问:
A confusion regarding the nature of elements in a field extension.
I read this statement in a well-known textbook: "Since $[W:F]\leq mn$, every element in $W$ satisfies a polynomial of degree at most $mn$ over $F$."

Let $F[a]$ be a field extension of $F$. Can $f\in F$ be algebraic over $F$, and how? I was under the impression that elements belonging only to $F[a]/F$ can be algebraic over $F$?

答:
Your question is confusing. The letter $f$ is usually reserved for a polynomial, but you are using it to mean an element of a field (you wrote $f\in F$). Also, any element $a\in F$ is clearly algebraic over $F$, because it is a root of the polynomial $g(x)=x-a\in F[x]$.

Perhaps it is worth emphasizing the definition of "algebraic element": given any field $F$, and any extension $L$ of $F$, then an element $a\in L$ is said to be algebraic over $F$ when there is a polynomial $g\in F[x]$ such that $g(a)=0$. Note that the field $L$ does not have to be equal to $F(a)$; that is, it can be the case that
$$\large L\supsetneq F(a)\supseteq F.$$
The element $a$ is obviously an element of $F(a)$. I'm not sure where you were going with this, so I don't know if I've addressed your concerns.

Let $F$ be a field, and let $L$ be an extension of $F$. For any $a\in L$, we have that
$$[F(a):F]=\deg(f)$$
where $f\in F[x]$ is the minimal polynomial of $a$. This polynomial is the monic polynomial in $F[x]$ of smallest degree such that $f(a)=0$. Consequently, if $g\in F[x]$ is any other polynomial with $g(a)=0$, we must have that $f\mid g$, and hence $\deg(f)\leq \deg(g)$.

Also, recall that degrees are multiplicative: if $K\supseteq L\supseteq F$ are fields, then
$$[K:F]=[K:L][L:F].$$
Thus, given the information that $[W:F]\leq mn$, and noting that $W\supseteq F(a)\supseteq F$ for any $a\in W$, we have that
$$mn=[W:F]=[W:F(a)][F(a):F]\geq [F(a):F]=\deg(f)$$
where $f$ is the minimal polynomial of $a$ over $F$. Thus, any $a\in W$ satisfies a polynomial in $F[x]$ of degree $\leq mn$.


问:
Showing linear independence using matrices.

If $\mathbf{v}_1,...,\mathbf{v}_m \in F^n$ are written as rows of an $m\times n$ matrix $A$ and $B$ is the row-reduced echelon form of $A$, then $\{\mathbf{v}_1,...,\mathbf{v}_m\}$ is linearly independant if and only if $B$ has no all zero rows

I thought that a good way of showing this may be to show that linear dependence implies row of zeroes in reduced row echelon form and row of zeroes in reduced row echelon form implies linear dependence.

So I start from assuming that $B$ has a row of zeroes. I'm able to show that each row of $B$ is a linear combination of the rows of $A$ by considering the elementary row operations which reduce $A$ to $B$, and so since the rows of $A$ are $\mathbf{v}_1,...,\mathbf{v}_m$, I have that each row of $B$ is a linear combination of $\mathbf{v}_1,...,\mathbf{v}_m$.

So if $B$ has a row of zeroes, then we have that the zero vector is a linear combination of $\mathbf{v}_1,...,\mathbf{v}_m$. However, my issue is that I am unsure whether this linear combination could possibly be  $0\mathbf{v}_1+0\mathbf{v}_2+...+0\mathbf{v}_m = \mathbf{0}$ as in this case, having a row of zeroes wouldn't imply linear dependence of $\mathbf{v}_1,...,\mathbf{v}_m$. Also, I don't know how I would prove in the opposite direction: That if $\mathbf{v}_1,...,\mathbf{v}_m$ are linearly dependent then $B$ has a row of zeroes.

Can anyone help me with this? Is there a better way to approach this question?

答:
Note that $A$ and $B$ have the same row space, as the operations of Gauss-Jordan elimination leave the row space invariant.

So the span of $v_1,\dots,v_m$ is the same as the span of the rows of $B$. Further, the non-zero rows of $B$ are independent, from the construction of the echelon form.

So $B$ has no non-zero rows iff $A$ has rank $m$, i.e. iff $v_1,\dots,v_m$ are linearly independent.

ContentLicense: CC BY-SA 4.0


问:
Uniqueness of finite fields with $p^n$ elements.
It's well known there exists only one finite field with $p^n$ elements, up to isomorphism, where $p$ is a prime and $n \geq 1$.

Let $n = m t$, where $m, t > 1$.

Then $F = Z_p[X]/(f(X))$ is a field with $p^n$ elements, provided $f$ is some irreducible polynomial of degree $n$ in $Z_p$.

By the same token, $G = Z_p[X]/(g(X))$ is a field with $m$ elements ($g$ irreducible of degree $m$). Then $G$ has $p^m$ elements.

Finally let: $H = G[X]/(h(X))$ where $h$ is an irreducible polynomial of degree $t$ with coefficients in $G$.

Now, in my understanding $F$ and $H$ have both $p^n$ elements. So my question is:

Are $F$ and $H$ isomorphic?

答:
This relies on a general fact about splitting fields.

Let $F$ be a field and $f(X)\in F[X]$ be a monic polynomial. An extension field $K$ of $F$ is a splitting field for $f$ if

$f(X)=(X-a_1)(X-a_2)\dots(X-a_k)$ in $K[X]$ (the roots need not be distinct);

$K=F(a_1,a_2,\dots,a_k)$

Theorem. If $K_1$ and $K_2$ are splitting fields of $f(X)\in F[X]$, then there exists a field isomorphism $\varphi\colon K_1\to K_2$ leaving $F$ pointwise fixed.

The proof is longish and can be found in any book on Galois theory, because it's a basic tool thereof.

Now consider the polynomial $X^{p^n}-X\in\mathbb{F}_p[X]$, where $\mathbb{F}_p$ is the $p$-element field (which is unique up to unique isomorphism).

Let $K$ be a splitting field of $f(X)$. Then $f(X)$ has $p^n$ distinct roots in $K$ (because the derivative of the polynomial is $-1$). On the other hand, the set of roots of $f(X)$ is a subfield of $K$: indeed, if $a,b$ are roots, then
$$
(a+b)^{p^n}-(a+b)=a^{p^n}+b^{p^n}-a-b=0
$$
so $a+b$ is a root of $f$. Analogously
$$
(ab)^{p^n}-ab=a^{p^n}b^{p^n}-ab=ab-ab=0
$$
and it's easy to check the reciprocals. Since also $0$ and $1$ are roots we're done.

Thus $K$ is the set of all roots of $f$ and therefore $|K|=p^n$.

Conversely, if $K$ is a field with $p^n$ elements, then the same argument as before shows that $X^{p^n}-X$ has $p^n$ distinct roots in $K$, so $K$ is a splitting field for $f(X)$.

Uniqueness up to isomorphism now follows from the theorem above.


问:
Find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ given that $a_1,a_2,a_3,a_4$ are roots of a biquadratic equation.
So the biquadratic equation is $x^4+(2-\sqrt3)x^2+2+\sqrt3=0$. Let $a_1,a_2,a_3,a_4$ be its roots. So we have to find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ .

My attempt:

So of we put $x^2=t$, and let the roots of the new quadratic equation be $a_1,a_2$. So we get that $a_1=-a_3;a_2=-a_4$. So the expression whose value we have to find will become $(1-a_1^2)(1-a_2^2)$. And now we multiply the expression to get, $1-(a_1^2+a_2^2)+(a_1a_2)^2=1-((a_1+a_2)^2-2a_1a_2)+(a_1a_2)^2$.

Now substituting the values of sum of roots and product of the same, I get the answer $5+10\sqrt3$. But the answer according to my question paper is $5$.

So please can someone point out the mistake and tell the answer?

ContentLicense: CC BY-SA 3.0

Body:
\begin{align}
f(x) &= (x-a_1)(x-a_2)(x-a_3)(x-a_4) \\
&= x^4 +(2−√3) x^2 +2 + √3 \\
f(1)& = 1 + 2 - √3 + 2 - √3 = 5
\end{align}


问:
Fourier Transform of $\frac{\sin(\|x\|)}{\|x\|}$
I am having a problem finding the Fourier Transform of $$f(x) = \frac{\sin(\|x\|)}{\|x\|}$$

I am interested in the cases where $x \in \mathbb{R}^n,  n= 2, 3$.

Any help much appreciated !

答:
Formally
$$\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}e^{-isx}e^{-ity}\,dx\,dy =\int_{-\pi}^{\pi}\int_{0}^{+\infty}\sin(\rho)e^{-is\rho\cos\theta}e^{-it\rho\sin\theta}\,d\rho\,d\theta $$
equals
$$ \int_{-\pi}^{\pi}\frac{d\theta}{(s\cos\theta+t\sin\theta)^2-1}=\int_{-\pi}^{\pi}\frac{d\phi}{(s^2+t^2)\sin^2\phi-1}=\frac{-2\pi}{\sqrt{1-(s^2+t^2)}} $$
and the Fourier transform of a radial function is a radial function, but there are issues in the application of Fubini's theorem since $\frac{\sin\|x\|}{\|x\|}$ does not belong to $L^1(\mathbb{R}^2)$ or $L^1(\mathbb{R}^3)$.


问:
Elementary conditional probability in dealing out a bridge hand
"In the card game bridge, the $52$ cards are dealt out equally to 4 players—called East,West, North, and South. If North and South have a total of $8$ spades among them, what is the probability that East has $3$ of the remaining $5$ spades?"

The book "A First Course in Probability" by Sheldon Ross states in its solution to the above question that the probability of South-North having a total of $8$ spades among them is $C(26,13)$. I can not understand where it came from. Any help would be appreciated.

答:
The total number of cases where N and S having a total of $8$ spades is not used in the solution, because it is independent from what the question is asking.

$26\choose 13$ is the number of ways to distribute the remaining $26$ cards between E and W. Out of all those distributions, ${5\choose3}$ ways for the E player to choose from the spades and ${21\choose 10}$ ways to choose from the non-spades.

Thus the final answer to the problem is ${5\choose3}{21\choose 10}\over{26\choose 13}$.

The probability does not depend on which of the $8$ spades are chosen for N and S and therefore it is not a conditional probability type problem.

The problem can be restated as "We have a deck of card consisting of $5$ spades and $21$ non-spade cards and we have two players, what's the probability first player gets $3$ spades?"

It becomes much easier to understand.


答:
From Bracewell, the n-dimensional Fourier Transform

$$F(s_1, \dots, s_n) = \int_{-\infty}^{\infty} \dots \int_{-\infty}^{\infty} f(x_1,\dots, x_n) e^{-2\pi i (x_1 s_1 + \dots + x_n s_n)} \space dx_1 \dots dx_n $$

when there is n-dimensional symmetry, can be manipulated into

$$F(q) = \dfrac{2\pi}{q^{\frac{1}{2}n-1}} \int_{0}^{\infty} f(r) J_{\frac{1}{2}n-1}(2\pi qr)\space r^{\frac{1}{2}n} \space dr$$

Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.

For the case of $n=2$, you need to compute

$$F(q) = 2\pi \int_0^\infty \dfrac{\sin(r)}{r}J_0(2\pi qr)\,r\, dr =  2\pi \int_0^\infty \sin(r)J_0(2\pi qr) \,dr$$

For the case of $n=3$ note that

$$J_{\frac{1}{2}}(x) = \sqrt{\dfrac{2}{\pi x}}\sin(x)$$

so you need to compute

$$F(q) = \dfrac{2\pi}{\sqrt{q}} \int_0^\infty \dfrac{\sin(r)}{r}\sqrt{\dfrac{2}{\pi (2\pi qr)}}\sin(2\pi qr)\,r\sqrt{r}\, dr = \dfrac{2}{q}\int_0^\infty \sin(r)\sin(2\pi qr)\, dr$$

Update

It is worth noting that the n-spherically symmetric Fourier transform (aka Hankel Transform) is its own inverse.  Also for the $n=3$ case

$$\begin{align*} f(r) &= \dfrac{2\pi}{\sqrt{r}} \int_0^\infty \dfrac{\delta\left(q-\frac{c}{2\pi}\right)}{2q}\sqrt{\dfrac{2}{\pi(2\pi rq)}}\sin(2\pi rq)q\sqrt{q} \, dq\\
\\
&= \dfrac{1}{r}\int_0^\infty \delta\left(q-\frac{c}{2\pi}\right)\sin(2\pi rq) \, dq\\
\\
&= \dfrac{\sin(cr)}{r}\\
\end{align*}$$

Title: Fourier Transform of $\frac{\sin(\|x\|)}{\|x\|}$

Body:
I am having a problem finding the Fourier Transform of $$f(x) = \frac{\sin(\|x\|)}{\|x\|}$$

I am interested in the cases where $x \in \mathbb{R}^n,  n= 2, 3$.

Any help much appreciated !


问:
What are the different types of "averages" and when to use each one?
So, the concept of an average truly is somewhat abstract. Most statisticians define it as a "measure of central tendency." Others say it is the "center of gravity" for a set of numbers.

I personally prefer a slightly more concrete explanation: A statistic that describes the "typical", or better yet, "representative" value of a data set. We humans get to decide what "representative" actually means. Is it the most common number? The number that falls "in the middle" of a set of numbers? Etc.

I am going to list a few types of averages and am wondering if someone could provide a data set where that particular average is the best choice for describing a "typical" or "representative" value for that data set.

