问:
How can something true follow from a false proposition?
Trying to wrap my head around conditional statements/implication and the respective truthtable in propositional logic. Read a number of the related posts on here. I understand that there is no causal relationship between $A$ and $B$, so from a false proposition anything can follow.

This made sense to me especially with the "promise" analogy: $A \Rightarrow B$ is a promise, that can only be broken when after a true condition $A$, $B$ is false (second row of the truth table). If the condition isn't met then the promise can't possibly be broken, no matter $B$. With most real life examples this makes perfect sense to me: "If you write an A in an exam, then you get a Dollar", "if you finish supper, then you get dessert" etc.

However one particular math example confuses it again for me:

$A:$ $x$ is an even number

$B:$ $x$ is divisible by two

How can $A \Rightarrow B$ be true when $A$ is false? An odd number is never going to be divisible by two. It's like saying an odd number is even. What am I missing here? What's my misconception? Am I understanding it generally wrong?

Also on a related note, what's the proper terminology for the "if-part" and the "then-part" of a conditional statement?

答: Your particular example confuses you because it's an "if and only if" statement. In fact, it's $x \text{ is an even number}  \Leftrightarrow x \text{ is divisible by two}$, meaning that if $x$ it's not even, it's not going to be divisible by $2$; if $x$ was odd and divisible by $2$, due to the other implication, it'd be even, and that's a contradiction. This is different from, let's say, "if you finish supper, then you get dessert", because this is $\text{you finish supper} \Rightarrow \text{ you get dessert}$, and in fact you may still get dessert even if you haven't finished your supper.


问:
$P_{n}(A)\xrightarrow{n \to \infty} P(A)$ for any $P$-continuity set $A$ iff $F_{n}(x)\xrightarrow{ n \to \infty} F(x)$ for all continuity points
For a while, I have been strunggling to find why Levy's continuity Theorem would imply that:

$P_{n} \xrightarrow{\text{distribution}} P\iff \phi_{n}(t)\xrightarrow{n \to \infty}\phi(t)\;  $

since Levy's Continuity Theorem is concerned with weak convergence rather than convergence in distribution. I thus attempted to find an equivalence between convergence in distribution and weak convergence. This leads me to the Portmanteau Lemma which shows many equivalent notions of weak convergence. So I am to attempt the following equivalence in the case of the measure being on $\mathbb R$:

$$ P_{n}\xrightarrow{\text{weakly}}P \iff P_{n}\xrightarrow{\text{distribution}}P$$

For "$\Rightarrow$" consider the equivalent notion of "weak convergence" from the Portmanteau Lemma, namely:

$P_{n}(A)\xrightarrow{n \to \infty} P(A)$ for any $P$-continuity set $A$.

Now let $x$ be a continuity point of $F$ in $\mathbb R$, then the set $(-\infty,x]$ is a continuity set under $P$ and hence: $F_{n}(x)=P_{n}((-\infty,x])\xrightarrow{n \to \infty} P((-\infty,x])=F(x)$ thus $F_{n}\xrightarrow{ \text{distribution}} F$.

I do not know what equivalent definition of weak convergence to prove for "$\Leftarrow$" direction. Continuity sets could take vastly different forms to $(-\infty,x]$ , thus I do not think I could use it. Any help would be greatly appreciated.

答:
I believe from your question that you take as definition of convergence in distribution that $\lim_n F_n(x) = F(x)$ where $x$ is a continuity point of $F$. Hence, I will show that this condition implies
$$\liminf_n P_n(G) \geq P(G)$$
for all non-empty open subsets $G$ (you already proved the converse). The latter condition is equivalent with weak convergence, by the Portmanteau lemma.

Fix such a non-empty open subset $G$. For any $x \in G$, the fact that $G$ is open allows us to select $\epsilon_x > 0$ with $]x-2\epsilon_x, x + 2\epsilon_2[\subseteq G$. Then $G= \bigcup_{x \in G} ]x-\epsilon_x, x + \epsilon_x[$. By a theorem of Lindelöf (every open cover has a countable subcover), we can select a sequence $(x_n)_n\subseteq G$ with $G = \bigcup_{n=1}^\infty ]x-\epsilon_{x_n}, x + \epsilon_{x_n}[$ and by our construction also $$]x_n-2\epsilon_{x_n}, x_n + 2 \epsilon_{x_n}[\subseteq G.$$
Since $F$ is monotone, the set $D(F)$ of points of discontinuity is at most countable. Hence, for any $n \geq 1$, we can find $a_n, b_n$ continuity points with
$$x_n - 2 \epsilon_{x_n} < a_n < x_n- \epsilon_{x_n}< x_n + \epsilon_{x_n}  < b_n < x_n + 2 \epsilon_{x_n}.$$
It follows that $G = \bigcup_{n=1}^\infty]a_n, b_n]$ where $a_n < b_n$ are continuity points for all $n \geq 1$. Put $$\mathcal{P}:= \{]a,b]: a < b, a,b \notin D(F)\}\cup\{\emptyset\}.$$
Then $\mathcal{P}$ is closed under finite intersections. If $C = ]a,b] \in \mathcal{P}$, then
\begin{align}\lim_n P_n(C) &= \lim_n \{P_n(]-\infty, b])- P_n(]-\infty, a])\}\\
&=\lim_n {F_n(b)- F_n(a)}\\ &= F(b)-F(a) = P(C).\end{align}
If $C_1, \dots, C_m \in \mathcal{P}$, then by inclusion-exclusion principle also
$$\lim_n P_n\left(\bigcup_{i=1}^m C_i\right) = P \left(\bigcup_{i=1}^m C_i\right).$$
Hence, for all $m \geq 1$,
$$\liminf_n P_n(G) \geq \liminf_n P_n\left(\bigcup_{i=1}^m ]a_i, b_i]\right) = P\left(\bigcup_{i=1}^m ]a_i, b_i]\right).$$
But $\bigcup_{i=1}^m ]a_i, b_i[ \nearrow G$, so by continuity of measure we obtain
$$\liminf_n P_n(G) \geq P(G).$$

Feel free to ask for clarification if something is not clear to you!

ContentLicense: CC BY-SA 4.0


问:
Asymptotic formula for roots of a polynomial-like function
Let $\sigma \in (0, 1)$ and $n \geq1$ be given.

How can one rigorously show that the positive root $x_0 \in (0, n)$ of

\begin{equation}
f(x) = (n^2-x^2-\sigma n^2 x)-(n^2-x^2+\sigma n^2x)e^{-4x}
\end{equation}
is of the form
\begin{equation}
x_0 = x^*-\varepsilon_n
\end{equation}
for some $\varepsilon_n$ such that $\varepsilon_n \rightarrow 0$ as $n\to+\infty$, and
where $$x^* := \frac12(n\sqrt{n^2\sigma^2+4}-n^2\sigma)$$ is the positive root of $f_1(x) = (n^2-x^2-\sigma n^2x)$?

This question is related and complementary to  Roots of polynomial/exponential function.

Body:
Let
$$g(x) = (n^2-x^2-\sigma n^2 x)-(n^2-x^2+\sigma n^2x)e^{-\frac{4}{\sigma}}.$$
Since $0 < x_0 < x^\ast = \frac{4n^2}{2(n\sqrt{n^2\sigma^2+4}+n^2\sigma)} < \frac{1}{\sigma}$,
we have $g(x_0) \ge f(x_0)$ that is
$$-(1 - e^{-\frac{4}{\sigma}} )x_0^2 - (1 + e^{-\frac{4}{\sigma}})\sigma n^2 x_0 + n^2(1 - e^{-\frac{4}{\sigma}}) \ge 0$$
which results in
$$x_0 \le -\frac{(1 + e^{-\frac{4}{\sigma}})\sigma n^2}{2(1 - e^{-\frac{4}{\sigma}})}
+ \frac{1}{2}\sqrt{\frac{(1 + e^{-\frac{4}{\sigma}})^2\sigma^2n^4}{(1 - e^{-\frac{4}{\sigma}})^2} + 4n^2}
=: x_u.$$

We have
$$\lim_{n\to \infty} (x^\ast - x_u) = \frac{2}{(1 + e^{\frac{4}{\sigma}})\sigma}.$$
Thus, $x^\ast - x_0$ does not converge to zero as $n\to\infty$.


问:
If $p$ is an odd prime then prove that there cannot exist a finite group $G$ such that ${\rm Aut}(G)\cong \mathbb{Z}_p$.

If $p$ is an odd prime then prove that there cannot exist a finite group $G$ such that ${\rm Aut}(G)\cong \mathbb{Z}_p$.

Can anyone tell me how to proceed in this question?

Here is my attempt:

If ${\rm Aut}(G)$ is cyclic then ${\rm Inn}(G)$ is also cyclic which implies $G/Z(G)$ is cyclic which implies $G$ is abelian.
If $G$ is abelian then for $\phi \in{\rm Aut}(G)$ such that $\phi(g)=g^{-1}$. then $\phi(\phi(g)) = \phi(g^{-1}) = g$. Hence $\phi$ is of order $2.$

Hence the order of ${\rm Aut}(G)$ must be a multiple of 2 and hence cannot have prime order.

Edit:- I cannot seem to proceed when $\phi$ becomes just the identity mapping. In that case every element of the group has order $2$ (except for the identity element of order $1$) .

答: Hint: yes $G$ is abelian and consider the map $\phi$: $G \rightarrow G$, defined by $\phi(g)=g^{-1}$. Prove that this is an isomorphism. What is the order of $\phi$?


问:
Example of the topology which has a dense closed point and has non-empty proper closed subset which does not contain any closed point
is there topology $X$ satisfying the following conditions ?

the set of closed point $H$ is a dense subset of $X$

there is non-empty proper closed subset $Z \subset X$ which does not contain any closed point

[My attempt]

I believe the $X$ exists but I can't take any example..

First, by condition 2, since   $ H \subset X \setminus Z $ so we know that $X \setminus Z $ is a dense open subset of $X$

Second, by condition 2, $X$ is not $T_1$ space because any singleton set of $T_1$ space is a closed subset.

I checked the zariski topology, the Sierpinski space, the overlapping interval topology , etc (the examples in wikipedia) ..
but they are not solutions.

答:
Let $X=H\cup Z$ where $H$ is infinite, $|Z|\ge2$, and $H\cap Z=\emptyset$, with the topology $\tau=\{\emptyset\}\cup\{H\setminus F:F\subseteq H,\ F\text{ finite}\}\cup\{X\setminus F:F\subseteq H,\ F\text{ finite}\}$.

$H$ is a dense subset of $X$, and every finite subset of $H$ is closed; $Z$ is closed but no nonempty proper subset of $Z$ is closed.

P.S. Here is an example where $X$ is a $T_0$ space.

Let $X=H\cup Z$ where $H$ and $Z$ are disjoint infinite sets, and let $\lt$ be a total ordering of $Z$ with no greatest element. Let $\tau$ be the collection of all sets $U\subseteq X$ satisfying the conditions:

(1) if $U\ne\emptyset$ then $H\setminus U$ is finite;

(2) if $a,b\in Z$, $a\lt b$, and $b\in U$, then $a\in U$.

Then $\tau$ is a $T_0$ topology on $X$; a singleton set $\{x\}$ is closed in $X$ (a "closed point") if and only if $x\in H$; the set $H$ of all closed points is an open dense subset of $X$; and $Z$ is a nonempty closed set containing no closed points.


问:
If $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert<\infty$, then $\{a_n\}_{n=1}^\infty$ has a convergent subsequence
Problem: Let $k\in\mathbb N$ be fixed and suppose that the series $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert$ converges. Prove that $\{a_n\}_{n=1}^\infty$ has a convergent subsequence.

My Thoughts: Using the triangle inequality we see that for any fixed $n\in\mathbb N$ with $n>k+1$ we have
\begin{align*}
\sum_{j=k+1}^{n}\vert a_{j+k}\vert-\sum_{j=1}^k\vert a_j\vert
&\leq \sum_{j=1}^n\vert a_{j+k}-a_j\vert\\
&\leq \sum_{j=1}^\infty\vert a_{j+k}-a_j\vert\\&\leq M,
\end{align*}
for some $M>0.$ Therefore we have the uniform bound
$$\sum_{j=k+1}^{n}\vert a_{j+k}\vert\leq M+\sum_{j=1}^k\vert a_j\vert,$$
and hence the series on the left-hand side converges. This implies that
$$\lim_{n\to\infty}a_{n+k}=0,$$
and we have our convergent subsequence.

Do you agree with the proof presented above?

Thank you for your time and feedback.

答:
Consider the sequence $(b_n) = (a_{kn})$.
From your hypothesis one can deduce that this sequence is a Cauchy sequence,
so that it must be convergent.

Let $\epsilon > 0$. Since $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert$ converges, there is a $N$ such that $\sum_{n=N}^{\infty} \vert a_{n+k}-a_n\vert < \epsilon.$
Since $k\geq 1,$ we also have $\sum_{n=kN}^{\infty} \vert a_{n+k}-a_n\vert < \epsilon.$

Now let $m_2>m_1>N.$ Then

$$
\vert b_{m_2} - b_{m_1} \vert = \vert a_{km_2} - a_{km_1} \vert
= \vert (a_{km_2} - a_{k(m_2-1)}) + \ldots + (a_{k(m_1+1)} - a_{km_1})\vert
\leq {\vert a_{km_2} - a_{k(m_2-1)} \vert + \ldots + \vert a_{k(m_1+1)} - a_{km_1}\vert}
\leq \sum_{n=kN}^{\infty} \vert a_{n+k}-a_n\vert < \epsilon.
$$

$(b_n)$ is therefore a Cauchy sequence and hence convergent.


答: Let $M= \sum\limits_{n=1}^{\infty} |a_{n+k}-a_n|$. Check that $ \sum\limits_{n=1}^{\infty} |a_{k(n+1)}-a_{kn}| \leq M$. The sequence $b_n=a_{k n}$ satisfies $\sum |b_{n+1}-b_n| <\infty$. This implies that $(b_n)$ is Cauchy.

Title: If $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert<\infty$, then $\{a_n\}_{n=1}^\infty$ has a convergent subsequence

Body:
Problem: Let $k\in\mathbb N$ be fixed and suppose that the series $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert$ converges. Prove that $\{a_n\}_{n=1}^\infty$ has a convergent subsequence.

My Thoughts: Using the triangle inequality we see that for any fixed $n\in\mathbb N$ with $n>k+1$ we have
\begin{align*}
\sum_{j=k+1}^{n}\vert a_{j+k}\vert-\sum_{j=1}^k\vert a_j\vert
&\leq \sum_{j=1}^n\vert a_{j+k}-a_j\vert\\
&\leq \sum_{j=1}^\infty\vert a_{j+k}-a_j\vert\\&\leq M,
\end{align*}
for some $M>0.$ Therefore we have the uniform bound
$$\sum_{j=k+1}^{n}\vert a_{j+k}\vert\leq M+\sum_{j=1}^k\vert a_j\vert,$$
and hence the series on the left-hand side converges. This implies that
$$\lim_{n\to\infty}a_{n+k}=0,$$
and we have our convergent subsequence.

