问:
non-equivalence of $p$-norms in $\ell_1$

For $x\in \mathbb{R}^{\mathbb{N}}, $ define $\|x\|_\infty := \sup_{i\in\mathbb{N}}|x_i|$ and for $1\leq p < \infty,$ the $p$-norm $\|x\|_p$ is given by $\|x\|_p := (\sum_{i} |x_i|^p)^{1/p}.$ The $\ell_p$ space is the set of all sequences with real entries with finite $p$-norm. Two norms $\|\cdot \|_a$ and $\|\cdot \|_b$ are equivalent on a real vector space $V$ if there are $m, M > 0$ so that $m\|x\|_a \leq \|x\|_b \leq M\|x\|_a$ for all $x\in V.$

Show that for $1\leq p < r \leq \infty,$ the norms $\|\cdot \|_p$ and $\|\cdot \|_r$ on $\ell_1$ are not equivalent.

I know how to show that $\ell_1\subseteq \ell_p$ for each $1\leq p\leq \infty.$ I think I should deal with the case where $r=\infty$ first as that might be easier. Taking the negation of the definition of equivalence, I think I should show that $\forall m, M > 0,$ there exists $x \in \ell_1$ so that $m\|x\|_p > \|x\|_r$ or $\|x\|_r > M\|x\|_p.$ However, I'm not sure how to find this particular $x$ for each choice of $m$ and $M.$

答:
Consider the all-ones vector $u^{(n)}$ on $n$ elements. For $1\leq p < r < \infty$, we have
$$\begin{align}
\|u^{(n)}\|_1 &= n
\\
\|u^{(n)}\|_p &= n^{1/p}
\\
\|u^{(n)}\|_r &= n^{1/r}
\\
\|u^{(n)}\|_\infty &= 1
\end{align}$$
The first shows that, indeed, $u^{(n)}\in\ell_1$ for all $n\geq 1$. The last 3 show what you want: those norms cannot be equivalent, as the ratios can be made arbitrarily large/small as $n\to\infty$.

If you want some intuition: this is basically the type of vectors that makes Holder's inequality tight (worst possible correspondence between $\ell_p$ norms).

Also, as an important comment: this will show that for $1\leq p<r\leq \infty$, the norms are not equivalent, by showing there cannot be a constant $M>0$ (depending only on $p$ and $r$) such that
$$
\|\cdot\|_p \leq M\|\cdot\|_r
$$
However, the other inequality will always hold with $m=1$: $\|\cdot\|_r \leq \|\cdot\|_p$. This follows from the monotonicity of $\ell_p$ norms.


问:
Finding trigonometric integral (challenging)
Integrate:

$$\displaystyle\int \dfrac{(1+\cos^2A\times\cos(2A))^2}{\cos(2A)\times(2\cos^4A+\sin^2A)}dA$$

Again my working with trigonometric identities has not take me anywhere useful, that is I haven't been able to break free of fractions or nasty products. But again, I'm a beginner to calculus and I only know how to integrate the 6 primary trigonometric functions. Perhaps there is a good way to bypass the tedious mess?

Despite the most off-putting look, the teacher claims that the integral is not anywhere as nasty as one would expect. This, of course, is not a part of class work but a challenge she threw at us.

Thanks in advance and as usual, all advice welcome.

ContentLicense: CC BY-SA 3.0

Body: The answer is $1/2\tan(A)/(\tan(A)^2+1)-1/2\ln(\tan(A)-1)+1/2\ln(\tan(A)+1)+1/2A.$


问:
If complex numbers are merely a definition with defined operations, how come they can help prove properties of real numbers?
It is often said that complex numbers were to some degree originally introduced by Rafael Bombelli, who together with Gerolamo Cardano was trying to find solutions to cubic equations.  In one YouTube series titled "Imaginary Numbers are Real", it is mentioned that Bombelli studied equations of the type $x^3=dx+c$, and found
$$x=\sqrt[3]{\frac{d}{2}+\sqrt{\frac{d^2}{4}-\frac{c^2}{27}}} + \sqrt[3]{\frac{d}{2}-\sqrt{\frac{d^2}{4}-\frac{c^2}{27}}}$$
to be one solution. Looking at the equation $x^3=15x+4$ and substituting $d=15,c=4$ one obtains:
$$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}$$
it is then said that Rafael essentially just assumed that $\sqrt{-1}$ was a normal number with the property that $\sqrt{-1}^2=-1$, and that $\sqrt{-a}=\sqrt{a}*\sqrt{-1}$, and then using some clever algebra and solving a system of equations, managed to eventually cancel out any $\sqrt{-1}$'s, and get:
$$x=4$$
And indeed one can check that $x=4$ is a solution to $x^3=15x+4$. When one introduces a new definition into mathematics, there are obviously certain restrictions, such as the definition having to be consistent with the rest of mathematics. At the end of the day however, it is just a definition, and we get to choose it and it's operations. For example, we could define complex numbers without much reference to the intuitive idea behind "square roots of minus 1" as:

A complex number $z$ is an ordered pair $(a,b)$ of two real numbers $a$ and $b$, often written with the following notational rule
$$(a,b): a+bi $$
plus the following shorthands:
$$\begin{eqnarray}
    && (0,b) & = & bi \\
    && (a,0) & = & a \\
    && (a,1) & = & a & + & i \\
    && (a,-1) & = & a & - & i \\
    && (0,1) & = & i  \\
    && (0,-1) & = - & i \\
    \end{eqnarray}$$
The equality relation $=$ for two complex numbers $z_1, z_2$
is defined as:
$$z_1 = z_2 \equiv (a_1=a_2 \wedge b_1=b_2)  $$
The addition operation $+$ is defined as:
$$z_1+z_2 = (a_1+a_2,b_1+b_2)  $$
The multiplicative operation $\cdot$ is defined as:
$$z_1 \cdot z_2 = (a_1 \cdot a_2 - b_1 \cdot b_2, a_1 \cdot b_2 + b_1 \cdot a_1)$$
Note that setting $a_1=a_2=0$ and $b_1=b_2=1$ and using the notation yields:
$$i \cdot i=(0 \cdot 0 - 1 \cdot 1)+(0 \cdot 1 + 1 \cdot 0)i = -1 $$
This is all fine, and one could define other operations such as raising complex numbers to real powers, real powers to complex numbers, taking logarithms of complex numbers and so on. But if this is merely definition, how do we know we can use it to prove facts about the real numbers as Rafael did? Was it merely intuition and luck on his side that gave him the final correct conclusion?

答:
What you wrote

At the end of the day however, it is just a definition, and we get to choose it and it's operations.

is technically correct. We could, in the same way,
define variants of the complex numbers such as
split-complex numbers
which appear with overa dozen different names. Another variant is the
dual numbers.
These are all mere variant definitions (they only appeared
in the 1800s) and have equal validity in some sense.
Complex numbers were not well understood (not quite "real")
by anyone until around 1800 when it was noticed that the
complex numbers could be identified with the complex plane.
That is, the points in a Cartesian coordinate system
could be interpreted as complex numbers along with
a geometric interpretation of complex addition and complex
multiplication. With this discovery, the complex
numbers were endowed with geometrical meaning and
no longer just an arbitrarily defined extension
of real numbers. There is a lot more to the history
of complex numbers but this is the very brief version.

Your question

But if this is merely definition, how do we know we can use it to prove facts about the real numbers as Rafael did? Was it merely intuition and luck on his side that gave him the final correct conclusion?

is a good one. The answer is mostly based
on the fact that the complex numbers are
the algebraic closure of the real numbers.
What this means is that in order to provide
all polynomials with real coefficients with
roots we need to introduce complex numbers.
From Wikipedia The fundamental theorem of algebra

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero.
Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.

We can apparently avoid this by restating
the fundamental theorem as

... it follows from the fundamental theorem of algebra that every non-constant polynomial with real coefficients can be written as a product of polynomials with real coefficients whose degrees are either 1 or 2

and in this version the complex numbers
appear implicitly in complex conjugate
pairs as the roots of the degree 2 factors.
Note that the other two complex number
variants that where mentioned earlier do
not provide any non real roots for
real polynomials.

The Wikipedia articles that I link to have
historical information that may be helpful
for you.

