问:
Group Structure on $\Bbb R$
$(\Bbb R,+)$ is a topological group. Is there any other group structure on $\Bbb R$ such that it is still a topological group and this group is not isomorphic to $(\Bbb R,+)$ ?

Refer to Non-isomorphic Group Structures on a Topological Group for the general problem. In that problem I have conjectured that there is no such group however I couldn't prove it.

答:
Incnis Mrsi's nice argument from 2014 has not actually been completed so let's complete it, since this came up recently, twice (actually a third time on MO, too).

We have a topological group $G$ with identity $e$ homeomorphic to $\mathbb{R}$. Incnis has established that this group is totally ordered with respect to the usual order on $\mathbb{R}$ (in a way compatible with the topology; that is, $G$ is totally ordered, the total order is isomorphic to $\mathbb{R}$, and the isomorphism is a homeomorphism also), and in particular torsion-free. Thanks to the total order it follows that for any positive integer $n$, the $n^{th}$ power function $g \mapsto g^n$ is strictly monotonic and hence injective.

Proposition: $g \mapsto g^n$ is unbounded from above and below.

Proof. If the image of $g \mapsto g^n$ contains some positive $L > e$ then it also contains $L^2 > L$, which is strictly larger. Similarly for negatives. $\Box$

Corollary: The function $g \mapsto g^n$ is bijective; equivalently, every element has a unique $n^{th}$ root.

Proof. By the above, the image of $g \mapsto g^n$ contains $e$ and is unbounded from above and below. Since it is also connected, the image is all of $G$, so $g \mapsto g^n$ is surjective. Since it is also strictly monotonic, it is injective (as observed above), hence bijective. $\Box$

Pick any $g > e$ in $G$ and define $f : \mathbb{Q} \to G$ by sending $\frac{p}{q}$ to the unique $q^{th}$ root of $g^p$. This is well-defined because if $\frac{p}{q} = \frac{r}{s}$ then $g^{ps} = g^{qr}$ has a unique $qs$-th root. It is a homomorphism for the same reason. And it is strictly monotonic, hence extends continuously to a homomorphism $F : \mathbb{R} \to G$ by a limiting argument. (This should all be very familiar if you've ever constructed the exponential $a^x$ in terms of limits of rational exponentials in a real analysis class, and in fact that argument is a special case of this one, applied to $G = (\mathbb{R}_{+}, \times)$.)

$F$ is strictly monotonic, hence injective. We will write $F(r) = g^r$.

Proposition: $r \mapsto g^r$ is unbounded from above and below.

Proof. If the image of $r \mapsto g^r$ contains any positive $L > e$ then it also contains $gL > L$, which is strictly larger. Similarly for negatives. $\Box$

Corollary: $r \mapsto g^r$ is surjective, hence an isomorphism of topological groups $G \cong (\mathbb{R}, +)$.

Proof. The image of $r \mapsto g^r$ is unbounded from above and below, contains $e$, and is connected, hence is all of $G$. So $r \mapsto g^r$ is surjective, hence a bijection. Since it's strictly increasing (hence continuous), its inverse is also strictly increasing (hence continuous), so it's a homeomorphism. $\Box$

Corollary: $G$ is abelian.

Title: Group Structure on $\Bbb R$


问:
Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation?

Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation?

Rank is defined to be $\dim(\text{im}(T))$ right? Every output of $T$ is a vector belonging to $\Bbb{R^2}$ and $\Bbb{R^2}$ is the image of $T$ so I assumed that $\text{rank}(T) = 2$. Turns out it's actually $1$, which I don't seem to understand... Every output of $T$ can be expressed as $c(1,2)$ but this is also just a tuple in $\Bbb{R^2}$ so how can the rank be $1$?

The nullity is defined to be the set of vectors for which $T(x) = 0$ so I have that $$(x+y,2(x+y)) = 0 \Longrightarrow x+y=0, 2x+2y=0 \Longrightarrow (x,y) = (0,0)$$

so the set of vectors for which $T(x) = 0$ is just $\{(0,0)\}$ so $\ker(T) = 1$?

答:
Hint:

The matrix for this linear map is
$$
\begin{pmatrix}
1&1\\
2&2
\end{pmatrix}
$$
which is singular.

Notes 1. $x+y=0$ and $2x+2y=0$ implies that $x=-y$, not $x=y=0$. So the kernel of your map is $\operatorname{span}\{(1,-1)\}$ and thus the nullity is $1$. Now you can apply the rank-nullity theorem to find the rank or directly tells it from the matrix.

Notes 2.

The rank of a linear transformation is the dimension of the image. That's correct.

Every output of $T$ is a vector in $\mathbb{R}^2$ does not mean that the image of $T$ is $\mathbb{R}^2$.

ContentLicense: CC BY-SA 4.0


问:
How do I find a third side of a triangle with two sides and a bisecting line segment?
I am using a laser range finder to calculate the height of a second story wall. I have a fixed point and three separate lengths hitting the top, the bottom, and an indeterminate point on the wall. With the exact lengths of all three line segments, how do I find the length of the wall? Note: I am unable to create a right triangle due to the height and the uneven wall underneath.

答:
Results from DeepSea's calculation:

Add the two segment lengths and simplify

$$ X=(1+\frac{r}{p}) \sqrt{\dfrac{p^3-pq^2+p^2r+q^2r-pr^2}{r}} $$

$$X =(1+\frac{r}{p}) \sqrt{\dfrac{p}{r}(p^2-q^2)+(p^2+q^2-pr)} $$

The wall need not be vertical, the scheme is even valid for three collinear points on a ceiling.

ContentLicense: CC BY-SA 3.0


问:
Euclids Algorithm for polynomials and a greatest common divisor
I have question about a problem I've encountered while attempting to solve an exercise (it's from an exercise in a homework series).

Suppose we have two polynomials $f$ and $g$ (presumably over the reals, although the field over which these polynomials are defined isn't mentioned explicitly). Assume that deg($g$) > 1. Then there exist polynomals $q_1,r_1$ such that $f = q_1 g + r_1$. We can now repeat the polynomial divison, which gives $g = q_2 r_1 + r_2$ for polynomials $q_3,r_3$ and $r_1 = q_4 r_2 + r_4$ for polynomials $q_4,r_4$, etcetera. At some point we get $r_{p-2} = q_{p}r_{p-1} + r_p$ and $r_{p-1} = q_{p+1} r_p$. I now want to show that $r_p$ is the greatest common divisor of $f$ and $g$.

I have already showed (I think) that $r_p$ divides both $f$ and $g$ by writing the equations in matrixform, giving $$\left(\begin{array}{c}f\\g\end{array}\right) = \Pi_{i=1}^{p+2}\left(\begin{array}{cc} q_{i} & 1 \\ 1 & 0 \end{array}\right) \left(\begin{array}{c}r_p\\0\end{array}\right)$$ But now I want to show that $r_p$ is the greatest divisor.  I think the best way to do it is to assume that there is some $\tilde{r}_p$ which also divides $f$ and $g$ but which has a degree larger than $r_p$. At this moment I'm not sure how to proceed (I'm a bit rusty on my abstract algebra). I feel that the solution is obvious but I'm not seeing it. How is this sort of problem best handled?

答: Those matrices are all invertible, which means that $r_p$ is a combination of $f$ and $g$. Hence every divisor of $f$ and $g$ also divides $r_p$.</p>


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
You can prove the tail sum formula for expectation using linearity of expectation. Will only prove in the discrete case. Let $X$ be a non-negative discrete R.V., and note that

$$X = \mathbb{I}\{X \ge 1\} + \mathbb{I}\{X \ge 2\} + \ldots$$

Use linearity of expectation on the above and you are done.

Somewhat circularly, you get back the definition of expectation from a property of expectation.


问:
Does the "$\neq$" in, for instance, "Prove that $(x+y)^2\neq x^2+y^2$" mean "never equal"? "not equivalent"?
I have the problem:

Prove that $(x + y)^2 \not= x^2 + y^2$

I have answered it with something like this:

$$(x+y)^2 = x^2 + 2xy + y^2$$

$$(x^2 + 2xy + y^2) - (x^2 + y^2) = 2xy$$

$$\therefore (x + y)^2 = x^2 + y^2 \Rightarrow 2xy = 0$$

$$\therefore (x+y)^2 = x^2 + y^2 \Rightarrow x=0 \lor y=0$$

(Apologies if I'm using symbols incorrectly here, I'm very new to this sort of thing, and as I'm teaching myself I have no one to point these sorts of things out to me)

My problem is that, as far as I can tell, this proves that $(x + y)^2 = x^2 + y^2$ for at least some combination of values for $x$ and $y$. So I suppose my question is, for this kind of proof question, am I being asked to prove that the two statements are not equivalent, or that they are never equal?

Here's a picture of the problem:

答:
You are correct. The statement ought to be $$\text{ Prove that } (x+y)^2\not\equiv x^2+y^2$$

For which it suffices to simply give a counterexample: $4=(1+1)^2\neq1^2+1^2=2$.

The technical distinction between $=$ and $\equiv$ is often overlooked, especially in high school courses, but it is very important. $=$ means it works for a specific value or set of values, $\equiv$ means it works for all values.


问:
Why does $\sum_{k=0}^{n} (-1)^k {n \choose k} = 0$ if $n\neq 0$ and 1 otherwise?
Basically, I've computed a few values by hand, and those followed the pattern in the question. Wolframalpha also claims that this function equals the Kronecker delta function which follows
$$\delta_{ij} = \begin{cases}0 &\text{if } i \neq j \\1 &\text{if }x=1\end{cases}$$
Why is this the case?

答: This counts the number of even subsets of a set of size $n$ minus the number of odd subsets. If $n=0$ there is only one subset, which is even ($\varnothing$); otherwise there are the same number of each (swapping whether or not $1$ is included gives a bijection from even to odd).


答:
Just saw a nice application in a lecture1 that fits neatly with the case of non-independent random variables you mentioned - let $S = \{(x_1, \ldots, x_n) \in \{0, 1\}^n | \sum_i x_i \leq k\}$. If $k \leq \frac{n}{2}$, then $|S| \leq 2^{n \cdot H_b\left(\frac{k}{n}\right)}$ where $H_b(p)$ is the binary entropy function.

We show this by letting $(X_1, \ldots, X_n)$ be distributed uniformly in $S$. Then $H(X_1, \ldots, X_n) = \log |S|$. Noticing that the $X_i$ are identically distributed: $$H(X_1, \ldots, X_n) \leq H(X_1) + \ldots + H(X_n) = n \cdot H(X_1)$$

$X_1$ takes values in $\{0, 1\}$, so we could calculate $H(X_1) = H_b(\mathbb{E}[X_1])$ and finish this if we can bound $\mathbb{E}[X_1]$. Now comes the use of linearity of expectation: we have $$n \cdot \mathbb{E}[X_1] = \mathbb{E}[X_1] + \ldots + \mathbb{E}[X_n] = \mathbb{E}[X_1 + \ldots + X_n] \leq k$$
so $\mathbb{E}[X_1] \leq \frac{k}{n}$. As $\frac{k}{n} \leq \frac{1}{2}$ and $H_b(p)$ is increasing on $[0, \frac{1}{2}]$, we have $H(X_1) \leq H_b\left(\frac{k}{n}\right)$, finishing the proof.

1: From https://ttic.uchicago.edu/~madhurt/courses/infotheory2017/l2.pdf

Title: Striking applications of linearity of expectation

Body:
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?


问:
Solving quadratic inequalities algebraically without graphing/common-sensing the equality signs
I've just been presented an inequality at GCSE (high school) level that, solving algebraically (the method used for linear inequalities), doesn't work correctly which has caught me by surprise. Research always leads to graphed or common-sensed solutions. Can anyone explain why this doesn't work and if there's an algebraic manipulation that leads to the right answers?

$$x^2 - 49 > 0$$
Using the quadratic identity
$$(x + 7)(x - 7) > 0$$
Divide both sides by (x+7)...
$$x - 7 > 0$$
$$x > 7$$
Divide both sides by (x-7)...
$$x + 7 > 0$$
$$x > -7$$

Alternatively,
$$x^2 - 49 > 0$$
$$x^2 > 49$$
$$x > \pm7$$

The negative solution has to be $$x < -7$$ but I don't know how that's arrived at through algebra.

答:
If you want a purely algebraic solution then note that the inequality $ab>0$ is only satisfied when $a>0$ and $b>0$, or when $a<0$ and $b<0$. So in your example,
$$
(x+7)(x-7) > 0 \, ,
$$
we need both $x+7$ and $x-7$ to be greater than $0$ (which gives us the solution $x>7$), or we need $x+7$ and $x-7$ to both be less than $0$ (which gives us the solution $x<-7$). However, I think you are much less prone to error if you solve this geometrically.


答:
Recall the binomial formula: $$(x+y)^n=\sum_{k=0}^n \binom nk x^k y^{n-k}$$

Then: $$0=0^n=((-1)+1)^n=\sum_{k=0}^n\binom nk (-1)^k (1)^k=\sum_{k=0}^n\binom nk (-1)^k$$

Title: Why does $\sum_{k=0}^{n} (-1)^k {n \choose k} = 0$ if $n\neq 0$ and 1 otherwise?