Do you agree with the proof presented above?

Thank you for your time and feedback.


问:
Behaviour of two distinct, arbitrary points of the Thue-Morse Subshift
Let $x^+$ denote the infinite Thue-Morse word, $$x^+ = 0110100110010110\ldots,$$ which is defined as being the only fixed point starting with $0$ of the morphism $S$ (defined over the words on the alphabet $\{0,1\}$) given by $S(0) = 01$ and $S(1) = 10$ - that is, $$x^+ = S^{\omega}(0) = \lim\limits_{n \geqslant 1} S^n(0);$$ this is only one among many different, equivalent definitions of such word (for instance, pick $A_0 = 0$ and let $A_{n+1} = A_n\,^\frown\overline{A_{n}}$ - where $^\frown$ denotes concatenation and $\overline{A_n}$ denotes the bitwise complement of $A_n$. Then $x^+ = \lim\limits_{n \in \mathbb{N}} A_n$). From now on, let $x^+ = (t_n)_{n \in \mathbb{N}}$.

It is known that the Thue-Morse word is uniformly recorrent (meaning, for every finite subword $w$ of $x^+$ there is some $n \in \mathbb{N}$ such that, for every $i \in \mathbb{N}$, the block $t_{i + 1}t_{i + 2}\ldots t_{i + n}$ contains some ocurrence of $w$) but it is $\mathbf{\textrm{not}}$ ultimately periodic (meaning, it is not true that there are $p \geqslant 1$ and $N \geqslant 0$ such that $t_i = t_{i + p}$ for all $i \geqslant N$).

The two-sided Thue Morse sequence $x = (x_i)_i \in \mathbb{Z}$ is defined as $x = x^{-}\,^\frown x^+$, meaning that $x_n = t_n$ for all $n \geqslant 0$ and $x_{-n} = t_{n - 1}$ for all $n \geqslant 1$.  So, $$x = \ldots 100101100.1101001 \ldots$$

Let $\sigma$ denote (as usual) the left shift map on $^\mathbb{Z}\{0,1\}$. The Thue-Morse subshift $X$ is the closure (in the Tychonoff topology) of the orbit $\{\sigma^n(x): n \geqslant 0\}$. Equivalently, $X$ is the family of all doubly infinite sequences $z$ such that every finite subword of $z$ is a subword of $x^+$. It is known that $X$ is a infinite, minimal subshift (that is, every point of $X$ has a dense orbit in $X$).

My question is:

Let $s = (s_i)_{i \in \mathbb{Z}}\,$ and $u = (u_i)_{i \in \mathbb{Z}}\,$ be two distinct points of the Thue-Morse subshift $X$ - that is, there is some integer $i \in \mathbb{Z}$ such that $s_i \neq u_i$. Consider  the set of integers given by $$\{i \in \mathbb{Z}: s_i \neq u_i \}.$$ Is it true that such set is, necessarily, an $\mathbf{\textrm{infinite}}$ set of integers ?

My conjecture is that the answer is "yes", but I wasn't able to produce a proof.

答:
Bonjour, it is first easy to observe that $S^n (0)$ is obtained from $S^n (1)$ changing the 0's in 1's and vice-versa. Suppose $E=\{ i\in \mathbb{Z}| s_i\not = u_i \}$ is finite and $u$ and $s$ distincts. Let say that $E$ is included in an interval of length $L$.
Let $N$ such that $2^N > L$.
By the recognizability result of Mossé for primitive substitutions, and shifting both sequences of the same amount if needed, there are two points $s'$ and $u'$ of the Thue-Morse shift such that $s=S^N(s')$ and $u=S^N(u')$ (this is not direct but I left it as an exercise). Then there exists $i$ such that for all $j\not \in \{ i, i+1\}$, $S^N(u'_j)=S^N (s'_j)$ and $S^N(u'_iu'_{i+1})\not =S^N (s'_is'_{i+1})$. Thus either $S^N(u'_i)\not =S^N (s'_i)$ or $S^N (u'_{i+1})\not =S^N (s'_{i+1})$. But in both case, from the first observation, it implies that the number of differences is at least $2^N$. This contradicts the facts that $L$ is strictly less than $2^N$.

ContentLicense: CC BY-SA 3.0


问:
What are the factors of the exterior product of this exterior differential algebra?
I have questions about [Risch 1979] (see the citation and reference below).

1.) What are the factors of the exterior product of the exterior differential algebra in the citation below?

I already know the terms exterior algebra and differential algebra.

2.) Is $\Omega^i(\mathfrak{D})$ the set of the $i$-forms on $\mathfrak{D}$?

"First some notation. We will work with fields $\mathfrak{D}$, $\mathfrak{E}$,... that are always
finitely generated over some algebraically closed field $\textit{K}$ of characteristic $0$ (...). If $x_1,...,x_t$, is a transcendence basis of $\mathfrak{D}/\textit{K}$, then all $\textit{K}$ derivations of $\mathfrak{D}$ are of the form $\sum_{i=1}^t\ g_i\partial/\partial x_i$, $g_i\in \mathfrak{D}$. $\Omega(\mathfrak{D})$ = (the exterior differential algebra formed from $\mathfrak{D}$) = $\sum_0^t\Omega^i(\mathfrak{D})$, with $\Omega^0(\mathfrak{D})=\mathfrak{D}$, $\Omega^1(\mathfrak{D})=\{\sum_{i=1}^tg_idx_i\colon g_i\in\mathfrak{D}, dx_i(\partial/\partial x_j)=\delta_j^i\}$ = dual space to the derivations of $\mathfrak{D}/\textit{K}$."

[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759

答:
Disclaimer: My exposure to these notions is rather small. Take everything with a grain of salt.

Short answer: Most likely, what's dubbed an exterior differential algebra here is actually the exterior algebra over the so called “Module of Kähler differentials of $\mathfrak D$ over $K$”, written $\Omega_{\mathfrak D /K}$, which in turn should have the structure of a differential (graded) algebra in a canonical way.

Long answer:
It appears that Eisenbud[1] provides the necessary basics here:

The context is a commutative ring $R$ and an $R$-Algebra $S$.
Given these and an $S$-Module $M$ (in our case, $R=K$ and $S=M=\mathfrak D$), we can define $\operatorname{Der}_R(S, M)$ to be all the $R$-linear maps $S\to M$ satisfying the Leibniz rule.
This has evidently the structure of an $S$-module by the usual pointwise defined action.

As it turns out, there is a universal object describing these:
Every such derivation corresponds uniquely to an $S$-linear map $\Omega_{S/R}\to M$, where $\Omega_{S/R}$ is the $S$-Module generated by symbols $\{d(f)\mid f\in S\}$ subject to the relations $d(fg) = fd(g) + gd(f)$ for all $f, g\in S$ and $d(\lambda f + g) f \lambda d(f) + d(g)$ for all $f, g\in S$ and $\lambda\in R$.
Even stronger, we have $$
    \operatorname{Der}_R(S, M)\simeq \operatorname{Hom}_S(\Omega_{S/R}, M),
$$ where the underlying isomorphism is $S$-linear and natural in $M$.

Now if we head to our case, setting $R = K, S = M = \mathfrak D$, the isomorphism gives us $$
    \operatorname{Der}_K(\mathfrak D) := \operatorname{Der}_K(\mathfrak D, \mathfrak D)\simeq \operatorname{Hom}_{\mathfrak D}(\Omega_{\mathfrak D/K}, \mathfrak D) =: \Omega_{\mathfrak D/K}{}^\ast,
$$ and assuming reflexivity, interpreting $\Omega(\mathfrak D)$ as the usual exterior algebra of $\mathfrak D$-modules $\bigwedge \Omega_{\mathfrak D/K}$ should work.

The reflexivity should hold due to the remark at the beginning, where $\operatorname{Der}_K(\mathfrak D)$ is identified as a free $\mathfrak D$-module over the  derivations $\partial/\partial x_i$, and free modules are obviously reflexive, since we can provide an explicit dual basis (here denoted by $dx_i$).

I don't know a proof of this off the top of my head, but I recommend looking at the simple case $K=\mathbb Q$ and one transcendental element. Intuitively, by the Leibniz rule, every differential should be given by its value at its transcendental element, and transcendental elements should be assignable to every value in the large field. But this is by no means formal.

This naked exterior algebra can now be equipped with a differential structure by means of the map $$
d^i (g d(f_1) \wedge d(f_2) \wedge \ldots \wedge d(f_i))
:= dg \wedge d(f_1) \wedge d(f_2) \wedge \ldots \wedge d(f_i),
$$ Turning it into a differential graded algebra.

[1] Chapter 16, “Modules of Differentials” in David Eisenbud - Commutative Algebra – with a View Toward Algebraic Geometry, 1995.


问:
Euclidean norm vs pythagorean theorem?
I recently stumbled upon an Euclidian norm. First I thought there are the powers and square root to deal with possible negative values (like in Standard deviation formula) but then I realized, the final number (sum of squares and square root of it) is not same as sum of absolute numbers.

So then I noticed it is more related to Pythagorean theorem (Euclidian distance).

But if looked in google for PT in 3d space i found the formula like

```<code>(a^2 + b^2 + c^2)^1/2
</code>```

http://www.math.brown.edu/tbanchof/Beyond3d/chapter8/section02.html

But in the Euclidian norm it is

```<code>SQRT(a^2 + b^2 + c^2)
</code>```

How comes, it is different?

Second question, how comes that the theorem works for any count of dimensions?

答:
The two formulas are indeed the same.

You can generalize it to $n$ dimensions by repeated application of Pythagoras:

$$\|(a,b,c)\|=\|(a,\|(b,c)\|)\|=\|(a,\sqrt{b^2+c^2})\|=\sqrt{a^2+(b^2+c^2)}$$ and so on.

(In this reasoning, you project $(a,b,c)$ to the plane $yz$, giving $(b,c)$ as an intermediate point, joining the origin to $(b,c)$, then $(b,c)$ to $(a,b,c)$.)

Title: Euclidean norm vs pythagorean theorem?


问:
Integral over null set is zero but integral of Dirac delta function is 1
We know integral of any function over a null set is zero.
But for Dirac delta function ($\delta=+\infty$ iff $x=0$ otherwise $\delta=0$)
$$
\int_{-\infty}^{+\infty}\delta =\int_0^0\delta =1.
$$
Is it a contradiction?</p>

答:
You can regard it as a measure, the Dirac measure, which gives mass $1$ to $\{0\}$ , and zero mass to other subsets of $\mathbb R$ which do not contain zero(let's say Borel sets). For a Borel set $A\subset \mathbb R$ define: </p>

<p>$$\mu (A)=\begin{cases}1,&0\in A\\0,&0\notin A\end{cases}$$
Therefore: $$\int _{-\infty }^\infty d\mu=\int_{\{0\}}1\cdot d\mu+\int_{\mathbb R\setminus\{0\}}d\mu=\mu(\{0\})+\mu(\mathbb R\setminus\{0\})=1+0=1$$</p>

Body:
You can regard it as a measure, the Dirac measure, which gives mass $1$ to $\{0\}$ , and zero mass to other subsets of $\mathbb R$ which do not contain zero(let's say Borel sets). For a Borel set $A\subset \mathbb R$ define:

$$\mu (A)=\begin{cases}1,&0\in A\\0,&0\notin A\end{cases}$$
Therefore: $$\int _{-\infty }^\infty d\mu=\int_{\{0\}}1\cdot d\mu+\int_{\mathbb R\setminus\{0\}}d\mu=\mu(\{0\})+\mu(\mathbb R\setminus\{0\})=1+0=1$$


问:
How to calculate this integral as following?
Can someone give me an idea, how to derive the integration?
$$\int_0^{\infty}e^{-e^{t}}t^{k}\mathrm{d}t~\text{for}~k\in \mathbb
N^{+}.$$
I found that the function may has a connection with the Gamma function or the Meijer G-function, but I can not know haow to solve this problem.

答:
$$I_k=\int_0^{\infty}e^{-e^{t}}\,t^{k}\,dt$$
As far as I know, there is no other representation of this integral beside the Meijer G hypergeometric function (except for $k=0$) provided $k$ to be an integer (if $k$ is not an integer, I do not know any formula).

The patterm is very simple
$$I_1=G_{2,3}^{3,0}\left(1\left|
\begin{array}{c}
 1,1 \\
 0,0,0
\end{array}
\right.\right)$$

$$I_2=2 \,G_{3,4}^{4,0}\left(1\left|
\begin{array}{c}
 1,1,1 \\
 0,0,0,0
\end{array}
\right.\right)$$

$$I_3=6 \,G_{4,5}^{5,0}\left(1\left|
\begin{array}{c}
 1,1,1,1 \\
 0,0,0,0,0
\end{array}
\right.\right)$$ that is to say
$$I_k=k! \,G_{k+1,k+2}^{k+2,0}\left(1\left|
\begin{array}{c}
 1,1,1,1,\cdots \\
 0,0,0,0,0,\cdots
\end{array}
\right.\right)$$

If you want a table of numerical values, let me know and I shall edit.


问:
Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $
The limit can be rewritten as

$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$

Recall the inequality,

$$ \cos x < \frac{\sin x}{x} < 1$$

holds for $ x\in (-\pi/2, \pi/2) $.

This provides the new inequalities

$$ 0 < \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] < \frac{1-\cos x}{x},  \text{ with } x>0 $$

$$ \frac{1-\cos x}{x} < \frac{1}{x} \left[1- \frac{\sin x}{x} \right]< 0,  \text{ with } x<0 $$

Applying the Squeeze Theorem, we get

$$ \lim_{x \rightarrow 0^{\pm}}  \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] = 0$$

Note:

$ \lim_{ x \rightarrow 0} \frac{1-\cos x}{x} = 0 $

答:
Your proof is right. But I would have used Taylor developments :

$$\frac{x-\sin x}{x^2} \underset{x\rightarrow 0}{=} \frac{x-(x-x^3/6+o(x^3))}{x^2}\underset{x\rightarrow 0}{=}\frac{x}{6}+o(x)\underset{x\rightarrow 0}{\longrightarrow} 0$$


答:
It is a fine and nice argument with trig functions.

Another way to proceed is to use asymptitics (see them as Taylor expansions at $x=0$.

Then you can write $$\lim_{x \rightarrow 0} \frac{-(\sin(x) -x)}{x^2} = \lim_{x \rightarrow 0} -(\frac{-x^3/3!}{x^2} )= \lim_{x \rightarrow 0} x/6 = 0$$

Title: Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $

Body:
The limit can be rewritten as

$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$

Recall the inequality,

$$ \cos x < \frac{\sin x}{x} < 1$$

holds for $ x\in (-\pi/2, \pi/2) $.

