问:
Sum of inverse gamma random variables with unity parameters? (challenging integral)
I came across a problem on this site (which I now cannot find) asking about the distribution of sums of i.i.d. inverse gamma random variables for paramters $\alpha=\beta=1$. Defining $Z=X+Y$ with $X,Y\sim\operatorname{InvGamma}(1,1)$ we can apply the usual formula to obtain the density
$$
f_Z(z)=\int_0^z(z-x)^{-2}\exp\left(-\tfrac{1}{z-x}\right)x^{-2}\exp\left(-\tfrac{1}{x}\right)\,\mathrm dx,\qquad z>0.
$$
This integral seems deceptively hard which is mainly due to the exponential terms. Now we do have the useful fact that
$$
\int(z-x)^{-2}\exp\left(-\tfrac{1}{z-x}\right)\,\mathrm dx=\exp\left(-\tfrac{1}{z-x}\right),
$$
however, I don't think integrating by parts gets us very far. Another observation is that the integrand is symmetric about $x=z/2$ so that
$$
f_Z(z)=2\int_0^{z/2}(z-x)^{-2}\exp\left(-\tfrac{1}{z-x}\right)x^{-2}\exp\left(-\tfrac{1}{x}\right)\,\mathrm dx.
$$
But again I'm not sure this accomplishes much. How can we deal with the exponential terms to put this integral in a more useful form for evaluating? Taylor series expansions are certainly out. Maybe a clever substitution? The integrand is quite interesting and reminds me of a bump function. Here it is for $z=1$:

答: Substitute $y=z^2/[4x(z-x)]$. Then $0<x<z/2$ maps to $1<y<\infty$, and $$f_Z(z)=\frac{8}{z^3}\int_1^\infty\left(1-\frac1y\right)^{-1/2}\exp\left(-\frac{4y}{z}\right)\,{\rm d}y=-\frac{4}{z^3}g'\left(\frac2z\right),$$ where $$g(a)=\int_1^\infty\frac{e^{-2ay}\,{\rm d}y}{\sqrt{y(y-1)}}\ \underset{y=(1+t)/2}{\phantom{\Big[}=\phantom{\Big]}}\ \int_1^\infty\frac{e^{-a(1+t)}}{\sqrt{t^2-1}}\,{\rm d}t=e^{-a}K_0(a)$$ uses the modified Bessel function $K_0$. Hence $f_Z(z)=(4/z^3)e^{-2/z}\big(K_0(2/z)+K_1(2/z)\big)$.


问:
Looking for classic intro to intermediate algebra books, similar to AOPS
I am looking for a classic intro to intermediate or intermediate algebra book OTHER than gelfand’s. I own gelfand algebra and while it’s great, it doesn’t go into logarithms and it is a little short. Preferably a book or books similar to AOPS Algebra - Intermediate Algebra books. I like AOPS books but I’ve used a few and I prefer the style of non US texts.

Recommendations for a classic intro to intermediate algebra book, preferably non American. Rigorous is good too

答:
I don't know what you mean by "intermediate algebra" (the term has a specific meaning in U.S. colleges, especially in community colleges, and there are very many texts with that phrase as their title), but this usage that I am familiar with is for books designed for remedial work, which is far removed from what AOPS is designed for. So I'll assume you mean advanced school level algebra. For advanced school level algebra, there are many older freely available texts that might be of interest to you, such as the following.

William Steadman Aldis -- A Text Book of Algebra

George Chrystal -- Algebra. An Elementary Text-Book (Part I and Part II)

George Chrystal -- Introduction to Algebra

Henry Burchard Fine -- A College Algebra

Hall and Knight -- Higher Algebra

Charles Smith -- A Treatise on Algebra

Isaac Todhunter -- Algebra for the use of Schools and Colleges

Webster Wells -- Advanced Course in Algebra

Wilczynski and Slaught -- College Algebra with Applications

The above happen to be on my bookshelves (obtained at various times and in various used bookstores during the past several decades) and simply by googling their titles/authors along with archive.org, I was able to find freely available digitized copies of all of them.

See also this list of online books.

ContentLicense: CC BY-SA 4.0


问:
How to Factorise these expressions?
I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$

答:
You need to find the factors of the term with the power of 2 and the last term (which can also be in the power of 2)

Let's take this one:

$4a^2 - 16ab - 9b^2$

We know that the factors of the coefficient of the term in $a^2$ are 1, 2 and 4 (we can take the all positives here and ignore the negative factors to make it easier) and for the coefficient of the term in $b^2$, they are 1, 3, 9, -1, -3 and -9.

My method (there are several) is to put the brackets and the variables:

$$(a\ \ \ b)(a\ \ \ b)$$

I try first with 1 and 4 for $a^2$:

$$(a\ \ \ b)(4a\ \ \ b)$$

```<code>                                    1  2 3  4
</code>```

Now, the only possible terms in $ab$ will be $4ab$ (#2 times #3) and $ab$ (#1 times #4), but we need to get $-16ab$ (see original expression), so that we need $-20ab$ to add to the current $4ab$. How can we get $-20$ from the factors 1, 3, 9, -1, -3 and -9, and $4ab$? Doesn't seem possible. So maybe putting #2 as $3b$, meaning #4 has to be $3b$ as well...

$$(a\ \ \ 3b)(4a\ \ \ 3b)$$

But this gives $12ab+3ab$, or if we take the negative, $-12ab-3ab$ not very useful... so probably 1 and 4 are not good. Let's try 2 and 2.

$$(2a\ \ \ b)(2a\ \ \ b)$$

```<code>                                    1  2 3  4
</code>```

Now, the only possible terms in $ab$ will be $2ab$ (#2 times #3) and $2ab$ (#1 times #4), but again we need to get $-16ab$, so that we need $-20ab$ to add to the current $4ab$. How can we get $-20$ from the factors 1, 3, 9, -1, -3 and -9, and $2ab$, $2ab$? Well, here, you can notice we can have $-9\times2ab + 1\times 2ab = -16ab$. So that:

$$(2a-9b)(2a+b)$$

If we expand it, we get: $4a^2 - 18ab + 2ab -9b^2 = 4a^2 - 16ab - 9b^2$

It takes time and a lot of practice to make all the process faster (especially the part where you have to see what numbers go where) and more natural, but if you keep at it, it will remain in your head for a long time!

ContentLicense: CC BY-SA 3.0

Title: How to Factorise these expressions?

Body:
You need to find the factors of the term with the power of 2 and the last term (which can also be in the power of 2)

Let's take this one:

$4a^2 - 16ab - 9b^2$

We know that the factors of the coefficient of the term in $a^2$ are 1, 2 and 4 (we can take the all positives here and ignore the negative factors to make it easier) and for the coefficient of the term in $b^2$, they are 1, 3, 9, -1, -3 and -9.

My method (there are several) is to put the brackets and the variables:

$$(a\ \ \ b)(a\ \ \ b)$$

I try first with 1 and 4 for $a^2$:

$$(a\ \ \ b)(4a\ \ \ b)$$

                                    1  2 3  4

Now, the only possible terms in $ab$ will be $4ab$ (#2 times #3) and $ab$ (#1 times #4), but we need to get $-16ab$ (see original expression), so that we need $-20ab$ to add to the current $4ab$. How can we get $-20$ from the factors 1, 3, 9, -1, -3 and -9, and $4ab$? Doesn't seem possible. So maybe putting #2 as $3b$, meaning #4 has to be $3b$ as well...

$$(a\ \ \ 3b)(4a\ \ \ 3b)$$

But this gives $12ab+3ab$, or if we take the negative, $-12ab-3ab$ not very useful... so probably 1 and 4 are not good. Let's try 2 and 2.

$$(2a\ \ \ b)(2a\ \ \ b)$$

                                    1  2 3  4

Now, the only possible terms in $ab$ will be $2ab$ (#2 times #3) and $2ab$ (#1 times #4), but again we need to get $-16ab$, so that we need $-20ab$ to add to the current $4ab$. How can we get $-20$ from the factors 1, 3, 9, -1, -3 and -9, and $2ab$, $2ab$? Well, here, you can notice we can have $-9\times2ab + 1\times 2ab = -16ab$. So that:

$$(2a-9b)(2a+b)$$

If we expand it, we get: $4a^2 - 18ab + 2ab -9b^2 = 4a^2 - 16ab - 9b^2$

It takes time and a lot of practice to make all the process faster (especially the part where you have to see what numbers go where) and more natural, but if you keep at it, it will remain in your head for a long time!


问:
Minimal polynomial of $x$ over $\mathbb{Q}(x+y+z, x^2+y^2+z^2, x^3+y^3+z^3)$
Let $K$ be the field of fractions of $\mathbb{Q}[x,y,z]$,
$$\alpha=x+y+z,\quad \beta=x^2+y^2+z^2,\quad \gamma = x^3+y^3+z^3\in K$$
and $M=\mathbb{Q}(\alpha, \beta, \gamma)$.

Is $x\in K$ algebraic over $M$? What is the minimal polynomial?

What is the degree of transcendence of $M$ over $\mathbb{Q}$?

For part 1 I really have no idea. I tried some combinations but it didn't get me anywhere. Is there a systematic way of approaching this kind of problems?

For part 2 I'm guessing the degree is 3, but I'm not sure how to prove it.

答:
Consider the polynomial $g(t)=(t-x)(t-y)(t-z)$ in $\mathbb{Q}(x,y,z)[t]$. Expanding it out yields
\begin{align}
g(t)&=t^3-(x+y+z)t^2 +\\
&=(yz+xz+xy)t-xyz.
\end{align} Since $x$ is certainly a root of $g$, to show that $x$ is algebraic over $\mathbb{Q}(\alpha,\beta,\gamma)$ it suffices to show that each of the coefficients of $g$ lies in $\mathbb{Q}(\alpha,\beta,\gamma)$. To see this, note:

$(x+y+z)=\alpha\in\mathbb{Q}(\alpha,\beta,\gamma)$.

$\alpha^2=\beta+2(yz+xz+xy)$, so $(yz+xz+xy)=(\alpha^2-\beta)\big/2\in\mathbb{Q}(\alpha,\beta,\gamma)$.

$\alpha^3=3\alpha\beta-2\gamma+6xyz$, so $xyz=(\alpha^3-3\alpha\beta+2\gamma)\big/6\in\mathbb{Q}(\alpha,\beta,\gamma)$.

This shows part $1$ of the problem, and in fact the analogous result holds for $\mathbb{Q}(x_1,\dots,x_n)$ for any $n\in\mathbb{N}$. For more on this I recommend reading about symmetric polynomials.

Now, to compute $\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(\alpha,\beta,\gamma)$, note that $g(t)$ witnesses that all of the generators of $\mathbb{Q}(x,y,z)$, and hence the field $\mathbb{Q}(x,y,z)$ itself, are algebraic over $\mathbb{Q}(\alpha,\beta,\gamma)$. What can you conclude about the respective transcendence degrees of $\mathbb{Q}(\alpha,\beta,\gamma)$ and $\mathbb{Q}(x,y,z)$ over $\mathbb{Q}$? Answer below, but try to figure it out yourself first!

 Suppose $S=\{m_1,\dots,m_k\}\subset\mathbb{Q}(\alpha,\beta,\gamma)$ is a transcendence basis for $\mathbb{Q}(\alpha,\beta,\gamma)$ over $\mathbb{Q}$. By definition this means (i) there is no polynomial in $\mathbb{Q}[t_1,\dots,t_k]$ satisfied by $S$, and (ii) $\mathbb{Q}(\alpha,\beta,\gamma)$ is algebraic over $\mathbb{Q}(S)$. Now we claim that $S$ is in fact a transcendence basis for $\mathbb{Q}(x,y,z)$. Indeed, condition (i) holds immediately, and condition (ii) follows from the fact that an algebraic extension of an algebraic extension is algebraic; we have shown above that $\mathbb{Q}(x,y,z)$ is algebraic over $\mathbb{Q}(\alpha,\beta,\gamma)$, so it follows that $\mathbb{Q}(x,y,z)$ is algebraic over $\mathbb{Q}(S)$, as desired. In particular, we have $$\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(\alpha,\beta,\gamma)=\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(x,y,z)=3,$$ just as you suspected. Note that this argument works much more generally, and we have, for any triple of fields $E\subseteq F\subseteq G$, if $G$ is algebraic over $F$, then $$\operatorname{tr}.\operatorname{deg}_E F=\operatorname{tr}.\operatorname{deg}_E G.$$


问:
Proof of $\lim_{x \to \infty}\tan x/x$ does not exist
There is an answer for this question as follows:

If we approach to infinity with the sequence $a_{n}=n\pi$ then limit is zero, on the other hand if we approach with the sequence $b_{n}=\pi/2+n\pi-1/n^2$ then limit is infinity.

I can not understand the process that is described.

答:
This is a very late response, but something else led me to the function $\frac{\tan{x}}{x}$ and its limit as $x \to \infty$. I came across this post only now.

We can, in fact, prove a more general result for the function $f(x) = \frac{\tan{x}}{x}$ than what is said originally in the post. We can prove the following:

Given any $l \in \mathbb{R}$, there exists a sequence of real numbers, $\langle b_n \rangle_{n \in \mathbb{N}}$, such that $\lim \limits_{n \to \infty} b_n = \infty$ and $\lim \limits_{n \to \infty} f(b_n) = l$.

To see this, let $\epsilon_n = \frac{1}{\alpha n}$ where $n \in \mathbb{N}$ and $\alpha \neq 0$ be any (fixed) real number. Then $\lim \limits_{n \to \infty} \epsilon_n = 0$. Now define $b_n = \frac{\pi}{2} + n \pi - \epsilon_n$ for $n \in \mathbb{N}$. Then $\lim \limits_{n \to \infty} b_n = \infty$, and you can prove that $\lim \limits_{n \to \infty} f(b_n) = \frac{\alpha}{\pi}$. Given any $l \in \mathbb{R}$, $l \neq 0$, if you now choose $\alpha = \pi l$, then it is clear that $\lim \limits_{n \to \infty} f(b_n) = l$.

When $l = 0$, let $b_n = n \pi$. Then $\lim \limits_{n \to \infty} f(b_n) = 0$.

Also, if you let $\epsilon_n = \pm \frac{1}{n^p}$ where $p > 1$, you can prove that $\lim \limits_{n \to \infty} f(b_n) = \pm \infty$.

