答:
Starting from your last inequation on x,y,z, you can factorize to get a cannonical expression :

$$\left(x+\frac{z}{2}\right)^2+\left(y+\frac{z}{2}\right)^2\leq\frac{1}{2}, 0\leq z\leq3$$

So you have :

$$0\leq z\leq3 \\ -\frac{z}{2}-\frac{1}{\sqrt{2}} \leq x \leq -\frac{z}{2}+\frac{1}{\sqrt{2}} \\-\frac{z}{2}-\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2}\leq y\leq -\frac{z}{2}+\sqrt{\frac{1}{2}-\left(x+\frac{z}{2} \right)^2} $$

Title: Finding the volume of $\{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$ by integration

Body:
I have been trying to solve this integral for some time now without success:

$$ T = \{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$$
$$ I = \int_{T} ~dx~dy~dz $$

I have tried simplifying the domain, by writing:
$$ (x+y+z)^2 + (x-y)^2 \leq 1 \to $$
$$ 2x^2 + 2y^2 + z^2 + 2xz + 2yz \leq 1 \to $$
$$ x^2 + y^2 + \frac{z^2}{2} + xz + yz \leq \frac{1}{2} $$
Now, I tried doing this change of variable: $ x = \rho \cos\theta$, $y = \rho \sin\theta$ and $z = \sqrt 2 w$
With $\rho \in (0,\infty)$, $\theta \in [0,2\pi]$. They are basically cylindrical coordinates.
I do some calculation and I have the following:
$$ \rho^2 + w^2 + (\rho \cos\theta + \rho \sin \theta) \cdot \sqrt{2}w \leq \frac{1}{2}$$
But I don't see how this is helping. I have also tried spherical coordinates substitution without success.

Can I have some hints?


问:
All solutions $(x, y, z) \in (\mathbb{N},\mathbb{N},\mathbb{N})$ for $3x^2 + 6y^2 = z^2$
My attempt:

$z^2 = 3(x^2 + 2y^2)$

so $3|z^2$ and thus $3|z$

letting $z = 3k$, we reduce the equation to $x^2+2y^2=3k^2$

Obviously $(n,n,n) $ represents infinitely many solutions, but not all since $(5,1,3)$ is a solution for example.

Also, if $x$ and $k$ are even then $4|2y^2$, then $y$ is even

If $x$ and $k$ are odd, then $x^2$ and $k^2$ are congruent to 1 mod 8 so $2y^2$ is congruent to 2 mod 8 so y is also odd.Therefore, it's sufficient to find the odd solutions.

But I don't know how to proceed from here.
Any hints on how to parametrize the odd solutions or a suitable mod $n$ to consider to simplify it further would be appreciated.

答:
There is a systematic way to solve such equation:  the idea is to parametrize the ellipse $C: 3X^2+6Y^2=1$ by a family of lines through a fixed rational point. (in other words, an ellipse is a rational curve).

Here $A=({1\over 3}, {1\over 3})$ is a rational point on this ellipse.  Note that a line through $A$ has equation $L_t:(X-1/3)=t(Y-1/3)$ must cut the ellipse in exactly one otherpoint which  is a rational fonction of $t.

It is preferable to write $C= (3.(X-1/3)^2+6.(Y-1/3)^2+6(X-1/3)+12(Y-1/3)=0$

Let $U=X-1/3, V=Y-1/3$
The intersection of $L_t\cap C$ is $U=tV, 3U^2+6V^2+6U+12V=0$

Or $3t^2V^2+6V^2+6tV+12V=0$

If we exclude the point A $(V=0)$ we get; $V(3t^2+6)=-6T-12$, $V=-{t+4\over t^2+3}, U=t V$

Now $t={a\over b}\in \bf Q$, ${x\over z}=U+1/3=-t{t+4\over t^2+3}+1/3$ and ${y\over z}=V+1/3=-{t+4\over t^2+3}$

${x\over z}=-{a^2+4ab\over a^2+3b^2}+1/3={-2a^2+3b^2-12ab\over a^2+3b^2}$

${y\over z}=-{a/b+4\over (a/b)^2+3}+1/3=$
$-{ab+4b^2\over a^2+3b^2}+1/3=$ $a^2-3ab-9b^2\over 3(a^2+b^2)$

So $x=-2a^2+3b^2-12ab$, $y=a^2-3ab-9b^2$, $z= 3(a^2+b^2)$ is a solution if $(a,b,c)\in \bf Z$.

Conversely, for every solution, the point  $ P=(x/z, y/z)$ is on the ellipse, so $(x,y,z)$ is proportional to the triple $(-2a^2+3b^2-12ab , a^2-3ab-9b^2,3(a^2+b^2))$, for $(a,b)$ the slope of the line $(A,P)$.

Note that a different choice for $A$   yields a different parametrization of the set of solution...

ContentLicense: CC BY-SA 4.0


问:
Roulette strategy
I came up with a strategy, which imitates the martingale (doubling) strategy , but it seems that there is a good probability to gain profit, which I want to compute.

So this is how we start:

We do not bet on black/red, but we chose columns, or dozens. Let's say we start with 100 chips.

Bet $1$ chip in e.g. a column. The probability to win is $12/37$. If you win, start over again, while if you lose:

Bet again $1$ chip to a column. Why again $1$? This is because we will still have profit if we win ( I will have won 3 chips, and lost 2 ). Again, if win go to step $1$, if lose:

Bet $2$ chips in a column. Each time we bet the lowest amount of funds, such that we have profit in case we win. In this case, if we win we will have played $4 $ chips , and won $6$, which gives the minimum  profit($2 $ chips).

Continuing with this strategy, one has to bet in columns/dozens, the following sequence of chips (each time he loses), in order to have the minimum profit :
$$1,1,2,3,4,6,9,14,21,31,47\dots$$
So, here is my question:

Using this strategy, and starting with $100$ chips, what is the probability of gaining $+50\%$, and what is the probability of doubling?
You continue until you have doubled, and stop if you have less chips than they are required to put the next bet.

*We have to consider that when we gain some profit, e.g. we have reached $140$ chips, we are able to bet more chips of the above sequence. With $100$ chips we can lose up to 9 times, and then bet $31$ chips. But with $140$ chips we can lose one more time , and then bet $46$.

Due to the complexity of the problem, I would also appreciate a numerical solution, but I do not have the knowledge to run a simulation.

Anyone can figure out a solution using probability theory?

Title: Roulette strategy

Body:
Assuming I didn't code a billion bugs, running a quick 100,000 samples gave me:

100 -> 150 case:  $0.55989 \pm 0.00156976$.

100 -> 200 case:  $0.37052 \pm 0.0015272$.

For a smaller case:

10 -> 15 case:  $0.56233 \pm 0.00156881$.

10 -> 20 case:  $0.39492 \pm 0.00154583$.

For comparison,

$12/37 = 0.324324$

ContentLicense: CC BY-SA 3.0


问:
Is it ok to use equivalence here where we have $\lim0\times(-\infty)$?
$$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))$$

Here to evaluate the limit I used the equivalence $\ln(u+1)\sim u$ twice  :

$$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))=\lim_{x\to 0^+}(e^x-1)\times(x-1)=(1-1)\times(0-1)=0$$

But the book I am reading write the expression as $\cfrac{\ln(\ln(1+x))}{\cfrac1{e^x-1}}$ then applied L'Hopital Rule and at the end it get $0$ too after relatively long calculation.

I wonder is my approach valid ?

答:
No we can't use equivalence in this way, we can proceed by standard limits using that

$$(e^x-1)\ln(\ln(1+x))=\frac{e^x-1}x \cdot x\ln(\ln(1+x)) \to 1 \cdot 0=0$$

indeed

$$x\ln(\ln(1+x))=\ln(1+x)\cdot\ln(\ln(1+x))\cdot \frac{x}{\ln(1+x)} \to 0 \cdot 1 =0$$

since for $t\to 0$ we have that $t\ln t \to 0$.

Refer to the related

How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

How can I deduce that $\lim\limits_{x\to0} \frac{\ln(1+x)}x=1$ without Taylor series or L'Hospital's rule?


问:
a proof for $\sum \limits_{n\le x} c_n f(n)=S(x)f(x)-\int \limits_{n_0}^x S(t)f'(t)dt$
I want to go through the book "An Introduction to Sieve Methods and their Application" by Cojocaru & Murty. On page 4 one can find the Theorem 1.3.1, that $\sum \limits_{n\le x} c_n f(n)=S(x)f(x)-\int \limits_{n_0}^x S(t)f'(t)dt$ is true with $S(x)=\sum \limits_{n\le x} c_n$ and $c_j =0$ for all $j<n_0$.

This formula looks like a discrete version of the chain rule and could be true, but if I take an example, it doesn't work. We can set $c_0=0$ and $c_n=1$ for all $n>0$ and $f(n)=n$. We know, that $\sum \limits_{n\le x} n=\frac{1}{2} x(x+1)$, but $S(x)f(x)-\int \limits_{n_0}^x S(t)f'(t)dt=\frac{1}{2}(x^2+1)$.

In the given proof of that theorem in the book I don't understand the step to the last line, thats why I took the example.

The proof in the book is very short, so I can copy it.
$\begin{align}
\sum \limits_{n\le x} c_n f(n)&=\sum \limits_{n\le x} S(n)f(n) -\sum \limits_{n\le x-1} S(n)f(n+1)\\
&=S(x)f(x)-\sum \limits_{n\le x-1} S(n)\int \limits_n^{n+1}f'(t)dt\\
&=S(x)f(x)-\int \limits_{n_0}^x S(t)f'(t)dt
\end{align}$

What is wrong here?

答:
There's nothing wrong.  With your example $c_n = 1$ for $n \geq 1$ and $f(n) = n$, you have $S(t) = \lfloor t \rfloor$, so the formula reads (assuming $x$ is an integer)
\begin{align*}
\sum_{n \leq x} n \cdot 1 &= S(x) f(x) - \int_1^x S(t) f'(t) \,dt = x \cdot x - \int_1^x \lfloor t \rfloor \cdot 1 \,dt \\
&= x^2 - (1+2+\dots+(x-1)) = x^2 - \frac{(x-1)x}{2} = \frac{x(x+1)}{2}
\end{align*}
The formula is correct, it's just not useful in this particular example, because the only way to evaluate the integral term is to already know the answer to the sum you're trying to evaluate.

As a remark, you should think of this formula more like a discrete version of integration by parts rather than a discrete version of the chain rule.


问:
What is the expected number of white balls, conditionally, in each row of this grid of white and black balls?
Suppose I have an $m\times n$ grid of black balls, $m,n \in \mathbb{N}$, and I randomly switch $k$ of them to white, where $1 \leq k \leq mn$. Every possible configuration of $k$ white balls is equally probable.

I want to compute the expected number of white balls in any row containing at least one white ball.

When $mn \gg k$, I can appeal to the fact that $p_0$, the probability that any given row contains no white balls, is approximately $\left(1-m^{-1}\right)^k$, hence the expected number of rows containing no white balls is $E[n_0] \approx mp_0 = m\left(1-m^{-1}\right)^k$, leaving $k$ white balls in the remaining $$E[n_{>0}] = m - E[n_0] = m - m\left(1-m^{-1}\right)^k$$ rows. The expected number of white balls in each of these rows can therefore be approximated $$E[n_{\text{white}}|n_{\text{white}} \geq 1] \approx k\left(m - m\left(1-m^{-1}\right)^k\right)^{-1}$$

However, as $\phi = \frac{k}{mn}$ increases towards $1$, this approximation becomes increasingly inadequate.

答:
There are $\binom{mn}{k}$ ways to select the cells with white balls, all of which we assume are equally likely.

First, what is the probability that row $i$ contains at least one white ball, for $1 \le i \le m$? It is $1$ minus the probability that the row contains no white balls, i.e.
$$P(\text{at least one white ball}) = 1 - \frac{\binom{mn-n}{k}}{\binom{mn}{ k}}$$
Second, what is the probability that row $i$ contains at least one white ball and cell $i,j$ contains a white ball?  This is the same as the probability that cell $i,j$ contains a white ball:
$$\frac{k}{mn}$$
Let's define $X_{ij} = 1$ if cell $i,j$ contains a white ball, given that row $i$ contains at least one white ball, $0$ otherwise. Then
$$E(X_{ij}) = P(X_{ij} =1) = \frac{k / mn}{1 - {\binom{mn-n}{k}} /{\binom{mn}{ k}}}$$
By linearity of expectation, the expected number of white balls in row $i$, given that the row contains at least one white ball, is
$$\begin{align}
E \left( \sum_{j=1}^n X_{ij} \right) &= \sum_{j=1}^n E(X_{ij}) \\
&= n E(X_{ij}) \\
& = \frac{k / m}{1 - {\binom{mn-n}{k}} /{\binom{mn}{ k}}}
\end{align}$$


答:
It is easy enough to get all rational solutions by stereographic projection around a single point. The trouble is that we want all integer solutions with $\gcd(x,y,k) = 1.$

There is a method for producing recipes of the Pythagorean Triple type. As far as I know, first written down in Fricke and Klein (1897).

$$  x = u^2 + 4 u v - 2 v^2 \; , \; \; y = u^2 - 2 uv - 2 v^2 \; , \; \; k = u^2 + 2 v^2  $$

In order to keep both $u,v$ positive, I just added in the effect of negating $v$ as:

$$  x = u^2 - 4 u v - 2 v^2 \; , \; \; y = u^2 + 2 uv - 2 v^2 \; , \; \; k = u^2 + 2 v^2  $$

The proof that this one recipe suffices is simply that forcing all to have gcd 2 or 3 leads to a new version of the same thing, when divided by that 2 or 3.
We stick with the first description and allow negative variables $u,v.$

If we demand $u = -2t,$  we arrive at
$$  \frac{-x}{2} = v^2 + 4tv - 2t^2 \; , \; \frac{-y}{2} = v^2 - 2tv - 2t^2 \; , \; \frac{k}{2} = v^2 + 2t^2  $$

