问:
prove by induction that every finite nonempty set of reals has a minimum.
Exercise on Mathematical Induction is from Mathematics for Computer Science by Albert R. Meyer, Eric Lehman, and Frank Thomson Leighton. It says,

Prove by induction that every nonempty finite set of real numbers has a minimum element.

I rephrased it in the following way to make induction variable explicit.

Let $S_n$ denote the set of $n$ real numbers.

   Theorem: Every nonempty $S_n$ has a smallest element.

Proof. By ordinary induction. Let induction hypothesis $P(n)::=$ in any set $S_n$, containing $n$ real numbers, $\exists m \in S_n: \forall s \in S_n: m \leq s$. Prove $P(n)$ holds for all $n \geq 1$.

 Base case ($n=1$): $P(1)$ is obviously true. If we choose $m$ to be that single element of $S_n$, then it's simply less than or equal to itself. Therefore, $P(1)$ holds.

  Inductive step: Suppose $P(n)$ holds for some $n \geq 1$. Show that $P(n) \Rightarrow P(n+1)$.

   Since $P(n)$ holds, we can conclude there exists $m \in S_n$ such that $\forall s \in S_n: m \leq s$. Since $P(n)$ holds for any set containing $n$ real numbers, it must be hold for the first $n$ elements of $S_n$. That is, there exists an element $m$ smaller than all of the first $n$ elements. We just need to compare $m$ to the (n+1)th element of $S_{n+1}$ and take the minimum of these two as new $m$, which proves $P(n+1)$.

  We can conclude by induction principle $p(n)$ holds for all $n \in N$.

Note: $N::= \{1,2,...\}$

 I've doubt on the proof and structure of this induction proof. I know its inductive part is a bit wordy. I didn't use set notations to write, instead tried to explain it in this way. Is this way of proving acceptable? I appreciate your answers.

答:
There are several problems.

“Let $S_n$ denote the set of $n$ real numbers.” It makes no sense, since there are infinitely many such sets.

“Theorem: Every nonempty $S_n$ has a smallest element.” The statement should be that every finite and non-empty set of real numbers has a smallest element.

You write as if you were working on one set $S_n$ with $n$ elements and one set $S_{n+1}$ with $n+1$ elements. You can do it as follows: let $S$ be a set with $n+1$ elements. Take $m\in S$. If $m$ is the smallest element of $S$, you're done. otherwise, let $S'=S\setminus\{m\}$. The $S'$ has $n$ elements and therefore, by the induction hypothesis, it has a smallest element $m'$. Since $m$ is not the smallest element of $S$, $m'\leqslant m$ and, for each $k\in S'$, $m'\leqslant k$. Therefore, $m'$ is the smallest element of $S$.


问:
Write vector $\vec{u}$ as a sum of vector parallel to $\vec{v}$ and perpendicular to $\vec{v}$
Write vector $\vec{u}$ as a sum of vector parallel to $\vec{v}$ and perpendicular to $\vec{v}$

$\vec{u}$ = $< 2,4,2 >$ and $\vec{v}$ = $< 1,2,-1 >$

I was pretty sure that the way to solve it was to find the parallel vector by doing

$
  u_1
= \left(\frac{|u|\cdot |v|}{|v| \cdot |v|}\right)v $

and then subtract that vector from $\vec{u}$ to get the perpendicular one.

That left me with $\vec{u}$ = $< 2,4,-2 > + < 0,0,4 >$ which can't be correct because the second vector isn't perpendicular to $\vec{v}$.

What am I doing wrong, or did I mess up my algebra?

答: The standard way is\begin{align}\vec u&=\left\langle\vec u,\frac{\vec v}{\left\|\vec v\right\|}\right\rangle\frac{\vec v}{\left\|\vec v\right\|}+\left(\vec u-\left\langle\vec u,\frac{\vec v}{\left\|\vec v\right\|}\right\rangle\frac{\vec v}{\left\|\vec v\right\|}\right)\\&=\left(\frac{4}{3},\frac{8}{3},-\frac{4}{3}\right)+\left(\frac{2}{3},\frac{4}{3},\frac{10}{3}\right).\end{align}

ContentLicense: CC BY-SA 4.0


问:
A simple yet hard task for (theoretically) Poisson distribution
Sorry if I don't use the words properly, I haven't learnt these things in English, only some of the words.
Anyway, I'm practicing to one of my exams and sadly this task seemed more challanging for me than it should be. Some kind of explain would help a lot!

10 meters of clothes have 6 holes in it.

a) What kind of distribution does the number of holes per meter follow? (I think it must be Poisson distribution)

b) What's the probability that there's more than 10 holes in 5 meters of clothes?

答:
The Poisson model is not unreasonable. So if $X$ is the number of holes in a $1$ meter piece of cloth, then it is reasonable to assume that $X$ has a more or less Poisson distribution with parameter $\lambda =\dfrac{6}{10}$.

Let $Y$ be the number of holes in a random $5$ meters of cloth. Then it is reasonable to assume that $Y$ has more or less Poisson distribution with parameter $\lambda=6\cdot\frac{5}{10}=3$.

The probability there are more than $10$ holes is $1$ minus the probability there are $10$ or fewer holes. The probability there are $10$ or fewer holes is
$$\sum_{i=0}^{10} e^{-3}\frac{3^i}{i!}.$$

The calculation is a little unpleasant. The last few terms are pretty small, and are not really terribly relevant. One should realize that the Poisson model is only a model. It fits reality modestly well, but one should not expect a  high accuracy fit.

ContentLicense: CC BY-SA 3.0


问:
How do I plug this function into the limit?
I've been having some trouble with these questions, specifically with plugging the function in.
Could somebody help me with the step by step on how to plug it in and simplify? I want to get an understanding of this.

ORIGINAL QUESTION:
$$\lim _{Δx\to 0}\left(\frac{f\left(x+Δx\right)-f\left(x\right)}{Δx}\right)$$

For the function

$$f(x)=x^2-3x+2$$

答:
Here it is a more general approach for the sake of curiosity.

Let $f(x) = ax^{2} + bx + c$, where $a\neq 0$. Then we have that

\begin{align*}
\lim_{h\to 0}\frac{f(x + h) - f(x)}{h} & = \lim_{h\to 0}\frac{a[(x+h)^{2} - x^{2}] + bh}{h}\\\\
& = \lim_{h\to 0}\frac{2axh + h^{2} + bh}{h}\\\\
& = \lim_{h\to 0}(2ax + b + h) = 2ax + b
\end{align*}

Such result is known as the derivative of the function $f(x) = ax^{2} + bx + c$ at the point $x\in\mathbb{R}$.

Hopefully this helps!


Body:
Write vector $\vec{u}$ as a sum of vector parallel to $\vec{v}$ and perpendicular to $\vec{v}$

$\vec{u}$ = $< 2,4,2 >$ and $\vec{v}$ = $< 1,2,-1 >$

I was pretty sure that the way to solve it was to find the parallel vector by doing

$
  u_1
= \left(\frac{|u|\cdot |v|}{|v| \cdot |v|}\right)v $

and then subtract that vector from $\vec{u}$ to get the perpendicular one.

That left me with $\vec{u}$ = $< 2,4,-2 > + < 0,0,4 >$ which can't be correct because the second vector isn't perpendicular to $\vec{v}$.

What am I doing wrong, or did I mess up my algebra?

Body:
You want that

$$(\vec{u}-\lambda \vec{v})•\vec{v}=0$$

or

$$((2,4,2)-\lambda (1,2,-1))•(1,2,-1)=0$$

So,

$$(2-\lambda)+2(4-2\lambda)-(2+\lambda)=0$$

thus

$$\lambda=\frac 43$$
and

$$\vec{u}=\lambda \vec{v}+ (\vec{u}-\lambda \vec{v})$$


Title: Write vector $\vec{u}$ as a sum of vector parallel to $\vec{v}$ and perpendicular to $\vec{v}$

Body:
You must have done some calculation mistake. By your formula the parallel vector is
$$v_{//}=\frac{(2,4,2)\cdot (1,2,-1)}{1^2+2^2+(-1)^2}(1,2,-1)=\frac{4}{3}(1,2,-1)$$
Then
$$v_{\perp}=u-v_{//}=\left(\frac{2}{3},\frac{4}{3},\frac{10}{3}\right)$$

And indeed $v_{\perp}\cdot v_{//}=\dfrac{4}{9}(2+8-10)=0$


问:
Express a irreducible polynomial $f(X)$ in the field of characteristic $p>0$ $F[X]$ as $g(X^{p^m})$
$F[X]$ is of the form $a_0+a_1X+...+a_nX^n$ where $a_0,a_1,...a_n\in F\text{ with characteristic p>0}.$
Express $f\in F[X]\ as\ g(X^{p^m})$, where the nonnegative integer m is a large as possible and $f$ is irreducible. Show that g is irreducible and separable.

This is a problem in section 3.4 of Basic Algebra by Robert Ash.

The question I had about is how to write $f(X)$ in form of $g(X^{p^m})$. I have tried this using the property $(a+b)^p=a^p+b^p$, since $F$ has characteristic p and the Frobenius Automorphism indicate if $\alpha \in F, \alpha = \beta ^p,$ for some $\beta \in F$. $g(X^{p^m})$ can be therefore written as $(b_0+b_1X+...+b_nX^n)^{p^m}.$ This is writing $g(X^{p^m})$ in in form of $f(X)$, for the other way around, is it just taking the $p^m$ root of the above expression:
$$f(X)=\sqrt[p^m]{g(X^{p^m})}$$

答:
The fact that $g$ is irreducible follows from the fact that $f$ is; i.e. if $g = h_1 h_2$ then $f = h_1(x^{p^m})h_2(x^{p^m})$.

Once we show existence, this proves that $g$ is separable too: an irreducible polynomial is separable iff $g' \neq 0$, and in characteristic $p$ the fact that $g' = 0$ in turn means that $g = \widetilde{g}(x^p)$ for some $\widetilde g$. We cannot have the latter, otherwise it would contradict the maximality of $m$.

Now, consider

$S = \{m \in \mathbb N_0 : f = g(x^{p^m}), \text{ for some $g$}\}.$

The set $S$ is non empty, because $0 \in S$, and it is bounded above by degree considerations. Hence there exists $m = \max S$ with an associated $g$, which is irreducible and separable by the previous remarks.


问:
Problem 17(i i ): Show $\Bbb R$ is regular

I will try to prove $\Bbb R$ is regular

Attempt (edited)

Let $\Bbb R$ be defined on usual topology

Let A be closed  and  A$\subset\Bbb R$ and x$\in \Bbb R$ and x$\notin A$

So if [a,b] is closed then it’s complement U=$\Bbb R\setminus [a,b]$ is open.

By Def 3.2.1 if x$\in\Bbb R $ then x$\in U\subset \Bbb R $ s.t (a,b)$\cap A=\emptyset$

And U=(-$\infty$,a)$\cup(b,\infty$) which is the union of open sets,so open

So U $\supseteq A$ and A and
(-$\infty$,a) and (b,$\infty$)are disjoint containing x and A respectively ?

Any help would be appreciated.

Am l missing anything?

Do l have to find to subsets within (-$\infty$,a) and (b,$\infty$)

答:
As I said in my comment, you are supposed to assume that $\Bbb R$ has its usual topology, not the finite-closed topology (also known as the cofinite topology), and most of what you’ve written suggests that you were in fact thinking of the usual topology. For instance, an interval $[a,b]$ is not finite unless $a=b$, so in general it is not closed in the finite-closed topology, but it is closed in the usual topology. However, it is very far from being the only kind of closed set: $\Bbb Z$ is a closed subset of $\Bbb R$ that is not of the form $[a,b]$, as is $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, to name just two.

Now let’s take a look at your argument. You start with a set $A\subseteq\Bbb R$ and a point $x\in\Bbb R$ such that $x\notin A$; since you’re trying to prove that $\Bbb R$ is regular, that’s a very reasonable place to start, except that you also want to assume that $A$ is closed. After that, though, things go a bit haywire. You let $p$ be a point of $A$, but why? You never mention $p$ again, so it serves absolutely no purpose.

Then you say that we can find ‘an open nbhd of $[a,b]$ of $x$ such that $[a,b]\cap A=\varnothing$’. I suspect that the first of wasn’t supposed to be there, and it was supposed to read ‘an open nbhd $[a,b]$ of $x$’, but $[a,b]$ isn’t open. Its complement $U=(\leftarrow,a)\cup(b,\to)$ is open in the usual topology, not the ‘closed topology’, what ever that is.

As you’ve set things up it is indeed true that $U\supseteq A$, and it is also true that $(\leftarrow,a)$ and $(b,\to)$ are disjoint, but $A$ could be a subset of either of those rays: there is no reason to think that $A\subseteq(b,\to)$. And since you apparently wanted $[a,b]$ to be a nbhd of $x$, $x$ must be in $[a,b]$, so clearly $(\leftarrow,a)$ cannot contain $x$.

Let’s go back to the beginning and start over. Let $A$ be a closed set in $\Bbb R$, and let $x\in\Bbb R\setminus A$. Then $\Bbb R\setminus A$ is an open set containing $x$, so there is an open interval $(a,b)$ such that $x\in(a,b)\subseteq\Bbb R\setminus A$. Since $x\in(a,b)$, we know that $a<x<b$, so there are $c,d\in\Bbb R$ such that $a<c<x<d<b$. Let $U=(c,d)$ and $V=(\leftarrow,a)\cup(b,\to)$; clearly $U$ and $V$ are open sets, $x\in U$, and $U\cap V=\varnothing$. I’ll leave it to you to complete the proof by showing that $A\subseteq V$.


