问:
$GL(n, \mathbb{C})$ is algebraically closed?
Let $GL(n,\mathbb{C})$ the group of non-singular matrices. Is it algebraically closed? For $GL(1,\mathbb{C})$ is it true; but if I take linear combinations of elements in $GL(n,\mathbb{C})$ with coefficients matrices in $GL(n,\mathbb{C})$ could I find solutions in $GL(n, \mathbb{C})$? We know that diagonalizable applications are dense: maybe it is useful. If $n=2$ can we use the isomorphis of $GL(2, \mathbb{C})$ with quatarnions?</p>

答:
One way (and the only one I know about) to define polynomials over a non-commutative ring $A$ is to consider them as elements of $\bigoplus_{\Bbb N} A$, i.e. as sequences $(a_0,a_1,\dotsc,a_n,\dotsc)$ with only a finite number of $a_i\in A$ not null. Beware, though, that polynomial functions over a non-commutative ring will probably behave in unexpected ways. For example, polynomial functions over $\Bbb H$ can have an infinite number of roots.

Please note that a polynomial is just a formal sequence $f\in A[X]$, while a polynomial function is a map $f(X)\colon A\to A$, defined by substituting an element of $A$ in place of the "variable" $X$. Moreover, in general we can define what a root is only for the latter.

There is a problem, though, in trying to define $GL_n(\Bbb C)[X]$, as $GL_n(\Bbb C)$ is not a ring. In particular, it doesn't have a $0$ element, so we cannot construct sequences as above.

But suppose, for the sake of argument, that we could somehow define polynomial functions with coefficients in $GL_n(\Bbb C)$. In particular, for any $i>0$ we could consider the monomials $X^i=I_nX^i$, where $I_n$ is the identity of $GL_n(\Bbb C)$. Let $A\in M_n(\Bbb C)$ be a matrix which is a root of $X^i=0$ for some $i>0$. Then it would satisfy $A^i=0$, i.e. it would be nilpotent (possibly $0$). Thus $A$ isn't invertible, so $A\notin GL_n(\Bbb C)$.


问:
Closed-form for $ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $ where $p\in\mathbb{N}$
Is there a closed form to the sum $$ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $$
where $p$ is a positive integer? I know that for $p = 1$ the series sums to $xe^x$, but not sure if there's a simple expression for $p > 1$. If there isn't a closed form, is this series related to any special functions? Thanks!

答:
Hint: here are the first few values for $p=0,\ldots,5$:
$$
e^x\cdot\left\{1,x,x^2+x,x^3+3 x^2+x,x^4+6 x^3+7 x^2+x,x^5+10 x^4+25 x^3+15 x^2+x\right\}
$$You may recognize the coefficients of these polynomials; they are the Stirling numbers of the second kind. The recurrence relation can be proved by induction using the power rule. So in summary,
$$
\sum_{n=0}^{\infty}\frac{x^n}{n!}n^p = e^x \cdot \sum_{k=1}^p \left\{
\begin{array}{cc}
 p \\ k
\end{array}
\right\} x^k
$$

ContentLicense: CC BY-SA 4.0


问:
Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$
Hi this is the question:

Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$

What I've tried:

I know that the range of $\csc(x)$ which is $R\setminus (-1,1) $, the range of $\sec(x)$ is $R\setminus (-1,1)$ too. And I've managed to simplify the expression to have in terms of $\sec x$ and $\csc x$ as such:
$$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$
But then, what do I do when I have 3 terms and not just $\sec$ or $\csc$?

