问:
If there is a 30% chance of rain for both Saturday and Sunday, what is the probability there would be rain on both days?
Can someone give me a formula for this question?

Wanda said there was a 30% chance of rain for both Saturday and Sunday. It rained on both days.

If her calculations were correct and there was a 30% chance of rain each day, what is the probability there would be rain on both days?

Please help!!!

答:
This is an appallingly constructed problem and is likely to lead to confusion and error for anyone trying to learn from it.

For two events $A$ and $B$ with probabilities $P(A)$ and $P(B)$, the probability of them both occurring - technically the intersection of the events $A \cap B$ is

$$P(A\cap B)=P(A)P(B|A)=P(B)P(A|B)$$

For the particular example, this says that the probability of it raining on both days is the probability of it raining on Saturday times the probability that it rains on Sunday given that it has rained on Saturday (or vice-versa).

Now, for independent events, that is that the fact that one has (or has not) occurred does not affect the other, $P(A|B)=P(A)$. However, the chance of rain on two consecutive days in the same location are not independent, not even close.

What the question wants you to do is multiply $0.3$ by $0.3$, but this is soooo wrong that it is frightening. People have been sent to goal (or jail if you prefer) for decades because so called experts cannot tell the difference between independent and correlated events.

Please note that the relationship between these events does not necessarily have to be causal. Rain is an effect of atmospheric conditions - these generally last longer than 24 hours so if it rains today it is more likely to rain tomorrow - this doesn't mean that today's rain caused tomorrows - they are simply different effects of another cause.

The correct answer to this problem is:

$$P(\text{rain on both days})=P(\text{rain Saturday})P(\text{rain Sunday}|\text{rain Saturday}) = 0.3 \times ?$$

You are not given enough information to go further!


问:
Finding isomorphism between structures
What is the best way to check isomorphism between structures? For example, checking isomorphism between the next structures:
$$\langle \mathbb{N},+\rangle, \langle \mathbb{Z},+\rangle,\langle \mathbb{R},+\rangle,\langle \mathbb{R}^{+},+\rangle,\langle (0,1),\cdot \rangle$$
By intuition, I can see how $\langle \mathbb{N},+\rangle$,$\langle \mathbb{Z},+\rangle$ and $\langle \mathbb{R},+\rangle$ aren't isomorfic, but intuition isn't always the best way for checking isomorphism. Therefore, I be glad to see how you suggest doing such questions, since I am putting a lot of time on them. Thank you!

答:
First, you need to find a bijection $\phi$ between the carrier sets.
In your example, $\Bbb N$ an $\Bbb Z$ are both denumerable and there is a bijection.
Likewise, there is a bijection between each to of the sets $\Bbb R$, $\Bbb R^+$, and $(0,1)$.

Second, you need to check if the equation $\phi(x+y)=\phi(x)\oplus\phi(y)$ for all $x,y$ holds, where $+$ is the operation in the pre-image structure and $\oplus$ is the operation in the image structure.

ContentLicense: CC BY-SA 4.0

Title: Finding isomorphism between structures


问:
If two antisymmetric and symmetric matrices are similar, why are they zero?
If two antisymmetric and symmetric matrices are similar, why are they zero?

My work :
Let $A$ be the antisymmetric matrix and $S$ the symmetric one.
$A$ is diagonalizable in $M_n(\mathbb C)$, we write $\Delta_A$ its diagonal matrix. $S$ is diagonalizable in $M_n(\mathbb R)$, we write $\Delta_S$ its diagonal matrix.

$\Delta_A$ and $\Delta_S$ are similar, so they must have the same eighten values : it is impossible because some are in $\mathbb R$, the others in $\mathbb iR$. So both matrixes must be zero

答:
Assuming $A$ and $S$ are both real matrices, you got the right idea, but the phrasing is a bit off. "It is impossible because some are in $\mathbb R$, the others are in $\mathrm i\mathbb R$" is wrong.

What is correct is that $A$ and $S$ have the same eigenvalues as similar matrices and that those of $A$ are in $\mathrm i\mathbb R$ and those of $S$ in $\mathbb R$. Hence, the eigenvalues are all in $\mathrm i\mathbb R\cap\mathbb R = \{0\}$, so all eigenvalues are $0$.

Finally, the only diagonalizable matrix with $0$ as its only eigenvalue is the zero matrix.


问:
Continuity and closure of sets
Let $X$ and $Y$ be two metric spaces and $f$ a mapping from $X$ to $Y$ .

Prove or disprove: Assume that $f$ is injective. Then $f$ is continuous iff for every subset $A$ of $X$, $f(\bar A) = \overline{f(A)}$

My attempt (Trying to prove the above):

I have already proved the fact that $f$ is continuous if and only if for every subset $A$ of X, $f(\bar A) \subset \overline{f(A)}$.

Suppose $$f(\bar A) = \overline{f(A)}$$
$$\implies \bar A = f^{-1}(\overline{f(A)})$$

As $\bar A$ and $\overline{f(A)}$ are closed sets, we can say that $f$ is continuous as the inverse image of the closed set is closed.

Now to show the converse part we assume that f is continuous and injective, we need to show $f(\bar A) = \overline{f(A)}$.

By using the above fact we can show that $f(\bar A) \subset \overline{f(A)}$, but I need help to show $\overline{f(A)} \subset f(\bar A)$?

答:
No, your argument for the first part is not correct. You have to start with an arbitary closed set $C$ in $Y$ and show that $f^{-1}(C)$ is closed. For this take $A=f^{-1}(C)$ in the hypothesis. You get $\overline {f(A)} =f(\overline A)$ so $f(\overline A) \subseteq  C$ since $f(A)=f(f^{-1}(C)) \subseteq C$ and $C$ is closed. Hence, $\overline A \subseteq f^{-1}(C)$ which means $\overline A \subseteq A$, so $A$ is closed.

The converse is not true. Define $f: [0,1) \to S^{1}$ by $f(x)=e^{2\pi ix}$. Let $A=[\frac 1 2 ,1)$. Then $A$ is closed but $f(A)$ is not closed, so we do not have $f(\overline A)=\overline {f(A)}$ even though $f$ is continuous and injective.

Title: Continuity and closure of sets


问:
Prove or Disprove Equations on Triangle

Given a triangle $ABC$ with $D,E,F$ respectively on $BC,CA,BA$. If $P$ is the intersection point of $AD$ and $EF$, is $(\frac{AB}{AF} \times DC) + (\frac{AC}{AE} \times DB) - (\frac{AD}{AP} \times BC)=0$?

If triangle $ABC$ is equilateral and $D,E,F$ respectively divide $BC,CA,BA$ into halves, I think it's easy to prove that statement is true. But what if the triangle is scalene and we don't know whether $D,E,F$ respectively divide $BC,CA,BA$ into halves?

答:
Proving the given statement is equivalent to proving that $(\frac {FB}{AF}\times DC)+(\frac {CE}{AE}\times BD)=(\frac {PD}{AP}\times BC)$

Extend $EF$ to meet $BC$ at point $Q$. Applying Menelaus' Theorem on $\triangle ABD$ and $\triangle ADC$ considering $FEQ$ as the transversal gives,

$\frac{FB}{AF}\cdot \frac{AP}{PD}\cdot \frac{DQ}{BQ}=1$ and $\frac{CE}{AE}\cdot \frac{AP}{PD}\cdot \frac{DQ}{CQ}=1$

$(\frac {FB}{AF}\times DC)+(\frac {CE}{AE}\times BD)$

$=\frac{PD}{AP}\{\left(\frac{FB}{AF}\times \frac{AP}{PD}\times DC\right)+\left(\frac{CE}{AE}\times \frac{AP}{PD}\times BD\right)\}$

$=\frac{PD}{AP}\{\left(\frac{BQ\times DC}{DQ}\right)+\left(\frac{CQ\times BD}{DQ}\right)\}$

Now, $\left(BQ\times DC\right)+\left(CQ\times BD\right)$ can be written as,

$\left(BQ\times DC\right)+\left(BQ-BC\right)\left(BC-DC\right)$

This simplifies to $BC\cdot DQ$.

Hence, $\left(\frac {FB}{AF}\times DC\right)+\left(\frac {CE}{AE}\times BD\right)$

$=\frac{PD}{AP}\left\{\left(\frac{BQ\times DC}{DQ}\right)+\left(\frac{CQ\times BD}{DQ}\right)\right\}=\frac{PD}{AP}\times BC$

Title: Prove or Disprove Equations on Triangle


问:
given a closed $1$-form $\alpha$, find $0$-form $f$ s.t. $df = \alpha$.

Let $\alpha$ be a closed $1$-form on a connected manifold $M$. We
assume that $M$ is simply connected, i.e. for all paths
$\gamma_0,\gamma_1$ with the same endpoints there exists a smooth
homotopy $h\colon[0,1]\times[0,1]\to M$ between $\gamma_0$ and
$\gamma_1$. Use the Poincare-Lemma to show that

$$\int_{\gamma_0}\alpha = \int_0^1\gamma_0^*\alpha =
\int_0^1\gamma_1^*\alpha = \int_{\gamma_1}\alpha$$

Use this
observation to construct a smooth function $f$ so that $df =  \alpha$.

I struggle to construct the desired function $f$.

So we are obviously looking for a $0$-form $f \in \Omega^0 M$ as the antiderivative $f = \int \alpha$ of our given $\alpha \in \Omega^1M$.

Since $\gamma_0,\gamma_1$ are homotopic, it does not depend over which $\gamma_t$ we integrate. Thus $$f = \int_{\gamma_1}\alpha = \int_0^1 \gamma^*\alpha$$

But how do i continue? I know the definition of the pullback of a differential form on a smooth manifold, but i do not know how to make use of it in this case.

Can anyone help me?

Thanks in advance for any advise!

Addendum:

For example, the solution to the problem starts with:

pick $\gamma_i(t)=tx^i$ in coordinates $x^i$

and then proceeds with

$ f = \int_\gamma \alpha = \int \alpha_{\gamma(t)}\frac{d\gamma}{dt}dt$

but i don't understand what's going on. I don't understand the coordinate-change nor the last integral.

答:
Fix a point $x_0\in M$, and then for a given point $x\in M$, define
$$f(x):=\int_\gamma \alpha$$
where $\gamma$ is any path from $x_0$ to $x$.


答:
An isomorphism between two magmas (or semigroups, or monoids, …) $(A,\bullet_A)$ and $(B,\bullet_B)$ is a map $\phi\colon A\to B$ satisfying
$$
\phi(x\bullet_A y) = \phi(x)\bullet_B \phi(y)
$$
(so it is a homomorphism) as well as being bijective (making it an isomorphism).

In particular, when $(A,\bullet_A)$ and $(B,\bullet_B)$ are isomorphic (meaning there exists an isomorphism), then they have the same cardinality as sets. This means that $(\mathbb N,+)$ and $(\mathbb Z,+)$ can't be isomorphic to any of the three others, since these are countable while the others aren't.

Now $(\mathbb N,+)$ and $(\mathbb Z,+)$ have the same cardinality (there is a set bijection $\mathbb N\to\mathbb Z$), but they are not isomorphic: both have a unique identity element, which is $0$ in both cases (so they are monoids) but in $(\mathbb Z,+)$ every element has an inverse while in $(\mathbb N,+)$ only $0$ has  an inverse. (I'm assuming $\mathbb N=\{0,1,2,\dots\}$, otherwise one doesn't even have an identity element.)

The same argument applies to $(\mathbb R,+)$ and $(\mathbb R^+, +)$, assuming that $\mathbb R^+$ denotes the non-negative real numbers.

Finally $((0,1),\,\cdot\,)$ can't be isomorphic to any of the others, since it is the only one having an element $a$ such that $a\cdot x=a$ for all $x$, namely $a=0$.

So, to answer your question: if you can find a property satisfied by one structure that isn't satisfied by the other, then they can't be isomorphic. If you can't find such a property, you might try to construct an isomorphism.

Title: Finding isomorphism between structures

Body:
What is the best way to check isomorphism between structures? For example, checking isomorphism between the next structures:
$$\langle \mathbb{N},+\rangle, \langle \mathbb{Z},+\rangle,\langle \mathbb{R},+\rangle,\langle \mathbb{R}^{+},+\rangle,\langle (0,1),\cdot \rangle$$
By intuition, I can see how $\langle \mathbb{N},+\rangle$,$\langle \mathbb{Z},+\rangle$ and $\langle \mathbb{R},+\rangle$ aren't isomorfic, but intuition isn't always the best way for checking isomorphism. Therefore, I be glad to see how you suggest doing such questions, since I am putting a lot of time on them. Thank you!


问:
$f(x)=x-x^2$ if $x$ is rational, $x+x^2$ if $x$ is irrational

$$f(x)= \left\{\begin{array}{ll}
  x-x^2 &\mbox{if $x$ is rational,}\\
  x+x^2 &\mbox{if $x$ is irrational.}
  \end{array}\right.$$
  Show that $f'(0)=1$ and yet there is no neighborhood $I$ of the point $0$ on which this function is monotonically increasing.

答:
To show that $f'(0)=1$, it suffices to prove that $\lim_{x\rightarrow 0}\frac{f(x)-x}x = 0$ (since $f(0) = 0$). Since $f(x)-x=\pm x^2$, it is always true that $|f(x)-x|=x^2$. Dividing by $x$ and making $x\rightarrow 0$ gives the answer.

As for the second part. Suppose such a neighborhood $I$ exists. Then let $x\in I$ be irrational. Since $\mathbb Q$ is dense in $\mathbb R$, there exists a sequence $r_n$ of rationals in $I$ such as $r_n\rightarrow x$ and $r_n > x$.

Since $r_n$ is rational, $f(r_n) = r-r_n^2$, on the other hand $f(x) = x+x^2$. $f$ is monotically increasing on $I$, so that $f(x)\leq f(r_n)$ for all $n$. In other words

$x+x^2\leq r_n-r_n^2$ for all $n$.

