问:
If a binary quadratic form primitively represents $n$ and $n^3$, must it be the identity form?
I've been searching numerically, and whenever a positive definite binary quadratic form can primitively represent an integer $n>1$ and $n^3$, it turns out to be equivalent to the identity form in that class group.

Is there some elementary way to see that this is only possible for identity forms?

Here is what initially gave me hope that it could be possible with other forms.

For any binary quadratic form $A x^2 + B xy + C y^2$, if it can represent integers $r$, $s$, $t$ then it can also represent $rst$.

This can be seen as follows:

For any form with coefficients $(A,B,C)$, in the class group its inverse is $(A,-B,C)$.

Both a form and its inverse can represent the same numbers (if $(x,y)$ with the former, use $(x,-y)$ with the later).

If a form can represent $r$ with $(x_1,y_1)$, and $s$ with $(x_2,y_2)$ then the identity form can represent $rs$ with some integers $(u,v)$ due to form composition:

$$
\begin{aligned}
rs &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 + B x_2 y_2 + C y_2^2) \\
   &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 - B x_2 (-y_2) + C (-y_2)^2) \\
   &= I(u,v) \\
\end{aligned}
$$

Futhermore, if the form can represent $t$ with $(x_3,y_3)$, then it can represent $rst$ by some integers $(x_4,y_4)$ due to form composition.

$$rst = I(u,v)(A x_3^2 + B x_3 y_3 + C y_3^2) = (A x_4^2 + B x_4 y_4 + C y_4^2)$$

It is definitely possible to primitively represent cubes, for example

```<code>a) 2 x^2 - xy + 6 y^2 = 3^3  with x=3,y=-1
b) 2 x^2 - xy + 6 y^2 = 8^3  with x=2,y=-9
</code>```

But in example 'a', it is not possible for that form to represent 3. While in example 'b', the form can represent 8, but not primitively (x=2,y=0).

This seems to be the case for any form not equivalent to an identity form.

While for an identity form, I've been able to find a lot of examples that work.

```<code>    x^2 + xy + 12 y^2 = 18     with x=2, y=1, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 18^3   with x=-76, y=7, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 118    with x=2, y=3, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 118^3  with x=964, y=207, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 63     with x=3, y=2, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 63^3   with x=-501, y=2, gcd(x,y)=1
</code>```

Can one prove that if a form can primitively represent $n$ and $n^3$, that is must be equivalent to the identity form?

UPDATE:

A related question On products of ternary quadratic forms $\prod_{i=1}^3 (ax_i^2+by_i^2+cz_i^2) = ax_0^2+by_0^2+cz_0^2$ shows a possible solution for any form with B=0 (all that is left is to analyze if the result is a primitive representation). Such forms are in the same equivalence class as its inverse in the class group, and are called "ambiguous forms".

Below Will Jagy gave an explicit solution for an ambiguous form which is not equivalent to the identity form, showing that the result can be a primitive representation.

The connection to what I tried is that the identity is always an ambiguous form. So that likely refines the question to:

If a form can primitively represents $n$ and $n^3$, is the form necessarily an ambiguous form?

答:
Cox treats $x^2 + 14 y^2$ in detail for the Hilbert Class Field; the outcome is pages 113-115 in the first edition, and Theorem 5.33 says a prime $p \neq 7$  is represented by $x^2 + 14 y^2$ if $(-14|p) = 1 $ and $(x^2 + 1)^2 \equiv 8 \pmod p $ has an integral solution.

The primes represented by $3x^2 + 2xy + 5 y^2 $ all the appropriate values $\pmod{56}$ Then each $p^3$ is also primitively represented by $3x^2 + 2xy + 5 y^2 .$  On the other hand, for any primitively represented composite $n,$ we know $n^3$ will be represented but perhaps not primitively. There are numerous ways to have a number $n$ be represented;  primes are restricted to some extent.

For hand calculation of composition, it helps to write the middle coefficients as $12,$  using Dirichlet's description. The product of first and last coefficients is then $30.$  The four classes become $\langle  1, 12, 50 \rangle \, , \; \; \; $ $\langle  2, 12, 25 \rangle \, , \; \; \; $ $\langle  5, 12, 10 \rangle \, , \; \; \; $ $\langle  10, 12, 5 \rangle \, . \; \; \; $
with direct calculation giving
$$\langle  5, 12, 10 \rangle^2 \, , \; \; \; = \langle  25, 12, 2 \rangle \approx \langle  2, 12, 25 \rangle  $$
since  $\langle  2, 12, 25 \rangle$ really is ambiguous

$$ \left( 5x^2 + 12 xy + 10 y^2 \right)^2 = 25 X^2 + 12 XY + 2 Y^2 $$
with
$$ X = x^2 - 2 y^2 \, , \; \; \; Y = 10 xy + 12 y^2  $$
As usual, some manipulation is necessary to write this as $2 u^2 + 7 v^2$

Next, it took some calisthenics, but we finally get
$$  \left(5 x^2 + 12 xy + 10 y^2\right)^3 = 5 p^2 + 12 pq + 10 q^2 $$
with
$$p = -5x^3 + 30 x y^2 + 24 y^3 \, , \; \; \;  q = 6x^3 + 15 x^2 y - 10 y^3 \, . \; \; \;   $$
I made explicit use of an improper automorphism to allow $\gcd(p,q) = 1$

Might as well type that in
$$
\left(
\begin{array}{rr}
1&0 \\
6&-1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
2&6 \\
6&25 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1&6 \\
0&-1 \\
\end{array}
\right) =
\left(
\begin{array}{rr}
2& 6\\
6&25 \\
\end{array}
\right)
$$

