问:
Similarity of real matrices over real and complex numbers
The problem

The goal is the show that two real matrices are similar over $\mathbb{R}$ if and only if they are similar over $\mathbb{C}$:

Let $ T \in GL_{n}\mathbb{C}$, write $T = T_{1} + iT_{2}$ with $T_{1}, T_{2} \in Mat(n, \mathbb{R})$. Prove that there exists a real number $c \in \mathbb{R}$ such that $T_{1} + cT_{2}$ is invertible.

Prove that two real matrices $A, B \in Mat(n, \mathbb{R})$ are similar over $\mathbb{R}$ if and only if they are similar over $\mathbb{C}$.

What I tried so far

I tried a proof by contradiction:

Suppose $\nexists c \in \mathbb{R}$ such that $T_{1}+cT_{2}$ is invertible, but $T_{1} +iT_{2}$ is invertible. This is the same as saying $\forall c \in \mathbb{R}: det(T_{1} + cT_{2}) = 0$. Then for $c = 0$ we can see that $det(T_{1}) = 0$, so $T_{1}$ is not invertible. From here on I wanted to show that $T_{1} + iT_{2}$ cannot be invertible if $T_{1}$ is not invertible, but a simple counter example proves this false:

$
T_{1} = \left( \begin{array}{rrr}
0 & 0 \\
1 & 1 \\
\end{array}\right),
T_{2} = \left( \begin{array}{rrr}
2 & 3 \\
1 & 2 \\
\end{array}\right)\\
$

$T_{1} + iT_{2}$ is invertible, but $T_{1} + cT_{2}$ is not for $c=1$. I am not sure how to go on from here.

I tried this:

$"\Rightarrow"$ Is trivial, if $\exists S \in Mat(n,\mathbb{R}): B = S^{-1}AS$, then $S \in Mat(n, \mathbb{C})$. Following the definition of matrix similarity, $A, B$ are similar over $\mathbb{C}$.

$"\Leftarrow"$ There exists $S \in GL_{n}(\mathbb{C})$ such that $B =  S^{-1}AS \Leftrightarrow SB = AS$. Write $S = S_{1} + iS_{2}$, with $S_{1},S_{2} \in Mat(n, \mathbb{R})$. Therefore $SB = S_{1}B + iS_{2}B = AS_{1} + iAS_{2} = AS$. Separating the real and imaginary parts of the equation gives us $S_{1}B = AS_{1}$ and $iS_{2}B = iAS_{2}$. We are only interested in the real part, which proves $A, B$ are similar over $\mathbb{R}$.

My question

First off, I am not sure wheter my work on 1. could lead somewhere. How would I prove this part? And second, is my proof for 2. correct?

答:
You are almost there, but as it stands, your proof isn't correct because $S_1$ and $S_2$ can be both singular. Your argument can be salvaged by observing that some linear combination of $S_1$ and $S_2$ must be nonsingular. See this answer for more details.

From a higher perspective, two real matrices are similar over $\mathbb R$ iff they are similar over $\mathbb C$ because every matrix has a unique rational canonical form and the similarity transform $A\mapsto P^{-1}AP$ that brings a matrix $A$ to its rational canonical form can be carried out by using an invertible matrix $P$ whose entries are taken from the field generated by the entries of $A$. See this answer for more details.


问:
Range of numbers not divisible by $2$ or $3$ or $5$
How many numbers lie in the range of $1331$ and $3113$ (Endpoints Inclusive) such that they are not divisible by $2$ or $3$ or $5$?

I came across this Number System question and was not able to figure out the way to get the answer. At first I thought to separate out the numbers which are divisible by the LCM of $2$, $3$ & $5$ but this will still leave some numbers which will be exclusively divisible by any one of $2$ or $3$ or $5$.
Another method I found out to solve these type of questions is by using the Euler's Totient method but I am not aware of this.
So can someone please explain a general way to solve these types of question?

General question can be : How many numbers lie in the range of $A$ and $B$ such that they are not divisible by $x$ or $y$ or $z$ and so on?

答:
Another approach:

An integer $n$ is not divisible by any of
$2$, $3$, and $5$ if and only if $n$ is relatively prime to $30$; and this in turn occurs if and only if $n$ leaves one of the eight remainders: $1, 7, 11, 13, 17, 19, 23, $ or $29$ upon division by $30$.

Because remainders upon division by $30$ occur cyclically: among any $30$ consecutive integers, exactly $8$ will not be divisible by any of $2, 3, 5$.

The interval $1331$ to $3100$ contains $1770=59\cdot 30$ integers. So this interval will contain $59\cdot 8=472$ integers not divisible by any of $2, 3, 5$.

From $3101$ to $3113$, the remainders upon division by $30$ are $11, 12, ..., 23$. Of these, the five with remainders of $11, 13, 17, 19, 23$ will not be divisible by any of $2, 3, 5$.

So the total in the desired interval is $472 + 5=477$ integers not divisible by any of $2, 3, 5$.

ContentLicense: CC BY-SA 4.0


问:
Error in book's definition of open sets in terms of neighborhoods?
The following is copied verbatim from a book (I. Protasov, Combinatorics of numbers, p. 14):

Suppose that to each point $x$ of a set $X$ a collection $\mathcal{B}(x)$ of subsets of $X$, which are called neighborhoods of $x$, is assigned so that the following conditions are satisfied:

(B1) $x\in U$ for every neighborhood $U \in \mathcal{B}(x)$;
(B2) if $U \subseteq V, U \in \mathcal{B}(x)$, then $V\in \mathcal{B}(x)$;
(B3) if $U_1, \dots, U_n \in \mathcal{B}(x)$, then $U_1 \cap \dots \cap U_n\in \mathcal{B}(x)$;
(B4) if $U\in \mathcal{B}(x)$, then there is a neighborhood $V\in\mathcal{B}(x)$ such that $U\in \mathcal{B}(y)$ for every $y\in V$.

A subset $A\subseteq X$ is defined to be open, if $A$ is a neighborhood of each of its points, i.e. $A\in\mathcal{B}(x)$ for every $x\in A$.  Evidently, open sets satisfy the following properties:

(O1) $X, \varnothing$ are open sets;
(O2) if $U_1, \dots, U_n$ are open sets, then $U_1 \cap \dots \cap U_n$ is an open set;
(O3) if $U_\alpha, \alpha\in J$, is a collection of open sets, then $\bigcup\{U_\alpha : \alpha \in J\}$ is an open set.

I don't see why (O1) is true.  More generally, I don't see what guarantees that $\mathcal{B}(x) \neq \varnothing$ for any $x \in X$.  In fact, if we set $\mathcal{B}(x) = \varnothing$ for all $x \in X$, conditions (B1)-(B4) are all satisfied vacuously.  In this case there could be no open sets, so (O1) could not hold.

Am I missing something?

If not, there must be some error in the book, and I'd like to know how to fix it such that what results agrees with the standard way of defining open sets in terms of neighborhoods.

答:
I think I finally got it.  All we need to do is replace B3 with

(B3) each collection $\mathcal{B}(x)$ is closed with respect to finite (including empty) intersections;

Then the empty intersection (namely $X$) will belong to every $\mathcal{B}(x)$.

(It could be argued that the original (B3) already implies this.)

ContentLicense: CC BY-SA 3.0


问:
Let $X = A\cup B$. Show that $V$ is an open subset of $X$.

Let $X = A\cup B$, $V \subset A \cap B$ and $V \cap A$ be an open subset of $A$. Also let $V \cap B$ be an open subset of $B$. Show that $V$ is an open subset of $X$.

This seems very confusing. I'm not sure why they have all these conditions. I'm not sure what definitions I could use here to show this... How can I start approaching this?

答:
Definitions first: $V \cap A$ is an open subset of $A$ means that there is an open set $O_1$ of $X$ such that $O_1 \cap A = V \cap A$. Similarly, $V \cap B$ is an open subset of $B$ means that there is an open set $O_2$ of $X$ such that $O_2 \cap B = V \cap B$.

Now, if $x \in V$, we know that $x \in A$ and $x \in B$ as $V \subseteq A \cap B$. So $x \in A \cap V = O_1 \cap V$, and so $x \in O_1$, and $x \in B \cap V = O_2 \cap V$ so $x \in O_2$.

So $V \subseteq O_1 \cap O_2$. We would be done if the reverse inclusion also held, as a finite intersection of open sets is open.
So let $x \in O_1 \cap O_2$. Then we have $x \in A$ or $x \in B$ as $X=A \cup B$.
In the first case $x \in O_1 \cap A= A \cap V \subseteq V$ and in the second $x \in O_2 \cap B = V \cap B \subseteq V$, so we're done showing the other inclusion.

So $$V = O_1 \cap O_2$$ is open in $X$. Check: we exactly used all the given assumptions.


问:
How to find an exponential variable?
What is the way to solve a simple equation of this type?

$$x ^ 4 = 1.25$$

Update: I should have asked, how to calculate N'th root in a calculator?

答:
Note that $\sqrt[4]{x}=x^{\frac 14}=\sqrt{\sqrt{x}}$ so just hit the square root button twice or use y^x with $x=0.25$.

For the n-th root $\sqrt[n]{x}=x^{\frac 1n}$ so you can use the y^x button with $x=\frac 1n$.

(it can be challenging if your calculator doesn't accept parenthesis to enter $\frac 1n$, in that case you calculate it in advance and enter the decimal value, e.g. $0.33333333333$ for $\frac 13$).

When $n$ is a power of $2$ then you can just hit square root as many times as desired, e.g. $8=2^3$ so $\sqrt[8]{x}=\sqrt{\sqrt{\sqrt{x}}}$ three times.

Title: How to find an exponential variable?


问:
Find probability that a gap greater than 7 seconds occurs between two consecutive arrivals
So I have a question that basically says:
Customers arrive at a facility at a mean rate of $12$ per minute.
Find the Probability that a gap of greater than $7$ seconds occurs between two consecutive arrivals.

I've started the question and found the mean time between arrivals ($\frac{1}{12}$ minutes $= 5$ seconds).
I also know I need to use the negative exponential distribution, I should use $e^{-λt}$, however, I'm not sure what my $\lambda$ and t value should be.
I think it might be quite simple and I'm overthinking it, any help on how I would obtain $\lambda$ and t would be great.

Thanks in advance

答:
If your time was in unit of minutes, your $\lambda$ would be $1/12$, but it is not, right, it is more convenient for us to use seconds and use $\lambda = 0.2$ per second.
Then we use the cumulative distribudion function to find the probability of being outside the interval $[0, 7]$ seconds.
So we may take $1 - $P(being in that interval).

The CDF of $\lambda e^{- \lambda t}$ is $1 - e^{-\lambda t}$.

This equals $1 - CDF[7]$ with the right $\lambda$ (rate, intensity).

$$ 1 - CDF[7] = 1 - (1 - e^{-\lambda 7} )= e^{-0.2*7} = 0.247$$


问:
Show that $g^{-1} \circ f \circ g$ is a translation, if $g$ is a rigid motion and $f$ is a translation
Question:

Let $g: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a rigid motion, and let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a translation.

Show that $g^{-1} \circ f \circ g$ is also a translation.

My attempt:

I know that for any finite dimensional vector space over $\mathbb{R}$, any rigid motion $g(x)$ can be written as: $g(x) = f_{g(0)} \circ T(x)$, where:

$f_{g(0)}$ is a translation given by $f_{g(0)}(w) = w + g(0)$

$T$ is an orthogonal and linear operator given by $T(x) = f_{-g(0)} \circ g(x)$

We note that both $f_{g(0)}$ and $T$ are invertible, and the inverse of $f_{g(0)}$ is $f_{-g(0)}

So then $g^{-1} \circ f \circ g = T^{-1} \circ f_{-g(0)} \circ f \circ f_{g(0)} \circ T$

I assume it's safe to say here that translations commute, so the above simplifies to:

$T^{-1} \circ  f  \circ T$

I'm not sure if my working up to this point has been correct, and I'm also not sure how to proceed from here. Any suggestions?

Thank you!

答:
Let $h = T^{-1}\circ f\circ T$.
Observe that $f(v) = v + f(0)$ and that $h(0) = T^{-1}(f(0))$.
We have

$$\begin{align}
h(v) - h(0)
&= T^{-1}\big(\,f(T(v))\,\big) - T^{-1}(f(0))
\\&= T^{-1}\big(f(T(v)) - f(0)\big) \tag{$T^{-1}$ is linear}
\\&= T^{-1}\big(T(v) + f(0) - f(0)\big) \tag{$f$ is a translation}
\\&= T^{-1}\big(T(v) \big) = v,
\end{align}$$

that is, $h(v) = v + h(0)$.
In other words, $h$ is a translation.


答: Suppose $f(x) = x + v$ is translation by an arbitrary vector $v$. Explicitly calculate $(g^{-1} \circ f \circ g)(x)$ using the formula for $f$ and the fact that $g$ and $g^{-1}$ are linear.

Title: Show that $g^{-1} \circ f \circ g$ is a translation, if $g$ is a rigid motion and $f$ is a translation

Body:
Question:

Let $g: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a rigid motion, and let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a translation.

