问:
Caution while dealing with finite field in Linear Algebra by Hoffman & Kunze
In the first section $1.1$ of the book Linear Algebra by Hoffman & Kunze, author says that

In the examples and exercises of this book, the reader should assume that the field involved is a subfield of complex numbers, unless it is expressly stated that the field is more general.

I have two questions :

$1.$ So does this mean all the theorems in the book are true for any fields.

$2.$ If I want to be a little careful about finite fields, which steps I should do carefully while doing examples. Like in a subfield of $\mathbb{C}$, $x.y=0 \Rightarrow x=0 \ \text{or} \  y=0$, but this is not true for finite fields. Are there any other such steps, which I might do in flow which are true for a subfield of $\mathbb{C}$, but not for finite field.

答:
"Linear algebra" is really a broad term that covers diverse topics such as matrix theory, systems of linear equations, vector geometry comes up, we often see some quadratic forms... so it is somewhat difficult to say if everything will still hold over a finite field.

But you will find that almost all of the standard "linear algebra" theorems are either still true over a finite field, or have a close analog. There are a few exceptions.

The most major exception is that we don't really have a notion of magnitude in a finite field (i.e. we can't say one finite field element is "larger than" another), and by extension, we don't consider finite field elements to be "positive" or "negative". So any statements about the distance between two vectors, or the length of a vector, or a quantity being positive or negative, are meaningless. This also means that we don't have a notion of an inner product (since it requires $\langle \bf{x}, \bf{x}\rangle \geq 0$ for all $\bf{x}$). Instead of inner products we typically deal with what are known as bilinear forms.

A second exception is that finite fields are, well, finite; so a finite-dimensional vector space over a finite field has a finite number of vectors. Obviously this means that systems of equations cannot have infinitely many solutions. This also leads to some nice results that can be proven using counting arguments.

A third exception has to do with fields of characteristic 2. In these fields, $1+1=0$ (really $a+a=0$ for any $a \in \mathbb{F}$). This can lead to some results that hold in most fields, but not those of characteristic 2; there are also some results that are still true over fields of characteristic 2, but require an alternate proof. The main example I can think of where characteristic 2 is an issue, is when looking at quadratic forms. You would typically represent a quadratic form $Q(x,y) = x^{2}+xy+y^{2}$ using a symmetric matrix
$$Q(x,y) = \begin{bmatrix} x & y\end{bmatrix}
\begin{bmatrix} 1 & \frac{1}{2}\\ \frac{1}{2} & 1 \end{bmatrix}
\begin{bmatrix} x\\y \end{bmatrix}$$
but this does not work in characteristic 2, so modifications have to be made, typically by using an upper triangular matrix instead of a symmetric matrix, like so:
$$Q(x,y) = \begin{bmatrix} x & y\end{bmatrix}
\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}
\begin{bmatrix} x\\y \end{bmatrix}.$$
It is worth noting that, unlike a symmetric matrix, an upper-triangular matrix cannot always be diagonalized.


问:
Inequality induction proof with multiple variables
For all $s ∈ N$ and all $k > 0$ it holds that $1+sk ≤(1+k)^s$

How would one use induction to prove this statement?

Im following the basic steps: First do base case then do induction step but
i dont know how to even set the base case up correctly.
Im setting up RHS and LHS to prove base case, but im not sure how to do it
when there is multiple variables.

答:
For $s=1$, LHS$=1+k$, and RHS$=(1+k)$. Therefore the statement is true. Assume that it is true for $s=s'$. Then

$$
(1+k)^{s'+1} = (1+k)(1+k)^{s'} \geq (1+s'k)(1+k) = 1+k+s'k+s'k^2 \geq 1+(1+s')k.
$$

In the last inequality use the fact that $k>0$.

ContentLicense: CC BY-SA 4.0


问:
Limit of this sequence $ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$?
How to find the limit of this sequence. I tried using the squeeze theorem but no result yet.

$ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$

答:
The limit is $\frac{4}{6} = \frac{2}{3}$. Prove that this is so.

HINT: Check that for each $\epsilon > 0$, there is an $n(\epsilon)$ such that for all $n>n(\epsilon)$, the denominator $2^n+3^n+4^n$ satisfies $$(1-\epsilon)4^n \le 2^n+3^n+4^n \le (1+\epsilon)4^n.$$ Likewise, there is an $n'(\epsilon)$ such that for all $n>n'(\epsilon)$, the numerator $5^n+6^n$ satisfies $$(1-\epsilon)6^n \le 5^n+6^n\le (1+\epsilon)6^n.$$

So for each $\epsilon > 0$ and for all $n > \max\{n(\epsilon), n'(\epsilon)\}$ the followign holds:

$$\sqrt[n]{\frac{1-\epsilon}{1+\epsilon}}\sqrt[n]{\frac{4^n}{6^n}} \le \sqrt[n]{\frac{5^n+6^n}{2^n+3^n+4^n}} \le \sqrt[n]{\frac{1+\epsilon}{1-\epsilon}}\sqrt[n]{\frac{4^n}{6^n}}$$

Can you finish from here.


问:
Problem related with boundary value problem and eigenvalue, eigenfunctions
I was looking at previous year exam papers and was stuck  on the following problem:

For the boundary value problem,   $\,\,y''+\lambda y=0; y(0)=0,y(1)=0, \,\,\exists$ an eigenvalue $\lambda$ for which there  corresponds an eigenfunction in $(0,1)$ that

does not change sign

changes sign

is positive

is negative

Now,I have to decide which of the aforementioned options are correct ?

My Attempt:

The general solution is :

$y(x)=a\cos(\sqrt \lambda x)+b\sin(\sqrt \lambda x).$

Then we apply the boundary values. From $y(0)=0$,we get $a=0.$

From $y(1)=0,a=0\,\,$ we get $\,\,0=b\sin(\sqrt \lambda )$.

We assume that $b \neq 0$ so that $\,\,0=b\sin(\sqrt \lambda )$.Then we have,$\sqrt \lambda=n \pi \implies \lambda =(n \pi)^2$.Let's write $\lambda_n=(n \pi)^2.$ Since we have $a=0,$ only sine term remains,so eigenfunctions are $y_n=\sin(\sqrt \lambda_n x)\,\,$ with eigenvalues $\lambda_n=(n \pi)^2,\,\,\,\, n=1,2,3, \dots$.

Now,when I look at the options ,I can not progress about which way to go. Can someone point me in the right direction keeping in mind for deciding which options are correct?

答: You've got it! What can you say about the function $y_1(x)=\sin(\pi x)$ on the interval $(0,1)$? What about scalar multiples for various scalars? And what about $y_2(x)$? P.S. I find the problem confusing. I interpret it as $4$ separate T/F questions.</p>

ContentLicense: CC BY-SA 3.0


问:
$\sum_{n=1}^{\infty}a_n b_n$ Convergence
If $\lim_{n \to \infty} a_n=0$ and $\lim_{n \to \infty} b_n = 0$, then does $\sum_{n=1}^{\infty}a_n b_n$ converge? I think this is false - I don't know about both $a_n$ and $b_n$ but I know that given $a_n=\frac{1}{x}$ the sum $\sum_{n=1}^{\infty}a_n$ does not converge.

答:
Remember that

$$\sum_{n=0}^{+\infty} s_n ~\text{converges} \Rightarrow \lim_{n \to +\infty} s_n = 0,$$

but not viceversa, i.e.

$$\lim_{n \to +\infty} s_n = 0 \not\Rightarrow \sum_{n=0}^{+\infty} s_n ~\text{converges} .$$

In your case, $s_n =a_n b_n$...


问:
Why is $\kappa({A^{-1}})$ important and what does it tell us?
We need the condition number of a matrix $\kappa(A)= \frac{\text{max}_{\vert\vert \vec{y} \vert\vert =1}\vert\vert A \vec{y} \vert\vert}{\text{min}_{\vert\vert \vec{y} \vert\vert =1}\vert\vert A \vec{y} \vert\vert}$ to help us understand how much the solution vector will change if there is a small change in the input matrix or output vector. It can also be used to see that there are rounding errors made in solving $A \vec{x} = \vec{b}$ when working with floating point numbers.

However, why would we need to know about the condition number of the inverse of $A$? Why is $\kappa({A^{-1}})$ important?

答:
However, why would we need to know about the condition number of the inverse of $A$? Why is $\kappa(A^{-1})$ important?

The question rather silly, given that by definition $\kappa(A)=\|A\|_2 \cdot \|A^{-1}\|_2 = \kappa(A^{-1})$.


问:
Can you check if it's a group monomorphism from the group presentations?
Let $G_{1}$, $G_{2}$ be two groups given by their presentations. Then if $f$ is a map defined on the generators of $G_{1}$ we can prove that $f$ is a homomorphism by proving that the set of the relations of $G_{1}$ is mapped through $f$ to relations that hold in $G_{2}$. Is there a similar way to prove that $f$ is 1-1 by using the fact that we know the presentation of both groups?

答: Nope. Showing that $f$ is 1-1 would amount to showing that for distinct $a$ and $b$ in $G_1$ that $f(a)$ and $f(b)$ are distinct in $G_2$. In general this is undecidable given just the presentation of $G_2$. See The word problem for groups.


Title: Limit of this sequence $ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$?

Body:
How to find the limit of this sequence. I tried using the squeeze theorem but no result yet.

$ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$

Body:
It is easy to see that $$\frac{1}{2}(\frac{4^n}{6^n})<\frac{2^n+3^n+4^n}{5^n+6^n}<3(\frac{4^n}{6^n}).$$ Therefore,
$$\frac{4}{6}\sqrt[n]{\frac{1}{2}}<(\frac{2^n+3^n+4^n}{5^n+6^n})^\frac{1}{n}<\sqrt[n]{3}\frac{4}{6}.$$ Since $\sqrt[n]{c}\to 1$ as $n\to\infty$, we have that $$\lim_{n\to\infty}(\frac{2^n+3^n+4^n}{5^n+6^n})^\frac{1}{n}=\frac{4}{6}=\frac{2}{3}.$$


问:
IEEE-754 General Issues
I read the documents of the IEEE-754 and I could not understand some things

1) Is it possible to represent any number using this method? I know there are a number of final states, and there are infinite possibilities then no, but if there is a more detailed explanation of why?

2)Why can not represent the number zero? And how I represent it?

3) Why I dont need to represent 1 in representation of floating point number, I know its normalized, but normalized mean only 1 in the left side of the number? or everynumber, for example $8.3131e^{-1}$

Thanks!

答:
1) Using $64$ bits, you can represent at most $2^{64}$ numbers exactly. Many more can be represented with sufficient accuracy for applications. Very very large numbers, such as $10^{1000}$ get represented as infinity (which may also be enough for applications).

2) $0$ is represented by all bits being zero.

3) I don't really understand this question, but you can use this conversion tool to find the IEEE-754 representation of $1$, or  other numbers.

Title: IEEE-754 General Issues


问:
Does the Lie derivative of a harmonic form with respect to a killing vector field vanish
On the wikipedia page for Killing vector fields under 'properties' it is stated that if $X$ is a Killing vector field and $\omega$ is a harmonic differential form then $\mathcal{L}_{X}\omega=0$. https://en.wikipedia.org/wiki/Killing_vector_field#Properties

It seems like this should be true but I can't find it stated anywhere else except for the online previews of a couple of books which I don't have access to. I have tried to show this using the formula for the Lie derivative of a form $\mathcal{L}_{X}\omega=d(i_{X}\omega)+i_{X}(d\omega)$ along with the co-closure condition on $\omega$ and Killing's equation but I'm not really sure where to go.

So, my question is: is this true and if so, what direction should I go in to try and show it?

答:
For completing the proof consider the following lemma:

Lemma: If $\psi$ is exact (i.e. $\psi=d\eta$) and coclosed (i.e. $\delta\psi=0)$ then $\psi =0$.

This is easy to prove: Consider the induced inner product $\langle\!\langle., .\rangle\!\rangle$ that $d$ and $\delta$ are adjoint (i.e. $\langle\!\langle dA, B\rangle\!\rangle= \langle\!\langle A, \delta B\rangle\!\rangle$) w.r.t. that inner product. Then compute $\|\psi\|^2=\langle\!\langle\psi, \psi\rangle\!\rangle$.

(Hint for rest of proof: ${\cal L}_X\omega=d(i_X(\omega))+0=d(\text{something})\implies$ ${\cal L}_X\omega$ it is exact.).


问:
Global expression of 1-forms on a manifold
Question: Any 1-form can be expressed as a finite sum $\displaystyle \sum_{i} f_i dg_i$ for some smooth functions $f_i$ and $g_i$.

My attempt:

Let $M$ be an $n$-dimensional smooth real manifold. Let $\{(U_i,\phi_i)\}_{i \in I}$ be an atlas for $M$. Let $\{\psi_i\}_{i \in I}$ be a partition of unity subordinate to $\{U_i\}_{i \in I}$. Let $\omega \in \Omega^1(M)$ be a $1$-form on $M$. Then, on $U_i$, $\omega$ can be expressed as $$\omega_i=\displaystyle \sum_{j}X^j dx_{i}^j,$$ where $\{X^j\}_{j=1}^n$ are smooth functions and $\{x_{i}^j\}_{j=1}^n$ are local coordinates on $U_i$. Now, globally, we can write $$\omega=\displaystyle \sum_{i} \psi_i \omega_i.$$

Then we have, $$\omega=\displaystyle \sum_{i}\left(\displaystyle \sum_{j}\psi_i X^j\right)dx_{i}^j.$$ Rewrite it as $$\omega=\displaystyle \sum_{i}\left(\displaystyle \sum_{j}\psi_i X^j\right)\left(\displaystyle \sum_{i}\psi_i dx_{i}^j\right).$$

Let $f_i=\displaystyle \sum_{j}\psi_i X^j$. If we have $dg_i=\displaystyle \sum_{i}\psi_i dx_{i}^j$, then we are done. So, we want to find such $g_i$.

