问:
What are the cohomology groups of Moore space?
I am using the following definition of Moore space:

Definition:

Let $G$ be an abelian group and $n$ an integer $\geq 2.$ A Moore space of type $(G,n)$ is $1-$connected, $CW-$complex $X$ such that:

$$ \widetilde{H}_i(X) = \begin{cases}
      0 & if \, i \neq n, \\
      G &  if   i = n.
   \end{cases} $$

And I want to find:

$1- H_*(M(\mathbb Z_{p^a}, 2); \mathbb Z).$

$2- H^*(M(\mathbb Z_{p^a}, 2); \mathbb Z).$

$3- H_*(M(\mathbb Z_{p^a}, 2) \mathbb Z_{p^t}).$

$4- H^*(M(\mathbb Z_{p^a}, 2); \mathbb Z_{p^t}).$

My answers are:

1-Using the given definition of Moore space, I can tell that  $H_0 = \mathbb{Z} \oplus \widetilde{H_0}$ and $H_i = \widetilde{H_i} \text{ for } i \geq 1.$Is that correct?

2- I do not know what are the cohomology groups of Moore space, could anyone tell me a source where I can find this information calculated? I know that it can be calculated using chain complex, but is not there any easier way of calculating it? I want to know what the final groups should be.

3- I know that, knowing $(1,)$ we can use the UCT for homology to calculate $H_*(M(\mathbb Z_{p^a}, 2); \mathbb Z_{p^t}).$ correct?

4- I know that, knowing $(2,)$ we can use the UCT for cohomology to calculate $H^*(M(\mathbb Z_{p^a}, 2); \mathbb Z_{p^t}).$ correct?

Could anyone help me in answering those questions, specifically question 2?

答:
Yes, it is correct.

2-4. Yes, the UCT makes it a purely algebraic exercise. But I think you misunderstand the UCT because in fact all 2-4 follow from (1) by it, in particular both (2) and (4). For (3) there will be $H_* \otimes -$ and $Tor(H_*, -)$ and for (2) and (4) there will be $Hom(H_*, -)$ and $Ext(H_*, -)$. Note that both times only homology groups are used.


问:
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.

That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4} \geq 0$, isn't?

I have also found the derivative $f'(x) = \frac{2x^3 + 2x^2 - 1}{x^2 + x}$ and that there is a solution of $f(x)$ in the interval of $\left[\frac12, \frac1{\sqrt 2}\right]$.

But from there I am stuck, I am not able to see how to link all the concepts.

Could someone give me a hint?

Thank you and Happy New Year 2021 :)

答:
Using a result from Some Logarithmic Inequalities, E.R.Love :
$$
\forall x>0 \quad ; \quad \ln (1+x) > \frac{2x}{2+x}
$$

Making the transformation $x \to \frac 1x$ gives us :
$$
\forall x > 0 \quad ; \quad \ln\left(1+\frac 1x\right) > \frac{\frac 2x}{2+\frac 1x} = \frac 2{2x+1}
$$

Thus if $g(x) = x^2 + \frac {2}{2x+1}$ then $g(x) \leq f(x)$ for $x>0$.

(Note : there may be easier bounds, but for this one the minima was the easiest to find)

The minimum of $g(x)$ is $\frac 54$ at $x = \frac 12$. We easily have $\frac {1+ 2 \ln 2}{4} \leq 1$ by using  $\ln 2 \leq 1$. Thus, we are done.

ContentLicense: CC BY-SA 4.0


问:
Help evaluating $\int_0^\infty \frac{1}{x^{1/2}(x+1)}dx$
I began solving this with U sub and partial fractions...first for $x^{1/2}$ and then for $x+1$  but neither of those methods got me the answer of $\pi$.

I know the indefinite integral should be $2\tan ^{-1} x^{1/2} + C$.

答:
The integral may also be evaluated using residue theory, using a keyhole contour $C$.  Consider

$$\oint_C dz \frac{z^{-1/2}}{z+1} = \int_{\epsilon}^R dx \frac{x^{-1/2}}{x+1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{R^{-1/2} e^{-i \theta/2}}{R e^{i \theta}+1}+ e^{-i \pi} \int_R^{\epsilon} dx dx \frac{x^{-1/2}}{x+1} \\ + i \epsilon\int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\epsilon^{-1/2} e^{-i \phi/2}}{\epsilon e^{i \phi}+1}$$

The second integral vanishes as $R \to \infty$, the fourth integral vanishes as $\epsilon \to 0$.  Thus,

$$\oint_C dz \frac{z^{-1/2}}{z+1} = 2 \int_0^{\infty} dx \frac{x^{-1/2}}{x+1} $$

This is equal to $i 2 \pi$ times the residue at $z=-1$, which may be computed by recognizing that, in the branch we have chosen, $-1=e^{i \pi}$.  Thus

$$2 \int_0^{\infty} dx \frac{x^{-1/2}}{x+1} = i 2 \pi e^{-i \pi/2} = 2 \pi $$

Therefore

$$\int_0^{\infty} dx \frac{x^{-1/2}}{x+1} = \pi $$

ContentLicense: CC BY-SA 3.0


问:
Investigate local maxima and minima
Question: $x^2 + 4xy + 4y^2 + x^3 +2x^2y+y^4 $ critical point $(0,0)$,  $F_{xx}*F_{yy} - (F_{xy})^2$ = $0$  now test failed to conclude any thing. But solution says it has minimum at the origin. Please guide me how to proceed. And lastly, I am unable to find its other critical point. Please help.

答:
Note that

$$
p(x,y) = x^2 + 4 x y + 4 y^2 + x^3 + 2 x^2 y + x^4=\left(\begin{array}{c}y\\ x\\ x^2\end{array}\right)^{\dagger}\left(\begin{array}{ccc}4& 0 & 0\\ 4 & 1 & 0\\ 2 & 1 & 1\end{array}\right)\left(\begin{array}{c}y\\ x\\ x^2\end{array}\right)
$$

Note also that matrix $M$

$$
M = \left(\begin{array}{ccc}4& 0 & 0\\ 4 & 1 & 0\\ 2 & 1 & 1\end{array}\right)
$$

has eigenvalues $\{4,1,1\}$ then it is positive definite hence $p(x,y)$ has a minimum at the origin.

Title: Investigate local maxima and minima


问:
Are all analytic continuations consistent with $1-1+1-1+\dots = \frac{1}{2}$?
Background:

This is a follow-up to this question, which is the same question except for a different series (and I suspect the answer may be different for this series—more on that later). In short, I am curious about whether the idea of associating divergent sums with analytic continuations could be used to rigorously define a summation method for divergent series.

Question:

Let $A$ and $B$ be open subsets of $\mathbb C$ with $A\subset B$ and $\{a_n\}_{n\in\mathbb N_0}$ be a family of analytic functions $a_n:B\to\mathbb C,z\mapsto a_n(z)$ such that:

$a_n(z_0)=(-1)^n$ for all $n\in\mathbb N_0$ and for some $z_0\in B$,

$f(z):=\sum_{n=0}^\infty a_n(z)$ is well defined and analytic on $A$.

Let $g$ be the analytic continuation of $f$ to the rest of $\mathbb C$. Must $g(z_0)=\frac{1}{2}$?

The linked question above asks the same thing for $1+2+3+\dots=-\frac{1}{12}$, and a comment links to another question that provides a counter-example. So the answer is “no” in that case. But I suspect it may be “yes” in this case because $1-1+1-\dots=\frac{1}{2}$ is better behaved in that it is Cesaro mean convergent.

Additionally there are already two examples of such analytic continuations that agree with this sum in the sense described above: (1) The Dirichlet eta function, with $a_n(z)=\frac{(-1)^n}{n^z},z_0=0$, and (2) The geometric series function, with $a_n(z)=z^n,z_0=-1$.

答:
Analytic continuation is a very fragile method that very heavily depends on the function you choose to continue.

Unlike those other major methods that usually give the same result, you often can arrive at what results you want with analytic continuation.

Possibly, it can be used to more rigorously define summation of divergent series but one should add additional conditions on the behavior of the functions they use to analytically continue. So far, I am unaware of any such work.

On the other hand, it seems, analytic continuation of Zeta function and other Dirichlet series works quite safe.

Addendum regarding notation

When you write (for instance) $1+1+1+\dots$, it is unclear what series do you mean, it can de different ones. For instance, it can be

