问:
(Tao's Ex 1.1.3) Uniqueness of elementary measures; confused how to proceed.
This is an exercise from Tao's Introduction to Measure Theory.

Exercise 1.1.3:($\textit{uniqueness of elementary measure}$) Let $d \ge 1$. Let $m':\mathcal{E}(\mathbb{R}^d)\to \mathbb{R}_{\ge 0}$ be a map from the collection $\mathcal{E}(\mathbb{R}^d)$ of elementary subsets of $\mathbb{R}^d$ to the non negative reals that obeys the non-negativity, finite additivity and translation invariance. Show that there exists a constant $c \in \mathbb{R}_{\ge 0}$ such that $m'(E)=c\,m(E)$ for all elementary sets $E$. In prticular, if we impose the additional normalisation $m'\left([0,1)^d\right)=1$ then $m' \equiv m$.

Attempt: Given hint: set $c:=m'([0,1]^d)$ and then compute $m'\left([0,\tfrac{1}{n})^d\right)$. For simplicity I've tried for $d=1$. Partitioning we get $E:=[0,1)=[0,\tfrac{1}{n})\cup [\tfrac{1}{n},\tfrac{2}{n})\cup \cdots \cup [\tfrac{n-1}{n},1)$ for some $n\in \mathbb{N}$. Now translating $[0,1)$ by $\tfrac{k}{n}$ we get $[\tfrac{k}{n},\tfrac{k+1}{n})$ where $k=1,\ldots,n-1$. Now given that $m'$ obeys translation invariance, then we write  $$m'\left([0,\tfrac{1}{n})\right)=m'\left([\tfrac{k}{n},\tfrac{k+1}{n})\right)$$ Now using the finite additivity property we get
\begin{align}
    m'([0,1))&=m' \left(\bigcup_{k=0}^{n-1}[\tfrac{k}{n},\tfrac{k+1}{n})\right)=\sum_{k=0}^{n-1}m' \left([\tfrac{k}{n},\tfrac{k+1}{n})\right)\nonumber \\&=\sum_{k=0}^{n-1}m'([0,\tfrac{1}{n})]=n\cdot m'([0,\tfrac{1}{n}))\nonumber
\end{align}
This gives us $m'([0,\tfrac{1}{n}))=\tfrac{1}{n}m'([0,1))=c\cdot m([0,\tfrac{1}{n})).$
I can do this for any $d$, also setting $m'([0,1)^d)=1$, i will have $m'\equiv m.$ But does intervals of type $[0,\tfrac{1}{n})$ exhaust all possibilities? Don't i need to consider also intervals of the form $[0,a)^d$ where $a\in \mathbb{R}_{\ge 0}$? I'm confused at this stage. Appreciate any help.

答:
Set $c=m'(Q_1)$ where $Q_1$ is the standard $1$-box. Note that $m(Q_1)=1$. Now divide $Q_1$ into $2^{nd}\ \ 2^{-n}$boxes which are necessarily disjoint. It follows that if $Q$ is a $2^{-n}$-box, then $2^{nd}m'(Q)=m'(Q_1)=cm(Q_1)=c2^{nd}m(Q)\Rightarrow m'(Q)=cm(Q).$ So the result holds for any $2^{-n}$-box.

In fact, we get even more from this: since every open set is a countable disjoint union of such sets, the result is true for any open set and so for any Borel set by regularity of the measures $m$ and $m'$.


问:
How to solve inequality equation with ceil operator product
I am trying to solve this inequality for $x$:

$$ \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil \leq N $$

With: $c_1, c_2, x, N \in \mathbb{N}$ and $c_1, c_2, N$ being just constants.

I could make my system work with a solution assuming that $ x \leq c_1, c_2 $ but I would like ot have a general solution if someone can help me. The context behind this equation is how to split an image of $[H,W]=[c_1, c_2]$ in square patches of size $[H_{patch}, W_{patch}]=[x,x]$.

What I have tried yet:

Solve the equation for the lower bound of each inequality: I does not work.

$$\frac{c_1c_2}{x^2} \leq N \rightarrow \sqrt{\frac{c_1c_2}{N}} \leq x$$

Solve the equation for the upper bound: I am stuck getting x out of this expression.

$$ (\frac{c_1}{x^2}+1) \cdot (\frac{c_1}{x^2}+1) =\frac{c_1c_2}{x^2} + \frac{c_1}{x} + \frac{c_2}{x} + 1\leq N$$

Forgive my possible mistakes, I am not a mathematician. Thank you for your help in advance.

答:
As my previous answer is quite long, I continue here with your attempt for the upper bound.

Let $k \in \mathbb{N}$, we have $\lceil{\frac{k}{x}} \rceil \leq \frac{k}{x} + 1 \Rightarrow (\frac{c_1}{x} + 1)(\frac{c_2}{x} + 1) \leq \lceil{\frac{c_1}{x} + 1} \rceil \lceil{\frac{c_2}{x} + 1} \rceil \leq N$

$(\frac{c_1}{x} + 1)(\frac{c_2}{x} + 1) \leq N$

$\frac{c_1c_2}{x^2} + \frac{c_1}{x} + \frac{c_2}{x} + 1 \leq N$

$c_1c_2 + c_1 x + c_2 x + x^2 \leq Nx^2$

$0 \leq (N - 1)x^2 - (c_1 + c_2)x - c_1c_2$

This is a quadratic for $x$, so using the quadratic formula we have

$x_{1, 2} = \frac{c_1 + c_2 \pm \sqrt{(c_1 + c_2)^2 + 4(N - 1)c_1c_2}}{2(N - 1)}, N \neq 1$,

If we subtract the square root, we have a negative numerator therefore a negative $x$, this means that we can only use addition, and that $0$ always falls between the two zeroes and at $x = 0$ the original quadratic is negative therefore between the two zeroes $\forall x$ the statement is false, so we have

$x \geq \frac{c_1 + c_2 + \sqrt{(c_1 + c_2)^2 + 4(N - 1)c_1c_2}}{2(N - 1)}, N \neq 1$.

Let us use $c_1 = 10, c_2 = 4$ again.

If $N = 8$, $x \geq 3.59... \Rightarrow x = 4$, this is correct, but as we have seen in the previous answer, $x = 3$ is the minimum value.

If $N = 7$, $x \geq 4 \Rightarrow x = 4$, this is correct, and as we have seen in the previous answer, it is the minimum value.

Again, I can not show if this is the method that always give the closest value to the minimum for $x$, but possibly it is.

ContentLicense: CC BY-SA 4.0


问:
A limit on binomial coefficients
Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.

What I can do is just use Stolz formula. But I could not proceed.

ContentLicense: CC BY-SA 3.0

Title: A limit on binomial coefficients

Body:
$$
\frac{1}{n^2}\sum_{k=0}^{n}ln\binom n k
=\frac{1}{n^2}\sum_{k=0}^{n}ln\frac{n!}{k!(n-k)!}
=\frac{1}{n^2}\sum_{k=0}^{n}(ln(n!)-ln(k!)-ln((n-k)!))
=\frac{1}{n^2}\sum_{k=0}^{n}ln(n!)-\frac{2}{n^2}\sum_{k=1}^{n}ln(k!)
=\frac{n+1}{n^2}ln(n!)-\frac{2}{n^2}\sum_{k=0}^{n}ln(k!)
$$

because ln(n!)~nln(n)-n,so

$$
\frac{1}{n^2}\sum_{k=0}^{n}ln\binom n k
=\frac{n+1}{n^2}(nln(n)-n)-\frac{2}{n^2}\sum_{k=1}^{n}(kln(k)-k)
$$

then I am going to prove $\sum_{k=1}^{n}(kln(k)) \rightarrow  \frac{n^2}{2}(ln(n)-1/2)$

$$
\int_{i=x-1}^{x}(iln(i))dx<xln(x)<\int_{i=x}^{x+1}(iln(i))dx
$$

and

$$
\int(xln(x))dx=x^2/2(lnx-1/2)
$$

so sum it up, we got
$$
\sum_{k=1}^{n}(kln(k)) \rightarrow  \frac{n^2}{2}(ln(n)-\frac{1}{2})
$$
then

$$
\frac{1}{n^2}\sum_{k=0}^{n}ln\binom n k
=\frac{n+1}{n^2}(nln(n)-n)-\frac{2}{n^2}\sum_{k=1}^{n}(kln(k)-k)
=\frac{n+1}{n}(ln(n)-1)-\frac{2}{n^2}(\frac{n^2}{2}(lnn-\frac{1}{2})-\frac{n(n+1)}{2})
=ln(n)-1-(ln(n)-1/2-1)=1/2
$$


问:
Proof of Inclusion Exclusion Principle
Show that $|A \cup B| + |A \cap B|= |A| + |B|$ for two finite sets $A$ and $B$.

Can you please check the proof below, and let me know if it's right? It makes me a bit uneasy for some reason, and I can't tell why.

My givens are:

$|A|$ is defined as $n$ if there is a bijection $f: A \to \{1,2,\dots,n\}$. The cardinality of an empty set is 0.

If there is some bijective $f: A \to B$ and a bijective $g: B \to C$, then there exists some $h: A \to C$ such that $h$ is also bijective.

Rudimentary results about sets from Chapter 1 of the text I'm using.

答:
We show this by decomposing $|A \cup B| + |A \cap B|$ into disjoints units.

First, $|A \cup B| = |A-B| + |B-A| + |A \cap B|$ where each segment of the RHS is disjoint. From this, we can write

\begin{eqnarray}
|A \cup B| + |A \cap B| &=& |A-B| + |B-A| + |A \cap B| + |A \cap B| \\
&=& (|A-B| + |A \cap B|) + (|B-A| + |A \cap B|) \\
&=& |A| + |B|\\
\end{eqnarray}

Title: Proof of Inclusion Exclusion Principle


问:
Solve limit using definite integration
I have to solve this limit using definite integral

$${\lim_{n\to \infty} \frac{1}{n} \cdot \sum_{i=1}^n \frac{1}{1+(\frac{i}{n})^i}}$$

Well, my development was:

I tried to give the Riemann integral definition form, using a regular partition such that:

$$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\cdot\sum_{i=1}f\left(a+i\frac{b-a}{n}\right)$$

So, i set $b=1, a=0$ for simplicity, then we have $f(\frac{i}{n})$ must be equal to ${\frac{1}{1+(\frac{i}{n})^i}}$.

Let ${x=\frac{i}{n}}$, then $${f(x)=\frac{1}{1+x^{nx}}}$$

The problem is that $n$ is a dummy variable that does not make sense outside the limit, that is, it does not make sense for the function $f$

So, I need to do some kind of algebraic transformation or variable change for $n$, in such a way that the function $f$ remains only in terms of $x$ and thus I can use it in the definite integral. However, I have not been able to find such a magical algebraic transformation or variable change for $n$.

答:
Denote the expression under the limit by $a_n$. Clearly $a_n<1$, so that $\limsup\limits_{n\to\infty}a_n\leqslant 1$.

Further, for a fixed $n$, and positive real values of $x$, the map $x\mapsto(x/n)^x$ has a global minimum at $x=n/e$. Thus, for $1\leqslant i\leqslant m$ we have $(i/n)^i\leqslant\max\{1/n,(m/n)^m\}$.

This gives an idea for a lower bound. Let $0<\varepsilon<1$. Then $$1\leqslant i\leqslant(1-\varepsilon)n\implies(i/n)^i\leqslant\max\{1/n,(1-\varepsilon)^{n(1-\varepsilon)}\}\color{gray}{\underset{n\to\infty}{\longrightarrow}0},$$ hence there exists $N_\varepsilon$ such that for $n>N_\varepsilon$ and $1\leqslant i\leqslant(1-\varepsilon)n$ we have $(i/n)^i<\varepsilon$, and $$a_n>\frac1n\times\lfloor(1-\varepsilon)n\rfloor\times\frac1{1+\varepsilon}\implies\liminf_{n\to\infty}a_n\geqslant\frac{1-\varepsilon}{1+\varepsilon}.$$

Since $\varepsilon$ is arbitrary, we have $\liminf\limits_{n\to\infty}a_n\geqslant 1$, giving finally $\lim\limits_{n\to\infty}a_n=1$.

A variant with Riemann sums (far-fetched): take a positive integer $m$, then for $n\geqslant m$ $$a_n:=\frac1n\sum_{i=1}^n\frac{1}{1+(i/n)^i}\geqslant\frac1n\sum_{i=1}^nf_m(i/n),\quad f_m(x):=\frac{1}{1+x^{mx}},$$ so that $\liminf\limits_{n\to\infty}a_n\geqslant\int_0^1 f_m(x)\,dx$ and, since $m$ is arbitrary, we may take $m\to\infty$: $$\liminf_{n\to\infty}a_n\geqslant\lim_{m\to\infty}\int_0^1 f_m(x)\,dx=1.$$

Title: Solve limit using definite integration


问:
Prove any consistent system with coefficient matrix A has exactly $p^{n-rank(A)}$ solutions in $\mathbb{Z_p}$.
"Prove that in the field $\mathbb{Z_p}$ ($_p$ being a prime), when we have a matrix A that is $m⨯n$. Any consistent system with coefficient matrix A has exactly $p^{n-rank(A)}$ solutions."

I got this question and I can't find the solution to it, I don't even know where to start. Does anyone know how to solve this? I feel really stupid.

P.S. How do you deal with anger when doing a maths question?

Thank you.

答:
Here is something that might work. First of all I am assuming that you are referring the field with $p$ elements (I would recommend you use the notation $\mathbb{Z}/p\mathbb{Z}$ and not $\mathbb{Z}_p$ in the future).