I am not able to proceed further. Any help is much appreciated.

答:
Let me assume that $M$ is compact. Take a local coordinate patch $U$ with coordinates $x^i$ taking values in unit ball. Let $V \subset U$ be the set of points with coordinates such that $\sum_i (x^i)^2 < \frac{1}{2}$. We can find functions $y^i$ which are equal to $x^i$ on $V$ and can be extended to whole $M$ (we can for example assume that they vanish for $\sum_i (x^i)^2 > \frac{3}{4}$). Thus $y^i$ are globally smooth and have the property that at every point of $V$ differential forms $dy^1,...,dy^n$ span the cotangent space. Now any point $p$ of $M$ has a neighbourhood $V_p$ which admits functions $y^i_p$ as above. By compactness, we can find $p_1,...,p_m$ such that $V_{p_1},...,V_{p_m}$ cover $M$. Then all $\{ y_{p_1}^1,...,y_{p_1}^n, ..., y_{p_m}^1,...,y_{p_m}^n \}$ are smooth functions defined on whole $M$. We will take them as our $g_i$. Clearly at every point of $M$ we have that $dg_i$ span the cotangent space. From here it should be easy to finish the proof using partitions of unity.

In the non-compact case I doubt you can always find a finite family of $f_i, g_i$ which works, but I'm pretty sure you can find one such that any point has a neighbourhood in which only finitely many $f_i$ and $g_i$ are nonzero. I will leave the exercise to you; I think you can proceed as I did but invoking paracompactness (instead of compactness) to construct the $g_i$. I have not worked out the details though.


问:
Let $(f_n)$ be a sequence of continuous functions such that $f_n\to f$ uniformly on $\mathbf{R}$. Show that $\lim_{n\to\infty}f_n(x_n)=f(x)$

Let $(f_n)$ be sequence of continuous functions such that $f_n\to f$ uniformly on $\mathbf{R}$. Show that $\lim_{n\to\infty}f_n(x_n)=f(x)$ for all sequence of points $x_n \in \mathbf{R}$ such that $x_n\to x$ as $n\to\infty$ for $x\in \mathbf{R}$. I would like to know if my proof holds and have a feedback, please.

As $f_n\to f$ uniformly, we have by definition that: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x \in \mathbf{R}$: $|f_n(x)-f(x)|<\epsilon/3$. As $x_n\in\mathbf{R}$, we have as well that $|f_n(x_n)-f(x_n)|<\epsilon/3$ $\forall \epsilon>0$.

Then, $f_n$ is a continuous sequence of functions. Let $a \in \mathbf{R}$. Thus, $\forall \epsilon>0 \ \exists\delta>0 \ \forall x \in \mathbf{R}$: $|x-a|<\delta\implies |f_n(x)-f_n(a)|<\epsilon/3$.

Moreover, $x_n$ is a converging seuquence. Therefore $\forall \epsilon>0 \ \exists N' \ \forall n\ge N': |x_n-x|<\epsilon$

We have then $\forall \epsilon>0 \ \forall n\ge n_0:=\max(N,N')$:

$|f_n(x_n)-f(x)|\le|f_n(x_n)-f_n(a)|+|f_n(a)-f_n(x)|+|f_n(x)-f(x)|=\epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$. So we showed that $\lim_{n\to\infty}f_n(x_n)=f(x)$

答:
Let $\varepsilon>0$ be given. Choose $N_{1}\in\mathbb{N}$ such that
$|f_{n}(y)-f(y)|<\varepsilon/2$ whenever $n\geq N_{1}$ and $y\in\mathbb{R}$.
Since $f$ is continuous at $x$ (recall that uniform limit of continuous
functions is also continuous), we may choose $\delta>0$ such that
$|f(y)-f(x)|<\varepsilon/2$ whenever $|y-x|<\delta$. Sine $x_n\rightarrow x$, we may choose $N_{2}\in\mathbb{N}$
such that $|x_{n}-x|<\delta$ whenever $n\geq N_{2}$. Let $N=\max(N_{1},N_{2})$.
Let $n\geq N$ be arbitrary. We have that
\begin{eqnarray*}
 &  & |f_{n}(x_{n})-f(x)|\\
 & \leq & |f_{n}(x_{n})-f(x_{n})|+|f(x_{n})-f(x)|\\
 & < & \varepsilon.
\end{eqnarray*}


问:
Explicite proof that complex power function are holomorphic
Consider the function  $f(z):=z^n$ and prove that $\dfrac{\partial}{\partial \bar{z}}f(z)=0$.

Here is my attempt:
\begin{equation*}
\begin{split}
    \dfrac{\partial}{\partial\bar{z}}f(z)&=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)(x+iy)^n=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)\left(\sum_{k=0}^n\binom{n}{k}x^{n-k}(iy)^k\right)\\
    &=\dfrac{1}{2}\sum_{k=0}^n\binom{n}{k}\left[\dfrac{\partial}{\partial x}\left(x^{n-k}(iy)^k\right)+i\dfrac{\partial}{\partial y}\left(x^{n-k}(iy)^k\right)\right]\\
    &=\dfrac{1}{2}\sum_{k=0}^n\binom{n}{k}\left[(n-k)x^{n-k-1}i^ky^k+ki^{k+1}x^{n-k}y^{k-1}\right]\\
    &=\dfrac{1}{2}\sum_{k=0}^n\binom{n}{k}\left(i^kx^{n-k-1}y^{k-1}\right)\left[(n-k)y+kix\right].
    \end{split}
\end{equation*}
Now, how can I show that it actually vanishes? Thank you.

答:
We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial\overline{z}}z^n}
&=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\,\frac{\partial}{\partial y}\right)(x+iy)^n\\
&=\frac{1}{2}\frac{\partial}{\partial x}(x+iy)^n
+\frac{i}{2}\frac{\partial}{\partial y}(x+iy)^n\\
&=\frac{1}{2}n(x+iy)^{n-1}+\frac{i}{2}in(x+iy)^{n-1}\\
&\,\,\color{blue}{=0}
\end{align*}


问:
Factoring $x^4 + 12x^3 + 46x^2 + 59x + 18$

How do I factorize the following?
$$x^4 + 12x^3 + 46x^2 + 59x + 18$$

I've tried looking for a root by trial and error to no avail.

The answer is
$$(x^2 + 5x + 2)(x^2 + 7x + 9)$$

答:
The first step is check for linear factors.  By the rational roots theorem, the only possible rational roots are $\pm 1, \pm 3, \pm 9, \pm 2, \pm 6, \pm 18$.  Once you have eliminated linear factors, the only possible factorization is as two quadratics.

With that out of the way, a good general technique to look for factorizations is to reduce mod n for various n.  If $f(x)=p(x)q(x)$, then reducing mod n yields a factorization for $f(x) \pmod n$.  If $n$ is bigger than any of the coefficients of $p, q,$ or $f$, then we can read off the actual factorization from the modular factorization.

To figure out what primes we might attempt to reduce modulo, let us compute $f(k)$ for small values of $k$, as a factor here will yield that $k$ is a root mod that factor.

Working mod 2, we have $f(x)=x^4+x=x(x+1)(x^2+x+1)$, the last factor being irreducible mod 2.  Working mod 3, $f(x)=x^4+x^2-x=x(x+1)(x^2-x-1)$, the last term also being irreducible mod 3. Computing $f(k)$ for small values of $k$ and factoring to find primes such that $f(k)$ will have at least one linear factor, we see that $(x+8)$ will be a factor mod 13 and 17, and $(x-3)$ will b a factor mod 13.  We get $f(x)\equiv (x-3)^3(x+8) \pmod{13}$ and $f(x)\equiv (x+3)(x-1)(x-6)^2 \pmod{17}$.

Thus, either $p(x)\equiv (x-3)^2 \pmod{13}$ and $q(x)\equiv (x-3)(x+8) \pmod{13}$ or vice versa.  Similarly, $p(x)\equiv (x-6)^2,(x-6)(x-1),(x-6)(x+3)$, or $(x-1)(x+3)\pmod{17}.$

Just using the mod 13 and 17 information is enough to find the factors mod 221, which is enough to specify the factors entirely.  The mod 2 and 3 information serves as a good double-check.  There are still a few combinations to check, but it makes the problem tractable.

The advantage of this method is that when a factorization does NOT exist, this can potentially show that too, by narrowing down what the shape of the factorization would have to be, and getting contradictory information.  For example, if you had $f(x)=x^5+15x^2+3$, then mod 3, this would either factor as $x(x^4)$ or $x^2(x^3)$, and since the constant term is not divisible by $9$, neither of these can happen.  This particular case can be extended to Eisenstein's criterion, but the general idea is useful more broadly.


问:
Inverting a diagonal (almost) linear relationship in a numerically stable way
I have a relationship of the form

$$y=Dx$$

where $D$ is a diagonal, square matrix with  non-negative entries. Some of the diagonal elements of $D$ may be zero. However, since $D$ is calculated by a computer, I don't know which ones are "really" zero, and which ones are just very small. That is, in practice none of them are exactly zero, just -1.34534634e-16 or something, but I believe that in the "true" matrix, were I to calculate it by hand, that entry would be a zero. Since I know from the theory that the matrix has non-negative entries, I can assume that tiny negative values are "really" zero, but it's less clear with a small positive value like 1.0234234e-05 or something.

I need to calculate $\tilde x$, where:

If $D_{i,i}>0$, then $\tilde x_i=x_i=D_{i,i}^{-1} y_i$.

If $D_{i,i}=0$, then $\tilde x_i=a_i$, where $a$ is some otherwise unrelated, constant vector.

How can I do this? Notice that if $D_{i,i}$ is very small, the two rules will give wildly different results depending on whether I assume it's actually zero or not.

For context: I have a linear ODE $u'=Du$, and I want to compute $u$ from $u'$. The vector $a$ mentioned above is in fact $u(0)$: when an entry of $D$ is zero I know that the corresponding entry in $u(t)$ doesn't change over time, so I can obtain it from $u(0)$. The reason I want to compute $u$ from $u'$ is because I have another ODE $v'=Au$, so by rewriting this as something like $v'=AD^{-1}u'$ I can integrate it to compute $v(t)$. $D$ is the diagonalization of a symmetric semi-postive-definite real matrix, as computed by some linear algebra package.

答:
Let's integrate $v' = AD^{-1} u'$. We get:
$$
v(t) = v(0) + AD^{-1} \left[u(t) - u(0)\right]. \tag{*}
$$
Note that for $D_{ii}$ that are close to zero the diagonal entries $D_{ii}^{-1}$ (which are large) are multiplied by $u_i(t) - u_i(0)$ which are small. But this large-small approach does not give an answer what is the value of the limit
$$
\lim_{D_{ii} \to 0} \frac{u_i(t) - u_i(0)}{D_{ii}} = {?}
$$

Let's derive the explicit expression for $v$ in a slightly other way:
$$
u' = Du \implies u = e^{tD} u(0)
$$
$$
v' = Au = Ae^{tD} u(0) \implies v = v(0) + A \left[\int_0^t e^{sD} ds\right] u(0)
$$
$$
\int_0^t e^{sD} ds = \int_0^t \sum_{k=0}^\infty \frac{s^k D^k}{k!} ds =
\sum_{k=0}^\infty \frac{t^{k+1} D^k}{(k+1)!}.
$$
when $D$ is not singular the series converge to $D^{-1} (e^{tD} - I)$ which perfectly matches (*).
When $D$ is singular we can pull the first term out obtaining
$$
\int_0^t e^{sD} ds = t I + t\sum_{k=1}^\infty \frac{t^k D^k}{(k+1)!}
$$
For $D_{ii}$ that are close to zero, the sum would have simply $t$ in the $i$-th diagonal element.

So the final expression for $v(t)$ takes the following forms:
$$
v(t) = v(0) +  tAu(0) + tA \left[\sum_{k=1}^\infty \frac{t^k D^k}{(k+1)!}\right] u(0) =
v(0) +  tA \left[\sum_{k=0}^\infty \frac{t^k D^k}{(k+1)!}\right] u(0).
$$
This gives you an answer for zero-diagonal entries, but requires summation for the rest.

The term $\sum_{k=0}^\infty \frac{t^k D^k}{(k+1)!}$ can be computed as matrix function $\phi(x) = \sum_{k=1}^\infty \frac{x^k}{(k+1)!} = \frac{e^x - 1}{x}$ applied to the matrix $tD$. Matlab has built in function funm for that based on this paper (beware of the paywall). Scipy also has scipy.linalg.funm.


问:
Understanding bicomplex numbers
I found by chance, the set of Bicomplex numbers. These numbers took particularly my attention because of their similarity to my previous personal research and question. I should say that I can't really understand the fact that $j^2=+1$ (and must of other abstract algebra) without using matricial interpretations. When I look a bit on the Bicomplex numbers, the think that surprised me a lot was the fact that $ij=ji=k$ and $k^2=-1$. Because using matricial representations, we get :
$$ij=\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}=\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}$$
$$ji=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
So, $ij$ should be different then $ji$. Then if we define $k=ij$ or $k=ji$, we get both way $$k^2= I$$
Looked at the Wikipedia article and did some internet search but couldn't find a matricial representation for the bicomplex $k$. So I wanna ask some clarifications about where from the relations $ij=ji$ and $k^2=-1$ come from.</p>

答:
The algebra you discribed in the question is tessarines.

