答:
I'm not sure what $a$ and $b$ are supposed to be, but I'll assume $a=1$ and $b=\infty$.  A simple example is $$f(x)=\cases{1,&$x\in \mathbb{N}$\\0,&$x\notin \mathbb{N}$}$$

You can modify this example to make $f$ continuous, by making it piecewise linear, or even $C^\infty$ by using bump functions.

Title: Is there a function $f(x)$ for which $\int_a^b f(x)dx$ converges but $\sum_{i=1}^\infty f(i)$ diverges?

Body:
I was recently learning about improper integrals, so I am not that advanced, and I asked myself whether there is some function $f(x)$ for which $\int_a^b f(x)dx$ with the limits being $a$ or $b$, it doesn't matter, converges while $\sum_{i=1}^\infty f(i)$ diverges.

I have found similar questions being asked like this one but there are usually the other way around, so the integral is divergent and the sum convergent, like with $f(x)=\sin(\pi x)$.


问:
Bounds for singular values of block matrices
Let
\begin{align}
M = \begin{pmatrix} A \\ v^T \end{pmatrix}
\end{align}
with $A \in \mathbb{R}_{\geq 0}^{n \times k}$, $n\geq k$, $\text{rank}(A) = k$, and $v^T \in \mathbb{R}_{\geq 0}^{1 \times k}$.
Moreover, I know bounds for the sum of every row $A_i$ of $A$ and for $v^T$, i.e.,
\begin{align}
 \varepsilon &\leq \sum_j A_{i,j} \leq \sqrt{2} \quad \forall i=1,...,n \\
 \varepsilon &\leq \sum_jv^T_j \leq \sqrt{2}
\end{align}

I'm interested in the condition number of $M$, i.e.
\begin{align}
\text{cond}(M) := \frac{\sigma_{max}(M)}{\sigma_{min}(M)},
\end{align}
where $\sigma_{max}(M)$ and $\sigma_{min}(M)$ represent the largest and smallest non-zero singular value of $M$, respectively.

Question

Is it possible to bound $\text{cond}(M)$ by information that I have regarding $A$ and $v^T$? It seems (numerical experiment) that
\begin{align}
\text{cond}(A) \not \leq \text{cond}{M},
\end{align}
which I hoped for at first. I guess it makes sense that this does not hold true; I'd have to include information regarding $v^T$.
The relation $\sigma_k(M) = \sqrt{\lambda_k(M^T M )} =  \sqrt{\lambda_k(A^T A + vv^T )} $ seems useful but didn't help me. Even with (I hope this is correct)
\begin{align}
 \sigma_{max}(M) \leq \sigma_{max}(A) + \sigma_{max}(v^T) = \sigma_{max}(A) + ||v^T||_2
\end{align}
I only have an upper bound for the enumerator, but no lower bound for the denominator (of the expression for the condition number).

答:
Your upper bound is correct, but we can do better. In particular, for any $x \in \Bbb R^k$, we have
$$
\|Mx\|_2^2 = \left\|\pmatrix{A \\ v^T} x \right\|_2^2 = \left\|\pmatrix{Ax \\ v^Tx} \right\|_2^2 = \|Ax\|_2^2 + \|v^Tx\|_2^2 \leq [\sigma_\max(A)^2 + \|v\|_2^2] \cdot \|x\|_2^2.
$$
With that, we can conclude that
$$
\sigma_\max(M) = \max_{x \neq 0} \frac{\|Mx\|_2}{\|x\|_2} \leq \sqrt{\sigma_\max(A)^2 + \|v\|_2^2}.
$$
We can apply a to get the lower bound $\sigma_\max(M) \geq \max\{\|v\|_2, \sigma_\max(A)\}$.

For lower bounds, we can apply similar ideas. Note that $\sigma_\min(M) = \min_{x \neq 0} \frac{\|Mx\|_2}{\|x\|_2}$. Using the ideas above, we can reach the conclusion that
$$
\sigma_\min(A) \leq \sigma_\min(M) \leq \sqrt{\sigma_\min(A)^2 + \|v\|_2^2}.
$$
Putting everything together, we can conclude that
$$
\frac{\sigma_\max(A)}{\sqrt{\sigma_\min(A)^2 + \|v\|_2^2}}\leq
\operatorname{cond}(M) \leq \frac{\sqrt{\sigma_\max(A)^2 + \|v\|_2^2}}{\sigma_\min(A)}.
$$

ContentLicense: CC BY-SA 4.0


问:
If $\beta = \left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)\left(2\; 4 \;10\right) \in S_{10}$, what is the smallest integer such that $\beta^n = \beta^{-5}$?
I manually found that $\beta^{-5} = \left(1\; 5 \;9 \;6 \;3 \;7 \;8\right)\left( 2 \;10 \;4\right)$. The nice thing is that since beta is composed of two disjoint cycles, so I know that $\beta^n = \left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)^n\left(2\; 4 \;10\right)^n$. Now I just tried to find an $n$ such that $\left(2\; 4 \;10\right)^n = \left( 2 \;10 \;4\right)$ after finding one I would check to see if $\left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)^n= \left(1\; 5 \;9 \;6 \;3 \;7 \;8\right)$. Clearly $n=2$ works and luckily $\left( 1 \;3 \;5 \;7 \;9 \;8 \;6\right)^2= \left(1\; 5 \;9 \;6 \;3 \;7 \;8\right)$. So I can conclude that $\beta^2 = \beta^{-5}$.

My question is how do I check is $n=2$ is indeed the smallest integer. All I showed so far is that $\beta^2 = \beta^{-5}$ and nothing about if 2 is the lowest value. One possibility is to consider the integers from -2 to 1 and try to show that those fail. But is there a better way to do a question like this.

答:
This is the smallest $n$ such that the order of $\beta$ divides $n+5$.

Since you have disjoint cycles of length $7$ and $3$, the order is $21$. Hence, $n = 16$.


问:
Coordinate Functions and Coordinate Bases
Let $M$ be a smooth manifold of n dimension. And let $(U,\varphi)$ and $(V,\psi)$ be two smooth charts on $M$.

Define the coordinate functions of $\varphi$ and $\psi$ as $(x^i)$ and $(\tilde{x}^i)$ respectively, i.e. $\forall p \in U\cap V$, $\varphi(p)=(x^1(p), ..., x^n(p)\,)$ and $\psi(p)=(\tilde{x}^1(p), ..., \tilde{x}^n(p)\,)$.

And then the author says that any tangent vector at $p \in U\cap V$ can be represented with respect to either basis $(\partial\,/ \, \partial x^i \vert _p)$ or $(\partial\,/ \, \partial x^i \vert _p)$. Later he also uses notation like $\frac{\partial}{\partial \widetilde{x}^j}$.

So I guess the author here identifies the coordinate bases with the coordinate functions. But how is such identification valid? I.e. Is there a one-to-one correspondence between the set of coordniate bases and the set of coordinate functions?

I'm really confused. Thanks for help.

Edit: What I really wanted to ask was that, if $f\in C^\infty[\varphi(U \cap V)]$, what does $\partial _{x^i} \vert _{\varphi(p)}f$ mean?

答: The set of tangent vector at $p$ noted $T_pM$ is a vector space of dimension $n$. You can have a lot of basis, and what is said is simply that you can use both basis to decompose any tangent vectors. Remember that $\partial_{x_i}$ is a notation: for $f$ smooth, we have $$\partial_{x_i}f=\partial_i(f\circ x^{-1})(x(p))$$ where $x=(x_i)$ is the diffeomorphism that have $x_i$ as coordinate functions.


问:
How many arrangements of red and blue balls are there so that, the number of red balls with: the ball immediately to the right is also red, is $9$.
The question is too long to fit in the title, but I tried.

$50$ balls: $23$ indistinguishable red balls; $27$ indistinguishable blue balls. The balls are arranged in a line. How many distinct arrangements of the balls are there so that the following property holds:

The number of red balls such that the ball immediately to the right of it is also red, is $9.\quad (1)$

?

Can I please get my following solution verified. Also, feel free to post alternative solution of course.

Let the number of red balls with property $(1)$ be $k$. If we start by placing all $23$ red balls next to each other, then $k=22.\ $ Now between each red ball, we imagine there is a bin. We don't yet have a bin to the left of the first red or a bin to the right of the last red. There are $22$ bins between the first and last red ball. Each time you put a blue ball into one of these bins, you reduce the number $k$ by $1$. So what we can first do is find the number of distinct ways to put $(22-9=)\ 13$ blue balls into $22$ bins so that no two balls are in the same bin. This is just $\binom{22}{13}.$

Now we have placed $13$ blue marbles into bins such that no two blue marbles are in the same bin, and we did this so that $k=9$, and so property $(1)$ holds. We now have $27-13=14$ blue marbles remaining to put into bins. Ah, but not $22$ bins because that would ruin property $(1)$, and not just $13$ bins: the ones that we have just put the blue balls into, but...

Now we introduce a bin to the left of the first red and another bin to the right of the last red. We put the remaining $(27-13=)\ 14 $ blue balls into $(1+13+1=)\ 15$ bins! How many ways are there of doing this? Well, it is a classic stars and bars problem, in particular see Theorem 2 on wikipedia. The number of ways of putting the remaining $14$ blue balls into $15$ bins is $\binom{14+15-1}{14} =\binom{28}{14}.$

I believe all possible arrangements can be done like this, and there are no repeats in this method. Therefore the final answer is:

$$ \binom{22}{13} \times \binom{28}{14} = 497420 \times 40116600 = 19954799172000. $$

答:
To summarize the discussion in the comments:

Here is a somewhat different way to do the count which confirms your calculation.

We distinguish between two cases, according to whether the rightmost ball is red or blue.

Case I (rightmost ball blue).  As there are $23-9=14$ reds which are not to the immediate left of another red we must have $14$ blocks of the form $R^{a_i}B$ where $a_i≥1$ and $\sum a_i=23$.  There are $\binom {22}{13}$ ways to form those blocks.  There are then $13$ unassigned blues and $15$ bins in which to place them, so there are $\binom {13+15-1}{15-1}=\binom {27}{14}$ ways to place the blues.  Thus case I contributes $$\binom {22}{13}\times \binom {27}{14}$$

Case II:  (rightmost ball red.)  Now we have $13$ blocks of the form $R^{b_i}B$ and a final block of the form $R^{b_{14}}$.  As before there are $\binom {22}{13}$ ways to form those blocks.  Now we have $14$ unassigned blues and $14$ bins in which to place them, and there are $\binom {14+14-1}{14-1}=\binom {27}{13}$ ways to do that.  Thus case II contributes $$\binom {22}{13}\times \binom {27}{13}$$

Finally we add to get $$\binom {22}{13}\times \binom {27}{14}+\binom {22}{13}\times \binom {27}{13}=\binom {22}{13}\times \binom {28}{14}$$

which confirms your result.


问:
How to desgin a fair raffle
I need to organize a raffle and it has to be fair.

Some people have a peice of land that lets say has 22 owners, and in that land they are planning to build 39 apartments (maybe 47 but that's not the point), so they have a contractor that will build it, and in order to decide which owner gets which apartment they want to have a raffle.

They wanted to number all the apartments from 1 to 39 and draw from a hat.

By the way they only get 22 or 24 apartments, the rest go to the contractor as his payment.

Problem is it's not fair, because the first to draw had a 1/39 chance the second has a 1/38 etc.

I taught about using a 2 phase raffle with random number pages.

Phase 1, we select the order in which people raffle using the random numbers, this phase is important in my opinion since not every apartment is the same there are 2 penthouses 2 garden apartments and other factors, so i order to keep everythibg fair, we also raffle the order of the raffle.

One important point about special apartments (penthouses and garden apartments) is that it may be agreed uopn in advance, that the contractor or an owner will get them WITHOUT a raffle, but that is ok since it was agreed upon.

After that is done using a w digit method we raffle for the apartments.

Is that fair enoguh? Is there a way to make a fully 100% fair raffle?

Thabks in advance

答: Your statement that the first has $\frac 1{39}$ chance to win and the second has $\frac 1{38}$ is not correct.  The second has $\frac 1{39}$ chance and the raffle is fair.  You are imagining that the first does not win, which improves the odds for the second one, but some of the time the first will win.  If you do the detailed calcultion you will find the second one has $\frac 1{39}$ chance to win.  Alternately, let each of the first two draw their numbers without looking.  Then have the second person look at his number first.  How can the order they look change the odds?

Title: How to desgin a fair raffle


问:
Does having real eigenvalues have a geometric meaning?
Is it possible to characterize the set of real matrices which have real eigenvalues geometrically? That is, is it possible to say that a linear map $T$ has real eigenvalues if and only if it satisfies some property ${\cal P}$ which has a geometric flavor?

I understand, of course, that different people will disagree on what ``geometric'' means. Still, I'm quite interested in seeing any possible answers. It seems plausible that properties of matrices which are independent of choice of basis have geometric interpretations.

Examples of the sort of thing I'm looking for:

A linear map $T$ has eigenvalues within the unit circle if and only if $\lim_k T^k x = 0$ for any $x$.

A linear map on $\mathbb{R}^n$ has all of its eigenvalues equal to zero if $T^n x = 0$ for any $x \in \mathbb{R}^n$.

A linear map $T$ has $1$ in its set of eigenvalues if and only if it has a fixed point.

A linear map $T$ will have a root of unity in its set of eigenvalues if and only if it has a periodic orbit.

答: $T^2$ embeds in a continuous 1-parameter real semigroup of transformations $(T^2)^\alpha $.  This is a dynamical interpretation, not purely geometric but close.</p>

ContentLicense: CC BY-SA 3.0


问:
Pull-back of differential forms
My professor gave me the definition:  Assuming $\phi : M \rightarrow N $ a smooth map between smooth manifolds M and N.