Passing to the limit, it is found $x+x^2\leq x-x^2$ which is absurd (if $x\neq 0$). Therefore $I$ does not exist.

ContentLicense: CC BY-SA 3.0


问:
Confusion in the equivalent statement of projective module
The following statements are equivalent

If $P$ is an $R-$module, and if it sastisfies the following equivalent properties then it is called a projective module.

$1$- Every exact sequence of the form $0 \to A \to B \to P \to 0$ splits.

$2$-There exists a $R-$module $Q$ such that $P \oplus Q$ is a free $R$-module.

Now if the sequence splits, then $B \cong A \oplus P$, Does that imply that $B$ is a free module? I do not think so, that $B$ is a free module?  if not then how do I show the existence of a module $R-$module $Q$ such that $P \oplus Q$

答:
No, if $0\to A\to B\to P\to 0$ is exact and $P$ is projective then $B$ is not necessarily free. But we know that all short exact sequences with $P$ at the end split, so we can just choose one where $B$ is free. So let $(x_i)_{I}$ be any generating system of $P$ and let $K$ be the kernel of $R^{(I)}\to P$, $e_i\mapsto x_i$. Then we have the short exact sequence $$0\to K\to R^{(I)}\to P\to0$$ which splits by assumption, i.e. $R^{(I)}\cong K\oplus P$.

(Here $R^{(I)}=\bigoplus_{i\in I}R$)


问:
If $\frak{g}$ is a semi-simple lie algebra then every homomorphic image of $\frak{g}$ is semisimple.
Could someone give me a suggestion to solve this problem?

PROBLEM:

If $\frak{g}$ is a semi-simple lie algebra then every homomorphic image of
$\frak{g}$ is semisimple.

I was trying to prove that if $\varphi: \frak{g} \rightarrow \frak{h}$ is a homomorphism of lie algebras then the radical of

$\varphi(\frak{g})$ is zero, but I could not.

答:
We know that $\phi(\mathfrak g)=\mathfrak g / ker \phi$.

Since $ker\phi$ is an ideal of $\mathfrak g$, we can write $\mathfrak g=ker\phi \oplus (ker\phi)^\perp$. Here $ker\phi $ and $(ker\phi)^\perp$ are both ideals of $\mathfrak g$. (This is proved on page 23 of 'Introduction to Lie Algebras and Representation Theory-Humphreys'.) So, $\phi(\mathfrak g)=\mathfrak g / ker \phi=(ker\phi)^\perp$. Since every ideal of semisimple is semisimple, $\phi(\mathfrak g)=(ker\phi)^\perp$ is semisimple.


问:
25th percentile of the sum of 500 samples coming from a given probability deistribution
I came across this problem that was modified from a business case study:

An environmental start-up recycles catalytic converters. It estimates that the worth of one catalytic converter has the following probability distribution: 1200\$ with probability 0.35, 1800\$ with probability 0.50, and 2400\$ with probability 0.15. The start-up is offered a load of 500 catalytic converters, and the start-up needs to assess the worth of the entire load to decide whether to purchase it (assuming that the worth of any one converter is independent of the worth of the others). To ensure a profit, the start-up would not pay a price higher than the 25th percentile of the probability distribution associated with the worth of the load. (In other words, they would agree to a price only if there is a $75 \%$ chance that the load is worth more than that price.) Following this rule, what is the maximum price that the start-up would be willing to pay for the load of 500 catalytic converters?

If we define $X$ to be the random variable for the worth of one converter and $Y$ to be the random variable for the worth of 500 converter, then $E[Y]=E[500X]=500E[X]=500(1200(0.35)+1800(0.5)+2400(0.15))=840000$ and $\sigma(Y)=\sigma(500X)=500\sigma(X)=203469.89$.

My first instinct to answer the question was use the observation that the 25th percentile for $X$ has value 1200 and multiply this to 500. However, this is probably wrong.

My second instinct is that $$\text{25th percentile of the total worth} =\text{expected value} + (z-\text{value of 25th percentile})(\text{standard deviation}).$$

However, I still think that this is wrong. Since we have a sample of 500 here, I think the central limit theorem must come into play somehow. How does one compute for the 25th percentile of the total worth of 500 converters?

答:
I will use thousands of dollars throughout, in order to keep numbers from
getting too large and to avoid writing more digits than precision supports.

Let $X$ be the value of a randomly chosen catalytic
converter. Then  $\mu_X = E(X) = 1.68$ thousand dollars and
$\sigma_X = SD(X) = 0.407$ thousand dollars. [Computations in R.]

```<code>kd = c(1.2, 1.8, 2.4)
pr = c(.35, .50, .15)
mu.x = sum(kd*pr); mu.x
[1] 1.68

va.x = sum(pr*(kd-mu.x)^2); va.x
[1] 0.1656
sg.x = sqrt(va.x);  sg.x
[1] 0.4069398
</code>```

Then the total value $T$ of 500 converters should
have $\mu_T=E(T) = 500\mu_X = 840$ thousand dollars and
$\sigma_T = SD(T) = \sqrt{500\sigma_X^2} =  9.0995$
thousand dollars.

```<code>mu.t = 500*mu.x;  mu.t
[1] 840
sg.t = sqrt(500*va.x);  sg.t
[1] 9.099451
</code>```

The random variable $X$ is highly discrete, but it
is probably OK to assume that $T$ is approximately
$\mathsf{Norm}(\mu_T = 840,
\sigma_T = 9.09945)$ in thousands of dollars.

Then the lower quartile of $T$ would be the lower
quartile of this normal distribution: 833,863 thousand dollars.

```<code>qnorm(.25, 840, 9.09945)
[1] 833.8625
</code>```

Simulation.

We can simulate 100,000 batches of 500 converters
to see how well results match with those above. With
100,000 iterations we can expect at at least two significant digits of accuracy, so results
match within the margin of simulation error.
In particular the normality assumption does seem OK,

```<code>set.seed(125)
t = replicate(10^5, sum(sample(kd, 500, rep=T, p=pr)))
mean(t);  sd(t);  quantile(t, .25)
[1] 840.0297
[1] 9.084199
25%
834
</code>```

```<code> hist(t, prob=T, col="skyblue2",
      xlab="Thousand Dollars", main="Simulated Batches")
  curve(dnorm(x, mean(t), sd(t)), add=T)
  abline(v = 834, col="red", lwd=2)
</code>```

Note: The simulated lower quartile 834 thousand dollars, might be
more accurate than the lower quartile 833.8625 thousand dollars of the theoretical
normal distribution. Even though the histogram does not show that the distribution of $T$ is discrete; among the 100,000
simulated values of $T$ there were only $124$ uniquely different values.

```<code> length(unique(t))
 [1] 124
</code>```

If this is a textbook exercise in means, variances, and
quartiles, then \$833,862.50 is likely the expected
answer. But if it's a real-life problem, in which the lower
quartile is set as protection that someone may have skimmed
many of the more valuable converters from some batches,
then it may be better to use the simulated value \$ 834,000.

Body:
I will use thousands of dollars throughout, in order to keep numbers from
getting too large and to avoid writing more digits than precision supports.

Let $X$ be the value of a randomly chosen catalytic
converter. Then  $\mu_X = E(X) = 1.68$ thousand dollars and
$\sigma_X = SD(X) = 0.407$ thousand dollars. [Computations in R.]

kd = c(1.2, 1.8, 2.4)
pr = c(.35, .50, .15)
mu.x = sum(kd*pr); mu.x
[1] 1.68

va.x = sum(pr*(kd-mu.x)^2); va.x
[1] 0.1656
sg.x = sqrt(va.x);  sg.x
[1] 0.4069398

Then the total value $T$ of 500 converters should
have $\mu_T=E(T) = 500\mu_X = 840$ thousand dollars and
$\sigma_T = SD(T) = \sqrt{500\sigma_X^2} =  9.0995$
thousand dollars.

mu.t = 500*mu.x;  mu.t
[1] 840
sg.t = sqrt(500*va.x);  sg.t
[1] 9.099451

The random variable $X$ is highly discrete, but it
is probably OK to assume that $T$ is approximately
$\mathsf{Norm}(\mu_T = 840,
\sigma_T = 9.09945)$ in thousands of dollars.

Then the lower quartile of $T$ would be the lower
quartile of this normal distribution: 833,863 thousand dollars.

qnorm(.25, 840, 9.09945)
[1] 833.8625

Simulation.

We can simulate 100,000 batches of 500 converters
to see how well results match with those above. With
100,000 iterations we can expect at at least two significant digits of accuracy, so results
match within the margin of simulation error.
In particular the normality assumption does seem OK,

set.seed(125)
t = replicate(10^5, sum(sample(kd, 500, rep=T, p=pr)))
mean(t);  sd(t);  quantile(t, .25)
[1] 840.0297
[1] 9.084199
25%
834

 hist(t, prob=T, col="skyblue2",
      xlab="Thousand Dollars", main="Simulated Batches")
  curve(dnorm(x, mean(t), sd(t)), add=T)
  abline(v = 834, col="red", lwd=2)

Note: The simulated lower quartile 834 thousand dollars, might be
more accurate than the lower quartile 833.8625 thousand dollars of the theoretical
normal distribution. Even though the histogram does not show that the distribution of $T$ is discrete; among the 100,000
simulated values of $T$ there were only $124$ uniquely different values.

 length(unique(t))
 [1] 124

If this is a textbook exercise in means, variances, and
quartiles, then \$833,862.50 is likely the expected
answer. But if it's a real-life problem, in which the lower
quartile is set as protection that someone may have skimmed
many of the more valuable converters from some batches,
then it may be better to use the simulated value \$ 834,000.


问:
What would be the expected number of targets which didn't get hit by any of the shooters?

Assume there are 5 shooters and 5 targets. Each shooter would choose a target randomly and when the signal is given all shooters will fire at the same time and hit their target. What would be the expected number of targets which didn't get hit by any of the shooters?

My attempts: Initially, there are $5^5$ different possible case. Now I assume a R.V $X$ be the number of unshot target. So $P(X=0)=5!/5^5, P(X=1)=5!/5^5,P(X=2)=C_2^5\times3!/5^5$ and so on but it seems that the probability I get is wrong but i am not sure which part goes wrong and how to get the correct one. Any solution or hints are welcome, thanks.

答:
Define random variables $X_1,X_2,\dots, X_5$ by $X_i=1$ if Target $i$ didn't get hit, and by $X_i=0$ otherwise. Then the number of targets hit is $X_1+\cdots+X_5$. By the linearity of expectation,
$$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_5).$$

To find $E(X_i)$, we only need to find the probability Target $i$ didn't get hit. This is (if the shooters choose their targets uniformly and independently) equal to $\left(\frac{4}{5}\right)^5$. This is $E(X_i)$.

Remark: Note that we do not need to know the distribution of $Y$ to calculate the expectation. The $X_i$ are not independent, but linearity of expectation always holds.  Also, the approach works just as smoothly with $k$ targets and $n$ shooters.


问:
A problem on inverse trigonometric function
How to sum the following series: $$S=\cot^{-1}2+\cot^{-1}8+\cot^{-1}18+\cot^{-1}32+\cdots+\cot^{-1}∞$$ My attempt better to say a solution was,\begin{align}S &=\sum_{n=1}^{∞}\cot^{-1}2n^2\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{1}{2n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{4n^2}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{2}{1+4n^2-1}\\
&=\sum_{n=1}^{∞}\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\\
&=\sum_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]\\
&=\tan^{-1}∞-\tan^{-1}1\\
&=\frac{π}{2}-\frac{π}{4}\\
&=\frac{π}{4}\end{align} Now one thing I have to confess that I knew this method previously,so I did and I don't understand from where this idea suddenly comes about telescoping? Can someone help me explain with any different approach is added as an answer. Thanks for your attention.

答: The only way one can expand an $\arctan$ is to find some expression like $\frac{a+b}{1-ab}$ and hence this method is to try to find a representation of the value we have in this form


问:
Where is my mistake in deriving this probability?
$X$ and $Y$ are independent Uniform(0,1) random variables. I want to find $\mathbb{P}(XY < \frac{1}{2})$

I proceeded like this:

$$\mathbb{P}(XY < \frac{1}{2}) = \mathbb{P}(X < \frac{1}{2Y}) $$

If $Y$ was discrete we could write

$$\mathbb{P}(X < \frac{1}{2Y}) = \sum_y \mathbb{P}(X < \frac{1}{2Y}|Y = y) \mathbb{P}(Y = y) $$

So in a continuous case this becomes an integral:

$$\mathbb{P}(X < \frac{1}{2Y}) = \int_{0}^1\frac{1}{2y}*1 dy$$

But the integral doesnt converge. Where did I go wrong? I would be most grateful if you show a mistake in my derivations instead of just solving the problem

答:
What happens when $Y$ is smaller than $1/2$?  Then the outcome of $X \in [0,1]$ does not matter, because we are guaranteed that $XY < 1/2$ and the probability of this event is $1$.  This is why your integral fails to converge, because in the expression $$\int_{y=0}^1 \frac{1}{2y} \, dy,$$ the integrand $1/(2y)$ becomes too large when $y$ is "close" to $0$.  You need to take this into account.

The correct expression is $$\Pr[XY < 1/2] = \Pr[X < 1/(2Y)] = \int_{y=0}^{1/2} \Pr[X < 1] f_Y(y) \, dy + \int_{y=1/2}^1 \Pr[X < 1/(2y)] f_Y(y) \, dy.$$  The first integral is just $1/2$.  The second integral is left as an exercise.


