问:
Description of Euler Characteristic Using $\operatorname{Ext}^i$'s
I am reading Nakajima's Lectures on Hilbert Schemes of Points on Surfaces, and I am confused about a detail at the top of page 13. He is sketching a proof that $X^{[n]}(=$ Hilbert scheme of $n$ points on a nonsingular surface $X$) is nonsingular. The strategy is to show that for $[Z]\in X^{[n]}$, $\dim T_{[Z]}X^{[n]}=2n$. The method of doing this is a fairly straightforward argument using
$$0\to \mathcal{I}_Z\to\mathcal{O}_X\to \mathcal{O}_X/\mathcal{I}_Z\to 0$$
and the long exact sequence for $\operatorname{Ext}^\bullet_{\mathcal{O}_X}(-,\mathcal{O}_Z)$. I need help understanding the following claim:

The Euler characteristic $\sum_{i=0}^2 (-1)^i \dim \operatorname{Ext}^i(\mathcal{I}_Z,\mathcal{O}_Z)$ is independent of $Z$.

I believe Nakajima means $\chi(X,\mathcal{O}_Z)=\sum_{i=1}^2(-1)^i\dim \operatorname{Ext}^i(\mathcal{I}_Z,\mathcal{O}_Z).$ I agree that for points $[Z]\in X^{[n]}$, $\chi(X,\mathcal{O}_Z)$ should be the same; namely equal to $n$. I just don't understand why the Euler Characteristic can be written in this fashion. It's probably easy, but escapes me!

答:
It is not true that the alternating sum of dimensions of $\mathrm{Ext}^i$ is equal to the Euler characteristic of $\mathcal{O}_Z$, but it is true that it is independent of $Z$. The point is that it only depends on the classes of arguments in the numerical Grothendieck group $K_0(X)$; indeed, this follows from the long exact sequence of $\mathrm{Ext}$-spaces.

Now, it remains to note that the classes of $\mathcal{O}_Z$ and $\mathcal{I}_Z$ do not depend on $Z$. Indeed, $\mathcal{O}_Z$ has a filtration with $n$ factors isomorphic to $\mathcal{O}_{x_i}$, the structure sheaves of points, hence its class is equal to $n[\mathcal{O}_x]$ for any point $x$ on $X$. Similarly, the class of $\mathcal{I}_Z$ is equal to $[\mathcal{O}_X] - n[\mathcal{O}_x]$.


问:
The difference between $\mathbb{Z}$ and $\mathbb{Z}^2$
I know that $\mathbb{Z}$ is the set of integers. But, what does $\mathbb{Z}^2$ mean? How is it different from $\mathbb{Z}$?

Thanks.

答: $ \mathbb{Z} $ is defined as the set of integers, and in general for any set A, we can define $ A^2=A\times A=\left\{(a,b): a,b\in A\right\}$. According to this definition, $ \mathbb{Z}^2=\mathbb{Z}\times\mathbb{Z}=\left\{(a,b): a,b\in \mathbb{Z}\right\}$, so it's basically the set of vectors with 2 coordinates, when every coordinate is an integer.</p>

ContentLicense: CC BY-SA 3.0


问:
A contradiction involved extending vector bundles from punctured affine plane.
Consider the punctured affine plane $\mathbb{A}^2\setminus \{(0,0)\}$ over a field. Every vector bundle on this is trivial. Especially every vector bundle on this extends to a vector bundle on $\mathbb{A}^2$. Every morphism of vector bundles on this also extends uniquely to a morphism of vector bundles on $\mathbb{A}^2$. (You can look at the morphism of vector bundles as a section of the $\text{Hom}$ vector bundle and the section of vector bundles extends uniquely to a section of the extended vector bundle. This is a property that holds for reflexive sheaves on an open that its complement has codimension greater than or equal to 2.).

More generally If $X$ is a smooth variety and $Y$ a closed subvariety of codimension $\geq 2$. Every reflexive sheaf on $X\setminus Y$ extends uniquely to a reflexive sheaf on $X$. Every section of a reflexive sheaf on $X\setminus Y$ extends uniquely to a section of the extended reflexive sheaves. You can see here. The extension functor is simply just the pushforward and restriction is just the pullback to the open.

Now back to our situation, on the punctured plane and affine plane, reflexive sheaves coincide with trivial vector bundles. So pushforward and pullback functors induce mutually inverse functors between the category of vector bundles on the plane and the punctured plane. This implies an equivalence of categories between those two. However this cannot be true because if it was then this implies the algebraic $K$-theory of the two are isomorphic. Looking at the localization sequence for a point, $\mathbb{A}^2$ and $\mathbb{A}^2\setminus \{(0,0)\}$ this implies that algebraic $K$-theory of point is trivial, which cannot be true.

What mistake am I making in my logic?

The pushforward functor probably is not exact but how it is possible for a functor that induces equivalence of categories to not to be exact? (its inverse pullback is exact)

答:
The categories of vector bundles are exact categories, and you need to use the structure of exact categories to define the $K$-theory. Now the point is that the functor you are considering is not compatible with the exact structures; indeed, the morphism
$$
\mathcal{O} \oplus \mathcal{O} \stackrel{(x,y)}\to \mathcal{O}
$$
is an admissible epimorphism on the punctured plane, but not on the plane itself.

ContentLicense: CC BY-SA 4.0


问:
The indifference curve of $U(x,y)=\min(x,y)^2+\max(x,y)$
I am trying to draw the indifference curves for $U(x,y)=\min(x,y)^2+\max(x,y)$. It should not come as a straight line, right? I tried to calculate it by setting one variable $= 0$ and the other variable greater than zero. But, it is coming as $(0,1)$ and $(1,0)$. Any help is appreciated. thank you.

答: You want to obtain level curves for $U(x,y)$. Those are only called indifference curves by economists when $U(·)$ is what they call a utility function. Fix $U(x,y)=k$ for some $k>0$ and the level curve is formed by all the $(x,y)$ that satisfy the resulting equation. For instance, if you choose $k=6$, any of the following will result in $U(x,y)=6$: $(0,6),(\frac{1}{2},\frac{23}{4}),(1,5),(\frac{3}{2},\frac{15}{4})$ and $(2,2).$


问:
Show $\frac{1}{\sum_{1}^{n}a^2}\leq \frac{n}{\left ( \sum_{1}^{n}a \right )^2}$
I want to show that $$\frac{1}{\sum_{1}^{n}a^2}\leq \frac{n}{\left ( \sum_{1}^{n}a \right )^2}$$

Until now, I only have $$\left (\sum_{1}^{n}a  \right )^2\leq \left (\sum_{1}^{n}a^2 \right )n$$

Please, any idea on how to procede?

答: This is equivalent to $\tfrac1n\sum_{i=1}^na_i^2\ge\left(\tfrac1n\sum_ia_i\right)^2$, i.e. that the mean of $a^2$ is at least the squared mean of $a$, note the difference is $a$'s variance.


问:
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$
To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?

答:
Using @Bernard solution
$$y=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}$$ using partial fractions
$$y=-\frac{1+i}{2(u-i)}-\frac{1-i}{2 (u+ i)}+\frac{\sqrt{2}+2 i}{2 \left(\sqrt{2} u+\sqrt{2}-2 i\right)}+\frac{\sqrt{2}-2 i}{2 \left(\sqrt{2} u+\sqrt{2}+2 i\right)}$$
$$\int y\,du=\left(-\frac{1}{2}-\frac{i}{2}\right) \log (-2 u+2 i)\left(-\frac{1}{2}+\frac{i}{2}\right) \log (2 u+2 i)+$$
$$\frac{\left(\sqrt{2}+2 i\right) \log \left(\sqrt{2} u+\sqrt{2}-2 i\right)}{2
   \sqrt{2}}+\frac{\left(\sqrt{2}-2 i\right) \log \left(\sqrt{2} u+\sqrt{2}+2 i\right)}{2
   \sqrt{2}}$$

Now, recombining everything
$$\int y\,du=\tan ^{-1}(u)-\sqrt{2} \tan ^{-1}\left(\frac{u+1}{\sqrt{2}}\right)+\frac 12 \log \left(\frac{u^2+2 u+3}{u^2+1}\right)$$


问:
Self-Adjointness of sum of $\ell^{2}$ operators using Kato Rellich Theorem
Consider the two operators on $\ell^{2}$: $T\left(x_{n}\right)_{n} = \left(nx_{n}\right)_{n}$ and $V\left(x_{n}\right)_{n} = \left(\sqrt{n}x_{n}\right)_{n}$ defined on $\text{dom}(T) = \left\{x \in \ell^{2} : (nx_{n})\in \ell^{2}\right\}$ and $\text{dom}(V)= \left\{x \in \ell^{2} : (\sqrt{n}x_{n})\in \ell^{2}\right\}$ respectively. I want to show that $T+V$ is self-adjoint using Kato-Rellich-Theorem. I could show that $T$ is s.a and $V$ is symmetric. However I run into troubles in showing that $V$ is $T$-bounded, with bound $<1$ i.e. that there are $a>0$ and $b \in (0,1)$ such that: $\|Vx\|^{2}\le a\|x\|^{2} + b\|Tx\|^{2}$. Can someone give me a hint on how to get this inequality?
$$\|Vx\|^{2} = \sum_{n}nx_{n}^2 \le ??$$
How can I split this up into two terms as above? Any ideas or help is greatly appreciated!

答:
We have
$$
|| V x ||^2
= \sum_{n} n x_n^2
\le \sum_n \left( \frac{1}{2} + \frac{n^2}{2}\right) x_n^2
= \sum_n \frac{n^2}{2} x_n^2
+ \sum_n \frac{1}{2} x_n^2
= \frac{1}{2} \| T x \|^2 + \frac{1}{2} \| x \|.
$$
Note that also all values of $a > \frac{1}{2}$ are possible.


问:
How to find the value of the constant $c$?
Given $X,Y$ random variables continuously I need to find the Probability density function of $X$
$$f_{X,Y}(x,y) = \begin{cases} c e^{-y} \cdot e^{-\frac{ (y-1-x)^2 }{2}} & x >0 \\ 0 & x \leq 0
\end{cases}$$

my plan is to find $c$ and than I will be able to calculate $f_{\small {X}}$. to find $c$ I need to use the fact that $$1=\int _{-\infty }^{\infty }\int _0^{\infty }ce^{-y}e^{−\left(y−1−x\right)^2}\:dxdy$$

I am not sure if the limits of $y$ is true here, and I am stuck on calculate this integral.

assume that I found $c$ I just need to write
$$f_{\small {X}}=\int _{-\infty }^{\infty }ce^{-y}e^{−\left(y−1−x\right)^2}\:dy$$
Is my plan correct? how do I find $c$?

答:
To find $c$ we need to solve:
$$1=\int _{-\infty }^{\infty }\int _0^{\infty }ce^{-y}e^{−\left(y−1−x\right)^2}\:dxdy$$

this is a bit easier to solve is we first switch the order of integration and do a bit of rearranging
$$\frac{1}{c}=\int _{0}^{\infty }\int _{-\infty}^{\infty } \exp[-(y+(y-1-x)^2)] \:dydx$$

expanding and completing the square for the expression inside
$$y+(y-1-x)^2 = y^2-y(1+2x)+(1+x)^2$$
$$=\left(y-\frac{1+2x}{2}\right)^2-\left(\frac{1+2x}{2}\right)^2+(1+x)^2$$
$$=\left(y-\frac{1+2x}{2}\right)^2 +x +\frac{3}{4}$$

Returning to our original integral now
$$\frac{1}{c}=\int _{0}^{\infty }\int _{-\infty}^{\infty } \exp[-(y+(y-1-x)^2)] \:dydx$$
$$ = \int _{0}^{\infty }\int _{-\infty}^{\infty } \exp \left[-\left(y-\frac{1+2x}{2}\right)^2 - x -\frac{3}{4}\right] dydx$$
$$ = e^{-\frac{3}{4}} \int _{0}^{\infty } e^{ - x} \int _{-\infty}^{\infty } e^{-\left(y-\frac{1+2x}{2}\right)^2} dydx$$

Notice that the pdf of a normal distribution with mean $\left(\frac{1+2x}{2}\right)$ and var $\frac{1}{2}$ is
$$\frac{1}{\sqrt{\pi}}e^{-\left(y-\frac{1+2x}{2}\right)^2}\:$$
therefore the integral of the above from $-\infty$ to $\infty$ is $1$

$$ \frac{1}{c} = \sqrt{\pi} e^{-\frac{3}{4}} \int _{0}^{\infty } e^{ - x} \int _{-\infty}^{\infty } \frac{1}{\sqrt{\pi}}e^{-\left(y-\frac{1+2x}{2}\right)^2} dydx$$
$$ = \sqrt{\pi} e^{-\frac{3}{4}} \int _{0}^{\infty } e^{ - x} = \sqrt{\pi} e^{-\frac{3}{4}}$$

Thus $c = \frac{e^{\frac{3}{4}}}{\sqrt \pi}$


答:
If $A$ is a set then $A^2$ is a shorthand for $A\times A$, which is the set of ordered pairs with elements from $A$. That is: $$A^2=\{\langle a,b\rangle\mid a,b\in A\}.$$

So $\Bbb Z^2$ is the set of all ordered pairs whose elements are integers.

Title: The difference between $\mathbb{Z}$ and $\mathbb{Z}^2$

Body:
I know that $\mathbb{Z}$ is the set of integers. But, what does $\mathbb{Z}^2$ mean? How is it different from $\mathbb{Z}$?

Thanks.


问:
Help construst a measurable set in $\mathbb{R} $
Let $0＜\theta＜1$. Construct a measurable set $E\subset\mathbb{R}$ such that

$\lim_{r\rightarrow 0}\frac{\lambda(E\cap (0,r))}{r}＝\theta$

I'm considering that since $\lambda(E\cap (0,r))$ is a continuous fuction of r. If we can choose E to let it be a polygonal with the form of
$\theta r＜\lambda(E\cap (0,r))＜\theta r+o(r)$ then the conclusion is obvious. But it seems that I can't construct a E with a form of $\bigcup[a_n,b_n]$.