问:
The topology of the free topological group
I don't understand something about the following result:

Claim: The topology of the group $F(X)$ is the finest topological group topology on $F_a(X)$ that generates on $X$ its original topology.

($X$ is a Tychonoff topological space, and $F(X)$ denotes the free topological group on $X$, and $F_a(X)$ denotes the free group on $X$).

Proof: We already know that $F(X)$ is algebraically isomorphic with $F_a(X)$ and the topology $\tau$ of $F(X)$ generates on X its original topology

Now, let $\tau '$ be a group topology on $F_a(X)$ such that $\tau '\big|_X=\tau _X$, where $\tau _X$ is the topology of $X$. Extend the identity mapping $i:X\to X$ to a continuous homomorphism $\overline{i}:F(X)\to F_a(X)$, where $F_a(X)$ is endowed with the group topology $\tau '$. Clearly, $\overline{i}$ is an algebraic isomorphism, and the continuity of $\overline{i}$ implies that $\tau '\subseteq \tau$.

My question is, why does the continuity of $\overline{i}$ imply that $\tau '\subseteq \tau$?

答: $\bar i$ continuous $\overset{def}\iff$ for all open $U\subseteq F_a(X)$, i.e., $U\in\tau'$ we have $\bar i^{-1}(U)$ is open in $F(X)$, so, as $i$ and by the group structure, $\bar i$ is also identity, $U=\bar i^{-1}(U)$, it means $U\in\tau$.</p>


问:
Compatibility of tensor product with subspace structure
Suppose that $W$ (resp. $W'$) is a subspace of the vector space $V$ (resp. $V'$).

My question is : how can we see the tensor product of $W$ and $W'$ (seen as vector spaces) as a subspace of $V\otimes V'$ ? More precisely :

As vector spaces themselves, we can define the tensor product $W\otimes' W'$ as the quotient
of the free vector space over $W\times W'$ by the space generated by the terms of the type $\lambda (v,w) - (\lambda v,w)$, $(v+v',w)-(v,w)-(v',w)$, $\lambda(v,w)-(v,\lambda w)$, and $(v,w+w')-(v,w)-(v,w')$, where $v,v'\in W$, $w,w'\in W'$ and $\lambda\in \mathbb{K}$.

I used the notation $\otimes'$ for the tensor product on $W\times W'$ ($W$ and $W'$ seen as vector spaces), to distinguish with the tensor product on $V\times V'$ (denoted $\otimes$).

There is a natural linear map $f:W\otimes' W'\to V\otimes V'$ such that $f(w\otimes' w')= w\otimes w'$. The question is : is that map $f$ injective ?
It seems to me that the answer is yes, but I haven't been able to find a precise reference of that fact.

答:
Hint: tensor products over vector spaces are convenient enough, in the sense that if $V$ and $V'$ have basis $B$ and $B'$, then $\{v \otimes v' : v \in B, v' \in B'\}$ is a basis for $V \otimes V'$.

Proof sketch: since tensor products commute with direct sums, we have

$$
V \otimes V' = \bigoplus_{v \in B} kv \otimes \bigoplus_{v' \in B'} kv' = \bigoplus_{v\in B, v' \in B'} kv \otimes kv'
$$

and $k v \otimes kv' \simeq k \otimes k \simeq k$. Hence $V \otimes V'$ is a sum of copies of $k$. Track each isomorphism to recover the explicit basis in term of elementary tensors. "$\square$"

Now, fix bases of $W$ and $W'$, extend them to $V$ and $V'$, and check what the image of $f$ on a basis of $W \otimes W'$ is.


问:
Does partitioning a sample space change probability of events?
Suppose I have 10 white balls and 1000 black balls.
Considering that picking each ball is equally likely, the probability of picking a white ball comes out to be $\frac{10}{1010}$, now if I keep all the white balls in box-1 and all the black balls in box-2, will this constitute as valid partitioning, as now the probability of selecting a white ball becomes $\frac{1}{2}$ i.e. equal to the probability of selecting box-1?
Although sample space and measure of that space (probability) are different concepts, I feel that this view of partitioning is wrong, as it violates the following equation*:

If U=[A1,...,An] is a partition of sample space,S and B is an arbitrary event then,
  P(B)=P(B|A1)P(A1)+...+P(B|An)P(An)

So according to this the probability of withdrawing a white ball should have remained the same and partitioning would have given us just another way to calculate it.

$*$ from page number 32, of Probability, Random Variables and Stochastic Processes ($4^{th} edition$) Book by Athanasios Papoulis and S. Unnikrishna Pillai.

答: @jflipp gave the correct answer, but I'll offer another perspective since this has confused me in the past as well.  When you physically divide the white and black balls into two separate boxes, you have gone way beyond partitioning the sample space.  You have actually created new sample space where the elements are Box 1 and Box 2.  Now the probability of selecting Box 1 = probability of selecting Box 2.  Since your sample space is now different, the outcome of the experiment is different.  Since you've essentially created a new "experiment", the outcomes will be as you stated (50% chance of picking white or black).</p>


问:
Proof and precise formulation of Welch-Satterthwaite equation
In my statistics course notes the Welch-Satterthwaite equation, as used in the derivation of the Welch test, is formulated as follows:

Suppose $S_1^2, \ldots, S_n^2$ are sample variances of $n$ samples, where the $k$-th sample is a sample of size $m_k$ from a population with distribution $N(\mu_k, \sigma_k^2)$. For any real numbers $a_1, \ldots, a_n$, the statistic
  $$
L = \frac{\nu}{\sum_{k=1}^{n} a_k\sigma_k^2}\sum_{k=1}^n a_kS_k^2
$$
  with
  $$
\nu = \frac{\left(\sum_{k=1}^m a_kS_k^2\right)^2}{\sum_{k=1}^m \frac{(a_kS_k^2)^2}{N_k - 1}}
$$
  approximately (sic) follows a chi-squared distribution with $\nu$ degrees of freedom.

Doing a quick Google search, most sources seem to follow a similar formulation.  There are a few questions I have though:

How can the number of degrees of freedom of a chi-squared distribution depend on the statistics $S_i$? Shouldn't the number of degrees of freedom be a constant?

What does 'approximately follow a distribution' mean? This is closely related to my third question:

How can one justify this approximation? All 'proofs' I have seen on the internet seem to assume that $L$ follows a chi-squared distribution and then show that $\nu$ must have the stated value, using basic properties of expected value and variance. Even those arguments are confusing to mee, since they seem to ignore the fact that the $S_i$ are itself stochastic variables and not known constants.

I found a link to the original papers by Welch and Satterthwaite, which I can't access freely. I wonder if there is a reasonably short answer to at least the first two of my questions. Us students are not expected to be able to prove the equation, so it's not that bad if there is no readily available proof, but I'd like to at least understand the statement itself.

答: See Loveland, Theorem 23, page 47. She seems to provide a really nice treatment!</p>


问:
Which notation to use for "large arguments" in a function of two or more variables?
While performing some calculations it has often come up the gaussian integral. I use the notation:

$ I(A,B) = \int\limits_{-\infty}^{+\infty} \exp\left( -\frac{A}{2}x^2 + Bx \right) \mathrm{d} x $

However this becomes cumbersome when A and B are themselves some expressions. For instance, in my case:

$ A = \frac{i}{\hbar}\frac{\varepsilon}{m} \qquad B = \frac{i}{\hbar}\left(\dot{x}_n^i - \frac{1}{2} f^i(x_{n+1},t) - \frac{1}{2} f^i(x_n,t)\right)\varepsilon$

which results in:

$ I(A,B) = I\left(\frac{i}{\hbar}\frac{\varepsilon}{m},\; \frac{i}{\hbar}\left(\dot{x}_n^i - \frac{1}{2} f^i(x_{n+1},t) - \frac{1}{2} f^i(x_n,t)\right)\varepsilon\right) $

I've also tried:

$ I(A,B)\bigg|_{A = \frac{i}{\hbar}\frac{\varepsilon}{m},\; B = \frac{i}{\hbar}\left(\dot{x}_n^i - \frac{1}{2} f^i(x_{n+1},t) - \frac{1}{2} f^i(x_n,t)\right)\varepsilon} $

but I think it's worse.

I'd rather not "interrupt" calculations by introducing new variables midway and the long expression may obscure the comma between the arguments. Is there some (relatively common and understandable) notation which prevents this problem (or make it less significant)?

答:
You could use the notation $I(A \mid B)$ instead of $I(A,B)$, resulting in
$$
   I(A,B) = I\left(\frac{i}{\hbar}\frac{\varepsilon}{m} \;\middle|\;\frac{i}{\hbar}\left(\dot{x}_n^i - \frac{1}{2} f^i(x_{n+1},t) - \frac{1}{2} f^i(x_n,t)\right)\varepsilon\right).
$$
That way the separator doesn't get lost. This is not common notation, but not completely made up. I've seen it used for the same purpose with inner products and other two-parameter expressions, though not with $I(A,B)$ in particular.

To typeset this in LaTeX: I prefer \mid over | for a vertical separator, but LaTeX doesn't let you stretch \mid, so the second expression above uses \middle| with \;'s around it for manual spacing.


问:
Why is the intersection of this indexed set a closed interval?
Let $I= \mathbb Z_+$. Let $B_i=(-1-\frac{1}{i},1+\frac{1}{i})$. The union of $\bigcup\limits_{i\in I} B_{i}=(-2,2)$ but why is $\bigcap\limits_{i\in I} B_{i}=[-1,1]$? How did we get to a closed interval?

答: Hint: the interval $[-1,1]$ is contained in each $B_i$, hence it should be contained in their intersection. But for any other point $x \not \in [-1,1]$, can't you find some $B_i$ such that $x \not \in B_i$? Maybe doing some drawings can help to get a mental picture of the phenomenon.


问:
Is pi lying on the ground, and on TV? - and on the sun?
Consider the leaves from a bunch of trees in a terraced plaza in the Autumn. It may well happen that the tiles of the terrace are squares whose length easily exceeds the length of the stem of the leaves (assuming leaves all of the same kind). Cannot this be regarded as a Buffon process? That is, if you took a photograph of it and counted the line crossing of the leaf-stems with the lines determined by the tiles, you would get Buffon's calculation for pi. Furthermore, you could do this even in the absence of the tiles, simply by imposing an appropriate tiling over the photograph later, right? And if we can impose a tiling later, what about imposing it on a TV program, with respect to some recurring object? Would that, properly done, not constitute a Buffon process as well?

edit: It might apply to sunspots as well, because sunspots, during solar min, typically have low latitudes, i.e., they  occur near the equator of the sun, which can be considered as approximately a flat surface, and each sunspot approximates a highly non-smooth Jordan curve, which therefore almost always is going to have a unique greatest diameter, which can be considered as the needle. Sunspots vary greatly in size, so we could restrict our attention to ones of a given size (most likely the smaller ones, so as to maximize the number of sunspots considered), and then consider the circles of latitude, starting at the equator, separated by a distance a little bit greater than the needle length (perhaps, such that the needle length is 5/6 that of the circle separations, in imitation of what Mario Lazzarini did in 1901.) Of course, since we’re considering solar min, we might want to consider all sunspots over, say, a 22-year period, so as to get a lot of data points (trial tosses of the needle). So, how good an approximation to pi would be get? – and is there any correlation between how good an approximation we get and certain solar events such as flares?

答:
Let's do an experiment! Here's a deck with leaves on it:

Counting the leaves, I see about 100 leaves. Now lets assume all the stems are the same size, even for the larger and smaller leaves. Also let's try to visualize stems on some leaves which don't seem to have stems. Interpolating, I see about 30 crossings.

Now the Buffon needle problem says that the probability of a crossing is

$$P(\mbox{crossing})=\frac{2L_{stem}}{L_{deck}\pi}$$

Now the rough ratio I'm seeing from the image is $L_{stem}/L_{deck}\approx 1/2$, which gives:

$$P(\mbox{crossing})\approx \frac{1}{\pi}$$

or with around 100, leaves, we'd expect to see $100/\pi\approx 32$ crossings. Not bad, since we counted 30, which gives our experimental approximation of $\pi$ as 3.33, and since I probably counted pretty poorly, I probably have a standard error of around three leaves, which puts $\pi$ within our experimental error. By the way, it can be shown that we used an unbiased estimator for $\pi$, whose mean-square error scales something like $1/n$, where $n$ is the number of leaves, so we should get a somewhat accurate guess of $\pi\approx 3.14$ past around 1000 leaves. I leave that experiment to you.


问:
parameterizing intersections of 2 planes
The curve C is the intersection of $x^2+y^2=1$ and $2x+y+z=4$

Finding the curve C according to the book is doing cylinder coordination such that, $x=cos\theta , y=sin\theta,z=4-sin(\theta)-2(cos(\theta)), 0 \leq \theta \ \leq2\pi$

but i don't they chose that approach to describe the intersections between the 2 planes

答:
$x^2+y^2=1$ is a cylinder and $2x+y+z=4$ is a plane

The intersection is an ellipse (in blue in the picture below) whose parametric equations are

$$(x=\cos \theta,y=\sin \theta,z=4-\sin  \theta-2 \cos \theta);\;\theta\in[0,2\pi]\tag{1}$$

Because $\cos^2\theta+\sin^2\theta=1$ so the curve $(1)$ satisfies the equation of the cylinder. Furthermore $z=4-y-2x$ is the explicit equation of the plane, so $(1)$ satisfies the equation of the plane, too.

$$...$$


问:
Find $q(x) \in \Bbb{Q}[x] $ such that $f(a)$ is its zero
Let $p(x) \in \Bbb{Q}[x] $ and $a \in \Bbb{Q}$ such that $p(a)=0$. Find $q(x) \in \Bbb{Q}[x] $ such that $f(a)$ is its zero, where
$p(x) =x^3 - 9x - 3$ and $f(a)=a^2-2 $.