$\phi^*$  : $\Lambda(N) \rightarrow \Lambda(M)$ the pullback defined by:

($\phi^*\omega)_x$ ( $v^1,...,v^p$) = $\omega_{\phi(x)}$ ($T_x \phi v^1,...,T_x \phi v^p$)

Where I think the $\Lambda(N)$ and $\Lambda(M)$ are exterior algebra on manifold N and M.

So I knew the definition of the pullback defined on the dual spaces.

But there I don't see what does $T_x \phi v^1$ means?

答:
Here $T_x\phi$ probably denotes the differential of $\phi$ at $x$. Since it is applied to $v^i$ you should think of $v^i$ as a tangent vector to $M$ at $x$. I.e. we take $v^1,\ldots, v^n\in T_xM$ and $T_x\phi(v^i)\in T_{\phi(x)}N$.

Some people prefer this notation as it is more "functorial."

Title: Pull-back of differential forms


问:
General formula of partial sum of series (non-geometric)
Like user71317 in his question I am struggling to understand how we arrive at the general formula of the partial sums of a series. In my case the following series:

$$\sum_{n=2}^{\infty} \frac{1}{n^2-1}$$

The general formula of the partial sums is:

$$s_n=\sum_{i=2}^{n} \frac{1}{i^2-1} = \frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$$

I understand the answer to the linked question where the answer is the formula for the sum of a finite geometric series. Since my series is not geometric; how do I go from the series to the partial sums in this case?

Update:

How is $\sum_{i=2}^{n} \frac{1}{i^2-1}$ equal to $\frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$?

答:
This series can be converted into a telescoping series. Using partial fraction decomposition, we can see that
$$\frac{1}{n^2-1}=\frac{1}{(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$
Therefore,
$$\begin{align}
2\sum_{r=2}^n\frac{1}{r^2-1}&=\sum_{r=2}^n\left(\frac{1}{r-1}-\frac{1}{r+1}\right)\\
&=\frac{1}{1}-\frac{1}{3}\\
&~+\frac{1}{2}-\frac{1}{4}\\
&~+\frac{1}{3}-\frac{1}{5}\\
&~~~~~~~\vdots~~~~~~~\vdots\\
&~+\frac{1}{n-2}-\frac{1}{n}\\
&~+\frac{1}{n-1}-\frac{1}{n+1}\\
\end{align}$$
And all the terms will cancel apart from $4$ of them, so we are left with
$$2\sum_{r=2}^n\frac{1}{r^2-1}=1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}=\frac{3}{2}-\frac{1}{n}-\frac{1}{n+1}$$
which means that $$\sum_{r=2}^n\frac{1}{r^2-1}=\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$$

We can take the limit as $n\to\infty$ to find that
$$\sum_{r=1}^\infty\frac{1}{r^2-1}=\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{r^2-1}=\frac{3}{4}$$


问:
Can an irrational number raised to an irrational power be rational?
Can an irrational number raised to an irrational power be rational?

If it can be rational, how can one prove it?

答:
Let me expand orangeskid's answer, both because I think it teaches us something useful, and it might be the easiest elementary proof for this question.

The proof that $ \sqrt{2} $ is irrational is well-known, so I will not repeat it here.

But there's a proof just as simple showing that $ \log 3 / \log 2 $ is irrational.  Suppose on contrary that $ \log 3 / \log 2 = p / q $ where p and q are integers.  Since $ 0 < \log 3 / \log 2 $, we can choose $ p $ and $ q $ both as positive integers.  The equality then rearranges to $ 3^q = 2^p $.  But here, the left hand side is odd and the right hand side is even, so we get a contradiction.

That gives a positive answer to the original question:
$$
\big(\sqrt{2}\big)^{2 \log 3 / \log 2} = 3
$$

Thanks to Rand al'Thor, who mentioned this problem in SE chat and thus inspired this answer.


答:
This is somewhat abuse of notation for the pushforward. Let $\phi : M \to N$, then for a $k$-form $\omega$ in $\Lambda^k TN$, we have that,
$$(\phi^*\omega)_x(X_1,\dots, X_k) = \omega_{\phi(x)}(\mathrm d\phi_x(X_1), \dots,\mathrm d\phi_x(X_k))$$

which is a form taking vectors on $M$, where the pushforward is $\mathrm d\phi_x : T_xM \to T_{\phi(x)} N$. If one thinks of a tangent vector as a derivation on functions, then $\mathrm d\phi_x(X)(f) = X(f\circ\phi)$.

Another way to see the pushforward: choosing a chart $U$ around $x$ and $V$ around $\phi(x)$, the pushforward is a map (using the same notation) $\phi : U \to V$ and then,
$$\frac{\partial}{\partial u^i} \mapsto \frac{\partial \phi^j}{\partial u^i} \frac{\partial}{\partial v^j}.$$

In practice this is how you compute it hands on: take for example $\omega = uv^3 \mathrm du \wedge \mathrm v$ with $u,v$ coordinates on $\mathbb R^3$ and we have a map $\phi : \mathbb R^3 \to \mathbb R^2$ given by $(x,y,z) \mapsto (x^2+yz,e^{xyz}).$

Then we have $\phi^* \omega = (x^2+yz)(e^{xyz})^3 \mathrm d(x^2+yz) \wedge \mathrm d(e^{xyz})$, where $\mathrm d(x^2+yz)$ for example is interpreted in the usual way, when taking the exterior derivative of a degree zero form. You can expand this out and you'll get a form in $\mathrm dx, \mathrm dy, \mathrm dz$.

You should convince yourself from the formal definition that in practice it means just substituting in when you have a smooth map like this.

Title: Pull-back of differential forms

Body:
My professor gave me the definition:  Assuming $\phi : M \rightarrow N $ a smooth map between smooth manifolds M and N.

$\phi^*$  : $\Lambda(N) \rightarrow \Lambda(M)$ the pullback defined by:

($\phi^*\omega)_x$ ( $v^1,...,v^p$) = $\omega_{\phi(x)}$ ($T_x \phi v^1,...,T_x \phi v^p$)

Where I think the $\Lambda(N)$ and $\Lambda(M)$ are exterior algebra on manifold N and M.

So I knew the definition of the pullback defined on the dual spaces.

But there I don't see what does $T_x \phi v^1$ means?


问:
calculating the Taylor series remainder of $\ln(2)$
I'm trying to calculate the remainder for the Taylor series expansion of $\ln(2)$, and would greatly appreciate your support.

I've learnt that when expanding the Taylor series:

$f(x) = f(a) + \frac{f'(a)}{1!}(x-a)... + \frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+R_n(x)$

Where $R_n(x)$ is the remainder, s.t. $R_n(x)=\frac{f^{(n)}(a+\theta(x-a))}{n!}(x-a)^n$, given $0 <\theta<1$

My assumption was that using  the series $\ln(1+x)$ and replacing it by $2$ would guide me to a solution.

$\ln(1+x) = x-\frac{1}{2}x^2+\frac{1}{3}x^3...$

Though, that's not the right expansion. How would I correctly expand this, thanks for the help.

答:
$$\sum_{k=0}^{n-1}(-x)^k=\frac1{1+x}-\frac{(-x)^n}{1+x}$$

and after integration the exact remainder is

$$\int_0^x\frac{t^n}{1+t}dt$$ (sign omitted). This is an incomplete Beta integral.


问:
GNS constructions of two faithful positive linear functional
If $\rho_1$ and $\rho_2$ are two faithful positive linear functionals on a unital $C^*$-algebra $A$. Then $\rho_i(x)=\langle x\omega_i,\omega_i \rangle$,
where $\omega_i=1+N_{\rho_i}$,
$N_{\rho_i}=\{x\in A: \rho_i(x^*x)=0\}(i=1, 2).$

Since each $\rho_i$ is faithful, we have $N_{\rho_i}=0$. Then each cycic vector $\omega_i=1(i=1, 2),$ $\rho_1$ and $\rho_2$ have the same GNS construction.

Does the above conclusion correct?

答: No, even if $\rho_1$ and $\rho_2$ are states. For instance take $A=\text{UHF}(2^\infty)$ and take $\rho_1$ to be the trace and $\rho_2$ to be the Powers' state $\phi_\lambda$. Then $\pi(A)''$ is the hyperfinite II$_1$-factor, while $\rho_2(A)$ is a type III$_\lambda$ factor.


问:
Find the matrix of a Transformation $\Bbb R^{3×3} \to\Bbb R^8$
Denote by $MQ$ the set of 3×3 magic squares and let $MQ_a$ be the set of magic squared with magic sum $a$. The sets $MQ$ and $MQ_0$ are vector spaces over $\Bbb R$ with matrix addition and scalar multiplication.

a) find a linear map $T:\Bbb R^{3×3}\to\Bbb R^8$ such that $\ker(T)$ is precisely the vector space $MQ_0$

b) let $e_{ij}$ be the matrix that has $1$ at entry $ij$ and $0$ elsewhere. What is the matrix of $T$ in the ordered basis $(e_{11}, e_{12}, e_{13}, e_{21}, e_{22}, e_{23}, e_{31}, e_{32}, e_{33})$ for $\Bbb R^{3×3}$ and the standard basis $(e_1,\dots,e_8)$ in $\Bbb R^8$?

For a) I think the linear map take a 3x3 matrix and outputs an 8x1 matrix where each row is the sum of a row, column or diagonal of the original 3x3, of which there are 8 of. Therefore, the $\ker(T)$ should be the vector space of $MQ_0$, as all of the rows, columns and diagonals of the original 3x3 sum to 0
My input is a 3x3 matrix ([x1 x2 x3],[x4 x5 x6],[x7,x8,x9]) reading across the rows. My output is then ([x1+x2+x3},[x4+x5+x6],[x7+x8+x9],[x1+x4+x7],[x2+x5+x8],[x3+x6+x9],[x1+x5+x9],[x3+x5+x7])

For b) I have found a matrix to describe $T$ by writing the columns as $T(e_{11})$ and so on but this forms an 8x9 matrix. I don't see how this can then be multiplied by a 3x3 matrix to get the 8x1 matrix. Perhaps I should rewrite my 3x3 as a 9x1?

Any help is much appreciated thank you!!

答:
Basically a matrix must take as its input a column vector and output a column vector.

Suppose I have a linear transformation whose input is a matrix. Let's say

$$T \left( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \right) = \begin{pmatrix}  a + b \\ c + d \\ a + c \\ b + d \end{pmatrix} $$

I convert that into a matrix equation by picking an ordering on the elements of the matrix. Reasonable choices here are $a,b,c,d$ and $a,c,b,d$ (either reading the matrix by row or by column).

Once the ordering is fixed, we can write this as a matrix equation:

$$ \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix}  a + b \\ c + d \\ a + c \\ b + d \end{pmatrix}. $$

Your task is to do the same but for a $3\times 3$ matrix and with the ordering given to you: $e_{11}, e_{12}, e_{13}$, etc. It will also depend on how you order the equations in your transformation $T$.


问:
Which algorithm will find the path closest to zero?
So I have a graph and I want to find (having the weight of each edge - which is either negative or positive) the path that reach zero (or is closest to it) by adding each weight.

I need that to calculate which path in a function cleans up the stack fully.

Which algorithm should I use - I have found about the Bellman–Ford algorithm because it allows negative indices but I'm not sure if it'll do the job.

答:
The question of whether a path with total absolute weight $\leq k$ exists is NP-complete. That it is in NP is hopefully clear (addition is fast); to see that it is NP-complete, let us reduce subset-sum to your problem. Let us first do $k=0$ (i.e., does a weight-zero path exist) for simplicity.

Suppose that we have an instance of subset-sum with set $S = \{s_1, \ldots, s_m\}$ and potential sum $n$. Now build a graph with $|S| + 1$ vertices: between the $i$th and $(i+1)$th vertex add an edge with weight $s_i$, as well as an edge with weight 0. (You can use an additional vertex with two edges in case your problem does not allow for multiple edges between vertices.) Finally, add an additional target vertex, connected only to the final vertex by an edge with weight $-n$. Then a path from the first vertex to the target vertex of weight 0 is precisely a subset of $S$ that adds up to $n$.

You can reduce to the case for general $k$ by simply multiplying all these weights in the constructed graph by $k+1$.

Therefore you should not expect there to be a fast algorithm in general for the problem you have described.


问:
Solving system if equations containing trigonometric functions with Ti-Nspire
In trying to solve the following system of equation:

$20000\times9.81+a\cos b=0$

$a\sin b=6.17\times20000$

Find $a$ and $b$ .

It gives me something containing "n2" in bold and I don't know why?

$a=-231780.07/(-1)^{n2}$

$b=(180*(n2*pi-0.56143476))/pi$

How do I get the right answer? Thanks

答:
Compare with standard solution of polar /cartesian  coordinate relations and interpret the differences. Let

$$a/20000= \alpha, 9.81= g;$$

$$ \left(g +\alpha \cos b=0, -6.17+ \alpha \sin b=0 \right)$$

Eliminate b ;

$$ \alpha= \sqrt{g^2+ 6.17^2}; $$

Multiply this by 20000 to get a ;

and

$$ b=\sin^{-1}\dfrac{6.17 }{\alpha}.$$


问:
$\pm 1$ valued vectors in an arbitrary subspace.
In a paper I have read on random Bernoulli matrices it was stated that for any subspace $V\subseteq \mathbb{R}^n$ of dimension at most $l$ one has:
$$|V\cap \{\pm 1\}^n|\leq 2^l$$
With no proof. A reference was provided.
However, the claim does not appear there.

How should this be proven?