答:
If you construct $E$ so that
$$\lambda\left(E\cap\left(\frac1{n+1},\frac1n\right)\right)=\frac\theta{n(n+1)},$$
then for $\frac1{n+1}\le r\le\frac1n$ you will have
$$\frac n{n+1}\cdot\theta\le\frac{\lambda(E\cap(0,r))}r\le\frac{n+1}n\cdot\theta.$$


问:
Prove that $\det (E + h A) = 1 + h \operatorname{tr} A + O(h^2)$ for $h \to 0$
Prove that for any square real matrix, the equality $\det (E + h A) = 1 + h \operatorname{tr} A + O(h^2)$ is true for $h \to 0$. What is the geometric meaning (in terms of oriented volumes) of this formula? What will change if we take a complex matrix instead of a real one?

答:
The matrix looks like this:
$$\begin{pmatrix}1+ha_{11}&ha_{12}&\dots&ha_{1n}\\
ha_{21}&\ddots&\ddots&\vdots\\
\vdots&\ddots&\ddots&ha_{n-1,1}\\
ha_{n1}&\dots&ha_{n,n-1}&1+ha_{nn}\end{pmatrix}$$
Taking its determinant, we see that it's definitely a polynomial expression in $h$, and we can try to identify where terms which are constant or linear in $h$ might come from. We're essentially taking all combinations of entries such that we have exactly one from each column and row, multiply those entries, and then add all the products we get this way. Notice that if we have any non-diagonal entry as a factor, we get a factor of $h$, so the only combination of entries that might contribute to the term without $h$ is all the diagonal entries. This is also the only combination which can contribute to the linear term, since if we have one non-diagonal entry in the combination, we need at least one more, otherwise we can't cover all rows and columns. You can calculate the product of the diagonal entries, and it will turn out to be
$$1+\sum_{k_1}^nha_{kk}+O(h^2)=1+h\operatorname{tr}A+O(h^2).$$
All other contributing products are $O(h^2)$, so the determinant is also
$$\det(E+hA)=1+\sum_{k_1}^nha_{kk}+O(h^2)=1+h\operatorname{tr}A+O(h^2).$$

This derivation didn't rely on any of the properties of the reals, so it's the same with complex entries. As far as geometric intuition is concerned: If $A$ is diagonalizable, then the trace is the sum of its eigenvalues. If we imagine $1+hA$ acting on a unit parallelepiped whose spanning vectors are the eigenvectors of $A$, then $E+hA$ stretches this parallelepiped in each direction with the corresponding eigenvalue (of $E+hA$, which has the same eigenvectors) as the stretching factor. We can imagine this stretching as adding or removing slices to or from the sides of the parallelepiped. These slices will overlap, but the overlap is kinda small if the eigenvalues are close to $1$ - which we can ensure by scaling $A$ with a small factor $h$. The parts which don't overlap scale linearly with the corresponding eigenvalue of $A$. Adding all the nonoverlapping parts, we get something that scales linearly with the sum of all eigenvalues - the trace. That's the $h\operatorname{tr}A$ contribution. The $O(h^2)$ contributions come from the overlapping parts of the slices.


问:
choosing two numbers $x_1,x_2$ out of $\{1 \dots n\}$ what is the probability that $x_1<x_2$?
Given that the two selections are independent:

My solution is quite combinatoric.

The sample space $\Omega $ is all possible selections, thus : $|\Omega| = n(n-1)$

The number of ways to choose two numbers such that $x_1<x_2$ is the number of ways to choose a set $X\subseteq \{1, \dots, n\}$ such that $|X| =2$, since after $X$ is chosen there is only one way to organize $x_1,x_2$ such that $x_1<x_2$.

This gives a final result of $\frac{\binom{n}{2}}{n(n-1)} = \frac{1}{2}$

According to the text book the final solution is $\frac{n-1}{2n}$

I can't seem to find my mistake.

Any help would be appreciated.

UPDATE:

It seems like they meant in the text book that $|\Omega| = n^2$ , means it is possible to choose the same number. Than my answer would be correct.

答: You did not pick $x_1$ and $x_2$ independently by disallowing $x_1=x_2$. One might argue that that "pick two numbers" is ambiguous enough to be interpreted as "pick two distinct numbers", but actually one should explicitly mention distinctness if it is desired.


问:
Is this composition of bounded operators self adjoint?
Let $\mathcal{H}$ be a Hilbert space. It seems obvious at first glace if $T: \mathcal{H} \to \mathcal{H}$ is bounded, self-adjoint operator and $P: \mathcal{H} \to M \subseteq \mathcal{H}$ is an orthogonal projection on a closed subspace of $\mathcal{H}$, then we should have

$$ \langle PTPx,y\rangle = \langle TPx,Py \rangle = \langle Px,TPy \rangle = \langle x,PTPy \rangle. $$
Am I overthinking this, or is there some counter-example I didn't notice?

答:
Your suspicion is not unfounded: indeed, it is not true in general that for self-adjoint $A, B$ their product will be self-adjoint as well, because taking the adjoint is an antiautomorphism - it reverses the order of products:

$$(AB)^* = B^* A^*$$

And for a product of many operators

$$(A_1 A_2 \dots A_n)^* = A_n^*\dots A_2^* A_1^*$$

In particular, this implies that if the product $A_1 A_2 \dots A_n$ is a palindrome ($A_1 = A_n$, $A_2 = A_{n-1}$, etc) and all $A_i$ are self-adjoint, their product will be self-adjoint as well. This is your case precisely, for the product $PTP$ is indeed a palindrome, and $P$ and $T$ are self-adjoint as given.


问:
How to reduce system of non linear equations to a polynomial root finding problem?
I have a set of 5 non-linear equations:

$w_0 + w_1 + w_2 = 1$,

$w_0v_0 + w_1v_1 + w_2v_2 = a$,

$w_0v_0^2 + w_1v_1^2 + w_2v_2^2 = a^2 + b^2$,

$w_0v_0^3 + w_1v_1^3 + w_2v_2^3 = a^3 + 3ab^2$,

$w_0v_0^4 + w_1v_1^4 + w_2v_2^4 = a^4 + 6a^2b^2 + 3b^4$,

Where $v_0 = a$.
I want to express the values of $w_0,w_1,w_2,v_1,v_2 $in terms of $a,b,$ and $c = w_1/w_2,$ where c is a root of a polynomial of c parametrized by a and b. I know I can do this for the case of $v_0 = 0$, but I can't manage to do it it for $v_0 = a$. Could anyone help me figure this out, or more generally to reduce the system to a root finding problem of a polynomial (that is not necessarily a polynomial of $c = w_1/w_2)$. Any tips would also be much appreciated.

答: MATLAB has isolate command which does this kind of operations. You can arrange your equations according to any variable by using this command.</p>


问:
Do I have to use T-Student to calculate this confidence interval?
Given a distribution $X~Pois(λ)$ with unknown parameter $λ=E(X)$ with $14$ random observations given from this distribution.

How do I calculate a confidence Interval at $95\%$ for $λ?$ Do I have to use T-Student or normal distribution? (I personally think I have to use T-Student, since $λ$ is unknown and n is small).

I tried it in the following way:

$λ =$ Mean([14 Values]) = 8

$pt(0.975, 13) = 0.8263$

Confidence interval: $(λ - 0.8263\sqrt{(λ/n)}, λ + 0.8263\sqrt{(λ/n)})$

Is my approach correct?

Thank you

答:
No, Student t distribution is used when you are estimating the mean of a Normal model with unknown variance. Here the model is not normal. It's a poisson one.

$n=14$ is great enough to use a Gaussian approximation for the interval.

Using CLT we know that, with $n$ great enough, $\overline{X}_n\dot{\sim} N(\lambda;\frac{\lambda}{n})$ thus the pivotal quantity is the following

$$\frac{\overline{X}_n-\lambda}{\sqrt{\frac{\lambda}{n}}}$$

First approximation

Estimate standard deviation with $\sqrt{\frac{\overline{X}_n}{n}}$

Then solving in $\lambda$

$$-1.96\leq \frac{8-\lambda}{\sqrt{\frac{8}{14}}}\leq 1.96$$

That is

$$\lambda \in[6.52;9.48]$$

Second approximation

Solve in $\lambda$ the following double inequality

$$-1.96\leq \frac{8-\lambda}{\sqrt{\frac{\lambda}{14}}}\leq 1.96$$

getting, after some algebraic calculations

$$8+\frac{1.96^2}{28}\pm\sqrt{\Bigg(\frac{1.96^2}{28}\Bigg)^2+8\frac{1.96^2}{14}}$$

That is

$$\lambda\in[6.65;9.63]$$


问:
Non-commutative rings without identity
I'm looking for examples (if there are such) of non-commutative rings without multiplicative identity which have the following properties:

1) finite with zero divisors

2) infinite with zero divisors

3) finite without zero divisors

4) infinite without zero divisors

I'll be grateful for any examples and hints. Thanks in advance.

答:
As indicated in Jared's comment, one may find easy examples of 1), 2), and 4) by taking non-unital subrings (such as left ideals or two-sided ideals) of rings with identity.

The most interesting request here is 3).  In fact, any finite nonzero associative ring $R$ (possibly without identity) without zero divisors is a field.

First, let's prove that $R$ in fact has an identity. Let $a \in R$ be a nonzero element. The function $f \colon R \to R$ defined by $\phi(x) = ax$ is injective, and since $R$ is finite, it's a bijection.  Again, because $R$ is finite, this bijection must have finite order.  Thus for some $d$, the function $\phi^d(x) = a^d x$ is the identity.  It follows that $a^d \in R$ is an identity element.

At this point, it's a standard exercise to show that a finite ring with identity and no zero divisors is a division ring. (Hint: think about the function above for any $a \in R$.)  And Wedderburn's little theorem states that any finite division ring is a field.

Title: Non-commutative rings without identity


答: There are $n^2$ ways to choose $x_{1}$ and $x_{2}$ and total $\binom{n}{2}=\frac{n(n-1)}{2}$ ways to choose two distinct numbers. Thus required probability  $$\frac{\binom{n}{2}}{n^{2}}=\frac{n(n-1)}{2n^{2}}=\frac{(n-1)}{2n}$$.Done.Hope it helps.

Title: choosing two numbers $x_1,x_2$ out of $\{1 \dots n\}$ what is the probability that $x_1<x_2$?

Body:
Given that the two selections are independent:

My solution is quite combinatoric.

The sample space $\Omega $ is all possible selections, thus : $|\Omega| = n(n-1)$

The number of ways to choose two numbers such that $x_1<x_2$ is the number of ways to choose a set $X\subseteq \{1, \dots, n\}$ such that $|X| =2$, since after $X$ is chosen there is only one way to organize $x_1,x_2$ such that $x_1<x_2$.

This gives a final result of $\frac{\binom{n}{2}}{n(n-1)} = \frac{1}{2}$

According to the text book the final solution is $\frac{n-1}{2n}$

I can't seem to find my mistake.

Any help would be appreciated.

UPDATE:

It seems like they meant in the text book that $|\Omega| = n^2$ , means it is possible to choose the same number. Than my answer would be correct.


问:
Maximum area of a triangle inscribed in a circle with radius r.
We want to find the maximum area of a triangle inscribed in a circle with radius r and with constant difference of two of its angles.

If $a, b, c$ are the angles of the triangle, if we set, wlog that $a>b$, we need to have:

$a-b = k$ (constant) and

$a+b+c=180$, so $a+b = 180-c$

I know that in general, without the restriction of the $2$ angles fixed difference, the largest triangle is the equilateral.
Any assistance is much appreciated (by the way this is not homework or anything; just challenge between friends).

答:
Hint: use this formula"

$S=\frac{abc}{4R}$

Now we want S be maximum for a constant R. We  apply sine rule:

$\frac a{ \sin \alpha}=\frac b { \sin \beta}=\frac c{\sin \gamma}=2R$

$\alpha +\beta=180-\gamma$

$\alpha - \beta =k$

These two relations give:

$\alpha = \frac {\pi}2+\frac {k-\gamma}2$

$\beta = \frac {\pi}2 -\frac{k +\gamma}2$

Plug these in S you get:

$S=2R^2 \sin{\left( \frac {\pi}2+\frac {k-\gamma}2 \right)} \sin { \left(\frac {\pi}2 -\frac{k +\gamma}2 \right)} \sin (\gamma)$

Now take derivative of S ,$\gamma$ is the variable, set ir to 0 find $\gamma$ and put it in S and find the value of S.
It can be seen (for example plotting S in wolfram) that for any k , S is maximum when $\gamma=90^o$, that is triangle must be right triangle to have maximum ares for any k.


问:
How many spanning subgraph of a graph G
If I have a graph $\mathbb G$ with $n$ vertices, $m$ edges and $c$ components, how can I count how many spanning subgraphs it has?

Thanks!

答: we have m edges. And by definition of Spanning subgraph of a graph G is a subgraph obtained by edge deletion only. If we make subsets of edges by deleting one edge, two edge, three edge and so on. As there are m edges so there are 2^m subsets. Hence G has 2^m spanning subgraphs.</p>

Title: How many spanning subgraph of a graph G


问:
How to check if a given subset is a subspace?
Given
$$ V = \Bbb R^{2 \times 2}$$

$$V_1 = \left\{A \in V:
A \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} -
\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}A =
 \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\right\}$$

$$V_2 = \left\{A \in V:
A \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} -
\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}A =
 \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}\right\}$$

I've been given these subsets of $V$ and I need to check if they are subspaces of $V.$

I'm aware of the rules of whether a subset is a subspace,

If it contains the zero vector,
if $V$ contains a vector $v$ multiplied by a scalar
and if $V$ contains the combination of $2$ vectors from within $V.$

I'm not sure what to do about this issue, specifically on how to manipulate the argument I have inside the $V1/V2$ brackets in order to give me the information I need.

Any tips will be happily accepted
Thanks in advance.

答:
Just to recap, if $V$ is a (real) vector space, then $U\subseteq V$ is a vector subspace of $V$ if

$0\in U$

$v+\lambda w\in U$ whenever $v$ and $w$ belong to $U$ and $\lambda\in \mathbb R$.

You can easily prove that the zero-matrix does not belong to $V_1$, hence $V_1$ cannot be a vector subspace.

Secondly, you can show that both of above conditions hold for $V_2$:

$0\in V$ belongs to $V_2$, as it satisfies the given equation.

