问:
About little o notation
I was wondering if the order between multiplication and little o can be exchanged, for example, $$x \times o\left(\frac{y}{x}\right) = o(y)?$$

I am a little confused.

In the example, I need to consider the case when $x$ goes to infinity, and don't know if eliminating $x$ will be a good idea. As far as I know, it is correct that $\lim_{x\rightarrow \infty} x \times o\left(\frac{y}{x}\right) =0$, but I can't get $\lim_{x\rightarrow \infty}  o(y) =0$. So I think $$x \times o\left(\frac{y}{x}\right) = o(y)$$  is not true?

Added:
 My case is that given $x \times o_{t\rightarrow \infty}(t^2)$ and $t=\sqrt{y/x}$, so I get $x \times o_{y/x \rightarrow \infty}(y/x)$. Now with this meaning and my consideration when x goes to infinity, is $x \times o_{y/x \rightarrow \infty}(y/x)$ same as $o_{y \rightarrow \infty}(y)$ or as $o_{y/x \rightarrow \infty}(y)$? Is it right that $\lim_{x\rightarrow \infty} f(x,y)=0, \forall f \in x \times o(y/x)$?

Thanks!

答:
You have every right to be confused; little-o notation itself is problematic, and this question highlights at least two of its problems.

Let's be completely precise. We say that $f(x) \in o_{x \to \infty}(g(x))$ (note the subscript) if $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$. This is the most precise form of little-o notation: I have not used an equals sign (because $o(g(x))$ is not a function), I have specified what variable the notation refers to, and I have even specified what its limit is. (Most people who use little-o or related notations don't bother to do any of these things, and the result can be quite confusing if you aren't aware of the context.)

Thus $f(y) \in o_{y \to \infty}(y)$ if $\lim_{y \to \infty} \frac{f(y)}{y} = 0$. In this situation, we might want to interpret

$$f(x, y) \in x o_{x \to \infty} \left( \frac{y}{x} \right)$$

to mean that

$$\frac{f(x, y)}{x} \in o_{x \to \infty} \left( \frac{y}{x} \right)$$

where $y$ is possibly a function of $x$. This is equivalent, of course, to $\lim_{x \to \infty} \frac{f(x, y)}{y} = 0$, so it is equivalent to the statement that

$$f(x, y) \in o_{x \to \infty}(y).$$

But your question is a little confusing, because you have two variables $x$ and $y$ and you didn't specify which one of them the little-o notation refers to, and you also didn't refer to $f$. Both of these are weaknesses of little-o notation itself.

Title: About little o notation


问:
Rationale behind truth values
I originaly asked a question on Programmers.SE to know why $0$ was consider $\text{false}$ and all the other [integral] values were considered $\text{true}$. That was a huge debate and many said it was a legacy from Boolean algebra where $0$ is indeed $\text{false}$ and $1$ is $\text{true}$.

Somebody suggested I go further and ask here why this is actually the case in Boolean algebra. So here is the question: what is the rationale for $0$ to be $\text{false}$ and $1$ to be $\text{true}$ and not the other way around in Boolean algebra?

答: I wouldn't be surprised if a version of this convention goes back to Boole himself, in his algebra of classes.  I believe he used $0$ for the empty class and $1$ for the class of "everything".  (This was before the set-theoretic paradoxes made people queasy about the class of everything.)  Under the natural correspondence between classes and functions to "true" and "false" (where a class $C$ corresponds to the function sending elements of $C$ to "true" and "everything" else to "false"), these would be the constant "false" function for $0$ and the constant "true" function for $1$.  So it was natural and convenient to identify the truth values with these numbers.</p>

ContentLicense: CC BY-SA 3.0

Title: Rationale behind truth values


问:
What does this $f(x) = \{$ mean?
Let $f:\mathbb{Z}\to\mathbb{Z}$ be defined by

$$f(x)=\begin{cases}\frac{x}{2}&x \text{ is even}\\
0&x \text{ is odd}\end{cases}$$

Does this mean for all values of $f(x) = x/2, x$ is even and for all values of $f(x) = 0,x$ is odd?

答: It means that if $x$ is even, hence $x=2k$ for some $k\in\mathbb{Z}$, then $f(x)=k$. Otherwise, if $x$ is odd then $f(x)=0$

ContentLicense: CC BY-SA 4.0

Title: What does this $f(x) = \{$ mean?


问:
Same number or each Sylow p-subgroup $\implies$ isomorphism?
I have a question about Sylow theory and isomorphic groups. I came with it while trying to solve some problems from my group theory course. The question is if this statement is true in general:

Being $G$, $H$ two groups of same order, $|G|=|H|=p_1^{\alpha_1}\cdots
> p_n^{\alpha_n}$ with each $p_i$ prime. Then, if the number of Sylow
$p_i$-subgroups in $G$ equals the number in $H$ for all $i=1,...,n$,
then $G\cong H$.

Is this true? Is it true just in certain cases? If not, why? Any help will be appreciated, thanks in advance.

答: This looks false to me. Take $Q_8$ and $D_8$ for instance. Then, they have the same number of Sylow 2-subgroups (since their orders are $8$, they both have one) but they are not isomorphic.


问:
Interesting rotating parabola proof
So, I am reading linear algebra and cartesian geometry by Juan de Burgos for the first time, and came across this problem that I just can't figure out. Goes like this:

Consider a parabola, whose distance from the focus to the directrix is a fixed p, and that rotates while keeping its focus fixed.
We also consider a certain fixed direction in the parabola's plane, and consider the tangent to the parabola that has that given direction.
Find the equation of the geometric space that all the tangency points describe.

The original question is in Spanish but I have tried to translate it the best I can. If anyone speaks Spanish I can send you the original question.

Edit: Ill keep working on it and try to write my work in here, sorry that I didnt do from the beginning

Ok, so first off the book gives the soultion to a problem in the back, but without any explanation or none of that. it says:

**Origin at the focal point, << x axis>> in the given direction;

directrix $xcos(\theta)+ysin(\theta)=p$.

Parabola $x^2 sin^2(\theta)+y^2 cos^2(\theta)-2xysin(\theta)cos(\theta)+2p(xcos(\theta)+ysin(\theta))=p^2$

The polar of (0,1,0) is $xsen^2(\theta)-ysin(\theta)cos(\theta)+pcos(\theta)=0$.

Eliminating $\theta$ we get $p^2(x^2+y^2)=4y^4$.**

I get the part of creating a new coordinates taking the focal point to be F(0,0). as so:
see image here
The directrix in a standard parabola where the vertix is in the origin has directrix x=p/2, so because we are taking the focal point it would be now x=p. Because of the rotation described for the parabola, i thought of using rotation of coordinate axes, as so:
see image
And thats why the directrix would be  $xcos(\theta)+ysin(\theta)=p$ as described in the solution given by the book.
(Thanks to a commenter here who clarified to me that  $xcos(\theta)+ysin(\theta)=p$ or  $xcos(\theta)+ysin(\theta)=-p$ would make no difference, I was confused on that too).

Now, how in the world did they get the parabola described in the solution and the stuff that comes after?

I have read the chapter like three times but I dont know why i cant understand this one problem

答:
I understood the problem in this way:

A parabola  (focus to directrix distance = 2p)  rotates about its focus  at origin. Find equation to the variably rotated parabola and its tangent as function of the rotation angle $\alpha$.

Answer:

I have tried to rotate the entire  parabola/tangent at vertex as one combination.

The given parabola equation:

$$ x^2 \sin ^2 t + y^2 \cos^2 t-2 x y \sin t \cos t+y^2 \cos ^2 t +2 p (x \cos t +y \sin t) = p^2 \tag 1 $$

For $t=0$ it reduces to
$$ 2 p x + y^2 = p^2 \tag 2 $$

Bring in parameter $t$ in a slightly different way

$$ x= \dfrac{p^2-t^2}{2p},   y = t  \tag 3$$

and then we try to  rotate the entire parabola through $\alpha$.

$$ x\to x \cos \alpha- y \sin \alpha , y\to x \sin \alpha +y \cos \alpha \tag 4 $$

Solving for $(x,y)$ from the rotated coordinates ( rather than directly)  with new parametrization $t$ included

$$
x \cos \alpha- y \sin \alpha =t, \; x \sin \alpha +y \cos \alpha = \dfrac{p^2-t^2}{2p} \tag 5 $$

Next put $t=0$ gives starting point $( x=0, y=p/2)$ that is the point of tangency of tangent $x= 2 p$ at parabola vertex is obtained by $\alpha$ rotation

$$ x \sin \alpha+y \cos \alpha = p/2 $$

are lines tangent to circle diameter $p$ not sketched for the  familiar tangent to circle tangent lines not to crowd up too much in the image.

Let the variable $ \cos \alpha = u $  for short.

$$ x u -y \sqrt{1-u^2}=t,\; x \sqrt{1-u^2} + y u =\dfrac{p^2-t^2}{2p} \tag 6$$

Solve for $(x,y)$  as functions of $(u,t)$ that respectively act upon to enable tangent point change as well as rotate the entire parabola.

$$ (x,y)=$$
$$ \left( \dfrac {t u + (p^2 - t^2) \sqrt{1 - u^2}} {2 p}, \dfrac {  u(p^2-t^2)  + 2 p t \sqrt {1 - u^2}}{2 p} \right) \tag 7 $$

The graph is drawn for $p=2$.

A non-vertex parameter $t$ point could be also taken for a parameter wlog.

Title: Interesting rotating parabola proof


答:
Yes: if, say, $P(x)$, $Q(x)$ and $R(x)$ are predicates in the variable $x$, then $$f(x)=\begin{cases}g(x)&\text{if }P(x)\\ h(x)&\text{if }Q(x)\\ i(x)&\text{if }R(x)\end{cases}$$

is standard notaton for a function $f$ such that $f(x)=g(x)$ for all $x$ such that $P(x)$, $f(x)=h(x)$ for all $x$ such that $Q(x)$, and $f(x)=i(x)$ for all $x$ such that $Q(x)$. And so on for more cases. This kind of presentation of a function is called piecewise definition, and the "if" is often omitted because that is the way the cookie crumbles. It is highly recommendend, if not mandatory, that the cases listed cover the entirety of the domain of $f$. It is good practice to also have the cases listed be mutually exclusive. Often, the last case listed may be substituted by the word "otherwise" instead of actually writing down the negation of the logical disjunction of all the other cases.

Title: What does this $f(x) = \{$ mean?

Body:
Let $f:\mathbb{Z}\to\mathbb{Z}$ be defined by

$$f(x)=\begin{cases}\frac{x}{2}&x \text{ is even}\\
0&x \text{ is odd}\end{cases}$$

Does this mean for all values of $f(x) = x/2, x$ is even and for all values of $f(x) = 0,x$ is odd?


答:
For any even integer number $x$ i.e $....,-2, 2 , 4 ,6 ..... $
the function divides $x$ by 2.

For any odd integer number $x$ i.e $....,-1, 1 , 3 ,5 ..... $
the function output is zero.

Title: What does this $f(x) = \{$ mean?


问:
sum of ten squares
You are given an unlimited supply of $1\times 1,2\times 2,3\times 3,4\times 4,5\times 5,6\times 6$ squares.Find a set of ten squares whose areas add up to $48$.If not the whole solution,even a little prod in the right direction would help.</p>

答:
One solution is obvious:

8 squares of 1x1

1 square of 2x2

1 square of 6x6

$$ Area = (8 \times 1) + (2 \times 2) + (6 \times 6) = 48$$

Title: sum of ten squares


问:
Relation between variance, standard deviation and mean
I have a survival function with multiple terms like:

$$ S(X) = {1 \over 3}e^{(-x/3)}+{2 \over 3}e^{(-x/6)}$$

I want to calculate the expected value and variance.

$$ E(X) = \int_0^{\infty} S(X) = 5 $$

For the variance I used:

$$ S(X) = 1 - F(X) = 1 - ({1 \over 3}e^{(-x/3)}+{2 \over 3}e^{(-x/6)})  $$

$$ f(x) = {1 \over 9 } e^{-x/3} + {2 \over 18 } e^{-x/6} $$

$$ \sigma^2 = 2 \int_0^{\infty} xS(X) dx - E(X)^2 = \int_0^{\infty} x^2f(x)dx -E(X)^2  = 54 - 25 = 29$$

For exponential distributions, the mean is $ 1 \over \lambda$ and the variance $1 \over \lambda^2 $, so that the $ E(X) = \sqrt{\sigma^2} = \sigma$. This is not the case for the formula above and somehow I think I understood the whole thing not completely.

The first question is, is my variance correct? If yes, how can I imagine it visually? I always thought about it as from 0 to $\sigma^2$ with the mean in the middle.

Thank you, WiPu

答: Your problem concerns a random sampling of one of two exponential distributions. So$$\Bbb EX=\tfrac133+\tfrac236=5,\,\Bbb EX^2=\frac132\cdot3^2+\frac232\cdot6^2=54,$$as per your calculations. More generally, if $X$ samples $\operatorname{Exp}(\lambda_i)$ with probability $p_i$ then$$\Bbb EX=\sum_i\frac{p_i}{\lambda_i},\,\Bbb EX^2=\sum_i\frac{2p_i}{\lambda_i^2}=\sum_{ij}\frac{2p_ip_j}{\lambda_i^2}\implies\sigma^2=\sum_{ij}\left(\frac{2p_ip_j}{\lambda_i^2}-\frac{p_ip_j}{\lambda_i\lambda_j}\right).$$


问:
Proving $n\text{P}(Y>n) \rightarrow 0$ implies $n\text{P}(Y> \epsilon n) \rightarrow 0$ for any $\epsilon > 0$ for
I am currently stuck at the following question: If we know that $Y$ is a random variable in $\text{R}$ that satisfies $n\ \text{P} (Y>n)\rightarrow 0$ as $n \rightarrow \infty$. Can we conclude that $n \  \text{P}(Y>n \epsilon)$ will also converge to zero for any $\epsilon >0$? If so, which theorem can we employ to prove this? Thanks for any hints or insights provided!

