答:
Let's first come up with a general solution for these type of problems, so that they don't keep coming up with different length values in this forum.

Always construct the equilateral triangle $\bigtriangleup APA'$ and connect $A'$ to $C$. Since $\bigtriangleup ABC$ is equilateral we should immediately notice that $\bigtriangleup BAP$ and $\bigtriangleup CAA'$ are congruent. Our goal is to apply the law of cosines in $\bigtriangleup ACP$. To do this, we first note that we should find $\cos(\alpha)$, which can be done by  applying the law of cosines in $\bigtriangleup A'CP$. Now, we can apply the law of cosines in $\bigtriangleup ACP$ since we know how to evaluate $\cos(\alpha + 60°)$.

$\hspace{5cm}$

Ex. Problem

What if $y^2=x^2+z^2-xz$ was given in the problem (instead of the given relationship); can you show that one of the angles around point $P$ is exactly $120°$ ?

Bonus

What if we select a random point $P$ in a square and give the distances to three of the vertices, can you find the relationship between these distances and the length of a side of the square?

 $(a^2+y^2-x^2)^2+(a^2+y^2-z^2)^2=(2ay)^2$

$\hspace{5cm}$

Title: Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$

Body:
$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?

If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.


问:
Why is an affine translation said to be a shear in a higher dimension?
I was doing a little refresher on matrix transformations, and started reading on affine transformations. After a while I kind of understood the purpose of utilizing 3x3 matrices for 2D transformations, but then I stumbled upon the affine transformation matrix for a translation:

Some of the sources I have found mention that what this is doing is basically performing a shear of the point in 3D space along the plane $z = 1$ (at least for this case of 2D transformations), but even after watching videos demonstrating what's happening underneath (like this excellent one on Wikipedia), I can't quite understand how that matrix is a meant to represent a shear of any sort, let alone one in 3D.

PS: I'm more accustomed to the simple shear transformation of the likes of $
\begin{bmatrix}
1 & h_x\\
h_y & 1
\end{bmatrix}
$, so I'm not very knowledgeable of the forms they can take and that may be the cause of my confusion.

答:
Your transformation looks like this.
$$
x'=x+t_xz;\quad y'=y+t_yz,\quad z'=z
$$
First of all, points move horizontally, since the $z$ is preserved. Also, observe that $(x,y)$ becomes $(x,y)+z(t_x,t_y)$ which means, for any fixed $z$ a translation  by the (horizontal) vector $z(t_x,t_y,0)$ which keeps its direction and has a length, proportional to $z$. That is exactly what a shear is: a translation on each plane parallel to $z$ by a vector, proportional to $z$.

ContentLicense: CC BY-SA 4.0


问:
Given a k-algebra A, representations $V, W$ of $A$, and homomorphism $\phi: V \rightarrow W,$ prove that ker$\phi$ is a subrepresentation of $A$
I think my issue here is wrapping my head around what is, exactly, a homomorphism of representations.

A representation is some vectorspace $S$ and a homomorphism $A \rightarrow$ End$(S)$.

So, to be specific, we have that $V$ is some vectorspace $X$ and a homomorphism $x: A \rightarrow$ End$(X)$. Then $W$ is a vectorspace $Y$ and a homomorphism $y: A \rightarrow$ End$(Y)$.

To make things easier... $V = (x, X)$ and $W = (y, Y)$.

Then... a homomorphism $\phi: V \rightarrow W$ is... what, exactly? It's somehow defined on both the morphisms $x,y$ and the vectorspaces $X,Y$, right? But what does that really "look like?" It  kind of feels like a really, really small functor, but I'm having a hard time putting that into words.

And then, like. Okay. The kernel of any morphism is, colloquially speaking, "the set of things which get mapped to $0$." But I don't know what's really being mapped, here, and what $0$ is in this context. The $0$ representation?

答:
Okay, so I read (part of) a book.

My confusion was coming from the fact that technically a representation is a 2-tuple, and I didn't really understand how the homomorphism was defined on those tuples.

But part of this issue can be solved simply by choosing which part of the representation to refer to. I'll choose the vectorspace part.

So then our homomorphism $\phi: V \rightarrow W$ is defined on the vectorspaces. In other words, $\phi: X \rightarrow Y$, such that $\phi(av) = a \phi(v)$ for all $v \in V_1$. Also, it's linear.

So, there. It's kind of like a "really, really small" functor. Because I say it is.

Now the question of the kernel becomes way mundane. It's all $v \in V_1$ such that $\phi(v) = 0$.

To show that this is a subrepresentation of $V,$ we just have to show that ker$\phi \subset V$ and that $\phi(av) = a\phi(v)$ for all $v \in$ ker$\phi$.

Will update this more after I eat.


问:
Given $\sigma$ a coprime automorphism of $G$ a finite group, must the centre of $G^\sigma$ be contained in the centre of $G$?
Let $G$ be a finite group, $\sigma$ an automorphism of order coprime to $|G|$. Let $K$ denote the subgroup of $\sigma$ fixed points of $G$. Then do we necessarily have that every element of $G$ commutes with every element of the centre of $K$?

I haven't been able to prove this or come up with a counterexample, but in the direction of trying to prove it, I have the following.

Bootstrapping from the prime power order case, we see that there is no fusion from $K$ into $G$ (Glaubermans lemma), so if $g\in G$ doesn't commute with $x\in Z(K)$, then $xgx^{-1}$ isn't in $K$. In a similar way, if $g$ isn't in $N_G(Z(K))=Z_G(Z(K))$, then $Z(K)^g\cap Z(K)=\{e\}$, so we have some trivial intersection behaviour.

答:
Let $p$ be an odd prime, and let $G$ be the nonabelian group of order $p^3$ and exponent $p$,
$$G=\langle x,y\mid x^p=y^p=[y,x]^p=[y,x,y]=[y,x,x]=1\rangle.$$
Let $\sigma$ be the automorphism obtained by exchanging $x$ and $y$, of order $2$.

Elements of $G$ can be written uniquely as $x^ay^b[y,x]^c$, $0\leq a,b,c\lt p$, with product
$$\Bigl( x^ay^b[y,x]^c\Bigr)\Bigl(x^{\alpha}y^{\beta}[y,x]^{\gamma}\Bigr) = x^{a+\alpha} y^{b+\beta}[y,x]^{c+\gamma+b\alpha}.$$
We have
$$\begin{align*}
\sigma(x^ay^b[y,x]^c) &= y^ax^b[x,y]^c\\
&= x^by^a[y,x]^{ab-c}.
\end{align*}$$
Thus, $x^ay^b[y,x]^c$ is fixed by $\sigma$ if and only if $a=b$ and $ab\equiv 2c\pmod{p}$.

For $p=3$, that means that $K$ consists exactly of three elements: $e$, $xy[y,x]^2$, and $x^2y^2[y,x]^2$. This is of course cyclic, so the center of $K$ is $K$ itself.

But $x$ does not commute with $xy[y,x]^2$:
$$\begin{align*}
x\bigl(xy[y,x]^2\bigr) &= x^2y[y,x]^2\\
\bigl(xy[y,x]^2\bigr)x &= x^2y[y,x]^{2+1}= x^2y.
\end{align*}$$
So the center of $K$ is not central in $G$.

For arbitrary $p$, you get $p$ elements (one for each possible value of $a$), so again $K$ is cyclic; and $x$ does not commute with $xy[y,x]^{1/2}$ (where $1/2$ is the solution to $2x\equiv 1\pmod{p}$, which exists since $\gcd(p,2)=1$):
$$\bigl[x,xy[y,x]^{1/2}\bigr] = [x,x][x,y][x,[y,x]]^{1/2} = [x,y]\neq e.$$


问:
Continuous Images of $GL_n(\mathbb{R})$
Which of the following is a continuous image of $GL_n(\mathbb{R})$?

The real line $\mathbb{R}$

The subspace $\{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ of $\mathbb{R}^2$

$\mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$

$\{(x_1,x_2,...,x_n):x_1^2+x_2^2+...+x_n^2 \le 1\}$ of $\mathbb{R}^n$

How do I proceed?

$GL_n(\mathbb{R})$ is not compact, not connected, so I cannot rule out some options using topological properties.

答:
Currently do not have enough reputation to comment, apologies.
How about just the trace function? One can always construct a matrix $M \in GL(n, \mathbb{R})$ such that $\mathrm{Tr}(M)=x, \forall x \in \mathbb{R}$

Trivial example, $M_{11} = x, M_{ij} = 1, i,j \neq (1,1)$.

Title: Continuous Images of $GL_n(\mathbb{R})$


问:
Equivalence class for a determinant
Show that the relation on the set of all 2 x 2 matrices defined by A ~ B if detA=det B is an equivalence relation. Describe the equivalence class.

I have determined that it is an equivalence relation. I am troubled by the equivalence class part. Determinants exist in the set of real numbers, right? How would I indicate the matrices that satisfy this?

答:
It's clear that the relation $A\sim B$ is an equivalence relation :

$\textbf{Reflexive:}$
$A\sim A$ is clear because $$det A=det A$$

$\textbf{Symmetric:}$ If $A\sim B$ then $B\sim A$ since
$$A\sim B \iff det A= det B \iff det B= det A\iff  B\sim A$$

$\textbf{Transitive:}$ Let $A$,$B$,$C$ be three matrices :

$A\sim B$ and $B \sim C$ $\implies$ $det A= det B$ and  $det B = det C$

$\implies det A=det C \implies A\sim C$

So it's an equivalence relation.

For the equivalence class let $detA=z\in \mathbb C$, we have :

$$\begin{align}
\bar{A}
&=\{B\in M_2(\mathbb C) / A\sim B\}\\
&=\{B\in M_2(\mathbb C) / det A= det B\}\\
&=\{B\in M_2(\mathbb C) / a_1a_4-a_3a_2=b_1b_4-b_3b_2=z\}\\
\end{align}
$$

Which gives $ b_1b_4-b_3b_2=z$, which has infinitely many solutions.

So we can only write

$$\bar{A}=\{B\in M_2(\mathbb C) / det B=z\}$$

Title: Equivalence class for a determinant


问:
Why is "for all $x\in\varnothing$, $P(x)$" true, but "there exists $x\in\varnothing$ such that $P(x)$" false?
There exists an $X\in A$ such that $P(X)$. When $A$ is the empty set then this statement is false because there is nothing in $A$ that when plugged in for $X$, makes $P(X)$ come out True.

However, when the quantifier is the universal one, meaning "For all $X\in A$" (which is the empty set), then the statement is true !!

How is that ? .. Because by the reasoning in the first statement, there are not any values of $X$ that makes $P(X)$ come out True.

I'm missing something !!

(I'm studying Electrical Engineering and self studying velleman's how to prove it and the quantifiers are already kicking my butt so simplify the answers for me if possible :))

答:
Let $A$ be the set of monkeys in the room, let $P(x)$ denote the sentence "x is playing chess".

$\forall x\in A, P(x)$ means "every monkey in the room is playing chess", which is true almost all the time, since most rooms are monkey-free.

$\exists x\in A, P(x)$ means "there is a monkey in the room playing chess", which is almost never true.

ContentLicense: CC BY-SA 3.0


问:
Proving the derivative of sin(x)
I was able to manipulate this identity: $$\sin{\left(x+\alpha\right)}-\sin{\left(x\right)}=2\sin{\left(\frac{\alpha}{2}\right)}\cos{\left(x+\frac{1}{2}\alpha\right)}$$ to be $$=\cos{\left(x+\frac{\alpha}{2}\right)}\cdot\frac{\sin{\left(\frac{\alpha}{2}\right)}}{\frac{\alpha}{2}}$$

Using the limit definition of the derivative, it is found that the derivative of this is $\cos(x)$. How does this prove the derivative of $\sin(x)$ is $\cos(x)$?

答:
You've done most of the hard work

The definition of the derivative of a function at a point $x $ is

$$f'(x) = \lim_{\alpha\to 0} \frac{f(x+\alpha) - f(x)}{\alpha}$$

Now, if you use your calculation and put it in the numerator, and use the fact that

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

You should get the proof

Title: Proving the derivative of sin(x)


问:
What is the point of discontinuity for the complex function $f(z)= \frac{2z-3}{z^2+2z+2}$?
Can someone help me find the point of discontinuity of a complex function? $$f(z)= \frac{2z-3}{z^2+2z+2}$$

答: Is it really a discontinuity point if the function is not even defined on it? For me, this "function" (you didn't specify the domain) is continuous.</p>


问:
Gradient of function that can be expressed as another function
Given unknown constants $a$ and $b$, let

$$f(x,y) := \frac{a}{2} x^2+ \frac{b}{2} y^2$$

meaning that
$$\nabla f(x,y) = \begin{bmatrix} ax\\ by\end{bmatrix}$$

Then I have $h(k) = f((1-ka)x_0, (1-kb)y_0)$, where $x_0$ and $y_0$ are constants.

(This is line search for gradient descent btw, we had the formula $h(k) = f(z_i + kp_i)$ where $z_i$ is the starting point we have ($(x_0, y_0)$ in this case) and $p_i$ is the negative gradient at $z_i$ ($-\nabla f(z_i)$))

I was asked to set the gradient of $h$ to $0$ and solve for $k$. Would this makes $\nabla h(k) = \nabla f((1-ka)x_0, (1-kb)y_0) = (a(1-ka)x_0, b(1-kb)y_0)$? In which case I would need $a(1-ka)x_0 = b(1-kb)y_0 = 0?$ Which is impossible since $a$ and $b$ might very well be not equal?

So the question I have is 1) did I misunderstand something/did a step wrong while doing the derivation and 2) is it actually possible to find a $k$ where gradient of $h(k)$ is $0$?

答:
Setting $u(k)=(1-ka)x_0,v(k)=(1-kb)y_0$, gradient of $h(k)$ is simply$$\begin{align*}h'(k)&=\frac{df(u(k),v(k))}{dk}\\&=f_uu'(k)+f_vv'(k)\\&=-ax_0f_u-by_0f_v\\&=-a^2ux_0-b^2vy_0\end{align*}$$

Substituting back $u(k),v(k)$ and equating this to zero gives$$k=\frac{a^2x_0^2+b^2y_0^2}{a^3x_0^2+b^3y_0^2}$$


答:
You seem a bit confused. We usually abuse notation and call both the representation and the underlying vector space the same thing. In your question's terminology, $V=X$.