For example, we get the prime $10037=5 x^2 + 12 xy + 10 y^2$ with $x=79, y= -36$ We get a representation of $10037^3=5 p^2 + 12 pq + 10 q^2$ with $p=-513419, q = 54654.$  Note $\gcd(p,q) = 1.$

```<code>./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
3 2 5
Discriminant  -56

Modulus for arithmetic progressions?
56

      3,      5,     13,     19,     59,     61,     83,    101,    131,    139,
    157,    173,    181,    227,    229,    251,    269,    283,    293,    307,
    349,    397,    419,    461,    467,    509,    523,    563,    587,    619,
    643,    661,    677,    691,    733,    773,    787,    797,    811,    829,
    853,    859,    941,    971,    997,   1013,   1021,   1069,   1091,   1109,

    3    5   13   19   27   45    mod 56

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./cube_of_prime_Primitively_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
3 2 5
Discriminant  -56

 apparent biggest legal target   766957636
Maximum number represented?
 766957636

 xbound  16550       ybound 12819
Sat Jan 16 11:14:10 PST 2021

          27 = 3^3
         125 = 5^3
        2197 = 13^3
        6859 = 19^3
      205379 = 59^3
      226981 = 61^3
      571787 = 83^3
     1030301 = 101^3
     2248091 = 131^3
     2685619 = 139^3
     3869893 = 157^3
     5177717 = 173^3
     5929741 = 181^3
    11697083 = 227^3
    12008989 = 229^3
    15813251 = 251^3
    19465109 = 269^3
    22665187 = 283^3
    25153757 = 293^3
    28934443 = 307^3
    42508549 = 349^3
    62570773 = 397^3
    73560059 = 419^3
    97972181 = 461^3
   101847563 = 467^3
   131872229 = 509^3
   143055667 = 523^3
   178453547 = 563^3
   202262003 = 587^3
   237176659 = 619^3
   265847707 = 643^3
   288804781 = 661^3
   310288733 = 677^3
   329939371 = 691^3
   393832837 = 733^3
   461889917 = 773^3
   487443403 = 787^3
   506261573 = 797^3
   533411731 = 811^3
   569722789 = 829^3
   620650477 = 853^3
   633839779 = 859^3

Sat Jan 16 11:16:21 PST 2021
</code>```

Body:
Cox treats $x^2 + 14 y^2$ in detail for the Hilbert Class Field; the outcome is pages 113-115 in the first edition, and Theorem 5.33 says a prime $p \neq 7$  is represented by $x^2 + 14 y^2$ if $(-14|p) = 1 $ and $(x^2 + 1)^2 \equiv 8 \pmod p $ has an integral solution.

The primes represented by $3x^2 + 2xy + 5 y^2 $ all the appropriate values $\pmod{56}$ Then each $p^3$ is also primitively represented by $3x^2 + 2xy + 5 y^2 .$  On the other hand, for any primitively represented composite $n,$ we know $n^3$ will be represented but perhaps not primitively. There are numerous ways to have a number $n$ be represented;  primes are restricted to some extent.

For hand calculation of composition, it helps to write the middle coefficients as $12,$  using Dirichlet's description. The product of first and last coefficients is then $30.$  The four classes become $\langle  1, 12, 50 \rangle \, , \; \; \; $ $\langle  2, 12, 25 \rangle \, , \; \; \; $ $\langle  5, 12, 10 \rangle \, , \; \; \; $ $\langle  10, 12, 5 \rangle \, . \; \; \; $
with direct calculation giving
$$\langle  5, 12, 10 \rangle^2 \, , \; \; \; = \langle  25, 12, 2 \rangle \approx \langle  2, 12, 25 \rangle  $$
since  $\langle  2, 12, 25 \rangle$ really is ambiguous

$$ \left( 5x^2 + 12 xy + 10 y^2 \right)^2 = 25 X^2 + 12 XY + 2 Y^2 $$
with
$$ X = x^2 - 2 y^2 \, , \; \; \; Y = 10 xy + 12 y^2  $$
As usual, some manipulation is necessary to write this as $2 u^2 + 7 v^2$

Next, it took some calisthenics, but we finally get
$$  \left(5 x^2 + 12 xy + 10 y^2\right)^3 = 5 p^2 + 12 pq + 10 q^2 $$
with
$$p = -5x^3 + 30 x y^2 + 24 y^3 \, , \; \; \;  q = 6x^3 + 15 x^2 y - 10 y^3 \, . \; \; \;   $$
I made explicit use of an improper automorphism to allow $\gcd(p,q) = 1$

Might as well type that in
$$
\left(
\begin{array}{rr}
1&0 \\
6&-1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
2&6 \\
6&25 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1&6 \\
0&-1 \\
\end{array}
\right) =
\left(
\begin{array}{rr}
2& 6\\
6&25 \\
\end{array}
\right)
$$

For example, we get the prime $10037=5 x^2 + 12 xy + 10 y^2$ with $x=79, y= -36$ We get a representation of $10037^3=5 p^2 + 12 pq + 10 q^2$ with $p=-513419, q = 54654.$  Note $\gcd(p,q) = 1.$

./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
3 2 5
Discriminant  -56

Modulus for arithmetic progressions?
56

      3,      5,     13,     19,     59,     61,     83,    101,    131,    139,
    157,    173,    181,    227,    229,    251,    269,    283,    293,    307,
    349,    397,    419,    461,    467,    509,    523,    563,    587,    619,
    643,    661,    677,    691,    733,    773,    787,    797,    811,    829,
    853,    859,    941,    971,    997,   1013,   1021,   1069,   1091,   1109,

    3    5   13   19   27   45    mod 56

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./cube_of_prime_Primitively_go
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
3 2 5
Discriminant  -56

 apparent biggest legal target   766957636
Maximum number represented?
 766957636

 xbound  16550       ybound 12819
Sat Jan 16 11:14:10 PST 2021

          27 = 3^3
         125 = 5^3
        2197 = 13^3
        6859 = 19^3
      205379 = 59^3
      226981 = 61^3
      571787 = 83^3
     1030301 = 101^3
     2248091 = 131^3
     2685619 = 139^3
     3869893 = 157^3
     5177717 = 173^3
     5929741 = 181^3
    11697083 = 227^3
    12008989 = 229^3
    15813251 = 251^3
    19465109 = 269^3
    22665187 = 283^3
    25153757 = 293^3
    28934443 = 307^3
    42508549 = 349^3
    62570773 = 397^3
    73560059 = 419^3
    97972181 = 461^3
   101847563 = 467^3
   131872229 = 509^3
   143055667 = 523^3
   178453547 = 563^3
   202262003 = 587^3
   237176659 = 619^3
   265847707 = 643^3
   288804781 = 661^3
   310288733 = 677^3
   329939371 = 691^3
   393832837 = 733^3
   461889917 = 773^3
   487443403 = 787^3
   506261573 = 797^3
   533411731 = 811^3
   569722789 = 829^3
   620650477 = 853^3
   633839779 = 859^3

Sat Jan 16 11:16:21 PST 2021


问:
Logic without the three traditional laws of thought
The wikipedia page on the laws of thought mentions the The three traditional laws: The Law of the Excluded Middle, Law of Non-contradiction and The Law of the Identity. These laws are embedded in a vast majority of logic systems, but not all.

In intuitionistic logics, the The Law of the Excluded Middle does not hold. In paraconsistent logics, the Law of Non-contradiction does not hold. As pointed out in this answer, in Schrodinger logics, the The Law of the Identity does not hold.

One interesting thing about these logics is that the removal of the laws has a larger ramification, it affects other inference rules. For example, in intuitionistic logic the law of double negation doesn't hold, and reduction ad absudum may only prove negative statements. Another example, as stated in stanford's entry on paraconsistent logic: "One feature of LP which requires some attention is that in LP modus ponens comes out to be invalid". The removal of these laws makes the system restrictive in more ways than just not being able to use that specific law.

So that raises the quesion, what about a system where none of them hold? Can such a system make any deductions at all?

Note that this question is mostly for exploration/amusement/curiosity. It does not require an example of an actual formal system where these laws don't apply, as a formal system without any of these would hardly be of any use, and therefore, it has probably never been developed. However, it still feels like an interesting question.

Is it a requirement for at least one of these to hold for a system to be logical, or is it possible to build some system, even if very restrictive, in which none of these laws hold, but that is still capable of producing sound arguments?

答:
There are plenty of inference rules in classical logic which do not rely on any of those three laws. Some examples include:

Modus ponens: $A, A\rightarrow B\vdash B$.

Left and right conjunction elimination: $A\wedge B\vdash A$ and $A\wedge B\vdash B$.

Absorption: if $A, A\vdash B$ is a valid sequent, then $A\vdash B$ is a valid sequent. (Actually this is just a special case of absorption, but oh well.)

These are simply unrelated to LEM/noncontradiction (negation doesn't appear) or to identity (they make sense in propositional logic which has no notion of equality in the first place).

So we can drop all three "laws of thought" and still have a nontrivial deductive system. Of course, this raises the question of whether such a system still counts as a "logic," and there is no accepted notion of what exactly a logic is. But I believe the current stance is a very pluralistic one, according to which even such weak systems still count as logics. (FWIW I hold such a pluralistic view.)

ContentLicense: CC BY-SA 4.0


问:
Proving diagonalizability
Suppose I have a SPD matrix, $A$, and a symmetric square matrix of appropriate size, $B$. I wish to show that $AB$ is diagonalizable, has all real eigenvalues (which are the same as $BA$), and that the eigenvectors of $AB$ are orthogonal.

My approach: It seems if I can prove that $AB = BA$ then I am done and I have proven everything at once. Is this correct? Assuming I am correct, this is where I am having trouble. My thought is since $A$ is SPD I can use its Cholesky factorization and try to get somewhere with that.

答:
Since $A$ is a real symmetric matrix, there exists an orthogonal matrix $U$ and a real diagonal matrix $\Lambda$ such that $A = U\Lambda U^T$. Since $A$ is symmetric positive definite the diagonal values of $\Lambda$ are strictly positive. It follows that $A$ has a nonsingular symmetric square root $A^{\frac{1}{2}}$ given by
$$
A^{\frac{1}{2}} = U \Lambda^{\frac{1}{2}} U^T.
$$
The square root of a diagonal matrix is obtained by taking the square root of the individual diagonal entries.

The matrix $C = AB$ is similar to the matrix $D = A^{-\frac{1}{2}} C A^{\frac{1}{2}} = A^{\frac{1}{2}}  B A^{\frac{1}{2}} $. It is clear that $D$ is symmetric. Hence there exists an orthogonal matrix $V$ and a real diagonal matrix $E$ such that $D = V E V^T$. It follows that
$$ A^{\frac{1}{2}}  B A^{\frac{1}{2}} = V E V^T$$
We conclude that
$$ AB = A^{\frac{1}{2}} (V E V^T) A^{-\frac{1}{2}} = (A^{\frac{1}{2}} V) E (A^{\frac{1}{2}} V)^{-1}.$$
This exhibits the eigenvalues and the eigenvectors of $AB$. The eigenvalues of $AB$ are real, but the eigenvectors of $AB$ are not necessarily orthogonal.

Title: Proving diagonalizability


问:
Avoidance principle of mean curvature flow at singularities
The avoidance principle for mean curvature flow states as follows.

$\textbf{Theorem.}$ Let $M_0$ and $N_0$ be two smooth closed surfaces and let $M_t$ and $N_t$ be their evolutions under mean curvature flow. Let $T > 0$ be such that $M_t$ and $N_t$ are both defined for all $t \in [0,T]$. Suppose that $M_0$ and $N_0$ are disjoint. Then $M_t$ and $N_t$ are disjoint for all $t \in (0,T]$.

My problem is, can we use this theorem if two surfaces converge to the same point at the same time? To be more specific, given two mean curvature flow $M_t$ and $N_t,$ if they both "shrink" to a point $p$ as $t\to T,$ can we use the avoidance principle to conclude that $M_t$ and $N_t$ intersects for each $t\in[0,T)?$

My concerns are:

I am not quite sure if in the setting above we could say that $M_t$ and $N_t$ are defined on $[0,T]$ (really including $T$?) to apply the theorem.

In the proof of the avoidance principle, it applys the maximum principle to say that if they intersect, then they should be identical. However, it doesn't seem to derive a contradiction since at that time they are really the same (point).

Any comments are welcome!

($\textbf{Edit.}$) When I say "applying the theorem to the point convergent case," I mean the following argument. Suppose $M_t$ and $N_t$ are disjoint for some $t\in[0,T).$ Then by the theorem (if it applies, which is what I would like to make sure in this question), $M_T$ and $N_T$ should be disjoint, contradicting to the assumption ($M_T=N_T=\{p\}$).

答:
There is a more general form of avoidance principle: not only that they remain disjoint, but also that the distance $d_t$ between them
$$ d_t = \min_{x \in M_t, y\in N_t} |x-y|$$
is non-decreasing with respect to $t$ (This is shown in the proof of the avoidance principle). Thus it is impossible that the two MCFs shrink to the same point.


问:
The j-invariant of an elliptic curve in Weierstrass normal form in terms of its coefficients
This question comes from Vakil's FOAG, page 530, right before exercise 19.9 E, self-study.

Suppose you have defined the $j$-invariant of an elliptic curve $(E, p)$ given by $y^2\,z = x^3 + a\,x\,z^2 + b\,z^3$ to be

$$j(\lambda) = 2^8\frac{(\lambda^2 - \lambda + 1)^3}{\lambda^2(\lambda - 1)^2}$$

where $\lambda$ is the cross-ratio of the three distinct roots of the right-hand side's affine form (ditch the $z$) and the point at infinity $p = [0, 1, 0]$.

How do you conclude from this that $j$ can also be given by

$$j(a, b) = 2^8\frac{3^3a^3}{4a^3 + 27b^2}?$$

Tools it seems we have at our disposal: the formula for $\lambda$ as a cross-ratio is, calling the 4 roots $(p, q_1, q_2, q_3)$, where $p$ is the point at infinity, assuming I derived it correctly

$$\lambda = \frac{(q_3-q_1)(q_2 - p)}{(q_3 - p)(q_2-q_1)}$$

I suppose we also have Vieta's formulas and the cubic formula if we need them.

I have been unable to synthesize these tools to produce the result, so it feels as if I am not realizing a key fact. Vakil mentions that this takes a little sweat, so an outline of the method should suffice; I can work it out myself once I know the general procedure.

答:
The exercise requires some elementary algebraic manipulation.
The cubic equation is
$$ x^3 \!+\! a x \!+\! b \!=\!
 (x\!-\!q_1)(x\!-\!q_2)(x\!-\!q_3)
 \!=\! 0 \tag{1} $$
where $\,q_1,q_2,q_3\,$ are its roots. From Vieta's formulas
$$ q_3 \!=\! -q_1\!-\!q_2, \quad
 a \!=\! q_1q_2 \!+\! q_1q_3 \!+\! q_2q_3, \quad
 b \!=\! -q_1q_2q_3. \tag{2} $$

The discriminant is defined by
$$ \Delta  := ((q_1-q_2)(q_1-q_3)(q_2-q_3))^2. \tag{3} $$
Note that $\,a = -(q_1^2+q_1q_2+q_2^2)\,$
which explains the minus sign in
$$ \Delta = -(4a^3 + 27b^2). \tag{4} $$
Let $\,\lambda\,$ be the cross ratio of the roots with infinity
$\,\{\infty,q_1,q_2,q_3\}\,$ following
Wikipedia Cross-ratio
$$ \lambda = (\infty,q1;q_2,q_3) :=
\frac{(q_2-\infty)(q_3 - q_1)}{(q_2 - q_1)(q_3-\infty)} \\ = \frac{(q_3-q_1)}{(q_2-q_1)}. \tag{5} $$
Now verify using elementary algebra and
following Wikipedia
Modular lambda function that
$$ j(\lambda) := 2^8\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2} = 2^8
\frac{(3a)^3}{4a^3+27b^2}. \tag{6} $$
Note that, depending on the order of the
elements of the cross-ratio, there are six
values of $\,\lambda\,$ but they all lead
to the same value for $\,j(\lambda).$

An alternative approach is to express
$\,j(\lambda)\,$ in terms of the
elementary symmetric polynomials
of the cubic roots. From Vieta's formulas
$$ e_1 := q_1 + q_2 + q_3 = 0, \\
  e_2 := q_1q_2+q_1q_3+q_2q_3 = a, \\
  e_3 := q_1q_2q_3 = -b. \tag{7} $$
The difficulty depends on identifying
the parts of the rational function
$\,j(\lambda)\,$ in terms of $\,e_1,e_2,e_3\,$
which is hard to do by manual calculations.
With Computer algebra systems
such as Mathematica, it can be automated.
For example the Wolfram language code

```<code>Divide @@ (Module[{L, q1,q2,q3, numden},
   L = (q3-q1)/(q2-q1);
   numden = {Numerator@#, Denominator@#}& /@ Factor[
       (L^2-L+1)^3 / (L^2(L-1)^2)];
   SymmetricReduction[#, {q1,q2,q3}, {0,a,-b}]& /@
   numden] //Transpose //First) //Factor //InputForm
</code>```

returns the result

(27*a^3)/(4*a^3 + 27*b^2)

which depends critically on the
SymmetricReduction function.

Body:
The exercise requires some elementary algebraic manipulation.
The cubic equation is
$$ x^3 \!+\! a x \!+\! b \!=\!
 (x\!-\!q_1)(x\!-\!q_2)(x\!-\!q_3)
 \!=\! 0 \tag{1} $$
where $\,q_1,q_2,q_3\,$ are its roots. From Vieta's formulas
$$ q_3 \!=\! -q_1\!-\!q_2, \quad
 a \!=\! q_1q_2 \!+\! q_1q_3 \!+\! q_2q_3, \quad
 b \!=\! -q_1q_2q_3. \tag{2} $$

The discriminant is defined by
$$ \Delta  := ((q_1-q_2)(q_1-q_3)(q_2-q_3))^2. \tag{3} $$
Note that $\,a = -(q_1^2+q_1q_2+q_2^2)\,$
which explains the minus sign in
$$ \Delta = -(4a^3 + 27b^2). \tag{4} $$
Let $\,\lambda\,$ be the cross ratio of the roots with infinity
$\,\{\infty,q_1,q_2,q_3\}\,$ following
Wikipedia Cross-ratio
$$ \lambda = (\infty,q1;q_2,q_3) :=
\frac{(q_2-\infty)(q_3 - q_1)}{(q_2 - q_1)(q_3-\infty)} \\ = \frac{(q_3-q_1)}{(q_2-q_1)}. \tag{5} $$
Now verify using elementary algebra and
following Wikipedia
Modular lambda function that
$$ j(\lambda) := 2^8\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2} = 2^8
\frac{(3a)^3}{4a^3+27b^2}. \tag{6} $$
Note that, depending on the order of the
elements of the cross-ratio, there are six
values of $\,\lambda\,$ but they all lead
to the same value for $\,j(\lambda).$

An alternative approach is to express
$\,j(\lambda)\,$ in terms of the
elementary symmetric polynomials
of the cubic roots. From Vieta's formulas
$$ e_1 := q_1 + q_2 + q_3 = 0, \\
  e_2 := q_1q_2+q_1q_3+q_2q_3 = a, \\
  e_3 := q_1q_2q_3 = -b. \tag{7} $$
The difficulty depends on identifying
the parts of the rational function
$\,j(\lambda)\,$ in terms of $\,e_1,e_2,e_3\,$
which is hard to do by manual calculations.
With Computer algebra systems
such as Mathematica, it can be automated.
For example the Wolfram language code

Divide @@ (Module[{L, q1,q2,q3, numden},
   L = (q3-q1)/(q2-q1);
   numden = {Numerator@#, Denominator@#}& /@ Factor[
       (L^2-L+1)^3 / (L^2(L-1)^2)];
   SymmetricReduction[#, {q1,q2,q3}, {0,a,-b}]& /@
   numden] //Transpose //First) //Factor //InputForm

returns the result

(27*a^3)/(4*a^3 + 27*b^2)

which depends critically on the
SymmetricReduction function.


问:
integral over a bounded domain of a bounded (from below) function
Let $\Omega$ be an open bounded domain and $f:\mathbb{R}\to\mathbb{R}$ be a continuous and bounded from below function.

It is true that a positive constant $C$ exists such that
$$\int_{\Omega} f(t) dt\geq C?$$
I think it is true, but how to prove it?

Could anyone please help?

Thank you in advance!

答:
No. Consider $f(x)=0$ on $x\in(0,1)$.

If you have a negative lower bound, cancellation could happen.

But if you have $f\ge 0$ (and continuous), then $\int_\Omega f>0$ if $f$ is not constantly zero.


问:
Is there a logic with quantifier "almost always".
I would like to describe a reasoning with quantifier "almost always".
For example, if probability of $Z$ is above $95$%, I would like to say that $Z$ is "almost always true".

