问:
Local Homeomorphism with constant fiber is a covering map
Let $f:X\to Y$ be a local homeomorphism between Hausdorff spaces such that $f$ is surjective and $f^{-1}(y)$ is compact $\forall y\in Y$. If $|f^{-1}(y_1)|=|f^{-1}(y_2)|,\forall y_1,y_2\in Y$ then show that $f$ is a covering map.

Attempt: I know $f^{-1}(y)$ is finite as it has a finite cover $\cup_{i=1}^nU_i\supseteq f^{-1}(y)$ with each $U_i$ open such that $f_{U_i}$ is a homeomorphism. I considered $V_i\subseteq U_i$ with $V_i\cap V_j=\emptyset$ but the problem is the preimage of $\cap_{i=1}^nf(V_i)$ is not necessarily in $\cup_{i=1}^nU_i$ otherwise I would have found an evenly covered neighborhood of $y$. I need to show that $f$ is closed so I can use the set $(\cap_{i=1}^nf(V_i))\backslash (f(X\backslash \cup_{i=1}^nU_i))$ instead or consider another approach by using the constant preimage assumption which I do not see how it fits there...

答:
Apparently the only remaining difficulty is to show that $f$ is closed, so let us concentrate in proving this.

So let $C$ be a closed subset of $X$ and pick
$$
  y\in \overline{f(C)}.
  $$
Write
$$
  f^{-1}(y)=\{x_1, x_2, \ldots , x_n\},
  $$
and observe that $n$ is therefore the constant number of elements of each fiber of $f$.

For each $k=1,\ldots ,n$, choose open sets $U_k\subseteq X$ and $V_k\subseteq Y$, such that $x_k\in  U_k$, $y\in  V_k$, and $f$ is a
homeomorphism from $U_k$ onto $V_k$.  By using that $X$ is Hausdorff and shrinking the $U_k$ appropriately we may assume
that the $U_k$ are pairwise disjoint.

Without loss of generality we may also assume that the $V_k$ are all the same since otherwise we may replace each $V_k$ by
$$
  V=\bigcap_{k=1}^n V_k,
  $$
while replacing each $U_k$ by $f^{-1}(V)\cap U_k$.

Since $y\in \overline{f(C)}$, we may find a net $\{c_i\}_{i\in I}\subseteq C$, such that $y=\lim_{i\in I}f(c_i)$.  By  discarding an initial segment of our net
we may assume that $f(c_i)\in  V$, for all $i$, so we may choose  $d^k_i\in U_k$, such that $f(d^k_i)=f(c_i)$.

Due to the fact that the $U_k$ are pairwise disjoint we have  that, for each $i$,  the elements
$$
  d^1_i, d^2_i,\ldots , d^n_i
  $$
are
all distinct, so there are precisely $n$ of them,  and since they all lie in $f^{-1}\big (f(c_i)\big )$ we deduce from the
hypothesis that
$$
  f^{-1}\big (f(c_i)\big ) =\{d^1_i, d^2_i,\ldots , d^n_i\}.
  $$

It  follows that
$$
  c_i\in \{d^1_i, d^2_i,\ldots , d^n_i\},
  $$
for every $i$, so the pigeonhole principle  implies that there is a
$k$ and a
co-final subset $I_0\subseteq I$, such that   $c_i=d^k_i$ for all $i\in  I_0$.  We then have that
$$
  x_k =   \lim_{i\in I}d^k_i =  \lim_{i\in I_0}d^k_i =  \lim_{i\in I_0}c_i \in  C,
  $$
so
$$
  y=f(x_k)\in  f(C),
  $$
whence $f(C)$ is closed.


问:
Proof that the maximum of $F(x)=\arctan(aTx)-\arctan(Tx)$ is $\arcsin(\frac{a-1}{a+1})$
$a$ and $T$ are constants.

I have plotted $F(x)$ and $y=\arcsin(\frac{a-1}{a+1})$ for different values of $a$ and found that it's true.

I tried using $\arctan(x)=\arcsin(\frac{x}{\sqrt{1+x²}})$.

I also tried calculating the derivative of $F(x)$ but I couldn't get to extract an $\arcsin$.

答:
Computing the derivative
$$F'(x)=\frac{a T}{a^2 T^2 x^2+1}-\frac{T}{T^2 x^2+1}$$ and we want it to be equal to $0$. This happens at
$$x_*=\pm \frac{1}{T \sqrt{a}}$$

Now, we need to check the second derivative
$$F''\left(\pm\frac{1}{T\sqrt{a} }\right)=\mp\frac{2 (a-1) a^{3/2} T^2}{(a+1)^2}$$ So, if $a>1$ the maximum is at $x_*=\frac{1}{T\sqrt{a}}$ and vice-versa.
$$F\left(+\frac{1}{T\sqrt{a} }\right)=\tan ^{-1}\left(\sqrt{a}\right)-\tan ^{-1}\left(\frac{1}{\sqrt{a}}\right)=\tan ^{-1}\left(\frac{a-1}{2 \sqrt{a}}\right)$$
$$F\left(-\frac{1}{T\sqrt{a} }\right)=\tan ^{-1}\left(\frac{1}{\sqrt{a}}\right)-\tan ^{-1}\left(\sqrt{a}\right)=\tan ^{-1}\left(\frac{1-a}{2 \sqrt{a}}\right)$$ and your result is true if you take the positive solution.

Now, play with a few trigonometric relations.

ContentLicense: CC BY-SA 4.0


问:
Chinese remainder theorem issue
Let's say I have the following equations:

$$x \equiv 2 \mod 3$$
$$x \equiv 7 \mod 10$$
$$x \equiv 10 \mod 11$$
$$x \equiv 1 \mod 7$$

And I need to find the smallest x for which all these equations are correct. So:

$N = 3\times10\times11\times7 = 2310$

$N_1 = \frac{2310}{3} = 770$

$N_2 = \frac{2310}{10} = 231$

$N_3 = \frac{2310}{11} = 210$

$N_4 = \frac{2310}{7} = 330$

Then:

$\gcd (3, 770) = 1 = 257\times3 - 1\times770; \qquad x_{1} = -1 \equiv 2 \mod 3$
$\gcd(10, 231) = 1 = -23\times10 + 1\times231; \quad x_{1} = 1$

Etc., and finally:

$x = 2\times2\times N_1 + 7\times1\times N_2 + ...$

I understand the whole method except for one thing - when I solved $\gcd(3, 770)$ I got $x_{1} = -1$...why is there $-1 \equiv 2 \mod 3$, from which the final result is 2?

ContentLicense: CC BY-SA 3.0

Title: Chinese remainder theorem issue

Body:
More simply: $ $ mod $3,11\!:\ x\equiv -1,\ $ so $ $ mod $33\!:\ x\equiv -1\ $ so $\ x = -1 + 33j.$

mod $7\!:\ 1 \equiv x \equiv -1 + 33j\ \Rightarrow\ j\equiv \frac{2}{33} \equiv \frac{2}{{-}2} \equiv -1\ $ so $\ j = -1 + 7k.$

Substituting:  $\ x = -1 + 33j = -1 + 33(-1 + 7k) = -34 + 231k.$

mod $10\!:\ 7 \equiv x \equiv -34 + 231k \equiv -4 + k\ $ so $\ k\equiv 7+4\equiv 1,\ $ i.e. $\ k = 1+10n.$

Substituting: $\ x = -34 + 231k = -34 + 231(1+10n) = 197 + 2310n.$


答:
Equating $F'(x)$ to zero,

$\begin{align}\frac{aT}{1+(aTx)^2}= \frac{T}{1+(Tx)^2} &\Rightarrow a+a(Tx)^2 = 1+ (aTx)^2 \\&\Rightarrow (a-1) = aT^2x^2(a-1) \\&\Rightarrow \boxed{x = \pm\frac{1}{T\sqrt a}}\end{align}$

$T$ can't be zero and $a$ can't be $1$, as these would make the function $F(x) = 0$.

Using these values of $x$, check if $F''(x)<0$ for maxima to occur.

For $x=  \frac{1}{T\sqrt a}$

$\begin{align}F(\frac{1}{T\sqrt a}) &= \arctan(\sqrt a) - \arctan\left(\frac{1}{\sqrt a}\right) \\\\&= \arctan\left(\frac{\sqrt a - \frac{1}{\sqrt a}}{1+1}\right) \\\\& = \arctan\left(\frac{a-1}{2
\sqrt a}\right) \\\\&= \arcsin\left(\frac{(a-1)/(2\sqrt a)}{\sqrt{1+\frac{(a-1)^2}{4a}}}\right) \\\\&= \arcsin \left(\frac{a-1}{a+1}\right)\end{align}$

Title: Proof that the maximum of $F(x)=\arctan(aTx)-\arctan(Tx)$ is $\arcsin(\frac{a-1}{a+1})$

Body:
$a$ and $T$ are constants.

I have plotted $F(x)$ and $y=\arcsin(\frac{a-1}{a+1})$ for different values of $a$ and found that it's true.

I tried using $\arctan(x)=\arcsin(\frac{x}{\sqrt{1+x²}})$.

I also tried calculating the derivative of $F(x)$ but I couldn't get to extract an $\arcsin$.


问:
Reference Needed: $m$-faced dice with $n$ options
I would like to investigate the following question and I am not sure if similar ideas have been investigated. I cannot really find a detail discussion about this matter on the Internet. It will be great if you can tell me what domain I should look into, or you may also give your view on the following questions.

Suppose we have one fair dice. We would like to choose an option randomly from 6 possible options. We roll the dice once then we can decide which option we should choose.

Suppose we now have 8 options. To choose an option randomly, different algorithms can be used.
For example, roll the dice once first.

If the result is odd, we focus on option 1 to 4. Then roll the dice again, if we get 1 to 4, we know which option to use. If it is a 5 or 6, repeat the second step again.

If the result is even, we focus on option 5 to 8. Then roll the dice again, if we get $n$, which $1\leq n\leq 4$, we choose option $n+4$. If it is a 5 or 6, repeat the second step again.

One can check the the probability of choosing each option is exactly $\frac{1}{8}$.

My question now is to investigate if it is possible that for whatever number of options we have, we can always use a dice to choose an option randomly. And is there a general algorithm to follow in order to choose an option?

My second question is more advance. Suppose we have a fair $m$-faced dice and we have $n$ total options. Can we choose an option randomly by using the dice for any $m$ and $n$?

Once again, you don't have to answer the above two questions. I am looking for hints or domain that I should look into. Thank you.

答:
Given an $m$-faced die, you can simulate an $n$-faced die, for any values of $m,n>1$.

Suppose the faces of your $m$-die are $0,1,\dots,m-1$. Then write out $0,1,\dots,n-1$ in base $m$. So each face of your $n$-die is associated with a base-$m$ string of length $\ell=\lceil\log_m(n)\rceil$.

Then roll your $m$-die a total of $\ell$ times consecutively, and record each die roll. This gives you a base-$m$ number with $\ell$ digits. If this matches up with one of $0,1,\dots,n-1$, great. If not, repeat the process again.

It's clear that each of the values $0,\dots,n-1$ are equally likely to show up. Furthermore, you will almost surely require a finite number of rolls. In fact, in expectation you will need $m^\ell/n<m$ rounds.

As a concrete example, let's take $m=2$ and $n=7$. Roll our $2$-sided die thrice, the list  of outcomes is $$000,001,010,011,100,101,110,111.$$The first $7$ of these correspond to the faces $0,\dots,6$ of the $7$-sided die. If we roll three $1$s, then we simply ignore the result and go again.


问:
How can I attack this definite integral - roots problem?
I have this problem:

Let $f(x)$ be continuous on the interval $[0,1] .$ Prove that if
$$
  \int_{0}^{1} f(x) d x=\int_{0}^{1} x f(x) d x=0
  $$
then $f$ has at least two different roots in (0,1)

Is it better to start with parts integration on second one? or triying
$$
  \int_{0}^{1} f(x) d x=\int_{0}^{1} x f(x) d x=0\Rightarrow \int_{0}^{1} f(x) -x f(x) d x=0 ?
 $$

If I use mean value Theorem I coul get
$$
\int_{0}^{1} (1-x) f(x) d x=(1-\xi)\int_{0}^{1}  f(x) d x=0
$$
With $\xi\in[0,1]$

Therefore
$$
  \int_{0}^{1} f(x) d x=0\qquad  or\qquad \xi =1
 $$

But doesn't seems to help, I already know
$$
  \int_{0}^{1} f(x) d x=0
 $$
and $\xi=1$ has no further use

So I'm kind of stuck right now

I'll appreciate any help or suggestion

Thanks in advance

答:
With $\int_0^1 f = 0$ we get that there is at least one root $x_0$. In fact, $f$ can't be always positive or always negative or the integral wouldn't be $0$ (i exclude the trivial case where $f=0$).

Let's suppose it is the only root. Then since $\int_0^1 f = 0$ and $\int_0^1 xf(x) dx = 0$, we have that $\int_0^1 \pm(x-x_0) f(x) dx = 0$. Let's say for instance that $f$ is positive before $x_0$, and negative after. Then $x\mapsto (x-x_0) f(x)$ is always positive. Since its integral is $0$, this means the function is $0$, which is contradictory.

So there are at least two roots.