ContentLicense: CC BY-SA 4.0


问:
Can't understand the answer to this combinatorics problem
```<code>Consider a deck of 40 cards (10 separate cards for each of the four suits).
Alice, Bianca, and Carlo draw from the deck respectively 5, 4, and 3 cards.
How many are the possible extractions in which Alice and Bianca have not drawn swords, while Carlo has
extracted at least 2?
</code>```

The answer is: $30\choose{5}$$\times$$25\choose{4}$$\times$$($$10\choose{2}$$\times$$21\choose{1}$+$10\choose3$$)$

But I can't understand why $10\choose3$ is inside the parentheses, something like:
$30\choose{5}$$\times$$25\choose{4}$$\times$$($$10\choose{2}$$\times$$21\choose{1}$$)$+$10\choose3$ it seemed more logical to me, because I'm choosing $A+B\iff A \ or \ B \iff (exactly\ 2 \ drawn \ swords) \ or \ (at\ least \ 2\ swords)$

However it's a little different than I thought.

答:
Break the question down into two cases: either Carlo draws exactly $2$ swords or $3$ swords. In the former case, the number of possible extractions is ${30 \choose 5}{25\choose 4}{10\choose 2}{21\choose 1}$, because Alice must draw $5$ non-swords, Bianca must draw $4$, and Carlo must draw $1$ non-sword, and $2$ swords.

In the latter case, the number of possible extractions is ${30 \choose 5}{25\choose 4}{10\choose 3}$, by similar reasoning. Adding these two together, and factoring out the common factor gives the desired answer.


问:
Confusion about notation in this proof
I am looking at the following problem from Munkres' topology and a proof previously asked on this forum.

If $\{\tau_\alpha\}$ is a family of topologies on $X$, show that $\cap \tau_\alpha$ is a topology on $X$.

This was the proof given by a user:

For all $\alpha$, $\varnothing \in \tau_\alpha$ and $X \in \tau_\alpha$. So, $\varnothing \in \cap \{\tau_\alpha\}$ and $X \in \cap \{\tau_\alpha\}$. Now, let $\{U_\beta\}_{\beta \in J}$ be an indexed collection of subsets of $\cap \{\tau_\alpha\}$. Then, for all $\alpha$, $\{U_\beta\}_{\beta \in J} \subseteq \tau_\alpha$. This implies that for all $\alpha$, $\cup\{U_\beta\}_{\beta \in J}\in\tau_\alpha$, since $\tau_\alpha$ is a topology. But then, $\cup\{U_\beta\}_{\beta \in J}\in\cap \{\tau_\alpha\}$.

Now, let $U_1, U_2, \ldots U_n$ be subsets of $\cap \{\tau_\alpha\}$. Then, for all $\alpha$, $U_i \in \tau_\alpha$. This implies that for all $\alpha$, $\cap_{i=1}^n U_i \in \tau_\alpha$, since each $\tau_\alpha$ is a topology. But then, $\cap_{i=1}^n U_i \in \cap \{\tau_\alpha\}$. Therefore, $\cap \{\tau_\alpha\}$ is a topology on $X$.

People told this user the proof was correct, but I am confused on the notation. Wouldn't it be more correct to make the following changes in the proof, or am I confused?

For all $\alpha$, $\varnothing \in \tau_\alpha$ and $X \in \tau_\alpha$. So, $\varnothing \in \cap \tau_\alpha$ and $X \in \cap \tau_\alpha$. Now, let $\{U_\beta\}_{\beta \in J}$ be an indexed collection of elements of $\cap \tau_\alpha$. Then, for all $\alpha$, $\{U_\beta\}_{\beta \in J} \subseteq \tau_\alpha$. This implies that for all $\alpha$, $\cup U_{\beta}$ in$\tau_\alpha$, since $\tau_\alpha$ is a topology. But then, $\cup U_{\beta}\in\cap \tau_{\alpha}$.

Now, let $U_1, U_2, \ldots U_n$ be elements of $\cap \tau_{\alpha}$. Then, for all $\alpha$, $U_i \in \tau_\alpha$. This implies that for all $\alpha$, $\cap_{i=1}^n U_i \in \tau_\alpha$, since each $\tau_\alpha$ is a topology. But then, $\cap_{i=1}^n U_i \in \cap \tau_{\alpha}$. Therefore, $\cap {\tau}_{\alpha}$ is a topology on $X$.

I am mostly confused as to why people giving this user feedback said it was completely correct, I feel like the placement of parenthesis in the unedited proof, make the user's proof very incorrect. Am I right? What is going on here?

答:
It seems your point is that the intersection and union of a family of sets should be written as $\cup U_\alpha$ instead of $\cup \{U_\alpha\}$.

The next paragraph is from the ZFC axioms page on wikipedia:

The axiom of union states that for any set of sets $\mathcal{F}$ there is a set $A$ containing every element that is a member of some member of $\mathcal{F}$:

$\forall \mathcal{F} \,\exists A \, \forall Y\, \forall x [(x \in Y \land Y \in \mathcal{F}) \Rightarrow x \in A]$.

That is, you can think of the union as a 'formula' that assigns to a family of sets the union of all the sets in that family. Seen this way, the notation $\cup \{\tau_\alpha;\alpha\in A\}$ makes sense. As in the question they just called the family $\{\tau_\alpha\}$, they also called it this way in the answer, although it would be better to write the index set.

The notation you use is also accepted and more widely used.

Title: Confusion about notation in this proof


答: If $\mathcal S$ is collection of sets, $\bigcap \mathcal S$ is the same as $\bigcap_{A \in \mathcal S} A$. So, in that problem are abusing the notation by writting $\{\tau_\alpha\}$ instead of $\{\tau_\alpha\}_{\alpha \in J}$ for some index set $J$. Thus, in that case $\bigcap \{\tau_\alpha\}$ refers to $\bigcap \{\tau_\alpha\}_{\alpha \in J}$, which is the same as $\bigcap_{\alpha \in J} \tau_\alpha$, as you say.

Title: Confusion about notation in this proof

Body:
I am looking at the following problem from Munkres' topology and a proof previously asked on this forum.

If $\{\tau_\alpha\}$ is a family of topologies on $X$, show that $\cap \tau_\alpha$ is a topology on $X$.

This was the proof given by a user:

For all $\alpha$, $\varnothing \in \tau_\alpha$ and $X \in \tau_\alpha$. So, $\varnothing \in \cap \{\tau_\alpha\}$ and $X \in \cap \{\tau_\alpha\}$. Now, let $\{U_\beta\}_{\beta \in J}$ be an indexed collection of subsets of $\cap \{\tau_\alpha\}$. Then, for all $\alpha$, $\{U_\beta\}_{\beta \in J} \subseteq \tau_\alpha$. This implies that for all $\alpha$, $\cup\{U_\beta\}_{\beta \in J}\in\tau_\alpha$, since $\tau_\alpha$ is a topology. But then, $\cup\{U_\beta\}_{\beta \in J}\in\cap \{\tau_\alpha\}$.

Now, let $U_1, U_2, \ldots U_n$ be subsets of $\cap \{\tau_\alpha\}$. Then, for all $\alpha$, $U_i \in \tau_\alpha$. This implies that for all $\alpha$, $\cap_{i=1}^n U_i \in \tau_\alpha$, since each $\tau_\alpha$ is a topology. But then, $\cap_{i=1}^n U_i \in \cap \{\tau_\alpha\}$. Therefore, $\cap \{\tau_\alpha\}$ is a topology on $X$.

People told this user the proof was correct, but I am confused on the notation. Wouldn't it be more correct to make the following changes in the proof, or am I confused?

For all $\alpha$, $\varnothing \in \tau_\alpha$ and $X \in \tau_\alpha$. So, $\varnothing \in \cap \tau_\alpha$ and $X \in \cap \tau_\alpha$. Now, let $\{U_\beta\}_{\beta \in J}$ be an indexed collection of elements of $\cap \tau_\alpha$. Then, for all $\alpha$, $\{U_\beta\}_{\beta \in J} \subseteq \tau_\alpha$. This implies that for all $\alpha$, $\cup U_{\beta}$ in$\tau_\alpha$, since $\tau_\alpha$ is a topology. But then, $\cup U_{\beta}\in\cap \tau_{\alpha}$.

Now, let $U_1, U_2, \ldots U_n$ be elements of $\cap \tau_{\alpha}$. Then, for all $\alpha$, $U_i \in \tau_\alpha$. This implies that for all $\alpha$, $\cap_{i=1}^n U_i \in \tau_\alpha$, since each $\tau_\alpha$ is a topology. But then, $\cap_{i=1}^n U_i \in \cap \tau_{\alpha}$. Therefore, $\cap {\tau}_{\alpha}$ is a topology on $X$.

I am mostly confused as to why people giving this user feedback said it was completely correct, I feel like the placement of parenthesis in the unedited proof, make the user's proof very incorrect. Am I right? What is going on here?


问:
Show that $\hat{\delta_h} = e^{-2\pi i h \cdot \xi}$
Show that $\hat{\delta_h} = e^{-2\pi i h \cdot \xi}$, where $\delta_h(\phi) = \phi(h)$

By definition, we know that $\hat{\delta_h}(\phi)=\delta_h(\hat{\phi})=\hat{\phi}(h).$

but $\hat{\phi}(h) = \frac{1}{(2\pi)^{N/2}}\int_{\mathbb{R}^N}\phi(x)e^{-ix\cdot h}dx.$ This is really all I got; I'm not sure how this is supposed to be equal to the desired expression.

Any thoughts would be greatly appreciated.

Thanks in advance.

答: There are two points to be noted. They are using a slightly different definition of FT: $\hat f (h)=\int f(\xi)e^{-2\pi ih.\xi} d\xi$. Once you realize this you can use the following: If $g$ is alocally integrable fuction then we an think of $g$ as  a distribution  which acts on a test fucnction $\phi$ to give the value $\int g(\xi) \phi (\xi) d\xi$. I hope this clear up your confusions.


问:
A curious infinite product
Let $R$  be the ring  $\mathbb{Z[q]}/(q^2)$ whose elements $a+bq$ satisfy $q^2=0.$

Define $g_n\in R$ by $g_1=1$, $g_{2^n}=1-2^{n-1}q$ for $n>0$ and $g_{2n}=0,$ $g_{2n+1}=-q$ else.

It seems that in $R$ the following identity holds in the sense of formal power series with coefficients in $R$:
$$1+x+(1-q)x^2+(1-2q)x^3+(1-3q)x^4+\dots =(1+g_{1}x)(1+g_{2}x^2)(1+g_{3}x^3)\dots.$$
For $q=0$ this reduces to the well-known identity
$$1+x+x^2+x^3+x^4+\dots =(1+x)(1+x^2)(1+x^4)(1+x^8)\dots.$$
Any idea how to prove this?

答:
Write $\epsilon = -q$, which will save us some minus signs. Start by substituting $x \mapsto \left( 1 + \frac{\epsilon}{2} \right) x$ into the well-known identity, which gives

$$\prod_{k \ge 0} (1 + (1 + 2^{k-1} \epsilon)) x^{2^k} = \frac{1}{1 - \left( 1 + \frac{\epsilon}{2} \right) x} = \sum_{n \ge 0} \left( 1 + \frac{n\epsilon}{2} \right) x^n.$$

This incorporates all the power-of-$2$ factors except that the factor for $k = 0$ is slightly off; above it's $(1 + \left( 1 + \frac{\epsilon}{2} \right) x)$ when you ask for it to be $1 + x$. We can correct this by multiplying by $\frac{1 + x}{1 + \left( 1 + \frac{\epsilon}{2} \right) x}$, which gives

$$\prod_{k \ge 0} (1 + g_{2^k} x^{2^k}) = \frac{1 + x}{1 - (1 + \epsilon) x^2} = \sum_{n \ge 0} \left( 1 + \left\lfloor \frac{n}{2} \right\rfloor \epsilon \right) x^{n}.$$

Now we need to incorporate the odd factors. The product of the odd factors only is, by direct expansion,

$$\prod_{k \ge 1} (1 + \epsilon x^{2k+1}) = 1 + \epsilon \sum_{k \ge 1} x^{2k+1} = 1 + \epsilon \frac{x^3}{1 - x^2} = \frac{1 - x^2 + \epsilon x^3}{1 - x^2}$$

and taking the product with the power-of-$2$ factors above gives

$$\prod_{n \ge 0} (1 + g_n x^n) = \frac{1 - x^2 + \epsilon x^3}{(1 - x^2 - \epsilon x^2)(1 - x^2)}.$$

This is not yet in the form we want it; we need to move the $\epsilon$ in the denominator to the numerator by multiplying by the "conjugate" $1 - x^2 + \epsilon x^2$, which gives

$$\begin{eqnarray*} \frac{(1 + x)(1 - x^2 + \epsilon x^3)(1 - x^2 + \epsilon x^2)}{(1 - x^2)^3} &=& \frac{(1 - x^2)^2 + \epsilon (x^2 + x^3 - x^4 - x^5)}{(1 - x)(1 - x^2)^2} \\
 &=& \frac{1}{1 - x} + \epsilon \frac{x^2}{1 - x} \end{eqnarray*}$$

which is the desired result. I don't have a good explanation for why there is so much cancellation at the end.

Title: A curious infinite product


问:
How Many Positive Integers Below $201$ Are Multiples of $4$ But Not Multiples of $10$?
I have recently been starting to work on counting problems, and currently, I am working on this one, but cannot figure it out:

How many positive integers less than 201 are multiples of 4 but not multiples of 10?

I was also wondering if there was a specific formula for this type of problem, so I can fully understand it in the future. Any help would be appreciated, and thanks in advance

答:
You could split this into two successive questions:

How many positive integers below $201$ are multiples of $4$?

How many of those numbers are multiples of $10$?

The first part is fairly easy - they are the numbers $\{4,8,12,16,20,24,\ldots, 196,200\}$.

Now also eliminating those divisible by $10$ is straightforward; the set to remove will look like $\{20,\ldots,200\}$

Hopefully you can work out the size of both these sets to answer the question.

With a little thought, this should also give you a method of calculating the answer in other questions of this sort. It's worth double-checking the end-points of the included/excluded numbers.


问:
Inequality for determinant of sum of two orthogonal matrices
My goal is to show that :

$$\forall A,B\in O_n(\mathbb{R}), |{\rm det}(A+B)| \le 2^n.$$

How can I access the determinant of the sum of two matrices ? I don’t know many ways to establish inequalities in Algebra, except with Bessel’s inequality and Cauchy Schwarz inequality, but they don’t seem to be very helpful here.

答:
Mechanodroid's solution is great. Here's a slightly different way to look at it.

Suppose $\lambda$ is an eigenvalue of $A+B$ with corresponding unit-length eigenvector $v$. Then,

$$|\lambda| = \|\lambda v\| = \|(A+B)v\| = \|Av+Bv\| \le \|Av\|+\|Bv\| = \|v\|+\|v\| = 2\|v\| = 2,$$

where we have used the fact that $\|Av\| = \|v\|$ and $\|Bv\| = \|v\|$ since $A,B$ are orthogonal. Since all $n$ eigenvalues of $A+B$ have magnitude at most $2$, the determinant of $A+B$ which is the product of the $n$ eigenvalues of $A+B$ is at most $|\det(A+B)| \le 2^n$.


问:
Show that a operator is self-adjoint in $\mathbf{R}^{3}$
Let $A:\mathbf{R}^{3}\rightarrow \mathbf{R}^{3}$ be given by $A(x,y,z)=(3x-z,2y,-x+3z)$. Show that $A$ is self-adjoint.