Is there a modal or fuzzy logic with such a quantifier? If not, is it even possible to define a logic with this quantifier? May be, it would contradict some assumptions about quantifiers in a meaningful logic?

答:
This logic exists
https://www.researchgate.net/publication/221551396_The_Modal_Logic_of_Probability

I just found it. They define modal operators L_a, meaning probability is above a.
I guess, modal operator is something used in propositional logic, and quantifiers are used with predicates with variables.


问:
How to calculate this integral: $\int_{C}{} (z^2-2z) dz$ where $C$ is a line segment starting at $1$ and ending in $i$
The task is as follows:
$$
\text{Calculate } \int_{C}{}f(z) dz \text{, where }
\\
\text{a) } f(z)=z^2-2z \text{, } C \text{ - line segment starting at } 1 \text{ and ending at } i \text{;}
\\
\text{b) }f(z)=\frac{\bar{z}}{z+i} \text{, } C \text{ - circle } |z+i| = 3 \text{.}
$$

I do not know how to calculate either of the integrals. Any help or push in the right direction would be much appreciated. Thank you.

Update:

I calculated the $a)$ integral to be equal to $\frac{7}{3}-\frac{i}{3}$. If someone could check if I did good, that would be great.

答:
Hint.

For (a), you can simply use the fundamental theorem of calculus (also true in complex analysis) since you have an entire function:
$$
\int_Cf(z)dz=F(B)-F(A)
$$
where $A$ is the initial point of $C$ and $B$ the final point, and $F$ is an antiderivative of $f$.

For (b), the first thing you need is to parameterize your path $C$, which means writing
$$
z=g(t),\quad t\in[0,1]
$$
for some function $g$.

Then
$$
\int_Cf(z)dz=\int_0^1 f(g(t))g'(t)dt.
$$


问:
Show the subspace of a set of vectors

If we consider the set $$\alpha = \{(1, 1, 0),(0, 0, 1),(-2,-2, 3)\} \ {\displaystyle \subset } \ \mathbb{R}^{3}  $$ a)  Find the subspace:  $$\ W =  \{ u \in \mathbb{R}^{3} \mid \ \left \langle \{(1, 1, 0),(0, 0, 1),(-2,-2, 3)\}  \right \rangle \}$$

I understand how to prove a subspace by definition and the "span method", but I don't understand how to use $ u $ here. I was thinking of using $a$, $b$ and $c$ with each vector trying to create a system but I have nothing to match these values to.

Maybe i'm confused. Any hints?

答:
Hint: sift the sequence of vectors in $α$.

The sifted sequence will have the same span as the original sequence (i.e., $W$), and will be linearly independent, hence a basis of $W$.

Title: Show the subspace of a set of vectors


问:
Convergence of $\int_{0}^{1} \frac{\ln^{2} x}{x^{2}+x-2} \ dx $
How do you show that $\displaystyle \int_{0}^{1} \frac{\ln^{2}x}{x^2+x-2} \ dx $ converges?

The singularity at $x=1$ is not an issue since it is removable.

But what about at $x=0$?

答:
You can show that
$$\int_0^1 dx\ \ln^n x$$
converges by changing variables (it's a gamma function). Now, your denominator is benign in a neighborhood of $x=0$, and you can do a simple comparison to show that your integrand is smaller in magnitude than the integrand mentioned above.</p>

ContentLicense: CC BY-SA 3.0


问:
Evaluating $\int_{0}^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \mathrm{d}x$

Does the integral $$\int_0^{\pi} \frac{1}{\sqrt{u^2+2u\cos x +1}} \text d x \hspace{30pt} (u \le 1) $$ has a closed form? If it has, how do we evaluate it?

I was solving a physics problem which I have asked on physics SE as well (here), and this integral popped out. How do I solve it?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi}{\dd x \over
\root{u^{2} + 2u\cos\pars{x} +1}}
\right\vert_{\, u\ \leq\ 1}}
\\[5mm] = &\
{1 \over 2}\int_{-\pi}^{\pi}{\dd x \over
\root{u^{2} + 2u\cos\pars{x} +1}}
\\[5mm] = &\
{1 \over 2}\int_{-\pi}^{\pi}{\dd x \over
\root{\pars{u + \expo{\ic x}}\pars{u +
\expo{-\ic x}}}}
\\[5mm] = &\
{1 \over 2}\,\Re\oint_{\verts{z} = 1}
{\dd z/\pars{\ic z} \over
\root{\pars{u + z}\pars{u + 1/z}}}
\\[5mm] = &\
{1 \over 2\root{u}}\,\Im\oint_{\verts{z} = 1}
{\dd z \over
\root{z}\root{z + u}\root{z + u^{-1}}}
\\[5mm] = &\
-{1 \over 2\root{u}}\ \times
\\[2mm] &
\Im\int_{-1}^{-u}
{\dd z \over
\pars{\root{-z}\expo{\ic\pi/2}}
\pars{\root{-z - u}\expo{\ic\pi/2}}\root{z + u^{-1}}}
\\[2mm] &
\,\,\,-{1 \over 2\root{u}}\ \times
\\[2mm] &\
\Im\int_{-u}^{0}
{\dd z \over
\pars{\root{-z}\expo{\ic\pi/2}}
\root{z + u}\root{z + u^{-1}}}
\\[2mm] &
\,\,\,-{1 \over 2\root{u}}\ \times
\\[2mm] &\
\Im\int_{0}^{-u}
{\dd z \over
\pars{\root{-z}\expo{-\ic\pi/2}}
\root{z + u}\root{z + u^{-1}}}
\\[2mm] &
\,\,\,-{1 \over 2\root{u}}\ \times
\\[2mm] & \Im\int_{-u}^{-1}
{\dd z \over
\pars{\root{-z}\expo{-\ic\pi/2}}
\pars{\root{-z - u}\expo{-\ic\pi/2}}\root{z + u^{-1}}}
\\[5mm] = &\
{1 \over \root{u}}\int_{0}^{u}
{\dd z \over
\root{z}
\root{u - z}\root{u^{-1} - z}} =
\bbx{2\on{K}\pars{u^{2}}} \\ &
\end{align}
$\ds{\on{K}}$ is the
$\ds{\on{K}}$ Elliptic Function.


问:
Is an argmax of products equal to a product of argmax's?
Given that $f(x) \geq 0,g(y) \geq 0,h(z) \geq 0 \ \forall \ x,y,z$, is this statement true?

$$
\DeclareMathOperator*{\argmax}{\arg\!\max}
\argmax_{x,y,z} f(x) \cdot g(y) \cdot h(z) = \left[\argmax_{x} f(x)\right] \cdot \left[\argmax_{y} g(y)\right] \cdot \left[\argmax_{z} h(z)\right]
$$

EDIT

I am asking this question in the context of belief propagation and the Viterbi algorithm. Consider instead this true statement:

$$
\sum_{x,y,z} f(x) \cdot g(y) \cdot h(z) = \left[\sum_{x} f(x)\right] \cdot \left[\sum_{y} g(y)\right] \cdot \left[\sum_{z} h(z)\right]
$$

If I replace the summations with argmax's, is this statement still true?

答: Yes, it's true.  Let $(x^*,y^*,z^*)$ denote the argmax on the right.  If you obtained anything else from the argmax on the left, say $(x',y^*,z^*)$, then replacing $x'$ with $x^*$ would only increase the LHS, never decrease it.


问:
Wrong/naive inductive hypothesis and tower of hanoi as an example
I am reading about how a wrong formulation of the tower of Hanoi and the inductive hypothesis can lead to a dead-end.

The example I am reading states the following:

The task is to move N discs from a specific pole to another
specific pole. Assume there are poles $A$, $B$ and $C$.   The base case is that when the number of discs is 0 then no steps are needed
to complete the tasks.      For the inductive step assume that we
can move $n$ discs from pole $A$ to pole $B$ and we are required to
show how to move $n + 1$ discs from $A$ to $B$

Then it highlights that this definition is a dead-end since the only 2 ways to use the induction hypothesis as set don't lead anywhere.

Specifically the options are:

Move the top $n$ discs from pole $A$ to $B$. After this point all possibilities of using the induction hypothesis have been exhausted
since $n$ discs are on pole $B$ and we do not have hypothesis about
moving discs from that pole.

Move the smallest disc from pole $A$ to $C$. Then move the remaining $n$ discs from $A$ to $B$. Once again we have exhausted all
possibilities of using the induction hypothesis, because $n$ discs are
now on pole $B$, and we have no hypothesis about moving discs from
this pole.

The reasoning for $1$ is clear to me. If we move the top $n$ discs i.e. the $n$ smaller discs to pole $B$ then at some point we would have to move them again since we would need to place the $nth + 1$ remaining largest disc bellow them. And there is no inductive hypothesis for that so I think I get it.

I need help with the second part. How I understand it is we move the smallest disc to pole $C$. Then we use the inductive hypothesis to move the remaining $n$ larger discs from pole $A$ to $B$. (At this part I am not sure how this could happen if pole $C$ is occupied by the smallest disc but I guess it is part of the logic of using induction in proof.)

Then at that state it seems to me that the only thing pending would be to move the smallest disc from $C$ to $B$ and finish the task.

Why does it state that we would need to move the $n$ discs from pole $B$ and that is not possible since we don't have an induction hypothesis?

Am I misunderstanding something about statements on induction here?

答:
You are correct: the author’s explanation of the problem with the second part is rubbish. He should instead have pointed to the problems that you saw:

The induction hypothesis is incomplete, since it pretty clearly must include a requirement that pole $C$ be available for the disks being transferred from $A$ to $B$.

If the transfer of the bottom $n$ disks from $A$ to $B$ actually were possible, we clearly could complete the job of transferring the whole stack by moving the smallest disk from $C$ to $B$.

The problem here isn’t really that the induction hypothesis leads to a dead end: it’s that it is too strong and allows a fallacious argument if this problem isn’t noticed in time.


问:
Double Eigenvalue by calculation matrix Exponential
Yea, I had no luck. Drawing the eigenvalues of $A = \left(\begin{array}{c} 1 \quad 3 \\ 0\quad1\end{array}\right)$ I get $\lambda_{1,2} = 1$. Hence $A -\lambda\,I = \left(\begin{array}{c} 0 \quad 3 \\ 0\quad0\end{array}\right)$. If I remember right from Calculus 1 the Kernel of this thing is having Dimension 1. That sounds inconvenient. Thus the Eigenspace would be 1 dimension...

What I'm trying to do: Calculation Matrix Exponential of A.

答:
To find $f(A)$ for a holomorphic function $f$ defined on some neighbourhood of $1$, you need to find a polynomial $p \in \Bbb{C}[x]$ of degree $< 2$ such that
$$p(1) = f(1), \quad p'(1) = f'(1)$$
and then $f(A)=p(A)$.
Trying out $p(x) = ax+b$, we find that $p(x) = ex$ is the desired polynomial if $f$ is the exponential function.

Therefore
$$e^A = p(A) = eA = \pmatrix{e & 3e \\ 0 & e}.$$


问:
The forgetful functor $U: \text{Con}(\mathcal C) \to \mathcal C$ has a left adjoint.
Let $\mathcal C$ a category. We define the category of cones $\text{Con}(\mathcal C)$ in the following way:

Objects: quadruples $(\mathcal Z, \underline{M}, M, m)$ consisting of a small category $\mathcal Z$, a functor $\underline{M}: \mathcal Z \to \mathcal C$ and a cone $(M, m)$ of $\underline{M}$ (i.e. $M$ is an object in $\mathcal C$ and $m = (m_Z)_{Z \in \mathcal Z}$ is a collection of morphism $m_Z: M \to \underline{M}(Z)$ satisfying a compatibility condition) .

Morphisms: a morphism $\underline{\phi}: (\mathcal Z, \underline{M}, M, m) \to (\mathcal Y, \underline{N}, N, n)$ is a triple $\underline{\phi}:( \phi^0, \phi^1, \phi^2)$ where $\phi^0: \mathcal Y \to \mathcal Z$ is a functor, $\phi^1: M \to N$ is a morphism in $\mathcal C$ and $\phi^2: \underline{M}\circ \phi^0 \to \underline{N}$ is a natural transformation such that $\phi^2_Y \circ m_{\phi^0(Y)} = n_Y \circ \phi^1$ for all $Y \in \mathcal Y$.

I am trying to show that the forgetful functor $U: \text{Con}(\mathcal C) \to \mathcal C$ has a left adjoint, i.e. there is a functor $S : \mathcal C \to \text{Con}(\mathcal C)$ such that
$$\text{Hom}_{\text{Con}(\mathcal C)}(S(X), Y) \cong \text{Hom}_\mathcal C (X, U(Y)).$$
The more natural way to associate an element $C \in \mathcal C$ to a quadruple in $\text{Con}(\mathcal C)$ is to send $C$ to $(\varnothing, F_\varnothing, C, m)$ where $F_\varnothing$ is the unique functor that goes from $\varnothing$ to $\mathcal C$ and $(C, m)$ is a cone for $F_\varnothing$ (a cone for this functor is just an element, there is no need to consider such an $m$).