问:
$A,B,C,B-C$ positive definite, prove the minimum eigenvalue of $BAB$ is greater that of $CAC$.
$A,B,C,B-C\in R^{n\times n}$ positive definite, prove the minimum eigenvalue of $BAB$ is greater that of $CAC$. At first I try to transform them to canoical forms like diagonal but since congruence transformation chances the order of eigenvalues and of course $B^2-C^2$ is not necessarily positive definite. I have difficulty with this problem

答:
Although it is true that
$\lambda_\min(B^{1/2}AB^{1/2})
=\lambda_\min(A^{1/2}BA^{1/2})
>\lambda_\min(A^{1/2}CA^{1/2})
=\lambda_\min(C^{1/2}AC^{1/2})$, your inequality is false in general. Here is a random counterexample:
$$
A=\pmatrix{5&2\\ 2&1},\ B=\pmatrix{4&5\\ 5&22},\ C=\pmatrix{2&0\\ 0&9}.
$$
One can easily verify by Sylvester's criterion that $A,B,C$ and $B-C$ are positive definite, but $\lambda_\min(BAB)=3.2248<3.3168=\lambda_\min(CAC)$.


问:
Probability that you win in rock scissors paper with 3 players?
Let's say person A, person B and person C are playing rock paper scissors.

Clearly $n(S)=3\times 3\times 3=27$. We want to find the probability that person A wins.

$n(\text{only A wins})=3$ (A wins with rock/scissors/paper).

$n(\text{A and one other person wins})=2\times 3=6$
($2$ ways to choose the other winner, $3$ ways to win (rock/scissors/paper))

$\therefore P(\text{A wins})=\frac{3+2\times3}{27}=\frac{1}{3}$, but this is flawed but this because it doesn't account for the probability where you draw. (i.e. if the probability of each person winning is $\frac{1}{3}$, then the total probability is already $1$, when we haven't accounted for the probability of drawing yet.)

What is the correct probability that you will win in a game of rock scissors paper with 3 players? Thanks.

答:
There is nothing wrong with your calculation; each player wins with probability $1/3$. The point is that these are not disjoint events, so $$P(A\text{ wins})+P(B\text{ wins})+P(C\text{ wins})\neq P(A\text{ or }B\text{ or }C\text{ wins}).$$
In fact, the sum of the probabilities is the expected number of players who win. (When there are no multiple winners, so the events are disjoint, this number is always $0$ or $1$ and so its expectation is the probability someone will win, but this is not true in general.)

Here $P(\text{no-one wins})=3\times\frac{1}{27}+\frac{3!}{27}=\frac13$, $P(2\text { people win})=3\times 3\times\frac1{27}=\frac13$, and $P(1\text{ person wins})=\frac13$, so the expectation is exactly $1$, consistent with your answer.


问:
The limit of an integral (with statistical backgrounds)
I would like to compute the following limit of an integral $$I=\lim_{n\to+\infty}\int_0^n\frac{x^n\text{e}^{-x}}{n!}\text{d}x$$
which I encountered in a statistics problem. I can easily do this using Gamma distribution and Central Limit Theorem:

 Consider a random variable $X_n$ with PDF $f_n(x)=\frac{x^n\text{e}^{-x}}{n!}$. From the properties of Gamma distribution, it can be decomposed into exponential distributions, $X_n=\sum_{k=1}^n X_k$ where $X_k\sim\text{Exp}(1)$. Now let $F_n(x)=\text{P}(\frac{X_n-n}{\sqrt n}\leq x)$; from Central Limit Theorem, $$\lim_{n\to+\infty}F_n(x)=\lim_{n\to+\infty}\text{P}\left(\frac{\sum_{k=1}^n X_k -n}{\sqrt n}\leq x\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\text{e}^{\frac{-t^2}{2}}\text{d}t$$ Plugging $x=0$ in gives $I=\frac 1 2$.

Nevertheless, I would like to know if there are any pure math methods for solving this (that don't involve anything statistical). The integral itself, from Wolfram|Alpha, is $$\int_0^n\frac{x^n\text{e}^{-x}}{n!}\text{d}x=\frac{\Gamma(n+1)-\Gamma(n+1,n)}{n!}$$
which doesn't seem to be very useful (it's just rewriting this integral in a Gamma function form). Any help would be appreciated.

答:
Not sure it is as pure math as you'd wish but here goes. Let:
$$
I_{m,n} = \int_0^m \frac{x^ne^{-x}}{n!}\text{d}x
$$
Using a simple integration by parts yields, for $n>0$:
$$
I_{m,n} -I_{m,n-1} = -\frac{m^ne^{-m}}{n!}
$$
Furthermore:
$$
I_{m,0} = \int_0^m e^{-x}\text{d}x = 1 - e^{-m}
$$
Then:
$$
I_{m,n} = I_{m,0} + \sum_{k=1}^n (I_{m,k} - I_{m,k-1})
$$

$$
I_{m,n} = 1 - e^{-m} - \sum_{k=1}^n \frac{m^ke^{-m}}{k!}
$$

Which, when setting $m$ to $n$:

$$
I_{n,n} = \int_0^n \frac{x^ne^{-x}}{n!}\text{d}x =  1 - e^{-n} \sum_{k=0}^n \frac{n^k}{k!}
$$

That was the rather easy part. You can then conclude using the result of this question (where most of the answers essentially go back the other way, so really you're just better off just picking up from where your integral appears).


问:
Help needed in proving an result about a function with only poles as singularity
This question was part of my assignment in complex analysis and I was unable to solve it and hence asking for help here.

Show that $~f(z)~$ has no singularities in extended plane other than poles if and only if $~f(z)~$ is quotient of $~2~$ polynomials.

Assuming $~f(z)~$ is quotient of $~2~$ polynomials I have proved that only singularities are poles but I am having trouble in proving the converse.

Let function be $~p(z)=  f(z) \left[1/(z-z_0)^{m_0}  + \cdots +1/(z-z_r)^{ m_r} \right]~,$ where $~f(z)~$ is analytic but how to prove that $~f(z)~$ must be a polynomial ?

Thank you!!

答:
Question: Show that $f(z)$ has no singularities in extended plane other than poles if and only if $f(z)$ is quotient of $~2~$ polynomials.

Answer: Let $f(z)$ be the quotient of two polynomials, say, $$f(z)=\dfrac{p(z)}{q(z)}~,\tag1$$ where
$~p(z) = α_0 + α_1z + α_2z^2 +\cdots + α_nz^n~$ and $~q(z) = β_0 + β_1z + β_2z^2 + \cdots + β_mz^m~.$
 Clearly, the polynomials $~p(z)~$ and $~q(z)~$ are analytic at all finite points on the complex plane.

Let us assume that $~p(z)~$ and $~q(z)~$ have no common zeros;
if they do have a common zero at $~z = z_0~,$ it is always possible to write $~f(z)~$ as the quotient of two polynomials with no common zeros by canceling a suitable number of the $~(z − z_0)$ factors.

Obviously, the only possible singularities of $~f(z)~$ are situated at the zeros
of $~q(z)~.$ Since the zeros of $~p(z)~$ do not coincide with those of $~q(z),~$ so $~f(z)~$ necessarily diverges at the zeros of $~q(z)~.$ Such points can be poles but not essential singularities in view of the Weierstrass theorem.
Thus all singularities of rational functions $~f(z)~$ are necessarily poles.

To prove the converse, suppose that all the singularities of an analytic function $~f(z)~$ are poles at the points $~a_1,~a_2,~\cdots ,~a_n~.$ The orders of these poles
are denoted by $~m_1,~m_2,~ \cdots ,~m_n~,$ respectively. In the vicinity of the point $~a_ν~,$ the function $~f(z)~$ has a Laurent series expansion of the form
$$f(z)=\dfrac{c_{-m_v}^{(v)}}{(z-a_v)^{m_v}}+\dfrac{c_{-m_v-1}^{(v)}}{(z-a_v)^{m_v-1}}+\cdots+\dfrac{c_{-1}^{(v)}}{(z-a_v)}+\sum_{\mu=0}^\infty c^{(u)}_\mu (z-a_v)^\mu~,$$where the superscripts $~(ν)~$ on $~c^{(ν)}~$ indicate that they are the coefficients that belong to the $ν$th poles, $~z = a_ν~.$ Denote the principal part by
$$g_v(z)=\dfrac{c_{-m_v}^{(v)}}{(z-a_v)^{m_v}}+\dfrac{c_{-m_v-1}^{(v)}}{(z-a_v)^{m_v-1}}+\cdots+\dfrac{c_{-1}^{(v)}}{(z-a_v)}$$and consider the expression $$h(z)=f(z)-g_1(z)-g_2(z)-\cdots-g_n(z)=f(z)-\sum_{v=1}^n g_v(z)~.$$ Since $~f(z) − g_ν(z)~$ is analytic at $~z = a_ν~,$ and $~g_ν(z)~$ is analytic everywhere
except at $~z = a_ν~,$ it follows that $h(z)$ is analytic at all points of the complex
plane, including the point at infinity.
In view of Liouville’s theorem such a
function is necessarily a constant. Thus we have identically $~h(z) ≡ γ_0~,$ whence
$$f(z)=\gamma_0+\sum_{v=1}^n g_v(z)$$which implies that $~f(z)~$ can be brought into the form equation $(1)$.
This completes the proof of our theorem.

Ref.: Higher Mathematics for Physics and Engineering: Mathematical Methods for Contemporary Physics by Tsuneyoshi Nakayama & Hiroyuki Shima


答:
It these cases, a drawing will result always very useful and self explaining.

$$\mathbb{P}\Bigg[XY<\frac{1}{2}\Bigg]=\mathbb{P}\Bigg[Y<\frac{1}{2X}\Bigg]$$

The requested probability is the colored area:

$$\mathbb{P}\Bigg[XY<\frac{1}{2}\Bigg]=0.5+\int_{0.5}^1 \frac{1}{2x}dx=0.5(1+\log2)$$

Title: Where is my mistake in deriving this probability?

Body:
$X$ and $Y$ are independent Uniform(0,1) random variables. I want to find $\mathbb{P}(XY < \frac{1}{2})$

I proceeded like this:

$$\mathbb{P}(XY < \frac{1}{2}) = \mathbb{P}(X < \frac{1}{2Y}) $$

If $Y$ was discrete we could write

$$\mathbb{P}(X < \frac{1}{2Y}) = \sum_y \mathbb{P}(X < \frac{1}{2Y}|Y = y) \mathbb{P}(Y = y) $$

So in a continuous case this becomes an integral:

$$\mathbb{P}(X < \frac{1}{2Y}) = \int_{0}^1\frac{1}{2y}*1 dy$$

But the integral doesnt converge. Where did I go wrong? I would be most grateful if you show a mistake in my derivations instead of just solving the problem


问:
Parachutist and Differential Equations Modeling Air Reistance
A parachutist whose mass is  65 kg drops from a helicopter hovering 2500 m above the ground and falls toward the ground under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant $b_1$= 20 N-sec/m when the chute is closed and  $b_2$= 90 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches  m/sec, after how many seconds will the parachutist reach the ground? Assume that the acceleration due to gravity is $9.81 m/s^2$.

The answer is apparently 345.06 seconds.

In my work, the subscripts 1 and 2 represent the quantities they are attached to before and after the chute is opened, respectively. Can someone point out where I went wrong?

$$\sum F_{1}=mg-b_1v_1=m\frac{dv_1}{dt}$$

$$\frac{dv_1}{dt}=g-\frac{b_1}{m}v_1;\frac{b_1}{m}=c_1=\frac{4}{13}$$

$$\int_{0}^{v}\frac{dv_1}{g-c_1v_1}=\int_{o}^{t_1}dt;c_1=\frac{b_1}{m}\Rightarrow -\frac{ln(g-c_1v_1)}{c_1}=-\frac{ln(g-c_1v_1)}{c_1}+\frac{ln(g)}{c_1}=t_1$$

$$ln(\frac{g}{g-c_1v_1})=c_1t\Rightarrow e^{c_1t}=\frac{g}{g-c_1v_1}\Rightarrow v_1=-\frac{g}{c_1e^{c_1t}}+\frac{g}{c_1}=-\frac{9.81}{\frac{4}{13}e^{\frac{4}{13}t}}+\frac{9.81}{\frac{4}{13}}$$

$$v_1(t)=\frac{dx_1}{dt}=-31.885e^{-\frac{4}{13}t}+31.8825$$

$$\sum F_2=m\frac{dv_2}{dt}=mg-b_2v_2$$

$$\frac{dv_2}{dt}=g-\frac{b_2}{m}v_2;\frac{b_2}{m}=c_2=\frac{18}{13}$$

$$\int_{25}^{v}=\frac{dv}{g-c_2v_2}=\int_{0}^{t_2}dt\Rightarrow -\frac{ln(g-c_2v_2)}{c_2}+\frac{ln(g-25c_2)}{c_2}=t_2$$

$$v_2(t)=17.915e^{-\frac{18}{13}}+7.085$$

From

$$x(t)=m\frac{d^2x_2}{dt^2}=mg-b_2\frac{dx_2}{dt}$$

I have

$$\frac{m(gt-v)}{b}+C=\frac{m(gt-(17.915e^{-\frac{18}{13}}+7.085))}{b}+C$$

答:
You could also just take the general ODE $v'=g-cv$, set $u=g-cv$, find $u'=-cu$ so that $u(t)=e^{-ct}u_0$ and substitute back for $v$.
$$
(g-cv(t))=e^{-ct}(g-cv(0))\implies v(t)=v(0)e^{-ct}+(1-e^{-ct})\frac{g}{c}
$$
Then insert the values for both phases, with $v_1(0)=0$. This shortens the computation considerably and thus reduces the probability of errors.

Next compute the time $t_o$ of the parachute opening, that is, solve $v_1(t)=v_o$ or
$$
e^{-c_1t_o}=\frac{g-c_1v_1(t_o)}{g-c_1v_1(0)}\iff
t_o=\frac{-\ln(1-\frac{c_1v_o}{g})}{c_1}
$$
Then apply the general equation with shifted origin to the second phase for initial condition $v_2(t_o)=v_o$
$$
g-cv_2(t)=e^{-c_2(t-t_c)}(g-c_2v_2(t_o))
\implies v(t_2)=v_oe^{-c_2(t-t_o)}+(1-e^{-c(t-t_o)})\frac{g}{c_2}
$$
To compute the distance to the ground you need to integrate the velocity
$$
v_1(t) = (1-e^{-c_1t})\frac{g}{c_1}\implies x_1(t)=(e^{-c_1t}-1+c_1t)\frac{g}{c_1^2}
$$
Then integrate $v_2$ with $x_2(t_o)=x_1(t_o)=(e^{-c_1t_o}-1+c_1t_o)\frac{g}{c_1^2}$.


