问:
How to solve Discrete Combination problem?
Translated from a native language to english.

A factory produces an item whose defect rate is 0.03 (meaning it's not defective 0.97 of the time).
Taking a sample of 500 items, what is the chance of finding 30 or more defective items?

Given directly from a teacher in a College Statistics class.

I can see that p=0.03 and q=0.97 have calculated that $P(x)= (0.03)^x*(0.97)^{500-x}$.I guess I could brute-force it by calculating $1-P(1)+P(2)+...+P(30)$ but that solution doesn't seem elegant. I don't understand what path I can take here. I don't think i can apply this to any formula like hypergeometric distribution or binomial distribution

答:
$P(x)$ has the form $P(x)=c r^x$ for some constants $c$ and $r$ I let you determine.
Now $\sum_{k=1}^{30}P(k)$ is a sum of consecutive terms of a geometric sequence. Can you take it from here ?


答:
You also need to subtract $P(0)$, and you need to remove the $P(30)$ term, but otherwise your solution is correct. You are correct in your suspicion that there is no simple expression for the final answer.

However, the problem might be asking you to do a normal approximation (instead of obtaining the exact probability). You will find that $30$ is about $4$ standard deviations above the mean of $15$ defective items, so there is virtually zero chance of seeing more than $30$ defective items.

ContentLicense: CC BY-SA 4.0

Title: How to solve Discrete Combination problem?

Body:
Translated from a native language to english.

A factory produces an item whose defect rate is 0.03 (meaning it's not defective 0.97 of the time).
Taking a sample of 500 items, what is the chance of finding 30 or more defective items?

Given directly from a teacher in a College Statistics class.

I can see that p=0.03 and q=0.97 have calculated that $P(x)= (0.03)^x*(0.97)^{500-x}$.I guess I could brute-force it by calculating $1-P(1)+P(2)+...+P(30)$ but that solution doesn't seem elegant. I don't understand what path I can take here. I don't think i can apply this to any formula like hypergeometric distribution or binomial distribution


问:
Points $E$ and $F$ are on sides $AB$ and $AC$ such that $DE$ and $DF$ are the angle bisectors of $\angle ADB$ and $\angle ADC$ respectively.
$\triangle ABC$ has $D$ a point on side $BC$. Points $E$ and $F$ are on sides $AB$ and $AC$ such that $DE$ and $DF$ are the angle bisectors of $\angle ADB$ and $\angle ADC$ respectively. Find the value of $\frac{AE}{EB} * \frac{BD}{DC} * \frac{CF}{FA}$ .

What I Tried: Here is a picture in Geogebra :-

I had suspected that, to find $\frac{AE}{EB} * \frac{BD}{DC} * \frac{CF}{FA}$, I probably have to show $AD,BF,CE$ are concurrent, because then it will be a direct apply of Ceva's Theorem and the answer will come out to be $1$. I realized that is true, since they are indeed concurrent when I make the triangle in Geogebra as you can see.

The real question is, how to prove it? I am finding no elemental methods to show them they are concurrent, like angle-chasing, similarity, areas and so on.

Can anyone help me? Thank You.

答:
In $\triangle ADB$, angle bisector DE divides AC in ratio of adjacent sides : $$ \dfrac{AE}{EB} = \dfrac{AD}{DB} $$

Similarly $\triangle ACD$ : $$ \dfrac{CF}{AF} = \dfrac{CD}{AD} $$

So that $$  \dfrac{BD}{DC}\cdot \dfrac{AE}{EB} \cdot \dfrac{CF}{FA} = \dfrac{BD}{DC}\cdot \dfrac{AD}{BD} \cdot \dfrac{CD}{AD} = 1$$


问:
If the result of my integral is of the form $ \ln(\infty)-ln(\infty) $ (they are the same functions), can i say that the integral evaluates to $0$?
$$ \int f(t)dt=ln(\infty)-ln(\infty)= 0 $$

Can this be true?

If this is true, then these terms fall out and i can evaluate my integral better.

答: Actually i just realized I can use Hospital's rule on $\ln(\frac{\infty}{\infty})$ to evaluate the 'combination' of the two infinities.


答: The number $X$ of defective items with defect rate $0.03$ is binomial-distributed, i.e. $X\sim \mathrm{Binom}(n=500,p=0.03)$. The probability in the question is $$\sum \limits_{x=30}^{500}{500\choose x}0.03^x 0.97^{500-x},$$ which is equal to the easier to compute $$1-\sum \limits_{x=0}^{29}{500\choose x}0.03^x 0.97^{500-x}.$$ To go on from here, I suggest you use a calculator, open up the binomial cumulative probability table.


答:
The situation you described is exactly descriped by a binomial distribution:

Insert $\textbf{P}(X\geq 30)=1-\textbf{P}(X<30)=1-\sum^{29}_{k=0}\binom{500} k p^k(1-p)^{500-k}.$

I don't think there is an easy way out for an exact solution. You can argue that $p$ is small and $n$ is large (allthough none of them is really as large/small as they should be) and use a poisson-approximation.


问:
Can someone please explain $e$ in layman's term?
I never really understood what $e$ means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that?</p>

答:
$e$, the constant, is the limit of $(1 + {1\over n})^n$ as n approaches infinity; As expressed by the series: $$e = 1 + {1\over 1} + {1\over 1\cdot2 } + {1\over 1\cdot2\cdot3} + {1\over 1\cdot2\cdot3\cdot4} + ... + {1\over 1\cdot 2\cdot...\cdot\infty}$$

It is the base of the natural logarithm.

ContentLicense: CC BY-SA 3.0


问:
How we form a group from the subset $\{5,29,53,77\}$ of the multiplicative group $\mathbb{Z}_{96}^*$ (extra challenge: $\{5,29,53,77,85\}$)?
Let $\mathbb{Z}_{96}^*$ the multiplicative group modulo $96$ (the group of units).
The subset $\{5,29,53,77\}$ contains those integers that all have a remainder $5$ when divided by $24$.
What is a common way to form a group from this set - no matter what the operation is? (Preferably the group will be multiplicative.)

And (maybe a bit more difficult) what if we take the set $\{5,29,53,77, 85\}$?

答:
The group must contain $1$, but just adding $1$ to the four-element set does not form a group because its order must divide that of $\mathbb Z_{96}^×$ which is $32$. Thus we repeatedly multiply elements together until closure is achieved, and get an $8$-element group $\{1,5,25,29,49,53,73,77\}$.

With an extra element $85$ we do the same operation and get a $16$-element group.


问:
Set of infinite binary sequence and Cantor's theorem
For each i ∈ N, let f_i : N → {0, 1}. Let A = {f_i : i ∈ N} and E = {n ∈ N : f_n (n) = 0}. Does there exist a f ∈ A such that E = {n ∈ N : f(n) = 1}.

I know that this question is related to set of binary sequence , but how can i answer the question it ask while relating it to Cantor's diagonal theorem?

答:
Hint: Here's a table to help you understand the definitions being given.
$$
\begin{array}{c|ccccc}
i & f_i(0) & f_i(1) & f_i(2) & f_i(3) & \cdots\\
\hline
0 & \color{red}0&1&0&0&\cdots\\
1 & 1&\color{red}1&0&1\\
2 & 0&0&\color{red}1&1\\
3 & 1&0&1&\color{red}0\\
\vdots &\vdots &&&&\ddots
\end{array}
$$
For this selection of functions $f_i$, we see that $0,3 \in E$ and $1,2 \notin E$. Correspondingly, the function $f$ being described would satisfy
$$
f(0) = 1, \quad f(1) = 0, \quad f(2) = 0, \quad f(3) = 1.
$$


问:
why $T^*=A_1T^×A_2^{-1}?$
I have  some confusion inKreyszig Book Functional analysis

My confusion given  below  marked in red box

My doubt :why $T^*=A_1T^×A_2^{-1}?$

My attempt : It seem like $A_2^{-1}$ goes from  $H_2 \rightarrow H_2^{'}$ so image of $A_2^{-1}$ is $H_2'$

again $T^×$ from $ H_2' \to H_1'$ so image will be $H_1'$

that  mean $A_1(H_2'H_1')=T^*$

答:
I cannot really make sense of what you wrote in your attempt. What would $H_2'H_1'$ be?

The map $A_1T^\times A_2^{-1}$ maps $H_2\to H_2'\to H_1'\to H_1$, which with the right domain and codomain for $T^*$.

Let $x\in H_1$, $y\in H_2$. If $g\in H_2'$ is the functional $z\longmapsto \langle z,y\rangle$, then $A_2g=y$, and
$$
x_0=A_1T^\times A_2^{-1}y=A_1 T^\times g
$$
is the element in $H_1$ such that
$$
g(Tx)=(T^\times g)x=\langle x,x_0\rangle.
$$
So
$$
\langle A_1T^\times A_2^{-1}y,x\rangle=\langle x_0,x\rangle=\overline{ \langle x,x_0\rangle}
=\overline{g(Tx)}=\overline{\langle Tx,y\rangle}=\langle y,Tx\rangle.
$$
As this occurs for all $x\in H_1$, $y\in H_2$, we get that $T^*=A_1T^\times A_2^{-1}$.

Title: why $T^*=A_1T^×A_2^{-1}?$


问:
Cosine rule formula proof problem with do product, where does positive sign come from?
I if want to prove the cosine rule with the vector summation (such as I have mentioned in the picture) I start with the $\vec{a}+\vec{b}=\vec{c}$

$\vec{a}=\vec{c}-\vec{b}$
then
$a^2=\vec{a}.\vec{a}=(\vec{c}-\vec{b}).(\vec{c}-\vec{b})=b^2+c^2-2bcCos(A)$
which is completely correct.

If I change the vector such as this picture

I tried to write
$\vec{a}+\vec{b}=-\vec{c}$

$\vec{a}=-\vec{c}-\vec{b}$
$a^2=\vec{a}.\vec{a}=(\vec{c}+\vec{b}).(\vec{c}+\vec{b})=b^2+c^2+2bcCos(A)$

The positive sign before cosine is my problem.

答: The angle $\theta$ you meant from the identity $\vec u\cdot \vec v =uv\cos (\theta)$ is to be measured between the two vectors when the tail is at the same point. In the figure you have shown the angle $A$ is not what you use. You need $\pi -A$. From this, you reduct to angle and would recover the missing minus sign.


问:
$A_n \cap A_m$ has exactly n + m elements
Find an infinite family $A$ of infinite subsets of $\mathbb{N}$ such that for each $A_n, A_m$ holds that $A_n \cap A_m$ has exactly n + m elements

My idea was to split $\mathbb{N}$ in such a way that it is fulfilled, but so far I have not managed to formalize this idea.

Body:
You have a countably infinite set of pairs $\{A_m,A_n\}$.  You also have an endless supply of numbers to put in the $A_n$.  My suggested sets are:
$$A_1=\{1,2,3,4,5,6,7,8,9,10,11,12\ldots\}   \\
A_2=\{1,2,3,13,14,15,16,17,\ldots\}   \\
A_3=\{4,5,6,7,13,14,15,16,17,\ldots\}  \\
A_4=\{8,9,10,11,12,\ldots\} $$
Then 18-23 go in $A_1$ and $A_5$, 24-29 in $A_2$ and $A_4$, and so on.


答:
The idea of $e$ may be explained through the concept of compound interest.  Let's say that you earn an annual interest rate $r$ on a principal $P$ dollars.  Let's say that the money is compounded annually.  Then after $n$ compounding periods, you  have $P (1+r)^n$ dollars.

Now lets say we compound semiannually; then after the same amount of time $n$ years, i.e., $2 n$ compounding periods, you will have

$$P \left ( 1+\frac{r}{2} \right )^{2 n}$$

dollars.  You can now imagine more frequent compounding events; let's say that there are $M$ such events per year.  Then after $n$ years you will have

$$P \left ( 1+\frac{r}{M} \right )^{M n}$$

dollars.

Now imagine that you are compounding continuously (like, every microsecond).  This corresponds to the limit as $M \to \infty$: after $n$ years, you will have

$$P \lim_{M \to \infty} \left ( 1+\frac{r}{M} \right )^{M n} =P\, e^{r n}$$

dollars.

Title: Can someone please explain $e$ in layman's term?

Body: I never really understood what $e$ means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that?</p>


问:
Integration of $\frac{1}{x(x+1)(x+2)...(x+m)}$
I came across this question,
$$\int \frac{1}{x(x+1)(x+2)...(x+m)} dx$$

I tried to split the rational function into partial fractions
$$ \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+...+\frac{Z}{x+m}$$
Not really sure how to proceed from here.

Can someone enlighten me with step by step explanations? I kinda get confused easily when certain steps are skipped.

Body:
Note
$$\frac{1}{x(x+1)(x+2)...(x+m)}=\sum_{k =0}^{m} \frac{a_k}{x+k}
$$
where $a_k$ are obtained as follows
\begin{align}
a_k &=\lim_{x\to -k} \frac{x+k}{x(x+1)(x+2)...(x+k)...(x+m)}\\
&= \frac{1}{[(-k)(1-k)(2-k)(-2)(-1)]\cdot[(1)(2)...(m-k-1)(m-k)]}\\
&=\frac1{(-1)^k k!(m-k)!}
\end{align}
Thus

$$\int \frac{dx}{x(x+1)(x+2)...(x+m)}
=\int \sum_{k =0}^{m} \frac{a_k}{x+k}dx
= \sum_{k =0}^{m} \frac{(-1)^k\ln|x+k|}{k!(m-k)!}+C
$$


问:
$\mathfrak{p}$-adic topology and the symbolic power topology
I'm studying symbolic powers from "Singularities and Foliations. Geometry, Topology and Applications".