More properly, a representation of a $k$-algebra $A$ is a pair $(V, \rho)$ where $V$ is a $k$-vector space and $\rho \colon A \to \mathrm{End}(V)$ is a $k$-linear ring homomorphism. However, we often abbreviate this by saying "$V$ is a representation of $A$", leaving $\rho$ implicit.

This convention makes more sense when you think of representations as actions: given $v \in V$, we can act on $v$ by $a \in A$ to get another element $a \cdot v \in V$, and this operation is appropriately linear and associative. Formally, $a \cdot v = \rho(a)(v)$ connects the two perspectives. In this sense, a representation "is" the vector space $V$, together with the extra data of an action, but we often just let actions come along for the ride without establishing extra notation for the action map, like scalar multiplication for vector spaces themselves.

A homomorphism of representations then must be a map between the vector spaces that's appropriately compatible with the actions: it's a $k$-linear map $\phi \colon V \to W$ such that $\phi(a \cdot v) = a \cdot \phi(v)$. In this guise it's pretty obvious that $\ker \phi$ is a subrepresentation since $\phi(v)=0$ implies $\phi(a \cdot v) = a \cdot \phi(v) = a \cdot 0 = 0$, giving closure.

You could write the action compatibility axiom more precisely as $\phi(\rho_V(a)(v)) = \rho_W(a)(\phi(v))$. Most people find that impractical.

Title: Given a k-algebra A, representations $V, W$ of $A$, and homomorphism $\phi: V \rightarrow W,$ prove that ker$\phi$ is a subrepresentation of $A$

Body:
I think my issue here is wrapping my head around what is, exactly, a homomorphism of representations.

A representation is some vectorspace $S$ and a homomorphism $A \rightarrow$ End$(S)$.

So, to be specific, we have that $V$ is some vectorspace $X$ and a homomorphism $x: A \rightarrow$ End$(X)$. Then $W$ is a vectorspace $Y$ and a homomorphism $y: A \rightarrow$ End$(Y)$.

To make things easier... $V = (x, X)$ and $W = (y, Y)$.

Then... a homomorphism $\phi: V \rightarrow W$ is... what, exactly? It's somehow defined on both the morphisms $x,y$ and the vectorspaces $X,Y$, right? But what does that really "look like?" It  kind of feels like a really, really small functor, but I'm having a hard time putting that into words.

And then, like. Okay. The kernel of any morphism is, colloquially speaking, "the set of things which get mapped to $0$." But I don't know what's really being mapped, here, and what $0$ is in this context. The $0$ representation?


问:
If a function f has no jump discontinuities, does it have the intermediate value theorem property?
I realized I was confused by this concept (while preparing for my exam).

If a function f has no jump discontinuities, does it have the intermediate value theorem property?

Facts, I know:
I know that continuity implies intermediate value theorem property.  However, intermediate value theorem property does not imply continuity.  All derivatives have the intermediate value theorem property, but we can have discontinuous derivatives.
Derivatives do not have jump discontinuities, or discontinuities of the first kind.

I don't know if a function has no jump discontinuities then it necessarily has the intermediate value theorem property.  I know this works for derivatives.  I was trying to construct a discontinuous function with discontinuities of the second kind, (one of left or right does not exist), with no jump discontinuities, that doesn't have intermediate value theorem property, but I couldn't think of any.

Thanks

答:
Another example: The characteristic function on the rationals:

$$\chi_{\mathbb{Q}} = \cases{1 & $x \in \mathbb{Q}$ \cr 0 & $x \notin \mathbb{Q}$}$$

At no point does either one sided limit exist (so no jump discontinuities), but the function assumes no value between 1 and 0.


问:
Is martingale problem interesting?
Why is the Martingale problem interesting, or useful to areas outside of math like economics, game theory, physics, etc.?

As a reminder, the Martingale problem is about finding a process so that when a given linear operator is applied to it, you get a martingale.

http://www.ma.utexas.edu/mediawiki/index.php/Martingale_Problem

Impreciseness is fine, I'm only looking for intuition and scope of knowledge here. (For instance, that website is too advanced for me to understand technically.)

I know Gaussian processes and Markov Chains are useful to physics.  Are Martingales also useful to physics?

答:
First of all, martingale problems often arise in the theory of Markov processes as they are related to the characterization. For example, let $(X_n)_{n\in \Bbb N}$ be a sequence of Markov processes with generators $\mathscr L_n$ and we would like to check whether they converge to a Markov process $X$ with a generator $\mathscr L$. In such a case, one can just check a certain convergence $\mathscr L_n\to\mathscr L$. However, to extend this result from $\mathscr L$ to $X$ one needs to show that a martingale problem for $\mathscr L$ has a unique solution. In general, one can define Markov processes as solutions of martingale problems.

Martingales themselves are not only (or almost not at all) about martingale problems (which themselves are more about operators). They often appear in optimal control problems thanks to their almost-constant structure, and are inevitable when dealing with SDEs or SPDEs. The latter often arise in evolutional dynamics and can have examples in all systems that evolve in time.

P.S. your reminder of what is a martingale problem, especially a linear operator applied to a process is misleading.

Title: Is martingale problem interesting?


问:
Probability of CDC data

All of this is my work, for (c) I believe its $1 - pbinom(144, 138, 2.9)$ which yields $.0192 $

For (d) I know that when D is negative, that would mean the woman is taller than the man, so I think it is just $ 69.2/63.8$ ?

Please correct me if I am wrong, Thank you in advance!

Title: Probability of CDC data

Body:
Use pnorm; pbinom is for binomial random variables.

> 1-pnorm(72,69.2,2.9)
[1] 0.1671429

Correct

$$P\left(Z\ge\frac{144-133}{\sqrt{16.82}}\right)=P(Z\ge2.726249)$$

> 1-pnorm(2.726249)
[1] 0.003202933

Correct

$$P\left(Z\le\frac{0-5.4}{\sqrt{16.82}}\right)=P(Z\le -1.316682)$$

> pnorm(-1.316682)
[1] 0.09397262


问:
Equivalence of $R$-linear additive categories are $R$-linear?
Let $R$ be a commutative ring. Let $C,D$ be $R$-linear categories (https://stacks.math.columbia.edu/tag/09MJ ) such that they are also additive categories. Let $F: C \to D$ be an equivalence of categories.

Then, is $F$ an $R$-linear functor? i.e., for every $X,Y\in  Ob(C)$, is it true that

$$F: \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X), F(Y))$$

is an $R$-module homomorphism?

My thoughts: Since $C,D$ are additive categories, so $F$ is an additive functor i.e. $F: \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X), F(Y))$
is an abelian group homomorphism. So I think we only have to check whether $F: \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X), F(Y))$

preserves multiplication by elements of $R$ or not.

答:
No.  For a really simple way of getting examples, note that if $\alpha:R\to R$ is any endomorphism and $C$ is an $R$-linear category, we can make a different $R$-linear category $D$ with the same underlying category as $C$ but with the action of $R$ modified via $\alpha$ (i.e., $r\cdot_D f=\alpha(r)\cdot_C f$).  Then the identity functor is an equivalence $C\to D$, but it is will rarely be $R$-linear.  In particular, if $\alpha$ is not the identity and the action of $R$ on $C$ is faithful (e.g., if $C$ is the category of $R$-modules), the identity functor is not $R$-linear.

Or for another really simple type of example (if you don't require the categories to be additive, as the original question did not), note that an $R$-linear category with one object is just an $R$-algebra.  Then you are asking whether if $A$ and $B$ are $R$-algebras and $f:A\to B$ is an isomorphism of multiplicative monoids, then $f$ is $R$-linear.  There are very easy counterexamples to this (for instance, when $R=A=B=\mathbb{Z}$).


问:
$\prod\limits_{n=1}^{x}\left(1+\frac{1}{n}\right)=x+1$. What is this formula called? Is it even useful?
So I was playing around with some equations and I got to this formula:

$$
\prod\limits_{n=1}^{x}\left(1+\frac{1}{n}\right)=x+1
$$

I checked it with particular values and it holds true when x is a positive integer.

I never saw this formula before. Is it just a useless formula? If not, does it have a name?

答:
This is an interesting observation. To see what's going on it helps to rewrite $(1 + 1/n)$ as $\frac{n+1}{n}$. Then for example if $x = 4$ we have
$$
\Pi_{n=1}^4 (1 + \frac1n) = \frac21 \cdot \frac32 \cdot \frac 43 \cdot \frac54 = 5.
$$
Everything cancels out except for the numerator in the final term in the product.

I'm not aware of any name for this formula.


答:
Let's try to understand the algorithm, and that should also answer your query. We have a convex scalar function in space $f:R^2 \to R$ and we would like to find the point where it achieves it's minimum/maximum value

To start the game, we need to pick a starting point - arbitrarily labelled $(x_0,y_0)$. The beauty of a convex function is that you could pick any starting point and it would eventually converge to your solution

Now, we need a direction to proceed to the next point - how to pick it? The gradient gives us a direction which represents the fastest rate of change - hence for a small step in that direction we would change the value the most. Seems like a nice choice of direction. (I'm adding the negative sign here but even if you don't the value of $k$ would be negative)

$$\vec{p_0} = -\nabla f(x_0, y_0)$$

Finally - we need a distance to proceed in that direction. This is where $k$ comes into the picture. How far along should we travel? A good way of doing it is to keep moving till $f$ is minimised - right? That would translate to

$$k^* = \arg\min h(k)$$

where $h(k) = f((x_0,y_0) - k\vec{p_0})$

In your case that simply means finding $k^*$ such that

$$\frac{dh(k)}{dk}|_{k^*} = 0$$

Once you get $k^*$ you can find the next point and repeat the process till it converges

Title: Gradient of function that can be expressed as another function

Body:
Given unknown constants $a$ and $b$, let

$$f(x,y) := \frac{a}{2} x^2+ \frac{b}{2} y^2$$

meaning that
$$\nabla f(x,y) = \begin{bmatrix} ax\\ by\end{bmatrix}$$

Then I have $h(k) = f((1-ka)x_0, (1-kb)y_0)$, where $x_0$ and $y_0$ are constants.

(This is line search for gradient descent btw, we had the formula $h(k) = f(z_i + kp_i)$ where $z_i$ is the starting point we have ($(x_0, y_0)$ in this case) and $p_i$ is the negative gradient at $z_i$ ($-\nabla f(z_i)$))

I was asked to set the gradient of $h$ to $0$ and solve for $k$. Would this makes $\nabla h(k) = \nabla f((1-ka)x_0, (1-kb)y_0) = (a(1-ka)x_0, b(1-kb)y_0)$? In which case I would need $a(1-ka)x_0 = b(1-kb)y_0 = 0?$ Which is impossible since $a$ and $b$ might very well be not equal?

So the question I have is 1) did I misunderstand something/did a step wrong while doing the derivation and 2) is it actually possible to find a $k$ where gradient of $h(k)$ is $0$?


问:
Jacobian matrix of a composition of functions with chain rule
Consider the differentiable functions $f:\mathbb{R}^{\ell}\to\mathbb{R}^{m}$, $g:\mathbb{R}^{m}\to\mathbb{R}^{n}$, and $h:\mathbb{R}^{n}\to\mathbb{R}^{o}$. Let $F:\mathbb{R}^{\ell}\to\mathbb{R}^{o}$ be the composition of these functions, i.e., $F=h\circ g\circ f$. We define the Jacobian matrix of $F$ evaluated at $\boldsymbol{x}\in\mathbb{R}^{\ell}$ as $J_{F}(\boldsymbol{x})\in\mathbb{R}^{o\times \ell}$, and similarly, we define the Jacobian matrices $J_{f}$, $J_{g}$, and $J_{h}$ for $f$, $g$, and $h$, respectively.

Using the chain rule, is it safe to say that $ J_F(x)=J_h(g\circ f(x))\cdot J_g(f(x))\cdot J_f(x) $ ?

If so, how can that be converted in matrix form ?

答:
$$
J_F(x)=J_h(g\circ f(x))\cdot J_g(f(x))\cdot J_f(x)
$$


问:
Continuity of derivative function
Let $f:X \to Y$ be a function such that it is differentiable on the interval $[x ,y] \subseteq X   (x < y).$ If $a \in (x , y)$ and if $\lim\limits_{z \to a^+}f'(z)$ and $\lim\limits_{z \to a^-}f'(z)$ both exists, then $f'$ is continuous at $a$.

I tried using Darboux's Theorem and the property of limit but was not able to bring a complete proof. Is there any hints to tackle this?

答: Argue by contradiction.  WLOG, assume $\lim_{y\uparrow a}f'(y)=m<f'(a)$.  By intermediate value property, for any $n>0$ we can find point $x_n\in(\max(a-\frac1n,x),a)$ with $f'(x_n)=\frac{m+f'(a)}{2}$.  So $\lim_n f'(x_n)=\frac{m+f'(a)}{2}\neq m=\lim_{y\uparrow a}f'(y)$, but $x_n\uparrow a$ by construction.

Title: Continuity of derivative function


问:
Calculate big-$\Theta$ for $T(x) = \log(x2x!)$
$T(x) = \log(x2x!)$

use the property of log, $\log(x2x!)$ is equivalent to $\log(2x) + \log(x!)$

My approach is to prove big-$O$ and big-$\Omega$ for $T(x)$,then big-$\Theta$ just follows.

If I want to calculate big-$O$ and big-$\Omega$ for $T(x)$, can I treat $\log(2x)$ as a constant and ignore it since it's growth rate is so slow compare to $\log(x!)$?