$\sum_{k=0}^\infty 1 =\frac12$
or
$\sum_{k=1}^\infty 1 =-\frac12$

You can test the both in Mathematica computer algebra system with Dirichlet regularization:

```<code>Sum[1,{x,0,Infinity},Regularization->Dirichlet]
Sum[1,{x,1,Infinity},Regularization->Dirichlet]
</code>```

In your case you use alternating signs, so the series is stable-method-summable, (for instance, with Cesaro or Abel), so it sums to $\frac12$, but again, please use exact notation.

Body:
Analytic continuation is a very fragile method that very heavily depends on the function you choose to continue.

Unlike those other major methods that usually give the same result, you often can arrive at what results you want with analytic continuation.

Possibly, it can be used to more rigorously define summation of divergent series but one should add additional conditions on the behavior of the functions they use to analytically continue. So far, I am unaware of any such work.

On the other hand, it seems, analytic continuation of Zeta function and other Dirichlet series works quite safe.

Addendum regarding notation

When you write (for instance) $1+1+1+\dots$, it is unclear what series do you mean, it can de different ones. For instance, it can be

$\sum_{k=0}^\infty 1 =\frac12$
or
$\sum_{k=1}^\infty 1 =-\frac12$

You can test the both in Mathematica computer algebra system with Dirichlet regularization:

Sum[1,{x,0,Infinity},Regularization->Dirichlet]
Sum[1,{x,1,Infinity},Regularization->Dirichlet]

In your case you use alternating signs, so the series is stable-method-summable, (for instance, with Cesaro or Abel), so it sums to $\frac12$, but again, please use exact notation.


问:
Alternate Proof of Unique Lifting Property of Covering Spaces
I proved one of Hatcher's propositions on my own and my proof is quite a bit different than his.

The Unique Lifting Property says: Given a covering space $p:\tilde{X} \rightarrow X$ and a map $f: Y \rightarrow X$, if $\tilde{f}_{1}, \tilde{f}_{2}: Y \rightarrow \tilde{X}$ are lifts of $f$ which agree at one point of $Y$ and $Y$ is path connected, then $\tilde{f}_{1}$ and $\tilde{f}_{2}$ agree at all points of $Y$.

My proof is as follows:

Let $A=\{y \in Y : \tilde{f}_{1}(y)=\tilde{f}_{2}(y)\}$ and $B=\{y \in Y : \tilde{f}_{1}(y)\neq \tilde{f}_{2}(y)\}$; then $Y=A \cup B$. Note that $A \neq \emptyset$ by hypothesis. Suppose $y \in A$ and consider $f(y) \in X$. Let $U$ be a neighborhood of $f(y)$ which is evenly covered. Let $\tilde{V}$ be the sheet above $U$ containing $\tilde{f}_{1}(y)=\tilde{f}_{2}(y)$. Let $G=\tilde{f}^{-1}_{1}(\tilde{V}) \cap \tilde{f}^{-1}_{2}(\tilde{V})$; suppose $t \in G$ then $p(\tilde{f}_{1}(t))=p(\tilde{f}_{2}(t))=f(t)$. Since $\tilde{f}_{1}(t),\tilde{f}_{2}(t)\in V$ and $p$ is injective on $V$, we have $\tilde{f}_{1}(t)=\tilde{f}_{2}(t)$. Thus $G$ is an open subset of $A$ containing $y$.
Suppose now that $y \in B$ and let $U$ be a neighborhood of $f(y)$ which is evenly covered. Let $\tilde{V}_{1}$ and $\tilde{V}_{2}$ be the sheets above $U$ containing $\tilde{f}_{1}(y)$ and $\tilde{f}_{2}(y)$, respectively. Note that $\tilde{V}_{1}\neq \tilde{V}_{2}$ as this would imply that $\tilde{f}_{1}(y)=\tilde{f}_{2}(y)$. Let $G=\tilde{f}^{-1}_{1}(\tilde{V}_{1}) \cap \tilde{f}^{-1}_{2}(\tilde{V}_{2})$. If $t \in G$ then $\tilde{f}_{1}(t) \in \tilde{V}_{1}$ and $\tilde{f}_{2}(t) \in \tilde{V}_{2}$ so they are not equal. Thus $G$ is an open subset of $B$ containing $y$. It follows that $B = \emptyset$ or this would be a separation of $Y$, contradicting that $Y$ is connected.

Does this proof seem correct?

答:
[I'm not allowed to comment due to missing reputation, but my post might be considered a comment.]

Your proof seems similar to the one given by Edwin H. Spanier in his book "Algebraic Topology". I'm including a screenshot of his proof for reference.


问:
$f$ continuous in a dense set $G_\delta$
If $f:X \to Y$  is a function between a metric space baire $X$ and a metric separable space $Y$ and if for each open $U$ of $Y$, $f^{-1}(U)$ is countable union of closed sets of $X$, then there exists a dense $G_{\delta}$ in $X$ such that $f$ is continuous.

My objective: Suppose that $Z$ is the set $G_{\delta}$, then there exists a family of dense openings $\{W_i : i \in \mathbb{N}\}$ of $X$ such that $Z= \cap_{i \in \mathbb{N}} W_i$, also, as for each open $U$ of $Y$ we have to $f^{-1}(U)= \cup_{i \in \mathbb{N}}B_i$ where $B_i$ es closed in $X$, i.e, $f^{-1}(U)$ is $F_\sigma$, and since $Y$ is separable, then $Y$ has a dense subset. My idea is to use the separability of $Y$ to build my $G_ {\delta}$, but I don't know how to do it. How could I use it? or any other ideas for this test? please thank you very much.

答:
Since $Y$ is a separable space it has a countable base $\mathcal B$. For each $B\in \mathcal B$ let $f^{-1}(B)$ be a union of a countable family $\mathcal C_B$ of closed subsets of $X$. Put $Z=X\setminus \bigcup_{B\in\mathcal B}\bigcup_{C\in\mathcal C_B} C\setminus\operatorname{int} C$. Then $Z$ is a $G_\delta$-subset of $X$. Since $X\setminus Z$ is a union of countable many nowhere dense subsets of a Baire space $X$,  $Z$ is dense in $X$, see, for instance, Theorem 1.13 in [HC].

We claim that a restriction $f|Z$ is continuous. Indeed, let $x\in Z$ be any point any $U$ be any neighborhood of $f(X)$. There exist a set $B’\in\mathcal B$ such that $f(x)\in B’\subset U$. Then
$$x\in Z\cap f^{-1}(B’)=\left(X\setminus \bigcup_{B\in\mathcal B}\bigcup_{C\in\mathcal C_B} C\setminus\operatorname{int} C\right)\cap \bigcup_{C\in\mathcal C_{B’}} C= Z\cap  \bigcup_{C\in\mathcal C_{B’}}\operatorname{int} C$$
is an open subset of $Z$ and $f\left(Z\cap f^{-1}(B’)\right)\subset B’\subset U$.

References

[HC] R. C. Haworth, R. C. McCoy, Baire spaces, Warszawa, Panstwowe Wydawnictwo Naukowe, 1977.


问:
Countable Chain Condition for separable spaces?
I'm trying to find a proof behind a small proposition. Recall that a topological space satisfies the countable chain condition if each disjoint collection of open sets is countable.

Why is it the case that every separable spaces satisfies the CCC, but the converse is not true?

Thanks.

答:
I've been trying to come up with a simpler example to answer your question, although the answes of course complete and great, I still fancied a simpler example. So here it is:

Every separable space is CCC. The converse is not true.

For example:

Let $X$ be a space with $|X|= \aleph_1$, let $\tau_X= \lbrace U: X\setminus U \text{ is countable } \rbrace$. This space is CCC, but not separable.

Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$.

$X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open.


问:
Solution or algorithm for linear diophantine equations with unit coefficients
I am interested in nonnegative integer solutions $\{x_{mq}\}$ to a particular type of linear system. The system can be broken into a "horizontal" and "vertical" subsystem; that is, the horizontal subsystem is as follows

$$x_{11} + x_{12} + \ldots + x_{1Q} = h_1 $$
$$x_{21} + x_{22} + \ldots + x_{2Q} = h_2 $$
$$\vdots$$
$$x_{M1} + x_{M2} + \ldots + x_{MQ} = h_M $$

whereas the vertical system is
$$x_{11} + x_{21} + \ldots + x_{M1} = v_1 $$
$$x_{12} + x_{22} + \ldots + x_{M2} = v_2 $$
$$\vdots$$
$$x_{1Q} + x_{2Q} + \ldots + x_{MQ} = v_Q $$

for constants $h_1, \ldots, h_M$ and $v_1, \ldots , v_Q$.

So the vertical system are simply summing restrictions formed from the "columns" of the horizontal system. In matrix form, the systems can be written as
$$
X 1_Q = h
$$
$$
X' 1_M = v
$$
where $1_K$ denotes the vector of ones in $\mathbb{R}^K$.

I doubt that this system has a closed form for the family of solutions (but would be very glad to be proved wrong!). Does a straightforward algorithm exist instead? The plan that I'm currently implementing is as follows:

Enumerate all solutions $x_{m1}, x_{m2}, \ldots , x_{mQ}$ to the horizontal system for each $m$.

Iterate through all combinations of the above solutions, eliminating combinations that violate the vertical solution.

As one might guess, this is not a terribly efficient algorithm. I can save much time by placing restrictions on the horizontal solutions I find in (1) by first eliminating any that have $x_{mq} > v_q$. Even so, with just $M = Q = 6$, I'd be iterating over a list with something on the order of tens of billions of elements.

答:
I'm exploring this, so for the beginning I assume, that the matrix $XX$ is a square-matrix, with #rows=#cols=n.

First we note, that $\sum_{k=1}^n h_k = \sum_{k=1}^n v_k $, so $v_n $ is not free to be set arbitrarily but is determined by the other given coefficients.

Keeping this in mind I begin with the matrix $XX=\{x_{r,c} \}_{r,c=1..n}=0$, and initially filling in $x_{1,1}=h_1$.
The sum of the first row gives now correctly $h_1$.

Now the first column needs a compensation to be able to sum up to $v_1$: in the second row $x_{2,1}$ I set $x_{2,1}=v_1-x_{1,1}$ .
From this follows the setting of $x_{2,2}=h_2-x_{2,1}$, then $x_{3,2}=v_2-x_{2,2}$     and so on with that simple recursive algorithm, down to the bottom right entry.

With an example of $n=4$ I get the following solution:
$$ \small\begin{array} {rrrr|r}
h_1&\cdot&\cdot&\cdot&h_1\\
v_1-h_1&h_1+h_2-v_1&\cdot&\cdot&h_2\\
\cdot&(v_1+v_2)-(h_1+h_2)&(h_1+h_2+h_3)-(v_1+v_2)&\cdot&h_3\\
\cdot&\cdot&(v_1+v_2+v_3)-(h_1+h_2+h_3)&(h_1+h_2+h_3+h_4)-(v_1+v_2+v_3)&h_4\\\hline
v_1&v_2&v_3&v_4&\cdot\\
\end{array}$$
Note that I've appended the rowsums at the right and the column-sums at the bottom.

Note also, that the last entry $x_{4,4}$ requires (or simply: expresses) the equality which I've mentioned at the beginning of this post.


问:
Prove $\prod_{r=1}^{(n-1)/2} 2\cos \frac{\pi r}{n}=1$
I would like to know how to prove that

$$\prod_{r=1}^{(n-1)/2} 2\cos\frac{\pi r}{n}=1~~~~~\text{where $n$ is odd}$$
I have tried induction and using trigonometric identities to attempt to transform the product into something more helpful, but nothing has worked.

It may be more helpful to think of the product as
$$\prod_{r=1}^{n} 2\cos\frac{\pi r}{2n+1}=1$$

Many thansk for your help.

Body:
Using this

$$\cos(2n+1)x=2^{2n}\cos^{2n+1}x+\cdots+(-1)^n\cos x$$

If $\cos(2n+1)x=-1=\cos\pi\implies(2n+1)x=(2r+1)\pi$

$\implies x=\dfrac{(2r+1)\pi}{2n+1}; 0\le r\le2n$

So, the roots of $$2^{2n}c^{2n+1}+\cdots+(-1)^nc+1=0$$ are $\cos\dfrac{(2r+1)\pi}{2n+1}; 0\le r\le2n$

Using Vieta's Formulas
$$\prod_{r=0}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=-\dfrac1{2^{2n}}$$

Exclude $r=n$

$$\prod_{r=0,\ne n}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac1{2^{2n}}$$

Now if $r_1+r_2=2n+1,\cos\dfrac{(2r_1+1)\pi}{2n+1}=\cos\dfrac{(2r_2+1)\pi}{2n+1}$

$$\implies\prod_{r=0}^{n-1}\cos^2\dfrac{(2r+1)\pi}{2n+1}=\dfrac1{2^{2n}}$$

Now $\dfrac{(2r+1)\pi}{2n+1}>\dfrac\pi2\iff4r+2>2n+1\iff r>\dfrac{2n-1}4$

So, if $m=n-1-\left\lceil \dfrac{2n-1}4\right\rceil$

$$\implies(-1)^m\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac1{2^n}$$

If $2r+1>\dfrac{2n+1}2,   0>\cos\dfrac{(2r+1)\pi}{2n+1}=-\cos\dfrac{2(n-r)\pi}{2n+1}$

$$\implies\prod_{m=1}^n\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^n}$$


问:
Question 10.17 (a) Apostol's Mathematical analysis
This question is from apostol  Mathematical analysis Chapter 10.

Assume that $f \in \mathbb{R}$ on $[a,b]$ for every $b>a>0$. Define $g$ by the equation $x g(x)=\int_{1}^{x} f(t) dt$ if $x>0$, assume that $\lim_{x\to \infty} g(x)$ exists and $a$ and $b$ are fixed positive integers. Prove that :

$$\int_a^{b} f(x)/x dx=  g(b) -g(a) +\int_{a}^b g(x)/xdx.$$

I tried using  $x g(x)=\int_{1}^{x} f(t) dt$ and trying to work with both LHS and RHS 1 at a time and trying to get other but failed.

I am not getting any other Idea.

So, How should I attempt this?

答:
Hint. Integrating by parts,
$$\int \frac{f(x)}{x}\,dx=\int (xg(x))'\cdot \frac{1}{x}\,dx=xg(x)\cdot \frac{1}{x}-\int xg(x)\left(\frac{1}{x}\right)'\, dx.$$


问:
Numbers to base $i$
Does it make sense to think of numbers to base $i$, i.e. $\sqrt{-1}$?

For conventional bases e.g. $N = 2, 8, 10, 16$ etc., we talk of using digits from $0$ through the base $N-1$ in the numeric representation. What digits would be permitted for base $i$?

What about other unconventional bases like $\pi$? $e$? Irrational numbers?

Just need a "frame of reference" to think through these things logically. Would appreciate pointers to papers or websites. I am not a Math PhD, just a hobbyist, so please keep explanations simple. Thank you.

答:
Just to clarify, it does make sense to think of numbers base $i$. If we have a string of digits $d_z$ for integers $z$, then this corresponds to $\sum_zd_zi^z$. However, unlike with normal number bases, there's no clear choice of what your set of digits should be (unlike $\{0,1,2,3,4,5,6,7,8,9\}$ for decimal), and if they have to be integers, then we can only produce numbers called Gaussian integers; those of the form $a+bi$ with $a$ and $b$ both integers. Furthermore, unlike with normal bases, this only works if only finitely many of the $d_i$ are zero. Otherwise we get something like the number $1.111111...$ representing
$$1\times i^0+1\times i^1+1\times i^2+1\times i^3+1\times i^4...=1\times1+1\times i+1\times-1+1\times-i=1+i-1-i...$$
which doesn't make converge (once you define convergence for sequences of complex numbers). Furthermore, we have $100000000=10000=1$, since $i^8=i^4=i^0=1$. Of course $0.9999...=1$ in base ten but at least with normal bases we only have non-unique expansions for a subset of rational numbers, each of these only has two expansions, and one of them has terminating expansion and the other doesn't.

Title: Numbers to base $i$


问:
Closed form for $(a \cdot x)\bmod n = r$
Given $a$, $n$ and $r$, is there a closed form for computing $x$?

$(a \cdot x)\bmod n = r$

$x$ in [0..n)

For example,

$(7 \cdot x)\bmod 10 = 1$

$x = 3, 13, 23,...$

$x = 3$ for $x$ in $[0..n)$

I rewrote the equation but not sure how to proceed:

$ x = \frac {qn + r}{a}$

答:
Not a closed formula, but an algorithm to find a solution, if there is one. Assume $a, n$ are not both zero, otherwise this is trivial.

You are looking for $x, y$ such that
$$
a x + n y = r.
$$
A necessary condition for this to happen is that
$$\tag{cond}
\gcd(a, n) \mid r
$$
Conversely, if (cond) holds, use the extended Euclidean algorithm to find $u, v$ such that
$$
a u + b v = \gcd(a, n),
$$
and then multiply by the integer $\dfrac{r}{\gcd(a, n)}$ to get
$$
a \left(u\cdot \dfrac{r}{\gcd(a, n)}\right) + b \left(v\cdot \dfrac{r}{\gcd(a, n)}\right) = r,
$$
so that
$$
x = u \cdot \dfrac{r}{\gcd(a, n)}
$$
is a solution. To find all the solutions, note that two solutions differ by a $t$ such that $a y \equiv 0 \pmod{n}$.


问:
Cutting point of $2$ functions
$y=\sqrt{x}$ and $y=\frac{\left(x-3\right)}{2}$

Cut each other in $x=9$

But if you see $\sqrt{x}=\frac{\left(x-3\right)}{2}$

$x^2-10x+9=0$

Gives us $x=1$ and $x=9$

So how do we know to deny $x=1$?

(without drawing the function of course)

Thank you.

答: Squaring results in some extra roots. In this case, if you put $x=1$ in the second equation you get $y=-1$ whereas if you put $x=1$ in the first equation you get $y=1$, and $1\ne-1$. So you neglect $x=1$.

Title: Cutting point of $2$ functions


问:
For the following monic polynomial,$f$ of even degree how to prove that that $lim_{|x|\to\infty }(\sqrt {f(x)}-g(x))=0$
For any monic polynomial $f \in \mathbb {Q[x]}$ of even degree,how to prove, there exists polynomial $g \in \mathbb {Q[x]}$ such that $lim_{|x|\to\infty }(\sqrt {f(x)}-g(x))=0$</p>

答:
Let $f$ have degree $2k$.

Solution 1: Write $f(x) = x^{2k}(1 + c_1x^{-1} + c_2x^{-2} + \cdots + c_{2k}x^{-2k}$. Then $\sqrt{f(x)} = x^k(1 + d_1x^{-1} + d_2x^{-2} + \cdots)$, where $1 + d_1x^{-1} + d_2x^{-2} + \cdots$ is the Maclaurin series for $\sqrt{1 + c_1x^{-1} + c_2x^{-2} + \cdots + c_{2k}x^{-2k}}$. Show that $g(x) = x^k (1 + d_1x^{-1} + d_2x^{-2} + \cdots + d_kx^{-k})$ works.

Solution 2: It suffices to find a monic polynomial $g(x)$ of degree $k$ such that $f(x)-g(x)^2$ has degree at most $k-1$. (To see this, bound $\sqrt{f(x)}-g(x) = (f(x)-g(x)^2)/(\sqrt{f(x)}+g(x))$ when $x$ is large.) Writing $g(x) = x^k + b_{k-1}x^{k-1} + \cdots + b_1x + b_0$, one can recursively solve for $b_{k-1},\dots,b_1,b_0$ in terms of the previous $b_j$ and the coefficients of $f$.

(The two solutions give the same polynomial $g(x)$; indeed, it's easy to show from first principles that such a $g(x)$ must be unique.)


问:
Different characterization of the subspace topology
Following task in a textbook:

Give a characterization of the subspace topology through:

a) closed sets

b) closure

c) neighborhoods

And I am not sure what actually is asked here. So what a suitable characterization should fulfill.
Of course it should be equivalent.

So let $(X,\tau)$ be a topological space, and $(Y,\tau_Y)$ a subspace with the subspace topology.
For a) I thought about this at first. Take $\{A\subseteq Y: A=Y\cap U^c~\text{with $U\in\tau$}\}$.

As $U^c$ is closed, this (somewhat) gives a characterization with closed sets. But this was not satisfying at all.

It is easy to show that a set $A\subseteq Y$ is closed (in $Y$) iff it exists a closed set (in $X$) $W\subseteq X$ with $A=Y\cap W$.

Which seems like a more suitable answer, and I could discribe the subspace topology in terms of closed sets in the obvious way.

But what about b)?

I know that if $A\subseteq Y$ then $\overline{A}\cap Y$ is the closure of $A$ in $Y$. But how would that lead to a characterization of the subspace topology?

It feels more like I am giving a characterization of the terms closed set, closure, neighborhood with regards to the subspace topology.

So I would say that $U\subseteq Y$ is a neighborhood of $y\in Y$ (with regards to the subspace topology), iff there is a neighborhood $V\subseteq X$ of $y$ with $U=Y\cap V$.

But I am not sure about the solution the task wants you to find.
Can you help me out. Are my thoughts correct?

Thanks in advance.

答:
Possible answers (proofs left to the reader, and not hard)

If $Y \subseteq X$ is a subset of a space $X$, then the closed sets of the subspace topology on $Y$ are exactly of the form $C \cap Y$ where $C$ is closed in $X$.

If $A \subseteq Y$ we can define the subspace closure operation in terms of the closure on $X$ by
$$\operatorname{Cl}_Y(A) = \operatorname{Cl}_X(A) \cap Y$$

And the neighbourhood system for $y \in Y$ in the subspace topology is given by $$\mathcal{N}_Y(y) = \{N \cap Y: N \in \mathcal{N}_X(y)\}$$

All of these characterise the subspace topology. You can define a topology by giving its open sets, its closed sets, its closure operation or its neighbourhood systems (there are other ways too). The above characterisations follow quite easily from knowing what the open sets are and how closure and niieghbourhood systems are defined in terms of open sets.


问:
Difficult Cauchy Problem
Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$

I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.

I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.

I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.

答:
$$\sum{a\over bc}=3$$

$$\sum{a\over b^2c^2}\times(a+b+c)\geq(\sum{a\over bc})^2$$ by Cauchy.

Title: Difficult Cauchy Problem


问:
$f(x-y)=f(x)f(y)-f(a-x)f(a+y)$

If $f(x-y)=f(x)f(y)-f(a-x)f(a+y), f(0)=1$

Show

a) $f(a)=0  $

b) $f(t)=f(-t)$

a) falls out easily setting $y=0, x=a$

I can get really close but I can't quite do this. I would appreciate a hint more than a full solution.

Setting $x=t, y=t$
$$f(0)=f(t)^2-f(a-t)f(a+t)$$
and $x=-t,y=-t$
$$f(0)=f(-t)^2-f(a+t)(a-t)$$

Then subtracting we find $f(t)= \pm f(-t)$. How do I eliminate the negative solution, or is there a better approach altogether?

Title: $f(x-y)=f(x)f(y)-f(a-x)f(a+y)$

Body:
$1) x=t, y=0$

$f(t)=f(t) f(0)-f(a-t) f(a)=f(t) f(0)$

because $ f(a)=0$

$2) x=0, y=t$

$f(-t)=f(t) f(0)-f(a-t) f(a)=f(t) f(0)$

From 1) and 2)
$\Rightarrow \quad f(t)=f(-t)$


问:
Proof regarding Weistrass theorem.

I'm looking for feedback on my work, since we've only just started talking about continuity and differentiability and I'm not 100% confident in my knowledge.. I'd love to know if there are any inaccuracies or general tips for improvement. :)

We know $f$ is continuous on $[0,1]$,

then by Weierstrass's  Theorem there exist $a ∈ [0,1]$ such that $f(a)
 ≤ f(x)$, for every $x ∈ [0,1]$.

Consider the fucntion $g: [0,f(a)]➜[0,f(a)]$, $\ g(x) = x$.

Note that by the algebra of continuous functions $g$ is continuous on
$[0,1]$.

Since $f(x)>0$ for every $x ∈ [0,1]$, and since $g(0) = 0$,$ \
 g(f(a))=f(a)$, then we have

$g(0) < \frac{f\left(a\right)}{2} < g(f(a))$.

Denote $\frac{f\left(a\right)}{2} = ϵ$.

Then we proved the existance of an ϵ > 0 such that f(x) > ϵ for every x
∈ [0,1].

I was originally planning to use the intermediate value theorem after finding the element between the two values of g, but then I noticed that I don't actually need to and I've already found such an element..
Is there anything I can improve?

答: I see no reason for introducing that function $g$. You defined $\varepsilon$ as $\frac{f(a)}2$. Then you can simply say that, for each $x\in[0,1]$,\begin{align}f(x)&\geqslant f(a)\\&>\frac{f(a)}2\text{ (since $f(a)>0$)}\\&=\varepsilon.\end{align}

Title: Proof regarding Weistrass theorem.


问:
$A$ and $A'$ have the same eigenvectors: Flaw in the proof?
Consider the  equation

$$Ax=\lambda x$$
where $x$ is non zero. Then, we have:
$$x'Ax=\lambda x'x$$

$$(x'Ax)'=x'A'x=(\lambda x'x)'=\lambda x'x$$
$$x'(A'x-\lambda x)=0$$
$$A'x=\lambda x$$ (since $x'$ is not zero).

Hence, this proves that A and A' have the same eigenvalues and the same eigenvectors as well.

However, my professor claims that in general, A and A' do not have the same eigenvectors.

What exactly then, is the flaw in my proof?

答: You cannot deduce from the fact that $x'$ is not $0$ together with $x'(A'x-\lambda x)=0$ that $A'x-\lambda x=0$. For instance, if $A$ is a $2\times2$ matrix, if $x'=(1,0)$ and if $A'x-\lambda x=(0,1)$, then $x'(A'x-\lambda x)=0$, but $A'x-\lambda x\ne0$.


问:
How to deal with the essential singularity when evaluating $\oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{z^2 + 9}dz$
I am trying to use the Cauchy integral formula and the residue theorem to show that

$$
\oint_{|z-i| = 3} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{z^2 + 9} dz= 2\pi i\cosh(1) - \pi i \cosh\left( \dfrac{9+6i}{13}\right)
$$

In preparation to use the Cauchy integral formula, my first attempt was to rewrite the integral:

$$
\oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{(z-3i)(z+3i)}dz = \oint_{C} \dfrac{\dfrac{z\cosh{\left( \frac{z}{z-2}\right)}}{z+3i}} {z-3i}dz
$$

However, the function $g(z) = \frac{z\cosh{\left( \frac{z}{z-2}\right)}}{z+3i}$ fails to be analytic at $z = 2$ which is an interior point of $C = \{z \in \mathbb C : |z-i| = 3  \}$. Therefore, Cauchy's integral theorem does not directly apply. Next, I decomposed one of the multiples into partial fractions:

$$
\dfrac{z}{\left(z - 3 i \right) \left(z + 3 i\right)}=\dfrac{1}{6 i + 2 z} - \dfrac{1}{6 i - 2 z} = \dfrac12\left[ \dfrac{1}{z + 3 i } - \dfrac{1}{3 i - z} \right]
$$

Which gives

$$
\oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{(z-3i)(z+3i)}dz = \dfrac12\left[ \oint_C \dfrac{\cosh{\left( \frac{z}{z-2}\right)}}{3 i + z} - \oint_C \dfrac{\cosh{\left( \frac{z}{z-2}\right)}}{3 i - z} \right]
$$

This, unfortunately, does not help with the singularity at $z=2$. In addition, the point $z=3i$ is not even in the region of integration. Thus, I decided to try and solve this using the residue theorem. There are three singular points for the initial function: $\{3i, -3i, 2\}$. Since $-3i \notin C$, we only need the residues at the remaining two points.

Residue at $z=3i$:

$$
\begin{align}
\lim_{z \to 3i} (z-3i)f(z) &= \lim_{z \to 3i} \dfrac{z^2 - 3zi}{z^2 + 9} \cosh\left(\frac{z}{z-2}\right)\\ &= \cosh\left(\frac{3i}{3i-2}\right)\lim_{z \to 3i} \dfrac{z^2 - 3zi}{z^2 + 9}\\ &=\cosh\left(\frac{3i}{3i-2}\right) \cdot \dfrac12
\end{align}
$$

I evaluated the last limit using L'Hopital's rule.

Residue at $z=2$:

The problem is that the limit
$$\lim_{z \to 2} (z-2)f(z)$$
does not exist since the left-hand limit is $-\infty$ whereas the right-hand limit is $+\infty$. Therefore, I cannot use the residue theorem. I am unsure what to try next. How do I deal with the singularity at $z=2$ and what other approaches can I try?

答:
Some ideas for you to evaluate the residue at $\;z=2\;$ of $\;\cosh\frac z{z-2}\;$ (Check carefully the calculations, though the algorithm is the important thing here: sometimes we need, and some even would say must, calculate Laurent series) :

Since $\;\cosh z=\cos(iz)\;$, we get that

$$\cosh z=\sum_{n=0}^\infty \frac{ z^{2n}}{(2n)!}\implies \cosh\frac z{z-2}=\sum_{n=0}^\infty \frac{ \left(\frac z{z-2}\right)^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{  z^{2n}}{(2n)!}\cdot\frac1{(z-2)^n}$$

But

$$z=2+(z-2)=2\left(1+\frac{z-2}2\right)\implies\cosh z=\sum_{n=0}^\infty \frac{2^n\left(1+\frac{z-2}2\right)^n}{(2n)!}\cdot\frac1{(z-2)^n}$$

Let us try to calculate the coefficient $\;c_{-1}\;$  of $\;(z-2)^{-1}\;$ , which is the residue of $\;\cosh\frac z{z-2}\;$ at $\;z=2\;$ . First we write

$$\left(1+\frac{z-2}2\right)^n\cdot\frac1{(z-2)^n}=\left(\frac1{z-2}+\frac12\right)^n=\sum_{k=0}^n\binom nk(z-2)^{-k}\cdot 2^{-n+k}$$