Body:
Basically, I've computed a few values by hand, and those followed the pattern in the question. Wolframalpha also claims that this function equals the Kronecker delta function which follows
$$\delta_{ij} = \begin{cases}0 &\text{if } i \neq j \\1 &\text{if }x=1\end{cases}$$
Why is this the case?


答:
Answering my own question:

This is simply a special case of the equality

$$(a+b)^n = \sum_{k=0}^n {n \choose k}a^{n-k}b^k$$

with $a=1$ and $b=-1$.


问:
Find generator of principal ideal
The ideal $(9, 2 + 2\sqrt{10})$ of $\mathbb{Z}[\sqrt{10}]$ is a principal ideal; it is generated by $1+\sqrt{10}$.

This is easy enough to check once it's been found, but can anyone tell me some way to arrive at this (or some other) generator? That could be done with pencil & paper?

Ordinarily I would use the Euclidean algorithm, but we don't have that here...

Title: Find generator of principal ideal

Body:
Hint: let $\ w = 1\!+\!\sqrt{10}.\ $ Then $\ ww' = -\color{#c00}9.\ $ Suppose $\ (9,2w) = (a).\ $

Then $\ (aa') = (a)(a') = (9,2w)(9,2w') = 9(9,2w,2w',-4) = (\color{#c00}9),\ \  {\rm by}\ \ \ (9,4) = (1).$


答:
Without further information there is a couple ways to interpret the problem statement.

Perhaps it is to prove that $(x+y)^2 \neq x^2 + y^2$ fails to be an identity. Which is to say there exists values for which it fails. Thus we prove it is not an identity by giving a counter-example to $(x+y)^2\neq x^2 + y^2$.

We could interpret it to be asking us to prove $(x+y)^2 > x^2 + y^2$ or $(x+y)^2 < x^2 + y^2$. However, this also is case dependent. So again would fail to be an identity.

My guess would be that it is asking for the first interpretation I have listed above.

Title: Does the "$\neq$" in, for instance, "Prove that $(x+y)^2\neq x^2+y^2$" mean "never equal"? "not equivalent"?

Body:
I have the problem:

Prove that $(x + y)^2 \not= x^2 + y^2$

I have answered it with something like this:

$$(x+y)^2 = x^2 + 2xy + y^2$$

$$(x^2 + 2xy + y^2) - (x^2 + y^2) = 2xy$$

$$\therefore (x + y)^2 = x^2 + y^2 \Rightarrow 2xy = 0$$

$$\therefore (x+y)^2 = x^2 + y^2 \Rightarrow x=0 \lor y=0$$

(Apologies if I'm using symbols incorrectly here, I'm very new to this sort of thing, and as I'm teaching myself I have no one to point these sorts of things out to me)

My problem is that, as far as I can tell, this proves that $(x + y)^2 = x^2 + y^2$ for at least some combination of values for $x$ and $y$. So I suppose my question is, for this kind of proof question, am I being asked to prove that the two statements are not equivalent, or that they are never equal?

Here's a picture of the problem:


问:
solve a linear congruence
I am studying for an exam and came upon the problem of solving the linear congruence $10 x = 15 \pmod{45}$.  I am stuck as the $10$ and $45$ are not coprime but have the $5$ as a common factor.

Title: solve a linear congruence

Body:
Clue:

$$45|10x-15\iff 9|2x-3$$


答: Note that $$T(x,y) = (x+y) (1,2),$$so the range of $T$ is spanned by $(1,2)$. But a non-zero vector by itself is linearly independent, so $\{(1,2)\}$ is a basis for the range, and thus ${\rm rank}(T) = 1$. By the rank-nullity theorem, $\dim \ker T = 1$ as well, and so any non-zero vector in the kernel of $T$ forms a basis for it, by the same argument above: clearly $(1,-1)$ fits the bill.

Body:
Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation?

Rank is defined to be $\dim(\text{im}(T))$ right? Every output of $T$ is a vector belonging to $\Bbb{R^2}$ and $\Bbb{R^2}$ is the image of $T$ so I assumed that $\text{rank}(T) = 2$. Turns out it's actually $1$, which I don't seem to understand... Every output of $T$ can be expressed as $c(1,2)$ but this is also just a tuple in $\Bbb{R^2}$ so how can the rank be $1$?

The nullity is defined to be the set of vectors for which $T(x) = 0$ so I have that $$(x+y,2(x+y)) = 0 \Longrightarrow x+y=0, 2x+2y=0 \Longrightarrow (x,y) = (0,0)$$

so the set of vectors for which $T(x) = 0$ is just $\{(0,0)\}$ so $\ker(T) = 1$?


问:
Proving a property of the Fourier transform
Let $0 < f ∈ L^1(\mathbb{R^2})$ almost everywhere.
Prove: $|\hat{f}(x)| < \hat{f}(0) \quad \forall x \in \mathbb{R^d}\backslash\{0\}$.

Could someone please help me with this question? I tried solving this with the step function but I don't seem to get anywhere. Any advice would be much appreciated. Thank you in advance.

Body: $$|\hat{f}(\xi)| < \int_{\mathbb{R}^{d}} |e^{-i x \xi}| f(x) dx=\int_{\mathbb{R}^{d}} f(x) dx=\hat{f}(0)$$


答:
Buffon's needle: rule a surface with parallel lines a distance $d$ apart. What is the probability that a randomly dropped needle of length $\ell\leq d$ crosses a line?

Consider dropping any (continuous) curve of length $\ell$ onto the surface. Imagine dividing up the curve into $N$ straight line segments, each of length $\ell/N$. Let $X_i$ be the indicator for the $i$-th segment crossing a line. Then if $X$ is the total number of times the curve crosses a line,
$$\mathbb E[X]=\mathbb E\left[\sum X_i\right]=\sum\mathbb E[X_i]=N\cdot\mathbb E[X_1].$$
That is to say, the expected number of crossings is proportional to the length of the curve (and independent of the shape).

Now we need to fix the constant of proportionality. Take the curve to be a circle of diameter $d$. Almost surely, this curve will cross a line twice. The length of the circle is $\pi d$, so a curve of length $\ell$ crosses a line $\frac{2\ell}{\pi d}$ times.

Now observe that a straight needle of length $\ell\leq d$ can cross a line either $0$ or $1$ times. So the probability it crosses a line is precisely this expectation value $\frac{2\ell}{\pi d}$.


问:
Earliest use of adjunction spaces?
An adjunction space is a topological space $Y= B \cup _f X$ determined by an inclusion $i : A \to X$ and a map $f: A \to B$; it can  be described briefly as the pushout  of the two maps $f,i$. The advantage of this notion is that it is behind the J.H.C. Whitehead notion of CW-complex, which is about inductively attaching cells $E^n$ by means of maps $S^{n-1} \to X_{n-1}$. Good properties of $A,X,B$ are often carried over to $Y$ (eg being Hausdorff).  The pushout property allows one to construct continuous functions on $Y$. { cf Topology and Groupoids).

The earliest paper I have on this notion is by Whitehead: "Note on a Theorem due to Borsuk",  Bull AMS (1948) 1125-1137. I would be grateful for information on earlier reference to this notion.

答:
Since I was asked to state an official answer, here it is, although it's just copy paste from Engelking's book:

According to Engelking, General Topology, Historical and bibliographic notes of section 2.4, adjunction spaces were defined by Borsuk in 1935 (for compact metric spaces)

Title: Earliest use of adjunction spaces?


问:
Change of variables inside conditional expectation
Let $X, Y$ be random variables, $g$ be some function on $\mathbb{R}$ that is invertible.

Suppose also that we know $\mathbb{E}[g(Y)|X]$, and the goal is to find $\mathbb{E}[Y|X]$. Is this possible in closed-form? I'm aware we could do a Taylor approximation here, but what about an exact solution?

答:
The conditioning on $X$ is immaterial here; the question boils down to

If I know $E[g(Y)]$ can I determine $E[Y]$?

Since $g$ is invertible, the question further reduces to

If I know $E[Y]$, can I determine $E[f(Y)]$ where $f$ is invertible?

This is not true: for exapmle there are many zero-mean random variables $Y$ such that $E[f(Y)]$ are different.

(Take for instance $Y_n / n \sim \text{Bernoulli}(1/2)$ and $f(x) = e^x$. Then $E[Y_n] = 0$ for all $n$, but  $E[f(Y_n)] = \frac{e^n + e^{-n}}{2}$ which depends on $n$.)


问:
Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$.
How would one solve for x in the following equation:

$\dfrac{(1+x)^{36} -1}{x} =20142.9/420$

I tried factorising the top but that didnt really help much.

$((1+x)^{18} - 1)((1+x)^{18}+1)$

Any help is appreciated thanks.

答:
Assuming we want a small positive value for $x$, we can approximate $(1+x)^{36}$ as $$1^{36}+{36 \choose 1} 1^{35}x^1 + {36\choose 2} 1^{34} x^2+O(x^3)$$

With this approximation, the equation becomes approximately $$\frac{36x+630x^2}{x}=47.959$$

This can be rearranged as $630x=47.959-36$, which has solution $x\approx 0.01898$.


问:
Strength of "Cofinite Choice"
Let $A$ be a set of sets such that $(\bigcup A) \setminus x$ is finite for every $x \in A$.

How strong is the variant of AC asserting that such an $A$ has a choice function?
I'd assume it to be weaker than countable choice, but ZF doesn't seem to prove it either.

答:
The axiom of cofinite choice is equivalent to the principle that every Dedekind-finite set is finite. Equivalently, for every infinite set $X$, there is an injective function $\omega\to X$. So as you suspected, it is (strictly) weaker than countable choice, but not provable in ZF.

Assume cofinite choice and let $X$ be an infinite set. Let $A$ be the set of all cofinite subsets of $X$, and let $g$ be a choice function for $A$. Then we can construct an injective function $\omega\to X$ by recursion. Let $f_0\colon 0\to X$ be the empty function. Given an injective function $f_n\colon n\to X$, the set $X_n = X\setminus \text{im}(f_n)$ is cofinite, and we can extend $f_n$ to $f_{n+1}\colon (n+1)\to X$ by defining $f_{n+1}(n) = g(X_n)$. Then $f = \bigcup_{n\in \omega} f_n$ is an injective function $\omega\to X$.

Conversely, assume that every Dedekind-finite set is finite. Let $A$ be a set of nonempty sets such that $(\bigcup A)\setminus x$ is finite for all $x\in A$. If $\bigcup A$ is finite, then $A$ is finite, and ZF proves that every finite set has a choice function. If $\bigcup A$ is infinite, let $f\colon \omega\to \bigcup A$ be an injective function. Then for any $x\in A$, since $(\bigcup A)\setminus x$ is finite, $x\cap \text{im}(f)$ is nonempty. So we can define a choice function $g$ on $A$ by defining $g(x) = f(n)$ for the minimal $n\in \omega$ such that $f(n)\in x$.

Title: Strength of "Cofinite Choice"


问:
$K[\alpha]$ is a field implies $\alpha$ is algebraic
Let $L$ be a field extension of $K$, and $\alpha \in L$. In this class I am taking, the following proof that $K[\alpha]$ is a field implies $\alpha$ is algebraic over $K$ is given:

If $\alpha$ is not algebraic, $\not \exists P \in K[x]$ s.t. $P(\alpha) = 0$.

This means $i \colon K[x] \to L$ given by $i(P) = P(\alpha)$ is injective (since if $P(\alpha) = 0$ then $P = 0$ by the above assumption that $\alpha$ is not algebraic). But $K[x]$ is not a field and the image of $i = K[\alpha]$ is a field, which is a contradiction.

I'm fine with everything up to this last statement. My thought process is that, since $\ker(i) = (0)$, then by the first ring isomorphism theorem $K[x]/(0) \cong K[\alpha]$, and since $K[x]/(0) \cong K[x]$ this implies $K[x] \cong K[\alpha]$, which is a contradiction since $K[x]$ is not a field.

Is there a more obvious explanation? Or is this obvious enough that it is ok to leave it out as the instructor did in the original proof.

答:
I think it is way simpler to observe that if $\;K[a]\;$ is a field (and assuming $\;a\neq0\;$ otherwise it is trivial), then

$$\frac1a\in K[a]\implies \exists \;0\neq p(x)\in K[x]\;s.t.\;\;\frac1a=p(a)\implies ap(a)-1=0$$

and certainly the polynomial $\;f(x)=xp(x)-1\in K[a]\;$ is non zero, thus $\;a\in L\;$ is algebraic over $\;K\;$ .


问:
How to calculate $\int_{0}^{\infty}y^{2}e^{-y(x+1)}dy$?
I want to evaluate the following integral in a question I had in Probability:
$$
\int_{0}^{\infty}y^{2}e^{-y(x+1)}dy
$$
But I'm stuck figuring out how to solve this integral. How to calculate this integral?

答:
Without using exponential distribution shortcuts,
$$\int_{0}^{\infty}y^2e^{-y(x + 1)}\text{ d}y = \lim_{t \to \infty}\int_{0}^{t}y^2e^{-y(x + 1)}\text{ d}y\text{.}$$
We then integrate by parts twice.