答: This is where Stirling's Approximation comes in handy. The result is that $$ \log(2x \cdot x!) = \Theta\left(x \log x + \log 2 + \log x\right) = \Theta\left(x \log x\right) $$</p>


问:
What is the right ball-park in a sub-space?
In a proof of the theorem, "A sub-space, Y, of a totally bounded space X is totally bounded",  a covering of X by a finite number of balls, B(x, r/2), is divided into balls that meet Y and those that do not.   From each ball that meets Y, a point y is chosen and another set of balls, B(y; r), is defined.  Since the first set covers X and therefore Y, the second finite set of twice the radius also covers Y.   The proof continues with their intersections with Y and asserts that the finite set of balls, B(y; r) ^ Y, lie within the subspace, and make Y totally bounded. QED.  Now, my perhaps trivial question:  "Ball" seems to imply something spherical, and
B(y; r) ^ Y is not.  The definition of totally bounded that I have reads "a finite covering by open balls", not "a finite covering by open sets", as for example,  B(y; r) ^ Y, which may not be spherical.  Am I slicing this too finely?</p>

答:
The general definition of totally bounded can be often phrased in terms of $\epsilon$-nets.
Given a metric space $X$, a subset $E$ and some $\epsilon>0$ We call $N^{\epsilon}$ an $\epsilon$-net of $E$ if for any $x \in E$ there exists some $y \in N^{\epsilon}$ such that $d(x,y)<\epsilon$.

A set is totally bounded if it admits a finite $\epsilon$-net for any $\epsilon >0$.

In your case, sure, the intersections may not be balls when seen from the perspective of the big space $X$, but for the subspace $Y$, by all means the intersections produce "balls" even if they aren't inherently spherical because of the weird structure of $Y$. Rephrasing totally bounded-ness in terms of $\epsilon$-nets not lets you see that it is the distances that really matter, and not quite the shape of the sets you are covering with (although, yes surely the definition you gave is equivalent to the one stated in my answer - and it should be possible to extract "balls" from your intersection that agree with my statement about $\epsilon$-nets). It may also be fruitful to see this as the inherited topology in some sense, to get more of an intuitive feeling.


问:
Method of characteristics problem
I'm trying to solve the equation $x_1 u_{x_1} + x_2 u_{x_2}=2u, u(x_1,1)=g(x_1)$
by the method of characteristics. I've found the characteristic equations:

\begin{cases}
      \dot{x_1}(s) =x_1(s) \\
      \dot{x_2}(s) =x_2(s) \\
      \dot{z}(s)=2z
   \end{cases}

Which gives:

\begin{cases}
      x_1(s) =c_1e^s \\
      x_2(s) =c_2e^s \\
      z(s)=c_3e^{2s}
   \end{cases}

The curve $x(s)=(x_1(s),x_2(s))$ should lie on the boundary $\Gamma=\{(x,y) \in \mathbb{R^2}: y=1 \}$ at $s=0$  ($x(0)=(x_0,1)$ ) and this gives:

\begin{cases}
      x_1(s) =x_0e^s \\
      x_2(s) =e^s \\
   \end{cases}

Similarly one should have $z(0)=u(x(0))= u(x_0,1)=g(x_0)$ so that
$z(s)=g(x_0)e^{2s}$.

If I've understood it rght I should solve $s$ for $x_1,x_2$ in $z(s)$ but I'm not sure how to do this. Could somebody explain to me how this is done?

答:
Another way we can solve this problem is by going into polar coordinates

$$x_1u_{x_1}+x_2u_{x_2} = ru_r = 2u \implies u = f(\theta)r^2$$

or in other words

$$u(x_1,x_2) = f\left(\frac{x_1}{x_2}\right)(x_1^2+x_2^2)$$

Then plugging in the initial condition gives us

$$u(x_1,1) = g(x_1) = f(x_1)(x_1^2+1) \implies f(x) = \frac{g(x)}{1+x^2}$$

which means the final answer is

$$u(x_1,x_2) = \frac{g\left(\frac{x_1}{x_2}\right)}{1+\frac{x_1^2}{x_2^2}}(x_1^2+x_2^2) = g\left(\frac{x_1}{x_2}\right)x_2^2$$

Title: Method of characteristics problem


问:
Computational complexity of this algorithm
Consider a function $f(n,k)$ for $n,k\in\mathbb{N}$ and an algorithm that implements that function. The structure of the algorithm is as follows:

do some calculations that take $O(n)$ time

define $k' = k \mod 2^n$

do some calculations that take $O(k')$ time

What is the computational complexity of this algorithm? Specifically, is it $O(n)$ or $O(2^n)$?

答:
@Listing is correct.

The question tests to see if one recognizes that

```<code>f(n,k)
= *O*(n) + *O*(1) + *O*(k')
= *O*(n) + *O*(1) + *O*(2^n)
= *O*(2^n)
</code>```

For example,

```<code>|f(n,k)|
<= M|n| + C|1| + N|k'| as x goes to infinity, and some constant M and N that satisfy:
    (a) |work in part one| <= M|n|
    (b) |definition of part two| <= C|1| (by assumption)
    (c) |work in part three| <= N|k'|
as given or assumed.

<= M|n| + C|1| + N|2^n| since |k'| <= 1|2^n| as x goes to infinity (all k' are bounded above by 2^n by the definition of mod)

<= N|2^n| since | M|n| + C|1| + N|2^n| | <= 2N|2^n| as x goes to infinity
</code>```

We note that if f(n,k) were to be O(n), then it would also be O(2^n); however, in this case, it is not O(n) since (as @Listing stated) k' can reach values much greater than n since k' takes on all values up to 2^n-1 (eg, when k=2^n-1, k'=2^n-1), and |2^n-1| > A|n| for all constants A as n goes to infinity.

See this page for more information on big O notation.

Body:
@Listing is correct.

The question tests to see if one recognizes that

f(n,k)
= *O*(n) + *O*(1) + *O*(k')
= *O*(n) + *O*(1) + *O*(2^n)
= *O*(2^n)

For example,

|f(n,k)|
<= M|n| + C|1| + N|k'| as x goes to infinity, and some constant M and N that satisfy:
    (a) |work in part one| <= M|n|
    (b) |definition of part two| <= C|1| (by assumption)
    (c) |work in part three| <= N|k'|
as given or assumed.

<= M|n| + C|1| + N|2^n| since |k'| <= 1|2^n| as x goes to infinity (all k' are bounded above by 2^n by the definition of mod)

<= N|2^n| since | M|n| + C|1| + N|2^n| | <= 2N|2^n| as x goes to infinity

We note that if f(n,k) were to be O(n), then it would also be O(2^n); however, in this case, it is not O(n) since (as @Listing stated) k' can reach values much greater than n since k' takes on all values up to 2^n-1 (eg, when k=2^n-1, k'=2^n-1), and |2^n-1| > A|n| for all constants A as n goes to infinity.

See this page for more information on big O notation.


问:
Closure of a subspace with respect to a inner product
I just have a question in general. If we are trying to show that a subspace of a vector space is a closed subspace, I know we need to prove that all convergent sequences in that subspace converge to a limit in that set. But if the subspace is defined in terms of the inner product, for example:

A = {f is a continuous function on C[0,1]: < f,sinh > = 0} and we knew that < f,sinh > is a continuous function of f, why would this imply A is a closed subspace?

答: Since $\phi(f)=<f,sinh>$ is a continuous function of $f$, $\phi^{-1}(0)=A$ will be closed.</p>


问:
$\int_0^{2 \pi} \sinh^{-1} \frac{h}{\sqrt{s^2 + R^2 -2sR \cos(\phi)}} \, d\phi $
The definite integral is the following one:

$$I_0 = \int_0^{2 \pi} \sinh^{-1} \frac{h}{\sqrt{s^2 + R^2 -2sR \cos(\phi)}} \, d \phi $$

Where $h, s, R$ are positive real quantities and $\sinh^{-1}$ is the arcsinh function. Basically, I have a cylinder of height $h$ and radius $R$ centered at the $xy$ plane origin. It goes up along the $z$ axis. I want to calculate the magnetic potential at a radial distance $s$ from the $z$ axis. And $\phi$ is the argument in respect to the $x$ axis. In other words, I am using the usual cylindrical coordinates, $s, \phi, z$.

This integral appeared when I was calculating the magnetic vector potential $A$. I tried to make substitutions with inverse functions ($\phi = \sin(x)$ and so on...) and also tried to substitute the whole argument of arcsinh so as to achieve an easier one. I also tried to use the approximation $\ln(2x) = \sinh^{-1}(x)$. But nothing worked out well. Also tried to expand in a power series the integrand itself and just the cosine function...

One of the results I found was:

$$I_0 = 4h^2 \int_{\sinh^{-1}(\frac{h}{s-R})}^{\sinh^{-1}(\frac{h}{s+R})} \frac{y \coth(y)}{\sqrt{-(\sinh^4(y)(s^2-R^2)^2-2(s^2+R^2)h^2 \sinh^2(y)
 +h^4  )}}dy$$

It appears to be a binomial expansion, but it isn't. It then becomes:

$$I_0 = 4h^2 \int_{\sinh^{-1}(\frac{h}{s-R})}^{\sinh^{-1}(\frac{h}{s+R})} \frac{y \coth(y)}{\sqrt{ (\sinh^2(y) 2sR)^2  - (\sinh^2(y)(s^2-R^2)-h^2)^2  }}dy$$

Any suggestions would be very much appreciated! Thanks!

MAJOR EDIT

It seems very unlikely that this integral above is easily solvable. That made me go back and realize the following:

$A(s, 0, 0) = \underbrace{\frac{u_0 nI h R}{4 \pi}}_{C} \displaystyle\int_0^{2 \pi} \displaystyle\int_{-1}^1 \frac{ \bar{\phi}}{\sqrt{s^2-2sR \cos(\phi) +R^2+h^2z^2}} dz d \phi$,

where $\bar{\phi}$ is the unit vector (cylindrical coordinates). Notice that, if we solve the integration in respect to $z$ we will achieve the nasty integral above ($I_0$). However, let's approach differently this time. Let's invert the order of integration and perform the following substitution: $\phi = \arccos(x)$. Then it becomes:

$A(s, 0, 0) = 2C \displaystyle\int_{-1}^1 \displaystyle\int_{-1}^1 \frac{\bar{\phi}}{ \sqrt{1-x^2} \cdot \sqrt{(s^2+R^2+h^2z^2)-2sRx}} dxdz$

Let's take both radicands to be non-negative and we then have:

$A(s, 0, 0) = 2C \displaystyle\int_{-1}^1 \displaystyle\int_{-1}^1 \frac{\bar{\phi}}{ \sqrt{ 2sRx^3 - (s^2+R^2+h^2z^2)x^2 - 2sRx + (s^2+R^2+h^2z^2) } } dxdz$

Which seems to be an elliptic function (if my quick search was correct). If these steps are correct, how should I proceed? Maybe residue theorem? Although I could recognize, I don't have the knowledge to solve it properly.

So far, I have found the following websites:

http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf ;
https://functions.wolfram.com/EllipticIntegrals/EllipticE/introductions/CompleteEllipticIntegrals/ShowAll.html ;
https://en.wikipedia.org/wiki/Elliptic_integral

Although they mentioned a similar integral to $A(s, 0, 0)$ and the need of elliptic integrals, they do not explain much.

Body:
I think that we can make approximations.

First of all, we have
$$\int_0^{2\pi} \sinh^{-1}(\text{argument})\,d\phi=2\int_0^{\pi} \sinh^{-1}(\text{argument})\,d\phi$$ What I think is that we could expand as (at least) two series : one around $\phi=0$ to compute the integral between $0$ and $\frac \pi 2$ and another one around $\phi=\frac \pi 2$ to compute the integral between $\frac \pi 2$ and $\pi$.

Using the same values as in @Ryan Howe's answer and using series to $O\left(t^{n+1}\right)$ and $O\left(\left(t-\frac{\pi }{2}\right)^{n+1}\right)$, I computed using these series
$$I_1=2\int_0^{\frac\pi 2} \sinh^{-1}(\text{argument})\,d\phi\qquad I_2=2\int_{\frac\pi 2}^\pi \sinh^{-1}(\text{argument})\,d\phi$$

$$\left(
\begin{array}{cccc}
n & I_1 & I_2 & I_1+I_2 \\
 3 & 0.33073804 & 0.29589435 & 0.62663239 \\
 4 & 0.33531563 & 0.29420164 & 0.62951727 \\
 5 & 0.33531563 & 0.29435349 & 0.62966912 \\
 6 & 0.33404741 & 0.29461198 & 0.62865939 \\
 7 & 0.33404741 & 0.29444469 & 0.62849210 \\
 8 & 0.33445144 & 0.29447519 & 0.62892663 \\
 9 & 0.33445144 & 0.29449613 & 0.62894757 \\
 10 & 0.33431268 & 0.29447737 & 0.62879005 \\
 11 & 0.33431268 & 0.29448269 & 0.62879537 \\
 12 & 0.33436278 & 0.29448433 & 0.62884711
\end{array}
\right)$$ while the "exact" values are $I_1=0.33434920$, $I_2=0.29448286$ and $I_1+I_2=0.62883206$.


问:
Prove that, there does not exist any $A ⊆ R$ satisfying $m(A ∩ I) = α l(I)$, for all intervals $I ⊂ R$
Let $0 < α < 1$. Prove that, there does not exist any $A ⊆ R$ satisfying $m(A ∩ I) = α l(I)$, for all intervals $I ⊂ R$. Here, $m$ denotes the Lebesgue measure on $R$ and $l$ denotes the length of the interval.

I have been asked to prove this, but I have found a counterexample instead.

$I=(1,3)$ and $α=\frac{1}{2}$, I have found an $A=(2,4)$ such that $m(A ∩ I) = 1 = α l(I)= \frac{1}{2}*2$

I have also been asked a follow up question:
Does there exist $A ⊆ R$ satisfying
$m(A ∩ I) = l(I)$, for all intervals $I ⊂ R?$ and I think a similar example can be constructed. Am I going wrong somewhere?

答:
The question clearly says 'for all intervals' and you are giving an example with one fixed interval.