The $n$th symbolic power of the ideal $I$ in a Noetherian domain $R$, is defined by $I^{(n)}= \bigcap_{\mathfrak{p}\in Ass_R(R/I)}(I^nR_{\mathfrak{p}}\cap R)$.

"There is a linear relationship between the $\mathfrak{p}$-adic topology and the symbolic power topology of the prime $\mathfrak{p}$: There exists a constant $h$ such that for all $n$, $\mathfrak{p}^{(hn)}\subset\mathfrak{p}^n$."

I couldn't find the definition of $\mathfrak{p}$-adic topology and the symbolic power topology of the prime $\mathfrak{p}$. If anyone introduces any reference that has defined them I would be really thankful

答: In general, any filtration (a descending chain) of ideals induces a topology on $R$. A basis for this topology is given by all translates of ideals appearing in the filtration. So, if our filtration is $\{I_i\}$, then the open basis consists of sets of the form $a+I_i$ for any $a \in R$ and any $i$. If $I$ is any ideal, one defines the $I$-adic filtration to be $\{I^n\}$ and the $I$-adic topology is the topology induced by this filtration. The Symbolic power topology considers instead the topology induced by the filtration $\{I^{(n)}\}$ of symbolic powers of $I$.


问:
What is the smallest group containing all finite cyclic groups?
What is the smallest group $S$(in terms of subgroups) for which every finite cyclic group is a subgroup?
I know that $\mathbb Q/\mathbb Z$ contain all finite cyclic groups as subgroups, is there a way to prove it doesn’t contain a non-isomorphic subgroup who contains all of them?
Can any of you help?

答:
This is a tough question because of the subjectivity of the word "smallest," but I can at least give you a smart-ass correct answer :) The group you're looking for will be the limit of the full subcategory of the category of abelian groups that consists only the (finite) cyclic groups. This isn't helpful except that it tells you such a groups exists since limits must exist in this category. Only considering finite cyclic groups you get something isomorphic to $\mathbf{Q}/\mathbf{Z}$ as folks have pointed out.

Idk how to describe the one where you consider the free cyclic group $\mathbf{Z}$ though. I'll leave this answer CW for anyone who wants to try to flush out details about this group.


问:
Integrating a sine function that is to an odd power
I've started the chapter in my book where we begin to integrate trig functions, so bear in mind I've only got started and that I do not have a handle on more advanced techniques.

$\eqalign{
  & \int {{{\sin }^3}x} dx  \cr
  &  = \int {\sin x({{\sin }^2}x} )dx  \cr
  &  = \int {\sin x({1 \over 2}}  - {1 \over 2}\cos 2x)dx  \cr
  &  = \int {{1 \over 2}\sin x(1 - \cos 2x)dx}   \cr
  &  = \int {{1 \over 2}\sin x(1 - (2{{\cos }^2}x - 1))dx}   \cr
  &  = \int {{1 \over 2}\sin x(2 - 2{{\cos }^2}x)dx}   \cr
  &  = \int {\sin x - {{\cos }^2}} x\sin xdx  \cr
  & y = {1 \over 3}{\cos ^3}x - \cos x + C \cr} $

I got the right answer but it seems like an awfully long winded way of doing things, have I made things harder than they should be with this method?

答: Using the formula $$\sin (3x)=3\sin x-4\sin^3 x$$ you can calculate the integral in just one step.</p>


问:
Probability of Precision Error With Uniform Random Variables
I ran into an interesting probability problem that I am not certain how to solve. The nonmathematical problem is this: a system has a level of precision $\epsilon$ such that if two points are within $\epsilon$, they are indistinguishable. If you need to distinguish between $n$ uniformly distributed points, what is the probability that you have at least 1 error (pair of points within $\epsilon$ of each other)? $k$ errors?

The problem is mathematically: given $X_1, \dots, X_n$ iid random variables distributed uniformly on $[0,1]$, what is the probability that $2k$ of the $X_i$s are within $\epsilon$ of another random variable for a fixed epsilon?

Equivalently, what is
$$P \left(\sum_{i = 1}^N \sum_{j = 1}^N \mathbb{1}_{|X_i - X_j| < \epsilon} \leq 2k \right) $$

which I'll denote (somewhat incorrectly) as:
$$P(\varphi(N) \leq 2k)$$

My intuition for solving this problem is essentially to build up to the general case above from a chain of conditional probabilities since I'm not sure how to solve for the above CDF in any closed form.

So for the sake of argument, let

$$\psi(n) = \begin{cases} 1& \varphi(n) > \varphi(n - 1)\\ 0 & \varphi(n) = \varphi(n - 1) \end{cases} $$

It's easy to see that
$$ P(\psi(n) = 1| \varphi(n - 1) = 0) = 2(n - 1)\epsilon $$

and

$$2\epsilon < P(\psi(n) = 1| \varphi(n - 1) = n - 1) < n\epsilon$$

Which then lends itself to general bounds

$$2(n - k)\epsilon <P(\psi(n) = 1| \varphi(n - 1) = k) < 2(n - k - 1)\epsilon + (k + 1)\epsilon$$

which are not very tight. We can easily find the probability of having at least one error since

$$P(\varphi(n) = 0) = \prod_{i = 1}^n (1 - 2(i - 1)\epsilon)$$

so for $n = 20$, $\epsilon = 0.0001$ we have the probability of failure is $1 - P(\varphi(n) = 0) \approx 0.0373$.

So for large $N$, $P(\varphi(N) > 0) \to 1$. The part that is tricky is handling overlap when we have many errors, as in one case an entire $\epsilon$ could be added to the region we want to avoid, in the other $0$ is added to the region we want to avoid (though both edge cases have measure 0).

This is just a problem for fun, so hints are encouraged.

答:
I was apparently not paying much attention when I first found this problem. Here is what I believe to be the answer:

Let $X \sim \text{Unif}(0,1)$ and $Y \sim \text{Unif}(-1,0)$ (i.e. $Y \sim -X$ but with $X, Y$ independent). The pdf of both $X$ and $Y$ is $1$. Let $Z = X + Y$ then

$$f_Z(z) = \int f_X(z - y) f_Y(y) \,dy $$

so

$$f_Z(z) = \int_{-1}^0 zy - y^2 \,dy = \frac{zy^2}{2} - \frac{y^3}{3} \bigg\rvert_{-1}^0 = \frac{z}{2} + \frac{1}{3}$$

and

$$P(-\epsilon < Z < \epsilon) = \int_{-\epsilon}^{\epsilon} \frac{z}{2} + \frac{1}{3} \,dz = \frac{z^2}{4} + \frac{z}{3} \bigg \rvert_{-\epsilon}^\epsilon = \frac{\epsilon^2}{4} + \frac{\epsilon}{3} - \frac{\epsilon^2}{4} + \frac{\epsilon}{3} = \frac{2\epsilon}{3}$$

Since the $X_i$ are iid we have this holds for each pair of $X_i$s. So we test for an error between each pair of random variables and have a "error"-"not error" distribution given as expected:

$$\xi \sim \text{Binomial}\left(k; {n\choose{2}}, \frac{2\epsilon}{3}\right)$$

So the probability of getting $k$ errors is

$$P(\xi = k) = {{n\choose{2}}\choose{k}} \left(\frac{2\epsilon}{3} \right)^k \left( 1 - \frac{2\epsilon}{3}\right)^{{n\choose{2}} - k} $$

Note that the first calculation of having no errors is different than if you calculate the same example here. This is likely an error in the conditional thinking of my first attempt.

Just for fun, with precision 0.001 and 20 samples we expect to have no error with probability: 0.88099

1 error: 0.00058

2 errors: $3.92074 \cdot 10^{-7}$

and the number of samples we'd need to almost guarantee an error with the same precision is: 187 (with probability of no error $< 10^{-5}$)


问:
Find the least number of moves required.

A move of a coin is defined as crossing any number of points in a straight line on the $4 \times 4$ grid (horizontally,vertically or diagonally). What is the lest number of moves in which a coin, starting from the indicated position, can cover all nine points within the marked square?

I did it in $5$ moves. The way I did it is as follows $:$

Is it the optimal one or there are more efficient moves exist? Any help regarding this will be appreciated.

Thanks for your time.

EDIT $:$ MY RE-ATTEMPT $:$

答:
This is a classic puzzle. It can be done in 4 moves.
From your marked starting position: Down 2; diagonal up/right 3; left 3; diagonal down/right 2 (or 3)</p>


问:
Door in an infinite wall problem

Given is a line $l$ and a start point $s$ on $l$

A target point $t$ is placed on $l$ at an unknown location

a point robot with start location $s$ is searching for the target point $t$. The robot can detect $t$ if it reaches $t$

the robot does not know whether t is located to the left or right of its current position

The question is

Suppose the target $t$ is located at distance $d = x + \epsilon$ (where $\epsilon$ > 0 and is really small)  to the
right of $s$ ($x$ is an integer)

The solution to this according the slids looks like this:

The robot walks 1 unit to the right along l and then back
to s

Then it walks 1 unit to the left along l and then back to s

Then it walks 2 unit to the right along l and then back to s

Then it walks 2 unit to the left along l and then back to s

...

Which results in the following formula:

And this is where I get lost

In the first line I don't understand why the algorithm has to walk $(x+1) + (x +1) + x + \epsilon$.
I do get that when we walk $x + x$ we will barely miss the target $t$ because it is located at  $x + \epsilon$. So in the next iteration we are now first walking $x+1$ to the right. Why are we not finding the target $t$ then already? Since it is located at $x+ \epsilon$  we must have found it by then already?

答:
The algorithm, which is based on a worst case scenario, is accurate.

WLOG, the robot always makes his first move to the right.

Consider two examples:

$x = 0$ and $t$ is located $\epsilon$ to the left.

First, the robot uses up two units going to the right and then returning.

Then, the robot travels $\epsilon$ to the left and finds the target.

The target is located $x + \epsilon$ and when the robot is  searching by the steps $x + x$, the robot begins by traveling towards $\epsilon.$ Here, the robot just misses the target.

Then, when the robot is searching by the steps $(x + 1) + (x + 1)$, the robot is looking in the wrong direction.  So the robot journeys $(x + 1)$ away from the target and then returns to the origin via the journey $(x + 1).$

Then, the robot starts on a search of (potentially) $(x + 2)$ out in the same direction where the target is located.  During this journey, the robot finds the target after traveling $(x + \epsilon).$

Title: Door in an infinite wall problem


问:
Given a convex angle and a line segment of length k, determine the locus of those points...
Got this problem last week on a geometry test, but didn't manage to solve. I have tried reflecting the points, but I didn't get anywhere, I am in 10th class, and I am familiar whith basic geometry tools.

Given a convex angle and a line segment of length k, determine the locus of those points inside the angle through which there exists a line cutting off a triangle of perimeter k from the angle.

答:
Let $O$ be the apex and $OA$, $OB$ be the two rays bounding the convex angle.

For any $P \in OA$, $Q \in OB$ such that $\triangle OPQ$ has perimeter $k$. Draw a circle tangent to $OA$, $OB$ and $PQ$ (within the convex angle, on different side of $PQ$ with respect to $O$). Let $A'$, $B'$ and $C'$ be the points on the circle tangent to rays $OA$, $OB$ and line segment $PQ$ respectively. Notice

$$PC' = PA'\quad\text{and}\quad QC' = QB'$$
We have
$$\begin{align} OA' + OB' &= (OP + PA') + (OQ + QB') = OP + OQ + (PC' + C'Q)\\ &= OP + OQ + PQ = k\end{align}$$
Since $OA' = OB'$, this implies $OA' = OB' = \frac{k}{2}$. This means the two tangent points $A', B'$ and hence the circle through $A',B',C'$ is independent of the choice of $P$, $Q$.

From this, it is easy to see the set of point $X$ where you
can find a line segment $PQ \ni X$ with $\triangle OPQ$ having perimeter $k$ is a curved triangular region.
Two sides of it lies on the rays $OA$, $OB$ with length $\frac{k}{2}$
(ie. $OA'$ and $OB'$). The remaining side is the circular arc $A'B'$ (the one facing $O$) on the circle mentioned above.


答: The smallest group that contains (copies of) all finite cyclic groups is the torsion subgroup of $\mathbb C^\times$, that is, the elements of finite order. It's the set of all complex roots of unity.

Title: What is the smallest group containing all finite cyclic groups?

Body:
What is the smallest group $S$(in terms of subgroups) for which every finite cyclic group is a subgroup?
I know that $\mathbb Q/\mathbb Z$ contain all finite cyclic groups as subgroups, is there a way to prove it doesn’t contain a non-isomorphic subgroup who contains all of them?
Can any of you help?


问:
How do I solve $\frac{d^2y}{dx^2} = (1+(\frac{dy}{dx})^2)^{3/2}$?
$$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$

My progress: I have used substituon $u = \frac{dy}{dx}$ and arrived at $u^2 = \frac{(x+c)^2}{1-(x+c)^2}$ equation with $c$ - constant. After that I was unsure on whether it is allowed to consider separate cases for $u$ or not: maybe for some values of $x$, $u$ will have positive sign, on other occasions negative sign. In fact, any help related to such sign issues would be welcomed.

答:
I think we need to be careful here, a maximal solution to this differential equation should be a function $\phi$ defined on some interval $I$, and circles are clearly not graphs of functions.