How do I approach this problem?

答:
We have that $a$ is a root of $p$, so $a^3-9a-3=0$. This is equivalent to

$$a(a^2-9)-3=0\Leftrightarrow a(a^2-2-7)-3=0\Leftrightarrow a(f(a)-7)-3=0$$

so $f(a)$ is a root of $q(x)=ax-7a-3=0$. Since $a\in Q$, we can guarantee that $q(x)\in Q[x]$.


答:
In the plane

$$\{(x,y)\in \Bbb R^\color{red}{2}\;:\; x^2+y^2=1\}$$

is a circle with $(0,0)$ as center and $ R=1 $ as radius.

In the space
$$\{(x,y,z)\in\Bbb R^\color{red}{3}\;:\; x^2+y^2=1\}$$
is a vertical cylinder which can be parametrized as

$$x=\cos(t)$$
$$y=\sin(t)$$
$$z=z$$

If a point $ (x,y,z) $ lies in the intersection, then
$$x=\cos(t)$$
$$y=\sin(t)$$
$$z=4-2x-y=4-2\cos(t)-\sin(t)$$
where $ t\in [0,2\pi]$.

Title: parameterizing intersections of 2 planes

Body:
The curve C is the intersection of $x^2+y^2=1$ and $2x+y+z=4$

Finding the curve C according to the book is doing cylinder coordination such that, $x=cos\theta , y=sin\theta,z=4-sin(\theta)-2(cos(\theta)), 0 \leq \theta \ \leq2\pi$

but i don't they chose that approach to describe the intersections between the 2 planes


问:
Show that $\frac{1}{(-q,q)_\infty}=(q,q^2)_\infty$
Show that $\frac{1}{(-q,q)_\infty}=(q,q^2)_\infty$

I found the following proof:
$\frac{1}{(-q,q)_\infty}=\frac{(1-q)(1-q^2)\cdots}{(1-q^2)(1-q^4)\cdots}=(q,q^2)_\infty$
However, this seems to lack formality since I am treating the infinite product as something that can  "cancel out" infinitely. I feel like I would have to make this argument for the partial product, but this method no longer works. Is there a way to formalize what I did here?

答:
We know the infinite product
\begin{align*}
(a,q)_{\infty}:=\prod_{k=0}^\infty\left(1-aq^k\right)
\end{align*}
is analytic in the interior of the unit disc which implies it can be expanded as convergent power series at the origin for $|q|<1$.

We show the following is valid:
\begin{align*}
\color{blue}{\frac{1}{(-q,q)_{\infty}}=\left(q,q^2\right)_{\infty}}\tag{1}
\end{align*}

We obtain for $|q|<1$:
\begin{align*}
\color{blue}{\frac{1}{(-q,q)_{\infty}}}&=\frac{(q,q)_{\infty}}{(-q,q)_{\infty}(q,q)_{\infty}}\\
&=\frac{\prod_{j=0}^\infty\left(1-q^{j+1}\right)}{\prod_{k=0}^\infty\left(1+q^{k+1}\right)\prod_{l=0}^\infty\left(1-q^{l+1}\right)}\\
&=\frac{\prod_{j=0}^\infty\left(1-q^{j+1}\right)}{\prod_{k=0}^\infty\left(1-q^{2k+2}\right)}\tag{2}\\
&=\frac{\lim_{N\to\infty}\prod_{j=0}^{2N+1}\left(1-q^{j+1}\right)}{\lim_{{M}\to\infty}\prod_{k=0}^M\left(1-q^{2k+2}\right)}\tag{3}\\
&=\lim_{N\to\infty}\frac{\prod_{j=0}^{2N+1}\left(1-q^{j+1}\right)}{\prod_{k=0}^{N}\left(1-q^{2k+2}\right)}\tag{4}\\
&=\lim_{N\to\infty}\prod_{j=0}^{N}\left(1-q^{2j+1}\right)\tag{5}\\
&=\prod_{j=0}^{\infty}\left(1-q^{2j+1}\right)\\
&\,\,\color{blue}{=(q,q^2)_{\infty}}
\end{align*}
and the claim (1) follows.

Comment:

In (2) we use $\left(1-q^{j+1}\right)\left(1+q^{j+1}\right)=\left(1-q^{2j+2}\right)$ and multiply two convergent infinite products.

In (3) we use the definition of the infinite product as limit of a sequence of finite products. This notation is admissible, since none of the factors is zero. Note, we conveniently set upper limits which is useful in the next steps.

In (4) we use one single limit instead of two limits which can be shown by recalling the $\varepsilon$ definition of the limits.

In (5) we cancel factors with even powers of $q$.


问:
Can a real plane curve have a singular complex point but no singular real points?
Let $p(x,y)\in\mathbb{R}[x,y]$. Is it possible that $\partial_xp(x,y)=\partial_yp(x,y)=p(x,y)=0$ has a solution $(x,y)\in\mathbb{C}^2$ but no solutions $(x,y)\in\mathbb{R}^2$?

In other word, can you see if a real curve is "smooth" just by looking at the real points?

Of course, one could ask similar questions about arbitrary varieties over an arbitrary field $k$ (with $\mathbb{C}$ replaced by the algebraic closure $\bar{k}$): can a variety $X$ over $k$ be smooth at all $k$-points but have a singular $\bar{k}$-point, and if so, are there nice conditions (on $k$ or $X$) which make this impossible?

答: Sure.  For a simple example, take $p(x,y)=(x^2+1)(y^2+1)$.  This has no real points at all, but has singular complex points at $(\pm i,\pm i)$.  (This example can immediately be generalized to any non-algebraically closed field, by letting $p(x,y)=q(x)q(y)$ where $q$ is a polynomial with no roots in the base field.  Or, if you want to require $p$ to be squarefree, this works as long as the field is not separably closed.  I don't know an irreducible example off the top of my head but I expect it is possible.)


问:
How to reparametize a distribution?
I have a hard time understanding what's the purpose of reparametization and how to do it. My prof briefly talked about it without giving us an example and asked us to reparametize the Log Normal distribution but I have no idea how to start it.

Thank you in advance for your help!

答:
A log-normal distribution is usually parametrised by the mean $\mu$ and variance $\sigma^2$ of the underlying normal distribution.

But you could do something else, for example by the actual mean $m$ and variance $v$ of the log-normal.  If you try this, you need to find $\mu$ and $\sigma^2$ in terms of $m$ and $v$, and then substitute that into the pdf or CDF.  You might start with

$m=\exp\left(\mu+\frac{\sigma^2}{2}\right)$

$v=\left(\exp(\sigma^2)-1\right) \exp\left(2\mu+\sigma^2\right)$

so $\exp(\sigma^2)=1+\frac{v}{\mu^2}$

and so on

Title: How to reparametize a distribution?


问:
Help with this double summation
I'm having trouble when the indexes are related in a double summation. For example, this problem:
$\sum^n_{i = 1} \sum^n_{j = i+1} j - i + 2$

How could I sum this and what's a general approach to this type of double/triple summations?

答:
It depends greatly on what you know. I might be inclined to let $k=j-i$ and evaluate it like this:

$$\begin{align*}
\sum_{i=1}^n\sum_{j=i+1}^n(j-i+2)&=\sum_{i=1}^n\sum_{k=1}^{n-i}(k+2)\\\\
&=\sum_{i=1}^n\left(\sum_{k=1}^{n-i}k+2(n-i)\right)\\\\
&=\sum_{i=1}^n\frac12(n-i)(n-i+1)+2\sum_{i=1}^n(n-i)\\\\
&=\sum_{i=1}^n\left(\binom{n-i+1}2+2(n-i)\right)\\\
&=\sum_{i=0}^{n-1}\left(\binom{i+1}2+2i\right)\\\\
&=\binom{n+1}3+2\binom{n}2\\\\
&=\binom{n}3+3\binom{n}2\;.
\end{align*}$$

Of course, you need to know some binomial coefficient identities in order to do that.

Or you could go at it straight up. First,

$$\sum_{i=1}^n\sum_{j=i+1}^n(j-i+2)=\sum_{i=1}^n\left(\sum_{j=i+1}^nj-\sum_{j=i+1}^n(i-2)\right)\;.$$

Now $\sum_{j=i+1}^nj$ is just the sum of an arithmetic progression, so it’s $\frac12(n+i+1)$. And $\sum_{j=i+1}^n(i-2)=(n-i)(i-2)$, since it’s the sum of $n-i$ identical terms. The double sum then reduces to a single sum over $i$ of a quadratic in $i$, which is straightfoward to evaluate if you know that $\sum_{i=1}^ni^2=\frac16n(n+1)(2n+1)$.

Title: Help with this double summation


问:
Is a well-ordered set always countable?
Suppose that $A$ is a well-ordered set. We can find $a_{0} = \min{A}$. Then, if $A \setminus \left\{a_{0}\right\} \neq \emptyset$, we can find $a_{1} = \min{\left(A \setminus \left\{a_{0}\right\}\right)}$, etc. Intuitively, this seems to be a counting process. Then my hypothesis is, every well-ordered set is countable. Is this a valid guess? In my understanding, if it is valid, then for finite well-ordered sets, we can use mathematical induction to prove it. Then what about infinite cases?

答: Every ordinal is well-ordered by $\in$. Ordinals are arbitrarily large. All the axiom of choice adds is that all sets biject with an ordinal, thereby being well-orderable. Even proper classes can be well-orderable; the class of ordinals is well-ordered by $\in$.

Title: Is a well-ordered set always countable?


问:
What is the value of a delta function?
So my understanding is that a delta function $\delta(t)$ has the value of $\infty$ at $t = 0$ and $\delta(t) = 0$ everywhere else. Essentially, it's just one big spike on a graph right?

In my signal processing lecture, I came across the following:

Here, we arrive at $X_1[k] = \frac{1}{2}\delta [k - 1] + \frac{1}{2}\delta [k + 1]$, which means  $X_1[k]$ only has non-zero value at $k - 1$ and $k + 1$. However, using my definition of a delta function, shouldn't $X_1[k = \pm 1] = \infty$?

This isn't the case when you look at the graph. It shows $X_1[k = \pm 1] = \frac{1}{2}$, so what's wrong? Is there some other delta function I just don't know?

答:
$\newcommand{\eqd}{\triangleq}$
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$\newcommand{\brp}[1]{{\left(#1\right)}}$
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$\newcommand{\deriv} [2]   {{\frac{\mathrm{d}#1}{\mathrm{d}#2} }}$
$\newcommand{\R}{\Bbb{R}}$
$\newcommand{\Z}{\Bbb{Z}}$
$\newcommand{\Rx} {{\R^\ast}}$
$\newcommand{\intcc} [2]  {{\left[#1:#2\right]}}$
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$\newcommand{\intoc} [2]  {{\left(#1:#2\right]}}$
$\newcommand{\intco} [2]  {{\left[#1:#2\right)}}$
$\newcommand{\ff}{\mathrm{f}}$
$\newcommand{\fy}{\mathrm{y}}$
$\newcommand{\fF}{\mathrm{F}}$
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$\newcommand{\du}{\mathrm{du}}$
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$\newcommand{\dt}{\mathrm{dt}}$
$\newcommand{\dtau}{\mathrm{d\tau}}$
$\newcommand{\kdelta}{\bar{\delta}}$
$\newcommand{\setu}{\cup}$
$\newcommand{\setn}[1]{{\left\{{#1}\right\}}}$
$\newcommand{\seqn}[1]{{\left[{#1}\right]}}$
$\newcommand{\inprod}[2]{\left\langle{#1}\,|\,{#2}\right\rangle}$

Building on what others have already said, the two functions are just that—two distinct functions.
One is the real-valued Kronecker Delta function $\kdelta:\Z\to\R$ defined as
$$\kdelta(n)\eqd\brbl{\begin{array}{cl}
  1 & \text{if $n=0$}\\
  0 & \text{otherwise}
\end{array}}$$
and the other is the Dirac Delta $\delta:\R\to\Rx$, which obviously is not a real-valued function at all.
This is not so much a problem in and of itself
(functions can be mappings to real numbers, complex numbers, colors, faces of a die, ...)
but in the case of $\delta$, it maps not into a well behaved set like the real numbers $\R$,
but rather into the set of
extended real numbers
$\Rx\eqd\R\setu\setn{-\infty,+\infty}$.
The set $\Rx$ is an Alice-in-Wonderland type world and when working therein requires great care
and proper safety equipment.
In $\Rx$ crazy things can happen like $\infty+\infty=\infty$.

Because $\delta(x)$ maps to $\Rx$, many would not call it a function at all, but rather an
extended function, or a distribution
(see for example Mallat Appendix A.7).
A distribution may be something of a loose cannon when left to its own devices,
but when constrained to the inside of an integral
(or more generally inside an inner product)
it becomes quite well-behaved, predictable, and useful.