答: I believe the hypothesis $n P(Y > n) \to 0$ implies $n \epsilon P(Y > n \epsilon) \to 0$, from which you can obtain the desired result.


问:
I can't find the exact limits to find the marginal probability density function (PDF) from a uniformly joint PDF.
For a uniform probability density function (PDF) of two random variables $X$ and $Y$, the PDF exists only in the square of length $\sqrt{2}$, whose vertices are $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$ (please see attached image below, where I also defined the equations of the lines).

Now, we know that the area of this square is $2$, and its PDF is $$f_{X,Y}(x,y)=\frac{1}{2}$$

Now, if I want to find the marginal PDF of $X$, for example, I want to sum the joint PDF over all values $Y$ can take. But I don't know what the limits of integration will be. Do I just sum from -1 to 1 for $y$?

答:
When you are finding the marginal pdf of $X$ from the joint pdf of $X$ and $Y$, you are calculating the integral
$$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y)\, \mathrm dy\tag{1}$$
in which $x$ is just a constant, say $0.123$ or $-0.456$. That is, you need to evaluate $(1)$ for uncountably infinitely many values of $x$. However, things are not as bad as this suggests, because we can figure out the value of $f(x)$ for huge chunks of values of $x$ if we just try to draw a few more lines on the figure that you have sketched.  For example, Eq. $(1)$ tells us that
for any choice of $x \in [-1,0)$,
\begin{align}f_X(x) &= \int_{-\infty}^\infty f_{X,Y}(x,y)\, \mathrm dy\\
&= \int_{-1-x}^{1+x} \frac 12\, \mathrm dy\end{align}
which you can figure out by drawing a vertical line at some choice of $x \in [-1,0)$ and checking where it intersects the edges of the joint pdf support. You even have the equations for the edges marked on your sketch!!
Similarly,  for any choice of $x \in [0,1]$,
\begin{align}f_X(x) &= \int_{-\infty}^\infty f_{X,Y}(x,y)\, \mathrm dy\\
&= \int_{-1+x}^{1-x} \frac 12\, \mathrm dy.\end{align}


答:
First, note that this function has as its domain and contradiction the set of integers.

The expression
$$f(x)=\begin{cases}\dfrac{x}{2}&x \text{ is even}\\[3pt]
0&x \text{ is odd}\end{cases}$$

It means that if $x$ is even, then the value of $f(x)$ will be $x/2$. For example, $f(10)=10/2=5$, $f(2)=2/2=1$, if $x$ is odd, then $f(x)$ is $0$, for example $f(1)=0, f(3)=0, f(21)=0$, ...

This is a piecewise function

Title: What does this $f(x) = \{$ mean?


问:
a question on $\epsilon$-chain sequences
Suppose $X$ is a compact metric space and $ f : X \to X$ is a homeomorphism.

By an $\epsilon$-chain from $x$ to $y$ in $X$ we mean a finite sequence $x_0 ,x_1 ... , x_k$ in $X$ with $x_0=x$ and $x_k=y$ such that :
\begin{align}
d(f(x_n) , x_{n+1})< \epsilon \; , \; n=0 ,..., k-1
\end{align}
a point $x \in X$ is said to be chain recurrent of $f$ , if for any $\epsilon>0$ there is an $\epsilon$-chain from $x$ to $x$.

Now suppose $f: [a,b] \to [a,b]$ is a strictly increasing homeomorphism which $a$ , $b$ are fixed points of $f$.

Now if $x \not\in Fix(f)$ then we may assume $f(x)>x$, take $\epsilon=\dfrac{|f(x)-x|}{2}$. I want to prove that there is no $\epsilon$-chain from $x$ to $x$ with this $\epsilon$.

By contradiction I suppose by this $\epsilon$ we have an $\epsilon$-chain so we have a finite sequence like $x=x_0 ,x_1 ... x_k=x$ such that :
\begin{align}
d(f(x_n) , x_{n+1})<\epsilon , \; n=0 ,..., k-1
\end{align}
I can't reach a statement which contradicts my assumption.

答: Suppose that $x=x_0,x_1,\ldots,x_k=x$ is an $\epsilon$-chain, where of course $k>0$. Clearly $x_1>f(x_0)-\epsilon=x+\epsilon$. In particular, $x_1>x$, so $$f(x_1)>f(x)=x+2\epsilon\,,$$ and $$x_2>f(x_1)-\epsilon>x+\epsilon\,.$$ In general, suppose that $x_i>x$ for some $i\in\{1,\ldots,k-1\}$; then $$f(x_i)>f(x)=x+2\epsilon\,,$$ so $$x_{i+1}>f(x_i)-\epsilon>x+\epsilon\,.$$ By induction it follows that $x=x_k>x+\epsilon$, which is absurd.


问:
Problem understanding fallacious argument on Courant/Robbins's "What is Mathematics"?
I've been reading Courant/Robbins' "What is Mathematics?"

The first time they mentions $a),b)$ is here:

In another section of the book, he points out there is an alternative formulation:

Now here:

I don't see what is the fallacy here: He showed that $A_1$ is true, that is: He showed that $b)$ is true. But when trying to show that $a)$ is true, it seems he did a weird thing in there that looks like $a)$ but isn't? It's not clear to me. Can you help?

答: $\alpha$ and $\beta$ don't necessarily satisfy the hypotheses of $A_r$.


答:
Admittedly, this would be more legible perhaps when written
$$ f(x)=\begin{cases}\frac x2&\text{if $x$ is even}\\0&\text{if $x$ is odd}\end{cases}$$
but the short is, well, shorter.

This is the unique function with the properties

If $x$ is even, then $f(x)=\frac x2$

If $x$ is odd, then  $f(x)=0$.

Note that $f(0)=0$ even though $0$ is even (cf. Numberphile).

It is important to note that for each $x$ in the domain, exactly one of the cases applies and hence a unique function value is defined, as is needed.

(Sometimes, you may encounter a piecewise definition with overlapping conditions, perhaps something like
$$ |x|=\begin{cases}x&\text{if }x\ge 0\\-x&\text{if }x\le 0\end{cases}$$
in that case, one has to verify that different branches do give the same function value where they overlap)

Title: What does this $f(x) = \{$ mean?


答:
There is nothing weird about what he’s doing in (a): he is simply trying to apply the induction hypothesis. The induction hypothesis is that $A_r$ is true: if $\max(a,b)=r$, then $a=b$. Now we assume that $a$ and $b$ are positive integers such that $\max(a,b)=r+1$. Then $\max(a-1,b-1)=r$, so by the induction hypothesis $a-1=b-1$, and therefore $a=b$.

The problem with this is that $A_r$ is the statement that if $a$ and $b$ are any two positive integers, and $\max(a,b)=r$, then $a=b$, and the fact that $a$ and $b$ are positive does not ensure that $a-1$ and $b-1$ are positive. If at least one of $a$ and $b$ is equal to $1$, we cannot apply the induction hypothesis $A_r$ to $a-1$ and $b-1$, because they are not both positive integers.

Title: Problem understanding fallacious argument on Courant/Robbins's "What is Mathematics"?

Body:
I've been reading Courant/Robbins' "What is Mathematics?"

The first time they mentions $a),b)$ is here:

In another section of the book, he points out there is an alternative formulation:

Now here:

I don't see what is the fallacy here: He showed that $A_1$ is true, that is: He showed that $b)$ is true. But when trying to show that $a)$ is true, it seems he did a weird thing in there that looks like $a)$ but isn't? It's not clear to me. Can you help?


问:
For $A\subset X$ a subset of a topological space, a universal property uniquely determines the quotient $X/A$
I have been working on a question with two parts.
PART ONE

Let $(X, \tau_x)$ and $(Y,\tau_Y)$ be topological spaces and $A\subset X$ a subset. Define an equivalence relation $\sim$ on $X$ by $x \sim y$ if and only if $x=y$ or $x,y, \in A$.

Let $X/A$ be the quotient $X/\sim$ equipped with the quotient topology, denoted simply by $\tau$ and let $p: X\to X/A$ be the resulting map. Let $f: X\to Y$ be a continuous map such that $f(A)=\{ f(a): a\in A\}$ has exactly one element. Show that there exists a unique continuous map:
$$g: X/A \to Y \quad \text{ with } f= g \circ p.$$

I derived here that the desired property essentially fixes our map (giving us uniqueness) for $x \in X$:
$$ f(x)= (g \circ p)(x)= g(p(x))= g([x]) \implies g([x])=f(x).$$
Well-definedness follows from the fact that $|f(A)|=1$, so if $x\sim x$ we have $g([x])=f(x)$ and $g([x'])=f(x')$.
Now by $x \sim x'$ either $x=x'$ or $x,x'\in A$ we see that since $|f(A)|=1$ in both cases $f(x)=f(x')$ hence the map is well-defined. (Now we also have existence, since by construction it satisfies the property).

I can then use a lemma in the book, since $f$ is continuous and $p$ is the canonical projection, hence surjective, we have that $g$ is continuous. END OF FIRST PART.

Now for the difficult part, part 2. This exercise proves a universal property. The claim is that this universal property uniquely determines the quotient.

We wish to show that this last property determines $X/A$ uniquely as follows:

Let $(Q, \tau_Q)$ be another topological space with a continuous map $q: X \to Q$ such that $q(A)$ has
exactly one element and the following property: for any continuous map $f : X \to
Y$ such that $f(A)$ has exactly one element, there exists a unique continuous map $g: Q \to Y$ such that $f = g \circ q$.

Show that there is a unique homeomorphism $\phi: Q \to X/A$ such that $\phi \circ  q = p$.

I understand that what we want to show is that in a way all the conditions fix $Q$ to be the same quotient  as $X/A$. To show that two spaces are homeomorphic we need to construct a map that is a bijection and also continuous with continuous inverse map. I cannot find a good way to construct this $\phi$ however.  Does anyone have a good candidate?

答: HINT: Take $Y=X/A$ and $f=p$ to see that there is a unique continuous $h:Q\to X/A$ such that $p=h\circ q$; this is your candidate for $\varphi$. Then take $Y=Q$ and $f=q$ to see that there is a unique continuous $\hat h:X/A\to Q$ such that $q=\hat h\circ p$. Note that $p=h\circ q=h\circ\hat h\circ p$, and $q=\hat h\circ p=\hat h\circ h\circ q$.


答:
It is obvious that $A_2$ is false. (The maximum being two doesn’t imply the two positive integers are equal.) Thus the argument in the induction step in part (a) must be invalid.

Where is the error? It’s in the tacit assumption that the maximum of $\alpha$ and $\beta$ is even defined (according to the precise definition given in the statement of the problem). The trouble is that it might not be: indeed it won’t be if $a$ or $b$ is equal to one, for then either $\alpha$ or $\beta$ is not positive. The maximum function, as defined in the problem, is defined only for positive integers.


问:
For which type of input does this operation repeat the outcome eventually and run forever?
This problem is a part of another (programming) problem on which I am working, the start of which requires us to identify a certain pattern, I believe.

We are given a pair of positive integers x and y. We are going to perform an operation on this pair repeatedly to obtain a pair with equal values. We could as well say that this operation tries to balance this pair of integers so as to make them equal. The balancing operation is as follows: If the pair has unequal values, take the minimum, add it to the smaller integer and subtract it from the larger one.
If the pair still has unequal values, repeat this operation on the new values.
The task is to find whether this operation will run forever or eventually stop(making the values equal)? Can this be determined by seeing the initial input itself?

Hint for remembering the operation easily: The smaller one gets doubled, the larger one diminishes by value equal to the smaller one. See examples below.

Example: Start with (1, 4). We see that (1, 4) --> (2, 3) --> (4, 1) --> (3, 2) --> (1, 4). At this point it is clear that this operation will not give equal values for the pair(1, 4). Also, at the point when we got the interchanged values (4, 1), we could say that they are going to be repeated (I guess).

Example: Start with (1, 1). We see that the numbers are already equal. So, this operation is not applied. Or, we could say the algorithm/operation terminated at the start itself.

Example: Start with (5, 8). We see that (5, 8) --> (10, 3) --> (7, 6) --> (1, 12) --> (2, 11) --> (4, 9) --> (8, 5) --> (3, 10). If ordering is ignored, the values look the same as before. They are in cycle. It will not give equal values ever.

Example: Start with (1, 3). We see that (1, 3) --> (2, 2). Operation terminated since the pair has equal values now.

答:
Since $x+y$ remains constant, this is essentially a 1D problem. 2D is just smoke and mirrors.

Let us denote $x+y=2n$ (because if the sum is odd, good luck trying to split it in equal halves). Let us suppose that on the first step one of the integers in the pair is $x$, because  er, well, that's what it is. What will it be on the next step? There are two options:

If it was the smaller number, it will turn into $2x$.

If it was the bigger one, it will become $x-y = x-(2n-x) = 2x-2n$, which still looks awfully like $2x\pmod{2n}$ to me.

By the same logic, on the next step we'll get $4x\pmod{2n}$, and so on.

So that's what is going on: we iterate through various remainders $2^k\cdot x\pmod{2n}$. We'll inevitably start going in cycles at some point or another, because there are only so many different remainders. Now the question is: will it be a cycle of size 1 (or rather, the cycle of size 1, because there is only one), or not? In other words, will our sequence ever hit $n\pmod{2n}$, or not?

Let's say $2^m$ is the greatest power of 2 that divides $n$. In other words, $n=2^m\cdot l$, where $l$ is odd. Now here is the deal: $\pmb x$ must be divisible by $\pmb l$. If it is, we'll hit $n$; if it isn't, we won't.

