问:
Normal subgroups: is the element $g^{-1}$ of the product $gng^{-1}$ required to be in the group?
Considering a group such as the integers, how's it possible to have normal subgroups if there are no rational numbers which are the the multiplicative inverse to conjugate with?

In particular the word is usually written as a product $gng^{-1}$ and the integers would have addition defined on it and not multiplying.

I just guess it is okay to do the calculation outside of the group setting yet that is the question.

答:
The group of integers $\Bbb Z$ is additive. Its operation is $+$. Thus the conjugation of $\Bbb Z$ by $g\in \Bbb Z$ is given by

$$g+\Bbb Z-g=\{ g+x-g\in \Bbb Z\mid x\in\Bbb Z\}.$$

Warning: one way to view the group $\Bbb Z$ is via the presentation $\langle a\mid \rangle$; that is, it is the free group on one generator. It is possible, then, to write $g\Bbb Z g^{-1}$, since, in this case, the elements of $\Bbb Z$ are understood as powers of the abstract symbol $a$.


问:
Finding the bound of a sequence sum
I am dealing with a sum of this form:
$S = l *[b^{-1}*a_1 + b^{-2}*a_1*a_2 + b^{-3}*a_1*a_2*a_3...+b^{-n}*a_1*a_2...*a_n]$, where $l$ is some positive constant, $b$ is a constant $>1$, and the sum of terms $a_1+a_2+a_3...+a_n = C$, where C is some constant $>n$. I initially was wondering if $S$ is maximized when $a_1=a_2=a_3...=a_n$, however this appears not to be the case. Is there a general way to calculate the upper bound of $S$ in terms of constants, or to identify the values of $a_1, a_2...$ such that $S$ is maximized?

答:
Take the simple case of maximizing $J = a_0 + a_0 a_1$ subject to $a_0 + a_1 = A$. The constraint tells us
$a_1 = A - a_0$ so $J = a_0 + a_0(A - a_0) = (A + 1)a_0 - a_0^2$, which has an extremum at $a_0 = (A + 1)/2$, $a_1 = (A - 1)/2$, its value there $(A + 1)^2/4$. Might help a bit, and it does give you a counterexample.

Thanks to @Mindlack for pointing an algebra misstep of mine, now fixed.

ContentLicense: CC BY-SA 4.0

Title: Finding the bound of a sequence sum


问:
What's the possible solutions for a 6 known parameters of 3x3 Orthogonal Matrix?
What is the possible solutions for an $3x3$ size orthogonal matrix with the known parameters of the first two columns which makes 6 parameters known?

Let's think of a rotation matrix which is an orthogonal matrix:

$R= \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$

The parameters of $r_{11},r_{21},r_{31}, r_{12},r_{22},r_{32}$ are known values. And it is desired to find the last (third) column with the parameters of $r_{13},r_{23},r_{33}$.

Is there a unique solution? What are/is the possible solution(s)?

I am confused with possible solutions, therefore I wanted to ask to be sure, how to find it.

Thanks in advance.

Edit: I take the square root of each raw which is equivalent to 1 unit vector.

$r_{11}^2+r_{12}^2+r_{13}^2=1$

$r_{21}^2+r_{22}^2+r_{23}^2=1$

$r_{31}^2+r_{32}^2+r_{33}^2=1$

Therefore we can get to the equations of

$r_{13}= \pm \sqrt[]{1-r_{11}^2-r_{12}^2}$

$r_{23}= \pm \sqrt[]{1-r_{21}^2-r_{22}^2}$

$r_{33}= \pm \sqrt[]{1-r_{31}^2-r_{32}^2}$

However, if there is one solution for the rotation matrix of R, how can I find that unique solution. Square root of any raw gives me two solutions.
Whichever (positive/negative square root of $r_{13},r_{23},r_{33}$) I would take, it would be still an orthogonal matrix:

$r_{13}^2+r_{23}^2+r_{33}^2=1$

答:
An ortogonal matrix is a matrix whose columns are an orthonormal basis of vectors. So, if you are given the first two columns, $(r_{11},r_{21},r_{31})$ and $(r_{12},r_{22},r_{32})$, the third column must ve a vector which forms an orthonormal base with the other two. That is, it is a vector perpendicular to the other two columns and of norm 1.  If you are told it is a rotation matrix, there will be just 1 possible vector. To calculate it you just have to take the cross product of the two other columns.

The cross product of two perpendicular unit vectors $v_1$, $v_2$ is the only vector $v_3$ such that $v_1,v_2,v_3$ form a orthonormal basis and the determinant of the matrix with columns $v_1,v_2,v_3$, in that order, is positive. So you have to take the third column to be the cross product of the first column and the second column.


问:
Existance of piecewise continious function if laplace transformation is given.
How to confirm if there exists a piecewise continuous function $f(t)$ whose Laplace transform $F(s)$ is given? I know the existence theorem for the existence of Laplace transform but don't know how to use it?

Say if $F(s)=e^{s^{2}+1}$, how can I confirm if there exist a piecewise continuous function $f(t)$ corresponding to the given $F(s)$?

Could you please explain it?

Thank you!

答:
Not an answer

I plotted numerical integrals for $t\in[-4,4]$
$$f(t)={\mathcal {L}}^{-1}\{e^{s^2+1}\}(t)={\frac {1}{2\pi i}}\int _{ -i \infty}^{i\infty}e^{st}e^{s^2+1}\,ds$$
There should exist an inverse


问:
What can be said about a linear transformation if there is a basis where its matrix is symmetric?
What can be said about a linear transformation if there is a basis where its matrix is symmetric?

And how about the opposite direction? That is, if we have a symmetric matrix, what can be said about it as a linear transformation.

Thank you!

答:
You can consider to start the set of Hermitian matrices. It will be relevant since a matrix that has only real entries is Hermitian if and only if it is symmetric, and a real and symmetric matrix is a large special case of Hermitian matrices and likewise a large special case of symmetric matrices.

We have the following results:

Every Hermitian matrix is a normal matrix.

The sum of any two Hermitian matrices is Hermitian.

The inverse of an invertible Hermitian matrix is Hermitian as well.

The finite-dimensional spectral theorem says that any Hermitian matrix can be diagonalized by a unitary matrix, and that the resulting diagonal matrix has only real entries. This implies that all eigenvalues of a Hermitian matrix A with dimension n are real, and that A has n linearly independent eigenvectors.

The product of two Hermitian matrices A and B is Hermitian if and only if $AB = BA$.

$A^{n}$ is Hermitian if A is Hermitian and n is an integer.

If A and B are Hermitian, then $ABA$ is also Hermitian.

The determinant of a Hermitian matrix is real.

There are even more things you can say about Hermitian matrices but now is a good time to stop.


问:
Solving this question using ln?
I know that ln is used to bring down exponents usually, so I was wondering where would I go from there ? I've looked at symbolab and don't really understand the implicit differentiation approach.

Here's the question $e^{(x^2)y}=2x+2y$

I was able to get to:

$$x^2y \ln(e) = \ln(2x)+ \ln(2y)$$

Where would i go from here ?

答:
I'd like to point out that you have made a common mistake. You seem to believe (correct me if I'm wrong) that
$$\ln(a+b)=\ln a+\ln b$$
This is wrong. In actual fact, the log law that you may be confusing this with is:

$$\ln(ab)=\ln a+\ln b$$
Note: this log law is not unique to the natural logarithm, it is a property of all logarithms.

About your question, your equation has infinitely many solutions. I assume that you want to differentiate it implicitly (as opposed to solving it), so this is how you'd do it. It is inconvenient in this case to use $\ln$ as we don't have a product on the right hand side of our original equation, so I will differentiate directly:

$$e^{x^2y}=2x+2y$$
Now, differentiating with respect to $x$, and using the fact that $\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$ we have
$$\begin{align}
&\frac{d}{dx}(e^{x^2y})=\frac{d}{dx}(2x)+\frac{d}{dx}(2y)\\
&\left(2xy+x^2\frac{dy}{dx}\right)e^{x^2y}=2+2\frac{dy}{dx}
\end{align}$$
Now rearrange to find $\frac{dy}{dx}$. If you need any more help please don't hesitate to ask.

Title: Solving this question using ln?


问:
Is node connectivity simply the smallest node degree in the network?
I was reading Network Science with Python and NetworkX, which has the following definition:

The node connectivity is the smallest min-cut over all node pairs. The edge connectivity is defined similarly.

It then proceeds to show that Zachary's Karate Club, as well as several other examples, has a node connectivity of 1. Does this simply means that there exists a node with a single edge (i.e. a degree of 1)?

In general, could the node connectivity ever exceed the smallest degree node in the network? Will it ever be less than that?

答:
Definitions are adopted from "Graph Theory" by Reinhard Diestel. All considered graphs are finite, simple and undirected.

A graph $G=(V,E)$ is $k$-connected (for $k \in \mathbb{N}$) if $|G| > k$ and $G-X$ is connected for every set $X \subseteq V$ with $|X| < k$. In other words, no two vertices of $G$ are separated by fewer than $k$ other vertices. We call the greatest integer $k$ such that $G$ is $k$-connected the connectivity of $G$ and denote it by $\kappa(G)$.

If $|G| > 1$ and $G-F$ is connected for every set $F \subseteq E$ of fewer than $l$ edges, then $G$ is $l$-edge-connected. The greatest integer $l$ such that $G$ is $l$-edge-connected is the edge-connetivity $\lambda(G)$ of $G$.

The following inequalities are due to Whitney.

Theorem: Let $G=(V,E)$ be a graph with minimum degree $\delta(G)$, then $\kappa(G) \leq \lambda(G) \leq \delta(G)$.

Proof Sketch: Consider a vertex $v$ of minimum degree $deg(v)=\delta(G)$. By removing $deg(v)$ edges that are incident to $v$, $G$ becomes disconnected. We need to show that $\kappa(G) \leq \lambda(G)$. Let $X \subset E$ be the set of edges incident to $v$. Find a set of at most $|X|$ vertices that disconnects $G$ (this is relatively straight forward by the way). Deleting a vertex, also deletes every edge that is incident to that vertex (using this, we can show that assuming $\lambda(G) < \kappa(G)$ leads to contradiction).


问:
Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Prompt:

Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question:

I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$.

What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane?

Is there any intuition to be gained here?

答: This equation can be restated as $(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$. With $z:=z_1-z_2$, some $\omega=(\exp\pm2\pi i/3)$ satisfies $z_2-z_3=\omega z,\,z_3-z_1=\omega^2z$. Since $\omega^3=1\ne\omega$, $\omega^2+\omega+1=\frac{\omega^3-1}{\omega-1}=0$. The LHS is then $z^2(1+\omega^2+\omega^4)=z^2(1+\omega^2+\omega)=0$.

ContentLicense: CC BY-SA 3.0


问:
Does the following converge uniformly?
Does $$f_n(x)=\frac{nx}{1+n(1+x^2)}$$ converge uniformly on $\mathbb{R}$?  To what?  Justify.</p>

答:
Note that $${f_n}(x) = \frac{x}{{\frac{1}{n} + 1 + {x^2}}}$$

Can you work something out now? Note that the pointwise limit is $$f(x)=\frac{x}{1+x^2}$$ and that $$\begin{align}
  \left| {f(x) - {f_n}(x)} \right| &= \left| {\frac{x}{{\left( {1 + {x^2}} \right)\left( {1 + n\left( {1 + {x^2}} \right)} \right)}}} \right| \\
   &\leqslant \frac{1}{2}\frac{1}{{1 + n\left( {1 + {x^2}} \right)}}\\
   &\leqslant \frac{1}{2}\frac{1}{{n\left( {1 + {x^2}} \right)}} \\
   &\leqslant \frac{1}{{2n}} \end{align} $$

because $\dfrac{x}{1+x^2}$ is at most $1/2$ and $\dfrac{1}{1+x^2}$ is at most $1$ over the real line.

Title: Does the following converge uniformly?


问:
Why is $(b-Ax)^T(b-Ax) = (b^Tb - 2b^T Ax + x^T A^T Ax)$?
Why is it true that
$$
(b-Ax)^T(b-Ax) = (b^Tb - b^T Ax - b^T Ax + x^T A^T Ax)
$$
Shouldn't it be
$$
(b-Ax)^T(b-Ax) = (b^Tb - b^T Ax - x^T A^T b + x^T A^T Ax)
$$?

So in why are there two $b^T Ax$? This must mean that $x^T A^T b = b^T Ax$. If true, why can you assume that? $b$ is a $M \times 1$ vector, $x$ is a $N \times 1$ vector, and $A$ is a $M \times N$ matrix.

答: It is true since $a^{T}b = b^{T}a$ (dot product is commutative). Simply take $a=Ax$.


问:
Multiplying two numbers is the average squared minus sum of odd numbers of k equal average minus smallest number. Why?
Multiplying two numbers is the average squared minus sum of odd numbers k equal average minus smallest number.

Why is that happening? What is the logic behind it? It is puzzling me.

For example (examples only for numbers that give integer averages, since for fractional averages you have to add the first number to the result):

```<code>a) 5 * 3 is (4*4)-1 (since difference between 4 and 5 is 1, its only the 1st odd number substracted)
b) 3 * 9 is (6*6)-1-3-5 (since difference between 6 and 3 is 3, its first 3 odd numbers substracted)
c) 21 * 35 is (28*28)-1-3-5-7-9-11-13 (difference between 28 and 21 is 7, its first 7 odd numbers substracted)
d) 21 * 36 is (28*28)-1-3-5-7-9-11-13+21 (fractional average, compute with *35 and add another 21 at the end)
and so on ... 233 * 53213 ... (too many odd numbers between 26723 and 233, but the result is sound)
</code>```

This is a basic code implemented to compute the formula

```<code>function multiply(no1,no2)

 no_computation = ( floor( (no1 + no2)/2) )^2;
 for k from 1 to abs( floor((nr1 + nr2)/2) - no1)
  no_computation = no_computation -  (k*2-1);
 end of for
 multiply_result = no_computation + modulo(no1+no2, 2)*no1;

end of function multiply
</code>```

Steps

The "no_computation" variable is initialized as the average value, squared. If the average is fractional then it takes the largest integer smaller than the average (ie, a no1 number is substracted in order to give an integer average).