Try viewing $A$ as a map from $(\mathbb{Z}/p\mathbb{Z})^n \to (\mathbb{Z}/p\mathbb{Z})^m$ and try to see how many elements are there in a subspace of given dimension. Now try to compute the number of elements in the solution space of the system using the rank nullity theorem.

Hope this helps.


问:
Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices?
My question might be very silly as I've only recently started to learn about Lie algebras.

Given a Lie algebra $\mathfrak{g}$, and a vector space basis $\{e_i|i=1 \ldots \dim \mathfrak{g}\}$, we can write the adjoint action $\text{Ad}_x: \mathfrak{g} \to \mathfrak{g}: y \mapsto [x,y]$ in matrix form, where the $i,j$th entry of the matrix $[\text{Ad}_x]$ equals $\lambda_i$ if $[x,e_j] = \sum_{k}\lambda_k e_k$. I am looking for necessary and sufficient conditions on our Lie algebra $\mathfrak{g}$ such that there exists a basis in which all of these matrices $\{ [\text{Ad}_x]| x \in \mathfrak{g} \}$ are skew-symmetric.

If the condition turns out to be either simplicity of compactness of the associated simply connected Lie group, I would be very happy.

答:
$\DeclareMathOperator{\tr}{tr}\DeclareMathOperator{\ad}{ad}$I will expand my comment here. I am assuming (though I am not sure if this is the case) that the OP is interested in real Lie algebras. If $\mathfrak{g}$ is the Lie algebra of a compact semisimple Lie group, then with respect to any orthonormal basis of $\mathfrak{g}$ with respect to (minus) the Killing form, the matrix representing $\operatorname{ad}_x \in \mathfrak{gl}(\mathfrak{g})$ is skew-symmetric, for any $x \in \mathfrak{g}$.

Define the Killing form $B(-,-): \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ by:

$$B(x,y) = \tr(\ad_x \circ \ad_y).$$

I claim that for any $z \in \mathfrak{g}$, $B([z,x],y) + B(x,[z,y]) = 0$. For short, it is common to say that the Killing form is $\ad$-invariant.

$$ \begin{align*} & B([z,x],y) + B(x,[z,y]) \\
    = & \,\tr(\ad_{[z,x]} \circ \ad_y)+
    \tr(\ad_x,\ad_{[z,y]}) \\
    = & \,\tr([\ad_z,\ad_x] \circ \ad_y) + \tr(\ad_x \circ \, [\ad_z, \ad_y]) \\
    = & \,0 \end{align*}$$

by the cyclic property of the trace ($\tr(BCA) = \tr(ABC)$). I have also used the fact that $\ad_{[x,y]} = [\ad_x,\ad_y]$.

There is another fact from Lie theory that we need, namely that if $\mathfrak{g}$ is the Lie algebra of a compact semisimple Lie group, then its Killing form is negative definite.

Let us now choose an orthonormal basis of $\mathfrak{g}$, say $x_1,\ldots,x_n$, with respect to minus the Killing form. Note that, for any $x \in \mathfrak{g}$, we have

$$ -B(\ad(x)(x_i),x_j) = -B([x,x_i],x_j) = B(x_i, [x,x_j]) = B(x_i, \ad_x(x_j)).$$

Hence the matrix representing $\ad_x$ with respect to the orthonormal basis $x_1,\ldots,x_n$ is skew-symmetric.


问:
Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$
so I stumbled upon this equation/formula, and I have no idea how to prove it. I don't know how should I approach it:
$$
\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\
=\
2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right)
$$

where $n\in\mathbb N$ and the square root sign appears $n$-times.

I thought about using sequences and limits, to express the LHS as a recurrence relation but I didn't get anywhere.

edit: Solved, thanks for your answers and comments.

答:
I know it is a bit old, but I'd like to elaborate a bit the answer here going step by step.
As commented above, the best hint is to use the cosine half-angle formula ($*=\cos\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$) and work by induction. Therefore, let's get to it:

Base case:

The first partial sum, $S(1)$, will be:
$$S(1)=\sqrt{2}$$
Using the provided formula, one sees that
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=2\frac{1}{\sqrt{2}}=2\cos\left(\frac{\pi}{4}\right)=2\cos\left(\frac{\pi}{2^2}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{2}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}{2}}=2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$$

If we start directly from the cosine and try to get to the partial sum, it is actually nicer imho:
$$2\cos\left(\frac{\pi}{4}\right)=2\cos\left(\frac{\pi}{2^2}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{2}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}{2}}=2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$$

If we go for the second partial sum, we have:
$$S(2)=\sqrt{2+\sqrt{2}}$$
Which is:
$$2\cos\left(\frac{\pi}{8}\right)=2\cos\left(\frac{\pi}{2^3}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{4}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{4}\right)}{2}}=2\sqrt{\frac{1}{2}+\frac{1}{2}\cos\left(\frac{\pi}{4}\right)}=2\sqrt{\frac{1}{2}+\frac{1}{2}\frac{\sqrt{2}}{2}}=2\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2}}=2\sqrt{\frac{2+\sqrt{2}}{4}}=\sqrt{2+\sqrt{2}}$$

So now you can already develop some intuition that eventually you will have a factor $\frac{1}{2}$ inside the sum that will be cancelled by the outer 2.

Induction step:

Assume $S(n)$ works, i.e.:

$$\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=2\cos\left(\pi\over2^{n + 1}\right)$$

which clearly implies:
$$**=\frac{1}{2}S(n)=\frac{1}{2}\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=\cos\left(\frac{\pi}{2^{n+1}}\right)$$

Let's find out if $$S(n+1)=2\cos\left(\frac{\pi}{2^{n+2}}\right)$$

$$2\cos\left(\frac{\pi}{2^{n+2}}\right)=2\cos\left(\frac{1}{2}\cdot\frac{\pi}{2^{n+1}}\right)\stackrel{*}{=}2\sqrt{\frac{1+\cos\left(\frac{\pi}{2^{n+1}}\right)}{2}}=2\sqrt{\frac{1}{2}+\frac{1}{2}\cos\left(\frac{\pi}{2^{n+1}}\right)}\stackrel{**}{=}2\sqrt{\frac{1}{2}+\frac{1}{2}\frac{1}{2}S(n)}=2\sqrt{\frac{1}{2}+\frac{1}{4}S(n)}=2\sqrt{\frac{2+S(n)}{4}}=\sqrt{2+S(n)}$$

Bottom line for other cases:

To me, if I would encounter the $S(n)$, trying to get a nice closed-form solution would be the most complicated thing to do. Once you realise it goes in the direction of cosines is quite straight forward, as you can see. In this case, the trick was to observe that the cosine formula they relate it to is the half angle: see how $\cos\left(\frac{\alpha}{2^{n+1}}\right)$ has this $2^{n+1}$. It should pop up as an alarm as saying "hey, you will eventually have to split this denominator".


问:
Are there real numbers that have no finite description?
Here's my argument:

Descriptions of finite length are countably infinite (we can enumerate through all descriptions of max length 1 with 1 distinct symbols, then max length 2 with 2 distinct symbols, etc.)

Some finite length descriptions specify a single real number including the axioms/definitions needed to define the real number.

Therefore the real numbers that have a finite length description are a subset of all finite length descriptions.

Since there are uncountable reals, then some reals have no finite length description.

Someone may argue, we cannot define which of those descriptions correspond to real numbers and which do not.

Agreed - due to Godel's incompleteness theorem, some of those descriptions will correspond to real numbers that cannot be proven to be real numbers from the axioms and definitions in the description.

Therefore, given a finite description of axioms and definitions and a candidate real number, we cannot decide whether or not it's a real number from the axioms and definitions in the description.

The enumeration of defineable real numbers, then is not defineable.

However, every defineable real number is an element of descriptions of finite length.

A similar question is here: Are there real numbers that cannot be uniquely expressed with a finite number of symbols?

I don't understand this part of this answer:

Externally, it is possible (but not necessary) that every real number can be uniquely expressed in the same metalanguage you use to describe the set theory you're using. Here is a related question asking whether, not just each individual real number, but every set can be defined.

Apparently there's a model in ZFC that can specify each set - and thus real number. How does that work with my argument? Does that model require an infinite length description? If so, then to fully describe those real numbers, you need to fully describe the model of ZFC, which requires an infinite description. That's compatible with my argument that says some real numbers require an infinite description.

答:
But how do you know that there are uncountably many reals? Skolem's paradox is addressing exactly that. If $M$ is a countable model of $\sf ZFC$, then $M$ contains countably many reals. Yet internally to that model, the reals are uncountable.

The math tea argument is, when presented correctly, the following:

Let $\cal L$ be the language of analysis, i.e. $0,1,+,\cdot,\exp,\int,\sin,\cos$ and the function symbols for a few other things we want, perhaps. Then in the "natural interpretation" of $\cal L$ in $\Bbb R$ (which is even second, or third order logic) there are only countably many real numbers that are definable, and therefore most reals are undefinable.

But if we start adding more things to the universe, more real numbers can turn up definable. For example, if $r$ is any real number that wasn't definable originally, we can add it as a constant symbol. Or maybe we added a subset $A\subseteq\Bbb R$ such that $A\cap [n,n+1)$ has exactly the number of elements as the $n$th decimal expansion of $\Bbb R$. This allows us to define $r$, and that's just one example.

But all of this happens inside a universe of a set theory, that is much larger and much more complex, so there's a lot more than can be defined there.

So the problem, really, is that "finite description" doesn't tell us what you are allowed to use for the description. And when you start allowing the description to come from the entire mathematical universe, you invariably start having more things that you can describe. Moreover, since the definitions given over the whole mathematical universe are not internal to the universe, they usual arguments simply break down.


问:
Line integral - Parametric representation
the question is: C is the following area:

$$C = \left\lbrace (x,y) \; \middle| \;
         x^2 + y^2=\frac{\pi^2}{4}, \;  x \geq 0  \right\rbrace$$

Begins in: $\displaystyle \left( 0,\frac{\pi}{2} \right)$ end in $ \displaystyle \left( 0, -\frac{\pi}{2} \right)$

we want to calculate the integral : $$I = \int_{C} \left( e^x\sin{y} \; + 24 y \right) \, dx + e^x \cos{y} \,dy$$

First I'm required to use green's theorem, - I completed C to a complete circle and calculated :
$$ \displaystyle
  J   =  \int_{C\cup C_1 }(e^x\sin{y} \; + 24 y) \,dx + e^x\cos{y} \, dy
      =  \iint 24 dxdy = {3\pi ^3} $$

Now I'm told to subtract the following line integral with $y(t) = t$:
$$
\int_{C_1} ( e^x \sin{y} \; + 24 y ) \, dx + e^x \cos{y} \, dy = {} \rule{60pt}{0pt}
$$

Where $C_1$ is the "rest of the circle"
So its:
$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} *something* dt$

I Can't figure out what that something should be.. help would be greatly appreciated.

答:
$\displaystyle
  J   =  \oint_{C\cup C_1 }(e^x\sin{y} \; + 24 y) \,dx + e^x\cos{y} \, dy
      =  \iint 24 \, dA = {6\pi ^3}$ (Area of the given circle is $\frac{\pi^3}{4}$).

(given it is clockwise orientation).

The part that you need to notice is that we can split the vector field into two parts to make it easier to work with -

$\vec{F} = \vec{F_1} + \vec {F_2} \, $ where

$\vec{F_1} = (e^x\sin{y}, e^x\cos{y}), \vec {F_2} = (24y, 0)$

Now the first vector field is a conservative vector field and is gradient of scalar function $e^x \, \sin y$. So the line integral of this vector field will be path independent and will only depend on start and end points. This part of the vector field would not contribute anything to the line integral over the full circle which we calculated earlier.

$\vec{F_1} = \nabla f$, where $f = e^x \sin y$.

Line integral of $\vec{F_1} \,$ over semicircular path $C_1$,

$J_1 = \displaystyle \int_{C_1} \vec{F_1} \cdot d\vec{r} = \int_{(0, \pi/2)}^{(0,3\pi/2)} (\nabla f) \cdot d\vec{r} = f(0,\frac{3\pi}{2}) - f(0,\frac{\pi}{2})$

Now let's calculate the line integral of the vector field $\vec{F_2}$ over $C_1$ oriented counterclockwise.

$\vec{r}(t) = (\frac{\pi}{2} \cos t, \frac{\pi}{2} \sin t) \,$ (parametric equation of the circle with radius $\frac{\pi}{2}$ and centered at the origin).

$\vec{r'}(t) = (-\frac{\pi}{2} \sin t, \frac{\pi}{2} \cos t)$

$J_2 = \displaystyle \int_{C_1} \vec{F_2} \cdot d\vec{r} = \int_{\pi/2}^{3\pi/2} \vec{F_2} \cdot \vec{r'}(t) \, dt$

The line integral you are looking for over the semicircle from point $(0, \frac{\pi}{2})$ to $(0,-\frac{\pi}{2})$ oriented clockwise should be $ = J + J_1 + J_2 \,$ as $J$ is line integral over the full circle oriented clockwise, $J_1$ is the line integral of a conservative vector field between given points and $J_2$ is the line integral over semicircle oriented counterclockwise.


问:
If there exist any $c, d$ so that $\frac{a_{new}}{b_{new}}\rightarrow \frac{c}{d}$ but never reaches $\frac{c}{d}$, I think

I have an algorithm to solve this problem  * given two valid simplifying fractions $\dfrac{a}{b}$ and $\dfrac{c}{d},$ each transformation is adding both $a$ and $b$ by $+1,$ simplifying it, again and again until we meet $\dfrac{c}{d}.$ Count the number of steps used, $0$ if none" so that $\dfrac{a}{b}, \dfrac{c}{d}< 1$ and $0< a< b\leq 10^{5}, 0< c< d\leq 10^{5}.$ Easily, I see that the steps only continue if $ad< bc.$ Because we have that $\dfrac{a}{b}< \dfrac{a+ 1}{b+ 1}\overset{\div gcd(a, b)}{\rightarrow}\dfrac{a_{new}}{b_{new}}\rightarrow \dfrac{c}{d}.$ That sounds good.