答:
First of all, we need

$\csc2x$ must be finite $\implies\sin2x\ne0\implies2x\ne n\pi\  \ \ \  (1)$ where $n$ is any integer

Now if $\sin x+\cos x+1=0,$

using double angle formula, $$\cos\dfrac x2\left(\sin\dfrac x2+\cos\dfrac x2\right)=0$$

$\cos\dfrac x2=0\iff x=(2r+1)\pi\  \ \ \  (2)$

and $\sin\dfrac x2+\cos\dfrac x2=0\implies x=2r\pi-\dfrac\pi2\  \ \ \  (3)$

Both $(2),(3)$ are been precluded by $(1)$

Let $\sin x+\cos x+1=k\implies k\ne0$

$\sin x\cos x=\dfrac{(\sin x+\cos x)^2-1}2=\dfrac{k^2-2k}2$

$f(x)=\dfrac{2k}{k(k-2)}=\dfrac2{k-2}$

As $k\ne0, f(x)\ne\dfrac2{0-2}\  \ \ \  (4)$

Now $-\sqrt2\le\sin x+\cos x\le\sqrt2\implies-\sqrt2-1\le k-2\le\sqrt2-1$

If $0< k\le\sqrt2-1, f(x)\ge\dfrac2{\sqrt2-1}=?$

If $0>k\ge-\sqrt2-1,  f(x)\le\dfrac2{-(\sqrt2+1)}=?$

But remember $(4)$


问:
Prove that if $f$ is continuous on an open interval, then there is a $c \in I$ such that $f(c)=r$
Let $D\subseteq \mathbb{R}$ be an open interval. Let $f \colon D \rightarrow \mathbb{R}$ be a function with these properties:

$f$ is continuous in $D$.

$\exists x_0 \in D$ such that $f(x_0)=0$.

There exists $a \in \mathbb{R}$ such that $a < 0$ and $\lim_{x \rightarrow \infty}f(x)=a$.

Can you give me any insight on how to prove that, for all real number $r$ s.t. $a<r \leq 0$, there is a $c \in D$ such that $f(c)=r$? Thank you.

答:
This can be done using Intermediate Value Theorem (IVT).

Since $f$ is continuous in $D$ (and $D$ is an interval), we can define the interval $D_2$ as a subset of $D$, where $D_2 = [x_0, \infty)$. We know that $$f(x_0) \neq a = \lim_{x \rightarrow \infty} f(x)$$

hence we can apply IVT. Choose arbitrarily $r \in \mathbb{R}$ where $a<r\leq 0$. By IVT, we know there exists a $c \in D_2$ (therefore $c \in D$) such that $f(c) = r$.


问:
Dot product and Cross product question
Let $f : \mathbb R → \mathbb R^3$ be a three-times differentiable vector-valued function.

Is this statement true?

$f '(u) \cdot [ f '(u)  \times  f ''(u)] = 0$

If so how can it be justified or explained? Is it correct to say that these vectors are perpendicular?

答:
$f'(u)\times f''(u)$ is perpendicular to $f'(u)$ (the cross product of vectors is perpendicular to them),

and the dot product of perpendicular vectors is $0$.

Title: Dot product and Cross product question


问:
Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$
Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.

I've already tried to write it like this:
$$\frac{\sqrt{1}}{1} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 3$$
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 4$$
but I don't know that to do next or if it's the right way prove this.

答:
Have you tried using induction?

Let's assume the following proposition is true for $n$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$

By induction we must prove it works for the case $n=1$ and for all the cases $n+1$.

$P(1)$:

The base case is trivial, since
$$\frac{1}{\sqrt{1}} = 1 \leq 3\sqrt{2} -3 = 3(\sqrt2 - 1)$$

Since $\sqrt2 > 1$ it is easy to see why it is true.

Now we must bring our proposition $p(n) := \sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$ to test for all $n+1$ and see if it holds.

$P(n) \Rightarrow P(n+1)$:

$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} + (\frac{1}{\sqrt{n+1}}) \leq 3\sqrt{n+1} -3 + (\frac{1}{\sqrt{n+1}})$$
$$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3 + \frac{1}{\sqrt{n+1}} \leq 3\sqrt{n+2} - 3 $$

Therefore

$$\sum_{k=1}^{n+1}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+2} - 3 $$

You can check that $3\sqrt{n+1} + \frac{1}{\sqrt{n+1}} \leq 3\sqrt{n+2}$ is true for all $n$. Hence we proved by induction that the inequality $P(n)$ is true.


问:
discrete pareto distribution?
I always thought, that Pareto distriubtion is continuous.