This seems to be a natural way to define the functor $S:\mathcal C \to \text{Con}(\mathcal C)$ as I did above but I do not really see how it goes for the morphism. If we consider a morphism $\underline{f} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), Y)$ (for $S(C)$ construct above and $Y = (\mathcal Y, \underline{N}, N, n)$ a quadruple in $\text{Con}(\mathcal C)$), $\underline{f}$ should be a triple $\underline{f} = (f^0, f^1, f^2)$ with $f^0: \mathcal Y \to \varnothing$ and $f^3: F_\varnothing \circ f^0 \to \underline{N}$. The problem is that $F_\varnothing \circ f^0 : \mathcal Y \to \mathcal C$ factors trough the empty set and therefore it has no image, $F_\varnothing \circ f^0$ is not even a functor.

My questions are the following: Is it really possible to consider a functor as $f^0$ such that $f^0: \mathcal Y \to \varnothing$ ? Is it the right way to construct the left adjoint $S$ ? I also tried to construct $S$ by sending $C$ to an other quadruble where the associate $\mathcal Z$ is not the empty set but it seems that in those cases there are many possiblities which are all very arbitrary (there is a unique functor $\varnothing \to \mathcal C$ but there are a lots a functor from $\mathcal Z = \underline{1} \to \mathcal C$, where $\underline{1}$ is the category with a unique element, for example).

答:
I've tried to show the statement for $S(C) = (\underline{1}, F_{C}, C, 1_C)$ (where $F_C$ is the functor $\{*\} \to C$) as recommended by Berci and I wanted to know everything is good in the justification, let me know if you guys see anything wrong.

For $\widetilde{Y} = (\mathcal Y, \underline{N}, N, n) \in \text{Con}(\mathcal C)$, we can consider $\underline{\phi} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), \widetilde{Y})$ so that
$$\underline{\phi} = (\phi^0, \phi^1, \phi^2)$$
where $\phi^0$ is the unique functor that goes from $ \mathcal Y$ to $\underline{1}$, $\phi^1: C \to N$ is any morphism in $\text{Hom}_\mathcal C(C, N) = \text{Hom}_\mathcal C(C, U(\widetilde{Y}))$ and $\phi^2$ is a natural transformation $F_C \circ \phi^0 \to \underline{N}$.

The natural transformation $\phi^2$ requires

$$\phi^2_Y = n_Y \circ \phi^1,$$
since $m_{\phi^0(Y)} = 1_C$. Therefore, $\underline{\phi} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), \widetilde{Y})$ is entirely determined by a morphism $\phi^1 \in \text{Hom}_\mathcal C(C, U(\widetilde{Y}))$ since $\phi^0$ is unique and $\phi^2_Y$ is equal to $n_Y \circ \phi^1$. We conclude that
$$\text{Hom}_{\text{Con}(\mathcal C)}(S(C), \widetilde{Y})\cong \text{Hom}_\mathcal C(C, U(\widetilde{Y})).$$
(The naturality in $C, \widetilde{Y}$ seems quite ugly so I'm gonna stop here).


问:
How do I write the generic series definition that can produce any term in the series expansion?
I'm trying to learn how to build the generic series definition for a series of numbers. For some reason I'm having a hard time pulling out this pattern. I'm always pulling out the wrong details for when it comes to building the generic series definition. Is there a more controlled and efficient approach to determining these patterns.

What does it mean when a geometric series can be infinite or finite? It seems to me that you can only find the sum of a series that are finite or when r < |1|. However, I'm understanding that I can use the Maclaurin Series expansion formula to solve for f(x) when x is within range.

I'm just not understanding the application or idea of finding the sum vs. solving for f(x). Are these the same approaches? Am I doing the same thing but deriving a formula to calculate sums for infinite series?

Can someone elaborate on this as I'm having quite the time understanding how to approach these applications.

答:
A finite geometric series can always be summed.  If the first term is $a$ and the ratio is $r$, so the series is $a, ar, ar^2, ar^3 \ldots ar^n$, the sum is $\frac {a(1-r^{n+1})}{1-r}$ [unless $r=1$, then it is $a(n+1)]$.  When we let $n \to \infty$ this only converges if $|r| \lt 1$

For a MacLaurin series, the radius of convergence depends on how quickly the terms decrease.  The above is the series for $\frac a{1-r}$.  The series for $e^z$ converges for any $z$ in the complex plane because the terms decrease so quickly.


问:
Absolute convergence of a double matrix and of $\frac{1}{\zeta(\frac{1}{2}+\epsilon)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{\frac{1}{2}+\epsilon}}$
According to Wikipedia the convergence of the right hand side of:
$$\frac{1}{\zeta(\frac{1}{2}+\epsilon)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{(\frac{1}{2}+\epsilon)}}$$

for $\epsilon$ an arbitrary small number, is equivalent to the Riemann hypothesis.

The logarithm of the left hand side with $s$ instead $\frac{1}{2}$:

$$\log \left(\frac{1}{\zeta \left(s+\epsilon\right)}\right)$$

The derivative of that is:

$$\frac{\partial \frac{1}{\zeta (s+\epsilon)}}{\partial s}=-\frac{\zeta '(s+\epsilon)}{\zeta (s+\epsilon)^2}$$

Notice that according to Mathematica:

$$-\frac{\zeta '(s+\epsilon)}{\zeta (s+\epsilon)^2}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s+\epsilon)}{\zeta (c+s+\epsilon-1)}-\zeta (c)\right)$$

Therefore:

$$-\frac{\zeta '(s+\epsilon)}{\zeta (s+\epsilon)^2}+\zeta(1+\epsilon)=\frac{\zeta (1+\epsilon) \zeta (s+\epsilon)}{\zeta (1+\epsilon+s+\epsilon-1)}$$

Now by joriki's and GH from MO's proofs, for $s>1$ and $c>1$:
$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{\varphi^{-1}(\gcd(n,k))}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$

where:
$$\varphi^{-1}(n)=\sum_{d \mid n}d\mu(d)$$

Therefore:
$$\frac{\zeta (1+\epsilon) \zeta (s+\epsilon)}{\zeta (1+\epsilon+s+\epsilon-1)} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s + \epsilon} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^{1+\epsilon}} = \sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{\varphi^{-1}(\gcd(n,k))}{n^{1+\epsilon} \cdot k^{s+\epsilon}}$$

We then set $s=\frac{1}{2}$ and $\epsilon=\frac{1}{p}$ and look for symmetry by solving:

$$\frac{\zeta \left(\frac{1}{2}+\epsilon\right) \zeta (1+\epsilon)}{\zeta \left(\frac{1}{2}+\epsilon+1+\epsilon-1\right)}=\frac{\zeta (x) \zeta (x)}{\zeta (x+x-1)}$$

for $x$. Unfortunately I can only solve this numerically in Mathematica for various values of smaller and smaller epsilons, and plot the result. But we know that $x>\frac{1}{2}$ and we can set a sufficiently high upper bound for $x$. In the program I use: $\frac{1}{2} < x \leq 120$

