问:
Computing continued fraction expansions
My question concerns the numerical accuracy of a continued fraction expansion.  A typical algorithm for computing a continued fraction can be written in Python as :

```<code>x0 = sqrt(2)

N = 40
a = [0]*N
u = [0]*N

x = x0
for k in range(N):
    a[k] = int(x//1)
    u[k] = x % 1      # Often replaced with x - a[k]  ???
    x = 1/u[k]

print(a)
</code>```

For $x=\sqrt{2}$, this produces

```<code>[1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 1, 3, 1, 1, 2, 1809, 1, 2, 5, 2, 2, 1, 2, 1]
</code>```

Or for $x = \pi$, I get :

```<code>[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 3, 3, 23, 1, 1, 7, 4, 35, 1, 1, 1, 2, 3, 3, 3, 3, 1, 1, 14, 6, 4, 5, 1, 7, 1, 5, 1]
</code>```

One observation is that round-off error has clearly crept into the calculation.  The expansion for $\sqrt{2}$ should be repeating 2s, and the approximation for $\pi$ is accurate up to the first appearance of 14 (12 terms after the leading $a_0=3$ term).

What I also found is that while round off error does creep in, the number of correct terms in the expansion is exactly what is needed to get convergents that agree with $x$ up to machine precision.  So for example,  convergent 20 of $\sqrt{2}$  is given by

```<code>54608393 / 38613965
</code>```

This convergent approximates $\sqrt{2}$ to $4.44 \times 10^{-16}$.  Convergent 21 agrees with $\sqrt{2}$ exactly (in finite precision arithmetic).  Interestingly, the expansion has exactly 20 correct terms (i.e. 2s) after the leading $a_0 = 1$.

Similarly, for $\pi$, the convergent 12 is given by

```<code>80143857 / 25510582
</code>```

which agrees with $\pi$ to within an error of $4.44 \times 10^{-16}$.  Convergent 13 agrees with $\pi$ exactly (in finite precision arithmetic).

In both cases, the numerators for the convergents agree with the OEIS. See A001333 and A002485.

The above observations led me to these questions :

Is there a standard stopping criteria that can be used to determine when the expansion has converged to a desired tolerance?

Is it always the case that one will have enough correct terms in the expansion to approximate the desired number to machine precision?

Is it possible to detect whether a continued fraction is periodic?  Or, if one knows it will be periodic (i.e. $x$ is a root of a quadratic with integer coefficients), is it possible to get the periodic sequence exactly?

It has also occurred to me that nobody would think of computing continued fraction expansions using finite precision arithmetic!  I would like to do this problem as an exercise in a beginning computational math course, and was hoping not to go into variable precision arithmetic (an area which is really outside of my wheel house).

答:
Here is a program and the protocol, using Pari/GP; but I hope the program is understandable. The stopping-criterion is, whether the error in the reconstruction of the intended irrational number is equal the machine-epsilon.

I set the Pari/GP precision to small number of digits and then run the loop. Using $\sqrt2$.

```<code>fmt(10)       \\ set very small ;-) internal precision (19 dec digits)
va=vector(50) \\ this takes the found coefficients, set length as you think serves well
w=b=sqrt(2)   \\ this is the initial value, shall have 18 or 19 dec digits correct
ia=0          \\ through iterating, this is the index for the va-vector
pq=[0,1;1,0]  \\ this captures the current convergents (p,q)
              \\  allows checking whether  w - p/q  <= machine-epsilon

</code>```

protocol:

```<code>\\  w=  1.41421356237 \\ protocol of w=b=sqrt(2), internal 18 or 19 dec digits correct
</code>```

program:

```<code>{for(k=1,50,  \\ set upper limit as you think serves well
    a=floor(b);
      ia++; va[ia]=a; \\ save the cf-digit (="cf-quotient") in vector
    b=1/(b-a);
    pq=pq*[0,1;1,a];  \\ compute current convergent
    p=pq[1,2];q=pq[2,2];
    print([k,a,err = w - p/q]); \\ document progress
    if(err==0.0,break());   \\ if machine-epsilon reached, stop iteration
    );}

</code>```

Protocol:

```<code> k  a  err             :: a=current quotient/digit, err=w - p/q
[1, 1, 0.414213562373]
[2, 2, -0.0857864376269]
[3, 2, 0.0142135623731]
[4, 2, -0.00245310429357]
[5, 2, 0.000420458924819]
[6, 2, -0.0000721519126192]
[7, 2, 0.0000123789411424]
[8, 2, -0.00000212390141476]
[9, 2, 0.000000364403551901]
[10, 2, -0.0000000625217745896]
[11, 2, 0.0000000107270403544]
[12, 2, -0.00000000184046916492]
[13, 2, 0.0000000003157745862]
[14, 2, -5.417835374 E-11]
[15, 2, 9.29553586 E-12]
[16, 2, -1.594861909 E-12]
[17, 2, 2.736349747 E-13]
[18, 2, -4.694840810 E-14]
[19, 2, 8.05490168 E-15]
[20, 2, -1.382117461 E-15]
[21, 2, 2.370426006 E-16]
[22, 2, -4.074183443 E-17]
[23, 2, 6.853781456 E-18]
[24, 2, -1.285552860 E-18]
[25, 2, 8.87489049 E-20]
[26, 2, -1.470440003 E-19]
[27, 1, 0.E-18]             \\ stopped at iteration 27
</code>```

Show continued-fraction quotients

```<code> va=vecextract(va,Str("1..",ia))} \\ remove superfluous trailing entries in va
 %630 = [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
            2, 2, 2, 2, 2, 2, 2, 2, 1]
 -----------------------------------^ last entry is likely wrong
</code>```

Using $w=\pi$ gives the following protocol

```<code>[1, 3, 0.141592653590]
[2, 7, -0.00126448926735]
[3, 15, 0.0000832196275291]
[4, 1, -0.000000266764189063]
[5, 292, 0.000000000577890634]
[6, 1, -0.0000000003316278063]
[7, 1, 1.223565329 E-10]
[8, 1, -2.914338504 E-11]
[9, 2, 8.71546711 E-12]
[10, 1, -1.610740232 E-12]
[11, 3, 4.040668867 E-13]
[12, 1, -2.214484967 E-14]
[13, 14, 5.78983025 E-16]
[14, 2, -1.641519209 E-16]
[15, 1, 7.80993605 E-17]
[16, 1, -1.932805900 E-17]
[17, 2, 3.079202631 E-18]
[18, 2, -5.73807725 E-19]
[19, 2, 0.E-18]
</code>```

and the continued fraction

```<code> %645 = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2]
 ------------------------------------------------------------------^
           (last quotient a=2 likely wrong)
</code>```

Title: Computing continued fraction expansions

Body:
Here is a program and the protocol, using Pari/GP; but I hope the program is understandable. The stopping-criterion is, whether the error in the reconstruction of the intended irrational number is equal the machine-epsilon.

I set the Pari/GP precision to small number of digits and then run the loop. Using $\sqrt2$.

fmt(10)       \\ set very small ;-) internal precision (19 dec digits)
va=vector(50) \\ this takes the found coefficients, set length as you think serves well
w=b=sqrt(2)   \\ this is the initial value, shall have 18 or 19 dec digits correct
ia=0          \\ through iterating, this is the index for the va-vector
pq=[0,1;1,0]  \\ this captures the current convergents (p,q)
              \\  allows checking whether  w - p/q  <= machine-epsilon

protocol:

\\  w=  1.41421356237 \\ protocol of w=b=sqrt(2), internal 18 or 19 dec digits correct

program:

{for(k=1,50,  \\ set upper limit as you think serves well
    a=floor(b);
      ia++; va[ia]=a; \\ save the cf-digit (="cf-quotient") in vector
    b=1/(b-a);
    pq=pq*[0,1;1,a];  \\ compute current convergent
    p=pq[1,2];q=pq[2,2];
    print([k,a,err = w - p/q]); \\ document progress
    if(err==0.0,break());   \\ if machine-epsilon reached, stop iteration
    );}

Protocol:

 k  a  err             :: a=current quotient/digit, err=w - p/q
[1, 1, 0.414213562373]
[2, 2, -0.0857864376269]
[3, 2, 0.0142135623731]
[4, 2, -0.00245310429357]
[5, 2, 0.000420458924819]
[6, 2, -0.0000721519126192]
[7, 2, 0.0000123789411424]
[8, 2, -0.00000212390141476]
[9, 2, 0.000000364403551901]
[10, 2, -0.0000000625217745896]
[11, 2, 0.0000000107270403544]
[12, 2, -0.00000000184046916492]
[13, 2, 0.0000000003157745862]
[14, 2, -5.417835374 E-11]
[15, 2, 9.29553586 E-12]
[16, 2, -1.594861909 E-12]
[17, 2, 2.736349747 E-13]
[18, 2, -4.694840810 E-14]
[19, 2, 8.05490168 E-15]
[20, 2, -1.382117461 E-15]
[21, 2, 2.370426006 E-16]
[22, 2, -4.074183443 E-17]
[23, 2, 6.853781456 E-18]
[24, 2, -1.285552860 E-18]
[25, 2, 8.87489049 E-20]
[26, 2, -1.470440003 E-19]
[27, 1, 0.E-18]             \\ stopped at iteration 27

Show continued-fraction quotients

 va=vecextract(va,Str("1..",ia))} \\ remove superfluous trailing entries in va
 %630 = [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
            2, 2, 2, 2, 2, 2, 2, 2, 1]
 -----------------------------------^ last entry is likely wrong

Using $w=\pi$ gives the following protocol

[1, 3, 0.141592653590]
[2, 7, -0.00126448926735]
[3, 15, 0.0000832196275291]
[4, 1, -0.000000266764189063]
[5, 292, 0.000000000577890634]
[6, 1, -0.0000000003316278063]
[7, 1, 1.223565329 E-10]
[8, 1, -2.914338504 E-11]
[9, 2, 8.71546711 E-12]
[10, 1, -1.610740232 E-12]
[11, 3, 4.040668867 E-13]
[12, 1, -2.214484967 E-14]
[13, 14, 5.78983025 E-16]
[14, 2, -1.641519209 E-16]
[15, 1, 7.80993605 E-17]
[16, 1, -1.932805900 E-17]
[17, 2, 3.079202631 E-18]
[18, 2, -5.73807725 E-19]
[19, 2, 0.E-18]

and the continued fraction

 %645 = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2]
 ------------------------------------------------------------------^
           (last quotient a=2 likely wrong)


问:
Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection
Clearly, the total number of subsets possible is $2^5$

For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.

Now there are far too many cases (IMO) for me to manually curate, so is their a shorter way, complementary probability perhaps?

Even in complimentary method there seem to be a lot of subsets, but here is what I got

$$\binom 51(\binom 41 +\binom 42 +\binom 43 +\binom 44) + \binom 52 (\binom 31 + \binom 32 +\binom 33).....+\binom 54 (\binom 11)$$
$$=5(2^4-1) +10 (2^3-1) +10(2^2-1) +5 (2-1)$$
$$=180$$

Which comes out to be greater than $2^5$, which isn’t possible. Where am I going wrong?

Edit: My reasoning

I found the cases where one set has an element, and then the other set does not have that element, although I now realise I missed the cases where they have 1 element common or 3 or 4 or 5, which effectively makes my attempt useless.

答:
Let's say instead of choosing two subsets from a single set, we choose subsets from the two sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5\}$, so number of subsets that can be chosen from $A$ are $2^{5}$ and similarly total number of subsets chosen from $B$ are $2^5$.

So, in a total of $2^5 \cdot 2^5=2^{10}$ ways, we can choose the subsets from $A$ and $B$.

Now, we need to find the cases, where there are exactly $2$ elements common to them.

If there are two elements in subset choosen from $A$, which can be done in $\displaystyle \binom{5}{2}$ ways, for two elements to be common, we can choose subsets in $B$ in $\displaystyle \binom{2}{2}$ ways, and for the rest of $3$ the elements, they can be chosen in $2^3$ ways.

If there are three elements in subset choosen from $A$, which can be done in $\displaystyle \binom{5}{3}$ ways, so choose $2$ elements from those three choosen for $A$, that is, $\displaystyle\binom{3}{2}$ and for rest choose them in $2^2$ ways.

If there four elements in subset choosen from $A$, there are $\displaystyle \binom{5}{4}$ ways. Choose two from those $4$, i.e $\displaystyle\binom{4}{2}$, choose rest in $2^1$ ways.

If all five elements are choosen in $A$, choose any two elements in $B$, for which there are $\displaystyle\binom{5}{2}$ ways.

So we have, probability that there is exactly one element common is

$$\dfrac{\binom{5}{2}\binom{2}{2}2^3+\binom{5}{3}\binom{3}{2}2^2+\binom{5}{4}\binom{4}{2}2^1+\binom{5}{5}\binom{5}{2}2^0}{2^{10}}=\dfrac{135}{512}$$

ContentLicense: CC BY-SA 4.0


问:
Tricky question about Ito's stochastic integral and continal law
Consider $B=(B_t)_{t\geq 0}$ real $\mathcal F_t$ - brownian motion  starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider a new real $\mathcal F_t$ - brownian motion $\tilde{B}=(\tilde{B_0})_{t\geq t}$ independent of $B$ as weel as a process $H=(H_t)_{t\geq 0}$ given by

$$ H_t := \frac{1}{\int _0^t f^2(B_s) ~ds}\int _0^t f(B_s) ~d \tilde B_s \mathbf 1_{\{ \int _0^t f^2(B_s) ~ds>0\}}, \ t\geq 0,$$

where $f \in \mathcal C^0(\mathbb R, \mathbb R)$ and $f \not\equiv 0$.

What is the conditional distribution of $H_t$ knowing $B$?

I dont have any idea on how to start to approach it. Any advice will be strongly appreciate. Thank's in advance.