But my test is not perfect, I wonder if there exist any $c, d$ so that $\dfrac{a_{new}}{b_{new}}\rightarrow \dfrac{c}{d}$ but never reaches $\dfrac{c}{d}.$ I need to your helps.

答:
Find $x$ such that $\frac{a+x}{b+x}=\frac cd$, i.e. $x=\frac{bc-ad}{d-c}$. If this is negative or not an integer, fail immediately.
Otherwise, of all prime divisors $p$ of $b-a$, find the one minimising $y:=p-(b \bmod p)$. If $y\ge x$, the answer is $x$, otherwise it is $y$ plus the result for $\frac{a+y}{b+y}$ (simplified) abd$\frac cd$.


答:
What you are looking at is the symmetry group of the square, which has $8$ elements. Four of them are generated by a quarter turn rotation (the identity the rotation itself, a half turn and the rotation in the opposite sense), and the other $4$ can be obtained from these by composing each with the same reflection (it does not matter which one, as long as you use the same one). Say if I is identity, R the rotation, and H a horizontal flip I,R,RR,RRR,  H,RH,RRH, RRRH.
One has identities IX=X=XI for all X, RRRR=I, HH=I, and HR=RRRH (so indeed, order does matter here), which allow any combination of operations to be brought into one of these $8$ forms. A vertical flip V is just V=RRH and so does not add anything new. If you really need to know more about this, you will have to study a bit of group theory.

Title: Which combinations of NxN matrix rotations and horizontal / vertical flips result in the same outcome?

Body:
I am doing an Advent of Code 2020 puzzle and currently trying to figure out which combinations of rotations and horizontal / vertical flips performed on an NxN matrix (2D array, really) result into a same MxM matrix.

For example, given this initial NxN matrix A:

```<code>#........#
........#.
#......#..
......#...
.....#....
....#.....
...#.....#
..#......#
.#.......#
#.#....###
</code>```

If we rotate it 90 degrees CW, we get B:

```<code>#......#.#
.#........
#.#.......
...#......
....#.....
.....#....
......#...
#......#..
#.......#.
####.....#
</code>```

