问:
Is this sextic equation solvable?
I have the equation $x^6 + 3 x^5 + 3 x^4 + 3 x^3 + 2 x^2 + 1 = 0$ and I want to find the roots in terms of radicals, if possible.

According to Wikipedia, "It follows from Galois theory that a sextic equation is solvable in term of radicals if and only if its Galois group is contained either in the group of order 48 which stabilizes a partition of the set of the roots into three subsets of two roots or in the group of order 72 which stabilizes a partition of the set of the roots into two subsets of three roots."

I've tried to figure out what Galois group my equation is in, without success. Galois theory is way over my head.

Is this equation solvable or not?

答:
You can compute Galois groups of polynomials using the open-source computer algebra system Sage. The documentation for Sage's tools for working with Galois groups can be found here. To give an example:

```<code>A.<x> = QQ[]   # defines A = Q[x] to be the polynomial ring in x over the rationals
poly = x^6 + 3*x^5 + 3*x^4 + 3*x^3 + 2*x^2 + 1
G = poly.galois_group()
G.is_solvable()
</code>```

If you want to do the computation by hand (or understand how computer implementations work), the book A Course in Computational Number Theory by Cohen (Graduate Texts in Mathematics 138, Springer-Verlag, 1993) is a good reference for efficient algorithms. In particular, section 6.3.5 deals with the case of determining the Galois group of a sextic polynomial. The method described there uses resolvent polynomials to distinguish between the 16 possible transitive subgroups of $S_6$.

Title: Is this sextic equation solvable?

Body:
You can compute Galois groups of polynomials using the open-source computer algebra system Sage. The documentation for Sage's tools for working with Galois groups can be found here. To give an example:

A.<x> = QQ[]   # defines A = Q[x] to be the polynomial ring in x over the rationals
poly = x^6 + 3*x^5 + 3*x^4 + 3*x^3 + 2*x^2 + 1
G = poly.galois_group()
G.is_solvable()

If you want to do the computation by hand (or understand how computer implementations work), the book A Course in Computational Number Theory by Cohen (Graduate Texts in Mathematics 138, Springer-Verlag, 1993) is a good reference for efficient algorithms. In particular, section 6.3.5 deals with the case of determining the Galois group of a sextic polynomial. The method described there uses resolvent polynomials to distinguish between the 16 possible transitive subgroups of $S_6$.


问:
Grasping the practical usage of Central Limit Theorem
I've been reviewing the Central Limit Theorem in statistics and I'm having trouble intuitively grasping how it can be practically used. The theorem tells us, subject to constraints, that the distribution of the sample mean will be approximately normal, even if the population distribution is not. From what I understand, this remarkable fact now allows us to use inferential statistical methods that apply to the normal distribution to answer questions about the population distribution. But I struggle with intuitively grasping why this is possible? That is, if I want to answer some question about a random variable X of unknown distribution, how does working with the sample mean of X get me to the answer about X? Is there a step missing that allows me to infer properties of X from properties of the sample mean of X?

答:
Here is an example which, if you work it through, will make you appreciate CLT.
Say, you toss a coin 10,000 and you got heads 5,600 times.
The question is this coin fair? Intuitively, number of heads and tails for a fair coin should be about fifty-fifth. In this case we have 5.600 and 4,400 so one would think ... yeah, there is nothing wrong with this coin.
But if you take a look at i.i.d random variables with probability of 1 set to $1/2$ and probability of -1 set to $1/2$ and estimate the probability that their sum is bigger or equal to 600 for "large N" (here N = 10,000) [This is were you would use CLT ] you will get that probability of such event is of magnitude 1% (or less).
In words, it is highly unlikely that a fair coin would produce 5,600 heads in 10,000 trails.
This is how one would use CLT in general.
You test a hypothesis by modeling a sequence of i.i.d, use CLT and then compare it with empirical data.

ContentLicense: CC BY-SA 4.0


问:
Calculate $\sum\limits_{k=0}^{n} \frac{1}{(k+1)(k+2)} \binom{n}{k}$
Any input on how to solve this?

$$\sum\limits_{k=0}^{n} \frac{1}{(k+1)(k+2)} \binom{n}{k}$$

I have tried using Pascal's triangle but wasn't able to get anywhere.

答:
Hint:

$$\frac{1}{(k+1)(k+2)}\binom{n}{k} = \frac{1}{(k + 1)(k + 2)}\frac{n!}{k!(n-k)!} = \frac{n!}{(k+2)!(n - k)!}$$

$$ = \frac{1}{(n+1)(n+2)}\frac{(n + 2)!}{(k + 2)!((n + 2) - (k + 2))!} = \frac{1}{(n + 1)(n + 2)}\binom{n + 2}{k + 2}$$

Notice that the part in front of the binomial coefficient is now independent of the summation variable, so you can pull it out of the summation.

Can you take it from here?


答:
You are quite right. In a way, the CLT smoothes out all characteristics of the distribution of $X$, as you always end-up with a normal. All information is lost but $\mu,\sigma$.

On another hand, the CLT is very powerful in studying the properties of the mean of a sample quantitatively, even without knowledge of the distribution. Like Chebyshev's inequality, it is universal.

Title: Grasping the practical usage of Central Limit Theorem

Body: I've been reviewing the Central Limit Theorem in statistics and I'm having trouble intuitively grasping how it can be practically used. The theorem tells us, subject to constraints, that the distribution of the sample mean will be approximately normal, even if the population distribution is not. From what I understand, this remarkable fact now allows us to use inferential statistical methods that apply to the normal distribution to answer questions about the population distribution. But I struggle with intuitively grasping why this is possible? That is, if I want to answer some question about a random variable X of unknown distribution, how does working with the sample mean of X get me to the answer about X? Is there a step missing that allows me to infer properties of X from properties of the sample mean of X?


问:
A triangle $ABC$ with $c=60,l_c=10$ and $P=125$
A triangle $ABC$ is given with $c=60,l_c=10$ and $P=125$, where $AB=c$, $l_c$ is the angle bisector of $\measuredangle C$ and $P$ is the perimeter. Find $a$ and $b$.

Here are my thoughts:

We have $\begin{cases}\dfrac{m}{n}=\dfrac{b}{a}\\l_c^2=ab-mn\end{cases}.$ From the first equation we can get $$\dfrac{m+n}{n}=\dfrac{a+b}{a}$$ or $$\dfrac{12}{n}=\dfrac{13}{a}.$$ Then we can substitute $l_c=10$ into the second equation to get $$100=ab-mn.$$ I am stuck here. Thank you in advance!
Katherine

答:
We are given the system of equations $$\begin{cases}m+n=60\\a+b=65\\m\cdot a=n\cdot b\\100=ab-mn\end{cases}$$

You can solve it, for instance, taking $m=60-n$ and $a=65-b$ into the third and fourth equations, which will yield two equations with two unknowns: $$\begin{cases}(60-n)\cdot (65-b)=n\cdot b\\100= (65-b)\cdot b-(60-n)\cdot n\end{cases}\iff \begin{cases}n=60 - \frac{12}{13}\cdot b\\100= (65-b)\cdot b-(60-n)\cdot n\end{cases}$$ Where the last equation is a quadratic in $b$ after substituting the first equation. This will yield $b$. You can solve for $a$ analogously.

I got

 $(a,b,m,n)=(13,52,48,12)$ or $(a,b,m,n)=(52,13,12,48)$


问:
What is the greatest number of baskets that Mikayla can make if she wants to put the same amount in each?
She had 60 apples and 72 pears and we want to split them equally into baskets.
I'm kinda confused because I'm in third grade and I'm learning this.</p>

答: Because the Greatest Common Divisor (GCD) of 60 and 72 is 12, Mikayla can make 12 baskets of apples and oranges, with each basket having 5 apples and 6 oranges.</p>


问:
Using Fourier Series of $|x|$ to find $\sum \frac{1}{n^2}$
I found the Fourier series of $f(x) = |x|$ to be $$f(x) = \frac{\pi}{2} + \sum_{n = 1}^\infty 2 \cos(nx) \frac{-1 + (-1)^n}{\pi n^2}$$

Taking $x = 0$, prove that a) $$ \sum_{n: odd \geq 1} \frac{1}{n^2} = \frac{\pi^2}{8}$$ and b) $$\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

Problem a) is easy is to show if you assume changing the index of $n$ does not change the equality (which it should). In that case you obtain $$0 = \frac{\pi}{2} + \sum_{n:odd \geq 1} \frac{-4}{\pi n^2} \rightarrow \sum_{n:odd \geq 1} \frac{1}{n^2} = \frac{\pi^2}{8} $$ But the Fourier series representation I found for $f$ has a sum over all $n \geq 1$ and not just the odd $n$.

For problem b) Im not sure what to do about the $(-1)^n$ term in the numerator.

答:
About a), you get only odd $n$'s because $-1+(-1) ^n$ is zero for even $n$!

About $(b) $, if you define

$$A= \sum_{n \ge 1 \text{ even}} \frac{1}{n^2} ,\ \ \  B =\sum_{n\ge 1 \text{ odd}} \frac{1}{n^2} $$

Then you have, setting $n=2m$ for $n$ even:

$$ A = \sum_{\ge 1 \text{ even}} \frac{1}{n^2} = \sum_{m\ge 1 } \frac{1}{4 m^2} = \frac{1}{4} ( A+B) $$

Can you continue from here?


问:
$\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$
Let $\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$,$s_k$ denotes the partial sum then could anyone tell me which of the following is/are correct?

$1.$ $s_k=0$ for infinitely many $k$

$2$. $s_k>0$ and $<0$ for infnitely many $k$

3.$s_k>0$ for all $k$

4.$s_k>0$ for all but finitely many $k$

if we take $a_n=(-1)^n{1\over n}$ then $\sum a_n$ is convergent but not absolutely convergent,but I don't know $\sum_{n=1}^{\infty} a_n=0$? so I am puzzled could any one tell me how to proceed?

答:
It is easy to see that the first statement is wrong. The idea for a counterexample is in 3.

For the second statement use Riemann's rearrengement theorem, which (if you unterstood the proof) gives you the existence of such a series.

For the third statement construct $a_k$ such that $s_k=(-1)^k \cdot \frac{1}{k}$.

The fourth look at the third.

On the other hand all of the things can be true.

For the first let $(b_n)_{n\in \mathbb{N}}$ be an arbitrary null sequence. Define $a_n$ via
\[a_n= \begin{cases} b_k & 2k=n\\
-b_k & 2k+1=n
\end{cases} \]
We see that
\[  \sum_{n=0}^\infty a_n = 0\]
and furthermore
\[\sum_{n=1}^{2N+1} a_n=0\]
for any $N\in \mathbb{N}$. If you chose a "slow" enough null sequence it won't be absolute convergent.

For the second to be true use the series constructed in the first part at 3.

For the third part make something that a subsequence of $s_k$ is $\frac{1}{k}$, and blow up the rest with $n$ terms with values $\pm \frac{1}{n}$ and let $n \to \infty$ while $k\to \infty$.

The fourth is solved by the third.

ContentLicense: CC BY-SA 3.0


问:
What does meromorphic continuation mean?
Sorry if this question is slightly out of context.

I've learned that $\zeta(k)=\sum\limits^{\infty}_{z=1} \frac{1}{z^k}$ has meromorphic continuation to $Re(s)>0$.

I know that, from convergence test, $\zeta(s)$ converges when $Re(s)>1$. Also, it is clear that $\zeta(k)$ diverges at $k=0.5$. Meromorphic continuation of $\zeta(s)$ to $Re(s)>0$ means that we would have some meromorphic function $f(s)$ that has the same value with $\zeta(s)$ at $Re(s)>1$ but $f(s)$ can be defined at $Re(s)>0$, with the "compensation" of having some poles. Is this correct?

答: Yes, but actually no such analytic (or even continuous) continuation exists, since$$\lim_{s\to1^+}|\zeta(s)|=\infty.$$But it turns out that it has an analytic (and even meromorphic) continuation to $\Bbb C\setminus\{1\}$; in particular, it has an analytic continuation to$$\{z\in\Bbb C\mid\operatorname{Re}(z)>0\}\setminus\{1\}.\tag1$$ Actually, it's not hard to prove that such an analytic continuation can be expressed as$$z\mapsto\frac{\sum_{n=1}^\infty(-1)^{n-1}n^{-z}}{1-2^{1-z}}$$for each $z$ in $(1)$ such that $2^{1-z}\ne1$.


问:
Legendre 3 Square problem
I got a little help on the way here yesterday but it seems like my question is dead.

Problem:

If $ n \in \mathbb{N} $ can be represented as $ n = n_1^2 + n_2^2 + n_3^2 ,\quad n_1, n_2, n_3 \in \mathbb{Z}, $ show that then $ n \neq 4^a(8m+7)$ for $a,m \in \mathbb{Z}.$

Solution:

$ 4^a(8m+7) \equiv 0 \mod 4, \forall a \geq 1.$ If $ n_i $ is an odd square, $ (n_i)^2 = (2k +1)^2 = 4(k^2+1) + 1 \equiv 1 \mod 4.$

So $n_1, n_2, n_3 $ must be even squares. So we can write $ (\frac{n_1}{2})^2 +(\frac{n_2}{2})^2 + (\frac{n_3}{2})^2 = \frac{n}{4} $, which yields to

$$ \frac{n}{4} = 4^{a-1}(8m+7). $$

Now I want to apply $ \mod 8 $ somehow I believe.

We got the easier parts: $ (8m+7) \equiv 7 \mod 8 ,\quad 4^{a-1} \equiv \begin{cases} 4 \quad a =2\\ 0 \quad a > 2\end{cases} $.

So $ 4^{a-1}(8m+7) \equiv 4  \quad \text{or} \quad 0 \mod 8. $

How do I proceed? $n$ is even, $ \frac{n}{4} \equiv  \quad ? \mod8$. Any help is muy appriciated! <3

Edit;

The squares in $ \mathbb{Z_8} $:

$ 0^2 = 0$

$ 1^2 = 1 $

$ 2^2 = 4 $

$3^2 = 1 $

$4^2 = 0 $

$5^2 = 1 $

$ 6^2 = 4 $

$7^2 = 1 $

With the sum of three squares, we can never reach $ 7 \mod 8$?

答:
What you have done so far is to prove that:

Let $a \geq 1$. If $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}4 = 4^{a-1}(8m+7)$.

Applying this observation $a$ times, one notices that if $n = 4^a(8m+7)$ can be written as a sum of three squares, then so can $\frac{n}{4^a} = 8m+7$. Thus, if we prove that no number of the form $8m+7$ can be written as a sum of three squares, then no number of the form $4^a(8m+7)$ can be written as a sum of three squares either.

So consider a number of the form $8m+7$. Modulo $8$, it is congruent to $7$.

As you've noticed, the squares modulo $8$ are $0$, $1$ and $4$. So if $8m+7 = x_1^2+x_2^2+x_3^2$ with $x_j \in \mathbb Z$, then

$$8m + 7 \equiv 7 \equiv x_1^2 + x_2^2 + x_3^2 \pmod 8,$$

which is not a solvable equation, since $7$ cannot be written as the sum of $0$, $1$ and $4$, using three summands (repetitions allowed).

Title: Legendre 3 Square problem


问:
Recurrence relation for Stirling number of first kind with a minimum value
As it states in Wikipedia: Stirling numbers of the first kind count permutations according to their number of cycles. And the recurrence relation is as follows:

$\left[{n+1\atop k}\right] = n \left[{n\atop k}\right] + \left[{n\atop k-1}\right]$

I was wondering how the recurrence relation will look like if there was a lower bound on the size of the cycle ( at least x nodes in a cycle ).

答:
This are called Associated Stirling numbers of the first kind, they are denoted by ${n\brack k}_{\geq x}$. The problem is then that you can not just put the ${n\brack k-1}$ because that corresponds to a fix point (cycle length = 1) so you have
$${n+1\brack k}_{\geq x}=n{n\brack k}_{\geq x}+{n+1-x\brack k-1}_{\geq x}\cdot \binom{n}{x-1}(x-1)!,$$
where the last multiplication corresponds to take $x-1$ elements less than $n+1$ and put them with $n+1$ in a cycle of length $x,$ the $(x-1)!$ takes care of the cyclic order.

Notice that if you take $x=1,$ you recover the normal Stirling numbers and if you set $x=2$ and add, you get the well known recurrence for derangements.


问:
Wilcoxon ranked sign test and the pseudomedian
I need help understanding the Wilcoxon signed rank test and the pseudomedian. More concisely about the confidence interval of the pseudomedian.

Say I have the observations $-2,3,6,10,20$.

I will find the pseudomedian for which I first need the Walsh averages $\frac{X_{i}+X_{j}}{2}$:

$-2,0.5,2,3,4,4.5,6,6.5,8,9,10,11.5,13,15,20$

From here it's clear that the pseudomedian is 6.5 (the measurement with equal distance from both ends). How can I find the confidence at $95\%$ intervals for this pseudomedian?

答:
With only $n = 5$ observations, you are right about the Walsh averages and the pseudomedian. Also, the (only approximately) 95% CI from the Wilcoxon signed rank test is $(-2, 20).$ In R:

```<code>x = c(−2,3,6,10,20)
wilcox.test(x, conf.int=T)

        Wilcoxon signed rank test

data:  x
V = 14, p-value = 0.125
alternative hypothesis:
  true location is not equal to 0
95 percent confidence interval:
 -2 20
sample estimates:
(pseudo)median
           6.5
</code>```

It is not clear whether you are more interested
in the theory and computations for Wilcoxon's signed
rank test, or whether you are more interested in
making sense of an actual small dataset. If the latter,
please provide actual data.

For the sample you give in your question, one kind of a 95% nonparametric bootstrap CI is $(-8,14),$ which may be
more useful than the CI from the Wilcoxon SR test.

```<code>set.seed(128)
h.obs = median(x); h.obs
[1] 6
d.re = replicate(5000, median(sample(x, 6, rep=T)) - h.obs)
UL = quantile(d.re, c(.975,.025))
h.obs - UL
97.5%  2.5%
-8    14
</code>```

Five observations is hardly enough for a reliable
bootstrap CI for the median and I would not bet much
on 95% coverage probability of the resulting CI. [Among the $B = 5000$ re-samples, there were only five uniquely different values of d.re.]

A 95% nonparametric bootstrap CI for the population mean
is $(0.4, 13.4).$ For various reasons, I have a little
more faith in this CI. One of the reasons is that your
sample of five observations is hardly symmetrical.

```<code>x = c(−2,3,6,10,20)
set.seed(128)
a.obs = mean(x); a.obs
[1] 7.4
d.re = replicate(10000, mean(sample(x, 5, rep=T)) - a.obs)
UL = quantile(d.re, c(.975,.025))
a.obs - UL
97.5%  2.5%
  0.4  13.4
</code>```

Body:
With only $n = 5$ observations, you are right about the Walsh averages and the pseudomedian. Also, the (only approximately) 95% CI from the Wilcoxon signed rank test is $(-2, 20).$ In R:

x = c(−2,3,6,10,20)
wilcox.test(x, conf.int=T)

        Wilcoxon signed rank test

data:  x
V = 14, p-value = 0.125
alternative hypothesis:
  true location is not equal to 0
95 percent confidence interval:
 -2 20
sample estimates:
(pseudo)median
           6.5

It is not clear whether you are more interested
in the theory and computations for Wilcoxon's signed
rank test, or whether you are more interested in
making sense of an actual small dataset. If the latter,
please provide actual data.

For the sample you give in your question, one kind of a 95% nonparametric bootstrap CI is $(-8,14),$ which may be
more useful than the CI from the Wilcoxon SR test.

set.seed(128)
h.obs = median(x); h.obs
[1] 6
d.re = replicate(5000, median(sample(x, 6, rep=T)) - h.obs)
UL = quantile(d.re, c(.975,.025))
h.obs - UL
97.5%  2.5%
-8    14

Five observations is hardly enough for a reliable
bootstrap CI for the median and I would not bet much
on 95% coverage probability of the resulting CI. [Among the $B = 5000$ re-samples, there were only five uniquely different values of d.re.]

A 95% nonparametric bootstrap CI for the population mean
is $(0.4, 13.4).$ For various reasons, I have a little
more faith in this CI. One of the reasons is that your
sample of five observations is hardly symmetrical.

x = c(−2,3,6,10,20)
set.seed(128)
a.obs = mean(x); a.obs
[1] 7.4
d.re = replicate(10000, mean(sample(x, 5, rep=T)) - a.obs)
UL = quantile(d.re, c(.975,.025))
a.obs - UL
97.5%  2.5%
  0.4  13.4


问:
Show that a vector that is orthogonal to every other vector is the zero vector
I have the following question, and I'd like to get some tips on how to write the proof. I know why it is, but I'm still not so great at writing it mathematically.

If $u$ is a vector in $\mathbb{R}^n$ that is orthogonal to every vector in $\mathbb{R}^n$, then $u$ must be the zero vector. Why?

I'm starting off like this, but I don't know if it's the right way to do it, or if it is and I just don't know how to continue.

\begin{align}
(\exists u\in\mathbb{R}^n)(\forall v\in\mathbb{R}^n)[\text{u is orthogonal to v}]&\iff u\cdot v=0\\
&\iff ?
\end{align}

From here, instinctively I want to divide both sides by $v$, but I don't know if there is such a thing as dividing a dot product.

答:
Hint: what would it mean for a vector to be orthogonal to itself?

If the problem says "every other vector" as in the title (but not the quoted text) you could instead consider what it means for a given vector $\vec{u}$ to be orthogonal to multiples of itself, e.g. to $-\vec{u}$.


问:
how to find bases for subspace span of $\mathbb{R}^3$
I have searched the internet, but I found contradicting answers.

Must the number of bases be exactly 3 if the space is $\mathbb{R}^3$? What if there is 1 redundant vector in the span? Will the bases then be 2 now?

For example, for subspace span{(1,2,3), (4,5,6), (7,8,9)} in $\mathbb{R}^3$, am I right to say the bases can be {(1,2,3),(4,5,6)} or perhaps {(1,2,3), (7,8,9)}?

But wouldn't that mean there are 2 bases, must there be 3 bases in $\mathbb{R}^3$?

答:
If we take the set of vectors you wrote $\mathcal A= \Bigg\{\begin{pmatrix}1\\2\\3 \end{pmatrix},\begin{pmatrix}4\\5\\6 \end{pmatrix},\begin{pmatrix} 7\\8\\9\end{pmatrix}\Bigg\}$ we can observe that
$$M=\begin{pmatrix}1&4&7\\2&5&8\\3&6&9 \end{pmatrix}\sim\begin{pmatrix}1&4&7\\0&-3&-6\\0&0&0 \end{pmatrix}\implies\text{ rk}(M)=2$$
and this shows that the three vectors are not linearly independent. They span a vector space of dimension $2$.

For example $\mathbb R^3\ni\begin{pmatrix} 0\\0\\1\end{pmatrix}\notin \langle \begin{pmatrix}1\\2\\3 \end{pmatrix},\begin{pmatrix}4\\5\\6 \end{pmatrix}\rangle$, so $\mathcal A$ is not a basis for $\mathbb R^3$.

An example of basis for $\mathbb R^3$ is
$$\mathcal C=\{e_1,e_2,e_3\}=\Bigg\{\begin{pmatrix}1\\0\\0 \end{pmatrix},\begin{pmatrix}0\\1\\0 \end{pmatrix},\begin{pmatrix}0\\0\\1 \end{pmatrix} \Bigg\}$$

For definition a set of vector $\{v_1,\dots,v_n\}$ is a basis for a vector space $V[\mathbb K]$ if:

$(i)$ $\{v_1,\dots,v_n\}$ is a system of generators of $V$;

$(ii)$ $v_1,\dots,v_n$ are linearly independent vectors;


问:
Conditional probability - voting
There is a population that consists of a mixture of "Unchangeables" and "Changeables". If you choose a person at random, then the probability that they're an Unchangeable is $p$ and the probability that they're a Changeable is $1-p$.

Every person is asked repeatedly to vote either "yes" or "no" for a proposal (which doesn't change). Unchangeables will always vote the same way on the same proposal. However, each time a Changeable votes, the probability that they will change their mind from their previous vote is $r$ and the probability that it will stay the same is $1-r$. (That is for example if a Changeable votes "yes" one time, then the probability that they vote "yes" next time is $1-r$ and "no" is $r$).

A randomly chosen person is noticed to have voted the same way on the proposal twice in succession. What is the probability that they will vote in the same way next time?

So far I've got that there are 4 possible categories of people, each combination of Unchangeable and Changeable that votes either twice the same or twice different in succession (at a particular instance), so this table should give the probabilities for getting each type of person if you pick someone randomly from the population.

twice same
twice different

Unchangeable
$p$
$0$

Changeable
$(1-r)(1-p)$
$r(1-p)$

That way, you'd expect to get $$P(\text{unchangeable } |\text{ voted same way twice}) = \frac{p}{(1-r)(1-p)+p}$$ and $$P(\text{changeable } |\text{ voted same way twice}) = \frac{(1-r)(1-p)}{(1-r)(1-p)+p}$$

and then combine these linearly to get $$P(\text{will vote the same next time}) = \frac{p}{(1-r)(1-p)+p} \times 1 + \frac{(1-r)(1-p)}{(1-r)(1-p)+p} \times (1-r). $$

Is this a correct approach? This seems to be correct to me, but I'm not completely sure that I haven't overlooked anything.

答:
I got a totally different result from yours.

Intro. 1: $$P(A \mid B) = P(A \mid BC) \cdot P(C \mid B) + P(A \mid B \overline C) \cdot P(\overline C \mid B)$$

Proof:
\begin{align}
P(A \mid B) P(B) &= P(AB) \\
&= P\left[AB(C \cup \overline C)\right] \\
&= P\left[(ABC) \cup (AB \overline C)\right] \\
&= P(ABC) + P(AB \overline C) \\
&= P(A \mid BC) \cdot P(BC) + P(A \mid B \overline C) \cdot P(B \overline C) \\
&= P(A \mid BC) \cdot P(C \mid B) \cdot P(B)  + P(A \mid B \overline C) \cdot P(\overline C \mid B) \cdot P(B) \\
&= P(B) \cdot \left[ P(A \mid BC) \cdot P(C \mid B) + P(A \mid B \overline C) \cdot P(\overline C \mid B) \right]
\end{align}

Intro. 2: $$P(A B \mid C) = P(A \mid C) \cdot P(B \mid AC)$$

Proof:
\begin{align}
P(AB \mid C) &= \frac{P(ABC)}{P(C)} \\
&= \frac{P(B \mid AC) \cdot P(AC)}{{P(AC)}/{P(A \mid C)}} \\
&= P(A \mid C) \cdot P(B \mid AC)
\end{align}

Say person A is the randomly selected one, whose proposals could be noted as $a_1, a_2, \ldots{}$ et cetera. $A_i$ stands for "yes" and $\overline A_i$ stands for "no".
If A is unchangeable, we say $X$; otherwise, $\overline X$.
If unchangeable A proposes "yes" every time, we say $Y$; otherwise, $\overline Y$. Obviously, $Y, \overline Y \subseteq X$, $Y \cup \overline Y = X$.

We can say $P(X) = p$, $P(\overline X) = 1 - p$.