问:
Frobenius element in cyclotomic extension
Let $K=\mathbb{Q}(\zeta_m)$. Then if $p\nmid m$ is any odd prime, how i can show that Frobenius map is
$(p,K/\mathbb{Q})(\zeta_m)=\zeta_m^p$.

We know, if $P$ is a prime above $p$

$$(p,K/\mathbb{Q})(x) \equiv x^p (\mbox{mod } P) \quad\forall x \in \mathbb{Z}_K$$

Thanks!

答:
In an arbitrary Galois extension $K/\mathbf Q$ with Galois group $G$, and an unramified prime $\mathfrak p$ of $K$ sitting over $p$, we know that the Galois group of the residual extension $\mathcal O_K/\mathfrak p$ over $\mathbf F_p$ identifies with the decomposition group $D_{\mathfrak p/p} = \{ \sigma \in G : \sigma(\mathfrak p) = \mathfrak p\} \subseteq G$. The Galois group of a finite extension of finite fields is cyclic, and comes equipped with a canonical generator : the $p$-th power map, also commonly known as the Frobenius. Under the identification of $D_{\mathfrak p/p}$ with $\text{Gal}((\mathcal O_k/\mathfrak p)/\mathbf F_p)$, the element of $D_{\mathfrak p/p}$ corresponding to the $p$-th power map is called the Frobenius at $\mathfrak p$ and is denoted $\sigma_{\mathfrak p/p}$.

If $\mathfrak p$ and $\mathfrak p'$ lie both above $p$, there is an element of the Galois group sending one to the other, and it follows that the decomposition groups $D_{\mathfrak p/p}$ and $D_{\mathfrak p'/p}$ are conjugate. Under this correspondence, the various Frobenius elements at $\mathfrak p$, when $\mathfrak p$ ranges over all the primes above $p$, determine no longer an element but a conjugacy class of $G$. This conjugacy class is written $\sigma_p$  and is called the Frobenius at $p$.

When $G$ is abelian, conjugacy classes are just elements of $G$, and therefore the Frobenius $\sigma_p$ is actually a well-determined element of $G$. Thus, for each unramified prime of an abelian extension, we obtain a canonical element of the Galois group attached to it.

In your case, you are looking at the cyclotomic extension $K=\mathbf Q(\zeta_m)$. This is an abelian extension, unramified away from the primes dividing $m$. The Galois group identifies canonically with $(\mathbf Z/m\mathbf Z)^\times$, by $\zeta_m \mapsto \zeta_m^a$. Now, look at the automorphism $\alpha : \zeta_m \mapsto \zeta_m^p$. Claim: we have $\alpha(x) \equiv x^p \mod p$. Indeed, since $\mathcal O_K = \mathbf Z[\zeta_m]$, we can write an arbitrary element as $x= \sum_{i}a_i \zeta_m^i$; since the binomial coefficients $p \choose i$ vanish for $1<i<m$, we have
$$x^p \equiv\sum_i a_i^p \zeta_m^{ip} \equiv \sum a_i\zeta_m^{ip} \equiv \alpha(x) \mod p$$
(using the fact that $a_i \equiv a_i^p$.) Therefore, $\alpha$ induces the $p$-power map on the residual extension so by definition of the Frobenius at $p$, $\alpha = \sigma_p$.

This fact is a primitive example of a "reciprocity law", which describes the splitting of primes in an extension in terms of arithmetic in the base.


问:
Are the integrals uniformly continuous and differentiable on $\mathbb{R}$?
I want to check if one of the following is uniformly continuous and differentiable on $\mathbb{R}$.

$\int_0^{x^3}\sin t\, dt$

$\int_0^x\cos (t^3)\, dt$

$\int_0^x[t]\, dt$

$\int_0^xe^{t^3}\, dt$

To  check that do we have to check if the derivative is Lipschitz continuous?

If yes I suppose that teh second one satisfies that property, since the cosine function is bounded.

答:
Your answer for 2) is correct.

For 1) note that the function is just $1-\cos (x^{3})$. Can you show that this is not uniformly continuous?  [Consider the points $(2n\pi)^{1/3}$ and $(2n\pi+\frac {\pi} 2)^{1/3}$.

For 3)use the fact that you can compute the integral explicitly by splitting the integral into $0$ to $1$, $1$ to $2$ etc. Now check that the values at $n$ and $n+\frac 1 n$ difer by $1$ which shows that the function is not uniformly continuous. I will leave differentiabilty to you.

For 4) here is  a hint: $0<x<y$ implies $\int_x^{y} e^{t^{2}} dt \geq e^{x^{2}} (y-x)$. Choose $x$ and $y$ appropriately to contradict uniformly continuity.


问:
Proving a statement: The set A={a,b,c} is a finite set with three elements.
It is from maths workshop course, we were asked to prove statement below.

The set $A=\{a,b,c\}$ is a finite set containing three elements.

I rewrote this as:

Let A={a,b,c} be a set $\iff$ A is a finite set conaining three elements.$\qquad (\times)$

And began to prove it as implication in two ways.

The lecturer said that my re-statement is incorrect, but that it should be instead:

Let A be a set containing three elements: a,b,c. Then A is finite set with three elements.

or equivalently:

$A=\{a,b,c\}$ is a set $\implies A$ is finite set with three elements.

I wasn't given reason why my restatement $(\times)$ is incorrect. Is lecturer correct by calling it incorrect, could anyone comment on this?

答:
I would say it is because you only stated that $A$ has three elements and these three elements could be anything and not only $\{a,b,c\}$.

So if you assume that A has three elements you cannot conclude that the three elements are $a,b$ and $c$.

However if you concluded that there exists a set $A=\{a,b,c\}$ then it would be more correct.


答:
First of all, it should be noted that $a,b,c$ are distinct, otherwise $A = \{a,b,c\}$ does not contain three elements.

Moreover, the problem in your restatement seems to lie in the direction $\Leftarrow$: it lacks quantifiers. One should write

$$A \text{ is a finite set with three elements} \implies \exists \text{ distinct elements } a,b,c \text{ such that } A = \{a,b,c\}.$$

In other words, you make it seem that $a,b,c$ are fixed from the beginning. If they are, then $\Leftarrow$ is incorrect. If they are not, well, then you'll need to quantify them.

Title: Proving a statement: The set A={a,b,c} is a finite set with three elements.

Body:
It is from maths workshop course, we were asked to prove statement below.

The set $A=\{a,b,c\}$ is a finite set containing three elements.

I rewrote this as:

Let A={a,b,c} be a set $\iff$ A is a finite set conaining three elements.$\qquad (\times)$

And began to prove it as implication in two ways.

The lecturer said that my re-statement is incorrect, but that it should be instead:

Let A be a set containing three elements: a,b,c. Then A is finite set with three elements.

or equivalently:

$A=\{a,b,c\}$ is a set $\implies A$ is finite set with three elements.

I wasn't given reason why my restatement $(\times)$ is incorrect. Is lecturer correct by calling it incorrect, could anyone comment on this?


问:
Is there an "addition theorem" of logarithm?
In trigonometry, there is a relation called the addition theorem:

$\begin{align} \sin(A+B) &= \sin{A}\cos{B}+\cos{A}\sin{B} \\ \cos{(A+B)} &= \cos{A}\cos{B} -\sin{A}\sin{B}\end{align}$

I wondered if there is a similar relation for logarithms. So what I want to know is that, is there a polynomial or a rational function $f(x,y)$ that satisfies
$\log(a+b)= f(\log(a),\log(b))$?(Assume $a>0,b>0$)

I think it is highly unlikely, but I haven't been able to prove that. I also would love to know if we can loosen the restriction on $f(x,y)$ and make it true.

答: if you could express log(x+y) in terms of log(x) and log(y) then it would mean that you could express log(x+1) in terms of log(1) and log(x). Since log(1)=0 then it means only in terms of log(x) suppose that there existed a polynomial in terms of log(x) which would be equal to log(x+1). if you differentiate both sides with respect to x, then you would get $\frac{1}{x+1}=$ a polynomial of $\frac{1}{x}$. But it is impossible. the same procedure works for a rational function. therefore you can't express log(x+y) in terms of a polynomial or a rational function in terms of log(x)  and log(y).


答:
Your approach is the hard way, but it can be done.

Consider the case of one missed target. There are $5$ ways to choose which target is missed. There are $\binom52$ ways to choose which $2$ of the $5$ shooters hit the same target, and there are $4$ ways to choose which target it is. There are then $3!$ ways to assign targets to the remaining $3$ shooters, so there are altogether

$$5\cdot\binom52\cdot4\cdot3!=\binom525!=10\cdot5!$$

ways to get $X=1$. Thus,

$$\Bbb P(X=1)=\frac{10\cdot5!}{5^5}=\frac{2\cdot4!}{5^3}\;,$$

not $\dfrac{5!}{5^5}$.

For two missed targets there are $\binom52$ ways to choose the missed targets, but the rest of the calculation is a bit more complicated, because there are two possibilities.

We can have $3$ shooters hitting one target. There are $\binom53$ ways to choose the shooters and $3$ ways to choose their target, and there are then $2$ ways to assign the other two targets to the other two shooters.

We can have two targets hit by two shooters each. There are $\binom32$ ways to choose which two targets will be hit twice. There are $\binom52$ ways to choose which $2$ shooters will hit the first of these, and there are then $\binom32$ ways to choose which $2$ will hit the other one.

The total count of ways to get $X=2$ is therefore

$$\binom52\left(\binom53\cdot3\cdot2+\binom32\binom52\binom32\right)=1500\;,$$

and

$$\Bbb P(X=2)=\frac{1500}{5^5}=\frac{12}{25}\;.$$

This analysis can be continued, but you can see that it will be quite a bit of work.

The easy approach is to use linearity of expectation on $X=\sum_{k=1}^5X_k$, where $X_k$ is $1$ if target $k$ is missed and is $0$ otherwise; I see that while I was writing up the messy bit, André has already covered this.

Body:
Assume there are 5 shooters and 5 targets. Each shooter would choose a target randomly and when the signal is given all shooters will fire at the same time and hit their target. What would be the expected number of targets which didn't get hit by any of the shooters?

My attempts: Initially, there are $5^5$ different possible case. Now I assume a R.V $X$ be the number of unshot target. So $P(X=0)=5!/5^5, P(X=1)=5!/5^5,P(X=2)=C_2^5\times3!/5^5$ and so on but it seems that the probability I get is wrong but i am not sure which part goes wrong and how to get the correct one. Any solution or hints are welcome, thanks.


问:
Finding a such that a parametric equation has solution
I've been trying to solve this parametric equation:
Find all $a$ for which there exist $x$ and $y$ such that $a(3x+2y) = y + \sqrt{2a^2(4x^2+y^2)} + \sqrt{2a^2x^2+2(a-1)^2y^2}$. I noticed x and y can scale, but trying to assume e.g. $x + y = 1$ has led me nowhere. I can't see any other substitution that would help me. Basic inequalities (AM-GM, Jensen, Cauchy-Schwarz) have also led me nowhere. After all this I have taken a look at the graph of the function on desmos and it looks like there are no such $a$. But how does one prove it?

答:
Hint.

Making $y = \lambda x$ and substituting into

$$
a (3 x+2 y)-\sqrt{2} \sqrt{a^2 \left(4 x^2+y^2\right)}-\sqrt{2 a^2 x^2+2 (a-1)^2 y^2}-y = 0
$$

we have

$$
a (2 \lambda +3) x-\sqrt{2} \left(\sqrt{a^2 \left(\lambda ^2+4\right) x^2}+\sqrt{x^2 \left(a^2+(a-1)^2 \lambda ^2\right)}\right)-\lambda  x=0
$$

or

$$
\left(a (2 \lambda +3)-\sqrt{2} \sigma(x)\left(\sqrt{a^2 \left(\lambda ^2+4\right)}+\sqrt{ \left(a^2+(a-1)^2 \lambda ^2\right)}\right)-\lambda  \right)x=0
$$

where $\sigma(x)$ is the sign function. Now discarding the trivial solution $x=0$, for $\sigma(x)=1$ the solution is $\lambda=a=0$ and for $\sigma(x)=-1$ the solution is $\lambda =a=2$


问:
How can I "see" that calculus works for multidimensional problems?
Let's say I have some function f(x) = x^2 + b. I can see what's going on, I can count the slope geometrically even without knowing the rules of derivatives.

When I need to minimize some cost function for linear problem (linear regression with gradient descent), I just need to picture 3d, and I can "see", and be quite confident why and how it works.

How can I "see" or get intuition that calculus works for multidimensional problems? Let's say I have a problem with many variables like:

f(area_km, bedrooms, ...) = theta + theta1 * area_km + theta2 * bedrooms etc

If I want to apply gradient descent, I know I need to calculate partial derivatives and multiply it with a learning rate etc. It works. But it's kinda magical that it works.

I am sorry this is a silly question, I am just beginning.

答:
Well, in one variable you need to solve $f'(x)=0$ and controll wheter it is maximum or a minimum. Such equation is called Euler equation and it holds also in more variables, but with the formulation:
$$\nabla f(x)=0$$
where $\nabla f(x)=(\frac{\partial f(x)}{\partial x_{1}},\dots,\frac{\partial f(x)}{\partial x_{n}})$. Now, also here you should controll if it is a minimum or not and this is done via checking on the Hessian of $f$. That is, if $\bar{x}$ is such that $\nabla f(\bar{x})=0$ then $\bar{x}$ is a minimum if $Hf(\bar{x})$ admits only positive eigenvalues, where $Hf$ is the matrix made of second order partial derivatives of $f$. Notice that we are assuming $\bar{x}$ belongs to the interior of a set, exactly as for functions with one variable.