答: By an elementary duality argument, there is a subset $A$ of $\{1,\dots,n\}$ such that the natural projection $\Pi : \mathbb{R}^n \to \mathbb{R}^A$ (that selects coordinates indexed by $A$) induces an isomorphism between $V$ and $\mathbb{R}^A$ (in particular, the cardinal of $A$ is $\ell$). Since $\Pi$ sends $\{\pm 1\}^n \cap V$ into $\{\pm 1\}^{A}$, and is injective on $\{\pm 1\}^n \cap V$ (because it is injective on $V$), we see that the cardinal of $\{\pm 1 \}^n \cap V$ is less than $2^\ell$.


问:
Causality in Vaccine versus Placebo Group
I am having a debate with some people on statistical significance of a certain event.

It is based on a real world situation, namely the AstraZeneca vaccine and the incidence of Transverse Myelitis.

As far as I know the trial invovled two groups with about 25,000 people in each group. (50,000 people altogether).

In the vaccine group there were two incidences of Transverse Myelitis, and in the placebo group there was one incidence.

Without going into the maths I can clearly see that there is no statistical significance of Transverse Myelitis in these two groups. The numbers are just too small.

Can anyone give me a basic mathematical breakdown of this situation? I would just like to back up my suspicions with some cold hard maths.

Thank you.

答:
Since we don't know what's the chance of catching TM is, we will use the placebo group as reference.  That is, the chance of catching it (within the duration of the study) is 1 in 25k.

Now imagine you choose from blind 25k people from a population where 1 in 25k has TM.  Then the possibility of picking up 2 or more TM cases is approximately $1-2e^{-1}\approx 26\%$ (the exact number is $1-(\frac{24999}{25000})^{25000}-(\frac{24999}{25000})^{24999}=0.264241\dots$).  Note this is not saying 26% chance of the vaccine does not cause TM or 74% chance of taking the vaccine will cause TM (analogy: the chance of getting a six on a fair die is 1/6, the fact that you rolled once and get a six doesn't mean the probability of this six happening not from randomness but from magic is 5/6).

That is, not counting publication bias.  If the vaccine group also had 1 case then there was nothing for the antivaxxer to winne about, and if there are none it also wouldn't make the news (nobody sane would use it to claim the vaccine also vaccinate against TM).


问:
Explanation to statement "If $f,g$ are uniformly continuous on $S$,then $fg$ may not be uniformly continuous on $S$"
If $f,g$ is uniformly continuous on some domain $S$,then $fg$ still uniformly continuous. I think i could find a counter example but when i try to explain why the statement fail by considering some property the uniform continuous functions have, it seems that i don't quite see(or don't quite get the big picture) why they fail. So my question is can it be explained by considering the property of the uniform continuity to explain the statement above

The definition of uniform continuity I am studying: $ \forall \epsilon>0, \exists \delta>0 s.t.\forall x,y \in S,d(x,y)<\delta \implies d(f(x),f(y))<\epsilon$

答:
Well, look at the function $f(x) = x$ over the whole real line. It is uniformly continuous but its product with itself is not. Uniform continuity is about how the function changes on the domain with respect to the independent variable. If the change is too steep, like in $f(x) = x^{2}$, you are not likely to get uniform continuity. If the change is reasonably slow, like in $f(x) = x$ or $f(x) = \sqrt{x}$ for example, you are likely to get uniform continuity.
Thus, while $f$ and $g$ themselves may change "reasonably" slowly, their product may change faster and hence may not be uniformly continuous.</p>


问:
Localization with respect to set that is NOT mutliplicatively closed
All rings are assumed to be commutative with unity. Usually denoted $R$

Intro: I understand that there are two definitions of localization (or possibly more, but these two are somewhat canonical for me):

1 Defining ring of fractions with elements from multiplicative $S$ as denominators.

2 For any set $S$ by universal property of localization: $\alpha \in Hom(R, S^{-1}R)$ such that a)$\alpha(s)$ is invertible and b)If $\beta \in Hom(R, T)$ and $\beta(s)$ invertible in $T$ then there is $\gamma \in Hom(S^{-1}R, T)$ such that these ring homomorphisms commute.

I understand the "fractions definition" and from this, I built some understanding of Categorical definition. We also had a proposition in my Alg. Geometry class showing that rings of fractions from the first definition (together with homomorphism $r->r/1$) are basically the same thing as  in the categorical definition.

Question However, the categorical definition needs not the assumption of $S$ being multiplicatively closed. How does a (categorical) localization with respect to $S$ not multiplicatively closed looks like?

My thoughts Of course the definition does not say that such localization exists. But what looks more natural to me is take $S'\supset S$ the smallest multiplicatively closed set containing $S$ and consider localization with respect to $S'$. I suppose that all elements from $S'$ would be invertible also in $S^{-1}R$ so maybe using universal property of localization should work?

答: That's correct: the localization with respect to $S$ is just the same as the localization with respect to the multiplicatively closed set $S'$ that $S$ generates.  This is because if $\alpha:R\to T$ is any ring homomorphism, then the set of $s$ such that $\alpha(s)$ is invertible is always multiplicatively closed, so it contains $S$ iff it contains $S'$.  So, the universal properties of localization with respect to $S$ and localization with respect to $S'$ are equivalent.


问:
A certain function expansion around x=0 with integral
given $f(x)= $$\int_1^x \frac{t^2} {5+t^2} \, dt $ expand the function around $x=0$
What I tried was:

I took $\frac{t^2} {5+t^2}=g(t)$ and then $\frac {1}{t^2}$*$\frac{t^2} {(\frac{5}{t^2}+1)}$ so we get $\frac {1}{(\frac{5}{t^2}+1)}$ after that I took $\frac{5}{t^2}$ and identified it by x, $\frac{5}{t^2}=x$

now it seems kind of better for the geometric expansion ${\sum_{k=0}^\infty x^k}$=$\frac {1}{1-x}$

in my case I did $\frac {1}{1-(-x)}$ because I have $\frac {1}{1+x}$ and therefore I got ${\sum_{k=0}^\infty (-x)^k}$=$\frac {1}{1-(-x)}$
after that I return the $\frac{5}{t^2}=x$
and we get  ${\sum_{k=0}^\infty (-\frac{5}{t^2})^k}$=$\frac {1}{(\frac{5}{t^2}+1)}$.

I could not continue from here as I got too lost with it.. the final answer in the book is  ${\sum_{k=0}^\infty \frac{(-1)^k}{5^{k+1}(2k+3)}*x^{2k+3}}$ but there is no explanation just a final answer which doesn't help much.

appreciate any help!

答:
Note that you have $$ f'(x) = \frac{x^2}{5+x^2} = \frac{x^2}{5}\frac{1}{1+\frac{x^2}{5}} = \frac{x^2}{5}\sum_{k=0}^\infty \left(\frac{-x^2}{5}\right)^k = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+2}}{5^{k+1}}$$
Then you just need to integrate it back
$$ f(x) = f(0) + \int_0^x f'(t)dt$$
Since $\int_0^x t^{2k+2} = \frac{x^{2k+3}}{2k+3}$ you'll get the result, up to the term $f(0)$, which I believe may be missing in the answer your book is giving.


问:
Why is the endomorphism ring of $\mathbb{Z}\times\mathbb{Z}$ noncommutative?
So I hear that the endomorphism ring of an abelian group is not always commutative. In particular, I'm looking at the abelian group $A=\mathbb{Z}\times\mathbb{Z}$, and considering $\text{End } A$. I can't find a counterexample to show that $\text{End } A$ is not commutative. Does anyone know of two such endomorphisms that don't commute?</p>

答:
Hint: try to show that the ring $M_2(\mathbb{Z})$ of $2 \times 2$ matrices with $\mathbb{Z}$-coefficients acts <s>effectively</s> faithfully on $\mathbb{Z}^2$ by endomorphisms.  This is a non-commutative ring, so this will answer your question.

More generally, if $A$ is an abelian group and $n \in \mathbb{Z}^+$, then one can show that $\operatorname{End}(A^n) = M_n(\operatorname{End}(A))$.  (It happens that the endomorphism ring of the group $\mathbb{Z}$ is canonically isomorphic to the ring $\mathbb{Z}$, so the previous paragraph is a special case of this.)  This gives many examples of non-commutative endomorphism rings.


问:
Determining Which Areas Under Curves Are Equal
I have the following question from my calculus 2 homework and I'm really stumped:

Three regions have areas $A$, $B$, $C$

The grey area $A$ is the area of a region bounded by the graph of the function $f(x) = \ln x$ and the $x$-axis over the interval $x \in[1, e]$.

The green area $B$ is the area of a region bounded by the graph of the function $f(x) = e^{\sin x}\sin 2x$ and the $x$-axis over the interval $x \in [0, \pi/2]$.

The orange area $C$ is the area of a region bounded by the graph of the function $f(x) = xe^x$ and the $x$-axis over the interval $x \in [0, 1]$.

Two of the three areas $A$, $B$, and $C$ are actually equal. Find out which areas are equal. Do not forget to write down the correct mathematical solution to this problem!

So far, all I've been able to do is setup the areas as integrals:

$A = \int_1^e\ln x\ dx$

$B = \int_0^{\pi/2}e^{\sin x}\sin2x\ dx$

$C = \int_0^1xe^x dx$

I'm thinking that in order to show a mathematical solution to the question, I have to solve or at least simplify the integrals. Here's where the problem lies.

I have no idea how to solve these. For example, for the area of $A$, it seems that I have to use integration by parts to solve it but we haven't been taught how to do that yet (I know for sure that we are not expected to read ahead). So far, we've only done basic substitution. So now I wonder, is there another way to solve these without using integration by parts?

答:
Hint:

Try making the substitution $u=\ln x$ in $A$, and consider the form of the resulting definite integral.


问:
Example of a striclty convex real valued function that's discontinuous
So I've been learning about convex functions, and I know it is the case that if $f$ is diffrentiable and $f'$ is increasing then $f$ is convex.

However is it possible for a strictly convex function to be discontinuous or even just non diffrentiable on an interval?

In the case of convex functions, I can think of the example of the function made from choosing various points on a convex function graph and "joining them up" with straight lines, which would be non-diffrentiable at any of the "links" (though it would still be continuous)

And for strictly convex functions I can think of the case of the lower unit semicircle, which is not differentiable at the end points of the interval.

However I would like to know of a less trivial counterexample for strictly convex functions (since the unit semicircle only loses differentiability at the boundary)

Also I would like to know of a strictly convex (or even just convex) function that is discontinuous (again not including the trivial case of at the endpoint of the interval)

答:
What about $f$ defined on $[-1,1]$ as

$$f(x)=\begin{cases}
1 & x = -1\\
x^2 - 1 & -1 \lt x \lt 1\\
1 & x = 1
\end{cases}$$

Note: a convex map is always continuous on an open interval.


问:
proof check: $f$ maps positive integers to positive integers. $f(m)f(n)=f(mn)$ and $m+n$ divides $f(m)+f(n)$
Find all possible functions $f$ such that $f$ maps positive integers to positive integers, $f(m)f(n)=f(mn)$ and $m+n$ divides $f(m)+f(n)$.

Can we safely say that $f(m)=m^k \cdot a$ (where $a$ is a constant number)? pOne of my students is arguing with me and he says we can.

答:
This is a partial answer,

$f(1)=f(1)^2$, so $f(1)=1$, since $f:\Bbb Z^+\rightarrow\Bbb Z^+$.

Since $m+n\mid f(m)+f(n)$ let $m=n$, then $m\mid f(m)$. We can factor $m=\prod_{i\in\Bbb N}p_i^{a_i}$ so that $$f(m)=\prod_{i\in\Bbb N}f(p_i)^{a_i}$$ is determined by where $f$ maps the prime numbers. We have $f(p_i)=p_ic_i$, where $c_i$ is some constant. Since $p_i+1\mid f(p_i)+1$ we must have $$\frac{p_ic_i+1}{p_i+1}\in\Bbb Z$$ which happens if and only if $c_i=1+k_i(p_i+1)$ for some constant $k_i$.

$f(m)=m^ka$ works if $a=1$, since $f(1)=1$, but fails for every even $k$ in general, since $$2m+m\not\mid 2^km^k+m^k$$ if $3\not\mid m$.


问:
How can I prove that $4^n-1$ is divisible by 3?
I was faced with the hypothesis that $4^n-1$ is always divisible by 3. I believe that this problem can be solved by a proof by induction.

As far as I understand, a proof by induction works by first testing whether or not this is true for a small value of n (e.g. 0, 1, etc.). Then we assume that it is true for an arbitrary number $n=k$, and prove how it is also true for $n=k+1$.

My first step

At first, I tried to prove it for a small number ($n=1$).

$$4^1-1 = 3$$

3 is clearly divisible by 3, so the first step should hold true. However, I am completely stuck now when trying to prove that the same holds true for $4^{k}-1$ and $4^{k+1}-1$.

答:
We have:

$$4^{n}-1 = (4-1)(4^{n-1}+4^{n-2}+\dots + 4 + 1) = 3(4^{n-1}+4^{n-2}+\dots + 4 + 1)$$

Which is clearly $\boxed{\text{divisible by }3.}$

Or, a proof by induction:

Note that $4^{1}-1 = 3$. Now, assume that the statement is true for $n=k$. Then, we have:

$$4^{k+1}-1 = 4(4^{k}-1)+3$$

However, the RHS is a multiple of $3$ plus $3$, so it must be divisible by $3$. Thus, the statement is true for $n = k+1$ if it is true for $n=k$, completing the proof by induction.


Title: How can I prove that $4^n-1$ is divisible by 3?

Body:
I was faced with the hypothesis that $4^n-1$ is always divisible by 3. I believe that this problem can be solved by a proof by induction.

As far as I understand, a proof by induction works by first testing whether or not this is true for a small value of n (e.g. 0, 1, etc.). Then we assume that it is true for an arbitrary number $n=k$, and prove how it is also true for $n=k+1$.