Show that $g^{-1} \circ f \circ g$ is also a translation.

My attempt:

I know that for any finite dimensional vector space over $\mathbb{R}$, any rigid motion $g(x)$ can be written as: $g(x) = f_{g(0)} \circ T(x)$, where:

$f_{g(0)}$ is a translation given by $f_{g(0)}(w) = w + g(0)$

$T$ is an orthogonal and linear operator given by $T(x) = f_{-g(0)} \circ g(x)$

We note that both $f_{g(0)}$ and $T$ are invertible, and the inverse of $f_{g(0)}$ is $f_{-g(0)}

So then $g^{-1} \circ f \circ g = T^{-1} \circ f_{-g(0)} \circ f \circ f_{g(0)} \circ T$

I assume it's safe to say here that translations commute, so the above simplifies to:

$T^{-1} \circ  f  \circ T$

I'm not sure if my working up to this point has been correct, and I'm also not sure how to proceed from here. Any suggestions?

Thank you!


问:
Triangle with perpendicular medians
In the triangle $ABC$ $AC=3,BC=4$ and the medians $AA_1$ and $BB_1$ are perpendicular at $M$.

The Pythagorean theorem gives $${AB}^2={AM}^2+{BM}^2\\{A_1B_1}^2={MA_1}^2+{MB_1}^2\\{AB_1}^2={AM}^2+{MB_1}^2\\{BA_1}^2={BM}^2+{MA_1}^2$$ If we multiply the third and fourth equality by $(-1)$ and then add we will get: $$AB^2+A_1B_1^2-AB_1^2-BA_1^2=0$$ From here we can get $AB=\sqrt{5}.$ ($A_1B_1=\dfrac12 AB$) Why can't we just substract the LHSs and RHSs of the equalities to get this equation? I mean can't we say that: $$AB^2-A_1B_1^2-AB_1^2-BA_1^2=AM^2+BM^2-MA_1^2-MB_1^2-AM^2-MB_1^2-BM^2-MA_1^2?$$ Why not? Thank you in advance!

答: The last equality holds, but that's not what we're searching for. The idea is to get rid of the segments with one endpoint at $H$.

Title: Triangle with perpendicular medians


问:
Prove that $\int_{\psi(Q^{N-1})} L \cdot \vec{n} \,dm_S=\int_{Q^{N-1}} (\vec{a}(\psi(t)) \cdot \vec{N}_\psi(t))\, dt$
Let $\psi:U' \to \mathbb{R}^{N}$ be a smooth function, where $U' \supset \overline{U} \supset U \supset Q^{N-1}$ with $$Q^{N-1}=\{(x_1,\dots,x_{N-1}: x_1 \geq 0, \dots, x_{N-1}\geq 0 \hbox{ and } \sum_{i=1}^{N-1}x_i \leq 1\}.$$ Suppose that $U$ is bounded, $\psi$ is injective and $\psi'(x)$ has rank $N-1$ for all $x \in U'$ and define $S\doteq \psi(U).$

Consider the map $\vec{N}_\psi(t_0)=\frac{\partial \psi}{\partial t_1}(t_0) \times \cdots \times \frac{\partial \psi}{\partial t_{N-1}}(t_0)$ and define
$$S \ni p \mapsto \vec{n}(p)=\frac{\vec{N}_\psi(t)}{|\vec{N}_\psi(t)|}, \quad \psi(t)=p.$$

Consider the $\sigma-$algebra of subsets of $S$ defined by
$$\mathcal{M}_S=\{\psi(A):A \subset U, A \hbox{ is Lebesgue measurable}\}$$
and the measure $m_S:\mathcal{M}_S \to [0,\infty)$ given by
$$m_S(\psi(A))\doteq \int_{A} |\vec{N}_\psi(t)|\, dm(t).$$

Let $L$ be the vector field given by $L=\sum_{j=1}^{N} a_j(x) \frac{\partial}{\partial x_j}$.

My question: How to prove that $$\int_{\psi(Q^{N-1})} L \cdot \vec{n} dm_S=\int_{Q^{N-1}} (\vec{a}(\psi(t)) \cdot \vec{N}_\psi(t))\, dt,$$
where $\vec{a}=(a_1,\dots,a_N).$

This seems to be related to the Radon-Nikodyn Theorem. I tried to develop the calculations but I didn't get it.

答:
I think I managed to prove that $$\int_{S} \phi dm_S=\int_{U} \phi(\psi(t)) |\vec{N}_\psi(t)|dm(t)$$ when $\phi$ is a simple function. Indeed, if $\phi=\sum_{j=1}^{m}a_j \chi_{A_j}$, $S=\bigcup_{j=1}^{m}A_j$. Then, $U=\bigcup_{j=1}^{m}B_j$, where $B_j=\psi^{-1}(A_j)$. Thus,

\begin{eqnarray}
\int_{S} \phi dm_S&=&\sum_{j=1}^{m} a_j m_S(A_j)\\
&=&\sum_{j=1}^{m} a_j m_S(\psi(B_j))\\
&=&\sum_{j=1}^{m} a_j \int_{B_j}|\vec{N}_\psi(t)|dm(t)\\
&=&\sum_{j=1}^{m} \int_{B_j}a_j|\vec{N}_\psi(t)|dm(t)\\
&=&\sum_{j=1}^{m} \int_{B_j}\phi(\psi(t))|\vec{N}_\psi(t)|dm(t)\\
&=&\int_{U}\phi(\psi(t))|\vec{N}_\psi(t)|dm(t),
\end{eqnarray}
since if $t \in B_j$ then $\psi(t) \in A_j$ which implies that $\phi(\psi(t))=a_j$. Now the idea is pass to the limit.


问:
Theorem on Spectrum, point spectrum and the Calkin algebra
Apparently it's a theorem by Weyl.

Let $H$ be an infinite dimensional, separable Hilbert space and let $A,B \in B(H)$.

Let $K(H)$ be the set of compact operators acting on $H$.

Show that if $A-B \in K(H)$ and if $\lambda \in \sigma(A)-\sigma_p(A)$, or if $\lambda$ is an eigenvalue of infinite multiplicity, then $\lambda \in \sigma(B)$.

Here $\sigma_p$ is the point spectrum, and it consists of the eigenvalues.

Hint: assume wlog $\lambda=0$(I'm not sure why we can do that) and write $A=B+(A-B)$.

Some thoughts:

We have two cases:

Suppose  $A-B \in K(H)$ and $\lambda \in \sigma(A)-\sigma_p(A)$.

Then we have $\lambda \in \sigma(B+(A-B))-\sigma_p(B+(A-B))$.

So $\lambda \in \sigma(B+ K(H))$ and $\lambda$ is not an eigenvalue of $(B+ K(H))$.

Now I feel like I have to use the compactness of $A-B$. Not sure to continue.

Suppose  $A-B \in K(H)$ and $\lambda$ is an eigenvalue of infinite multiplicity. Then we have $\lambda \in \sigma_p(A-B)$.

Again.. I'm missing something about compactness.

Any help will be appreciated.

Thank you in advance!

答:
An operator $T\in B(H)$ is said to be a Fredholm operator if the null space of $T$ is finite dimensional and the range of
$T$ has finite co-dimension.  In this case the index of $T$ is defined by
$$
  \text{ind}(T) = \text{dim}(\text{Ker}(T)) - \text{codim}(\text{Ran}(T)).
  $$

Among  the standard results of the theory of Fredholm opertors one has:

Theorem.  If $T$ is Fredholm and $K$ is compact,  then $T+K$ is also Fredholm and
$\text{ind}(T) = \text{ind}(T+K)$.

Letting $A'=A-\lambda $,  and $B'=B-\lambda $,  as suggested by @Berci,   we need to prove that,  if $A'-B'$ is compact,  and $A'$
fails to be invertible either because:

its null space is infinite dimensional,  or

$A'$ is injective,  but not surjective,

then $B'$ is also not invertible.

In order to prove this,  suppose by contradiction that $B'$ is invertible.  Then $B'$ is clearly Fredholm,  with
$\text{ind}(B') = 0$,
and
hence $A'$ is also Fredholm with $\text{ind}(A') = 0$.  However this is incompatible with either (1) or (2), so we have
arrived at a contradiction, proving that $B'$ is not invertible, and hence that $\lambda \in \sigma (B)$.


问:
Prove that the algebraic sum of the ordinates of intersection of a circle and a parabola is 0
I consider two curves $x^2=4ay$ and $x^2+y^2=\lambda^2$

So
$$y^2+4ay-\lambda^2=0$$
And $$y_1+y_2\not =0$$

I just want to whether the question itself is right or not

答:
The equation $y^2+4ay-\lambda^2=0$ has the solutions

$$y_{1/2}=-2a \pm \sqrt{4a^2+\lambda^2}.$$

Hence

$$y_1+y_2=-4a.$$


答:
For $x^2=4ay$, sum of abscissae will be zero due to symmetry about y-axis.

For $y^2=4ax$, sum of ordinates will be zero due to symmetry about x-axis.

(Perhaps the question has been mixed up? Or was it originally a true/false question?)

Title: Prove that the algebraic sum of the ordinates of intersection of a circle and a parabola is 0

Body:
I consider two curves $x^2=4ay$ and $x^2+y^2=\lambda^2$

So
$$y^2+4ay-\lambda^2=0$$
And $$y_1+y_2\not =0$$

I just want to whether the question itself is right or not


问:
Example of presentation of a group.
I’m reading up on presentations of a group. In here, the example is the dihedral group $D_8$, with generators rotation $r$ of order $8$, and flip $f$ of order $2$. In the construction of the presentation of $D_8$ from the free group $F$ generated by those two letters $r$ and $f$, it’s fairly obvious why $r^8 = 1$ and $f^2 = 1$ are the necessary relations. But what about $rfrf = 1$? Where does it come from? What happens if we drop it from the presentation, i.e. if we just have $\langle r,f \mid r^8 = 1, f^2 = 1 \rangle$?

Also, just to be sure, when the article says that $D_8 \cong F/N$, does the argument implicitly involves a homomorphism $\phi: F \to D_8$, with $N$ as the kernel of $\phi$?

答:
If you drop $rfrf=1$, you get the free product

$$\Bbb Z_8\ast\Bbb Z_2\cong\langle r,f\mid r^8, f^2\rangle$$

of the presentations $\Bbb Z_8\cong\langle r\mid r^8\rangle$ and $\Bbb Z_2\cong\langle f\mid f^2\rangle$.

Indeed, with a rotation of order two, we get the infinite dihedral group

$$D_\infty\cong \Bbb Z_2\ast\Bbb Z_2\cong\langle r,f\mid r^2, f^2\rangle.$$

Title: Example of presentation of a group.


答:
The group $\langle r,f \mid r^8 = 1, f^2 = 1 \rangle$ is infinite, and the dihedral group $D_8 = \langle r,f \mid r^8 = 1, f^2 = 1, (rf)^2 = 1 \rangle$ is finite. And the comment of @rogerl explains the geometric reason for that relation $(rf)^2=1$.

To get to the heart of the matter, here's how to prove that for any word in the free group $F = \langle r,f \rangle$, it's image in $D_8$ is equal to the image of an element of the form $r^a f^b$ where $a \in \{0,1,2,3,4,5,6,7\}$ and $b \in \{0,1\}$. Start with any word in $r$, $r^{-1}$, $f$, $f^{-1}$. By combining consecutive $f$'s into one power of $f$, and similarly for $r$, one obtains a word that alternates between powers of $f$ and powers of $r$. Using $f^2=1$, one can replace each power of $f$ with either the empty word or $f$ itself. Using $frf=r^{-1}$, which together with $f=f^{-1}$ implies that $fr^af=r^{-a}$ one can then reduce the number of $f$'s by $2$, eventually reaching a word of one of the forms $r^a f r^b$ or just $r^a$. To simplify the first form further, using $fr=r^{-1}f$, one can move the $f$ to the end, and one gets $r^a f$ or just $r^a$. Using $r^8=1$, one can reduce $a$ modulo $8$. Thus, group element is equal to one of the form $r^a f^b$ where $a \in \{0,1,2,3,4,5,6,7\}$ and $b \in \{0,1\}$.

Regarding your last paragraph, what you write is essentially correct, and moreover it holds in great generality. To say that a group $G$ has a presentation $G \cong \langle s_i \mid r_j\rangle$ means that there is a surjective homomorphism $\phi : F \langle s_i \rangle \to G$ defined on the free group $F = F\langle s_i \rangle$ with free basis $\{s_i\}$, such that the kernel $N$ of $\phi$ is normally generated by the subset $\{r_j\} \subset F$, and therefore $G \cong F / N$.

Title: Example of presentation of a group.

Body:
I’m reading up on presentations of a group. In here, the example is the dihedral group $D_8$, with generators rotation $r$ of order $8$, and flip $f$ of order $2$. In the construction of the presentation of $D_8$ from the free group $F$ generated by those two letters $r$ and $f$, it’s fairly obvious why $r^8 = 1$ and $f^2 = 1$ are the necessary relations. But what about $rfrf = 1$? Where does it come from? What happens if we drop it from the presentation, i.e. if we just have $\langle r,f \mid r^8 = 1, f^2 = 1 \rangle$?