My work, (I consider the inner product like dot product in $\mathbf{R}^{3}$):

Let $w,z\in \mathbf{R}^{3}$ where $w=(x_{1},y_{1},z_{1})$ and $z=(x_{2},y_{2},z_{2})$ and $x_{i},y_{i},z_{i} \in \mathbf{R}$ for $i=1,2$, so:
\begin{eqnarray*}
\langle Aw,z \rangle &=& \langle (3x_{1}-z_{1},2y_{1},-x_{1}+3z_{1}),(x_{2},y_{2},z_{2}) \rangle \\
&=&3x_{1}x_{2}-z_{1}x_{2}+2y_{1}y_{2}-x_{1}z_{2}+3z_{1}z_{2}
\end{eqnarray*}
By other way:
\begin{eqnarray*}
\langle w,Az \rangle &=& \langle (x_{1},y_{1},z_{1}),(3x_{2}-z_{2},2y_{2},-x_{2}+3z_{2}) \rangle \\
&=&3x_{1}x_{2}-z_{2}x_{1}+2y_{1}y_{2}-x_{2}z_{1}+3z_{1}z_{2}
\end{eqnarray*}
But if $A$ is self-adjoint then $\langle Aw,z\rangle=\langle w,Az \rangle$ and I can't see this.

Body:
Your calculations are correct, you just need to reorder the terms. Does this help?

\begin{eqnarray*}
\langle Aw,z \rangle &=& \langle (3x_{1}-z_{1},2y_{1},-x_{1}+3z_{1}),(x_{2},y_{2},z_{2}) \rangle \\
&=&3x_{1}x_{2}-\color{blue}{z_{1}x_{2}}+2y_{1}y_{2}-\color{red}{x_{1}z_{2}}+3z_{1}z_{2}
\end{eqnarray*}
\begin{eqnarray*}
\langle w,Az \rangle &=& \langle (x_{1},y_{1},z_{1}),(3x_{2}-z_{2},2y_{2},-x_{2}+3z_{2}) \rangle \\
&=&3x_{1}x_{2}-\color{red}{z_{2}x_{1}}+2y_{1}y_{2}-\color{blue}{x_{2}z_{1}}+3z_{1}z_{2}
\end{eqnarray*}


答:
Well Let $A = $ the number of positive multiples of $4$ that are less than $201$.

 One out of ever $4$ is a multiple of $4$ there are $200$ positive integers less than $201$ (they are $1,2,3,......, 200$ so there are $\frac {200}{4} = 50$ that are multiples of $4$.  $A = 50$.

Suppose $n$ is both a multiple of $4$ and of $10$ what can we say about $n$?

 If an number is a multiple of $4$ and $10$ then it is a multiple of $\operatorname{lcm}(4,10)=\operatorname{lcm}(2^2, 2\times 5)=2^2\times 5=20$.

Let $B$ be the number of integers that can be the thing in the hint above, that is a multiple of both $4$ and $10$.

  Every $20$ numbers is a multiple of $20$.  so there are $\frac {200}{20} = 10$ multiples of $20$.  So $B=10$.

Then the number of multiples of $4$ that aren't multiples of $10$ will be all the mulitples of $4$ less the multiples of $4$ that are multipls of $10$ so that number is $A-B$.

So your answer is $A-B$.

  $A-B = 50 -10 = 40$.

Title: How Many Positive Integers Below $201$ Are Multiples of $4$ But Not Multiples of $10$?

Body:
I have recently been starting to work on counting problems, and currently, I am working on this one, but cannot figure it out:

How many positive integers less than 201 are multiples of 4 but not multiples of 10?

I was also wondering if there was a specific formula for this type of problem, so I can fully understand it in the future. Any help would be appreciated, and thanks in advance


问:
about embedding or proper embedding
let $f:(0,1] \rightarrow \mathbb{R^2} , f(t)=\left(t \cos \left( \frac{1}{t}\right), t \sin \left(\frac{1}{t}\right)\right )$

is $f$ an immersion, embedding or proper embedding?

$f$ is an immersion because $f^\prime(t)\neq 0$ for all $t \in (0,1]$ and $f$ is one to one.for embedding we must show $f:(0,1] \rightarrow im(f)$ is open map .

答:
Let $D^2 = \{(x,y) \in \mathbb R^2 \mid \lVert (x,y) \Vert \le 1\}$ denote the closed unit disk and define $n : D^2 \setminus \{0\} \to (0,1], n(x,y) =  \lVert (x,y) \rVert$.

We have $\lVert f(t) \rVert =\sqrt{(t \cos( \frac{1}{t}))^2, (t \sin (\frac{1}{t})^2} = t \in (0,1]$, thus $f((0,1]) \subset D^2 \setminus \{0\}$. Then $n(f(t)) = t$ which shows that $f$ is injective (if $f(t) = f(t')$, then $t = n(f(t)) = n(f(t')) = t'$).

Thus $\bar f : (0,1] \stackrel{f}{\to} f((0,1])$ is a continuous bijection; we want to show that it is a homeomorphism (which is equivalent to $\bar f$ being an open map). Let $g = n \mid_{f((0,1])} : f((0,1]) \to (0,1]$. This is a continuous map such that $g \circ \bar f = id_{(0,1]}$ Let $\bar f^{-1} : f((0,1]) \to (0,1]$ denote the inverse of $\bar f$ (at the moment we only know that is a function between sets). Then $\bar f^{-1} = id_{(0,1]} \circ \bar f^{-1} = (g \circ \bar f) \circ f^{-1} = g \circ (\bar f \circ f^{-1}) = g \circ id_{f((0,1])} = g$ which means that $\bar f^{-1}$ is continuous.

Title: about embedding or proper embedding


问:
How to compute $a \ast b \ast c$ or more generally $a_1^{b_1} a_2^{b_2} a_3^{b_3} \cdots$?
I'm currently a self-learner on Abstract Algebra and Group theory. The book that I'm currently reading is A First Course In Abstract Algebra by John B. Fraleigh. So I know this question might truly be a basic question which every student would be taught in class and there is confusion about it. However, maybe it is me who is not careful in reading the book that raises in my mind this question. So given a binary operation $\ast $ on a set $S$, how can we compute:

$$a \ast b \ast c$$
Actually, on page 23 of the book, the author does mention this but I still can not get it. I don't know if we should compute this from left to right or reversely or we will just leave it there if $\ast$ is not associative. Moreover, if this binary operation is associated, is it true to define that:
$$a \ast b \ast c = \left(a \ast b \right)\ast c$$
Then, after learning about Cyclic group there is a problem of which can be stated as below:

Let $a$ and $b$ be elements of a group $G$. Show that if $ab$ has finite order $n$, then $ba$ also has order $n$.

I know that we should show that $(ba)^n = e$ which can be driven from $(ab)^n = e$. Unfortunately, since $G$ might not be an Abelian group so what I had gone so far before looking for a solution is just manipulating $(ab)^n = e$ by using the fact that $(ab)^{-1} = b^{-1} a^{-1}$ since I don't know how to move further after splitting $(ab)^n = (ab)(ab)^{n-1}$. I also get confused whether $(ab)^n = (ab)(ab)^{n-1}$ or $(ab)^n = (ab)^{n-1} (ab)$. So I checked up for the solution and this is the way they manipulate it that I found: Show that if ab has finite order n, then ba also has order n. - Fraleigh p. 47 6.46.
So I completely have no idea about this can be infered:

\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies
\color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e
\end{align}
I also tried to check another link in order to understand the solution: same question And the asker states that:
$$(ab)^n=\underbrace{(ab)(ab)(ab)\cdots(ab)}_{n~\text{copies}}$$
$$=a\underbrace{(ba)(ba)(ba)\cdots(ba)}_{n-1~\text{copies}}b$$
$$=a(ba)^{n-1} b$$
It is here that the number of terms which we're dealing with is $n$ now. I don't have any idea how could he manipulates from the first line to the second line so I think that I need some helps on this. In summary, how can we compute binary operation of more than 3 elements? Pleas explain the problems that I state above and give some other examples if possible. Thank you so much.

答: Conjugation by any element of the group is an automorphism, and thus preserves order.  All that is left is to note that $ab$ and $ba$ are conjugates:  $b(ab)b^{-1}=ba$.


答:
You can do this problem "abstractly".

Observe that the linear map $A$ is the following matrix:
$$
A=\begin{pmatrix}
3&0&-1\\
0&2&0\\
-1&0&3
\end{pmatrix}
$$
and it is symmetric.

For any (row) vector $v\in V:=\mathbb{R}^3$, $A(v)=vA$ where the right-hand side is matrix multiplication.

So now you have
$$
\begin{align}
\langle A(w),z\rangle
&=(wA)z^T\\
&=((wA)z^T)^T&(\textrm{since you have a 1-by-1 matrix})\\
&=z(wA)^T\\
&=z(A^Tw^T)=(zA^T)w^T\\
&=(zA)w^T=\langle w,A(z)\rangle
=\langle A(z),w\rangle
\end{align}
$$

Title: Show that a operator is self-adjoint in $\mathbf{R}^{3}$

Body:
Let $A:\mathbf{R}^{3}\rightarrow \mathbf{R}^{3}$ be given by $A(x,y,z)=(3x-z,2y,-x+3z)$. Show that $A$ is self-adjoint.

My work, (I consider the inner product like dot product in $\mathbf{R}^{3}$):

Let $w,z\in \mathbf{R}^{3}$ where $w=(x_{1},y_{1},z_{1})$ and $z=(x_{2},y_{2},z_{2})$ and $x_{i},y_{i},z_{i} \in \mathbf{R}$ for $i=1,2$, so:
\begin{eqnarray*}
\langle Aw,z \rangle &=& \langle (3x_{1}-z_{1},2y_{1},-x_{1}+3z_{1}),(x_{2},y_{2},z_{2}) \rangle \\
&=&3x_{1}x_{2}-z_{1}x_{2}+2y_{1}y_{2}-x_{1}z_{2}+3z_{1}z_{2}
\end{eqnarray*}
By other way:
\begin{eqnarray*}
\langle w,Az \rangle &=& \langle (x_{1},y_{1},z_{1}),(3x_{2}-z_{2},2y_{2},-x_{2}+3z_{2}) \rangle \\
&=&3x_{1}x_{2}-z_{2}x_{1}+2y_{1}y_{2}-x_{2}z_{1}+3z_{1}z_{2}
\end{eqnarray*}
But if $A$ is self-adjoint then $\langle Aw,z\rangle=\langle w,Az \rangle$ and I can't see this.


问:
Does the factorization operation for composite knots applicable for any projection of the knot?
Let $K$ be a knot in 3-space. We know that if $K$ is not prime (composite) then $K$ can be represented as a connected sum of two knots, both are non-trivial. My question is : does this apply to any arbitrary projection of $K$ in a plane? In other words, if I take any projection of $K$, then I can decompose it into two non-trivial knots by the decomposition operation or in some projections, we have to apply some Reidemeister moves first before applying the decomposition operation.

答:
I will make an assumption about what you mean by "decompose it into two non-trivial knots by the decomposition algorithm." Let's call a diagram composite if there is a simple closed curve in the plane intersecting the diagram exactly twice away from the crossings such that both the interior and exterior of that simple closed curve contain crossings of the diagram. I will interpret your question as follows:

Rephrased question: Are all diagrams of composite knots composite?

If I've interpreted your question correctly, then the answer is no. Here is a counterexample of the connected sum of two trefoils.


问:
How to approximate the integral $\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx$ for a constant $\mu \gg 1$
I think this integral has no exact solution, but I've found that you can approximate it for large $\mu$ as:

$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx \approx \sqrt{\frac{\pi \mu}{2}} e^{-\mu}$$

This is used a couple of times in a book I'm reading. I've checked it numerically and it works, but I wanted to get the symbolic expression.

I've tried expanding the square root:
$$\sqrt{x^2-\mu^2}=\mu\sqrt{-(1-x^2/\mu^2)}\approx i\mu \left(1-\frac{1}{2}\frac{x^2}{\mu^2}-\frac{1}{8}\frac{x^4}{\mu^4}-...\right)$$

That means that I get an imaginary solution as far as I know, because the first term is a simple logarithmic integral:

$$\int_{\mu}^{\infty} \frac{i\mu}{e^x-1}dx=-i\mu\ln{|e^{-\mu}-1|}$$

And the following terms are integrals that I don't know how to solve. So I guess that this is not the way to proceed.

Any ideas on how to do this approximation, or a more exact solution? Appreciate your help.

答:
With the substitution $z=x-\mu$, the integral becomes

$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1}\,dx=\int_0^\infty \frac{\sqrt{z^2+2\mu z}}{e^{\mu+z}-1}\,dz=\frac{\sqrt{2\mu}}{e^\mu }\int_0^\infty\frac{\sqrt{z+z^2/(2\mu)}}{e^z-e^{-\mu}}\,dz$$

Taking the limit $\mu\to\infty$ on the integrand and substituting $w=\sqrt{z}$, the integral becomes  $$\int_0^\infty \sqrt{z}\,e^{-z}\,dz=\int_0^\infty 2w^2 e^{-w^2}\,dx=\int_{-\infty}^\infty w^2 e^{-w^2}\,dw.$$
This is a standard variation on the Gaussian integral and evaluates to $\sqrt{\pi}/2$. Hence $$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1}\,dx\sim \frac{\sqrt{2\mu}}{e^\mu}\frac{\sqrt{\pi}}{2}=\sqrt{\frac{\pi u}{2}}e^{-\mu}$$ in the $\mu\to\infty$ limit as claimed.


问:
Sudoku puzzle with exactly 3 solutions
While published sudoku puzzles typically have a unique solution, one can easily conceive of a sudoku puzzle with two solutions. However, is it possible to construct a sudoku puzzle with exactly 3 different solutions?

Inspired by https://puzzling.stackexchange.com/a/6.

答: I think it's possible. This adaptation of Andrew Stuart's tough example has two unique rectangles interacting via a triple. Let me know if this doesn't work! starting sudoku grid</p>

Title: Sudoku puzzle with exactly 3 solutions


问:
On the equality of two generating functions related to plane partitions
I'd like to prove
$$\prod_{(i,j,k)\in\mathcal{B}(r,s,t)}\frac{1-q^{i+j+k-1}}{1-q^{i+j+k-2}}=\prod_{i=1}^r\prod_{j=1}^s\frac{1-q^{i+j+t-1}}{1-q^{i+j-1}},$$
where
$$\mathcal{B}(r,s,t)=\{(i,j,k):1\leq i\leq r,1\leq j\leq s,1\leq k\leq t\}.$$
I'm guessing that we could find a bijection between products of terms from the left-hand side and terms from the right-hand side, but how I am not sure how. Of course there might be something easy/obvious that I am missing. Any help or general hints on how to approach problems like this would be appreciated!</p>

答:
The left hand side is the same as

$$\prod_{i=1}^r\prod_{j=1}^s\prod_{k=1}^t\frac{1-q^{i+j+k-1}}{1-q^{i+j+k-2}}$$

Now just consider

$$\prod_{k=1}^t\frac{1-q^{i+j+k-1}}{1-q^{i+j+k-2}}=\frac{1-q^{i+j+t-1}}{1-q^{i+j+1-2}}$$

because a lot of the factors cancel each other out.


问:
Is a rational and projectively normal curve a rational normal curve?
I am confused about the notion of "rational normal curve".