Take $A,B\in V_2$ and $\lambda\in \mathbb R$. As $A,B$ belong to $V_2$, they satisfy the corresponding equation. Now, just check that $(A+\lambda B)$ also satisfies the equation. It is a direct computation, no fancy tricks involved.


问:
When does equality holds in $A\subseteq P(\cup A)$
Note: $P$ is power set. It's easy to prove that this inclusion holds. But when is other inclusion true? I can't even think of one example...</p>

答:
Here is one example: $A=\{\varnothing\}$.

Also is $A=V_{\alpha+1}$ for any ordinal $\alpha$, then $A=P(V_\alpha)$, therefore $\bigcup A=V_\alpha$, and the equality holds.


问:
A definition of defect groups
In Rouquier's 2006 ICM talk (see here), he gives the following definition of the defect group of a block $b$ (definition 2.1.2)

A subgroup $D$ of $G$ is a defect group of the block $b$ if it is minimal among subgroups such that the restriction functor $D^b(\mathcal OGb)\to D^b(\mathcal OD)$ is faithful. $D$ is then unique up to conjugacy and it's a $p$-subgroup of $G$

(where we are in modular characteristic $p$)

This is not the usual definition that one finds, and in fact the definition in the paper does not come with a reference. My question is :

Are there some references that define defect groups in that way, and work with this definition ? Or at least some references that prove that this definition is equivalent to the usual ones ?

Ideally I would prefer references that do the first thing, i.e. just define defect groups the same way Rouquier does, and prove their first properties from this definition, and then at some point prove the other characterizations - but the second one is fine as well.

答:
I doubt that there are references that do the first thing, and I don't know any that do the second. Rouquier's definition is a basically a cute and slightly idiosyncratic reformulation of the standard definition of a defect group as a minimal subgroup $D$ such that every $\mathcal{O}Gb$-module is relatively $D$-projective.

I know this is not quite what you asked for, but here is a sketch of why Rouquier's definition is equivalent to standard definitions.

First, if $D$ is a defect group (or contains a defect group) then the standard theory of defect groups gives that the natural epimorphism (given by multiplication)
$$\mathcal{O}Gb\otimes_{\mathcal{O}D}\mathcal{O}Gb\to\mathcal{O}Gb$$
of $\mathcal{O}Gb$-bimodules splits.

So for any objects $X$, $Y$ of the derived category, the restriction map
$$\operatorname{Hom}_{D^b(\mathcal{O}Gb)}(X,Y)\to
\operatorname{Hom}_{D^b(\mathcal{O}Gb)}(\mathcal{O}Gb\otimes_{\mathcal{O}D}X,Y)\cong
\operatorname{Hom}_{D^b(\mathcal{O}D)}(X,Y)$$
is a (split) monomorphism of $\mathcal{O}$-modules. So the restriction functor between derived categories is faithful.

Second, if $D$ does not contain a defect group, then there is some $\mathcal{O}Gb$-module $M$ such that the natural epimorphism $\mathcal{O}Gb\otimes_{\mathcal{O}D}M\to M$ does not split as a map of $\mathcal{O}Gb$-modules. However, it does split as a map of $\mathcal{O}D$-modules. Taking the kernel, we get a short exact sequence
$$0\to L\to \mathcal{O}Gb\otimes_{\mathcal{O}D}M\to M\to0$$
of $\mathcal{O}Gb$-modules that splits over $\mathcal{O}D$, but not over $\mathcal{O}Gb$.

In other words we have a nonzero element of the kernel of the restriction map
$$\operatorname{Hom}_{D^b(\mathcal{O}Gb)}(M,L[1])
=\operatorname{Ext}^1_{\mathcal{O}Gb}(M,L)
\to\operatorname{Ext}^1_{\mathcal{O}D}(M,L)
=\operatorname{Hom}_{D^b(\mathcal{O}D)}(M,L[1]).$$
So the restriction functor beween derived categories is not faithful.

Title: A definition of defect groups


问:
Find Find all maxima and minima on interval
I am trying to understand the solution.

Why isn't $f^{\prime}(x)$ only $= 1-sin x $. What am I missing?

$f(x)=x+\cos x$ on $[-\pi, 2 \pi]$ $f^{\prime}(x)=1-\sin x \geq 0$ and
$\sin x \leq 1 ;$ this is true for any real $x$.

The maximum value > is $f(2 \pi)=2 \pi+\cos (2 \pi)=2 \pi+1$ and the minimum value is
$f(\pi)=\pi+\cos (\pi)=\pi-1$

Thanks.

答:
It's true that $f'(x)=1-\sin x$. Then, as David comments, the solution says that $f'(x)\geqslant0$ for all $x\in\mathbb{R}$, because $\sin x\leqslant1$, so
$$\sin x\leqslant1\Longrightarrow-\sin x\geqslant-1\Longrightarrow f'(x)=1-\sin x\geqslant0.$$
We can conclude that $f$ is always increasing and that its minimun value is attained at the lower end of the interval. ¿$\pi$ or $-\pi$? (something is wrong there). And its maximun value can be found at the higher end, $x=2\pi$.


问:
$3$ and $5 $cent coins
Prove that any amount of more that $7$ cents can be represented by $3$
and $5$ cent coins. (Assume $3$ cent coins exist.)

Let P(n) be true if we can find $n$ cents with $3$ and $5$ cent coins.

My basis step is:

$P(8)$ is true since $3 + 5 = 8$

Would this be correct?

As for the inductive step, I'm not sure how to show it.

I've looked around online and it seems to be:

$P(k) = 8 ≤ k ≤ n$

How would I prove this to be true?

答:
proof by induction

p(8): 3.(1)+5.(1)=8 holds

Assume p(k) holds => 3.a + 5.b = k

we must prove that p(k+1) holds.

p(k+1): 3x+5y = k+1=> 3x+5y = 3a+5b+1=> 3(x-a) + 5(y-a)

thus suffices to prove that 3z+5ω =1 has an integer solution. That is true because gcd(3,5)=1 | 1 (1 divides 1)

Calculating gcd(3,5) gives: 5 = 1*3 + 2

3 = 1*2 + 1

2 = 2*1 + 0

Then applying the Extended Euclidean Algorithm:

1 = (1 * 3) + (-1 * 2) = (-1 * 5) + (2 * 3)=> z0=2, ω0=-1 => 3.2+5.(-1) = 1=> p(k+1) holds as well

Title: $3$ and $5 $cent coins


问:
Roots of polynomial/exponential function
Let $n\in \mathbb{N}$ and $\sigma \in (0,1]$ be fixed.
I am interested in finding the roots of the function
\begin{equation}
f(x) = \big(x^2-n^2-\sigma n^2 x\big)e^{x} - \big(x^2-n^2+\sigma n^2 x\big)e^{-3x};
\end{equation}
on $[0,n^2]$, namely solving $f(x) = 0$ for $x \in [0,n^2]$.

I would be happy with an approximation formula, which clearly indicates the dependence on $n \in \mathbb{N}$.

Clearly one root is $x=0$.
A way to proceed for finding the other root(s?) could of course be to factor the polynomials appearing in $f$; namely,
\begin{equation}
f(x) = \big(x-\big(n^2\sigma-n\sqrt{n^2\sigma^2+4}\big)\big)\big(x-\big(n^2\sigma+n\sqrt{n^2\sigma^2+4}\big)\big)e^x - \big(x-\big(-n^2\sigma-n\sqrt{n^2\sigma^2+4}\big)\big)\big(x-\big(-n^2\sigma+n\sqrt{n^2\sigma^2+4}\big)\big)e^{-3x}.
\end{equation}
Though, I don't see clearly how one may proceed from here.

答:
There is a formal analytical solution to the equation.

Rewrite it as
$$e^{-4x}=\frac {x^2-\sigma n^2 x-n^2 } {x^2+\sigma n^2 x-n^2 }=\frac{(x-a)(x-b) }{(x-c)(x-d)}$$ and the solution is given in terms  of the generalized Lambert function (have a look at equation $(4)$ in the paper).


问:
Proof that $ \operatorname{det}\left(A^{-1}-\frac{1}{\lambda} I\right)=0 $
Here is my problem: $$\\$$
Let $ A $ is the reversible matrix, $ \lambda \in R$ satisfy $\operatorname{det}(A-\lambda I)=0 $
where I is the identity matrix.
Proof that:
$$
\operatorname{det}\left(A^{-1}-\frac{1}{\lambda} I\right)=0
$$
Thank you for your help.

答:
Hint:

$\det B = 0 \iff $ $B$ is singular $\iff Bv=0$ for some $v\ne0$

$Av = \lambda v \implies \frac{1}{\lambda} v = A^{-1}v$


问:
A problem on Division and Distribution of objects.
Find Number of ways to distribute $4$ identical red balls , $4$ identical  green balls, $4$ identical white balls, $4$ identical  black balls among four people so that each person receives at least one ball but no one gets $4$ red balls.

There is a really similar question here ; Number of ways of distributing $4$ identical red balls,$1$ green ball,$1$ black ball among $4$ persons . But in my book, it was written : Number of ways of distributing $4$ identical red, green, black, white balls among $4$ persons so that each person receives at least one ball but no one gets $4$ red balls.

So i interpreted the question as the question i am posting now and i just couldn't solve it. So i looked it up on MSE and found the question (link attached). So i am wondering if it is even possible to solve that question because I can not stop thinking about it but i just can't solve it.

Thanks!

I THINK THIS PROBLEM IS WRONG

答:
Begin by distributing the green, white, and black balls arbitrarily. Let $A_k$ be the set of distributions where exactly $k\in[4]$ persons get $\geq1$ balls. The $|A_k|$ can be calculated by distributing  arbitrarily to $4$, $3$,$2$, $1$ persons and using inclusion exclusion. E.g., there are ${7\choose3}^3$ arbitrary distributions to $4$ persons, then ${6\choose2}^3$ arbitrary distributions to $3$ persons, etcetera.

For each $k\in[4]$ compute the number $n_k$ of ways to finally distribute the $4$ red balls such that in all an admissible distribution results. E.g., when $k=3$, give the  person $P$ having no ball yet a red ball, and distribute the remaining three red balls arbitrarily among the $4$ persons, except all three of them to $P$. The final number $N$ then is
$$N=\sum_{k=1}^4 n_k\,|A_k|\ .$$


答:
If $\lambda$ is a non-zero eigen value of $A$ and $A^{-1}$ exist, then
$$\det|A-\lambda I|=0 \implies \det|A(I-\lambda A^{-1}) \implies \det|A^{-1}|\det|I-\lambda A^{-1}|=0$$ $$ \implies \det|A^{-1}-\frac{I}{\lambda}|=0$$

Title: Proof that $ \operatorname{det}\left(A^{-1}-\frac{1}{\lambda} I\right)=0 $

Body:
Here is my problem: $$\\$$
Let $ A $ is the reversible matrix, $ \lambda \in R$ satisfy $\operatorname{det}(A-\lambda I)=0 $
where I is the identity matrix.
Proof that:
$$
\operatorname{det}\left(A^{-1}-\frac{1}{\lambda} I\right)=0
$$
Thank you for your help.


问:
How can I find all the solutions of $\sin^5x+\cos^3x=1$

Find all the solutions of $$\sin^5x+\cos^3x=1$$

Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.

答:
Use what you know about the magnitudes of $\sin x$ and $\cos x$.

$$
\begin{aligned}
\sin^5x+\cos^3x &\le |\sin^5x+\cos^3x| \\
&\le |\sin^5 x| + |\cos^3 x| \\
&\le |\sin^2 x| + |\cos^2 x| \\
&= \sin^2 x + \cos^2 x\\
&= 1
\end{aligned}
$$

The inequality where the exponents are changed is only satisfied if the individual terms are equal: $\sin x$ and $\cos x$ must both be $0$ or $1$. Put that into the original equation, and you get $\sin x = 1$ or $\cos x = 1$. So,  $x= \frac\pi2 + 2n\pi$ or $x = 2m\pi$.


问:
Find the max and min values of $\alpha$ for which $x^3-5x^2+8x+\alpha$ has 3 real roots.

Find the max and min values of $\alpha$ for which $x^3-5x^2+8x+\alpha$ has 3 real roots.

I noticed that if we have 3 different roots then the set of solutions for $\alpha$ is open i.e no min/max.

I'm not sure how to approach the problem when also considering roots with degree $> 1$.

Maybe we could define $F(x,y,z,w,\alpha)= \alpha$

And using lagrange multipliers look for an ext under constraints:

$x^3-5x^2+8x+\alpha = 0$,

$(x-y)(x-z)(x-w)=0$

But this seems very unpleasant...

Would appreciate hints.

答:
If $f(x)=x^3-5x^2+8x+\alpha$, then $f'(x)=3x^2-10x+8$ and therefore $f'(x)>0$ if $x<\frac43$ or $x>2$, and $f(x)<0$ if $x\in\left(\frac43,2\right)$. That means that $f$ is strictly increasing on $\left(-\infty,\frac43\right]$ and on $[2,\infty)$ and it is strictly decreasing on$\left[\frac43,2\right]$. So, $f$ has $3$ distinct real roots if and only if $f\left(\frac43\right)>0$ and $f(2)<0$. This means that $\alpha\in\left(-4-\frac4{27},-4\right)$

Edit: If you're after those $\alpha$'s for which $x^3-5x^2+8x+\alpha$ has $3$ real roots when we count them with their multiplicities, then the inequalities that you should solve are $f\left(\frac43\right)\geqslant0$ and $f(2)\leqslant0$.


问:
prove that error order of backward euler method is $o(h)$
The question is to prove that error order of backward euler method is $o(h)$
We know that in backward euler method $$f_i’=\frac{f_i-f_{i-1}}{h}$$
By using taylor seris we can get $$f(x)=f(x_i)+(x-x_i)f’(x_i)+\frac{(x-x_i)^2}{2!}f’’(x_i)+....$$

By putting $x=x_{i-1}$ i get :
$$\frac{f_i-f_{i-1}}{h}+f’i=\frac{h}{2!} f’’_i+\frac{h^2}{3!}f’’’_i+...$$
Which is not what i want to get
Cause the error is $$\frac{f_i-f_{i-1}}{h}-f’_i$$

Any help ?
I could prove that $o(h)$ is the order of error of forward Euler method by using $x=x_{i+1}$
But for the backward method it seems it doesn’t work

答:
You also need to take into account that $x-x_i$ at $x=x_{i-1}$ has the value $-h$.