Thus, for each $\;n=1,2,....\;$, the coefficient of $\;(z-2)^{-1}\;$ is given when $\;k=1\;$ (not for $\;n=0\;$...!), and thus it is given by

$$\sum_{n=1}^\infty\binom n1\frac1{2^{n-1}}=\sum_{n=1}^\infty\frac{n}{2^{n-1}}=4\;$$

For the last series' value: differentiate $\;\cfrac1{1-z}\;$ and substitute the correct value of $\;z$


问:
Writing the inclusion relation on the power set $P(\{a,b,c\})$ in matrix notation.
Here is the question. As far as I know, I understand what a power set is, but the last half of the question confused me, and I have no idea what it means.

Write the inclusion relation on the power set $P(\{a,b,c\})$, namely $$(X,Y)\in R \quad\text{iff}\quad X \subseteq Y$$
in the matrix notation.

答:
Your attempt is off, I am not sure what you are doing.

For the answer, the graph is created by the vertices being the subsets of $\{a,b,c\}$ and two subsets being directly connected if the former is a subset of the latter. Therefore, we take our vertex set in the order : $\{\emptyset , \{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}, \{a,b,c\}\}$. So the row for $\{a\}$ will look like $[0,1,0,0,1,1,0,1]$ because that is the subsets which contain $a$ according to our order. Therefore , creating the matrix gives :
$$
\color{red}{
\begin{pmatrix}
1&1&1&1&1&1&1&1 \\
0&1&0&0&1&1&0&1 \\
0&0&1&0&1&0&1&1 \\
0&0&0&1&0&1&1&1\\
0&0&0&0&1&0&0&1\\
0&0&0&0&0&1&0&1 \\
0&0&0&0&0&0&1&1 \\
0&0&0&0&0&0&0&1
\end{pmatrix}}
$$

which looks correct and beautiful as well.


问:
Rectangle is divided into 14 identical smaller rectangles of width $x$. What is its perimeter in terms of $x$?
Rectangle region ABCD show in below partitioned into 14 identical small rectangles, each of which has width x.

What is perimeter of ABCD in terms of x?

I have used rectangle perimeter formula but answer getting wrong . what is method to find answer in shortest way ?

答: If each small rectangle has width $x$, then we can see that the length of the small rectangles is 7 widths of the small rectangle, thus $7x$. Therefore, sides $AP$, $DQ$ and $BC$ have length $7x$. Finally, $AD$, $QC$, and $PB$ have length $7x$. Thus, we have six sides with length $7x$, so the perimeter has length $6\cdot 7x = 42x$.</p>


问:
Expected Value for Arrows game

In a game Daniel shoot arrows on a circular target with a radius of $R$.

Daniel's arrows hit the target uniformly distributed over the target.

Daniel shoots one arrow on a circular target with radius R, that is
split into $10$ circles each with similar size $((R^2\pi)/10)$ such
that the innermost circle gives $10$ points, the second innermost gives $9$ and so on till the outermost gives $1$ point. What's the expected value for the points he got?

Let D be a random variable that measures the distance from the center.

Previously I found with your help that $E(D)=(2R)/3$ from there I think I should find the circle within that radius and multiply with points. But I'm not sure how to do this formally or find that circle.

答: It is given that the arrow lands on the target uniformly. This means the probability of landing on a circle is proportional to the area of the circle. In this case we have $10$ circles of equal area, so the probability  that the arrow lands on any circle is $1/10$. The expected value of the points is$$\frac1{10}[1+2+\cdot\cdot\cdot+10]=\frac{11}2.$$

Title: Expected Value for Arrows game


问:
Angle Chasing in geometry in triangle
In triangle $\Delta ABC$, $\angle CAB = 30^{\circ}$ and $\angle ABC = 80^\circ$. The point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$
and $\angle MCA = 30^\circ$. Find the value of $180^\circ – \angle BMC$ in degrees .

答:
Hint:

Extend lin segment $AM$ and draw line segment $BF$ such that $\angle CBF = 10^0$.

Can you see $ABFC$ is cyclic with $\angle ACF = \angle ABF = 90^0$?

That leads to center of the circle on $AF$.

If $\angle ABG = 20^0$ then can you see $AG = BG = CG$?

You can then show $\triangle BCG$ is equilateral.

Now, as $\angle MGC = \angle MCG = 20^0$, $\angle GBM = \angle CBM = 30^0$.

You also know $\angle BCM = 40^0$.

And adding both $\angle CBM$ and $\angle BCM$, you get $(180^0 - \angle BMC)$.

Title: Angle Chasing in geometry in triangle


问:
Can a unit set $C(K)$ form a commutative subdring?
I was asked to prove that the set $C(K)=\{a\in K| a\cdot r=r \cdot a, \forall\; r \in K\}$ forms a commutative subring inside the ring $K$.

However unit sets do not contain the element $0$, so they are not closed under addition. Therefore, $C(K)$ cannot be a subring.

I want to know if there is any possibility for $C(K)$ to form  a ring.

答:
You are confusing the set of invertible elements of $K$, which are the elements $a\in K$ which satisfy
$$ \exists b\in K, ab=ba=1 $$
and its center (which you write $C(K)$ but is more commonly written $Z(K)$), which consists of the elements $a\in K$ such that
$$\forall b\in K, ab=ba.$$

They are completely different sets that have nothing to do with each other. The center is indeed a subring, and you can see that $0$ fits the definition ($0$ commutes with every element). The set of invertible elements (also called "units" to my great despair) is not a subring, but it is a subgroup for the multiplication.


问:
The equivalent of multiplying by $-1$ in $GF(4)$?
In $\mathbb{R}$, we have the notion of multiplying an element by $-1$. For example:

$12 \in \mathbb{R}$, then $12 \times (-1) = -12 \in \mathbb{R}$,

$7 \in \mathbb{R}$, then $7 \times (-1) = -7 \in \mathbb{R}$.

Now, assume we are in $GF(4)$ and its definition is a standard definition with elements $\{0, 1, \omega, \bar{\omega}\}$. We will label them as $\{0, 1, 2, 3\}$, respectively:

$$
\begin{array}{|c|cccc|}\hline
    + & 0 & 1 & 2 & 3\\\hline
    0 & 0 & 1 & 2 & 3\\
    1 & 1 & 0 & 3 & 2\\
    2 & 2 & 3 & 0 & 1\\
    3 & 3 & 2 & 1 & 0\\\hline
\end{array}\qquad
\begin{array}{|c|cccc|}\hline
    - & 0 & 1 & 2 & 3\\\hline
    0 & 0 & 1 & 2 & 3\\
    1 & 1 & 0 & 3 & 2\\
    2 & 2 & 3 & 0 & 1\\
    3 & 3 & 2 & 1 & 0\\\hline
\end{array}\qquad
\begin{array}{|c|cccc|}\hline
    \times & 0 & 1 & 2 & 3\\\hline
    0 & 0 & 0 & 0 & 0\\
    1 & 0 & 1 & 2 & 3\\
    2 & 0 & 2 & 3 & 1\\
    3 & 0 & 3 & 1 & 2\\\hline
\end{array}\qquad
\begin{array}{|c|cccc|}\hline
    \div& 0 & 1 & 2 & 3\\\hline
    0 & - & 0 & 0 & 0\\
    1 & - & 1 & 3 & 2\\
    2 & - & 2 & 1 & 3\\
    3 & - & 3 & 2 & 1\\\hline
\end{array}\qquad
$$

It is usually understood that $-x$ is the  additive inverse so that for any element $x \in GF(4)\,\,$, $x + (-x) = 0$.

I am interested in multiplying by $-1$ in $GF(4)$ defined as the same mathematical meaning as ordinary multiplying by $-1$ in $\mathbb{R}$. How would I go about doing something like this?

In the case where you would like more context, this deals with finding determinant of a matrix with elements in $GF(4)$ and multiplying by $-1$ is a necessary condition.

答:
Well, $GF(4)$ has $GF(2) = \{0,1\}$ as a subfield.
In this subfield, $-1=1$ as you work modulo 2.


问:
Different definitions of regular map (between affine varieties)?
On Robin Hartshone's Algebraic Geometry book, on page 15, the concept of a regular function is given as follows:

A function $f$ from a affine variety $Y$ to the base field $k$ is regular at a point $x$ if there exists an open neighborhood $U$ of $x$ and $p,q\in k[x_1,x_2,\ldots , x_n]$ such that $f=p/q$ on $U,$ and $q\neq 0$ on $U.$

Now, the definition on Wikipedia looks a bit different. Let me summarise it as follows:

A regular map $Y \to \mathbb A^1(=k)$ is called a regular function. So, if we write out the entire definition, $f:Y \to k$ is regular at point $x$ if in a neighborhood $U$ of $x$, $f=p,$ where $p\in k[x_1,\ldots,x_n] /I(Y).$ $I(Y)$ is the ideal such that $Y=Z(I(Y)).$ (The quotient "$/Y(I)$" on the ring does not actually make a difference to the definition and can be ignored)

Why the definition of Wikipedia doesn't involve quotient polynomials of any sort, but the definition on the books allows the quotient $p/q$?

答:
Let $k$ be a field and let $\mathbb{A}^n(k)$ denote the set of all $n$-tuples
$(a_1,..,a_n)$ with $a_i\in k$ for all $i=1,..,n$.

Let $X:=Z(I)\subseteq \mathbb{A}^n(k)$ be the zero set of an ideal $I\subseteq k[x_1,..,x_n]$. Hence $X$ is by definition the set of $n$-tuples $a:=(a_1,..,a_n)\in \mathbb{A}^n(k)$ with $f(a)=0$ for all polynomials $f\in I$.

Let $Y:=Z(J)\subseteq \mathbb{A}^m(k)$ with $J\subseteq k[y_1,..,y_m]$.

Definition 1: Let $U\subseteq X \subseteq \mathbb{A}^n(k)$ be an open set (the complement of a Zariski closed set), and let $f: U \rightarrow k$ be a map of sets. The map $f$
is a "regular map" iff for any point $x\in U$ there is an open set $x\in U(x) \subseteq U$ and polynomials $p,q\in k[x_1,..,x_n]$ such that $f_{U(x)}=\frac{p}{q}$ on $U(x)$.

Here the notation $f_{U(x)}$ means "the restriction of $f$ to the open subset $U(x)$".

If $U\subseteq X$ is a Zariski open set, let $\mathcal{O}_X(U)$ denote the set of regular maps $f:U \rightarrow k$ in the sense of Definition 1.

Lemma. The set $\mathcal{O}_X(U)$ is a $k$-algebra for any open set $U\subseteq X$.
If $V\subseteq U\subseteq X$ are open sets, there is a canonical restriction map

$\rho_{U,V}: \mathcal{O}_X(U) \rightarrow \mathcal{O}_X(V)$.

This defines a sheaf of $k$-algebras $\mathcal{O}_X$ on $X$.

Proof. This holds because the definition of "regular map" in Definition 1 is local.

You may define $\mathcal{O}_{X,x}:= lim_{x\in U}\mathcal{O}_X(U)$ and it follows $\mathcal{O}_{X,x}$ is a local ring. The pair $(X, \mathcal{O}_X)$ is a locally ringed space.

Hartshorne defines a map $\phi: X \rightarrow Y$ to be a "morphism of algebraic varieties" iff for every open set $V\subseteq Y$ and every regular map $f:V\rightarrow k$, it follows the map

$f\circ \phi: \phi^{-1}(V) \rightarrow k$ is a regular map. This defines a map

$\phi^{\#}(V):\mathcal{O}_Y(V) \rightarrow \mathcal{O}_X(\phi^{-1}(V)):=\phi_*\mathcal{O}_X(V)$

by $\phi^{\#}(V)(f):=f \circ \phi$. This again corresponds to a map of sheaves

$\phi^{\#}: \mathcal{O}_Y \rightarrow \phi_*\mathcal{O}_X$.

Hence Hartshorne is implicitly defining a map of locally ringed spaces in chapter I in his book.

Your comment: "A function f from a affine variety Y to the base field k is regular at a point x if there exists an open neighborhood U of x and p,q∈k[x1,x2,…,xn] such that f=p/q on U, and q≠0 on U."

When Hartshorne defines "regular map", he implicitly defines the structure sheaf $\mathcal{O}_X$ on $X$. In this language you may speak of a morphism of varieties

$\phi: X \rightarrow \mathbb{A}^1(k)$. When $k$ is algebraically closed Hartshorne proves in Theorem I.3.5 the isomorphism

$Hom_{var}(X,Y) \cong Hom_{k-alg}(A(Y), A(X))$

where $A(X), A(Y)$ are the coordinate rings of $X,Y$. Hence in the case when $Y=\mathbb{A}^1(k)$ it follows

$Hom_{var}(X,Y)=Hom_{k-alg}(k[x], A(X)) \cong A(X)$.

Your question: "Why the definition of Wikipedia doesn't involve quotient polynomials of any sort, but the definition on the books allows the quotient p/q?"

Answer: You should not use wikipedia as a reference, instead you should use a book that has been in use "for a while". Most books contain errors, but the fact that a book is popular and has been reprinted several times is a sign it is a good book.

It seems to me you confuse the notion "regular map" and "morphism of algebraic varieties" when you write the equality $\mathbb{A}^1 (=k)$, and this confusion may be because of wikipedia.
Anyone can write on wikipedia, and there is noone checking the quality of what is written there.

Example: What happens when the base field $k$ is not algebraically closed?

Let $k:=\mathbb{R}$ be the field of real numbers and consider $X:=\mathbb{A}^1(k)$. What is the ring of global regular maps $\Gamma(X, \mathcal{O}_X)$ from Definition 1 in this case? Is it the polynomial ring $k[x]$?

Let $z:=a+ib$ with $a,b \in k$ and $b\neq 0$ and define

$p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2 \in k[x]$.

Let $S$ be the multiplicatively closed set of functions on the form $p_{z_1}(x)\cdots p_{z_l}(x)$ with $z_i \in \mathbb{C}-\mathbb{R}$. It follows we may construct the localization $B:=S^{-1}(k[x])$. Choose an element $f(x):=\frac{p(x)}{q(x)}\in B$. It follows the function $f(x)$ defines a regular map

$f(x): k \rightarrow k$

since $q(x)\neq 0$ on $X$. Hence there is a canonical inclusion

$B \subseteq \Gamma(X, \mathcal{O}_X):=A(X)$.

Hence the global sections $\Gamma(\mathbb{A}^1(k), \mathcal{O}_{\mathbb{A}^1(k)})$ is much larger than $k[x]$ when $k$ is not algebraically closed.

If you define the spectrum $Y:=Spec(\mathbb{R}[x])$, its topological space has as points prime ideals in $A:=\mathbb{R}[x]$ and the prime ideals in $A$