Also, just to be sure, when the article says that $D_8 \cong F/N$, does the argument implicitly involves a homomorphism $\phi: F \to D_8$, with $N$ as the kernel of $\phi$?


问:
How to prove the optimality of greedy algorithm in a change problem with binary coin values?
Consider the change making problem, the input to this problem is an integer L. The output should be the minimum cardinality collection of coins required to make L cents of change (that is, you want to use as few coins as possible). In Binaryland, the coins are worth $1$,$2$,$2^2$,$2^3$,...,$2^{1000}$ cents. Assume that you have an unlimited number of coins of each type. Prove or disprove that the greedy algorithm (that takes as many coins of the highest value as possible) solves the change problem in Binaryland.

I'm new to proofs and couldn't get far enough with induction or contradiction, so I'd really appreciate your help.

答:
For a proof with contradiction, you could use the following: Assume that you do not use a coin that has the highest possible value. Then, you will have to take a combination of coins that sums up to that amount exactly (this is the case since you use $1,2,2^2,..$) and the fact that you use more than one coin to get to the exact same amount concludes the proof since this contradicts the fact that you have a solution with a few coins as possible.

Hope this helps!


问:
One-point compactification. The local compactness and para compactness is the concept but I unable to understand the basic thought of this problem
Prove that the one-point compactification of $R^n$ is homeomorphic to $S^n$.

答:
It’s well-known that the stereographic projection shows that $\Bbb S^n\setminus \{p\} \simeq \Bbb R^n$ (where $p$ is the north pole of the sphere, typically) and then apply

Theorem If $X$ is compact Hausdorff and $Y$ is locally compact Hausdorff and for some point $p \in X$ we have that $Y \simeq X\setminus \{p\}$, then $X$ is homeomorphic to the one-point compactification of $Y$.


问:
In How Many Ways Hiring Can Be Made?
I've been trying to solve this question, but no luck. I would really appreciate your help.

In the first place I thought, 8x5=40 for the first question and 8x4=32 for the second question.

```<code>There are 8 positions to be filled in the company and 5 applicants for each position.
Only 1 applicant to each of 8 positions has a background.

a) In how many ways these positions can be filled?

b) In how many ways these positions can be filled
only with people which do not have any background?
</code>```

The link of the screenshot of the question itself,

https://i.stack.imgur.com/Sm6oQ.png

And also the error I got when I entered 390625,

https://i.stack.imgur.com/OEu9i.png

答:
For the first question, you have 8 positions and 5 options per position, this means that you can select 5 candidates for the first, 5 candidates for the second position, and so on. Now, by taking $5 \times 5 \times ... \times 5 = 5^8 = 390625$ you will find all possible combinations since for each position you will have to consider all other possibilities for the other positions.

The second question is solved similarly, here you only have 4 candidates per position, hence $4^8$ = 65536 options.

Hope this helps!

Title: In How Many Ways Hiring Can Be Made?


问:
If a forgetful functor between varieties preserves coproducts, does it have a right adjoint?
Suppose that $C,D$ are two varieties (equational classes) and that $U:C\rightarrow D$ is the forgetful functor. Is it true that if this functor preserves coproducts and initial objects then it has a right adjoint?

I remember someone once told me this fact but I do not remember whom it was to ask for details. I just remember that I have been told that it follows from Freyd's criterion and some properties of varieties, but whenever I look at Freyd's criterion, it only talks about the existence of left adjoints and not right adjoints. Maybe this is easy, but I did not study the subject properly but need to use this result but I want to make sure it is true and understand some of its backgrounds.

I would appreciate any clarification/reference where I can find proof for this fact.

答:
Let me give a more conceptual proof using Lawvere theories.

Let $\mathcal{C}$ be a category of algebras of a variety.
Suppose the language only has operations of arity $< \kappa$, where $\kappa$ is an infinite regular cardinal.
Let $\mathcal{C}_\mathrm{F}$ be full subcategory of $\mathcal{C}$ spanned by the free algebras generated by each cardinal $< \kappa$.
There is a fully faithful functor $\mathcal{C} \to [\mathcal{C}_\mathrm{F}^\textrm{op}, \textbf{Set}]$ defined by $X \mapsto \mathcal{C}(-, X)$.
It is clear that $\mathcal{C}_\mathrm{F}$ has coproducts of families of $< \kappa$ objects and that this functor $\mathcal{C} \to [\mathcal{C}_\mathrm{F}^\textrm{op}, \textbf{Set}]$ has image contained in the full subcategory of functors $\mathcal{C}_\mathrm{F}^\textrm{op} \to \textbf{Set}$ that send coproducts of families of $< \kappa$ objects in $\mathcal{C}_\mathrm{F}$ to products in $\textbf{Set}$.
The fundamental theorem regarding Lawvere theories is that this is precisely the essential image of $\mathcal{C} \to [\mathcal{C}_\mathrm{F}^\textrm{op}, \textbf{Set}]$.

Let $\mathcal{D}$ is a category of algebras of a variety.
Suppose there is a forgetful functor $U : \mathcal{C} \to \mathcal{D}$ that preserves coproducts.
(More precisely: suppose the language of $\mathcal{D}$ is a reduct of the language of $\mathcal{C}$ and that the forgetful functor $\mathcal{C} \to \mathcal{D}$ preserves coproducts.)
This restricts to a functor $U : \mathcal{C}_\mathrm{F} \to \mathcal{D}$.
Thus, for any object $Z$ in $\mathcal{D}$, we obtain a functor $\mathcal{D} (U -, Z) : \mathcal{C}_\mathrm{F}^\textrm{op} \to \textbf{Set}$ that sends coproducts of families of $< \kappa$ objects in $\mathcal{C}_\mathrm{F}$ to products in $\textbf{Set}$.
By the fundamental theorem, this is representable by some object $G Z$ in $\mathcal{C}$, i.e. $\mathcal{C} (X, G Z) \cong \mathcal{D} (U X, Z)$ naturally in $X$, for $X$ in $\mathcal{C}_\mathrm{F}$.
This defines a functor $G : \mathcal{D} \to \mathcal{C}$.
Since $U$ preserves coproducts, the earlier isomorphism $\mathcal{C} (X, G Z) \cong \mathcal{D} (U X, Z)$ extends to all free algebras $X$.
We don't yet know that $U$ preserves coequalisers in general, but it is straightforward to check that $U$ preserves the special reflexive coequalisers of the kind appearing in the precise monadicity theorem, and this is enough to extend the isomorphism $\mathcal{C} (X, G Z) \cong \mathcal{D} (U X, Z)$ to all algebras $X$.
Thus $G : \mathcal{D} \to \mathcal{C}$ is a right adjoint of $U : \mathcal{C} \to \mathcal{D}$, and therefore $U : \mathcal{C} \to \mathcal{D}$ indeed preserves coequalisers.

Let me also give a proof dealing with coequalisers first and then constructing the right adjoint by abstract nonsense.

It is well known that the category of algebras of a variety is exact in the sense of Barr, i.e. every morphism factors as a regular epimorphism followed by a monomorphism, pullbacks preserve regular epimorphisms, and every congruence is a kernel pair.
Moreover, the forgetful functor to the category of sets respects these properties.
Thus, any forgetful functor between categories of algebras also respects these properties.

There are basically two ways to construct coequalisers in the category of algebras of a variety, corresponding to the two ways of constructing the smallest congruence extending a binary relation on an algebra.
The first way is impredicative: the collection of all congruence relations extending a given binary relation is a set and is closed under arbitrary intersections, so there is indeed a smallest one.
The second way is iterative: first, make the binary relation reflective and symmetric; next, if $a \sim b$ and $b \sim c$ then add in $a \sim c$ too; then close it under the algebraic operations; repeat these steps until the construction stabilises.
If the language only has operations of arity $< \kappa$ and $\kappa$ is an infinite regular cardinal then the construction will stabilise after at most $\kappa$ steps.

Now back to category theory.
Both constructions of the smallest congruence extending a binary relation can be translated into categorical language.
For the purposes of this question, it is the iterative construction that we should think about.
I claim:

Proposition.
Let $\mathcal{C}$ be a category with finite limits, finite coproducts, $\kappa$-filtered colimits, exact in the sense of Barr, with $\kappa$-filtered colimits preserving finite limits.
Then $\mathcal{C}$ also has coequalisers.
Moreover, if $\mathcal{D}$ is a category of the same kind and $U : \mathcal{C} \to \mathcal{D}$ is a functor that preserves finite limits, finite coproducts, $\kappa$-filtered colimits, and regular epimorphisms, then $U : \mathcal{C} \to \mathcal{D}$ also preserves coequalisers.

Proof.
The hypotheses are precisely what we need to translate the iterative construction of the smallest congruence extending a binary relation.
Given a parallel pair of morphisms $X \rightrightarrows Y$ in $\mathcal{C}$, take the image of the induced morphism $X \to Y \times Y$: this defines a binary relation on $Y$, so we may form the smallest congruence $R \rightrightarrows Y$ extending it.
Every congruence is a kernel pair, so $R \rightrightarrows Y$ has a coequaliser.
It is straightforward to verify that the coequaliser of $R \rightrightarrows Y$ is the coequaliser of $X \rightrightarrows Y$.
All of the structures used in this construction are preserved by $U : \mathcal{C} \to \mathcal{D}$, so we may conclude that coequalisers are also preserved.　■

Corollary. If a forgetful functor between categories of algebras preserves coproducts and $\kappa$-filtered colimits for some infinite regular cardinal $\kappa$, then it also preserves coequalisers.

Next, regarding the existence of adjoints: if there is an infinite regular cardinal $\kappa$ such that all the operations in the language have arity $< \kappa$, then the category of algebras is a locally $\kappa$-presentable category.
It is known that any colimit-preserving functor between locally $\kappa$-presentable categories has a right adjoint.
Thus, in this situation, preserving coproducts is enough to imply the existence of a right adjoint.


问:
Two questions about implications between $\mathsf{DC}, \mathsf{BPI}$ and $\mathsf{AC}_\omega$
Does the implication $\mathsf{DC} \implies \mathsf{BPI}$ hold?

And does the implication  $\mathsf{BPI} \implies \mathsf{AC}_\omega$ hold?

I checked with Howard/Rubin's "Consequences of the Axiom of Choice" in part V where they list relations between forms, in particular, I could not find any mention of either of the above implications on page 324 and 326 respectively.

答:
No implication whatsoever.

In Cohen's first model $\sf BPI$ holds, while both $\sf AC_\omega$ and $\sf DC$ fail.

In Shelah's model where every set has the Baire property, $\sf BPI$ fails, and $\sf DC$ holds (and consequently $\sf AC_\omega$ holds). One can also use for this Solovay's model or models of $\sf AD+\it V=L(\Bbb R)$, if one is willing to take large cardinals for granted.


问:
How do I obtain the inequality for the two points of a triangle that fullfill a condition?
This problem states that on $△ABC$, $AB=3$, $AC=1$, $AD=1$ $∠BAC=θ$. It says that there are two points that fullfill the condition $DP_0=\frac{1}{3}BC$ and asks for the range of $\cos \theta$ that is needed to fullfill the condition.
These images are given:

Image 1, Image 2

The problem says that by looking at image#2, it can be known that the conditions for P to exist are $0\leq AP<AP_0$ and the explanation of the problem says that this is because $P$ is between $A$ and $P_0$ but I don't understand this. How can I get the conditions for $P$ to exist only by looking at the second image?

答:
Applying law of cosine in $\triangle ABC$,

$BC^2 = 3^2 + 1^2 - 2 \cdot 3 \cos \theta = 10 - 6 \cos \theta$

Now in $\triangle ADP$, if $AP = x$,

$DP^2 = \frac{BC^2}{9} = 1^2 + x^2 - 2 x \cos \theta$

i.e $ \ \displaystyle \frac{10 - 6 \cos \theta}{9} = 1 - \cos^2 \theta + (x- \cos \theta)^2$

$(1 - 3 \cos \theta)^2 = 9 (x-\cos\theta)^2$

$1 - 3 \cos \theta = \pm (3x - 3 \cos \theta)$

So, $x = \frac{1}{3} $ or $x = \frac{6\cos\theta - 1}{3}$

Range of $\theta$ so we have two points on line segment $AC$ meeting the condition $DP = \frac{BC}{3}$,

we must have $0 \lt x \lt 1 \implies \frac{1}{6} \lt \cos\theta \lt \frac{2}{3}$.

Also note that when $\cos \theta = \frac{1}{3}$, we have only one point $P$ where $DP = \frac{BC}{3}$.


问:
$L$-function of elliptic curves expansion into Dirichlet series
Let $E/\mathbb{Q}$ be an elliptic curve. The $L$-function of $E$ is defined to be the Euler product
$$
L_E(s) = \prod_{\text{ bad }p} (1 - a_p p^{-s})^{-1} \prod_{\text{ good }p} (1 - a_p p^{-s} + p^{1-2s})^{-1},
$$
where $a_p = p + 1 - \#E(\mathbb{F}_p)$ if $p$ is a good prime and $a_p \in \{ 0, \pm 1\}$ if $p$ is a bad prime. A lot of textbooks I've read then says that then $L_E(s)$ can be written as a Dirichlet series
$$
L_E(s) = \sum_{n=1}^\infty a_n n^{-s},
$$
by using unique factorization in $\mathbb{Z}$. The problem is, I don't see how to get there.

I understand that there are only finitely bad primes, so we can just focus on the good primes. The good primes factor in $L_E(s)$ can be expanded using the standard power series identity $1/(1-z) = 1 + z + z^2 + \cdots$, so we get
\begin{align}
\frac{1}{1 - (a_p p^{-s} - p^{1-2s})} = \sum_{i=0}^\infty (a_p p^{-s} - p^{1-2s})^i.
\end{align}
But then I don't know how to proceed anymore. I've tried binomial expansion but that doesn't seem to get me nearer to the solution. Any hints in expanding the Euler product of the $L$-function? I apologize if this is a stupid question.

答:
For each individual $p$, you may view $p^{-s}$ as a formal variable, say $T$.

Thus e.g. for a good $p$, the Euler factor is $(1 - a_p T + pT^2)^{-1}$.

This can then be expanded as a formal power series in $T$. We have:
$$(1 - a_p T + pT^2)^{-1} = 1 + a_p T + (a_p^2 - p)T^2 + (a_p^3 - 2pa_p)T^3 + \dots$$

Now for every $p$ you get a series $\sum_r a_{p^r} p^{-rs}$ and the product gives you a Dirichlet series $\sum_n a_n n^{-s}$.


问:
Discrepancy in Calculating Expectation of a Dependent Random Variable
Suppose $X$ a is random variable with pdf $$\begin{cases} f(x), & x\ge 0 \\ 0, & x\lt 0 \end{cases} $$ The random variable $Y$ is defined as $$Y= \begin{cases} A, & X\lt k \\ A+BX, & X\ge k \end{cases} $$ where $A,B,k$ are positive constants. If I’m not wrong, the pdf of $Y$ is then $$ \begin{cases} \int_0^k f(x) dx, & y=A \\ f\left(\frac{y-A}{B}\right), & y\gt A \\ 0, & \text{otherwise}\end{cases} $$
as the first case corresponds to $P(X\lt k)$ and the second case corresponds to $P\left(X=\frac{y-A}{B} \right)$, and so $$E(Y)=A\int_0^k f(x) dx \ + \int_A^{\infty} yf\left(\frac{y-A}{B} \right) dy $$

But at the same time, I could also look at this expectation as $$E(Y) = A + B\int_k^{\infty} xf(x) dx $$ as $Y$ will always be a certain amount more than $A$, which depends on $X$.

Which of the two is the correct expression? Am I making a mistake somewhere?

答:
The two methods lead exactly to the same expectation.

Your second expression is

$$ \bbox[5px,border:2px solid red]
{
\mathbb{E}(Y)=A+B\int_k^{\infty} xf_X(x)dx
\qquad (1)
}
$$

and it is correct!

For the first case, the rv is not absolutely continuous anyway we can derive its CDF (that always exists) in the following way:

$$ F_Y(y) =
\begin{cases}
0,  & \text{if $y<A$ } \\
\int_0^k f_X(x)dx,  & \text{if $y=A$ } \\
F_X\left(\frac{y-A}{B}\right), & \text{if $y\geq A+Bk$}
\end{cases}$$

Thus the expectation is

$$\mathbb{E}(Y)=A\int_0^kf_X(x)dx+\int_{A+Bk}^{\infty}\frac{1}{B}yf_X\left(\frac{y-A}{B}\right)dy$$

Now substitute $\frac{y-A}{B}=x$ and get

$$\mathbb{E}(Y)=A\int_0^kf_X(x)dx+\int_{k}^{\infty}(A+Bx)f_X(x)dx$$

Which is the same as (1)

Here is an useful drawing

As you can see,  Y-domain is X-image thus

$$Y \in A \cup [A+Bk;\infty)$$


问:
Intuitive Difference between posterior variance $V(Y|X)$ and variance of conditional expectation $V(E(Y|X))$
Intuitive Difference between posterior variance $V(Y|X)$ and variance of conditional expectation $V(E(Y|X))$

From Eves theorem we have as expectation is non-negative,

$V(Y) \geq V(E(Y|X))$

It means after knowing information X, my prior variance V(Y) has reduced

Is posterior variance  $V(Y|X)$ or $V(E(Y|X))$. Both look similar to me. Please elaborate intuitive difference /relation between these 2. I am a beginner. Request your patience.

答:
You can prove that if $Y\in L^2(\Omega, \mathcal{F}, \mathbb{P})$, then

$Var(Y)=Var(\mathbb{E}(Y\mid X))+\mathbb{E}(Var(Y\mid X))$.

Note that $Var(Y\mid X)$ is a random variable and $Var(\mathbb{E}(Y\mid X))$ is a number.


答:
As the other answers have mentioned, if you drop that relation you get the infinite dihedral group. Check out the groups of order 8, and note that conjugation gets written into the relations as $frf^{-1} = r^{-1}$, so $r \mapsto r^{-1}$, for the dihedral group of order $8$, and $frf^{-1} = r$, so $r \mapsto r$ for the direct product $Z_4 \times Z_2$.

This might clear up why you need that particular relation.

And yes, I would agree they are using the first isomorphism theorem as you suggest.

Title: Example of presentation of a group.


问:
Finding the angle between the radius and vector
I'm trying to find the angle between the radius vector produced and the tangent to the curve whose equation is $r = 4\cos\theta$ at the point where $\theta = \frac{\pi}{4}$.

My assumption was that I take the derivative, such that $\frac{dr}{d\theta}=-4\sin\theta$. Then substitute in the value for $\theta$.

Though, this isn't the right answers, for I expect $\frac{3\pi}{4}$.

I would appreciate your help, and if possible a pictorial example to help me understand this. The book i'm using is fairly terse and vague.

答:
As $\theta$ increases the point on the curve moves from $(4,0)$ to $(2,2)$ counterclockwise.

The tangent and the produced radius vector form an angle $=135^{\circ}$.


问:
A computer's memory is finite, so how can there be languages more powerful than regular?
A computer has a finite memory. There are no computers with infinite memory. Therefore the only languages that a computer can process are those whose member strings are finite. As I recall, the computational power required for any finite language is small -- a deterministic finite state automata is sufficient to process finite languages, I think.

So there is no such thing as context-free languages or recursively enumerable languages, right? Programming languages are not context-free, they are just simple regular languages, right? The idea that a programming language is a context-free language ... well, that's an illusion, right?

Why do we even bother to talk about context-free languages, recursive languages, recursively enumerable languages since inside the (finite) computer everything is just a regular language?

Obviously I am playing devils-advocate here, but I would really like to understand the fallacy of my argument.

Summary of Responses

Thank you very much to all who responded. I have studied your responses carefully. There are so many pearls of wisdom in your responses that I decided to summarize them. If my summary is incorrect, unclear, or missing an important concept, please let me know.