问:
Is it possible $\sin\alpha=\cos\alpha=1?$
Is it possible $\sin\alpha=\cos\alpha=1?$

We have defined trigonometry using a unit circle. The point $X(1;1)$ does not lie on the trig circle, right? Does this mean that $\alpha$ does not exist?

答: $\sin ^2 x + \cos ^2 x = 1$ for all $x$, so if you had $\cos \alpha = \sin \alpha = 1$, then $\sin ^2 \alpha + \cos ^2 \alpha$ would be equal to $2$. This is absurd.


问:
Indefinite integral $\displaystyle \int 1 / (p + q \sin a x)^2 \, \mathrm d x$
I am evaluating:

$$\int \frac {\mathrm d x} {(p + q \sin a x)^2}$$

which the book gives me as:
$$\frac {q \cos a x} {a (p^2 - q^2) (p + q \sin a x) } + \frac p {p^2 - q^2} \int \frac {\mathrm d x} {p + q \sin a x}$$

This is Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.361$.

I am using a Weierstrass substitution: $u = \tan \dfrac {a x} 2$ which yields $\mathrm d x = \dfrac 2 a \dfrac {\mathrm d u} {1 + u^2}$ and $\sin a x = \dfrac {2 u} {u^2 + 1}$.

After algebra, this yields me the primitive:

$$\frac 2 a \int \frac {(u^2 + 1) \mathrm d u} {(p u^2 + 2 q u + p)^2}$$

I split this up into two separate primitives:

$$\frac 2 a \int \frac {u^2 \mathrm d u} {(p u^2 + 2 q u + p)^2} + \frac 2 a \int \frac {\mathrm d u} {(p u^2 + 2 q u + p)^2}$$

both of which are obtained via standard (though unwieldy) results:

$$\frac 2 a \left({\frac {(4 q^2 - 2 p^2) u + 2 p q} {p (4 p^2 - 4 q^2) (p u^2 + 2 q u + p) } + \frac {2 p} {4 p^2 - 4 q^2} \int \frac {\mathrm d u} {p u^2 + 2 q u + p} }\right)$$

and:

$$\frac 2 a \left({\frac {2 p u + 2 q} {(4 p^2 - 4 q^2) (p u^2 + 2 q u + p) } + \frac {2 p} {4 p^2 - 4 q^2} \int \frac {\mathrm d u} {p u^2 + 2 q u + p} }\right)$$

After some straightforward cleaning up, I get:

$$\frac {2 q (q u + p) } {a p (p^2 - q^2) (p u^2 + 2 q u + p)} + \frac p {(p^2 - q^2) } \left({\frac 2 a \int \frac {\mathrm d u} {p u^2 + 2 q u + p} }\right)$$

The term on the right is in completely the correct format, returning me $\displaystyle \frac p {p^2 - q^2} \int \frac {\mathrm d x} {p + q \sin a x}$ after I put $u$ back.

But my left hand term has gone astray. In order to return $\dfrac {q \cos a x} {a (p^2 - q^2) (p + q \sin a x) }$, I really need to get it into the form:

$$\frac {q (1 - u^2)} {a (p^2 - q^2) (p u^2 + 2 q u + p)}$$

but instead I have:
$$\frac {2 q (q u + p) } {a p (p^2 - q^2) (p u^2 + 2 q u + p)}$$

I have gone through my working carefully several times, but I can't put my finger on where I have gone astray -- it could be at any of the above steps.

答:
Write the numerator as

$$q(1-u^2)=-\frac q p (pu^2+2qu+p) +\frac{2q}p( qu +p)
$$

where the first term cancels the denominator, becoming a non-essential constant, and second term produces what you have.


问:
Find all finite subsets of $\mathbb{C}$ closed to multiplication.
Find all finite subsets of $\mathbb{C}$ such that they are closed to multiplication.

Hint: calculate the powers of $\sqrt2+\sqrt3$ until they become a linear combination of two square roots.

I know that since $|a \cdot b| = |a| \cdot |b|,$ these subsets are all subsets of the unit circle. Also, I've concluded that these subsets consist of numbers $e^{i \phi}$ where $ 2 \pi | \phi$ because $e^{i \phi_1} \cdot e^{i \phi_2} = e^{i (\phi_1+\phi_2)}$ so these numbers always stay in a closed set.

I'm not sure how to formally prove this statement and if these are only possible solutions. Also, I'm not sure how the given hint helps me.

Any hints would be appreciated!

答:
Hint The group generated by $e^{iu}$ is dense if $u$ is different of ${p\over q}\pi$ where $p,q$ are integers. This follows from the fact a subgroup of the real line  generated  by a real number $a$ is dense if $a$ is not rational.

You have to add $0$ to the finite groups of the unit circle eventually.


问:
Linear algebra - dot product - orthogonal
Question is from Gilbert 5th 2.1 B:

x+3y+5z=4

x+2y-3z=5

2x+5y+2z=8

The vector (1,1,-1) is orthogonal to all three columns of A but not to B.

So, take the dot product of each column of A ( and also b) with y = (1,1,-1). How do those dot products show that no combination of columns equals b?

(Sol)
The dot product of each column of A with y = (1,1,-1) is zero.

My question is below:

The dot product should be (1,3,5) * (1,1,-1) = -1, (1,2-3) * (1,1,-1) = 6, and (2,5,2) * (1,1,-1) = 5.
column picture should be (1,1,2) * 1 + (3,2,5) * 1 + (5,-3,2) * -1

I can't get zero.

I believed that I probably misunderstand somewhere.

答:
$b$ is a linear combination of the columns of the matrix if there exist $x,y,z$ such that $xC_1+yC_2+zC_3=b.$ Let's denote $u=(1,1,-1).$ Then

$$(xC_1+yC_2+zC_3)\cdot u=xC_1\cdot u +yC_2\cdot +zC_3\cdot u=0$$ while $$b\cdot u\ne 0.$$ So, $b$ is not a linear combination of the columns of the matrix (or in other words, the system is inconsistent).


问:
Roots of a polynomial modulo some prime

Let $p$ be a prime number not dividing $x$, let $n>1$. Consider the equations
$$\tag{1} x^n-1 \equiv 0 \pmod{p}$$
and
$$\tag{2} x^n+1\equiv 0 \pmod{p}$$
What is the meaning (or the implication) of saying that equation (1) or equation (2) has at most $1$ solution $x$ modulo $p$, counting multiplicity?

I know that such equations have at most $n$ solutions $x$ modulo $p$, counting multiplicity by a theorem of Lagrange. Here $n$ is not necessarily equal to 2.

答: If $x^n-1\equiv0\pmod{p}$ has at most one solution, it has precisely one solution, because $x=1$ is a solution. That it has no other solution means that $x=1$ is the only element of $(\Bbb{Z}/p\Bbb{Z})^{\times}$ of order dividing $n$. This in turn implies that $\gcd(n,p-1)=1$. Similar statements hold for the other equation.

Title: Roots of a polynomial modulo some prime


问:
Given a list of points on a rectilinear path, identify the corners

You are provided list of co-ordinates in form of pair as shown below:

$$li=[(576, 64), (576, 192), (448, 192), (320, 192), (320, 320), (192, 320), (192, 448), (192, 576), (320, 576), (448, 576), (576, 576), (704, 576), (832, 576), (960, 576)]$$

After modifying it according to goal.

Goal is to first plot all points in co-ordinate plane then join them now carefully we need to find out all corner points.

$$li=[(576, 64), (576, 192),(320, 192), (320, 320), (192, 320), (192, 576), (960, 576)]$$

Please refer the image I need the points which are encircled by a blue rectangle, where as red circled points are points that are provided in the problem.

I tried brute force approach it worked but yes please I need something better than that.

答:
Converting a comment (with some adjustments) to an answer, as OP is satisfied with its validity ...

For each index $i$ (starting with the second, and ending with the next-to-last), compare $p_{i-1}$ to $p_{i+1}$: if (and only if) their $x$-coordinates differ and their $y$-coordinates differ, then (and only then) $p_i$ is a corner.

This algorithm ignores the first and last points, but those are already "corners" by declaration.

As OP has noted, the algorithm has complexity $O(n)$, as it simply steps through the list once.


问:
Finding the equation for $ E(x) $
Here is the question

Let $E$ be the differential function and satisfying $E(u + v) = E(u)E(v)$ for all $u$ and $v$. Find the equation for $E(x)$ Hint: First find $E^\prime (v)$.

I was figured it that the function $E(x) = e^x$. If I put $x = u + v$, then it should be satisfying that

$\begin{align} E(u+v) &= e^{(u+v)} \quad \text{using this property: } e^{a+b} = e^a e^b \\ &= e^u e^v \\ &= E(u)E(v) \end{align}$

And I though it was just only $e^x$, but I'm very confused that the hint tell me that I have to find the $E^\prime (v)$ and I don't really get it.

Please, correct me if my work is wrong.

答:
By deriving the expression with the variable $u$, we get $E'(u+v) = E'(u)E(v)$. And with $u=0$, we get $E'(v)=E'(0)E(v)$. If we note $c=E'(0)$ and integrate the relation, we get $E(v)=c e^{ax}$ for some $c$.

Then verify it works only for $c=1$.

Title: Finding the equation for $ E(x) $


问:
Confused with the 2 different ideas on derivatives
The professor explained two different things about derivatives:

It is related to a line connecting 2 points on a function

It can be calculated using rules

I can understand visually the first of those claims. Even when it requires infinitesimals. But the second one, the formulas, were derived algebraically.

Is there any way to build an intuition, or use another method other than infinitesimals to find out what a derivative of a function is, or this concept and results only arises when we use infinitesimals?

Any article for reading is welcome as long as it's not too technical.

答:
First of all, why do you think that these are "different things"? This is what is normal in mathematics: You define some concept, and then you use that definition to figure out various theorems and formulas related to that concept.

A right triangle is defined as three line segments sharing end points, with two of the segments being perpendicular to each other (which itself is defined as "if you extend one of the segments, the two angles formed by the other segment and the two sides of the extended segment are congruent"). There is no mention in the definition of any relationship between the lengths of the three sides. Yet still we are able to prove $a^2 + b^2 = c^2$.

The problem with derivatives is that the idea behind the derivative is not something we can directly calculate. But it is something we can approximate as closely as we like. And we leverage that ability to approximate it into indirect tools that we can calculate.

Calculus is the mathematics of refining approximations. And though it is generally held to have been invented by Leibniz and Newton, its roots are far older. They date back more than 2400 years to the Method of Exhaustion of Antiphon and Eudoxus. Though the method of exhaustion was originally developed as a tool for area calculation, the basic idea is more general:

If you have a quantity $M$ that you can only approximate, and you want to show it is equal to some number $m$, do this: For every $a < m$, approximate $M$ closely enough to show that it must be $> a$. For every $b > m$, approximate $M$ closely enough to show it must be $< b$. Now you've shown that $M$ is greater than every number less than $m$ and $M$ is less than every number greater than $m$. The only possibility left is that $M = m$.

A concrete example: Find the slope of the tangent line to $y =x^2$ at $x = 1$. The point of tangency is $(1,1)$. if I choose $x = z \ne 1$, I get a second point $(z, z^2)$ and can form the secant line passing through the two points. That line has slope:
$$m = \dfrac{z^2-1}{z-1} = \dfrac{(z+1)(z-1)}{z-1} = z + 1$$
I cannot have $z = 1$ in this equation because the fraction is not defined there. I can only say that $m = z+1$ when $z \ne 1$. If I were allowed to have $z = 1$, $2$ would be the obvious choice for the corresponding $m$. So how to show that $m$ actually is $2$ at $z = 1$? The idea is that the derivative is the value that the slopes approximate when $z$ is close to $1$, with the approximation improving as $z$ gets closer to $1$.

Suppose $a < 2$. I can choose a value for $z$ with $a - 1 < z < 1$, for example, the average $z = \frac {(a-1) + 1}2 = \frac a2$. Then $$m(z) = z + 1 > (a-1) + 1 = a$$
In particular, for any $z$ closer to $1$ than $a-1$, $m(z)$ is closer to $2$ than $a$. Thus the desired value of $m(1)$ cannot be $a$. The same is true when $a > 2$. Picking $1 < z < a-1$ gives $2 < m(z) < a$, so $m(1)$ cannot be $a$ either. Therefore the derivative $m(1)$ at $z = 1$ cannot be less than $2$ or greater than $2$. It must be $2$.

Notice that there is no mention of limits here. I've dodged around the language a bit to avoid getting into certain issues that would dramatically lengthen this post, but the whole concept of the derivative can be defined without actually discussing the idea of a limit. This is somewhat of a cheat, because the idea behind the limit is still there. But discussions of infinitesimals or of epsilons and deltas are not. You just eliminate all values but one from being a possibility, and define the derivative to be the one that is left. "Limits" themselves are just a way of cleaning up the process.

Notice also that even though I started woking with just a vague idea of approximating the tangent slope with secant slopes, and produced my answer only from that idea, I found a nice formula in the middle that gave me the answer. The rest of the argument was devoted to showing that yes, it really was the answer. This is why the derivative is defined in terms of limits of secant lines, but then we have all these formulas for calculating it. The formulas follow from the definition.


问:
Pick two random people in the world, what's the probability of them knowing each other?
I have one question popped in my head out of nowhere at 3AM.

If we pick two random people in the world, what's the probability of them knowing each other?