Types of averages:

Arithmetic Mean: The number that you could use in place of each of the values of a data set, and still have the same sum. Formula: $$\overline{X}=\frac1N\sum\limits_{i=1}^{N} X_i$$

Mode: The most common value in a data set. No formula that I know of.

Median: The literal middle number of a data set where the values are listed in ascending order. No formula that I know of.

Root Mean Square: Don't really know how to describe this average in a physical sense. Maybe the average that gives larger numbers in the data set more "weight" or "significance?" Formula: $$X_{rms}=\sqrt{\frac1N\sum\limits_{i=1}^{N} X_i^2}$$

Mean Root Square: I just made this one up, but it seems to work well when I applied it to some random data sets. It seems to do the opposite of the RMS and gives smaller numbers in the data set more "weight" and "significance." Formula:
$$X_{mrs}=\frac1N\sqrt{\sum\limits_{i=1}^{N} X_i^2}$$
EDIT: Turns out this should not be considered an average because it fails to describe a data set that consists of only one number, unlike all the other averages.

Geometric mean: The number that you could use in place of each of the values of a data set, and still have the same product. Formula:
$$GM=\sqrt[N]{\prod\limits_{i=1}^{N} X_i}$$

Feel free to add in any other popular or useful types of averages and when to use them!

答:
In the data set (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) "6' is the median.

There are 5 numbers before it, (1, 2, 3, 4, 5), and 5 numbers after it (6, 7, 8, 9, 10, 11)

In the data set, (1, 22, 37, 45, 51, 66, 79, 84, 90, 107, 3035), "66" is the 'median'.
This doens't tell you much about the OTHER members of this set of numbers, but it is still the 'median'.


问:
Integral of Legendre polynomials with tangent
I have encountered the following relationship$^{[1][2]}$, stated without proof both times

$$\int_0^\gamma dt \tan(t/2)\cdot [P_n(\cos(t))+P_{n-1}(\cos(t))]=\frac{1}{n}[P_{n-1}(\cos(\gamma))-P_{n}(\cos(\gamma))]    $$

Where $P_n(x)$ is the $n^{th}$ Legendre polynomial, and $\gamma<\pi$.

I've tried integration by parts, substitution of $u=\cos(t)$, and looked in Gradshteyn. Mathematica doesn't integrate it (at least, not in the current form). I've verified that the relationship is correct for modest values of $n$. I suspect that there is some property of the Legendre polynomials that facilitates this, but I don't see what it is.

[1] Mixed boundary value problems, Dean G. Duffy, eq. 3.1.80, p95

[2] This paper, eq. A8

答:
Since $\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}$, by letting $\cos(\gamma)=T\in[-1,1]$ your identity can be written in the simplified form

$$ \int_{T}^{1}\frac{dx}{1+x}(P_n(x)+P_{n-1}(x))\,dx = \frac{1}{n}(P_{n-1}(T)-P_n(T)). $$

The identity clearly holds at $T=1$. By differentiating both sides wrt $T$ we find Bonnet's identity and we are done.


答:
All elements $f\in F$ are algebraic over $F$, that is: there is
some polynomial $p\in F[x]$ s.t $p(f)=0$. Namely $$p(x)=x-f\in F[x]$$

Note 1: $F\subseteq F[a]$ so if this isn't any contradiction to statement
that all elements of $F[a]$ are algebraic over $F$.

Note 2: Not only the elements of $F[a]$ are algebraic over $F$.
For example consider $$F=\mathbb{Q},a=\sqrt{2},b=\sqrt{3}$$ then
$$F[a]=\mathbb{Q}(\sqrt{2})$$ and $$b\not\in F[a]$$ but is algebraic
over $F$.

Title: A confusion regarding the nature of elements in a field extension.

Body:
I read this statement in a well-known textbook: "Since $[W:F]\leq mn$, every element in $W$ satisfies a polynomial of degree at most $mn$ over $F$."

Let $F[a]$ be a field extension of $F$. Can $f\in F$ be algebraic over $F$, and how? I was under the impression that elements belonging only to $F[a]/F$ can be algebraic over $F$?


问:
Define the relation $\mathscr{R}$ on $\Bbb Z$ by $x \mathscr{R} y$ iff $xy >0$. I wish to examine several properties of this relation.
Considering on $\mathbb{Z}$ the following relation:

$$\mathscr{R} = \left \{ (x,z) \in \mathbb{Z} \times \mathbb{Z} \quad \mbox{ such that } \quad xz > 0  \right \}$$

i.e. $\forall x,z \in \mathbb{Z}, x \mathscr{R} z \iff \mbox{ the product } \, \, x \cdot z > 0 $

Check if the relation $\mathscr{R}$ is:

i) reflexive;

ii) symmetrical;

iii) transitive;

iv) anti-symmetrical;

v) of order;

vi) of equivalence.

What I have done is the following:

i) is it a reflexive relation?

$\forall x,x \in \mathbb{Z}, \quad x \mathscr{R} x \iff x\cdot x = x^2 > 0 $
since a square is always $> 0$ the relation is reflexive.

ii) is it a symmetrical relation?

$\forall x,z \in \mathbb{Z}, \quad x \mathscr{R} z \iff x \cdot z > 0$ by hypothesis,

since the operation of multiplication is commutative on $\mathbb{Z}$ the following is true:

$\forall x,z \in \mathbb{Z}, \quad z \mathscr{R} x \iff z \cdot x > 0$

hence, the relation is symmetrical.

iii) is it a transitive relation?

$\forall x,z \in \mathbb{Z}, \quad x \mathscr{R}z \iff x \cdot z > 0$ by hypothesys,

if we take another ordered pair

$\forall z,m \in \mathbb{Z}, \quad z \mathscr{R} m \iff z \cdot m > 0$

hence,

if $x \mathscr{R}z, \quad z \mathscr{R} m \Rightarrow x \cdot m > 0$

i.e. if $x \cdot z > 0, \quad z \cdot m > 0 \Rightarrow x \cdot m > 0$

so the relation is transitive.

iv) is it a anti-symmetrical relation?

No. Because it is not true that $z \mathscr{R}x \iff x=z$, as we have seen in the symmetrical relation.

v) is it a order relation?

No. Because it is not anti-symmetrical.

vi) is it a equivalence relation?

Yes, because it is reflexive, symmetrical and transitive.

What do you think about it?

Please, can you give me any suggestions? Many thanks!

答:
$
\newcommand{\RR}{\mathscr{R}}
\newcommand{\Z}{\mathbb{Z}}
$
Note, intuitively, what we have. This relation $\RR$ on $\Z$ essentially means $x \RR y$ if and only if $x,y$ have the same sign and are nonzero. (Same sign ensures nonnegative, nonzero ensures positive.) Hopefully this makes the intuition clearer.

Onto the errors with your approach:

You were wrong on reflexivity. Yes, $x^2 > 0$ for any $x \in \Z$, but only if $x \ne 0$.

Your justification for transitivity is unclear, in my opinion.

Your justification for antisymmetry seems to amount to "it's symmetrical and thus not antisymmetrical." The properties are not necessarily opposed; a relation can be both, or neither. You're right in that this relation is not antisymmetrical, though, but even so. (In particular, a relation on $A$ of the form $\{(x,x) \mid x \in X\}$ for some $X \subseteq A$ is both antisymmetric and symmetric.)

Onto the overall right ideas:

Reflexivity: This does not hold, because $(0,0) \not \in \RR$. This is because $0 \cdot 0 = 0 \not > 0$.

Symmetry: This does hold under commutativity of integer multiplication. Assume $(x,z) \in \RR$. Then $xz > 0$. But then $xz = zx > 0$. Thus, $(z,x) \in \RR$ on the premise $(x,z) \in \RR$.

Transitivity: This does hold. It's easiest to refer back to the intuitive idea: $(a,b) \in \RR$ if and only if $a,b$ are nonzero and share the same sign. Suppose $(a,b),(b,c) \in \RR$. Then $a,b$ share the same sign, and are nonzero; likewise, the same is true of $b,c$. Thus, $a,b,c$ are nonzero and have the same sign. Thus, $a,c$ have the same sign and are nonzero, and thus $(a,c) \in \RR$.
Of course, this can be verified algebraically too. Since $(a,b),(b,c) \in \RR$, then $ab>0$ and $bc>0$. Multiply the inequalities and you have $ab^2c > 0$. Since $b \ne 0$ (otherwise neither inequality would hold) and $b^2 >0$ as a result, we may divide both sides by $b^2$ and see $ac > 0$. Thus, $(a,c) \in \RR$.

Antisymmetry: Anti-symmetry requires that, when $(a,b) \in \RR$ and $(b,a) \in \RR$, that $a=b$. However, commutativity of multiplication allows them to be distinct. More specifically: $(1,2),(2,1) \in \RR$ since $1 \cdot 2 = 2 = 2 \cdot 1 > 0$, but obviously $1 \ne 2$.

Order: An order relation requires antisymmetry, so we don't have this.

Equivalence: Since we don't have reflexivity, we don't have equivalence.


问:
Conditional PMF of two independent Poisson r.v.s
Problem:

My solution: $$P(X=k\mid X+Y=n)=P(Y=n-k)=e^{-\lambda}\frac{\lambda^{n-k}}{(n-k)!}$$
Official Solution:

My question: How does the first equality even work. I suppose they are using the independence of $X$ and $Y$ to say something like:

$$P(X=k \cap X+Y=n)=P(X=k)P(Y=n-k)$$

But I don't think this is right because in this specific probability you do not know that $X=k$.

My question(2): Why is my answer wrong. These two expressions are definitely not equivalent, but I cannot see where I went wrong.

答:
The event $\{X=k\} \cap \{X+Y = n\}$ is equal to $\{X = k\} \cap \{Y = n-k\}$. Then you can use independence.

You can see that the official solution is your answer multiplied by $\frac{P(X=k)}{P(T=n)}$. The error in your solution attempt is the first equality: you haven't justified why $P(X=k \mid X+Y=n) = P(Y=n-k)$. Your intuition is correct that somehow the event $Y=n-k$ plays a role, but you need to go back to the definition of the conditional probability $P(X=k \mid X+Y=n) = \frac{P(X=k, X+Y=n)}{P(X+Y=n)}$ to see how the event $Y=n-k$ enters the calculation.

Side note: what is true is $P(X=k \mid X+Y = n) = P(Y = n-k \mid X+Y = n)$. But $P(Y = n-k \mid X+Y = n) \ne P(Y = n-k)$.


问:
How to calculate $\lim_{n \to \infty} \int_{[n,n+1)} f(x) dx$ if $f\in L^1(\mathbb{R})$
Let $f \in L^1(\mathbb{R})$. I need to calculate the following integral: $$\lim_{n \to \infty} \int_{[n,n+1)} f(x) dx$$ but I don't really how to begin with...

I would be really thankful if someone could give me a hint.

I think that  $$\lim_{n \to \infty} \int_{[n,n+1)} f(x) dx$$ should be equal to  $$ \int_{\mathbb{R}} f(x) dx$$ but I don't get to a solution... I know it's value is finite because $f$ is an integrable function. Is it possible to say anything else about this integral?

答: Here's an alternative answer not using dominated convergence. Recall that $S \mapsto \int_S |f|$ is a measure so by disjoint additivity, $$\sum_{n = 1}^\infty \int_{[n, n + 1)} |f(x)| = \int_{[1, \infty)} |f(x)| \leq \int_{\mathbb{R} } |f(x)| < \infty $$ which implies that $\lim_{n \to \infty} \int_{[n, n +1)} |f(x)| = 0$ since otherwise, the sum would diverge.


问:
What is the difference between "expectation", "variance" for statistics versus probability textbooks?
It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $X$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like,
$$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$. The variance is,
$$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide  (def 2)? How do I know which mean the person is talking about?/

答:
The general expression for arithmetic mean is $\frac{\sum\limits_{i= 1}^n w_i x_i}{\sum\limits_{i= 1}^n w_i}$, or even more generally, $\frac{\int w(t) f(t)dt}{\int f(t)dt}$ (there are then ways to recover the discrete case from that).

If you set all $w_i$ to $1$, or really to anything as long as it's constant, you get $\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$. This is referred to as an "unweighted" average, although technically it's still weighted, it's just that you're multiplying everything by $1$, so you don't notice it. If you set $p_i = \frac{w_i}{ \sum\limits_{k= 1}^n w_k}$, and interpret $p_i$ as the probability of the $i$th event, then you get the average weighted by probability, which is also known as expected value.

You have to be careful about "unweighted" averages, as they often actually are weighted, but by weights that you didn't want. For instance, suppose you want the average income over the US, and you have the average income for each state individually. You could just add all of those averages together, and then divide by $50$. People will often call this the "unweighted" or "simple" average, but you're actually weighting people by the reciprocal of their states population; the fewer people there are in a state, the more each person's income affects that state's average, and so the more they affect the total "unweighted average". As a result, to get the actual overall average income from the individual states' averages, you have to multiply each state's average by its population to get its total income, add all of those together, and then divide by the total population.

A common weighting you'll see is frequency weighting. This is where you multiply each value by the number of times it appears. For instance, if you measure something once a month for a year, and the only values you get are $0$, $1$, and $2$, taking the simple average of those values gives you $1$. But that's probably not the real average. To get a more meaningful average, you should take each of these values, weight them by how many months they appear for, and then take the average.