Furthermore, we have
$P(a_{k+1} = a_k \mid X) = 1$,
$P(a_{k+1} \neq a_k \mid X) = 0$,
$P(a_{k+1} \neq a_k \mid \overline X) = r$,
$P(a_{k+1} = a_k \mid \overline X) = 1 - r$.

Suppose $P(A_1 \mid \overline X) = q$, $P(\overline A_1 \mid \overline X) = 1 - q$.

Suppose $P(A_1 \mid X) = s$, $P(\overline A_1 \mid X) = 1 - s$.

Now we want to calculate:
\begin{align}
P &= P(A_3 \mid A_1 A_2) + P(\overline A_3 \mid \overline A_1 \overline A_2) \\
&= P(A_3 \mid A_1 A_2 X) \cdot P(X \mid A_1 A_2) + P(A_3 \mid A_1 A_2 \overline X) \cdot P(\overline X \mid A_1 A_2) \\
&+ P(\overline A_3 \mid \overline A_1 \overline A_2 X) \cdot P(X \mid \overline A_1 \overline A_2) + P(\overline A_3 \mid \overline A_1 \overline A_2 \overline X) \cdot P(\overline X \mid \overline A_1 \overline A_2)
\end{align}

\begin{align}
P(A_1 A_2 \mid X) &= P(A_1 \mid X) \cdot P(A_2 \mid A_1 X) = P(A_1 \mid X) \cdot P(a_{k+1} = a_k \mid X) = s \\
P(A_1 A_2 \mid \overline X)
&= P(A_1 \mid \overline X) \cdot P(A_2 \mid A_1 \overline X)
= P(A_1 \mid \overline X) \cdot P(a_{k+1} = a_k \mid \overline X)
= q (1-r) \\
P(A_1 A_2) &= P(X) \cdot P(A_1 A_2 \mid X) + P(\overline X) \cdot P(A_1 A_2 \mid \overline X) = ps + (1-p)q(1-r) \\
P(X \mid A_1 A_2) &= \frac{P(X) \cdot P(A_1 A_2 \mid X)}{P(A_1 A_2)} = \frac{ps}{ps + (1-p)q(1-r)} \\
P(\overline X \mid A_1 A_2) &= \frac{P(\overline X) \cdot P(A_1 A_2 \mid \overline X)}{P(A_1 A_2)} = \frac{(1-p)q(1-r)}{ps + (1-p)q(1-r)} \\
P(\overline A_1 \overline A_2 \mid X) &= P(\overline A_1 \mid X) \cdot P(\overline A_2 \mid \overline A_1 X) = 1 - s \\
P(\overline A_1 \overline A_2 \mid \overline X) &= P(\overline A_1 \mid \overline X) \cdot P(\overline A_2 \mid \overline A_1 \overline X) = (1-q)(1-r) \\
P(\overline A_1 \overline A_2) &= P(X) \cdot P(\overline A_1 \overline A_2 \mid X) + P(\overline X) \cdot P(\overline A_1 \overline A_2 \mid \overline X) = p(1-s) + (1-p)(1-q)(1-r) \\
P(X \mid \overline A_1 \overline A_2) &= \frac{P(X) \cdot P(\overline A_1 \overline A_2 \mid X)}{P(\overline A_1 \overline A_2)} = \frac{p(1-s)}{p(1-s)+(1-p)(1-q)(1-r)} \\
P(\overline X \mid \overline A_1 \overline A_2) &= \frac{P(\overline X) \cdot P(\overline A_1 \overline A_2 \mid \overline X)}{P(\overline A_1 \overline A_2)} = \frac{(1-p)(1-q)(1-r)}{p(1-s)+(1-p)(1-q)(1-r)}
\end{align}

Therefore,
\begin{align}
P &= \frac{ps}{ps+(1-p)q(1-r)} + (1-r)\frac{(1-p)q(1-r)}{ps+(1-p)q(1-r)} + \frac{p(1-s)}{p(1-s)+(1-p)(1-q)(1-r)} + (1-r)\frac{(1-p)(1-q)(1-r)}{p(1-s)+(1-p)(1-q)(1-r)} \\
&= \frac{ps + (1-p)q(1-r)^2}{ps + (1-p)q(1-r)}+\frac{p(1-s)+(1-p)(1-q)(1-r)^2}{p(1-s)+(1-p)(1-q)(1-r)}
\end{align}

Title: Conditional probability - voting


问:
Show that $M_2(\mathbb R)AM_2(\mathbb R)=M_2(\mathbb R)$
Let $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Show that $M_2(\mathbb R)AM_2(\mathbb R)=M_2(\mathbb R)$

Since $M_2(\mathbb R)AM_2(\mathbb R)$ is an ideal is it sufficient to show that it contains an invertible matrix (a unit) and is therefore equal to the whole ring?

答:
Here is an abstract answer:

It is a general fact that if $R$ is a ring, then the ideals of $M_n(R)$ are of the form $M_n(I)$ where $I$ is an ideal in $R$. Thus, $M_2(\mathbb{R})A M_2(\mathbb{R})$ is of the form $M_2(I)$ for an ideal $I$ of $\mathbb{R}$. However, the only ideals in $\mathbb{R}$ are $0$ and $\mathbb{R}$ itself, because every non-zero element is invertible. It thus follows that $M_2(\mathbb{R})A M_2(\mathbb{R})$ is either $M_2(0) = 0$ or $M_2(\mathbb{R})$. But $M_2(\mathbb{R})A M_2(\mathbb{R})$ contains the matrix $A$, so that it cannot be zero and thus it must be equal to all of $M_2(\mathbb{R})$.

Basically, I used that if $R$ is a simple ring, then $M_n(R)$ is also a simple ring.

Here is a concrete answer:

Let $rE_{i,j}$ be the elementary matrix that is $r$ on position $(i,j)$ and $0$ elsewhere. Note that $A= E_{1,1}$. It suffices to show that $M_2(\mathbb{R})A M_2(\mathbb{R})$ contains all these elementary matrices, because one can take their sums to obtain any other matrix. You can write every $r E_{i,j}$ in the form $X E_{1,1} Y$ for matrices $X,Y$, as an easy exercise in matrix calculations shows. For example
$$\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}0 & r \\0 & 0\end{pmatrix}= \begin{pmatrix}0 & r \\ 0 & 0\end{pmatrix}.$$


问:
Finding standard deviation given mean and probability of a sample (algebraically).
I am trying to solve for the standard deviation of a normal distribution given the probability of a sample and mean:

$P(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp(-0.5(\frac{x-\mu}{\sigma})^2)$

So given $P(x)=a, x=b, \mu=c$ is it possible to find a closed-form solution of $\sigma$?

I tried reducing the problem to a form where it could be solved using the Lambert W function with no luck.

答:
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Indeed, this case is suitable for solution in terms of the Lambert $\W$ function as follows:

\begin{align}
\frac1{\sqrt{2\pi}\sigma}\exp\left(-\frac{(b-c)^2}{2\sigma^2}\right)
&=a
\\
\frac1{2\pi\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)
&=a^2
\\
\frac1{\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)
&=2\pi a^2
\\
\frac{-(b-c)^2}{\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)
&=-2\pi(b-c)^2 a^2
\\
\W\left(\frac{-(b-c)^2}{\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)\right)
&=\W\left(-2\pi(b-c)^2 a^2\right)
\\
\frac{-(b-c)^2}{\sigma^2}
&=\W\left(-2\pi(b-c)^2 a^2\right)
\\
\sigma^2&=
\frac{(b-c)^2}{-\W\left(-2\pi(b-c)^2 a^2\right)}
\\
\sigma&=
\frac{|b-c|}{\sqrt{-\W\left(-2\pi(b-c)^2 a^2\right)}}
.
\end{align}

Or, using identity $\W(x)=x\exp(-\W(x))$, we can get
an equivalent, but more convenient form that includes the case $b=c$
without resort to the limits,

\begin{align}
\sigma&=
\frac1{a\sqrt{2\pi}}\,\exp\left(\tfrac12\,\W\left(-2\pi a^2(b-c)^2\right)\right)
.
\end{align}

Note that if $(-2\pi a^2(b-c)^2)\in(-\tfrac1{\e},0)$ there are two real solutions for $\sigma$:

\begin{align}
\sigma_0&=
\frac1{a\sqrt{2\pi}}\,\exp\left(\tfrac12\,\Wp\left(-2\pi a^2(b-c)^2\right)\right)
\\
\text{and }\quad
\sigma_1&=
\frac1{a\sqrt{2\pi}}\,\exp\left(\tfrac12\,\Wm\left(-2\pi a^2(b-c)^2\right)\right)
,
\end{align}

where $\Wp$ and $\Wm$ are the two real branches of the Lambert $\W$ function.

For example, setting $a=0.75,\ \mu=0.5$, $x=0.2$ (or $x=0.8$)
corresponds to two curves, one with
$\sigma=\sigma_0\approx0.403437$
and another
with $\sigma=\sigma_1\approx0.234237$, see image below:

$\endgroup$


问:
Determining the equation of a line parallel to two planes.
**The task is to find the equation of a line that intersects lines $\textbf{p1}$ and $\textbf{p2}$, and is parallel to the planes $\textbf{T1}$ and $\textbf{T2}$ the equations are:

$$\textbf{p1:} \quad \frac{x+5}{2} = \frac{y-3}{-4} = \frac{z+1}{3} $$

$$\textbf{p2:} \quad \frac{x-3}{-2} = \frac{y+1}{3} = \frac{z+2}{4}$$

$$\textbf{T1:}\quad  3x+12y-3z-5=0$$

$$\textbf{T2:} \quad 3x-4y+9z+7=0$$

If anyone has an idea for this task and similar tasks please share!

**

答:
Ok so we know this line is parallel to both T1 and T2. which means it must be perpendicular to both of their normal vectors, given by

$\begin{bmatrix}
3 & 12 & -3
\end{bmatrix}$
and
$\begin{bmatrix}
3 & -4 & 9
\end{bmatrix}$

if you find the cross product of these two vectors you will get the directional vector of the line and then by using P1 or P2 we can find an initial point.


答:
Length of angle bisector is given by

$\displaystyle l_c^2 = \frac{ab}{(a+b)^2} ((a+b)^2 - c^2)$

$a+b = 125-60 = 65, c = 60$

Leads to $\displaystyle 10^2 = \frac{ab}{65^2}(65^2-60^2)$

i.e., $ab = 676 \implies a(65-a) = 676$

We get a quadratic $a^2 - 65a + 676 = 0 \implies a = 13, 52$

So $a = 13, b = 52$ or $a = 52, b = 13$

Title: A triangle $ABC$ with $c=60,l_c=10$ and $P=125$

Body:
A triangle $ABC$ is given with $c=60,l_c=10$ and $P=125$, where $AB=c$, $l_c$ is the angle bisector of $\measuredangle C$ and $P$ is the perimeter. Find $a$ and $b$.

Here are my thoughts:

We have $\begin{cases}\dfrac{m}{n}=\dfrac{b}{a}\\l_c^2=ab-mn\end{cases}.$ From the first equation we can get $$\dfrac{m+n}{n}=\dfrac{a+b}{a}$$ or $$\dfrac{12}{n}=\dfrac{13}{a}.$$ Then we can substitute $l_c=10$ into the second equation to get $$100=ab-mn.$$ I am stuck here. Thank you in advance!
Katherine


答:
Hint:

The direction of the line is given by the cross-product of the normals to the planes. If you project the two lines parallelly to this direction to some plane, their 2D intersection is the projection of the searched line.

Title: Determining the equation of a line parallel to two planes.

Body:
**The task is to find the equation of a line that intersects lines $\textbf{p1}$ and $\textbf{p2}$, and is parallel to the planes $\textbf{T1}$ and $\textbf{T2}$ the equations are:

$$\textbf{p1:} \quad \frac{x+5}{2} = \frac{y-3}{-4} = \frac{z+1}{3} $$

$$\textbf{p2:} \quad \frac{x-3}{-2} = \frac{y+1}{3} = \frac{z+2}{4}$$

$$\textbf{T1:}\quad  3x+12y-3z-5=0$$

$$\textbf{T2:} \quad 3x-4y+9z+7=0$$

If anyone has an idea for this task and similar tasks please share!

**


问:
Generalizing the Leibniz rule
Given the Leibniz rule:

$$\frac{d}{dy}\int^a_b f(x,y) dx = \int_b^a \frac{\partial f}{\partial y} (x,y) dx$$

How do I prove a more general case using the chain rule and the above:

$$\frac{d}{dy} \int_{g_1(y)}^{g_2(y)} f(x,y) dx =?$$

From the fundamental theorem of calculus we have that: $$\frac{d}{dt} \int_{f_2(t)}^{f_1(t)} g(s) ds = g(f_1(t))f'_1(t) - g(f_2(t))f'_2(t)$$

But when I just apply the fundamental theorem of calculus I get an incorrect answer (because I also need to use Leibniz rule itself...).

答:
You're very, very close. The last term is the Leibnitz rule as you are familiar with it itself. That is,
$$\dfrac{d}{dt}\int_a^bf(x,t)dx=f(b,t)b'-f(a,t)a'+\int_a^b\dfrac{\partial f(x,t)}{\partial t}dx.$$ This is because our integral

$$=F(b,t)-F(a,t)$$ which when differentiated gives
$$\dfrac{\partial F(b,t)}{\partial b}b'+\dfrac{\partial F(b,t)}{\partial t}-\dfrac{\partial F(a,t)}{\partial b}a'-\dfrac{\partial F(a,t)}{\partial t}$$ which afterwards it becomes obvious that
$$\dfrac{\partial F(b,t)}{\partial b}b'-\dfrac{\partial F(a,t)}{\partial a}a'=f(b,t)b'-f(a,t)a'$$ and that because we took the partial derivatives of F as if $b$ and $a$ were independent of $t$, we obtain that the last term is
$$\dfrac{\partial F(b,t)}{\partial t}-\dfrac{\partial F(a,t)}{\partial t}=\dfrac{\partial F(x,t)}{\partial t}\Bigg|_{x=a}^{x=b}=\dfrac{\partial}{\partial t}\int_a^b f(x,t)dx=\int_a^b \dfrac{\partial}{\partial t}f(x,t)dx$$ Done!