You can assume that there are 7 billion people in the world and the idea of Six Degrees of Separation is relevant, or maybe you can assume one person has a circle of friends up to 100 people. You can provide any other assumptions to make it simple.

Please help answer my 3AM thought so I can sleep peacefully.

答:
Suppose there are $n$ people in the world. Each individual is identified with a number $i$, $1 \le i \le n$. Let $k_i$ be the number of persons known by $i$, in a reciprocal way (i.e. if $i$ knows $j$ then $j$ knows $i$). The number of couples that know each other is:

$$\frac{1}{2}\sum_{i=1}^n{k_i}$$

because we count the couple $(i, j)$ twice, once for $i$ and once for $j$. Therefore the probability you are looking for is:

$$\frac{\frac{1}{2}{\sum_{i=1}^n{k_i}}}{{n \choose 2}}=\frac{\frac{1}{n}{\sum_{i=1}^n{k_i}}}{n-1}$$

i.e. it can be viewed also as the average number of people known (reciprocally) by an individual divided by all the other people ($n-1$).

Suppose you would like to compute the probability including one way knowing relationships. Each person $i$ will thus know $v_i$ "VIPs" or "influencers". Suppose the number of influencers is $m$ and that influencer $j$, $1 \le j \le m$, has $f_j$ followers. Then the above probability increases like this:

$$\frac{\frac{1}{2}{\sum_{i=1}^n{k_i}+\sum_{i=1}^n{v_i}}}{{n \choose 2}}=\frac{\frac{1}{n}{\sum_{i=1}^n{k_i}}}{n-1}+\frac{\sum_{i=1}^n{v_i}}{{n \choose 2}}=\frac{\frac{1}{n}{\sum_{i=1}^n{k_i}}}{n-1}+\frac{\sum_{j=1}^m{f_j}}{{n \choose 2}}$$


问:
Redundant equations in the definition of jointly independent events
The events $A_1, A_2, \ldots A_n$ are said to be jointly independent if for every $2 \leq k \leq n$ and $1 \leq i_1 < i_2 < \ldots < i_k \leq n$ we have

$$ P(A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k}) = P(A_{i_1}) P(A_{i_2}) \ldots P(A_{i_k}). $$

A set of jointly independent events are always pairwise independent, but for $n \geq 3$ the converse is false, which means the equations for $k=2$ only do not necessarily imply joint independence.

In this sense it is natural to ask the following question: What is the minimum number of equations of the type

$$ P(A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k}) = P(A_{i_1}) P(A_{i_2}) \ldots P(A_{i_k}) $$

which already give a sufficient condition for joint independence? I.e. how many redundant equations are there in the definition of jointly independent events?

答:
Claim: The answer is $0$. We can't get rid of any equations.

Lemma: for every $n$ there are $n+1$ events, which are not jointly independent, but any $n$  items of them are independent.

Proof of lemma: let $\xi_{i}$ be independent $Bern(\frac12)$ r.v., $1 \le i \le n$. Put $\xi_{n+1} = I_{\sum_{i=1}^n \xi_i \text{ is odd}}$. Finally, put $A_i = I_{\xi_i = 1}$.

Proof of claim: it's sufficient to prove that for any $j_1 < j_2 < \ldots < j_m$ there are $A_1$, $A_2, \ldots, A_n$ such that $$ P(A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k}) = P(A_{i_1}) P(A_{i_2}) \ldots P(A_{i_k})$$
for any $\{i_1, \ldots, i_k\}$ except $\{j_1, j_2, ..., j_m\}$.

Take $A_{j_1}, A_{j_2}, \ldots, A_{j_m}$ from lemma and put $A_i = \varnothing$ for all $i \notin \{j_1, j_2, ..., j_m\}$. Thus if $i_{\alpha} \in  \{j_1, j_2, ..., j_m\}$ we have $$ P(A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k}) =  0= P(A_{i_1}) P(A_{i_2}) \ldots P(A_{i_k}).$$

If $\{i_1, i_2, ..., i_k\} \subset \{j_1, j_2, ..., j_m\}$ and $k < m$ then it follows from lemma that $$ P(A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k}) =  P(A_{i_1}) P(A_{i_2}) \ldots P(A_{i_k}).$$

Q.e.d.


问:
Given N points in the Cartesian plane, how many unique sequences are formed by ordering them by distance from any arbitrary test point?
Let $P$ be a set of $N$ distinct points in the Cartesian plane $C$, and let $S_P$ be the set of all $N$-sequences of these Cartesian points; a particular sequence $s$ has elements $s_1$ through $s_N$.

Consider the subset of all sequences that that ordered with respect to some point in the Cartesian plane:
$$D(P) = \{ s \in S_P \mid \textrm{for some } x \in C, \left| s_{k} - x \right| \leq \left| s_{k+1} - x \right| \textrm{for all } 1 \leq k < N  \}$$
My question is, is there an upper bound on the cardinality of $D$? Or, is there some asymptotic behavior for
$$ D_N = \max_{\{ P \subset C : \left| P \right| = N\}} \left| D(P) \right| $$
that is stronger than $O(N!)$?

[EDIT: to rule out some cases where a point is equally near all points in the plane and trivially $\left| D(P) \right| = N!$, as pointed out by @HagenvonEitzen, this definition can be made more interesting by restricting $D(P)$ to uniquely ordered sequences:
$$D'(P) = \{ s \in S_P \mid \textrm{for some } x \in C, \left| s_{k} - x \right| < \left| s_{k+1} - x \right| \textrm{for all } 1 \leq k < N  \}$$.
The set of points $x$ forming sequences in $D(P)$ but not $D'(P)$ must be equidistant to at least two points in $P$ and are therefore constrained to lie on some collection of lines, and thus have measure zero.
]