Crossposted on overflow.

答:
Answered on overflow by The Bridge. </p>

<p>$$ \mathcal L(H_t| B) = \mathcal N (0,1)$$
 since the integral which defines $H$ is a Wiener integral if  we know the all path of $B$.</p>

ContentLicense: CC BY-SA 3.0

Body:
Answered on overflow by The Bridge.

$$ \mathcal L(H_t| B) = \mathcal N (0,1)$$
 since the integral which defines $H$ is a Wiener integral if  we know the all path of $B$.


问:
Equivalent condition of convergence in measure
Suppose $E \subset \Bbb R^n$, Lebesgue mesure $m(E) < +\infty$.
$f(x), f_1(x), \cdots, f_n(x), \cdots$ are measurable functions on $E$.

Prove that

$f_n$ convergence to $f$ in measure $\iff$ $\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}=0$.

($\implies$) $\forall \varepsilon > 0$, $\alpha<\frac{\varepsilon}{2}$, there exists $N$ s.t. for $n \ge N$, $m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\frac{\varepsilon}{2}$, so $\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\varepsilon$ and we can take the infimum.

However, for the ($\impliedby$) part, $\alpha$ appears in  form $\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})$, so it's difficult to single out the $m(\{x \in E: |f_n(x)-f(x)|>\delta\})$ term. So how can we proceed?

答:
For the ($\implies$) part, your argument is correct. Let us now prove:

$f_n$ convergence to $f$ in measure $\iff$ $\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}=0$.

Proof:

($\implies$) As your wrote:  $\forall \varepsilon > 0$, $\alpha<\frac{\varepsilon}{2}$, there exists $N$ s.t. for $n \ge N$, $m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\frac{\varepsilon}{2}$, so $\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\varepsilon$ and we can take the infimum.

($\impliedby$) Let us prove by the counter-positive. Suppose  $f_n$ does not convergence to $f$ in measure. Then, there is $\alpha_0 >0$ and  there is $\varepsilon_0>0$, such that, for all $N\in \Bbb N$, there is $n>N$ such that
$$m(\{x \in E: |f_n(x)-f(x)|>\alpha_0\})>\varepsilon_0$$

So, for all $N\in \Bbb N$, there is $n>N$ such that, for all $\alpha>0$,

if $\alpha > \alpha_0$, then
$$\alpha + m(\{x \in E: |f_n(x)-f(x)|>\alpha\}) \geqslant \alpha > \alpha_0 \geqslant \min(\alpha_0, \varepsilon_0)$$

if $\alpha \leqslant \alpha_0$, then $m(\{x \in E: |f_n(x)-f(x)|>\alpha\}) \geqslant m(\{x \in E: |f_n(x)-f(x)|>\alpha_0\})$. So
\begin{align*}
\alpha +  m(\{x \in E: |f_n(x)-f(x)|>\alpha\} & \geqslant m(\{x \in E: |f_n(x)-f(x)|>\alpha\}) \geqslant \\
& \geqslant m(\{x \in E: |f_n(x)-f(x)|>\alpha_0\})> \\
& > \varepsilon_0 \geqslant \min(\alpha_0,\varepsilon_0)
\end{align*}

So, for all $N\in \Bbb N$, there is $n>N$ such that, for all $\alpha>0$,

$$\alpha + m(\{x \in E: |f_n(x)-f(x)|>\alpha\})> \min(\alpha_0, \varepsilon_0)$$

So, for all $N\in \Bbb N$, there is $n>N$ such that,

$$\inf_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\} \geqslant \min(\alpha_0, \varepsilon_0)$$

So,
$$\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}\neq 0$$


问:
Intersection of group $\mathbb{Z}_p \times \mathbb{Z}_p$ and $\mathbb{Z}_{p^2}$
Consider the finite groups, $\mathbb{Z}_p \times \mathbb{Z}_p$ and $\mathbb{Z}_{p^2}$ for prime $p$. What is the intersection of these two groups?

My naive guess is $\mathbb{Z}_p \subset \mathbb{Z}_p \times \mathbb{Z}_p$ and $\mathbb{Z}_p \subset \mathbb{Z}_{p^2}$, so at least $\mathbb{Z}_p$ is in both of them.  Am I right?  How I can prove this in more rigorously?

答:
In retrospect this is an overkill answer, but anyway here it is.

Intersection is an extrinsic notion in the sense that for abstract sets $X$ and $Y$ it does not make sense to define $X\cap Y$ unless $X$ and $Y$ come equipped with injections $i_1:X\to Z$ and $i_2:Y\to Z$ where $Z$ is another set which we think of as containing $X$ and $Y$. Even so, $i_1(X)\cap i_2(Y)$ depends on the choice of embeddings.

In the case you are considering, the only reasonable way to pose the question would be to find a group $G$ into which $\Bbb{Z}/p^2\Bbb{Z}$ and $\Bbb{Z}/p\Bbb{Z}\times \Bbb{Z}/p\Bbb{Z}$ both have injective homomorphisms (equivalent  to existence of subgroups of $G$ isomorphic to these groups) and to then analyze the intersection.

However, this is still not well-posed because copies of these groups in $G$ can intersect with cardinality $1$ or $p$ depending on the group $G$. Indeed, $\Bbb{Z}/p\Bbb{Z}\subseteq \Bbb{Z}/p^2\Bbb{Z}$ as $\{[np]:0\le n \le p-1\}$, and so in $\Bbb{Z}/p^2\Bbb{Z}\times \Bbb{Z}/p\Bbb{Z}$ there is a copy of $\Bbb{Z}/p\Bbb{Z}\times \Bbb{Z}/p\Bbb{Z}$ intersecting $\Bbb{Z}/p^2\Bbb{Z}\times \{0\}\cong \Bbb{Z}/p^2\Bbb{Z}$ with cardinality $p$, while in $$\Bbb{Z}/p^2\Bbb{Z}\times(\Bbb{Z}/p\Bbb{Z}\times \Bbb{Z}/p\Bbb{Z})$$
we have $(\Bbb{Z}/p^2\Bbb{Z}\times \{0\})\cap (\{0\}\times (\Bbb{Z}/p\Bbb{Z}\times \Bbb{Z}/p\Bbb{Z}))=0.$


问:
Background knowledge to prove that 'if set S in R is closed and bounded and non-empty, then sup S is in S.'
I am new to Real Analysis and Metric Space.
While searching for proof of this, I found that I am not fully understanding that if the nonempty set S is closed and bounded, then there must exist a subsequence that converges to supS.
I would like to know why the subset S of R must have a subsequence that converges to supS so that supS is in S.
Please help.</p>

答: If $\alpha = \sup S$, then for any $n \in \mathbb{N}$, $(\alpha-1/n, \alpha+1/n) \cap S$ is nonempty. Let $a_n \in  (\alpha-1/n, \alpha+1/n) \cap S$ and note $a_n \to \alpha$ as $n \to \infty$. Hence $\alpha \in S$ since $S$ is closed.


答:
HINT

Here I propose a sketch of an alternative proof.

If the set $S\subset\mathbb{R}$ is bounded and it is not empty, then it admits an infimum and a supremum.

Once it is closed, it contains all of its adherent points.

Given that the infimum and the supremum are adherent points, the result holds.

Title: Background knowledge to prove that 'if set S in R is closed and bounded and non-empty, then sup S is in S.'

Body:
I am new to Real Analysis and Metric Space.
While searching for proof of this, I found that I am not fully understanding that if the nonempty set S is closed and bounded, then there must exist a subsequence that converges to supS.
I would like to know why the subset S of R must have a subsequence that converges to supS so that supS is in S.
Please help.</p>


问:
Show the unilateral forward shift is essentially normal
Let $H$ be a complex, infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward shift with $Te_n= e_{n+1}$.

We say that $T \in \mathcal{B(H)}$ is essentially normal if the image $\pi(T)$ of $T$ under the canonical quotient map $\pi: \mathcal{B(H)} \rightarrow {\mathcal{B(H)}}/{\mathcal{K(H)}}$ in a sense that $\pi(T)\pi(T^*)= \pi(T^*)\pi(T)$

Show that $T$ is essentially normal.

What I've tried:

Since the quotient map is linear, I want to show that

$\pi(T)\pi(T^*) - \pi(T^*) \pi(T)=\pi(TT^*) - \pi(T^*T)= \pi(TT^*- T^*T)=0$

But not I'm confused because isn't that equivalent to showing that $T$ is normal?
i.e. $TT^* - T^*T = 0$.

Also, I'm not sure how to find $TT^* - T^*T$.

Any help will be appreciated!

Thank you in advance!

答: $\pi$ is not injective. The equality $\pi(T^*T-TT^*)=0$ does not tell you that $T^*T-TT^*=0$, but that $T^*T-TT^*$ is compact.


问:
Complex-Orthogonal Similarity
First, a quick definition: I say that a matrix $M$ with complex entries is complex-orthogonal if $MM^T = I$.  Note that $T$ here refers to the transpose, not the conjugate transpose.  That is, we are taking the adjoint with respect to a bilinear (as opposed to sesquilinear) dot-product over $\Bbb C$.  A complex-orthogonal matrix need not be unitary, and vice versa (unless of its entries are in $\Bbb R$).

Consider the following claim:

Claim: For any $A \in \Bbb C^{n \times n}$, there exists a complex-orthogonal $U$ such that $UAU^T$ has equal diagonal entries.

My Question: Does this claim hold?

It may be helpful to take a look at this paper which proves that for any $A$, there is an $S$ such that $SAS^{-1}$ has equal diagonal entries.  It goes on to show that such an $S$ may be taken to be a unitary matrix.  Another proof of the same fact is presented in p. 77 of R.A. Horn and C.R. Johnson’s Matrix Analysis (1985/7, Cambridge Univ. Press).

Notably, both of these proofs exploit facts for which an analog fails to exist in my case.  For Horn and Johnson's proof, we need the fact that the unitary matrices form a compact subset of $\Bbb C^{n \times n}$.  For the proof linked, we need the numerical range to have certain "nice properties", which I believe fail in my case.

Any insight here is appreciated.

答:
The claim is false. As a counterexample to the claim consider
$$
A = \begin{pmatrix}
i & 1
\\
1 & -i
\end{pmatrix}.
$$

To see this, note that the $2 \times 2$ complex orthogonal matrices can be parameterised as
$$
O =
\begin{pmatrix}
\cos z & \mp \sin z
\\
\sin z & \pm \cos z
\end{pmatrix}
$$
for $z \in \mathbb{C}$. This yields
$$
B = O A O^T = \mathrm{e}^{ \pm 2 i z}\begin{pmatrix}
i & \pm1
\\
\pm1 & -i
\end{pmatrix}.
$$
thus we always have $B_{11} = - B_{22}$, the only possible solution to $B_{11} = B_{22}$ is $B_{11} = 0 = B_{22}$, which is approached only in the limit $\mathrm{Im}(z) \to \pm \infty$, a limit in which every element of $O$ tends to infinity and $O$ is no longer invertable.

Title: Complex-Orthogonal Similarity


问:
How can I find the conditional probability given one of two satisfy a condition?
Prompt: According to a survey, only 43% of Americans have a net worth above \$20,000. This varies by age. Out of Americans below the age of 35, only 37% have a net worth above \$20,000. Where as 52% of Americans over the age of 35 have a net worth above \$20,000.

Question: If I pick one person under 35, and another over the age of 35, given that only one of them has a net worth above \$20,000, what is the probability it is the older person?

I see that it is a conditional probability question.

If we let $P(A)$ = over the age of 35 (however it is not stated that half of people are under 35 and the other half are over 35 so this is also confusing).

and $P(B)$ = one of the two has a net worth over \$20,000

then the solution would be $P(A|B)$, but how can I find $P(B)$?

答:
You are setting this up wrong.  We know that one person is under $35$ and the other is over $35$.  This is not a matter of probability.

Let $A$ be the event that the younger person's net worth is over $\$20000$ and let $B$ be the event that the older person's net worth is over $\$20000$.  We are given that exactly one of these events occurs, that is $(A\cap \overline{B})\cup(\overline{A}\cap B)$.

We are asked to compute $\Pr(B|(A\cap \overline{B})\cup(\overline{A}\cap B))$

Can you do it now?


问:
$\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$
I try to prove if de groups $\mathbb{Z}_{p^{2}}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$ are isomorphic.
I was using the fact that $ \mathbb{Z}_ {mn} $ is isomorphic to $ \mathbb{Z}_m \times \mathbb{Z}_n $ if and only if $ (m, n) = 1 $ taking $ m = n = p $,in this case as $ p $ is a prime greater than 1, then $ (p, p) = p \neq 1 $ and so I don't have the isomorphism, but I don't know if this is true.

答: It is.  The statement $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{mn}$ is equivalent to $(m,n)=1$.


问:
How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$
$\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.</p>

答:
$$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots \sqrt{2^{2^{n-1}}}}}}\\&<\sqrt{1+\sqrt{2+\sqrt{2^2+\cdots }}}\\&=\sqrt{1+\sqrt{2}\cdot\sqrt{1+\sqrt{1+\cdots }}}\\&=\sqrt{1+\sqrt{2}\phi}\\&<2\end{aligned}$$

We can get a tighter bound for the limit by breaking the pattern a little further down the line:

$$\begin{aligned}\sqrt{1+\sqrt{2+\sqrt{3+\cdots \sqrt{n}}}}&<\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2^2+\cdots}}}}\\&=\sqrt{1+\sqrt{2+\sqrt{3+2\sqrt{1+\cdots }}}}\\&=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{5}}}}\approx 1.7665398\end{aligned}$$

$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots }}}\approx 1.7579327$$


问:
$p_{i + k} - p_i \neq \text{const}$ for any $k \geq 1$ where $p_i = i$th prime number.
Let $p_i$ be the $i$th prime number.  This should be simple to prove:

$$
\forall k \geq 1, c \in \Bbb{Z}, \\
p_{i + k} - p_i \neq c, \\
$$
for some $i \geq 2$.  But for example:

$$
11 - 5 = 6 \\
13 - 7 = 6 \\
17 - 11 = 6 \\
19 - 13 = 6 \\
23 - 17 = 6 \\
$$

So long strips do exist, I'm saying the whole diagonal strip (a diagonal line in the table for $p_i - p_j$) cannot be constant.