Then the "for" function determines how many odd numbers there are between the average and the first number, and for each time it substracts the next odd number and refreshes the value of "no_computation"

the multiply_result is the value "no_computation" with the added twist that if the average was fractional then it'll correct the result by adding the value of number no1 (since in the 1st step above a no1 number was substracted at the "floor" operation). This is done by "modulo" function.

What arithmetic property is involved in this computation?

答:
The sum of the first odd integers is the square of the quantity of numbers we added
$$\sum _{k=1}^{n} (2 k-1)=n^2$$
You subtract from the square of the mean $$\left(\frac{a+b}{2}\right)^2$$
the quantity of odd integers given by the difference between the mean and the minimum between $a$ and $b$
$$\sum _{k=1}^{\frac{a+b}{2}-\min (a,b)} (2 k-1)=\frac{1}{4} \left(-4 a \min (a,b)+4 \min (a,b)^2-4 b \min (a,b)+a^2+2 a b+b^2\right)=$$
$$=\frac{1}{4} \left(4 a \min (a,b)-4 \min (a,b)^2+4 b \min (a,b)-a^2-2 a b-b^2\right)+\frac{1}{4} (a+b)^2=$$
$$=a \min (a,b)- \min (a,b)^2+b \min (a,b)+\left[-\frac{1}{4} a^2-\frac{1}{2} ab-\frac{1}{4} b^2+\frac{1}{4} a^2+\frac{1}{2} ab+\frac{1}{4} b^2\right]=$$
the terms in the square bracket cancel
$$=a \min (a,b)+b \min (a,b)-\min (a,b)^2=\ldots$$
now suppose that $\min (a,b)=a$. We get
$$a \min (a,b)+b \min (a,b)-\min (a,b)^2=a^2+ab-a^2=ab$$
if $\min (a,b)=b$, we get
$$a \min (a,b)+b \min (a,b)-\min (a,b)^2=ab+b^2-b^2=ab$$

and that's why your intuition works.


问:
Covariance of order statistics (uniform case)
Let $X_1, \ldots, X_n$ be uniformly distributed on $[0,1]$ and $X_{(1)}, ..., X_{(n)}$ the corresponding order statistic. I want to calculate $Cov(X_{(j)}, X_{(k)})$ for $j, k \in \{1, \ldots, n\}$.

The problem is of course to calculate $\mathbb{E}[X_{(j)}X_{(k)}]$.

The joint density of $X_{(j)}$ and $X_{(k)}$ is given by $$f_{X_{(j)}, X_{(k)}}=\binom{n}{k}\binom{k}{j-1}x^{j-1}(y-x)^{k-1-j}(1-y)^{n-k}$$ where $0\leq x\leq y\leq 1$. (I used the general formula here.)

Sadly, I see no other way to calculate $\mathbb{E}[X_{(j)}X_{(k)}]$ than by $$\mathbb{E}[X_{(j)}X_{(k)}]=\binom{n}{k}\binom{k}{j-1}\int_0^1\int_0^yxyx^{j-1}(y-x)^{k-1-j}(1-y)^{n-k}\,dx\,dy.$$

But this integral is too much for me. I tried integration by parts, but got lost along the way.

Is there a trick to do it? Did I even get the limits of integration right?

Apart from that, I wonder if there's a smart approach to solve the whole problem more elegantly.

答:
I don't know if there is a smart way to solve the problem, but the standard way is the following:

For $1\le j< k\le n$, the joint density function should be
$$
f_{X_{j},X_{_k}}=\frac{n!}{(j-1)!\,(k-j-1)!\,(n-k)!}\,x^{j-1}\,(y-x)^{k-1-j}\,(1-y)^{n-k}
$$
where $0<x<y<1.$

Now to compute $E[X_{j}X_{_k}]=\int_0^1\mathbb{d}y\int_0^y xy\cdot f(x,y)\,\mathbb{d}x$, you need to be familiar with Beta function. Let
$$C=\frac{n!}{(j-1)!\,(k-j-1)!\,(n-k)!},$$
so that
$$\begin{eqnarray}
E[X_{j}X_{_k}]
&=&C\int_0^1\mathbb{d}y\int_0^y xy\cdot x^{j-1}(y-x)^{k-1-j}(1-y)^{n-k}\,\mathbb{d}x \\
&=&C\int_0^1\mathbb{d}y\int_0^y \left[\frac{x}{y}\right]^{j}\left[1-\frac{x}{y}\right]^{k-1-j}y^k(1-y^{n-k})\,\mathbb{d}x\\
&=&C\cdot B(j+1,k-j)\int_0^1y^{1+k}(1-y)^{n-k}\,\mathbb{d}y \\
&=&C \cdot B(j+1,k-j)\cdot B(k+2,n-k+1)\\
&=& \frac{j\cdot (k+1)}{(n+1)(n+2)}.
\end{eqnarray}$$

Finally, $$
\begin{eqnarray}\operatorname{Cov}[X_{j}X_{_k}]&=&E[X_{j}X_{_k}]-E[X_{j}]\cdot E[X_{_k}]\\ &=& \frac{j\cdot (k+1)}{(n+1)(n+2)}-\frac{j\cdot k}{(n+1)^2} \\ &=& \frac{j\cdot (n+1-k)}{(n+1)^2(n+2)}.
\end{eqnarray}$$


答: I assume you are using vectors from $ \mathbb{R}^n $. In real euclidian spaces the scalar product is symmetric. So here you have $ x^T\cdot A^T\cdot b=(A\cdot x)^T\cdot b=\langle A\cdot x, b \rangle=\langle b,A\cdot x \rangle = b^T\cdot A\cdot x$

Title: Why is $(b-Ax)^T(b-Ax) = (b^Tb - 2b^T Ax + x^T A^T Ax)$?

Body:
Why is it true that
$$
(b-Ax)^T(b-Ax) = (b^Tb - b^T Ax - b^T Ax + x^T A^T Ax)
$$
Shouldn't it be
$$
(b-Ax)^T(b-Ax) = (b^Tb - b^T Ax - x^T A^T b + x^T A^T Ax)
$$?

So in why are there two $b^T Ax$? This must mean that $x^T A^T b = b^T Ax$. If true, why can you assume that? $b$ is a $M \times 1$ vector, $x$ is a $N \times 1$ vector, and $A$ is a $M \times N$ matrix.


问:
$|f(1)^2-f(0)^2|\le 2$
Let H the set of the functions $f:[0,1]\to\mathbb{R}$ differentiable, with $f'(x)$ continuous and $\int^1_0 (f(x)^2)\,dx =\int^1_0 (f'(x)^2)\,dx=1$
Prove that
$|f(1)^2-f(0)^2|\le 2$ for any $f\in H$.
I did not had any idea at it other than using the mean value theorem but I do not think it is helpful.

Title: $|f(1)^2-f(0)^2|\le 2$

Body:
Because $(f^2)'=2ff'$, you have
$$f^2(1)-f^2(0) =  2\int_0^1 f(x)f'(x)dx $$

So
$$|f^2(1)-f^2(0)| =  2\left|\int_0^1 f(x)f'(x)dx \right| \leq 2 \left(\int_0^1 f^2(x) dx\right)^{1/2}\left(\int_0^1 f'^2(x) dx\right)^{1/2}$$

by Cauchy-Schwarz, so because
$$\int_0^1 f^2(x) dx=\int_0^1 f'^2(x) dx=1$$ you deduce
$$|f^2(1)-f^2(0)| \leq 2$$


问:
Proof of triangle inequality
I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)</p>

答:
Since the proof of CW is necessary and similar I will do both

Start both proofs with the fact that a vector dotted with itself is greater than or equal to 0

for CW substitute vector = x-ty, for triangle inequality vector = x+y

for CW, after dotting x-ty with itself let t = (x.y)/(y.y), for triangle ineq. after dotting x+y with itself and getting a quadratic equation with a dot product in the middle, use CW to show that this quadratic is less than or equal to the same quadratic with the moduluses of the vectors of the dot product in the equation on the left

after rearranging both sqrt both sides

Title: Proof of triangle inequality


问:
Minimum number of vertices in an r-regular graph with no triangles and squares

Given a graph $G = (V,E)$, $r-\text{regular}$, with no triangles/squares (no cycles of length $3$ or $4$,) prove that $|V(G)| \geq r^2 +1 $

I'll be more than happy to get a direction, thanks!

答:
Since the OP has solved the problem, I'll post a solution.

Pick a vertex $a$. If $x$ and $y$ are two neighbors of $a$, then $a$ is their only common neighbour (or we'd have a 4-cycle). So for each neighbor $x$ of $a$ we get $k-1$ vertices at distance two from $a$, and these sets of size $k-1$ are pairwise disjoint (or we'd have a 4-cycle).

So we have $a$, its $k$ neighbours, and $k(k-1)$ vertices at distance two from $a$, adding up to $1+k+k^2-k=1+k^2$ vertices as a lower bound.


问:
Is the Legendre transform of a differentiable function also differentiable?
Let $f=f(x)$ be a real-valued differentiable convex function. The Legendre transform $g$ of $f$ is a function of $p$ defined by:
$$g(p) = \max_{x}(px - f(x))$$

Question: Is $g$ differentiable?

答: Not always. Take for example a linear function $f(x)=ax+b$ in one dimension. Then $f^*(p)=+\infty$ for $p\neq a$ and $f^*(a)=b$. If $f$ is differentiable and strictly convex then $f^*$ is convex and differentiable.


问:
$1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^\times$.
I have seen this problem:

Let $p$ be an odd prime and let $n$ be a positive integer. Use the Binomial Theorem to show that $(1+p)^{p^{n-1}}\equiv 1 \mod p^n$ but $(1+p)^{p^{n-2}}\not\equiv 1 \mod p^n$. Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^\times$.

be posted but no one has given a complete answer for the last part:

Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^\times$.

Can someone give an explanation for how this comes from the previous part?

答:
Claim 1: Let $r_1$ and $r_2$ be any two positive integers such that $(1+p)^{r_1} \equiv_{p^n} (1+p)^{r_2} \equiv_{p^n} 1$. Now let $r=\gcd(r_1,r_2)$. Then $(1+p)^r \equiv_{p^n} 1$ as well.

Claim 2: Let $r$ be an integer satisfying $(1+p)^r \equiv_{p^n} 1$. Then $(1+p)^{mr} \equiv_{p^n} 1$ for any integer $m$ as well.

Let us suppose that there is a smaller positive integer $r_1 < p^{n-1}$ that satisfies
$(1+p)^{r_2} \equiv_{p^n} 1$. Now $r_2|(p-1)p^{n-1}$ [recall that $|(\mathbb{Z}/p^n\mathbb{Z})^x| = (p-1)p^{n-1}$] so $r_2$ has to be either of the form
$r= cp^j$ for some integer $j < n-1$ and some integer $c$ that divides $p-1$; $c$ may be 1. But then in either case, as $p$ is prime,
$\gcd(r_1,r_2) = p^j \doteq r$. So Claim 1 gives $(1+p)^{p^j} \equiv_{p^{n}} 1$.
Then as $j < n-1$, Claim 2 gives $(1+p)^{p^{n-2}} \equiv_{p^n} 1$ as well. But this contradicts what you discovered earlier, that  $(1+p)^{p^{n-2}} \not \equiv_{p^n} 1$ after all.


