问:
In what vector space do the arrows we know from basic geometry live? Do they form the domain of the dot product?
Lots of people say that they are simply elements of $\mathbb R^3$, but I think that this is not the entire story. My guess would be that it is a three dimensional real vector space and thus can be identified with $\mathbb R^3$ once we have chosen a basis.

Q: Is this something that gets formalized if one studies geometry? If yes, could you please give me some insight and/or a good reference? I have to talk about the cross product and the dot product and it bugs me that I can't really say what their domain is (I assume that this is strongly correlated to my question).

答:
You can't add arrows based at different points.  Well, you may hear people say vectors in $\mathbf R^3$ that are parallel translates of each other are "the same", but you would not agree that the gravitational force vectors on a big rock are the same if it is dropped over one of your feet or dropped some distance away from your feet.  So there is that compelling reason not to consider vectors to be the same if they are not genuinely in the same place.

The arrows based at a common point in $\mathbf R^3$ form the tangent space to $\mathbf R^3$ at that point, and the (disjoint!) union of all those tangent spaces at different points are the tangent bundle of $\mathbf R^3$.  That is a common space where all the different "vector arrows" in $\mathbf R^3$ live.


问:
$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$
Given $a,b,c\geq 0$, then this following is true
$$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$$

I tried to prove it by using some cases, i.e.

Case 1: if $\,\,a\leq b \leq c\,\,$ and $\,\,a\leq c \leq b,\,\,$ it's clear that
\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq a+a=2a
\end{align*}
Case 2: if $\,\,b\leq a\leq c\,\,$ and $\,\,c\leq a \leq b,\,\,$ we have

\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq b+a
\end{align*}
and
\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq a+c
\end{align*}
Case 3: $\,\,c\leq b\leq a\,\,$ and $\,\,b\leq c \leq a,\,\,$

I am not sure about this case, because although $a$ is the biggest from $b$ and $c$, it can satisfies $a\leq b+c$ or $b+c\leq a$. Is the third case must be done with two cases again? i.e for $a\leq b+c$ and $b+c\leq a$?

or there is another way to prove $\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$?

Any help will be appreciated. Thanks.

答:
There's another way. Consider the function $f(a)=\min(a,b)+\min(a,c)-\min(a,b+c)$. It's piece-wise linear, so to prove it's non-negative, we only need to check its value at each endpoint of the intervals (including $+\infty$):

$f(0)=0+0-0=0;$

$f(b)=b+\min(b,c)-\min(b,b+c)=\min(b,c) \ge 0.$

By symmetry $f(c)=\min(b,c)$.

$f(b+c)=b+c-(b+c)=0.$

And lastly, when $a\to \infty, f(a)=b+c-(b+c)=0.$

Some other problems can be solved this way as well:

Show $\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$ : This one you just check the function $f(b)=2|a-b|-|a|$ at $b=\frac{|a|}{2} - \varepsilon, -\frac{|a|}{2}+ \varepsilon$, with $\varepsilon>0$ arbitrarily small.

Prove that $|a|\leq \max\{|b|,|c|\}$ if $b\leq a \leq c$ : This one you check $a=0, b, c$ (even though $0$ may be outside of $[b,c]$).

Even equations: Why does the equality $\max\{a,\min\{b,c\}\}=\min\{\max\{a,b\},\max\{a,c\}\}$ hold? where you check $a=\min(b,c), b,c, -\infty, +\infty$.

Or even this one Reverse Triangle Inequality Proof albeit not the best way.

ContentLicense: CC BY-SA 4.0


问:
A continuous map that is not a proper map
I know that if a map is proper then it is continuous. But to disprove the converse I am trying to think of a counter example of a continuous map that is not proper.</p>

答:
Let $\Bbb Z$ have the discrete topology, and let

$$f:\Bbb Z\times\Bbb Z\to\Bbb Z:\langle n,m\rangle\mapsto n$$

be the projection to the first factor. Clearly $f$ is continuous, since its domain has the discrete topology, but $f^{-1}[\{n\}]=\{n\}\times\Bbb Z$ is not compact for any $n\in\Bbb Z$, so $f$ is not proper.


问:
Question on continuity and optimization
Let    $f:\mathbb{R}^2\to\mathbb{R}$ be a continuous function on $\mathbb{R}^2$.

Is it true that, if    $$\lim_{||(x,y)||\to{\infty}}{f(x,y)}=-{\infty}$$ then $f$ has a global maximum on $\mathbb{R}^2$?

I think that is true, but I don't know how to prove it. Can someone give me a proof or a counter-example?

答:
It is true.

As $\lim_{||(x,y)||\to{\infty}}{f(x,y)}=-{\infty}$, there is $N>0$ such that if $\|(x,y)\|>N$, then $f(x,y)<-|f(0,0)|-1$.

Now, observe that $K:=\{(x,y)\in\Bbb R^2:\ \|(x,y)\|\le N\}$ is compact, so, since $f$ is continuous, Weierstrass' Theorem guarantees the existence of a maximum on $K$. Let $M:=\max_{x\in K} f(x)$.

If $f(0,0)\ge 0$, then $M\ge f(0,0)>-|f(0,0)|-1$, so $M$ is teh global Maximum of $f$.

If $f(0,0)<0$, then $M\ge f(0,0)=-|f(0,0)|>-|f(0,0)|-1$, and, again, $M$ is the global maximum of $f$.

Title: Question on continuity and optimization


答:
Here is an example that is not using the discrete topology. The inclusion map
$$
i: (-1,1) \to \mathbb{R}
$$
is continuous but not proper, as $i^{-1}\left([-1,1] \right) = (-1,1)$ is not compact.

This example shows that even an embedding may not be proper.

Title: A continuous map that is not a proper map

Body: I know that if a map is proper then it is continuous. But to disprove the converse I am trying to think of a counter example of a continuous map that is not proper.</p>


问:
How to integrate $\int \sqrt{x^2+a^2}dx$
$a$ is a parameter. I have no idea where to start</p>

答:
Integrating by parts

$$I=\int \sqrt{x^2+a^2}\cdot1dx$$

$$=\sqrt{x^2+a^2}\int dx-\int\left( \frac{d \sqrt{x^2+a^2}}{dx}\int dx\right)dx$$

$$=x\sqrt{x^2+a^2}-\int\frac{x^2 dx}{\sqrt{x^2+a^2}}$$

Now, $$\int\frac{x^2 dx}{\sqrt{x^2+a^2}}=\frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx=I-a^2\int\frac{dx}{\sqrt{x^2+a^2}}$$

Put $x=a\tan \theta $ in

$$\int\frac{dx}{\sqrt{x^2+a^2}}$$

ContentLicense: CC BY-SA 3.0


问:
How can this theorem regarding limits of infinite sequences be proven.
Theorem: Assume that f is a function that is continuous at c, and assume that $\lim\limits_{x \to \infty}$$(s_{n})$
= c , where all the terms
$s_{n}$ are in the domain of f. Then $\lim\limits_{x \to \infty}$$(f(s_{n}))$ = f(c).

How can this theorem be proven, using the precise limit definitions for functions and infinite sequences?

答:
Proposition

Let $f:X\to Y$ be a real-valued function over $X\subseteq\mathbb{R}$. Then $f$ is continuous at $x_{0}\in X$ if and only if $f(x_{n})\to f(x_{0})$ whenever $x_{n}\in X$ converges to $x_{0}$.

Proof

We shall prove the implication $(\Rightarrow)$ first.

Let us suppose that $f:X\to Y$ is continuous at $x_{0}\in X$. Then for every $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} > 0$ such that for every $x\in X$ one has that
\begin{align*}
|x - x_{0}| \leq \delta_{\varepsilon} \Rightarrow |f(x) - f(x_{0})| \leq \varepsilon
\end{align*}

Moreover, since $x_{n}\to x_{0}$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ such that
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow |x_{n} - x_{0}| \leq \delta_{\varepsilon}
\end{align*}
Gathering both results, we deduce that for every $\varepsilon > 0$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ such that
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow |x_{n} - x_{0}|\leq\delta_{\varepsilon} \Rightarrow |f(x_{n}) - f(x_{0})|\leq\varepsilon
\end{align*}
whence we get that $f(x_{n})\to f(x_{0})$, and we are done.

Let us prove the implication $(\Leftarrow)$ this time.

Let us assume that $x_{n}\in X$ converges to $x_{0}$ and $f$ is not continuous at $x_{0}$.

This means there exists an $\varepsilon > 0$ such that for every $\delta$ there corresponds a $x_{\delta}\in X$ satisfying

$$(|x_{\delta} - x_{0}| \leq \delta)\wedge(|f(x_{\delta}) - f(x_{0})| > \varepsilon).$$

In particular, if we take $\delta = 1/n$, there corresponds a $x_{n}\in X$ such that $|x_{n} - x_{0}| \leq 1/n$.

Taking the limit in the last expression, we conclude through the sandwich theorem that $x_{n}\to x_{0}$.

But $f(x_{n})\not\to f(x_{0})$ since $|f(x_{n}) - f(x_{0})| > \varepsilon > 0$, and we are done.

Hopefully this helps!


问:
Perfect graphs, and perfectly orderable graphs
I am currently interested in learning more about perfect graphs, and perfectly orderable graphs, however some of the basic concepts are escaping me, and I am simply unable to wrap my head around some of the definitions and examples I am facing. I understand that, for example, a perfectly orderable graph is one where its vertices may be ordered in such a way that allows it to be optimally colored via a greedy coloring algorithm - https://en.wikipedia.org/wiki/Perfectly_orderable_graph.

Here is another definition:

"Given an ordered graph (G, < ), the ordering < is called perfect (Chvátal, V., Perfectly ordered graphs, Perfect graphs, Ann. Discrete Math. 21, 63-65 (1984). ZBL0559.05055.) if for each induced ordered subgraph (H, < ) the greedy algorithm produces an optimal
colouring of H. The graphs admitting a perfect ordering are called perfectfy
orderable. An obstruction in an ordered graph is a chordless path with four
vertices abcd such that a <b and d <c. It is easily seen that a perfectly ordered
graph has no obstruction. Chvatal has shown that this condition is also sufficient:
a graph is perfectly orderable if and only if it admits an obstruction-free ordering" (Hoàng, Chín T.; Maffray, Frédéric; Olariu, Stephan; Preissmann, Myriam, A charming class of perfectly orderable graphs, Discrete Math. 102, No. 1, 67-74 (1992). ZBL0776.05091.)

Nevertheless, I am struggling to understand how to determine that a graph is perfectly orderable (or not perfectly orderable). For example a perfectly orderable graph would be (with a perfect ordering, for example of c<d<e<b<a, and the only two induced subgraphs of 4 vertices, being: abcd & aedc):

And a non-perfectly orderable graph, which does not exhibit a perfect ordering:

In short, why is the first one perfectly orderable but not the other?

答:
For the first one, we can check that the ordering given is perfect: $c<d<e<b<a$. This can be done directly by checking all induced subgraphs (only subgraphs on at least $4$ vertices need to be checked, and there's $6$ of those). Or, we could look at the chordless paths in the graph, and check that they don't form obstructions:

$a,b,c,d$ doesn't because $a>b$ in the ordering (also because $d>c$).

$a,e,d,c$ doesn't because $a>e$ in the ordering.

No other paths are chordless.

For the second one, we want to show that no ordering is perfect, which is easiest to do by showing that there's always an obstruction. In principle, there's $5!$ orderings to check, but we can save on duplicated effort by making use of the symmetry.

Without loss of generality, $a$ is first in the ordering (because the vertices are all identical, and one of them has to be first). Then:

If $c > d$, then $a,b,c,d$ is an obstruction, because $a<b$ and $d<c$.

If $c < d$, then $a,e,d,c$ is an obstruction, because $a<e$ and $c<d$.

So either way, there is an obstruction.


问:
A doubt regarding uniqueness of IVP.
Consider IVP$$y’=f(x,y), y(x_0)=y_0$$  where $f$ being continuous and Lipschitz on a rectangle $$|x-x_0|\leq a, |y-y_0|\leq b.$$ Then by existence and uniqueness theorem, the above differential equation has a unique solution in some  neighbourhood of $x_0$ (say) $ (x_0-h, x_0+h)$ where $h$ is as given in uniqueness theorem. Now let if I solved the problem as usual method( like variable separable etc.) and got a solution (say) $y$ with domain as $D_y$ . Clearly ( as I think ) $$(x_0-h, x_0+h)\subseteq D_f$$

Now my question is that can I say that $y$ is unique solution of above ordinary differential equation in the domain $D_f ?$ or not guarantee outside $(x_0-h, x_0+h)$ even if differential equation has solution outside it .
I am trying by taking example like $$y’=y^{2},y(0)=1$$ by solving as I got solution $y=\frac{1}{1-x}$ and I am not seeing  any other solution in $(-\infty, 1)$. Please suggest . Thanks .

答:
It's not clear what you mean by $D_f$. Given the right side $f$ of an ODE and an initial point $(x_0,y_0)\in D_f$ one can ask for solutions to such an IVP. When the assumptions to the existence and uniqueness theorem are fulfilled at $(x_0,y_0)$ are fulfilled then there is indeed an $h>0$ such that there is a unique solution $x\mapsto y(x)$ of the IVP, defined in an interval $(x_0-h,x_0+h)$. One then wants to "continue this solution analytically onto a maximal $x$-interval".

To do this we need the assumptions of the basic theorem in a global setting. Otherwise it may happen that further away from $x_0$ we reach bad zones where the uniqueness is no longer guaranteed. Consider the following example:
$$y'= 3|y|^{2/3},\quad y(-1)=-1\ .\tag{1}$$
This IVP has in the open $x$-interval $(-2,0)$ the unique solution $x\mapsto y(x)=x^3$. When this solution arrives at $(0,0)$ it can be continued by $y(x)=x^3$ for $x>0$, but also by $y(x)\equiv0$ for $x>0$.