I found a paper, that states that $$P(X=c)=\frac{1}{c^{\alpha}} - \frac{1}{{c+1}^{\alpha}}$$

for $c=1,2,3,...$, where $X$ is a Pareto random variable.

Is that right?

Maybe they meant an other distribution?

答: This seems like a natural discretization of the continuous Pareto distribution. I agree it's not what people typically mean. $P(X=c)$ in that paper is simply the probability under the true Pareto distribution that $c\le X\le c+1$. The key characteristic of the Pareto distribution in many cases is that $P(X>c)=O(1/c^\alpha)$, which holds here.

Title: discrete pareto distribution?


问:
Does anyone recognize the function: ${ F }_{ \delta }(x)=\frac { 1 }{ 2\delta } \int _{ -\delta }^{ \delta }{ f(x+t)dt } $
I was given as homework a question related with this "smoothing" function:
 ${ F }_{ \delta  }(x)=\frac { 1 }{ 2\delta  } \int _{ -\delta  }^{ \delta  }{ f(x+t)dt } $

Where $f(x)$ is a continuous function.

Does it have some formal name?
I'm trying to read some more about it but I wasn't able to find useful info.

Thanks!

答:
If you generalize it a bit, you get a well known and much used construction. The integral can be written the form $$F_\delta(x)=\frac1\delta\int_{-\infty}^{\infty}f(x-t)\varphi\Bigl(\frac t\delta\Bigr)\,dt$$
where, in this case, $$\varphi=\frac12\chi_{[-1,1]}$$
(where, in general, $\chi_A$ is the function which takes the value $1$ in the set $A$ and $0$  outside it).
You can replace $\varphi$ by any nonnegative function which vanishes outside the interval $[-1,1]$ and has $\int_{-\infty}^\infty\phi(t)\,dt=1$. Then $F_\delta\to f$ when $\delta\to0$. This is most useful when $\varphi$ is infinetely differentiable, for then $F_\delta$ too becomes infinitely differentiable.

The function $\varphi$ is called a mollifier, and the process of moving from $f$ to $F_\delta$ is called mollification. It is a much used technique in real analysis.

ContentLicense: CC BY-SA 3.0


问:
$A$ is the moore penrose of $A^+$: $(A^+)^+ = A$
$A^+ =VS^+ U^H$
but I think rather than using the definition, I shall use this: $AA^+A = A$ and $A^+=A^+AA^+$
where $A$ is $mn$ matrix.
however my trying ended up  complete mess. I start with $(A^+)^+$ and end up proving that $(A^+)^+=(A^+)^+$.

same with proving $(A^+)^T=(A^T)^+$
I'm stuck in a circle. please any hints will greatly help!

答:
According to Wikipedia, the definition of $A^+$ is the unique matrix $B$ that satisfies:

$ABA = A$,

$BAB = B$,

$AB$ is Hermitian,

$BA$ is Hermitian.

Let $B = A^+$ and $C = B^+ = (A^+)^+$. Then $C$ is the unique matrix that satisfies:

$BCB = B$,

$CBC = C$,

$BC$ is Hermitian,

$CB$ is Hermitian.

Note that $C = A$ makes statement 5 equivalent to 2, 6 equivalent to 1, 7 equivalent to 4, and 8 equivalent to 3. That is, $C = A$ satisfies the conditions, and hence $(A^+)^+ = A$.


问:
Local injectivity of a function defined as $f: \Bbb R^2 \to \Bbb R^2$

Determine if the function defined as $f: \Bbb R^2 \to \Bbb R^2$, $f(x,y) = (x^2, x^2+y)$ is locally injective.

Computing the partials one has $\frac{\partial f_1}{\partial x}=2x, \frac{\partial f_1}{\partial y}=0, \frac{\partial f_2}{\partial x}=2x, \frac{\partial f_2}{\partial y}=1.$ Thus the Jacobi is as follows $$J_f = \begin{bmatrix} 2x && 0 \\2x && 1\end{bmatrix}.$$

And from here $\det(J_f) = 2x \ne0,$ when $x \ne 0.$

However, I'm not sure I entirely understand the definition of local injectivity and how to continue from here. The definition I have deals with topological spaces which I haven't been studying yet.