For the first time, let $u_1 = y^2$ and $\text{d}v_1 = e^{-y(x+1)}\text{ d}y$, so that
$$
\begin{align}
\int_{0}^{t}y^2e^{-y(x + 1)}\text{ d}y &= \left.\dfrac{y^2}{-(x+1)}e^{-y(x+1)}\right|_{0}^{t} - \int_{0}^{t}\dfrac{e^{-y(x+1)}}{-(x+1)}\cdot 2y\text{ d}y \\
&= \dfrac{-t^2}{x+1}e^{-t(x+1)}+\dfrac{2}{x+1}\int_{0}^{t}ye^{-y(x+1)}\text{ d}y
\end{align}$$
Now, for the integral $\int_{0}^{t}ye^{-y(x+1)}\text{ d}y$, we set $u_2 = y$ and $\text{d}v_2 = e^{-y(x+1)}\text{ d}y$, so that
$$\begin{align}
\int_{0}^{t}ye^{-y(x+1)}\text{ d}y &= \left.\dfrac{y}{-(x+1)}e^{-y(x+1)}\right|_{0}^{t} - \int_{0}^{t}\dfrac{e^{-y(x+1)}}{-(x+1)}\text{ d}y \\
&= \dfrac{t}{-(x+1)}e^{-t(x+1)}+\dfrac{1}{x+1}\int_{0}^{t}e^{-y(x+1)}\text{ d}y \\
&= \dfrac{t}{-(x+1)}e^{-t(x+1)}+\dfrac{1}{(x+1)^2}\left[1-e^{-t(x+1)}\right]
\end{align}$$
Thus we obtain
$$\int_{0}^{t}y^2e^{-y(x+1)}\text{ d}y = \dfrac{-t^2}{x+1}e^{-t(x+1)}+\dfrac{2}{x+1}\left\{ \dfrac{t}{-(x+1)}e^{-t(x+1)}+\dfrac{1}{(x+1)^2}\left[1-e^{-t(x+1)}\right]\right\}$$
and taking the limit as $t \to \infty$,
$$\int_{0}^{\infty}y^2e^{-y(x+1)}\text{ d}y = \dfrac{2}{x+1}\left[\dfrac{1}{(x+1)^2}\right] = \dfrac{2}{(x+1)^3}\text{.}$$

With exponential distribution shortcuts, write
$$\int_{0}^{\infty}y^2e^{-y(x+1)}\text{ d}y = \int_{0}^{\infty}z^2e^{-z(x+1)}\text{ d}z = \dfrac{1}{x+1}\int_{0}^{\infty}z^2 \cdot (x+1)e^{-z(x+1)}\text{ d}z$$
and one will recognize that the above integral has a part which is the PDF of an exponential distribution with mean $\frac{1}{x+1}$: namely $f(z) = (x+1)e^{-z(x+1)}$ for $z > 0$.

So,
$$\int_{0}^{\infty}z^2 \cdot (x+1)e^{-z(x+1)}\text{ d}z = \int_{0}^{\infty}z^2f(z)\text{ d}z$$
is $\mathbb{E}[Z^2]$, where $Z$ follows an exponential distribution with mean $\mathbb{E}[Z] = \dfrac{1}{x+2}$.

Thus, one may recall that $\text{Var}(Z) = \dfrac{1}{(x+2)^2}$, so
$$\mathbb{E}[Z^2] = \text{Var}(Z) + (\mathbb{E}[Z])^2 = \dfrac{1}{(x+2)^2} + \dfrac{1}{(x+2)^2} = \dfrac{2}{(x+2)^2}$$
thus
$$\int_{0}^{\infty}y^2e^{-y(x+1)}\text{ d}y = \dfrac{1}{x+1} \cdot \mathbb{E}[Z^2] = \dfrac{2}{(x+1)^3}$$


问:
Triple volume integral of divergence
S is given by $z = \sqrt{x^2+ y^2} \ $ for $1 ≤ z ≤ 4$, then calculate $\nabla \cdot \vec{F} \ $ for the vector field $\vec{F} = (xz, −zy, x^3)$.

Evaluate $I = \iiint \ (\nabla \cdot \vec{F}) \ dV$ ,
where $V$ is the volume enclosed by the surface S and the two discs
{(x, y, z) : x^2+y^2 ≤ 1 , z = 2} and {(x, y, z) : x^2+y^2 ≤ 9 , z = 3}$.

I got a negative answer it thet right?

答:
Nothing forbids you from having a negative flux because it’s just an integral.

Through that surface you have portions where the field is outgoing, hence it has the same direction of the external local normal versor of the surface, and other portions where the field is incoming, hence it has the opposite direction of the external local normal versor.

The net balance is the value measured by the integral.

You are always calculating the flux as a outgoing flux, so if it’s positive it means that it’s mainly outgoing, while if it’s negative then it is mainly incoming.

Title: Triple volume integral of divergence


问:
Normalized "probability functions" such that the marginal are not probability functions
This question is inspired from Quantum Optics. There one has a nonnegative normalized function, "Husimi function" such that

$$
\int\int dx dp Q(x,p) = 1,
$$

where the integrals are done over the whole complex plane. However, in the literature it is insisted time and again that this is only a quasi-distribution, in the sense that the marginals are not obtained from it, e.g.

$$
\int dp Q(x,p) \neq P(x),
$$

where $P(x)$ is the probability function over the $x$-axis. However, it might be that $\int dp Q(x,p)= R(x) $ such that $\int R(x)=1$, positive definite and at the end $R(x)$ "resembles" a probability density function in its own right.

With this $R(x)$, is it still possible to say that $Q(x,p)$ is a probability function instead of a "quasi-probability"? Does it exist a probability function $Q'(x,p)$ nonnegative and normalized to one such that $\int dp Q'(x,p) = R'(x)$ but $\int R(x)\neq 1$?

答:
It might help you to refer to this PSE question and its answer. Your texts should have emphasized why all these phase-space distributions, semi-definite-positive or not,
cannot be probability distributions.

The reason is they all fail Kolmogorov's third axiom that different points of the plane represent mutually exclusive logical alternatives, which, you might know from physics, contiguous points within an $\hbar$ distance do not, by dint of the uncertainty principle.

The positive semi-definiteness itself of Q is a red herring, ritually confusing the novice. Suppose one has a function f(x), whose expectation value one wishes to find. Your texts should have described why
$$
\langle f(x)\rangle \neq \int\int\!\!dpdx ~ Q(x,p) f(x) = \int\!\! dx ~  R(x) f(x).
$$

The easiest way to see this is to refer Q to the Wigner distribution W, (the Cartesian coordinate system of these distributions, so to speak!), whose marginal is a bona-fide probability distribution P(x),
$$
\langle f(x)\rangle = \int\int\!\!dpdx ~ W(x,p) f(x) = \int\!\! dx ~  P(x) f(x).
$$

You thus utilize the fact that the Hussimi is the Weierstrass transform of the Wigner distribution,
$$
 Q(x,p) = \exp \left({ 1\over 4}(\partial_x^2
           +\partial_p^2 ) \right) W(x,p)     \\
 \propto \int \int dx' dp'
\exp \left( -{(x'-x)^2-(p'-p)^2   } \right) ~ W(x',p') ,
$$
to see that its marginal
$$
\int  \int \int  dp dx' dp'
\exp \left( -{(x'-x)^2-(p'-p)^2   } \right) ~ W(x',p') ,
$$
is not the probability distribution P(x) we already saw.

Instead, you'd need something like

$$
\langle f(x)\rangle =  \int\!\!dx ~ \left (\int dp \exp \left(-{ 1\over 4}(\partial_x^2 + \partial_p^2)
              \right)  Q(x,p)\right )~~ f(x) = \int\!\! dx ~  P(x) f(x).
$$

Damning the torpedoes to go full speed ahead, and using plain Q s without Weierstrass kernels produces Wehrl quasi-entropies , uncontrolled approximants of the von Neumann entropy.