One property of weighted averages is that multiplying all of the weights by a constant number doesn't change the final result (you'll just divide it out again when you divide by the total of the weights). So weighting by the frequencies is equivalent to weighting by the percentage of cases each value is. That is, if $0$ is the value for $5$ months, $1$ is the value for $4$ months, and $2$ is the value for $3$, the weighting of $5,4,3$ is equivalent to the weighting of $\frac 5 {12},\frac 4{12},\frac 3{12}$.

So if you have a probability distribution where one thing has a $60$% chance of happening, and something else has a $40$% chance of happening, the expected value is simply the frequency weighted average.

For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?

The expected value of a random variable is the expected value of that random variable. It is a property of the distribution. If you're asked for the expected value of a random variable, you find the expected value of that random variable. If you aren't asked for the expected value of a random variable, you don't go looking for a random variable to take the expected value of. How you know whether you take the expected value of random variable is if there is a random variable to take the expected value of. Even if values come from a distribution, the average of those values is the average of those values, not the average of the distribution they came from.

If people are talking rigorously, they will explicitly say they want the expected value. You many, however, see people asking for the "mean" or "average", when they really want the expected value, but you can recognize those cases by there being a random variable. For instance, if someone asks "What's the average payout of this slot machine?", the context suggests that you should take the expected value of the distribution of payouts, and not simply take the set of different possible payouts and take the simple average. There could be some ambiguity as to whether "the payout" refers to the random process that pays money out, in which case you should take the expected value of the distribution (population mean), or the actual money paid out, in which case you should take the average of all the actual payouts made by the machine (sample mean), but in the latter case, you still should take the frequency weighted average.


问:
Number of schedules that do not repeat a matchup

Background

There are n teams such that n is even. A matchup is a pair of two distinct teams where order does not matter so that (A,B) and (B,A) are equivalent. A schedule is comprised of all n teams in $\frac n2$ matchups. For example, the following is a possible schedule of 8 teams labeled A ... H:

{(A,B) (C,D) (E,F) (G,H)}

Question

From the set of all possible schedules, one is chosen at random. How many schedules in the set of all possible schedules share no matchups with this random schedule?

Progress I've Made

The size of the set of schedules is (n-1)(n-3)(n-5)...(3)(1). For n = 6 there are 15 possible schedules. I've listed these by hand. By inspection, 8 of the 15 possible schedules would share no matchup with a random schedule. However, I have not been able to hit upon an equation that would describe this for all n.

Areas of Inspection

I've attempted to use derangements and other variation of the principle of inclusion-exclusion, but either this is the wrong approach or I'm not using the principles correctly. I read about orthogonal latin squares but am unsure if they apply to the problem.

答:
Use Inclusion-Exclusion.

For example  for six people, three pairs:

Start with $1×3×5=15$

Subtract $3\choose1$ times $1×3$, which is nine.

Add $3\choose2$ times $1$, which is three.

Subtract $3\choose3$ times $1$, which is one.

Total $15-9+3-1=8$


问:
Deriving another formula for least square
In a proof of a theorem in this paper(page 26), they started with what is shown below

The $\|\cdot\|_*$ is the trace norm and I do understand SVD decomposition but I don't understand how they decompose $||Z-W||_F^2$ to what is between brackets?

答:
The Frobenius norm is induced by the inner product $\langle A, B\rangle_F := \text{Tr}(A'B)$ (where $A,B \in \mathbb{R}^{m \times n}$), in the sense that $\|A\|_F^2 = \langle A, A \rangle_F$.

A useful fact is that if $U$ and $V$ are orthogonal matrices of the appropriate dimension, then $\|UAV\|_F = \|A\|_F$. You can prove this easily using the above inner product formulation of the Frobenius norm.

You can show that

$$\|Z-W\|_F^2 = \|Z\|_F^2 - 2 \langle Z, W \rangle_F + \|W\|_F^2.$$

Note that $\|Z\|_F^2 = \|D\|_F^2 = \sum_{i=1}^n d_i^2$.

Note that $\langle Z, W\rangle_F = \langle UDV', W \rangle_F
= \text{Tr}(VD'U'W) = \text{Tr}(D'U'WV)$. Think through the definition of trace and matrix multiplication to show that this equals $\sum_{i=1}^n d_i u_i' W v_i$.

After correcting the typo in the paper ($\|Z\|_F^2$ should be $\|W\|^2_F$), this leads to the desired expression in the paper.


问:
Find $a$ so a quadratic expression describes a straight line (problem from a Swedish 12th grade ’Student Exam’ from 1931)

$x$ and $y$ are coordinates in an orthogonal coordinate system. If the constant $a$ is suitably chosen the equation
$$9x^2-3xy+ay^2+15x-16y-14=0$$
describe two straigth lines. Find the value of $a$ and the equations for the two lines.

I tried to complete the square (using Mathematica) but the resulting expression was not helpful’. Any idea where to start? TIA.

(N.B.: The problem is quoted from a Swedish 12th grade ’Student Exam’ from 1931. I find these problems rather fun to solve in the evenings.)

答:
The quadratic expression describes a conic, and the symmetric matrix associated to the conic is:
$A=\begin{bmatrix}
9 & -3/2  & 15/2\\
-3/2 & a  & -8 \\
15/2 & -8 & -14
\end{bmatrix}$

If the rank of the matrix $A$ is not equal to $3$ (which means that its determinant is equal to $0$), then the quadratic form describes a degenerate conic: the union of two distinct lines (which could also be, for example, two conjugate complex lines) or a double line.

We have that: $det(A)=\frac{-729}{4}a-\frac{729}{2}$. So $A$ is singular if and only if $a=-2$. In fact $9x^2-3xy-2y^2+15x-16y-14=0$ describes two incident lines in the affine real plane.


问:
Is every element of $\mathbb{R}$ a member of $\mathbb{Q}$ adjoined with finitely many members of its transcendence basis?
Recently I've been interested in creating somewhat non-constructive solutions to problems using the concept of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, which exists assuming the Axiom of Choice but I only know some basic Field Theory. As part of my increasing understanding, I ask:

Let $W$ be the transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$. Is it true that $$\mathbb{R} = \bigcup_{w\subset W, \;w \text{ finite}}\mathbb{Q}(w)$$? What if we replace "finite" with "countable"?

答:
Edit. This answer is incorrect. I read "transcendence basis" as "vector space basis". I think @AndreasCaranti 's answer is correct. I will leave mine up so no one else makes the same mistake.

Yes, since every element of $\mathbb{R}$ is a finite $\mathbb{Q}$-linear combination of basis elements. That means it's in the union of the corresponding extensions.


问:
Invert Trigonometric System of Equations
In my research, I have a system of trig equations that I would like to invert. I know $\alpha$ and $\beta$ ($\theta$ is a constant) in the following system.

$$\cos \alpha  = \cos \theta \cos x + \sin \theta \sin x \cos y$$
$$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$

I would like to find an inversion in the form:
$$\text{trig}(x) =\; ...$$
$$\text{trig}(y) = \;...$$

How can this be done?

答:
$$\cos \alpha  = \cos \theta \cos x + \sin \theta \sin x \cos y$$
$$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$
you set $\cos x=t;\;\sin x=u;\;\cos y=v;\;\sin y=w$  and get
$$\cos \alpha  = \cos \theta\, t + \sin \theta\, u v$$
$$\tan \beta = \frac{t +\cos \theta \cos \alpha}{\sin \theta\, u w} $$

add the conditions $$t^2+u^2=1;\;v^2+w^2=1$$

If you send the constant values I can try to solve it for you.

It is too complicated to solve it in general.


问:
Proving that length of a curve is $\infty$
Let $f$ be a differentiable and continious function in $(0,1]$ and $lim_{x\to 0^+}f(x)=\infty$. Prove that the length of the curve on (0,1] is $\infty$.
Steps I tried: $L=\int_0^1 \sqrt{1+f\prime(x)^2}dx\geq\int_0^1\sqrt{f\prime(x)^2}=\int_0^1f\prime(x)=f(1)-lim_{x\to0^+}f(x)=-\infty$ (and I had to prove $\int\geq\infty$ and not $-\infty$).</p>

答:
I think your question is wrong, because in order for $f$ to be continuous at 0, it has to be bounded near 0, but your claim that $\lim_{x \rightarrow 0^+}f(x)=\infty$ means that $f$ is unbounded at 0, contradiction. I think you mean that $\lim_{x \rightarrow 1^-}f(x)=\infty$. Then, you get the limit as $\infty$ from your computation.

P.S. From one of your steps, $\sqrt{f'(x)^2} \neq f'(x)$, but it is $|f'(x)|$. So I think you just need to change that equality into inequality

Edit in response to edited question:
Then, once we have $\int_0^1 \sqrt{f'(x)^2}=\int_0^1 |f'(x)| \geq |\lim_{t \rightarrow 0^+}\int_t^1 f'(x)|=|f(1)-\lim_{x \rightarrow 0^+}f(x)|=|-\infty|=\infty$

This corrects and finishes the proof.


问:
embedded submanifold in euclidean space derivative
Let $y$ be the coordinate function on $\mathbb{R}^2$. Consider $\mathbb{S}^1$ as an embedded submanifold. If $p\in (1,0)$, how do I show that $\frac{\partial}{\partial{y}}|_p\in T_p\mathbb{S}^1$? It clearly is in $T_p\mathbb{R}^2$.

My attempt: Since $\mathbb{S}^1$ is defined by the smooth function $F(x,y)=x^2+y^2-1$,

$T_p\mathbb{S}^1=Ker(F_{*,p})$ for all $p\in \mathbb{S}^1$.
Hence if $h\in C^{\infty}$, $F_{*,p}(\frac{\partial}{\partial{y}}|_p)h=\frac{\partial{h\circ F}}{\partial{y}}|_p=0$

where the last equality holds because $\frac{\partial{F}}{\partial{y}}|_p=0$

Question 2: In Loring Tu's book, does the differentiable structure as constructed in Proposition 9.4, coincide with the differentiable structure as constructed in example 5.16 for the sphere?

答:
Yes, that argument works. Perhaps it wouldn't hurt to make the use of the chain rule explicit. Also note that $\frac{\partial F}{\partial y}\Big\vert_p=0$ holds true precisely because $p=(1,0)$, this would fail at other points (convince yourself of this).

Alternatively, you don't need to be this abstract. Since we're working in $\mathbb{R}^2$, we can identify all tangent spaces with subspaces of $\mathbb{R}^2$. Then $\frac{\partial}{\partial y}\Big\vert_p$ is just the vector $(0,1)$ and, for all $p\in S^1$, the tangent space $T_pS^1$ is precisely the subspace of $\mathbb{R}^2$ orthogonal to $p$. And, of course, $(1,0)$ is orthogonal to $(0,1)$ (this is what's illustrated in Fig. 19.1. in Tu's book).


答:
Perhaps I am missing something, but, quoting for instance from this MSE post:

a set $T$ of elements of an extension field $k/F$ is a transcendence basis if

for all $n$, and distinct $t_{1}, \dots, t_{n} \in T$, there is no nonzero polynomial $f(X_1,\dots,X_n)\in F[X_1,\dots,X_n]$ such that $f(t_1,\dots,t_n)=0$;

$k$ is algebraic over $F(T)$.

So an element like $\sqrt{2}$ will not be in any of your $\mathbb{Q}(w)$.

Title: Is every element of $\mathbb{R}$ a member of $\mathbb{Q}$ adjoined with finitely many members of its transcendence basis?

Body:
Recently I've been interested in creating somewhat non-constructive solutions to problems using the concept of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, which exists assuming the Axiom of Choice but I only know some basic Field Theory. As part of my increasing understanding, I ask:

Let $W$ be the transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$. Is it true that $$\mathbb{R} = \bigcup_{w\subset W, \;w \text{ finite}}\mathbb{Q}(w)$$? What if we replace "finite" with "countable"?


问:
Prove $\lim\limits_{x\to0} xf(x) = 0$?
I am confused about the following proof on a textbook I'm reading.
Suppose $f:(0,b] \to \Bbb R$ is continuous, positive, and integrable on $(0,b]$. Suppose further that as $x \to 0$ from the right, $f(x)$ increases monotonically to $+\infty$. Prove that $\lim\limits_{x\to0} xf(x) = 0$.

It would be nice if I could write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$. Then for all $\epsilon > 0$ there exists a $\delta$ such that $0<x<\delta$ implies $0<f(x)x \le \int_a^{a+x} f(t)dt <\epsilon$. But this is not true since $f$ increases monotonically as $x\to0$. Instead of writing $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$, I should probably write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(a+x)dt = f(a+x)x$. But that way my proof does not work. Can someone give me a hint as to how to proceed?

EDIT: Thanks for the comments and I realized that the monotonicity of $f$ should be one of the conditions.

答:
In the following proof, we'll suppose that $ f $ is $\underline{\textbf{decreasing}}$ on $ \left(0,b\right] $. (Otherwise the statement is false)

Let $ \varepsilon >0 $.

Since $ f $ is positive and integrable on $ \left(0,b\right] $, then $ \lim\limits_{x\to 0^{+}}{\int_{x}^{b}{f\left(y\right)\mathrm{d}y}}=\ell\in\mathbb{R} $, which means there exists $ \eta\in\left(0,b\right) $, such that $ \left(\forall x\in\left(0,\eta\right)\right),\ \left|\int_{x}^{b}{f\left(y\right)\mathrm{d}y}-\ell\right|<\frac{\varepsilon}{4} $.