```<code>> A computer has a finite memory. There are no computers
> with infinite memory. Therefore the only languages that
> a computer can process are those whose member strings
> are finite.
</code>```

Responses

A computer's memory is finite, so in its memory it can recognize only finite languages. However, if you would pass an external input to the computer, then it could recognize some infinite languages.

Example: a*b is an infinite language and any string in that language can be recognized, even if the string exceeds the size of the computer's memory. Here's how: pass the string into the computer character by character. If the first character that is passed in is not an "a" then go to an error state. Otherwise, stay in the "a" state until a non-"a" character is input. If the character is not a "b" then go to the error state. Otherwise, go to an accept state. For any characters beyond "b" go to the error state.

Not all infinite languages can be recognized using this technique of feeding the input strings in as external input. In fact, the technique can be used only with regular languages. And even of the regular languages, only some of them can be recognized.

Example: For illustration purposes, suppose the computer has a ridiculously small memory size, say, 2 bytes. Suppose 1 byte is required for each state of a finite state automaton. Then the computer will not be able to recognize this regular language: a*b*c* because at least three byes are needed – one byte for the state that consumes all the a's, a second byte for the state that consumes all the b's, and a third byte for the state that consumes all the c's.

Any finite state automaton that requires more states than there are bytes in the computer's memory cannot be recognized.

Finally, with this technique of feeding arbitrarily long strings into the computer as external input, we would need to find a way around certain technical limitations such as a continuous power supply.

```<code>> A computer has a finite memory. Therefore the only languages
> that a computer can process are those whose member strings
> are finite … So there is no such thing as context-free languages
> or recursively enumerable languages, right?
</code>```

Responses

Consider an infinite language. It consists of strings of arbitrary length. Many strings will be longer than the size of the computer's memory. It is incorrect to say that because a computer can hold in its memory only those strings that are of a certain finite length, the language has only finitely many strings.  It is true that any finite subset of a language is regular. But that does not mean that the original language must have been regular.

```<code>> A computer has a finite memory. Therefore, in general, a computer
> can only process a finite subset of an infinite language. A finite
> subset of a language is regular. So why don't we simply treat all
> languages used in computers as regular?
</code>```

Responses

The reason to treat, say, a programming language as context-free is that the context-free grammar tells how to parse the language. If you considered just the subset of programs with length no more than 232, that would be regular but the regular expression would likely consist of millions of individual cases and wouldn't be helpful for parsing the programs.

答:
I think, Carl Mummet's answer hits the point but I'd like to add the practical view because Carl doesn't mention the essence of the practical meaning of (non-)regular languages.

Remember: practice never equals theory. Different from theory, it makes more sense in practice to define language types by their efficency of memory use (to recognize a language), not the possible structure of infinite languages (infinite languages don't even exist in the real world, just think of planned obsolescence of technology, limited life spans etc., finite particles universe, so it won't make even sense to consider inifinite languages in real situations).

After all, no useful programming language is just a context-free language (it wouldn't allow for variables, although regular programming languages with turing-complete semantics exist, for example Iota and Jot or other esoteric programming languages). Useful programming languages are typically context-sensitive needed for productive semantics like namespaces. Of course, you could enumerate all possible warning-free compilable C programs which fit into 1KiB but storing them would take massively more memory, significantly more programs than particles in our universe. This is pratically impossible.

By this argument, regular languages cannot practically achieve the same like non-regular languages, even when limited to finite memory. (Parsers of) Languages of certain size cannot fit into your memory, if you try to recognize them as a regular language since the regular expression will grow large, in the worst case (for totally random words) almost linearly with the number of words in the language! Whereas infinity in practice means finity in theory. But with something like a Linear bounded automaton, you can use finite memory to recognize languages containing exponentially (or more?) words. Of course, regular languages are FAST but without a certain efficiency of memory use you won't be able to detect languages at all in practice!

And practice also tells us, that computers cannot process infinite languages with certain complexity. If the program requires unboundedly growing memory to recognize the language. Except for control systems which are LTI-filters or digital hardware automata, it concerns most interesting infinite languages since with finite memory, you could only compress infinite languages into finite space which is possible for finitely bounded languages as well. You really need infinite memory to make use of the power of parsing infinite languages but then what should be the use for us human of an infinite result, even if it's effectively infinite by the notion of infinity due to very long processing time?

So actually, the set of infinite languages that a computer could really recognize in practice are very limited at all. It can be even doubtful, if any reasonable or good computer software should parse infinite languages since maintenance is a relevant factor (everyhwere) as well as limited lifetime of the CPU.


问:
Volume by disk perpendicular to $x$ axis
Find the volume of the solid formed when region enclosed by $y=x^{1/2},\quad y=6-x$ and $y=0$ revolves around $x$ axis.

Can I use the equation for finding solids by splitting the solid into 2 parts?

enter image description here

答:
$y=x^{1/2},\quad y=6-x$  , so put $x^{1/2} =6-x$

square both sides and get to $x^2 -13x +36 =0$

solutions are 4 and 9, use $x=4$ and $x=0$  for the limits in the volume formula $V=\int \pi y^2 dx$ for curve $y=x^{1/2}$.

Then similar for the line equation, between 4 and 6. Do separately then add.

Draw a diagram to see the $x=6$ point and why $x=4$ was used, not $x=9$.


问:
Calculate the expectation of a random variable with known PDF
This question came up while I was doing a manual black-level calibration for an image sensor. To do this, you cover the sensor so that no light hits it and take lot's of raw images. The average of the images gives the black level for each pixel. Out of curiosity I plotted the histogram of one pixel and it looked like this:
histogram (the orange line is the mean).

I wondered if there are better ways to calculate the expectation if the PDF is known. So I came up with the following example.

Let's say we have a random variable. The PDF of the random variable has two equal peaks spaced apart by the distance $d$. The expectation of the RV is obviously in the middle between the two peaks.
image of the pdf (please excuse my poor drawing skills).

A number of samples are taken from the RV. The expectation can of course be estimated by calculating the mean of the samples taken. But since the cdf is known we can do much better.

First of all we look at the distance between the samples and put them into two groups. We add $\frac{d}{2}$ to the lower group and substract $\frac{d}{2}$ from the upper group. After that we calculate the mean. The resulting estimate for the expectation is much better than just calculating the mean.