Title: A doubt regarding uniqueness of IVP.


问:
How to find the present value of an annuity-due after a certain period p while interest also varies for each year t
Given:

i) X is the current value at the end of year two of a 20-year annuity-due of 1 per annum.

ii) The annual effective rate for year t is: $$i_t = \frac {1}{8+t}$$

Calculate X.

$$ a(t) = (1+i_t) = \frac {9+t}{8+t}$$

From this point, I honestly have no idea how to evaluate the annuity...

答:
At the end of year $2$, there are $18$ payments of $1$ still to come. The first one come immediately, so its present value is $1$.  The second one comes at the end of year $3$, in which the interest rate is $1/11$ so the present value is $$\left(1+\frac1{11}\right)^{-1}=\frac{11}{12}$$  The third comes at the end of year $4$, so the present value is
$$\left(1+\frac1{11}\right)^{-1}\left(1+\frac1{12}\right)^{-1}=\frac{11}{12}\frac{12}{13}=\frac{11}{13}$$

I'm sure you see how to continue.  We have to discount each payment by the rate in every year from the end of yer $2$ to the payment date, so we get for the present value of the future payments
$$1+11\left(\frac1{12}+\frac1{13}+\cdots+\frac1{28}\right)$$

We must add to this the accumulated value of the past payments, of which there have been $2$.  The payment at the beginning of year $1$ has grown by $1+\frac1{10}=\frac{11}{10}$ and the payment at the beginning of year $1$ has grown by $\left(1+\frac19\right)\frac{11}{10}=\frac{11}9$ so the current value is $$11\left(\frac19+\frac1{10}+\cdots+\frac1{28}\right)=\sum_{t=9}^{28}\frac{11}t$$


问:
Show that the following series $\sum_{n=1}^\infty f'(n)=\sum a_n, \sum_{n=1}^\infty \frac{f'(n)}{f(n)}=\sum b_n$ converge or diverge together.
Let $f: (0,\infty) \to (0,\infty)$ be a $C^2$-function (i.e. twice continuously differentiable) such that $f''(x) < 0$ for all $x \in (0,\infty)$.

Show that the following series $A$ and $B$ either both converge or both diverge
$$ A = \sum_{n=1}^\infty f'(n)=\sum a_n, \quad B = \sum_{n=1}^\infty \frac{f'(n)}{f(n)}=\sum b_n.$$

If $B$ converges then $\lim_{n \to \infty}\frac{a_n}{b_n}=\lim_{n \to \infty}f(n)=0$ and hence $A$ converges.

If $A$ converges then notice that $f''<0$. So $f'$ is a strict decreasing function. Lower Riemann Sum is Right Riemann Sum, Hence we have:
$$\sum_{n=1}^{\infty}f'(n)=f'(1)+\sum_{n=2}^{\infty}f'(n)<f'(1)+\int_{1}^{\infty}f'(x)=f'(1)+f(\infty)-f(1)=f'(1)-f(1)$$

Also Upper Riemann Sum is left Riemann Sum, Hence we have:
$$\sum_{n=1}^{\infty}f'(n)>\int_{1}^{\infty}f'(x)=f(\infty)-f(1)=-f(1)$$

By mixing two inequalities:
$$-f(1)<A<f'(1)-f(1)$$
$$-A<f(1)<f'(1)-A$$

But this is not helping me as I can't conclude for $B$ and don't know whether $\frac{f'(n)}{f(n)}$ is increasing or not!

答:
$f$ is concave and positive, and therefore an increasing function. It follows that $L = \lim_{x \to \infty} f(x)$ exists as a finite value or $+ \infty$.

$f'$ is decreasing, so that $A = \sum_{n=1}^\infty f'(n)$ converges if and only if
$$
 \int_1^\infty f'(x) \, dx = \lim_{x \to \infty} f(x) - f(1) < \infty \, ,
$$
i.e. if $L$ is finite.

Also
$$
 \left( \frac{f'(x)}{f(x)} \right)' = \frac{f''(x)}{f(x)} - \left( \frac{f'(x)}{f(x)} \right)^2 < 0
$$
so that the same argument can be applied to $B = \sum_{n=1}^\infty \frac{f'(n)}{f(n)}$: The series converges if and only if
$$
\int_1^\infty \frac{f'(x)}{f(x)} = \lim_{x \to \infty} \ln f(x) -  \ln f(1) < \infty \, .
$$

So both series converge if and only if $L = \lim_{x \to \infty} f(x)$ is finite.


问:
A special kind of sieve. Find lower bound for cardinality of a given set.
Let:

$A=\{s,s+1,s+2,...,t\}$. Where $s,t$ are some fixed natural numbers

$A(d,a,b)=\{m\in A: d|(m-a) \vee d|(m-b)\}$, where $d,a,b\in \mathbb{N}$

$C$ be a finite subset of prime numbers

$B =A\setminus \bigcup_{p\in C}A(p,a,b)$

Calculate/find formula for $|B|$. If it is impossible, the find nontrivial lower bound.

Here is what i have tried:

$$|B|=|A|-\sum_{i=1}^{|C|}(-1)^i\sum_{\{i_1,...,i_s\}\subset C}|A(i_{1},a,b)\cap ...\cap A(i_{s},a,b)|$$

I have used here an inclusion-exclusion formula.

If we put $a=b=0$, then $A(d,a,b)=A(d,0,0)=\{m\in A:d|m\}$

And we see that for relativiely prime $Q$ and $W$ we get $A(Q\cdot W, 0,0)=A(Q,0,0)\cap A(W, 0,0)$.

Furthermore $|B|=\sum_{d|D}\mu(d)A(d,0,0)$, where $D$ is a product of all the elements from $C$, and $\mu$ is a Möbius function

It seems, that we can't make a similar result for any other $a, b$

Please correct me if i am wrong.

Regards

答:
Not sure if it is what's expected but

$D =\prod_{p\in C} p$
$$|B|=\sum_{m\in [s,t],\ \gcd(m-a,D)=1 \ and \ \gcd(m-b,D)=1}1
$$ $$=\sum_{m\in [s,t]}\ \ \sum_{k| \gcd(m-a,D)}\mu(k)\sum_{l| \gcd(m-a,D)}\mu(l)$$ $$
 =\sum_{k| D}\sum_{l| D}\mu(k)\mu(l)\sum_{m\in [s,t], k | m-a \ and \ l|m-b}1
$$

If $\gcd(k,l)$ doesn't divide $a-b$ then  $$\sum_{m\in [s,t], k | m-a \ and \ l|m-b}1=0$$ otherwise $$\sum_{m\in [s,t], k | m-a \ and \ l|m-b}1=\frac{t-s}{lcm(k,l)} +O(1)$$
(here we can precompute the unique residue class $\bmod lcm(k,l)$ which satisfies $m\equiv a\bmod k,m\equiv b\bmod l$ if we want an exact formula)

Thus we are left with

$$|B|=\sum_{k| D,l| D, \gcd(k,l)|a-b}\mu(k)\mu(l) (\frac{t-s}{lcm(k,l)}+O(1))$$
$$ =(t-s)\eta +O(4^{|C|})$$


问:
Evaluating $\lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{1/n}$ using Stirling's approximation
Although it is a very simple question, I am not able to get similar results using Stirling's approximation as obtained using Integration. Here is what I have attempted.

$$L=\lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{1/n}$$

$$\begin{aligned}L&=\lim_{n\to\infty}\left(\frac{\sqrt{2\pi n}}{e^{n}}\right)^{1/n}\\&=\lim_{n\to\infty}\frac{1}{e}n^{1/2n}\end{aligned}$$

I am not able to reason out as to how this will evaluate to $1/e$. Any hints are appreciated. Thanks.

答: $\lim_n \sqrt[n]{a_n}=\lim_n \frac{a_{n+1}}{a_n}$ if the last limits exists. So $$\lim_n (\sqrt{2\pi n})^{1/n}=\lim_n \frac{\sqrt{2\pi (n+1)}}{\sqrt{2\pi n}}=\sqrt{\lim_n \frac{n+1}{n}}=1$$


问:
Laurent series of $\frac{1}{(1-z)}$
So I see this in my book:

So the first one I understand is the power series representation of a geometric series right?

But what is the second? Is the second just showing that this is an alternative representation of $\frac{1}{(1-z)}$ with negative powers and that's what makes it a laurent series? Since there are only negative powers in the second representation, that's what makes it a Laurent series right?

The two are different representations of the same series correct?

答:
Basically correct.  The two series have different regions of convergence:  the first on $|z|\lt1$, the second $|z|\gt1$.  They are both Laurent series.  The first happens to be a Taylor series,  as well.  They are both around $z=0$.

I wouldn't say  they are representations of the same series, but rather the same function on different regions.

Title: Laurent series of $\frac{1}{(1-z)}$


问:
About integrating product of two sinc function using Fourier transform
So the problem is

which I think is pretty straight-foward by using Fourier transform and convolution property of two sinc functions and evaluating the convolution at 5. However, I got sinc(t) for the convolution result(So the answer is sinc(5)?). But when I use matlab to check, it says that the result is 100sinc(t)(again 100sinc(5))?  I'm really confused and got stuck here.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{%
\int_{-\infty}^{\infty}\on{sinc}\pars{t}\on{sinc}\pars{t - 5}\,
\dd t}
\\[5mm] = &\
\int_{-\infty}^{\infty}
\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic\omega t}\dd\omega}
\bracks{{1 \over 2}\int_{-1}^{1}
\expo{-\ic\nu\pars{t - 5}}\,\,\dd\nu}\dd t
\\[5mm] = &\
{\pi \over 2}\int_{-1}^{1}\expo{5\nu\ic}\
\overbrace{\int_{-1}^{1}\
\underbrace{\int_{-\infty}^{\infty}\expo{\ic\pars{\omega - \nu}t}
\,\,{\dd t \over 2\pi}}_{\ds{\delta\pars{\omega - \nu}}}\
\,\dd\omega}^{\ds{=\ 1}}\ \,\dd\nu
\\[5mm] &\
{\pi \over 2}\int_{-1}^{1}\expo{5\nu\ic}\,\dd\nu =
\pi\int_{0}^{1}\cos\pars{5\nu}\,\dd\nu =
\pi\,{\sin\pars{5} \over 5}
\\[5mm] = &\
\bbx{\pi\,\on{sinc}\pars{5}} \approx -0.6025 \\ &
\end{align}


问:
How to show that this language is decidable
The question :

The language is $L=$ { $< G,w > $ : $G$ is a grammar in normal form of Chomsky and $w$ is a word on the terminal alphabet that can be derivated in $G$ by at least 2 differents derivations trees. }

In other words, find an algorithm that when given a grammar that is in normal form of Chomsky and a word $w$, it concludes after a finite number of steps if yes or no $w$ can be derivated by two derivations trees.

My reflexion and attempt:

Since the grammar is in Chomsky normal form, we know that it will take exactly $2|w|-1$ steps of derivation. So everything is done in a finite time space.

So we have to take our string and be able to derivate it with two derivation tree.

The part i am not sure in the algorithm is how to say that we are switching the derivations rules used ( so we don't get the same derivation tree).

Here's my attempt but it seems odd:

Entry : < G, w >;

Try every derivations in $G$ of length $2|w| -1$.

If two derivations generate w stop and accept < G,w >. Else stop and reject < G,w >.

Thanks a lot for your help.

答:
I’m afraid the “try every derivations” part of your attempt is a bit imprecise.  The idea is there, but how would you do it?  Well, there is a well-known dynamic programming algorithm that “tries every derivation” in time $O(n^3)$: the CYK algorithm.

The algorithm, from Wikipedia, is as follows:

```<code>let the input be a string I consisting of n characters: a1 ... an.
let the grammar contain r nonterminal symbols R1 ... Rr, with start symbol R1.
let P[n,n,r] be an array of booleans. Initialize all elements of P to false.

for each s = 1 to n
    for each unit production Rv → as
        set P[1,s,v] = true

for each l = 2 to n -- Length of span
    for each s = 1 to n-l+1 -- Start of span
        for each p = 1 to l-1 -- Partition of span
            for each production Ra → Rb Rc
                if P[p,s,b] and P[l-p,s+p,c] then set P[l,s,a] = true

if P[n,1,1] is true then
    I is member of language
else
    I is not member of language
</code>```

However, in this basic form, the CYK algorithm is a mere recognizer – it will only tell you if a word belongs to the grammar or not.  To get what we’re after, we need to do some bookkeeping.  Instead of booleans, we can store the parse trees in the $P[l, s, v]$ array.  Here is a modification to the pseudo-code:

```<code>let the input be a string I consisting of n characters: a1 ... an.
let the grammar contain r nonterminal symbols R1 ... Rr, with start symbol R1.
let P[n,n,r] be an array of lists. Initialize all elements of P to the empty list.
let TerminalNode(Rv,as) denote a terminal node with production Rv → as
let NonTerminalNode(Ra,Rb,Rc) denote a nonterminal node with production Ra → Rb Rc

for each s = 1 to n
    for each unit production Rv → as
        append TerminalNode(Rv,as) to P[1,s,v]

for each l = 2 to n -- Length of span
    for each s = 1 to n-l+1 -- Start of span
        for each p = 1 to l-1 -- Partition of span
            l_cell = P[p,s,b]
            r_cell = P[l-p,s+p,c]
            for each production Ra → Rb Rc
                l_nodes = [node in l_cell | node.symbol = Rb]
                r_nodes = [node in r_cell | node.symbol = Rc]
                for each l_node in l_nodes
                    for each r_node in r_nodes
                        append NonTerminalNode(Ra,l_node,r_node) to P[l,s,a]
</code>```

Now, $P[n, 1, 1]$ tells you the entire parse forest that was used to derive the string.  From here, you can easily determine if the derivation was ambiguous or not.

Body:
I’m afraid the “try every derivations” part of your attempt is a bit imprecise.  The idea is there, but how would you do it?  Well, there is a well-known dynamic programming algorithm that “tries every derivation” in time $O(n^3)$: the CYK algorithm.