If such  a set exists with finite measure  then $\frac  1 {b-a} \int_a^{b} \chi_A (x) dx =\alpha$ whenever $a <b$. Letting $b \to a$ we get, using Lebesgue's differentiation theorem, $\chi_A(x)=\alpha $ almost everuywhere. But $\chi_A(x)$ takes only the values $0$ and $1$. Hence, such a set does not exist.

For the case when $A$ does not have finite measure consider $A \cap (-1,1)$ and use intervals $I$ conatined in $(-1,1)$ in above proof.
For the second question take $A=\mathbb R$.

Ref: https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem


答:
If you know topology, by the definition the map $\phi$ is continuous if the preimage $\phi^{-1}(A)$ is open given that the set $A$ is open. Equivalently, $\phi^{-1}(B)$ is closed whenever $B$ is closed. In your case
$$
  \phi(f) = \langle f,\sin \rangle
$$
is a continuous function from $C([0,1])$ to $\Bbb R$, and since $B = \{0\}$ is a closed set, your $A = \phi^{-1}(B)$ is closed as well. Clearly, it is also a linear space, so it is a closed subspace.</p>

Title: Closure of a subspace with respect to a inner product

Body:
I just have a question in general. If we are trying to show that a subspace of a vector space is a closed subspace, I know we need to prove that all convergent sequences in that subspace converge to a limit in that set. But if the subspace is defined in terms of the inner product, for example:

A = {f is a continuous function on C[0,1]: < f,sinh > = 0} and we knew that < f,sinh > is a continuous function of f, why would this imply A is a closed subspace?


问:
Finding the absolute maximum and minimum of a function in a disk?
Find the absolute maximum and minimum of the function $ \ e^{-x^2-y^2}(x^2+2y^2) \ $ in the disk $ \ x^2 \  + \  y^2 \ \le \ 4 \ $.

I found the partial derivatives but am unable to find the critical points and proceed further

答:
As has already been pointed out, the function $ \ f(x,y) \ = \ (x^2+2y^2)·e^{-x^2-y^2} \ = \ (x^2+2y^2)·e^{-(x^2+y^2)} \ $ is non-negative; since the exponential ("decaying") factor is always positive, this function is only equal to zero at the origin, so that is the location of the absolute minimum.

What we may also say about this function is that it has "four-fold" symmetry $ \ [ \ f(-x,y) \ = \ f(x,-y) \ = \ f(-x,-y) \ = \ f(x,y) \ ] \ $ , so that any critical points not on a coordinate axis will appear in "sets of four" and those on an axis will form pairs.  If the function were simply $ \ (x^2+y^2)·e^{-(x^2+y^2)} \ , \ $ the function would have "radial symmetry", and we know from a frequently-posed problem that the maximum of $ \ r^2·e^{-r^2} \ $ occurs at radial distance  $ \ r \ = \ 1 \ $ , thus at all points on a unit circle centered at the origin.  For our function, the factor $ \ (x^2+2y^2) \ $ has the effect of "stretching" the radial function "away from" the $ \ y-$ axis, turning the circle into an ellipse with its major axis on the $ \ x-$ axis.  This "breaks" the radial symmetry down to the four-fold symmetry of $ \ f(x) \ $ .  Since the distance-squared $ \ (x^2+y^2) \ $ is increased for points on or outside the unit circle but not on the $ \ y-$ axis, the exponential factor decreases "more rapidly" than $ \ (x^2+2y^2) \ $ increases, so we might intuit that $ \ f(x) \ $ is only maximal at the points where the unit circle intercepts the $ \ y-$ axis.

We can check this by exploiting the four-fold symmetry and considering "slices" through the function on the lines $ \ y  =  \pm  \ mx \  $  to produce $ \ \phi(m,x) \  = \ [ \ (1+2m^2)x^2 \ ]·e^{-[ \ (1+m^2)x^2 \ ]} \  $ .  (This is a variation on the idea of using polar coordinates.)  On a particular such slice, fixing $ \ m \ $  , the extremization calculation gives us

$$ [ \ 2 \ (1+2m^2)x \ ]·e^{-[ \ (1+m^2)x^2 \ ]} \ - \ [ \ 2 \ (1+m^2)x \ ]·[ \ (1+2m^2)x^2 \ ]·e^{-[ \ (1+m^2)x^2 \ ]} \ \ = \ \ 0 $$

$$ \Rightarrow \ \ \ x \  \ = \ \  (1+m^2) \ x^3 \ \ \ \Rightarrow \ \ \ x \ = \ 0  \ \ , \ \ x^2 \ = \ \frac{1}{1 \ + \ m^2} \ \ . $$

We already know about $ \ x \ = \ 0 \ $ ; inserting our second result into $ \ \phi(m,x) \ $ tells us that the maximum value on the slice is

$$  \left(  \frac{1+2m^2}{1+m^2} \right)·e^{- \left( \frac{1+m^2}{1+m^2} \right)} \ \ = \ \ \left(  \frac{1+2m^2}{1+m^2} \right)· \frac{1}{e} \ \ . $$

For the (two) critical points on the $ \ x-$ axis ( $  m = 0  \ \ \Rightarrow \ \ x \ = \ \pm 1 $ ) , the maximum of the function is $ \ \frac{1}{e} \ $ .  We see that
$$ \ \lim_{m \ \rightarrow \ \infty} \ \left(  \frac{1+2m^2}{1+m^2} \right)· \frac{1}{e} \ \ = \ \ \frac{2}{e} \ \ , $$

so our expectation that the maxima of $ \ f(x,y) \ $ lie on the $ \ y-$ axis ( $  \lim_{m \ \rightarrow \ \infty} \ \frac{1}{1 \ + \ m^2} \ = \ 0 $ ) was correct. The absolute maxima are confirmed as $ \ f(0,1) \ = \  f(0,-1) \ = \ \frac{2}{e} \ \ $  .


问:
What's the intuition behind the Co-Area formula?
I mainly work in statistics and I know only basic measure theory. I was trying to understand the Co-Area formula by Federer.

If $f:\mathbb{R}^M\to \mathbb{R}^N$ is a Lipschitz function with $M \geq N$ then
$$
\int_A J_N f(x) d\mathcal{L}^M x = \int_{R^N} \mathcal{H}^{M-N} (A\cap f^{-1}(y)) d\mathcal{L}^N y
$$

Could anyone please give me an intuition of what the formula implies and what it means practically? In particular, this is a simple example I am trying to wrap my head around: suppose we are in 2 dimensions and there is a curve defined by a function

What does the co-area formula tell us in this context?

Here are some practical questions:

My understanding is that we have some space of dimension $M$ and there a manifold in it of dimension $M-N$ and we would like to compute the area of this manifold using the Hausdorff measure. Is this what is happening?

I am completely lost as to why we need $A\cap f^{-1}(y))$ what's the intuition behind this?

On the right hand side we have $\mathcal{H}^{M-N}$. I am familiar with the Lebesgue measure $d\lambda$ however I am lost as to why we seem to have both the Lebesgue and the Hausdorff measure on the right hand side. I thought we should only have the Hausdorff measure on the RHS?

答:
This is most easily understood when $N = 1$. Suppose $A \subset \mathbb{R}^m$ is a bounded domain and $f: A \rightarrow \mathbb{R}$ is a smooth function whose gradient is everywhere nonzero.

Its level sets are non-intersecting hypersurfaces. Suppose first that you want to compute the volume of the $A$ in terms of the surface areas of the hypersurfaces. Suppose $a$ is the inf of $f$ on $A$ and $b$ is the sup. You can divide the interval $[a,b]$ into equal sized subintervals $I_1 = [t_0,t_1], \dots, I_N = [t_{N-1},t_N]$ of size $\delta = (b-a)/N$. The volume of $A$ is the sum of the volumes of $f^{-1}(I_k)$. On the other hand, each $f^{-1}(I_k)$ is a shell with varying thickness. At each point, the thickness is roughly $\Delta t/|\nabla f|$. So the volume of the shell is roughly
$$
V(f^{-1}(I_k)) \simeq \Delta t\int_{f^{-1}(t_k)}\frac{dA}{|\nabla f|},
$$
where $dA$ is the $(M-1)$-dimensional Hausdorff measure on $f^{-1}(t_k)$.
Adding up the volumes of the shells and taking the limit $N \rightarrow \infty$, we see that the volume of $A$ is
$$
V(A) = \int_{-\infty}^{\infty} \int_{f^{-1}(t)} \frac{dA_t}{|\nabla f|}\,dt,
$$
where $dA_t$ is the $(M-1)$-dimensional Hausdorff measure of $f^{-1}(t)$.

If, on the other hand, you integrate $|\nabla f|$ on $A$ using the same approach, you get
$$
\int_{f^{-1}(I_k)} |\nabla f|\,dx \simeq \Delta t\int_{f^{-1}(t_k)}|\nabla f|\frac{dA}{|\nabla f|},
$$
and therefore
$$
\int_A |\nabla f|\,dx = \int_{-\infty}^{\infty} \int_{f^{-1}(t)}\,dA_t\,dt
$$
The $N>1$ case is similar, except you chop $\mathbb{R}^N$ into small rectangular pieces and use the fact that the cross section of the inverse image of each piece is roughly a parallelogram whose volume is $(J_Nf)^{-1}$.
and therefore
$$
\int_A J_Nf\,dx = \int_{\mathbb{R}^N} \int_{f^{-1}(y)}\,dA_y\,dy,
$$
where $dA_y$ is the $(M-N)$-dimensional Hausdorff measure on $f^{-1}(y)$.

The general formula is
$$
\int_A \phi(x)\,dx = \int_{\mathbb{R}^N} \int_{f^{-1}(y)} \phi(x)\frac{dA_y}{J_Nf}\,dy
$$


问:
Regular polygons that touching to a sphere surface
What is  the possible number of n sided polygons(every face is the same regular polygon) that touching their corners to sphere surface and also touching each other ?
I would like to know the relation between n sides polygon  and possible placing number.

And also I would like to find a (one side of polygon )  depends on r, n (number of sides), m (total number of polygons that placed on the sphere).

Examples:
Can I place 17 regular triangles on a sphere surface?

Can I place 7 squares on a sphere surface?

Can I place  10 regular hexagons on a sphere surface?

I do not know the related tag about my question. Please feel free to retag it. Thanks a lot for answers.

答:
every face is the same regular polygon

A strictly convex solid where every face of the solid is the same regular polygon is, by definition, a Platonic solid, one of the 5 Platonic solids.

What is the possible number of n sided polygons(every face is the same
regular polygon) that touching their corners to sphere surface and
also touching each other ?

A strictly convex solid made up of entirely of n-sided regular polygons is Johnson solid, which includes all 14 Archimedean solids and all 5 Platonic solids.
The 14 Archimedean solids and the 5 Platonic solids all have a sphere that touches every corner of the solid.
Many (but not all) of the remaining Johnson solids also have a sphere that touches every corner of the solid.

...

Can I place 10 regular hexagons on a sphere surface?

The truncated icosahedron, more commonly known as "the soccerball shape", is an Archimedean solid that includes 20 regular hexagons, all touching each other and all with their corners touching the surface of a sphere. (The remaining surface is covered by 12 regular pentagons, none of which touch each other).

Can I place 7 squares on a sphere surface?

The rhombicuboctahedron is an Archimedean solid that includes 18 square faces, all touching each other and all with their corners touching the surface of a sphere. (The remaining surface is covered by 8 regular triangles, none of which touch each other).

Have fun!


问:
Given an arbitrary quadratic polynomial $ax^2 + bx + c$, how do I find the path the vertex of the parabola travels upon when b is varied?
https://www.desmos.com/calculator/ysuklqqwfe

I've made a desmos to illustrate the problem. If you press play on b you'll see what I mean.

I'm pretty sure the arc is parabolic because the y coordinate varies quadratically with respect to x. But I'm not sure how to go about proving it, or to find the equation of the path.

Any insights?

答:
Let $y = ax^2 + bx + c$.  For the time being, suppose $a$, $b$, and $c$ are fixed, with $a \ne 0$.

The vertex of the parabola is located where the parabola reaches an extremum, either a minimum (when $a > 0$) or a maximum (when $a < 0$).  To see where this occurs, we complete the square:

$$\begin{align}
y &= a\left(x^2 + \frac{b}{a} x\right) + c \\
&= a \left(x^2 + 2 \cdot \frac{b}{2a} x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c \\
&= a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\
&= a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}.
\end{align}
$$

As a function of $x$, keeping in mind $a, b, c$ are fixed, the only term that changes is the first; moreover, because the square of a real number is never negative, the minimum value that $(x + \frac{b}{2a})^2$ can attain is $0$, when $x = -\frac{b}{2a}$.  And as stated previously, this choice leads to a minimum when $a > 0$ and a maximum when $a < 0$.  Therefore, this is the $x$-value for the vertex, and the $y$-value is the second term, $-\frac{b^2-4ac}{4a}$; thus the vertex has coordinates
$$(x,y)_{\text{vertex}} = \left(-\frac{b}{2a}, -\frac{b^2-4ac}{4a}\right).$$
Now for a fixed $a$ and $c$, where $b$ is allowed to vary, what is the locus of the vertex--what shape does the vertices of the resulting family of parabolas trace out?  This can be realized by expressing the $y$-coordinate of the vertex in terms of the $x$-coordinate, in turn as a function of $b$.  That is to say, if $x = -b/(2a)$, then $$y = -\frac{b^2 - 4ac}{4a} = -a \left(-\frac{b}{2a}\right)^2 + c = -ax^2 + c,$$ hence the locus of the vertex is the equation $$y = -ax^2 + c,$$ also a parabola, but with the linear term equal to $0$, and facing the opposite direction to the original family of parabolas. This of course should be obvious in retrospect.


问:
Finding torque of a force

A force $\vec F=2\hat i+3\hat j$ is acting on a point $(1,0,3)$. Find the torque of this force about the axis along $y=3x$.

I know 3-D geometry. So, I can handle equations of lines in 3-D too.

Definition of torque: $$\tau(\text{torque})= \vec r \times\vec F $$

where $\vec r$ is the position vector of point of application of force about the axis.