Of course there is the possibility that only one group arises. In this case we calculate the mean over the group and add $\frac{d}{2}$. There is a $50%$ chance that we hit the lower group and get it right.

I have simulated this in MATLAB from 1 to 200 samples taken from the RV. For each number of samples and method the results were averaged over 1000 tests. The results show clearly that this method converges way faster than the mean and that much less samples are needed to achieve the same quality of estimate (diagram).

The question is now:

Is there a general way to do this for a RV with an arbitrary cdf?

Is there an even better way to estimate the expectation?

答:
Maximum likelihood estimation

I don't know why I didn't think about this in the first place, but you can use Maximum likelihood estimation to estimate a parameter of a pdf. For the example stated here we have the random variable $X$ and, assuming the peaks are normally distributed, the pdf

$ P(X=x) = \frac{1}{2\sqrt{2\pi}\sigma} e^{-\frac{1}{2}(\frac{x-\mu-0.5}{\sigma})^2} + \frac{1}{2\sqrt{2\pi}\sigma} e^{-\frac{1}{2}(\frac{x-\mu+0.5}{\sigma})^2}$.

The likelihood formula looks like this:

$L(\vartheta)=\prod_{i=1}^{n} f_{\vartheta}(x_i)$

with $\vartheta$ being the unknown parameter, $n$ being the number of samples and $x_i$ being the sample teken from the RV. In this case the expectation $\mu$ is the unknown parameter. The pdf is then inserted into the likelihood formula like this:

$L(\mu)=\frac{1}{2\sqrt{2\pi}\sigma}[e^{-\frac{1}{2}(\frac{\prod_{i=1}^{n}x_i-\mu-0.5}{\sigma})^2}+e^{-\frac{1}{2}(\frac{\prod_{i=1}^{n}x_i-\mu+0.5}{\sigma})^2}]$

The resulting function is the likelihood for a given $\mu$ and given samples ${x_1,x_2,...,x_n}$. To get the 'best' $\mu$ we need to maximize the function. For that the derivative is taken and solved for zero. This is not that easy in this case since the function has multiple solutions depending on the number of samples $n$. Using numerical methods I compared it to my handcrafted method in the question and it performs even better.

TLDR: If samples and the pdf of a random variable are known, use maximum likelihood estimation to get a much better estimation of the expectation than the mean.


问:
$10$ identical black balls, $5$ identical red balls, $2$ identical white balls. What is the probability of choosing three black balls from this box?
I have $10$ identical black balls, $5$ identical red balls, $2$ identical white balls in my box. What is the probability of choosing three black balls from this box?

My Attempt: We have to first consider each ball is different and the points in the sample space is ${17} \choose 3$.

We can choose three black balls in ${10} \choose 3$  ways. So the probability is $\frac{{10} \choose 3}{{17} \choose 3}$

Can anyone please check my attempt ?

答:
The answer is correct, if you draw the three balls without putting them back. To be even clearer, you can also write the solution this way:
$$\frac{\binom{10}{3}\binom{5}{0}\binom{2}{0}}{\binom{17}{3}} = \frac{\binom{10}{3}}{\binom{17}{3}}$$


答:
We can write:$$Y=A+BX1_{[k,\infty)}(X)$$

Then:$$\mathbb EY=\mathbb E\left[A+BX1_{[k,\infty)}(X)\right]=A+B\mathbb EX1_{[k,\infty)}(X)=A+B\int_k^{\infty}xf(x)dx$$

In cases like this it is not handsome to figure out what the PDF (if it exists) of $Y$ looks like.

Actually it is possible here that $Y$ does not even have a PDF. For that note that $P(Y=A)$ might well be  positive. If so then $Y$ is not a continuous random variable (hence has no PDF).

Title: Discrepancy in Calculating Expectation of a Dependent Random Variable

Body:
Suppose $X$ a is random variable with pdf $$\begin{cases} f(x), & x\ge 0 \\ 0, & x\lt 0 \end{cases} $$ The random variable $Y$ is defined as $$Y= \begin{cases} A, & X\lt k \\ A+BX, & X\ge k \end{cases} $$ where $A,B,k$ are positive constants. If I’m not wrong, the pdf of $Y$ is then $$ \begin{cases} \int_0^k f(x) dx, & y=A \\ f\left(\frac{y-A}{B}\right), & y\gt A \\ 0, & \text{otherwise}\end{cases} $$
as the first case corresponds to $P(X\lt k)$ and the second case corresponds to $P\left(X=\frac{y-A}{B} \right)$, and so $$E(Y)=A\int_0^k f(x) dx \ + \int_A^{\infty} yf\left(\frac{y-A}{B} \right) dy $$

But at the same time, I could also look at this expectation as $$E(Y) = A + B\int_k^{\infty} xf(x) dx $$ as $Y$ will always be a certain amount more than $A$, which depends on $X$.

Which of the two is the correct expression? Am I making a mistake somewhere?


问:
Let a, b ∈ R with 0 < a  0 and a < sqrt(r) <b.
Density theorem:
If x,y ∈ R and x<y then here exists r ∈ Q st. x<r<y

From the density theorem, it is clear that a<r<b. I am having trouble getting the sqrt(r). I do not really know where to begin with that.

答: Hint: If $0<a<b$ then $a^2<b^2$. Apply the density theorem now and see what you get.

Title: Let a, b ∈ R with 0 < a < b. Prove using the Density Theorem that there exists r ∈ Q such that r > 0 and a < sqrt(r) <b.


问:
What is this limit equal to? $\lim_{n\to\infty}( \frac{f(1)}{2} + ... + \frac{f(n)}{n+1})^{\frac{1}{n}}$
$f:(0,\infty)\rightarrow R$ given by $f(x)=(x+1)\ln(1+\frac{1}{x})$

$\lim_{n\to\infty}( \frac{f(1)}{2} + \frac{f(2)}{3} + ... + \frac{f(n)}{n+1})^{\frac{1}{n}}$

I have calculed the derivative and established that the function is monotonically decreasing, and I tried to calculate $\lim_{n\to\infty}( \frac{f(1)}{2} + \frac{f(1)}{3} + ... + \frac{f(1)}{n+1})^{\frac{1}{n}}$ and $\lim_{n\to\infty}( \frac{f(n)}{2} + \frac{f(n)}{3} + ... + \frac{f(n)}{n+1})^{\frac{1}{n}}$ to maybe pin the results   between  two other limits but I have not been successful

答: HINT: Note that $$\frac{f(1)}{2} + \frac{f(2)}{3} +\cdots+ \frac{f(n)}{n+1}=\ln\left(2\right)+\ln\left(\dfrac32\right)+\cdots+\ln\left(\dfrac{n+1}{n}\right)$$ and $\ln(a)+\ln(b)=\ln(ab)$ for any $a, b$ positive reals.


问:
How to get ${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$
I found this in my test book, any hints? Given $${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$$ Then find the value of x and y in n. According to the answer provided on last pages of that book, it's $2n \choose n$. What i dont understand is how to get the answer (there's no explanation written there). All i said is thank you so much for every activity to me and i'm sorry for  my bad english (english is not my native language) and my messy post.</p>

答:
Let $E=\{a_1,\ldots a_n,b_1,\ldots,b_n\}$ a set with $2n$ elements. There's $2n\choose n$ subsets of $E$ with $n$ elements: $k$ elements from $\{a_1,\ldots,a_n\}$ and $n-k$ elements from $\{b_1,\ldots,b_n\}$ for some $k=0,\ldots,n$ hence we have
$${2n\choose n}=\sum_{k=0}^n{n\choose k}{n\choose n-k}=\sum_{k=0}^n{n\choose k}^2$$</p>


问:
A linear subspace is dense if and only if the annihilator of it is zero.
A similar question has been posted here: Dense subspace of linear space and functional equal to $0$.

Let $(V, \|\ \cdot\ \|)$ be a Banach space on $\mathbb{C}$ and let $V^{*}$ be its dual, endowed with the dual norm $$\|f\|_{V^{*}}=\|f\|_{op}=\sup_{\|x\|_{V}\leq 1}|\langle x,f \rangle_{V, V^{*}}|.$$ For $M\subset V$, we define the annihilator of $M$ as $$M^{\bot}:=\{f\in V^{*}:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ x\in M\}\subset V^{*}.$$ and similarly for $N\subset V^{*}$, we define $$^{\bot}N:=\{x\in V:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ f\in N\}=\bigcap_{f\in N}\ker(f)\subset V.$$

I am working on a problem stating as follows:

Let $W\subset V$ be a vector subspace. Show that $W$ is dense in $V$ if and only if $W^{\bot}=\{0\}$.

Using Hahn-Banach geometric form, I have proved that $\overline{W}=\ ^{\bot}(W^{\bot})$. This gives us the direction $(\Leftarrow)$, as follows:

Suppose $W^{\bot}=\{0\}$, then $$\overline{W}=\ ^{\bot}(W^{\bot})=\ ^{\bot}\{0\}=\{x\in V: \langle x,0\rangle_{V,V^{*}}=0\}=V,$$ since $\langle x,0\rangle_{V,V^{*}}=0$ is always true for all $x\in V$. So, we have $\overline{W}=V$, which means that $W$ is dense in $V$.

However, I got stuck in the direction $(\Rightarrow)$. Suppose $W$ is dense in $V$, and suppose $W^{\bot}\supset \{0\}$, i.e. there exists $f\in V^{*}$ that is not identically zero such that $\langle x, f\rangle_{V,V^{*}}=0$ for all $x\in W$.

I tried to use the density to continue, but by density we can only $y\in V$ such that $\|x-y\|_{V}<\epsilon$, where $\epsilon>0$ was arbitrarily fixed. I don't know how to continue then...

Is there a way for me to get contradiction? Thanks!

答:
Assume there is an $f\ne 0$ which vanishes on $W$. Then at some point $v\in V$ we have $f(v)\ne 0$. Since $W$ is dense, we find a sequence $(w_n)$ inside $W$ converging to $v$. By continuity,
$$
f(v)=f(\lim w_n)=\lim f(w_n)=\lim 0=0\ .
$$A contradiction.

$\square$


问:
Reduce the number of trigonometry calls in $ f(x,y,n)=2^{n-1}(1-\cos{\frac{x\pi}{2^n}})(1-\cos{\frac{y\pi}{2^n}})$ (and $\sum_{n=0}^bf(x,y,n))$
Using trigonometric identities rearrange the following function to have a single trigonometry call:

$$
f(x,y,n)=2^{n-1}\left(1-\cos{\frac{x\pi}{2^n}}\right)\left(1-\cos{\frac{y\pi}{2^n}}\right)
$$

I'm looking for help in optimizing it for a computer program I'm working on. Bonus points: run this entire summation with a single trigonometry reference. How can I get the $n$ out of the trigonometry call?
$$
\sum_{n=0}^b{f(x,y,n)}
$$

答:
Presumably, $x$ and $y$ are unrelated and arbitrary, and you do have to compute the two cosines.

Second bad news, I see no way to avoid the full evaluation of the terms in the summation.

To reduce the number of trigonometric calls drastically, you can rely on the recurrence relation

$$\cos(2^{-(n-1)}\alpha)=2\cos^2(2^{-n}\alpha)-1$$ or $$c_{n-1}=2c_n^2-1.$$

Anyway, this is numerically unstable when $c_n\approx1$, and instead,

$$1-c_{n-1}=2(1-c_n)(2-(1-c_n)).$$

Finally, note that the terms decrease roughly like $32^{-n}$ so that you don't necessarily need to accumulate all of them, contrary to my initial opinion.


问:
Eigenvectors of hermitian matrix are not coming out to be orthogonal
I have the following hermitian matrix before me:
\begin{pmatrix}2 &1+i\\ 1-i& 3\end{pmatrix}
I calculated its characteristic polynomial as $k^2-5k+4$ which has $1$ and $4$ as its roots.
Eigenvectors corresponding to these two eigenvalues are \begin{pmatrix}-(1+i) &1 \end{pmatrix} and \begin{pmatrix}1 &1-i\end{pmatrix}
But these are not orthogonal. Why is it so? Am I doing something wrong?

答:
Check Eigen values are $4,1$ and eigenvectors are $V_1=\begin{bmatrix} 1+i \\2 \end{bmatrix}$ and $V_2=\begin{bmatrix} -1-i \\ 1 \end{bmatrix}$, then
$V_1^{\dagger} V_2=0$, $\dagger$ means conjugate and transpose.


问:
The definition of Sobolev spaces
I have some confusion about the definition of Sobolev space for a real positive real $r>0$.

I know that for a positive integer $m>0$ and $p\geq 1$ and any subset $\Omega$ of $\mathbb{R^n}$, we define the Sobolev space denoted by $W^{m,p}$ by
$$W^{m,p}=\{f\in L^p : \text{the partial derivatives up to order m are in }   L^p   \}.$$
I am struggling with the definition of a Sobolev space by the summation with the Fourier coefficients. And I want also to know the general definition of Sobolev space for any real positive $r >0.$

答: As explained here https://en.wikipedia.org/wiki/Sobolev_space and here Is $B^s_{p,p} = W^{s,p}$?, there are several extensions of Sobolev spaces for non integer spaces: the Sobolev-Bessel spaces $H^{s,p} = F^s_{p,2} = (1-\Delta)^{-s/2} L^p$ and the Sobolev-Slobodeckij spaces $W^{s,p} = F^s_{p,p} = B^s_{p,p}$. More generally, the Besov spaces $B^s_{p,q}$ and the Triebel-Lizorkin spaces $F^s_{p,q}$ are intermediate spaces between Sobolev spaces of integer order.

