问:
A point $p$ of a variety $X$ is smooth if and only if it is smooth in $U\subset X$ an open sub variety.
I want to show the following:

Let $X$ be an algebraic variety, let $U$ be an open subvariety of $X$ containing $p$. Then $X$ is smooth at $p$ if and only if $U$ is smooth at $p$.

Here is my definition of being smooth at a point

Definition: Let $X$ be an algebraic variety. We say $X$ is smooth at $p$ if there is an open subvarity $U$ of $X$ containing $p$ and an isomorphism $\varphi: U \rightarrow Z(f_1,..,f_{n-d})\subset \mathbb A^n$ for some $d\leq n$ for some $f_1,..,f_{n-d}$ such that the rank of $$\left(\frac{\partial f_i}{\partial x_j}(\varphi(p)\right)$$
is maximal.

Attempt:

($\Leftarrow$)
Let $U$ be smooth at $p$, then there is some $\overline{U}$ an open subvariety of $U$ and $\varphi:U\rightarrow Z(f_1,...,f_{n-d})$ satisfying the smoothness criterion. Since $U$ is a subvariety of $X$ so to is $\overline{U}$. Again this $\varphi:U\rightarrow Z(f_1,...,f_{n-d})$ still satisfies the smoothness crieterion so we are done.

($\Rightarrow$)
Let $X$ be smooth at $p$, then there is some $\overline{U}$ an open subvariety of $X$ and an isomorphism $\varphi:\overline{U}\rightarrow Z(f_1,..f_{n-d}$ satisfying the smoothness criterion.

We want an open subvariety of $U$, define $V=U\cap\overline{U}$ this certainly is an open subvariety of $U$. And $\varphi\mid_V$ is an isomorphism onto its image. But how do we know its image is of the form $Z(f_1,...,f_{n-d})$ and how do we know that the Jacobian of the restriction still has full rank?

答:
The other answer has some valuable content, but it's long, not an easy read to my eyes, and could be more self-contained. Here I present what I feel is a more direct solution.

One direction is easy: if $p\in X$ is smooth as a point of $U\subset X$, then the open neighborhood $U'\subset U$ obtained from the definition of smoothness is also an open neighborhood of $p\in X$ satisfying the same required properties.

For the other direction, suppose $p\in X$ is a smooth point and $U$ is an open neighborhood of $p$. Let $U'$ be the open neighborhood of $p$ guaranteed by the definition of smoothness. If $U'\subset U$, we're done. Else, $U'\cap U$ is an open subset of $U'$, with closed complement $V\subset U'$.

As $V$ is a closed subvariety of the affine variety $U'\cong Z(f_1,\cdots,f_{n-d})\subset \Bbb A^n$ which is disjoint from $p$, we can find a function $f$ on $U'$ vanishing on $V$ and nonvanishing on $p$. Then $D(f)\cap U'$ is an open subset of $U\cap U'$ and also an affine variety, cut out in $\Bbb A^{n+1}$ by the equations $(fx_{n+1}-1,f_1,\cdots,f_{n-d})$.

To see that the Jacobian matrix of this system has rank $n+1-d$, it suffices to check that the entry corresponding to $\frac{\partial (fx_{n+1}-1)}{\partial x_{n+1}}$ is nonzero: every other term of the form $\frac{\partial f_i}{\partial x_{n+1}}$ is zero because $f_i$ doesn't depend on $x_{n+1}$. But $\frac{\partial (fx_{n+1}-1)}{\partial x_{n+1}}=f$, which is nonvanishing on $D(f)\cap U'$. So our Jacobian matrix looks like $$\begin{pmatrix} \frac{\partial(f_1,\cdots,f_{n-d})}{\partial(x_1,\cdots,x_n)} & 0 \\ * & f \end{pmatrix}$$ and this has rank $n+1-d$, so we're done.


问:
Prove that $\dim(T(H))\leq \dim(H)$.

Let $V$ and $W$ be finite-dimensional vector spaces and $T$ be a linear transformation $T:V\to W$.

Let $H$ be a non-zero subspace of $V$, and let $T(H)$ be the set of images of vectors in $H$. Prove that $\dim(T(H))\leq \dim(H)$.

Proof:

Let $B=\{b_1,...,b_n\}$ be a basis for $H$.

Because $B$ is a basis and $T$ is linear, for any $x\in T(H)$, we have

$$x=T(c_1b_1+...+c_nb_n)=c_1T(b_1)+...+c_nT(b_n)$$

Therefore, $\text{span}\{T(b_1)...T(b_n)\}=T(H)$.

If $\{T(b_1)...T(b_n)\}$ is linearaly independent, then $\text{dim}(T(H))=\text{dim}(H)$ and we are done.

Otherwise, $\{T(b_1)...T(b_n)\}$ is linearaly dependent, and there exist $T(b_i)$ that can be expressed as a linear combination of the other vectors in this set.

So let $T(b_i)=d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)$.

Now let $y\in T(H)$,then
$$y=c_1T(b_1)+..+c_iT(b_i)+..+c_nT(b_n)=c_1T(b_1)+...+d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)+...+c_nT(b_n)=(c_1+d_1)T(b_1)+..+(c_{i-1}+d_{i-1})T(b_{i-1})+(c_{i+1}+d_{i+1})T(b_{i+1})+...+(c_n+d_n)T(b_n)$$

So, $\text{span}\{T(b_1)...T(b_n)\}=\text{span}\{T(b_1)...T(b_{i-1}),T(b_{i+1}),...,T(b_n)\}=T(H)$

If this new set is linearly independent then $\text{dim}(T(H))=n-1<\text{dim}(H)$.

Otherwise, we are linearly dependent and a vector can be removed without changing the span. If we get down to one remaining vector and that vector is not the $0$ vector then  $\text{dim}(T(H))=1<\text{dim}(H)$. Otherwise, it is the $0$ vector and $\text{dim}(T(H))=0<\text{dim}(H)$.

Thank you for reading the proof. Please provide feedback on the validity of my proof. Have I assumed anything I shouldn't have assumed? Any leaps in logic? I appreciate any tips suggestions and insights.

答:
This question can be found here as well.

Now about your proof, she is correct. But I probably instead of writing,

Otherwise, $\{T(b_1),...,T(b_n)\}$ is linearaly dependent, and there exist $T(b_i)$ that can be expressed as a linear combination of the other vectors in this set.

I would write,

Otherwise, $\{T(b_1),...,T(b_n)\}$ is linearaly dependent, so there is at least one vector $T(b_i)$, that can be expressed as a linear combination of the other vectors in this set.

Also note that when you write $y$, the vector $T(b_i)$ is being multiplied by $c_i$, then $y$ would be written as,

$$y=c_1T(b_1)+...+c_{i-1}T(b_{i-1})+c_i \big(\underbrace{d_1T(b_1)+...+d_{i-1}T(b_{i-1})+d_{i+1}T(b_{i+1})+...+d_nT(b_n)}_{T(b_i)}\big)+...+c_nT(b_n)$$

So in the next equality, instead of just, for example, $(c_1+d_1)$, you need to write, $(c_1+c_id_1)$, and then the proof follows as you wrote it.

ContentLicense: CC BY-SA 4.0

Title: Prove that $\dim(T(H))\leq \dim(H)$.


问:
Orthogonal idempotents in $M_{4\times 4}(\mathbb{C})$.
If $e_1,\cdots,e_n \in M_{4\times 4}(\mathbb{C})$ are $n$ distinct, nonzero, $4\times 4$ matrices with complex entries that satisfy $e_ie_j = e_je_i = 0$ and $e_i^2 = e_i$ for all $1\leq i \leq n$, then I want to show that $n \leq 4$.

My initial instinct was to look at the Jordan Canonical forms of the matrices $e_i$, and note that the minimal polynomial of $e_i$ must divide $x^2-x$, and hence must be $x,(x-1)$ or $x(x-1)$. The minimal polynomial can't be $x$ (since the $e_i$'s were assumed nonzero) and it can't be $x-1$, since then $e_i$ is the identity matrix which is only orthogonal to the zero matrix.
Thus, then minimal polynomial of $e_i$ is $x(x-1)$, and the invariant factor decomposition of $e_i$ must be one of
\begin{align*}
&(x,x,x(x-1)) &(x-1,x-1,x(x-1))& &(x(x-1),x(x-1)).
\end{align*}
However, at this point I realized that this method does not seem to be on the right track, since  there is no reason why the $e_i$'s shouldn't be similar, so knowing their JCF doesn't seem to help me reason why the total number of such matrices should be less than $4$.

Any thoughts or hints would be greatly appreciated.

答:
By this minimal polynomial condition we see that $e_1$ is diagonalizable, so there's a basis in which $$e_1=\begin{bmatrix}I_d & 0 \\ 0 & 0\end{bmatrix}$$ for some $1\leq d\leq 4$. Now we want to use trivial product relations and this basis to analyse the remaining matrices.

Continue for spoiler:

 Since $e_1e_i=0$ for $i\geq 2$, the columns of $e_i$ are contained in the kernel of $e_1$. So in this basis, the columns of $e_i$ must have the first $d$ entries equal to $0$. Similarly, since $e_ie_1=0$ for $i\geq 2$, the rows of $e_i$ must have the first $d$ entries equal to $0$ (in the given basis). So the $e_i$ really live in the space of $(n-d)\times(n-d)$ matrices. Then this problem reduces to a lower dimension and you just keep going until you hit dimension $1$ where the problem is trivial.


问:
How is Brownian motion predictable?
Could someone please explain how Brownian motion is predictable? My understanding is that a predictable process is one that depends on information up to time t say but not t itself, therefore W_t has to be observed at time t to determine its value.</p>

答: Brownian motion is continuous in time so you can determine the value of $W_t$ as a limit from the left: $W_t=\lim_{s\rightarrow t^-}W_s$. </p>

ContentLicense: CC BY-SA 3.0

Title: How is Brownian motion predictable?


问:
Eilenberg-Watt's theorem reference.
I'm looking for a reference (with proof!) for the following result:

Let $A$ and $B$ be unital, associative rings.

(Eilenberg-Watt) Let $F: {}_{A}\mathrm{Mod} \to {}_{B}\mathrm{Mod}$ be a functor that is
additive and preserves direct sums and quotients. Then there is a
natural isomorphism $$\alpha: P \otimes_A \bullet \to F$$ where
$P=F(A)$ has the natural right $A$-module structure.

In short, an additive functor that preserves direct sums and quotients coincides with the functor $P \otimes_A \bullet$.

Please check if all assumptions match, because I have seen plenty of variations of this theorem but not this exact one.

答:
If we require that for any map $\varphi : X \to Y$ then $F(coker \varphi) \simeq coker F(\varphi) $, then the result is true.

Let $M$ be a module, and let $\{m_i\}_{i\in I}$ be a set of generators. This defines a map $\psi: A^{\oplus I} \to M$ sending the 1 in the $i$-th component to $m_i$. Now for any relation $\sum a_{ij} m_i =0$ in $M$, let $r_j\in A^{\oplus I}$ be the element $\sum a_{ij}$. This defines a map $\varphi: A^{\oplus J}\to A^{\oplus I}$, and you can check that $M \simeq coker \varphi $. Let me drop the $\oplus$ above the $A$'s. We can use the cokernel commutativity to write

$$ F(M) \simeq F(coker \varphi ) \simeq coker F(\varphi) $$

Note also that
$$F(A^I) \simeq F(A) ^I \simeq F(A) \otimes A^I$$
and that $F$ being additive implies also
$$F(\varphi) = F(A) \otimes \varphi$$

The latter formula means to tensor with the identity on $F(A) $. Also, recall that tensoring with a module is a left adjoint functor, thus preserve cokernels. In particular we have

$$ coker F(\varphi) \simeq coker F(A) \otimes \varphi \simeq F(A) \otimes coker \varphi \simeq F(A) \otimes M $$

And voilà. Maybe this was the result you were looking for?

Title: Eilenberg-Watt's theorem reference.


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答: Visual Complex Analysis by Tristan Needham</p>


问:
closed measurable rectangles generates $\mathscr{B}(\mathbb{R}^2) ? $
Given the following proposition : X,Y random variables are indipendent $\iff$ $\forall \hspace{0.1cm} f,g \in C_b(\mathbb{R})$ it holds that $\mathbb{E}[f(X)g(Y)] = \mathbb{E}[f(X)]\mathbb{E}[g(Y)]$

I think the proof of $[\Rightarrow]$ is using the Fubini-Tonelli theorem properly.

Conversely I'm not sure how to conclude from the followin : Let's take $f_n,g_n \in C_b(\mathbb{R})$ such that $f_n \uparrow I_C,g_n \uparrow I_C'$ pointwise, where $C,C'$ are closed in $\mathbb{R}$, it follows that using monotone convergence

$$\mathbb{P}(X \in C, Y \in C') = \mathbb{E}[I_C I_C'] = \mathbb{E}[\lim\limits_{n \to +\infty}f_n(X)g_n(Y)]$$ $$ = \lim\limits_{n \to +\infty} \mathbb{E}[f_n(X)g_n(Y)] = \lim\limits_{n \to +\infty} \mathbb{E}[f_n(X)] \mathbb{E}[g_n(Y)] = \mathbb{P}(X \in C)\mathbb{P}(Y \in C')$$

Then I'd like to conclude for a criterion of coincidence of measures, but I'm not sure since I'm supposed to affirm that the closed measurable rectangles, i.e. $A \times B : A \in \mathscr{B}(\mathbb{R}), B \in \mathscr{B}(\mathbb{R})$ with $A,B$ closed in $\mathbb{R}$, generate $\mathscr{B}(\mathbb{R}^2)$. Is this true ? If so, there is a nice way to prove it ?

答:
I'll focus on the question if the closed measurable rectangles generate the Borel algebra on $\mathbb{R}^2$. I.e. I will show that
$$\mathcal{C}:=\sigma(\{A \times B\mid A , B \text{ closed subsets of $\mathbb{R}$}\})= \mathcal{B}(\mathbb{R}^2).$$

Note first that if $A$ and $B$ are closed subsets of $\mathbb{R}$, then $A \times B$ is a closed subset of $\mathbb{R}^2$. In particular, $A \times B \in \mathcal{B}(\mathbb{R}^2)$. Hence, $\mathcal{B}(\mathbb{R}^2)$ is a $\sigma$-algebra containing all rectangles of closed sets, so by minimality of $\mathcal{C}$ we get $\mathcal{C}\subseteq \mathcal{B}(\mathbb{R}^2)$.