are the following: The zero ideal $(0)$ and maximal ideals. A maximal ideal
is on the form $(p(x))$ where $p(x)\in A$ is an irreducible polynomials
and $p(x)$ is irreducible iff $p(x)=(x-r)$ for $r\in \mathbb{R}$
or $p(x)=p_z(x)$ for $z\in \mathbb{C}-\mathbb{R}$. The ring of global sections $\Gamma(Y, \mathcal{O}_Y)=\mathbb{R}[x]$, hence in this case you "recover" the ring $\mathbb{R}[x]$ when calculating global sections of $\mathcal{O}_Y$. Hence the structure sheaf of the spectrum $Spec(\mathbb{R}[x])$ has fewer global sections than the structure sheaf of $\mathbb{A}^1(k)$. When studying "real algebraic geometry" one uses the construction in Definition 1 and the space $\mathbb{A}^n(k)$, since functions on the form $\frac{f(x)}{p_{z_1}(x)\cdots p_{z_l}(x)}$ with $f(x)\in \mathbb{R}[x]$ and $z_j \in \mathbb{C}-\mathbb{R}$ are algebraic functions.

Example. You may construct "real projective space" $\mathbb{P}^n(k)$ and its structure sheaf $\mathcal{O}:=\mathcal{O}_{\mathbb{P}^n(k)}$ using Definition 1. With this definition there is a closed embedding of "real algebraic varieties"

$j:\mathbb{P}^n(k) \rightarrow \mathbb{A}^d(k),$

hence "real projective space" is isomorphic to an affine real algebraic variety.

You may construct $\mathbb{P}^n_k:=Proj(k[x_0,..,x_n])$ using Hartshorne, and it follows $\mathbb{P}^n(k)$ is the topological subspace of $\mathbb{P}^n_k$ consisting of $k$-rational points. Hence there is a continuous inclusion of topological spaces

$u: \mathbb{P}^n(k) \rightarrow \mathbb{P}^n_k$.

But the structure sheaves differ much. If we let $U_i \subseteq \mathbb{P}^n(k), V_i\subseteq \mathbb{P}^n_k$ be the open subset where $x_i\neq 0$ it follows

$U_i\cong \mathbb{A}^n(k)$ and $V_i\cong \mathbb{A}^n_k:=Spec(k[y_1,..,y_n])$. And there is an inclusion of rings

S1. $S^{-1}k[w_1,..,w_n] \subseteq \Gamma(U_i, \mathcal{O}_{\mathbb{P}^n(k)})$

where $S$ is the multiplicative subset containing the functions $p_{z_1}(w_1)\cdots p_{z_n}(w_n)$ with $z_i\in \mathbb{C}-\mathbb{R}$.

There is an equality

S2. $\Gamma(V_i, \mathcal{O}_{\mathbb{P}^n_k})=k[y_1,..,y_n]$.

Hence the space $\mathbb{P}^n(k)$ has more local sections in its structure sheaf. The Whitney embedding theorem says that any (real) smooth manifold $M$ of dimension $n$ can be smoothly embedded in $\mathbb{R}^{2n}$. Hence if we view real projective space $\mathbb{P}^n(k)$ with its smooth structure, we can embed $\mathbb{P}^n(k)$ smoothly into $\mathbb{A}^{2n}(k)$. Hence the locally ringed space $(\mathbb{P}^n(k), \mathcal{O}_{\mathbb{P}^n(k)})$ - even though it is "algebraic" - "behaves" like a real smooth manifold in the sense that it can be (algebraically) embedded into affine real space.

Your title was "Different definitions of regular map (between affine varieties)?"

This post illustrates that when you change the notion "regular map" you may algebraically embed real projective space into real affine space, realizing real projective space as an affine algebraic variety of finite type over the real numbers. More generally if $X\subseteq \mathbb{P}^n(k)$ is any algebraic subvariety of real projective space, it follows we may realize $X$ as an affine algebraic sub-variety of real affine space $\mathbb{A}^d(k)$. This is not true if you consider an arbitrary
closed subscheme $Y \subseteq \mathbb{P}^n_k$.

The field "real algebraic geometry" is an independent field. By the above argument you may view real projective space $\mathbb{P}^n(k)$ as a topological subspace of $\mathbb{P}^n_k$ with an "enlarged" structure sheaf and a larger set of regular maps. This larger set of regular maps can be used to realize any real projective variety as a real affine variety.

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


问:
For what $n$ and $m$ is this number a perfect square?
Find all positive integers $n$ and $m$ such that $$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3$$ is a perfect square.

Of course $n=m=1$ is a trivial solution and if you put $n=m+1$ (assume WLOG that $n \geq m$), the above expression is equal to $4n^4$ which is a perfect square - But I don't know if this is the only solution. I would really appreciate any help.

答:
Given the OP's form,

$$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3 = w^2\tag{1}$$

ccorn and Alyosha showed it is equivalent to the more aesthetic forms,

$$\tfrac{1}{4}\big((2m-1)^2+1\big)\big((2n-1)^2+1\big)-1 =w^2$$

$$(m+n-2mn-1)^2+(n-m)^2-1 =w^2$$

respectively. There are infinitely many positive integer solutions to (1) other than $n=m+1$ as we can use a Pell equation $x^2-dy^2=1$ to solve it. In fact, we can use infinitely many $d$. For any positive integer $m>1$, the solution is then,

$$\begin{aligned}
&n =(2m-3)x^2+4(m-1)^2xy-(m-2)\\
&w = 4(m-1)^2x^2+(2m-3)\big((2m-1)^2+1\big)xy-2(m-1)^2
\end{aligned}$$

where $x,y$ satisfy the Pell equation,

$$x^2-\big((2m-1)^2+1\big)y^2=1$$

For the simplest case when $m=1$, we have the alternative and simpler form,

$n = (x+1)/2,\;\; w=y,\;\; \text{where}\;\;x^2-2y^2=1.\tag{2}$

(Note: Since the $x$ of (2) is odd, then $n$ is an integer.)


问:
>Let $X_1,\ldots,X_n$ indepent variable RV $X_n\sim Bern(1/n)$ Does the $X_n\underset{a.s}{\to}0$

Let $X_1,\ldots,X_n$ indepent variable RV $X_n\sim Bern(1/n)$

Does the $X_n\underset{a.s}{\to}0$

I tried to use Borel-Cantelli but I get that $\sum_{i=1}^{\infty}\mathbb{P}(A_n^{\epsilon})=\infty$

where $A_n^{\epsilon}=\{|X_n-X|\geq\epsilon\}$

any hint please

答: $\sum P(X_n=1)=\sum \frac  1 n =\infty$. By independence and  Borel Cantelli Lemma this implies $P(X_n =1 i.o.) =1$. Hence it is not true that $X_n \to 0$ almost surely.


问:
Identity for categorical product is terminal?
It is well-known that for a category $\mathbf{C}$ with binary products and terminal object $\mathbf{1} \in \mathbf{C}$, we have $\mathbf{1} \times A \cong A$ naturally in $A$. I want to know if the converse holds, i.e. if $B \times A \cong A$ for all objects $A$, does this mean $B$ is terminal?

答:
If we strengthen the condition, we can get an affirmative answer.

Namely, first assume that specifically the second projection $\pi_2:B\times A\to A$ is invertible for all objects $A$ (implying that the isomorphism is natural in $A$).

Then we can conclude that $B$ is a terminal object:

For any object $A$ we have an arrow $A\overset{\pi_2^{-1}}\to B\times A\overset{\pi_1}\to B$, and this is unique, since for any $f:A\to B$, the arrow $(f,1_A):A\to B\times A$ is a right inverse of $\pi_2$, as $\pi_2\circ(f,1_A)=1_A$, hence it is its inverse now, and thus
$$f=\pi_1\circ (f,1_A)=\pi_1\circ\pi_2^{-1}\,.$$

Secondly, if we only have a natural isomorphism $\tau_A:A\to B\times A$, the conclusion still holds:

Let $\alpha_A:=\pi_2\circ\tau_A:A\to A$, this is still natural in $A$, and consider $u:=\tau_A^{-1}\circ (f,1_A):A\to A$ with an arbitrary $f:A\to B\ $ (say, $f=\pi_1\circ\tau_A$), then by naturality of $\alpha:{\rm id}_{\bf C}\to{\rm id}_{\bf C}$, we obtain
$$\alpha_A\circ u = u\circ\alpha_A$$
where we have $\alpha_A\circ u=\pi_2\circ (f,1_A)=1_A$, so $\alpha_A$ and hence also $\pi_2$ is invertible, and $(f,1_A)=\pi_2^{-1}$ with every $f$, showing its uniqueness as above.


问:
Indefinite integral of $\int \frac {\cosh^{-1} (x/a) \, \mathrm d x} {x^2}$
Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.655$ gives:

$$\int \frac {\cosh^{-1} (x/a) \, \mathrm d x} {x^2} = \dfrac {-\cosh^{-1} (x/a) } x \mp \dfrac 1 a \ln \left({\dfrac {a + \sqrt {x^2 + a^2} } x}\right)$$

(negative for $\cosh^{-1} (x/a) > 0$, positive for $\cosh^{-1} (x/a) < 0$)

Let's ignore the fact for now that $\cosh^{-1}$ is AFAIK defined as being the $+$ve branch of the function, and roll with what we've got.

It is assumed at this stage that $a > 0$.

Integration by parts:

$u = \cosh^{-1} (x/a) \implies \dfrac {\mathrm d u} {\mathrm d x} = \dfrac 1 {\sqrt {x^2 - a^2} }$

$\dfrac {\mathrm d v} {\mathrm d x} = \dfrac 1 {x^2} \implies v = \dfrac {-1} x$

allows us to assemble:

$$ \int \frac {\cosh^{-1} (x/a) \,\mathrm d x} {x^2} = \frac {-\cosh^{-1} (x/a)} x + \int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } + C$$

But then we've got this standard integral:

$$\int \frac {\mathrm d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \operatorname {arcsec} \left|{\frac x a}\right| + C$$

In order to return to Spiegel's result, the denominator of the integrand on the RHS of the above would need to be $\displaystyle \int \dfrac {\mathrm d x} {x \sqrt {a^2 - x^2} }$.

But I can't work out how to make it be so. The derivative of $\cosh^{-1} (x/a)$ is pretty well established as being $\dfrac 1 {\sqrt {x^2 - a^2} }$, and you would expect this to be the case, as $\cosh^{-1}$ is defined only on $x > a$ in the first place.

Am I correct in assuming that Spiegel has made a mistake?

答:
I simply differentiate the second part of Schaum's result and realized their mistake:

for some reason they integrated
$$\int\frac{1}{x \sqrt{x^2 + a^2}}\,dx$$
instead of the correct
$$\int \frac{1}{x \sqrt {x^2 - a^2} }\,dx$$
Thus the correct integral is
$$\int \frac{1}{x^2} \operatorname{arcosh} \left(\frac{x}{a}\right) \,dx = - \frac{1}{x} \operatorname{arcosh} \left(\frac{x}{a}\right) + \frac{1}{a} \operatorname{arcsec} \left(\frac{x}{a}\right) + C $$


问:
Finding the vertex of $y = x^2 - 6x + 1$. My solution doesn't match the book.
I'm studying on how to find the vertex of a Parabola and got stuck with this question.

Determine the vertex of $y = x^2 - 6x + 1$

I found the vertex is $V=(3,-8)$ (see work below), but my workbook showed it as $V=(3,10)$. Unfortunately, it does not have a section where it shows how it got to that answer so I'm doubting if my understanding is incorrect. If I use online algebra calculators, it matches with mine but I just want to make sure that I understand how signs and the formula work.

Formula I used to look for the Vertex:
$$V=\left(-\frac{b}{2a} , \frac{4ac-b^2}{4a}\right)$$

My solution:

$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{-32}{4} = \frac{-8}{1}=-8
\end{align}$$

$$V=(3,-8)$$

What I think was done on the book:

$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{40}{4} = \frac{10}{1}=10
\end{align}$$

$$V=(3,10)$$

Is it correct that I should've added $4-36$ since they are both positive numbers or subtract it?

Any explanation is appreciated. Thanks!

答:
Any general parabola can be written as $L_1^2=4AL_2$, if $L_1$ and $L_2$ are Eq. of non-parallel lines.

If $L_1$ and $L_2$ are perpendicular and $L_1$ and $L_2$ are normalized, life is simple. The length of latus rectum (LR) is 4A. The Eq. of axis of the parabola is $L_1=0$ Tangen at vertex is $L_1=0$. Vertex is found by solving $L_1=0$ and $L_2=0$. The Eq. of directrix is $L_2=-A$, Eq. of LR is $L_2=A$, the focus F come by solving $L_1=0$ & $L_2=A$.

So in your case $y=x^2-6x+1 \implies y+8= (x-3)^2$ Then $L_1=x-3, L_2=y+8, A=1/4$ The vertex is given by $L_1=0, L_2=0$ which is $x=3, y=-8$. the vertex is (3,-8)$.


答:
$4-36$ is indeed $-32$. Whenever you have $a - b$ and $b > a$, you should interpret it as $a - b = -1 \cdot (b - a)$. So your calculation seems to be right. And yes, the vertex of the parabola is at $(3, -8)$, this can be verified by differentiating and setting $y' = 0$.

$y' = 2x - 6 = 0 \implies x = 3$ and then $y(x=3) = 3^2 - 6 \cdot 3 + 1 = 9 - 18 + 1 = -8$. Your method is correct, I just mentioned the differentiation method to confirm the answer.

Title: Finding the vertex of $y = x^2 - 6x + 1$. My solution doesn't match the book.

Body:
I'm studying on how to find the vertex of a Parabola and got stuck with this question.

Determine the vertex of $y = x^2 - 6x + 1$

I found the vertex is $V=(3,-8)$ (see work below), but my workbook showed it as $V=(3,10)$. Unfortunately, it does not have a section where it shows how it got to that answer so I'm doubting if my understanding is incorrect. If I use online algebra calculators, it matches with mine but I just want to make sure that I understand how signs and the formula work.

Formula I used to look for the Vertex:
$$V=\left(-\frac{b}{2a} , \frac{4ac-b^2}{4a}\right)$$

My solution:

$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{-32}{4} = \frac{-8}{1}=-8
\end{align}$$

$$V=(3,-8)$$

What I think was done on the book:

$$\begin{align}
x &=-\frac{-6}{2(1)} = -\frac{-6}{2} = -\frac{-3}{1} = 3 \\[6pt]
y &=\frac{4(1)(1)-(-6^2)}{4(1)} = \frac{4-36}{4} = \frac{40}{4} = \frac{10}{1}=10
\end{align}$$

$$V=(3,10)$$

Is it correct that I should've added $4-36$ since they are both positive numbers or subtract it?

Any explanation is appreciated. Thanks!


问:
Boundary regularity of the domain in the use of Poincare Inequality