问:
Show that stopping time of a Brownian Motion is independent of the particular BM
Let $\{B_t\}$ and $\{\tilde{B}_t\}$ be two Brownian motions starting at $0$. Define the two random variables

$$\tau_a = \inf\{t: B_t =a\}\quad \text{and}\quad \tilde{\tau}_a = \inf\{t: \tilde{B}_t=a\}$$

I want to show that $\mathbb{E}(\tau_a) = \mathbb{E}(\tilde{\tau}_a)$.

My attempt:

The functions $\max_{s\leq t}B_s(\omega)$ and $\max_{s\leq t}\tilde{B}_s$ are clearly measurable by continuity of paths and $\{\tau_a\leq t\} = \{\max_{s\leq t}B_s\geq a\}$ so these functions are measurable. We have that

$$\mathbb{E}(\tau_a) = \int_\Omega\tau_a(\omega)\mathrm{d}\mathbb{P}(\omega) = \int_\mathbb{R}\mathbb{P}(\tau_a>t)\mathrm{d}t=\int_\mathbb{R}\mathbb{P}(\max_{s\leq t}B_s\geq a)\mathrm{d}t=?=\int_\mathbb{R}\mathbb{P}(\max_{s\leq t}\tilde B_s\geq a)\mathrm{d}t = \mathbb{E}(\tilde \tau_a)$$
Now if the max wasn't there I would be done, but I don't see how to continue. I basically need that $\mathbb{P}(\max_{s\leq t}B_s\geq a) = \mathbb{P}(\max_{s\leq t}\tilde B_s\geq a)$ to conclude the argument. Is this a viable method? Could someone give a tip on how to proceed?

答:
Here is an answer based on the hint by @Brian Moehring:

We first show that $\mathbb{P}(\sup_{s\leq t} B(s)\geq a) = \mathbb{P}(\sup_{s\leq t}\tilde{B}(s)\geq a)$. We do this by proving that $\mathbb{P}(\sup_{s\leq t} B(s)\geq a) \leq \mathbb{P}(\sup_{s\leq t}\tilde{B}(s)\geq a)$. Observe that if $\{s_k\}$ is a dense sequence of $[0,t]$ then $\sup_{s\leq t}B(s) = \sup_kB(s_k)$ by the continuity of paths and hence these functions are $\mathscr{A}$-measurable.

We begin by proving that $\mathbb{P}(\max(B_s,B_t)\geq a) = \mathbb{P}(\max(\tilde B_s,\tilde{B}_t)\geq a)$. To see this we let $\epsilon>0$ be arbitrary and begin by choosing $\{r_{k}\}_{-\infty}^{k=0}$ such that $r_k<r_{k+1}<r_k+\epsilon$, $r_0 = a$ and $(-\infty,a) = \bigcup_{-\infty}^{k=-1}(r_k,r_{k+1})$. Assuming that $s<t$ we obtain
\begin{align*}
   \mathbb{P}(\max(B_s,B_t)\geq a)& = \mathbb{P}\Big[(B_s\geq a)\cup \Big((B_t\geq a)\cap (B_s<a)\Big)\Big] \\
   & = \mathbb{P}(B_s\geq a)+ \mathbb{P}\Big((B_t\geq a)\cap (B_s<a)\Big) \\
   & = \mathbb{P}(B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((B_t\geq a)\cap B_s\in (r_k,r_{k+1})\Big) \\
   & \leq \mathbb{P}(B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((B_t-B_s\geq a-r_{k+1})\cap B_s\in (r_k,r_{k+1})\Big) \\
   & = \mathbb{P}(B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((B_t-B_s\geq a-r_{k+1})\Big) \mathbb{P}\Big(B_s\in (r_k,r_{k+1})\Big) \\
   & = \mathbb{P}(\tilde B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((\tilde B_t-\tilde B_s\geq a-r_{k+1})\Big) \mathbb{P}\Big(\tilde B_s\in (r_k,r_{k+1})\Big) \\
   & = \mathbb{P}(\tilde B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((\tilde B_t\geq a+r_k-r_{k+1})\cap(\tilde B_s\in (r_k,r_{k+1}))\Big) \\
   & \leq \mathbb{P}(\tilde B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((\tilde B_t\geq a-\epsilon)\cap(\tilde B_s\in (r_k,r_{k+1}))\Big) \\
   & = \mathbb{P}(\tilde{B}_s\geq a)+\mathbb{P}\Big((\tilde{B}_t\geq a-\epsilon)\cap (\tilde{B}_s<a)\Big)\\
   & \xrightarrow{\epsilon \rightarrow 0} \mathbb{P}(\tilde{B}_s\geq a)+\mathbb{P}\Big((\tilde{B}_t\geq a)\cap (\tilde{B}_s<a)\Big)\\
   & = \mathbb{P}(\max(\tilde B_s,\tilde B_t)\geq a).
  \end{align*}
As mentioned before this shows that $\mathbb{P}(\max(B_s,B_t)\geq a)=\mathbb{P}(\max(\tilde B_s,\tilde B_t)\geq a)$.

Now if $\{t_1,\dotsc,t_n\}$ is a finite sequence with $t_1<t_2<\dotsc<t_n$ we observe that $B_{t_k}-B_{t_{k-1}}$ is independent from $\max_{1\leq j \leq k-1}B_{t_j}$ and hence the above argument suitably modified gives an inductive procedure of showing that
\begin{equation*}
   \mathbb{P}\left(\max_{1\leq j \leq n}B_{t_{j}}\geq a\right) = \mathbb{P}\left(\max_{1\leq j \leq n}\tilde B_{t_{j}}\geq a\right).
  \end{equation*}
Choosing a dense sequence of $[0,t]$ we are thus able to conclude via the continuity of paths property that
\begin{equation*}
   \mathbb{P}\left(\max_{0\leq s\leq t}B_s\geq a\right) = \mathbb{P}\left(\max_{0\leq s\leq t}\tilde B_s\geq a\right)
  \end{equation*}
this in fact implies that $\max_{0\leq s\leq t}B_s$ and $\max_{0\leq s\leq t}\tilde B_s$ have the same distribution. Since $\{\omega: \tau_a(\omega)\leq t\} = \{\omega: \max_{0\leq s\leq t}B_s\geq a\}$ we conclude that $\tau_a$ and $\tilde{\tau}_a$ share distribution.


答:
No, it's the other way round:

$$\exp(a+b)=\exp(a)\exp(b)$$

or

$$\log(ab)=\log(a)+\log(b).$$

In fact, via the complex numbers, the first identity directly leads to the trigonometric ones.

We can refute the hypothesis $$\log(a+b)=f(\log(a)+\log(b))$$ for rational functions with algebraic coefficients.

Indeed,

$$\log(2)=\log(1+1)=f(\log(1),\log(1))=f(0,0)$$ is a transcendental number, but cannot be the result of a rational function with algebraic coefficients applied to algebraic arguments.

Title: Is there an "addition theorem" of logarithm?

Body:
In trigonometry, there is a relation called the addition theorem:

$\begin{align} \sin(A+B) &= \sin{A}\cos{B}+\cos{A}\sin{B} \\ \cos{(A+B)} &= \cos{A}\cos{B} -\sin{A}\sin{B}\end{align}$

I wondered if there is a similar relation for logarithms. So what I want to know is that, is there a polynomial or a rational function $f(x,y)$ that satisfies
$\log(a+b)= f(\log(a),\log(b))$?(Assume $a>0,b>0$)

I think it is highly unlikely, but I haven't been able to prove that. I also would love to know if we can loosen the restriction on $f(x,y)$ and make it true.


问:
Prove that $a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$ for positive reals $a+b=1$

Problem 1: Let $\lambda = \frac{\ln (7 + 3\sqrt{5})}{\ln 2} - 1 \approx 2.77697$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$$
with equality if and only if $a = b = \frac{1}{2}$

Problem 2(a weaker version of Problem 1): Let $\lambda_1 = \frac{25}{9}$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda_1 b} + b^{\lambda_1 a} + a^{\lambda_1 b^2} + b^{\lambda_1 a^2} \le 2.$$

Background Information: In Proposition 5.2 in [1], Vasile Cirtoaje gives the following result:

Problem 3: If $a, b$ are nonnegative real numbers satisfying $a + b = 1$, then $a^{2b} + b^{2a} \le 1$.

In Conjecture 5.1 in [1], Vasile Cirtoaje proposes the following conjecture:

Let $a, b$ be nonnegative real numbers satisfying $a + b = 1$. If $k \ge 1$, then $a^{(2b)^k} + b^{(2a)^k} \le 1$.

The case $k=2$ has been proved:

Problem 4: Let $a, b$ be positive real numbers with $a + b = 1$. Prove that $a^{4b^2} + b^{4a^2} \le 1$.

See If $a+b=1$ so $a^{4b^2}+b^{4a^2}\leq1$

I combine Proposition 5.2 and Problem 4 to come up with the problems.

I may use appropriate bounds to prove Problems 3 and 4 (something like
Inequality $a^{2b}+b^{2a}\leq \cos(ab)^{(a-b)^2}$). Now the problems are more difficult.

Reference

[1] Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938

答:
working the weaker case and using $b=1-a$, we have
$$f(a)=a^{\frac{25}{9} (1-a)}+(1-a)^{\frac{25}{9}a  }+a^{\frac{25}{9} (1-a)^2}+(1-a)^{\frac{25 }{9}a^2}$$
The derivative cancels close to $a=\frac 14$. In fact, the first iterate of Newton method is extremely close to $\frac 6 {25}$

$$f\left(\frac{1}{4}\right)\sim 1.94050 \qquad \text{and} \qquad
f\left(\frac6{25}\right)\sim 1.94029$$


问:
Coordinate change in a vector space
Suppose $A = \{x_1, x_2\}$ is a basis for some vector space $V$, and that $T: V \rightarrow V$ is a linear transformation. Let $B = \{y_1, y_2\}$ be another basis for $V$ with $y_1 = x_1 + x_2$ and $y_2 = 2x_1 + 3x_2$. If $$T_A = \begin{bmatrix} 3 & 1 \\ -2 & 4\end{bmatrix}$$ I wish to find $T_B$. How would I go about doing this? Thanks.

答:
We only have to "read" the images of the vectors $x_1$ and $x_2$ and then write them in function of $y_1$ and $y_2$.

$T(x_1)=3x_1-2x_2$ and $T(x_2)=x_1+4x_2$. Since the map is linear we can now observe that
$$T(y_1)=T(x_1+x_2)=T(x_1)+T(x_2)=4x_1+2x_2$$
$$T(y_2)=T(2x_1+3x_2)=2T(x_1)+3T(x_2)=6x_1-4x_2+3x_1+12x_2=9x_1+8x_2.$$
Now we have $T(y_1)=8y_1-2y_2$ and $T(y_2)=11y_1-y_2$, therefore
$$\mathcal M_B^B(T)=T_B=\begin{pmatrix} 8 && 11\\-2 && -1\end{pmatrix}.$$

In order to find $T(y_i)$ I solved the systems
$$T(y_1)=\alpha y_1+\beta y_2\text{ and }T(y_2)=\alpha y_1+\beta y_2$$

Title: Coordinate change in a vector space


问:
Show that the solution of this (nonlinear) ODE cannot remain bounded as $t\to\infty$
Preliminary properties: Let the state vector $x(t)=[x_1(t),\dots,x_n(t)]^T\in\mathbb{R}^n$ be constrained to the dynamical system
$$
\dot{x} = Ax +
\begin{bmatrix}
\phi_1(x_1) \\
\vdots \\
\phi_n(x_1) \\
\end{bmatrix}, \ \ \ \ x(0) = x_0
$$
where $A$ is defined by:
$$
A =
\begin{bmatrix}
\lambda_1 & 1 & 0 &\cdots& 0\\
0 & \lambda_2 & 1 &\ddots&\vdots\\
\vdots&\ddots&\ddots&\ddots&0\\
0&\cdots&0&\lambda_{n-1}& 1\\
0&\cdots&0&0&\lambda_n
\end{bmatrix}
$$
with $\lambda_i>0$, and $\phi_i(x_1) = \beta_i |x_1|^{\alpha_i}\text{sign}(x_1), \beta_i>0$, $0<\alpha_i<1$.

Question: Is it possible to show that for any initial condition $x_0\neq 0$, the solution $x(t)$ either converge to the origin, or $
\lim_{t\to\infty}\|x(t)\| = +\infty
$, but cannot remain in a bounded trajectory different from staying at the origin?

Concretelly, what additional structure or conditions on the system or the initial condition do we require to show this?

In case you find this useful, here are my attempts to understand/solve the problem.

Attempt 1: I was trying to use results such as the ones from here which can conclude what I want, but require to find a Lyapunov-like function (not necesarilly positive definite) for which $\ddot{V}\neq 0, x\neq 0$. However, I haven't been able to come up with a suitable such function.

Attempt 2: The differential equation have "explicit" solution (not precisely explicit but can be expressed as)
$$
x(t) = e^{At}x_0 + e^{At}\int_0^se^{-As}\Phi(x_1(s))ds
$$
where $\Phi(x_1) = [\phi_1(x_1),\dots,\phi_n(x_1)]^T$. So I wanted to proceed by contradiction: assume that there exists $b,B>0$ and $T>0$ such that $b\leq \|x(t)\|\leq B$ for all $t\geq T$. Hence,
$$
b\leq \left\|e^{At}x_0 + e^{At}\int_0^se^{-As}\Phi(x_1(s))ds\right\|\leq B
$$
And noticing that in this case there should be $c,C>0$ such that $0<c\leq\|\Phi(x_1(t))\|\leq C $, for all $t\geq T$. Thus, try to obtain a contradiction, for example by using $C\geq\|\Phi(x_1(t))\|$ to show that $B\leq\|x(t)\|$. But unfortunately I haven't obtained anything positive in this direction neither.