Building on what @Anixx and @Thomas-Andrews have already pointed out,
the Dirac Delta distribution can be defined, together with $\int\dx$, as an
operator (a mapping from one
linear space to another/same linear space)
with the following definition:
$$\inprod{\ff(x)}{\delta(x)} \eqd \int_{\R}\ff(x)\delta(x)\dx \eqd \ff(0)$$

So the Dirac Delta distribution $\delta(x)$ and the Kronecker delta function are distinctly different.
But that being said (that they are very different in form), they are very similar in function.
In particular, note the following:

(1) Both "sum" to 1:
\begin{align*}
  \boxed{\int_{\R} \delta(x)\dx}
    &= \int_{\R} 1\cdot\delta(x)\dx
  \\&\eqd \boxed{1}
    && \text{by definition of $\delta$}
  \\
  \\
  \boxed{\sum_{n\in\Z}\kdelta(n)}
    &= \kdelta(0)
    && \text{by definition of $\kdelta$}
  \\&= \boxed{1}
\end{align*}

(2) Both have a similar time-shift property:
\begin{align*}
   \boxed{\int_{\R} \delta\brp{x-a}\ff(x)\dx}
     &= \int_\R \delta\brp{u}\ff(u+a)\du
     && \text{where $u=x-a$ $\implies$ $\dx=\du$}
   \\&= \boxed{\ff(a)}
  \\
  \\
  \boxed{\sum_{n\in\Z}\kdelta(n-k)y(n)}
    &= \sum_{n\in\Z}\kdelta(m)y(m+k)
    && \text{where $m=n-k$ $\implies$ $n=m+k$}
  \\&= \kdelta(0)y(0+k)
    && \text{by definition of $\kdelta$}
  \\&= \boxed{y(k)}
\end{align*}

(3) Both induce a projection operator.
A projection operator $P$ is a linear operator such that
$P^2=P$:
\begin{align*}
  \boxed{P^2}\ff(x)
    &= PP\ff(x)
  \\&\eqd P\int_\R \delta(x)\ff(x) \dx
    && \text{where here $P$ is the Dirac Delta operator}
  \\&\eqd P\ff(0)
    && \text{by definition of $\delta$}
  \\&\eqd \int_\R \delta(x)\ff(0)\dx
    && \text{where here $P$ is the Dirac Delta operator}
  \\&= \ff(0) \int_\R \delta(x)\dx
    && \text{by linearity of the integral operator}
  \\&= \ff(0)
    && \text{because $\int_{\R} \delta(x)\dx=1$}
  \\&= \int_\R\delta(x)\ff(x)\dx
    && \text{by definition of $\delta$}
  \\&= \boxed{P}\ff(x)
    && \text{where here $P$ is the Dirac Delta operator}
  \\
  \\
  \boxed{P^2}\fy(n)
    &= PP\fy(n)
  \\&\eqd P\sum_{n\in\Z} \kdelta(n)\fy(n)
    && \text{where here $P$ is the Kronecker Delta operator}
  \\&\eqd P\fy(0)
    && \text{by definition of $\kdelta$}
  \\&\eqd \sum_{n\in\Z} \kdelta(n)\fy(0)
    && \text{where here $P$ is the Kronecker Delta operator}
  \\&= \fy(0) \sum_{n\in\Z} \kdelta(n)
    && \text{by linearity of the summation operator}
  \\&= \fy(0)
    && \text{because $\sum_{n\in\Z} \kdelta(n)=1$}
  \\&= \sum_{n\in\Z}\kdelta(n)\fy(n)
    && \text{by definition of $\kdelta$}
  \\&= \boxed{P}\fy(n)
    && \text{where here $P$ is the Kronecker Delta operator}
\end{align*}

(4) Both can be used for sampling.
The Dirac can be used—inside an integral—as a projection operator to map
a function $\ff(t)$ to a single point $\ff(a)$ for some value $t=a$.
That is, it can be used to sample $\ff(t)$ at a given time $t=a$.
In the field of Digital Signal Processing
(DSP) a continuous time function $\ff(t)$ can be transformed (mapped) to a

sequence
(a function with domain $\Z$)
$\seqn{\ldots, x_{n-1}, x_n, x_{n+1}, \ldots}$,
where each element $x_k$ of this sequence is
$$x_k \eqd \int_\R \ff(t)\delta(t-kT)\dt = \ff(kT)$$
Likewise, the Kronecker delta can be used to sample (OK, maybe a bit of a stretch here?)
a sequence $\seqn{\ldots, x_{n-1}, x_n, x_{n+1}, \ldots}$
in the sense
$$y(n)= \cdots + x_{-2}\kdelta(n+2) + x_{-1}\kdelta(n+1) + x_0\kdelta(n) + x_1\kdelta(n-1) + x_2\kdelta(n-2) + \cdots$$

Title: What is the value of a delta function?


问:
If a two variable smooth function has two global minima, will it necessarily have a third critical point?
Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a  $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$, and then imagine an endless rain pouring with water level rising steadily on the entire plane.

Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima.

Those two points are connected by a compact line segment $K$. As a continuous function, $f$ attains a maximum value $M$ on the set $K$. This means that when the water level has reached $M$, the two lakes will have been merged.

The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$.

It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$. But, can we prove that one always exists?

Follow-ups:

Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ($S^1\times S^1$) or the surface of a sphere ($S^2$). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima?

Similarly, what if we have local minima instead of global?

If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded - not sure that would be at all relevant, but who knows).

答:
With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is
$$
f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, .
$$
$f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$.

If the gradient
$$
\nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \,  2e^y(e^y-x^2) \bigr)
$$
is zero then $e^y =x^2$ and $x(x^2-1) = 0$. $x= 0 $ is not possible, so that the gradient is zero only if $x=\pm1$ and $y=0$, that is only at the global minima.

The construction is inspired by Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?. We have $f(x, y) = g(\phi(x, y))$ where:

$g(u, v) = (u^2-1)^2 + v^2$ has two global minima, but also an additional critical point at $(0, 0)$, and

$ \phi(x, y) = ( x , e^y-x^2)$ is a diffeomorphism from the plane onto the set $\{ (u, v) \mid v > -u^2 \}$. The image is chosen such that it contains the minima of the function $g$, but not its critical point.

With respect to the “connected lakes” approach: The level sets
$$
L(z) = \{ (x, y) \mid f(x, y) \le z \}
$$
connect the minima $(-1, 0)$ and $(1, 0)$ exactly if $z > 1$. The infimum of such levels is therefore $m=1$, but $L(1)$ does not connect the minima (it does not contain the y-axis). Therefore this approach does not lead to a candidate for a critical point.

The above approach can also be used to construct a counterexample with bounded derivatives. Set $f(x, y) = g(\phi(x, y))$ with

$g(u, v) = \frac{(u^2-1)^2}{1+u^4} + \frac{v^2}{1+v^2}$, which has two global minima at $(\pm 1, 0)$, one critical point at $(0, 0)$, and bounded derivatives.

$\phi(x, y) = (x, \log(1+e^y) +1 -\sqrt{1+x^2} )$, which is a diffeomorphism from $\Bbb R^2$ with bounded derivatives onto the set $\{ (u, v) \mid v > 1- \sqrt{1+v^2} \}$, which contains the points $(\pm 1, 0)$ but not the point $(0, 0)$.


问:
Closed sets as $A\cap B = \varnothing$ such as $\inf \{\|a − b\| \mid a\in A, b\in B\} = 0$
Let $E$ be a normal vector space, $A$, $B\subset E$ closed, nonempty subsets.

We denote by $c = \inf\{\|a − b\| \mid a\in A, b\in B\}$.

I gotta find $2$ closed sets $A$, $B$ such that $c = 0$ and $A\cap B = \varnothing$.

And then, for these sets, give an example of sequences $\{a_{n}\}$ and $\{b_{n}\}$ such that $\|a_{n} - b_{n}\| \to 0$.

I have the idea of  $A = N - \{0\}$ and $B = \{(n+1)/n \mid n \in\mathbb{N}_{>=2}\}$ for my closed sets.

Is it okay? I'm kinda lost about the sequences

Thank You for your help

答: I think that there is a solution over $\mathbb{R}^2$: take the function $x\mapsto 0$ and for example $x\mapsto\frac{1}{x}$ because for continuous functions the set ${(x,f(x))}$ is closed and in this case the graphs are disjoint


问:
Poisson distribution upper bound
We assume $X \sim Poi(\lambda)$, and want to prove that for every $i<\lambda$, $P(X\leq i)\leq e^{-\lambda}\frac{(e\lambda)^i}{i^i}$.
I tried first assuming $i>0$ and following definition:

$P(X\leq i)=\sum_{{k<i}}e^{-\lambda}\frac{\lambda^k}{k!}$ and bounding by $\sum_{{k<i}}e^{-\lambda}\frac{\lambda^i}{k!}$.

So we get $P(X\leq i)=\sum_{{k<i}}e^{-\lambda}\frac{\lambda^k}{k!}\leq e^{-\lambda}\lambda^i\sum_{{k<i}}\frac{1}{k!}$, which can be then bounded by just taking the sum over all the natural numbers, meaning $$P(X\leq i)\leq e^{-\lambda}\lambda^i\sum_{{k<i}}\frac{1}{k!}\leq e^{-\lambda}\lambda^ie$$, but that's the best bound I could find. What am I missing here?

答:
It is not true in general:

Take $\lambda = 2, i =5$ then $P(X \le 5) \approx 0.98344$ and yet $e^{-2}\frac{(2e)^5}{5^5} \approx 0.205675$.

EDIT: I just noticed OP had put $i<\lambda$

Title: Poisson distribution upper bound


答: To clarify exactly what's going on, your argument shows there is a countably infinite subset of any infinite well-ordered ordinal. The reason it doesn't show any well-ordered set is countable is that there's no reason your process should exhaust the set A. If we consider the set $\mathbb{N}\cup\{\infty\}$, where we add a point larger than any natural, we can check this is still a well-ordered set. But the process you describe will never "count" the point at infinity. Of course this set is still countable, but the idea is instead of adding 1 point we can add uncountably many (a priori it's not clear how to order these points, this is just supposed to show why we can get uncountable ordinals).

Title: Is a well-ordered set always countable?

Body: Suppose that $A$ is a well-ordered set. We can find $a_{0} = \min{A}$. Then, if $A \setminus \left\{a_{0}\right\} \neq \emptyset$, we can find $a_{1} = \min{\left(A \setminus \left\{a_{0}\right\}\right)}$, etc. Intuitively, this seems to be a counting process. Then my hypothesis is, every well-ordered set is countable. Is this a valid guess? In my understanding, if it is valid, then for finite well-ordered sets, we can use mathematical induction to prove it. Then what about infinite cases?


问:
Probability that a character kills the boss in a 3 versus 1 fight
In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.

Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

答:
Extra assumptions:

attack damage > HP left means kill.

all damage ranges include both edge values, so e.g. for Character A the damage value $d_A \in [25, 50]$.

With problems like this, if you want to obtain a definite answer, you probably want to start from the end and proceed by backward induction. Here, this means that you begin by considering HP values in the "kill range" of the current attacker. I am going to sketch out the main points of a precise solution, but I will not give you a closed-form answer. I hope that my "sketch" will be clear enough for you to come to the answer you want.

Suppose that it is Character A's move, and the boss's HP is $x \leq 50$. What is the probability of dealing enough damage? It's simply $$\Pr(\text{A kills} \,|\, \text{it's A's turn and } HP = x) = \frac{50+1-x}{50-25+1} = \frac{51-x}{26}.$$ Similarly, for B and C we get, for $x \leq 70$ and $x \leq 80$ respectively, $$\Pr(\text{B kills} \,|\, \text{it's B's turn and } HP = x) = \frac{70+1-x}{70-30+1} = \frac{71-x}{41}$$ and $$\Pr(\text{C kills} \,|\, \text{it's C's turn and } HP = x) = \frac{80+1-x}{80-10+1} = \frac{81-x}{71}.$$

By now we can already write out the answer… in an implicit form:
$$\Pr(\text{A kills}) = \sum_{i=1}^{50} \Pr(\text{it's A's turn and } HP = i) \times \Pr(\text{A kills} \,|\, \text{it's A's turn and } HP = i) =\\= \sum_{i=1}^{50} \Pr(\text{it's A's turn and } HP = i) \times \min\left\{ 1, \max\left\{ 0, \frac{51-x}{26} \right\} \right\},$$
and similarly for the other two characters. The $\min\left\{ 1, \max\left\{ 0, \frac{51-x}{26} \right\} \right\}$ term comes from the fact you want the probability to stay in between 0 and 1.

So the missing part now is the $\Pr(\text{it's X's turn and } HP = i)$ terms. We can write down a recurrent formula for it, no matter the character $X \in \{A,B,C\}$ — as long as the current HP is low enough for X to have had at least one attack before:
$$\Pr(\text{it's X's turn and } HP = i) = \sum_{s=25+30+10}^{50+70+80} \Pr(\text{it's X's turn and } HP = i + s) \times \Pr(\text{total damage over the last round} = s).$$

You see that for the expression above to become useful, we need to know the distribution of damage per round, $\Pr(\text{total damage over the last round} = s)$, for $s \in [65,200]$ (it is zero outside that interval). This probability is $$\Pr(\text{total damage over the last round} = s) = \sum_{d_A = \max\{25,s-70-80\}}^{\min\{50,s-30-10\}} \sum_{d_B = \max\{30,s-d_A-80\}}^{\min\{70,s-d_A-10\}} \frac{1}{26} \cdot \frac{1}{41} \cdot \frac{1}{71}.$$ The fraction product up there is simply the probability of a particular damage value triplet $(d_A,d_B,d_C)$ materializing; the sum limits are trickier — let me explain them.

You have the task to deal exactly $s$ damage in three consecutive hits by character A, B, and C. Suppose that A deals $d_A$ damage. What can $d_A$ be? On the one hand, it has to be at least enough so that B's and C's strongest attacks put the total at $s$, hence $d_A \geq s - 70 - 80$. On the other hand, $d_A$ has to be low enough so that B's and C's weakest attacks do not throw you above $s$, therefore $d_A \leq s - 30 - 10$. And then you also have the limits $25 \leq d_A \leq 50$ from the rules. Applying both means that, at each of the edges, whichever limit is stricter is the one that rules. Similarly, after A's attack you have exactly $(s-d_A)$ more damage to deal, so you get $s-d_A - 80 \leq d_B \leq s-d_A - 10$ in addition to the existing rule that $30 \leq d_B \leq 70$. Finally, after both A and B have hit, C must deal exactly $d_C=s-d_A-d_B$ damage; all that remains is to check that $d_C$ is within the $[10,80]$ interval, but that always holds by construction of the sum limits.