问:
Graphing divergence of a 3D vector field
I want to be able to compute and graph the divergence of a three dimensional vector field. I have tried
the following and the resulting graph is as attached.

```<code>clc
syms x y z
f = input("Enter the 3D vector function as [f1, f2, f3]");
div(x, y, z) = divergence(f, [x, y, z]);
p(x, y, z) = f(1); q(x, y, z) = f(2); r(x, y, z) = f(3);
x = linspace(-4, 4, 20); y = x; z = x;
[X, Y, Z] = meshgrid(x, y, z);
U = p(X, Y, Z); V = q(X, Y, Z); W = r(X, Y, Z);
hold on
quiver3(X, Y, Z, U, V, W, 1.5);
axis on
hold off
title("Vector Field of F(x, y, z) = [f1, f2, f3]");
</code>```

and the figure output for $(x, 2y^2, 3z^3)$ -

Now obviously I would like for the figure to be represented in 3D, and I did use quiver3 for the purpose.

Surely there's something iffy with my code. An elaborate explanation about my mistakes here would be much appreciated.

I must state that I copied the code from what was taught to us for computing the divergence of 2D functions and simply went about making changes to accommodate the third variable. Also, I'm a total noob at MATLAB. My professor doesn't help my situation. Their forums have been helpful in understanding.

答:
Your function is $f({\bf x}) = (x, 2 y^2, 3 z^3)$ so $\nabla f({\bf x}) = (1, 4 y, 9 z^2)$.

Here is a plot of the divergence (as a scalar color) atop the vector function (rendered in Mathematica):


答:
This is a standard example of an equation
that does not have an explicit formula for its root.
It can only be solved numerically.

The important thing is to get a good first approximation to the root.

For example, if you want to solve
$\frac{(1+x)^n-1}{x} = a$,
and you think that there might be a root close to $0$,
you can approximate $(1+x)^n$
by $1+nx+n(n-1)x^2/2$
(the first terms of the binomial theorem)
to get
$n + n(n-1)x/2 = a$
or $x = 2(a-n)/(n(n-1))$.

Then apply Newton's iteration.

Title: Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$.

Body:
How would one solve for x in the following equation:

$\dfrac{(1+x)^{36} -1}{x} =20142.9/420$

I tried factorising the top but that didnt really help much.

$((1+x)^{18} - 1)((1+x)^{18}+1)$

Any help is appreciated thanks.


问:
Prove that products of permutations $ab$ and $ba$ have the same cycle length
For example:
Let $a,b \in S_3$.

$a = (1,2,3), b = (1,2)(3)$,
then $ab = (1,3)(2) $ and $ba = (2,3)(1) $.

These are two different permutations, but their cycle type is the same.
How could we prove this for a general case?

I have some idea of what conjugate cycles are and that they might be required to prove this, but if there is a proof without them, that would be even better.

答:
I'm sorry but I'm gonna give you a proof using conjugation since it is just the easiest. Like you I would indeed be interested in seeing a different proof.

Lemma Let $(x_1, x_2, \ldots, x_n)$ be a cycle (so each $x_i$ is an integer between 1 and $N$ for some $N \geq n$), and let $a$ be a permutation in $S_N$. Then the permutation $a(x_1, x_2, \ldots, x_n)a^{-1}$ can also be written as a cycle of length $n$.

proof The easiest way to show that it can be written in this form is by doing it. For this we need a notational convention: I denote by $a(x_i) \in [1, \ldots, N]$ the number to which $x_i$ is sent by permutation $a$. (So I view $a$ as a function from $[1, \ldots, N]$ to itself.)

Now the permutation $a(x_1, \ldots, x_n)a^{-1}$ equals the cycle $(a(x_1), \ldots, a(x_n))$.

To see that just check where both permuations send a given number $z$. If the two answers are the same for every $z$ we know the permuations are equal.

We know that $z$ equals $a(y)$ for some (unique) $y$. We consider two cases: $y$ is one of the $x_i$ or it is not.

In the first case let's see what $a(x_1, \ldots, x_n)a^{-1}$ does do to $z$. First we let $a^{-1}$ act that sends it to $y = z_i$. Then we have the cycle that sends $z_i$ to $z_{i+1}$ (or to $z_1$ if $i = n$). Finally we have $a$ that sends $x_{i+1}$ to $a(x_{i+1})$. So the answer is $a(x_{i+1})$.

On the other hand what does the permuation $(a(x_1), \ldots, a(x_n))$ do with $z$? Well since $z = a(x_i)$ by assumption, this cycle sends it so $a(x_{i+1})$. So the answers are equal as expected.

In other case, when $y$ is not one of the $x_i$ what do both permuations do? $a(x_1, \ldots, x_n)a^{-1}$ acts by first letting $a^{-1}$ send $z$ to $y$. Then $y$ gets sent to itself by the cycle and finally $a$ sends $y$ to $a(y) = z$, so that $z$ is sent to itself. It is clear that the permutation $(a(x_1), \ldots, a(x_n))$ also sends $z$ to itself when $z$ is not one of the $a(z_i)$ so also in this case the two answer coincide. End proof.

Corollary Let $p$ be any permutation, then $p$ and $apa^{-1}$ have the same cycle type.

proof Let $p = c_1 c_2 \cdots c_n$ with each $c_i$ a cycle. We know from the lemma that $ac_ia^{-1}$ is a cycle of the same length as $c_i$ for each $i$. But now we can just write
$$a^{-1}pa = a^{-1}c_1(aa^{-1})c_2(aa^{-1})c_3(aa^{-1}) \cdots (aa^{-1})c_na = (a^{-1}c_1a)(a^{-1}c_2a)\cdots (a^{-1}c_na)$$ which is a product of cycles of the same length as the $c_i$ appearing in the cycle decompostion of $p$. End proof.

I love this trick.

Now how does this relate to your question?

$$ba = a^{-1}(ab)a$$

So we can apply the corollary with $p = ab$.


答:
You can do a Taylor approximation here. Let's denote $Z=g(Y)$ then $Y = g^{-1}(Z)$. For simplicity's sake, denote also $f(Z)=g^{-1}(Z)$ and $a(X)=E[Z|X]$.
We have
$$f(Z)=f(E[Z|X]) + \sum_{i=1}^{+ \infty}(Z-E[Z|X])^n\frac{d^n}{dz^n}f(E[Z|X])$$
$$f(Z)=f(a(X)) + \sum_{i=1}^{+ \infty}(Z-E[Z|X])^n\frac{d^n}{dz^n}f(a(X))$$
This implies that
$$E[Y|X]=E[f(Z)|X] = f(a(X)) + \sum_{i=1}^{+ \infty}E[(Z-a(X))^n |X]\frac{d^n}{dz^n}f(a(X))$$
But if you know only $E[Z|X]$ and not the variable $Z$ (or $Y$) itself, the term $E[(Z-a(X))^n|X] = E[(g(Y)-a(X))^n|X]$ is undefined.

The conclusion: it is unable to determine $E[Y|X]$ from the sole information $E[g(Y)|X]$.

Title: Change of variables inside conditional expectation

Body:
Let $X, Y$ be random variables, $g$ be some function on $\mathbb{R}$ that is invertible.

Suppose also that we know $\mathbb{E}[g(Y)|X]$, and the goal is to find $\mathbb{E}[Y|X]$. Is this possible in closed-form? I'm aware we could do a Taylor approximation here, but what about an exact solution?


问:
Ordinary/Partial Differentiation Confusion
So, I've seen from different sources that $\frac{d}{dx}$ and $\frac{\partial}{\partial x}$ function as basically the same thing. But then, I saw that if $$\frac{d}{dx}(z(x, y(x)) = 0$$ then $$ z(x, y(x)) = c$$
where c is a constant, because "if the ordinary (not partial…) derivative of something is zero, that something must have been a constant to start with."

Related to this I think is the notion that in implicit differentiation, taking the $\frac{d}{dx}$ of a given $g(y)$ gets $\frac{d}{dy}\frac{dy}{dx}$, while in other cases I've heard that taking $\frac{d}{dx}$ of $g(y)$ simply equals zero since y can be treated as a constant.

What are the distinctions here?

答:
$\frac{d}{dx} z(x, y(x))$ is unambiguous: we have a function $f(x) = z(x, y(x))$ of $x$ obtained by composing two other functions $z$ and $y$, and we want to compute its derivative. But you might interpret the partial derivative

$$\frac{\partial}{\partial x} z(x, y)$$

to refer to the partial derivative of $z$ in the first variable only, holding $y$ fixed, and ignoring the fact that $y$ is a function of $x$. You might do this in order to state the chain rule in this setting, for example, which can be written

$$\frac{d}{dx} z(x, y(x)) = \frac{\partial}{\partial x} z(x, y) + \frac{\partial}{\partial y} z(x, y) \frac{dy}{dx}.$$

This gets at a very ambiguous thing about partial derivative notation, which is that any partial derivative is implicitly measuring the change in one variable while "keeping all other variables fixed," and it can get quite ambiguous what "all other variables fixed" means; in particular, although the notation only refers to a single variable it actually depends on a choice of many other variables (the "fixed" ones), which is omitted from the notation and to be "understood from context."

This can get very confusing and in my opinion is a sign that partial derivative notation should be replaced by something more precise, although it's tricky to see how to do it without making everything excessively verbose. I think Sussman and Wisdom's Structure and Interpretation of Classical Mechanics may do something like this but I haven't read it.  They at least discuss the example of the Euler-Lagrange equation in the preface.


问:
How many different equivalence relations with exactly two different equivalence classes are there on a set with $n$ elements
I came across with this topic. It looks straight forward for $5$ elements, but what if I want to find how many different equivalence relations with exactly two different equivalence classes are there on a set with $n$ elements?

It says that the set of equivalence relations on a set are in direct bijection with the set of partitions on a set. This made me read about Bell numbers but I'm not sure how help us here. How to solve this problem?

答: Since equivalence relations bijectively correspond to partitions on the same set (via their equivalence classes) we see that the number of equivalence relations with exactly two different equivalence classes is the number of partitions of a set with $n$ elements into two non-empty sets. There are in total $2^n$ different subsets of which $2^n-2$ are non-empty and not the whole set. Therefore the number of these partitions is $\frac{2^n-2}{2}=2^{n-1}-1$ (we need to divide by $2$ since otherwise we would count each partition twice)


问:
Uniqueness of payout of Nash equilibrium in a 2-player zero-sum game
I'm convinced that in a 2 player zero sum game, any two NE strategies must have the same expected payout for both players. I'm even more convinced that this is not true for 3 or more players.

Where could I find a reference for this?
Is this still true if the game is not zero sum?
(you could add a dummy player to make it zero sum, so I suppose not)

答:
The couple of strategies $s_1^e$, $s_2^e$  is an equilibrium point (saddle point ) according to Von Neumann concept if the two following conditions hold:

$g\left(s_1^e, s_2^e \right) \ge \ g\left(s_{1,i}, s_2^e \right) \forall s_{1,i} \in S_1$

$g\left(s_1^e, s_2^e \right) \le \ g\left(s_1^e, s_{2,j} \right) \forall s_{2,j} \in S_2$

The definition of equilibrium has been formulated by v. Neumann for zero sum games.
The definition of equilibrium according to Von Neumann was given for zero-sum games and it can be seen as the "translation" of the concept of equilibrium in the sense of Nash given later for non-cooperative games. The following shows how to pass from the concept of Von Neumann equilibrium to the concept of Nash equilibrium: just remember that for a zero-sum game we have:

$g_1 = g$ and $g_2 = -g$

so it can be written that

$g_1 \left(s_1^e, s_2^e \right) \ge \ g_1 \left(s_{1,i}, s_2^e \right) \forall s_{1,i} \in S_1$

The first condition of equilibrium in the sense of Von Neumann is therefore the first condition of equilibrium in the sense of Nash referred to player 1.

For player 2 it becomes:

$-g_2 \left(s_1^e, s_2^e \right) \le \ -g_2 \left(s_1^e, s_{2,j} \right) \forall s_{2,j} \in S_2$
that is

$g_2\left(s_1^e, s_2^e \right) \ge \ g_2\left(s_1^e, s_{2,j} \right) \forall s_{2,j} \in S_2$

E. Burger, Introduction to the Theory of Games, Prentice-Hall, Inc., 1959

Non-constant sum games are implicitly non-zero-sum games; while all constant-sum games can be considered as zero-sum games without altering the outcome of the game (dummy player).

References:

J. von Neumann, Contributions to the theory of games. Vol. IV, Annals. of Mathematics Studies, no.40, Princeton Univ. Press, 1959

John von Neumann, Oskar Morgenstern (1944): Theory of Games and Economic Behavior, Princeton University Press; ed. 1953


问:
Factoring property for a surjective morphisms of sheaves
Let $X \to Y$ be a surjective morphism of sheaves on the site $(\operatorname{Sch}/S)_{et}$ and suppose that $X$ is a scheme.   Consider the cartesian diagram
$\require{AMScd}$
\begin{CD}
F  @>{}>> W \\
@VVV @VVV\\
Y  @>{\triangle_Y}>> Y \times_S Y
\end{CD}
where $\triangle_Y$ is the diagonal map and $F:= W \times_{Y \times_S Y, \triangle_Y} Y$ is the fiber product.

Let us assume that $S$ and $W$ are affine schemes.

It is stated in Olsson's Algebraic spaces and stacks in subsection 5.2.9 that: Since $X \to Y$ is a surjective morphism of sheaves, there exists an etale cover $W' \to W$ such that the composition
$$W' \to W \to Y \times_S Y$$
factors through $X \times_S X$.

What about the fact that $X \to Y$ is a surjection implies the existence of such an etale cover factoring through $X \times_S X$?

答:
I might be making a mistake here but I think this comes down to the definition of a surjective morphism in the category of sheaves on $(\operatorname{Sch}/S)_{et}$. Namely, $\phi : X \to Y$ is surjective/an epimorphism if for all objects $T \to S$ any element $f \in Y(T)$ can be pulled back to an element of $X(T)$ after possibly taking an etale cover. This means $\exists g: T' \to T$ etale such that the element $f \circ g : T' \to T \to Y$ comes from an element of $X(T')$ under $\phi$. Of course, this is equivalent to $T' \to T \to Y$ factoring through $\phi : X \to Y$. I assume the claim then follows by observing the product map $X \times_S X \to Y \times_S Y$ is also surjective.

See Tag 00WL for definition of surjective morphisms of sheaves on sites.


问:
Evaluate a sum involving n choose r
Evaluate: $\sum_{k=0}^6 (-1)^k \binom{6}{k}$ where $\binom{n}{r}= \frac{n!}{r!(n-r)!}$.

I'm unsure how to compute the part with $\binom{6}{k}$, it should be something along the lines of $\frac{6!}{k!(6-k)!}$ can we simplify this down?

I think there was an expansion method for factorials I could used in the denominator that was somewhere along the lines of $(6-k)(5-k)(4-k)(3-k)(2-k)(1-k)$ ?

答:
Hints:

$$nCr:=\binom{n}r\,,\;\;\text{and by the Binomial Theorem (Newton), we get :}$$

$$(a+b)^n=\sum_{k=0}^n\binom nka^{n-k}b^k$$

Substitute above now with $\,a= 1 \;,\;b = -1\;,\;n=6$ ...

Title: Evaluate a sum involving n choose r


问:
Calculating an improper integral.
There is an improper integral which has to be calculated and I have no idea about it. The problem is in the argument of cosine being just $x$ and not $ax$. If it were $ax$, the problem would have been trivial. Here it is: $$\int_{0}^{+\infty} \frac{a x \cos(x)-\sin(ax)}{x^2} \, dx, \qquad a>0$$
No complex analysis is allowed to use, only calculus knowledge, for example, differentiating integrals depending on a parameter.

答:
If $\text{Ci}(x)$ is an antiderivative of $\cos(x)/x$, an antiderivative of your integrand is $$ F(x) = \frac{\sin(ax)}{x} - a\; \text{Ci}(ax) + a\; \text{Ci}(x)$$
Now $\text{Ci}(t) \sim \text{constant} + \ln(t)$ as $t \to 0+$, while
$\text{Ci}(t) \to \text{constant}$ as $t \to +\infty$.

So the answer is $a \ln(a) - a$.

Title: Calculating an improper integral.


问:
Covariance as double summation over all probability indices and other times as a single summation over joint probability indices.
When reading about covariance, I've seen covariance represented sometimes as a double summation, and other times as a single summation.

An example of the single summation version can be seen in Wolfram's definition:

$Cov(X,Y) = \sum\limits_{i} p_{i} (x_i- \mu_X) (y_i-\mu_Y)$

An example of the double summation can be seen on pg 545 of Strang's Introduction to Linear Algebra 5th Edition or in this math exchange post:

$Cov(X,Y) = \sum\limits_{i}\sum\limits_{j} p_{ij} (x_i- \mu_X) (y_j-\mu_Y)$

There doesn't seem to be an intuitive way to derive one from the other, and no definition of the covariance that I've seen bothers to explain or compare both formulas. How should I interpret these formulas?

答:
This question has been answered as part of another question I asked by the accepted answer (by David K) in this post: https://math.stackexchange.com/a/3984564/333560.

To summarize his explanation, the single summation gets used when only a specific sequence of values, corresponding to the summation indices, have a non-zero probability. Technically the other combination of values exist (as in the double summation), but the probability is zero we can omit them. This is the case when we formulate $N$ discrete random numbers as a a vector in $N$-dimensional space.


问:
What form of choice is "every Dedekind-finite set is finite" equivalent to?
Halmos in his Naive Set Theory proves that every infinite set has a subset equivalent to $\omega$ using the axiom of choice with its full power. And this leads to the corollary that a set is infinite if and only if it is equivalent to some proper subset of it, which leads to each Dedekind-finite set being finite.

But I've also seen a proof (on Wikipedia) that this can also be proven with just countable choice. However Wikipedia also states that this result is strictly weaker than countable choice.

Question: It is clear that we do require some form of choice, not just ZF, to prove this result.$^1$ But it is even weaker than the countable choice. Can we explicitly state the form of this choice which is equivalent to this result?

$^1$ I've come across the fact that there exists a model of ZF (whatever that means (sorry I've not done any model theory; this is just for your reference)) in which every infinite set is Dedekind-infinite, and yet the countable choice fails.

答:
While answering this question: Strength of “Cofinite Choice”, I discovered that "every Dedekind-finite set is finite" is equivalent to the following "axiom of cofinite choice":

Let $A$ be a set of non-empty sets such that $(\bigcup A)\setminus X$ is finite for all $X\in A$. Then $A$ has a choice function.

See the linked answer for a proof. This seems to me to be a fairly natural choice principle.


答:
Assuming $x+1 > 0$, do a change of variables $y = t/(x+1)$ and use the definition of the
Gamma function.

Title: How to calculate $\int_{0}^{\infty}y^{2}e^{-y(x+1)}dy$?

Body:
I want to evaluate the following integral in a question I had in Probability:
$$
\int_{0}^{\infty}y^{2}e^{-y(x+1)}dy
$$
But I'm stuck figuring out how to solve this integral. How to calculate this integral?


问:
How to set boundaries when approximating a discrete RV with a normally distributed RV?
I am approximating a random variable $S_n \sim \text{Bin}(n,0.02)$ with $Z_n=\frac{S_n-\mu n}{\sqrt{np(1-p)}} \sim \mathcal{N}(0,1)$ for high enough $n$ (by central limit theorem).

My problem is how to set interval boundaries when calculating probabilities with that approach.

For example when calculating the chance of $2$ or more (edited the value, but not important) successes in $100$ trials, what is the interval I would like $S_n$ to be in? Is it $[2,100]$ (meaning $S_n\geq2$) or $(1, 100]$ (meaning $S_n>1$), or something inbetween like $[1.5, 100]$?

Calculation for $S_n \in (1,100]$:
$$P(S_n \in (1,100])=1-P(S_n \notin (1,100])=1-P\Big[\frac{S_{100}-0.02\cdot100}{\sqrt{100\cdot0.02\cdot0.98}}\leq \frac{1-0.02\cdot100}{\sqrt{100\cdot0.02\cdot0.98}}\Big]=1-\Phi(-0.71)=\Phi(0.71)\approx0,76$$

Calculation for $S_{100} \in [2,100]$:
$$P(S_{100} \in [2,100])=1-P(S_{100} \notin [2,100])=1-P\Big[\frac{S_{100}-0.02\cdot100}{\sqrt{100\cdot0.02\cdot0.98}}\leq \frac{2-0.02\cdot100}{\sqrt{100\cdot0.02\cdot0.98}}\Big]=1-\Phi(0)=0.5$$

Both results are not exactly good approximations of the exact result $0.6$ calculated with the binominal distribution. I strongly suspect it is due to bad boundary setting.

答:
It seems the task is to find $P(X > 2) = 1-P(X \le 2)$ when $X \sim \mathsf{Binom}(n = 100, p=0.02).$ So the computation reduces to finding $P(X \le 2) = 0.6767,$ to four places. Exact computation in R is as below: first using the R binomial CDF function pbinom; second, using R as a calculator to compute the necessary three terms of the binomial PDF formula.