So, there are two matrical representations of tessarines.

For a tessarine $z=w_1+w_2i+w_3j+w_4ij$ the matrix representation is as follows:

$$\left(
\begin{array}{cccc}
 {w_0} & -{w_1} & {w_2} & -{w_3} \\
 {w_1} & {w_0} & {w_3} & {w_2} \\
 {w_2} & -{w_3} & {w_0} & -{w_1} \\
 {w_3} & {w_2} & {w_1} & {w_0} \\
\end{array}
\right)$$

One has to use the 4x4 matrix. After performing the operations, the coefficients of the resulting tessarine can be extracted from the first column of the resulting matrix.

Alternatively, one can use a 2x2 matrix of the form $z=a+bj=\left(
\begin{array}{cc}
 a & b \\
 b & a \\
\end{array}
\right)$

where both $a$ and $b$ are complex numbers. Since complex numbers are embedded in most computer algebra systems, I prefer the second method.

Title: Understanding bicomplex numbers


问:
Sketching a polar curve
Continued off the question I asked earlier, I also have to sketch the curve.

$r^2=−4\sin(2\theta)$

So I have to set up a table of values I'm assuming. How do I know what values to choose for $\theta$. Then I'm just lost all together trying to plot the theta and $r$ values to get the curve.

Oh and one quick other question. If it also asks me to find all the points of intersections and to set up the integral to evaluate the area inside after sketching the curve, what does that mean? Do I have to pick a top and bottom or something? Confused about that.

答:
To get started, you might try values of $\theta$ where it's easy to calculate stuff, like $\theta=0,\pi/12,\pi/4,\pi/2,\dots$. Maybe even better to note first that since $r^2\ge0$ it only makes sense to take $\theta$ with $\sin2\theta\le0$. </p>

<p>You won't get very far with the integral until you have a good sketch of the curve. </p>

Title: Sketching a polar curve

Body:
To get started, you might try values of $\theta$ where it's easy to calculate stuff, like $\theta=0,\pi/12,\pi/4,\pi/2,\dots$. Maybe even better to note first that since $r^2\ge0$ it only makes sense to take $\theta$ with $\sin2\theta\le0$.

You won't get very far with the integral until you have a good sketch of the curve.


问:
Question on cardinal arithmetic
How do I go about showing:

$\aleph_{\omega}^{\aleph_{1}}$=$\aleph_{\omega}^{\aleph_0}$.$2^{\aleph_1}$

My first thought was to apply Hausdorff's formula directly, but that doesn't quite work. The book provides a hint using products and sums that I don't quite understand. Can you please help?

The question is from Introduction to Set Theory by Hrbacek and Jech.

答:
That we have a product on the right-hand side is an indication that there are two cases to be distinguished. Note that it will clearly suffice to prove $\aleph_\omega^{\aleph_1} \le \aleph_\omega^{\aleph_0}2^{\aleph_1}$.

Suppose first that for some $\kappa < \aleph_\omega$, we have $\kappa^{\aleph_1} \ge \aleph_\omega$. Then $\kappa^{\aleph_1} \le \aleph_\omega^{\aleph_1} \le (\kappa^{\aleph_1})^{\aleph_1} = \kappa^{\aleph_1}$. I leave it to you to use Hausdorff's formula to establish $\kappa^{\aleph_1} = 2^{\aleph_1}$.

Now suppose that $\kappa^{\aleph_1} < \aleph_\omega$ for all $\kappa <\aleph_\omega$. Then since $\aleph_1 \ge \operatorname{cf}\, \aleph_\omega = \aleph_0$, it is known that $\aleph_\omega^{\aleph_1} = \aleph_\omega^{\operatorname{cf}\aleph_\omega} = \aleph_\omega^{\aleph_0}$.

This latter fact appears as Theorem 5.20(iii)(b) in Jech's Set Theory. I can add a proof of this if you want, but I guess it's in Hrbacek-Jech as well.

Title: Question on cardinal arithmetic


问:
"Extremely low probability events happen all the time" - Does this allow us to draw any conclusions?
"Probability of getting any specific hand in poker is the same as getting Royal Flush" - Statements of these kinds seem fairly intuitive to interepret. But recently I thought of the following statement - "There are 10 towns each with its own distinct specialty, like - Food, Landscape, Culture, etc. Now Bob visits exactly one of these towns for his winter vacation."

Now not knowing the fact the bob has no preference whatsoever, I conclude that since he visited town X he must like the town's culture, food, etc. Since the probability of him choosing that specific culture he very less. But this does not make sense as I can say that about any town bob chooses.

Could we or should we draw conclusions like these in a scenario like this? What kind of conclusions we can draw from such an event? What conditions should an event like this satisfy so that meaningful conclusions can be derived?

Thanks in Advance.

答:
All statistics tells you about is correlations -- these two events occur together with greater probability than expected by chance; these two events occur together with lesser probability than expected by chance.

Your specific inference types

Now not knowing the fact the bob has no preference whatsoever, I conclude that since he visited town X he must like the town's culture, food, etc. Since the probability of him choosing that specific culture he very less. But this does not make sense as I can say that about any town bob chooses.

are about causation.  Correlations do not imply causation.  Inferring causation requires more than statistics and probability, one has to go on and do (root) cause analysis and other forms of science.

Statistics helps you find which correlations are explainable by chance, so you do not invest further time/effort in attempting to establish a causal relationship.  Statistics also help you find which correlations are not adequately explained by chance, so you can invest your time/effort investigating apparent causes/relationships.


问:
Question on a proof about invariant atomic measures.
The statement is let $T: X \to X$ be a continuous transformation on a compact metrizable space $X$.  If $\mu$ is a $T-invariant$ borel probability measure that is also purely atomic then it is a (possible infinite) convex combination of measures of the form $\frac{1}{N} \sum_{i=0}^{N-1} \delta_{T^i(x)}$.

Proof:  $\mu$ is invariant iff $\int f \circ T d \mu = \int f d \mu$ for every continuous f.  Now suppose $\mu=\sum_{i=1}^{\infty} p_i \delta_{x_i}$ with $\sum p_i =1$.  We can assume that the $p_i$ are decreasing.  From the above integral condition we get $\sum p_i f(T^{i+1}(x))=\sum p_i f(T^{i}(x))$ so that $T(x_1)$ is an atom with measure $p_1$.  From this we get that $T^{N_1}(x_1)=x_1$ and each atom $T^i(x)$ has the same measure.  The desired result follows from simple induction.

I don't get why you can conclude that $T(x_1)$ is an atom with measure $p_1$ or any of the statements after that.  Any help would be appreciated.

答:
I don't get the insistence on arguing through continuous functions. Really what is being said is that

$$
\mu(\{T(x_1)\})=T^*\mu(\{T(x_1)\})=\mu(T^{-1}(T(x_1))=\sum_{i=1}^{\infty} 1_{T(x_i)=T(x_1)} p_i\geq p_1
$$
so $T(x_1)$ is an atom. Furthermore, since the $p_i$ are decreasing, we must have $\mu(T(x_1))=p_1$. By induction, $\mu(T^n(x_1))=p_1$ for all $n$. Furthermore, we get that the restriction of $T$ to the orbit of $x_1$, that is to say $\mathcal{O}(x_1):=\{T^n(x_1)|n\in \mathbb{N}\}$, must be injective (why?).

Hence if $x_1, T(x_1),...,T^n(x_1)$ are distinct, then
$$
1\geq \mu(X)\geq \mu(\{x_1,...,T^n(x_1)\})=(n+1)p_1
$$
Hence, there must be some $n<m\leq \frac{1-p_1}{p_1}$ such that $T^n(x_1)=T^m(x_1)=T^n(T^{n-m}(x_1))$. However, since $T$ is injective on the orbit of $x_1$, this implies that $x_1=T^{m-n}(x_1)$.

Now $\mu(\cdot)=\mu(\cdot |\mathcal{O}(x_1))\mu(\mathcal{O}(x_1))+\mu(\cdot|X\setminus \mathcal{O}(x_1))(1-\mu(\mathcal{O}(x_1)))$ and we just proved that
$$
\mu(\cdot |\mathcal{O}(x_1))=\frac{1}{m-n}\sum_{i=0}^{m-n-1} \delta_{T^{-i}(x_1)}
$$
Can you see how to continue from here?


问:
Non-isomorphic locally ringed spaces which represent isomorphic functors $\mathsf{CommRing} \to \mathsf{Set}$.
It's well known that the restricted Yoneda functor $よ : \mathsf{Schemes} \to \operatorname{Fun}(\mathsf{CommRing},\mathsf{Set})$ is an embedding, so that (in particular) if $X$ and $Y$ are schemes such that $よ(X) \cong よ(Y)$, then $X \cong Y$. This is the key idea of the "functor of points approach" to algebraic geometry (which then proceeds to classify precisely which co-presheaves on commutative rings are in the essential image of $よ$).

It's my understanding that the restricted Yoneda $よ : \mathsf{LRS} \to \operatorname{Fun}(\mathsf{CommRing},\mathsf{Set})$ is no longer an embedding, where $\mathsf{LRS}$ is the category of locally ringed spaces. Moreover, I expect there to exist non-isomorphic locally ringed spaces $X$ and $Y$ such that $よ(X) \cong よ(Y)$.

Does anyone know a simple example of this failure of essential injectivity?

Edit: Just realized I messed up the variance at first, hopefully I got it right now.

答:
If $A$ is a local ring, let $X_A$ denote the locally ringed space with one point with $\mathcal{O}_{X_A}(X_A)=A$.  If $B$ is a ring, then a morphism of ringed spaces $\operatorname{Spec} B\to X$ is just given by a homomorphism $f:A\to B$.  In order for this to be a morphism of locally ringed spaces, $f(m)$ must be contained in every prime ideal of $B$, where $m$ is the maximal ideal of $A$.  That is, $f(m)$ must be contained in the nilradical of $B$.  If $m$ is finitely generated, this means $f$ must factor through the quotient $A/m^n$ for sufficiently large $n$.

In particular, this means that morphisms of locally ringed spaces $\operatorname{Spec} B\to X$ actually depend only on the completion $\hat{A}$ of $A$ with respect to $m$, since $\hat{A}/(m\hat{A})^n\cong A/m^n$ for all $n$.  So, the morphism $X_{\hat{A}}\to X_A$ becomes an isomorphism after applying the functor $よ : \mathsf{LRS} \to \operatorname{Fun}(\mathsf{CommRing},\mathsf{Set})$.

Or, let $A$ be a valuation ring whose value group $\Gamma$ is strongly nonarchimedean in the sense that for each positive $x\in\Gamma$, there is a positive $y\in\Gamma$ such that $ny<x$ for all $n\in\mathbb{N}$.  (For instance, $\Gamma$ could be the ordered ring $\mathbb{Z}[x]$ where $x$ is infinitesimal.)  Note that then if $I\subset A$ is any ideal whose radical is the maximal ideal $m$, then $I$ must be equal to $m$, since $I$ must contain elements of arbitrarily small positive valuation.  This means that if $f:A\to B$ is any homomorphism such that $f(m)$ is nilpotent, $f(m)$ must actually be $0$.  By the first paragraph above, this means that morphisms $\operatorname{Spec} B\to X$ are just in bijection with homomorphisms $A/m\to B$.  In other words, the natural morphism $\operatorname{Spec} A/m\to X_A$ becomes an isomorphism after applying $よ$.

Here is a rather different sort of example, inspired by a now-deleted answer by Aragogh.  Let $X$ be a $T_1$ locally ringed space with the following property: for each $x\in X$, there is a neighborhood $U$ of $x$ and a section $a\in \mathcal{O}_X(U)$ that vanishes at $x$ but at no other point of $U$ (where by "vanishing at a point" I mean its image in the stalk at that point is in the maximal ideal).  For instance, $X$ could be a nice (say, metrizable) topological space equipped with the sheaf of continuous real-valued functions.  I claim that if $Y$ is an affine scheme and $f:Y\to X$ is a morphism of locally ringed spaces, then the underling map of topological spaces is locally constant.

To prove this, suppose $y\in Y$ and $f$ is not locally constant at $y$. We know that $f$ induces a local homomorphism $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$.  Moreover, each prime ideal $p$ of $\mathcal{O}_{Y,y}$ corresponds to a generization of $y$ which $f$ must map to the same point (since $X$ is $T_1$), and the corresponding homomorphism $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,p}$ is just obtained by composing with the localization map $\mathcal{O}_{Y,y}\to \mathcal{O}_{Y,p}$.  Since all of these compositions with localizations are local homomorphisms, our homomorphism $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$ must actually map the maximal ideal of $\mathcal{O}_{X,f(y)}$ into the nilradical of $\mathcal{O}_{Y,y}$ (since its image must be contained in every prime ideal).

Now take a neighborhood $U$ of $f(y)$ and a section $a\in \mathcal{O}_X(U)$ that vanishes at $f(y)$ and nowhere else.  Since the image of $a$ in $\mathcal{O}_{Y,y}$ is nilpotent, we may replace $a$ with a power of itself and assume there is some neighborhood $V\subseteq f^{-1}(U)$ of $y$ such that the image of $a$ in $\mathcal{O}_Y(V)$ is $0$.  Since $f$ is not locally constant at $y$, there exists some $z\in V$ such that $f(z)\neq f(y)$.  But then $a$ is invertible in a neighborhood of $f(z)$, so the image of $a$ in $\mathcal{O}_{Y,z}$ is invertible.  This is a contradiction, since the map $\mathcal{O}_X(U)\to\mathcal{O}_{Y,z}$ factors through the map $\mathcal{O}_X(U)\to\mathcal{O}_Y(V)$ that maps $a$ to $0$.