I can't find $\vec r$ correctly. Can someone help me out?

答:
To find out $\vec r$ we need two point. From  $y=3x$ we can say that (0,0,0) is situated on it and $\vec F$ is working on $(1,0,3)$ so . $$ \vec r=(1-0) i+(0-0)j+(3-0)K $$
$$\vec r=i+3k $$</p>

Title: Finding torque of a force


问:
Show limit point of product topology and sequence converging to $0$
I have been really struggling with the product topology. May I have a solution verification and get some pointers on how I should approach these product topology questions?

Let $\mathbb{R}^{\omega}$ be the countable product of copies of $\mathbb{R}$.So every point in $\mathbb{R}^{\omega}$ is a sequence $(x_1,x_2,...)$.Let $A \subset \mathbb{R}^{\omega}$ be the set consisting of all points with positive coordinates. Show that in the product topology $0=(0,0,...)$ is a limit point of the set $A$ and there is a sequence of points in $A$ converging to $0$. Then show that in the box topology, $0=(0,0,...)$ is a limit point of the set $A$, but there is no sequence of points in $A$ converging to $0$.

Attempt:
Let $U$ be a basic open set containing $(0,0,...)$. Then $U=\prod U_i$.If $U_i \neq \mathbb{R}$ since $U_i$ is open in $\mathbb{R}$ for each $U_i \neq \mathbb{R}$ and $\epsilon_i>0$ there is an interval $(-\epsilon_i,\epsilon_i) \subset U_i$. So $U_i \cap R^{+}\neq \varnothing$. Choose $\epsilon=\text{min}\{\epsilon_i\}$. Then $(-\epsilon,\epsilon)\subset U_i$ for each $i$ and so $U \cap A \neq \varnothing$, and $(0,0,...)$ is a limit point of $A$.

Consider the sequence of points in $A$ given by $x_n=(\frac{1}{n},\frac{1}{n},...)$.Let $U=\prod U_i$ be a basic open set containing $(0,0,...)$ if $U_i \neq \mathbb{R}$ there is an $N_i \in \mathbb{N}$ such that $\frac{1}{n} \in U_i$ for each $n \geq N_i$.Choose $N=\text{max}\{N_i\}$ Then $n \geq N \implies x_n \in U$, so $x_n \rightarrow (0,0,...)$.

In the box topology $(0,0,...)$ is a limit point of $A$ since if $U=\prod U_i$ is an open set containing $(0,0,...)$ for all $i$ there is an $\epsilon>0$ such that $(-\epsilon,\epsilon) \subset U$. So $U_i \cap R^{+} \neq \varnothing$ for each $i$ and $(0,0,...)$ is a limit point of $U$.

Comment:I do not know if any of this is correct. I used some help I had learned previously to try and give this solution a shot. However I really think this is incorrect, but would like to know how it is supposed to be done properly, or if I am in the right direction. Also I do not understand why there is no sequence converging to $(0,0,...)$ in the box topology, so I omit it. Does it have anything to do with not being able to take the maximum of the $N_i's$ since there might be an infinite number of $U_i's$ that are not equal to $\mathbb{R}$

答:
Your proof that the zero sequence is a limit point of $A$ in the product topology is correct, and you correctly found a sequence in $A$ converging to the zero sequence. In proving that it does so, however, you could be clearer. Let $I=\{i:U_i\ne\Bbb R\}$, and for $i\in I$ define $N_i$ as you did; then when you take $N=\max_{i\in I}N_i$, it’s explicit that you’re taking the maximum of a finite set of positive integers.

In your argument that the zero sequence is also a limit point of $A$ in the box topology it would be better to say that for each $i$ there is an $\epsilon_i>0$ such that $(-\epsilon_i,\epsilon_i)\subseteq U_i$, to emphasize that the choice of $\epsilon_i$ really does depend on the coordinate factor involved. In fact, this is basically why in the box topology the zero sequence is not the limit of a sequence in $A$.

To show this, let $\sigma=\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ be any sequence of points in $A$, where $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$ for each $n\in\Bbb N$. For each $k\in\Bbb N$ let $r_k=\frac12x_k^{(k)}$, and let $U_k=(-r_k,r_k)$. Let $U=\prod_{k\in\Bbb N}U_k$, and show that in the box topology $U$ is an open nbhd of the zero sequence that contains no points of the sequence $\sigma$. (The intuitive basis for this construction is that we can construct $U$ to exclude the $k$-th term of $\sigma$ on the $k$-th coordinate.)


问:
The context free grammar for language $L = \{a^nb^mc^k \mid k = |n - m|, n≥0,m≥0,k≥0\}$ is
The context free grammar for language $L = \{a^nb^mc^k \mid k = |n - m|, n≥0,m≥0,k≥0\}$ is:

$S→S_1S_3, S_1→aS_1c |S_2|λ, S_2→aS_2b|λ, S_3→aS_3b|S_4| λ, S_4→bS_4c|λ$

$S→S_1S_3, S_1→aS_1S_2c |λ, S_2→aS_2b|λ, S_3→aS_3b|S_4| λ, S_4→bS_4c|λ$

$S→S_1|S_2, S_1→aS_1S_2c|λ, S_2→aS_2b|λ, S_3→aS_3b|S_4| λ, S_4→bS_4c|λ$

$S→S_1|S_3, S_1→aS_1c |S_2|λ, S_2→aS_2b|λ, S_3→aS_3b|S_4| λ, S_4→bS_4c|λ$

My attempt:

None of these with given options

$S_3\implies aS_3b\implies aS_4b \implies abS_4cb\implies abcb$

Best way to by make either $n=k+m$ or $m=k+n$

Seems typo's with option $(3)$ too.

Can you explain it, please?

It also CFG for the language $L = \{(a^n)(b^m)(c^k) \mid k = |n – m|, n,m,k \geqslant 0\}$ asked there, but need more explaination, please.

Is it option $(4)$ true?

答:
Option (4) is correct as first part has #a = #b+#c and second part has #b = #a+#c, which is required for given language.

First part, for $n=k+m$ :
$$S_1→aS_1c |S_2|λ, S_2→aS_2b|λ$$
Second part, for $m=k+n$ :
$$S_3→aS_3b|S_4| λ, S_4→bS_4c|λ$$

Thus, $n=k+m$ or $m=k+n$ :
$$S→S_1|S_3, S_1→aS_1c |S_2|λ, S_2→aS_2b|λ, S_3→aS_3b|S_4| λ, S_4→bS_4c|λ$$

Language of above grammar would be inherently ambiguous.


问:
Does the following necessarily converge to a normal random variable in distribution?
Suppose $X_1, \dots, X_n$ are i.i.d. random variables with $E(X_1) = 0$ and $\operatorname{Var}(X_1) = 1$. Let
$$
S_n = \frac{1}{n}\sum_{i=1}^n\sqrt{i}X_i.
$$
Does $S_n$ converge to a normal random variable?

Originally, I attempted to use Lindeberg CLT to prove this. However, I ran into a wall because I can't figure out a way to check the Lindeberg condition that for $\forall \varepsilon > 0$
$$
\sum_{i=1}^nE(|Y_i|^2\mathbf{1}(|Y_i| > \varepsilon)) \to 0
$$
where $Y_i = \sqrt{i}X_i/s_n$ and $s_n^2 = \sum_{i=1}^ni\operatorname{Var}(X_i) = \frac{n(n+1)}{2}$. If I can prove this, then I can use Slutsky, then we are done. But I have no idea what $X_i$ actually is so I don't know how to verify the condition.

Then I tried using characteristic functions and try to do expansion and approximation. However, I also hit a wall due to the changing index of $i$.

I also tried finding counterexample, but nothing came up.

Can anyone provide some hint? Thank you!

答:
For $c_n=\sqrt n$, observe that $\frac{\max_{1\le k\le n}c_k^2}{\sum_{k=1}^n c_k^2}=\frac{n}{n(n+1)/2}=\frac{2}{n+1}\to 0$ as $n\to \infty%$.

As $X_i$'s are i.i.d, by Hajek-Sidak's CLT,

$$\frac{\sum_{k=1}^n c_k X_k}{\sqrt{\sum_{k=1}^n c_k^2}}=\sqrt{\frac{2}{n(n+1)}}\sum_{k=1}^n \sqrt k X_k \stackrel{d}\longrightarrow N(0,1)$$

That is, $$\sqrt{\frac{2n}{(n+1)}}S_n\stackrel{d}\longrightarrow N(0,1)$$

Hajek-Sidak's CLT can be shown using Lyapounov's condition (which implies Lindeberg's condition) under the additional assumption $E|X_1|^3<\infty$. But I am not aware of the general proof.


问:
What is the geometric difference between the eigenvectors and eigenspace of a 3x3 matrix?
I am trying to understand the geometric difference between a transformations:

"Eigenvectors"

"Eigenspace"

"Maximal set of nonzero orthogonal eigenvectors"

Let's say we have this symmetric 3x3 matrix:
$$B_1 =\begin{bmatrix}
11 & -8 & 4 \\
-8 & -1 & -2 \\
4 & -2 & -4
\end{bmatrix}
$$

It has the following eigenvalues: $\lambda_1 = -5, \lambda_2 = 16$.

Now using $\lambda_2$, I found one eigenvector, $v_1 = (4,-2,1)$ with this eigenvalue. From what I understand, an eigenvector is a vector that stays in the same span (stretches/squished by a scalar) following a transformation. Therefore, the span of this eigenvector is:

$$ span(v_1) = \{a\begin{bmatrix}
4\\
-2 \\
1
\end{bmatrix},  a\in\mathbb R \}$$

Question 1: Is this span the same as the eigenspace?

Now using $\lambda_1$, I end up getting an eigenvector, $u_1$, that seems to be a linear combination of two other vectors:

$$u_1 = b*m + c*n  = b\begin{bmatrix}
0.5\\
1\\
0
\end{bmatrix} + c\begin{bmatrix}
-0.25\\
0 \\
1
\end{bmatrix} = \begin{bmatrix}
0.25\\
1\\
1
\end{bmatrix}, b = 1, c= 1$$

This is where I get really confused.

Question 2: Are $u_1$, $m$, and $n$ all eigenvectors that stay on their respective lines after a transformation?

Question 3: Do $m$ and $n$ form the basis vectors that span a plane of vectors (one of which is $u_1$)? Does this mean the plane is stretched or squished by the eigenvalue? Is this plane also an "eigenspace"?

Question 4: What is the "maximal set of orthogonal eigenvectors"? Is that simply the $span(v_1)$ + $span(m,n)$?

答:
First let's note that any non-zero vector of the form
$$ b\begin{bmatrix}
0.5\\
1\\
0
\end{bmatrix} + c\begin{bmatrix}
-0.25\\
0 \\
1
\end{bmatrix} $$
is an eigenvector for this value.

Q1. Yes. The eigenspace is spanned by all the eigenvectors corresponding to the eigenvalue. In this case is just the span of this eigenvector.

Q2 and Q3. Yes this is correct.

Q4. You just need an eigenvector corresponding to $16$ and two orthogonal vectors from the eigenspace for eigenvalue $-5$.


问:
3D line integral question
Let $C$ be the curve of intersection of the cylinder $x^2 + y^2 = 1$ and the surface $z = xy$, oriented counterclockwise around the cylinder. Compute the integral $\int_C y\,dx + z\,dy + x\,dz$.</p>

答:
Parameterize as follows:

$$x=\cos\theta;\space y=\sin\theta;\space z=\cos\theta\sin\theta=\frac{1}{2}\sin2\theta$$

$$\vec{r}(\theta)=\langle{\cos\theta,\sin\theta,\frac{1}{2}\sin2\theta}\rangle,0\le\theta<2\pi$$

$$\oint_Cy\space{dx}+z\space{dy}+x\space{dz}=\int_{0}^{2\pi}(-\sin^2{\theta}+\frac{1}{2}\sin{2\theta}\cos{\theta}+\cos{\theta}\cos{2\theta})\space d\theta$$

From there you can make use of trig identities to evaluate the integral.

Title: 3D line integral question


答:
(a) For the part related to product topology.

Claim 1: $0$ is an accumulation point of $A$: Clearly $\vec{0}\notin A$.
Let $U$ be an open neighborhood of $0$, then there exists open subset
$U_{i}\subseteq\mathbb{R}$ such that $U_{i}=\mathbb{R}$ excepts
for at most finitely many $i$ and $0\in\prod_{i=1}^{\infty}U_{i}\subseteq U.$
Choose $N\in\mathbb{N}$ such that $U_{i}=\mathbb{R}$ whenever $i>N$.
Since $U_{i}$ is an open neighborhood of $0$, we may choose $x_{i}\in U_{i}$
with $x_{i}>0$. Let $\vec{x}=(x_{1},x_{2},\ldots)$, then $\vec{x}\in A\cap\prod_{i=1}^{\infty}U_{i}\subseteq A\cap U$.
This shows that $A\cap U\neq\emptyset$ and hence $\vec{0}$ is an
accumulation of $A$.

Claim 2: There exists a sequence $(\vec{x}_{n})$ in $A$ such that
$\vec{x}_{n}\rightarrow0:$ For each $n$, let $\vec{x}_{n}=(\frac{1}{n},\frac{1}{n},\ldots)\in A$.
We assert that $\vec{x}_{n}\rightarrow\vec{0}$. Let $U$ be an arbitrary
open neighborhood of $0$. Then there exist open sets $U_{i}$ such
that $\vec{0}\in\prod_{i=1}^{\infty}U_{i}\subseteq U$ and $U_{i}=\mathbb{R}$
excepts for at most finitely many $i$. Choose $N$ such that $U_{i}=\mathbb{R}$
whenever $i>N$. For $i=1,\ldots,N$, $U_{i}$ is an open neighborhood
of $0\Rightarrow$ there exists $\delta_{i}>0$ such that $(-\delta_{i},\delta_{i})\subseteq U_{i}$.
Let $\delta=\min(\delta_{1},\delta_{2},\ldots,\delta_{N})>0$. Choose
$n_{0}\in\mathbb{N}$ such that $\frac{1}{n_{0}}<\delta$. Clearly,
for any $n\geq n_{0}$, $\vec{x}_{n}\in\prod_{i=1}^{\infty}U_{i}$.
It follows that $\vec{x}_{n}\rightarrow\vec{0}$.