问:
Formula for each angle of a polygon
This link (https://www.mathsisfun.com/geometry/interior-angles-polygons.html) says that the formula for interior angle is (n-2)*180/n. When I tried to use it in my Python program then the pentagon didn't draw correctly (in fact what was drawn wasn't even a polygon). I had to change the formula to 360/n. Did Is the link incorrect.

import turtle

```<code>def polygon(sides, length):
    bob = turtle.Turtle()
    #print(bob)
    #angle = ((sides-2)*180)/sides --> doesn't work
    angle = 360/sides #works
    #print("angle is "+angle)
    for i in range(sides):
        bob.fd(100)
        bob.lt(angle)

polygon(5,100)
turtle.mainloop()
</code>```

答:
The interior angle of a poligon is $\frac{(n-2)\cdot180}{n}$. The exterior angle is $\frac{360}{n}$. You probably mixed them up. You can check that this formula works for a few poligons or search a proof in the internet.

You have to write $\frac{360}{n}$ because the turtle module of Python uses exterior angles, not interior ones.

Title: Formula for each angle of a polygon


问:
What is the relation between a matrix as a linear function versus the same matrix as a bilinear function?
Given an $n \times n$ matrix $A$, we can define a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by $T(x)=Ax$.

We could also define a bilnear function $T: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ by $T(x,y) = y^TAx$.

Is there a relation between these two uses of the matrix?

Also, we could do the same thing with an $n \times m$ matrix and get a linear function $\mathbb{R}^m \rightarrow \mathbb{R}^n$ and a bilinear function $\mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}$. Is the relationship between using $A$ to define a linear function versus a bilinear function the same in this case?

答:
Any bilinear form $b\colon \mathbb R^n\times\mathbb R^n\to\mathbb R$ naturally corresponds to a linear map $\Phi_b\colon \mathbb R^n\to(\mathbb R^n)^*$, where $(\mathbb R^n)^*=\operatorname{Hom}(\mathbb R^n,\mathbb R)$ is the dual space. This correspondence is given by
\begin{align}
b \quad\longmapsto\quad \Phi_b\colon\mathbb R^n&\to(\mathbb R^n)^*,\\
x &\mapsto b(x,-).
\end{align}
Here $b(x,-)$ denotes the linear map $\mathbb R^n\to\mathbb R$ sending $y$ to $b(x,y)$.

Considering the elements of $\mathbb R^n$ as column vectors, we can consider the elements of $(\mathbb R^n)^*$ as row vectors: for $\rho\in(\mathbb R^n)^*$ and $x=(x_1,\dots,x_n)^T$ we have
\begin{align*}
\rho(x) &= \rho(x_1 e_1+\cdots+x_n e_n) = x_1 \rho(e_1) +\cdots + x_n \rho(e_n)
\\&= \begin{pmatrix} \rho(e_1) & \rho(e_2) & \dots & \rho(e_n)\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix}.
\end{align*}
Using this identification, there is an isomorphism ${}^T\colon(\mathbb R^n)^*\to\mathbb R^n$ given by $y\mapsto y^T$.

Now starting with $b(x,y) = y^TAx$ this yields $\Phi_b(x)=(y\mapsto y^TAx)$ which we identified with the row vector $\Phi_b(x)=(Ax)^T$. Using the above isomorphism we then obtain a linear map $\mathbb R^n\to\mathbb R^n$:
\begin{align}
\mathbb R^n &\xrightarrow{\Phi_b}\ (\mathbb R^n)^*\ \xrightarrow{T} \mathbb R^n,\\
x &\longmapsto (Ax)^T \mapsto Ax.
\end{align}

This is how starting with the bilinear map defined by $A$ we obtain the linear map defined by $A$. You can of course go in the other way as well.

More abstractly you can think of this in terms of the tensor-hom adjunction together with the (basis dependent!) isomorphism $(\mathbb R^n)^*\cong(\mathbb R^n)$:
\begin{align}
\{\text{bilinear maps $\mathbb R^n\times\mathbb R^n\to\mathbb R$}\}
&\leftrightarrow
\operatorname{Hom}(\mathbb R^n\otimes\mathbb R^n,\mathbb R)
\\&\cong
\operatorname{Hom}(\mathbb R^n,\operatorname{Hom}(\mathbb R^n\to\mathbb R))
\\&=
\operatorname{Hom}(\mathbb R^n,(\mathbb R^n)^*)
\\&\cong
\operatorname{Hom}(\mathbb R^n,\mathbb R^n).
\end{align}


答:
Redefining symbols to avoid ambiguity: $T: \mathbb{R}^n \to \mathbb{R}^n$ is the linear map defined as $T(x) = Ax$ and $S: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ is the bilinear map defined as $S(x, y) = y^T A x$.

Constructing bilinear functions from linear functions using inner product

One way to understand $S$ is as composition of $T$ with the standard inner product $\phi: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ defined as $\phi(x, y) = y^T x$, namely

$$
S(x, y) = \phi(T(x), y).
$$

This view allows us to notice some properties of $S$ based on properties of $T$ and known properties of $\phi$. For example, since $\phi$ is known to be non-degenerate, $S$ is non-degenerate if and only if $T$ is an isomorphism.

The construction is readily generalized to the $n \times m$ case by composing $T: \mathbb{R}^m \to \mathbb{R}^n$ with $\phi$.

Change of basis

For a fixed matrix $A \in \mathbb{R}^{n \times n}$ the construction above yields two functions: a linear function $T_A: \mathbb{R}^n \to \mathbb{R}^n$ and a bilinear function $S_A: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$. The construction takes place in a fixed basis, but the resulting functions $S$ and $T$ are independent of basis, so it is natural to ask how their matrix representation changes under basis transformations.

It easy to see that the matrix representing a linear function transforms differently than the matrix representing a bilinear function. Let $B$ denote an invertible matrix describing a change of basis. Then

$$
T_A = T_{A'}
$$

whenever $A' = B^{-1}AB$, i.e. the matrices representing a fixed linear map are similar. On the other hand,

$$
S_A = S_{A^{''}}
$$

whenever $A^{''} = B^TAB$, i.e. the matrices representing a fixed bilinear map are congruent.

This shows that care must be taken when using matrix representations of linear and bilinear functions. Even when a linear function $T$ and a bilinear function $S$ are represented by the same matrix in one basis, it does not imply that they are represented by the same matrix in another basis (unless the basis transformation is orthogonal).

Title: What is the relation between a matrix as a linear function versus the same matrix as a bilinear function?

Body:
Given an $n \times n$ matrix $A$, we can define a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by $T(x)=Ax$.

We could also define a bilnear function $T: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ by $T(x,y) = y^TAx$.

Is there a relation between these two uses of the matrix?

Also, we could do the same thing with an $n \times m$ matrix and get a linear function $\mathbb{R}^m \rightarrow \mathbb{R}^n$ and a bilinear function $\mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}$. Is the relationship between using $A$ to define a linear function versus a bilinear function the same in this case?


问:
Differential operator; its algebraic manipulation in deriving the Laplacian ($\nabla ^2$) formula for polar coordinates
I tried out using $r = \sqrt{x^2+y^2}$ and $\tan \theta = y/x$, then getting the formula for $\partial/\partial x$ using the chain rule, and then obtaining the second partial derivative operator without differentiating any further but through purely algebraic means, i.e., expanding.

In the fourth line of the below derivation, I could recognise that something was going wrong because the terms consisting $\partial^2/\partial x^2$ should have indeed been five, i.e.,

$$\frac{\partial ^2}{\partial  x^2} = \cos^2\theta \frac{\partial ^2}{\partial x^2} - \frac{2 \sin \theta \cos \theta}{r} \frac{\partial ^2}{\partial r \partial \theta} + \frac{\sin ^2 \theta}{r} \frac{\partial }{\partial r} + \frac{2 \sin \theta \cos \theta }{r^2} \frac{\partial }{\partial \theta} + \frac{\sin ^2 \theta}{r^2} \frac{\partial ^2}{\partial  \theta ^2}$$, but there are only three terms in my derivation.

I heard in quantum mechanics classes that treating linear (differential) operators algebraically is perfectly fine, but this seems to be a counterexample.

If anyone wants to see how I tried out further derivations please see below. Thanks.

答:
$\def\th{\theta}$Manipulation of differential operators in this fashion is perfectly fine and is in fact just shorthand for more cluttered notation.
(We leave out the function being operated on.)
The rules of manipulation may seem strange, but if one simply considers what is being represented they are straightforward to work out and remember.

We begin with an example from calculus of one variable.
Note that
$$(f g)' = f'g + f g'.$$
This can be represented in operator notation as
\begin{align*}
D f = (D f)  + f D,\tag{1}
\end{align*}
where $D=d/dx$.
This is product rule in operator notation.
It is understood in (1) that $D f$ on the left-hand side is meant to act on another function and that $(D f)$ on the right-hand side is simply $f'$.
Think of a differential operator as "acting to the right" in the following way:

\begin{align*}
D f g &= f'g+fg' \\
(D f)g &= f'g \\
(f D)g &= f D g = fg'.
\end{align*}
The uncertainty principle in quantum mechanics essentially follows from the fact that
$$xf' - (xf)' = xf' - (f+xf') = -f,$$
or
\begin{align*}
[x,D] &\equiv x D - D x \\
&= x D - (1 + x D) \\
&= -1.
\end{align*}

Regarding the original problem, let $D_x = \partial/\partial x,r_x = \partial r/\partial x$, etc.
We can work out $D_x^2$ as follows:
\begin{align*}
D_x &= r_x D_r + \th_x D_\th \\
D_x^2 &= (r_x D_r + \th_x D_\th)(r_x D_r + \th_x D_\th) \\
&= (r_x D_r)^2
    + (r_x D_r)(\th_x D_\th) + \ldots \\
&= r_x r_{xr}D_r + r_x^2 D_r^2 \\
    &\quad + r_x \th_{xr}D_\th + r_x \th_x D_{\th r} \\
    &\quad + \ldots
\end{align*}
Using that
\begin{align*}
r_{xr} &= 0 \\
r_{x\th} &= -\sin\th \\
\th_{xr} &= \frac{\sin\th}{r^2} \\
\th_{x\th} &= -\frac{\cos\th}{r}
\end{align*}
one will arrive at the correct relation for $D_x^2$.


问:
What is the meaning of symmetry of equalities?
For the equation:

$$a+\frac1b=b+\frac1c=c+\frac1a=t$$

According to @FundThmCalculus's Answer  We have symmetry of equalities. I wonder what does that mean. in other word when equations are symmetric in mathematics?

答:
This is known to some of us as a cyclic symmetry, with the group of symmetries being $A_3$ (also the cyclic group $C_3$, but here named as a normal subgroup of $S_3$, the full group of symmetries on three objects).

The rest of this answer probably takes you well away from the context of your original problem, but it does attempt to illustrate something of mathematical symmetry - which is expressed very often in group theory.

Full symmetries and cyclic symmetries remain interesting as the number of variables increases, but the algebraic study of symmetry (study of algebraic symmetries) is founded more naturally around Galois theory. Sometimes the full group of symmetries will be involved, and the study of symmetric functions of $n$ variables invariant under the full group is a place to begin. Every (suitably defined) symmetric function can be expressed in terms of certain elementary symmetric functions - and that is often useful to know when embarking on practical calculations, though cases with three variables are soon learnt. It is as the number of variables becomes larger that things start to get interesting. Here is a rough outline of the kind of thing which can happen.

Many systems $S$ are symmetric under subgroups of the symmetric group (rather than fully symmetric under the whole symmetric group - cyclic symmetry is an example of this), and we sometimes want to analyse subsystems $T$ of $S$. Galois theory exploits the fact that, if $T$ sits nicely in $S$, very often normal subgroups will be involved when we examine the relevant groups of symmetries. In fact for each normal subgroup of an appropriate $S$ we hope to find a nice sub-system $T$, and each nice subsystem will relate to a normal subgroup.

You might want to look at the process of solving cubic and quartic equations in a single variable, which is where this starts to take root. Note, for example, that the three expressions $ad+bc, ac+bd, ab+cd$ are permuted amongst themselves under any permutation of the four letters $a, b, c, d$. It is this symmetry which is essentially exploited in a key step of solving the quartic as it takes us from the original four letters (the roots of the equation) down to three expressions. And this in turn relates to the fact that the symmetric group on four letters $S_4$ has a normal subgroup $V$ of order $4$.

The fact that the quintic equation relates to the symmetric group $S_5$, together with the fact that the alternating group $A_5$ is simple, means that $S_5$ has no small normal subgroups, and that means there are no sub-systems of roots small enough to make the same kind of reduction as in the case of the quartic. Thus we have the famous Abel-Ruffini Theorem that the quintic equation cannot be solved "by radicals". Obviously there is a fair amount of detail missing, but it is an illustration of how exploiting our knowledge of symmetry can show us what is, and what is not possible.


问:
Deriving derivatives with differentials
What does it mean to derive the rule $d(x^3)=3x^2dx$ using differentials? Are differentials derivatives? In that case what is the question?

答:
There are some pedagogical tools for describing derivatives which date back to the presentation given by Leibniz and which form the foundation of Non-Standard Analysis.

These techniques differ from the standard epsilon delta approach to derivatives and can be formalized by a set of algebraic type rules.

H. Jerome Keisler, David Tall, and some other educators maintain that the use of infinitesimals is more intuitive and more easily grasped by students than the "epsilon–delta" approach to analytic concepts.

As a result in some early calculus courses they introduce a subset of these techniques as a pedagogical tool and they are presented as allowed shortcuts which can be used when taking derivatives.

I recommend checking your notes or book to see what is allowed.

I will take a guess based on what I have seen before:

$\frac{d(x^{3})}{dx} = 3x^{2}$, so $d(x^{3}) = 3x^{2}dx$.

Title: Deriving derivatives with differentials


问:
Largest consecutive integers with no prime factors except $2$, $3$ or $5$?
The number $180$ has a special property. Its prime factors are only $2$, $3$, and $5$. However the number $220$ does not have this special property because one of its prime factors is $11$.

In the first one-hundred natural numbers, the numbers with this property are plentiful. They cover almost half of that range. However, into the thousands, or tens of thousands, they get increasingly rare, and you might go hundreds of integers without finding one. As far as I know, $80$ and $81$ are the two largest consecutive numbers that both have this property — $80 = 2^4 \times 5$ and $81 = 3^4$. I have tested numbers up to one million, and have not found another pair of numbers like this.

But, is it possible to mathematically prove this? Has anyone ever tried to do it before? All evidence points to the fact that $80$ and $81$ are the largest pair of integers with this property, but has it ever been proven before?

答:
Only one of the two consecutive numbers can be divisible by $2$. Similarly, only one is divisible by $3$ and only one is divisible by $5$. Depending on how we pair the prime powers we therefore have six possibilities.

The cases $p^Aq^B=r^C-1$ can all be solved in the same way.

$2^A3^B=5^C-1 $

If $A\ge4$ then modulo $16$ we have $C\equiv 0\pmod 4$ and then $5^C-1$ has a factor of $13$.

If $B\ge2$ then modulo $9$ we have $C\equiv 0\pmod 6$ and then $5^C-1$ has a factor of $31$.

The remaining values of $A,B$ give the solutions $\{4,5\},\{24,25\}$.

$2^A5^B=3^C-1 $

If $A\ge5$ then modulo $32$ we have $C\equiv 0\pmod 8$ and then $3^C-1$ has a factor of $41$.

If $B\ge2$ then modulo $25$ we have $C\equiv 0\pmod {20}$ and then $3^C-1$ has a factor of $11$.

The remaining values of $A,B$ give the solutions $\{2,3\},\{8,9\},\{80,81\}$.

$3^A5^B=2^C-1 $

If $A\ge2$ then modulo $9$ we have $C\equiv 0\pmod 6$ and then $2^C-1$ has a factor of $7$.

If $B\ge2$ then modulo $25$ we have $C\equiv 0\pmod {20}$ and then $2^C-1$ has a factor of $31$.

The remaining values of $A,B$ give the solutions $\{1,2\},\{3,4\},\{15,16\}$.

The cases $p^Aq^B=r^C+1$.

$2^A\times3^B=5^C+ 1$

Modulo $4$ we have $2^A\times3^B\equiv 2\pmod 4$. Therefore $A=1$ and $2\times3^B=5^C+ 1.$

If $B\ge2$ then modulo $9$ we have $C\equiv 3\pmod {6}$ and then $5^C+1$ has a factor of $7$.

The remaining values of $B$ give the solution $\{5,6\}$.

$2^A\times 5^B=3^C+ 1$

Modulo $4$ we have $2^A\times3^B\equiv 2\pmod 4$. Therefore $A=1$ and $2\times5^B=3^C+ 1.$

If $A\ge3$ then modulo $8$ we have $3^C\equiv -1\pmod {8}$ which is impossible.

If $B\ge2$ then modulo $25$ we have $C\equiv 10\pmod {20}$ and then $3^C+1$ has a factor of $1181$.

The remaining values of $A,B$ give the new solution $\{9,10\}$.

$3^A\times 5^B=2^C+ 1$

The cases $AB=0$ have been covered in previous equations. So we can suppose $AB>0$ when the equation is impossible modulo $15$.


问:
Solution to the Friedmann Equations
In cosmology I have seen three kinds of energy densities: that of matter, radiation, and dark energy. I have seen solutions to the First Friedmann equation when the universe contains either one of these + curvature. Energy densities have an equational form: $\rho = \rho_0 a^{-3(1 + \omega)}$, where matter has $\omega = 0$, radiation has $\omega = 1/3$ and dark energy $\omega = -1$. Suppose there was a new kind of energy density with $-1/5 < \omega < 1$ with $\omega \neq 0$ and consider an universe with only this energy and curvature $\kappa$. The first Friedmann equation is: $\dot{a}^2 = \frac{8\pi G\rho}{3}a^2 - \kappa = \frac{8\pi G\rho_0a^{-3(1+\omega)}}{3} - \kappa = \beta a^{-3(1+\omega)} - \kappa$. From this I got the following integral equation

$\int_0^{a}da\sqrt{\frac{a^{3\omega + 1}}{\beta - a^{3\omega + 1}\kappa}} = \pm\int_{0}^{t}dt$. When $\kappa = 0$, I can solve this equation as $a(t) = (\pm 1.5(\omega + 1)t)^{1/(3\omega + 3)}$ and so I have an expanding universe. However when the universe is curved, I am unable to determine the scale factor as a function of time. Can anyone tell me how I might try to solve this problem I created? Will such a universe expand or collapse? Is there an easier way I can approximate $a(t)$ if this cannot be solved exactly? Please let me know?

答:
As a matter of preference, I'll call
\begin{align}
\frac{8\pi G\rho_0}{3}=\text{H}_0^2
\end{align}
for short. For the $\kappa=0$ case, and evaluating today it simplifies down to the definition of $\textrm{H}_0$. I'll also call $\kappa=\Omega_c/\textrm{H}_0^2$ for brevity.

You've lost your $a^2$ term,
\begin{align}
\dot{a}^2=\frac{\textrm{H}_0^2}{\rho_0}\rho a^2-\kappa=\textrm{H}_0^2a^{-3(1+w)}\color{red}{a^2}-\kappa=\textrm{H}_0^2\color{red}{a^{-(1+3w)}}-\kappa.
\end{align}
So,
\begin{align}
\dot{a}^2=\textrm{H}_0^2(a^{-(1+3w)}-\Omega_c)\\
\frac{1}{\sqrt{a^{-(1+3w)}-\Omega_c}}\frac{\mathrm da}{\mathrm dt}=\textrm{H}_0.
\end{align}
As for integrating, I'm afraid you've now been introduced to the Gauss hypergeometric function.

https://www.wolframalpha.com/input/?i=int+1%2Fsqrt%28a%5E%28b%29-c%29+da

Title: Solution to the Friedmann Equations


问:
Let $G$ be a group in which $aba = b \; \forall a,b \in G$. Prove that $G$ is Abelian.
I was wondering if someone could tell me if my method is correct. I feel like the substation $b=e$ is a stretch and that there is a better way to do it.

We know that $G$ is Abelian $\iff (ab)^{-1} = a^{-1}b^{-1}$. So I know that we will have to manipulate $aba = b$ into $(ab)^{-1} = a^{-1}b^{-1}$.

But this is where I get stuck.
$$aba = b \implies ab = ba^{-1} \implies (ab)^{-1} = (ba^{-1})^{-1} \implies (ab)^{-1}=ab^{-1}$$
This is close, but not what I want. One thing that I did notice is that if $b=e$, then $aba=b \implies a^2 = e \implies a=a^{-1}$. Since $a^{-1}$ is a unique element we know that $a=a^{-1}$ regardless of what $b$ is. So if we make that substitution into out previous equation we get
$$(ab)^{-1} = a^{-1}b^{-1}$$
Therefore $G$ is Abelian.

答:
You're correct.

If $a^2=e$ for all $a\in G$, then for any $g,h\in G$,

$$\begin{align}
(gh)^2&=ghgh\\
&=e\\
&=ee\\
&=g^2h^2\\
&=gghh,
\end{align}$$

from which it follows that $gh=hg$. Hence $G$ is abelian.


答:
You can't prove something that is not true. If you use in particular $f(x) = e ^x $ then
$$ \int_0^1 {f(x)^2} dx = \int_0^1 e^{2x} dx = \frac{e^2 - 1}{2} $$
$$ \int_0^1 {f'(x)^2} dx = \int_0^1 e^{2x} dx = \frac{e^2 - 1}{2} $$
So we can say:
$$ \int_0^1 {f(x)^2} dx =  \int_0^1 {f'(x)^2} dx $$

But:

$$ \vert{f(1)^2 - f(0)^2 }\vert = \vert e^2 - 1 \vert = 6.38 > 2 $$

Title: $|f(1)^2-f(0)^2|\le 2$

Body:
Let H the set of the functions $f:[0,1]\to\mathbb{R}$ differentiable, with $f'(x)$ continuous and $\int^1_0 (f(x)^2)\,dx =\int^1_0 (f'(x)^2)\,dx=1$
Prove that
$|f(1)^2-f(0)^2|\le 2$ for any $f\in H$.
I did not had any idea at it other than using the mean value theorem but I do not think it is helpful.


问:
Reference for infinite exponentiation
I'm trying to find a book or a paper on infinite exponentiation: more precisely, it should be proving its full interval of convergence on the positive real line, i.e. if $x\in\mathbb{R}^+$, then $x^{x^{x^\cdots}}$ converges if and only if $e^{-e}\le x\le e^{1/e}$.

While I didn't manage to find any books about it, there are these two papers from The American Mathematical Monthly:

Knoebel's Exponentials Reiterated

Mitchelmore's A Matter of Definition

Also, there's this paper on arXiv:

Moroni's The strange properties of the infinite power tower

Every one of them is problematic.

Paper #1 doesn't make sense towards the end of page 242 and is rather incomprehensible there.

Paper #2 shows only the convergence for $1\le x\le e^{1/e}$ and the conclusion is

DEFINITION. For all real $x$ such that $0\le x\le e^{1/e}$, $x^{x^{x^\cdots}}$ is the real solution for $t$ of the equation $x^t=t$, and in case this equation has two solutions, then the lesser one.

That is not satisfactory at all.

Paper #3 makes an incorrect assumption on page 15, which is the following:

If the first first derivative is $|z'(y_1^*)|\lt 1$ (that is $-1\le z'(y_1^*)\lt 0$...

There shouldn't be $-1\le$ but $-1\lt$, the author probably does so to avoid the analysis of the case when $x=e^{-e}$, but at the same time he further in the text states that the infinite tetration for $x=e^{-e}$ is convergent. So that one is incomplete.

答: In Ioannis Galidakis' A Collection of References For Infinite Exponentials and Tetration, see items #7, 19, 88, 110, 111, 166. See also On the Convergence of Iterated Exponentiation—I by Creutz/Sternheimer (1980) and On ${x^{x^{x^{{\cdot}^{{\cdot}^{\cdot}}}}}}$ by Louis A. Talman (written in 1999, I think).

Title: Reference for infinite exponentiation


问:
number of different sequences
He, let consider Basketball. We can make throw for 1, 2 or 3 points.

For example for 5 points we have throws sequences like:

```<code>{3, 2}, {2, 3}, {3, 1, 1}, {1, 1, 3} {1, 3, 1}, {2, 2, 1}, {2, 1, 2}, {1, 2, 2}, {2, 1, 1, 1}, {1, 2, 1, 1}, {1, 1, 2, 1}, {1, 1, 1, 2}, {1, 1, 1, 1, 1}.
</code>```

How can compute it for any n?

答:
For $r \in \mathbb{R},$ let $\lfloor r\rfloor$ denote the floor of $r$ (i.e. the largest integer $\leq r).$

Compute $$A = \left\lfloor \frac{n}{3}\right\rfloor.$$

For $a \in \{0,1,2,\cdots, A\}$, let $T_a$ denote the set of all sequences that use exactly $a$ occurrences of 3.

The following algorithm will let you specify $T_a$.

For each $a$, compute $$B_a = \left\lfloor \frac{n - 3a}{2}\right\rfloor.$$

Then for each $b$ in $\{0,1,2,\cdots, B_a\}$, you can form a complete sequence consisting of

$a$ 3's

$b$ 2's

$(n - 3a - 2b)$ 1's.

Note:

Although not requested, if instead you are looking for a count of the number of distinct sequences:

$$\sum_{a = 0}^A (B_a + 1).$$

Edit

I just realized that the OP is construing $(3,2)$ and $(2,3)$ as different sequences.  The remedy for the above work is as follows:

Given a sequence of $a$ 3's, $b$ 2's, and

$c = (n - 3a - 2b)$ 1's, let $t = a+b+c.$

Then the number of sequences that can be permuted from the combination of $a$ 3's, $b$ 2's and
$c$ 1's is:

$$\binom{T}{a} \times \binom{T-a}{b}$$.

Consequently, the formula for the count of the number of permuted sequences is $\{$setting $f(a,b) = a + b + (n - 3a - 2b)\}$:

$$\sum_{a = 0}^A \sum_{b = 0}^{B_a}\left[\binom{f(a,b)}{a} \times \binom{f(a,b)-a}{b}\right] .$$

Title: number of different sequences


问:
Is the statement still true?
Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a entire function. Suppose that there is $M$ and $n\in \mathbb{N}$ such that $\left|f(z)\right|<M\left|z\right|^n$ for all $z \in \mathbb{C}$ with $\left|z\right|\geq 1$. It is well known that in this case $f$ is a polynomial of degree at most $n$. Now if I suppose to have have it just for all $z \in \mathbb{C}$, can I conclude the same?

I would say yes. We will just not get the if and only if. So if $f$ is a polynomial of degree n than we can not bound it like this.

答:
If your condition holds for all $C$ then it holds for the subset that you needed to apply the theorem and so the theorem still applies.

I suppose you need only worry about the fact that you might have made the condition overly restrictive so that no nonzero function will satisfy it. However you need not worry since take $K<M$

$$|Kz^{n}|=|K||z|^{n} = K |z|^{n} < M |z|^{n}$$

So $f(z)=Kz^{n}$ will satisfy your inequality.

Title: Is the statement still true?


问:
What is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ; n \in \mathbb{N_{0}} \}$ of relation $\sqsubseteq$?
On the set of polynomials with real coefficients we define $\sqsubseteq$ as:

$p \sqsubseteq  q \iff p=q\ \vee \exists n \in \mathbb{N_{0}}:(p(n) <q(n)\ \wedge \forall m \in \mathbb{N_{0}}:(m<n \Rightarrow p(m)=q(m)))$

First part of the question was to prove that this is total order, which I had no problem with. Then it also asks what is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ;  n \in \mathbb{N_{0}} \}$?

My initial though was to take $\prod_{i=0}^{\infty}(x-i)$, which would have all natural numbers as its zeroes, but if I remember correctly, a polynomial can't have infinitely many zeroes unless it is constant. Now I have two questions:

Why can't polynomial have infinitely many zeroes, as does $\prod_{i=0}^{\infty}(x-i)$?

If not $\prod_{i=0}^{\infty}(x-i)$, then what is $\text{inf} \{ \prod_{i=0}^{n}(x-i) ;  n \in \mathbb{N_{0}} \}$ equal to?

答:
For $n\in\Bbb N_0$ let $p_n(x)=\prod_{i=0}^n(x-i)$. Clearly for each $n\in\Bbb N_0$ we have $p_n(k)=0$ for all $k\in\Bbb N_0$ such that $k\le n$, while $p_n(k)>0$ for $k\in\Bbb N_0$ such that $k>n$. It follows that if $m,n\in\Bbb N_0$, and $m<n$, then $p_m(k)=p_n(k)=0$ for all $k\in\Bbb N_0$ such that $k\le m$, while $p_m(m+1)>0=p_n(m+1)$, so $p_n\sqsubseteq p_m$ (and in fact $p_n\sqsubset p_m$).