First of all, if a projective curve is rational and normal, then using the Zariski's Main Theorem, we can conclude that it is isomorphic to $\mathbf{P}^1$. Is there an easier way to see this? Or we really need to translate the argument in ZMT to get an elementary proof?

As for the rational normal curve, I am using the standard definition (See Wikipedia page https://en.wikipedia.org/wiki/Rational_normal_curve). It says: "The term "normal" refers to projective normality, not normal schemes." Projectively normality is necessary because Hartshorne II Exercise 3.18(b) provided such example: $(x, y) \to (x^4, x^3y, xy^3, y^4)$. Is it a sufficient condition? In other words, is every rational and projectively normal curve in $\mathbf{P}^n$ comes from an $d$-uple embedding: $\mathbf{P}^1 \hookrightarrow \mathbf{P}^d \subseteq \mathbf{P}^n$?

In this wikipedia page: https://en.wikipedia.org/wiki/Normal_scheme, there is a stronger statement: "linearly normal ... is the meaning of "normal" in the phrases rational normal curve and rational normal scroll." Is every linearly normal and rational curve a rational nomal curve? I try to find this result in standard textbook, but none of Hartshorne, Shafarevich or Harris explain the terminology. Maybe it is hided in some Exercise...

========

After some search, I saw two results which might count as a proof. In this question Definitions of "linearly normal" variety, it is said that "$V$ as projective variety cannot be obtained by an isomorphic linear projection from a projective space of higher dimension, except in the trivial way of lying in a proper linear subspace." On the other hand, by Shafarevich I P158, "any rational map from $\mathbf{P}^m$ is obtained by composing the Veronese map with a projection". Combining the two facts, we see that a linear normal rational curve is a rational normal curve. Is there a standard reference for the first fact?

答: Yes, so you can argue like this: First of all projective normality implies normality, but for a curve that amounts to smoothness, hence this curve is isomorphic to $\mathbb{P}^{1}$, but the only linear series on $\mathbb{P}^{1}$ are those $\mathcal{O}(n)$. But now projective normality says that the embedding is given by the complete linear series (equivalent to saying that it does not come from a projection) for those $n>0$, which is the definition of rational normal curve.


问:
How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$?
This is a claim in a textbook that I'm using, although it was not proven.

When trying this for small $n$'s it is clear.

\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}

What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?

答:
In binary, this is

$$1+1+10+100+...+1\underbrace{00...00}_{n-2\text{ times}}=1+\underbrace{11...11}_{n-1\text{ times}}=1\underbrace{00...00}_{n-1\text{ times}}$$

which in decimal is $2^{n-1}$.


答:
Let me suggest 4 ways:

$ a^{n}-b^{n}=(a-b)(a^{n-1} + ba^{n-2}+ ... + b^{n-1}) \Rightarrow 2^{n} = 2^{n} - 1 + 1 = 1+2+ ... +2^{n-1}  +1 $

Considering all subsets from set with $ n>1 $ elements and considering map between subsets and binary numbers

Mathematical Induction

Newton's binomial theorem

Title: How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$?

Body:
This is a claim in a textbook that I'm using, although it was not proven.

When trying this for small $n$'s it is clear.

\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}

What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?


答:
It's because it's a geometric progression. Since

$$1+p_i+{p_i}^2+··· +{p_i}^{a_i} =  \dfrac{{p_i}^{a_i+1}-1}{p_i-1}. $$

you just add an extra one at the start of the progression.


问:
Can all naturals can be expressed as sums of prime powers of $\{3,5,7,13\}$?
Define the set $$S=\left\{(p,k) \mid p\in\{3,5,7,13\} \land k \in\mathbb Z^{0+}\right\}.$$

The question is whether every natural number $n$ can be expressed as the sum of some subset $s\subseteq S$ of distinct pairs $(p,k)$ such that $$n=\sum_{(p,k)\in s}{p^k}.$$ To be clear, any pair in $S$ may be used at most once in the sum, though $k=0$ is allowed as an exponent. As an example, when using only the primes $\{3,5,7\}$, $24$ has two valid representations:

$$ 3^1+3^2+5^1+7^1 = 3^0+3^2+5^0+5^1+7^0+7^1 = 24. $$

Another way to describe the problem is to ask whether every natural can be described as the sum of four numbers, one each in bases 3, 5, 7, and 13, using only $0$ and $1$ digits. So equivalently, in that formulation, $$110_3+10_5+10_7=101_3+11_5+11_7=24.$$

Using $\{3,5,7,13\}$ gives 20 representations for $24$, which seemed a little spammy to include.

This post is a direct follow-up to this one, which posed the same question for $p\in\{3,5,7\}$. A nice proof was given that that set isn't sufficient to cover the naturals; the bottom line is that this arises because $$\frac{1}{3-1}+\frac{1}{5-1}+\frac{1}{7-1}=\frac{11}{12}<1.$$

This leaves open the interesting edge case of whether it would be possible for $\{3,5,7,13\}$, as $$\frac{1}{3-1}+\frac{1}{5-1}+\frac{1}{7-1}+\frac{1}{13-1}=1.$$

Note: To explain comments below, I originally thought I had a counterexample, but that was a rounding error. I'm leaving this question up as it seems like it still has merit. Clever arguments as to why $\{3,5,7,13\}$ should or should not work are also welcome.

Update

It actually seems like you may be able to limit this to proper prime powers after all, i.e. disallowing $p^0$ terms, for all $n\geq 113$. The only numbers I can find at first glance that require them are $\{1, 2, 4, 6, 11, 26, 31, 107, 112\}$ (though this is not proven!) So if you confine yourself to $n \geq 113$, each term $p^k$ in the sum must be distinct, no caveats required.

This seems like a cleaner formulation of the problem, but it's unclear to me which version would is more likely to be tractable or of greater interest. Input on this welcome.

答:
I cannot confirm your counterexample. Here are the details so nobody else has to duplicate this effort. Following the method from Brian Moehring's answer, we want to find a value of $n$ such that $\log_5(3^n - 1), \log_7(3^n - 1), \log_{13}(3^n - 1)$ all have fractional parts close to $1$. If their integer parts are $n_5, n_7, n_{13}$, it then suffices to show that

$$\sum_{p = 3, 5, 7, 13} \left( \sum_{k=0}^{n_p} p^k \right) = \sum_{p = 3, 5, 7, 13} \frac{p^{n_p + 1} - 1}{p - 1} < 3^n - 1.$$

With $n = 338993010578$ we get (noting that $3^n$ is never a power of any other prime so $\lfloor \log_p(3^n - 1) \rfloor = \lfloor \log_p(3^n) \rfloor = \lfloor 3 \log_p n \rfloor$) we get (computations done in WolframAlpha)

$$n_3 = n - 1 = 338993010577$$
$$n_5 = \lfloor n \log_5 3 \rfloor = 231398728907$$
$$n_7 = \lfloor n \log_7 3 \rfloor = 191386990490$$
$$n_{13} = \lfloor n \log_{13} 3 \rfloor = 145196584918$$

and now it remains to show the desired inequality. If we try to compute the sums above directly we may run into rounding errors; we can do the following instead. First, since $\sum_{p = 3, 5, 7, 13} \frac{1}{p - 1} = 1$, we can add $1$ to both sides to convert the desired inequality into the slightly simpler form

$$\sum_{p= 3, 5, 7, 13} \frac{p^{n_p + 1}}{p - 1} < 3^n.$$

Second, we can divide both sides by $3^n$ to convert the inequality into the form

$$\sum_{p=3,5,7,13} \frac{3^{(n_p+1) \log_3 p - n}}{p - 1} < 1.$$

From here we see that to get the desired inequality it suffices (again using the fact that $\sum_{p=3,5,7, 13} \frac{1}{p-1} = 1$) that $3^{(n_p+1) \log_3 p - n} \le 1$ for each $p$ and that the inequality is strict for at least one $p$. Taking logs, equivalently we want $(n_p + 1) \log_3 p \le n$ or

$$n \log_p 3 \ge n_p + 1.$$

Unfortunately this is false for $p = 5, 7, 13$ since by definition we have $n \log_p 3 < n_p + 1$ (and we have equality for $p = 3$). So $3^{(n_p + 1) \log_3 p - n} > 1$ for $p = 5, 7, 13$ which implies that the inequality in Brian Moehring's answer cannot hold in this case for any value of $n$. The method can be salvaged if you show that one of the largest powers can't appear in the sum but this seems tricky.


问:
Cusp containment (Shimura Proposition 2.4)
In Shimura's Introduction to Arithmetic Theory of Automorphic Functions, the following is given:

Proposition 2.4: Let $\Gamma,\Gamma'<SL_2(\mathbb{R})$ be Fuchsian and $\alpha\in GL_2^+(\mathbb{R})$ such that $\alpha\Gamma\alpha^{-1}<\Gamma'$ with finite index. Then $f\mapsto f\mid[\alpha]_k$ gives a $\mathbb{C}$-linear injections $A_k(\Gamma')\hookrightarrow A_k(\Gamma)$, $G_k(\Gamma')\hookrightarrow G_k(\Gamma)$, and $S_k(\Gamma')\hookrightarrow S_k(\Gamma)$, which are surjective if $\Gamma'=\alpha\Gamma\alpha^{-1}$.

Proof: Let $C$ (resp. $C'$) denote the set of cusps of $\Gamma$ (resp. $\Gamma'$). Then $\alpha(C)=C'$ and our assertion follows immediately from the definition.

The containment $\alpha(C)\subset C'$ is relatively straightforward to show, but where does the reverse containment $\alpha(C)\supset C'$ come from without using the equality at the end of the Proposition statement? (Or does the finite index have something to do with this?)

Additionally, it seems that the non-surjection case can be proved without the reverse containment. Is it the case that that containment is only needed for the surjections?

Thanks in advance for any answers.

答:
The cusps of $\Gamma'\backslash\Bbb{H}$ are the same as those of $\alpha \Gamma \alpha^{-1}\backslash\Bbb{H}$, except that the orbits under the action of $\Gamma'$ on those cusps are larger than the orbits under the action of $\alpha \Gamma \alpha^{-1}$.

The cusps of $\Gamma\backslash\Bbb{H}$ correspond to those of $\alpha \Gamma \alpha^{-1}\backslash\Bbb{H}$ through $z\to \alpha(z)$.


问:
Series of logarithms $\sum\limits_{k=1}^\infty \ln(k)$ (Ramanujan summation?)
I had this question earlier, so to say as a "standalone" problem, but now it pops up in context of an analysis with the lngamma-function. As well as we can convert the question of sums of like powers $Su_p(n)=1^p+2^p+3^p+\cdots+n^p$ in terms of the Hurwitz-zeta $Su_p(n) = \zeta(p,1)-\zeta(p,n+1)$ (and solve using the Bernoulli-polynomials) I try to express this for sums of logarithms (and powers of logarithms) : $$ Sl_p(n) = \ln(1)^p+\ln(2)^p+\cdots+\ln(n)^p $$

I have seemingly proper coefficients for power series which allow that computations/approximations. The result is always found by the difference of $ Sl_p(n)-Sl_p(1) $ . Here the power series has zero as constant term.

However, to have the analogue to the Hurwitz-zeta I should have the constant term the "value" for the infinite sum $ Sl_p(\infty) $ instead. (This cancels properly if I formulate the sum of (powers of) logarithms always as that difference $ Sl_p(n)-Sl_p(1) $) And also, I'd like that this agreed conceptually more to the notion:
$$ \begin{array} {rll} Sl_p(n)-Sl_p(1) &=& T_p(0)-T_p(n) \\\
 & =&  (\ln(1)+\ln(2)+\ln(3)+\cdots) \\\
 & & - (\ln(n+1)+\ln(n+2)+\ln(n+2)+\cdots) \end{array} $$

as this agrees more with the idea of the Hurwitz-zeta-difference.

But this requires, that $T_p(0)$ represents the (infinite) sum of the $p$'th powers of the logarithms, and the constant in the power series of $T_p(0)$ must contain just such a value. Let's talk about the first powers of the logs $p=1 $ first and denote the assumed sum as $L_1$ : $ L_1 = \lim_{n\to \infty} T_1(0) $

[update 2]

I think, thanks to the hint of J.M., I can answer Q1 myself now; only Q2 remains somehow vague - besides a simple empirical heuristic I did still not get the formally correct approach to the constant term/integral-definition in the Ramanujan summation - but this is now only a side problem here (however it would be nice to get help also for this question).

With the help of the knowledge about the derivatives of the zeta at zero the relevant part of the problem could now satisfyingly be solved, so I put it here in an answer to my own question, see that answer below....

Final remark/conclusion: it is interesting, that the power series for the lngamma pops up here "automatically" - we need no other uniqueness criterion for the argument, that the (Eulerian) gamma-function gives "the correct" interpolation for the factorial problem. It is just the result of the construction of an operator for the "indefinite summation" (which was the comceptual goal from where the problem/question arose initially)

In the following I keep the rest of the original question although the approach to the Ramanujan summation contains an error

[end Update 2]

I tried to give sense to that divergent series by replacing the powers of $x$ in the Mercator series for $ \ln(1+x) $ by appropriate zetas at negative integer arguments. If I understand things correctly then this is similar to the method of Ramanujan summation, where the Bernoulli numbers are just replaced by the according zeta-values (appropriately scaled). With this I got then an approximation of $$L_1 \approx -0.0810614667953 $$

which seems a rather "random" value...