How so?

First the negative part. Any number which equals $n\pmod{2n}$ should be divisible by $l$, but if $x$ isn't, then neither is any of the $2^k\cdot x$, so we're out of luck.

Now the positive part. Let's say $x$ is some odd number times $2^p$. Obviously, $k<m$ (otherwise $x$ would be divisible by $l$ and by $2^m$, which means it would be divisible by $n$, which means it would be $n$ and the journey would end right where it started). Well, on the next step it would be $p+1$ instead of $p$, and so on, until we reach $m$. At that point, our $2^k\cdot x$ is divisible by $l$ and by $2^m$, which means it is divisible by $n$, but not by $2n$, which means... see above.

As a particular case, if $n$ is a power of 2, then any $x$ will do.

So it goes.


问:
Problem on inequality with power sum
Let $m$ and $n$ are positive integer

Can it be shown that, For every $m\ge 5$

$$\sum_{i=1}^ni^m-m^i>0\iff n=2,3,...,m$$

Example: let $m=5$, choose any $n$ between $2$ to $5$, now let $n=2$ then $\sum_{i=1}^2i^5-5^i=(1^5-5^1)+(2^5-5^2)=-4+7=3>0$

More on observation, $\sum_{i=1}^ni^m-m^i=0$ holds only for $(m,n)=\{(1,1),(2,3),(2,4)\}$