What this argument shows is that you can "adjust" the limit of $f(b_n)$ to anything you want. Thus there is no unique limit.


问:
A linear Equation in 4D space corresponds to the whole 3D space
This equation is said to be a plane in 4D space $u+ v+ w+ z=6$

Does that plane correspond to the whole real 3D space?

I think so. Below is my attempt

Say we think of the three real space axis as given by $u ,v, w$ ,then fixing $z=d$ ($d $ is some arbitrary number ),gives me $u+ v+ w=6-d$ which is a plane in 3D space. This plane is a set of solution.

Now I change  value of $z$ and I again get another plane as a solution , continuing this way I can fill all of 3D space with the solution planes.

Therefore we can say that a plane in 4D is the whole 3D space.

答:
If at least one of $u, v, w, z$ is nonzero, then $u + v + w + z = 6$ is a system of $1$ equation in $4$ variables with rank $1$. Thus the nullity is $3$, so the system $u + v + w + z = 0$ is like a copy of 3D space. Since the solution space $S$ of the system $u + v + w + z = 6$ is just a translation$^*$ of the solution space of $u + v + w + z = 0$, you can also think of $S$ as a copy of 3D space.

$^*$ You can verify that the solution space $S$ of a consistent system $Ax = b$ is a translation of the solution space of $Ax = 0$. In other words, $S = p + \ker A$ for some point $p$.


问:
Check my work: $\lim a_n = 0 \Rightarrow \lim \sqrt{a_n} = 0 $? (for $a_n$ positive)
I'm trying to prove, as "properly" as possible the following:$$\left[ \lim z_n = z \right] \iff \left[ \lim x_n = x \quad \wedge \quad \lim y_n = y \right]$$

where $z_n = x_n + i y_n$ and $z=x+iy$.

For the converse direction, I have $\lim |x_n - x|=0$ and $\lim |y_n - y|=0$, then I show this implies $\lim |x_n - x|^2=0$ and $\lim |y_n - y|^2=0$ so that $$\lim |x_n - x|^2 +|y_n - y|^2 =0 $$

Now I want to say this implies $\lim \sqrt{|x_n - x|^2 +|y_n - y|^2 }=0 = \lim |z_n-z|$ because of the fact that $f:x \rightarrow \sqrt{x}$ is continuous at $x=0$. But is this justified, given that $f$ is undefined to the left of $0$? Does that matter at all?

答: If $0\leq a<\varepsilon^2$, then $0\leq\sqrt{a}<\varepsilon$.  </p>


问:
Integration using digamma functions
I'm trying to prove that for $x,y,r>0$ the following identity (gotten via CAS) holds:

$$\int_0^x \frac{x^r}{x^r+y^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1+r}{2r}\right)-\psi\left(\frac{1}{2r}\right)\right),$$

where $\psi(\cdot)=\frac{\Gamma'(\cdot)}{\Gamma(\cdot)}$ denotes the Digamma function. Does anyone has a clue, how integration can be done this way?

As an additional information following identity, holds also to be true:
$$\int_0^x \frac{y^r}{y^r+x^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1}{2r}+1\right)-\psi\left(\frac{1}{2r}+\frac{1}{2}\right)\right),$$
Which was obtained using:
$$\psi(x)=H_{x-1} - \gamma,$$
where $\gamma$ denotes the Euler-Mascheroni constant and $H_x$ denotes the $x^{th}$ harmonic number.