My first step

At first, I tried to prove it for a small number ($n=1$).

$$4^1-1 = 3$$

3 is clearly divisible by 3, so the first step should hold true. However, I am completely stuck now when trying to prove that the same holds true for $4^{k}-1$ and $4^{k+1}-1$.

Body: Hint: $4^{k+1}-1=4\cdot 4^k-1=3\cdot 4^k+(4^k-1)$.


答: As $$4 \equiv 1 [3]$$ you get by induction $$4^n \equiv 1 [3]$$ for all $n \in \mathbb N$ and therefore the desired result.


问:
$n^5-n$ is divisible by $10$?
I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!</p>

答:
Without modular arithmetic? How about induction?

Base case: it is true for $n = 0$ since $0^5 - 0 = 0$. It is also true for $-1$ and $1$ since $-1^5 + 1 = 0$, and $1^5 - 1 = 0$. And it is true for $-2$ and $2$: $-2^5 + 2 = -30$ and $2^5 - 2 = 30$.

Inductive hypothesis: if it is true for $k$, it can be shown to be true for $k + 1$ or vice versa. We can work it from $k + 1$ to $k$, avoiding any "trap door" steps, so that all derivation works both ways.

$$(k + 1)^5 - (k + 1) = 10q$$

$$k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - (k + 1) = 10q$$

Rearrange terms, cancel 1 and -1:

$$k^5 - k + 5k^4 + 10k^3 + 10k^2 + 5k = 10q$$

Isolate $k^5 - k$:

$$k^5 - k  = 5k^4 + 10k^3 + 10k^2 + 5k + 10q$$

Now we need to show that the right hand side is divisible by ten. We can do this as follows. First, rearrange some terms:

$$k^5 - k  = 5k^4 + 10k^2 + 5k + 10k^3 + 10q$$

Now note that $10k^3$ and $10q$ are divisible by 10, the latter having come from our inductive hypothesis. So let us focus on the remaining terms, which comprise this formula:

$$5k^4 + 10k^2 + 5k$$

We can show that this is divisible by 10 by factoring out $5k$:

$$5k(k^2 + 2k + 1)$$

$$5k(k + 1)(k + 1)$$

But $k(k + 1)(k + 1)$ is an even number, which, multiplied by 5 is divisible by 10.   To show that $k(k + 1)(k + 1)$ we divide into cases. If we suppose that $k$ is odd, then we have odd x even x even, which is even. If we suppose that $k$ is even, then we have even x odd x odd, which is even again.

So the inductive hypothesis is true. If $(k + 1)^5 - (k + 1)$ is  divisible by $10$, then so is $k^5 - k$, and vice versa. By induction from the base case in the positive and negative directions, it is true for all $k \in \mathbb{Z}$.

Modular arithmetic wasn't used, but one basic argument which was used is linked to modular arithmetic. Namely, the argument that some $N$ {is/isn't} is divisible by $10$, then $N + 10k (k \in \mathbb{Z})$ likewise {is/isn't} divisible by $10$. This is equivalent to the modular concept that $N$ is congruent to $N + 10k$ modulo $10$, but without the formal trappings. Furthermore the argument about the evenness of $k(k + 1)(k + 1)$ also relies on congruences in disguise. Division into even/odd cases is nothing more than division into the two symbols of the mod 2 congruence.
We cannot really even discuss divisibility without invoking ties to congruences. Divisibility by 10 means congruence to 0 mod 10.

Title: $n^5-n$ is divisible by $10$?


问:
Formal argument for why $(\mathbb{Z}/(4))_{(2)} = \mathbb{Z}/(4)$
I'm learning about localizations, and I came across this statement:
$$ \left(\mathbb{Z}/(4)\right)_{(2)} = \mathbb{Z}/(4) $$
Now, this statement makes sense because inverting all of the odd elements have no effect since the images of odd elements in $\mathbb{Z}/(4)$ are all invertible. But I'm struggling to write this down formally, what exactly can I write down to convince myself of this fact beyond a reasonable doubt?

答: The complement of $(2)$ is the set of all units of $\mathbb{Z}/(4)$.


问:
Show that $a = \sup \{ a_n | n \in \mathbb{N} \} = \limsup\limits_{n \to \infty} a_n$

Let $a \in \mathbb{R}$ be an accumulation value of the real sequence $(a_n)_{n \in \mathbb{N}}$ and an upper bound of the set $\{a_n | n \in \mathbb{N}\}$. Show that
$$a=\sup\{ a_n | n \in \mathbb{N} \} =  \limsup_{n \to \infty} a_n$$

I think I understand the equation but I do not know how to prove it in a rigorous way.

答:
Since $a$ is an upper bound of $\{a_n\}_{n=1}^\infty$, therefore
\begin{equation}
L\le a\tag1\label{1}
\end{equation}
where $\displaystyle L:=\limsup_{n\to\infty}a_n$ (we know that limsup always exists, finitely or infinitely). Also, since $a$ is an accumulation point, for any $\varepsilon>0$, $a-\varepsilon$ has infinite number of points of the sequence above it. Hence,
\begin{align}
&a-\varepsilon\le L\qquad\forall\varepsilon>0\\
\implies &a\le L\tag2\label{2}
\end{align}
From \eqref{1} and \eqref{2}, we get the desired result.


问:
Question about refining finite cover of a compact k in a TVS E by translates of neighboroods of 0
In a TVS E with countable basis (U_n) of open neighborood of 0 , if K is a compact K of E, it is always possible to find fot each integer n a finite set T_n such that K is covered by the open sets t + U_n where t is an element of T_n.

I am asking , with the supplementary conditions that the U_n are balanced and decreasing for inclusion and such that, for each integer n, U_(n+1) + U_(n+1) is contained in U_n,

if we can impose for each integer n, T_(n+1) is contained in the set  T_n + U_(n+1) (the sum of the sets T_n and U_n). This is not evident for me, so I need a prooof of this fact if there is one because I found this implicitely affirmed in a proof of the following lemma:

let f : E -> F linear continuous and surjective (E and F TVS being metrisable and complete) and K a compact subset of F, then there is a compact subset L of E such that f(L) = K

答: The answer is very simple : for each n, the compact K is covered by a finite number of open sets of the form t+ U_(n+1) where t belongs to the finite set T_n . Then it is obvious that we have, for each n, T_(n+1) contained in T_n + U_(n+1).</p>


问:
Composition of distance and continuous function
Let (M, $d$) a metric space and $h$ a continuous function $h : \mathbb R_+ \to \mathbb R_+$.
We suppose that :

(i) $h(0) = 0$ ;

(ii) $h$ is differentiable on $]0, +\infty[$ ;

(iii) $h'(x)> 0$ on $]0, +\infty[$ ;

(iv) and $h'$ is decreasing on $]0, +\infty[$.

Show that $f = h ∘ d$ is also a distance on M.

I struggle to show the triangular inequality $f(x,y) \leq f(x,z) + f(z,y)$. I was told to use mean value theorem.

I have a few attempts : suppose we have $y \leq x$, we would have a $c \in ]y,x[$ such that $h(x)-h(y)=h'(c)(x-y)\leq h'(y)(x-y)$, we also have an $u \in ]0,y[$ such that $h(y)-h(0)=h(y)=h'(u)y$. (I think we want to show that $h(x+y) \leq h(x) + h(y)$).

答: We know that $d(x,y) \leq d(x,z) + d(z,y)$ because $d$ is a distance. $h'$ is decreasing so h is concave which means $h(d(x,y)) \leq h(d(x,z) + d(z,y)) \leq h(d(x,z)) + h(d(z,y))$


问:
calculation $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}$.
calculation$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}.$$
I try to use $|\sin x-x|\leqslant\frac{x^2}{2}$, but I don't feel like it's going to work. Is there any other way?

答:
This is essentially a Riemann sum of the function $$f(x) = \frac{\sin \pi x}{x}$$ over the unit interval $[0,1]$.
The partition seems to be $x_k = k/(n+1)$ for $0 \le k \le n+1$ but only $n$ points are being sampled ($x_1$ through $x_n$) instead of $n+1$. But the difference is negligible, and we have
$$ \sum_{k=1}^n f(x_k) \cdot \Delta x_k = \sum_{k=1}^n \frac{n+1}{k} \sin\left(\frac{\pi k}{n+1}\right) \cdot \frac{1}{n+1} \overset{n\to\infty}{\longrightarrow} \int_0^1 \frac{\sin \pi x}{x} \, dx = \mathrm{Si}(\pi) = 1.851\ldots$$


问:
Given f(x) in k[x], is it possible to find two field extensions K/k and K'/k such that f has two different factorizations as linear polynomials?
By Kronecker's theorem, given a field $k$ and $f(x)$ a polynomial with coefficients in $k$, there exists a field $K$ containing $k$ as a subfield and with $f(x)$ a product of linear polynomials in $K[x]$;
\begin{equation*}
f(x)=a(x-\alpha_{1})\cdots (x-\alpha_{n})
\end{equation*}
with $\alpha_{i} \in K$ for all $i$ and $a \in k$
(Rotman, Advanced Modern Algebra, p. 191).

Since $K$ is a field, $\alpha  \in K$ is a root of $f$ (i.e. $f(\alpha)=0$) iff the polynomial $x- \alpha$ divides the polynomial $f(x)$ (in $K[x]$), then the above can be restated as "$K$ contains all the roots of $f$".

My question is if it can happen to exist a second field extension $K'/k$ such that $f$ has a factorization in $K'[x]$ of the form
\begin{equation*}
f(x)=b(x-\beta_{1})\cdots (x-\beta_{n})
\end{equation*}
with $\beta_{j} \in K'$ for all $j$ and $b \in k$, with at least one $\beta_{j} \neq \alpha_{i} $ for some $i,j$.

if I'm not mistaken it would mean then that the polynomial $f$ has two different collections of $n$ roots.

[Obviously if $K \subseteq K'$ such thing cannot happen, since it would imply $f$ to be a polynomial with coefficients in $K´$ of degree $n$ with more than $n$ roots].

I started recently to study field extensions then I don't know if this is a silly question or not, sorry in advance if it is :-)

答:
You can have $K$ and $K'$ constructed as distinct sets, and thus in the strictest sense no element of $K$ is equal to any element of $K'$.

However, the smallest subfield of $K$ that contains all the roots of $f$ (i.e. $k(\alpha_1,\ldots,\alpha_n)$) is necessarily isomorphic to the smallest subfield of $K'$ that contains all the roots of $f$ (i.e. $k(\beta_1,\ldots,\beta_n)$). And there is an isomorphism that sends each $\alpha_i$ to some $\beta_j$.

So up to isomorphism, there is only one splitting field of $f$.


答:
Be careful when talking about whether two things are equal in different extensions. Of course, they might just be the same thing with different names so they are not strictly equal.

However, you can take the subfield generated by the roots. This is known as the splitting field and is unique up to isomorphism. See Theorem 3.13 (https://courses.maths.ox.ac.uk/node/view_material/52996)

Title: Given f(x) in k[x], is it possible to find two field extensions K/k and K'/k such that f has two different factorizations as linear polynomials?

Body:
By Kronecker's theorem, given a field $k$ and $f(x)$ a polynomial with coefficients in $k$, there exists a field $K$ containing $k$ as a subfield and with $f(x)$ a product of linear polynomials in $K[x]$;
\begin{equation*}
f(x)=a(x-\alpha_{1})\cdots (x-\alpha_{n})
\end{equation*}
with $\alpha_{i} \in K$ for all $i$ and $a \in k$
(Rotman, Advanced Modern Algebra, p. 191).

Since $K$ is a field, $\alpha  \in K$ is a root of $f$ (i.e. $f(\alpha)=0$) iff the polynomial $x- \alpha$ divides the polynomial $f(x)$ (in $K[x]$), then the above can be restated as "$K$ contains all the roots of $f$".

My question is if it can happen to exist a second field extension $K'/k$ such that $f$ has a factorization in $K'[x]$ of the form
\begin{equation*}
f(x)=b(x-\beta_{1})\cdots (x-\beta_{n})
\end{equation*}
with $\beta_{j} \in K'$ for all $j$ and $b \in k$, with at least one $\beta_{j} \neq \alpha_{i} $ for some $i,j$.

if I'm not mistaken it would mean then that the polynomial $f$ has two different collections of $n$ roots.

[Obviously if $K \subseteq K'$ such thing cannot happen, since it would imply $f$ to be a polynomial with coefficients in $K´$ of degree $n$ with more than $n$ roots].

I started recently to study field extensions then I don't know if this is a silly question or not, sorry in advance if it is :-)


问:
How to solve an optimisation equation with unknown parameters?
Given an example equation:

$$ z = Mx + Ny $$

where $M$, $N$ are unknown parameters and $x, y, z$ are features of a dataset.

My initial guess is to use gradient descent and the least squares error to obtain $M$ and $N$ from the dataset.

After that, we construct the inequality equations and apply these with the new equation (known $M$ and $N$) to a Lagrange multiplier to minimise $z$.

Is this a correct approach to the problem?

答:
As I understood, you have a data set $D=\{x_i,y_i,z_i\},\ \ i = 1,2\cdots,n$

so you can solve

$$
\min_{M,N}\sum_{i=1}^n\left(M x_i+N y_i-z_i\right)^2
$$

a typical least squares problem.


问:
$S/I$ is an integral extension of $R/R\cap I$
I have some probelems with the following statement: Let $S, R$ be rings. Suppose $S$ is an integral extension of $R$ and $I$ is an ideal in $S$ then $S/I$ is an integral extension of $R/(R\cap I)$.

