问:
Finding $\frac {d(\tan \theta)}{d\theta}$
I was reading Needham's Visual Complex Analysis, and could not figure out how we get length $Ld\theta$ here:

答:
By "ultimately similar" Needham means the following (see pages 20–21):

We stress here the words "ultimately equal" and "infinitesimal" are being
used in delicate, technical senses; in particular, "infinitesimal" does not
refer to some mystical, infinitely small quantity7. More precisely,
if two quantities $X$ and $Y$ depend on a third quantity $\delta$, then
\begin{align*}
\lim_{\delta \to 0} \frac{X}{Y}\quad & \iff \quad \text{"$X = Y$  for infinitesimal $\delta$".}\\
&\iff \quad \text{"$X$ and $Y$ are ultimately equal as $\delta$ tends to zero".}
\end{align*}
It follows from the basic theorems on limits that "ultimate equality" inherits many of the properties of ordinary equality.

7For more on this distinction, see the discussion in Chandrasekhar [1995].

Since two triangles are similar if and only if they have the same set of (internal) angles when Needham says that "the black triangle is ultimately similar [exercise] to the shaded triangle" he probably means that the ratio of any angle of the black triangle with the corresponding angle of the shaded triangle tends to $1$ as $d\theta$ tends to zero.

Let the shaded triangle be labelled $\triangle \mathit{ABC}$, where $\angle A = \theta$ and $\angle B = \pi/2$. The black triangle $\triangle\mathit{CDE}$ is obtained by constructing the perpendicular to the line segment $\mathit{AC}$ at $C$ and taking the point $D$ to be where it meets the straight line that is at an angle of $\theta + d\theta$ to the base $\mathit{AB}$, and by extending the segment $\mathit{BC}$ until it meets the same line at the point $E$.

Let us compute the angles of $\triangle \mathit{CDE}$. Examining $\triangle \mathit{ABE}$, we have $\angle E = \pi/2 - \theta - d\theta$ in $\triangle \mathit{CDE}$. Examining $\triangle \mathit{ACD}$, we have $\angle D$ in this triangle to be $\pi/2 - d\theta$. Hence, $\angle D$ in $\triangle \mathit{CDE}$  equals $\pi/2 + d\theta$. Thus, we must have $\angle C$ in $\triangle \mathit{CDE}$ to be equal to $\theta$ (this could also have been seen directly from our construction of $\triangle \mathit{CDE}$).

So, what do the ratios of corresponding angles between $\triangle \mathit{ABC}$ and $\triangle{\mathit{CDE}}$ look like in the limit as $d\theta \to 0$?
\begin{align*}
&\lim_{d\theta \to 0} \frac{\angle \mathit{CAB}}{\angle \mathit{DCE}} = \lim_{d\theta \to 0} \frac{\theta}{\theta} = 1.\\
&\lim_{d\theta \to 0} \frac{\angle \mathit{BCA}}{\angle \mathit{CED}} = \lim_{d\theta \to 0} \frac{\pi/2 - \theta}{\pi/2 - \theta - d\theta} = 1.\\
&\lim_{d\theta \to 0} \frac{\angle \mathit{ABC}}{\angle \mathit{EDC}} = \lim_{d\theta \to 0} \frac{\pi/2}{\pi/2 + d\theta} = 1.\\
\end{align*}

This proves that the black triangle and shaded triangle in Needham's diagram are ultimately similar. (Of course, we could have just done this for any two angles, since the third angle of a triangle is uniquely determined once the other two are specified.)

Now, call $\lvert \mathit{CE} \rvert$ to be $dT$. What can we say about $\lvert \mathit{CD} \rvert$?
Let $F$ be the point on the segment $\mathit{AD}$ such that $\lvert \mathit{AF} \rvert = \lvert \mathit{AC} \rvert = L$. Then, certainly $\lvert \mathit{CF} \rvert < L\, d\theta < \lvert \mathit{CD} \rvert$. So, to show that $L\, d\theta$ and $\lvert \mathit{CD} \rvert$ are ultimately equal, it suffices to show that $\lvert \mathit{CF} \rvert$ and $\lvert \mathit{CD} \rvert$ are ultimately equal.

Applying the cosine law on $\triangle \mathit{ACF}$, we get
$$
\cos(d\theta) = \frac{L^2 + L^2 - \lvert \mathit{CF} \rvert^2}{2^{\vphantom{2}}\!\;\! L^2} \implies \lvert \mathit{CF} \rvert = 2L\sin\left(\frac{d\theta}{2}\right).
$$
Examining $\triangle \mathit{ACD}$, we have
$$\tan(d\theta) = \frac{\lvert \mathit{CD} \rvert }{ L } \implies \lvert \mathit{CD} \rvert = \frac{2L\tan
\left(\frac{d\theta}{2}\right)}{1^\vphantom{2}\! - \tan^2 \left(\frac{d\theta}{2}\right)}.
$$
Thus,
$$
\frac{\lvert \mathit{CF} \rvert}{\lvert \mathit{CD} \rvert} = \cos\left(\frac{d\theta}{2}\right) \cdot \left(1 - \tan^2\left(\frac{d\theta}{2}\right)\right).
$$
As $d\theta$ tends to $0$ the RHS tends to $1$, so $\lvert \mathit{CF} \rvert$ and $\lvert \mathit{CD} \rvert$ are ultimately equal.

To summarise, the ratio $\dfrac{\lvert \mathit{CE} \rvert}{\lvert \mathit{CD} \rvert}$ is ultimately equal to $\dfrac{\lvert \mathit{AC} \rvert}{\lvert \mathit{AB} \rvert}$ because $\triangle \mathit{CDE}$ is ultimately similar to $\triangle \mathit{ABC}$, and $\lvert \mathit{CD} \rvert$ is ultimately equal to $L\, d\theta$. Thus, $\dfrac{dT}{L\, d\theta}$ is ultimately equal to $\dfrac{L}{1}$, so we get
$$
\frac{dT}{d\theta} = L^2 = 1 + T^2
$$
in the limit as $d\theta \to 0$.

The reference to Chandrasekhar [1995] mentioned in footnote 7 in the quote at the beginning is listed on page 574 as follows: S. Chandrasekhar (1995), Newton's Principia for the Common Reader. Clarendon Press, Oxford.


问:
How do i prove that $\frac{1}{\pi} \arccos(1/3)$ is irrational?
How do i prove that $\frac{1}{\pi} \arccos(1/3)$
 is irrational?</p>

答: Let’s call this angle $\Theta$. If it were rational, then there would be an $N$ such that $\cos(N\Theta)=1$. That would say that $\cos\Theta$ was a root of $T_N-1$, where $T_N$ is the appropriate Čebyšev polynomial, which it can’t be ’cause it’s transcendental.</p>

ContentLicense: CC BY-SA 3.0


问:
closed set contained in the rationals between $0$ and $1$.
Let $E= [0,1] \setminus \mathbb{Q}$. Let $\epsilon>0$ be given. Construct a closed set contained in $E$, call it $F$ such that we have have $m_*(E \setminus F)<\epsilon$ where $m_*$ is exterior measure not Lebesgue measure. Any hints?

答:
Note that $E \setminus F = [0, 1] \setminus \mathbb Q \setminus F$ is Lebesgue measurable, so it makes no difference if we use exterior or Lebesgue measure. Also if for any $\epsilon > 0$, $m_*(X) < \epsilon$, then just $m_*(X) = 0$.

There is no such set: as $F$ is closed, $(0, 1) \setminus F$ is open, also $\frac{1}{2} \in (0, 1) \setminus F$, so for some $\epsilon$, $\left( \frac{1}{2} - \varepsilon; \frac{1}{2} + \epsilon\right) \cap F = \varnothing$. So we have $\left( \frac{1}{2} - \epsilon; \frac{1}{2} + \epsilon\right) \setminus \mathbb Q \subset E \setminus F$. And as $m_*(\left( \frac{1}{2} - \epsilon; \frac{1}{2} + \epsilon\right) \setminus \mathbb Q) = 2\epsilon$, $m_*(E \setminus F) \geqslant 2\epsilon$.

If you switch order of quantifiers - ie $\forall \epsilon \exists F\, m_*(E \setminus F) < \epsilon$, it's standard construction: let $q_m$ be some enumeration of rationals in $[0, 1]$ and let $F = [0, 1] \setminus \bigcup\limits_{m=1}^\infty (q_m - \epsilon\cdot 2^{-m - 1}; q_m + \epsilon\cdot 2^{-m - 1})$.

ContentLicense: CC BY-SA 4.0


问:
Integration with $\int_0^\infty \frac{\cos (ax)}{x^2}dx$
I am trying to integrate $$\int_0^\infty \frac{\cos (ax)}{x^2}dx$$

I get $$-a\operatorname{Si}(ax)-\frac{\cos(ax)}{x}+C$$ as indefinite integral and when I put limits, I understand how to get $-\frac{{\pi}a}{2}$ for the Sine integral. But I don't understand how to put limits for the second term, it seems to me that term goes to infinity. Any help?

Edit:
Sorry for not being clear,
I am trying to integrate this integral $$D_1=\frac4\pi\int_0^\infty\frac{d\lambda}{\lambda^2}\sin\lambda q_1\sin\lambda q_2\sin\lambda q_3\sin\lambda q_4,\tag{25}$$
which I converted to sum of cosines
$$\sum\ \eta_1\eta_2\eta_3\eta_4\frac{\cos ((\eta_1q_1+\eta_2q_2+\eta_3q_3+\eta_4q_4)\lambda)}{16}, \eta_{1,2,3,4}= \pm1$$

I am supposed to get these results,

For $q_1+q_2>q_3+q_4$ and $q_1+q_4>q_2+q_3$, $$D_1=\frac12(q_2+q_3+q_4-q_1)\tag{30}$$

For $q_1+q_2>q_3+q_4$ and $q_1+q_4<q_2+q_3$, $$D_1=q_4\tag{33}$$

I am trying to reproduce calculations given in the Appendix A of this paper.

答:
Whilst the specified integral does not converge, the following does:
$$I(a)=\int_0^\infty\frac{\cos(ax)-1}{x^2}dx$$
$$I'(a)=-\int_0^\infty\frac{\sin(ax)}{x}dx$$
now make the substitution:
$$u=ax\Rightarrow dx=\frac{du}a$$
and you will see that:
$$I'(a)=-\int_0^\infty\frac{\sin(x)}{x}dx$$
and this is called the Dirichlet integral so we know its value:
$$I'(a)=-\frac\pi2$$
Now we can integrate to get:
$$I(a)=-\frac\pi2 a+C$$
now since we know that $I(0)=0$ we can say $C=0$ which finally gives us:
$$\int_0^\infty\frac{\cos(ax)-1}{x^2}dx=-\frac{\pi a}{2}$$


问:
A natural transformation
On the page $9$ here I have a very basic question: in the definition $3.1$ $\alpha$ never appears in the condition

"such that if $(U_1f)a\in U_2 HB$ then $a\in U_2 HA$ for each $f:A\to B$ in ${\cal K}_1$".

Also, I cannot syntactically verify this condition.

答:
$\alpha$ is a monic natural transformation of functors to Set, so the authors identity the components of $\alpha$ with the subset relation. That is, for all $A\in \mathcal{K}_1$, the authors suppress the injective function $\alpha_A\colon U_2HA\hookrightarrow U_1A$ and instead pretend $U_2HA\subseteq U_1A$. This can be done formally at the expense of replacing $U_1$ or $U_2$ with a naturally isomorphic copy.

So the condition "if $(U_1f)a\in U_2HB$, then $a\in U_2HA$ for each $f\colon A\to B$ in $\mathcal{K}_1$" could be rewritten using $\alpha$ as "if $(U_1f)a\in \text{im}(\alpha_B)\subseteq U_1B$, then $a\in \text{im}(\alpha_A)\subseteq U_1A$ for all $f\colon A\to B$ in $\mathcal{K}_1$".

Title: A natural transformation


问:
"Inverse" moment generating function of standard normal distributed random variable
This is just a trivial question maybe but, is the Moment generating function for $X$ the same as for $-X$ for a normally distributed random variable, so $E(e^{tX})=E(e^{-tX})$? If not, what is the difference between them?

答:
For any random variable $X$, letting $M(t)=E(e^{tX})$ then $E(e^{-tX})= M(-t)$ so your question is essentially: when does $M(t)=M(-t)$, i.e. when is the moment generating function an even function?