This provides the new inequalities

$$ 0 < \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] < \frac{1-\cos x}{x},  \text{ with } x>0 $$

$$ \frac{1-\cos x}{x} < \frac{1}{x} \left[1- \frac{\sin x}{x} \right]< 0,  \text{ with } x<0 $$

Applying the Squeeze Theorem, we get

$$ \lim_{x \rightarrow 0^{\pm}}  \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] = 0$$

Note:

$ \lim_{ x \rightarrow 0} \frac{1-\cos x}{x} = 0 $


问:
What is $2 - 1 + 1$?
$2-1+1$; a fairly straightforward question, but I (well, not me, but Henry Reich) found something strange.

Most people would evaluate it as $2+(-1)+1 = 2$; however, this goes against the famed, and fairly standard B.E.D.M.A.S./P.E.D.M.A.S., which states that addition goes first, and then subtraction.

If this is the case, then the answer is $2 - (1 + 1) = 2-2 =0$.

Which is the correct answer, and why is the conventional way (B.E.D.M.A.S./P.E.D.M.A.S.) so ambiguous?

答:
Forget garbage like PEMDAS.  Learn mathematics instead.

(And I would take "AS" to mean "addition and subtraction", not "addition, then subtraction".  I learned in 6th and 7th grades that those are done from left to right.  If someone taught you a rule that addition should precede subtraction in such instances, they are ignorant.)

Title: What is $2 - 1 + 1$?


问:
Polynomial Tricky Function
Here is a problem that is a bit hard for me. I can understand the question but can't make out how to approach. Please post a complete solution to the given question. I started with solving $P(P(1)) = P(P(2)) = 0$. But it didn't helped

$P(x) = x^2 + ax + b$

$P(P(1)) = P(P(2)) = 0$

$P(1) \ne P(2)$

Fro the given conditions, find the value of $\textbf{-2P(0)}$.

答:
We can observe from substitution that $P(1)=a+b+1$ and $P(2)=2a+b+4$. These are the roots of the polynomial $P$. By Vieta's formulae, we then have $a=-(P(1)+P(2))=-(3a+2b+5)$, and we also have $b=P(1)P(2)=(a+b+1)(2a+b+4)$.

From the first equation, we get $a=\frac{-(2b+5)}{4}$. When substituting this in the second equation:
$$8b=(-2b-5+4b+4)(-2b-5+2b+8)=3(2b-1)$$
$$8b=6b-3 \implies b=-\frac{3}{2}$$

Hence, we have:
$$-2P(0) = -2b = -2 \bigg(\frac{-3}{2} \bigg)=3$$

Title: Polynomial Tricky Function


问:
Prove that one circle contains the other circle's radius when six circles meet at one point in a plane
The Problem is:

Prove that one circle contains the other circle's radius when six circles meet at one point in a plane.

Comment:

This problem is one of exercises in my textbook on chapter pigeonhole principle.
To get some hint, I tried drawing identical circles one by one and I could get it by picture. But did not reach the key hint for explaining it by pigeonhole principle. I thought the angle approach could work but didn't find the way to go.

答: I assume the circles are equal. There are six rays $PO_i$, where $P$ is the common points. Two of them will make an angle smaller than $\frac{2\pi}{6}$, so the distance between the centers will be at most the radius.


答:
Since absolute convergence implies convergence,

$\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert<\infty \implies \sum_{n=1}^\infty\left( a_{n+k}-a_n\right)<\infty.$

Next, if it weren't true that $ \sum\limits_{n=1}^{\infty} \left(a_{k(n+1) + i}-a_{kn+i} \right) $ converges for all $i \in \{0,...,k-1 \}$, then the original sum $\sum_{n=1}^\infty\left( a_{n+k}-a_n\right)$ would not converge, a contradiction.

So we choose one $i \in \{0,...,k-1 \}$, and consider the subsequence $\{b_n\} = \{ a_{kn+i} \} $ of $\{a_n\}.$

We already have that $ \sum\limits_{n=1}^{\infty} \left(b_{n+1}-b_n\right) = L$, so I'll show that {$b_n$} is a convergent sequence.

By supposition, $ \sum\limits_{n=1}^{\infty} \left(b_{n+1}-b_n\right) = L \in \mathbb{R}.$

By cancellation of terms, this is the same as:

$lim_{n \to \infty} \left(b_{n+1}-b_1\right) = L$

$\implies lim_{n \to \infty} \left(b_{n}\right) = L + b_1.$


答:
$P(1)$ and $P(2)$ are two distinct roots of $P$, and we know that $P$ has only two roots (at most). So we have these two roots : $P(1)=1+a+b$ and $P(2)=4+2a+b$.

So $$P(X) = (X-P(1))(X-P(2)) = (X-1-a-b)(X-4-2a-b)$$

So $-2P(0) = -2P(1)P(2)$.

But we know more about the roots in fact we know they are $\frac{-a+\sqrt{a^2-4b}}{2}$ and $\frac{-a-\sqrt{a^2-4b}}{2}$. So the product of both is $b$.

So $-2P(0) = -2b$.

Let's determine $b$.

First the sum of the two roots is $-a$ and also $5+3a+2b$. So $4a+2b+5=0$. And with the difference of the two roots we get (by taking the square) $a^2-4b=(3+a)^2$ so $6a+4b+9=0$.

By multiplying the first equation by $2$, the second by $3$, and taking the difference, we get $$2b=-3$$

So $$-2P(0) = 6$$

Title: Polynomial Tricky Function

Body:
Here is a problem that is a bit hard for me. I can understand the question but can't make out how to approach. Please post a complete solution to the given question. I started with solving $P(P(1)) = P(P(2)) = 0$. But it didn't helped

$P(x) = x^2 + ax + b$

$P(P(1)) = P(P(2)) = 0$

$P(1) \ne P(2)$

Fro the given conditions, find the value of $\textbf{-2P(0)}$.


Title: Polynomial Tricky Function

Body:
Hint:

$$P(2)-P(1)=3+a$$

$$P(P(2))-P(P(1))=(3+a)(5+4a+2b)$$


答:
The usual convention is to read $a-b$ as an abbreviation for $a+(-b)$, that is, "add the opposite".

This makes possible to associate or permute the terms in the expression getting always the same result:
$$
2-1+1=(2-1)+1=2+(-1+1)=
2-1+1=2+1-1=-1+2+1=1-1+2
$$
This should explain why this convention is used instead of a more difficult one. Just forget about subtraction, which has less pleasant properties than addition.

The same is for "division": we avoid it preferring "multiplying by the inverse". This avoids ambiguities.

How would you parse
$$
3-1+2-3+4
$$
with Henry Reich's convention? With the standard convention it is
$$
3+(-1)+2+(-3)+4=5
$$
That's all: mathematics is based on conventions.

Title: What is $2 - 1 + 1$?

Body:
$2-1+1$; a fairly straightforward question, but I (well, not me, but Henry Reich) found something strange.

Most people would evaluate it as $2+(-1)+1 = 2$; however, this goes against the famed, and fairly standard B.E.D.M.A.S./P.E.D.M.A.S., which states that addition goes first, and then subtraction.

If this is the case, then the answer is $2 - (1 + 1) = 2-2 =0$.

Which is the correct answer, and why is the conventional way (B.E.D.M.A.S./P.E.D.M.A.S.) so ambiguous?


问:
Amy has a cuboid with volume at most 8

Amy has a clump of iron with volume $8$. She melts it and make a cuboid (she might have iron left over). Her boyfriend Ben increased the length, width and height of the cuboid by $1$ to form a new cuboid. Show that the surface area of the new cuboid is at least two times of the volume of the new cuboid.

I made this problem, so I know a solution (using inequality). Any solution is accepted as long as it makes sense. There might be a geometric solution that I don't know! I will post my solution if no one posts it.

答:
Let V and A be the area and volume of the new cuboid. Let L, W, H be the dimensions of the original. Then:

\begin{align*}V &= \left(L+1\right)\left(W+1\right)\left(H+1\right) \\ &= LWH + LW + WH + HL + L + W + H + 1 \\ &= LWH + \frac{LWH}{H} + \frac{LWH}{L} + \frac{LWH}{W} + L + W + H + 1 \\ &= LWH + LWH \left(\frac{1}{H} + \frac{1}{L} + \frac{1}{W}\right) + \left(L + W + H\right) + 1\end{align*}

Meanwhile:

\begin{align*}A &= 2\left(L+1\right)\left(W+1\right) + 2\left(W+1\right)\left(H+1\right) + 2\left(L+1\right)\left(H+1\right) \\ &= 2 \left(LW + WH + LH + 2L + 2W + 2H\right) + 6 \\ &= 2 \left(\frac{LWH}{H} + \frac{LWH}{L} + \frac{LWH}{W}\right) + 4\left(L + W + H\right) + 6 \\ &= 2 LWH \left(\frac{1}{H} + \frac{1}{L} + \frac{1}{W}\right) + 4\left(L + W + H\right) + 6\end{align*}

Let $$V_0 = LWH$$ $$Q = \left(\frac{1}{H} + \frac{1}{L} + \frac{1}{W}\right)$$ $$R = L + W + H$$
so then: $$V = V_0 + V_0 Q + R + 1 \hspace{2.54cm} A = 2V_0Q + 4R + 6 \:,$$ or $$\frac{A}{V} = \frac{V_0 \left(2Q\right) + \left(6 + 4R\right)}{V_0 \left(1+Q\right) + \left(1 + R\right)}$$

Still not totally clear to me, so looking at a few extreme cases (I haven't used the 8 or the inequality yet)...

For $Q$ very small, this means $R$ is very large, in which case the ratio limits to $A/V \approx 4$. Okay.

For $R$ very small, $Q$ must be very large, in which case the ratio limits to $A/V \approx 2$, but still greater. So far so good.

The in-between cases should be covered by the top-heavy fraction.

As a non-surprising sanity check, the special case $V_0=8$ with $L=W=H=2$, gives exactly $2$, not greater. $$\frac{A}{V} = \frac{8 \left(2 \cdot 3/2\right) + \left(6 + 4 \cdot 6\right)}{8 \left(1 + 3/2\right) + \left(1 + 6\right)} = \frac{24+6+24}{20+7} = \frac{54}{27} = 2$$

(I now see why you chose 8 for the original volume.)

Title: Amy has a cuboid with volume at most 8


问:
Geometric sequence and series problem
Hello i need help please,
$$
\begin{cases}
u_0&= 1\\
u_{n+1}&=\frac{3u_n + 2v_n}{5}
\end{cases}
\qquad
\begin{cases}
    v_0 = 2\\
v_{n+1} = \frac{2u_n+3v_n}{5}
\end{cases}
$$

1.  Calculate $u_1$, $u_2$, $v_1$, and $v_2$.

2.  We consider the sequence $(d_n)$ defined for any natural number $n$ by $d_n = v_n-u_n$.

a.  Show that the sequence $(d_n)$ is a geometric sequence of which we will give its common ratio and its first term.

b.  Deduce the expression of $d_n$ depending on $n$.

3.  We consider the sequence (s_n) defined for any natural number $n$ by $s_n = u_n+v_n$.

a.  Calculate $s_0$, $s_1$, and $s_2$.  What can we guess?

b.  Show that, for all $n \in \mathbb{N}$, $s_{n+1} = s_n$.  What can we deduce? Deduce an expression of $u_n$ and $v_n$ depending on $n$.

5.  Determine depending on $n \in \mathbb{N}$,

a.  $T_n = u_0 + u_1 + \dots + u_n$.

b.  $W_n = v_0 + v_1 + \dots + v_n$.

Actually i answered the first question and the second, and i found the common ratio is $1/5$ and the first term 1 so $d_n = 1 \cdot 1/5^n$
For the 3. I found $3$
And i cant resolve the 4 and 5  can you help me

答:
Hints. Since you have successfully found $d_n=v_n-u_n$ and $s_n=v_n+u_n$ as functions of $n,$ then you can find expressions for $u_n$ and $v_n$ simply by subtraction and addition of the above equations, so that you have that $$2v_n=s_n+d_n$$ and $$2u_n=s_n-d_n.$$

From here it should be easy to continue.

Title: Geometric sequence and series problem


问:
Simplify the expression $\log(2^x+2^y)$ for large values of $x,y$
Given two numbers $x,y$, I need to calculate $\log(2^x+2^y)$.

Is there a way to simplify this function, in a way that would allow to calculate the complete expression without going through calculating $2^x$ and $2^y$?

Assume that $x,y$ are larger than 1500 and I'm trying to calculate the value using a computer, so it can't really handle $2^{1500}$, but eventually I want to stay in the logarithmic scale so I don't really want to calculate $2^x$.

A (good) approximation is also fine (if exact simplification doesn't exist).

答:
If $x\simeq y$, then we have $\log (2^x + 2^y) \simeq \log (2^{x+1}) = (x+1)\log 2$ (which is equal to $(x+1)$ if we are in base $2$).

If $x\gg y$, then we have $\log (2^x + 2^y) \simeq \log(2^x) = x\log 2$ (equal to $x$ if we are in base $2$).

More precisely , if $h=x-y >0$, then$$\ln(2^x+2^y)=x\ln2+\ln(1+2^{-h})\in x\ln2+2^{-h}-O(4^{-h}).$$


问:
Find the expected value of $\frac{1}{X+1}$ where $X$ is binomial
The problem:

X is a binomial random variable, find $E[\frac{1}{X+1}]$
n and p are not given

PDF for a binomial distribution is $\binom{n}{k}p^k(1-p)^{n-k}$

Expected value is
$\sum{x_ip(x_i)}$

But this is where I get stuck, I'm really rusty on my statistics and I'm not sure exactly how to structure it in the next step? I think I want to get the form of the following out of the summation

$\sum _{k=0}^{n} \binom{n}{k}p^k(1-p)^{n-k} = (p + 1 - p)^n = 1$

But I'm not sure if it should look like

$\sum \frac{1}{xp(x)+1} $

and if it should where to go from here?

Body:
we know :$\frac{1}{X+1}=\int_{0}^{1}s^{X}ds$

$E(\frac{1}{X+1})=\int_{0}^{1}E(s^{X})ds$

$MGF=E(e^{tx})=(q+pe^t)^n;(p+q=1)$

Also

$E(s^{X})=(q+ps)^n$

Thus

$E(\frac{1}{X+1})=\int_{0}^{1}(q+ps)^nds=\dfrac{(q+ps)^{n+1}}{p(n+1)}\bigg]_{0}^{1}=\dfrac{(q+p)^{n+1}}{p(n+1)}-\dfrac{q^{n+1}}{p(n+1)}$
$=\dfrac{1}{p(n+1)}-\dfrac{q^{n+1}}{p(n+1)}$


问:
How can I solve an integral by using hypergeometric function of the second kind?
This is an article that I've been studing for a while. I came across this integral equation:

$$\mathrm dm=\frac{\mathrm dr}{{r}^{z+1}\left[1-{\left(\dfrac{r_h}{r}\right)}^{n+z-1}\right]}$$

in which $n$, $z$ and $r_h$ are parameters and $r$ and $m$ are my variables. The general solution for $n + z > 1$ is:

$\qquad m=\frac{-{r}^{-z}}{z} _2F_1\big[1,\frac{z}{n+z-1};1+\frac{z}{n+z-1};{(\frac{r_h}{r})}^{n+z-1}\big]$

I want to know how that integral leads to a hypergeometric function of the second kind.