Score Pari/GP

```<code>for(m=5,50,for(n=1,50,if(sum(i=1,n,i^m-m^i)>0,print([m,n]))))
</code>```

Title: Problem on inequality with power sum

Body:
Lemma 1: $2^m> m^2+m-1, \forall m>5$.

When $m=5, 2^5=32>29=5^2+5-1$. If $2^m>m^2+m-1$ then $$2^{m+1}-[(m+1)^2+(m+1)-1]\\>2(m^2+m-1)-(m+1)^2-(m+1)+1 = m(m-1)-3 >0.$$

Lemma 2: $i^j < j^i, \forall i>j>e.$

We check the first derivative of the function $f(x)=x^\frac 1x$:
$$f'(x)=x^{\frac 1x -2}(1-\ln x) < 0, \forall x>e.$$

First we prove $\Leftarrow$.

From the lemmas
$$\sum_{i=1}^n (i^m-m^i)=1+2^m-m-m^2+\sum_{i=3}^n(i^m-m^i)>0.$$

Next we prove $\implies$.

If $n=1$, it's trivial.

If $n>m$,
$$\sum_{i=1}^n (i^m-m^i)=\sum_{i=1}^{m+1} (i^m-m^i) + \sum_{i>m+1} (i^m-m^i) \le \sum_{i=1}^{m+1} (i^m-m^i) \text{ via Lemma 2}$$

Note that $e^{-\frac km} \ge 1-\frac km \implies \left(1-\frac km \right)^m \le e^{-k}, k=0, 1, \cdots, m-1$

Hence
$$
\sum_{i=1}^{m+1} (i^m-m^i) = \sum_{i=1}^{m+1} i^m - \sum_{i=1}^{m+1} m^i = m^m\left(\left(1+\frac 1m\right)^m + \sum_{k=0}^{m-1} \left(1-\frac km\right)^m \right) - \frac{m^{m+2} - m}{m-1}\\
< m^m \left(e+\sum_{k=0}^\infty e^{-k} - \frac{m^2-m^{1-m}}{m-1}\right) < m^m \left(e+\frac{1}{1-e^{-1}} - m\right) \approx m^m(4.3003-m) <0.\blacksquare
$$


问:
Approximation of matrix in 2-norm

Given a rank-$k$ matrix $A$, find a rank-$j$ matrix $B$, where $j<k$, such that $\|A-B\|_2$ is minimal.

My idea was to choose, if

$$A=P \operatorname{diag}(\sigma_1,\ldots,\sigma_k,0,\ldots) Q^H$$

then

$$B=P \operatorname{diag}(\sigma_1,\ldots,\sigma_j,0,\ldots) Q^H$$

Is this approach correct? If so, then I would try to prove that this is actually the best approximation.

答: Yes, your approach is correct and is a standard result on Singular value decomposition. The topic is termed as Low Rank Approximation. It should be there in any standard matrix theory book talking about SVD. </p>

Title: Approximation of matrix in 2-norm


问:
Generalising norms over arbitrary fields
I am curious about the structure we require on arbitrary fields if we want to generalise the notion of norms on vector spaces over fields other than either $\mathbb{R}$ or $\mathbb{C}$. Specifcally, given a vector space $V$ over $\mathbb{F}$, what structure must be imposed on $\mathbb{F}$, such that we have a map $\|\cdot\|: V \to \mathbb{F}$ s.t. for any $a \in \mathbb{F}$ and any $\bar u, \bar v\in V$,

$$\text{(i) }\|\bar u+ \bar v\| \leq \|\bar u\|+\|\bar v\|$$
$$\text{(ii) }\|\bar v\| = e \to \bar v = \boldsymbol{0}, \text{where } e \text{ is the additive identity of } \mathbb{F}$$
$$\|a\bar u\| = |a|\|\bar u\|$$

From this question here, it seems that positive definiteness requires that we have an ordered field. The accepted answer implies this is enough structure to generalise the inner product.

Where I am getting somewhat confused is how the generalisation works for norms. Specifically, is it really enough to have just have an ordered field in order to satisfy the norm property of $\|a\bar x\| = |a|\|\bar x\|$?

I know that having an ordered field allows us to generalise (i) and (ii), but  how do we define what the modulus of $a$ is? I think the linked question is relevant since if we generalise the inner product over an arbitrary field, and we get a norm out of those inner products, we have made some progress.

How does ordering the field allow us to define the modulus of scalars in a vector space over an ordered field? Do we need to specify additional structure?

答:
The trivial real-valued absolute value $$|a|_{tr}=\cases{1 \ if \ a\ne 0\\ 0 \ otherwise}$$ works over any field $F$. We can then consider the trivial real-valued norm $$\| v\| = \sup_j |v_j|_{tr}, \qquad v\in F^n$$
When $char(F)=0$ you can think to $\|.\|$ as being $F$-valued, taking its values in the ordered subfield $\Bbb{Q}$.


答: The method is always the same for all kinds of universal properties: The universal property of $X/A$ applied to the map $q:X\longrightarrow Q$ yields a unique map $X/A\longrightarrow Q$ with special properties. The universal property of $Q$ applied to the map $p:X\longrightarrow X/A$ gives a unique map $Q\longrightarrow X/A$, also with those properties. It then turns out that these two induced maps are inverses of each other, which I leave to you to prove, so they are isomorphisms (in the case of continuous maps: homeomorphisms).

Title: For $A\subset X$ a subset of a topological space, a universal property uniquely determines the quotient $X/A$

Body:
I have been working on a question with two parts.
PART ONE

Let $(X, \tau_x)$ and $(Y,\tau_Y)$ be topological spaces and $A\subset X$ a subset. Define an equivalence relation $\sim$ on $X$ by $x \sim y$ if and only if $x=y$ or $x,y, \in A$.

Let $X/A$ be the quotient $X/\sim$ equipped with the quotient topology, denoted simply by $\tau$ and let $p: X\to X/A$ be the resulting map. Let $f: X\to Y$ be a continuous map such that $f(A)=\{ f(a): a\in A\}$ has exactly one element. Show that there exists a unique continuous map:
$$g: X/A \to Y \quad \text{ with } f= g \circ p.$$

I derived here that the desired property essentially fixes our map (giving us uniqueness) for $x \in X$:
$$ f(x)= (g \circ p)(x)= g(p(x))= g([x]) \implies g([x])=f(x).$$
Well-definedness follows from the fact that $|f(A)|=1$, so if $x\sim x$ we have $g([x])=f(x)$ and $g([x'])=f(x')$.
Now by $x \sim x'$ either $x=x'$ or $x,x'\in A$ we see that since $|f(A)|=1$ in both cases $f(x)=f(x')$ hence the map is well-defined. (Now we also have existence, since by construction it satisfies the property).

I can then use a lemma in the book, since $f$ is continuous and $p$ is the canonical projection, hence surjective, we have that $g$ is continuous. END OF FIRST PART.

Now for the difficult part, part 2. This exercise proves a universal property. The claim is that this universal property uniquely determines the quotient.

We wish to show that this last property determines $X/A$ uniquely as follows:

Let $(Q, \tau_Q)$ be another topological space with a continuous map $q: X \to Q$ such that $q(A)$ has
exactly one element and the following property: for any continuous map $f : X \to
Y$ such that $f(A)$ has exactly one element, there exists a unique continuous map $g: Q \to Y$ such that $f = g \circ q$.

Show that there is a unique homeomorphism $\phi: Q \to X/A$ such that $\phi \circ  q = p$.

I understand that what we want to show is that in a way all the conditions fix $Q$ to be the same quotient  as $X/A$. To show that two spaces are homeomorphic we need to construct a map that is a bijection and also continuous with continuous inverse map. I cannot find a good way to construct this $\phi$ however.  Does anyone have a good candidate?


问:
Characteristic function - how to guess distribution
I have random variable $Z$ with characteristic function of the form $\cos^3(t)$. How to calculate characteristic function of $Y=Z^3+2Z^2+1$

Body: Since $Z$ has characteristic function$$((e^{it}+e^{-it})/2)^3=\tfrac18(e^{3it}+3e^{it}+3e^{-it}+e^{-3it}),$$$P(Z=1)=P(Z=-1)=\frac38$ while $P(Z=3)=P(Z=-3)=\frac38$, so$$P(Y=4)=P(Y=2)=\tfrac38,\,P(Y=46)=P(Y=-8)=\tfrac18.$$Hence $Y$ has characteristic function $\frac18(e^{-8it}+3e^{2it}+3e^{4it}+e^{46it})$.


问:
expected value of sum of random variables

Suppose $U_i\sim U(0,1)$ for $1\leq i\leq n$ independently. Define the random variable $T = \min \{t : \sum_{i=1}^t U_i \geq c\}$, where $c>0$. What is the expected value of $T$? What does this value approach as $k\to \infty$?

I think $P(T > t) = P(\sum_{i=1}^t U_i < c)$ and $\sum_{t=0}^\infty P(T > t) = \sum_{t=0}^\infty \sum_{x=t}^\infty P(T =x) = \sum_{x=0}^\infty \sum_{t=0}^x P(T = x) = \sum_{x=0}^\infty xP(T = x) = E(T).$

Also, by the central limit theorem, $\sum_{i=1}^t U_i$ approaches the $N(0,t)$ normal distribution as the $U_i$ are independent. So would it be correct to say that $E(T) = \sum_{t=0}^\infty P(\sum_{i=1}^t U_i<c)$ approaches $\sum_{t=0}^\infty P(0 \leq Z < \frac{c}{\sqrt{t}})$ for $c$ large? If so, how could one simplify this and if not, is there an easier method?

答:
If we denote $S_t=\sum_i^tU_i$, we can rewrite your summation as:

$$
\mathbb{E}T=f(c)=\sum_{t=0}^\infty P(S_t<c).
$$

Let's find out $f(c)$ when $0<c<1$. Condition $\sum U_i<c$ and $U_i>0$ gives us $t$-dimensional simplex in the metric space $(U_1,\ldots U_t)$. Since $c<1$, $U_i<1$ are true for all points of the simplex. And thus $P(S_t<c)$ is equal the ratio of volumes of simplex and cube $[0,1]^t$, which is $c^t/t!$ Thus,
$$
P(S_t<c) = \sum\frac{c^t}{t!}=e^c,\qquad\text{when}\quad 0<c<1.
$$

If $c>1$, then with one jump we surely don't reach $c$. So we can perform one jump by $x$ and then we will need to find the expected number of jumps to reach $c-x$:
$$
f(c) = 1 + \mathbb{E}_x f(c-x) = 1 + \int_0^1f(c-x)dx=1+\int_{c-1}^c f(z) dz
$$

If we denote as $f_{n}(c)=f(c)$ when $c\in[n, n+1]$, $n\in\mathbb{N}$, then after differentiation by $c$:
$$
f'_n(c)=f_n(c)-f_{n-1}(c-1),
$$
with continuity condition that $f_n(n) = f_{n-1}(n)$.

Since $f_0(c)=e^c$, we can make an educated guess to have $f_n(c)=P_0(c)e^c+P_1(c)e^{c-1}+\ldots P_n(c)e^{c-n}$, where $P_n(c)$ is a polynomial of power $n$. Indeed:

$$
f'_n(c) = \left(P_0(c)e^c+P_1(c)e^{c-1}+\ldots P_n(c)e^{c-n}\right) + \left(P'_0(c)e^c+P'_1(c)e^{c-1}+\ldots P'_n(c)e^{c-n}\right) =
\left(P_0(c)e^c+P_1(c)e^{c-1}+\ldots P_n(c)e^{c-n}\right) -
\left(P_0(c-1)e^{c-1}+P_1(c-1)e^{c-2}+\ldots P_{n-1}(c-1)e^{c-n}\right),\\
P'_1(c)e^{-1}+\ldots P'_n(c)e^{-n} = -\left(P_0(c-1)e^{-1}+\ldots P_{n-1}(c-1)e^{-n}\right).
$$

If we let $P_n(c)=\frac1{n!}(n-c)^n$,
then
$$
P'n(c) = \frac n{n!}(n-c)^{n-1} = -\frac 1{(n-1)!}\Big((n-1)-(c-1)\Big)^{n-1} = -P_{n-1}(c-1).
$$

Thus, final formula is:
$$
f(c) = \sum_{n=0}^{\lceil c\rceil}\frac{(n-c)^n}{n!}e^{c-n}.
$$

When $c\to\infty$, we can notice that for $t<c$, $P(S_t<c)=1$ and for $t>c$ $S_t\sim N(t/2, t/12)$, thus $P(S_t<c)=1-\Phi(\frac6t(2c-t))$. So the relative number of $t$, where $P(S_t<c)$ is significantly different from 1 or 0 goes down as $1/\sqrt{t}$, and finally $f(c)\approx 2c$


答:
Since there was a discussion under my previous answer, for consistency, I am adding this as a new answer for the cases where order of the group has more than one distinct prime factors:

Take $\mathbb{Z}_6\times \mathbb{Z}_2$ and $\mathbb{Z}_{12}$ for instance. Then, since they are abelian, Sylow 2-subgroups and Sylow 3-subgroups are normal in these groups. This means, $n_2=n_3=1$ for both but they are not isomorphic.

Title: Same number or each Sylow p-subgroup $\implies$ isomorphism?

Body:
I have a question about Sylow theory and isomorphic groups. I came with it while trying to solve some problems from my group theory course. The question is if this statement is true in general:

Being $G$, $H$ two groups of same order, $|G|=|H|=p_1^{\alpha_1}\cdots
> p_n^{\alpha_n}$ with each $p_i$ prime. Then, if the number of Sylow
$p_i$-subgroups in $G$ equals the number in $H$ for all $i=1,...,n$,
then $G\cong H$.

Is this true? Is it true just in certain cases? If not, why? Any help will be appreciated, thanks in advance.


问:
Given $x_1=1,x_2=2,x_{n+2}=3x_{n+1}-x_n\forall n\in\mathbb N$. Find $x_n$.

Given $x_1=1,x_2=2,x_{n+2}=3x_{n+1}-x_n\forall n\in\mathbb N$. Find $x_n$.

I tried to find ways to telescope, but failed. Please help. Thank you.

答:
The set of sequences of complex numbers (actually any field, but let's keep things simple) $(x_n)$ satisfying
$$
\begin{cases}
x_0=a\\
x_1=b\\
x_{n+2}=rx_{n+1}+sx_{n}
\end{cases}
$$
can be made into a vector space by defining addition componentwise and multiplication by scalars as multiplying each term by the scalar. Since a sequence is determined as soon as the $0$ and $1$ term are fixed, this vector space has dimension $2$.

Let's try finding two linearly independent sequences in it, by first looking at the simplest kind, that is, geometric progressions. So, when does the sequence $(t^n)$ belong to the vector space? It should satisfy
$$
t^{n+2}=rt^{n+1}+st^n
$$
for all $n$, in particular for $n=0$, that is,
$$
t^2-rt-s=0.
$$
Conversely, if $t$ satisfies this equation, the corresponding geometric progression belongs to the vector space.

If the polynomial $X^2-rX-s$ has distinct roots $\alpha$ and $\beta$, we want to show that the sequences $(\alpha^n)$ and $(\beta^n)$ are linearly independent. If
$$
p(\alpha^n)+q(\beta^n)=(0)
$$
where $(0)$ denotes the constant zero sequence, we have, in particular
$$
\begin{cases}
p+q=0\\
p\alpha+q\beta=0
\end{cases}
$$
and the determinant of the system is
$$
\det\begin{bmatrix}1 & 1\\\alpha&\beta\end{bmatrix}=\beta-\alpha\ne0,
$$
so the system has a unique solution, with $p=q=0$.

Therefore, for any sequence $(x_n)$ in the vector space there are coefficients $p$ and $q$ such that
$$
x_n=p\alpha^n+q\beta^n \qquad \text{for all $n\ge0$}
$$
and $p$ and $q$ can be uniquely determined by plugging in the given values for $x_0$ and $x_1$.

If the polynomial has only one (double) root $\alpha$, then we can see that the sequences $(\alpha^n)$ and $(n\alpha^n)$ are linearly independent: the linear system to look at is
$$
\begin{cases}
p=0\\
p\alpha+q\alpha=0
\end{cases}
$$
that surely is satisfied for $p=q=0$. Notice that $\alpha\ne0$, unless the recurrence is the trivial one $x_{n+2}=0$.

Of course this can be generalized to any linear recurrence equation.


问:
For sets A and B, where A is countable and B is uncountable, what would $A \setminus B$ be?
Since A is countable and B is uncountable then this would be undefined as B is larger than A right? Or am I missing something here?</p>

答: It depends on the relation between $A$ and $B$. For example, if $A = \mathbb{N}$ and $B = \mathbb{R}$, then $A \setminus B = \varnothing$. However, if you take $A = \mathbb{N}^{+}$ and $B = \mathbb{R}^{-}$, then $A \setminus B = A$.


答:
Let $A=\{-1\}$, $B=\mathbb{R}^+$. Clearly, $A$ is countable and $B$ is uncountable. However, $A \backslash B=A$ is countable.

In fact, for countable $A$ and uncountable $B$, $A \backslash B$ is necessarily countable since $A \backslash B \subseteq A$, and subsets of countable sets are necessarily countable.

Title: For sets A and B, where A is countable and B is uncountable, what would $A \setminus B$ be?

Body: Since A is countable and B is uncountable then this would be undefined as B is larger than A right? Or am I missing something here?</p>


问:
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$
Number Theory : Solve the system of congruence

(1)    $28x+17y\equiv 18 \pmod{41}$

(2)   $31x+11y\equiv 35\pmod{41}$

Attempt :

we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic

lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.

(1)    $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$

(1) $868x+527y\equiv 558\pmod{41}$

(2)   $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$

(2) $868x+308y\equiv 980\pmod{41}$

So finally we can subtract equation (1) from (2) we get:

$219y\equiv -422\pmod{41}$

lets check the $\gcd(219,41)$ by Euclidian algorithm :

$219 = 41\cdot 5 + 14$

$41= 14\cdot 2 + 13$

$14= 13\cdot 1 + 1$

$\gcd(219,41)=1$

Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.

$219y\equiv 1\pmod{41}$

$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.

I don't know how to continue from here .

答:
Starting from your result of

$$219y \equiv -422 \pmod{41} \tag{1}\label{eq1A}$$

the first thing I recommend doing is reducing the coefficients to smaller values since these are generally easier to deal with, plus sometimes you may be able to find simplifications you can use, such as common factors relatively prime to $41$ so they can be "removed" using their multiplicative inverses. Doing this, and using that $\gcd(2, 41) = \gcd(7, 41) = 1$, gives

$$\begin{equation}\begin{aligned}
14y & \equiv -12 \pmod{41} \\
7y & \equiv -6 \pmod{41} \\
7y & \equiv 35 \pmod{41} \\
y & \equiv 5 \pmod{41}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Using this result in your first congruence equation, along with $\gcd(28, 41) = 1$, gives

$$\begin{equation}\begin{aligned}
28x + 85 & \equiv 18 \pmod{41} \\
28x + 3 & \equiv 18 \pmod{41} \\
28x & \equiv 15 \pmod{41} \\
28x & \equiv 56 \pmod{41} \\
x & \equiv 2 \pmod{41}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

In general, though, if you can't fairly easily simplify the results as I did above, you can instead use something like either the constructive proof or direct proof methods described in Wikipedia's "Chinese remainder theorem" article.


问:
How to write absolute value as a "true" function
Here's the basic absolute value ... what?

\begin{align*}
 |x| = \left\{ \begin{array}{r@{\quad \mathrm{if} \quad}l}
 x  &  x > 0, \\
 0  &  x = 0, \\
 \!\! -x  &  x < 0. \end{array} \right.
 \end{align*}

Is this a function, an "expression," or just some short-hand statement? How would you categorize this? Could it be a function if I wrote it like this?

\begin{align*}
 f(x) = \left\{ \begin{array}{r@{\quad \mathrm{if} \quad}l}
 x  &  x > 0, \\
 0  &  x = 0, \\
 \!\! -x  &  x < 0. \end{array} \right.
 \end{align*}

or is this not really a proper algebra function format?

答:
By definition, a function consists of three data:  (1) a domain, (2) a codomain, and (3) a way of identifying elements of the domain with elements of the codomain.  So, for example, the function
$$ f: \mathbb{R} \to \mathbb{R} : x \mapsto x^2 $$
is the function which takes real numbers to real numbers by squaring them. We often think of a function as being a "machine" or "rule" which takes an input (an element of the domain) and produces an output (an element of the codomain).  With this point of view in mind, standard notation is to write $f(x)$ to denote the element of the codomain which corresponds to the element $x$ in the domain.  Thus $f$ is a function, while $f(x)$ is a value in the codomain of the function.  In the case of the function described above, for any $x$ in the domain of $f$, $f(x) = x^2$.

When we write something like
$$ f(x) = \begin{cases} x & \text{if $x \ge 0$, and} \\
-x & \text{if $x < 0$,} \end{cases}$$
we are indicating that $f$ is a function which has the property that the value $f(x)$ which corresponds to some $x$ in its domain is determined by the given piecewise formula.  Again, $f(x)$ is "just" some element of the codomain (or, really, the range) of the function $f$.

However, we often elide the "name" of a function when we define that function in terms of some formula, or introduce some other notation instead of a name.  The absolute value function is an example of this.  When we write
$$ |x| = \begin{cases} x & \text{if $x \ge 0$, and} \\
-x & \text{if $x < 0$,} \end{cases} \tag{1}$$
we are saying that the value $|x|$ (called the absolute value of $x$) is defined by the formula on the right.  Personally, I would say that the mathematical object at (1) is an "identity"—we are identifying the notation $|x|$ (which denotes a number) with the expression on the right.

This identity defines a function, i.e. the absolute value function, but $|x|$ is not that function.  If you want to refer to the function itself (rather than a value taken by the function), you are going to have to introduce some new notation (or just use a lot of words).  For example, if you want to refer to the absolute value function (rather than the absolute value of an input), you could write
$$|\cdot|, \qquad \operatorname{abs}, \qquad\text{or}\qquad d(\cdot,0),$$
where the last expression denotes the distance from zero.  The variable $x$ is absent from all of these notations, as the goal is to refer to the function, and not the values of the function.

There are other times when the standard notation for a function consists of some notation, rather than a letter (or letters) which name that function.  For example

$\|\cdot\|$ is used to denote the norm of a vector,

$\lfloor \cdot \rfloor$ and $\lceil \cdot \rceil$ are often used to denote the floor and ceiling functions,

$(\cdot,\cdot)$ or $\langle \cdot, \cdot \rangle$ are often used to denote an inner product, or (possibly) even

$+$, which is a function on $\mathbb{R}^2$ which maps the ordered pair of real numbers $(a,b)$ to the real number $a+b$.


问:
Probability about event.
It is found that in manufacturing a certain article, defects of one type occur with probability 0.1 and defects of a second type with probability 0.05. What is the probability that:
(a) an article does not have both kinds of defects?
(b) an article is defective?
(c) an article has only one type of defect, given that it is defective?

think I understood how to solve through formulas but not through numbers, this for the first 2 points.
the first point should be P(AandB) complement=1-P(AandB) if these two events are independent I find the intersection through the multiplication law but if they are not? also how do I understand if they are independent not having the intersection?
the second point should be P(AorB)=P(A)+P(B)-P(AandB)
for the third point I have no ideas.

答:
Let $A$ be the event that an article has the first type of defect, and $B$ be the event that an article has the second type of defect. We are given $P(A) = 0.1$ and $P(B) = 0.05$.

IF WE ASSUME INDEPENDENCE

a) $P\left(\lnot (A \land B)\right) = 1 - P(A \land B) = 1 - P(A)P(B) = 1 - 0.01 \cdot 0.5 = 0.995$ so 99.5% of articles have fewer than two defects.

b) $P(A \lor B) = P\left(\lnot(\lnot A \land \lnot B)\right) = 1 - P(\lnot A \land \lnot B) = 1 - (1 - P(A))(1 - P(B)) = 1 - 0.99 \cdot 0.95 = 0.0595$ so 5.95% of articles have at least one defect.

c) $P(A \triangle B | A \lor B) = P(A \land \lnot B|A \lor B) + P(\lnot A \land B|A \lor B) = (P(A \land \lnot B) + P(\lnot A \land B)) / P(A \lor B) = (P(A)(1 - P(B)) + (1 - P(A)) P(B)) / P(A \lor B) = (0.01 \cdot 0.95 + 0.99 \cdot 0.05)/0.0595 \approx 0.9916$ so approximately 99.16% of defective articles have only one defect.

IF WE DO NOT ASSUME INDEPENDENCE

Make a Venn diagram with two circles A and B. We are given that circle A occupies 10% of the space and circle B occupies 5% of the space. How much could be in their overlap $P(A \land B)$?

They might not overlap at all. Or circle B could lie entirely within circle A. Or any amount in between. So $0 \leq P(A \land B) \leq 0.05$ is all we can say.

If the amount of overlap is EXACTLY 0.5% of the space, then the events are independent (this is the formal definition of indenpendence.)

The answers to a), b), and c) will thus be ranges. For example:

a) $P(\lnot (A \land B)) = 1 - P(A \land B)$ so $0.95 \leq P(\lnot (A \land B)) \leq 1$ is the best we can do.

You should get confirmation as to whether independence of the defects is assumed before proceeding further.

Title: Probability about event.


答:
$A \setminus B$ is defined even if $B$ is not a subset of $A$. It's the set of everything that is in $A$ but not in $B$.

Since $A \setminus B$ is a subset of $A$, and $A$ is countable, it follows that $A \setminus B$ is countable (possibly finite or empty).

Example where $A \setminus B$ is countably infinite: $A = \mathbb Q$, $B = \mathbb R \setminus \mathbb Q$, then $A \setminus B = \mathbb Q$.

Example where $A \setminus B$ is finite but not empty: $A = \mathbb Q$, $B = \mathbb R \setminus \{0\}$, then $A \setminus B = \{0\}$.

Example where $A \setminus B$ is empty: $A = \mathbb Q$, $B = \mathbb R$. Then $A \setminus B = \emptyset$.


问:
Solving recurrence relation of the form $T(n)=T\left(\frac{nb}{a}\right)+T\left(\frac{(n-b)c}{a}\right)+n$
How do we solve a recurrence of the form:

$T(n)=T\biggl(\dfrac{nb}{a}\biggr)+T\biggl(\dfrac{(n-b)c}{a}\biggr)+n$ ?

I tried substituting $q=\dfrac{a}{b},\ r=\dfrac{a}{c}, \ s=\dfrac{bc}{a}$ to obtain:

$T(n)=T\left(\dfrac{n}{q}\right)+T\left(\dfrac{n}{r}-s\right)+n$

but could not proceed further.

This question solves a somewhat similar problem but it has a convenient form which makes for a neat substitution.

答:
Assume $a,b,c\neq0$ and $\dfrac{b}{a}\neq1$ to maintain the key meaning of this question.

$T(n)=T\biggl(\dfrac{nb}{a}\biggr)+T\biggl(\dfrac{(n-b)c}{a}\biggr)+n$

$T(n)-T\biggl(\dfrac{nb}{a}\biggr)-T\biggl(\dfrac{(n-b)c}{a}\biggr)=n$

Getting the particular solution part is very easy when $a-b-c\neq0$ .

Let $T_p(n)=An+B$ ,

Then $An+B-\biggl(\dfrac{Anb}{a}+B\biggr)-\biggl(\dfrac{A(n-b)c}{a}+B\biggr)\equiv n$

$\dfrac{(a-b-c)An}{a}+\dfrac{bcA}{a}-B\equiv n$

$\therefore\begin{cases}\dfrac{(a-b-c)A}{a}=1\\\dfrac{bcA}{a}-B=0\end{cases}$

$\begin{cases}A=\dfrac{a}{a-b-c}\\B=\dfrac{bc}{a-b-c}\end{cases}$

$\therefore T_p(n)=\dfrac{an+bc}{a-b-c}$

But getting the particular solution part is not easy when $a-b-c=0$ .

For getting the complementary solution part, we should handle the equation $T_c(n)-T_c\biggl(\dfrac{nb}{a}\biggr)-T_c\biggl(\dfrac{(n-b)c}{a}\biggr)=0$

Even let $T_c(n)=\int_{-\infty}^\infty e^{nt}K(t)~dt$ ,

Then $\int_{-\infty}^\infty e^{nt}K(t)~dt-\int_{-\infty}^\infty e^{\frac{nbt}{a}}K(t)~dt-\int_{-\infty}^\infty e^{\frac{(n-b)ct}{a}}K(t)~dt=0$

$\int_{-\infty}^\infty e^{nt}K(t)~dt-\int_{-\infty}^\infty e^{\frac{bnt}{a}}K(t)~dt-\int_{-\infty}^\infty e^{\frac{cnt}{a}}e^{-\frac{bct}{a}}K(t)~dt=0$

$\int_{-\infty}^\infty e^{nt}K(t)~dt-\text{sgn}\left(\dfrac{a}{b}\right)\int_{-\infty}^\infty e^{nt}K\left(\dfrac{at}{b}\right)d\left(\dfrac{at}{b}\right)-\text{sgn}\left(\dfrac{a}{c}\right)\int_{-\infty}^\infty e^{nt}e^{-bt}K\left(\dfrac{at}{c}\right)d\left(\dfrac{at}{c}\right)=0$

$\int_{-\infty}^\infty e^{nt}K(t)~dt-\int_{-\infty}^\infty\dfrac{a}{b}\text{sgn}\left(\dfrac{a}{b}\right)e^{nt}K\left(\dfrac{at}{b}\right)dt-\int_{-\infty}^\infty\dfrac{a}{c}\text{sgn}\left(\dfrac{a}{c}\right)e^{nt}e^{-bt}K\left(\dfrac{at}{c}\right)dt=0$

$\int_{-\infty}^\infty\left(K(t)-\dfrac{a}{b}\text{sgn}\left(\dfrac{a}{b}\right)K\left(\dfrac{at}{b}\right)-\dfrac{a}{c}\text{sgn}\left(\dfrac{a}{c}\right)e^{-bt}K\left(\dfrac{at}{c}\right)\right)e^{nt}~dt=0$

$\therefore K(t)-\dfrac{a}{b}\text{sgn}\left(\dfrac{a}{b}\right)K\left(\dfrac{at}{b}\right)-\dfrac{a}{c}\text{sgn}\left(\dfrac{a}{c}\right)e^{-bt}K\left(\dfrac{at}{c}\right)=0$

But still very difficult to solve in general.


问:
what is the percentage chance that I will invoke the spear every 5 strokes?
if I have a 3% chance to summon a magic spear every 1 strokes, what is the percentage chance that I will invoke the spear every 5 strokes?

Attemp:
I don't think this is correct; if I'm not mistaken, the question is asking the probability that only one spear is invoked every 5 strokes. Then, the probability would be $\binom{5}{1} \cdot \frac{3}{100} \cdot (\frac{97}{100})^4 = 0.13279...$ so the answer is about $\boxed{\text{13.3 percent}}$.

答:
Much of the work of a probability problem like this is translating the question into terms that are more "probabilistic".

If each stroke has a $3\%$ chance to summon a spear,
you can treat the summoning of a spear in one stroke as a Bernoulli variable with $p=0.03$, that is, probability $0.03$ of success.

If you have a sequence of $5$ strokes, and each has an independent chance to summon a spear, the number of spears summoned is a binomial variable $X$ with
$n=5,$ $p=0.03$.

You appear to have correctly intuited this, because the formula
$\binom{5}{1} \cdot \frac{3}{100} \cdot (\frac{97}{100})^4 \approx 0.13279$
is the correct formula for the probability that $X=1$ (you summon exactly one spear)
if $X$ is defined as above.

To summon at least one spear, notice that the only way you will not summon at least one spear is if you summon no spears at all.
So compute the probability that $X=0$ (no spears).
Then when you subtract that probability from $1,$ what is left will be the probability that $X \neq 0,$ which is the probability that you get at least one spear.

As you probably know, $P(X=0) = (\frac{97}{100})^5$ in this example.


问:
If $f$ is continous on $J_1\times J_2$ and $g$ is Riemann integrable on $J_1$, then $F(t)=\int_a^bf(x,t)g(x)dx$ is continous
If $J_1=[a,b], J_2=[c,d]$, and if $f$ is continous on $J_1\times J_2$ to $\mathbb{R}$ and $g$ is Riemann integrable on $J_1$, then the function $F$, defined on $J_2$ by $$F(t)=\int_a^bf(x,t)g(x)dx$$ is continuous on $J_2$.

My approach:

I know have to show that for $t_0\in [c,d]$ $\forall$, $\exists \delta > 0$, such that for all $t$, $|t-t_0|<\delta\quad$ then $\quad |\int_a^bf(x,t)g(x)dx-\int_a^bf(x,t_0)g(x)dx|<\epsilon$

I can see that the continuity of $F$ comes from the continuity of $f$. And that $g$ is Riemann Stieltjes integrable with respect to $x$.

I've been struggling with this proof for a while. Any suggestions would be great!

答:
$f$ being continuous on a compact set ($J_{1} \times J_{2}$) implies uniform continuity.

$$ \begin{aligned} | F(t) - F(t_0) | & = \left| \int_{a}^{b} (f(x,t)-f(x,t_0)) g(x) \text{ d}x \right| \\ & \leqslant \int_{a}^{b} \underbrace{|f(x,t) - f(x,t_0)|}_{< \ \varepsilon \text{ by uniform continuity}} |g(x)| \text{ d}x \\ & \leqslant \varepsilon \int_{a}^{b} |g(x)| \text{d}x \end{aligned} $$

$g$ being Riemann integrable implies that $g$ is bounded (and in particular $|g|$ is bounded too).


问:
sum of ten squares
You are given an unlimited supply of $1\times 1,2\times 2,3\times 3,4\times 4,5\times 5,6\times 6$ squares.Find a set of ten squares whose areas add up to $48$.If not the whole solution,even a little prod in the right direction would help.</p>

答:
These are all the solutions:

```<code>1       2       3       4       5       6
-----------------------------------------
3       5       1       1       0       0
4       2       4       0       0       0
6       2       1       0       1       0
7       0       1       2       0       0
8       1       0       0       0       1
</code>```

If you only want one solution, you can try the greedy approach of trying to find a solution with one $6 \times 6$ piece. You're then left with the problem of expressing $12=48-36$ with nine pieces, which leads you to the answer by alex23 (the last row in the table).

This combination of greedy and dynamic programming finds all the solutions, but there is probably a lot of book keeping to do. (I just wrote a simple program :-)

Title: sum of ten squares

Body:
These are all the solutions:

1       2       3       4       5       6
-----------------------------------------
3       5       1       1       0       0
4       2       4       0       0       0
6       2       1       0       1       0
7       0       1       2       0       0
8       1       0       0       0       1

If you only want one solution, you can try the greedy approach of trying to find a solution with one $6 \times 6$ piece. You're then left with the problem of expressing $12=48-36$ with nine pieces, which leads you to the answer by alex23 (the last row in the table).

This combination of greedy and dynamic programming finds all the solutions, but there is probably a lot of book keeping to do. (I just wrote a simple program :-)


问:
A GRE algebra question
GRE Problem

Got this GRE question from my friend and we both stuck on it.

I think it's quite conceptual and we don't really need calculation.
It's still obvious that the slope between $P$ and $Q$ is $1$.

The answer is $B$, not sure why it's always true when the line $l$ tilted a little bit from $PQ$ then the slope is less than $1$.

Highly appreciated for any response.

答: Note that the slope of the line $PQ$ is $$\frac{y_2 - y_1}{x_2-x_1} = \frac{6-1}{3-(-2)}=\frac{5}{5} = 1.$$ Assuming the diagram is accurate, the slope of $l$ is smaller than that of $PQ$ (to see why, try shifting $l$ to the right so that $P$ lies on $l$, and note that $l$ crosses $y=6$ -- the $y$-coordinate of $Q$ -- later than $PQ$), hence (b) 'Quantity $B$ is greater' is the correct answer.

Title: A GRE algebra question


问:
Approximation of the gradient of a log-likelihood function implies the approximation of the log-likelihood function.
Suppose that I have the log-likelihood (scalar) function $\ln(p(\mathbf{v};\mathbf{\theta}))$, where $\mathbf{v}$ is some random vector and $\mathbf{\theta}$ is a vector of parameters, and suppose it is given that:

$$
\frac{\partial \ln(p(\mathbf{v};\mathbf{\theta}))}{\partial \mathbf{\theta}} = -\mathbb{E}_{p(\mathbf{h}|\mathbf{v})}\left[\frac{\partial E(\mathbf{v},\mathbf{h};\mathbf{\theta})}{\partial \mathbf{\theta}} \right] + \mathbb{E}_{p(\mathbf{v},\mathbf{h})}\left[\frac{\partial E(\mathbf{v},\mathbf{h};\mathbf{\theta})}{\partial \mathbf{\theta}} \right]
$$

Where:

$\mathbf{h}$ is another random vector.

$E(\mathbf{v},\mathbf{h};\theta)$ is some scalar function of $\mathbf{v}$, $\mathbf{h}$, and $\theta$.

$\mathbb{E}_{p(\mathbf{h}|\mathbf{v})}$ is an expectation with respect to the conditional probability distribution $p(\mathbf{h}|\mathbf{v})$.

$\mathbb{E}_{p(\mathbf{v},\mathbf{h})}$ is an expectation with respect to the conditional probability distribution $p(\mathbf{v},\mathbf{h})$.

The expectations can be approximated using Monte Carlo integration such that:

$$
\frac{\partial \ln(p(\mathbf{v};\mathbf{\theta}))}{\partial \mathbf{\theta}} \approx - \frac{1}{N} \sum_{i = 1}^{N} \left[\frac{\partial E(\mathbf{v},\mathbf{h}_i;\mathbf{\theta})}{\partial \mathbf{\theta}} \right] + \frac{1}{M} \sum_{j=1}^{M} \left[\frac{\partial E(\mathbf{v}_j,\mathbf{h}_j;\mathbf{\theta})}{\partial \mathbf{\theta}} \right]
$$

Where:

$\mathbf{h}_i$ is sampled from $p(\mathbf{h}|\mathbf{v})$.

$\mathbf{v}_j$ and $\mathbf{h}_j$ are sampled from $p(\mathbf{v},\mathbf{h})$.