What do we have at this point? We know the probabilities of killing the boss when within range; we also know the probabilities of ending up in a particular game situation (active character + HP count) at any time that is certainly past the first round, i.e. for $HP \leq 1000 - 50 - 70 - 80 = 800$. What remains to be calculated are the probabilities of the game situations for higher HP levels. The problem with that range is, not every combination of a character and an HP level is attainable there, so one has to be careful. Here I would probably switch to analysing the characters move by move rather than round by round as I did before.

Consider $\Pr(\text{it's C's turn and } HP = i)$. Since character C is in play, it must have been B's move right before. Therefore, we have $$\Pr(\text{it's C's turn and } HP = i) = \sum_{d_B = 30}^{\min\{70,1000-25-i\}} \frac{1}{71} \times \Pr(\text{it's B's turn and } HP = i + d_B).$$ (We're talking $i \leq 1000-25-30 = 945$ as anything higher is unattainable, and the respective probability is zero.) By the same token, $$\Pr(\text{it's B's turn and } HP = i) = \sum_{d_A = 25}^{\min\{50,1000-i\}} \frac{1}{26} \times \Pr(\text{it's A's turn and } HP = i + d_A).$$ Here $i \leq 1000-25 = 975$, otherwise $\Pr(\text{it's B's turn and } HP = i) = 0$. Finally, $$\Pr(\text{it's A's turn and } HP = i) = \sum_{d_C = 10}^{\min\{80,1000-25-30-i\}} \frac{1}{72} \times \Pr(\text{it's C's turn and } HP = i + d_C)$$ with, importantly, $i \leq 1000-25-30-10 = 935$; for $i \in [935,999]$ $\Pr(\text{it's A's turn and } HP = 0$, and for $i = 1000$ it is one — that's the start of the game. This concludes the solution.

Remark: I could have left out the part that involves using the distribution of total damage per round and gone for the move-by-move induction rightaway, but I think that, first, it is useful for learning how to solve such problems, and second, it should probably save up quite some computations.

Another remark: my answer seems similar, in terms of formulae, to @leonbloy's. I would also like to reinforce the statement in @Karl's reply: the more rounds, in expectation, the boss takes to kill, the closer the result would be to the straightforward approximation $37.5/(37.5+50+45)$.


问:
Calculate $\sum_{k=1}^{\infty}\frac{|\sin(k x)|}{1+k^2}$
In https://math.stackexchange.com/a/4015346/198592 it was shown that the sum

$$s(x) = \sum_{k=1}^{\infty}\frac{\sin(k x)}{1+k^2}$$

is exactly expressible in terms of the hypergeometric function.

I wonder what happens íf we replace the sine by its absolute value, i.e. ask for the properties of the function $a(x)$ defined by the sum

$$a(x) = \sum_{k=1}^{\infty}\frac{|\sin(k x)|}{1+k^2}$$

Questions

(1) can you find a closed expression for $a(x)$ if $x$ is a rational multiple of $\pi$?

(2) is there a closed expression for $a(x)$?

Here is a plot of the two functions $a(x)$ and $s(x)$

答:
Here I provide a "semi-closed" form expression for $a(x)$ in case of $x\in \mathbb{Q}$ (rational numbers). Perhaps from it, we can obtain the closed form expression for $a(x)$ for $x\in \mathbb{R}$.

As $a(x)$ is a periodic function, it suffices to calculate $a(x)$ for $x\in (0,\pi)$.

Suppose  $x\in \mathbb{Q}$, then $x=\frac{p}{q}\pi$ with $p,q\in \mathbb{N}$, $p<q$ and $(p,q)=1$.
\begin{align}
a(x) &= \sum_{k=1}^{\infty}\frac{|\sin(k x)|}{1+k^2} \\
&= \sum_{k=1}^{\infty}\frac{|\sin(k\frac{p}{q}\pi)|}{1+k^2} \\
&= \sum_{k=1}^q \left(\left|\sin \left(k\frac{p}{q}\pi \right)\right| \sum_{n=0}^{\infty}\frac{1}{1+(k+nq)^2} \right) \tag{1} \\
\end{align}
Denote
$$u(k,q)=\sum_{n=0}^{\infty}\frac{1}{1+(k+nq)^2}$$
We have
\begin{align}
u(k,q) &=\frac{i}{2q} \sum_{n=0}^{\infty}\frac{-2\frac{i}{q}}{(n+\frac{k}{q})^2+\frac{1}{q^2}} \\
&=\frac{i}{2q} \sum_{n=0}^{\infty} \left(\frac{1}{n+\frac{k}{q}+\frac{i}{q}}-\frac{1}{n+\frac{k}{q}-\frac{i}{q}}  \right) \tag{1} \\
\end{align}

We know that the digamma function for complex number is calculated as
$$\psi(z)=-\gamma + \sum_{n=0}^{\infty}\frac{1}{n} - \sum_{n=0}^{\infty}\frac{2}{n+z}=\int_0^{+\infty} \left( \frac{e^{-t}}{t} -\frac{e^{-zt}}{1-e^{-t}}  \right)dt \tag{3}$$

From (2) and (3), we can easily deduce

$$u(k,q)=\int_0^{+\infty} \frac{e^{-\frac{k}{q}t}}{1-e^{-t}}\frac{\sin \left( \frac{t}{q} \right)}{q}dt$$

Hence, from (1) we have
\begin{align}
a(x) &=\sum_{k=1}^q \left(\left|\sin \left(k\frac{p}{q}\pi \right) \right| \int_0^{+\infty} \frac{e^{-\frac{k}{q}t}}{1-e^{-t}}\frac{\sin \left( \frac{t}{q} \right)}{q}dt\right) \tag{4} \\
\end{align}

The formulas (4) can be considered to be the "semi-closed form expression" for $a(x)$ in case of $x\in \mathbb{Q}$ (rational numbers).

Find the closed-form expression for $a(x)$ in general case ($x\in \mathbb{R}$) is equivalent to find the closed-form expression of $a(\frac{p}{q}\pi)$

We have this identity from Fourier transform (link)
$$\left|\sin x\right| = \frac{2}{\pi}-\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}\cos(2j x)$$

Then
\begin{align}
a \left(\frac{p}{q}\pi \right)&=\sum_{k=1}^q \left(\left(\frac{2}{\pi}-\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}\cos \left(2j k\frac{p}{q}\pi\right) \right) \int_0^{+\infty} \frac{e^{-\frac{k}{q}t}}{1-e^{-t}}\frac{\sin \left( \frac{t}{q} \right)}{q}dt\right) \tag{5}\\
\end{align}

Remark: It's possible that from (5), we can transform $a(x)$ to a function $f\left(\frac{p}{q}\right)$. If the function $f\left(\frac{p}{q}\right)$ exists, it suffices then replace $\frac{p}{q}$ by $r \in \Bbb R$.

We can use the formula $(5)$ by noticing that $\sin(x)\cos(y) =\frac{1}{2}(\sin(x+y)+\sin(x-y))$ and

$$\sum_{i=1}^n \sin(nx) = \frac{\sin\left(\frac{nx}{2}\right)\sin\left(\frac{(n+1)x}{2}\right)}{\cos\left(\frac{x}{2}\right)}$$
We have
\begin{align}
a \left(\frac{p}{q}\pi \right)&=\sum_{k=1}^q \left(\left(\frac{2}{\pi}-\sum_{j = 1}^\infty \frac{4}{\pi(4j^2-1)}\cos \left(2j k\frac{p}{q}\pi\right) \right) \int_0^{+\infty} \frac{e^{-\frac{k}{q}t}}{1-e^{-t}}\frac{\sin \left( \frac{t}{q} \right)}{q}dt\right) \\
&=\frac{2}{\pi}\sum_{k=1}^q  \int_0^{+\infty} \frac{e^{-\frac{k}{q}t}}{1-e^{-t}}\frac{\sin \left( \frac{t}{q} \right)}{q}dt - \sum_{j = 1}^\infty \left( \frac{4}{\pi(4j^2-1)} \int_0^{+\infty} \frac{\sum_{k=1}^q  e^{-\frac{k}{q}t}\cos \left(2j k\frac{p}{q}\pi \right) \sin \left( \frac{t}{q} \right)}{(1-e^{-t})q}dt \right)\\
&= I_1 - I_2 \\
\end{align}
The first term $I_1$ is equal to
$$I_1 = \frac{2}{\pi} \int_0^{+\infty} \frac{\sin \left( \frac{t}{q} \right)}{q(e^{\frac{t}{q}}-1)}dt = \frac{2}{\pi} \int_0^{+\infty} \frac{\sin (t)}{e^t-1}dt$$
The second term $I_2$ is equal to
\begin{align}
I_2 &=\sum_{j = 1}^\infty \left( \frac{4}{\pi(4j^2-1)} \int_0^{+\infty} \frac{\sum_{k=1}^q  e^{-\frac{k}{q}t}\cos \left(2j k\frac{p}{q}\pi \right) \sin \left( \frac{t}{q} \right)}{(1-e^{-t})q}dt \right)\\
&=\sum_{j = 1}^\infty \left( \frac{4}{\pi(4j^2-1)} \int_0^{+\infty}
\frac{
\sin(\frac{t}{q})(-e^{-t}+\cos(2jpt)+e^{t+\frac{t}{q}}\cos(2j\frac{p}{q}t)-e^{\frac{t}{q}}\cos(2j(\frac{p}{q}+p)t))}{q(e^t-1)(1+e^{\frac{2t}{q}} -2e^{\frac{t}{q}}\cos(2j\frac{p}{q}t))
}
\right)\\
\end{align}

Denote $r = \frac{p}{q}$
\begin{align}
I_2 &=\sum_{j = 1}^\infty \left( \frac{4}{\pi(4j^2-1)} \int_0^{+\infty}
\frac{
\sin(\frac{t}{q})(-e^{-t}+\cos(2jrqt)+e^{t+\frac{t}{q}}\cos(2jrt)-e^{\frac{t}{q}}\cos(2jr(1+q)t))}{q(e^t-1)(1+e^{\frac{2t}{q}}+-2e^{\frac{t}{q}}\cos(2jrt))
}
\right)\\
\end{align}

We can transform the integrand into a function depended on only $r$ and $t$. But it's too calculating. I must stop here.


问:
Domain of a function quick question
$$f(x)=\arcsin\left(\frac{x-1}{2x}\right)$$
We need to find the domain of this function.

My try:

$$-1\leq \frac{x-1}{2x} \leq 1 $$

We can split this into
$$-1\leq \frac{x-1}{2x} \quad\text{and} \quad\frac{x-1}{2x} \leq 1$$
My idea is to solve for this two inequalities and then take the intersection of them both.

$$-2x\leq x-1\\
-3x\leq -1\\
x\geq\frac{1}{3}$$

$$x-1 \leq 2x\\
-1 \leq x$$

By doing the intersection I get the wrong answer which is: $x\geq\frac{1}{3}$

What have I done wrong?

答:
Hint:
You forgot the case when x is negative and then, you can't multiply the inequality with $2x$.You need to change the direction of the inequality.

So take by the 2 cases:

1.$x>0$

2.$x<0$

If you still don't get it to work, tell me and I will come up with a complete solution

Title: Domain of a function quick question


问:
Directed graphs, chains, paths...
First of all - what's a directed graph? Is this just some normal graph but with no loops or edges? I don't really understand definitions found in the web...

What is a path? I mean, as far as I know, it is a directed chain, where the chain is a "path between two vertices, connected anyhow". Ok, I understand what the chain is, but what is the DIRECTED chain?

答:
A directed graph is a graph where the edges (i.e., the "connections" between vertexes) have a direction. In diagramms, such edges are usually drawn as arrows.

You use such graphs if you want to represent relationships between nodes (i.e., the vertexes) which aren't symmetric. Say, for example, your graph represents the debt between the members of a group of people. The vertices are the individuals, and edges represent debts. Now, debt isn't a symmetric relationship - there's a difference between you owing me 30 dollars, and me owing you 30 dollars. So just knowing that person $A$ and $B$ have some non-zero debt balance isn't sufficient - you need to know who owes whom. Thus, when drawing such a graph, you draw an arrow from the lending to the borrowing party to indicatie the direction of the debt.

Now, for directed graphs, when looking at chains (or paths, or whatever you call them), you aren't usually interested in just any connection between two nodes, but in connections with a certain direction as well. In the debt-graph situation from above, you might ask yourself "Does $A$, directly or indirectly, owe $B$ money?". This is the case if there's either an arrow from $B$ to $A$, or if there are intermediate parties $C_1,C_2,\ldots,C_n$ such that there are arrows $$
  B\to C_1 \to C_2 \to \ldots \to C_n  \to A
$$

Note that this is more specific than asking for just any connection between $A$ and $B$. For example, if both $A$ and $B$ lent money to $C$, then you will have $$
  A \to C \leftarrow B
$$
Yet this doesn't imply that $A$ owes anything to $B$, nor that $B$ owes anything to $A$.

Title: Directed graphs, chains, paths...


问:
Moment of Inertia of a Lamina Around the Center of Mass

I have a two-dimensional lamina in the $xy$-plane, and I need to calculate the moment of inertia around the center of mass.
I know that the moment of inertia around the $x$-axis is $I_x = \int\int y^2 \delta(x,y) \mathrm{d}A$ and $I_y = \int\int x^2 \delta(x,y) \mathrm{d}A$ for the $y$-axis, but I'm not sure how to apply that to / find the formula for the vertical axis going through the center of mass...