Now, consider a maximal solution $(\phi,I)$. The fact that, for $x\in I$, we have
$$\phi’’(x)=(1+(\phi’(x))^2)^{3/2}>0$$
implies that $\phi$ is convex on $I$, or equivalently, that $\phi’$ is increasing on I. Further,
$$\forall x\in I,\quad\left(\frac{\phi’(x)}{\sqrt{1+\phi’^2(x)}}\right)’=1$$
Thus, there exists some constant $a$ such that

$$\forall x\in I,\quad\frac{\phi’(x)}{\sqrt{1+\phi’^2(x)}}=x-a$$
In particular, for all $x\in I$ we have $x-a\in (-1,+1)$ that is $I\subset (a-1,a+1)$. Furthermore the sign of $\phi’(x)$ is the same as the sign of $x-a$. It follows that
$$\forall x\in I,\quad\phi’(x)=\frac{x-a}{\sqrt{1-(x-a)^2}}$$
A further integration shows that there exists a real constant  $b$ such that
$$\forall x\in I,\quad\phi(x)=-\sqrt{1-(x-a)^2}+b.$$
Conversely, for any real numbers $a$ and $b$ the function
$$\phi:(a-1,a+1)\to\mathbb{R}, \phi(x)=-\sqrt{1-(x-a)^2}+b$$ is a maximal solution to the proposed ode.$\qquad\square$

Remark. So, we note that the graphs of our solutions are in fact half unit circles.


问:
Extend an integral inequality from $C[0,1]$ to the whole $L^{2}([0,1])$ [related to operator norm].
Consider the following integral defined for $f\in L^{2}[0,1]$ and $x\in [0,1]$, that $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$

This shows that for $f\in C[0,1]$ and for any $x\in [0,1]$, we have $$|(T^{n}f)(x)|\leq \dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}|f(r)|dr\leq \dfrac{\|f\|_{\infty}}{n!}.$$ By definition of operator norm, this implies that the operator norm of $T^{n}$ $$\|T^{n}\|\leq \dfrac{1}{n!},$$ if the operator is on $C[0,1]$.

However, I want to prove the same bound of this operator norm on the whole $L^{2}([0,1])$.  Is there anyway for me to extend this proof to all of $f\in L^{2}([0,1])$? and to conclude $\|T^{n}\|\leq\frac{1}{n!}$

My attempt was to use density of continuous function of compact support in $L^{2}$ to find $g$ such that $f=g+\epsilon$ where $\epsilon>0$ is arbitrarily fixed. Then we bound the integral in the same way, and you will have a summation, the second term in the summation will go to $0$ when $\epsilon\searrow 0$. Therefore, we again have $$|(T^{n}f)(x)|\dfrac{\|g\|_{\infty}}{n!}.$$  But this cannot say anything to the operator norm $\|T\|$.

What should I do? Thank you!

Edit:

The overall purpose is to derive that $\|T\|\leq\frac{1}{n!}$. So if there is any other way to find this, it will also be really good.

Edit2: Proof

Okay, I confused myself in the first place. The operator norm has nothing to do with the sup-norm since $T:L^{2}\longrightarrow L^{2}$, so what we should do is to use Cauchy-Schwarz and $\|f\|_{L^{2}}$. See below my own answer to my post for details.

Also, Ruy gave a general theorem about how to compute the operator norm for Hilbert-Schmidt operator, and my computation will show you the idea about the proof. Basically, the absolute value of operator will be bounded by $\|f\|_{L^{2}}$ multiplication of the $L^{2}-$integral of the kernel, so we should expect such a theorem.

Thank you so much for all the users who helped me!!

答:
Given  $k\in L^2([0, 1]\times[0, 1])$,
one may define an operator $A_k$ on $L^2([0, 1])$ by
$$
  A_k(f)|_x  = \int_0^1 k(x, y)f(y)\,  dy, \quad \forall f\in  L^2([0, 1]), \quad \forall x\in [0,1].
  $$

This is called an integral operator and $k$ is said to be its integral kernel.

Besides being bounded, $A_k$ is Hilbert-Schmidt, and its Hilbert-Schmidt norm $\|A_k\|_2$ may be computed as
$$
  \|A_k\|_2 = \left(\int_0^1 \int_0^1 |k(x, y)|^2 \, dx\,  dy\right)^{1/2}.
  $$
See [1,Theorem VI.23] for details.

Since the operator norm is always bounded by the Hilbert-Schmidt norm, one may use the above to estimate the former.

The operator $T^n$ mentioned by the OP is clearly an example of the above, where
$$
  k(x, r)=\frac{1}{(n-1)!}(x-r)^{n-1}[r\leq x],
  $$
where $[r\leq x]$ takes the values 1 or 0 according to whether $r\leq x$ or not.

If my calculations are correct this method provides an even better estimate for the norm of $T^n$.

[1] Reed, Michael; Simon, Barry, Methods of modern mathematical physics. I: Functional analysis. Rev. and enl. ed, New York etc.: Academic Press, A Subsidiary of Harcourt Brace Jovanovich, Publishers, XV, 400 p. $ 24.00 (1980). ZBL0459.46001.

EDIT:
OK, prompted by Calvin Khor's challenge, here is the full calculation based on the Hilbert-Schmidt norm and leading to a slightly better estimate.

I will temporarily ignore the coefficient $\frac{1}{(n-1)!}$ since it doesn't play much of a role in the computation, so
let us first
deal with
$$
  h(x, r)=(x-r)^{n-1}[r\leq x].
  $$
We then have
$$
  \|A_h\|_2^2 =
  \int_0^1 \int_0^1 |h(x, r)|^2 \, dr\,  dx =
  \int_0^1 \int_0^1 |(x-r)^{n-1}[r\leq x]|^2 \, dr\,  dx = $$$$ =
  \int_0^1 \int_0^x (x-r)^{2n-2} \, dr\,  dx =
   \int_0^1 \frac{-(x-r)^{2n-1}}{2n-1} \Big|_0^x\,  dx = $$$$ =
   \int_0^1 \frac{x^{2n-1}}{2n-1}\,  dx =
   \frac{x^{2n}}{2n(2n-1)}\Big|_0^1 =
   \frac{1}{2n(2n-1)}.
  $$
This implies that
$$
  \|A_k\| \leq
  \|A_k\|_2 =
  \frac{1}{(n-1)!}  \|A_h\|_2 =
  \frac{1}{(n-1)!\sqrt{2n(2n-1)}}.
  $$
So we indeed get a better estimate than just   $\frac{1}{n!}$ because
$$
  \sqrt{2n(2n-1)} \geq    \sqrt{(2n-1)^2} = 2n-1 \geq  n.
  $$

Remark.  Since the Hilbert-Schmidt norm is usually much bigger than the operator norm, there is still a lot of room for
improvement.  Any bidders? :-)


答:
Euler's Number, $e\approx2.718281828$, is a special constant in calculus because of several important properties it holds.

First, $y=e^{x}$ (also written as $y=\exp(x)$ due to the exponential growth of the function) and constant multiples thereof is the only function for which:

$$\frac{dy}{dx}=y$$

Which means in simple language that the gradient of the curve at a point is proportional to the value of the curve at that point. For instance, at the point $(1,e)$, the gradient of the tangent (shown as blue in the graph below) to the curve $y=e^{x}$ (shown in red) is $e$.

A proof that $y=C\exp(x)$ is the only real function for which this rule applies can be found here.

Euler's Number is also the number which the expression $\left(1+\frac{1}{x}\right)^{x}$ approaches as $x$ gets closer and closer to infinity. Formally:

$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$$

Euler's number also can be found as the unique positive real number $e$, such that:

$$\int_{1}^{e}\frac{dt}{t}=1$$

Which in common language means the area under the curve $y=\frac{1}{x}$ between $1$ and $e$ has unit area (or area equal to 1).

Also, interestingly, $e$ can be found by applying Stirling's approximation to the factorial in reverse to get:

$$e=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}$$


问:
Is this a typo in the retraction formula?
In the book, optimization on matrix manifolds by Absil, Chapter 3 pg 59, it is given that retraction on the orthogonal manifold $O_n$, using the Cayley's transform, is given by $$
R_{X}(X \Omega)=X\left(I-\frac{1}{2} \Omega\right)^{-1}\left(I+\frac{1}{2} \Omega\right)
$$ I am trying to derive to this from first principles but stuck mid-way.

My attempt:

Let $Q \in O_n$ and $\Omega$ be a skew symmetric matrix. By Cayley's transform, for each $\Omega$ we can find a $Q$ such that $$Q = \left( I - \Omega \right) \left( I + \Omega \right)^{-1}$$ Let $\phi$ be the mapping from $[Q^{-1},I - \Omega] \rightarrow I+\Omega$. Therefore, we can write $\pi_{1} \circ \phi^{-1}$ as matrix that takes in $I+\Omega$ and outputs $Q^{-1}$.

\begin{align}
R_{X}(X+X\Omega) &:=\pi_{1}\left(\phi^{-1}(X+X\Omega)\right) \\
& = X \pi_{1}\left(\phi^{-1}(I+\Omega)\right) \\
& = X Q^{-1} \\
& = X \left( I + \Omega \right) \left( I - \Omega \right)^{-1} \\
& = X \left( I - \Omega \right)^{-1} \left( I + \Omega \right)  \qquad \quad \text{as the matrices commute}\\
\end{align}

I am not sure how to get the the $\frac{1}{2}$ as mentioned in the textbook. Can you please help. Thank you.

Update: One of the answers says that this is a typo in the book. Can others please confirm?

Below is a picture of the relevant text and symbols.

答:
An errata for Optimization Algorithms on Matrix Manifolds can be found at https://sites.uclouvain.be/absil/amsbook/errata.html. So far, the Cayley retraction formula on top of page 59 has been considered to be correct.

I doubt that Proposition 4.1.2 can be used here. The first step would be to clarify what $\mathcal{N}$ is, and to make sure that $\phi$ has the required domain and codomain.

Hint: A way to prove that
$$
R_{X}(X \Omega)=X\left(I-\frac{1}{2} \Omega\right)^{-1}\left(I+\frac{1}{2} \Omega\right)
$$
is a retraction on the orthogonal group is to confirm that it satisfies the properties given in Definition 4.1.1. The hardest part is arguably (ii) therein. Using the product rule, the chain rule and (A.3) should yield the sought result.


问:
How to solve (step-by-step) this trig. equation below?
$$\cos{2x}\cdot\cos{\left(x-\frac{\pi}{6}\right)}=\sin{2x}\cdot\sin{\left(x-\frac{\pi}{6}\right)}$$

The root of my struggle with this problem is that I know (said by Wolframalpha) that it can be solved, I just can't find a "nice" way to write it down, step-by-step...

答:
First recognize that subtracting $$\sin{2x}\sin{\left(x-\frac{\pi}{6}\right)}$$from both sides will show the recognizable equation of $$\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}=\cos\left(\alpha+\beta\right)$$ now that you see this equation you can then realize that when you subtracted $$\sin{2x}\sin{\left(x-\frac{\pi}{6}\right)}$$ from both sides the equation became: $$\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}=0$$ This means that $$\cos{\left(\alpha+\beta\right)}=0$$ This is recognizable to the basic cosine $$\cos{90^\circ}=0$$ Therefore all you have to do is solve $$2\mathit{x}+\left(x-\frac{\pi}{6}\right)=\frac{\pi}{2}+n\pi$$
Where $n$ is any integer. Solving this algebraically shows that $$\mathit{x}=\frac{2\pi}{9}+n\frac{\pi}{3}$$ Tell me if I should clarify further.


答:
As Mike points out, "smallest" is not a precise terminology. However, since we already have a model $\mathbb{Q}/\mathbb{Z}$, one can prove what razivo suggests at least, i.e. there's no proper subgroups of $\mathbb{Q}/\mathbb{Z}$ satisfying the desired property. That's basically the existing question:Q/Z has a unique subgroup of order n for any positive integer n?. In the process, one actually shows each element of $\mathbb{Q}/\mathbb{Z}$ lies in some cyclic group of order $n$ (Note: $n$ might not be unique though). Thus if any element is taken away, there must be some cyclic group hurt.

PS. The torsion subgroup of $\mathbb{C}^\times$, which lhf suggests, is isomorphic to $\mathbb{Q}/\mathbb{Z}$ through the exponential map.

EDIT: I would like to present a possibly "fair" way to understand "smallest" in this context. For example, we hope that the smallest group containing $\mathbb{Z}/3\mathbb{Z}, \mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z}/8\mathbb{Z}$ should be $\mathbb{Z}/24\mathbb{Z}$. When we add more groups, for example, $\mathbb{Z}/5\mathbb{Z}$ and $\mathbb{Z}/10\mathbb{Z}$, we could embed all the groups mentioned above into $\mathbb{Z}/120\mathbb{Z}$, which is again smallest. Now it's inspiring to consider the diagram consisting of all cyclic groups $\mathbb{Z}/n\mathbb{Z}$, with arrows the emdeddings from $\mathbb{Z}/d\mathbb{Z}$ into $\mathbb{Z}/n\mathbb{Z}$ if and only if $d|n$. The colimit of this (directed) diagram seems to be the precise characterisation of "smallest" in our sense. Not surprisingly, one can prove this colimit is exactly $\mathbb{Q}/\mathbb{Z}$.