Mathematica 8.0.1:

```<code>(*start*)
Clear[x, p, epsilon, a, s, c];
Monitor[TableForm[a = Table[
    epsilon = 1/p;
    c = x /.
      FindRoot[
       Zeta[1/2 + epsilon]*
         Zeta[1 + epsilon]/Zeta[1/2 + epsilon + 1 + epsilon - 1] ==
        Zeta[x]*Zeta[x]/Zeta[x + x - 1], {x, 2 - 1/2, 1/2, 120},
       WorkingPrecision -> 60], {p, 5, 100}]], p]
ListLinePlot[a, PlotRange -> {1/2, 3/2}]
(*end*)
</code>```

The plot of the function $x(p)$, with $\epsilon=\frac{1}{p}$, looks like this:

which clearly tends to $1$ as $p \rightarrow \infty$, and which would mean that we have absolute convergence for at least the second row and the second column in the now symmetric $\varphi^{-1}(\gcd(n,k))$ matrix and probably also all the other columns and rows since they are similar to alternating series like the second column and second row.

Question 1: Does the solution $$\lim_{\epsilon \to 0 } \, x(\epsilon)=1$$ tend to
$1$ as the graph suggests?

In the the book "Convergence Methods for Double Sequences and Applications" by M. Mursaleen and S.A. Mohiuddine, Springer India 2014, there is on page 150 in the last chapter, the following Lemma 9.1:

"A double series $\sum_{k,l} a_{k,l}$ is absolutely convergenct if an only if the following [two] conditions hold:"

(i) There are $(k_{0},l_{0}) \in \mathbb{N} \times \mathbb{N}$ and $\alpha_{0} > 0$ such that:
$$\sum\limits_{k=k_0}^{m} \sum\limits_{l=l_0}^{n}|a_{k,l}|\leq \alpha_{0} \text{ for all } (m,n) \geq (k_{0},l_{0})$$

(ii) Each row-series and each column-series are absolutely convergent.

My comment above was that (ii) probably already holds. So left is to show that (i) also holds.
On page 156 the authors list Raabe's test applied to double series:

Theorem 9.11 (Raabe's test) Let $(a_{k,l})$ be a double sequence.

(i) Suppose that each row-series and each column-series corresponding to $\sum_{k,l} |a_{k,l}|$ are convergent. If there is $p>1$ such that

$$|a_{k,l+1}| \leq \left(1-\frac{p}{l}\right)|a_{k,l}| \text{ and } |a_{k+1,l}| \leq \left(1-\frac{p}{l}\right)|a_{k,l}|$$ whenever $k$ and $l$ are large, then $\sum_{k,l} |a_{k,l}|$ is convergent.

(ii) If $|a_{k,l+1}| \geq (1-\frac{1}{l})|a_{k,l}|>0$ for some $k \in \mathbb{N}$ and all large $l \in \mathbb{N}$, or if $|a_{k+1,l}| \geq \left(1-\frac{1}{k}\right)|a_{k,l}|>0$ for some $l \in \mathbb{N}$ and all large $k \in \mathbb{N}$, then $\sum_{k,l} |a_{k,l}|$ is divergent.

Part (i) in Raabes test above, I have computed as a limit for the second row and the second column, in Mathematica, and it seems to make sense. Part (ii) I have yet to explore.

Question 2: Does Raabe's test apply to all rows and columns in
$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{\varphi^{-1}(\gcd(n,k))}{n^c \cdot k^s}$, for $c>1$ and $s>1$?

答: Can you find the asymptotic of the LHS of $\frac{\zeta \left(\frac{1}{2}+\epsilon\right) \zeta (1+\epsilon)}{\zeta \left(\frac{1}{2}+\epsilon+1+\epsilon-1\right)}=\frac{\zeta (x) \zeta (x)}{\zeta (x+x-1)}$ as $\epsilon\to 0$ ? For $\epsilon$ and the equality to hold it is immediate that $x$ must be close to $1$.


问:
Notation concatenation of tuples
Is there any common notation for concatenating two tuples?

For instance, let's assume we have two tuples $X:=(x_1, \ldots ,x_n )$ and $Y:=(y_1, \ldots ,y_m)$. I want the resulting tuple to be $Z:=(x_1, \ldots ,x_n , y_1, \ldots ,y_m)$.

For sequences, I have read here that one possibility is to use $\frown$ (\frown).

Is this used for tuples as well, such that $Z:=X\frown Y$ would work?

答:
Tuples and sequences are the same thing, so it seems to me that this question is basically a duplicate of the previous one. But in any case I disagree with the accepted answer. I have never seen $\frown$ used in this way. In mathematics I would use $\times$ for the cartesian product here. I would notate individual tuples with lowercase letters and reserve uppercase letters for sets, so $x = (x_1, \dots x_n)$ and $y = (y_1, \dots y_m)$ give $x \times y = (x_1, \dots x_n, y_1, \dots y_m)$.

If the $x_i$ and $y_i$ are vectors in vector spaces you can also use $\oplus$ for the direct sum.

Title: Notation concatenation of tuples


问:
Does ECDSA work without cyclic groups
I'm currently working on the different signature schemes, here is my question :

Does ECDSA work without cyclic groups ? Is it secure ?

For example, if I use a non cyclic group, are there some cases in which the signature will not work ? If it always work, is it secure ? why / why not ?

Thank you

答:
Being cyclic adds a property to a group $G$ such that there is an element $g \in G$ so that $\langle g \rangle = G$. Instead of adding a property, cryptographers look the other way, the Generic Group Model;

The generic group model is an idealized cryptographic model, where the adversary is only given access to a randomly chosen encoding of a group, instead of efficient encodings, such as those used by the finite field or elliptic curve groups used in practice.

The model includes an oracle that executes the group operation. This oracle takes two encodings of group elements as input and outputs an encoding of a third element.

In 2016, there is an interesting work that showed the security of the GenericDSA then they indicated that forging signatures in (EC)DSA is as hard as solving discrete logarithms.

On the Provable Security of (EC)DSA Signatures

based on this we can say that (EC)DSA is secure when the discrete log is secure.

While choosing a non-cyclic group, be careful, since all elements have some finite order in the finite groups. So, if you choose your element to have a small order then DLog will be easy.

Since you are asking  on ECDSA here a non-cyclic ECC example from Stinson's book;

Suppose that $p > 3$ is an odd prime, and $a,b$ is an element of $\mathbb Z_p$. Further, suppose that the equation $x^3 + ax + b$ is congruent to $0 \bmod p$) has 3 distinct roots in $\mathbb Z_p$. Prove that the corresponding elliptic curve group $(E, +)$ is not cyclic.

Still, you will still work on a cyclic-subgroup of the Curve.

In some constructions, the signature may fail, need to be looked for cases.

Title: Does ECDSA work without cyclic groups


问:
Showing that a sequence weakly converges but no subsequence converges strongly
I have a sequence $(f_k)$ in $L_p(R)$ and I want to show that it weakly converges to $0$ but no subsequence converges strongly to $0$.
The sequence is $$f_k = k^{1/p}1_{[0,1/k]}$$ and while I'm able to show weak convergence, I don't see how I can show that no subsequence $f_{k_j} \to 0$. I'm looking for a hint as to what theorem might be applicable.

答:
Because
$$\Vert f_k \Vert_p^p = \int_0^{1/k} |k^{1/p}|^p dx = \int_0^{1/k} k dx = 1$$
so
$$\Vert f_k - 0 \Vert_p = \Vert f_k \Vert_p = 1 \not \to 0 ;$$
in particular, for any subsequence $\{ f_{k_i} \}_{i\in \Bbb{N}}$, the elements $f_{k_i}$ are still such that
$$\Vert f_{k_i} \Vert_p = 1,$$
so we have the same conclusion.


问:
Matrix Equation, Solving for Variables.
I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:

$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$

What I have done so far is:

$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$

And the solving for variables,

$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$

$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$

$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$

$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$

Am I going about this correctly? Or am I just doing this completely incorrect?

答:
Error 1: $a-b=8$ is not the same as $a = \frac{8}{-b}$ but $a = 8 + b$

Error 2: $b+c=1$ is not the same as $b = \frac{1}{c}$ but $b = 1-c$

Error 3: $3d+c=7$ ... same as errors 1 and 2

Error 4: $2a-4d=6$ Incorrect substitution because of previous errors. Fix the previous errors and then subsititue to fix this error

Title: Matrix Equation, Solving for Variables.


问:
Volume using triple integral with spherical and cylindrical coordinates
I had to solve this triple integral and I tried to solve by cylindrical and spherical coordinates but couldn't get anywhere. I was hoping someone could help me in this problem.

Solve $\int \int \int _{E} \frac{1}{x^2+y^2+z^2 }dV, E=\{(x,y,z) \epsilon  \mathbb{R}^3 : z\geq \sqrt{\frac{x^2+y^2}{3}}, 2z\leq x^2+y^2+z^2 \leq 9\} $

答:
You can do this in both cylindrical and spherical coordinates but it is more straightforward in spherical coordinates.

Based on inequalities, the region $E$ we need to integrate over is the volume of cone $z \geq  \sqrt{\frac{x^2+y^2}{3}} \ $ inside the sphere $x^2 + y^2 + z^2 \leq  9 \ $ but outside the other sphere $x^2+y^2+z^2 \geq 2z$.

In spherical coordinates, $x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi, x^2 + y^2 + z^2 = \rho^2$

So the maximum polar angle $\phi$ from the equation of the cone,

$\rho \cos \phi = \frac{\rho \sin \phi}{\sqrt3} \implies \phi = \frac{\pi}{3}$

The equation of the second sphere can be re-written as

$\rho^2 = 2 \rho \cos \phi \implies \rho = 2 \cos \phi, 0 \leq \phi \leq \frac{\pi}{2}$

So the integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi/3} \int_{2\cos \phi}^3 \sin \phi \ d\rho \ d\phi \ d\theta$

(Please note that the integrand is $\frac{1}{\rho^2}$ which cancels out with $\rho^2$ in the Jacobian).


问:
How to solve the linear congruence $ 25x -4 \equiv 9x +4 \pmod 8$?
I have the following problem

Argue if the following linear congruence is solvable, and if it is, list all the incongruent solutions.
$$ 25x -4 \equiv 9x +4   \pmod 8$$

My attempt

I know that if I have a linear congruence of the form
$$
ax +b\equiv 0 \pmod m \tag{1}
$$
then the above system is solvable if and only if $ax + my =b$ has integer solutions for $x$ and $y$, so my idea is to transform the congruence into something like $(1)$. Knowing that $a \equiv a \pmod k$ I get that
$$
4 - 9x \equiv4 -9x \pmod 8
$$
so I get
\begin{align*}
(25x -4) + (4 - 9x) &\equiv (25x -4) +(4 -9x) \pmod 8\\
16 x &\equiv 0 \pmod 8
\end{align*}
but since $16 x = 8(2x)$, the above equivalence holds true for any integer $x$. Now, since $\gcd(16,8) = 8$, we know that we have $8$ incongruent solutions (modulo $8$). Since a particular solution to the diophantine equation
$$
16x + 8y = 0
$$
is $x_0 = 1, y_0 = -2$, I know that the incongruent solutions will be
$$
x_0,\ x_0 + \frac{8}{\gcd(8,16)}, \ x_0 + \color{blue}{2}\frac{8}{\gcd(8,16)},  ... ,\ x_0 + \color{blue}{7}\frac{8}{\gcd(8,16)}
$$
and since $\frac{8}{\gcd(8,16)} = 1$ the incongruent solutions turn out to be
$$
1, 1 + 1, 1+2, ... 1+7 = 1,2,3..., 7,8
$$

I'm not sure if I solved the above congruence correctly. Could anyone tell me if my solution is correct? Or alternatively, can someone tell me where I made a mistake and how to correct it? Thank you!

答:
Your solution is correct. I will outline a simpler method below.
$$ 25x -4 \equiv 9x +4   \pmod 8$$
Note that $25x=3\times8x+x\equiv 0+x\equiv x\pmod8$ and similarly, $9x=8x+x\equiv 0+x\equiv x\pmod8$.
So in fact your equation is
$$x-4\equiv x+4\pmod 8\iff-4\equiv4\pmod8$$
which is true for all integer values of $x$.


问:
Determine which of the folowing statements hold: $f(x) = O(g(x))$, $f(x) = o(g((x))$, $g(x) = o(f(x))$, $g(x) = O(f(x))$ when $x → ∞$
Determine which of the folowing statements hold: $f(x) = O(g(x))$, $f(x) = o(g((x))$, $g(x) = o(f(x))$, $g(x) = O(f(x))$ when $x → ∞$:

a) $f(x) = e^x$, $g(x)= e^\sqrt{x}$

b) $f(x) =e^{log_2^2 x}$ , $g(x) = x^2$.

I know that for a, only $g(x) = o(f(x))$ holds, and for b, $f(x) = O(g(x))$, $f(x) = o(g(x))$ and $g(x) = O(f((x))$ hold as $x → ∞$ using a calculator.

I wonder how we prove these points using the formal definition of big-O-notation (and small-O-notation), that is:

(i) $f(x) = O(g(x))$ when there exists $C > 0$ such that $f(x) ≤ Cg(x)$ when $x → ∞$ ($X = ∞$ here)

( (ii) $f(x) = o(g(x))$ when $\lim_{x \to +\infty} {f(x) \over g(x)} =0)$

答:
Firstly we need exact definition (considering $x \to +\infty$ for non negative functions):
$$O(g)=\{\phi \colon, \exists C>0, \exists \delta >0, \forall x> \delta, \phi(x) \leqslant C g(x)  \}$$
so, if we consider $f(x)=e^x$ and $g(x)=e^\sqrt{x}$, then having $\sqrt{x} \leqslant x$ for $x>1$, immediately gives $e^\sqrt{x} \in O(e^x)$ for $C=\delta=1$.

For little-$o$
$$o(g)=\{\phi \colon, \exists \varepsilon(x), \lim\limits_{x \to +\infty}\varepsilon(x)=0 , \exists \delta >0, \forall x> \delta, \phi(x) = \varepsilon(x)g(x)  \}$$
now using representation $e^\sqrt{x} =e^{-x+\sqrt{x}} e^x$ and knowing $\lim\limits_{x \to +\infty}e^{-x+\sqrt{x}}=0$ we have $e^\sqrt{x} \in o(e^x),x \to +\infty$.

Of course, as rightly indicated in  the  Greg Martin's comment, you can obtain first from second, because $\lim\limits_{x \to +\infty}\varepsilon(x)=0$  gives $\exists C>0$ such that $\varepsilon(x) \leqslant C$.


问:
Numbers that can be expressed as $ab + a + b$
How many positive integers $n$ less than $100$ can be expressed in the form $ab + a + b,$ where $a,b$ are positive integers?

I wasn't quite sure how to approach this problem besides simply finding values for all the numbers or making a table, so is there a slicker solution? Thanks.

答: Hint: $ab+a+b=(a+1)(b+1)-1$ so the numbers $n$ that can be expressed in that form are exactly those for which $n+1$ is not prime.


问:
Existence of a certain basis?
Let V be a finite dimensional k-vector space and let f: V → V be a k-linear map. Show that there exists a basis $\{e_i\}$ such that

\begin{equation}
    f(e_i)= \sum_{j=i}^{n} \alpha{_i}{_j} e_j
\end{equation}

答:
This is not true as stated. Let $\Bbb{k} = \Bbb{R}$ and consider the linear map $f : \Bbb{R}^2 \to \Bbb{R}^2$ given by $$f(x,y) = (y,-x)$$

and assume there is a basis $\{e_1, e_2\}$ such that $$f(e_1) = \alpha_{11} e_1 + \alpha_{21}e_2, \quad f(e_2) = \alpha_{22}e_2.$$