答:
Assuming $x\neq0$, substitute $z=y/x$:
$$
\int_0^x\frac{x^r}{x^r+y^r}dy=x\int_0^1\frac{dz}{1+z^r}
$$
Then, using Mathematica as an integral-table lookup device:

```<code>Integrate[1/(1 + z^r), {z, 0, 1}]
</code>```

$$
\frac{\psi\left(\frac{1+r}{2 r}\right)-\psi\left(\frac{1}{2 r}\right)}{2 r}
\qquad \text{if $\left(\Re\left((-1)^{\frac{1}{r}}\right)\geq 1\lor \Re\left((-1)^{\frac{1}{r}}\right)\leq 0\lor (-1)^{\frac{1}{r}}\notin \mathbb{R}\right)\land \Re(r)>0$}
$$
The conditions on $r$ are always satisfied for $r>0$.

Title: Integration using digamma functions

Body:
Assuming $x\neq0$, substitute $z=y/x$:
$$
\int_0^x\frac{x^r}{x^r+y^r}dy=x\int_0^1\frac{dz}{1+z^r}
$$
Then, using Mathematica as an integral-table lookup device:

Integrate[1/(1 + z^r), {z, 0, 1}]

$$
\frac{\psi\left(\frac{1+r}{2 r}\right)-\psi\left(\frac{1}{2 r}\right)}{2 r}
\qquad \text{if $\left(\Re\left((-1)^{\frac{1}{r}}\right)\geq 1\lor \Re\left((-1)^{\frac{1}{r}}\right)\leq 0\lor (-1)^{\frac{1}{r}}\notin \mathbb{R}\right)\land \Re(r)>0$}
$$
The conditions on $r$ are always satisfied for $r>0$.


问:
Evaluate : $\int\limits_0^{\pi/2} \cot x \ln(\sec x) dx$
I want to solve this integral :
$$\int_0^{\pi/2} \cot x \ln(\sec x) dx$$
I tried the following substitution : $\ln(\sec x)=t$ which means $dt=\tan x dx$
$$I=\int_0^\infty \frac{\cot x}{\tan x}tdt=\int_0^\infty \frac{t}{\tan^2 x}dt$$
I'm really disturbed by the $\tan^2 x$, I tried also to substitute $\sec x =t$ but it's not helpful either. Any helpful approach to solve this problem ?

答:
Another possibility, using the Basel problem formula $\sum_{k\geq 1}k^{-2}=\pi^2/6$.

We have
$$
I=-\int_0^{\pi/2}\frac{\cos(x)}{\sin(x)}\ln\cos(x)dx=
-\frac 12\int_0^{\pi/2}\frac{\cos(x)}{\sin(x)}\ln(1-\sin^2 x) dx
$$
$$
=\frac 12\sum_{k\geq 1}\frac 1k\int_0^{\pi/2}\cos(x)\sin^{2k-1}(x) dx
$$
The integral in the summation is evaluated as
$$\int_0^{\pi/2}\cos(x)\sin^{2k-1}(x) dx=\frac 1{2k}\int_0^{\pi/2}d(\sin^{2k}(x))dx=\frac 1{2k}$$
Hence
$$
I=\frac 14\sum_{k\geq 1}\frac 1{k^2}=\frac{\pi^2}{24}.
$$


问:
$63^{63^{63}} \mod 100$
I need to find $63^{63^{63}} \bmod 100$.

This is what I've got so far:

Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$

Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$

Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$

I'm stuck here and don't know what to do next, what could I do now?

答:
$(63)^7 = (60+3)^7 = (60^7 + {7\choose 1} 60^6\cdot 3 + \cdots + {7\choose 6} 60\cdot 3^6 + 3^7 $

All the terms to the left of the last 2 are equivalent to 0 modulo 100, and can be dropped.

Leaving $7\cdot 60\cdot 3^6 + 3^7\pmod {100}$

We only need to worry about the one figure for the first term since we know it will end in 0.

$3^4 = 81, 3^6\equiv 9 \pmod {10}$

$6\cdot 7 \cdot 9 \cdot 10 + 81\cdot 27\equiv 80 + 87 \equiv 67 \pmod {100}$

Title: $63^{63^{63}} \mod 100$


问:
Foundation of Formal Logic
All the books that I’ve read about formal logic either starts assuming the existence of set theory (they talk of countable sets of symbols and define formulae as sequences of symbols), or follow an axiomatic approach that assumes known the meaning of words like symbol, formula, substitution etc. Is there a way to formalize these concepts (with no reference to set theory, I consider set theory a product of formal logic)?

We could assume as primitive the notion of symbol, but could the concepts of string, formula, composition of formulae be rendered in more precise way that don’t rely on intuition? A book or paper reference on this matter would be appreciated.

答:
Here is a possible approach.  Consider the minimalist language Brainfuck. This extremely stripped-down language comprises only eight commands and an instruction pointer.  The first step is to train yourself to be able to execute Brainfuck programs.  This does require some amount of intuition; you need the ability to read and write, at least well enough to be able to read and understand a description of the Brainfuck language.  You need to be able to recognize a Brainfuck program when you see one, parse the symbols, and remember how to execute the commands using whatever writing method you have mastered.   But you don't have to understand what a "symbol" is in general or what a "rule" is in general. You just have to master a very specific set of symbols and rules.

If you get this far, then in principle you have more than enough power to reproduce any formal logical proof, since Brainfuck is Turing complete. Of course, someone still has to code up logical axioms and rules in Brainfuck, but that's their problem, not yours. If they do their part, you can do your part by executing their Brainfuck programs.  In this way, you can formally reproduce all of logic and mathematics, with a minimal reliance on non-formal intuition.  (You won't understand anything, of course, but that doesn't seem to be your question.)

Title: Foundation of Formal Logic


问:
The subring test
This is how the Wikipedia article on subring defines the subring test

The subring test states that for any ring $R$, a nonempty subset of $R$ is a subring if it is closed under addition and multiplication, and contains the multiplicative identity of $R$.

When you follow the link for the subring test, it is stated as follows

In abstract algebra, the subring test is a theorem that states that for any ring, a nonempty subset of that ring is a subring if it is closed under multiplication and subtraction. Note that here that the terms ring and subring are used without requiring a multiplicative identity element.

My question is, is the first statement of subring test correct? This is also how a subring is "defined" in Atiyah-Macdonald. It seems incorrect to me as $\mathbb{R}_+$ satisfies those conditions and is not a subring unless I am missing something.

Looking at the responses I feel I should further clarify my question. Closure under subtraction and multiplication (with the added provision that the given subset contain the identity depending on how you define your rings), guarantees a subring, as in the second statement. I am comfortable with this statement as I know that closure under subtraction for a subset of a group (written additively) gives a subgroup. My question is whether the first statement is correct - is closure under addition and multiplication enough?

答:
It seems that the problem lies in what it means to be closed under addition. My interpretation of being closed under addition is that if you restrict the binary operation of addition to the subset that you want to study, then you get a well defined function.

Say, if you have a ring $(R, +, \cdot)$, then if we have a subset $S \subseteq R$, I would interpret $S$ being closed under the addition inherited from $R$ as meaning that if $a, b \in S$ then $a + b \in S$, or that the image of the map

$$+ : S \times S \longrightarrow R$$

that results from restricting the addition to the elements of $S$ lies in $S$, that is, that $+(S \times S) \subseteq S$ (however weird that notation may seem). So if this is what is what it means for $S$ to be closed under addition, then certainly $\mathbb{R}_{+}$ would satisfy the requirements in the first formulation of the "subring test" that you give, but it will not be a subring of $\mathbb{R}$ since it will not contain the additive inverses.

The same thing would happen when considering $\mathbb{N} \subseteq \mathbb{Z}$.

I just checked my copy of Atiyah-Macdonald and indeed they define a subring in this way, so maybe it is just a misunderstanding or maybe a lack of care when defining a subring.

Title: The subring test


问:
Calculation of a hard integral $\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx $
Calculate
$$\int^{1/2}_0\frac{1}{x^2-x+1} \cdot \ln\frac{1+x}{1-x}\,dx $$
I tried to subsitute $x=1/2-t$ so $dx=-dt$ but I just complicated more my problem.

答:
Mathematica gives:

$$\frac{\pi  \log \left(\frac{9}{8}\right)-3 i \left(\text{Li}_2\left(-\frac{1}{2}
   (-1)^{2/3}\right)+\text{Li}_2\left(\frac{1}{1-\sqrt[3]{-1}}\right)-\text{Li}_2\left(\frac{1}{1+\sqrt[3]{-1}}\right)+\text{Li}_2\left(\frac{1}{1-(-1)^{2/3}}\right)-\text{Li}
   _2\left(\frac{1}{1+(-1)^{2/3}}\right)+\text{Li}_2\left(\frac{3}{4}-\frac{i
   \sqrt{3}}{4}\right)-\text{Li}_2\left(\frac{1}{4}+\frac{i
   \sqrt{3}}{4}\right)-\text{Li}_2\left(\frac{3}{4}+\frac{i \sqrt{3}}{4}\right)\right)}{3
   \sqrt{3}}$$

which suggests you have a lot of work ahead of you if you want to do this by hand.


问:
Show that every rank-deficient matrix has a full rank matrix arbitrarily close to it
I came across the following exercise in preparation for a test.

Let $A \in \mathbb{R}^{n \times m}$ with rank $r < \min\{n,m\}$. Use SVD of $A$ to show that for every $\epsilon >0$ (no matter how small), there existsa full-rank matrix $A_{\epsilon} \in \mathbb{R}^{n\times m}$ such that $|| A - A_\epsilon ||_{2} < \epsilon$.

I am really puzzled on how to approach this exercise. From my understanding
$$ A - A_\epsilon = \underset{n\times n}{U} \ \underset{n \times m}{\Sigma} \ \underset{m\times m}{V^{T}} - \underset{n\times n}{U} \ \underset{n \times m}{\Sigma_\epsilon} \ \underset{m\times m}{V^{T}} = U(\Sigma - \Sigma_\epsilon)V^{T}.$$
Then $\Sigma$ is diagonal matrix with singular values ranging from $\sigma_1$ to $\sigma_{\min\{n,m\}}$ and $\Sigma_\epsilon$ having singular values from $\sigma_1$ to $\sigma_m$ since $A_\epsilon$ is of rank $m$.

Then since $\min\{n,m\} \le m$
$$ \Sigma - \Sigma_\epsilon = \begin{pmatrix} \sigma_1 &  &  & \\
& \ddots &  \\
 & & \sigma_n &  \\
& & & 0\end{pmatrix} - \begin{pmatrix} \sigma_1 &  &  & \\
& \ddots &  \\
 & & \sigma_{\min\{n,m\}} &  \\
& & & 0\end{pmatrix} =  \begin{pmatrix} 0&  &  & \\
& \ddots &  \\
 & & \sigma_? &  \\
& & & 0\end{pmatrix}$$
So since a singular value exists for the matrix difference above, $|| A - A_\epsilon ||_2 = \sigma_?$ which has to be small since $\sigma_1 > \sigma_2 > \dots > \sigma_r$, and here this $\sigma_?$ is among among the last in the sequence.

I am most likely writing non-sense, I am aware of that, I don't know how to proceed, can anyone give me some hint or idea ?

答:
Note that in the case of square matrices $(n\times n$), there is maybe a shorter proof.

$p(x)=\det(A-xI)$ is a polynomial in $x$ and has at most $n$ roots in $\mathbb C$.

So let's take $A_\epsilon=A-\epsilon I$, we have $\det(A_\epsilon)=\det(A-\epsilon I)=p(\epsilon)$

The number of roots of $p$ being finite, we can always choose $\epsilon$ such that $p(\epsilon)\neq 0$, making $A_\epsilon$ invertible.


答:
You might use Chinese Remainder Theorem.  $100 = 4 \cdot 25$, and

$$63^7 \equiv (-1)^7 \equiv 3 \pmod{4}.$$

Then $$63^7 \equiv 9^7 7^7 \equiv 729\cdot 729 \cdot 9 \cdot 49^3\cdot 7
\equiv 4\cdot 4 \cdot 9 \cdot (-1)^3 \cdot 7 $$

$$\equiv 144(-7) \equiv (-6)(-7) \equiv 42 \pmod{25}.$$

So you just have to add or subtract various multiples of $25$ to $42$ until you get a number that's $3 \pmod{4}.$  And $42+25 = 67$ works.

Title: $63^{63^{63}} \mod 100$

Body:
I need to find $63^{63^{63}} \bmod 100$.

This is what I've got so far:

Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$

Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$

Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$

I'm stuck here and don't know what to do next, what could I do now?


问:
QR Factorization definiton
We know that for any complex $(m \times n)$-matrix $A$
with linearly independent columns, there exists an $(m \times n)$-matrix $Q$ and an $(n \times n)$-matrix $R$ such that $A = QR$, $Q^* Q = I$, $R$ is a invertible upper triangular matrix, and the columns of $Q$ are an orthonormal basis for $\operatorname{Im}(A)$.

But I wonder if the columns of $A$ can form a orthogonal basis of $\operatorname{Im}(A)$?

答: The columns of $A$ are orthogonal if and only if the upper-triangular matrix $R$ is diagonal.

Title: QR Factorization definiton


答:
Use the fast exponentiation algorithm:

\begin{array}{rrrr}
n& x^{2^k} && P\bmod 100 \\\hline
7 & 63 && 63 \\
3 & 69 &&69\cdot 63\equiv 47 \\
1 & 61 && 61\cdot 47\equiv \color{red}{67} \\\hline
\end{array}

Title: $63^{63^{63}} \mod 100$


问:
First-order recursion problem with a special cubic equation: by Ji_Chen

Given a natural number $n$ and a real number $d_{n},$ for the least $d_{n+1}$ so that $32d_{n}^{5}\!\left ( d_{n}+ 2 \right )\!=$
$$= 2\left ( 5d_{n}^{4}+ 8d_{n}^{2}+ 8d_{n}+ 8 \right )d_{n+ 1}^{3}+ 4d_{n}\left ( 5d_{n}^{4}+ 11d_{n}^{3}- 2d_{n}^{2}+ 14d_{n}- 4 \right )d_{n+ 1}^{2}$$
$+ 4d_{n}^{3}\left ( 5d_{n}^{3}+ 2d_{n}^{2}+ 24d_{n}- 16 \right )d_{n+ 1}.$ Prove that if $d_{2}= \frac{8}{5},$ then $\left \{ d_{n} \right \}_{n> 1}$ is rational and
$$\lim\frac{n^{2}}{\ln n}\left ( \frac{2}{n}- d_{n} \right )= 2$$
Source: AoPS/@Ji_Chen (still unsolved)

I'm into the related one here https://Artofproblemsolving.com/community/c6h318726p1713923 _I also see $d_{2}= \!\frac{8}{5}.$ Maybe they are same so I used discriminant to define the concrete value for $d_{n}$ but unsuccessfully, I need to the help.

答:
Define a sequence $\{x_i\}_i$ so that $x_1=1/2$ and
$$x_{n+1}=\frac{x_n^2+1}{2}.$$
We claim that
$$d_n=\frac{1-x_{n-1}}{x_{n-1}x_n}.$$
We show this by induction on $n$, with the base case $n=2$ true since $x_2=5/8$. For the inductive step, we show that, if $d=\frac{2(1-x)}{x(x^2+1)}$, then
$$\frac{}{}$$
is the largest root of the given polynomial. This is via the factorization of
$$2(5d^4+8d^2+8d+8)y^3+4d(5d^4+11d^3-2d^2+14d-4)y^2+4d^3(5d^3+2d^2+24d-16)y-32d^5(d+2)$$
with $d=\frac{2(1-x)}{x(x^2+1)}$ as