Conversely, note that every open set in $\mathbb{R}^2$ is the countable union of sets of the form $]a,b[\times ]c,d[$ (these are called open cubes). Any union can be made countable for example by using a density argument. Next, note that $]a,b[ \times ]c,d[$ is in $\mathcal{C}$, since it is the closed cube $[a,b] \times [c,d]$ in which the sides are removed (one can explicitely describe the sides as a finite union of sets of the form $A \times B$ where either $A$ or $B$ is a singelton). It follows that every open set is in $\mathcal{C}$ and thus $\mathcal{B}(\mathbb{R}^2)\subseteq \mathcal{C}$.


问:
Arc length equation question
If you were given the following and were told to find $\dfrac{\text{d}y}{\text{d}x}$.

$y^2=2(x+4)^3$

How do you solve the problem when there is a $y^2$? When I tried solving the problem I got

$\dfrac{\text{d}y}{\text{d}x}= \dfrac{3x^2+24x+48}{y}$

I was trying to put the answer into the equation for finding the length of a curve which doesn't make sense because the equation is $\displaystyle \int_{a}^{b} \sqrt{1+\left(\dfrac{\text{d}y}{\text{d}x}\right)^2}\text{d}x$

My answer is a fraction with a $y$ variable on the bottom, but the formula at the end has a $\text{d}x$? How do you solve for $\dfrac{\text{d}y}{\text{d}x}$ so that there isn't a $y$ on the bottom fraction?

Title: Arc length equation question

Body:
Solve wrt $y$

$y=\pm \sqrt{2} (x+4)^{3/2}$

differentiate

$y'=\pm \frac{3\sqrt 2}{2}(x+4)^{1/2}$

Integrate
$$\int_a^b \sqrt{\frac{9 (x+4)}{2}+1} \, dx=\left[\frac{4}{27} \left(\frac{9 x}{2}+19\right) \sqrt{\frac{9 x}{2}+19}\right]_a^b$$


问:
Geometrically, why are complex numbers both linear maps that perform rotation/dilation, and points on the plane?
The standard geometric interpretation of complex numbers is to draw the complex plane by identifying the horizontal axis as the real axis and the vertical axis as the imaginary axis, so that the complex number $z = x + iy$ is identified with the point $(x, y)$. That is to say, a complex number is geometrically a point on the plane.

But we also know from the polar form of complex numbers that when we multiply two complex numbers $z=r(\cos \alpha + i \sin \alpha)$ and $w=s(\cos \beta + i \sin \beta)$ that we can “add the angles and multiply the distances” to get $zw=rs(\cos(\alpha + \beta) + i \sin(\alpha + \beta))$. That means we can view one of the complex numbers, say $z$, as being a linear map that performs a rotation and/or dilation on the other number (here $w$) which is viewed as a point, to return a new point.

In linear algebra though, a point is a vector $v \in V$, and a linear map is a function $T: V \to W$. These are two different kinds of objects, and it would be a type error to try to identify $v$ with $T$ somehow. So my question is something like, why does it work out that complex numbers can be viewed in two different ways even though that seems like a type error?

If I stopped here though I feel like people would just give me examples of other things in math where it's possible to view a single object through multiple geometric "lenses". (Maybe that is what I want, but I feel like there is still something to my original question.) So let me go a little further: Of course there are many situations where the same objects can be visually represented in different ways, e.g. a real number can be both a length and an area and a volume and so forth. And even in the case of real numbers, we can say that we start with some distance $x$, then multiply by a distance $y$ to get a distance $yx$. In that case, isn’t $y$ acting as a “linear map” $x \mapsto yx : \mathbf R \to \mathbf R$? That’s true, but we can also use similar triangles (figure from this book, p. 3) to show that $y$ can be seen as a length as well, so that $x$, $y$, and $yx$ are all lengths in a single figure. It’s this extra step that seems to be missing from the case of complex number multiplication.

答:
Perhaps a slight change of perspective could prove useful: we think of complex numbers just as that, points in the complex plane. However, we can prove that the functions which just multiply by some fixed complex number

$f_c: \mathbb{C} \to \mathbb{C},    f_c(z) := cz$

are exactly all of the linear maps which perform rotation and dilation. Since every such function arises from exactly one $c \in \mathbb{C}$ and vice versa, it's natural to identify them (rather than directly defining complex numbers as being these maps). So the solution to this "type error" (which I think is a nice way to put it) is that we do indeed have two different sets (the set of vectors in the complex plane, and the set of maps as already described) which just happen to be very naturally bijective.


问:
Probability and supremum
Problem:

Suppose $X_1,......,X_n.....$ are independent random variables on $(\Omega, \mathcal{F}, \mathbb{P})$.

Suppose $\mathbb{P}(sup_nX_n<\infty)=1$. Show that $sup_nX_n(x)<\infty$ for all $x\in \Omega$

This is a lemma I have constructed, but i'm not sure how to prove it or if a counter example exists. Intuitively, it means that the probability of the set $\{$ $sup_nX_n<\infty \}$ occuring is 100%, which would make intuitive sense to then imply that $sup_nX_n(x)<\infty$. May someone help?

答:
This is very very false. You can always change the values of $X_i$'s on set of measure $0$ to find a point where the supremum is $\infty$.

For example let $(X_n)$ be i.i.d coin tossing random variables. Define $Y_n(x)=X_n(x)$ if $X_n(x) \neq 1$ for some $n$ and $Y_n(x)=n$ if $X_n(x) =1$ for all $n$.

Title: Probability and supremum


答:
Consider $I_i=\sum_{j\ne i}e_jR$, where $R=M_{4\times4}(\mathbb{C})$. If $x\in e_iR\cap I_i$, then
$$
x=\sum_{j\ne i}e_jr_j
$$
and also $x=e_ir$, from which $e_ix=e_ie_ir=e_ir=x$, so
$$
x=e_ix=\sum_{j\ne i}e_ie_jr_j=0
$$
We conclude that
$$
R=\bigoplus_{1\le i\le n}e_iR
$$
as a right $R$-module.

Now note that $R$ is a simple artinian ring, so it has a unique minimal right ideal up to isomorphism. Since the right ideal formed by matrices of the form $[u\ 0\ 0\ 0]$ (only the first column being possibly nonzero) is minimal, it has dimension $4$ over $\mathbb{C}$.

Each $e_iR$ is a nonzero right ideal of $R$, so it contains a minimal right ideal. Thus the number of direct summands cannot exceed $4$ or dimension $16$ of $R$ would be exceeded as well.

Title: Orthogonal idempotents in $M_{4\times 4}(\mathbb{C})$.

Body:
If $e_1,\cdots,e_n \in M_{4\times 4}(\mathbb{C})$ are $n$ distinct, nonzero, $4\times 4$ matrices with complex entries that satisfy $e_ie_j = e_je_i = 0$ and $e_i^2 = e_i$ for all $1\leq i \leq n$, then I want to show that $n \leq 4$.

My initial instinct was to look at the Jordan Canonical forms of the matrices $e_i$, and note that the minimal polynomial of $e_i$ must divide $x^2-x$, and hence must be $x,(x-1)$ or $x(x-1)$. The minimal polynomial can't be $x$ (since the $e_i$'s were assumed nonzero) and it can't be $x-1$, since then $e_i$ is the identity matrix which is only orthogonal to the zero matrix.
Thus, then minimal polynomial of $e_i$ is $x(x-1)$, and the invariant factor decomposition of $e_i$ must be one of
\begin{align*}
&(x,x,x(x-1)) &(x-1,x-1,x(x-1))& &(x(x-1),x(x-1)).
\end{align*}
However, at this point I realized that this method does not seem to be on the right track, since  there is no reason why the $e_i$'s shouldn't be similar, so knowing their JCF doesn't seem to help me reason why the total number of such matrices should be less than $4$.

Any thoughts or hints would be greatly appreciated.


问:
Proving that $\operatorname{Span}(u, v + lu) \subseteq \operatorname{Span}(u, v)$
Let $u, v$ be elements of a real vector space $V$, and let $l\in\mathbb R$.

a) Prove that $\operatorname{Span}(u, v + lu) \subseteq \operatorname{Span}(u, v)$.

b) Prove that $\operatorname{Span}(u, v) \subseteq \operatorname{Span}(u, v + lu)$?

I'm really struggling and not sure how to prove part a) and part b). Any answers or where to start? Any help would be much appreciated.

答:
For part a), think about what a generic element $w \in$ span$(u, v + lu)$ looks like, just from the definition of spanning:

$w = c_1 u + c_2 (v + lu) \ \ \ (c_i \in \mathbb{R})$

Can you prove that this is also element of span$(u,v)$? If so, you've proven that span$(u,v+lu) \subseteq $ span$(u,v)$ (do you see why?).


问:
Space of embedded surfaces with a common point
Consider the space of all embedded orientable surfaces in $ R^3 $ of constant mean curvature (the minimal case is included) passing through the origin. I'm asking if there exists a topology on this space such that it is sequentially compact (and Haussdorf) and if $ M_n $ is convergent to $ M $ then $ H_n \rightarrow H $ and $ |A_n(0)| \rightarrow |A(0)| $ where $ H_n,H $ are the mean curvatures and $ A_n,A $ are the second fundamental forms.

Thanks

答:
Presumably you want a bound on the mean curvature, otherwise the  sequence of spheres $(x-1/n)^2+y^2+z^2=1/n^2$ will be problematic: it has no convergent subsequence in any reasonable sense.

Already  the case $H_n\equiv 0$ is very complicated. To begin with, think of a sequence of catenoids with neck size shrinking to zero. If the point that you fixed sits on the neck, you will see both principal curvatures blowing up: one to $+\infty$, the other to $-\infty$. This survey by Minicozzi lists other things that can happen with a sequence of minimal surfaces.

To suggest a direction in which a positive result may be found: up to rotation, the surface is given near the origin by equation $z=f(x,y)$ where $f(0,0)=0$ and  $f$ is real-analytic. Impose a uniform lower bound on the radius of disk $x^2+y^2<r^2$ in which $f$ is defined. Also impose a uniform upper bound on $|f|$. Then the interior regularity for mean curvature equation (Corollary 16.7 in the book by Glbarg & Trudinger) yields a uniform upper bound on Hessian of $f$ at $0$. This allows you to select a subsequence for which the second fundamental form at $0$ converges.


问:
Let $U,W \subseteq \mathbb{R}^n$ be subspaces. Assume that $U \oplus W = \mathbb{R}^n$. Prove that $U^\perp \oplus W^\perp=\mathbb{R}^n$.
I started by writing out the definitions of $U^\perp$ and $W^\perp$ as
$$U^\perp=\{x \in \mathbb{R}^n:\langle x,u \rangle =0, \forall u\in U\} \qquad W^\perp=\{y \in \mathbb{R}^n:\langle y,w \rangle =0, \forall w \in W\}$$
Sadly I dont really know how to progress from this point (if you can even call this a start). I thought about writing any vector of $\mathbb{R}^n$ as $r$ and then letting $r = u \oplus w$. However, I do not see how this helps.

Any tips?

答:
All you have to prove that $W^{\bot} \cap U^{\bot} = \{0\}$ and then you use some dimension argument like $$\dim U+\dim U^{\bot }=n$$
for every subspace $U$.

Let $w_1,w_2,...,w_k$ be a basis for $W$ and let $u_1,u_2,...,u_m$ be a basis for $U$. Then all these vector form a basis for $\mathbb{R}^n$ so $k+m=n$, since $U \oplus W = \mathbb{R}^n$.

If $v\in W^{\bot} \cap U^{\bot}$ then $v\cdot w_i=0$ and $v\cdot u_j=0$ for all $i,j$. So $v$ is in $(\mathbb{R}^n)^{\bot} = \{0\}$ so $v=0$.

Since $W^{\bot} \oplus U^{\bot} \leq \mathbb{R}^n$ we have  $$n=(n- k)+(n-m)= \dim W^{\bot} +  \dim U^{\bot} =\dim (W^{\bot} \oplus U^{\bot})\leq n$$

So $\dim (W^{\bot} \oplus U^{\bot}) = n$ so $W^{\bot} \oplus U^{\bot} = \mathbb{R}^n$


问:
Integral question with a constant function resulting in solving a confusing equation
The question is: Let $f(x)= \int_0^{2a^3} \sin(\frac{x}{a})dx$, where $a$ is a positive constant less than 4. How many values of $a$ are there such that $f(x)^2=2f(x)$? (no calculator)

My problem comes from the "no calculator part". I think if I integrate $f(x)$ correctly I get $a-a\cos(2a^2)$. But if I set it equal to 2 ($a-a\cos(2a^2)-2=0$)  since $f(x)$ can equal either 0 or 2, I have no clue how to solve it without a calculator. Could someone help me figure out how to evaluate this?

答:
If you calculate the integral you get
$$f(x)=a\Big(1-\cos(2a^2)\Big) \mbox.$$
Now you want to solve:
$$a^2\Big(1-\cos(2a^2)\Big)^2=2a\Big(1-\cos(2a^2)\Big)$$
which becomes
$$a\cos^2(2a^2)+(2-2a)\cos(2a^2)+(a-2)=0$$
and solving for $\cos(2a^2)$ with the quadratic formula you get
$$\cos(2a^2)=\frac{2a-2\pm 2}{2a}$$
So the two solutions are: $1$ and $\frac{a-2}{2}$.

(i) If $\cos(2a^2)=1$, then $2a^2=2\pi k$ with $k\in \mathbb{Z}$,so $a=\sqrt{\pi k}$; but only when $a\le 4$, which means only when $\pi k \le 16$ (by hand you can see that $3,15\cdot 5 = 15,75$), so 5 solutions.

(ii) Now if $\cos(2a^2)=\frac{a-2}{a}$, you can graph the two functions of variable $a$ and see that they have only one solution and it is between $0$ and $2$.

In total 6 solutions.