I specified $i \geq 2$ because I'm only interested in odd primes, but lowering to $i \geq 1$ should also work.

It should be easy to prove, but I'm stuck on it.

If $p_{i + k} - p_i = c$ for all $i \geq 2$ then in particular $3 + cz$ is prime for all $z \geq 0$, which would include $z \in 3 \Bbb{Z}$?  Does that work as a proof?

My next question is how can we generalize the statement? I got a little too specific with the indices.

I think I've got it now.

Let $b,c,k \in \Bbb{N}$ be positive numbers.  Then:

$$
p_b + c = p_{b + k} \\
p_{b + k} + c = p_{b + 2k} \\
p_{b + 2k} + c = p_{b + 3k} \\
\vdots
$$

would imply
$$
p_b + c = p_{b + k} \\
p_b + 2c = p_{b + 2k} \\
p_b + 3c = p_{b + 3k} \\
\vdots
$$
or in general $p_b + cz = p_{b + zk}$ for all $z \geq 0$.  Thus choose any $z \in p_{b}\Bbb{N}$.  For more tightness we'd only have to choose the minimal composite in $w \in p_b + c \Bbb{N}$, and then take $z = \dfrac{w - p_b}{c}$

Thus to generalize, $b + zk$ was in the place of $i + k$ so:

$$
b + zk = i + k = b + \dfrac{w - p_b}{c} k \implies \\
\exists i, b \leq i \leq b + k(\dfrac{w - p_b}{c} - 1) : \\
p_{i + k} - p_i \neq c \\
$$

答:
If I understand your question,
the fact that there there are,
for any $n$,
more than $n$ consecutive
composite integers
should prove what you want
by taking $n > c$.


问:
On eigenvalues of complex-orthogonal matrices
Suppose that $A$ is a complex matrix satisfying $A^TA = I$ (so $A$ is the entrywise transpose, not the conjugate transpose).  What can be said about the eigenvalues of $A$, if $A$ is "complex-orthogonal" in this sense?

Of course, for any eigenpair $(\lambda,x)$, we have
$$
x^Tx = x^TA^TAx = (Ax)^TAx = \lambda^2 (x^Tx)
$$
which allows us to conclude that $\lambda^2 = 1$... so long as $x^Tx \neq 0$.  Can anything else be said?  Does the case in which $A$ has real entries allow us to conclude that $|\lambda| = 1$?

答:
At the risk of stating the obvious, as it hasn't been explicitly stated here, I add that:

Each eigenvalue is either (a) one of a pair of eigenvalues $(\lambda, \lambda^{-1})$, or (b) $\lambda = \pm 1$.

The product of all the eigenvalues satisfies $\prod_i \lambda_i = \pm 1$.

Both of these claim are straightforward to show:

Let $A x = \lambda x$ define the left eigenvector $x$, then by taking the transpose it follows that $x^T A = x^T \lambda^{-1}$, i.e. $x^T$ is a right eigenvector with eigenvalue $\lambda^{-1}$. If (a) $x^T x = 0$ then these are distinct eigenvalues and $\lambda$ may take any value, if (b) $x^T x \neq 0$ then they are the same eigenvalue, and we require $\lambda = \lambda^{-1}$, i.e. $\lambda = \pm 1$.

Note $1 = \det(I) = \det(A A^T) = \det(A)\det(A^T)= \det(A)^2$. This implies $\det(A) = \prod_i \lambda_i = \pm 1$.


问:
Estimation Theorem on a straight line in complex plane
Apply the Estimat Theorem (ML inequality) to show that
$$\bigg| \int_{[0, 1+i]}(z^2+1)^{-1}\bigg|\leq \sqrt 2$$

Note that $\gamma(t)=t(1+i), t\in [0, 1]$ and $\gamma'(t)=1+i$. Let $f(z)=(z^2+1)^{-1}$, so that
\begin{align*}
\bigg|\int_{\gamma}f(z)\ dz\bigg| \leq \int_{0}^{1}|f(\gamma(t))||\gamma'(t)|\ dt = \int_{0}^{1} \bigg|\frac{1+i}{2it^2+1}\bigg|\ dt = \int_{0}^{1}\frac{\sqrt{2}}{\sqrt{1+4t^4}}\ dt
\end{align*}
I am stuck at this point because I not sure I can just take the $\sqrt{2}$ and that is. I am also not sure it is integrable. Any help will be great. Thanks in advance.

答:
You are making things unnecessarily complicated.

The length of $[0,1+i]$ is $\sqrt{2}$ and $(1+z^2)^{-1}$ has magnitude $\leq 1$ (achieved only at $z=0$).  So ML-inequality gives the answer claimed.


问:
What subset of the reals have unique prime factorizations if you allow rational exponents?
By the fundamental theorem of arithmetic we know that all positive integers have a unique prime factorization. So if $n$ is some positive integer, then
$$n = \prod_{i\in\mathbb{N}}{p_i^{e_i}}$$
where $p_i$ is the $i$th prime number and $\{e_i\}$ is a sequence of natural (I'm including $0$) numbers, all but finitely many of which are $0$.

If we allow the elements of $\{e_i\}$ to be integers and not just naturals, then the subset of $\mathbb{R}$ with "prime factorizations" is $\mathbb{Q}$, but what happens when we allow the exponents to be rationals? What reals have a unique "prime factorization" then? It's some subset of the algebraic numbers, but I don't know if there's any other characterization of this set.

Also, in the case of integer or natural number exponents, it was necessary for all but finitely many exponents to be 0 for the product to converge, but given that rationals are dense in the reals, it seems like you could have products of the form
$$\prod_{i\in\mathbb{N}}{p_i^{e_i}}$$
that converge without needing only finitely many non-zero exponents so long as they converged to $0$ quickly enough. What subset of the reals have unique "factorizations" in this sense? It seems plausible to me that these sorts of factorizations might not be unique too.

答:
For the case where the factorizations are required to contain only finitely many nonzero exponents:

Let $\alpha$ be a nonzero element of $\mathbb{R}$ with a "rational exponent prime factorization": $$\alpha = \pm\prod\limits_{i \in \mathbb{N}} p_i^{r_i/s_i}$$

where $p_i$ are the positive primes, $r_i$ and $s_i$ are coprime integers, all the $s_i > 0$ and only finitely many of the $r_i$ are nonzero.

Then we may choose $d$ as the LCM of the finitely-many $s_i$ corresponding to these nonzero $r_i$.

We then find that $\alpha^d$ has a factorization where each exponent is an integer, so $\alpha^d$ must be rational.

Therefore, the subset of real numbers with such "rational exponent prime factorizations" consists of the real numbers that are $d$-th roots of rational numbers, for some $d \in \mathbb{N}$. Conversely, any $d$-th root of a rational number has an obvious factorization of the desired form.

The factorization in the form given above is unique. The result for positive and negative integer exponents is well-known (Fundamental Theorem of Arithmetic and properties of multiplication in $\mathbb{Q}$), and the case of rational exponents can be reduced to the previous case after raising to the $d$-th power.

To flesh the argument out in more detail... Consider two candidate factorizations: $$\alpha = \pm\prod\limits_{i\in\mathbb{N}}p_i^{r_i/s_i} = \pm\prod\limits_{i\in\mathbb{N}}p_i^{r_i'/s_i'}$$

The signs are uniquely defined ($+$ if $\alpha$ is positive, $-$ if $\alpha$ is negative), so we can take absolute values.

We can again define $d$ as the LCM of the $s_i$ associated with nonzero $r_i$, and similarly $d'$ as the LCM of the $s_i'$ associated with nonzero $r_i'$. We then raise to the $dd'$-th power:

$$|\alpha|^{dd'} = \prod\limits_{i\in\mathbb{N}}p_i^{r_id'\frac{d}{s_i}} = \prod\limits_{i\in\mathbb{N}}p_i^{r_i'd\frac{d'}{s_i'}}$$

By the choice of $d$ and $d'$, both products only contain integer exponents. By uniqueness of the factorization for integer exponents, we see that $r_id'\frac{d}{s_i} = r_i'd\frac{d'}{s_i'}$ for all $i \in \mathbb{N}$, from which it follows that $r_i/s_i = r_i'/s_i'$ for all $i \in \mathbb{N}$ (note that $d, d' \gt 0$).

Not all real algebraic numbers are of this form, however. The set of real numbers with "rational exponent prime factorization" can be described as the subset $S \subset \bar{\mathbb{Q}} \cap \mathbb{R}$: $$S = \{\alpha \in \mathbb{R}: \text{there exists some }d \in \mathbb{N}\text{ such that } \alpha^d \in \mathbb{Q}\}$$

So in particular this only includes elements in radical extensions of $\mathbb{Q}$, and it is well-known from Galois theory that not all real algebraic extensions of $\mathbb{Q}$ are radical (this is probably overkill for the current purposes, but it's enough to show that $S$ is a proper subset of the real algebraic numbers).

The case of infinitely many nonzero exponents in the factorization looks more difficult.

One issue you can run into is that lots of products converge to $0$, simply by choosing lots of negative exponents, for example $\prod\limits_{i\in\mathbb{N}}p_i^{-1} = 0$.

Besides this issue, it might also be possible that the factorizations of nonzero real numbers are not unique, but it requires finding sequences $a_i$ of rational numbers, not all zero, with the following property:

$$\prod\limits_{i\in\mathbb{N}}p_i^{a_i} = 1$$

This can also be rewritten as an infinite series:

$$\sum\limits_{i\in\mathbb{N}} a_i\log p_i = 0$$

So you would need to pick the sequence of rational exponents in such a way that the "weighted average" of the $\log p_i$ tends to $0$. I'm not sure at the moment whether this is possible or not. I think it is possible, the idea would be that you alternate positive and negative values for $a_i$, and you tweak their magnitudes in such a way that the partial sums always oscillate around $0$ with diminishing amplitude. If this idea works, it should also prove that all real numbers have an infinite prime factorization with rational exponents. But I haven't checked the details yet.

Here's a fully fleshed-out argument showing that all real numbers have a rational exponent prime factorization if you allow infinite products. Moreover, there are infinitely many distinct factorizations. This follows from the first claim by using nontrivial factorizations of $1$.

Given a real number $\alpha \in \mathbb{R}$, we wish to find a sequence of rational numbers $a_i \in \mathbb{Q}$ such that $\alpha = \pm\prod\limits_{i \in \mathbb{N}} p_i^{a_i}$. The case where $\alpha = 0$ was already discussed, and the case of negative $\alpha$ is trivially reducible to the remaining case of positive $\alpha$ by choosing the sign, so we assume $\alpha \gt 0$.

We may then take logarithms and define $L = \log \alpha$. Now the problem becomes, as mentioned before, that of finding a sequence of $a_i \in \mathbb{Q}$ such that $$\sum\limits_{i \in \mathbb{N}}a_i\log p_i = L$$

The sequence will be chosen in such a way that for every $i \in \mathbb{N}$, we satisfy $$|s_i - L| < \frac{1}{i+1}$$ from which it follows that the sum converges to $L$. Here $s_i$ is the sequence of partial sums, so $s_1 = a_1\log p_1$ and $s_i = s_{i-1} + a_i\log p_i$ for all $i \geq 2$.

The sequence is defined inductively. First choose some rational number $a_1$ such that $$\frac{L - \frac{1}{2}}{\log p_1} < a_1 < \frac{L + \frac{1}{2}}{\log p_1}$$

This makes sure that $|s_1 - L| < \frac{1}{2}$.

Now assume that for some $i \geq 2$, the values $a_1, \ldots, a_{i-1}$ were chosen so that $|s_{i-1} - L| < \frac{1}{i}$. There are two cases to consider for the choice of $a_i$.

$L - \frac{1}{i} < s_{i-1} \leq L$:

Then we choose $a_i$ to be a rational number satisfying $$\frac{1}{i(i+1)\log p_i} < a_i < \frac{1}{(i+1)\log p_i}$$

For $i \geq 2$, this defines a nonempty open interval, and since $\mathbb{Q}$ is dense in $\mathbb{R}$, this means that such a choice is always possible.

$L < s_{i-1} < L + \frac{1}{i}$:

Then we choose $a_i$ to be a rational number satisfying $$-\frac{1}{(i+1)\log p_i} < a_i < -\frac{1}{i(i+1)\log p_i}$$

And the choice is again possible for $i \geq 2$.

By induction, the sequence of rational numbers $a_i$ is constructed in such a way that $|s_i - L| < \frac{1}{i+1}$ for all $i \in \mathbb{N}$, from which the result follows.


答: I claim that $\mathbb{Z}/p^2$ is not isomorphic to $\mathbb{Z}/p\times \mathbb{Z}/p$. By definition, the former is the cyclic group of order $p^2$, which implies that it has a generating element of order $p^2$. Any element of $\mathbb{Z}/p\times \mathbb{Z}/p$ is of form $(a,b)$, with $a,b\in \mathbb{Z}/p$. But observe that $p(a,b)=(pa, pb)=(0,0)$, so the order of any such $(a,b)$ is at most $p$. Since isomorphisms preserve order, no such isomorphism is possible.