$$\frac{16}{x^6(x^2+1)^6}\big(y(x^2+1)(x^4+2x^2+5)-8(1-x^2)\big)\big(a(x)y^2+b(x)y+c(x)),$$
times $a(x)y^2+b(x)y+c(x)$, where
\begin{align*}
a(x)&=x^2 (x^2 + 1) (x^4 - 2 x^3 + 4 x^2 - 4 x + 2) (x^4 + 2 x^3 - 2 x + 1)\\
b(x)&=2 (x - 1)^2 x (x^7 + x^6 + 6 x^5 - 5 x^4 + 9 x^3 - 2 x^2 - 6 x + 4)\\
c(x)&=16(x-1)^4(x^2-x+1).
\end{align*}
It is not hard to see (for example, by graphing them) that $a(x),b(x),c(x)\geq 0$ as long as $x>0$; since $0<x_n<1$ for all $n$, this gives that
$$\frac{8(1-x^2)}{x^6+3x^4+7x^2+5}$$
is the largest root of the polynomial, so
$$d_{n+1}=\frac{8(1-x_{n-1}^2)}{(x_{n-1}^2+1)(x_{n-1}^4+2x_{n-1}^2+5)}=\frac{16(1-x_n)}{2x_n(4x_n^2+4)}=\frac{1-x_n}{x_nx_{n+1}},$$
as desired.

This shows that $d_n$ is rational; we only need to show that
$$\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-d_n\right)=2.$$
Define
$$t_n=\left(x_{n-1}-\left(1-\frac 2n\right)\right)n^2.$$
Then
\begin{align*}
t_{n+1}&=1-n^2+(n+1)^2x_n\\
&=1-n^2+(n+1)^2\left(\frac{\left(1-\frac2n+\frac{t_n}{n^2}\right)^2+1}{2}\right)\\
&=1-n^2+(n+1)^2\left(1-\frac2n+\frac2{n^2}+\frac{t_n}{n^2}-\frac{2t_n}{n^3}+\frac{t_n^2}{2n^4}\right)\\
&=\frac2n+\frac2{n^2}+t_n\left(1-\frac2{n^3}-\frac3{n^2}\right)+\frac{t_n^2}{2n^4}.
\end{align*}
So, we may show by induction on $n$ that $\{t_n-2H_{n-1}\}_n$ is bounded, where
$$H_k=\sum_{i=1}^k\frac1i$$
is the $k$th harmonic number, and so
$$x_{n-1}=1-\frac2n+\frac{2\ln n}{n^2}+O\left(\frac1{n^2}\right),$$
since $H_n\sim \ln n$. Now
\begin{align*}
\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-d_n\right)
&=\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-\frac{1-x_{n-1}}{x_{n-1}x_n}\right)\\
&=\lim_{n\to\infty}\frac{n^2}{\ln n}\left(\frac2n-\frac{\frac2n-\frac{2\ln n}{n^2}+O\left(\frac1{n^2}\right)}{x_{n-1}x_n}\right)\\
&=\lim_{n\to\infty}\frac{n^2}{x_{n-1}x_n\ln n}\left(\frac{2x_{n-1}x_n}{n}-\frac2n+\frac{2\ln n}{n^2}+O\left(\frac1{n^2}\right)\right)\\
&=\lim_{n\to\infty}\frac{2n(x_{n-1}x_n-1)}{x_{n-1}x_n\ln n}+\frac{2}{x_{n-1}x_n}+O\left(\frac{1}{x_{n-1}x_n\ln n}\right)\\
&=2-2\lim_{n\to\infty}\frac{n(1-x_{n-1}x_n)}{\ln n}.
\end{align*}
Since
$$1-x_{n-1}x_n=1-x_{n-1}+x_{n-1}(1-x_n)=O\left(\frac1n\right),$$
this limit is just $2$, as desired.


答:
you are not getting other people are saying
people are using this formula

$$(a-b)^2=a^2+b^2-2ab$$

means you have to make terms in square form like in case of

$9m^2 - 24mn + 16n^2\implies(3m)^2+(4n)^2-2\times3m\times 4n\implies (3m-4n)^2$

try yourself first one.And one more thing bracket do not affect this question

Title: How to Factorise these expressions?

Body:
I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$


问:
Using epsilon-delta to prove a multivariable limit
I'm trying to find $\lim_{x,y \to 0,0} f(x,y)$ where $f(x,y)=\frac{e^\frac{-1}{\sqrt{x^2+y^2}}}{e^{x^2+y^2}-1}$ I've tried converting to polar coordinates but it doesn't seem to be a helpful simplification. Graphically, it seems that the limit is 0 from all possible paths, and I tried to construct an epsilon-delta proof to prove this. I'm just trying to figure out if I'm on the right track.

Let $\epsilon>0$ be given, find a $\delta>0$ such that if $|{(x,y)-(0,0)}|<\delta$ implies $|{f(x,y)-L}|<\epsilon$, Assuming $L=0$. We know that $\sqrt{x^2+y^2}<\delta\rightarrow x^2+y^2<\delta^2$
which implies $\frac{1}{e^{x^2+y^2}-1}<\frac{1}{e^{\delta^2}-1}$
$|{f(x,y)-0}|=|{\frac{e^\frac{-1}{\sqrt{x^2+y^2}}}{e^{x^2+y^2}-1}-0}| =
|{\frac{1}{e^{x^2+y^2}-1}*e^\frac{-1}{\sqrt{x^2+y^2}}}|$.
Since ${e^\frac{-1}{\sqrt{x^2+y^2}}}<1$ for all $x,y\neq(0,0)$, we get $|{f(x,y)-0}|< |\frac{1}{e^{\delta^2}-1}*1|$. Setting $\delta=\sqrt{ln(\frac{1}{\epsilon}-1)}$ We can plug in to the inequality to get $|{f(x,y)-0}|<|\frac{1}{e^{\delta^2}-1}*1|<\epsilon$. Thus the limit is equal to $0$

答:
$0<\sqrt {x^2 + y^2} = r < \delta$

$0<\frac {e^{\frac {-1}{r}}}{e^{r^2} + 1} < \frac 12 e^{\frac {-1}{r}}<\epsilon$

We can say this because $e^{r^2} > 1$ and $e^{r^2} + 1 > 2$

$-\frac {1}{r} < \ln 2\epsilon\\
r< -\frac {1}{\ln 2\epsilon}$

For any $\epsilon > 0,$ when $\delta < -\frac {1}{\ln 2\epsilon}$ then  $|f(x,y)| < \epsilon$


问:
Confidence Level - Could anyone help me figure out where I went wrong?
I have been working on this practice problem for a while, and worked it out in two ways, and following from inputting the answers in the system, it notified me that the answers for both of them was wrong.

Could anyone help me figure out where I went wrong in my procedure? And help me work it out the correct way?

The question goes as follows:

The first method

The second method

Part (a)

$$E = \frac{0.2319-0.1270}{2}= 0.05245$$
$$p = 0.2314-0.05245 = 0.17945$$
$$\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.17945(1-0.17945)}{195}}= 0.0275$$

Therefore, the critical $z$ value is,
$$z = \frac{E}{\sigma} = \frac{0.05245}{0.0275} = 1.9073 \approx 1.91$$

From the standard table,
$$P ( -1.91 < z < 1.91) = 0.9719 $$

Confidence Level: 97.19%

Part (b)

Confidence Level computed as:
$$\frac{19}{20}= 0.95$$

From the standard table,
$$P(z < 1.645) = 0.95$$

For a point estimate of proportion here as $p = 0.17945$, the margin of error is,
$$E = z \times \sqrt{\frac{p(1-p)}{n}}$$
$$n = \frac{z^2 \times p(1-p)}{E^2}= \frac{1.645^2 \times 0.17945(1-0.17945)}{0.05245^2}= 144.84$$

Required sample size: 145

答:
You need to use a finite population correction factor for the interval estimate, because the batch has size $N = 2750$ and you are sampling $n = 195$ axles without replacement from that batch.  So a $100(1-\alpha)\%$ asymptotic confidence interval will have the form
$$\hat p \pm z_{\alpha/2}^* \sqrt{\frac{\hat p (1 - \hat p)}{n} \frac{N - n}{N - 1}},$$ where $\hat p$ is the point estimate of the proportion of failed axles, and $z_{\alpha/2}^*$ is the upper $\alpha/2$ quantile of the standard normal distribution, i.e. $\Pr[Z \le z_{\alpha/2}^*] = 1 - \alpha/2$.  In your case, $$\hat p = \frac{0.1270 + 0.2319}{2},$$ hence the margin of error is $$ME = z_{\alpha/2}^* \sqrt{\frac{\hat p (1 - \hat p)}{n} \frac{N - n}{N - 1}} = \frac{0.2319 - 0.1270}{2}$$ and the critical value is $$z_{\alpha/2}^* = \frac{0.2319 - 0.1270}{2} \sqrt{\frac{195}{(0.17945)(1 - 0.17945)} \cdot \frac{2750 - 1}{2750 - 195}}.$$  This gives $1 - \alpha/2 = \Phi(1.97984)$ or $\alpha = 0.0477213$.

For the second part, what they are telling you is that the desired margin of error is $ME = 4\% = 0.04$, and the desired confidence level is $19/20 = 95\%$, corresponding to $\alpha/2 = 0.025$.  If we assume the point estimate from part (a) stays the same, then the required sample size is $$n = \frac{m}{1 + \frac{m-1}{N}}, \quad m = \frac{(z_{\alpha/2}^*)^2 \hat p (1 - \hat p)}{ME^2}.$$  This gives $n = 314$, meaning $314 - 195 = 119$ more axles need to be sampled.


答:
I'm achieving your approach :

Using the Identity I mentioned :
$$\tan^2 x=\sec^2 x-1$$
And :
$$\ln(\sec x) =t\Longleftrightarrow \sec x=e^t$$
So $\tan^2 x$ in function of $t$ is :
$$\tan^2 x=e^{2t}-1$$
And your integral is :
$$\int_0^\infty \frac{t}{e^{2t}-1}dt=\frac{1}4\int_0^\infty \frac{t}{e^t-1}dt$$
Note that :
$$\zeta(s)\Gamma(s)=\int_0^\infty \frac{t^{s-1}}{e^t-1}dt$$
In our case :
\begin{align}
\zeta(2)\Gamma(2)&=\int_0^\infty  \frac{t}{e^t-1}dt\\
\frac{\pi^2}{6}\times 1&=\int_0^\infty  \frac{t}{e^t-1}dt\\
\end{align}
Therefore :
\begin{align}
I&=\frac{1}4 \int_0^\infty \frac{t}{e^t-1}dt\\
&=\frac{1}4\times \frac{\pi^2}{6}\\
&=\frac{\pi^2}{24}
\end{align}
Which is the result @DavidG.Strock said the answer Mathematica gives.

$\zeta(s)$ denotes Riemann Zeta function, $\Gamma(s)$ denotes Euler Gamma function.
$$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
$$\Gamma(2)=(2-1)!=1$$

Title: Evaluate : $\int\limits_0^{\pi/2} \cot x \ln(\sec x) dx$

Body:
I want to solve this integral :
$$\int_0^{\pi/2} \cot x \ln(\sec x) dx$$
I tried the following substitution : $\ln(\sec x)=t$ which means $dt=\tan x dx$
$$I=\int_0^\infty \frac{\cot x}{\tan x}tdt=\int_0^\infty \frac{t}{\tan^2 x}dt$$
I'm really disturbed by the $\tan^2 x$, I tried also to substitute $\sec x =t$ but it's not helpful either. Any helpful approach to solve this problem ?


问:
Complexity in Mathematics
How does one decide whether a theorem is qualitatively complex rather than hard?

Namely, how does one know when the proof of the theorem is irreducibly complex up to the least amount of computational steps required for a theorem's proof to Q.E.D.

答:
Proof complexity is mainly studied for propositional calculus. To extend the notion to typical "real" math theorems that require the expressiveness of first-order logic, you could define the complexity of a theorem to be the number of steps in its shortest natural deduction proof.

I don't think there's a practical way to compute this, though; you'll essentially have to just enumerate all possible proofs in order of length, checking to see if they prove the given sentence. If you don't already know that the given sentence (or its negation) is provable (i.e. that the sentence is not independent), there's no way to know when to stop searching.

More precisely, there's no computable function $L$ from strings to numbers such that whenever $\varphi$ is provable, $L(\varphi)$ is an upper bound on the length of the shortest proof of $\varphi$. If there were, then we could solve the halting problem by letting $\varphi$ be the statement that a given program halts and then checking all possible proofs up to length $L(\varphi)$ to see if $\varphi$ is provable.

In other words, as we consider longer and longer sentences, we encounter theorems that require "very long" proofs: any upper bound on proof complexity grows faster than any computable function.

Title: Complexity in Mathematics


问:
Infinitesimal Generator of a Group Action
This question is from IV.3, Exercise 6 in Boothby's Introduction to Differentiable Manifolds.  Given the action $$\theta(t; x, y) = (xe^{2t}, ye^{-3t})$$ from $\mathbb{R}\times\mathbb{R}^2 \rightarrow \mathbb{R}^2$ (treating $\mathbb{R}$ as your group and $\mathbb{R}^2$ as the manifold), determine the infinitesimal generator $X_p$ (herein "inf gen") and show that it is $\theta$-invariant.

He generically defines the infinitesimal generator as $$X_p f = \sum^n_{i=1} \dot{h^i}(0,x)\left(\frac{\partial \hat{f}}{\partial x^i}\right)_{\phi(p)}.$$

Here $x^i = {h^i}(0,x)$ and $\hat{f}(x^1, \ldots, x^n)$ is a local expression for the function $f\in C^\infty$.  The overdot indicates a derivative with respect to $t$.

Thank you in advance for any help.

[I originally asked for hints and geometric intuition.  See below for the solution attempt.]

答:
The answer per the hint is given as follows.

We have $\theta(t; x, y) = (xe^{2t}, ye^{-3t})$.  On inspection, we see that it is an action, as $\theta(0; x, y) = (x, y)$ and $\theta(t+s; x, y) = \theta(s+t; x, y)$.  Per the note, we compute

\begin{equation} \label{eq1}
\begin{split}
\dot{\theta_1} & = 2xe^{2t} \\
        & = 2\theta_1 \\
\dot{\theta_2} & = -3ye^{-3t} \\
        & = -3\theta_2. \\
\end{split}
\end{equation}

This means that the infinitesimal generator $X_{(x,y)} = 2x \frac{\partial}{\partial x} -3y \frac{\partial}{\partial y}$.  By Theorem 3.4 in Boothby IV.3, $X$ is $\theta$-invariant.  Without citing this, we can check if $X$ is $\theta$-invariant, i.e. if $\theta_{t*}(x,y) = X_{\theta_(x,y)}$.

The left-hand side has $X_{(x,y)}$ as above, with $\theta_t(x,y) = (xe^{2t}, ye^{-3t})$, which implies that $X_{\theta_t(x,y)} = 2xe^{2t}\frac{\partial}{\partial x} - 3ye^{-3t}\frac{\partial}{\partial y}$.

The right-hand side requires the computation of the pushforward $\theta_{t*}$.  We consider this $\theta_{t}$ as a function of $x$ and $y$.  Generally, for any $v \in T_pM$ with $v = v_1 \frac{\partial}{\partial x} +v_2 \frac{\partial}{\partial y}$, $$\theta_{t*} (v) =
\begin{pmatrix}
    \frac{\partial \theta_{t,1}}{\partial x}  &  \frac{\partial \theta_{t,1}}{\partial y}      \\
    \frac{\partial \theta_{t,2}}{\partial x}  &  \frac{\partial \theta_{t,2}}{\partial y}
\end{pmatrix}
\begin{pmatrix}
    v_1 \\
    v_2
\end{pmatrix}
$$

where the basis for $v$ (the matrix on the right) is implicitly $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$.  Computing the Jacobian, we have

$$\theta_{t*} =
\begin{pmatrix}
   e^{2t}  &  0    \\
   0  &  e^{-3t}
\end{pmatrix}.$$

So $\theta_{t*}(X_{(x,y)}) = \theta_{t*}(2x \frac{\partial}{\partial x} -3y \frac{\partial}{\partial y})$, and substituting into the above, we have that

$$\theta_{t*}(X_{(x,y)}) =
\begin{pmatrix}
   e^{2t}  &  0    \\
   0  &  e^{-3t}
\end{pmatrix}
\begin{pmatrix}
    2x \\
    -3y
\end{pmatrix}
=
\begin{pmatrix}
    2xe^{2t} \\
    -3ye^{-3y}
\end{pmatrix} .$$

Here again the basis is understood to be $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$.  So we are done.


答:
Substitute $t=\frac{1-x}{1+x}$

\begin{align}
&\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx \\
=&-2\int_{\frac13}^1 \frac{\ln t}{1+3t^2}dt
\overset{\sqrt3t\to t}= -\frac2{\sqrt3} \int_{\frac1{\sqrt3}}^{\sqrt3} \frac{\ln t-\ln\sqrt3}{1+t^2}dt\\
=& \frac{\ln 3}{\sqrt3} \int_{\frac1{\sqrt3}}^{\sqrt3} \frac{1}{1+t^2}dt = \frac{\pi\ln3}{6\sqrt3}
\end{align}
Note $\int_{\frac1{\sqrt3}}^{\sqrt3} \frac{\ln t}{1+t^2}dt \overset{t\to\frac1t} =0$

Title: Calculation of a hard integral $\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx $

Body:
Calculate
$$\int^{1/2}_0\frac{1}{x^2-x+1} \cdot \ln\frac{1+x}{1-x}\,dx $$
I tried to subsitute $x=1/2-t$ so $dx=-dt$ but I just complicated more my problem.


问:
Computing the gradient knowing directional derivatives