Thank you in advance

答:
Not a full answer (so please don't downvote), but this figure should help.  The black arrow is the $z$ axis through the center of mass, and the blue arrow represents $r$.  Integrate over all possible $r$s of

$$I_z = \iint\limits_{\Omega} r^2\ dm = \iint\limits_{\Omega} (x^2 + y^2)\ dm$$

where $\Omega$ is the laminar region (area).


答:
The two inequalities can be handled at once as following without inviting mistakes :

For $x \neq 0$
$$\frac{x-1}{2x} \in [-1,1]$$
$$\frac{1}{2} - \frac{1}{2x} \in [-1,1]$$
$$-\frac{1}{2x} \in [\frac{-3}{2},\frac{1}{2}]$$
$$\frac{1}{x} \in [-1,3]$$
Taking reciprocal (and changing the direction),
$$x \in \; ]-1,\frac{1}{3}[$$

That is $$x \le -1 \vee x \ge 1/3$$

Title: Domain of a function quick question

Body:
$$f(x)=\arcsin\left(\frac{x-1}{2x}\right)$$
We need to find the domain of this function.

My try:

$$-1\leq \frac{x-1}{2x} \leq 1 $$

We can split this into
$$-1\leq \frac{x-1}{2x} \quad\text{and} \quad\frac{x-1}{2x} \leq 1$$
My idea is to solve for this two inequalities and then take the intersection of them both.

$$-2x\leq x-1\\
-3x\leq -1\\
x\geq\frac{1}{3}$$

$$x-1 \leq 2x\\
-1 \leq x$$

By doing the intersection I get the wrong answer which is: $x\geq\frac{1}{3}$

What have I done wrong?


问:
Can I get some help with a line integral? $\int_\Gamma (x^2 + 5y + 7z)dx + (y^2+z+5x)dy + (z^2+7x + y)dz)$
A(0,10, 0), B($3 \sqrt 10, 0, \sqrt10$), C(2, 8, 0) with $\int_\Gamma (x^2 + 5y + 7z)dx + (y^2+z+5x)dy + (z^2+7x + y)dz)$

$\Gamma$  is made from the following segments:

A->B {$ (x,y,z)∈ \Bbb R^3 / x^2+y^2+z^2=100,x-3z=0, x≥0, y≥0$}

[BA] line

From what I understand the A B portion is a quarter of a sphere with the radius of 10 and a plane that goes through the origin and cuts the quarter sphere? So I think I should apply Stokes' Theorem on the first part then I substract from that the second part somehow

Sorry for my bad english and formatting. Thank you for the future insight

答:
Please note that the vector field $\vec F = (x^2 + 5y + 7z, y^2+z+5x, z^2+7x + y)$ is a conservative vector field and its potential function is

$\vec F = \nabla f(x,y,z)$ where $\displaystyle f(x,y,z) = \frac{x^3+y^3+z^3}{3} + 5xy + yz + 7zx$.

So the line integral is path independent and it should simply be

$I = f(c) - f(a)$ where $a$ is the starting point and $c$ is the end point.

Can you take it from here?


问:
If $f:S^1\to S^1$ doesn't have any fixed point then it is homotopic to the identity
How to show that every continuous function $f:S^1\to S^1$ without fixed points is homotopic to the identity? (without using homology nor the concept of degree). </p>

答:
Here is a slightly longer argument that doesn't go through the antipodal map:

Suppose $f:S^1\rightarrow S^1$ has no fixed points. Identify $S_1$
with $\mathbb{R}/\mathbb{Z}$ and let $\pi:\mathbb{R}\rightarrow S_1$
be the quotient map. Let
$\gamma:[0,1] \rightarrow S^1$ be the loop $t \mapsto f(t+\mathbb{Z})$,
starting at $f(\bar{0})$.
Choosing $x_0 \in \pi^{-1}(f(\bar{0}))\cap [0,1)$, we have a
unique lift $\overline{\gamma}:[0,1]\rightarrow \mathbb{R}$
which starts at $x_0$.  Now we can define the function
\begin{align*}
        F:S^1 \times [0,1] &\longrightarrow S^1 \\
        (t+\mathbb{Z}, \lambda) &\longmapsto \pi \left( \lambda \cdot \overline{\gamma}
        (t) + (1-\lambda)\cdot t \right)
    \end{align*}
where $t$ is chosen such that $t\in [0,1)$. This is clearly
continuous if $t\neq 0$. Remains to check continuity at $\bar{0}$.

By the definition of a lift, we know that $\overline{\gamma}(1) \in
    \{x_0+n\;|\; n\in \mathbb{Z}\}$. Claim for all $t$, we have $t < \overline{\gamma}(t) <
    t + 1 $. If not, suppose (wlog) the lower bound doesn't hold. But we
chose $f(0)=x_0>0$, so by the
intermediate value theorem we have $\overline{\gamma}(t)=t$ for
some $t$ -- we have $\pi\circ \overline{\gamma}(t) =
    \pi(t)$ i.e. $f(\bar{t}) = \bar{t}$, giving us a fixed point of
$f$ [a contradiction].
This shows $\overline{\gamma}(1) = x_0 + 1$, hence
$\lim_{t\rightarrow 1} F(t+\mathbb{Z}, \lambda) = \lim_{t\rightarrow 0}
    F(t+\mathbb{Z},\lambda)$ as required. Thus (F) is a homotopy $f\simeq \text{id}$.


问:
Prove that there exist infinitely many positive integers $n$ such that $a_{n}$ is divisible by $n$
Let $\{a_{n}\}$be a sequence defined by $$a_{0}=0,a_{1}=a_{2}=1,a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$$
Prove that there exist infinitely many positive integers $n$ such that $a_{n}$ is divisible by $n$

I try:let $a,b.c$ be the set of all roots of $X^3-X^2-X-1\in \Bbb{Q}[x]$, we can take $(s,t,u)\in \Bbb{C}^3$, such that $$a_{n}=sa^n+tb^n+uc^n$$ for all $n\in \Bbb{Z}$

答:
$\DeclareMathOperator*{\lcm}{lcm}$We will need a few theorems from "Some properties of a generalized Fibonacci sequence modulo $m$", M.E. Waddill (1978). Let's denote $h(x)$ the Trisano period of the $a_n$ sequence taken modulo $$. To begin with, a few empirical observations:
$$\begin{aligned}
 h(3) &= 13\\
 h(13) &= 168
\end{aligned}$$
Also
$$\begin{align*}
a_{13\cdot 3} &\equiv 0 \pmod{h(3)}\\
a_{13\cdot 3} &\equiv 0 \pmod{h(13)}
\end{align*}$$
This allows us to derive
$$\begin{align*}
a_{13\cdot 3} &\equiv 0 \pmod{\lcm(h(3),h(13))}\\
&\equiv 0 \pmod{h(13\cdot 3)} &\text{(see Waddill Theorem 4)}
\end{align*}$$
And $39|a_{39}$ follows.

The general form for $13\cdot 3^{1+6k}$ with $k$ non-negative integer (as conjectured in the question's comments) poses no major difficulties. We know that

$$a_{13\cdot 3^{1+6k}} \equiv 0 \pmod{h(3^{1+6k})} \tag{1}$$
is true because
$$\begin{aligned}
h(3^{1+6k}) &= 3^{6k}h(3) &\text{(see Waddill Theorem 8)}\\
&= 13\cdot 3^{6k} &\text{which divides $13\cdot 3^{1+6k}$.}
\end{aligned}$$
Taking into account that $3^m$ has period $6$ when taken modulo $168$, we also have
$$a_{13\cdot 3^{1+6k}} \bmod{h(13)} = a_{13\cdot 3} \bmod{h(13)} = 0\tag{2}$$
So we can claim that
$$\begin{align*}
a_{13\cdot 3^{1+6k}} &\equiv 0 \pmod{\lcm(h(3^{1+6k}),h(13))}\\
&\equiv 0 \pmod{h(13\cdot 3^{1+6k})}
\end{align*}$$
and conclude that
$$13\cdot 3^{1+6k} | a_{13\cdot 3^{1+6k}}$$
for all non-negative integer $k$.

On the topic of finding $h(p)$ for $p$ prime: Trisano periods exist for all $p$ (that's Waddill's Theorem 1), however they are no more predictable than Pisano periods are for Fibonacci sequences. One can only compute as many terms as needed to find the first reoccurrence of the first terms. For small primes, using a spreadsheet is fast and simple:

Column A for the indices, column B for the sequence itself, cell D1 for the modulus. Then you copy A4:B4 down a few hundred times, and you will find a new occurrence of $0,1,1$ at index $168$.


答:
I'd like to point out why your intuitive argument is wrong. Your sequence starts with the smallest element of $A$, then moves to the next smallest element of $A$ - the successor of $a_0$. By induction (as per the Peano axioms), this process will eventually reach every natural number if we apply it to $\mathbb N$. By the way, it's typical to write $\omega = \mathbb N$ when referring to it as an ordinal. The thing is, your process being exhaustive would require that every element of $A$ other than the smallest one is a successor of some other element. I'll give you an example of a well ordered set where this does not hold.

As I said, we have $\omega = \mathbb N$. Let's define a new set, called $\omega + 1$, as $\omega \cup \{\infty\}$, where $\infty$ is just some symbol not in $\omega$. We'll keep the ordering on $\omega$ and assert that $\infty$ is strictly bigger than every element of $\omega$. This makes $\omega + 1$ into a well ordered set. Essentially, if a nonempty subset contains $\infty$ then its minimum is either $\infty$ itself or removing $\infty$ does not change the minimum. Now, if you tried to apply this process to $\omega + 1$, you would get $a_0 = 0$, $a_1 = 1$,... , $a_n = n$. However, you would never reach $\infty$ as there is no $n \in \omega$ whose successor is $\infty$.

I'd like to point you to the notion of transfinite induction, however. This is a method that allows you to make inductive arguments on any well ordered set, as you have tried to do. The difference is that you have to account for these $\infty$ esque elements which are not a successor of anything (called limit ordinals). If you can prove a property $P$ holds for all limit ordinals in $A$ and that $P(\alpha) \implies P(succ(\alpha))$ then $P$ holds for all $\alpha$ in $A$.

I'd also like to point out that the more standard notation is $\omega + 1 = \omega \cup \{\omega\}$. This definition of the successor ordinal allows for a very clean theory. Here, we have that $\omega \in \omega + 1$ is a limit ordinal, so doing induction on $\omega + 1$ requires treating this case specially.

Title: Is a well-ordered set always countable?


问:
Does exists a matrix $B$ such that $A^TA=A^TB+B^TA$? with $B^TB$ being a diagonal matrix and $A$ an incidence matrix
$A$ is a incidence matrix for some undirected graph.

$A^TA$ is a positive definite matrix, so I know that we can factorize it as $A^TA = C + C^T$

There exists always a matrix $C$ such that $C = A^TB$?
Satisfying the next requirement, $B^TB = cI$ is a diagonal matrix, being $I$ the identity matrix and $c$ a scalar, and $B$ also is related to the incidence matrix $A$ in the sense that for each edge, we select one node.

$A$ might not be square (more edges than nodes).

For instance,

$
A = \begin{bmatrix}1 & 0 & -1\\-1 & 1 & 0\\0 &-1 & 1\end{bmatrix} \quad
B = \begin{bmatrix}0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 0\end{bmatrix},
$
satisfies $A^TA=A^TB+B^TA$ and $B^TB = I$

Note that here I have taken for $B$ all the $-1$ from $A$.

Can I find such $B$ for instance for complete undirected graphs? what about for more general undirected grahps?

答:
If you think of this in terms of columns vectors, write $A = \begin{bmatrix} v_1 & \dots v_n \end{bmatrix}$ and $B = [ b_1 /2,
\dots, b_n/2 ]$ (the one-half factor is just to normalize stuff), so that
$$
A^{\top}A = [(v_i \, | \, v_j)], \qquad A^{\top} B + B^{\top} A = [(v_i \, | \, b_j/2)] + [ (b_j/2 \, | \, v_i) ] = [(v_i \, | \, b_j)].
$$
where $( \cdot \, | \, \cdot )$ denotes inner product. So you want to find $n$ column vectors $b_1,\dots, b_n$ ($n$ is the number of edges in your graph, the vectors lie in $\mathbb R^k$ where $k$ is the number of vertices) such that
$$
(v_i \, | \, v_j) = (v_i \, | \, b_j).
$$
In particular, if $k \le n$ and $\{ v_1, \dots, v_k \}$ forms a basis of $\mathbb R^k$, your vectors $b_j$ are completely determined by the first $k$ vectors, i.e. any set of edges which gives rise to $k$ linearly independent vectors in $\mathbb R^k$ (perhaps there is a graph-theoretic interpretation to such a constraint) determines if there exists such a matrix $B$, and what it is if it exists.

This puts a lot of constraints on the possible graphs which would acheive this, so I don't expect you to be able to find such a matrix $B$ for every graph if it has more edges than vertices in general. Otherwise these equations tell you how to compute the vectors $b_j$ ; it reduces to solve a linear system given your matrix $A$.

Hope that helps,


问:
Can the idea of a 'function of a variable' be made rigorous?
Suppose that $y=x^2$. Very often people describe this relationship by saying that '$y$ is a function of $x$'. It seems that there are several logical problems with this statement:

A function is simply a set of ordered pairs of numbers. It does not matter how you denote the input of a function, and there is no doubt that $x \mapsto x^2$ and $y \mapsto y^2$ are the same function. Indeed, in both cases $y$ and $x$ are simply dummy variables used to illustrate what happens when you plug in an arbitrary number into the function.