The algorithm, from Wikipedia, is as follows:

let the input be a string I consisting of n characters: a1 ... an.
let the grammar contain r nonterminal symbols R1 ... Rr, with start symbol R1.
let P[n,n,r] be an array of booleans. Initialize all elements of P to false.

for each s = 1 to n
    for each unit production Rv → as
        set P[1,s,v] = true

for each l = 2 to n -- Length of span
    for each s = 1 to n-l+1 -- Start of span
        for each p = 1 to l-1 -- Partition of span
            for each production Ra → Rb Rc
                if P[p,s,b] and P[l-p,s+p,c] then set P[l,s,a] = true

if P[n,1,1] is true then
    I is member of language
else
    I is not member of language

However, in this basic form, the CYK algorithm is a mere recognizer – it will only tell you if a word belongs to the grammar or not.  To get what we’re after, we need to do some bookkeeping.  Instead of booleans, we can store the parse trees in the $P[l, s, v]$ array.  Here is a modification to the pseudo-code:

let the input be a string I consisting of n characters: a1 ... an.
let the grammar contain r nonterminal symbols R1 ... Rr, with start symbol R1.
let P[n,n,r] be an array of lists. Initialize all elements of P to the empty list.
let TerminalNode(Rv,as) denote a terminal node with production Rv → as
let NonTerminalNode(Ra,Rb,Rc) denote a nonterminal node with production Ra → Rb Rc

for each s = 1 to n
    for each unit production Rv → as
        append TerminalNode(Rv,as) to P[1,s,v]

for each l = 2 to n -- Length of span
    for each s = 1 to n-l+1 -- Start of span
        for each p = 1 to l-1 -- Partition of span
            l_cell = P[p,s,b]
            r_cell = P[l-p,s+p,c]
            for each production Ra → Rb Rc
                l_nodes = [node in l_cell | node.symbol = Rb]
                r_nodes = [node in r_cell | node.symbol = Rc]
                for each l_node in l_nodes
                    for each r_node in r_nodes
                        append NonTerminalNode(Ra,l_node,r_node) to P[l,s,a]

Now, $P[n, 1, 1]$ tells you the entire parse forest that was used to derive the string.  From here, you can easily determine if the derivation was ambiguous or not.


问:
How to directly show that Figure 8 injective immersion is not a monomorphism
I'm working in the category of smooth manifolds. The injective immersion that takes the open unit interval $(0,1)$ to the figure 8 is a well-known example of an injective immersion that is not an embedding. Now embeddings are the monomorphisms in our category. So the figure 8 immersion (let's call it $f$) must not be a monomorphism. Hence, there must exist two distinct maps $g_1,g_2:X\rightarrow (0,1)$ such that $f\circ g_1 = f\circ g_2$. What is an example of such maps?</p>

答:
Let $f : X \to Y$ be any smooth map whatsoever. Suppose $f (x) = f (x')$. Then $x$ and $x'$, considered as smooth maps $1 \to X$, have the property that $f \circ x = f \circ x'$. So if $f$ is monic in the category of manifolds it must be injective.

Conversely, suppose $f : X \to Y$ is an injective smooth map. Then, for any smooth maps $g, g' : T \to X$, if $f \circ g = f \circ g'$, then $g = g'$ (because smooth maps are determined by their action on points). Thus any injective smooth map is monic. In particular the figure-8 injective immersion is monic.


问:
How to determine confidence interval using central limit theorem?
How to determine the confidence interval for the unknown theta parameter of the Uniform($[0, \theta]$)-distribution using the central limit theorem, considering that the significance level is given and $\theta > 0$?

When generating the sample, theta should be considered as known and afterwards should be checked whether or not it's in the interval.

I know it can somehow be solved by generating a uniform sample, calculating the value for z and then using a polynomial with coefficients $a, b$ and $c$, then generating a $\delta$ and the roots will be the interval's endpoints, but I don't understand how I can get those coefficients.
I have a sample of size $N$ and I calculated the mean and the value for z but I'm stuck at this point. How do I calculate those coefficients for the polynomial?

答:
For a random sample with large $n$ you could use the CLT to get a less-than-optimal CI for $\theta.$
Here is an outline:

Notice that $E(\bar X_n) = \theta/2.$ So you might consider the unbiased estimator
$\hat \theta = 2\bar X_n.$

By the CLT, $\hat \theta$ is nearly normal. With what mean and standard deviation?

Then you can use an approximate 95% CI of the form
$\hat\theta \pm 1.96\, \widehat{\mathrm{SE}}(\hat\theta),$ where  $\widehat{\mathrm{SE}}(\hat\theta)$ estimates $\mathrm{SD}(2\bar X_n).$

Note: The estimator of @tommik (+1) will tend to be shorter than the intervals above. For $n=20,$ 95% CIs above are about 2 units long, whereas
95% CIs based on the maximum are about 0.612 units long.

Here is a relevant simulation for the case $n = 20, \theta=4.$ (Vertical lines on plots of histograms suggest typical lengths of 95% confidence intervals for $\theta$ made from sample means and sample maximums, respretively.)

```<code>set.seed(1128)
m = 10^5;  n = 20;  th = 4
x = runif(m*n, 0, th)
MAT = matrix(x, nrow=m)    # mxn matrix, each row is sample of n
th.est = 2*rowMeans(MAT)
mean(th.est);  sd(th.est)
[1] 4.002066               # aprx E(th.est) = 4
[1] 0.5160839
hist(th.est, prob=T, col="skyblue2")
curve(dnorm(x, mean(th.est), sd(th.est)), add=T, col="red", lwd=2)
pm=c(-1,1)
abline(v = mean(th.est)+pm*1.96*sd(th.est))
</code>```

```<code>mx = apply(MAT, 1, max)  # vector of maximums
mx.unb = 21*mx/20
mean(mx.unb)             # aprx E(unb MLE) = 4
[1] 4.000818
mean(mx/.05^(1/20) - mx)
[1] 0.6156929            # aprx length of 95% CIs based on max
hist(mx.unb, prob=T, col="skyblue2", xlim=c(2,4.5))
 abline(v = c( mean(mx), mean(mx/.05^{1/20})))
</code>```

Body:
For a random sample with large $n$ you could use the CLT to get a less-than-optimal CI for $\theta.$
Here is an outline:

Notice that $E(\bar X_n) = \theta/2.$ So you might consider the unbiased estimator
$\hat \theta = 2\bar X_n.$

By the CLT, $\hat \theta$ is nearly normal. With what mean and standard deviation?

Then you can use an approximate 95% CI of the form
$\hat\theta \pm 1.96\, \widehat{\mathrm{SE}}(\hat\theta),$ where  $\widehat{\mathrm{SE}}(\hat\theta)$ estimates $\mathrm{SD}(2\bar X_n).$

Note: The estimator of @tommik (+1) will tend to be shorter than the intervals above. For $n=20,$ 95% CIs above are about 2 units long, whereas
95% CIs based on the maximum are about 0.612 units long.

Here is a relevant simulation for the case $n = 20, \theta=4.$ (Vertical lines on plots of histograms suggest typical lengths of 95% confidence intervals for $\theta$ made from sample means and sample maximums, respretively.)

set.seed(1128)
m = 10^5;  n = 20;  th = 4
x = runif(m*n, 0, th)
MAT = matrix(x, nrow=m)    # mxn matrix, each row is sample of n
th.est = 2*rowMeans(MAT)
mean(th.est);  sd(th.est)
[1] 4.002066               # aprx E(th.est) = 4
[1] 0.5160839
hist(th.est, prob=T, col="skyblue2")
curve(dnorm(x, mean(th.est), sd(th.est)), add=T, col="red", lwd=2)
pm=c(-1,1)
abline(v = mean(th.est)+pm*1.96*sd(th.est))

mx = apply(MAT, 1, max)  # vector of maximums
mx.unb = 21*mx/20
mean(mx.unb)             # aprx E(unb MLE) = 4
[1] 4.000818
mean(mx/.05^(1/20) - mx)
[1] 0.6156929            # aprx length of 95% CIs based on max
hist(mx.unb, prob=T, col="skyblue2", xlim=c(2,4.5))
 abline(v = c( mean(mx), mean(mx/.05^{1/20})))


问:
Hash function and one near hard example?
Example: Suppose $H:${$1,...,n$} $\rightarrow ${$1,..,n$} be a uniform hash function. for input $x$, $z$ is equal to number of trailing zero in the right side of $H(x)$. for $0 \leq c \leq 1$ what is the order of probability $ z \geq c \log_2 n$? $C$ is constant here.

Answer: $O(1/n^c)$

How  this this is can be achieved?

Update:

The Logarithm base is $2$ not $10$.

答:
Since the hash function is uniform, to find $P(z \geq c \log n)$ we can simply count the number of elements in $S = \{1,\ldots,n\}$ that satisfy $z \geq c \log n$.

Take $n=2^p$ for some $p\in\mathbb{N}$. Then $n/2$ elements in $S$ end on one zero, $n/4$ elements end on two zeros, etc. So, $P(z \geq c \log n) = (n/2)/n = 1/2$ for $c\log n =1$, $1/4$ for $c\log n=2$, $1/8$ for $c\log n=3$, etc.

So $P(Z \geq c \log n) = (1/2)^{c \log n} = (1/2^{\log n})^c = (1/n)^c$.


问:
Application more than actual Question I am answering?
I have this hypothetical thought experiment is to find two arbitrary functions with $\mathrm{i}$ starting point \begin{equation}\sum_{n=i}^\infty \frac{f(n)}{g(n)} \end{equation} Converges, and that the following converges as well: \begin{equation} \sum_{n=i}^\infty \frac{f'(n)}{g'(n)} \end{equation}
My question is how would I go about doing so in a coordinated, pragmatic way? They don't have to converge to the same number. I started by testing algebraic functions, and trigonometric, and transcendental. It's ok if this is not possible as well, I would like to just for the gain of the skill of developing a rationale for math intuition. This is not a homework exercise.

答: Hint. The best way to develop intuition is to look for simple examples. I think that counts as pragmatic. In this case think about what happens if $f$ and $g$ are each polynomials, or even just powers of $x$. Think about their degrees.


问:
For which values of $x$ are the terms of this sequence always positive?
Let the sequence $a_n$ be defined as such:
$$\begin{cases}a_0=1\\ a_1=x\\a_n=xa_{n-1}-a_{n-2},\ n \in \mathbb{N}\setminus\{1,2\} \end{cases}$$
I am looking for the $x$ for which the sequence is always positive.

For $x\ge 2$, the sequence is always positive, since it is the sequence of natural numbers for $x=2$.Then, a sequence for $x>2$ is always greater or equal to the sequence for which $x=2$.  I tried finding the limit $\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}$, and since $x>1$ ($a_2=x^2-1$), this limit only exists for $x\ge 2$, as $L=\frac{x\pm \sqrt(x^2-4)}{2}$, so for $2>x>1$ the sequence can't converge. (Since this limit is always 1 for positive sequences that converge.)

Also, if for $2>x>1$ $\ \ \ a_n\le a_{n-1}\Leftrightarrow a_{n+1}\le(x-1)a_{n-1}\Rightarrow a_{n+1}<a_{n-1}$  for positive $a_n$. Meaning that the sequence has to be stricly increasing or at some point it will be negative, because it can't converge since the limit above (of the ratio of $a_{n+1}$ and $a_n$) doesn't exist. But now I'm stuck. How can I show that for all $1<x<2$ the sequence isn't strictly increasing? Or can you find some value for which it is? (I tested experimentally and it seems that for all values under 2 the sequence decreases at some point, but one can never be sure.)

P.S. I found a way to write the general expression of the sequence, but it didn't help me much.
$a_n = x^n - {n-1 \choose 1}x^{n-2}+{n-2 \choose 2}x^{n-4}-{n-3 \choose 3}x^{n-6}+\cdots  $

答:
$a_n=xa_{n-1}-a_{n-2};\;a_0=1,a_1=x$

has characteristic equation

$\lambda^2-x\lambda+1=0$

$$\lambda=\frac{x\pm\sqrt{x^2-4}}{2}$$

therefore the solution of the recurrence are

$$a_n=\alpha\left(\frac{x+\sqrt{x^2-4}}{2}\right)^n+\beta\left(\frac{x-\sqrt{x^2-4}}{2}\right)^n$$
setting the initial values
$$a_0=\alpha+\beta=1;\;a_1=\alpha\,\frac{x+\sqrt{x^2-4}}{2}+\beta\,\frac{x-\sqrt{x^2-4}}{2}=x$$
$$\alpha=\frac{1}{2}+\frac{x}{2 \sqrt{x^2-4}};\;\beta=\frac{1}{2}-\frac{x}{2 \sqrt{x^2-4}}$$
Finally the solution
$$a_n=\left(\frac{1}{2}+\frac{x}{2 \sqrt{x^2-4}}\right)\left(\frac{x+\sqrt{x^2-4}}{2}\right)^n+\left(\frac{1}{2}-\frac{x}{2 \sqrt{x^2-4}}\right)\left(\frac{x-\sqrt{x^2-4}}{2}\right)^n$$

$a_n>0$ for all $n$ if $x-\sqrt{x^2-4}>0\to x>2$


问:
Two circles $C_1$ and $C_2$ of radii $10$ cm and $8$ cm are respectively are tangent to each other internally at a point $A$.

Two circles $C_1$ and $C_2$ of radii $10$ cm and $8$ cm are respectively are tangent to each other internally at a point $A$. $AD$ is the diameter of $C_1$ and $P$ and $M$ are points on $C_1$,$C_2$ respectively such that $PM$ is tangent to $C_2$ . If $PM = \sqrt{20}$  and $\angle PAD = x^\circ$, find $x$.

What I Tried: Here is a picture :-

I absolutely have no good ideas to start this. Let $O$ and $K$ be the centers of the $2$ circles. The only thing which I could do was to find $PO$ using Pythagoras Theorem, which will come out to be $\sqrt{84}$.

