问:
angle determined by tangent ratios
I'm new here and not a math student so I apologize if this question is too dumb but I don't know where to begin to solve it.
Given the image below, I'm looking for the red "α" angle such that the ratio between the 2 red segments is equal to the ratio between the 2 blue segments. That's all you need to know. Every other angle is a known variable.

I'd lose my pride if someone told me the full answer but I need at least a starting point because my trigonometry skills are failing me.

答:
Assign names to points along the black tangent line: $A$ at the point of tangency, $R$ at the intersection with the red line, $B$ at the next intersection with a black line, $C$ at the intersection after that, and $D$ at the final intersection.
Also, let $O$ be the center of the circle.

I suppose the locations of the points $O$, $A$, $B$, $C$, $D$ are known.
Hence the angles $\theta = \angle AOB$ and $\beta = \angle BOC$ can be found and so can the ratio
$$ r = \frac{AB}{BD}. $$

The location of $R$ is initially unknown, and so is the angle
$\alpha = \angle ROB,$
but you know that the ratio of the red segments must equal the ratio of the blue segments.
The length of the lower red segment is $OA \times \tan\alpha$,
the total length of the two red segments is $OA \times \tan(\alpha + \beta)$,
and the length of the second red segment is the difference of those two lengths,
so the ratio of those two segments is
$$
\frac{OA \times \tan\alpha}{OA \times \tan(\alpha + \beta) - OA \times \tan\alpha}
= \frac{\tan\alpha}{\tan(\alpha + \beta) - \tan\alpha}.
$$

Since it is given that this ratio equals the ratio of the blue segments,
$$
\frac{\tan\alpha}{\tan(\alpha + \beta) - \tan\alpha} = r.
$$

Note that there is a formula (sometimes called the sum identity for tangent) that expresses $\tan(\alpha+\beta)$ in terms of $\tan\alpha$ and $\tan\beta$.

After using the sum identity there is not much more need for trigonometry.
That may be enough of a hint to solve the problem.

Title: angle determined by tangent ratios


问:
Integer solution to $B(J-T)+1 = J(T-B), 31 \leqslant B < T < J \leqslant 49$
I found the following equation in some old notes:
$$
B(J-T)+1=J(T-B) \tag 1
$$
I don't remember where I got the problem but it is something like this: There are three brothers, James, Ted and Bob. James the eldest is in his forties, and Bob the youngest is at least 31 years old. The product of James's age and the (positive) difference between the other two brothers' is one more than the product of Bob's age and the (positive) difference between James's and Ted's. What are their ages?

The solution I came up with was to apply Simon's favorite factoring trick on (1) and I got
$$
T^2 - 2 = (2B-T)(2J-T) \tag 2
$$

Note that $J>T>B$, and $32 \le T \le 48, 14\le 2B-T\le 46, 41 \le 2J-T \le 66$. So I just manually (via Excel) checked all factors of the following 17 numbers: $32^2-2, 32^2-2, \ldots, 48^2-2$, and found only one solution $J=44, T=39, B=35$.

I would like to know:

If there's a better approach (James gives this as an interview question to someone looking for a job in his firm);

If anybody could find a source of the problem.

I appreciate your help.

答:
There are infinitely many solutions. If you have one $J>T>B>0$ you get a new one  (bigger)

There is more variety: we can begin with column vector
$$
\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right)
$$
and multiply on the left by a square matrix

as in
$$
\left(
\begin{array}{ccc}
45 & -1 & 28 \\
40 & -1 & 24 \\
36 & -1 & 21
\end{array}
\right)
\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right) =
\left(
\begin{array}{c}
44 \\
39 \\
35
\end{array}
\right)
$$

```<code>  6     -1      1 transposed        6      4      3
  4     -1      0 transposed       -1     -1     -1
  3     -1      0 transposed        1      0      0

 15     -1      6 transposed       15     12     10
 12     -1      4 transposed       -1     -1     -1
 10     -1      3 transposed        6      4      3

 20     -6      1 transposed       20     15     12
 15     -4      1 transposed       -6     -4     -3
 12     -3      1 transposed        1      1      1

 28     -1     15 transposed       28     24     21
 24     -1     12 transposed       -1     -1     -1
 21     -1     10 transposed       15     12     10

 35     -6      6 transposed       35     21     15
 21     -4      3 transposed       -6     -4     -3
 15     -3      2 transposed        6      3      2

 42    -15      1 transposed       42     35     30
 35    -12      1 transposed      -15    -12    -10
 30    -10      1 transposed        1      1      1

 45     -1     28 transposed       45     40     36
 40     -1     24 transposed       -1     -1     -1
 36     -1     21 transposed       28     24     21
</code>```

ContentLicense: CC BY-SA 4.0

Body:
There are infinitely many solutions. If you have one $J>T>B>0$ you get a new one  (bigger)

There is more variety: we can begin with column vector
$$
\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right)
$$
and multiply on the left by a square matrix

as in
$$
\left(
\begin{array}{ccc}
45 & -1 & 28 \\
40 & -1 & 24 \\
36 & -1 & 21
\end{array}
\right)
\left(
\begin{array}{c}
1 \\
1 \\
0
\end{array}
\right) =
\left(
\begin{array}{c}
44 \\
39 \\
35
\end{array}
\right)
$$

  6     -1      1 transposed        6      4      3
  4     -1      0 transposed       -1     -1     -1
  3     -1      0 transposed        1      0      0

 15     -1      6 transposed       15     12     10
 12     -1      4 transposed       -1     -1     -1
 10     -1      3 transposed        6      4      3

 20     -6      1 transposed       20     15     12
 15     -4      1 transposed       -6     -4     -3
 12     -3      1 transposed        1      1      1

 28     -1     15 transposed       28     24     21
 24     -1     12 transposed       -1     -1     -1
 21     -1     10 transposed       15     12     10

 35     -6      6 transposed       35     21     15
 21     -4      3 transposed       -6     -4     -3
 15     -3      2 transposed        6      3      2

 42    -15      1 transposed       42     35     30
 35    -12      1 transposed      -15    -12    -10
 30    -10      1 transposed        1      1      1

 45     -1     28 transposed       45     40     36
 40     -1     24 transposed       -1     -1     -1
 36     -1     21 transposed       28     24     21


问:
Find $\angle ADB$ where $\angle BAC = 18^\circ$, $\angle ABC = 12^\circ$ and $AB = CD$.
Let $\triangle ABD$ be a triangle and $C$ be a point on $BD$ such that $ {AB} ={CD}$, $\angle BAC = 18^\circ$ and $\angle ABC = 12^\circ$. The question is to find $x$, where $\angle ADC = x^\circ$.

I think this is a pretty classic problem but I am not able to find a solution using the techniques which were used to solve a similar kind of problem (only the angles were different and they were $30^\circ$ and $24^\circ$ respectively). The idea was to create an equilateral triangle by drawing a perpendicular line segment from $B$ to $AC$ with the same length as $AB$. However, that technique does not work here and I cannot find another way. Any help is appreciated (I forgot to add this but an answer which uses "geometrical" methods like drawing line segments or forming new triangles etc. would be great), thank you. The figure is as follows:

答:
I will insert a slow geometric solution. Doing it in steps may allow a pause after any step, the ambitious reader may want to find her or his own path to the solution. This may be also a good preparation for an olympiad or related challenge... The reader in hurry may chose to consume one of the two solutions listed in (4) and/or (5), which can be understood independently.

(1) Some picture:

In such situations it may be useful to use the pattern of an existing symmetry inside a regular polygon. The given triangle $\Delta ABC$ should be realized with vertices among the vertices of the polygon. (But the imagination of the solver should not stop here, sometimes using the center or some related intersection of diagonals may be also fruitful.) In our case we search a constellation with $A,B,C$ among the vertices of the regular polygon $P_{30}$.
If the vertices are $0,1,2,3,\dots,14, 15=-15, 16=-14,\dots, 29=-1$, denoted using the elements of $\Bbb Z/30$, then a possible choice is $B=0$, $C=3$, $A=5$.

We draw first a circle $(O)$ centered in some point $O$.
Since the angles with measures $12^\circ$ and $18^\circ$ are given, we expect to "see" also an inscribed angle in $(O)$ with measure $6^\circ$, so we decide to draw a $30$-gon inscribed in $(O)$. We fix the position of $A$, make then the one or the other choice of the position of $B$ and obtain the following picture:

In the figure, we have also constructed the point $D$ as follows.
Let $(A)$ be the circle centered in $A$ with radius $AB$. Let $S,T$ be the two points of intersection of $(A)$ with the parallel trough $A$ to $BC$. Here $S$ is the point making $\Delta ABS$ to have angles $<90^\circ$, so the angles are $12^\circ$ in $A$, and $84^\circ$ in $B$ and $S$.

The triangle $\Delta ABT$ has an angle with measure $180^\circ-12^\circ$ in $A$, and two $6^\circ$ angles in $B$ and $T$. Then $D$ is determined by constructing the parallelogram $TACD$.

Note that the circle $(A)$ passes through $O$.

We draw $DA$ and ask for the angle between the lines $AD$ and $BCD$. The picture suggests the next step, try to pause consuming the solution at this point...

(2) Some more observations and another picture:

In the regular polygon $P_{30}$ recall that the vertices are indexed via $\Bbb Z/30$. Then $BA=05$ or in general a segment $[k,k+5]$ is the side of a regular hexagon inscribed in $(O)$. So $BA$ is also equal to the radius $OA=OB$ in $(O)$. The OP mentions the need to find an equilateral triangle, well, here is one...