Let $p(x)$ be a real polynomial. It is clear that if $p\sqsubseteq p_n$ for all $n\in\Bbb N_0$, then $p(k)\le 0$ for all $k\in\Bbb N_0$: if some $p(k)>0$, then $p_k\sqsubset p$. The simplest real polynomial satisfying this requirement is the constant function $z(x)=0$, and indeed $z\sqsubset p_n$ for each $n\in\Bbb N_0$. It is also clear that if $p$ is any real polynomial such that $p(k)\le 0$ for all $k\in\Bbb N_0$, and there is a $k_0\in\Bbb N_0$ such that $p(k_0)<0$, then $p\sqsubset z$, and hence $p\ne\inf\{p_n:n\in\Bbb N_0\}$. The only possible candidate for $\inf\{p_n:n\in\Bbb N_0\}$ besides $z$ would therefore be some real polynomial $p$ such that $p(k)=0$ for each $k\in\Bbb N_0$. But it is a well-known consequence of the fundamental theorem of algebra that a non-zero real polynomial of degree $n$ has at most $n$ real roots, so there is no such $p$: the zero polynomial $z$ is the only real polynomial taking the value $0$ at each non-negative integer, and it follows that $\inf\{p_n:n\in\Bbb N_0\}=z$.

marty cohen has already pointed out that $\prod_{i\ge 0}(x-i)$ is undefined for all $x\in\Bbb R\setminus\Bbb N_0$: unless $x$ is a non-negative integer, all of the factors are non-zero, exactly two have absolute values less than $1$, and there are factors with arbitrarily large absolute value. It is also not a polynomial, so it is irrelevant here: the infimum is taken in the domain of the order $\sqsubseteq$, i.e., in the set of real polynomials.


问:
Prove that an uncountable set with the cofinite topology is Fréchet–Urysohn
A space $X$ is said to be Fréchet–Urysohn give that,

$$x\in \overline{A} \text { if and only if there is a sequence } \langle x_n \rangle \text{ in } A \text{ such that } \langle x_n \rangle\to x.$$

Lemma A: Let $X$ be any uncountable set with the cofinite topology. Let $x\in X$ and $\langle x_n \rangle$ a sequence in $X$. Then $\langle x_n \rangle$ converges to $x$ if and only if for every $y\neq x$ we have $x_n=y$ for only finitely many $n\in\Bbb N$.

Using Lemma A, how can I prove that $X$, an uncountable set with the cofinite topology, is Fréchet–Urysohn?

I know that if $A\subseteq X$ is finite then $\overline{A}=A$, whereas if $A\subseteq X$ is infinite then $\overline{A}=X$.

答: ... so if $\lvert A\rvert\ge\aleph_0$, then consider an injective sequence $x_\bullet:\Bbb N\to A$: that sequence converges to every point of $X$ by the lemma.


问:
I'm struggling with a particle motion with one degree of freedom question.
Q) Given that y(x) = log(1 + cos x) on the interval (0, π)

Taking signed arclength s=0 at x=0 show that:

x(s) = 4arctan(tanh(s/4))

Attempt)

I tried using (ds/dx)^2= 1 + (dy/dx)^2 rule to get dx/ds = (cosx+1)/root(2cosx+2)

Then integrating dx/ds gives me x(s) =  2sin(x/2)+ c

Im really stuck on the next step on what to do next to get x(s) = 4arctan(tanh(s/4)) and how do I calculate y(s) from this?

答:
The problem is a bit obscure. Recalling what you did, for the sake of public understanding:

$$\ell = \frac{\text{d}s}{\text{d}x} = \sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2}$$

$$y(x) = \ln(1 + \cos(x)) \longrightarrow 1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = \frac{2}{1 + \cos(x)}$$

$$s(x) = \int \sqrt{\frac{2}{1 + \cos(x)}}\ \text{d}x = 2 \ \text{arctanh}\left(\sin \left(\frac{x}{2}\right)\right) + C$$

On the other hand we have too:

$$\frac{\text{d}x}{\text{d}s} = \frac{1}{\frac{\text{d}s}{\text{d}x}} = \frac{1}{\sqrt{\frac{2}{1 + \cos(x)}}} = \cos\left(\frac{x}{2}\right)$$

Yet from here I cannot understand the statement of the problem and most of all how it gets to that result. Do you have a copy of the homework, for even finding the inverse function I do not think there is a way to come to that result...


答:
I'm not sure what your background is. But the term differential often refers to the differential operator $d$ in the so called de Rham complex

$$0 \to \Omega^0(\mathbb{R}) \xrightarrow{d} \Omega^1(\mathbb{R}) \xrightarrow{d} \cdots$$

where $\Omega^k(\mathbb{R})$ is the space of (differential) $k$-forms. In degree $0$, the space $\Omega^0(\mathbb{R})$ is simply the space of smooth (infinitely differentiable) real valued functions.

Therefore, given the smooth function $x^3 \in \Omega^0(\mathbb{R})$, its differential is the $1$-form $d(x^3) \in \Omega^1(\mathbb{R})$ given by $$d(x^3) = 3x^2dx.$$

$1$-forms are simply the usual differential of differentiable functions we are already familiar with.

Does this help you?

Title: Deriving derivatives with differentials

Body: What does it mean to derive the rule $d(x^3)=3x^2dx$ using differentials? Are differentials derivatives? In that case what is the question?


问:
Is it permitted to use $\tan 90^\circ$ to prove an equation/identity even if it is not defined? defined?
Domain of tangent function limits to all real numbers except odd integral multiples of $\frac{\pi}{2}$.

I noticed when proving an equality that using $\tan 90^\circ$ yielded the exact same result. Here what I did:

For a triangle $\mathrm{ABC}$, prove that $$\tan \frac{A}{2}\tan\frac{ B}{2 }+\tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2}=1$$

Since $\mathrm{A+B+C = \pi}$
$$\mathrm{\frac{A+B+C}{2} = \frac{\pi}{2}}$$
I applied identity for sum of angles
$$\tan{(A/2 + B/2 + C/2)}=\frac{\tan \frac{A}{2} +\tan \frac{B}{2} +\tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{1-\tan \frac{A}{2}\tan\frac{ B}{2 }-\tan \frac{B}{2} \tan \frac{C}{2} - \tan \frac{C}{2}\tan \frac{A}{2}}$$
Then I reciprocated both side to get L.H.S. to be equal to 0 and I was able to prove the given equation.

Why did I arrive at the right conclusion by using the value out of domain for tangent function?

答:
Just like there is an identity for the tangent of a sum of three angles, there is an identity for the cotangent of a sum of three angles.
Not surprisingly, you can write a version of it just by taking the reciprocal of the formula for the tangent of the sum:

$$
\cot\left(\alpha + \beta + \gamma\right)
= \frac{1 -\tan \alpha \tan\beta - \tan\beta \tan\gamma - \tan\gamma \tan\alpha}
       {\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha \tan\beta \tan\gamma}.
$$

Plug in $\alpha = \frac A2,$ $\beta = \frac B2,$ $\gamma = \frac C2$:
$$
\cot\left(\frac A2 + \frac B2 + \frac C2\right)
= \frac{1 - \tan\frac A2 \tan\frac B2 - \tan\frac B2 \tan\frac C2 - \tan\frac C2 \tan\frac A2}
{\tan\frac A2 + \tan\frac B2 + \tan\frac C2 - \tan\frac A2 \tan\frac B2 \tan\frac C2}.
$$

Now just use the fact that $\cot\left(\frac\pi2\right) = 0.$

Intuitively, this seems to be what you were trying to do.
The main difference in the methods is that $\cot\left(\frac\pi2\right) = 0$
is a true equation with well-defined quantities on each side, wherease
$1/\tan\left(\frac\pi2\right) \stackrel?= 0$ is not.


问:
Minimum distance travelled in order to reach a known maximum acceleration given a constant jerk value?

Problem

Given the following:

$A$ = maximum acceleration.

$J$ = constant jerk (change in acceleration with respect to time).

Initial velocity and acceleration of 0.

How might one determine the minimum distance that must be travelled in order to reach the maximum acceleration?

Attempt

In the same way that one might solve distance $D$ given a maximum velocity $V$ and constant acceleration $A$ like so:

$D = \frac{V^2}{2A}$

I thought I might be able to apply the same to my problem by swapping out maximum velocity with maximum acceleration, and the constant acceleration for constant jerk, before realising that this seems to give the velocity at which the maximum acceleration is reached, and not the distance.

$V = \frac{A^2}{2J}$

I'm as yet unsure how to apply what I know so far to determine the distance travelled before the maximum acceleration is reached. Any help would be greatly appreciated!

Application Context

I'm currently attempting to improve a linear motion profile for some linear actuators that are present in an art installation that I'm working on in a small team.

Previously, the linear actuator was driven using a constant acceleration selected by us. We could achieve fast movement nicely with high acceleration, and we could achieve slow movement nicely with slow acceleration, however slow movement with a high acceleration resulted in a lot of "chatter" from the motor.

The idea is to use a constant jerk rather than a constant acceleration to achieve smoother behaviour at both low and high speeds. The implementation seems to be almost complete, however I'm struggling to determine the distance at which I should begin decreasing deceleration in order to arrive at the target position with a velocity and acceleration of 0. It occurs to me that this is the same as solving the distance required to reach the maximum acceleration, hence the framing of this question.

答:
If $\dot{a}=J$, then repeated integration yields

$$ a(t)=Jt+C_1 \xrightarrow{a(0)=0} a(t)=Jt$$
$$ v(t)=\frac{Jt^2}{2}+C_2 \xrightarrow{v(0)=0} v(t)=\frac{Jt^2}{2}$$
$$ x(t)=\frac{Jt^3}{6}+C_3 \xrightarrow{x(0)=0}x(t)=\frac{Jt^3}{6}$$

Solving the last equation for $t$

$$ t=\left(\frac{6x}{J}\right)^{1/3}$$

Substitute in $a(t)$

$$ a=J^{2/3}(6x)^{1/3}$$

Solve for $x$

$$ x(a)=\frac{a^3}{6J^2}$$


答:
If $A$ is finite, and $x \in X$ we have two cases: $x \notin A$, then $X\setminus A$ is a neighbourhood of $x$ that misses all of $A$, so no sequence from $A$ can converge to $x$.
If $x \in A$, a constant sequence will show already that $x$ is a sequential limit from $A$. So $A$ equals the set of sequential limits from $A$, and also this equals $\overline{A}$.

If $A$ is infinite, and $x \in X$, then any sequence fom $A$ that has all terms distinct (this can be done easily  when $A$ is infinite) converges to $x$. (Lemma A is not really needed for this, it's an easy observation). So $X$ equals the set of all sequential lmits from $A$ and this is also $\overline{A}$.

So in all cases sequential closure of $A$ is closure of $A$ and $X$ is Fréchet-Urysohn.

Title: Prove that an uncountable set with the cofinite topology is Fréchet–Urysohn

Body:
A space $X$ is said to be Fréchet–Urysohn give that,

$$x\in \overline{A} \text { if and only if there is a sequence } \langle x_n \rangle \text{ in } A \text{ such that } \langle x_n \rangle\to x.$$

Lemma A: Let $X$ be any uncountable set with the cofinite topology. Let $x\in X$ and $\langle x_n \rangle$ a sequence in $X$. Then $\langle x_n \rangle$ converges to $x$ if and only if for every $y\neq x$ we have $x_n=y$ for only finitely many $n\in\Bbb N$.

Using Lemma A, how can I prove that $X$, an uncountable set with the cofinite topology, is Fréchet–Urysohn?

I know that if $A\subseteq X$ is finite then $\overline{A}=A$, whereas if $A\subseteq X$ is infinite then $\overline{A}=X$.


问:
What values of alpha guarantee the convergence of $\sum _{n=0}^{\infty }\left( \frac {\alpha n} {n+1}\right) ^{n}$?
I have managed to use the root test and get as far as getting the limit to equal to $1$. But, I still need clarification on why the answer for the limit to this questions is $\alpha < 1$.

\begin{align}
\sum _{n=0}^{\infty }\left( \frac {\alpha n} {n+1}\right) ^{n}\end{align}

\begin{align}
\lim _{n\rightarrow \infty }\left( \left( \frac {\alpha n} {n+1}\right) ^{n}\right) ^{1/n}
=\left| \frac {\alpha n} {n+1}\right| =\alpha \lim _{n\rightarrow \infty }\left| \frac {n} {n+1}\right|=1.\end{align}

答:
Notice that $$\lim_{n\to \infty} \alpha^n\left(\frac{n}{n+1}\right)^n=\lim_{n\to\infty}\alpha^n\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n=\frac{1}{e}\lim_{n\to\infty}\alpha^n$$ the only way for this to converge to zero is that $|\alpha|<1$. Otherwise the sum will diverge. You can use this idea to check the series, for example: if $\alpha\geq 0$, then since $$\left(\frac{n}{n+1}\right)^n\leq 1,\ \forall n $$ then $$\sum_{n=0}^\infty \alpha^n\left(\frac{n}{n+1}\right)^n\leq \sum_{n=0}^\infty \alpha^n<\infty\  \mathrm{iff} \ \alpha<1$$ notice as well that for $\alpha=1$ your series it's
$$\sum_{n=0}^\infty \left(\frac{n}{n+1}\right)^n=\infty $$ and the reason behind it is the limit previously computed.


问:
Serving Slices of Pizza
Can someone please double check my work? Thanks!

There are 15 slices of pizza: 5 cheese, 5 cheese with pepperoni, and 5 no cheese.

Suppose the pizzas are served in a certain order from 1 to 15.

(1) What is the probability that all cheese with pepperoni slices are served within the first 10?

My work: I first choose 5 spots out of the first 10 to put them in. They can be ordered in 5! ways. The remaining 10 spots can be ordered in 10! ways. Total number of ways is 15! Answer is 8.39%.

(2) What is the probability that after serving 10 slices, only two types remain?
Kinda stuck on this one, but my thinking is the following:
Choose one type which can be done 3 ways. Place 5 of that type in the first 10. Those can be ordered 5! ways. Choose two spots in the final five which can be done 10 ways and those can be arranged in 2! ways. Arrange the remaining in 8! ways. Answer is 5.6%.