答:
Making $a=r_h$, you want to compute
$$I=\int  r^{-z-1} \left(1-\frac{a}{r}\right)^{-n-z+1}\,dr$$ Make $\frac{a}{r}=t$
$$I=-a^{-z}\int t^{z-1} (1-t)^{-n-z+1}\,dt=-a^{-z}\int \frac{t^{z-1} } {(1-t)^{n+z-1} }\,dt$$ which would give
$$I=-a^{-z} t^z \left(\frac{\, _2F_1(z,n+z;z+1;t)}{z}-\frac{t \,
   _2F_1(z+1,n+z;z+2;t)}{z+1}\right)$$ which is probably more pleasant that the one in your post.

OwnerDisplayName: Q Tara


问:
Is $f(x)=x^2\cos\left(\frac{1}{x}\right)$ of bounded variation on $[-1,1]$?

Is $f(x)=x^2\cos\left(\frac{1}{x}\right)$ with $f(0)=0$ of bounded variation on $[-1,1]$?

I know that $g(x)=x\cos\left(\frac{1}{x}\right)$ and $h(x)=x^2\cos\left(\frac{1}{x^2}\right)$ are both NOT of bounded variation, so I'm guessing that $f$ isn't either...? I just don't know which partition to take, since the proofs of the ones for $g$ and $h$ involved expressing the Variation into a harmonic series, which diverges when the number of points in the partition increases, but I don't know how to do the same for $f$. Thanks!

答:
I know that I am going to answer an old question, but I think the way I used to answer here is new and easy. So you can show this by using any one of the following ways:

Result$~\bf 1$: If $~f~$ is a differentiable function having bounded derivatives on $[a,b].$ Then $~f~$ is of bounded variation on $[a,b].$

Here $~f(x)=x^2\cos\left(\frac{1}{x}\right),~$ then $~f'(x)=2x\cos\left(\frac{1}{x}\right)+\sin\left(\frac{1}{x}\right)~$ for all $x\ne0~.$

So $~|f'(x)|=\left|2x\cos\left(\frac{1}{x}\right)+\sin\left(\frac{1}{x}\right)\right|\le 2\left|x\cos\left(\frac{1}{x}\right)\right|+\left|\sin\left(\frac{1}{x}\right)\right|\le 2~|x|+1~=2\cdot 1+1=3,$
as $~|\sin x|\le 1~$ and $|\cos x|\le 1~,~~\forall ~x~.$

Hence $~|f'(x)|\le 3~~,~\forall ~x\in[-1,1].$

Clearly, by the previous result, $~f~$ is of bounded variation on $[-1,1]~.$

Result$~\bf 2$: If $~f(x)=x^\alpha\cos\left(\frac{1}{x^\beta}\right),~~\forall~x\ne 0~,$ then $~f~$ is of bounded variation if and only if $~\alpha\gt\beta~.$

So in this case $~\alpha=2~$ and $~\beta=1~$.
Therefore the condition  $~\alpha\gt\beta~$ satisfied and hence $~f~$ is of bounded variation on $[-1,1]~.$


问:
$a^b = c$, is it possible to express $b$ without logarithms?
$ a^b = c $

is it possible to express b without logarithms?

答: Not without more information, since finding $b$ in this case is what logarithms are.  Maybe in some special cases something can be said.</p>


问:
Expressing $ \frac{\cos x + \sin x}{\cos x - \sin x} $ as a single trigonometric function
For a while now, I have been trying to get this expression into 1 term instead of the 2 term fraction.
$$
\frac{\cos x + \sin x}{\cos x - \sin x}
$$
The most I have gotten down to is get it to a single ratio, ie,
$$
\frac{1 + \tan x}{1 - \tan x}
$$

I have been looking for an identity or formula of some sort to factorize $ \cos x + \sin x $ and $ \cos x - \sin x $.

Any help will be appreciated.
Thanks

答:
$
\dfrac{\cos x + \sin x}{\cos x - \sin x} = \dfrac{1+ \tan x}{1 - \tan x}
$

Now, $\tan (x+y) = \dfrac{\tan x+ \tan y}{1 - \tan x \tan y}$

Therefore,  $\tan (x+ \frac{\pi}{4}) = \dfrac{\tan x+ 1}{1 - \tan x}$

Therefore, $\dfrac{\cos x + \sin x}{\cos x - \sin x} = \tan(x+ \frac{\pi}{4})$

Is this what you meant by 1 term?.


答: You may write it as $\tan(\pi/4+x)$, which is a single term.

Title: Expressing $ \frac{\cos x + \sin x}{\cos x - \sin x} $ as a single trigonometric function

Body:
For a while now, I have been trying to get this expression into 1 term instead of the 2 term fraction.
$$
\frac{\cos x + \sin x}{\cos x - \sin x}
$$
The most I have gotten down to is get it to a single ratio, ie,
$$
\frac{1 + \tan x}{1 - \tan x}
$$

I have been looking for an identity or formula of some sort to factorize $ \cos x + \sin x $ and $ \cos x - \sin x $.

Any help will be appreciated.
Thanks


问:
Parametric family of distributions that is closed under mixture
Suppose $F$ and $G$ are two cumulative distribution functions on $\mathbb{R}$. Let $\alpha F + (1 - \alpha) G$ be their convex combination for some $\alpha \in [0, 1]$. We know it is also a cdf.

Is there any well known parametric family of distributions that is "closed" under mixture? That is, if both $F$ and $G$ belong to this family, then so is any $\alpha F + (1 - \alpha) G$ for $\alpha \in (0, 1)$?

Thank you!

答:
I don't have a complete answer to your question, but clearly, one of the consequences of such closure would be that any member of this parametric family would need to be expressible as $$F = \sum_{i=1}^\infty \alpha_i F_i$$ subject to the constraint $$\forall i, \alpha_i \ge 0, \quad \sum_{i=1}^\infty \alpha_i = 1.$$

In fact, we can use this to construct a family from a single parametric distribution; e.g., use the standard normal and construct the family of distributions that are mixtures of the standard normal, parametrized by the mixture weights themselves. Then this family is closed under additional mixing.

For further thought.  Suppose $X_i \sim \operatorname{Bernoulli}(p_i)$ with CDF $F_i(x)$.  Then what is the mixture of $F_1$ with $F_2$?  What is the $n$-fold mixture of $F_1, F_2, \ldots, F_n$?


问:
Product of CDFs at different points
Suppose that $X$ is a continuous random variable with finite moments, and $\mathbb a=(a_i) \in \mathbb R^n$. In my calculations, I have reached an expression as

$$ f(\mathbb a )=\prod_i \Pr \{ X\le a_i\} $$

but I am not sure what to do with it to make it a simple expression (i.e. a CDF of a univariate random variable or a limiting theorem that make the whole thing tractable at least asymptotically, etc.). Let's suppose that $n$ is large enough.

One idea was to take another random variable $ Y_i = X - a_i$ , then $$ f(\mathbb a )= \prod_i \Pr \{ Y_i\le 0\} = \Pr \{ \max_i (Y_i)\le 0\} $$ and try to use results from Extreme Value Theory but I only know this theory for i.i.d. case (here we have different first moments).

答:
I can only help you with a special case: Let's X is Gumbel-distributed

$$\mathrm{Pr}\{X\leq a_{i}\}=e^{-e^{-\frac{x-a_{i}}{\beta}}}$$

with the same variance $\beta$. Then

$$f(a)=\prod_{i=1}^{n}\mathrm{Pr}\{X\leq a_{i}\}=\prod_{i=1}^{n}e^{-e^{-\frac{x-a_{i}}{\beta}}}=e^{-e^{-\frac{x}{\beta}}\left(\sum_{i=1}^{n}e^{\frac{a_{i}}{\beta}}\right)}=e^{-e^{-\frac{x-\beta\log\left(\sum_{i=1}^{n}e^{\frac{a_{i}}{\beta}}\right)}{\beta}}}=\mathrm{Pr}\left\{ X\leq \bar{a}\right\} $$

where the new mean

$$\bar{a}=\beta\log\left(\sum_{i=1}^{n}e^{\frac{a_{i}}{\beta}}\right)$$

is often called the mellowmax.

Title: Product of CDFs at different points


问:
Multiplication of matrix: ABAB
I was doing this Putnam exercise:

Let $A$, $B$ be $n \times n$ matrices such that $ABAB = 0$. Can we conclude that $BABA = 0$?

I realized that this is not always true, it can fail to non-invertible matrices with an arbitrary n order. But, when I searched in google to confirm my answer, I read an interesting answer from another user here that says:
"If $n=2$ and $AB$ is nilpotent, so as the $BA$ and this affirmation is always true"

Okay, it does not disprove my answer but it was interesting to know

For $n = 2$: The characteristic polynomials of $AB$ and $BA$ are the
same (for any $n$). Therefore, if $AB$ is nilpotent then $BA$ is also
nilpotent. So, we necessarily have $BABA = (BA)^2 = 0$.

I can see the characteristic polynomial are equal by the equation $t^2 - \operatorname{tr}(AB)t + \det(AB)$, but what does this imply? I mean, my knowledge about characteristic polynomials is restricted to find the eigenvalues and eigenvectors.

答:
Since ${\rm trace}(AB) = {\rm trace} (BA)$ and $\det (AB)=\det (BA)$ it means that for $n=2$ matrices $AB$ and $BA$ have the same characteristic polynomial so they have the same eigenvalues.

But in this case you don't need that. Since $ABAB = 0$ we get $BABABA =0$ so $(BA)^3=0$ which means that $BA$ is also nilpotent but with order $3$ or less, so it eigenvalues are all $0$.

Title: Multiplication of matrix: ABAB


答:
Think of each number as an item;
$2−1+1$ evaluates to: $$(+2)+(-1)+(+1)$$ Thus, represented in succession, it evaluates to:

$$n = 0$$
$$n = n+2 = +2$$
$$n = n-1 = +1$$
$$n = n+1 = +2$$
$$n = 2$$

Title: What is $2 - 1 + 1$?


问:
A problem on Angle bisectors of a triangle and perpendiculars.

From $A$ perpendiculars $AX, AY$ are drawn to the bisectors of the exterior angles of $B$ and $C$ of $\Delta ABC$. Prove that $XY || BC$.

Prove that the feet of the four perpendiculars dropped from a vertex of a triangle upon the four bisectors of the other two angle are collinear.

In triangle $\Delta ABC, X $and $Y$ be the feet of perpendiculars from vertex $A$ to the internal angle bisector of $∠B$ and $∠C$ respectively. Line $XY$ meets $AB$ at $P$ and $AC$ at $Q$. If $AB = 7$ cm, $BC = 8$ cm and $CA = 5$ cm then find $PQ$ and $XY$.

These were 3 questions with almost similar configurations; as I wasn't able to solve the first one, I couldn't solve the next two either (I don't know if it's just psychological or something but anyway).

So, could someone please provide the SOLUTION for the first one and just hints for the next two

A solution using just elementary geometry would be ideal but any solution would be really helpful,

Thanks

答:
Here is a nice solution using trigonometry; I will name the points according to your diagram (i.e., using $AD$ and $AE$ instead of $AX$ and $AY$). So let $F$ and $G$ be the feet of the perpendiculars from $D$ and $E$ onto $BC$, respectively. It suffices to show that $d(BC, D)=|DF|=|EG|=d(BC, E)$.

Notice that $\Delta DGB$ and $\Delta ADB$ are similar and hence

$$\frac{|DF|}{|AD|}=\frac{|DB|}{|AB|}\Leftrightarrow |DF|=\frac{|AD||DB|}{|AB|}$$

However, using $\alpha, \beta, \gamma$ to denote the interior angles of the triangle, as usual, we have $|AD|=|AB|\cos(\beta/2)$ and $|DB|=|AB|\sin(\beta/2)$, so

$$|DF|=|AB|\sin(\beta/2)\cos(\beta/2)=\frac{1}{2}|AB|\sin(\beta)$$

Finally, the sine rule in $\Delta ABC$ then gives

$$|DF|=\frac{1}{2}|BC|\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}$$

and by symmetry, the same argument for $E$ instead of $D$ gives

$$|EG|=\frac{1}{2}|BC|\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}$$

and we are done.

For problem 2, call the feet onto the internal bisectors $H$ and $I$. Then a similar computation should give exactly the same values for $d(H, BC)$ and $d(I, BC)$ as above and you are done.

For problem 3, use that the length of the altitude from $A$ in $\Delta ABC$ is given by $|BC|\frac{\sin(\beta)\sin(\gamma)}{\sin(\alpha)}$.


答:
I don't see why the characteristic polynomial should be involved here.

Instead note that $(BA)^3=B(AB)^2A=0$ so $BA$ is indeed nilpotent, although possibly only for the third power. When $n=2$ any nilpotent matrix may be reduced to the form
$$ M = \pmatrix{ 0 & 1 \\ 0  & 0}, $$
so the order of milpotency must be 2. You may construct counterexamples for $n\geq 3$, e.g.:
$$ A = \pmatrix{ 0 & 1 &0\\ 0 &0&0\\ 0 & 0 &1}, \ \
B = \pmatrix{ 1 & 0 &0\\ 0 &0&1\\ 0 & 0 &0}, \ \
$$
for which you check that $(AB)^2=0$ but $(BA)^2\neq 0$.

Title: Multiplication of matrix: ABAB

Body:
I was doing this Putnam exercise:

Let $A$, $B$ be $n \times n$ matrices such that $ABAB = 0$. Can we conclude that $BABA = 0$?

I realized that this is not always true, it can fail to non-invertible matrices with an arbitrary n order. But, when I searched in google to confirm my answer, I read an interesting answer from another user here that says:
"If $n=2$ and $AB$ is nilpotent, so as the $BA$ and this affirmation is always true"

Okay, it does not disprove my answer but it was interesting to know

For $n = 2$: The characteristic polynomials of $AB$ and $BA$ are the
same (for any $n$). Therefore, if $AB$ is nilpotent then $BA$ is also
nilpotent. So, we necessarily have $BABA = (BA)^2 = 0$.

I can see the characteristic polynomial are equal by the equation $t^2 - \operatorname{tr}(AB)t + \det(AB)$, but what does this imply? I mean, my knowledge about characteristic polynomials is restricted to find the eigenvalues and eigenvectors.