My feeling is that $D_N \sim O(N^2)$; I might have heard or read this somewhere before but can't recall the source. For a particular point $x$, let the set of sequences of points ordered by distance from $x$ be denoted $x(P)$ (below I'll gloss over the fact that there can be more than one when distances are equal). The set $P$ determines a Voronoi diagram in the plane, with each point $x$ in a cell with its nearest neighbor $p \in P$. For $x$ close enough to $p$, $x(P) = p(P)$. Once $x$ gets slightly further away, its next nearest neighbor may be distinct from $p$'s nearest neighbor $q$. A simple 1 dimensional example:

```<code>q---p-x---r
</code>```

The point $x$ is two units away from $p$ and four from $r$, but $p$ is closer to $q$ than $r$. I'm sure I could draw up cases where $x$'s third nearest neighbor differ's from $p$'s second, and so on; however, it seems difficult to continue this forever and create a situation where $x(P)$ differs from $p(P)$ for its entire length (after the initial element where they must agree). Once the ordering reaches far enough away neighbors $w$, it seems that $\left|p-x\right|$ should (usually) be small compared to $\left|p-w\right|$. Of course, there are scenarios where that doesn't hold (e.g. if one Voronoi cell is much larger than all others), so this isn't a precise statement that I'm able to work into a rigorous argument.

It would also be interesting to know how this generalizes to other geometries.

答:
With the definition of $D(P)$ in the OP, we can let $P$ be a set of $n$ points on a circle; then thr centre of the circle witnesses that $D(P)$ is the set of all $n!$ permutations of $P$. Thus $$D_n = n!.$$

Let us adjust $D$ to allow only points in general position, i.e.,
$$D(P) = \{ s \in S_P \mid \textrm{for some } x \in C, \left| s_{k} - x \right| < \left| s_{k+1} - x \right| \textrm{for all } 1 \leq k < N  \}.$$
Now we can show
$$ D_n=O(n^4)$$
(and can perhaps be improved).
Indeed, we must pick $x$ in any of the regions that the plane is cut into by the $m={n\choose 2}$ perpendicular bisectors between points in $P$, and it is well-known that the maximal number of regions obtainable with $m$ lines is ${m+1\choose 2}+1$. As we are not free in the choice of our $m$ lines, I suspect that lowering the exponent is still possible.


问:
If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$
Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded.
Show that $f'(x)\to0$ as $x\to\infty$.
(This is inspired by a comment/answer to a different question)</p>

答:
Let $M$ be a bound for $f''$. Then $|f'(x+h)-f'(x)|\le M|h|$ for all $x,h$.
Let $\epsilon>0$ be given. We have to show that $|f'(x)|<\epsilon$ for all sufficiently big $x$.
As $f(x)\to0$, there is an $x_0$ such that $|f(x)|<\frac{\epsilon^2}{4 M}$ for all $x>x_0$.
Consider $x>x_0$ and assume $f'(x)> 0$.
Then $f'(x+h)\ge f'(x)-Mh$ for $h\ge 0$ and hence
$$ \begin{align}f\left(x+\frac {f'(x)}{M}\right)-f(x)&=\int_0^{\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\ge \int_0^{\frac {f'(x)}{M}}(f'(x)-M|h|)\,\mathrm dh\\&=\frac{(f'(x))^2}{2M}.\end{align}$$
If on the other hand $f'(x)<0$, we similarly have  $f'(x+h)\le f'(x)+Mh$ for $h\ge 0$ and hence
$$ \begin{align}f\left(x-\frac {f'(x)}{M}\right)-f(x)&=\int_0^{-\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\le \int_0^{-\frac {f'(x)}{M}}(f'(x)+M|h|)\,\mathrm dh\\&=-\frac{(f'(x))^2}{2M}.\end{align}$$
In both cases we find
$$ \left|f\left(x+\frac {|f'(x)|}{M}\right)-f(x)\right|\ge \frac{(f'(x))^2}{2M}$$
and as $x+\frac {|f'(x)|}{M}>x_0$, we conclude $\frac{(f'(x))^2}{2M}< 2\cdot  \frac{\epsilon^2}{4M}$ and hence $|f'(x)|<\epsilon$ as was to be shown.$_\square$</p>


问:
Simplifying an algorithm to evaluate $\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n(x_i\&x_j)(x_j|x_k)$ for non-negative integers
Let $x_1, x_2, \ldots, x_n$ sequence of non-negative integers. I need to calculate $$\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}(x_i \& x_j)(x_j | x_k)$$
Here "$\&$" stands for bitwise AND, and "$|$" stands for bitwise OR

I know how to write an algorithm that runs ~ $O(n^3)$. How can you get rid of the triple amount?

答:
Here's a hint. Let $f(a,b) = x_a \wedge x_b$ and $g(a,b) = x_a \vee x_b$. You are asked to compute
$$
\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n f(i,j) g(j,k) = \sum_{j=1}^n \left(\sum_{i=1}^n \sum_{k=1}^n f(i,j) g(j,k) \right).
$$

The straightforward way is to do $O(n^3)$ ANDs, ORs, multiplications and additions.
How can you  simply speed up the computation just by manipulating the above formula?


问:
Does being able to factor semiprimes entail being able to factor any number?
I'm asking this question with specific reference to Shor's Quantum Factorization Algorithm, which uses the fact that finding a non-trivial square root of $1 \pmod N$ for some integer $N$ yields factors of $N$. The plausibility of this setup (i.e. the fact that we don't have to hope to be overly lucky for this to work) relies on the following claims:

Let $x$ be chosen randomly from $0 \leq x < p$ is prime. We know that there exists a generator of $\mathbb{Z}_p$, call it $g$, so that $g^k \pmod p = x$. If $k$ is odd, then $\text{ord}(x)$, defined to be the minimum $r$ such that $x^r \pmod p = 1 \pmod p$, is even. Consequently, the probability that $\text{ord}(x)$ is even for a random $x$ is at least $1/2$.

If $N = pq$ for primes $p$ and $q$, then with probability at least $3/8$ the order $r$ of a randomly chosen element $x \in \mathbb{Z}_{pq}$ is even, and furthermore, $x^{r/2} \pmod {pq} \neq \pm 1 \pmod {pq}$. The proof of this is roughly as follows:

Choosing a random $x$ in $\mathbb{Z}_{pq}$ is equivalent to choosing some $a \in \mathbb{Z}_p$ and $b \in \mathbb{Z}_q$. Namely, given some $a = x \pmod p$ and $b \pmod q$, there is exactly one element in $\mathbb{Z}_{pq}$ satisfying these relations, by the Chinese Remainder Theorem. Letting $r_a = \text{ord}(a)$ (with respect to $\mathbb{Z}_p$) and $r_b = \text{ord}(b)$ (with respect to $\mathbb{Z}_q$), clearly we can see that $r_a, r_b ~|~ r$ (where $r = \text{ord}(x)$ with respect to $\mathbb{Z}_{pq}$). From result one, $r_a$ and $r_b$ are even with probability at least $1/2$, so $r$ will be even with probability at least $3/4$.

In addition, we claim that if $r$ is even, the probability that $x^{r/2} \pmod {pq} = \pm 1$ is at most $1/2$. Since $x^{r} = 1 \pmod {pq}$ if and only if $x^r = 1 \pmod p$ and $x^r = 1 \pmod q$, and these in turn imply that $x^{r/2} = \pm 1 \pmod p$ and $x^{r/2} = \pm 1 \pmod q$. We can form four systems of two equations with these, and they will each have unique solutions $x \in \mathbb{Z}_{pq}$. Two of them are $\pm 1$, and thus the other two will thus be non-trivial. The result then follows.

What I fail to see is how this can extend to factoring numbers in general. What if the number were of the form $p^2 q^3$, or $pqr$ (where $p, q, r$ are prime)? Is it either possible to tweak one of these results and its proof so that we can find a non-trivial square root for any $N$ (not just those which are semiprime)? Or alternatively, is there some simple explanation for why factoring just reduces to factoring a semiprime? Thanks!

UPDATE: It appears that the following is true (Theorem 5.3 in Nielsen and Chuang, Quantum Computation and Quantum Information):

Suppose $N$ is odd and has $m$ distinct prime factors. If we choose $x$ uniformly under the constraint $\gcd(x, N) = 1$ and $r$ is the order of $x$ modulo $N$, then
$$\mathbb{P}(r \text{ is even and } x^{r/2} \neq -1 \pmod N) \geq 1 - 2^{-m}$$

But no proof or proof reference is given. If someone can prove this assertion, that would answer my question as well.

答:
I've found the proof of this statement in Nielsen and Chuang, Quantum Information and Quantum Computing, Appendix 4 (thanks to @DaftWullie on QC Stack Exchange). The claim which I believe is correct is slightly different:

Suppose $N$ is odd and has $m$ distinct prime factors. If we choose $x$ uniformly under the constraint $\gcd(x, N) = 1$ and $r$ is the order of $x$ modulo $N$, then
$$\mathbb{P}(r \text{ is even and } x^{r/2} \neq -1 \pmod N) \geq 1 - 2^{1-m}$$

Let $\mathbb{Z}_{N}^*$ be the multiplicative group of integers less than $k$ that are relatively prime to $k$. The proof of the statement relies on the following lemma:

Let $p$ be an odd prime. If $2^d$ is the largest power dividing $\varphi(p^a) = p^{a-1}(p-1)$, then with probability $1/2$ a randomly chosen element of $\mathbb{Z}_{p^a}^*$ will have order divisible by $2^d$.

Proof: $\mathbb{Z}_{p^a}^*$ is cyclic, so there exists a generator $g$ for it, i.e. every element is expressible as $g^k \pmod {p^a}$ for some $k < \varphi(p^a)$. Suppose $x = g^k \pmod {p^a}$ and the order of $x$ modulo $p^a$ is $r$. Consider two cases. If $r$ is odd, then $g^{kr} \pmod {p^a} = 1$, so $\varphi(p^a)$ divides $kr$ (since the order of $g$ must be $\varphi(p^a)$). Since $k$ is odd, $2^d ~|~ r$. Now if $k$ is even, then
$$x^{\varphi(p^a) /2} \pmod {p^a} = g^{\varphi(p^a) k/2} \pmod {p^a} = 1^{k / 2} \pmod {p^a}$$
so it follows that $r$ divides $\varphi(p^a) / 2$ and thus, $2^d \nmid r$. But $k$ is odd with probability exactly half, so we are done. $\square$

From here, we note that if $N = p_1^{a_1} \cdots p_k^{a_k}$, then choosing a random element $x$ from $\mathbb{Z}_N^*$ is equivalent to choosing random elements $x_i$ from $\mathbb{Z}_{p_i^{a_i}}^*$ and mandating that $x = x_i \pmod {p_i^{a_i}}$ for all $i$. Let $r_i$ be the period of $x_i$ modulo $p_i^{a_i}$, and define $d_i$ to be such that $2^{d_i}$ is the largest power of two dividing $r_i$.

Let $r$ be the order of $x$ modulo $N$. The crucial claim, then, is that if $r$ is odd or $x^{r/2} \pmod N = -1$, then the $d_i$ must all be the same. Indeed, we can observe that $r_i \mid r$ for every $i$, because
$$x^r = 1 \pmod N \implies x^r \pmod {p_i} = x_i^r \pmod {p_i} = 1 \pmod {p_i}$$
Now if $r$ is odd, then all the $r_i$ must be odd and thus $d_i = 0$ for all $i$. If $x^{r/2} \pmod N = -1$, then by the Chinese Remainder Theorem we must have $x_i^{r/2} = -1 \pmod {p_i^{a_i}}$ for all $i$. This implies that $r_i$ does not divide $r / 2$, for every $i$. In particular, this means that all the $d_i$ must be the same. If this weren't the case, say $d_j \geq d_k + 1$ for some $j$ and $k$, then since $2^{d_j}$ divides $r$, it would have to follow that $r = z \cdot 2^a \cdot 2^{d_k}$ for some integer $z$ and $a = d_j - d_k \geq 1$. Hence, it would follow that $2^{d_k}$ divides $r / 2$, a contradiction.

From here, we note that the probability that all of the $d_i$ are the same is worst (i.e. largest) when the same number $d$ is such that $2^d$ divides $p_i^{a_i}$ for every $i$. In this case however, the lemma dictates that a randomly chosen element from $\mathbb{Z}_{p_i^{a_i}}^*$ has order divisible by $2^d$ with probability half. Now the orders of $x_i$ modulo $p_i^{a_i}$ are chosen independently from one another, so this probability in the worst case is the probability that all of the orders $r_i$ are equivalent to $r_1$, which is $2^{1-m}$ (recall $m$ is the number of distinct prime factors of $N$). The result then follows. $\square$

In particular, this means that Shor's algorithm as commonly written can be applied to any number, and with constant probability of success, it will return a non-trivial factor of it! The probability of success is worst when $N$ is a semiprime (the lower bound on the probability of success is $1/2$) and tends arbitrarily closer to $1$ as the number of prime factors of $N$ increases.


问:
Change of coordinate system on a sphere
This might take a while to explain, so bear with me:

I've got a perfect sphere. I've set up an arbitrary longitude/latitude ("angle") coordinate system on it (imagine an equator around the middle, with an arbitrary (0,0) on the equator). The coordinates are angles (longitude from $-\pi$ to $\pi$, and latitude from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$).

In order to visualize what I want to do, I came up with this way of thinking about it: Imagine a sheet of graph paper pinned (through the middle) to the sphere at the origin (0,0). The graph paper cannot be rotated. The lines on the graph paper form a "grid".

The graph paper can be wrapped around the equator with no distortion (forming a cylinder), but wrapping it completely around the sphere will force distortion around the poles (imagine the graph paper scrunching up to fit at the poles). This is expected. It turns out (I think) that the graph paper can be wrapped around any equator (with any arbitrary origin), so long as that equator includes the pinned point. Of course, wrapping it around a different equator will yield distortion at different poles (those relative to that equator), with no distortion at all along that equator.

Now, pick any point on the sphere (other than (0,0)) and call it the "local origin". I can wrap the graph paper around the sphere along the "global equator" and this gives me a "global coordinate system". I can also wrap the graph paper along the (unique) "local equator" that crosses both the global origin and local origin, yielding a "local coordinate system". Both have the same amount of distortion, but in different places.

The graph paper is in arbitrary units representing a fixed number of radians in the global coordinate system.

I've got a specific point which is expressed in the local coordinate system. Now, what I want to do is get the location of that point expressed in the global coordinate system. I tried a change of basis in the grid (yay!), but it turns out this doesn't work properly (aww) because the grid is warped differently in each coordinate system, so the grid units aren't equivalent in size anywhere except on the local equator (where there's no distortion).