Attempt 3: Can Bendixon's/Dulac criterion (see Theorem 11  here) be used to conclude something for this system? It is easy to verify that if we write this system as $\dot{x} = f(x)$, we obtain $\nabla\cdot f(x)>0$.

I know that neither my attempts nor my exposition here are perfect. However, I'm looking for suggestions/references or any idea which might help me understand more this problem.

答:
Inspired by the answer of open problem one can say a bit more in general when considering only the $\alpha_i=1$ cases. Although, it is stated that $0 < \alpha_i < 1$, so technically these cases would just barely violate the considered domains for each $\alpha_i$. In these cases the dynamics is linear and can be described with $\dot{x} = M\,x$, with

$$
M =
\begin{bmatrix}
\lambda_1 + \beta_1 & 1 & 0 & \cdots & 0 \\
\beta_2 & \lambda_2 & 1 & \ddots & \vdots \\
\vdots & 0 & \ddots & \ddots & 0 \\
\beta_{n-1} & \vdots & \ddots & \lambda_{n-1} & 1 \\
\beta_n & 0 & \dots & 0 & \lambda_n
\end{bmatrix}. \tag{1}
$$

These kind of systems can have non-zero bounded trajectories if $M$ has at least one eigenvalue of zero. One necessary condition for this would be that $\det(M) = 0$, since the determinant of a matrix is equal to the product of its eigenvalues.

It can be shown that in general the determinant of $(1)$ is equal to

$$
\det(M) = \prod_{k=1}^n \lambda_k + \sum_{k=1}^n \left((-1)^{k+1} \beta_k \prod_{m = k+1}^n \lambda_m\right). \tag{2}
$$

Even though it holds that $\lambda_i,\beta_i > 0$ for all $i = 1, \cdots, n$, due to the minus signs inside $(2)$ it is possible to have that $\det(M) = 0$ for $n \ge 2$.

For example for $n = 2$ with $\lambda_1,\lambda_2,\beta_1 = 1$ and $\beta_2 = 2$ yields

$$
M =
\begin{bmatrix}
2 & 1 \\
2 & 1
\end{bmatrix}, \tag{3}
$$

which has the eigenvalues $0$ and $3$ and thus can have non-zero bounded trajectories if the initial condition $x(0)$ is chosen such that it doesn't excite the unstable mode associated with the eigenvalue $3$.

Another sufficient condition for a counter argument would be if a certain system satisfying your description has other equilibria besides the origin. It can be noted that for the linear cases the mode, whose associated eigenvalue is zero, gives a line of equilibria. For all choices for $\alpha_i$ and using $x_1 = 1$ yields $\phi_i(x_1) = \beta_i$. Therefore, a non-zero equilibrium can be constructed by solving $\dot{x} = 0$. In order to split the knowns from the unknowns I define $x' = \begin{bmatrix}x_2 & \cdots & x_n\end{bmatrix}^\top$, such that $\dot{x} = 0$ can be split into $\dot{x}_1 = 0$ and $\dot{x}' = 0$. Substituting $x_1 = 1$ into those two expressions yields

$$
\lambda_1 + x_2 + \beta_1 = 0, \tag{4}
$$

$$
A'\,x' + B = 0, \tag{5}
$$

with $B = \begin{bmatrix}\beta_2 & \cdots & \beta_n\end{bmatrix}^\top$ and

$$
A' =
\begin{bmatrix}
\lambda_2 & 1 & 0 & \cdots & 0 \\
0 & \lambda_3 & 1 & \ddots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 \\
0 & \cdots & 0 & \lambda_{n-1} & 1 \\
0 & \cdots & 0 & 0 & \lambda_n
\end{bmatrix}. \tag{6}
$$

Solving $(5)$ for $x'$ yields $x' = - A'^{-1} B$. Therefore $B$ can be chosen to ensure that $\beta_i > 0$ for $i=2,\cdots,n$. However, this doesn't ensure that $\beta_1 > 0$. Namely, solving $(4)$ for $\beta_1$ yields $\beta_1 = -\lambda_1 - x_2$, where $x_2$ can be obtained from the solution for $x'$. It can be noted that scaling $B$ by a positive scalar $\gamma$ also scales $x'$ by the same scalar. Therefore, if for some valid $B$ one obtains a negative value for $x_2$ one could always find a large enough $\gamma$ such that after scaling $\beta_1$ would become positive. The inverse of $A'$ from $(6)$ can shown to be equal to

$$
A'^{-1}_{ij} = \left\{
\begin{array}{ll}
\frac{(-1)^{j-i}}{\prod_{k=i}^j \lambda_k} & \text{if}\ j \geq i \\
0 & \text{otherwise}
\end{array}
\right., \tag{7}
$$

where $X_{ij}$ denotes the element of matrix $X$ at its $i$th row and $j$th column. Given that each element of $B$ is positive the expression for $x_2$ would thus be a sum of alternating negative and positive terms. Therefore, by choosing some of the odd terms of $B$ sufficiently large would guarantee that the associated solution for $x_2$ would be negative, which thus ensures that $\beta_1$ can be made positive.

For example for $n = 2$ with $\lambda_1,\lambda_2 = 1$ and $\beta_2 = 2$ yields $x_{eq} = \begin{bmatrix}1 & -2\end{bmatrix}^\top$ as equilibrium for every possible $\alpha_i$. It can be noted that due to the fact the the expression for $\dot{x}$ is odd in $x$ also implies that $-x_{eq}$ (thus $\begin{bmatrix}-1 & 2\end{bmatrix}^\top$) would be an equilibrium as well.

However, I am not sure if for $n \geq 2$ these systems always have multiple equilibrium points for any arbitrary choice for $\lambda_i,\beta_i > 0$. But at least I have shown there exists systems that satisfy your description that violate your postulated limits.


问:
Family of norms ordered by a peculiar relation
Let $X$ be a vector space together with a family $\left( \left \Vert \cdot \right \Vert_k \right)_{k \in \mathbb{N}}$ of norms.
For what I am working on, I have proved that I need a kind of ordering property on the norms such that $\forall k,l \in \mathbb{N} \; \forall x \in X :$

\begin{equation}
k \le l \implies B^k_{\left \Vert x \right \Vert_k} \left( 0 \right) \subseteq B^l_{\left \Vert x \right \Vert_l} \left( 0 \right)
\end{equation}

where $B^k_r \left( 0 \right)$ denotes the open ball of radius $r$ in the $\left \Vert \cdot \right \Vert_k$ norm around $0$.

I have never seen such a property before and so I wonder whether such families of norms have been studied before and if so what name is given to them and perhaps what alternative characterizations might exist.

答:
The given inclusion is equivalent to $\|z\|_k\le\|x\|_k\implies\|z\|_l\le\|x\|_l$ for all $x,z$.

By symmetry this implies $\|z\|_k=\|x\|_k\implies \|z\|_l=\|x\|_l$.

Hence $B^k_r=\alpha B^l_r$, since suppose $\|x\|_l/\|x\|_k=\alpha$ for some fixed $x$. Then for any other $y$ such that $\|y\|_k=r$, $\|y\|_k=\|\frac{rx}{\|x\|_k}\|_k$ implies $\|y\|_l=r\|x\|_l/\|x\|_k=r\alpha$.

So the  norms are equal in the sense that $\|x\|_k=f(k)\|x\|_1$.


问:
What are some examples of subtle logical pitfalls?
Here's an example:

Demonstrating that the assumption $A=B$ leads to a true statement is a vacuous truth.
In order the show that $A=B$, prove that the difference $\Delta =A-B$ is zero.  The subtle change being that $\Delta$ is not assumed to be zero.

What are some other examples of subtle logical pitfalls that the amateur Mathematician should be aware of?

Here is an specific argument that shows how assuming $A=B$ leads to absurdity.

$$
\begin{eqnarray}
2&=&1\\
2-1&=&1-1\\
1&=&0\\
\end{eqnarray}
$$

$$
\begin{eqnarray}
a+b&=&a+b\\
a+1*b&=&1*a+b\\
a+0*b&=&0*a+b\\
a&=&b
\end{eqnarray}
$$

A falsity implies anything. Assuming that the false statement is true implies that the two undefined objects $a$ and $b$ are equal, absurd. However, if we define the difference as $\Delta$, then a true statement is forced.

$$
\begin{eqnarray}
2-1&=&\Delta\\
\Delta &=&1\\
2&=&\Delta +1\\
2&=&2
\end{eqnarray}
$$

答:
I think this is probably the subtlest point I have yet to encounter in set theory.

Let's assume the universe of sets satisfies $\sf ZFC$.

Fact I: For every $T_0$, a finite subset of $\sf ZFC$, we can prove in $\sf ZFC$ that $T_0$ has a model.

Fact II: The compactness theorem is provable from $\sf ZFC$, so if a theory $T$ is such that every finite fragment has a model, then $T$ has a model.

Fact III: If $\sf ZFC$ is consistent, then $\sf ZFC$ cannot prove its own consistency. In particular, if we assume that the universe of sets satisfies $\sf ZFC$, then it is impossible to prove that there is a set which is a model of $\sf ZFC$.

On first sight it seems that the first two facts should contradict the third! But the truth is that the first fact is proved in the meta-theory. And we cannot apply compactness arguments in the meta-theory, and have the results in the theory itself!

This pitfall is $\large\sf\text{confusing between meta-theory and theory}$. It is very easy to fall for it when making your first steps in set theory and consistency results.


问:
Temperature distribution in an infinite cylinder
the Problem
An infinite Cylindrical:
$$\frac{\partial T}{\partial t} =2-3r+\nabla^2T$$
B.C/I.C:
$$\frac{\partial T}{\partial r}(1,\theta,t)=0$$
$$T(r,\theta,0) =1-3r^2+2r^3$$
Hint: Assume axial symmetry to simplify derivations

Well as i see the problem is Z axis independent, can i assume that from the fact  the cylindrical is infinite long? the hint?

Is $\theta$ is variable which need to be considered?
in cylinder which subjected to wind or any natural convection, which perpendicular to the cylinder surface, this is the case because, with different angle, there is temp variation. Here i don't see it happening.

答:
Hint: Consider the definition of $\nabla^2$ in terms of cylindrical coordinates and simplify the PDE based on the symmetry of the problem.

See here for the Laplacian in cylindrical coordinates.
https://en.wikipedia.org/wiki/Laplace_operator


问:
A different group associated with a Lie algebra
Let $\mathfrak g$ be a Lie algebra over a field $k$. Everybody loves to study its automorphism group,

$$\mathrm{Aut}(\mathfrak g) := \{\alpha \in Aut_k(\mathfrak g): [\alpha(x),\alpha(y)] = \alpha([x,y]) \text{ for all } x,y \in \mathfrak g\}$$

which, if $\mathfrak g$ is sufficiently nice, can be an algebraic group over $k$, and for $k \in \{\mathbb R, \mathbb C\}$ might even be a Lie group.

But the following is also a group:

$$\mathrm{F}(\mathfrak g) := \{\phi\in Aut_k(\mathfrak g): [\phi(x),\phi(y)] = [x,y] \text{ for all } x,y \in \mathfrak g\}$$

Note that $-id \in \mathrm{F}(\mathfrak g)$ and hence $\mathrm{F}(\mathfrak g)$ is never trivial for $\mathfrak g \neq 0$ (unless $char(k)=2$, which we shall exclude from now on).

I wondered if it has been studied, and why I have never heard of it. Of course the suspicion is: Because it is "somewhat trivial" or "something known in disguise".

Indeed if we tried something like that for a unital associative $k$-algebra $R$,

$$\mathrm{F}(R) := \{\phi\in Aut_k(R): \phi(x) \cdot \phi(y) = x \cdot y \text{ for all } x,y \in R \}$$

it follows pretty easily that $\mathrm{F}(R) = \{\pm id\}$ is not interesting.

Well:

$\mathfrak g$ is abelian if and only if $\mathrm{F}(\mathfrak g) = \mathrm{Aut}(\mathfrak g)$.

Actually, all $ f \in \mathrm{F}(\mathfrak g) \cap \mathrm{Aut}(\mathfrak g)$ act as the identity on all commutators, hence on the derived subalgebra $[\mathfrak g, \mathfrak g]$. In particular, if $\mathfrak g$ is perfect, $\mathrm{F}(\mathfrak g) \cap \mathrm{Aut}(\mathfrak g) = \{id\}$.

<s>Further, when restricted to $[\mathfrak g, \mathfrak g]$, all $\phi \in \mathrm{F}(\mathfrak g)$ commute with all $\alpha \in \mathrm{Aut}(\mathfrak g)$.</s> (I realised my reason for this was flawed, and I am actually not sure about this.)

A hands-on calculation shows that $\mathrm{F}(\mathfrak{sl}_2(k)) = \{\pm id\}$ and still I suspect that this is the case for a large class of Lie algebras.

Question 1: I suspect from the last point, using something like Jacobson density, or a Lie algebra version of von Neumann's bicommutant theorem, or something else that I just don't see, that for (semi)simple $\mathfrak g$, we have that $\mathrm{F}(\mathfrak g)$ contains only scalars, i.e. actually is as trivial as it can be, $\{\pm id\}$. Is that the case? What is the centraliser of $\mathrm{Aut}(\mathfrak g)$ in $\mathrm{Aut}_k(\mathfrak g)$ for (semi)simple $\mathfrak g$?

However, there is a lot of room between semisimple and abelian Lie algebras. E.g. for the non-abelian two dimensional Lie algebra $\mathfrak g$ with basis $x,y$ and $[y,x] =x$, written in that basis, we have

$$\mathrm{Aut}(\mathfrak g) = \{ \pmatrix{a&b\\0&1} :a \in k^\times, b \in k \}$$

$$\mathrm{F}(\mathfrak g) = \{ \pmatrix{a&b\\0&a^{-1}} :a \in k^\times, b \in k \}$$

in particular $\mathrm{F}(\mathfrak g) \cap \mathrm{Aut}(\mathfrak g) = \{ \pmatrix{1&b\\0&1} : b \in k \}$ is unipotent and (as mentioned before) acts trivially on the derived subalgebra $k \cdot x$; but also, $\mathrm{F}(\mathfrak g)$ and $\mathrm{Aut}(\mathfrak g)$ together generate the group $\{ \pmatrix{a&b\\0&c} :a, c \in k^\times, b \in k \}$ of all $k$-linear automorphisms of $\mathfrak g$ which respect the derived series.