By searching in other online sources I got the suggestion (OEIS), that this is also $$ \int_1^2 \ln(\Gamma(t)) \; dt \approx -0.08106146679532725821967026 $$

With this my power series for $T_1(x)$ begins like

$ \qquad \small \begin{array} {l}
 - & 0.0810614667953 \\
 - & 0.577215664902x \\
 + & 0.533859200973x^2 \\
 + & 0.325578788221x^3 \\
 + & 0.125274140308x^4 \\
 + & 0.0337256506589x^5 \\
 + & 0.00685935357296x^6 \\
 + & 0.00117260810356x^7 \\
 + & O(x^8)
 \end{array}
$

The other coefficients occur also in the power series for $f(x)=\ln(\Gamma(\exp(x)) $

Also, $L_1$ seem to satisfy the following expression:
$$ \exp(L_1) = {\sqrt{(2\pi)} \over e} \approx 0.9221370088957891168791517 $$

The questions are:
$\qquad$   Q1: Is that value $-0.0801\ldots$ a meaningful (or even correct) representation for the infinite sum of logarithms?

$\qquad$   Q2: Did I reproduce the Ramanujan summation correctly here?

[update 1]: J.M. mentions the relation to $ y= - \zeta'(0) \approx 0.918938533205 $ where the representation of the derivative of zeta at 0 equals formally just the sum of logarithms. Now my $L_1$ and the $y$ are related by $L_1 = y - 1$. So I expect some error in my derivation which leads just to that unit difference...

答:
I think, thanks to the hint of J.M. I can answer Q1 myself now; only Q2 remains somehow vague - besides a simple empirical heuristic I did still not get the formally correct approach to the constant term/integral-definition in the Ramanujan summation - but this is now only a side problem here (hoewever it would be nice to get help also for this question).

With the help of the knowledge about the derivatives of the zeta at zero the relevant part of the problem could now satisfyingly be solved, so I put it here in an answer to my own question.

Final remark/conclusion: it is interesting, that the powerseries for the lngamma pops up here "automatically" - we need no other uniqueness criterion for the argument, that the (Eulerian) gamma-function gives "the correct" interpolation for the factorial problem.

Since the infinite series of like powers of logarithms can be expressed by the derivatives of $ \zeta(n) $ at $n=0$ we have expressions for $L_p$ and thus for $T_p(0)$ .

$$  L_p = T_p(0) = (-1)^p \zeta^{(p)}(0) = \ln(1)^p + \ln(2)^p + \ln(3)^p + \ldots $$

To actually compute that derivatives I use the powerseries representation of zeta using the Stieltjes-constants:
$$ \zeta(z) =  {1 \over z-1} + \sum_{k=0}^{\infty} (-1)^k \cdot \gamma_k \cdot { (z-1)^k \over k!} $$
where the $\gamma_k$ are the Stieltjes constants, beginning with $\gamma_0$ and the values $ \approx [0.5777...,-0.072..., -0.096...] $

The derivatives are then expressed by termwise differentation, giving
$$ \begin{eqnarray} \zeta^{(p)}(z) &=&
 ({1 \over z-1})^{(p)} &+& \sum_{k=p}^{\infty} \binom kp \cdot p!\cdot (-1)^k \cdot \gamma_k \cdot { (z-1)^{k-p} \over k!} \\
&=& - (-1)^p {p! \over (z-1)^{p+1} }& +& (-1)^p \sum_{k=0}^{\infty} (-1)^k \cdot \gamma_{k+p} \cdot { (z-1)^k \over k!} \\
&=& {p! \over (1-z)^{p+1} }& +& (-1)^p \sum_{k=0}^{\infty} \gamma_{k+p} \cdot { (1-z)^k \over k!} \\
\end{eqnarray} $$

Thus
 $$  \begin{eqnarray}
 L_p = T_p(0) &=&  (-1)^p*( {p! \over 1^{p+1} }&
             +& (-1)^p \sum_{k=0}^{\infty} \gamma_{k+p} \cdot { 1^k \over k!} ) \\
&=& (-1)^p*p! & +& \sum_{k=0}^{\infty} {\gamma_{k+p} \over k!}
\end{eqnarray} $$

The explicite sum in the last expression converges very well and can be computed to as many digits as the accuracy of the Stieltjes numbers allows. With this I get the logsums $L_0 = T_0(0) $ to $L_{23}=T_{23}(0) $ (read from left to right):

$ \qquad \small \begin{array} {rr}
 -0.500000000000 & 0.918938533205 \\
 -2.00635645591 & 6.00471116686 \\
 -23.9971031880 & 120.000232908 \\
 -720.000936825 & 5039.99915018 \\
 -40320.0002324 & 362880.000331 \\
 -3628799.99946 & 39916800.0004 \\
 -479001600.000 & 6227020800.00 \\
 -87178291200.0 & 1.30767436800E12 \\
 -2.09227898880E13 & 3.55687428096E14 \\
 -6.40237370573E15 & 1.21645100409E17 \\
 -2.43290200818E18 & 5.10909421717E19 \\
 -1.12400072778E21 & 2.58520167389E22
 \end{array} $

where the sequence of the deviation from the signed factorials is

$ \qquad \small \begin{array} {rr}
 0.500000000000 & -0.0810614667953 \\
 -0.00635645590858 & 0.00471116686225 \\
 0.00289681198629 & 0.000232907558455 \\
 -0.000936825130051 & -0.000849823765002 \\
 -0.000232431735512 & 0.000330589663612 \\
 0.000543234115780 & 0.000375493172907 \\
 -0.0000196035362810 & -0.000407241232563 \\
 -0.000570492013282 & -0.000393927078981 \\
 0.0000834588058255 & 0.000660943729629 \\
 0.00102622728654 & 0.000865575776779 \\
 0.0000192936717837 & -0.00135690605213 \\
 -0.00269215645875 & -0.00305138562124 \\
 -0.00142429184942 & 0.00270778921289 \\
 0.00860288096928 & 0.0135561620310 \\
 0.0127785133267 & 0.000576026175993 \\
 -0.0264657041471 & -0.0641102553488 \\
 -0.0940806022511 & -0.0801887691867 \\
 0.0263889614064 & 0.263594454732 \\
 0.608778480504 & 0.904597405553 \\
 0.786784764403 & -0.316856127028 \\
 -2.99127389406 & -7.27888475259 \\
 -11.6472353086 & -11.6164312194 \\
 1.20065709453 & 37.8116918598 \\
 105.106316717 & 189.660315159
 \end{array} $

(I displayed more numbers here because the divergence of the sequence cannot be seen from the leading numbers only!)

A/the matrix $\Lambda$ contains then in each column the coefficients of the powerseries to compute $T_p(x)$ by the dot-product of a vandermonde rowvector of the type $V(z)=[1,z,z^2,z^3,...]$ with the column $p$ of $\Lambda$:

$ \qquad V(\ln(1+x))~ * \Lambda = \sum_{k=x}^\infty V(\ln(1+k))~ = [T_0(x),T_1(x),T_2(x),\ldots] $

and the sum of consecutive powers of the logarithms are taken by the difference of two $T_p(n)$-computations:

$ \qquad Sl_p(n) = T_p(0) - T_p(n) = \sum_{k=1}^n \ln(k) $

One of the reasons for that long derivations was, that the n are not required to be integer here, analoguosly to the solution for the sum of like powers by means of the Bernoulli polynomials.

Appendix:

Here is the top-left-segment of the matrix $\Lambda$:

$ \qquad \small
\begin{bmatrix}
 -0.5000000 & 0.9189385 & -2.006356 & 6.004711 & -23.99710 & \ldots \\
 -1.000000 & 0.5772157 & -0.1456317 & -0.02907109 & 0.008215338  & \ldots \\
 -0.5000000 & -0.5338592 & 0.6345700 & -0.1858111 & -0.06050253  & \ldots \\
 -0.1666667 & -0.3255788 & -0.3868589 & 0.6713043 & -0.2104926  & \ldots \\
 -0.04166667 & -0.1252741 & -0.2407114 & -0.3127569 & 0.6958103  & \ldots \\
 -0.008333333 & -0.03372565 & -0.09916203 & -0.1918968 & -0.2670350  & \ldots \\
 -0.001388889 & -0.006859354 & -0.02847304 & -0.08146653 & -0.1603119  & \ldots \\
 -0.0001984127 & -0.001172608 & -0.005923793 & -0.02452507 & -0.06891184  & \ldots \\
 -0.00002480159 & -0.0001831833 & -0.0009884023 & -0.005280337 & -0.02143180  & \ldots \\
 -0.000002755732 & -0.00002257650 & -0.0001620035 & -0.0008613236 & -0.004778981 & \ldots \\
 \ldots & \ldots  & \ldots  & \ldots  & \ldots  & \ldots
 \end{bmatrix}
$


问:
Can I compare the cardinality of two uncountable sets
If I have an infinite set, say $B=(0,1)$, and I have $A \subsetneqq B$, can I say $|A|<|B|$, where $|\cdot|$ denotes cardinality?

On one hand, for every element in $A$, we can find a corresponding element in $B$; but the inverse is not true. Thus, it seems to suggest that $|A|<|B|$.

On the other hand, it also seems that we cannot use such intuitive methods to compare the cardinality of these two sets because they are uncountable.

I hope someone can help me clarify my doubts. The reason I'm wondering this is that I hope to know whether I can write a rigorous mathematical expression to describe $A$ as the largest subset of $B$ that satisfies some constraints (let's just call the constraint $f(A)=c$ here) as the following

$$
A = \arg\max_{A\subset B, f(A)=c} |A|
$$

答:
No. Two sets have the same cardinality iff there is a bijection between them. Let $A=(0,1/2)\subsetneq B$. There is a bijection $$f:A\to B:x\mapsto 2x.$$

Even for countable set, this intuition is also false. Let $S$ be the set of all even integers. Then $S\subsetneq\mathbb{Z}$ but $|S|=|\mathbb{Z}|$ with the same map $n\mapsto 2n$.

That is also why infinite sets are more confusing that finite sets. In fact, some definition of infinite sets also says that

A set $A$ is infinite if there is a proper subset $B$ of $A$ such that $|B|=|A|$.


Body: Let $$x = \sum_{m=2}^{n}2^{m-2},$$ and note that $$1 + x = 1 + 1 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{1}.$$  Then, multiplying $(1)$ by $2$, we have $$2+2x = 2 + 2 + 4 + \dotsb + 2^{n-2} + 2^{n-1}\tag{2}.$$  We can rewrite $(1)$ as $$1 + x = 2 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{3}.$$  Subtracting $(3)$ from $(2)$ gives us $$1 + x = 2^{n-1},$$ which is the same as $$2^{n-1} = 1 + 1 + 2 + 4 + \dotsb + 2^{n-2}.$$


问:
Can someone please check my answers for these questions regarding improper integrals?
a) Evaluate $$\int^{1}_{0.5} \frac{1}{x^2}dx$$

my answer: $-1+\frac{1}{0.5}=1$

b) Explain why  $$\int^{-1}_{1} \frac{1}{x^2}dx$$ is undefined.

my answer: since the function $\frac{1}{x^2}$ is undefined at $x=0$, the integral is undefined.

c) Evaluate $$\int^{k}_{1} \frac{1}{x^2}dx$$ for some positive number $k$

my answer: $\int^{k}_{1} \frac{1}{x^2}dx = \frac{-1}{x} |^{k}_{1} = \frac{-1}{k}+1$

d) Take the limit of your answer in part c as $k\rightarrow \infty $ What value would you assign $\int^{\infty}_{1} \frac{1}{x^2}dx$?

my answer: $\lim_{k\to\infty} [\frac{-1}{k}+1]=1$

$\frac{-1}{x} |^{\infty}_{1}=1$

答:
a) Fine.

b) I would add that the function isn't bounded.

c) You switched the order of the integration limits by mistake.

d) error follows from previous part.


问:
The solution to a Cauchy-Euler differential equation
Problem statement: Assume that $y=x^2 \ln (x)$ is a solution to the differential equation
$ x^2y'' +axy' +by=0$. What is the value of $a+2b$?

Solution attempt: It is known that a Cauchy-Euler equation has a solution of the form $y=cx^m\ln(x)$ if the characteristic polynomial of the equation has a repetitive root. The polynomial is:
$$m(m-1) + am +b= m^2 + (a-1)m +b= 0. $$
To force repetitive roots, the so-called delta has to equal $0$. That is,
$$ (a-1)^2-4b=0\Rightarrow (a-1)^2 = 4b \Rightarrow a-1 = \pm 2b. $$
The last equation in particular implies $a+2b = 1. \square$

However, my solution manual says that $a+2b = 5$ without providing the details. Can you help me understand this?

Body:
$$ m^2 + (a-1)m +b= 0.$$
Since $m=2$ is a solution we deduce that:
$$b=-2(1+a)$$
The disctiminant is zero $\Delta=0 $:
$$(a-1)^2-4b=0$$
$$(a-1)^2+8(a+1)=0$$
$$(a+3)^2=0$$
$$ \implies a=-3$$
$$(a,b)=(-3,4)$$
And
$$a+2b=5$$


答:
What you suggest basically seems to work - I'm going to change the leading value into $n$ rather than $n-1$ for ease of reading:

First, $2^1 = 1+1$ and $2^2 = 1+1+2 = 1+2^0+2^1$

Then using induction, assuming that the summation holds for lower $n$, we can use your observation $2^n = 2^{n-1} + 2^{n-1}$ and change the first $2^{n-1}$ into the summation form, giving  $2^{n} = 1+1 + 2 + \cdots + 2^{n-2} + 2^{n-1}$ as required.


问:
Minimal polynomials over the rationals and the reals
Find the minimal polynomial over $\mathbb Q$ and $\mathbb R$ for ...$\sqrt[3]{3}$, $1- i\sqrt{3}$, $2 + i$, $i\sqrt[3]{3}$

Sorry for my sqrt formulas .. I'm new here, hope to learn really fast to write a correct question in math symbols.

I have to find both minimal polynomial over $\mathbb Q$, as well over $\mathbb R$ ... It's kind of homework ... Any good explanations will be nice and very appreciated. Thanks

答:
I'll do the first one for you as hint.

Let $\alpha = \sqrt[3]{3}$

Let's get rid of that pesky radical by cubing both sides:

$\alpha^3 = 3$

Then move 3 over to the other side:

$\alpha^3 - 3 = 0$

Let $f = x^3 - 3$. Now f is irreducible in $\mathbb Q[x]$  by Eisenstein's Criterion (p=3) and Gauss' lemma. So the minimal polynomial for $\sqrt[3]{3}$ over $\mathbb Q$ is $m_\sqrt[3]{3} = x^3 - 3$.

The rest follow a very similar procedure.


问:
What is the sample space, outcomes, event space, random variables in machine learning?
Reading through materials of machine learning problems, I see people treating things like they are doing with probability.

Particularly, consider linear regression. I cannot figure out what is the sample space, outcomes, events, random variables. In what manner they are using the word "probability measure" in the field of machine learning?

For instance, please take a look at this article.

Moreover, in general, why do we need probability for machine learning? I don't see how we are going to calculate any probability of something in machine learning. While the need of linear algebra is obvious, since we are working on lists matrix of numbers. What properties of probability make it essential for machine learning?

答:
I'd like to answer your question in the context of computer vision. An example would be image classification, where the labels of each image are mutually exclusive (one label per image). Lets say there are $n$ classes.