For any $ x\in\left(0,\eta\right] $, we have : \begin{aligned} \left|xf\left(x\right)\right|=\left|2\int_{\frac{x}{2}}^{x}{f\left(x\right)\mathrm{d}y}\right|\leq 2\left|\int_{\frac{x}{2}}^{x}{f\left(y\right)\mathrm{d}y}\right|&=2\left|\int_{\frac{x}{2}}^{b}{f\left(y\right)\mathrm{d}y}-\int_{x}^{b}{f\left(y\right)\mathrm{d}y}\right|\\ &\leq 2\left|\int_{\frac{x}{2}}^{b}{f\left(y\right)\mathrm{d}y}-\ell\right|+2\left|\int_{x}^{b}{f\left(y\right)\mathrm{d}y}-\ell\right|\\ &\leq\varepsilon \end{aligned}

Thus : $$ \lim_{x\to 0^{+}}{x f\left(x\right)}=0 $$

Title: Prove $\lim\limits_{x\to0} xf(x) = 0$?


问:
2 equivalent definitions of finite cover property
I have 2 different definitions of the finite cover property, f.c.p. a property of first order theory, which I would like to prove
equivalent. I was trying to prove this equivalence but I got stuck.

答:
Definition 2.1 defines what it means for a theory $T$ to not have f.c.p. Negating the statement, we find that it says:

A theory has f.c.p. if and only if there exists a formula $\varphi(x;y)$ with the following property, which I'll call $(\star)$: For every integer $n$, there exists a set $A$ in a model of $T$ and a set $p\subseteq \{\varphi(x;a),\lnot\varphi(x;a)\mid a\in A\}$ such that $p$ is inconsistent and every subset of $p$ of size less than $n$ is consistent.

So to show that the two definitions are equivalent, we just need to show that $T$ has a formula with $(\star)$ if and only if $T$ has a formula with f.c.p. according to Definition 4.1. But $(\star)$ and f.c.p. essentially say the same thing. The only substantial difference is that $(\star)$ allows instances of both $\varphi$ and $\lnot \varphi$, while f.c.p. requires just instances of $\varphi$. This can be fixed by the standard trick of coding finitely many formulas (in this case $\varphi$ and $\lnot\varphi$) in a single formula.

Here are the details.

Claim 1: If $\varphi(x;y)$ has f.c.p., then it has $(\star)$.

This is the easy direction. Suppose $\varphi(x;y)$ has f.c.p. Let $n$ be an integer. By f.c.p., there is some $N\geq n$ and $a^0,\dots,a^{N-1}$ such that $$\models \lnot \exists x\, \bigwedge_{k<N}\varphi(x;a^k)$$ and for every $\ell<N$, $$\models \exists x\, \bigwedge_{k<N,k\neq \ell}\varphi(x;a^k).$$ Let $A$ be the set of all elements appearing in the tuples $a^0,\dots,a^{N-1}$, and let $p = \{\varphi(x;a^k)\mid k<N\}$. Then $p$ is inconsistent, but every subset of $p$ of size $(N-1)$ is consistent. Since $N\geq n$, every subset of $p$ of size less than $n$ is contained in a subset of size $(N-1)$, and hence is consistent. So $\varphi(x;y)$ has $(\star)$.

Claim 2: If $\varphi(x;y)$ has $(\star)$, then there is a formula $\psi(x;y,z_1,z_2)$ with f.c.p.

Let $\psi(x;y,z_1,z_2)$ be the following formula: $$((z_1 = z_2)\land \varphi(x;y))\lor ((z_1\neq z_2)\land \lnot \varphi(x;y))$$
Note that $\psi(x;a,c_1,c_2)$ is equivalent to $\varphi(x;a)$ if $c_1 = c_2$ and equivalent to $\lnot\varphi(x;a)$ if $c_1\neq c_2$.

Let $N\geq 3$ be a natural number. By $(\star)$, there exists a set $A$ in a model of $T$ and a set $p\subseteq \{\varphi(x;a),\lnot\varphi(x;a)\mid a\in A\}$ such that $p$ is inconsistent and every subset of $p$ of size less than $N$ is consistent. Since $p$ is inconsistent, by compactness some finite subset of $p$ is inconsistent. Let $q\subseteq p$ be a finite subset of minimal size which is inconsistent, and note that $|q| \geq N$. Let $n = |q|$ (and note that by taking $N$ arbitrarily large, we can make $n$ arbitrarily large). Then we can enumerate $q$ as $$\varphi(x;a^0), \dots \varphi(x;a^{m-1}), \lnot \varphi(x;a^{m}),\dots,\lnot \varphi(x;a^{n-1}).$$

Technical point: A model of $T$ must have at least two elements. Otherwise, there would be at most one tuple which is substitutable for $y$, and hence $q$ would contain at most two formulas, but we took $N\geq 3$. So we can fix arbitrary elements $c_1 \neq c_2$.

Define $b^i = (a^i,c_1,c_1)$ for each $0\leq i < m$, and define $b^i = (a^i,c_1,c_2)$ for each $m\leq i < n$. Then:
$$\models \lnot \exists x\, \bigwedge_{k<n}\psi(x;b^k),$$ because $q$ is inconsistent, and for every $\ell<n$, $$\models \exists x\, \bigwedge_{k<n,k\neq \ell}\psi(x;b^k),$$ because no proper subset of $q$ is inconsistent.


问:
Maximum of the ratio of the incircle radius to the circumcircle radius for an arbitrary triangle
For an arbitrary triangle, let $r$ = incircle radius and $R$ = circumcircle radius. Consider the ratio $r/R$? What is its maximum value? For an equilateral triangle this ratio is $1/2$. For very skewed triangles, this ratio approaches $0$. I am wondering if the maximum value is $1/2$.

This ratio is used as a measure of the quality of a triangulated surface mesh.

答:
Your claim is correct.

We use the well-known area formulae $\Delta=rs=\frac{abc}{4R}=\sqrt{s(s-a)(s-b)(s-c)}$ to get
$$\frac{r}{R}=\frac{4\Delta^2}{abcs}=\frac{4(s-a)(s-b)(s-c)}{abc}=\frac{4xyz}{(y+z)(z+x)(x+y)},$$
where $x=s-a, y=s-b,z=s-c$ are all positive. But $y+z\geq 2\sqrt{yz}$, $z+x\geq2\sqrt{zx}$, $x+y\geq2\sqrt{xy}$, so we get the result.

This result is called Euler's inequality. It is a corollary of the fact that $IO^2=R(R-2r)$, where $I$ is the incentre, $O$ the circumcentre.


问:
How to approximate $\pi$ using the Maclaurin series for $\sin(x)$
We have that
$$\sin(x)=x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
Now plugging in $x=\pi$,
$$0=\pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + \cdots$$

Is there a way we can use this beautiful result to calculate better and better approximations to $\pi$?

EDIT: This is a problem from a Mathologer video here titled Euler's real identity NOT e to the i pi = -1 at minute 7:06. I think this problem is meant to be doable, so please try it and don't just say that it is impossible and pointless, because if that were the case, Mathologer wouldn't have put it in his video (hopefully?)

答:
We can write $\pi$ like this:
$\pi = 4\cdot \frac{\Large\pi}{\Large4}=4\cdot\arctan(1)$

And use Maclaurin Series for $\arctan(x)$:
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+~...~+(-1)^n\cdot\frac{x^{2n+1}}{(2n+1)!}+~...$$
Hope my answer will help you =)


问:
Prove that $3^n>n^4$ if $n\geq8$
Proving that $3^n>n^4$ if $n\geq8$

I tried mathematical induction start from $n=8$ as the base case, but I'm stuck when I have to use the fact that the statement is true for $n=k$ to prove $n=k+1$.  Any ideas?

Thanks!

答: It can be proved by induction. The n+1 can be solved in the following way (assuming the claim is correct for n): need to be prove $3^{n+1}\geq (n+1)^4$. but we assumed that $3^n\geq n^4$ so the claim also can be written as $3*3^n\geq3*n^4\geq(n+1)^4$ but taking the 4th square on each side (both sides are non-negative), $3^\frac{1}{4}n\geq n+1\rightarrow 0.31n\geq 1$ which is correct since we prove the claim for $n\geq 8$.</p>

Title: Prove that $3^n>n^4$ if $n\geq8$


问:
Does convergence in distribution imply "convergence of supports"?
Let $P_{n}(n=1,2, \ldots)$ on $([0,1], \mathcal{B}([0,1]))$, where $\mathcal{B}([0,1])$ is the Borel $\sigma$-algebra on $[0,1]$, be a sequence of measures which converges weakly to the finite measure $P$ (i.e. $P_{n} \Rightarrow P$). My question is, for every $x \in supp\;P$, does there exist an $N$ such that for all $n \geq N$ $x \in supp\;P_n$?

My approach: $x \in supp\;P$, therefore $P((x-\delta,x+\delta))>0$ for all $\delta >0$. If $x \in C(P)$, i.e. in the continuity set of $P$, there exists $K_{x,\delta}$ such that for all $n \geq K_{x,\delta}$, $P_n((x-\delta,x+\delta))>0$. But it may so happen that for all such $n$ there exists some $\delta_n \in (0,\delta)$ such that $P_n((x-\delta_n,x+\delta_n))=0$. How do we know that there exists some $N$ such that for all $\delta >0$, $P_n((x-\delta,x+\delta))>0$ for all $n \geq N$? And this is just for points of continuity of $P$. I'm not sure how to argue for other points in the support of $P$ (such as the ones with atoms, if any).

If not "convergence" (I'm using this a bit informally, of course supports are sets and so the term convergence does not directly apply), can we say something weaker about what happens to the supports as a sequence of measures converges? They must be "somehow coming closer" but I'm not able to describe exactly how.

答:
The answer is negative.

Consider Dirac measures $P_n = \delta_{\frac{1}n} \to P = \delta_0 $. Then $supp P_n = \{ \frac1{n} \}$ and $supp P = \{ 0 \}$.


问:
Reflexive and Irreflexive
I've seen an example that goes as follows:

Let $R$ be an irreflexive, anti-symmetric, and transitive relation on a set $A$.

$A = \{1,2,3\}$

$R = \{(1,1)(1,2),(2,3)(1,3)\} $

How can the relation $R$ include $(1,1)$? Wouldn't that just make it neither reflexive or irreflexive.

I thought for relations to hold they were required to hold for all $x \in A$. Therefore it isn't irreflexive anymore because the condition $(x,x) \notin R$ is broken by $(1,1)$.

答:
As noted in the comments, you are correct. For $R$ a relation on $A$ to be irreflexive, we would require that $\forall a \in A$, $(a,a) \not \in A$. However, that is broken by $a=1$, so $R$ as given is not irreflexive.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.

Title: Reflexive and Irreflexive


Title: Find $a$ so a quadratic expression describes a straight line (problem from a Swedish 12th grade ’Student Exam’ from 1931)

Body:
$x$ and $y$ are coordinates in an orthogonal coordinate system. If the constant $a$ is suitably chosen the equation
$$9x^2-3xy+ay^2+15x-16y-14=0$$
describe two straigth lines. Find the value of $a$ and the equations for the two lines.

I tried to complete the square (using Mathematica) but the resulting expression was not helpful’. Any idea where to start? TIA.

(N.B.: The problem is quoted from a Swedish 12th grade ’Student Exam’ from 1931. I find these problems rather fun to solve in the evenings.)

Body:
$$a y^2+9 x^2-3 x y+15 x-16 y-14=0$$
collect $x$
$$9 x^2+x (15-3 y)+ay^2-16 y-14=0$$
discriminant is
$$\Delta=(15-3y)^2-4\cdot 9(a y^2-16y-14)=9(y^2-4 a y^2+54 y+81)$$
In order to  give the equation of straight lines, $\Delta$ must be  a perfect square:
$$y^2-4 a y^2+54 y+81=(3y+9)^2\text{ if } a=-2$$
the solutions are
$$x=\frac{-(15-3y)\pm 3(3y+9)}{18}$$
that is
$$x=\frac{2 (y+1)}{3};\;x=\frac{1}{3} (-y-7)$$
and finally
$$3 x-2 y-2=0;\;3 x+y+7=0$$


问:
How are we able to calculate specific numbers in the Fibonacci Sequence?
I was reading up on the Fibonacci Sequence, $1,1,2,3,5,8,13,\ldots $ when I noticed some were able to calculate specific numbers. So far I've only figured out creating an array and counting to the value, which is incredibly simple, but I reckon I can't find any formula for calculating a Fibonacci number based on it's position.

Is there a way to do this? If so, how are we able to apply these formulas to arrays?

答:
There's a perfectly general way for this sort of problem.  I used it to look for instances of $p$, such that if $p$ divided a Fibonacci number, so did $p^2$. There are none under twenty million twelftywise $20.00.00.00.00_{120}=4147200000.$

This method can be used with modular arithmetic into the hundreds of millions.

These are a number of different series, the fibonacci, Lucas series, the side and diagonal of the square approximate, and Heron's triangle.  It's written in the REXX programming language.

The actual process is to be found in the 'isoquad' routine, which extends an isoseries (ie a form $T_{n+1}= a.T_n - T_{n-1}$).  The actual equation as written gives $T_p$, given in order $a, p, T_0 ,T_1$.  It relies on that any isoseries, one can take steps of m, and thus take $T_{m+n} = a\text{^^}m \cdot T_n - T_{n-m}$.