Title: The definition of Sobolev spaces


答:
To calculate area from known shapes

$$ 4\cdot2\cdot\dfrac23+2\cdot 2\cdot\dfrac12 $$

Title: Volume by disk perpendicular to $x$ axis

Body:
Find the volume of the solid formed when region enclosed by $y=x^{1/2},\quad y=6-x$ and $y=0$ revolves around $x$ axis.

Can I use the equation for finding solids by splitting the solid into 2 parts?

enter image description here


问:
Correctly Guessing The Coin Toss

In a football game, the sides in the soccer field are determined on the basis of a single coin flip by the referee and the team which correctly guesses the flip result gets its desired side. What is the probability that a team will have to wait until its 8th game to get its desired side?

I am stuck with this question, any help would be much appreciated!

I thought, $\frac{1}{2}^8$ but it is wrong.

答: it depends. if they play against a team who usually picks the side that 'our' team wants again and again, than they are more likely to wait until their 8th game, but then they might lose the toss again. the other way around is the exact opposite, because if both teams want different sides, than they will probably get their chosen side first game. it's most likely to happen if each team who wins the toss chooses randomly, in which case the chances are still as low as 1/2^16.</p>

Title: Correctly Guessing The Coin Toss


问:
Probability: Rotating circular board
Let there be a circular board that is divided into $2^k$ sectors (similar to a dartboard). Each sector can be red with probability $p$ and green with probability $1- p$. If we rotate the board, when it comes to rest, I am interested in an event that a red sector covers the 6 o'clock direction.

Now let $X$ denote the reward that one gets when the above event occurs. The random variable $X$ takes a value according to the following rule: $X = 1/t$, when $1 \leq t\leq 2^k$ sectors are red and one of them covers the 6 o'clock direction, i.e., the reward of $1$ is shared among those red sectors. I am interested in the probability mass function of $X$. When none of the sectors that are red covers the 6 o'clock direction, the reward is $0$, i.e., $X = 0$.

My attempt: There are $2^k + 1$ combinations depending on how many of the $2^k$ sectors are red, e.g., $0$, $1$, $2$, $\dotsc$, $2^k$. Let $p_n$ denote the probability that $n$ sectors are red before rotating the board. Then we have
$$p_n = {2^k \choose n}p^{n}(1-p)^{2^k-n}, \,\,\,\, 1 \leq n \leq 2^k.$$
Now, for the case when there are $1 \leq t \leq 2^k$ sectors are red, the probability that a red sector covers the 6 o'clock direction is $$q_t = \frac{t}{2^k}.$$

So, one has
$$
X =
\begin{cases}
1, \quad \text{w.p.}~~ p_1q_1 \\
\frac{1}{2}, \quad \text{w.p.}~~ p_2q_2\\
\vdots \\
\frac{1}{2^k}, \quad \text{w.p.}~~ p_{2^k}q_{2^k} \\
0, \quad \text{w.p.}~~(1-p)^{2^k}.
\end{cases}
$$

Then the probability that a red sector covers the 6 o'clock direction is
$$P_{c} = \sum_{n = 1}^{2^k} p_n q_n.$$
The probability that no red sector covers the 6 o'clock direction is
$$P_{n,c} = (1-p)^{2^k}.$$

Then the following equation should hold:
$$P_{c} + P_{n,c} = 1.$$
But it does not hold true. To check this, let us consider the case of $k = 1$. In this case,

$$P_{c} = \frac{1}{2}2p(1-p) + \frac{2}{2}p^2$$
and
$$P_{n,c} = (1-p)^2.$$
Here, $P_c + P_{n,c} \neq 1$.

What am I missing here?

答:
Where your attempt goes awry is in equating the probability, $P_{n,c}$, that no red sector covers the 6 o'clock position with $(1-p)^{2^k}$. The latter is the probability that there are no red sectors at all. But it's entirely possible to have a mix of red and green sectors with the 6 o'clock position covered by green. So $P_{n,c}$ lies somewhere between $(1-p)^{2^k}$ and $1-p^{2^k}$ (since $p^{2^k}$ is the probability that all sectors are red).

In fact, the probability $P_c$ and $P_{n,c}$ are simply $p$ and $1-p$ respectively. That's because the 6 o'clock position is necessarily covered by some sector, and whichever sector that is will be red with probability $p$ and green with probability $1-p$.

Title: Probability: Rotating circular board


问:
GCSE: Find the number of person that will go to a party
Looking please for any help!

John emailed 128 people to invite them to the restaurant's inauguration party. He asked these guests to send the email to 5 others in exchange for participation in a lottery where a free meal for two is won. Based on the assumptions made below, how many of those who received the email are expected to come to the party?

Assumptions
•half of the 128 people emailed the other 5.
•1/4 of these sent the email to 5 others.
•1/8 of these sent the email to 5 others
•1/16 of these sent the email to 5 others.
• No one received the email more than once.
• 10% of those who received the email will go to the party.

My attempt was the following:

128: 2= 64    64+5=69

128:4= 32      32+5=37

128: 8= 16    16+5=21

128:16= 8      8+5=13

10% of 140= 14 person

14 person will go to the restaurant’s inauguration party.

答:
Not sure if this is the answer to some kind of test you need to make...

But note that in 'step' 1, 128 people receive an email. For step 2, we have that out of these 128 persons, 1/2 forwarded it to 5 people, hence #persons receiving an email = $(\tfrac{1}{2}*128)*5$. For the third step, we have that out of these $(\tfrac{1}{2}*128)*5$ people, 1/4th forwards this to 5 new persons, i.e. the number of individuals receiving this email is: $\left(\tfrac{1}{4}\left[(\tfrac{1}{2}*128)*5\right]\right)*5$, and so forth to the $n$th step. Thus we can describe the number of people actually receiving the email with
$$\sum_{i=0}^n \frac{128\cdot 5^i}{2^{(\sum_{j=0}^i j )}}.$$
With $n$ the 'step' number. Your answer is 10% of the above with $n=4$.

(this results in percentages of persons btw)


答:
I haven't read your calculation to look for an error.

The probability that a random position on the wheel is red is just $p$, the probability that each sector is red.

Edit in response to the edited question, that introduces the reward for red.

The reward information is independent of the error.
The argument still relies on a calculation of

the probability that a red sector covers the 6 o'clock direction

that relies on the sum over the possible numbers of red sections. But that probability is still just $p$. So I suspect an error somewhere in that calculation.

Title: Probability: Rotating circular board

Body:
Let there be a circular board that is divided into $2^k$ sectors (similar to a dartboard). Each sector can be red with probability $p$ and green with probability $1- p$. If we rotate the board, when it comes to rest, I am interested in an event that a red sector covers the 6 o'clock direction.

Now let $X$ denote the reward that one gets when the above event occurs. The random variable $X$ takes a value according to the following rule: $X = 1/t$, when $1 \leq t\leq 2^k$ sectors are red and one of them covers the 6 o'clock direction, i.e., the reward of $1$ is shared among those red sectors. I am interested in the probability mass function of $X$. When none of the sectors that are red covers the 6 o'clock direction, the reward is $0$, i.e., $X = 0$.

My attempt: There are $2^k + 1$ combinations depending on how many of the $2^k$ sectors are red, e.g., $0$, $1$, $2$, $\dotsc$, $2^k$. Let $p_n$ denote the probability that $n$ sectors are red before rotating the board. Then we have
$$p_n = {2^k \choose n}p^{n}(1-p)^{2^k-n}, \,\,\,\, 1 \leq n \leq 2^k.$$
Now, for the case when there are $1 \leq t \leq 2^k$ sectors are red, the probability that a red sector covers the 6 o'clock direction is $$q_t = \frac{t}{2^k}.$$

So, one has
$$
X =
\begin{cases}
1, \quad \text{w.p.}~~ p_1q_1 \\
\frac{1}{2}, \quad \text{w.p.}~~ p_2q_2\\
\vdots \\
\frac{1}{2^k}, \quad \text{w.p.}~~ p_{2^k}q_{2^k} \\
0, \quad \text{w.p.}~~(1-p)^{2^k}.
\end{cases}
$$

Then the probability that a red sector covers the 6 o'clock direction is
$$P_{c} = \sum_{n = 1}^{2^k} p_n q_n.$$
The probability that no red sector covers the 6 o'clock direction is
$$P_{n,c} = (1-p)^{2^k}.$$

Then the following equation should hold:
$$P_{c} + P_{n,c} = 1.$$
But it does not hold true. To check this, let us consider the case of $k = 1$. In this case,

$$P_{c} = \frac{1}{2}2p(1-p) + \frac{2}{2}p^2$$
and
$$P_{n,c} = (1-p)^2.$$
Here, $P_c + P_{n,c} \neq 1$.

What am I missing here?


问:
sum of coefficients of polynomial?
If $\sqrt{2+(\sqrt3 +\sqrt5)}$ is root of polynomial of eighth degree then, the sum of absolute values of coefficients of polynomial is?

I found this question on https://brilliant.org/assessment/s/algebra-and-number-theory/1974729/
I want to know is there any simple way to solve it?

答:
Call your number $x$. That means that we have the relation $x = \sqrt{2+(\sqrt3 +\sqrt5)}$

It follows by squaring both sides that $x^2 - 2 = \sqrt{3} + \sqrt{5}$. Further squaring and subtracting leftover integers yields
$$
((x^2 - 2)^2 - 8)^2 - 60 = 0
$$
and you have your polynomial. Now all you need to do is multiply out the parentheses, and you're more or less there.

Note that to get a unique answer, we have to require the polynomial to be monic. I assume that this is the case.

Title: sum of coefficients of polynomial?


问:
Doubts about the slice category
Given a Poset $(X, \leq)$ and $c \in X$, could you illustrate how to see that the slice category $c/X$ is $[c,\rightarrow]$. I understand that I can identify every object $f: c \rightarrow x$ there with x, but I can't see how to identify the morphisms and their composition with the structure of $[c, \rightarrow]$.

Edit: In general, what is the intuitive notion of slice category $c/C$ of $C$ under $c$?

答: The slice category has objects the morphisms $f:c\to x$ and morphisms $g\to f$ commutative diagrams $c\to_g y \to_f x.$ Translating this into the poset language, an object $f$ is just an object $x$ of $X$ with $c\le x$ and a morphism is a relation $c\le y\le x.$ So the slice category $c/X$ is just the set of elements of $X$ which are larger than $c.$

Title: Doubts about the slice category


答:
In a posetal category morphisms are unique in the sense that $\hom_X(a,b)$ is a singleton set if $a\le b$ and empty otherwise. Therefore the compositions are also the unique possible maps.

Specifically, the arrow $c\to x$ as object of $c/X$ corresponds to $x\in [c,\to]=\{x\in X:c\le x\}$, and the (necessarily commutative) triangle $c\to x\to y,\ c\to y$ as morphism in $c/X$ corresponds the unique arrow $x\to y$ in the posetal category of $[c,\to]$ which exists because $x\le y$ (because there's an arrow $x\to y$ in the posetal category of $X$).

Title: Doubts about the slice category

Body:
Given a Poset $(X, \leq)$ and $c \in X$, could you illustrate how to see that the slice category $c/X$ is $[c,\rightarrow]$. I understand that I can identify every object $f: c \rightarrow x$ there with x, but I can't see how to identify the morphisms and their composition with the structure of $[c, \rightarrow]$.

Edit: In general, what is the intuitive notion of slice category $c/C$ of $C$ under $c$?


问:
Find the number of persons who will go to the party
John emailed 128 people to invite them to the restaurant's inauguration party. He asked these guests to send the email to 5 others in exchange for participation in a lottery where a free meal for two is won. Based on the assumptions made below, how many of those who received the email are expected to come to the party?

Assumptions •half of the 128 people emailed the other 5. •1/4 of these sent the email to 5 others. •1/8 of these sent the email to 5 others •1/16 of these sent the email to 5 others. • No one received the email more than once. • 10% of those who received the email will go to the party.

All help/ solutions appreciated!

答:
See my answer at: this equivalent question

(copy-paste)

Not sure if this is the answer to some kind of test you need to make...

But note that in 'step' 1, 128 people receive an email. For step 2, we have that out of these 128 persons, 1/2 forwarded it to 5 people, hence #persons receiving an email = $(\tfrac{1}{2}*128)*5$. For the third step, we have that out of these $(\tfrac{1}{2}*128)*5$ people, 1/4th forwards this to 5 new persons, i.e. the number of individuals receiving this email is: $\left(\tfrac{1}{4}\left[(\tfrac{1}{2}*128)*5\right]\right)*5$, and so forth to the $n$th step. Thus we can describe the number of people actually receiving the email with
$$\sum_{i=0}^n \frac{128\cdot 5^i}{2^{(\sum_{j=0}^i j )}}.$$
With $n$ the 'step' number. Your answer is 10% of the above with $n=4$.