I hope no mistakes were made.


问:
Complex Discrete Form
I was studying mechanical vibrations and had a doubt at the following differential equation as mentioned in the figure below

Here the above equation has the exact solution in the form of  $ cos(wt) $ but the discrete solution will have an additional  $ sin(wt) $ term also. How is this exact complex discrete form prooved?

I have tried to represent $ cos(wt) $ in terms of hyperbolic exponential terms, to get a discrete form, but when I tried to take the increments in "n", the solution seems to be wrong.

答: I general, the solution of the differential equation $y'' + \omega^2 y=0$ is a linear combination of two independent components: $cos(\omega x)$ and $sin(\omega x)$. For this particular problem, the constraint $y'(0)=0$  rules out $sin(\omega x)$ from the solution.

Title: Complex Discrete Form


问:
How do I reduce the index of a radical $\sqrt[9]{64a^6b^{12}}$?
$\sqrt[9]{64a^6b^{12}}$

It doesn't have to be the answer, just tell me what I should do

Body: $64= 2^6$ so this could be written $(2^6a^6b^{12})^{1/9}= 2^{6/9}a^{6/9}b^{12/9}= 2^{2/3}a^{2/3}b^{4/3}= b\sqrt[3]{2^2a^2b}$.


问:
How to establish (formally) equality of sets?
The axiom of extensionality establishes that "if two sets have the same elements, then they are identical", but not the second implication (the one I will use to prove that two sets are equal).

If we try to solve this by defining "A and B are equal sets iff they have the same elements", then why do we need to establish the axiom of extensionality?

答:
Preliminary note : in order to prove that two sets are identical, you will use " if A and B have the same elements, they are equal". You will not use " if A = B, they have the same elements" for you cannot use as a starting point what you're supposed to prove. You need your target sentence as the consequent of the conditional you are going to use.

The extensionality principle ( a biconditional statement) says two things:

(1) being coextentional ( having the same elements) is a necessary
condition for a set A and a set B to be identical ( if A = B, then A
and B have the same elements).

(2) being coextentional is a sufficient condition for them to be
identical ( if A and B have the same elements, then A = B).

Assertion (1) is no surprise , as is clearly shown above. it follows from logic alone.

But assertion (2) is really a thesis with genuine content. This is why the whole statement has to be stated as an axiom.

Assertion (2) expresses the decision to consider as totally identical "two"  collections on the sole ground that they have the same members. Ths is not trivial.

Consider the following  collections : $C_1$ = " the set of all extraterrestrial bodies on which man has landed before 1970 "  and $C_2$ " the set of all natural satellites of the Earth".

"These two collections are not the same, are they? Their definitions do not mean the same thing at all..."

The extensionality axiom part (2) says : yes,  they are perfectly identical, since they have exactly the same elements, namely a unique element in each case :

$C_1 = \{The \space Moon\}= C_2$.

Note : part (2) of the axiom amounts to the decision not to take into account intensional or conceptual differences between collections, but only their extensional aspect; the intension / extension distinction being inherited from traditional logic.


答:
To reduce $(a)^{(1/b)}$ where $a,b \,\in \,\mathbb{Z^+}$ and $1 < a,b$ the first step is to try to factor $b$.

Your goal is to try to find positive integers $r,s$ such that the following is true:

$(r \times s) = b.$

There exists a positive integer $c$ such that $a = c^r$.

Then, you have that

$$(a)^{(1/b)} = \left(c^r\right)^{\frac{1}{(r\times s)}} = \left\{\left(c^r\right)^{1/r}\right\}^{1/s} = (c)^{1/s}.$$

Title: How do I reduce the index of a radical $\sqrt[9]{64a^6b^{12}}$?

Body:
$\sqrt[9]{64a^6b^{12}}$

It doesn't have to be the answer, just tell me what I should do


问:
Show that the limit sequence $f$ is in $\mathcal{L}^1$
For $\Omega = \mathbb{N}$ with counting measure is $\mathcal{L}^1$ complete.
Let $(f_k)_{k \in \mathbb{N}}$ be Cauchy sequence in $\mathcal{L}^1$ such that  every $f_k$ is a sequence $(f^n_k)_{n \in \mathbb{N}}$.

I already showed that a sequence $f=(f^n)_{n \in \mathbb{N}}$ exists such that $f_k$ converges pointwise to $f$. How do I show that the limit sequence $f$ is in $\mathcal{L}^1$?

答:
Attention: Since the functions $f_k$ are functions defined on $\mathbb{N}$, it is clearer to write $f_k(n)$ and $f(n)$ instead of $f_k^n$ and $f^n$.

Let us prove the result.

Since $(f_k)_{k \in \mathbb{N}}$ be Cauchy sequence in $\mathcal{L}^1(\mathbb{N},\#)$ (where $\#$ is the counting measure), we have that, for every $\varepsilon >0$, there is $M \in \mathbb{N}$, such that, for all $r,s >M$,
$$ \sum_{i \in \mathbb{N}} |f_r(i) -f_s(i)| = \int |f_r -f_s| d\# < \varepsilon/2$$

So, for every $\varepsilon >0$, there is $M \in \mathbb{N}$, such that, for all $r,s >M$ and for all $p\in\mathbb{N}$,
$$ \sum_{i=0}^p |f_r(i) -f_s(i)| \leq \sum_{i \in \mathbb{N}} |f_r(i) -f_s(i)| = \int |f_r -f_s| d\# < \varepsilon/2$$

Now, consider a fixed $p\in\mathbb{N}$, since $f_r$ converges pointwise to $f$, we have, for all $s >M$
\begin{align*}
\sum_{i =0}^p |f(i) -f_s(i)|&= \sum_{i =0}^p  | \lim_{r \to +\infty}f_r(i) -f_s(i)| =\\
&=\sum_{i =0}^p \lim_{r \to +\infty} |f_r(i) -f_s(i)|=\\
&=\lim_{r \to +\infty} \sum_{i =0}^p |f_r(i) -f_s(i)|\leq\varepsilon/2
\end{align*}

So, for any $p\in\mathbb{N}$, for all $s >M$
$$\sum_{i =0}^p |f(i) -f_s(i)| \leq\varepsilon/2$$
So,  for all $s >M$,
$$ \int |f -f_s| d\#=\sum_{i \in \mathbb{N}} |f(i) -f_s(i)| = \sup_p \left(\sum_{i =0}^p |f(i) -f_s(i)|\right) \leq\varepsilon/2<\varepsilon \tag{1} $$

It also follows that, for all $s >M$,

$$\left | \int |f| d\# - \int |f_s| d\#\right | \leq \int | \, |f|-|f_s| \, | d\# \leq \int |f -f_s| d\# \leq\varepsilon/2<\varepsilon \tag{2}$$

So, by $(2)$ we see that $\int |f| d\# <+\infty$, so $f \in \mathcal{L}^1(\mathbb{N},\#)$.

By $(1)$, we have that $f_s$ converges to $f$ in $\mathcal{L}^1(\mathbb{N},\#)$.

And, by $(2)$ again, we see that $\int |f_s| d\#$ converges to $\int |f| d\#$. That means, $\|f_s\|_1$ converges to $\|f\|_1$.


Body:
Notice that, $\sqrt[n]{x}=x^{\frac{1}{n}}$, in your case, $n = 9$, and $64=2^6$, so

$$\sqrt[9]{64a^6b^{12}}=\sqrt[9]{2^6a^6b^{12}}=\sqrt[9]{(2a)^6b^{12}}$$

Using the above property,

$$\sqrt[9]{(2a)^6b^{12}}=\Big((2a)^6b^{12}\Big)^{\frac{1}{9}}=(2a)^{\frac{6}{9}}b^{\frac{12}{9}}=(2a)^{\frac{2}{3}}b^{\frac{4}{3}}=\Big((2a)^2b^{4}\Big)^{\frac{1}{3}}$$

Therefore,

$$\sqrt[9]{(2a)^6b^{12}}=\Big((2a)^2b^{4}\Big)^{\frac{1}{3}}=\sqrt[3]{(2a)^2b^{4}}$$

It is possible to simplify a little more, noting that $b^4=b\cdot b^3$, so

$$\sqrt[3]{(2a)^2b^{4}}=\sqrt[3]{(2a)^2b\cdot b^3}=b\sqrt[3]{(2a)^2b}$$


问:
Show that $Y_t = (W_t + t)\exp\{-W_t - \frac{1}{2}t\}$ is a martingale.
Let $W$ be standard (one-dimensional) Brownian motion and define the process $Y = \left\{ Y_t, \mathcal{F}_t; 0 \le t < \infty \right\}$ by
$$ Y_t = (W_t + t)\exp\left\{-W_t - \frac{1}{2}t \right\}. $$
I am trying to show that $Y$ is an $\left\{ \mathcal{F}_t \right\}$-martingale using Ito's Formula.

Here's my approach:

For each $t \ge 0$, put $G_t := W_t + t$, $M_t := -W_t - \frac{1}{2}t$, and $H_t := \exp\{M_t\}$. Then $Y_t = G_t H_t$, and $M$ is an Ito process with $(u_t, v_t) = (-1/2, -1)$. Now, let $f(t,w) = f(w) = e^w$. Then $f(M_t) = f'(M_t) = f''(M_t) = H_t$; and so, by Ito's Formula,
$$ d H_t = \left[0 - \frac{1}{2}f'(M_t) d M_t + \frac{1}{2}f''(M_t)\right]dt - f'(M_t) dW_t = 0 - f'(M_t) dW_t = - H_t dW_t. $$
This means that
$$ H_t = 1 - \int_0^t H_s \, dW_s, \qquad t \ge 0;$$
and so, the process $H := \left\{H_t, \mathcal{F}_t; 0 \le t < \infty \right\}$ is a martingale.

Similarly, $G$ is an Ito process with $(u_t, v_t) = (1,1)$; i.e.,
$$ d G_t = d t + d W_t. $$
Now, using the integration-by-parts formula,
\begin{align*}
d Y_t &= G_t \, dH_t + H_t \, dG_t + dG_t \, dH_t \\
 &= -G_t H_t d W_t + H_t\left(d t + d W_t \right) - \left(d t + d W_t \right)\left(H_t d W_t \right)  \\
 &= 0 + (H_t - G_t H_t)d W_t \\
 &= (1 - t - W_t)H_t d W_t.
\end{align*}
Thus,
\begin{align*}
Y_t &= \int_{0}^t (1 - s - W_s)H_s d W_s.
\end{align*}
So, if we put $Q_t := (1 - t - W_t)H_t$, for each $t \ge 0$, then it clear that the process $Q := \left\{Q_t, \mathcal{F}_t ; 0 \le t < \infty \right\}$ is measurable and $\left\{\mathcal{F}_t \right\}$-adapted. Now, we just need to show that
\begin{align*}
\mathbb{E}\left[\left(\int_{0}^T Q_s^2 \, ds\right)^{1/2} \right] &= \mathbb{E}\left[\left(\int_{0}^T (1 - s - W_s)^2H_s^2 d s \right)^{1/2}\right] \\
&= \mathbb{E}\left[\left(\int_0^T (s^2H_s^2 + 2sW_sH_s^2 - 2sH_s^2 + W_s^2H_s^2 - 2W_sH_s^2 + H_s^2) ds \right)^{1/2}\right]
\end{align*}
is finite for all $T > 0$. Naturally, this is where I am stuck. Given the ugliness of this integrand, I can only I imagine I've done something wrong here (or that I am overlooking some "slicker" approach)...

答: Actually you need to show something stronger, that $$\mathbb{E}[\int_0^T Q_s^2ds]< +\infty.$$ This is much easier (and actually implies the thing that you wanted to show, because of Holder). You can do this exchanging the integrals and using the fact that you know the law of $W_s$ for any given $s$.


问:
Surjectiveness of $e^{x^2}$
Let $f: \mathbb R \to \mathbb R^+$ be given by $f(x)=e^{(x^2)}$. I am not sure whether this function is surjective or not.

If I am not wrong, I can see surjectivity for codomain $[1,\infty)$.
Let $y \in [1,\infty)$ then $\sqrt{\ln y} \in \mathbb R^+$. Say $\sqrt{\ln y}=x$. $\Rightarrow \ln y=x^2 \Rightarrow e^{(x^2)}=y=f(x)$. Could you please fix me if I made mistake?

However, I am not sure about $y<1$ case. I need help on this.

Very thanks in advance for any help.

答:
If $0 < y < 1$ then $\ln y< 0$.  So if $e^{x^2} = y< 1$ then $x^2  = \ln y < 0$.

Which... is not possible.

So $f(x) =e^{x^2} = y <1$ is not possible and $f$ is not surjective.

Title: Surjectiveness of $e^{x^2}$


问:
Need help with this proof by induction
I don't understand a step in the following proof by induction.

I have to prove that:
$$A(n):\quad2^n>n+1$$
By induction:

The base clause is given by $n=2$, which yields $2^2=4$ and $2+1=3$, which is true because $4>3$.

The proof starts as follows:
$$2^{k+1}=2\cdot 2^k.$$
Due to
$$A(n):\quad2^n>n+1$$
it follows that:
$$2^{k+1}>2\cdot (k+1).$$

The rest of the proof is as follows:
$$ =2k+2$$
$$ = k+2+\underbrace{k}_{\geq2} $$
$$ > k+2 $$
$$ = (k+1) +1$$

Where does the $k$ come from in the second line?

答:
I am a bit confused on what you wrote, so I will just prove this with how induction usually goes. First, find a base case:
$$4=2^2>2+1=3$$
Now we have to show that $2^k>k+1\implies 2^{k+1}>(k+1)+1=k+2$. Assume $2^k>k+1$, then
$$2^k>k+1\implies 2(2^k)>2(k+1)\implies 2^{k+1}>2k+2>k+2$$
and we are done.