The actual process is like this.  The example is to find T(37)  Numbers in brackets are the members of the original T(n).  We halve out the T(n), and increase the step constant to $T(2x) = T(x)^2-2$,  This is the difference between c0, c1, c2, c3.

The values in the second and third columns are as one divides 37 into binary.

```<code> diff          c0   c1   c2   c3   (role, by col 3)     (regular powers)
 (1)  18  1 : (0)  (1)  (2)  (3)   keep odd places       18 1 (1)   (1)
 (2)   9  0 : (1)  (3)  (5)        keep even places       9 0 (2)   (1) = 1+0*2
 (4)   4  1 : (1)  (5)  (9)  (13)  keep odd places        4 1 (4)   (5) = 1+1*4
 (8)   2  0 : (5)  (13) (21)       keep even places       2 0 (8)   (5) = 5+0*8
 (16)  1  0 :  (5)  (21) (37)       keep even places      1 0 (16)  (5) = 5+0*16
       0  1:   (5)  (37)        c2 = 0, so answer in c1.  0 1 (32)  (37) = 5+1*37
</code>```

The actual method is similar to finding $a^m$ by binary methods, and runs at 2/3 of the speed of said process.

By adjusting the nature of the multiplication to be a modulus, one can test to see if something like $f_{14400}$ is a multiple of 14401 or not.  The heron series, with p set to large powers of 2, is used to find fermat primes, by this very method.

```<code>fibon: ; procedure  ;parse arg t0; return isoquad(3,div(t0+2,2),mod(t0,2),1+mod(t0,2))
lucas: ; procedure  ;parse arg t0; return isoquad(3,div(t0+2,2),2-mod(t0,2),3+mod(t0,2))
qside: ; procedure  ;parse arg t0; return isoquad(6,div(t0+2,2),mod(t0,2),2+mod(t0,2)*3)
qdiag: ; procedure  ;parse arg t0; return isoquad(6,div(t0+2,2),  1  ,3+mod(t0,2)*4)
heron: ; procedure  ;parse arg t0; return isoquad(4,t0,2,4)

div:   ; procedure  ;parse arg a, b; c = a % b; if b < a*c then c = c-1 ; return c
mod:   ; procedure  ;parse arg a, b; c = a // b; if c<0 then c=c+b; return c

isoquad:  procedure
parse arg a2, a0, a3, a4
if a0 = 0 then do; a4 = a3; a0 = 1; end
if a0 < 0 then do; a0 = -a0; a4 = a3*a2-a4; end
do forever; if a0 = 1 then leave
a1 = a0//2; a5 = a4*a2 - a3;
if a1 = 0 then a4 = a5
else do; a3 = a4; a4 = a5*a2 - a3; end
a0 = a0 % 2; a2 = a2 * a2 - 2; end
return a4
</code>```

Body:
There's a perfectly general way for this sort of problem.  I used it to look for instances of $p$, such that if $p$ divided a Fibonacci number, so did $p^2$. There are none under twenty million twelftywise $20.00.00.00.00_{120}=4147200000.$

This method can be used with modular arithmetic into the hundreds of millions.

These are a number of different series, the fibonacci, Lucas series, the side and diagonal of the square approximate, and Heron's triangle.  It's written in the REXX programming language.

The actual process is to be found in the 'isoquad' routine, which extends an isoseries (ie a form $T_{n+1}= a.T_n - T_{n-1}$).  The actual equation as written gives $T_p$, given in order $a, p, T_0 ,T_1$.  It relies on that any isoseries, one can take steps of m, and thus take $T_{m+n} = a\text{^^}m \cdot T_n - T_{n-m}$.

The actual process is like this.  The example is to find T(37)  Numbers in brackets are the members of the original T(n).  We halve out the T(n), and increase the step constant to $T(2x) = T(x)^2-2$,  This is the difference between c0, c1, c2, c3.

The values in the second and third columns are as one divides 37 into binary.

 diff          c0   c1   c2   c3   (role, by col 3)     (regular powers)
 (1)  18  1 : (0)  (1)  (2)  (3)   keep odd places       18 1 (1)   (1)
 (2)   9  0 : (1)  (3)  (5)        keep even places       9 0 (2)   (1) = 1+0*2
 (4)   4  1 : (1)  (5)  (9)  (13)  keep odd places        4 1 (4)   (5) = 1+1*4
 (8)   2  0 : (5)  (13) (21)       keep even places       2 0 (8)   (5) = 5+0*8
 (16)  1  0 :  (5)  (21) (37)       keep even places      1 0 (16)  (5) = 5+0*16
       0  1:   (5)  (37)        c2 = 0, so answer in c1.  0 1 (32)  (37) = 5+1*37

The actual method is similar to finding $a^m$ by binary methods, and runs at 2/3 of the speed of said process.

By adjusting the nature of the multiplication to be a modulus, one can test to see if something like $f_{14400}$ is a multiple of 14401 or not.  The heron series, with p set to large powers of 2, is used to find fermat primes, by this very method.

fibon: ; procedure  ;parse arg t0; return isoquad(3,div(t0+2,2),mod(t0,2),1+mod(t0,2))
lucas: ; procedure  ;parse arg t0; return isoquad(3,div(t0+2,2),2-mod(t0,2),3+mod(t0,2))
qside: ; procedure  ;parse arg t0; return isoquad(6,div(t0+2,2),mod(t0,2),2+mod(t0,2)*3)
qdiag: ; procedure  ;parse arg t0; return isoquad(6,div(t0+2,2),  1  ,3+mod(t0,2)*4)
heron: ; procedure  ;parse arg t0; return isoquad(4,t0,2,4)

div:   ; procedure  ;parse arg a, b; c = a % b; if b < a*c then c = c-1 ; return c
mod:   ; procedure  ;parse arg a, b; c = a // b; if c<0 then c=c+b; return c

isoquad:  procedure
parse arg a2, a0, a3, a4
if a0 = 0 then do; a4 = a3; a0 = 1; end
if a0 < 0 then do; a0 = -a0; a4 = a3*a2-a4; end
do forever; if a0 = 1 then leave
a1 = a0//2; a5 = a4*a2 - a3;
if a1 = 0 then a4 = a5
else do; a3 = a4; a4 = a5*a2 - a3; end
a0 = a0 % 2; a2 = a2 * a2 - 2; end
return a4


问:
How to find the profit earned from getting right a set of logic statements?
The problem is as follows:

Diana offers her sister Maria $3$ bitcoins for each true statement and $4$ bitcoins for each false that she finds correctly in the statements shown, where $\mathbb{Z}$ is the universal set for the variables $x$ and $y$.

I. $\forall y; \forall x: x^2+y^2>0$

II. $\exists x; \exists y: x+y<2xy$

III. $\forall x; \exists y: x+y=0$

How many bitcoins did Maria receive, after answering all of them correctly?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{11 bitcoins}\\
2.&\textrm{12 bitcoins}\\
3.&\textrm{10 bitcoins}\\
4.&\textrm{9 bitcoins}\\
\end{array}$

How exactly this problem should be approached?. I'm stuck at the part where it comes the inequalities, from then on I don't know what to do. Can someone help me?.

The thing which I'm getting is that the set from where $x$ and $y$ belong to is $\mathbb{Z}$ hence we're talking about integers but I don't know how to use this information to further solve this problem. Can someone help me from this part in the most detailed way as possible and step by step?.

答:
To start, I found it helpful to write the terminology out in more details:

When dealing with ∀ (for all), you need only one case where it is false, since that means not all integers work

When dealing with ∃ (there exists), you only need one case where it is true, since that means there exists at least one valid answer

When combination, you are back to only needing one case where it is false but it is a bit simpler because it must be a case where the ∀ variable is the one that makes the statement impossible to resolve.

Since you mentioned you had an issue with the inequalities, let's look at techniques for those. I've used a lot of spoiler tags since it can be helpful to try to guess yourself before seeing the next conclusion.

As general tool, it helps to list out the properties of integers, such as (not an exhaustive list):

Any positive integer is greater than any negative integer

A positive times a negative is negative; a positive times a positive (including squaring) is a positive; a negative times a negative (including squaring) is positive;

Multiplying by zero gives you zero

All positive integers have one equivalent negative integer such that x + y = 0

Any integer divided by or multiplied by 1 remains the same ($1x=x$,$y/1=y$)

For ∀ and an inequality, I sometimes flip the inequality and try to prove that there exists ∃ a case for which it is true since I find ∃ simpler to deal with. In this case, that means evaluating

 $∃y;∃x:x^2+y^2≤0$

Then breaking that into two parts:

 The inequality portion, $x^2 + y^2 < 0$, is always false, because squaring any non-zero integer, positive or negative, will  give you a positive number.

 However, the equality case can be resolved to a true statement, since $x^2 + y^2 = 0$ is true if $x,y=0$.

Because there exists (∃) a case for which $x^2 + y^2 ≤ 0$, it cannot be true that for all (∀) $x,y$ the condition $x^2 + y^2 > 0$ is satisfied.

For the second inequality $∃x;∃y:x+y<2xy$, I would go back to the properties of integers. In this case,

A positive integer is always greater than a negative integer, multiplying two negative integers will always yield a positive, and adding two negative integers will results in a negative integer.

Therefore,

Taking all those into account, if x,y are any two negative integers, 2xy will be positive and x+y will be negative, and thus ∃ two integers such that $x+y<2xy$


问:
Is the relation on $\Bbb R \times\Bbb Z$ given by $(a,b) \sim (c,d) \iff (\exists t \ne0)(\exists n \in \Bbb Z)(c = at \land b^2 = 3n+d^2)$ symmetric?

Check if $R$  defined on $\mathbb R \times \mathbb Z$ is symmetric, when
$$(a,b) \sim (c,d) \iff (\exists t \ne0)(\exists n \in \mathbb Z)(c = at \land b^2 = 3n+d^2)$$

I have attempted to solve this but I got stuck very quickly. If $R$ was symmetric, then it would imply that

\begin{align*}
&(\exists t\ne0)(\exists n \in \mathbb Z)(c = at \land b^2 = 3n+d^2) \\
\implies &(\exists t_2 \ne 0)(\exists m \in \mathbb Z)(a = ct_2 \land d^2 = b^2 + 3m)
\end{align*}
Is enough to set $t_2 = \frac 1 t$ and $3n = -3m$?

答:
Ultimately, yes, that's enough. Let's rephrase this in a bit clearer of a fashion: assume the following is true:

$$(a,b) \sim (c,d), \text{ i.e. } (\exists t \in \Bbb R \setminus \{0\})(\exists n \in \mathbb Z)(c = at \land b^2 = 3n+d^2)$$

We want to show, on the premise of the above, the following is true:

$$(c,d) \sim (a,b), \text{ i.e. } (\exists t' \in \Bbb R \setminus \{0\})(\exists n' \in \mathbb Z)(a = ct' \land d^2 = 3n'+b^2)$$

To show the existence of $t',n'$, we can simply define them and observe that it works. Namely,

$$c = at \iff \frac 1 a = \frac t c \iff a = \frac 1 t c$$

which suggests the choice of $t' = 1/t$. Similarly,

$$b^2 = 3n + d^2 \iff \frac{b^2-d^2}{3} = n \iff \frac{d^2-b^2}{3} = -n$$

which suggests the choice of $n' = -n$. Of course, since $t \ne 0$, then $1/t \ne 0$; likewise, since $n \in \Bbb Z$, then $-n \in \Bbb Z$. It is also obvious by construction that these definitions of $t',n'$ otherwise satisfy the conditions, validating them.

Thus your relation is indeed symmetric.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
How to measure the width
In the attached picture, RS can be determined as follows:
RS is obtained from fitting the function $$0.5(1+\operatorname{erf}((I\%-50)/RS))$$ where $I\%$ is the current pulse amplitude as percentage of threshold and $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x}e^{-\lambda^2}d \lambda$.
It is about determining the spread of a curve.

答:
Yes, the relative spread as defined by the equation
$$0.5(1+\operatorname{erf}((I\%-50)/RS))$$
is proportional to the coefficient of variation. Your alternative formulation also leads to the coefficient of variation; a more direct approach to obtaining RS is to construct the tangent line at the point where the probability is 50%. See this answer for a comprehensive discussion.

Title: How to measure the width


问:
Application of Induction in the analyze of the convergence a sequence defined recursive.

Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad  n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.

My approach: Note that  $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly  I would like to find $a_{n}$. Now, I was trying to find a  pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.

How can find $a_{n}$?

Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.

答:
If the sequence converges to $L$, taking the limit on both sides of the recurrence shows that

$$L=\frac1{4-3L}\,,$$

or $3L^2-4L+1=0$. The quadratic factors nicely: $(3L-1)(L-1)=0$, so the only possible limits are $L=\frac13$ and $L=1$.

Clearly the sequence is undefined if $a_1=\frac43$ and constant if $a_1=\frac13$ or $a_1=1$.

If $a_k<1$, then $1<4-3a_k$, and $0<a_{k+1}<1$.

If $a_k>\frac43$, then $a_{k+1}<0$, so $0<a_{k+2}<1$.

If $1<a_k<\frac43$, let $r=a_k-1$; then $0<3r<1$, so $$a_{k+1}=\frac1{4-3a_k}=\frac1{1-3r}=\sum_{n\ge 0}(3r)^n>1+3r>a_k\,.$$ The sequence cannot have a limit in $\left(1,\frac43\right]$, so either it hits $\frac43$ and dies, or $a_\ell>\frac43$ for some $\ell>k$, and then $a_n\in(0,1)$ for all $n\ge\ell+2$.