Suppose that the directional derivatives of f(x, y) are known at a given point in two nonparallel directions given by unit vectors u and v. Is it possible to find ∇f at this point?

I would like to know if the following is a valid way to solve this:

I was thinking of computing the tangent plane(assuming f is differentiable, which I believe is assumed in the problem) to the given point. Since the the tangent lines to the function in the direction of u and v must be contained in the plane, could I just take the cross product of

(u1,u2,directional derivative in the direction of u) x  (v1,v2,directional derivative in the direction of v)

and the resulting normal vector will be (a,b,c)
where

a= partial derivative with respect to x

b= partial derivative with respect to y

c = -1
Finally, the gradient would be --> (a,b)

Maybe, this is totally wrong, but I was trying to come up with a solution where the tangent plane was involved.
Thanks for your help!

答:
Just solve the linear system of equation
\begin{align}
a=\mathbf{u}\cdot \nabla f = f_x(a, b)u_x+f_y(a, b)u_y\\
b=\mathbf{v}\cdot \nabla f = f_x(a, b)v_x + f_y(a, b) u_y
\end{align}
where you know $a, b, \mathbf{u}$ and $\mathbf{v}$. That is
\begin{align}
\nabla f = [\mathbf{u}, \mathbf{v}]^{-1}
\begin{bmatrix}
a\\
b
\end{bmatrix}.
\end{align}


问:
Show that a sequence is mutually orthogonal.
I found this test which was given to math majors in a local University and one of the question(problem) says

Show that $\bigg\{\sin\bigg(\dfrac{n \pi x}{L}\bigg) \bigg\}_{n=1}^{\infty}$ is mutually orthogonal on $-L \leq x \leq L$.

I could not understand what mutually orthogonal means for a sequence. Can anyone elaborate this? Thanks in advance.

答: Hint: It means you have to show: For $m \neq n \implies \displaystyle \int_{-L}^L \sin\left(\frac{m\pi x}{L}\right)\cdot \sin\left(\frac{n\pi x}{L}\right)dx = 0$. Use: $2\sin A\sin B = \cos(A-B) - \cos(A+B)$ to rewrite the integrand.


问:
Solve $2\csc^2 x - \csc x - 1 = 0$ for $0 ≤ x ≤ 2π$
Solving some trigonometric equations and I'm getting a little thrown off by this one as it's the first one I've seen with the secondary trig ratios used instead of the primary ones. Could use a little help with this one please and thanks.</p>

答:
Try to not let the choice of function throw you; the approach is the same.  This is a quadratic equation.  Make the substitution to make this look more familiar.

Let $u = \csc x$

$2u^2 -  u - 1 = 0$

Factor

$(2u+1)(u-1) = 0$
$u = 1, u = -\frac 12$

Now we can think about these trig functions.

$|\csc x| \ge 1$ so we can disregard one of our values of $u.$

$\csc x = 1\\
\sin x = 1\\
x = \frac {\pi}{2}$


问:
Radius of convergence of power series (complex)
I don't know if my reasoning is right on this exercise:
If the power series $\sum a_n z^n$ has radius of convergence $R$, which is the radius of convergence of the series $\sum a_n^2 z^n$ and $\sum a_n z^{2n}$?

For the first one I am thinking: $\displaystyle \lim_{n \to \infty } \left|\dfrac{a^2_{n+1}}{a^2_n}\right|=\lim_{n \to \infty } \left|\dfrac{a_{n+1}}{a_n}\right|^2$= $R^2$.

And for the second, I believe it must have the same radius $R$, because it has the same coefficients, but we are only taking the pair powers, is this right?

答:
Hint: The radius of convergence $R$ of the series $\sum\limits_na_nz^n$ is characterized by the fact that $|a_n|r^n\to0$ when $r\lt R$ and $|a_n|r^n\not\to0$ when $r\gt R$ (note how one carefully avoids to specify what happens when $r=R$, with reason since several different behaviours are possible).

Thus, $|a_n^2|r^n=(|a_n|(\sqrt{r})^n)^2\to0$ if $\sqrt{r}\lt R$ and $|a_n^2|r^n\not\to0$ if $\sqrt{r}\gt R$. These conditions can be rewritten as $r\lt R^2$ and $r\gt R^2$, respectively, hence the radius of convergence of  $\sum\limits_na_n^2z^n$ is $R^2$.

Surely you can proceed likewise for $\sum\limits_na_nz^{2n}$.


问:
Value of $x$ in a failure rate problem
An electronic component is known to have a useful life which follows a probability
distribution with failure rate of $10^{-5}$
failures per hour. What is the probability that the
component would fail before the mean life of expected life?

This is Poisson Distribution right? Here the mean is $10^{-5}$ right but what is the value of $x$?

答:
A Poisson distribution is a counting distribution; it counts the random number of failures occurring in a fixed period of time.  Conversely, an exponential distribution is a failure time distribution; it measures the lifetime (time until failure) of a single component.

As such, a component that has a failure rate of $10^{-5}$ failures per hour translates (under the exponential lifetime model) to what average lifetime?  Think of it this way.  If on average we observe $10^{-5}$ failures per hour, then in $10^5$ hours, we would expect to observe $1$ failure, so the average lifetime of one component is $10^5$ hours.

Now, given this, what is the probability that the component happens to fail before its expected lifetime?  In other words, if $$X \sim \operatorname{Exponential}(\mu = 10^5)$$ where $\mu = \operatorname{E}[X] = 10^5$ hours, what is the probability $$\Pr[X \le \mu]?$$

Title: Value of $x$ in a failure rate problem


问:
If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?
If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?

I tried to solve it as follows:

I state that $p=ab$

$p^2=(2b+15)(2a+15)$

$p^2=4ab+30(a+b)+225$

$p^2=4p+30(a+b)+225$

and this is where I got stuck. I don't know how to get over this hurdle. could you please explain to me how to solve the question?

答:
Let $ a + b = c$.

Then, the quadratic $x^2 - 2 (c-x) -15 = x^2 +2x + (-15-2c) = 0 $ has distinct roots $a , b$.

Hence, these are the roots to the quadratic.

Thus, by Vieta's formula, $ a+b = - 2 \Rightarrow  c = -2$ and $ ab = -15 - 2c = -11$.


问:
Quick Confusion Regarding Expectation of Sample Variance in Poisson Distribution
Could somebody please quickly clear up a confusion of mine regarding the following?

Let there be a random sample from the Poisson distribution with parameter $\lambda$ as well as an an unbiased estimator of $\lambda$, $\tilde{\lambda}=S^2$. How is $\tilde{\lambda}$ unbiased?

More specifically, knowing that $\lambda=\mathbb{E}X=var X$, how is the expectation of the variance equivalent to the variance?

Thanks a bunch and apologies in advance if this seems trivial.

答:
how is the expectation of the variance equivalent to the variance?

Not exactly. The expectation of the sample variance is equal to the population variance

$$\mathbb{E}[S^2]=\mathbb{E}\left[\frac{1}{n-1} \Sigma_i(X_i-\overline{X}_n)^2    \right]=\frac{1}{n-1}\mathbb{E}\left[ \Sigma_i(X_i-\mu)^2-n(\overline{X}_n-\mu)^2    \right]=$$

$$=\frac{1}{n-1}\left[ \Sigma_i\mathbb{E}(X_i-\mu)^2-n\mathbb{E}(\overline{X}_n-\mu)^2    \right]=\frac{1}{n-1}\left[n\sigma^2-n\mathbb{V}[\overline{X}_n]\right]=$$

$$=\frac{1}{n-1}\left[n\sigma^2-n\frac{\sigma^2}{n}\right]=\sigma^2=\lambda$$


答:
This is quit easy if you are not afraid to use complex numbers. I will focus on the indefinite integral first. First note that $$x^2-x+1=(x-e^{i\pi/3})(x-e^{-i\pi/3})$$ (two purely complex roots of $-1$) and hence $$\dfrac{1}{x^2-x+1}=\dfrac{\sqrt{3}i}{3(x-e^{-i\pi/3})}-\dfrac{\sqrt{3}i}{3(x-e^{i\pi/3})}.$$ Since $$\ln\left(\dfrac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$$ We can easily split the given integral into four simple integrals of the form $$\displaystyle\int\dfrac{\ln (1\pm x)}{x-e^{\pm i\pi/3}}\,dx$$ and you can easily use dilogarithm to solve each of them respectively.

Added: In general, $$\displaystyle\int\dfrac{\ln(x-a)}{(x-b)}\, dx=\operatorname{Li}_2\left(\dfrac{a-x}{a-b}\right)+\ln(x-a)\ln\left(\dfrac{x-b}{a-b}\right)+C.$$


问:
How to prove that $\sum_{i=0}^\infty \int_{i\pi}^{(i+1) \pi}\frac{\sin x}{x}$ converges?
In short, I am trying to find a counterxample to Theorem 10.49b in Apostol's 'Mathematical Analysis' which is a claim that misses some assumptions related to this post: If $a_n=\int_n^{n+1} f,$ is it true that $f$ is Lebesgue integrable over $[1,\infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges?. My candidate is $\frac{\sin x}{x}$ which is not Lebesgue integrable over $\mathbb{R}_{\geq 0}$. To finish my counterxample I am left to prove that the series above convrges. But the only bounds I can think of are that $\frac{\sin x}{x} \leq \frac{1}{i\pi}$ on $[i\pi, (i+1)\pi)$. But of course, using this bound shows that the series is bounded as $$\sum_{i=0}^\infty \int_{i\pi}^{(i+1) \pi}\frac{\sin x}{x} \leq \sum_{i=1}^\infty \int_{i\pi}^{(i+1)\pi}\frac{1}{\pi} = \sum_{i=1}^\infty \frac{(i+1)\pi - i\pi}{i\pi} = \sum_{i=1}^\infty \frac{1}{i} \to \infty$$
which is useless. What am I missing?

答: This is an alternating series with decreasing terms, so yes, it converges.</p>


问:
$63^{63^{63}} \mod 100$
I need to find $63^{63^{63}} \bmod 100$.

This is what I've got so far:

Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$

Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$

Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$

I'm stuck here and don't know what to do next, what could I do now?

Title: $63^{63^{63}} \mod 100$

Body:
$63^7 = (60 +3)^7= ..... + {7\choose 2}60^2\cdot 3^5 + 7\cdot 60\cdot 3^6 + 3^7 \equiv 20\cdot 3^6 +3^7\pmod{100}$.

And as $2\cdot 3^6 \equiv 2\cdot 3^2 \equiv 8 \pmod {10}$ and as $3^7 = 9^3\cdot 3 = (10-1)^3 \cdot 3 = 3(100^3 - 3\cdot 100 + 3\cdot 10 - 1)\equiv 90-3 \equiv 87 \pmod {100}$

So have $63^7\equiv 20\cdot 3^6 + 3^7 \equiv 80 + 87 \equiv 67 \pmod{100}$.

.....

Also nothing wrong with  successive squaring.  $7$ is a small number.

$63^2 = 3969 \equiv 69$.

So $63^3 \equiv 69\cdot 63 = 4347 \equiv 40 + 80 + 27\equiv 47\pmod {100}$.

So $63^6 \equiv 47^2 \equiv 2209\equiv 9\pmod{100}$ and so

$63^7 \equiv 9\cdot 63 \equiv 567 \equiv 67 \pmod {100}$


答:
If $s = 1/\sin x $ the reciprocal trig relation can be

$$\dfrac{2}{x^2}-\dfrac{1}{x}-1=0,\to 2s^2-s-1 =0, \;(s-1)((2s+1)=0; $$

First root is $ 1/x=\pi/2 + 2 k \pi$

Second root is $ 1/x=3 \pi/2 \pm \pi/3 + 2 m \pi$

where $(k,m)$ are integers and we take their reciprocals once again.

Title: Solve $2\csc^2 x - \csc x - 1 = 0$ for $0 ≤ x ≤ 2π$

Body: Solving some trigonometric equations and I'm getting a little thrown off by this one as it's the first one I've seen with the secondary trig ratios used instead of the primary ones. Could use a little help with this one please and thanks.</p>


答:
This is a straightforward application of the alternating series test.

Define $a_i = \int\limits_{i\pi}^{(i + 1)\pi} \frac{\sin x}{x} dx$. Then $|a_i| = \int\limits_0^\pi \sin(x) \frac{1}{x + \pi i} dx$. Then clearly, $|a_i|$ is a decreasing sequence, since $|a_i| - |a_{i + 1}| = \int\limits_0^\pi \sin(x) (\frac{1}{x + \pi i} - \frac{1}{x + \pi(i + 1)}) dx > 0$.

And clearly, $a_i$ is an alternating sequence. And it's easy to show that $\lim\limits_{i \to \infty} a_i = 0$.

So by the alternating series test, your sum converges.

Title: How to prove that $\sum_{i=0}^\infty \int_{i\pi}^{(i+1) \pi}\frac{\sin x}{x}$ converges?

Body:
In short, I am trying to find a counterxample to Theorem 10.49b in Apostol's 'Mathematical Analysis' which is a claim that misses some assumptions related to this post: If $a_n=\int_n^{n+1} f,$ is it true that $f$ is Lebesgue integrable over $[1,\infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges?. My candidate is $\frac{\sin x}{x}$ which is not Lebesgue integrable over $\mathbb{R}_{\geq 0}$. To finish my counterxample I am left to prove that the series above convrges. But the only bounds I can think of are that $\frac{\sin x}{x} \leq \frac{1}{i\pi}$ on $[i\pi, (i+1)\pi)$. But of course, using this bound shows that the series is bounded as $$\sum_{i=0}^\infty \int_{i\pi}^{(i+1) \pi}\frac{\sin x}{x} \leq \sum_{i=1}^\infty \int_{i\pi}^{(i+1)\pi}\frac{1}{\pi} = \sum_{i=1}^\infty \frac{(i+1)\pi - i\pi}{i\pi} = \sum_{i=1}^\infty \frac{1}{i} \to \infty$$
which is useless. What am I missing?


问:
normal tables for a random variable which is the sum of N standard normal random vars?
Are there normal tables for a random variable which is the sum of N standard normal random vars?

https://youtu.be/rYefUsYuEp0?t=776

The lecture here says we can find $\xi'$ such that this probability $\alpha$ is $5$%.

But how do we do this really?

$H_0$ is the hypothesis that the $n$ variables are $X_i \sim N(0,1)$.

$H_1$ is the hypothesis that the $n$ variables are $X_i \sim N(1,1)$.

The $N$ variables are independent and identically distributed.

So if e.g. $N=5$ how do I find such $\xi'$ so that the tail (of the distribution of the sum of 5 standard normal random variables) has probability 5%. I thought only the standard normal distribution is tabulated really.

I mean, is there some way to do this without (too much) computer help?

Say I want to do this in an old-fashioned way with paper, pencil,

and some books and tables that I have in my hands. How?

答:
It is a matter of verifying a system of simple hypothesis (both simple) thus Neyman Pearson's lemma can be applied.

At the end you evaluate

$$P(\overline{X}_n\geq k|\mu=0)=0.05$$

$k$ can be found using Z-table

If $\overline{X}_n\geq k$ you decide for $H_1: \mu=1$ otherwise you decide for the mean to be zero

Speaking about numbers, using a random sample with size $n=5$, you set

$$P(\overline{X}_n\sqrt{5}\geq k)=0.05$$

$k=1.64$ is given by the z-table thus you reject $H_0$, say you decide for $H_1$ if and only if

$$\overline{X}_n\geq 0.7356$$


问:
Characterization of initial locally injective maps.