Title: Need help with this proof by induction


问:
Comparing Rate of Convergence of Two Square Root Algorithms
Source: Rudin's "Principles of Mathematical Analysis" 3.17(d)

Problem: Compare the rapidity of convergence of the process

$$x_{n+1}=\frac{\alpha+x_n}{1+x_n}=x_n+\frac{\alpha-x_n^2}{1+x_n}$$
with $\alpha>1$ and $x_1>\sqrt{\alpha}$ with the process
$$x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right).$$
In 16b, Rudin lets $\epsilon_n = x_n-\sqrt{\alpha}$ and we show that with $\beta=2\sqrt{\alpha}$,
$$\epsilon_{n+1}<\beta\left(\frac{\epsilon_1}{\beta}\right)^{2^n}.$$

My work:
For finding the rate of convergence in 16, we must find the $$\alpha\geq 1: \lim\limits_{n\rightarrow\infty} \frac{|\epsilon_{n+1}|}{|\epsilon_n|^\alpha}<\infty.$$ After 16b, we can deduce that with $\alpha=2$, the limit converges and so the rate of convergence is quadratic. For the latter, we just go through the same process as in 16b,c. I think $\beta=\frac{(\sqrt{\alpha}-1)x_1-1+\sqrt{\alpha})}{2x_1+1+\alpha}$ gives us a similar inequality as in 16b. Then from here,
$$\lim\limits_{n\rightarrow\infty}\frac{\left|\beta\left(\frac{\epsilon_1}{\beta}\right)^n\right|}{\left|\beta\left(\frac{\epsilon_1}{\beta}\right)^{n-1}\right|^\alpha}<\infty\iff \lim\limits_{n\rightarrow\infty}n-(n-1)\alpha<\infty\iff \alpha=1.$$
Hence it's rate of convergence is linear at best.

My Question: I was hoping to get some feedback on if my work is sufficient/good.

答:
For the first one,

$\begin{array}\\
x_{n+1}
&=\dfrac{a+x_n}{1+x_n}\\
\text{so}\\
x_{n+1}-\sqrt{a}
&=\dfrac{a+x_n}{1+x_n}-\sqrt{a}\\
&=\dfrac{a+x_n-\sqrt{a}(1+x_n)}{1+x_n}\\
&=\dfrac{a+x_n-\sqrt{a}-\sqrt{a}x_n}{1+x_n}\\
&=\dfrac{a-\sqrt{a}+x_n-\sqrt{a}x_n}{1+x_n}\\
&=\dfrac{\sqrt{a}(\sqrt{a}-1)-x_n(\sqrt{a}-1)}{1+x_n}\\
&=\dfrac{(\sqrt{a}-1)(\sqrt{a}-x_n)}{1+x_n}\\
\text{so}\\
\dfrac{x_{n+1}-\sqrt{a}}{\sqrt{a}-x_n}
&=\dfrac{\sqrt{a}-1}{1+x_n}\\
\text{or}\\
\dfrac{x_{n+1}-\sqrt{a}}{x_n-\sqrt{a}}
&=\dfrac{1-\sqrt{a}}{1+x_n}\\
\end{array}
$

so the convergence is linear.

For Newton,

$\begin{array}\\
x_{n+1}
&=\dfrac{1}{2}\left(x_n+\dfrac{a}{x_n}\right)\\
\text{so}\\
x_{n+1}-\sqrt{a}
&=\dfrac{1}{2}\left(x_n+\dfrac{a}{x_n}\right)-\sqrt{a}\\
&=\dfrac{1}{2}\left(x_n-2\sqrt{a}+\dfrac{a}{x_n}\right)\\
&=\dfrac{1}{2}\left(\dfrac{x_n^2-2x_n\sqrt{a}+a}{x_n}\right)\\
&=\dfrac{1}{2}\left(\dfrac{(x_n-\sqrt{a})^2}{x_n}\right)\\
\end{array}
$

so the convergence is quadratic.


问:
$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$

Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
 1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
 {\rm d}t=\dfrac{\pi }{\sqrt{2}}$.

We can obtain
$$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$
but how to go on ?

Body:
Finally I was able to handle this problem. It is just tedious calculation. Let $u=\frac{f(t)}{f(x)}$ and then
\begin{eqnarray}t&=&1-\frac{(1-x)(x-1+\sqrt{(1-x)^2+8u(1+x)}}{2u(1+x)},\\
dt&=&-\frac12\frac{ (x-1) \left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)}{u^2(1+x)\sqrt{(1-x)^2+8u(1+x)}}du.
\end{eqnarray}
So
\begin{eqnarray}
&&\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t\\
&=&\lim_{x \to 1}\int_{\frac1{f(x)}}^1\frac{\sqrt{t}\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
   }{u \sqrt{(1-u)
   (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}{\rm d}u\\
&=&\int_0^1\frac{1}{\sqrt{2u(1-u)}}du\\
&=&\frac{\pi}{\sqrt 2}.
\end{eqnarray}
Here
$$ \lim_{x \to 1}\frac1{f(x)}=0, \lim_{x \to 1}t=1,  $$
and it turns out that when $x=1$, the integrand
$$ \frac{\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
   \sqrt{\frac{(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)}{2 u (x+1)}+1}}{u \sqrt{(1-u)
   (x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}\bigg|_{x=1}=\frac{1}{\sqrt{2u(1-u)}}$$


问:
Showing that $K[x,y,z]/(y^2 - xz) \simeq K[x,y,\frac{y^2}{x}]$
Let $K$ be a field and consider the ideal $(y^2 - xz) \lhd K[x,y,z]$. Intuitively, it seems clear to me that $K[x,y,z]/(y^2 - xz)$ should be isomorphic to the subring $K[x,y,\frac{y^2}{x}]$ of $K(x,y) = \text{Frac}(K[x,y])$. The only way I can think of showing this is to take the obvious map $\varphi: K[x,y,z] \rightarrow K[x,y,\frac{y^2}{x}]$ and show that $\ker\varphi = (y^2 -xz)$. Of course $(y^2 -xz) \subseteq \ker\varphi$ is clear, but I have problems with showing containment in the other direction. More precisely, I probably could just "brute-force" it, but I feel that there must be a more elegant way of showing this, given that the conclusion seems so obvious.

Is there a "nice" way to show either $\ker \varphi \subseteq (y^2 - xz)$, or directly that $K[x,y,z]/(y^2 - xz) \simeq K[x,y,\frac{y^2}{x}]$? I would be particularly happy to learn general techniques that can be used in similar situations.

答: We can write $$f(x,y,z)=(z-y^2/x)g(y,z)+r(y)$$ with $g\in k(x)[y,z]$ and $r\in k(x)[y]$. Multiplying by $xh(x)$ we get $$xh(x)f(x,y,z)=(xz-y^2)h(x)g(y,z)+xh(x)r(y),$$ where $h\in k[x]$ is chosen with the property $h(x)g(y,z),h(x)r(y)\in k[x,y,z]$. Since $f(x,y,y^2/x)=0$ we have $xh(x)r(y)=0$, so $r=0$. Thus $xh(x)f(x,y,z)\in(xz-y^2)$. But the ideal $(xz-y^2)$ is prime, and $x,h(x)\not\in(xz-y^2)$, so $f\in(xz-y^2)$.


问:
Dual basis and annihilator problem
I think they're fairly simple but I really don't know where to start, the problems are these:

First one:

$V$ a vector space of dimension $n$ and $\phi \in V^* \setminus \{0\}$.

Prove that $\dim \ker \phi = n-1$.

Second one:

Let $B = \begin{bmatrix}2 & -2 \\ -1 & 1\end{bmatrix}$ and $W = \{A \in \operatorname{Mat}_{\mathbb R}(2\times2) : AB = 0\}$. Suppose $f \in W^o$ (the annihilator of $W$) such that $f(\operatorname{Id}_{2\times2}) = 0$ and $f\left({\scriptstyle\begin{bmatrix}0 & 0 \\ 0 &1\end{bmatrix}}\right) = 3$. Calculate $f(B)$.

For this is one I don't really know when I have to use that $AB = 0$.

答:
For the first question, recall the Rank-Nullity Theorem: for any linear operator $T:V\to W$ between two finite dimensional vector spaces the following equality holds: $\dim\text{ker}\ T+\dim\text{Im}\ T=\dim V$. If $\phi\in V^*\setminus\{0\}$, then $\phi:V\to\mathbb{F}$, where $\mathbb{F}$ is the field over which you consider $V$ and $W$. The dimension of $W=\mathbb{F}$ over $\mathbb{F}$ is 1. Now use the theorem.

Concerning the second question, notice that $W$ is the subspace of matrices of the form:

$$\begin{bmatrix}{}
a & 2a\\
c & 2c
\end{bmatrix}$$

where $a,c\in\mathbb{R}$. Thus $\dim W=2$. As $f(Id)=0$, $Id\not\in W$ and $f\neq0$, we have that $\ker\ f=W\oplus\text{span}\{Id\}$. Now we have:

$$f(B)=f\begin{bmatrix}{}
2 & -2\\
-1 & 1
\end{bmatrix}=f\begin{bmatrix}{}
-1 & -2\\
0 & 0
\end{bmatrix}+f\begin{bmatrix}{}
0 & 0\\
-1 & -2
\end{bmatrix}+f\begin{bmatrix}{}
3 & 0\\
0 & 3
\end{bmatrix}$$

The first and the second matrices from RHS are in W, the third one is the multiplication of $Id$. So $f(B)=...$

Title: Dual basis and annihilator problem


问:
Does the ratio of the $x-$values of two continuous functions $f(x), g(x) \in \mathbb{R^+}$ tell you anything about the ratio between their $y-$values?
Does the ratio of the $x-$values of two continuous functions $f(x), g(x) \in \mathbb{R^+}$ tell you anything about the ratio between their $y-$values?

In general, obviously not. But suppose the following is also true:

$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0$,

$\lim_{x \to \infty} \frac{f(x)}{g(x)}$ converges to $L_1$ and $\lim_{y \to 0+} \frac{f^{-1}(y)}{g^{-1}(y)}$ converges to $L_2$.

Then what can we say about $L_2$ in terms of $L_1$? e.g. If $L_1 = \infty$ then does $L_2 = \infty$? If $L_1$ is finite then is $L_2$ finite?

答:
If $\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0$ and
\begin{equation}
  \lim_{x \to \infty} \frac{f(x)}{g(x)}=L_1 \text{ and }
\lim_{y \to 0+} \frac{f^{-1}(y)}{g^{-1}(y)}=L_2\tag{1}
\end{equation}
then if $L_1>1$, $L_2\geq 1$ and if $L_1<1$, $L_2\leq 1$.  Since
$L_1>1$ implies that for sufficiently large $x$, $f(x)>g(x)$, for
sufficiently small $y$, $f^{-1}(y)>g^{-1}(y)$ ($x$ must be bigger for
$f$ to shrink to $y$). Thus, $L_2\geq 1$. The second conclusion is
similarly shown. Too many examples:

Let $L_1$ and $L_2$ be arbitrary positive real numbers either
both greater than one or both less than one. Set
$\alpha=\log{L_2}/\log{L_1}$ and note that $\alpha>0$. Let
$f(x)=L_1x^{-1/\alpha}$ and $g(x)=x^{-1/\alpha}$ so that
$f^{-1}(y)=L_2y^{-\alpha}$ and $g^{-1}(y)=y^{-\alpha}$, and hence (1) holds.

Let $L_1=1$ and $L_2$ be an arbitrary positive real number. Let
$a=-\log{L_2}$ and set $f(x)=(\log x+a)^{-1}$ and $g(x)=(\log x)^{-1}$.
Since $f^{-1}(y)=L_2e^{1/y}$ and $g^{-1}(y)=e^{1/y}$,  (1) holds.

Let $L_1=1$ and $L_2=\infty$. Set
$f(x)=(\log\log x-\log 2)^{-1}$ and $g(x)=(\log\log x)^{-1}$.  Since
$f^{-1}(y)=e^{2e^{1/y}}$ and $g^{-1}(y)=e^{e^{1/y}}$, (1) holds.

Let $L_1=1$ and $L_2=0$. Set
$f(x)=(\log\log x+\log 2)^{-1}$ and $g(x)=(\log\log x)^{-1}$.  Since
$f^{-1}(y)=e^{\frac12 e^{1/y}}$ and $g^{-1}(y)=e^{e^{1/y}}$, (1) holds.

Let $L_1=\infty$ and $1<L_2<\infty$. Set
$f(x)=e^{-x/L_2}$ and $g(x)=e^{-x}$.  Since
$f^{-1}(y)=-L_2\log y$ and $g^{-1}(y)=-\log y$, (1) holds.

Let $L_1=0$ and $0<L_2<1$. Set
$f(x)=e^{-x/L_2}$ and $g(x)=e^{-x}$.  Since
$f^{-1}(y)=-L_2\log y$ and $g^{-1}(y)=-\log y$, (1) holds.

Let $L_1=0$ and $L_2=0$. Set
$f(x)=x^{-2}$ and $g(x)=x^{-1}$.  Since
$f^{-1}(y)=y^{-1/2}$ and $g^{-1}(y)=y^{-1}$, (1) holds.

Let $L_1=\infty$ and $L_2=\infty$. Set
$f(x)=x^{-1}$ and $g(x)=x^{-2}$.  Since
$f^{-1}(y)=y^{-1}$ and $g^{-1}(y)=y^{-1/2}$, (1) holds.


问:
Given any sequence of real numbers $a_n$, show that there exists a subsequence $a_{n_k}$ of $a_n$....
Given any sequence of real numbers $a_n$, show that there exists a subsequence $a_{n_k}$ of $a_n$ and a real number $b$ such that $\sum\limits_{k\geq1}\mid b-\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}\mid$ is convergent.

I tried like this. For any subsequence  $a_{n_k}$ of $a_n$ , the sequence $\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}$ is always a bounded sequence. Hence it contains a convergent subsequence. Now from this I got stuck. please help

答:
Since there is a $b$ and $(n_k)|_{k=1}^{\infty}$ such that
$\dfrac{a_{n_k}}{1+\mid a_{n_k}\mid}
\to b
$,
we can choose a subsequence
$(n_{k_m})|_{m=1}^{\infty}$ of $n_k$
such that
$\left|\dfrac{a_{n_{k_m}}}{1+\mid a_{n_{k_m}}\mid}- b\right|
\lt \dfrac1{2^m}
$,
and the sum of these converges.