(3) What is the probability that two of each type are among the first six to be serviced?
Place all six in the first six slots. Order them in 6! ways. Order the rest in 9! ways. Answer is 0.02%.

答:
On question 2, if the interpretation is that at most two types remain after $10$ slices are served then

For one type of pizza, the probability that it finishes in first $10$ serving is $\frac{12}{143}$ as you found in question $1$. But it also includes probability that two types would have finished in first $10$ which is

$\displaystyle  \frac {10 \choose 10}{15 \choose 10} = \frac{1}{3003} $ for any two. So applying inclusion-exclusion,

$\small P(A \cup B \cup C) = P(A) + P(B) + P(C) - P (A \cap B) - P (B \cap C) - P(A \cap C) + P(A \cap B \cap C) \ $

where $A, B, C$ are events of a type of pizza finishing in first $10$ serving.

So, $\small P(A \cup B \cup C) =  3 \times \frac{12}{143} - \frac{3}{3003}$

If the question is about exactly two types to remain in last $5$, then we have all of one type served in first $10$ and for other two types, number of slices serves in first $10$ are $(4, 1), (3,2), (2,3), (1,4)$.

So, Probability $ \displaystyle = \frac{ {3 \choose 1} \times 2 \times \big({5 \choose 4}  {5 \choose 1} + {5 \choose 3} {5 \choose 2}\big)} {15 \choose 10} = \frac{250}{1001}$

On question 1, while your working is correct, you can simply look at it as

$\displaystyle  = \frac{10 \choose 5}{15 \choose 10} = \frac{12}{143}$

which is in the first $10$, we choose $5$ out of $5$ cheese with pepperoni pizza slices and $5$ out of remaining $10$ slices vs. any $10$ of $15$ slices.

For question 3, think number of ways of choosing $2$ out of $5$ pizza slices of each type vs. any $6$ slices out of $15$.

Title: Serving Slices of Pizza


问:
Find a normal subgroup $N$ of $(\Bbb Z_{10}, +)$.
I am to find a normal subgroup $H$ of $(\Bbb Z_8, +)$ and a normal subgroup $N$ of $(\Bbb Z_{10}, +)$.

The definition of a subgroup is given by:

A normal subgroup $F$ of the group $G$ has $gfg^{-1}\in F$ for $\forall g\in G$ and $\forall f\in F$. Every subgroup of an abelian group is normal since we have that $gfg^{-1}=gg^{-1}f=ef=f$.

And the following theorem defines a subgroup:

The subgroup $F$ of the group $G$ is normal if and only if one of the following conditions are met:

$aF=Fa$ for $\forall a\in G$

$F$ is equal to its conjugate, that is, $aFa^{-1}=F$ for $\forall a\in G$

$F$ is invariant under all inner automorphics of $G,$ that is, $aFa^{-1}=F$ for $\forall a\in G$

Now, I've solved the first part of the question by writing:

$\Bbb Z_8$ is an abelian group since it is a product of abelian groups, and also because the center $Z(G)$ of $G$ is equal to $G$. Thus, any subgroup of $\Bbb Z_8$ is normal. Therefore, we choose an arbitrary subgroup of $Z_8$ as $H$.
$$H=\{0,2,4,6\}.$$

答:
Each group $G$ is a normal subgroup of itself. The trivial subgroup is also normal in every group.

However, I'm guessing you're looking for a nontrivial, proper subgroup that is normal for each $\Bbb Z_8$ and $\Bbb Z_{10}$.

It is useful to note (and rather easy to show) that these two groups are abelian, so any subgroup of each is normal in that group.

For $\Bbb Z_8$, take the subgroup $\langle [2]_8\rangle$ generated by $[2]_8$; for $\Bbb Z_{10}$, consider $\langle [2]_{10}\rangle$.


问:
What is the relation between a matrix as a linear function versus the same matrix as a bilinear function?
Given an $n \times n$ matrix $A$, we can define a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by $T(x)=Ax$.

We could also define a bilnear function $T: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ by $T(x,y) = y^TAx$.

Is there a relation between these two uses of the matrix?

Also, we could do the same thing with an $n \times m$ matrix and get a linear function $\mathbb{R}^m \rightarrow \mathbb{R}^n$ and a bilinear function $\mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}$. Is the relationship between using $A$ to define a linear function versus a bilinear function the same in this case?

Body:
Given
$$
{\bf y}_{\,1}
  = \left( {\matrix{ {y_{1,1} }  \cr  \vdots \cr {y_{h,1} } \cr } } \right)
 = {\bf A}\,{\bf x}_{\,1}
$$
then
$$
{\bf Y}
 = \left( {\matrix{ {y_{1,1} } &  \cdots  & {y_{1,h} }  \cr \vdots  &  \ddots  &  \vdots   \cr
   {y_{h,1} } &  \cdots  & {y_{h,h} }  \cr  } } \right)
 = {\bf A}\,{\bf X}
$$

Therefore
$$
{\bf Y}^{\,T} {\bf Y} = {\bf Y}^{\,T} {\bf A}\,{\bf X}
 = {\bf X}^{\,T} {\bf A}^{\,T} {\bf A}\,{\bf X}
 = \left( {\matrix{  {{\bf y}_{\,1}  \cdot {\bf y}_{\,1} } &  \cdots
  & {{\bf y}_{\,1}  \cdot {\bf y}_{\,h} }  \cr     \vdots  &  \ddots  &  \vdots   \cr
   {{\bf y}_{\,h}  \cdot {\bf y}_{\,1} } &  \cdots  & {{\bf y}_{\,h}  \cdot {\bf y}_{\,h} }  \cr
 } } \right)
$$


问:
what are the last 2 digits of $n^n$ without computing the formula
Computing this exponential is costly for number like $n=10^{400}$

Would rather find a way to find the answer without any exponentiation.

答:
Last 2 digits of the result depend only of the last 2 digits of the original number.

Thus do all operations using modulo 100. Details about why we can simplify multiplication in modular multiplication see in Wikipedia.

result1 = n mod 100

result2 = (result1 * result1) mod 100

result3 = (result2 * result1) mod 100

result4 = (result3 * result1) mod 100

result5 = (result4 * result1) mod 100

And so on, repeat it $n$ times. Thus you get the result very quickly.

Also you can write a simple Python script, that would be literally a few lines of code.


问:
Evaluating particular solution
I'm trying to find the solution to the following difference equation:

$$y[n] - \frac{1}{4} y[n-1] - \frac{1}{8} y[n-2] =3x[n] $$

with $x[n]=(\frac{1}{2})^nu[n]$  where $u[n]$ is a step signal defined as :

$$ u[n]= 1 \forall n\ge 0$$
$$= 0 \forall n<0 $$

but somehow my solution does not match the given solution.

The complementary solution or homogeneous solution is : $k_1(\frac{1}{2})^n+k_2 (\frac{-1}{4})^n]u[n]$

And after that I take the particular solution of the form $$y_p[n]=kn(\frac{1}{2})^nu[n]$$ since the input is of the same form as the roots of the characteristic equation.

and get k as equal to 2.

and evaluate the coefficients of complementary solution as
$$y[n]=\frac{8}{3} (\frac{1}{2})^n u[n]+\frac{1}{3} (\frac{-1}{4})^n u[n]+2n (\frac{1}{2})^n u[n]$$

but the solution given is $$\frac{1}{3} (\frac{1}{4})^n u[n] + 4(n+1)(\frac{1}{2})^{(n+1)} u[n+1] + \frac{2}{3} (\frac{1}{2})^n u[n]$$

I don't where the anomaly in my evaluation is? Or is the given solution wrong?
I'm new to the concept of difference equations.

I wish to solve this difference equation without using z transform.

答:
A resolution method not based on the $z$ transform.

Making $y_1(k) = y(k), y_2(k) = y_1(k+1)$ and $Y = (y_1, y_2)^{\dagger}$ we have

$$
Y_{k+1}= M Y_k + B x_k
$$

considering now $M = T\Lambda T^{-1}$ we have

$$
T^{-1}Y_{k+1}=\Lambda T^{-1} Y_k + T^{-1}B x_k
$$

making now $Z_k = T^{-1} Y_k,\ \ \ V = T^{-1}B $ we follow with

$$
Z_{k+1} = \Lambda Z_k + V x_k
$$

the homogeneous solution is direct

$$
Z_k^h = \Lambda^k C_0
$$

now assuming a particular solution $Z_k^p = \Lambda^k C_k$ and substituting into the complete recurrence

$$
\Lambda^{k+1}C_{k+1} = \Lambda^{k+1}C_k+V x_k
$$

gives the recurrence

$$
C_{k+1}-C_k = \Lambda^{-k-1}V x_k
$$

then

$$
C_k = \sum_{j=0}^{k-1}\Lambda^{-j-1}V x_j
$$

and finally

$$
Z_k=Z_k^h + Z_k^p = \Lambda^k C_0+\Lambda^k\sum_{j=0}^{k-1}\Lambda^{-j-1}V x_j
$$

hence

$$
Y_k = T Z_k
$$

NOTE

$$
M = \left(
\begin{array}{cc}
 0 & 1 \\
 \frac{1}{8} & \frac{1}{4} \\
\end{array}
\right),\ \ T = \left(
\begin{array}{cc}
 2 & -4 \\
 1 & 1 \\
\end{array}
\right),\ \ \Lambda = \left(
\begin{array}{cc}
 \frac{1}{2} & 0 \\
 0 & -\frac{1}{4} \\
\end{array}
\right),\ \ B = \left(
\begin{array}{c}
 0 \\
 1 \\
\end{array}
\right),\ \ V = \left(
\begin{array}{c}
 \frac{2}{3} \\
 \frac{1}{3} \\
\end{array}
\right)
$$

Title: Evaluating particular solution


问:
Minimal polynomial matrix
I want to show that $ x^n-1$ is the minimal polynomial of the permutation matrix $P:=(e_2,e_3,....,e_n,e_1)$ where $e_i$ is the i-th unit vector written as a column vector. </p>

<p>And now I have to show that over an arbitrary field $x_n-1$ is the minimal polynomial. This seems to be tough. I have already proved that $P^n-Id=0$. Does anybody of you know how to do this or has a hint for me? </p>

答:
Since you observed that $P$ is annihilated by $X^n-1$, it only remains to check that no nonzero polynomial of degree $\leq n-1$ annihilates $P$ to conclude that the minimimal polynomial is $X^n-1$. Since the latter has degree $n$ and divides the characteristic polynomial which is also monic of degree $n$, it will also follow that the characteristic polynomial is $X^n-1$.

So assume that a polynomial of degree $\leq n-1$ annihilates $P$, that is
$$
a_0I_n+a_1P+\ldots+a_{n-1}P^{n-1}=0.
$$
Applying the latter to the first vector $e_1$ of the canonical basis, we get
$$
a_oe_1+a_1Pe_1+\ldots+a_{n-1}P^{n-1}e_1=a_0e_1+a_1e_2+\ldots+ a_{n-1}e_n=0.
$$
That is a linear combination of the basis, whence, by linear independence of the latter, $a_0=a_1=\ldots=a_{n-1}=0$. So no nontrivial polynomial of degree $\leq n-1$ annihilates $P$, which concludes the proof.

Note: you might want to note that this is a special case of a companion matrix $C(q)$, for which in general the characteristic and the minimal polynomials are both equal to $q$.

Title: Minimal polynomial matrix


问:
Why is $(b-Ax)^T(b-Ax) = (b^Tb - 2b^T Ax + x^T A^T Ax)$?
Why is it true that
$$
(b-Ax)^T(b-Ax) = (b^Tb - b^T Ax - b^T Ax + x^T A^T Ax)
$$
Shouldn't it be
$$
(b-Ax)^T(b-Ax) = (b^Tb - b^T Ax - x^T A^T b + x^T A^T Ax)
$$?

So in why are there two $b^T Ax$? This must mean that $x^T A^T b = b^T Ax$. If true, why can you assume that? $b$ is a $M \times 1$ vector, $x$ is a $N \times 1$ vector, and $A$ is a $M \times N$ matrix.

答:
Let $A=\left(\begin{matrix}-A_1^T-\\-A_2^T-\\...\\-A_M^T-\end{matrix}\right)$

$b^TAx=\left(\begin{matrix}b_1&b_2&...&b_M\end{matrix}\right)\left(\begin{matrix}A_1^Tx\\A_2^Tx\\...\\A_M^Tx\end{matrix}\right)=\sum_{j=1}^Mb_jA_j^Tx$

$x^TA^Tb=x^T\left(\begin{matrix}|&|&&|\\A_1&A_2&...&A_M\\|&|&&|\end{matrix}\right)\left(\begin{matrix}b_1\\b_2\\...\\b_M\end{matrix}\right)=\left(\begin{matrix}x^TA_1&x^TA_2&...&x^TA_M\end{matrix}\right)\left(\begin{matrix}b_1\\b_2\\...\\b_M\end{matrix}\right)=\sum_{j=1}^Mx^TA_jb_j$

But $x^TA_j=A_j^Tx$. Further, $b_jA_j^Tx=A_j^Txb_j$ as this is multiplying two scalars.

Alternatively, $b^TAx=b^T(Ax)=b^Ta=(b^Ta)^T=a^Tb=x^TA^Tb$


问:
Prove theorem about bases
I was attempting to prove this theorem, and got lost on the second part of the proof.

Theorem:

Let $(X,\tau)$ be a topological space, and let $\mathcal{B}$ be a collection of subsets of $X$. Then $\mathcal{B}$ is a basis for $\tau$ if and only if

$(1) \mathcal{B} \subset \tau$ and

$(2)$ for each set $U$ in $\tau$ and point $p$ in $U$ there is a set $V$ in $\mathcal{B}$ such that $p \in V \subset U$.