A convenient proof may use the disjoint-union characterisation of directed colimits, based on which we notice: once we identify $\mathbb{Z}/n\mathbb{Z}$ as $\frac{1}{n}\mathbb{Q}/\mathbb{Z}$, all the equivalent relations required in the characterisation of colimits (see the link above) coincide with those in $\mathbb{Q}/\mathbb{Z}$. For example, $4$ in $\mathbb{Z}/10\mathbb{Z}$ is equivalent with $6$ in $\mathbb{Z}/15\mathbb{Z}$, since they both correspond to $12$ in $\mathbb{Z}/30\mathbb{Z}$. This corresponds exactly to the fact $\frac{4}{10} = \frac{6}{15}$ in $\mathbb{Q}/\mathbb{Z}$ (since they both equal to $\frac{12}{30}$, by elementary arithmetic). Thus the directed colimit coincides with $\mathbb{Q}/\mathbb{Z}$.


问:
Extract a maximal independent subsystem and complete this to a basis in $\mathbb{R}^3$

I have $4$ vectors $v_1=(0,-1,2),$ $v_2= (1,0,1),$ $v_3= (-3,-2,1),$ $v_4= (1,-2,5)$ which form a dependent system and $S=\text{span}\{v_1,v_2,v_3,v_4\}.$ Find a maximal independent subsystem of $S$ and complete this subsystem to a basis in $\mathbb{R}^3$.

My idea is to find a maximal subsystem is to find $3$ vectors and to prove that the determinant isn't $0.$ But can $4$ vectors form a basis in $\mathbb{R}^3?$

答:
Note that $(1,0,1) = (1,-2,5) - 2\cdot(0,-1,2)$ so $v_2 = v_4-2\cdot v_1.$

Therefore $$\langle v_1,v_2,v_3,v_4 \rangle = \langle v_1,v_3,v_4 \rangle.$$

But $(0,-1,2) = \frac{1}{8}((-3,-2,1) + 3\cdot(1,-2,5)),$ so $v_1 = \frac{1}{8}(v_3 + 3\cdot v_4).$

Hence

$$\langle v_1,v_3,v_4 \rangle = \langle v_3,v_4 \rangle.$$

Since $v_3 \neq \lambda \cdot v_4, \forall \lambda \in \mathbb{R},$ we conclude that $A = \{v_3,v_4\}$ is linearly independent.

To form a basis of $\mathbb{R}^{3},$ first note the $\dim{\mathbb{R}^3} = 3$ but $\dim{A} = 2.$

So $A$ itself doesn’t form a basis of $\mathbb{R}^3.$ But (by Steinitz Theorem) we can form a basis of $\mathbb{R}^3$ that contains $A$ by adding one more vector.

But this extra vector has to be chosen carefully, note that this one, when reunited with $A$ must “preserve” the linearly independence of $A.$

We can pick this vector, for example, from the standard basis of $\mathbb{R}^3,$ which is

$$((1,0,0),(0,1,0),(0,0,1)).$$

Take, for example, $(1,0,0).$ Then

$$((-3,-2,1),(1,-2,5),(1,0,0))$$

form a basis of $\mathbb{R}^3.$


问:
Does a bounded well-ordered set have a maximum?
Suppose $(A,\preccurlyeq)$ is a well-ordered chain, i.e., every subset $B\subseteq A$ has a least element under the order $\preccurlyeq$.

We say that a set $B\subseteq A$ is bounded if there exists $s\in A$ such that $b\preccurlyeq s$ for all $b\in B$.

Question: Does every bounded subset of $A$ have a maximum?

If $B$ is finite this is obviously true, since we can start with the minimum $m=\min B$, and keep applying the successor operation
$$x^+=\min\textstyle_{\preccurlyeq}(B\smallsetminus\{b\in B: b\preccurlyeq x\})$$
and stop when we end up with a singleton $\{M\}$, this will be the maximum of $B$. But what if $B$ is infinite?

I can't think of an easy counterexample to this, it feels like it should be true.

答: If we let $A = \{-\frac{1}{n} \mid n \in \mathbb{N}_+\} \cup \{0\}$ with the usual order, then $A$ is well-ordered. Now take $B = A \setminus\{0\}$ which is bounded, as $0$ is an upper bound for $B$, but as $0$ is not an element of $B$, and no other element of $A$ provides an upper bound for $B$, it follows that $B$ has no maximum.


问:
Building a tower using colorful blocks
How many possibilities are there to build a tower of n height, using colorful blocks, where:

white block is of 1 height

black, red, blue, green, yellow, pink blocks are equal to 2 heights

I need to find the generating function formula for this.

So, for n = 1 I get 1 possibility,

for n = 2 I get 2 possibilities,

for n = 3 I get 3 possibilities

for n = 4 I get > 4 possibilities etc.

The generating function at this moment would be $1 + 2x + 3x^{2} + ...$. But I have no idea how can I find the general formula calculating this.

Could you give me any suggestions, or solutions (;-)) ?

答:
It looks to me like for $n=2$ you have $7$ possibilities, either two white or one of any of the $6$ colors.  Similarly for $n=3$ you can have three white, or a white and a color, or a color and a white, for $13$ possibilities.

To make the recurrence, if there are $T(n)$ ways to make a tower $n$ high, we can either put a white block on a tower one shorter or put a colored block (6 ways) on a tower two shorter, so $$T(n)=T(n-1)+6T(n-2), T(1)=1, T(0)=1$$

This can be solved by the usual techniques-find the characteristic polynomial, factor, etc.

Added:  if color doesn't matter but order does, change the $6$ to $1$ above.  If color and order don't matter, the only choice to make is how many $2$ size blocks to use, which can range from $0$ to $\lfloor \frac n2 \rfloor$, for a total of $1+\lfloor \frac n2 \rfloor$ ways.

Title: Building a tower using colorful blocks


问:
Let $F(x)=\int_0^x\left(t\int_1^tf(u)\,du\right)\,dt$. Show that $F''(x)$ is increasing on $(0, 1)$.
Question:

Suppose that $f(x)$ is continuous and increasing on $[-1, 2]$ with $f(x)>0$. Let $F(x)=\int_0^x\left(t\int_1^tf(u)\,du\right)\,dt$. Show that $F''(x)$ is increasing on $(0, 1)$.

This is an exercise about the Fundamental Theorem of Calculus. Below is my attempt.

My idea is to show that $F'''(x)\geq0$ for all $x\in(0, 1)$.

$$\begin{align}F(x)=\int_0^x\left(t\int_1^tf(u)\,du\right)\,dt&\implies F'(x)=x\int_1^xf(u)\,du\\
&\implies F''(x)=\int_1^xf(u)\,du+xf(x)\\
&\implies F'''(x)=2f(x)+xf'(x)
\end{align}$$

Here I want to say that $f'(x)\geq0$ for all $x\in(0, 1)$ since $f(x)$ is increasing on $(0, 1)$. However, I am not sure whether $f'(x)$ exists for all $x\in(0, 1)$. The question only states that $f(x)$ is "continuous", not "differentiable".

How should I continue? Thanks in advance.

答:
Let $x \geq y$ then observe that
$$xf(x)-yf(y) = x(f(x)- f(y)) + f(y)(x - y) \geq 0 $$ since $f$ is increasing. So $xf(x)$ is also increasing.

Now $$F^{''}(x) - F^{''}(y) = \int_{y}^{x} f(u) \, du + xf(x) - yf(y) > 0$$ since $f > 0$ and $xf(x)$ is increasing.

So we conclude that $F^{"}(x) $ is increasing.


问:
Is the structure $(\mathbb{R}, *)$ "weakly minimal" in this precise sense?
This is a follow-up to my previous question about definability in the structure $(\mathbb{R}. *)$ The definition of a minimal structure is that every parameter-definable subset of that structure is finite or cofinite. Now, certainly the structure $(\mathbb{R}, *)$ is not minimal. However, I conjecture that it does have the following weaker property: that every parameter-definable set is either finite, cofinite, the positive reals modulo a finite set, or the negative reals modulo a finite set. By modulo a finite set, I mean the set with some finite number of elements added and/or removed. Is this conjecture true? I would be very surprised if it was false.

答:
$\mathcal{R}=(\mathbb{R};*)$, while not itself minimal, has a strongly minimal subset which really captures the whole: namely, $\mathcal{R}^+=(\mathbb{R}_{>0},*)$. $\mathcal{R}$ is basically two copies of $\mathcal{R}^+$, plus a definable singleton, glued together in a definable way, so we can often derive results about $\mathcal{R}$ from results about $\mathcal{R}^+$.

In particular, we have via an appropriate interpretation of $\mathcal{R}$ in $\mathcal{R}^+$:

If $X$ is a definable-with-parameters subset of $\mathcal{R}$, then $\{\vert c\vert: c\in X\}$ is a definable-with-parameters subset of $\mathcal{R}^+$.

Now unlike $\mathcal{R}$, the substructure $\mathcal{R}^+$ actually is minimal: it's just a $\mathbb{Q}$-vector space (think about the map $x\mapsto e^x$). This lets us rule out counterexamples to the conjecture in the OP as follows. Given a definable-with-parameters $X$ in $\mathcal{R}$, let $X_+=X\cap\mathbb{R}_{>0}$ and $X_-=X\cap\mathbb{R}_{<0}$. By the facts above, both $X_-$ and $X_+$ must be either finite or cofinite, so we're done.

More snappily, this is a general feature of any "finite cover" of a minimal structure in the appropriate sense.


问:
Number of divisors of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form $4n+1,n\in N$?
In this question, the OP asks to find the number of divisors of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ which are of the form $4n+1,n\in N$. The top answer points out that the required divisor is of the form $$3^a\cdot 5^b\cdot 7^c$$ with $0\leq a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $a+c$ being even. The answer therefore is, apparently, $(4 \cdot 4 \cdot 6)/2=48$.

But this is wrong according to my book: the correct answer is $47$. Obviously, one case has been overcounted, but which? As far as I know, the person who wrote the top answer employed a fairly standard approach and should have arrived at the correct answer.

答: As pointed out in the comments by Daniel Fischer and lulu, my book considers $0 \notin N$, so discounts the case where $a=b=c=0$, i.e., $4(0)+1=1$.


问:
Find max of $x^7+y^7+z^7$

Find max of  $x^7+y^7+z^7$ where $x+y+z=0$ and $x^2+y^2+z^2=1$

I tried to use the inequality:$$\sqrt[8]{\frac {x^8+y^8+z^8} 3}\ge\sqrt[7]{\frac {x^7+y^7+z^7} 3}$$ but stuck

答:
Here is an elementary way to proceed, which may not be wholly convenient, but which illustrates some interesting techniques.

If $x,y,z$ are the roots of a cubic $t^3-s_1t^2+s_2t-s_3=0$, then we have$$s_1=x+y+z=0$$$$2s_2=2xy+2yz+2zx=(x+y+z)^2-(x^2+y^2+z^2)=-1$$
Thus we have (for some $u$) $$2t^3-t-u=0$$

This gives also (multiply by $t^r$) $$2t^{r+3}-t^{r+1}-ut^r=0$$

Now let $a_r=x^r+y^r+z^r$, and substitute $x, y, z$ successively into the last equation and add the three together to obtain:$$2a_{r+3}-a_{r+1}-ua_r=0 \text{ with }a_0=3, a_1=0, a_2=1$$

Then you can express $a_7$ in terms of $u$ using the recurrence. The constraint on $u$ is that all the roots of the cubic must be real. Combining these two pieces of information you can get a handle on your question.

Title: Find max of $x^7+y^7+z^7$


问:
How to prove that $a^3 + b^3 \geq a^2b + ab^2$?
Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--

For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$

I have no idea how to begin but this is what I did.

By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)

Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$

Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.

(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)

Body:
Note
$$\frac{a^3+b^3}{a^2b+ab^2}=\frac{a^2-ab+b^2}{ab}=\frac{(a-b)^2}{ab}+1\ge 1
$$
Thus
$$a^3+b^3\ge a^2b+ab^2$$


问:
How to find sum of $\frac{1}{2}+\frac{3}{8}+\frac{15}{48}+\frac{105}{384}+\cdots$?

Sum $S_n$ of this series is given by, $S_n=\frac{1}{2}+\frac{3}{8}+\frac{15}{48}+\frac{105}{384}+...$

Express $r^{th}$ term ($U_r$) in this series and hence show that,$$\frac{U_r}{U_{r-1}}=\frac{2r-1}{2r}$$
Considering the above relation find $f_{(r)}$such that $U_{r-1}=f_{(r)}-f_{(r-1)}$. Here $f_{(r)}$shall be expressed interms of $U_r.$
Hence deduce $\sum_{r=1}^nU_r=(2n+1)U_n-1$

My Try:

I was able to find $U_r=\frac{(2r-1)!!}{(2r)!!}$

Simplifying further I got the next relation,

$U_r=\frac{(2r-1)(2r-3)!!}{2r(2r-2)!!}=\frac{(2r-1)}{(2r)}U_{r-1}$

But how do I proceed to find the summation. Please help me. Thanks in advance!

Body:
Hint:$$\frac{U_{r}}{U_{r-1}}=\frac{2r-1}{2r}$$

Rearrange $$U_{r-1}=\color{red}{2(r)U_{r}}-\color{blue}{2(r-1)U_{r-1}}=f_{r}-f_{r-1}$$

here $f_r=2rU_r$

Now telescope....


问:
$\mathbb{Q}/\mathbb{Z}$ has cyclic subgroup of every positive integer $n$?
I would like to know whether $(\mathbb{Q}/\mathbb{Z},+)$ has

$1$. Cyclic subgroup of every positive integer $n$?

$2$. Yes, unique one.

$3$. Yes, but not necessarily unique one.