I've thought of transforming the point from the local coordinate system into a (local) 3D $(x,y,z)$ point, then applying a transformation that would rotate it into its place in global 3D space, then working backward from that point to get the global angle coordinates. But that's kind of ugly (plus, I'm trying to keep the number of operations down since I'm using fixed-precision arithmetic).

Surely there's a simple way to go from local (angle) coordinates to global (angle) coordinates?

Thanks!

Update: Nope, I really don't believe there's a simpler way than converting to local 3D Cartesian coordinates, applying a transform to get global 3D coordinates, then going back to global angle coordinates. I could, of course, be wrong.

答:
I think, I had the same problem. I also have not found any simpler solution so far. But after I saw an answer from a mathematics professor from ETH, I am now pretty sure, there is not. The solution:

Transforming from one spherical coordinate system to another

The simplifications should be tan, cot and csc appearing.

PS: Another one: https://stackoverflow.com/questions/5278417/rotating-body-from-spherical-coordinates

Title: Change of coordinate system on a sphere


问:
Is it true that if $Y$ is infinite and $|X|<|Y|$, then $|Y^X| \leq |Y|$?
I know that for any finite $n$, $|\mathbb{N}^n| \leq |\mathbb{N}|$, in other words, the set of all $n$-tuples of natural numbers is no bigger than the set of natural numbers and $|\mathbb{R}^\mathbb{N}| = |\mathbb{R}|$, the set of sequences of real numbers is no bigger than the set of real numbers.

答: $|X|=\aleph_0,\,|Y|=\aleph_1$ then $|Y|\le2^{\aleph_0}$ in $\mathsf{ZFC}$, so that $|Y^X|=2^{\aleph_0}$, giving a counterexample in $\mathsf{ZFC+\neg CH}$. More generally, in $\mathsf{ZFC+\neg GCH}$ some ordinal $\alpha$ satsifies $2^{\aleph_\alpha}>\aleph_{\alpha+1}$, so $|X|=\aleph_\alpha,\,|Y|=\aleph_{\alpha+1}\le2^{\aleph_\alpha}$ provides a similar counterexample. By contrast, in $\mathsf{ZFC+GCH}$ ordinals $\alpha<\beta$ satisfy $|X|=\beth_\alpha,\,|Y|=\beth_\beta,\,|Y^X|=|Y|$. Therefore, the conjecture is equivalent in $\mathsf{ZFC}$ to $\mathsf{GCH}$.


问:
Reference recommendation: Mackey Topological in terms of Compactness and Completeness and Dunford-Pettis on Topological groups
Unfortunately, I have to work with Mackey Topology on a topological group under the duality theory. My primary reference is Topological Vector Space, but it is concise. I have searched for literature; it is scattered and seems not so much. I am interested in completeness, compactness, and, most importantly, Dunford-Pettis properties. Any help is needed.</p>

答:
Köthe's Toplogical Vector Spaces I is a pretty good book, where $\S21$ seems to cover the Mackey topology fairly well.  It is a little old though and I doubt it covers anything on Dunford-Pettis.

Bogachev is another good author for these kinds of things, so you could try Topological Vector Spaces and Their Applications by him and Smolyanov though you might find some of the results you are interested in relegated to the exercises.

I don't have it to hand to check, but I also seem to recall Treves's Topological Vector Spaces being comprehensive, so assuming you're checking all these in a library, that might also be worth a look.


问:
Conceptual doubt in reading the domain of a relation
I can already see the heavy downvotes but any help will be appreciated.

Let there be a relation R and the domain of R is given by
$$\{x:(x,y)\in R\}$$

How should I convey this set if I am to read it out to someone?

My try was :"the set of all x such that x comma y belongs to R"

Is this right it does not make alot of sense to me .

Thanks alot in advance.

答:
The definition of the domain of $R$ isn't quite right. It should be $$\{x:\color{red}{\exists y}((x,y)\in R)\}.$$ That is, the domain of $R$ is the set of things which are $R$-related to something; or, perhaps more smoothly, $x$ is in the domain of $R$ iff there is some $y$ such that $x$ is $R$-related to $y$. The definition you've given suggests that we're looking at some specific $y$; this would describe the $R$-preimage of the particular element $y$ (or of the particular set $\{y\}$ - there's some abuse of terminology here), but that's not the whole domain of $R$ in general.

(This is a situation where natural language is a bit clunky, unfortunately.)


问:
How to compute next/previous representable rational number?
An (approximate) non-negative rational number representation is a pair of natural numbers each not greater than some fixed limit M (and of course denominator being non-zero).

With this condition there is finite number of representable rational numbers. This means that for each such number we can name previous and next number in the set (of course except for smallest 0 and largest 1). How to compute them?

In other words lets have a set of numbers

$$
R(M)=\{\frac{p}{q} : p,q\in\mathbb{N},q\neq0,p\leq q\leq M\}
$$

where $M\in\mathbb{N}_+$

and given number $\frac{p_1}{q_1}\in R(M)$.

I want to find numbers $\frac{p_S}{q_S}\in R(M)$ and $\frac{p_L}{q_L}\in R(M)$ (if they exist) such that

$$
\frac{p_S}{q_S}<\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_S}{q_S}<\frac{p}{q}<\frac{p_1}{q_1}
$$

and smilarly

$$
\frac{p_L}{q_L}>\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_1}{q_1}<\frac{p}{q}<\frac{p_L}{q_L}
$$

答:
As I mentioned in a closely related question, this has a very beautiful geometrical interpretation employing Pick's Area Theorem: $\rm\: a/b < c/d\:$ are two adjacent terms in a Farey sequence iff the triangle $\rm T$ with vertices $\rm (0,0), u = (b,a), v = (d,c)$ contains no lattice points except for vertices. By Pick's formula this is true iff  $\rm\ area\ T = $ #interior_points $\ +\ 1/2\ $ #boundary_points $\:-\ 1\ =$ $\ 0 + 3/2 - 1 = 1/2\:.$ But by basic analytic geometry $\rm\: area\ T\: =\: |det(u,v)|/2 = |ad-bc|/2\:.\:$ Hence $\rm\:a/b,\:c/d\:$ are meighbors in some Farey sequence iff $\rm\:|ad-bc| = 1\:.\:$

Therefore the problem reduces to solving the linear Diophantine equation $\rm\:ax+by = 1\:.\:$ As is well-known, such equations may be solved by employing the extended Euclidean algorithm, which is conveniently implemented by performing the Euclidean algorithm in parallel on an identity-augmented system of equations. For an example worked-out in detail see my post here.

In fact it deserves to be much better known that Pick originally applied his area formula in a similar way to give a beautiful geometric proof of the Bezout linear representation of the gcd.


问:
Vector Field Generating Variation Along Curve
I'm learning a proof of the fact that length extremising curves are geodesics of the Levi-Civita connection, and have found something I don't understand. The argument states the following.

Suppose $\gamma$ is an arbitrary curve in a manifold $M$ and $N$ is an arbitrary vector field along $\gamma$ which vanishes at the endpoints. Then there must exist a variation $\tilde{\gamma}$ of $\gamma$ such that $\gamma'$ and $N$ are the pushforwards of the coordinate vector fields on the domain of the variation.

I can't see how this is true, because it would immediately imply that all vector fields commute (I think!). Am I missing something or is it false? If it's wrong, does anyone know how to prove that statement that length extremizing implies geodesic.

The relevant lectures notes from which I've taken this argument are here, page 32. Thanks in advance!

答: To answer your point in the third paragraph: No, it only implies that vector fields that are push-forwards of (constant multiples of) coordinate vector fields commute. True enough, two linearly independent vector fields can, in appropriate coordinates, be coordinate vector fields if (and only if) their Lie bracket vanishes.


问:
Is it possible $\sin\alpha=\cos\alpha=1?$
Is it possible $\sin\alpha=\cos\alpha=1?$

We have defined trigonometry using a unit circle. The point $X(1;1)$ does not lie on the trig circle, right? Does this mean that $\alpha$ does not exist?

答:
The answer you already gave is the clearest in my opinion. The trigonometric functions can be defined using the unit circle, so that $\sin(\theta)$ represents the $y$-coordinate when you have travelled $\theta$ degrees/radians anticlockwise around the unit circle, starting from the point $(1,0)$. Likewise, $\cos(\theta)$ represents the $x$-coordinate when you have travelled $\theta$ degrees/radians around.

Since the equation of a unit circle is $x^2+y^2=1$, and $(x,y)=(\cos(\theta),\sin(\theta))$, we obtain the identity
$$
\cos(\theta)^2+\sin(\theta)^2=1 \, ,
$$
which math's answer references. This is one way to answer your question. More directly, we know that the point $(1,1)$ does not lie on the unit circle, since the distance from the origin to that point is $\sqrt{2}$. Hence, it is never true that $\sin(\theta)=\cos(\theta)=1$.


问:
equivalent statements about fully idempotent
Let $R$ be a ring with identity. Show that the following are equivalent:

(a) $R$ is fully idempotent (that is, $I^2 = I$ for every ideal $I$ in $R$);

(b) for ideals $I$ and left ideals $A$ in $R$, we have $A \cap I \subset AI$;

(c) for ideals $I$ and left ideals $A$ in $R$ with $A \subset I$, we have $A \subset AI$.

(b) $\Rightarrow$ (c) is clearly.

(c) $\Rightarrow$ (b) Let $I$ be an ideal and $A$ a left ideal in $R$. Then $A \cap I$ is a left ideal in $R$ with $A \cap I \subset I$. It follows from (c) that
$$A \cap I \subset (A \cap I)I \subset AI.$$

(c) $\Rightarrow$ (a) Let $I$ be an ideal in $R$. It follows from (c) that $I \subset I^2$. Since always $I^2 \subset I$, we have $I = I^2$. So $R$ is fully idempotent.

But I can't prove that (a) $\Rightarrow$ (b). Can somebody help me?

Thank you.

答:
If $A\subseteq I$, then $AR\subseteq I$. Now $AR$ is an ideal. Assuming (a), $A\subseteq AR=(AR)^2\subseteq ARI=AI$

This shows (a) implies (c), completing your equivalence.


问:
Shortest distance and calculus of variation.
Every answer to the question "show that the shortest path between two points on $\mathbb{R}^2$ is the line segment joining them", seems to rely on some version of calculus of variation.

My problem with that approach is that most of them make the assumption that the path is the graph of a smooth function, rather than a generic curve. Okay, we may exclude the possibility of self-intersection, but how can we eliminate the possiblity that the curve may not be the graph of a function or not (piecewise) smooth?

答:
Not true. See: Wikipedia article: we normally define the length of a curve as a (potentially infinite) supremum of all the lengths (in the "standard" sense) of polygonal lines joining the beginning and the end of the curve and "inscribed" in the curve. We say that the curve is "rectifiable" if this supremum is finite.

With that definition, the fact that the straight line is the shortest follows pretty much immediately from the definition. Pretty much, the only slightly nontrivial bit is to prove that the length of the straight line (in this new sense) is the same as the length of the straight line (in the standard sense). This is true because, for every polygonal line "inscribed" in a straight line, the (standard) length of it is exactly the same as the (standard) length of the original straight line.

This consideration works in any $\mathbb R^n$. However, it does not scale well to other manifolds, because those may be locally homeomorphic with $\mathbb R^n$ but not locally isometric with $\mathbb R^n$. The common way of adding metric to manifolds (which is: adding a metric tensor) requires differentiability of the maps between charts in the atlas (i.e. the manifold must be differentiable), in order for the metric tensor to allow for a change in co-ordinates when switching from one chart to another. That approach scales a lot better, but requires differentiability, and in that approach, the length of the shortest path is calculated using calculus of variations. This is probably what you had in mind in your question.


问:
Is a diagonalization of a matrix unique?
I was solving problems of diagonalization of matrices and I wanted to know if a diagonalization of a matrix is always unique? but there's nothing about it in the books nor the net.

I was trying to look for counter examples but I found none.

Any hint would be much appreciated

Thanks!

答: Actually it is not unique. Even without changing the order of eigenvalues. For example, the matrix $$A=\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0& -1 \\ 1 & 0 & -1 \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}1&1& 0 \\ 0 & 0 & 1 \\ -1&1&0\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{2}}&0& -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}& 0 \\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\end{pmatrix}.$$


问:
linear transformation- cant continue the proof
as part of studying for an exam i encountered a proof question:

Let vector spaces $U,V$ over a field $F$ and Linear maps $T,S:U\rightarrow V$ be given.

Prove that there exist bases $B_1$,$B_2$ of $U$ and $C_1,C_2$ of $V$ such that $[T]$ from $B_1$ to $C_1$ = $[S]$ from $B_2$ to $C_2$  if and only if $rank(T)=rank(S)$.
Here the rank of a linear map is defined as the dimension of its image: $rank(T)=dim(Im(T))$.

I know that in order to prove this i need to show both ways.