We consider closer the relation of the circles centered in $O,A,C$ having as radius this distance
$$
OA=OB=AB=AS=AT=CD=CO\ .
$$
Consider also the regular congruent $30$-gons on the circles $(O)$, $(A)$, $(C)$. (Note that main diagonals in the one or the other $30$-gon pass through vertices of the other $30$-gons, angle argument.)

Observation: We obtain the $30$-gon inside $(C)$ from the $30$-gon inside $(A)$ by a translation with the vector $\vec{AC}=\vec{TD}$.

This picture is of course too much and not needed, for the OP we want to know the angles in the paralellogram $ACDT$ formed by the diagonal $DA$ with the sides.
Let $U$ be the point with $\vec{AC}=\vec{SU}$. The parallelograms $TDCA$ and $ACUS$ are congruent.

(3) An other observation and yet an other picture:

In (2) the way to reason was as follows. The circles $(A)$ and $(C)$ have the same radius, thus they are congruent. Consider the vector $\vec{AC}$, it translates $(A)$ into $(C)$, and in particular also a specific $30$-gon inscribed in $A$ into a specific $30$-gon inscribed in $(C)$ if we insure that (at least) one vertex of the first $30$-gon goes to one vertex of the other $30$-gon.

The same applies also with the pair of circles $(A)$ and $(O)$, and/or with the pair of circles $(C)$ and $(O)$ correspondingly instead.

A similar observation can be done with rotations (instead of translations). For example... The circles $(A)$ and $C$ are passing through $O$ (and an other point). Consider the rotation centered in $O$ which maps $A$ into $C$. Then it maps the $30$-gon inscribed in $A$ with one vertex in $O$ into the $30$-gon inscribed in $C$ with one vertex in $O$. A picture suggesting the movement of the vertices by drawn circles of guidance would be:

Again, this is too much for our purposes, but we are only collecting ideas.

When searching for a solution, the idea is to use some pointed points under the many points constructed above.

(4) Solution I:

Let $O$ be a point and $(O)$ a circle centered in $O$ with radius $R$. Let $O(30)$ denote a fixed $30$-gon inscribed in $(O)$, its vertices are denoted by $0,1,2,\dots\in\Bbb Z/30$, and for some points we use alternative notations.

The first three points we want to evidentiate are $B=0$, $C=3$, and $A=5$.

Let $(A)$ and $(C)$ be the circles centered in $A$, respectively $C$ passing through $O$. They also have radius $R$. We draw the $30$-gons $A(30)$, and respectively $C(30)$ inscribed in these two circles having $O$ as one vertex in common, and denote the vertices in them by
$A_0$, $A_1$, $A_2$, and so on,
respectively
$C_0$, $C_1$, $C_2$, and so on. The indices are also living in $\Bbb /30$.

The starting point when counting is $O$, so $(A)$ and $(C)$ intersect in $O=A_0=C_0$, and in $A_{13}=C_{-13}$.

$\Delta ABO$ is equilateral, its sides are $R$.

We thus know the angle $\widehat{OBC}=60^\circ+12^\circ$, so $BC$ intersects $(C)$ for the second time in $D=C_{-9}$.

The radius $AB$ sits on the diameter $A_5A_{-10}$. We know the angle  $\widehat{ABC}=12^\circ$, so $BC$ intersects for the second time $(A)$ in $A_{-12}$.

Similar knowledge of angles shows that the parallel through $A$ to the line $BCDA_{-12}$ is the line $TC_{-8}AA_4C_5$.

Now start the solution, and show the equality of lines $(5,-5)$ and $C_3C_{-9}$. In $(O)$, the line between $5$ and $-5$ is the perpendicular bisector of the radius $O0=OB$, as a visual help consider the equilateral triangles $\Delta OB5=\Delta OBA$ and $\Delta OB-5$. The translation of the plane with vector $\vec{OC}$ transforms circles $(O)\to(C)$, and regular polygons $O(30)\to C(30)$. The chord $(5, -5)$ is mapped to the chord $C_{-13}C_7$. Note also that the perpendicular in $0=A_5$ to $(O)$ has known angle w.r.t. $AB=AA_5$, wo we can identify the direction perpendicular to $OB$ also in the circle $(A)$, it is $A_5A_{15}$. From this we obtain a parallelism
$$(5,-5)\| C_{-13}C_7\|C_{-13+4}C_{7-4}=C_{-9}C_3\|\dots
\\
\dots\| A_5A_{-15}\|A_4A_{-14}\|A_3A_{-13}\|A_2A_{-12}\|A_1A_{-11}\|OA_{-10}\ .$$
The diameter through $C_{-9}=D$ in $C$ is $C_{-9}C_6$. Reflecting the line
$$
(l) =C_{-9}CC_6=DCB=\color{blue}{A_{-12}A_5}
$$
w.r.t. to the line $\color{red}{C_{-9}C_3}$ gives the line
$$
(l')=C_{-9}C_{3-3}=DO=D\color{blue}{A_0}\ .
$$
And yes, there is exactly one isosceles trapez $\color{blue}{A_{-12}A_0A_5A_?}$ which is symmetric w.r.t. a direction matching the above parallelism, it is
$\color{blue}{A_{-12}A_0A_5A_{-13}}$, and its symmetry axis passes through the center $A$ of the circle $(A)$. So the parallel lines
$\color{red}{C_{-9}C_3}$ and $\color{red}{-5\;A}$ both contain $A$, so they are equal. This shows the colinearity of the points
$$
D=C_{-9}\ ,\ A=5\ ,\ C_3\ ,\ -5\ .
$$
The needed angle is for instance half of the angle between the diagonals of the above trapez, so it is
$$
\widehat{ADB}
=\frac 12\widehat{ODB}
=\frac14(\overset\frown{A_{-12}A_{-13}} + \overset\frown{A_0A_5})
=\frac14(1\cdot 12^\circ + 5\cdot 12^\circ )
=18^\circ\ .
$$

$\square$

(5) Solution II:

We use the same notations as above, else this would make 60% of the solution.

Using the rotation $\mathcal R$ centered in $O$ that maps $C=3$ to $A=5$ we can ask for the angle in $A_{-9}$ in $\Delta A_{-9}57=\mathcal R\Delta C_{-9}35$ instead of the angle in $A_{-9}$ in $\Delta C_{-9}35=\Delta DCA$ asked in the OP, and thus we completely remove from the picture $(C)$ and $C(30)$.

We know the angle
$\widehat{A_{-9}AO}
=\widehat{C_{-9}CO}$, or use the colinearity $(A_{-9}52)=\mathcal R(C_{-9}30)$, to  obtain that $A_{-9}52A_6$ is a line (supported on the diameter $A_{-9}A_6$.

The angles between the one fixed line $O,10$ and $OA_{-9}$ and $O8$ are equal, so $O, 8,A_{-9}$ are colinear. The two angles

between the chords $8,O,-7$ and $7,-3$ in $(O)$

between $A_{-9}A_0$ and $A_{-9}A_3$ in $(A)$

are equal, $4-1=3$, and two of the above chords are on the same line, so we obtain the colinearity of points
$$
A_{-9},\ A_3\ ,\ 7\ ,\ -3\ .
$$
The needed angle is thus
$$
\widehat{CDA} =
\widehat{3C_{-9}5} \overset{\mathcal R}=
\widehat{5A_{-9}7} =
\widehat{A_6A_{-9}A_3} =
\frac 12\overset\frown{A_6A_3}
=\frac 12\cdot3\cdot 12^\circ
=18^\circ
\ .
$$
$\square$


问:
Finding variance of X, the random variable that represents the number of isolated nodes in a random graph
Assuming a random n-node graph that is undirected, can't have multi-edges or loops, consider a scenario in which each edge exists independently with probability p. Set a random variable X to be the number of isolated nodes.

I found $E[X] = n(1−p)^{n-1}$

I know that $Var[X] = E[X^2] + E[X]^2$

Would I be correct in saying $Var[X] = (n(1−p)^{n-1})^2 + n$ because of the proposition that for the random variable $Sn$ defined as $S_{n} = X_1 + X_2 +... + X_n$, $E[S_n^2] = n$? Because I'm setting $X_i$ to be an indicator variable, where it's one if the i'th node is isolated and 0 otherwise, and $X = X_1 + X_2 +...+X_n$ is essentially the $S_n$

Or is this invalid?

答:
When $(X_i)$ is the sequence of indicator functions for isolation, then $S_n=\sum_{i=1}^n X_i$ will count the isolated nodes.

Now the expectation, $\mathsf E(X_i)$, is the probability that node $i$ is isolated from each of the $n-1$ remaining nodes, which is: $(1-p)^{n-1}$.

Since each $X_i$ has Bernoulli distribution, the expectation of ${X_i}^2$ equals the expectation of $X_i$.$$\begin{align}\mathsf{Var}(X_i)&=\mathsf E({X_i}^2)-\mathsf E(X_i)^2\\[1ex]&=(1-p)^{n-1}\,(1-(1-p)^{n-1})\end{align}$$

So indeed.

$$\begin{align}\mathsf E(S_n) &=\sum_{i=1}^n\mathsf E(X_i)&&\text{linearity of expectation}\\[1ex]&= n\,\mathsf E(X_1)&&\text{identical distributions}\\[1ex]&=n\,(1-p)^{n-1}&&\text{.}\end{align}$$

However for the expectation of the square of the sum, things are not linear. The square of a sum is not the sum of squares :  $~(a+b)^2=a^2+b^2+2ab$ .

$$\begin{align}\mathsf E\left({S_n}^2\right)&=\mathsf E\left(\sum_{i=1}^nX_i\cdot\sum_{j=1}^n X_j\right)\\&=\sum_{i=1}^n \mathsf E\left({X_i}^2\right)+2\sum_{1\leqslant i<j\leqslant n}\,\mathsf E(X_iX_j)\\[1ex] &= n\mathsf E({X_1}^2)+ n(n-1)\,\mathsf E(X_1X_2)\\[1ex]&= n~(1-p)^{n-1}+ n(n-1)\,\mathsf E(X_1X_2)\end{align}$$

That leaves you to evaluate the probability that two particular nodes are both isolated, $\mathsf E(X_1X_2)$.


问:
UMVU Estimator of Bernoulli Distribution
I am working on this question, and I got stuck.
Let $X_1, . . . , X_n$ be i.i.d. Bernoulli variables with success probability p.

$\hat{p}_n= \frac{(X_1 + ... +X_n)}{n} $

I was able to show that $\sqrt{n} (\hat{p}^2_n - p^2) \rightarrow N(0, 4p^3(1-p))$ by using the delta method; $\sqrt{n} (\hat{p}_n - p) \rightarrow N(0, p(1-p))$.

Now I am trying to find the UMVU estimator $\delta_n$ of $\sigma^2 = 4p^3(1-p))$. From what I understand the UMVU is going to be the bias corrected version of the MLE. However, I am not sure how to proceed. I would appreciate any tips.

答:
Using Rao - Blackwell & Lehmann - Scheffé together,  the UMVUE is

$$\mathbb{E}[T|S]$$

Where $T$ is an unbiased estimator for $p$ and $S$ is a Complete and Sufficient Statistic.

Let's set $ T=X_1$ and $S=\Sigma_i X_i$

$$\mathbb{E}[T|S]=\mathbb{P}[X_1=1|\Sigma_i X_i=s]=\frac{\mathbb{P}[X_1]\mathbb{P}[\Sigma_i X_i=s|X_1=1]}{\mathbb{P}[\Sigma_i X_i=s]}  = $$

$$=\frac{\mathbb{P}[X_1]\mathbb{P}\Big[\Sigma_{i=2}^n X_i=s-1]}{\mathbb{P}[\Sigma_i X_i=s]}=\frac{p\binom{n-1}{s-1}p^{s-1}(1-p)^{n-s}   }{\binom{n}{s}p^s(1-p)^{n-s}}=\frac{s}{n}=\overline {X}_n$$


问:
Prime number generator, how to make
Can anybody point me an algorithm to generate prime numbers, I know of a few ones (Mersenne, Euclides, etc.) but they fail to generate much primes...

The objective is:
    given a first prime, generate the 'n' next primes. But thanks for the link ;-)