$4$. Does not have cyclic subgroup of every positive integer $n$.

What I know about the given group is infinite group but every element has finite order. Please help how to proceed.

答:
As mentionned here, $\mathbb{Q} / \mathbb{Z} \simeq \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]$ (its canonical decomposition as a divisible group).

Let $p_1^{n_1} \dots p_r^{n_r}$ be the prime decomposition of $n$. For each $1 \leq j \leq r$, let $\displaystyle \zeta_j=\exp \left(i \frac{2k_j \pi}{p_j^{n_j}} \right)$ be an element of $\mathbb{Z}[p_j^{\infty}]$ with $k_j$ not divisible by $p_j$. Set $\zeta=\zeta_1 \dots \zeta_r$.

Now, $\text{ord}(\zeta)=\prod\limits_{i=1}^r \text{ord}(\zeta_i)= \prod\limits_{i=1}^r p_i^{n_i}=n$ and $\langle \zeta \rangle = \bigoplus\limits_{i=1}^r \langle \zeta_i \rangle$.

Of course, there is a choice for each $k_j$, but it doesn't change $\langle \zeta_j \rangle$. Indeed, let $\zeta_1=\exp \left(i \frac{2k_1\pi}{p^n} \right)$ and $\zeta_2= \exp \left( i \frac{2k_2\pi}{p_n} \right)$ where $k_1$ and $k_2$ are not divisible by $p$. Using Bezout equality, there are $u,v \in \mathbb{Z}$ such that $uk_1+vp^{n}=1$ hence $k_2=uk_1k_2+vk_2p^n$. So $\zeta_2= \zeta_1^{uk_1}$, ie. $\zeta_2 \in \langle \zeta_1 \rangle$. By symmetry, $\zeta_2 \in \langle \zeta_1 \rangle$. Therefore, $\langle \zeta_1 \rangle=\langle \zeta_1 \rangle$.

Finally, one deduce that $\mathbb{Q}/ \mathbb{Z}$ has a unique subgroup of order $n$.


问:
need help with how to Construct Binary Tree from Inorder traversal
Question : we have the following list  J, R, D, G, T, E, M, H, P, A, F, Q
which is inserted in order in an empty binary tree.

This is the inorder traversal :

A, D, E, F, G, H, J, M, P, Q, R, T

construct the binary tree.

The answer is :

I understand that inorder is : You visit the node after you visited his left sub-tree and then you visit the right sub-tree.

I understand how to do it also when u have the given sequence for pre-order and in-order at the same time but i need help understanding what to do when u only have the in-order sequence.

I have no clue why we say that J is the root, i have no clue why F is in the right branch of E and the list goes on...

It seems to me that i am missing out some basic concept to help me understand this, if anyone could explain step by step the process i would greatly appreciate it.

Thanks a lot.

答:
Solution

You are given two clues for how to solve this problem: The result of the Inorder-Tree-Walk, and the order the values are inserted. As another clue, you may use that if a node is inserted into a binary tree, it will always become the root and remain so until it is removed. If a node $x$ is inserted into a tree containing just a single root $y$, then $y$ will become the parent of $x$ and stay so.

Given what you already know from the inorder walk, you should now be able to reconstruct the tree. Since J is inserted first, you know it must be the root. Next, you know that R will be a child of J. To see if it is the left or the right child, you can simply examine the order in the inorder walk.

More information

This sort of tree is called a binary search tree. Let $u$ be a node in the tree, and define $u.\mathit{key}$ to be the value stored in the node (e.g. J or D). A binary search tree has the property that for any given node $u$, then if $x$ is a node in the left subtree of $u$, $x.\mathit{key} \leq u.\mathit{key}$. If $x$ is a node in the right subtree of u, $x.\mathit{key} \geq u.\mathit{key}$.

There is an algorithm for inserting nodes into a binary search tree which makes sure this property is maintained, but you actually don't need it to solve your problem. If, however, you are interested, you can find more information in the excellent book Introduction to Algorithms by Cormen et. al.


问:
Continuous bijective mapping between $\mathbb{C}$ and $\mathbb{C}^*$
Is there a continuous bijective mapping from $\mathbb{C}$  to   $\mathbb{C}^* = \mathbb{C}\setminus\{0\}$?

答:
Hint: one  is simply connected, the other is not.

EDIT: Why in the world was this answer deleted?

EDIT: Suppose $f$ is such a map.  Let $x \in \mathbb C$, and $V$ a closed disk centred at $x$.  Since $V$ is compact, $f$ is a homeomorphism of $V$ onto $f(V)$.  The image of the interior of $V$ is a subset of $\mathbb C^*$ homeomorphic to an open disk and containing $f(x)$, so it is a neighbourhood of $f(x)$.  Thus $f$ is actually a homeomorphism of $\mathbb C$ onto $\mathbb C^*$.


问:
Prove subspace existence
I am trying to solve the following problem:

Given the following equations system

$
\begin{equation}
\begin{cases}
20a + b + c + d + e = 0 \\
a + 20b + c + d + e = 0 \\
a + b + 20c + d + e = 0
\end{cases}
\end{equation},
$

How to prove that its solutions set are a $\mathbb{R}^5$ subspace? How to describe it?

I am trying to solve the system, but once it demands a $\mathbb{R}^5$ subspace (I have never worked with it), I really don't know how to deal with it.

答:
Equivalently, subtract the second equation from the first to get 19a- 19b= 0 or a= b.  With that the equations become  21a+ c+ d+ e= 0 and 2a+ 20c+ d+ e= 0.  Subtract the second of those from the first to get 19a- 19c= 0 or a= c.

a= b= c, d, and e satisfy all three equations for any c, d, and e so a solution can be written {c, c, c, d, e}.  To show that this is a subspace you need to show that it is "closed" under addition and scalar multiplication.  You can do that by showing that xu+ yv, for u and v members of that set and x and y real numbers, is also of that form.

That is, x(c1, c1, c1, d1, e1)+ y(c2, c2, c2, d2, e2)= (xc1+ yc2, xc1+ yc2, xc1+ yc2, xd1+ yd2, xe1+ yd2) which is of the form (c, c, c, d, e) with c= xc1+ yc_2, d= xd1+ yd2, and e= xe1+ ye2.

Title: Prove subspace existence


问:
$A$ be $n \times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of $A$
I am trying assignment questions in Linear Algebra and this question could not be solved by me. So, I thought of posting it here.

Let A be an $n\times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A. Then which of the following are true:

All eigenvalues of A are equal

All eigenvalues of A are distinct

$A=\lambda I $ for some $\lambda \in \mathbb{C}$ , where I is the n times n identity matrix

The minimal and characterstic polynomial for A are equal.

I tried by taking eigenvectors to be distinct and the hypothesis but couldn't solve any options. Can you please give hints how to start the problem!

Thank you!

答: Hint: if $u$ is an eigenvector for eigenvalue $\alpha$ and $v$ an eigenvector for eigenvalue $\beta \ne \alpha$, can $u+v$ be an eigenvector?


问:
Does "normal" induction work in any well-ordered set?
This is a follow up from this question.

I want to prove whether the usual "induction theorem" works in a general  well-ordered set $A$:

If $(A,\preccurlyeq)$ is a well-ordered chain, $\Phi(\min A)$ is true, and for all $x\in A$, $\Phi(x)\Rightarrow \Phi(x^+)$,  then $\Phi(a)$ is true for all $a\in A$.

For $\mathbb N$ it works because apart from the successor, you can also evaluate the "precursor"
$$x^-=\max\{a\in A : \text{$a\preccurlyeq x$ and $a\neq x$}\}.$$

Indeed, one can proceed by contradiction, noting that if it is false, $\min\{a\in A : \lnot\Phi(a)\}$ exists (call it $a_0$), but then you get that $\Phi(a_0^-)$ is true, so $\Phi((a_0^-)^+) = \Phi(a_0)$ is true, a contradiction.

(In the case of $\mathbb N$, the maximum always exists because the corresponding set is bounded above and finite, hence my original question.)

When I see this stated for the general case, it is always phrased in the "strong induction" way, and the proof depends on that fact. Here is an example:

So is my (normal) induction statement true? I suspect not now that I think about it, since if we have order type $\omega+1$ (say) we never establish $\Phi(x)$ for the last element by chaining implications together as induction usually does for $\mathbb N$ (i.e., $\Phi(1) \Rightarrow\Phi(2)\Rightarrow\cdots$).

答:
No, your "normal induction statement" is not true.

Consider the following well ordered set $A$, $\{0,1\}\times\omega$, with the lexicographic order. That is, $(a,b)\lt (c,d)$ if and only if $a\lt c$, or $a=c$ and $b\lt d$. Think of it as "green" and "yellow" natural numbers, with all the green natural numbers going first, and all yellow natural numbers going later. The successor funtion in this set is $(a,b)^+ = (a,b^+)$.

This is well ordered: given any nonempty subset $X$ of $A$. If $X\cap \{0\}\times\omega$ is nonempty, then the preimage of the least element of the projection of this subset onto $\omega$ is the least element; otherwise, the least element of $X$ is the smallest "yellow" natural number that it contains. (That is, if $X$ has green natural numbers in it, then its least element is the smallest green natural number in it; otherwise, take the smallest yellow number in it).

Now consider the property $P(a,n)$ to mean "$a=0$" (i.e., "$n$ is green"). Clearly, $P(a,n)$ does not hold for all elements of $A$.

However: $P(0,0)$ holds. And if $P(a,b)$ holds, then $P( (a,b)^+) = P(a,b^+)$ holds, since $a=0$. Thus, your "normal induction statement" holds for $P$ in $A$. Yet $\forall x(x\in A\implies P(x))$ is false.

Note that your statement for "strong induction" can be simplified. In fact, you only need to prove that
$$\forall a\in A\Bigl( \bigl(\forall a' (a'\lt a\implies P(a'))\bigr)\implies P(a)\Bigr)$$
and that will prove $P(a)$ for all $a\in A$, since for $\min(A)$, the premise $\forall a' (a'\lt \min(A)\implies P(a'))$ is true by vacuity. So you don't need to separate out the least element of $A$, and then explicitly exclude it from the inductive step. See comments here and here. It's just that one usually needs to do "special cases" for the least element, but sometimes for other values instead.


问:
Notion of cusp in algebraic geometry
This relates to problem 3.14(b) of Hartshorne, wherein I am required to show that the projection of the twisted cubic curve (viewed as a subset of $\mathbb P^3$) from the point $(0,0,1,0)$ onto the hyperplane $\mathbb P^2$ defined by the equation $z=0$ is a cuspidal cubic curve. I have been able to show that the said projection is the zero set of the homogenous polynomial $f(x,y,z):= y^3-x^2z \in k[x,y,z]$ so it is a cubic curve. From some searching, I learnt that a cuspidal curve is one where all singularities are cusps and by using the characterization of singular points in projective curves defined by a single equation as points where all partial derivatives vanish, I have been able to show that the only singularity of $f$ occurs at the point $(0, 0, 1) \in \mathbb P^2$. Now I am not sure how to show that this point is a cusp, because I require a rigorous definition of "cusp" in this context which is easy to verify.

Most references I came across give a definition to the effect that a "cusp" is a point where two "branches" have a common semi-tangent, and the notion of "branch" is either handwaved or explained only for Euclidean spaces. The only rigorous definition I have found after searching of a cusp $P$ on a variety $Y$ involves completions of $\mathcal O_{P, Y}$. Now I am not familiar with the idea of completion in this context and would really appreciate a definition which is rigorous and can be used in the aforementioned context.

Edit: Here (https://arxiv.org/pdf/1511.02691.pdf) is a reference, where "cusp" is defined on pg. 6, but it uses the notion of a "branch", which has not been defined. I would really like to know a simple rigorous definition of that terminology.

答:
The plane curve $f(x,y)=x^2(x+1)-y^2=0\subset \Bbb{C}^2$ doesn't have a cusp at $(0,0)$ because we have  a parametrization $(x,x\sqrt{1+x}),(x,-x\sqrt{1+x})$ of two distinct "analytic curves" around $(0,0)$.

In contrary $x^2-y^3=0$ has a cusp because $(y\sqrt{y},y)$ is not an analytic curve around $(0,0)$.

Over $\Bbb{C}$ we have the ring of all functions that are analytic in some neighborhood of $x_0=0$, but this doesn't generalize well to arbitrary fields.

We may try to replace it by the ring of formal power series in $x$, which is $k[[x]]=\varprojlim k[x]/(x^n)$,
or in $y$, which is $k[[y]]=\varprojlim k[y]/(y^n)$, but
it appears that the most insightful ring is $\varprojlim k[x,y]/(f(x,y))/(x,y)^n=k[[x,y]]/(f(x,y))$.

If $char(k)\ne 2$ then $\sqrt{1+x}=\sum_{k\ge 0} {1/2\choose k}x^k$ is in this ring and $(y-x\sqrt{1+x}),(y+x\sqrt{1+x})$ are two distinct prime ideals. $k[[x,y]]/(f(x,y))$ is not an integral domain as $(y-x\sqrt{1+x})(y+x\sqrt{1+x})=0$. In contrary to $k[[x,y]]/(x^2-y^3)$ which is an integral domain.

For an arbitrary variety this generalizes to $\varprojlim O_P / \mathfrak{p}^n$ where $O_P$ is the ring of rational functions regular at $P$ (or the localization thereof for a non-closed point) and $\mathfrak{p}$ is the prime ideal of those vanishing at $P$ (its unique maximal ideal).