Let $F : X → Y$ be a continuous function between topological spaces, and let $a ∈ X$.
We say that $F$ is locally injective (or locally one-to-one) at $a$ if there exists a
neighborhood $U$ of a such that $F|_U$ is injective.

How does the fact that the Jacobian determinant is nonzero help here? Since the Jacobi can be interpreted as a matrix and it's invertible iff the determinant is nonzero it seems that we're striving for a stronger statement that $f$ would actually be invertible? This raises another question, if $f$ is invertible is it immediately locally injective?

答:
In any neighborhood of a point of the form  $(0,y_0)$ you may find distinct points $(a,y_0)$ and $(-a,y_0)$ by choosing
$a$ sufficiently small.  Since
$$
  f(a,y_0) = (a^2, a^2 + y_0) = f(-a,y_0),
  $$
we see that  $f$ is not injective on any neighborhood of $(y_0,0)$.  Because of this $f$ is not locally injective at $(0, y_0)$.

On the other hand, regarding a point of the form $(x_0, y_0)$,  with $x_0\neq 0$,  one has that $\text{det}(J_f)(x_0,y_0)\neq 0$, so
the Inverse Function Theorem implies that there exists a neighborhood $U$  of $(x_0, y_0)$ such that $f|_U$ is (a
diffeomorphism and hence) injective.  So $f$ is locally injective at $(x_0, y_0)$.

Finally,  if $f$ is (globally) injective,  it is also locally injective since,  given any $(x_0, y_0)$,  you can
choose the neighborhood $U=\mathbb R^2$,  where $f$ is injective.


问:
Finding a periodic orbit
I have given the following system of ODEs:
\begin{cases} x'= -y + x(r^4-3r^2 + 1) \\ y'= x + y(r^4-3r^2 + 1) \end{cases}
where $r^2=x^2+y^2$.
I have to find the equilibrium and ,using the Poincaré-Bendixson theorem, I have to show that there is a periodic orbit in the region $\{x \in \mathbb{R}^2|0<|x|<1\}$.

Now I showed that the only equilibrium is the origin (and from the Poincaré-Bendixson it follows that there is a periodic orbit) and in case I wrote the system in polar coordinates:
\begin{cases} \theta'= 1 \\ r'= r(r^4-3r^2 + 1) \end{cases}
How can I procede?

答:
Consider your dynamical system in polar coordinates:

$$\theta'= 1$$
$$r'= r(r^4-3r^2 + 1)$$

In theory, you can immediately discern the existence of a periodic orbit for $r = \frac{1}{2}(\sqrt{5} - 1)$ as the second root of that expression for $r'$, but we want to use the Poincaré-Bendixson theorem to show it.

Consider the system at two $r$'s that are a little bit larger & smaller than $\frac{1}{2}(\sqrt{5} - 1)$; say, at $r = \frac{1}{2}$ and $r = 1$.

For $r = 1/2$, we find that $r' = 5/32$, so trajectories at this radius tend to move away from the origin. For $r = 1$, we find that $r' = -1$, so trajectories tend to move towards the origin.

Therefore, the annulus region defined between these two radii is a trapping region, and there must be a periodic orbit somewhere inside of it as per the Poincaré-Bendixson theorem.

Title: Finding a periodic orbit


问:
Question 13.3(a) Tom Apostol Introduction to Analytic number theory
I am self studying analytic number theory from Tom M Apostol and  got struck on question 3(a) of Chapter-13.

It's image:

In (a) I thought of using Perron's Formula (statement here)

by taking identity $log\zeta(s) =\sum_{n=2}^{\infty}\frac{\Lambda(s) }{n^s log n}$ .

But the problem is that I am not sure what will occur on the RHS ie I am not able to simplyfy $\sum_{n=2}^{\infty}\frac{\Lambda(n) }{n^s log n}$.