For the normal distribution, $M(t) = e^{\mu t + \frac{1}{2}\sigma^2 t^2}$ so $M(t)$ is even only when $\mu = 0$.


问:
Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$

Prove that for all integers $n$, $n \geq 1$,
  $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$

How would I go about proving this?

答:
A nice version of the inductive step is the following:

Using induction, the problem comes down to establishing the result that

$$
n^2=(n-1)^2+(2n-1)
$$

Now it is easy to establish this by multiplying out the brackets, but I want to show you another way that I thought of one day.  As an example, let $n-1=5$.  Then:

\begin{align}
5^2+5 &= 5\times5+5=6\times5\\
6\times5+6&=6\times6
\end{align}

So $6^2=5^2+5+6$.  In general:

\begin{align}
(n-1)^2+(n-1)&=(n-1+1)(n-1)=n(n-1)\\
n(n-1)+n&=n(n-1+1)=n^2
\end{align}

So $n^2=(n-1)^2+(n-1)+n=(n-1)^2+(2n-1)$.

Note: This answer is not intended to be particularly useful or even mathematically revealing.  It is certainly not the most beautiful way of establishing the identity (the use of induction disqualifies it immediately), but I thought it was a nice way of showing the inductive step, without having to multiply out the brackets.


问:
Motivation for vector spaces
For instance I can make a narrative for defining some familiar sets, leading up to complex numbers, as follows:

Natural numbers are obvious. They are "ordered sets" useful for counting things, and we can define addition and multiplication on them in a useful way.

Integers were defined because we wanted additive inverses, e.g. $5+x=3$. This also an "ordered set"

Rationals were defined because we wanted multiplicative inverses, e.g. $3x=2$, and now we have a "field" and since they are ordered, an "ordered field."

Reals were defined because we wanted to solve $x\cdot x = 2$, these are also an "ordered field."

Complex numbers were defined because we wanted $x^2=-1$. We defined multiplication and addition specifically so that they satisfy the field axioms, which makes complex numbers also a "field." (We lose "ordered" because we can't define it in a consistent way, but the reals are a subfield which we can order so we're still ok.)

Ok, so now we have ordered sets and fields and ordered fields. Is there a similar narrative for defining "vector spaces?"

Most textbooks simply start with "here are fields and here are the axioms for vector spaces over fields." e.g. what is the sequence of logic that led us to invent $\mathbb{R}^N$ or $\mathbb{C}^N$, along with the addition/scaling/identity/closure axioms that go with it? Why didn't we introduce the concept of multiplication (I don't mean inner/outer/cross product) into the definition?

答:
Here is one possible thought process.

It seems natural to attempt to generalize calculus to functions that take a list of numbers as input and return a list of numbers as output. This leads us to introduce $\mathbb R^n$. Since the fundamental strategy of calculus is to approximate a nonlinear function locally by a linear function, we must define what it means for a function to be "linear" in this new setting. This leads us to introduce the ideas of vector addition, scalar multiplication, and linear transformations (and matrices that describe linear transformations).

Now having become familiar with $\mathbb R^n$, we might want to extend calculus to handle functions that take as input, say, a matrix, or a symmetric matrix. And we might notice that the same fundamental operations (addition and scalar multiplication) also make sense when considering, say, the set of all $m \times n$ matrices, or the set of all $n \times n$ symmetric matrices. This motivates us to generalize the linear algebra ideas that we developed in $\mathbb R^n$ to more general sets where addition and scalar multiplication make sense, leading to the idea of a vector space.

A fantastic motivating example for introducing vector spaces is the set of all possible solutions to an $n$th-order homogeneous linear ordinary differential equation. That is a wonderful vector space. The set of all solutions to the ODE $y'(t) = A y(t)$ is another great example of a vector space. What is its dimension? What is a basis for it? Arguably, the idea of a vector space was introduced first in order to understand these spaces of solutions to linear differential equations. (I don't know the true history though.)

When understanding the effect of applying a linear transformation $T$ to a vector $v \in \mathbb R^n$, a reasonable strategy is to first seek out special vectors $v$ which have the nice property that $T(v)$ is just a scalar multiple of $v$. While $T$ might seem complicated, it is at least simple to understand what $T$ does to $v$. And then if some other vector $w$ can be written as a linear combination of these special "eigenvectors", then perhaps understanding what $T$ does to $w$ is not so hard either. But, when seeking eigenvectors for $T$, we often find that the entries of $w$ need to be complex. This motivates us to introduce $\mathbb C^n$ and to do linear algebra in $\mathbb C^n$.

Regarding multiplication, it seems that there is not a natural way to multiply two vectors in $\mathbb R^n$, resulting in a new vector in $\mathbb R^n$. So such a concept was not introduced despite efforts along those lines. (You can read the history of how mathematicians introduced quaternions, but had to give up commutativity of multiplication when doing so; and then octonions were introduced, sacrificing even associativity.)

Title: Motivation for vector spaces


问:
Evaluate $\int \frac{(1 + e^x)^2}{e^x} dx$
I am struggling with this integral: $$\int \frac{(1 + e^x)^2}{e^x} dx$$

I tried u substitution setting u = 1+ e^x but that was not fruitful. I also tried expanding $(1+e^x)^2$ and got $1 + 2e^x + e^{2x}$. I don't know where to go from there.

答:
Based on the fact that $(1 + e^{x})^{2} = 1 + 2e^{x} + e^{2x}$, we have that
\begin{align*}
\int\frac{(1+e^{x})^{2}}{e^{x}}\mathrm{d}x & = \int\frac{1 + 2e^{x} + e^{2x}}{e^{x}}\mathrm{d}x\\\\
& = \int(e^{-x} + 2 + e^{x})\mathrm{d}x\\\\
& = -e^{-x} + 2x + e^{x} + c
\end{align*}

and we are done.


问:
Show that the single matrix that commutes with all invertible matrices is the scalar matrix
Show that the scalar matrix $\lambda E$, with $E$ being the identity matrix and $\lambda \in K$ a constant, is the only matrix $A\in K^{n\times n}$ that commutes with all invertible matrices.

I am not sure how to tackle this question. I would appreciate if you could help me. Thanks!

答:
I assume your matrices are over $\mathbb{R}$, although this approach can be modified using algebraic geometry to work over general fields.

There is a nice two-step approach: first show that a matrix that commutes with all invertible matrices in fact commutes with all matrices, and then show that the only matrices that do that are scalar multiples of the identity.

Now for step 1 there is a really nice surprising trick: use topology!

The key is that the set of invertible matrices is dense in the set of all matrices, meaning that for every matrix B, no matter how restrictive your notion of 'tiny' is, you can always find a tiny matrix $C$ such that $B + C$ is invertible. This is intuitively clear once you realize the same is true fore the determinants of $B, C$ and $B + C$.

Now given your matrix $A$ that commutes with all inverible matrices we have that the map that sends matrix $B$ to matrix $AB - BA$ is continuous and equals the zero matrix almost everywhere (on the dense subset of invertible $B$ that is) so that it follows that it equals $0$ everywhere.

Step 2 doesn't involve any surprise trick like this, so I leave that to you


问:
The unfair subway
This is problem 24, "The Unfair Subway", in Mosteller's Fifty Challenging Problems in Probability with Solutions

Marvin gets off work at random times between 3 and 5 P.M. His mother lives uptown, his girl friend downtown. He takes the first subway that comes in either direction and eats dinner with the one he is first delivered to. His mother complains that he never comes to see her, but he says she has a 50-50 chance. He has had dinner with her twice in the last 20 working days. Explain.

The accompanying solution says that it's because the uptown train always arrives one minute after the downtown train, which in turn arrives nine minutes after the uptown train, in this time span. So there's a nine-to-one chance that Marvin will get on the downtown train and not the uptown one.

Huh? Then what happened to the "50-50 chance" part of the problem?

The problem seemed to be posed as a probabilistic inference problem, i.e. one where the goal is to calculate:
$$\binom{20}{2} (0.5)^2 (1-0.5)^{18} \approx 0.00018$$
but it turns out it was a statistical inference problem (one based on maximum likelihood estimates at that) that contradicts information in the problem itself.

So my question is: is this a valid problem in probability? Am I missing something that would make this a valid problem?

答: The 50-50 chance is a red herring.  As the solution points out, from the fact that there are the same number of trains in each direction, you cannot infer that there is a 50% chance that the next train will be going in each direction.  If you randomly selected a train from the list and then met it, you would have a 50% chance of going either direction.  I would say it is a valid problem, but not one that requires any calculation at all.  The problem is just to figure out how the next train can not be 50-50 if there are the same number of trains in each direction.</p>

Title: The unfair subway


问:
How do I show $eIe = I \cap eRe$?

Let $R$ be an associative ring with $1_R$ and $e \in R$ be an
idempotent element such that $e^2 = e$. Show that

(a) $eRe$ is a ring in itself with identity $e$

(b) For any two sided ideal $I \subset R$, $I \cap eRe = eIe$

$(a)$ Do I just show all the properties or is there a faster way? For example what I mean is show

(i) Abelian in $+$, for any $x,y \in eRe$, x+y = er_1e + er_2e = e(r_1e +r_2e) =e(r_1 + r_2)e = e(r_2 + r_1)e = y + x. The third equality uses $R$ is a ring.

(ii) Assoc Multiplication. $(xy)z = (er_1e er_2 e)(er_3e) =\dots = x(yz)$

(iii) Distributive property $x(y + z) = \dots = xy + xz$ so forth and so forth.

(b) I only have to show set equality right?

So if $x \in I \cap eRe$, then $x = ere$ for some $r \in R$. But $exe = ere$ and $exe \in eIe$. Therefore $x = ere = exe \in eIe$ so $I \cap eRe \subset eIe.$

All I can deduce for the other equality is that for $y \in eIe$, we have $y = exe \in eIe$ and clearly $eIe \subset eRe$. Here is what I got stuck.

答:
(a) You show all the ring properties yes, but things like associativity or distributivity immediately follow from the corresponding properties in the ring $R$.

(b) In fact, the inclusion you are stuck at is trivial: If $y = exe \in eIe$ with $x \in I$, then clearly $y = exe \in eRe$ and since $I$ is an ideal also $y = exe \in I$, so $eIe \subseteq I \cap eRe.$

Title: How do I show $eIe = I \cap eRe$?


问:
The set of scalar function can be structured to a field.
I ask if it is possible to equip the set of scalar functions $\Bbb F^X$ between a set $X$ and a field $\Bbb F$ with a field structure. Indeed if $\Bbb F$ is a field and if $X$ is a set then for any $f,g\in\Bbb F^X$ I naturally thought to define an addition and multiplication in $\Bbb F^X$ through the following two identities
$$
(f+g)(x):=f(x)+g(x)\\
(f\cdot g)(x):=f(x)\cdot g(x)
$$
for any $x\in X$ and this seems to me that make the set $\Bbb F^X$ to a field but I'd like to know from otherone if this is effectively true. So could someone help me, please?

答:
No this does not work, because we will always have zero divisors when $|X| > 1$ (which cannot exist in a field).

Indeed, fix distinct elements $x\neq y$ in $X$. Define $f: X \to \mathbb{F}$ to be $1$ in $x$ and $0$ elsewhere. Similarly, define $g: X \to \mathbb{F}$ to be $1$ in $y$ and $0$ elsewhere. Then $fg = 0$ yet $f \neq 0$ and $g \neq 0$. Hence, $\mathbb{F}^X$ is never a field, unless $|X| = 1$ in which case $\mathbb{F}^X\cong \mathbb{F}.$


问:
An speical generating element of the ring of integers in local fields
See the accepted answer to this question Can a generator of the ring of integers of local fields can be chosen so that it is also a uniformizer at the same time?

"$O_L=O_K[\pi_L]$ iff $f(L/K)=1$ and $O_L=O_K[\zeta_{q_L-1}]$ iff $e(L/K)=1$." I can see why this sentence is true.