Title: Generalizing the Leibniz rule


问:
Is this type of function continuous?
Pic of the function

Is this function continius along the x axis?

答: This type is called as a rational function. It being a ratio of 2 polynomials, will be continuous everywhere, except at those points where the denominator goes to $0$, called the zeros of the function. It can have at most $m$ discontinuities, where $m$ is the degree of the polynomial in the denominator.

Title: Is this type of function continuous?


问:
$V$ is the direct sum of its weight spaces
If $\mathfrak{g}$ is a semisimple Lie algebra, with Cartan subalgebra $\mathfrak{h}$, and irreducible representation $(V, \rho)$, where $V$ has a non-zero weight-space, then $V$ is the direct sum of its weight spaces.

I see why is that the intersection of two distinct weight spaces is $\{0\}$, indeed let $0 \neq v \in V_\lambda \cap V_\mu,   \ $  then: $\forall \ h \in \mathfrak{h}, \ \ \lambda(h)\cdot v = h \cdot v =\mu(h)\cdot v \Rightarrow \forall \ h \in \mathfrak{h},\ (\lambda-\mu)(h)=0, \ \text{ as } \ v \neq 0 \ \Rightarrow \ \mu =\lambda.$

But it's not clear to me why the sum of the weight spaces must equal the entire space.

答:
You can express $\mathfrak g$ itself as $\mathfrak h\oplus\bigoplus_{\alpha\in R}\mathfrak g_\alpha$, whre $R$ is the set of roots.

Now, consider the sum $S$ of the weight spaces; you want to prove that $S=V$. You know that $S\ne\{0\}$ , since you are working over $\Bbb C$, which is algerbaically closed, and therefore the set of wights is not empty. On the other hand, by the definition of $S$, if $H\in\mathfrak h$ and $v\in S$, then $Hv\in S$. And if $X\in\mathfrak g_\alpha$, for some root $\alpha$, and if $v\in V_\beta$, for some weight $\beta\in\mathfrak h^*$, then $Xv\in V_{\alpha+\beta}\in S$. But then $S$ is a submodule which is not $\{0\}$. So, since $V$ is irreducible, $S=V$.


问:
Derivative of Euclidean distance in 3D
I am currently working on a problem involving the gradient of a composite function with respect to a single particle $k$, following

$$
\nabla_k u(r_{k,i})
$$
for which $r_{k,i} = |\mathbf{r}_k - \mathbf{r}_{i}|$. This far I have worked out

$$
\nabla_k u(r_{i,k}) = \frac{\partial u(r_{i,k})}{\partial r_{i,k}} \frac{\partial r_{i,k}}{\partial \mathbf{r}_k}
=
u'(r_{i,k})\frac{\partial}{\partial \mathbf{r}_k}
|\mathbf{\mathbf{r_k}} - \mathbf{r_i}|
$$
Could I treat the last partial differential operation like one would treat an absolute value function in 1D? Or would I have to expand the radial variable into its Cartesian coordinates and work from there?

Many thanks in advance.

答:
By taking the partial derivatives, the gradient of

$$\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$$ is

$$\frac{(x-x_0,y-y_0,z-z_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}=\frac{\mathbf r-\mathbf r_0}{|\mathbf r-\mathbf r_0|}.$$

Title: Derivative of Euclidean distance in 3D


问:
Integral of product log and square root $ \int_{-2}^2 \log(a-bx)\sqrt{4-x^2} dx $
I need to compute the following integral as a function of the parameters $a$ and $b$
$$
\int_{-2}^2 \log(a-bx)\sqrt{4-x^2} dx
$$

But, I don't know how I should deal with it.

Thanks in advance!

答:
With $\int_{-1}^1 \sqrt{1-t^2} dt=\frac\pi2$, rewrite the integral as

$$
I = \int_{-2}^2 \log(a-bx)\sqrt{4-x^2} dx
\overset{x= 2t}= 2\pi \log a + 4J\left(\frac{2b}a\right)
$$
where

$$J(p)= \int_{-1}^1 \log(1-pt )\sqrt{1-t^2} dt
$$
$$J’(p)= -\int_{-1}^1 \frac{ t\sqrt{1-t^2}}{ 1-pt} dt=-\frac{\pi}{2p^3}\left(1-\sqrt{1-p^2}\right)^2
$$
Thus
\begin{align}
I & = 2\pi \log a + 4\int_0^{\frac{2b}a}J’(p)dp\\
&= 2\pi \log a + 2\pi \left(\frac1{\sqrt{1-p^2}+1}+\ln\left(1+\sqrt{1-p^2}\right)\right)\bigg|_0^{\frac{2b}a}\\
 &= 2\pi \left( \frac{a}{\sqrt{a^2-4b^2}+a} +\ln \left({\sqrt{a^2-4b^2}+a}\right) -\frac12 -\ln2\right)
\end{align}


问:
Please explain, "Asymmetric is stronger than simply not symmetric".
In some textbook I found a statement like, "Asymmetric is stronger than simply not symmetric".

But as I try to perceive this statement, both appear to be same to me.

For example, parentof is an asymmetric relation. If $A$ is a parentof $B$, $B$ can not be parentof $A$. We arrive at the same conclusion if we call this relation to be not symmetric.

(Liyang Yu. A Developer’s Guide to the Semantic Web.
2nd. Springer, 2014. p 226, last sentence of second paragraph from bottom.)

答:
What the book said is wrong.

Consider such a case: Let <∅, ∅> be an empty set with an empty relation defined on it. Then the empty relation on the empty set must be reflective, irreflective, symmetric, antisymmetric, asymmetric, and transitive simultaneously. Since the relation is both asymmetric and symmetric, it follows that asymmetry cannot deduce non-symmetry.

Indeed, what the book said is right mostly. However, it may fail in some extreme scenarios, like the example presented above.


答:
Note that
$$\mathrm{Tr}\left[(A-A^T)^2\right]=-2\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2.$$
Therefore the identity can be rewritten as
$$2(1-\mathrm{Tr}\,A)^2-\mathrm{Tr}\left[(A-A^T)^2\right]=8.$$
Since $\mathrm{Tr}\,AA^T=\mathrm{Tr}\,1=3$, this is equivalent to
$$\left(\mathrm{Tr}\,A\right)^2-\mathrm{Tr}\,A^2-2\mathrm{Tr}\,A=0.$$
The left side of the last relation can be rewritten in terms of eigenvalues $\lambda_{1,2,3}$ of $A$:
$$(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_2^2+\lambda_2^2+\lambda_3^2)-2(\lambda_1+\lambda_2+\lambda_3).$$
Using that $\lambda_1\lambda_2\lambda_3=\mathrm{det}\,A=1$, we reduce the problem to showing that
$$\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=\lambda_1+\lambda_2+\lambda_3$$
But this follows from $\mathrm{Tr}\,A^{-1}=\mathrm{Tr}\,A^T=\mathrm{Tr}\,A$. $\blacksquare$

In case someone would like to have an "explanation" of the very first identity. The matrix $A-A^T$ can be seen as an element $a$ of Lie algebra $\mathfrak{so}(3)$, which can in turn be seen as a three-dimensional vector. The squared length of this vector coincides with Cartan-Killing norm of $a$ up to a numerical factor.

Title: How prove that $\;(1-\mathrm{Tr}\,A)^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4\;\;?$

Body:
Let $A=\begin{bmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}
\end{bmatrix}$ be an orthogonal matrix with $a_{i,j}\in \mathbb R$, where $\det(A)=1$

Show that
$$(1-\mathrm{Tr}\,A)^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4$$


问:
Solving this problem without Taylor series ( I made an extra assumption)
I want to show that the derivative of a (one-time differentiable) function $f(x)$ is given by $$f'(a)= \lim \limits_{h \to 0} \dfrac {1}{h} \displaystyle \int_{-1}^{1} \dfrac{3u}{2} f(a+uh) \ du$$

I used the Taylor's expansion about the point $x=a$ and the conventional $h$ as $uh$ [i.e. $f(x+h)= f(x) + h f'(x) + \cdots$] and got the answer. However, I made an additional assumption that the function $f(x)$ possesses a convergent series about $uh$ and this includes the assumption that the function is differentiable more than once.

Tried applying the Leibniz rule, but could not get to the RHS. How can I prove the result without using the Taylor series?

答:
Hint.

Try the change of variables $y=uh$ to the integral:
$$
\frac32\cdot\frac{1}{h^3}\int_{-h}^{h}yf(a+y)dy
$$
Then you can apply the L'Hopital rule and the fundamental theorem of calculus to get

$$
\lim_{h\to 0 }\frac{h(f(a+h)-f(a-h))}{2h^2}=
\lim_{h\to 0 }\frac{f(a+h)-f(a-h)}{2h}
$$
Now use the differentiability of $f$ at $a$.


问:
Can a finitely generated $\mathbb{Z}$-algebra contain $\mathbb{Q}$?
Is there a ring between $\mathbb{Q}$ and $\mathbb{R}$ that is finitely generated as an algebra over $\mathbb{Z}$? My guess is there isn't.

I can see that it would have to be finitely generated over $\mathbb{Q}$ as well, and I think I can deal with algebraic generators. But if there are algebraically dependent transcendentals, I don't see how to exclude some rational. Why couldn't there be $\alpha$, $\beta$ transcendental, such for every prime $p$, $1/p$ is given by some integer polynomial in $\alpha,\beta$?

答: The residue fields of a finitely generated $\mathbb{Z}$-algebra (i.e. the quotients by maximal ideals) are finite fields (see here). There is no homomorphism from $\mathbb{Q}$ to a finite field.


问:
Help with this simple example about projection of linear spaces
I'm studying the book Linear Algebra by Hoffman and Kunze and I'm trying to find a concrete example in $\mathbb R^2$ of this fact:

If $R$ and $N$ are subspaces of $V$ such that $V=R \oplus N$, there is one and only one projection operator $E$ which has range $R$ and null space $N$. That operator is called the projection on $R$ along $N$.

I'm thinking about the linear space

$$V=\langle(1,1)\rangle$$

being $E:V\to V$ the projection given by

$$E(x,y)=(x,0)$$

Following my calculations we should have $R=\langle(1,0)\rangle$ and $N=\langle (0,1)\rangle$. I don't understand why

$$R\oplus N=\mathbb R^2\neq V$$

答: Note that neither $R$ nor $N$ are actually subspaces of $V$ in your example and your suggested projection, considered from $V\to\mathbb{R}^2$ has trivial kernel and range of $R$. As @hardmath noted in their comment, this result is much more interesting to think about in the case where $V$ has dimension $>1$.


问:
Plot a multivariable function in Matlab
Let $f:\mathbb{R}^2 \to \mathbb{R}_+$ be the function defined piecewise by
$$
f(x,y)
        :=
        \begin{cases}
            \sqrt{1 - x^2 - y^2} &\text{, if } x^2+y^2 \leq 1 \text{ and } y \geq 0, \\
            0 &\text{, otherwise.}
        \end{cases}
$$

I would like to plot this function in MATLAB. I tried the following code, but only recieve error notifications... (I suspect mistakes in the definition of f and/or the line with Z);

```<code>f = @(x,y) piecewise(x^2+y^2<=1, piecewise(y>=0,sqrt(1-x^2-y^2),y<0,0), x^2+y^2>1, 0);

xarray = linspace(-5,5,51);
yarray = linspace(-5,5,51);
[X,Y] = meshgrid(xarray,yarray);
Z = f(xarray,yarray);

% Plot the function %
figure
surfc(X,Y,Z)
xlabel('x')
ylabel('y')
zlabel('f(x,y)')
colorbar
</code>```

Thanks in advance for any help! :)

答:
Two different naive approaches (they are naive because they don't rigorously treat discontinuities):

```<code>xmin = -1.5; xmax = -xmin; xpto = 500;                               % creating 2D domain for your function..
ymin = -1.5; ymax = -ymin; ypto = 500;                               % creating 2D domain for your function..
[X,Y] = meshgrid(linspace(xmin,xmax,xpto),linspace(ymin,ymax,ypto)); % creating 2D domain for your function..

FXY_1 = sqrt(1-X.^2-Y.^2); % f1 values..
FXY_2 = 0*X;               % f2 values..

FXY_1_CONDITION = ( X.^2 + Y.^2 ) <= 1 & Y >= 0; % f1 domain..
FXY_2_CONDITION = ~FXY_1_CONDITION;              % f2 domain..