If we demand $u = s +2t, v = s - t,$  so that $u \equiv v \pmod 3,$  we arrive at
$$  \frac{x}{3} = s^2 + 4st - 2t^2 \; , \; \frac{-y}{3} = s^2 - 2st - 2t^2 \; , \; \frac{k}{3} = s^2 + 2t^2  $$

```<code>Thu Oct 29 08:38:25 PDT 2020
    1              1    1    1
    1              1    1    1  fricke
    1              1    1    1  fricke
    3              5    1    3
    3             -5    1    3  fricke
    9              1   11    9
    9              1  -11    9  fricke
   11             19    1   11
   11             19    1   11  fricke
   11              5   13   11
   11             -5   13   11  fricke
   17             23   13   17
   17            -23   13   17  fricke
   17             25   11   17
   17             25  -11   17  fricke
   19             29   11   19
   19            -29  -11   19  fricke
   19              5   23   19
   19             -5  -23   19  fricke
   27             43   13   27
   27             43   13   27  fricke
   33             23   37   33
   33            -23   37   33  fricke
   33             47   23   33
   33            -47  -23   33  fricke
   41             25   47   41
   41             25  -47   41  fricke
   41             71    1   41
   41            -71    1   41  fricke
</code>```

Title: All solutions $(x, y, z) \in (\mathbb{N},\mathbb{N},\mathbb{N})$ for $3x^2 + 6y^2 = z^2$

Body:
My attempt:

$z^2 = 3(x^2 + 2y^2)$

so $3|z^2$ and thus $3|z$

letting $z = 3k$, we reduce the equation to $x^2+2y^2=3k^2$

Obviously $(n,n,n) $ represents infinitely many solutions, but not all since $(5,1,3)$ is a solution for example.

Also, if $x$ and $k$ are even then $4|2y^2$, then $y$ is even

If $x$ and $k$ are odd, then $x^2$ and $k^2$ are congruent to 1 mod 8 so $2y^2$ is congruent to 2 mod 8 so y is also odd.Therefore, it's sufficient to find the odd solutions.

But I don't know how to proceed from here.
Any hints on how to parametrize the odd solutions or a suitable mod $n$ to consider to simplify it further would be appreciated.

Body:
It is easy enough to get all rational solutions by stereographic projection around a single point. The trouble is that we want all integer solutions with $\gcd(x,y,k) = 1.$

There is a method for producing recipes of the Pythagorean Triple type. As far as I know, first written down in Fricke and Klein (1897).

$$  x = u^2 + 4 u v - 2 v^2 \; , \; \; y = u^2 - 2 uv - 2 v^2 \; , \; \; k = u^2 + 2 v^2  $$

In order to keep both $u,v$ positive, I just added in the effect of negating $v$ as:

$$  x = u^2 - 4 u v - 2 v^2 \; , \; \; y = u^2 + 2 uv - 2 v^2 \; , \; \; k = u^2 + 2 v^2  $$

The proof that this one recipe suffices is simply that forcing all to have gcd 2 or 3 leads to a new version of the same thing, when divided by that 2 or 3.
We stick with the first description and allow negative variables $u,v.$

If we demand $u = -2t,$  we arrive at
$$  \frac{-x}{2} = v^2 + 4tv - 2t^2 \; , \; \frac{-y}{2} = v^2 - 2tv - 2t^2 \; , \; \frac{k}{2} = v^2 + 2t^2  $$

If we demand $u = s +2t, v = s - t,$  so that $u \equiv v \pmod 3,$  we arrive at
$$  \frac{x}{3} = s^2 + 4st - 2t^2 \; , \; \frac{-y}{3} = s^2 - 2st - 2t^2 \; , \; \frac{k}{3} = s^2 + 2t^2  $$

Thu Oct 29 08:38:25 PDT 2020
    1              1    1    1
    1              1    1    1  fricke
    1              1    1    1  fricke
    3              5    1    3
    3             -5    1    3  fricke
    9              1   11    9
    9              1  -11    9  fricke
   11             19    1   11
   11             19    1   11  fricke
   11              5   13   11
   11             -5   13   11  fricke
   17             23   13   17
   17            -23   13   17  fricke
   17             25   11   17
   17             25  -11   17  fricke
   19             29   11   19
   19            -29  -11   19  fricke
   19              5   23   19
   19             -5  -23   19  fricke
   27             43   13   27
   27             43   13   27  fricke
   33             23   37   33
   33            -23   37   33  fricke
   33             47   23   33
   33            -47  -23   33  fricke
   41             25   47   41
   41             25  -47   41  fricke
   41             71    1   41
   41            -71    1   41  fricke


问:
What's the probability for the balls and bins problem?
There are 5 bins.

You randomly throw 7 balls into them.
 What is the probability that at least one of the bins has precisely 3 balls in it?

How to approach this problem? How to find the probability of any 2 bins has precisely 3 balls in it?

答:
Approach for first problem:

Number the bins and for $i=1,2,3,4,5$ let $B_i$ denote the event that bin $i$ has precisely $3$ balls in it.

Then to be found is: $$P\left(\bigcup_{i=1}^5B_i\right)$$

Applying inclusion/exclusion and symmetry we find that:$$P\left(\bigcup_{i=1}^5B_i\right)=5P(B_1)-10P(B_1\cap B_2)$$


问:
If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90 ^\circ$.
If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90^\circ$.

I have proved it going the other way. When given that $C$ is $90^\circ$, I can show that the Lemoine point lies on the altitude, but I cannot figure out how to go backward or incorporate the $BC = AC$ part.

I have tried using the median from vertex $C$, which I know is an isogonal conjugate of the symmedian from $C$. But then from there I don't know where to go.

Thanks for your help.

答:
The coordinates of the symmedian point $X_6$ and
the foot of the altitude $H_c$
in terms of vertices $A(A_x,A_y),\ B(B_x,B_y),\ C(C_x,C_y)$ and side lengths $a,b,c$
can be found as
\begin{align}
X_6&= \frac{a^2\,A+b^2\,B+c^2\,C}{a^2+b^2+c^2}
\tag{1}\label{1}
,\\
H_c&=
\tfrac12\,(A+B)+\frac{a^2-b^2}{2c^2}\,(A-B)
\tag{2}\label{2}
.
\end{align}

Condition $X_6\in CH_c$ is equivalent to

\begin{align}
\operatorname{Im}
\left(
\frac{H_c-C}{X_6-C}
\right)
&=0
\tag{3}\label{3}
.
\end{align}

Let $C=(0,0)$, then the condition \eqref{3} simplifies to

\begin{align}
(b^2-a^2)(a^2+b^2-c^2)(A_x B_y-A_y B_x)
&=0
\tag{4}\label{4}
,
\end{align}

which holds either if $a=b$,
or $c^2=a^2+b^2$.
The third option, $A_x B_y=A_y B_x$,
corresponds to the degenerate case,
when all the vertices of $\triangle ABC$
are collinear.


问:
Uniformly distributed random variable on unit circle.
Let the random vector $(x,y)$ be uniformly distributed within the unit circle $B = \{ (x,y) : x^2 + y^2 \leq 1 \}$. Then the probability density function is

$$ f(x,y) = \begin{cases} \frac{1}{\pi} \quad &(x,y) \in B \\ 0 \quad &\text{otherwise} \end{cases}$$

Now, the marginal distributions are

$$\tilde{f}(x) = \int_{\mathbb{R}} f(x,y) \, dy = \frac{2}{\pi} \sqrt{1-x^2}$$

and similar for $\tilde{f}(y)$.

Question is:

Let $y = y_0 > 0$ be fixed. What is the probability distribution function for the random variables $x^4$, $|x|$ and $-x$?

Any ideas?

答:
When they say:"Let $y=y_0>0$ be fixed" it is evident to me that they are interested in the conditional distribution as fixing y the support of the marginal $X$ changes.

The problem is trivial because the conditional distribution $X|Y=y$ is uniform.

$$f_{X|Y=y_0}(t)=\frac{1}{2\sqrt{1-y_0^2}}\mathbb{1}_{\Big[-\sqrt{1-y_0^2};\sqrt{1-y_0^2}\Big]}(t)$$

Clarified this, I think you will have no problems to continue by your own

Example 1:

In the case $Z=|X|y_0|$ the distribution function is

$$F_Z(z)=F_{X|Y}(z)-F_{X|Y}(-z)=\frac{z}{\sqrt{1-y_0^2}}$$

...better

$$F_Z(z)=\frac{z}{\sqrt{1-y_0^2}}\mathbb{1}_{\Big[0;\sqrt{1-y_0^2}\Big]}(z)+\mathbb{1}_{\Big[\sqrt{1-y_0^2};+\infty\Big)}(z)$$

(again an Uniform Distribution)

Example 2

$Z=X^4$

$$F_Z(z)=\mathbb{P}[X^4\leq z]=\mathbb{P}\Big[- z^{\frac{1}{4}}\leq X\leq z^{\frac{1}{4}}\Big]=F_X\Big(z^{\frac{1}{4}}\Big)-F_X\Big(-z^{\frac{1}{4}}\Big)$$


问:
How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$
Somebody please help me to solve this Differential equation.

$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $

答: It is an exact DE, solution is $x^3\tan{y}+y^4+\frac{y^3}{x^2}+C=0$


问:
Showing integral exists
I am having a hard time seeing how the following is calculated:

Consider $f:[1,\infty] \to \mathbb{R}$ defined as $f(x) = \frac{(-1)^n}{n}, \text{ for } n \leq x \leq n+1.$

Then $$\lim_{n\to\infty}\int_{1}^{n}f = \sum_{k=1}^{\infty}\frac{(-1)^k}{k}.$$

I considered defining the following:

$$f_n := \sum_{k=1}^{n}\frac{(-1)^k}{k}\chi_{[k,k+1]}.$$

Then using the Monotone Convergence Theorem, however, the $\{f_n\}$ are not non-negative...

Title: Showing integral exists

Body: You are looking at $$\int_1^nf=\sum_{k=1}^{n-1}\int_k^{k+1}f=\sum_k^{n-1}\frac{(-1)^k}{k}\,.$$


问:
Does the complex series $s = \sum_{n=1}^\infty i^n$ converge?
How would one proof if the complex series
$s_N = \sum_{n=1}^N a_n$ with
$a_n=i^n$ converges or diverges?

$s = \sum_{n=1}^\infty i^n$ where $i^2=-1$.

Thanks in advance for all tips, links and references!

答:
A (real or complex) sequence $b_n$ is defined to "converge to a limit $b$" iff for every choice of $\epsilon>0$ there exists a choice of $N_0$ such that the implication $n>N_0 \implies |b_n - b|<\epsilon$ holds true.

By negating the condition:

A (real or complex) sequence $b_n$ is defined to "diverge" iff for every proposed limit $b$ there exists a choice of $\epsilon > 0$ such that for every choice of $N_0$ there exists some $n>N_0$ such that $|b_n-b|\geq \epsilon$

In lay terms, that is to say that a sequence converges if it gets closer and closer to a limit and that for any desired closeness there is some point in the sequence such that from that point on all terms in the sequence are at least that close.

On the other hand, a sequence diverges if regardless how far we look in the sequence there are always at least some terms which are farther away than we would like.

For your specific series, we look at the sequence of partial sums.  We see that the sequence of partial sums goes $i,-1+i, -1, 0, i, -1+i, -1, 0, i, -1+i, -1, 0,\dots$ repeating this pattern ad infinitum.

We can see that by setting $\epsilon = \frac{1}{4}$ for instance that regardless what proposed limit we take, at least one of $0$ or $-1$ will be further away from the limit than $\frac{1}{4}$ and for any choice of $N_0$ there will always be some $n>N_0$ such that the term is equal to whichever of $0$ or $1$ is further than $\frac{1}{4}$ of the proposed limit.

Thus, the sequence diverges.

Note, the choice of $\frac{1}{4}$ here was entirely arbitrary.  I could have used $\frac{1}{10}$ or $\frac{1}{\pi}$ or $0.42$ or whatever else so long as it was small enough.  I just needed to pick something such that being within that range of one of the values in the sequence necessarily implied it was outside of that range of another value in the sequence.

As for taking shortcuts, I emphasize that given that you asked this question you should not be worrying about them yet.  Focus on your basic definitions.  That said, there are many ways you can see this series diverges very quickly.  The most basic of which is the "basic divergence test."

You can prove that every series who converges must necessarily have the summands approach zero.  The contrapositive of this statement is that if you have a series whose summands do not approach zero, then the series necessarily must diverge.  Your summands here are all of absolute value equal to $1$ and do not approach zero.


Title: How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$

Body:
Somebody please help me to solve this Differential equation.

$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $

Body:
$$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0$$
$$\left(\tan (y)dx^3+ y^3d\dfrac 1 {x^2}\right)+ \left(dy^4 +{x^3}d{\tan y} + \frac{dy^3}{x^2}\right) =0$$
$$d(\tan (y)x^3)+ d\dfrac {y^3} {x^2}+ dy^4  =0$$
Integrate.
$$x^3\tan (y)+ \dfrac {y^3} {x^2}+ y^4  =C$$


答:
You can even go beyond the limit composing Taylor series one piece at the time
$$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$
$$\log(\log(1+x))=\log (x)-\frac{x}{2}+O\left(x^2\right)$$
$$e^x-1=x+\frac{x^2}{2}+O\left(x^3\right)$$
$$(e^x-1)\log(\log(1+x))=\left(x+\frac{x^2}{2}+O\left(x^3\right) \right) \left(\log (x)-\frac{x}{2}+O\left(x^2\right) \right) $$
$$(e^x-1)\log(\log(1+x))=x \log (x)+\frac{1}{2} x^2 (\log (x)-1)+O\left(x^3\right)$$ which for sure shows the limit but also how it is approached.

Title: Is it ok to use equivalence here where we have $\lim0\times(-\infty)$?

Body:
$$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))$$

Here to evaluate the limit I used the equivalence $\ln(u+1)\sim u$ twice  :

$$\lim_{x\to\ 0^+}(e^x-1)\ln(\ln(1+x))=\lim_{x\to 0^+}(e^x-1)\times(x-1)=(1-1)\times(0-1)=0$$

But the book I am reading write the expression as $\cfrac{\ln(\ln(1+x))}{\cfrac1{e^x-1}}$ then applied L'Hopital Rule and at the end it get $0$ too after relatively long calculation.

I wonder is my approach valid ?


问:
Artin-Wedderburn: Decomposition of a semisimple Dual Hopf algebra
1. Context

My lecture notes prove that any cocommutative finite-dimensional Hopf algebra over a field $k$ of characteristic zero is semisimple and cosemisimple.
They try to argue from there that any finite-dimensional, cocommutative Hopf algebra over a field of characteristic zero is isomorphic to a group algebra:

Since $H^*$ is semisimple, it is, as an algebra, isomorphic to $H^* \cong k \times. . . \times k$ by the Artin-Wedderburn theorem. The projection $p_i$ to the $i$-th factor is a morphism of algebras or, put differently, a grouplike element in $H^{**} \cong H$. All projections give a basis of $H$ consisting of grouplike elements. Thus $H$ is a group algebra of a finite group.