Question 2: Does this "polarisation" phenomenon generalise to something interesting for a bigger class of Lie algebras, i.e. what can one say in general about the intersection $\mathrm{F}(\mathfrak g) \cap \mathrm{Aut}(\mathfrak g)$ and the product $\mathrm{F}(\mathfrak g) \cdot \mathrm{Aut}(\mathfrak g)$ in $\mathrm{Aut}_k(\mathfrak g)$?

答:
Not an answer, but yes, this has been studied, at least on the Lie algebra level. Consider the Lie algebra of the Lie group ${\rm Aut}(\mathfrak{g})$, i.e., ${\rm Der}(\mathfrak{g})$ and similarly the Lie algebra $f(\mathfrak{g})$ of $F(\mathfrak{g})$. Then $f(\mathfrak{g})$ is a Lie algebra of generalized derivations. There is a lot of literature on such spaces. One of the most important references there is the following paper by Leger and Luks:

Leger, Luks: Generalized Derivations of Lie Algebras, Journal of Algebra ($2000$).

For applications see also our paper Post-Lie algebra structures on pairs of Lie algebras.


问:
Conjectural closed-form representations of sums, products or integrals
What are some examples of infinite sums, products or definite integrals that have conjectural closed form representations that were confirmed by numerical calculations up to whatever maximum precision anybody tried but still remain unproved?

I am also interested in values of special functions at certain points that have conjectural representations in terms of simpler functions (e.g. special values of hypergeometric functions, Meijer G-function or Fox H-function that representable in terms of elementary functions and well-known constants like $\pi$, $e$, Catalan, Euler–Mascheroni, Glaisher–Kinkelin or Khinchin).

To give some examples:

Gourevitch conjecture mentioned at MathOverflow:
$$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7\stackrel{?}{=}\frac{32}{\pi^3}.$$

Riemann hypothesis (in an unusual form, also mentioned at MathOverflow)
$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt\stackrel{?}{=}\frac{\pi(3-\gamma)}{32}.$$

Another conjecture from MathOverflow attributed to J. M. Borwein, D. H. Bailey, Mathematics by Experiment: Plausible Reasoning in the 21st Century:
$$\frac{\displaystyle\int_{\pi/3}^{\pi/2}\log\left|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right|dt}{\displaystyle\sum_{n=1}^\infty\left(\frac n7\right)\frac{1}{n^2}}\stackrel{?}{=}\frac{7\sqrt{7}}{24},$$
where $(\frac{n}{7})$ denotes the Legendre symbol.

答:
Some conjectural formulas for $\pi$ are given in:

J. Guillera, "Kind of proofs of Ramanujan-like series"

J. Guillera, "Mosaic supercongruences of Ramanujan-type"

J. Guillera, "About a new kind of Ramanujan type series"

J. Guillera, "Some challenging formulas for $\pi$"

For example (I expressed an infinite sum given in the paper in terms of hypergeometric functions):
$$224 \,
   _5F_4\left(\frac{1}{3},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{2}{3
   };1,1,1,1;8235 \sqrt{5}-18414\right)\\-100 \sqrt{5} \,
   _5F_4\left(\frac{1}{3},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{2}{3
   };1,1,1,1;8235 \sqrt{5}-18414\right)\\-1655540 \,
   _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
   };1,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\+740380
   \sqrt{5} \,
   _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
   };1,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\-1237563
   \,
   _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
   };2,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\+553455
   \sqrt{5} \,
   _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3
   };2,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\stackrel{?}{=}\frac{4}{\pi^2}$$

You can also look at:

J. Borwein, D. Bradley, "Empirically determined Apéry-like formulae for ζ(4n + 3)"

And one more from J. Guillera homepage:
$$\sum_{n=0}^\infty\frac{(5418n^2+693n+29)(6n)!}{(-23887872000)^n n!^6}\stackrel?=\frac{128\sqrt5}{\pi^2}.$$


问:
Possible to cover all vertices of a tri-partite graph with triangles?
I'm designing fictional fruits. Each fruit has three attributes; color, taste and smell. Also, each of the values of the attributes have some compatibility with the values of the other attributes. So, we get a tri-partite graph. An example is shown below. Here, 1 and 2 are color attributes, 3,4 and 5 are taste attributes and 6 and 7 are smell attributes. Also, color-1 is compatible with taste-3 and smell-6 and so on.

I want to tell if its possible to design a certain number of fruits in a way that all levels of all three attributes are covered. For instance, in the example graph shown above, this isn't possible. The only way to cover color-2 is to connect it with taste-5. But then, 5 connects with smell-6 while 2 connects with smell-7. So, we can't cover color-2 with any fruit while satisfying the constraints dictated by the graph above.

The next natural question once we determine feasibility is of course to actually find the fruits (minimal required). I asked a related question on this here: https://cs.stackexchange.com/questions/134766/triangles-covering-all-vertices-of-a-tri-partite-graph

答:
The feasibility question is easy to answer: you just have to check whether every vertex is in a triangle. This entails checking for every vertex whether the neighbourhood of the vertex is an independent set, which can of course be done in polynomial time.

Finding a minimum triangle cover is hard. More specifically, in the special case when the three color classes have the same cardinality $k$, it is NP-hard to decide whether there is a triangle cover of size $k$, see here. (Note that such a triangle cover must necessarily consist of vertex-disjoint triangles.)

The question is then whether you want an algorithm that can solve small instances in a reasonable time, or one that can give an approximate solution for large instances efficiently. A simple approximation algorithm would be just to take a triangle for each vertex in the graph. Let $OPT$ denote the size of a minimum triangle cover and let $x,y,z$ denote the sizes of the color classes in the graph. The resulting triangle cover has size at most $x + y + z \leq 3 \cdot \max(x,y,z) \leq 3 \cdot OPT$.

You can do better by deleting the edges that are not in a triangle, covering the first color class by $x \leq OPT$ triangles and then finding a minimum edge cover of the bipartite graph induced by the other two color classes. Since every edge is in a triangle, the edge cover gives rise to a set of triangles that covers the last two color classes. Notice that a minimum triangle cover also gives an edge cover of size at most $OPT$, so the minimum edge cover has size at most $OPT$ as well. It follows that the triangle cover we constructed has size at most $2 \cdot OPT$.

For better algorithms (be it approximate or exact), you might want to look up the vertex clique cover problem. Note that a triangle cover gives a vertex clique cover of the same size after deleting overlaps, and conversely, after deleting the edges of the graph that are not in a triangle, any vertex clique cover can be made into a triangle cover of the same size. Finding a minimum clique cover is equivalent to optimally coloring the complement graph, and of course there is a vast literature on (the algorithmic aspects of) graph coloring.


问:
Inclusion-exclusion principle problem.
I currently faced a problem I couldn't solve: given sets A, B and C, s.t. $|A \setminus B|=24,|B \setminus C|=17, |C \setminus A|=11, |A \cap B \cap C|=11$ find the cardinality of $A\cup B\cup C$.

What I tried is $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C|$ but I'm stuck here. Thanks in advance.

答:
Hint

$$|A\setminus B| +  |B\setminus C| + |C\setminus A| = ?$$

Title: Inclusion-exclusion principle problem.


问:
Under group action, can we define the stabilizer for a set of points?
Let a group $G$ act on a space $S$. If a point $p$ in the space remains invariant under the action of a subgroup of $G$, we call the subgroup the stabilizer of $p$.

$\text{Stab}(p) = \{g: g \in G, \, p\in S, \, g.p = p\}$

Can the notion of stabilizer be extended to a set of points (say $P$) in the space rather than a single point? i.e. can we define a subgroup in the following way?

$\{g: g \in G, \, P\subset S, \, g.p_i = p_j \, \forall \, p_i, p_j \in P\}$

If so, what is such a group called?

答:
For a subset $S'$ of $S$, define $Stab_1(S') = \bigcap_{p\in S'} Stab(p)$ which is the point-wise stabilizer of $S'$. This is a subgroup of $G$ since the intersection of subgroups is itself a subgroup.

Another set to consider is $Stab_2(S') = \{g\in G\mid g\cdot S'\subseteq S'\}$ which stabilizes the subset $S'$. Can you check whether this set is a subgroup of $G$?


问:
Convergence of sequence of inner product with orthogonal basis
Suppose X is a Seperable Hilbert space and E={$u_{n}$ : $n\in $N} is denumerable orthonormal basis.Let ($x_{n}$) be a sequence in X and $x\in X$.Show that if $\langle x_{n}$,$u_{m} \rangle$ converges to $\langle x$,$u_{m} \rangle$ for all $m\in N $ , then $\langle x_{n},y \rangle$ converges to $\langle x$,$y \rangle$ for all $y\in X$

My attempt:

$\langle x_{n}-x$,$y \rangle$ = $\lim_{m\to\infty}\langle S_{m}$,$y \rangle$ where $S_{m}=\sum_{i=1}^{m}\langle x_{n}-x$,$u_{i} \rangle u_{i}  $

Now, $\lim_{n\to\infty}\langle x_{n}-x$,$y \rangle  = \lim_{n\to\infty}\lim_{m\to\infty} \langle S_{m}$,$y \rangle $

If limits above can be exchanged  , since  $\lim_{n\to\infty} \langle S_{m}$,$y \rangle= 0 $ ,we can conclude the proof. Im not sure if this is correct or when the limits can be exchanged.Does the seperability condition comes in play anywhere?

答:
This is not correct. Let $E=\ell^2(\mathbb{N})$ and $u_m(k)=\delta_{m,k}$. The sequence given by $x_n(k)=n\delta_{n,k}$ satisfies $\lim_{n\to\infty}\langle x_n,u_m\rangle=0$ for every $m\in\mathbb{N}$, but if we take $y(k)=\frac 1 k$, then
$$
\langle x_n,y\rangle=\sum_{k=1}^\infty n\delta_{n,k}\frac 1k=1.
$$


问:
Order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$
Back to easier problems for a bit...

I have been told that it is possible to find the order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ without considering automorphisms of $\mathbb{Z}_{720}$.  Here's what I have:

We have that $\mathrm{Aut}(\mathbb{Z}_{720}) \cong U(720)$.  Then since $720=2^4 \cdot 3^2 \cdot 5$ we have $U(720)=U(16) \oplus U(9) \oplus U(5)$.  Some simple computations show that $U(16)$ is not cyclic, while $U(9)$ and $U(5)$ are, so $U(720) \cong U(16) \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_4$.  Since $\gcd(6,4) \ne 1$, $\mathbb{Z}_6 \oplus \mathbb{Z}_4$ is not cyclic, so this, coupled with the fact that the noncyclic $U(16)$ of order 8 would have a cyclic subgroup of order 4 at the most, gives us that the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ is of order $6$.

Is this correct?

I didn't look at automorphisms of $\mathrm{Aut}(\mathbb{Z}_{720})$ per se, but I did consider elements of $\mathrm{Aut}(\mathbb{Z}_{720})$ up to isomorphism...  Is there a way to derive this result just from looking at the structure of $\mathbb{Z}_{720}$?

Thanks.

答:
If we find an element of maximal order in $Aut({\mathbb Z_{720}})$, then this element will be a generator of the maximal order cyclic subgroup of $Aut({\mathbb Z_{720}})$.

First, we use Thm: $Aut({\mathbb Z_n})\cong U(n)$. You had the first step:

$$Aut({\mathbb Z_{720}})\cong U(720)=U(16\cdot 9\cdot 5)\cong U(16)\oplus U(9)\oplus U(5)\cong {\mathbb Z_2}\oplus{\mathbb Z_4}\oplus{\mathbb Z_6}\oplus{\mathbb Z_4}
$$

We see immediately that the orders of the elements of $U(720)$ can only be $1,2,3,4,6,$ and $12$, since an element from ${\mathbb Z_2}\oplus{\mathbb Z_4}\oplus{\mathbb Z_6}\oplus{\mathbb Z_4}$ has the form $(a,b,c,d)$, where $|a|=1$, $|b|\in \{1,2,4\}$, $|c|\in\{1,2,3,6\}$, and $|d|\in\{1,2,4\}$. Since the maximum order of an element in this form $(a,b,c,d)$ is the least common multiple of the orders of each element, it is easy to see that we see that $12$ is the maximum order of an element in $U(720)$. Hence the maximal order cyclic subgroup of $Aut({\mathbb Z_{720}})$ has order $12$.

We can arrive at the same conclusion considering the properties of $Aut({\mathbb Z_{720}})$ alone:

Let $\phi\in Aut({\mathbb Z_{720}})$. Then the mapping $\phi$ is completely determined by $\phi(1)$, which must be relatively prime to $720$. Note that ${\mathbb Z_{720}}\cong {\mathbb Z_{16}}\oplus {\mathbb Z_9}\oplus {\mathbb Z_5}$. Now the image under isomorphism $\phi(1)=(a,b,c)$, where $a,b$, and $c$ are multiplicative units of ${\mathbb Z_{16}}$,  is a multiplicative unit of ${\mathbb Z_{16}}$, ${\mathbb Z_9}$, and ${\mathbb Z_5}$, respectively. But these groups of units are isomorphic to ${\mathbb Z_2}\oplus {\mathbb Z_4}, {\mathbb Z_6}$, and ${\mathbb Z_4}$ respectively. The order of $\phi$ is the least common multiple of $(|a|,|b|,|c|)$, and therefore is at most $12$.