So, every morphism from an affine scheme to $X$ is locally constant.  Now let $D(X)$ be $X$ with the discrete topology, equipped with the sheaf of local rings that has the same stalks as $X$.  The identity function $i:D(X)\to X$ has an obvious structure of a morphism of locally ringed spaces, and a locally constant morphism to $X$ is just one that factors (uniquely) through $i$.  So, $よ(i)$ is an isomorphism.

This gives rise to a bunch of other related examples.  For instance, if $M$ and $N$ are two nonempty manifolds of the same dimension $d>0$, equipped with the sheaves of continuous (or smooth) real-valued functions, note that $D(M)$ and $D(N)$ are isomorphic: they are each just a disjoint union of $2^{\aleph_0}$ singletons whose local ring is the ring of germs of real-valued functions at a point of $\mathbb{R}^d$.  It follows that $よ(M)$ and $よ(N)$ are isomorphic (though this isomorphism does not come from a map between $M$ and $N$).

In fact, we can go further.  If $X$ is a topological space, $x\in X$, and $(A,m)$ is the local ring of germs of real-valued continuous functions at $x$, then $A$ has the following property.  For each nonzero $a\in m$, there exists a nonzero $b\in m$ such that $a$ is divisible by $b^n$ for all $n\in\mathbb{N}$.  Indeed, you can take $b=g(a)$ where $g:\mathbb{R}\to\mathbb{R}$ is a continuous function such that $g(0)=0$ and as $t\to 0$, $g(t)$ approaches $0$ more slowly than $|t|^{1/n}$ for each $n$.  This property means that the only ideal of $A$ whose radical is $m$ is $m$ itself.  As in the third paragraph above, this means the canonical morphism $\operatorname{Spec} \mathbb{R}\to X_A$ becomes an isomorphism after applying $よ$.  It follows that if $X$ is a nice space as above equipped with its sheaf of continuous real-valued functions, then actually the canonical morphism $\coprod_{x\in X}\operatorname{Spec}\mathbb{R}\to X$ becomes an isomorphism after applying $よ$, since $よ$ can't see the difference between the points of $D(X)$ and copies of $\operatorname{Spec}\mathbb{R}$.  In particular, this means that if $X$ and $Y$ are two metrizable spaces of the spaces of the same cardinality equipped with their sheaves of continuous real-valued functions, then $よ(X)\cong よ(Y)$.

(In fact, you can generalize this a bit further: all of this applies to spaces on which real-valued continuous functions separate points, not just metrizable spaces.  To sketch the proof, the argument above shows that if $Y$ is an affine scheme and $f:Y\to X$ is a morphism of locally ringed spaces, then every continuous real-valued function on $X$ actually pulls back to a locally constant real-valued function on $Y$.  If $f$ were not locally constant, then its image would be infinite, and then since functions separate points on $X$, you can conclude that the functions on $X$ pull back to an infinite-dimensional algebra of locally constant functions on $Y$, which must come from an infinite subalgebra of the Boolean algebra of clopen subsets of $Y$.  But then you can construct a uniformly convergent infinite sum of such functions which is not locally constant, and this sum will still come from a continuous function on $X$, giving a contradiction.)


问:
Extrapolating an abstract algebra proof, arriving upon an incorrect conclusion.
Could you kindly point out what is wrong with my reasoning?

EDIT: What I have unintendedly proven through my reasoning is that every field can only have one automorphism- the identity mapping. Hope this helps in navigating the mess below

Let $k_{1},k_{2},\dots k_{n}$ be an arbitrary number of distinct, non-zero elements $\in\mathbb{K}$. $\mathbb{K}$ is a field of $0$ characteristic.

Also, $\sigma_{1},\sigma_{2},\dots \sigma_{m}$ are distinct automorphisms of $\mathbb{K}$.

I will show that $m\leq n$.

Let us assume that $m>n$. We can form the following system of linear equations:

$$\sigma_{1}(k_{1})x_{1}+\sigma_{2}(k_{1})x_{2}+\dots \sigma_{m}(k_{1})x_{m}=0$$ $$\sigma_{1}(k_{2})x_{1}+\sigma_{2}(k_{2})x_{2}+\dots \sigma_{m}(k_{2})x_{m}=0$$ $$\dots$$$$\dots$$$$\sigma_{1}(k_{n})x_{1}+\sigma_{2}(k_{n})x_{2}+\dots \sigma_{m}(k_{n})x_{m}=0$$

As $m>n$, we are assured a non-zero solution for $(x_{1},x_{2},\dots x_{m})$. Let it be $(a_{1}, a_{2},\dots a_{m})$, not all zero.

Then, taking the first linear equation into consideration, $$\sigma_{1}(k_{1})a_{1}+\sigma_{2}(k_{1})a_{2}+\dots \sigma_{m}(k_{1})a_{m}=0, a_{i}\in\mathbb{K}$$

This is not possible (I can give a proof, but it would unnecessarily lengthen the discussion).

Hence, on selecting an arbitrary ($n$) number of elements from $\mathbb{K}$, this result shows that the number of automorphisms $\mathbb{K}$ can have $\leq n$. This directly implies that the number of automorphisms $\mathbb{K}$ has is $1$- the identity automorphism.

It is quite obvious that this conclusion is wrong. What is erroneous in my thought process though?  Thanks in advance!

答:
It looks to me from the comments that you have based your proof a situation in which $\mathbb K$ is an extension of the ground field $\mathbb F$, and the author is examining the automorphisms of $\mathbb K$ which fix $\mathbb F$ in order to get make a comparison with the degree of the extension.

If you have, then your proof essentially assumes $\mathbb F = \mathbb K$, the degree of the extension is $1$, and the automorphisms you are considering are assumed to fix $\mathbb K$. The only automorphism of $\mathbb K$ which fixes $\mathbb K$ is the identity, and what you have done is to prove a special case of the general theorem.

You have not given enough information to show where the assumption about the fixed field comes in - it would be in the part of the proof that you haven't given.


问:
Find the mean and standard deviation of the distribution
Hypothetically seen if I had to find the mean and standard deviation of the distribution for this question. How would I get to that?

“A basketball player is traditionally a 73% foul shooter. In a season, he takes 428 foul shots. Find the mean and standard deviation of the distribution”

I've been trying to figure it out but I seem to be getting the wrong answer every time

答:
The mean and variance of a binomial distribution, ${\cal B}(n,p)$, for $n$ samples with success rate $p$ is a classic problem in statistics:

${\rm mean} = n p = 312.44$

${\rm variance} = n (1-p) p = 84.3588$, and the standard deviation is merely the square root of the variance, $9.1847$.


问:
Trying to show a certain map is a contraction
I'm having trouble understanding each step of the inequality shown below:
$$\Large\begin{eqnarray}
\left\|F(x)-F(x')\right\|
&=&  \left\|\:\int_0^1 DF_{(x-x')t+x'}      \cdot      (x-x') \:dt\: \right\|
\\[1.5ex]
&\leq& \int_0^1 \left\|DF_{(x-x')t+x'}\right\|\: \left\|x-x'\right\| \: dt
\\[1.5ex]
&\leq& \left(\sup_{z \in B_R} \left\|DF_z\right\|\right) \left\|x-x'\right\|
\end{eqnarray}$$
[Transcribed from this image.]

To give some context, this is an inequality that will be used in the proof of the inverse function theorem.  (The full document can be viewed here.)

Here, $F$ is a continuously differentiable function on $\mathbb{R}^n$ to $\mathbb{R}^n$, and $B_R$ denotes the closed ball of radius $R >0$ centered at the origin.  The norms involved are the Euclidean norm for the vectors and the operator norm for the linear operator $DF_z$.

My attempt: I have an idea that the mean-value theorem in several variables is involved, but I can't apply it to $F$ directly since it is a function to $\mathbb{R}^n$.  I believe I can finish it off if I can prove the first equality, $\| F(x) - F(x') \| = \| \int_{0}^{1}DF_{(x-x')t+x'}(x-x') dt \|$ and this one: $\| \int_{0}^{1}DF_{(x-x')t+x'}(x-x') dt \| \leq \int_{0}^{1} \| DF_{(x-x')t+x'} (x-x') \| dt$, but the trouble I am having is that the integral is computed componentwise, so I'm unsure how to go about this.

答:
First line to second line: triangle inequality for integrals.  See https://math.stackexchange.com/a/103783/123905 for a quick discussion.

Second line to third line: This is the ML inequality (maximum of integrand times length of integration interval) together with the definition of the norm as the maximum on the unit ball.


问:
Miller–Tucker–Zemlin formulation of asymmetric TSP
The Wikipedia article https://en.wikipedia.org/wiki/Travelling_salesman_problem#Integer_linear_programming_formulations states two conditions including inequalities about some Uis. I understand how any solution satisfies these two conditions. However, I don't realize Why any assignment that satisfies these two conditions consists of edges that form a Hamiltonian cycle, not a disjoint union of 2-regular graphs. (That's actually the reason we set these two conditions.) Can anybody help?</p>

答:
The article gives a proof by contradiction.  If you sum the constraints along a $k$-cycle $C$ that does not contain the depot (city 1), you obtain
$$\sum_{(i,j)\in C}(u_i-u_j+n x_{i,j}) \le \sum_{(i,j)\in C} (n-1) \tag1$$
The $u_i$ variables cancel in pairs, and $\sum_{(i,j)\in C} x_{i,j} = k$,
yielding
$$nk \le k (n-1),$$
which is a contradiction.


答:
Actually, this isn’t the mean value theorem, it really is simpler than that.

Consider the function $t \in [0,1] \longmapsto F((1-t)x+tx’)\in \mathbb{R}^n$. It is differentiable enough, and its derivative at $t$ is $DF_{(1-t)x+tx’}\cdot (x’-x)$. By the fundamental theorem of calculus (it works when a finite-dimensional vector space instead of $\mathbb{R}$ or $\mathbb{C}$ is the range and the derivative is continuous – basically just apply it to each coordinate) $\int_0^1{DF_{(1-t)x+tx’}\cdot (x’-x)dt}=F(x’)-F(x)$. To get the first equality, just apply the norm to each side.

For the first inequality, you need to prove that if $f: [0,1] \rightarrow \mathbb{R}^n$ is continuous, $\|\int_0^1{f(t)dt}\|\leq \int_0^1{\|f(t)\|dt}$. The simplest way to see this (in finite dimension at least) is to check that for any continuous linear form $L$ of norm at most one, $L\left(\int_0^1{f(t)dt}\right) \leq \int_0^1{\|f(t)\|dt}$. This is easier because you can convince yourself easily with Riemann sums that LHS can also be written as $\int_0^1{L(f(t))dt}$.

The last inequality stems from the fact that the line segment $[x,x’]$ is a subset of $B_R$ (as long as $R$ is chosen so that $x,x’ \in B_R$).

Title: Trying to show a certain map is a contraction

Body:
I'm having trouble understanding each step of the inequality shown below:
$$\Large\begin{eqnarray}
\left\|F(x)-F(x')\right\|
&=&  \left\|\:\int_0^1 DF_{(x-x')t+x'}      \cdot      (x-x') \:dt\: \right\|
\\[1.5ex]
&\leq& \int_0^1 \left\|DF_{(x-x')t+x'}\right\|\: \left\|x-x'\right\| \: dt
\\[1.5ex]
&\leq& \left(\sup_{z \in B_R} \left\|DF_z\right\|\right) \left\|x-x'\right\|
\end{eqnarray}$$
[Transcribed from this image.]

To give some context, this is an inequality that will be used in the proof of the inverse function theorem.  (The full document can be viewed here.)

Here, $F$ is a continuously differentiable function on $\mathbb{R}^n$ to $\mathbb{R}^n$, and $B_R$ denotes the closed ball of radius $R >0$ centered at the origin.  The norms involved are the Euclidean norm for the vectors and the operator norm for the linear operator $DF_z$.

My attempt: I have an idea that the mean-value theorem in several variables is involved, but I can't apply it to $F$ directly since it is a function to $\mathbb{R}^n$.  I believe I can finish it off if I can prove the first equality, $\| F(x) - F(x') \| = \| \int_{0}^{1}DF_{(x-x')t+x'}(x-x') dt \|$ and this one: $\| \int_{0}^{1}DF_{(x-x')t+x'}(x-x') dt \| \leq \int_{0}^{1} \| DF_{(x-x')t+x'} (x-x') \| dt$, but the trouble I am having is that the integral is computed componentwise, so I'm unsure how to go about this.


问:
Relations between hypergeometric functions
I am trying to find a relation between hypergeometrics $${}_2F_1(a,b,c;z)\,\,\text{and}\,\,{}_2F_1(a+1,b+1,c+1;z)$$

I can see that $$\frac{\partial}{\partial z}{}_2F_1(a,b,c;z) = \frac{ab}{c}{}_2F_1(a+1,b+1,c+1;z)$$ but I was wondering if I can write a relation between them that admits the form $${}_2F_1(a+1,b+1,c+1;z) = f(a,b,c,z) \cdot {}_2F_1(a,b,c;z) + \,\,\text{some other hypergeometric} $$

I know there is a hypergeometric identities page on Wolfram with a plethora of identities, but I couldn't see such an identity for the case I am considering. I am not sure, however, if this list is exhaustive.