```<code>for example : primes( 17, 50 ) -> Generate 50 consecutive primes starting at 17
</code>```

and do not fail any prime in this 50... no holes!

答:
Looking at this from a computer science like perspective, I would use the following algorithm to generate my list of primes.

```<code>input = (17,50)
    if 17 is prime:
    counter = 1
    from_seed = seed
      while counter <= 50
        if from_seed is prime
         print from_seed
         counter+1
         from_seed+1
        else
       from_seed+1
</code>```

To check if the number is prime then I would do the very simple

```<code>boolean isPrime (n)
   if n <= 2
     return false
   else
     for i .. sqrt(n)
       if n % i == 0
         return false
   return true
</code>```

This will very easily generate what you want. Below I have the working Java code, if you wish to implement it.

```<code>public class PrimeGeneration_Seed {

    public static void main(String[] args) {
        prime_Generation(17,50);
    }

    public static void prime_Generation (int seed, int number_of_primes) {
        if(!isPrime(seed)) {
            System.out.println("Seed is not prime"); // simple out statement saying that "seed" is not prime
        }else {
            int primes = 1;
            int from_seed = seed;
            while(primes <= number_of_primes) {
                if(isPrime(from_seed)) {
                    System.out.println(from_seed);
                    primes++;
                    from_seed++;
                }else {
                    from_seed++;
                }
            }
        }

    }

    // simple primality test - can be modified to be more complex and more efficient
    public static boolean isPrime(int n) {
        if(n <= 2){
            return false;
        }else {
            for(int i = 2;i<=Math.sqrt(n);i++) {
                if(n % i == 0) {
                    return false;
                }
            }
        }
        return true;
    }

}
</code>```

You can access this link to see the program in action.

ContentLicense: CC BY-SA 3.0

Title: Prime number generator, how to make

Body:
Looking at this from a computer science like perspective, I would use the following algorithm to generate my list of primes.

input = (17,50)
    if 17 is prime:
    counter = 1
    from_seed = seed
      while counter <= 50
        if from_seed is prime
         print from_seed
         counter+1
         from_seed+1
        else
       from_seed+1

To check if the number is prime then I would do the very simple

boolean isPrime (n)
   if n <= 2
     return false
   else
     for i .. sqrt(n)
       if n % i == 0
         return false
   return true

This will very easily generate what you want. Below I have the working Java code, if you wish to implement it.

public class PrimeGeneration_Seed {

    public static void main(String[] args) {
        prime_Generation(17,50);
    }

    public static void prime_Generation (int seed, int number_of_primes) {
        if(!isPrime(seed)) {
            System.out.println("Seed is not prime"); // simple out statement saying that "seed" is not prime
        }else {
            int primes = 1;
            int from_seed = seed;
            while(primes <= number_of_primes) {
                if(isPrime(from_seed)) {
                    System.out.println(from_seed);
                    primes++;
                    from_seed++;
                }else {
                    from_seed++;
                }
            }
        }

    }

    // simple primality test - can be modified to be more complex and more efficient
    public static boolean isPrime(int n) {
        if(n <= 2){
            return false;
        }else {
            for(int i = 2;i<=Math.sqrt(n);i++) {
                if(n % i == 0) {
                    return false;
                }
            }
        }
        return true;
    }

}

You can access this link to see the program in action.


问:
Find an increasing function in $\mathbb{R}$ such that its set of discontinuity is $\mathbb{Q}$.
Question: Find an increasing function in $\mathbb{R}$ such that its set of discontinuity is $\mathbb{Q}$.

Attempt: Suppose $(q_n)_{n \in \mathbb{N}}$ be an enumeration of the rationals, and define
$$f(x) := \sum_{k=1}^\infty 2^{-k} \chi_{(q_k, \infty)}(x).$$

Then the function is increasing by construction.

Now, I am stuck at showing that $f$ is discontinuous at $x \in \mathbb{Q}$.

In particular, I am having trouble showing that if for some $n \in \mathbb{N}$ such that $$x=q_n < y < \max_{q_i > q_n,i =1 \ldots, n} q_i\, ,$$ then we have
$f(x) + 2^{-n} \leq f(y)$.

Any help would be appreciated.

答:
Another old example is: (i) $f(0)=1.$ (ii) If $0\ne m\in \Bbb Z$ and $n\in \Bbb N$ with $\gcd(m,n)=1$ then $f(m/n)=1/n.$ (iii) If $x\in \Bbb R$ \ $\Bbb Q$ then $f(x)=0.$

For $n\in \Bbb N$ let $S(n)=\{m/d:m\in \Bbb Z \land n\ge d\in \Bbb N\}.$ If $b,c$ are distinct members of $S(n)$ then $|b-c|\ge 1/n^2$ so $S(n)$ is closed. So if $(y_j)_{j\in \Bbb N}$ is a sequence of rationals converging to $x\not \in \Bbb Q$ then for any $n\in \Bbb N$ the set $\{j: f(y_j)\ge 1/n\}$ is a subset of the $finite$ set $\{j: y_j\in S(n)\}$; Hence $\lim_{j\to\infty}f(y_j)=0=f(x).$

More generally, let $(X,d)$ be a  metric space. (i'). The set of discontinuities of any $f:X\to \Bbb R$ is an $F_{\sigma}$ subset of $X.$ (ii'). If $X$ has no isolated points and $Y$ is an $F_{\sigma}$ subset of $X$ then there is an $f:X\to [0,1]$ such that $f^{-1}(0,1]=Y$ and $Y$ is the set of discontinuities of $f.$


问:
If$|f(x)-f(y)|\le (x-y)^2$, prove that $f$ is constant
(Baby Rudin Chapter 5 Exercise 1)

Let $f$ be defined for all real $x$, and suppose that
\begin{equation}\tag{1}
    |f(x)-f(y)|\le (x-y)^2
\end{equation}
Prove that $f$ is constant.

My attempt:

Let $f$ be defined for all real-valued inputs. Let $x \in \mathbb{R}$ and $y \in \mathbb{R} \smallsetminus \{ x \}$, and suppose that (1) holds.
Then, we have:
\begin{align*}
        \left| \dfrac{f(x)-f(y)}{x-y}\right| \le (x-y)
    \end{align*}
As $x\to y,  \lim\limits_{x \to y}\left| \dfrac{f(x)-f(y)}{x-y}\right| \le 0$. Since it cannot be that $\left|f'(y)\right| < 0$, we have that $\left|f'(y)\right| = 0 \implies f'(y) = 0$.

Can someone please read over my proof and let me know if it is correct?

答:
Late answer, but here is a slight generalization of this problem.

Let $X,Y\subseteq\mathbb{R}$ and suppose $f:X\to Y$ is a function. Suppose $f$ is $\alpha$-Holder continuous with $\alpha\in\mathbb{R}$ and $\alpha>1$. Then, there exists a $K\in\mathbb{R}$ such that for all $x,y\in X$, we have $|f(x)-f(y)|\leq K|x-y|^{\alpha}$. Now, consider $$0\leq \lim_{x\to y}\bigg|\frac{f(x)-f(y)}{x-y}\bigg|\leq \lim_{x\to y}K|x-y|^{\alpha-1} $$ Since $\alpha>1$, then $\alpha-1>0$, which implies $\lim\limits_{x\to y}|x-y|^{\alpha-1}=0$. By the squeeze theorem, we have $\lim\limits_{x\to y}\bigg|\frac{f(x)-f(y)}{x-y}\bigg|=0$ which implies $f'(x)=0$. Since the only functions whose derivatives are identically zero are the constant functions, then we must have $f(x)=c$ for some $c\in\mathbb{R}$.


问:
What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$?
We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$

What is the limit of this function as $n \rightarrow \infty$?

My idea:

$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$

Is this correct?

答:
Another method, using Riemann sums applied to $f:x\mapsto x$:
$$
\frac{\sum_{k=1}^n k }{n^2} = \frac{1}{n}\sum_{k=1}^n \frac{k }{n}  \xrightarrow[n\to\infty]{}
\int_{0}^{1} xdx
$$

As for your question, each term goes to $0$, but the number of terms grows; you are not in the case where you have the sum of "constantly many terms, each of them converging".


问:
How do we know if an integral converges or diverges from a limit?
If we know that the limit of $f(x)$ approaches infinity equals some value like $5$, how can I show that $$\int_0^\infty f(x)\text{ }dx$$ converges or diverges?

答:
We can prove that it diverges. Here's how:

If $\lim_{x\to\infty}f(x)=5$, then there exists a $\delta>0$ such that for all $x>\delta$, $|f(x)-5|<1$. This latter inequality is equivalent to $4<f(x)<6$, showing that $f(x)>4$ for all $x>\delta$.

It is clear that the integral $\int_{\delta}^{\infty}4\text{ }dx$ diverges, so $\int_{\delta}^{\infty}f(x)\text{ }dx$ diverges by the integral comparison test. Adding a finite integral like $\int_{0}^{\delta}f(x)\text{ }dx$ doesn't affect convergence/divergence, so $\int_{0}^{\infty}f(x)\text{ }dx$ is also divergent.


问:
Maximum Likelihood Estimation for Bernoulli distribution
I'm just confused about the purpose of MLE.

What I've read about MLE for normal distribution is that it is a method that will find the values of parameters (μ and σ in this case) that result in the curve that best fits the data, or in simpler version maximize the probability of observing our data.

This makes sense when it comes to normal distribution, but I can't imagine a best 'curve' for Bernoulli distribution, what is the point of having MLE in this case?

答:
I can't imagine a best 'curve' for Bernoulli distribution,

The likelihood is a function of the parameter, considering $\mathbf{x}$ as given data. Thus for bernulli distribution

$$L(\theta)=\theta^k(1-\theta)^{n-k}$$

Where $k=\Sigma_i X_i$ and $\theta \in (0;1)$

To maximize L it is easy use logL

$$logL=klog\theta+(n-k)log(1-\theta)$$

Derivating in $\theta$ and setting =0 you get

$$\hat{\theta}=\frac{k}{n}=\overline {X}_n$$


问:
Why is $P(X>m+n) = P(X>m)\cdot P(X>n)$?
$P$ is probability, $X$ is a random variable, $m$, $n$ are two scalars. I know that $P(X>m+n)$ is not $P(X>m\cap X>n)$, and think that this does make some sense. I just need some more detailed explanation. Thanks!

答:
It is not generally true, unless you are dealing with a memoryless distribution (such as geometric or exponential), and so for positive $m,n$: $$\mathsf P(X>m+n\mid X>m)=\mathsf P(X>n)$$

Thus, since for positive $m,n$ it follows that $X>m+n$ entails $X>m$. $$\begin{align}\mathsf P(X>m+n) &=\mathsf P(X>m \cap X>m+n)\\&~~\vdots\\&=\mathsf P(X>m)\mathsf P(X>n)\end{align}$$

Title: Why is $P(X>m+n) = P(X>m)\cdot P(X>n)$?


问:
Can we say $C_G(x)=C_G(x^2)$
Let $(G,.)$ be a group with $e$ as an identity element. If $x\in G$ and $C_G(x)=\left\{g \in G: gx=xg\right\}$ and $|x|=5$. In the process of proving $C_G(x)=C_G(x^3)$, i proved $C_G(x)=C_G(x^2)$ as follows:

Let $g \in C_G(x)$

So $gx=xg$

Now $gx^2=g(xx)=(gx)x=(xg)x=x(gx)=x(xg)=(xx)g=x^2g$

So all $g \in C_G(x)$ satisfies $gx^2=x^2g$

So can we conclude $$C_G(x)=C_G(x^2)$$

答:
Yes. Here is a lemma. Let $G$ be a group, and $g \in G$. Then for any $m$,
$$
C(g) \subset C(g^m).
$$
Proof: If $y \in C(g)$, then $yg^m = gyg^{m-1} = \ldots = g^my$, so $y \in C(g^m)$.

Now in your group, you have an element $x$ of order $5$. The lemma shows that
$$
C(x) \subset C(x^2).
$$

But $x = (x^2)^3$, so

$$
C(x^2) \subset C(x),
$$
again by the lemma.

Title: Can we say $C_G(x)=C_G(x^2)$


问:
Good books to Learn General topology
I am quite interested in learning topology from the basics, I would like to find a book that is quite detailed and has a large number of examples.

What are the best books to easily understand topology?

Thanks.

答:
"Topology" by James Dugundji has the best of many worlds: formality and richness in content without being too obnoxious or cryptic for someone who has no idea of topology.

"Topology" by James Munkres is (I think) an excellent introductory book, very digestible and easy to handle. The main drawback I see from this one is that it can be a bit slow, so if you are in a hurry to learn topology go for Dugundji instead.

"General Topology" by Stephen Willard is an excellent book that not only teaches you the concepts, but also the reason behind them. Unfortunately, Willard leaves too much to the reader (for my taste), and it can be a bit lacking in problems.

Last but not least, "Topology" by Ryszard Engelking is somewhat considered a bible among students and teachers. It is an extremely complete book, but it can be a tough read for a beginner. I don't recommend starting with this one, but I wanted to mention it since it can be a good reference in the future.

Unfortunately, I have yet to find THE perfect book for an introductory course to Topology, but I feel that the combination of this four books is an excellent way to dip your toes in topology. My main recommendation is to start with Munkres, if you feel it's too easy and slow go for Dugundji while complementing with Willard, and if everything still feels too basic and obvious, go for Engelking.

Title: Good books to Learn General topology


答:
I'll put $n$ for $20$
and start the sum at $i=1$
since the inner sum is not well defined
at $i=0$.

The sum is

$\begin{array}\\
s(n)
&=\sum_{i=1}^n \dfrac{i}{n/2}\sum_{j=i}^n \dfrac1{j}\\
&=\dfrac{2}{n}\sum_{i=1}^n i\sum_{j=i}^n \dfrac1{j}\\
&=\dfrac{2}{n} \sum_{j=1}^n \sum_{i=1}^ji \dfrac1{j}
\qquad\text{since } j \ge i, i \le j\\