答:
Hint:

Rewrite

$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C|$

As

$|A\cup B\cup C|=|A|-|A\cap B| +|B|-|B\cap C| + |C| - |A\cap C|+ |A\cap B \cap C|$

Now $|A| - |A \cap B| = |A \setminus B|$. Similarly check other pairs.

Title: Inclusion-exclusion principle problem.

Body:
I currently faced a problem I couldn't solve: given sets A, B and C, s.t. $|A \setminus B|=24,|B \setminus C|=17, |C \setminus A|=11, |A \cap B \cap C|=11$ find the cardinality of $A\cup B\cup C$.

What I tried is $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C| + |A\cap B \cap C|$ but I'm stuck here. Thanks in advance.


答:
By drawing a Venn diagram this can be solved. You will find that all you need to do is add the numbers.

Note that $|A\backslash B|$ denoted the number of elements in $A$ without $B$ and so that includes all the elements in $|A|$ except $|A\cap B|$. Therefore $|A\backslash B|=|A|-|A\cap B|$

Similarly, $|B\backslash C|=|B|-|B\cap C|$ and $|C\backslash A|=|C|-|C\cap A|$

Replace these in your formula and so you only need to add the numbers and the final answer would be $63$.

Title: Inclusion-exclusion principle problem.


问:
Can I divide two equal vectors and assign 1?
I was reading a mechanics textbook where the following 3-step derivation is not convincing.

In the left-hand side of the second step equation, both $\frac{\partial V}{\partial \vec{r}}$ and $\vec{\dot{r}}$ are vectors right? There should have been a dot or cross there.

Assuming there was dot, can I cancel between $d\vec{r}$ in $\frac{\partial V}{\partial \vec{r}}$ and $\frac{d\vec{r}}{dt}$ thus getting $\frac{dV}{dt}$ ?

If my approach was totally wrong, please help me understand how to get the second equation.

Thanks in advance!

答:
No, it is the dot product, and what is happening is as follows

Suppose you are in an $N$ dimensional space, where $V: R^n \to R$ and $\vec{r} = (r_1, r_2,...,r_n)$

Hence, by definition, $$\frac{\partial V}{\partial \vec{r}} = \left(\frac{\partial V}{\partial r_1}, \frac{\partial V}{\partial r_2}, ..., \frac{\partial V}{\partial r_n}\right)$$

Now,

$$\frac{\partial V}{\partial \vec{r}} \cdot \dot{\vec{{r}}} = \sum \frac{dr_i}{dt}\cdot\frac{\partial V}{\partial r_i}$$

We then use the chain rule to simplify this to

$$\frac{\partial V}{\partial \vec{r}} \cdot \dot{\vec{{r}}} = \frac{dV}{dt}$$


问:
A problem about quadrilateral inscribed in circle divide the segment.
Quadrilateral ABCD is inscribed in circle O.

AB and DC intersects at point E, AD and BC intersect at point F.

The area of traingle ABC ,traingle ADC are equal, prove that EH = HF.

I wonder whether this problem can be solved by pencil of conics that pass A,B,C,D.

Which I suppose EF is the X-axis, then EH , FH can be the two root of the formula.

I don't know how to take advantage of the given condition that the area of triangles are equal.

Any help is appreciated! Thanks:)

答:
If we denote the intersection point of $BD$ and $AC$ by $L$, then $L$ will be the midpoint of $BD$.

Then, $\frac{BC}{AD}=\frac{CL}{LD}=\frac{CL}{BL}=\frac{CD}{AB}$

$\triangle EBC\sim \triangle EDA$

$\Rightarrow  \frac{CE}{AE}=\frac{BC}{AD}$

$\Rightarrow   \frac{CE}{AE}=\frac{CD}{AB}$

$\Rightarrow   \frac{CE}{CD}=\frac{AE}{AB}$

Applying Menelaus' Theorem on $\triangle AED$ considering $BCF$ as the transversal gives,

$\frac{AB}{BE}\cdot \frac{EC}{CD}\cdot \frac{DF}{AF}=1$

$\Rightarrow   \frac{AB}{BE}\cdot \frac{AE}{AB}\cdot \frac{DF}{AF}=1$

$\Rightarrow   \frac{AE}{BE}=\frac{AF}{DF}$

Hence, $BD\parallel EF$.

Since $AC$ bisects $BD$, $AC$ must bisect $EF$ as well. Thus, $H$ is the midpoint of $EF$.


问:
maximal/minimal element on >= relation
The definition I found on a book is

```<code>If (A, R) is a poset, then an element x ∈ A is called a maximal element of A
if for all a ∈ A, a != x => x !R a.
</code>```

The contrapositive statement is

```<code>x is a maximal element if for each a ∈ A, x R a => x = a.
</code>```

If it apply to >= relation on A = {1,2} where R = {(2,1),(2,2),(1,1)} is a poset and based on this definition 1 is maximal element. Is this correct? Am I missing something?

答:
Yes, but the book uses the (usual) convention that $(a, b) \in \mathcal R$ (or, with another notation, $a \mathcal R b$) should be understood as "$a$ is smaller than $b$". Of course, it does not really matter, but the definition of "maximal element" that you use makes more sense if you understand the order in this direction.

Now, if we want to reformulate the definition of the book, the second one says that $x$ is a maximal element if for each $a \in A$, $x \leq a \Rightarrow x = a$, meaning that we cannot find an element $a$ greater than $x$ - which is exactly the definition of a maximal element.

Title: maximal/minimal element on >= relation


问:
How to estimate E$[\alpha-\alpha']$ and Var$[\alpha-\alpha']$ conditioned on $\alpha>\alpha'>0$ where $\alpha$ and $\alpha'$ are iid standard normal?
Let $\alpha$ and $\alpha'$ be independent standard normal random variables, how to estimate the conditional mean $\mathbb{E}[\alpha-\alpha'|\alpha > \alpha' >0]$ and variance Var$[\alpha-\alpha'|\alpha > \alpha' >0]$? (a bound would be ok) It's clear that, due to $\alpha > \alpha'$, $\alpha-\alpha'$ is a normal random variable sampled from the positive part, but I am not sure how to deal with the condition $\alpha'>0$.

Body:
Let's rewrite the problem with $X$ and $Y$ instead of $\alpha$ and $\alpha'$.
We need to calculate $E(X-Y|X>Y>0)$ and $E((X-Y)^2|X>Y>0)$  (because $\text{Var}(X-Y|X>Y>0) = E((X-Y)^2|X>Y>0) - (E(X-Y|X>Y>0))^2 $).

We have
$$
\begin{align}
E(X-Y|X>Y>0) &= \frac{E((X-Y)\mathbb{I}_{\{X>Y>0\}})}{P(X>Y>0)}\\
&=8E((X-Y)\mathbb{I}_{\{X>Y>0\}}) \\
&=8\int_{x>y>0}{\frac{x-y}{2\pi}\exp{(-\frac{x^2+y^2}{2}})}dxdy \\
\end{align}
$$
Make a change of varibles $x=r\cos(t)$ and $y=r\sin(t)$ then

$$
\begin{align}
E(X-Y|X>Y>0) &= 8\int_{r>0,0<t<\frac{\pi}{8} }{\frac{r(\cos(t)-\sin(t))}{2\pi}\exp{(-\frac{r^2}{2}})}rdrdt \\
&= \frac{4}{\pi}\int_{r>0}r^2\exp{(-\frac{r^2}{2}})dr \int_{0<t<\frac{\pi}{8}}(\cos(t)-\sin(t))dt \\
&= \frac{4}{\pi}\int_{r>0}r^2\exp{(-\frac{r^2}{2}})dr \int_{0<t<\frac{\pi}{8}}(\cos(t)-\sin(t))dt \\
&= \frac{4}{\pi}*\sqrt{\frac{\pi}{2}} *(-1 + \frac{\sqrt{2 - \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2}) \\
\end{align}
$$

Now, we work with the second term $E((X-Y)^2|X>Y>0)$, we have
$$E((X-Y)^2|X>Y>0) = \frac{E((X-Y)^2\mathbb{I}_{\{X>Y>0\}})}{P(X>Y>0)} = 8E((X-Y)^2\mathbb{I}_{\{X>Y>0\}})$$
But by symmetry, we have $E((X-Y)^2\mathbb{I}_{\{X>Y>0\}}) = \frac{1}{8}E((X-Y)^2)$, then $$E((X-Y)^2|X>Y>0) = E((X-Y)^2)$$
It's obvious that
$$E((X-Y)^2) = E(X^2) +E(Y^2) = 1+1=2$$
Hence, $E((X-Y)^2|X>Y>0) = 2$.


问:
Evaluate the I?
This  is question is inspired from this

find the value   of $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}\sqrt{ \frac{u^4}{a^2} +u^2} du dv$$

My attempt : $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du dv$$

after that im not able to solved this

答:
For

$$\int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du $$

use the substitution $t= \frac{u^2}{a^2}+1$. Then you get the integral

$$\frac{a^2}{2}\int_1^3 \sqrt{t} dt,$$

which is independent of $v$, hence

$$\int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du = 2 \pi \cdot \frac{a^2}{2}\int_1^3 \sqrt{t} dt.$$

Title: Evaluate the I?


问:
Seeing a positive trend line on chart with Negative R^2. How can the trend be positive, but the correlation negative?

Hi! I have the data and scatterplot as shown. I fit a line to it.

The fit line looks positive and yields a function of:

y=0.125715x+(84.663831)

However, when I calculate the r squared I get -29.3.
This is true if I calculate by hand or using a function.

How do I interpret this result? I agree that the chart looks noisey enough to not have a strong r squared, but the strong trend line is still there.

Is it possible this is a polynomial distribution and my r2 is not fit for this?

答: The answer was a mistake in my use of the function. I was not using the model's predicted output, and instead reusing the initial data!</p>


问:
When does l'Hospital's rule work for series?
This question stems from a response a colleague of mine received as a (flawed) solution to a problem on his calculus exam. The student was to determine convergence of a series of the form $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$. The student instead considered the series  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ and thought that this series and the original must behave the same way. This, of course, is false in general.

What conditions on $f$ and $g$ are required so that the series  $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$ converges iff  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ converges?

答:
In my oppinion any such result would be artificial and make very little mathematical sense (excepting of course for students).

How would we use such a result? Given the series $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$, since $f,g$ are defined only for integers, what does $f'(n)$ and $g'(n)$ mean?

Backwards same problem, how do we integrate a function defined only on positive integers.

Of course, most of the exercises the students meet involve only elementary functions, so for students is pretty natural to derivate/integrate such sequences; yet in reality any function defined on the integers has uncountably many differentiable extensions to the real numbers.
Any such test would either need to work for all of them (I highly doubt it), or to excluse most of them.

Last but not least, keep in mind given any function of the type $\frac{f(n)}{g(n)}$ defined on the integers, it is easy to extend it (probably in countably many ways) to differentiable functions $f,g$ on ${\mathbb R}$ so taht $f'(n)=0$ and $g'(n)\neq 0$....
And this can be done most of the times so that basic properties of $f,g$ and $\frac{f}{g}$ such as monotony are presearved....

Just to include a simple example to emphasise the problem:

Look to the series $\sum \frac{1}{n^2}$.

What would the derivative series be be? Obvious right $\sum \frac{0}{2n}$...

Or since our series is also the same as  $\sum \frac{1+\sin(2n \pi)}{n^2}$ then the derivative series should be $\sum \frac{2 \pi \cos(2n \pi)}{2n}=\sum \frac{ \pi }{n}$.

These two derivatives have completely different behavior... And if you think this is easy, if $h(x)$ is any differentiable function then our originar series is also:

$$\sum \frac{1+\sin(2n \pi)h(n)}{n^2} \,.$$

and now things really get messy....

Added

Asking for $f$, $g$ to be monotonic doesn't change too much, is just that counterexamples are a little harder to build. Lets say that $f$ is decreasing.

Fix an $n$. Then the function $h: [n,n+1]$ defined by

$$h(x)= \frac{f(n)+f(n+1)}{2}+ \frac{f(n)-f(n+1)}{2}\cos((x-n)\pi) \,.$$

is decreasing, and $h(n)=f(n), h(n+1)=f(n+1)$ wile $h'(n)=h'(n+1)=0$.

Doing this over each interval, you can replace any decreasing $f$ by a new function, which is monotonic and whose derivative is identically zero on the integers...

And more complicated examples can be constructed so the derivative is any negative sequence, and I am pretty sure this can be done to be even $C^\infty$.


答:
If $\gcd(n,m)=1$, note that $$\mathbb Z_n\oplus\mathbb Z_m\cong\mathbb Z_{nm}.$$
Therefore $$\mathbb Z_4\oplus\mathbb Z_6\cong \mathbb Z_4\oplus\mathbb Z_2\oplus\mathbb Z_3\cong \mathbb Z_2\oplus\color{red}{\mathbb Z_{12}}.$$</p>

Title: Order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$

Body:
Back to easier problems for a bit...

I have been told that it is possible to find the order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ without considering automorphisms of $\mathbb{Z}_{720}$.  Here's what I have:

We have that $\mathrm{Aut}(\mathbb{Z}_{720}) \cong U(720)$.  Then since $720=2^4 \cdot 3^2 \cdot 5$ we have $U(720)=U(16) \oplus U(9) \oplus U(5)$.  Some simple computations show that $U(16)$ is not cyclic, while $U(9)$ and $U(5)$ are, so $U(720) \cong U(16) \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_4$.  Since $\gcd(6,4) \ne 1$, $\mathbb{Z}_6 \oplus \mathbb{Z}_4$ is not cyclic, so this, coupled with the fact that the noncyclic $U(16)$ of order 8 would have a cyclic subgroup of order 4 at the most, gives us that the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ is of order $6$.

Is this correct?

I didn't look at automorphisms of $\mathrm{Aut}(\mathbb{Z}_{720})$ per se, but I did consider elements of $\mathrm{Aut}(\mathbb{Z}_{720})$ up to isomorphism...  Is there a way to derive this result just from looking at the structure of $\mathbb{Z}_{720}$?

Thanks.