问:
Graphing a tangent function with an amplitude
Because of the all real values for the range of a tangent graph, the amplitude seems to have no effect on the maximum or minimum of the graph, but to change its steepness. i just want to seek reassurance whether this is true or not?</p>

答: If you add a coefficient in front of $\tan{x}$, you do not alter the asymptotes. So yes, all it will do is change the steepness. One way you could measure/observe this is the slope at $x=2πn$. But since it isn't a wave, I don't think amplitude is the right word.


答:
Assume $x\ge y$. Because:

$$\log(2^x+2^y)=\log(2^x(1+2^{-(x-y)}))=x\log(2)+\log(1+2^{-(x-y)})$$

and, as you may know, $\log(1+h)\approx h$ for $h$ small, then:

If $2^{-(x-y)}$ is small, then take $\log(2^x+2^y)\approx x\log(2)+2^{-(x-y)}$

If $2^{-(x-y)}$ is large, leave it as is: $\log(2^x+2^y)=x\log(2)+\log(1+2^{-(x-y)})$.

The point here is that the error you get when you replace $\log(1+h)$ with $h$ is given by the error in Taylor expansion of $\log(1+h)$ around $h=0$, which is:

$$\log(1+h)=h-\frac{h^2}{2(1+\xi)^2}$$

for some $\xi, 0\lt\xi\lt h$, and so the error is bounded by $h^2/2$. Thus, I suggest you use that estimate of an error to choose for how small $h=2^{-(x-y)}$, i.e. for how big difference $x-y$ you are happy with the above approximation.

Example: Suppose you want the result to be correct up to two decimal points, i.e. the error to be smaller than $0.005$. By solving $h^2/2\le 0.005$ you get $h\le 0.1$, i.e. the approximation above is good whenever $x-y\ge\log_2(0.1)\approx 3.32$.

Altogether, luckily, $2^{-(x-y)}$ converges very quickly (exponentially) to zero as $x-y$ grows. This means you will probably accept the approximation for all but a very few differences $x-y$, for which you will need to use the exact formula with $\log(1+2^{-(x-y)})$. For those small $x-y$, if $x$ and $y$ are integers, you can even pre-calculate a table of the values of $\log(1+2^{-(x-y)})$.

Title: Simplify the expression $\log(2^x+2^y)$ for large values of $x,y$

Body:
Given two numbers $x,y$, I need to calculate $\log(2^x+2^y)$.

Is there a way to simplify this function, in a way that would allow to calculate the complete expression without going through calculating $2^x$ and $2^y$?

Assume that $x,y$ are larger than 1500 and I'm trying to calculate the value using a computer, so it can't really handle $2^{1500}$, but eventually I want to stay in the logarithmic scale so I don't really want to calculate $2^x$.

A (good) approximation is also fine (if exact simplification doesn't exist).


问:
When two elements of a group generate the same subgroup
Let $G$ be a group and $x,y$ two elements from $G$. Suppose that $\langle xy \rangle = \langle yx \rangle$ , i.e. they generate the same subgroup of $G$. I want to find a counterexample in which $xy \neq yx$ and $xyyx \neq 1$. Maybe in a permutation group such a counterexample exists, but I haven't found one yet.

A counterexample for $xy \neq yx$ can be easily found in dihedral group $D_6$, for example $x=\rho^2\sigma$, $y=\rho\sigma$, where $\sigma^2=\rho^6=1$, but, in this case, $xyyx=1$.

Thank you in advance!

答: Suppose $3<n<p$ and $\Bbb F_p^+$ is the additive group of the finite field $\Bbb F_p:=\frac{\Bbb Z}{p\Bbb Z}=\{\mathbf 0,\mathbf 1\,\dots,\mathbf{p-1}\}$ of size $p$. Let $$x=(u,\mathbf 1)\;\;\;\;\;y=(v,\mathbf 1)$$ in the product group $D_{n}\times\Bbb F_p^+$ where $u,v\in D_n$ are reflections and $\langle uv\rangle$ is the entire $D_n$ subgroup $R_n$ of rotations. Now, $$\langle yx\rangle=\langle xy\rangle=\langle (uv,\mathbf 2)\rangle =R_n\times\Bbb F_p^+$$ Since $\text{ord}(\mathbf 2)=p$ is odd we must have $$(uv,\mathbf 2)=xy\neq (yx)^{-1}=(uv,\mathbf 2^{-1})$$ and finally $xy\neq yx$ because $uv\neq vu$.


问:
Entropy of dyadic Toeplitz system
I have failed to find the topological entropy of dyadic Toeplitz system. Do you know what this entropy is?

Dyadic Toeplitz system is a subshift of $\{0,1\}^{\mathbb{Z}}$, i.e. it is an orbit closure of point $x$ constructed as follows:
on every second coordinate we place zero, we get sequence $(...*0*0*0*0*...)$, then instead of every second * we place one, we get $(...010*010*010...)$ and so on. In other words, coordinates of element $x$ can be decomposed into arithmetic progressions, on which $x$ is constant.
Thank a lot in advance!

答: Bonjour, your sequence is the fixed point of the primitive substitution 0-->01, 1-->00 thus its entropy is 0:its number of different words of length $n$ should be bounded by 2n or something like that but it is sublinear as all primitive substitutive sequences. You can find more information in a paper of Cassaigne-Karhumaki:1997 or Koskas:1998.

Title: Entropy of dyadic Toeplitz system


问:
Is there a closed expression for $\sum_{n=0}^\infty \frac{n^2a^n}{n!}$, where $a \in (0,\infty)$?
Is there a closed expression for $\displaystyle\sum_{n=0}^\infty \frac{n^2a^n}{n!}$, where $a \in (0,\infty)$?

I have concluded that the sum converges using the ratio test:

$\displaystyle\frac{(n+1)^2a^{n+1} n!}{n^2a^n(n+1)!}=\frac{(n+1)a}{n^2} \rightarrow 0$ as $n\rightarrow \infty$.

Also, by comparison the following "similar expression" has a closed form:

$$\sum_{n=0}^\infty \frac{na^n}{n!} = a\sum_{n=0}^\infty \frac{na^{n-1}}{n!}=ae^a$$

Body:
Hint. Note that for $n\geq 2$,
$$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}.$$


问:
Prove that M is an extension of N of degree 2

Let $L$ be a field of rational functions in one variable over $\mathbb{C}; i.e$

$$L = \mathbb{C}(y) = \{ \frac{f(y)}{g(y)} | f(y), g(y) \in \mathbb{C}(y)\}$$

Let $M = \mathbb{C}(\beta)$ with $\beta = y^5$ and let $N = \mathbb{C}(\alpha)$ with $\alpha = -\frac{(y^5-1)^2}{y^5} \in M.$ Notice that now we have

$$\mathbb{C} \subseteq N = \mathbb{C}(\alpha) \subseteq M = \mathbb{C}(\beta)
\subseteq L = \mathbb{C}(y)$$
a)Prove that M is an extension of N of degree 2.

My attempts

$\alpha = -\frac{(y^5-2)^2}{y^5}$ with $\beta = y^5$

Here Polynomial f(x) = $x^2+(\alpha-2)x+1=0$ has a root $\beta,\frac{1}{\beta}$

and we know minimal polynomial $m_{\alpha}(x)$ | f(x). So, deg $m_{\alpha}(x) =$ 1 or 2.

but how can we eliminate the case with deg $m_{\alpha}(x) =1$.

I am struggling to prove that $\beta$ or $\frac{1}{\beta} \notin \mathbb{C}(\alpha)$. It would be great if someone can give hint in this direction. Thanks in advance

答: We have the identity $2-\alpha= \beta + \beta^{-1}$ and therefore $\mathbb{C}(\alpha)=\mathbb{C}(\beta + \beta^{-1})$. Now assume $\beta\in\mathbb{C}(\beta + \beta^{-1})$. The equation $\beta\cdot g(\beta + \beta^{-1}) = f(\beta + \beta^{-1});\ f,g\in \mathbb{C}[X]$ will lead to a contradiction when we compare coefficients.


问:
Find the envelope of the family of straight lines $x \cos α + y \sin α = a \sec α$ , where $α$ is the parameter.
Ok, so here's what I have done I have divided both the sides cosα and then simplified it to

$(x-a) + y\tan \alpha - a\tan^2 \alpha = 0$

I am stuck as to what I should do here... Some help needed.

答:
Write $$a t^2-y t+(a-x),~ t=\tan \alpha~~~~(1)$$
For the envelope (the fixed curve which is touched by the given family of lines(1)), demand  $B^2=4AC$, in this quadratic. Then we get
$$y^2=4a(a-x)~~~~(2)$$
which is the required envelope.


答:
$$\log_2(2^x+2^y)=\max(x,y)+\log_2(2^{x-\max(x,y)}+2^{y-\max(x,y)})
\\=\max(x,y)+\log_2(1+2^{-|x-y|}).$$

The second term is a gentle function close to zero and not exceeding $1$. You can compute it very accurately if the function log1p is available in your environment (for large values of $|x-y|$, say $>50$, just ignore it).


答:
Start with
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
Differentiate both sides:
$$e^x=\sum_{n=0}^\infty\frac{nx^{n-1}}{n!}$$
Multiply by $x$:
$$xe^x=\sum_{n=0}^\infty\frac{nx^n}{n!}$$
Differentiate both sides:
$$(x+1)e^x=\sum_{n=0}^\infty\frac{n^2x^{n-1}}{n!}$$
Multiply by $x$:
$$x(x+1)e^x=\sum_{n=0}^\infty\frac{n^2x^n}{n!}$$

Title: Is there a closed expression for $\sum_{n=0}^\infty \frac{n^2a^n}{n!}$, where $a \in (0,\infty)$?

Body:
Is there a closed expression for $\displaystyle\sum_{n=0}^\infty \frac{n^2a^n}{n!}$, where $a \in (0,\infty)$?

I have concluded that the sum converges using the ratio test:

$\displaystyle\frac{(n+1)^2a^{n+1} n!}{n^2a^n(n+1)!}=\frac{(n+1)a}{n^2} \rightarrow 0$ as $n\rightarrow \infty$.

Also, by comparison the following "similar expression" has a closed form:

$$\sum_{n=0}^\infty \frac{na^n}{n!} = a\sum_{n=0}^\infty \frac{na^{n-1}}{n!}=ae^a$$


问:
What is the analogue of the rotation matrix in polar coordinates?
In cartesian coordinates, a rotation around the z axis is represented as:

$$\left( \begin{array}{ccc}
\cos\phi & \sin\phi & 0\\
-\sin\phi & \cos\phi & 0\\
0&0&1
\end{array} \right)$$
What is the analogue in spherical coordinates?

EDIT
My change of coordinates is:

$x=r \sin\theta\cos\phi$

$y=r \sin\theta\sin\phi$

$z=r \cos\theta$

答: We wish to rotate a vector $\vec{r}$ to $R\vec{r}$ for a suitable rotation matrix $R$. I should make an important conceptual correction: the correct representation of $\vec{r}$ is $r\hat{r}$, not $r\hat{r}+\theta\hat{\theta}+\phi\hat{\phi}$ or whatever you were thinking of. In particular$$\begin{align}\hat{r}&=\sin\theta\cos\phi\hat{i}+\sin\theta\sin\phi\hat{j}+\cos\theta\hat{k},\\\hat{\theta}&=\cos\theta\cos\phi\hat{i}+\cos\theta\sin\phi\hat{j}-\sin\theta\hat{k},\\\hat{\phi}&=-\sin\phi\hat{i}+\cos\phi\hat{j}\end{align}$$is our orthonormal basis. We can write these equations as $\hat{S}_A=T_{Aa}\hat{C}_a$ (I sum over repeated indices), with $\hat{C}_a$ ($\hat{S}_A$) the Cartesian (spherical) basis vectors, e.g. $T_{ri}=\sin\theta\cos\phi$. Now you just want to compute $R_{AB}=T_{Aa}R_{ab}(T^{-1})_{bB}$, or in matrix terms $R^\text{spherical}=TR^\text{Cartesian}T^{-1}$; I'll leave that to you.


问:
product of end equal to end of product
In the theory of ends, is it true in general that the end of a product is equal to the product of the ends? If not, then what are sufficient conditions for it to be true?

For instance, if $F,G,H,I\in[\mathcal{C}^{\text{op}},\text{Set}]$, is it true that

$$\int_Y \text{Set}(F(Y),G(Y))\times \int_Y\text{Set}(H(Y),I(Y)) \cong
\int_Y \text{Set}(F(Y),G(Y))\times\text{Set}(H(Y),I(Y))$$
I seem to have seen steps like this done in a few places but haven't found a justification for it. Does one prove this from 'first principles' i.e. the definitions of ends and products, or is there a more immediate way to see this is true?

答:
Limits commute with limits and ends are particular kinds of limits. But you have to take into account the correct formula for this, that is, given a functor $L : {\cal I}\times{\cal J}\to \cal C$, the limits
$$\textstyle
\lim_{\cal I}\lim_{\cal J} L\quad \lim_{\cal J}\lim_{\cal I} L\quad \lim_{\cal I\times J} L
$$ are all isomorphic (even more precisely, one exists iff all three exist and they are canonically isomorphic objects).

So, in general no, it is false; the integration variable is mute, so the red $Y$ and the blue $Y$ in the product of integrals
$$
\int_{\color{red} Y} \text{Set}(F({\color{red} Y}),G({\color{red} Y}))\times \int_{\color{blue} Y}\text{Set}(H({\color{blue} Y}),I({\color{blue} Y}))
$$ have to be treated as different. The best you can say thus is that
$$
\int_{\color{red} Y} \text{Set}(F({\color{red} Y}),G({\color{red} Y}))\times \int_{\color{blue} Y}\text{Set}(H({\color{blue} Y}),I({\color{blue} Y}))\cong\int_{\color{red} Y\color{blue} Y} \text{Set}(F({\color{red} Y}),G({\color{red} Y}))\times \text{Set}(H({\color{blue} Y}),I({\color{blue} Y}))
$$ to further reduce this to be equal to
$$
\int_{\color{green} Y} \text{Set}(F({\color{green} Y}),G({\color{green} Y}))\times \text{Set}(H({\color{green} Y}),I({\color{green} Y}))
$$ you need further assumptions on $\cal C$.

Title: product of end equal to end of product


问:
A simple probability reasing to predict rain fall
A friend told me the following about whether it will rain tomorrow (or not):

The probability that it will rain tomorrow is $1/2$ since it will either happen or not. But -even as a non mathematician- I do not find this argument a convincing one, because the same reasoning can be used to predict that I will be struck by a bolt of lighting in five minutes.

Can someone tell me what is mathematically wrong about this argument?

答:
The argument is based on the premise that the chance of rain or no rain is equally likely.

There may be places in the world where this is true in general - i.e. of the 365 days in the year it rains on 182.5 of them - somewhere in the tropics perhaps?