(this results in percentages of persons btw)


问:
Given a 2D polygon in 3D space identify all points that are not a part of the polygon
I do not have a background in mathematics. I am a developer attempting to process point cloud data, but I believe this to be the correct place to post for this type of problem.

Given points of a planar polygon shape based on point cloud data in 3D space, how can all points that are not on this 2D plane be identified?

e.g.:

The red points represent the values that we know, the orange is the interference. None of the points of the interference are known.

The expected result is an approach to identify all of the points in the same 3D space which are not a part of this 2D shape, based on the points known.

Is there an algorithm or mathematical approach available for solving this type of problem?

I hope that this explanation is clear enough, feel free to comment for further clarification if required.

Many thanks.

答:
If we know the red points $p_1,p_2,p_3,p_4$ then we know a close plane to them. Calling this plane $\pi\to a x+b y +c z +d=0$ or $p\cdot \vec n + d=0$ with $p=(x,y,z),\ \ \vec n = (a,b,c)$ we calculate the distances $\lambda_i$ from the points $p_i$ to the plane $\pi$ as follows

$$
(p_i+\lambda_i\vec n)\cdot \vec n + d = 0\Rightarrow \lambda_i = |\frac{d+p_i\cdot\vec n}{\|\vec n\|^2}|
$$

rejecting the points with $\lambda_i \gt \lambda_{ref}$

```<code>SeedRandom[1]
a = 1; b = 1; c = 1; d = 1;
n = {a, b, c};
p = {x, y, z};
pi = p.n + d;
np = 100;
selected = {};
dataxy = RandomReal[{-2, 2}, {np, 2}];
data = Table[{dataxy[[i, 1]], dataxy[[i, 2]], -(dataxy[[i, 1]] a + dataxy[[i, 2]] b + d)/c } + RandomReal[{-1, 1}, 3], {i, 1, np}];
lambdaref = 0.2;

For[i = 1, i <= np, i++,
 lambda = Abs[d + data[[i]].n]/n.n;
 If[lambda <= lambdaref,
   AppendTo[selected, data[[i]]]
 ]
]

cloud = Table[Graphics3D[{Red, Sphere[data[[i]], 0.04]}], {i, 1, np}];
grselected = Table[Graphics3D[{Red, Sphere[selected[[i]], 0.04]}], {i, 1, Length[selected]}];
gr0 = ContourPlot3D[pi == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, Mesh -> False, ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]]];
Show[gr0, cloud, PlotRange -> All]
Show[gr0, grselected, PlotRange -> All]
</code>```

Body:
If we know the red points $p_1,p_2,p_3,p_4$ then we know a close plane to them. Calling this plane $\pi\to a x+b y +c z +d=0$ or $p\cdot \vec n + d=0$ with $p=(x,y,z),\ \ \vec n = (a,b,c)$ we calculate the distances $\lambda_i$ from the points $p_i$ to the plane $\pi$ as follows

$$
(p_i+\lambda_i\vec n)\cdot \vec n + d = 0\Rightarrow \lambda_i = |\frac{d+p_i\cdot\vec n}{\|\vec n\|^2}|
$$

rejecting the points with $\lambda_i \gt \lambda_{ref}$

SeedRandom[1]
a = 1; b = 1; c = 1; d = 1;
n = {a, b, c};
p = {x, y, z};
pi = p.n + d;
np = 100;
selected = {};
dataxy = RandomReal[{-2, 2}, {np, 2}];
data = Table[{dataxy[[i, 1]], dataxy[[i, 2]], -(dataxy[[i, 1]] a + dataxy[[i, 2]] b + d)/c } + RandomReal[{-1, 1}, 3], {i, 1, np}];
lambdaref = 0.2;

For[i = 1, i <= np, i++,
 lambda = Abs[d + data[[i]].n]/n.n;
 If[lambda <= lambdaref,
   AppendTo[selected, data[[i]]]
 ]
]

cloud = Table[Graphics3D[{Red, Sphere[data[[i]], 0.04]}], {i, 1, np}];
grselected = Table[Graphics3D[{Red, Sphere[selected[[i]], 0.04]}], {i, 1, Length[selected]}];
gr0 = ContourPlot3D[pi == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, Mesh -> False, ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]]];
Show[gr0, cloud, PlotRange -> All]
Show[gr0, grselected, PlotRange -> All]


问:
Write the definition of prime number using symbols and no words
I'm trying to write the definition of prime numbers using only symbols.
Here is what I have:
A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers.

Note:

N$^{*}$ is the set of all natural numbers except for $0$.

So the set of prime numbers is S = {x $\in$ N$^{*}$, x $\neq$ 1 $|$ ... [I'm stuck here]}

I don't know how to explain. I was just introduced to proofs and sets so I do not want to use complex symbols. Can someone suggest me a simple way for the "such that" part, please?
Thank you!

答: $$S = \{ x \in \mathbb{N}, x \neq 1 | \forall p,q \in \mathbb{N},\  pq=x \implies p,q \in \{1, x\} \}$$


问:
Series of Cyclotomic polynomials
How can I show that if $\Phi$ is a Cyclotomic polynomial, $$\Phi_n(x)=\prod_{\substack{1\leq k\leq n\\(n,k)=1}}(x-\zeta_n^k)$$
With $\frac{d}{dx}\Phi_n(x)=\Phi'_n(x)$

Then, $$\sum_{n=1}^\infty\frac{1}{n^4}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{45\zeta(3)}{\pi^4e^{2\pi}}+\frac{7}{4\pi e^{2\pi}}$$

$$\sum_{n=1}^\infty\frac{1}{n^8}\frac{\Phi'_n(e^{2\pi})}{\Phi_n(e^{2\pi})}=\frac{4725\zeta(7)}{\pi^8e^{2\pi}}+\frac{19}{12\pi e^{2\pi}}$$

答:
$\Phi'_n(x)/\Phi_n(x) = \sum 1/(x - \zeta_n^k) =  x^{-1} \sum 1/(1 - \zeta_n^kx^{-1}) = x^{-1} \sum_{d \ge 0} \zeta_n^{kd} x^{-d}$.

Let $S(n,d) = \sum \zeta_n^{kd}$. We easily have $S(1,d) = 1, S(n,0) = \phi(n)$ and $S(n,1) = \mu(n)$ (the Möbius function).

If $\gcd(d,n) = 1$ then $\zeta_n^k \mapsto \zeta_n^{kd}$ is a permutation of the roots, so $S(n,d) = S(n,1) = \mu(n)$.

If not, then write $d = gd'$ and $n = gn'$ where $g = \gcd(d,n)$. We have $\zeta_n^d = \zeta_{n'}^{d'}$, and so $S(n,d) = \phi(n)S(n',d')/\phi(n')$. Then, since $\gcd(n',d') = 1$, $S(n,d) = \phi(n)\mu(n',d')/\phi(n')$.

Now, $\Psi(s,x) = \sum n^{-s}\Phi'_n(x)/\Phi_n(x) = x^{-1}\sum_{d \ge 0} (\sum_{n \ge 1} S(n,d)n^{-s})x^{-d}$.

Since $\phi$ and $\mu$ are multiplicative functions, so is $S(n,d)$ when $d$ is fixed.

If $v_p(d) = m$, with the convention that $v_p(0) = \infty$ and $p^{-\infty} = 0$, we see that $$\sum_{k \ge 0} S(p^k,d)p^{-ks} = \sum_{0 \le k \le m} \phi(p^k)p^{-ks} + \sum_{k \gt m} \phi(p^k)\mu(p^{k-m})p^{-ks}/\phi(p^{k-m})
\\ = 1 + \sum_{1 \le k \le m} (p-1)p^{-1+k(1-s)} - p^{-s+m(1-s)} = (1 - p^{-s})\sum_{0 \le k \le m} p^{k(1-s)}
\\ = (1 - p^{-s})\sigma_{1-s}(p^m)$$

Hence, $$\sum_{n \ge 1} S(n,d)n^{-s} = \prod (1-p^{-s}) \sigma_{1-s}(p^{v_p(d)}) = \sigma_{1-s}(d)/\zeta(s)$$

Notice that if $d=0$, then $\sigma_{1-s}(0) = \zeta(s-1)$, and if $d>0$, then $\sigma_{1-s}(d) = \sigma_{s-1}(d)/d^{s-1}$, which proves that :
$$\Psi(s,x) = \frac 1 {x\zeta(s)}\left(\zeta(s-1) + \sum_{d\ge 1} \frac{\sigma_{s-1}(d)}{d^{s-1} x^d} \right)$$

Title: Series of Cyclotomic polynomials


问:
How to integrate this $\int \left[(u+a)(u+b)\right]^{n-1} du$?
I having some problems with the following integral

$$\int \left[(u+a)(u+b)\right]^{n-1} du$$

I try to use $u$-substituion and integration by parts but is not working. Can someone help me?

Body: Write $(u+a)(u+b)=x^2-c^2$ with $x:=u+\tfrac{a+b}{2},\,c:=\frac{a-b}{2}$ so for $n\ge1$ your integral is$$\begin{align}\int(x^2-c^2)^{n-1}dx&=\int\sum_{k=0}^{n-1}\binom{n-1}{k}(-c^2)^{n-1-k}x^{2k}dx\\&=\sum_{k=0}^{n-1}\binom{n-1}{k}(-c^2)^{n-1-k}\frac{x^{2k+1}}{2k+1}+C\\&=\sum_{k=0}^{n-1}\binom{n-1}{k}\left(-\left(\tfrac{a+b}{2}\right)^2\right)^{n-1-k}\frac{(u+\tfrac{a+b}{2})^{2k+1}}{2k+1}+C\end{align}$$by the binomial theorem.


问:
Direct sum, product, sum and intersection of ideals
Is is true that in general the direct sum of an ideal of a ring with itself is equal to the square of that ideal (i.e. the product with itself). More generally, is $I \oplus J = IJ$ for I,J ideals?

Moreover, what can be said in general about the relation of the direct sum, the sum of ideals and their intersection?

In the particular case of $R = \mathbb{Z}[\sqrt{-5}]$ and the ideal $I = (2,1+\sqrt{-5})$, I know that $I^2 = (2)$, but I think one can also show that $I^2 = R^2$ as $R$-modules. Is this true?

答:
There's no reason for the equality $I+J=IJ.$ If you want a specific example, consider $I=J=(2)\subset \mathbb Z.$ Then $I+J=(2)$ whereas $IJ=(4).$ In general we have the relationship $IJ\subset I\cap J\subset I+J.$ The containment $IJ\subset I\cap J$ is actually an equality if $I$ and $J$ are comaximal, that is, if $I+J=R$ is the unit ideal (as long as $R$ is commutative).

In your particular example it is true that $I^2=(2).$ However, $1\in R,$ so $1=1^2\in R^2.$ Thus $R^2=R$ as ideals.

This is all contained in chapter 7 of Dummit and Foote's book.


问:
In a multinomial distribution, compute $\mathbb E[I_i \cdot I_j]$ where $I_i = 1$ when event $i$ doesn't occur at all
Suppose there are $n$ independent trials each of which can take the values :  $n_1,n_2, \cdots, n_r$ with probabilities : $p_1,p_2,\cdots,p_r$.

Let the Indicator variable $I_j = 1 $ when the event $j$ never occurs in the $n$ trials and $0$ otherwise.

I am trying to prove that $\mathbb E[I_i \cdot I_j] = ( 1 - p_i - p_j)^n$ where $i \ne j , 1 <= i <= r , 1 <=j <= r ; ~i,j$ are integers

Attempt: From a previous problem, I know that $I_i, I_j$ are not independent.

Now  $\mathbb E[I_i \cdot I_j] = 1 \cdot P(I_i \cdot I_j = 1)$

$P(I_i \cdot I_j = 1) = P(I_i = 1 ~\cap~I_j = 1) = P(I_i = 1 ~\cup~I_j = 1) - P(I_i = 1) - P(I_j = 1)$

How can I evaluate this further? So, neither $i$th or $j$th event happens with probability $(1 - P(\text{ith event or jth event}))^n$.  What I am a little confused is : As per the intended result, how is $P(\text{ith event or jth event})) = p_i+p_j$. These events are not mutually exclusive!

Help is much appreciated!

答:
$I_i I_j=1$ if $I_i=1$ and $I_j=1$, meaning that neither event $i$ nor event $j$ ever occurs, otherwise it is zero. Therefore

$$E[I_i I_j]=1 \cdot P(\text{neither } i \text{ nor } j \text{ ever occurs}) + 0 \cdot (1-P(\text{neither } i \text{ nor } j \text{ ever occurs})).$$

Can you proceed from there?


问:
About $\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$
How to show $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$ ?

I've found this identity in this post. Following the comment by rnorris in that post "use integration by parts, starting from the right-hand side. Let $u=\sin^2(x)$, and $dv=x^{-2}dx$. Look at trig identities that involve the product $\sin(x)\cos(x)$."

I arrived to $$-\dfrac{\sin^2(x)}{x}\Big\vert_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\dfrac{\sin(2x)}{x}\mathrm dx$$ where I used the trig identity $\sin(x+y)=\sin(x)\cos(x)+\cos(x)\sin(x)$ to change  $\sin(x)\cos(x)$.

Could someone tell me what can be done next?

I don't see how I could  modify the integral to have $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x $

Also, the trig identity $\sin(x+y)=\sin(x)\cos(x)+\cos(x)\sin(x)$, was that the one to be applied in the first place?

Thanks in advance.

答:
Instead of using the trig identity you should do a change of variable.

Let $u = 2x$.


问:
Expected number of intersections within a circle
Consider a circle. We perform random walks on the boundary of this circle. Without loss of generality, consider the circle to be centered at the origin, and our random walk starts $0^{\circ}$. You return to the starting point after $n$ steps.

We will make $n$ total steps. Each step you randomly traverse to some point on the boundary. There is a straight line that connects the points for each step. So for example, if you traverse to $90^{\circ}$ on the first step, there will be a straight line going from $0^{\circ}$ to $90^{\circ}$, and then if you go to $155^{\circ}$ afterwards, there will then be another line going from $90^{\circ}$ to $155^{\circ}$.

What is the expected number of intersections created by these line segments?

I think this question may be somewhat similar to Expected number of intersection points when $n$ random chords are drawn in a circle, but here all the $n$ line segments are connected. Which means two consecutive line segments cannot intersect one another. So this reduces the maximum number of intersections from $\frac{n(n-1)}{2}$ to $\frac{(n-1)(n-2)}{2}$, I believe. The probability of an intersection between these pairs is still $1/3$.