$x$ and $y$ are often described as variables that are related in some way; in this case, with $y$ being the square of $x$. However, in this Math Overflow post  Mike Shulman states that the idea of a variable is 'not a standard part of modern formalizations of mathematics'. Strangely, the idea of a dummy variable makes much more sense to me. For instance, when we say 'consider the function $f$ defined by $f(x)=x^2$ for all $x$', it is clear that the only purpose of the letter $x$ is to declare that the second entry of the ordered pair is the square of the first.

If $y=x^2$ and $x \geq 0$, then we just as well might write $x=\sqrt{y}$. This flexibility is not really allowed when speaking of functions: $x \mapsto x^2$ and $x \mapsto \sqrt{x}$ are certainly not the same function. Perhaps it is possible evade this by treating $x$ as the independent variable and $y$ as the dependent variable.

Some authors get around this ambiguity by saying that 'the function $y=x^2$' is just a shorthand for 'the function $y(x)=x^2$', which in turn is just a shorthand for 'the function $y$ defined by $y(x)=x^2$ for all x'. However, this doesn't seem to align with how people treat the relationship between $y$ and $x$ in practice. Even if we accept that $y=x^2$ is simply a shorthand for $y(x)=x^2$, it still seems that there is a tendency towards treating $x$ as an independent variable representing the input of $y$, as opposed to simply a dummy variable that can be replaced by any other letter.

So I ask, in formal mathematics, is it possible to interpret $y=f(x)$ in such a way that $x$ and $y$ are variables representing the inputs and outputs of a function? And if so, how should the statement '$y$ is a function of $x$' be understood?

答:
When we say “$y=f(x)$”, we are stating a typographical convention that we are going to adopt in the present context: Namely, whenever the letters $x$ and $y$ appear, the values to which they refer are connected by the functional relationship $f$. The terminology “$y$ is a function of $x$” is confusing. While it is still employed by many people who use mathematics, it is often avoided by present-day mathematicians, who are aware that $y$ and $f$ refer to quite different types of mathematical object.

For example, consider the function $$f:\Bbb R\to\Bbb R_{\geqslant0}:x\mapsto f(x):= x^2.$$ In this case, $f$ may be identified with a certain subset of $\Bbb R\times\Bbb R_{\geqslant0}$. Quite separately, and in addition, we may adopt the naming convention of using $y$ instead of $f(x)$. But $y$ is just some element of $\Bbb R_{\geqslant0}$, albeit dependent on $x$, while $f$ is a quite particular subset of  $\Bbb R\times\Bbb R_{\geqslant0}$ (which determines the relationship between $x$ and $y$).

In this context, the notation “$y=y(x)$” is often used, which further reinforces the confusion between $y$ and $f$. This sort of notation may be convenient for (say) practical engineering calculations, but it's not a good place to start when you want to maintain a clear mathematical concept of function.

In the above, for simplicity, I have adopted the convention of identifying a function with its graph. In some branches of mathematics this is inconvenient, and to specify a function it is necessary to specify a codomain for it (of which the range of the function is a subset).


问:
Show that $\frac{1}{(-q,q)_\infty}=(q,q^2)_\infty$
Show that $\frac{1}{(-q,q)_\infty}=(q,q^2)_\infty$

I found the following proof:
$\frac{1}{(-q,q)_\infty}=\frac{(1-q)(1-q^2)\cdots}{(1-q^2)(1-q^4)\cdots}=(q,q^2)_\infty$
However, this seems to lack formality since I am treating the infinite product as something that can  "cancel out" infinitely. I feel like I would have to make this argument for the partial product, but this method no longer works. Is there a way to formalize what I did here?

答:
The simplest way to formalize seems to use
the idea of
formal power series. Suppose we have a sequence $\,a_n\,$
of formal powers series where $\,a_n = O(x^{n+1}).\,$ For an
example, suppose $\,a_n = q^n.\,$ Then the
sum $\,\sum_{n=1}^\infty a_n\,$ is a well-defined formal
power series. That is, the sum converges in the natural
topology of formal power series (See my note at the end).

Define another sequence
$\, b_n := a_n - a_{2n} = O(x^{n+1}).\,$
Then the sum $\,\sum_{n=1}^\infty b_n = \sum_{k=1}^\infty
a_{2k-1}\,$ is a well-defined formal power series for the
same reason. The $\,a_{2n}\,$ get subtracted out and what
is left are the odd indexed terms. More
precisely, $$ \sum_{k=1}^n b_k =
\sum_{k=1}^n a_k -
\sum_{k=1}^{\lfloor n/2\rfloor} a_{2k}
+ O(x^{n+1}). $$

In the case of infinite products of formal power series,
where we have $\,a_n = 1-q^n\,$, and
$\,b_n = \frac{(1-q^n)}{(1-q^{2n})},\,$
we can take logarithms and the previous reasoning applies.
For the case of convergent complex power
series, use similar reasoning
to prove convergence. The
proof is essentially the same, but the  details are a bit more involved.

NOTE: the Wikipedia article states

In mathematics, a formal power series is a generalization of a polynomial, where the number of terms is allowed to be infinite, with no requirements of convergence.

The emphasis is my own, but this is not strictly true.
Later it states

The topology has the useful property that an infinite summation converges if and only if the sequence of its terms converges to $0$, which just means that any fixed power of $X$ occurs in only finitely many terms.

Thus, there is a natural topology on formal powers series which
makes them into convergent series in that topology.


答:
Here it is another approach for the sake of curiosity.

I would start with squaring both sides as it is done next:

\begin{align*}
\left|\frac{x-1}{2x}\right| \leq 1 & \Longleftrightarrow \left(\frac{x-1}{2x}\right)^{2} \leq 1\\\\
& \Longleftrightarrow \frac{x^{2} - 2x + 1 - 4x^{2}}{4x^{2}} \leq 0\\\\
& \Longleftrightarrow \frac{3x^{2} + 2x - 1}{4x^{2}} \geq 0
\end{align*}

We may now study the behavior of the function at the numerator.

To begin with, let us rearrange its expression:
\begin{align*}
3x^{2} + 2x - 1 & = \left(3x^{2} + 2x + \frac{1}{3}\right) - \frac{4}{3}\\\\
& = \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} - \frac{4}{3}
\end{align*}

Consequently, the solution set to problem proposed is given by
\begin{align*}
3x^{2} + 2 x - 1 \geq 0 & \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} \geq \frac{4}{3}\\\\
& \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}} \geq \frac{2}{\sqrt{3}}\right)\vee\left(\sqrt{3}x + \frac{1}{\sqrt{3}} \leq -\frac{2}{\sqrt{3}}\right)\\\\
& \Longleftrightarrow \left(x \geq \frac{1}{3}\right)\vee\left(x \leq -1\right)
\end{align*}

Since both intervals does not contain the value $x = 0$, we conclude the sought domain is given by
\begin{align*}
D_{f} = (-\infty,-1]\cup\left[\frac{1}{3},+\infty\right)
\end{align*}

Hopefully this helps!

Title: Domain of a function quick question


问:
Basic proof that Isogeny = group homomorphism without Riemann Roch?
It is a well known fact that an isogeny $f:E_1\rightarrow E_2$ between elliptic curves (i.e., where $f(O)=O$) is a group homomorphism under the natural group structure of the elliptic curves.

Every proof that I have found (usually based on Silverman's book) uses  Riemann-Roch theorem. Of course, it is a very important theorem, so why not.

But I was wondering if it is possible to prove it directly, without first going to $\operatorname{Pic}^0(E_i)$. Does anyone know of any such proof? (The simple the better, of course).

Cheers.

答:
This is proven for abelian varieties (of any dimension, the 1-dimensional case being elliptic curves) as Corollary 1 in section 4 of Mumford's Abelian Varieties. It is an immediate corollary of the "rigidity lemma" stated immediately prior:

Rigidity Lemma (Form I). Let $X$ be a complete variety, $Y$ and $Z$ any varieties, and $f\colon X \times Y \to Z$ a morphism such that for some $y_0 \in Y$, $f(X \times \{y_0\})$ is a single point $z_0$ of $Z$. Then there is a morphism $g\colon Y \to Z$ such that if $p_2\colon X \times Y \to Y$ is the projection, $f = g \circ p_2$.

In other words, if $f$ is constant on one fiber of the projection map $p_2$, then $f$ is constant on every fiber of $p_2$.

The proof is brief, so I'll just reproduce it here:

Proof. Choose any point $x_0 \in X$, and deﬁne $g\colon Y \to Z$ by $g(y) = f(x_0, y)$. Since $X \times Y$ is a variety, to show that $f = g \circ p_2$, it is sufficient to show that these morphisms coincide on some open subset of $X \times Y$. Let $U$ be an affine open neighbourhood of $z_0$ in $Z$, $F = Z - U$, and $G = p_2(f^{-1}(F))$; then $G$ is closed in $Y$ since $X$ is complete and hence $p_2$ is a closed map. Further $y_0 \notin G$ since $f(X \times \{y_0\}) = \{z_0\}$. Therefore $Y - G = V$ is a non-empty open subset of $Y$. For each $y \in V$, the complete variety $X \times \{y\}$ gets mapped by $f$ into the affine variety $U$, and hence to a single point of $U$. But this means that for
any $x \in X$, $y \in V$, $f(x, y) = f(x_0, y) = g \circ p_2(x, y)$, and this proves
our assertion.

In particular, if $f \colon X \to Y$ is a morphism of abelian varieties such that $f(0_X) = 0_Y$, then the morphism $\phi\colon X \times X \to Y$ defined for all $x_1, x_2 \in X$ by
$$\phi(x_1, x_2) = f(x_1 + x_2) - f(x_1) - f(x_2)$$
is constant with value $0_Y$ on $X \times \{0_X\}$ and $\{0_X\} \times X$. Applying the rigidity lemma, $\phi$ is constant everywhere on $X \times X$, which means that $f$ is a homomorphism.

The analogous relative statement for abelian schemes (over any base scheme) is Corollary 6.4 in Mumford, Fogarty, and Kirwan's Geometric Invariant Theory. The general idea of that proof is similar.

References:

Mumford, D. Abelian Varieties. Corrected reprint of the second (1974) edition. Tata Institute of Fundamental Research Studies in Mathematics, vol. 5. Hindustan Book Agency, New Delhi, 2008. ISBN: 978-81-85931-86-9. [MR 2514037]

Mumford, D., Fogarty, J., and Kirwan, F. Geometric Invariant Theory. Third edition. Ergebnisse der Mathematik und ihrer Grenzgebiete, vol. 34. Springer-Verlag, Berlin, 1994. ISBN: 3-540-56963-4. [MR 1304906]


问:
Matrices (Hermitian and Unitary)
Hey! I have some short proofs I’m quite stuck on below. I know the definitions but get stuck on how to use them to prove what’s required. If possible, can you please explain how to apply the definition to get these proofs so that I can try again? Thanks!

Q1. Show that if there exists a unitary matrix P such that $P*AP = D$ where $P*$ is the conjugate transpose of $P$ and $D$ is a diagonal matrix then $A$ is a normal matrix. Show that the columns of $P$ form an orthornormal basis of $Cn$ if and only if $P$ is unitary.

A1. If $P$ is unitary, $PP*=P*P=I$. A is normal if $AA*=A*A$. I'm not sure where to go from there.

Q2. Show that eigenvectors of a Hermitian matrix corresponding to distinct eigenvalues are orthogonal.

A2. Hermitian if $A=A*$. Eigenvalues of $A$ are given by the equation: $Av=(\lambda)v$. Again, what do I do with this?

Q3. Show that the eigenvalues of a Hermitian matrix are real, and that the eigenvectors corresponding to different eigenvalues are orthogonal.

A2. Similar to the above.

Q4. What is meant to say that a matrix is unitarily diagonalisable? Prove that if $P$ is a unitary matrix then all of the eigenvalues of $P$ have modulus equal to one. Further, prove that the column vectors of $P$ form an orthornormal set (with respect to the Euclidean inner product).

A3. Is it unitarily diagonalisable when it can be expressed as a matrix with orthornormal vectors? I’m not sure about what the rest of the question even means! :(

答:
Here are partial answers to your question:

Q'1. Show that if there exists a unitary matrix P such that PAP = D where P is the conjugate transpose of P and D is a diagonal matrix then A is a normal matrix.

A'1. We have $P^*AP=D$. By applying $()^*$ to both sides, we get $P^*A^*P=D^*$.

Now, $(P^*AP)(P^*A^*P)=DD^*$. On the other hand, $(P^*A^*P)(P^*AP)=D^*D$. But $DD^*=D^*D$, since $D$ is diagonal. Hence $(P^*AP)(P^*A^*P) = (P^*A^*P)(P^*AP)$. Therefore $(P^*AA^*P) = (P^*A^*AP)$. Lastly, premultiply both sides by $P$ and postmultiply both sides by $P^*$ to get $AA^*=A^*A$.

Q3. Show that eigenvectors of a Hermitian matrix are real and that eigenvectors corresponding to distinct eigenvalues are orthogonal.

A'3. First, let $\lambda \in \mathbb{C}$ be an eigenvalue with associated eigenvector $v \neq 0$. Look at the inner product $\langle v,v \rangle$. We have:

$\lambda \langle v,v \rangle = \langle Av,v \rangle = \langle v,A^*v \rangle = \langle v,Av \rangle = \langle v, \lambda v \rangle = \overline{\lambda} \langle v,v \rangle$.

Since $\langle v,v \rangle \neq 0$, we must have $\lambda = \overline{\lambda}$.

This shows that the eigenvalues of a Hermitian matrix are real.

Now let $\lambda$ and $\mu$ be distinct eigenvalues, with associated eigenvectors $v$ and $w$ respectively.

Look at the inner product $\langle v,w \rangle$. We have:

$\lambda \langle v,w \rangle
= \langle \lambda v,w \rangle
= \langle Av,w \rangle
= \langle v,A^*w \rangle
= \langle v,Aw \rangle
= \langle v,\mu w \rangle
= \mu \langle v,w \rangle$

But $\lambda \neq \mu \implies \langle v,w \rangle =0$.