I thought of Perron's Formula due to the setting in LHS in which question was asked.

So, can you please tell how to simplify RHS if question can be simplified by Perron's Formula?

答:
To solve (a), you are correct to use the Perron's formula. Namely, we get that

\begin{align*}
\frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty}\log(\zeta(s))\frac{x^s}{s}ds&=\frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty}\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s\log(n)}\frac{x^s}{s}ds\\
&=\sum_{n<x}\frac{\Lambda(n)}{\log(n)}
\end{align*}

We now note that there are $\pi\left(x^{1/k}\right)$ powers $p^k$ of primes, and at powers $p^k$ we have

$$\frac{\Lambda(n)}{\log(n)}=\frac{\log(p)}{\log(p^k)}=\frac{1}{k}$$

Summing over all $k$ and counting the number of elements of each group yields the result.

To solve (b), we note that $\pi(x)<x$ and so since $x^{1/\log_2(x)}=2$ we get that

\begin{align*}
\sum_{k=2}^{\infty}\frac{\pi(x^{1/k})}{k}&=\sum_{k=2}^{\log_2(x)}\frac{x^{1/k}}{k}\\
&\leq \frac{1}{2}\log_2(x)\sqrt{x}\\
&=o(x/\log(x))
\end{align*}

and thus this will only be assymptotic to $x/\log(x)$ if $\pi(x)$ is which completes our proof.


问:
using the $sin^2x=1-cos^2x$,show that |axb| = $\sqrt{a^2.b^2−(a.b)^2}$
using $sin^2x=1-cos^2x$ in $|v|=|a|.|b|.sinx$, where  $v=a x b$ (vector product)

show that  |axb| = $\sqrt{a^2.b^2−(a.b)^2}$

ı dont understand what ı have to do exactly.
according to equation, must be Zero.

,so,Do I need to find the vectors what makes equation 0 ?
if I do,where ı have to begin the  solution ?

Body:
$$|a\times b|=|a||b||sin{x}|=|a||b|\sqrt{1-\cos^2x}=$$
$$=\sqrt{a^2b^2-(|a||b|\cos{x})^2}=\sqrt{a^2b^2-(a\cdot b)^2}.$$


问:
Proof $(A-\{a\}) \sim (B - \{b\}) $
I must prove the following:

"let $A$ and $B$ two set, with $A \neq \emptyset$ and $B \neq \emptyset$ and $A \sim B$, and $a \in A$ and $b \in B$, then $(A-\{a\}) \sim (B - \{b\})$

The symbol $\sim$ is used for "equipotency"!!

Thanks in advance!

答:
Let $f$ be a bijection between $A$ and $B$. If $f(a)=b$ then this would be wonderful, since the restriction of $f$ to $A\setminus\{a\}$ would be a bijection onto $B\setminus\{b\}$.

But if $f$ is not such a bijection, we can make it into such:

Let $c\in A$ such that $f(c)=b$ and $d\in B$ such that $f(a)=d$. What could be a good solution for our problem? If there was a slightly modified version of $f$, say $\hat f$, such that $\hat f(c)=d$ and $\hat f(a)=b$, and otherwise $\hat f(x)=f(x)$, then we would be done.

But we actually just described the solution. Let $\hat f$ be the following: $$\hat f=\big(f\setminus\{\langle c,d\rangle,\langle a,d\rangle\}\big)\cup\{\langle a,b\rangle,\langle c,d\rangle\}$$

I leave it to you to verify that $\hat f$ is a bijection from $A$ onto $B$ and $\hat f(a)=b$. Now return to the first sentence of this answer, and finish.

Title: Proof $(A-\{a\}) \sim (B - \{b\}) $


问:
problems with a delta function identity
I want to prove $x\delta'(x)=-\delta(x)$. What I did was integrating the right side around 0 (since both sides are equals when $x \neq 0$ trivially):

$$\int_{-\varepsilon}^{\varepsilon}x\delta'(x)dx=\left[x\delta(x)\right]_{-\varepsilon}^{\varepsilon}-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx=0-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx=-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx$$

I see the right and left sides are equal under this integration, but how does that tell me anything about their value at 0? Couldn't the integral be equal around 0 in  "as small interval as we wish" yet still the original functions get different values at zero?