2. Question

Why does the isomorphism $H^* \cong k \times. . . \times k$ exist? Where is the Artin-Wedderburn theorem used?

The Artin-Wedderburn theorem gives an isomorphism $H^* \cong \prod M_{n_i}(D_i)$ where the $n_{i}$ are natural numbers, the $D_i$ are finite dimensional division algebras over $k$ and $M_{n_i}(D_i) $ is the algebra of $n_i \times n_i $matrices over $D_i$. If $k$ were algebraically closed we would even know that $H^* \cong \prod M_{n_i}(k)$ holds. How to proceed? I am not familiar with Artin-Wedderburn, I guess. So any hint would be appreciated.

答:
Since you confirmed it in the comment above:

It sounds like the intended context was an algebraically closed field, and that $H^\ast$ is a commutative semisimple ring.

Using the Artin-wedderburn theorem it is easy to see why all the matrices must have side length $1$ in order to be commutative. That makes a product of fields, each one a finite field extension of $k$.  But adding algebraiclly closed means that these fields are in fact just $k$.


问:
System with $x + ay = 3 , 2x - y = b$
Quesstion about this system:
$x + ay = 3 , 2x - y = b$

Where $a, b \in\mathbb R$. The system admits as solution $(1, 2)$.
I'm struggling with the logic of getting a final result. Is there an efficient way of calculating this

答: The point $(1,2)$ is a solution to the system iff $1+a\cdot2=3$ and $2\cdot1-2=b$ are both true statements.  From here $a=1$ and $b=0$.

Title: System with $x + ay = 3 , 2x - y = b$


答:
There is no room to apply it twice.

$$\log(\log(1+x))\sim\log(x)$$ and $x$ is not $1+x$ !

Tinkering with $x=1+(x-1)$ does not work because $x-1$ is not close to $0$.

On the opposite, $$e^x-1\sim x$$ will work and finally

$$(e^x-1)\ln(\ln(1+x))\sim x\log(x)\to0.$$


答:
Set $a_n:=\int_1^n f$ for all $n$, so $a_1=0, a_2=-1, a_3=-1+\frac{1}{2},$ etc. If you have to show only the existence of the limit, it is sufficient to see that
$$
\forall n \ge 1, \quad
\int_{2n}^{2n+2}f=a_{2n+2}+a_{2n+1}=-\frac{1}{2n+1}+\frac{1}{2n} \in \left[0,\frac{1}{n^2}\right].
$$
In particular, you can apply the Monotone Convergence Theorem at the sequence of functions $(f \chi_{[1,2n]}: n\ge 1)$.

Title: Showing integral exists

Body:
I am having a hard time seeing how the following is calculated:

Consider $f:[1,\infty] \to \mathbb{R}$ defined as $f(x) = \frac{(-1)^n}{n}, \text{ for } n \leq x \leq n+1.$

Then $$\lim_{n\to\infty}\int_{1}^{n}f = \sum_{k=1}^{\infty}\frac{(-1)^k}{k}.$$

I considered defining the following:

$$f_n := \sum_{k=1}^{n}\frac{(-1)^k}{k}\chi_{[k,k+1]}.$$

Then using the Monotone Convergence Theorem, however, the $\{f_n\}$ are not non-negative...


问:
Unit before and after Fourier transform not the same
Suppose the functions $e(x)$ and $h(x)$ with units $\left[\frac{V}{m}\right]$ and $\left[\frac{A}{m}\right]$ and their Fourier transforms $E(k)$ and $H(k)$ with units $\left[V\right]$ and $\left[A\right]$. The time avaverage A of $e(x)h(x)$ in function of their Fourier transforms $E(k)$ and $H(k)$ can be written as
\begin{align}
A &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} e(x)h(x) dx \\
&= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} \int\limits_{-\infty}^{\infty} E(k)e^{ikx} dk  \int\limits_{-\infty}^{\infty} H(k')e^{ik'x} dk' dx \\
&=  \int\limits_{-\infty}^{\infty}  \int\limits_{-\infty}^{\infty}E(k) H(k')\left[ \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X}e^{i(k+k')x} dx \right] dk dk'  \\
&=  2 \int\limits_{-\infty}^{\infty}  \int\limits_{-\infty}^{\infty}E(k) H(k')\delta(k'+k) dk dk' \\
&=  2 \int\limits_{-\infty}^{\infty}  E(k) H(-k)dk \\
\end{align}

The question: the unit of the time average A is $\left[\frac{VA}{m^2}\right]$ whereas the unit of the final result is $\left[\frac{VA}{m}\right]$. How can this be and how to interpret it?

Example: if you know the Fourier transforms $E(k)$ and $H(k)$, and you want to calculate the average A, then you will use $A=2 \int\limits_{-\infty}^{\infty}  E(k) H(-k)dk$ but this will give another unit as expected? So, then you have the wrong result?

答:
Your error is in the integral converging to the Dirac delta. Normally it should be $$\lim_{X\to\infty} \int_{-X}^X e^{i(k+k')x}\,\mathrm d x = \delta_{k+k'}$$
so without dividing by $X$. This is at this step that you are loosing a $m$.


问:
Best proof for the number of independent components of a totally (anti) symmetric tensor
I'm asking this question for future reference. Given a totally antisymmetric and a totally symmetric tensor having $k$ indices living in $n$ dimensions. Denote the antisymmetric and symmetric tensor by $A_{i_1\dots i_k}$ and $S_{i_1\dots i_k}$ respectively. Without these symmetries would have $n^k$ independent components. Because of their (anti) symmetricity each of the components can be written such that the indices appear in ascending order. For example $A_{321}=-A_{123}$. The independent components can be written as a list where the indices of each element are ascending. The antisymmetric tensor has the additional constraint that no two indices can be the same.
\begin{align}
\text{indep}(A)&=\left\{A_{i_1\dots i_k}\middle|1\leq i_1<\dots<i_k\leq n\right\}\\
\text{indep}(S)&=\left\{S_{i_1\dots i_k}\middle|1\leq i_1\leq\dots\leq i_k\leq n\right\}
\end{align}
Now it is known that the number of independent components are given by
\begin{align}
\#(A)\equiv\big|\,\text{indep}(A)\,\big|&=\pmatrix{n\\k}\\
\#(S)\equiv\big|\,\text{indep}(S)\,\big|&=\pmatrix{n+k-1\\k}
\end{align}
My question is what are nice ways to proof this last statement (both for the symmetric/antisymmetric case). With  nice I mean either easy to remember, intuitive or elegant. The antisymmetric case has been proven here and both cases have also been proven in this paper. The reason I'm asking this question is to have these facts nicely together and also to get some more proofs to make it easier to remember. Also I found a nice proof myself.

答:
The antisymmetric case can be rephrased as 'I have $n$ digits. In how many ways can I group them in groups of $k$'. This has the answer $\pmatrix {n\\k}$.

The symmetric case is also in the form of an binomial coefficient so this invites us to the idea of writing the components of a symmetric $n,k$ tensor as an antisymmetric $n+k-1,k$ tensor. Let's call it $A'$. The difference between the antisymmetric/symmetric components is that the symmetric components can have repeating indices while the antisymmetric components can't. Let's construct the following map from $\text{indep}(S)\rightarrow\text{indep}(A')$: if $S_{i_1\dots i_k}$ has no repeating indices then map it directly to $A'_{i_1\dots i_k}$. If $S$ does have repeating indices we record the places of repeating indices. Then delete all repeating indices, leaving us with a couple indices to be filled. Fill these empty indices with the locations but add  $n$ to each of them.

For example let's take a look at $S_{1,1,1,5,9}$ where $k=5,n=9$. In $S_{1,1,1,5,9}$ we have repeating indices at location 1 and 2. If we delete all repeating indices we get $A'_{1,5,9,\cdot,\cdot}$ with two spots left. We fill these spots with $9+1=10$ and $9+2=11$ to 'remember' where we deleted indices. This gives $A'_{1,5,9,10,11}$. Another example: $S_{5,5,7,7,7}\rightarrow A'_{5,7,10,12,13}$. This new tensor has dimension $n+k-1$ since $k-1$ is the largest position at which a double index can occur $n+k-1$ is the largest index that will occur. It also stricly increasing indices so it can be mapped to an antisymmetric tensor and correspondingly it has $\pmatrix{n+k-1\\k}$ components. For any antiysymmetric tensor we can reverse this map so it is a bijection. So we can conclude that our original symmetric tensor has the same number of components as this new tensor. So $\#(S)=\pmatrix{n+k-1\\k}$


问:
is Root a minus number undefined
Assume that we have a function $f$ such that $f(x) = \sqrt x$
Is a domain of $-1$ undefined.
I thought it wouldn’t because the answer would be $i\sqrt x$.
However, apparently, it is undefined. Can someone explain to me why?

答:
Both.

The real square root function is $$\mathbb R^+\to\mathbb R^+:x\to\sqrt x.$$

$-1$ is not in the domain and $\sqrt{-1}$ undefined.

The complex square root is

$$\mathbb C\to\mathbb C:z\to\sqrt z$$ and requires to choose a specific branch (usually, $\sqrt{-1}=i$).

Title: is Root a minus number undefined


问:
Upper bound of sum of powers of lowest divisors
If $n = p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k}$, where $p_1$ is lowest prime divisor.
Let $d(n) = \alpha_1$.

Is there any upper bound for $\displaystyle \sum_i^{n} d(i)$?
We can roughly estimate it as $O(n \log n)$, because $d(n) \in O(\log n)$, but maybe exists better estimate for this sum? For example $O(n \log \log n)$?

答:
We get an $O(n \log \log n)$ bound from the obvious estimate
$$d(n) = \alpha_1 \leqslant \alpha_1 + \ldots + \alpha_k = \Omega(n)$$
and the rather well-known (e.g. Theorem 430 in Hardy/Wright)
$$\sum_{n \leqslant x} \Omega(n) = x\log \log x + B_2x + o(x)$$
which follows from Mertens' second theorem.

We can get an asymptotically smaller bound, $O(n)$, by using a slightly more conservative bound. With $\omega(n)$ denoting the number of distinct prime factors of $n$, i.e. $\omega(n) = k$ if $n = p_1^{\alpha_1}\ldots p_k^{\alpha_k}$, we can easily see
$$d(n) = \alpha_1 \leqslant 1 + (\alpha_1 - 1) + \ldots + (\alpha_k - 1) = 1 + \Omega(n) - \omega(n)\,.$$
Thus, since (see also Theorem 430 in Hardy/Wright)
$$\sum_{n \leqslant x} \omega(n) = x\log \log x + B_1 x + o(x)$$
we have
\begin{align}
\sum_{i = 1}^{n} d(i)
&\leqslant \sum_{i = 1}^{n} \bigl(1 + \Omega(i) - \omega(i)\bigr) \\
&= n + \sum_{i = 1}^{n} \Omega(i) - \sum_{i = 1}^{n}\omega(i) \\
&= n + (B_2 - B_1)n + o(n)\,.
\end{align}
Since trivially $d(i) \geqslant 1$ for $i > 1$ it follows that
$$\sum_{i = 1}^{n} d(i) \asymp n\,.$$


问:
Example for Riemann integrable function such that $\int^{a+1}_a f(x)dx=0$ but $ f(x)\neq 0$
I'm looking for Riemann integrable function $f:\mathbb{R}\to\mathbb{R}$ with $\int^{a+1}_a f(x)dx=0$ for all $a\in\mathbb{R}$,  but $ f(x)\neq 0$.

I suspect that floor function involves here, if so, then how?

Thank you all!

To clarify: $f$ must not be identically equal to $0$, and it should be integrable over any finite interval.

答: What about the characteristic function of a singleton?</p>


问:
Find an enclosed formula for $a_n$ where $a_n$ is the $n$th term of the sequence $1,0,1,1,0,0,1,1,1,0,0,0,1,\dots$
I have reached that the formula should look something like this
$$a_n=\frac{1+(-1)^{f(n)}}{2}$$
where $f: \mathbb{N} \rightarrow \mathbb{N}$ is a function which outputs an even number if the $n$th term of the sequence is $1$ and an odd number if the $n$th term is $0$.

I tried manipulating the sequence a bit but haven't reached any conclusion. Any ideas?

答:
If this is one $1$ followed by one $0$ followed by two $1$s followed by two $0$s followed by three $1$s followed by three $0$s ... beginning with $a_0=1$, then it is OEIS A118175 and can be written with ceiling and rounding functions $$a_n=1 - \lceil \sqrt{n+1} \,\rceil + [\sqrt{n+1}\,]$$

but there are other possibilities for what it is, including $16$ listed at OEIS


问:
Functor that does not preserve monic and epic
In Jacobson’s book BAII, he gave two exercises to show that there exists functor that does not preserve monic or epic.

Ex-1.Let M and N be monoids as categories with a single object. Show that in this identification,a functor is a homomorphism of M into N.

Ex-2.Use Ex-1 to construct a functor $F$ and a monic(epic) $f$ such that $F(f)$ is not monic(epic)

I have solved Ex-1. However, I don’t know how to construct concrete example to solve Ex-2.

I do hope someone can give me some hints. Thank you very much!

答:
Hint: Let $M$ be a free monoid.

 A simple example is with $M=(\Bbb N,+),\ \,N=(\{0,1\},\max)$ and $F(m)=\min(1,m)$.

 $1$ is cancellable in $M$ but not in $N$.


答:
Looks to me like $n$ $1$'s followed by $n$ $0$'s, for $n = 1,2,3, \ldots$.
If the indexing starts from $0$, the $n$ $1$'s are in positions $n^2+n$ to $(n+1)^2-1$.  Thus $a_k$ is the indicator function of $\{k \in \mathbb N:k \ge \lfloor \sqrt{k} \rfloor^2 + \lfloor \sqrt{k} \rfloor\}$

Title: Find an enclosed formula for $a_n$ where $a_n$ is the $n$th term of the sequence $1,0,1,1,0,0,1,1,1,0,0,0,1,\dots$

Body:
I have reached that the formula should look something like this
$$a_n=\frac{1+(-1)^{f(n)}}{2}$$
where $f: \mathbb{N} \rightarrow \mathbb{N}$ is a function which outputs an even number if the $n$th term of the sequence is $1$ and an odd number if the $n$th term is $0$.

I tried manipulating the sequence a bit but haven't reached any conclusion. Any ideas?


问:
A Number theory & polynomial question
Prove that if $p$ is a some polynomial with integer coefficients, $M$ is an integer such that $p(n)$ is prime for every $n\ge M $. Then $p$ is constant.

Please help , I have no clue how to prove it.

答:
Suppose $p$ is not constant. We know that $p(M)=q$ with $q$ prime. Since $p$ is not constant $\lim_{n \to \infty} |p(M+nq)|=\infty$. That is, for $n$ big enough, $|p(M+nq)|>q$. But notice that
$$
p(M+nq) \equiv p(M) \equiv 0 \ mod \ q.
$$
So $M+nq \geq M$ and $p(M+nq)$ is a composite number.

Title: A Number theory & polynomial question


答:
Consider any integrable periodic function with period $1$, which integrates to $0$, i.e., let $g(x)$ be any function defined on interval $[0,1]$. Then consider