However, even if this is the case, there will be seasonal variances to the rainfall pattern i.e. rain is more likely in the monsoon season than the dry season.

Alternatively in the desert of Antarctica it almost never rains. In the Sahara it rains a little more.

In general, the best predictor of rain tomorrow is if it is raining today. In Sydney, Australia you will be right with this about 75% of the time. This would translate to probabilities of (very roughly) $\frac{3}{4}$ if it is raining and $\frac{1}{4}$ if it isn't.


问:
If $W$ is a neighborhood of $0$ in $X$ then there is a neighborhood $U$ of $0$ which is symmetric and which satisfies $U+U\subset W$
What is shown below is a reference from the text Functional Analysis by Walter Rudin.

If $X$ s a vector space over the field $\Bbb F$ then  the following notation well be used
$$
x+A=\{x+a:a\in A\}\\
x-A:=\{x-a:a\in A\}\\
A+B:=\{a+b:a\in A\wedge b\in B\}\\
\lambda A:=\{\lambda a:a\in A\}
$$
for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.

Definition

If $\tau$ is a topology on a vector space $X$ such that

every point of $X$ is a closed set

the vector space operation are continuous with respect to $\tau$

we say that $X$ is a topological vector space equipped with the vector topology $\tau$.

So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.

Proposition

Let be $X$ a topological vector space over the field $\Bbb F$ and for some $v\in X$ and $\lambda\in\Bbb F$ we define the $v$-traslation and $\lambda$-multiplication functions through the condition
$$
T_v(x):=x+v\,\,\,\text{and}\,\,\,M_\lambda(x):=\lambda x
$$
for any $x\in X$. So the above two defined functions are homeomorphisms.

Proof. Omitted.

One consequence of the above proposition is that every vector topology $\tau$ is translation-invariant, that is a set $E\subseteq X$ is open if and only if each of its translates $v+E$ is open and thus $\tau$ is completely determined by any local base so that the open sets of X are then precisely those that are unions of translates of memebrs of any local base.

Lemma

Let be $X$ a topological vector space over the field $F$. So if $W$ is a neighborhood of $0$ in $X$ then there is a neighborhood $U$ of $0$ which is symmetric (that is $U=-U$) and which satisfies $U+U\subseteq W$.

Proof. To see this, note that $0+0=0$ and that addition is continuous and that $0$ therefore has neighborhoods $V_1$ and $V_2$ such that $V_1+V_2\subseteq W$. So if we put
$$
U=V_1\cap V_2\cap (-V_1)\cap(-V_2)
$$
the $U$ has the required properties.

So I ask to explain why effectively $0$ has neighborhoods $V_1$ and $V_2$ such that $V_1+V_2\subseteq W$ and why the neighborhood $U$ above defined has the required properties. So could someone help me, please?

答:
Let $P: X \times X \rightarrow X$ denote the addition. It’s continuous at $(0,0)$ so $P^{-1}(W)$ is a neighborhood of $(0,0)$. In particular, it contains a “basic” open subset (for the product topology) which itself contains $(0,0)$. Such basic open subsets are the $V_1 \times V_2$ where $V_1,V_2$ are open subsets of $X$ with $0 \in V_1 \cap V_2$.
Then, $U=V_1 \cap (-V_1) \cap (V_2) \cap (-V_2)$ is an open subset of $X$ containing $0$, clearly it’s symmetric, and $U+U=P(U \times U) \subset P(V_1 \times V_2) \subset W$, since $U \subset V_1$ and $U \subset V_2$.


问:
Question about Lyapunov stability for a nonlinear system with inputs
I am studying Lyapunov stability for systems of the type:

$\dot{x}=f(x)$

and I think I have well understood the theory.

But, suppose now I have a system with an input, so of the type:

$\dot{x}=f(x)+g(x)u$

Can I still use the Lyapunov analysis for analizing the stability of this system?

答:
No, but there are extensions that allow you to do so.

You can look up input to state stability for example. But you will need some assumptions for $u$, for example that the input is bounded.

For example, the system

$$
\dot{x} = -x^3 + xu
$$

is input to state stable if $|u| < C$ for any finite positive $C$ because as $x$ gets large the stabilizing $-x^3$ "beats" the $xu$.

But the system

$$
\dot{x} = -x^3 + x^3u
$$

is not input to state stable for any bounded input, because with the input $u = 2$ you get

$$
\dot{x} = x^3
$$

which is unstable.

This is of course not very rigorous but there exists a rigorous theory and a lot of literature for this topic. However it gets quickly much more complicated if you have more complicated systems.


问:
Succinct notation for finite sets
I have sometimes somewhere see a very concise way to express sets of finite size whose elements are really not important, only that they are different, but I'm not sure what of the following variants it was (or maybe there is even better notations):

$$
S=\{s_i\}\\
S=\{s\}_i\\
S=\{s_i\}_1^n
$$

I hope it's the first notation $S=\{s_i\}$, because you give the same information as in the other methods but write much less; you express the index you will use whenever you refer to elements in $S$, you don't need to iterate from $1$ to $n$ because in maths, indexes usually start at $1$, and the last index, or number of elements, can always be written as $\#S$ or $|S|$, so taking the symbol $n$ is unnecesary.

As a related question, when you want to express set of pairs, what you would write, assuming one of the above notations are correct?

$$
S=\{(a, b)_i\}, \text{ or}\\
S=\{\{a, b\}_i\}, \text{ or even}\\
S=\{a, b\}_i
$$

答:
Without context, I would interpret your first three definitions of $S$ as:

A set consisting of a single element called $s_i$,

A set of elements $s$, all identical, indexed by $i$ which is probably, but not definitely, infinite, and

A finite set of $n$ elements $s_1, s_2, \ldots, s_n$ which may be all different or may contain repeats.

In cases $1$ and $2$ I would look for further context nearby where the symbols are introduced and worry if I couldn't find it; for $3$ I would feel I already understand from what is written.

The succinct form of the set is $S$, in fact, so it is important to define what you mean when $S$ is introduced, and then you can refer back to just $S$ whenever you need it.  To specify what you're after: a finite set of distinct elements, I would suggest:
$$ S := \{ s_i \mid s_i \not= s_j, 1\leq i,j \leq N \}$$
where we assume that the reader is not about to assume that $s_1 \not= s_1$ just to be difficult.

If you need to refer to several sets of this kind, I would set that up as a notational paragraph at the start; something like:

We define $S_N, T_N$, etc. to be any set of $N$ distinct elements from an alphabet/universe $\cal U$, specifically
$$ S_N := \{ s_i \mid s_i \not= s_j, 1\leq i,j \leq N \}$$

If using the notation you've described is really important to you, then again, define what you mean by it and then probably use $\{s_i\}_{i \leq N}$ as the short form.

Title: Succinct notation for finite sets


问:
Sitting younger, older alternately.
In a group of eight friends, everyone is of different age. Everyone decided to take a seat by an eight-person round table. The youngest of them sat at seat 1. Later they sat: older, younger, older, younger, etc. [seat 2 is older than seat 1, seat 3 is younger than seat 2, ..., seat 1 is younger than seat 8] How many ways can they sit?

Actually, I feel this is the hardest of all the problems I have posted recently.

My first fhought was that you have to split them evenly into the "younger" and "older" groups, but you can find solutions that satisfy the requirements but don't follow this rule.

What is true though, is you cannot the currently youngest person as the "older" one or currently olders person as the "younger" one.
Since it is an archival local math contest, I know the answer is supposed to be 272.

Title: Sitting younger, older alternately.

Body:
Here is an alternative approach which does not require identifying this problem as "calculating $A_7$" and then either copying its value from somewhere or proving any of the recursive formulas for up-down numbers on one's own: List all of the possible ways of the people sitting down explicitly:

Let's enumerate people as $1,2,3,4,5,6,7,8$, going from youngest to oldest. Then what we want is to find sequences $a_1a_2\cdots a_8$ with the following properties:

$a_1=1$

$a_j>a_{j-1}$ if $j$ is even

$a_j<a_{j-1}$ if $j$ is odd.

Such a sequence gives us a way of people sitting down: $a_j$ sits at the $j$-th place.

The following Octave code generates all such sequences in a file "list.txt":

P=perms([2 3 4 5 6 7 8]);

text='';

for i=1:size(P)(1)
    if P(i,1)>P(i,2) && P(i,2)<P(i,3) && P(i,3)>P(i,4) && P(i,4)<P(i,5) && P(i,5)>P(i,6) && P(i,6)<P(i,7)
        text=[text sprintf('\n') int2str(1)];
        for j=1:7
            text=[text ', ' int2str(P(i,j))];
        endfor
    endif
endfor

filetowrite=fopen(['list.txt'],'w');
fprintf(filetowrite,text);

fclose(filetowrite);