Thanks!

答:
Little late to the game here but your request is possible by using the contigious relations of the hypergeometric function. Using the notation in the linked paper
$$
F\left({a_1+1,a_2+1\atop a_3+1};x\right)=\mathcal A_1\mathcal A_2\mathcal A_3 F\left({a_1,a_2\atop a_3};x\right).
$$
There are a number of operators you can derive from the contigious relations that will be equivalent to $\mathcal A_1\mathcal A_2\mathcal A_3$. For example, equation $(17)$ gives $\mathcal A_1\mathcal A_2\mathcal A_3=-\frac{a_3}{a_2 z}(\mathcal I-\mathcal A_1)$ so that
$$
\begin{aligned}
F\left({a_1+1,a_2+1\atop a_3+1};x\right)%
&=-\frac{a_3}{a_2 z}(\mathcal I-\mathcal A_1) F\left({a_1,a_2\atop a_3};x\right)\\
&=-\frac{a_3}{a_2 z}F\left({a_1,a_2\atop a_3};x\right)+\frac{a_3}{a_2 z}F\left({a_1+1,a_2\atop a_3};x\right),
\end{aligned}
$$
which is in the form presented in the question.

If any of the parameters $a_1$, $a_2$, or $a_3$ are integers it is likely that repeated application of these contigious relations can reduce the second hypergeometric term above to some "simpler" product of elementary functions.


问:
Need help verifying understanding of convexity (concavity) + quasi convexity(quasi concavity)
I wanted to verify the definition of quasi concavity/convexity. In my textbooks it says a function is quasi concave if the upper contour set is convex. It also says a function is quasi concave if f(λx+(1−λ)y) ≥ min {f(x), f(y)}.
If I use f(x)=|x| to show this, we can see visually that the upper contour set is obviously convex, suggesting quasi concavity but if you take values x and y on the domain and use the definition f(λx+(1−λ)y) ≥ min {f(x), f(y)}, we can see that a part of f(λx+(1−λ)y) makes this statement untrue.</p>

答: Hint: So actually $f(x)=|x|$ is convex (and quasiconvex), not concave (or quasiconcave). If you try verifying this definition for $f(x)=-|x|$, you may have more luck :-)


问:
Understanding pairwise congruent modulo $n$
I have the following small theorem in my Number Theory textbook and I need some help understanding it,

Let $a$ and $n$ be natural numbers with $gcd(a,n)=1$ and $k=ord_n(a)$. Then the numbers $a^1,a^2,...,a^k$ are pairwise incongruent modulo $n$

Where I'm confused is the line phrase "pairwise incongruent modulo $n$". What information is this phrase telling me? What literal arithmetic is being performed in something that is "pairwise incongruent"?

Furthering this, what approach would I used to prove this theorem?

答:
Two integers $a,b$ are incongruent mod $n$ means that $a \not\equiv_n b$. So, $a,a^2,...,a^k$ are pairwise incongruent modulo $n$ means that $a^i \not\equiv_n a^j$, for every $1\leq i <j \leq k$.

To prove the result, note that $a$ and $n$ are coprime means that there exist integers $x,y$ such that $ax+ny=1$, i.e., $ax\equiv_n1$ and one can "divide by a" when working modulo $n$.

That being said, suppose that $a^i \equiv_n a^j$, for some $1\leq i <j \leq k$. Then, $1 \equiv_n a^i x^i \equiv_n a^{j-i} a^i x^i \equiv_n a^{j-i}$. But $k$ is the order of $a$ mod $n$, and that means that $k$ is the least positive integer such that $a^k \equiv_n 1$. But $0<j-i<k$, which is a contradiction with the minimality condition on $k$.

Therefore, the hypothesis that $a^i \equiv_n a^j$ for some $1\leq i <j \leq k$ is false, i.e., $a,a^2,...,a^k$ are pairwise incongruent modulo $n$.


问:
Implicit Euler integration scheme for DAEs
Consider following DAE

\begin{equation}
\begin{bmatrix}
a_{11} & a_{12} & a_{13} & 0\\
a_{21} & a_{22} & a_{23} & 0\\
0 & 0 & 0 & 0
\end{bmatrix}\frac{\mathrm{d}}{\mathrm{d}t}
\begin{bmatrix}
y_{1}\\
y_2 \\
y_3 \\
y_4
\end{bmatrix}
+
\begin{bmatrix}
b_{11} & b_{12} & b_{13} & b_{14}\\
b_{21} & b_{22} & b_{23} & b_{24}\\
b_{31} & b_{32} & b_{33} & b_{34}\\
b_{41} & b_{42} & b_{43} & b_{44}\\
\end{bmatrix}
\begin{bmatrix}
y_{1}\\
y_2 \\
y_3 \\
y_4
\end{bmatrix}
=
\begin{bmatrix}
\gamma_{1}(y_{1},y_2,y_3,y_4)\\
\gamma_2(y_{1},y_2,y_3,y_4) \\
\gamma_3(y_{1},y_2,y_3,y_4) \\
\gamma_4(y_{1},y_2,y_3,y_4)
\end{bmatrix}
\end{equation}
RHS vector $\gamma$ is composed from possibly nonlinear functions.
How can I setup implicit Euler method scheme for this problem? The part which confuses me is the algebraic one. Can you rewrite it in terms of
\begin{equation}
y_i^{k + 1} = y^k + \Delta t f(t^{k+1},y^{k+1})
\end{equation}

How do I incorporate those algebraic constraints in scheme such that I can call some NR solver for each time step $k$?

答:
I believe the first matrix misses a row.

Anyway, applying implicit Euler for ODE and for DAE is straightforward: just take everything except time derivative from the $k+1$-st time step and replace $\frac{dy_i}{dt}$ with
$\frac{y^{k+1}_i - y^k_i}{\Delta t}$.

In matrix form you DAE is
$$
A \frac{d\mathbf y}{dt} + B \mathbf y = \mathbf \Gamma(\mathbf y)
$$
with matrix $A$ being singular. The corresponding implicit Euler would be
$$
A \frac{\mathbf y^{k+1} - \mathbf y^k}{\Delta t} + B \mathbf y^{k+1} = \mathbf \Gamma(\mathbf y^{k+1}).
$$
Separating knowns from unknowns gives a nonlinear system of equations
$$
A \mathbf y^{k+1} + \Delta t B \mathbf y^{k+1} - \Delta t \mathbf \Gamma(\mathbf y^{k+1}) = A \mathbf y^k.
$$
You may now apply Newton-Rhapson to this system and hope that the jacobian matrix
$$
A + \Delta t B - \Delta t \frac{\partial \mathbf \Gamma}{\partial \mathbf y}
$$
is not singular in the neighborhood of your solution. This depends mainly on $\mathbf \Gamma$ so little may be said in advance.


问:
Formula to get a control point closest for a given point what belongs to this quadratic curve
We have a quadratic bezier curve, with control point A (red start), control point B (red end), and yellow point X what belongs to the curve and what you actually "drag" - so it should be the closest point to the missing control point C (blue) what forms the curve. What is a formula getC(A,B,X) to calculate it?

drawing http://www.freeimagehosting.net/newuploads/d89e6.gif

A popular solution is when we think that X corresponds to t=1/2 for calculating C is not giving a nice result.

The concept is explained on this thread. But still I can't figure out the final formula for calculating C from it - having difficulty to understand the "orthogonality relation"

Thanks for any help!

答:
Actually the bisector thing is a beautiful found I did at 2012 also at my paper that cover that theme “Quadratic bezier through three points” and something related of what I call “equivalent quadratic bezier”. Hope you found it interesting

https://microbians.com/math/Gabriel_Suchowolski_Quadratic_bezier_through_three_points_and_the_-equivalent_quadratic_bezier_(theorem)-.pdf

also had some more info on quadratic offseting and examples here https://microbians.com/mathcode


问:
Principle of Mathematical Induction Problem with Mathematical Series
Can anyone help me with this problem? Been strugling with it for a few hours now. My finite induction principle is quite rusty.

Consider $n \in \mathbb{N}^{*}$.

For $P(n):9+13+...+(4n+5)=\frac{4n^2+14n+1}{2}$, show that $P(k)\Rightarrow P(k+1)$ is true $\forall k \in \mathbb{N}^{*}$.

Does the procedure in 1. demonstrate that the proposition is true for every $n \in \mathbb{N}^{*}$? Justify your answer.

Determine the subset of $\mathbb{N}^{*}$ that satisfies the proposition $P(n)$. Justify your answer.

Thank you guys.

答:
Welcome to MSE!

#1,2 In essence you can think of this as dominos falling, $P(k) \Rightarrow P(k+1)$ for all $k \in \mathbb{N}^*$ means the dominos are all lined up so that if you tip one, the rest thereafter will fall... you just need one to go. But none will ever go because $P(k)$ is false for all $k \in \mathbb{N}^*$.

$n \in \mathbb{N}^*$ so we start $n$ at $1$ not $0$.

Check:

Let $n=1$, then

$P(1) = 9 \not= \frac{4+14+1}{2} = \frac{19}{2}$, False.

$P(k)$ is not true $\forall k \in \mathbb{N}^*$ because $4n^2 +14n+1 \equiv 1 \pmod2$

However if you start with $P(k) = 9 + \cdots +(4k+5) = \frac{4k^2+14k+1}{2}$ true, then $$P(k+1)= \frac{4(k+1)^2+14(k+1)+1}{2} = \frac{4k^2+14k+1}{2} + 4k+9 $$
$$= \frac{4k^2+14k+1}{2} + 4(k+1)+5 = P(k)+4(k+1)+5$$ therefore $P(k+1)$ would be true by the definition of $P$.

Therefore
the induction step is true: if $P(k)$, then $P(k+1)$, $\forall k \in \mathbb{N}^*$, however the statement is always false because we can never satisfy the if, so I think the goal of this exercise was to show you the importance  of testing a base case, finding at least one tipping of a domino.

#3
The subset of $\mathbb{N}^*$ which satisfies $P(n)$ is the empty subset $\varnothing$ because as mentioned above $2 \not|\;\;4n^2 + 14n+1 $ for any $n \in \mathbb{N}^*$, so $9+ \cdots + (4k+5)$ is always an integer on the LHS while $\frac{4n^2+14n+1}{2} = 2n^2+7n+\frac{1}{2}$ is never an integer on the RHS, so they can never be equal.


问:
If we randomly pick 7 balls from 49 balls numbered from 1 to 49, what is the probability that we get 5 or more balls with number less than 10?

If we randomly pick 7 balls from 49 balls numbered from 1 to 49, what is the probability that we get 5 or more balls with number less than 10?

I know that there are totally 49*48*47*46*45*44*43 cases of drawing 7 balls.

Please help. Thank you.

答:
Hint: How many ways are there to pick 7 balls? Does order matter to you?

Hint: How many ways are there to pick 5 balls with number less than 10 (and 2 balls with number greater than 10)?

Hint: How many ways are there to pick 6 balls with number less than 10 (and 1 balls with number greater than 10)?

Hint: How many ways are there to pick 7 balls with number less than 10 (and 0 balls with number greater than 10)?


问:
Algebraic function field of one variable as a scheme
Let $F|K$ be an algebraic function field of one variable. How can it be considered as a scheme? And how can one define geometric and arithmetic genus for it?!

答:
There exists $t\in F$ transcendental over $K$ such that $F/K(t)$ is finite. Let $A$ (resp. $B$) be the integral closure of $K[t]$ (resp. $K[1/t]$) in $F$. Note that $B_{1/t}$ and $A_t$ are equal, so we can glue the spectra of $A$ and $B$ along $D(t) = D(1/t)$, where the LHS is an open subset of the spectrum of $A$ and the latter is an open subset of the spectrum of $B$.

The scheme $X$ which is obtained is naturally an integral $K$-scheme with function field $F$ and even has a natural morphism to $\mathbb{P}^1$ corresponding to the rational function $t$. Usual scheme theory applies to define a genus.

When $F/K(t)$ is separable, $A$ and $B$ are Dedekind rings and finitely generated modules over $K[t]$ and $K[1/t]$, which entails that $X$ is a noetherian Dedekind scheme hence is regular. So you may easily enough define a notion of arithmetic and geometric genus.


问:
Does $\sum_{n=1}^{\infty}\left(\frac{\sin x}{x}\right)^n$ converges?
Does $\sum_{n=1}^{\infty}\left(\frac{\sin x}{x}\right)^n$ converge for all values of x, except for $x=0$? I tried some values and it seems to make sense. Any proofs for this?

答:
Note that $-1<\frac{\sin x}{x}<1$

let $u=\frac{\sin x}{x}$ then,

$$\begin{align}\sum_{n=1}^{\infty}\left(\frac{\sin x}{x}\right)^n&=\sum_{n=1}^{\infty}(u)^n\\
&=\lim \limits_{n \to \infty}\sum_{i=1}^{n}(u)^i\\
&=\lim \limits_{n \to \infty}\frac{1-u^{n+1}}{1-u}\\
&=\frac{\lim \limits_{n \to \infty}(1-u^{n+1})}{\lim \limits_{n \to \infty}(1-u)}\\&=\frac{1}{1-u}\end{align}$$


答: There are 9 balls less than 10, and only 5 of the picked balls have to be less than 10. That means that there are ${9 \choose 5}=126$ combinations of balls that can be picked among the 5 restricted balls. After those 5 balls are picked, only 44 balls are left. We have two balls left to pick, and the other two balls can be any of the other 44 balls, which is $44 \choose 2$. The total combinations are $49 \choose 5 $We end up with $$\frac{{9 \choose 7}{44 \choose2}}{49 \choose 7}$$

Body:
If we randomly pick 7 balls from 49 balls numbered from 1 to 49, what is the probability that we get 5 or more balls with number less than 10?