"In general it is $O_L=O_K[\zeta_{q_L-1}+\pi_L]$." I can not see it. I can just see that $O_L=O_K[\zeta_{q_L-1}, \pi_L]$, and $O_K[\zeta_{q_L-1}+ \pi_L] \subseteq O_K[\zeta_{q_L-1}, \pi_L]$

答:
I have a proof but this may not be the same one as suggested by reuns in the link.
Let me try to restate the claim:

Let $l,k$ be residue fields of finite extension of local fields $L/K$ and let $\mathfrak{q},\mathfrak{p}$ be corresponding maximal ideals of $\mathcal{O}_L,\mathcal{O}_K$. Then $l/k$ is extension of finite fields, hence $l=k(\zeta_{q^n-1})$ where $q=|k|$ is a prime power, $q^n=|l|$ and $\overline{g}=x^{q^n-1}-1$ minimal polyonomial of $\zeta_{q^n-1}$.  We have $\mathcal{O}_L=\mathcal{O}_K[\zeta_{q_L-1}+\pi_L]$ where here we interpret $\zeta_{q_L-1}$ as any lift of $\zeta_{q^n-1}\in l$ to $\mathcal{O}_L$.

Let $a_0$ be any lift of $\zeta_{q^n-1}\in l$ to $\mathcal{O}_L$ with minimal polynomial $g\in \mathcal{O}_K[x]$ whose taking mod $\mathfrak{p}$ is $\overline{g}$. We let $a=\begin{cases}a_0 & v_{\mathfrak{q}}(g(a_0))=1 \\ a_0+\pi_L & v_{\mathfrak{q}}(g(a_0))>1 \end{cases}$ and we will show that $\mathcal{O}_L=\mathcal{O}_K[a]$.

First, the choice of $a$ gives $v_{\mathfrak{q}}(g(a))=1$. Indeed, consider when $v_{\mathfrak{q}}(g(a_0))>1$, $a=a_0+\pi_L$. We can expand $g(x)=g(a_0)+(x-a_0)g'(x)+h(x)(x-a_0)^2$ for $h\in \mathcal{O}_K[x]$ (here $g'$ is the formal derivative of $g$) then $g(a)=g(a_0)+\pi_L g'(a)+h(a)\pi_L^2$. Hence, we just need to show $v_{\mathfrak{q}}(g'(a))=0$. Assume otherwise, then $0\equiv g'(a)\equiv g'(a_0)\pmod{\mathfrak{q}}$ so this implies $\overline{g}$ and its derivative $\overline{g}'$ has $\zeta_{q^n-1}$ as common root, a contradiction since $\overline{g}=x^{q^n-1}-1$ is separable.

Now, with $[L:K]=m=ne$ then we want to show $1,a,\ldots, a^{m-1}$ generate $\mathcal{O}_L$ as $\mathcal{O}_K$-module. By Nakayama's lemma (Version 3), it suffices to show $1,a,\ldots, a^{m-1}$ spans $\mathcal{O}_L/\mathfrak{p}$ as $k=\mathcal{O}_K/\mathfrak{p}$-vector space. As $\mathfrak{p}\mathcal{O}_L=\mathfrak{q}^e$ so every element of $\mathcal{O}_L/\mathfrak{p}$ can be written as $b_0+b_1\pi_L+b_2\pi_l^2+\ldots + b_{e-1}\pi_L^{e-1}$ where $b_i\in l=\mathcal{O}_L/\mathfrak{q}$ and $\pi_L$ uniformiser of $\mathcal{O}_L$, which we can choose to be $g(a)$ because $v_{\mathfrak{q}}(g(a))=1$. Because $\pi_L=g(a)$ and $l=k(\zeta_{q^n-1})$ where $a\pmod{\mathfrak{q}}=\zeta_{q^n-1}$. Hence, each $b_i$ can be written as linear combination of $a\pmod{\mathfrak{q}}$ and $\pi=g(a)$, a polynomial in $a$. We are done.

Observe that when $e=1$ then we don't need the condition $v_{\mathfrak{q}}(g(a))=1$ to show $\mathcal{O}_L=\mathcal{O}_K[a]$, hence in this case, $a$ can be any lift of $\zeta_{q^n-1}$.


问:
Existence of continuous function using fixed point theorem
Using the fixed point theorem, the problem asked me to prove there is a continuous function $x(t):[0,1]\to\mathbb{R}$ (maybe unique??)satisfying the equation $$x(t)=\frac{1}{3}\int_0^t s^2\sin(x(s))ds $$

What I did (bounding in $[0,1]$):
$$|x(t)-x(v)|=\frac{1}{3}\left|\int_v^t s^2\sin(x(s))ds\right|\leq \frac{1}{3}|t-v| $$  so $x$ is Lipschitz and uniformly continuous in $[0,1]$. Indeed, $x'(t)=\frac{1}{3} t^2\sin(x(t))\geq 0$ this is because $\sin(\cdot)$ is positive in $[0,1]$ so this means $x$ non decreasing  $x(t)\leq x(1)\leq \frac{1}{3}$.   Accordingly to fixed point theorem if $x:X\to X$, and $d(x(t),x(v))\leq \lambda d(x,y)$ then there is a unique $x(t_0)=t_0$. But how does this imply the existence of the solution? for my problem and its uniqueness?

答:
Let $F:C[0,1] \to C[0,1]$ be defined by $(Fx)(t) = {1 \over 3}\int_0^t s^2 \sin x(s) ds $ then you have shown that $\|Fx\| \le { 1\over 3} \|x\|$ and
so is a contraction. Hence there exists a unique fixed point satisfying
$Fx = x$.


答: Apply Fixed Point Theorem to the space $C[0,1]$ of continuous functions on $[0,1]$ with the sup norm. Define $F(x)(t)=\frac1 3 \int_0^{t} s^{2} \sin (x(s))ds$. Then you get $\|F(x)-F(y)\|\leq \frac 1 3 \|x-y\|$. Hence there exists $x\in C[0,1]$ such that $F(x)=x$ and this gives the solution to the given integral equation.

Title: Existence of continuous function using fixed point theorem

Body:
Using the fixed point theorem, the problem asked me to prove there is a continuous function $x(t):[0,1]\to\mathbb{R}$ (maybe unique??)satisfying the equation $$x(t)=\frac{1}{3}\int_0^t s^2\sin(x(s))ds $$

What I did (bounding in $[0,1]$):
$$|x(t)-x(v)|=\frac{1}{3}\left|\int_v^t s^2\sin(x(s))ds\right|\leq \frac{1}{3}|t-v| $$  so $x$ is Lipschitz and uniformly continuous in $[0,1]$. Indeed, $x'(t)=\frac{1}{3} t^2\sin(x(t))\geq 0$ this is because $\sin(\cdot)$ is positive in $[0,1]$ so this means $x$ non decreasing  $x(t)\leq x(1)\leq \frac{1}{3}$.   Accordingly to fixed point theorem if $x:X\to X$, and $d(x(t),x(v))\leq \lambda d(x,y)$ then there is a unique $x(t_0)=t_0$. But how does this imply the existence of the solution? for my problem and its uniqueness?


问:
Explaining the difference between two definitions for recursively enumerable languages.
I'm a little confused about the definition for recursively enumerable languages in my script. A recursively enumerable language is defined as:

A language $A \subseteq \Sigma^*$ is called recursively enumerable, if $A = \emptyset$ or there exists a total and computable function $f: \mathbb{N} \to \Sigma^*$ with $A=\{f(0), f(1), f(2), ...\} := f(\mathbb{N})$. Then $f$ enumerates $A$.

However, the definitions that I found on Wikipedia and Youtube videos is rather different. For example, according to Wikipedia:

A recursively enumerable language is a formal language for which there exists a Turing machine (or other computable function) that will halt and accept when presented with any string in the language as input but may either halt and reject or loop forever when presented with a string not in the language.

The second definition makes more intuitive sense to me and also seems more widely accepted. What's the connection between the two definitions? From what I understand, the first definition essentially "numbers" each alphabet in $A$. But I'm not sure.

答:
Call the first concept enumerable and the second one semi-decidable. You want to show that $A\subseteq\Sigma^*$ is enumerable if and only if $A$ is semidecidable.

Assume first that $A$ is enumerable; we have to find a Turing machine which matches the "semi-decidable" description (this is the simple direction). If $A=\emptyset$, take the machine which always rejects. If $A\neq\emptyset$, consider the machine which does the following: Given an input $x$, start computing $f(0),f(1),f(2),\ldots$ If you encounter an $i$ such that $f(i)=x$, then accept. Note that the computation will never halt if $x\notin A$.

Assume now that $A$ is semidecidable, as witnessed by some Turing machine $T$. We want to find a computable function $f$ enumerating $A$. If $A$ is finite, you can simply write down a (non-injective) enumeration function, so assume $A$ is infinite. Let $x_1,x_2,\ldots$ be an enumeration of $\Sigma^*$, and let $n$ be the input for the function we are going to define. $f(n)$ is computed as follows: Using the machine $T$, perform one computation step on $x_1$. Then perform a second computation step on $x_1$ and a first computation step on $x_2$. Then do a third computation step on $x_1$, a second computation step on $x_1$ and a first computation step on $x_3$. Continue in this fashion, so that you are computing in parallel more and more steps on more and more inputs. When you get an "accept" answer on some input, do not continue the computation on that input. Wait until you get the $(n+1)$-th "accept", say on input $x_i$. Then output $f(n):=x_i$. (Note that since $A$ is infinite, you must eventually get an $(n+1)$-th accept, so the computation of $f$ on $n$ terminates.)


问:
Proving an identity using formal power series

4.

(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.

(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
  $$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$

Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?

答:
For the first, $$ \frac{1}{1 + \frac{x^2}{1 - x}} = \frac{1 - x}{1 - x + x^2} = \frac{(1 + x)(1 - x)}{(1 + x)\left(1 - x + x^2\right)} = \frac{1 - x^2}{1 + x^3} $$

For the second, consider the Maclaurin Series of $ \frac{1 - x^2}{1 + x^3} $ which is $$ \left( 1 - x^2\right)\sum_{k = 0}^\infty \left(-x^3\right)^n = \sum_{k = 0}^\infty a_nx^n$$ where $$ a_n = \begin{cases} &0 & x \equiv 1 \mod 3 \\ -&1 & x \equiv 2 \mod 6 \vee x \equiv 3 \mod 6 \\ &1 &x \equiv 5 \mod 6 \vee x \equiv 0 \mod 6 \end{cases} $$

Meanwhile, he Maclaurin Series for $ \frac{1}{1 + \frac{x^2}{1 - x}} $ is $ \displaystyle \sum_{n = 0}^\infty f(n) x^n $ where $$ f(n) := \sum_{k\ge 0 } (-1)^k  \dbinom{n -k -1}{n - 2k} $$Because the two functions are equivalent near $ x = 0 $, their Maclaurin series are the same and the equality follows.

The hard part is proving that definition of $ f(n) $. It involves expanding $ \frac{x^2}{1 - x} $ as its own Maclaurin series and then using a geometric series on $ \frac{1}{1 - \left(-\frac{x^2}{1 - x}\right)}.


问:
Couples are equally likely to have 1 or 2 children. How likely does a randomly chosen person have a sibling?
Assume that every couple can only have exactly 1 child or two children, with those outcomes being equally likely. Ignore any silly extra factors (e.g. 1 child dying, another being alive).

If I choose a human on the face of the earth randomly, what is the probability that they had a sibling?

As a follow-up, what would be the general answer if couples had $i$ children with probability $\pi_i$?

My thought for the simple case is either 1/2 or 2/3, and I can think of compelling reasons for both.

Siblings get counted twice, and only-children get counted once, leading to 2/3.

A given person's parents were equally likely to give him a sibling or not, leading to 1/2.

This kind of problem occurred to me one day. A similar problem occurs to me when I think about divorce rates. (e.g. if the divorce rate is 50%, is the probability a person is divorced 50% or 66.6%?) Any help is appreciated!

答:
The answer is 2/3 and more generally the probability of having $i$ siblings is:

$$\frac{i\pi_i}{\sum_{k} k\pi_k}$$

where $\pi_i$ is the probability of a couple having $i$ kids.

One way to see this is assume there are 100 couples. Then we would expect 50 of them to have 1 kid and 50 to have 2 kids so we have 150 kids total and 100 siblings which is 2/3. You can do a similar calculation to see why the general equation holds.


问:
What is this fallacy called?
I tried to show that $\frac{1}{x} \neq 0$ by contradiction:

$$\begin{align} \frac{1}{x} &= 0\ \text{(multiplty by $x$)} \\[6pt]
1 &= 0\end{align}$$

Btw I know other ways to prove this (i.e. proving this is not the goal of my question)

Someone pointed out that you cannot time by $x$. I know why one cannot divide by $x$.

Say $x +1 = 2$. If I times by $x$, I get $x^2 -x = 0$ which leads to the solutions $x = 0,-1$. Obviously $x=0$ is extraneous and can be ruled out, also, this step was unnecessary but this isn't fallacious though- or is it?