Thus, if $a_1$ actually generates an infinite, non-constant sequence, that sequence ends up in $(0,1)$. What happens there?

If $\frac13<a_k<1$, let $r=a_k-\frac13$. Then $$a_{k+1}=\frac1{4-3a_k}=\frac1{3(1-r)}=\frac13\sum_{n\ge 0}r^n\,,$$ so $$a_{k+1}-\frac13=\frac13\sum_{n\ge 1}r^n=\frac{r}3\sum_{n\ge 0}r^n=ra_{k+1}<r=a_k-\frac13\,,$$
and $a_{k+1}<a_k$. In this case the sequence must converge to $\frac13$.

If $0<a_k<\frac13$, let $r=\frac13-a_k$. Then $$a_{k+1}=\frac1{4-3a_k}=\frac1{3(1+r)}=\frac13\sum_{n\ge 0}(-1)^nr^n\,,$$ so $$\begin{align*}\frac13-a_{k+1}&=\frac13-\left(\frac13+\frac13\sum_{n\ge 1}(-1)^nr^n\right)=\frac13\sum_{n\ge 0}(-1)^nr^{n+1}\\&=\frac{r}3\sum_{n\ge 0}(-1)^nr^n=ra_{k+1}<r=\frac13-a_k\,,\end{align*}$$ and $a_{k+1}>a_k$. Again the sequence converges to $\frac13$.

We have now shown that $a_1=1$ yields the constant sequence $a_k=1$ for all $k\ge 1$, and every other initial value yields either a sequence converging to $\frac13$ or one that eventually dies because some $a_k=\frac43$. It only remains to determine for which initial values some $a_k=\frac43$.

Solving $y=\frac1{4-3x}$ for $x$, we find that $x=\frac{4y-1}{3y}=\frac43-\frac1{3y}$. Let $b_1=\frac43$, and for $k\ge 1$ let $b_{k+1}=\frac{4b_k-1}{3b_k}$. It’s easy to show by induction on $k$ that $a_k=\frac43$ if and only if $a_1=b_k$, so $\{b_k:k\ge 1\}$ is the set of initial values that do not yield a convergent sequence, and it only remains to find a closed form for the numbers $b_k$.

If we write $b_k$ as a fraction $\frac{c_k}{d_k}$, then

$$b_{k+1}=\frac{\frac{4c_k}{d_k}-1}{\frac{3c_k}{d_k}}=\frac{4c_k-d_k}{3c_k}\,,$$

so $c_{k+1}=4c_k-d_k$, and $d_{k+1}=3c_k$, with initial conditions $c_1=4$ and $d_1=3$. Then $c_{k+1}-d_{k+1}=c_k-d_k$, so by induction $c_k-d_k=c_1-d_1=1$ for all $k\ge 1$. It follows that $c_{k+1}=d_{k+1}+1=3c_k+1$. Solving the recurrence $c_{k+1}=3c_k+1$ with initial value $c_1=4$ by any standard method, we find that

$$c_k=\frac{3^{k+1}-1}2$$

and hence that

$$d_k=\frac{3^{k+1}-3}2\,,$$

so that

$$b_k=\frac{3^{k+1}-1}{3^{k+1}-3}\,.$$


答: Another way to see this is to actually produce a slice chart about $p=(1,0).$ Note that if $f(x,y)=x^2+y^2-1$ then $\partial f_x(1,0)=2\neq 0$ so if we set $\phi(x,y)=(f,y)$ then, we get a slice chart in a small neighborhood $p\in U\mathbb \subseteq R^2.$ That is, $(\phi|_{S^1},U\cap S^1)$ is a chart about $p$ in $S^1$. But $\phi|_{S^1}(f,y)=(0,y)\cong y$ and so $T_pS^1=\operatorname {span} \{\partial_y\}.$

Title: embedded submanifold in euclidean space derivative

Body:
Let $y$ be the coordinate function on $\mathbb{R}^2$. Consider $\mathbb{S}^1$ as an embedded submanifold. If $p\in (1,0)$, how do I show that $\frac{\partial}{\partial{y}}|_p\in T_p\mathbb{S}^1$? It clearly is in $T_p\mathbb{R}^2$.

My attempt: Since $\mathbb{S}^1$ is defined by the smooth function $F(x,y)=x^2+y^2-1$,

$T_p\mathbb{S}^1=Ker(F_{*,p})$ for all $p\in \mathbb{S}^1$.
Hence if $h\in C^{\infty}$, $F_{*,p}(\frac{\partial}{\partial{y}}|_p)h=\frac{\partial{h\circ F}}{\partial{y}}|_p=0$

where the last equality holds because $\frac{\partial{F}}{\partial{y}}|_p=0$

Question 2: In Loring Tu's book, does the differentiable structure as constructed in Proposition 9.4, coincide with the differentiable structure as constructed in example 5.16 for the sphere?


问:
Calculate probability from marginal probability
Suppose $X,Y$ two random variables following the uniform distribution $U(0,1)$. The cdf $F(x,y)$ is known and equal to $\max {(x+y-1,0)}$.
I want to calculate the probability $P(XY \leq a)    \text{   }\forall a\in(0,1)$.

From the cdf $F(x,y)$, we can deduce the probability $P(X \leq u,Y = v)$ as following
$$P(X \leq u,Y = v) = \frac{\partial P(X \leq u,Y \leq v)}{\partial y}=\frac{\partial F(u,v)}{\partial y} =\mathbb{I}_{(u+v-1 \geq 0)} $$
where $\mathbb{I}_{(\mathbf{A})}$ is the indicator function of the event $\mathbf{A}$.
So,
$$
\begin{equation}
\begin{split}
P(XY \leq a) & = \int_{y\in [0,1]}P(XY\leq a,Y=y)dy \\
             & = \int_{y\in [0,1]}P(X\leq \frac{a}{y},Y=y)dy \\
             & = \int_{y\in [0,1]}\mathbb{I}_{(\frac{a}{y}+y-1 \geq 0)}dy \\
\end{split}
\end{equation}
$$

For $a \geq \frac{1}{4}$, we have $\frac{a}{y}+y-1 \geq 2\sqrt{\frac{a}{y}y}-1 \geq 2\sqrt{\frac{1}{4}}-1 =0$, so
$$P(XY \leq a) = \int_{y\in [0,1]}\mathbb{I}_{(\frac{a}{y}+y-1 \geq 0)}dy = \int_{y\in [0,1]}1dy = 1$$

In particular, $P(XY<0.5) = 1$  !!!

Where is my error? And how we calculate correctly the probability $P(XY<a) ,  \forall a\in (0,1)$

Thank you in advance.

答:
As kimchi lover pointed out, you don't have an error.

It may be easier to recognize that the CDF implies $Y=1-X$. Then the probability you seek is $P(X(1-X) \le a)$. This event is equivalent to $X^2 - X + a \ge 0$, which is equivalent to $(X-\frac{1}{2})^2 \ge \frac{1}{4} - a$. (Note that rearranging your inequality $\frac{a}{y} + y - 1 \ge 0$ leads to the same quadratic.)

When $a \ge 1/4$, then any $X \in [0,1]$ satisfies this, so indeed the probability is $1$.

When $0 \le a \le 1/4$, then $X \notin \left[\frac{1}{2} - \sqrt{\frac{1}{4} - a},\frac{1}{2} + \sqrt{\frac{1}{4} - a}\right]$ works, so the probability is $1 - 2 \sqrt{\frac{1}{4} - a}$.


问:
To what ring is $\mathbb{Z}[X,Y,Z]/(X-Y, X^3-Z)$ isomorphic?
The problem:

Let $(\mathbb{Z}[x,y,z],+,\cdot)$ be the ring of polynomials with coefficients in $\mathbb{Z}$ in the variables $x$, $y$ and $z$ and the obvious operations $+$ and $\cdot$. Let $(x-y, x^3-z)$ be the ideal generated by $x-y$ and $x^3-z$. Find the ring to which $\mathbb{Z}[x,y,z]/(x-y,x^3-z)$ is isomorphic.

A similar question has been asked before: Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?. Using the same approach, we set $x=y$ and $x^3=z$ and substitute for $y$ and $z$. Then we find $\mathbb{Z}[x,x,x^3]$ which is just $\mathbb{Z}[x]$. I'm trying to find out why this approach would work, i.e., formalise it in a proper proof. My idea is to use the first isomorphism theorem, but I get stuck near the end. I would very much appreciate if someone could help fill in this gap, check my proof for mistakes or provide additional (intuitive) information as to why the above method works.

My proof:

Consider the map $\varphi : \mathbb{Z}[x,y,z] \to \mathbb{Z}[x] : p(x,y,z) \mapsto p(x,x,x^3)$.

This is a homomorphism, since it obeys the relations $\varphi(p) + \varphi(q) = \varphi(p+q)$ and $\varphi(p) \cdot \varphi(q) = \varphi(p \cdot q)$ for all $p,q \in \mathbb{Z}[x,y,z]$ (verifying these is quite trivial).

Now take any $q \in \mathbb{Z}[x]$ and define a $p \in \mathbb{Z}[x,y,z]$ such that $p(x,y,z) = q(x)$. Since $\varphi(p(x,y,z)) = p(x,x,x^3) = q(x)$, $\varphi$ is surjective and it follows that $\mathbb{Z}[x] \subseteq \varphi(\mathbb{Z}[x,y,z])$. It is obvious that $\varphi(\mathbb{Z}[x,y,z]) \subseteq \mathbb{Z}[x]$ and, therefore, we must have $\varphi(\mathbb{Z}[x,y,z]) = \mathbb{Z}[x]$.

Lastly, we must show that $\ker(\varphi) = (x-y, x^3-z)$. For brevity's sake I will denote $I := (x-y,x^3-z).$

Let $p \in I$. Then there must exist $q_1, q_2 \in \mathbb{Z}[x,y,z]$ such that $p = q_1(x-y) + q_2(x^3-z)$. It follows that:
\begin{align*}\varphi(p(x,y,z)) &= \varphi(q_1(x,y,z)(x-y) + q_2(x,y,z)(x^3-z))\\ &= q_1(x,x,x^3)(x-x) + q_2(x,x,x^3)(x^3-x^3) = 0.\end{align*}
Thus $p \in \ker(\varphi)$ and $I \subseteq \ker(\varphi)$, since $p$ was chosen arbitrarily in $I$.

Now, take a $q \in \ker(\varphi)$. Then $\varphi(q(x,y,z)) = q(x,x,x^3) = 0$. Somehow we must get that $q \in I$ so that we can conclude $\ker(\varphi) \subseteq I$ and thus $I = \ker(\varphi)$, but I seem to get stuck on every path. Proof by contradiction or by contraposition also seems fruitless. What am I missing?

With all of the above, we may apply the first isomorphism theorem to conclude that:
$$\mathbb{Z}[x,y,z]/I = \mathbb{Z}[x,y,z]/\ker(\varphi) \cong \varphi(\mathbb{Z}[x,y,z]) = \mathbb{Z}[x]$$

答: Hint: Note that $q$ is congruent to $q(x,x,x^3)$ modulo $I$.


问:
degree field extension by a power of transcendental element
Basic field theory exercise:

Let $K$ be a field and $x \in K$ transcendental.
Firstly I need to prove that $x^n$ is transcendental too. I proved that the map $ev_{x^n}:K[T] \longrightarrow K[x]$ is injective as a consequence of the injectivity of $ev_x$.

Now I need to show that $[K(x):K(x^n)]=n$

I'm still a bit confused by the definition of $K(x):=\{ \frac{f(x)}{g(x)}:f(T),g(T) \in K[T] \} \cong \{ \frac{f(T)}{g(T)}: f,g \in K[T] \}=K(T)$.
I see that every element of $K(x):=\{ \frac{f(x^n)}{g(x^n)}:f(T),g(T) \in K[T] \}$ is "associated to an element of degree multiple of $n$ in K(T)".

I believe that basis of $K(x)$ over $K(x^n)$ looks like $B:=\left(\frac{f(x)}{g(x)}, x \cdot \frac{f(x)}{g(x)},x^2 \cdot \frac{f(x)}{g(x)},..., x^{n-1} \cdot \frac{f(x)}{g(x)} \right)$. Since $x^k=x^{k'} \iff k=k'$ linear independence should be clear. Showing that $=K(x)$ looks more complicated but would anyway follow from $|B|=n$.

If I'm not mistaken the solution is pretty much there but I'm still quite confused so I would really appreciate any hint on how to make it more formal.

答:
Recall that if $F/E$ is an extension such that $F = E(\alpha)$ for some $\alpha \in F$, then $F/E$ is a finite extension whenever $\alpha$ is algebraic over $E$; moreover, if $p(T) \in E[T]$ is the minimal polynomial of $\alpha$ over $E$, then $[F:E] = m := \deg p(T)$ and $1,\alpha,\dots,\alpha^{m-1}$ is a $E$-basis for $F$.

Now, notice that $K(x) = K(x^n)(x)$,  and so it suffices to find an irreducible monic polynomial $p(T) \in K(x^n)[T]$ with $n$-degree such that $p(x) = 0$. In fact, you can consider $p(T) := T^n-x^n$, and once you have shown that $p(T)$ is irreducible, it follows that a $K(x^n)$-basis for $K(x)$ is given by $1,x,\dots,x^{n-1}$.