for the first way, i know that if $[T]=[S]$ than it doesnt matter in in what basis the matrixes are, the ranks of the matrixes will be equal, therefore $rank(T)=rank(S)$.

for the other way, i know that $rank(T)=rank(S)$ $\Rightarrow$ $dim(Im(T))=dim(Im(S))$ $\Rightarrow$ from the rank-nullity theorom i will know that $dim(ker(T))=dim(ker(S))$.

from the point that the dimensions of the kernels are equal i dont know how to continue the proof. i would like someone to continue the proof for me. and i dont know if the first way is true as well so if someine can tell me if it is true

答:
Possibly overkill if you don't know the SVD:

By the singular value decomposition (SVD), there exists orthonormal bases $B_1, B_2$ of $U$ and $C'_1, C'_2$ such that $[T]_{B_1, C'_1}$ and $[S]_{B_2, C'_2}$ are [rectangular] diagonal with nonnegative entries on the diagonal, called singular values. These diagonal entries can be assumed to be nonincreasing, and the number of nonzero entries is the rank of the matrix.

By scaling the basis vectors of $C'_1$ and $C'_2$ by the corresponding singular values, you can get new (no longer orthonormal) bases $C_1, C_2$ for $V$ such that $[T]_{B_1, C_1}$ and $[S]_{B_2, C_2}$ are diagonal matrices whose diagonal entries are $1$s and $0$s (in nondecreasing order). Since the number of $1$s is the rank of each matrix, the two matrices must be the same.

Sketch of alternate solution:

Applying row operations to a matrix (e.g., Gaussian elimination) is equivalent to modifying a basis for the codomain $V$. In particular, if you know Gaussian elimination can bring a matrix into RREF form, then you immediately know that there exists a basis $C_1$ for the codomain $V$ for which the matrix of $T$ is in RREF form. Similarly, column operations are equivalent to modifying a basis for the domain $U$. You can apply column operations to the RREF matrix to obtain a rectangular diagonal matrix with only $1$s and $0$s, where the number of $1$s is the rank of $T$, so there exists a basis $B_1$ for $U$ (along with the basis $B_2$ for $V$) such that the matrix of $T$ is this kind of matrix. Doing the same for $S$ will bring you to the same matrix since their ranks are the same.


问:
Avoid Loss of significance error
I want to compute $f(x)$to avoid loss of significance$$f(x)=\frac{1-\cos x}{x^2}$$

One way is writing Taylor series approximation for $\cos x$ about $x=0$. what about this approach:

$$f(x)=\frac{2\sin^2(\frac x2)}{x^2}$$

Is this ok ?

答:
when you say "compute" I assume you mean without using standardised functions, so how would you then continue from this $\sin^2$ term for which the series will not be as pleasant. In terms of using the first series we know that:
$$1-\cos(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}x^{2n}$$
and so:
$$f(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n)!}x^{2(n-1)}$$
However if you mean to take values where $x$ is very small I think the second method is good as approximations will show the value at $x=0$

Title: Avoid Loss of significance error


问:
Prove that $f(x)$ is everywhere finite and everywhere locally unbounded.
Consider the following. If $f(x)$ is rational function so that $x=\frac{m}{n}$, where $m$ and $n$ are integers such that the fraction $\frac{m}{n}$ is in lowest terms and $n>0$, then $m$ and $n$ are uniquely determined. Therefore the following function is well-defined:

$$f(x)= \left\{ \begin{array}{11} n & x \in \mathbb{Q}, \,\, x=\frac{m}{n} \,\, \text{in lowest terms}, n>0 \\ 0 & x \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right.$$

Prove that this function is everywhere finite and everywhere locally unbounded.

Hint: Start the second part by defining "locally bounded."

I would appreciate if anyone could define what everywhere finite means. It seems to me that this function is everywhere finite since it only outputs values of $n$ when $n>0$ and $0$ otherwise. We know since $\frac{m}{n}$ is in lowest terms, then a lot of fractions will repeat.

I then defined a locally bounded function $f(x)$. Locally bounded function $f(x)$ means for any $x$, $\exists$open set $O$ such that $x \in O$ such that $f(O)$ is a bounded set. In other words, for some number $M$, $|f(x)| \leq M \,\, \forall x \in O$.

I tried using the Archimedean Property to show that now function value can be bounded.

答:
First approach:

Assume $f$ is bounded in open $\varepsilon$-neighbourhood of $a$, then obviously the denominators of all rational numbers in this neighbourhood must be bounded (i.e. $n$ for $x=\frac mn$). This implies that the numerators must also be bounded (i.e. $m$ for $x = \frac mn$), otherwise $\frac mn$ would not lie in $\varepsilon$-neighbourhood of $a$ for large values of $m$. But this would mean there are only finitely many rational numbers in that neighbourhood of $a$. This contradiction proves the claim. $\square$

Second approach:

Let's look at the Thomae's functions first:
$$g(x) = \begin{cases}\frac 1n, \ \ x = \frac mn \\ 0, \ \ x \in \mathbb R \setminus \mathbb Q \end{cases}$$ You can find a proof at the Internet that the limit of Thomae's function vanish at all points (here is one at mathcounterexamples). Therefore,
$$f(x) = \begin{cases}\frac{1}{g(x)}, \ \ x = \frac mn \\ 0, \ \ x \in \mathbb R \setminus \mathbb Q \end{cases}$$ is unbounded at any point.


问:
Understanding blow up of curves
I am trying to desingularize the curve $V=V(y^n-x^d)$, where $n>d$ and $gcd(n,d)=g$. It is singular at $P(0,0)$ so I am trying a blow-up.

My intuition is that a blow-up is locally like $(x,y)\mapsto (xy, y)$ or $(x,y)\mapsto (x, xy)$. So I will obtain still singular curves: $y^n-(xy)^d$ that is $y^{n-d}-x^d$, and in the same way I will arrive to $y^{n-d_1d}-x^d$ with $d_1$ the entire divisor of $d$ and $n$,  so $n-d_1 d=r_1$ is the rest. Like in the Euclid's algorithm, I will come to something like $y^g-1$.

I am stuck and I need some help as a hint or a reference

答:
$g=\gcd(n,d)$ and $n>d>0$.

The blow up of $V(y^n-x^d)$ at $(0,0)$

(the Zariski closure of $\{((x,y), [u:v]), xu+yv=0,y^n-x^d=0,(x,y)\ne (0,0)\}$)

is isomorphic to $V(Y^{n-d}-X^d)$ (where $(x,y)=(XY,X)$)

repeating several times we'll arrive at $V(W^g-Z^g)$.

(in characteristic $0$) $V(W^g-Z^g)$ is smooth if $g=1$, otherwise it is an union of several smooth curves intersecting at $(0,0)$.

Title: Understanding blow up of curves


答:
Yes, your formulation is perfect for the job.  While the Taylor series approach is also good, using the half-angle sine is easier to program and equally accurate.

As a further exercise, apply the Taylor series expansion

$\sin(x/2)=(x/2)-(x/2)^3/3!+(x/2)^5/5!-...$

and square the results through the $x^6$ term.  Compare this result with what the direct Taylor series expansion for $1-\cos x$ would give.

Title: Avoid Loss of significance error

Body:
I want to compute $f(x)$to avoid loss of significance$$f(x)=\frac{1-\cos x}{x^2}$$

One way is writing Taylor series approximation for $\cos x$ about $x=0$. what about this approach:

$$f(x)=\frac{2\sin^2(\frac x2)}{x^2}$$

Is this ok ?


问:
An elementary problem on function
The problem is:

let $p(x)$ be a polynomial with real Coefficients that is non-negative for all $x$. Prove that for some $k$, there are polynomials $f_1 (x),\cdots f_k (x)$ such that $p(x)=\sum_{i=0}^k (f_i (x))^2$.

I have tried a lot but don't able to make any substantial progress so that I can state it, I will be happy if anyone help how to approach this kind of problem or with any possible solutions. Thanks.

Edit  : thanks to those who tried to help with this $i=2$ version but I don't able to convert it into solution so ask for further help or any references.

答:
Since $P$ is positive, all its real roots have an even multiplicity.

So we can write $P = \prod (X-\gamma_i)(X-\overline{\gamma_i}) \prod (X-x_k)^2$, where the $\gamma_i$ are the complex roots, and the $x_k$ the real (eventually equal).

Then if we call $A = \prod (X-\gamma_i)$, we have $\overline{A} = \prod (X-\overline{\gamma_i})$. So that if we note $C = \prod (X-x_k)$, we have $$ P = AB\cdot \overline{A}B$$

Since $B$ is a real polynomial, $B=\overline{B}$.

So $P=AB\cdot\overline{AB}$.

If we note $AB = R+iQ$ with $R$ and $Q$ real polynomials, we have : $$P=R^2 + Q^2$$

So for the exercise, we can notice that $k=1$ or $2$.

Title: An elementary problem on function


问:
$+$- and $\Sigma$-types in modal type theory
I am reading Bierman & de Paiva's paper on intuitionistic modal logic. They offer a typed version on page 401. I wonder whether the result generalizes to $\Sigma$- and $+$-types in Martin-Löf type theory. In particular, I wonder whether the T-rule $\Box A\rightarrow A$ generalizes to the two mentioned types. In other words, can we have the following rules?

$$\Box (A+B)\rightarrow (A+B),\quad\quad\quad\Box\Sigma x:A.B(x)\rightarrow\Sigma x:A.B(x).$$

答:
There's quite a bit of research on modalities in type theory, but I'm uncertain what exactly what you're looking for.

I think some of the oldest stuff would be by Frank Pfenning and Rowan Davies. I believe they develop modal type theory with a view toward staged computataion, where $\square A$ is like an internal classification of closed terms of type $A$. I'm not sure if any of their work has $Σ$ and $+$ specifically, but once that is your mindset, it's not much of a surprise that $\square A → A$ works for them (you get an $A$ value by actually reducing the $\square A$ term).

If you're looking to read about modal type theory from the ground up, I'd look into their work, because it's probably the simplest to start with. A Judgmental Reconstruction of Modal Logic might be a good precursor. A Modal Analysis of Staged Computation should get into some of the ideas I was talking about above. I'm afraid I'm not familiar with all their work on the subject, so there might be additional/better papers I'm not aware of.

If you merely want to see something like MLTT with modalities, I found this. It uses a different presentation, though, and looks quite a bit more complex. Of course, some of that complexity is due to questions that necessarily arise with dependent types and modalities (like, how should equality/identity act for values of modal type). If you get really interested in modal type theory, there is also this which seems like a pretty comprehensive treatment of certain sorts of modal type theories (unfortunately, 'modality', is a vague enough term that it's impossible to concretely describe a generalized framework encompassing modal theories; also unfortunately, I haven't gotten around to reading it thoroughly yet, but it seems great).

Hopefully that's enough to get you started.


答:
WLOG $p$ is monic. First, $\deg p$ is even. So we can pair off the roots in complex conjugate pairs (note that any real root has even multiplicity as $p\geq0$).

So we can write $p(x)=\prod (x-a_k)(x-\bar{a_k})=q(x)\overline{q(x)}$, where $q(x)=\prod(x-a_k)$. Then there exist real polynomials $f,g$ such that $q=f+ig$. Then $$p=(f+ig)(f-ig)=f^2+g^2.$$

Title: An elementary problem on function

Body:
The problem is:

let $p(x)$ be a polynomial with real Coefficients that is non-negative for all $x$. Prove that for some $k$, there are polynomials $f_1 (x),\cdots f_k (x)$ such that $p(x)=\sum_{i=0}^k (f_i (x))^2$.

I have tried a lot but don't able to make any substantial progress so that I can state it, I will be happy if anyone help how to approach this kind of problem or with any possible solutions. Thanks.

Edit  : thanks to those who tried to help with this $i=2$ version but I don't able to convert it into solution so ask for further help or any references.


问:
Thesis in Machine Learning BUT...

Context

I am a final year Bachelor of Mathematics student and next semester I will write my Bachelor thesis.

My interests are in Machine Learning (ML) and I will do a master in ML next year. More specific sub-fields I like are

Deep Learning

Computer Vision

Natural Language Processing

Reinforcement Learning

And my interests outside of ML and mathematics include

Self-driving cars (e.g. Tesla)

Rocket and space exploration

more vaguely, I find tech interesting as a whole.

Question

I am looking for a thesis which would bring me as close as possible to the field of ML. Do you have topic recommendations ?

BUT my Bachelor is in Mathematics therefore I shall not write a thesis in Computer Science as it would not be accepted by my study director.

Some thoughts

I know some people who were in my case. One of them for instance discovered and proved some convergence results in the context of Gradient Descent. Maybe this will inspire you with your answers.

Thanks in advance!

答:
Great question. I can think of a couple of topics. (1) Humans exploit symmetry in a pretty big way to make learning vastly more efficient. The canonical example is that objects "are the same" whatever their size and rotation. Can you somehow tie symmetry into the learning in a fundamental way? I'm sure that people have already worked on that, so you'll to see what's been done already and see how you can take it farther.

(2) Bayesian inference is a pretty general framework for reasoning under uncertainty, but the application of it is hampered by the need for approximations for multidimensional integrals, which are typically very expensive to compute. Can you expand the class of models for which exact results are known? For example, can you derive some exact results for the posterior distribution of the parameters of a system which is governed by differential equations.

Good luck and have fun!

Title: Thesis in Machine Learning BUT...


问:
Limit of a function with an undefined feature