Check also the other signs, the Taylor terms should be alternating.


问:
Characterize solutions to Laplace's and Poisson's equation in the unit square with periodic boundary conditions
So I am studying for a qualifying examination and there was this problem from an old exam.

(a) Does the problem $\Delta u = 0$ in the unit square in the plane with u and and all of its partial derivatives satisfying periodic boundary conditions have a unique solutions? If not, characterize all solutions.

(b) For smooth f, does the problem $\Delta u = f$ in the unite square, with u and all of its partial derivatives satisfying periodic boundary conditions generally have a solution or are there conditions on f? Give a proof of your conclusion.

I feel like I can approach (a) to some degree. However, I am having a hard time feeling like I am characterizing $all$ solutions. ie, separation of variables leads to the set of ODE's $P''(x) + \lambda^2P(x) = 0$ and $Q''(y) + \lambda^2Q(y)= 0$ and the periodic boundary conditions lead to $P(0)Q(y)=P(1)Q(y)$ and $P(x)Q(0) = P(x)Q(1)$. To illustrate my confustion in the simplest matter take $\lambda = 0$. Then both $P,Q$ are linear. If at least one is constant than it satisfies the equation and boundary conditions. But if they are both linear, it is not clear to me how to show that a solution is, or is not possible. That is I get if $P(x) = ax + b, Q(y) = cy + d$ then Bndry Conditions implies $b(cy+d) =(a+b)(cy+d)$ and $d(ax+b)=(ax+b)(c+d)$

Moreover, I want to translate the bndry conditions to imply $P(0) = P(1), Q(0) = Q(1)$ but it seems like this is only true if $Q(y),P(x) \ne 0$ for some $0 \le x,y \le 1$ and it certainly could be.

So yeah, I'm unsure of a real affective approach

For part (b) things are even worse. I'm not sure what to require except it seems reasonable that we would want f to periodic on the boundary (at least the limit as it approaches the boundary). But besides that I don't know. The most general starting point I can think of would be the general solution:

$u(x) = \int_U \psi(x-y)f(y)dy - \int_{\partial U} g(y)\frac{\partial \psi(x-y)}{\partial b} - \psi(x-y)\frac{\partial u(y)}{\partial n}dS(y)$

Anywho, any suggestions would be much appreciated! Thanks for your time!

答:
The issue here is, the problem seems to set you up to use certain results, but those theorems are irrelevant in the problem as actually stated.

So long as the boundary conditions lie in the appropriate trace space ($H^{-\frac{1}{2}}$ of the boundary), and $f$ lies in $H^{-1}$ of the unit square, you can solve the equation on the square in the weak sense; this is just the Lax-Milgram theorem, which you can look up in any textbook on PDEs.

However, this has nothing to do with periodicity, which I don't know if the problem is hinting at. If you did not transcribe the problems faithfully, this may matter (no shame if you did not or you are working from another language - this happens all the time in math, and while it causes confusion, people get over it quickly).

If you want a bounded periodic solution to Laplace's equation in the plane (which is to say, it solves the equation everywhere, not just in tiles of the unit square), then you can invoke the Liouville theorem, which says that bounded harmonic functions are constants only.

Similarly, if you want a bounded periodic function $u$ that solves $\Delta u = f$ everywhere with $f$ periodic, you need to have $\int_S f = 0$ where $S$ is the unit square, you see this by considering the integral
$$ 0 = \int_{\partial S} u_\nu = \int_S \Delta u = \int_S f$$
where the first equality follows from the periodicity of $u$ and the remainder are integration by parts.


问:
How did $e^x$ get equal to $e^{2x}$?
$$  \begin{vmatrix}
e^x \, \cos(2x) & e^x \, \sin(2x) \\
e^x \, \cos(2x)- 2 e^x \, \sin(2x) & e^x \, \sin(2x)+ 2 e^x \, \cos(2x) \\
\end{vmatrix}  $$

I am finding Wronskian examples to work with and in this particular. I'm getting a different answer which is $2e^{2x}$. However, the example used $2e^x$. I rechecked it and it still gives me the same answer.  I don't know where i went wrong somewhere I know this is basic for some but if  $2e^{2x}$ is correct then is there an explanation for this? Can $e^x$ be used in place of $e^{2x}$? Thank you.

答:
The determinant is linear in its rows (or its columns). This means that you can extract a common factor of each single row or each single column. For a scaled matrix this gives one factor for each row, that is
$$
\det(cA)=c^n\det(A)
$$
when $A$ is a $n\times n$ matrix. Here $n=2$ and $c=e^x$ with no other exponential terms in the remaining matrix, so that the result has a factor $e^{2x}$.

Title: How did $e^x$ get equal to $e^{2x}$?


答: Imagine the graph of the polynomial $f(x)$. Here $f'(x)$ is equal to $0$ for two real points and let them be at $a$ and $b$. Observe that the $f(a).f(b)<0$ iff it has three distinct real roots as then $f(a)$ and $f(b)$ have opposite signs(visualize). Also $f(a).f(b)=0$  implies multiple real roots.  So the range of alpha can be derived from $f(a)f(b)\leqslant0$.

Body:
Find the max and min values of $\alpha$ for which $x^3-5x^2+8x+\alpha$ has 3 real roots.

I noticed that if we have 3 different roots then the set of solutions for $\alpha$ is open i.e no min/max.

I'm not sure how to approach the problem when also considering roots with degree $> 1$.

Maybe we could define $F(x,y,z,w,\alpha)= \alpha$

And using lagrange multipliers look for an ext under constraints:

$x^3-5x^2+8x+\alpha = 0$,

$(x-y)(x-z)(x-w)=0$

But this seems very unpleasant...

Would appreciate hints.


问:
A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$
$$\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx~~~~(1)$$
Several difficult integral can be handled using (1) easily. For instance we know that $$\int_{-\infty}^{\infty} \text{sech}^2 x~dx=2~~~~~~~~~~~~(2)$$ but what is interesting is that one may verify numerically that $$\int_{-\infty}^{\infty} \text{sech}^2(x-a/x)~dx=2, a>0.~~~~~~~~~(3)$$ This interesting property (1) follows from Glasser's Master Theorem. See one more interesting integral and its solutions in MSE post:s
https://en.wikipedia.org/wiki/Glasser%27s_master_theorem.

In (1) what is also interesting is that LHS is independent of $a(>0)$. I have tried proving (1) or even (3) by the ordinary rules of integration ( Real Analysis) without a success. Can someone help me?

EDIT: One more friendly (doable by hand) twin is
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^2}=\pi=\int_{-\infty}^{\infty} \frac{dx}{1+(x-a/x)^2} dx , a >0$$

答:
Since the integrand is even, we have
$$\mathcal{I} \stackrel{def}{=}\int_{-\infty}^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} dx
= 2\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})}dx$$
In one copy of the integral on RHS, change variables to $y = \frac{a}{x}$, one get
$$\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} =
\int_0^\infty \frac{1}{\cosh^2(\frac{a}{y} - y)} \frac{a}{y^2}dy$$
Renaming $y$ back to $x$ and add back to another copy of integral on RHS, we get
$$\mathcal{I} = \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\left(1 + \frac{a}{x^2}\right)dx
= \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\frac{d}{dx}\left(x - \frac{a}{x}\right) dx$$
Change variable to $z = x - \frac{a}{x}$, this becomes
$$\mathcal{I} = \int_{-\infty}^\infty \frac{dz}{\cosh^2(z)}$$
i.e. transform your integral $(3)$ to integral $(2)$, the one you already know.

Let's look at the more general identity $(1)$.

As you increases $x$ from $-\infty$ to $0$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ once. If one further increases $x$ from $0$ to $\infty$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ the second time.
For any $t \in \mathbb{R}$, let $x_1(t) < 0$, $x_2(t) > 0$ be the two roots of
$$x - \frac{a}{x} = t \equiv x^2 - tx - a = 0$$
This is a quadratic equation in $x$. By Vieta's formula, we have

$$x_1(t) + x_2(t) = t \implies x_1'(t) + x_2'(t) = 1$$

If you change variable to $t = x - \frac{a}{x}$ for both $(-\infty,0)$ and $(0,\infty)$, we obtain:

$$\begin{align}
 \int_{-\infty}^\infty f(x - \frac{a}{x}) dx
&= \left(\int_{-\infty}^0 + \int_0^\infty\right) f(x - \frac{a}{x})dx\\
&= \int_{-\infty}^{\infty} f(t) x'_1(t) dt +
 \int_{-\infty}^{\infty} f(t) x'_2(t) dt\\
&= \int_{-\infty}^\infty f(t) (x'_1(t) + x'_2(t))dt\\
&= \int_{-\infty}^\infty f(t)dt
\end{align}$$

This is the identity $(1)$ we seek.


问:
Showing that a polynomial in $k[x_1, \dots, x_n]$ is irreducible
Let $p\in k[x_1, \dots, x_n]$ such that p does not involve the variable $x_i$ and is not square of any polynomial. Then we want to show that the polynomial $f = x_i^2-p$ is irreducible.

I tried assuming otherwise, assuming $f=gh$ for some non-constant polynomials $g,h\in k[x_1, \dots, x_n]$. Then differentiating both sides with respect to $x_i$, we get

\begin{equation} \frac{\partial f}{\partial x_i} = \frac{\partial (x_i^2-p)}{\partial x_i} = 2x_i = \frac{\partial g}{\partial x_i}h +g\frac{\partial h}{\partial x_i}
\end{equation}
This should somehow lead to a contradiction with the fact that $p$ is not square of any polynomial but I couldn't get further from here on.

答:
Down to earth argument. Assume that $f=gh$. View $f,g,h$ as elements of $R[x_i]$, where $R=k[x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n]$.

Then $\deg(f)=2=\deg(f)+\deg(h)$, so $\deg(g)=0,1,2$ and $\deg(h)=2,1,0$.

Note that comparing leading terms shows that the leading terms of $g,h$ are in $R^\times=k$, hence we can assume wlog that $g,h$ are monic.

In this case, if $g$ or $h$ is constant, it is equal to $1$, hence invertible, and we are done.

So we may asume that $g,h$ are monic of degree $1$. Write $g=x_i-a,h=x_i-b, a,b\in R$.

Then $f=x_i^2-p=gh=x_i^2-(a+b)x_i+ab$, so $b=-a$, and $-p=ab=-a^2$, so $p=a^2$, contradiction.

Commutative algebra argument.  $R=k[x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n]$ is a UFD, and $f$ is monic in $x_i$, so $f$ is irreducible over $R$ if and only if it is irreducible over $K=Frac(R)$.

Since $f$ has degree $2$, $f$ is irreducible over $K$ if and only if it has a root in $K$, that is, if and only if $p$ is a square in $K$.

So we have an equality $pv^2=u^2$, where $u,v\in R$ are coprime. If $v$ is not a unit of $R$,  it is divisible by an irreducible element $\pi$,  but then $\pi$ divides $u^2$, so it divides $u$, contradiction. Hence $v$ is a unit, and $p$ is a square in $R$, contradiction again.


问:
Finding point on a plane closest to a point in $\mathbb{R}^n$ using least squares method?
Suppose S is a 2d plane in $\mathbb{R}^3$ s.t. it is the set of all vectors in $\mathbb{R}^3$ with $ax_1+bx_2=0$ (a,b not equal to 0). Let $b=(x_1,x_2,x_3)^T$ be any vector in $\mathbb{R}^3$. How can I use least squares fitting to find the point in S that's closest to b? I understand it will be the projection, but how does least squares  come into play? Do I have to use formula $x_{min}=(A^TA)^{-1}A^Tb$?

答:
Let us first consider the case when it is required to find the point on the
$$\tag{1}
P:\;ax+by+cz=d
$$
closest to the origin. The coordinates of the point $Q$ on the plane described by (1) nearest to the origin are equal to the minimal norm solution of (1):
$$\tag{2}
\left(\begin{array}{c}
x_Q\\y_Q\\z_Q
\end{array}\right)=
\left(\begin{array}{ccc}
a& b &c
\end{array}\right)^+\cdot d
$$
Now consider the case when it is required to find the point on the plane
(1)
closest to some arbitrary point $M(x_M,y_M,z_M)$. After the change of variables
$$
x'=x-x_M,\quad y'=y-y_M,\quad z'= z-z_M
$$
$M$ tranforms into the origin and (1) transforms into
$$
a(x'+x_M)+b(y'+y_M)+c(z'+z_M)=d
$$
or
$$
ax'+by'+cz'=d-ax_M-by_M-cz_M.
$$
According to (2), we obtain
$$
\left(\begin{array}{c}
x'_Q\\y'_Q\\z'_Q
\end{array}\right)=
\left(\begin{array}{ccc}
a& b &c
\end{array}\right)^+\cdot (d-ax_M-by_M-cz_M)
$$
and
$$
x_Q= x'_Q+x_M,\quad y_Q= y'_Q+y_M,\quad z_Q= z'_Q+z_M.
$$
Since $\left(\begin{array}{ccc}
a& b &c
\end{array}\right)$ has linearly independent rows,
$$
\left(\begin{array}{ccc}
a& b &c
\end{array}\right)^+=
\left(\begin{array}{ccc}
a& b &c
\end{array}\right)^T
(\left(\begin{array}{ccc}
a& b &c
\end{array}\right)\left(\begin{array}{ccc}
a& b &c
\end{array}\right)^T)^{-1}
=
\frac1{a^2+b^2+c^2}
\left(\begin{array}{c}
a\\b\\c
\end{array}\right),
$$
thus,
$$
\left(\begin{array}{c}
x_Q\\y_Q\\z_Q
\end{array}\right)=
\frac{d-ax_M-by_M-cz_M}{a^2+b^2+c^2}
\left(\begin{array}{c}
a\\b\\c
\end{array}\right)+
\left(\begin{array}{c}
x_M\\y_M\\z_M
\end{array}\right).
$$


问:
Non-Noetherian ring with a single prime ideal

What are the most simple examples of a commutative ring $R$ satisfying both of the following two properties:

  1. $R$ is not Noetherian.