答:
Your argument is correct. It is always the case that
$$a=b$$
implies
$$f(a)=f(b)$$
for any function $f$ applied to both sides. In particular,
$$\frac{1}x=0$$
implies
$$1=0$$
since we can take $f$ to be "multiply by $x$" - which is a perfectly good function, even if $x$ were zero (note that the same cannot be said of "divide by $x$"). Of course, $1\neq 0$, so this is a contradiction and thus it cannot be that $\frac{1}x=0$ for any $x$ since that implies a contradiction. (Or, perhaps more precisely we might say that if $y\cdot x=1$ then $y\neq 0$, in order to avoid possibly writing $\frac{1}0$ by accident)

The danger of multiplying by $x$ is that the statements $a=b$ and $f(a)=f(b)$ might not be equivalent - which allows for extra solutions to sneak in. In your example, if you start with
$$x-1=0$$
then multiplying by $x$ gives
$$x^2-x=0.$$
It's certainly true that if $x-1=0$ then $x^2-x=0$. The extraneous solution comes from the fact that $x^2-x=0$ doesn't imply that $x-1=0$ - but your argument isn't trying to use this reverse direction, so it's not an issue that extraneous solutions might pop in.

This is a point that sometimes gets overlooked, since algebra is most often done using functions such as "add $2$" or "subtract $x$" which are invertible (or at least "injective"), which allows us to prove the reverse implication - formally, if for the functions $f$, there's another $g$ so that $g(f(x))=x$, we can find that $f(a)=f(b)$ implies $g(f(a))=g(f(b))$ which then implies $a=b$ - meaning that then $a=b$ and $f(a)=f(b)$ are equivalent. Working with functions that aren't like this - such as squaring both sides or multiplying by some $x$ that might be zero - feels different, but as long as you keep track of the direction of the implication, there is absolutely no problem in doing so.

Title: What is this fallacy called?


答: To your question, “How should I expand a fraction into a power series?”, the hand-computational answer is this. You write numerator and denominator down with powers ascending to the right and then proceed exactly as you did in high school when you divided one polynomial into another.

Title: Proving an identity using formal power series

Body:
4.

(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.

(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
  $$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$

Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?


问:
Charts for the complex projective line .
Let $\mathbb{C}P^1$ be the complex projective line quotient of $\mathbb C^2 \setminus \{0\}$ by the equivalence relation :

$$(z_1,z_2) \sim \;(z_1',z_2') \iff \exists \rho \in \mathbb C^*, \quad  (z_1',z_2')=\rho (z_1,z_2) $$
For $i=1,2$, let $U_i=\{[z_1,z_2] \in \mathbb{C}P^1 : z_i \neq0\}$ and $\phi_i : U_i \to \mathbb C$ defined by :
$$\phi_i([z_1,z_2])=\frac{z_2}{z_1} \text{ and } \phi_2([z_1,z_2])= \frac{z_1}{z_2}$$
I want to find $\phi_i ^{-1} $ for $i=1,2.$
I didn't know how to find it. Thanks in advance.

答: Verify that $\phi_1^{-1}(z)=[1,z]$, and similarly for $\phi_2^{-1}$.


问:
$\mathcal{E}'(\mathbb{R^n}) \subset \mathcal{S}'(\mathbb{R^n})$?
Can I have pls the proof of why is the space of distribution with compact support $\mathcal{E}'(\mathbb{R^d})$ subset in the set of tempered distribution $\mathcal{S}'(\mathbb{R^n})$?

答:
I'm assuming here that your $\mathcal E$ is $C^\infty$. Say $\mathcal S$ is the Schwarz space. Define $i:\mathcal S\to\mathcal E$ by $if=f$.

Show $i$ is bounded.

Calculate $i^*:\mathcal  E'\to\mathcal S'$.


问:
Normal rings and being integrally closed
Let $A$ be a commutative ring. I am thinking about the condition of being integrally closed in relation to being normal. How do the following conditions relate to each other?

$A$ is normal

$A/\mathfrak p$ is integrally closed in $\text{Frac}(A / \mathfrak{p})$ for each prime $\mathfrak p$

$A_{\mathfrak{p}}/\mathfrak{q}$ is a valuation ring for each prime $\mathfrak{p}$ in $A$ and each prime $\mathfrak{q}$ in $A_{\mathfrak{p}}$.

I have found in my own thinking that (2) and (3) are equivalent, and that they imply (1), but I cannot get the other direction. Inn particular, is there an integrally closed domain $A$ such that $A/p$ is not integrally closed for some $p$?

答:
Yes, there are tons of integrally closed domains $A$ with a prime ideal $\mathfrak{p}$ so that $A/\mathfrak{p}$ isn't integrally closed. Take any non-normal affine variety, like $V(x^2-y^3)\subset \Bbb A^2_k$: $k[x,y]$ is a UFD, hence normal, but $k[x,y]/(x^2-y^3)\cong k[t^2,t^3]$ which is not integrally closed in its field of fractions $k(t)$. So (1) does not imply (2).

Let me also point out that the condition in (3) of $A_\mathfrak{p}/\mathfrak{q}$ being a valuation ring for all $\mathfrak{p,q}$ is incredibly uncommon: even when $A$ is the coordinate ring of a smooth variety over a field, the local ring of any point $\mathfrak{p}$ of codimension greater than one with $\mathfrak{q}=0$ provides a counterexample. Condition 3 should be equivalent to $A$ being regular of dimension one.


问:
Prove that $\int 1/x\, \text{d}x=\ln |x|$
What is the proof that $$\int \frac{1}{x}\, \text{d}x=\ln |x|?$$

Please note that similar questions refer to $\ln x$, and not $\ln |x|$ as I have seen in Calculus textbooks.

Given the formula that $$\frac{\text{d}}{\text{d}x}\ln x=\frac{1}{x},$$ it is obvious that for $x>0$, $\int 1/x\, \text{d}x=\ln x$; however I am unsure how the above definition is proven over all $x\neq 0$.

答:
On $(0,\infty)$, we have $\log|x|=\log x$ and, as you know, $\log'(x)=\frac1x$.

And, on $(-\infty,0)$, $\log|x|=\log(-x)$ and, by the chain rule, if you differentiate $\log(-x)$, you get $-\frac1{-x}$, which is equal to $\frac1x$.


问:
Differential equation which has following solution $y=A+B\tanh(\frac{x-D}{C})$
How can I find the ODE from the following solution:

$$y=A+B\tanh\left(\frac{x-D}{C}\right)$$

$A,B,C,D$ are constants.

These are some examples of ODE with that solution:
$$Ay''-B(y-1)y'=0$$

Other
$$Ay''+2yy'=0$$
Other
$$y'=1-y^2$$

and other

$$\frac{dy}{dx}= ay^2+ by+ c$$

I need to solve this equation to get something with $\tanh$.
I have started with an idea but have not been able to complete it.

Note that:
$$z:=\frac{y-A}{B}=\tanh\left(\frac{x-D}{C}\right) \text{ satisfies } z''=\frac{2}{c}zz'=\frac{2}{c}(1-z)(1+z)z$$

答:
Eliminate one constant after the other. You get rid of $A$ by just taking the derivative
$$
y'=\frac{B}{C}\cosh\left(\frac{x-D}C\right)^{-2}.
$$
Now a logarithmic derivative will remove $B$
$$
\frac{y''}{y'}=-\frac{2}{C^2}\tanh\left(\frac{x-D}C\right)
$$
Now insert this back into the first equation to eliminate the transcendental function. Unfortunately that revives $A$ and $B$
$$
y=A-\frac{BC^2}2\frac{y''}{y'}.
$$
Start again with a simple derivative
$$
y'=-\frac{BC^2}2\frac{y'''y'-y''^2}{y'^2}
$$
and continue again with a logarithmic derivative
$$
\frac{y''}{y}=\frac{y''''y'-y'''y''}{y'''y'-y''^2}-2\frac{y''}{y'}
\\~\\
y''(y'-2y)(y'''y'-y''^2)=yy'(y''''y'-y'''y'')
$$
This is a 4th order ODE as required for 4 integration constants. While terms on the left and right cancel, I do not see how that could simplify the equation.


答:
When $x\lt 0$, $\ln |x|=\ln(-x)$, but then:

$$\frac{d}{dx}\ln(-x)=\frac{1}{-x}\frac{d}{dx}(-x)=\frac{1}{-x}(-1)=\frac{1}{x}$$

so $\ln|x|$ is the antiderivative of $\frac{1}{x}$ for $x\lt 0$ too.

Title: Prove that $\int 1/x\, \text{d}x=\ln |x|$

Body:
What is the proof that $$\int \frac{1}{x}\, \text{d}x=\ln |x|?$$

Please note that similar questions refer to $\ln x$, and not $\ln |x|$ as I have seen in Calculus textbooks.

Given the formula that $$\frac{\text{d}}{\text{d}x}\ln x=\frac{1}{x},$$ it is obvious that for $x>0$, $\int 1/x\, \text{d}x=\ln x$; however I am unsure how the above definition is proven over all $x\neq 0$.


答:
I believe the easiest solution is to show that $\frac{d}{dx} \ln \lvert x \rvert = \frac{1}{x}$, and since you know that $\ln(x) = \ln \lvert x \rvert$ for $x >0$ you can treat this as an extension of the "normal" natural logarithm.

You can do this by using that $\lvert x \rvert = \sqrt{x^2}$ and using the chain rule. You get
\begin{align*}
\frac{d}{dx} \ln \lvert x \rvert & = \frac{d}{dx} \ln \left( \sqrt{x^2}\right)\\
& = \frac{1}{ \sqrt{x^2}}\cdot \frac{d}{dx}  \sqrt{x^2}\\
& = \frac{1}{ \sqrt{x^2}}\cdot \frac{1}{2\sqrt{x^2}}\cdot\frac{d}{dx}  x^2\\
& = \frac{1}{ \color{purple}{\sqrt{x^2}}}\cdot \frac{1}{\color{blue}{2}\color{purple}{\sqrt{x^2}}} \cdot \color{blue}{2}x\\
& = \frac{1}{\color{purple}{x^2}} \cdot x\\
&= \frac{1}{x}
\end{align*}


问:
Why does $(13)(47)(13)(23)=(47)(23)$?
Given $a=(3714)$ and $b=(123)$. Compute $a^2b$. I got $(13)(47)(13)(23)$ as my answer, but why does $(13)(47)(13)(23)=(47)(23)$? I got the left hand side of this equation, but I don't understand why the $(13)$ terms cancel out. Any help is greatly appreciated.

答: Becouse this commutes: $(13)(47)(13)$; also if you check exactly where the elements go, you get that it is the same as the right side.

Title: Why does $(13)(47)(13)(23)=(47)(23)$?


答:
The key insight that you need is that disjoint cycles commute. This means that:

$$ (1 3)(4 7)(1 3)(2 3) = (1 3)(1 3)(4 7)(2 3) = (4 7)(2 3) $$

Title: Why does $(13)(47)(13)(23)=(47)(23)$?

Body: Given $a=(3714)$ and $b=(123)$. Compute $a^2b$. I got $(13)(47)(13)(23)$ as my answer, but why does $(13)(47)(13)(23)=(47)(23)$? I got the left hand side of this equation, but I don't understand why the $(13)$ terms cancel out. Any help is greatly appreciated.


问:
Motivation for vector spaces
For instance I can make a narrative for defining some familiar sets, leading up to complex numbers, as follows:

Natural numbers are obvious. They are "ordered sets" useful for counting things, and we can define addition and multiplication on them in a useful way.

Integers were defined because we wanted additive inverses, e.g. $5+x=3$. This also an "ordered set"

Rationals were defined because we wanted multiplicative inverses, e.g. $3x=2$, and now we have a "field" and since they are ordered, an "ordered field."

Reals were defined because we wanted to solve $x\cdot x = 2$, these are also an "ordered field."

Complex numbers were defined because we wanted $x^2=-1$. We defined multiplication and addition specifically so that they satisfy the field axioms, which makes complex numbers also a "field." (We lose "ordered" because we can't define it in a consistent way, but the reals are a subfield which we can order so we're still ok.)

Ok, so now we have ordered sets and fields and ordered fields. Is there a similar narrative for defining "vector spaces?"