问:
Computing 3D Zernike Moments on 3D Point Clouds
I'm trying to understand the formulation for 3D Zernike descriptors used in A Comparative Study of Object Classification Methods Using 3D Zernike Moment on 3D Point Cloud. For context, I'm looking for a translation invariant method for comparing point cloud representations of molecular surfaces. In researching this, the subject of 3D Zernike moments comes up a lot (for example). I'm having trouble finding a clean algorithm to actually compute these values.

The "3D Zernike Moment on 3D Point Cloud" paper promises a fast computation of these moments, but quite give enough to go on. Here's what the paper says:

\begin{align}
Z_{l,m,n}(X) = \sum_{v=0}^{k} Q_{k,m,n} |X|^{2v}e_{m,n}(X)
\end{align}

\begin{align}
Q_{k,m,n} = \frac{(-1)^{k}}{2^{2k}}  \sqrt{\frac{2m+4k+3}{3}} \binom{2k}{k} (-1)^{v} \frac{\binom{k}{v}\binom{2(k+m+v)+1}{2k}}{\binom{k+m+v}{k}}
\end{align}

Where

\begin{align}
l \in [0, \text{Max}], \quad m\in[0,l], \quad n\in[-m,m], \quad k=\frac{l-m}{2}
\end{align}

The notation of $|X|^{2v}e_{m,n}(X)$ is not defined in the paper.

The paper references this algorithm for computing the moments, but I can't seem to match the notation between papers.

I found aa Python implementation here, but this implementation uses meshes (point clouds with triangular faces) rather than just point clouds, as described in the first paper. This method also requires iterating over all faces, which is very slow. The paper describes computation in milliseconds to seconds.

Is anyone familiar with how to compute these values?

答:
Summary: One definition of $\mathbb{Z}$ is in spherical coordinates, the other in Cartesian coordinates. $\mathbb{R}$ is used for any 3D point in spherical coordinates, and $X$ for any point in Cartesian coordinates.

In https://doi.org/10.1155/2012/353406 formulae 2.1 - 2.9 explicitly define these:

3D Zernike polynomials are defined as
$$\mathbb{Z}_{n, \ell, m}(r, \theta, \phi) = R_{n, l}(r) Y_{\ell, m}(\theta, \phi) \tag{2.1}\label{2.1}$$
Any function $f(r, \theta, \phi)$ defined within the unit ball can be described in terms of the 3D Zernike polynomials via
$$f(r, \theta, \phi) = \sum_{n=0}^{\infty} \sum_{\ell=0}^{n} \sum_{m=-n}^{n} \Omega_{n, \ell, m} \mathbb{Z}_{n, \ell, m} \tag{2.2}\label{2.2}$$
where $\Omega_{n, \ell, m}$ are the 3D Zernike moments, and determined (mathematically) by
$$\Omega_{n, \ell, m} = \int_{0}^{1} d r \int_{0}^{\pi} d \theta \int_{0}^{2 \pi} d \phi ~ \overline{\mathbb{Z}_{n, \ell, m}(r, \theta, \phi)} ~ f(r, \theta, \phi) ~ r^2 \sin \theta \tag{2.3}\label{2.3}$$
where the overline refers to complex conjugation, $\overline{~ a + i b ~} = a - i b$, where $i$ is the imaginary unit, $i = \sqrt{-1}$.

The Cartesian-to-Spherical coordinate system mapping used here is
$$\begin{aligned}
x &= r \sin \theta \sin \phi \\
y &= r \sin \theta \cos \phi \\
z &= r \cos \theta
\end{aligned} \tag{2.4}\label{2.4}$$

In Cartesian coordinates,
$$\mathbb{Z}_{n, \ell, m}(x, y, z) = \sum_{v=0}^{k} Q_{k, \ell, v} ~ \lvert x^2 + y^2 + z^2 \rvert^v ~ e_{\ell, m}(x, y, z) \tag{2.5}\label{2.5}$$
with $k \in \mathbb{N}$ and $k = (n - \ell) / 2$, and where
$$Q_{k, \ell, v} = \frac{(-1)^k}{2^{2 k}} ~ \sqrt{\frac{2\ell + 4 k + 3}{3}} ~ {{2 k} \choose k} ~ (-1)^v ~ \frac{{k \choose v} ~ {{2 ( k + \ell + v ) + 11} \choose {2 k }}}{{k + \ell + v} \choose k} \tag{2.6}\label{2.6}$$
and
$$\begin{aligned}
e_{\ell, m}(x, y, z) ~ = & ~ C_{\ell, m} ~ \sqrt{x^2 + y^2 + z^2}^\ell ~ \left(\frac{i x - y}{2}\right)^m ~ (x + i y)^{\ell - m} ~ \cdot \\
~&~ \sum_{\mu = 0}^{\lfloor (\ell - m) / 2 \rfloor} ~ {\ell \choose \mu} ~ {{ \ell - \mu } \choose { m + \mu }} ~ \left(-\frac{x^2 + y^2}{4 (x + i y)^2}\right)^\mu  \\
\end{aligned}\tag{2.7}\label{2.7}$$
where and the $\cdot$ (multiplication) is a reminder that the expression is a product split on two lines.  In $\eqref{2.7}$,
$$C_{\ell, m} = \frac{\sqrt{(2\ell + 1) ~ (\ell + m)! ~ (\ell - m)!}}{\ell !} \tag{2.8}\label{2.8}$$
(note, factorial ($!$) has precedence over multiplication) so $C_{-\ell, m} = C_{\ell, m}$, and for $\eqref{2.7}$,
$$e_{\ell, -m}(x, y, z) = (-1)^m ~ \overline{e_{\ell, m}(x, y, z)} \tag{2.9}\label{2.9}$$
where the overline again indicating complex conjucation.

As a reminder, $\lfloor ~ \rfloor$ indicate floor (round towards zero) operation, the binomial coefficient is
$${n \choose k} = \frac{n!}{k! ~ (n-k)!}$$
and factorial $n! = 1 \cdot 2 \cdot \ldots \cdot (n-1) \cdot n$.


问:
Limit Involving a Tetration of the Variable
We wish to evaluate this challenging limit
$$
\lim_{x\to 1}\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}
$$

My Observation : As an initial guess, I shall consider $f_{n}(x)=x^{x^{x{^{.^{.^{.}}}}}}$ with $n$ being the number of $x$. To simplify this problem I considered the case of $n=1$, it is easy to see that the limit is $1$, to be sure of my assertion I further considered the case for $n=2$, we receive :
$$
\left(\frac{x^{x}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}
$$
We thus simplify this expression :
\begin{align*}
\left(\frac{x^{x}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{\exp(x\ln(x))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{\exp(x(x-1+o(x-1)))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{1+x(x-1)+o(x(x-1))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(x+o(x)\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}(x-1+o(x-1))\right) \\
&=\exp(\sqrt{x-1}+o(\sqrt{x-1}))
\end{align*}
Where :$$\displaystyle\lim_{x\to 1}\exp(\sqrt{x-1}+o(\sqrt{x-1}))=1$$

Now the big question is how can we extend this for the general case that is for $n\geq3$? Any help would be much appreciated.

答:
$\textbf{Hint}:$ In the limit, the following relationship holds

$$x^{x^{\cdots}} = y \implies x^y = y$$

$$ 1 = yx^{-y} = ye^{-y\log x} $$

which means

$$y = \frac{-W(-\log x)}{\log x}$$

where $W$ is the principal branch of the Lambert-W or product log function (the principal branch is chosen because the limit takes $\log x \to 0$, and the two branches would either make $y$ a $\frac{0}{0}$ or $\frac{\infty}{0}$ situation). Can you take it from here?


问:
Set theory question about countable sets
Prove that $\omega\cup\omega^{2}\cup$... is countable

References

...a set is called denumerable (countable)if it is in
one-to-one correspondence with ω.

6.1 Definition By the set of the natural numbers we mean the intersection of all the successor sets. The set of the natural numbers is designated by the symbol ω; every element of ω is called a natural number.

$\approx$= equipotent

7.20 Theorem ω × ω ≈ ω.

7.10 Theorem If A ≈ B and C ≈ D, then A × C ≈ B × D.
————————————————————————————————

I believe it has to be done  by induction  by repeatedly using 7.20,and 7.10 and the fact a union of countable sets is countable.

I think AoC comes into too.
Is my reasoning correct? The only thing I
can’t get is mapping for something continuous like this

Base case

By 7.20 and the fact that countable union of countable sets are countable $\omega \cup \omega^{2}$ is countable

Induction step:
We want to show that $\bigcup\omega^{n^{+}}$
is countable...

Any help would be appreciated.

答:
Given that $\omega^n$ is countably infinite, 7.10 and 7.20 tell you that $\omega^{n+1}$ is countably infinite, so by induction $\omega^n$ is countably infinite for each $n\in\Bbb Z^+$. If you’ve already proved that the union of countably many countable sets is countable, you’re done at that point. Otherwise, you’ll have to prove that the union is countable. For instance, for $n\in\Bbb Z^+$ you could let $f_n:\omega\to\omega^n$ be a bijection and then define

$$h:\omega\times\omega\to\bigcup_{n\in\Bbb Z^+}\omega^n:\langle m,n\rangle\mapsto f_{m^+}(n)$$

and show that it is a bijection.


问:
Find the angle between the 2 points (50.573,-210.265) and (117.833,-80.550)
I am attempting to find the angle between the 2 points (50.573,-210.265) and (117.833,-80.550).