These are my thoughts so far.
Let $i$ be embeddings and $\pi$ natural projections the following diagram commutes (right?)
\begin{matrix}
R & \overset{\pi _1}{\rightarrow } & R/(I\cap R)\\
i\downarrow &  & i\downarrow\\
 S& \overset{\pi _2}\rightarrow & S/I
\end{matrix}
For any $\overline{s}\in S/I$, there exists a monic polynomial $p(x)\in S[x]$ with $p(s)=0$, clearly $\pi _1 (p(s))=0$ in $R/(I\cap R)$ and $\pi _1(p(x))$ is still monic. Furthermore $\pi _2(i(p(x)))=i(\pi _1(p(x)))$ since $p(x)$ has coefficents in $R$.

I feel like one have to show that if $\overline{t}= \overline{s}$ then $p(t)\in R$.

If $p(t)\in R$ then $p(t)-p(s)\in (I\cap R)$ and hence $t$ is also a root of $\pi _1(p(x))\in R/(I\cap R)[x]$. Is this not needed? How can one explicit argue.

答:
Let $\bar{s} \in S/I$ with $s \in S$. By hypothesis, $s$ is integral over $R$, hence there are $a_i \in R$ such that $s^n+a_{n-1}s^{n-1}+\cdots+a_1 s + a_0=0$. Now this equation modulo $I$ becomes $\bar{s}^n+\bar{a}_{n-1}\bar{s}^{n-1}+\cdots+\bar{a}_1 \bar{s} + \bar{a}_0=0$ where we view the coefficients $\bar{a}_i$ as elements of the $R$-module $(R+I)/I$. Now by the third isomorphism theorem we have $(R+I)/I \cong R/(R\cap I)$ and we are done.

Alternatively and more consistently with your notation, let $\pi:S \rightarrow S/I$ be the natural projection. For $s \in S$ there exists monic $p(x) \in R[x]$ such that $p(s)=0$. Now $p(s)$ is an element of $S$ and its image under $\pi$ is $p^{\pi}(\pi(s))=0$, where $p^{\pi}$ denotes the monic polynomial that we obtain from $p(x)$ be applying $\pi$ to its coefficients. Now use the third isomorphism theorem to see that actually $p^{\pi}(x) \in R/(R\cap I)[x]$.


答: In the join of $K''$ of $K$ with $K'$ we'd have two distinct factorisations of the same polynomial, a contradiction since $K''$[x] is a UFD. You can have two distinct factorisations in irreducible non-linear polynomials though, for intermediate fields, coming from distinct "rearrangements" of the linear factors.


问:
Solve ODE with log and absolute value
Let $f(u)=-u\ln(\lvert u \rvert)$ if $u \neq 0$ and $f(0)=0$. Solve:

$$
\begin{cases}
u'(t)=f(u(t)), t \in \mathbb{R} \\
u(0)=u_0
\end{cases}
$$

For $u > 0$, I found:

$$u(t)=\exp(\exp(-t+\ln(\ln(u_0)))$$

For $u < 0$, I found:

$$u(t)=\exp(\exp(-t+\ln(\ln(-u_0))$$

All in all, for $u_0 \neq 0$, $$u(t)=\exp(\exp(-t+\ln(\ln(\rvert u_0\lvert))$$

Does it look correct to you? I am always troubled by absolute values in ODE. Furthermore, I don't really know what happens when $t=0$ or $u=0$.

答:
To accommodate both positive and negative values of $u_0$ the solution should be in the form

$$ u(t)=\frac{u_0}{|u_0|}\exp(\exp(-t+\ln(\ln(|u_0|))) $$

Graphing the function illustrates that $u_0\in(-\infty,-1)\cup(1,\infty)$ so there is no concern for what happens when $u_0=0$.

Furthermore, letting $t=0$ gives, simply, $u(0)=u_0$ which is as it should b.

Title: Solve ODE with log and absolute value


问:
Simple proof that there exist that you can obtain difference of 1 between multiples of coprimes $a$ and $b$
Given two coprimes $a$ and $b$ (assume wlog that $a < b$), there are non-negative integers $n_a$ and $n_b$ such that $n_b \cdot b = n_a \cdot a + 1$. Easy to prove using Bézout's identity, but is there a simpler, more "intuitive" way of proving this?

Another way of putting it is that there exists $n \in \mathbb{N}$ such that the remainder of the Euclidean division of $nb$ by $a$ is $1$.

答: $\gcd(a,b)=1$ implies $\bar b\in\mathbb Z/a\mathbb Z$ is invertible, so there exists $n_b\in\mathbb Z$ such that $n_bb\equiv1\pmod a$. Hence there exists $n_a\in\mathbb Z$ such that $n_aa=n_bb-1$, i.e. $n_bb=n_aa+1$.


问:
Angle between lines joining tetrahedron center to vertices
What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?

I'm trying to make these:

I need to know what angles to bend the metal.

答:
Another possible solution:

Consider the equilateral triangle, edge length x, formed by any 3 tips of a regular tetrapod (a face on it's corresponding tetrahedron).

The 3 medians of this equilateral triangle section it into 6 congruent right triangles with hypotenuse length y, and angles of 60° between lines around the centroid.

sin60° = (1/2x) / y

therefore

y = x / 2sin60°

Now consider a line running orthogonal to the previously identified triangle, from its centroid to the tip of the opposing tetrapod leg. This line is coincident with that opposing tetrapod leg and forms a right triangle with the previously identified y, having a hypotenuse equal in length to x.

Back to our tetrapod, any two legs form an isosceles triangle with base length x. Therefore, the angle between any two legs, A, can be expressed:

A = 180° - 2

where

 = arcsin (y / x)

 = arcsin ([x / 2sin60°] / x)

 = arcsin (1 / 2sin60°)

therefore

A = 180° - 2[arcsin (1 / 2sin60°)]

A ≈ 109.47°

A more spatial approach, and certainly not the most direct, but it got me out of a pinch during a grade 12 chemistry test years ago. We were supposed to have the angles of molecular structures memorised and of course I blanked mid-test so I found myself deriving something to this effect in the margins.

This was on mobile so please forgive the formatting and lack of diagrams which surely would have made this explanation easier to follow.


问:
What is "group of graph"?
I'm reading some old article and I have one small question: what in general is the group of a graph? By the article, definition should be in Harary's Graph Theory, but unfortunately I don't have any access to that book.

答:
The group of $G$ just means the automorphism group of $G$, i.e. the collection of all isomorphisms of $G$ with itself, the group operation being composition.

Here is the relevant section from Harary's Graph Theory:

Title: What is "group of graph"?


问:
Minimal polynomial of $\alpha^3 - 14 \alpha$ over $\mathbb{Q}$
I am trying to solve a problem that goes as follows.

Show that the polynomial $f(X) = X^4 - 16X^2 + 4$ is irreducible over $\mathbb{Q}$. Let $\alpha$ be a root of $f(X)$ in some field extension of $\mathbb{Q}$. Determine the minimal polynomials of $\alpha^2$ and of $\alpha^3 - 14 \alpha$ over $\mathbb{Q}$.

I have shown that $f(X)$ is irreducible over $\mathbb{Q}$ and found the minimal polynomial of $\alpha^2$. I am now stuck on the very last part of this problem, to find the minimal polynomial of $\alpha^3 - 14\alpha$. I know that $0 = \alpha^4 - 16 \alpha^2 + 4$ but I can not proceed from here. To find the minimal polynomial of $\alpha^2$ was easy enough, just set $X := \alpha^2$ in the equation and simplify. However, with $\alpha^3 - 14\alpha$ I can't seem to find a way to use substitution in this way. Any hints?

答:
To find the minimal polynomial of $\gamma=\alpha^3 - 14\alpha$, we may start by looking at the powers of $\gamma$.

Luckily, $\gamma^2=48$ (WA), and we're done. (But there is one crucial point to prove. Find it!)


问:
why $n^{\log{\log{n}}}=\log{n}^{\log{n}}$?
I was reading the solution of one exercise on my book on algorithms.
What I need to do in short is order some function from fast to slow.
This is the link of the solution: site.

At some point I get the following property:
$$n^{\log{\log{n}}}=\log{n}^{\log{n}}$$
but how do you prove that?

I've tried to use the properties of logarithm, but I wasn't able to get that result.
Can you give me some hints?

Body: $$n^{\log{\log{n}}} = n^{(\log{\log{n}})/(\log{n})(\log n)} = n^{(\log_n \log n)(\log{n})} = \left(n^{\log_n \log n}\right)^{\log{n}} = (\log n)^{\log{n}}$$


问:
Taylor expansion of a (nearly) gaussian integral
My goal is to expand the function $$\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}$$ from this question asked by me: Taylor expansion of a gaussian integral.

However, this time I want to expand for large $x\gg1$. I should get an answer of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^{\frac{-(2n+1)}{4}}.
\end{align*}
I have tried expanding the exponential in the integrand using the classical taylor expansion of $e$, but this lead me to nowhere. I have no clue how to begin this expansion so any hint to get me started would be very helpful :).

答:
To get an expansion for $x>>1$, let's first make substitution $y = x^{-\frac14}t$ in the integral:
\begin{align} f(x) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^4}{24} - \frac{t^2}{2\sqrt{x}}} \frac{dt}{\sqrt[4]{x}} = \frac{x^{-\frac14}}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{t^4}{24}} \sum_{n=0}^\infty \frac{1}{n!}\left(- \frac{t^2}{2\sqrt{x}}\right)^n dt \\
&= \sum_{n=0}^\infty \frac{(-1)^nx^{-\frac14-\frac{n}{2}}}{\sqrt{2\pi} 2^n n!} \int_{-\infty}^\infty t^{2n}e^{-\frac{t^4}{24}}  dt \end{align}
For these integrals we have, with substitution $t = \sqrt[4]{24u} $
\begin{align} \int_{-\infty}^\infty t^{2n}e^{-\frac{t^4}{24}}  dt &= 2 \int_0^\infty t^{2n}e^{-\frac{t^4}{24}}  dt = \\
&= 2 (24)^{\frac{2n+1}{4}}\int_0^\infty u^{\frac{2n-3}{4}}e^{-u}  du = \\
&= 2 (24)^{\frac{2n+1}{4}} \Gamma(\frac{2n+1}{4}) \end{align}
where $\Gamma$ is the Euler's gamma function. In total, we get
$$ f(x) = \sum_{n=0}^\infty \frac{(-1)^n \Gamma(\frac{n}{2}+\frac14)}{\sqrt{2\pi} 2^{n-1} n!} (\frac{x}{24})^{-\frac{2n+1}{4}} $$
It can be shown that this series is convergent for all $x>0$.


答:
Let $ n\in\mathbb{N}^{*} $, we have the following : \begin{aligned}\sum_{k=1}^{n}{\frac{1}{k}\sin{\left(\frac{k\pi}{n+1}\right)}}&=\frac{2}{n+1}\int_{0}^{\frac{\pi}{2}}{\sum_{k=1}^{n}{\cos{\left(\frac{2kx}{n+1}\right)}}\,\mathrm{d}x}\\ &=\frac{2}{n+1}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{\left(\frac{nx}{n+1}\right)}}{\sin{\left(\frac{x}{n+1}\right)}}\,\mathrm{d}x}\end{aligned}

For all $ x\in\left(0,\frac{\pi}{2}\right] $ : \begin{aligned}\left\vert\frac{\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}}-\frac{\sin{x}}{x}\right\vert&=\left\vert\frac{\left(x-\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}\right)\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)x\sin{\left(\frac{x}{n+1}\right)}}+\frac{\sin{\left(\frac{nx}{n+1}\right)}-\sin{x}}{x}\right\vert\\ &\leq\frac{\sin{\left(\frac{nx}{n+1}\right)}}{x\sin{\left(\frac{x}{n+1}\right)}}\left\vert\frac{x}{n+1}-\sin{\left(\frac{x}{n+1}\right)}\right\vert+\frac{\left\vert\sin{x}-\sin{\left(\frac{nx}{n+1}\right)}\right\vert}{x}\\ &\leq\frac{x^{2}\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)^{3}\sin{\left(\frac{x}{n+1}\right)}}+\frac{2\sin{\left(\frac{x}{2\left(n+1\right)}\right)}\cos{\left(\frac{\left(2n+1\right)x}{2\left(n+1\right)}\right)}}{x}\end{aligned}

Combining the inequalities : $ \left(\forall y\in\left[0,\frac{\pi}{2}\right]\right),\ \frac{2}{\pi}x\leq\sin{x}\leq x $, $ \left(\forall y\in\mathbb{R}\right),\ \cos{y}\leq 1 $ and $ \left(\forall y\in\left[0,\frac{\pi}{2}\right]\right),\ \sin{y}\leq 1 $, we get : \begin{aligned}\left\vert\frac{\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}}-\frac{\sin{x}}{x}\right\vert\leq\frac{x^{2}\times 1}{\left(n+1\right)^{3}\times\frac{2x}{\pi\left(n+1\right)}}+\frac{2\times\frac{x}{2\left(n+1\right)}\times 1}{x}&=\frac{\pi x}{2\left(n+1\right)^{2}}+\frac{1}{n+1}\\ &\leq\frac{n+1+\frac{\pi^{2}}{4}}{\left(n+1\right)^{2}}\end{aligned}

Thus : $$ \frac{1}{n+1}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{\left(\frac{nx}{n+1}\right)}}{\sin{\left(\frac{x}{n+1}\right)}}\,\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{x}}{x}\,\mathrm{d}x}=\frac{1}{2}\int_{0}^{\pi}{\frac{\sin{x}}{x}\,\mathrm{d}x} $$

Which means : $$ \lim_{n\to +\infty}{\sum_{k=1}^{n}{\frac{1}{k}\sin{\left(\frac{k\pi}{n+1}\right)}}}=\mathrm{Si}\left(\pi\right) $$

Title: calculation $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}$.