Since the gradient is a linear operator, then the above approximation can be re-arranged such that:

$$
\frac{\partial \ln(p(\mathbf{v};\mathbf{\theta}))}{\partial \mathbf{\theta}} \approx \frac{\partial}{\partial \mathbf{\theta}} \left(\frac{1}{M} \sum_{j=1}^{M} \left[E(\mathbf{v}_j,\mathbf{h}_j;\mathbf{\theta}) \right] - \frac{1}{N} \sum_{i = 1}^{N} \left[E(\mathbf{v},\mathbf{h}_i;\mathbf{\theta}) \right] \right)
$$

Does this approximation then imply that:

$$
\ln(p(\mathbf{v};\theta)) \approx \frac{1}{M} \sum_{j=1}^{M} \left[E(\mathbf{v}_j,\mathbf{h}_j;\mathbf{\theta}) \right] - \frac{1}{N} \sum_{i = 1}^{N} \left[E(\mathbf{v},\mathbf{h}_i;\mathbf{\theta}) \right]
$$

答:
Let's assume some regularity conditions such that we can exchange the partial derivative $\partial/\partial\theta$ with the expectations in the first equation. This allows us to perform a line integral and find:

$$\begin{align}
\ln p(\mathbf v;\theta)&=\ln p(\mathbf v;\theta_0)+\int_{\theta_0}^\theta\frac{\partial \ln p(\mathbf{v};\mathbf{\theta})}{\partial \mathbf{\theta}}\cdot d\theta\\
&=\ln p(\mathbf v;\theta_0)+\int_{\theta_0}^\theta\left\{-\mathbb{E}_{p(\mathbf{h}|\mathbf{v})}\left[\frac{\partial E(\mathbf{v},\mathbf{h};\mathbf{\theta})}{\partial \mathbf{\theta}} \right] + \mathbb{E}_{p(\mathbf{v},\mathbf{h})}\left[\frac{\partial E(\mathbf{v},\mathbf{h};\mathbf{\theta})}{\partial \mathbf{\theta}} \right]\right\}\cdot d\theta\\
&=\ln p(\mathbf v;\theta_0)+\int_{\theta_0}^\theta\left\{-\frac{\partial}{\partial\theta}\mathbb{E}_{p(\mathbf{h}|\mathbf{v})}\left[E(\mathbf v,h;\theta)\right] + \frac{\partial}{\partial\theta}\mathbb{E}_{p(\mathbf{v},\mathbf{h})}\left[E(\mathbf{v},\mathbf{h};\mathbf{\theta}) \right]\right\}\cdot d\theta\\
&=-\mathbb{E}_{p(\mathbf{h}|\mathbf{v})}\left[E(\mathbf v,h;\theta)\right] + \mathbb{E}_{p(\mathbf{v},\mathbf{h})}\left[E(\mathbf{v},\mathbf{h};\mathbf{\theta}) \right]+C\\
\end{align}$$

Where $\theta_0$ is an arbitrary vector in the parameter space. $C$ is a constant given by $C=\ln p(v;\theta_0)-\mathbb{E}_{p(\mathbf{h}|\mathbf{v})}\left[E(\mathbf v,h;\theta_0)\right] + \mathbb{E}_{p(\mathbf{v},\mathbf{h})}\left[E(\mathbf{v},\mathbf{h};\mathbf{\theta_0}) \right]$.

Now, your Monte Carlo estimator converge to their expected value with an errors $O_p\left(\frac{1}{\sqrt M}\right)$ and $O_p\left(\frac{1}{\sqrt N}\right)$, where $O_p$ denotes stochastic boundness. If we define $n=\min(M,N)$, the error associated to the sum of these estimators is bounded by $O_p\left(\frac{1}{\sqrt n}\right)$.

Summing up, using these Monte Carlo estimators you described will give you a good approximation, up to the additive constant $C$. If your objective is to find maximum likelihood estimators or posterior distributions, that constant might not be relevant, as it would cancel out in later calculations. But it might matter depending on what exactly you are trying to do.

Hope it was helpful!


答:
In order to define a norm in an $F$-vector space for a field $F$, you need to have a modulus defined on $F$. Otherwise, $\lVert \alpha v\rVert=\lvert \alpha\rvert\lVert v\rVert$ does not make much sense.

If you want the familiar identities to hold, you should also want the codomain of $\lvert\cdot\rvert$ and $\lVert\cdot\rVert$ to be the same and be equipped with multiplication, addition and an order, since you want $\lvert\cdot\rvert$ to be multiplicative and subadditive, you want the modulus of $0$ to be $0$, etc.

In general, you can consider $\lvert\cdot\rvert\colon F\to K$ for an ordered field $K$, making $F$ into a "$K$-normed field", and then you can consider "$K$-normed vector spaces" defined in the obvious way. If you have a linear order on $F$, then you have a canonical choice of $\lvert\cdot\rvert$, but I believe that is the extent of it.

This can be further generalised by requiring $F,K$ to be only rings, possibly non-commutative, etc. The main issue with using $F$ and $K$ distinct from $\mathbf R,\mathbf C$ is that if $K$ is not the field of reals (or a subfield), then the resulting topology will be (I think always, but definitely sometimes) non-metrisable, and even if $K$ is the reals, then $F$ might be incomplete (e.g. if it is the rational with the standard modulus) or non-Archimedean (e.g. if it is the p-adics with the p-adic modulus), which immediately transfers to properties of finite-dimensional vector spaces you might consider pathological.

For example, I believe there should be no funny business happening if you consider real-normed vector spaces over the quaternions (beyond the non-commutativity itself, of course), or real-normed vector spaces over the p-adics (beyond the usual non-Archimedeanness). And as pointed in the other answer, you can also consider the trivial discrete norm on any field and all its vector spaces, which will give you the trivial topology on all of them.

There is also a related notion of a valued field and a valued vector space, in which you have maps into an ordered abelian group (plus infinity or 0, depending on how you order your value groups), which are, to my knowledge, much more common than normed fields with norms taking values outside of the reals.

Title: Generalising norms over arbitrary fields

Body:
I am curious about the structure we require on arbitrary fields if we want to generalise the notion of norms on vector spaces over fields other than either $\mathbb{R}$ or $\mathbb{C}$. Specifcally, given a vector space $V$ over $\mathbb{F}$, what structure must be imposed on $\mathbb{F}$, such that we have a map $\|\cdot\|: V \to \mathbb{F}$ s.t. for any $a \in \mathbb{F}$ and any $\bar u, \bar v\in V$,

$$\text{(i) }\|\bar u+ \bar v\| \leq \|\bar u\|+\|\bar v\|$$
$$\text{(ii) }\|\bar v\| = e \to \bar v = \boldsymbol{0}, \text{where } e \text{ is the additive identity of } \mathbb{F}$$
$$\|a\bar u\| = |a|\|\bar u\|$$

From this question here, it seems that positive definiteness requires that we have an ordered field. The accepted answer implies this is enough structure to generalise the inner product.

Where I am getting somewhat confused is how the generalisation works for norms. Specifically, is it really enough to have just have an ordered field in order to satisfy the norm property of $\|a\bar x\| = |a|\|\bar x\|$?

I know that having an ordered field allows us to generalise (i) and (ii), but  how do we define what the modulus of $a$ is? I think the linked question is relevant since if we generalise the inner product over an arbitrary field, and we get a norm out of those inner products, we have made some progress.

How does ordering the field allow us to define the modulus of scalars in a vector space over an ordered field? Do we need to specify additional structure?


问:
The sum of $(1+i)^x + (1-i)^x$ in a proof
My friend sent me an interesting question:
For all non-negative integers $x$, show that:

$$(1+i)^x + (1-i)^x = 2 \sum_{j=0}^{\lfloor x/2\rfloor}\binom{x}{2j}(-1)^j$$

I've tried induction on $x$, but that didn't seem to do the trick. Some hints as to how to approach this proof would be greatly appreciated.

Body:
Hint: Use
$$ (1+x)^n=\sum_{j=0}^n\binom{n}{j}x^j. $$
Using
$$ i^{2j}+(-i)^{2j}=2(-1)^j, i^{2j+1}+(-i)^{2j+1}=0 $$
one has
\begin{eqnarray}
&&(1+i)^x + (1-i)^x \\
&=& \sum_{j=0}^x\binom{x}{j}\bigg(i^j+(-i)^j\bigg)\\
&=&2\sum_{j=0}^{\lfloor x/2\rfloor}\binom{x}{2j}(-1)^j
\end{eqnarray}


问:
Needing Some Help With this Recurrence Problem
Ishaan only cooks 1 of 4 meals: pesto chicken, eggplant parmesan, peanut noodles, or
cauliflower wings. His only rule is that he never makes peanut noodles two or more days in
a row.

Let $a_n$ represent the number of ways Ishaan can choose meals across $n$ days (where $n\ge 0$).

My Thinking:

From the past day we can say there is three ways to choose what Ishaan ate, so $3a_{n-1}$. Then the day before that there is three ways to choose what Ishaan ate, so $3a_{n-2}$.

Then all together I have: $a_n = 3a_{n-1}+3a_{n-2}$, but I don't think this is right.

I was wondering if someone could help me with getting the correct recurrence formula and inital conditions. I am lost...

答:
Your recurrence is correct, but your explanation for it doesn’t make much sense.

On day $n$ Ishaan can cook pesto chicken, eggplant parmesan, or cauliflower wings no matter what he made on day $n-1$; that accounts for $3a_{n-1}$ possible sequences of $n$ meals. He can also cook peanut noodles if he did not cook them on day $n-1$. No matter what he cooked on day $n-2$, he could have cooked any one of the other three dishes on day $n-1$, so there are $3a_{n-2}$ sequences of $n-1$ meals that do not end with peanut noodles on day $n-1$. Each of these sequences allows him to cook peanut noodles on day $n$, so each of them contributes one more possible sequence of $n$ meals, for a total of $a_n=3a_{n-1}+3a_{n-2}$ possible sequences of $n$ meals.

The initial conditions are $a_0=1$ and $a_1=4$: there is only one empty sequence of meals, and on day $1$ he can cook any one of the $4$ meals.


答:
Another way to understand this question is to draw vertical lines through $P$ and $Q$, intersecting the given line $\ell$ at points $P' = (-2, y_p')$ and $Q' = (3, y_q')$, respectively.  We know the $x$-coordinates of $P'$ and $Q'$ are the same as the $x$-coordinates of $P$ and $Q$ because lines $PP'$ and $QQ'$ are vertical.

We also can observe from this diagram that $y_p' > 1$ and $y_q' < 6$.  Therefore, the slope of $\ell$ is $$m = \frac{y_q' - y_p'}{3-(-2)} = \frac{y_q' - y_p'}{5}.$$
$y_q' < 6$ implies $y_q' - y_p' < 6 - y_p'$, and $y_p' > 1$ implies $6 - y_p' < 6 - 1$.  Therefore,
$$m < \frac{6 - 1}{3+2} = 1.$$

Title: A GRE algebra question

Body:
GRE Problem

Got this GRE question from my friend and we both stuck on it.

I think it's quite conceptual and we don't really need calculation.
It's still obvious that the slope between $P$ and $Q$ is $1$.

The answer is $B$, not sure why it's always true when the line $l$ tilted a little bit from $PQ$ then the slope is less than $1$.

Highly appreciated for any response.


问:
Calculus: A given sequence $(a_n)_{n=1}^{\infty}$ with the property that $a_n\cdot a_{n+1}=1$ .
Calculus: A given sequence $(a_n)_{n=1}^{\infty}$ with the property that  $a_n\cdot a_{n+1}=1$ .

Find all the partial limits of the series $(a_n)_{n=1}^{\infty}$.

Given $(a_n)_{n=1}^{\infty}$ a converges. Find the limit of the series.

Attempt:

Unfortunately, I have no idea

$a_n\cdot a_{n+1}$ need to be equal to $1$

Option 1: if all the sequence equal to $1$ ?

答:
That's pretty close to trivial isn't it?  Let $a_1= a$.  Then $a_1\cdot a_2= a(a_2)= 1$ so $a_2= \frac{1}{a}$ $a_2\cdot a_3= \left(\frac{1}{a}\right)a_3= 1$ so $a_3= a$.
$a_3\cdot a_4= \left(a\right)a_4= 1$ so $a_4= \frac{1}{a}$.

See a pattern? You can conjecture that $a_n= a_1$ for n odd and $a_n= \frac{1}{a_1}$ for n even.  Can you prove that for all n?


问:
limit of Lebesgue integrable functions
Given two sequences of integrable functions $\{f_{n}\}, \{g_{n}\}$ with limits $f$ and $g$ both also integrable. Does this always hold

$$\lim_{n}(f_{n}-g_{n})=\lim_{n}f_{n}-\lim_{n}g_{n}=f-g$$

I mean what if  for some point x, $f(x)=\infty$ and $g(x)=\infty$ that would make $f-g=\infty-\infty$. Then what happened at that point? or in order for a sequence of, in this case, integrable functions one should have that the limit is different to $\infty$ at every point.
thanks for the answers beforehand

答:
In the Lebesgue theory, one generally deals with equivalence classes of functions, rather than with individual functions; maybe a better way of saying this is that when one appears to be referring to a function $f$, one is really referring to the class of functions equivalent to $f$. Functions that differ at just one point are equivalent (indeed, functions that differ at a set of measure zero are equivalent) so it doesn't matter if there's one point $x$ where $f(x)=g(x)=\infty$. </p>

<p>Alternatively, if you insist on dealing with functions, not equivalence classes, then you can't say $f$ is integrable if there's an $x$ with $f(x)=\infty$, because that means $f$ isn't even defined on its putative domain. </p>

Title: limit of Lebesgue integrable functions

Body:
In the Lebesgue theory, one generally deals with equivalence classes of functions, rather than with individual functions; maybe a better way of saying this is that when one appears to be referring to a function $f$, one is really referring to the class of functions equivalent to $f$. Functions that differ at just one point are equivalent (indeed, functions that differ at a set of measure zero are equivalent) so it doesn't matter if there's one point $x$ where $f(x)=g(x)=\infty$.

Alternatively, if you insist on dealing with functions, not equivalence classes, then you can't say $f$ is integrable if there's an $x$ with $f(x)=\infty$, because that means $f$ isn't even defined on its putative domain.


问:
Show that $\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

Given $a_1,a_2,...,a_n$ be positive reals. Show that $\displaystyle\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.