Is my calculation correct because a program is giving me a different answer? It says the angle is 27'24'27.27 DMS

```<code>dx = x2 - x1;
dy = y2 - y1;
angle = Atan2(dy,dx) * 180 / PI;
angle = 62.59242
</code>```

My result is 62.59242 degrees but the program says its 27'24'27.27 DMS. Which is correct?

Also if the points were 3d would that affect the angle? Or does the z axis not impact on the horizontal angle?

答: I agree with your calculation.  It appears they are defining it with the arguments switched.  The fact that you answer is $90-$theirs supports this. Note that there is no angle between two points. What you have is the angle between the line through those points and the $x$ axis.</p>


问:
Is there a clever solution to Arnold's "merchant problem"?
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

答:
Huh. I must be a 5 or 6 year old since I thought this was completely trivial. I have noticed though that I tend to reason visually far more often than algebraically.

Before:

After:

I mean, it has to be! No matter whether it's a spoonful, or a pinch, or whether you moved stuff back and forth 3 or 4 times, or what have you ... in the end, whatever amount of wine gets replaced by tea must have ended up in the cup.


问:
A problem pertaining to functional equations with conditions $P(2020) = 2019$ and (b) $(P(x) + 1)^2 = P(x^2 + 1)$ for all $x\in\mathbb{R}$

Determine all polynomials $P(x)$ with real coefficients satisfying the
following conditions:

$P(2020) = 2019$

$(P(x) + 1)^2 = P(x^2 + 1)$ for all
$x\in\mathbb{R}$

I wrote down $p(x$) as $a_nx^n +\ldots+a_0$ and then started applying substituting $2020$ and tried to get awkward results. I know this is absolutely wrong. Kindly provide a rigorous answer

答:
►Degree $0:\hspace0.5cm P(x)=c$. One must have $(c+1)^2=c\iff c^2+c+1=0$ so  $c$ is not real and there are not solutions.

►Degree $1:\hspace0.5cm Px)=ax+b$. One must have $(ax+b+1)^2=a(x^2+1)+b$ for all $x$. Then
$$a^2-a=0,\hspace0.5cm2a(b+1)=0,\hspace0.5cm(b+1)^2-a-b=0$$ whose only solution with $a\ne0$ is  $(a,b)=(1,-1)$ so $P(x)=x-1$ could be a solution. Since $P(2020)=2019$ we have a solution of the first degree $\boxed{P(x)=x-1}$.

►Degree $2:\hspace0.5cm P(x)=x^2+bx+c$ (for degree greater than or equal to two, clearly the leading coefficient should be equal to $1$ because $a_n^2 = a_n$). One must have $(x^2+bx+c+1)^2=(x^2+1)+b(x^2+1)+c$ for all $x$. Then $$x^4+b^2x^2+(c+1)^2+2bx^3+2(c+1)x^2+2b(c+1)x=x^4+2x^2+1+bx^2+b+c$$ so $b=0$ and $2(c+1)=2$ then $c=-1$. Thus $P(x)=x^2-1$ but then $P(2020)\ne2019$ so there are not solutions for quadratic polynomials.

►Degree $n\gt2:\hspace0.5cm P(x)=x^n+Q(x)$ where $Q(x)$ is of degree at most $n-1$. One must have $$(x^n+Q(x)+1)^2=(x^2+1)^n+Q(x^2+1)\text { for all x}.$$
$$\left(x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+(a_0+1)\right)^2=(x^2+1)^n+\cdots+a_1(x^2+1)+a_0$$
Since all powers in RHS are even, all coefficient of all odd power in LHS must be null. We get
$$a_1(a_0+1)=0\\a_3(a_0+1)+a_1a_2=0\\a_5(a_0+1)+a_1a_4+a_2a_3=0\\a_7(a_0+1)+a_1a_6+a_2a_5+a_3a_4=0\\a_9(a_0+1)+a_1a_8+a_2a_7+a_3a_6+a_4a_5=0\\.......................\\.......................\\a_{2n-1}(a_0+1)+a_1a_{2n-2}+a_2a_{2n-3}+\cdots+a_{n-1}a_{n}=0 \text{ where $a_n=1$}$$ from which it is easily deduced $a_1^2a_2=0,\space a_1a_2^2a_3=0,\space a_1a_2a_3^2a_4=0,......$

Do these necessary conditions lead to $P(x)=x^n-1$ in which case $P(2020)\ne{2019}$?

We see the cubic case separately and leave the general case which offers some difficulty in handling the coefficients for whoever wants to solve it, mainly for those who have suitable electronic calculators. It is not unlikely that there is any artifice to resolve this problem in an easier way.

On the other hand,it seems very likely that the only solution is $P(x)= x-1$.

►Degree $3:\hspace0.5cm P(x)=x^3+bx^2+cx+d$.
$$(x^3+bx^2+cx+d+1)^2=(x^2+1)^3+b(x^2+1)^2+c(x^2+1)+d$$ Equating to $0$ the coefficients of $x$ and $x^3$ on the left side  and the coefficients of $x^2$ on both sides, we have the system
$$\begin{cases}d+1+bc=0\\c(d+1)=0\\c^2+2bd-c=3\end{cases}$$ whose solutions are
$$(a,b,c)=(-1.5,0,-1),(0,\frac{1+\sqrt{13}}{2},-1),(0,\frac{1-\sqrt{13}}{2},-1)$$ and it is verified that none of the three resulting cubic polynomials satisfies $P(2020)=2019$. Thus there is no solution for degree $3$.


问:
inner product space and gram matrix
I have a question from my proffesor that I can not figure it out.

V will be inner product system above R2.

Let E some basis with the gram matrix (E={v1,v2})
This is the gram matrix:

\begin{pmatrix}
  2 & -1\\
  -1 & 1\\
  \end{pmatrix} =
\begin{pmatrix}
  {(v_{1},v_{1})} & {(v_{1},v_{2})}\\
  {(v_{2},v_{1})} & {(v_{2},v_{2})}\\
  \end{pmatrix}

Let T:V->V be a linear map with a matrix represent T according to basis E

\begin{pmatrix}
  1 & 2\\
  2 & 1\\
  \end{pmatrix}

is T is self adjoint ?

well first of all according to gram matrix it is very easy to see that E is not a orthonormal set and therefore T* is necessarily equal to the matrix of T with a transpose and conjunction.

So where can I go from here because I can not assume that the inner product is the standard one and therefore I do not know How to look for it, if exists ?

and if it is not exist how can I proof that if I do not know inner product and by that I can not find the adjoint

答:
Let us call $P$ the matrix of the basis $E$ expressed in an orthonormal basis, so that
$$
P^*P=\pmatrix{2&-1\\-1&1}=Q
$$
and, if $A$ denotes the matrix of $T$ in the same orthonormal basis,
$$
P^{-1}AP=\pmatrix{1&2\\2&1}=B.
$$
So the question amounts to: do we have $A^*=A$? Now, note that $B^*=B$ and
$$
A^*=A\;\iff\;(PBP^{-1})^*=PBP^{-1}\;\iff\; (P^*)^{-1}BP^*=PBP^{-1}\;\iff\;BQ=QB.
$$
I let you check whether the latter is true.</p>

Title: inner product space and gram matrix


问:
If $g(x)$ is a polynomial in $GF(q)$, then $[g(x)]^q$ = $g(x^q)$.
If $g(x)$ is a polynomial in $GF(q)$, then $[g(x)]^q$ = $g(x^q)$.

I came across this statement as lemma 2.1 in Factoring polynomials over large finite fields, ER Berlekamp.

The author also mentions that this Lemma is discussed among the others in two other publications. However, I could not find adequate proofs there either. The books just state them as fact.

Is there something obvious here that I am missing that can be used to prove this statement?

答:
Hint:

A finite field $\mathbf F_q$, where $a=p^n$ is a prime power, is the splitting field of the polynomial $X^q-X$, and the map from $\mathbf F_q$ to itself defined by $x\longmapsto x^q$ is the $n$-iterate of the Frobenius automorphism $x\longmapsto x^p$.


问:
Are compositions of rotations a rotation?
The problem asks:

Suppose that $R_\theta$ is the counter-clockwise rotation of any vector in $\mathbb R^2$ by an angle $\theta$ $\epsilon$ $(0,\pi]$. Suppose that $F \circ G$ denotes the composition of the two functions, i.e,  $F \circ G(x)$ = $F(G(x))$. Is $R_\theta\circ R_\phi$ a rotation? What is the matrix $M(R_\theta \circ R_\phi)$?

Clearly, it is a rotation but I don't know how to prove how. Any videos or solutions can help.
Thank you!

Body: $\pmatrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta}\pmatrix{\cos\phi&-\sin\phi\\\sin\phi&\cos\phi}=\pmatrix{\cos(\theta+\phi)&-\sin(\theta+\phi)\\\sin(\theta+\phi)&\cos(\theta+\phi)}$


问:
Given iid random variables, expected number of times that a random variable is greater than all previous variables.
I'm given $X_1, X_2, \dots$ iid with continuous cdfs, and let $A_1 = \Omega, A_n=\{\omega:X_n
(\omega) > X_i(\omega) \textrm{ for } i < n\}$.

I'd like to show that $\lim_{n\rightarrow \infty}{\frac{1}{\log(n)}\sum_{j=1}^{n}{1_{A_j}}} = 1$ almost surely.

This is a pretty interesting result, but I'm pretty unsure of how I might go about showing this.

答:
Let me briefly sketch a proof. The first step should be to note that the $A_n$'s are independent and  $\mathsf{P}(A_n)=1/n$ for all $n$ (do you see why?). From Kolmogorov's inequality, for any $\epsilon>0$

$$\mathsf{P}\left(\max_{2^{n^2}\le k\le 2^{(n+1)^2}} \left|\sum_{j=1}^k \left(1_{A_j}-1/j\right)\right|> \epsilon n^2\right)\lesssim \frac{1}{\epsilon^2 n^2}.$$

Now we use Borel-Cantelli lemma to argue that almost surely, for $n$ large enough

$$\left|\sum_{j=1}^n \left(1_{A_j}-1/j\right)\right|\le \epsilon\log n.$$

What we just proved is

$$\forall \epsilon>0, \mathsf{P}\left(\exists n_0(\epsilon), \forall n>n_0(\epsilon), \left|\frac{1}{\log n}\sum_{j=1}^n 1_{A_j}-\frac{1}{\log n}\sum_{j=1}^n 1/j\right|\le \epsilon\right)=1.$$

As the intersection of a countable number of almost sure events is almost sure,

$$\mathsf{P}\left(\forall \epsilon\in\mathbb{Q}_+^*, \exists n_0(\epsilon), \forall n>n_0(\epsilon), \left|\frac{1}{\log n}\sum_{j=1}^n 1_{A_j}-\frac{1}{\log n}\sum_{j=1}^n 1/j\right|\le \epsilon\right)=1.$$

Now notice that if we take $n_0(\epsilon)$ large enough, from the asymptotics of the harmonic sum, it holds that for all $n>n_0(\epsilon)$, $\left|\frac{1}{\log n}\sum_{j=1}^n 1/j-1\right|\le \epsilon$. Hence:

$$\mathsf{P}\left(\forall \epsilon\in\mathbb{Q}_+^*, \exists n_0(\epsilon), \forall n>n_0(\epsilon), \left|\frac{1}{\log n}\sum_{j=1}^n 1_{A_j}-1\right|\le 2\epsilon\right)=1.$$

That's exactly the definition of almost sure convergence of $\sum_{j=1}^n 1_{A_j}/\log n$ to $1$.


问:
Need help for gre questions
Of the $240$ employees at a certain company, $75$ percent attended the company's $25$th anniversary receptions. Of the employees who attended the reception, $80$ percent brought one guest each and the rest each brought either no guests or $2$ guests. Which of the following could be the total number of company's employees and their guests who attended the reception?

A. $300$

B. $330$

C. $350$

D. $380$

E. $410$

F. $430$

I'm not sure whether my interpretation is wrong or the answer is wrong.

My approach as since we have $75$ percent of member attended the reception, then the total number who attended the reception is $75\% \times 240 = 180$.

Within $180$ members, there are at least $144$ guests since $80\%$ of $180$ members brought at least $1$ guest, and out of $36$ member brought either no guests or $2$ guests.

So the range of total members and guests who attended the reception is

$180+ 144 + 36 \times 0= 324$  to $80 + 144 + 36 \times 2 = 396$

The answer I have should be BCD, but the given answer is EF.

I probably missed something, any suggestions?

答:
In short, it appears as though you are correct and the problem must be mislabeled (or you've mis-transcribed something). Here is a formal algebraic solution:

Variables

$A = $ attendees

$E = $ employees

$E_A = $ employees in attendance

$G_A = $ guests in attendance

$f_{EA} = $ fraction of the total employees that are in attendance

$f_i = $ fraction of the employees in attendance that brought i guests, where $i \in [0,2]$

$g_i = $ guests per employee in attendance, where $i \in [0,2]$

Algebra

$A = \left[ E_A \right] + \left[ G_A \right]$

$A = \left[ E_A \right] + \bigl[ \left(f_0 \times E_A \times g_0\right) + \left(f_1 \times E_A \times g_1\right) + \left(f_2 \times E_A \times g_2\right) \bigr]$

factoring $E_A$ out of the expressions on the right and acknowledging that $g_0 = 0$, $g_1 = 1$, and $g_2 = 2$:

$A = E_A \bigl[ \left[ 1 \right] + \left(0\right) + \left(f_1\right) + \left(2 \times f_2 \right) \bigr]$

$A = E_A \bigl[ 1 + f_1 + 2 f_2 \bigr]$

Where $E_A = f_{EA} \times E$:

$A = f_{EA} \times E \times \bigl[ 1 + f_1 + 2 f_2 \bigr]$

Where $f_{EA} = 0.75$, $E = 240$, and $f_1 = 0.8$:

$A = 0.75 \times 240 \times \bigl[ 1 + 0.8 + 2 f_2 \bigr]$

$A = 180 \times \bigl[ 1.8 + 2 f_2 \bigr]$

Where $f_2 \in [0, 0.2]$

$A_{min} = 180 \times \bigl[ 1.8 + 2 \times 0 \bigr]$

$A_{min} = 180 \times \bigl[ 1.8 \bigr]$

$\therefore A_{min} = 324$

$A_{max} = 180 \times \bigl[ 1.8 + 2 \times 0.2 \bigr]$

$A_{max} = 180 \times \bigl[ 2.2 \bigr]$

$\therefore A_{max} = 396$

Final Note: Deceptive English

In theory, it is possible that more than 75% of the employees were in attendance. It would be outrageous for the GRE to be that deceptive, but, in theory, if 100% of the employees were in attendance, it would also be true that 75% of the employees were in attendance. It is pretty straightforward to show that, in that case, the maximum number of attendees would be 528. Even more ridiculously, it could be true that 80% of the employees in attendance brought more than one guest, but it would also be true that 80% of the employees brought one guest. If you are actually studying for the GRE, I highly recommend you disregard this entire paragraph, assume the correct answer was supposed to be BCD, and move on.

Title: Need help for gre questions


问:
Solving $x^2 + py^2 = 4z^3$ with a fixed $z$
Given an odd prime $p$ and a positive integer $z$, I'd like to find all positive integers $x$, $y$ such that
$$
x^2 + p y^2 = 4 z^3.
$$

Because the left is positive definite there is only a finite number of solutions.

I could solve by incrementally guessing values of $y$ and checking if $4z^3-py^2$ is a square, until $py^2> 4 z^3$. But this is slow, and many times the answer just turns out to be there are no solutions, or wasted time after finding the first solution which turns out to be unique.

I've learned the left hand side is called a binary quadratic form. Which has a class size $h(-4p)$. If it  somehow helps, assume I know this value.

Is there an easy way to determine that, if a solution exists, it is unique?

Is there an easy way to find all the solutions?

答:
I found a necessary condition for there to be a solution which may be useful as a way to rule out some solutions. The condition is: Divide out the highest power of $p$ diving $z$, then what is left must be a quadratic residue mod $p$.