I also did, for example, checked this in Geogebra and I found $\angle PAD$ to be $60^\circ$, so I suppose that is the answer, but why is it coming so? I didn't find any ideas using Geogenra either.

Can anyone help me? Thank You.

答:
Note $2x =180^\circ-\angle PKO$ and apply the cosine rule
$$\cos 2x=-\cos\angle PKO
= - \frac{KO^2+KP^2 - PO^2}{2\cdot KO\cdot KP}
$$
Plug in $KO=2$, $KP=10 $ and $OP^2={8^2+20}=84$ to get
$\cos 2x =-\frac12$, yielding $x =60^\circ$.


问:
What is the definition of a ring element raised to the zeroth power?
I have a fairly straightforward question about defining exponentiation in rings: Given any element $a \in R$ of any ring $R$, what is $a^0$ defined as? Is it the additive identity or the multiplicative identity (if it exists)?

In the context of groups we define $a^0 = e$ for whatever the identity of that one operation is. But in the context of rings, $a^n$ would refer to the ring product, so I would assume $a^0$ to be the multiplicative identity, but this doesn't always exist.

Is there any convention, or is it simply left undefined/context dependent?

Edit: In regards to being a duplicate of the $0^0$ question, I had already read through both the question and some answers, but I thought this was a 'fairly separate' question. In particular, this is about rings in general, and while there is a relationship, most of the discussion in the other question pertains explicitly to $\mathbb{R}$. For example discussion of non-integer powers, and limits as a motivating value for the definition, and the fact $\mathbb{R}$ has unity.

答:
Loosely speaking, for $a\in R$, $a^n$ means the product $a\cdots a$ consisting of $n$ factors. Rigorously, we could establish a right $\mathbb{N}^*$-pseudomodule over $(R,\cdot)$ whose module operation is $^\wedge:R\times \mathbb{N}^*\to R$. Here, a pseudomodule is a generalization of a semimodule that does not require the monoid to be commutative.

The properties of $R_{\mathbb{N}^*}$ are then

$a^{n+m} = a^n\cdot a^m$

$a^1 = a$

When $R$ is unitary, then we can use the right $\mathbb{N}$-pseudomodule which includes the property that
$$a^0=1_R$$
for $a\neq 0_R$. Thus in a unitary ring, a zeorth power is indeed the multiplicative identity of the ring.

If $R$ is a division ring, then we upgrade to a right $\mathbb{Z}$-pseudomodule and enable negative exponents. If $R$ is commutative then we can go to a $\mathbb{N}^*$-semimodule. If $R$ is a field, then we have the familiar $\mathbb{Z}$-semimodule of the rational numbers and the real numbers. Rational and then irrational exponents require further properties.


问:
Why is $\text{Var(}\text{min}(X, Y)) = \text{Var(}\text{max}(X, Y))$ for $X, Y \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0,1)$?
I intuitively get that there is some sort of symmetry at play that makes $\text{Var}(\text{max}(X, Y)) = \text{Var}(\text{min}(X, Y))$ for $X, Y \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0,1)$ since, generally, there is symmetry for two continuous, i.i.d random variables $A, B$ in that $P(A > B) = P(A < B)$.

What I don't quite understand is this justification:

By symmetry of the Normal, $(X, Y)$ has the same
distribution as $(-X, -Y)$, so $\text{Var}(\text{max}(X, Y)) = \text{Var}(\text{min}(X, Y))$. (Source)

This post begins to tease this out a little, but I'm still missing why the fact that $-X = X$ and $-Y=Y$ means it's necessarily so that the dispersion of the minimum and maximum of two i.i.d standard normal random variables is the same.

答:
Because $\max(x, y) = -\min(-x, -y)$, you have that
$$\text{Var}(\max(X, Y)) = \text{Var}(-\min(-X, -Y))$$

But since $\text{Var}(-X) = \text{Var}(X)$, this simplifies to $\text{Var}(\min(-X, -Y)$. And then since the distribution of $X$ and $-X$ are the same, this simplifies to $\text{Var}(\min(X, -Y))$. Finally because the distribution of $Y$ and $-Y$ are the same, this simplifies to $\text{Var}(\min(X, Y))$.

In general, if $X$ and $Y$ are chosen from symmetric distributions (it doesn't necessarily have to be the same distribution for each), then $$\text{Var}(\max(X, Y)) = \text{Var}(\min(X, Y))$$


问:
Determining a matrix M that represents $\phi$ relative to a basis B
This is for a specific example.

Let V be a vector space of functions q : $R^2$ $\to$ $R$ where f(x,y) = $a_1$$x^2$ + $a_2$$xy$ + $a_3$$y^2$ + $a_4$$x$ + $a_5$$y$ + $a_6$

$a_i$ $\in$ $R$ and $\phi$ : V $\to$ V be linear map given by

$\phi$(f) = $\partial$/$\partial$x $\int$ f(x,y) dy

Question: Determine the matrix M that represents $\phi$ relative to the basis $B$ = { $x^2$, $xy$, $y^2$, $x$, $y$, 1 }

Do you get $\phi$($x^2$) = 2$xy$, $\phi$($xy$) = $y^2$/2, $\phi$($y^2$) = 0, $\phi$($x$) = $y$, $\phi$($y$) = 0, $\phi$(1) = 0 then represent it as a matrix so the 1st column would be "0 2 0 0 0 0 ", 2nd "0 0 1/2 0 0 0", 3rd "0 0 0 0 0 0", 4th "0 0 0 0 1 0" and last two columns are all Os?

Or do you get $\phi$(f(x,y)) = 2$a_1$$xy$ + $a_2$$y^2$/2 + $a_4$y then use that?

I'm not really sure what I'm looking for. The next parts ask for the minimal polynomial and Jordan normal form so i assume the main diagonal in M has non-zero entries.

Any help would be appreciated, thanks.

答:
The general process for representing a linear map $\phi: V \to V$ as a matrix with respect to a basis $B$ is very simple:

Compute the images of the elements of $B$ under $\phi$.

Write each of those images as a sum of elements of $B$.

Use the coefficients of those terms as the entries of your matrix.

In this case, you have already done the first two steps, so you need only write down the matrix

$$M = \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
2 & 0 & 0 & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

Or do you get $\phi(f(x,y)) = 2a_1xy + a_2y^2/2 + a_4y$ then use that?

You could also do it this way: you'd get the exact same answer.

The next parts ask for the minimal polynomial and Jordan normal form so i assume the main diagonal in M has non-zero entries.

Neither of those requries the main diagonal to have non-zero entries.


问:
Why is $e^\pi - \pi$ so close to $20$?
$e^\pi-\pi\approx 19.99909998$

Why is this so close to $20$?

答:
Probably the simplest possible answer is in looking at

$$f(x,y)=x^y-y$$

around $(x,y)=(3,3)$ because both $\pi$ and $e$ are close to $3$

Taylor to first degree would be

$$f(3+x,3+y)=3^3-3+(y-3)(3^3\log(3)-1)+3^3(x-3)=24+(y-3)(27\log(3)-1)+27(x-3)$$

Notice that a very close expression (without $\log(3)$)

$$24+(y-3)(27-1)+27(x-3)=24+26(y-3)+27(x-3)$$

for rational approximations $e=\frac{19}{7}$ and $\pi=\frac{22}{7}$ is giving

$$24+(\frac{22}{7}-3)(27-1)+27(\frac{19}{7}-3)=20$$

So we are dealing with two functions that are very close and linear in the region which we are interested in, as you can see in the graph.

We want to arrive from the black dot at the blue line to the black dot at the orange line. But look, the situation is very symmetrical and quite linear so it is not surprise that we do not move too much from targeted $20$ when we use the intended full arguments not their rational approximations.

Let us swap the arguments to make this more obvious. Notice that both functions behave very linearly in this small space.

$$f_1(x,y)=x^y-y-20$$

$$f_2(x,y)=24+26(y-3)+27(x-3)-20$$

$$f_1(\frac{19}{7},\frac{22}{7})=(\frac{19}{7})^{\frac{22}{7}}-\frac{22}{7}-20=-0.079$$
$$f_2(\pi,e)=24+26(\pi-3)+27(e-3)-20=0.075$$

So we expect that swapping arguments will cancel the effect and keep the value intact, so we will get close to $20$ again, even with $f(x,y)=x^y-y$ and $(x,y)=e,\pi$

To make it totally clear create two implicit functions:

$$g_1(x,y):(\frac{19}{7}+x)^{\frac{22}{7}-y}-(\frac{22}{7}-y)=20$$

$$g_2(x,y):24+26(\pi+y-3)+27(e-x-3)=20$$

The difference between these two functions in the region we are interested in $(e-\frac{19}{7},\frac{22}{7}-\pi)$ is just about the precision we have in the expression which we are observing that is $0.0001$. As we push $x,y$ towards the value in $g_2$ that coincide with rational approximations, $g_1$ goes towards the intended values of $e$ and $\pi$.

Together this magic is consisting of three points:

Having rational approximations of $\pi$ and $e$ with a small denominator, actually the small one, $7$, with sufficient precision. Sure, this makes it easy to conjure up quite some integer connections, the simplest of which would be

$$\frac{e}{19}-\frac{\pi}{22}=0.0002...$$

Having $\pi$ and $e$ sufficiently close to $3$

A specific function $x^y-y$ that is comparable with another closely related linear function whose evaluation with rational approximations of $e$ and $\pi$ is an integer.

I am quite certain that just based on these principles of low but sufficiently precise rational approximations, one can invent tons of similar "surprising" examples. The choice of function is just that - a choice. The fact that $e$ and $\pi$ have the same small denominator in rational approximation might be something that would require some slightly deeper theory, but I doubt we would need much there.

Still, it is kind of a small theory, not just an incident I would say. It is just a couple of connected general properties.

Using the above system you can derive this "marvel":

$$\frac{1}{8}(e^4+\pi^4)=19.000905$$

Add-on:

After checking many similar examples, almost all of them follow the same pattern as long as the incigers (ha, did you see what I have done there) are relatively small.

I will show the system for $\pi$ but it is all the same for any other number and combination as long as the rational approximation is sufficiently good. For $\pi$ we set a reference to be $3$ but it could be any other rational number close to $\pi$.

Create an equation

$$a_0 + a_1(x-3)$$

where $a_0$ and $a_1$ are not extremely large, so that $m$ is an integer when $x$ is a rational approximation of $\pi$, $x_{\pi}$ like $\frac{22}{7}$.

$$a_0 + a_1(x_{\pi}-3)=m$$

Find a function $f(x)$ that is somewhat linear around $3$ so that

$$f(3) \approx a_0$$

and

$$f'(3) \approx a_1$$

if not equal. Then we expect $f(\pi) \approx m$ as well. Use this over $x+x^2$ and you will find not perfect but a good starter $\pi+\pi^2=13.01...$

Title: Why is $e^\pi - \pi$ so close to $20$?


答: A slight variant on @VarunVejalla's approach: let $\Phi$ denote the $\mathcal{N}(0,\,1)$ CDF so$$P(\min(X,\,Y)\le z)=1-(1-\Phi(z))^2=1-\Phi(-z)^2$$(the last $=$ uses symmetry), and $P(\max(X,\,Y)\le z)=\Phi^2(z)$. So $\min(X,\,Y)$'s distribution is that of $-\max(X,\,Y)$, whence$$\operatorname{Var}\min(X,\,Y)=\operatorname{Var}[-\max(X,\,Y)]=\operatorname{Var}\max(X,\,Y).$$

Title: Why is $\text{Var(}\text{min}(X, Y)) = \text{Var(}\text{max}(X, Y))$ for $X, Y \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0,1)$?

Body:
I intuitively get that there is some sort of symmetry at play that makes $\text{Var}(\text{max}(X, Y)) = \text{Var}(\text{min}(X, Y))$ for $X, Y \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0,1)$ since, generally, there is symmetry for two continuous, i.i.d random variables $A, B$ in that $P(A > B) = P(A < B)$.

What I don't quite understand is this justification:

By symmetry of the Normal, $(X, Y)$ has the same
distribution as $(-X, -Y)$, so $\text{Var}(\text{max}(X, Y)) = \text{Var}(\text{min}(X, Y))$. (Source)

This post begins to tease this out a little, but I'm still missing why the fact that $-X = X$ and $-Y=Y$ means it's necessarily so that the dispersion of the minimum and maximum of two i.i.d standard normal random variables is the same.


问:
Cone equations given its vertex and a sphere circumscribe in it
In a solved problem it appears the following equations describing the points in a cone with vertex $v=(a,b,c)$ wich circumscribe the sphere $x^2+y^2+z^2=1$ :
$$\mathcal{C}: (ax+by+cz-1)^2=(x^2+y^2+z^2-1)(a^2+b^2+c^2-1)$$
I want to know where this equation comes from.

答:
Let:

$$X_1^T=(a,b,c,1) \ \ \text{and} \ \ \ X^T=(x,y,z,1)$$

(homogeneous coordinates).

The given equation has the classical form

$$T^2=SS_1 \ \iff \ (X^TAX_1)^2 = (X^TAX)(X_1^TAX_1)\tag{1}$$

of a cone with vertex $(a,b,c)$ circumscribed to any quadric surface with equation $X^TAX=0$ where $A$ is the associated $4 \times 4$ matrix (here $A=diag(1,1,1,-1)$).

Here is a simple proof:

Evidently, $X=X_1$ verifies (1).

Consider a ray $X_1+\lambda U$ issued from $X_1$.