FXY_1(FXY_2_CONDITION) = NaN; % ignoring f2 domain in f1 (NaN values are ignored by plot function)...
FXY_2(FXY_1_CONDITION) = NaN; % ignoring f1 domain in f2 (NaN values are ignored by plot function)...

figure('color','white');
hold on;
mesh(X,Y,FXY_1); % plot f1..
mesh(X,Y,FXY_2); % plot f2..
colorbar;
</code>```

.

```<code>xmin = -1.5; xmax = -xmin; xpto = 500;                               % creating 2D domain for your function..
ymin = -1.5; ymax = -ymin; ypto = 500;                               % creating 2D domain for your function..
[X,Y] = meshgrid(linspace(xmin,xmax,xpto),linspace(ymin,ymax,ypto)); % creating 2D domain for your function..

FXY = NaN .* X;            % initializing f..
FXY_1 = sqrt(1-X.^2-Y.^2); % f1 values..
FXY_2 = 0 .* X;            % f2 values..

FXY_1_CONDITION = ( X.^2 + Y.^2 ) <= 1 & Y >= 0; % f1 domain..
FXY_2_CONDITION = ~FXY_1_CONDITION;              % f2 domain..

FXY(FXY_1_CONDITION) = FXY_1(FXY_1_CONDITION); % passing f1 values to f..
FXY(FXY_2_CONDITION) = FXY_2(FXY_2_CONDITION); % passing f2 values to f..

figure('color','white');
mesh(X,Y,FXY); % plot f..
colorbar;
</code>```

Title: Plot a multivariable function in Matlab

Body:
Two different naive approaches (they are naive because they don't rigorously treat discontinuities):

xmin = -1.5; xmax = -xmin; xpto = 500;                               % creating 2D domain for your function..
ymin = -1.5; ymax = -ymin; ypto = 500;                               % creating 2D domain for your function..
[X,Y] = meshgrid(linspace(xmin,xmax,xpto),linspace(ymin,ymax,ypto)); % creating 2D domain for your function..

FXY_1 = sqrt(1-X.^2-Y.^2); % f1 values..
FXY_2 = 0*X;               % f2 values..

FXY_1_CONDITION = ( X.^2 + Y.^2 ) <= 1 & Y >= 0; % f1 domain..
FXY_2_CONDITION = ~FXY_1_CONDITION;              % f2 domain..

FXY_1(FXY_2_CONDITION) = NaN; % ignoring f2 domain in f1 (NaN values are ignored by plot function)...
FXY_2(FXY_1_CONDITION) = NaN; % ignoring f1 domain in f2 (NaN values are ignored by plot function)...

figure('color','white');
hold on;
mesh(X,Y,FXY_1); % plot f1..
mesh(X,Y,FXY_2); % plot f2..
colorbar;

.

xmin = -1.5; xmax = -xmin; xpto = 500;                               % creating 2D domain for your function..
ymin = -1.5; ymax = -ymin; ypto = 500;                               % creating 2D domain for your function..
[X,Y] = meshgrid(linspace(xmin,xmax,xpto),linspace(ymin,ymax,ypto)); % creating 2D domain for your function..

FXY = NaN .* X;            % initializing f..
FXY_1 = sqrt(1-X.^2-Y.^2); % f1 values..
FXY_2 = 0 .* X;            % f2 values..

FXY_1_CONDITION = ( X.^2 + Y.^2 ) <= 1 & Y >= 0; % f1 domain..
FXY_2_CONDITION = ~FXY_1_CONDITION;              % f2 domain..

FXY(FXY_1_CONDITION) = FXY_1(FXY_1_CONDITION); % passing f1 values to f..
FXY(FXY_2_CONDITION) = FXY_2(FXY_2_CONDITION); % passing f2 values to f..

figure('color','white');
mesh(X,Y,FXY); % plot f..
colorbar;


问:
How many different codes can be formed such that the code consists of two letters followed by four digits?
I have a question as follows:

A small design company gives each of its products a code that consists of two letters
followed by four digits. How many different codes can be formed such that the letters
are taken from $\{A,B,C,D\}$ and may include repetition, and the four digits are taken
from $\{1,2,3,4,5,6,7,8,9\}$ and must be all different and arranged in decreasing order?

My idea: we know that the options for the first two letters are $4^2$. But how to arrange the four digits in decreasing order? I feel like that the options of "all different" are $\tbinom{9}{4}$.

答: You need to break everything down. $\tbinom{9}{4}$ means the number of all unique subsets with four elements of a set of 9 elements. Now a set of 9 elements of $\{1,2,3,4,5,6,7,8,9\}$ remains the same even if you shuffle the numbers, because you can still choose from the same values. This is also true for its subsets: a unique subset must have all of its values so that none of the other subsets have the same values, irrespective of ordering. Based on the comment, $\{2, 9, 3, 4\}$ is the same subset as $\{9, 4, 3, 2\}$, though the latter is sorted. Now you need to see if you can have multiple options from a given subset by sorting, or possibly less. If you have sorted elements like $\{9, 4, 3, 2\}$, as each element is unique, it means that an element and its smaller neighbour (if has) have a difference of at least 1, therefore no neighbours can be swithced, as that would break ordering. From this follows that no any 2 elements can be switched. So a unique subset has exatly one sorting order, therefore the sorting condition does not increase the number of options. It can also not decrease, as subsets are unique which means per above not all the numbers are the same for any 2 subsets, and after sorting this attribute will still be true as no value has changed, and therefore this meets the "must be all different" condition. So the question is as you correctly feel: how many unique subsets with four elements of a set of 9 elements has?


问:
Number of terms in a Polynomial Expansion
For a binomial $(a + b)^n$, the number of terms is n + 1.

For a trinomial $(a + b + c)^n$, the number of terms is $\frac{(n+1)(n+2)}{(2)}$.

For a multinomial $(a + b + c +d)^n$, the number of terms is $\frac{(n+1)(n+2)(n+3)}{(6)}$.

I'm guessing that for $(a + b + c + d + e)^n$, the number of terms formula would include $(n+1)(n+2)(n+3)(n+4)$ on the numerator but I don't know what should be its denominator.

Question:

What is the number of terms for $(a + b + c + d + e)^n$?

Do we have a general formula for the number of terms of a polynomial expansion?

What if the given is $(a^2 + a + b)^n$, can I still use the formula $\frac{(n+1)(n+2)}{(2)}$.when 2 terms in the expansion has the same variable?

What if the given is $(a + b + Constant)^n$, would the constant affect the number of terms?

Body: for (a+b+c+d+e)^n it would be [(n+4)(n+3)(n+2)(n+1)]/24


答:
The map $E$ you defined doesn't map to your $V$, it's rather a map $\Bbb R^2\to\Bbb R^2$, which would justify $R\oplus N=\Bbb R^2$.

On the proof of the claim: by the direct sum property, every vector $v\in V$ can be uniquely decomposed as $v=r+n$ with $r\in R,\ n\in N$. Now define
$$P(r+n):=r$$
By the uniqueness of the decomposition, you can verify that it's linear, and obviously one has $P^2(r+n)=r=P(r+n)$, hence $P^2=P$.

On the other hand, if $P^2=P$ and $\ker P=N,\ {\rm range\,}P=R$, then we must have $Pn=0$ for all $n\in N$ and for any $r\in R$ there's an $x\in V$ with $P(x)=r$, then
$$P(r)=P(P(x))=P^2(x)=P(x)=r\,,$$
so, $P$ must satisfy $P(r+n)=r$.

Title: Help with this simple example about projection of linear spaces

Body:
I'm studying the book Linear Algebra by Hoffman and Kunze and I'm trying to find a concrete example in $\mathbb R^2$ of this fact:

If $R$ and $N$ are subspaces of $V$ such that $V=R \oplus N$, there is one and only one projection operator $E$ which has range $R$ and null space $N$. That operator is called the projection on $R$ along $N$.

I'm thinking about the linear space

$$V=\langle(1,1)\rangle$$

being $E:V\to V$ the projection given by

$$E(x,y)=(x,0)$$

Following my calculations we should have $R=\langle(1,0)\rangle$ and $N=\langle (0,1)\rangle$. I don't understand why

$$R\oplus N=\mathbb R^2\neq V$$


问:
total cost of membership over period at set growth rate per renewal
I need to calculate a membership cost.

The first year's membership will be £530, the second year will be £530 + £25, so £555.Every year, the renewal will be an additional £25, so year one = 530, year two = 555, year 3 = 580 and so on.

I need to calculate, based on how long the membership will be, wat the total cot of that membership will be i.e. if it were a 3 year membership it would be 530 + 555 + 580.

Shy of working out each year's cost, then adding them all together, is there a formula I can use to work this out? I'll need to program this into a query, so need the specific equation.

答:
Hint-1: Suppose the first year's membership was $0$ and the membership rose by $1$ each year.  What would the cost be after $n$ years.

Related question: What is the formula for $1 + 2 + \cdots + (n-1)$?

Hint-2: Suppose the first year's membership was $0$ and the membership rose by $k$ each year.  How has that affected the answer given to Hint-1?

Hint-3: Suppose the first year's membership was $n$ and the membership rose by $k$ each year.  How has that affected the answer given to Hint-2?


问:
Calculate $\sum\limits_{k=0}^{n} \frac{1}{(k+1)(k+2)} \binom{n}{k}$
Any input on how to solve this?

$$\sum\limits_{k=0}^{n} \frac{1}{(k+1)(k+2)} \binom{n}{k}$$

I have tried using Pascal's triangle but wasn't able to get anywhere.

Body:
Another way:

$$(1+x)^n= \sum_{k=0}^{n} {n \choose k} x^k$$
Integrate w.r.t. $x$
$$\frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^{n} {n \choose k} \frac{x^{k+1}}{(k+1)}+C \implies C=\frac{1}{n+1}$$
Again integrate w.r.t. $x$ in $(0,1)$ to get
$$\frac{(1+x)^{n+2}}{(n+1)(n+2)}|_{0}^{1}=\sum_{k=0}^{n}{n \choose k}\frac{1}{(k+1)(k+2)}+C$$
$$\implies \sum_{k=0}^{n}{n \choose k}\frac{1}{(k+1)(k+2)}=\frac{2^{n+2}-n-3}{(n+1)(n+2)}$$


问:
"Reducing" spliting fields
Let $f(X)=X^n+a_{n-1}X^{n-1}+\ldots a_{1}X+a_{0} \in \Bbb Q[X]$, irreducible in $\Bbb Q[X].$ Let $K_{1},K_{2},\ldots,K_{n}$ be the $n$ roots of $f(X)$ in $\mathbb{C}$. Why $\Bbb Q[K_{1},K_{2},\ldots,K_{n}] = \Bbb Q[K_{1},K_{2}, \ldots, K_{n-1}]?$

I think that i need to use the multiplicative property of dimensions of field extensions but i dont know if i'm right. Any help is appreciated!

答: Note that $f=(X-K_1)\dots(X-K_n)$. The field $F=\Bbb Q(K_1,\dots,K_{n-1})$ contains the roots $K_1,\dots,K_{n-1}$ and hence the polynomial $(X-K_1)\dots(X-K_{n-1})$ is defined over $F$. We may therefore divide $f$ by $(X-K_1)\dots(X-K_{n-1})$ and get a polynomial over $F$ again. But $$\frac{f}{(X-K_1)\dots(X-K_{n-1})}=X-K_n$$ hence $K_n\in F$.

Title: "Reducing" spliting fields


问:
Burnside's Lemma necklace with a clasp
How many necklaces with clasps can be made with 6 beads in one colour, 4 in second and 3 in the third?

Use Burnside's Lemma I've got:

Let $X=\{\text{necklaces of length 13 beads = 6 white, 4 black and 3 blue}\}$, $|X|=\frac{13!}{6!4!3!}=60060$.
Now G in my case is $|G|=id + \text{symmetry}=2$

So I apply Burnside's Lemma with specified number of beads.

Count fixed points for each $g\in G$:

Identity $g_{id}$ sends any bead back to itself $|fix(g_{id})|=|X|=60060$

Then I can consider flipping the necklace.
There is only 1 flip with axis through the centre - the middle bead. Permutation that represents it has 6 orbits of size 2 (vertices on both sizes of symmetry axis) and 1 orbit of size 1 (vertex that is immutable):
$$s \rightarrow (1\,13)(2\,12)(3\,11)(4\,10)(5\,9)(6\,8) $$ last, $(7\,7)$, has to stay immutable so was omitted.
To make such colouring with fixed vertex being in the colour of uneven order (blue) and assigning all 3 colours to rest of them: there is 1 way of choosing middle bead (only blue) 20 ways of choosing 3 out of 6 monochromatic pairs(first let's choose white) and 3 ways of choosing 2 out of 3 pairs(this time black), the last one can be chosen in 1 way(2 blue beads). So $20\cdot 3\cdot1^2=60$.

Since $|G|=2$ we get:

$$N=\frac{1}{\left|G\right|}\sum_{g\in G} \left|fix(g)\right|$$
here it would be
$\frac{1}{2}\sum_{g\in G} \left |fix(g)\right|=\frac{1}{2}\cdot(60060+20\cdot 3) = 30060$

答:
Well done! I presume you wanted this use of Burnside checked and it is very good.

One could quibble that for an example where there is only one relevant symmetry, applying Burnside seems overly sophisticated. You could just say:

There are $\frac{13!}{6!4!3!}=60060$ arrangements in total, of which $\frac{6!}{3!2!1!}=60$ are symmetrical.

The number of necklaces is therefore $60 + 30000=30060$.

However, as a test of understanding Burnside it was a good easy example to use.

Title: Burnside's Lemma necklace with a clasp


答:
Well, you have $f=(x-K_1)\cdots (x-K_{n-1})(x-K_n)$ over $\Bbb C$.
Thus if you have the first $n-1$ roots $K_1,\ldots,K_{n-1}$, the remaining root is given by division of $f$ into $(x-K_1)\cdots (x-K_{n-1})$.

Title: "Reducing" spliting fields

Body:
Let $f(X)=X^n+a_{n-1}X^{n-1}+\ldots a_{1}X+a_{0} \in \Bbb Q[X]$, irreducible in $\Bbb Q[X].$ Let $K_{1},K_{2},\ldots,K_{n}$ be the $n$ roots of $f(X)$ in $\mathbb{C}$. Why $\Bbb Q[K_{1},K_{2},\ldots,K_{n}] = \Bbb Q[K_{1},K_{2}, \ldots, K_{n-1}]?$

I think that i need to use the multiplicative property of dimensions of field extensions but i dont know if i'm right. Any help is appreciated!


问:
Question about proving ceiling function equivalence relation
I read a proof about ceiling function equivalence relation as linked here.
The theorem to prove is:

Let $\mathcal R$ be the relation defined on $\mathbb{R}$ such that: $\forall x, y \in \mathbb{R}: \left({x, y}\right) \in \mathcal R \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$
where $\left \lceil {x}\right \rceil$ is the ceiling of $x$.

Then $\mathcal R$ is an equivalence.

Proving reflexitivity and symmetry is trivial, but I have questions about proving transitivity.

According to definition, we should just proof that if $\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$ and $\left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$, then $\left \lceil {x}\right \rceil = \left \lceil {z}\right \rceil$. In the proof, the author took extra steps to prove this. But isn't $\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$ trivial? Are there any reason why the author went those extra steps?

答: Yes, it is trivial. Actually, it is so trivial that it has nothing to do with the ceiling function. If $f\colon X\longrightarrow Y$ is any function and if you define a binary relation $\sim$ on $X$ by $x\sim y$ if and only if $f(x)=f(y)$, then $\sim$ is an equivalence relation.


问:
Prove that $(X,d)$ is complete
Let $(X,d)$ - metric space such that $A,B \subset X$ and $ X=A\cup B$ where $(A,d_{|A\times A}), (B,d_{|B\times B})$ are complete subspaces. Prove that $(X,d)$ is complete.

I think it should be obvious that the sum of the complete subspaces is complete, but nevertheless a few years ago this task was in an exam so I would like to know how to prove it.

答: Let $(x_n)_n$ be a Cauchy sequence in $X$. We want to show that $x_n$ converges. As $X=A\cup B$ infinitely many elements of $(x_n)_n$ and hence a subsequence have to be in one of the sets $A,B$. Without loss of generality assume that there is a subsequence $(x_{n_k})_k$ of $(x_n)_n$ contained in $A$. As subsequences of Cauchy sequences are again Cauchy, we see that $(x_{n_k})_k$ is a Cauchy sequence. By assumption $A$ is complete, hence $(x_{n_k})_k$ is convergent. This then implies that $(x_n)$ is convergent (In general a Cauchy sequence with a convergent subsequence is itself convergent). Hence $X$ is complete.

Title: Prove that $(X,d)$ is complete


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
You might know only $\Pr[A\mid B]$ and not $\Pr[B\mid A]$, not because someone "adversarially told you the wrong one", but because one of those is a natural quantity to compute, and the other is a natural quantity to want to know.

I am about to teach Bayes' theorem in an undergraduate course in probability. The general setting I want to consider is when:

We have several competing hypotheses about the world. (Several candidates for $B$.)

If we assume one of these hypotheses, then we get a nice and easy probability problem where it's easy to find the probability of $A$: some observations that we've made. (Outside undergraduate probability courses, "nice and easy" is a relative term.)

We want to figure out which hypothesis is likelier.

The mammogram example might be natural, but it's less obviously natural because we have to track down where the numbers that are given to us come from, and ask why we couldn't be given the other quantities in the problem. So here are some examples where we have fewer numbers coming to us out of thin air.

Suppose you are communicating over a binary channel which flips bits $10\%$ of the time. (This part is given to us out of nowhere, but it's the natural quantity to ask about first.) Your friend has several possible messages they might send you: these are the hypotheses $B_1, B_2, \dots, B_n$. You receive a message: that's the observation $A$. Then $\Pr[A \mid B_i]$ is just $(0.1)^k (0.9)^{n-k}$ if $B_i$ is an $n$-bit message that differs from the one you received in $k$ places. On the other hand, $\Pr[B_i \mid A]$ is the quantity we want: it will tell us how likely it is that your friend sent each message.

You have a coin, and you don't know anything about its fairness. One possible assumption is that it lands heads with probability $p$, where $p \sim \text{Uniform}(0,1)$, but we could vary this. Then you flip the coin $n$ times and see $k$ heads. There are infinitely many hypotheses $B_p$, one for each possible $p$; under each of them, $\Pr[A \mid B_p]$ is just a binomial probability. Knowing the conditional PDF of $p$, which is what Bayes' theorem tells us, tells us more about how likely the coin is to land heads.


问:
Proving that $\frac{x^2}{2(1+x)}<x-\ln(1+x)<\frac{x^2}{2}$, $x>0$. Using mean value theorem only

Using applications of the MVT, prove that $\frac{x^2}{2(1+x)}<x-\ln(1+x)<\frac{x^2}{2}$, $x>0$.

I am not able to choose the required function for the mean value theorem. I have to use only the differential form of the mean value theorem. I don't know how I am supposed to do this. Any help or any other method you could offer would be appreciated.

答:
Take $$f(x)=x-\ln(1+x)-x^2/2 \implies f'(x)=\frac{-x^2}{1+x}<0, ~if~x>0$$
So $f(x)$ is decreasing for $x>0$. Then $x>0 \implies f(x)<f(0)=0.$
This proves that $$x-\ln(1+x)<\frac{x^2}{2}$$
Next take $g(x)=x-\ln(1+x)-\frac{x^2}{2(1+x)} \implies g'(x)=\frac{x^2}{2(1+x)^2}>0.$
So $g(x)$ is increasing function, then $x>0 \implies g(x) >g(0)=0$, we prove that $$x-\ln(1+x)>\frac{x^2}{2(1+x)}.$$


问:
No cardinal between that of a finite set and the set plus one element
How can we show that there is no set with cardinality strictly between a finite set S and the set $S \bigcup\{S\}$ ?

答:
Generally, we say that $|A|\leq|B|$ iff there exists an injection from $A\to B$, and $|A|<|B|$ if $|A|\leq|B|$ and there is no bijection between $A,B$.

Suppose there exists some $X$ such that $S\to X\to S\cup\{S\}$ is a sequence of strict injections. We know that $f:S\to X$ is not a bijection, so there exists some $x\in X$ such that there is no $s\in S$ where $f(s)=x$. So, we can construct an injection $g:S\cup\{S\}\to X$ given by $g(s)=\begin{cases}x&s=S\\f(s)&s\neq S\end{cases}$. This, by C-S-B, shows there exists a bijection between $S\cup\{S\},X$. This is a contradiction.

Note that I haven't used $S$ being finite, since the claim doesn't require it.


问:
Interpreting ETCS in ZFC
I am currently doing a project on foundation of mathematics. Set theory and category theory are two different approaches to describe the core principles. ZFC is an axiomatic system which is inspired by set theory that is commonly accepted as the foundation of maths. Then there's also ETCS axiomatisation which has categorical origins.

I am looking into category theory as of now to have a better undertstanding
of the context and motivation behing ETCS and looking to interpret ETCS in ZFC. However, i am very overwhelmed and not sure where to start my research into category theory and struggling to find any good reasources that shows the connection between the two axiomatic systems. So i would greatly appreciate if you guys could let me know some good introductory resources.

Thank you in advance.

答:
The most introductory material related to ETCS is Trimble’s informal notes starting here: https://ncatlab.org/nlab/show/Trimble+on+ETCS+I

However, this may not be introductory enough, depending on your background, since you say you’re quite new to category theory. Leinster’s book Basic Category Theory is where’d I’d usually send an undergraduate (my best guess at your level, do let me know if I’ve misjudged) wanting to learn some category theory proper, which is likely to be necessary to understand much about ETCS. You would also likely find Lawvere and Rosebrugh’s book Sets for Mathematics a useful and initially gentle introduction to both category theory and how it thinks about set theory.

It might be worth noting a warning here that ETCS is not explicitly used as a logical foundation by any (as far as I know) working mathematicians in the way that ZFC is. Rather it purports to better explicate how such mathematicians actually assume sets to behave. Thus the purposes of the two systems are not perfectly comparable. Penelope Maddy’s papers on “believing the axioms” might be a good starting point if you hope to clarify your thoughts about what exactly a foundation is for.

Title: Interpreting ETCS in ZFC


答:
The sum of the roots $K_1+\dots+K_{n-1}+K_n$ is $-a_{n-1}\in Q\subseteq Q[K_1,\dots, K_{n-1}]$.

Also  $K_1+\dots+K_{n-1} \in Q[K_1,\dots, K_{n-1}]$.

So $K_n\in Q[K_1,\dots, K_{n-1}]$.

Title: "Reducing" spliting fields


问:
Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$

How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$

Attempt:

$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$

Any hints would be appreciated.

Edit: Testing it with different values of $n$, it seems like the integral evaluates to be $\frac{n \pi^2}{6(n+1)}$

答:
I don't know how to evaluate
$$\int_0^1\frac{\ln(1-x^{n+1})}{x}dx$$
(sorry) but I can do the other one:
$$\int_0^1-\frac{\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$
This result follows when you consider the Maclaurin expansion of the integrand and on recalling the solutions to the Basel problem, as shown below:
$$\begin{align}\int_0^1-\frac{\ln(1-x)}{x}dx&=\int_0^11+\frac{x}{2}+\frac{x^2}{3}+\frac{x^4}{4}+\cdots dx\\
&=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\\
&=\frac{\pi^2}{6}\end{align}$$


问:
how to show that this is isomorphic to a certain quotient group
Let $A,B,C$ be normal subgroups of a group $G$. Suppose that in turn, $A$ is a normal subgroup of $B$.

How do I show that $B/(A(B \cap C))$ is isomorphic to a quotient group of $B/A$ ?

答:
BIG HINT

The comments are directing you to use the following Theorem.

If $N$ and  $K$ are normal subgroups of G such that  $N\subseteq K\subseteq G$, then $G/K$ is isomorphic to the quotient group $(G/N)/(K/N)$ .

All you need to do is to apply this to an appropriate choice of $N,K,G$.

Can you now see what this choice must be?


问:
How many ways can we write a word of 4 letters from the group of {1,2,3,4} without the sequence 12 and 23?
How many ways can we write a word of 4 letters from the group of {1,2,3,4} without sequence 12 and 23?
the options are:

16
2.256
3.172
4.24

I think we can repeat on the same letter so I tried to this:
count all of the words without any constraint: $4^4$
count the illegal words and then to do: $4^4$- illegal--> to get what I want.
but I didn't get any of these options.
thank you for your help!

答:
In retrospect, this response is nothing but a re-hash of the comments of J Moravitz, which preceded it.

Hints:

When employing Inclusion-Exclusion in this problem, things to note:

Superficially, you would surmise that the number of ways that the sequence "12" can occur at least once is $\binom{3}{1} \times 4^2 = 48.$  This is wrong, since it over-counts the sequence "1212".  Therefore, the correct enumeration here is $47.$

When enumerating the number of ways that the sequences "12" and "23" can both occur, for the add-back, re Inclusion-Exclusion, you have to count separately : "X123", "123X", "1223", and "2312".


问:
If for every continuous distribution function $F$, $E[F(X_n)] \to E[F(X)]$ then $X_n \Rightarrow X$
$(X_n)_n$ and $X$ are random variables.

Show that $(X_n)_n$ converges in distribution to $X$ if and only if $$\lim_nE[F(X_n)]=E[F(X)]$$ for every continuous distribution function $F.$

$\implies$ is trivial, Any ideas for the converse?

答:
We know that $X_n\Rightarrow X$ iff $F_{X_n}(x) \to F_X(x)$ in all continuity points of $F_X$, so it is enough to prove the former convergence. This is done in quite a standard way; actually, the standard proof that the weak convergence implies such pointwise convergence uses monotone continuous functions.

Specifically, for any fixed $x<y$ define $$
F_{x,y} (z) = \begin{cases} 0,& z<x,\\
\frac{z-x}{y-x},& x\le z\le y\\