  2. $R$ has exactly one prime ideal.

Body: Try $k[x_1, x_2, ...]/(x_i x_j)$ for all $i, j$.


问:
Probability distribution, finding Cumulative distribution function
I have a quick question.
If
$$
P(X=n)=\frac{1}{2^{n+1}}
$$
and I want to find CDF should I start with $n=0$? I think that then this probability is going to be closer to $0$, and if $n<0$ then the limit would be $1$. I don't know nice way of writing this formally though.

答:
Your pmf is the following

$$\mathbb{P}[X=n]=\Big(\frac{1}{2}\Big)^n\cdot\frac{1}{2}$$

We immediately recognize a Geometric distribution counting the failures before the first success, thus the support is $n \in\{0;1;2;...\}$

The CDF is the following

$$\mathbb{P}[X\leq n]=\frac{1}{2}\sum_{x=0}^n\Big(\frac{1}{2}\Big)^x=\frac{1}{2}\cdot\frac{1-\Big(\frac{1}{2}\Big)^{n+1}}{1-\frac{1}{2}}=1-\Big(\frac{1}{2}\Big)^{n+1}$$


问:
How to show the second fundamental form of a flat minimal surface is zero?
The surface is flat if its Guassian curvature vanishes everywhere.
The surface is minimal if ist mean curvature vanishes everywhere.
Therefore, for a function f(x,y), its corresponding surface's Gaussian curvature and mean curvature are both 0.

My question is how to show its second fundamental form
Ldu^2 + 2Mdudv + Ndv^2 = 0?

答:
HINT:

If you meant $r(u,v)$ then due to orthogonality between normal N and tangent vectors we have

$$ N \cdot r_1=0 $$

Differentiate w.r.t. $u$ we have by Chain Rule $ \; N\cdot r_{11} + N_1 \cdot r_{1} =0 $

and  since $ (N_1, r_1) $ are orthogonal their dot product vanishes leaving $ (r_{11}\cdot N) =L=0$. Differentiation w.r.t. $v$ leaves  $M=0$.

Similarly $ N \cdot r_2=0 $, again differentiate w.r.t $v$ in $ N\cdot r_{22} +N_2\cdot r_{2}   $ second  term vanishes  leaving the first term $ (M= r_{22}\cdot N=0 )$  so that

$$ L  du^2+ 2 M du dv + N dv^2 =0 ;\ $$

Moreover in the plane i.e., principal scalar curvatures  are all zero

$$ \{ k_1k_2=0, k_1+k_2=0\} \to \{k_1=0,k_2=0\} .$$


问:
Essential singularity of $h = f \cdot g$.
I want to proof the following claim

Let $a \in \mathbb{C}$. Consider a function $f$, where $f$ is
holomorphic on $\mathbb{C}\setminus\{a\}$ and $a$ being an essential
singularity of $f$. Moreover let $g$ be an entire function such that
$g(a) \neq 0$. Then $h = f\cdot g$ has an essential singuarity at $z = a$.

My attempt:

$f$ has an essential singularity if $\lim_{z \to a} |(z-a)^k\cdot f(z)| = +\infty$ does not exist for every $k \in \mathbb{N}$. Otherwise we would have a pole of order $k$, if $k$ is the smallest number such that the limit exists.

Then
\begin{align}
\lim_{z \to a} |(z-a)^k \cdot h(z)| &= \lim_{z \to a} |(z-a)^k \cdot f(z) \cdot g(z)|\\ & = \lim_{z \to a} |(z-a)^k \cdot f(z)| \cdot \lim_{z \to a} |g(z)| \\ &= \underbrace{\lim_{z \to a} |(z-a)^k \cdot f(z)|}_{\infty} \cdot \underbrace{|g(a)|}_{\neq 0} \\ &= +\infty
\end{align}
for every $k \in \mathbb{N}$. Hence, $h$ has to have an essential singularity at $z = a$.

I am unsure about whether that is enough to proof the claim or not. Can someone confirm my attempt or give a hint how the statement can be proven?

Edit:

I am particulary unsure about the second step where I write

\begin{align}
\lim_{z \to a} |(z-a)^k \cdot f(z) \cdot g(z)| = \lim_{z \to a} |(z-a)^k \cdot f(z)| \cdot \lim_{z \to a} |g(z)|.
\end{align}
Am I allowed to do this here?

答:
Your solution seems fine.

The step
$$\lim_{z\to L}f(z)g(z)=\lim_{z\to L}f(z)\lim_{z\to L}g(z)$$
is viable whenever the right hand side leads you to a determinable expression, which is the case here.


问:
Plucker's Formula proof
I just read a proof of Plucker's Formula in Rick's Miranda book that says

A smooth projective plane curve of degree $d$ has genus $\frac{(d-1)(d-2)}{2}$.

To do this he defines map $\pi : X \rightarrow P^1$ such that $\pi[x:y:z]=[x:z]$ and he claims the map has degree $d$, and this is where i have my doubt, i know this is true if $X$ is given by a a function $F$ of the form $x^d+y^d+z^d$ so maybe this is true because we can do some kind of change of variable and assume this ?  Also i have tried to see that its degree $d$ by computing it but got nowhere, since we know $\pi^{-1}[1:0]=[1:y:0]$ such that $F[1:y:0] = 0$ we will have $d$ solutions to this and we know that a point is a ramification point for this map iff $\frac{\partial F}{\partial y}(p)=0$, and we know that $F[x:y:0]=\frac{1}{d}(\frac{\partial F}{\partial x} + y\frac{\partial F}{\partial y})$ but I cant seem to find a reason why $\frac{\partial F}{\partial y} \neq 0$ so any help is welcomed. Or is it because by changing coordinates we can assume that the smooth projective curve will be given by $x^d+y^d+z^d=0$?

Also another thing I dont fully understand, when proving that the degree of a smooth projective curve $X$ will be the degree of an hyperplane divisor we assume that the hyperlane divisor will be given by $G(x,y,z)=x$ and that $[0:0:1]$ is not in $X$, my question is why can we assume both of these things ? I get that by changing the coordinates we could get one but I dont see how we could get the two withouse losing some generality. Thanks in advance!

答:
To find the degree of $\pi$, consider a general point $[\lambda:1] \in \mathbb{P}^1$ i.e. a non-branch point. The preimages of $[\lambda:1]$ would be $[\lambda:y:1]$ such that $F(\lambda,y,1)=0 $. Notice that the point $[0:1:0]$ has been removed from $X$ by a change of coordinates, otherwise the map $\pi$ wouldn't make sense.

Hence, $F(\lambda,y,1)$ is a degree $d$ polynomial in $y$ and therefore has $d$ roots. Let $f(y) = F(\lambda,y,1)$. Note that
$$ \frac{\partial F}{\partial y}(\lambda,y,1) = \frac{\partial f}{\partial y}(y)$$
and, $y_0$ is a repeating root of $f \implies \partial f/\partial y(y_0)=0 \implies \partial F/\partial y(\lambda,y_0,1)=0 \implies [\lambda:y_0:1]$ is a ramification point of $\pi$.

Hence, all the roots of $f$ are distinct and therefore $\pi$ has degree $d$.

For the second part, you can first make a coordinate transformation so that $G(x,y,z)=x$ and then follow up with another transformation of just the $yz$ plane to remove the point $(0,1)$ from the roots of $F(0,y,z)$.

Title: Plucker's Formula proof


问:
Limiting distributions of non-overlapping sums are independent?
I'm being stuck in the following statement in Billingsley's Convergence of Probability Measures. Let $[・]$ denote the integer part of the argument. Assume that $\{\xi_n\}$ is a sequence of iid r.v.'s with mean zero and variance finite. Define $S_n = \xi_1 + \cdots + \xi_n.$ Define also $X_t^n$ as $$ X_t^n =  \frac{1}{\sqrt{n}}S_{[nt]} + (nt - [nt])\xi_{[nt] + 1}.$$ Then, we have the following: for $s \leq t$,  $$(X_s^n, X_t^n - X_s^n) \Rightarrow (N_1, N_2) \ \ (n \to \infty),$$ where $N_1$ and $N_2$ are independent and normal.

I have two questions:

Why are $N_1$ and $N_2$ independent?

Why can we say that they jointly converge in distribution?

I notice that $X_s^n$ and $X_t^n - X_s^n$ are non-overlapping partial sums (up  to the interpolating term) of the independent sequence, so that they are independent for all $n$. I know from the CLT that each of the partial sums converges in distribution. But I also know that even if $\{X_n\}$ and $\{Y_n\}$ are two independent sequences(i.e., $X_n$ and $Y_n$ are independent for all $n$), their limiting distributions are not necessarily independent. So I can't see why $N_1$ and $N_2$ are independent under the above conditions.

I'm also wondering about the joint convergence of $X_s^n$ and $X_t^n - X_s^n$. I have only seen the Lindeberg-Levy CLT states that a single partial sum converges in distribution. Or can we derive the joint convergence of non-overlapping partil sums?

Thanks a lot !

Edit: I think question 2 will be solved if the weak limits of $X_s^n$ and $X_t^n - X_s^n$ are independent. So, an answer only to question 1 will be of great help.

答:
As you guessed, the fact that we obtain at the limit a vector of independent random variables comes from this special setting. To see this, we use the Cramer-Wold device: we have to show that for each real numbers $a$ and $b$, $aX_s^n+b(X_t^n-X_s^n)$ converges in distribution to $aN_1+bN_2$, where $N_1$ and $N_2$ are independent
normal. Since $N_1$ and $N_2$ are Gaussian and independent, $aN_1+bN_2$ has a normal distribution with mean zero and variance $a^2s +b^2(t-s) $.

One can show that $aX_s^n+b(X_t^n-X_s^n)$ behave like $Y_n:=\frac1{\sqrt n}\left(aS_{[ns]}+b(S_{[nt]}-S_{[ns]})\right)$ and a use of the central limit theorem under Lindeberg's conditions for an array of rowwise independent random variables shows that the former expression converges to a Gaussian random variable whose limit is the limit of the variance of $Y_n$, which is indeed $a^2s +b^2(t-s) $.


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

答:
This is an overkill solution, but this is an immediate application of Cauchy-Schwarz:

$$\left( 1 \cdot \sqrt{1001}+1 \cdot\sqrt{999}\right)^2 \leq (1+1)(1001+999)=4000\,.$$


答:
$u=x-\frac ax\Rightarrow du=(1+\frac a{x^2})dx$ and so:
$$\int_{-\infty}^\infty f(x-a/x)dx=\int_{-\infty}^\infty\frac{f(u)}{1+\frac a{x^2}}dx$$
now we need to get this expression on the bottom in terms of $u$.
$$\frac{f(u)}{1+a/x^2}=\frac{x^2f(u)}{x^2+a}=f(u)-\frac{a^2}{x^2+a}f(u)$$
So now the challenge we face is showing that:
$$\int_{-\infty}^\infty\frac{a^2}{x^2+a}f(u)du=0$$
we can start by trying to undo the substitution:
$$\int_{-\infty}^\infty\frac{a}{x^2+a}f(x-a/x)(1+a/x^2)dx=a\int_{-\infty}^\infty\frac{1+a/x^2}{x^2+a}f(x-a/x)dx$$
$$=a\int_{-\infty}^\infty\frac1{x^2}\frac{x^2+a}{x^2+a}f(x-a/x)dx=a\int_{-\infty}^\infty\frac{f(x-a/x)}{x^2}dx$$
if we try letting $x\to-x$ and combine we get:
$$I_1=\frac a2\int_{-\infty}^\infty\frac{f(-x+a/x)+f(x-a/x)}{x^2}dx=\frac a2\int_{-\infty}^\infty\frac{f(-(x-a/x))+f(x-a/x)}{x^2}dx$$
now we can try and make our substitution again:
$$u=x-a/x$$
$$I_1=\frac a2\int_{-\infty}^\infty\frac{f(u)+f(-u)}{x^2}\frac{du}{1+a/x^2}dx=\frac a2\int_{-\infty}^\infty\frac{f(u)+f(-u)}{x^2+a}du$$
Now notice how this function is symmetric i.e. $I_1=-I_1\therefore I_1=0$

Title: A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$

Body:
$$\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx~~~~(1)$$
Several difficult integral can be handled using (1) easily. For instance we know that $$\int_{-\infty}^{\infty} \text{sech}^2 x~dx=2~~~~~~~~~~~~(2)$$ but what is interesting is that one may verify numerically that $$\int_{-\infty}^{\infty} \text{sech}^2(x-a/x)~dx=2, a>0.~~~~~~~~~(3)$$ This interesting property (1) follows from Glasser's Master Theorem. See one more interesting integral and its solutions in MSE post:s
https://en.wikipedia.org/wiki/Glasser%27s_master_theorem.

In (1) what is also interesting is that LHS is independent of $a(>0)$. I have tried proving (1) or even (3) by the ordinary rules of integration ( Real Analysis) without a success. Can someone help me?

EDIT: One more friendly (doable by hand) twin is
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^2}=\pi=\int_{-\infty}^{\infty} \frac{dx}{1+(x-a/x)^2} dx , a >0$$


答:
Your proof is fine because $\lim_{z \to a} |g(z)| = |g(a)| \ne 0$. It might be simpler to argue in the opposite direction. If $h$ has a non-essential singularity at $z=a$ then
$$
c = \lim_{z \to a} (z-a)^k \cdot h(z)
$$
exists for some non-negative integer $k$. It follows that
$$
(z-a)^k \cdot f(z) = \frac{(z-a)^k \cdot h(z)}{g(z)} \to \frac{c}{g(a)}
$$
for $z \to a$, so that $f$ has a non-essential singularity at $z=a$, contrary to the assumption.

Title: Essential singularity of $h = f \cdot g$.

Body:
I want to proof the following claim

Let $a \in \mathbb{C}$. Consider a function $f$, where $f$ is
holomorphic on $\mathbb{C}\setminus\{a\}$ and $a$ being an essential
singularity of $f$. Moreover let $g$ be an entire function such that
$g(a) \neq 0$. Then $h = f\cdot g$ has an essential singuarity at $z = a$.