Body:
other nice methods
$$\Longleftrightarrow \left(\dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}\right)+\left(\dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}\right)+\cdots+\left(\dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}\right)\ge 0$$
$$\Longleftrightarrow \left(\dfrac{a_{1}}{\sqrt{a_{2}}}-\sqrt{a_{2}}\right)^2+\cdots+\left(\dfrac{a_{n}}{\sqrt{a_{1}}}-\sqrt{a_{1}}\right)^2\ge 0$$


问:
Demonstrate the statement in section 3.I.3 on the decomposition of the Möbius transformations!!!
press here to view textbook question

I'm wondering how to do this question. I believe the question wants to decompose the following statements together from the text book to find the Mobius function

Thank you in advance (:

答:
They basically gave you the solution. Let $f, g, h, i$ are the functions in that order.

Then notice what $i(h(g(f(z))))$ is:

$i(h(g(f(z)))) = i(h(g(z+d/c))) = i(h(\frac{1}{z + \frac{d}{c}})) = i(h(\frac{c}{cz+d})) = i(-\frac{ad-bc}{c^2}\frac{c}{cz+d}) = i(-\frac{ad - bc}{c(cz+d)}) = \frac{a}{c} - \frac{ad-bc}{c(cz+d)} = \frac{az+b}{cz+d}$.

So $M = ihgf(z)$ as desired.


问:
Find a property of groups that is inherited by quotients but not by subgroups.
This is Exercise 4.6a of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.

The Details:

From p.116 of Roman's book,

Definition: Let $\mathcal{P}$ be a property of groups. [ . . . ] We write $G\in \mathcal{P}$ if $G$ has property $\mathcal{P}$. [ . . . ] A property $\mathcal{P}$ of groups is inherited by subgroups if $$G\in \mathcal{P}\text{ and }H\le G\implies H\in \mathcal{P},$$ and $\mathcal{P}$ is inherited by quotients if $$G\in \mathcal{P}\text{ and }H\unlhd G\implies G/H\in \mathcal{P}.$$

The Question:

Find a property of groups that is inherited by quotients but not by subgroups.

Thoughts:

My first guess - I don't know why - was to let $\mathcal{P}$ be "nonabelian", which fell quickly in the light that $G/G=\{G\}$ is abelian for all groups $G$.

Having $\mathcal{P}$ be "abelian" won't  work, since it is inherited by both quotients and subgroups; <s>out goes "cyclic" along with it.</s>

"Finite" is not an example, obviously.

"Solvable" is inherited by both quotients and subgroups. This can be checked on Wikipedia. (I haven't done the Maths, ironically; but with such a standard property of a group, I reckon it's more reliable here than not).

<s>All nilpotent groups are solvable, so they're out; the same goes for supersolvable groups and, in turn, metacyclic groups. Metabelian groups are also solvable.</s>

"Simple" appears not to work.

It would be fiendish if Roman had not defined a group property as required prior to the exercise, so, if all else fails, I could sieve through the definitions in the first 144 pages; this is cheating though (and quite a daunting task!).

Please help :)

答:
Being finitely generated is such a property.

Let $F_n$ denote the free group on $n$ generators. If $F_n\to G$ witnesses that $G$ is f.g. then $F_n\to G\to G/H$ witnesses that the quotient also is, while $F_2$ contains a subgroup isomorphic to $F_\infty$.


问:
Calculus 1, True / False question
The statement is:
If the function $f$ continuous in $[a,b]$. so there is a point $a<c<b$ such as:
$f(c) = f(a) + (f(b) - f(a)) / 3$.

Now there is 3 answers:

1.This statement is False by given an example of function which there is no point such as..

2.This statement is True and can be proven by Weierstrass (Extreme value theorem).

3.This statement is True and can be proven by intermediate value theorem.

So, I'm pretty sure this statement is true but I cant find how to prove it.
I though first of the intermediate value theorem because it's look kind of similar but nothing comes to my head.

any help will be appreciated.

答:
The "intemediate value theorem" says that if f is continuous on [a, b] and F lies between f(a) and f(b) then there exist c, between a and b, such that f(c)= F.

To apply that here, show that f(a)+ (f(b)- f(a))/3 lies between a and b. Of course, f(a)+ (f(b)- f(a))/3= (2/3)f(a)+ (1/3)f(b).

Title: Calculus 1, True / False question


问:
Extreme Points of Disc Algebra
I know that the extreme points of the closed unit ball of $\mathcal{H}^\infty(\mathbb{D})$, the space of all bounded holomorphic functions on the unit disc are the functions $f\in\mathcal{H}^\infty(\mathbb{D})$ such that $\int\limits_{-\pi}^\pi\log(1-|f(e^{i\theta})|)d\theta=-\infty$. What are the extreme points of the closed unit ball of the disc algebra, i.e. all functions continuous on $\overline{\mathbb{D}}$ and holomorphic on $\mathbb{D}$?

It is clear that all the extreme points of the closed unit ball of $\mathcal{H}^\infty(\mathbb{D})$, that are in the disc algebra are extreme points of the closed unit ball of the disc algebra. But are these the only extreme points?

Are there any extreme points of the closed unit ball of $\mathcal{H}^\infty(\mathbb{D})$ that is not in the disc algebra?

Is there any extreme point of the closed unit ball of the disc algebra that is not a finite Blaschke product?

答:
$\newcommand\TT{\mathbb T}$ $\newcommand\AA{\mathbb A}$ $\newcommand\abajo{\\[0.3cm]}$ $\newcommand\e{\varepsilon}$ $\newcommand\DD{\mathbb D}$

The same condition characterizes the extreme points. Since $\AA\subset \mathcal H^\infty$, every $f$ satifying the integral condition is an extreme point.

Conversely, suppose that $\log(1-|f(e^{i\theta})|)$ is integrable. The argument is roughly the same as in $H^\infty$, with a twist to make the function $g$ stay in $\AA$. Let $w\in C([0,2\pi])$ with $w(2\pi)=w(0)$, $0\leq w(t)\leq1-|f(e^{it})|$, with $\log w\in L^1([0,2\pi])$, and $e^{iw}$ continuously differentiable on the open arcs in the set $\{|f|\ne1\}\cap\TT$ (details at the end). Define
$$
g(z)=\exp\bigg[\frac1{2\pi}\,\int_0^{2\pi}\frac{e^{it}+z}{e^{it}-z}\,\log w(t)\,dt\bigg]
$$
By the Schwarz formula, the expression inside the brackets is analytic on $\DD$ and has real part $\log w(t)$ when $z=e^{it}$; then $|g(e^{it})|=w(t)$ is continuous on $\TT$.

Since $|g(e^{it})|\leq  1-|f(e^{it})|$ we have, via the maximum modulus principle, that  $\|f\|_\infty+\|g\|_\infty\leq 1$. Then $g\in\AA$,  $f\pm g\in \AA$, and
$$
f=\tfrac12\,(f+g)+\tfrac12\,(f-g)
$$
is not extreme.

How to construct the function $w$: (not claiming that this is a smart way to do it)

The set $\{|f|\ne1\}\cap\TT=|f|^{-1}(-\infty,1)\cap\TT$ is open in $\TT$, so it is a countable union of disjoint arcs. Say (the union might be finite, but that only simplifies the argument)
$$
\{|f|\ne1\}\cap\TT=\bigcup_n \exp(i\,(a_n,b_n)),
$$
with $0<a_n<b_n<a_{n+1}<b_{n+1}<2\pi$, for all $n$. Let $h_n\in C[0,1]$ be continuously differentiable, with $0\leq h_n\leq1$, $h_n(0)=h_n(1)=0$,  and $\int_{h_n<1}|\log t|<2^{-n}$ (this condition is achievable due to $\log t$ being integrable in $[0,1]$).
Define
$$
w(t)=(1-|f(e^{it})|)\,\sum_n h_n\big(\frac{t-a_n}{b_n-a_n}\big)\,1_{(a_n,b_n)}.
$$
The sum is trivially defined because only one term is nonzero at a time; for the same reason, $w$ is continuously differentiable on each $(a_n,b_n)$. Now
\begin{equation*}
\begin{split}
\int_0^{2\pi}|\log w(t)|\,dt
&=-\sum_n\int_{a_n}^{b_n}\log h_n\big(\frac{t-a_n}{b_n-a_n}\big)\,dt
=-\sum_n\int_{0}^{1}\log h_n\big(t\big)\,dt\abajo
&=-\sum_n\int_{h_n\ne1}\log h_n\big(t\big)\,dt\abajo
&\leq\sum_n2^{-n}<\infty
\end{split}
\end{equation*}

Title: Extreme Points of Disc Algebra


问:
Prove that the setwise limit of signed meaures is a signed measure
Let $(\mathcal{X} , \mathcal{B})$ be a measurable space. Let $(\nu_n)_{n=1}^\infty$ be a sequence of signed measures on $\mathcal{B}$ which converge setwise to a set function $\nu$. Suppose there exists a finite measure $\mu$ on $\mathcal{B}$ such that $|\nu_n(B)| \leq \mu(B)$ for every $B \in \mathcal{B}$ and $n \in \mathbb{N}$. Prove that $\nu$ is a signed measure.

I've been banging my head against the wall on this for a while, and so far this is the only progress I've made:

Let $\bigcup_{i=1}^\infty A_i$ be a union of pairwise disjoint members of $\mathcal{B}$. To show $\nu$ is a signed measure, it is enough to show that $\nu(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \nu(A_i)$. Notice that each $\nu_n$ is absolutely continuous with respect to $\mu$, so we can write $d \nu_n = f_n d\mu$ for some integrable function $f_n : \mathcal{X} \to \mathbb{R}$ (this is a result from a previous exercise/Radon-Nikodym theorem). So:

$\nu(\bigcup_{i=1}^\infty A_i) = \lim_{n \to \infty} \int_{\bigcup A_i} f_n d\mu = \lim_{n \to \infty} (\sum_{i=1}^\infty \int_{A_i} f_n d\mu)$

$\sum_{i=1}^\infty \nu(A_i) = \sum_{i=1}^\infty (\lim_{n \to \infty} \int_{A_i} f_n d\mu)$

So I've managed to reduce this question to the interchange of this limit and infinite sum. The most natural thing to try is Dominated Convergence Theorem, but I can't seem to find an integral dominator for all the $f_n$'s. Could someone please point me in the right direction here?

答: Note that $|\nu_{n}(B)|\leq\mu(B)$ for all $B$ in fact implies that $|\nu_{n}|(B)\leq\mu(B)$. Indeed, $\nu_{n}^{+}(A)$ is the supremum of all $\nu_{n}(B)$'s for $B\subseteq A$, thus $\nu_{n}^{+}(A)\leq\mu(A)$ for all $A$, and the same holds for $\nu_{n}^{-}$, so the same holds for $|\nu_{n}|$. Then now you can show that $|f_{n}|$ is bounded by $1$.


问:
Show integrability of convex (and increasing) function of martingale (submartingale)

I want to prove the above lemma. How can I show (c)? There is a hint: for $f:I\to \mathbb R$ convex and increasing, there exist $a \in \mathbb R$ and $b \in \mathbb R_+$ such that $a+bx\leq f(x)$. I showed that the hint is correct, but how can I apply it in this context?

答: The negative part $Y_t^-$ of $Y_t$ is integrable because $f$ is bounded below by an affine function, per the hint. Meanwhile, by Jensen, if $0\le  s<t$, then $E[Y_t^+|\mathcal F_s]\ge f(E[M_t|\mathcal F_s])^+=f(M_s)^+=Y_s^+$ (the positive part of $Y_s$), even on $\{E[Y_t|\mathcal F_s]=+\infty\}$. Consequently, $E[Y^+_t]\ge E[Y_s^+]$, so the integrability of $Y_t$ implies that of $Y_s$ for $0\le s<t$.


问:
Using Lagrange multiplier to find maximum value.
The maximum value of the function $f(x, y) = xy$, and subject to condition $x^2+y^2=1$:

So do I apply Lagrange's Multiplier method to find the maximum value?
I tried to find the numbers just by applying few numbers but it doesn't seem to work.

答:
If you wanted to use Lagrange Multipliers, you could do the following:

$$f(x,y) = xy\\
g(x,y) = x^2+y^2 = 1$$

So

$$\nabla f(x,y) = < f_x,f_y>=<y,x>, \,\mathrm{and}\, \nabla g(x,y) = <g_x,g_y> = <2x,2y>$$

Now for the method of Lagrange Multipliers, we must find all values of $x$, $y$, $\lambda$ such that

$$\nabla f(x,y) = \lambda \nabla g(x,y)\\
x^2 + y^2 = 1$$

So
$$f_x = \lambda g_x \,\,\mathrm{and}\,\, f_y=\lambda g_y$$

Thus we must solve
$$
y=2\lambda x\\
x=2\lambda y \\
x^2+y^2=1
$$

If we rearrange the second equation for $y$, we get

$$ y = \frac{x}{2\lambda} $$

So

$$ 2 \lambda x = \frac{x}{2\lambda} \\
4\lambda^2x=x\\
4\lambda^2x-x=0\\
x(4\lambda^2-1)=0\\
$$

So $x=0$, $\lambda=\frac{1}{2}$

If
$x=0$,

$y=2\lambda (0) = 0$

BUT if $x=y=0$, then the condition $x^2+y^2=1$ does not hold. So $x$,$y\neq 0$

If $\lambda = \frac{1}{2}$, $y=x$

So
$$x^2+x^2 = 1\\
2x^2=1\\
x^2=\frac{1}{2}\\
x=\pm \frac{1}{\sqrt{2}}\\
$$

Now $y=x$, so $y=\frac{1}{\sqrt{2}}$ when $x=\frac{1}{\sqrt{2}}$ and $y=-\frac{1}{\sqrt{2}}$ when $x=-\frac{1}{\sqrt{2}}$

Thus
$$f(x,y) = xy = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)\\
f(x,y) = \frac{1}{2}$$

That was a lot of work just to get the same answer as @Amire...