So does the solution simply just become $\frac{1}{6}(n-1)(n-2)$? I saw some discussions offline that it could be $\frac{1}{6}n(n-3)$.

答:
Because you end at the starting point, there are $$\binom{n}{2} - n = \frac{n(n-3)}{2}$$ pairs of non-adjacent chords, and each pair of these chords intersects with probability $\frac13$ (here you use that the points are sampled uniformly, as in Expected number of intersection points when $n$ random chords are drawn in a circle). By linearity of expectation, the total expected number of intersections is $\frac{n(n-3)}{6}$ (coinciding intersections, i.e., three or more chords passing through the same point, happen with probability $0$).

Your formula $\frac{(n-1)(n-2)}{6}$ would be correct if the walk would not necessarily end at the starting point.
Note that for $n=3$ your formula gives $\frac13$, whereas the answer should be $0$ because the walk ends at the starting point.


问:
Binary search complexity
In sorted array of numbers binary search gives us comlexity of O(logN).
How will the complexity change if we split array into 3 parts instead of 2 during search?</p>

答: Same. You will get a running time only differing by constant factor ($\log_2 3=\frac{\ln 3}{\ln 2}$).</p>

Title: Binary search complexity


答:
Let $ax=t$
$$I=\int_{-\infty}^{\infty} \frac{\sin ax}{x} dx=\int_{-\infty}^{\infty} \frac{\sin t}{t} dt$$
$I$ is independent of $a$.

Title: About $\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$

Body:
How to show $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$ ?

I've found this identity in this post. Following the comment by rnorris in that post "use integration by parts, starting from the right-hand side. Let $u=\sin^2(x)$, and $dv=x^{-2}dx$. Look at trig identities that involve the product $\sin(x)\cos(x)$."

I arrived to $$-\dfrac{\sin^2(x)}{x}\Big\vert_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\dfrac{\sin(2x)}{x}\mathrm dx$$ where I used the trig identity $\sin(x+y)=\sin(x)\cos(x)+\cos(x)\sin(x)$ to change  $\sin(x)\cos(x)$.

Could someone tell me what can be done next?

I don't see how I could  modify the integral to have $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x $

Also, the trig identity $\sin(x+y)=\sin(x)\cos(x)+\cos(x)\sin(x)$, was that the one to be applied in the first place?

Thanks in advance.


问:
Why, when we impose restrictions such as the ones of our physical world, new properties arise?
Unfortunately I am not and expert mathematician nor a philosopher so I don't have the right words to phrase the concept, but the following example should be able to make the question clear:

Take the complex numbers. We can't impose an order relationship between them (>, <). But as we impose a new constrain by saying that $\sqrt{-1}$ doesn't exist, we find that we can now tell which number is bigger than another.

How can we rigorously explain what's going on?

EDIT: Second example concerning Achilles and the tortoise removed since multiple users pointed out that it contained numerous errors. Minor corrections.

答:
Turning my comment above into an answer:

As Don Thousand commented above, narrower classes of objects have more guaranteed properties a priori. You're focusing on tameness properties, but it's also worth noting that we can get guaranteed pathologies this way too. For example, here are a couple ways in which adding a natural requirement to a starting class of structures can limit our ability to do algebra:

When we shift from the class of all fields (which includes both $\mathbb{R}$ and $\mathbb{C}$) to the class of ordered fields (which includes $\mathbb{R}$ but not $\mathbb{C}$), we are guaranteed to lose the ability to solve the equation $x^2+1=0$.

When we shift from the class of all rings (which includes both $\mathbb{Z}$ and $\mathbb{Q}$) to the class of discrete rings (which includes $\mathbb{Z}$ but not $\mathbb{Q}$), we are guaranteed to lose the ability to solve the equation $2x=1$.

More restrictions make things more decisive, but even very "nice" restrictions do not necessarily make things better.


问:
Finding the angle between the radius and vector
I'm trying to find the angle between the radius vector produced and the tangent to the curve whose equation is $r = 4\cos\theta$ at the point where $\theta = \frac{\pi}{4}$.

My assumption was that I take the derivative, such that $\frac{dr}{d\theta}=-4\sin\theta$. Then substitute in the value for $\theta$.

Though, this isn't the right answers, for I expect $\frac{3\pi}{4}$.

I would appreciate your help, and if possible a pictorial example to help me understand this. The book i'm using is fairly terse and vague.

答:
Angle enclosed in a semicircle, other two for given condition are shown in isosceles right triangle.

Actually two correct answers. The angle radius vector makes to the outer tangent $\psi$ is $ 3 \pi/4$ but considering inner tangent it could be also $ \pi/4$ as shown... even with a consistent CCW convention.

The differential equation  determines the tangent. Its ode is $ \sin \psi =r/4 $. At the given radius, accordingly there are two arcsin $\psi$ values


问:
Magic squares and matrices
A $3 × 3$ magic square with entries in $\mathbb{R}$ is a $3 × 3$ matrix $A = (a_{ij} )$ $\in$ $\mathbb{R}^{3× 3}$such that the sums of the numbers in each row, each column, and both main
diagonals are the same. That common sum is called the magic sum of the
magic square. Denote by MQ the set of $3 × 3$ magic squares, and for $a ∈ \mathbb{R}$,
denote by $MQ_a \subset MQ$, the subset of those magic squares with magic sum
a.

How do I find the magic square in $MQ_3$ and also the general formula for $MQ_a$?

Any help is much appreciated.

答:
Start by considering the constraints of the magic square. I denote the sum by $s$ rather than $a$ to avoid confusion. The sum of diagonals is
$$ a_{11}+a_{13}+a_{31}+a_{33}+2 a_{22}=2 s$$
The sum of the first and last row minus the second column is
$$ (a_{11}+a_{12}+a_{13})+(a_{31}+a_{32}+a_{33})-(a_{12}+a_{22}+a_{32})=a_{11}+a_{13}+a_{31}+a_{33}-a_{22} =s$$
From these equations you get $a_{22}= s/3$, in your case $a_{22}=1$.

Fix now two entries that are not on the same diagonal, let say $a_{11}$ and $a_{13}$. Given the constraints, all other entries are now fixed.

EDIT

We have $a_{22}=s/3$ and we fix $a_{11}=b$, $a_{13}=b'$ ($b$ and $b'$ are names that i give to recall that these entries are fixed).

The entry $a_{33}$ is fixed due to the constraint on the diagonal
$$a_{11}+ a_{22}+a_{33}= s\Rightarrow  a_{33} = s-\frac s3-b=\frac{2s}{3}-b$$
Similarly, $a_{31}=\frac{2s}{3}-b'$

We now have all the corner entries. This fixes all the other entries. For example,
$$a_{12}=s-a_{11}-a_{13}=s-b-b'$$
Similarly,
$$a_{21}=\frac{s}{3}-b+b', \quad a_{23}=\frac{s}{3}+b-b',\quad a_{32}=b+b'-\frac{s}{3}$$

(I let you verify that all the constraints are indeed verified.)

Title: Magic squares and matrices


问:
If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$
I have come across the following comment in the lecture:

If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$.

I do not see why this is true. I attempted to prove it by rewriting $q-2$ as $q-2 = 8k+3$ for some integer $k$ and tried to see what happens when I divide it by an arbitrary prime. However, this did not help. Could you please give me a hint?

答: Hint: If all prime factors of $n$ are congruent to  $\pm 1 \bmod 8$, then $n \equiv \pm 1 \bmod 8$.


问:
How to derive test statistic under composite null hypothesis?
Suppose I have a multiple linear regression of the form:

$$y_i = \beta_0 + \beta_1 x_{i1} + \beta_2 x_{i2} +\epsilon$$

Then suppose I am interested in testing this set of hypotheses:

$$H_0: \beta_1 = 0 \text{ OR } \beta_2 = 0$$
$$H_A: \beta_1 \neq 0 \text{ AND } \beta_2 \neq 0$$

I am interested in deriving a test statistic for this hypothesis.  I understand that I need to derive this under the null hypothesis.  However, it is not as straightforward as just plugging in $\beta_1 = 0 \text{ and } \beta_2 = 0$, since our null is that either/or is zero, not both.

Any input would be much appreciated, thank you!

答:
Your null hypothesis is equivalent to
$$
H_0: \min \{ |\beta_1|, |\beta_2| \}= 0
$$

In general, you are interested in the following problem:
$$
H_0 : \theta \in \bigcup_{i=1}^k \Theta_i, \qquad \text{vs.} \qquad H_1 : \theta \in \bigcap_{i=1}^k \Theta_i^C
$$

If we let $R_i, i=1,\dots,k$ be the rejection regions for the local tests $H_{i0}: \theta \in \Theta_i$ versus $H_{i1}=\theta \in \Theta_i^C$, then (under some conditions) the optimal test to use is the IUT (Intersection Union Test), which states that the rejection region of the global test is the one with $R = \cup_{i=1}^k R_i$.

This is covered in the following paper, see section 1.3 for a description of a very similar problem to yours.

So to perform the test in your question, the approach is to consider the individual tests:

$$
T_1: \quad H_{10}: \beta_1 = 0 \qquad \text{vs.}\qquad H_{11}: \beta_1 \neq 0,
$$
$$
T_2: \quad H_{20}: \beta_2 = 0 \qquad \text{vs.}\qquad H_{21}: \beta_2 \neq 0,
$$

Find a rejection region for each test using the standard theory, then take an intersection to find a rejection region for the global test.


答:
The "paradox" of Achilles and the Tortoise has nothing to do with uncountable infinities. It can be easily be phrased within the (countable) rationals, since it is about
$$\tag1
\sum_{n=1}^\infty\frac1{2^n}=1.
$$
And in fact it doesn't require  infinity to be phrased, as it could be written as
$$\tag2
\forall \varepsilon>0,\ \exists N\in\mathbb N:\ \ \ \ 1-\varepsilon<\sum_{n=1}^N\frac1{2^n}<1
$$
and proven within the field of rational numbers.
That said, the only paradox in the "paradox" lies in the mistake of not noticing that in the setup of the paradox time is also subject to $(2)$.

As for your example with the complex numbers, I fail where you want to go, but that kind of thinking can be made in scores of situations:

In going from $\mathbb R$ to $\mathbb Q$ you "gain" that now every number is a fraction.

In going from $\mathbb Q$ to $\mathbb Z$ you "gain" that every element has a predecessor and a successor.

In going from $\mathbb Z$ to $\mathbb N$ we "gain" that every subset has a first element.

It is easy to argue that there lots of "gains" going the other way, too.

Title: Why, when we impose restrictions such as the ones of our physical world, new properties arise?

Body:
Unfortunately I am not and expert mathematician nor a philosopher so I don't have the right words to phrase the concept, but the following example should be able to make the question clear:

Take the complex numbers. We can't impose an order relationship between them (>, <). But as we impose a new constrain by saying that $\sqrt{-1}$ doesn't exist, we find that we can now tell which number is bigger than another.

How can we rigorously explain what's going on?

EDIT: Second example concerning Achilles and the tortoise removed since multiple users pointed out that it contained numerous errors. Minor corrections.


问:
getting line-line intersection if the axis intersections are given
A,B,C and D points are given, how can I calculate the E point?
I know this method, but I would like to know if there is any other solution exists, that does not need the equalation of both lines.

Maybe there is some trick using Pythagorean theorem or Intercept theorem or anything else.

Thank you for any advice!

答: Hint: Use Thales' Theorem, e.g. in the triangle A0B with E