$$f(x) = \begin{cases} g(x) - \underbrace{\int_0^1 g(x) dx}_b & \text{if }x\in[0,1]\\ g(\{ x\}) - b & \text{else}\end{cases}$$</p>

Title: Example for Riemann integrable function such that $\int^{a+1}_a f(x)dx=0$ but $ f(x)\neq 0$

Body:
I'm looking for Riemann integrable function $f:\mathbb{R}\to\mathbb{R}$ with $\int^{a+1}_a f(x)dx=0$ for all $a\in\mathbb{R}$,  but $ f(x)\neq 0$.

I suspect that floor function involves here, if so, then how?

Thank you all!

To clarify: $f$ must not be identically equal to $0$, and it should be integrable over any finite interval.


问:
Finding the volume of $\{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$ by integration
I have been trying to solve this integral for some time now without success:

$$ T = \{(x,y,z) \in \mathbb{R}^3 \vert (x+y+z)^2 + (x-y)^2 \leq 1, 0 \leq z \leq 3 \}$$
$$ I = \int_{T} ~dx~dy~dz $$

I have tried simplifying the domain, by writing:
$$ (x+y+z)^2 + (x-y)^2 \leq 1 \to $$
$$ 2x^2 + 2y^2 + z^2 + 2xz + 2yz \leq 1 \to $$
$$ x^2 + y^2 + \frac{z^2}{2} + xz + yz \leq \frac{1}{2} $$
Now, I tried doing this change of variable: $ x = \rho \cos\theta$, $y = \rho \sin\theta$ and $z = \sqrt 2 w$
With $\rho \in (0,\infty)$, $\theta \in [0,2\pi]$. They are basically cylindrical coordinates.
I do some calculation and I have the following:
$$ \rho^2 + w^2 + (\rho \cos\theta + \rho \sin \theta) \cdot \sqrt{2}w \leq \frac{1}{2}$$
But I don't see how this is helping. I have also tried spherical coordinates substitution without success.

Can I have some hints?

答:
Lt us integrate by slicing. The slice at height $z$ is defined by:

$$\left(x+\frac{z}{2}\right)^2+\left(y+\frac{z}{2}\right)^2\leq\frac{1}{2}$$

This inequation defines a disk with radius $\sqrt{\frac12}$

Therefore the area of the slice at height $z$ is constant, equal to:

$$S(z)=\pi \frac{1}{2}$$

giving a volume:

$$V=\int_0^3 S(z)dz=\frac{3\pi}{2}$$


问:
My Laplace Transform vs. My Desmos laplace Transform
If I were to calculate the Laplace Transform of 1 I would do the following:
\begin{equation}\mathcal{L}\{1\}=\int_{0}^{\infty}e^{-st}dt \end{equation}
From here I would go on evaluate the improper integral:\begin{align}&=-\frac{1}s[e^{-st}]\Bigg\vert_0^\infty \\&=\frac{1}s[e^{-s\cdot0}-e^{-\infty \cdot s}]\end{align}
So the end result would be that I would get $\mathcal{L}\{1\}=\frac{1}s$.
I plotted this on desmos. I used a really big upper limit for the approximation and saw it matches the positive x axis part of only of $\frac{1}x$. Why is that? Does this have heavy implications differential equations and/or calculus if so please let me know?

答:
Your evaluation of this improper integral is valid only for $s>0$.  For $s \le 0$ the integral diverges to $\infty$.  By replacing $\infty$ by some large number $R$, the integral becomes
$$ \frac{1}{s} \left(1 - e^{-Rs}\right)$$
When $s< 0$ the $e^{-Rs}$ is a very large number.


问:
Find all functions $f$ such that $f(f(x, y), z) = f(x, yz)$
Find all functions $f:\mathbb{R}_{\geq{0}} \times \mathbb{R}_{>{0}} \to \mathbb{R}_{\geq{0}}$ such that for all $x \in \mathbb{R}_{\geq{0}}$ and all $y, z \in \mathbb{R}_{>{0}},$
$$ f(f(x,y),z)=f(x,yz)$$
and
$$\forall x\in\mathbb{R}_{\geq{0}}: f(x,1)=x$$
where the function $f$ is differentiable over its entire domain (at least once).

I found a solution assuming that the function is analytical. Then by unfolding into Taylor's series and comparing the coefficient, I got
$$f(x,y) = y^{\alpha}x$$
where $\alpha$ is any real number.

I don't know how to prove that this solution is unique and how to circumvent the proof without the additional requirement of analytical function.

Edit

I already know why I had a problem with the ambiguity of the function. I forgot a very important condition.

Function $f$ is homogenous with respect to $x$:
$$f(\lambda{}x,y)=\lambda{}f(x,y)$$
Then there is only one unique solution: $f(x,y)=xy^\alpha$.

Thank you all for your advice, primarily @Max for his guidance to Transaltion function as general solution without this homogenous condition.

答:
Let $F(x,y)=\ln f(e^x, e^y)$ . Then

$$\begin{aligned} F(F(x,y), z)&=\ln f(e^{\ln f(e^x, e^y)}, e^z)= \\&\ln f( f(e^x, e^y), e^z)=\ln f(e^x, e^{y+z})\\&=F(x, y+z). \end{aligned}$$

This is the translation equation (see https://en.wikipedia.org/wiki/Abel_equation#Equivalence). Pick arbitrary continuous and strictly monotone $h$ with range all of $\mathbb{R}$ (and differentiable at least once with positive derivative, to get corresponding smoothness of $f$). Then $F(x, y)=h(h^{-1}(x)+y)$ solves it. Then $f(x,y)=e^{F(\ln x, \ln y)}=e^{ h(h^{-1}(\ln x)+\ln y)}$

Your solution is obtained when $h(u)=\alpha u$, i.e. $f(x,y)=e^{\alpha (\frac{1}{\alpha} \ln x+\ln y)}=x y^{\alpha}$.

The solution $f(x,y)=x^y$ is almost obtained from $h(u)=e^{u}$: if one ignores the fact that $\ln \ln x$ is not defined for $x\leq 1$ we have $f(x,y)=e^{e^{\ln \ln x+ \ln y}}=e^{y\ln x}=x^y$.

There is a discussion of translation equation in Section 6.1 of Aczel's book on functional equations. It shows that (under mild continuity conditions) if $F(x,\cdot)$ is non-constant for every $x$ then the solution is in fact of the form above. This suggests that $f(x,y)=x^y$ is in some sense exceptional.


问:
Basic book about mathematical proofs
I'm looking for an understandable book about mathematical proofs. </p>

<p>As an engineer/computer scientist major I'm used to do higher math on a daily base, but whenever I'm asked in an exercise to "show" something I get the "where should I start? and what exactly do they want me to do?" syndrome. </p>

答: You might also like this: How to Solve it: Modern Heuristics. I've read the first few pages and it's not bad at all.</p>

Title: Basic book about mathematical proofs


答: If you mean "Riemann integrable on every finite interval", try $f(x)=\sin{2\pi x}$. If it needs to be non-zero everywhere, you may redefine it to be $1$ for $2x\in\mathbb Z$.</p>


问:
Is this definition of distance between two points correct?
I came across this definition

but it's confusing to me.

Shouldn't $A$ be $(x_1,y_1)$ instead of $(x_1,x_2)$? Same for $Y$. And even if $A$ was correct like that then shouldn't the formula for the distance be written as $$d(X,Y) =\sqrt{(y_1-x_1)^2+(y_2-x_2)^2}$$

What am I getting wrong?

答:
Indeed there is a typo and your expression

$$d(X,Y) =\sqrt{(y_1-x_1)^2+(y_2-x_2)^2}$$

is the correct one by direct application of pythagorean theorem.


问:
Catalan identity to calculate the determinant of a matrix.
I want to show that the determinant of the matrix $A$ of order $n\times n$ with entries $a_{ij}=C_{i+j-2}$ is $1$, where $C_m$ is the $m$th term of the Catalan sequence. To solve this problem, the hint is defining a decomposition $LU$ of $A$.

Set $\,u_{ij}\,$ to be
$$ u_{ij} = \frac{2i+1}{i+j+1}{2j \choose j-i}, $$
and let $\,U_n = (u_{ij})_{0\le i,j\le n-1},\,$ which is an
$\,n\times n\,$ upper triangle matrix whose diagonal entries
are $\,1.\,$ Put $\,L_n = \,^{t}U_{n}.\,$ Then we can show
that $\,C_n^0 = L_nU_n,\,$ which immediately implies that
$\,\det C_n^0 = 1.\,$ We omit the detail but one can easily
prove that these $\,LU$-decompositions reduce to the
following identity:
$$ \sum_{k\ge 0} \frac{(2k+1)^2}{(i+k+1)(j+k+1)}
   {2i \choose i-k}{2j \choose j-k} = \frac1{i+j+1}
   {2i+2j \choose i+j}. $$

My question is regarding the U matrix, I don't know why it is triangular, but in case such how can I prove the mentioned identity?

Original image: https://i.stack.imgur.com/xcDM3.png

Body:
 Introduction

The identity

$$\sum_{k\ge 0} \frac{(2k+1)^2}{(p+k+1)(q+k+1)}
{2p\choose p-k} {2q\choose q-k} =
\frac{1}{p+q+1} {2p+2q\choose p+q}$$

is identical to

$$\sum_{k=0}^{\min(p,q)} (2k+1)^2
{2p+1\choose p+k+1} {2q+1\choose q+k+1} =
\frac{(2p+1)(2q+1)}{p+q+1} {2p+2q\choose p+q}$$

or

$$\sum_{k=0}^{\min(p,q)} (2k+1)^2
{2p+1\choose p-k} {2q+1\choose q-k} =
\frac{(2p+1)(2q+1)}{p+q+1} {2p+2q\choose p+q}.$$

The LHS is

$$S= [z^p] (1+z)^{2p+1} [w^q] (1+w)^{2q+1}
\sum_{k=0}^{\min(p,q)} (2k+1)^2 z^k w^k.$$

The two coefficient extractors enforce the upper limit of the sum:

$$[z^p] (1+z)^{2p+1} [w^q] (1+w)^{2q+1}
\sum_{k\ge 0} (2k+1)^2 z^k w^k
\\ = [z^p] (1+z)^{2p+1} [w^q] (1+w)^{2q+1}
\frac{z^2 w^2 + 6 z w + 1}{(1-zw)^3}
\\ = - [z^p] \frac{1}{z^3} (1+z)^{2p+1} [w^q] (1+w)^{2q+1}
\frac{z^2 w^2 + 6 z w + 1}{(w-1/z)^3}
\\ = - [z^{p+3}] (1+z)^{2p+1} [w^q] (1+w)^{2q+1}
\frac{z^2 w^2 + 6 z w + 1}{(w-1/z)^3}.$$

The coefficient extractor in $w$ is

$$\mathrm{Res}_{w=0} \frac{1}{w^{q+1}} (1+w)^{2q+1}
\frac{z^2 w^2 + 6 z w + 1}{(w-1/z)^3}.$$

 Residue at infinity

Now residues sum to zero and the residue at infinity is given by

$$-\mathrm{Res}_{w=0} \frac{1}{w^2}
w^{q+1} \frac{(1+w)^{2q+1}}{w^{2q+1}}
\frac{z^2/w^2 + 6 z/w + 1}{(1/w-1/z)^3}
\\ = -\mathrm{Res}_{w=0}
\frac{(1+w)^{2q+1}}{w^{q+2}}
\frac{z^2 w + 6 z w^2 + w^3}{(1-w/z)^3}
\\ = -\mathrm{Res}_{w=0}
\frac{(1+w)^{2q+1}}{w^{q+1}}
\frac{z^2 + 6 z w + w^2}{(1-w/z)^3}.$$

Next applying the coefficient extractor in $z$ we find

$$\mathrm{Res}_{z=0}
\frac{(1+z)^{2p+1}}{z^{p+4}}  \mathrm{Res}_{w=0}
\frac{(1+w)^{2q+1}}{w^{q+1}}
\frac{z^2 + 6 z w + w^2}{(1-w/z)^3}
\\ = \mathrm{Res}_{z=0}
\frac{(1+z)^{2p+1}}{z^{p+2}}  \mathrm{Res}_{w=0}
\frac{(1+w)^{2q+1}}{w^{q+1}}
\frac{1 + 6 w/z + w^2/z^2}{(1-w/z)^3}
\\ = \mathrm{Res}_{z=0}
\frac{(1+z)^{2p+1}}{z^{p+2}}  \mathrm{Res}_{w=0}
\frac{(1+w)^{2q+1}}{w^{q+1}}
\sum_{k\ge 0} (2k+1)^2 \frac{w^k}{z^k}
\\ = \sum_{k\ge 0} (2k+1)^2
{2p+1\choose p+k+1} {2q+1\choose q-k} = S.$$

This means that $S$ is minus  half the residue at $w=1/z$, substituted
into the coefficient extractor in $z.$

 Residue at $w=1/z$

The residue at $w=1/z$ is

$$\mathrm{Res}_{w=1/z} \frac{1}{w^{q+1}} (1+w)^{2q+1}
\frac{z^2 w^2 + 6 z w + 1}{(w-1/z)^3}
\\ = \mathrm{Res}_{w=1/z} \frac{1}{w^{q+1}} (1+w)^{2q+1}
\left(\frac{8}{(w-1/z)^3} + \frac{8z}{(w-1/z)^2} +
\frac{z^2}{w-1/z}\right).$$

Evaluating the three pieces in turn we start with

$$8\frac{1}{2}\left(\frac{ (1+w)^{2q+1}}{w^{q+1}}\right)''
= 4 (q+1)(q+2)\frac{(1+w)^{2q+1}}{w^{q+3}}
\\ - 8(q+1)(2q+1)\frac{(1+w)^{2q}}{w^{q+2}}
+ 4 (2q+1)(2q)\frac{(1+w)^{2q-1}}{w^{q+1}}.$$

Evaluate at $w=1/z$ to get

$$4(q+1)(q+2) \frac{(1+z)^{2q+1}}{z^{q-2}}
\\ - 8 (q+1)(2q+1) \frac{(1+z)^{2q}}{z^{q-2}}
+ 4 (2q+1)(2q) \frac{(1+z)^{2q-1}}{z^{q-2}}.$$

Substituting into the coefficient extractor in $z$ we find

$$- 4(q+1)(q+2) {2p+2q+2\choose p+q+1}
\\ + 8 (q+1)(2q+1) {2p+2q+1\choose p+q+1}
- 4 (2q+1)(2q) {2p+2q\choose p+q+1}.$$

Continuing with the middle piece we have

$$8z\left(\frac{ (1+w)^{2q+1}}{w^{q+1}}\right)'
= -8z (q+1) \frac{(1+w)^{2q+1}}{w^{q+2}}
+8z (2q+1) \frac{(1+w)^{2q}}{w^{q+1}}.$$

Evaluate at $w=1/z$ to get

$$-8(q+1) \frac{(1+z)^{2q+1}}{z^{q-2}}
+ 8(2q+1) \frac{(1+z)^{2q}}{z^{q-2}}.$$

The coefficient extractor now yields

$$8(q+1) {2p+2q+2\choose p+q+1} - 8(2q+1) {2p+2q+1\choose p+q+1}.$$

The third and last piece produces

$$\frac{(1+z)^{2q+1}}{z^{q-2}}$$

which when substituted into the coefficient extractor yields

$$-{2p+2q+2\choose p+q+1}.$$

 Collecting the three pieces

We get

$$-(2q+1)^2 {2p+2q+2\choose p+q+1}
+ 8q (2q+1) {2p+2q+1\choose p+q+1}
- 8q (2q+1) {2p+2q\choose p+q+1}
\\ = -(2q+1)^2 {2p+2q+2\choose p+q+1}
+ 8q (2q+1) {2p+2q\choose p+q}
\\ = - 2 (2q+1)^2 {2p+2q+1\choose p+q}
+ 8q (2q+1) {2p+2q\choose p+q}
\\ = - 2 (2q+1)^2 \frac{2p+2q+1}{p+q+1} {2p+2q\choose p+q}
+ 8q (2q+1) {2p+2q\choose p+q}
\\ = -2 \frac{(2p+1)(2q+1)}{p+q+1} {2p+2q\choose p+q}.$$

Halve this value and flip the sign to obtain the coveted

$$\bbox[5px,border:2px solid #00A000]{
\frac{(2p+1)(2q+1)}{p+q+1} {2p+2q\choose p+q}.}$$


问:
Spectrum (eigenvalues) and their relation to Projection Valued Measures(PVM)
I'm going to ask 2 questions but they're (I believe) related.

1) Let $z_0$ be an eigenvalue and $\psi$ a corresponding normalized eigenvector. Compute $μ_{\psi}$.

This is the whole question but we could assume operator $A$ which we know has eigenvalue $z_0$ is self-adjoint and $\mu_{\psi}(\Omega)$ is given by $\langle\psi,P_A(\Omega)\psi\rangle$ where $P_A$ is associated PVM to $A$.

2) Show that $z_0$ is an eigenvalue if and only if P({$z_0$})$\neq0$. Show that $Ran(P(\{z_0\}))$is the corresponding eigenspace in this case. Again we could assume $A$ is self-adjoint.

I've already (kind of) answered both of the questions but reason for why i'm asking these questions is both questions seems very intuitive like the moment i saw the first question i immediately said if $\Omega$ contains $z_0$ it is $||\psi||^2$ if not it is $0$. And i would like to answer them more formally \

3) Under the shadow of these 2 questions i would like to ask another question.We know spectrum $$\sigma(A)=\{z\in\mathbb{R} : P_A(z-\epsilon,z+\epsilon)\neq0 \enspace \forall\epsilon>0\}$$ but this does not differentiate between different types of spectrums pure point,point embedded in continuum and purely continuous.Clearly eigenvalues are in point part of the spectrum and above results shows us $P(\{z_0\})$ is an eigenspace but what happens if $z_0\in$ purely cont.?Could someone give me an example(maybe example from physics like quantum harmonic oscillator etc.)?

Related links Show that eigen-vectors belong in range of projection valued measure. Spectral measure associated to eigenvector of self-adjoint operator

Any hint and solution is appreciated THANKS!

答:
I'm not sure if there's a way to see this directly from the definition of $P_A$, but here is a proof using the resolvent, which is quite natural in view of the proof of the spectral theorem. With $R_A(z)=(A-zI)^{-1}$, we know that