The input features are the image/s itself (which can be preprocessed in a multitude of ways such as rgb channel normalization, resizing, various data augmentation schemes etc.). The features are transformed through the convolutional layers of a backbone structure (such as VGG-$16$, Alexnet, or Resnet), and then fed through a Multi-Layer Perceptron to a Sigmoid layer to ultimately get a probability vector say $p$ with $n$ entries, one per class (sometimes $n+1$ for a background class, depending on if there are "empty" images).

During training, using a cross-entropy loss function, we compute for a single image (ignoring batching for now)

$\sum_{i=1}^{n} - g_{i} log(p_i)$ where $g_i$ is $1$ for an image with ground truth label $i$, and $0$ otherwise. As $p_i$ approaches $1$, we get a smaller loss, and hence a good prediction. As $p_i$ approaches $0$, we get a large loss, and hence a bad prediction. The network is learning to model the true probability distribution $p(Class_{i}|Image)$ which is $1$ for images with a classification label belonging to class $i$ and $0$ otherwise.

The optimizer (Stochastic Gradient Descent, Adam, etc.) then takes gradients of the model variables in the network that require gradient updates with respect to the loss, and updates the weights according to the optimizer algorithm.

For some other computer vision uses of probability...

In object detection, confidence scores (probabilities) are assigned to the detections that are output by a model, in addition to bounding boxes. This is done in order to threshold which detections will be evaluated in the computation of mean average precision or mean average recall, which are the evaluation metrics commonly used.

For a detection whose bounding box IOU is larger than some threshold, say $0.5$, for a given ground truth box with category label $c$, the model should learn to have high confidence values reflecting how certain the prediction is. The probability thresholds yield a trade-off between precision and recall. Set it too low? You will have a less precise model. Set it too high? You will miss predictions.

In semantic segmentation, for an image of size $(600, 600)$, with say $C$ classes (including the background), your model should get a feature map tensor of size $(600, 600, C)$ where by choosing at each pixel the channel value ($C$ value) that has maximal output, we choose which class the pixel belongs to for our inference prediction.


问:
Are proofs by contradiction really logical?
Let's say that I prove statement $A$ by showing that the negation of $A$ leads to a contradiction.

My question is this: How does one go from "so there's a contradiction if we don't have $A$" to concluding that "we have $A$"?

That, to me, seems the exact opposite of logical. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true".

答:
The best way to get a perspective on the issue - as already seen by many of the replies here - is to pull back from classical logic to something more general and to address the matter in the more general setting. The reason is that classical logic flattens out important distinctions that lend the needed perspective on the matter and clarify what is going on. So, we need to consider a more general setting where those distinctions are not conflated.

In reality, what you're actually asking - in reverse - is what kind of logic permits the rule you cited. When you ask "why", you also need to have something on the playing field where the "why not" is also the case: and that's what makes it necessary to step back. Otherwise, the answer would simply be "the rule is a rule in classical logic, because that's just what classical logic is" and that would be that.

But actually, even a foray into intuitionistic logic isn't enough. To get a better view, still, of the matter, you should also broaden the scope to bi-intuitionist logic. I will show you, first, a comparison of the three and where they fit in terms of the question and see where everything fits together.

There are actually two senses of the word "not" that are in common use; and people do actually make a distinction between the two. The weaker version of "not A" is "A is not true", while the strong one of "not A" is "A is false". The most familiar example is seen in political discourse, where respected adversaries disagree not by saying the other person's point "is false", but only that it is "not true".

In classical logic, no such distinction exists: they're the same.

In intuitionistic logic, only "is false" sense can be formulated: formalizing the "is false" sense of "not A" as ¬A = A⊃⊥. This is expressed in terms of the "only if" connective ⊃ (i.e. A⊃B means "A only if B", or equivalently "if A then B") and the "false" statement ⊥. So, let's call the stronger version of not "refute".

A key property - the defining property in fact - of the conditional is that there is a one-to-one correspondence between inferences C∧A ⊢ B (that is: inferences of B from the conjunction "C and A") and inferences C ⊢ A⊃B (i.e. inferences of the conditional A⊃B from C alone). The two are equivalent.

Applied to the "refute" operator, this expresses a one-to-one correspondence between C∧A ⊢ ⊥ and C ⊢ ¬A ... voilá! There's part of what you're looking for: the positive version of proof by contradiction. If C and A together yield a contradiction then from C alone we may infer the strong "A is false" version of "not A" - that we: we may refute A, from C.

If you use this version of "not", then the best you can achieve from this is to start with the inference C∧¬A ⊢ ⊥ and conclude that C ⊢ ¬¬A. In classical logic, this is good enough, because you have the double-negative rule. But this doesn't really answer your question: it only shifts the question over to the double-negative rule and puts the spotlight on that, instead. And, so you don't get any real insight on the matter other than "it holds in classical logic, but not in intuitionistic logic ... because they're different."

You actually want the negative version of proof by contradiction, that whenever the inference (the context of Other Stuff) and (not A) ⊢ ⊥ holds, then we also have the inference (the context of Other Stuff) ⊢ A. That's what your question is alluding to.

Collecting all the "Other Stuff" into C, this reads as the rule that given the inference C∧(not A) ⊢ ⊥ we should also have the inference C ⊢ A.

Do we? Well, that's where we get into bi-intuitionistic logic; because that's actually a characterization of the other, weaker version of "not"!

Bi-intuitionistic logic is a form of logic that lies between intuitionistic and classical logic. It includes the dual versions of the "not" and "only if" operators; which we will denote here respectively: A⊂B for "A unless B", ⨽B = ⊤⊂B, for the "B is not true" version of "not B". We'll call this the "denial" operator.

Here, the property that characterizes the "unless" connective ⊂ is that there is a one-to-one correspondence between inferences A ⊢ B∨D and A⊂B ⊢ D. That is: if we can infer the disjunction B∨D (i.e. B or D) from A, then by excluding B, we may infer D, alone, i.e. from "A unless B".

When applied to the denial operator, this leads to the rule that if we can infer ⊤ ⊢ B∨D (that is: if we can actually assert that B or D is true), the we can equivalently infer ⨽B ⊢ D: i.e. we can infer D alone from the denial of B. Now you're almost where you need to be.

In bi-intuitionistic logic, a denial may be inferred from a refutation. That is: ¬A ⊢ ⨽A is a valid inference: "is not true" follows from "is false".

The instant you assert the converse as a rule, you have classical logic. Classical logic is equivalent to bi-intuitionistic logic plus the rule ⨽B ⊢ ¬B that equates denial with refutation. And that's where important distinctions start to collapse.

The threshold from intuitionistic or bi-intuitionistic logic to classical logic is actually a watershed. If any of a large number of rules is added to either form of logic, the result is equivalent to classical logic. And every single rule in that watershed is a rule that has had a long controversial history attached to it.

My favorite one is the "continuation rule", which generalizes the equivalence rule for conditionals by attaching a "continuation" (∨D) to the right. It says that the inference C∧A ⊢ B∨D is equivalent to the inference C ⊢ (A⊃B)∨D. If you add that rule to intuitionistic logic, you get classical logic.

In many cases, the watershed rules have reflections expressed in terms of the dual operators that are valid in bi-intuitionistic logic, but don't cross the watershed. So, we can begin to see the distinctions that classical logic flattens out.

First, from the trivial valid inference ¬A ⊢ ¬A, we have ¬A∧A ⊢ ⊥. The "rule of contradiction" applies to the "refute" operator. In contrast, if the same rule is expressed in terms of the denial operator B∧⨽B ⊢ ⊥, then it's actually one of the above-mentioned watershed rules that pushes you into classical logic.

Second, from the trivial valid inference ⨽B ⊢ ⨽B, we have ⊤ ⊢ B∨⨽B. The "excluded third" rule holds for the "denial" operator. But if we try to impose the same rule on the "refute" operator ⊤ ⊢ ¬A∨A, then the result (once again) is another watershed rule and we've landed in classical logic.

Bi-intuitionism plus the rule B∧⨽B ⊢ ⊥ equals classical logic.

Intuitionism plus the rule ⊤ ⊢ ¬A∨A also equals classical logic.

Your version of proof by contradiction rule (where expressed in terms of the "refute" operator) ... as we saw above ... is tantamount to the double-negative rule; and that's one of the watershed rules that pushes you into classical logic.

In contrast, when your version of the proof by contradiction rule is expressed in terms of the "denial" operator, the result is a weaker rule that actually does hold in bi-intuitionistic logic, but can't be expressed in intuitionistic logic, because it is using the extra "denial" and "unless" connectives.

To see this, suppose that taking C and the denial of B together yields a contradiction. That is: suppose we have the inference C∧⨽B ⊢ ⊥. Then, we may switch the order of the conjunction to rewrite this as ⨽B∧C ⊢ ⊥ and then apply the conversion for the "refute" operator to rewrite this equivalently as ⨽B ⊢ ¬C and apply the conversion for the "denial" operator to write this, in turn, equivalently as ⊤ ⊢ B∨¬C.

So, now let's add the supposition of C, itself, and attach it as a conjunction. Then we have C ⊢ C∧⊤ ⊢ C∧(B∨¬C) ⊢ (C∧B)∨(C∧¬C), where we apply distributivity in the last step.

Since we already have the rule of contradiction ¬C∧C ⊢ ⊥, then switching the order of the conjunction, we have the inference C∧¬C ⊢ ⊥. Applying this to the above result, we get C ⊢ (C∧B)∨(C∧¬C) ⊢ (C∧B)∨⊥ ⊢ C∧B ⊢ B.

So, the answer to your question is: Yes. Your rule is logical - but only with the weaker "denial" operator in bi-intuitionistic logic. And the answer to the implied, underlying, question (what logic is the rule valid in) is: bi-intuitionistic logic.

Many of the "watershed" rules that push you from either intuitionism or bi-intuitionism over into classical logic have weaker versions that are written by replacing one version of the "not" operator with the other. So, we can actually segregate the rules by which "not" they work with and draw a very finely-graded distinction between each of the watershed rules.

One further example I will give here is that the double-negative rule holds in one direction A ⊢ ¬¬A for the "refute" operator and in the other direction ⨽⨽B ⊢ B for the "deny" operator. The converse direction only holds for negatives; i.e. ¬¬¬A ⊢ ¬A and ⨽B ⊢ ⨽⨽⨽B are both valid.

(Correspondingly, we may then say that in intuitionistic logic, the "regular" statement - i.e. the statements equivalent to their double-negatives - are one and the same as the "negative" statements; and that classical logic is what results from intuitionistic logic by allowing only negative statements; or more briefly: classical logic is pessimistic).

There is actually an entire hierarchy of statements that may be generated from a single statement C, by repeatedly applying different combinations of ⨽ and ¬ to C; so the word "not" acquires a very nuanced meaning in bi-intuitionistic logic. So, there is plenty of room to draw many distinctions between different nuances of each of the watershed rules, as we just did above.


问:
Pole order of $\frac{4}{27}\frac{\left(\lambda^2-\lambda+1\right)^3}{\lambda^2\left(1-\lambda\right)^2}\left(=j(\tau)\right)$
Concerning the relation
$$j=\frac{4}{27}\frac{\left(\lambda^2-\lambda+1\right)^3}{\lambda^2\left(1-\lambda\right)^2},$$
I understand, that the RHS is an element of $\mathbb{C}(j)$, and thus the LHS can be expressed as a polynomial in $\lambda$. I also understand, that if the order of the pole of the RHS at infinity is equal to 1, the highest degree of the polynomial can only be 1.