Title: $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$

Body:
I try to prove if de groups $\mathbb{Z}_{p^{2}}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$ are isomorphic.
I was using the fact that $ \mathbb{Z}_ {mn} $ is isomorphic to $ \mathbb{Z}_m \times \mathbb{Z}_n $ if and only if $ (m, n) = 1 $ taking $ m = n = p $,in this case as $ p $ is a prime greater than 1, then $ (p, p) = p \neq 1 $ and so I don't have the isomorphism, but I don't know if this is true.


问:
Probability of second ball being blue
There are $3$ boxes. One box has $2$ blue balls. One box has $2$ red. The last has one blue and one red.

You randomly pick a box and take $1$ ball out. You see that it's blue. What is the probability that the other ball in the box is blue.

The answer is $\frac{2}{3}$ and I confirmed this with Bayes rule. However, I am confused why the answer is not $0.5$. I know that if we aren't given the information that the first ball taken out is blue, then for sure, the second ball being blue has a probability of $0.5$.

However, without referring to Bayes rule, if I account for the fact that the first ball is blue, I can essentially eliminate the box with $2$ reds from my consideration. Because it is surely not possible that we chose that box. That leaves us with $2$ boxes. Then my thought process was that well, we could remove a blue ball from each box (because we just drew one and no longer need to consider it), so now the $2$ boxes each has $1$ ball. Blue in $1$ box and red in another, with each having the same probability of being the second ball, which would make the answer $0.5$, but apparently that's not correct. What is wrong with this thought process?

One thing that I am pondering is, if we know the first ball is blue, then we can remove the box with $2$ reds, but I think knowing this information might also make it so that there's a higher probability that we chose the box with $2$ blues, than the box with $1$ blue and $1$ red. In these $2$ boxes, there are $3$ blue balls total, but because $2$ of these belong in a single box, then that box should have a $2/3$ chance of being chosen conditioned on us knowing that we observed $1$ blue already. Is this right? If so, then I think that answers my previous question.

答:
One thing that I am pondering is, if we know the first ball is blue, then we can remove the box with 2 reds, but I think knowing this information might also make it so that there's a higher probability that we chose the box with 2 blues, than the box with 1 blue and 1 red. In these 2 boxes, there are 3 blue balls total, but because 2 of these belong in a single box, then that box should have a 2/3 chance of being chosen conditioned on us knowing that we observed 1 blue already. Is this right?

Yes.

You had an equal probability (no bias) for obtaining each individual blue ball; and two among the three were in the jar that held two blue balls.

So when given that you draw a blue ball there is a probability of $2/3$ that there was another blue ball left in that jar.

Title: Probability of second ball being blue


问:
Diameter metric space
Suppose $(A, d)$ is a metric space with a non-empty subset $E \subseteq A$. Let $S = \{d(a, b) : a, b \in E\}$. Define $\phi (E)$ be the diameter of $E$ so that $\phi (E) = \sup S$, and $\phi(E) = \infty$ if $S$ is unbounded. Prove the following:

(a) If $E \subseteq A$ is non-empty then $\phi(E) = \phi(\overline{E})$ (the diameter of $E$ is equal to that of the closure of $E$).

(b) If $\phi (E) < \epsilon$ for some $\epsilon > 0$ and $E \cap B_{\epsilon}(x) \neq \emptyset$ for some $x$, then $E \subseteq B_{2\epsilon}(x)$.

For (a) I came up with a proof by contradiction but I am curious as to whether there are other proofs. After much time spent, I still cannot figure out a proof for (b). Any assistance or alternatively proofs are appreciated.

答: For part (b): Following the comments of Martin R and myself, we pick a point $y \in E \cap B_\varepsilon(x)$, since the latter is non-empty.  Then for any point $z \in E$, we have $d(z, y) \leq \varepsilon$ (because $\phi(E) = \varepsilon$), and we also have $d(y, x) \leq \varepsilon$ (since $y \in B_\varepsilon(x)$).  Therefore, by the triangle inequality, we must have $d(z, x) \leq d(z, y) + d(y, x) \leq 2\varepsilon$.

Title: Diameter metric space


问:
Incomplete Beta Funtion and Regularized Incomplete Beta Function
How could you solve this statement :

$1 - \frac{1}{2} I_{\frac{a}{a+x^2}}(\frac{a}{2},\frac{1}{2}) = \frac{\sqrt{a}}{2} (\sqrt{I_{\frac{x}{x+1}}(\frac{a}{2},\frac{a}{2})} - \frac{1}{\sqrt{I_{\frac{x}{x+1}}(\frac{a}{2},\frac{a}{2})}}) $

with Hint :

$I_x(a,a) = 1 - \frac{1}{2}I_{4x(1-x)}(a,\frac{1}{2}) ,$ where 1/2<= x <= 1 .

Some definitions : $I_x(a,b) = \frac{B(x;a,b)}{B(a,b)} $ ,where B(x;a,b) is incomplete beta function and B(a,b) is beta function.
I tried to set left side minus right side to be a function f. Then I considered $\frac{df}{dx}$ and f(1), if both are equal $0$ then it is proved. But it doesn't go any further. Could anyone help me ?

答:
As writen, I have the strange feeling that this is not an identity.

Consider the function
$$f(x,a)=\text{lhs - rhs}$$ Consider the constant term in the series expansion of $f(x,a)$ around $x=1$. For $a=1$, it is
$$\frac{\left(4+\sqrt{2}\right) \pi -2
   B_{\frac{1}{2}}\left(\frac{1}{2},\frac{1}{2}\right)}{4 \pi }\neq 0$$ and, beside this specific value, it is an increasing function of $a$.

Doing the same around $x=\frac 12$, the constant term of the expansion is an increasing function of $a$ and, for $a=1$, this constant is $\neq 0$.

Now, making explicit the function, we have
$$f(x,2)=\frac{1}{2} \left(\sqrt{\frac{x^2}{x^2+2}}+\frac{\sqrt{2}}{\sqrt{x}
   \sqrt{x+1}}+1\right)$$

Similarly, $f'(1,a)$ is not a constant. It is $0$ only for $a=1$.


问:
If $b = c \times a$ and $c = a \times b$, and length $b$ = length $c$, $a$ is a unit vector.
If $\vec b = \vec c \times \hat a\,$  and $\,\vec c = \hat a \times \vec b\,$, and  $|\vec b|$ = $|\vec c|$.

Assuming $\vec b \ne 0$.

I have managed to prove $\vec a$, $\vec b$ and $\vec c$ are orthogonal, but not much else.

Help appreciated! Thanks.

答:
We have
\begin{align*}
  b \times c &= b \times (a \times b)\\
     &= \bigl(b \cdot b\bigr)a - \bigl((c\times a) \cdot a\bigr)b\\
     &= |b|^2 a
\end{align*}
and
\begin{align*}
  b \cdot c &= (c \times a)\cdot c\\
            &= 0,
\end{align*}
hence $|b \times c| = |b||c| = |b|^2$ and therefore
\[
  |a| = \frac{|b \times c|}{|b|^2} = 1.
\]</p>


问:
Are real numbers defined as Banach spaces?
I was listening to this lecture by Norman Wildberger, where he explains that real numbers are constructed as equivalence classes of Cauchy sequences. He never says that these equivalence classes are Banach spaces, but I was left wondering if that is what he means.</p>

答:
This is unfortunately a nonsense question.

There is a set of Cauchy sequences. This set has a relation on it ($(a_n) \sim (b_n)$ if $\lim_n (a_n - b_n) = 0$). There is therefore a quotient set or set of equivalence classes.

None of this ever passes through anything with more structure than sets. (If you like, it passes through rings: the set of Cauchy sequences is a ring, because I know how to add and divide in it, and the relation respects these operations, so that the quotient inherits them.)

In fact "$V$ is a Banach space" does not even make sense until you invent the real numbers. A Banach space is a pair $(V, N)$, where $N: V \to \Bbb R_{\geq 0}$ is a function satisfying certain properties.


问:
How do I output a line as a function of $x$?
This might be really dumb, but I want to know how I can output a line after inputting an $x$ value. For example, how would I write a function that would give me the tangent line of a graph at a certain $x$ value?

答:
First of all - it is not "really dumb". You are exploring and thinking and that’s great!

I'll work through the tangent line example since you explicitly mentioned that in your question.

Given a differentiable function $f(x)$ we can construct the general tangent line at a point $a$, we will call this $T_{a}(x)$. We know that $T_a = mx+b$ (since it is a line). It is common knowledge that the derivative at a point gives the slope of the tangent line at that point, so now we have $T_a(x) = f'(x)x + b$. We also want the tangent line to go through the point $(a,f(a))$. The most concise way to do this is to apply the horizontal shift $(x-a)$ and then set $b$ to $f(a)$ yielding $$T_a(x) = f'(x)(x-a) + f(a).$$

The first time I worked through this myself when I was younger and exploring just like you, I didn't think of the horizontal shift, and instead I solved $$f'(a)a + b = f(a)$$ which yields $$b = f(a) - f'(a)a$$ yielding $$T_a(x) = f'(a)x +f(a)-f'(a)a$$ which is also a valid expression for the tangent line.

For specific function $f$, you can plot the function and the tangent line in desmos to see them visually. By adding a slide for the value $a$ you can toggle $a$ around and see the tangent line move! Here is a screen shot of a demonstration using the function $f(x) = x^x$ which is differentiable for $x > 0$.


答:
The real numbers can be defined by the process of completion with respect to the usual norm $\lvert\:\cdot\:\rvert:\Bbb{Q}\to \Bbb{Q}$. The norm $\lvert \:\cdot\:\rvert$ defines a metric on $\Bbb{Q}$ by $d(x,y)=\lvert x-y\rvert$. We note that $\Bbb{Q}$ is not complete with respect to this metric because there are Cauchy sequences which do not converge (this is the definition of not being complete): for instance the sequence
$$
3,3.1,3.14,3.141,3.1415,\ldots
$$
is "trying" to converge to $\pi= 3.1415...$ but $\pi$ is not rational so this does not make sense. [Note that one does not really know what $\pi$ is before defining $\Bbb{R}$ so this reasoning is sort of bad, but it is a good source of motivation - you could do the same thing with $\sqrt{2}$ or any infinite decimal expansion you wish.] The completion of $\Bbb{Q}$ with respect to this norm is called $\Bbb{R}$ and is defined to be the set of Cauchy sequences in $\Bbb{Q}$ modulo the relation that two Cauchy sequences $(a_n)$ and $(b_n)$ are called equivalent if
$$
\lvert a_n-b_n\rvert\to 0
$$
as $n\to\infty$. Maybe this is already familiar to you. Anyway, the sequences themselves are not Banach spaces in any reasonable sense, but it is true that $\Bbb{R}$ has a norm defined by extending the norm $\lvert\:\cdot\:\rvert$ on $\Bbb{Q}$ from before. It is also true that the completion of a normed space with respect to a norm is complete (in that all Cauchy sequences converge). In particular, $\lvert\:\cdot\:\rvert$ extended from $\Bbb{Q}$ endows $\Bbb{R}$ with the structure of a complete normed vector space where we add Cauchy sequences by
$$
(a_n)+(b_n)=(a_n+b_n)
$$
and define scaling in the obvious way. So, $(\Bbb{R},\lvert\:\cdot\:\rvert)$ is a Banach space.

Title: Are real numbers defined as Banach spaces?

Body: I was listening to this lecture by Norman Wildberger, where he explains that real numbers are constructed as equivalence classes of Cauchy sequences. He never says that these equivalence classes are Banach spaces, but I was left wondering if that is what he means.</p>


问:
What are 'contexts' actually called?
Consider the following argument by contradiction.

\begin{array}{|l}
\mbox{We wish to deduce A.} \\
{\begin{array}{|l}
\mbox{Suppose not A.} \\
\hline \\
\mbox{Then B. Thus C. Therefore, contradiction. }
\end{array} } \\
\mbox{Thus, A.}
\end{array}

So to actually perform the argument by contradiction, we had to open a new ‘context’ wherein additional assumptions hold. What are these ‘contexts’ actually called? And which deductive systems actually utilize them?

答: Natural deductive systems often use such a context often for a few different rules.  Say we want to prove that Cpr (p implies r) in some proof.  Then we'll open such a new context, take p as a supposition and show that r follows.  Then we infer Cpr outside of the context.  Also, you can open a new context supposing p, show that a contradiction follows, and then infer Np (the logical negation of p) outside of the context.  The new contexts I've seen called "subproofs".

Title: What are 'contexts' actually called?


问:
How to contextualize an inverse function for interpolation purposes?
I have a function that describes the growth of a population in a logistic term:
$$N_t = rN_0(1-\frac{N_0}{\kappa})$$
I would like to interpolate the values so that, given a known $y$ I can get the related $x$. To make it easier, I linearized the equation by taking log($N_t$). I removed the point that were not in a straight line and the obtained a linear model:
$$log(N_t) = y = st + i = mx + b$$
Using R, I can interpolate x given the model y~x , as in this picture:

But what I need to do is the opposite:

Since in R is not possible to get y from the model, and it is not correct to invert the model as x~y, I found the inverse function as:
$$log(y) = sx+i$$ (where $s$ and $i$ are the slope and intercept of the first model
$$log(x) = sy+i$$
$$y = \frac{log(x) -i}{s}$$
But now I am lost on what to do next. Shall I use the inversed function on the original x or y values? How do I convert these values into the original space's one? In other words, I get something like this:

How do I use the inverted function?
Manually, I have found that if $y=1000000$, then $x=16.6$. The parameters are: $s=0.022$, $i=0.328$. If I solve, I get:
$$log(x) = s \cdot 10^6 -i$$
$$x = 10^{s \cdot 10^6 -i}= ∞$$ OR
$$x = 10^{s \cdot 6 -i}= 0.6$$
My apologies but I am not a mathematician.