Then again, we could do a series of other operations that also result in matrix B:

```<code>Initial
#........#
........#.
#......#..
......#...
.....#....
....#.....
...#.....#
..#......#
.#.......#
#.#....###

Rotate 90 (90)
#......#.#
.#........
#.#.......
...#......
....#.....
.....#....
......#...
#......#..
#.......#.
####.....#

Rotate 90 (180)
###....#.#
#.......#.
#......#..
#.....#...
.....#....
....#.....
...#......
..#......#
.#........
#........#

Rotate 90 (270)
#.....####
.#.......#
..#......#
...#......
....#.....
.....#....
......#...
.......#.#
........#.
#.#......#

Horz flip
#.#......#
........#.
.......#.#
......#...
.....#....
....#.....
...#......
..#......#
.#.......#
#.....####

Vert flip
#......#.#
.#........
#.#.......
...#......
....#.....
.....#....
......#...
#......#..
#.......#.
####.....#
</code>```

So what I want to know is that which set of operations I have to do, to have gone through all the possible transformations of the initial matrix aka. which sets of rotate|flipVert|flipHorz transformations result in the same outcome and therefore can be left out?

Edit: Also, now that I think of it, does the order of these operations matter?

P.S I am a programmer, not a mathematician. I am guessing this probably has something to do with matrix transformations / linear transformations and perhaps even congruent shapes?


问:
Apply complex residue theory to non even function improper integration
I am considering the integral

$$\int_{-\infty}^{\infty}\frac{t-5}{(t - 3+i)(t - 3-i)(t-2i)(t+2i)} dt.$$

The integrand, say $f(t)$, is readily not even, right? How do we use residues to solve this? The common technique uses the interior of large $[-R,R]$ upper-half plane semicircle and line segment combo. Only two singularities of the integrand are inside that interior. The rest is as usual.

Similar example is Complex analysis: Improper integral of a non-even function.

However, in this question and in that link as well, we will only get the value of Cauchy principle value $\lim_{R\rightarrow\infty} \int_{-R}^{R} f(t)\text{ }dt$, right?

If $f(t)$ is even, then we finally can get the value we want. Since here it is not, how do we continue? Similarly, why in that link, the answer can conclude the value too?

Thanks for any suggestion! Much appreciated!

答: In this case, you have the quotient of a polynomial function with degree $1$ by a polynomial function with degree $4$, without real roots. So, if $R(t)$ is the quotient, it is automatically true the the integrals $\int_0^\infty R(t)\,\mathrm dt$ and $\int_{-\infty}^0 R(t)\,\mathrm dt$ converge (compare them with $\frac1{t^3}$) and therefore$$\int_{-\infty}^\infty R(t)\,\mathrm dt=\lim_{R\to\infty}\int_{-R}^RR(t)\,\mathrm dt.$$So, go ahead and apply the method of residues.


问:
Proof that we can find rational numbers arbitrarily close to $\sqrt{2}$: direct approach.
The statement of the proposition:

Proposition. For every rational number $\epsilon > 0$, there exists a non-negative rational number $x$ such that $x^{2} < 2 < (x+\epsilon)^2$.

The most common approach to proving the proposition is by using contradiction (1,2).

My question is: is it possible to prove the proposition directly? More concretely, is it possible to find a function $f: \mathbb Q^+\rightarrow \mathbb Q^+$ such that for arbitrary positive rational $\epsilon$, we have

$$f(\epsilon)^2 < 2 < (f(\epsilon) + \epsilon)^2 $$

?

答: Using the fact that between any two real numbers there exists a rational number, given any rational $\varepsilon$ such that $4\sqrt{2}>\varepsilon>0, \exists x \in \mathbb{Q}$ such that $x \in (\sqrt{2} - \frac{\varepsilon}{2}, \sqrt{2})$, which gives: $x^2 < 2$ and $(x+\varepsilon)^2 > 2.$


问:
Counting Possible combinations
I do not now how to count the number of possibilities. Perhaps you can help.

I have 12 sided dice.
The sides are colored.
4 yellow
3 blue
2 green
1 red
1 black
1 white

If I roll 2 dice there are 16 possible combinations of two yellow sides showing out of a possible 144 combinations.

If I add dice, how do I count the number of possible yy combinations?

To complicate matters, the black dice are sometimes wild.
When they are, then I want to count the number of yellow and black.

If I have two dice, then I want to count the possible combinations of yy, yb, by, and bb. I can make a simple table and observe the number of instances. In this case, there are 25 possible combinations out of 144 possibilities.

When I have more than 2 dice how many yy combinations are possible given x dice?  And if I add the black dice how many yy, by, yb, bb combinations are possible given x dice?

Is there a formula that I can use to count the various combinations?  Tables become impractical when there are more than 2 dice.  I am unable to find formulas that help in counting the number of specific combinations our of the total number of combinations. In most examples, the identification of subset# out of total# seems to be done manually.

答:
This following is really a longish comment, not a solution, because it's not really feasible for large numbers of dice, but I think it's interesting to note that if we define
$$P=4 y + 3 b + 2 g + r + b + w$$
then if we expand $P^2$ we have counted all the possible outcomes of rolling two dice:
$$P^2 = 16 b^2+16 b g+8 b r+8 b w+32 b y+4 g^2+4 g r+4 g w+16 g y+r^2+2 r w+8 r y+w^2+8 w
   y+16 y^2$$
I.e. there are $16$ ways to roll blue-blue, $16$ ways to roll blue-green, $8$ ways to roll blue-red, etc.  And in general, expanding $P^n$ counts all the outcomes when rolling $n$ dice.  This process is only feasible for fairly small values of $n$, although it a computer algebra system can expand the range of application beyond the limits of pencil and paper computation.

Title: Counting Possible combinations


Body:
(From OMITS Indonesian $2015$)

Known that $f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$, if $x_1, x_2, x_3,..., x_{2015}$ is the roots of $f(x)$, Find :

$\sum_{n=0}^{2014}$ $\sum_{k=1}^{2015}$ $(x_k)^n$

$a. -2016$

$b. -2015$

$c. -2014$

$d. -2013$

$e. -2012$

What should i do? and where i can learn more about it?

Body:
First notice that $$\sum_{n=0}^{2014}\sum_{k=1}^{2015}(x_k)^n = \sum_{k=1}^{2015}\frac{(x_k)^{2015}-1}{x_k-1}$$
since $x_k$ is always $\ne1$.

Now we will compute the sums $S_1 = \sum\frac1{x_k}$ and $S_2 = \sum\frac1{x_k^2}$ over all roots. Consider the function $f(1/x)$ for $x \ne 0$. $f(1/x) = 0 \iff 2016x^{2015}+2015x^{2014}+\ldots+2x+1 = 0$. Let $x_1', \ldots, x_k'$ be the roots of this last equation. By Vieta's formulas $x_1'+\ldots+x_k' = -2015/2016$. Moreover, the sum of the squares of the roots is $x_1'^2+\ldots x_k'^2 = (x_1'+\ldots x_k')^2-2\sum_{0\le i < j \le k}x_i'x_j' = (-\frac{2015}{2016})^2-2\frac{2014}{2016}$.
It's quite obvious now that since each $x_k \ne 0$ the roots $x_1', \ldots, x_k'$ must be exactly $1/x_1, 1/x_2, \ldots 1/x_k$. Overall we find that $$S_1 = -\frac{2015}{2016}\ \ \text{and}\ \ S_2=\left(-\frac{2015}{2016}\right)^2-2\frac{2014}{2016}$$
Finally the comment by J.G. tells us that $$2016-2017x_k+x_k^{2017} = 0 \implies x_k^{2015} = -\frac{2016}{x_k^2}+\frac{2017}{x_k}\tag{1}$$and$$\frac{2016-2017x_k+x_k^{2017}}{1-x_k} = 0\implies 2016+\frac{x_k^{2017}-x_k}{1-x_k} = 0\implies \frac{x_k^{2016}-1}{x_k-1}=\frac{2016}{x_k}\tag{2}$$
Thus using $(1)$ and $(2)$ the solution is $$\sum_{k=1}^{2015}\frac{(x_k)^{2015}-1}{x_k-1} = \sum_{k=1}^{2015}\frac{(x_k)^{2016}-1}{x_k-1}-\sum_{k=1}^{2015}x_k^{2015} = -\sum_{k=1}^{2015}\frac1{x_k}+2016\sum_{k=1}^{2015}\frac1{x_k^2}=$$ $$= \frac{2015}{2016}+\frac{2015^2}{2016}-2\cdot 2014 = -2013$$


问:
Representation of Frechet derivative
A linear operator $\mathcal{L}_A : \mathcal{S}^n\to \mathbf{R}^{n\times n}$ is represented as $\mathcal{L}_AX:=AX+XA^{\top}$. Find a representation for Frechet derivative of $f(X) = AX+XA^{\top}$ and $g(X) = A^2X+2AXA^{\top}+X(A^{\top})^2$.

I am not sure if I understand the question clearly. What do I need to do exactly? Calculate the Frechet derivative of the given $2$ functions and then somehow represent them in terms of the linear operator? What does it exactly mean to represent in terms of an operator? I am new to this field of mathematics, so I am not sure how to go about the problem.

答:
The definition of Frechet derivative is the following

$\boldsymbol{Definition:}$ Let $\boldsymbol{V}$, $\boldsymbol{W}$ be vector spaces with norms $||.||_V, ||.||_W$. A function $f:\boldsymbol{V} \rightarrow \boldsymbol{W}$ is called Frechet differentiable at $\hat{x} \in \boldsymbol{int\ dom}f$, if $\exists$ a bounded linear operator $Df(\hat{x}): \boldsymbol{V} \rightarrow \boldsymbol{W}$ s.t.,
$$\lim_{x\rightarrow \hat{x}\\x \in \boldsymbol{int\ dom}f\\x \neq \hat{x}} \frac{||f(x) - f(\hat{x}) - Df(\hat{x})(x-\hat{x})||_W}{||x - \hat{x}||_V} = 0$$

If $f:\boldsymbol{V} \rightarrow \boldsymbol{W}$ is Frechet differentiable at $\hat{x}$, then $Df(\hat{x})$ is called the Frechet derivative of $f$ at $\hat{x}$.

In the light of this definition, we can find the Frechet derivative of the given functions.

$\boldsymbol{Claim}:$ Let $f:\boldsymbol{S}^n \rightarrow \boldsymbol{R}^{n \times n}$ be the given function such that $f(X) = AX +XA^T$. Then, Frechet derivative $Df(\hat{X})$ of $f$ is the linear operator given by $Df(\hat{X}) = f$, i.e., $$ Df(\hat{X})(X) = AX + XA^T$$

$\boldsymbol{Proof}:$ Let $ Df(\hat{X})(X) = AX + XA^T$, and consider the following limit,
$$\lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||f(X) - f(\hat{X}) - Df(\hat{X})(X-\hat{X})||}{||X - \hat{X}||} = \lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||(AX + XA^T) - (A\hat{X} + \hat{X}A^T) - Df(\hat{X})(X-\hat{X})||}{||X - \hat{X}||} = $$
$$\lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||[A(X-\hat{X}) + (X-\hat{X})A^T] - [A(X-\hat{X}) + (X-\hat{X})A^T]||}{||X - \hat{X}||} = \lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||0||}{||X - \hat{X}||} = 0$$
Since the above limit goes to zero with the given operator, it is the Frechet derivative of the function $f$ at $\hat{X}$, i.e., $Df(\hat{X}):\boldsymbol{S}^n \rightarrow \boldsymbol{R}^{n \times n}$ such that $Df(\hat{X})(X) = AX + XA^T$. $\blacksquare$

For the second function $g(X) = A^2X + 2AXA^T+X(A^T)^2$, the Frechet derivative at $\hat{X}$ is $Dg(\hat{X}) = g$, i.e., $$Dg(\hat{X})(X) = g(X) = A^2X + 2AXA^T+X(A^T)^2$$
The above proof can be applied to prove that.

$\boldsymbol{Remark}:$ The reason why $Df(\hat{X}) = f$, and $Dg(\hat{X}) = g$ is that the both functions are linear in $X$.

Title: Representation of Frechet derivative


问:
Let $B:=A^2-A-id$, where $A$ is the left shift operator. Find $dim(\ker(B))$, basis for $\ker(B)$ and if $B$ is surjective.
Let $V$ be the infinite dimensional $\mathbb{R}$-vector space of sequence of real numbers $(a_0, a_2, ...)$. Let $A\in \operatorname{End}(V)$ (endomorphism) be the left shift operator, so that $(a_1, a_2, ...) \to (a_2, a_3, ...) $ and define:
\begin{equation}
B:=A^2-A-\text{id} \in \operatorname{End}(V)
\end{equation}
Show that B is surjective. Determine also the dimension of $\ker(B)$ and find a Basis of $\ker(B)$.

I am confused about the concepts of surjectivity (particularly, how to prove it), how to express kernels formally, and how to build basis. My approach to the given problem was the following:

a) As $$B(a_1, a_2, ...) = (a_3, a_4, ...) - (a_2, a_3, ...) - (a_1, a_2, ...) = (a_3-a_2-a_1, a_4-a_3-a_2, ...)$$
I think it is surjective, as all elements from $\operatorname{img}(B)$ are the result of $B(a_1, a_2, ...)$, but how should I express or prove this in a more formal way? (please, correct me if I am wrong)

b) First, I want to express $B(a_1, a_2, ...)$ as follows (am I right to do it this way?):
$$Ba = (a_{k+2}-a_{k+1}-a_{k}|k\in \mathbb{N}\text{ and }a_k\in\mathbb{R})$$.

I suppose that I can express the kernel as follows:
$$\ker(B)=(x_k|x_{k+2}=x_{k+1}+x_{k})$$

As each $x_k$ is determined by another two elements, then follows $\dim(\ker(B))=2$. Am I right? I also think there should be a better way to express this statement (or to calculate the dimension in general).

c) How should I calculate the Basis of $\ker(B)$?

Thank you for your help.

答:
Let me adress surjectivity first. You start with a sequence $b=(b_1, b_2, \dots)$ and you want to find some $a=(a_1, a_2, \dots)$ such that $B(a)=b$. Now we can express this in terms of equations for every entry. Namely, for the first entry we get
$$ b_1 =a_3 - a_2 - a_1.$$
We can for example pick $a_1=0=a_2$ and $a_3=b_1$. For the second entry we get
$$ b_2 = a_4 - a_3 - a_2.  $$
With our previous choice we get
$$ b_2 = a_4 - b_1 $$
and hence, we have to choose $a_4= b_1+b_2$. Now we could go on. We just recursively define
$$ a_{n+2} = b_n + a_{n+1} + a_n $$
and $a_1=0=a_2$.

Now let us consider the kernel. This is the same thing, but with $b=(0, 0, \dots).$ In every entry we have the equation
$$ 0 = a_{n+2} - a_{n+1} - a_n. $$
Thus, we get $a_{n+2}=a_{n+1}+ a_{n}$. This means, $a$ is uniquely determined by $a_1$ and $a_2$ and hence, the kernel has dimension $2$. For example we could take $a_1=1=a_2$, then we get the Fibonacci numbers patched together. We could complete the vector with $a=(0,1,1,2, \dots)$, i.e. the previous vector shifted to the right and with an additional zero.


问:
How to compute Frechet Derivatives that involve matrices.
Suppose I have a linear operator which maps $S^n\to\mathbb R^{n \times n}$. It is represented as $LX:= XA^T + AX.$ How can I find the Frechet derivative of $f(X) = XA^T + AX.$

答:
I will give the same proof included in https://math.stackexchange.com/a/3956281/864940.

The definition of Frechet derivative is the following

$\boldsymbol{Definition:}$ Let $\boldsymbol{V}$, $\boldsymbol{W}$ be vector spaces with norms $||.||_V, ||.||_W$. A function $f:\boldsymbol{V} \rightarrow \boldsymbol{W}$ is called Frechet differentiable at $\hat{x} \in \boldsymbol{int\ dom}f$, if $\exists$ a bounded linear operator $Df(\hat{x}): \boldsymbol{V} \rightarrow \boldsymbol{W}$ s.t.,
$$\lim_{x\rightarrow \hat{x}\\x \in \boldsymbol{int\ dom}f\\x \neq \hat{x}} \frac{||f(x) - f(\hat{x}) - Df(\hat{x})(x-\hat{x})||_W}{||x - \hat{x}||_V} = 0$$

If $f:\boldsymbol{V} \rightarrow \boldsymbol{W}$ is Frechet differentiable at $\hat{x}$, then $Df(\hat{x})$ is called the Frechet derivative of $f$ at $\hat{x}$.