Most textbooks simply start with "here are fields and here are the axioms for vector spaces over fields." e.g. what is the sequence of logic that led us to invent $\mathbb{R}^N$ or $\mathbb{C}^N$, along with the addition/scaling/identity/closure axioms that go with it? Why didn't we introduce the concept of multiplication (I don't mean inner/outer/cross product) into the definition?

答:
Your "narratives" 1–3 are okay, but 4 and 5 are not quite right and can not define the real and complex number fields:

You can also solve the equation $x^2=2$ in the field $\mathbf{Q}(\sqrt{2})$, not only in $\mathbf{R}$.

Similarly, you can solve the equation $z^2=-1$ in $\mathbf{Q}(i)$,
not only in $\mathbf{C}$.

Instead, you might want to say that

the reals are the minimal complete metric space that contains a copy of $\mathbf{Q}$

and the complex numbers are the minimal algebraically closed field
that contains a copy of the reals.

So (informal) "narratives" can be used for your own "understanding" of (motivation of) mathematical objects, but they can not be substitutions of definitions.

For vector spaces, here is an excerpt from an excellent set of lecture notes in linear algebra by Terence Tao. This excerpt gives a very good "motivation narrative" for what a vector is. (It is of course not intended for a rigorous definition.)

Informally, a scalar is any quantity which can be described by a single number. An example is mass: an object has a mass of $m$ kg for
some real number $m$. Other examples of scalar quantities from physics
include charge, density, speed, length, time, energy, temperature, volume, and pressure. In finance, scalars would include money, interest
rates, prices, and volume. (You can think up examples of scalars in
chemistry, EE, mathematical biology, or many other fields).

The set of all scalars is referred to as the field of scalars; it is usually
just $\mathbb{R}$, the field of real numbers, but occasionally one likes to work
with other fields such as $\mathbb{C}$, the field of complex numbers, or $\mathbb{Q}$, the field of rational numbers.

Informally, a vector is any
member of a vector space; a vector space is any class of objects which
can be added together, or multiplied with scalars. (A more popular,
but less mathematically accurate, definition of a vector is any quantity
with both direction and magnitude. This is true for some common
kinds of vectors - most notably physical vectors - but is misleading or
false for other kinds). As with scalars, vectors must obey certain rules
of algebra.

Why didn't we introduce the concept of multiplication (I don't mean inner/outer/cross product) into the definition?

We do have the notion of algebra over a field where you can multiply vectors.

Title: Motivation for vector spaces


问:
Help me find the triple improper integral
I don't understand how to calculate the integral
$$\int\limits_{ \mathbb{R}^{3} }\exp\left(-\frac{x^{2}_{1} + x^{2}_{2} + x^{2}_{3} }{ a }\right)\operatorname{arctg}\left( x_{1}+x_{2}+x_{3} + 1\right)dx_{1}dx_{2}dx_{3}$$

Help please, thanks :)

答:
Convert to spherical coordinates

$$\int_0^{2\pi}\int_0^\pi\int_0^\infty r^2\sin\theta\exp\left(-\frac{r^2}{a}\right)\tan^{-1}(r\sin\theta(\cos\phi+\sin\phi)+r\cos\theta+1)drd\theta d\phi$$

Now let $\sqrt{a}t = r$

$$a^{\frac{3}{2}}\int_0^{2\pi}\int_0^\pi\int_0^\infty t^2\sin\theta \:e^{-t^2}\tan^{-1}(\sqrt{a}t\sin\theta(\cos\phi+\sin\phi)+\sqrt{a}t\cos\theta+1)dtd\theta d\phi$$

By dominated convergence the limit $a\to0^+$ is boring and goes to $0$, but the limit of $a^{-\frac{3}{2}}\int df$ is given by

$$\frac{\pi}{4}\int_0^{2\pi}\int_0^\pi\int_0^\infty t^2\sin\theta \:e^{-t^2}\:dtd\theta d\phi = \frac{\pi^{\frac{5}{2}}}{4}$$


答:
Because disjoint cycles commute, as others have already said. But to convince yourself of this, it might help to follow what happens to each of the elements $1,2,3,4,7$ as they "go through the mill". For instance, look what happens to $3$:
$$(13): 3\to 1$$
$$(47): 1\to 1$$
$$(13): 1\to 3$$
$$(23): 3\to 2$$
So $3$ gets mapped to $2$.

If you do this for yourself, you will find that the result is the same as $(47)(23)$.

Title: Why does $(13)(47)(13)(23)=(47)(23)$?


问:
Show that a complete ternary tree of depth d has path width >= d-1
Show that $pw(T_{d}) \geq d- 1$. Feel free to suggest a completely different approach, if you know of a better one.

A complete ternary tree $T_{d}$ is a rooted tree in which each vertex has a 3 descendants and has depth $d$. Depth $d$ indicates the distance from the root vertex to a vertex at the deepest level of a tree. For example the complete ternary tree $T_{1}$ is a root node with 3 descendants (a 3 claw tree).

A path decomposition of a tree is a tree decomposition, in which the resulting tree is a path. Path width is defined as the bag with the largest amount of vertices minus 1.

My attempt:

I have attempted the problem through induction on $d$. The claim holds trivially for $d=1$ and $d=2$, as any connected graph must have $pw \geq 1$. Any edge in a connected graph must be in some bag, meaning there exists a bag with 2 vertices. Suppose the claim holds for depth $<d$. The tree $T_{d}$ is formed of a root vertex $v_{0}$ and three connected components, all of which are complete ternary trees of depth $d-1$. In other words the tree $T_{d}$ is formed of 3 tress $T_{d-1}$ and a root vertex $v_{0}$. By the induction hypothesis the path width of $T_{d-1}$ must be $\geq d-2$.

How can I calculate the path width of $T_{d}$ given that I know that I have 3 subtrees of path width $d-2$?

答:
Here's a sketch: The key idea is that a path decomposition which covers all three principal subtrees of $T_d$ must pass through the subtrees in some order, and it has to "thicken" the middle one a bit.

Call the three subtrees of $T_d$ adjacent to the root $a$, $b$, and $c$, and let $P=(\mathcal{B},E)$ be a tree decomposition of $T_d$ which is a path, having bags $B$ and edges $E$.  The induction step follows from these observations:

$P$ contains tree decompositions of each of $a,b,c$.  By the induction hypothesis, there are corresponding bags $B_a, B_b, B_c \in \mathcal{B}$ having at least $d$ vertices of the respective subtree.

$P$ also has bags $B_a', B_b', B_c' \in \mathcal{B}$ which contain the root of $T_d$ (called $v_0$ in the OP) and the respective roots of $a, b, c$.

We have a dichotomy.  Either:

One of $B_a, B_b, B_c$ (say $B_a$) lies in between two of $B_a', B_b', B_c'$ in the path $P$, so that by the properties of tree decompositions, we must have the root vertex $v_0\in B_a$.  Thus $|B_a|\ge d+1$, and we are done.

Or:

If the preceding does not hold, then one of $B_a, B_b, B_c$, say $B_a$, must lie in between $B_b$ and $B_b'$ or between $B_c$ and $B_c'$ in the path $P$.  WLOG suppose $B_a$ lies between $B_b$ and $B_b'$.  Then $B_a$ must contain a vertex of subtree $b$ (otherwise there would a vertex of $b$ on both sides of $B_a$ in $P$ with no connecting path).  So again $|B_a|\ge d+1$, as desired.


问:
How do i prove that $\frac{1}{\pi} \arccos(1/3)$ is irrational?
How do i prove that $\frac{1}{\pi} \arccos(1/3)$
 is irrational?</p>

答:
Let $\theta = \arccos\dfrac13$ so that $\cos\theta=\dfrac13$.

If $\theta$ is a rational multiple of $\pi$, say $\theta=\dfrac mn \pi$, then $\cos(n\theta)=\pm1$.  Now $\cos(n\theta)=T_n(\cos\theta)$, where $T_n$ is the $n$th-degree Chebyshev polynomial.  Using mathematical induction and some trigonometric identities, you can show that the leading coefficient in the $n$th-degree Chebyshev polynomial is $2^{n-1}$ for $n\ge2$.  We have
$$
\pm1=T_n\left(\frac 13\right) = 2^{n-1}\left(\frac13\right)^n+\text{lower-degree terms}.
$$
Multiplying both sides by $3^n$, we get
$$
\pm3^n = 2^{n-1} + \text{terms divisible by $3$}.
$$
And that says a positive power of $2$ is a multiple of $3$, which violates uniqueness of prime factorizations.


问:
What is the probability that the hand contains more than one ace, given that it contains an ace of spades?
A poker hand of 5 cards is dealt from a normal deck of 52 cards. What is the probability that the hand contains more than one ace, given that it contains an ace of spades?

I'm assuming that I need to use conditional probability to solve this question.  Specifically, let X be the number of aces in a five-card hand. Then, I need to solve

Pr( X > 1 | Ace of Spades) = Pr(X > 1 and Ace of Spades) / Pr(Ace of Spades)

Here is how I tried to find the Pr(Ace of Spades)

Pr(Ace of Spades) = 4C1 * 48C4 / 52C5

I'm not sure how to go about solving for Pr(X > 1 and Ace of Spades).

答:
Your formula for $\mathbb P(A\spadesuit)$ looks a little off -- it should be $\frac{\binom{51}{4}}{\binom{52}{5}}=\frac{5}{52}$. This is because if we want the probability of drawing the ace of spades, then we need to allow the other four cards to be anything.

Your expression looks like the probability of drawing exactly one ace (not necessarily the ace of spades).

Anyhow, it's probably easier to think of the probability as $$\mathbb P(X>1\mid A\spadesuit)=1-\mathbb P(X=1\mid A\spadesuit),$$
But the probability of drawing exactly one ace, conditional on drawing the ace of spades, is $$\frac{\mathbb P(X=1\cap A\spadesuit)}{\mathbb P(A\spadesuit)}=\frac{\binom{48}{4}/\binom{52}{5}}{\binom{51}{4}/\binom{52}{5}}=\frac{3243}{4165}\implies\mathbb P(X>1\mid A\spadesuit)=\frac{922}{4165}.$$


问:
Coloring an intersection graph of lines and a circle

Given a finite set of lines in the plane with no three meeting at a common point, and a circle that contain all intersection points of the lines in its interior, form a graph G whose vertices are the intersections of the pair of lines and the intersections of the lines with the circle with two vertices adjacent if they appear consecutive on one of the lines or consecutive on the circle. Prove that χ(G) ≤ 3.

I was leaning towards the idea of greedy coloring. If we remove the circle then the plane in rotatable in such a way that no 2 points have the same $x$ coordinate, then it's possible to sort the vertices according to their $x-$ coordinate and each one has at most 2 earlier neighbors in the ordering, hence $G$ is 3- colorable. The circle introduces constraints that cannot be solved by rotating the plane since I could always have a vertex that has 3 earlier neighbors in that ordering so I don't see how to go on using greedy coloring. One other idea was list coloring, but I got stock there too.
Any ideas?

答:
First $3$-colour all intersection points of lines as you describe.

Each intersection point on the circle has only one already coloured neighbour. So each point, $v$, on the circle has one colour, $c'(v)$, which it cannot use.

Let two adjacent points $v,w$ on the circle have $c'(v)\ne c'(w)$ and then colour $v$ with $c(w)$. Then go round the circle away from $w$ successively colouring points since each has at most two 'prohibited' colours. This will still apply when we reach $w$.

That leaves just one case, when every $c'(v)$ is the same colour. However, there are an even number of intersection points on the circle so, in that case, all the points on the circle can be coloured alternately with the two 'non-prohibited' colours.


问:
How does one know how to structure the unknown when factorizing using the remainder theorem?
When $x^3+x^2-7$ is divided by $(x-3)$, my book states that because the polynomial is cubic and the divisor is linear then the quotient must be quadratic and the remainder must be a constant. Basically, it sets put the following:
$x^3+x^2-7=(Ax^2+Bx+C)\cdot(x-3)+D$

In another example, the book divides $x^4+x^3+x-10$ by $x^2+2x-3$, and states that because the divisor is quadratic and because the polynomial is quartic then quotient must be a quadratic and the remainder must be a linear expression.

I want to understand what logic they use here, in order to decide the morphology of the equation whose coefficients they will begin to find. I have been going at this question for a long time. No online teacher seems to tackle it, although it seems like a bog issue.

Another reason why I really want to understand this is because it arises in partial fractions, such as the partial fraction involved in the first order differential equation in question 7 of this booklet (https://madasmaths.com/archive/maths_booklets/further_topics/integration/1st_order_differential_equations_substitutions.pdf). The logic is even more baffling. Someone please help, I have a further maths exam coming up in 3 days and this isn't difficult, but just obscure.