Body:
calculation$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}.$$
I try to use $|\sin x-x|\leqslant\frac{x^2}{2}$, but I don't feel like it's going to work. Is there any other way?


问:
Show that a local ring is equicharacteristic iff it contains a subfield
A local ring $(A,\mathfrak m)$ is equicharacteristic if $\operatorname{char} A=\operatorname{char}  \kappa (m)$.

Need hints to solve the following question:

A local ring is equicharacteristic iff it contains a subfield.

答:
$\kappa (\mathfrak m):=A/\mathfrak m$ and $\operatorname{char} A=\operatorname{char} \kappa (m)$ iff $A$ contains a field.

If $A$ contains a field $k$, then we have $\operatorname{char} A=\operatorname{char} k$ and $k\subseteq A\to A/\mathfrak m$, so $\operatorname{char}A/\mathfrak m=\operatorname{char}k$.

If $\operatorname{char} A=\operatorname{char} \kappa(\mathfrak m)$ and $\operatorname{char} \kappa(\mathfrak m)=p>0$, then obviously $\mathbb Z/p\mathbb Z\subseteq A$. If $\operatorname{char} \kappa(\mathfrak m)=0$, then $\mathbb Z\subseteq A$ and every non-zero integer remain non-zero in $\kappa(\mathfrak m)$. It follows that every non-zero integer is invertible in $A$, so $\mathbb Q\subseteq A$.


答:
We can simplify the expression $\alpha^3 - 14 \alpha$. This occurs because $\alpha^2$ satisfies $\alpha^4 = 16\alpha^2 - 4$, therefore $$\alpha^3 = 16\alpha - 4\alpha^{-1} $$

and so $\theta = \alpha^3 - 14\alpha = 2\alpha - 4\alpha^{-1}$. (note that $\alpha \neq 0$ so is invertible)

Let us square this term :
$$
\theta^2 = 4\alpha^2 - 16\alpha^{-2} -16 = \frac{4\alpha^4 - 16}{\alpha^2} - 16 = 4\frac{\alpha^4 - 4}{\alpha^2} -16
$$

but $\alpha^4 - 4 = 16 \alpha^2$! So substitute that in and get $4 \times 16 - 16 = 48 = \theta^2$.

Thus, the minimal polynomial should be $x^2-48$ as long as you see that is irreducible.

We are lucky that such a manipulation worked out. In general, however, we'd either have to visit higher powers, or need good manipulative intuition to find the minimal polynomial. In some cases, knowing a root can be helpful, so you can directly compute the expression for that root.

For example, the polynomial given is in fact the minimal polynomial of $\sqrt 3 + \sqrt 5$, so one can try the manipulation with this as $\alpha$ to see that $\alpha^3 - 14 \alpha = 4\sqrt 3$ (in case that no intelligent techniques work, this is brute force).

Title: Minimal polynomial of $\alpha^3 - 14 \alpha$ over $\mathbb{Q}$

Body:
I am trying to solve a problem that goes as follows.

Show that the polynomial $f(X) = X^4 - 16X^2 + 4$ is irreducible over $\mathbb{Q}$. Let $\alpha$ be a root of $f(X)$ in some field extension of $\mathbb{Q}$. Determine the minimal polynomials of $\alpha^2$ and of $\alpha^3 - 14 \alpha$ over $\mathbb{Q}$.

I have shown that $f(X)$ is irreducible over $\mathbb{Q}$ and found the minimal polynomial of $\alpha^2$. I am now stuck on the very last part of this problem, to find the minimal polynomial of $\alpha^3 - 14\alpha$. I know that $0 = \alpha^4 - 16 \alpha^2 + 4$ but I can not proceed from here. To find the minimal polynomial of $\alpha^2$ was easy enough, just set $X := \alpha^2$ in the equation and simplify. However, with $\alpha^3 - 14\alpha$ I can't seem to find a way to use substitution in this way. Any hints?


问:
Is there by chance an analytic solution of the following non-linear ODE: $x'(t) = a\, x(t) + b\, x^3 (t)$?
I'm in fact interested in a PDE for which I try to get some intuition (roughly, I interpret my PDE as a function of time with value in a space of functions of space variables). Does someone by luck know a solution to the ODE in the title (wandering around this site I did see some incredible analytic solutions to some ODEs).

As for my line of thoughts so far, I see that $\enspace x:0< t\mapsto \frac{C}{\sqrt{t}}$ satisfies $x' = -\frac{1}{2}\, x^3$. The ODE being non-linear, I guess it is not very helpful to look at solutions of $x' = b x^3$  alone. I also don't think that Laplace or Fourier transfo will work. Series solution (haven't checked yet...)? My last idea is to use the trick of writing the equation in integral form and insert it within itself...

Follow up question: the next step is to consider a system of ODE of the form
$$\dot{\mathbf{X}} = A\cdot \mathbf{X} + B\cdot \mathbf{X}^3 $$
where here $\mathbf{X}^3$ denotes a vector with $i^{\text{th}}$-components $x_i^3$. Let us assume also that $A$ is diagonalizable but not $B$ (or at least no in the same basis). The problem boils down for each component of $\mathbf{X}$ to solving
$$ x_i'(t) = a\, x_i(t) + b\, x_i^3 (t) + f(t)$$

答:
$$x'(t) = a\, x(t) + b\, x^3 (t)$$
This is Bernoulli's differential equation.
$$\dfrac {x'(t)}{x^3} = \dfrac a { x^2(t)} + b$$
$$-\dfrac {u'}{2}=  au + b$$
Where $u=\dfrac 1 {x^2}$. The DE is now linear of first order.


问:
The axiom of choice for a category
I am currently studying the counterpart of axiom of choice in ETCS which is the axiom that every surjective function has a right inverse.

In category of sets the surjective functions are epimorphsims however, in general categories not every epimorphism has a right inverse. An example of this is the category of partially ordered set. Every arrow in this category is an epimorphism and non identity arrows in this category does not have a right inverse.

I recently saw in an article that we can reformulate the axiom of choice in any category as the following statement:

$\bf{The\ Axiom\ of\ Choice\ for\ a\ Category}$: every epimorphism is split.

The definition i am using for a split epimorphism is as follow:

A ${split\ epimorphism}$ is a morphism $f: X \rightarrow Y$ where there exists a morphism $g : Y \rightarrow X$ such that the composite of $f$ and $g$ is the identity morphism.

But how can the axiom of choice for a category be true when we can prove that not every epimorphism had a right inverse?

答:
As a footnote to Asaf's answer, note that $\mathsf{ETCS}$ is not supposed to be true in every category - it's a set of axioms intended to describe the category Sets specifically (remember that it stands for elementary theory of the category of sets).

The idea of categories as possible "universes" for mathematics is an important one, but it needs to be treated carefully. Not every category results in an interesting mathematical universe. The system $\mathsf{ETCS}$ pins down a particular class of categories (= those categories in which the $\mathsf{ETCS}$ axioms, appropriately interpreted, are true) which provide interesting mathematical universes. If the $\mathsf{ETCS}$ axioms held in every category, they wouldn't be able to implement much mathematics - because they wouldn't be able to rule out "uninteresting" categories.

Title: The axiom of choice for a category


问:
Moment generating function. A miner is trapped in a mine containing 3 doors.
A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 hours of travel. The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after 7 hours. If we assume that the miner is at all times equally likely to choose any one of doors, what is the expected length of time until he reaches safety?

I did it normally but I have been asked to solve through moment generating function. I don't know how to proceed

答:
Let $X$ be the random variable denoting the length of time to reach safety. Let's name the events corresponding to the visits to the doors as $D_3$ and $D_5$ and $D_7$, respectively, then any sequence leading to safety can be denoted by the regular expression "$(D_5|D_7)^*D_3$", i.e., any number of visits (including $0$) to doors 5 or 7, followed by the last visit to door 3.

The following figure shows a DFA which we can be used to conceptualize / visualize the state transitions for above problem (with intermediate / transient states $D_5$, $D_7$ and final / absorbing state $D_3$ corresponding to the safe exit, start state can be any one of the states):

We have the following pmf of X (keeping in mind that at any point in time choosing the doors $D_3$, $D_5$ and $D_7$ are equally likely with probability $\frac{1}{3}$):

$P(X=3)=\frac{1}{3}$ (1-step exit probability, corresponding to $D_3$)

$P(X=5+3)=P(X=7+3)=\frac{1}{3^2}$ (2-step exit probability, $D_5D_3$ and $D_7D_3$)

$P(X=2.5+3)=P(X=2.7+3)=\frac{1}{3^3}$ (for $D_5D_5D_3$ and $D_7D_7D_3$) and $P(X=5+7+3)=\frac{2}{3^3}$ (for $D_5D_7D_3$ and $D_7D_5D_3$) (3-step exit probability)

$P(X=3.7+3)=P(X=3.5+3)=\frac{1}{3^4}$ and $P(X=5+2.7+3)=P(X=7+2.5+3)=\frac{3}{3^4}$ (4-step exit probability)

$P(X=4.7+3)=P(X=4.5+3)=\frac{1}{3^5}$, $P(X=5+3.7+3)=P(X=7+3.5+3)=\frac{4}{3^5}$ and $P(X=2.5+2.7+3)=\frac{6}{3^5}$ (5-step exit probability)

$\ldots$

Now, let's generalize the expression for the $k$-step exit probability.

Obviously, such a sequence will be of length $k$, consisting of total $k-1$ length-$5$ or length-$7$ paths, followed by a single length $3$ path.

(1) The number of possible length-$(k-1)$ sequences consisting of $5$s or $7$s is $2^{k-1}$ and total number of 5s (= total number of 7s, by symmetry) combining all such sequences is $\frac{k-1}{2}$, leading to the travel of sum total length $2^{k-1}.\frac{k-1}{2}.(5+7)=2^{k-1}(6k-6)$.

(2) Also, each of this $2^{k-1}$ sequences will end with a path of $3$ (the last door to be visited, leading to safety), so it will contribute $2^{k-1}.3$ to the sum.

Adding (1) and (2), the total length (in time, hrs) of travel corresponding to $k$-step exit is $2^{k-1}(6k-6)+2^{k-1}.3=2^{k-1}.(6k-3)$, with $k$ state transition probability as $\left(\frac{1}{3}\right)^k$. Hence, the total probability for a $k$-step exit $=\frac{2^{k-1}.(6k-3)}{3^k}$

Finally, the m.g.f = $m_X(t) = E[e^{tX}] = \sum_\limits x P(X=x).e^{tx}$

$=\frac{1}{3}e^{3t}+\frac{1}{3^2}(e^{8t}+e^{10t})+\frac{1}{3^3}(e^{13t}+e^{17t}+2.e^{15t}) +\frac{1}{3^4}(e^{18t}+e^{24t}+3.e^{20t}+3.e^{22t}) +\frac{1}{3^5}(e^{23t}+e^{31t}+3.e^{25t}+3.e^{29t}+6.e^{27t}) + \ldots$

E[X] will be the coefficient of t in the power series representation of $m_X(t)$, or equivalently, $m_X^{\prime}(0)$, which will be

$=\frac{3}{3}+\frac{8+10}{3^2}+\frac{13+17+2.15}{3^3}+\frac{18+24+3.20+3.22}{3^4}+\frac{23+31+4.25+4.29+6.27}{3^5} + \ldots $

$=\frac{3}{3}+\frac{18}{3^2}+\frac{60}{3^3}+\frac{168}{3^4}+\frac{432}{3^5}+\ldots$

$=\sum\limits_{k=1}^\infty\frac{2^{k-1}(6k-3)}{3^k}$ (as shown above)

$=\sum\limits_{k=1}^\infty\frac{2^{k-1}(2k-1)}{3^{k-1}}$

$=2\sum\limits_{k=1}^\infty k\left(\frac{2}{3}\right)^{k-1}-\sum\limits_{k=1}^\infty \left(\frac{2}{3}\right)^{k-1}$