$\boldsymbol{Claim}:$ Let $f:\boldsymbol{S}^n \rightarrow \boldsymbol{R}^{n \times n}$ be the given function such that $f(X) = AX +XA^T$. Then, Frechet derivative $Df(\hat{X})$ of $f$ is the linear operator given by $Df(\hat{X}) = f$, i.e., $$ Df(\hat{X})(X) = AX + XA^T$$

$\boldsymbol{Proof}:$ Let $ Df(\hat{X})(X) = AX + XA^T$, and consider the following limit,
$$\lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||f(X) - f(\hat{X}) - Df(\hat{X})(X-\hat{X})||}{||X - \hat{X}||} = \lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||(AX + XA^T) - (A\hat{X} + \hat{X}A^T) - Df(\hat{X})(X-\hat{X})||}{||X - \hat{X}||} = $$
$$\lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||[A(X-\hat{X}) + (X-\hat{X})A^T] - [A(X-\hat{X}) + (X-\hat{X})A^T]||}{||X - \hat{X}||} = \lim_{X\rightarrow \hat{X}\\X \in \boldsymbol{int\ dom}f\\X \neq \hat{X}} \frac{||0||}{||X - \hat{X}||} = 0$$
Since the above limit goes to zero with the given operator, it is the Frechet derivative of the function $f$ at $\hat{X}$, i.e., $Df(\hat{X}):\boldsymbol{S}^n \rightarrow \boldsymbol{R}^{n \times n}$ such that $Df(\hat{X})(X) = AX + XA^T$. $\blacksquare$


答:
First, we note that $f(x)-f(x)/x=\frac{-2016}{x}+\sum_{k=0}^{2015} x^k$  Dividing by $x$ and rearranging, we get
$$\sum_{k=0}^{2014} x^k=f(x)/x-f(x)/x^2+2016/x^2 -1/x$$
Evaluating at the roots of $f(x)$ and summing yields

$$\sum_i\sum_{k=0}^{2014} x_i^k=\sum_i 2016/x_i^2 -1/x_i.$$

This reduces the problem to finding the sum of the reciprocals of the roots, and finding the sum of the squares of the reciprocals of the roots.  This is an exercise in rewriting symmetric functions in terms of elementary symmetric polynomials to rewrite the expression in terms of the coefficients of $f(x)$.

Title: $f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$


问:
Dot product of vector valued function
I know that a dot product $p\cdot q$ measures the projection of parts $p$ on $q$, so that if $p$ and $q$ are orthogonal, $p\cdot  q$ = 0. However, I have been wondering about whether this applies to the dot product of any arbitrary vector valued function ( with vectors as inputs) of $p$ and $q.$ Would, in general, $f(p)\cdot f(q)$ be 0 when $p$ and $q$ are orthogonal? (where $f(p)$ and $f(q)$ are any arbitrary vector valued function with vectors as inputs evaluated at $p$ and $q$)

答: No. For example, if $p,\,q$ are $n$-dimensional vectors and $A,\,B$ are $n\times n$ matrices then $Ap\cdot Bq=p\cdot(A^TB)q$ depends on $A^TB$. In particular, we could choose $A=I$, then choose $B$ so $Bq\parallel p$, so $Ap\cdot Bq\ne0$ (unless $Bq=0$).

Title: Dot product of vector valued function


问:
Is there any specific terminology to refer to an initial sequence of a sequence?
Lets say you have a sequence  $S = (0, 1, 2, 3, 4, 5, 6, 7, 8)$

And another sequence $T = (0, 1, 2, 3)$

Is there any specific mathematical term that defines the relationship between $S$ and $T$, that specifically says that $S$ starts with $T$?

I thought $T$ would be called an initial sub-sequence, but this is incorrect because a sub-sequence seems to be a any subset of the elements of the sequence in the same order of the sequence (so even $(2, 4, 6, 8)$ would be a sub-sequence, while I want the prefix sub-sequence part, i.e. a sub-sequence that the sequence starts with)

答:
I usually use "initial segment", and "tail segment" or "end segment" or "final segment" to denote the last part of a sequence.

If you want to indicate the segment is not everything, then "proper initial segment" should suffice.


问:
Covering lattice edges with Ls
I recall seeing (approximately this) problem in an old issue of "Quant", but I cannot recall anymore how it was solved:

Suppose that the edges of a $n\times n$ lattice are covered with Ls. Each L is a figure consisting of two adjacent lattice edges, and an L may point in any of the four directions: upwards left, upwards right, downwards left, downwards right. If all the lattice edges are covered with Ls and no two Ls overlap, prove that there are equally many Ls pointing upwards left as those pointing downwards right (and the same for upwards right and downwards left pointing Ls).

Any hint?

答:
Hint: restrict you attention to a single diagonal line of lattice vertices, going from vertex on the upper border or left border and going down to the right. Show there are an equal number number upper-left Ls and downward right Ls whose corner is on that line.

Further hint:

 If you rotate the grid so this diagonal is horizontal, then one type of L looks like an open parenthesis, and the other like a closed parenthesis.

Title: Covering lattice edges with Ls


问:
Why doesn't the Rational Root Theorem work on $x^3-6x^2+4x-5=0$?
I tried using the Rational Root Theorem on $x^3-6x^2+4x-5=0$. However, I could not find a rational root of the cubic. When I plugged the cubic into Wolfram Alpha, it yielded a very messy real solution:

How do you tell whether the RRT doesn't work on a particular cubic of the form $ax^3+bx^2+cx+d=0$? How do you tell whether $ax^3+bx^2+cx+d=0$ doesn't have any rational roots?

答:
After some extensive testing of my own on another polynomial that had zeros but could not be determined by the rational roots test I came to the following conclusion:

It appears, that the RRT will only give you $\mathit rational$ numbers as answers. Meaning, a polynomial might very well have zeros but the RRT will not give you any results. That is because said zeros are real numbers which cannot be represented by rational numbers. $(x\in\Bbb R, x \notin\Bbb Q)$

For OP's polynomial $x^3-6x^2+4x-5=0$, doing the RRT quickly leads one to believe that there are no zeros for it ;(it is left to the reader to confirm this as an exercise). It does however have one. Using an online zeros calculator one finds that the zero in question is $x=5.43316142415711$. This number is real (as in $\Bbb R$) and cannot be represented by a fraction.

In conclusion: The RRT will only give you results that are part of $\Bbb R$. There may however be polynomials that do very well possess zeros but those may not be part of $\Bbb R$. I hope this answers the question in the title.


问:
Since $i$ can be used for $90^\circ$ rotations, and $\sin$ and $\cos$ are orthogonal, why is it that $\sin t\neq i\cos t$ and $i\sin t\neq-i\cos t$?
As I know imaginary unit number $i$ can be used for $90^\circ$ rotation. For example, between $5$ and $5i$ there are $90^\circ$.

Also $sin$ and $cos$ are orthogonal too.

So why $\sin \theta \neq i \cos\theta $
or  $ i \sin\theta \neq -i\cos\theta$?

I think I found my answer in below equation of Euler:

$$e^{inx} = cos(nx) + i sin(nx)$$
so
$$ cos(nx) = e^{inx}-isin(nx)$$
It is not exactly how I imagined but yes It gives me the answer.

also as @José Carlos Santos stated in comments

quote:

They are orthogonal with respect to the inner product

$$ \langle f,g\rangle=\frac1{2\pi}\int_0^{2\pi}f(t)g(t)\,\mathrm dt.$$
This has nothing to do with 90∘ rotations in the plane

答:
Maybe the following is helpful to know. Since Euler we know
$$ e^{ix}=\cos(x)+i\sin(x). $$
This implies by multiplication
$$ i\,e^{ix}=i(\cos(x))+i(i\sin(x))=-\sin(x)+i\cos(x). $$
Thus, in some sense, it seems that
$\,i\cos(x)=-\sin(x)\,$ and $\,i\sin(x)=\cos(x)\,$ if you are
not looking at the equation carefully, but it will
give rise to false results.

As for orthogonality, as functions or signals, the
functions $\,\sin(x)\,$ and $\,\cos(x)\,$ are $\,90^\circ$
out of phase. This may be considered as a reason to state that they
are orthogonal, and there are other definitions of orthogonality.


问:
How does a calculator calculate the sine, cosine, tangent using just a number?

Sine $\theta$ = opposite/hypotenuse

Cosine $\theta$  = adjacent/hypotenuse

Tangent $\theta$  = opposite/adjacent

In order to calculate the sine or the cosine or the tangent I need to know $3$ sides of a right triangle. $2$ for each corresponding trigonometric function. How does a calculator calculate the sine, cosine, tangent of a number (that is actually an angle ?) without knowing any sides?

答:
Calculators either use the Taylor Series for $\sin / \cos$ or the CORDIC algorithm. A lot of information is available on Taylor Series, so I'll explain CORDIC instead.

The input required is a number in radians $\theta$, which is between $-\pi / 2$ and $\pi / 2$ (from this, we can get all of the other angles).

First, we must create a table of $\arctan 2^{-k}$ for $k=0,1,2,\ldots, N-1$. This is usually precomputed using the Taylor Series and then included with the calculator. Let $t_i = \arctan 2^{-i}$.

Consider the point in the plane $(1, 0)$. Draw the unit circle. Now if we can somehow get the point to make an angle $\theta$ with the $x$-axis, then the $x$ coordinate is the $\cos \theta$ and the $y$-coordinate is the $\sin \theta$.

Now we need to somehow get the point to have angle $\theta$. Let's do that now.

Consider three sequences $\{ x_i, y_i, z_i \}$. $z_i$ will tell us which way to rotate the point (counter-clockwise or clockwise). $x_i$ and $y_i$ are the coordinates of the point after the $i$th rotation.

Let $z_0 = \theta$, $x_0 = 1/A_{40} \approx 0.607252935008881 $, $y_0 = 0$. $A_{40}$ is a constant, and we use $40$ because we have $40$ iterations, which will give us $10$ decimal digits of accuracy. This constant is also precomputed1.

Now let:

$$ z_{i+1} = z_i - d_i t_i $$
$$ x_{i+1} = x_i - y_i d_i 2^{-i} $$
$$ y_i = y_i + x_i d_i 2^{-i} $$
$$ d_i = \text{1 if } z_i \ge 0 \text{ and -1 otherwise}$$

From this, it can be shown that $x_N$ and $y_N$ eventually become $\cos \theta$ and $\sin \theta$, respectively.

1: $A_N = \displaystyle\prod_{i=0}^{N-1} \sqrt{1+2^{-2i}}$


问:
A function vanishing in infinity with zero integral is integrable?
Suppose we have a continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ s.t.:

$f(x) \rightarrow 0$ as $||x|| \rightarrow \infty$;

$\iint_{\mathbb{R}^2}f(x)dx = 0$;

Would that mean that $\iint_{\mathbb{R}^2}|f(x)|dx < \infty$? It seems like it can be true by some form of generalization of alternating series test
 to integrals. If not, a counterexample would be appreciated. Thanks in advance!

答:
Hint: On $\mathbb R^1,$ function

$$f(x) =\frac{\sin x}{x} - \frac{1}{1+x^2}$$

satisfies your two conditions, yet

$$\int_{\mathbb R}|f(x)|\,dx = \infty.$$


问:
How do we arrive at the following modular arithmetic equation?
For ABC E - Throne, which says the following:
We have
N
chairs arranged in a circle, one of which is a throne.

Takahashi is initially sitting on the chair that is
S
chairs away from the throne in the clockwise direction. Now, he will repeat the move below.

Move: Go to the chair that is
K
chairs away from the chair he is currently sitting on in the clockwise direction.

After how many moves will he be sitting on the throne for the first time? If he is never going to sit on it, report -1 instead.

The solution says, He can sit on the throne after
x
moves if and only if
S
+
x
K
≡
0
mod
N. Why is this true?

答:
Number the chairs from $0$ to $N-1$ in the clockwise order. Assume that chair $0$ is the throne. This means that Takahashi sits on chair $S$ initially. Now, he moves to the chair that is $K$ chairs away in the clockwise direction. So the chair number would be $S+K$, but if this number goes past $N-1$ we should only look at the remainder when $S+K$ is divided by $N$(this is the essence of modular arithmetic where we wrap around the numbers in a circle). This means that in the first move, his position will be $S+K\pmod N$. In the second move, it will be $S+2K\pmod N$. His position after $x$ moves will be $S+xK\pmod N$.

Since chair numbered $0$ is the throne, he sits on it after $x$ moves if and only if $S+xK\equiv 0\pmod N$. The smallest $x$ for which this holds gives you the number of moves when he'll sit on the throne for the first time.


问:
Given rank-$1$ matrix $A$, how to compute $A^{100}$?

$$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$.

I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?

答:
Every matrix is annihilated by its characteristic polynomial. Since the characteristic polynomial is $$P(x)=x^2-2x,$$ we deduce $$A^2=2A.$$
Then $$A^{100}=(A^2)^{50}=(2A)^{50}=2^{50}(A^2)^{25}=\dots = 2^{99}\cdot A$$


问:
Are there non-homogeneous topological spaces which are "almost homogeneous" in Rudin's sense?
Let $X$ be a nonempty topological space and $G$ the group of all homeomorphisms $g:X\to X$. Recall that $X$ is homogeneous if $G$ acts transitively on $X$. In

"Homogeneity Problems in the Theory of Čech Compactifications", The Mathematical Legacy of Eduard Čech (1993), pages 81-92,

Walter Rudin considers the following condition on $X$: for all point $x\in X$ and all nonempty open subset $U\subset X$ there is a $g\in G$ such that $g(x)\in U$. He calls such a space almost homogeneous.

I suspect that there are simple examples of almost homogeneous non-homogeneous spaces, but I haven't been able to find any. Let me ask the implicit question formally, together with an obvious follow-up question:

Question 1. Are there almost homogeneous non-homogeneous spaces?

Question 2. Assuming the answer to Question 1 is Yes, what is the least possible cardinality of an almost homogeneous non-homogeneous space?

[I added the quotient-spaces tag because, in the above notation, $X$ is almost homogeneous if and only if the quotient topology of $G\backslash X$ is coarse.]

答:
Let $X_n$ be the Cayley graph of the free product $G_n=\frac{1}{2^n}\mathbb{Z}*\frac{1}{2^n}\mathbb{Z}$ (with respect to the two obvious generators).  Call edges of $X_n$ that come from the first generator horizontal edges and edges that come from the second generator vertical edges.  The inclusion map $G_n\to G_{n+1}$ induces an inclusion map $X_n\to X_{n+1}$.  Let $G=\mathbb{Z}[1/2]*\mathbb{Z}[1/2]$ be the colimit of the groups $G_n$, and let $X$ be the colimit of the spaces $X_n$.  Note that $G_n$ acts on $X_n$ for each $n$, and these actions glue to give an action of $G$ on $X$.  There is also an automorphism of $G_n$ that swaps the two generators which induces an automorphism of $X_n$ that swaps the horizontal and vertical edges, and these glue to give an automorphism $\sigma$ of $X$.