But no matter how I approach the RHS, the poleorder always seems to be two. For instance: If $\tau \to i\infty$, then $\lambda(\tau)\to 0$, thus we have a $0^2$ in the denominator. Thus the poleorder should be 2.

I suspect, that if I can express $\lambda$ as a 1 periodic Fourie series, that the problem may resolve, but my skills seem to be failing me.

Any help would be greatly appreciated.

答: No $j(\tau)$ is not a polynomial in $\lambda(\tau)$. You already have a rational function expression in $\lambda(\tau)$, it is unique.


问:
Curve of number of attempts to gain a result with probability p
Recently, I've been reinvigorated to get a better intuition for probability and statistics by its several simultaneously vexing and intriguing paradoxes. One of which I've discovered through my interest in programming.

I have a program that will generate a random number between 0 and 1 over and over again until it is greater than some number N, counting the number of times it does this. It then keeps a list of each possible number of tries it could have taken and counts them up. In essence, it's repeating random chance 1-N until it gets the result and telling me how many times the program was able to get it in X attempts.

Now, originally I was expecting a bell curve around the number 1/p where p would be 1-N, or just the probability. However, I was completely surprised to find out that the most common result was only 1, and it trails off in probability from there! The average number of attempts is still 100, I suppose, but it's simply baffling to me that the graph comes out this way. For an example of why it seems paradoxical to me, take the case of N = 0.99, so P = 0.01 or 1 in 100. It will be more likely that you have a result before fifty attempts than it will be that you have a result between 75 to 125 attempts.

I don't know if anyone here is able to shed light on the topic, as it may just come down to poor intuition, but I felt like it was striking enough to post a question regarding it. Below I will post a graph of my results from 10,000,000 trials where X is the number of attempts before getting the result, and Y is how many times it took exactly that number of attempts.

答:
Your $N$ is presumably in $[0,1)$, i.e. $0 \le N \lt 1$, making $p=1-N$ be the probability of exceeding $N$, so $0 \lt p \le 1$

What you have in $X$ is a geometrically distributed random variable as the number of attempts until you exceed $N$.  The probability mass function for $X$ is $$\mathbb P(X=x) = (1-p)^{x-1}p$$ where $x \in \{1,2,3,\ldots\}$ since you have to initially fail $x-1$ times and then succeed $1$ time.  You get the sort of exponentially decreasing curve you illustrated from your simulation. You can extend this to the cumulative distribution function $$\mathbb P(X\le x) = 1- (1-p)^{x}$$

Your thought that it will be more likely that you have a result before $50$ attempts than it will be that you have a result between $75$ and $125$ attempts is basically correct since the latter involves failing $75$ times and then succeeding in the next $49$ attempts and that is $$\mathbb P(75 \lt X \lt 125) = \mathbb P(X \le 124) - \mathbb P(X \le 75) \\=(1- (1-p)^{124})-(1-(1-p)^{75})= (1-p)^{75}-(1-p)^{124}  = (1-p)^{75}(1-(1-p)^{49}) \\= \mathbb P(X\ge 75)  \mathbb P(0 \lt X \lt 50) \\ \lt \mathbb P(0 \lt X \lt 50)$$ i.e. less than just having to to succeed in the first $49$ attempts, and I find this an intuitively sensible result.


问:
Combinatorial/ Binomial Identity
I am finding difficulty in showing the following identity:

$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1} = \frac{2}{n+1}.$$

Mainly because of the inverse part.

答:
Recall that we have the beta integral

$$\int_0^1 t^a (1 - t)^b \, dt = \frac{1}{(a + b + 1) {a+b \choose b}}.$$

Substituting $b = k, a = 2n-k-1$ gives that the sum can be rewritten

$$2n \sum_{k=1}^n {n-1 \choose k-1} \int_0^1 t^{2n+k-1} (1 - t)^k \, dt = 2n \int_0^1 \left( \sum_{k=1}^n {n-1 \choose k-1} t^{2n-k-1} (1 - t)^k \right) \, dt.$$

The sum in the integrand can be reindexed  via the substitution $k \mapsto k+1$ to give

$$\sum_{k=0}^{n-1} {n-1 \choose k} t^{2n-k-2} (1 - t)^{k+1} \, dt = t^{n-1} (1 - t) (t + (1 - t))^{n-1} = t^{n-1} - t^n$$

which is easily integrated, giving

$$2n \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{2}{n+1}.$$

Title: Combinatorial/ Binomial Identity


问:
Payoff from an option contract
In period 1 the consumer of type $\theta$ selects an option contract consisting of an up-front fee, $B>0$, and exercise price, $\bar{R}$. The consumer pays $B$ at the end of the first period. In period 2, he realises his valuation, $\theta$, distributed on $[0,1]$ by CDF $G(\theta)$ with density $g>0$. His expected payoff, from choosing contract $(B, \bar{R})$, should thus be $-B+\int_{\bar{R}}^{1}(\theta-\bar{R})g(\theta)d(\theta).$ However, in the paper I am told that it is instead: $-B+\int_{\bar{R}}^{1}(1-G(\theta))d\theta.$ Are these two expressions equivalent? What does it mean to integrate a CDF in this fashion? Thank you.

答:
The CDF satisfies $G(0) = 0$, $G(1) = 1$ and $G'(\theta) = g(\theta)$.

Integrating by parts, with $u = 1- G(\theta)$ and $dv = d\theta$, we have $du = -G'(\theta)\,d\theta = - g(\theta)\,d\theta$, $v = \theta$, and

$$\int_{\bar{R}}^1(1- G(\theta)) \, d\theta = \left.(1- G(\theta))\theta\right|_{\bar{R}}^1 + \int_{\bar{R}}^1 \theta\,g(\theta) \, d\theta = -(1- G(\bar{R}))\bar{R} + \int_{\bar{R}}^1 \theta\, g(\theta) \, d\theta\\=-\bar{R} \int_{\bar{R}}^1 g(\theta)\, d\theta+ \int_{\bar{R}}^1 \theta\, g(\theta) \, d\theta = \int_{\bar{R}}^1 (\theta-\bar{R}) \, g(\theta) \, d\theta$$

Title: Payoff from an option contract


问:
Question about eigenvalues
I have this :

i dont understand why  they write $\lambda=m^2 , m\in \mathbb{N}\cup\lbrace0\rbrace$ ,

it's right that $\lambda=m^2$ is the eigenvalues of $(P_0)$ ,but $0$ is not an eigenvalue !.

after that why they  write $\mathbb{N}\cup \lbrace0\rbrace$ 0 already belongs to $\mathbb{N}$ !

Please, help me

Thank you.

答:
Zero is an eigenvalue. Let $x(t)=1$ to see this.

Whether or not zero is a natural number depends on your definition of $\mathbb N$ which is not standardized. Either convention can be used.

Title: Question about eigenvalues


问:
Let $X$ be a continuos r.v and X > 0 is this true? $\mathbb{E}(\frac{1}{X}) = \frac{1}{\mathbb{E}(X)}$
Let $X$ be a continuos r.v and X > 0 is this true?
$\mathbb{E}(\frac{1}{X}) = \frac{1}{\mathbb{E}(X)}$

答:
Not in general:

$E\left(x\right)\:=\:\sum _{i=1}^n\:\left(x_i\cdot p_i\right)$

$E\left(\frac{1}{x}\right)\:=\:\sum _{i=1}^n\:\left(\frac{1}{x_i}\cdot p_i\right)$

Now, what happens when you plug numbers into that?
Say X = {1,2,3,4} and p = {1/4, 1/4, 1/4, 1/4}. Then:
E(X) = 2.5 whilst E(1/X) = 25/48.

However, that ws a mistake for omitting the key to the question: CONTINUOUS RV. Thus:

$E\left(x\right)=\int _{-\infty }^{\infty }\left(x\cdot f\left(x\right)\right)\:dx;\:E\left(\frac{1}{x}\right)=\int _{-\infty \:}^{\infty }\left(\frac{1}{x}\cdot \:f\left(x\right)\right)\:dx$

The intuition is retained from above. Most any other example will prove that E(x) doesn't equal E(1/x), except (maybe) for a some coincidences that may appear here and there.


Body:
Let $S:=1+2+4+\dots+2^{n-2}$.

Then, since it's a geometric series, we have $S=\dfrac{1-2^{n-1}}{1-2}=2^{n-1}-1\implies S+1=2^{n-1}$.


问:
Principal normal vector of a parabolic path is not orthogonal
I have a parameterized equation of a path that looks like a parabola, $\textbf{r}(t) = \langle t, t^2 \rangle $. The principal normal vector is $\textbf{N}(t) = \frac{\textbf{T}'(t)}{|\textbf{T}'(t)|}$. Without normalizing, which doesn't change the direction of the vector, we can find the tangent vector $\textbf{T}(t) \propto \langle 1, 2t \rangle$, and the normal vector $\textbf{N}(t) \propto \langle 0, 2 \rangle$. At any time t, the normal vector $\textbf{N}(t)$ and the tangent vector $\textbf{T}(t)$ are not orthogonal to each other. I read somewhere that it has to do with curvature. Is there an easier way to understand why this simple path's tangent and normal vectors are not orthogonal to each other?

The theorem in the Briggs Calculus textbook states:

Let $\textbf{r}$ describe a smooth parameterized curve with unit tangent vector $\textbf{T}$ and principal unit normal vector $\textbf{N}$. Then, $\textbf{T}$ and $\textbf{N}$ are orthogonal at all points of the curve; that is, $\textbf{T} \cdot \textbf{N} = 0$ at all points where $\textbf{N}$ is defined.

I would argue that this path satisfies the conditions to this theorem, so not sure why the two vectors wouldn't be orthogonal.

答:
While it's true that the vector $\langle{1,2t}\rangle$ is in the same direction as the unit vector $\textbf{T}(t)$, the scalar multiple, $c(t)$ say, varies as a function of $t$.

But in your attempt to find a vector in the same direction as $\textbf{T}'(t)$ by taking the derivative of the vector $\langle{1,2t}\rangle$, you are implicitly assuming that $c(t)$ is truly constant (i.e., independent of $t$).

To verify the orthogonality of $\textbf{T}(t)$ and $\textbf{N}(t)$ for the curve $\textbf{r}(t)=\langle{t,t^2}\rangle$ we can proceed as follows . . .$\\[10pt]$
\begin{align*}
\textbf{r}'(t)&=\langle{1,2t}\rangle\\[2pt]
|\textbf{r}'(t)|&=\sqrt{1+4t^2}\\[16pt]
\textbf{T}(t)&=\frac{\textbf{r}'(t)}{|\textbf{r}'(t)|}\\[2pt]
&=
\left\langle
\frac{1}{\sqrt{1+4t^2}}
,
\frac{2t}{\sqrt{1+4t^2}}
\right\rangle
\\[16pt]
\textbf{T}'(t)
&=
\left\langle
-\frac{4t}{(1+4t^2)^{\frac{3}{2}}}
,
\frac{2}{(1+4t^2)^{\frac{3}{2}}}
\right\rangle
\\[2pt]
|\textbf{T}'(t)|
&=
\frac{2}{1+4t^2}
\\[16pt]
\textbf{N}(t)&=\frac{\textbf{T}'(t)}{|\textbf{T}'(t)|}\\[2pt]
&=
\left\langle
-\frac{2t}{\sqrt{1+4t^2}}
,
\frac{1}{\sqrt{1+4t^2}}
\right\rangle
\\[16pt]
\textbf{T}(t){\,\cdot\,}\textbf{N}(t)
&=
\left(\frac{1}{\sqrt{1+4t^2}}\right)
\left(-\frac{2t}{\sqrt{1+4t^2}}\right)
+
\left(\frac{2t}{\sqrt{1+4t^2}}\right)
\left(\frac{1}{\sqrt{1+4t^2}}\right)
\\[3pt]
&=
\;0
\\[2pt]
\end{align*}


问:
Actuarial Problem Help
So this is an official Exam P practice problem:

The density of a manufacturer's loss, X,  is given by

$$
f_{X}(x) = \frac{2.5(0.6)^{2.5}}{x^{3.5}}
$$
for x > 0.6 and 0 otherwise.

The manufacturer has an insurance policy with a deductible of 2, after which point all loss is covered. Calculate the expected value of all loss not covered by the insurance company, Y.

I understand how to solve this problem; you split up the integral as such:

$$
E(Y) = \int_{0.6}^{2} xf_X(x)dx + \int_{2}^{\infty}2f_X(x)dx
$$

and you end up getting 0.9343, the correct answer to the problem.

What I'm having trouble doing is connecting this problem back to the theory of conditional expectation that I know.  Y is definitely conditional on X. Y|(0.6<X<2) = x and Y|(X>2) = 2 and 0 otherwise.  How would you define the conditional density of Y given X, and how would you make sense of the conditional expectation formula in the context of this problem?

答:
The conditional distribution of $\ Y\ $ given $\ X\ $ is always concentrated at a single atom, namely $\ X\ $ if $\ X<2\ $, or $2$ if $\ X\ge2\ $. It therefore doesn't have a density (unless you allow  generalized functions as "densities"), and
$$
E(Y|X)=\cases{X&if $\ X<2$\\
              2&if $\ X\ge2\ $.}
$$
You then have
\begin{align}
E(Y)&=E(E(Y|X))\\
&=\int_{0.6}^\infty E(Y|X=x)f_X(x)dx\\
&=\int_{0.6}^2 xf_X(x)dx+\int_2^\infty2f_X(x)dx\ ,
\end{align}
as you've already found.

Title: Actuarial Problem Help


问:
Showing that $ZFC$ can't prove the existence of Strongly Inaccessible Cardinals. Why the extra steps?
It is possible I am missing something exceedingly obvious. We know that $V_{\kappa} \vDash ZFC$ when $\kappa$ is strongly inaccessible.

Pretty much every proof of $ZFC \not \vdash "\exists$ inaccessible cardinals" goes like:

Suppose $ZFC \vdash "\exists$ strong inaccessibles" (I'll shorten this from now one as $\exists I$). Then pick the least such one, $\kappa$. $V_{\kappa} \vDash "\exists I"$. By standard absoluteness, what $V_{\kappa}$ thinks are strongly inaccessible will indeed be strongly inaccessibles, contradicting the choice of $\kappa$ as the smallest such cardinal.

And this argument is fine to me. But I don't see why we have to go through the trouble of estabilishing absoluteness, picking the smallest one and everything. Why can't we simply do:

Suppose $ZFC \vdash "\exists I"$. Since we may formalize model theory, the truth predicate for set models, consistency statements, etc, we have that $ZFC \vdash "\exists I" \rightarrow Con(ZFC)$ (by exhibiting a set model of $ZFC$). Then we have by basic modus ponens that $ZFC \vdash Con(ZFC)$. This contradicts Godel's incompleteness theorem and then we're done.

This argument seems fine to me, and more natural too. However, most arguments I can see seem to follow the first argument instead. And the first one isn't hard per se, but it makes me wonder why we're doing all these extra steps.

Am I missing something? Does the second argument not work?