(b) For the part related to box topology.

Claim 3: $0$ is an accumulation point of $A$ (with respect to the
box topology): Clearly $\vec{0}\notin A$. Let $U$ be an open neighborhood
of $0$, then there exist open sets $U_{i}$ such that $\vec{0}\in\prod_{i=1}^{\infty}U_{i}\subseteq U.$
For each $i$, $U_{i}$ is an open neighborhood of $0$, so we can
choose $x_{i}>0$ such that $x_{i}\in U_{i}$. Define $\vec{x}=(x_{1},x_{2}\ldots)$,
then $\vec{x}\in A\cap\prod_{i=1}^{\infty}U_{i}\subseteq A\cap U$.
This shows that $A\cap U\neq\emptyset$ and hence $\vec{0}$ is an
accumulation point of $A$.

Claim 4: There is no sequence in $A$ that converges to $\vec{0}$
(with respect to the box topology): Prove by contradiction. Suppose the contrary that there exists a sequence $(\vec{x}_{n})$ in $A$ such that $\vec{x}_{n}\rightarrow\vec{0}$
with respect to the box topology. For each $n$, denote $\vec{x}_{n}=(x_{n1},x_{n2},\ldots)$.
For each $i=1,2,\ldots$, define $U_{i}=(-\frac{1}{2}x_{ii},\frac{1}{2}x_{ii})$.
Clearly $U_{i}$ is an open neighborhood of $0$, so $U:=\prod_{i=1}^{n}U_{i}$
is an open neighborhood of $\vec{0}$ (of course with respect to the
box topoogy). For that open neighborhood, there exists $N$ such that
$\vec{x}_{n}\in U$ whenever $n\geq N$. In particular $\vec{x}_{N}\in U$.
Hence, $\vec{x}_{N}(N)\in U_{N}$, i.e., $x_{NN}\in(-\frac{1}{2}x_{NN},\frac{1}{2}x_{NN})$,
which is a contradiction.

Title: Show limit point of product topology and sequence converging to $0$

Body:
I have been really struggling with the product topology. May I have a solution verification and get some pointers on how I should approach these product topology questions?

Let $\mathbb{R}^{\omega}$ be the countable product of copies of $\mathbb{R}$.So every point in $\mathbb{R}^{\omega}$ is a sequence $(x_1,x_2,...)$.Let $A \subset \mathbb{R}^{\omega}$ be the set consisting of all points with positive coordinates. Show that in the product topology $0=(0,0,...)$ is a limit point of the set $A$ and there is a sequence of points in $A$ converging to $0$. Then show that in the box topology, $0=(0,0,...)$ is a limit point of the set $A$, but there is no sequence of points in $A$ converging to $0$.

Attempt:
Let $U$ be a basic open set containing $(0,0,...)$. Then $U=\prod U_i$.If $U_i \neq \mathbb{R}$ since $U_i$ is open in $\mathbb{R}$ for each $U_i \neq \mathbb{R}$ and $\epsilon_i>0$ there is an interval $(-\epsilon_i,\epsilon_i) \subset U_i$. So $U_i \cap R^{+}\neq \varnothing$. Choose $\epsilon=\text{min}\{\epsilon_i\}$. Then $(-\epsilon,\epsilon)\subset U_i$ for each $i$ and so $U \cap A \neq \varnothing$, and $(0,0,...)$ is a limit point of $A$.

Consider the sequence of points in $A$ given by $x_n=(\frac{1}{n},\frac{1}{n},...)$.Let $U=\prod U_i$ be a basic open set containing $(0,0,...)$ if $U_i \neq \mathbb{R}$ there is an $N_i \in \mathbb{N}$ such that $\frac{1}{n} \in U_i$ for each $n \geq N_i$.Choose $N=\text{max}\{N_i\}$ Then $n \geq N \implies x_n \in U$, so $x_n \rightarrow (0,0,...)$.

In the box topology $(0,0,...)$ is a limit point of $A$ since if $U=\prod U_i$ is an open set containing $(0,0,...)$ for all $i$ there is an $\epsilon>0$ such that $(-\epsilon,\epsilon) \subset U$. So $U_i \cap R^{+} \neq \varnothing$ for each $i$ and $(0,0,...)$ is a limit point of $U$.

Comment:I do not know if any of this is correct. I used some help I had learned previously to try and give this solution a shot. However I really think this is incorrect, but would like to know how it is supposed to be done properly, or if I am in the right direction. Also I do not understand why there is no sequence converging to $(0,0,...)$ in the box topology, so I omit it. Does it have anything to do with not being able to take the maximum of the $N_i's$ since there might be an infinite number of $U_i's$ that are not equal to $\mathbb{R}$


问:
Can we say that this matrix is in row reduced echelon form?
Can we say that this matrix is in row reduced echelon form?
\begin{bmatrix}1&0&0&3&1\\0&0&1&0&0\\0&0&0&0&1\\0&0&0&0&0\end{bmatrix}
I know that, it has leading numbers as 1 and other rows are zeros.
Is there a rule to have columns which have all zeros need to occur the beginning?
Your help is highly appreciated. Thank you.

答:
No, columns of 0s don't have to appear at the beginning. There is no such rule as that. You are more interested in the rows. The only two rules are

the zero rows are at the bottom

the leading coefficient of a nonzero row is to the right of the one above it, if they were placed in the same row

Your matrix satisfies these two and thus is in row echelon form. But to get it in reduced row echelon form, it must be that

Any column that contains a pivot has 0 in all other entries.

the pivots are all 1's

So, to put your matrix in rref you subtract row 3 from row 1

$$\begin{bmatrix}1&0&0&3&0\\0&0&1&0&0\\0&0&0&0&1\\0&0&0&0&0\end{bmatrix}$$


问:
2000 people play chess, need to find the maximal number of games during the marathon if…
2000 people play chess, each two play once. For every 2 players (which play in the same game) at least one plays not more than 30 games during the marathon. What is the maximal number of games in that marathon?

I've done this:
Let's divide those 2000 player in two groups.
In the first group are those people who played less than 31 games during the marathon. (Let's call them "F" people)
And second group: Those people who played more than 30 times. (Calling them "M" people)
Let's make a graph with 2000 vertexes and every two are connected if 2 players played chess together.
Let's look at the vertex which has the most number of edges.
If it's F vertex, then all vertexes are "F" types. So in that case the maximal number of games will be $2000×30/2$.
If it's M vertex, well, then all people who played with him will be "F"s. let'ss suppose our first vertex  has the maximal number of edges that are possible, in that case, the maximal number of games will be $1999+(1998×29+28)/2=30984>30000$(case of absence of "M"s)
If we had two M vertexes (which would be connected to the others as well) the number of games would be $30969$ and it's less than 30984. Similarly if we add M vertexes, the sum decreases every time. So 30984 should be the maximum.

But they're many shortcomings as we haven't talked about such cases when "M"s aren't connected to  all vertexes. Also, is it possible to say that if the pattern (of decrease) is visible, this means that we can judge the value of number?
So I think there's more effective and simple-idea solution for this problem.

答: You can't have any games between $M$ people.  You should be able to convince yourself that you want few games between the $F$ people.  This suggests a bipartite graph.  Let there be $m\ M$ people and $f \ F$ people.  We have $m+f=2000$.  From the $F$ side there are $30f$ games.  Each $M$ person can play all the $F$ people, so there are $mf$ games from that side.  This gives $m=30, f=1970$ and there are $30 \cdot 1970=5910$ games.


问:
Divergence theorem: compute triple integral over a paraboloid between two planes

Let $S$ be the surface of equation $z=x^2 + y^2$, which is between the planes $z=1$ and $z=x+y+3$. Let $F = (x^2,y,z)$. Compute the flux of $F$ across $S$

I think I should use divergence theorem, hence I only need to compute $$\int \int \int_V \operatorname{div}(F)dxdydz$$ but I don't know how to describe the volume enclosed by $S$.

I'd like to integrate this way:

$$\int \int_{D} \Big[\int_{1}^{x+y+3}2x + 2 dz\Big] dxdy$$

But I don't know how to define the domain $D$ where the coordinates $x$ and $y$ are. I think I should the two conditions on the planes, but I can't. Any help is highly appreciated.

答:
If you are applying divergence theorem, you will have to close the  surface with disks in planes $z=1$ and $z = x+y+3$. You will need to separately calculate flux through the disks and subtract from the flux through the closed surface.

Now good thing is that within the paraboloid domain, plane $z = x + y + 3$ is throughout above plane $z=1$. Also note that the intersection of the given circular paraboloid and the given plane can be seen in $XY$ plane as -

$x^2+y^2 = x + y + 3 \implies (x-0.5)^2 + (y-0.5)^2 = 3.5$

So at the minimum, it makes sense to use cylindrical coordinates with

$x = 0.5 + r \cos \theta, y = 0.5 + r \sin\theta, 0 \leq r \leq \sqrt{3.5}$

$1 \leq z \leq 4 + 0.5 \cos\theta + 0.5 \sin\theta, \ 0 \leq \theta \leq 2\pi$

So the integral to find flux becomes

$\displaystyle \int_0^{2\pi} \int_0^{\sqrt {3.5}} \int_1^{4+r\cos\theta+r\sin\theta} r(3+ 2r\cos\theta) \ dz \ dr \ d\theta$

Now please find flux through disks at $z=1$ and $z=x+y+3$ and subtract from the above. That should give you the flux through paraboloid surface.

Just as a side note: if you want to simplify the limits of integration further, you can in fact rotate the paraboloid by $\frac{\pi}{4}$. There is no change to Jacobian just like in the first case. It will simplify the equation of the plane $z = x + y + 3$.

$x = 0.5 + r \cos (\frac{\pi}{4} + \theta), y = 0.5 + r \sin(\frac{\pi}{4} + \theta), 0 \leq r \leq \sqrt{3.5}$

Equation of plane becomes $z = x+y+3 = 4 + r \cos(\frac{\pi}{4} + \theta) + r \sin(\frac{\pi}{4} + \theta) = 4 + r \sqrt2 \cos\theta$.


问:
Given a set of m numbers {1,2,...m} we can always select a subset having sum of [1,m(m+1)/2]
Given a set s of m numbers {1,2,3,4,....m} i need to prove that given any number x in the interval [1,m(m+1)/2] i can select a subset s' from the set s such that sum of elements in s' is equal to x.

My approach was to consider two cases:
1) 1<=x<=m
In this case s' is just a singleton set containing x

2)m<x<= (m*(m+1))/2

In this case i first find the greatest integer 'i' such that x - [m+(m-1)+(m-2)+(m-3).....(m-i)] is positive. Let this quantity be c = x - [m+(m-1)+(m-2)+(m-3).....(m-i)] . Now because of the way we have defined 'i' , if we subtract m-i-1 from c we will get a non positive value.
 This means that c-(m-i-1)<=0.
This means that  c<=(m-i-1).
Also 0<=c.

if c=0 we have,

c = x - [m+(m-1)+(m-2)+(m-3).....(m-i)] = 0.
This means x = m+(m-1)+(m-2)+(m-3).....(m-i)
And we can select s' containing elements m,(m-1),(m-2),(m-3).....(m-i).

but if c is not equal to 0,

Then we know that 1<=c<=(m-i-1), We can select this value of c.
This means we can select s' as m,(m-1),(m-2),(m-3).....(m-i), c

Is this proof correct??

How can this be proved using induction??

答:
An induction proof.

For $m=1$ the result is trivial.

Suppose the result is true for $S=\{1,2, ... , k\}$ and now consider the numbers up to $k+1$. If a sum up to $\frac{k(k+1)}{2}$ is required then you can just use a subset of $S$. So suppose you require a total sum, $T$, greater than this.  Then $$\frac{k(k+1)}{2}<T\le \frac{(k+1)(k+2)}{2}.$$
Then $$T-(k+1)\le \frac{(k+1)(k+2)}{2}-(k+1)=\frac{k(k+1)}{2}$$ and so a subset of $S$ has total sum $T-(k+1)$ and then adding $k+1$ to this subset gives a set with total $T$.


问:
Find the Laplace $\mathcal{L}(y(t))$ if $y'(t)-y(t)+\int_0^t y(u)e^{\alpha(t-u)}du=\alpha e^{\alpha t}$
How to find in terms of $y(0)$ and $\alpha$, the  Laplace transform $\mathcal{L}\{y(t)\}(s)$ if $$y'(t)-y(t)+\int_0^t y(u)e^{\alpha(t-u)}du=\alpha e^{\alpha t}$$

Comment: By the properties of Laplace transform, applying directly with $Y(s)=\mathcal{L}\{(y(t))\}(s) $

$$sY(s)-y(0)-Y(s)+Y(s)\mathcal{L}(e^{\alpha t})=\alpha\mathcal{L}(e^{\alpha t}) $$ here I'm using the convolution property, is it well applied? I mean is $$ \mathcal{L}\left(\int_0^t y(u)e^{\alpha(t-u)}du \right)=Y(s)\mathcal{L}(e^{\alpha t}) ?$$

答:
Note that if $f$ and $g$ be continous function on $\mathbb{R}^{+}$, so we can define the convolution between the functions $f$ and $g$ as $$f(t)\ast g(t)=\int_{0}^{t}f(u)g(t-u)\operatorname{du}.$$
Let's to make $f(u):=y(u)$ and let $g(t):=e^{at}$, so we have that $g(t-u)=e^{a(t-u)}$. Therefore, $$\color{blue}{y(u)\ast e^{at}}=\int_{0}^{t}y(u)e^{a(t-u)}\operatorname{du}.$$

On the other hand, by convolution's theorem we know that: if $f$ and $g$ be piecewise continous function on $\mathbb{R}^{+}$ and $f$ and $g$  be exponential order, so $$\mathcal{L}\{f\ast g\}=\mathcal{L}\{f(t)\}\cdot \mathcal{L}\{g(t)\}=F(s)G(s)$$
Then you can see that $$\color{red}{\mathcal{L}\left\{ \int_{0}^{t}y(u)e^{a(t-u)}\operatorname{du} \right\}}=\mathcal{L}\{y(u)\ast e^{at}\}\color{red}{=Y(s)\mathcal{L}\{e^{at}\}}.$$

Reviewing your comment, you may need to know what $$\color{blue}{e^{at}=\mathcal{L}^{-1}\left\{ \frac{1}{s-a}\right\},  \quad s\not=a}.$$
Finally, solve algebraically $Y(s)$ from the algebraic equation and apply $$\color{blue}{f(t)=\mathcal{L}^{-1}\{F(s)\}}.$$


问:
Geometric Multiplicity, Eigenspace of Matrix
If $A$ is an $r$ x $s$ matrix and B is an $s$ x $r$ matrix.