&=\dfrac{2}{n} \sum_{j=1}^n\dfrac1{j} \sum_{i=1}^ji \\
&=\dfrac{2}{n} \sum_{j=1}^n\dfrac1{j}\dfrac{j(j+1)}{2} \\
&=\dfrac{1}{n} \sum_{j=1}^n (j+1)\\
&=\dfrac{1}{n} (\dfrac{n(n+1)}{2}+n)\\
&=\dfrac{1}{n} \dfrac{n(n+3)}{2}\\
&=\dfrac{n+3}{2}\\
\end{array}
$

For $n=20$ this is
$\dfrac{23}{2}
=11.5
$.

Title: Evaluate $\sum_{i=0}^{20} \frac{i}{10}(\frac{1}{i} + \frac{1}{1+i} + ...+ \frac{1}{20})$

Body: Would the answer be 11.5? I simplified the above expression to $\frac{1}{10}\{1(1) + \frac{1}{2}(1+2) +. .. + \frac{1}{20}(1+2+...+20)\}$ =$\frac{1}{10}\sum_{k=1}^{20} \frac{k+1}{2}$ and thus got 11.5


问:
Combinatorics- to place rooks!
Find the number of ways to place 3 rooks on a 5 × 5 chess board so
that no two of them attack each other
The way I tried this question was that there are $5C_3$ ways to choose a row and a column in which 3 rooks can be placed so I got $5C_3^2$ but it is wrong.
The second way I thought was that 1st rook can be placed in any of the first 5 blocks of a row and 2nd rook in 4 ways so total comes out to be $5!$ ways but thats wrong please point out my mistakes

答:
Your partial work of $~\left[\binom{5}{3}\right]^2 = 100$

is the right starting point.

In effect, you have determined that there are 100 possible ways of choosing the pertinent (3 rows) and (3 columns).

Now, you have to consider how many ways are there of placing 3 rooks in a 3 $\times$ 3 chessboard, so that they don't attack each other.

Let $R_i$ denote the rook placed in row $i$.

Then, there are three choices (i.e. columns) that $R_1$ can be placed in, and then two (remaining) columns for $R_2.$

Final answer:

$$\left[\binom{5}{3}\right]^2 \times 3! = 600.$$

Title: Combinatorics- to place rooks!


问:
Galois extensions of degrees $p$ and $p^{n-1}$ given a Galois extension of $p^n$
Suppose $K$ is a Galois extension of a field $F$ of degree $p^n$ for a $p$ a prime.

I want to see if there are Galois extensions of degrees $p$ and $p^{n-1}$ over $F$.

If $G=\text{Gal}(K/F)$, then $|G|=p^n$. If $G$ is abelian, I know there are subgroups of order $p^i$ for $0\leq i\leq n$, so there are subgroups of orders $p^{n-1}$ and $p$, and their corresponding fixed fields are of degrees $p$ and $p^{n-1}$ over $F$, and are Galois extensions since the subgroups are obviously normal in $G$.

But if $G$ is not abelian, is this still true?

答:
I recognize this question was asked 9 years ago, but I was doing this problem (in Dummit and Foote) and found what I believe to be a simpler solution.  I'll write it here to brag and to help future students of Dummit and Foote.

Using the class equation (as in the accepted answer here) find a normal subgroup of order $ p $, $ \mathbb{Z}_p \trianglelefteq \mathrm{Gal}(K/\mathbb{Q}) $.  Its fixed field $ K_1 \subseteq K $ is indeed a Galois extension of $ \mathbb{Q} $ of degree $ p^{n-1} $.  Now repeat this argument with $ K_1 $, finding again $ \mathbb{Z}_p \trianglelefteq \mathrm{Gal}(K_1/\mathbb{Q}) $ with fixed field $ K_2 \subseteq K_1 $.  We now have a Galois extension of order $ p^{n-2} $.  Continuing by the same argument for $ n $ steps gives $ \mathbb{Q} \subseteq K_{n-1} \subseteq \ldots \subseteq K_1 \subseteq K $ where each $ K_i $ is Galois over $ \mathbb {Q} $ of degree $ p^{n-i} $.


问:
order of kernel of surjection between curves is equal to separate degree
Let $Φ:C→D$ be a non constant map of smooth curves.
Silverman says in AEC

for all but finitely many $y∈D$,
$$\#\phi^{-1}(y) = \deg_s(\phi)$$

How can I prove this? Silverman also say to prove this, see Hartshorne Ⅱ6.8. But I can find no connection between them.

Some remarks in mathstacks I see use the ideal correspondence and Hilbert's Nullstellensatz, but curves $C$ and $D$ in AEC might not be defined over algebraically closed field, so I need to show the proposition over arbitrary fields.

答:
I assume you mean Proposition II.2.6(b) in AEC. The curves must be defined over an algebraically closed field, or else the statement is not true. An easy counterexample in the non-algebraically closed case is $K = \mathbb{R}$, with curves $C = D = \mathbb{A}^1(\mathbb{R})$, and $f(x) = x^2$. Then $\# \phi^{-1}(y) = 0$ for infinitely many $y$ (namely, all $y < 0$), but $0 \neq \deg_s\phi = 2$.

In order to understand how II.2.6(b) incorporates a hypothesis of an algebraically closed field, it is necessary to refer to the Definition of smooth curve (on page 17), which refers to the Definition of projective variety (on page 9), which refers to the Definition of projective algebraic set (on page 7), which refers to the definition of $V_I$ (on page 7), which refers to the Definition of $\mathbb{P}^n$ (on page 6), which refers to the Definition of $\mathbb{A}^n$ (on page 1). We finally see that $\mathbb{A}^n$ by definition means $\mathbb{A}^n(\bar{K})$, so we are indeed working over an algebraic closure.

In order to avoid a notation conflict with $L$ and $K$ below, I will use $F$ to denote the (algebraically closed) scalar field over which $C$ and $D$ and $\phi$ are defined.

To prove the result, use the fact that over an algebraic closure, points on curves are in 1-1 correspondence with nonzero prime ideals in the coordinate ring. The splitting of prime ideals in finite extensions of global fields then proves the desired result. Specifically, take $K = \phi^* F(D)$ to be the function field of $D$, and $L = F(C)$ to be the function field of $C$, and consider the formula
$$
[L:K] = \sum_{j=1}^g e_i f_i
$$
from the aforementioned page, where $e_i$ and $f_i$ denote ramification index and inertial degree. The residue fields at every point must all be equal to $F$, since $F$ is algebraically closed, meaning that no further nontrivial finite extensions of this residue field are possible. Hence $f_i = 1$ for all $i$, and the formula then reduces to the simpler expression
$$
[L:K] = \sum_{j=1}^g e_i.
$$
(Note that this expression is identical to the statement of AEC II.2.6(a).) Since $[L:K] = \deg \phi$ by definition of $\deg \phi$, this equation is exactly the statement that $\deg \phi = \# \phi^{-1}(y)$ for all $y \in D$, where the points in $\phi^{-1}(y)$ are counted with multiplicity (the respective values of $e_i$ denoting the multiplicity of each element of $\phi^{-1}(y)$). In order to obtain II.2.6(b) in the separable case, simply observe that $\deg_s\phi = \deg \phi$, and that there are only finitely many points in $C$ where $e_i \neq 1$.

In the inseparable case, you will have $e_i = d$ where $d$ denotes the inseparable degree of $[L:K]$. Dividing both sides of the equation by $d$ then yields the result.

I also highly recommend that you read Remark II.2.8 in AEC since this remark adds some further information to this discussion.


问:
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$

Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.

How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.

答:
Well, in the worst case when we are stuck we can do cases:

Case 1: $x+y\ge 0$ and $x-y \ge 0$ then

$x\ge -y$ and $x \ge y$ so $x \ge \max(y,-y) = |y|$.

And $|x+y| + |x-y| = (x+y)+(x-y) = 2x = 2$ so $x = 1$ and $|y| \le x=1$ so $-1 \le y \le 1$.

$x^2 - 6x +y^2 = -5+y^2$ and the max that $y^2$ can be is $1$ so the max of $x^2 -6x +y^2$ is $-4$ when $y = \pm 1$ if was assume $x+y\ge 0$ and $x-y \ge 0$.

(Likewise the max of $x^2 + y^2 + 10y = 1+y^2 + 10y$ is $12$ if $y=1$.)

Case 2:

$x +y < 0$ and $x-y \ge 0$ then

$x < -y$ and $x \ge y$ so $y \le x < -y$ so $y < 0$.

And $|x+y| + |x-y| = -x -y + x-y = -2y = 2$ so $y = -1$ and $-1\le x < 1$.

$x^2 -6x + y^2 = x^2 -6x +1$ and $x^2 \le 1$ and $x \ge -1$ so $-6x \le 6$ so the very max that $x^2-6x +y^2$ is $1+6 +1 = 8$ if $x=-1$.

(Likewise $x^2 +y^2 + 10y = x^2 + 11$ an the max that can be is $12$ if $x = -1$.)

Case 3: $x +y \ge 0$ and $x -y < 0$ then

$x \ge -y$ and $x < y$ so $-y \le x < y$ so $y > 0$.

$|x+y| + |x-y| = x+y+y-x = 2y =2$ so $y =1$ and $-1 \le x < 1$ and

the max value of $x^2-6x +y^2 = x^2-6x+1$ and $x^2 \le 1$ and $-6x \le 6$ so $x^2 -6x+1 \le 8$ and the max value is $8$ when $x=-1$.

case 4: $x+y < 0$ and $x-y < 0$ so $x< -y$ and $x < y$ so $x< \min (y,-y) = -|y|$

And $|x+y| + |x-y| = -x -y +y -2 =-2x = 2$ so $x = -1$ and $-|y| > x=-1$ so $|y|< 1$.

so $x^2 -6x +y^2 = 7 + y^2< 8$.  but $x^2-6x +y^2 = 8$ never occurs; it's an unacheived least upper bound.

Of those four cases the max value of $x^2 -6x +y^2 = 8$.

....

that's probably not an efficient or clever way but it is a complete hit the ground running systematic way.


问:
Greatest common divisor of Characteristic Polynomials
I am stuck with the following proof about matrices:

Let $A\in \textbf{M}_{n}(k),B\in \textbf{M}_{m}(k)$ and $M\in \textbf{M}_{n\times m}(k)$ such that $AM=MB$. Prove that the degree of gcd{$ P_{A} ,P_{B} $} is larger or equal to $r=rankM$.  (with $P_{M}$ I mean characteristic polynomial)

I know that if $v$ is an eigenvector of $B$ with eigenvalue $\lambda$, then $AMv=MBv=M\lambda v= \lambda Mv$, so if $Mv$ is nonzero i could say that $\lambda$ is an eigenvalue of A, though I'm not sure if this information will help in any way to solve the problem. Any help would be great!

答:
$AM=MB$ gives $f(A)M=Mf(B)$ for any polynomial $\in k[x]$, and $M\ker(f(B))\subset \ker(f(A))$.

We can assume that $k$ is algebraically closed.

Let $d=\max(n,m)$.

The multiplicity of $\lambda$ in $\det(B-xI)$ is $\dim(\ker( (B-\lambda I)^d)$.

Moreover $k^m=\sum_\lambda \ker( (B-\lambda I)^d)$.

Whence $M k^m=\sum_\lambda M\ker( (B-\lambda I)^d)$.

This space has dimension $rank(M)$ and $\lambda$ has multiplicity $\dim\ker( (A-\lambda I)^d)\ge \dim(M\ker( (B-\lambda I)^d))$ in $\det(A-xI)$.


问:
Number of partitions of an integer with extra conditions
In how many ways can you write number 15 as a sum of positive integers if number 1 can appear 3 times at most in the sum and number 3 can only appear even number of times?

I've tried doing this as $$(1+t+t^2+t^3)(1+t^2+t^4+t^6+t^8+t^{10}+t^{12}+t^{14})(1+t^6+t^{12})\cdots(1+t^{15})$$ and finding the coefficient of $t^{15}$, but it's taking too much time. Is there an easier way to do this? I think the result should be around 81. (I found the number of partitions of number 15 with no limitations, then ruled out those that do not satisfy the condition.)

Body:
Via inclusion-exclusion, where $p_n$ is the number of unrestricted partitions of $n$:
\begin{align}
& p_{15}
- \left[p_{15-4\cdot1} + (p_{15-1\cdot3} - p_{15-2\cdot3}) + (p_{15-3\cdot3} - p_{15-4\cdot3}) + p_{15-5\cdot3} \right]
+ \left[(p_{15-4\cdot1-1\cdot3} - p_{15-4\cdot1-2\cdot3}) + p_{15-4\cdot1-3\cdot3} \right] \\ &= p_{15}
- \left[p_{11} + (p_{12} - p_9) + (p_6 - p_3) + p_{0} \right]
+ \left[(p_8 - p_5) + p_{2}\right] \\
&= 176
- \left[56 + (77 - 30) + (11 - 3) + 1 \right]
+ \left[(22 - 7) + 2\right] \\
&= 176 - 112 + 17 \\
&= 81
\end{align}


问:
Find the greatest common divisor of the following polynomials in Z5[x].

After I do the division algorithm, I get
$x^4 + x^3 +3x= x (x^3 + x^2 -2) +5x$

Since $5$ is congruent to $0$ $(mod$  $5)$, does it imply that the remainder is $0$?
So the gcd is $b(x)$?

答:
Yes, or you could say simply $b(x)=[1]_5x^3+[1]_5x^2+[3]_5$,

and then it's obvious that $a(x)=xb(x)$.


问:
About the induced vector measure of a Pettis integrable function
In what follows, $X$ stands for a Hausdorff LCTVS and $X'$ its topological dual.

Let $(T,\mathcal{M},\mu)$ be a finite measure space, i.e., $T$ is a nonempty set, $\mathcal{M}$ a $\sigma$-algebra of subsets of $T$ and $\mu$ is a nonnegative finite measure on $\mathcal{M}$.

A function $f:T\to X$ is called scalarly integrable iff for each $x'\in X'$, the composition map
$$x'\circ f:T\to \mathbb{R}$$
is Lebesgue integrable.

A function $f:T\to X$ is said to be Pettis-integrable iff it is scalarly integrable and for each $E\in \mathcal{M}$, there exists $x_E\in X$ such that
$$x'(x_E)=\int_E(x'\circ f)d\mu$$
for all $x'\in X'$. In this case, $x_E$ is called the Pettis integral of $f$ over $E$ and is denoted by
$$x_E=\int_E fd\mu.$$

Question. How can we prove that if $f:T\to X$ is Pettis integrable then the map
$$m_f:\mathcal{M}\to X$$
defined by
$$m_f(E)=\int_E fd\mu, \quad E\in \mathcal{M}$$
is a vector measure on $\mathcal{M}$ which is countably additive?

答:
Let $\{E_n:n\in\mathbb{N}\}\subset\mathcal{M}$ be a sequence of disjoint sets. By $E$ we denote their union. Fix $x'\in X^*$, then
$$
x'(m_f(E))
=x'\left(\int_E fd\mu\right)
=\int_E(x'\circ f)d\mu
=\sum\limits_{n=1}^\infty\int_{E_n}(x'\circ f)d\mu
=\sum\limits_{n=1}^\infty x'\left(\int_{E_n} fd\mu\right)
=\sum\limits_{n=1}^\infty x'\left(m_f(E_n)\right)
=x'\left(\sum\limits_{n=1}^\infty m_f(E_n)\right)
$$
Using Orlicz-Pettis theorem for locally convex spaces we can be sure that the series $\sum\limits_{n=1}^\infty m_f(E_n)$ is convergent.
Since $x'\in X^*$ is arbitrary by corollary of Hahn-Banach theorem
$$
m_f(E)=\sum\limits_{n=1}^\infty m_f(E_n)
$$</p>


问:
Let $x_0 = 2003$, and let $x_{n+1} = \frac{1 + x_n}{1 - x_n}$ for $n >= $1. Compute $x_{2019}$