Is an initial and locally injective map between topological spaces an embedding? By initial I mean the topology of the domain is the initial topology. It seems easy to prove that such a map is a local embedding, but I fail to prove it is injective. On the other hand I can't find a counterexample to prove that the assertion is false.</p>

答: Let $f:X\to Y$ be initial and not injective. There are $x_0,x_1\in X$, $x_0\ne x_1$ such that $f(x_0)=f(x_1)=y$. Let $U$ be any open nbhd of $x_0$; then $U=f^{-1}[V]$ for some open nbhd $V$ of $y$, so $x_1\in U$. Thus, $f$ is not injective on any nbhd of $x_0$ and hence is not locally injective.


问:
Question about invertible matrix
Let V be a vector space, $dim(V) = n$, and $T: V \rightarrow V$ is an isomorphism. Suppose there is an invertible $n \times n$ matrix $P$. Does there exist a basis $\beta$ of $V$ such that $[T]_{\beta} = P$?

答:
Not necessarily. Take the special case where $T$ is the identity map. Then no matter what basis you pick, the matrix representation of $T$ will always be the identity matrix.

One particular counterexample would be

$P = \begin{pmatrix}
1 & 1 \\
0 & 1 \end{pmatrix}$

$n = 2$

$T = $ the identity map.

Title: Question about invertible matrix


问:
Integration using digamma functions
I'm trying to prove that for $x,y,r>0$ the following identity (gotten via CAS) holds:

$$\int_0^x \frac{x^r}{x^r+y^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1+r}{2r}\right)-\psi\left(\frac{1}{2r}\right)\right),$$

where $\psi(\cdot)=\frac{\Gamma'(\cdot)}{\Gamma(\cdot)}$ denotes the Digamma function. Does anyone has a clue, how integration can be done this way?

As an additional information following identity, holds also to be true:
$$\int_0^x \frac{y^r}{y^r+x^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1}{2r}+1\right)-\psi\left(\frac{1}{2r}+\frac{1}{2}\right)\right),$$
Which was obtained using:
$$\psi(x)=H_{x-1} - \gamma,$$
where $\gamma$ denotes the Euler-Mascheroni constant and $H_x$ denotes the $x^{th}$ harmonic number.

答:
First, substitute away the $x$ as in @Roman's answer. Then,
$$\int_0^1 \frac{dz}{1+z^r} = \int_0^1 \sum_{k=0}^\infty (-1)^k z^{kr} dz = \sum_{k=0}^\infty (-1)^k \frac{z^{kr}}{kr+1}\Bigg|_0^1 =  \sum_{k=0}^\infty \frac{(-1)^k}{rk+1}.$$
The result follows from the total convergence of the geometric series in an appropriate domain of $z$ and from the identity
$$\sum_{k=0}^\infty \frac{(-1)^k}{rk+1} = \frac{1}{2r} \left( \psi\left(\frac{r+1}{2r} \right) - \psi\left( \frac{1}{2r}\right)\right),\tag{$*$} $$
which is listed in the MathWorld page for $\psi$. Identity $(*)$ can be proven through the following series representation of $\psi$, which links it to the harmonic series:
$$\psi(s)=-\gamma+\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+s}\right);$$
through this, the RHS of $(*)$ becomes
$$\sum_{k=0}^\infty\left( \frac{1}{2kr+1}-\frac{1}{2kr+r+1}\right), $$
which turns out to yield the series on the LHS. (Try it!)

This procedure concerns the first integral, but can in principle be used to calculate the second integral as well. I wouldn't advise it, though, seeing as there is a much faster way: notice that summing the two integrals (omitting the $x$ in front) yields
$$\int_0^1 \frac{dz}{1+z^r} + \int_0^1 \frac{z^rdz}{1+z^r} = \int_0^1 \frac{1+z^r}{1+z^r} dz = 1. $$

Title: Integration using digamma functions


问:
Are consecutive products in the form $2^k-1$ divisible by the first such consecutive product?
Is it true that $\frac{(2^m-1)(2^{m-1}-1)\cdots(2^{m-k+1}-1)}{(2^k-1)(2^{k-1}-1)\cdots1}$ is always an integer, for integers $m\ge k\ge1$?

I've noticed that if $x$ divides $y$, then $2^x-1$ divides $2^y-1$, and since both the numerator and denominator have $k$ terms we might be able to prove it like in the proof that binomial coefficients are integers, but I'm not sure how. I've tried induction over $k$ and $m$ but it hasn't led anywhere, so how else can we prove that the answer is yes?

答:
Actually more is true: for all $m\geq k\geq 1$ the rational function
$$
\frac{(x^m-1)\cdots(x^{m-k+1}-1)}{(x^k-1)\cdots(x-1)}=\frac{(x^m-1)\cdots(x-1)}{(x^k-1)\cdots(x-1)(x^{m-k}-1)\cdots(x-1)}
$$
is a polynomial in $x$. Indeed, let us recall that
$$
x^n-1=\prod_{d|n}\Phi_d(x)
$$
where $\Phi_d(x)$ are the cyclotomic polynomials, defined by
$$
\Phi_d(x):=\prod_{j\in\{1,...,d\},\ (j,d)=1}(x-\exp(2\pi i d/j)).
$$
It then follows that
$$
(x-1)\cdots(x^K-1)=\prod_{j=1}^K\Phi_j^{\left\lfloor K/j\right\rfloor}.
$$
Hence the original rational function is
$$
\frac{(x^m-1)\cdots(x-1)}{(x^k-1)\cdots(x-1)(x^{m-k}-1)\cdots(x-1)}
=
\frac{\prod_{j=1}^m\Phi_j^{\left\lfloor m/j\right\rfloor}}
{\left(\prod_{j=1}^k\Phi_j^{\left\lfloor k/j\right\rfloor}\right)\left(\prod_{j=1}^{m-k}\Phi_j^{\left\lfloor {m-k}/j\right\rfloor}\right)}
$$
from which it should be possible to conclude (the degree of each cyclotomic polynomial in the denominator is lower than its corresponding degree in the numerator).


问:
How does Riesz representation theorem give us the left invariance of the Radon measure from a left invariant positive linear functional on $C_c(G)$?
I am trying to understand the end of the (canonical) proof of the existence of Haar measure on a locally compact group by Riesz representation theorem.

Let $G$ be a locally compact group and $C_c(G)$ denote the set of all compactly supported function on $G$. Suppose $I$ is a left-invariant positive linear functional on $C_c(G)$ (Here, by left-invariant we mean $I(L_x f)=I(f)$ for any $x\in G$, where $L_x f(y)=f(xy)$ ).

By Riesz representation theorem, there exists a unique Radon measure $\mu$ on $G$ such that

$$I(f)=\int_Gfd\mu, \forall f\in C_c(G).$$

From here I can use the Urysohn's lemma, monotone convergence theorem, etc. to conclude that $$\mu(xU)=\mu(U)$$

for any $x\in G$ and open subsets $U$ of $G$. But I wonder how to prove that

$$\mu(xE)=\mu(E)$$

for any $x\in G$ and Borel subset $E$ of $G$ (so that $\mu$ is a left invariant Haar measure on $G$).

Note that the latter doesn't follow from the former immediately.

答:
Use the uniqueness part of the theorem.  For each $x\in G$, the measure $\mu_x(E)=\mu(xE)$ satisfies $$\int f d\mu_x=\int L_{x^{-1}}f d\mu=I(L_{x^{-1}}f)=I(f)$$ for all $f\in C_c(G)$.  By the uniqueness of $\mu$, this implies $\mu=\mu_x$.

Or more directly, the fact that $\mu$ (and hence also $\mu_x$) is a Radon measure implies that it is uniquely determined by its values on open sets, by outer regularity.


问:
Factorize $x^4-4$ in $\mathbb{Q}[x]/(x^4+1)[x]$

What are the irreducible factors of $x^4-4$ in $K[x]$ where $K=\mathbb Q[x]/(x^4+1)$?

Clearly $x^4-4=(x^2+2)(x^2-2)$, but from here I'm not sure what to do. I tried showing that $\sqrt{2}\in K$ or $\sqrt{2}\not\in K$ but couldn't do either.

答: $K=\mathbb Q[x]/(x^4+1)=\mathbb Q(a)$, where $a^4+1=0$. This shows that $a$ is a primitive $8$th root of unity, so we may take $a=\frac{1}{\sqrt 2}(1+i)$. Show that $K=\mathbb Q(\sqrt 2,i)$. Now finish the proof.


问:
Let $(a_d)_{d\in D}$ be a real net such that $\lim_{d\in D}a_d=∞$. Is there a cofinal set $D'\subset D$ such that $(a_d)_{d\in D'}$ is increasing?
Let $(D, \geq)$ be a directed set, and let
$(a_d)_{d\in D}$ be a real-valued net satisfying
$$
\lim_{d\in D}a_d =+\infty.
$$
Can we find a cofinal subset $D'\subset D$ such that the restricted net $(a_d)_{d\in D'}$ is increasing, i.e., $a_{t}\geq a_d$ whenever $t,d\in D'$ and $t\geq d$?

The result is easily proved under the additional assumption that there exists a cofinal sequence $(t_n)_{n\in \mathbb N}$ in $D$, but I would like to avoid this assumption.

答:
This is a partial answer offered in the hope that it  might  help those searching for a counter example:

The claim is that, if $D$ contains a linearly ordered,  cofinal  subset, then the answer is positive.

In order to prove it we may of course assume that $D$ itself is linearly ordered.

It is then easy to recursevely   build an increasing sequence
$$
  i_0<i_1<\cdots <i_n<\cdots
  $$
in $D$ such that, for each $k\in {\mathbb N}\setminus\{0\}$, one has that
$$
  i\geq i_k \Rightarrow a_i> \max\{k, a_{i_{k-1}}\}.
  \tag 1
  $$

Letting $D'=\{i_0, i_1,  i_2, \ldots \}$, it is obvious that the restriction of the net to $D'$ is (strictly) increasing, so it suffices to
prove that $D'$ is cofinal.
For this, pick any $i$ in $D$, and let $k$ be any integer such that $k> a_i$.
Observe that one cannot have that $i\geq i_k$  or, otherwise,  (1) would imply that $a_i>k$.  The assumption that $D$ is
linearly ordered then gives $i<i_k$, hence proving $D'$ to be cofinal.

In conclusion,  should one be looking for a counter-example,  it will be necessary to consider directed sets $D$ not
possessing any linearly ordered,  cofinal subset.

One such example is the directed set
$\mathscr P_{\text{fin}}({\mathbb R})$
consisting of all finite subsets of ${\mathbb R}$,  under the
order of inclusion.   If $\mathscr L$ is a linearly ordered subset of
$\mathscr P_{\text{fin}}({\mathbb R})$,  one has that
$$
  \bigcup_{X\in \mathscr L}X
  $$
is countable,  so $\mathscr L$ cannot be cofinal.


问:
$\sum_{\rho}\frac{1}{|\rho|^2}=\sum_\rho\frac{1}{\rho(1-\rho)}$
Riemann Hypothesis(RH) is equivalent to the statement "
https://mathoverflow.net/questions/91280/is-this-sum-of-reciprocals-of-zeta-zeros-correct "$$\sum_{\rho}\frac{1}{|\rho|^2}=\sum_\rho\frac{1}{\rho(1-\rho)}$$ where $\rho$ denotes non trivial zeros of Riemann zeta function.

Question Prove that RH is equivalent to $$\sum_{\rho}\frac{1}{|\rho|^2}=\sum_\rho\frac{1}{\rho(1-\rho)}$$
Attempt
Assume RH then $\Re(\rho)=\frac{1}{2}$ ,so $1-\rho=\overline{\rho}$
$$\sum_\rho\frac{1}{\rho(1-\rho)}=   \sum_{\rho}\frac{1}{|\rho|^2} $$
Conversely, Assume,$$\sum_\rho\frac{1}{\rho(1-\rho)}=   \sum_{\rho}\frac{1}{|\rho|^2} $$
$$\sum_\rho [\frac{1}{\rho(1-\rho)}-\frac{1}{\rho\overline{\rho}}]=0$$
$$\sum_\rho \left(\frac{2\Re(\rho)-1}{\rho \bar{\rho}(1-\rho)}\right)=0$$
How to conclude that $$\Re(\rho)=\frac{1}{2}?$$

答:
Note that if $1/2<\sigma<1, t \ne 0, \rho=\sigma+it$ we have $|\rho|^2=\sigma^2+t^2>(1-\sigma)^2+t^2=|1-\rho|^2$ so $\frac{2\sigma -1}{|\rho|^2} < \frac{2\sigma -1}{|1-\rho|^2}$ (since $2\sigma-1>0$) and by rearranging a little we get:

$\frac{2\sigma}{|\rho|^2} +\frac{2(1-\sigma)}{|1-\rho|^2}< \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$

But now the expression above is symmetric in $\sigma, 1-\sigma$ so by interchanging them the inequality holds for $\sigma < 1/2$ too

Assume $\sum_\rho\frac{1}{\rho(1-\rho)}=   \sum_{\rho}\frac{1}{|\rho|^2}$ and note that both sides are absolutely convergent hence we can regroup at will.

In particular, we rearrange the terms so we group $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ together. If $\Re \rho =1/2$ this grouping has only two terms and appears twice (once for $\rho$ and once for $1-\rho$) while the sum is $\frac{1}{|\rho|^2}=\frac{1}{|1-\rho|^2}$ so the corresponding terms on RHS are equal.

If RH is false there is $\rho=\sigma+it, \sigma \ne 1/2, t > 0$ and the grouping above has $4$ terms each appearing twice (this time for $\rho, 1-\bar \rho$) with sum $\frac{1}{\rho(1-\rho)}+\frac{1}{\bar \rho(1-\bar \rho)}= \frac{2\sigma}{|\rho|^2} +\frac{2(1-\sigma)}{|1-\rho|^2}$ However in this case the corresponding terms on RHS are $\frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$ and by the result in the beginning we get a strict inequality.

Since by the grouping above we showed that LHS is a sum of positive terms, some equal with the corresponding ones on RHs and some strictly less we get the contradiction $\sum_\rho\frac{1}{\rho(1-\rho)} <  \sum_{\rho}\frac{1}{|\rho|^2}$ so our supposition that RH is false cannot hold. Done!


问:
The probability to pick a spade and then a heart.
From a standard deck of playing cards, you pick a card and without returning the card you pick a second card.  What is the probability that you picked a spade and then a heart?</p>

答: There are 4 suits and each suit has 13 cards. In your first draw, you will get a spade with probability 1/4 (each suit is equally likely). Now that you have a spade, there are 51 cards (12 spades and 13 each of the other 3 suits). Now your probability of getting a heart is 13/51. So the probability of getting a spade followed by a heart is $1/4 \times 13/51$.


问:
Show that the set $A=\{(x,y)\in\mathbb{R^2}: |x-y| \leq 2\}$ is not bounded.
im new on math and I'm trying to prove that the following set $A=\{(x,y)\in\mathbb{R^2}: |x-y| \leq 2\}$ is not bounded. This is my argument and I just need someone else's aprove.

My definition is

Let $A\subseteq \mathbb{R^2}$, we say that, for $A\not = \emptyset$, $A$ is bounded if $\exists M>0$ s.t. $\forall \bar{a} \in A$, $||\bar{a}|| \leq M$.

Proof.

Let $M\in \mathbb{N}$ such that $M>0$. Let $\bar{a}=(2M, 2M)$ thus $|2M-2M|=0 \leq 2$, then $\bar{a} \in A$ and $A \not = \emptyset$. We have that $||\bar{a}|| = \sqrt{4M^2+4M^2} = \sqrt{8}M$ which is clearly greater than $M$, that is to say $||\bar{a}||>M$.

Therefore, $A$ is a non bounded set.

答:
Very good demonstration! ButThe only detail I see is that:
\begin{equation*}
||\bar{a}|| = \sqrt{4M^2 + 4M^2} = 2\sqrt{2}M,
\end{equation*}
but anyway it is true that $||\bar{a}|| > M.$


问:
Find $\bigcup\limits_{n∈N} A_{n}$
$\bigcup\limits_{n∈N} A_{n}$ for $A_{n} = \{x∈R|x^2-nx>0\}$

From my understanding:

$x^2-nx=0$

$x(x-n)=0$

$x=0$

$x=n$

So:
$x^2-nx>0$ when $x<0$ or $x>n$

$x∈(-∞;0) U (n;+∞)$

So the union of these sets should be the set with the most elements? So since $x∈N$, the smallest x could be 0.0001 or 0.0000001. This is the part I'm having trouble with.