Now I claim $X$ is almost homogeneous.  Let $x\in X$ and let $U\subseteq X$ be a nonempty open subset.  Then there is some $n$ such that $x\in X_n$ and $U$ contains an open interval of some edge of $X_n$.  Applying the automorphism $\sigma$ if necessary, we may assume that $x$ is in a horizontal edge of $X_n$ and that $U$ also contains an open interval of a horizontal edge of $X_n$.

Now note that each edge of $X_n$ gets split in half to become two edges in $X_{n+1}$.  So for some $N\geq n$, the open interval that $U$ contains in $X_n$ actually contains an entire horizontal edge of $X_N$.  But now the action of $G_N$ is transitive on the horizontal edges of $X_N$, so some element of $G_N$ maps $x$ into the horizontal edge of $X_N$ that $U$ contains.  This gives an automorphism of $X$ that maps $x$ into $U$.

On the other hand, $X$ is not homogeneous.  To see this, note that if $x\in X$ is a vertex of $X_n$ for some $n$, then $X\setminus \{x\}$ has $4$ connected components (the four "branches" of the tree coming out from the vertex), whereas if $x\in X$ is not a vertex in any $X_n$, then $X\setminus\{x\}$ has only $2$ connected components (the two sides of the tree coming off of the edge that $x$ is in the middle of in each $X_n$).

This example has cardinality $2^{\aleph_0}$; I don't know whether you can get a smaller example but I suspect it is possible to get a countably infinite one.  It is easy to see that an almost homogeneous finite space must be homogeneous (consider the minimal nonempty open sets; by almost homogeneity every point must be in one and they all must have the same cardinality, which means the space is a disjoint union of indiscrete spaces of the same cardinality).


答:
If $f$ is a constant non-zero function, then $f(p)\cdot f(q)$ is a constant non-zero scalar for any $p$ and $q$, orthogonal or not.

There are many examples that are less extreme. The more interesting question would be what functions preserve the orthogonality property for all pairs of vectors.

Title: Dot product of vector valued function

Body: I know that a dot product $p\cdot q$ measures the projection of parts $p$ on $q$, so that if $p$ and $q$ are orthogonal, $p\cdot  q$ = 0. However, I have been wondering about whether this applies to the dot product of any arbitrary vector valued function ( with vectors as inputs) of $p$ and $q.$ Would, in general, $f(p)\cdot f(q)$ be 0 when $p$ and $q$ are orthogonal? (where $f(p)$ and $f(q)$ are any arbitrary vector valued function with vectors as inputs evaluated at $p$ and $q$)


问:
Contest Problem - Darts
Wendy is playing darts with a circular dartboard of radius 20. Whenever she throws a dart, it lands
uniformly at random on the dartboard. At the start of her game, there are 2020 darts placed randomly
on the board. Every turn, she takes the dart farthest from the center, and throws it at the board again.
What is the expected number of darts she has to throw before all the darts are within 10 units of the
center?

Let us call the area within 10 units of the center be I-Inner and the area outside be O-Outer.  There are 2020 darts altogether and the probability that it is in I is $\frac{1}{4}$ and that it is in O is $\frac{3}{4}$. If she takes a dart from O with probability $\frac{3}{4}$ and throws back at the board again, it lands in I with prob $\frac{1}{4}$ and in O with $\frac{3}{4}$. Thus I can think of this as a markov chain and represent this by the following transition matrix.

\begin{pmatrix} &I & O\\
I & 1 & 0\\
O & \frac{1}{4} & \frac{3}{4}
\end{pmatrix}

$A= \pmatrix{1&0\cr \frac{1}{4}&\frac{3}{4}}$

$I= 1$

Find the inverse of $(I-A)^{-1} = F$

$(I-A)= 1-\frac{3}{4}=\frac{1}{4}$

$F=4$

From F we can get to find the expected number of turns for the dart to reach the absorbing state is $4$ for one dart On an average there are $\frac{3}{4}\times 2020 = 1515$ darts in O and by linearity of expectation for 1515 darts, it is $4*1515 = 6060$.

Is this the correct approach of solving the problem in a different way than the solution outlined in the link ( problem number 13 of the Guts round):
HMMO 2020.

Could someone verify if this is a valid approach?

Thanks

答: For each individual dart, the expected number of throws until it ends up forever in the inner ring, is 4. However, this includes the initial placement throw! Hence in total, we expect (by additivity) $4\cdot2020$ throws, but have to subtract the $2020$ "preparation“ throws for a final answer of $6060$.

Title: Contest Problem - Darts


Title: Given rank-$1$ matrix $A$, how to compute $A^{100}$?

Body:
$$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$.

I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?

Body: Since $\;A=x^ty\;$ where $\;y=(6,4)\;$ and $\;x=(1,-1)\;$ then $A^{100}=(x^ty)^{100}=x^t(yx^t)^{99}y\;$. Since $\;yx^t=2\;$ (it is an inner product) you have $x^t(yx^t)^{99}y=2^{99}x^ty=2^{99}A\;$


问:
Limit as $x$ approaches $1$ from the right of $\frac{1}{\ln x}-\frac{1}{x-1}$
$$
\lim_{x\rightarrow 1^+}\;\frac{1}{\ln x}-\frac{1}{x-1}
$$

So I would just like to know how to begin to solve this limit, or what topic does this problem fall under so that I can search for examples online or in text.

I know the answer is 1/2 and I have tried plugging in numbers  smaller or greater than one but as I assumed that was the wrong way to go.

答:
No need to restrict to the right limit. By variable change $u=x-1$, this is
$$
\lim_{u\rightarrow 0}\;\frac{1}{\ln (1+u)}-\frac{1}{u}=\lim_{u\rightarrow 0}\;\frac{u-\ln (1+u)}{u\ln(1+u)}.
$$

1) L'Hospital: if we differentiate once top and bottom, we get
$$
\frac{u}{(1+u)\ln(1+u)+u}.
$$
Once more, we find
$$
\frac{1}{\ln(1+u)+2}\longrightarrow \frac{1}{2}.
$$
So the initial limit is indeed $\frac{1}{2}$ by a double application of L'Hospital.

2) Taylor: we can use that $\ln(1+u)=u-\frac{u^2}{2}+O(u^3)$. Then
$$
\frac{u-\ln(1+u)}{u\ln(1+u)}=\frac{\frac{u^2}{2}+O(u^3)}{u^2+O(u^3)}=\frac{\frac{1}{2}+O(u)}{1+O(u)}\longrightarrow \frac{1}{2}.
$$


问:
Is a broken (Euclidean) ball still convex?
Consider the Euclidean ball $B^n(x,r)$ in $\mathbb{R}^n$ given by:
$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_2 \leq r\}$$ with centre $x\in\mathbb{R}^n$ and radius $r\geq 0$.
Now we break this ball by removing an arbitrary subset from the boundary of $B^n(x,r)$. Is the resulting set still convex?

My intuition says Yes, and it's easy to visualise this in $\mathbb{R}^2$ or $\mathbb{R}^3$. I'm wondering how I can prove it generally, i.e. for $\mathbb{R}^n$ by considering an arbitrary subset $S\subseteq B^n(x,r)$? If $S = B^n(x,r)$ then the resulting set is empty, which is convex - so we can ignore that situation. We assume $S\subset B^n(x,r)$. What's next?

I consider two points $y_1,y_2 \in B^n(x,r)\backslash S$, and want to show that for $t\in [0,1]$ we have $ty_1 + (1-t)y_2 \in B^n(x,r)\backslash S$. How do I take it from here? I'm stuck.

Thanks!

Addendum:

What if we consider the $p$-norm in general, instead of the Euclidean norm? How do things change with $p$ if the ball is defined as follows?
$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_p \leq r\}$$
It sounds fun breaking balls in higher dimensions under different norms!

答: Along every (parametrised by $t$) line, the squared Euclidean norm is a quadratic function in $t$ and does not attain a value $\ge1$ between distinct points where it is $\le1$.


问:
If $E(X_n) \to E(X)$, does this imply convergence in $L^1$?
My question is basically if there exists any counterexample where we have that $\mathbb{E}(X_n)\to \mathbb{E}(X)$ as $n\to \infty$, but $X_n$ does not converge to $X$ in $L^1$. If not, then it should be true that $\mathbb{E}(X_n)\to \mathbb{E}(X)$ as $n\to \infty$ implies that $X_n \overset{L^1}{\rightarrow}X$ but I cannot really get to prove that one.

Any tips on how to approach this problem? Or any counterexample you could maybe think of?

答: Let $X$ be 1 with probability 0.5 and 0 with probability 0.5. Then let $X_n$ be copies of $X$. Let $Y = 1-X$. Then $\mathbb E[X_n] = 0.5 = \mathbb E[Y]$ but $\lVert X_n - Y \rVert_{L^1} = \Vert X - Y \rVert_{L^1} = 1$ for all $n$, so we don't have $L^1$ convergence.


答:
Note that

$$A= 2B \text{ with } B=\begin{bmatrix} 3 & 2\\ -3 & -2\end{bmatrix} \text{ and } B^2 = B$$

Hence,

$$A^{100} = 2^{100}B = 2^{100}\begin{bmatrix} 3 & 2\\ -3 & -2\end{bmatrix}$$


问:
For $\triangle ABC$ with circumradius $R$ and nine-point center $E$, prove that $EA+EB+EC\le3R$

Given a triangle $ABC$ with the circumradius $R$, the centroid $G$, and the nine-point center $E$. Prove that $$EA+EB+EC\le 3R$$

Note. I'm trying to use the vector equation below  to prove this inequality, however, I'm stuck: $$\overrightarrow{EA} + \overrightarrow{EB} + \overrightarrow{EC} = 3\overrightarrow{EG}$$

答:
We need the help of two triangle inequalities to prove the inequality mentioned in OP’s statement. We provide the proof of one of them while leaving the other for OP to try his/her hand at.

Consider the scalene triangle $UVW$ shown in $\mathrm{Fig.\space 2}$. Its sides and one of its medians (i.e. $WM$) have lengths $u$, $v$, $w$, and $m$. We shall write,
$$u^2=m^2+\frac{w^2}{4}+m\times w\times \cos\left(\omega\right)\qquad\left(\mathrm{for}\space \triangle MVW\right), \tag{1}$$
$$v^2=m^2+\frac{w^2}{4}-m\times w\times \cos\left(\omega\right)\qquad\left(\mathrm{for}\space \triangle UMW\right). \tag{2}$$

When we add (1) to (2), we get,
$$u^2+v^2 =2m^2+\frac{w^2}{2}\quad\rightarrow\quad 4m^2= 2u^2+2v^2-w^2. \tag{3}$$

As shown below, equation (3) can be transformed into the first of the two sought triangle inequalities, which is, in fact, the mainstay of the proof of OP’s inequality.

$$4m^2= 2u^2+2v^2-w^2= u^2+v^2+\left(u^2+v^2-w^2\right) = u^2+v^2+2uv\cos\left(\phi\right)$$

Since thelargest value of $\cos\left(\phi\right)$ is 1, we cam write,
$$4m^2\le u^2+v^2+2uv=\left(u+v\right)^2 \quad\rightarrow\quad   u+v\ge 2m. \tag{4}$$

Here, the equality holds only for the degenerated triangle. The purpose of the numerical values shown in the diagram is to convince you that this inequality holds.

Now, pay your attention to the $\mathrm{Fig.\space 1}$, which depicts the configuration described by OP and auxiliary points and segments introduced by us. $H$ and $O$ are the orthocenter and circumcenter of the triangle $ABC$ respectively. We assume that OP is aware of the fact that the center of the nine-point circle $E$ is the midpoint of the segment $HO$. We denote the radius of the circumcircle $ABC$ as $R$.

Consider the triangle $AOH$, where $AE$ is one of the medians. We can apply the inequality (4) to this triangle to obtain $2AE\le AO+AH$. Since $AO=R$, this becomes $2AE\le R+AH$. In a similar vein, we can state $2BE\le R+BH$ and $2CE\le R+CH$ by considering the triangles $BHO$ and $CHO$ respectively. When we add these three inequalities together, we have,
$$2\left(AE+BE+CE\right)\le 3R+\left(AH+BH+CH\right). \tag{5}$$

Let us try to express the segment $AH_b$ as a function of $R$. Consider the isosceles triangle $ABO$. There, we have $AB=2R\sin\left(\hat{C}\right)$. Next, consider the right-angled triangle $ABH_b$. It is easy to see that $$AH_b =AB\cos\left(\hat{A}\right)= 2R\sin\left(\hat{C}\right) \cos\left(\hat{A}\right). \tag{6}$$

Finally, we consider the right-angled triangle $AHH_b$ to obtain
$$AH_b =AH\sin\left(\hat{C}\right) \tag{7}.$$

Now, we have two expressions for $AH_b$, i.e. (6) and (7). From them, it follows,
$$AH_b =2R\sin\left(\hat{C}\right) \cos\left(\hat{A}\right)= AH\sin\left(\hat{C}\right)\qquad\rightarrow\qquad AH=2R\cos\left(\hat{A}\right).$$

Similarly, we are able to express both $BH$ and $CH$ as $BH=2R\cos\left(\hat{B}\right)$ and $CH=2R\cos\left(\hat{C}\right)$ respectively. When we substitute these values in (5), we get,
$$2\left(AE+BE+CE\right)\le 3R+2R\left(\cos\left(\hat{A}\right)+\cos\left(\hat{B}\right)+\cos\left(\hat{C}\right)\right). \tag{8}$$

It is up to OP to prove using trgonometry, for a given triangle,
$$ \cos\left(\hat{A}\right)+\cos\left(\hat{B}\right)+\cos\left(\hat{C}\right)\le \frac{3}{2}.$$

Once OP has done that, he/she can insert it in (8) to get the sought inequality, i.e.
$$AE+BE+CE\le  3R, $$
where equality holds if and only if $\triangle ABC$ is an equilateral triangle, This is because, in an equilateral triangle, the two centers in question coincide.


Body:
$$|\lambda I - A| = \lambda^2 - 2\lambda$$
A matrix satisfies its characteristic equation:
$$A^2 -2A =0$$
$$A^2 = 2A$$
$$A^4 = 4A^2 = 8A$$
$$A^8 = 16A^4= 2^7 A$$
$$A^{16} = (A^4)^4 = 2^{12}A^4=2^{15}A$$
$$A^{32}= (A^16)^2 = 2^{30}A^2 = 2^{31}A$$
$$A^{64} = (A^16)^4 = 2^{60} A^4 = 2^{63}A$$
$$A^{100}= A^{64+32+4} = 2^{63+31+3}A^3 = 2^{97}A^3=2^{99}A$$
$$A^{100} \approx \pmatrix{3.8&2.5\\-3.8&-2.5}\cdot10^{30}$$


问:
Why Spin(V) is a connected group?
I want to know why $Spin(V)$ is connected. I am watching $Spin(V)$, where $V$ is a real vector space with a positive defined metric, as the products of elements of the Clifford algebra $Cl(V)$ of the form $v_{1}...v_{r}$ where $v_{i} \in Cl(V)$ has norm $1$ and $r$ an even integer.

Why is $Spin(V)$ connected? (At the moment, I amnot supposed to know that is the covering space of any space).

答:
So $r=2n$. Let $\gamma_i:[0,1]\to V$ be a path connecting $v_i$ to a fixed vector $v_0\in V$ with $|v_0|=1$, and such that $|\gamma_i(t)|=1$ for all $t$ (this exists because the unit sphere in $V$ is path connected). Then $\Gamma:[0,1]\to Spin(V)$ defined by $\Gamma(t)=\prod_i \gamma_i(t)$ connects $\prod_i v_i$ to $(v_0)^{2n}={-1}^n \in Spin(V)$.

Now take a path  $\gamma_0$ connecting $v_0$ to $-v_0$ in the unit sphere of $V$. Then $v_0\gamma_0(t)$ connects $-1$ to $1$ in $Spin(V)$.

All together, any element of $Spin(V)$ is connected by a path to either $1$ or $-1$, and those two are connected by a path. So the whole $Spin(V)$ is (path) connected.

Title: Why Spin(V) is a connected group?


问:
Is $[0,5]$ open in $[-5,5]$?
I'm writing examples about relatively open sets. Can an open subset $U\subset\Bbb{R}$ be found such that $[0,5]=U \cap[-5,5]$, or is this maybe a nonsensical question because of some theorem?

答:
Bear with me

we want  $[0,5] \subset U\cap [-5,5]$. So that would mean $[0,5]\subset U$ and $0 \in U$.

And we want $U$ to be open in $\mathbb R$.  That would mean $0$ is interior point, or that there is some  $\epsilon > 0$ so that $(-\epsilon, \epsilon)\subset U$. If we let $\rho = \min(\epsilon, 5)$ then $(-\rho, \rho) \subset (-\epsilon,\epsilon)\subset U$.  And $(-\rho,\rho)\subset U\cap [-5,5]$.