[Note: please note that I don't want some complex mathemtical proof. I just want know the logic so that I can, with reasonable simplicity, be able to explain to someone else how to decide upon the morphology of the the right hand side when using the factor theorem and factorising a polynomial by inspection, in order to find the coefficients.]

答:
Work it through.

Suppose $P(x) = \sum_{k=0}^n a_k x^k; a_n\ne 0$ is an $n$ degree polynomial. And $Q(x)= \sum_{k=0}^m b_k x^k; b_m\ne 0; m < n$ is an $m$ degree polynomial.

To say $P(x) = K(x)Q(x) + R(x)$ with $R(x)$ having degree less than $m$ would be the same as saying $P(x)- K(x)Q(x) = R(x)$.

For that to happen $K(x)Q(x)$ must be a $n$ degree polynomial with all coefficients $j: m\le j \le n$ all equal to $a_j$.

We can claim

For that to happen $K(x)$ must be degree $n-m$.

Pf:  If $K(x)=\sum_{k=0}^w c_k x^k; c_w\ne 0$ then $K(x)Q(x) = c_wb_{m}x^{w+m} + (c_wb_{m-1} + c_{w-1}b_m x^{w+m+1} + ........ + (c_1b_0 + b_0c_1)x^1 + c_0b_0$ will be a polynomial of degree $w+m$ with a leading coefficient of $c_wb_m$.

If $w+m< n$ then $P(x) - K(x)Q(x)$ will have a leading $a_n$ coefficient and will be a polynomial of degree $n > m$.

If $w+m > n$ then $P(x) -K(x)Q(x)$ will have a leading coefficient of $-c_wb_m$ and will be a polynomial of degree $w+m > n >m $.

So $w+m =n$ or $w= n-m$ is the only option.

But $K(x)Q(x)$ has leading of $c_{n-m}b_m$ so $P(x)-K(x)C(x)$ shouuld have a leading coefficient of $a_n - c_{n-m}b_m$ and if that is not equal $0$ then  $P(x) -K(x)C(x)$ will have a degree of $n$.

So we can claim

If $P(x) -K(x)Q(x)=R(x)$ is a polynomial of degree less than $n$, then we must have $c_{n-m} = \frac {a_n}{b_m}$ and as $b_m\ne 0$ we can do this.

The next coefficient of $K(x)Q(x)$ is $c_{n-m}b_{m-1} + c_{n-m-1}b_m$ and the next coefficient of $P(x) - K(x)Q(x) = a_{n-1} - (c_{n-m}b_{m-1} + c_{n-m-1}b_m)$.  If we want $P(x)-K(x)Q(x)$ to be a degree less than $m \le n-1$ we must have $a_{n-1} - (c_{n-m}b_{m-1} + c_{n-m-1}b_m)=0$ and therefore

$c_{n-m-1} = \frac {a_{n-1} - c_{n-m}b_{m-1}}{b_m}$ and we can do that as $b_m \ne 0$.

If $n = m+1$ were are done.  $K(x)Q(x) = a_nx^n + a_{n-1=m} x^{n-1=m} + \sum_{k=0}^{n-2}(\sum_{j=0}^kc_{n-m-k}b_k) x^k$ so $P(x) - K(x)Q(x)=R(x) = \sum_{k=0}^{n-2}(a_k -\sum_{j=0}^kc_{n-m-k}b_k) x^k$, a uniquely determined polynomial of degree less than $m$.

However if $n > m+1$ we can keep repeating each step.

We will need $a_{n-k} - (c_{n-m}b_{m-k} + c_{n-m-1}b_{m-k+1} + ...... + c_{n-m-k}b_m) = 0$ and we have already determined what $c_{n-m}, c_{n-m-1},....., c_{n-m-k + 1}$ are.

That uniquely determines  that we must have $c_{n-m-k} = \frac {a_{n-k} - (c_{n-m}b_{m-k} + c_{n-m-1}b_{m-k+1} + ...... +c_{n-m-k+1}b_{m-1}}{b_m}$.

And we do that until we unique determined what $c_0$ must be.  And as $K(x)$ we uniquely determined $R(x) = P(x) - K(x)Q(x)$ is uniquely determined.

It might help to do and example to convince yourself:

Suppose we wanted to solve $P(x) = x^8 + 2x^7 + 3x^6 + 4x^5 + 5x^4 + 6x^3+7x^2+8x+ 9$ and $Q(x)  = (4x^3 + 3x^2 + 2x + 1)$.  And we want $P(x)= K(x)Q(x) + R(x)$ where  $R(x)$ has degree less than $3$.

Well, If $K(x)$ has degree $w$ then $K(x)Q(x)$ has degree $3+w$.  And unless $3+w=8$ we will have a polynomial of degree $8$ minus a polynomial of degree $3+w$ will be of degree $\max(8,3+w) > 3$.  So we must have $w = 5$.

So $K(x) = (ax^5 + bx^4 + cx^3 + dx^2 + ex + f)$.

And so $K(x)Q(x) = (ax^5 + bx^4 + cx^3 + dx^2 + ex + f)(4x^3 + 3x^2 + 2x + 1) =$

$4ax^8 + (3a+4b)x^7 + (2a+3b+4c)x^6+(1a+2b+3c+4d)x^5 + (1b+2c + 3d +4e)x^4 + (1c + 2d+3e+4f)x^3 + (1d+2e + 3f)x^2 + (1e + 2f)x + f$.

And for $P(x) - K(x)Q(x)= R(x) =$

$(1 -4a)x^8 + (2 - (3a+4b))x^7 + (3- (2a+3b+4c))x^6+(4- (1a+2b+3c+4d))x^5 + (5- (1b+2c + 3d +4e))x^4 + (6-(1c + 2d+3e+4f))x^3 + (7- (1d+2e + 3f))x^2 + (8 - (1e + 2f))x + (9- f)$

And if that is to have degree less than $3$ and we must have

$1-4a =0$

$2-(3a+4b) = 0$

$3-(2a+3b+4c) =0$

$4- (1a+2b+3c+4d=0$

$5- (1b+2c + 3d +4e)=0$

$6-(1c + 2d+3e+4f)=0$

So six linear equations and six unknowns.  There is one uniqe set of solutions to that:

$1-4a =0; a = \frac 14$

$2-(3a+4b) = 0; b=\frac {2-3a}4= \frac {2-\frac 34}{4}=\frac {\frac 54}{4}=\frac 5{16}$

.... and so on......


问:
Prove $\frac{1}{n^{p+1}}<\frac{1}{p}\left[\frac{1}{(n-1)^p}-\frac{1}{n^p}\right]$
Problem: Prove the inequality following
$$\dfrac{1}{n^{p+1}}<\dfrac{1}{p}\left[\dfrac{1}{(n-1)^p}-\dfrac{1}{n^p}\right],\ \forall p>0, n \in \mathbb{N}.$$

My attempt: Since the form of this inequality, I tried to use the Lagrange theorem with $f(p)=\dfrac{1}{n^p},\ p>0$ but it did not work.

答:
Give $p \in (0,+\infty)$.

Consider $f(x)=\dfrac{1}{x^p},\ x>1$. According to the Lagrange theorem, for $x>1$, exist $c \in (x-1,x)$ that $$\dfrac{f(x-1)-f(x)}{x-1-x}=f'(c).$$
This yields $$\dfrac{1}{(x-1)^p}-\dfrac{1}{x^p}=\dfrac{p}{c^{p+1}} > \dfrac{p}{x^{p+1}}.$$
Therefore $$\dfrac{1}{x^{p+1}} < \dfrac{1}{p}\left[\dfrac{1}{(x-1)^p}-\dfrac{1}{x^p}\right] (*).$$
Replace $x=n \ (n \in \mathbb{N})$ in (*) we have
$$\dfrac{1}{n^{p+1}} < \dfrac{1}{p}\left[\dfrac{1}{(n-1)^p}-\dfrac{1}{n^p}\right].$$


问:
Follow-up question on conditionally convergent series.
This is a follow-up from this question.

Hypothesis: If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series with $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, then is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?

Maybe I'm wrong but I believe this isn't answered in the comments to the answer, because not only does it not allow for zeros to be part of the sequence, but also it's not clear to me how interweaving an absolutely convergent sequence answers this question.
$$$$
Actually I think I've found a sequence that disproves my hypothesis. For the following sequence, it doesn't matter what $a_0, a_1$ are: $a_0 =-1,\ a_1 =-1,\ a_2 = \frac{1}{2},\ a_3 = \frac{1}{3},\ a_4 = -\frac{1}{4},\ a_5 = -\frac{1}{5},\ a_6 = \frac{1}{6},\ a_7 = \frac{1}{7},\ a_8 = -\frac{1}{8},\ a_9 = -\frac{1}{9}, ...$

It's an alternating sequence of decreasing terms (which $\to 0$ ) when put into pairs $(a_0 + a_1, a_2+a_3, a_4+a_5,... ),$ so I think the corresponding series $\sum\limits_{n=0}^{\infty} a_n\ $ is conditionally convergent.
Also, the series $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ is $\geq $ the sum of the pairs: $\sum\limits_{ \text{n is even} }^{\infty} |a_n+a_{n+1}|\ \geq \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + ...,$ which famously diverges. So I believe this sequence disproves my hypothesis(but I'd like confirmation please).
$$$$
Assuming this is all correct, this then begs the modification to the original question:

If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series with $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, then is $\sum\limits_{n=0}^{\infty} a_n+a_{n+1}\ $ convergent?

But for this question, we can even get rid of the $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$ condition and the answer is still yes, because

$\sum\limits_{n=0}^{\infty} a_n+a_{n+1}\ = lim_{n \to \infty} \left(a_0+2\left(a_1+a_2+a_3+...a_{n-1}\right) + a_n\ \right) = 2\sum\limits_{n=0}^{\infty} a_n\ - a_0, $ which converges because $\sum\limits_{n=0}^{\infty} a_n\ $ converges because it is conditionally convergent.

答:
Conditional convergence means the sum converges, but not absolutely.

That is, $\sum_{n\in \Bbb N} a_n $ converges, but $\sum_{n\in\Bbb N} |a_n|$ does not.

Now: $\limsup_{n\to\infty} |\frac{a_{n+1}}{a_n}|=1$ tells us that $|a_{n+1}|$ and $|a_n|$ are very close in value as $n\to\infty$, which implies we can take: $$\sum_{n\in\Bbb N} |a_n+a_{n+1}|\approx\sum_{n\in\Bbb N} |2a_n|=2\sum_{n\in\Bbb N} |a_n|$$
which does not converge.

For the same reason: $$\sum_{n\in\Bbb N} a_n+a_{n+1}\approx \sum_{n\in\Bbb N} 2a_n=2\sum_{n\in\Bbb N} a_n$$ whenever $a_n$ and $a_{n+1}$ are sufficiently close, this implies convergence. While $\limsup|\frac{a_{n+1}}{a_n}|=1$ is a sufficient condition for this, it is  not necessary.


问:
How do I change between cartesian and cylindrical coordinate systems?
I have a vector $\textbf{D}=(x,3,5)$ in cartesian coordinates $(x,y,z)$ that I want to express in cylindrical coordinates $(r,\phi,z)$. Do I just plug the values for x,y,z into the expressions to get the new coordinates and thats it?
$$ \begin{align*}  \phi &= \arctan \frac{x}{y} \\ r &= \sqrt{x^2+y^2} \\ z &= z \end{align*}  $$

答: That is basically correct although you have to be careful with arctan. Cylindrical coordinates is the same thing as converting the $x,y$ plane to polar and keeping the $z$-axis unchanged. This wikipedia article explains the subtleties of the arctan conversion: https://en.wikipedia.org/wiki/Polar_coordinate_system#Converting_between_polar_and_Cartesian_coordinates.


问:
Let $A$ be a set and $R,S$ are relations on $A$. Then....
Let $A$ be a set and $R,S$ are relations on $A$. Define the relation $R+S$ on $A$ as $$R+S = (R-S)\cup(S-R)$$
Now show that

If $A$ is countable then so does $R+S$

Give and example of a set $A$ that is uncountable but $R+S$ is countable.

Prove or disprove: If $R$ and $S$ are partial orders on A then so does $R+S$

For the first part, I am thinking that a relation gives us subset of the set on which it is defined, which is in our case $A$ so if $A$ is a countable set then its subset is countable as well.