Any help or tip would be appreciated!

答:
First note if $x$ is negative,  $x^2-nx>0$ is always true for $n\in\mathbb N$.

And if $x$ is positve $$x^2-nx=x\cdot (x-n)>0$$ is equivalent to $$x-n>0.$$
Thus for any $n\in\mathbb N$ you get,

$A_n=\mathbb R^{-}\cup \{x\in\mathbb R^{+}\colon x-n>0\}=\mathbb R^{-}\cup \{x\in\mathbb R^{+}\colon x>n\}$.

Can you make it from here? Try to compute $A_1,A_2,A_3$. Now you should see $A_n\subset A_1$ for any $n\in\mathbb N$ and therefore $\bigcup_n A_n=A_1$

Title: Find $\bigcup\limits_{n∈N} A_{n}$


答:
You have come quite far on your own. The final stretch is concluding how the set must look like once forming the union over all n:

Let $n\in\mathbb{N}_{+}$, then $x^{2}-nx=x(x-n)>0 \iff (x>n \lor x<0)$, therefore $A_{n}=(-\infty,0)\cup(n,\infty)$. Hence $\forall n\in\mathbb{N}_{+} : A_{n+1} \subseteq A_{n}$ and obviously $\forall n\in\mathbb{N}_{+} : A_{1}\supseteq A_{n}$. If $A_{1}$ is the first set and contains all the others, then it is invariant to the union with later sets.

Thus, $\bigcup_{n\in\mathbb{N}_{+}}A_{n}=A_{1}=\mathbb{R}\setminus{[0,1]}$

Title: Find $\bigcup\limits_{n∈N} A_{n}$

Body:
$\bigcup\limits_{n∈N} A_{n}$ for $A_{n} = \{x∈R|x^2-nx>0\}$

From my understanding:

$x^2-nx=0$

$x(x-n)=0$

$x=0$

$x=n$

So:
$x^2-nx>0$ when $x<0$ or $x>n$

$x∈(-∞;0) U (n;+∞)$

So the union of these sets should be the set with the most elements? So since $x∈N$, the smallest x could be 0.0001 or 0.0000001. This is the part I'm having trouble with.

Any help or tip would be appreciated!


问:
Composition of the a linear transformation to equal another linear transformation
I'm trying to solve this question:

Consider a linear transformation $T: \mathbb{R}^2 \rightarrow
 \mathbb{R}^2$ given by:

$T(x,y) = (10x - 6y, -6x +10y)$

Decide if exists or not a linear transformation $F: \mathbb{R}^2
 \rightarrow \mathbb{R}^2$ such as $F\circ F = T$ and justify it.

I have no idea how to start solving this.

答:
$$T=\begin{pmatrix} 10 & -6 \\ -6 &10 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
Note that $T$ is symmetric, so this means it will have real eigenvalues and can be diagonalized in an orthonormal basis.
$$T = \begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix} \begin{pmatrix} 16 & 0 \\ 0 & 4  \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix}^{-1} = \begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 2  \end{pmatrix}^2 \begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix}^{-1}$$
$$=\begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 2  \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix}^{-1} \begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix}\begin{pmatrix} 4 & 0 \\ 0 & 2  \end{pmatrix} \begin{pmatrix} -1 & 1 \\ 1 &1  \end{pmatrix}^{-1}$$


答:
Another practical way is to take an unbounded sequence of $A$, say,
$$(x_n,y_n)=(n,n), \quad n\in \mathbb{N}.$$

Title: Show that the set $A=\{(x,y)\in\mathbb{R^2}: |x-y| \leq 2\}$ is not bounded.

Body:
im new on math and I'm trying to prove that the following set $A=\{(x,y)\in\mathbb{R^2}: |x-y| \leq 2\}$ is not bounded. This is my argument and I just need someone else's aprove.

My definition is

Let $A\subseteq \mathbb{R^2}$, we say that, for $A\not = \emptyset$, $A$ is bounded if $\exists M>0$ s.t. $\forall \bar{a} \in A$, $||\bar{a}|| \leq M$.

Proof.

Let $M\in \mathbb{N}$ such that $M>0$. Let $\bar{a}=(2M, 2M)$ thus $|2M-2M|=0 \leq 2$, then $\bar{a} \in A$ and $A \not = \emptyset$. We have that $||\bar{a}|| = \sqrt{4M^2+4M^2} = \sqrt{8}M$ which is clearly greater than $M$, that is to say $||\bar{a}||>M$.

Therefore, $A$ is a non bounded set.


问:
Solve ODE $xy+(3xy^2+x^2)y'=-1$
How to solve $xy+(3xy^2+x^2)y'=-1$ ?

I have tried all the possible ways to solve an ODE but could not solve this one.

答:
Write equation as
$$(xy+1)dx+(3xy^2+x^2)dy=0$$
Then integrating factor is
$$\mu=\frac{1}{x}$$
and solution
$$y^3+xy+\ln x=C$$

Title: Solve ODE $xy+(3xy^2+x^2)y'=-1$


答:
A matrix $M$ representing $T$ in any basis must satisfy the same polynomials that $T$ does: in particular its minimal polynomial must be the same.  So e.g. $M = \pmatrix{1 & 1\cr 0 & 1\cr}$ has minimal polynomial $(x-1)^2$, and can only represent $T$ if
$(T-I)^2 = 0$ and $T \ne I$.

Title: Question about invertible matrix

Body: Let V be a vector space, $dim(V) = n$, and $T: V \rightarrow V$ is an isomorphism. Suppose there is an invertible $n \times n$ matrix $P$. Does there exist a basis $\beta$ of $V$ such that $[T]_{\beta} = P$?


问:
Counting the number of partitions such that every part is divisible by $k$.
I would like to do this using generating functions. I'm comfortable with the generating function for the number of partitions of $n$: $$(1+x+x^2+\dots)(1+x^2+x^4+\dots)(1+x^3+x^6+\dots)\cdot\dots,$$ although I'm not entirely sure how to adapt it to deal with certain restrictions, such as asking that every part is divisible by $k$, or divisible by some pair of numbers, say $k$ and $j$.

I was thinking something like $$(1+x^k+x^{2k}+\dots)(1+x^{2k}+x^{4k}+\dots)(1+x^{3k}+x^{6k}+\dots)\cdot\dots.$$

Any advice or hints to get be going would be greatly appreciated!

答:
What you have is correct.  If $$f(x)=\sum_{n\ge 0} p(n) x^n= \prod_{j\ge 1} \frac{1}{1-x^j}$$ is the generating function for all partitions, then
$$\prod_{\substack{j\ge 1:\\ j\pmod k=0}}\frac{1}{1-x^j}
=\prod_{m\ge 1}\frac{1}{1-x^{km}} = f(x^k)
$$
is your desired generating function.


问:
Zero, a mystery.

Is zero even?

Is zero a multiple of ANY number?

I did some reading, and I found the following:

for 1) I see a whole Wiki article saying that this is true, but I just can't understand why?

for 2) I think it is, but I don't have a precise definition, and the only thing I remember is that in the junior classes, I used to write that the smallest multiple of any number is the number itself and it was approved by my teachers...so a proof on 1) and 2) would be highly appreciated...
Thanks a Ton!

答:
Remember that $k$ is even if and only if there is some $m$ such that $k=m+m$. No one said that $m$ cannot be equal to $k$. For $k=0$ we indeed take $m=0$ and we have $0=0+0$.

Similarly $0=0\cdot0=0\cdot1=0\cdot2=\ldots$, and as the definition of "$k$ is a multiple of $m$" says that there exists $n$ such that $k=n\cdot m$, we can take any $m$ whatsoever and $n=0$.

Title: Zero, a mystery.


问:
Does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists?
Let $k>2$, is a natural number, does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? If $k \to \infty$, then I think no, because $\ln k$ will go to infinite. And $\sum \frac{1}{n}$ diverge too, and so it's $\infty - \infty$. But someone say the limit exists? and it's positive. How?

答:
$$\sum_{n=1}^{k}\frac{1}{n}$$
is the sequence of harmnonic numbers (of order 1).

This is a plot of the starting of the minuends/subtrahents under consideration:

This plot shows the starting of the difference:

This looks by graphical representation pretty convergent.

The limit can be representated as the riemannian sum limit of the difference of two integrals the second one as riemannian sum, the as easy integral:
$$\lim_{k\rightarrow\infty}(-ln(k)+\sum_{n=1}^{k}\frac{1}{n})=\int_{1}^{\infty}(\frac{1}{x}-\frac{1}{\lfloor x\rfloor})=\gamma=0.577216$$
$\lfloor x\rfloor$ represents the floor function.
This page show a lot more information about the Euler Mascheroniconstant and that this identity stems from L. Euler himself.
The proof is rather easy because the integral is over this

A step size below 1 does not change anything. It is a stair function in the denominator. The riemannian sum approximation is exact for stair functions.

Despite the exact representation this is not a formlar for calculating the Euler-Mascheroni-constant depsite up to some decimals. The name of the differnce is convention and the numerical value an reals with only approcimate represenations.

The convergence is really slow but this picture is one the is the really famous signature of the convergence as example between 7030 and 7100 is looks still nice convergent

This is exact to the 5th decimal.

Mind the the first term in the middle of my equation chain does not exist alone. So this is indeed an instruction to calculate the Euler-Mascheroni-constant numerical. The riemannian theory of integrals gave it a new meaning centuries later.

Both term of the difference are each divergent but with a finite value at infinity. That is really strange despite it is clear by the graphics and by the continous decline of the difference and the lower limit given by the x-axis. The sequence of the difference is strictly monotonous decrease and has a lower limit, so is it convergent. Rules valid in general found here, Limit of a sequence.

$$\frac{1}{k+1}+ln(k)-ln(k+1)<0    :    \forall k>2$$

$$\frac{1}{k+1}>-ln(k)+ln(k+1)=ln(\frac{k+1}{k})=ln({1+\frac{1}{k})   :    \forall k>2$$

$ln(x)$ is divergent and there monotonous increasing. Therefore is always $ln(k+1)>ln(k)$ for all integer $k$. $\frac{1}{x}$ in monotonous decrease and converges towards $0$. To add this to $ln(k)$ does not enlarge the value large enought to get bigger than $ln(k+1)$.

This is initially verifiable by

The sequence is elementwise positive. This can be infered by the first plot of my answer and the fact that both a monotonously increasing. All increase are strict. From sentences about steady functions with one know order at point and the knowledge that both are strictly increasing follows the condition if fullfilled.

That is all needed. This can be put into statements with quantors.


Title: Prove $7^{71}>75^{32}$

Body:
My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$

I did the first question: \begin{align}\frac{6^9}{10^7}&=\frac{4}{5}\times\frac{27^3}{25^3}\\&=0.8\times1.08^3\\&>0.8\times(1+3\times0.08+3\times0.08^2)\\&>0.8\times(1+3\times0.086)\\&>0.8\times1.25=1.\end{align}

But I can't work out the second,  I calculated it out on my computer, $\frac{7^{71}}{75^{32}}=1.000000949\cdots$

Body:
$\color{green}{\textbf{Just now!}}$

$$\ln\dfrac{75^{32}}{7^{71}}=\ln\left(\dfrac{20}{21}\left(\dfrac{2400}{2401}\cdot\dfrac{225}{224}\right)^{14}\dfrac{78125}{78732}\right)$$
$$=\ln\left(\dfrac{41-1}{41+1}\left(\dfrac{4801-1}{4801+1}\cdot\dfrac{449+1}{449-1}\right)^{14}\dfrac{156857-607}{156827+607}\right)$$
$$= 2\sum\limits_{k=1}^\infty\dfrac1{2k-1}\left(-\dfrac1{41^{2k-1}}-\dfrac{14}{4801^{2k-1}}+\dfrac{14}{449^{2k-1}}-\dfrac{607^{2k-1}}{156827^{2k-1}}\right)$$
$$< 2\left(-\dfrac1{41}-\dfrac1{4\cdot41^3}-\cdot\dfrac{14}{4801}+\dfrac{14}{449}-\dfrac{607}{156827}\right)$$
$$-\dfrac1{6\cdot41^3}+ \dfrac1{48\cdot449^2}\left(1+\dfrac{5}{3\cdot 449^2}+\dfrac{5^2}{3^2\cdot 449^4}+\dots\right) <0.$$
Proved!

$\color{brown}{\textbf{Previous version.}}$

Equivalent inequality is
$$71\ln 7 > 32\ln3+64\ln 5.$$

Taking in account numeric equalities
$$7^5=16807,\quad 3^3\cdot 5^4=16875,\quad 5^6 = 15625,\quad 3^7\cdot7 = 15309,$$
one can denote
$$2\ln5-3\ln3=\ln\dfrac{25}{27} = \ln\dfrac{1-x}{1+x},\quad x = \dfrac1{26},$$
$$3\ln3+4\ln5-5\ln 7= \ln \dfrac{3^3\cdot5^4}{7^5} = \ln\dfrac{1+y}{1-y},\quad y = \dfrac{34}{16841},$$
$$7\ln3+\ln7-6\ln5 = \ln \dfrac{3^7\cdot7}{5^6} = \ln\dfrac{1-z}{1+z},\quad z = \dfrac{158}{15467}.$$

Easily to check expression for the goal sum:
$$S=32\ln3+64\ln 5-71\ln7 = \ln\dfrac{1-x}{1+x}+14\ln\dfrac{1+y}{1-y}-\ln\dfrac{1-z}{1+z}.$$
If $\;x,y,z,t>0\;$ then
$$S = -\ln\dfrac{1+x}{1-x}+14\ln\dfrac{1+y}{1-y}+\ln\dfrac{1+z}{1-z},$$
$$\ln\dfrac{1+t}{1-t} = \ln(1+t)-\ln(1-t) = 2t +\dfrac23t^3+\dfrac25t^5+\dfrac27t^7+\dots,$$
$$S_1=-\ln\dfrac{1+x}{1-x} < -2x-\dfrac23x^3 = -\dfrac{2x(3+x^2)}{3}$$
$$S_1<-\dfrac{2029}{26364}<-0.076\,961\,007,$$
$$S_2=14\ln\dfrac{1+y}{1-y}<28y(1+y^2+y^4+\dots)=\dfrac{28y}{1-y^2},$$
$$S_2<\dfrac{28\cdot34\cdot16841}{16841^2-34^2}<0.056\,528\,941,$$
$$S_3=\ln\dfrac{1+z}{1-z}\le2z+\dfrac23z^3(1+z^2+z^4+\dots) =2z+\dfrac23\,\dfrac{z^2}{1-z^2} =\dfrac{2z}{3}\dfrac{3-2z^2}{1-z^2},$$
$$S_3 <\dfrac{316}{46401}\dfrac{3\cdot15467^2-2\cdot158^2}{15467^2-158^2}<0.020\,431\,305$$
$$S=S_1+S_2+S_3 < -0.000\,000\,761 < 0.$$

Proved!


问:
Concrete Filtration at specific time
Consider random variables $X(t) \geq 0$ for $t=0,1,2,3$ defined on a filtered  probability space with $\Omega=\{u,d\}^3$, where $u>d$ and $u,d \geq 0$.

I holds  $$X(t)(\omega)=w_t$$ for any $\omega=(\omega_1,.., \omega_3)$

The filtration at $t$ is generated by $X(t)$, meaning : $\mathcal{F}(t)=\sigma(X(1),..,X(t))$

For $t=0$ define $\mathcal{F}(0)=\{ \emptyset, \Omega\}$.

I want to compute $\mathcal{F}(1),\mathcal{F}(2)$.

Since I know at $t=1$, which state $X(1)$ has taken, I would conclude: $$\mathcal{F}(1)=\{ (u,u,u), (u,d,d), (u,u,d), (u,d,u)\}$$
I assumed $X(1)=u$.
In similar way I would argue for $t=2$:$$ \mathcal{F}(2)=\{ (u,u,d), (u,u,d)\}$$
assuming $X(1)=X(2)=u$.

What do you think?

答:
The meaning of a filtration

See, the idea of a filtration is of storing accumulated information.

What does this mean? Well, imagine you are travelling on a bus, on a route you've never been on. You note down the stops that come your way : when you are at a certain time, you know everything about a the place you are at and everything before it, but you don't quite know what comes afterwards. That's the point of a filtration : to store the list of events that you know have happened within a certain time, or are decided by a certain list of random variables, you use a filtration.