Let $\Omega $ be a bounded, connected, open subset satisfying some boundary regularity. For any $p \in [1,\infty]$, there is $C(n,p,\Omega)$ such that
$$
\lVert u - u_{\Omega} \rVert_{L^p(\Omega)} \leq C \lVert \nabla u \rVert_{L^p(\Omega)} \text{ for any $u \in W^{1,p}(\Omega)$. }
$$
Here $u_{\Omega} = \dfrac{1}{|\Omega|} \displaystyle\int_{\Omega}u$. In Evan's book and Trudinger's book, they tell me that $\partial \Omega$ should be $C^1$. However from Wiki, it required that $\partial \Omega$ is Lipschitz. So may I know that what is the least condition about the boundary regularity of $\Omega$ for Poincare Inequality being valid?

答:
I am not aware of a characterisation of domains supporting the Poincaré inequality of the above type, but there are several interesting results in this direction.

(1) We say $\Omega$ satisfies the cone condition, which asserts that there is a finite cone $C$ of the form $\{x \in \Bbb R^n : x_1^2+\cdot x_{n-1}^2 \leq bx_n^2, 0 < x_n < a\}$ for constants $a,b>0$ such that for all $x \in \Omega,$ there is a cone $C_x$ congruent to $C$ (i.e. obtained by translating and rotating) with vertex at $x,$ such that $C_x \subset \Omega.$

It is a classical result that the cone condition implies the Poincaré inequality of the above form. This is proved for instance in Sobolev Spaces by Adams and Fournier (section 6 via Rellich-Kondrachov), and in Sobolev Spaces by Maz'ya (section 1.1.11 via local representation formulae). It's worth noting that Lipschitz domains do satisfy the cone condition, but this is more general as it includes punctured domains also (e.g. the unit ball with the origin removed).

(2) The above is a special case of the more general result, which asserts that the Poincaré-Sobolev inequality
$$\left( \int_{\Omega} |u-u_{\Omega}|^{\frac{np}{n+p}} \,\mathrm{d} x \right)^{\frac{n+p}{np}} \leq C \left( \int_{\Omega} |\nabla u|^p \,\mathrm{d} x \right)^{\frac1p} $$
holds if $\Omega$ is a John domain; I'll omit the precise definition but by a result of Martio these are domains where the cone is replaced by images of balls by Lipschitz mappings which are of "fixed size." For a detailed discussion see the paper:

Bojarski, B., Remarks on Sobolev imbedding inequalities, Complex analysis, Proc. 13th Rolf Nevanlinna-Colloq., Joensuu/Finl. 1987, Lect. Notes Math. 1351, 52-77 (1988). ZBL0662.46037.

Moreover a partial converse was proved by Buckley and Koskela, asserting that under mild topological conditions (which holds if $\Omega$ is simply connected), $\Omega$ must be a John domain to satisfy the above Poincaré-Sobolev inequality. The reference is below.

Buckley, S.; Koskela, P., Sobolev-Poincaré implies John, Math. Res. Lett. 2, No. 5, 577-593 (1995). ZBL0847.30012.

This precisely captures the regularity of the boundary; note that John domains are weaker than Lipschitz, but it shows that Hölder regularity is not sufficient - it is known that Sobolev embedding fails for domains with Hölder continuous boundary (see for instance 8.7 in Analysis by Lieb and Loss), and the above refines this fact.

(3) Going back to the Poincaré inequality it is known that you need some regularity; in Section 1.1.4 of Mazy'a's book, it is shown that the inequality fails for a domain constructed by Nikodym. Here a function $u$ is furnished such that $\nabla u \in L^2(\Omega)$ but $u \not\in L^2(\Omega),$ but the example is easy to modify to disprove the validity of an estimate of the above type.

However I am not aware of further results in either direction; whether you can relax the John domain assumption (I imagine there are easy examples that say you can, but general results?), or if there are nicer domains (e.g. domains with Hölder continuous boundary) where the inequality fails - if anyone knows any references in this direction please let me know. I hope my answer does give some insight into results of this type however, and that it is a fairly involved problem.


问:
Irreducibility of certain polynomials
Let $n \geq 2$ and we work over the rationals.

Question 0: Is a factorisation into irreducibles of polynomial of the form
$p_{n,i}(x)=x^n+x^i+1$ known, where $0<i<n$?

Question 1:Is it true that the $n$ such that all $p_{n,i}(x)$ are irreducible is given by the sequence https://oeis.org/A038754? This is true for $n \leq 55$.

Question 2: In case $p_{n,i}(x)$ is irreducible, what is its Galois group?

答:
Concerning Question 1:

Let $n=3^rps$, where $p$ is a prime, $p\ge5$, and $s$ is not a multiple of $3$. Let $m=3^rt$ where $ps>t>0$ and $ps+t$ is a multiple of $3$. Let $\zeta=e^{2\pi i/3^{r+1}}$. Then $\zeta^n+\zeta^m+1$ is the sum of the three cube roots of unity, so it's zero, so $x^n+x^m+1$ is divisible by the minimal polynomial for $\zeta$. That polynomial has degree $2\times3^r$, which is less than $n$, so $x^n+x^m+1$ is reducible.

Now let $n=4t$ for some $t$. Then $$x^n+x^{n/2}+1=x^{4t}+x^{2t}+1=(x^{2t}+x^t+1)(x^{2t}-x^t+1)$$ so $x^n+x^{n/2}+1$ is reducible.

This only leaves $n$ of the form $3^r$ and $2\times3^r$ to consider. Suppose $n$ is of one of these forms, and consider $x^n+x^m+1$, $0<m<n$. At this point, we have to bring in the big result of the papers cited at mathoverflow.net/questions/56579/about-irreducible-trinomials. This says that $x^n+x^m+1$ has at most one non-cyclotomic factor, where by cyclotomic factor I mean a polynomial whose zeros are all at roots of unity. That is, $x^n+x^m+1$ is either $P(x)$ or $Q(x)$ or $P(x)Q(x)$, where $P(x)$ is a cyclotomic factor, and $Q(x)$ is an irreducible non-cyclotomic factor. If it's $Q(x)$, then we're done – we've proved it's irreducible, as requested. So, we assume it has a cyclotomic factor $P(x)$, which has a root $\zeta$, which is a root of unity. Then $\zeta^n+\zeta^m+1=0$, a vanishing sum of three roots of unity, which can only be the sum of the three cube roots of unity. From this, I want to conclude that we must have $n=2\times3^r$, $m=3^r$, and $x^n+x^m+1$ is the minimal polynomial for $\zeta$, hence, irreducible, and we're done, but I'm not seeing it right now. I'll try to come back to finish this up in a day or two.

Title: Irreducibility of certain polynomials


答:
Ah, well, both you and the book got the same $x$ coordinate for the vertex: $x=3$. So let's substitute $x=3$ into $y=x^2-6x+1$ and see the $y$ coordinate. It comes out as $y=-8$, so your solution is right.

(This shows there isn't much value in remembering the formula $y=\frac{4ac-b^2}{4a}$ - you can always just remember the formula for $x$ and then substitute $x$ into the original equality $y=ax^2+bx+c$ - the result will be the same.)


问:
Bijection between $ (0,1) ^ n $ and $ (0,1) $.
Let $ z \in (0,1) ^ n $.  We can express this "$ z $" as a $ n $ -tuple of real numbers $ z = (z_1, z_2, \ldots ,z_n) $, where each $ z_i $ will have its infinite decimal expansion.

We can write:

$\begin{align*}z_1&=0,z_{11}z_{12} \ldots z_{1n} \ldots\\ z_2&=0,z_{21}z_{22}\ldots z_{2n}\ldots
 \\ \ldots \\ z_n&=0,z_{n1}z_{n2}\ldots z_{nn}\ldots
 \end{align*}$

If we now take the decimals of this arrangement according to the Cantor diagonal method, we obtain $$y=0,z_{11}z_{21} z_{12}z_{13}z_{22} \ldots \in (0,1)$$ Having that function, how is it shown that this is a bijection?  To have $ | (0,1) ^ n | = | (0,1) | $

答:
I suppose $(0,1)=[0,1[$.

The application is clearly a surjection.

If $y=0,z_{11}z_{21}z_{12}...=t_{11}t_{21}t_{12}...$, then you clearly have the equality of $z_1,...z_n$ with $t_1,...,t_n$.

The only problem (which is not specific to your function) comes from the fact that some numbers have two different decimal expansion : think of $0,099999...$ and $0,100...$ for instance, wich are the same numbers.

This is why we usually decide to take only one of these decimal expansion, for instance only the one that is not stationary at $9$.

Then since you have a bijection, you have that the two sets are (by definition) equipotent, which can be written $| (0,1) ^ n | = | (0,1) |$.


问:
The orbit space $\dfrac{\mathbb{R}^n}{\mathbb{R}^+}$ has an open subset homeomorphic to $S^{n-1}$
Let $X$ be the orbit space $\dfrac{\mathbb{R}^n}{\mathbb{R}^+}$, where the action of $\mathbb{R}^+$ is given by the following lemma. I want to show that $X$ has an open subset homeomorphic to $S^{n-1}$, and a point that belongs to every nonempty closed subset. Could anyone help me show this?
It's problem $2$ of the chapter $21$ of Introduction of smooth manifolds john lee.

here is the lemma :
For any continuous action of a topological group $G$ on a topological
space $M$; the quotient map $\pi : M \to \dfrac{M}{G}$ is an open map.

Recall: Suppose we are given an action of a group $G$ on a topological space $M$; which we
write either as $\theta: G \times G \to M$ or as $(g,p) \to g.p $. (For definiteness, let us assume
that $G$ acts on the left; similar considerations apply to right actions.) Recall that the
orbit of a point $p \in M$ is the set of images of $p$ under all elements of the group:
\begin{align}
G.p=\{g.p : g \in G\}
\end{align}
Define a relation on$M$ by setting $p \sim q$ if there exists $g \in G$ such that $g.p=q$.
This is an equivalence relation, whose equivalence classes are exactly the orbits of
$G$ in $M$. The set of orbits is denoted by $\dfrac{M}{G}$; with the quotient topology it is called
the orbit space of the action

答:
Hint 1: Show that $(\mathbb{R}^n - \{0\})/\mathbb{R}^+$ is homeomorphic to $S^{n-1}$.

Hint 2: Recall quotient topology. A set in the quotient space is closed iff. its inverse image under the quotient map is closed. Now, I claim that any nonempty subset of the quotient space which does not contain the image of the origin, cannot be closed. (This is because its inverse image under the quotient map will contain points arbitrarily close to the origin but not the origin. Hence the inverse image is not closed).

UPDATE: I will recall the following basic theorem from point-set topology (which is labeled as "Corollary 22.3" in Munkres's Topology).

Theorem.  Let $X$, $Y$ be topological spaces, and $f : X \to Y$ be a continuous surjective map. Let $X^\ast$ denote the fiber space of
$f$, ie., $X^\ast = \{f^{-1}(y) : y \in Y\}$. Let $q : X \to X^\ast$
be the quotient map and equip $X^\ast$ with the quotient topology.
Then the induced map $\widetilde{f} : X^\ast \to Y$ is a homeomorphism
iff. $f$ is a quotient map.

Put $X = \mathbb{R}^n-\{0\}$, $Y = S^{n-1}$ and $f : x \mapsto x/\|x\|$. It is easy to show that the fibers of this map are in one-to-one correspondence with the orbits of $\mathbb{R}^n -\{0\}$ under the $\mathbb{R}^+$ action. So $X^\ast$ is this case is $(\mathbb{R}^n -\{0\})/\mathbb{R}^+$.

Now, all that we need is to show that $f$ is a quotient map. I claim that $f$ is actually an open map. Here is a simple argument. In one of your comments you suggested that you have a homeomorphism $h : \mathbb{R^n}-\{0\} \to S^{n-1}\times\mathbb{R}^+$. Let $p_1 : S^{n-1}\times\mathbb{R}^+ \to S^{n-1}$ be the projection onto the first factor. Then it is easy to show that our map $f$ is the composition $p_1\circ h$. So, being a composition of open maps, $f$ itself is open. Hence $f$ is a quotient map (if a map is continuous, surjective and open then it is a quotient map). And so, by the theorem the induced map $\widetilde{f} : (\mathbb{R}^n -\{0\})/\mathbb{R}^+ \to S^{n-1}$ is a homeomorphism.


问:
Find the $x^2$ coefficient by induction of $\frac{\sin x}{x}$
In one of Wikipedia's proofs of the solution to the Basel problem, they state that the $x^2$ coefficient of $$\begin{align*}
\frac{\sin(x)}{x} &=
\left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right)\cdots
\end{align*}$$
can be proven inductively to be
$$-\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}\cdots\right)$$

I do not see any way of proving this by induction. I haven't ever come across this use of induction before actually. Could someone point me in the right direction of how to prove this using induction? I don't even know what the inductive hypothesis would be.

I can see this will certainly be the case though as the only way to have $x^2$ terms is by having each $x^2$ term in each bracket only multiplied together by the $1$'s inside the other brackets.

Thanks for your help.

答:
There's a footnote in the Wikipedia page, did you read it?
It's a formula for the coefficient of $x^2$ of the partial product. That's what can be proved by induction


问:
Inverse CDF for generating random points from U[0,1]
I read online that:

"The inverse CDF technique for generating a random sample uses the fact that a continuous CDF, F, is a one-to-one mapping of the domain of the CDF into the interval (0,1). Therefore, if U is a uniform random variable on (0,1), then X = F–1(U) has the distribution F."