In a first step, saying that this ray is tangent to the quadric amounts it has a single common point with it, i.e., the following quadratic equation in $\lambda$ has a double root:

$$(X_1+\lambda U)^TA(X_1+\lambda U)=0$$

$$\lambda^2U^TAU+2 \lambda U^TAX_1+X_1^TAX_1=0$$

Otherwise said, its discriminant must be $0$:

$$(U^TAX_1)^2-(U^TAU)(X_1^TAX_1)=0 \tag{2}$$

Setting

$$p:=X_1^TAX_1, \ \ \ q:=U^TAX_1  \ \ \ \text{and} \ \ \ r:=U^TAU,\tag{3}$$

relationship (2) can be written:

$$q^2=pr \tag{4}$$

Now, in the second step, let us prove that, with $U$ as above, any

$$X=X_1+\mu U, \tag{5}$$

for any $\mu$ (this is the subtle part: taking a general $\mu$, independent of the $\lambda$ above...) verifies equation (1).

Indeed, plugging (5) into (1) gives:

$$((X_1+\mu U)^TAX_1)^2 = ((X_1+\mu U)^TA(X_1+\mu U))(X_1^TAX_1)$$

which, by expansion, using notations (3), becomes

$$(p+\mu q)^2 = (p+2q\mu+r\mu^2)p \tag{6}$$

Taking (4) into account, (6) appears as an identity, valid for any $\mu$, proving that the whole ray $X_1+\mu U$ belongs to the locus.

Remark: A proof for the 2D equivalent expression can be found here.


问:
Algebra step for Noether theorem
In Tong’s preliminaries for QFT, he makes the following step while showing Noether’s theorem:

$(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta \dot q) = \left(\frac{\partial L}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q} \right)\right)\delta q + \frac{\partial L}{\partial \dot q}\delta q$

How do you get algebraically from the left to the right side there?

答: The equality you asked about is wrong, and I will show why. Since $$\frac{\partial L}{\partial \dot{q}}\delta \dot{q} =\frac{\partial L}{\partial\dot{q}}\left ( \frac{d}{dt}\delta q \right )$$ we can write using the product rule: $$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}}\delta q \right ) = \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}$$ and so $$\frac{\partial L}{\partial \dot{q}}\delta \dot{q} = \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}}\delta q \right )-\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\delta q$$ and so the equality would be: $$\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q} = \left ( \frac{\partial L}{\partial q} - \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\right )\delta q + \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \delta q\right )$$ which is the same as your equality, except it has a time derivative on the last term of the RHS. Notice that the first two terms in the RHS vanish by the Euler-Lagrange equation.

Title: Algebra step for Noether theorem


问:
Help to resolve a Double Integral
I'm doing a workout guide about double integrals and I came across an exercise that I could not resolve for a while.

$$\int_0^2\int_1^2 \frac{x}{\sqrt{1+x^2+y^2}} \,\mathrm dx\,\mathrm dy$$

I guess that the easier order of integration is $dxdy$, because if I try to integrate respect to $y$ first, I would have to deal with the integral of a root of $1+x^2+y^2$, while outside of the root there's no $y$.

I tried Integration by substitution (with $u = 1 + x^2 + y^2$ and $du = 2xdx$) to solve the inner integral, but then I felt I couldn't solve the outside integral.

Any hints you can give me?.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2}\int_{1}^{2}
{x \over \root{1 + x^{2} + y^{2}}}\,\dd x\,\dd y} =
\int_{0}^{2}\int_{1}^{2}
\nabla\times\pars{\root{1 + x^{2} + y^{2}}\,\hat{y}}
\cdot\dd\vec{S}
\\[5mm] = &\
\oint_{\partial\bracks{\pars{1,2}\times\pars{0,2}}}\,\,\,
\root{1 + x^{2} + y^{2}}\,\hat{y}\cdot\dd\vec{r}
\\[5mm] = &\
\int_{0}^{2}\root{1 + 2^{2} + y^{2}}\,\dd y +
\int_{2}^{0}\root{1 + 1^{2} + y^{2}}\,\dd y
\\[5mm] = &\
\underbrace{\int_{0}^{2}\root{5 + y^{2}}\,\dd y}
_{\ds{3 + {5 \over 4}\ln\pars{5}}}\ -\
\underbrace{\int_{0}^{2}\root{2 + y^{2}}\,\dd y}
_{\ds{\root{6} + \ln\pars{\root{3} + \root{2}}}}
\\[5mm] = &\
\bbx{3 - \root{6} + {5 \over 4}\,\ln\pars{5} -
\ln\pars{\root{3} + \root{2}}} \approx 1.4161 \\ &
\end{align}

Title: Help to resolve a Double Integral


问:
Borel sigma algebras of subsets of $\mathbb{R}$
I'm trying to prove a result that seems to be used routinely, but the proof is skipped over.

Let $\mathbb{R}$ have the Euclidean metric, and this defines a topology. Let $A \subset \mathbb{R}$ be equipped with the subspace topology. Then,

$\mathcal{B}(A) = \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$

If $A \in \mathcal{B}(\mathbb{R})$, then $\mathcal{B}(A) = \{ B \in \mathcal{B}(\mathbb{R}) : B \subset A\}$.

The proof for the $\subset$ direction of 1. is simple; note that $\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$ is a sigma algebra, which contains all the open sets of $A$. There are also other questions on stackexchange which address this direction of the proof. However, I am stuck on the $\supset$ direction. Does anyone have a nice proof for this?

(2. is an immediate consequence of 1., as far as I can see)

答:
You have already proved that
$\mathcal{B}(A) \subseteq  \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} $, and your prooof is correct: $\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$ is a sigma algebra, which contains all the open sets of $A$. So $\mathcal{B}(A) \subseteq  \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} $.

Let us prove, in a simple way, that
$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq
\mathcal{B}(A)$$

Proof: Let $\tau$ be the collections of open sets in $\mathbb{R}$.

Let $$K = \{ E \cup F : E  \in \mathcal{B}(A) \textrm{ and } F \in \mathcal{B}(A^c) \}$$

(note that $E$ and $F$ are disjoint). It is easy to see that $K$ is a $\sigma$-algebra. For all $O \in \tau$, $O =(O\cap A) \cup (O\cap A^c)$, $(O\cap A)\in\mathcal{B}(A)$ and  $(O\cap A^c)\in\mathcal{B}(A^c)$. So $O \in K$.  It follows that $\mathcal{B}(\mathbb{R}) \subseteq K$. So we have

$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \{ G \cap A : G \in K\}$$

But $\{ G \cap A : G \in K\}=\mathcal{B}(A)$. So we have:
$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq
\mathcal{B}(A)$$

Remark: A second way to do this proof (which is just a variant) is:

Proof 2: Let $\tau$ be the collections of open sets in $\mathbb{R}$.

Let $$K = \{ G \subseteq \mathbb{R} :  G\cap A  \in \mathcal{B}(A)  \}$$

It is easy to see that $K$ is a $\sigma$-algebra. For all $O \in \tau$, $(O\cap A)\in\mathcal{B}(A)$. So $O \in K$. It follows that $\mathcal{B}(\mathbb{R}) \subseteq K$.

$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \{ G \cap A : G \in K\}$$

But $\{G \cap A : G \in K\} \subseteq \mathcal{B}(A)$. So we have:
$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq
\mathcal{B}(A)$$


问:
$\mathbb{Z}_{2}G$ as the class of finite subsets of $G$
Let $G$ be a group. Then the group ring $RG$ is defined as $\sum_{g\in G} R$ (a copy of $R$ for every $g\in G$) and a typical element is of the form $  \sum_{i=1}^{n} r_{g_{i}}g_{i} $. Now if $R= \mathbb{Z}_{2}$ we can think $\mathbb{Z}_{2}G$ as the class of finite subsets of $G$. Why is that? I'm having trouble understanding it.

答:
Because each element in $\Bbb Z_2$ is either $0$ or $1$.

If you interpret $0$ as "not in the subset" and $1$ as "in the subset", then a finite sum $\sum_{g\in G} r_g g$ defines a finite subset and this gives a bijection between elements in $\Bbb Z_2 G$ and finite subsets of $G$.

Note that this gives a structure of group ring to the set of finite subsets of $G$.


问:
Show that Fourier series of $f(x)=|x|$ converges pointwise and uniformly to $f(x)$ on $[-\pi,\pi]$.
The Fourier series of $f(x)=|x|$ on $[-\pi,\pi]$ can be written as $$f(x)\sim \dfrac{\pi}{2}+2\sum_{n=1}^{\infty}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx}.$$

This post Show absolute and uniform convergence of a Fourier series has shown that this Fourier series converges uniformly on $[-\pi,\pi]$. However, I want to show that this Fourier series converges uniformly to $f$ on the whole $[-\pi,\pi]$.

To this end, I define the partial sum $$S_{N}:=\dfrac{\pi}{2}+2\sum_{n=1}^{N}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx},$$ and then try to estimate $|f(x)-S_{N}(x)|$. I have some preliminary result, but it does not give me $$\sup_{x\in[-\pi,\pi]}|f(x)-S_{N}(x)|\longrightarrow 0,$$ when $N\rightarrow\infty$.  Certainly, I can use the $\epsilon-N$ definition, but my estimate outcome is not that computation friendly either for me to get $N$ for each $\epsilon$.

The following is my estimate:

As $|e^{-inx}|=1$, we can have the following estimate
\begin{align*}
|f(x)-S_{N}(f)(x)|=\Bigg|f(x)-\dfrac{\pi}{2}-2\sum_{n=1}^{N}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx}\Bigg|&\leq |f(x)|+\dfrac{\pi}{2}+2\sum_{n=1}^{N}\Bigg|\dfrac{(-1)^{n}-1}{\pi n^{2}}\Bigg|\\
&=|x|+\dfrac{\pi}{2}+2\sum_{n=1}^{N}\Bigg|\dfrac{(-1)^{n}-1}{\pi n^{2}}\Bigg|\\
&\leq \dfrac{3\pi}{2}+\dfrac{4}{\pi}\sum_{n=1}^{N}\dfrac{1}{n^{2}}\\
&\leq \dfrac{3\pi}{2}+\dfrac{4}{\pi}\Bigg(\dfrac{\pi^{2}}{6}-\dfrac{1}{N+1}\Bigg)\\
&=\dfrac{3\pi}{2}+\dfrac{2\pi}{3}-\dfrac{4}{\pi(N+1)}.
\end{align*}

Then I got stuck. We have a good outcome that the bound does not depend on $x$, but solve the bound $<\epsilon$ seems really complicated. Is there any way for me to make this nicer? The best case would be that I can directly conclude the bound goes to $0$ when $N\rightarrow\infty$.

Thank you!

Edit: Pointwise Convergence

As the answer of "Mostafa Ayaz" suggested, we need to firstly prove that the Fourier series converges to $f(x)$ pointwise on the interval $[-\pi,\pi]$.

In fact, the reason why I directly proved the uniform convergence was that I did not know how to prove the pointwise convergence.

I mean, it is straightforward to prove that the series convergences, but how to prove it convergent pointwise to $f(x)$ on the whole $[-\pi,\pi]$?

Edit 2:

Okay. I just recalled that $f(x)=|x|$ is Holder continuous, so the partial sum must converge pointwise.

答:
Your approach is correct, but using the triangle inequality at the first step is a bit overkill and leads to irrelevant bounds. Just do the following
$$
|f(x)-S_{N}(f)(x)|{=\Bigg|f(x)-\dfrac{\pi}{2}-2\sum_{n=1}^{N}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx}\Bigg|
\\=
\Bigg|\dfrac{\pi}{2}+2\sum_{n=1}^{\infty}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx}-\dfrac{\pi}{2}-2\sum_{n=1}^{N}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx}\Bigg|
\\=
\Bigg|2\sum_{n=N+1}^{\infty}\dfrac{(-1)^{n}-1}{\pi n^{2}}e^{-inx}\Bigg|
\\\le
\Bigg|2\sum_{n=N+1}^{\infty}\dfrac{(-1)^{n}-1}{\pi n^{2}}\Bigg|
\\\le
\sum_{n=N+1}^{\infty}\dfrac{4}{\pi n^{2}}
}
$$
From now, it is very straightforward.


问:
Composite of Continuous Real Functions at Point -- not necessarily continuous?
Let us recall the definition of continuity of a real function at a point.

$f(x)$ is continuous at $a$ iff:

$\forall \epsilon \in \mathbb R_{>0}: \forall \delta \in \mathbb R_{>0}: \forall x \in \mathbb R: |x - a| < \delta \implies |f(x) - f(a)| < \epsilon$

Let $f(x)$ denote the function:

$\forall x \in \mathbb R: f(x) = \begin{cases} 0 & : x \text { irrational} \\ \dfrac 1 q & : x = \dfrac p q: p, q \in \mathbb Z, q > 0, \gcd (p, q) = 1 \end{cases}$

This is Thomae's function, and is well-known for its pathological property of being continuous at all $x$ irrational but discontinuous at all $x$ rational.

Let $g(x)$ denote the function:

$\forall x \in \mathbb R: g(x) = \begin{cases} 0 & : x \le 0 \\ 1 & : x > 0 \end{cases}$

This is continuous everywhere except where $x = 0$.

Now consider the function:

$\forall x \in \mathbb R: h(x) = g(f(x))$

If you work this out, you establish that $h(x)$ is the function defined as:

$\forall x \in \mathbb R: h(x) = \begin{cases} 0 & : x \text { irrational} \\ 1 & : x \text { rational} \end{cases}$

which is also a famous pathological function: it is a Dirichlet function, which has the delightful property of being discontinuous everywhere.

Hence we have the composite of 2 functions, each of which is continuous at certain points (i.e. irrationals for $f$, non-zero for $g$) but itself is not continuous anywhere.

But then, you turn the page in your Topology textbook (in this case "Introduction to Metric and Topological Spaces" by W.A. Sutherland), and you read:

"If $T_1$, $T_2$ and $T_3$ are topological spaces, and $f: T_1 \to T_2$, $g: T_2 \to T_3$ are continuous mappings, then $g \circ f: T_1 \to T_3$."

The language of continuity is defined in topology in terms of inverse mappings on open sets:

$\forall U \in \tau_2: f^{-1} [U] \in \tau_1$

where $\tau_1$ and $\tau_2$ are the topologies of $T_1$ and $T_2$ respectively.