```<code>pbinom(2, 100, .02)
[1] 0.6766856

sum(dbinom(0:2, 100, .02))
[1] 0.6766856

.98^100 + 100*.98^99*.02 + choose(100,2)*.98^98*.02^2
[1] 0.6766856
</code>```

For $n$ as large as $100$ and $p$ as small as $0.02$ one
can get a reasonable approximation to  $\mathsf{Binom}(100, 0.02)$
by using $\mathsf{Pois}(\lambda = 2).$ The approximation gives
$0.6767,$ to four places.

```<code>ppois(2,2)
[1] 0.6766764
exp(-2)*(2^0 + 2 +2^2/2)
[1] 0.6766764
</code>```

I do not believe a normal approximation is appropriate for
this problem. The n0rmal distribution with matching mean
and standard deviation is $Y \sim\mathsf{Norm}(\mu = 2, \sigma = 1.4).$ With a continuity correction we would have $P(Y \le 2.5) = 0.6395,$
which does not give even two place accuracy. Standardizing
and using printed normal CDF tables would introduce some
rounding error, even with some interpolation.

```<code>sqrt(100*.02*.98)
[1] 1.4
pnorm(2.5, 2, 1.4)
[1] 0.6395076

(2.5-2)/1.4
[1] 0.3571429
pnorm(.357)    # in table, interpolate btw .35 & .36 ...
[1] 0.6394541  # ... to get about .639 or .640, NOT .6767
</code>```