问:
Do I need to prove a function between two topological spaces $X, Y$ is onto to prove it induces a function on $Y$?
If there is a mapping $f: X\to Y$ between two topological spaces and $X$ has a topology $T_X$ then $Y$ will have a topology $T_f$ induced by $X$ and function $f$. But recently I got confused by something my teacher said while explaining identification functions:

"If I have a mapping from $X$ to $Y$ and $X$ has a topology and if I were able to prove the mapping was onto then I can say there is a topology $T_f$ induced on $Y$ by $T_X$ and $f$ and then we can complete our prove for identification function".

So my question is, why do I need the function to be onto in order to have an induced topology on $Y$? Now I know I need onto condition to prove identification function but I am asking for the induced topology itself why do i need onto condition for that?

答:
Another way to see why surjectivity of the function is involved is that if you have a function $$f:X \rightarrow Y$$ you can form an equivalence relation given by the function identifying $x \sim y$ if and only if $f(x) = f(y)$. Then you see immediately that it is an equivalence relation and so you can form the quotient set $X /\sim$ and so you have the projection onto the quotient that is a surjective function $\pi : X \rightarrow X /\sim$.

As Hitman Reborn correctly pointed out, this is sufficient to define the identification topology via its definition. Nevertheless, note that to have a surjective function is not necessary; it only gives a natural definition to induce a topology on the quotient set, that is to define a topology on it that makes the projection map continuous.


问:
Proof by induction for a sequence, given $a_1$ and $a_2$
Question:  prove by induction that this statement holds.
Define  the sequence $(a_{k}),  k∈\mathbb N ,$
by $a_1 = 2, a_2 = 5 $
and
$a_{k+2} = 5a_{k+1} − 6a_{k} $ for all $k≥1$. Then $a_{k} =2^{n−1} +3^{n-1}$ for all $n≥1 $.

What I did was find the base case -- $ P(1)=2 $ -- holds
Then made an assumption that $P(k)$ holds.  Thus we can assume $P(k+1)$ and I got that
$= 5×2^{k−1} +5×3^{k−1} −3×2^{k−1} −2×3^{k−1}.$

Where do I go from here?

Body:
Say $P(n)$ is $a_n=2^{n-1}+3^{n-1}$.

Base:  $P(1):  a_1=2=2^0+3^0$, $P(2):  a_2=5=2^1+3^1$.

Now assuming $P(k)$ and $P(k+1)$, we have

$P(k+2):  a_{k+2}=5a_{k+1}-6a_k=5(2^k+3^k)-6(2^{k-1}+3^{k-1})$

$=(5-3)2^{k}+(5-2)3^{k}=2^{k+1}+3^{k+1}.$


问:
Why does the addition of generic ultra-filters in forcing not lead to inconsistent theories?
Apologies if this is a dumb question - I’m trying to dig into the intricacies of set-theoretic forcing at the moment and am struggling with a step.

In particular, given our poset P we need to show that our generic ultra-filter G does not already exist in our transitive model M. I  think I understand the argument around this - we can show the subset P\G is also dense and clearly G cannot intersect this, so no such G exists in our model on pain of inconsistency

My (probably naive) confusion is about why, when we build our extended model M[G], can we not run a similar argument to show there is a dense subset of P that G does not intersect? My intuition is that the “generic-ness” of G is not a property we require G to have in M[G], but rather a property we require G to have to guarantee it does not exist in M (and that it has certain filter properties etc), and once we extend M to include G we no longer care whether it intersects every dense subset of P in M[G]

Is this intuition correct or am I missing something fundamental in the argument?

答:
Yes, $G$ meets every dense set in $M$ but not in $M[G]$. So $G$ is $\mathbb{P}$-generic over $M$, but not $\mathbb{P}$-generic over $M[G]$, and that's fine.

(Incidentaly, it's "generic filter," not "generic ultrafilter," in the context of forcing-via-posets; ultrafilters emerge when we use the Boolean algebras approach to forcing.)


问:
Free abelian group $F_i$ with free basis $B_i$
I have to prove this statement:

Let free abelian group $F_i$ with free basis $B_i$ for $i=1,2$ then $F_1\cong F_2$ iff $\lvert B_1 \rvert = \lvert B_2 \rvert$.

I prove it by the idea that element of basis is linear independent and they generate whole group. So we have injection $F_1= \langle B_1 \rangle \rightarrow \langle B_2\rangle=F_2$, WLOG $\lvert B_1 \rvert \leq \lvert B_2 \rvert$.

So now is trivial both implication.

My question is can I do use this idea for proving?

答:
If you accept that vector spaces over a field have a well-defined dimension, we can prove this using that result as follows:

Theorem. Let $F$ be a free abelian group, and let $B$ be a basis for $F$. Let $G=F/2F$; then $G$ is a vector space over $\mathbb{F}_2$, the image of $B$ is a basis for $G$ as a vector space, and the restriction of the canonical map $\pi\colon F\to G$ to $B$ is one-to-one. In particular, $|B|=\dim_{\mathbb{F}_2}(G)$.

Proof. $G$ is certainly an abelian group of exponent $2$; that means it is a vector space over $\mathbb{F}_2$. If $b_1,b_2\in B$ and $\pi(b_1)=\pi(b_2)$, then there exists $x\in G$ such that $b_1=b_2+2x$. Expressing $x$ in terms of the basis $B$ and using uniqueness of the representation we conclude that $x=0$ and $b_1=b_2$, so that $\pi|_{B}$ is injective.

Now, $B$ spans $F$, hence $\pi(B)$ spans $G$. To show that $\pi(B)$ is linearly independent over $\mathbb{F}_2$, let $b_1,\ldots,b_n\in B$ be pairwise distinct, and let $a_1,\ldots,a_n\in\mathbb{F}_2$ be such that
$$a_1\beta_1+\cdots+a_n\beta_n = 0,$$
where $\beta_j=\pi(b_j)$. We want to show that $a_1=\cdots=a_n=0$. Let $r_i=0$ if $a_i=0$, and let $r_i=1$ if $a_i\neq 0$. Then $r_1b_1+\cdots+r_nb_n\in F$ is such that $\pi(r_1b_1+\cdots+r_nb_n)=a_1\beta_1+\cdots+a_n\beta_n$. Since the image is trivial, there exists $x\in F$ such that $r_1b_1+\cdots+r_nb_n=2x$. Again, expressing $x$ in terms of $B$ and using the uniqueness of the representation we conclude that all $r_i$ must be even, and hence equal to $0$. Thus, $a_i=0$ for all $i$, so $\beta_1,\ldots,\beta_n$ are linearly independent. This proves that $\pi(B)$ is linearly independent, and hence is a basis. $\Box$

Now the result follows easily: $F_1/2F_1\cong F_2/2F_2$ as abelian groups, hence as vector spaces. Thus, $|B_1|=\dim_{\mathbb{F}_2}(F_1/2F_1) = \dim_{\mathbb{F}_2}(F_2/2F_2) = |B_2|$. The converse is trivial.


问:
Identities with quadratic forms
Let $y_1,\dots, y_n$ be vectors in $\mathbb{R}^p$ and let $\mu$ be an additional vector in $\mathbb{R}^p$. Let $M\in\mathbb{R}^{p\times p}$ be a symmetric positive definite matrix. I have some difficulties into prove the following identity
\begin{equation}\sum_{i=1}^n (y_i-\mu)'M^{-1}(y_i-\mu)=n(\bar{y}-\mu)'M^{-1}(\bar{y}-\mu)+\sum_{i=1}^n (y_i-\bar{y})'M^{-1}(y_i-\bar{y})\end{equation}
where
\begin{equation}\bar{y}\triangleq \frac{1}{n}\sum_{i=1}^n y_i\end{equation}
I've tried to expand the generic product $(y_i-\mu)'M^{-1}(y_i-\mu)$ but this seems to me useless to prove the previous identity.

答:
Hint You could replace $\mu$ by $t\mu$. It is clear that the coefficients of $t$ and $t^2$ are the same on both sides of the equality. It remains to prove the equality of the constant terms (w.r.t $t$), that is to say the same equality without $\mu$.

The equality that is left resembles the equation $E(X^2) = E(X)^2 + \text{Var}(X)$ in probability. The proof is similar by using $\frac{1}{n}\sum_{i=1}^n$ instead of $E$.

Title: Identities with quadratic forms


问:
Bound on subsequent terms of inductively defined sequence
Let $(a_n)$ be the sequence defined by $a_0=1$ and $a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}$. Show that $|a_{n+1}-a_n|\leq(\frac{1}{2})^{n+1}$.

Using the induction hypothesis I can show that

$$|a_{n+1}-a_n|\leq \frac{|a_n-a_{n-1}|}{2} + |\frac{1}{a_{n-1}}- \frac{1}{a_n}|\leq (\frac{1}{2})^{n+1} + |\frac{1}{a_{n-1}}- \frac{1}{a_n}|.$$

I cannot get rid off the extra term and I have tried different approaches, but this one seems like the most promissing one because the bound appears naturally. I think I am missing something obvious.

Thank you.

Body:
Assume you want $|a_{n+1}-a_n| \le \frac{1}{2^{n+1}}$

First $a_n\ge 0$. If $n\ge 1$ then $a_n \ge 2\sqrt{\frac{a_{n-1}}{2} \cdot \frac{1}{a_{n-1}}} = 2\sqrt{\frac 12}=\sqrt 2$ via AM-GM.

$a_{n+1} - a_n = \frac{1}{a_n} - \frac{a_n}{2}=\frac{2-a_n^2}{2a_n} \le 0$

Lastly,
$$a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}, a_n=\frac{a_{n-1}}{2} + \frac{1}{a_{n-1}}\\
\implies a_{n+1} - a_n = (a_n-a_{n-1}) \left( \frac 12 - \frac{1}{a_n a_{n-1}}\right) \\
\implies |a_{n+1} - a_n | \le \frac 12|a_n-a_{n-1} | \le \frac{1}{2^n}|a_1-a_0|=\frac{1}{2^{n+1}}$$

Note: we have $0 \le \frac 12 - \frac{1}{a_n a_{n-1}} \le \frac 12$. LHS because $a_n\ge \sqrt 2 $, RHS because $a_n>0$


问:
Does convergence in the $r$th mean for all $r$ imply almost sure convergence?
Almost sure convergence of a sequence of random variables and convergence in the $r$th mean (for any $r$) each imply convergence in probability. However, (I believe) neither of them implies the other for fixed $r$.

What about convergence in the $r$th mean for all $r$? This implies almost sure convergence, correct?

答:
The answer is negative.

Consider $\Omega = [0,1]$ with Lebesgue measure. Put $A_1 = [0,1]$, $A_2 = [0, \frac12]$, $A_3 =  [\frac12, 1]$, $A_4 = [0, \frac14]$. $A_5 = [\frac14, \frac12]$, $A_6 = [\frac12, \frac34]$, $A_7 = [\frac34,1]$, $A_8 =[0,\frac18]$ and so on. Put $f_n = I_{A_n}$. Hence $f_n \to 0$ in every $L_r$, $r \ge 1$ and $f_n$ doesn't converge to $0$ (or anywhere else) a.s.


答:
This is not true, because of the following lemma (cf. Wikipedia):

If $(X_n)_n$ converges in probability to $X$, and if $\mathbb{P}\{|X_n| \leq b\} = 1$ for all $n$ and some $b$, then $(X_n)_n$ converges in $r$th mean to $X$ for all $r \geq 1$.

In particular, this applies to the bounded sequence $(X_n)_n$ with $X_n$ being a Bernoulli with parameter $1/n$. But converges in probability to $0$, yet not almost surely.

Title: Does convergence in the $r$th mean for all $r$ imply almost sure convergence?

Body:
Almost sure convergence of a sequence of random variables and convergence in the $r$th mean (for any $r$) each imply convergence in probability. However, (I believe) neither of them implies the other for fixed $r$.

What about convergence in the $r$th mean for all $r$? This implies almost sure convergence, correct?


问:
Corrections to the $\delta$ limit of the Cauchy distribution
The Cauchy distribution
$$
C(x) = \frac{1}{\pi} \frac{\gamma}{(x-\omega)^2 + \gamma^2}
$$
approaches the Dirac delta distibution as $\gamma \to 0$. What is the $O(\gamma)$ correction to this? I am thinking it should involve a Cauchy principal value, based on considering two cases:

Case 1: $x \ne \omega$

Off resonance, $C(x)$ has a convergent series expansion for $|\gamma| < |x - \omega|$
$$
C(x) = \frac{1}{\pi} \frac{\gamma}{(x-\omega)^2} + O(\gamma^3)
$$

Case 2: $x = \omega$

On resonance, $C(x)$ diverges as $\gamma \to 0$ as
$$
C(x) \sim \frac{1}{\pi\gamma}
$$
But this is the same divergent behavior that leads us to identify the Dirac delta distribution as the $\gamma \to 0$ limiting behavior.

These two cases lead me to want to write
$$
C(x) \sim \delta(x-\omega) + \frac{\gamma}{\pi} P \frac{1}{(x-\omega)^2}
$$
where $P$ means to take the Cauchy principal value. But if I use this in expansion in real calculations, will I recover correct $O(\gamma)$ behavior? Or have I made some subtle mistake?