I know that there are totally 49*48*47*46*45*44*43 cases of drawing 7 balls.

Please help. Thank you.


问:
Find my mistake: Conditional probability
A box contains 5 balls numbered 1-5.

Two balls are drawn randomly.
Calculate the probability of selecting balls numbered 1 and 2.

Answer via counting: 1/5C2

Attempt with conditional probability:

A= ball numbered 1 drawn
B= ball numbered 2 drawn

P(draw 1 and 2)=P(A and B)=P(A)P(B|A)=(1/5)(1/4).

My calculation: Given A the updated sample space is:
{1,2},{1,3},{1,4},{1,5}

So P(B|A)= 1/4.

What mistake am I making?

答:
Your second calculation gives the probability of drawing ball 1 first and then ball 2. Your first calculation gives the probability of drawing the two balls in either order.

If you want the second calculation to allow either order, then $P(A)$ should be $\frac25$, not $\frac15$.


问:
What is the meaning of $X = \sum_{i}^n X_i$ in the context of binomial random variable $X$?
I have been reading some of the problems and solutions for my Statistiscs course. I saw the following in the solution of a problem.

If $ X\in Bi(n, p),$ than $X = \sum_{i}^n X_i$

I dint really understand what this means.
Can someone help me build some intuition?

答:
Bernoulli distribution

The simplest distribution possible. Let's call it $f_{BE}$. Informally, it is used to represent a coin (fair or unfair), i.e. it is used to represent a random phenomenon where only two events are allowed. If the coin/random variable is $X$, and the coin can assume values $0$ or $1$ (or $H$ and $T$, or whatever you like, but just two events only!), then:

$$f_{BE}(x) =\mathbb{P}(X = x) = \begin{cases}
a & \text{if}~ x = 0 \\
1-a & \text{if}~ x = 1
\end{cases},$$

where $a \in (0, 1)$. If $X$ represents a fair coin, then $a = \frac{1}{2}$, otherwise not. Besides the trivial example of the coin, we can use the Bernoulli r.v. for representing for example the success of a candidate during an exam, or the victory of a football (or other sports!) team in match...

Binomial distribution

Suppose that you are concerned to find the probability that your favorite football team will win exactly $2$ matches in a streak of $3$. Each match is represented by $X_i$, with $i \in \{1, 2, 3\}$. Each $X_i$ has a Bernoulli distribution with parameter $a$ fixed. Moreover, each match is independent, does not matter what happened before, or what will happen later.

If there is a victory, $X_i = 1$, and $X_i= 0$ otherwise. The total number of victories in the streak is:

$$X = \sum_{i=1}^{3} X_i.$$

You want to find $\mathbb{P}(X = 2).$

It can be proven that $X$ has a binomial distribution with parameters $n=3$ and $p=a$. If the binomial distribution is denoted by $f_{BI}$, then:

$$\mathbb{P}(X = 2) = f_{BI}(2).$$


答:
Picking $5$ balls numbered less than $10$: $\binom{9}{5}\cdot\binom{40}{2}$ (choose $5$ and another $2$ with higher numbers)

Picking $6$ balls numbered less than $10$: $\binom{9}{6}\cdot\binom{40}{1}$

Picking $7$ balls numbered less than $10$: $\binom{9}{7}$

So the number of ways to choose at least $5$ balls numbered less than $10$ is $$\binom{9}{5}\cdot\binom{40}{2}+\binom{9}{6}\cdot\binom{40}{1}+\binom{9}{7}$$
and the number of ways to draw $7$  balls is $\binom{49}{7}$

(I assumed the order of the drawing is not important). So, the required probability will be:...

Title: If we randomly pick 7 balls from 49 balls numbered from 1 to 49, what is the probability that we get 5 or more balls with number less than 10?


问:
$f (x) = 1$ if the digit $9$ appears in the decimal representation of $x$ and $f (x) = 0$ otherwise — show that $f$ is continuous

The function $f : [0, 1] \to \Bbb R$, where $f (x) = 1$ if the digit $9$ appears in the decimal representation of $x$ and $f (x) = 0$ otherwise. We use a decimal representation that does not end in repeating $9$s.

 Show that $f$ is continuous at $x$ if and only if $f (x) = 1$.

 (Use only $\varepsilon$, $\delta$ definition of continues. You are not allowed use limit).

I intuitively know that if $x$ is a decimal which appears $9$ then we can find another decimal in any neighbourhood with the same property. if not then again we can find another decimal in any neighbourhood which appears $9$. But how write it formally I couldn't. After several days I gave up. Actually I understand the basic idea. But I couldn't write formally anything. Please help me.

答: The set of numbers with a $9$ in the decimal representation is dense in $[0, 1]$ (just replace the $n$-th digit by $9,$ for some high $n.$) This shows that the function is not continuous if it is $0.$ On the other hand, the set of numbers which do not have a nine is not dense (a number which has a nine in the $k$-th position will have it if perturbed by something smaller than $10^{-k-2}.$)


问:
Trouble with textbook problem and questionable proof of midpoint formula.

I need help with problem 33. I don’t really know how to go about solving it. Whenever I try to express the vertices as variables, everything just ends up being really messy.

For the guy to proof of the midpoint formula given in problem 35, the proof seems a bit wishy-washy that’s not rigorous. Wouldn’t it be better to use the distance formula to show that the segment addition postulate and definition of midpoint from geometry hold?

答:
I don’t understand your second question at all: there is no proof at all in Problem $35$, so there certainly isn’t one that’s not rigorous. The problem is for you to provide proofs of those five statements. How you do so will depend on what postulates and earlier theorems you have available. I can only guess what your segment addition postulate and definition of midpoint are, but it seems likely that they are the tools that you’re expected to use.

Once you know the result in part e of Problem $35$, Problem $33$ is just a matter of using the given information to set up a couple of systems of simultaneous equations. If the vertices are $A(x_A,y_A)$, $B(x_B,y_B)$, and $C(x_C,y_C)$, the information on the midpoints tells you that $7=\frac12(x_A+x_B)$, $3=\frac12(y_A+y_B)$, $10=\frac12(x_B+x_C)$, $9=\frac12(y_B+y_C)$, $5=\frac12(x_A+x_C)$, and $5=\frac12(y_A+y_C)$. That gives you three equations involving the three $x$-coordinates:

$$\begin{align*}
x_A+x_B&=14\\
x_B+x_C&=20\\
x_A+x_C&=10\,.
\end{align*}$$

This system is easy to solve: subtract the last equation from the first to find that $x_B-x_C=4$, add that to the second to get $2x_B=24$ and therefore $x_B=12$, and then use the first two equations to find $x_A$ and $x_C$. Finding the $y$-coordinates is equally easy.


问:
What is the inverse element of $u=i+j+k$ with respect to the multiplication?
What is the inverse element of $u=i+j+k$ with respect to the multiplication?

When I have $z = x + iy$ inverse is $\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}$ but what happens in my case?

答:
In tessarines, it is

$$\frac1{i+j+ij}=-\left(\frac{2}{5}+\frac{i}{5}\right)+\left(\frac{3}{5}-\frac{i}{5}\right)j$$


问:
Given $I(a,b)=\int_{0}^{1}x^a(1-x)^bdx$, find truth/falsity

Given $I(a,b)=\int_{0}^{1}x^a(1-x)^bdx$, where $a,b$ are integers, with $a \geq 1$, $b \geq 1$. Is the following true: $I(2,2)=2I(1,1)$?

I understood this to be whether this integral is linear, like $I(c(a,b))=c(I(a,b))$, where $c\in \mathbb{R}$. I declare this to be false. If you substitute $a=1, b=1$, then the integral evaluates to $2\int_{0}^{1}0dx$, which is clearly not the case with $a=2, b=2$.

Next, is $I(a,b)=I(b,a)$? This is of course a property of symmetry, if there ever was one. This time there's no way to verify whether it holds. Shall I try a few examples to see if it disproves it?

答:
In fact your integral is a version of the so-called "Beta" function, with a simple closed form expression for positive integer exponents:

$$I(a,b)=\int_0^1 x^{a} (1-x)^b dx= \dfrac{a! b!}{(a+b+1)!}\tag{1}$$

This gives an answer to your second question: yes, $I(a,b)=I(b,a).$

For the first one, one obtains from (1):

$$I(2,2)=\dfrac{2 \times 2}{120}=\dfrac{1}{30} \ \ \text{and} \ \ I(1,1)=\dfrac{1 \times 1}{6}=\dfrac{1}{6}$$

Thus the exact answer to your first question is: no, we have

$$I(2,2)=\dfrac15 I(1,1).$$


答:
A more elementary approach,

First of all let $f(x)= 0$ then i.e. $x = \sum_{i=1}^{\infty} \frac{x_i}{10^i}$ where $x_{i}\neq 9$, $\forall \ i$. Then for every $\epsilon >0 $, there exists large enough $i_0$ such that, $y := \sum_{i=1}^{i_0 - 1} \frac{x_i}{10^i} + \frac{9}{10^{i_0}} + \sum_{i=i_0}^{\infty} \frac{x_i}{10^i}$, is $\epsilon$-close to $x$, but $f(y)=1$. So indeed $f$ is not continuous at $x$ where $f(x)=0$.

Now let $f(x)=1$. Then $9$ appears in its decimal representation and consider $i_1\geq 1$ such that $9$ appears for the first time in its decimal expansion i.e.  $x = \sum_{i=1}^{i_1 - 1} \frac{x_i}{10^i} + \frac{9}{10^{i_1}} + \sum_{i=i_1}^{\infty} \frac{x_i}{10^i}$, where $x_i\neq 9$ for all $1\leq i < i_1$. Then for $\epsilon_1 < \frac{9}{10^{i_1 +1}}$ we have that every $y$ that is $\epsilon_1$-near $x$ has $9$ in its decimal expansion and more precisely $y_i = x_i$ for (at least) all $1\leq i \leq i_1$

Title: $f (x) = 1$ if the digit $9$ appears in the decimal representation of $x$ and $f (x) = 0$ otherwise — show that $f$ is continuous

Body:
The function $f : [0, 1] \to \Bbb R$, where $f (x) = 1$ if the digit $9$ appears in the decimal representation of $x$ and $f (x) = 0$ otherwise. We use a decimal representation that does not end in repeating $9$s.

 Show that $f$ is continuous at $x$ if and only if $f (x) = 1$.

 (Use only $\varepsilon$, $\delta$ definition of continues. You are not allowed use limit).

I intuitively know that if $x$ is a decimal which appears $9$ then we can find another decimal in any neighbourhood with the same property. if not then again we can find another decimal in any neighbourhood which appears $9$. But how write it formally I couldn't. After several days I gave up. Actually I understand the basic idea. But I couldn't write formally anything. Please help me.


问:
Computing the residues for $\frac{1}{z^3-1}$
I'm struggling to follow this example in my lecture notes where they compute the residue for $\frac{1}{z^3-1}$. I understand how to get to the singularities $c=1$, $c=\exp^{2i\pi/3}$ and $c=\exp^{4i\pi/3}$.

I also understand how the steps go from L'Hopital's rule to obtain the limit as $z$ tends to $c$, which gets $\frac{1}{3c^2}$. But the notes rather helpfully stop here.

This hasn't found the residues, right?

I had a look on Mathematica, the singularity $c=1$ gives a residue of $0$, so why if you were to "sub-in" $c=1$ into $\frac{1}{3c^2}$ would you get $\frac{1}{3}$? Is it the Mathematica code I'm doing wrong or am I missing something completely?