Thank you

答:
since $a_n$ convergent to $a$, we multiple $a_n$ to your sequence and plus an extra term $b_n^{2}/2$, ignore whether $b_n$ is convergent or not at first.

then product your sequence to be $\frac{(aa_n)^2-2aa_nb_n+b_n^{2}}{2}-b_n^{2}/2=(aa_n-b_n)^2/2-b_n^{2}/2$. since $a\in{(0,1]}$, so $aa_n\in{(0,1]}$ be bounded (avoid squares ?), the above terms can be merge to $aa_n$ and $aa_n-2b_n$, only depends one term for $b_n$. now your conclusion seems reliable.


问:
Find the marginal PDF
I am having a hard time trying to find the marginal of the following joint pdf.

$$f_{U,V}(u,v) =2 \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1}$$
where $ u<v<\frac{u+\theta+1}{2}$ and $\theta < u <\theta+1$.

This was my attempt

$\begin{align*}
f_V (v) &= \int_{\theta}^{\theta + 1} \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\
&=  \frac{n!}{\left( \frac{n}{2}-1 \right)!\left( \frac{n}{2}-1 \right)!}\int_{\theta}^{\theta + 1}\left[(u-\theta)(\theta + 1-2v+u)\right]^{\frac{n}{2}-1} \cdot 2 du\\
\end{align*}$

答:
Your support is: $\theta\lt u< \theta +1$ and $u< v< (u+\theta+1)/2$.

That is equivalently: $\theta\leq v<\theta+1$ and $\max(\theta,2v-\theta-1)\leq u<v$

Title: Find the marginal PDF


问:
How to prove that $G$ is semidirect product?
If $ N\triangleleft G $ and $ G / N $ is a free group, I want to show that $ G $ is a semi-direct product of $ N $ by $ G / N $, but I don't see how to prove it, I know if $ N $ is normal to $ G $, then $ G $ is a semi-direct product of $ N $ and $ G / N $ if and only if there is a homomorphism $s: G / N \to G$ such that $v\circ s: G/N \to G/N = id$, where $v: G\to G/N$ is the natural projection. but I don't know where to put the hypothesis that $ G / N $ is a free group.

答:
Basically, free groups are projective, so every surjection onto a free group has a section. Explicitly:

If $G/N$ is a free group, say it is (isomorphic to) a free group on the set $X$. For each $x\in X$, let $g_x\in G$ be such that $\pi(g_x)=x$, where $\pi\colon G\to G/N$ is the canonical projection.. By the universal property of the free group, there exists a group homomorphism $f\colon G/N\to G$ such that $f(x)=g_x$ for each $x\in X$. Since $\pi\circ f$ is the identity on $X$, the universal property of the free group tells us that $\pi\circ f=\mathrm{id}_{G/N}$, and so it follows that $f$ is one-to-one. Therefore, $G$ contains a subgroup isomorphic to $G/N$.

Now verify that $f(G/N)\cap N=\{e\}$ and that $G$ is generated by $N$ and $f(G/N)$.


问:
Breakdown of Inverse Transform from Contour Integral at short times
I need to invert a Laplace transform to obtain the temperature variation in a particular problem in time and (1-D) space. In the frequency domain, I have

$$F(s)=\frac{1}{s}e^{H(s)\xi}$$

where

$$H(s)=\frac{1}{C}\Big[s + Q\sqrt(s)\frac{K_1(\sqrt{s})}{K_0(\sqrt{s})}\Bigr]$$

and where $K_0()$ and $K_1()$ are the Modified Bessel Functions of the Second Kind, of orders $0$ and $1$, respectively. The function $H(s)$ has a branch point at $s=0$, but a closed contour can be generated with a branch cut as shown in the figure below

Noting that the integrand of the Bromwich integral vanishes on the large semicircle as its radius tends to $\infty$, and with the use of the Cauchy Integral Theorem, the temperature distribution is (skipping details) given by the integral in real space:

$$\theta(\tau,\xi) = 1-\frac{2}{\pi}
\int_0^\infty \frac{1}{u} e^
{-\Bigl[ u^2\Big(\tau-\frac{\xi}{C}\Big) +\frac{Q}{C}uA(u)\xi\Bigr]}
\sin\Big(\frac{Q}{C}uB(u)\xi\Big) \text{d}\xi$$

where $\tau$ and $\xi$ are the dimensionless temporal and spatial coordinates in the physical domain, and where $A(u)$ and $B(u)$ are functions involving the Modified Bessel functions. The term  $C$ can be thought of as the wave speed in the underlying first order PDE.

For large times, the solution matches that obtained from a Gaver-Stehfest function-sampling algorithm, although I do not necessarily state this as proof of correctness.

The problem however is that for smaller times, when $\xi/C > \tau$, the solution breaks down since the exponential argument becomes unbounded as $u\rightarrow \infty$ which is evident from the solution above.  At small times $s\rightarrow\infty$ and the ratio of the Modified Bessel functions tends to unity in the expression for $H(s)$. But this will still not resolve the problem caused by the condition $\xi/C > \tau$. Is it the case that the solution is only expected to be valid for $\tau > \xi/C >$? It appears to be that the spatial location cannot be ahead of the distance traversed by the thermal front at the wave speed in the elapsed time.

I am really interested in the short term behaviour, so it's a pity if this is the case, after all of the hard work that I put in working out the contour integral! Any suggestions therefore for how to work around this issue would be greatly appreciated.

答: I believe that I may have answered my own question within it, and that the solution is indeed valid only for locations that are upstream of and at the thermal front propagating at its wave speed. For all downstream locations, the solution remains at the initial condition.</p>


问:
Do $L^p$ integrals over fixed radius balls satisfy this reverse Hölder-like inequality?
For measurable $f:\mathbb R^n \to \mathbb R$, define$$M_qf(x) =\left(\int_{B_{1/2}(0)} |f(x+y)|^{q} dy  \right)^{1/q} = \|f\|_{L^q(B_{1/2}(x))} $$
and define the (inhomogeneous) dyadic annuli $ D_0 := B(0,1)$, $D_j = \{x\in\mathbb R^n: 2^{j-1} \le |x| \le 2^j\} $.

Hölder’s inequality trivially gives for $1\le q\le q’ \le \infty $ the pointwise bound $M_qf(x) \le M_{q’}f(x)$.

I’m interested in proving the following inequality, for all $q\in[1,\infty]$ and some $C>0$, (which I have on good authority is true): setting $\tilde D_j = D_{j-1}
\cup D_j \cup D_{j+1}$,
$$\fbox{$x\in D_j \implies M_q f(x)\le C\int_{
\tilde D_j} M_q f$}$$
and I think $C$ can be taken independent of $j$? Note that this is equivalently asking for $\|M_q f\|_{L^\infty(D_j)} \le C\|M_q f\|_{L^1(\tilde D_j)} $ which looks like a reverse of the relation for $q,q’$ above, but you need to take a second norm, and it’s at the cost of changing the set slightly. (In fact, this proves after an interpolation argument that $\|M_q f\|_{L^{p’}(D_j)} \le C\|M_q f\|_{L^p(\tilde D_j)} $ for all $1\le p\le p’ \le \infty$.)

(There was an attempt here, but I've removed it because it was long and I don't feel it is useful. The interested can see the edit history)

答:
I got some help, so I'll post a self-answer. My $C=C(n)$ is indeed independent of $j$ but it is rather large for large $n$; I would like to know if the constant can be taken independent of $n$, but I kind of doubt it. I believe the result is false for $N=2^n$. Anyway:

Let $\alpha\in(0,1)$ be arbitrary. First note that there exists $x_0\in D_j$ such that
$$ \left(\int_{ B_{1/2}(0)} |f(x_0+y)|^{q} dy  \right)^{1/q} = M_q f(x_0) \ge \alpha \| M_q f\|_{L^\infty(D_j)}.$$
Covering $ B_{1/2}(0)$ with a large number $N=N(n)$ (to be chosen shortly) of balls $B_k$ of equal radius $O(1/N^{1/d})$ (their total volume is therefore $O(1)$ which far exceeds  $|B_{1/2}(0)|=C2^{-d}$), we obtain by an easy contradiction argument that at least one of the sets $S=B_{1/2}(0)\cap B_k$ satisfies$$ \left(\int_{ S} |f(x_0+y)|^{q} dy \right)^{1/q} \ge \frac\alpha N \|  M_q f\|_{L^\infty(D_j)}$$
Now, choosing  $N$ large enough,   we have that $ S\subset  B_{1/2}(x)$ for each fixed $x\in S$. Thus,
$$  \frac  \alpha N \|  M_q f\|_{L^\infty(D_j)} \le \left(\int_{ B_{1/2}(x)}|f(x_0+y)|^q dy\right)^{1/q} =  M_qf(x_0+x)$$
Since all such points $x_0+x$ as $x$ varies over $S$ belong to $\tilde { D}_j :=  D_{j-1}\cup  D_{j} \cup  D_{j+1}, $ we obtain
$$ \frac{\alpha |S| }{N} \|M_q f\|_{L^\infty(D_j)} \le \int_{S} M_q f(x_0+x)dx \le \int_{\tilde D_j} M_q f(z)dz$$
Since  we can arrange all such sets $S$ to satisfy $|S|\gtrsim 1/N$  the result holds with some $C<\infty$. The above constructed $C$ satisfies
$C \lesssim N^2 $.


问:
Chebyshev-Gauss quadrature with $\tan(x)$
With Chebyshev-Gauss quadrature, solve $\int_0^{\pi/4}x\cdot tan^2(x)$, for $n=3$.

Needs first to determine the change in the integral, to change the limits of integrals and then reduce in form integrate $$\int_{-1}^1{\dfrac{1}{(1-x^2)^{1/2}}}$$
http://upload.wikimedia.org/math/3/7/3/3739c7537ace93cb3bc05e3957a44ff3.png

Thanks a lot, this is really important for me.

Greetings,
Tanya

答:
Hint: make the substitution $$4x - \frac{\pi}{2} = \arcsin(t).$$

You will need to simplify some horrific looking trig. functions in the integrand but this substitution should give you the correct limits of integration and a factor of $\sqrt{1 - t^2}$ in the denominator.


问:
Find the last two digit of $5^{121}*3^{312}$
The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$

and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of  $5^{143}*3^{312}$ are $25$.

Don't know why  $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.

Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$

答:
From my point of view, your book may contain some typographical errors. I think what the book really wants to express is as below.
First of all, note that $\forall k \in \mathbb N^* \ \ \ 25^k\equiv 25\mod 100$

Besides, $5^2*3^4\equiv 5^2\mod100$

Note that $5^2$ appears on both sides of the equation, which means that we can use "Recursion" here.

For any $k\in \mathbb N$, $5^2*3^{4*k}=5^2*3^4*3^{4*(k-1)}\equiv 5^2*3^{4*(k-1)}\mod 100$, so keep doing this, we will find that $5^2*3^{4*k}\equiv5^2\mod 100$

Wow, $312=4*78$, so we get $5^2*3^{312}\equiv 25\mod100$

The rest is easy. Just multiply $5^m$($m=143-2$  or  $m=121-2$) on both sides, and you will still get 25 on the right side.


问:
Finding the analytic root of a multivariate function that involves trigonometric functions
I am currently stuck in a problem, where
I need to find an analytic expression $x_0=f(\alpha,\beta)$.
Let me first try to explain what $x_0$ is:

Let $x$ be a real-valued variable, which must obey the following relation:
\begin{equation}
x^2 + \left[a_1(\omega) \alpha + a_2(\omega) \beta \right] x + \left[ a_3(\omega) + a_4(\omega) \alpha \beta \right] \leq 0
\end{equation}
with $\omega \in [0,\pi]$ and both $\alpha$ and $\beta$ are positive and real-valued.
The $a_i(\omega)$ terms are defined as follows:
\begin{equation}
a_1(\omega) = \frac{4 \left( cos(\omega) - 1\right)}{sin^2(\omega) + 2 \left(\cos(\omega) - 1\right)}
\end{equation}
\begin{equation}
a_2(\omega) = \frac{-8}{cos(2\omega)-1}
\end{equation}
\begin{equation}
a_3(\omega) = \frac{3-cos(\omega)}{cos(\omega)-1}
\end{equation}
\begin{equation}
a_4(\omega) = \frac{-16}{ sin^2(\omega) \left(cos(\omega)-1\right)}
\end{equation}
The equation for $x$ can be written in terms of
\begin{equation}
x^2 + p x + q \leq 0
\end{equation}
with
\begin{equation}
p=a_1(\omega)\,\alpha + a_2(\omega)\,\beta
\end{equation}
\begin{equation}
q=a_3(\omega) + a_4(\omega)\,\alpha\,\beta.
\end{equation}

If $4 q < p^2$ is true, one can obtain a possible solution for $x$:
\begin{equation}
\frac{1}{2}\left(-\sqrt{p^2-4q} - p\right) \leq x \leq \frac{1}{2}\left(\sqrt{p^2-4q} - p\right)
\end{equation}

Now, I am especially interested in finding the maximum of the upper bound of this solution regarding $\omega$, which I call $x_0(\alpha,\beta)$:
\begin{equation}
x_0(\alpha,\beta) = \max_{\omega} \frac{1}{2}\left(\sqrt{p^2-4q} - p\right)
\end{equation}

To estimate an expression for $x_0(\alpha,\beta)$, I thought I could calculate the first derivative of $(\sqrt{p^2-4q}-p)/2$, setting it zero and solve for $\omega$. This should give me an expression like $\omega_0 = f(\alpha,\beta)$, where $\omega_0$ is the $\omega$-value at which the maximum $x_0$ is located.
This expression could then be inserted into $(\sqrt{p^2-4q}-p)/2$ and should result in an expression for $x_0=f(\alpha,\beta)$.

Of course, it should also be checked, if $x_0$ is actually a maximum via the second derivative.
However, I checked the shape of $(\sqrt{p^2-4q}-p)/2$ for some $\alpha$ and $\beta$ values I am interested in and it didn't seem to have local minimas in that range.