For the second part, Lets say $A=\mathbb{R}$ relation $S$ gives us rational numbers (or its subset) and $R$ gives us Natural numbers (or its subset) then their symmetric difference which is $R+S$ becomes a countable set however $\mathbb{R}$ is uncountable.

For third part, I am thinking of relations $R=S=\{(a,a)|a\in \mathbb{Z} \}$ which holds both reflexive and transitive property but I am not sure about anti symmetry. Also this is just an example. How can we prove (if it is right) or disprove (Give counter example)

Please help

答:
A relation $R$ on a set $A$ is not a subset of $A$: it is a subset of $A\times A$. That means that your answer to the second question is wrong from the start: relations on $\Bbb R$ are not subsets of $\Bbb R$. You could, however, let $A=\Bbb R$ and let $R=S$ be any relation on $\Bbb R$, say $\{\langle x,x\rangle:x\in\Bbb R\}$, the identity relation on $\Bbb R$. Then

$$R+S=(R\setminus S)\cup(S\setminus R)=\varnothing\cup\varnothing=\varnothing\,,$$

which is certainly countable.

If the source of this problem misuses countable to mean countably infinite, you’ll have to work a little harder. You could, for instance, let $$R=\{\langle n,n+1\rangle:n\in\Bbb N\}$$ and $$S=\{\langle n,n-1\rangle:n\in\Bbb Z\}\,;$$ then $$R+S=(R\setminus S)\cup(S\setminus R)=R\cup S\,,$$ which is countably infinite.

Something very much like your idea, but with a correct understanding of what a relation on $A$ is, can be used to answer the first question. If $A$ is countable, then so is $A\times A$, and therefore so is every subset of $A\times A$, i.e., every relation on $A$.

Your example for the third question will work: your relation is a partial order, and $R+S=R+R=\varnothing$ is not reflexive, so it is not a partial order.


问:
Existence of distribution for a Moment Generating Function
Question : Does a distribution exist for which $M_X(t)=\frac{t}{(1-t)},|t|<1 ? \text{ If yes, find it. If no, Prove it.}$

Answer : Since the mgf is defined as $M_X(t)=\mathbb{E}e^{tX}$, We necessarily have $M_X(0)=\mathbb{E}e^{0}=1.$ But $\frac{t}{1-t}$ is zero at $t=0$, therefore it cannot be an mgf.

Could you please explain the answer kindly ?

答:
It is based on the definition of MGF. $M_x(t)=E(e^{tx}) \\$

$M_x(0)=E(e^{0})=E(1)=1$

and for this case, $M_x(0)=\frac{0}{1}=0$


答: Hint:  $(13)$ and $(47)$ commute, because they are disjoint.  Also, $(13)$ has order two.

Title: Why does $(13)(47)(13)(23)=(47)(23)$?


问:
What does "well defined" mean?
On page 153 of Linear Algebra Done Right the second edition, it says:

Define a linear map $S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$ by:

7.43:   $S_1 (\sqrt{T^* T}v)=Tv$

First we must check that $S_1$ is well defined. To do this, suppose $v_1, v_2 \in V$ are such that $\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$. For the definition given by 7.43 to make sense, we must show that $Tv_1=T v_2$.

It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?

Thanks

答:
When authors claim that something needs to be proven to be well defined that is because the definition given is under suspicion of having some flaw: it might not define anything at all. For example:

(Fake) Definition.  Let $x$ be the smallest strictly positive real number.

Since there is no smallest positive real number, the definition above doesn't define anything.

Most of the time, this occurs in the literature when one is attempting to define a function, as lavishly explained in several other answers, but whenever there is reason for suspicion, a definition must be proven to be ok.

Title: What does "well defined" mean?


问:
Is it possible/meaningful to have a right triangle with sides $0$, $1$, and $i$?
If $i^2 = -1$, does that mean you could have a triangle with side lengths $0$, $1$, and $i$?  But if you draw such as triangle on a complex number grid, the hypotenuse length is $\sqrt 2$, not $0$.  Does the pythagorean theorem only apply to real numbers?

答:
No.

That wouldn't be possible because distance is inherently a nonnegative real number, usually -- you can see this in the formal definition of a metric, which is how we define distance in a mathematically rigorous and formal sense.

Within that framework, then, it's not possible to have negative distances, much less complex ones. For instance, we can define the distance between the complex numbers $0,i,1$ by the usual "Euclidean distance" or "$2$-metric", the distance you're familiar with. But for complex numbers, this ties back somewhat to definition of magnitude. We define the distance $d(z_1,z_2)$ between complex numbers $z$ (under this metric) to be

$$d(z_1,z_2) = |z_1-z_2|$$

where $| \cdot |$ denotes magnitude. Notice how this is a real number (magnitudes of complex numbers always are). In fact, if $z_i = x_i + i y_i$ for $i=1,2$, then

$$d(z_1,z_2) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$$

What this all says is that it is nonsense to say that a side length can have length "$i$". This is because what you're ultimately, truly meaning is "what is the distance between $0$ and $i$ in the complex plane (under a certain metric)?" Yes, $0+i=i$, but that distance is meant to be measured in terms of nonnegative, real numbers -- otherwise, the distance between $-2$ and $0$ is $-2$, which doesn't make sense.

Or in a different sense, perhaps bear in mind the difference between "distance/length" and "displacement."


问:
Show that $2\le U_n \le 3$ for all $n$
Given $U_n = (1 + 1/n)^n$ , $n = 1,2,3,\ldots\;.$

Show that $2 \le U_n \le 3$ for all n

This is what I've done. Can anyone help?

$$\begin{align*}
a_n=\left(1+\frac1n\right)^n&=\sum_{r=0}^n{^nC_r}(1)^r\left(\frac1n\right)^{n-r}\\
&=\sum_{r=0}^n{^nC_r}\left(\frac1n\right)^{n-r}\\
&=1+\frac{n}n+\frac1{2!}\frac{n(n-1)}{n^2}+\frac1{3!}\frac{n(n-1)(n-2)}{n^3}\\
&\quad\,+\ldots+\frac1{n!}\frac{n(n-1)\ldots(n-n+1)}{n^n}
\end{align*}$$

Since $\forall k\in\{2,3,\ldots,n\}$: $\frac1{k!}<\frac1{2^k}$, and $\frac{n(n-1)\ldots\big(n-(k-1)\big)}{n^k}<1$,

$$\begin{align*}
a_n&<1+\left(1+\frac12+\frac1{2^2}+\ldots+\frac1{2^{n-1}}\right)\\
&<1+\left(\frac{1-\left(\frac12\right)^n}{1-\frac12}\right)<3-\frac1{2^{n-1}}<3
\end{align*}$$

答:
Here is another way to look at it that might be helpful, if the question is why for all $n$ it holds. Consider $f(x)=(1+\frac{1}{x})^x$. It is an increasing function of $x$:

$$ f'(x)=\frac{d}{dx}e^{x \ln(1+\frac{1}{x})}
= \left[\ln\left(1+\frac{1}{x}\right) - \frac{1}{1+x}\right]\left(1+\frac{1}{x}\right)^x >0, \forall x>0$$

Now:

$f(1)=2$, and $\lim_\limits{n\rightarrow+\infty}f(n)=e<3$, so $2\leq f(n) < e < 3, \forall n$

Title: Show that $2\le U_n \le 3$ for all $n$


问:
Diagonalizing a block diagonal matrix
I have a matrix of $2 \times 2$ blocks which looks like

\begin{equation}
M =
\begin{pmatrix}
A & B & B & \cdots \\
B & A & B & \cdots \\
B & B & A & \cdots \\
\vdots &\vdots& \vdots& \ddots
\end{pmatrix}
\end{equation}
with $dim M = 2N$ (or $N \times N$ blocks). I can put this into a nice block-diagonal form with Mathematica:

\begin{equation}
M =
\begin{pmatrix}
A + (N-1)B &  &  &  \\
 & A-B &  &  \\
 &  & A-B &  \\
\, &\,& \,& \ddots
\end{pmatrix}
\end{equation}
but I don't know where exactly this form comes from. This may just be due to the form of the blocks, which for completeness are
\begin{equation}
A =
\begin{pmatrix}
1 & 1 \\
a & a + 1
\end{pmatrix}, \; \; \; B =
\begin{pmatrix}
0 & 0\\
\eta & 0
\end{pmatrix}
\end{equation}
but I'd like to know how to see this diagonalization explicitly.

答:
First, let's review the problem where $A,B$ are merely scalars. Then we can express our matrix as $$M=(a-b)I_N+buu^\top$$ where $u$ is a vector of 1s. We take advantage of this by noting that the eigenbasis of the rank-one matrix $uu^\top$ is simply $u_1:=u$ itself (with eigenvalue $N$) and the $N-1$ eigenvectors $u_2,u_3,\ldots u_N$ where $u_k:=e_{k}-e_{1}$ (with eigenvalue $0$). Note that that the latter vectors are all orthogonal to $u$.

To make this more explicit, we introduce $$U=\begin{pmatrix} u_1 & u_2 & \cdots & u_N\end{pmatrix} =
\begin{pmatrix}
1 & -1 & -1 & \cdots &  -1 \\
1 & 1 & 0 &\cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 &    0 & \cdots & 1
\end{pmatrix}$$
and observe that $$uu^\top U=\begin{pmatrix} u(u^\top u_1) & u(u^\top u_2)& \cdots& u(u^\top u_N)\end{pmatrix}=\begin{pmatrix} Nu & 0 & \cdots & 0\end{pmatrix}=U \Lambda$$ where $\Lambda=\operatorname{diag}(N,0,\ldots,0)$. Hence $uu^\top$ is diagonalized as $uu^\top=U^{-1}\Lambda U$, and therefore $M$ is as well:

\begin{align}
UMU^{-1}
&=(a-b)I_N+ b U^{-1}uu^\top U  \\
&=(a-b)I_N+b U^{-1}U\Lambda \\
&= (a-b)I_N+N b \Lambda\\
&= \operatorname{diag}(a+(N-1)b,a-b,\ldots, a-b)
\end{align} in agreement with Mathematica's proposition.

That's fine for the case of scalar $A,B$, but what about 2-by-2 matrices? To deal with this, we write $M$ using the Kronecker product $\otimes$  as $$M=I_N\otimes (A-B)+ uu^\top\otimes B.$$ We can now diagonalize $M$ as before, but now using the matrix

$$U\otimes I_2=
\begin{pmatrix}
I_2 & -I_2 & -I_2 & \cdots &  -I_2 \\
I_2 & I_2 & 0_2 &\cdots & 0_2\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
I_2 & 0_2 &    0_2 & \cdots & I_2
\end{pmatrix}$$

where $0_2$ is a 2-by-2 matrix of zeros. We have

\begin{align}
(U\otimes I_2)M(U\otimes I_2)^{-1}
&=(U \otimes I_2)\Big[I_N\otimes (A-B)+uu^\top \otimes B\Big ](U^{-1} \otimes I_2)\\
&=(UI_N U^{-1})\otimes(I_2(A-B)I_2)+(Uu^\top U^{-1})\otimes(I_2 B I_2)\\
&=I_N\otimes (A-B)+\Lambda \otimes B
\end{align}
which is exactly the block-diagonal form given by Mathematica. So we can indeed block-diagonalize $M$ in the way proposed by Mathematica, in a way that's entirely analogous to the construction for scalar $A,B$. (Indeed, it's straightforward to do this for square matrices $A,B$ of arbitrary dimension.)

A closing observation: While the first column of $U$ was determined explicitly (up to a multiplicative constant), the remaining columns were chosen to have $N-1$ vectors which were orthogonal to $u$. These were far from unique: Not only would any permutation suffice, but we could also have picked for instance the vectors $$e_1-e_2,e_2-e_3,\ldots,e_N-e_{N-1}.$$ As such, there's no one way to diagonalize $M$ in the desired form. The form chosen above is simply what seemed easiest to write out explicitly.

Title: Diagonalizing a block diagonal matrix


问:
If $X$ has mean $\mu_x$, variance $\sigma_x^2$, and $Y$ has mean $\mu_y$, variance $\sigma_y^2$, what are the mean and cov of $(X, Y)$?
If r.v. $X$ has mean $\mu_x$, variance $\sigma_x^2$, and r.v $Y$ has mean $\mu_y$, variance $\sigma_y^2$. I'm interested in the expectation and covariance of the random vector $\begin{bmatrix} X \\ Y \end{bmatrix}$. Without any additional information about the joint distribution, is it always true that

$$\mathbb{E}\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix}$$
$$Cov\begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \sigma_x^2 & \sigma_{xy} \\ \sigma_{xy}& \sigma_y^2 \end{bmatrix}$$

where $\sigma_{xy}$ is the covariance $Cov(X, Y)$?