From: Article

But I'm failig to undertand this, in the specific case that I have:

$\displaystyle F_{X}( x) =\frac{e^{x}}{1+e^{x}}$, and therefore $\displaystyle F^{-1}_{X}( x) =\log\frac{x}{1-x}$.

From the derativate I was able to get the PDF: $\displaystyle f_{X}( x) =\frac{e^{x}}{\left( 1+e^{x}\right)^{2}}$.

How are the two functions mapping one-to-one?

If I have a random point generator $\displaystyle U$ between $\displaystyle [ 0,1]$. How exactly should I be able to generate a random point in $\displaystyle f_{X}( x)$? This is what I'm not getting, like what should I do exactly, and why it works.

答: With $U\sim \mathcal{U}(0,1)$, you have $\log \frac{U}{1-U} \sim \mathrm{logistic}(0,1)$. That is, you generate $U$ from a uniform on unit interval, transform it according to $\log \frac{U}{1-U}$, to obtain a random draw from a standard logistic distribution.


问:
Viewing schemes as varieties
So this question is motivated from example 6.5.2 in Hartshorne Chapter 2. There he looks at $X=$Spec$(A)$ where $A=k[x,y,z]/(xy-z^2)$ and he calls $X$ a cone in $k^3$ which to me is the variety of that polynomial as defined in Chapter 1.

From section 4 there is a correspondence between varieties and integral separated schemes over a field and since we're working with divisors all schemes we are dealing with have these properties (plus others), so I can sort of see the relationship he is using here. However to what extent can we treat $X$ as a variety (ie. $Z(xy-z^2)$)? As he goes through the example he uses other variety-esque constructions (eg. $Y=\{y=z=0\}$ he claims is a prime divisor of $X$- how can you show that $Y$ is a closed integral subscheme of codim 1?) so how is this all justified?

答:
Question: "However to what extent can we treat X as a variety (ie. Z(xy−z2))? As he goes through the example he uses other variety-esque constructions (eg. Y={y=z=0} he claims is a prime divisor of X- how can you show that Y is a closed integral subscheme of codim 1?) so how is this all justified?"

Answer 1. When Hartshorne writes $Y: y=z=0$ he means the closed subscheme defined by the ideal $I:=\{\overline{y},\overline{z}\}$ in $Spec(A)$ where $A:=k[x,y,z]/(xy-z^2)\cong k[\overline{x}, \overline{y}, \overline{z}]$. By definition this is the scheme $Spec(B)$ where $B:=A/I \cong k[x]$.

Answer 2: $dim(X)=2$ and $dim(Y)=1$ hence $codim(Y,X)=1$. Since $Y\cong Spec(k[x])$ it follows $Y$ is an integral scheme since the polynomial ring $k[x]$ is an integral domain.

Note that an affine scheme $X:=Spec(A)$ is an "integral scheme" iff the ring $A$ is an integral domain: If $X$ is an integral scheme it follows by definition that $A$ is an integral domain. Conversely if $A$ is an integral domain, we must prove that $\mathcal{O}_X(U)$ is an integral domain for all open  sets $U$. Since $\mathcal{O}_X(U) \subseteq K(A)$ where $K(A)$ is the quotient field of $A$, it follows $\mathcal{O}_X(U)$ is an integral domain.

Title: Viewing schemes as varieties


问:
Calculating the value of $f'(x)$ if it exists.

Note that I am not looking for a solution, just a general advice on how to approach these questions and practical tips.

We've just started differentiability and I'm completely lost. I'm asked to calculate the value of $f'(x)$ at $x_0$ or prove it doesn't exist. But we haven't done much practice and only focused on theorems so far so I'm not sure how to approach these questions.

Are there any special tricks regarding calculating differential function limits or is it basically the same as ordinary limits?

$f\left(x\right)\ =\ \left|x^{2}-3x-4\right|,\ \ \ x_{0}=4$

$\lim _{x\to 4}\frac{\left(f\left(x\right)-f\left(4\right)\right)}{x-4} =\ \lim _{x\to 4}\frac{\left|x^2-3x-4\right|-0}{x-4}$

$\lim _{x\to 4^+}\frac{x^2-3x-4}{x-4} =\ \lim _{x\to 4^+}\frac{\left(x-4\right)\left(x-3\right)+4x-16}{x-4} =\ \lim _{x\to 4^+}\frac{\left(x-4\right)\left(x-3\right)+4\left(x-4\right)}{x-4} $

$=\ \lim _{x\to 4^+}\left(x-3\right)+4 =\ \lim _{x\to 4^+}\left(4-3\right)+4 = 5$

$\lim _{x\to 4^-}\frac{-x^2+3x+4}{x-4} =\ \lim _{x\to 4^-}\frac{\left(x-4\right)\left(3-x\right)-4x+16}{x-4} =\ \lim _{x\to 4^-}\frac{\left(x-4\right)\left(3-x\right)-4\left(x-4\right)}{x-4} $

$=\ \lim _{x\to 4^-}\left(3-x\right)-4 = \left(3-4\right)-4 = -5$

Therefore the limit doesn't exists, and f is not differentiable at 4.

b.

$f\left(x\right)\ =\ \sqrt[3]{x^{2}-\left|x\right|},\ \ x_{0}\ =0$

$\lim \:_{x\to \:0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =\ \lim \:_{x\to \:0}\frac{\left(\sqrt[3]{x^2-x}-0\right)}{x}$

$\lim _{x\to 0^+}\frac{\sqrt[3]{x^2-x}}{x}\cdot \frac{\sqrt[\frac{2}{3}]{x^2-x}}{\sqrt[\frac{2}{3}]{x^2-\:x}}=\lim \:_{x\to 0^+}\frac{\left(x^2-x\right)}{x\left(\sqrt[\frac{2}{3}]{x^2-\:x}\right)}=\lim \:_{x\to \:0^+}\frac{x-1}{\sqrt[\frac{2}{3}]{x^2-\:x}}=\frac{0-1}{\sqrt[\frac{2}{3}]{0^+}}=-∞$

A one sided limit doesn't exist, therefore f is not diffirintable at 0.

c.

$f\left(x\right)=
 \begin{cases}
      \frac{\sin x^{2}}{x} & if \ x ≠ 0\\
       0 & if \ x = 0 \\
   \end{cases}
\
, \ x_{0}\ =\ 0$

$\lim \:\:_{x\to \:\:0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =\ \lim \:\:_{x\to \:\:0^{ }}\frac{f\left(x\right)}{x}
$

$\lim _{x\to 0^+}\frac{\left(\frac{sin\left(x^2\right)}{x}\right)}{x}=\lim \:_{x\to 0^+}\left(\frac{sin\left(x^2\right)}{x^2}\right)=1$

$\lim _{x\to 0^-}\frac{\left(\frac{sin\left(x^2\right)}{x}\right)}{x}=\lim \:_{x\to 0^-}\left(\frac{sin\left(x^2\right)}{x^2}\right)=1$

Therefore the limit exists and f is differentiable at 0.

答:
Do the first one by writing the left and right pieces as
$f(x)=(x^2+x)^{1/3}, x<0; f(x)(x^2-x)^{1/3}, x \ge 0$
So the left derivative id $L f'(x)=\frac{1}{3}(x^2+x)^{-2/3}(2x+1).$ So $Lf'(0)=\infty$ so left derivative doe not exist, similarly right one also is infinite so it  also does not exist. So the derivative does not exist at $x=0$

Second one you do by $h$-method:
$$Lf'(0)=\lim_{ h \to 0} \frac{f(0-h)-f(0)}{-h}=\lim_{h \to 0}\frac{\frac{\sin^2 h}{- h}-0}{-h}=\lim_{h\to 0}(\frac{\sin h}{h})^2=1 $$
$$Rf'(0)=\lim_{ h \to 0} \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{\frac{\sin^2 h}{h}-0}{h}=\lim_{h\to 0}(\frac{\sin h}{h})^2=1 $$
Both being finite and equal $f(x)$ is differentiable at $x=0$.


问:
When $x=c^2$, we write $c= \pm \sqrt{x}$. For higher powers, do we still write $\pm$ only?
When we write $x=c^2$ , then we write $c= \pm \sqrt{x}$

What is the case when it is cube root or $4$th power of root ? Do we still write $\pm$ only ?

答: If $x^4=c$, then the solutions are $c= \sqrt[4]{x},-\sqrt[4]{x},i\sqrt[4]{x},-i\sqrt[4]{x}$, where $i$ is the imaginary unit and a primitive 4th root of unity. This works if you work over the reals or complex numbers.


问:
$X$ RV $X(S)=s$. Prove $X_n\underset{a.s}{\to}X$.

Let $([0,1],\mathcal{B}_{[0,1]},\lambda)$ probabilty space and let $X_n(s)=\frac{n}{n+1}s+(1-s)^n$ a sequence for RVs for every $s\in[0,1]$. and let $X$ RV $X(S)=s$.

Prove $X_n\underset{a.s}{\to}X$.

I've tried using Borel-Cantelly and I got

$$\sum\mathbb{P}(|X_n-x|\geq\epsilon)=\sum(1-\mathbb{P}(|X_n-s|\leq\epsilon)=\sum(1-\mathbb{P}(|-\frac{s}{n+1}+(1-s)^n|<\epsilon)$$

Couldn't figure how to continue from here.

答:
No theorem is needed. Just the definition of almost sure convergence.

For every $s \in (0,1]$ we have $\lim X_n(s)=(1)(s)+0=s=X(s)$. And $\lambda (\{0\})=0$. This proves that $X_n \to X$ almost surely.


问:
Closed form for $(a \cdot x)\bmod n = r$
Given $a$, $n$ and $r$, is there a closed form for computing $x$?

$(a \cdot x)\bmod n = r$

$x$ in [0..n)

For example,

$(7 \cdot x)\bmod 10 = 1$

$x = 3, 13, 23,...$

$x = 3$ for $x$ in $[0..n)$

I rewrote the equation but not sure how to proceed:

$ x = \frac {qn + r}{a}$

答:
Closed form for $ax\equiv r\mod n$ with $gcd(a,n)=1$ is

$x\equiv a^{-1}r\mod n$.

The condition $gcd(a,n)=1$ is sufficient and necessary that $a$ is invertible modulo $n$.


答:
An easier way to do this is to take a derivative, solve for the minimum, this will give you, your x-value. The substituting that value back into the original equation we can get the y-value. For example here you have:
y=(x^2)-6x+1
derivative: y'=(2x-6)
Solving for x-value (solving for when y'=0) : 2x-6=0 -> x=3   <-X-value
Now that we have the x value we can use this value in our original equation
y=3^2 -6*3 +1 We find that y=-8
This gives us the coordinates for the vertex. This is probably the fastest approach here as taking the derivative of this function is pretty basic.. Hopefully this helps.</p>


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答:
It might be too obvious, but Gödel's incompleteness theorems are certainly landmarks, not only because of their historical significance and "popularity" outside mathematics (where usually they are quoted in a wrong way, in a wrong context, in order to claim a wrong claim) but also because it's ideas are actually the roots of Computability theory - and understanding both the theorems and their connections to computability might take some time indeed (but I wouldn't say a couple of months...)

In complexity theory I'd say the Cook-Levin theorem, which leads to the entire theory on NP-completeness, is a good representative. But then again, the theorem itself is mainly technical and not very illuminating - it's the implications that are important and need to be studied.

Yet another main landmark of modern complexity theory is the PCP theorem (actually PCP theorems - there are many versions). This is a very surprising and powerful result.

Title: Landmarks of subjects of mathematics


问:
Conditional expectation of a squared random variable
I wanted to ask how to work with a conditional expectation of a squared random variable.

For instance I want to find E[Y^2|X].

The information I have given is: Y|X,Z is normally distributed with mean 0 and variance 1+Z and X and Z are both normally distributed.

I tried to apply the law of iterated expectations such that I have E[Y^2|X]=E[E[Y^2|X,Z]|Z] but I don't know how to solve this. For simple Y I have no problems, but I don't know how to work with Y^2.

Thank you a lot in advance.

答:
$Y|X, Z \sim \mathcal{N}(0, 1+Z)$ hence $Var[Y|X, Z] = \mathbb{E}[Y^2|X, Z] = 1+Z$.
By law of iterated expectation, $\mathbb{E}[Y^2|X] = \mathbb{E}[\mathbb{E}[Y^2|X, Z]|X] = 1+\mathbb{E}[Z|X]$


问:
Percentage Problem: Can someone help me with finding the right percentage?
In this article, we have the data that police officers have around 50 million interactions with the public and 1000 Fatal shootings but later they say "... those interactions led to fatal shootings about 0.00002 percent of the time."

How exactly does it work? Shouldn't it be:
(1000/50Mil)*100 = 0.002% instead?

答:
$1000$ over $50000000$ is $0.00002$, but just like my students do, they forget, hopefully just because of math ignorance, that percentage is over $100$

It's like saying that $150$ over $300$ students are girls, thus the percentage of girls is $0.5\%$

The correct percentage of fatal shootings is $0.002\%$, like you said correctly in your post.

Remark. I find tons of math mistakes on the papers. The most recent? On a national famous journal  in an article about smoking they said that "[...] yearly  17 millions euro go up in smoke." A mistake is a MISTAKE if you miss 3 orders of magnitude...


问:
Weighted median of the set $\{1,2,3,4,5\}$ with weights $\{0.1, 0.1, 0.1, 0.1, 0.6\}$?
The weighted median is described in Wikipedia in the following way: link.

What is the weighted median of the following set of numbers $\{1,2,3,4,5\}$ with each number having weights $\{0.1, 0.1, 0.1, 0.1, 0.6\}$, respectively?

Note that, the sum of weights is $\displaystyle \sum_{i=1}^{5}w_{i}=1$, but there is no element $x_{k}$, such that:

$$\sum_{i=1}^{k-1}w_{i}\leq 0.5\qquad\text{and}\qquad \sum_{i=k+1}^{5}w_{i}\leq 0.5.$$

Namely:

For $x_{2}=2$, we have

$$\sum_{i=1}^{1}w_{i}=0.1\leq 0.5\qquad\text{but}\qquad \sum_{i=3}^{5}w_{i}=0.8.$$

For $x_{3}=3$, we have

$$\sum_{i=1}^{2}w_{i}=0.2\leq 0.5\qquad\text{but}\qquad \sum_{i=4}^{5}w_{i}=0.7.$$

For $x_{4}=4$, we have

$$\sum_{i=1}^{3}w_{i}=0.3\leq 0.5\qquad\text{but}\qquad \sum_{i=5}^{5}w_{i}=0.6.$$

答:
I choose $x_5$.  Note
$$
\sum_{i=1}^4 w_i = 0.4,\qquad
\sum_{i=6}^5 w_i = 0,
$$
the last is an empty sum.


答:
1
2
3
4
5

0.1
0.1
0.1
0.1
0.6

0.1
0.2
0.3
0.4
1

The last row is the cumulated weighs.

Then you can see that the median is $5$

Title: Weighted median of the set $\{1,2,3,4,5\}$ with weights $\{0.1, 0.1, 0.1, 0.1, 0.6\}$?

Body:
The weighted median is described in Wikipedia in the following way: link.

What is the weighted median of the following set of numbers $\{1,2,3,4,5\}$ with each number having weights $\{0.1, 0.1, 0.1, 0.1, 0.6\}$, respectively?