$=\frac{2}{(1-2/3)^2}-\frac{1}{1-2/3}$

$=15$ hrs

With simulation in R (the result holds by LLN):

```<code>set.seed(111)
mean(replicate(100000,
                {
                    tot <- 0
                    cur <- sample(c(3,5,7),1)
                    while (cur == 5 | cur == 7) {
                        tot <- tot + cur
                        cur <- sample(c(3,5,7),1)
                    }
                    tot <- tot + cur
                    tot
                }))
#[1] 15.0603
</code>```

Body:
Let $X$ be the random variable denoting the length of time to reach safety. Let's name the events corresponding to the visits to the doors as $D_3$ and $D_5$ and $D_7$, respectively, then any sequence leading to safety can be denoted by the regular expression "$(D_5|D_7)^*D_3$", i.e., any number of visits (including $0$) to doors 5 or 7, followed by the last visit to door 3.

The following figure shows a DFA which we can be used to conceptualize / visualize the state transitions for above problem (with intermediate / transient states $D_5$, $D_7$ and final / absorbing state $D_3$ corresponding to the safe exit, start state can be any one of the states):

We have the following pmf of X (keeping in mind that at any point in time choosing the doors $D_3$, $D_5$ and $D_7$ are equally likely with probability $\frac{1}{3}$):

$P(X=3)=\frac{1}{3}$ (1-step exit probability, corresponding to $D_3$)

$P(X=5+3)=P(X=7+3)=\frac{1}{3^2}$ (2-step exit probability, $D_5D_3$ and $D_7D_3$)

$P(X=2.5+3)=P(X=2.7+3)=\frac{1}{3^3}$ (for $D_5D_5D_3$ and $D_7D_7D_3$) and $P(X=5+7+3)=\frac{2}{3^3}$ (for $D_5D_7D_3$ and $D_7D_5D_3$) (3-step exit probability)

$P(X=3.7+3)=P(X=3.5+3)=\frac{1}{3^4}$ and $P(X=5+2.7+3)=P(X=7+2.5+3)=\frac{3}{3^4}$ (4-step exit probability)

$P(X=4.7+3)=P(X=4.5+3)=\frac{1}{3^5}$, $P(X=5+3.7+3)=P(X=7+3.5+3)=\frac{4}{3^5}$ and $P(X=2.5+2.7+3)=\frac{6}{3^5}$ (5-step exit probability)

$\ldots$

Now, let's generalize the expression for the $k$-step exit probability.

Obviously, such a sequence will be of length $k$, consisting of total $k-1$ length-$5$ or length-$7$ paths, followed by a single length $3$ path.

(1) The number of possible length-$(k-1)$ sequences consisting of $5$s or $7$s is $2^{k-1}$ and total number of 5s (= total number of 7s, by symmetry) combining all such sequences is $\frac{k-1}{2}$, leading to the travel of sum total length $2^{k-1}.\frac{k-1}{2}.(5+7)=2^{k-1}(6k-6)$.

(2) Also, each of this $2^{k-1}$ sequences will end with a path of $3$ (the last door to be visited, leading to safety), so it will contribute $2^{k-1}.3$ to the sum.

Adding (1) and (2), the total length (in time, hrs) of travel corresponding to $k$-step exit is $2^{k-1}(6k-6)+2^{k-1}.3=2^{k-1}.(6k-3)$, with $k$ state transition probability as $\left(\frac{1}{3}\right)^k$. Hence, the total probability for a $k$-step exit $=\frac{2^{k-1}.(6k-3)}{3^k}$

Finally, the m.g.f = $m_X(t) = E[e^{tX}] = \sum_\limits x P(X=x).e^{tx}$

$=\frac{1}{3}e^{3t}+\frac{1}{3^2}(e^{8t}+e^{10t})+\frac{1}{3^3}(e^{13t}+e^{17t}+2.e^{15t}) +\frac{1}{3^4}(e^{18t}+e^{24t}+3.e^{20t}+3.e^{22t}) +\frac{1}{3^5}(e^{23t}+e^{31t}+3.e^{25t}+3.e^{29t}+6.e^{27t}) + \ldots$

E[X] will be the coefficient of t in the power series representation of $m_X(t)$, or equivalently, $m_X^{\prime}(0)$, which will be

$=\frac{3}{3}+\frac{8+10}{3^2}+\frac{13+17+2.15}{3^3}+\frac{18+24+3.20+3.22}{3^4}+\frac{23+31+4.25+4.29+6.27}{3^5} + \ldots $

$=\frac{3}{3}+\frac{18}{3^2}+\frac{60}{3^3}+\frac{168}{3^4}+\frac{432}{3^5}+\ldots$

$=\sum\limits_{k=1}^\infty\frac{2^{k-1}(6k-3)}{3^k}$ (as shown above)

$=\sum\limits_{k=1}^\infty\frac{2^{k-1}(2k-1)}{3^{k-1}}$

$=2\sum\limits_{k=1}^\infty k\left(\frac{2}{3}\right)^{k-1}-\sum\limits_{k=1}^\infty \left(\frac{2}{3}\right)^{k-1}$

$=\frac{2}{(1-2/3)^2}-\frac{1}{1-2/3}$

$=15$ hrs

With simulation in R (the result holds by LLN):

set.seed(111)
mean(replicate(100000,
                {
                    tot <- 0
                    cur <- sample(c(3,5,7),1)
                    while (cur == 5 | cur == 7) {
                        tot <- tot + cur
                        cur <- sample(c(3,5,7),1)
                    }
                    tot <- tot + cur
                    tot
                }))
#[1] 15.0603


问:
Sobolev spaces - about smooth aproximation
Consider $\Omega $ a open and bounded set of $\mathbb R^n$ . Let $u \in H^{1}(\Omega)$ a bounded function. I know that there exists a sequence $u_m \in C^{\infty} (\Omega) $   where $u_m \rightarrow u$ in $H^{1}(\Omega)$ . I can afirm that exists $C>0$ where $|u_m (x)| \leq C , \forall \ x , \forall m  ?$

I dont have idea how to prove this or how to show a counter example ...

my english is terrible, sorry ..

答:
Yes, it is possible to find a uniformly bounded approximating sequence. Let  $\phi:\mathbb R\to\mathbb R$ be a bounded $C^\infty $ function such that

$\phi(x)=x$ for $x\in [-M,M]$ where $M=\operatorname{ess\,sup}|u|$

$ 0\le \phi'(x)\le 1$ for all $x$. (As a consequence, $|\phi(x)|\le |x|$ for all $x$.)

For every $m$, the function $\phi\circ u_m$ is smooth and satisfies $\|\phi\circ u_m\|_{H^1}\le \|u_m \|_{H^1}$. Since the sequence $(u_m)$ converges in the $H^1$ norm, it is bounded in the $H^1$ norm. Therefore, the sequence $(\phi\circ u_m)$ is bounded in the $H^1$ norm. It follows that $(\phi\circ u_m)$ has a subsequence that converges in the weak topology of $H^1(\Omega)$. On the other hand, $\phi\circ u_m\to \phi\circ u=u$ in $L^2(\Omega)$. Therefore, the aforementioned weak limit is indeed $u$. Furthermore, it is the strong limit because $\limsup\|\phi\circ u_m\|_{H^1}\le \limsup\| u_m\|_{H^1}=\|u\|_{H^1}$.


问:
$0 \to A \to B\to C$ is left exact if and only if $f= \ker (g)$.
We work in an abelian category.

Consider the sequence
$$0 \to A \stackrel{f}\to B \stackrel{g}\to C $$
where $gf=0$. I want to show that if this sequence is left exact, then $f$ is a kernel of $g$.

My definition of exactness:

A sequence $A \stackrel{f}\to B \stackrel{g}\to C$ is called exact in $B$ if $H_B=0$, where $H_B= \text{coker}(f')$ where $f'$ is the unique morphism $f': A \to \text{Ker}(g)$ satisfying $\ker(g)f' = f$.

My attempt:

I managed to show that $f'$ is an epimorphism. I tried to directly verify the universal property of the kernel, but given a morphism $T \to B$, I don't see how to construct a morphism $A \to T$.

答:
If we can show that $f'$ is a monomorphism, then from your observation $f'$ is both a monomorphism and an epimorphism, which means that $f'$ is an isomorphism. From there, the uniqueness of the kernel up to isomorphism is enough to tell us that $\ker g \circ f' = f$ is a kernel of $g$.

To that end, show that $f$ is a monomorphism using the exactness of $0 \to A \overset{f}{\to}B$. From there, it follows that for any morphisms $\alpha,\beta$ to $A$,
$$
f' \alpha = f' \beta \implies \ker (g)f'\alpha = \ker(g)f'\beta  \implies f\alpha = f \beta \implies \alpha = \beta.
$$


问:
Symmedian and orthic triangle properties
I am trying to prove the following lemma that may be useful for junior-level international contests:

Given $\Delta ABC$ an acute triangle and $(BE)$ and $(CF)$ its altitudes. Let's consider $(AM)$ the median line, $M \in (BC)$ and $AM \cap EF = \{N\}$. If $NP \perp BC$, $P \in (BC)$, then $AP$ is the $A$-symmedian in the triangle $\Delta ABC$.

Attempt. I have tried using the following metric relations in a triangle:

$s_a = \frac{2bc}{b^2 + c^2} m_a$

$m_a^2 = \frac{2(b^2 + c^2 - a^2)}{4}$ (Stewart)

$\frac{BD}{DC} = {AB^2}{AC^2}$ if and only if $AD$ symmedian (Steiner)

Drawing $AD$ height, I proved that $\frac{PM}{DM} = \frac{a^2}{b^2 + c^2}$. I could not prove that $AP$ symmedian, although I have tried to calculate ratios and lengths using Pythagorean Theorem or trigonometric relations.

答:
Let $D$ be the mid-point of $EF$

Since $MF=ME=BC/2$, it follows that $MD\perp EF$. So $MNDP$ is a cyclic quadrilateral, and

$$\angle NDP = 180^\circ - \angle NMP = 180^\circ - \angle ADE.$$

Thus $A, D, P$ are colinear and it follows that $AP$ is the symmedian.


问:
Finding conjugate of a complex number using some algebra
it is given that z is a complex number satisfying

$z^3-iz+1=0$

I'm supposed to find argument of  $ z+z \bar z + \bar z$

The only thing I could do was

$(z+1)(z^2+1-z)=iz$

Any help is appreciated

答:
Hint:

$$z+z\bar z+\bar z=2\Re(z)+|z|^2.$$

You still have to find out the sign.


答:
I will go with 3x3 example cause it's just easier to write it.

I thought it would be a good idea to take the $(A)^t$ but then I am stuck again because i don't know how to calculate the $det(A)^t-tI_{10})$

$(A)^t-tI_3=\begin{pmatrix}
-t& 0 & a\\
 1&  -t&0 \\
 0& 1 & -t
\end{pmatrix}$ so if this was a $10x10$ matrix how should I calculate the determinant ?

Title: Find the characteristic polynomial of a $10\times 10$ matrix

Body:
Let  $$A=\begin{pmatrix}
0 & 1 & ... & &  &  &  &  &  ...& 0\\
0 &0  &1  &. .. &  &  &  &  &  ...&0 \\
 0& 0 &0  &1  &...  &  &  &  &  ...&0 \\
0 & 0 & 0 &0  &1  & ... &  &  &  ...&0 \\
 .&  &...  &  &  &  &  &  & ... & \\
 .&  & ... &  &  &  &\ddots  &  & ... & \\
 .&  &...  &  &  &  & &  & ... &\\
 .&  &...  &  &  &  &  &  & ... & \\
 0&0  &0  &...  &  &  &  &  &  ...& 1\\
 10^{10}& 0 & 0 & 0 & 0 &0  &  0& 0 &0  &0
\end{pmatrix}$$

Find the characteristic polynomial. This probably needs a trick I don't know. Does someone know it and want to share it with me?


问:
Expected number of vertices for a given random graph
So I have to find out the expected value of the number of vertices of degree 1.
Let's the given graph if G(n,p). Then the $$\mathbb{P}(deg(v)=1) = {n \choose 1} (p)^{1}(1-p)^{n-1}$$

And $$\mathbb{E}X = (n){n\choose 1} p(1-p)^{n-1} = n^2p(1-p)^{n-1}$$

So I do not know if I am doing it right. Can one simplify it further? I am stuck here.

Thanks in advance.

答: You've made one mistake.  There are only $n-1$ possible neighbors of a given vertex. So you should have $$n(n-1)p(1-p)^{n-2}$$


问:
There is a rectangle in 3-dimensional space. If I am only given the $x$ and $y$ coordinates, how do I find the $z$ coordinates?
We are given the $x$ and $y$ coordinates of the four vertices of a rectangle.
How do I find the $z$-coordinates?

I know that this 3 dimensional rectangle must obey all the usual conditions of rectangle.
Opposite sides are equal.
The angles between two adjacent sides are 90 degrees.

答:
(Rewritten after missing an entire degree of freedom!)

Let's say the vertices are in (clockwise or counterclockwise order) $\vec{A} = (x_A, y_A, z_A)$, $\vec{B} = (x_B, y_B, z_B)$, $\vec{C} = (x_C, y_C, z_C)$, and $\vec{D} = (x_D, y_D, z_D)$.

In order for the quadrilateral to be a parallelogram, pairs of opposing sides must be equal:
$$\left\lbrace\begin{aligned}
\vec{B} - \vec{A} &= \vec{C} - \vec{D} \\
\vec{D} - \vec{A} &= \vec{C} - \vec{B} \\
\end{aligned}\right. \quad \iff \left\lbrace\begin{aligned}
x_A - x_B + x_C - x_D &= 0 \\
y_A - y_B + y_C - y_D &= 0 \\
z_A - z_B + z_C - z_D &= 0 \\
\end{aligned}\right. \tag{1}\label{G1}$$
which means we can ignore $\vec{C}$ completely, because it is defined exactly by the three other points (via $\vec{C} = \vec{B} + \vec{D} - \vec{A}$), if it is a parallelogram.  (And rectangles and squares are always also parallelograms.)

First, check if the coordinates are valid for any rectangle that is perpendicular to the $z$ axis. If
$$(x_D - x_A)(x_B - x_A) + (y_D - y_A)(y_B - y_A) = 0 \tag{2a}\label{G2a}$$
then all real $z_A = z_B = z_C = z_D$ are valid solutions, as are $z_A = z_B \ne z_C = z_D$ and $z_A = z_D \ne z_B = z_C$.  (The latter are solutions because changing $z_A \to z_A + h$ and $z_B \to z_B + h$, or $z_A \to z_A + h$ and $z_D \to z_D + h$ is still a rectangle, increasing $h$ in magnitude just makes the rectangle taller or wider.)