Let $x_0 = 2003$, and let $x_{n+1} = \frac{1 + x_n}{1 - x_n}$ for $n >= $1. Compute $x_{2019}$

I tried expressing $x_2$, $x_3$, $x_4$.. in terms of previous terms to see if there is a pattern but $n >= 1$ is troubling me. Please help.

Body:
HINT: Note that

$$\begin{align*}
x_{n+2}&=\frac{1+x_{n+1}}{1-x_{n+1}}\\
&=\frac{1+\frac{1+x_n}{1-x_n}}{1-\frac{1+x_n}{1-x_n}}\\
&=\frac{1-x_n+1+x_n}{1-x_n-1-x_n}\\
&=-\frac1{x_n}\,.
\end{align*}$$


问:
How do I find the least number of employees in a supermarket?
The problem is as follows:

7-11 (Seven-eleven) has numerous branches across Taipei. The newest
manager knows that in $k$ stores where a total of $10k$ employees
work. The latest report also indicates that in none of those stores
works less than $k+2$ neither more than $3(k+1)$ employees. If the
greatest number of employees in four stores is $43$. Find the value of
$k$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{6 employees}\\
2.&\textrm{5 employees}\\
3.&\textrm{7 employees}\\
4.&\textrm{9 employees}\\
\end{array}$

What sort of logic is used to solve this problem?. What I attempted so far was doing this:

$k+2<k(10k)<3(k+1)$

From this I'm getting:

$k+2<10k^2<3k+3$

I don't know if the intended meaning of this problem is that those less than or greater than consider the boundaries, in other words to be $\leq$ or $\geq$.

Does it exist any word in the problem which specify that any of these conclusions should be considered?. I don't know when to use $\leq$ or $\geq$.

More over what to do next?. My book states that the answer is $7$, but I don't know how to get there.

Can someone help me with a step by step solution?.

答:
$k$ denotes the number of stores.

$10k$ employees work in these stores, so the average number of employees in each store is 10.

Therefore $~(k + 2) ~\leq ~10 ~\leq ~3(k+1).$

Therefore $~k ~\leq ~8 ~~\text{and} ~~k \geq 3.$

Further, $~10 k ~\geq ~43 ~\implies ~k ~\geq ~5.$

Examine each of these cases individually:

$\underline{\text{Case 1:} ~k = 5}$

Minimum in each store is 7 and maximum is 18.

50 total employees in all stores combined.

At least one group of 4 stores has 43 employees.

No group of 4 stores has more than 43 employees.

$\{7, 10, 11, 11, 11\}~$ works.

$\underline{\text{Case 2:} ~k = 6}$

Minimum in each store is 8 and maximum is 21.

60 total employees in all stores combined.

At least one group of 4 stores has 43 employees.

No group of 4 stores has more than 43 employees.

$\{8, 9, 10, 11, 11, 11\}~$ works.

$\underline{\text{Case 3:} ~k = 7}$

Minimum in each store is 9 and maximum is 24.

70 total employees in all stores combined.

At least one group of 4 stores has 43 employees.

No group of 4 stores has more than 43 employees.

$\{9, 9, 9, 10, 11, 11, 11\}~$ works.

$\underline{\text{Case 4:} ~k = 8}$

Minimum in each store is 10 and maximum is 27.

80 total employees in all stores combined.

At least one group of 4 stores has 43 employees.

No group of 4 stores has more than 43 employees.

This won't work because:

If one group of 4 stores has 43 employees, the total number of
employees in the other 4 stores must be 37.  This contradicts the requirement that the minimum that work in each store is 10.

Therefore, you have 3 possible values of $k$ that will satisfy the requirements.  This is the point where I would meta-cheat.  The way that the problem is worded, and the way that the choices are given, as a mathSE reviewer, the status of the problem at this point is unacceptable.  Therefore, I have only two choices:

Request that I would like to buy a vowel.

Attempt telepathy.

Forgoing buying a vowel, look at the title of the query:

How do I find the least number of employees in a supermarket?

My telepathy suggests that you are supposed to identify all possible satisfying values of $k$, and then choose the minimum.
Based on this:

Final answer:

$$k = 5.$$


问:
Sobolev injections in $\mathbb{R}$
a) Let $a<b$. Prove there exists an universal contant $C>0$ such that
$$
\|u\|_{L^{\infty}(a, b)} \leq C\|u\|_{H^{1}(a, b)} \quad \forall u \in H^{1}(a, b)
$$

b) Let $a<b$. Prove that the injection $H^{1}(a, b) \rightarrow L^{\infty}(a, b)$ is compact.

c) Let $\Omega \subset \mathbb{R}^{2}$ a bounded open set with boundary of class $C^{1} .$ Prove that there no exists a constant $C>0$ such that
$$
\|u\|_{L^{\infty}(\Omega)} \leq C\|u\|_{H^{1}(\Omega)} \quad \forall u \in H^{1}(\Omega)
$$

So I proved the part a) indeed the constant is independet from $p$ in this case $p=2.$ For b) I have no idea because I saw the same problem with the compact $[a,b]$ but for that problem the solution used the Arzelá-Ascoli Theorem so compactness is important, so I have no idea how to prove b).
And c) is also a mystery.

答:
You said you can do b) with compact [a,b]. This approach can be salvaged because you can extend an $H^1$ function from $(a,b)$ to $[a,b]$ as a (Hölder) continuous function.

For c) try to construct a radially symmetric singular function whose $H^1$ norm is finite.

Title: Sobolev injections in $\mathbb{R}$


答:
$I=\int_{0}^{\infty} f(x) dx$ converges if in $f(x)$ is finite and continuous (or piece-wise continuous) in $[0,\infty)$ and $|f(x)|$ vanishes faster that $|x|^{-1}$ as $x \to \infty$. For instance
$I_1=\int_{0}^{\infty} \frac{dx}{1+x} dx$ diverges, $I_2=\int_{0}^{\infty} \frac{dx}{1+x^2} dx$ conveges, $I_3=\int_{0}^{\infty} \frac{dx}{(1+x^4)^{1/4}}$ diverges because the integrand when $x \sim \infty$ behaves as $f(x)\sim  \frac{1}{|x|^\beta}, \beta=1$. For convergence, $\beta >1$.

Title: How do we know if an integral converges or diverges from a limit?

Body: If we know that the limit of $f(x)$ approaches infinity equals some value like $5$, how can I show that $$\int_0^\infty f(x)\text{ }dx$$ converges or diverges?


问:
Let $f:[a, b]\rightarrow\mathbb{R}$ be differentiable. If $f'(a)=f'(b)$, then exists a $c \in (a, b)$, such that $f'(c) = \frac{f(c) - f(a)}{c - a}$
In the book (Curso de Análise, volume 1,Elon Lages), there is a suggestion that helps a lot.

First, consider that $$f'(a) =f'(b)=0$$ Then, consider the function $g:[a,b] \rightarrow \mathbb{R}$, where $g(x) = \frac{f(x) - f(a)}{x - a}$ and $g(a) = 0$. Show that $g$ reaches it's maximum or minimum in a point $c \in (a,b)$. For the general case, consider $$g(x) = f(x) - xf'(a)$$

I can see why the first case: if a take the derivative of g, I end up with something like:

$$g'(x) = \frac{1}{x - a} \left( f'(x) - \frac{f(x) - f(a)}{x - a} \right)$$

So, by being continuous (from the differentiable hypothesis) on a compact set, by Weierstrass's Theorem, we have that $g$ must have it's maximum/minimum on $c \in [a,b]$. By being a critical point, we must have $g'(c) = 0$, and assuming $c \neq a$, we have our first conclusion.

But (1) I can't see why it must be a interior point (seriously, I've been on this question for 4 days), and (2) the second suggestion isn't that clear for me.

Any other ideas for solutions will be of great help to me.

答:
There are several steps involved in the first part and here are some hints:

If $g$ does not have  a local minimum or a local maximum in $(a,b)$ then $g$ is necessarily strictly monotonic. This implies that $f$ is concave or convex (depending on whether $g$ is increasing or decreasing). Hence $f'$ is monotonic. The fact that $f'(a)=f'(b)=0$ shows that $f'\equiv 0$ and $f$ is a constant. So any $c \in (a,b)$ is good enough.

The second part is straightforward. Just apply the first part to $g(x)=f(x)-xf'(a)$. Since $g'(a)=g'(b)=0$ this is possible). Simplify the equation $g'(c)=\frac {g(c)-g(a)} {c-a}$ to finish the proof.


问:
Is set x+y where x$\in G_1$ and $y \in G_2$ open or closed or neither( Description of open and closed sets)
The following question was part of topology quiz and I was unable to think on how should I approach this question. A similar question was also asked in another quiz and I could also not answer it.

Let $G_1 $ and $G_2$ be two subsets of $\mathbb{R}^2$ and $f: \mathbb{R}^2 \to \mathbb{R}^2$ be a function. Then if $G_1$ is open and $G_2$ is closed then $G_1 +G_2$ ={x+y : $x\in G_1$ , $y\in G_2$} is neither open nor closed.

The problem is i am unable to see how $G_1 +G_2$ looks like( I mean how to write an open set and closed set in it) . What does + means to elements of set $G_1 $ and $G_2$?

That's why I could not solve this problem and require your help.

答:
Suppose $y \in G_2$. Then since $G_1$ is an open set, we have $G_1 +y = \{x + y : x \in G_1 \}$ is an open set too. Now take union over $y \in G_2$. What can you say from here?

 In my opinion, $G_1$ $+$ $G_2$ should be an open set


问:
How to solve the recurrence relation $a_{1}=2, a_{n}=\frac{a_{n-1}+2}{2 a_{n-1}+1}(n \geq 2)$ with generating functions?
There's already a way to solve it, called "fixed point method", that is, from the relation we define its characteristic equation as $x=\dfrac{x+2}{2x+1}$，then we have $x_1=1,x_2=-1$. So the following relation established:
$$
\frac{a_{n}-1}{a_{n}+1}=\frac{\frac{a_{n-1}+2}{2 a_{n-1}+1}-1}{\frac{a_{n-1}+2}{2 a_{n-1}+1}+1}=-\frac{1}{3} \cdot \frac{a_{n-1}-1}{a_{n-1}+1}
$$
It is obvious that $\displaystyle \frac{a_{n}-1}{a_{n}+1}=\frac{1}{3} \cdot\left(-\frac{1}{3}\right)^{n-1}$, and then we have $a_{n}=\dfrac{3^{n}-(-1)^{n}}{3^{n}+(-1)^{n}}$.

My question is, how to solve this kind of recurrence relations with generating functions? Also, "fixed points" can be applied to solving recurrences like $a_{n+1}=\dfrac{a_{n}^{2}+b}{2 a_{n}+d}$, which seems impossible to solve using generating functions.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{%
\left\{\begin{array}{rcl}
\ds{a_{1}} & \ds{=} & \ds{2}
\\
\ds{a_{n}} & \ds{=} &
\ds{{a_{n - 1} + 2 \over 2 a_{n - 1} + 1}\,,\quad n \geq 2}
\end{array}\right.}}$

$\ds{\large Another\ Method}$: Lets $\ds{a_{n} \equiv x_{n}/y_{n}}$ such that
\begin{align}
&\bbox[5px,#ffd]{a_{n}  = {x_{n} \over y_{n}} =
{x_{n - 1}\,/y_{n - 1} + 2 \over 2x_{n - 1}\,/y_{n - 1} + 1} = {x_{n - 1} + 2y_{n - 1} \over 2x_{n - 1} + y_{n - 1}}}
\end{align}
and sets $\ds{\quad x_{n} = x_{n - 1} + 2y_{n - 1}\,,\quad
y_{n} = 2x_{n - 1} + y_{n - 1}}$ which is equivqlent to
\begin{align}
\pars{\begin{array}{c}\ds{x_{n}} \\ \ds{y_{n}}\end{array}}
& =
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}
\pars{\begin{array}{c}
\ds{x_{n - 1}} \\ \ds{y_{n - 1}}
\end{array}} =
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}^{2}
\pars{\begin{array}{c}
\ds{x_{n - 2}} \\ \ds{y_{n - 2}}
\end{array}}
\\[5mm] & = \cdots =
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}^{n - 1}
\pars{\begin{array}{c}
\ds{x_{1}} \\ \ds{y_{1}}
\end{array}}
\end{align}
The above matrix has eigenvalues $\ds{\lambda_{1} = 3}$ and $\ds{\lambda_{2} = -1}$ with,
respectively, orthonormal eigenvectors
$\ds{{\bf u}_{1} = {1 \over \root{2}}{1 \choose 1}}$ and
$\ds{{\bf u}_{2} = {1 \over \root{2}}
{-1 \choose \phantom{-}1}}$. Then,
\begin{align}
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}} & =
\sum_{j = 1}^{2}\lambda_{j}\,{\bf u}_{j}\,{\bf u}_{j}^{T}
\\[5mm] \mbox{and}\
\pars{\begin{array}{cc}
\ds{1} & \ds{2}
\\
\ds{2} & \ds{1}
\end{array}}^{n - 1} & =
\sum_{j = 1}^{2}\lambda_{j}^{n - 1}\,\,
{\bf u}_{j}\,{\bf u}_{j}^{T}
\\[2mm] & =
{ 1 \over 2}\pars{\begin{array}{cc}
\ds{3^{n -1} - \pars{-1}^{n}} & \ds{3^{n -1} + \pars{-1}^{n}}
\\
\ds{3^{n -1} + \pars{-1}^{n}} & \ds{3^{n -1} - \pars{-1}^{n}}
\end{array}}
\end{align}
Therefore,
\begin{align}
a_{n} & =
{\bracks{3^{n - 1} - \pars{-1}^{n}}\ \overbrace{x_{1}/y_{1}}^{\ds{= a_{1} = 2}}\ +\ 3^{n - 1} + \pars{-1}^{n} \over
\bracks{3^{n - 1} + \pars{-1}^{n}}x_{1}/y_{1} + 3^{n - 1} - \pars{-1}^{n}}
\\[5mm] & =
\bbx{3^{n} - \pars{-1}^{n} \over 3^{n} + \pars{-1}^{n}} \\ &
\end{align}


问:
Topologies and manifolds
This question might seem philosophical a bit:

in a standard manifolds introductory course. when one talks about open , closed sets in $\mathbb{R}^n$ it's always the standard euclidian topology that we think about. Is there a more general framework in which one defines manifolds in different topologies in a way that generalize the standard setting that is closer to the physical world of course.

Thank you in advance for your answers

答:
The definition of a manifold is reasonably nice topological space that locally looks like Euclidean space. More formally, $M$ is a topological manifold of dimension $n$ if $M$ is second countable and Hausdorff and for each $p \in M$ there exists an open set $U \ni p$ (in whatever topology $M$ has) such that there is a homeomorphism $U \stackrel{\approx}{\to} V$, where $V \subset \mathbb R^n$, open.

It sounds like you've either only seen manifolds as subsets of Euclidean space or you're wondering if we can replace the homeomorphism with subsets of $\mathbb R^n$, with the usual topology, with subsets of $\mathbb R^n$ with a different topology. In either case, hopefully understanding the formal definition helps. We only call something a manifold if it locally looks like $\mathbb R^n$ with the usual topology.

Title: Topologies and manifolds