Note that, the sum of weights is $\displaystyle \sum_{i=1}^{5}w_{i}=1$, but there is no element $x_{k}$, such that:

$$\sum_{i=1}^{k-1}w_{i}\leq 0.5\qquad\text{and}\qquad \sum_{i=k+1}^{5}w_{i}\leq 0.5.$$

Namely:

For $x_{2}=2$, we have

$$\sum_{i=1}^{1}w_{i}=0.1\leq 0.5\qquad\text{but}\qquad \sum_{i=3}^{5}w_{i}=0.8.$$

For $x_{3}=3$, we have

$$\sum_{i=1}^{2}w_{i}=0.2\leq 0.5\qquad\text{but}\qquad \sum_{i=4}^{5}w_{i}=0.7.$$

For $x_{4}=4$, we have

$$\sum_{i=1}^{3}w_{i}=0.3\leq 0.5\qquad\text{but}\qquad \sum_{i=5}^{5}w_{i}=0.6.$$


问:
Question about the Souslin property
I've been working on problems dealing with the Souslin property. A topological space $X$ has the Souslin property if every pairwise disjoint family of non-empty open subsets of $X$ is countable.

I came across this problem, and I am having trouble with formulating a proof.

A metrizable space $X$ has the Souslin property if and only if it has a countable base.

Can anyone help? Thanks in advance!

答:
I've been trying to come up with a simpler proof to answer your question( Brain's answer is of course completed and great but I still fancied a simpler proof). So here it is:

If $X$ has countable base, then it is easy to see that $X$ has Souslin property.
  Now let $X$ be a metric space with Souslin property. By the Bing metrization theorem $X$ has a $\sigma$-discrete base $\mathscr{B}=\bigcup_n\mathscr{B}_n$, where each $\mathscr{B}_n$ is discrete. Since $X$ has Souslin property , any discrete family of open sets in $X$ is countable, so each of the families $\mathscr{B}_n$ is countable, and therefore the base $\mathscr{B}$ is countable as well.

Title: Question about the Souslin property


问:
Show that C is a circle and find its radius and centre.

A transformation $T$ from the $z$ plane to the $w$ plane is given by$$w=\frac1{z+1},z\ne-1$$Show that $T$ maps the imaginary axis in the $z$ plane onto a circle $C$ in the $w$ plane and find its centre and radius.

This is from a June 2019 FP2 IAL edexcel paper. Not sure about my answer. Please help!

答:
Let $z=it$ where $t$ is a real number  we get $$w=\frac{1}{1+it} = \frac{1}{1+t^2}-\frac{it}{1+t^2}$$

With $$x= \frac{1}{1+t^2}$$ and $$y=-\frac{t}{1+t^2}$$ we have $$x^2 + y^2 = x$$ which is a circle with center at $\big(\frac{1}{2}, 0\big)$ and radius of $\frac{1}{2}.$


问:
Find all positive integer solutions for the following equation:
Find all positive integer solutions for the following equation:

$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$

I've tried simplifying the equation and then refactoring but I can't find any solutions.

答:
Using wolfram alpha, all the solutions are given by the expression : $$\begin{cases}y=\frac{x-1}{x+1}, x\neq -1 \\y = \frac {x+3}{x+1}, x\neq -1 \end{cases}$$
For example, the solutions for $x=1$ are $y=0$ and $y=2$.
You could try to find these solutions by hand solving the quadratic equation in respect to $y$. If you expand the original equation you get:
$$x^2 y^2 - 2 x^2 y + x^2 + 2 x y^2 + 2 x + y^2 - 2 y + 1 = 4 x y + 4$$ $$\Leftrightarrow (x^2 +2x+1) y^2 + (-2x^2-4x-2)y + (x^2+2x+1-4)=0 $$

Which you can solve using the quadratic formula.


答:
We write the given equation equivalently:
$$
\begin{aligned}
0 &= x^2y^2 - 2x^2y + 2xy^2 + x^2 - 4xy + y^2 + 2x - 2y - 3\ ,\\
0 &= x^2(y-1)^2 + 2x(y-1)^2 + (y-1)^2-4\ ,\\
4 &= (x+1)^2(y-1)^2\ .
\end{aligned}
$$
Now consider all possible ways to write $4$ as a product of two perfect squares.

Body:
Find all positive integer solutions for the following equation:

$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$

I've tried simplifying the equation and then refactoring but I can't find any solutions.


Title: Find all positive integer solutions for the following equation:

Body:
Expansion:

$$x^2y^2+x^2+y^2+1+2x-2y+2xy^2-2x^2y-4xy=4$$

Re-factoring:

$$(x^2+2x+1)(y^2-2y+1)=4$$

Simplify

$$(x+1)^2(y-1)^2=4$$

I will leave you to check the cases.


答:
First of all, one needs to understand that $\pm$ is a shorthand in writing. "$c=\pm\sqrt{x}$" means "$c=\sqrt{x}$ or $c=-\sqrt{x}$".

Second, I presume you are working with real numbers. In that case, when $x$ is negative, then $c^2=x$ has no solutions in $c$, and if $x\ge 0$, then $\sqrt{x}$ is well defined, and the solutions for $c$ are precisely $\pm\sqrt{x}$.

Which brings us to the question: (a) Which other equations are you solving, and (b) In which set of numbers.

Real number case

Start with $c^3=x$. If you are working with real numbers, then this has a unique solution for every $x$, and this solution is given by $c=\sqrt[3]{x}$. No $\pm$. The story is the same for all odd exponents ($c^5=x, c^7=x$ etc.)

For even exponents, e.g. $c^4=x$, the situation is the same as for the square. If $x\lt 0$ there is no solution. If $x\ge 0$, the solutions are $c=\pm\sqrt[4]{x}$ (and similarly for $c^6=x, c^8=x$ etc.)

Complex number case

I was about to write a quite long story here, as the situation is, erm, a lot more complex. (Pun intended.) It involves talking about roots of $1$ and about trigonometry (the "polar form" of a complex number).

If $x\ne 0$ there are always $n$ different solutions in $c$ for $c^n=x$, and if $n=2$, the solutions are still given by $c=\pm\sqrt{x}$.

However, the symbol $\sqrt{x}$, and even more the symbol $\sqrt[n]{x}$, is ambiguous.

However, there are ways to resolve this ambiguity by adopting a certain convention. ("Choosing a branch of a multi-valued complex function" is what this is called.)

However, in teaching, different people adopt different conventions :( and so the whole thing is a mess.

Let's not get there in this post.

Title: When $x=c^2$, we write $c= \pm \sqrt{x}$. For higher powers, do we still write $\pm$ only?

Body:
When we write $x=c^2$ , then we write $c= \pm \sqrt{x}$

What is the case when it is cube root or $4$th power of root ? Do we still write $\pm$ only ?


答:
Real numbers

Every positive number $x$ has $2$ square roots, $1$ cube root, $2$ fourth roots, $1$ fifth root, etc. This means that if $x^4=5$, for instance, then $x=\pm \sqrt[4]{5}$. The symbol $\sqrt[4]{}$ denotes the positive fourth root of a number. So, if we were solving $x^4=10$ but we knew $x>0$, then we would write $x=\sqrt[4]{10}$. If $x$ is a negative number, then it has no square roots, $1$ cube root, no fourth roots, $1$ fifth root etc.

Complex numbers

Let $w$ denote a non-zero complex number. The equation $z^n=w$ for $n\in\mathbb{Z^+}$ has exactly $n$ solutions, i.e. every complex number number has $n$ $n$-th roots. This is a direct consequence of the fundamental theorem of algebra:
$$
z^n-w=0
$$
is an $n$-th degreee polynomial, meaning that it has $n$ roots.


问:
Asymptotic behaviour of quantity related to an ODE
Consider the differential equation

$$u''(t) + u(t) + u^3(t) = f(t) u(t) + g(t),\,\, u(0) = u_0, \,u'(0) = u_1$$

where $u :\mathbb{R}^+ \to \mathbb{R}$ is a bounded solution  and $f, g: \mathbb{R}^+ \to \mathbb{R}$ are given continuous functions such that $$\lim_{t\to \infty} f(t) = \lim_{t\to \infty} g(t) = 0.$$
Define the quantity $E(t) := \frac 12 (u(t))^2 +\frac 12 (u'(t))^2 + \frac 14 (u(t))^4.$
I want to study the asymptotic of $E$ when $t \to \infty.$  First, I considered many examples for $f, g$ using mathematica (See the attached images where I plotted $t\mapsto E(t)$). I'm interested in the case where
$\liminf_{t\to \infty}E(t) >0$ or roughly $\lim_{t\to \infty} E(t) >0.$ I wonder if this is the general case or it depends on the initial values $u_0, u_1$ and $f,g.$ In all the examples that I tested numerically, I found that $\lim_{t\to \infty} E(t) >0.$
My attempt started by differentiating $E$ with respect to $t.$ It seems that in the case where
$u'$ is bounded, we have $$\lim_{t\to \infty} E'(t) =0,$$
since $E' = (f u + g) u'.$ But this does not imply necessarily that $\lim_{t\to \infty}E(t) >0.$
The limit might be zero.

Thank you for any hint.

答:
I'd be grateful if you give more explicit details about the decaying force in the harmonic oscillator

You can do it as I said or you can do it in an even simpler way and put $f=0$, $g=u^3-u'$ (effectively eliminating the non-linear term and adding the damping). Then you just have a linear equation $u''+u'+u=0$, which has an exponentially decaying solution, so your $g$, indeed, tends to $0$ at infinity. The more interesting question seems to be

Is there any pair $f,g$ tending to $0$ at $\infty$ such that all (or, at least, many) solutions decay to $0$ at infinity?

though, perhaps, I'm missing something simple here.


问:
Proof of " dyadic rational numbers are dense in $\mathbb R$"
The set $\left\{\frac{m}{2^n}:m \in \mathbb Z,n \in \mathbb N^+\right\}$ is called the set of dyadic rational numbers,show that the set is dense in $\mathbb R$.
The proof is as follows:

Consider two arbitrary real numbers $a,b$ with $a< b $,By the Archimedean Property there exists $n \in \mathbb N$ such that $$0< \frac{1}{n} <b-a \;\;\text{ which implies} \;\; 0<\frac{1}{2^{n}}<\frac{1}{n}<b-a$$
Thus we have $1<b2^n-a2^n$,As the distance between $a2^n$ and $b2^n$ is is greater than $1$, there exist
$m \in \mathbb N$ such that $a2^{n}<m<b2^{n}$

I don't understand where does such $m$ come from,why such $m$ should exists?

which implies that $a<\frac{m}{2^{n}}<b$,since $a,b$ were arbitrary so the claim is proved.

答:
Any interval $(a,b)$ with $b-a>1$ contains an integer.
Indeed, let
$$m=\max(\{k\in\mathbb{Z}: k\leq a\}),$$
then
$$a<m+1<\underbrace{a}_{\geq m}+\underbrace{b-a}_{>1}=b.$$
Therefore the integer $m+1$ belongs to the interval $(a,b)$.


答:
This is not correct. $z=x+iy$ is a variable point, so $x,y$ are variable. $\left(\frac1{1-y^2},0\right)$ does not describe a fixed centre neither does $\frac1{1-y^2}$ describe a fixed radius.

To solve your question, equate the real and imaginary parts of the equality you obtained:$$iuy-vy=(1-u)-iv\implies uy=-v\wedge vy=u-1$$Isolate $y$ from the first equation and substitute for it in the second equation:$$uy=-v\implies uvy=-v^2\implies u(u-1)+v^2=0$$Thus you have shown that the image of the imaginary axis in the $z$ plane is a subset of the circle $C:|w-1/2|=1/2$ in the $w$ plane.

For proving that the image is onto, you will need to prove $\forall w\in C$, there exists a purely imaginary $z$ such that $T(z)=w$. You have already figured out the inverse transformation. For $w\ne0$:$$wz=1-w\iff z=\frac1w-1=\frac{-[u(u-1)+v^2]-iv}{u^2+v^2}$$Can you verify that $\mathfrak{Re}(z)=0$ whenever $|w-1/2|=1/2$?

As a final step, notice that $0\in C$. What purely imaginary $z$ satisfies $T(z)=0$?

Title: Show that C is a circle and find its radius and centre.

Body:
A transformation $T$ from the $z$ plane to the $w$ plane is given by$$w=\frac1{z+1},z\ne-1$$Show that $T$ maps the imaginary axis in the $z$ plane onto a circle $C$ in the $w$ plane and find its centre and radius.

This is from a June 2019 FP2 IAL edexcel paper. Not sure about my answer. Please help!


问:
$f(z)$ has a pole if $\lim_{z\to a}\, f(z) = \infty$
here is the statement that I've found in 2 books:

If $\lim_{z\to a}f(z)=\infty$, the point $a$ is said to be a pole of $f(z)$, and we set $f(a) = \infty$

Intuitively I understand about pole. For example $\frac{1}{z-2}$ has a simple pole at the point $2$. But I don't know how do we get $\infty$ when we take the limit $z$ goes to $2$ since in the real plane this function has no limit cz the left and the right side of limit aren't the same. That's my point. So, how does it have $\infty$ as its limit if that function itself has a pole in the complex plane? So, What's the meaning of that statement actually?

It's been a year I didn't study about complex analysis. Hope you can help me to understand this.

Thanks.

答: $\lim_{z\to a}f(z)=\infty$ means $\lim_{z\to a}|f(z)|=+\infty$.


答: As $f$ is non-zero in a punctured neighbourhood of $a$, we can conclude $$\lim_{z\to a}\frac 1{f(z)}=0.$$

Title: $f(z)$ has a pole if $\lim_{z\to a}\, f(z) = \infty$

Body:
here is the statement that I've found in 2 books:

If $\lim_{z\to a}f(z)=\infty$, the point $a$ is said to be a pole of $f(z)$, and we set $f(a) = \infty$

Intuitively I understand about pole. For example $\frac{1}{z-2}$ has a simple pole at the point $2$. But I don't know how do we get $\infty$ when we take the limit $z$ goes to $2$ since in the real plane this function has no limit cz the left and the right side of limit aren't the same. That's my point. So, how does it have $\infty$ as its limit if that function itself has a pole in the complex plane? So, What's the meaning of that statement actually?

It's been a year I didn't study about complex analysis. Hope you can help me to understand this.

Thanks.