For nontrivial solutions, we need all four angles of the parallelogram to be right angles:
$$\left\lbrace\begin{aligned}
\bigl( \vec{D} - \vec{A} \bigr) \cdot \bigl( \vec{B} - \vec{A} \bigr) &= 0 \\
\bigl( \vec{A} - \vec{B} \bigr) \cdot \bigl( \vec{C} - \vec{B} \bigr) &= 0 \\
\bigl( \vec{B} - \vec{C} \bigr) \cdot \bigl( \vec{D} - \vec{C} \bigr) &= 0 \\
\bigl( \vec{C} - \vec{D} \bigr) \cdot \bigl( \vec{A} - \vec{D} \bigr) &= 0 \\
\end{aligned}\right.$$
Because $\vec{C} = \vec{D} + \vec{B} - \vec{A}$, these all simplify to the same rule,
$$(x_B - x_A)(x_D - x_A) + (y_B - y_A)(y_D - y_A) + (z_B - z_A)(z_D - z_A) = 0$$
i.e.
$$(z_B - z_A)(z_D - z_A) = - (x_B - x_A)(x_D - x_A) - (y_B - y_A)(y_D - y_A)$$
which we can solve for $z_D$ (noting that we have already covered the $z_A = z_B$ cases in $\eqref{G2a}$):
$$z_D = z_A - \frac{ (x_B - x_A)(x_D - x_A) + (y_B - y_A)(y_D - y_A) }{z_B - z_A}, \quad z_A \ne z_B \tag{2b}\label{G2b}$$

meaning we can choose any pair of reals $z_A \ne z_B$, and $\eqref{G2b}$ and
$$z_C = z_D + z_B - z_A$$
will yield a valid rectangle.

Let's test this with
$$\vec{A} = \left[\begin{matrix} 3 \\ 3 \\ z_A \end{matrix}\right], \quad
\vec{B} = \left[\begin{matrix} 10 \\ 5 \\ z_B \end{matrix}\right], \quad
\vec{C} = \left[\begin{matrix} 11 \\ 8 \\ z_B + z_D - z_A \end{matrix}\right], \quad
\vec{D} = \left[\begin{matrix} 4 \\ 6 \\ z_D \end{matrix}\right]$$
(which yields $13 \ne 0$ for $\eqref{G2a}$, so $z_A=z_B=z_C=z_D$ are not valid rectangles).

Rule $\eqref{G2b}$ simplifies to
$$(z_D - z_A)(z_B - z_A) = -13 \quad \iff \quad z_D = z_A - \frac{13}{z_B - z_A}, z_A \ne z_B$$
Here are some randomly generated solutions:
$$\begin{array}{cc|cccc}
z_A & z_B & \vec{A} & \vec{B} & \vec{C} & \vec{D} \\
\hline
0 & 1 & (3, 3, 0) & (10, 5, 1) & (11, 8, -12) & (4, 6, -13) \\
1 & 0 & (3, 3, 1) & (10, 5, 0) & (11, 8, 13) & (4, 6, 14) \\
1 & -1 & (3, 3, 1) & (10, 5, -1) & (11, 8, 5.5) & (4, 6, 7.5) \\
-3 & 10 & (3, 3, -3) & (10, 5, 10) & (11, 8, 9) & (4, 6, -4) \\
\end{array}$$


问:
Calculate the sum of A.
Let $\omega=\mathrm{e}^{\mathrm{i} \frac{2 \pi}{7}}$. Calculate
$$
A=\omega+\omega^{2}+\omega^{4}
$$

Since $\omega=e^{i\frac{2\pi}{7}}$ it mean that $\omega$ is the 7th roots of unity. So it have known that $1+\omega+\omega^2+\cdots+\omega^6=0$.
But the induction want to calculate $A=\omega+\omega^2+\omega^4$. I'm not quite sure how to do this. Please kindly give a help for me or something recommendations for me. Thank is advanced!

答:
On the one hand, observe that
$$A = \omega+\omega^{2}+\omega^{4} \quad \text{and} \quad B = \omega^3+\omega^{5}+\omega^{6}$$
are conjugate numbers since $AB = 2$. On the other hand, as you noticed, $A + B = -1$. Therefore, $A$ and $B$ are solutions of the polynomial $x^2 + x + 2 =0$.

Then, you can solve the equation and find $A$ and $B$. To know which solution correspond to $A$, notice that $\text{Im}(A) > 0$.

Title: Calculate the sum of A.


问:
If $Gf$ is connected and locally-connected then $f$ is continuous.
I encountered this proof in the H. Herrlich's Topology book. There are several steps there that are unclear, perhaps someone would have an idea.

Prove that:

$f: \underline R \rightarrow \underline R\ $ is a map and $GF = \{(x, f(x)) \ | \ x \in R)\}$ it's graph. Then $f$ is continuous if $GF$ is connected and locally-connected.

The author starts by stating that if $f$ were discontinuous then WLOG there would exist such $x \in R$ and $r > 0$ and monotone icreasing $x_n \rightarrow x$.

It would hold then that $f(x_n) > f (x) + r, \forall n \in N$. Then the set U = $\{ (a,b) \in Gf \ |\ b < f(x) + r \}$ would be an open neighbourhood of $(x, f(x))$.

Since $GF$ is locally-connected, there is such open and connected $V \subset U$.
If one defines $V_n = \{(a,b) \ \in V | \ a < x_n\}$ then each $V_n$  has to be empty. Therefore, from $a < x_n$ it follows that $(a, f(a)) \notin V$.

Hence, $V$ is an open neighbourhood of $(x,f(x))$, such that it does not intersect $A = \{ (a,b) \in Gf \ | \ a < x \}$.

$A$ would be therefore both open and closed, which contradicts the connectedness of $Gf$.

I've marked unclear parts using bold font.

In order to prove that $A$ is closed the author needs to show that any point not from $A$ has an open neighbourhood that does not intersect $A$. He finds it by defining $V$ for $(x, f(x))$. The way I see it $(x, f(x))$ is not an arbitrary point. However, in the beginning, the author states that it is. By writing WLOG, it looks as if the author assumes that the map is discontinuous at every point. Why is that?

Why does it follow that each $V_n$ must be empty? I see why $V_n$ is supposed to be closed and open set and since $V$ is connected then $V_n$ must be either $\emptyset$ or $V$ itself. Why is the latter not possible and only the $\emptyset$ is considered?

Why does it follow that $(a, f(a)) \notin V$? I guess it has to hold for each $V_n$ but how does one transition from there to $V$? $V$ could be bigger than the union of all $V_n$'s and thus contain $(a,f(a))$ in that gap.

答:
Let us begin with some notation. For $x \in \mathbb R$ let $p(x) = (x,f(x)) \in Gf$, $H_+(x) = \{ (a,b) \in \mathbb R^2 \mid a > x \},  H_-(x) = \{ (a,b) \in \mathbb R^2 \mid a < x \}$. The sets $H_\pm(x)$ are open in $\mathbb R^2$, hence the sets $Gf_\pm(x) = Gf \cap H_\pm(x)$ are open in $Gf$.

W.l.o.g. :

If $f$ is not continuous, then there exists $x$ and $r > 0$ and a sequence $(x_n)$ converging to $x$ such that $x_n \ne x$ and $\lvert f(x_n) - f(x) \rvert > r$. Thus $f(x_n) - f(x)  > r$ for infinitely many $n$ or $f(x_n) - f(x)  <  -r$ for infinitely many $n$. We may assume that $f(x_n) - f(x)  > r$ for infinitely many $n$, the other case can be treated similarly. So w.l.o.g. all  $f(x_n) - f(x)  > r$ (otherwise take a subsequence). We may also assume that the sequence $(\lvert x_n - x \lvert)$ is stricty decreasing (otherwise take a subsequence). But infinitely many $x_n < x$ or infinitely many $x_n > x$. We may assume that infinitely many $x_n < x$, the other case can be treated similarly. So w.l.o.g. all  $x_n < x$ (otherwise take a subsequence). This means that $(x_n)$ is strictly increasing.

$V_n$ has to be empty :

Note that $f(x_n) > f(x) + r$, thus $p(x_n) \notin U$ and hence $p(x_n) \notin V$. Therefore $V = (V \cap Gf_+(x_n)) \cup (V \cap Gf_-(x_n))$, where $V \cap Gf_\pm(x_n)$ are disjoint open subsets of $V$. Clearly $V \cap Gf_+(x_n) \ne \emptyset$ because $p(x) \in V$ and $p(x) \in Gf_+(x_n)$ because $x > x_n$. Since $V$ is connected, we conclude that $V_n = V \cap Gf_-(x_n)$ must be empty. Now assume that $V \cap Gf_-(x) \ne \emptyset$. Then there exists $(a,b) \in V$ such that $(a,b) \in Gf_-(x)$, i.e. $a < x$. There exists $n$ such that $a < x_n < x$ which implies $(a,b) \in Gf_-(x_n)$. Thus $(a,b) \in V \cap Gf_-(x_n)$, a contradiction.

$A = Gf_-(x)$ is open and closed in $Gf$ :

It is open by definition. Moreover $Gf = Gf_-(x) \cup V \cup Gf_+(x)$. We have $Gf_-(x) \cap (V \cup Gf_+(x)) = Gf_-(x) \cap V \cup Gf_-(x) \cap Gf_+(x)) = \emptyset$. Therefore $Gf_-(x)$ is the complement of the open subset $V \cup Gf_+(x) \subset Gf$.


问:
Complete metric on $\mathcal C([0,1],\mathbb S^1)$
On $\mathcal C([0,1],\mathbb R^n)$ we have for example a norm, namely $\sup_{s\in [0,1]}\|f(s)-g(s)\|_{\mathbb R^n}$ that make $\mathcal C([0,1],\mathbb R^n)$ complete. Is there such a metric on $\mathcal C([0,1],\mathbb S^1)$, the space of continuous function that map $[0,1]$ on the unit circle ? My idea would be to consider $g$ being the tensor metric of $\mathbb S^1$, and thus, to consider $\sup_{s\in [0,1]}\|f(s)-g(s)\|_g$ where $\|f(t)\|_g:= g(f(t),f(t))$. But in somehow, I have the impression that it doesn't make so much sense, since depending on which chart we are, the quantity $\sup_{0\leq s\leq 1}\|f(s)\|_g$ may be not well defined... Any idea ?

答:
The $\sup$-metric (let us call it $d$) still works: Let $(f_n)_{n \in \mathbb{N}} \subseteq \mathcal{C}([0, 1], \mathbb{S}^1)$ be a Cauchy-Sequence in the $\sup$-metric. Since $\mathcal{C}([0, 1], \mathbb{R}^n)$ is complete, there exists some $f \in \mathcal{C}([0, 1], \mathbb{R}^n)$ such that
$$
\sup_{s \in [0, 1]} \lVert f_n(s) - f(s) \rVert = d(f_n, f) \overset{n \rightarrow \infty}{\longrightarrow} 0.
$$
Therefore $f_n$ converges pointwise, which implies that $\lVert f_n(x) \rVert$ converges to $\lVert f(x) \rVert$ for all $x \in [0, 1]$. By assumption, for arbitrary $x \in [0, 1]$ we have
$$
1 = \lVert f_n(x) \rVert
$$
and as $n \rightarrow \infty$
$$
1 = \lVert f(x) \rVert.
$$
So, $f \in \mathcal{C}([0, 1], \mathbb{S}^1)$ and we are done.

In general: If $(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $K \subseteq X$ is compact, then $\mathcal{C}(K, Y)$ is complete w.r.t. to the metric $\displaystyle d(f, g) := \sup_{x \in K} d_Y(f(x), g(x))$ iff $Y$ is complete.


问:
Why is the endomorphism ring of $\mathbb{Z}\times\mathbb{Z}$ noncommutative?
So I hear that the endomorphism ring of an abelian group is not always commutative. In particular, I'm looking at the abelian group $A=\mathbb{Z}\times\mathbb{Z}$, and considering $\text{End } A$. I can't find a counterexample to show that $\text{End } A$ is not commutative. Does anyone know of two such endomorphisms that don't commute?</p>

答:
An endomorphism of $\mathbb{Z}\times\mathbb{Z}$ can be represented as a $2\times2$ matrix with entries in $\mathbb{Z}$, and all such matrices give endomorphisms of $\mathbb{Z}\times\mathbb{Z}$.  (The ones with determinant 1 give automorphisms.)

So any two $2\times 2$ matrices that don't commute will provide you with an example.


答:
Let
$$x = \log n^{\log n}.$$ Then, $$ e^{\log x} = e^{\log\log n^{\log n}} = e^{\log n\log\log n} = n^{\log\log n}.$$

Title: why $n^{\log{\log{n}}}=\log{n}^{\log{n}}$?

Body:
I was reading the solution of one exercise on my book on algorithms.
What I need to do in short is order some function from fast to slow.
This is the link of the solution: site.

At some point I get the following property:
$$n^{\log{\log{n}}}=\log{n}^{\log{n}}$$
but how do you prove that?

I've tried to use the properties of logarithm, but I wasn't able to get that result.
Can you give me some hints?


问:
$\pm 1$ valued vectors in an arbitrary subspace.
In a paper I have read on random Bernoulli matrices it was stated that for any subspace $V\subseteq \mathbb{R}^n$ of dimension at most $l$ one has:
$$|V\cap \{\pm 1\}^n|\leq 2^l$$
With no proof. A reference was provided.
However, the claim does not appear there.

How should this be proven?

答:
Consider the solid $n$ dimensional hypercube $Q$ with side length $4$, with vertices $\{\pm 2\}^n$. We can partition this into $2^n$ sub-cubes with side length $2$, with a cube $Q_i$ centered at vector $i$ for each $i\in \{\pm 1\}^n$.

Let $S=Q\cap V$. For each $j\in V\cap \{\pm1\}^n$, let $S_j=Q_j\cap V$. Note that $S_j$ and $S$ are both sections of an $l$-dimensional plane through the center of a cube, the difference being that the cube for $S_j$ has half the side length for the cube for $S$. Letting  $\def\vol{\text{vol}}\vol$ denote $l$-dimensional volume, this proves

$\sum_{j\in  V\cap \{\pm1\}^n} \vol(S_j)\le \vol(S)$

$\vol(S_j)=2^{-l}\vol(S)$, because $S_j$ is congruent to $S$ scaled down by a factor of $2$.

Combined, these prove $|V\cap \{\pm1\}^n|\le 2^l$.

I believe this volume argument can be modified to prove the same result is true even when $V$ is an arbitrary affine subspace, not just a linear one.