$$\langle\psi,R_A(z)\psi\rangle=\langle \psi,\frac{1}{z_0-z}\psi\rangle=\frac{1}{z_0-z}||\psi||^2$$
This is in turn, by the spectral theorem, equal to $F(z):=\int_{\mathbb{R}}\frac{1}{\lambda-z}\,d\mu_{\psi}(\lambda)$, which is the Borel transform of $\mu_\psi$. We may recover a measure from its Borel transform via the Stieljes inversion formula:
$$\mu_\psi(\lambda)=\lim_{\delta\downarrow0}\lim_{\epsilon\downarrow 0}\frac{1}{\pi}\int_{-\infty}^{\lambda+\delta}\text{Im}(F(t+i\epsilon))\,dt$$
I suggest your try this computation, but if you cannot get it I am happy to include details. The key point is that the integrand has a singularity as $\epsilon\rightarrow 0$ precisely at $t = z_0$.

If $\psi$ is an eigenvector corresponding to $z_0$ then
$$0\neq\langle \psi,\psi\rangle=\int_{\mathbb{R}}\,d\mu_\psi=\int_{\{z_0\}}\,d\mu_\psi=\langle \psi,P(\{z_0\})\psi\rangle$$
where we have used the fact that $\mu_\psi$ is a point-mass from $1.$ Conversely, if $P(\{z_0\})\neq 0$, we may find $\psi$ such that $P(\{z_0\})\psi=\psi$ because $P(\{z_0\})$ is a projection. This also means that $P(\mathbb{R}\setminus\{z_0\})x=0$. Now, the result follows from DisintegratingByParts answer here. They are proving the backwards implication (more or less), but if you look at the argument it works to prove what we want by tracing it in reverse.

The canonical example of an operator with purely ac spectrum is the free energy $-\Delta$, where $\Delta$ is the Laplacian. For this fact, see Theorem 7.8 in Mathematical Methods in Quantum Mechanics by Teschl. Teschl's book is an excellent reference for this material and you may find it free online. While extremely important, this example is a little hard to make sense of because $-\Delta$ is an unbounded operator. For a simpler example, you may try to show that $f(x)\mapsto xf(x)$ on $L^2([0,1])$ has purely ac spectrum.


问:
Is there a concise way to denote two sets that do not contain each other but have non-empty intersection?
Let $A$ and $B$ two sets. Is there any concise way to write these three conditions together?

$$A\not\subseteq B \quad\text{and}\quad B\not\subseteq A \quad\text{and}\quad A\cap B \neq \emptyset$$

答:
The easiest way I can think of to write it is $B\setminus A\neq \emptyset \neq A\setminus B \neq A$, though I'm not sure this is better than what you proposed.

Another way would be to use symmetric difference (defined by $X\mathrel{\bigtriangleup}Y=(X\cup Y)\setminus (X\cap Y)$) and say $A\mathrel{\bigtriangleup}B\neq A\cup B\not\in\{A,B\}$.


问:
Conditional probability involving Gambler's Ruin
Consider a gambler whose initial fortune is $2$ dollars, and who repeatedly makes fair bets in which his fortune either increases or decreases by $1$ dollar. He keeps betting until he runs out of money. Given that the gambler got broke on his $10th$ bet, what is the probability that his wealth trajectory went upto $6$ dollars and then crashed down to zero?

From what I understand, isn't there just one possible case in which this can happen? $+2, +3... +6, +5, ...0$? So shouldn't the probability just be $\frac{1}{1024}$? ($1/2$ for each state change?) I know I am missing something but I am unable to figure it out. It has something to do with conditional probability I think but I don't know what exactly is wrong here.

答:
You are asked for the conditional probability that the one case you mentioned occurred, given that the gambler went broke on the $10$'th bet.  By definition,
$$ \mathbb P(A \mid B) = \dfrac{\mathbb P(A \cap B)}{\mathbb P(B)}$$
You have found $\mathbb P(A \cap B)$, but you need to divide by $\mathbb P(B)$.

EDIT: Let $f(n,b)$ be the probability that it takes exactly $n$ bets to go broke starting with $b$ dollars.  Thus $f(0,0)=1$, $f(n,0) = 0$ if $n \ge 1$, $f(n, b) = 0$ if $b > n$, $f(n,b) = (f(n-1,b-1) + f(n-1,b+1))/2$ for $1 \le b \le n$.  You want $f(10, 2)$. Compute $f(1,1)$, $f(2,2)$, $f(3,1)$, $f(3,3)$, etc.


问:
Finding the Inverse Laplace Transform
I am having trouble finding the inverse Laplace transform of: $$\frac{1}{s^2-9s+20}$$

I tried writing it in a different way:

$L^{-1}\{\frac{1}{(s-\frac{9}{2})^2-(\frac{1}{2})^2}\}=2L^{-1}\{\frac{\frac{1}{2}}{(s-\frac{9}{2})^2-(\frac{1}{2})^2}\}$

Can someone help ?

Title: Finding the Inverse Laplace Transform

Body:
I have done it:
$$\begin{array}{l}
 X(s) = \frac{1}{{{s^2} - 9s + 20}} = \frac{1}{{(s - 4)(s - 5)}} = \frac{{ - 1}}{{s - 4}} + \frac{1}{{s - 5}} \\
  =  > x(t) =  - {e^{4t}} + {e^{5t}} \\
 \end{array}$$


问:
Analytic function on $\mathbb{D}$ belongs to $H^2$ iff $\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)<\infty$

Analytic function on $\mathbb{D}$ belongs to Hardy space $H^2$ iff $\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)<\infty$ where $dA$ is the area measure.

By definition, $f\in H^2$ if $$\|f\|^2_{H^2}=:\sup_{0<r<1}\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2d\theta<\infty$$
Moreover we know that $\|f\|^2_{H^2}=\sum_{k=0}^{\infty}|a_k|^2$ where $\{a_k\}\in \ell^2$.

Then, we have, $$\begin{align*}\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)&=\int_0^{2\pi}\int_0^1|f'(re^{i\theta})|^2(1-r^2)rdrd\theta\\
&=\int_0^{2\pi}|f'(re^{i\theta})|^2d\theta\int_0^1r(1-r^2)dr\\
&=\frac{1}{4}\sum k^2|a_k|^2r^{2n-2}\end{align*}$$

I think the calculation is a little messed up, and I'm not sure how I should go about proving the statement.

答:
\begin{align*}I=\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)&=\int_0^1\int_0^{2\pi}|f'(re^{i\theta})|^2(1-r^2)rd\theta dr\\
&=2\pi\sum_{k \ge 1} k^2|a_k|^2\int_0^1r^{2k-2}r(1-r^2)dr
\end{align*}

(by applying orthogonality in $\theta$ and switching sum and integral by absolute convergence)

Integrating in $r$ one gets that:

$I=2\pi\sum_{k \ge 1} k^2|a_k|^2 (1/(2k)-1/(2k+2))=\pi\sum_{k \ge 1} \frac{k}{k+1}|a_k|^2$ so obviously $I$ coneverges or diverges precisely when $\sum_{k \ge 1}|a_k|^2$ does so hence we have the required if and only if ($f(0)$ has no bearing on the convergence/divergence of $\sum |a_k|^2$)


问:
Annual variance from monthly data (returns)
I can't determine how we establish equality between the two bracketed expressions in the image below. The square of the sum is being equated with the sum of the squares, which in general isn't true. What am I missing?
Thank you.

答:
We can write the squared sum of residual returns as

$$\left(\sum_{i=1}^{12} (r_i - \bar{r})\right)^2 = \sum_{i=1}^{12} (r_i - \bar{r})\sum_{j=1}^{12} (r_j - \bar{r}) = \sum_{i=1}^{12} \sum_{j=1}^{12}(r_i - \bar{r}) (r_j - \bar{r}) \\ = \sum_{i=1}^{12} (r_i - \bar{r})^2 + \underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r}) $$

Taking the expected value, we get

$$\sigma_y^2 = \mathbb{E} \left(\sum_{i=1}^{12} (r_i - \bar{r})^2 + \underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r}) \right)  \\= \mathbb{E}\left(\sum_{i=1}^{12} (r_i - \bar{r})^2\right) + \mathbb{E}\left(\underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r})\right)  $$

For the second term on the right-hand side, we have

$$\mathbb{E}\left(\underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r})\right) = \underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}\mathbb{E}\left((r_i - \bar{r}) (r_j - \bar{r})\right)$$

You are given "... the monthly returns ... are mutually uncorrelated," which means

$$\mathbb{E}\left((r_i - \bar{r}) (r_j - \bar{r})\right) = 0 \,\, \text{ for } i \neq j$$

Thus,

$$\sigma_y^2 = \mathbb{E}\left(\sum_{i=1}^{12} (r_i - \bar{r})^2\right) = \sum_{i=1}^{12} \mathbb{E}\left((r_i - \bar{r})^2\right) = 12 \sigma^2$$


问:
How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$
Somebody please help me to solve this Differential equation.

$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $

答:
Claim:  This is an exact first order ordinary differential equation.

To verify this claim, we impose some standard notation, then perform a simple test.  Let
\begin{align*}
M(x,y) &= 3x^2 \tan(y) - \frac{2y^3}{x^3}  \text{,}  \\
N(x,y) &= 4 y^3 + \frac{x^3}{\cos^2 y} + \frac{3y^2}{x^2}
\end{align*}
(Notice that, for a while, we will not treat $y$ as a dependent variable.)
We wish to find an $f(x,y)$ such that both
\begin{align*}
\frac{\partial}{\partial x} f(x,y) &= M(x,y)  \\
\frac{\partial}{\partial y} f(x,y) &= N(x,y)
\end{align*}
If so, the equation is exact.  We perform the test:
$$  \frac{\partial}{\partial y} M(x,y) \overset{?}= \frac{\partial}{\partial x} N(x,y)  $$
(This test verifies $\frac{\partial^2}{\partial y \, \partial x} f(x,y) = \frac{\partial^2}{\partial x \, \partial y} f(x,y)$, which is true under mild assumptions about the various partial derivatives of $f$.)

So we calculate \begin{align*}
\frac{\partial}{\partial y} M(x,y) &= \frac{-6y^2}{x^3} + 3x^2 \sec^2(y)  \text{,}  \\
\frac{\partial}{\partial x} N(x,y) &= \frac{-6y^2}{x^3} + 3x^2 \sec^2(y)  \text{.}
\end{align*}
These are equal, so the equation is exact.

Since the equation is exact, such an $f(x,y)$ exists and the general solution is
$$  f(x,y) + C  \text{.}  $$
We know two partial derivatives of $f$, so we can (partially) recover $f$ by integration.
\begin{align*}
f_1(x,y) &= \int M(x,y) \,\mathrm{d}x  \\
    &= \int 3x^2 \tan(y) - \frac{2y^3}{x^3} \,\mathrm{d}x  \\
    &= \frac{y^3}{x^2} + x^3 \tan(y) + C_1(y)  \text{, and}  \\
f_2(x,y) &= \int N(x,y) \,\mathrm{d}y  \\
    &= \int 4 y^3 + \frac{x^3}{\cos^2 y} + \frac{3y^2}{x^2} \,\mathrm{d}y  \\
    &= \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C_2(x)  \text{.}
\end{align*}
Here, $C_1(y)$ is a function of integration, analogous to the constant of integration.  Notice that if $f(x,y)$ has terms that depend only on $y$ or are constant, those terms are sent to zero in $\frac{\partial}{\partial x} f(x,y)$, so the general antiderivative must represent all possible $f$s differing only by a function of $y$ (and constants).  Similarly, $C_2(x)$ is a function of integration.

Notice that we actually can use $f_2$ to find out about $C_1(y)$.  $C_1(y) = y^4 + C$, where $C$ is a constant of integration.  And reasoning the similarly, $C_2(x) = C$.  (Some describe this process of merging the terms from the two solutions into a single solution that matches both partial derivatives.  Perhaps a better way to describe this is that we are antidifferentiating both sides of the original equation, each with respect to the variable in the differential already present, and then matching the resulting functions of integration.)  So we have
$$  f(x,y) = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C  \text{.}  $$
Then $f(x,y) = C$ is the general solution, so,
$$  C = \frac{y^3}{x^2} + x^3 \tan(y) + y^4 + C  $$
or, what is the same thing (because the range of "an arbitrary number minus another arbitrary number" is an arbitrary number),
$$  C = \frac{y^3}{x^2} + x^3 \tan(y) + y^4  $$
is the general solution to the given equation.

We can, of course, verify that this is the case using implicit differentiation.  First, we stop pretending that $y$ is an independent variable.  Next, we implicitly differentiate the claimed general solution
$$  0 = \frac{x^2 (3y^2 y') - y^3 (2x)}{x^4} + 3x^2 \tan(y) + x^3 \sec^2(y) y' + 4y^3 y'  \text{.}  $$
Then, we solve for $y'$.  \begin{align*}
\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y) &= \frac{x^2 (3y^2 y')}{x^4} + x^3 \sec^2(y) y' + 4y^3 y'  \\
\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y) &= \left( \frac{x^2 (3y^2)}{x^4} + x^3 \sec^2(y) + 4y^3 \right) y'  \\
\frac{\frac{y^3 (2x)}{x^4} - 3x^2 \tan(y)}{\frac{x^2 (3y^2)}{x^4} + x^3 \sec^2(y) + 4y^3} &= y'  \\
\frac{\frac{2 y^3 }{x^3} - 3x^2 \tan(y)}{\frac{3y^2}{x^2} + x^3 \sec^2(y) + 4y^3} &= \frac{\mathrm{d}y}{\mathrm{d}x}  \\
\left(\frac{2 y^3 }{x^3} - 3x^2 \tan(y) \right)\mathrm{d}x &= \left(\frac{3y^2}{x^2} + x^3 \sec^2(y) + 4y^3 \right) \mathrm{d}y  \text{,}
\end{align*}
which should be easy enough to see is the given equation.


问:
Why are those inequalities always true?
Here are two inequalities that i don't know how to explain.

$\text{For }n\in\mathbb N^* \text{ and } (a_1,...,a_n)\in(\mathbb R^{+})^n\text{ , }
\sqrt{\sum_{k=1}^{n}k^2a_k}\le\sum_{k=1}^{n}\sqrt{\sum_{s=k}^{n}a_s}$

$\text{For }n\in\mathbb N^* \text{ , } (a_1,...,a_n)\in\mathbb R^n\text{ and }(b_1,...,b_n)\in\mathbb R^n\text{ such that }a_1\le...\le a_n \text{ and } b_1\le...\le b_n , \left(\frac{1}{n}\sum_{k=1}^{n}a_k\right)\left(\frac{1}{n}\sum_{k=1}^{n}b_k\right)\le\left(\frac{1}{n}\sum_{k=1}^{n}a_kb_k\right)$

Why are those inequalities always true?

答:
For the first inequality, let

$x_k=\sqrt{\sum_{s=k}^n a_s}$, RHS becomes $\sum_{k=1}^n x_k$

It's easy to show that LHS $= \sqrt{\sum_{k=1}^n (2k-1)x_k^2}$

We now prove that $LHS^2 <= RHS^2$

Note that $x_1 \ge x_2 \ge \cdots \ge x_n$

$x_1^2=x_1^2$

$(2k-1)x_k^2 = (2k-1) x_k \cdot x_k \le \left(\sum_{s=1}^k x_s + \sum_{s=1}^{k-1} x_s \right) x_k = \left( \sum_{s=1}^k x_s \right)^2 - \left( \sum_{s=1}^{k-1} x_s \right)^2, \forall k\ge 2$

Therefore $LHS^2 \le x_1^2 + \sum_{k=2}^n \left( \left( \sum_{s=1}^k x_s \right)^2 - \left( \sum_{s=1}^{k-1} x_s \right)^2 \right) =\left(\sum_{s=1}^n x_s \right)^2 = RHS^2$. Q.E.D.

Title: Why are those inequalities always true?


问:
Is there a discrete version of de l'Hôpital's rule?
When considering asymptotics of runtime functions, you often have to find limits of quotients of discrete functions, e.g.

$\displaystyle\qquad \lim\limits_{n \to \infty} \frac{4^n}{\binom{2n}{n}\sqrt{n}}.$