答:
Hint:

Let $x_r=\tan(y_r)$

$$x_{n+1}=\tan(y_{n+1})=\tan\left(\dfrac\pi4+y_n\right)$$

$$\implies\tan(y_{n+r})=\tan(y_n)\text{  if }4\mid r$$

and $$\tan(y_{n+r})=-\dfrac1{\tan(y_n)}\text{  if }r\equiv2\pmod4$$

Body:
Let $x_0 = 2003$, and let $x_{n+1} = \frac{1 + x_n}{1 - x_n}$ for $n >= $1. Compute $x_{2019}$

I tried expressing $x_2$, $x_3$, $x_4$.. in terms of previous terms to see if there is a pattern but $n >= 1$ is troubling me. Please help.


问:
On Solution of FLT for exponent 3
There is a simple proof that $a^3$ + $b^3$ + $c^3 = 0$ has no solutions in integers, is this really the same thing for the actual case of $a^3$ + $b^3 = c^3$, what is the connection between the two equations, how does a no-solution for the first form translates into a no-solution for the second form?

答:
$$a^3 + b^3 = c^3 \iff a^3 + b^3 - c^3 = 0$$
$$\iff a^3 + b^3 + (-c)^3 =0 $$
so (a,b,c) is a solution of the second equation if and only if $(a,b,-c)$ is a solution of the second one, and this gives  a bijection between solutions to the two equations over any ring.

Title: On Solution of FLT for exponent 3


答: I assume for both equations the assumption is that $a, b, c \neq 0$, otherwise both equations have solutions. Alex's solution is right: $(a, b, c)$ satisfies the first equation iff $(a, b, -c)$ satisfies the second. I just want to remark that FLT asserts that there are no solutions to $a^3 + b^3 = c^3$ where $a, b, c \in \mathbb{N}$. However, this is equivalent to asserting that there are no solutions to $a^3 + b^3 = c^3$ where $a, b, c \in \mathbb{Z} \setminus \{0\}$. Note that if all $a^3, b^3, c^3$ have the same sign, then this is equivalent to FLT. If they don't all have the same sign, then one of them has the opposite sign as the others. Then simply re-arrange so that the one with the different sign is isolated on one side of the equation, and by renaming variables we will have arrived at the exact same statement as FLT with the natural numbers.

Title: On Solution of FLT for exponent 3

Body: There is a simple proof that $a^3$ + $b^3$ + $c^3 = 0$ has no solutions in integers, is this really the same thing for the actual case of $a^3$ + $b^3 = c^3$, what is the connection between the two equations, how does a no-solution for the first form translates into a no-solution for the second form?


问:
Am I missing something, or does this simple PDE lack an explicit solution due to the nature of its (also simple) boundary conditions?
Short version:
Below, I present a simple PDE with simple boundary conditions (BCs), which has a simple solution. I then modify one of the BCs, and end up with a transcendental equation for the wave number parameter, preventing me from going further. This surprises me, since the resulting problem doesn't seem too complicated. My main question: Am I right to conclude that an explicit solution is impossible to find in this case?

Long version:
Consider the equation (EQ1) (Note: I'm starting from the heat conduction equation here but omitting some constants for clarity, consider all variables and constants dimensionless)
$$ \theta_{xx} = \theta_t, \qquad 0 < x < a,$$
where $\theta = \theta(x,t)$ and the subscripts denote partial derivatives. The boundary conditions are (BC1 and BC2, resp.)
$$ \theta_x = 0, \qquad x = 0, $$
$$ \theta_x = 0, \qquad x = a. $$
The initial condition IC is
$$ \theta = \theta_0(x), \qquad t = 0. $$

The solution is obtained by a straightforward application of separation of variables: $\theta(x,t) = X(x)T(t) \Rightarrow$
$$ \frac{X''(x)}{X(x)} = \frac{T'(t)}{T(t)} = -\lambda^2 \Rightarrow $$
$$ T(t) = C_1\exp({-\lambda^2t}), $$
$$ X(t) = C_2\sin(\lambda x) + C_3\cos(\lambda x). $$
Now, BC1 $\Rightarrow C2 = 0$ and BC2 $\Rightarrow$
$$ -C_3\lambda\sin(\lambda a) = 0 \Rightarrow \sin(\lambda a) = 0 \Rightarrow $$
$$ \lambda_n = \frac{n\pi}{a}. $$
It follows that
$$ X_n(x) = C_3\cos(\frac{n\pi x}{a}) $$
and finally
$$ \theta(x,t) = \sum_{n=0}^{\infty}c_n\cos(\frac{n\pi x}{a})\exp({-\frac{n^2\pi^2 t}{a^2}}). $$
Now, the coefficients $c_n$ can be obtained by applying the initial condition and using the trigonometric orthogonality property. The latter requires knowledge of the spatial period of the solution.

Now, consider replacing BC2 with
$$ \theta_x = \theta, \qquad x = a. $$
Then, instead of the $\lambda_n$ given above, we would obtain
$$ -C_3\lambda\sin(\lambda a) = C_3\cos(\lambda a) \Rightarrow $$
$$ -\lambda\tan(\lambda a) = 1. $$
This is a transcendental equation in $\lambda$, so no explicit expression for $\lambda$ can be obtained.

Does this mean that we can't go any further? That more or less complicated numerical methods are the only way to proceed? Or am I missing something here?

答:
No, you do not miss anything: no explicit $\lambda$s can be found. However, the general theory tells us:

there are infinitely many lambda's

there is minimal $\lambda$ and there is no maximal one

the eigenfunctions corresponding to lambdas are orthogonal on the interval

Hence we can always write the solution as an infinite series. And if you want to actually calculate the function, you will need a numerical routine to find your eigenvalues (this is really simple in your case)

Added: About the point, raised by @Lost1 in comments. Note that your boundary conditions imply (physically) that the left side of the bar is isolated, and at the right side of the bar the flux of temperature is equal to the temperature itself. This means that there will be always incoming flux of the energy in your bar, which means obviously that the temperature has to approach infinity, which implies that you have to assume that you constant of separation is positive. Take not $-\lambda^2$ but $\mu^2$ and see the implications. And this also means that $\cos$ are not the only eigenfunctions in your problem.


问:
Solve for $x$: $a_1\ln(1+b_1x)+a_2\ln(1+b_2x)=0$, maybe with non-elementary function?
Given: $a_i>0$, $b_1>0$, $b_2<0$

Solve for: $a_1\ln(1+b_1x)+a_2\ln(1+b_2x)=0$

My try: The equation is equivalent to $(1+b_1x)^{a_1}=(1+b_2x)^{-a_2}$,

Which seems to be solved with only non-elementary functions...

Is it possible to prove that no elementary function can solve this and then find a non-elementary one to solve this?

答:
You want to find the zero's of function
$$f(x)=a_1\log(1+b_1x)+a_2\log(1+b_2x)$$ This is an highly transcendental equation and there is no hope to find an explicit solution if $a_1 \neq a_2$; this means that you will need some iterative numerical methods and then a reasonable starting guess.

For the first derivatives, we have
$$f'(x)=\frac {a_1b_1 }{1+b_1x }+\frac {a_2b_2 }{1+b_2x }$$
$$f''(x)=-\frac {a_1b_1^2 }{(1+b_1x)^2 }-\frac {a_2b_2^2 }{(1+b_2x)^2 }\quad <0 \quad \forall x$$

The first derivative cancels at
$$x_*=-\frac {a_1b_1 +a_2b_2 } {(a_1+a_2)b_1b_ 2}$$ which may exist (the requirement is $x < -\frac 1{b_2}$ because of the logarithm). If it does and $f(x_*)>0$, there will two roots since, by the second derivative test, $x_*$ corresponds to a maximum. But $x=0$ being a solution, we need the other one in the reduced range
$$-\frac 1 {b_1} < x <  x_*$$

Make a series expansion of $g(x)$ around $x_*$
$$f(x)=f(x_*)+\frac{1}{2} f''(x_*)(x-x_*)^2 +O\left((x-x_*)^3\right)$$ and solve it to get, as an approximation,
$$x_0=x_*- \sqrt{-2 \frac{f(x_*)}{f''(x_*)}}$$

Let us try with $a_1=1$, $b_1=3$, $a_2=5$, $b_2=-7$. These conditions give
$$x_*=-\frac{16}{63} \qquad \text{and} \qquad f(x_*)=5 \log \left(\frac{25}{9}\right)-\log \left(\frac{21}{5}\right) >0$$  but is gives
$$x_0=-\frac{16}{63}-\frac{5}{63} \sqrt{\frac{5}{3} \left(5 \log
   \left(\frac{25}{9}\right)-\log \left(\frac{21}{5}\right)\right)}\approx -0.450338$$ which is outside the range $\left(-\frac 13,-\frac{16}{63} \right)$; so, this does not work but we reduced significantly the range to be explored since now we know that the solution is such that
$$-\frac 13 <x<-\frac{16}{63}$$

So, at least for the time being, what I would suggest is to divide the new interval (say $100$ equally spaced points)  and pick as a starting guess the last value of $x$ for which $f(x) <0$ (application of Darboux theorem).

For the worked example, we shall have to compute the $f(x_i)$
$$x_i=-\frac 13+\frac 1 {100}\left(-\frac{16}{63}+\frac 13\right) i=\frac{i}{1260}-\frac{1}{3}$$ and $x_1=-\frac{419}{1260}$ becomes our starting point. Now, using (as the simplest) Newton method, the iterates will be
$$\left(
\begin{array}{cc}
 n  & x_n \\
 0 & -0.3325396825 \\
 1 & -0.3325166882 \\
 2 & -0.3325163490
\end{array}
\right)$$

What we could try is one iteration of  the quintic order of Newton-like method starting with $x_0=x_*$; this order is required being the first which makes the function value  appearing in denominator.

It will write
$$x_1=x_*-\frac{4 \,f(x_*)\, f'''(x_*)}{6 f''(x_*)^2-f(x_*) \,f''''(x_*)}$$

Applied to the worked problem, it gives
$$x_1=\frac{32 \left(45+220 \log \left(\frac{25}{9}\right)-44 \log
   \left(\frac{21}{5}\right)\right)}{567 \left(10+35 \log
   \left(\frac{25}{9}\right)-7 \log \left(\frac{21}{5}\right)\right)}\approx -0.326529$$


问:
Prime ideals of the ring $\mathbb{Q}[X]/ (X)^3$
The following question was part of my abstract algebra quiz and I am unable to solve it.

Let A denote the ring $\mathbb{Q}[X]/ (X)^3$ . Then is there only 1 prime ideal of A?

Ideals of $A$  are $x/\langle x^3 \rangle$ , $x^2 / \langle x^3 \rangle$  and $1$ and $ 1/\langle x^3 \rangle$ .

I proved $x/\langle x^3 \rangle$  prime. THen I tried with $x^2 / \langle x^3 \rangle$. Here any element of $A/(x^2 / \langle x^3 \rangle)$ would be of the form $a_0 + a_1 x$ but I am neither able to prove it prime or not prime.

As other ideals would be of the same form so I am looking for you help to do this part.

答:
$A/((X)/(X^3))=(\mathbb{Q}[X]/(X^3))/((X^2)/(X^3))\cong \mathbb{Q}[X]/(X^2)$.

Now $X\neq 0\in \mathbb{Q}[X]/(X^2)$ and $X\cdot X=0\in \mathbb{Q}[X]/(X^2)$,
so $\mathbb{Q}[X]/(X^2)$ is not integral domain.


问:
Find an explicit formula for $ \sum_{n=0}^{\infty}{x^{n^2}} ,\quad \forall x \in (0,1) $
My question askes me to find an explicit formula for
$$ \sum_{n=0}^{\infty}{x^{n^2}} \quad\left(\forall x \in (0,1)\right)$$
And I feel it kind of interesting to find an appropriate f(x) that satisfies
$$ \lim_{x \to 1^{-} }\frac{\sum_{n=0}^{\infty}{x^{n^2}}}{f(x)}=1 $$
Both questions make me feel puzzled, and I tried
$$ \sum_{n=0}^{\infty}{x^{n^2}}<\sum_{n=0}^{\infty}{x^n}=\frac{1}{1-x} $$
Hence I guess maybe
$$ \sum_{n=0}^{\infty}{x^{n^2}}\sim\frac{C_{1}}{\sqrt{1-x}} \quad \text{when} \quad x \to 1^{-}$$
or at least $$ \sum_{n=0}^{\infty}{x^{n^2}} \sim \frac{C_{2}}{(1-x)^\alpha} \quad \text{for} \quad \text{some} \quad \alpha$$
Any help or recommendation related would be greatly appreciated.

Thanks a lot for your focus.

答: $$\sum_{n=0}^{\infty}{x^{n^2}} =\frac{1}{2} (\vartheta _3(0,x)+1)$$ This is the definition of the theta function.


问:
Clear my doubt regarding this probabililty concept.
A bag contains 2 red and 3 green balls. 2 balls are draw at random. Find the probability that the ball is red and green.What is the correct solution to the problem.

Solution 1- 2/5 * 3/4 = 3/10 i.e probability of getting red multiplied by probability of getting green from the remaining balls.

Solution 2- 2C1 * 3C1/ 5C2 = 2*3/10=3/5 Using combinations.

I cant understand the difference between the two. In many books answer is given as solution 1, simple way. Whereas is some others for the same kind of questions, they are doing using combinations.
In which type of questions can we use the simple method and when we have to take combinations for finding the probability.

答:
It seems that the question is asking the probability of getting a red ball first and then green ball, while you are calculating the probability that you get a red ball and a green ball, not caring about the order.

Notice that if you divide your answer by 2, you get the textbook solution. This is because you must divide by 2! = 2 to get only the case where you get red ball first and then the green ball.

So basically first solution requires first red ball and second green ball, while the second solution does not require order.


问:
cardinality of a set with repeating elements?
What is the cardinality of a set which has repeating elements ?
For example $S = \{1,1,1,2,2\}$
Is each individual element counted?
Please quote a reference text if possible. </p>

答:
The set $A = \{1,1,1,2,2\}$ is identical to the set $B = \{1, 2\}$ because one can show $A\subseteq B$ and $B\subseteq A$ and therefore $A=B$. Both have cardinality 2.

For example, in this answer I quoted Halmos:

The ordered pair of a and b… is the set $(a, b)$ defined by:
  $$(a, b) = \{\{a\}, \{a,b\}\}.$$
  …
  We note first that if $a$ and $b$ happen to be equal, then the ordered pair
  $(a, b)$ is the same as the singleton $\{\{a\}\}$.