Now there does exist a definition of continuity at a point in the topological context:

The mapping $f$ is continuous at $x$ (with respect to the topologies $\tau_1$ and $\tau_2$) iff:

for every neighborhood $N$ of $f(x)$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f [M] \subseteq N$.

And in this context, the continuity or not of a composite function at a point is not well explored. It seems to be taken for granted that the composite of two continuous functions at a point is itself composite, but (as seen in the specific instance of Thomae's function) this is specifically contradicted by certain pathological examples.

(You can see how this applies to the $\epsilon-\delta$ definition of continuity on Thomae's function: the closer you get to an irrational number, the larger and larger get the denominators of the rational numbers near it, hence the smaller and smaller get the images of those rational numbers. So you have an open set around an irrational $c$, say, which is a teeny-tiny open real interval containing rational numbers with large denominators, which is the preimage of another teeny-tiny open real interval containing reciprocals of those large denominators, and so the topological definition of a continuous mapping applies.)

When you map this image interval, containing $0$ and $+$ve rationals in it, to $\mathbb R$ using $g$, the subsequent image is $\{0, 1\}$. Going back the other way, I am having difficulty interpreting the "neighbourhood" definition of continuity in order to investigate the reason why the continuity of a composite function at a point in a general topological space.

Can somebody help me through this? I need to be guided through the exploration of continuity of the composite of mappings which are continuous at a point ''from the topological perspective''.

答:
You have the following result: Let $f : T_1 \to T_2, g: T_2 \to T_3$. If $f$ is continuous at $x \in T_1$ and $g$ is matchingly continuous at $f(x) \in T_2$, then $g \circ f : T_1 \to T_3$ is continuous at $x \in T_1$.

The problem with your $f : \mathbb{R} \to \mathbb{R}$, Thomae's function and $g : \mathbb{R} \to \mathbb{R}$, the step function at $0$, is that the continuities don't match as above. More specifically, $f$ is continuous at all irrational $j$. However, it maps those points to $0$: $f(j) = 0$, precisely where $g$ fails to be continuous. Thus, the result above does not allow you to conclude the continuity of $g \circ f$ at irrational $j$.


问:
How many zeros does $x^2 + y^2 \pmod d$ have on $[0, d-1]^2$?
I have been doing some work on Pythagoras's Theorem with my Year 8 Maths class (7th Grade in US speak). I had them investigating what values were unobtainable for the square on the hypotenuse of right-angled triangles with integer lengths. In other words, the values that $x^2+y^2$ can take for integers $x,y$.

One student decided to start looking at the factors of $x^2+y^2$. They came up with the following question (in slightly less formal language):

Given a positive integer $d$, when is $x^2+y^2 \equiv 0\pmod d$, for integers $x$ and $y$?

I created this Geogebra resource to visualise where the zeros occur (unexpectedly creating some beautiful patterns).

Some values of $d$, like $d=17$ and $d=25$, give loads of zeros. Others, like $d=19$, give zeros only when $xy \equiv 0 \pmod {19}$.

The following table gives the number of zeros ($N$) of $x^2+y^2 \pmod d$ for integer $(x,y) \in [0,d-1]^2$, for $d=1,...,20$:

... and this scatter graph plots $N/d$ against $d$ for $d=1,...,40$:

I am now stuck with the following questions:

Given $d$, is it possible to predict $N$? (The values of $N$ seem dependent upon whether $d$ is a square number and whether $d$ occurs in a Pythagorean Triple, but with some weird exceptions like $15$ and $25$.)

Is there some intuitive explanation of the wavy patterns that are generated by plotting $x^2+y^2 \pmod d$, as in the Geogebra app?

Any guidance appreciated! Thank you.

$\small{\textrm{(Image: the contrast of each coordinate is proportional to $x^2+y^2 \pmod {40}$, with zeros highlighted pink.)}}$

答:
Some information copied over from oeis.org/A086933, but in the OP's notation.

$N(d)$ is a multiplicative function. That is, if $c, d$ are relatively prime, then $N(cd) = N(c)N(d)$.
Also

$N(2^e) = 2^e$

$N(p^e) = ((p-1)e+p)p^{e-1}$ for primes $p \equiv 1 \mod 4$

$N(p^e) = p^{e-(e \bmod 2)}$ for primes $p \equiv 3 \mod 4$.

For example, $300 = 2^2\cdot3\cdot5^2$. Now, $3 \equiv 3\mod 4$ and $5\equiv 1 \mod 4$, so

$N(2^2) = 2^2 = 4$,

$N(3) = N(3^1) = 3^{1-(1\bmod 2)} = 3^{1 - 1} = 3^0= 1$,

$N(5^2) = ((5-1)\cdot2 + 5)\cdot 5^{2-1} = 13\cdot 5 = 65$.

Therefore $N(300) = 4\cdot1\cdot 65 = 260$.

From this we can see that $N(d) = 1$ if and only if $d$ is a product of distinct primes congruent to $3\bmod 4$.

Another formula for $N(d)$ is:
$$N(d) = \sum_{r|d, r \text{ odd}} (-1)^{(r-1)/2}{\phi(r)}\frac dr$$


问:
Geometry - Equilateral triangle covered with five circles
I have to cover an equilateral triangle (whose sides are 1m long) with 5 identical circles: what's the minimum radius of the circles?</p>

答:
The minimum radius $r$ of the circle is $1/4$.

Hans Melissen showed that $r\le 1/4$.

Now let $A$ be a subset of the triangle, consisting of its vertices and the middles of its edges. Then the distance between each pair of different points of $A$ is at least $1/2$. Since in each cover of the triangle by five circles there is a circle covering two different points of $A$, its radius is at least $1/4$.


答:
The characteristic equation has two roots: $r=\frac{x+\sqrt{x^2-4}}{2}, s=\frac{x-\sqrt{x^2-4}}{2}$.

Note that $a_n-(r+s)a_{n-1}+rsa_{n-2}=0$

Case 1: $x>2$, $r, s$ are real and $r>s>0$.

$$
a_{n+1} - s a_n = r(a_n-sa_{n-1})
\Rightarrow a_{n+1} - s a_n =  r^n(a_1-sa_0) =r^n (x-s) =r^{n+1}\tag1\\
$$
$$a_{n+1} - r a_n = s(a_n-ra_{n-1})
\Rightarrow a_{n+1} - r a_n = s^n(a_1-ra_0)= s^n (x-r) = s^{n+1}\tag2
$$

(1)-(2),
$$a_n=\frac{r^{n+1}-s^{n+1} }{r-s} >0  \tag 3$$

(Note that $a_n=\frac{r^{n+1}-s^{n+1} }{r-s}$ as long as $r\neq s$, even if they are complex).

Case 2: $x=2$.

$$a_n-2a_{n-1}+a_{n-2}=0 \implies a_n-a_{n-1}=a_{n-1}-a_{n-2} \\
\implies a_n-a_{n-1}=a_1-a_0=1 \implies a_n=n$$

So $a_n>0, \forall n$.

Case 3: $x<2$, $r$ and $s$ are conjugate complex numbers.

Let $r=\exp\{i\theta\}, s=\exp\{i(-\theta)\}$ (since $rs=1$ we know $|r|=|s|=1$).

Now $$a_n=\frac{r^{n+1}-s^{n+1} }{r-s}=\frac{2\sin((n+1)\theta)}{2 \sin \theta} = \frac{\sin((n+1)\theta)}{\sin \theta}$$

Note that $\theta=\tan^{-1}\frac{4-x^2}{x} \in (0,\frac{\pi}{2}) \implies \lambda :=\frac{\theta}{\pi} \in (0,\frac 12)$.

If $a_n>0, \forall n$, then $\lfloor n \lambda \rfloor$ is always even. Let $k$ be the smallest integer such that $k\lambda \ge 2$, then $(k-1)\lambda < 1, \lambda = k\lambda - (k-1)\lambda > 2-1=1$, a contradiction.

So for some $n$, $a_n \leq 0$.

Conclusion: The sequence is always positive if and only if $x\geqslant 2$.

Title: For which values of $x$ are the terms of this sequence always positive?

Body:
Let the sequence $a_n$ be defined as such:
$$\begin{cases}a_0=1\\ a_1=x\\a_n=xa_{n-1}-a_{n-2},\ n \in \mathbb{N}\setminus\{1,2\} \end{cases}$$
I am looking for the $x$ for which the sequence is always positive.

For $x\ge 2$, the sequence is always positive, since it is the sequence of natural numbers for $x=2$.Then, a sequence for $x>2$ is always greater or equal to the sequence for which $x=2$.  I tried finding the limit $\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}$, and since $x>1$ ($a_2=x^2-1$), this limit only exists for $x\ge 2$, as $L=\frac{x\pm \sqrt(x^2-4)}{2}$, so for $2>x>1$ the sequence can't converge. (Since this limit is always 1 for positive sequences that converge.)

Also, if for $2>x>1$ $\ \ \ a_n\le a_{n-1}\Leftrightarrow a_{n+1}\le(x-1)a_{n-1}\Rightarrow a_{n+1}<a_{n-1}$  for positive $a_n$. Meaning that the sequence has to be stricly increasing or at some point it will be negative, because it can't converge since the limit above (of the ratio of $a_{n+1}$ and $a_n$) doesn't exist. But now I'm stuck. How can I show that for all $1<x<2$ the sequence isn't strictly increasing? Or can you find some value for which it is? (I tested experimentally and it seems that for all values under 2 the sequence decreases at some point, but one can never be sure.)

P.S. I found a way to write the general expression of the sequence, but it didn't help me much.
$a_n = x^n - {n-1 \choose 1}x^{n-2}+{n-2 \choose 2}x^{n-4}-{n-3 \choose 3}x^{n-6}+\cdots  $


问:
Show that $\{1, \cos x, \sin x, \cos(2x), \sin(2x), \dots, \cos(nx), \sin(nx)\}$ is an orthogonal set of V .
Suppose $V := CR([0, 2\pi])$ is the collection of real valued functions on $[0, 2\pi]$
with inner product given by
$$\langle f,g\rangle:= \int_0^{2\pi}f(x)g(x)\,dx$$

Show that $$S_n := \{1, \cos x, \sin x, \cos(2x), \sin(2x), \dots , \cos(nx), \sin(nx)\}$$ is an orthogonal set of V.

Find the orthogonal projection of a function $f \in V$ on $span(S_n).$ Particularly find the orthogonal projections of the function $x$  on $span(S_2)$ and $span(S_3).$

(And we know for $n \to \infty,$ the orthogonal projection of a function $f$ on $S_n$ is nothing but the Fourier series expansion of the function $f$).

答:
$$\cos (a x) \sin (b x)=\frac{1}{2} (\sin (a x+b x)-\sin (a x-b x))$$
Integrating
$$\int_0^{2\pi} \cos (a x) \sin (b x)\,dx=\int_0^{2\pi}  \frac{1}{2} (\sin (a x+b x)-\sin (a x-b x))\,dx=$$
$$=\left[\frac{\cos (x (a-b))}{2 (a-b)}-\frac{\cos (x (a+b))}{2 (a+b)}\right]_0^{2\pi}=\\=\frac{\cos (2 \pi  (a-b))}{2 (a-b)}-\frac{\cos (2 \pi  (a+b))}{2 (a+b)}-\left(\frac{1}{2 (a-b)}-\frac{1}{2 (a+b)}\right)=0,\;\forall a,b\in\mathbb{Z}$$
Edit

For the other products $\sin(ax)\sin(bx);\;\cos(ax)\cos(bx)$ a similar procedure leads to the same result.


问:
Proving the injective part of the First Isomorphism Theorem
I'm reviewing set theory, and am trying to work out part of the proof of the First Isomorphism Theorem.

Let $f$ be a function from sets $X$ to $Y$. I'd like to prove that $f$ induces an injection from $X/\mathrm{Ker}(f)$ to $\mathrm{Im}(f)$.

Let $R_1,R_2\in X/\mathrm{Ker}(f)$, and let $g:X/\mathrm{Ker}(f)\rightarrow X$ be a choice function such that $x_1 = g(R_1)$ and $x_2 = g(R_2)$. Now if $f(x_1)=f(x_2)$ then $(x_1,x_2)\in\mathrm{Ker}(f)$, according to the definition of the kernel of $f$.

But now since $R_1 = \{y\in X:(y,x_1)\in\mathrm{Ker}(f)\}$, both $x_1$ and $x_2$ are members of $R_1$. The same can be said for $R_2$, and so $R_1$ must be the same as $R_2$, since $R_1$ and $R_2$ are part of a partition (they must be identical or otherwise mutually exclusive).

Therefore, we have that $f(x_1)=f(x_2)\rightarrow R_1 = R_2$, so the mapping induced by $f$ is injective.

Does this seem correct? Do I really need to invoke choice functions? It really seems there must be a simpler way to do this.

答:
The first step is to define a mapping from $\varphi:X/\ker f\to\operatorname{im}f$: you can’t prove that it’s injective until you know what it is. For each $x\in X$ let

$$[x]=\{y\in X:f(y)=f(x)\}\in X/\ker f\,;$$

the natural thing to try is

$$\varphi:X/\ker f\to\operatorname{im}f:[x]\mapsto f(x)\,.$$

The first step is to show that $\varphi$ is well-defined. That is, we need to make sure that if $x_0,x_1\in R\in X/\ker f$, so that $[x_0]$ and $[x_1]$ are both names for the same set $R$, the value of $\varphi(R)$ does not depend on which of these names we use. This isn’t hard: if $x_0,x_1\in R\in X/\ker f$, then $f(x_0)=f(x_1)$, and therefore

$$\varphi([x_0])=f(x_0)=f(x_1)=\varphi([x_1]),,$$

as desired.