Basically, my problem is the step, where I want to solve for $\omega$ when setting the first derivative to zero:
\begin{equation}
\frac{\partial}{\partial \omega} \frac{1}{2}\left(\sqrt{p^2-4q} - p\right) = 0
\end{equation}
My problem is, that $\omega$ occurs only in the trigonometric functions in the $a_i(\omega)$-terms.
I tried to solve the problem with with a symbolic programming library (sympy), but failed at this point to estimate a solution.

Has anyone a hint for me, how such a problem can be solved or in which direction I could go?
Maybe there is some kind of trick I can use to simplify the problem?

Thank you all in advance!

答:
I haven't double checked to make sure there are no typing errors but something like that will do the trick. The main idea was to write a polynomial system in order to compute the Groebner basis in lexicographic order.

```<code>from sympy import *

a, b, w = symbols("a b w")

a1 = (4*(cos(w)-1))/(sin(w)**2+2*(cos(w)-1))

a2 = (4)/(sin(w)**2)

a3 = (3-cos(w))/(cos(w)-1)

a4 = (-16)/(sin(w)**2*(cos(w)-1))

p = a1 * a + a2 * b

q = a3 + a4 * a * b

x, r, s = symbols("x r s")

EQN_1 = cancel(x**2 + x*p + q)

EQN_2 = cancel(diff(q,w) + x*diff(p,w))

num_EQN_1, den_EQN_1 = fraction(EQN_1)

num_EQN_2, den_EQN_2 = fraction(EQN_2)

EQNS = Matrix([[num_EQN_1, num_EQN_2, r**2+s**2-1]])
EQNS = EQNS.subs(sin(w),r)
EQNS = EQNS.subs(cos(w),s)

"""
SOL = groebner(EQNS, s, r, x, order='lex')
SOL = groebner(EQNS, x, s, r, order='lex')
"""
SOL = groebner(EQNS, x, r, s, order='lex')

"""
TESTING for alpha = beta = 1
"""

SOL_1 = SOL
SOL_1 = SOL_1.subs(a,1)
SOL_1 = SOL_1.subs(b,1)
sym_cos_w = solve(SOL_1[2],s)
num_cos_w = sym_cos_w

for i in range(0,len(sym_cos_w)):
    num_cos_w[i] = N(num_cos_w[i])
    if ( abs(im(num_cos_w[i])) < 1e-6 ) and ( abs(re(num_cos_w[i])) < ( 1 + 1e-6 ) ):
        num_sin_w = solve(SOL_1[1].subs(s,num_cos_w[i]),r)
        for j in range(0,len(num_sin_w)):
            num_x = SOL_1[0].subs(s,num_cos_w[i])
            num_x = solve(num_x.subs(r,num_sin_w[j]))
            for k in range(0,len(num_x)):
                print('cos_w = ', num_cos_w[i], 'sin_w = ', num_sin_w[j], 'x = ', num_x[k])
</code>```

EDIT:

The answer was updated because the line EQN_2 = cancel(2*diff(q,w) + x*diff(p,w)) was wrong. The correct assignment is EQN_2 = cancel(diff(q,w) + x*diff(p,w)).

Thank you, FloSe, for pointing it out.

Body:
I haven't double checked to make sure there are no typing errors but something like that will do the trick. The main idea was to write a polynomial system in order to compute the Groebner basis in lexicographic order.

from sympy import *

a, b, w = symbols("a b w")

a1 = (4*(cos(w)-1))/(sin(w)**2+2*(cos(w)-1))

a2 = (4)/(sin(w)**2)

a3 = (3-cos(w))/(cos(w)-1)

a4 = (-16)/(sin(w)**2*(cos(w)-1))

p = a1 * a + a2 * b

q = a3 + a4 * a * b

x, r, s = symbols("x r s")

EQN_1 = cancel(x**2 + x*p + q)

EQN_2 = cancel(diff(q,w) + x*diff(p,w))

num_EQN_1, den_EQN_1 = fraction(EQN_1)

num_EQN_2, den_EQN_2 = fraction(EQN_2)

EQNS = Matrix([[num_EQN_1, num_EQN_2, r**2+s**2-1]])
EQNS = EQNS.subs(sin(w),r)
EQNS = EQNS.subs(cos(w),s)

"""
SOL = groebner(EQNS, s, r, x, order='lex')
SOL = groebner(EQNS, x, s, r, order='lex')
"""
SOL = groebner(EQNS, x, r, s, order='lex')

"""
TESTING for alpha = beta = 1
"""

SOL_1 = SOL
SOL_1 = SOL_1.subs(a,1)
SOL_1 = SOL_1.subs(b,1)
sym_cos_w = solve(SOL_1[2],s)
num_cos_w = sym_cos_w

for i in range(0,len(sym_cos_w)):
    num_cos_w[i] = N(num_cos_w[i])
    if ( abs(im(num_cos_w[i])) < 1e-6 ) and ( abs(re(num_cos_w[i])) < ( 1 + 1e-6 ) ):
        num_sin_w = solve(SOL_1[1].subs(s,num_cos_w[i]),r)
        for j in range(0,len(num_sin_w)):
            num_x = SOL_1[0].subs(s,num_cos_w[i])
            num_x = solve(num_x.subs(r,num_sin_w[j]))
            for k in range(0,len(num_x)):
                print('cos_w = ', num_cos_w[i], 'sin_w = ', num_sin_w[j], 'x = ', num_x[k])

EDIT:

The answer was updated because the line EQN_2 = cancel(2*diff(q,w) + x*diff(p,w)) was wrong. The correct assignment is EQN_2 = cancel(diff(q,w) + x*diff(p,w)).

Thank you, FloSe, for pointing it out.


问:
Do you know of any functions which resemble this function?
I am looking for a function which is non $0$ between two $x$ coordinates and $0$ for all other coordinates. I have found one function like this, which is $\frac{1}{x^{\infty}+1}$: see image. However, it is not ideal, as $1^{\infty}$ is indeterminate. Do you know of any other such functions? I have also played around with equations involving square roots, however this doesn't seem to work as then the function is undefined for all values not between the x coordinates, which is something that I don't want.

答: $$f(x) = \begin{cases}1 \text{ if } |x|<1\\0 \text{ otherwise}\end{cases}$$


答: What about smooth bump functions?</p>

Title: Do you know of any functions which resemble this function?

Body: I am looking for a function which is non $0$ between two $x$ coordinates and $0$ for all other coordinates. I have found one function like this, which is $\frac{1}{x^{\infty}+1}$: see image. However, it is not ideal, as $1^{\infty}$ is indeterminate. Do you know of any other such functions? I have also played around with equations involving square roots, however this doesn't seem to work as then the function is undefined for all values not between the x coordinates, which is something that I don't want.

Body: What about smooth bump functions?


问:
Proving $f: \mathbb{R}^n \to \mathbb{R}^n$ defined by $f(x) = |x|x$ is smooth
Given the function $f: \mathbb{R}^n \to \mathbb{R}^n$ defined by $f(x) = |x|x$, I am trying to prove that $f$ is smooth.

So far, I took its partial derivatives, all of which are polynomials. I know that polynomials are continuous and, intuitively, it must be   case that they are smooth, and if all the components are smooth, the function is smooth. I've only proved that a function is differentiable if its components are differentiable, but not smoothness. I assume it also holds, but I don't know how to prove it. Is induction it?

答:
If it was smooth, its first component $f_1(x)=|x|x_1$ would be smooth when its domain was restricted to the line $$\{ (t,0,\dots,0): t\in\mathbb R\}$$
and on this line, $f_1((t,0,\dots,0)) = g(t):= |t|t$. But $g$ is not smooth; $g$ is merely once differentiable at $t=0$. Indeed,
$$ g(t)=\begin{cases}t^2 & t\ge 0, \\ -t^2 & t< 0\end{cases}$$
so (and one checks $t=0$ carefully)
$$ g'(t)=\begin{cases}2t & t\ge 0, \\ -2t & t< 0\end{cases} = 2|t| $$
and $|t|$ is not differentiable at $t=0$. (But $|t|$ is continuous, so $g$ belongs to the slightly better space $C^1(\mathbb R)$.)

We see a posteriori that the partials are not polynomials. (In fact, the above is an explicit calculation of a partial.) However, if you consider restricting the domain to a set that avoids the origin, then $f$ is indeed smooth.


问:
ZF: Regularity axiom or axiom schema?
I have seen the axiom system ZF for set theory described including a single axiom of regularity (aka "foundation"), namely
$$\forall x\neq\emptyset \, \exists y\in x \ y\cap x = \emptyset$$
and also including regularity as an infinite axiom schema, with an axiom for every formula $\varphi(x,x_1,..,x_n)$:
$$\forall x_1,..,x_n \,\exists x \left(\varphi \rightarrow \exists x \, \left( \varphi \land \forall y\in x \ \neg \varphi\frac{y}{x}\right)\right)$$

The second version states that each non-empty class has an $\in$-minimal element, while the first one states that every non-empty set has an $\in$-minimal element.
Is the second one stronger? Is it needed?

答:
Let $\phi(x,x_1, \ldots, x_n)$ be a given formula. For given $x_1, \ldots, x_n$, suppose there is an $x$ such that $\phi(x, x_1, \ldots, x_n)$ holds. Let $X$ be the transitive closure of $\{x\}$ (which is a set) and

\[ z = \{y \in X \mid \phi(y, x_1, \ldots,  x_n) \} \]
Then $z$ is non-empty, by regularity, $z$ has an
$\in$-minimal element $x'$. Let $y \in x'$, then $y \in X$ (as $X$ is transitive) and $y \not\in z$ (as $x'$ is $\in$-minimal), so $\neg\phi(y,x_1,\ldots, x_n)$. That is, $x'$ is an $\in$-minimal element of the class $\phi$.

So the schema follows from the other axioms of $\mathsf{ZF}$.

Title: ZF: Regularity axiom or axiom schema?


答:
$3^{2m}\equiv1\pmod4$

$\implies5^r3^{2m}\equiv5^r\pmod{4\cdot5^r}$

$\equiv5^r\pmod{100}$ for $r\ge2$

Again for $r\ge2,$

$$5^r-25=25(5^{r-2}-1)\equiv0\pmod{100}$$ using Why is $a^n - b^n$ divisible by $a-b$?

Title: Find the last two digit of $5^{121}*3^{312}$

Body:
The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$

and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of  $5^{143}*3^{312}$ are $25$.

Don't know why  $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.

Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$


答:
The first sentence says that whatever power greater than $1$ you pick, $5$ to that power ends in $25$.  Note that the powers of $5$ are $5,25,125,625,3125\ldots $  The end digits are always $25$ after the first one because when you multiply some number of hundreds by $5$ you still have some number of hundreds and when you multiply $25$ by $5$ you get $125$.  The $1$ goes into the hundreds and you are left with $25$.

Then from $3^{40} \equiv 1 \pmod {100}$ you get $3^{280} \equiv 1 \pmod {100}$ by taking the seventh power.  We now want $5^{121} \cdot 3^{32}$ or $5^{143} \cdot 3^{32} \pmod {100}$.  Noting that $5^2 \cdot 3^4 \equiv 1 \pmod {100}$ is new information, it is not derived from the preceding.  Now we can say $(5^2 \cdot 3^4)^8=5^{16} \cdot 3^{32} \equiv 1 \pmod {100}$  We multiply by the remaining factors of $5$, whether $105$ or $127$ and get $25 \pmod {100}$


问:
Show that the angle that a tangent line of a hyperbola forms with the intersection with a vertical line, and a focus is a right angle.
Question:
Let $P(p, q)$ be a first quadrant point on $x^2/a^2 −y^2/b^2 = 1$. Let $D$ be the point where the
tangent at $P$ meets the line $x =a^2c$. $F_2$ is the focus $(c, 0)$.

Show that $∠DF_2P$ is a right angle.

I believe the way to solve this problem is to confirm that the slope of $DP$ is the negative reciprocal of the slope of $DF_2$.But the points being $(p,q)$, $(a^2,y)$. and $(c,0)$ doesn't seem to get me there.

答:
Hint: Try using parametric coordinates $( a\sec \theta ,b\tan \theta )$ for the point P.

Henceforth calculate the intersection of the tangent line
$\frac{x\sec \theta }{a} -\frac{y\tan \theta }{b} =1$ with the line $x=a^{3} e$, assuming the focus of the hyperbola is (ae,0), where e is the eccentricity.

This intersection point turns out to be:
$\left( a^{3} e,\frac{\left( a^{2}\sec \theta -1\right) b}{\tan \theta }\right)$.

So, we have P, D and F.
Can you now confirm that the slope of DP is the negative reciprocal of the slope of DF?


问:
Is there exist a hollow module which is not Artinian
A module is said to be hollow if, its every proper submodule is small, that is, for any two proper submodule $N,K$ of $M$, $N+K\neq M$. A module is said to be Artinian if it satisfies descending chain condition on its submodules.

I am seeking for an example which is hollow but not Artinian.