答:
They both work. However, I don't actually think that the second one is more natural: it involves first proving the second incompleteness theorem, which is nontrivial and not obviously relevant to the specific problem at hand.

By contrast, in the first approach we give a method for building models of $\mathsf{ZFC}$ + "There are no inaccessibles" from models of $\mathsf{ZFC}$. Moreover, this method "preserves niceness" in many senses, e.g. if the starting $\mathsf{ZFC}$-model is well-founded then the $\mathsf{ZFC\neg I}$-model is well-founded as well. Note that this is not something incompleteness can do for us alone - e.g. there is no nice way to take a well-founded model of $\mathsf{ZFC}$ and produce a well-founded  model of $\mathsf{ZFC+\neg Con(ZFC)}$ since there are no well-founded models of $\mathsf{ZFC+\neg Con(ZFC)}$ in the first place.

So I think the first approach actually is the better one.


问:
$R_0$ and normal implies completely regular
We say a space ($X, \tau$) is $R_0$ if for each $x,y \in X$ such that $\overline{\{x\}}\neq\overline{\{y\}}$ there exists open sets with $x\in U-V$ and $y \in V-U$.

If for each $F, K \subseteq X$ disjoint closed sets there exists disjoints open sets $U$ and $V$ such that $F \subseteq U, K \subset V$ we say that the space is normal.

And if for each $x \in X$ and $F \subset X$ closed such that $x \notin F$ there exists $f:X \rightarrow [0,1]$ continuous with $f(x)=0$ and $f[F]=\{1\}$ we say that the space is completely regular.

I want to show:

Theorem: If a space is $R_0$ and normal, then is completely regular.

I've tried to show that with this conditions the space must be $T_1$.  If  $\overline{\{x\}} \neq \overline{\{y\}}$ for any pair of distinct points then I'm done, this is equivalent to be $T_0$, but I don't know if this is always true?

答:
The space need not be $T_1$. Let $X=\{0,1,2\}$ and $\tau=\big\{\varnothing,\{0\},\{1,2\},X\big\}$; then $\langle X,\tau\rangle$ is $R_0$ and normal but not $T_0$ (and hence not $T_1$). And it is completely regular but not Tikhonov.

HINT: To prove the theorem, use Urysohn’s lemma: if $F$ and $K$ are disjoint closed set in $X$, there is a continuous $f:X\to[0,1]$ such that $f[F]=\{0\}$ and $f[K]=\{1\}$. Suppose that $x\in X$, $F\subseteq X$ is closed, and $x\notin F$. Let $K=\operatorname{cl}\{x\}$. If $K\cap F=\varnothing$, we can immediately apply Urysohn’s lemma to get the desired result.

If not, let $y\in K\cap F$. $X\setminus F$ is an open nbhd of $x$ that does not contain $y$, so $x\notin\operatorname{cl}\{y\}$, and therefore $\operatorname{cl}\{x\}\ne\operatorname{cl}\{y\}$. Now use the hypothesis that $X$ is $R_0$ to get a contradiction, showing that in fact $K\cap F=\varnothing$.


问:
Is the projective dimension of modules with finite projective dimension bounded?
I know that a Noetherian local ring $R$ has finite global dimension if and only if $R$ is regular in which case $\mathrm{gldim}{R} = \dim{R}$. Therefore, for regular rings, every module has projective dimension at most $\dim{R}$. When $R$ is not regular there must exist modules with infinite projective dimension.

My question is:

For $R$ Noetherian local but not regular, do there exist modules with finite but arbitrarily large projective dimension or are the projective dimensions of modules with $\mathrm{pd}(M) < \infty$ bounded? If so, are they bounded by $\dim{R}$?

答:
I'll assume all rings are commutative. In this case, your question was answered precisely by Auslander and Buchsbaum in "Homological dimension in local rings". Link!

The answer is yes: these finite projective dimensions are bounded by the Krull dimension of $R$!


问:
Simplifying Formulae
I'm doing an 8th grade math text book and I came across this simple problem:

$$T = 2\pi  \sqrt{\frac{l}{g}} \text{ then }T\text{ is _________  proportional to }\sqrt l$$

I had actually solved this, but now I forgot how I solved this.
I came till $\frac{T \sqrt g}{2\pi} = \sqrt l$, but I could not conclude.

please help

答:
$$T=2π\sqrt{\frac{l}{g}}\quad \text{is the given equation.}$$
$$\text{Since }\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$
$$\text{We can write }\sqrt{\frac{l}{g}} \text{ as } \frac{\sqrt{l}}{\sqrt{g}}$$
$$\text{therefore we can write  } T=2π\frac{\sqrt{l}}{\sqrt{g}}$$
$$\text{Multiplying by } \sqrt{g} \text{ and dividing by 2π, we get your answer of } \frac{T\sqrt{g}}{2π}=\sqrt{l}$$

Title: Simplifying Formulae


问:
equality in the Boolean algebra of the power set of the power set
Let $A$ be a Boolean algebra, let $p$ be a probability measure an $A$.
Let $N = \{1,2,3,...,n\}$ for a fixed $n \in \mathbb{N}$, and let $\epsilon$ be a function from the set of all subsets of $N$ (the power set $2^N$ of $N$) into real numbers $\epsilon: 2^N \rightarrow \mathbb{R}$ with the assumption that $\epsilon(\emptyset) = 0$.

For each such function define a function $p_{\epsilon}$ from the cartesian product $A^n$ into the real numbers: $p_{\epsilon}(A_1, \dotsc, A_n):= \sum\limits_{S \subseteq N} \epsilon (S) p \left(\cap_{i \in S}A_i\right)$ for all $A_1, \dotsc, A_n \in A$. ($\cap$ denotes the Boolean operation of the meet in $A$)

Now choose for $A$ the Boolean algebra of all the subsets of the set $2^N$ (this algebra has $2^{2^n}$ elements). The elements $S \subseteq N$ are the atoms of $A$.

Take $A_i = \cup_{\{i\}\subseteq S \subseteq N}S, \, i = 1, 2, ..., n$.

Then the following equation holds:

$p_{\epsilon}\left(A_1, \dotsc, A_n\right) = \sum\limits_{T \subseteq N}\left(\sum\limits_{S \subseteq T} \epsilon(S)\right)p(X_T)$, $\quad $where $X_T = T$ by the definition of $A$.

My question now is how you can see that the equality does hold?

Besides, how does the Boolean algebra $A$ look like? I mean, it seems to be the power set of the power set of $N$. What do its elements and the operations of meet, join and complement look like? Moreover, if $S \subseteq N$, then why is $S$ an atom of $A$? Wouldn't it have to be of the form $\{S\}$? (I'm familiar with the power set as a Boolean algebra, but not with the power set of the power set.)

答:
There is nothing unusual about the Boolean algebra $A$; it is just the usual power set Boolean algebra of the set $2^N$, with union and intersection and set complement as its operations.  The fact that $2^N$ happens to itself be a power set is irrelevant.

It appears that there is either a typo or a rather unfortunate abuse of notation that is confusing you: when it refers to $S$ as being an atom of $A$, it should indeed be $\{S\}$ instead.  Similarly, the definition of $A_i$ should be $$A_i = \bigcup_{\{i\}\subseteq S \subseteq N}\{S\}.$$  In other words, $A_i$ is just the set of all subsets of $N$ that contain $i$.

So, this means that for $S\subseteq N$, $\bigcap_{i\in S}A_i$ is just the set of all $T\subseteq N$ such that $S\subseteq T$.  We can write this as the union of the singletons $\{T\}$ for each such $T$.  So, since $p$ is additive, $$p\left(\bigcap_{i\in S}A_S\right)=\sum_{T\supseteq S}p(\{T\})$$ and thus $$p_\epsilon(A_1,\dots,A_n)=\sum_{S\subseteq N}\epsilon(S)\sum_{T\supseteq S}p(\{T\})=\sum_{T\subseteq N}\sum_{S\subseteq T}\epsilon(S)p(\{T\}).$$


问:
Showing the existence of a function $M \rightarrow L.$
Let $R$ be a commutative ring. For $R-$modules $L,M,N$ show that the following conditions are equivalent.(all functions are $R-$ module homomorphisms.)

a- $M \cong_{R} L \oplus N.$

b- There exists a left-split short exact sequence $0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$

Here is my trial:

$a \implies b.$

Let $R$ be a commutative ring. And let $L,M,N$ be $R-$modules.

Assume that $M \cong_{R} L \oplus N.$ We want to show that there exists a left-split short exact sequence $$0 \rightarrow{L} \rightarrow{M} \rightarrow{N} \rightarrow{0.}$$ i.e., $\exists$ a retraction $r: M \rightarrow{L}$ s.t.$$r \circ i = id_L \quad (1)$$ Where $i: L \rightarrow M $ and it is injective. So we need to find the functions $r,i$ that satisfies $(1)$ above.

So, since $M \cong_R L \oplus N,$ we can say that we have $0 \rightarrow L \rightarrow L \oplus N  \rightarrow N \rightarrow 0,$ with $i:  L \rightarrow L \oplus N$ the embedding $i(l) = (l,0)$ and  $p:  L \oplus N \rightarrow N$ the projection $p(l,n) = n.$

If we define $r:  L \oplus N  \rightarrow N $ with $r(l,n) = l$ then we have $(r\circ i) (l) = r((l,0)) = l$ as required.

Now, we want to verify that $r$ is an $R-$module homomorphism:

1- $r((l_{1},n_{1}) + (l_{2}, n_{2})) = r((l_{1} + l_{2}, n_{1} + n_{2})) = l_{1} + l_{2} = r((l_{1}, n_{1})) + r((l_{2},n_{2}))$ where the first equality is by definition of addition in direct sum.

2- Let $r' \in R$ and let $m \in M,$ then $r(r'm) = $ but then I do not know how to complete. could anyone help me in this step please? also, is my solution correct?

Also I do not know how to prove the reverse direction, could anyone help me in this please?

答: Given $r' \in R$ and $(l,n) \in L \oplus N$, $$r(r'(l,n)) = r((r'l,r'n)) = r'l = r'r((l,n)),$$ and this shows that $r(r'm) = r'r(m)$ for every $r' \in R$ and $m \in L \oplus N$. Now, for $b \!\implies\! a$, tag the left-split short exact sequence $0 \to L \to M \to N \to 0$ as $$0 \longrightarrow L \stackrel{i}{\longrightarrow} M \stackrel{p}{\longrightarrow} N \longrightarrow 0$$ and let $r : M \to L$ be the $R$-module homomorphism such that $r \circ i = \operatorname{id}_L$. Show that the induced homomorphism $(r,p) : M \to L \oplus N$ given by $(r,p)(m) := (r(m),p(m))$ is an isomorphism using the fact that $r$ and $p$ are surjective (why?) and that $\ker r \cap \operatorname{im} i = 0$.


问:
If τ is the finite-closed topology, then the semi-open sets are precisely the open sets
If τ is the finite-closed topology, then the semi-open sets are precisely the open sets.

We can assume the set is infinite, otherwise the proof is trivial. I can easily show that open sets are semi-open. Any open set A are element of X without a finite number of points. If we remove another point, it is still an open set, O, and is a subset of A. The closure of O is the entire set because that is the smallest closed set that contains O since every closed set is finite. So every open set fits the criterion for being semi open. Now I need to show that if there are any other semi open sets in the topology, is is just part of the open sets. I am not sure how to show this.

答: Let $X$ be an infinite space with the cofinite topology. Let $A\subseteq X$ be semi-open, meaning that there is an open $U$ in $X$ such that $U\subseteq A\subseteq\operatorname{cl}U$. If $U=\varnothing$, then $\operatorname{cl}U=\varnothing$, so $A=\varnothing$, which is open. If $U$ is non-empty, then $U=X\setminus F$ for some finite $F\subseteq X$, and $\operatorname{cl}U=X$, so $X\setminus F\subseteq A\subseteq X$. But then $X\setminus A\subseteq F$, so $X\setminus A$ is finite, and $A$ is open.


问:
Can someone please check my answers for these questions regarding improper integrals?
a) Evaluate $$\int^{1}_{0.5} \frac{1}{x^2}dx$$

my answer: $-1+\frac{1}{0.5}=1$

b) Explain why  $$\int^{-1}_{1} \frac{1}{x^2}dx$$ is undefined.

my answer: since the function $\frac{1}{x^2}$ is undefined at $x=0$, the integral is undefined.

c) Evaluate $$\int^{k}_{1} \frac{1}{x^2}dx$$ for some positive number $k$

my answer: $\int^{k}_{1} \frac{1}{x^2}dx = \frac{-1}{x} |^{k}_{1} = \frac{-1}{k}+1$

d) Take the limit of your answer in part c as $k\rightarrow \infty $ What value would you assign $\int^{\infty}_{1} \frac{1}{x^2}dx$?

my answer: $\lim_{k\to\infty} [\frac{-1}{k}+1]=1$

$\frac{-1}{x} |^{\infty}_{1}=1$

答:
For any expression $h(x)$,

let $\lim_{x \to k^+} h(x)$ denote the limit of $h(x)$ as $x$ approaches $k$, from above.

Similarly,
let $\lim_{x \to k^-} h(x)$ denote the limit of $h(x)$ as $x$ approaches $k$, from below.

b) Explain why  $$\int^{-1}_{1} \frac{1}{x^2}dx$$ is undefined.

Let $f(x)$ denote $\frac{1}{x^2}$, which implies that $f(x)$ is well defined throughout $(-1,1)$, except at $x=0$.

Therefore, from the definition of improper integrals,

$$I = \int_1^{-1} f(x)dx = I_1 + I_2$$

where

$$I_1 = \lim_{b \to 0^+} \int_1^b f(x)dx$$

and

$$I_2 = \lim_{a \to 0^-} \int_a^{(-1)} f(x)dx.$$

This means that in order for $I$ to exist (i.e. be well defined) it is necessary (and sufficient) that each of $I_1$ and $I_2$ be (separately) well defined.  Failure with respect to either $I_1$ or $I_2$ is sufficient to conclude that $I$ is not well defined.

Therefore, the problem reduces to showing that $I_1$ is not well defined (i.e. that the corresponding limit does not exist).

$$\int_1^b f(x)dx = F(b) - F(1)$$

where

$$F(x) = \frac{-1}{x}.$$

Therefore,

$$\int_1^b f(x)dx = \frac{-1}{b} - \frac{-1}{1} = \frac{b-1}{b}.$$

Therefore,

$$I_1 = \lim_{b \to 0^+} \frac{b-1}{b}.$$

As you can see, as $b \to 0^+,$ the above numerator goes to $-1$, and the above denominator goes to $0$, from above.  Therefore, as $b \to 0^+, I_1 \to -\infty.$

Therefore, $I_1$ is not well defined.

Therefore, $I$ is not well defined.