If $e_2 = (x,y)$ then $$(y,-x) = f(x,y) =\alpha_{22}(x,y) \implies \begin{cases} y= \alpha_{22}x,\\ \alpha_{22}y = -x \end{cases}$$
so $\alpha_{22}^2 x = -x$ which implies $x = 0$ since $\alpha_{22} \in \Bbb{R}$. Hence $y = \alpha_{22}x = 0$ as well so $e_2= 0$ which is a contradiction. Therefore there is no such basis.

The background is that the matrix $$\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$$
is not similar to a real lower triangular matrix $$\begin{bmatrix} \alpha_{11} & 0 \\ \alpha_{21} & \alpha_{22} \end{bmatrix}$$
since the diagonal elements would have to be the eigenvalues $\pm i$.

Title: Existence of a certain basis?


问:
Probability of Non-Consecutive Cards
I've got a deck of 36 cards... 3 are labeled with #1, 3 with #2, and so on up to 3 with #12. I'm drawing 3 cards from the deck & wondering what the probability is that I will end up with NONE of the cards being within 2 of one another (including wrap-arounds--meaning a #12 is allowed to wrap-around to a #1). For example, {1,3,5} or {4,10,12} would qualify, but {10,12,1} would not.

I know that it's 38.9% WITH replacement as I actually drew the whole thing out on a spreadsheet with every possible outcome (and got 56/144), but I know that the number will increase slightly WITHOUT replacement, and I'm just not sure how to come up with the formula to figure it out.

I came up with this formula WITH replacement (using the inverting method)... 1 - {9/36 + (27/36 * 9/36) + (25/36 * 9/36)}, but I don't know how to adjust it for WITHOUT replacement. And I may have the wrong formula altogether. I'm new at this.

Is there anybody out there who wouldn't mind helping me out on this one? I've been at it for weeks with no luck!

答:
With replacement

With replacement, you can choose any card you like as the first card.
After that there are three cases:

If the second card is your number or your number plus or minus $1$ you have a bad set; this can happen $9$ ways out of $36$.

If the second card is your number plus or minus $2$ (which has $6$ of $36$ ways to happen) then there are $15$ out of $36$ cards that will make a bad set if they are drawn as the third card.

For any other second card ($21$ of the $36$ cards) there are $18$ out of $36$ possibilities for the third card that will make a bad set.

The three cases are mutually exclusive, so altogether the chance of a bad set is found by adding the three cases together,
$$ \frac{9}{36} + \left(\frac{6}{36}\right)\left(\frac{15}{36}\right)
    + \left(\frac{21}{36}\right)\left(\frac{18}{36}\right) = \frac{11}{18}, $$
and therefore the probability of a good set is
$1 - \frac{11}{18} = \frac{7}{18} \approx 0.38889.$

Without replacement

Without replacement, there are only $8$ ways out of $35$ for the first case to occur;
the second case occurs for $6$ of $35$ second cards but only $13$ of $34$ third cards;
and the third case occurs for $21$ of $35$ second cards but only $16$ of $34$ third cards.
Now the sum of the three cases is
$$ \frac{8}{35} + \left(\frac{6}{35}\right)\left(\frac{13}{34}\right)
    + \left(\frac{21}{35}\right)\left(\frac{16}{34}\right) = \frac{49}{85}, $$
and the probability of a good set is
$1 - \frac{49}{85} = \frac{36}{85} \approx 0.423529.$

Without replacement - alternative method

A second way to compute the probability without replacement uses the fact that a good set cannot contain two matching numbers.
That is, you must choose three different numbers from the set
$\{1, 2, 3, \ldots, 12\}$ without choosing two numbers that would be adjacent on a clock face.

For the case where $12$ is one of the numbers chosen,
you can choose any two distinct numbers $x < y$ from the set
$\{1, 2, 3, \ldots, 8\}$; then use $x + 1$ as your first number
(so $2\leq x\leq 9$ and $x$ is not too close to $12$),
and $y+2$ as your second number (so $x + 1 < y \leq 10$ and $y$ is not too close to either $x$ or $12$).
There are $\binom82$ choices of $x$ and $y$ in this case.

To cover all other cases you can choose any three distinct numbers $x < y < z$ from the set $\{1, 2, 3, \ldots, 9\}$; then use $x$ as your first number,
$y + 1$ as the second number, and $z + 2$ as the third number;
then $x + 1 < y < z - 1$, so the numbers are not too close in sequence,
and the $12,1$ case also is eliminated.
There are $\binom93$ choices of $x$, $y$, and $z$ in this case.

Note that for any set of three distinct numbers that does not include "adjacent" numbers (that is, for every good set), you can reverse the procedures in the previous two paragraphs in order to express the set either as $\{x,y+1,12\}$ or as $\{x,y+1,z+2\}.$
That is, the combinations in the preceding two paragraphs generate every good set of numbers once and only once.

So there are a total of
$\binom82 + \binom93 = 28 + 84 = 112 = 2^4 \times 7$
good sets of three numbers.

For each good set of three numbers, there are three choices of card for each number, $3^3 = 27$ choices altogether.

That means there are $112 \times 27 = 3024 = 2^4 \times 3^3 \times 7$ good sets of three cards.

But there are
$\binom{36}{3} = \frac{36\times 35\times 34}{3\times 2\times 1} = 7140$
ways to choose three cards without replacement from your $36$ cards.

As a result, the chance of drawing a good set of cards is

$$
\frac{3024}{7140}
= 2^4\times 3^3\times 7\times \frac{3\times 2\times 1}{36\times 35\times 34}
= \frac{36}{85}.
$$

Without replacement - second alternative

Another way to count the number of "good" sets of three numbers chosen from
$\{1, 2, 3, \ldots, 12\}$ is to choose any of those $12$ numbers first.
This leaves a set of $9$ numbers that are consecutive around a clock face;
choose $x < y$ from the set $\{1, 2, 3, \ldots, 8\}$,
then choose the $x$th and $y+1$st of the $9$ consecutive numbers
as the other two members of your set of three numbers;
there are $\binom82 = 28$ choices for that step.
This is guaranteed not to produce any two numbers that are adjacent around a clock face, but it produces each set three times (because any of the three numbers in the set can be the first one chosen).
So we have a total of
$$ \frac{12 \times 28}{3} = 112 $$
possible "good" sets of three numbers.
For the rest of the calculation, follow the part of "without replacement - alternative method" after the step where we find that there are $112$ good sets of numbers.

With replacement - alternative method

With replacement, you are really just doing three repetitions of choosing a number uniformly from the set $\{1, 2, 3, \ldots, 12\}.$
Each number has a $\frac1{12}$ chance of being chosen each time,
independent of the other two choices.
So we can say the second card "ruins" the set in just $3$ ways out of the $12$ possible choices of the second card.
Another way to put this is, when working out the probabilities with replacement,
instead of writing $\frac{9}{36}$ we can write $\frac{3}{12}$,
instead of writing $\frac{15}{36}$ we can write $\frac{5}{12}$,
and so forth.

Title: Probability of Non-Consecutive Cards


问:
$n-$section formulas in goniometry without the calculus
We know that the formulas of bisection of an angle $\alpha$ it is:

$$\sin^2\left(\frac{\alpha}2\right)=\frac{1-\cos(\alpha)}{2}, \qquad \cos^2\left(\frac{\alpha}2\right)=\frac{1+\cos(\alpha)}{2}$$

In particular

$$\sin\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}, \qquad \cos\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$$

My question is:

For the formulae of $n-$section of an angle $\alpha$, i.e.

$$\sin\left(\frac{\alpha}n\right), \qquad \cos\left(\frac{\alpha}n\right)$$

Do I must necessarily use the development in series of Taylor or the general formulas can be obtained with methods that do not use the derivatives, or the numerical methods?

My doubts arise for the students of an high school that not use the derivate.

答:
This is not a complete answer but I think it will be too long for a comment and moreover it gives you a explicit formula for the case when $n$ is a power of $2$:

For a given $\alpha$, let's call $\beta=\alpha/2$. Then your formula applied to $\beta/2$ reads

$$
\sin\left(\frac{\beta}{2}\right) = \pm\sqrt{\frac{1-\cos(\beta)}{2}}.
$$

But $\beta=\alpha/2$ so we can apply agian the same formula:

$$
\sin\left(\frac{\alpha}{4}\right) = \pm\sqrt{\frac{1-\cos\left(\frac{\alpha}{2}\right)}{2}}= \pm\sqrt{\frac{1\mp\sqrt{\frac{1-\cos(\alpha)}{2}}}{2}} =\pm\sqrt{\frac{1\mp\sqrt{\frac{1-\cos(\alpha)}{2}}}{2}} .
$$

It can be simplified a little as

$$
\sin\left(\frac{\alpha}{4}\right) = \pm\sqrt{\frac{\sqrt 2 \mp\sqrt{1-\cos(\alpha)}}{2\sqrt 2}} .
$$

You can play the same game for $\gamma=\beta/2$. Undoing the change of variable, and if I am not mistaken, the formula should be

$$
\sin\left(\frac{\alpha}{8}\right) = \pm\sqrt{\frac{2\sqrt2 \mp\sqrt{\sqrt2\mp \sqrt{1-\cos(\alpha)}}}{2\sqrt{2\sqrt2}}} .
$$

To get the general formula, it is better to denote square roots by $2^{1/2}$, so that $\sin(\alpha/8)$ becomes

$$
\sin\left(\frac{\alpha}{8}\right) = \pm\sqrt{\frac{2^{1+1/2} \mp\sqrt{2^{1/2}\mp \sqrt{2^0-\cos(\alpha)}}}{2^{1+1/2+1/4}}} .
$$

And there you go:

$$
\sin\left(\frac{\alpha}{2^m}\right) = \pm\sqrt{\frac{2^{2^{1-(m-1)}\left(2^{m-1})-1\right)} \mp\sqrt{2^{2^{1-(m-2)}\left(2^{m-2})-1\right)} \mp \sqrt{\cdots \sqrt{1-\cos(\alpha)}}}}{2^{2^{1-m}\left(2^m-1\right)}}} .
$$

Remark. I am using the fact that

$$
\sum_{k=0}^{m-1} \frac{1}{2^k} = 2^{1-m}\left(2^m-1\right)
$$


问:
Final step in proof that embedded lie subgroups are closed
A Lie subgroup $H$ of a Lie group $G$ is defined to be an embedded submanifold of $G$ that is also a subgroup. The following exercise helps the reader in showing that every Lie subgroup is closed.

Let $G$ be a Lie group and $H$ a Lie subgroup.

(1) Let $\overline H$ be the closure of $H$ in $G$. Show that $\overline H$ is a subgroup in $G$.

(2) Show that each coset $Hx$ , $x\in\overline H$, is open and dense in $\overline H$.

(3) Show that $H=\overline H$, that is, every Lie subgroup is closed.

Using continuity of the multiplication map, (1) follows readily. Also, since for any fixed $x\in\overline H$, multiplication by $x$ is a diffeomorphism, (2) can be shown relatively easily too. However, I'm stuck at (3). Clearly I should be using (2) but I don't see how. Any hints?

答: Let $x\in \bar H$, write $x=lim_nx_n, x_n\in H$, there exists $N$ such that $n>N$ implies that $x_n\in Hx$ since $Hx$ is open and contains $x$, we deduce that $x_n=hx,$  and $h\in H$ for $n>N$ and $x=h^{-1}x_n\in H$.


问:
How many whole rectangles can you catch in a grid?
There are $n$ rectangles packed in a square; all of them are axes-parallel.
You are allowed to partition the square into a grid of cells, with $1$ or more rows and $1$ or more columns. You score a point for each cell that contains at least one whole rectangle.

What is the maximum score $s(n)$ you can get, for the worst-case initial arrangement of $n$ rectangles?

Here are two examples for $n=4$. Here:

your score is 4 (the maximum), since there are 4 cells (out of 6) that contain a whole rectangle.
However, here:

your score is only 3, since there are only 3 cells and each of them contains a whole rectangle (a cell with more than one rectangle is worth only one point).
Moreover, here:

In any grid containing more than 2 cells (e.g. $1\times 3$ or $3\times 1$ or $2\times 2$), at least 2 rectangles are cut, so at most 2 cells contain whole rectangles. This example proves that $s(4)\leq 2$.

On the other hand, it is obvious that you can always score at least $2$ by just taking any two rectangles and creating a $2\times 1$ grid. Therefore $s(4)=s(3)=s(2)=2$.

The above example can be generalized by partitioning each rectangle into $n/4$ rectangles, such that each horizontal line cuts at least $n/4$ vertical rectangles and vice-versa.
It gives an upper bound of $s(n) \leq \lceil n/2 \rceil$.

Is it always possible to score at least $\lceil n/2 \rceil$?
If not, what is the maximum possible score in the worst case?

NOTE: This question is related, but it considers a single cut rather than a grid.

答:
Proposition 5. For each natural $k\ge 4$, $s(k^2)\le k^2/4+k$.

Proof. Consider the following configuration of $k^2$ squares of size $a$ contained in a big square.

Let the catching grid consists of $h$ horizontal and $v$ vertical lines, so it catches at most $(h+1)(v+1)$ squares. Removing empty grid cells, we can assume that the distance between any two parallel grid lines it at most $a$ and $h,v\le k$. This provides all squares crossed by parallel grid lines are distinct. It is easy to check that each grid line crosses at least $k-1$ squares, so the grid crosses at least $h(k-1)+v(k-1)-hv$ squares ($hv$ is subtracted because of the double count), and so it catches at most $k^2-(k-1)(h+v)+hv$ squares.
Put $h+v=2x\le 2k$. By the inequality between arithmetic and geometric means, $hv\le x^2$. So the grid cathes at most $\min\{x^2+2x+1, k^2-2(k-1)x+x^2\}$ squares. The following cases are possible:

If $0\le x\le (k-1)/2$ then $x^2+2x+1\le (k^2+2k+1)/4$.

If $k/2\le x\le k-1$ then $k^2-2(k-1)x+x^2\le k^2-2(k-1)(k/2)+(k/2)^2=k^2/4+k$.

If $x=k-1/2$ then $k^2-2(k-1)x+x^2=2k-3/4$.

If $x=k$ then $k^2-2(k-1)x+x^2=2k$. $\square$

Corollary 6. For each natural $n\ge 16$, $s(n)\le n/4+3\sqrt{n-1}/2+1$.

Proof. Pick the smallest $k$ such that $k^2\ge n$. Then $(k-1)^2\le n-1$ and $s(n)\le k^2/4+k\le 4+3\sqrt{n-1}/2+1.$ $\square$.

Proposition 7. When the input rectangles are all unit squares, $s(n)\geq n/4$.

Proof. Represent the squares by their bottom-left coordinates, $(x_i,y_i)$ for $i\in[n]$.  Partition the squares into four groups:

$\lfloor x_i \rfloor$ is odd and $\lfloor y_i \rfloor$ is odd;

$\lfloor x_i \rfloor$ is odd and $\lfloor y_i \rfloor$ is even;