While this particular case can easily be dealt with by Stirling's formula, I have been wondering. Mathematicians often like to use de l'Hôpital's rule, but it can obviously not be applied to the discrete case immediately (no mean value theorem). If—as in this case—you are lucky, you might find nice and well-studied continuations on the reals.

What to do in general, though? Is there a discrete version/relative of de l'Hôpital's rule, maybe using difference quotients?

答:
The discrete version of L'Hôspital's rule, in my opinion, is Abelian theorems, including the L'Hôspital's rule, Silverman-Toeplitz theorem and its sepcial case, Stolz-Cesàro theorem.

On de Bruijn's Asymptotic methods in analysis, it's said that

A theorem which derives asymptotic information about some kind of average of a function from asymptotic information about the function itself, is called an Abelian theorem. If one can find a supplementary condition under which the converse of an Abelian theorem holds, then this condition is called a Tauberian condition, and the converse theorem is called a Tauberian theorem.

The amount you gave, as far as I've tried, couldn't be easily solved with these theorems.

Let $$a_n=\frac{4^n}{\binom{2n}n\sqrt n}$$
we have
$$\ln a_{n+1}-\ln a_n=\frac12\ln(n+1)-\ln(n+\frac12)+\frac12\ln n=-\frac1{4n^2}+O\left(\frac1{n^3}\right)\tag1$$
Therefore $\ln a_n$ converges as $n\to\infty$. However, the preceding equation, the asymptotic behavior of the difference, is not enough to determine the limit value, even if the result is refined. Such efforts are generally unsuccessful.

However, if $S=\lim_{n\to\infty}\ln a_n$, we could determine the asymptotic behavior of $a_n-S$ through (1) easily, since $\ln a_n=S-\sum_{k\ge n}(\ln a_k-\ln a_{k+1})$.

Remark
One could determine $S$ through Stirling's formula. There's another approach, more elementary, I think:

$$a_n^2=\left(\frac{(2n-1)!!}{(2n)!!}\right)^2(2n+1)\cdot\frac n{2n+1}\to\frac1\pi$$

by Wallis product, therefore $S=1/\sqrt\pi$.


问:
Proof that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} < \ln n < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}$
Please help me to prove this inequality

$$\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} < \ln n < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}$$

答:
Hint:

$$\lfloor x\rfloor\le x\le\lceil x\rceil\implies \frac1{\lceil x\rceil}\le \frac1x\le \frac1{\lfloor x\rfloor}.$$

Integrate in unit intervals and sum.

$$\frac1n\le\log(n)-\log(n-1)\le\frac1{n-1}$$ which telescopes to $$H_{n}-1\le\log(n)-\log(1)\le H_{n-1}.$$


问:
Find the extremal point of $x^2 + 2xy + 4y^2 + 6$ and show it is a minimum point by completing the square
I'll admit that my elementary algebra sucks. This winter break I want to work through Algebra: An Elementary Textbook by G. Chrystal. The question at hand is confusing me; I hope someone can walk me through it.</p>

答:
I'm not showing the calculus part as that's direct calculation.

The minimum occurs at $ x = 0, y = 0 $, and has value 6.

In terms of completing the square, we have

 $$ (x+y)^2 + 3y^2 + 6 $$

Hence, the minimum is 6.

Another way of completing the square is

 $$ (2y + \frac{x}{2} ) ^2 + \frac{3}{4} x^2 + 6 $$

Again, the minimum is 6.


问:
If the sequence $s_{n}$ converges to $l$, does $\frac{1}{s_{n}}$ converges to $\frac{1}{l}$?
I'm not sure if that's true, in our textbook there is only an affirmation that if $s_{n}$ converges to $l$ then $\frac{1}{s_{n}}$ is bounded, but I think that it also should converge to $\frac{1}{l}$ (when $l \neq 0$ obviously). I tried to prove it by definition of convergence but didn't get anywhere. Thanks in advance :)

答:
Since

$l \ne 0, \tag 1$

there is an open set $N$ containing $l$ with

$0 \notin N; \tag 2$

then the function

$f(x) = \dfrac{1}{x} \tag 3$

is continuous on $N$; thus if

$s_n \to l \tag 4$

we have

$\dfrac{1}{s_n} = f(s_n) \to f(l) = \dfrac{1}{l}. \tag 5$


问:
Let V = C[0,1] and W = {$f \in C[0,1] | f(1) \leq 0$} is W a subspace of the vector space V
I've shown that W is closed under addition and contains the 0 function but I'm a bit confused about how scalar multiplication works for C[0,1] so I'm not sure whether W is closed under it.

If f(1) = -1 and the scalar c = -1 then cf(1) > 0 so cf(x) is not contained in W. But can the scalar be a negative number in this context or does it need to be between 0 and 1 because f is in C[0,1]?

答: No, the scalar does not need to be between $0$ and $1$. As you have found, the set $W$ fails to be a vector subspace of $V$ because it is not closed under scalar multiplication.


问:
Probability distribution with $p(x) = p(1/x)$
I am looking for a probability distribution defined on $x\in(0,\infty)$ that satisfies, for all $x$:

$$p(x) = p(1/x)$$

Naturally, we must have:

$$\int_0^\infty p(x) \, dx= 1$$

and thereore:

$$
\int_0^1 p(x) \, dx + \int_1^\infty p(x) \, dx = 1
$$

Does such a probability distribution exist?

答:
It is not true that $\int^1_0 p(x)\, dx$ and $\int^\infty_1 p(x) dx$ necessarily both need to be equal to $0.5$, just that their sum needs to be equal to $1$.

As easy example could be $$p(x) = \begin{cases} \exp(-\alpha/x)&\text{for $0<x<1$},\\
\exp(-\alpha x) & \text{for $x \geq 1$},\end{cases}$$

where $\alpha$ is a parameter such that
$$\left[\int^1_0\exp(-\alpha/x)\, dx\right] + \left[\int^\infty_1 \exp(-\alpha x) \, dx\right] = 1.$$
A quick numerical check yields $\alpha \approx 0.68$.


问:
Lagrange Multiplier in multivariable calculus
I want to ask when we apply Lagrange multiplier, we will get several critical points. And is it sufficient that the global extremum will exactly come from the value evaluated by these found critical points? In other words, is it possible for all the critical points found to be saddle points or only maximum but no minimum for example?  If not, could you please give me some examples?
What's more, is it possible to get only one critical points after applying Lagrange multiplier?</p>

答:
Assume we have $f, g:\mathbb{R}^n \rightarrow \mathbb{R}$ differentiable and we want to find the maximum and minimum of $f$ on the set
$$
\mathcal{D} := \lbrace x\in \mathbb{R}^n: g(x) = 0 \rbrace.
$$
Then one statement (and one statement only) is always true. If $x^* \in \mathcal{D}$ is a local minimum/maximum of $f$ on $\mathcal{D}$, then there is some $\lambda \in \mathbb{R}$ such that
$$
\nabla f(x^*) + \lambda \nabla g(x^*) = 0.
$$

We call those points critical points.

What you usually do is decide if $\mathcal{D}$ is compact or not. If it in fact is, we know that a global maximum and minimum must exist. Of course, they are among the critical points. You can find them by evluating $f$ at the critical points. The lowest value is the minimum and the greatest is the maximum.

First example:
$f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x, y) = x+y$,  $g(x, y) = x^2+y^2 - 1$

Obviously, $\mathcal{D}$ is compact. The only critical points are $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)^\top$ with $\lambda = \frac{1}{\sqrt{2}}$ and $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)^\top$ with $\lambda = -\frac{1}{\sqrt{2}}$. I think it is easy to decide which one is the minimum/maximum.

Second example: $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x, y)= \exp(-x^2-y^2)$, $g(x, y) = y$.

The only critical point is $(0, 0)$ with $\lambda = 0$. One can immediately see that it is a global maximum because of $\displaystyle \max_{t \in (-\infty, 0]} \exp(t) = \exp(0)$.

Third example: $f:\mathbb{R}^3 \rightarrow \mathbb{R}$, $f(x, y) = x^3 + y^3$, $g(x, y) = x-y$

The only critical point is $(0, 0)$ with $\lambda = 0$. It is a saddle point which you can see by looking at $f(t, 0) = t^3$. If $t>0$, $f(t, 0)$ is positive and if $t<0$, $f(t, 0)$ is negative.

(Of course other cases are possible)


问:
Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots

Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.

Obviously as soon as you factor this equation to $$(4x^2+8x+a^2)(x^2+ax+1)=0$$ then you have finished.
However, how am I supposed to think of factoring the equation in such a manor? That you are supposed to factor it, is obvious, but how can you find out which are its factors, aside from blind luck?

答:
Observing and categorizing the coefficients can be effective here.

Rewrite the equation into

$$(4x^4+8x^3+a^2x^2)+(4ax^3+8ax^2+a^3x)+(4x^2+8x+a^2)=0\\\Longrightarrow x^2(4x^2+8x+a^2)+ax(4x^2+8x+a^2)+(4x^2+8x+a^2)=0\\\Longrightarrow (x^2+ax+1)(4x^2+8x+a^2)=0$$

Or in this way:

$$(4x^4+4ax^3+4x^2)+(8x^3+8ax^2+8x)+(a^2x^2+a^3x+a^2)=0\\\Longrightarrow 4x^2(x^2+ax+1)+8x(x^2+ax+1)+a^2(x^2+ax+1)=0\\\Longrightarrow (4x^2+8x+a^2)(x^2+ax+1)=0$$

Does this help? I'm not sure if you are asking for a general way or  just for this one problem.


答:
Let $$x^2+2xy+4y^2+6=z$$, see it as quadratic of $y$, treat $x,z$ as constants. Find $y$:

$$y=\frac{-2x\pm \sqrt{4x^2-16(x^2+6-z)}}{8}$$ For $y$ to be real $$-12x^2-96+16z\ge 0 \implies 16z-96>12x^2 \ge 0 \implies 16z\ge 96 \implies z\ge 6.$$ The minimum of $z$ equal to 6 occurs when $x=y=0$.

One may also write $$z=(x+y)^2+3y^2+6 \implies z \ge 6$$

Title: Find the extremal point of $x^2 + 2xy + 4y^2 + 6$ and show it is a minimum point by completing the square

Body: I'll admit that my elementary algebra sucks. This winter break I want to work through Algebra: An Elementary Textbook by G. Chrystal. The question at hand is confusing me; I hope someone can walk me through it.</p>


问:
Local injectivity of a function defined as $f: \Bbb R^2 \to \Bbb R^2$

Determine if the function defined as $f: \Bbb R^2 \to \Bbb R^2$, $f(x,y) = (x^2, x^2+y)$ is locally injective.

Computing the partials one has $\frac{\partial f_1}{\partial x}=2x, \frac{\partial f_1}{\partial y}=0, \frac{\partial f_2}{\partial x}=2x, \frac{\partial f_2}{\partial y}=1.$ Thus the Jacobi is as follows $$J_f = \begin{bmatrix} 2x && 0 \\2x && 1\end{bmatrix}.$$

And from here $\det(J_f) = 2x \ne0,$ when $x \ne 0.$

However, I'm not sure I entirely understand the definition of local injectivity and how to continue from here. The definition I have deals with topological spaces which I haven't been studying yet.

Let $F : X → Y$ be a continuous function between topological spaces, and let $a ∈ X$.
We say that $F$ is locally injective (or locally one-to-one) at $a$ if there exists a
neighborhood $U$ of a such that $F|_U$ is injective.

How does the fact that the Jacobian determinant is nonzero help here? Since the Jacobi can be interpreted as a matrix and it's invertible iff the determinant is nonzero it seems that we're striving for a stronger statement that $f$ would actually be invertible? This raises another question, if $f$ is invertible is it immediately locally injective?

答:
You can say that your function is injective on U if and only if :

$$ f(x,y)=f(x',y') \Rightarrow (x,y)=(x',y')$$

So we are trying to find a subset U where F is injective. So first you can take $(x,y,x',y')\in\mathbb{R}^4, f(x,y)=f(x',y')$ Then you have :
$$x^2=x'^2, x^2+y=x'^2+y' \\x^2=x'^2, y=y' $$

In this case, you can easily see that f will be injective if $x\in \mathbb{R}^+$.
So you can say that f is injective on the subset $ U=\mathbb{R}^+\times\mathbb{R}$


问:
Subgroup transitive on the subset with same cardinality
Maybe there is some very obvious insight that i miss here, but i've asked this question also to other people and nothing meaningful came out:

If you have a subgroup G of $S_n$(the symmetric group on n elements), you can consider the natural action of G on the subset of $\{1,...,n\}$; my question is this:what are the G that act transitively on the subsets of the same cardinality; that is, whenever $A,B \subseteq \{1,...,n\}$ with $|A|=|B|$, there is  $g \in G$ so that $gA=B$.

答:
Derek Holt's answer works well for $n\geq 10$, but there are some exceptions for small values of $n$.  Using $\mathsf{GAP}$'s library of transitive permutation groups, I found the complete list of exceptions:

The subgroup $\big\langle(1\;2\;3\;4\;5),(1\;2\;4\;3)\big\rangle$ of $S_5$ acts transitively on $k$-element subsets of $\{1,2,3,4,5\}$ for every value of $k$.  This is the
Frobenius group of order 20 (i.e. $\mathbb{Z}_5 \rtimes \mathbb{Z}_4$).

The subgroup $\big\langle(1\;2\;3\;4\;6),(1\;2)(3\;4)(5\;6)\big\rangle$ of $S_6$ acts transitively on $k$-element subsets of $\{1,\ldots,6\}$ for every value of $k$.  This subgroup is isomorphic to $S_5$, and can be obtained as the action of $S_5$ on the six two-element subsets of $\{1,\ldots,5\}$.  This group can also be described as $\mathit{PGL}(2,5)$ acting on the six one-dimensional subspaces of $\mathbb{F}_5^2$.

The subgroup $\big\langle(1\;9)(2\;3)(4\;5)(6\;7),(1\;2\;4\;3\;6\;7\;5),(2\;5)(3\;6)(4\;7)(8\;9)\big\rangle$ of $S_9$ acts transitively on $k$-element subsets of $\{1,\ldots,9\}$ for every value of $k$.  This is the projective special linear group $\mathit{PSL}(2,8)$ (of order 504) acting on the nine one-dimensional subspaces of $\mathbb{F}_8^2$.

The subgroup of $S_9$ generated by $\mathit{PSL}(2,8)$ above as well as the permutation $(2\;4\;6)(3\;5\;7)$ also acts transitively on $k$-element subsets of $\{1,\ldots,9\}$ for every value of $k$.  This group has order 1512, and can be described as a semidirect product $\mathit{PSL}(2,8)\rtimes \mathbb{Z}_3$.  Presumably the $\mathbb{Z}_3$ comes from the action of the Frobenius automorphism of $\mathbb{F}_8$ on $\mathbb{F}_8^2$.

Incidentally, here is the $\mathsf{GAP}$ code that prints out the list of exceptions:

```<code>for n in [3..10] do
   for k in [1..NrTransitiveGroups(n)-2] do
      if ForAll([1..n],
          i -> Size( Orbits(TransitiveGroup(n,k),
                  Combinations([1..n],i), OnSets) ) = 1)
      then
         Print( "TransitiveGroup(", n, ",", k, ")\n" );
      fi;
   od;
od;
</code>```

Edit: This edit is in response to Igor Rivin's question below.  According to this website, the following polynomials over $\mathbb{Q}$ have the above permutation groups as their Galois groups:

The first group listed above (usually denoted 5T3 or F(5)) is the Galois group of
$$x^5-9x^3-4x^2+17x+12.
$$

The second group listed above (usually denoted 6T14, PGL(2,5), L(6):2, or $S_5(6)$) is the Galois group of
$$x^6 + 3x^4 - 2x^3 + 6x^2 + 1.$$

The third group listed above (usually denoted 9T27, L(9), or PSL(2,8)) is the Galois group of
$$x^9+x^7-4x^6-12x^4-x^3-7x^2-x-1.$$

The fourth group listed above (usually denoted 9T32, L(9):3, or P|L(2,8)) is the Galois group of
$$x^9-x^8-4x^7+28x^3+26x^2+9x+1.$$

Body:
Derek Holt's answer works well for $n\geq 10$, but there are some exceptions for small values of $n$.  Using $\mathsf{GAP}$'s library of transitive permutation groups, I found the complete list of exceptions:

The subgroup $\big\langle(1\;2\;3\;4\;5),(1\;2\;4\;3)\big\rangle$ of $S_5$ acts transitively on $k$-element subsets of $\{1,2,3,4,5\}$ for every value of $k$.  This is the
Frobenius group of order 20 (i.e. $\mathbb{Z}_5 \rtimes \mathbb{Z}_4$).

The subgroup $\big\langle(1\;2\;3\;4\;6),(1\;2)(3\;4)(5\;6)\big\rangle$ of $S_6$ acts transitively on $k$-element subsets of $\{1,\ldots,6\}$ for every value of $k$.  This subgroup is isomorphic to $S_5$, and can be obtained as the action of $S_5$ on the six two-element subsets of $\{1,\ldots,5\}$.  This group can also be described as $\mathit{PGL}(2,5)$ acting on the six one-dimensional subspaces of $\mathbb{F}_5^2$.

The subgroup $\big\langle(1\;9)(2\;3)(4\;5)(6\;7),(1\;2\;4\;3\;6\;7\;5),(2\;5)(3\;6)(4\;7)(8\;9)\big\rangle$ of $S_9$ acts transitively on $k$-element subsets of $\{1,\ldots,9\}$ for every value of $k$.  This is the projective special linear group $\mathit{PSL}(2,8)$ (of order 504) acting on the nine one-dimensional subspaces of $\mathbb{F}_8^2$.

The subgroup of $S_9$ generated by $\mathit{PSL}(2,8)$ above as well as the permutation $(2\;4\;6)(3\;5\;7)$ also acts transitively on $k$-element subsets of $\{1,\ldots,9\}$ for every value of $k$.  This group has order 1512, and can be described as a semidirect product $\mathit{PSL}(2,8)\rtimes \mathbb{Z}_3$.  Presumably the $\mathbb{Z}_3$ comes from the action of the Frobenius automorphism of $\mathbb{F}_8$ on $\mathbb{F}_8^2$.

Incidentally, here is the $\mathsf{GAP}$ code that prints out the list of exceptions:

for n in [3..10] do
   for k in [1..NrTransitiveGroups(n)-2] do
      if ForAll([1..n],
          i -> Size( Orbits(TransitiveGroup(n,k),
                  Combinations([1..n],i), OnSets) ) = 1)
      then
         Print( "TransitiveGroup(", n, ",", k, ")\n" );
      fi;
   od;
od;

Edit: This edit is in response to Igor Rivin's question below.  According to this website, the following polynomials over $\mathbb{Q}$ have the above permutation groups as their Galois groups:

The first group listed above (usually denoted 5T3 or F(5)) is the Galois group of
$$x^5-9x^3-4x^2+17x+12.
$$

The second group listed above (usually denoted 6T14, PGL(2,5), L(6):2, or $S_5(6)$) is the Galois group of
$$x^6 + 3x^4 - 2x^3 + 6x^2 + 1.$$

The third group listed above (usually denoted 9T27, L(9), or PSL(2,8)) is the Galois group of
$$x^9+x^7-4x^6-12x^4-x^3-7x^2-x-1.$$

The fourth group listed above (usually denoted 9T32, L(9):3, or P|L(2,8)) is the Galois group of
$$x^9-x^8-4x^7+28x^3+26x^2+9x+1.$$


问:
I need help finding algebraic area and perimeter, please
Could you help me solve some or all of the problems please? I can't understand the procedure for solving them...

solve the following problems determines algebraically the perimeter and area of the following questions

sorry if i did something wrong i'm new and i don't understand english very well

Thank you for your attention, it would be a pleasure if you could answer my question

答:
Question 1

note that we know how to solve the area and perimeter of a rectangle, a square being a unique version of this solution. I.e. $A=Width * Height , P=2*Width + 2*Height$

Rectangle 1: (note that all squares are rectangles, so technically this is correct)

$p= 2*3m+2*3m = 6m+6m = 12m $ , $A = 3m*3m = 9m^2$

Rectangle 2:

$p= 2*2n+2*3m = 4n+6m$ , $A= 2n*3m $

Rectangle 3:
$p=2*2n+2*2n=8n$ , $A=2n*2n = 4n^2$

Rectangle 4 (the large unmarked square) Is using the information on the previous examples
to formulate the same type of answer (note that you could also use the expressions that we have created to calculate the perimeter and area, I shall use this way but it is worth you trying to form it like I did for the smaller squares to see if you get the same answer)

$P=\frac{r_1+2r_2+r_3}{2} = \frac{12m+2(4n+6m)+8n}{2} = \frac{24m+16n}{2} = 12m+8n$

$A=r_1+2r_2+r_3 = 9m^2+2(2n*3m)+ 4n^2 = 9m^2+(4n*6m)+4n^2$
$A=(3m+2n)^2$

I would suggest looking at the last "collection' of shapes to see if you can apply the same reasoning as I did for the first three. If you can, have a think why I was able to do it differently.
I hope that answers your question


问:
Obtaining the three torus via Dehn surgery
It is a well known theorem from the '60 (Lickorish-Wallace) that any closed orientable three dimensional smooth manifold can be obtained performing a sequence of integral Dehn surgeries along knots in $\mathbb{S}^3$.
The most common examples found in any book are $\mathbb{S}^3, \mathbb{S}^2\times \mathbb{S}^1 $ and  the Lens spaces $L(p,q)$.
Curiosly,  I can't find how to get the three torus $\mathbb{T}^3 = \mathbb{S}^1\times\mathbb{S}^1 \times \mathbb{S}^1$ which is a quite ubiquitous $3$-manifold in Geometry.

How can I obtain  $\mathbb{T}^3$ via  (rational and integral) Dehn surgery from $\mathbb{S}^3$?

答:
Here's a way not to solve this. I don't know if that's something StackExchange goes for, but I feel like it's still instructive.

How to recognize $T^3$ as a manifold? The most obvious thing in my view is that it's a $T^2$ bundle over $S^1$ with trivial monodromy. So I was hoping to find a genus 1 fiber surface for the Borromean rings complement with trivial monodromy that gives 0-slopes on each boundary component. Then the filled manifold would fill in that fiber surface with three disks to get a torus, and again the filled manifold would be fibered with trivial monodromy.