Let's call $x=p^a x'$, $y=p^b y'$ and $z=p^c z'$ respectively, the ' meaning it is the part relatively prime to $p$. So now we are to show that $z'$ must be a quadratic residue for there to be a solution.

$$p^{2a}x'^2 + p^{2b+1}y'^2 = 4p^{3c}z'^3$$

Either $2a$ or $2b+1$ is smaller, since they can't be equal. Whichever the lowest one is, it must be equal to $3c$. This means we can divide out by $p^{3c}$ and one of the two terms on the left is divisible by $p$ at least once. That means we have,

$$x'^2 = 4z'^3 \mod p$$ or $$y'^2 = 4z'^3 \mod p$$

In both cases we must have that $z'^3$ is a quadratic residue mod $p$, which must mean $z'$ is a quadratic residue. If it isn't then we have a contradiction.


问:
Showing a diffeomorphism extends to the neighborhood of a submanifold
Does anyone have a proof of problem 14, on page 56 of Guillemin and Pollack? I meant to do it as an exercise (I'm teaching myself the subject) but I'm struggling with the last step. Suggestions?

Suppose that the derivative of $f: X \rightarrow Y$ is an isomorphism whenever $x$ lies in the submanifold $Z \subset X$, and assume that $f$ maps $Z$ diffeomorphically onto $f(Z)$. Prove that $f$ maps a neighborhood of $Z$ diffeomorphically onto a neighborhood of $f(Z)$. Essentially, this is the inverse function theorem but without assumptions on the compactness of $Z$.

So far I have local inverses $g_i : U_i \rightarrow X$ where $\{U_i\}$ is a locally finite collection of open subsets of $Y$ covering $f(Z)$. Define $W = \{y \in U_i : g_i(y) = g_j(y)\mbox{ whenever }y \in U_i \cap U_j\}$. We then have a smooth inverse $g: W \rightarrow X$.

How do I show that $W$ then contains an open neighborhood of $f(Z)$ (using local finiteness of $\{U_i\}$, presumably)?

答:
Eric: Here's the idea. A good example I've found helpful when presenting this is the spiral $f\colon\mathbb R^2\to\mathbb R^2$, $f(t) = (e^x\cos y,e^x\sin y)$, with $Z$ the diagonal. So $f(Z)$ is the spiral $\{(e^t\cos t,e^t\sin t): t\in\mathbb R\}$. You can imagine overlapping but non-adjacent little boxes on which $f$ is a local diffeomorphism, so that the local inverses don't quite glue.

As G&P suggest, cover $f(Z)$ with open sets $U_i$ on which we have a local inverse $g_i$ of $f$. The key point is that we may assume this is a locally finite open cover. (In our example, note that it's crucial that $0\notin f(Z)$.) Let $V_{ij} = \overline{\{v\in U_i\cap U_j: g_i(v)\ne g_j(v)\}}$. Then let $\tilde U_i = U_i - \cup_j V_{ij}$, and let $W=\cup_i \tilde U_i$. Check that $W$ works.

Let me know if this works better than what G&P have.


问:
Equivalent conditions involving (co)primary module (Proposition 3.9 from Eisenbud)
Let $R$ be a Noetherian ring and $M$ a f.g. $R$-module. Let $P$ be a prime ideal of $R$. I'm trying to understand why TFAE:

(a) $M$ is $P$-coprimary (i.e. $Ass(M)=\{P\}$)

(b) $P$ is minimal over $Ann(M)$ and every element not in $P$ is a nonzerodivisor on $M$

(c) A power of $P$ annihilates $M$ and every element not in $P$ is a nonzerodivisor on $M$

There is a proof in Eisenbud, but some points are unclear.

$(a)\implies (b)$ Here he refers to Theorem 3.1a that says that $Ass M$ is finite and nonempty set of primes, each containing $Ann(M)$, and $Ass(M)$ includes all primes minimal among primes containing $Ann(M)$. This should imply that $P$ is minimal over $Ann(M)$. But how? If I knew that primes minimal over $Ann(M)$ exist, then I could conclude that they lie in the set $Ass M=\{P\}$, so $P$ would be itself minimal over $Ann(M)$. Do they always exist?

$(b)\implies (c)$ Here he says that it suffices to prove the statement after localizing at $P$ since the elements in $R-P$ are nonzerodivisors. Why is it sufficient? He then says we can assume $R$ is a local ring with maximal ideal $P$; then $P=rad(Ann(M))$ since $P$ is minimal over $Ann(M)$. How is locality of $R$ and maximality of $P$ used here?

$(c)\implies (a)$ This is clear, I think.

答:
$(a)\implies (b)$ Here he refers to Theorem 3.1a that says that $Ass M$ is finite and nonempty set of primes, each containing $Ann(M)$, and $Ass(M)$ includes all primes minimal among primes containing $Ann(M)$. This should imply that $P$ is minimal over $Ann(M)$. But how? If I knew that primes minimal over $Ann(M)$ exist, then I could conclude that they lie in the set $Ass M=\{P\}$, so $P$ would be itself minimal over $Ann(M)$. Do they always exist?

In a paragraph after the statement of Theorem $3.1$, Eisenbud shows that primes minimal over a given ideal exist in any ring.

Furthermore, Proposition $3.4$ shows that $\operatorname{Ass} M$ is nonempty.

$(b)\implies (c)$ Here he says that it suffices to prove the statement after localizing at $P$ since the elements in $R-P$ are nonzerodivisors. Why is it sufficient? He then says we can assume $R$ is a local ring with maximal ideal $P$; then $P=rad(Ann(M))$ since $P$ is minimal over $Ann(M)$. How is locality of $R$ and maximality of $P$ used here?

Since every element not in $P$ is a nonzero divisor on $M$, then the canonical map $M \to M_P$ is injective (see Prop $2.1$).

Now what Eisenbud actually proves is that $PR_P = \operatorname{rad}(\operatorname{ann}M_P)$ which implies that $(PR_P)^n \subset \operatorname{ann}M_P$. But since $P \subset PR_P$ and $\operatorname{ann}M_P \subset \operatorname{ann}M$, then we have
$$P^n \subset (PR_P)^n \subset \operatorname{ann}M_P \subset \operatorname{ann}M.$$


问:
Prove using mathematical induction
Sequence $b_1, b_2,...$ where $b_1=b_2=1$ and for every natural number $k$ bigger than $2$, $b_k = b_{k-1}+b_{k-2}$.

Prove using mathematical induction that, for every $k \in \mathbb N, b_k \le (\frac{7}{4})^{k-1} $

I started with if  $k=3$, $k_3=1+1=2$ and $(\frac{7}{4})^{k-1} =(\frac{7}{4})^2=3 \frac{1}{16}$, $2<3 \frac{1}{16}$. This is the induction base?

Step:
Let us assume that S(m) is true, $b_m \le (\frac{7}{4})^{m-1}$ for every m $\in$ {3, ..., k}.

We have to show that  $b_{k+1} \le (\frac{7}{4})^k$.
Using assumption and $b_{k+1} = b_{k}+b_{k-1}$ I got
$$b_k+b_{k-1}\le (\frac{7}{4})^{k+1}+(\frac{7}{4})^{k}$$
I don't know what to do next.

答:
I don't know what to do next.

$ \left(\dfrac{7}{4}\right)^{k+1}+\left(\dfrac{7}{4}\right)^{k}=\left(\dfrac74\right)^k\left(\dfrac74+1\right)=\left(\dfrac74\right)^k\left(\dfrac{11}4\right)=\left(\dfrac74\right)^k\left(\dfrac{44}{16}\right)$

$<\left(\dfrac74\right)^k\left(\dfrac{49}{16}\right)=\left(\dfrac74\right)^k\left(\dfrac{7}{4}\right)^2=\left(\dfrac74\right)^{k+2}$

Title: Prove using mathematical induction


问:
Evaluating the volume of $4z^2=x^2+y^2$
I know the volume of the cone
$$4z^2=x^2+y^{2}$$
for $0\leq z\leq3$ is equal to $36\pi$ since its height is 3 and its base radius is 6. So, In the first octant, the volume should be
$$36\pi\div4=9\pi$$
But I can't find this value using integrals. I tried to write using spherical coordinates: it's clear that
$$0\leq\rho\leq6$$
and
$$0\leq\theta\leq\frac{\pi}{2}$$
and on the plane $yOz$ to find the interval for $\phi$, I realized that, if $x=0$, then
$$4z^2=y^2\Rightarrow z=\frac{y}{2}$$
so, for $z=3,y=6$. So,
$$\tan(\phi)=\frac{6}{3}=2\Rightarrow 0\leq\phi\leq\arctan(2)$$
However, Wolfram says the integral
$$\int_{0}^{\pi/2}\int_{0}^{\arctan(2)}\int_{0}^{6}\rho\sin^{2}(\phi)d\rho d\phi d\theta=\frac{9}{10}\pi(5\arctan(2)-2)$$
What am I doing wrong?

答:
The top of your cone is flat, yet this isn't reflected in your integral. Constant bounds on all the variables gives you regions shaped like spheres. That's why they are called spherical coordinates.

You will need $\rho$ to go from $0$ to $\dfrac{3}{\cos\phi}$.

Also, it's $\iiint\rho^2\sin\phi\,d\rho\, d\phi\, d\theta$, not $\iiint\rho\sin^2\phi\,d\rho \,d\phi \,d\theta$.

Title: Evaluating the volume of $4z^2=x^2+y^2$


答:
I started with if  $k=3$, $k_3=1+1=2$ and $(\frac{7}{4})^{k-1} =(\frac{7}{4})^2=3 \frac{1}{16}$, $2<3 \frac{1}{16}$. This is the induction base?

No. Since the series starts with $b_1$, you should start with $1$, not $3$. That is, for your base, show that $b_k \leq (\frac{7}{4})^{k-1}$ for $k=1$ ... and since the recursive definition of the series refers to two previous terms, you should also include the case $k=2$ in your base.

For your step you then show that for some arbitrary $k \geq 3$: if the claim is true for $k-2$ and for $k-1$, then it is true for $k$

Title: Prove using mathematical induction

Body:
Sequence $b_1, b_2,...$ where $b_1=b_2=1$ and for every natural number $k$ bigger than $2$, $b_k = b_{k-1}+b_{k-2}$.

Prove using mathematical induction that, for every $k \in \mathbb N, b_k \le (\frac{7}{4})^{k-1} $

I started with if  $k=3$, $k_3=1+1=2$ and $(\frac{7}{4})^{k-1} =(\frac{7}{4})^2=3 \frac{1}{16}$, $2<3 \frac{1}{16}$. This is the induction base?

Step:
Let us assume that S(m) is true, $b_m \le (\frac{7}{4})^{m-1}$ for every m $\in$ {3, ..., k}.

We have to show that  $b_{k+1} \le (\frac{7}{4})^k$.
Using assumption and $b_{k+1} = b_{k}+b_{k-1}$ I got
$$b_k+b_{k-1}\le (\frac{7}{4})^{k+1}+(\frac{7}{4})^{k}$$
I don't know what to do next.


问:
What is the equivalent of a normal subgroup in Linear Algebra?
In group theory, the quotient group G/H is a group only if H is a normal subgroup because you can be sure the group operations are well defined.

Now does linear algebra have something similar? For example, the linear map of the quotient space T(V/U, W) where U is a subspace of V. Is this a well defined linear transformation or does the subspace U need to meet some criteria for the linear map to be well defined?

答: For any subspace $U$ of a vector space $V$, the quotient $V/U$ has the structure of a vector space. So yeah, in this sense every subspace of a vector space is normal.


问:
Writing a Lebesgue Measurable Set as a Union of a Measure Zero Set and a Countable Union of Compact Sets
I'm trying to showing that using the notion of Lebesgue inner measure, i.e., using the fact that for a measurable set $E$, $m(E)=$Sup$\{m(K): K \;$is compact$\; \land \;K\subseteq E\}$, one can write $E$ as follows: $$E=A\cup\bigcup_{n=1}^{\infty}C_{n}$$ where $m(A)=0$ and each $C_{n}$ is compact.

I've generated a sequence of compact sets $\{K_{n}\}_{n=1}^{\infty}$ with each $K_{n}\subseteq E$ and $m(K_{n}) \to m(E)$ by the definition of the supremum, but I'm stuck at this point.

答:
Consider $C_n= K_1 \cup \ldots\cup K_n$. Then
$$m(K_n) \le m(C_n) \le m(E)$$
and since $m(K_n) \to m(E)$ we have also $m(C_n) \to m(E)$.  Now, since $C_n$ is an increasing sequence of measurable sets ( compacts) we have
$$m(C_n) \to m(\cup_n C_n) $$

Now, $\cup_n C_n \subset E$ and $m(\cup_n C_n) = m(E)$.

If we have also $m(E)< \infty$ we get $m(E\backslash \cup_n C_n) = 0$, so we can take $A=E\backslash \cup_n C_n$.

Otherwise, write $E= \cup_m E_m$, $E_m$ measurable, of finite measure. For every $m$ consider $L_m$ a finite union of compacts inside $E_m$, such that $m(E_m \backslash L_m) = 0$. Now take $A= E \backslash \cup_m L_m$.


问:
$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$

Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
 1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
 {\rm d}t=\dfrac{\pi }{\sqrt{2}}$.

We can obtain
$$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$
but how to go on ?

答:
I found a possible method here which is straightforward.  I will just carry it here.

\begin{aligned}
\lim _{x \rightarrow 1} u(x) &=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t} \cdot \frac{1+t}{(t-1)^{2}}}{\sqrt{\frac{1+x}{(x-1)^{2}}-\frac{1+t}{(t-1)^{2}}}} d t \\
&=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t}(1+t)(1-x)}{(1-t) \sqrt{(x-t)(3-t-x-x t)}} d t \\
&=\lim _{y \rightarrow 0} \int_{y}^{1} \frac{(2-s) \sqrt{1-s} y}{s \sqrt{(s-y)(2 s+2 y-s y)}} d s \quad(\text { Let } s=1-t, y=1-x)
\\
&=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y} \cdot y \cdot y}{u y \sqrt{(u y-y)\left(2 u y+2 y-2 u y^{2}\right)}} d u \quad(\text { Let } s=u y) \\
&=\lim _{y \rightarrow 0} \int_{1}^{\frac{1}{y}} \frac{(2-u y) \sqrt{1-u y}}{u \sqrt{(u-1)(2 u+2-u y)}} d u\\
&=\sqrt{2} \int_{1}^{+\infty} \frac{1}{u \sqrt{u^{2}-1}} d u \\
&=\sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-m^{2}}} d m \quad\left(\text { Let } m=\frac{1}{u}\right) \\
&=\frac{\sqrt{2}}{2} \pi
\end{aligned}


问:
Does the set $\{1, i , −1\}$ form a group?

Does the set $\{1, i , −1\}$ form a group?

This is exactly what the question in my book asks.

I am not sure how to answer this, I would wait it says to me if the group is defined under ordinary multiplication or in another way. Of course if it is the former is no, but what about the latter?

Am i missing something?

答:
It does not make grammatical sense, mathematically, to ask whether a set is a group, precisely because no operation has been specified. Many people do this and many people are being sloppy. The question that makes sense is to ask whether a subset of a group is a subgroup; in this case $\{ 1, i, -1 \}$ is a subset of, say, $\mathbb{C}^{\times}$, which is not a subgroup because it's not closed under (ordinary) multiplication.

Another question that makes sense is to equip a set $S$ with a binary operation $\ast : S \times S \to S$ (giving a magma) and ask whether that operation satisfies the group axioms. But again no operation has been specified here. A third question that makes sense is to ask whether a set $S$ admits some group operation, but obviously every finite set has this property, and in fact every set does (this is equivalent to the axiom of choice).

Most likely the intended question is whether $\{ 1, i, -1 \}$ is a subgroup of $\mathbb{C}^{\times}$.


问:
Proving that $\operatorname{Span}(u, v + lu) \subseteq \operatorname{Span}(u, v)$
Let $u, v$ be elements of a real vector space $V$, and let $l\in\mathbb R$.

a) Prove that $\operatorname{Span}(u, v + lu) \subseteq \operatorname{Span}(u, v)$.