My attempt:

$f$ has an essential singularity if $\lim_{z \to a} |(z-a)^k\cdot f(z)| = +\infty$ does not exist for every $k \in \mathbb{N}$. Otherwise we would have a pole of order $k$, if $k$ is the smallest number such that the limit exists.

Then
\begin{align}
\lim_{z \to a} |(z-a)^k \cdot h(z)| &= \lim_{z \to a} |(z-a)^k \cdot f(z) \cdot g(z)|\\ & = \lim_{z \to a} |(z-a)^k \cdot f(z)| \cdot \lim_{z \to a} |g(z)| \\ &= \underbrace{\lim_{z \to a} |(z-a)^k \cdot f(z)|}_{\infty} \cdot \underbrace{|g(a)|}_{\neq 0} \\ &= +\infty
\end{align}
for every $k \in \mathbb{N}$. Hence, $h$ has to have an essential singularity at $z = a$.

I am unsure about whether that is enough to proof the claim or not. Can someone confirm my attempt or give a hint how the statement can be proven?

Edit:

I am particulary unsure about the second step where I write

\begin{align}
\lim_{z \to a} |(z-a)^k \cdot f(z) \cdot g(z)| = \lim_{z \to a} |(z-a)^k \cdot f(z)| \cdot \lim_{z \to a} |g(z)|.
\end{align}
Am I allowed to do this here?


问:
In the real quaternion ring, for any $t$ where $t^2=-1$, there exists a $s$ such that $sts^{-1} = i$
While studying the notions of Extension fields and Splitting fields, I came across to handle the following question:

Consider the real quaternion ring $R$. Then for any $t \in R$ satisfying $t^2=-1$, prove that we can find a $s \in R$ such that $sts^{-1} = i$.

First I have found that, every $t \in R$ being a square of $-1$ in $R$ is pure imaginary unit quaternion; that is
\begin{gather*}
t = a_1 i + a_2 j + a_3 k, \quad a_1^2 + a_2^2 + a_3^2 =1
\end{gather*}
for any real $a_1, a_2, a_3 \in \mathbb{R}$. And I also have found that we can isomorphically identify the real quaternions with the elements in $M_2(\mathbb{C})$ as
\begin{gather*}
a_0 + a_1 i + a_2 j + a_3 k \iff \begin{pmatrix} a_0+a_1i & a_2+a_3i \\ -a_2+a_3i & a_0-a_1i\end{pmatrix}.
\end{gather*}
Then I merely tried to make a matrix calculation if what kind of element would $s^{-1}is$ be. By identifying $i \in R$ with $\begin{pmatrix} i &0 \\ 0 & -i\end{pmatrix}$,  I came to conclude that $t$ satisfying $t=s^{-1}is$ must be $-i$. But, this is definitely not the question wanted. So I have some doubts; whether I am misunderstanding the question or there is another brilliant way to tackle the problem.

If it was to find some $t$ (such that $t^2=-1$) and $s \in R$ satisfying $sts^{-1} =i$, the problem would be too easy. I like to get some advice. Please help!

答:
It was a good start that you figured out that we must have
$$
t=a_1i+a_2j+a_3k
$$
for some triple $(a_1,a_2,a_3)\in\Bbb{R}^3$ such that $a_1^2+a_2^2+a_3^2=1$. To find a suitable $s$ we do the following.

Case 1. If $a_2=a_3=0$ the task is easy. We must have $t=\pm i$. If $t=i$ we can use $s=1$. If $t=-i$ we can use $s=j$ as $jij^{-1}=-i$.

Case 2. Either $a_2\neq0$ or $a_3\neq0$. In this case the vectors $i$ and $t$ are linearly independent. Therefore their cross product $i\times t$ is a non-zero vector.
Furthermore, the angle $\theta$ between $i$ and $t$ is in the interval $\theta\in(0,\pi)$. Here we can use the following trick (well known to people using quaternions when implementing 3D-rotations in e.g. computer graphics, see also runway44's comment).
Let $u$ be the vector $i\times t$ scaled to have unit length. Then
$$
s=\cos(\theta/2)-u \sin(\theta/2)
$$
will work. This is because the linear transformation $v\mapsto svs^{-1}$ is the rotation about the axis $u$ by angle $\theta$. Remark: More often the sources uses
$$
s^{-1}=\cos(\theta/2)+u\sin(\theta/2),
$$
but then the rotation would map $i$ to $t$ rather than the other way arouns as we want here.

Example. Let $t=j$. We know that $i\times j=k$, so $u=k$. We know also that the angle between $i$ and $j$ is $\theta=\pi/2$. The recipe gives
$$
s=\cos\frac{\pi}4-k\sin\frac{\pi}4=\frac{1-k}{\sqrt2}.
$$
This has unit norm, so we know that
$$
s^{-1}=\overline{s}=\frac{1+k}{\sqrt2}.
$$
We can verify that
$$
\begin{aligned}
sjs^{-1}&=\frac1{\sqrt2}(1-k)j\frac1{\sqrt2}(1+k)\\
&=\frac12(j-kj)(1+k)\\
&=\frac12(j+i)(1+k)\\
&=\frac12(j+jk+i+ik)\\
&=\frac12(j+i+i-j)\\
&=i
\end{aligned}
$$
as prescribed.


问:
When does equality holds in $A\subseteq P(\cup A)$
Note: $P$ is power set. It's easy to prove that this inclusion holds. But when is other inclusion true? I can't even think of one example...</p>

答: If equality holds, then $A=\mathcal P(X)$ for some set $X$, namely $X=\bigcup A$. Conversely, note that if $A=\mathcal P(X)$, then $\bigcup A=X$. That is: $A=\mathcal P(\bigcup A)$ iff $A$ is the power set of some set $X$, in which case $X=\bigcup A$.</p>


问:
Inclusion Map $i : A \to X $ i(a) =a is continous for all a $a\in A$
While trying my assignment questions on Topology I am struck on the following assignment problem .

Let $(A, T_A) $ be a subspace of topological space (X,T). Prove that inclusion Map $i : A \to X $ i(a) =a is continuous for all a $a\in A$.

Well, The definition of continuity  is defined for open set and not for any element as given in the question. So, I am confused on how to prove continuos as defining open set by the help of element has onlly one meaning( in my opinion) that I took an open set A' whose element is a' ( There could be many open sets , I chose any 1) , now I have to prove that $i^{-1} A' \in A'$ but by i(a)=a , as i is 1-1 and for each a there exists a st i(a) =a , in my case there exists an a' st i(a') =a' and a' will belong to A'. So, for each A'$\subseteq X$, there exists an A'$\subseteq X$  and hence i is continuous.

this was my thinking. kindly tell what is my mistake in case i am wrong .

答: If you take an open subset O of X (i.e. a set from T)  you can disjointly devide it into $O\cap A$ and $O\cap A^c$. Then the latter part doesn't have a preimage, thus $i^{-1}(O\cap A^c)$ equals the empty set. Since $O\cap A$ is an open set (by definition of the subspace topology), and $i^{-1}(O\cap A)=O\cap A$, you know that $i^{-1}(O)=\emptyset\cup (O\cap A)$ is an open set.


问:
Give an example of a local domain with principal maximal ideal, which is not a valuation ring

Give an example of a local domain with principal maximal ideal, which is not a valuation ring.

If we had Noetherianness on top of it, we'd have that it's a DVR; but without it, I have no idea how to proceed. This might be a potential example, but I can't make it work.

答:
$$R=\Bbb{Z}[x,x/p,x/p^2,\ldots,y,y/p,y/p^2,\ldots]$$
Then $(p)=(p,x,x/p,x/p^2,\ldots,y,y/p,y/p^2,\ldots)$ is a principal maximal ideal and the unique one of $R_{(p)}$.

In $R_{(p)}\subset \Bbb{Q}[x,y]_{(x,y)}$,

$(x)\not\subset (y)$ and $(y)\not\subset (x)$.


问:
Renewal function with discrete interarrival times evaluated at non-integer points

Of this question, how do I do approach part d)? I do not recall any continuity results which I could use for this..

答:
Since the interarrival time $X_i$ are positive integers, there will be jumps on integral time point only, and thus

$$ N_{t+\varepsilon} = N_t $$

To write this out more explicitly, note that the relationship between the counting process and interarrival time

$$ N_t \leq n \iff \sum_{i=1}^{n+1} X_i > t $$

Therefore

$$ N_{t+\varepsilon} \leq n \iff \sum_{i=1}^{n+1} X_i > t + \epsilon $$

Moreover,

$$ \Pr\left\{\sum_{i=1}^{n+1} X_i > t\right\}
= \Pr\left\{t < \sum_{i=1}^{n+1} X_i \leq t + \varepsilon \right\} +
\Pr\left\{\sum_{i=1}^{n+1} X_i > t + \varepsilon \right\}$$

As $X_i$ are integer, $\displaystyle \sum_{i=1}^{n+1} X_i$ is also an integer, So for $t \in \mathbb{N}$,

$$ \Pr\left\{t < \sum_{i=1}^{n+1} X_i \leq t + \varepsilon \right\} = 0$$

As a result, for $t \in \mathbb{N}$,

$$ \Pr\left\{\sum_{i=1}^{n+1} X_i > t\right\} = \Pr\left\{\sum_{i=1}^{n+1} X_i > t + \epsilon\right\} $$

$$ \Pr\{N_t \leq n\} = \Pr\{N_{t+\epsilon} \leq n\}$$

$$ E[N_t] = E[N_{t+\epsilon}] $$


答:
I feel like what you should prove is that, for every open set $O\in T$, its preimage $i^{-1}(O)\in T_A$.

This is almost trivial.

Let $O\in T$. Then $i^{-1}(O)=O\cap A$. Because $O$ is open in $X$, by the definition of subspace topology, we know $i^{-1}(O)$ is open in $A$.

Title: Inclusion Map $i : A \to X $ i(a) =a is continous for all a $a\in A$

Body:
While trying my assignment questions on Topology I am struck on the following assignment problem .

Let $(A, T_A) $ be a subspace of topological space (X,T). Prove that inclusion Map $i : A \to X $ i(a) =a is continuous for all a $a\in A$.

Well, The definition of continuity  is defined for open set and not for any element as given in the question. So, I am confused on how to prove continuos as defining open set by the help of element has onlly one meaning( in my opinion) that I took an open set A' whose element is a' ( There could be many open sets , I chose any 1) , now I have to prove that $i^{-1} A' \in A'$ but by i(a)=a , as i is 1-1 and for each a there exists a st i(a) =a , in my case there exists an a' st i(a') =a' and a' will belong to A'. So, for each A'$\subseteq X$, there exists an A'$\subseteq X$  and hence i is continuous.

this was my thinking. kindly tell what is my mistake in case i am wrong .


问:
Formal definition of a monotonically increasing function
My professor is working with the definition that a function is monotonically increasing when $x_1<x_2. \implies f(x_1)<f(x_2)$. Is this definition correct?

And most importantly, if we already know that a function is increasing, and we know that $f(x_1)<f(x_2)$, can we prove that $x_1<x_2$?

Intuitively, we should be able to show this, but I lack the mathematical maturity to do it. Can somebody please help me?

Thanks.

答:
The most widely accepted definitions for an increasing function are the following:

A (monotonically) increasing function, which says that $x_1 \leq x_2 \implies f(x_1) \leq f(x_2)$.

A strictly (monotonically) increasing function, which says that $x_1 < x_2 \implies f(x_1) < f(x_2)$.

These two definitions are not the same, as any constant functions are increasing but not strictly increasing. Your definition would fit into the second one.

As for your second claim, indeed it holds. The contrapositive of $x_1 < x_2 \implies f(x_1) < f(x_2)$ asserts that $f(x_1) \geq f(x_2) \implies x_1 \geq x_2$. If $f(x_1) < f(x_2)$ is known, then we must have $x_1 \leq x_2$ by the contraposition, i.e. $x_1 = x_2$ or $x_1 < x_2$. But $x_1 = x_2$ can't be true, for otherwise we have that $f(x_1) = f(x_2)$, so we indeed have $x_1 < x_2$.

EDIT: OP asked a follow-up question in the comment on why in this case we are able to prove the bi-implication from just the implication alone here, but not in general. To illustrate this, we first note that more precisely the definition of a strictly increasing function should say:
$$
\forall x_1 \in \mathbb{R}, \; \forall x_2 \in \mathbb{R}, \; x_1 < x_2 \implies f(x_1) < f(x_2)
$$
The reason we can prove the reverse implication stems from the following two facts:

For any real numbers $\mathbb{R}$, either $x_1 < x_2$, $x_2 < x_1$ or $x_1 = x_2$. That is, $<$ is a connex relation.

There are two arbitrary variables involved in the definition above. Thus, $x_1$ and $x_2$ are just dummy variables that can be replaced by any other symbols.

With the contraposition, we see that we obtain:
$$
\forall x_1 \in \mathbb{R}, \; \forall x_2 \in \mathbb{R}, \; f(x_2) \leq f(x_1) \implies x_2 \leq x_1
$$

Replacing the dummy variables give:
$$
\forall x_1 \in \mathbb{R}, \; \forall x_2 \in \mathbb{R}, \; f(x_1) \leq f(x_2) \implies x_1 \leq x_2
$$

The two steps done above do not work for any general statements but are rather very special property of the relation $<$.


问:
Proof for an integral identity
Is it true that $\int_0^A dx \int_0^B dy f(x) f(y) = 2 \int_0^A dx \int_0^x dy f(x) f(y)$ ? If so, can this be proved?</p>

答:
My answer is essentially the same as gt6989b's, but I will leave it in case the difference in presentation is useful.