I'm really confused how to get to this last step! Thanks in advance for any tips :)

答:
I tried this example in Mathematica and got the answer $\frac13$:

```<code> Residue[1/(z^3-1),{z,1}]
</code>```

This is of course not surprising because the answer $\frac1{3c^2}$ is correct.

Body:
I tried this example in Mathematica and got the answer $\frac13$:

 Residue[1/(z^3-1),{z,1}]

This is of course not surprising because the answer $\frac1{3c^2}$ is correct.


问:
What is the derivative of the derivative?
If we consider the derivative as a function from a function space to function space, then does it make sense to talk about the derivative of the derivative? In particular, if we consider

$$D:C^\infty[\mathbb{R}]\to C^\infty[\mathbb{R}]$$

and arbitrary smooth function $f\in C^\infty[\mathbb{R}]$, then can we reasonably ask if there is some other function on smooth real functions given by the following?

$$\frac{d}{df}[Df]$$

I've tried to do this myself with limits, where we take an arbitrary smooth function $u$ that approaches the constant function of $0$. In particular, I found that

$$\lim_{u\to 0}\frac{D(f+u)-D(f)}{u}=\lim_{u\to 0}\frac{f'+u'-f'}{u}=\lim_{u\to 0}\frac{u'}{u}.$$

But I'm not sure if that approaches any particular value independent of $u$'s path toward $0$. However, plugging in $u = 0$ does indeed result in an indeterminant form, which leads me to suspect that there is a way of solving this. Instinctually, I think that this value should be $1$ both due to $D$ being considered linear, and applying L'Hopital's rule infinitely many times would result in

$$\lim_{u\to 0}\frac{u'}{u} = \lim_{u\to 0}\lim_{n\to\infty}\frac{u^{(n+1)}}{u^{(n)}}=\lim_{u\to 0}\lim_{n\to\infty}\frac{u^{(n)}}{u^{(n)}}=\lim_{u\to 0}\lim_{n\to\infty}1=1$$

But this does not seem particularly rigorous. Does my question even make sense and if it is, is there a rigorous way of finding the solution?

Edit: To add to the confusion, if we treat deriving with respect to a function as we do in $\mathbb{R}$, we find that
$$ \frac{d}{df}[Df] = \frac{df'}{dx}\frac{dx}{df}=\frac{f''}{f'}$$
Which I'm pretty sure is not quite the same as differentiating with respect to the function itself, but I don't fully understand why it would be different.

答:
This was a question that also bothered me when I was younger - and I was much surprised to find that the only satisfactory answer is that the derivative of the operator $D$ is itself. It's worth understanding why this must be - and not to bog ourselves down with too much analysis (especially since the main hurdle to such analysis is getting the derivative operator to behave well enough to do analysis on at all).

The most important fact about derivatives is that they represent linear approximations near a point. When we speak, in finite dimensions, of a derivative of a function $f:\mathbb R^n\rightarrow\mathbb R^m$, we are really just saying "near a point $p$, $f(v)$ is close to $f(p)+L(v-p)$ for some linear function $L$" - where by "near" we usually mean that the error is $o(|p-v|)$ - meaning that for any $\varepsilon > 0$ there is some $\delta$ so that if $|v - p| < \delta$ then $|f(v) - (f(p) + L(v-p))| \leq \varepsilon|p-v|$. If you're used to working with coordinates, that linear function $L$ is represented by the Jacobian matrix - and the existence of such a matrix is precisely how "differentiability" is usually defined for multiple variables. Sometimes this is stated by saying that $L$ is such that
$$\lim_{v\rightarrow p}\frac{f(v)-(f(p)+L(v-p))}{|v-p|} = 0$$
which is another notation for the same thing - note that we can't divide directly by $v-p$ because it's just some vector; it doesn't make sense to try to divide by, for instance "5 miles northeast" - even if that makes perfect sense as a position on a two dimensional grid. Instead, we have to suppose we already have a guess at what the derivative is, and are just seeking to verify it.

There's a notable special case here: if $f$ was already linear, this is trivial: the Jacobian, at every point, is just the matrix that represented $f$ to begin with. For instance, the map
$$f:\mathbb R^2\rightarrow\mathbb R$$
$$f(x,y)=2x+y$$
has its Jacobian at every point being, well, the matrix taking $(x,y)$ to $2x+y$ - because, near any point, this function tells us exactly how much a change to the input changes the output - we're not even approximating anything here because the error term is not only sublinear, it's zero.*

The derivative as a map falls into this special case - where "derivatives are linear" just means that $(f+g)' = f'+g'$ and $(\alpha f)'=\alpha(f')$ where $\alpha$ is a scalar (i.e. a real/complex number, depending on the context). Saying "the derivative of $D$ is itself" just says that if we have some value $f$ for which we know $Df$ and want to shift the input $f$ by some vector $g$, the resulting change in the output (which should be approximately the derivative of $D$ applied to the vector $g$) is "about" $Dg$ - where, by "about" we mean "exactly." As a result, under every definition of the derivative I'm aware of that allows us to even pose the question, the answer will be that the derivative of $D$ is $D$.

This can be a bit confusing because you seem tempted to want the derivative to be
$$\lim_{u\rightarrow 0}\frac{f'+u'-f'}{u}$$
which kinda makes sense, since $u$ is a function so it kinda makes sense to divide by it, but the notion of a "derivative" treats $u$ as if it were just some vector, not as if it's something we can do algebra on - it's just a confounding coincidence that we could write this expression, not a useful definition of a derivative. Note especially that $u$ could be zero at various points, which basically ensures that this limit does not make sense (even before we run into other problems with the definition - like it taking a limit of a whole function without specify how). Luckily for us, this isn't how derivatives are defined in multiple dimensions - and in terms of linear approximations, it's unavoidable that $D$ is its own derivative.

(*To catch a very reasonable confusion: when we're saying "the derivative of something is itself" it feels like we're discussing exponential functions - but that's a slightly different idea. The one dimensional analog of this is to imagine that $f(x)=2x$ has this property that near say $x$ we can approximate $f(x+\delta)\approx f(x)+2\delta$ - which is what a derivative means. What we mean to note is that this linear approximation $2\delta$ is just $f(\delta)$. This is distinct from saying that $g(x)=e^x$ has that $g(x)=g'(x)$, which instead expands to say that $g(x+\delta)$ is about $g(x)+g'(x)\delta$ - where, of course, the term $g'(x)\delta$ giving the (approximate) change over a small $\delta$ is not $g(\delta)$. This is not the sense we mean when we say $D$ is its own derivative - and is a sense that is limited to one dimension anyways)


问:
Limit does not exist or limit does exist? What does the unevaluated output mean?
From this short Mathematica program while investigating the convergence of the Dirichlet series for the Möbius function:

```<code>Clear[a, b, s, x];
s = 1/2 + 100*I;
Limit[1 - Sum[1/a^s, {a, 2, x}] +
  Sum[Sum[1/(a*b)^s, {a, 2, x}], {b, 2, x}], x -> Infinity]
</code>```

I get the output:

where Out[186]= is left unevaluated.

Does that mean that the limit exists but it is not known in the form of standard functions?

答:
First,  we consider

```<code> s = 1/2 + 100*I; Sum[1/a^s, {a, 2, x}]
</code>```

-1 + HarmonicNumber[x, 1/2 + 100 I]

Second (The crucial tool is an assumption.),

```<code>Sum[Sum[1/(a*b)^s, {a, 2, x},Assumptions -> b \[Element] PositiveIntegers], {b, 2, x}]
</code>```

(-1 + HarmonicNumber[x, 1/2 + 100 I])^2

At last,

```<code>Limit[1 + (-1 + HarmonicNumber[x, 1/2 + 100 I])^2 - (-1 +
HarmonicNumber[x, 1/2 + 100 I]), x -> Infinity]
</code>```

ComplexInfinity

Addition. The OP additionally requires  a*b<=x. In fact, this is not any restriction because the limit as x->Infinity is taken. The sum of a double series does not depend on an exhaustion. Since

```<code>ComplexExpand[1/Abs[(a*b)^(\[Sigma] + I*t)]]
</code>```

(Sqrt[a^2] Sqrt[b^2])^-\[Sigma] E^(t Arg[a b])

, the convergence depends only on \[Sigma]. It's more or less clear, that \[Sigma] should be greater than 1 to this end (I don't go into math deep.).
Therefore,

```<code>Sum[Sum[1/(a*b)^\[Sigma], {a, 2, Infinity}, Assumptions -> \[Sigma] > 1 &&
 b \[Element] PositiveIntegers], {b, 2, Infinity}, Assumptions -> \[Sigma] > 1]
</code>```

1 - 2 Zeta[\[Sigma]] + Zeta[\[Sigma]]^2

The same with Sum[1/a^s, {a, 2, x}]. Making use of Mathematica, we obtain the same answer as in MSE. The true convergence of a series is its absolute convergence. In view of it the series from the question are not absolutely convergent for s = 1/2 + 100*I.

OwnerDisplayName: Mats Granvik

Body:
First,  we consider

 s = 1/2 + 100*I; Sum[1/a^s, {a, 2, x}]

-1 + HarmonicNumber[x, 1/2 + 100 I]

Second (The crucial tool is an assumption.),

Sum[Sum[1/(a*b)^s, {a, 2, x},Assumptions -> b \[Element] PositiveIntegers], {b, 2, x}]

(-1 + HarmonicNumber[x, 1/2 + 100 I])^2

At last,

Limit[1 + (-1 + HarmonicNumber[x, 1/2 + 100 I])^2 - (-1 +
HarmonicNumber[x, 1/2 + 100 I]), x -> Infinity]

ComplexInfinity

Addition. The OP additionally requires  a*b<=x. In fact, this is not any restriction because the limit as x->Infinity is taken. The sum of a double series does not depend on an exhaustion. Since

ComplexExpand[1/Abs[(a*b)^(\[Sigma] + I*t)]]

(Sqrt[a^2] Sqrt[b^2])^-\[Sigma] E^(t Arg[a b])

, the convergence depends only on \[Sigma]. It's more or less clear, that \[Sigma] should be greater than 1 to this end (I don't go into math deep.).
Therefore,

Sum[Sum[1/(a*b)^\[Sigma], {a, 2, Infinity}, Assumptions -> \[Sigma] > 1 &&
 b \[Element] PositiveIntegers], {b, 2, Infinity}, Assumptions -> \[Sigma] > 1]

1 - 2 Zeta[\[Sigma]] + Zeta[\[Sigma]]^2

The same with Sum[1/a^s, {a, 2, x}]. Making use of Mathematica, we obtain the same answer as in MSE. The true convergence of a series is its absolute convergence. In view of it the series from the question are not absolutely convergent for s = 1/2 + 100*I.


答: Yes, if $c\in\left\{1,e^{2\pi i},e^{4\pi i}\right\}$, then$$\operatorname{res}_{z=c}\left(\frac1{z^3-1}\right)=\frac1{3c^2}.\tag1$$That's because\begin{align}\lim_{z\to c}(z-c)\frac1{z^3-1}&=\lim_{z\to c}\frac1{\frac{z^3-1}{z-c}}\\&=\frac1{\lim_{z\to c}\frac{z^3-1}{z-c}}\\&=\frac1{3c^2}.\end{align}It follows from this that, near $c$, you can write $\frac1{z^3-1}$ as$$\frac1{3c^2(z-c)}+a_0+a_1(z-c)+\cdots;$$in particular, $(1)$ holds.

Title: Computing the residues for $\frac{1}{z^3-1}$

Body:
I'm struggling to follow this example in my lecture notes where they compute the residue for $\frac{1}{z^3-1}$. I understand how to get to the singularities $c=1$, $c=\exp^{2i\pi/3}$ and $c=\exp^{4i\pi/3}$.

I also understand how the steps go from L'Hopital's rule to obtain the limit as $z$ tends to $c$, which gets $\frac{1}{3c^2}$. But the notes rather helpfully stop here.

This hasn't found the residues, right?

I had a look on Mathematica, the singularity $c=1$ gives a residue of $0$, so why if you were to "sub-in" $c=1$ into $\frac{1}{3c^2}$ would you get $\frac{1}{3}$? Is it the Mathematica code I'm doing wrong or am I missing something completely?

I'm really confused how to get to this last step! Thanks in advance for any tips :)


问:
Solving differential equation $y y' = x^2 y + (a^2 - 1) x$
I'm trying to solve the differential equation for $y=y(x)$:
$$
y y' = x^2 y + (a^2 - 1) x .
$$
I know that this the Abel equation of second kind. But, I could not solve it analytically. Any suggestion would be appreciated.

答:
Divide both sides by $y,$ you get

$$y^\prime = x^2 + (a^2-1)x/y.$$

Solve the homogeneous part (easy), then use indetermined coefficients.


问:
Adjoint Matrix Invertibility
If matrix A is invertible, is its adjoint matrix also always invertible?

The book "Elementary Linear Algebra" by Anton and Rorres use the terms adjucate matrix, adjoint matrix, and classical adjoint matrix interchangeably. Is there a difference between them? From  my limited experience, I think there is. In this problem I am referring to the adjoint matrix as the transpose of the cofactor matrix from A.

答: By Cramer's rule, the adjoint of an invertible matrix equals its inverse times its determinant. Now, is that invertible?</p>

Title: Adjoint Matrix Invertibility


问:
IEEE-754 General Issues
I read the documents of the IEEE-754 and I could not understand some things

1) Is it possible to represent any number using this method? I know there are a number of final states, and there are infinite possibilities then no, but if there is a more detailed explanation of why?

2)Why can not represent the number zero? And how I represent it?

3) Why I dont need to represent 1 in representation of floating point number, I know its normalized, but normalized mean only 1 in the left side of the number? or everynumber, for example $8.3131e^{-1}$

Thanks!

答:
1) No. Eg. if you want to represent $10^{50} + 1$ and $10^{50} + 2$ they're going to be the same in terms of floating points with not enough bits.

2) You have numbers in from $(-1)^{sign} (1 + \frac{mantisa}{2^{bitsofmantisa}}) 2^{exponent-bias}$ how would you express 0? You have to have special value of exponent for this.

3) For 1 you just don't have any positive bits in mantisa...

Title: IEEE-754 General Issues


问:
Rectangular Fractal name and Area
I would like to know the name of this fractal, and how to obtain the n-th step area. Thanks!

答:
Let us consider that the initial area is $1$.

The area diminishing ratio between a step and its successive step is $$r=\dfrac{13}{16}<1.$$

Therefore the area remaining at the $n$-th step is $r^{n-1}$ which tends to zero when $n$ tends to infinity, just like the Sierpinski triangle fractal.

Title: Rectangular Fractal name and Area


问:
Probability combined with permutations.
Suppose we select a permutation of the sequence 1,...,n, uniformly at random, where n ≥ 2 is even. What is the probability that none of numbers 1, . . . , n/2 stays in its place? Justify your answer.

I am struggling as to where to start any suggestions would be appreciated.

答:
Comments: For $n=10,$ here is an approximate answer $(0.5988 \pm 0.0005)$ from
a simulation in R. With a million iterations, one
expect two (maybe three) place accuracy.