How is it possible to define $\lim_{x \rightarrow a} f(x)$ if $f(x)$ is undefined at $a$? There is an infinite amount of $x$-values on either side of $x=a$, and without a boundary or a strategically-placed, removable discontinuity, isn't the limit unknown?

I am using this definition of a limit [1].

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and
only if, for each positive number $\epsilon$ there is another,
$\delta$, such
that whenever $0 < |x-a| < \delta $ we have $|f(x) - A|< \epsilon$.
That is, when $x$ is near $a$ (within a distance $\delta$
from it), $f(x)$ is near $A$ (within a distance $\epsilon$ from it).
In symbols we write $\lim_{x \rightarrow a} f(x) = A$.

[1] David V. Widder. Advanced Calculus. Dover 1989.

答:
The limit of $f$ at a point $a$ is defined as
\begin{equation}
\lim_{x\to a}f\left(x\right) = L \longleftrightarrow \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-a \rvert < \delta \longrightarrow \lvert f\left(x\right) - L \rvert < \epsilon.
\end{equation}
You can see that the condition $0 < \lvert x-a \rvert < \delta$ does not require the definition of $f$ at $a$. Whenever you would like to tell if $L$ exists, you may first consult the definition. If it is not straigtforward to tell from the definition, you may use the Cauchy theorem of limit.

Your question is based on intuitive understanding of math, which is the type of math you learned at high school. Unfortunately, contemporary math is based on bottom-level logical systems. Using rigorous logical language is the gap between math in high school and math in college.


答:
Yes, this is okay as the sine close to zero is computed as $xh(x^2)$, so that computing $\frac{\sin x}{x}$ is (almost) as exact as the implementation of $h(x^2)$. Plotting shows that indeed the original formula has large errors for $x\approx 10^{-8}$.

The difference plot to the 8th degree Taylor polynomial

shows that both alternative formulas have errors in the magnitude of the machine precision.

Title: Avoid Loss of significance error


答:
A limit uses the values of $f(x)$ everywhere except at $x=a$ ! Whether $f(a)$ is defined or not and whether $f(a)$ coincides with the limit or not is irrelevant. This is expressed in the definition by $0<|x-a|$.

The goal of a limit is precisely to "guess" what $f(a)$ is should be.

Title: Limit of a function with an undefined feature

Body:
How is it possible to define $\lim_{x \rightarrow a} f(x)$ if $f(x)$ is undefined at $a$? There is an infinite amount of $x$-values on either side of $x=a$, and without a boundary or a strategically-placed, removable discontinuity, isn't the limit unknown?

I am using this definition of a limit [1].

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and
only if, for each positive number $\epsilon$ there is another,
$\delta$, such
that whenever $0 < |x-a| < \delta $ we have $|f(x) - A|< \epsilon$.
That is, when $x$ is near $a$ (within a distance $\delta$
from it), $f(x)$ is near $A$ (within a distance $\epsilon$ from it).
In symbols we write $\lim_{x \rightarrow a} f(x) = A$.

[1] David V. Widder. Advanced Calculus. Dover 1989.


问:
Proving that $\mathscr B$ is a tautology iff $\lnot \mathscr B'$ is a tautology using induction.
I have the following logic question about duality:(Introduction to mathmatical Logic by elliot mendelson , exercise $1.30$)

If $\mathscr B$ is a statement form involving only $\lnot$ , $\land$, and $\lor$, and $\mathscr B'$ results from  $\mathscr B$  by replacing each $\land$ by $\lor$ and each $\lor$ by $\land$ , show that $\mathscr B$ is a tautology if and only if $\lnot \mathscr B'$ is a tautology.

Now , at first , I had some confusion about the question itself which I resolved with another post of mine. After that , While trying to prove it , I got stuck for a while until I found another post which has the same question If $\phi$ is a tautology then dual $\phi$ is a contradiction.

There , They defined a couple of things:

$1$.In general, for any two-place logical operator $\times$, the dual expression of $\phi \times \psi$ is $\phi^d \times^d \psi^d$, where $\times^d$ is the dual operator of $\times$, which you can find by: $\phi \times^d \psi \Leftrightarrow \neg (\neg \phi \times \neg \psi)$.

$2$.Define the complement statement $\phi'$ of a statement $\phi$ to be that statement that puts a negation in front of every atomic statement in $\phi$. Recursively:

$A' = \neg A$ for any atomic $A$

$(\neg \phi)' = \neg (\phi)'$

$(\phi \times \psi)' = \phi' \times \psi'$

Also the answer states:"you should be able to prove (by induction of course) that $\phi^d\Leftrightarrow \neg \phi'$.  And from that, the desired result follows fairly quickly."

I was able to prove that $\phi^d\Leftrightarrow \neg \phi'$ .But I stuck on figuring out how i will use this  corollary to prove that " $\phi$ is a tautology iff  $\phi^d$ is a contradiction" .Is it something really trivial I am missing out?

Edit: Here is the question again from mendelsons book but I rewrote some symbols in the context of the symbols used in this post so that it is easier to understand,

If $\phi$ is a statement form involving only $\lnot$ , $\land$, and $\lor$, and $\phi^d$ results from  $\phi$  by replacing each $\land$ by $\lor$ and each $\lor$ by $\land$ , show that $\phi$ is a tautology if and only if $\lnot \phi^d$ is a tautology.

答:
As per the post that you have linked, the Duality Theorem for boolean algebra is different:

Let $A'$ be the result of interchanging ∨ and ∧ in $A$, and replacing $P$ by $¬P$ for each atom $P$. Then $A' \Leftrightarrow ¬A$.

The proof is by structural induction.

The above proof does not apply sic et simpliciter to Mendelson's transformation.

We have to note that $P' = P$, for every atom $P$, and neither $P$ nor $P'$ are tautologies. Thus, the result holds trivially.

The same for a formula with a number of nagation whatever, but without occurrences of $\lor$ and $\land$.

Thus, the formula $A$ must contain at least one occurrence of a binary connective and one of negation.

A very simple example (maybe the simplest) of Mendelson's transformation is with formula $A = P \lor \lnot P$.

We have that $\lnot A' = \lnot (P \lor \lnot P)' = \lnot (P' \land \lnot P')=\lnot P \lor P$.

So, in the "basic case" the property holds.

If instead we consider a formula $B=P \land \lnot Q$, which is not a tautology, we have that $\lnot B'=\lnot (P \land \lnot Q)'= \lnot(P' \lor \lnot Q')=\lnot P \land Q$, and neither it is a tautology.

IMO, the result holds considering that for $A$ whatever, we may rewrite it equivalently in CNF, i.e. as a finite conjunction $C_1 \land \ldots \land C_n$ of formulas that are in turn disjunctions: $C_i= D_{i1} \lor \ldots \lor D_{im}$ of atoms or negated atoms.

We have that the CNF is a tautology iff every $C_i$ contains an atom and it negation, i.e. a "basic disjunction" $P \lor \lnot P$.

Consider now $A$ in CNF; what happens with Mendelson's transform? That $A'$ will be in DNF: a disjunction of conjuncts.

What happens when we prefix $\lnot$ to $A'$? That the and's and or's will be exchanged again when we move inside the negation sign and the new formula will be again in CNF.

What happened to the "basic disjunctions"? The first step changes $P \lor \lnot P$ into $P \land \lnot P$, while the second step changes it into $\lnot P \lor \lnot \lnot P$.

Thus, if every disjunct $D_i$ of the CNF corresponding to $A$ contains a "basic disjunction", this will be so also for the corresponding $D'_i$.


问:
Ehrenfest model limiting distribution
In my textbook (Dobrow) p.89. The stationary distribution for the Ehrenfest model with transtition matrix for $\{0,1,...,N\}$

$$P_{ij}=\begin{cases}\frac{i}{N}, &\text{if }j=i+1\\
\frac{N-i}{N}&\text{if }j=i-1\\
0 &\text{otherwise}
\end{cases} $$

is computed and shown to binomial with parameters N and 1/2. It then says without justification that there is no limiting distribution. I would like to know why this is so. Thanks.

What I've done so far:

I think I need to show that $lim_{n \to\infty}P^n_{0,0}$ does not exist. (It is also sufficient to show it for any other choice of states)

I know that $P^{2k+1}_{0,0}=0$ for all k>0. I need to show that $lim_{k \to \infty} P^{2k}_{0,0}$ either oscillates or converges to something else. I am having trouble with this. Working on small cases. $P^4_{0,0}=\left(P^2_{0,0}\right)^2+P^2_{0,0}P^2_{1,1}$. $P^6_{0,0}=\left(P^2_{0,0}\right)^3+P^4_{0,0}P^2_{0,0}+P^4_{0,0}P^2_{2,2}$

答:
$P^k_{0,0} = 0$ for any and all $k$ because in the Ehrenfest model, marbles (or fleas) can only exchange urns (or dogs) and can only increase or decrease by one. The transition matrix is linked, and no matter many times we propagate the transition matrix $P_{ii} = 0$ for any $i$.

Ehrenfest Model Transition Matrix

Title: Ehrenfest model limiting distribution


答:
I present you with the grap of the function $\;f(x)=\cfrac{\sin x}x\;$ :

Clearly the functions isn't defined at $\;x=0\;$ , yet the limit if the function when $\;x\to0\;$ is $\;1\;$ .

In words: the limit $\;\lim\limits_{x\to x_0}f(x)\;$ doesn't depend at all on whether $\;f(x_0)\;$ is defined or what its possible value could be, but only depends on the behavior of the function on an open punctured neighborhood of $\;x_0\;$ , meaning: without the point $\;x_0\;$ itself.


答:
You know that $p \in \mathbb{R}[X]$, so by factorizing it with irreducible polynomials you can always write it as :

$$p = \lambda \prod_{i=1}^r(X-x_i)^{\alpha_i}\prod_{j=1}^s(X^2+b_jX+c_j)^{\beta_j}$$

With all $x_i$ that are real and distinct, and $\forall j \in  \{1,\dots,q \}, (b_j,c_j)\in \mathbb{R}^2$ verifying $b_j^2-4c_j<0$.

If one of the $\alpha_i$ was an odd number so you could check that $p$ would change of sign near $x_i$, it implies that all $\alpha_i$ are even numbers (because you know $p$ has always the same sign). You can now check that $\lambda \geq 0$.

Now let's factorize $\displaystyle\prod_{j=1}^s(X^2+b_jX+c_j)^{\beta_j}$ in $\mathbb{C}[X]$ :
$$\prod_{j=1}^s(X^2+b_jX+c_j)^{\beta_j} = \prod_{j=1}^s(X-z_j)^{\beta_j}\prod_{j=1}^s(X-\overline{z_j})^{\beta_j}$$

But you can write $\displaystyle \prod_{j=1}^s(X-z_j)^{\beta_j} = g_1+ig_2$ where $g_1,g_2\in \mathbb{R}[X]^2$, therefore:

$$\prod_{j=1}^s(X-\overline{z_j})^{\beta_j} = \overline{\prod_{j=1}^s(X-z_j)^{\beta_j} } = \overline{g_1+ig_2}=g_1-ig_2$$

Thus:
$$\displaystyle \prod_{j=1}^s(X^2+b_jX+c_j)^{\beta_j} = \prod_{j=1}^s(X-z_j)^{\beta_j}\prod_{j=1}^s(X-\overline{z_j})^{\beta_j} = (g_1+i g_2)(g_1-ig_2) = g_1^2+g_2^2$$

Finally by taking $f_1 = \sqrt{\lambda} \left(\displaystyle \prod_{i=1}^r (X-x_i)^{\alpha_i/2}\right)g_1$ and $f_2 =\sqrt{\lambda} \left(\displaystyle \prod_{i=1}^r (X-x_i)^{\alpha_i/2}\right)g_2$, you have:

$$\boxed{f_1,f_2 \in \mathbb{R}[X] \text{ such that } p = f_1^2+f_2^2}$$

Title: An elementary problem on function


问:
Existence of solutions for the following PDE
We have that $a:\mathbb{R}^+\times \mathbb{R}\to \mathbb{R}$ be a given function such that $|a(t,x)|+|\partial_x a(t,x)|\leq M$. $\rho_0$ is the initial data where $$\rho_{|t=0} = \rho_0 \in L^1\cap L^{\infty}(\mathbb{R}),\quad \rho_0\geq 0.$$

Show that the iterative scheme
$$ \left\{ \begin{array}{lcc}
             \rho^{(0)}\in L^2((0,T)\times \mathbb{R}),\quad \partial_x \rho^{(0)}\in L^2((0,T)\times \mathbb{R}) \\
             \\ \partial_t \rho^{(n+1)} - \partial_{xx}^2 \rho^{(n+1)} = -\partial_x (a\rho^{(n)}) =   -\rho^{(n)}\partial_x a - a \partial_x \rho^{(n)}\\
             \\ \rho^{(n+1)}_{|t=0} = \rho_0
             \end{array}
   \right.$$
is well defined.

Show that it is contracting on $C^0 ([0; T]; L^2 (\mathbb{R}))$ endowed with norm
$$N(u)^2 = \sup_{0\leq t\leq T< \infty}\left(e^{-\gamma t}\int_\mathbb{R}|u(t,x)|^2 dx\right)$$
for a well chosen value of $\gamma> 0$.

My fisrt question is, what is the meaning of iterative schema well defined? and how to test it?

For the second item, I have the following idea:

I worked with
$$\small \partial_t (\rho^{(n+1)}-\rho^{(n)}) - \partial_{xx}^2( \rho^{(n+1)} - \rho^{(n)}) = -(\rho^{(n)}-\rho^{(n-1)})\partial_x a - a \partial_x (\rho^{(n)}-\rho^{(n-1)})$$
then, multiplying by $(\rho^{(n+1)}-\rho^{(n)})$ and integrating with respect to $x$ we have that
$$\scriptsize \frac{d}{dt}\int_\mathbb{R}\frac{|\rho^{(n+1)}-\rho^{(n)}|^2}{2}dx + \int_\mathbb{R}|\nabla(\rho^{(n+1)}-\rho^{(n)})|^2 dx = -\int_\mathbb{R} (\rho^{(n+1)}-\rho^{(n)})(\rho^{(n)}-\rho^{(n-1)})\partial_x a dx -\int_\mathbb{R} a(\rho^{(n+1)}-\rho^{(n)})\partial_x(\rho^{(n)}-\rho^{(n-1)}) dx$$

and from this expression, we will find majorities for the expression on the right until we can express a Gronwall lemma inequality.

答:
Concerning (1), I think the question is asking you to check that if $\rho^{(n)} \in L^{2}((0,T) \times \mathbb{R})$ and $\partial_{x} \rho^{(n)} \in L^{2}((0,T) \times \mathbb{R})$, then $\rho^{(n + 1)}$ is well-defined (i.e. there's a solution of the equation) and has the same property.  (I won't give the details, but it looks correct to me.)

Concerning (2), notice that, given $u \in C([0,T]; L^{2}(\mathbb{R}))$, $N_{\gamma}(u)$ can be written as
\begin{equation*}
N_{\gamma}(u)^{2} = \sup \left\{ \int_{-\infty}^{\infty} |u_{\gamma}(t)|^{2} \, dx \, \mid \, t \in [0,T] \right\}.
\end{equation*}
Here $u_{\gamma}$ is defined by $u_{\gamma}(t) = e^{-\gamma t/2} u(t)$.  Hence one way to proceed is to study $\rho_{\gamma}$ instead of $\rho$.

Rather than the scheme, let's show that the solution map $S : \mathcal{S} \to \mathcal{S}$ is a contraction.  Here $\mathcal{S}$ is the space of possible solutions