Define
$$
F(x)=\int_0^xf(y)\,\mathrm{d}y
$$
Then $f(x)=F'(x)$ and
$$
\begin{align}
2\int_0^A\int_0^x f(x)f(y)\,\mathrm{d}y\,\mathrm{d}x
&=2\int_0^Af(x)F(x)\,\mathrm{d}x\\
&=2\int_0^AF'(x)F(x)\,\mathrm{d}x\\
&=F(A)^2\\
&=\left(\int_0^Af(x)\,\mathrm{d}x\right)^2
\end{align}
$$
However,
$$
\int_0^A\int_0^B f(x)f(y)\,\mathrm{d}y\,\mathrm{d}x
=\left(\int_0^Af(x)\,\mathrm{d}x\right)\left(\int_0^Bf(y)\,\mathrm{d}y\right)
$$
Thus, your equation is true if
$$
\int_0^Af(x)\,\mathrm{d}x=\int_0^Bf(y)\,\mathrm{d}y
$$
or if
$$
\int_0^Af(x)\,\mathrm{d}x=0
$$

Title: Proof for an integral identity


答: First of all, the standard definition of (non-strict) monotonicity allows for equality, hence the definition becomes $$x_1\leq x_2 \implies f(x_1) \leq f(x_2).$$ Second of all, to answer your question, the reverse implication is actually true as well. This is immediate from the definitions: suppose the above statement holds for some $f$, and take two $x_1,x_2$ such that $f(x_1)\leq f(x_2)$. Now suppose for contradiction that $x_2 < x_1$. But then it follows from our definition of monotonicity that $f(x_2)\leq f(x_1)$. It couldn't be the case that $x_1=x_2$, so we get a contradiction. Therefore $$f(x_1)\leq f(x_2) \implies x_1 \leq x_2.$$

Title: Formal definition of a monotonically increasing function

Body:
My professor is working with the definition that a function is monotonically increasing when $x_1<x_2. \implies f(x_1)<f(x_2)$. Is this definition correct?

And most importantly, if we already know that a function is increasing, and we know that $f(x_1)<f(x_2)$, can we prove that $x_1<x_2$?

Intuitively, we should be able to show this, but I lack the mathematical maturity to do it. Can somebody please help me?

Thanks.


问:
Prove that a convolution product $(f\ast g)(x)$, with $f,g\in\mathcal{L}^2(\mathbb{R})$, goes to $0$ as $|x|$ goes to $\infty$.
I want to proof that the convolution product $(f\ast g)(x)$ of two square-integrable functions $f,g\in\mathcal{L}^2(\mathbb{R})$ goes to $0$ as $|x|$ goes to $+\infty$. The convolution product here denotes

$$(f\ast g)(x)=\int_\mathbb{R}f(y)g(x-y)dy.$$

I know that, using Hölder's inequality,

$$|(f\ast g)(x)|\leq \int_\mathbb{R}|f(y)g(x-y)|dy\leq ||f||_2||g||_2$$

but I do not immediately see how this can help me to finalise my proof.

答:
If we have:
$$I(x)=\lim_{n\to\infty}\int_{-n}^nf(y)g(x-y)dy$$
this denotes our convolution. Now we can say that we want $|x|\to\infty$ so we can write this as:
$$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{-n}^nf(y)g(x-y)dy$$
we can make the substitution $u=x-y,du=-dy$ and so:
$$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x+n}^{x-n}f(x-u)g(u)du=-\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(x-u)g(u)du$$
also:
$$I_{\infty}=\lim_{x\to\infty}\lim_{n\to\infty}\int_{-n}^nf(-y)g(x+y)dy$$
$$I_\infty=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(u-x)g(u)du=\lim_{x\to\infty}\lim_{n\to\infty}\int_{x-n}^{x+n}f(-(x-u))g(u)du$$
this suggests that $f(-(x-u))=-f(x-u)$ and since we are talking about $|x|\to\infty$ we can say that $f(-)=-f(+)$. If $g$ is an even function then we have an odd integrand over the real domain which will equal zero.


答:
To do this question, you'd better know some density theorem of $L^p$ space. We will use a theorem which shows for an $L^2(\mathbb{R})$-function, there exists a sequence of simple measurable functions $f_k$ with bounded support such that $f_k\to f$ in $L^2(\mathbb{R})$ and $f_k$'s can be chosen to be in $L^2(\mathbb{R})$.

Since simple function with bounded support is dense in $L^2(\mathbb{R})$, there exists $f_k\to f$ in $L^2$, where $f_k$'s are simple functions with bounded support in $L^2$. This shows
$$|(f*g)(x)|\leq \lVert f-f_k\rVert_2\lVert g\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g(y)\;dy\right|$$
Similarly, we can find a sequence of simple function $g_k$ with bounded support and $g_n\to g$ in $L^2$. Then,
$$\left|\int_{\mathbb{R}^n}f_k(x-y)g(y)\;dy\right|\leq \lVert f_k\rVert_2\lVert g-g_n\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g_n(y)\;dy\right|$$
Note that $f_k(x-y)g_n(y)=0$ for large enough $|x|$. This is because if the radius of the support of $f_k$ is $r_k$ and the radius of support of $g_n$ is $R_n$, then if $|x|>r_k+R_n$, either $|y|>R_n$ or $|x-y|>r_k$, so either $f_k(x-y)=0$ or $g_n(y)=0$. This implies that
$$|(f*g)(x)|\leq\lVert f-f_k\rVert_2\lVert g\rVert_2+\lVert f_k\rVert_2\lVert g-g_n\rVert_2+\left|\int_{\mathbb{R}^n}f_k(x-y)g_n(y)\;dy\right|$$
First take $|x|\to\infty$ on both sides, we have
$$\lim_{|x|\to\infty}|(f*g)(x)|\leq \lVert f-f_k\rVert_2\lVert g\rVert_2+\lVert f_k\rVert_2\lVert g-g_n\rVert_2$$
Then take $n\to\infty$ on both sides, since LHS is independent of $n$, we have
$$\lim_{|x|\to\infty}|(f*g)(x)|\leq\lVert f-f_k\rVert_2\lVert g\rVert_2$$
Finally, take $k\to\infty$ on both sides, since LHS is independent of $k$, we obtain $\lim_{|x|\to\infty}|(f*g)(x)|\leq 0$, i.e., $(f*g)(x)\to 0$ as $|x|\to \infty$.

Title: Prove that a convolution product $(f\ast g)(x)$, with $f,g\in\mathcal{L}^2(\mathbb{R})$, goes to $0$ as $|x|$ goes to $\infty$.

Body:
I want to proof that the convolution product $(f\ast g)(x)$ of two square-integrable functions $f,g\in\mathcal{L}^2(\mathbb{R})$ goes to $0$ as $|x|$ goes to $+\infty$. The convolution product here denotes

$$(f\ast g)(x)=\int_\mathbb{R}f(y)g(x-y)dy.$$

I know that, using Hölder's inequality,

$$|(f\ast g)(x)|\leq \int_\mathbb{R}|f(y)g(x-y)|dy\leq ||f||_2||g||_2$$

but I do not immediately see how this can help me to finalise my proof.


问:
Calculate $\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}$
Please help me to calculate the $$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}$$.

My attemp is:
$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}\cdot\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{\sqrt[3]{1+7x}-\sqrt{1+3x}}\cdot\frac{\sqrt[3]{1+7x}+\sqrt{1+3x}}{\sqrt[3]{1+7x}+\sqrt{1+3x}}$, but I do not know if I did well, can anyone help me

Body:
Hint:

$$\dfrac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}=\dfrac p{x-1}-\dfrac q{x-1}$$

where $\sqrt[3]{1+7x}-2=p\implies x-1=?$

and $\sqrt{1+3x}-2=q,x-1=?$


问:
Discontinuous function on the sphere
In $\mathbb{R}^n$ endowed with the euclidean norm , Let $\mathbb{S}^{n-1}$ be the unit sphere

we consider the function $f :\mathbb{S}^{n-1} \rightarrow \mathbb{R}_+$ not necessairly contiunous such that :

for each $u\in \mathbb{S}^{n-1}$,  $f(u)< +\infty $

Can we deduce that $f$  is bounded  ? ( i.e $Sup_{u\in \mathbb{S}^{n-1}}(f(u))<\infty$)

I know if $f$ is contiuous this is true (Extreme value theorem)

but what if $f$ is not continuos ?

答: This is clearly false as written. Take any bijection from $\mathbb{S}^{n-1}$ to $\mathbb{R}$. If you'd like to make it continuous almost everywhere, distinguish one point $\ast$ on $\mathbb{S}^{n-1}$. Then $\mathbb{S}^{n-1} \setminus \{\ast\}$ is homeomorphic to $\mathbb{R}^{n-1}$, and define your $f$ to one of those homeomorphisms away from $\ast$, and $f(\ast) = 0$.

Title: Discontinuous function on the sphere


答:
As a counter example consider any fixed point $s \in \mathbb{S}^{n-1}$ and define

$$f(u):=\begin{cases} \frac{1}{||u-s||} & \text{for $u\neq s$} \\ 0 & \text{for $u=s$}\end{cases}.$$

Title: Discontinuous function on the sphere

Body:
In $\mathbb{R}^n$ endowed with the euclidean norm , Let $\mathbb{S}^{n-1}$ be the unit sphere

we consider the function $f :\mathbb{S}^{n-1} \rightarrow \mathbb{R}_+$ not necessairly contiunous such that :

for each $u\in \mathbb{S}^{n-1}$,  $f(u)< +\infty $

Can we deduce that $f$  is bounded  ? ( i.e $Sup_{u\in \mathbb{S}^{n-1}}(f(u))<\infty$)

I know if $f$ is contiuous this is true (Extreme value theorem)

but what if $f$ is not continuos ?


问:
No. Of LI Solutions to a Non homogeneous system of equations
Let A be an n order matrix with Nulity m . I know the number of linearly independent solutions to the homogeneous equation is  m .
But what will be the number of linearly independent solutions if the system is nonhomogeneous system .
I thought it would be same m
Since the solution set is nothing but a translation or a parallel of the homogeneous system.
But my teacher said It is m+1

Please provide your views

答:
IF the equation $Ax=b$ has a solution, i.e. $b$ is in the column space of $A$, then the set of solutions is indeed an $m$ dimensional affine subspace $W$, specifically $W=x_0+\ker A$ where $x_0$ is any single solution.

If $b\ne 0$ and $v_1,\dots,v_m$ is a basis for $\ker A$, then as $x_0\notin\ker A$ it must be linearly independent from $v_j$, and thus the vectors
$$x_0,\ x_0+v_1,\ \dots,\ x_0+v_m$$
are all solutions and are linearly independent.

Since $W\subseteq{\rm span}(x_0,\ker A)$ which has dimension $m+1$, there can't be more linearly independent vectors in $W$.


问:
Example of non-separable subspace of a separable Hausdorff space
The following question was asked in my topology assignment and I was unable to solve it, so I am looking for help here.

Example of non separable subspace of a separable hausdorff space.

I couldn't construct such example and I looked for an answer by searching mathstackexchange and found this answer :Density character of a subspace of a topological space.

But this is beyond my understanding as I  have just started 1st course in topology (2 weeks back) and I am unable to understand that terminology.

So, kindly give a answer which is for a person that has just started learning topology. I am following Lecture notes and Assignment sheets of the course and not any textbook.

Thank you!

答:
Consider the set $\mathbb R$ with the following topology $\tau$: a set $A\subseteq\mathbb R$ is open if and only if, for each irrational number $a\in A$, there is a positive number $\varepsilon\gt0$ such that $\mathbb Q\cap(a-\varepsilon,a+\varepsilon)\subseteq A$.

You can easily verify that $\tau$ is a topology. It is Hausdorff because every set which is open in the usual topology of $\mathbb R$ is also open in the topology $\tau$. It is separable because $\mathbb Q$ is a countable dense set: there is no nonempty open set containing only irrational numbers. Finally, the subspace $\mathbb P$ consisting of the irrational numbers is not separable because it's an uncountable set with the discrete topology: for each irrational number $a$, the set $\mathbb Q\cup\{a\}$ is open in $(\mathbb R,\tau)$, so $\{a\}$ is open in $\mathbb P$ with the subspace topology.

P.S. I thank Henno Brandsma for informing me that this example is called the Michael line.


问:
$\lim_{n\to \infty}{\int_{0}^1 (1-x^2)^n} = 0$
$\lim_{n\to \infty}{\int_{0}^1 (1-x^2)^n} = 0$

My Attempt : Now $(1-x^2)^n$   is uniformly convergent to $0$ when $x \in (0 , \delta)$. So $\lim_{n\to \infty}{\int_{\delta}^1 (1-x^2)^n} = 0$. Here $0 < \delta < 1$.
IS it sufficient to prove $\lim_{n\to \infty}{\int_{0}^1 (1-x^2)^n} = 0$ ?

Can anyone please help me ??

答:
$$I_n=\int_0^1(1-x^2)^ndx$$
$u=1-x^2,du=-2xdx=-2\sqrt{1-u}dx\Rightarrow dx=-2\frac{du}{\sqrt{1-u}}$ and so we have:
$$I_n=2\int_0^1\frac{u^n}{\sqrt{1-u}}du$$
now on the lower part we have an issue as $u\to1$ and also on the top $\to1$ but for the rest of the interval it is clear that:
$$\lim_{n\to\infty}\frac{u^n}{\sqrt{1-u}}\to0\,\,\,\,u\ne1$$


问:
$\int \frac{a^2}{(a^2-x^2)^{\frac{3}{2}}}dx$
How can I solve the following integral?

$$\int \frac{a^2}{(a^2-x^2)^{\frac{3}{2}}}dx$$

答:
Let, $x=a\sin\theta\Rightarrow \frac{dx}{d\theta}=a\cos\theta\Rightarrow dx=(a\cos\theta)\,d\theta.$

$$\int \frac{a^2a\cos\theta}{(a^2-a^2\sin^2\theta)^\frac{3}{2}}d\theta$$

Thus, by solving this integral we get $\int(\sec^2\theta)\, d\theta = \tan\theta + C$

Since,$$x=a\sin\theta\Rightarrow \sin\theta = \frac{x}{a} \Rightarrow \cos\theta = \sqrt{1-\frac{x^2}{a^2}}$$

Since, $\tan\theta = \frac{\sin\theta}{\cos\theta}=\frac{x}{\sqrt{a^2-x^2}}$

So, the final result is:

$$\int \sec^2\theta \,d\theta = \tan\theta + C = \frac{x}{\sqrt{a^2-x^2}} + C$$

Note: $C$ is the integration constant.