which are

1, 8, 6, 7, 4, 5, 2, 3
1, 8, 6, 7, 3, 5, 2, 4
1, 8, 6, 7, 3, 4, 2, 5
1, 8, 6, 7, 2, 5, 3, 4
1, 8, 6, 7, 2, 4, 3, 5
1, 8, 5, 7, 4, 6, 2, 3
1, 8, 5, 7, 3, 6, 2, 4
1, 8, 5, 7, 3, 4, 2, 6
1, 8, 5, 7, 2, 6, 3, 4
1, 8, 5, 7, 2, 4, 3, 6
1, 8, 5, 6, 4, 7, 2, 3
1, 8, 5, 6, 3, 7, 2, 4
1, 8, 5, 6, 3, 4, 2, 7
1, 8, 5, 6, 2, 7, 3, 4
1, 8, 5, 6, 2, 4, 3, 7
1, 8, 4, 7, 5, 6, 2, 3
1, 8, 4, 7, 3, 6, 2, 5
1, 8, 4, 7, 3, 5, 2, 6
1, 8, 4, 7, 2, 6, 3, 5
1, 8, 4, 7, 2, 5, 3, 6
1, 8, 4, 6, 5, 7, 2, 3
1, 8, 4, 6, 3, 7, 2, 5
1, 8, 4, 6, 3, 5, 2, 7
1, 8, 4, 6, 2, 7, 3, 5
1, 8, 4, 6, 2, 5, 3, 7
1, 8, 4, 5, 3, 7, 2, 6
1, 8, 4, 5, 3, 6, 2, 7
1, 8, 4, 5, 2, 7, 3, 6
1, 8, 4, 5, 2, 6, 3, 7
1, 8, 3, 7, 5, 6, 2, 4
1, 8, 3, 7, 4, 6, 2, 5
1, 8, 3, 7, 4, 5, 2, 6
1, 8, 3, 7, 2, 6, 4, 5
1, 8, 3, 7, 2, 5, 4, 6
1, 8, 3, 6, 5, 7, 2, 4
1, 8, 3, 6, 4, 7, 2, 5
1, 8, 3, 6, 4, 5, 2, 7
1, 8, 3, 6, 2, 7, 4, 5
1, 8, 3, 6, 2, 5, 4, 7
1, 8, 3, 5, 4, 7, 2, 6
1, 8, 3, 5, 4, 6, 2, 7
1, 8, 3, 5, 2, 7, 4, 6
1, 8, 3, 5, 2, 6, 4, 7
1, 8, 3, 4, 2, 7, 5, 6
1, 8, 3, 4, 2, 6, 5, 7
1, 8, 2, 7, 5, 6, 3, 4
1, 8, 2, 7, 4, 6, 3, 5
1, 8, 2, 7, 4, 5, 3, 6
1, 8, 2, 7, 3, 6, 4, 5
1, 8, 2, 7, 3, 5, 4, 6
1, 8, 2, 6, 5, 7, 3, 4
1, 8, 2, 6, 4, 7, 3, 5
1, 8, 2, 6, 4, 5, 3, 7
1, 8, 2, 6, 3, 7, 4, 5
1, 8, 2, 6, 3, 5, 4, 7
1, 8, 2, 5, 4, 7, 3, 6
1, 8, 2, 5, 4, 6, 3, 7
1, 8, 2, 5, 3, 7, 4, 6
1, 8, 2, 5, 3, 6, 4, 7
1, 8, 2, 4, 3, 7, 5, 6
1, 8, 2, 4, 3, 6, 5, 7
1, 7, 6, 8, 4, 5, 2, 3
1, 7, 6, 8, 3, 5, 2, 4
1, 7, 6, 8, 3, 4, 2, 5
1, 7, 6, 8, 2, 5, 3, 4
1, 7, 6, 8, 2, 4, 3, 5
1, 7, 5, 8, 4, 6, 2, 3
1, 7, 5, 8, 3, 6, 2, 4
1, 7, 5, 8, 3, 4, 2, 6
1, 7, 5, 8, 2, 6, 3, 4
1, 7, 5, 8, 2, 4, 3, 6
1, 7, 5, 6, 4, 8, 2, 3
1, 7, 5, 6, 3, 8, 2, 4
1, 7, 5, 6, 3, 4, 2, 8
1, 7, 5, 6, 2, 8, 3, 4
1, 7, 5, 6, 2, 4, 3, 8
1, 7, 4, 8, 5, 6, 2, 3
1, 7, 4, 8, 3, 6, 2, 5
1, 7, 4, 8, 3, 5, 2, 6
1, 7, 4, 8, 2, 6, 3, 5
1, 7, 4, 8, 2, 5, 3, 6
1, 7, 4, 6, 5, 8, 2, 3
1, 7, 4, 6, 3, 8, 2, 5
1, 7, 4, 6, 3, 5, 2, 8
1, 7, 4, 6, 2, 8, 3, 5
1, 7, 4, 6, 2, 5, 3, 8
1, 7, 4, 5, 3, 8, 2, 6
1, 7, 4, 5, 3, 6, 2, 8
1, 7, 4, 5, 2, 8, 3, 6
1, 7, 4, 5, 2, 6, 3, 8
1, 7, 3, 8, 5, 6, 2, 4
1, 7, 3, 8, 4, 6, 2, 5
1, 7, 3, 8, 4, 5, 2, 6
1, 7, 3, 8, 2, 6, 4, 5
1, 7, 3, 8, 2, 5, 4, 6
1, 7, 3, 6, 5, 8, 2, 4
1, 7, 3, 6, 4, 8, 2, 5
1, 7, 3, 6, 4, 5, 2, 8
1, 7, 3, 6, 2, 8, 4, 5
1, 7, 3, 6, 2, 5, 4, 8
1, 7, 3, 5, 4, 8, 2, 6
1, 7, 3, 5, 4, 6, 2, 8
1, 7, 3, 5, 2, 8, 4, 6
1, 7, 3, 5, 2, 6, 4, 8
1, 7, 3, 4, 2, 8, 5, 6
1, 7, 3, 4, 2, 6, 5, 8
1, 7, 2, 8, 5, 6, 3, 4
1, 7, 2, 8, 4, 6, 3, 5
1, 7, 2, 8, 4, 5, 3, 6
1, 7, 2, 8, 3, 6, 4, 5
1, 7, 2, 8, 3, 5, 4, 6
1, 7, 2, 6, 5, 8, 3, 4
1, 7, 2, 6, 4, 8, 3, 5
1, 7, 2, 6, 4, 5, 3, 8
1, 7, 2, 6, 3, 8, 4, 5
1, 7, 2, 6, 3, 5, 4, 8
1, 7, 2, 5, 4, 8, 3, 6
1, 7, 2, 5, 4, 6, 3, 8
1, 7, 2, 5, 3, 8, 4, 6
1, 7, 2, 5, 3, 6, 4, 8
1, 7, 2, 4, 3, 8, 5, 6
1, 7, 2, 4, 3, 6, 5, 8
1, 6, 5, 8, 4, 7, 2, 3
1, 6, 5, 8, 3, 7, 2, 4
1, 6, 5, 8, 3, 4, 2, 7
1, 6, 5, 8, 2, 7, 3, 4
1, 6, 5, 8, 2, 4, 3, 7
1, 6, 5, 7, 4, 8, 2, 3
1, 6, 5, 7, 3, 8, 2, 4
1, 6, 5, 7, 3, 4, 2, 8
1, 6, 5, 7, 2, 8, 3, 4
1, 6, 5, 7, 2, 4, 3, 8
1, 6, 4, 8, 5, 7, 2, 3
1, 6, 4, 8, 3, 7, 2, 5
1, 6, 4, 8, 3, 5, 2, 7
1, 6, 4, 8, 2, 7, 3, 5
1, 6, 4, 8, 2, 5, 3, 7
1, 6, 4, 7, 5, 8, 2, 3
1, 6, 4, 7, 3, 8, 2, 5
1, 6, 4, 7, 3, 5, 2, 8
1, 6, 4, 7, 2, 8, 3, 5
1, 6, 4, 7, 2, 5, 3, 8
1, 6, 4, 5, 3, 8, 2, 7
1, 6, 4, 5, 3, 7, 2, 8
1, 6, 4, 5, 2, 8, 3, 7
1, 6, 4, 5, 2, 7, 3, 8
1, 6, 3, 8, 5, 7, 2, 4
1, 6, 3, 8, 4, 7, 2, 5
1, 6, 3, 8, 4, 5, 2, 7
1, 6, 3, 8, 2, 7, 4, 5
1, 6, 3, 8, 2, 5, 4, 7
1, 6, 3, 7, 5, 8, 2, 4
1, 6, 3, 7, 4, 8, 2, 5
1, 6, 3, 7, 4, 5, 2, 8
1, 6, 3, 7, 2, 8, 4, 5
1, 6, 3, 7, 2, 5, 4, 8
1, 6, 3, 5, 4, 8, 2, 7
1, 6, 3, 5, 4, 7, 2, 8
1, 6, 3, 5, 2, 8, 4, 7
1, 6, 3, 5, 2, 7, 4, 8
1, 6, 3, 4, 2, 8, 5, 7
1, 6, 3, 4, 2, 7, 5, 8
1, 6, 2, 8, 5, 7, 3, 4
1, 6, 2, 8, 4, 7, 3, 5
1, 6, 2, 8, 4, 5, 3, 7
1, 6, 2, 8, 3, 7, 4, 5
1, 6, 2, 8, 3, 5, 4, 7
1, 6, 2, 7, 5, 8, 3, 4
1, 6, 2, 7, 4, 8, 3, 5
1, 6, 2, 7, 4, 5, 3, 8
1, 6, 2, 7, 3, 8, 4, 5
1, 6, 2, 7, 3, 5, 4, 8
1, 6, 2, 5, 4, 8, 3, 7
1, 6, 2, 5, 4, 7, 3, 8
1, 6, 2, 5, 3, 8, 4, 7
1, 6, 2, 5, 3, 7, 4, 8
1, 6, 2, 4, 3, 8, 5, 7
1, 6, 2, 4, 3, 7, 5, 8
1, 5, 4, 8, 6, 7, 2, 3
1, 5, 4, 8, 3, 7, 2, 6
1, 5, 4, 8, 3, 6, 2, 7
1, 5, 4, 8, 2, 7, 3, 6
1, 5, 4, 8, 2, 6, 3, 7
1, 5, 4, 7, 6, 8, 2, 3
1, 5, 4, 7, 3, 8, 2, 6
1, 5, 4, 7, 3, 6, 2, 8
1, 5, 4, 7, 2, 8, 3, 6
1, 5, 4, 7, 2, 6, 3, 8
1, 5, 4, 6, 3, 8, 2, 7
1, 5, 4, 6, 3, 7, 2, 8
1, 5, 4, 6, 2, 8, 3, 7
1, 5, 4, 6, 2, 7, 3, 8
1, 5, 3, 8, 6, 7, 2, 4
1, 5, 3, 8, 4, 7, 2, 6
1, 5, 3, 8, 4, 6, 2, 7
1, 5, 3, 8, 2, 7, 4, 6
1, 5, 3, 8, 2, 6, 4, 7
1, 5, 3, 7, 6, 8, 2, 4
1, 5, 3, 7, 4, 8, 2, 6
1, 5, 3, 7, 4, 6, 2, 8
1, 5, 3, 7, 2, 8, 4, 6
1, 5, 3, 7, 2, 6, 4, 8
1, 5, 3, 6, 4, 8, 2, 7
1, 5, 3, 6, 4, 7, 2, 8
1, 5, 3, 6, 2, 8, 4, 7
1, 5, 3, 6, 2, 7, 4, 8
1, 5, 3, 4, 2, 8, 6, 7
1, 5, 3, 4, 2, 7, 6, 8
1, 5, 2, 8, 6, 7, 3, 4
1, 5, 2, 8, 4, 7, 3, 6
1, 5, 2, 8, 4, 6, 3, 7
1, 5, 2, 8, 3, 7, 4, 6
1, 5, 2, 8, 3, 6, 4, 7
1, 5, 2, 7, 6, 8, 3, 4
1, 5, 2, 7, 4, 8, 3, 6
1, 5, 2, 7, 4, 6, 3, 8
1, 5, 2, 7, 3, 8, 4, 6
1, 5, 2, 7, 3, 6, 4, 8
1, 5, 2, 6, 4, 8, 3, 7
1, 5, 2, 6, 4, 7, 3, 8
1, 5, 2, 6, 3, 8, 4, 7
1, 5, 2, 6, 3, 7, 4, 8
1, 5, 2, 4, 3, 8, 6, 7
1, 5, 2, 4, 3, 7, 6, 8
1, 4, 3, 8, 6, 7, 2, 5
1, 4, 3, 8, 5, 7, 2, 6
1, 4, 3, 8, 5, 6, 2, 7
1, 4, 3, 8, 2, 7, 5, 6
1, 4, 3, 8, 2, 6, 5, 7
1, 4, 3, 7, 6, 8, 2, 5
1, 4, 3, 7, 5, 8, 2, 6
1, 4, 3, 7, 5, 6, 2, 8
1, 4, 3, 7, 2, 8, 5, 6
1, 4, 3, 7, 2, 6, 5, 8
1, 4, 3, 6, 5, 8, 2, 7
1, 4, 3, 6, 5, 7, 2, 8
1, 4, 3, 6, 2, 8, 5, 7
1, 4, 3, 6, 2, 7, 5, 8
1, 4, 3, 5, 2, 8, 6, 7
1, 4, 3, 5, 2, 7, 6, 8
1, 4, 2, 8, 6, 7, 3, 5
1, 4, 2, 8, 5, 7, 3, 6
1, 4, 2, 8, 5, 6, 3, 7
1, 4, 2, 8, 3, 7, 5, 6
1, 4, 2, 8, 3, 6, 5, 7
1, 4, 2, 7, 6, 8, 3, 5
1, 4, 2, 7, 5, 8, 3, 6
1, 4, 2, 7, 5, 6, 3, 8
1, 4, 2, 7, 3, 8, 5, 6
1, 4, 2, 7, 3, 6, 5, 8
1, 4, 2, 6, 5, 8, 3, 7
1, 4, 2, 6, 5, 7, 3, 8
1, 4, 2, 6, 3, 8, 5, 7
1, 4, 2, 6, 3, 7, 5, 8
1, 4, 2, 5, 3, 8, 6, 7
1, 4, 2, 5, 3, 7, 6, 8
1, 3, 2, 8, 6, 7, 4, 5
1, 3, 2, 8, 5, 7, 4, 6
1, 3, 2, 8, 5, 6, 4, 7
1, 3, 2, 8, 4, 7, 5, 6
1, 3, 2, 8, 4, 6, 5, 7
1, 3, 2, 7, 6, 8, 4, 5
1, 3, 2, 7, 5, 8, 4, 6
1, 3, 2, 7, 5, 6, 4, 8
1, 3, 2, 7, 4, 8, 5, 6
1, 3, 2, 7, 4, 6, 5, 8
1, 3, 2, 6, 5, 8, 4, 7
1, 3, 2, 6, 5, 7, 4, 8
1, 3, 2, 6, 4, 8, 5, 7
1, 3, 2, 6, 4, 7, 5, 8
1, 3, 2, 5, 4, 8, 6, 7
1, 3, 2, 5, 4, 7, 6, 8

and there are 272 of them.


问:
Definition/statistics/calculations for "evenly spaced numbers"?
It seems like this ought to have a name like "smoothness" or something similar, but smoothness relates to something entirely different, and I can't find terminology that applies to the following problem:

Consider a set $A$ containing 50 unique integers, with $1 \leq (a \in A) \leq 100$.

If the set contained all the even (or odd) integers between $1$ and $100$, then a reasonable observer would call the numbers evenly spaced within the interval. On the other hand, if the set were all the integers between $51$ and $100$ (or $1$ to $50$), this would be the worst possible "even spacing" on the interval.

Determining "perfect" even spacing for sets with cardinalities not a divisor of the interval is more difficult. How, for instance, do we space out a set of $31$ integers? It looks like we need $24$ gaps of $3$ and $7$ gaps of $4$--but how are those gaps spaced?

We can generalize this to any set of integers and any length of interval. My suspicion/guess is that we could define the "perfect spacing" as, say, the nearest-integer of the zeroes of

$$f(x) = 1 - \cos(\frac{|A| \tau}{L}x)$$

where $|A|$ is the cardinality of $A$ and $L$ is the length of the interval. But then, given a set of integers that aren't "perfectly spaced," how do we determine how close they are to evenly-spaced?

I imagine it's possible to map "how close to evenly spaced" to a value between $0$ and $1$. I also imagine I'm not the first to ask this sort of question. Is there a proper terminology for these sorts of questions? Is there a specific distribution I should be looking at?

答:
I think @DavidDStork's comment is an idea worth considering; I would use the standard deviation instead of the variance. [See the note at the end for another possibility.]

If you have only odd numbers this 'index of evenness'
will be zero. For less 'even' sequences this measure will be greater. [Computations in R.]

```<code>x = seq(1, 100, by= 2);  x
 [1]  1  3  5  7  9 11 13 15 17 19 21 23 25
[14] 27 29 31 33 35 37 39 41 43 45 47 49 51
[27] 53 55 57 59 61 63 65 67 69 71 73 75 77
[40] 79 81 83 85 87 89 91 93 95 97 99
diff(x)
 [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[21] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[41] 2 2 2 2 2 2 2 2 2
sd(diff(x))
[1] 0
</code>```

If the 50 numbers are chosen at random without replacement,
you might wonder about the distribution of this measure.
How small does the measure have to be in order for the
sequence to be called 'relatively even'?

The following simulation shows the distribution for
100,000 such random choices:

```<code>set.seed(2020)
sd.dif = replicate(10^5, sd(diff(sort(sample(1:100, 50)))))
summary(sd.dif)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.8165  1.2410  1.3534  1.3680  1.4777  2.9226

hist(sd,dif, prob=T, col="skyblue2")
</code>```

.

So if you get an 'index of evenness' below  $1$ you
can be pretty sure the sequence of 50 wasn't randomly
chosen.

Here are three randomly generated sequences and their
indexes of smoothness:

```<code>a = sort(sample(1:100, 50)); a
 [1]   1   2   4   5   8   9  13  17  19  22
[11]  26  27  28  29  30  32  33  34  35  40
[21]  44  49  50  52  53  57  60  61  63  64
[31]  68  70  71  72  74  78  80  82  84  85
[41]  86  87  89  93  94  96  97  98  99 100
sd(diff(a))
[1] 1.266389

b = sort(sample(1:100, 50)); b
 [1]   1   2   3   4   5  10  13  14  15  16
[11]  17  21  25  27  28  33  34  35  36  38
[21]  41  42  43  44  48  55  57  59  60  62
[31]  63  64  67  68  69  70  71  72  75  76
[41]  79  84  86  92  93  94  95  96  99 100
sd(diff(b))
[1] 1.547711

c = sort(sample(1:100, 50)); c
 [1]   1   4   5   6   9  10  15  17  19  21
[11]  22  24  25  27  29  32  33  36  38  39
[21]  40  42  43  44  47  48  50  52  56  57
[31]  60  63  64  65  68  69  73  75  79  80
[41]  81  82  84  85  86  96  97  98  99 100
sd(diff(c))
[1] 1.561113
</code>```

By Stork's criterion, sequence a is 'smoother' than c. If we plot the sequences
below, then the diagonal line is smoother for a. [Note the big jump in the plot of c.]

```<code>par(mfrow=c(1,2))
 plot(a, pch=20); abline(0, 2, col="green2")
 plot(c, pch=20); abline(0, 2, col="green2")
par(mfrow=c(1,2))
</code>```

Note: The plots may show another useful way to define
'smoothness'. How nearly to they match a straight line?
One way to answer that is by looking at the correlation
of the sequence with the vector of numbers from 1 through 50. A correlation nearer to $1$ indicates better fit to
a straight line. By this criterion, c is smoother than
a. [Note the wobble in the plot of a.]