For a curve, when a parametrization $(x,h(x))$ exists with $h $ a formal power series then in fact $h$ is algebraic over $k[x]$ so that it can be seen as a rational function on a ramified covering of our curve.

Title: Notion of cusp in algebraic geometry


问:
Categorical characterisation of fields and groups
Given a monoidal category $\mathcal{V}$, is there a categorical characterisation of an object in $\mathcal{V}$ such that it is a group if $\mathcal{V} = Set$ and a field if $\mathcal{V} = Ab$? Or does anybody know of useful references? This would be much appreciated!

Ofcourse I do not expect it by simple means of extra data as groups cannot be descriped operadically, but rather by universal properties or something similar.

EDIT: As noted in the comments: I am looking for something different than group objects, as they do not encompass fields. So I am looking for some categorical framework in which both can fit.

答:
For groups, it's quite doable: a group object in a cartesian monoidal category is a particular kind of monoid. Now every object in a cartesian monoidal category is canonically a comonoid, where the comultiplication is the diagonal $\Delta:G\to G\times G$ and the counit is the unique map $!:G\to 1$. In fact if $G$ is a monoid, then this comonoid structure makes it a bimonoid, because $\Delta$ and $!$ are always monoid homomorphisms.

This suggests that we want to think about a kind of bimonoid, in a general symmetric monoidal category (the symmetry is necessary to make the tensor product of monoids a monoid.) So let $G$ in $\mathcal V$ be a bimonoid, equipped with a comultiplication $\Delta:G\to G\otimes G$ and a counit $\epsilon:G\to I$ as well as a multiplication $\mu$ and unit $\eta$. Then we can write down the inverse axiom as for a group: introduce an endomorphism $s:G\to G$ (called in this context an "antipode"), such that $\mu\circ(1\otimes s)\circ \Delta=\eta\circ \epsilon$. This is the definition of a Hopf monoid, and uniqueness of comonoid structures in Cartesian monoidal categories implies that a Hopf monoid in Sets is precisely a group.

I'm not so sure what to do about fields, since these aren't even algebraically defined--note that we need the notion of the complement of zero to get a field, so they naturally fit in something closer to a topos rather than in a monoidal category.


问:
Colored balls probability - color of remaining balls
A bowl contains 5 white and 8 black balls. We proceed to pick out balls one by one (without putting any back in the bowl) until only same-color balls remain in. What should I think to calculate the probability of only white balls remaining in the bowl?</p>

答:
Write the arrangements of $BBBBBBBBWWWWW$ out such that each one ends with a $W$, and count these.

To avoid repetition, consider the endings of $BW$, $BWW$, $BWWW$, $BWWWW$, $BWWWWW$, and then arrange the remaining balls in front of them.

So in total:

$$\binom{11}{4}+\binom{10}{3}+\binom{9}{2}+\binom{8}{1}+\binom{7}{0}$$

$$=330+120+36+8+1$$

$$=495$$

The probability is then $\frac{495}{\binom{13}{5}}=\frac{5}{13}$

which is $\frac{5}{5+8}$ because the numerator is $\binom{12}{4}$ by the hockey-stick identity, and the denominator is $13\cdot\binom{12}{4}\cdot\frac15$.


问:
Integral of Dirac delta function evaluated at a differential function
I am having some troubles in evaluate the following:
\begin{align}
\int_{a}^{b} \delta \left( f \left(x \right)\right) f'\left( x\right) dx,
\end{align}
where $f: \left[a,b \right] \to \mathbb{R}$ is differentiable, $f\left( a\right)\neq 0, f\left(b \right)\neq0$ and if $f\left( x\right) = 0$, then $f'\left( x\right) \neq 0$. How can I proceed? Thank you!

答:
using the substitution $y=f(x)$ you get:
$$I=\int_{f(a)}^{f(b)}\delta(y)dy$$
and this would be $I=1$ if $0\in[f(a),f(b)]$ or $I=0$ if $0\notin[f(a),f(b)]$

To answer your comment, remember that:
$$\int_a^b\delta(x)dx=1,\,0\in[a,b]$$
since the function $\delta$ has one peak for every zero for the function inside of it. So for a function $f(x)$ with $n$ roots $i=[1,n],\,\alpha_i\in[a,b]$ we can say that there will be $n$ delta peaks within the region $[a,b]$ and so:
$$\int_a^b\delta(f(x))dx=\int_a^b\sum_{i=1}^n\delta(x-\alpha_i)dx=n$$

As for the second part, notice that if we use the same substitution as in the original question we get:
$$\int_{f(a)}^{f(b)}\delta(y)\frac{|f'(x)|}{f'(x)}dy=\int_{f(a)}^{f(b)}\delta(y)\operatorname{sgn}(y')dy$$
Now think about what this represents


问:
Can someone please explain $e$ in layman's term?
I never really understood what $e$ means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that?</p>

答:
On can show that there are functions $f(x)$ with the property
$$\tag1f'(x)=f(x)$$ for all $x$. If one additionally requires $f(0)=1$, the solution is unique and called $\exp(x)$. From this property alone, it follows that the general solution to $f'(x)=f(x)$ is $f(x)=c\exp(x)$ with $c=f(0)$. And also that $$\tag2\exp(x+y)=\exp(y)\exp(x)$$ for all $x,y$
because $x\mapsto \exp(x+y)$ is also a solution of $(1)$. Especially, $\exp(n)=\exp(1)\exp(1)\ldots\exp(1)=(\exp(1))^n$. In effect, because of $(2)$, the function $x\mapsto \exp(x)$ suspiciously behaves like exponentiation and is therefore usually suggestively written like exponentiation, i.e. we write $e^x$ instead of $\exp(x)$. Of course for this to make sense, we must define $e=\exp(1)$. It turns out that $\exp(1)\approx 2.71828$.


问:
Finding closed formula for sequence
How can I find the closed formula for the sequence $$S = 2^0n + 2^{1}(n-1) + 2^{2}(n-2) + ... + 2^{n-1}(1) + 2^{n}(0).$$
Also, what kind of sequence is this because it looks both geometric and arithmetic.

Title: Finding closed formula for sequence

Body:
$$S_{n}-S_{n-1} =2^0 + 2^1+\cdots + 2^{n-1}=2^n-1$$

$$S_n-S_1 = (S_n-S_{n-1}) + (S_{n-1}-S_{n-2}) + \cdots + (S_2-S_1) \\
= (2^n-1)+(2^{n-1}-1)+ \cdots + (2^2-1)=2^{n+1}-2^2 - (n-1)$$

$$\implies S_n = 2^{n+1} - n-2. \blacksquare $$


问:
T-test vs Z-test

It is known that the diameter of a particular type of automobile tyre follows a normal distribution with mean and variance being $600$ millimetres and $3$ millimetres respectively. Ten samples are collected from a large population of manufactured tyres as :

It is believed that the manufactured lot has a diameter different from $600$ mm. Which hypotheses test can be used here? What conclusion can be drawn on $H_0$.

Select one:

T-test, Failed to reject $H_0$

Z-test, Failed to reject $H_0$

Z-test, Reject $H_0$

F-test, Reject $H_0$

Note: I'm confused between T-test and Z-test. When one should go for T-tset and when for Z test. In this particular senario which one is best and why?

答:
@Heropup (+1) is correct. Suppose the data were as entered into a recent release of Minitab, as follows:

```<code> 595.0   598.5   599.6   603.2
 601.5   584.6   604.0   597.5   598.1
</code>```

Then results of the appropriate one-sample z test
are as follows:

```<code>One-Sample Z: x

Test of μ = 600 vs ≠ 600
The assumed standard deviation = 3

Variable  N    Mean  StDev  SE Mean       95% CI           Z      P
x         9  598.00   5.78     1.00  (596.04, 599.96)  -2.00  0.046
</code>```

You reject this two-sided test at the 5% level.

You should find the sample mean $\bar X$ of your data and compute the $Z$ statistic by hand, for an exact answer.

Title: T-test vs Z-test

Body:
@Heropup (+1) is correct. Suppose the data were as entered into a recent release of Minitab, as follows:

 595.0   598.5   599.6   603.2
 601.5   584.6   604.0   597.5   598.1

Then results of the appropriate one-sample z test
are as follows:

One-Sample Z: x

Test of μ = 600 vs ≠ 600
The assumed standard deviation = 3

Variable  N    Mean  StDev  SE Mean       95% CI           Z      P
x         9  598.00   5.78     1.00  (596.04, 599.96)  -2.00  0.046

You reject this two-sided test at the 5% level.

You should find the sample mean $\bar X$ of your data and compute the $Z$ statistic by hand, for an exact answer.


问:
Complexification the real inner product space
Let $V$ be a real inner product space. If $W=V\times V$ with the operations
$(u_1,v_1)+(u_2,v_2)=(u_1+u_2,v_1+v_2)$ and $(\alpha +i\beta)(u,v)=(\alpha u-\beta v,\alpha   v+\beta u)$, where $u, u_1,u_2,v,v_1,v_2\in V$ and $\alpha,\beta$ are scalars.

Show that $W$ is a complex inner product with the inner product
  $$\langle (u_1,v_1),(u_2,v_2)\rangle=\langle u_1,u_2\rangle+\langle v_1,v_2\rangle+i\langle v_1,u_2\rangle-i\langle u_1,v_2\rangle$$

I do not know how to connect these operations, plz help

答:
Maybe too late, $\langle u_1+iv_1,u_2 + iv_2\rangle $ works as in $\mathbb{C}=\mathbb{R}+i\mathbb{R}$ i.e.,

$ (u_1+iv_1)\overline{(u_2+iv_2)}= (u_1+iv_1)(u_2-iv_2)=u_1u_2+v_1v_2+i(v_1u_2-u_1v_2)$. This definition inherits all the good properties from the original inner product $\langle\cdot,\cdot\rangle_V$.


问:
Find $\lim\limits_{x \to +\infty}\frac{f(x)}{x}.$
Suppose $f(x)$ be a periodical function defined over $\mathbb{R}$, bounded nearby $x=0$ and satisfy $\lim\limits_{x \to +\infty}[2f(x)-f(2x)]=0.$ Find $\lim\limits_{x \to +\infty}\dfrac{f(x)}{x}.$

Let $T$ be a positive period of $f(x)$. Then $2f(x)-f(2x)=2f(x+nT)-f[2(x+nT)]$, where $n=1,2,\cdots.$ Hence $\lim\limits_{n \to \infty}[2f(x)-f(2x)]=\lim\limits_{n \to \infty}\left[2f(x+nT)-f(2(x+nT))\right]=0,$ which implies $2f(x)=f(2x)$ holding for any $x\in \mathbb{R}$. Further, $f(x)=\dfrac{1}{2}f(2x)=\cdots=\dfrac{1}{2^n}f(2^nx).$ How to go on?

答:
You have most of what you need:  you have deduced for all $x$
$$ f(x) = 2^{-n} f(2^n x). $$
Replace $x$ with $2^{-n}x$ and you obtain for any $x \in [0,T]$,
$$ f(x) = 2^n f(2^{-n}x ). $$
Now use the boundedness of $f$ in the neighbourhood of $x=0$.  There exists $b > 0 $ and $\delta >0$ such that $|f(x)| \leqslant b $ whenever $|x| < \delta$.  Because $f$ is periodic, we can find a fixed integer $n$ such that $0 \leqslant 2^{-n} x  < \delta$ for all $x \in [0,T]$: simply choose $n > \log (T / \delta) / \log 2$.  Then, for every $x \in [0,T]$
$$
|f(x)| = |2^n f(2^{-n}x)| \leqslant 2^n b$$
and by periodicity, it follows $f(x)$ is bounded by $2^nb$ for all $x$.  Then
$$ \left\lvert \frac{f(x)}{x} \right\rvert \leq \frac{ 2^nb}{|x|} \to 0 \quad\text{as}\quad  x \to \infty.$$


问:
The probability of the dart hitting the smaller ring is?

A dart is randomly thrown at a circular board on which two concentric rings of radii $R$ and $2R$ having the same width (width much less than $R$) are marked. The probability of the dart hitting the smaller ring is:

$(1)$ Twice the probability that it hits the larger ring.

$(2)$ Half of the probability that it hits the larger ring.

$(3)$ Four times the probability that it hits the larger ring.

$(4)$ One-fourth the probability that it hits the larger ring.

My attempt $:$ Let $S$ be the radius of the circular board. Then the probability
that the dirt will hit the larger ring and the smaller ring would be  respectively $\frac {4R^2} {S^2}$ and $\frac {R^2} {S^2}.$ So the probability  of the dart hitting the smaller square will be one-fourth the the probability that it hits the larger square. So according to me option $(4)$ is correct. But the answer key suggests that the correct option is $(2).$ Am I missing something? Any help in this regard will be appreciated.

Thanks for your time.

Source $:$ CSIR NTA NET DECEMBER $2019.$

答:
Let's say the common width was $W$.  Then the area occupied by a ring of radius $r$ would be  $$\pi (r+W)^2-\pi r^2=2\pi r W+\pi W^2$$

Thus the probability that the dart will land in that ring is $$\frac {2\pi r W+\pi W^2}{\pi S^2}=\frac {2rW}{S^2}+\frac {W^2}{S^2}\approx \frac {2rW}{S^2}$$  where we have used the fact that $W$ is much smaller than $S$.

It follows that the ratio you want is approximately $$\frac {2RW}{S^2}\Bigg /\frac {4RW}{S^2}=\frac 12$$

As desired.