Q'4. Prove that if P is a unitary matrix then all of the eigenvalues of P have modulus equal to one.

A'4. Let $\lambda$ be an eigenvalue of $P$  with associated eigenvector $v \neq 0$. Then:

$\langle v,v \rangle = \langle PP^*v,v \rangle = \langle Pv,Pv \rangle = \langle \lambda v, \lambda v \rangle = \lambda \overline{\lambda} \langle v,v \rangle$.

Hence $(1-\lambda \overline{\lambda})\langle v,v \rangle=0$. Since $\langle v,v \rangle \neq 0$, $\lambda \overline{\lambda} = 1 \implies |\lambda|^2=1 \implies |\lambda|=1$.

Title: Matrices (Hermitian and Unitary)


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
I think it depends on people. Beyond practicing, it's about logic, and sometimes about abstraction. I think a lot of people would find hard to put effort into it, because abstracion is a totally irregular way of thinking for an everyday people.

Just like anything else, math can be learnt anytime by anyone. Just have to begin it!

However, talking about it without any subjective opinions is hard for me, because even though I hate practicing all the time, basically I love logic things, such as math (but I prefer programming) - overall, feel free to downvote me if I'm wrong.


问:
Stars and Bars with bounds but order matters
This question is related to Stars and Bars with bounds

I have what I thought is a fairly simple problem: Count non-negative
integer solutions to the equation

$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$

such that $0 \leq x_1 \leq 9$.

The accepted answer uses nCr such that order does not matter. But if order does matter can the  function be replaced from
\begin{align*}
[x^j](1+x)^n=\binom{n}{j}
\end{align*}
to then be

\begin{align*}
[x^j](1+x)^n=n^j
\end{align*}

答:
My confusion was with why the "Choose" function was used rather than the "Permute" function when order matters. According to @user by the nature of the stars and bars and by extension, generating function it takes the order of the bins into account despite using the choose function as a basis.

This offers the best example for how this happens: https://youtu.be/zjtxJSkjejc?t=308


问:
Unsure how to proceed in showing that sin(x/n)n converges uniformly
Currently I am trying to solve essentially the same problem as presented in uniform convergence of n sin(x/n) with the exception that the interval is restricted to [0, 1]. I have not made any progress beyond $\left|n\sin(x/n) - x\right| \leq \max_{x \in [0, 1]}\left|n\sin(x/n) - x\right|$ as the derivative $\frac{n\sin(x/n) - x}{\left|n\sin(x/n) - x\right|}$ does not provide anyting useful and I do not know any upperbound for the expression inside the absolute value. Also, I did not get much out of the two answers in the aforementioned link as (to me) it seems that one answers shows that the function does not converge uniformly and one shows that it does.

How should I proceed?

答:
Recall the basic inequalities $y- \frac{y^3}{6} \leqslant \sin y \leqslant y$ for $y>0$. If you are not familar with the LHS inequality, it is easily derived from the RHS inequality by integrating twice.

We have

$$x- x \sin \frac{x}{n}  = \begin{cases}x \left(1 - \frac{\sin \frac{x}{n}}{\frac{x}{n}}\right), & 0 < x \leqslant 1 \\ 0 , & x = 0\end{cases}$$

Using the basic inequalities, we obtain

$$\frac{x}{n} \geqslant \sin \frac{x}{n} \geqslant \frac{x}{n} - \frac{x^3}{6n^3},$$

and by rearranging, we get for $0 < x \leqslant 1$,

$$0 \leqslant x \left(1 - \frac{\sin \frac{x}{n}}{\frac{x}{n}}\right) \leqslant \frac{x^3}{6n^2}\leqslant \frac{1}{6n^2}$$

Thus,  for all $x \in [0,1]$,

$$\left|x \sin \frac{x}{n} - x \right| \leqslant \frac{1}{6n^2},$$

and it follows that $n \sin \frac{x}{n} \to x$ uniformly on $[0,1]$.


问:
How do we calculate $\cos (3\pi/4+k\pi)$?
Can someone help me to compute $\cos(\frac{3 \pi}{4}+\pi k)$?

I know that this is equal to $\cos(\frac{3 \pi}{4}) \cos(\pi k)-\sin(\frac{3 \pi}{4})\sin(\pi k)$, which equals to $-\frac{1}{\sqrt{2}} \cos(\pi k)-0 \sin(\pi k)$.

How do I proceed? Thank you.

Body:
Hint: observe that if $k = 0,  2, 4 ...$
$$cos(\pi k) = 1 $$

else if $k = 1, 3, 5, ...$

then:

$$cos(\pi k) = -1 $$


问:
Closed sets as $A\cap B = \varnothing$ such as $\inf \{\|a − b\| \mid a\in A, b\in B\} = 0$
Let $E$ be a normal vector space, $A$, $B\subset E$ closed, nonempty subsets.

We denote by $c = \inf\{\|a − b\| \mid a\in A, b\in B\}$.

I gotta find $2$ closed sets $A$, $B$ such that $c = 0$ and $A\cap B = \varnothing$.

And then, for these sets, give an example of sequences $\{a_{n}\}$ and $\{b_{n}\}$ such that $\|a_{n} - b_{n}\| \to 0$.

I have the idea of  $A = N - \{0\}$ and $B = \{(n+1)/n \mid n \in\mathbb{N}_{>=2}\}$ for my closed sets.

Is it okay? I'm kinda lost about the sequences

Thank You for your help

答:
Your set $A$ is OK, but $B$ has to be adapted to $\{n + \frac{1}{n} \mid n =2,3,4,,\ldots\}$. The set $B=\{\frac{n+1}{n}\mid n=2,3,4,\ldots\}$ that you wrote is not closed, as $1 \in \overline{B}\setminus B$

Then $a_n = n$ and $b_n = n+ \frac{1}{n}$ $(n \ge 2$) have the property that $|a_n - b_n|  = \frac1n \to 0$.

Another similar set can be found in the plane: $A= \{(x,0)\mid x \in \Bbb R\}$ and $ B = \{(x,\frac1x)\mid x \in \Bbb R\setminus \{0\}\}$, where $a_n=(n,0)$ and $b_n = (n,\frac1n)$ have similar properties.


问:
Power series of $\log(1+x)\log(1-x)$
On the bottom of page 7 of the paper Logarithmic Integrals by Morshed (arXiv link) there is an "interesting generating function":
$$ \log(1+x) \log(1-x) = \sum_{n=1}^\infty \left(H_n - H_{2n} - \frac{1}{2}\right) \frac{x^{2n}}{n}$$
but it doesn't seem to be correct. The function on the left has the series
$$-x^2 - \frac{5}{12}x^4 - \frac{47}{180}x^6 - \ldots$$
while the function on the right is
$$-x^2 - \frac{13}{24}x^4 - \frac{67}{180} - \ldots$$
These do not match. Not knowing how to actually find the power series for $\log(1+x)\log(1-x)$, I decided instead to try to identify the function on the right, using the identity
$$\sum_{n=1}^\infty H_n x^n = \frac{-\log(1-x)}{1-x}$$
as a starting point.
In the end I was able to obtain
$$\sum_{n=1}^\infty \left(H_n - H_{2n} - \frac{1}{2}\right) \frac{x^{2n}}{n} = \log(1+x)\log(1-x) + \tfrac{1}{2}\left(\mathrm{Li}_2(x^2) + \log(1-x^2)\right).$$
But the power series of $\mathrm{Li}_2(x^2) + \log(1-x^2)$ is very simple (it's $\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n^2} - \frac{1}{n}\right) x^{2n}$) and so
$$\log(1+x)\log(1-x) = \sum_{n=1}^\infty\left(H_n - H_{2n} - \frac{1}{2\color{red}{n}}\right) \frac{x^{2n}}{n} \tag{*}$$
which means the paper had a typo.

Question. How would you derive (*) without knowing it had anything to do with harmonic numbers?

I'm dissatisfied with my derivation because I started with a "guess" that was "close" (by virtue of it being a typo) and was able to go from there. I'd be especially interested in a "power series proof" (multiplying together the series for $\log(1-x) = -x - x^2/2 - x^3/3 - \ldots$ and $\log(1+x) = x - x^2/2 + x^3/3 - \ldots$ and simplifying the coefficients).

答:
We obtain
\begin{align*}
\color{blue}{\log}&\color{blue}{(1+x)\log(1-x)}\\
&=\left(\sum_{j=1}^\infty (-1)^{j+1}\frac{x^j}{j}\right)\left(-\sum_{k=1}^\infty\frac{x^k}{k}\right)\\
&=\sum_{n=2}^\infty\left(\sum_{{j+k=n}\atop{j,k\geq 1}}\frac{(-1)^j}{j}\,\frac{1}{k}\right)x^n\\
&=\sum_{n=2}^\infty\left(\sum_{j=1}^{n-1}\frac{(-1)^j}{j}\,\frac{1}{n-j}\right)x^n\\
&=\sum_{n=2}^\infty\left(\sum_{j=1}^{n-1}\frac{(-1)^j}{j}+\sum_{j=1}^{n-1}\frac{(-1)^j}{n-j}\right)\frac{x^n}{n}\tag{1}\\
&=\sum_{n=2}^\infty\left(\sum_{j=1}^{n-1}\frac{(-1)^j}{j}+\sum_{j=1}^{n-1}\frac{(-1)^{n-j}}{j}\right)\frac{x^n}{n}\tag{2}\\
&=\sum_{n=1}^\infty\left(\sum_{j=1}^{2n-1}\frac{(-1)^j}{j}\right)\frac{x^{2n}}{n}\tag{3}\\
&=\sum_{n=1}^\infty\left(-\sum_{j=1}^{n}\frac{1}{2j-1}+\sum_{j=1}^{n}\frac{1}{2j}-\frac{1}{2n}\right)\frac{x^{2n}}{n}\tag{4}\\
&=\sum_{n=1}^\infty\left(-H_{2n}+2\sum_{j=1}^{n}\frac{1}{2j}-\frac{1}{2n}\right)\frac{x^{2n}}{n}\tag{5}\\
&\,\,\color{blue}{=\sum_{n=1}^\infty\left(-H_{2n}+H_n-\frac{1}{2n}\right)\frac{x^{2n}}{n}}\\
\end{align*}
and the claim follows.

Comment:

In (1) we use the identity $\frac{1}{j(n-j)}=\frac{1}{n}\left(\frac{1}{j}+\frac{1}{n-j}\right)$.

In (2) we change the order of the right-hand inner sum $j\to n-j$.

In (3) we cancel odd terms of $n$.

In (4) we split the inner sum in odd and even part.

In (5) we use the identity $H_{2n}=\sum_{j=1}^{2n}\frac{1}{j}=\sum_{j=1}^{n}\frac{1}{2j}+\sum_{j=1}^n\frac{1}{2j-1}$.

Title: Power series of $\log(1+x)\log(1-x)$


答:
We have $\cos k\pi = (-1)^{k}$ for integer $k$.

So it simplifies to$$\cos (k\pi+3\pi/4)=\frac{(-1)^{k+1}}{\sqrt{2}}$$

Title: How do we calculate $\cos (3\pi/4+k\pi)$?

Body:
Can someone help me to compute $\cos(\frac{3 \pi}{4}+\pi k)$?

I know that this is equal to $\cos(\frac{3 \pi}{4}) \cos(\pi k)-\sin(\frac{3 \pi}{4})\sin(\pi k)$, which equals to $-\frac{1}{\sqrt{2}} \cos(\pi k)-0 \sin(\pi k)$.

How do I proceed? Thank you.


问:
Limit of a function defined at a single point
Let's suppose we have a function defined at a single point $f : \left\{ 1 \right\} \to \left\{ 1 \right\}$ defined by $f(x) = x$. Its graph is, therefore, composed of a single point $(1,1)$.

Does the following limit exist?

$$\lim_{x \to 1} \ f(x)$$

I've read this post but it addresses the question in a topology perspective. I'd like to answer this considering a typical first semester Calculus course.

Any help is highly appreciated.

答:
It depends on the definition, but here is one answer.

You can definite limits in a very general setting: for any topological space $X$, we say that a sequence $\{x_n\}$ converges to $x\in X$ if, for any open set $U\subset X$ containing $x$, there exists $N\in\Bbb N$ for which $n>N$ implies $x_n\in U$.

For a function $f\colon X\to Y$, we can then define $\lim\limits_{x\to c}f(x)=y$ to mean "if $\{x_n\}$ is a sequence of points in $X\setminus\{c\}$ which converges to $c$, then $\{f(x_n)\}$ converges to $y$."

So for the function $f\colon \{1\}\to\{1\}$, there are no sequences at all with values in $\{1\}\setminus\{1\}$. So it is vacuously true that, for every such sequence $\{x_n\}$, the corresponding $\{f(x_n)\}$ converges to $1$.

In my first answer, I really was writing the definition of the statement "$f$ is continuous at $1$ and $f(1)=1$." As in the familiar setting of real functions, continuity implies the existence of the limit.


问:
Why is an alternating multilinear map called "alternating"?
I know it's just a small detail, but it's kind of bothering me... A multilinear map is called "alternating" if having two equal arguments makes it zero.

Why use the word "alternating" then? What precisely is alternating?

答:
As mentioned in the comment, if the characteristic of the coefficient field is different from $2$, then the following statements are equivalent:

$f(x, y, \dots) = -f(y, x, \dots)$ for any $x, y$;

$f(x, x, \dots) = 0$ for any $x$.

The proof is an easy exercise.

In your question, you take the second statement as definition. But the first one is more suitable for the name "alternating".