$\lfloor x_i \rfloor$ is even and $\lfloor y_i \rfloor$ is odd;

$\lfloor x_i \rfloor$ is even and $\lfloor y_i \rfloor$ is even.

Pick the largest group; it contains at least $n/4$ squares. Suppose w.l.o.g. that it is group 1. Then cut at $x=1, 3, 5, ...$ and $y = 1, 3, 5, ...$; the cuts do not harm any square of group 1.

Currently, the best lower bound for arbitrary rectangles is $s(n)\geq \sqrt{n}$.
I conjecture that there should be a linear lower bound similar to $n/4$, but so far could not prove it.

EDIT: In the pattern below there are 40 squares, and it seems that the maximum score is 8, which implies an asymptotic upper bound $s(n)\leq n/5$ (not sure how to verify this):


问:
Prove that if a graph $G$ has an independent vertex subset $X \subseteq V G$ such that $|X| > |N (X)|$ then $G$ is non-Hamiltonian.
Prove that if a graph $G$ has an independent vertex subset $X \subseteq V G$ such that $|X| > |N (X)|$ then $G$ is non-Hamiltonian.

I have tried to delete m vertices in order to produce m component, but it doesn't work. can someone please help!

答:
@bof

Choose an orientation for the cycle, making it a directed cycle. Then you can match each vertex of the independent set X with its neighbor in the forward direction, which shows that $|X|≤|N(X)|$


答:
I doubt that there are such elementary general formulas. I will derive a possible (restricted) formula as an infinite series using De Moivre's formula, though this may not satisfy your requirement of calculus free derivation. One value of the multivalued expression $\cos\left(\dfrac{\theta}{n}\right)+i\sin\left(\dfrac{\theta}{n}\right),\quad n\ge2$ is given by
\begin{align}
\left(e^{i\theta}\right)^{1/n} &
= \left(\cos\theta+i\sin\theta\right)^{1/n} \\
 & = \left(\cos\theta\right)^{1/n}\left(1+i\tan\theta\right)^{1/n}
\end{align}

Now, assuming $0\lt\theta\lt\pi/4,$ we can expand $\left(1+i\tan\theta\right)^{1/n}$ via binomial series as $$1+\dfrac{1}{n}i\tan\theta +\dfrac{(n-1)}{2!n^2}\tan^2\theta -\dfrac{(n-1)(2n-1)}{3!n^3}i\tan^3\theta -\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$

Hence we have two infinite series $$\dfrac{\sin\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=\dfrac{1}{n}\tan\theta - \dfrac{(n-1)(2n-1)}{3!n^3}\tan^3\theta+ \dfrac{(n-1)(2n-1)(3n-1)(4n-1)}{5!n^5}\tan^5\theta+$$ and
$$\dfrac{\cos\left(\dfrac{\theta}{n}\right)}{\sqrt[n]{\cos\theta}}=1+\dfrac{(n-1)}{2!n^2}\tan^2\theta-\dfrac{(n-1)(2n-1)(3n-1)}{4!n^4}\tan^4\theta +\cdots.$$

Title: $n-$section formulas in goniometry without the calculus

Body:
We know that the formulas of bisection of an angle $\alpha$ it is:

$$\sin^2\left(\frac{\alpha}2\right)=\frac{1-\cos(\alpha)}{2}, \qquad \cos^2\left(\frac{\alpha}2\right)=\frac{1+\cos(\alpha)}{2}$$

In particular

$$\sin\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}, \qquad \cos\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$$

My question is:

For the formulae of $n-$section of an angle $\alpha$, i.e.

$$\sin\left(\frac{\alpha}n\right), \qquad \cos\left(\frac{\alpha}n\right)$$

Do I must necessarily use the development in series of Taylor or the general formulas can be obtained with methods that do not use the derivatives, or the numerical methods?

My doubts arise for the students of an high school that not use the derivate.


问:
Is $\cos(x^2)$ a solution of a 2nd order linear ODE defined on all of $\mathbb R$?
I've been asked in an assignment whether the function $\cos(x^2)$ can be the solution of a 2nd order homogeneous linear equation which satisfies the existence and uniqueness theorem on all of $\mathbb R$. On one hand I cannot find a reason why it wouldn't be a solution of such an equation (unlike $\sin(x^2)$, which was easy) since there is no point where both the function and its derivative vanish, on the other hand I cannot for the life of me find an example of such an equation. Assistance would be appreciated.

答: Your function is a solution of $$xy''-y'+4x^3y=0.$$


问:
Find the polynomial that goes through points B and C and is tangent to $f(x) = (2/3)x+4$ at A, Interpolation
I have 3 points:

$$A = (0,4) \\
B=(-5,0) \\
C=(5,0)$$

I need to find a polynomial that goes through B and C, and is tangent to $f(x) = (2/3)x+4$ at A.

I know that tangent means it must be equal to the derivative of f(x) at that point.

This is probably wrong, but I did the interpolation using points B, C and $(0,2/3)$. I got

Help?

答:
You have four conditions
$$\begin{align}f(0)&=4\\f(-5)&=0\\f(5)&=0\\f'(0)&=\frac23\end{align} $$
and so we should be looking for four unknowns, i.e., $f$ is of degree $3$, or
$$ f(x)=ax^3+bx^2+cx+d$$
and the four conditions translate into linear equations in the unknown coefficients:
$$\begin{align}d&=4\\-125a+25b-5c+d&=0\\125a+25b+5c+d&=0\\c&=\frac23\end{align} $$
As $c$ and $d$ are available immediately, you are essentially left with a simple system of two very user-friendly linear equations in $a$ and $b$.

Your interpolation considers only the first three givens.


问:
Let $G$ be a $p$-group : $|G| = p^r$. Prove that $G$ contains a normal subgroup of order $p^k$ for every $k \leq r$.
I am reading Aluffi chapter 0. This is an exercise in it. Since the book hasn't stated Sylow theorem or Cauchy theorem yet, I don't want to use it in my solution, though I am stuck.

Let $p$ be a prime number, and let $G$ be a $p$-group : $|G| = p^r$. Prove that $G$ contains a normal subgroup of order $p^k$ for every $k \leq r$.

I know that the center isn't trivial by the class formula.

答:
If $G$ is abelian, then this result follows from the theorem:

Theorem. Let $G$ be a finite abelian group of order $n$. Then $G$ has a subgroup of order $d$ for every divisor $d$ of $n$.

You can see the proofs from this link. The OP used Cauchy's theorem, but  the proof of Cauchy's theorem for abelian groups is much easier than for non-abelian ones. Here I give a proof for that

Proof for Cauchy's theorem on Abelian Groups. Here we can prove it by induction on $n$. The case for $n=1$ is trivial. Now we let $a\in G$ be an element of order $k$.

If $p\mid k$, say $k=p\ell$, then $a^\ell$ has order $p$.

Suppose $p\nmid k$. Since $G$ is abelian, we have $N:=\langle a\rangle\unlhd G$. Now we consider $G/N$, which has order $n/k$. Since $p\mid n$ but $p\nmid k$, we must have $p\mid(n/k)$. By induction hypothesis, there is an element $bN\in G/N$ of order $p$. Clearly, the order of $b$ is a multiple of $p$, so everything reduces to the first case. $\square$

As for the followings, you can check that link.

Thus, we assume that $G$ is non-abelian and prove it by induction on $r$. The case for $r=1$ is trivial as $G$ is cyclic in this case.

As for the inductive step, the center $Z(G)$ of $G$ has size $p^\ell$ where $1\leq\ell<r$. We then consider two cases:

If $k\leq\ell$, then we can find a subgroup $N$ of $Z(G)$ of order $p^k$ as $Z(G)$ is abelian. We can show that $N$ is a normal subgroup of $G$. (Hint. $N$ is contained in the center of $G$)

If $k>\ell$, we consider the quotient group $G/Z(G)$, which has order $p^{r-\ell}$. By induction hypothesis, there is a normal subgroup $\overline{N}$ of $G/Z(G)$ of order $p^{k-\ell}$. Besides, by the correspondence theorem (4-th isomorphism theorem), we have $\overline{N}=N/Z(G)$ for some $N\unlhd G$. As we can see, $$|N|=|N/Z(G)||Z(G)|=p^{k-\ell}p^\ell=p^k.$$


问:
Abstract Algebra automorphisms inner automorphisms - Are my answers correct?
Hope you are well.

Im posting the following as a favor on behalf of a friend who is not a native speaker, she just wants to confirm her answer is indeed correct.

Image of Questions:

Answer of 1st Question.

Answer of 2nd Question:

If possible please post any answers in the most simple form for easy understanding.

Any input would be greatly appreciated, Thanks in advance!

答:
Both questions seem fairly vague. For (ii), I'll assume that $(\mathbb{Q},+)$ refers to the underlying additive group of $\mathbb{Q}$. In that case, $\operatorname{Inn}(\mathbb{Q})$ refers specifically to those automorphisms generated by conjugations, but since $\mathbb{Q}$ is abelian, conjugations are trivial, and $\operatorname{Inn}(\mathbb{Q})$ is therefore trivial as well. Contrast this with $\operatorname{Aut}(\mathbb{Q})$: For every nonzero rational $q$, multiplication by $q$ is a nontrivial group automorphisms. This results in a homomorphism $\mathbb{Q}^{\times} \to \operatorname{Aut}(\mathbb{Q})$. In fact, this map is an isomorphism. To prove this, argue that, for any automorphism $f$, one has $f(n) = n\cdot f(1)$ for all integers $n$, and $f(1/n) = f(1) / n$ for all (nonzero) $n$ as well.

As for question (iii), I agree with lulu that the question is not clear. When they say 'group with multiplication', do they just mean the ring, and if so, is $\operatorname{Aut}(\mathbb{Q}[\sqrt{2}])$ the group of ring automorphisms? In that case, the provided answer seems correct: the underlying $\mathbb{Q}$ can only be mapped to itself trivially, and it brings us to the question what happens to $\sqrt{2}$ under an automorphism. Well, $\sqrt{2}$ is a root of $x^2 - 2$, and so must $f(\sqrt{2})$ be for any automorphism $f$, and there are two such roots, namely $\pm \sqrt{2}$. Thus, $\operatorname{Aut}(\mathbb{Q}[\sqrt{2}])$ consists of two elements, namely the identity and $a + b\sqrt{2} \mapsto a - b\sqrt{2}$.


问:
Interchange Laplacian transform and integral
I wonder under what condition the Laplacian transform is interchangeable with integrals.

Specifically, let $f(t, w)$ be continuous with repsect to both $w$ and $t$ and $g(t) = \int_0^1 f(t, w) \mathrm{d}w$. Then, when do we have
\begin{align*}
g(s) = \mathcal{L}(g(t)) &= \mathcal{L} \left(\int_0^1 f(t, w) \mathrm{d}w \right) \\ &= \int_0^1 \mathcal{L} (f(t, w)) \mathrm{d} w
= \int_0^1 f(s, w) \mathrm{d}w,
\end{align*}
where $g(s)$ and $f(s, w)$ are Laplacian transforms of $g(t)$ and $f(t, w)$, respectively?

答: You can always do this (given the conditions satisfied by your function), since this is nothing more than a change in order of integration. That is, you want $$\int_0^{+\infty} e^{-st}\int_0^1 f(t,w)dw dt=\int_0^1 \int_0^{+\infty} e^{-st}f(t,w)dt dw,$$ or in other words $$\int_0^{+\infty} \int_0^1 e^{-st} f(t,w)dw dt=\int_0^1 \int_0^{+\infty} e^{-st}f(t,w)dt dw.$$


问:
How to simplify a partial sum obtained by Legendre polynomials
Context

I am working on an electrostatics problem. I have undertaken Fourier analysis. By $k$, I denote a natural number $k=0,1,2,\ldots$. I have obtained the following partial sum in terms of the even number $2k$.

$$S{(2k)}=\frac1{2^{2k}}\sum_{m=0}^{k}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\,\frac{1}{2k-2m+1}.$$

The boundary problem that this question originates from has odd symmetry. Therefore, I expect all even Fourier coefficients--except potentially $2k=0$--to be null. This expectation is verified by the correct answer below.

Question

Can the above expression be simplified further?

答:
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}

We obtain for $k>0$
\begin{align*}
\color{blue}{S(2k)}&=\color{blue}{\frac{1}{2^{2k}}\sum_{m=0}^k(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\frac{1}{2k-2m+1}}\\
&=\frac{1}{k2^{k+1}}\sum_{m\geq 0}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k-2m+1}\tag{2}\\
&=\frac{1}{k2^{k+1}}\sum_{m\geq 0}(-1)^m\binom{2k}{m}[z^{2k-2m+1}](1+z)^{4k-2m}\tag{3}\\
&=\frac{1}{k2^{k+1}}[z^{2k+1}](1+z)^{4k}\sum_{m\geq 0}(-1)^m\binom{2k}{m}\left(\frac{z}{1+z}\right)^{2m}\tag{4}\\
&=\frac{1}{k2^{k+1}}[z^{2k+1}](1+z)^{4k}\left(1-\left(\frac{z}{1+z}\right)^2\right)^{2k}\tag{5}\\
&=\frac{1}{k2^{k+1}}[z^{2k+1}](1+2z)^{2k}\tag{6}\\
&\,\,\color{blue}{=0}\tag{7}
\end{align*}

Comment:

In (2) we use the binomial identity $\binom{4k-2m}{2k}\frac{1}{2k-2m+1}=\binom{4k-2m}{2k-2m}\frac{1}{2k-2m+1}
=\binom{4k-2m}{2k-2m-1}\frac{1}{2k}$.

In (3) we use the coefficient of operator according to (1). We also set the upper limit of the sum to $\infty$ which doesn't change the sum since we are adding zeros only.

In (4) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

In (5) we apply the binomial theorem.

In (6) we do some simplifications.

In (7) we select the coefficient of $z^{2k+1}$.


问:
Splitting $n^2$ objects into $n$ unordered groups

Is there a way to prove that the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$ is $(((n^2)!)/((n!)^{(n+1)}))$

Question arises from: https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_25

I tried to solve it using summation but that didn't work.
Any ideas?

答:
One way to do such a split up is to fill an $n\times n$ matrix with the $n^2$ objects in random order and then take the rows as subsets (and clearly every split up into $n$ groups of $n$ is obtainable this way).
There are $(n^2)!$ different orderings of the $n^2$ objects, but many lead to the same split-up: We can reorder each row in itself without changing the outcome, thus dividing the count by $n!$. As this applies to $n$ rows, we divide $n$ times, i.e., we divide by $n!^n$. Finally, the order of the rows is unimportant, i.e., we need to divide by yet another factor of $n!$. Final result:
$$\frac{(n^2)!}{n!^{n+1}}. $$


问:
Rank of a matrix
If a $3 \times 3$ matrix has determinant zero, then is it possible that its rank could be $3$? I think it only could be $2$ or less. I am right or wrong? Please explain.</p>

答: A matrix does have full rank iff it is invertible. The determinant of a matrix is zero iff it is not invertible. Thus if a $3 \times 3$ matrix has a determinant of zero it can not have the rank $3$.</p>

Title: Rank of a matrix


问:
What is a "distinguished subset"?
I don't know if this is another word for something I already know or if it is something altogether different. I'm reading my textbook in CS about distributed algorithms, and this came up.