And finally we also want $[0,5]$ to be open in $U\cap [-5,5]$.  But that means $0$ is an interior point of $[0,5]$ in $U\cap [-5,5]$  which means there is an $\omega>0$ so that $(-\omega,\omega)\cap [U\cap[-5,5]]$ is a subset of $[0,5]$.

But $(-\omega,0) \not \subset [0,5]$.  In fact $(-\omega,0)$ is disjoint from $[0,5]$.  So to have  $(-\omega,\omega)\cap [U\cap[-5,5]]\subset [0,5]$ we must have $[0,\omega)\subset U$ but $(-\omega,0)$ disjoint from $U$.

And that assures that $U$ is not open in $\mathbb R$ (because $(-\rho, \rho) \not \subset U$ after all!)

So it's possible to find a set $U$ where $[0,5]$ is open in $U\cap [-5,5]$ (all we need is that some $(-\omega,0)$ be disjoint from $U$ while $[0,5]\subset U$) but such a set can never be open in $\mathbb R$.

Title: Is $[0,5]$ open in $[-5,5]$?


问:
Book 2 of Visual Complex Functions
I am having a lot of fun in reading Visual Complex Functions by Professor Wegert. (It is a very interesting read and well-recommended by me.)

Inside it, he regularly lets things be and postpones until part 2 or book 2. I have asked Professor Wegert himself about his plans with it (or a title), but unfortunately he seems to be busy at the moment.

That's why I am asking the community: Does anyone know of the book 2 mentioned in Visual Complex Functions, or when it is due ?

答:
In personal correspondence with the author back in 2018, he said that the plan still exists for a second volume, but he cannot say when it will appear.

I am looking forward to any news too. If I hear anything, I will try to remember to post an update here.

Title: Book 2 of Visual Complex Functions


问:
Repeated application of L'Hopital's rule to evaluate $\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$
I came across an interesting limit while trying to draw the graph of the function $f(x)=\frac{3^{\ln x}}{x}$.

What is $$L=\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$

Since $\ln(0^+)\to -\infty$ and $3^{-\infty}=0$ we have the limit in indeterminate form of $\frac{0}{0}$.
So By L'Hopital's rule we get
$$L=\lim_{ x \to 0^+}\frac{3^{\ln x}\ln 3 \times  \frac{1}{x}}{1}=\ln 3\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$
So how to find this limit?

答:
Writing $x=e^{-t}$ and considering $t\to +\infty$ you get

$$\frac{3^{\ln x}}{x} = \left(\frac{3}{e}\right)^{-t}=\left(\frac{e}{3}\right)^{t}\stackrel{t\to +\infty}{\longrightarrow}0$$


问:
Why is this equivalence only for integers?
I read about the following rule:

$m < n$ is equivalent to $m + 1 \leq n$, for $m, n \in \mathbb Z$

Why is it only for integers though? I can't think of a counter example e.g. for real numbers.

答:
This holds for subsets of real numbers where the gap of consecutive numbers is at least $1$ and the gap doesn't even have to be uniform.

For example, it holds even for $m, n \in  \mathbb{Z}+\frac12$, which is technically just a translation of the integers.

If $m < n$, then since we know the  gap is at least $1$, we know that $m+1$ would not exceed $n$.


问:
Dot Product of Two vectors is commutative
I was reading this book by Oleg D. Jefimenko on Electricity and Magnetism which had a chapter on Vector Analysis. In this section, he defined dot product as $\bar A \cdot \bar B$ as |$ {\bar A} $||$\bar B$| $cos \theta$ and also as $A_xB_x+A_yB_y+C_xC_y$ and he stated that dot product is commutative.
And then he went used dot product to prove cosine law for triangles:

$$\bar C= \bar A+\bar B$$
$${\bar C}^2= (\bar A+\bar B)\cdot(\bar A+\bar B)$$
$${\bar C}^2= (\bar A \cdot \bar A)+(\bar A \cdot \bar B)+(\bar B \cdot \bar A)+(\bar B \cdot \bar B)$$
$${\bar C}^2= {\bar A}^2+{\bar B}^2+(\bar A \cdot \bar B)+(\bar B \cdot \bar A)$$

and then it went ahead and did something that left me confounded:

It said: $$ \bar A \cdot \bar B = |\bar A| |\bar B| cos \theta$$
$$ \bar B \cdot \bar A = |\bar B| |\bar A| cos (180^o -\theta)=-|\bar B| |\bar A| cos (\theta)$$
and then went ahead with the proof.
Isn't this contradictory to the part where it said that dot product of commutative.

I know this law can be proved using alternate means by choosing to reorient the sides of the triangle and rewriting $\bar C$ as $\bar C = \bar A - \bar B$.

But I want to know how did it manage to disregard the commutative property of dot product.

By the way this is on page 26 of the second edition of the book.

Edit:

答:
You are misunderstanding what he is saying. Note that he converts

$\mathbf b \cdot \mathbf c = bc\cos(\mathbf b, \mathbf c)$

$\mathbf c \cdot \mathbf b = cb\cos(\mathbf c, \mathbf b)$

And in the very next sentence, he clearly states:
$$\angle (\mathbf b, \mathbf c) =\angle (\mathbf c, \mathbf b)$$
Which along with commutivity of the multiplication $bc = cb$ still leaves us with
$$\mathbf b \cdot \mathbf c = \mathbf c \cdot \mathbf b$$

What he is saying is that neither of those angles is $\theta$. Instead they are both equal to $180^\circ - \theta$.

$\theta$ itself is the angle between $\mathbf c$ and $(-\mathbf b)$, the vector of the same length pointing in the opposite direction of $\mathbf b$. Remember that angles are calculated between vectors based at the same point. But in the picture, $\mathbf c$ has been slid so that its base is at the end of $\mathbf b$.


问:
Complex root of a number
Let us suppose I have a number $x$ such that $x<0$. If I want to write the roots of the $x^{1/n}$. How can we write the roots of this number. I thought we can write

$$|x|^{1/n}e^{i\pi\theta}$$ for $\theta = \frac{2l + 1}{n}$ and $l = 0,1,2$ etc.

Is this correct ?

Similary If I wanted to write $x^{m/n}$, I should I write

$$|x|^{m/n}e^{mi\pi\theta}$$ for $\theta = \frac{2l + 1}{n}$ ?

答:
What you have seems okay but i think $\theta$ in your case might be alittle off.

One way of doing this is through De Moivre's theorem, here it is:

$$|x|^\frac{1}{n}(cos(\theta)+isin(\theta))^{\frac{1}{n}}=|x|^{\frac{1}{n}}[cos(\frac{2k\pi + \theta}{n})+isin(\frac{2k\pi + \theta}{n})]$$

Where $k=0,1,2,3...,n-1$

So all your doing here is converting your complex number to polar form and applying De Moivre's theorem.

And then if you really need it in your form you can take the argument out of the trig form and use it in your form.

Im new to answering questions so any feedback is welcome thanks :)

Title: Complex root of a number


答:
It is generally easier to evaluate these problems if you can express the limit as one fraction. In this case, we have
$$
\frac{1}{\ln x} - \frac{1}{x-1} = \frac{x-1-\ln x}{\ln x(x-1)}.
$$
Applying L'Hopital's rule, we have
$$
\lim_{x\to1^+} \frac{1-\frac{1}{x}}{\frac{1}{x}(x-1) + \ln x}.
$$
We still have a limit of the form $\frac{0}{0}$, so again apply the rule again, and obtain
$$
\lim_{x\to1^+}\frac{\frac{1}{x^2}}{\frac{-1}{x^2}(x-1) + \frac{1}{x} + \frac{1}{x}} = \lim_{x\to1^+} \frac{x}{2x^2} = \frac{1}{2}.
$$</p>

Title: Limit as $x$ approaches $1$ from the right of $\frac{1}{\ln x}-\frac{1}{x-1}$

Body:
$$
\lim_{x\rightarrow 1^+}\;\frac{1}{\ln x}-\frac{1}{x-1}
$$

So I would just like to know how to begin to solve this limit, or what topic does this problem fall under so that I can search for examples online or in text.

I know the answer is 1/2 and I have tried plugging in numbers  smaller or greater than one but as I assumed that was the wrong way to go.


问:
Sum of nilpotent matrices with certain property is nilpotent
I want to prove that if $A$ and $B$ are nilpotent matrices such that $[A, [A,B]] = [B, [A,B]] = 0$ then $A + B$ is a nilpotent matrix as well.

I know that it must follow from some general theorem from Lie algebras theory. I am sure, though, that in this particular case it can be done by  simple methods.

My attempt uses rather the language of the Lie theory than its results  themselves. I noticed that such a matrices (if $dim \langle A, B, [A,B] \rangle = 3$) is a (faithful) representation if the Heisenberg algebra.  And since it is true for this algebra it must be true in any faithful representation. I think. I could't prove it.

I am looking for hints for proving it my way and for the solution via elementary methods.

答:
The statement is not true in general. E.g. consider
$$
A=\pmatrix{1&1\\ 1&1}
\text{ and } B=\pmatrix{0&1\\ 0&0}
$$
over $GF(2)$. We have $A^2=B^2=0,\,[A,B]=I_2$ and $[A,[A,B]]=[B,[A,B]]=0$, but $A+B$ is nonsingular.

Yet, the statement is true if $[A,B]$ is also nilpotent. (This occurs, for instance, when $A,B\in M_n(F)$ where $\operatorname{char}(F)=0$ or $\operatorname{char}(F)>n$, by Jacobson's lemma.) Since $A$ commutes with $[A,B]$, they share a common eigenvector $x$. Therefore $Ax=[A,B]x=0$. Hence we can prove by mathematical induction that $AB^kx=0$ for every nonnegative integer $k$, where in the inductive step we have
\begin{aligned}
AB^{k+1}x
&=ABB^kx\\
&=\left([A,B]+BA\right)B^kx\\
&=[A,B]B^kx\\
&=B^k[A,B]x\\
&=0.
\end{aligned}
However, since $B$ is nilpotent, there exists a nonnegative integer $k$ such that $B^kx\ne0=B^{k+1}x$. Therefore $B^kx\in\ker(A)\cap\ker(B)$, i.e. $A$ and $B$ share a common eigenvector. Proceed recursively, $A$ and $B$ can be simultaneously triangularised. Hence $A+B$ is nilpotent.

(Note that we don't need the condition $[B,[A,B]]=0$ in the above proof.)


问:
How to integrate $\int_{0}^{\infty} t^{-\frac{3}{2}} e^{-at - \frac{b}{t}}dt$ without gamma function?
How to solve this type of integral
\begin{equation}
\int_{0}^{\infty} t^{-\frac{3}{2}} e^{-at - \frac{b}{t}}dt
\end{equation}
where $a$ and $b$ are positive.

Please help me out with the integration without using the gamma function.

答:
Substitute $t=\sqrt{\frac ba    }\frac1{x^2 }$

\begin{align}\int_{0}^{\infty} t^{-\frac{3}{2}} e^{-at - \frac{b}{t}}dt
&=\sqrt[4]{\frac ab  } e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab }\>(x-\frac1{x})^2}dx\\
&= \sqrt[4]{\frac ab } e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab }\>x^2}dx\\
&=\sqrt[4]{\frac ab  } e^{-2\sqrt{ab}}\cdot\frac{\sqrt\pi}{\sqrt[4]{ab}}= \sqrt{\frac\pi b} e^{-2\sqrt{ab}}
\end{align}
where $\int_{-\infty}^\infty e^{-cx^2}dx=\sqrt{\frac \pi c}$ and $
\int_{-\infty}^{\infty} f\left(x-\frac 1x \right) dx
=\int_{-\infty}^{\infty} f(x) dx
$
are used.


问:
sine vs Sine: understanding the differences
I was using the textbook A History in Mathematics by Victor J. Katz. I saw a theorem from Nasir al-Din al-Tusi. The way the theorem is written in the book is like this:

In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB:AC=sin(angle ACB):sin (angle ABC). [Note that since we are considering a ratio it is irrelevant whether we use Sines or sines.)

This theorem is about the law of sines. My question is about the last sentence in parenthesis.

What is the difference between Sine and sine?

答:
Six pages earlier, at the beginning of 9.6.1, the first subsection on Islamic trigonometry, a parenthetical remark notes:

The Islamic sine of an arc, like that of the Hindus, was the length of a particular line in a circle of given radius $R$. We will keep to our notation of "Sine" to designate the Islamic sine function

Additionally, in 8.7.1 on Indian trigonometry, the book explains:

In what follows, we generally use the word "Sine" (with a capital S) to represent the length of the Indian half-chord, given that the half-chord is a line in a circle of radius $R$, where $R$ will always be stated. We reserve the word "sine" (with a small s) for the modern function (or, equivalently, when the radius of the circle is 1). Thus, $\mathrm{Sin}\,\theta=R\sin\theta$.

I have not seen this convention outside of this book. As noted in the comments, in modern mathematics, some might use a capital letter to denote a restriction of  the domain of the sine function.


答:
You can apply l'Hôpital's rule twice:
$$ \lim_{x\rightarrow 1^+}\frac{1}{\ln x}-\frac{1}{x-1} = \lim_{x\rightarrow 1^+}\frac{x-1-\ln x}{\ln x(x-1)} = \lim_{x\rightarrow 1^+}\frac{1-1/x}{1/x(x-1)+\ln x} = \lim_{x\rightarrow 1^+}\frac{x-1}{x-1+x\ln x} = \lim_{x\rightarrow 1^+}\frac{1}{1+\ln x +1} = \frac{1}{2}.$$</p>


问:
Formula to calculate how much should be added to the x and y to move to a new point (On a rotated square)

Hey!

(please see the diagram before reading the question)

i have this square ABCD which is 10x10cm and it is rotated 45 degrees clockwise

i need to move it 25 cm (so to square EHGF) over the red lines lines (AF and CG)

but i cant slide it over those lines, instead i need to move it an X amount to the left and an Y amount up to get to the square EHGF

is there a formula to calculate how much X and Y are in this case?

i am looking for a formula that always works no matter how much the square is rotated

All help is very appreciated!

答:
There are various ways to do this.

If you know the angle of rotation, you can use trigonometry to calculate the $x$ and $y$ components of the desired motion.
In this example, before rotation you would have started with $B$ directly above $A$
(at the same $x$ coordinate), and you then rotated $45$ degrees clockwise to get to the position shown.

In general, suppose you rotate by an angle $\theta$ clockwise,
and then you want to move by a distance $d$ in the rotated direction.
(In the figure it appears that $d$ is $15$. You say $25$ in the question.
Using a symbol $d$ for the distance means you can make the distance anything
you want; just make sure you set $d$ to the desired distance from $A$ to $E$.)
Then the amount to add to $y$ is
$$ d \cos(\theta) $$
(assuming positive $y$ is upward in the figure,
the direction from $C$ to $B$, as implied in the question)
and the amount to add to $x$ is
$$ -d \sin(\theta) $$
(assuming positive $y$ is leftward in the figure, the direction from $D$ to $A$, as implied in the question).
If adding a positive number to $x$ will move you to the right (as in the usual Cartesian coordinates on a graph) then the $x$ formula is reversed and becomes $d \sin(\theta).$

You can verify the formulas with a few test values, such as zero and a right angle. Just be careful to use the correct angle measurements when passing arguments to the sine and cosine functions; most computer math libraries expect these arguments to be measured in radians rather than degrees.

If you already have the coordinates of all the vertices of the square at the time you have to compute the $x$ and $y$ amounts, however, you can avoid the need to use trigonometric functions.