$E_\mu(C)$ is the eigenspace of square matrix C with eigenvalue $\mu ≠ 0$.

Proof:

$dimE_\mu(AB) = dimE_\mu(BA)$

I imagine one would need to find the basis of eigenvectors for both spaces, but I'm not sure how...

答: Hint: If $v$ is an eigenvector for $AB$, then $Bv$ is an  ...</p>


问:
Percentage increase when adding a specific amount
Is it possible to calculate $X$ if $100$ has been added to $X$ as a result of a $234$% increase?

So, I want to calculate what I started with from the statement "up \$$100$ ($+234$%) in the past $6$ months."

Is there a general formula for this, perhaps?

答: If $X$ gets increased by $234\%$, then it becomes $X+2.34X=3.34X$. And if you add $100$ to $X$, you get $X+100$. So, $X+100=3.34X$. Can you take it form here?


答:
HINT

Increasing $X$ by $234$ per cent is the same as multiplying it by 3.34.

So you have an equation $X+100=3.34X$.

Title: Percentage increase when adding a specific amount

Body:
Is it possible to calculate $X$ if $100$ has been added to $X$ as a result of a $234$% increase?

So, I want to calculate what I started with from the statement "up \$$100$ ($+234$%) in the past $6$ months."

Is there a general formula for this, perhaps?


问:
CDFs of indicator random variables

Consider two events $A$ and $B$ in the probability space and let $X = 1_A$ and $Y = 1_B$ be
their indicator random variables. Find their joint CDF.

So here's what I know and have tried:

A random variable is a function from the sample space $\Omega$ to $\Bbb R$, so $X(\omega) = 1$ if $\omega\in A$ and $0$ otherwise (it's the indicator). Similarly for $Y$. The joint CDF is supposed to look like this:
$$P(X\le x, Y\le y) = P((X\le x)\cap(Y\le y))$$
Now what's $P(X\le x)$?

$X$ can take only two values, i.e. $0,1$. I would expect $P(X\le x) = 0$ if $x < 0$. Also $P(X\le x) = 1$ if $x\ge 1$. What happens in between, it's quite confusing!

One thing that comes to mind is that the random variable $Z = 1_{A\cap B}$ would indicate when $A$ and $B$ occur together. Does that help?

Thank you!

答:
$X$ is a random variable taking values $0$ and $1$ with probabilities $P(A^{c})$ and $P(A)$ respectively. The CDF $F_X$ of $X$ is given by $F_X(x)=P(X \leq x)=0$ if $x <0$, $F_X(x)=P(A^{c})$ if $F_X(x)=0\leq x <1$ and $1$ if $x \geq 1$. Similarly for $F_Y$.

$F_{X,Y} (x,y)=P(X \leq x, Y\leq y)$ and this is

$0$ of $x <0$ or $y <0$,

$1$ if $x \geq 1$ and $y\geq 1$,

$P(A^{c}\cap B^{c})$ if $0 \leq x <1$ and $0\leq y <1$,

$P(A^{c})$ if $0 \leq x <1$ and $y \geq 1$ and

$P(B^{c})$ if $0 \leq y <1$ and $x \geq 1$.

Title: CDFs of indicator random variables


问:
open balls cover function space
Part A (which I understand)

If for every $\epsilon>0$, a metric space $X$ can be covered by a finite number of $\epsilon$-balls,

Then every point $x\in X$ is inside at least one $\epsilon$-ball.

Part B (which I am confused about)

If for every $\epsilon>0$, a continuous function space $C(X,Y)$ can be covered by a finite number of $\epsilon$-balls: $B(f_1,\epsilon),B(f_2,\epsilon),\dots$

Then every function $f\in C(X,Y)$ is inside at least one ball $B(f_i,\epsilon)$. I doubt this conclusion because I can imagine a counterexample: $f$ is partly covered by $B(f_i,\epsilon)$ and partly by $B(f_j,\epsilon)$, while neither of them cover $f$ entirely.

答:
The "counterexample" is nonsense: $B(f_i, \varepsilon)$ is a set of functions and a function is in it or not. You cannot be "partially" in it.

On an abstract level B is the same as A. It's just about points and sets of points that form a cover of the space they are points from. The fact that the points in B are functions internally does not change that at all.

Title: open balls cover function space


问:
Linear operator that commutes with projection
Let $T, \pi$ be linear operators on a vector space $V$. If $\pi$ is a projection and $T\pi = \pi T$, does it follow that $\pi \in F[T]$, i. e. $\pi$ can be written as a polynomial in $T$?

I know that the kernel and range of $\pi$ would be $T$ invariant, but can't figure out where to go from here.

答: No. Let $I:V\rightarrow V$ be the identity map. Let $\pi$ be a non-trivial projection (we assume that $\mbox{dim}(V) \geq 2$). Then $\pi$ commutes with $I$. However, $\pi$ cannot be written as a polynomial of $I$.


答:
If $X$ increases $100$%, it becomes $2X$. So, by similar logic, if $X$ is increased by $234$%, then it becomes $3.34X$. However, we also notice that $100$ is added to $X$ due to this increase, thus $3.34X=X+100$.

More generally, if $X$ is increased by $m$%, and this increases $X$ by $n$ amount, then this relationship can be describe by this formula $X(1+\frac{m}{100}) = X +n$.


问:
Does inverse property of a mathematical group require different elements X and Y?
I'm reading a textbook that talks about mathematical groups, and I'm currently solving some problems. The textbook asks if these infinite sets are groups. For this set below -- rational numbers under multiplication -- I was confused by the explanation.

(j) rational numbers, multiplication

i. Does it form a group?

The rational numbers under multiplication do not form a group because if the identity element is $1$, then  $1\cdot0=0\ne1$, so it cannot satisfy invertibility.

It says that the group doesn't satisfy invertibility, which is where X operation Y = I. From my understanding, 1 * 1 = 1. Doesn't that mean it satisfies invertibility? Or do they have to be different elements? Is the solution wrong?

I'm really confused because the dihedral group 3 (otherwise known as reflection/rotation group of an equilateral triangle) has I as the identity element, and I's inverse is also I. So the definition of invertibility must not require different elements, right?

答:
It says that the group doesn't satisfy invertibility, which is where X operation Y = I. From my understanding, 1 * 1 = 1. Doesn't that mean it satisfies invertibility?

Yes, $1$ is an invertible element.

Or do they have to be different elements? Is the solution wrong?

The final claim is not wrong. The explanation is somewhat misleading though (probably a typo). They don't claim that $1$ is not invertible. They claim that $0$ is not invertible. That's because $x\cdot 0=0$ regardless of $x$ and so $0\cdot x=1$ has no solution. There seems to be a mistake (typo?) when they write "$1\cdot 0=0\neq 1$". It should be "$x\cdot 0=0\neq 1$".


问:
Distributing 4 distinct balls into 3 distinct bins($bin_1,bin_2,bin_3$) , such that $bin_2$ always has a ball

Distributing 4 distinct balls into 3 distinct
bins($bin_1,bin_2,bin_3$) , such that $bin_2$ always has a ball

My approach to this problem is that without any conditions the total way of distributing the balls is $3^4 = 81$.

Now we deduct all the ways in which $bin_2$ doesn't have a ball i.e all the balls are distributed into bins : $bin_1$ and $bin_2$. This gives two options for each ball therefore $2^4 = 16$ ways.

Therefore total number of ways of distributing 4 distinct balls into 3 distinct bins such that $bin_2$ always gets a ball is $3^4-2^4 = 65$.

Is this approach and this answer correct ??

答:
Yes, I think your method is spot on. Instead of bins I like to call them colours, so that we colour 4 balls, numbered 1 to 4 using three colours $c_1,c_2,c_3$. That is the same thing, really. The total number of colourings is $3^4$ as we are making $4$ independent choices for colours of the balls, with three options per choice, so $3 ^4 = 81$ colourings.

The ones where $c_2$ is not used is $2^4$ for the same reasons (just two options, $4$ times), so $16$ "wrong" colourings, leaving $81 - 16 = 65$ correct colourings.


答:
An Increase of $234 \%$ is equal to $\$100$.

Meaning the original base quantity $X$ (which can also be viewed as $100 \% X$ is simply $\$ \frac {100}{234} (100) = \$ \frac{100}{2.34} \approx \$ 42.74$ to the nearest cent.

The simple rule of thumb is divide the percentage increase by $100$, get that value. Then take the dollar amount increase and divide by the value. That will give you the original amount.


问:
Reduction from 3SAT to NAE3SAT
I have seen two reductions -

 1. $(x, y, z) => (x, y, a) (a’, z, 0)$

 2. 3SAT to NAE4SAT to NAE3SAT

For (2) the initial transform is $(x, y, z) => (x, y, z, a) (x’, y’, z’, a’)$.

This works for $x = y = z = 1$ but not for $x = y = z = 0$. With the latter if $a = 1$, then the NAE4SAT expression is satisfied but the initial 3SAT expression is not.

What is the correct transform for the reduction via NAE4SAT?

答: the transform for (2) should be $(x,y,z)=>(x,y,z,a)(x′,y′,z′,a)$. This way even the NAE4SAT is not satisfiable if the initial 3SAT is not satisfiable.

Title: Reduction from 3SAT to NAE3SAT


问:
Find the range of $f:(-4,6) \rightarrow R$ , $f(x):x^2 -4x +1$
When i work over finding the range of a function , my textbook gives me a solution way such that if $f:(-3,2] \rightarrow R$ , $f(x):x^2 -2$ then its range can be found such that $-3<x \leq 2 \rightarrow 0 \leq x^2 < 9 \rightarrow -2 \leq x^2 -2 <  7$. It is fine , but when i want to apply it another question , it gave me wrong answer.

Question: Find the range of $f:(-4,6) \rightarrow R$ , $f(x):x^2 -4x +1$

My work : $-4<x<6 \rightarrow 0 \leq x^2 < 36$  , $-4<x<6 \rightarrow-24 < -4x < 16 $, and $-23 < x^2 -4x +1 < 53$

However , the answer is $[-3,33)$ .Why didnt the previous solution way work ? Why isnt the answer $(-23,53)$ ?

答: Note that$$x^2-4x+1=(x-2)^2-3.$$So,\begin{align}-4<x<6&\iff-6<x-2<4\\&\iff0\leqslant(x-2)^2<36\\&\iff-3\leqslant(x-2)^2-3<33\\&\iff-3\leqslant x^2-4x+1<33.\end{align}Also, note that here I have only used equivalences. Therefore, I get the actual range and not a set that could be larger than the range.


答:
Understand the quadratic $f(x)=x^2- 4x + 1$ on $\Bbb R$ first: it's a parabola with a minimum, and this minimum lies at $\frac{-b}{2a} = \frac{4}{2} = 2$ and has value $f(2)= 4 - 8 + 1 = -3$. Then on both sides of $x=2$ the function will be increasing to infinity (symmetrically around $x=2$). This should be included in your knowledge of parabolas.

Now the actual domain is $(-4,6)$, so we truncate the values on both sides (with domain $\Bbb R$ the range would have been $[-3, \rightarrow)$, because of the minimum and the shape of the graph). And $f(-4) = 16 + 16 + 1 = 33$ and $f(6) = 36 - 24 + 1 = 13$, so the graph reaches its highest point (almost) at $(-4,33)$ which is just on the boundary of the domain (which is open ended).

All values inbetween $-3$ inclusive and $33$ are assumed on $(-4,2]$ already, then some values are duplicated again on $[2,6)$ as well...

So range of $f$ is $[-3,33)$

Body:
When i work over finding the range of a function , my textbook gives me a solution way such that if $f:(-3,2] \rightarrow R$ , $f(x):x^2 -2$ then its range can be found such that $-3<x \leq 2 \rightarrow 0 \leq x^2 < 9 \rightarrow -2 \leq x^2 -2 <  7$. It is fine , but when i want to apply it another question , it gave me wrong answer.

Question: Find the range of $f:(-4,6) \rightarrow R$ , $f(x):x^2 -4x +1$

My work : $-4<x<6 \rightarrow 0 \leq x^2 < 36$  , $-4<x<6 \rightarrow-24 < -4x < 16 $, and $-23 < x^2 -4x +1 < 53$

However , the answer is $[-3,33)$ .Why didnt the previous solution way work ? Why isnt the answer $(-23,53)$ ?


问:
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

答:
First observe that since $5 \le n$, we have:
$$
4(n+1) =4n+4 < 4n+5 \le 4n+n = 5n \le (n)n = n^2
$$
Hence, since $4(n+1)<n^2 \Rightarrow \boxed{n+1<\dfrac{1}{4}n^2}$ (and $f(x)=\sqrt{x}$ is monotonically increasing and $n\ge0$), we have:

$$
\sqrt{n+1}+\sqrt{n} \le \sqrt{n+1}+\sqrt{n+1} = 2\sqrt{n+1}<2\sqrt{\dfrac{1}{4}n^2}=2\left(\dfrac{1}{2}n\right) = n
$$

as desired.