问:
Proving $\triangle DFG$ is equilateral

In this figure $\Delta ADC,\Delta BED$ are equilateral. $F$ is the midpoint of $AE$,$G$ is the midpoint of $CB$.Prove that $\Delta DFG$ is also equilateral.

My Try

Easily we can show that $$\Delta ADE \cong \Delta CDB  \tag{SAS}$$ hence  $$AE=BF$$
$$AF=FE=CG=BG$$  The rest  is getting complicated...

答:
Nope, the rest is elementary.

We have $\triangle ADF \cong \triangle CDG$ (SAS), since $AF = CG, AD = DC$ and $\angle EAD = \angle BCD$ from the pair of congruent triangles you showed. This gives $DF = DG$.

Finally $\angle FDG = \angle CDG - \angle CDF = \angle ADF -\angle CDF = 60^\circ$.

(Essentially, $\triangle CDG$ is $\triangle ADF$ rotated by $60^\circ$.)

Two equal sides with an included angle of $60^\circ$ produces an equilateral triangle.

Title: Proving $\triangle DFG$ is equilateral


问:
Building a tower using colorful blocks
How many possibilities are there to build a tower of n height, using colorful blocks, where:

white block is of 1 height

black, red, blue, green, yellow, pink blocks are equal to 2 heights

I need to find the generating function formula for this.

So, for n = 1 I get 1 possibility,

for n = 2 I get 2 possibilities,

for n = 3 I get 3 possibilities

for n = 4 I get > 4 possibilities etc.

The generating function at this moment would be $1 + 2x + 3x^{2} + ...$. But I have no idea how can I find the general formula calculating this.

Could you give me any suggestions, or solutions (;-)) ?

答:
An alternative approach is to note that the generating function for the number towers made out of exactly $k$ blocks of any size can be written as $$(x+6x^2)^k$$ So the total generating function must be:

$$\sum_{k=0}^\infty (x+6x^2)^k$$

which is just a geometric sequence for which a closed form can easily be found.

If you want to ignore the colors of the bigger blocks, then the generating function for exactly $k$ blocks is $$(x+x^2)^k$$ and the total generating function is $$\sum_{k=0}^\infty (x+x^2)^k$$

It turns out in this case the sequence will look familiar.

Title: Building a tower using colorful blocks


问:
finding bijections
Find bijections between A and B and proof their bijectivity

$A=\{2m| m \in \mathbb{Z}\}$, $B=\{2n-1| n \in \mathbb{Z}\}$.

Here I have $f: A \rightarrow B$,
$f(x)=x-1$. It is easy to prove that it is one to one, and that it is onto.

$A=\{3m| m \in \mathbb{Z}\}$, $B=\{3n-1| n \in \mathbb{Z}\} \cup \{3k-2|k \in \mathbb{Z}\}$

The elements which are in $B$ are the ones $1$ mod $3$ and $2$ mod $3$, but I can not come up with any solutions for (2). Should the function be piecewise there?

Could someone check whether 1) is correct and help me solving 2)?

答:
$1.$ is correct.  For $2.$, an injection from $A$ to $B$ would be

$f(a)=\dfrac a2+1$ if $a$ is even and $\dfrac{a+1}2$ if $a$ is odd.

Note that $\dfrac{1+(-1)^a}2$ is $1$ when $a$ is even and $0$ when $a$ is odd,

and $\dfrac{1-(-1)^a}2$ is $0$ when $a$ is even and $1$ when $a$ is odd,

so $f(a)=\dfrac{1+(-1)^a}2\left(\dfrac a2+1\right)+\dfrac{1-(-1)^a}2\left(\dfrac {a+1}2\right)$ works,

and that simplifies to $f(a)=\dfrac{2a+3+(-1)^a}4$.

Title: finding bijections


问:
Good books to Learn General topology
I am quite interested in learning topology from the basics, I would like to find a book that is quite detailed and has a large number of examples.

What are the best books to easily understand topology?

Thanks.

答: Gustavo Rubiano's book, introduction to Topology (Victor Anatolyevich Vassiliev).</p>

Title: Good books to Learn General topology


问:
How can I solve this continuity question?
Basically, I am trying to understand this problem, but I still have not gotten how to solve it, nor the steps to lead me to the answer. Could anyone help me out? Thanks.
Question: What value should be assigned to K to make f continuous at x = 0 Here is the problem.</p>

答:
A function is continuous at some point $a$ if $\displaystyle \lim_{x\to a} f(x)=f(a)$. Just compute, in your case, the previous limite, compare it to $f(0)$ and draw your conclusions.

Also, I must say that this forum is not the proper way of dealing with such a basic question. If you are not able to solve this question, you must really study and look at some examples before anything else.


答:
To flesh out the hint in the comments:

Imagine that you draw out all the balls, even though they might be mono-chromatic well before you are done.

The probability that you want is just the probability that the last ball drawn is white so the answer is $$\boxed {\frac {5}{5+8}=\frac 5{13}\approx .384615}$$

In general, starting with $W$ white balls and $B$ black, the answer is $$\frac {W}{W+B}$$

Title: Colored balls probability - color of remaining balls

Body: A bowl contains 5 white and 8 black balls. We proceed to pick out balls one by one (without putting any back in the bowl) until only same-color balls remain in. What should I think to calculate the probability of only white balls remaining in the bowl?</p>


答:
Rotate everything with respect to $D$ for $60^\circ$ in counterclockwise direction.

Since $\triangle BDE$ is equilateral, $B$ get rotated into $E$.

Since $\triangle DCA$ is equilateral, $C$ get rotated into $A$.

Since rotation preserve midpoint-ness, $G$ (midpoint of $BC$) get rotated into $F$ (midpoint of $EA$). This implies $DG = DF$ and $\angle GDF = 60^\circ$. So $\triangle DFG$ is equilateral.

Title: Proving $\triangle DFG$ is equilateral

Body:
In this figure $\Delta ADC,\Delta BED$ are equilateral. $F$ is the midpoint of $AE$,$G$ is the midpoint of $CB$.Prove that $\Delta DFG$ is also equilateral.

My Try

Easily we can show that $$\Delta ADE \cong \Delta CDB  \tag{SAS}$$ hence  $$AE=BF$$
$$AF=FE=CG=BG$$  The rest  is getting complicated...


问:
What happens to $dV$ and $dt$ in the proof of the logistic equation?
I understand that the following antiderivative is correct.
\begin{align}
\int \left(\frac{1}{V}+\frac{1}{N-V}\right)\,dV&=\int r\,dt\\
\ln{\left|V\right|}+c_1-\ln{\left|N-V\right|}+c_2&=rt+c_3\\
\end{align}
However, can someone explain what happens to $dV$ and $dt$ when going from the first equation to the second? Why is the left-hand side not multiplied by $V$ but $r$ on the right-hand side is multiplied by $t$? Sorry, I'm a bit confused!

答: It is integration, not multiplication. Although, your weird rule of multiplying by the variable of integration does work for the integral $\int kdt=kt+C$.


答:
Remember that $dV$ and $dt$ tell you what the integration variable is.

In the first case we are trying to find the antiderivative of $\frac{1}{V}+\frac{1}{N-V}$ with respect to $V$.  That, as you've said, is $$\ln|V| - \ln|N-V| + C$$ because when you differentiate with respect to $V$ you get back to $\frac{1}{V}+\frac{1}{N-V}$.  In the second case, notice that $r$ is a constant with respect to $t$, so when you find the antiderivative of $r$ with respect to $t$ you get $rt + C$.  This is because if you differentiate $rt + C$ with respect to $t$ you get back to $r$.

Title: What happens to $dV$ and $dt$ in the proof of the logistic equation?

Body:
I understand that the following antiderivative is correct.
\begin{align}
\int \left(\frac{1}{V}+\frac{1}{N-V}\right)\,dV&=\int r\,dt\\
\ln{\left|V\right|}+c_1-\ln{\left|N-V\right|}+c_2&=rt+c_3\\
\end{align}
However, can someone explain what happens to $dV$ and $dt$ when going from the first equation to the second? Why is the left-hand side not multiplied by $V$ but $r$ on the right-hand side is multiplied by $t$? Sorry, I'm a bit confused!


问:
Domain of a function defined with another function
Suppose $f(x)=\frac{1}{1+\frac{1}{x}}$, and $g(x)=\frac{1}{1+\frac{2}{f(x)}}$. To find the points not in the domain of $g(x)$, we obviously have to check the $x-$values where $f(x)$ becomes $0$ and $-2$, and count them first.

But where is the intuition behind considering the points for which $f(x)$ itself is undefined? On substituting $f(x)$ in $g(x)$, you get $g(x)=\frac{x}{3x+2}$. Why do we count the additional points where $f$ is discontinuous (in this case, the only extra one being $x=-1$, as $x=0$ is already necessary for $g(x)$'s denominator to be defined)?

答: We want to find when $g$ is undefined and this is when the denominator, $1+ \frac {2}{f(x)} = 0 $ which is only true when $f(x) = -2$. So then we check $-2 = \frac {1}{1+\frac{1}{x}} \implies 1+\frac {1}{x} = -\frac{1}{2}\implies x=-\frac {2}{3}.$ So $g(-\frac{2}{3})$ is not defined and $x = -\frac {2}{3}$ is not in the domain of $g$. Notice that $g$ is also not defined when $f(x) = 0$, but $f(x)$ never equals $0$ so this is not a problem. We must check where $f(x)$ is undefined now. This is when $1+ \frac{1}{x} = 0 \implies x = -1$. so then $g(-1)$ is not defined either. Also $f(0)$ is undefined since $\frac {1}{0}$ is undefined which means $g(0)$ is undefined. Hence the points not in the domain of $g$ are $x = -1, -\frac {2}{3},0$.


问:
Have I gone about this proof in a clear way?
Prove that if $f$ is uniformly* continuous on a bounded set $S \in \mathbb{R}$, then $f$ is a bounded function on $S$.

Assume that $f$ is not continuous on a bounded set $S \in \mathbb{R}$. By the Bolzano-Weierstrass theorem, $f$ has a convergent subsequence because it is bounded. Since $f$ is bounded and has a convergent subsequence, it must be Cauchy. That is,

$$f(x_{n_k}) \to L \implies x_{n_k} \to L_0$$

Since $x_{n_k}$ is Cauchy, it must be continuous, which is a contradiction to our assumption.  $\square$

* corrected

答: As far as I understood, you are trying to prove that $f$ is a bounded function on $\mathbb{R}$. Then why did you start with the statement $f$ is not continuous on a bounded set $S$?


问:
Is it possible to find the difference between two sets in set-builder form?
For instance, if there is a set $A$ in set-builder form and a set $B$ in set-builder form, what will be the difference $A-B$?

For example, if $A=\{x  \in  \mathbb R: 0<x<3\}$ and $B=\{x  \in\mathbb  R: 1\le x\le 5\}$, then how would I find $A-B$ or $B-A$? Is it necessary to first convert them to roster form and then solve them? Is their any way, so that the sets can be directly solved in set-builder form? However here it is about real numbers and you cannot list all of the real numbers. It's in the interval, but it is possible for, sets like where $x \in \mathbb N$ or $x \in \mathbb Z$

答: The result $A-B$ can be written as, in set-builder form, $\{x\in\mathbb{R}\mid 0<x<3 \wedge  \neg (1\leq x\leq 5)\}$. Other set operations such as union, intersection, and complement can also be done in this way. However, when you are going to further use this set, it's kind of obliged for you to solve the condition and write down precisely the elements.


问:
Path-connectedness and compactifications
Is the compactification of a path-connected space path-connected? Why or why not? </p>

<p>(I came across this question in my notes while studying for finals and I have no idea.) </p>

答: Let $\Gamma = \{(x, \sin(1/x)) \mid x>0 \}$ and $X=\Gamma \cup \{0\} \times [-1,1]$. Then $\Gamma$ is homeomorphic to $\mathbb{R}$ and $X$ is clearly a compactification of $\Gamma$. However, $X$ is not path connected.</p>


问:
Integrability of generalization of Thomae's function on $[0,1] \times [0,1]$
Let $f:\left[0,1\right]\times\left[0,1\right]\to \mathbb{R}$ be defined by
$$f\left ( x,y \right )=\begin{cases}
 0& ,x\in\mathbb{I} \\
 0& ,x\in\mathbb{Q},\hspace{2mm} y\in\mathbb{I}\\
 \frac{1}{q}& ,x\in \mathbb{Q},\hspace{2mm} y=\frac{p}{q}\hspace{2mm} \text{in lowest terms}
\end{cases}$$
Show that $f$ is integrable and

$$\int_{\left[0,1\right]\times\left[0,1\right]}f=0$$

That was my tried. We know that $L(f,P)=0$ for any partition of $[0,1]\times[0,1]$ (This is easy to proof since the density of $\mathbb{I}$ on $\mathbb{R}$). Therefore, is enough to proof that given $\varepsilon$ there exist a partition $P$ such that $U(f,P)<\varepsilon$. Then for any $\varepsilon$ there exist $n\in \mathbb{N}$ st $\frac{1}{n}<\frac{\varepsilon}{2}$. Let $P$ the following partition

$$P=(\{0,\frac{1}{n},\dots, \frac{n-1}{n},1\},\{0,1\})$$. The volume $v(S)$ of each rectangle $S$ of the partition $P$ is $\frac{1}{n}$, but i don't how to calculate $M_S(f)=\sup\{f(x,y):(x,y)\in S\}$.

This exercise was taken from Calculus on Manifolds - Michael Spivak, Chapter 3. Integration, exercise 3-7.