I googled for a definition but couldn't find one.

The state set $Q$ contains a distinguished subset of initial states.

答: It just means a particular subset that we're going to pay special attention to. It does not mean that the subset has any special properties.</p>

Title: What is a "distinguished subset"?


问:
Solving $\cos2x-2\sin x-\cos^2x=-3$ with identities
I need to solve
$$\cos2x-2\sin x-\cos^2x=-3$$
I'm trying to find possible solutions in the interval $[0,2\pi]$, but I need to simplify the equation using trig identities first. I just can't seem to find the right ones to use.

答:
$$\cos2x-2\sin x-\cos^2x=-3$$
Substitute $\cos 2x =\cos^2x-\sin^2 x$

$$\cos^2x-\sin^2 x-2\sin x-\cos^2x=-3\to \sin^2x+2\sin x-3=0$$
$$\sin x=1;\;\sin x=-3$$
$$\sin x=1\to x=\frac{\pi}{2}$$
The second result is discarded because $|\sin x|\le 1$ so $\sin x =-3$ is impossible.


答:
Alternative solution for the case without replacement.

First we compute the probability of getting $3$ different cards:

$$P_1 = \frac{\binom{12}{3} \cdot 3^3}{\binom{36}{3}} = \frac{99}{119}$$

Next we compute the probability of getting a good triple from $12$ different cards. If all three are consecutive there are $12$ cases. If only two are consecutive there are $12 \cdot 8=96$ cases, therefore

$$P_2 = 1- \frac{12+96}{\binom{12}{3}}=\frac{28}{55}$$

Finally, the probability of getting a good triple from $36$ cards without replacement is
$$P_1 \cdot P_2 = \frac{99}{119} \cdot \frac{28}{55}=\frac{36}{85}$$

Title: Probability of Non-Consecutive Cards

Body:
I've got a deck of 36 cards... 3 are labeled with #1, 3 with #2, and so on up to 3 with #12. I'm drawing 3 cards from the deck & wondering what the probability is that I will end up with NONE of the cards being within 2 of one another (including wrap-arounds--meaning a #12 is allowed to wrap-around to a #1). For example, {1,3,5} or {4,10,12} would qualify, but {10,12,1} would not.

I know that it's 38.9% WITH replacement as I actually drew the whole thing out on a spreadsheet with every possible outcome (and got 56/144), but I know that the number will increase slightly WITHOUT replacement, and I'm just not sure how to come up with the formula to figure it out.

I came up with this formula WITH replacement (using the inverting method)... 1 - {9/36 + (27/36 * 9/36) + (25/36 * 9/36)}, but I don't know how to adjust it for WITHOUT replacement. And I may have the wrong formula altogether. I'm new at this.

Is there anybody out there who wouldn't mind helping me out on this one? I've been at it for weeks with no luck!


问:
Well ordered set, Binary relation

My attempt: A is well ordered, hence every subset of A has a minimal element;

Since ⟨A,≤1⟩ is WO, it is isomorphic to an ordinal β, which is finite. But α is an ordinal as well. So we can define a function g: β-> α;

If we define the function f to map the min element a1 of an arbitrary subset of A,say A1, to f(a1), then for every a' in A1 we have that a1<=a', because of the order preserving.

Also f(a1)<=f(a').
If β and α are order isomorphic, then for every b in A (or after my assumption we can say β as well) we have f(b)=g(b).

And since A has an initial element a1, then this a1 will satisfy f(a1)<g(a1).

Any help is appreciated!

答:
Either I'm misreading the question or the following is a counterexample. Let $A=\omega$ with the usual ordering and let $\alpha=2$. For each natural number $n$, define $f_n:\omega\to2$ by
$$
f_n(k)=\begin{cases}0&\text{if }k<n\\1&\text{if }k\geq n. \end{cases}
$$
Then $f_0\succ f_1\succ f_2\succ\dots$, so $\preceq$ is not a well-ordering.

Title: Well ordered set, Binary relation


答: A distinguished subset is one that is chosen out of all the subsets.  In this case, some of the states are chosen to be "initial states".</p>

Title: What is a "distinguished subset"?

Body:
I don't know if this is another word for something I already know or if it is something altogether different. I'm reading my textbook in CS about distributed algorithms, and this came up.

I googled for a definition but couldn't find one.

The state set $Q$ contains a distinguished subset of initial states.


问:
Let $N_1\leq N_2\leq N_3\leq\cdots$ be normal subgroups of $G$. Show that $\bigcup_{i=1}^\infty N_i$ is a normal subgroup of $G$.

Let $N_1 \leq N_2\leq N_3\leq \cdots$ be normal subgroups of $G$. Show that  $\bigcup_{i=1}^{\infty}N_i$ is a normal subgroup of $G$.

Please help me to verify my attempt, in case I have left any details.

My attempt: Since $N_1 \leq N_2\leq N_3\leq \cdots$, we have that $N_1 \subseteq N_2\subseteq N_3\subseteq \cdots$, so $\bigcup_{i=1}^{\infty}N_i$ is a subgroup of $G$. (Justification)

Now if $a\in N_i$ and $g\in G$, then $g^{-1}ag\in N_i$. Thus $g^{-1}ag\in \bigcup_{i=1}^{\infty} N_i$ and we are done.

答:
It seems fine to me, but your "justification" is a little off (as it only proves the result for two subgroups); try this instead.

Oh, and you've overused $i$ in your last paragraph. You can remedy this as follows. Suppose $a\in H:=\bigcup_{i=1}^\infty N_i$. Then $a\in N_j$ for some $j$. Since $N_j$ is normal, $gag^{-1}\in N_j$ for all $g\in G$. But then $gag^{-1}\in H$. Hence $H\unlhd G$.


问:
Clarification on the definition of derivative/differentiablility
I'm having a hard time understanding where to use:

$$\lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}\qquad\text{and}\qquad \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$$

My textbook seems to use either one of these in similar exercises and from what I've tried they produce the same result. Any help?

答: You can use either one, they mean the same thing.  To see this, we can start with $$\lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$$ and set $h = x - a$.  Then we can say that $x\to a\implies h\to 0$, so we can make the substitution $x = a+h$: $$\lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{h\to 0}\frac{f(a+h) - f(a)}{h}.$$


答:
They are the same expression under the condition $x=a+h$:
$$
\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{a+h\to a}\frac{f(a+h)-f(a)}{a+h-a}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.
$$
Thus it is equivalent to use either one!

Title: Clarification on the definition of derivative/differentiablility

Body:
I'm having a hard time understanding where to use:

$$\lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}\qquad\text{and}\qquad \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$$

My textbook seems to use either one of these in similar exercises and from what I've tried they produce the same result. Any help?


问:
A continuous and derivable decreasing function s.t. the derivative is $0$ for $t>t_0$?
The question may be silly, but I am stucked with this thought...

I ask for a nonconstant continuous and derivable nondecreasing function s.t. the derivative approaches $0$ for $t\to\infty$, but more than that I might have $f'(t_0)=0$ for some $t_0$ and also for all $t>t_0$.

I really cannot think this example. For instance, $f(t)=\text{log}(t)$ does not mind, because $f'$ approaches to $0$, but $f'(t)\neq0\forall t$. And so on.

But because this question?

Because I have an equation on a model of form $\dfrac{d f}{d t}=g(t)$ with $g(t)$ nonpositive and I might have $g(t)=0$ for all $t>t_0$ for some $t_0>0$, but I cannot realize this.

Thank you so much.

答:
Consider the function $$f(x) = \int_{1}^{x}\frac{sin^{2}t}{t}dt$$
Since integral measures area below the graph and since the function being integrated is non-negative $f(x)$ is increasing. Of course, derivative of $f(x)$ is $\frac{sin^{2}x}{x}$ and it non-negative and is zero at $k*\pi$.


问:
Prove that $\frac{1}{x} = O\left(\frac{\pi}{2} - \arctan(x)\right)$ as $x\to\infty$
Prove that $\frac{1}{x}  = O\left(\frac{\pi}{2} - \arctan(x)\right)$ as $x\to\infty$.

I am confused how we used the formal definition of big-$O$-notation to prove this. Notice that this is just $\frac{1}{x}$.

答:
For $x>1,$

$$0<\pi/2 - \arctan x = \int_x^\infty \frac{1}{1+t^2}\,dt > \int_x^\infty \frac{1}{2t^2}\,dt =\frac{1}{2x}.$$

This implies $\dfrac{1}{x} < 2(\pi/2 - \arctan x)$ and proves the result.


问:
Finding the ellipsoid of maximum area inside a square - and extension to $\mathbb{R}^n$
In an answer here, I read that:

Take a max area ellipse. Apply an affine transform to make it a circle; then the problem becomes to show that a minimal area parallelogram containing a circle is a square. It is easy to see that both the height of the parallelogram and its base are at least the diameter. Q.E.D.

I understand the second half of the reasoning here, but I don't understand why the original problem is equivalent to its reformulation in terms of a min-area parallelogram. Could someone help me achieve this reformulation, please?

Some comments read:

Comment 1: By the way, the argument seems easy to adapt for cubes of any dimension.

Comment 2: When you go to higher dimensions, you might as well prove Auerbach's lemma in general, since the proof is elementary and simple, and remark that Auerbach's lemma for the Euclidean ball gives John's theorem for the cube.

From my understanding, this is how this argument adapts to $n$-dimensions: Perhaps we'll be finding the min-volume parallelepiped that contains the $n$-ball of radius $1$, and noticing that the base of the parallelepiped along the $k$th dimension (for all $k$) is at least as large as twice the radius of the ball should complete the proof.

However, what is Comment 2 saying?

答:
Your understanding of your opening question and of comment 1 now seems fine. So we have a Theorem and Proof along the following lines. (Which I will state in the 3D form for simplicity.)

Theorem An ellipsoid of maximal volume inscribed in a cube is a sphere.

Proof Consider an ellipsoid of maximal volume inscribed in a cube and transform it into a sphere, of diameter $d$ say, by directional scalings. Then the cube is transformed into a parallelepiped containing the sphere. The parellelepiped has volume at least $d^3$ and so, by minimality it is a cube.

The point of Comment 2 is that the above theorem even in $n$-dimensions can be generalised further. In particular, we can replace "inscribed in a cube" by "inscribed in a centrally symmetrical convex body".

n-dimensions

The crux of the argument comes at the end of the proof summarised above. For any direction, the $n$ -dimensional parallelepiped will lie between two planes perpendicular to that axis and a distance $d$ apart. So it's $n$ -dimensional volume is at most $d^n$.


问:
Convergence of $\int_{-\infty}^\infty \frac{1}{1+x^6}dx$
Okay, so I am asked to verify the convergence or divergence of the following improper integrals:

$$\int_{-\infty}^\infty \frac{1}{1+x^6}dx$$

and

$$\int_1^\infty \frac{x}{1-e^x}dx$$

Now, my first attempt was to use comparison criterion with $$\int \frac{1}{x^2}$$

and conclude that both of the improper integrals converge given that they are smaller than the general term $\frac{1}{x^2}$.

Is it the right path? Also, are the antiderivatives of the improper integrals given easy to find?

Thanks.

答: For the second, you have two problems. One is  at $\infty$, which your comparison will take care of.  At $1$ the denominator goes to zero.  Once you show it goes to a finite value you have shown convergence.</p>


问:
Why Not Fewer Axioms for Probability Theory?
From what I understand, modern probability theory assumes the following axioms:

$0 \le P(E) \le 1$.

$P(S) = 1$.

$P(\bigcup_{i=1}^\infty E_i) = \sum_{i=1}^\infty P(E_i)$ where $E_i \cap E_j=\emptyset$ for $1 \le i < j$.

Clearly, the inclusion–exclusion principle reduces to axiom 3 whenever events are mutually exclusive. Moreover, the inclusion–exclusion principle may be proven by induction, without any help from axiom 3 whatsoever. Why then is axiom 3 even an axiom at all?

答: Contra your claim, axioms $(1)$ and $(2)$ alone are extremely weak. For example, taking $S=[0,1]$ for concreteness, let $P_*(X)=1$ iff $0\in X$ or $1\not\in X$, and $P(X)=0$ otherwise. This $P_*$ is absolutely horrible: beyond merely failing to satisfy $(3)$, it's not even monotonic, since e.g. $P_*([{1\over 2}, 1))=1$ but $P_*([{1\over 2}, 1])=0$. For that matter, it also has $P_*(\emptyset)=1$.


问:
Reasons for $\lambda \mathbb{I} - T$ to be non-invertible.
Given a Hilbert space $\mathcal{H}$ an operator $T\in \mathcal{L}(\mathcal{H})$, it's spectrum is
$$\sigma(T):= \{ \lambda \in \mathbb{C} : \lambda \mathbb{I} - T \mbox{ is not invertible } \}.$$
Now to actually compute the elements in the spectrum we find out that there are only 3 reasons for which the operator $\lambda \mathbb{I} - T$ is not invertible:

(i) it is not injective;

(ii) it is not surjective;

(iii) it is injective and surjective, but the "inverse" (which is linear) isn't continuous.

But at this point we construct the whole theory on the fact that the case (iii) is not possible; I found a theorem called "Open Mapping Theorem" which would explain it since it would imply that $T$ is an open map, but I can't find a proof.

Could somebody direct me to one or anyways be able to prove it? Also, if what I asked is a problem, I would still be very happy to read an intuitive explanation of how one could prove it.

答:
What you are looking for is the bounded inverse theorem, which says that a bounded linear bijection between Banach space has bounded inverse. You can find a proof in virtually any book that contains some functional analysis. Here are some examples:

[1] Rudin, "Real and complex analysis", theorem 5.10 (p100).

2 Axler, "Measure, integration and real analysis", theorem 6.83 (p188).

[3] Conway; "A course in functional analysis", theorem 12.5 (p91)

The book 2 is very accessible and can be accessed online for free. See here.