1, &z>y.
\end{cases}
$$
Then, $\mathrm{P}(X_n>x)\ge \mathrm{E}[F_{x,y}(X_n)]$, so noting that $F_{x,y}$ is a continuous cdf and letting $n\to\infty$ we obtain
$$
\liminf_{n\to\infty} \mathrm{P}(X_n>x)\ge \mathrm{E}[F_{x,y}(X)] \ge \mathrm{P}(X>y).
$$
Now letting $y\to x+$ gives
$$
\liminf_{n\to\infty} \mathrm{P}(X_n>x)\ge \mathrm{P}(X>x).
$$
Similarly, for any $t<x$, from $\mathrm{P}(X_n>x)\le \mathrm{E}[F_{t,x}(X_n)]$, we get that
$$
\limsup_{n\to\infty} \mathrm{P}(X_n>x)\le \mathrm{P}(X\ge x).
$$
Consequently,
$$
\lim_{n\to\infty} \mathrm{P}(X_n>x)= \mathrm{P}(X>x)
$$
whenever $\mathrm{P}(X = x)=0$, and this is easily seen to be equivalent to the desired statement.


问:
Probability - Discrete Maths

Donal, Joy and Padraig run a dental practice in collaboration. The
  following table shows the percentage of llings that they carry out,
  and the percentage of those llings that they each carried out that
  later on turn out to be faulty:

Donal completed 32% of fillings and 5% of these were faulty. Joy
  completed 41% of fillings and 2% of these were faulty. Padraig
  completed 27% of fillings and 7% of these were faulty.

Let D denote the event that Donal did the filling, J that Joy did the filling and P that Padraig did the filling. Compute P(D), P(F|J), P(F) AND P(P|F).
  Blockquote

If someone could help me out with this question, it would be great. Thank you very much.

答:
Judging from the text in the question, I'll assume that the event $F$ is that the filling turns out to be "faulty".

If that is the case, then here are a few hints to get you started.

(1) For $P(D)$ we want the probability that Donald did the filling. In this context, another way of asking this question is what percent of the fillings were done by Donald?

(2) For $P(F|J)$ we want the probability that a filling is faulty given that it was done by Joy. Here we only want to focus on fillings done by Joy, so we ignore any info about Donald and Padraig. Then, what percent of Joy's fillings turn out to be faulty?

(3) For $P(F)$ we want the probability that a filling turns out to be faulty. Essentially we want to add up the following probabilities:

$$P(Faulty) = P(\text{Faulty and done by Donald}) + P(\text{Faulty and done by Joy}) + P(\text{Faulty and done by Padraig})$$

Here we use the fact that, for instance, $P(F \,and\, D)=P(F|D)P(D)$, so we really need

$$P(F) = P(F|D)P(D) + P(F|J)P(J) + P(F|P)P(P)$$

and these are all probabilities you can find using the hints above.

Hope that helps get you started!

Title: Probability - Discrete Maths


问:
Analyse convergence of: $\displaystyle \sum^{\infty}_{n=3} \left (\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right)$

Analyse convergence of:
$$\displaystyle \sum^{\infty}_{n=3} \left(\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right )$$

I know that all elements are negative, so I need to take $-1$ out and I will have only positive elements and then I can use comparison criterion. However, I don't know how to do the first step. I need to make it rational.

答:
If you are able to prove that both series$$\sum_{n=3}^\infty\sqrt[3]{n^3+2}-n\quad\text{and}\quad\sum_{n=3}^\infty\sqrt{n^2+7}-n\tag1$$converge, or that one of them converges and the other one diverges, then you're done.

Now, note that\begin{align}\sqrt[3]{n^3+2}-n&=\sqrt[3]{n^3+2}-\sqrt[3]{n^3}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\sqrt[3]{n^3}+\sqrt[3]{n^3}^2}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\,n+n^2}\end{align}and therefore$$\lim_{n\to\infty}\frac{\sqrt[3]{n^3+2}-n}{\frac1{n^2}}=\frac23$$and therefore the first series from $(1)$ converges.

Can you take it from here?


问:
Formula for arc length of a curve using limit only (instead of integration.)
I know that formula of a length of arc of a curve $y=f(x)$ from abscissae a to b is given by $$\int\limits_{a}^{b} \sqrt{1+\{f^\prime (x)\}^2}\;\mathrm{d}x$$
But according to my college professor, he says that formula to determine the length of arc is
$$\lim\limits_{h\rightarrow 0} \sum\limits_{r=1}^{\frac{b-a}{h}} \sqrt{\left\{ f\Big(a+rh\Big)-f\Big(a+(r-1)h\Big)\right\}^2 +h^2}$$

He also asked us to find the length of circumference of the unit circle at origin $(x^2+y^2=1)$ using this limit formula only, but I don't know how I can compute this. So please can anyone help me to solve this in this particular way using limits.

答:
The $y=f(x)$ in this case becomes $y=\pm\sqrt{1-x^2}$ and using symmetry across the $x$- and $y$-axes we can take $x\in[0,1]$, so with your limit formula we have

$$4\lim_{h\to 0} \sum_{r=1}^{\frac1h}\sqrt{\left\{\sqrt{1-(rh)^2}-\sqrt{1-((r-1)h)^2}\right\}^2 +h^2}\\
=4\lim_{h\to 0}\sum_{r=1}^{\frac 1h}\sqrt{2+2rh^2-2r^2h^2-2\sqrt{(1-r^2h^2)(1-r^2h^2+2rh^2-h^2)}}\\
=4\sqrt 2\lim_{h\to 0}\sum_{r=1}^{\frac 1h}\sqrt{1+rh^2-r^2h^2-\sqrt{(1+rh^2-r^2h^2)^2-h^2}}\\
r\in\left[1,\frac 1h\right]\to 1+rh^2-r^2h^2\in[h,1],\frac 1h\to n, \frac 1{1+rh^2-r^2h^2}\to t^2, t\in[1,\sqrt n]\\
=4\sqrt 2\lim_{n\to\infty}\frac 1n\sum_{t=1}^{\sqrt n}\sqrt{\frac 1{t^2}-\sqrt{\frac 1{t^4}-\frac 1{n^2}}}\\
=4\sqrt2\lim_{n\to\infty}\frac1{n}\sum_{t=1}^{\sqrt n}\frac {\sqrt{1-\sqrt{1-\frac{t^4}{n^2}}}}t\\
=4\sqrt2\lim_{n\to\infty}\frac 1n\sum_{t=1}^{\sqrt n}t\sqrt{\frac{\frac1{n^2}}{1+\sqrt{1-\frac{t^4}{n^2}}}}\\
=4\sqrt2\lim_{n\to\infty}\frac1{n^2}\sum_{t=1}^{\sqrt n}\frac t{\sqrt{1+\sqrt{1-\frac{t^4}{n^2}}}}$$

which is still a little ways from something reasonable.

We can use one approach as described in https://math.stackexchange.com/a/217251/86846 by considering the area of the same circle, especially if we take areas of triangles having height sides equal to the radius (i.e., $1$), and their bases as the same length calculation as used for our arc length, so each area is $\frac 12bh$, or

$$\frac 12\cdot 1\cdot\sqrt{\left(\sqrt{1-r^2h^2}-\sqrt{1-(r-1)^2h^2}\right)^2+h^2}$$

giving our total area formula (converging from overestimates) as

$$2\lim_{h\to 0} \sum_{r=1}^{\frac1h}\sqrt{\left\{\sqrt{1-(rh)^2}-\sqrt{1-((r-1)h)^2}\right\}^2 +h^2}$$

This approach only gives us that the circumference of the unit circle is twice the area, but at least it's a minor bit of progress.

For an alternate approach, consider taking a cue from https://tutorial.math.lamar.edu/classes/calcii/arclength.aspx on this formula and using one more "symmetry fold", allowing us to rewrite the initial limit as

$$8\lim_{h\to 0}\sum_{r=1}^{\frac 1{\sqrt2h}}h\sqrt{1+\left(\frac{-rh}{\sqrt{1-r^2h^2}}\right)^2}\\
=8\lim_{h\to 0}\sum_{r=1}^{\frac 1{\sqrt2h}}h\sqrt{1+\frac{r^2h^2}{1-r^2h^2}}\\
=8\lim_{h\to 0}\sum_{r=1}^{\frac 1{\sqrt2h}}\frac h{\sqrt{1-r^2h^2}}\\
=8\lim_{n\to\infty}\sum_{r=1}^\frac{\sqrt 2n}2\frac 1{\sqrt{n^2-r^2}}$$

but this fails to meet the professor's strict instructions...


问:
Matrix inverses over finite fields with composite moduli
I know that over a field $F$, a matrix is invertible if and only if its determinant is nonzero. And I understand why this is true, at least in the case where the field is $\mathbb{R}$.

But I do not understand how to actually compute the inverse over a finite field. In this question, there is a suggestion to use the formula involving the adjugate matrix, that works over $\mathbb{R}$. Here, another user suggests row reduction and provides a helpful step-by-step walkthrough, but the modulus is prime so everything is invertible.

He writes:

For extra credit review the algorithm, and observe that, all the operations and laws of algebra that the algorithm depends on, hold in all fields.

But I can't see why this is true. For instance, what if the determinant of the matrix (which it seems we usually end up dividing by when we take the inverse) has no multiplicative inverse in $F$?

答: By definition, every nonzero element in a field has a multiplicative inverse. When you're working over a non-field like $\mathbb Z_6$, then your worries come true, and you can have two matrices with non-zero determinant whose product is the zero matrix. Can you give some $2\times 2$ examples?


答:
Hint Use that when $\alpha\not = 0$,
\begin{equation}
(1 + x)^\alpha - 1 \sim \alpha x
\end{equation}
when $x\longrightarrow 0$.

Title: Analyse convergence of: $\displaystyle \sum^{\infty}_{n=3} \left (\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right)$

Body:
Analyse convergence of:
$$\displaystyle \sum^{\infty}_{n=3} \left(\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right )$$

I know that all elements are negative, so I need to take $-1$ out and I will have only positive elements and then I can use comparison criterion. However, I don't know how to do the first step. I need to make it rational.


问:
Prove this formula $\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}= \sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)$
I wish to use geometric power series of $\dfrac{1}{1-x}$ to prove this formula by Euler

$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...=1+\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)$$

In this thread, I have understood how to prove the same formula for $\cos(nx)$

Please see the answer of A-Level Student to see, his answer is:

Let
$$\begin{align}
C&=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\
S&=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\
\end{align}$$
Then
$$\begin{align}
C+iS&=1+r(\cos x+i\sin x)+r^2(\cos 2x+i\sin2x)+r^3(\cos3x+i\sin3x)+\cdots\\
&=1+re^{ix}+(re^{ix})^2+(re^{ix})^3+\cdots\\
&=\frac{1}{1-re^{ix}}=\frac{(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{1-r\cos x+ri\sin x}{1-2r\cos x+r^2}
\end{align}$$
Hence, equating real and imaginary parts we obtain
$$\begin{align}
C&=\frac{1-r\cos x}{1-2r\cos x+r^2}\\
S&=\frac{r\sin x}{1-2r\cos x+r^2}\\
\end{align}$$

I wish to use a similar method to prove the above mentioned series for $\sin(nx)$, but I am not sure value which series for of $C$ and $S$, should I choose.

I have chosen $C=1+r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...$

$S=rcos(x)+r^2cos(2x)+r^3cos(3x)+...$

And I let $1 + S + iC$, with $i$ being imaginary number.

So I have

$1 +S + iC = 1+r(\cos(x)+i\sin(x))+r^2(\cos(2x)+i\sin(2x)+...$

= $1 + S +iC = 1+ re^{ix} + (re^{ix})^2+ (re^{ix})^3+ (re^{ix})^4+ (re^{ix})^5+...$

And the $1+ re^{ix} + (re^{ix})^2+ (re^{ix})^3+ (re^{ix})^4+ (re^{ix})^5+... =  \dfrac{1}{1-e^{ix}}$

It is here that I am stuck, I don't think my choice of $S$ and $C$ is correct. I know that the Euler identity for $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$, but I don't know how to apply this formula.

答:
Use infinite GP:
$$S=\sum_{n=1}^{\infty} r^n \sin nx=\Im \sum_{n=1}^{\infty} r^ne^{inx}=\Im \frac{re^{ix}}{1-re^{ix}}$$
$$\implies S=\Im \frac{re^{ix}(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{r\sin x}{1-2r\cos x+r^2}$$
This result has been check with Mathematica numerically. For instance for $r=1/2,x=\pi/3$, we get $S=1/\sqrt{3}$


问:
Is there any analytic method for solving $\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$
Hi i am trying to solve the following two equations:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that  is $\phi(\theta)$. Is there any way to do this?

答:
$$\frac{d^2\phi}{dt^2} + 2\cot(\theta)\frac{d\theta}{dt}\frac{d\phi}{dt} =0$$
This separable ODE is directly integrable.
$$\ln(\frac{d\phi}{dt})+2\ln(\sin(\theta))=\text{constant}$$
$$\sin^2(\theta)\frac{d\phi}{dt}=c_1\quad\implies\quad \frac{d\phi}{dt}=\frac{c_1}{\sin^2(\theta)}$$
Second equation :
$$\frac{d^2\theta}{dt^2}-\sin(\theta)\cos(\theta)\left(\frac{d\phi}{dt}\right)^2=0$$

$$\frac{d^2\theta}{dt^2}-\sin(\theta)\cos(\theta)\left(\frac{c_1}{\sin^2(\theta)}\right)^2=0$$
$$2\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}-2(c_1)^2\frac{\cos(\theta)}{\sin^3(\theta)}\frac{d\theta}{dt}=0$$
$$\left(\frac{d\theta}{dt}\right)^2+(c_1)^2\frac{1}{\sin^2(\theta)}=(c_2)^2$$
$$\frac{d\theta}{dt}=\pm\sqrt{(c_2)^2-(c_1)^2\frac{1}{\sin^2(\theta)}}$$
$$t=\int \frac{\sin(\theta)}{\pm\sqrt{(c_2)^2\sin^2(\theta)-(c_1)^2 }}d\theta$$
$$t=\pm \sin^{-1}\left(\frac{c_2}{(c_1)^2+(c_2)^2}\cos(\theta) \right)+c_3$$
$$\frac{c_2}{(c_1)^2+(c_2)^2}\cos(\theta)=\pm\sin(t-c_3)$$
$$\theta(t)=\cos^{-1}\left(\pm\frac{(c_1)^2+(c_2)^2}{c_2}\sin(t-c_3) \right)$$


问:
Evaluate $E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] $, where $\{B(t); t\ge 0\}$ is a standard Brownian Motion