This was supposed to be done using the following definition:

Let $\tau$ be a topology on a set $X$, and let $\mathcal{B} \subset \tau$. Then $\mathcal{B}$ is a basis for the topology $\tau$ if and only if every open set in $\tau$ is the union of elements of $\mathcal{B}$. If $B \in \mathcal{B}$, we say $B$ is a basis element. Note that $B$ is an element of the basis of $\mathcal{B}$, but a subset of the set $X$.

Attempt:

($\Rightarrow$)Obviously $(1)$ is given in the definition, so it suffices to prove $(2)$. Since $\mathcal{B}$ is a basis, any open set $U$ can be expressed as a union $\bigcup\limits B_x$ of basis elements. Then $U=\bigcup\limits B_x$ . Let $p \in U$. Then $p \in B_x$ for some $x$ and $B_x \subset U$, proving $(2)$.

($\Leftarrow$)

For this direction can I just pick an arbitrary $U \in \tau$, then by condition $(2)$, for each $x \in U$, there exists a $B_x$ containing $x$, with $x \in B_x \subset U$, and then $U= \bigcup\limits B_x$? So since each open set is a union of sets satisfying $(1)$ and $(2)$, this collection of sets is a basis?

答:
Yes, you can prove it that way, but we can avoid making any choices : given $U \in \tau$ we can define $$\mathcal{B}_U = \{B \in \mathcal{B}\mid B \subseteq U\}$$

Then I claim that $\bigcup \mathcal{B}_U = U$.

If $x \in U$, we apply (2) to get a $V \in \mathcal{B}$ such that $x \in V \subseteq U$. But then $V \in \mathcal{B}_U$ by definition and so $x \in \bigcup \mathcal{B}_U$, and the reverse inclusion is obvious: all $B \in \mathcal{B}_U$ are subsets of $U$, hence so is their union.

So any $U \in \tau$ is a union of base elements and we're done.

Title: Prove theorem about bases


问:
Jordan–Bareiss algorithm
Jordan–Bareiss Algorithm is an algorithm that find determinant of $n \times n$ square matrix $M$. Its algorithm is given below:

Initialize the sign $s := 1$ and set $c := m^{(-1)}_{00}
 = 1;$

loop over rows $M_k$ of $M$ for $k ∈ \{1, \dots , n\}$:

a) find the pivot $m_{ik} \neq 0$ with $i ≥ k$;

b) if no such $m_{ik}$ exists, then return $0$ and quit;

c) if $i \neq k$, then swap the rows $M_k$ and $M_i$ and update the sign, setting $s := −s$.

d) replace every row $M_i$ with $i ∈ \{k + 1, \dots , n\}$ by an expression

$\hspace{2.4cm}$ $\frac{(m_{kk}·M_i − m_{ik}·M_k)}{c}$

e) update the divisor $c$ setting $c:= m_{kk}$;

return $\det M = s·m_{nn} $.

Here is my trial using SageMath but its giving wrong answer.

```<code>def JordanBarreiss(M):
    assert M.is_square(), "It must be a square matrix"
    n=M.dimensions()[0]
    s=1
    c=1
    for i in [1,..,n-1]:
        for k in [1,..,n-1]:
            if M[i,k] == 0 and i < k:
                return 0
            if i != k:
                M.swap_rows(i,k)
                s=-s
                for j in [k+1,..,n-1]:
                    M[j] = (M[k,k]*M[j] - M[j,k]*M[k])/c
        c = M[k,k]
    return M[n-1,n-1]
</code>```
```<code>A = matrix(QQ, [[5,1,-9,0],[-1,3,-1,1],[8,-2,1,-3],[2,2,1,2]]);
JordanBarreiss(A);

out[110]  0

The expected output M = 5  1  -9   0
                        0  16 -16  5
                        0  0   196 -30
                        0  0   0   475
where M[n-1,n-1] = 475 (det(M))

</code>```

Please, it will be highly appreciated if anyone could help to fix this error, Python is also allowed. thanks.

答:
Now I get it, thanks all</p>
```<code>def JordanBarreiss(M):
    """
     JordanBareiss: Function to compute determinant of nxn square matrix M of size this algorithm

     INPUT:

         - A square matrix M

    OUTPUT:

        - det(M) 'determinant of the given matrix'

    """
    assert M.is_square(), "Only Square matrix allowed"
    n=M.dimensions()[0]
    c=1
    for k in range(n-1):
        for i in range(k+1,n):
            for j in [k+1,..,n-1]:
                M[i,j] = (M[i,j]*M[k,k] - M[i,k]*M[k,j])/c
        c=M[k,k]
    return M[n-1,n-1]

<span class="math-container">```</span>
</code>```

Title: Jordan–Bareiss algorithm

Body:
Now I get it, thanks all

def JordanBarreiss(M):
    """
     JordanBareiss: Function to compute determinant of nxn square matrix M of size this algorithm

     INPUT:

         - A square matrix M

    OUTPUT:

        - det(M) 'determinant of the given matrix'

    """
    assert M.is_square(), "Only Square matrix allowed"
    n=M.dimensions()[0]
    c=1
    for k in range(n-1):
        for i in range(k+1,n):
            for j in [k+1,..,n-1]:
                M[i,j] = (M[i,j]*M[k,k] - M[i,k]*M[k,j])/c
        c=M[k,k]
    return M[n-1,n-1]

```


问:
Let $R$ be an equivalence relation on $X$ and $S$ an equivalence relation on $X/R$.....

Let $R$ be an equivalence relation on $X$ and $S$ an equivalence relation on $X/R$. Find an equivalence relation $T$ on $X$ such that $(X/R)/S$ is in one-to-one correspondence with $X/T$ under the mapping $[[x]R]S\mapsto[x]T$.

Not much to add. I found this question in Elements of Abstract Algebra by Allen Clark. Any help would be appreciated. Cheers!

答:
HINT: Use the relationship between equivalence relations on a set and the associated partitions of that set. $R$ induces a partition $\mathscr{P}_R$ of $X$. $S$ then induces a partition $\mathscr{P}_S$ of $X/R$. Each member of $\mathscr{P}_S$ is a set whose elements are members of $\mathscr{P}_R$. In other words, the equivalence relation $R$ lumps the members of $X$ into sets — the $R$-equivalence classes, which are also the members of $\mathscr{P}_R$ — and the equivalence relation $S$ then lumps those sets into collections of sets that are the $S$-equivalence classes, i.e., the members of $\mathscr{P}_S$.

It’s as if you were to take a box, $X$, of objects, and begin by sorting those objects into small boxes sitting inside $X$; those small boxes are the $R$-equivalence classes, the members of the partition $\mathscr{P}_R$. Then you sort those smaller boxes into medium-sized boxes still sitting inside $X$; those medium-sized boxes are the $S$-equivalence classes, the members of the partition $\mathscr{P}_S$.

Now you want to find a single partition of $X$ whose parts are in one-to-one correspondence with $\mathscr{P}_S$. How can you sort the objects in $X$ into boxes that correspond naturally to the medium-sized boxes?


问:
number of different sequences
He, let consider Basketball. We can make throw for 1, 2 or 3 points.