Now that we know that $\varphi$ really does define a function, we can try to show that this function is injective. Suppose that $\varphi([x_0])=\varphi([x_1])$. Then

$$f(x_0)=\varphi([x_0])=\varphi([x_1])=f(x_1)\,,$$

so $x_0\in[x_1]$, $[x_0]=[x_1]$, and $\varphi$ is indeed injective.


问:
Calculating Integrals with infinity
I'm trying to calculate this integral :
$$
\int_{0}^{+\infty}{e}^{-\left(\mathrm{i}w+2\right)t} dt
$$
Without using the limits => $\int f(t)dt$, I got this result below :

$\dfrac{\mathrm{e}^{-\left(\mathrm{i}w+2\right)t}}{-\mathrm{i}w-2} + C$

So to calculate the integral in this interval : $[0,+\infty]$ I just have to calculate the limit for this function $\dfrac{\mathrm{e}^{-\left(\mathrm{i}w+2\right)t}}{-\mathrm{i}w-2} + C$
in $0$ and $+\infty$ .
My question is how do we calculate the limit of such function in $+\infty$ .

答:
Assuming $w$ is real (which you didn't say),
$$
\left|e^{-(iw+2)t}\right| = e^{-2t}
$$
This goes to $0$ much more rapidly than any polynomial, so the limit at $t=\infty$ is $0$.

Title: Calculating Integrals with infinity


问:
Lebesgue measurable homework problem
Let $X \subseteq \mathbb{R}$. A subset $E \subseteq \mathbb{R}$ is called a hull of $X$ if

$E$ is measurable

$X \subseteq E$

If $F$ is any measurable set such that $X \subseteq F$, then $E$\ $F$ is a zero set

A hull of $X$ should be thought of as a sort of "smallest" measurable set containing $X$.
This is made precise in the problem.

(a) Prove that every subset of $\mathbb{R}$ has a hull

(b) Prove that if $E_{1}$ and $E_{2}$ are both hulls of $X$, then $E_{1}$\ $E_{2}$ and $E_{2}$\ $E_{1}$ are zero sets. Thus the hull of $X$ is unique up to zero sets

(c) Let E be a hull of X. Prove that $m(E)=m^{*}(X)$

The hint says to consider the case $m^{*}(X)< \infty$, then consider the case let $R_i=(i,i+1)$ for $i \in \mathbb{Z}$. Let $X_i=X \cap R_i$. Then $X=\cup_{i \in \mathbb{Z}} X_{i} \cup \mathbb{Z}$. I need help to prove this.

答:
I know this is an old question, but I keep waking up in the middle of the night with the sense that I've only dreamed the existence of measurable hulls and eventually find my way to this thread, so I would like to provide a more general answer so that I can finally get some sleep.

Let $(X,\Sigma,\mu)$ be a finite measure space, i.e. $\mu(X)<\infty$ (please note that I have made $X$ the space here, not the set we want the hull of, because $X$ is the typical symbol for a general measure space; I hope this doesn't create any confusion). Let $A\subset X$ be any subset, measurable or non-measurable. We want to find a hull of $A$, i.e. a measurable set $E$ containing $A$ with property 3 in the question.

Since $\mu(X)<\infty$,
$$
M:= \inf_{S \in \Sigma,\ A\subset S} \mu(S) < \infty.
$$
Then we can find measurable set $S_1, S_2,\dots$ containing $A$ such that $\mu(S_n)$ decreases to $M$. Let
$$
E = \bigcap_n S_n.
$$
Then $E$ is measurable, $A\subset E$, and since this measure space is finite, we can see that $\mu(E)$ is equal to the infimum. Lastly, suppose $F\supset A$ is also measurable. Then $E\cap F = E \setminus (E\setminus F)$ is also a measurable set containing $A,$ and is also a subset of $E$. Thus $M \leq \mu(E\cap F) \leq \mu(E) = M,$ so it follows that $\mu(E\setminus F) = 0$. Therefore, $E$ satisfies all the desired properties of a measurable hull.

This can be extended to $\sigma$-finite spaces, which are measure spaces $(X,\Sigma,\mu)$ in which $X$ written as a countable union of sets $Y_n$ of finite measure, even if $\mu(X)=\infty$. To find a hull in this case, we simply intersect $A$ with each $Y_n$ $X$, and a hull for the intersection in each $Y_n$, and then take the union. To check that property 3 is satisfied, remember that a countable union of sets of measure $0$ has measure $0$.

$\mathbb{R}$ with the Lebesgue measure is $\sigma$-finite (think along the lines of the hint in the problem), so this concludes the proof. Notice, though, that I never mentioned the Lebesgue outer measure. This might be disconcerting, because with a measure constructed from an outer measure, it seems just as good to define your hull to be a measurable set $E$ containing $A$ such that $\mu^*(A) = \mu(E)$, and with the properties above (in this case, property 3 is needed only to deal with sets of infinite outer measure). However, consider the way in which the hull I found and the outer measure are both defined using infima. This should show you how to prove that the definitions in this case are equivalent.

Title: Lebesgue measurable homework problem


问:
Show the finiteness of number of singularities.
Let $\Gamma$ be a bounded region in $\mathbb C$ and let $\Sigma$ be a closed set contained in $\Gamma$.
Suppose $f$ is a function that is holomorphic on $\Gamma$ apart from singularities where
it has poles. Show that the number of singularities within $\Sigma$ is finite.

I know that the singularities are isolated point but I can't see the finiteness of them;

is there a charitable soul who can help me?

答: Assume for contradiction that there infintely many poles within $\Sigma$. Then we could make a convergent sequence of points within $\Sigma$ which were singularities of the function (why?). But then, $f$ would not be meromorphic since the limit $z_0$ of this sequence would be a singularity, but not an isolated one since every neighborhood of $z_0$ would contain infinitely many singularities.


问:
Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$
If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$

So $x+y+z=0$

We need to minimize $$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{z}\right)^{4}+\left(\frac{z}{x}\right)^{4}$$
$\implies$
$$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{x+y}\right)^{4}+\left(\frac{x+y}{x}\right)^4$$
Letting $\frac{x}{y}=t$ we get:
$$E=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)^4}{t^4}$$
Now i tried using derivatives, but very tedious.

Body:
We assume $a\ne b\ne c$.
We see that if $$E(t)=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)}{t^4},$$
then
$$E'(t)=4\frac{(t^2+t+1)^2}{t^5(t+1)^5}p(t),$$
where $$p(t)=t^9+3t^8+t^7-3t^6+t^5+2t^4-10t^3-13t^2-6t-1.$$
From our assumption that $a\ne b\ne c$, we have that $t=\frac{a-b}{b-c}\not\in\{0,-1\}$, and since $a,b,c\in\Bbb R$, $t^2+t+1\ne0$. Thus the critical points of $E(t)$ are the real roots of $p(t)$. There are only three real roots of $p(t)$. Call them $\alpha,\beta,\gamma$, with $\alpha<\beta<\gamma$. We have
$$\begin{align}
\alpha&\approx -1.7212312143130658672\\
\beta&\approx -0.41902052920932262495\\
\gamma&\approx 1.3865179156901112246.
\end{align}$$
Thus
$$\begin{align}
E(\alpha)&\approx 12.5037895937134016209875697987512072072\color{red}{22979907033168192287594834}\\
E(\beta)&\approx 12.5037895937134016209875697987512072072\color{green}{14601727355467488179851200}\\
E(\gamma)&\approx 12.5037895937134016209875697987512072072\color{blue}{10427320909272240776493390}.
\end{align}$$
Thus it appears as if $E(\alpha)\ge E(\beta)\ge E(\gamma)>12+\frac{10803682224985055382}{21444830063582275811}$, so that
$$\left(\frac{a-b}{b-c}\right)^4+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^4>12+\frac{10803682224985055382}{21444830063582275811}.$$


问:
Does this linear system even have a solution? 2×3
$$\begin{aligned} x + 2y + kz &= 1 \\ 2x + ky + 8z &= 3\end{aligned}$$

The assignment is to find values of $k$ that make the system have a unique solution (which is impossible, so that's okay), have no solution at all and have infinite solutions. I don't even know what logic to use in this scenario and had no luck with research. I'd be very thankful to whoever helps me with this.

答:
$k = 4$ gives no solutions. Every other value of $k$ gives infinitely many solutions, because the matrix
$$ \begin{bmatrix} 1 & 2 & k \\ 2 & k & 8 \end{bmatrix} $$
has rank $1$ for $k = 4$, and rank $2$ for $k \neq 4$.

Another way to look at it: solve the first equation for $x$ to obtain
$$ x = 1 - 2y - kz. $$
Then substitute this into the second equation. You get
$$ (k - 4)y + (8-2k)z = 1. $$
Observe that if $k = 4$, this second equation makes no sense. But if $k \neq 4$, the second equation just gives the equation of some line on the $y-z$ plane.


问:
Cardinal Arithmetic Property Proof
The following proposition is "obvious" but I am having trouble obtaining a rigorous proof of it:
If the set $X$ is finite and $Y$ is a subset of $X$, then $Y$ is finite and #$(Y)\leq$#$(X)$. It is easy to show that if #$(Y)\leq$#$(X)$ then $Y$ is finite, but with regards to proving that #$(Y)\leq$#$(X)$, I have tried a whole bunch of ways to show this given the hypotheses and have come to no satisfactory proof (e.g. I tried a proof by contradiction of assuming #$(Y)>$#$(X)$, but was not able to derive any obvious contradiction).

答:
Assume that the claim is false. Without loss of generality, this implies that there is some $m,n\in\mathbb N$, $m<n$, such that there is an injection $f$ from $[n]\to[m]$ ($[n]$ is notation for $\{1,...,n\}$).

Also without loss of generality, we can assume that $n$ is the smallest natural for which this property holds.

If $n=1$, the claim is clearly false (you can write out the case yourself).

If $n>2$, let $x_0=1$, and let $x_{i+1}=f(x_i)$ for all naturals $i$.

I'll leave it to you to prove that if $X=\{x_i\}_{i\in\mathbb N}$, then $f_X$, the restriction of $f$ to $X$, is a bijection $f_X:X\to X$. Also, note that $X\subset[m]\subsetneq[n]$, so this implies that $f_{[n]\setminus X}$ is an injection from $[n]\setminus X$ to $[m]\setminus X$. This implies that the property holds for $n-|X|<n$, contradicting minimality of $n$.

Title: Cardinal Arithmetic Property Proof


问:
Convergence of a series involving $\arctan (1/n)$
For what positive values of $a$ does the series $\sum (1/n -\arctan(1/n))^a $ converge ?

All the terms in the series are non negative and the terms are monotonic decreasing. Can I say for all positive $a$, the series is convergent?

答:
Using Taylor expansion we have
$$\frac1n-\arctan\frac1n=\frac{1}{3n^3}+O(n^{-5}).$$
Then you can compare the series with $\sum n^{-3a}$.


问:
How to integrate $\int \sqrt{x^2+a^2}dx$
$a$ is a parameter. I have no idea where to start</p>

答:
Substitue $x=a \sinh t$, so $t=\text{arcsinh} \frac{x}{a} $ $dx=a\cosh t dt$ and $x^2+a^2=a^2(1+\sinh^2t)$. Your integral becomes $$\int|a|\cosh t a\cosh t dt=a|a|\int \cosh^2tdt.$$

Use the identity $\cosh^2t=\frac{1}{2}(1+\cosh2t)$ to get $$\frac{a|a|}{2} \int(1+\cosh2t)dt=\frac{a|a|}{2} \left(t+\frac{\sinh 2t}{2} \right)+C $$

If we go back to $x$, we get: $$\frac{a|a|}{2} \left(\text{arcsinh }\frac{x}{a}+\frac{\sinh2 \text{arcsinh } \frac{x}{a} }{2} \right) .$$

This can be further simplified using the identities $\sinh 2t=2\sinh t \cosh t$ and $\cosh \text{arcsinh t}=\sqrt{1+t^2}$


答:
We have
$$
\arctan(\frac 1 n)=\frac{1}{n}-\frac{1}{3n^3} + o(\frac{1}{n^3})
$$
so
$$
\frac{1}{n}-\arctan(\frac 1 n)=\frac{1}{3n^3} + o(\frac{1}{n^3})
$$
and
$$
(\frac{1}{n}-\arctan(\frac 1 n))^\alpha=\frac{1}{3^{\alpha}n^{3\alpha}}( 1+ o(1)).
$$
Thus
$$
(\frac{1}{n}-\arctan(\frac 1 n))^\alpha \sim \frac{1}{3^{\alpha}n^{3\alpha}}
$$
and the series converges iff $3\alpha > 1$ iff $\alpha > \frac{1}{3}$.

Title: Convergence of a series involving $\arctan (1/n)$

Body:
For what positive values of $a$ does the series $\sum (1/n -\arctan(1/n))^a $ converge ?

All the terms in the series are non negative and the terms are monotonic decreasing. Can I say for all positive $a$, the series is convergent?


问:
Is Language of All Binary-Digit Strings Not Containing Substring 0100 or Suffix 010 Context-Free?
Let $\Sigma$ be the alphabet consisting of the symbols {0,1}, let $\Sigma^{*}$ be its Kleene closure, and let $R$ be defined by

$R = \{w \in \Sigma^{*} \mid (0100 \text{ is not a substring of }w) \wedge (010 \text{ is not a suffix of }w)\}$.

Is $R$ a context-free language, and if so, what would be a grammar or FSM describing it?

Thank you...

答:
$R$ is not just context-free: it’s regular. Here is a the transition table for a DFA that recognizes $R$; $q_0$ is the initial state, and $q_0,q_1$, and $q_2$ are the acceptor states.

$$\begin{array}{c|c|c}
q&a&\delta(q,a)\\\hline
q_0&0&q_1\\
q_0&1&q_0\\
q_1&0&q_1\\
q_1&1&q_2\\
q_2&0&q_3\\
q_2&1&q_0\\
q_3&0&q_4\\
q_3&1&q_2\\
q_4&0&q_4\\
q_4&1&q_4
\end{array}$$

Draw the automaton and check that it is at $q_3$ if and only if the current input ends in $010$, and it reaches $q_4$ if and only if the input has a substring $0100$.