答:
Yes, the expectation is calculated elementwise, so the expected value of the random vector $(X, Y)^T$ it is simply column vector of $\mu_x$ and $\mu_y$, i.e., $\mu = (\mu_x, \mu_y)^T$. For the covariance, denote $Z = (X, Y)^T$, thus by definition
$$
Cov(Z) = E[(Z-\mu)(Z-\mu)^T]
$$
write $(Z-\mu)(Z-\mu)^T$ explicitly
$$
(Z-\mu)(Z-\mu)^T =
\begin{pmatrix}
(X-\mu_x)(Y-\mu_y) \quad (X-\mu_x)(Y-\mu_y)\\
(Y-\mu_y)(X-\mu_x) \quad (Y-\mu_y)(Y-\mu_y)
\end{pmatrix},
$$
and then apply expectation elementwise, so you got for the entries $\Sigma_{12}$ , and $\Sigma_{21}$ the definition of covariance between $X$ and $Y$, i.e., $E[(X-\mu_x)(Y-\mu_y)] = Cov(X,Y) = \sigma_{xy}$.


问:
A short exact sequence of groups and their classifying spaces
Suppose that we have a short exact sequence of topological groups:
$$1 \to H \to G \to K \to 1.$$
I have found some papers mentioning that the above sequence induces a fibration:
$$BH \to BG \to BK.$$
Here $B$ assigns to each (topological) group its classifying space.

My question:

How can we show the fibration?

Are there any good books/papers explaining the categorical properties of the classifying space functor $B$?

Note:

I found a relevant question at MO. But I can not see that $EK \times_K (EG/H)$ has the same homotopy type as $BG$.

At the same page at MO, the paper "Cohomology of topological groups" (by Segal) is suggested. But it is not available for me. So I am looking for other papers.

答:
I think there's a typo in that MO answer: $BG$ is modeled by $EG \times_G E(G/H)$ since this is just $BG \times E(G/H)$ and $E(G/H)$ is contractible.

Now we have a natural map $EG \times_G E(G/H) \rightarrow B(G/H)$ which is a fibration with fiber $(EG)/H \cong BH$.

For a great reference on all this stuff written in a very friendly style see:

http://www.math.washington.edu/~mitchell/Notes/prin.pdf


问:
What is an analytic coefficient? (Kashiwara)
In Kashiwara's thesis he uses $\mathcal{D}$-modules to investigate (systems of) linear PDEs with analytic coefficients. There are also mentions of

analytic manifolds

analytic linear PDEs

analytic $\mathcal{D}$-modules

analytic sheaves

analytic spaces

and others...

I understand what an analytic function is.

My question is what is an analytic coefficient?

Is it a constant? Is it a an analytic polynomial?

答: It seems like an analytic coefficient means that that the coefficient can be any analytic function only depending on the independent variable.</p>


问:
Different representations of hyperbolic space
Two equivalent representations (not quite sure if that's the right word) of hyperbolic space $\mathcal{H}^n$ are

the zero set of $- x_0^2 + x_1^2 + \ldots x_n^2 =-1 $

the quadratic space: vector space $\mathbb{R}^n$ with quadratic form $q= - x_1^2 +x_2^2 + \ldots x_n^2$

( For example, see https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.70.5236&rep=rep1&type=pdf and https://arxiv.org/abs/1910.07112)

However, with spheres $\mathcal{S}^n$ we take

the zero set of $ x_0^2 + x_1^2 + \ldots x_n^2 =1$

the quadratic space: the vector space $\mathbb{R}^n$ with quadratic form $q= x_1^2 + \ldots + x_n^2$

What is weird to me is that in the first representation of $\mathcal{H}^n$, the quadratic form is equal to a negative one. Although writing $ x_0^2 - x_1^2 - \ldots- x_n^2 =1 $, would have a lot of negative signs, it seems to be in better correspondence with the spherical case. However, if we do this, shouldn't the quadratic form in the second representation change and be $q = x_1^2 -x^2_2 - \ldots - x_n^2$? This would then change the signature, right?

So my questions are:

A) Why do we define hyperbolic space as the zero set of $- x_0^2 + x_1^2 + \ldots x_n^2 =-1 $ instead of the zero set of $ x_0^2 - x_1^2 - \ldots- x_n^2 =1 $?

B) If we define it as the zero set of $ x_0^2 - x_1^2 - \ldots- x_n^2 =1 $, shouldn't we say that the quadratic form (for the second representation) is then $q=  x_1^2 - x_2^2 - \ldots- x_n^2$?

答:
The first important feature in both cases --- $\mathcal H^n$ and $\mathcal S^n$ --- is that the equation defining that surface can also be used to define a quadratic form on $\mathbb R^{n+1}$. Namely, the surface is a level set of the quadratic form. (You mention quadratic forms on $\mathbb R^n$; see my last paragraph for remarks on that issue).

The second and key feature is that when that quadratic form on $\mathbb R^{n+1}$ is restricted to the tangent space at each point of the surface, the restricted form on that plane is positive definite (of rank $n$), and hence defines a Riemannian metric on the surface.

In particular, when you restrict $-x_1^2+x_2^2+...+x_n^2$ to each tangent space of the hyperplane $-x_1^2+x_2^2+...+x_n^2=-1$ then the result is positive definite of rank $n$. For example, at the point $x_1=1$, $x_2=...=x_n=0$, the tangent hyperplane is given by the equation $x_1=0$; and the restricted quadratic form on that plane is $x_2^2+...+x_n^2$ which is positive definite. With a bit of analytic geometry work you can verify that the restricted form is positive definition on the tangent hyperplane of every point.

But when you restrict the quadratic form $-x_1^2+x_2^2+...+x_n^2$ to the tangent spaces of the hyperplane $-x_1^2+x_2^2+...+x_n^2=1$, you will get a form of type $(-1,1,...,1)$ (with $n-1$ 1's), and hence is indefinite and does not define a Riemannian metric on the surface. For example, at the point $x_2=1$ and $x_1=x_3=...=x_n=0$, the tangent hyperplane is $x_2=0$, and the restricted quadratic form is $-x_1^2+x_3^2+...+x_n^2$ which has signature $(-1,1,..,1)$.

Remarks: Let me just add, it's not clear to me why you associate a quadratic form on $\mathbb R^n$ with the $\mathcal H^n \subset \mathbb R^{n+1}$, and it is similarly not clear to me why you associate a quadratic form on $\mathbb R^n$ with $\mathcal S^n \subset \mathbb R^{n+1}$. Instead, what you should be associating to each of $\mathcal H^n$ and $\mathcal S^n$ is a quadratic form on $\mathbb R^{n+1}$, which is how the earlier portions of my answer are written.


问:
Proving an identity using formal power series

4.

(a) Prove that $\dfrac{1-x^2}{1+x^3}=\dfrac{1}{1+\frac{x^2}{1-x}}$.

(b) By expanding each side of the identity in (a) as a power series, and considering the coefficient of $x^N$, prove that
  $$\left|\sum_{k\geq0}(-1)^{k}{N-k-1 \choose N-2k}\right|=\begin{cases}
0 & \text{ if }N\equiv1\pmod3\\
1 & \text{ otherwise }
\end{cases}$$

Apparently I need to use formal power series. However, I can't seem to derive anything useful by attempting to divide $\frac{1-x^2}{1+x^3}$. How should I proceed? More generally, how should I go about proving identities using formal power series, and how should I expand a fraction into a power series?

答:
As DonAntonio points out, the first part is just a matter of factoring and doing some algebra. For the second part, note first that

$$\begin{align*}
\frac{1-x^2}{1+x^3}&=(1-x^2)\cdot\frac1{1-(-x^3)}\\
&=(1-x^2)\sum_{n\ge 0}(-1)^nx^{3n}\\
&=\sum_{n\ge 0}(-1)^nx^{3n}-\sum_{n\ge 0}(-1)^nx^{3n+2}\;,
\end{align*}$$

in which the coefficient of $x^{3n}$ is $(-1)^n$, the coefficient of $x^{3n+1}$ is $0$, and the coefficient of $x^{3n+2}$ is $(-1)^n$. Clearly, then, the absolute value of the coefficient of $x^N$ is $0$ if $N\equiv 1\pmod 3$ and $1$ otherwise.

Similarly,

$$\begin{align*}
\frac{1}{1+\frac{x^2}{1-x}}&=\frac1{1-\left(-\frac{x^2}{1-x}\right)}\\
&=\sum_{n\ge 0}(-1)^n\left(\frac{x^2}{1-x}\right)^n\\
&=\sum_{n\ge 0}(-1)^nx^{2n}\frac1{(1-x)^n}\\
&=\sum_{n\ge 0}(-1)^nx^{2n}\sum_{k\ge 0}\binom{n+k-1}kx^k\;,\tag{1}
\end{align*}$$

where the last step uses the standard power series expansion of $\dfrac1{(1-x)^n}$.

The coefficient of $x^N$ in $(1)$ is

$$\sum_{n\ge 0}(-1)^n\binom{N-n-1}{N-2n}\;;$$

to get this I simply write $N=2n+k$ and sum over $n$, so that $k=N-2n$. (The binomial coefficient is $0$ when $n$ is too large, so there’s no need to specify an upper bound.)

The result is now immediate.


答:
Hint Let
$$U_n = (1 + 1/n)^n \\
V_n=U_n = (1 + 1/n)^{n+1}$$

It is clear that $U_n < V_n$ for all $n$. Use Bernoulli inequality to show that

$$\frac{U_{n+1}}{U_n} \geq 1 \\
\frac{V_{n}}{V_{n+1}} \geq 1$$

Find some $m$ so that $V_m \leq 3$.
Deduce from here that
$$U_1 \leq U_n \leq V_m$$ for all $n$.

Title: Show that $2\le U_n \le 3$ for all $n$

Body:
Given $U_n = (1 + 1/n)^n$ , $n = 1,2,3,\ldots\;.$

Show that $2 \le U_n \le 3$ for all n

This is what I've done. Can anyone help?

$$\begin{align*}
a_n=\left(1+\frac1n\right)^n&=\sum_{r=0}^n{^nC_r}(1)^r\left(\frac1n\right)^{n-r}\\
&=\sum_{r=0}^n{^nC_r}\left(\frac1n\right)^{n-r}\\
&=1+\frac{n}n+\frac1{2!}\frac{n(n-1)}{n^2}+\frac1{3!}\frac{n(n-1)(n-2)}{n^3}\\
&\quad\,+\ldots+\frac1{n!}\frac{n(n-1)\ldots(n-n+1)}{n^n}
\end{align*}$$

Since $\forall k\in\{2,3,\ldots,n\}$: $\frac1{k!}<\frac1{2^k}$, and $\frac{n(n-1)\ldots\big(n-(k-1)\big)}{n^k}<1$,

$$\begin{align*}
a_n&<1+\left(1+\frac12+\frac1{2^2}+\ldots+\frac1{2^{n-1}}\right)\\
&<1+\left(\frac{1-\left(\frac12\right)^n}{1-\frac12}\right)<3-\frac1{2^{n-1}}<3
\end{align*}$$


问:
Optimal game strategy New Zealand IMO selection problem
The numbers 1, . . . , 1000 are written on the board. Two players take
turns erasing a number; a number x may be erased if x = 1, or x − 1 has
been erased, or x is even and x/
2
has been erased. The player to erase
1000 wins.
Which player has a winning strategy?

The first player wins if the number of numbers removed  is odd once 1000 has been removed. I think that regardless of what number is removed the opponent can always choose to remove some number other than 500 or 999 until only these numbers and 1000 remains, and therefore the first player wins. This is because whichever number the opponent chooses to remove no numbers are eliminated from consideration, since each player will keep trying to avoid removing 500 or 999 until they and 1000 only remain and the opponent can pick a number 1 greater than the number they choose to erase. Assuming optimal play each player will avoid their loss as long as possible so in all 999 numbers will be removed at the end of the game, since either 500 or 999 may not be removed.

Is my approach correct?

答:
No :

$501$ cannot be removed before $500$

so there is a losing position when faced with $500,501,999,1000$ remaining