For example for 5 points we have throws sequences like:

```<code>{3, 2}, {2, 3}, {3, 1, 1}, {1, 1, 3} {1, 3, 1}, {2, 2, 1}, {2, 1, 2}, {1, 2, 2}, {2, 1, 1, 1}, {1, 2, 1, 1}, {1, 1, 2, 1}, {1, 1, 1, 2}, {1, 1, 1, 1, 1}.
</code>```

How can compute it for any n?

答:
Call $S_n$ the number if sequences of $n$ points.

For $1$ point, you have just one sequence, $(1)$.

For $2$ points, you have two sequences, $(2)$ and $(1,1)$.

For $3$ points, you have 4 sequences, $(3),(2,1),(1,2),(1,1,1)$.

So, $S_1=1, S_2=2, S_3=4$. Now let $n\geq4$. A sequence of $n$ points follows one of these cases:

$\bullet$ It begins by 1 and it´s followed by a sequence of $n-1$ points. ($S_{n-1}$ sequences)

$\bullet$ It begins by 2 and it´s followed by a sequence of $n-2$ points. ($S_{n-2}$ sequences)

$\bullet$ It begins by 3 and it´s followed by a sequence of $n-3$ points. ($S_{n-3}$ sequences)

Thus, we have that $S_n=S_{n-1}+S_{n-2}+S_{n-3}$. Thus, your the number of sequences of $n$ points is given by the  recursion:

$S(1)=1;S(2)=2;S(3)=4;S(n)=S(n-1)+S(n-2)+S(n-3)$.

This is a standard linear recurrence relation that can be solved for a explicit formula. If you want to try and solve it explicitly (the formula you obtain won´t be nice, to say the least), there is more info here.

Title: number of different sequences


问:
Suppose $(A, \le)$ is a partially ordered set. Define a relation $\preceq$ on $A\times A$ by $(a,b)\preceq (c,d)$ if
if and only if

1) $a\le c$ and $a \ne c$

2) $a = c$ and  $b\le d$

Prove that $(A\times A,\preceq)$ is also a partially ordered set.

So to prove this I would start with trying to find 1) Reflexivity, 2) anti-symmetricy and 3) Transitivity right?

How would I do that, does the $A\times A$ mean anything special in this case?

For Reflexivity would i say 1) since $\le$ is reflexive then that means $A\le A$ ... I just dont know how to begin this?

答:
The order $\preceq$ on $A\times A$ is known as the lexicographic order induced by $\le$. $A\times A$ is the Cartesian product of $A$ with itself: $A\times A=\{\langle a,b\rangle:a,b\in A\}$. To show that $\preceq$ is a partial order on $A\times A$, you do indeed have to show that it’s reflexive, transitive, and antisymmetric.

Reflexive: You must show that for every $\langle a,b\rangle\in A\times A$, $\langle a,b\rangle\preceq\langle a,b\rangle$. This is completely straightforward: just check the definition of $\preceq$.

Transitive: You must show that if $\langle a_0,b_0\rangle,\langle a_1,b_1\rangle,\langle a_2,b_2\rangle\in A\times A$, $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$, and $\langle a_1,b_1\rangle\preceq\langle a_2,b_2\rangle$, then $\langle a_0,b_0\rangle\preceq\langle a_2,b_2\rangle$. Since the definition of $\preceq$ has two clauses, you’ll have to consider several cases. I’ll do one of them to get you started. Suppose that $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$ because $a_0\le a_1$ and $a_0\ne a_1$. We know that $a_1\le a_2$ (why?), so $a_0\le a_2$ and $a_0\ne a_2$, and therefore $\langle a_0,b_0\rangle\preceq\langle a_2,b_2\rangle$, as desired.

Antisymmetry: You must show that if $\langle a_0,b_0\rangle,\langle a_1,b_1\rangle\in A\times A$, $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$, and $\langle a_1,b_1\rangle\preceq\langle a_0,b_0\rangle$, then $\langle a_0,b_0\rangle=\langle a_1,b_1\rangle$. Here again I’ll get you started. Suppose that $\langle a_0,b_0\rangle\preceq\langle a_1,b_1\rangle$ because $a_0\le a_1$ and $a_0\ne a_1$; then $a_1\not\le a_0$ (why?), and it is therefore impossible to have $\langle a_1,b_1\rangle\preceq\langle a_0,b_0\rangle$. Thus, we must have $a_0=a_1$ and $b_0\le b_1$. Now finish the argument by showing that $b_0=b_1$.


问:
Tree-width and Maximum Cliques of Triangulation - Confusion over proof.
I am confused over a line in the proof of a theorem which is part of some Graph Theory notes I am working through.

The statement of the theorem is as follows:

$tw(G)=min\{\omega(H)-1:$ $H$ is a triangulation of $G$$\}$, where $tw(G)$ is the tree-width of $G$ and $\omega(H)$ is the size of a maximum clique in $H$.

The proof starts by proving the following claim:

G is chordal (triangulated) if and only if it has a tree-decomposition with each bag (part) being a clique.

This part of the proof I understand. The proof then continues as follows:

Let $H$ be a triangulation of $G$. Since $G$ is a subgraph of $H$, any tree-decomposition of $H$ is also a tree-decomposition of $G$, and hence $tw(G) \leq tw(H)$. Since every clique of $H$ is contained in one of the bags of its tree-decomposition, we know that $tw(H) \leq \omega(H)-1$.

And so forth...

Now I understand that any tree-decomposition of $H$ is a tree-decomposition of $G$, since I proved this as an exercise. I therefore understand that $tw(G) \leq tw(H)$. I also understand that every clique of $H$ is contained in one of the bags of its tree-decomposition since I also proved this result as an exercise. But why does that imply that $tw(H) \leq \omega(H)-1$? Surely every tree-composition of $H$ must have a bag that contains the largest clique in $H$, hence the width of any tree-decomposition of $H$ must be at least $\omega(H)-1$, and hence $tw(H) \geq \omega(H)-1$? Am I missing something?

Many thanks

答: You are right: $tw(H) \geq \omega(H) - 1$. On the other hand, since $H$ is chordal, it has a tree-decomposition where every "bag" is a clique. Such a tree-decomposition must clearly have width $\omega(H) - 1$, which implies $tw(H) \leq \omega(H) - 1$. This shows that for a chordal graph $H$, $tw(H) = \omega(H) - 1$. Without seeing more of the proof it is difficult to say why this is given only as an upper bound; I suppose that the proof doesn't use the fact that equality holds.


问:
Translating logical statements in to English?
I've studied a little bit about logical statements, but apparently not enough. Can you recommend a good reference on how to translate logical statements in English?

For instance, when reading the Axiom of Extensionality on Wikipedia, there is a translation into plain English. Like below,

$$\forall A \, \forall B \, ( \forall X \, (X \in A \iff X \in B) \implies A = B)$$

"Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B."

After $$\forall A \, \forall B $$ how does the next group of parenthesis, $$ ( \forall X \, (X \in A \iff X \in B) \implies A = B)$$

relate to first two quantifiers?

I get confused on how to read these statements correctly with all the parenthesis and brackets, why is there a "such that" when there is no symbol for such that, etc...

For instance, in the Axiom of Pairing, there is a "such that," like below,

$$\forall A \, \forall B \, \exists C \, \forall D \, [D \in C \iff (D = A \lor D = B)]$$

"Given any set A and any set B, there is a set C such that, given any set D, D is a member of C if and only if D is equal to A or D is equal to B."

Or, perhaps my question should be, "how do you read these statements?"

Thank you!

答:
With logic, we want to go from the outside in, so addressing

$(∀X(X∈A⟺X∈B)⟹A=B)$

Let's ignore the innermost parentheses for a second - then this statement is essentially saying

$(∀X($Something$)⟹A=B)$

The forward implication symbol $⟹$ denotes an if-then statement, so we can translate this to:
"If, for all X, (Something), then A = B.

Now we can unpack the statement in the innermost parenthesis, which is saying that $X \in A$ if and only if $X \in B$. So putting this back into our last statement, we have "If, for all $X$, $X \in A$ if and only if $X \in B$, then $A = B$.

What does this have to do with our $\forall A \forall B$ before? well this is telling us that this above statement is true for ANY sets $A$ and $B$

Actually looking to see if this makes sense helps, essentially we're saying that if we pick any two sets at random, we're confirming that these two sets are equal, and this happens if every element of $A$ is an element of $B$ and vice versa, which is what the $\forall X(X\in A \Leftrightarrow X \in B)$ statement is saying.

When it comes to the $\exists X$ statement, as far as I am aware, linguistically translates to "There exists X such that...", whose conditions follow. Without the "such that", $\exists$ really means nothing at all.


问:
How to compute Trigonometric functions based on Inverse Square Root
I am writing a rational number library for performing fast math on integer-only microcontrollers. So far, I have all the basic operations, as well as a first-step estimation function for square root. It takes a positive integer input and returns a rational number approximating the square root (accurate to roughly 4 bits of precision; to be refined further by Newton-Raphson or similar method). Obviously, inverse (reciprocal) square root comes for free since the answer is a fraction that can be trivially inverted.

I read somewhere that other functions (e.g. trigonometric functions) can be calculated in terms of inverse square root (can't find the link now), but I am having a hard time finding any algorithms since search results are almost entirely about inverse trigonometric functions (mostly high school help stuff).

Does anyone have a handy resource where I can find efficient methods for calculating sin, cos, tan, atan, etc. in terms of square root or inverse square root? Is it possible or am I imagining things?

I know this is not a "code" StackExchange but I believe this is fundamentally a math question. If you're interested in my code, it's here. The sqrt_est() function is somewhat remarkable in that it is very short, uses only shifts and adds, no looping or iteration, or even multiplies or divides! (core implementation is on lines 39-45.)

答:
Square roots are key to all sorts of computations because they let us compute arbitrary powers. If you want to compute $2^{1/7}$, for example, first write $\frac17$ in binary as $0.001001001001001\dots$, which tells us that
$$
   2^{1/7} = 2^{1/8} \cdot 2^{1/64} \cdot 2^{1/512} \cdot 2^{1/4096} \cdot 2^{1/32768} \cdot \dots
$$
All of these are repeated square roots: $2^{1/2} = \sqrt 2$, $2^{1/4} = \sqrt{2^{1/2}}$, $2^{1/8} = \sqrt{2^{1/4}}$, and so on. So we can take as many square roots as we need to, and multiply them together.

Square roots are key to trig functions because those are just powers. If you want to compute $\cos \frac\pi7$, for example, then it's enough to compute $(-1)^{1/7} = \cos \frac\pi7 + i \sin \frac\pi7$. This is done in the same way as above:
$$
   (-1)^{\frac17} = (-1)^{1/8} \cdot (-1)^{1/64} \cdot (-1)^{1/512} \cdot (-1)^{1/4096} \cdot (-1)^{1/32768} \cdot \dots
$$
Computing iterated square roots $(-1)^{1/2} = \sqrt{-1}$, $(-1)^{1/4} = \sqrt{(-1)^{1/2}}$, and so on is a bit trickier, but comes down to real square roots in the end.

But if the $(-1)^x$ argument seems a little bit disingenuous to you, then we can rephrase this in terms of trig identities. Start with a known $(\cos x, \sin x)$ pair: it's convenient to start with $(\cos \frac\pi2, \sin \frac\pi2) = (0,1)$, but we can start anywhere if we want to. From $(\cos x, \sin x)$ we can compute $(\cos \frac x2, \sin \frac x2)$ by the pair of identities
$$
    \cos \frac x2 = \pm \sqrt{\frac{1 + \cos x}{2}}, \qquad \sin \frac x2 = \pm \sqrt{\frac{1 - \cos x}{2}}.
$$
The angles are determined by the quadrant, which in this case is always going to be positive for both $\cos$ and $\sin$. This lets us compute $(\cos \frac{x}{2^n}, \sin \frac{x}{2^n})$ for many $n \ge 0$, which we can actually save in a table.

Then, if you want to compute $\cos \theta$ for an arbitrary angle $\theta \in [0,2x]$, where $x$ is our reference angle, first write $\frac{\theta}{x}$ in binary. This is telling us $\theta$ as a sum of some subset of $\{x, \frac x2, \frac x4, \frac x8, \frac x{16}, \frac x{32}, \dots\}$, all of whose sines and cosines we know. Angle addition formulas tell us $\cos \theta$ and $\sin \theta$.

This is very similar in flavor to CORDIC algorithm also mentioned in the other answer, but that one is more efficient (and also more complicated).


问:
Let $G$ be a group in which $aba = b \; \forall a,b \in G$. Prove that $G$ is Abelian.
I was wondering if someone could tell me if my method is correct. I feel like the substation $b=e$ is a stretch and that there is a better way to do it.

We know that $G$ is Abelian $\iff (ab)^{-1} = a^{-1}b^{-1}$. So I know that we will have to manipulate $aba = b$ into $(ab)^{-1} = a^{-1}b^{-1}$.

But this is where I get stuck.
$$aba = b \implies ab = ba^{-1} \implies (ab)^{-1} = (ba^{-1})^{-1} \implies (ab)^{-1}=ab^{-1}$$
This is close, but not what I want. One thing that I did notice is that if $b=e$, then $aba=b \implies a^2 = e \implies a=a^{-1}$. Since $a^{-1}$ is a unique element we know that $a=a^{-1}$ regardless of what $b$ is. So if we make that substitution into out previous equation we get
$$(ab)^{-1} = a^{-1}b^{-1}$$
Therefore $G$ is Abelian.

答: The hypothesis implies an exponent of two for the group.  But then it is abelian, since $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$, for any two elements of the group $a$ and $b$.


问:
Solving for theta
I am trying to solve some a math problem in a 3D software project.

I know that:

$$\cos(\theta)\sin(75°)a+\cos(75°)b+\sin(\theta)\sin(75°)c=0$$

How do I write an equation to solve for $\theta$? Could someone go through all the steps please? Because I have to solve many similar problems.

I already know that there may be zero, one, or (usually) two solutions.

Thanks.

[edit]

By the way, what would the solutions for $\theta$ be in this case if a = 4, b = 5, and c = 6? (Just so I can work it out on my own.)

答:
HINT: Notice the common factor of $\sin(75^\circ)$. Write
$$a\cos\theta + c\sin\theta = \sqrt{a^2+c^2}\left(\frac a{\sqrt{a^2+c^2}} \cos\theta + \frac c{\sqrt{a^2+c^2}}\sin\theta\right).$$
There is an angle $\tau$ with
$$\cos\tau = \frac c{\sqrt{a^2+c^2}} \quad\text{and}\quad \sin\tau=\frac a{\sqrt{a^2+c^2}}.$$
Now you can use the addition formula for $\sin$ to express the term in parentheses as $\sin(\theta+\tau)$.

Title: Solving for theta


问:
An issue identifying Clifford Algebra with Endomorphism ring.
I am trying to work through pages 303-305 of Fulton and Harris and have ran into a problem. I will first give a little bit of setup to try and make this post self contained.

Let $V$ be an even dimensional complex vector space and $Q$ the standard non-degenerate symmetric bilinear form on $V$. We form the Clifford algebra $\mathcal{Cl}(V,Q)$. This algebra splits up into even and odd parts: $\mathcal{Cl}(V,Q)=\mathcal{Cl}(V,Q)^{\text{even}}\oplus\mathcal{Cl}(V,Q)^{\text{odd}}$, as does the exterior algebra: $\wedge^{\star}W=\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}}$.

With this choice of dimension and bilinear form, $V$ splits up into maximal isotropic subspaces: $V=W\oplus W'$. Following Fulton and Harris I have then been able to show that we have an isomorphgism of algebras:
$$
\mathcal{Cl}(V,Q)\cong\text{End}\left(\wedge^{\star}W\right)
$$

Fulton and Harris claim that from this decomposition and the above isomorphism that we have:

$$
\mathcal{Cl}(V,Q)^{\text{even}}\cong\text{End}\left(\wedge W^{\text{even}})\oplus\text{End}(\wedge W^{\text{odd}}\right)
$$

and this is where I get stuck. I proceeded as follows:

$$
\mathcal{Cl}(V,Q)
\cong
\text{End}\left(\wedge^{\star}W\right)
\cong
\text{End}(\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}})
\cong
\text{End}\left(\wedge W^{\text{even}})\oplus\text{End}(\wedge W^{\text{odd}}\right)
$$

This is obviously not correct, since it would imply that $\mathcal{Cl}(V,Q)\cong \mathcal{Cl}(V,Q)^{\text{even}}$. I would really appreciate it if anybody could point out where I've gone wrong.

EDIT.

Thanks to David Hill I have realised thast the identification:
$$
\text{End}(\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}})
\cong
\text{End}(\wedge W^{\text{even}})\oplus\text{End}(\wedge W^{\text{odd}})
$$
was incorrect. This can be seen by either noting that $\text{dim}\text{End}(\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}})=2^{2n}$, while $\text{dim}\text{End}(\wedge W^{\text{even}})\oplus\text{End}(\wedge W^{\text{odd}})=2^{2n-1}$, or by noting the existence of odd homomorphisms such as $f_{w}:\wedge^{\text{even}}W\to\wedge^{odd}W$, $u\mapsto wu$ for $w\in W$.

This means that so far I have only established the following isomorphism:
$$
\mathcal{Cl}(V,Q)^{\text{even}}\oplus\mathcal{Cl}(V,Q)^{\text{odd}}
\cong
\text{End}(\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}})
$$

I still need to somehow "split up", $\text{End}(\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}})$, and "project out", the odd part of the Cifford algebra, but I am not sure how to go about this. Fulton and Harris claim that this can be done by noting that $\mathcal{Cl}(W)^{\text{even}}$ respects the splitting $\wedge^{\star}W=\wedge W^{\text{even}}\oplus\wedge W^{\text{odd}}$, but I am not sure exactly what that means. Any pointers would be greatly appreciated.

答:
Okay, I think I have figured this out. First of all, by decomposing an endomorphism into block digonal and block anti-diagonal parts one can show that for $n$-dimensional algebras $V_{1},V_{2}$:

$$
\text{End}(V_{1}\oplus V_{2})
\cong
\text{End}(V_{1})
\oplus
\text{End}(V_{2})
\oplus
\text{Hom}(V_{1},V_{2})
\oplus
\text{Hom}(V_{2},V_{1})
$$

and so:

$$
\mathcal{Cl}(V,Q)^{\text{even}}
\oplus
\mathcal{Cl}(V,Q)^{\text{odd}}
\cong
\left(
\text{End}(\wedge^{\text{even}}W)
\oplus
\text{End}(\wedge^{\text{odd}}W)
\right)
\oplus
\left(
\text{Hom}(\wedge^{\text{even}}W,\wedge^{\text{odd}}W)
\oplus
\text{Hom}(\wedge^{\text{odd}}W,\wedge^{\text{even}}W)
\right)
$$

Let $\phi$ be the above isomorphism (it is given explicitly in Fulton and Harris). On can then show directly that $\phi$ respects this direct sum decomposition. Explicitly we have:
$$
\phi|_{\mathcal{Cl}(V,Q)^{\text{even}}}:\mathcal{Cl}(V,Q)^{\text{even}}\to\text{End}(\wedge^{\text{even}}W)
\oplus
\text{End}(\wedge^{\text{odd}}W)
$$
and
$$
\phi|_{\mathcal{Cl}(V,Q)^{\text{odd}}}:\mathcal{Cl}(V,Q)^{\text{odd}}\to\text{Hom}(\wedge^{\text{even}}W,\wedge^{\text{odd}}W)
\oplus
\text{Hom}(\wedge^{\text{odd}}W,\wedge^{\text{even}}W)
$$
And from here, since $\phi=\phi|_{\mathcal{Cl}(V,Q)^{\text{even}}}\oplus\phi|_{\mathcal{Cl}(V,Q)^{\text{odd}}}$, we see that both restrictions must be isomorphisms.

I think that my main difficulty with this problem was a findamental misunderstanding of the Endomorphism ring as pointed out by David Hill. Beyond that, working out an explicit example for $V=\mathbb{C}^{4}$ helped me realise that $\phi$ respects the $\mathbb{Z}_{2}$ grading of the Clifford algebra.


问:
How is this notation read?
How is this notation read $$a_{0}^{2^{i}}??$$

Here is also an image of the full reference:

答:
The number $a_0$ is the input to the function that does the exponentiation. The notation
$$a_0^{2^i}$$
then refers to the number $a_0$ to the power of $2^i$.

Title: How is this notation read?