问:
How many combinations are possible in this question?
I was curious how many combinations are possible in this scenario:

The characters allowed to make up the combination are so:

Letters a-z (lowercase only)

Numbers 0-9

Underscores _

Periods .

The combination is 4 digits

The combination cannot end with a period (.)

Just a side note, this can have repeating characters.

I also appreciate it a lot if someone were to answer this.

答: We have $26+10+1+1=38$ possible characters. We want $4$ characters and the combination cannot end with a period so we have $38^3 \times 37$ possible combinations.


问:
The euclidean space $\Bbb R^n$ is orientable as a manifold.
I know that

The euclidean space $\Bbb R^n$ is orientable as a manifold.

I think that it is orientable because it has a nowhere vanishing $n$-form.

But I am not sure.

Please can you explain me more formally and mathematically?

Thank you:)

答: You know that $\mathbb{R}^n$ has a natural set of global coordinates. Use these to write down a pointwise basis for $\Lambda^1_x(\mathbb{R}^n)$. Now use those to write down a basis for $\Lambda^n_x(\mathbb{R}^n)$. The coordinates are global. What does that tell you about your basis of $\Lambda^n_x(\mathbb{R}^n)$?</p>


问:
How can I further simplify this equation?
[1]: https://i.stack.imgur.com/0MnZo.jpg the original equation

as for the image provided, is there a way to simplify?
I have reduced the last two portions (the division) down to $$\frac{6}{a+2}$$

So the final equation is the following:
$$\frac{2(a+(a+1)+(a+2)+...+2a)}{a^2+3a+2}+\frac{6}{a+2}$$

Can I reduce it even further? the only thing bothering me is the series part. I don't know what to do with that part.

Thanks for any help.

Body:
Note $a^2+3a+2=(a+2)(a+1)$, so you might expect the term $(a+1)$ to appear in the numerator after simplifications so cancellation occurs.

We have $$a+(a+1)+(a+2)+...+(2a-2)+(2a-1))+2a$$
$$=a+(a+1)+(a+2)+...+(a+(a-2))+(a+(a-1))+a+a$$
$$=\underbrace{a+a+...+a}_{a+1 \space\space\text{times}}+(1+2+3+...+a-1+a)$$
$$=a(\underbrace{1+1+...+1}_{a+1 \space\space\text{times}})+\frac{1}{2}a(a+1)$$
$$=a(a+1)+\frac{1}{2}a(a+1)=\frac{3}{2}a(a+1)$$

Or note that the $n$-th term is $b_n=a+n$ so we have $$a+(a+1)+(a+2)+...+(2a-2)+(2a-1))+2a$$
$$=\sum_{n=0}^{a}b_n=\sum_{n=0}^{a}a+\sum_{n=0}^{a}n$$
$$=\frac{3}{2}a(a+1).$$

Or as mentioned by @kerf in the comments $$a+(a+1)+(a+2)+...+(2a-2)+(2a-1))+2a$$
$$=(1+2+3+...+a+...+2a)-(1+2+3+...+(a-2)+(a-1))$$
$$=a(2a+1)-\frac{1}{2}a(a-1)$$
$$=\frac{3}{2}a(a+1)$$


问:
Use that $a^5\equiv a\pmod 5$ and if $(a,5) = 1$ then prove $a^4\equiv 1\pmod 5$

Use that $a^5\equiv a\pmod 5$ and if $(a,5) = 1$ then prove $a^4\equiv 1\pmod 5$

I´ve been trying to solve this but I don’t get any where

答: $a^{5} \equiv a \pmod 5 \implies 5|(a^{5}-a)=a(a^{4}-1)\implies a(a^{4}-1)=5k%$ for some integer $k$. Since $\gcd(a,5) =1,$ and by the uniqueness of prime factorizations, $5$ is a prime factor of the LHS, but not a prime factor of $a$. Hence, $5$ must be a prime factor of $a^{4} -1$ so that $5|(a^{4}-1) \implies a^{4} \equiv 1 \pmod 5.$


问:
Method of Moment
In this lecture on method of moment, we have:

why is gradient of psi inverse a dxd matrix?
K-th moment $m_k$is defined as $ \mathbb E[X^k] $ and can be estimated by the average using Law of Large Numbers which here is represented by $\hat m_k$
My understanding is that the inverse function $\psi^-1 $  takes the vector of moments of size d, so why isn't the gradient of size d?

答: I have found the answer. Normally, for a function $\psi$ that goes from $\mathbb R^d$ to $\mathbb R$ then you have to take the derivative of the function in regards to each of the d parameters and you will have a vector of size d. In this case $\psi$ is a function from $\mathbb R^d$ to $\mathbb R^d$ therefore you will end up with a d-dimensional array for each of d elements in the "y" side.

Title: Method of Moment


问:
What is the purpose and dimensionality of expressing a representation of a matrix group as a block diagonal matrix?
The question relates to the following definition:

A decomposition of a representation $R:G \to GL(n, \mathbb C)$ is a splitting of $\mathbb C^n = V_1 \oplus \cdots \oplus V_k$ where each $V_i \subseteq \mathbb  C^n$ is a subrepresentation of $\mathbb C^n$. Which means that $R(g) v \in V_i$ whenever $v\in V_i.$ And $R(g)$ can be written as a block diagonal matrix:

$$R(g)=\begin{bmatrix}R(g)\vert_{V_i}&&\\
                       &\ddots& \\
                       &&R(g)\vert_{V_k}\end{bmatrix}$$
Or, equivalently, that $R=R\vert_{V_1}\oplus\cdots \oplus R\vert_{V_k}.$

Leaving aside possible technical misunderstandings, the idea is that the internal direct sum is supposed to change the span of subspaces involved, but not the dimension of the ambient space.

But each block in the matrix above is $n \times n,$ which would make the whole matrix $(kn)\times (kn),$ and inconsistent with a matrix acting on $\mathbb C^n$ vectors.

Is this a correct view of the dimensionality of the block matrix, and if so, what is the purpose of the block matrix expression of this direct sum? In particular, I find it confusing when it comes the the expression $R(g) v,$ which seems to say 'the representation applied to the vector $\vec v$', as in matrix multiplication, which would call for $[n\times n][n\times 1]$ dimensions.

答:
$R(g)$ is still a map that sends a vector $v \in \mathbb{C}^n$ to a vector $R(g) v \in \mathbb{C}^n$.

However, $R(g)|_{V_i}$ is not $n \times n$; it is $\dim(V_i) \times \dim(V_i)$. Since $\sum_i \dim(V_i) = n$, there is no discrepancy in the size of the matrix. The block diagonal form of $R(g)$  is assuming that we have chosen a basis for $\mathbb{C}^n$ by collecting together bases for each $V_i$. For example, vectors in $V_1$ would have zeros in all entries except the first $\dim(V_1)$ entries, so the restriction $R(g)|_{V_1}$ of $R(g)$ to $V_1$ can be represented as a $\dim(V_1) \times \dim(V_1)$ matrix.


答:
You can use this basic result:

An non-homogeneous linear system $Ax=b$ has solutions if & only if the matrix $A$ and the augmented matrix $(A|x)$ have the same rank. Furthermore, in this case, the common rank is the codimension of the (affine) space of solutions.

Title: Does this linear system even have a solution? 2×3

Body:
$$\begin{aligned} x + 2y + kz &= 1 \\ 2x + ky + 8z &= 3\end{aligned}$$

The assignment is to find values of $k$ that make the system have a unique solution (which is impossible, so that's okay), have no solution at all and have infinite solutions. I don't even know what logic to use in this scenario and had no luck with research. I'd be very thankful to whoever helps me with this.


问:
Triangular inequality of complex numbers
Let $0<a, b<1$ such that $a+b=1$. If $u,v \in\mathbb{C}$, with $u\neq v$, and $|u|=1=|v|$, then
$|au+vb|<1$.

I have tried to prove it absurd but I do not reach a contradiction.
Let $z:=x+iy$ and $w:=c+id$ such that
$|a(x+iy)+b(c+id)|=1$

$\iff$ $\sqrt{(ax+bc)^2+(ay+bd)^2}=1$

$\iff$ $(ax+bc)^2+(ay+bd)^2=1$...

I finally have to

$xc+yd=1$,

but I don't see the contradiction.

答:
It isn't true.  Let $u=2,v=2i$.  And $a,b=1/2$.  Then we get $|au+bv|=|1+i|=\sqrt2\gt1$.  In fact, we can take $u=r,v=ri$, and get $|au+bv|=\sqrt2r/2$.

On the other hand after your edit it's true, because when the angle between $z$ and $w$ is greater than zero, $|z+w|\lt |z|+|w|$.  So we get $|au+bv|\lt a|u|+b|v|=a+b=1$.


问:
Sum of all elements in a matrix
The trace is the sum of the elements on the diagonal of a matrix. Is there a similar operation for the sum of all the elements in a matrix?

答:
The term "grand sum" is commonly used, if only informally, to represent the
sum of all elements.

By the way, the grand sum is a very important quantity in the contexts of
Markovian transition matrices and other probabilistic applications of linear
algebra.

Regards,
Scott

Title: Sum of all elements in a matrix


答:
Let's consider the square of the modulus
$$
|au+bv|^2=(au+bv)(a\bar{u}+b\bar{v})=a^2+b^2+ab(u\bar{v}+\bar{u}v)
$$
Since $a^2+b^2=(a+b)^2-2ab=1-2ab$, the inequality to prove becomes
$$
1+ab(u\bar{v}+\bar{u}v-2)<1
$$
Since $ab>0$, this becomes
$$
u\bar{v}+\bar{u}v<2
$$
(note that this makes sense because the left-hand side is real).

Let $w=uv^{-1}=u\bar{v}$. Then also $|w|=1$ and the inequality is $w+\bar{w}<2$, which amounts to saying that
$$
\operatorname{Re}(w)<1
$$
which is true because $w\ne1$ by assumption.

Title: Triangular inequality of complex numbers

Body:
Let $0<a, b<1$ such that $a+b=1$. If $u,v \in\mathbb{C}$, with $u\neq v$, and $|u|=1=|v|$, then
$|au+vb|<1$.

I have tried to prove it absurd but I do not reach a contradiction.
Let $z:=x+iy$ and $w:=c+id$ such that
$|a(x+iy)+b(c+id)|=1$

$\iff$ $\sqrt{(ax+bc)^2+(ay+bd)^2}=1$

$\iff$ $(ax+bc)^2+(ay+bd)^2=1$...

I finally have to

$xc+yd=1$,

but I don't see the contradiction.


问:
Let $A, B:V \to V$ be two linear operators in a vector space with inner product. If $A$ and $B$ commutes then so does $A^\ast$ and $B^\ast$
$A^\ast$ and $B^\ast$ are adjoint operators

This is a proof that is for an undergraduate linear algebra class. This is what I tried but I don't know if it is enough or rigorous enough.

$\langle Av, w\rangle = \langle w, A^\ast w\rangle$

$\langle ABv, w\rangle =\langle w, A^\ast B^\ast w\rangle$

$AB = BA$

then

$\langle BAv, w\rangle = \langle w, B^\ast A^\ast w\rangle$.

答: Hint: What is the adjoint of $ST$, for operators $S,T$?


问:
Proving two equations involving the greatest common divisor

Show or prove that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m),$$
  and that
  $$\gcd \left(\frac{a^{2m + 1}+1}{a+1} , a + 1\right)=\gcd(a + 1 , 2m + 1).$$

答:
Let integer $d\ne0$ divides $a+1$ i.e., $a+1=c\cdot d$(say)  where $c\ne0$ is some integer

$\implies a=c\cdot d-1$

$$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )$$
$$=\gcd \left(\frac{(c\cdot d-1)^{2m}-1}{c\cdot d} ,c\cdot d\right )$$
$$=\gcd \left( (c\cdot d)^{2m-1}-\binom{2m}1(c\cdot d)^{2m-2}+ \binom{2m}2(c\cdot d)^{2m-3}+\cdots+\binom{2m}{2m-2}(c\cdot d)-\binom{2m}{2m-1},c\cdot d\right)$$

$$=\gcd\left( c\cdot d\{(c\cdot d)^{2m-2}-\binom{2m}1(c\cdot d)^{2m-3}+ \binom{2m}2(c\cdot d)^{2m-4}+\cdots+\binom{2m}{2m-2}\}-2m,c\cdot d\right)\text{ as } \binom{2m}{2m-1}=\binom{2m}{2m-(2m-1)}=\binom{2m}1=2m$$

$$=\gcd(-2m, c\cdot d)\text{ as }\gcd(p+kq,q)=\gcd(p,q)$$

$$=\gcd(2m, a+1)$$

Can you solve the second one now?


问:
meaning of "$\wedge$" in set theory
I have been reading Analysis I by Terence Tao and this symbol, $\wedge$, keeps popping in proof's in this forum as well as from other sources, for example, Power Set Axiom. Terence Tao' analysis 3.4.6. This symbol is not described in the book so I do not understand what this means, could anyone explain it please? Thank you.

答:
$X^Y$ where $X,Y$ are sets means the set of functions from $Y$ to $X$.

When $Y$ is finite, it resembles Cartesian product. For example, $\mathbb{R}^n$ means the set of $n$-tuples of reals (implying ordering).

What's an $n$-tuple? Each instance of $\mathbb{R}^n$ means: when you query an index number $i$ where $1\leq i\leq n$, you get a real number in return. And that is the same as a function mapping from indices $\{1, 2, 3,...,n\}$ to $\mathbb{R}$.

Collectively, the set $\mathbb{R}^n$ is equivalent as the set of functions mapping from $\{1, 2, 3,...,n\}$ to $\mathbb{R}$. It works in similar ways for infinite sets.

Title: meaning of "$\wedge$" in set theory