So you need to think about which criterion best matches
your personal view of 'smoothness'.

```<code>cor(1:50, a)
[1] 0.9974012
cor(1:50, c)
[1] 0.9980791
</code>```

Body:
I think @DavidDStork's comment is an idea worth considering; I would use the standard deviation instead of the variance. [See the note at the end for another possibility.]

If you have only odd numbers this 'index of evenness'
will be zero. For less 'even' sequences this measure will be greater. [Computations in R.]

x = seq(1, 100, by= 2);  x
 [1]  1  3  5  7  9 11 13 15 17 19 21 23 25
[14] 27 29 31 33 35 37 39 41 43 45 47 49 51
[27] 53 55 57 59 61 63 65 67 69 71 73 75 77
[40] 79 81 83 85 87 89 91 93 95 97 99
diff(x)
 [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[21] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[41] 2 2 2 2 2 2 2 2 2
sd(diff(x))
[1] 0

If the 50 numbers are chosen at random without replacement,
you might wonder about the distribution of this measure.
How small does the measure have to be in order for the
sequence to be called 'relatively even'?

The following simulation shows the distribution for
100,000 such random choices:

set.seed(2020)
sd.dif = replicate(10^5, sd(diff(sort(sample(1:100, 50)))))
summary(sd.dif)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.8165  1.2410  1.3534  1.3680  1.4777  2.9226

hist(sd,dif, prob=T, col="skyblue2")

.

So if you get an 'index of evenness' below  $1$ you
can be pretty sure the sequence of 50 wasn't randomly
chosen.

Here are three randomly generated sequences and their
indexes of smoothness:

a = sort(sample(1:100, 50)); a
 [1]   1   2   4   5   8   9  13  17  19  22
[11]  26  27  28  29  30  32  33  34  35  40
[21]  44  49  50  52  53  57  60  61  63  64
[31]  68  70  71  72  74  78  80  82  84  85
[41]  86  87  89  93  94  96  97  98  99 100
sd(diff(a))
[1] 1.266389

b = sort(sample(1:100, 50)); b
 [1]   1   2   3   4   5  10  13  14  15  16
[11]  17  21  25  27  28  33  34  35  36  38
[21]  41  42  43  44  48  55  57  59  60  62
[31]  63  64  67  68  69  70  71  72  75  76
[41]  79  84  86  92  93  94  95  96  99 100
sd(diff(b))
[1] 1.547711

c = sort(sample(1:100, 50)); c
 [1]   1   4   5   6   9  10  15  17  19  21
[11]  22  24  25  27  29  32  33  36  38  39
[21]  40  42  43  44  47  48  50  52  56  57
[31]  60  63  64  65  68  69  73  75  79  80
[41]  81  82  84  85  86  96  97  98  99 100
sd(diff(c))
[1] 1.561113

By Stork's criterion, sequence a is 'smoother' than c. If we plot the sequences
below, then the diagonal line is smoother for a. [Note the big jump in the plot of c.]

par(mfrow=c(1,2))
 plot(a, pch=20); abline(0, 2, col="green2")
 plot(c, pch=20); abline(0, 2, col="green2")
par(mfrow=c(1,2))

Note: The plots may show another useful way to define
'smoothness'. How nearly to they match a straight line?
One way to answer that is by looking at the correlation
of the sequence with the vector of numbers from 1 through 50. A correlation nearer to $1$ indicates better fit to
a straight line. By this criterion, c is smoother than
a. [Note the wobble in the plot of a.]

So you need to think about which criterion best matches
your personal view of 'smoothness'.

cor(1:50, a)
[1] 0.9974012
cor(1:50, c)
[1] 0.9980791


问:
Multivariable Chain Rule Formula doesn't make sense to me
Consider $z=f(x(t),y(t))$, then its chain rule formula is:

$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$

What I cannot make sense of, are the terms $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

$\frac{\partial f}{\partial x}$ is defined as the rate of change of $f$ with respect to $x$, when $y$ is held constant.

Now, in our original definition, both $x$ and $y$ depend on $t$. So if we hold $y$ constant, we'll have to hold $t$ constant, and thus $x$ will automatically be held constant. How, then, can we compute $\frac{\partial f}{\partial x}$ if we cannot hold only $y$ as constant?

答:
What is confusing you is that the letters $x$ and $y$ are used for two purposes: (a) As names of the functions $x(t), y(t)$; (b) As names of the "first argument" and "second argument" of $f$. Sadly, this notation is very usual in mathematics, and there is no suitable alternative.

If I could convince to write partial derivatives differently, e.g. instead of $\frac{\partial f}{\partial x}$ you write $f_1'$ ("derivative on the first parameter"), and instead of $\frac{\partial f}{\partial y}$ you write $f_2'$ ("derivative on the second parameter"), then the formula becomes a lot more palatable:

$$z'(t)=\frac{d}{dt}f(x(t),y(t))=f_1'(x(t),y(t))x'(t)+f_2'(x(t),y(t))y'(t)$$

(I also write $x'(t)=\frac{dx}{dt}, y'(t)=\frac{dy}{dt}, z'(t)=\frac{dz}{dt}$, if that helps.)

Now it is obvious we are in no way modifying $x(t)$ or $y(t)$ when calculating $f_1'$ and $f_2'$. Instead, we are keeping constant the second argument of $f$/the first argument of $f$, respectively. After having calculated $f_1'$ and $f_2'$ - both end up as functions of two arguments - you will substitute those arguments with $x(t), y(t)$.

Sadly, this notation I used above is completely non-standard, and for historical reasons we are stuck with the notation that is confusing for you.


答:
Maybe an example would help. Say we take $z=x^2y$ where as you said $x,y$ are functions of $t$ so we will define them as $x=t^2$ and $y=t^3$.

Our formula is:

$$\frac{dz}{dt}=\frac{\partial z}{\partial x} \frac{dx}{dt}+\frac{\partial z}{\partial y} \frac{dy}{dt}$$

Which is then:

$$\frac{dz}{dt}=(2xy)(2t)+(x^2)(3t^2)$$

Subbing for $x,y$:

$$\frac{dz}{dt}=(2t^5)(2t)+(t^4)(3t^2)$$

$$\frac{dz}{dt}=7t^6$$

I hope this makes sense. I think the way you are getting confused is with the $f$ and $z$ and what you are differentiating with respect to. If you follow this example closely you should see!

Title: Multivariable Chain Rule Formula doesn't make sense to me

Body:
Consider $z=f(x(t),y(t))$, then its chain rule formula is:

$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$

What I cannot make sense of, are the terms $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

$\frac{\partial f}{\partial x}$ is defined as the rate of change of $f$ with respect to $x$, when $y$ is held constant.

Now, in our original definition, both $x$ and $y$ depend on $t$. So if we hold $y$ constant, we'll have to hold $t$ constant, and thus $x$ will automatically be held constant. How, then, can we compute $\frac{\partial f}{\partial x}$ if we cannot hold only $y$ as constant?


问:
Fixed point result for a differentiable function.
Let $f$ be a differentiable function on $[0, 1]$ with $0$ and $1$ be its fixed points. Also, it is given that $$f'(0)>1, f'(1)>1.$$ Then prove that

(i) $f$ has at least one fixed point in $(0, 1)$.

(ii) $f$ has a fixed point $\xi\in (0, 1)$ such that $f'(\xi)\leq 1$.

My effort: I started by considering $g(x)=f(x)-x$. Clearly, $g$ is a differentiable function which means $g'(x)=f'(x)-1$. This means $g'(0)>0, g'(1)>0$. But I am not able to conclude anything. Please help.

答: Use first the definition of $g'(0)$ and $g'(1)$ to show that $g$ assumes a strictly positive and a strictly negative value on $(0,1)$.


答:
You also have $g(0) = 0$ and $g(1) = 0$ because $0, 1$ are fixed points of $f$. Now, because $g'(0) > 0$, you must have $g(x_1) > 0$ for some $x_1 \in (0, 1)$. Otherwise, for a contradiction, if $g(x) \leq 0$ were instead true for all $x \in (0, 1)$, then $g(x) - g(0) = g(x) - 0 \leq 0$ always. Hence, the fraction
$$
\frac{g(x) - g(0)}{x - 0} \leq 0, \ \forall x \in (0, 1) \implies g'(0) = \lim_{x \to 0, x \in (0, 1)}\frac{g(x) - g(0)}{x - 0} \leq 0
$$ contrary to $g'(0) > 0$.

For a similar reason, there must also be some $x_2 \in (0, 1)$ such that $g(x_2) < 0$. Obviously $x_1 \neq x_2$. So now, you can conclude via IVT.

Title: Fixed point result for a differentiable function.

Body:
Let $f$ be a differentiable function on $[0, 1]$ with $0$ and $1$ be its fixed points. Also, it is given that $$f'(0)>1, f'(1)>1.$$ Then prove that

(i) $f$ has at least one fixed point in $(0, 1)$.

(ii) $f$ has a fixed point $\xi\in (0, 1)$ such that $f'(\xi)\leq 1$.

My effort: I started by considering $g(x)=f(x)-x$. Clearly, $g$ is a differentiable function which means $g'(x)=f'(x)-1$. This means $g'(0)>0, g'(1)>0$. But I am not able to conclude anything. Please help.


答:
From the definition of derivative $$g'(0) = \lim_{h\to 0^+} \frac{g(h) - g(0)}{h} = \lim_{h\to 0^+} \frac{g(h)-0}{h} > 0$$ Since $h$ goes to $0$ from the positive side, $g(x)$ must be positive $(*)$  in some right neighbourhood of $0$ to have a positive limit.
$$g'(1) = \lim_{h\to0^-} \frac{g(1+h) - g(1)}{h} = \lim_{h\to0^-} \frac{g(1+h) -0}{h} > 0$$
Since $h$ goes to $0$ from the negative side $g(1+h) < 0$, which means $g(x)$ must be negative $(**)$ in some neighbourhood of $1$ to have a positive limit. Hence, combining $(*)$ and $(**)$ there exists $c \in (0,1)$ such that $g(c) = f(c)-c=0$


问:
Can we write $a \leq a$
I am proving some inequality and one of the steps i took is that i wrote $n+1 \leq n+1$ is it correct mathematically to write such thing ?

Note :
i was try to prove that $2^{n-1} \leq n! \leq n^{n} \space \space \forall n \in \mathbb{N^*}...P_n $

so i did it by induction and i got to $2^{n-1} \leq n! \leq (n+1)^n$

then i wrote that $2 \leq n+1 \leq n+1$ and i multiplied the inequalities side by side to get to $P_{n+1}$

答:
$a\leqslant b$ means $a <b$ or $a=b$.

So $a\leqslant a$.

Title: Can we write $a \leq a$


答:
Well, the ''less than equal'' relation on $\Bbb N_0$ is defined as

$a\leq b \;:\Longleftrightarrow \;\exists c\in\Bbb N_0: a+c=b.$

So $a\leq a$ is satisfied for $c=0$.

Title: Can we write $a \leq a$

Body:
I am proving some inequality and one of the steps i took is that i wrote $n+1 \leq n+1$ is it correct mathematically to write such thing ?

Note :
i was try to prove that $2^{n-1} \leq n! \leq n^{n} \space \space \forall n \in \mathbb{N^*}...P_n $

so i did it by induction and i got to $2^{n-1} \leq n! \leq (n+1)^n$

then i wrote that $2 \leq n+1 \leq n+1$ and i multiplied the inequalities side by side to get to $P_{n+1}$


答: When we write $a\leq b$ we mean that $a$ is either equal or greater than $b$. Every number is obviously equal to himself, so $a\leq a$ for every number $a$.

Title: Can we write $a \leq a$


答:
Yes, $a\leq a$ is correct.

$x\leq y$ means $x$ is not bigger than $y$.

Title: Can we write $a \leq a$


问:
In what sense does Homotopy type theory "model types as spaces"
I've read at multiple points (e.g. here) that homotopy type theory "models" types as spaces. I can understand informally that we can "think of" types as spaces, in a vague sense. But what is formally meant by a type theory "modelling" something?

When I was taught a course on formal logic (using first order logic and a form of natural deduction) using the book "Mathematical logic by Ebbinghaus, Flum, Thomas", there was a formal semantics of logical propositions, in terms of structures and interpretations. An "interpretation of a first-order logical formula" was formally defined as a set, with mathematical functions corresponding to the symbols classified as "function symbols", similarly for mathematical relations and constants. i.e. "interpretation" has a formal definition (which can be found in that book).

Is there similarly a formal definition of "interpretation of a type theory" written down somewhere?

答:
Type theories are algebraic theories. This means that for every type theory (there are multiple variants of type theory, with different type formers and constructions), a model of the theory is a particular collection of sets and operations, satisfying some equations. The initial (freely generated) model of a type theory is called the syntax of the type theory. Initiality entails that for any model, there is a morphism (a family of structure-preserving interpretation functions) from the syntax to the model.

The situation is exactly the same as with other algebraic structures; the difference is that models of type theories are significantly larger and more complicated gadgets than well-known algebraic structures.

For example, let's compare groups and models of a type theory.

A group is a set together with an identity element and an associative, unital and invertible binary function.

A model of the type theory is a category with a terminal object (objects represent typing contexts, the terminal object the empty context and morphisms parallel substitutions), together with "family" structure representing types and terms, and data for specific type formers. There are multiple equivalent ways to specify all of this, for example categories with families, comprehension categories or natural models. All of these are multi-sorted, since already in a category there are multiple sets (for objects and arrows).

The initial group is the trivial group. There is a unique group homomorphism from the trivial group to any group. A homomorphism is a function between underlying sets which preserves all structure, e.g. it maps the identity element to the identity element.

The initial model of the type theory is the syntax. There is a unique model morphism from the syntax to any model, which consists of interpretation functions, mapping syntactic contexts to contexts in the model, syntactic types to types in the model and so on, and these functions preserve all structure. For example, the interpretation function maps the syntactic empty context to the semantic empty context, and syntactic type/term formers to corresponding semantic type/term formers.

There are different formal classes of algebraic theories, based on how flexible/general the specification is. For example, we may talk about single-sorted or multi-sorted algebraic structures. Type theories require a fairly general notion of algebraic signatures to describe. There are again multiple ways to specify such signatures. All of the following systems are sufficient to specify type theories:

Essentially algebraic theories

Generalized algebraic theories

Quotient inductive-inductive types

Representable map categories

I remark that the notion of model that falls out from specifying theories in the algebraic fashion, is much more general than the conventional notion of model in the study of first-order logical theories. In logic, only the extra-logical structure can vary over models (interpretations of sorts, operations and axioms), the interpretation of logical structure (variables, quantifiers, logical connectives, sometimes the equality symbol as well) is fixed. In contrast, the algebraic specification allows every syntactic construction to be interpreted differently in different models.

It is possible to use algebraic model theory for first-order logic as well. Hyperdoctrines are one way to do this. It's also possible to specify the algebraic syntax of first-order logic in a more direct type-theoretic flavor, by reusing formal notions of binding, variables and substitutions to specify logical syntax, and adding connectives and quantifiers as certain proposition formers.

In what sense does Homotopy type theory “model types as spaces”?

To model types as spaces, we have to define a model of HoTT, such that closed types in the model are mathematical structures which can be viewed as spaces. There's again some wiggle room here, because there are multiple notions of spaces in mathematics.

For example, cubical sets is one way to define spaces. In the cubical set model of HoTT, we define a particular model, such that typing contexts and closed types are cubical sets, parallel substitutions are natural transformations, types are a certain kind of "dependent" cubical sets, and every type and term former has an interpretation, satisfying $\beta\eta$ equations.


答: The "$\leq$" sign means less than or equal to. As $a=a$, or $n+1=n+1$, the equals sign can be correctly replaced by the less than or equal to sign ($\leq$).

Title: Can we write $a \leq a$