答: Being perfect is such an example. If $G$ is perfect, then $G/N$ is also perfect for all normal subgroups $N$ of $G$ since in general $(G/N)' = G'N/N$. But perfectness is clearly not preserved by subgroups.

Body:
This is Exercise 4.6a of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.

The Details:

From p.116 of Roman's book,

Definition: Let $\mathcal{P}$ be a property of groups. [ . . . ] We write $G\in \mathcal{P}$ if $G$ has property $\mathcal{P}$. [ . . . ] A property $\mathcal{P}$ of groups is inherited by subgroups if $$G\in \mathcal{P}\text{ and }H\le G\implies H\in \mathcal{P},$$ and $\mathcal{P}$ is inherited by quotients if $$G\in \mathcal{P}\text{ and }H\unlhd G\implies G/H\in \mathcal{P}.$$

The Question:

Find a property of groups that is inherited by quotients but not by subgroups.

Thoughts:

My first guess - I don't know why - was to let $\mathcal{P}$ be "nonabelian", which fell quickly in the light that $G/G=\{G\}$ is abelian for all groups $G$.

Having $\mathcal{P}$ be "abelian" won't  work, since it is inherited by both quotients and subgroups; <s>out goes "cyclic" along with it.</s>

"Finite" is not an example, obviously.

"Solvable" is inherited by both quotients and subgroups. This can be checked on Wikipedia. (I haven't done the Maths, ironically; but with such a standard property of a group, I reckon it's more reliable here than not).

<s>All nilpotent groups are solvable, so they're out; the same goes for supersolvable groups and, in turn, metacyclic groups. Metabelian groups are also solvable.</s>

"Simple" appears not to work.

It would be fiendish if Roman had not defined a group property as required prior to the exercise, so, if all else fails, I could sieve through the definitions in the first 144 pages; this is cheating though (and quite a daunting task!).

Please help :)


答: The property of having no nontrivial quotient groups (aka being simple or the trivial group).</p>

Title: Find a property of groups that is inherited by quotients but not by subgroups.


答:
Let $d =  f(a) + \frac {f(b)-f(a)}3 $

If you can prove that $\min (f(a), f(b)) \le d \le \max (f(a),f(b))$ then the result follows directly from the intermediate value theorem. $f$ is continuous, and $d\in [\min(f(a),f(b)), \max(f(a), f(b))]$ so there exists a $c\in [a,b]$ where $f(c) = d =f(a) + \frac {f(b)-f(a)}3$

So let's prove  $\min (f(a), f(b)) \le f(a) + \frac {f(b)-f(a)}3 \le \max (f(a),f(b))$.

$f(a) + \frac {f(b) -f(a)}3 = \frac 23 f(a) + \frac 13 f(b)$ and so

$$\frac 23 \min(f(a), f(b)+\frac 13\min(f(a),f(b)) \le  \\\frac 23 f(a) + \frac 13 f(b)\le \\\frac 23 \max(f(a), f(b)+\frac 13\max(f(a),f(b))$$ and so

$\min(f(a), f(b)) \le \frac 23 f(a) + \frac 13 f(b)\le  \max (f(a),f(b))$.

Title: Calculus 1, True / False question

Body:
The statement is:
If the function $f$ continuous in $[a,b]$. so there is a point $a<c<b$ such as:
$f(c) = f(a) + (f(b) - f(a)) / 3$.

Now there is 3 answers:

1.This statement is False by given an example of function which there is no point such as..

2.This statement is True and can be proven by Weierstrass (Extreme value theorem).

3.This statement is True and can be proven by intermediate value theorem.

So, I'm pretty sure this statement is true but I cant find how to prove it.
I though first of the intermediate value theorem because it's look kind of similar but nothing comes to my head.

any help will be appreciated.


问:
Finite transcendence degree implies that every endomorphism is an automorphism?
In our last lecture, our algebra professor stated this result as a commentary, but i'm having some difficulties to prove it, mainly because i am still learning this subject:

Let $E/K$ be an extension, with $E$ an algebraically closed field with finite transcendence degree over $K$. Then, every $K$-endomorphism of $E$ is an automorphism.

I would appreciate any help.

答: Suppose $f:E\to E$ is a $K$-endomorphism.  Then we have a tower of field extensions $K\subseteq f(E)\subseteq E$ and $$\operatorname{trdeg}(E/K)=\operatorname{trdeg}(E/f(E))+\operatorname{trdeg}(f(E)/K).$$  But $\operatorname{trdeg}(f(E)/K)=\operatorname{trdeg}(E/K)$ since $f:E\to f(E)$ is a $K$-isomorphism, and since this common value is finite, the equation above implies $\operatorname{trdeg}(E/f(E))=0$.  That is, $E$ is algebraic over $f(E)$.  But $f(E)$ is algebraically closed, so this implies $E=f(E)$.  That is, $f$ is surjective and hence an automorphism.


问:
Showing this set is open in the topological space $\mathbb{R}\times\mathbb{R}$
Let $\pi_1 : \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ be the projection on the first coordinate. Let $A$ be the subspace of $\mathbb{R}\times \mathbb{R}$ consisting of all points $(x,y)$ for which either $x\geqslant 0$ or $y=0$ (or both) and let $q:A\to \mathbb{R}$ be obtained by restricting $\pi_1$

To prove that the function $q$ is continuous I have to show that for an open set $U$ in $\mathbb{R}$ that the set below is open in $\mathbb{R}\times\mathbb{R}$.

$\left((U \cap [0,\infty)\right)\times \mathbb{R}) \cup (U\times \{0\})$

The function is continuous and so the set should be open. However, $[0,\infty)$ and $\{0\}$ are not open sets in $\mathbb{R}$ and so I don't see how this set is in the box topology of $\mathbb{R}\times\mathbb{R}$.

答: $q^{-1}(U)=(U \times \mathbb{R}) \cap A$. Since $U \times \mathbb{R}$ is a product of open sets, it is open in $\mathbb{R}\times \mathbb{R}$ and so the intersection with $A$ is open in the subspace $A$ by definition of the subspace topology.


答:
An example of a first-order property which is inherited by quotients but not by subgroups is $\forall x\exists y[y^2=x]$. It is inherited by quotients because it's a positive sentence; it's not inherited by subgroups because it holds in $(\mathbb Q,+)$ but not in $(\mathbb Z,+)$.

Nobody asked about products but let me also remark that this property, being defined by a Horn sentence, is also preserved by (finite or infinite) direct products, and even reduced direct products.


问:
How do we prove that $\lim_{x\to\pi/4}\cos(2x)/(1-\sqrt{2}\sin(x)) = 2$ without L'Hôpital?
How can I prove the following limit equals two without LHopital?

Through L'Hôpital, I reached this conclusion:

$$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos 2 x}{1-\sqrt{2} \sin x}=2$$

Body:
HINT

\begin{align*}
\cos(2x) = \cos^{2}(x) - \sin^{2}(x) = 1 - 2\sin^{2}(x) = (1 - \sqrt{2}\sin(x))(1+\sqrt{2}\sin(x))
\end{align*}


问:
Isomorphism of group rings
I have to prove that $G=\mathbb{Z}/6\times\mathbb{Z}/2$ and $H=\mathbb{Z}/12$ are not isomorphic, but the group rings $\mathbb{C}[H]$ and $\mathbb{C}[G]$ are isomporphic. First part was quite easy. Second part is a bit trickier for me. I know that $\mathbb{C}[H]=\{\sum_{h\in H}Z_hh|Z_h\in\mathbb{C}\}$ and similarly $\mathbb{C}[G]=\{\sum_{g\in G}Z_gg|Z_g\in\mathbb{C}\}$. I tried defining a general isomorphism from $\phi:\mathbb{C}[G]\rightarrow\mathbb{C}^{|G|}$ by $\phi(\sum_{\in G}Z_gg)$ is the vector of coefficients, but I don't think it works. I thought that this has something to do with the fact that the field of complex numbers is algebraiclly closed, but I feel like I'm missing some basic properties of such group rings.

We did define the dual group $G^*$ the fourier transfrom of finite abellian groups, but I don't know if this has anything to do with the question itself.
Any hint would help, even some formal properties that might help me find the  solution myself.

答: By Maschke's Theorem both algebras are semisimple. Now Wedderburn's Theorem tells us that as $\mathbb{C}$ is algebraically closed any abelian group $G$ with $n$ elements must give $\mathbb{C}(G)\cong M_{n_1}(\mathbb{C})\times...\times M_{n_k}(\mathbb{C})\overset{G \text{ abelian}}{\cong}\underset{n \text{ times}}{\mathbb{C}\times...\times\mathbb{C}}$ as the $n_i$ represent the number of  elements in each conjugacy class of $G$.

Title: Isomorphism of group rings


问:
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$
Number Theory : Solve the system of congruence

(1)    $28x+17y\equiv 18 \pmod{41}$

(2)   $31x+11y\equiv 35\pmod{41}$

Attempt :

we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic

lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.

(1)    $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$

(1) $868x+527y\equiv 558\pmod{41}$

(2)   $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$

(2) $868x+308y\equiv 980\pmod{41}$

So finally we can subtract equation (1) from (2) we get:

$219y\equiv -422\pmod{41}$

lets check the $\gcd(219,41)$ by Euclidian algorithm :

$219 = 41\cdot 5 + 14$

$41= 14\cdot 2 + 13$

$14= 13\cdot 1 + 1$

$\gcd(219,41)=1$

Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.

$219y\equiv 1\pmod{41}$

$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.

I don't know how to continue from here .

答:
Alternatively, use linear algebra. The system is
$$
\begin{pmatrix} 28 & 17 \\ 31 & 11 \end{pmatrix}
\begin{pmatrix} x \\ y\end{pmatrix}
=
\begin{pmatrix} 18 \\ 35 \end{pmatrix}
$$
The inverse of the matrix is
$$
-\frac{1}{219}\begin{pmatrix} \hphantom-11 & -17 \\ -31 & \hphantom-28 \end{pmatrix}
$$
It remains to find the inverse of $219$ mod $41$, which is $3$ by the extended Euclidean algorithm.


问:
Jacobian of trace of matrix product
I would like to compute the following Jacobian with respect to $x$:

$$\frac{\partial trace[A(x)A(x)^{T}]}{\partial x}$$

where $A(x) \in \mathbb{R}^{n\times m}$, $x \in \mathbb{R}^{m\times 1}$. The result should be a vector in $\mathbb{R}^{m\times 1}$. So far I have something like:

$$\frac{\partial trace[A(x)A(x)^{T}]}{\partial x} \\
= \frac{\partial A(x)}{\partial x} \cdot \frac{\partial trace[A(x)A(x)^{T}]}{\partial A(x)} \\
= \frac{\partial A(x)}{\partial x} \cdot 2A(x)$$

But the dimensions aren't working out to give a result in $\mathbb{R}^{m\times 1}$. I'm definitely missing something. Any suggestions?

答:
Make sure you understand the meaning of differentiating a function by something that isn't a scalar. What you wrote is "basically correct", but the correct interpretation of $\frac{\partial A(x)}{\partial x}$ is not something you probably think is.

I don't think your notation $\frac{\partial A(x)}{\partial x}$ makes a lot of sense. The most "sane" interpretation of it might be the operator taking $x$ and then giving the directional derivative of $A$ along the direction $x$. Hence, here you need two inputs: the point you are evaluating the derivative, and the direction. A closely related (via so called "duality") way of interpretation is probably to think $\frac{\partial A(x)}{\partial x}$ as the "$m$-tuple of matrices", where the $j$th matrix is the partial derivative of $A(x)$ with respect to the $j$th component $x_{j}$ of $x$.

The same applies to the derivative of $\operatorname{tr}(AA^{T})$ with respect to $x$ and the derivative with respect to $A$.

If you are not sure what you mean by a certain symbol, then the best way to avoid confusion is to try to be explicit.

Okay, so let me give the answer. First, note that $\operatorname{tr}(AA^{T})$ is the squared sum of all entries of $A$. And let's define $\frac{\partial \operatorname{tr}A(x)A(x)^{T}}{\partial x}$ to mean the $m$-tuple of partial derivatives:
$$
\frac{\partial \operatorname{tr}A(x)A(x)^{T}}{\partial x}
=\begin{bmatrix}
  \frac{\partial \operatorname{tr}A(x)A(x)^{T}}{\partial x_{1}} \\
  \frac{\partial \operatorname{tr}A(x)A(x)^{T}}{\partial x_{2}} \\
  \vdots \\
  \frac{\partial \operatorname{tr}A(x)A(x)^{T}}{\partial x_{m}}
\end{bmatrix}
$$
So we are differentiating a scalar-valued function with respect to a scalar variable, thus we are fine. As noted earlier, $\operatorname{tr}(AA^{T})$ is the squared sum of all entries of $A$, so its partial derivative with respect to $x_{j}$ is nothing but
$$
\sum_{i=1}^{n}\sum_{j=1}^{m}2A_{ij}(x)\frac{\partial A_{ij}(x)}{\partial x_{j}}
=\operatorname{tr}\left(2A(x)\frac{\partial A(x)^{T}}{\partial x_{j}}\right)
$$
Thus what you are looking for is the vector
$$
\frac{\partial \operatorname{tr}A(x)A(x)^{T}}{\partial x}
=\begin{bmatrix}
  \operatorname{tr}\left(2A(x)\frac{\partial A(x)^{T}}{\partial x_{1}}\right) \\
  \operatorname{tr}\left(2A(x)\frac{\partial A(x)^{T}}{\partial x_{2}}\right) \\
  \vdots \\
  \operatorname{tr}\left(2A(x)\frac{\partial A(x)^{T}}{\partial x_{m}}\right)
\end{bmatrix}.
$$

Title: Jacobian of trace of matrix product