\begin{equation*}
\mathcal{S} = \{\rho \in C([0,T]; L^{2}(\mathbb{R})) \, \mid \, \rho \restriction_{t = 0} = \rho_{0}\}
\end{equation*}
and, given $\eta \in C([0,T];L^{2}(\mathbb{R}))$, we define $\rho = S\eta$ as the solution of the equation
\begin{equation*}
\left\{ \begin{array}{r l}
\partial_{t} \rho - \partial^{2}_{xx} \rho = - \partial_{x}(a \eta) & \text{in} \, \, (0,T) \times \mathbb{R} \\
\rho \restriction_{t = 0} = \rho_{0} & \text{on} \, \, \{0\} \times \mathbb{R}
\end{array} \right.
\end{equation*}
We want to show that, for an appropriate choice of $\gamma > 0$, there is an $\alpha_{\gamma} \in (0,1)$ such that $N_{\gamma}(S\eta - S\beta) \leq \alpha_{\gamma} N_{\gamma}(\eta - \beta)$.

Fix $\eta, \beta \in \mathcal{S}$.  Let $\rho^{\eta} = S\eta$ and $\rho^{\beta} = S \beta$ and let $\mu = \rho^{\eta} - \rho^{\beta}$ be the error.  If we set $\gamma > 0$, then $\mu_{\gamma}(t) = e^{-\gamma t/2} \mu(t)$ solves the equation
\begin{equation*}
\left\{ \begin{array}{r l}
\partial_{t} \mu_{\gamma} - \partial^{2}_{xx} \mu_{\gamma} + \frac{\gamma}{2} \mu_{\gamma} = - e^{-\gamma t/2} \partial_{x}(a (\eta - \beta)) & \text{in} \, \, (0,T) \times \mathbb{R} \\
\mu \restriction_{t = 0} = 0 & \text{on} \, \, \{0\} \times \mathbb{R}
\end{array} \right.
\end{equation*}
If we multiply the equation by $\mu_{\gamma}$ and integrate-by-parts in the right-hand side, we find
\begin{equation*}
\frac{1}{2} \frac{d}{dt} \left\{ \int_{-\infty}^{\infty} |u_{\gamma}(t)|^{2} \, dx \right\} + \int_{-\infty}^{\infty} |\partial_{x} u_{\gamma}(t)|^{2} \, dx = e^{-\gamma t/2} \int_{-\infty}^{\infty} a (\eta - \beta) \partial_{x} \mu_{\gamma} \, dx - \frac{\gamma}{2} \int_{-\infty}^{\infty} |u_{\gamma}(t)|^{2} \, dx
\end{equation*}
By Young's inequality, given $\nu> 1/2$, we have
\begin{equation*}
e^{-\gamma t/2} \int_{-\infty}^{\infty} a (\eta - \beta) \partial_{x} \mu_{\gamma} \, dx \leq \frac{1}{2 \nu} \int_{-\infty}^{\infty} |\partial_{x} \mu_{\gamma}|^{2} \, dx + \frac{\nu}{2} e^{-\gamma t} \int_{-\infty}^{\infty} a^{2} |\eta - \beta|^{2} \, dx
\end{equation*}
and, thus, the function $F_{\gamma}(t) = \int_{-\infty}^{\infty} |u_{\gamma}(t)|^{2} \, dx$ satisfies
\begin{equation*}
F_{\gamma}'(t) \leq - \gamma F_{\gamma}(t) + \frac{\nu M}{2} e^{-\gamma t} \int_{-\infty}^{\infty} |\eta(t) - \beta(t)|^{2} \, dx.
\end{equation*}
Therefore, we can apply Gronwall's Lemma to find
\begin{equation*}
F_{\gamma}(t) \leq \frac{\nu M}{2} \int_{0}^{t} e^{- \gamma(t - s)} \cdot e^{-\gamma s} \int_{-\infty}^{\infty} |\eta(s) - \beta(s)|^{2} \, dx \, ds \leq \frac{\nu M}{2} \int_{0}^{t} e^{-\gamma (t - s)} \, ds \cdot N_{\gamma}(\eta - \beta)^{2}.
\end{equation*}
which gives
\begin{equation*}
N_{\gamma}(S\eta - S \beta)^{2} = \sup \left\{ F_{\gamma}(t) \, \mid \, t \in [0,T] \right\} \leq \frac{\nu M}{2 \gamma} N_{\gamma}(\eta - \beta)^{2}.
\end{equation*}
Hence we get $\alpha_{\gamma}^{2} = \frac{\nu M}{2 \gamma} < 1$ as soon as $\gamma > \frac{2}{\nu M}$.

Given the above, notice that $N_{\gamma}(\rho^{(n + 1)} - \rho^{(n)}) \leq \alpha_{\gamma} N_{\gamma}(\rho^{(n)} - \rho^{(n - 1)})$ and, thus, $N_{\gamma} (\rho^{(n + 1)} - \rho^{(n)}) \leq \alpha_{\gamma}^{n} N_{\gamma}(\rho^{(1)} - \rho^{(0)})$.  This implies $\{\rho^{(n)}\}_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathcal{S}$ and it approaches its limit exponentially fast.


问:
Find $e^{At}$ where $A=\left(\begin{smallmatrix} -7&8&2\\-4&5&1\\-23&21&7 \end{smallmatrix}\right)$
Let $$A=\begin{pmatrix}
-7&8&2\\-4&5&1\\-23&21&7
\end{pmatrix}$$

Find $e^{At}$.

I've found that $p(\lambda)=(\lambda-1)(\lambda-2)^2$

$\underline{\lambda=1}$

$$\begin{pmatrix}
-8&8&2\\-4&4&1\\-23&21&6
\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\to v_1=\begin{pmatrix}3\\1\\8\end{pmatrix}$$

$\underline{\lambda=2}$

$$\begin{pmatrix}
-9&8&2\\-4&3&1\\-23&21&5
\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\to v_2=\begin{pmatrix}2\\1\\5\end{pmatrix}$$

$\begin{pmatrix}
-9&8&2\\-4&3&1\\-23&21&5
\end{pmatrix}=\begin{pmatrix}2\\1\\5\end{pmatrix}\to v_3=$some vector

setting $C=\begin{pmatrix}|&|&|\\v_1&v_2&v_3\\|&|&|\end{pmatrix}$

Then the solution is $$e^{At}=C\begin{pmatrix}e^{t}&0&0\\0&e^{2t}&te^{2t}\\
0&0&e^{2t}\end{pmatrix}C^{-1}$$

My question is if $\begin{pmatrix}e^{t}&0&0\\0&e^{2t}&te^{2t}\\
0&0&e^{2t}\end{pmatrix}$ has the right form?

答:
You cannot set $v_3$ to be arbitrary, you must pick $v_3$ to be such that conjugating gives you a matrix of the form

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}.$$

But if you do, you will have found the Jordan Normal Form of the matrix.  If $A$ is a sum of Jordan blocks, then $e^{At}$ will also be a sum of blocks, and to figure out what those blocks are, it suffices to understand what $e^{At}$ is when $A=J_n(\lambda)$. is a single Jordan block.  There is a nice trick to this: When can write $J_n(\lambda)=\lambda I + N$ where $N$ is the matrix of $1$s just above the diagonal.  Since $\lambda I$ and $N$ commute, we can actually apply the bionomial theorem to compute powers of $J_n(\lambda)$ (which does not in general work for calculating powers of sums of matrices).

Before we can do this, we need a small calculation: $N^k$ will be the matrix which is all $0$ except for $1$s along the diagonal $k$ above the main diagonal (or $0$ everywhere if $k\geq n$).

I will not work out the general case here, but let us see what it gives in your particular case, where $\lambda=2$ and $N^2=0$.

$$(2I+N)^n=\sum_{i+j=n}\binom{n}{i}2^iN^j=2^nI+n2^{n-1}N.$$

Then

$$\begin{align}e^{t(2I+N)}&=\sum_{n\geq 0} \frac{t^n}{n!}(2I+N)^n \\
             &=\sum_{n\geq 0} \frac{t^n}{n!}(2^nI+n2^{n-1}N)\\
             &=\sum_{n\geq 0} \frac{(2t)^n}{n!}I+\sum_{n\geq 0} \frac{n2^{n-1}t^n}{n!}N\\
             &=e^{2t}I+te^{2t}N.
\end{align}$$

The upshot of this calculation is that yes, your answer is correct (if you make the proper choice of $v_3$). It is a worthwhile exercise to do this calculation for a $3\times 3$ block.


问:
How to do integral $\int\frac1{x^3+x+1}dx$?
I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$

答: Since $x^3+x+1=(x-a)((x-b)^2+c^2)$ with$$\begin{align}a&:=\sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{2/3}{\sqrt{93}-9}},\\b&:=\frac{1}{\sqrt[3]{12(\sqrt{93}-9)}}-\sqrt[3]{\frac{\sqrt{93}-9}{144}},\\c&:=\sqrt{3}\left(\frac{1}{\sqrt[3]{12(\sqrt{93}-9)}}+\sqrt[3]{\frac{\sqrt{93}-9}{144}}\right),\end{align}$$the integrand is as follows (double-check my partial fractions):$$\frac{1}{(x-a)((x-b)^2+c^2)}=\frac{1}{(b-a)^2+C^2}\left(\frac{1}{x-a}+\frac{(b-a)-(x-b)}{(x-b)^2+c^2}\right).$$So the most general antiderivative is$$\frac{1}{(b-a)^2+C^2}\left(\ln\frac{|x-a|}{\sqrt{(x-b)^2+c^2}}+\frac{b-a}{c}\arctan\frac{x-b}{c}\right)+C,$$where the locally constant function $C$ can have different values either side of $x=a$.


答:
Decomposing in simple fractions,

$$\frac1{x^3+x+1}=\frac a{x-r}+\frac{bx+c}{x^2-2px+p^2+q^2}
\\=\frac a{x-r}+b\frac{x-p}{(x-p)^2+q^2}+\frac{bp+c}{(x-p)^2+q^2}.$$

The antiderivatives of the terms are

$$a\log(x-r)$$

$$\frac b2\log((x-p)^2+q^2)$$

and

$$\frac{bp+c}q\arctan\frac{x-p}q.$$

All coefficients are functions of the roots.

Title: How to do integral $\int\frac1{x^3+x+1}dx$?

Body:
I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$


问:
Arens-Eells spaces
I started reading Nik Weaver's book on Lipschitz Algebras, where Arens-Eells spaces are identified as the pre-dual of the space of Lipschitz functions.  The weak* convergence in $Lip_0(X)$ is then identified with point-wise convergence.  The book also gives specific examples of Arens-Eels spaces, for example that $Æ(\mathbb{R})$ is isometric with $L^1(\mathbb{R})$, which is a separable space.

My question is: what can one say about the separability of $Æ(X)$, for example, for "simple" $X$, such as compact metric spaces (or even closed and bounded subspaces of $\mathbb{R}^n$)?  The reason for asking is the following observation.  On the one hand, we have the weak* compactness of the unit ball in $Lip_0(X)$ owing to Banach-Alaoglu.  On the other hand, we have compactness (hence sequential compactness) of the unit ball of $Lip_0(X)$ as a subset of $C(X)$ owing to Arzela-Ascoli, and consequently the weak* sequential compactness of the unit ball of $Lip_0(X)$, since weak* convergence is equivalent to point-wise convergence.

答:
If $X$ is separable then so is $A\!E(X)$.

Let $x_n$ be a sequence in $X$ with $x_n\to x$. Then
$$\|x_n-x\|≤ d(x_n,x)$$
and $x_n -x \to 0$ in $A\!E(X)$. So if $X$ is separable with $\{y_n\mid n\in\Bbb N\}$ and let

$$m=\sum_i^n \lambda_i \ x_i$$
be a molecule. Suppose that $y_{n_i}$ is arbitrarily close to $x_i$ then
$$\left\|m - \sum_i^n \lambda_i \ y_{n_i}\right\| ≤ \sum_i^n |\lambda_i| \,\left\|x_i -y_{n_i}\right\| $$
and note that the molecules with coefficients in $\{y_n\mid n\in\Bbb N\}$ are dense in the space of molecules hence in all of $A\!E(X)$. It follows that the rational combinations of $\{y_n\mid n\in\Bbb N\}$ are countable and dense in $A\!E (X)$.

Title: Arens-Eells spaces


问:
Proof of injective functions
Let $f:\mathbb{N}\to\mathbb{N}$ and $N_k=\{1,2,3,...,k\}$ for all $k\in\mathbb{N}$.

I have to prove two statements. If $f_{\big|N_k}$ is injective function for all $k\in\mathbb{N}$, then $f$ is injective function(one to one) and second if $f[N_k]=N_k$ for all $k\in\mathbb{N}$, then $f$ is identity function.

Note that $f_{\big|N_k}$ is restricted domain of function and $f[N_k]=N_k$ is image of function.

I prove the first statement, let $x_1,x_2\in\mathbb{N}$ and assume that $f(x_1)=f(x_2)$. If $x_1,x_2\in{N_k}$, then we are done but if not ?
Next, the second statement I have to show that $f(x)=x$ for all $x\in\mathbb{N}$ but how to show that ?

Tanks for someone to help me.

答:
If $x_1,x_2\in\Bbb N$, take some natural number $k$ such that $x_1,x_2\leqslant k$. Then $x_1,x_2\in N_k$ and so, since $f|_{N_k}$ is injective, if $f(x_1)=f(x_2)$, then $x_1=x_2$.

Concerning the other question, suppose the $f$ is not the identity function. Then there is a smallest natural $k$ such that $f(k)\ne k$. The number $k$ cannot be $1$, since $f(N_1)=N_1$, which means that $f(\{1\})=\{1\}$, or $f(1)=1$. So, $k>1$. But $f(N_{k-1})=N_{k-1}$ and $f(k)\ne k$ implies that $f(N_k)\ne N_k$.

Title: Proof of injective functions


问:
Block diagonal matrix diagonalizable
I am trying to prove that:

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

If $A\in GL(\mathbb{C}^n)$ and $B\in GL(\mathbb{C}^m)$ are diagonalizable, then is easy to check the $C\in GL(\mathbb{C}^{n+m})$ is diagonalizable. But if I suppose that $C$ is diagonalizable, then exists $S = [S_1, S_2, \ldots, S_{n+m}]$, $S_i\in\mathbb{C}^{m+n}$, such that $S^{-1}CS = \mbox{diag}(\lambda_i)$ . Now $CS_i = \lambda_iS_i$, and  if $S_i = \left(\begin{smallmatrix}x_i\\y_i\end{smallmatrix}\right)$, $x_i\in\mathbb{C}^n$ and $y_i\in\mathbb{C}^m$, then
$$Ax_i = \lambda_ix_i\quad\mbox{ and }\quad By_i = \lambda_iy_i.$$
So, if I can justify that $\{x_1,\ldots,x_{n+m}\}$ have exactly $n$ linear independent vectors and $\{y_1,\ldots,y_{n+m}\}$ have $m$ linear independent vectors, I will prove that $A$ and $B$ are diagonalizables, but I don't know how to prove that? Please, anyone have an idea? Thanks in advance.

答:
You can also define $f_C$ as a linear operator between $V=\mathbb{C}^{n+m}$ sending $x$ to $C\cdot x$. Then, $C$ is diagonalizable iff $f_C$ is diagonalizable iff there exists a basis of eigenvectors of $f_C$.

Suppose $f_C$ is diagonalizable. Denote the basis of eigenvectors as $v_1,...,v_{n+m}$. Let $U=\operatorname{span}(e_1,...,e_n)$ and $W=\operatorname{span}(e_{n+1},...,e_{n+m})$, which are two subspaces of $V$. Note that $V=U\oplus W$ and $U, W$ are $f_C$-invariant space. Thus, $v_i$ can be written as $u_i+w_i$ with $u_i\in U$ and $w_i\in W$. By the property of direct sum, $\operatorname{span}(u_1,...,u_{n+m})=U$, $\operatorname{span}(w_1,...,w_{n+m})=W$, $f_C(u_i)=\lambda_i u_i$ and $f_C(w_i)=\lambda_i w_i$ hold. Therefore, $f\upharpoonright_U$ and $f\upharpoonright_W$ are both diagonalizable, which means that $A$ and $B$ are diagonalizable.

Here is this statement in the form of linear operator, which can be proved exactly the same as above: suppose $V$ is a vector space over field $\mathbb{F}$ and $f:V\to V$ is a linear operator. $U,W$ are $f$-invariant subspaces of $V$ s.t. $V=U\oplus W$. Then, $f$ is diagonalizable iff $f\upharpoonright_U$ and $f\upharpoonright_W$ are diagonalizable.

Title: Block diagonal matrix diagonalizable


问:
Is a function differentiable at a point discontinuity?
I recently learnt the proof that if a function is differentiable then it is continuous and that a function cannot be differentiable at a point discontinuity .
I am confused however take the function
$$
f(x)=\begin{cases}
-2x & x<4, \\
4 & x=9.
\end{cases}
$$
Isn't this function differentiable at $x=9$ as the left hand derivative exsists and the right hand derivative does not need to exsist as the function is not even defined for values of x greater than $9$ so what is the function differentiable and if not why?

答: This function is continuous but not differentiable at 9, this all comes down to the function's domain which is $(-\infty, 4)\cup \{9\}$. Here, 9 is an isolated point, and by the definition of continuity, every function is continuous at the isolated points of its domain. Moreover, as 9 is an isolated point, it is not an accumulation point, so again by definition you cannot calculate the limit $\lim_{h\rightarrow 0}\dfrac{f(9+h)-f(9)}{h}$. In fact, most books only define the derivative at a point when it is an interior point of the domain.