b) Prove that $\operatorname{Span}(u, v) \subseteq \operatorname{Span}(u, v + lu)$?

I'm really struggling and not sure how to prove part a) and part b). Any answers or where to start? Any help would be much appreciated.

答:
$Span(u,v+lu)$ is the set of all linear combinations of $u$ and $v+lu$,

thus all elements of the form $a \cdot u + b \cdot (v+lu)$ for $a,b \in \mathbb{R}$.

However, you can group together like terms: $(a+lb)\cdot u + b \cdot v$ and $a$ can be rewritten as $a= a’ -l$, and we get $ a’ \cdot u + b\cdot v$. (You can skip this step if you like, $a+lb$ is already a real number so technically we’d be done as soon as we brought together like terms, but it’s more for visuals).

So it turns out to be a linear combination of $u$ and $v$ also. $Span(u,v+lu) \subseteq Span(u,v)$.

Next, given $Span(u,v)$, all linear combinations of the form $a\cdot u + b\cdot v$ for $a,b\in \mathbb{R}$ can be written as $(a-bl+bl)\cdot u + b\cdot v = (a-bl)\cdot u + bl\cdot u + b\cdot v = (a-bl)\cdot u + b\cdot(v+lu),$ because we are adding $0=bl-bl$ to the coefficient $a$. Now $a,b,l \in \mathbb{R}$ thus so is $a-bl$, and you have a linear combination of the form $a’ \cdot u + b\cdot (v+lu)$ for $a’,b \in \mathbb{R}$.
$Span(u,v) \subseteq Span(u,v+lu).$


答:
Not sure it fully answers to your question, but note that we have a short exact sequence
$$0\longrightarrow U\longrightarrow V\longrightarrow V/U\longrightarrow 0,$$
and the $\operatorname{Hom}$ functor is left exact, so we deduce an exact sequence
$$0\longrightarrow\operatorname{Hom}(V/U,W)  \longrightarrow\operatorname{Hom}(V,W) \longrightarrow\operatorname{Hom}(U,W), $$
which shows the linear maps from $V/U$ to any vector space $W$  are obtained from the  linear maps from $V$ to $W$ which vanish on $U$.

Title: What is the equivalent of a normal subgroup in Linear Algebra?

Body:
In group theory, the quotient group G/H is a group only if H is a normal subgroup because you can be sure the group operations are well defined.

Now does linear algebra have something similar? For example, the linear map of the quotient space T(V/U, W) where U is a subspace of V. Is this a well defined linear transformation or does the subspace U need to meet some criteria for the linear map to be well defined?


答:
No (and yes${}^\dagger$).

No operation is defined for it, so it makes no sense to ask.

Assuming multiplication, since $-1\times i=-i\notin\{1,i,-1\}$, it is not closed, so it cannot be a group.

$\dagger$: If you want, you could pick any element of $\{1,i,-1\}$ to be the identity element $e$ with respect to concatenation, one of the remaining two elements to be the square of the other, call them $a$ and $a^2$, with $a^3=e$, so that, under this operation, the set is a group (isomorphic to $\Bbb Z_3$).

Body:
Does the set $\{1, i , −1\}$ form a group?

This is exactly what the question in my book asks.

I am not sure how to answer this, I would wait it says to me if the group is defined under ordinary multiplication or in another way. Of course if it is the former is no, but what about the latter?

Am i missing something?


问:
Sublattice of a power set, where elements are either empty or countably infinite
I was developing a new logic system that deals with undecidability (like Continuum Hypothesis is to ZFC), and this question has emerged.

Let $X$ be a countably infinite set. Endow $\mathcal{P}(X)$ with the usual lattice, i.e. Order as $\subset$, join as $\cup$, meet as $\cap$. I need a sublattice $\mathcal{L} \subset \mathcal{P}(X)$ such that:

$\emptyset \in \mathcal{L}$ (Empty set is a member)

$\forall (Y \in \mathcal{L} \setminus \{\emptyset\}) \quad |Y| = \aleph_0$ (Nonempty members are countably infinite)

$\forall (Y \in \mathcal{L}) \quad X \setminus Y \in \mathcal{L}$ (Closure under complement)

$\forall (\mathcal{M} \subset \mathcal{L}) \quad \bigcup\mathcal{M} \in \mathcal{L} $ (Closure under arbitrary join)

Note that closure under arbitrary meet is not a requirement, though closure under finite meet is.

The trivial example is $\mathcal{L} = \{\emptyset, X\}$. But the presence of undecidable statements means that this is not the case.

Up to isomorphism, I need $\mathcal{L}$ to be the biggest. Is such $\mathcal{L}$ unique? In that case, I can call $X$ a house, and can call its members residents.

<s>Edit

I thought about it, and found some examples of $\mathcal{L}$:

For some fixed $n \in \mathbb{Z}^+$, let $X = \mathbb{Z}$, and let $\mathcal{L_n}$ be the topology generated by a subbasis $\{\langle n \rangle + m : m \in \mathbb{Z} / n \mathbb{Z} \}$.

But $\bigcup_{n \in \mathbb{Z}^+} \mathcal{L}_n$ isn't closed under complement. Does this mean there is no biggest $\mathcal{L}$?</s>

Edit 2

It turns out that $X$ needs another requirement to be a house:

$\forall (Y \in \mathcal{L} \setminus \{\emptyset\}) \quad |\{Y \cap Z : Z \in \mathcal{L}\}| = 2^{\aleph_0}$

Or in other words, a sublattice of $\mathcal{L}$ bounded above by a countably infinite member $Y$ must be uncountable. Note that it also must satisfy if $Y = X$.

答:
Self-Answered

I think I found that such $\mathcal{L}$ doesn't exist.

Pick a nonempty $Y_0 \in \mathcal{L}$. Pick $x \in Y_0$. Pick a nonempty $Y_1 \in \mathcal{L}$ such that $x \in Y_1 \subsetneq Y_0$. Such $Y_1$ exists due to the requirement in Edit 2. Repeat.

Let $Y_\omega = \bigcap_{n=0}^\infty Y_n$. Then $x \in Y_\omega$, so $Y_\omega$ is countably infinite. Now we can continue this to arbitrary countable ordinals. We now have an uncountably long chain between $X$ and $\emptyset$. This contradicts against that $X$ is countable.

This arises a new question, but that should be another post.


问:
the coefficient of Adam Bashforth method
When using Adams-Bashforth method, after integrating the polynomial of k-th order, we obtain a linear combination of $f(t_n,y_n),f(t_{n+1},y_{n+1}),...,f(t_{n+k},y_{n+k})$.
However, I am curious about why the sum of all the coefficients is always 1.
Below is the formulae for k<4 from wikipedia. So please tell me why we always get 1 after summing up the coefficients.

答: Think about what the method should look like for a constant f, e. g. f(t,y)=1 :)</p>


问:
What is the minimum DFA of this?
I've been struggling for hours with this DFA, what I want to do is that get the minimum DFA of this.
But whenever I'm trying to minimize this then I get a DFA the same as this.

答:
The DFA is already minimal. To see why, we will show that for every two states $q_1$ and $q_2$, there is a word $x$ that separates between them. That is, there is a word $x$ such that one of the states $\delta^*(q_1, x)$ and $\delta^*(q_2, x)$ is accepting and the other is rejecting. To begin with, the rejecting and accepting states are separated via the empty word $\epsilon$ as $\delta^*(q, \epsilon) = q$, for every state $q$. So, the states $\{ A, C, D, G\}$ are separated from the states $\{ B, E, F, H\}$. The following holds:

$A$ is separated from $C, D, G$: $x = 1$ separates between $A$ and $C$. $x = 11$ separates between $A$ and $D$. Also, $x = 1$ separates between $A$ and $G$.

$C$ is separated from $A, G, D$: we've seen in item 1 that $C$ and $A$ are separated. $x = 011$ separates between $C$ and $G$. Also, $x = 01$ separates between $C$ and $D$.

$D$ is separated from $A, C, G$: we've seen in item $1$ that $D$ and $A$ are separated, and we've seen in item $2$ that $D$ and $C$ are separated. Finally, $D$ and $G$ are separated by $x = 1$.

$G$ is separated from $A, C, D$: we have seen that in the previous items.

$B$ is separated from $E, F, H$: $x = 0$ separates between $E$ and $B$. $x= 10$ separates between $F$ and $B$. Also, $x = 1$ separates between $B$ and $H$.

$E$ is separated from $B, F, H$: we have seen in item 5 that $E$ and $B$ are separated. $E$ and $F$ are separated by $x=1$. Also, $E$ and $H$ are separated by $x=0$.

$F$ is separated from $B, E, H$: we have seen in items 5 and 6 that $F$ is separated from $B$ and $E$. Finally, $x = 1$ separates between $F$ and $H$.

$H$ is separated from $B, E, F$: we have seen that in the previous items.

Title: What is the minimum DFA of this?


问:
Arithmetic operation
Given
$$9 ∆ 4=20,$$
$$7 ∆ 3=12,$$
$$10 ∆ 8=16,$$
find $m ∆ n$.

I have tried so many ways to get this but I may be missing something.

答:
As @lulu said in comments, it's mainly a guessing idea.

I noticed that the second number was divisible by the answer. $\frac{20}{4}=5, \frac{12}{3}=4$, et cetera. So one would think that a multiplication of $n$ would be involved. From there, it's just a matter of what to multiply by. Trying it out leads to $$(m-n)\cdot n.$$

Title: Arithmetic operation


答:
The 4-element set $\{1,-1,i,-i\}$ does form a group under multiplication--indeed they are the quartic roots of 1, and the 3 elements in $\{1,i, -1\}$ generate that 4-element group.

But I agree with the others--sloppily written. And perhaps a typo too, e.g., the author meant to include $-i$ as well?

Title: Does the set $\{1, i , −1\}$ form a group?


答:
Don't use spherical coordinates, but still use radial in $(x,y)$.

The area of the segment of a circle radius:
$$A = \frac{r^2\theta}{2}$$

Radius squared:
$$r^2 = 4z^2$$

Using area.

$$\int_0^3\int_0^{\pi/2} 2z^2 d\theta dz$$

$$\int_0^3 \pi z^2 dz = 9\pi $$

Title: Evaluating the volume of $4z^2=x^2+y^2$

Body:
I know the volume of the cone
$$4z^2=x^2+y^{2}$$
for $0\leq z\leq3$ is equal to $36\pi$ since its height is 3 and its base radius is 6. So, In the first octant, the volume should be
$$36\pi\div4=9\pi$$
But I can't find this value using integrals. I tried to write using spherical coordinates: it's clear that
$$0\leq\rho\leq6$$
and
$$0\leq\theta\leq\frac{\pi}{2}$$
and on the plane $yOz$ to find the interval for $\phi$, I realized that, if $x=0$, then
$$4z^2=y^2\Rightarrow z=\frac{y}{2}$$
so, for $z=3,y=6$. So,
$$\tan(\phi)=\frac{6}{3}=2\Rightarrow 0\leq\phi\leq\arctan(2)$$
However, Wolfram says the integral
$$\int_{0}^{\pi/2}\int_{0}^{\arctan(2)}\int_{0}^{6}\rho\sin^{2}(\phi)d\rho d\phi d\theta=\frac{9}{10}\pi(5\arctan(2)-2)$$
What am I doing wrong?


问:
Analytic solution to 2 dimensional PDE
I am trying to solve the 2D PDE
$$\frac{\partial p(\mathbf{x},t)}{\partial t}=-{f}(\mathbf{x})\frac{\partial }{\partial x_1}p(\mathbf{x},t) -{g}(\mathbf{x})\frac{\partial }{\partial x_2}p(\mathbf{x},t)+a\frac{\partial ^2}{\partial x_1^2}p(\mathbf{x},t)+b\frac{\partial ^2}{\partial x_2^2}p(\mathbf{x},t)$$
so that I can test the accuracy of a numerical solver that I wrote. Here, $a$ and $b$ are real valued, nonzero, scalars.

I learned how to solve this when $f$ and $g$ are constants, but I am not sure if it is possible to analytically solve this when they are not constant (Or if one is nonconstant and the other is constant). I have been searching in the literature but am having trouble finding solutions.

Are there any known solutions to this that I can use to test my code in this situation that either $f$ or $g$ or both are nonconstant? Even for a specific case would be helpful.

Thank you!

答:
Search for the method of manufactured solutions.
I would be very surprised if you can solve this analitical for arbitrary f,g.</p>

Title: Analytic solution to 2 dimensional PDE


问:
If $P$ minimal over $\operatorname{ann} M$, and every element not in $P$ is a nonzero divisor on $M$, then a power of $P$ annihilates $M$.
Let $P$ be a prime ideal of the Noetherian ring $R$, and let $M$ be a finitely generated $R$-module.

Proposition: If $P$ minimal over $\operatorname{ann} M$, and every element not in $P$ is a nonzero divisor on $M$, then a power of $P$ annihilates $M$.

I don't understand why proving the special case also shows the general statement we want to prove.

Why is it sufficient to localize?

答:
Showing that a power of $P$ annihilates $M$ is just showing that $P^n\subseteq\operatorname{ann}_R(M)$ for some $n\in\mathbb{N}$. So, suppose that the result holds for local rings, and let $R_P$ be the localization of $R$ at $S:=R\setminus P$ and $M_P$ the localization of $M$ at $S$.

We claim the localization map $M\rightarrow M_P$ is injective. Indeed, recall that $m$ lies in the kernel of this map if and only if there is an element $s\in S$ such that $s\cdot m=0$. Since every element of $S$ is a non-zero divisor on $M$ by hypothesis, this will happen if and only if $m=0$.

Now, by the result for local rings, there is some $n\in\mathbb{N}$ such that $P_P^n\subseteq\operatorname{ann}_{R_P}(M_P)$. We claim that this implies $P^n\subseteq\operatorname{ann}_R(M)$. Indeed, let $p\in P^n$ and $m\in M$. We have $p\big/1\in P_P^n$, so in particular $p\big/1\cdot m\big/1=0\big/1$; ie, $p\cdot m\big/1=0\big/1$. But now injectivity of the map $M\rightarrow M_P$ implies $p\cdot m=0$, as desired.


问:
Conditions for continuous map
today in my topology class we had discussed about continuos map in topological spaces. And I have this question, for $(X,\tau)$ and $(X',\tau')$ two topological spaces and $f\colon X\to X'$, if $f(A)\subseteq X'$ is an open set $\forall A\in\tau$ It implies that $f$ is continuous? or what is necessary to get the continuity of the map?

答:
It is neither necessary nor sufficient for continuity. A function that satisfies this condition is called an “open map”.

For an example of a continuous function that does not satisfy this property, consider the real numbers with the usual topology, and a constant function $f\colon \mathbb{R}\to\mathbb{R}$. This is continuous, but the only open set whose direct image is open is the empty set; all other sets have image a singleton, which is not open.

For an example of an open mapping that is not continuous, consider $X=X’=\{a,b\}$, $\tau=\{\varnothing, X\}$, and $\tau’=\{\varnothing, \{a\}, \{b\}, X’\}$. Then the identity map is open, but not continuous. Other examples can be found here.

Title: Conditions for continuous map