This isn't precisely what you want, but it is a consequence of it: if $a$ and $b$ happen to be equal then the set $S = \{\{a\}, \{a,b\}\}$ is $\{\{a\}, \{a, a\}\}$, and the repeated $a$ in the $\{a,a\}$ element can be ignored, since $\{a,a\} = \{a\}$.  Then $S = \{\{a\}, \{a\}\}$ and the repeated $\{a\}$ can be ignored, so $S$ is really just $\{\{a\}\}$.

That thread has a number of other references, particularly in this post by Brian Scott.


问:
Are the minimal polynomials of $A$ and $A^T$ related?
I'm wondering if the minimal polynomials of $A, A^T\in\mathbb{M}_{n\times n}(\mathbb{F})$ are somehow related? Are they the same?

If $m_A(t)$ is the minimal polynomial of $A$, then $m_A(A) = 0$. So, $(m_A(A))^T = 0$. If two matrices $P,Q$ commute then $(PQ)^T = P^TQ^T$. Using this, we can see that $m_A(A^T) =0$.

Is this enough to conclude that $m_A(t)$ is also the minimal polynomial for $A^T$?

P.S.
Does this depend on whether or not the minimal polynomial splits over $\mathbb{F}$?

答:
$m_A$ and $m_{A^T}$ are the minimal polynomials for $A$ and $A^T$ respectively.

So we have $m_A(A) = 0$ and $m_{A^T}(A^T) = 0$ by definition. Let $m_A(t) = p_1(t)p_2(t)...p_k(t)$ where $p_i(t)\forall 1\leq i\leq k$ are the irreducible factors of $m_A(t)$.

$m_A(A) = 0 \implies (m_A(A))^T = 0$, which is just $(p_1(A)p_2(A)...p_k(A))^T = 0$. Since polynomials in $A$ commute, this is just $(p_1(A))^T(p_2(A))^T...(p_k(A))^T = 0 \implies p_1(A^T)p_2(A^T)...p_k(A^T) = 0$. So $m_A(A^T) = 0$, which means that $m_{A^T}\vert m_A$.

Starting with $m_{A^T}(A^T) = 0$, we can similarly prove that $m_A\vert m_{A^T}$.

So $m_A(t) = m_{A^T}(t)$


问:
About the rank of Jordan blocks

What can be said about the rank of Jordan blocks of size $n$? When is rank $J_n(\lambda) = n$?

From my observation, it seems that whenever $\lambda\neq 0$, the corresponding Jordan block has full rank (i.e. $n$). This seems clear from the fact that $J_n(\lambda)$ is upper triangular, and if we select any two columns $c_1,c_2$ and put $ac_1 + bc_2 = 0$, then we must have $a=b=0$. However this fails when $\lambda = 0$, and the matrix becomes strictly upper triangular.

I would appreciate any hints on how to be able to make generalized comments about the rank of Jordan blocks. Thank you!

答:
I am adding this answer just for the sake of completeness, it is definitely trivial and unnecessary:

$$\text{rank }J_n(\lambda)= n-1 \text{ if }\lambda=0$$ and $$\text{rank }J_n(\lambda)=n \text{ if }\lambda\neq 0$$

Title: About the rank of Jordan blocks


问:
Example of Finite Group and Homomorphism
I'm new to Math Stack Exchange.

I saw a problem about a finite group, $G$, and a homomorphism that maps an element of $G$ onto another element of $G$ (the mapping doesn't necessarily have to be to a distinct element).

What exactly might this look like? Could someone give a super trivial example of what this would be?

答:
Take $G = \{0\}$, the trivial group. Define $f \colon G \rightarrow G$ by $f(0) = 0$.

Take $G = {0,1}$ $0+0 = 0$, $1+0 = 0+1 = 1$, $1+1 = 0$. Take $f \colon G \rightarrow G$ by $f(0) = 0$, $f(1) = 1$.


问:
Understanding equivalence classes
I am struggling to conceptualize and answer the following question. Consider the relation $R$ that consists of all pairs $(x,y)$ such that $x$ and $y$ are bit strings of length three or more that agree except perhaps in their first three bits. What are the equivalence classes, and how many are there?

I know that $R$ is an equivalence relation, as it is Reflexive (all bits of $x$ are the same as the bits of itself), symmetric (all bits of $x$ (except perhaps the first three) are the same as all bits of y (except perhaps the first three)), and Transitive.

Any help is appreciated!

答:
The elements of a particular equivalence class all share the bits after the 3rd bit in common which means that to every equivalence class can be associated a unique bit-string.

For example: 010 11011 and 011 11011 belong to the same equivalence class, so this equivalence class corresponds to the bit-string 11011 uniquely.

In particular, this means that the equivalence classes of $R$ are bijective with the set of bit-strings itself (so their number is infinite for example).

This is the bijection: $a \mapsto \{000a, 001a, 010a, 011a, 100a, 101a, 110a, 111a\}$

Title: Understanding equivalence classes


问:
Show that $n | \sum_{k=1}^n m^{\textrm{gcd}(k, n)}$ where $n, m \geq 1$
I must show that $n$ divides $\sum_{k=1}^n m^{\textrm{gcd}(k, n)}$ where $n, m \geq 1$.

When $n$ is prime, Fermat's theorem gives us the answer easily because then $\sum_{k=1}^n m^{\textrm{gcd}(k, n)} = m^n + (n-1)m \equiv 0  [n]$.

In the general case, how to show this? I would like to use mostly integers/arithmetics arguments, not too high level theories.

答:
In order to give a fully arithmetic answer to the problem, let us introduce a couple of notations:

we denote by $\mathbb{P}$ the set of all strictly positive prime integers, in other words the set of what is commonly understood as prime numbers (from the strict algebraic point of view, $0$ and $-2$ are both prime elements of the ring $\mathbb{Z}$).

for given $r \in \mathbb{Z}$ we denote by $\mathrm{D}(r)\colon=\{k \in \mathbb{N}|\ k \mid r\}$ the set of all its natural divisors.

for given $r \in \mathbb{Z}$ we introduce the set $\Pi(r)\colon=\{p \in \mathbb{P}|\ p\mid r\}$ of all its prime divisors.

for given $n \in \mathbb{N}$ we define the set $\mathrm{P}(n)\colon=\{k \in \mathbb{N}|\ 1 \leqslant k \leqslant n \wedge (k; n)=1\}$ of all nonzero natural numbers less than $n$ and coprime with it. We have also tacitly used the notation $(r; s)$ to indicate the greatest common divisor of given integers $r, s \in \mathbb{Z}$ (the role of the semicolon in the notation being to prevent confusion with the syntax for ordered pairs). When $n \in \mathbb{N}^{\times}\colon=\mathbb{N} \setminus \{0\}$, we have by definition that $|\mathrm{P}(n)|=\varphi(n)$.

for any two integers $r, s \in \mathbb{Z}$, we denote by $[r, s]$ the integer and not the real interval between $r$ and $s$.

for any $n \in \mathbb{N}^{\times}$ we introduce the polynomial $P_n\colon=\displaystyle\sum_{k=1}^n X^{(k; n)} \in \mathbb{Z}[X]$.

With the above notation, our goal is to show that for any $n \in \mathbb{N}^{\times}$ and any $m \in \mathbb{Z}$ the relation $n \mid P_n(m)$ holds, or more succinctly put that the subset:
$$M\colon=\left\{n \in \mathbb{N}^{\times}|\ \{P_n(m)\}_{m \in \mathbb{Z}} \subseteq n\mathbb{Z}\right\}$$
is equal to its entire ambient set $\mathbb{N}^{\times}$.

Before moving on to the core of the proof, a preliminary result is needed:

Proposition. For every $n \in \mathbb{N}^{\times}$ we have the expression $P_n=\displaystyle\sum_{d \in \mathrm{D}(n)}\varphi\left(\frac{n}{d}\right)X^d=\displaystyle\sum_{d \in \mathrm{D}(n)}\varphi(d)X^{\frac{n}{d}}$.

Proof. The second equality is a simple consequence of the fact that the map:
$$\begin{align*}
\mathrm{D}(n) &\to \mathrm{D}(n)\\
d &\mapsto \frac{n}{d}
\end{align*}$$
is an involutive permutation of the set $\mathrm{D}(n)$ of summation indices. To establish the first one, let us start by remarking that the map:
$$\begin{align*}
\delta \colon [1, n] &\to \mathrm{D}(n)\\
\delta(k)&=(k; n)
\end{align*}$$
is clearly a surjection whose fibre at every element in the codomain is given by $\delta^{-1}[\{d\}]=d\mathrm{P}\left(\frac{n}{d}\right)$. This means that $\left(d\mathrm{P}\left(\frac{n}{d}\right)\right)_{d \in \mathrm{D}(n)}$ is a partition of $[1, n]$ and we thus have:
$$P_n=\sum_{d \in \mathrm{D}(n)}\sum_{k \in d\mathrm{P}\left(\frac{n}{d}\right)}X^{\delta(k)}=\sum_{d \in \mathrm{D}(n)}\sum_{k \in d\mathrm{P}\left(\frac{n}{d}\right)}X^d=\sum_{d \in \mathrm{D}(n)}\left|d\mathrm{P}\left(\frac{n}{d}\right)\right|X^d=\sum_{d \in \mathrm{D}(n)}\varphi\left(\frac{n}{d}\right)X^d,$$
concluding our proof. $\Box$

We organise the proof in the following two steps:

first, we show that for any $m, n \in M$ such that $(m; n)=1$ we also have $mn \in M$.

second, we show that for any prime $p \in \mathbb{P}$ and for any exponent $k \in \mathbb{N}$ we have $p^k \in M$.

Establishing the two assertions above is sufficient, for it allows us to prove that $\mathbb{N}^{\times} \subseteq M$ by induction on $|\Pi(n)|$ for arbitrary nonzero natural $n$ (argument which of course relies on the Fundamental Theorem of Arithmetic; note that the base case of this induction corresponds to the claim $1 \in M$, which is clearly true). Let us now justify each of the two claims above:

Here we begin by establishing another auxiliary result:

Proposition. Let $m, n \in \mathbb{N}^{\times}$ be coprime, $(m; n)=1$. Then the map:
$$\begin{align*}
\mathrm{D}(m) \times \mathrm{D}(n) &\to \mathrm{D}(mn)\\
(k, l) &\mapsto kl
\end{align*}$$
is a bijection.

Proof. Consider first two pairs of divisors $(k, l), (k', l') \in \mathrm{D}(m) \times \mathrm{D}(n)$ such that $kl=k'l'$. Since $l, l' \mid n$ it is obvious that $(m; l), (m; l') \mid (m; n)=1$ and therefore $(m; l)=(m; l')=1$. Thus, from one of the fundamental properties of the g.c.d operation it follows that $(m; kl)=(m; k)=k$ and analogously $(m; k'l')=(m; k')=k'$. Since $kl=k'l'$ we immediately deduce that $k=k'$ and subsequently $l=l'$ (for $0 \notin \mathrm{D}(n)$ as long as $n\neq0$). This shows that the map introduced above is injective. As to its surjectivity, consider an arbitrary $h \in \mathrm{D}(mn)$ and define $k\colon=(m; h) \in \mathrm{D}(m)$. If we further introduce $l\colon=\frac{h}{k}$ we gather that $\left(\frac{m}{k}, l\right)=1$, by virtue of another fundamental property of the g.c.d operation. From $h \mid mn$ we infer that $l \mid \frac{m}{k}n$ and taking into account that $l$ and $\frac{m}{k}$ are relatively prime we gather that $l \mid n$. Thus, $l \in \mathrm{D}(n)$ and we have expressed $h=kl$ as a product of the divisors $k \in \mathrm{D}(m)$ and $l \in \mathrm{D}(n)$, which means that our map is also surjective. $\Box$

We continue by putting to use the result above in order to remark that given coprime nonzero natural numbers $m$ and $n$ we have:
$$\begin{align*}
P_{mn}&=\sum_{h \in \mathrm{D}(mn)}\varphi\left(\frac{mn}{h}\right)X^h\\
&=\sum_{\substack{k \in \mathrm{D}(m) \\ l \in \mathrm{D}(n)}}\varphi\left(\frac{mn}{kl}\right)X^{kl}\\
&=\sum_{\substack{k \in \mathrm{D}(m) \\ l \in \mathrm{D}(n)}}\varphi\left(\frac{m}{k}\frac{n}{l}\right)X^{kl}\\
&=\sum_{k \in \mathrm{D}(m)}\sum_{l \in \mathrm{D}(n)}\varphi\left(\frac{m}{k}\right)\varphi\left(\frac{n}{l}\right)X^{kl}\\
&=\sum_{k \in \mathrm{D}(m)}\varphi\left(\frac{m}{k}\right)\sum_{l \in \mathrm{D}(n)}\varphi\left(\frac{n}{l}\right)\left(X^k\right)^l\\
&=\sum_{k \in \mathrm{D}(m)}\varphi\left(\frac{m}{k}\right)P_n\left(X^k\right)
\end{align*}$$
together with the analogous relation (obtained by considering $\mathrm{D}(n)$ as the outer domain of summation):
$$P_{mn}=\sum_{l \in \mathrm{D}(n)}\varphi\left(\frac{n}{l}\right)P_m\left(X^l\right).$$
The coprimeness of $m$ and $n$ is essential in order to justify the relation $\varphi\left(\frac{m}{k}\frac{n}{l}\right)=\varphi\left(\frac{m}{k}\right)\varphi\left(\frac{n}{l}\right)$, bearing in mind that $\left(\frac{m}{k}; \frac{n}{l}\right) \mid (m; n)=1$.

Let us now consider arbitrary $r \in \mathbb{Z}$. If $m, n \in M$ we have $m \mid P_m(s)$ and similarly $n \mid P_n(s)$ for any $s \in \mathbb{Z}$. From the above relations we obtain in particular that:
$$\begin{align*}
P_{mn}(r)&=\sum_{l \in \mathrm{D}(n)}\varphi\left(\frac{n}{l}\right)P_m\left(r^l \right) \in m\mathbb{Z}\\
P_{mn}(r)&=\sum_{k \in \mathrm{D}(m)}\varphi\left(\frac{m}{k}\right)P_n\left(r^k \right) \in n\mathbb{Z},
\end{align*}$$
hence $P_{mn}(r) \in m\mathbb{Z} \cap n\mathbb{Z}=mn\mathbb{Z}$, the latter relation between ideals of $\mathbb{Z}$ also being valid by virtue of $(m; n)=1$. This means that $mn \in M$ and establishes our first claim.

Here again we proceed by induction on the exponent $k$ in order to show that $p^k \in M$ for all $k \in \mathbb{N}$, the base case $k=0$ being equivalent to $1 \in M$ and thus trivial by definition of $M$. To this end, we first remark the general relation:
$$P_{p^k}=\sum_{i=0}^k \varphi\left(p^i\right)X^{p^{k-i}}=X^{p^k}+\sum_{i=1}^k p^{i-1}(p-1)X^{p^{k-i}},$$
based on the fact that the map:
$$\begin{align*}
[0, k] &\to \mathrm{D}\left(p^k\right)\\
i &\mapsto p^i
\end{align*}$$
is a bijection and we employ this general relation to derive the recursive relation:
$$\begin{align*}
P_{p^{k+1}}&=X^{p^{k+1}}+\sum_{i=1}^{k+1}p^{i-1}(p-1)X^{p^{k+1-i}}\\
&=X^{p^{k+1}}+\sum_{i=0}^k p^i(p-1)X^{p^{k-i}}\\
&=X^{p^{k+1}}+(p-1)X^{p^k}+\sum_{i=1}^k p^i(p-1)X^{p^{k-i}}\\
&=X^{p^{k+1}}+(p-1)X^{p^k}+p\sum_{i=1}^k p^{i-1}(p-1)X^{p^{k-i}}\\
&=X^{p^{k+1}}-X^{p^k}+p\left(X^{p^k}+\sum_{i=1}^k p^{i-1}(p-1)X^{p^{k-i}}\right)\\
&=X^{p^{k+1}}-X^{p^k}+pP_{p^k}.
\end{align*}$$
Consider now an arbitrary $m \in \mathbb{Z}$. Since by the inductive assumption we are given that $p^k \in M$, we infer $P_{p^k}(m) \in p^k\mathbb{Z}$ and subsequently $pP_{p^k}(m) \in p^{k+1}\mathbb{Z}$. Thus, in order to show that $P_{p^{k+1}}(m) \in p^{k+1}\mathbb{Z}$ for arbitrary $m \in \mathbb{Z}$ and hence conclude that $p^{k+1} \in M$, it suffices to establish the following “higher order” version of Fermat’s little theorem:
$$(\forall k, m)\left(k \in \mathbb{N} \wedge m \in \mathbb{Z} \Rightarrow m^{p^{k+1}} \equiv\ m^{p^k} \left(\mathrm{mod}\ p^{k+1}\right)\right),$$
for fixed prime $p \in \mathbb{P}$.

Let us first remark that for fixed $k \in \mathbb{N}$ and any multiple $m \in p\mathbb{Z}$ of $p$ we have the obvious relation $p^{p^{k}} \mid m^{p^k}\left(m^{p^k(p-1)}-1\right)=m^{p^{k+1}}-m^{p^k}$ and since $p \geqslant 2$ we can also infer that $p^k \geqslant 2^k \geqslant k+1$ (it is a fundamental Bernoulli-type inequality that $2^n \geqslant n+1$ for any $n \in \mathbb{N}$). It then follows that in the particular case $p \mid m$ we therefore have $p^{k+1} \mid p^{p^k} \mid m^{p^{k+1}}-m^{p^k}$, which establishes the higher order claim.

In order to prove this higher order version for non-multiples of $p$ we once again resort to induction on $k$. The base case $k=0$ is none other than Fermat’s little theorem. Assuming the claim to be valid for arbitrary $k \in \mathbb{N}$ we establish it for $k+1$. Let $m \in \mathbb{Z} \setminus p\mathbb{Z}$. We have the factorisation:
$$\begin{align*}
m^{p^{k+2}}-m^{p^{k+1}}&=m^{p^{k+1}}\left(m^{p^{k+1}(p-1)}-1\right)\\
&=m^{p^{k+1}}\left(\left(m^{p^k(p-1)}\right)^p-1\right)\\
&=m^{p^{k+1}}\left(m^{p^k(p-1)}-1\right)\sum_{l=0}^{p-1}\left(m^{p^k(p-1)}\right)^l\\
&=m^{p^k(p-1)}\left(m^{p^{k+1}}-m^{p^k}\right)\sum_{l=0}^{p-1}\left(m^{p^k(p-1)}\right)^l,
\end{align*}$$
from which we gather that, in conjunction with the induction hypothesis according to which $m^{p^{k+1}}-m^{p^k} \in p^{k+1}\mathbb{Z}$, in order to show that $m^{p^{k+2}}-m^{p^{k+1}} \in p^{k+2}\mathbb{Z}$ it suffices to show that $p \mid \displaystyle\sum_{l=0}^{p-1}\left(m^{p^k(p-1)}\right)^l=\sum_{l=0}^{p-1}\left(m^{p^kl}\right)^{p-1}$. However, since by hypothesis $p \nmid m$ we also have that $p \nmid m^{p^kl}$ for any $l \in [0, p-1]$, whence by virtue of Fermat’s little theorem we infer that $\left(m^{p^kl}\right)^{p-1} \equiv\ 1 \left(\mathrm{mod}\ p\right)$ and finally that:
$$\sum_{l=0}^{p-1}\left(m^{p^kl}\right)^{p-1} \equiv\ \sum_{l=0}^{p-1}1=p \equiv\ 0 \left(\mathrm{mod}\ p\right),$$
which fully establishes our claim and brings our argument to an end.


问:
Prove that $\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))$ for $0\leq x <1$
It's inspired by a general result proposed in a book (Dictionary of inequalities second edition by Peter Bullen p.101) related to composition of  function :

In fact we have $x\in[0,1)$:

$$\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))\quad (1)$$

To prove it I have introduced a function :

$$f(x)=\arctan(\operatorname{arctanh}(x))-\operatorname{arctanh}(\arctan(x))$$

We differentiate to get:

$$f'(x)= -\frac{1}{((x^2-1)(\operatorname{arctanh}^2(x)+1))}+\frac{1}{((x^2+1)(\arctan^2(x)-1))}$$

And then I use :

Let $0\leq x <1$ then we have :

$$\operatorname{arctanh}(x)\geq x \geq \arctan(x)$$

Then I cannot conclude directly .

Question :

How to prove $(1)$ ?

Thanks in advance !

Body:
Fact 1: $\frac{-3x^3+12x}{-x^3-x^2+12x+12}\sqrt{\frac{1+x}{1-x}} \ge \operatorname{arctanh}(x)$ for all $x \in [0, 1)$.

(Hint: Use $u^2 = \frac{1+x}{1-x}$ and then take derivative.)

Fact 2: $\frac{64x^5+735x^3+945x}{225x^4+1050x^2+945} \ge \arctan x$ for all $x \in [0, 1)$.

(Hint: Take derivative.)

Let $f(x) = \arctan(\operatorname{arctanh}(x)) - \operatorname{arctanh}(\arctan(x))$.

By Facts 1-2, we have, for all $x$ in $[0, 1)$,
\begin{align}
f'(x) &= \frac{1}{(1-x^2)(\operatorname{arctanh}^2(x)+1)} - \frac{1}{(x^2+1)(1 - \arctan^2(x))}\\
&\ge \frac{1}{(1-x^2)(\frac{(-3x^3+12x)^2}{(-x^3-x^2+12x+12)^2}\frac{1+x}{1-x}+1)}
- \frac{1}{(x^2+1)(1 - (\frac{64x^5+735x^3+945x}{225x^4+1050x^2+945})^2)}\\
&\ge 0.
\end{align}
Also, $f(0) = 0$. Thus, $f(x) \ge 0$ for all $x$ in $[0, 1)$.

We are done.


问:
Let $x_0 = 2003$, and let $x_{n+1} = \frac{1 + x_n}{1 - x_n}$ for $n >= $1. Compute $x_{2019}$

Let $x_0 = 2003$, and let $x_{n+1} = \frac{1 + x_n}{1 - x_n}$ for $n >= $1. Compute $x_{2019}$

I tried expressing $x_2$, $x_3$, $x_4$.. in terms of previous terms to see if there is a pattern but $n >= 1$ is troubling me. Please help.

答:
If you look here, ou will find how to solve the first-order rational difference equation in the more general case.

For $x_0=a$, this would give after simplifications

$$x_n=\frac{a \cos \left(\frac{\pi  n}{4}\right)+\sin \left(\frac{\pi
   n}{4}\right)}{\cos \left(\frac{\pi  n}{4}\right)-a \sin \left(\frac{\pi    n}{4}\right)}$$


问:
Find the number of all ordered pair (a, b) such that a ∈ A, b ∈ B, a is not equal to b and a + b is even [Read Full Q. below]
Let A denote the set of all numbers between 1 and 700 which are divisible by 3 and let B denote the set of all numbers between 1 and 300 which are divisible by 7. Find the number of all ordered pair (a, b) such that a ∈ A, b ∈ B, a is not equal to b and a + b is even

My Approach

Okay so looking at the question, I can tell that I need to use the inclusion-exclusion principle but before applying it, I must dissect the sets. So a + b needs to be even this means that both of them (a,b) must be either even or odd. We can create a subset $p$ from $A$ which are all the even multiples of 3 till 300 and a subset $q$ for all odd multiples of 3. Then do the same for the set $B$ for the multiples of 7 and create subsets $m$, $n$ respectively.

Now we have 4 sets:

$p$ = {even multiples of 3 till 300}

$q$ = {odd multiples of 3 till 300}

$m$ = {even multiples of 7 till 300}

$n$ = {odd multiples of 7 till 300}

After which I apply the inclusion-exclusion principle to these sets $p ∪ q ∪ m ∪ n$ for which I got $84 - 252 + 189 - 84$ but that makes no sense since the answer would be in negative. There maybe a calculation error but anyway, if you have an easier approach please answer

答:
I’ll get you most of the way and leave it to you to finish.

The largest and smallest multiples of $3$ in $A$ are $699=3\cdot233$ and $3=3\cdot1$, respectively, so there are $233$ multiples of $3$ in $A$; $116$ of them are even, and $117$ of them are odd.

The largest and smallest multiples of $7$ in $B$ are $294=7\cdot42$ and $7=7\cdot1$, respectively, so there are $42$ multiples of $7$ in $B$; $21$ of them are even, and $21$ of them are odd.

If $\langle a,b\rangle\in A\times B$, $a+b$ is even if and only if either $a$ and $b$ are both even, or $a$ and $b$ are both odd. There are $116\cdot21$ pairs with $a$ and $b$ both even and $117\cdot21$ with $a$ and $b$ both odd, and from this you can easily find the number of pairs such that $a+b$ is even.

However, that includes some pairs for which $a=b$, e.g., the pair $\langle 42,42\rangle$. How many such pairs are there?


问:
How can you be sure that an integral does not exist, if it has no indefinite integral?
Say you have the integral $\displaystyle\int_1^\infty{\frac{1}{x^{1+\frac{1}{x}}}}\;\mathrm{d}x$

This integral cannot be completed. Not that it goes to infinity, but it physically just cannot be completed. How can you realize this if you encounter it? How can you prove it?

答:
@KaviRamaMurthy and @player2326 have answered this question.
The comparison test can be employed to solve this question.

Additional references:

How to see this improper integral diverges?

Checking whther the integral $\int_1^∞ \frac{1}{x^{\frac{1}{x}+1}} dx$ convergent


问:
Why is showing that $ker(T_A) = \vec 0$ the same as showing that a linear transformation is one-to-one?
Determine if the standard matrix $A$ representing the linear transformation is one-to-one.

$A = \begin{bmatrix}1&-1\\2&0\\3&-4\end{bmatrix}$

Answer: Yes

My question: When determining if multiplication by A is a one-to-one linear transformation, is it equivalent to determine if $ker(A)=\vec 0$ ? In other words, is the following method always a valid way to determine if $T_A$ is a one-to-one linear transformation?

$T_A$ is one-to-one if $ker$($T_A$)$=\vec 0$ , i.e, $A\vec x=\vec 0$ only has the trivial solution $\vec x=\vec 0$

$\begin{bmatrix}1&-1&0\\2&0&0\\3&-4&0\end{bmatrix}$ $\rightarrow$ $x=0,y=0$ $\rightarrow$ $\vec x=\vec 0$

So $T_A$ is one-to-one.

I am confused about this method because if $ker$($T_A$)$=\vec 0$ then that means the only vector that is mapped to $\vec 0$ after the linear transformation is $\vec 0$. So this only shows that the there is one input vector for the output vector  $\vec 0$. However, a one-to-one linear transformation means that each unique vector in the domain maps to only one vector  in the range. So why don't I have to show all vectors have this property and just show that zero vector has this property?

So my question is why is showing that the kernel of the transformation is $\vec 0$ enough to show that the transformation is one-to-one?

答:
First of all, if by one-to-one you mean bijective, then your matrix A is evidently not bijective as is represents a transformation from $\mathbb{R}^3\to\mathbb{R}^2$ which you can see dimensionnaly is never one-to-one, even if you add a column of zeros. In particular, if your add a column of zeros, the kernel of your matrix will no longer be trivial as any vector of the form $$(0,0,\alpha ),$$ will be sent to $0$.

Now having a trivial kernel does make a matrix injective which implies directly that it is surjective if the matrix is square -> it is one-to-one (you can try to see it dimensionnaly, or with the rank theorem). However this is not the case here.


答:
Recall the Jacobi triple product:
$$\prod_{n\ge1}(1-x^{2m})(1-y^2x^{2m-1})(1-y^{-2}x^{2m-1})=\sum_{n\in\Bbb Z}x^{n^2}y^{2n}.$$
Setting $y=1$, $$\sum_{n\in\Bbb Z}x^{n^2}=\left(\prod_{m\ge1}(1-x^{2m-1})\right)\prod_{m\ge1}(1-x^m),$$
because $\prod_{m\ge1}(1-x^{2m})(1-x^{2m-1})=(1-x^2)(1-x)(1-x^4)(1-x^3)(1-x^6)(1-x^5)\cdots=\prod_{m\ge1}(1-x^m)$. Similarly we may write $$\prod_{m\ge1}(1-x^{2m-1})=\prod_{m\ge1}\frac{(1-x^{2m-1})(1-x^{2m})}{1-x^{2m}}=\frac{\prod_{m\ge1}(1-x^m)}{\prod_{m\ge1}(1-x^{2m})}.$$
Writing everything in terms of the $q$-Pochhammer symbol
$$(a;q)_\infty=\prod_{k\ge0}(1-aq^k),$$
we have
$$\sum_{n\in\Bbb Z}x^{n^2}=\frac{(x;x)_\infty^2}{(x^2;x^2)_\infty}.$$
Since $(-n)^2=n^2$, we have that the left side is
$$\sum_{n\in\Bbb Z}x^{n^2}=1+2\sum_{n\ge1}x^{n^2},$$
so
$$\sum_{n\ge1}x^{n^2}=\frac12\left(\frac{(x;x)_\infty^2}{(x^2;x^2)_\infty}-1\right).$$
This is, as Claude mentioned,
$$\sum_{n\ge1}x^{n^2}=\frac12(\vartheta_3(0,x)-1),$$
where $\vartheta_3$ is the Jacobi theta function.

Title: Find an explicit formula for $ \sum_{n=0}^{\infty}{x^{n^2}} ,\quad \forall x \in (0,1) $

Body:
My question askes me to find an explicit formula for
$$ \sum_{n=0}^{\infty}{x^{n^2}} \quad\left(\forall x \in (0,1)\right)$$
And I feel it kind of interesting to find an appropriate f(x) that satisfies
$$ \lim_{x \to 1^{-} }\frac{\sum_{n=0}^{\infty}{x^{n^2}}}{f(x)}=1 $$
Both questions make me feel puzzled, and I tried
$$ \sum_{n=0}^{\infty}{x^{n^2}}<\sum_{n=0}^{\infty}{x^n}=\frac{1}{1-x} $$
Hence I guess maybe
$$ \sum_{n=0}^{\infty}{x^{n^2}}\sim\frac{C_{1}}{\sqrt{1-x}} \quad \text{when} \quad x \to 1^{-}$$
or at least $$ \sum_{n=0}^{\infty}{x^{n^2}} \sim \frac{C_{2}}{(1-x)^\alpha} \quad \text{for} \quad \text{some} \quad \alpha$$
Any help or recommendation related would be greatly appreciated.

Thanks a lot for your focus.


答:
Suppose You say that KerT ={0}

Then Suppose For two vectors u and v
.T(u) = T(v)
Implies that T(u) -T(v) = 0 Since T is Linear map so it
Implies T(u-v)=0.

But Ker T ={ 0 }
So u-v=0

Hence u= v

Then T is one one for any vector

Title: Why is showing that $ker(T_A) = \vec 0$ the same as showing that a linear transformation is one-to-one?

Body:
Determine if the standard matrix $A$ representing the linear transformation is one-to-one.

$A = \begin{bmatrix}1&-1\\2&0\\3&-4\end{bmatrix}$

Answer: Yes

My question: When determining if multiplication by A is a one-to-one linear transformation, is it equivalent to determine if $ker(A)=\vec 0$ ? In other words, is the following method always a valid way to determine if $T_A$ is a one-to-one linear transformation?

$T_A$ is one-to-one if $ker$($T_A$)$=\vec 0$ , i.e, $A\vec x=\vec 0$ only has the trivial solution $\vec x=\vec 0$

$\begin{bmatrix}1&-1&0\\2&0&0\\3&-4&0\end{bmatrix}$ $\rightarrow$ $x=0,y=0$ $\rightarrow$ $\vec x=\vec 0$

So $T_A$ is one-to-one.

I am confused about this method because if $ker$($T_A$)$=\vec 0$ then that means the only vector that is mapped to $\vec 0$ after the linear transformation is $\vec 0$. So this only shows that the there is one input vector for the output vector  $\vec 0$. However, a one-to-one linear transformation means that each unique vector in the domain maps to only one vector  in the range. So why don't I have to show all vectors have this property and just show that zero vector has this property?

So my question is why is showing that the kernel of the transformation is $\vec 0$ enough to show that the transformation is one-to-one?