答: Let $K$ be a field and let $V$ be a $K$-vector space. The ring $R=K\oplus V$, where elements $(k_1,v_1)$ and $(k_2,v_2)$ are added and multiplied by the rules $(k_1,v_1)+(k_2,v_2)=(k_1+k_2,v_1+v_2)$ and $(k_1,v_1)\cdot(k_2,v_2)=(k_1k_2,k_1v_2+k_2v_1)$, is a commutative local ring with maximal ideal $0\oplus V$. If you consider $R$ as an $R$-module, written ${}_RR$, then its submodule lattice is the ideal lattice of the ring $R$, which is isomorphic to the subspace lattice of $V$ with a new top element adjoined. So ${}_RR$ is hollow. It is Artinian only if $V$ is finite dimensional.


问:
If $b = c \times a$ and $c = a \times b$, and length $b$ = length $c$, $a$ is a unit vector.
If $\vec b = \vec c \times \hat a\,$  and $\,\vec c = \hat a \times \vec b\,$, and  $|\vec b|$ = $|\vec c|$.

Assuming $\vec b \ne 0$.

I have managed to prove $\vec a$, $\vec b$ and $\vec c$ are orthogonal, but not much else.

Help appreciated! Thanks.

答:
Note also that $u \times (v\times w)=(u\cdot w)v-(u\cdot v)w$

And $a\cdot c=a\cdot (a\times b)=0$

Apply this to $c=a\times b=a\times (c\times a)=|a|^2c-(a\cdot c)a=|a|^2c$

It looks as though the comment by Ali is the solution intended, as it can be carried through without knowing the identity for the triple product, but does use the information that $b$ and $c$ have the same length, which turns out to be unnecessary for the triple product solution.

Note that $c$ is non-zero since $b$ is non-zero and $b = c\times a$.


问:
If $E\subset Y\subset X$ and $X$ is a metric space then give the example & prove the Theorem stated below.
Suppose $E\subset Y\subset X$, where $X$ is a metric space.

Give an example such that $E$ is closed relative to $Y$, but not closed relative to $X$.

Show that $E$ is closed relative to $Y$ iff $E=Y\cap F$ for some closed set $F$ of $X$.

This question came to me when I was reading Sec. 2.32 (Compact Sets) from Baby Rudin. The questions are not explicitly stated in the book. So if any of them is false, then sorry and provide proof. But I suspect both the questions are true.

答:
The open balls in $(X,d)$ are of the form $B_X(a,r):=\{x \in X: d(a,x)<r\}$, then the open balls in $(Y,d)$ (let's write $d|_{Y\times Y}=d$) should be $B_Y(a,r):=\{x \in Y: d(a,x)<r\}$ but then $B_Y(a,r)=B_X(a,r)\cap Y$ and this is true for any open ball. The same is true for any open set too. Let $O\subseteq Y$ be open relative to $Y$, since $O$ is an open neighborhood of each of its points (relative to Y), for each $x_\alpha \in O,~~~\exists~~~r_\alpha>0$ such that $x_\alpha \in B_X(x_\alpha,r_\alpha)\cap Y\subseteq O$. So $\cup_\alpha (B_X(x_\alpha,r_\alpha)\cap Y)=O$ (you can show if $x_\alpha \in O$ then $x_\alpha \in B_X(x_\alpha,r_\alpha)\cap Y$) this implies $(\cup_{\alpha}B_X(x_\alpha,r_\alpha))\cap Y=O$. $B_X(x_\alpha,r_\alpha)$'s are open in $X$ and arbitrary union of open sets are open so our claim is proved.\

Now we prove 2. $\Rightarrow$   Let $E$ is closed relative to $Y$ then $Y\setminus E$ is open relative to $Y$, there exists an open set $O$ in $X$ such that $Y\setminus E = Y \cap O$ from which you can show  that $E=Y \cap (X \setminus O)$ where $X \setminus O=F$.\

$\Rightarrow$   Let $F$ be closed in $X$ and $E=Y \cap F$. We will show any $y \in Y \setminus E$ is not a limit point of $E$ relative $Y$. Note that $y \notin F$, there exists an open ball $B_X(y,r)$ in $X$ such that $B_X(y,r) \cap F=\phi $ which implies $(Y \cap B_X(y,r))\cap (Y \cap F)=\phi$. Thus $E$ is closed in $Y$.\

Take $X=\mathbb{R},~Y=(0,1),~E=[\frac{1}{2},1)$ then $E=Y \cap [\frac{1}{2},\frac{3}{2}]$.


问:
When I graph $y=x^{1/2}$ why does it only show the positive y values.
I understand the reason for $y=\sqrt{x}$. I've been told that the radical symbol gives out the positive answer. But $y=x^{1/2}$ doesn't use a radical symbol, and it still only shows the positive y values.

答:
$x^{\frac{1}{2}}$ is basically the same function as $\sqrt{x}$

So for example, take $9$, $9^{\frac{1}{2}}$ is the same as $\sqrt{9}$, which is 3


问:
Area of triangle within an ellipse and hyperbola that have the same foci

Question: An ellipse and a hyperbola have the same foci, $F_1$ and $F_2$. These curves cross at 4
points - let $P$ be one of the points. These curves also intersect the line $\overleftrightarrow{F_1F_2}$ at 4 points labelled $Q$, $R$, $S$ and $T$ in that order. If $RS$ = 20, $ST$ = 14 and $∆PF_1$ $F_2$ is isosceles,
compute the area of $∆PF_1$ $F_2$ .

I have no idea how to go about solving this. I know that the easiest way of getting the area is probably by SSS and Heron's Formula, and I did get that the major axis is $14 + 20 + 14=48$ so the $PF_1+PF_1=48$ and probably that $PF_1$ is congruent to $F_1F_2$

答:
If the length of transverse axis of hyperbola is $2a$ and of conjugate axis is $2b$ AND the length of major axis of ellipse is $2d$ and of minor axis is $2e$, $RS = 2a = 20 \implies a = 10$.

Also, $d = 10 + 14 = 24$.

As point $P$ is on both the hyperbola and the ellipse,

$PF_1 - PF_2 = 20$

$PF_1 + PF_2 = 48$

So $PF_1 = 34, PF_2 = 14$.

As $\triangle PF_1 F_2$ is isosceles triangle, either $F_1F_2 = PF_1$ or $F_1F_2 = PF_2$.

If distance of foci from the center is $c$, $c = \frac{F_1F_2}{2}$ and we know $c^2 = d^2 - e^2 = a^2 + b^2$. From this, we confirm that the only possibility is $F_1F_2 = PF_1 = 34$.

So the sides of the isosceles triangle $\triangle PF_1F_2$ are $34, 34$ and $14$. You can find altitude and then the area or use Heron's formula.


问:
Show that $GL(2, \Bbb Q)$ is isomorphic to a subgroup of $GL(3, \Bbb Q)$
What should be my approach for this particular question and general approach to prove isomorphism of one group to another ?</p>

答:
Just consider $\mathcal i: GL(2,\Bbb Q) \hookrightarrow GL(3,\Bbb Q)$ given by

$\begin{pmatrix} a_{11} &a_{12} \\ a_{21} &a_{22}\end{pmatrix} \mapsto \begin{pmatrix} 1 &0 &0\\ 0 &a_{11} &a_{12} \\ 0 &a_{21} &a_{22}\end{pmatrix}$


问:
Relationship between Mean Value Theorem and the maximum norm
I am seeking assistance with the following application of the Mean Value Theorem:

Let $x \in \Omega$ and construct an associated neighbourhood $N_x = (a, a+ \sqrt{\epsilon})$, such that $x \in N_x$ and $N_x \subset \Omega$. Then, by the Mean Value Theorem, for some $y \in\overline{N}_x $,

$$ \frac{u(a + \sqrt{\epsilon}) - u(a)}{\sqrt{\epsilon}} = u'(y)...........(1) $$

It follows that

$$ |u'(y)| \le 2\epsilon^{-1/2}||u||_{N_x}  \le C\epsilon^{-1/2}............(2)$$

My problem is that I do not quite follow how expression (2) is obtained from expression (1).

答:
We have $\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|_{N_x}}$
\begin{align*}
  \abs{u'(y)} &= \abs{\frac{u(a+ \sqrt \epsilon) - u(a)}{\sqrt \epsilon}}\\
         &\le \epsilon^{-1/2} \cdot \bigl(\abs{u(a + \sqrt \epsilon)} + \abs{u(a)}\bigr)\\
         &\le \epsilon^{-1/2} \cdot \bigl( \norm u + \norm u\bigr)\\
         &= 2 \epsilon^{-1/2} \norm u.
\end{align*}</p>


问:
Question about proof that $n! = \omega(2^n)$
I'm reading CLRS and have came across this sub-question:

Proof that $n! = \omega(2^n)$

I have seen several solutions that are more or less the same as the one stated here. The solution simply says

For all $n \ge 4$, $$\begin{align}n!&= n\cdot(n - 1) \cdot \ldots \cdot 2\cdot 1 \\&> 2 \cdot 2\cdot\ldots \cdot 2\cdot 2\\ &= \omega(2^n)\end{align}$$

Now, I do understand what the above says.

However, what I am confused about is that by the formal definition of $\omega$, we need to prove that

$$\forall C> 0 ~~\exists n_0 > 0 ~~ \forall n \ge n_0~~~~0 \le C\cdot2^n  < n!$$
The solution has proved the above case for $C = 1$: For $C = 1$, there exists $n_0 > 0$ (namely $n_0 = 4$) such that for all $n \ge 4$, we have $$0 \le 1\cdot 2^n < n!$$

but this is only for the case $C = 1$, not for all $C > 0$.

Am I right to say that the solution I quoted is incomplete, and that it is necessary to prove the inequality for all values of $C$ instead of just $C = 1$?

答:
I think the previous answer that is currently ticked is wrong please read this link

c's value must be a constant and independent of n, but may depend on n0 which in the link is referred to as k.
So, it can be 1.

(If this is wrong please delete this)

To prove f(n)=ω(g(n)) we must prove that

$\lim_{x \to \infty}\frac{ f(n)}{g(n)}$=$\infty$

which is proved by this link that was given in the question also.


问:
Unclear question on threshold probability
I found this question:
Consider the binomial subset of $\Gamma = \{1,...,N\}$. Find a threshold probability for
containing a tuple $x<y<z$ such that $x + y = z$.

I am really confused about what the author means by the 'binomial subset' here. Could anyone please elaborate on the question? Thanks for the help.

答:
The words "binomial subset" are pretty much nonsense, but from context, it's likely that the author meant: "Choose a subset of $\{1, 2, \dots, N\}$ by choosing each element independently at random with probability $p$."

A threshold probability means that we consider the limit as $N \to \infty$, letting $p$ depend on $N$. The threshold probability $p^*$, also depending on $N$, should satisfy the following:

if $p \ll p^*$, the desired property (having a tuple $x<y<z$ with $x+y=z$) holds with a probability tending to $0$ as $N \to \infty$.

if $p \gg p^*$, the desired property holds with a probability tending to $1$ as $N \to \infty$.


问:
Fixed Point Iteration and order of convergence
Consider fixed point iterations $x_{n+1} = 2 \sin x_n+ \tan x_n − 3x_n $, where $g(x) =
2 \sin x + \tan x − 3x$. Find its order of convergence at fixed point $\alpha = 0$. (Hint: Using Tayor’s
series might be easier.)

What have I tried?

I tried the idea of taking the derivative of g(x).

Because this is an fixed point iteration, $g(\alpha)$ will affect the convergence of the iteration. If $g(\alpha) < 1$, the iteration will converge with linear order. If $g(\alpha) = 1$, we have no clue whether it converges or not, and if it converges, it will converge very slow. if $g(\alpha) = 0$, it will converge with higher order.

I realize that $\alpha=0$ is the fixed point. So I tried to calculate $g', g'', g''', g^{(4)}$.

I came up with $g'(\alpha) = 0, g''(\alpha) = 0, g'''(\alpha)=g^{(4)}(\alpha) = 0$, and $g^{(5)}(\alpha) \neq 0$. So I guess the order of convergence is 5. But it take a lot of steps to evaluate until the fifth derivatives of $g(x)$. According to the hint, there might be a better and faster solution by using the Taylor' series.

Body:
$\sin x = x-\frac {x^3}{3!}+\frac {x^5}{5!} -\frac{x^7}{7!}+...$

$\tan x = x+\frac {x^3}{3}+\frac {2x^5}{15}+\frac {17x^7}{315}+...$

Therefore,

$g(x) = 2(x-\frac {x^3}{3!}+\frac {x^5}{5!}+-\frac{x^7}7+...) + (x+\frac {x^3}{3}+\frac {2x^5}{15}+\frac {17x^7}{315}+...)-3x $

$g(x) =  x^5(\frac {1}{5!}+\frac{2}{15})+x^7(\frac{-1}{7!}+\frac{17}{315})+...$

The leading term of the Taylor's expansion of $g(x)$ is $x^5$ which means $g'(\alpha)=g''(\alpha)=g'''(\alpha)=g^{(4)}(\alpha)=0$, $g^{(5)}(\alpha) \neq 0$,where $\alpha$ is the fixed point of $x=g(x)$

By applying Taylor's theorem to $g(x_n)$, we have:

$g(x_n)=g(\alpha) +\frac {g^{(5)}(c)}{5!}(x_n-\alpha)^5 $, where $c\in[\alpha,x_n]$

$g(x_n)\approx g(\alpha) +\frac {g^{(5)}(\alpha)}{5!}(x_n-\alpha)^5 $

$g(x_n)-g(\alpha)\approx \frac {g^{(5)}(\alpha)}{5!}(x_n-\alpha)^5 $

$|\alpha - x_{n+1}|\approx\frac {g^{(5)}(\alpha)}{5!}|\alpha-x_n|^5$

Therefore, the convergence order is 5


问:
Showing that the order of the zero of the $j$ function is $3$