For example  by time $1$ you are expected to have seen $X_1$, and therefore $\mathcal F(1)$ is the list of events that you know for sure have occurred or not, by only looking at $X_1$.

Note that an event is a subset of the sample space i.e. $\Omega$. So $\mathcal F(1)$ is a subset of $\mathcal P(\Omega)$, the set of subsets of $\Omega$. Every element of $\mathcal F(1)$ is a subset of $\Omega$. You seem to have written just one event when you wrote down $\mathcal F(1)$ ; that event turns out to be a good example though.

One can similarly speak of $\mathcal F(2)$.

Key point : The occurence of an event in $\mathcal F(1)$ depends $\color{green}{\text{only on } X_1}$, and every such event will belong to $\mathcal F(1)$.

Why do you think that $\mathcal F(0) = \{\emptyset , \Omega\}$? This is because these are the two events you can be certain have occured or not without knowing anything : for any other event, you will have to know something about the process to know if it occurred or not.

Examples

We will take examples, and in the process point out a stunning commonality of the affirmative cases, with the help of which I will mention a general pattern that one can use to form the filtrations in the blink of an eye.

Let's take the event $E_1 = \{(u,u,d)\}$. Is this event in $\mathcal F(1)$? Can I decide using just $X_1$ if $E_1$ has occurred or not? The answer is no, because if $X_1 = u$ occurs, I'd still have to check if $X_2 = u$ and $X_3 = d$ happen, only then can $E$ happen. So I need to go beyond $X_1$ to check if $E_1$ occurs or not.

Let's take $E_2 = \{(u,u,d),(u,d,u),(d,d,u)\}$. Is this event in $\mathcal F(1)$? Once again, the answer is no : for example, if $X_1 = u$ then I can't conclude that $E_2$ has occurred : what if $X_2= X_3 = d$ happens? Then $E_2$ won't happen. So $E_2$ very much depends upon $X_2,X_3$.

Let's take $E_3 = \{(u,u,d),(u,d,u),(u,d,d),(u,u,u)\}$. Is this event in $\mathcal F(1)$? The answer is YES. Here's why : if $X_1 = u$ then regardless of what $X_2$ and $X_3$ are, the event $E_3$ occurs. On the other hand, if $X_1 = d$ then once again the event certainly does not occur. In conclusion, $E_3$ depends only on $X_1$.

Let's take $E_4 = \{(u,u,d),(u,d,u),(u,d,d),(u,u,u) , (d,u,u)\}$. Is this event in $\mathcal F(1)$? The answer is NO, because if $X_1 = d$ then it is still possible that $X_2 = d$ so then $E_4$ won't occur.

Now we are beginning to see things : what's happened in $E_3$ is that after $X_1 = u$, all possible combinations of $X_2$,$X_3$ were covered in $E_3$. Because of that , the event $E_3$ said : I don't care about $X_2$ or $X_3$, just tell me if $X_1 = u$ or not. The other events did not . $E_4$ said :  if $X_1 = u$ then I occur, but if $X_1 =d$ then you'll need to check $X_2$ and $X_3$  to see if I occur or not.

This idea of covering all possible future combinations is the key to forming sets that belong to filtrations. Including all possible combinations says "I don't care about those coordinates" and that's what you want, isn't it?

Observing the pattern

The pattern is thus noticed as follows : if you want an event to depend only on a certain bunch of random variables, make sure that the event includes all possible combinations of the rest of the variables.

Now, what does that mean in the context of $\mathcal F(1) = \sigma(X_1)$? Well, you need all possible combinations of $X_2,X_3$ in any event in $\mathcal F(1)$.

Of course both the empty set and $\Omega$ belong to $\mathcal F(1)$. There are only two other possible events in $\mathcal F(1)$. Why? Well, let's imagine an event $E \in \mathcal F(1)$. If it contains $(u,*,*)$ for any value of $*$ then it must contain all possible combinations so that the value of those $*$s don't matter for $E$. In particular, it must contain $E_3 = \{(u,u,d),(u,d,u),(u,u,u),(u,d,d)\}$.

Similarly, if $E$ contains $\{(d,*,*)\}$ for any value of $*$ then it must contain all possible combinations ,so it must contain $\{(d,u,u),(d,u,d),(d,d,u),(d,d,d)\}$.

That gives us only these sets, which form $\mathcal F(1)$ :
$$
\mathcal F(1) =\Big\{\emptyset , \Omega , \{(u,u,d),(u,d,u),(u,u,u),(u,d,d)\} , \{(d,u,u),(d,u,d),(d,d,u),(d,d,d)\} \Big\}
$$

If you take any other set, it will depend on $X_1$. Try it!

A different way of understanding $\mathcal F(1)$

That is through the following lemma :

$\mathcal F(1)$ is precisely the set of all events that can be obtained as preimages of $X_1: \Omega \to \mathbb R$.

So what we need to do, is characterize the preimages of $X_1$. Let $S \subset \mathbb R$. Then $\{\omega : X_1(\omega) \in S\} = \{\omega : \omega_1 \in S\}$. But then, this event doesn't depend upon $\omega_2$ and $\omega_3$.  Now see the magic :

$$
\emptyset =  \{\omega : X_1(\omega) \in \emptyset\}\\
\Omega =  \{\omega : X_1(\omega) \in \{u,d\}\} \\
\{(u,u,d),(u,d,u),(u,u,u),(u,d,d)\}  = \{\omega : X_1(\omega) \in \{u\}\} \\
\{(d,u,u),(d,u,d),(d,d,u),(d,d,d)\} = \{\omega : X_1(\omega) \in \{d\}\}
$$

now do you understand the beauty of the lemma? We obtain the sets we need as preimages of $X_1$ under certain well chosen sets.

Onto $\mathcal{F}(2)$

For $\mathcal F(2)$ we again use free choice for the last coordinate $X_3$. So if an event in $\mathcal F(2)$ contains $(u,d,*)$ it must contain all values of $*$ so that the third coordinate stops mattering for that event.

With that, we get four fundamental sets , corresponding to the first two combinations of $X_1,X_2$ as $(u,u,*),(u,d,*),(d,u,*)$ and $(d,d,*)$:
$$
F_1 = \{(u,u,d),(u,u,u)\} \\
F_2 = \{(u,d,d),(u,d,d)\} \\
F_3 = \{(d,u,d),(d,u,u)\} \\
F_4 = \{(d,d,d),(d,d,u)\}
$$

Any event in $\mathcal F(2)$ is a union of the events $F_1,F_2,F_3,F_4$. For example $F_1 \cup F_2 \in \mathcal F(2)$, $F_3 \cup F_4 \cup F(1) \in \mathcal F(2)$. So $\mathcal F(2)$ is precisely given by the set $$\mathcal F(2) = \{\cup_{i \in \mathcal I} F_i : \mathcal I \subset \{1,2,3,4\}\}$$

(Note : the empty union gives $\emptyset$ and $\Omega = F_1 \cup F_2 \cup F_3 \cup F_4$, so they very much belong to the filtration).

And how to get this using the preimage method? Some examples are given from which I'd assume you'd be able to get the general idea :
$$
F_1 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(u,u)\}\} \\
F_3 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(d,u)\}\} \\
F_2 \cup F_4 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(u,d), (d,d)\}\}
F_1 \cup F_2 \cup F_4 = \{\omega : (X_1(\omega),X_2(\omega)) \in \{(u,u), (u,d), (d,d)\}\}
$$

Try to see that $\mathcal F(1) \subset \mathcal F(2)$ (hint : try to write each element of $\mathcal F(1)$ as a union of the $F_i$).

Title: Concrete Filtration at specific time


问:
Newton's root finding method that copes with stationary points
I am using Newton's method to solve an equation of the form $y=f(x)$ for increasing values of $y$. The value of $x$ found for some $y$ is used for the next $y$ as a starting approximation and this works fine.

$f$ is increasing but may have stationary points, and I do have situations where the range of interest of $y$ starts or ends on an extremum or very close. This causes the algorithm to experience a division by zero or yield very inaccurate roots.

Is there a simple way to cope ? Is there a known modified Newton's method able to deal with this situation ?

I know that double roots can be handled by the modified iterations $$x\leftarrow x-2\frac{f(x)}{f'(x)}$$ but this still does not work at the root, and one needs to know when to activate the factor $2$.

For information, my function $f$ is a cubic polynomial, but I want to avoid the explicit computation of the roots by Cardano's formulas.

Update:

If we restate Newton's method at a stationary point, we get to the next term of Taylor's development,

$$f(x+h)-y\approx f(x)-y+\frac{h^2}2f''(x),$$

so that $$h\approx\sqrt{-\frac{f(x)-y}{f''(x)}}.$$

In fact, the approximation

$$f(x+h)-y\approx f(x)-y+h f'(x)+\frac{h^2}2f''(x)$$ always holds, giving the general formula

$$h\approx \frac{f'(x)\pm\sqrt{f'^2(x)-2f''(x)(f(x)-y)}}{f''(x)}.$$

Due to the increased complexity, I don't want to use that formula all the time.

答:
Momentum-based algorithms are good at passing through stationary points while retaining fast convergence.

$$m_i=(1-\beta)f'(x_i)+\beta m_{i-1}=m_{i-1}-(1-\beta)(m_{i-1}-f'(x_i))$$
$$x_{i+1}=x_i-(1-\beta^i)\frac{f(x_i)}{m_i}$$

The $1-\beta^i$ factor serves as bias correction, supposing one takes $m_0=0$, since we get $m_i=(1-\beta^i)f'(x_i)$ if $x_i$ is fixed.

As $\beta\to0$, you get Newton's method and faster convergence, while for larger $\beta$ you will get better "resilience" against stationary points.

For your purposes, you know that $f'(x)\ge0$, so rather than setting $m_0=0$, you may want to initialize it to a small positive value, or perhaps even use the same $m_i$ sequence for every $y$, since it is likely you will converge to the solution for $y=f(x)$ for a given $y$ while $m_i$ is sufficiently large that starting from where it left off works well for the next $y$.

If you go with the latter approach, you'll want to choose $\beta$ based on the density of your $y$ around the stationary points. If $\beta$ is too small and you have many $y$ near a stationary point, then $m_i$ will decay rapidly, which leaves you with the original issue.


问:
Prove that there exists a path in G of length K

Let $G=(V,E)$ be a nonempty graph and let $d(v)\geq k$ for all $v\in V$. Prove that there exists a path in $G$ of length $k$.

Alright so my first intuition is
$$\sum_{i=1}^{n} d(v_i) \geq nk \quad(n=|E|)$$

So a path contains each vertex once and has no cycle. Seeing as each vertex has a degree $\geq k$, in the worst situation (because this limits the length of our path) we have:
$$\sum_{i=1}^{n} d(v_i) = nk$$

We also have $\sum_{v\in V} d(v_i) = 2|E|$, so
$$
2|E| = nk.
$$

This seems logical but I cant translate this into the length of a path. Any tips?
edit
maybe we can say because we have $\frac{nk}{2}$ edges, and $n$ vertices, we could maybe use induction?

答:
Hint. Pick a vertex $v_1$. It is connected to $k$ vertices. Pick one of them $v_2$, this is connected to $k$ vertices, and $k-1$ of them are not in the path (only $v_1$ can be).

$v_2$ is connected to at least $k-2>0$ vertices not in the path. Pick one of them, $v_3$ repeat....

Induction by $k$ is probably the  simplest way to write the solution....


问:
Show that if a map between manifolds has a linear coordinate representation at each point, then it has constant rank
I would like to prove the following corollary from the book ISM by John M. Lee, and the proof concerning the direction from (a) to (b) is confusing me. I can't understand the meaning of the sentence underlined with red. As far as I know, the rank of a linear transformation $T$ (between vector spaces) is defined as the dimension of the range of $T$, which is a number unique to $T$. So, why does Lee say that every linear map has constant rank? Does he mean that a linear map can have many ranks? Thank you.

答:
There are two different ways one might interpret the "rank" of a linear map $F:V\to W$ between finite-dimensional vector spaces.

In the linear-algebraic sense, the rank of $F$ is just the dimension of its image. This is just a nonnegative integer.

We can also consider $F$ as a smooth map between manifolds. In this context, its rank at a point $p\in V$ is the rank (in sense #1) of the linear map $dF_p\colon T_pV\to T_{F(p)}W$, so in principle $F$ might have different ranks at different points. But the differential of a linear map is the linear map itself (after we make the canonical identifications $T_pV\cong V$ and $T_{F(p)}W\cong W$), so when considered as a smooth map, every linear map has constant rank.


问:
Square root of a $3×3$ matrix
I was trying to find square root of this $3×3$ matrix
$$\begin{pmatrix}
12&8&31\\
7&14&28\\
9&35&6\\
\end{pmatrix}$$
I searched Wikipedia and some site for helpful information and also try to
$${ \begin{pmatrix}
a&b&c\\
d&e&f\\
g&h&i\\
\end{pmatrix} }^3 =
\begin{pmatrix}
12&8&31\\
7&14&28\\
9&35&6\\
\end{pmatrix}$$
Though I know that a similar question has been posted for a $2×2$ matrix, the workings are different so I'll need some help on this
Thanks

Title: Square root of a $3×3$ matrix

Body:
There are eight (complex) square roots of this matrix, since there are three distinct, nonzero eigenvalues.  The closed-form expressions for these will be quite unpleasant, as the characteristic polynomial is an irreducible cubic.  A numerical approximation for one of them is

$$
\left[\begin{array}{ccc} 2.994275474+ 0.1518364930 \;{i} &  1.197978251+ 2.057588565 \;{i} &  3.000734902- 2.183155336 \;{i}
\\
  0.7775140113+ 0.1253357713 \;{i} &  3.460596511+ 1.698468161 \;{i} &  2.737777769- 1.802119185 \;{i}
\\
  0.9410435414- 0.2015472331 \;{i} &  3.344874508- 2.731235902 \;{i} &  2.789009650+ 2.897912794 \;{i} \end{array}\right]
$$


问:
How to show that there is no positive rational number a such that $a^3$ = 2?
I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—the book answer says that $m^3$ becomes $m^3 = 2(4p)^3$. And $2n^3 = 2(4p)^3$. I'm super confused here and can't really understand what's going on here.

Can someone tell what exactly is going on here and how should I prove this?

答:
Note that $m^3 = 2n^3$ implies that $m$ is even (is it clear why?). So you can take $m=2p$ for some $p$. Therefore:
$$
m^3 = 2n^3 \implies (2p)^3 = 2n^3 \implies 4p^3 = n^3.
$$
Because of the same reason, $n=2q$ for some $q$. Finally, use that $m$ and $n$ do not have common factors (you did not mention it in your question, but that is a key point).