Title: Show that $n \ge \sqrt{n+1}+\sqrt{n}$


问:
Rank-Nullity Possible Pairs
Let $T : \mathbb{R}^3 → \mathbb{R}^3$ be a linear map of vector spaces over the reals. Assume that $T(1,2,0) = (2,3,0)$ and $T(−1,1,0) =(1, 1,0)$. What are the possible pairs of Rank(T) and Nullity(T)?

I know that Rank(T)+Nullity(T)=3. How do I find specific values of Rank(T) and Nullity(T)?
Any guidance is much appreciated

答: Since $T(\Bbb R^3)$ contains $(2,3,0)$ and $(1,1,0)$, which are linearly independent, $\operatorname{rank}(T)$ is at least $2$. If $T(0,0,1)=(0,0,0)$, then $T(\Bbb R^3)=\operatorname{span}\bigl(\{(2,3,0),(1,1,0)\}\bigr)$ and therefore $\operatorname{rank}(T)=2$ and $\operatorname{null}(T)=1$. And if $T(0,0,1)=(0,0,1)$, then $T(\Bbb R^3)=\Bbb R^3$. So, $\operatorname{rank}(T)=3$ and $\operatorname{null}(T)=0$.

Title: Rank-Nullity Possible Pairs


问:
Finding the eigenvalue of F(V)=(V⋅W)W+V where W=(1 , 1 , 2) and f:R^3 → R^3
I have a linear algebra test in 3 days and there was this problem that i was having difficulties with! Found this in some of the practice quizzes in my university website! So the Problem goes like this :

Let W=(1 , 1 , 2).Let R^3 be the endomorphism defined by: V ∈ R^3 such that F(V)=(V⋅W)W+V
The answer to this problem is : ""W is an eigenvector of eigenvalue 7""

Here is a picture of the problem as well:

答:
Take $\underline w=(1,1,2)$ and given your map $F:\mathbb R^3\to\mathbb R^3$, you simply have to calculate the image of the vector.

$$F(\underline w)=(\underline w\cdot\underline w)\underline w+\underline w=\lVert \underline w\rVert^2\underline w+\underline w=(1^2+1^2+2^2)\underline w+\underline w=6\underline w+\underline w=7\underline w$$.
For definition, the eigenvalue $\lambda$ associated to the vector $\underline w$ is equal to $7$, since $\underline w\overset{F}{\mapsto}7\underline w$.


答:
$$x^2-4x+1 = (x-2)^2-3$$
That is a  parabola open upward with a vertex at the point $(2, -3)$

Thus the minimum value of your function on $(-4, 6)$ is attained at $x=2$ and the supremum is $f(6)=33$

Thus $$ -3\le f(x) < 33$$

Title: Find the range of $f:(-4,6) \rightarrow R$ , $f(x):x^2 -4x +1$


问:
Does $(AB)^m = I_n$ imply $(BA)^m = I_n$?
The problem is to prove or disprove that $(AB)^m = I_n$ imply $(BA)^m = I_n$ where $A, B \in M_n(\mathbb{R})$ and $m \in \mathbb{Z}$.

I know that it's true for $m = 1$, and I tried to use the induction on $m$, but could not prove the statement.

答:
For negative $m$, note that $(AB)^m=I$ implies $[(AB)^m]^{-1}=(AB)^{-m}=I$.

Since $A,B$ are invertible, we get $B\underbrace{(AB)(AB)...(AB)}_{|m|\text{ times}}B^{-1}=BIB^{-1}=I$ and using the associativity of matrix multiplication we get $(BA)^{|m|}=I$.

For negative $m$, this implies $(BA)^{-|m|}=(BA)^m=I$.


问:
Kernel of a morphism between coherent sheaves.
Throughout this book, http://books.google.co.kr/books/about/Algebraic_Geometry_and_Arithmetic_Curves.html?id=uaLKdA0PxS4C&redir_esc=y, kernel of a morphism between coherent sheaves on a locally Noetherian scheme is also coherent.

More precisely, let $X$ be a locally Noetherian scheme and  $u:\mathcal{F}\to\mathcal{G}$ be a morphism of coherent sheaves on $X$. Then $\textrm{Ker }u$ is also coherent.

The author says, this is trivially true. However, I cannot prove it. Moreover, I do not know whether it is really true or not.

Even though I do not have any explicit example, I think this would be false in following sense; when $X$ is locally Noetherian scheme and $\mathcal{F}$ is a coherent sheaf on $X$, it is known that $\mathcal{F}(U)$ is finitely generated $\mathcal{O}_X(U)$-module for any affine open subset $U$ of $X$. If $\textrm{Ker }u$ is coherent, then the sequence $\textrm{Ker }u(U)\to\mathcal{F}(U)\to\mathcal{G}(U)$ of $\mathcal{O}_X(U)$-modules is an exact for any affine open $U$. However, in general, a submodule of a finitely generated module is not finitely generated. So, maybe there would be an example that arises form the sequence $\textrm{ker} f \to L\xrightarrow{f}M$ where $L,M$ are finitely generated modules and $\textrm{ker }f$ cannot be generated by any finite set. Is this reasonable?

Actually, we can find the same statement for $X$ is Noetherian case in Hartshorne's book. If the above statement is true, how can we expand the case?

答: Hint: A ring $R$ is noetherian iff submodules of f.g. $R$-modules are again f.g.</p>


答:
It's correct that by the dimension theorem we have that $$\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim(\mathbb{R}^{3})=3.$$
Now, you need to use the first hypothesis of the problem: we need that $T$ satisfy:

$T(1,2,0)=(2,3,0)$.

$T(-1,1,0)=(1,1,0)$.

Now, note that $$B=\left\{\color{red}{\begin{pmatrix} 1\\ 2 \\ 0 \end{pmatrix}}, \color{red}{\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix}}, \begin{pmatrix} 0\\ 0 \\ 1 \end{pmatrix} \right\}\subset \mathbb{R}^{3}$$
is a linearly independent set and generates $\mathbb{R}^{3}$; that is, $B$ is a basis for $\mathbb{R}^{3}$ (the starting vector space). We know the transformation of $(1,2,0)$ and $(-1,1,0)$, but we have added a vector but do we know the transformation of that vector? Sure not, but we know we only have two options:

$T(0,0,1)=(0,0,0) \implies (0,0,1)\in \operatorname{ker}(T)\implies \operatorname{nullity}(T)=1.$

$T(0,0,1)\not=(0,0,0) \implies (0,0,1) \in \operatorname{im}(T) \implies \operatorname{rank}(T)=3. $

Therefore, the answer is:

$(\operatorname{rank}, \operatorname{nullity}(T))=(2,1).$ (First choice).

$(\operatorname{rank}, \operatorname{nullity}(T))=(3,0).$ (Second choice).

Also, note that always we have that $$2+1=3, \quad \text{and} \quad 3+0=3.$$
no matter what choice we make.

Title: Rank-Nullity Possible Pairs

Body:
Let $T : \mathbb{R}^3 → \mathbb{R}^3$ be a linear map of vector spaces over the reals. Assume that $T(1,2,0) = (2,3,0)$ and $T(−1,1,0) =(1, 1,0)$. What are the possible pairs of Rank(T) and Nullity(T)?

I know that Rank(T)+Nullity(T)=3. How do I find specific values of Rank(T) and Nullity(T)?
Any guidance is much appreciated


答:
$$Rank(T)+Nullity(T)=3$$

The span of the transformation already contains $\left\{\begin{pmatrix}2\\3\\0\end{pmatrix},\begin{pmatrix}1\\1\\0\end{pmatrix}\right\}$

Thus if there exists any vector $v=\begin{pmatrix}a\\b\\c\end{pmatrix}$ such that $T(v)=\begin{pmatrix}a'\\b'\\c'\end{pmatrix},c'\ne0$ then Rank(T)=3 and Nullity(T)=0. Otherwise, if $T(v)=\begin{pmatrix}a'\\b'\\0\end{pmatrix}$ for all v where a' and b' change, then Rank(T)=2 and Nullity(T)=1.

Title: Rank-Nullity Possible Pairs


问:
Need help solving this log equation $\log_{5}(x-4) = \log_{10}(x+5)$
I need help solving for $x$:

$$\log_{5}(x-4) = \log_{10}(x+5)$$

答:
Using natural logarithms and simplifying, you are looking for the zero's of function
$$f(x)=\log (10) \log (x-4)-\log (5) \log (x+5)$$ Its first derivative
$$f'(x)=\frac{\log (10)}{x-4}-\frac{\log (5)}{x+5}$$ cancels at
$$x_*=\frac{4 \log (5)+5 \log (10)}{\log (5)-\log (10)}\sim -25.8974$$ which is lower than $4$. So, the first derivative is always positive and we have an increasing function.

Now, if you plot or use inspection, you will notice that the single root is close to $x=11$. Use it as the $x_0$ for Newton method which will generate the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 11.00000000 \\
 1 & 10.91979756 \\
 2 & 10.92036742 \\
 3 & 10.92036745
\end{array}
\right)$$


问:
Given a set of m numbers {1,2,...m} we can always select a subset having sum of [1,m(m+1)/2]
Given a set s of m numbers {1,2,3,4,....m} i need to prove that given any number x in the interval [1,m(m+1)/2] i can select a subset s' from the set s such that sum of elements in s' is equal to x.

My approach was to consider two cases:
1) 1<=x<=m
In this case s' is just a singleton set containing x

2)m<x<= (m*(m+1))/2

In this case i first find the greatest integer 'i' such that x - [m+(m-1)+(m-2)+(m-3).....(m-i)] is positive. Let this quantity be c = x - [m+(m-1)+(m-2)+(m-3).....(m-i)] . Now because of the way we have defined 'i' , if we subtract m-i-1 from c we will get a non positive value.
 This means that c-(m-i-1)<=0.
This means that  c<=(m-i-1).
Also 0<=c.

if c=0 we have,

c = x - [m+(m-1)+(m-2)+(m-3).....(m-i)] = 0.
This means x = m+(m-1)+(m-2)+(m-3).....(m-i)
And we can select s' containing elements m,(m-1),(m-2),(m-3).....(m-i).

but if c is not equal to 0,

Then we know that 1<=c<=(m-i-1), We can select this value of c.
This means we can select s' as m,(m-1),(m-2),(m-3).....(m-i), c

Is this proof correct??

How can this be proved using induction??

答:
Notation: $[m]=\{1,2,\dots,m\}$; for $S\subseteq[m]$ we write $\sum S=\sum_{n\in S}n$; in particular $\sum[m]=1+2+3+\cdots+m=m(m+1)/2$.

Let $m$ be a positive integer. Let $P(x)$ denote the statement: there is a set $S\subseteq[m]$ with $\sum S=x$. I will show by induction that $P(x)$ holds for every integer $x$ such that $1\le x\le\sum[m]$.

$P(1)$ is clear; take $S=\{1\}$.

Suppose $P(x)$ holds for some fixed $x\lt\sum[m]$; I have to show that $P(x+1)$ holds. Choose a set $S\subseteq[m]$ such that $\sum S=x$. Let $n$ be the least positive integer that does not belong to $S$. Since $\sum S\lt\sum[m]$, we must have $n\le m$. If $n=1$ let $S'=S\cup\{1\}$; if $n\gt1$ let $S'=(S\setminus\{n-1\})\cup\{n\}$. In either case, $S'\subseteq[m]$ and $\sum S'=\sum S+1=x+1$, so $P(x+1)$ holds.


Title: Does $(AB)^m = I_n$ imply $(BA)^m = I_n$?

Body:
The problem is to prove or disprove that $(AB)^m = I_n$ imply $(BA)^m = I_n$ where $A, B \in M_n(\mathbb{R})$ and $m \in \mathbb{Z}$.

I know that it's true for $m = 1$, and I tried to use the induction on $m$, but could not prove the statement.

Body:
From $(AB)^m=I_n$, one obtain $A^{-1}=B(AB)^{m-1}$.

Hence $(BA)^m=B(AB)^{m-1}A=I_n$.


问:
Is there a way to prove that $A \setminus C = (A \setminus (B \cup C)) \cup ((A \cap B) \setminus C)$?
Drawing out the Venn Diagrams I know this is true.

$(A \setminus (B \cup C))$ is just all the values that are solely in A.

$((A \cap B) \setminus C)$ is just all the values in both A and B but not in C.

And taking the union of these two leads us to the values that are either only in A or in only A and B (but not C).

I am unsure how to rigorously prove this, however. Would starting off by letting $x \in (A \setminus (B \cup C))$ be a good start?

Venn Diagram

答: To prove sets $S,\,T$ are equal, i.e. $x\in S\iff x\in T$, you should try to prove both directions of the $\iff$ simultaneously by only using $\iff$ steps. To wit:$$\begin{align}x\in A\setminus C&\iff x\in A\land x\notin C\\&\iff(x\in A\land x\notin B\land x\notin C)\lor(x\in A\land x\in B\land x\notin C)\\&\iff(x\in A\land x\notin B\cup C)\lor(x\in A\cap B\land x\notin C)\\&\iff(x\in A\setminus(B\cup C))\lor(x\in(A\cap B)\setminus C)\\&\iff x\in(A\setminus(B\cup C))\cup((A\cap B)\setminus C).\end{align}$$


问:
Solving $y'' = -y \cdot (y')^2$ analytical
I was trying to solve the following differential equation:

$$y'' = -y \cdot (y')^2$$

But I always have some trouble solving nonlinear ODE. I used wolfram alpha to check the solution but it used complex number and the $\text{erf}$ function and I'm not very used to work with it. Is that the only way of solving this analytical? If so, how Can I solve this equation?

答:
$$y'' + y \, (y')^2=0$$

Switch variables to make
$$-\frac{x''}{[x']^3}+\frac y{[x']^2}=0\implies x''=y \,x'$$ Reduction of order $p=x'$ gives
$$p=c_1 e^{\frac{y^2}{2}}$$ Now, another integration
$$x+c_2=\sqrt{\frac{\pi }{2}} c_1 \text{erfi}\left(\frac{y}{\sqrt{2}}\right)$$

Do not try to get y(x).