So here are two diagrams for the Borromean rings. I like the second one better because it makes it easier to draw a Seifert surface.

If I want a connected Seifert surface that gives 0-slopes on each boundary component, probably just start with the the obvious three disks and resolve their singularities in some way, right? Here's what that looks like:

This is certainly a genus 1 Seifert surface, as can be verified with a little Euler characteristic calculation. But it's not a fiber surface, and least not with the trivial monodromy. (I verified this with the criterion given in this paper by S. Baader and C. Graf. Specifically their Cor 2.5.) Even worse, this surface doesn't give the 0-framing on the 'clasp' components!

Then my friend pointed out that the Borromean rings are hyperbolic, hence they can't be fibered with trivial monodromy (else they would have e.g. essential tori)! So alas, this approach seems to just not work. I'll give another answer with the real solution once I find it.


答:
Set $x = e^{-y}$ for $y \in \Bbb R$. Thus, we look for probability distribution functions $p(x) = f (\ln x)$ for positive $x$ such that
$f = p \circ \exp$
is an even function.
Now, let us check that
$$
1 = \int_0^\infty p(x)\, dx = \int_0^1 p(x)\, dx + \int_1^\infty p(x)\, dx = \int_0^1 p(x) \left(1+ x^{-2}\right) dx \, ,
$$
where we have used the change of variable $z=1/ x$. This condition can be rewritten as
$$
1 = \int_{0}^{\infty} f(y)\left(e^{-y}+e^{y}\right) dy \, .
$$
For instance, we note that the function defined by $f(y) = e^{-(1+\sqrt{2}) |y|}$ is a suitable choice, and we have $$
p(x) = e^{-(1+\sqrt{2}) |\!\ln x|} \Bbb{I}_{x>0} = \left\lbrace
\begin{aligned}
&x^{(1+\sqrt{2})} && 0<x<1\\
&x^{-(1+\sqrt{2})} && x \geq 1
\end{aligned}
\right.
$$
in this case (see figure below). Thus there are obviously various possibilities, even if we restrict the study to continuous $p$.

NB. The function $p(x) = \exp(-\alpha e^{|\!\ln x|}) \Bbb{I}_{x>0}$ proposed by @molarmass corresponds to $f(y) = \exp(-\alpha e^{|y|})$.

Title: Probability distribution with $p(x) = p(1/x)$

Body:
I am looking for a probability distribution defined on $x\in(0,\infty)$ that satisfies, for all $x$:

$$p(x) = p(1/x)$$

Naturally, we must have:

$$\int_0^\infty p(x) \, dx= 1$$

and thereore:

$$
\int_0^1 p(x) \, dx + \int_1^\infty p(x) \, dx = 1
$$

Does such a probability distribution exist?


问:
Proof average codeword length of a prefix code $C$ is the sum of probability of all leaf vertices that are descendant
Let $I$ represent the set of all interior vertices in a binary tree. Define the probability of vertex $v \in I$ is $P(v)$, which is the sum of the probabilities of all leaf vertices that are descendants of $v$. Show that (or given a counter-example)

$$L(C) =  \sum_{v\in \mathcal{I}}^{} P(v) .$$

答:
Suppose that $u$ is a leaf at depth $d(u)$; then there are $d(u)$ interior vertices in the path from $u$ to the root, so $u$ contributes $d(u)P(u)$ to $\sum_{v\in I}P(v)$. Thus, if $L$ is the set of leaves,

$$\sum_{v\in I}P(v)=\sum_{v\in L}d(v)P(v)\,.$$

On the other hand, $d(u)$ is the length of the code word corresponding to a leaf $u$, so $\sum_{v\in L}d(v)P(v)$ is the mean length of a code word of $C$.


问:
Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$
Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.

I've already tried to write it like this:
$$\frac{\sqrt{1}}{1} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 3$$
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 4$$
but I don't know that to do next or if it's the right way prove this.

答:
By induction we have

base case: $n=1 \implies 1 \le 3\sqrt 2-3$

inductive step: we assume true

$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3 \tag 1$$

and we need to prove that

$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt{n+1}} \stackrel{(1)}\leq 3\sqrt{n+1} - 3+\frac{1}{\sqrt{n+1}} \stackrel{?}\le 3\sqrt{n+2} - 3$$

and last inequality is true indeed

$$ 3\sqrt{n+1} +\frac{1}{\sqrt{n+1}}\le 3\sqrt{n+2} $$

$$ 3(n+1) +1\le 3\sqrt{(n+2)(n+1)} $$

$$ 3n+4\le 3\sqrt{(n+2)(n+1)} $$

$$ 9n^2+24n+16\le 9n^2+27n+18 $$


问:
Cusps of parallel/offset curves
Found this remark in Wikipedia: https://en.wikipedia.org/wiki/Parallel_curve

When parallel curves are constructed they will have cusps when the distance from the curve matches the radius of curvature. These are the points where the curve touches the evolute.

I would like to prove it, or at least have a better explanation of why this happens.

Thanks in advance!!

答:
Since I don't recall seeing this question on this site before and because it may be of interest to other people, I'm going to sketch the computation to which I alluded in the comments.

Let's consider the parallel curve $\beta(s)=\alpha(s)+rN(s)$ with $r=1/\kappa_0$ (here the subscript denotes the value at $s=0$). The local canonical form (see, e.g., p. 17 of my text) for $\alpha$ at $s=0$ is
$$\alpha(s)=\big(s-\tfrac16\kappa_0^2s^3 + o(s^3)\big)T_0 + \big(\tfrac12\kappa_0s^2 + \tfrac16\kappa'_0s^3 + o(s^3)\big)N_0.$$
Differentiating, we see that
$$\alpha''(s) = \kappa(s)N(s) = \big({-}\kappa_0^2s + o(s)\big)T_0+\big(\kappa_0 + \kappa'_0s + o(s)\big)N_0,$$
and so
$$\frac1{\kappa_0}N(s) = \frac{\big({-}s+o(s)\big)T_0 + \big(\frac1{\kappa_0}+\frac{\kappa'_0}{\kappa_0^2}s+o(s)\big)N_0}{\sqrt{\left(1+\frac{\kappa'_0}{\kappa_0}s\right)^2+\kappa_0^2s^2+o(s^2)}}.$$
Using the usual first-order approximation  $\frac1{\sqrt{1+x}}\approx 1-\frac12x$, we work out that
$$\beta(s) = \big({-}\tfrac16\kappa_0s^3+O(s^3)\big)T_0 + \big(\tfrac1{\kappa_0} + \tfrac12\kappa_0 s^2 + O(s^2)\big)N_0.$$
This makes it clear that $\beta$ has a cusp at the point $\frac1{\kappa_0}N_0$.

Title: Cusps of parallel/offset curves


问:
Non Empty Set Dense In The Topological Subspace
Suppose that $A$ is a non empty subset of a topological space $(X, T)$, then prove that $A$ is a dense subset of the subspace $\overline{A}$.
$$$$Please Check My Proof
$$$$Now as a subspace of a given topological space is formed by the relative topology. Now the subspace is the topological space $(\overline{A}, T')$ with the topology $T'$ which is the collection of sets formed by the intersection of the class of sets in $T$ with $\overline{A}$. Now any subset of $\overline{A}$ open under $T$ lies in $T$ and hence lies in $T'$. Also any subset of $\overline{A}$ open under $T'$ lies in $T'$ and hence lies in $T$. Hence any subset of $\overline{A}$ is open under $T$ if and only if it is open under $T'$. Now to prove the above statement we need to show that the closure of $A$ under $T'$ say $\overline{A'}$ is equal to closure of $A$ under $T$, that is $$\overline{A'}=\overline{A}$$. Now any superset of $A$ lying in $\overline{A}$ which is closed under $T'$ is also closed under $T$ and by the definition of $\overline{A}$ we see that $\overline{A}$ lies in that set and hence $\overline{A}$ is equal to that set. So the only superset of $A$ closed under $T'$ lying in $\overline{A}$ is itself $\overline{A}$ and hence $$\overline{A'}=\overline{A}$$. So we get that $$\overline{A'}=\overline{A}$$ which proves that $A$ is dense in the topological subspace $(\overline{A}, T')$.

答:
Let $U\in T'$ with $U\cap A=\varnothing$.

Then $U=V\cap\overline A$ for some $V\in T$ so that $U\cap A=\varnothing$ implies that $V\cap A=\varnothing$ or equivalently that $A\subseteq V^{\complement}$.

Because $V^{\complement}$ is closed it follows that $\overline A\subseteq V^{\complement}$ or equivalently $U=\varnothing$ .

Proved is now that: $$\forall U\in T'[U\cap A=\varnothing\implies U=\varnothing]$$

We can reformulate this as: $$\forall U\in T'[U\neq\varnothing\implies U\cap A\neq\varnothing]$$

This is exactly the statement that $A$ is dense in space $(\overline A,T')$.


Title: Proof that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} < \ln n < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}$

Body:
Please help me to prove this inequality

$$\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} < \ln n < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}$$

Body:
there is a simple picture for this,

if we have $g(x) > 0$ but $g'(x) < 0,$  then
$$ \int_a^{b+1} \; g(x) \;  dx \;  < \; \sum_{j=a}^b \;  g(j) \; < \; \int_{a-1}^b \;  g(x) \; dx   $$

You need two versions, with $a=2,b=n$

$$ \sum_{j=2}^n \; \frac{1}{j} \; < \; \int_{1}^{n} \;  \frac{1}{x} \; dx = \log n  $$

$a=1 , b= n-1$
$$  \log n = \int_{1}^{n} \;  \frac{1}{x} \; dx  <   \sum_{j=1}^{n-1} \; \frac{1}{j}$$


问:
An interesting integral $\int{\dfrac{x^m}{x^{2m}+1}dx}$
I have, for a long period of time, tried to evaluate the indefinite integral $\int{\dfrac{x^m}{x^{2m}+1}dx}$ for different values of $m$, such as $m=2,3,4,6...$, and I have recently thought of generalizing this integral for all values of $m\in\mathbb{N}$. Since evaluating the integral using partial fractions for high values of $m$ by myself is very time consuming, and it happens that I can't find any direct solution on the internet, I tried to look for patterns of the integrals in WolframAlpha. After some comparisons, I found the below expression:
$$\int{\dfrac{x^m}{x^{2m}+1}dx}=\sum_{n=1}^{m}(-1)^{n-1}(\dfrac{1}{2m}\sin(\dfrac{(2n-1)\pi}{2m})\ln{|x^2+2\cos(\dfrac{(2n-1)\pi}{2m})x+1|}$$$$+\dfrac{1}{m}\arctan(\csc(\dfrac{(2n-1)\pi}{2m})(x+\cos(\dfrac{(2n-1)\pi}{2m})))\cos(\dfrac{(2n-1)\pi}{2m}))$$
How could I have found the above expression without comparing the integrals?

P.S. It seems that the equivalence $x^{2m}-2x^ma^m\cos(m\theta)+a^{2m}\equiv\displaystyle\prod_{r=0}^{m-1}(x^2-2ax\cos(\theta+\dfrac{2r\pi}{m})+a^2)$ for $a\in\mathbb{R}^+, \theta\in\mathbb{R}, m\in\mathbb{N}$ can be used, yet I don't know how I can evaluate partial fractions with it.

Body:
Factorize $x^{2n}+1= \prod_{k=1}^{2n}(x-x_k)$ with $x_k= e^{i a_k},\>a_k=\frac{(2k-1)\pi}{2n}$. Then
$$\frac{x^n}{x^{2n}+1}=-\frac1{2n}\sum_{k=1}^{2n}\frac{x_k^{n+1}}{x-x_k}$$
and integrate
\begin{align}
&\int \frac{x^n}{x^{2n}+1}dx \\=& -\frac1{2n}\sum_{k=1}^{2n}\int \frac{x_k^{n+1}}{x-x_k} dx
= -\frac1{2n}\sum_{k=1}^{2n} x_k^{n+1} \ln(x-x_k)\\
= &-\frac1{2n} \sum_{k=1}^{2n} [\cos(n+1)a_k +i \sin(n+1)a_k]
\ln(x-\cos a_k -i\sin a_k)\\
= &-\frac1{4n} \sum_{k=1}^{2n} \cos(n+1)a_k \ln(x^2-2x\cos a_k+1)+2\sin(n+1)a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k}
\end{align}

where only the real part survives.


问:
Example of uniform but not normal convergence
Let's consider a function series defined as follows.
For every $n\in \mathbb N, n\gt 1$, let $ f_n(x):= \frac{1} {n} $ if $x=\frac{1} {n}$ and $f_n(x)=0$ $x\neq\frac{1} {n}$.
Prove, using the definition of uniform convergence and the Weierstrass M-test, that $$\sum_{n=1}^{+\infty }f_n(x) $$ converges uniformly but not normally in $\mathbb R$.
(I assume a series is normally convergent if $\sum_{n=1}^{+\infty } sup \{ \vert f_n(x) \vert \} $ converges)

My attempt.

For every $n$, $sup \{ \vert f_n(x) \vert; x \in \mathbb R \}= \frac{1} {n} $ and the harmonic series diverges. So we don't have normal convergence.
Though, I'm not sure how to do with the uniform convergence. It may seem a trivial question, but I got lost with the epsilons and the supremums. I suspect that it converges to the 0 function, but $ sup \{ \vert \sum_{k=1}^{n} f_k(x) \vert; x \in \mathbb R\} \ge 1=f_1(1)=\sum_{k=1}^{n} f_k(1)$, so it does not tend to 0.

Any help, using the definition or the epsilon?

答:
My initial attempt mixed up.  I was looking at the right sum, but drew the wrong conclusion.  I must not be feeling well today, because it's the second answer I've screwed up.

Let $$S=\left\{1/n\mid n\in \mathbb{Z}^+\right\}$$  Then it's obvious that the sum converges pointwise to $$f(x)=\begin{cases}x,&x\in S\\0,&\text{otherwise}\end{cases}$$ because for each $x$, $f_k(x)\neq0$ for at most one value of $k$.  Then we have $$f(x)-\sum_{k=1}^nf_k(x)=\begin{cases}x,&x\in S, \frac1x>n\\0,&\text{otherwise}\end{cases}$$

For all $n>\frac1\varepsilon$, the above difference is $<\varepsilon$, so the convergence is uniform.

In answer to your comment, note that
$$\left\lvert f(x)-\sum_{k=1}^nf_k(x)\right\rvert=\left\lvert\sum_{k=n+1}^nf_k(x)\right\rvert$$ and uniform convergence means that the last expression is uniformly small.