The following figure shows the exact binomial probabilities (bars),
approximate Poisson probabilities (centers of open circles), and the
approximating normal density curve. At the scale of this
graph it is difficult to see the slight differences
between exact binomial and approximate Poisson probabilities. The normal curve is a very poor fit.

```<code>k = 0:10;  PDF = dbinom(k, 100, .02)
PDF.pois = dpois(k, 2)
hdr = "BINOM(100, .02) with Poisson & Normal Aprx."
plot(k, PDF, type="h", xlim=c(-1.5,10), lwd=2,
   col="blue", main=hdr)
 points(k, PDF.pois, col="red")
 curve(dnorm(x, 2, 1.4), add=T, lwd=2)
 abline(h=0, col="green2")
</code>```

Body:
It seems the task is to find $P(X > 2) = 1-P(X \le 2)$ when $X \sim \mathsf{Binom}(n = 100, p=0.02).$ So the computation reduces to finding $P(X \le 2) = 0.6767,$ to four places. Exact computation in R is as below: first using the R binomial CDF function pbinom; second, using R as a calculator to compute the necessary three terms of the binomial PDF formula.

pbinom(2, 100, .02)
[1] 0.6766856

sum(dbinom(0:2, 100, .02))
[1] 0.6766856

.98^100 + 100*.98^99*.02 + choose(100,2)*.98^98*.02^2
[1] 0.6766856

For $n$ as large as $100$ and $p$ as small as $0.02$ one
can get a reasonable approximation to  $\mathsf{Binom}(100, 0.02)$
by using $\mathsf{Pois}(\lambda = 2).$ The approximation gives
$0.6767,$ to four places.

ppois(2,2)
[1] 0.6766764
exp(-2)*(2^0 + 2 +2^2/2)
[1] 0.6766764

I do not believe a normal approximation is appropriate for
this problem. The n0rmal distribution with matching mean
and standard deviation is $Y \sim\mathsf{Norm}(\mu = 2, \sigma = 1.4).$ With a continuity correction we would have $P(Y \le 2.5) = 0.6395,$
which does not give even two place accuracy. Standardizing
and using printed normal CDF tables would introduce some
rounding error, even with some interpolation.

sqrt(100*.02*.98)
[1] 1.4
pnorm(2.5, 2, 1.4)
[1] 0.6395076

(2.5-2)/1.4
[1] 0.3571429
pnorm(.357)    # in table, interpolate btw .35 & .36 ...
[1] 0.6394541  # ... to get about .639 or .640, NOT .6767

The following figure shows the exact binomial probabilities (bars),
approximate Poisson probabilities (centers of open circles), and the
approximating normal density curve. At the scale of this
graph it is difficult to see the slight differences
between exact binomial and approximate Poisson probabilities. The normal curve is a very poor fit.

k = 0:10;  PDF = dbinom(k, 100, .02)
PDF.pois = dpois(k, 2)
hdr = "BINOM(100, .02) with Poisson & Normal Aprx."
plot(k, PDF, type="h", xlim=c(-1.5,10), lwd=2,
   col="blue", main=hdr)
 points(k, PDF.pois, col="red")
 curve(dnorm(x, 2, 1.4), add=T, lwd=2)
 abline(h=0, col="green2")


问:
Proof Check: $4x^2+3x+17$ is continuous by using a $\epsilon$, $\delta$ Argument
Can anyone double check my following epsilon delta proof.

I want to prove that the following function is continuous with an Epsilon Delta Argument.
$$ f: x \in \mathbb R \mapsto (4x^2+3x+17) \in \mathbb R $$

So I started with
$$\left\lvert f(x)-f(y) \right\rvert =$$
$$= \left\lvert (4x^2+3x+17) - (4y^2+3y+17) \right\rvert$$
$$= \left\lvert (4x^2+3x)-(4y^2+3y) \right\rvert$$
$$= 3\left\lvert (4x^2+x)-(4y^2+y) \right\rvert$$
$$= 3\left\lvert (\frac {4}{3}x^2+x)-(\frac {4}{3}4y^2+y) \right\rvert $$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
(Triangel inequality)

$$\le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)$$

Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)$$

$$= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)$$

$\delta \le 1$
$$\le 3 \left(\frac {4}{3}\left\lvert x + y \right\rvert \delta + \delta \right)$$
Adding zero in form of y - y
$$= 3 \left(\frac {4}{3}\left\lvert x + y - y + y \right\rvert \delta + \delta \right)$$
$$= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)$$
Triangel inequality
$$\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
$$= 4\delta^2+4\left\lvert 2y \right\rvert \delta + 3\delta $$
$$= \delta(4\delta+4\left\lvert 2y \right\rvert + 3) $$
Note that $\delta \le 1$
$$\le \delta(4 +4\left\lvert 2y \right\rvert + 3) $$
$$\le \delta(\left\lvert 8y \right\rvert + 7) = \epsilon$$
Therefore $$\delta = \frac {\epsilon}{(\left\lvert 8y \right\rvert + 7)} \;\; with \; \delta \le1$$

Thanks in advance for the help, I really appreciate it. :)

答:
Your solution looks good. The following steps could be simplified a bit to make it more readable.

\begin{align}
\left\lvert f(x)-f(y) \right\rvert
&=
3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert
\quad (\textrm{why bother factoring out $3$?})\\
& \le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)\\
& \le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)
\quad (\color{red}{\textrm{for $|x-y|<\delta$} })\\
&= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)\\
&= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)\\
&\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)
\quad (\textrm{triangle inequality})\\
&\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)\\
&=4(\delta +\left\lvert 2y \right\rvert) \delta + 3\delta
\\
&\le
4(1+\left\lvert 2y \right\rvert) \delta + 3\delta
\quad (\delta\le 1)\\
&\le (7+|8y|)\delta
\end{align}

Set $$\delta =\min(\frac {\epsilon}{\left\lvert 8y \right\rvert + 7} ,1).$$


问:
$\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$ is not valid for non-monotone function.
$\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$ is not valid for non-monotone function. How to find such a counterexample?

As is well-known, for monotone function $f$, we have $\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$. then for non-monotone function, is there any simple example such that $\int_0^1 f(x)dx$ exists (or for any $\delta>0$, $f$ is integrable on $[\delta, 1]$, and $\int_0^1 f(x)dx=\lim_{\delta\to 0}\int_\delta^1 f(x)dx$ exists), but $\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$ is not valid.

答:
For a counterexample, take $f:[0,1] \to \mathbb{R}$ such that for $m = 0,1,2,\ldots$

$$f(x) = \begin{cases}2^m,&x = 2^{-m}\\0, &\text{otherwise} \end{cases}$$

Then

$$S_n=\sum_{k=1}^{2^n}f\left( \frac{k}{2^n}\right)\left(\frac{k}{2^n}- \frac{k-1}{2^n}\right) = 2^{-n}\sum_{k=1}^{2^n}f\left( \frac{k}{2^n}\right) = 2^{-n}\sum_{r=0}^{n}\frac{2^n}{2^r}= 2 - \frac{2}{2^{n+1}},$$

and

$$\lim_{n \to \infty}S_n = 2 \neq 0 = \int_0^1f(x) \, dx$$


答: Hint: The binomial theorem tells that that: $$(x+y)^n=\sum_{i=0}^n\left(\begin{array}{c} n \\ i \end{array}\right)x^iy^{n-i}$$where $\left(\begin{array}{c} n \\ i \end{array}\right)$ is your nCi. Maybe some clever choices for $x$, $y$, and $n$ will make your summation easier.</p>

Title: Evaluate a sum involving n choose r

Body:
Evaluate: $\sum_{k=0}^6 (-1)^k \binom{6}{k}$ where $\binom{n}{r}= \frac{n!}{r!(n-r)!}$.

I'm unsure how to compute the part with $\binom{6}{k}$, it should be something along the lines of $\frac{6!}{k!(6-k)!}$ can we simplify this down?

I think there was an expansion method for factorials I could used in the denominator that was somewhere along the lines of $(6-k)(5-k)(4-k)(3-k)(2-k)(1-k)$ ?


答:
Pick an element $a\in S$. Then we have an obvious bijection between $$\{\text{equivalence relations on }S\text{ with exactly two equivalence classes}\}$$ and $$\{\text{proper subsets of }S\setminus\{a\}\,\}$$
given by
$$ {\sim}\mapsto \{\,x\in S\setminus\{a\}\mid x\sim a\,\}$$

Title: How many different equivalence relations with exactly two different equivalence classes are there on a set with $n$ elements

Body:
I came across with this topic. It looks straight forward for $5$ elements, but what if I want to find how many different equivalence relations with exactly two different equivalence classes are there on a set with $n$ elements?

It says that the set of equivalence relations on a set are in direct bijection with the set of partitions on a set. This made me read about Bell numbers but I'm not sure how help us here. How to solve this problem?


问:
Time period of a nonuniform oscillator
I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the variables for $\eqref{eq1}$ to get
$$\frac{1}{\omega-a\sin(\theta)}d\theta=dt.$$
I know that for the time to increase from $0$ to $T$, then the oscillator has gone from $0$ to $2\pi$. Therefore
$$\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta=\int_0^Tdt$$
and so
$$T=\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta.$$
This is where the fun begins. I use the Weierstrass substitution of $u=\tan\left(\frac{\theta}{2}\right)$ to evaluate this which changes my integral to
$$T=\int_{\theta=0}^{\theta=2\pi}\frac{2}{\omega u^2+\omega-2ua}du.$$
I take the factor of $2$ outside of the integral and complete the square of the quadratic in the denominator to arrive at
$$T=2\int_{\theta=0}^{\theta=2\pi}\frac{1}{\omega\left(\left(u-\frac{a}{\omega}\right)^2+1-\frac{a^2}{\omega^2}\right)}du.$$
I factor out $\frac{1}{\omega}$ from the integral, and use the substitutions $x=u-\frac{a}{\omega}$ and $y=1-\frac{a^2}{\omega^2}$ to make this
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{1}{x^2+y}dx.$$
I then use the substitution $x=\sqrt{y}\tan(v)$, meaning $x^2=y\tan^2(v)$ and $dx=\sqrt{y}\sec^2(v)dv$. This makes my integral
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{\sqrt{y}\sec^2(v)}{y\tan^2(v)+y}dv.$$
I take out a factor of $\frac{\sqrt{y}}{y}$ and find the following:
$$T=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\tan^2(v)+1}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\sec^2(v)}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}1 dv=\frac{2\sqrt{y}}{\omega y}\left[v\right]^{\theta=2\pi}_{\theta=0}.$$
Then I start the process of re-inserting my many substitutions. First $v=\tan^{-1}\left(\frac{x}{\sqrt{y}}\right)$, then $x=u-\frac{a}{\omega}$, then $u=\tan\left(\frac{\theta}{2}\right)$, and finally $y=1-\frac{a^2}{\omega^"}$. This gets me
$$T=\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}\left[\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)\right]^{2\pi}_{0}.$$
I can simplify this slightly. Firstly,
$$\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}=\frac{2}{\omega\sqrt{1-\frac{a^2}{\omega^2}}}=\frac{2}{\omega\sqrt{\frac{\omega^2-a^2}{\omega^2}}}=\frac{2}{\sqrt{\omega^2-a^2}},$$
and secondly
$$\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)=\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right).$$
So now I have
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$
I feel like I'm so very close to $\eqref{eq2}$ now, but this is where I become stuck. I know that $\tan\left(\frac{2\pi}{2}\right)=\tan\left(\frac{0}{2}\right)=0$, which means when I evaluate the square brackets between $0$ and $2\pi$ they just cancel out and I'm left with
$$T=0.$$

Could someone please tell me where I have gone wrong, or what I am missing here? Been stuck on this one all day to no avail!

答:
Taking on board the comment of @Hans Ludmark, I believe I have found the answer but I'm not entirely sure if it is rigorous. Considering the discontinuity at $\theta=\pi$ I now re-examine the equation

$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$

I have split the brackets to be evaluated into two. I now write this as
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{\pi}_{0}+\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{\pi}\right),$$
where the bracket on the left is to the left of $\theta=\pi$ and the bracket on the right is to the right of $\theta=\pi$.
I then write this as
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^--\tan^{-1}\left(\frac{\omega\tan\left(\frac{0}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)+\tan^{-1}\left(\frac{\omega\tan\left(\frac{2\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)-\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^+\right)$$
where a '$-$' superscript denotes the term on the left of $\pi$ and a '$+$' superscript denotes the term on the right of $\pi$. The terms $$-\tan^{-1}\left(\frac{\omega\tan\left(\frac{0}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)$$ and $$+\tan^{-1}\left(\frac{\omega\tan\left(\frac{2\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)$$ cancel since $\tan\left(\frac{0}{2}\right)=\tan\left(\frac{2\pi}{2}\right)=0$ This leaves
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^--\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^+\right)$$
For the term on the left of $\pi$, we can take the limit to find
$$\lim\limits_{x\to\frac{\pi}{2}^-}\tan(x)=\infty,$$
and for the term on the right of $\pi$ we take the limit to find
$$\lim\limits_{x\to\frac{\pi}{2}^+}\tan(x)=-\infty.$$
This reduces our equation to
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\tan^{-1}(\infty)-\tan^{-1}(-\infty)\right).$$
Taking limits again we find that
$$\lim\limits_{x\to\infty}\tan^{-1}(x)=\frac{\pi}{2}$$
and
$$\lim\limits_{x\to-\infty}\tan^{-1}(x)=-\frac{\pi}{2}.$$
Hence:
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\frac{\pi}{2}--\frac{\pi}{2}\right)=\frac{2\pi}{\sqrt{\omega^2-a^2}}.$$

Could someone let me know if this is a sound method?

Title: Time period of a nonuniform oscillator


问:
Another form of the ratio test
Let $a_k≠0, \forall k \in \mathbb{N}$. If $$\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_k}\right|<1$$ then the series $\sum_{k=1}^\infty a_k$ is absolutely convergent.

I am really interested into seeing a proof of this but I cannot find one. For the original ratio test, there are many proofs but this one is different. Has anyone an idea? Thanks.

答:
To follow leoli1's comment, you can do it as follows.

By the assumption and directly applying the ratio test you can show that the following two series are convergent:
$$
\sum_{k=1}^\infty |a_{2k}|,\quad \sum_{k=1}^\infty |a_{2k+1}|
$$

Now the partial sum of the original (absolute) series has the following estimate by the triangle inequality
$$
\sum_{k=1}^n|a_k|
\le
\sum_{k=1\\k\textrm{ odd}}^n|a_k|
+\sum_{k=1\\k\textrm{ even}}^n|a_k|\tag{1}
$$
Taking $n\to\infty$, we are done.

Notes. One actually has equality in (1) and thus the triangle inequality is not needed. (Thanks to hardmath's comment below.)

Title: Another form of the ratio test


问:
How to calculate $\int_{0}^{\infty}y^{2}e^{-y(x+1)}dy$?
I want to evaluate the following integral in a question I had in Probability:
$$
\int_{0}^{\infty}y^{2}e^{-y(x+1)}dy
$$
But I'm stuck figuring out how to solve this integral. How to calculate this integral?

Body:
Let $\alpha=(x+1)y$, then
$$\int_0^{\infty}y^2 e^{-(x+1)y}dy = \dfrac{1}{(x+1)^3}\int_0^{+\infty}\alpha^2e^{-\alpha}d\alpha=\dfrac{2}{(x+1)^3}$$


问:
When do we have $\log(\prod^{\infty}_{i=0}a_i)=\sum^{\infty}_{i=0}\log(a_i)$?
This shows up in convergence of the Euler product and I don't know in what condition this holds. Here $a_i$ is a complex sequence. Is this an equivalent condition?

答:
This is straightforwardly true if the $a_i$ are all positive reals; just argue by induction that

$$\log (\prod_{i=0}^n a_i) = \sum_{i=0}^n \log a_i$$

and then take the limit as $n \to \infty$ (and use the continuity of $\log$).