I'm aware that the $j$-invariant can be defined as a function on the upper half-plane $\Pi^{+}$(as seen in the following wikipedia article: https://en.wikipedia.org/wiki/J-invariant):

$$ j(\tau)=1728\frac{(g_{2}(\tau))^{3}}{(g_{2}(\tau)^{3}-27g_{3}(\tau)^{2})} $$

I have to show that the zero of $j(\tau)$ at $p= e^{\frac{2 \pi i}{3}}$ has order 3. Differentiating in a "naive" way gives:

$$ j'(\tau) = \frac{3g_{2}(\tau)^{2}((g_{2}(\tau)^{3}-27g_{3}(\tau)^{2}))- g_{2}(\tau))^{3}(3g_{2}(\tau)^{2})-54g_{3}(\tau))}{(g_{2}(\tau)^{3}-27g_{3}(\tau)^{2})^{2}} $$, this implies that $j'(e^{\frac{2 \pi i}{3}})=0$

In this way, I can show that $$ j''(p) = 0$$, but that $$j'''(p) \neq 0 $$, so the zero of the function is of order 3.

But, I'm not sure if this style of differentiation is correct for a function like the $j$-invariant. Can someone tell me if I'm doing this in an appropriate manner?

答:
$z\to e^{2i\pi /3}z$ is an automorphism of $\Bbb{Z+e^{2i\pi /3}Z}$ so that $g_2(e^{2i\pi 3})=(e^{2i\pi 3})^{-4} g_2(e^{2i\pi 3})\implies g_2(e^{2i\pi 3})=0$.

On the other hand $g_3(e^{2i\pi /3})\ne 0$ and $g_2'(e^{2i\pi /3})\ne 0$ (numerical check).


问:
Strong law of large numbers and the central limit theorem
The strong law of large numbers and the central limit theorem use different modes of convergence. Is it nevertheless true that the strong law of large numbers can be shown from the central limit theorem?

It seems that, if random variables $\sum_i X_i$ converge to the gaussian distribution in probability, then due to the fact we normalized them for this to happen, that if we did not normalize them, then they would converge almost surely to a dirac delta. But the details elude me.

答:
For example, suppose $X_n$ are iid standard normal random variables, and $U$ uniform on $[0,1]$ independent of the $X_n$. Let $T_n = n$ if $U$ is in an interval $[a_n, a_n+1/n] \subset [0,1]$ where the $a_n$ are arranged so every point of $[0,1]$ is in infinitely many of the intervals.  Let $Y_n = X_n + T_n - T_{n-1}$ (with $Y_1 = X_1 + T_1$).  Then the partial sums $S_n = \sum_{i=1}^n Y_n = T_n + \sum_{i=1}^n X_n$ satisfy the conclusion of the Central Limit Theorem in that $S_n/\sqrt{n}$ converges in distribution to the standard
normal distribution, but does not satisfy the conclusion of the Strong Law of Large Numbers in that  $S_n/n$ does not converge a.s. to $0$, indeed almost surely  $\limsup_{n \to \infty} S_n/n \ge 1$.


问:
Convergence of sums of $\sin(kY)$ where Y is uniform
Let $Y$ be uniformly distributed on $[0,2\pi]$. Let $X_k=\sin(kY)$.

I want to show that $\frac{X_1+X_2+\cdots+X_n}{n}\to 0$ almost surely.

I thought of using the strong law of large numbers but it requires the $X_k$'s to be identically distributed.

答: You can compute the geometric sum  $e^{iy}+e^{2iy}+..._+e^{iny}$ and show that $\frac 1 n (e^{iy}+e^{2iy}+..._+e^{iny}) \to 0$ for all real $y$ such that $e^{iy} \neq 1$. Take the imaginary part.  Can you finish?


问:
solving an equilibrium equation
I have the following example:

$ pa_1=a_0\\ pa_0+qa_1+pa_2=a_1\\qa_0+qa_2=a_2 $

where p+q=1.

I can see how to get $a_1=(1/p)a_0 $  but from there they say from the third equation they produce $a_2 = (q/p)a_0$. Then the values are substituted into the normalising equation of $a_0 + a_1 + a_2 =1$. With a final answer of $a_0 =p/2 \ , a_1=1/2  \ , \ a_2=q/2$

Update: I can see how they get $a_2 = (q/p)a_0 $

$qa_0+qa_2=a_2 \ -> qa_0=a_2 - qa_2\ -> qa_0=a_2(1-q)\ −> a_2 = qa_0/(1-q)   \  $
and as p+q=1 p=1-q hence $a_2=(q/p)a_0$

答:
$ p\ a_1=a_0\\ p\ a_0+q\ a_1+p\ a_2=a_1\\q\ a_0+q\ a_2=a_2 $

Now replacing $a_1$ from the first equation gives:

$$p\left[ \left(1+\frac{q-1}{p^2}\right)a_0+a_2\right]=0\\
q\ a_0=\left(1-q\right)a_2$$

Using the fact that $p=1-q$, the third equation gives(which is the same as what you might have got from the second equation):

$$a_2=\frac{q}{p}a_0$$

Now replacing $a_1$ and $a_2$ in the relation$a_0+a_1+a_2=1$, we get:

$$a_0+a_1+a_2=a_0+\frac{1}{p}a_0+\frac{q}{p}a_0=\frac{a_0}{p}\left(p+1+q\right)=\frac{2}{p}a_0=1\\ \Rightarrow a_0=\frac{p}{2},a_1=\frac{1}{2},a_2=\frac{q}{2}$$

Title: solving an equilibrium equation


问:
Closed form of minimum value of A
I want the minimum value of A. (closed form)

$$A = \int_{0}^{p} \sin(x) dx + \int_{p}^{\frac{\pi}{2}} (\sin(x)-\sin(p))dx$$
$$0 \leq p \leq \frac{\pi}{2}$$

So...
$$A = p \cdot \sin(p) - \frac{\pi}{2}\sin(p) + 1$$

$$\frac{d}{dp} A = \sin(p) + p \cdot \cos(p) - \frac{\pi}{2}\cos(p) = 0$$

$$\tan(p) + p - \frac{\pi}{2} = 0$$

I stopped here.
I know that:

$$A = 1 - \tan(p) \cdot \sin(p)$$

$$\sin(p) = \cos(\tan(p))$$
$$\cos(p) = \sin(\tan(p))$$
$$\tan(p) = \cot(\tan(p))$$
$$\csc(p) = \sec(\tan(p))$$
$$\sec(p) = \csc(\tan(p))$$
$$\cot(p) = \tan(\tan(p))$$

$$p \approx 0.71046$$
$$A \approx 0.4389$$

EDIT:

$$\tan(p) + p - \frac{\pi}{2} = 0$$

$$\tan(p) = \frac{\pi}{2} - p$$

$$-cot\left(p-\frac{\pi}{2}\right) = \frac{\pi}{2} - p$$

$$cot\left(p-\frac{\pi}{2}\right) = p - \frac{\pi}{2}$$

$$B:=p-\frac{\pi}{2}$$

$$cot(B) = B$$

"This kind of equations which mix polynomial and trigonometric terms do not, in general, show solutions which have a closed form expression..." as said by Claude Leibovici here.

答:
Starting from @vitamin d's answer, we need to solve for $p$ the equation
$$\sin (p)+\left(p-\frac{\pi }{2}\right) \cos (p)=0$$ for which we cannot expect a closed form (this is already the case for $x=\cos(x)$).

First, for the fun of it, I shall use two approximations, $1,400$ years old, which are valid for the range under consideration; they are
$$\sin(p)\simeq \frac{16 (\pi -p) p}{5 \pi ^2-4 (\pi -p) p} \qquad \text{and} \qquad \cos(p) \simeq\frac{\pi ^2-4p^2}{\pi ^2+p^2}$$

Unfortunately, this will lead to a quintic polynomial
$$32p^5-16(3 \pi-2)p^4+16 \pi  (3 \pi -2)p^3+8 (4-\pi ) \pi ^2 p^2-2 \pi ^3 (16+7 \pi )p+5 \pi ^5=0$$ the numerical solution of which being $0.710393$.

On the other side, we can make better and better explicit approximations of the solution performing one single iteration of high order method.

As a function of the order [$n=2$ means Newton, $n=3$ Halley, $n=4$ Householder, $n=5$ has no name], I give the expression obtained for $p$, its numerical representation and the value of the minimum value of $A$.

$$\left(
\begin{array}{cccc}
 n & p_{(n)} & p_{(n)} \sim & A_{\text{min}} \\
 2 & \frac{-16+12 \pi +\pi ^2}{4 (8+\pi )} & 0.70835286 & 0.43890829 \\
 3 & \frac{-256+208 \pi +24 \pi ^2+3 \pi ^3}{704+64 \pi +12 \pi ^2} & 0.71064251
   & 0.43890370 \\
 4 & \frac{-8448+6464 \pi +832 \pi ^2+124 \pi ^3+11 \pi ^4}{20480+3136 \pi +352
   \pi ^2+44 \pi ^3} & 0.71045640 & 0.43890366 \\
 5 & \frac{-327680+236800 \pi +43776 \pi ^2+6944 \pi ^3+656 \pi ^4+57 \pi
   ^5}{820224+147456 \pi +24960 \pi ^2+1920 \pi ^3+228 \pi ^4} & 0.71046310 &
   0.43890366 \\
\cdots & \cdots& \cdots & \cdots \\
\infty & &0.71046274 & 0.43890366
\end{array}
\right)$$

Title: Closed form of minimum value of A


问:
In the proof of Morse's Lemma, why can it happen that $h_{ij} \neq h_{ji}$?
The following is the first step in the proof of Morse's Lemma:

Without loss of generality, we can assume that $\overline x$ is the origin in $\mathbb R^N$ (in the sense that the chart takes $\overline x$ to the origin) and that $f(\overline x) = f(0) = 0$ (trivial).
Note that we can write, for $x$ in a neighborhood of the origin,
\begin{align*}
f(x)
& = \int_0^1 \sum_{i = 1}^N \frac{\partial f}{\partial x_i} (tx) x_i \ dt \\
& = \sum_{i,j = 1}^N \left(\int_0^1 \int_0^1\frac{\partial^2f}{\partial x_i \partial x_j} (stx) \ ds \ dt\right)x_ix_j \\
& = \sum_{i, j = 1}^N h_{ij}(x) x_i x_j.
\end{align*}

Then, Milnor argues that we can assume $h_{ij} = h_{ji}$ by taking $\overline h_{ij} = \frac 12(h_{ij} + h_{ji})$ if necessary. My question is, isn't $h_{ij} = h_{ji}$ already, since the function $f$ is smooth?

答:
Milnor applies the original lemma to the functions $g_i(x)=\displaystyle\int_0^1 \dfrac{\partial f}{\partial x_i}(tx)\,dt$ rather than trying to give the formula explicitly. Thus, he is not a priori guaranteed any symmetry.

I suggest that you did not correctly differentiate under the integral sign when you gave your explicit formula. Indeed, let's try a simple example, even with $n=1$. Take $f(x)=x^2$. Then $g_1(x)=\int_0^1 f'(tx)\,dt = \int_0^1 2tx\,dt = x$, as of course it should be. Now your formula gives
$\int_0^1\int_0^1 f''(stx)\,ds\,dt = \int_0^1\int_0^1 2\,ds\,dt = 2$, which does not recover the function $f$. We're off by a numerical factor. You need to go back and compute the derivative $\dfrac{\partial g_i}{\partial x_j}$ more carefully. (Examples are your friends!!)

That said, I think that when you make your explicit integral formula correct, you will in fact end up proving what you believed in the first place, i.e., that $h_{12}=h_{21}$ when $f$ is $C^2$.


问:
Prove that if $p_1,\dots,p_k$ are distinct odd primes then 1 has $2^k$ square roots $\mod m$ where $m$ is the product of the primes.
I think I am most of the way through this proof but I am stuck. Here was my approach: I looked at the square roots of $1$ mod $105$, and noticed that each one corresponded to one less than an integer multiple of a subset of the prime factors of $105$, i.e. we have for instance $71=24\cdot3-1$ or $29=2\cdot(3\cdot 5)-1$ and so on. I conjectured that given a subset of the prime factors of $m$, I could construct a unique square root of $1$ mod $m$, which would show that there are (at least) $2^k$ of them.

So I did that and I think that it worked. Pick some set of the prime factors of $m$, say $p_{a_1},\dots,p_{a_j}$. Claim: we can find a corresponding $n$ such that $n\prod_{i=1}^jp_{a_i}-1$ is a square root of $1$ mod $m$.

Proof: We are looking for a solution to

$$\left(n\prod_{i=1}^jp_{a_i}-1\right)^2\equiv 1 \pmod m$$
which may be rewritten as

$$\left(\prod_{i=1}^jp_{a_i}\right)^2n^2 - 2\prod_{i=1}^jp_{a_i}n\equiv 0 \pmod m$$

Now since $\prod_{i=1}^jp_{a_i}|m$ we may write this as $$\prod_{i=1}^jp_{a_i}n^2-2n \equiv 0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$$

or

$$n\left(\prod_{i=1}^jp_{a_i}n-2\right) \equiv 0 \mod {\frac{m}{\prod_{i=1}^jp_{a_i}}}$$

This has solutions when $n\equiv0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$ or when $\prod_{i=1}^jp_{a_i}n-2\equiv 0 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$. In the first case we end up with the same root for any choice of $p_{a_i}$'s so that's sort of a degenerate one. But in the second case we have $\prod_{i=1}^jp_{a_i}n\equiv 2 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$, which has a unique and nonzero solution for $n$. Plugging that $n$ back into $n\prod_{i=1}^jp_{a_i}-1$ thus gives a square root of $1$ mod $m$ which corresponds uniquely to your choice of prime factors, thus showing that there are (at least) $2^k-1$ square roots of 1. Then we add 1 in which corresponds to the empty set of prime factors and we get the full $2^k$.

My problem is ruling out the possibility of more square roots of 1, i.e. if $x^2\equiv1\mod m$ I need to show that it corresponds uniquely to something of the form $n\prod_{i=1}^jp_{a_i}-1$. This is proving to be difficult if not impossible.

I put a lot of work into going the one way and I really don't want it all to be for nothing. Can anyone see a way to go the other direction?

答: Since $x^2=1 \mod (p)$ has $2$ solutions, applying then the Chinese remainder theorem.</p>