答:
Let $\varphi$ be a test function. Then
$$
\int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\gamma}{(x-\omega)^2 + \gamma^2} \varphi(x) \, dx
= \{ x=y+\omega \}
= \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\gamma}{y^2 + \gamma^2} \varphi(y+\omega) \, dy
\\= \{ y=\gamma z \}
= \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\gamma}{\gamma^2z^2 + \gamma^2} \varphi(\gamma z+\omega) \, \gamma\,dz
= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \varphi(\gamma z+\omega) \,dz
\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \left(\varphi(\omega)+\gamma z\varphi'(\omega)+O((\gamma z)^2)\right) \,dz
\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \varphi(\omega) \, dz
+ \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \gamma z\varphi'(\omega)\,dz +O(\gamma^2)
.
$$

The first term is the $\delta$ term:
$$
\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \varphi(\omega) \, dz
= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \, dz \, \varphi(\omega)
= \varphi(\omega)
$$
since $\int_{-\infty}^{\infty} \frac{1}{z^2+1} \, dz = \pi.$

The term linear in $\gamma$ is
$$
\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \gamma z\varphi'(\omega)\,dz
= \frac{\gamma}{\pi} \int_{-\infty}^{\infty} \frac{z}{z^2+1}\,dz\, \varphi'(\omega)
= 0
$$
since $\int_{-\infty}^{\infty} \frac{z}{z^2+1}\,dz=0.$


问:
Factorize in R[x]
I have the polynomial $x^8+1$, I know that there's no root for solve this in $\Bbb R[x]$ but i want to factorize this to the minimal expression. This is possible or this is irreducible?</p>

答:
The answer depends on $R$.

Example $1$: $R=\mathbb{C}$ and you know that any polynomial, including
the one in question, splits into linear factors

Example $2$: $R=\mathbb{Q}$ denote $$f(x)=x^{8}+1$$ then $$f(x+1)=(x+1)^{8}+1=x^{8}+8x^{7}+28x^{6}+56x^{5}+70x^{4}+56x^{3}+28x^{2}+8x+2$$
is irreducible by Eisenstein with $p=2$ hence $f(x)$ is irreducible.

Title: Factorize in R[x]


问:
Intuitive explanation of tangent space of an open set of $\mathbb{R^n}$
My differential geometry professor said that the tangent space of an open set $\Omega$ of $\mathbb{R^n}$ at a point $P$ coincides with $\mathbb{R^n}$ , that is

$T_P(\Omega)=\mathbb{R^n}$

He gave an intuitive example for n=2:  if $\Omega$ is any region in $\mathbb{R^2}$, let'say a disc centerend in the origin, and $P$ any point in the disc, the tangent space if the set of all tangent vectors to  the curves whose traietories pass by $P$ and are contained in the plane. cleary the $T_P(\Omega)$ consides with the plane $\mathbb{R^2}$ itself.

I tried to do my own example for $\mathbb{R^3}$,  according to that is not true that $T_P(\Omega)=\mathbb{R^3}$. If $\Omega$ is half a spheric surface,  and $P$ any point on the surface, It is absurd to say that the tangent plane to the sphere is $\mathbb{R^3}$, since is it a PLANE! as it is obvious form the picture,
so $T_P(\Omega)=\mathbb{R^2}$

What I am getting wrong? Please refrain from talking about manifolds , varities or technical stuff. These are not part of my course

答:
You have embedded a two dimensional object into a three dimensional space however when you move to a higher dimensional space open sets in the lower dimensional space may not be open anymore. A simple example would be the interval $(0,1)$ in $\mathbb{R}$ which is open as it contains all of its interior points, however a line segment is never open in $\mathbb{R}^2$ because it has no interior points.

In your example the open sets would be on the surface of the sphere which are not open in the ambient space. This is for the same reason, every neighborhood around every point will contain points not on the sphere and so it has no interior points, and is therefore not open. As you've correctly deduced this detail is non-trivial otherwise your get a mismatch in dimension.


问:
How to solve the equation $x^2-2y=z^2$?
Consider the following equation
$$x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can help me. Thanks in advance(
I tried to rewrite the equation as follows
\begin{align}
2y=x^2-z^2 \Rightarrow 2y=(x-z)(x+z)\Rightarrow y=\frac{(x-z)(x+z)}{2}
\end{align}
but it did not work).

答:
The original equation suggests that $x,y,z$ are components $C,B,A$ of a Pythagorean triple except the y-term is not generated directly. We will accept $2y=B^2$.
$$x^2-2y=z^2\implies x^2=2y+z^2$$
There are infinite solutions. One method of generating them is with Euclid's formula (modified):

$$ z=A=m^2-k^2\qquad y=\frac{B^2}{2}=\frac{(2mk)^2}{2}=2(mk)^2\qquad x=C=m^2+k^2$$
where $m>1$ and $m>k$. Since we do not care about primitives, we can ignore the normal restriction of $GCD(m,k)=1$.

What this means is that, for any Pythagorean triple such as $(A,B,C)=(3,4,5)$, the solution we seek will be

$(C,\frac{B^2}{2},A)=(5,8,3)\quad$ or $\quad (5,12,13)\rightarrow (13,72,5)\quad$ or $\quad (15,8,17)\rightarrow (17,32,15)$

Another formula which generates only non-trivial positive $A,B,C$ components is

$$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$

This one works with any pair of natural numbers $n,k$ and generates only primitives and odd square multiples of primitive Pythagorean triples. The translation is the same for either formula: $(x,y,z)=(C,\frac{B^2}{2},A)$

Here is a list of $x,y,z$ where $x<100$

$$
(5,8,3)\\
(13,72,5)\quad
(17,32,15)\quad
(25,288,7)\quad
(29,200,21)\quad
(37,72,35)\quad
(41,800,9)\\
(45,648,27)\quad
(53,392,45)\quad
(61,1800,11)\quad
(65,128,63)\quad
(65,1568,33)\\
(73,1152,55)\quad
(85,3528,13)\quad
(85,648,77)\quad
(89,3200,39)\quad
(97,2592,65) $$

Given Pythagorean triple properties, we can expect $z$ to be a multiple of any odd number greater than one. Also, since $B$ can be any odd multiple of $4, B=4r\implies y=8r^2, r\in\mathbb{N}$. Not all combos will work but those that do work will be of this form.

Title: How to solve the equation $x^2-2y=z^2$?


问:
How to define a group object given an object $a$ in a finitely complete category $C$
Let $C$ be a finitely complete category and $a\in C$. Since $C$ is finitely complete, there exists (i.e. pullback) unique morphisms $m,n:a\times a\rightarrow a$ such that $m\circ1_a=n\circ 1_a$: i.e. $m=n$. We can take this $m$ to be our "multiplication". There is also a unique morphism $e:a\rightarrow1$ where $1$ is our terminal object; this can serve as our "multiplicative identity". How can I derive an $i:a\rightarrow a$ so that $(a,m,i,e)$ forms a group object in $C$?

I have used the fact that $C$ allows pullbacks to derive $m$. On the other hand, I have used the fact that $C$ possesses a terminal object to find $e$. What property gives me $i$?

答:
As in the comments, you've messed up the direction of the unit; it's a map $e : 1 \to a$, and such a map need not exist in a finitely complete category. An easy example is to take the category of $G$-sets for $G$ a nontrivial group; there a map $1 \to X$ for a $G$-set $X$ corresponds to a fixed point, so we can take for example $X = G$ being acted on by left multiplication, which has no fixed points and hence no possible unit map.

In most categories with finite products (this is all you need to define group objects, and actually with a little fiddling you don't even need this), most objects don't have a group object structure; it's a quite special feature of $\text{Set}$ that (assuming the axiom of choice) every set has a group structure. For example, in $\text{Top}$ a group object is a topological group and the underlying topological space is heavily constrained, e.g. it must be homogeneous, its fundamental group must be abelian, etc.


问:
Simplifying this expression $(e^u-1)(e^u-e^l)$
Is it possible to write the following

$$(e^u-1)(e^u-e^l)$$

as

$$e^{f(u,l)}-1?$$

答:
Yes, it is possible. Define
$$f(u,l) = \log(1+(e^u-1)(e^u-e^l))$$
We then have
$$e^{f(u,l)}-1 = (e^u-1)(e^u-e^l)$$

EDIT

The above is valid/defined provided $(e^u-1)(e^u-e^l) > -1$.


问:
The Geometric Mean as a Minimum.
This is problem 8.5 from the book "Cauchy Schwarz Master Class":

Problem: Prove that the geometric mean has the representation $$\left(\prod_{k=1}^n a_k \right)^{1/n} = \min\left\{\frac{1}{n} \sum_{k=1}^n a_kx_k:(x_1,\cdots,x_n)\in D\right\},$$
where $a_k \geq 0$ and $D$ is the region of $\mathbb{R}^n$ defined by $$D = \left\{(x_1,\cdots,x_n): \prod_{k=1}^n x_k =1, x_k \geq 0, k = 1,2, \cdots, n\right\}.$$
Question: The author proceeded to use the AM-GM inequality to show that
$$\left(\prod_{k=1}^n a_k \right)^{1/n}  = \left(\prod_{k=1}^n a_kx_k \right)^{1/n} \leq \frac{1}{n}\sum_{k=1}^n a_kx_k,$$and showed that equality occurs when $a_kx_k$ does not depend on $k$. But then the author said that picking $x_k = \frac{a_k}{(\prod\limits_{k=1}^n a_k)^{1/n}}$ yields equality in the AM-GM bound. However, when I plugged this in, $a_kx_k$ is constant iff $a_k$ is constant, which is not given. Am I missing anything here?

答:
Looks like the  equality-yielding definition for $x_k$ is upside down, or an exponent $-1$ is missing on one side. The assertion $a_kx_k=c$ for all $k=1,\ldots n$ leads to
$$
\prod_{k=1}^n a_k\prod_{k=1}^n  x_k = c^n.
$$ Since $\prod_{k=1}^n  x_k=1$, this gives $$c=\left(\prod_{k=1}^n  a_k\right)^{1/n}$$ and
$$
x_k = \frac c{a_k}=\frac{(\prod\limits_{k=1}^n a_k)^{1/n}}{a_k}.$$

Title: The Geometric Mean as a Minimum.


答:
In the usual AM-GM inequality $$\left(\prod_{k=1}^n b_k \right)^{1/n} \le \frac{1}{n} \sum_{k=1}^n b_k$$ (where I have used $b_k$ to avoid confusion), equality occurs if and only if $b_k = b$ for some constant $b \ge 0$.  This is what the author means by "$a_k x_k$ does not depend on $k$."  But that does not imply that the sequence $\{a_k \}_{k=1}^n$ is itself constant.  It implies the sequence of the products $\{a_k x_k \}_{k=1}^n$ is constant, where the $x_k$ can be thought of as "weights" on the $a_k$.  The $a_k$ are fixed; the goal is to find a corresponding sequence of weights $x_k$ that results in equality.

To illustrate, let us consider a concrete example.  Let $n = 5$ and $a_k = k^2$ for $1 \le k \le n$.  Then the usual AM-GM statement is $$(1 \cdot 4 \cdot 9 \cdot 16 \cdot 25)^{1/5} \le \frac{1 + 4 + 9 + 16 + 25}{5},$$ which of course is true, and the inequality is strict.  Now can we find a sequence $x_k$ such that $$(1 \cdot 4 \cdot 9 \cdot 16 \cdot 25)^{1/5} = \frac{1x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5}{5}?$$  Certainly.  In fact, there are many possible solutions, but we want one in which the product $x_1 x_2 x_3 x_4 x_5 = 1$.  To do this, we observe that when this product of weights equals $1$, the LHS can be written $$(1 x_1 \cdot 4 x_2 \cdot 9 x_3 \cdot 16 x_4 \cdot 25 x_5)^{1/5}.$$  And now, by AM-GM applied to the sequence of prodcuts, equality is attained when $1x_1 = 4x_2 = 9x_3 = 16x_4 = 25x_5$; this gives us $$x_k = \frac{c}{a_k}$$ for some constant $c$.  Then since we require this to equal $1$, we have $$c^5 = a_1 a_2 a_3 a_4 a_5,$$ hence $$x_k = \frac{(a_1 a_2 a_3 a_4 a_5)^{1/5}}{a_k}.$$

Thus, the author has incorrectly taken the reciprocal of the required weight, and you can see the incorrect choice leads to an expression on the RHS of the form $\frac{1}{c^5} (a_1^2 + a_2^2 + \cdots + a_n^2),$ which clearly does not work.

Title: The Geometric Mean as a Minimum.

Body:
This is problem 8.5 from the book "Cauchy Schwarz Master Class":

Problem: Prove that the geometric mean has the representation $$\left(\prod_{k=1}^n a_k \right)^{1/n} = \min\left\{\frac{1}{n} \sum_{k=1}^n a_kx_k:(x_1,\cdots,x_n)\in D\right\},$$
where $a_k \geq 0$ and $D$ is the region of $\mathbb{R}^n$ defined by $$D = \left\{(x_1,\cdots,x_n): \prod_{k=1}^n x_k =1, x_k \geq 0, k = 1,2, \cdots, n\right\}.$$
Question: The author proceeded to use the AM-GM inequality to show that
$$\left(\prod_{k=1}^n a_k \right)^{1/n}  = \left(\prod_{k=1}^n a_kx_k \right)^{1/n} \leq \frac{1}{n}\sum_{k=1}^n a_kx_k,$$and showed that equality occurs when $a_kx_k$ does not depend on $k$. But then the author said that picking $x_k = \frac{a_k}{(\prod\limits_{k=1}^n a_k)^{1/n}}$ yields equality in the AM-GM bound. However, when I plugged this in, $a_kx_k$ is constant iff $a_k$ is constant, which is not given. Am I missing anything here?