If the $a_i$ are complex then both the LHS and the RHS are ambiguous and one must specify a branch of the logarithm. In that context it's true if one takes the principal branch of the logarithm and the sum of the arguments of the $a_i$ remains bounded in absolute value by $\pi$. In general if you're not careful about branches then the LHS and the RHS may differ by an integer multiple of $2 \pi i$, so you just need to set things up to make sure this doesn't happen.


问:
What does a "half derivative" mean?
I was looking at fractional calculus on Wikipedia, specifically this section and came across the half derivative of the function $y=x$ which is $\frac{d^{1/2}y}{dx^{1/2}}=\frac{2\sqrt{x}}{\sqrt{\pi}}$ . The derivative tells the slope at any point on the curve, but what does the "half derivative" mean - it's obviously not $\frac{1}{2}$ the derivative of $y=x$ which would be just $\frac{1}{2}$.

I do not have a very deep understanding of calculus - I have just taken Calc 1 & 2 out of a 4 series, but anything helps!

Also, I have checked similar questions, but they did not seem to answer my question that I have bolded.

答:
Often with new mathematical objects, it's not that the object has no meaning, but that we're asking the wrong question, or from the wrong context.  It makes sense to double my apples, but it makes no sense to multiply my apples by i.  That's asking the wrong question.  One way of understanding multiplication by a complex number is that it scales for real values and rotates for imaginary values, so if we want a physical example that makes sense for multiplying imaginary numbers, allowing for rotation in the context would enable us to use this fact.  Numbers of apples doesn't work, but perhaps modifying the velocity vector of my car would.  We aren't accustomed to saying, "Double our velocity" or "i-ble" our velocity (turn 90 degrees), but it would make some sense, if we're asking the right question.  We definitely could talk about multiplying by imaginary numbers whenever we rotate ourselves, even though we commonly don't.

What question, what context, could make sense for a half derivative?  We have one famous context that I know of, namely the Riemann Hypothesis and the functions that surround it.  There are probably other contexts.

In the Riemann Hypothesis, we have essentially a Fourier analysis of prime numbers, and the F(s) frequency space function is related somehow to the Riemann zeta function.  (Question for the audience for my benefit: IS the frequency space function equivalent to the Re(z)=1/2 slice?).  Hopefully my ignorance won't hinder the discussion here too much.

The Riemann Zeta function is an interesting one, which for $Re(z) > 1$, can be simply defined like this:

$$
\zeta(z) = \sum_{k=1}^\infty \frac{1}{k^z}
$$

Since this combines both exponentiation and addition, this makes things tricky for moving around the graph of $\zeta(z)$.  How would we consider shifting by $z+2$, for example?  For integer amounts of shift, we might be able to rely on binomial expansion or something similar, but the number of terms becomes unwieldy quickly.

Differentiation and integration of a polynomial provides a convenient alternative, since they are linear operators, and modify the exponents.  If we could differentiate $\zeta(z)$ in respect to k, the exponent would change, providing every term with a multiplicative factor but we can deal with that.

Confirmation of differentiation being somehow related to the values of $\zeta(z)$ can be found in the equations that link the Riemann zeta function and the polygamma function $\psi_n(z)$, where different values of $z$ for the Riemann zeta function correspond to different amounts of derivative in the polygamma function:

$$
\psi_n(1) = (-1)^{n+1} n! \zeta(n+1)
$$

where $\psi_n(z)$ is the nth derivative of the digamma function.  But for fractional derivatives, and using the gamma function instead of the factorial, this can be made valid for non-integer n.

But the points in question for the Riemann zeta function are well defined for Re(z) for integer $z$, whereas the points in question are all on the Re(z) = 1/2 line.  We could probably use half derivatives of $\psi_n(z)$ to get there, and mayble we could use imaginary derivatives to traverse imaginary-wise along the critical line.  I would wager that this impetus is what led Riemann and Cauchy to developing the beginnings of fractional calculus in the first place.

Other contexts might deal with exponential functions, which are eigenfunctions for derivatives and calculus.  Repeated differentiation will lead to different eigenvalues.  If the eigenvalues can be real valued instead of just integer valued, and if those eigenvalues have physical meaning, then fractional calculus will also find meaning there.

Title: What does a "half derivative" mean?


问:
Understanding the derivation of Euler's Basel formula.
I was looking at the derivation of Euler's Basel formula and I had a question.

Since the zeroes of $\sin(x)$ are $0, ±π, ±2π,...$ , we can factorise $\sin (x)$ :

$$\sin(x) = ax\cdot(x-π)\cdot(x+π)\cdot(x-2π)\cdot(x+2π)\cdots$$

and the place where I found this equation gives the next simplification as :
$$\sin(x) = x\cdot(1-x/π)\cdot(1+x/π)\cdot(1-x/2π)\cdot(1+x/2π)\cdots$$
Since the limit as $x$ tends to $0$ of $\sin (x)/x = 1$

In the above equation shouldn't it be $x/π - 1$ and so on for the other such expressions and not the other way around and also how do you arrive at that expression? Is it by dividing each term of the factor with constant in it like $π, 2π$? but would not that change the value of the expression?

答:
I have a bunch of links which should clarify things (mainly the first one):

How was Euler able to create an infinite product for sinc by using its roots?

https://en.wikipedia.org/wiki/Basel_problem

How do we know that the Sine function has no Non-Real Roots?


答:
The condition
$$\limsup_{k\to\infty}\left\vert\frac{a_{k+2}}{a_k}\right\vert<1$$ implies that it exists $N \ge 1$ and $0 \le K \lt 1$ such that

$$\left\vert\frac{a_{k+2}}{a_k}\right\vert \le K$$ for all $k \ge N$. Therefore

$$\left\vert\frac{a_{N+2p}}{a_N}\right\vert \le K^p, \left\vert\frac{a_{N+2p+1}}{a_{N+1}}\right\vert \le K^p$$ for all $p \ge 0$ and

$$\begin{aligned}\sum_{k \ge N} \vert a_k \vert &= \sum_{k \ge N\\k\textrm{ odd}} \vert a_k \vert + \sum_{k \ge N\\k\textrm{ even}} \vert a_k \vert\\
&\le \vert a_{N+1} \vert\sum_{k \ge 0} K^{p} + \le \vert a_{N} \vert\sum_{k \ge 0} K^{p}\\
&\le \frac{2}{1-K} \max(\vert a_N \vert, \vert a_{N+1} \vert)
\end{aligned}$$

Proving that $\sum a_k$ converges absolutely.

Title: Another form of the ratio test

Body:
Let $a_k≠0, \forall k \in \mathbb{N}$. If $$\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_k}\right|<1$$ then the series $\sum_{k=1}^\infty a_k$ is absolutely convergent.

I am really interested into seeing a proof of this but I cannot find one. For the original ratio test, there are many proofs but this one is different. Has anyone an idea? Thanks.


问:
Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation?

Let $T : \Bbb{R^2} \to \Bbb{R^2}$ and $T(x,y) = (x+y, 2(x+y))$. What is the rank and nullity of this transformation?

Rank is defined to be $\dim(\text{im}(T))$ right? Every output of $T$ is a vector belonging to $\Bbb{R^2}$ and $\Bbb{R^2}$ is the image of $T$ so I assumed that $\text{rank}(T) = 2$. Turns out it's actually $1$, which I don't seem to understand... Every output of $T$ can be expressed as $c(1,2)$ but this is also just a tuple in $\Bbb{R^2}$ so how can the rank be $1$?

The nullity is defined to be the set of vectors for which $T(x) = 0$ so I have that $$(x+y,2(x+y)) = 0 \Longrightarrow x+y=0, 2x+2y=0 \Longrightarrow (x,y) = (0,0)$$

so the set of vectors for which $T(x) = 0$ is just $\{(0,0)\}$ so $\ker(T) = 1$?

答:
When calculating the rank of a linear transformation you're interested in the dimension of its image. Then nullity is the dimension of the kernel, that is all the vectors that are mapped to the zero vector by the transformation. The rank plus the nullity will be the dimension of the domain.

Lets consider a simple example first, the zero transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y)=(0,0)$. Here all the vectors are transformed into the zero vector, which is a zero dimensional space. Another simple example is the identity $T(x,y)=(x,y)$ and the image is clearly the entire plane, so it has rank $2$ and nullity $0$.

Now lets look at one that is almost your example with $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y)=(x+y,x+y)$. If we set $a=x+y$ we see that these points are all of the form $(a,a)$, which is diagonal in the plane. This line is one dimensional and so the image of the transformation is one dimensional, giving it rank $1$. Since we know that the dimension of the image and kernel must sum to $2$ this also tells us that the nullity is $1$. See if you can find the line through the origin that is the kernel of this transformation.

Finally your transformation is $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $T(x,y)=(x+y,2(x+y))$. Again setting $a= x+y$ we now have points of the form $(a,2a)$ which is again a line and so the image has rank $1$ and we calculate that the nullity is $1$ as well. Using some trial and error see if you can find the kernel, which is another line in the plane. You can test them by ensuring $T(x,y)=0$.

Now in all these cases you could set up the matrix and do row reduction as well to find the dimension. The number of non-zero pivots is the dimension of the image. All matrices has four fundamental subspaces, the two we discussed being the image and kernel, and the same two of its transpose.


问:
General solutions of $ty''(t) -(t+1)y’(t) + y(t) = t^2$ given that $y_1(t)=e^t$ and $y_2(t)=t+1$
So I know the answer is $y(t)= c_1e^t + c_2(t+1) -t^2 -2t -2$
(According to answer attached to the question)
But what don’t understand is that I thought $-2t$ and $-2 $ shouldn’t be there because it’s same power as $c_2t$ and $c_2$.
I don’t really understand the theory behind this as my professor only taught the process.

答:
Since that the ordinary differential equation is of order $2$, so the general solution for your problem is of the form $$\color{blue}{\boxed{\operatorname{solution-general}=\operatorname{homogeneous-solution}+\operatorname{particular-solution}}}$$ I suppose that the homogenous solution for your problem is $y_{1}(t)$ and $y_{2}(t)$,  and the particular solution you can find using $y_{1}(t)$ or $y_{2}(t)$ and the Abel's formula.  Asuming that the solution particular is $$y_{p}(t)=-t^{2}-2t-2$$
and that the homogenous solution is $$y_{h}(t)=c_{1}e^{t}+c_{2}(t+1), \quad c_{1},c_{2}\in \mathbb{R}$$
So, the general solution is $$y(t)=c_{1}e^{t}+c_{2}(t+1)-t^{2}-2t-2$$

Now, about your question, if you take $c_{1}=1$ and $c_{2}=-2$, so you have one solution for fixed values of the constants $$y(t)=e^{t}-2t-1-t^{2}-2t-1=e^{t}-t^{2}-4t-2$$
for that same reason you can fixed $c_{1}=0,1,2,3,\frac{1}{2},\ldots$ and $c_{2}=1,2,3,\frac{1}{2},\ldots$ and you can obtain more solutions  for  your problem.

It's important to see that $y(t)=c_{1}e^{t}+c_{2}(t+1)-t^{2}-2t-2$ contain a families of solutions for the ODE. For example here in Wolfram Mathematica you can see a nice simulating about this topic. Families of Solutions for ODEs.


问:
Solving the equation $\dfrac{(1+x)^{36} -1}{x} =20142.9/420$ for $x$.
How would one solve for x in the following equation:

$\dfrac{(1+x)^{36} -1}{x} =20142.9/420$

I tried factorising the top but that didnt really help much.

$((1+x)^{18} - 1)((1+x)^{18}+1)$

Any help is appreciated thanks.

答:
Make a change in variables: $y = x + 1$. Then the equation becomes:
$$\dfrac{y^{36} - 1}{y-1} = 20142.9/420$$
The numerator $y^{36}-1$ is equal to $(y-1)(y^{35}+y^{34}+\cdots+1)$ (use the formula for the sum of a geometric series). Therefore dividing by $y-1$ gives:

$$y^{35}+y^{34}+\cdots+1 = 20142.9/420$$

This is a polynomial of degree 35 in $y$. It happens to be that for polynomials of degree $5$ or higher there is no general formula to solve for $y$ (1). You can find an approximate solution using a tool like Wolfram Alpha. Try inputting "(y^(36) - 1)/(y-1) = 20142.9/420" and see what answers you get for $y$ (there may be more than one). Then use $x = y - 1$ to get approximate answers for $x$.


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
As lulu mentioned in a comment, the fact that a uniformly random permutation $\pi\colon\{1,2,\dots,n\}\to\{1,2,\dots,n\}$ has in expectation one fixed point is a quite surprising statement, with a one-line proof.

Let $X$ be the number of fixed points of such a uniformly random $\pi$. Then $X=\sum_{k=1}^n \mathbf{1}_{\pi(k)=k}$, and thus
$$
\mathbb{E}[X] = \mathbb{E}\left[\sum_{k=1}^n \mathbf{1}_{\pi(k)=k}\right]
= \sum_{k=1}^n \mathbb{E}[\mathbf{1}_{\pi(k)=k}]
= \sum_{k=1}^n \mathbb{P}\{\pi(k)=k\}
= \sum_{k=1}^n \frac{1}{n}
= 1\,.
$$


答: Note that since $$\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots,$$ Euler concluded that the product of the roots of $\frac{\sin x}{x}$ is $1.$ Therefore $a$ in your expansion should be equal to $$\frac{1}{\prod_{k} (kπ)^2}.$$

Title: Understanding the derivation of Euler's Basel formula.

Body:
I was looking at the derivation of Euler's Basel formula and I had a question.

Since the zeroes of $\sin(x)$ are $0, ±π, ±2π,...$ , we can factorise $\sin (x)$ :

$$\sin(x) = ax\cdot(x-π)\cdot(x+π)\cdot(x-2π)\cdot(x+2π)\cdots$$

and the place where I found this equation gives the next simplification as :
$$\sin(x) = x\cdot(1-x/π)\cdot(1+x/π)\cdot(1-x/2π)\cdot(1+x/2π)\cdots$$
Since the limit as $x$ tends to $0$ of $\sin (x)/x = 1$

In the above equation shouldn't it be $x/π - 1$ and so on for the other such expressions and not the other way around and also how do you arrive at that expression? Is it by dividing each term of the factor with constant in it like $π, 2π$? but would not that change the value of the expression?


问:
Adding the number of unique decision variables in the objective function.
In a linear program formulation, my objective function requires to add the number of unique decision variable it took. (in my original problem, unique decision variable may reduce cost).

Let's say I have 3 decision variables $x_1, x_2, x_3$. They are all integers and between 1 to 5.

The objective function will be:

$$
min\ ax_1+bx_2+cx_3 + D
$$

Here I need this term $D$ to add the number of unique decision variable. For example, if we are checking $3,5,4$ as $x_1, x_2, x_3$ then $D = 3$. However for $3,4,4$, $D$ will be 2.

Is there any way I can incorporate this in my objective function?

答:
For $i\in\{1,2,3\}$ and $j\in\{1,\dots,5\}$, let binary variable $y_{i,j}$ indicate whether $x_i=j$.
For $j \in \{1,\dots,5\}$, let binary variable $z_j$ indicate whether $x_i=j$ for some $i$.  The problem is to minimize $$a x_1 + b x_2 + c x_3 + \sum_{j=1}^5 z_j$$ subject to linear constraints
\begin{align}
\sum_j y_{i,j} &= 1 &&\text{for all $i$} \\
\sum_j j y_{i,j} &= x_i &&\text{for all $i$} \\
y_{i,j} &\le z_j &&\text{for all $i$ and $j$}