答:
For any $\epsilon > 0$, choose $N \in \mathbb{N}$ such that $1/N < \epsilon/2$.

The set $A_N = \left\{1,\frac{1}{2},\frac{1}{3},\frac{2}{3}, \ldots, \frac{1}{N}, \ldots, \frac{N-1}{N} \right\}$ contains a finite number of elements $m = \#(A_N)$. If $x\in \mathbb{Q} $ and $y \in A_N$ then $f(x,y)\geqslant 1/N$, and  if $x\in \mathbb{Q} $ and $y \not\in A_N$ then $f(x,y)<  1/N < \epsilon/2$

Take a partition $P$ of $[0,1] \times [0,1]$ with subrectangles $R_{jk} = [x_{j-1},x_j]\times [y_{k-1},y_k]$ $(j,k = 1,\ldots ,n)$ and   $\max_{1 \leqslant k\leqslant n}(y_k - y_{k-1}) < \frac{\epsilon}{4m}$.  Denote $M_{jk} = \sup_{(x,y) \in R_{jk}} f(x,y)$.

We have

$$U(f,P) = \sum_{R_{jk}}M_{jk}\,vol(R_{jk}) = \sum_{[y_{k-1},y_k]\cap A_N = \emptyset}M_{jk}\,vol(R_{jk})+ \sum_{[y_{k-1},y_k]\cap A_N \neq \emptyset}M_{jk}\,vol(R_{jk})$$

For the first sum on the RHS,  we have $M_{jk} < 1/N < \epsilon/2$ and

$$\sum_{[y_{k-1},y_k]\cap A_N = \emptyset}M_{jk}\,vol(R_{jk}) \leqslant \frac{\epsilon}{2}\sum_{[y_{k-1},y_k]\cap A_N = \emptyset}\,vol(R_{jk}) < \frac{\epsilon}{2} \cdot 1 = \frac{\epsilon}{2}$$

For the second sum, there are at most $n \cdot 2m$ subrectangles that contribute because no more than $2m$ intervals $[y_{k-1},y_k]$ can contain the $m$ points in $A_N$. As $M_{jk} \leqslant 1$ on any subrectangle, we have

$$\sum_{[y_{k-1},y_k]\cap A_N \neq \emptyset}M_{jk}\,vol(R_{jk}) \leqslant \sum_{[y_{k-1},y_k]\cap A_N \neq \emptyset}\,vol(R_{jk}) \leqslant 2m \cdot \max_{1\leqslant k \leqslant n}(y_{k} - y_{k-1}) \sum_{j=1}^n (x_j- x_{j-1}) \\ < 2m \cdot  \frac{\epsilon}{4m} \cdot 1 = \frac{\epsilon}{2}$$

Therefore, $U(f,P) < \epsilon$ and $f$ is integrable.


问:
Showing that a set is not homeomorphic to a given space (Cantor set)
Recall the definition of the Cantor set. We start with $Y_0 := [0, 1]$ (with the Euclidean topology). Let $Y_1$ be obtained from $Y_0$ by removing the open interval $(1/3,2/3)$. Let $Y_2$ be obtained from $Y_1$ by removing the open intervals $(1/9,2/9)$ and $(7/9,8/9)$. Recursively, let $Y_n$ be obtained from $Y_{n−1}$ by removing the open middle third of each of the $2^{n−1}$ components of $Y_{n−1}$. The Cantor set is the intersection $Y := \cap_{n=0}^∞ Y_n$.

$(a)$ Use compactness to argue that $Y$ is non-empty.

$(b)$ Show that $Y$ is homeomorphic to the space $X :=$ {$0, 1$}$^{\mathbb N}$ with the product topology.

Hint. Show that $ϕ : X → Y$ defined by $ϕ(x) := \sum_{n=1}^∞(2/3)^nx_n$ is a homeomorphism. You can understand $ϕ$ by thinking of $x$ as an “address” for $ϕ(x)$.

$(c)$ The above representation resembles the binary representation of the points in $[0, 1]$, where a point $z ∈ [0, 1]$ is represented by its binary expansion $0.z1z2z3 · · · $, so that $z_k ∈$ {$0, 1$} and $z = \sum_{n=1}^∞(1/2)^nz_n$. Explain why a similar argument as in the previous part would not show that $[0, 1]$ is homeomorphic to $X$.

I already proved parts $(a)$ and $(b)$. But how to explain part $(c)$ as similar arguments to part $(b)$?

In part $(b)$, I showed that $ϕ$ is one-to-one, i.e. if $x=y \in X$, then $ϕ(x)=ϕ(y)$. I also showed that $ϕ$ is onto, i.e., if $y \in Y$, and $y=\sum_{n=1}^∞(2/3)^ny_n$, then $ϕ(\sum_{n=1}^∞(2/3)^ny_n = y$. Which means that we have a unique pre-image for every element of $Y$ in $X$. Moreover, $ϕ$ is continuous, cause from part $(a)$ we see that each point of Cantor set is a limit point. And also, we know thag Cantor set is uncountable. Finally, $ϕ^{-1}$ is continuous since Cantor set is closed and bounded, and so inverse of $ϕ$ is closed and bounded and hence contains all of its limit points.

So how to show by a similar argument that if $z = \sum_{n=1}^∞(1/2)^nz_n$, then $[0, 1]$ is not homeomorphic to $X$. (part $(c)$).

答: The map $(z_1, z_2, \dots) \mapsto \sum_{n=1}^{\infty} 2^{-n} z_n$  is continuous but it's not injective, so it can't be a homeomorphism.  For example the sequences $(1,0,0,0\dots)$ and $(0,1,1,1,\dots)$ both map to $1/2$.  Of course this only proves that this particular map isn't a homeomorphism, not that $[0,1]$ and the Cantor set aren't homeomorphic in general.  But to prove they can't be homeomorphic at all, one only needs to observe (for example) that $[0,1]$ is connected and the Cantor set is not.


问:
$A$ be $n \times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of $A$
I am trying assignment questions in Linear Algebra and this question could not be solved by me. So, I thought of posting it here.

Let A be an $n\times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A. Then which of the following are true:

All eigenvalues of A are equal

All eigenvalues of A are distinct

$A=\lambda I $ for some $\lambda \in \mathbb{C}$ , where I is the n times n identity matrix

The minimal and characterstic polynomial for A are equal.

I tried by taking eigenvectors to be distinct and the hypothesis but couldn't solve any options. Can you please give hints how to start the problem!

Thank you!

答:
I assume our OP Avenger means "every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of A" instead of
"every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A"; otherwise, the problem makes no sense.

If every non-zero vector $w$ is an eigenvector, each with eigenvalue $\mu_w$,

$Aw = \mu_w w \tag 1$

by choosing two linearly independent vectors $u$ and $v$ we have

$\mu_u u + \mu_v v  = Au + Av = A(u + v)$
$= \mu_{u + v}(u + v) =  \mu_{u + v}u + \mu_{u + v}v; \tag 2$

linear independence of $u$ and $v$ then forces

$\mu_u = \mu_{u + v} = \mu_v; \tag 3$

thus all eigenvalues are equal to some

$\mu \in \Bbb C, \tag 4$

that is, for any vector $w$,

$Aw = \mu w, \tag 5$

or

$A = \mu I; \tag 6$

$A$ is a scalar multiple of $I$.

The characteristic polynomial of $A$ is

$\det(A - xI) = \det (\mu I - xI)$
$= \det((\mu - x)I) = (\mu - x)^n; \tag 7$

however, $A = \mu I$ also satisfies its minimal polynomial

$m(x) = \mu - x, \tag 8$

and

$m(x) \ne (\mu - x)^n \tag 9$

unless $n = 1$.

So we have items (1) and (3) in the OP's question are true, but (2) and (4) are false.


问:
What is the correct notation to indicate that a function is taking a particular value for one of its parameters?
Consider a function which has both variables and parameters, such as
$$
y(x) = 3x^2 + ax + b,
$$
where $a$ and $b$ are the paramters (constants) for any one realisation, but $x$ is a variable. What is the correct notation to use to indicate the function with $a$, for example, taking a certain value?

Should it be something like the "evaluated at" symbol, like this:
$$
y(x)\rvert_{a=2} = 3x^2+2x+b
$$

or would it be written as an argument like this:
$$
y(x,a=2) = 3x^2+2x+b.
$$

Or is there another accepted way that I am not aware of?

Thank you.

答: If you plan on specifying the parameter $a$, it might be best to choose beforehand a notation that reflects that intention. For example, either of the notational choices $y(x,a)=3x^2+ax+b$ or $y_a(x)=3x^2+ax+b$ makes it easy to show what you want to show. (And similar notational choices can be used for $b$, with or without $a$.)


问:
The right "weigh" to do integrals
Back in the day, before approximation methods like splines became vogue in my line of work, one way of computing the area under an empirically drawn curve was to painstakingly sketch it on a piece of graphing paper (usually with the assistance of a French curve) along with the axes, painstakingly cut along the curve, weigh the cut pieces, cut out a square equivalent to one square unit of the graphing paper, weigh that one as well, and reckon the area from the information now available.

One time, when we were faced with determining the area of a curve that crossed the horizontal axis thrice, I did the careful cutting of the paper, and made sure to separate the pieces above the horizontal axis from the pieces below the horizontal axis. Then, my boss suddenly scooped up all the pieces and weighed them all.

I argued that the grouped pieces should have been weighed separately, and then subtract the weights of the "below" group from the "above" group, while my boss argued that we were calculating the total area, and thus, not separating the pieces was justifiable.

Which one of us was correct?

答: I guess we speak of the area "under a curve" when the piece of curve is above the $x$ axis. Sign does matter in the definition of the proper integral, so I would say you were right, and your boss was wrong. Doing integrals by "weighing" things should allow for negative weights.</p>

ContentLicense: CC BY-SA 2.5

Title: The right "weigh" to do integrals


问:
Application of Galois theory
i have a question regarding roots of equation,
find all $a$,such that the cubic polynomial $x^3-bx+a=0$ has three integer roots,
how can you solve these by using galois theory,what does the reducible polynomials,splitting fields,field extensions have to do with these,
explain each of them,in detail,because it serves as a introduction to galois theory,
ok for eg take $b$ to be 3 and list all $a$ such that equation has three integer roots</p>

答:
Your problem is a classical diophantine problem that Fermat or Newton could have studied, long before Galois.

In the case you mention, you don't really need Galois Theory.

You can use symmetric polynomials of the roots to derive constraints on the coefficients.
EDIT: (see Quiaochu's answer for a simple and classical way to start doing this)

there is a complete algebraic solution of the 3rd degree equation giving the roots in term of the coefficients.

You might enjoy this representation of solutions for small $b$s and $a$s.

with red squares indicating combinations of $b$ and $a$ where the polynomial has 3 integer solutions and pink squares where there is only one (in this case the polynomial can be factorized with an integer linear factor, the two other solutions being complex conjugates).  You can guess with this image (but need to prove) that there is no solution for negative b, that the solutions are symmetric around the vertical axis and that solutions arise as intersections of straight lines representing one parameter families of equations with at least one integer root.

Galois Theory is not about whether roots of polynomial equations are integers. It has been developed and applied initially to study whether roots can be finitely expressed by addition, substraction, multiplication, division, n-th power and n-th root extraction starting from given quantities, including the coefficients of the polynomials, that is whether these solutions are algebraic in terms of the coefficients and other quantities.

The theory provides among other things a deep and far-reaching understanding of what happens for polynomial equations with degree greater or equal to 5: the individual solutions cannot always be expressed as an algebraic expressions of the coefficients, while some expressions involving simultaneously all the roots can.

Of course, if you can prove with an application of Galois Theory that the roots of a polynomial cannot be expressed algebraically in terms of the coefficients and if the coefficients are integer or rational, then you can deduce that the roots cannot be integers, but this is not what you are asking for. Galois Theory will not give you all the necessary and sufficient conditions on the coefficients for the roots to be integers. For polynomials of degree inferior to 5, Galois Theory is not needed and not sufficient for the result your looking for.

Title: Application of Galois theory


问:
Solve trig-equation
Can someone solve the equation $3\sin(x) + 2\sin^2(x) = 1$? I know that one answer is $x = 1/2\pi + k\cdot2\pi$ with $k \in \mathbb{Z}$ but can't seem to find the other one. </p>

答:
To answer the new question posted in the comments:
$$3\sin{x}-2\sin^{2}{x}=1$$
Let:
$$\sin{(x)}=\beta$$
Thus:
$$-2\beta^{2}+3\beta=1$$
Dividing everything by -2
$$\beta^{2}-\frac{3}{2}\beta=-\frac{1}{2}$$
Completing the square:
$$(\beta-\frac{3}{4})^{2}=\frac{1}{16}$$
Taking the square root:
$$\beta-\frac{3}{4}=\pm\frac{1}{4}$$
$$\beta=\pm\frac{1}{4}+\frac{3}{4}$$
$$\beta=1$$
$$\beta=\frac{1}{2}$$
Solving for sin(x):
$$1=\sin{x}$$
$$\frac{1}{2}=\sin{x}$$
Recognizably the sin(90)=1, therefore, x=90. Recognizing sin(30)=.5, thus, x=30. And the reference angle of 150 is 30, and as such$$\sin{150}=\sin{30}$$ So your solutions are:

x=90

x=30

x=Any number with a reference angle of 30, or 90, which includes 150.

In radians:
$$\mathit{x}=90=\frac{\pi}{2}$$
$$\mathit{x}=30=\frac{\pi}{6}$$

Title: Solve trig-equation