问:
Fourier Transforms of shifted sinc funtions
I would like to calculate the Fourier transform of the following functions:

$$\left(\dfrac{\sin(\pi x\pm\pi n/2)}{\pi x\pm\pi n/2}\right)^2$$

$$\dfrac{\sin(\pi x+\pi n/2)}{\pi x+\pi n/2}\cdot\frac{\sin(\pi x-\pi n/2)}{\pi x-\pi n/2}$$

with $n \in\mathbb{N}$

Any help will be highly appreciated!

答:
Hint 1: the transform of $\sin (x\pi)/(x\pi)$ is a rectangle with height 1 between $-\pi$ and $\pi$.

Hint 2: multiplication in one domain corresponds to convolution in the other domain.

Hint 3: shifting a function by $x_0$ corresponds to multiplication of its Fourier transform with $e^{\pm ix_0\omega}$.

EDIT: For solving (2.) we have the Fourier transforms of the two functions:
$$e^{\pm i\omega n/2}\text{rect}(\omega,\pi)$$

So the convolution becomes

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\Omega n/2}\text{rect}(\Omega,\pi)
e^{-i(\omega-\Omega) n/2}\text{rect}(\omega-\Omega,\pi)d\Omega=\\
=\frac{e^{-i\omega n/2}}{2\pi}\text{rect}(\omega,2\pi)\int_{\max (-\pi,\omega-\pi)}^{\min (\pi,\omega+\pi)}e^{j\Omega n}d\Omega$$

For $-2\pi<\omega < 0$ we get

$$\frac{e^{-i\omega n/2}}{2\pi}\int_{-\pi}^{\pi+\omega}e^{j\Omega n}d\Omega$$

and for $0<\omega<2\pi$ we have

$$\frac{e^{-i\omega n/2}}{2\pi}\int_{\omega-\pi}^{\pi}e^{j\Omega n}d\Omega$$

I trust you can take it from here.

EDIT 2:
OK, some more help and the final solution. The first integral (for $-2\pi<\omega < 0$ ) evaluates to

$$\frac{e^{-i\omega n/2}}{2\pi}\frac{(-1)^n}{in}\left (e^{in\omega} -1\right )=
\frac{(-1)^n}{n\pi}\sin\frac{n\omega}{2}$$
and the second integral for positive $\omega$ is the same, just with the opposite sign. So the final result is

$$\frac{(-1)^{n+1}}{n\pi}\text{sign}(\omega)\sin\frac{n\omega}{2},\quad
|\omega|<2\pi,\quad \text{(zero otherwise)}$$


答:
Put $Z_n = \int_0^n{(1-x^2)^n dx}$. We have $\forall \epsilon >0$
$$Z_n= \int_0^n{(1-x^2)^n dx} = \int_0^{\frac{\epsilon}{2}}{(1-x^2)^n dx}+\int_{\frac{\epsilon}{2}}^{1}{(1-x^2)^n dx} \leq \frac{\epsilon}{2} + \int_{\frac{\epsilon}{2}}^{1}{(1-(\frac{\epsilon}{2})^2)^n dx} \leq \frac{\epsilon}{2} + (1-\frac{\epsilon^2}{4})^n $$

And we have $\forall n> \frac{\ln (\frac{\epsilon}{2})}{\ln (1-\frac{\epsilon^2}{4})}$, $(1-\frac{\epsilon^2}{4})^n  \leq \frac{\epsilon}{2}$.

So, $\forall \epsilon >0, \exists N= \frac{\ln (\frac{\epsilon}{2})}{\ln (1-\frac{\epsilon^2}{4})}$ so that $\forall n>N , Z_n < \frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon $. We can conclude  $Z_n \longrightarrow 0 $

Body:
$\lim_{n\to \infty}{\int_{0}^1 (1-x^2)^n} = 0$

My Attempt : Now $(1-x^2)^n$   is uniformly convergent to $0$ when $x \in (0 , \delta)$. So $\lim_{n\to \infty}{\int_{\delta}^1 (1-x^2)^n} = 0$. Here $0 < \delta < 1$.
IS it sufficient to prove $\lim_{n\to \infty}{\int_{0}^1 (1-x^2)^n} = 0$ ?

Can anyone please help me ??


Title: Calculate $\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}$

Body:
Please help me to calculate the $$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}$$.

My attemp is:
$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}\cdot\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{\sqrt[3]{1+7x}-\sqrt{1+3x}}\cdot\frac{\sqrt[3]{1+7x}+\sqrt{1+3x}}{\sqrt[3]{1+7x}+\sqrt{1+3x}}$, but I do not know if I did well, can anyone help me

Body: Since $f^\prime(1):=\lim_{x\to1}\frac{f(x)-f(1)}{x-1}$, we have$$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-2}{x-1}=\left.\frac73(1+7x)^{-2/3}\right|_{x=1}=\frac{7}{12},$$and similarly $\lim_{x\to1}\frac{\sqrt{1+3x}-2}{x-1}=\frac34$, so your limit is $-\frac16$.


问:
Running time that is neither polynomial nor exponential
I have to find an example of a running time that is neither polynomial nor exponential. At first I was thinking about $n!$ but I think that this is bounded below by exponential functions for large n (which would imply exponential running time). Could someone help me out with this question? Thanks in advance

答: Your initial intuition is correct, factorial time is certainly not polynomial, and it is not bounded above by exponential time (if by exponential time you mean $a^n$ for some $a$). Note, however, that there is another category called the superexponential time, of the form $n^n$, which does indeed outgrow $n!$. See this question for more details.


问:
Mathematical induction $3$ and $5$ cent coins
Use mathematical induction to show that for all integers, any price equal or greater than $8$, can be paid for by $3c$ and $5c$ coins.</p>

答:
We define a recursive function $\mu = (\mu_1,\mu_2): \{8,9,10,\dots\} \to \Bbb N \times \Bbb N$ as follows:

$\quad \mu(8) \,\;= (1,1)$

$\quad \mu(9) \,\;= (3,0)$

$\quad \mu(10) = (0,2)$

$\quad$ If $\mu(k)$, $\mu(k+1)$, and $\mu(k+2)$ are defined then set $\mu(k+3) = \bigr(\mu_1(k)+1,\, \mu_2(k)\bigr)$

It is easy to prove by induction that

$\tag 1 \mu_1(k) \cdot 3 + \mu_2(k) \cdot 5 = k \quad \text{ for all } k \ge 8$


答:
Let change $x=1+u$ with $u\to 0$ then expand $(1+\alpha u)^\beta=1+\alpha\beta\, u+o(u)$

First order is enough for this limit:

$\require{cancel}
\dfrac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}=
\dfrac{(8+7u)^\frac 13-(4+3u)^\frac 12}u=
\dfrac 2u\left((1+\frac 78u)^\frac 13-(1+\frac 34u)^\frac 12\right)=
\dfrac 2u\left(\cancel{1}+\frac 7{24}u+o(u)-\cancel{1}-\frac 38u+o(u)\right)=-\dfrac 16+o(1)\to-\dfrac 16$


Body:
Just use  the definition of the derivative:
$$\dfrac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}=\underset{\substack{\downarrow\\\bigl(\sqrt[3]{1+7^{\vphantom{l}}x}\bigr)'_{x=1}}}{\dfrac{\sqrt[3]{1+7x}'\Bigl|_{x=1}-2}{x-1}}{}-\underset{\substack{\downarrow\\\bigl(\sqrt{1+3^{\vphantom{l}}x}\bigr)'_{x=1}}}{\frac{\sqrt{1+3x}-2}{x-1}}=\cdots$$


问:
The letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

Dr. Lizardo writes down a sequence of $n$ letters, where each letter is A, B, or C, and the letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

What I have tried so far

The last letter is predefined in terms of "A or not" based on the first $n-1$ letters. If there are an odd number of As so far, the last one has to be A, otherwise, B or C. Therefore, there are $3^{n-1}$ ways to write the first $n-1$ letters. However, the last letter can either have 1 choice or 2 choices.

I am thinking maybe we need to do $S_n = 3^{n-1} + S_{n-1}$. However, I need a closed form, not a recurrance. How can I do this?

Thank you!

Note: $S_n = 3^{n-1} + S_{n-1}$ seems to be working for the first few values of $n$ (I tested through $n=4$), so finding a closed form of that recurrance would work as well.

答:
Consider two cases.

Case 1. The sequence is all C's. There is just one such sequence, and it has an even number of A's, namely zero.

Case 2. The sequence is not all C's. There are $3^n-1$ such sequences, and exactly half of them, or $\frac{3^n-1}2$, have an even number of A's. To see this, look at the leftmost letter in the sequence that's not a C; toggling it between A and B, we see that there's a one-to-one correspondence between sequences with an even and an odd number of A's.

Combining the results from Case 1 and Case 2, we get a grand total of
$$1+\frac{3^n-1}2=\frac{3^n+1}2.$$

More generally, over an alphabet of $k$ letters, the number of sequences of length $n$ in which a certain letter appears an even number of times is
$$(k-2)^n+\frac{k^n-(k-2)^n}2=\frac{k^n+(k-2)^n}2;$$
see my answer to this question.


问:
Is is always true that $\sum_{1}^{n}\left ( a_{i}-\bar{a} \right )^2\leq \sum_{1}^{n}a_{i}^2$
Is it always true that
$$\sum_{1}^{n}\left ( a_{i}-\bar{a} \right )^2\leq \sum_{1}^{n}a_{i}^2$$

where $\bar{a}$ is  the mean of all $a_{i}$'s.

If yes, how to prove it?

答: $$\sum a_i^2-\sum(a_i-\bar a)^2=\sum2a_i\bar a-\sum\bar a^2$$$$=-n\bar a^2+2\bar a\sum a_i=-n\bar a^2+2\bar a(n\bar a)=n\bar a^2\geq0$$


问:
Maximize area of rectangle inside a triangle
Find the maximum area of a rectangle that is inside of the triangle forms by the x-axis and the lines y=-3x+12 and y=3x+12. The base of the triangle is on the x-axis and the two upper verticies are on the lines y=-3x+12 and y=3x+12.

With this question. Obviously the are of the rectangle is lw or xy... I have figured out that x = (8-2x) and y=(12-y). Im not sure of the next step.. Could someone help?

答:
The two lines are symmetric across the $y$-axis. Therefore, just take twice the area in the 1st quadrant.

Thus the area of the rectangle (width times height) is $2x(-3x+12) = -6x^2+24x$, where $0 ≤ x ≤ 4$.

Since the quadratic is concave up, notice that the maximum is at $x = -\frac{b}{2a} = -\frac{24}{2(-6)} = 2$, which gives the maximum area as $-6(2)^2 + 24(2) = \boxed{24}$.


问:
How to show that the only affine automorphism of $\mathbb{R}$ is the identity, using fundamental thorem of affine geometry?
Let the set of real numbers $\mathbb{R}$ be endowed with the natural affine structure (that means the group action is the usual addition). The question is how to prove that the only affine automorphism in affine space $\mathbb{R}$ is the identity, using the fundamental theorem of affine geometry. The affine automorphism is an affine map which is an automorphism. The fundamental theorem of affine geometry states as follows:

Let $X,X'$ be affine spaces of the same affine dimension $\ge2$. Let $f:X\to X'$ be a set-theoretical bijection which takes any three collinear points in $X$ into collinear points in $X'$. Then $f$ is semiaffine.

I have no idea know how to relate the fundamental theorem of affine geometry to the problem in question. In particular, the affine space $\mathbb{R}$ has affine dimension $1$, not $\ge2$ as stipulated in the theorem. What contradiction can I get if I assume there is a non-identity automorphism in affine space $\mathbb{R}$? I would appreciate your help greatly if you can direct me how to use this powerful fundamental theorem.

PS, for reference, the screenshot of the proposition and the fundamental theorem is copied below:

答:
The statements about $\mathbb{R}$ and $\mathbb{C}$ in the Proposition are about field automorphisms. The point is that

It is well known that the only field automorphism of $\mathbb{R}$ is the identity (the fact that the state this is what tells you they are talking about field automorphisms, since there are many examples of affine bijections of the real number line that are not field automorphisms).

We can view $\mathbb{C}$ as a 2-dimensional affine geometry over $\mathbb{R}$.

We can use the fact that a "continuous automorphism" of $\mathbb{C}$ must be an affine automorphism, that is, it is both an affine map in the sense of the fundamental theorem of affine geometry, and also a field automorphism.

This lets you use the Fundamental Theorem (which is about affine maps) to classify the continuous field automorphisms of $\mathbb{C}$. Notice that there is no claim about the transformations of $\mathbb{R}$ as an affine geometry; indeed the Fundamental Theorem does not apply here because $\mathbb{R} = \mathbb{R}^{1}$ is one-dimensional.


问:
Find the altitude of the triangle given the equation of sides

The sides of a triangle are on the lines $2x+3y+4=0$, $ \ \ x-y+3=0$, and $5x+4y-20=0$. Find the equations of the altitudes of the triangle.

Should I find the vertices first? Or is there a direct way? Actually, I tried finding the vertices, using the substitution method but I find it hard to turn it into an equation.

答:
$$r:2x+3y+4=0,\;s:x-y+3=0,\;t:5x+4y-20=0$$
The set of all lines passing through the intersection point $r\cap s$ is
$$\mathcal{F}:a(2x+3y+4)+b(x-y+3)=0\tag{1}$$
$$\mathcal{F}:(2a+b)x+(3a-b)y+4a+3b=0$$
To find the one perpendicular to $t:5x+4y-20=0$ we use the condition
$$5(2a+b)+4(3a-b)=0\to b=-22a$$
plug this value in the equation $(1)$ and get

$a(2x+3y+4)-22a(x-y+3)=0$

simplify $a$ and get the first height equation
$$20 x-25 y+62=0$$
In a similar way we can get the other two equations
$$27x-18y+46=0;\;7x+7y-16=0$$


问:
Which topics of mathematics should I study?
I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.

Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.

If possible, could you also give me advice on books/references which I should study from?

Thanks in advance.

Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.

答: Here's a link to a page which in turn links to documentation about the first three years of the Cambridge Mathematics Tripos (look at the Guides to Part IA, Part IB and Part II). This will tell you what one famous course thinks is pretty essential (in the first two years) and then -- as you begin to specialize -- what the next steps might be. Of course, there can be heated arguments about what should go in undergraduate courses when: but these documents, and similar ones from other places with top-ranked mathematics courses, should give you some very useful pointers.</p>