\end{align}


问:
Solving recurrence relations using generating functions
I have to solve given recurrence relation: $ a_{n+2} = 2*a_{n+1}^2 * a_{n}^3$, where: $ a_{1} = a_{0} = 1$

I have no idea how to do it honestly, I have calculated that first values of $a_{n}$ are (1, 1, 2, 8, 1024...). $a_{5}$ is so big that I will not put it down here.

I've tried using generating functions to solve it but with no success. I don't know how to split given sequence in order to receive couple of generating functions so that I can solve them. Maybe we have to use auxiliary sequence for that?

Thanks in advance! Any help is appreciated.

答:
First transform your recurrence: Let $b_n = \log a_n$, and:

$\begin{align*}
\log a_{n + 2}
  &= \log 2 + 2 \log a_{n + 1} + 3 \log a_n \\
b_{n + 2}
  &= 2 b_{n + 1} + 3 b_n + \log 2
\end{align*}$

Now use generating functions. Define $B(z) = \sum_{n \ge 0} b_n z^n$, multiply by $z^n$ and sum over $n \ge 0$, recognize the resulting sums:

$\begin{align*}
  \sum_{n \ge 0} b_{n + 2} z^n
    &= 2 \sum_{n \ge 0} b_{n + 1} z^n
         + 3 \sum_{n \ge 0} b_n z^n
         + \log 2 \sum_{n \ge 0} z^n \\
  \frac{B(z) - b_0 - b_1 z}{z^2}
     &= 2 \frac{B(z) - b_0}{z}
          + 3 B(z) + \frac{\log 2}{1 - z} \\
\end{align*}$

Use the initial values $b_0 = b_1 = 0$, solve for $B(z)$ and express as partial fractions:

$\begin{align*}
  B(z)
    &= \frac{z^2}{1 - 3 z - z^2 + 3 z^2} \log 2 \\
    &=  \frac{\log 2}{8 (1 - 3 z)}
         + \frac{\log 2}{8 (1 + z)}
         - \frac{\log 2}{4 (1 - z)}
\end{align*}$

Everything in sight is just geometric series:

$\begin{align*}
   b_n
     &= \frac{\log 2}{8} \cdot 3^n
           + \frac{\log 2}{8} \cdot (-1)^n
           - \frac{\log 2}{4} \\
   a_n
     &= 2^{3^n / 8 + (-1)^n / 8 - 1 / 4}
\end{align*}$


问:
Would $ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$ be equal to zero?
Suppose a limit of the form

$$ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$$

where $f(x)$ is an integrable function.

I think that this limit would be zero, since we would be taking an integral at a single point, even if the argument was divergent. Is this correct?

答: With $x=y\sigma$ it becomes $\lim_{\sigma\to0^+}\int_{-1}^1f(y\sigma)dy$, which if we can use dominated convergence is$$\int_{-1}^1\lim_{\sigma\to0^+}f(y\sigma)dy=\int_{-1}^1f(0)dy=2f(0)$$for $f$ continuous at $0$.


问:
Simplicial Complex and Euler characteristic
I find myself confused with this problem:

"For each integer $n$ find a simplicial complex with Euler's characteristic $n$".

In the case of positive integers, the simplicial complex would be $n$ linearly independent vertices. Is this answer possible?
And, for negative integers, I have been doing several cases where adding in an extra $0$-simplex and two $1$-simplices, I got negative integers as Euler's characteristic for example

$K_2 = \{[v0, v1], [v1, v2], [v0, v2], [v1, v3], [v2, v3], [v0], [v1], [v2], [v3] \}$

$K_3 = \{[v0, v1], [v1, v2], [v0, v2], [v1, v3], [v2, v3], [v1, v4], [v3, v4], [v0] , [v1], [v2], [v3], [v4]\}$

the $\chi(K_2) = -1$ and $\chi(K_3) = -2$
But I don't know if this is the way to solve the problem, or if it is only through a formula, I need help. Thank you

答:
Once you have simplicial complexes with $\chi = -1,0,1$, you could in principle 'cheat' and take their disjoint unions, since simplicial complexes need not be connected. A point has $\chi = 1$, a circle has $\chi = 0$, and you seem to have found an example with $\chi = -1$ yourself, though I haven't checked what complex it is, though alternatively a figure-eight should work.

If your simplicial complexes have to be connected, you may instead wish to invoke wedge sums. Using the fact that $\chi(X \vee Y) = \chi(X) + \chi(Y) - 1$, you can get connected simplicial complexes with arbitrary Euler characteristic. For instance, the wedge of $n$ circles should have $\chi = 1 - n$, and if $X$ is a connected simplicial complex, then $\chi(X \vee S^2) = \chi(X) + 1$.


问:
Are commutative squares that induce iso on cokernels cocartesian?
Does the converse of this question hold true? Explicitly, is a commutative square
$$\begin{matrix}
A & \stackrel a \to & X\\\llap{\scriptstyle a'}\downarrow&&\downarrow\rlap{\scriptstyle b'}\\Y & \stackrel b \to &  B
\end{matrix}$$
cocartesian if the induced maps $\operatorname{coker}a \to \operatorname{coker}b$ and $\operatorname{coker}a' \to\operatorname{coker}b'$ are isomorphisms?

Actually, I see no reason why this should be true, but maybe someone has a nice counter example and/or can elaborate on the problem.

答:
Here is an example: in the diagram below, the top and left arrows send $1 \mapsto 1$; then all cokernels are $0$, but the pushout $\mathbb{Z}/2 \sqcup_{\mathbb{Z}} \mathbb{Z}/2$ is $\mathbb{Z}/2$.
$\require{AMScd}$
\begin{CD}
\mathbb{Z} @>>> \mathbb{Z}/2 \\
@VVV @VVV\\
\mathbb{Z}/2 @>>> 0
\end{CD}


问:
Formula for quadratic equation
Consider the following expression

$$x^T (A+B)^{-1}x,\tag{1}$$

where $x$ is a column vector, and both $A$ and $B$ are invertible ($A$ is diagonal if this helps).

Is there some formula such that $(1)$ is reformulated into

$$f(A) + g(B),$$

where $f$ and $g$ are some functions?

The motivation is to decompose $(1)$ into the addition of two parts, where one part depends only on $A$ (and $x$), while the other term depends only on $B$ (and $x$).

答:
No. If such $f(A,x)$ and $g(B,x)$ exist, then $x^T(A+B_1)^{-1}x-x^T(A+B_2)^{-1}x$ does not depend on $A$, because it is equal to $g(B_1,x)-g(B_2,x)$. However, this is not the case in reality. E.g. when $x$ is a unit vector,
\begin{aligned}
&x^T(I+0)^{-1}x-x^T(I+I)^{-1}x=\frac12,\\
\ne\ &x^T(2I+0)^{-1}x-x^T(2I+I)^{-1}x=\frac16.\\
\end{aligned}

Title: Formula for quadratic equation


问:
Find if the function is surjective $f(x) = (x^2+1)/(x^2 + 2)$
I found that the following function from $f:\Bbb R \to\Bbb R$ is not one to one:

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

I found answer online that it's not surjective as the range is not equal to the codomain, where the range is $\left\{y\mid\frac12 \le y \lt 1 \right \} = \left [\frac12, 1\right)$. Can you please explain how we can get this range?

答: Since $f'(x)=\frac{2x}{(x^2+2)^2}$, $f$ is decreasing on $(-\infty,0]$ and increasing on $[0,\infty)$. Besides, $\lim_{x\to\pm\infty}f(x)=1$, and $f(0)=\frac12$. So, indeed, $f(\Bbb R)=\left[\frac12,1\right)$.


答:
Write $f(x)$ as
$$f(x)=1-\frac1{x^2+2}.$$
From this we see that the supremum of $f(x)$ is $1,$ though this is never achieved.

Further, the function is even, so we may only consider positive values. We note that $0\le x<y$ implies $\frac1{x^2+2}>\frac1{y^2+2},$ so $f(x)<f(y),$ so $f(x)$ is increasing over $\mathbb R_{\ge0}.$ It will achieve its minimum at $f(0)=\frac12.$

Title: Find if the function is surjective $f(x) = (x^2+1)/(x^2 + 2)$

Body:
I found that the following function from $f:\Bbb R \to\Bbb R$ is not one to one:

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

I found answer online that it's not surjective as the range is not equal to the codomain, where the range is $\left\{y\mid\frac12 \le y \lt 1 \right \} = \left [\frac12, 1\right)$. Can you please explain how we can get this range?


问:
Proof of Egoroff's Theorem
Let $\{f_n \}$ be a sequence of measurable functions, $f_n \to f$ $\mu$-a.e. on a measurable set $E$, $\mu(E) < \infty$. Let $\epsilon>0$ be given. Then $\forall \space n \in \mathbb{N} \space \exists A_n \subset E$ with $\mu(A_n) <\frac{\epsilon}{2^n}$ and $\exists N_n$ such that $\forall \space x \notin A_n$ and $k \ge N_n \space |f_k(x) - f(x)| < \epsilon$. That is: if we define $A = \cup_{n=1}^{\infty} A_n$ with $\mu(A) < \epsilon $ then ${f_n}$ converges uniformly on $E \setminus A$.

$\mathbf{Proof}$: (taken from Royden's Real Analysis)

Let $A = \cup_{n=1}^{\infty} A_n \Rightarrow A \subset E$ and $\mu(A) < \sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon. \mathbf{Q1}$.

choose $n_0$ such that $\frac{1}{n_0} < \epsilon$. If $x \notin A$ and $k \ge N_{n_0}$ we have $\space |f_k(x) - f(x)| < \frac{1}{n_0} < \epsilon \space$. $ \square$

$\mathbf{Q1}$: First off, I don't see how $\sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon$. It's a geometric series: $ \epsilon \sum_{n=1}^{\infty} \frac{1}{2^n} = \epsilon \frac{1}{1-\frac12} = 2 \epsilon$. Am I wrong?

$\mathbf{Q2}$: The idea behind Egoroff is in order to turn almost sure convergence into uniform convergence on $E$ we only need to take away a really small set, right? Interestingly, as $\epsilon \to 0$ (that is $f_n$ is getting closer to $f$), the measure of the set $A$ is getting proportionally smaller ($\mu(A) \to 0$). So are we ultimately taking away a set of zero measure?

答:
A1: we begin the sum at $1$, so the result is $\varepsilon$ (even if we had got $2\varepsilon$ it wouldn't matter).

A2: the problem comes from the dependence of the set $A$ and $\varepsilon$. Things are more concrete when we think about $X=[0,1]$ and $f_n(x)=x^n$.

Title: Proof of Egoroff's Theorem


问:
If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series, then is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?
If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series, then is the sum of the average of consecutive terms necessarily convergent?

In other words, is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?

For the most standard example, the alternating harmonic series $\sum\limits_{n=0}^\infty {(-1)^n\over n+1},$ we get that $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ is half the sum of the reciprocal triangular numbers, which does converge.

But does every conditionally convergent (but not necessarily alternating sign) series have this property, or is there one that diverges?

I feel like I'm missing some obvious triangle inequality trick, but I just don't see it.

答:
No. Take your example, and interweave it with the all-zero sequence:
$$
a_{n} = \begin{cases}
\frac{(-1)^{n/2}}{n+1} &\text{ if } n \text{ even}\\
0 &\text{ if } n \text{ odd}
\end{cases}
$$


问:
What is a more rigorous method of solving modulus function inequalities?
I have the inequality

$$\vert x-3\vert < 3x-4$$

To solve this I first break the modulus containing function into the two cases using the critical value

$$\vert x-3 \vert  \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.  $$

For the case where $x>3$:

$$x-3<3x-4$$
$$x>\frac{1}{2}$$

For the case where $x<3$:

$$3-x<3x-4$$
$$x>\frac{7}{4}$$
However it is apparent that these two statements are contradictory and by simply looking at the graph of the two functions it is clear that the value $x>\frac{1}{2}$ does not hold.

The value does however seem to hold for the function $f(x)=x-3$

My question is this, why has my method given a false answer and what improvements can be made to this method to make it such that these false answers no longer appear?

答: Note that the inequality is true whenever $3x-4\le 0.$ Therefore assume $3x-4>0$ and square both sides to obtain $$(x-3)^2>(3x-4)^2,$$ or $$(x+3)^2-(3x-4)^2>0,$$ which gives $$(x+3-3x+4)(x+3+3x-4)>0,$$ simplifying to $(-2x+7)(4x-1)>0,$ or $$(x-3.5)(x-0.25)<0.$$ This implies $0.25<x<3.5$ for $x>4/3,$ or in short $$4/3<x<3.5,$$ so that taking the union of $x\le 4/3$ and $4/3<x<3.5$ gives the result $$(-\infty, 3.5).$$


问:
Whats the relation between the Veronese embedding and the isomorphism $\operatorname{Proj}A\cong\operatorname{Proj}A^{(d)}$?
In the book Introduction to Schemes, G. Ellingsrud and J. Ottem show (in section 6.7) that if $A$ is a graded ring and
$$A^{(d)}:=\bigoplus_{n\geq 0} A_{nd}$$
then the natural inclusion $A^{(d)}\to A$ induces an isomorphism $\operatorname{Proj}A\cong\operatorname{Proj}A^{(d)}$. They call it the Veronese embedding. While I understand what they did, I don't understand why they called it this way. Isn't the Veronese embedding the inclusion $\operatorname{Proj} k[x_0,\dotsc,x_n]_d \to \mathbb{P}_k^{N-1}$, where $N=\binom{n+d}{d}$?

答:
They are the same thing. Let $A_*$ be a finitely generated graded ring. Like you say, you can consider the subring $A_{d*}$, something called the $d$-th Veronese subring. There is a natural map $\operatorname{Proj} A_* \to \operatorname{Proj} A_{d*}$ which is an isomorphism of schemes.

Here's how this then leads to the Veronese embedding: Take $A_*$ to be the graded ring $k[x_0,\ldots,x_n]$. Compose the isomorphism $\operatorname{Proj} A_* \to \operatorname{Proj} _{d*}$ with the natural embedding of $\operatorname{Proj} A_{d*}$ into $\mathbb{P}^{N-1}$, where $N$ is the dimension of the space of homogeneous degree-$d$ polynomials. The composition is what precisely what we call classically the Veronese embedding.

An example with low $n$ and $d$ may help. If $n = 1$ and $d = 2$, then $\operatorname{Proj} A_{2*} = \operatorname{Proj} k[x^2,xy,y^2]$, embedding naturally in $\mathbb{P}^2_k$. Can you match that with more classical descriptions of the Veronese embedding?


答:
I really love this proof of Bickel and Lehmann (Ann. Math. Stat. 1969), Unbiased Estimation in Convex Families: we can use linearity of expectation to prove unbiased estimators for certain quantities are impossible.
$\newcommand{\PP}{\mathbb{P}}
\DeclareMathOperator{\E}{\mathbb{E}}$

Let $T(\PP)$ be some property of a distribution you wish to estimate, where $\PP$ is a distribution in some class $\mathcal P$. Let $\PP_0 \ne \PP_1$ both in $\mathcal P$ such that $\PP_\alpha = (1-\alpha) \PP_0 + \alpha \PP_1$, a weighted mixture of the two distributions, is also in $\mathcal P$ for any $\alpha \in [0, 1]$.
We want to know whether there exists any estimator $\hat T(X_1, \dots, X_n)$ such that
$$\E_{X_i \stackrel{iid}{\sim} \PP} \hat T(X_1, \dots, X_n) = T(\PP).$$
Assume that there is some such estimator. Then
\begin{align*}
R(\alpha)
  &= T(\PP_\alpha)
\\&= \int_{x_1} \cdots \int_{x_n} \hat T (x_1, \dots, x_n) \,\mathrm{d}\PP_\alpha(x_1) \cdots \mathrm{d}\PP_\alpha(x_n)
\\&= \int_{x_1} \cdots \int_{x_n} \hat T(x_1, \dots, x_n) \,\left[ (1-\alpha) \, \mathrm{d}\PP_0(x_1) + \alpha \, \mathrm{d}\PP_1(x_1) \right] \cdots \left[ (1-\alpha) \, \mathrm{d}\PP_0(x_n) + \alpha \, \mathrm{d}\PP_1(x_n) \right]
\\&= (1-\alpha)^n \E_{X_i \stackrel{iid}{\sim} \PP_0}[ \hat T(X_1, \dots, X_n)]
+ \dots
+ \alpha^n \E_{X_i \stackrel{iid}{\sim} \PP_1}[ \hat T(X_1, \dots, X_n)]
,\end{align*}
and so $R(\alpha)$ must be a polynomial in $\alpha$ of degree at most $n$.

Now, for example, let $T$ be the standard deviation, take $\PP_0 = \mathcal N(0, 1)$ [the standard normal distribution] and $\PP_1 = \mathcal N(0, 2^2)$. Then by the law of total variance,
$$
R(\alpha) = T(\PP_\alpha)
= \sqrt{(1-\alpha) \cdot 1 + \alpha \cdot 4 + \operatorname{Var}[0]}
= \sqrt{1 + 3 \alpha}
,$$
which is not polynomial in $\alpha$ – and hence no unbiased estimator of the standard deviation exists, at least on any class of distributions containing two-component Gaussian mixtures.

You can also easily use this same technique to show, e.g., the non-existence of unbiased estimators for Hellinger distance or integral probability metrics (Theorem 3 here).