Suppose the $(x,y)$ coordinates of $A$ are $(a_x, a_y)$ and
the $(x,y)$ coordinates of $B$ are $(b_x, b_y)$.
Let $s$ be the length of the side of the square;
you say in the question that $s = 10,$ but if you don't know the size of the square it is easy to compute it from the coordinates:
$$ s = \sqrt{(b_x - a_x)^2 + (b_y - a_y)^2}. $$

Now the amount to add to the $x$ coordinate is just
$$ \frac ds(b_x - a_x) $$
and the amount to add to the $y$ coordinate is just
$$ \frac ds(b_y - a_y). $$

That is, you know that adding $b_x - a_x$ to the $x$ coordinates and $b_y - a_y$ to the $y$ coordinates will move you $s$ units in the desired direction;
to move $d$ units in that direction instead of $s$ units, you just have to move $d/s$ times as far.

By the way, if the side of the square is really $10$ cm and the points that appear to be at the intersections of grid lines really are at intersections of grid lines, then your grid lines are approximately
$4.714$ cm apart.
When an actual $10$-cm square is rotated $45$ degrees on a $5$-cm grid,
keeping one vertex on an intersection of grid lines,
the diagonally opposite vertex is visibly below the third grid line above the fixed vertex.


答:
Edit: it appears that the notation used in the book does not correspond to how I have seen $\sin$ and $\mathrm{Sin}$ being used elsewhere. However, I will keep this answer for the benefit of other readers.

$\sin$ is a function that (generally speaking) maps real numbers to real numbers. Here is how it is defined on Wolfram Mathworld:

Let $\theta$ be an angle measured counterclockwise from the $x$-axis along an arc of the unit circle. Then $\sin(\theta)$ is the vertical coordinate of the arc endpoint.

There are other equivalent definitions of sine, but however you choose to define it, it is not a one-to-one function:

Notice that $\sin(0)$ equals $0$, but so does $\sin(\pi)$, $\sin(2\pi)$, etc. (I assume you are familiar with radians.) This means that the $\sin$ function does not have an inverse. To get around this, we often restrict the domain of $\sin(\theta)$ by requiring that $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$:

Notice that along this restricted domain, there are no two values of $\theta$ that produce the same output. This means that the inverse exists. However, because a function is defined not just by its values but by its domain, when we restrict the domain of $\sin$ it becomes a new function that we often denote $\DeclareMathOperator{\Sin}{Sin} \Sin$. The inverse of $\Sin$ is called $\Sin^{-1}$. You will often see $\Sin^{-1}$ being written simply as $\sin^{-1}$, but this is really an abuse of notation. The function $\sin$ does not have an inverse—only $\Sin$ does. A more popular alternative to $\Sin^{-1}$ is $\arcsin$. This notation has the advantage of being more 'correct' than $\sin^{-1}$, while also avoiding the $\sin$/$\Sin$ confusion.

Title: sine vs Sine: understanding the differences

Body:
I was using the textbook A History in Mathematics by Victor J. Katz. I saw a theorem from Nasir al-Din al-Tusi. The way the theorem is written in the book is like this:

In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB:AC=sin(angle ACB):sin (angle ABC). [Note that since we are considering a ratio it is irrelevant whether we use Sines or sines.)

This theorem is about the law of sines. My question is about the last sentence in parenthesis.

What is the difference between Sine and sine?


问:
How to prove $\det(tI-BA^T)=\det(tI-B^TA)t^{m-n}$?

If $A,B\in M_{m \times n}$. Show that $\det(tI-BA^T)=\det(tI-B^TA)t^{m-n}$

I want to solve this question and I have an idea probably useful which is from Sylvester's determinant identity. If I can find some matrices such that $$P^{-1}\begin{bmatrix}I&\\&\ BA^T\end{bmatrix}P=\begin{bmatrix}I&\\&\ A^TB\end{bmatrix}$$, then two matrices are similar and have the same characteristic polynomial. Hope someone can give me a hint how to construct these four matrices.

答:
Hint: Use these two facts:

$1.$ $\det(tI - BA^T) = \det(tI - AB^T)$ since for any $C$, $\det(C) = \det(C^T)$

$2.$ The algebraic multiplicity of the non-zero eigenvalues of $AB^T$ and $B^TA$ are the same (see here.)


答:
$$ L = (\ln3)\cdot L $$

Dividing both sides by $L$ (if $L\ne0$) yields $1 = \ln 3,$ which is false.

So the only way the first line above can be true is if $L=0.$

Title: Repeated application of L'Hopital's rule to evaluate $\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$

Body:
I came across an interesting limit while trying to draw the graph of the function $f(x)=\frac{3^{\ln x}}{x}$.

What is $$L=\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$

Since $\ln(0^+)\to -\infty$ and $3^{-\infty}=0$ we have the limit in indeterminate form of $\frac{0}{0}$.
So By L'Hopital's rule we get
$$L=\lim_{ x \to 0^+}\frac{3^{\ln x}\ln 3 \times  \frac{1}{x}}{1}=\ln 3\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$
So how to find this limit?


答:
$\dfrac{3^{\log x}}{e^{\log x}} = (3/e)^{\log x};$

Take the limit $x\rightarrow 0^+.$


答:
To simplify notation, I'll assume, without loss of generality, that the center of the ball is at the origin.
Suppose, toward a contradiction, that you have a counterexample in some high dimension. Say your set contains $p$ and $q$ but not some point $tp+(1-t)q$ on the line segment joining $p$ to $q$. Then this is also a counterexample in the $2$-dimensional subspace spanned by (the vectors from the origin to) $p$ and $q$. So, once you know the result for $2$ dimensions, it follows for all higher dimensions.

Title: Is a broken (Euclidean) ball still convex?

Body:
Consider the Euclidean ball $B^n(x,r)$ in $\mathbb{R}^n$ given by:
$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_2 \leq r\}$$ with centre $x\in\mathbb{R}^n$ and radius $r\geq 0$.
Now we break this ball by removing an arbitrary subset from the boundary of $B^n(x,r)$. Is the resulting set still convex?

My intuition says Yes, and it's easy to visualise this in $\mathbb{R}^2$ or $\mathbb{R}^3$. I'm wondering how I can prove it generally, i.e. for $\mathbb{R}^n$ by considering an arbitrary subset $S\subseteq B^n(x,r)$? If $S = B^n(x,r)$ then the resulting set is empty, which is convex - so we can ignore that situation. We assume $S\subset B^n(x,r)$. What's next?

I consider two points $y_1,y_2 \in B^n(x,r)\backslash S$, and want to show that for $t\in [0,1]$ we have $ty_1 + (1-t)y_2 \in B^n(x,r)\backslash S$. How do I take it from here? I'm stuck.

Thanks!

Addendum:

What if we consider the $p$-norm in general, instead of the Euclidean norm? How do things change with $p$ if the ball is defined as follows?
$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_p \leq r\}$$
It sounds fun breaking balls in higher dimensions under different norms!


问:
Frobenius coin problem induction help
I need some help with the actual Induction proof of a Frobenius coin problem. This is the exact problem:
The government of Elbonia has decided to issue currency only in 5 and 9 cent denominations. Prove that there is largest value that Elbonians cannot pay with this denomination

And a later question says to prove that all values over this found value is payable with 5 and 9 cent coins.

First of all, I've found the largest value that can't be paid is 31 cents. I've found this by just writing out the combinations of coins until I got 5 in a row, and then each of those numbers can just have 5 added to them to continue forever, starting at 32. However I'm not sure if that could be considered "proof" and if I need to show this in a more official way.

I've started on the induction proof anyway, but I'm having some trouble as where to go. I know to prove this I need to show that for S(n): where n is the amount payable with 5 or 9 cent pieces, show that
S(n) -> S(n+5),
S(n+1) -> S(n+6),
S(n+2) -> S(n+7),
S(n+3) -> S(n+8),
S(n+4)-> S(n+9),

So for my base case I've let n=32, and shown that
32 = 3(9) + 3(5),
33= 2(9) + 3(5),
34 = 1(9) + 5(5),
35 = 7(5),
36 = 4(9)

So I've shown my base cases can be paid with 5 and 9 cent pieces, but now I'm stuck. What exactly do I assume for my inductive assumption? That S(k), s(k+1).. etc is true for some kEZ? Normally when we are given induction questions for the inductive step there is a way to rearrange it to make your assumption show up somewhere to help you prove it, but I can't see how to do that for the inductive step here.

Any help on this would be awesome, sorry for the long question! Thanks!

答:
Let $a$ and $b$ both be greater than or equal to $2$ and coprime.

Let

$\quad S = \{n \in \Bbb N \mid (\forall \, x,y \in \Bbb N)\; n \ne xa + yb \}$

Clearly $S$ is a nonempty set and it is the complement of integers with a Frobenius coin solution.

The OP is seeking a proof that $S$ is a finite set and we'll address that now.

It is easy to see that if Frobenius solutions can be found for an interval segment of integers of length, say $a$, starting at $s$ then

$\quad S \subset \complement \, [s,+\infty)$

and the argument that $S$ is finite would be complete.

Lemma: Let $s = (a-1)\cdot b$. Then Frobenius solutions exist on the interval $[s,s+a-1]$.

Proof Sketch

If $0 \le k \lt a-1$, consider the expression

$\quad k + (z)b   \quad \text{where } z = a - 1$

and, in order to 're-balance the accounting books', set $y = z - u$ so that

$\quad k + (z)b = k + bu + (y)b$

and consider

$\tag 1 k + bu \equiv 0 \pmod{a} \; \text{ iff } \; bu \equiv -k \pmod{a}$

Recall that there is one and only one solution of $\text{(1)}$ for $[u]$.

So it must be possible to write

$\quad s + k = (x)a+ (y)b, \text{ with } x,y \in \Bbb N$

In fact, there is exactly one such representation for $s+k$ (see also this). $\quad \blacksquare$

So the Frobenius number is indeed well-defined and is equal to the largest number in the set $S$.

Applying the above theory to the OP's specific problem we'll find the exact solutions on the interval

$\quad [(5-1)\cdot9,(5-1)\cdot9 + 5-1] = [36, 40]$

Due to the bijective correspondence, it is simplest to 'borrow and fit':

$\quad \text{Borrow } 0: \quad \quad\quad\quad\quad\quad\quad\quad\quad\;\, 36 = (0)5 + (4)9$

$\quad \text{Borrow } 1: \quad 36 = 9 + (3)9 \;\;\;\, \text{ iff } \; 37 = (2)5 + (3)9$

$\quad \text{Borrow } 2: \quad 36 = 18 + (2)9 \; \ \text{ iff } \; 38 = (4)5 + (2)9$

$\quad \text{Borrow } 3: \quad 36 = 27 + (1)9 \;\, \text{ iff } \; 39 = (6)5 + (1)9$

$\quad \text{Borrow } 4: \quad 36 = 36 + (0)9 \;\, \text{ iff } \; 40 = (8)5 + (0)9$

Title: Frobenius coin problem induction help


问:
Showing that $(\frac{\sin3A}{\sin A})^2 - (\frac{\cos3A}{\cos A})^2 = 8\cos2A$

Show that $$\left(\frac{\sin3A}{\sin A}\right)^2-\left(\frac{\cos3A}{cosA}\right)^2=8\cos2A$$

Solution

$$\begin{align}
\left(\frac{\sin3A}{\sin A}\right)^2-\left(\frac{\cos3A}{\cos A}\right)^2
&=\left(\frac{\sin3A}{\sin A}+\frac{\cos3A}{\cos A}\right)\left(\frac{\sin3A}{\sin A}-\frac{\cos3A}{\cos A}\right) \\[4pt]
&=\left(\frac{\sin3A\cos A+\sin A\cos3A}{\sin A\cos A}\right)\left(\frac{\sin3A\cos A-\sin A\cos3A}{\sin A\cos A}\right) \\[4pt]
&=\left(\frac{\sin(3A+A)}{\sin A\cos A}\right)\left(\frac{\sin(3A-A)}{\sin A\cos A}\right) \\[4pt]
&=\left(\frac{\sin4A}{\sin A\cos A}\right)\left(\frac{\sin2A}{\sin A\cos A}\right)
\end{align}$$

What do I do next?

The textbook from which the above question is taken hasn't introduced the double angle formulas yet. So I am trying to find a way to do this question without using the double angle formulas.

答: Since $\sin2x=2\sin x\cos x$,$$\frac{\sin4A}{\sin A\cos A}\frac{\sin 2A}{\sin A\cos A}=\frac{2\sin 4A}{\sin 2A}2=4\cos2A\cdot2=8\cos2A.$$


问:
Confusion regarding Ends
I am trying to understand the definition of Ends and CoEnds.

I was looking at some examples where I found the following case because of which I have become more confused.

So, here is the example:

Let $G \in [\textbf{C}^{op} \to \textbf{Set}]$, i.e. Let G be a presheaf.
Now, the example talks about: $\int_{B}GB$.

This does not make sense to me. From what I know, Ends are defined on functors from $\textbf{C}^{op} \times \textbf{C}$ to another category D which in this case will be Set.  How can one even write $\int_{B}GB$?!

I have tried thinking about this for a while and I thought maybe we can apply Yoneda's Lemma on $GB$. So, we will be able to write: $GB \simeq \hat{\textbf{C}}(\textbf{C}(-, B), G)$. So, we will be able to write: $\int_{B}GB \simeq \int_{B}\hat{\textbf{C}}(\textbf{C}(-, B), G)$.

Now, is $\hat{\textbf{C}}(\textbf{C}(-, B), G)$ a functor from $\textbf{C}^{op} \times \textbf{C}$ to Set?

I think I am missing something very simple or basic here. Any help with this would be great.

Thank you.

答:
$\int_BGB$ stands for $\int_B\tilde G(B,B)$ where $\tilde G:{\bf C}^{op}\times {\bf C}\to{\bf Set}$ is defined as
$$\tilde G(A,B):=GA\ \text{ and }\ \tilde G(f,g):=Gf\,.$$
One readily sees that it's indeed a functor, e.g. by observing that $\tilde G=G\circ\pi_1$ where $\pi_1:{\bf C}^{op}\times {\bf C}\to{\bf C}^{op}$ is the projection on the first coordinate.

As commented by Fosco, one can prove that $\int_B\tilde G(B,B)$ is just the limit of $G$ in $\bf Set$.

Title: Confusion regarding Ends


问:
$C^1[0,1]$ is not banach using the closed graph theory
Show that $C^1[0,1]$ is not a banach space using the closed graph theory with the maximum norm.
First, look at the derivative operator:
$D:C^1[0,1]\to C[0,1]$, $D(f)=f'$.

We can check that $D$ is linear and not bounded (by taking an example such as a polynomial $x^{n+1}$).
Thus $D$ is not continuos.
I'm not sure, if it is possible to show that $D$ has a closed graph (a linear map $T:X\to Y$ has a closed graph if $x_n\subset X$ such that $x_n\to x$ and $T_{x_n}\to y$ then $Tx=y$).So if by contradiction, we assume that $C^1[0,1]$ is banach with the sup norm, then get by the closed graph theory  that $D$ is continuos, which is not true according to what we've said.

答: $D$ has a closed graph: Let $u_n \to  u$ and $D(u_n) = u_n' \to v$ uniformly. We want to show that $v=D(u)=u'$. To this end, notice that  for all $t \in [0,1], \ \int_0^t u_n'(s) ds \to \int_0^t v(s) ds$ and thus $ u_n(t) - u_n(0) \to \int_0^t v(s) ds$. In conclusion, $u(t) =u(0) + \int_0^t v(s) ds$  and so $ u \in C^1[0,1]$ and $u'=v$.