```<code>set.seed(2021)
stay = replicate(10^6, sum(1:5==sample(1:10)[1:5]))
mean(stay)
[1] 0.498965     # aprx n/2 = 5
mean(stay==0)
[1] 0.598818     # aprx answer
sd(stay==0)/1000
[1] 0.000490138  # aprx 95% margin of sim error for ans
table(stay)/10^6
stay
       0        1        2        3        4        5
0.598818 0.315199 0.075046 0.010108 0.000795 0.000034
</code>```

Notes: The most likely outcome is that none of the
first five 'stay' in place.

It is trivial to see that the answer for $n=2$ is $1/2.$
And enumeration is within reach for $n=4.$

Using indicator variables one can see that the mean number staying in place
among $1,2 \dots n/2$ is $n/2.$ (But indicator variables
are not independent, so getting the variance is not so easy.)

I suppose an analytic solution to your question for general $n$ would require use of the inclusion-exclusion rule.

Title: Probability combined with permutations.

Body:
Comments: For $n=10,$ here is an approximate answer $(0.5988 \pm 0.0005)$ from
a simulation in R. With a million iterations, one
expect two (maybe three) place accuracy.

set.seed(2021)
stay = replicate(10^6, sum(1:5==sample(1:10)[1:5]))
mean(stay)
[1] 0.498965     # aprx n/2 = 5
mean(stay==0)
[1] 0.598818     # aprx answer
sd(stay==0)/1000
[1] 0.000490138  # aprx 95% margin of sim error for ans
table(stay)/10^6
stay
       0        1        2        3        4        5
0.598818 0.315199 0.075046 0.010108 0.000795 0.000034

Notes: The most likely outcome is that none of the
first five 'stay' in place.

It is trivial to see that the answer for $n=2$ is $1/2.$
And enumeration is within reach for $n=4.$

Using indicator variables one can see that the mean number staying in place
among $1,2 \dots n/2$ is $n/2.$ (But indicator variables
are not independent, so getting the variance is not so easy.)

I suppose an analytic solution to your question for general $n$ would require use of the inclusion-exclusion rule.


问:
Proving that $\mathcal{S} = \{E\subset X: E \text{ is countable or } X\setminus E \text{ is countable}\}$ is the smallest $\sigma$-algebra

Suppose $X$ is a set and $\mathcal{A}$ is the set of subsets of $X$ that consist of exactly one element: $$\mathcal{A} = \{\{x\}:x \in X\}.$$ Prove that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $E^c$ is countable.

To prove this, I first want to show that $\mathcal{S}$ is a $\sigma$-algebra where $$\mathcal{S} = \{E\subset X: E \text{ is countable or } X\setminus E \text{ is countable}\}.$$

My question: I've already shown that $\mathcal{S}$ is a $\sigma$-algebra. Now, I think I need to show that $\mathcal{A}$ is contained in $\mathcal{S}$. How can I show this (since both $\mathcal{A}$ and $\mathcal{S}$ are sets of sets)? So, how can $\mathcal{A}$ be contained in $\mathcal{S}$? Also, lastly, I think I need to show that $\mathcal{S}$ is the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{A}$. Can someone show how this can be proven?

答:
As you said you have already proved that
$$\mathcal{S} = \{E\subseteq X: E \text{ is countable or } X\setminus E \text{ is countable}\}$$
is in fact a $\sigma$-algebra.

Now, we have $\mathcal{A} = \{\{x\}:x \in X\}$ and let us prove that the smallest $\sigma$-algebra containing $\mathcal{A}$  is equal to $\mathcal{S}$.

Since for each $x \in X$ , $\{x\}$ is countable, we have that $ \mathcal{A} \subseteq \mathcal{S}$.

Since $\mathcal{S}$ is a $\sigma$-algebra, it follows that
$$ \sigma(\mathcal{A}) \subseteq \mathcal{S} \tag{1}$$
where $\sigma(\mathcal{A})$ is the smallest $\sigma$-algebra containing $\mathcal{A}$ (the $\sigma$-algebra generated by $\mathcal{A}$).

Now let us prove that $ \mathcal{S} \subseteq \sigma(\mathcal{A}) $.

We will begin by proving the claim:

Every countable set $F\subseteq X$ is in $\sigma(\mathcal{A})$.

Proof of the claim:
Since for every $x \in X$ , $\{x\} \in \mathcal{A} \subseteq \sigma(\mathcal{A})$ and every countable set $F\subseteq X$ is a countable union of single-point sets, we have that $F \in \sigma(\mathcal{A})$ (because $\sigma(\mathcal{A})$ is a $\sigma$-algebra). So, every countable subset of $X$ is in $\sigma(\mathcal{A})$ . $\square$

Now, for any $E \in \mathcal{S}$, we have two possibilities:

$E$ is countable. In this case, it follows directly from the claim that $E \in \sigma(\mathcal{A})$.

$X\setminus E$ is countable. In this case,  it follows directly from the claim that $X\setminus E \in \sigma(\mathcal{A})$. Since  $\sigma(\mathcal{A})$ is a $\sigma$-algebra, we have $X\setminus(X\setminus E) \in \sigma(\mathcal{A})$. But $X\setminus(X\setminus E) =E$, so we have $E \in \sigma(\mathcal{A})$.

So, in both cases, we have that  $E \in \sigma(\mathcal{A})$.
So, we have:

$$ \mathcal{S}= \{E\subseteq X: E \text{ is countable or } X\setminus E \text{ is countable}\} \subseteq \sigma(\mathcal{A})  \tag{2}$$

From $(1)$ and $(2)$,
$$ \sigma(\mathcal{A}) = \mathcal{S} $$
It means, the smallest $\sigma$-algebra containing $\mathcal{A}$  is equal to $\mathcal{S}$.


问:
How to understand the definition of divergence of a vector field on a Riemannian manifold as ${\rm d}( i_X {\rm d}V_g )=( {\rm div} X ) {\rm d}V_g$?
I am trying to read some basic constructions on Riemannian manifolds, and I came across the definition of the divergence of a vector field. It is given by the formula,

$$
d( i_{X} d V_g ) = ( \text{div} X ) dV_g.
$$

Here $i_{X} d V_g$ denotes the interior multiplication of the Riemannian volume form with $X$, giving us a $(n-1)$ form. $d$ denotes taking the exterior derivative, giving us a $n$ form, and the divergence is specified to be the smooth function multiplying with the resulting $n$ form.

Can anyone she some shed on this formula? I am finding it very hard to motivate this coordinate independent definition.

答:
This is a sketch rather than a complete answer. You did not specify what is for you the intuitive meaning of divergence (usually some version of "flux per unit of volume") so it is a bit hard to answer. But I can think of three possibly helpful things

write the expression in local coordinate and convince yourself that you get the same thing as the divergence you are familiar with - essentially you are going to get $\nabla_\mu X^\mu\  \mathrm{vol}$.

since the volume element is closed, the expression you wrote is equal to $L_X\  \mathrm{vol}$. So, what you have is how the volume element changes when Lie transported along $X$, which is kinda of the dual viewpoint of the flux ingoing/outgoing a fixed volume element perspective.

By Stokes' theorem
\begin{equation}
\int_V d(i_X\  \mathrm{vol})=\int _{\partial V}i_X \ \mathrm{vol}.
\end{equation}
Now you have to convince yourself that $i_X \ \mathrm{vol}= *X$ is the flux of $X$ across $\partial V$. Something along the following lines shoud work. Write $\mathrm{vol}=n \wedge W$ with $n$ a 1-form such that the associated vector field is unit normal to $\partial V$ and $W$ is a volume form on $\partial V$. This way $i_X\  \mathrm{vol}$ once restricted to $\partial V$ is equal to $\langle X,n \rangle W $.


问:
proving how many permutations exist.
Show that the number of permutations of n numbers, for n ≥ 2, with two cycles is at most
(n−1)!logn.

Unsure as to were to start any help would be appreciated.

答:
The number of cycle permutations on $k$ letters is $(k-1)!$ If your cycles are of lengths $m$ and $n-m,$ you have $\binom{n}{m} (m-1)! (n-m-1)!$ possibilities, so
your total is
$$\sum_{m=1}^{n-1}\binom{n}{m} (m-1)! (n-m-1)! = n!\sum_{m=1}^{n-1} \frac{1}{m(n-m)}.$$
Can you finish from here?

Title: proving how many permutations exist.


答:
You can't represent every number as the language is countable and there are uncountably many numbers.  More to the point, you can't even represent $1/10$ since
$$\frac{1}{10} = \frac{1}{2} \sum _{n=1}^{\infty } \frac{3}{16^n}.$$
This is exactly why computations like $0.1 + 0.2 - 0.3$ tend to yield results around $10^{-16}$, rather than exactly zero.
More generally, you cannot exactly represent a number with an infinite binary expansion.

You can represent zero. The fraction and exponent fields must contain all zeros and the sign field can be zero or one.

Of course, 1 needs (and has) a representation.

Title: IEEE-754 General Issues

Body:
I read the documents of the IEEE-754 and I could not understand some things

1) Is it possible to represent any number using this method? I know there are a number of final states, and there are infinite possibilities then no, but if there is a more detailed explanation of why?

2)Why can not represent the number zero? And how I represent it?

3) Why I dont need to represent 1 in representation of floating point number, I know its normalized, but normalized mean only 1 in the left side of the number? or everynumber, for example $8.3131e^{-1}$

Thanks!


问:
Prove that the orbital projection of a $G$-space $X$ is a functional cover
What shown to follow is a reference from the text Curso de Topología by Sergey A. Antonyan. I traslated it from spanish and thus for sake of completeness I put here a link where you can find the original text: in particular to follow I summarise the pages $234$-$236$ and the definition given at the page $258$.

Definition

Let be $G$ a group and let be $X$ a topological space. If we equipe $G$ with the discrete topology then an action of $G$ on $X$ is a continuous function $\theta:G\times X\rightarrow X$ such that

$\theta(g\cdot h,x)=\theta\big(g,\theta(h,x)\big)$ for any $g,h\in G$ and for any $x\in X$

$\theta(e,x)=x$ for any $x\in X$ where $e$ is the identity element of $G$.

So we call $X$ with the action $\theta$ $G$-space.

To follow we are indicating $\theta(g,x)$ as $gx$ and thus for any $g\in G$ we define a function $\tilde g:X\rightarrow X$ by letting that
$$
\tilde g(x)=gx
$$
for any $x\in X$ and we call it translation.

Proposition

Let be $X$ a $G$-space. For any $g\in G$ the translation is a hoemomorphism.

Proof. Omitted.

So we define a function $\Gamma$ from $G$ to the group $\hom(X)$ of homeomorphisms in $X$ by letting that
$$
\Gamma(g):=\tilde g
$$
for any $g\in G$.

Proposition

The function $\Gamma$ is a homomorphism.

Proof. Omitted.

Definition

An action $\theta$ is said effective if the homomorphism $\Gamma$ is injective, that is if $\tilde g=\text{id}_X$ then $g=e$.

Definition

The action of $G$ in $X$ is said independent if for any $g\in G\setminus\{e\}$ there exist $x\in X$ such that $gx\neq x$.

Clearly any independent action if effective.

Definition

If $X$ is a $G$-space then the set
$$
G(x):=\{gx:g\in G\}
$$
for any $x\in X$ is said orbit of $x$.

Proposition

If $G(x)\cap G(y)\neq\emptyset$ then $G(x)=G(y)$ and so the orbits determine a partition of $X$.

Proof. Omitted

The function $p:X\rightarrow X/G$ given by
$$
p(x):=G(x)
$$
for any $x\in X$ is surjective and it is called orbital projection and in particular if we equipped the set $X/G$ with the quotient topology we call it orbital space.

Proposition

The orbital projection of a $G$-space is a continuous and open function.

Proof. Omitted.

Definition

A $G$-space $X$ is said discontinuous if any $x\in X$ has a neighborhood $U$ such that $gU\cap hU=\emptyset$ if $g$ and $h$ are two distinct elements of $G$. In particular the neighborhood $U$ is said lock neighborhood.

Definition

Let be $X$ and $Y$ two topological spaces and let be $f:X\rightarrow Y$ a continuous and surjective funciton. So we say that $f$ is a functional cover if for any $y\in Y$ there exists a neighborhood $V_y$ and a collection $\mathcal U:=\{U_i:i\in I\}$ of open and disjoint sets in $X$ such that $f^{-1}[V]$ is disjoint union of these sets and such that the function $f|_{U_i}$ is a homeomorphism.

Theorem

The orbital projection of a $G$-space $X$ is a functional cover.

Proof. Let be $x$ a poin of $X$ and thus let be $U$ a lock neighborhood. So since $p$ is open then $p[U]$ is an open set of $G(x)=p(x)\in X/G$. Then
$$
p^{-1}\big[p[U]\big]=\bigcup_{g\in G}gU
$$
so that $\mathcal G:=\{gU:g\in G\}$ is a collection of open and disjoint set of $X$ whose union is $p^{-1}\big[p[U]\big]$ and clearly the function $p|_{gU}$ is for any $g$ a hoemomorphsim.

So I did not understand why $p|_{gU}$ is for any $g$ a hoemomorphsim and thus I ask to explain it: in particular I know that it is open and continuous so that the statement follows showing that it is injective but unfortunately I do not be able to do it. So could someone help me, please?

答:
So if $p(x)=p(y)$ then $y\in G(x)$ and thus there exist $h\in G$ such that $h\ast y=y$ but $x\in gU$ so that there must exit $\xi\in U$ such that $y=g\xi$ and thus $y=hx=h(g\xi)$ that means $y\in(hg)U$ so that
$$
gU\cap(hg)U\neq\emptyset
$$
and clearly this is impossible unless $h=e$.