Consider a standard Brownian motion $\{B(t); t\ge 0\}$ (zero mean and $\sigma^{2} = 1 $) at times

$0 < u < u+v < u+v+w <  u+v+w+x $, where $u, v, w, x > 0$

Evaluate the product moment:

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] $$

I tried using the same trick employed when solving $E[B(t)B(s)] = \sigma^{2} min\{s,t\}$.

Namely for $0<s \le t$,

$$E[B(t)B(s)] = E[B(s)\{B(t)-B(s)+B(s)]=E[B(s)^{2}]+E[B(s)]E[B(t))-B(s)]=\sigma^{2}s$$

Using the same method on the above product moment and starting with the 4th term I got

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)]\\ = \\ E[B(u)B(u+v)B(u+v+w)\{B(u+v+w+x)-B(u+v+w)+B(u+v+w)\}]$$
$$\\ = \\ E[B(u)B(u+v)B(u+v+w)^{2}] + E[B(u)B(u+v)B(u+v+w)\{B(u+v+w+x)-B(u+v+w)\}]$$

I then used the same trick with the other elements eventually I managed to end up with

$$E[B(u)^{2}B(u+v+w)^{2}] + E[B(u)\{B(u+v)-B(u)\}B(u+v+w)^{2}] \\+E[B(u)B(u+v)^{2}\{B(u+v+w+x)-B(u+v+w)\}] \\+ E[B(u)^{2}\{B(u+v+w) -B(u+v)\}\{ B(u+v+w+x)-B(u+v+w)\}] \\+ E[B(u)\{B(u+v)-B(u)\}\{B(u+v+w) -B(u+v)\}\{ B(u+v+w+x)-B(u+v+w)\}]$$

But I've gotten lost and not sure if I am going the right way, And I don't know how I would evaluate elements such as $E[B(u)^2B(u+v+w)^{2}]$ or  $E[B(u)B(u+v)^{2}\{B(u+v+w+x)-B(u+v+w)\}]$. I Believe the very last term is equal to zero by the definition of Brownian motion?

Any help would be appreciated.

答:
So I managed to figure it out so I thought I would post the solution in case anyone else is having difficulty with the same question.

Firstly one needs to use the law of total Expectation i.e. $E[E[X|Y]]=E[X]$ and apply it iteratively. So starting from the top.

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] =
E[B(u)B(u+v)B(u+v+w)E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]$$
Where $\mathcal{F}_{u}:= \sigma (\{B(t);t \le u\}) $.

Using $E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]=B(u+v+w)$ we get,

$$E[B(u)B(u+v)B(u+v+w)E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]= E[B(u)B(u+v)B^{2}(u+v+w)]$$

Applying the law of total probability again we have

$$E[B(u)B(u+v)B^{2}(u+v+w)] = E[B(u)B(u+v)E[B^{2}(u+v+w)|\mathcal{F}_{u+v}]]$$

Where $E[B^{2}(u+v+w)|\mathcal{F}_{u+v}] = B^2(u+v)+(u+v+w)-(u+v) = B^2(u+v)+w$ , so we have,

$$E[B(u)B(u+v)E[B^{2}(u+v+w)|\mathcal{F}_{u+v}]]= E[B(u)B(u+v)[B^2(u+v)+w]]$$

Expanding the right hand side we have

$$ E[B(u)B(u+v)[B^2(u+v)+w]] = E[B(u)B^3(u+v)] + wE[B(u)B(u+v)]$$

Applying the Law of total probability again to both parts of the RHS we end up with

$$E[B(u)B^3(u+v)] + wE[B(u)B(u+v)] = E[B(u)E[B^3(u+v)| \mathcal{F}_{u}]] + wE[B(u)E[B(u+v)| \mathcal{F}_{u}]] $$

where, $E[B^3(u+v)| \mathcal{F}_{u}] = B^3(u) +3B(u)((u+v)-u))=B^3(u)+3B(u)v$ insertiing this into the first term on the RHS we get

$$E[B(u)E[B^3(u+v)| \mathcal{F}_{u}]] =E[B(u)(B^3(u)+3B(u)v)] = E[B^4(u)]+3vE[B^2(u)] = 3u^2+3uv$$

Looking at the second term on the RHS we have

$$ wE[B(u)E[B(u+v)| \mathcal{F}_{u}]]=wE[B^2(u)]=wu$$

Where we used the result $E[B(u+v)| \mathcal{F}_{u}]=B(u)$

Hence we end up with the full answer as

$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] = 3u^2+3uv+uw $$


问:
What is the complement of empty set?
We have learned that for universal set $U$ and  for set $A$ belonging to $U$,
$A^c=U-A$ but what about for the empty set?</p>

答: But U will still contain the empty set, since all sets contain it. It seems nonsensical to talk about U without the null set, since the complement set still contains it. Help.</p>

Title: What is the complement of empty set?


答:
A short supplement to A-Level-Student's answer:

The first integral
$$\int_0^1\frac{\ln(1-x^{n+1})}{x}\mathbb dx$$
can be represented by the second one
$$I:=-\int_0^1\frac{\ln(1-x)}{x}\mathbb dx$$
by the substitution $u:= x^{n+1}$ and thus $1/u\cdot\mathbb du=(n+1)/x\cdot\mathbb dx$
$$\int_0^1\frac{\ln(1-x^{n+1})}{x}\mathbb dx=\frac{1}{(n+1)}\int_0^1\frac{\ln(1-u)}{u}\mathbb du=-\frac{I}{(n+1)}.
$$

Title: Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$

Body:
How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$

Attempt:

$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$

Any hints would be appreciated.

Edit: Testing it with different values of $n$, it seems like the integral evaluates to be $\frac{n \pi^2}{6(n+1)}$


问:
Why does calculating the return on a portfolio differ when calculated at a stock level rather than at a portfolio level?
I have a portfolio of three assets A, B and C, each with a beginning value of 100. I trade the portfolio over two days. On day 1, asset A returns 50, B returns 20, and C returns -10. On day two, A returns 0, B returns 0, and C returns 30. These are dollar returns.

When I calculate the returns of each stock over each day, the contribution to return as a percentage of the portfolio's beginning value is right. So, on day 1, A returns 50% (50/300), B returns 20% and C returns -3.3%. And on day two, A returns 0%, B gives 0% and C gives -8.3% (-30/(150+120+90)).

The return of the portfolio for two days is 10%. The profit in dollar terms is 30, and the daily return for the portfolio is 20% on day 1 and -8.3% on day two, which compounds to 10% using the time weighted rate of return. But, if I calculate the returns for each stock and then sum those returns (as a percentage), it does not add up. So, for stock A the return is 16.7%, B gives 6.7%, and C gives -11.4%, and these sum to 11.9%. Why does 11.9% differ from 10%, and what is the right way to adjust my stock level returns to match the 10% of the portfolio? Please see below the screenshot of the daily calculation. Essentially, the cells highlighted in green should sum up to the blue cell. In the picture below, Opening NAV = value of each stock at the beginning of day, Weight = weight of each stock in the portfolio at beginning of day, PnL = profit of each stock for the day, %ctr 1 = the stock's contribution to return of the portfolio (PnL divided by the value of the portfolio at the beginning of day).
Portfolio returns over two days

Edit: Please see a clarification of the problem below.

So at the end of day 1 the market value of A is 150, B is 120, and C is 90. Then on day two A has a profit of zero (the stock price doesn't move), B has a profit of zero, and C has a profit of -30. So the market values of A,B, and C at the end of day two are 150, 120, and 60 respectively. The returns of A,B and C for day two are 0% (0/360), 0% (0/120), and -8.3% (-30/360). So, over two days, what is the return of the portfolio? The return of the portfolio on day 1 is 60/300 (20%), and on day two it is -30/360 (-8.3%). So the return for the portfolio for the entire period is (1+20%)(1-8.3%)-1 = 10%. This is the time weighted rate of return formula and makes sense because your closing NAV is 330 and you began with 300 so you made 10%. BUT, AT A STOCK LEVEL, WHAT WAS YOUR RETURN OVER THE TWO DAY PERIOD? For stock A, I calculate (1+16.7%)(1+0%)-1 = 16.7%, For B, I calculate (1+6.7%)(1+0%)-1 = 6.7%, and then for C I calculate (1-3.3%)(1-8.3%)-1 = -11.4%. If I sum 16.7%+6.7%-11.4% = 12%. This differs from my portfolio return of 10%!

Edit: The key part of the question is to link the stock level returns to the portfolio return over multiple periods. So, if I have a portfolio total return of 10% over multiple periods, how do the stocks contribute to that return ie stock A contributed 9%, B contributed 5% and c contributed -4% ie 9% + 5% - 4% = 10%.

答:
Here you have a portfolio of 3 assets ($A,B,C$) and $2$ days. Firstly you calculate the $\textrm{growth factor}$ of the first day of the portfolio, from time $0$ to time $1$:

$$1+r_{p,01}=\frac{X_1^A+X_1^B+X_1^C}{X_0^A+X_0^B+X_0^C}=\frac{150+120+90}{100+100+100}=\frac{360}{300}=\frac{12}{10}=1.2,$$

where $X_t^A,X_t^B,X_t^C$ are the net asset values at time $t$.  Next you calculate the growth factor of the second day of the portfolio, from time $1$ to time $2$:

$$1+r_{p,12}=\frac{X_2^A+X_2^B+X_2^C}{X_1^A+X_1^B+X_1^C}=\frac{150+120+60}{360}=\frac{330}{360}=\frac{11}{12}\approx 0.9167$$

Then the growth factor for the portfolio in two days is $1+r_{p,02}=(1+r_{p,01})\cdot (1+r_{p,12})$ $=\frac{12}{10}\cdot \frac{11}{12}=1.1$. Thus the growth rate of the portfolio in two days is $r_{p,02}=1.1-1=10\%$

And the average daily growth rate of the portfolio is $\sqrt{1+r_{p,02}}-1=\sqrt{1.1}-1\approx 4.88\%$

Update

The contribution (in percent) of asset A at day 1 to the return of the portfolio is $r_{A1}^c=\frac{150-100}{300}=\frac16=16.\overline 6\%\approx 16.7\%$

The contribution (in percent) of asset C at day 1 to the return of the portfolio is $r_{C1}^c=\frac{90-100}{300}=\frac{-1}{30}=-3.\overline 3\%\approx -3.3\%$

The contribution (in percent) of asset C at day 2 to the return of the portfolio is $r_{C2}^c=\frac{60-90}{360}=\frac{-1}{12}=-8.\overline 3\%\approx -8.3\%$

Finally the ontribution (in percent) of asset C from time $t=0$ to time $t=2$ to the return of the portfolio in that period is

$$(1+r_{C1}^c)\cdot (1+ r_{C2}^c)-1=\left(1-\frac{1}{30}
\right)\cdot \left(1-\frac{1}{12}\right)-1$$ $$\approx 0.9666666\cdot 0.9166666-1=-0.11388901444444\approx -11.4\%$$

$\underline{\textrm{Important to mention}}$: Consecutive growth rates $r_i$ of $n$ periods  are never summed up. You multiply the the growth factors $(q_i=1+r_i)$ and  subtract 1 to obtain the growth rate in $n$ periods:

$$r_{0,n}=\prod\limits_{i=1}^n (1+r_i)-1=\prod\limits_{i=1}^n q_i-1$$


答:
Ok we have:
$$
I = \int_0^1 \frac{\ln(\sum_{k=0}^n x^n)}{x}dx
$$
Using:
$$
S_n = \frac{1-r^{n+1}}{1-r}
$$
$$
I = \int_0^1 \frac{\ln\left( \frac{1-x^{n+1}}{1-x} \right)}{x}dx
$$
Using the fact that $\ln(a/b) = \ln(a) - \ln(b)$
$$
I = \int_0^1 \frac{1}{x} \ln\left(1-x^{n+1} \right)dx - \int_0^1 \frac{1}{x} \ln(1-x) \, dx
$$
We now use $\ln$'s taylor series:
$$
\ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k}
$$
We obtain
$$
I = \int_0^1\sum_{k=1}^\infty \frac{x^{(n+1)k-1}}{k} dx - \int_0^1 \sum_{k=1}^\infty \frac{x^{k-1}}{k} dx
$$
Switching the bounds with the summation because of monotone convergence and integrating:
$$
I =  \left[ \sum_{k=1}^\infty \frac{x^{k(n+1)}}{(n+1)k^2} \right]_0^1 -\left[ \sum_{k=1}^\infty \frac{x^k}{k^2} \right]_0^1
$$
$$
I= \frac{\zeta(2)}{n+1} - \zeta(2)
$$
$$
I = \left(\frac{-n}{n+1}\right) \frac{\pi^2}{6}
$$


问:
Suppose $f,g$ are two absolutely continous functions in an interval $[a,b]$. Show $fg \in AC[a,b]$.

Suppose $f,g$ are two absolutely continous functions in an interval $[a,b]$. Show $fg \in AC[a,b]$.

My try:
$$
\begin{align}
\int_a^x (fg)'(t)dt & \stackrel{(1)}= \int_a^x \Big[f'(t)g(t) + f(t)g'(t)\Big]dt \\[8pt]
& = \int_a^x f'(t)g(t)dt  + \int_a^xf(t)g'(t)dt \\[8pt]
& \stackrel{(2)}= f(x)g(x) - f(a)g(a) - \int_a^xf(t)g'(t)dt + \int_a^xf(t)g'(t)dt \\[8pt]
& = f(x)g(x) - f(a)g(a)
\end{align}
$$
This should conclude the proof but I'm not very sure if I can use the product rule in $(1)$ and if I can integrate by parts the first integral in $(2)$.

Is there any other way to prove this fact?

答:
We can prove directly.

Choose $M>0$ such that $|f(x)|\leq M$ and $|g(x)|\leq M$ for all
$x\in[a,b]$. Let $\varepsilon>0$ be given. Since $f$ and $g$ are
absolutely continuous, there exists $\delta>0$ such that for any
(fintely many) pairwisely disjoint subintervals $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$
of $[a,b]$, if $\sum_{i=1}^{n}(y_{i}-x_{i})<\delta$, then $\sum_{i=1}^{n}|f(y_{i})-f(x_{i})|<\frac{\varepsilon}{2M}$
and $\sum_{i=1}^{n}|g(y_{i})-g(x_{i})|<\frac{\varepsilon}{2M}$. Now,
let $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$ be pairwisely disjoint subintervals
of $[a,b]$ with $\sum_{i=1}^{n}(y_{i}-x_{i})<\delta$. For each $i$,
observe that
\begin{eqnarray*}
 &  & \left|f(y_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
 & \leq & \left|f(y_{i})g(y_{i})-f(x_{i})g(y{}_{i})\right|+\left|f(x_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
 & \leq & M\left|f(y_{i})-f(x_{i})\right|+M\left|g(y_{i})-g(x_{i})\right|.
\end{eqnarray*}
It follows that
\begin{eqnarray*}
 &  & \sum_{i=1}^{n}\left|f(y_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
 & \leq & M\sum_{i=1}^{n}\left|f(y_{i})-f(x_{i})\right|+M\sum_{i=1}^{n}\left|g(y_{i})-g(x_{i})\right|\\
 & < & \varepsilon.
\end{eqnarray*}
Therefore $fg$ is absolutely continuous.