答:
Answering your question requires that we answer the following: if we have a (generalized) function $d(x)$, then how do we know that $d(x)$ is the delta-function? In other words, what is the definition of the delta function?

One definition is that corresponding to the sifting property: $\delta(x)$ is the unique function $d(x)$ for which $\int_{-\infty}^\infty f(x) d(x) = f(0)$ holds for every function $f$. With this definition, if we want to show that $d(x) = -x\delta'(x)$ is the delta function, then it suffices to note that
$$
\begin{align}
\int_{-\infty}^\infty f(x)d(x)\,dx &=
\int_{-\infty}^\infty f(x)(-x\delta'(x))\,dx
\\ & = -\int_{-\infty}^\infty (xf(x))\delta'(x)\,dx
\\ & = \int_{-\infty}^\infty (xf(x))'\delta(x)\,dx
\\ & = \int_{-\infty}^\infty (xf'(x) + f(x))\delta(x)\,dx
= 0\cdot f'(0) + f(0) = f(0).
\end{align}
$$
It follows that $d(x) = -x \delta'(x)$ is indeed equal to the delta function.

Note that in the above computation, I use the defining feature of the "function" $\delta'$: for any function $f(x)$, $\int_{-\infty}^\infty f(x)\delta'(x)dx = -f'(0)$.

Title: problems with a delta function identity


问:
What is this notation called and how does it work?
I came across this formula in a Theory of Interest class. I've never seen this notation before and can't seem to find out how it is meant to be interpreted. What I do know is that it is used for ordinary annuities.

If someone could tell me what its called and how to interpret it that would be much appreciated.

答:
The notation
$$
s_{\overline n | i}
$$
is shorthand for the expression
$$
\frac{(1+i)^n-1}i$$
which is the amount of an annuity of $1$ per payment interval for $n$ intervals. It is usually read as "$s$-angle-$n$ at $i$".

For example of usage, see page 245 of this excerpt.


问:
Please help me to solve the differential equation $y' = (y/2) \cos x + (1/y) \sin (2x)$
Please help me to solve this differential equation: $y' = \frac{y}{2}\cos (x) + \frac{1}{y}\sin (2x)$?

答:
Multiply both sides by $y$ we get
$$
yy'+\frac{y^2}{2}\cos x=\sin 2x
$$
or
$$
\frac{1}{2} (y^2)'+\frac{y^2}{2}\cos x=\sin 2x
$$
we write $u=\frac12 y^2e^{\int^x\cos x} =\frac12 y^2e^{\sin x} $ ($e^{\int^x\cos x}$  is an integrating factor prescribed for such an equation) we get
$$
u'e^{-\sin x}=\sin2x
$$
Which is solved to give
$$
u=\int^x\sin 2x e^{\sin x}+c=2e^{\sin x}(\sin x-1)+c_1
$$
Thus
$$
y^2=4e^{-\sin x}(e^{\sin x}(\sin x-1)+c_1)
$$


答:
As in @BenGrossman's comments and answer, your computation does capture the essence of the issue, though your concern about pointwise values is heading off into trouble.

For that matter, although the manipulation of integrals (as in your heuristic or in BenGrossman's explanation) tells the story, it is also possible to write an analogous, but/and absolutely rigorous argument, with fewer symbols: for a test function $f$, by unabashedly thinking of distributions as (continuous, linear) functionals on test functions:
$$
(-x\delta')(f) \;=\; -\delta'(xf) \;=\; \delta((xf)')
\;=\; \delta(f + xf') \;=\; f(0) + (xf')(0) \;=\; f(0)+0
\;=\; \delta(f)
$$

Title: problems with a delta function identity

Body:
I want to prove $x\delta'(x)=-\delta(x)$. What I did was integrating the right side around 0 (since both sides are equals when $x \neq 0$ trivially):

$$\int_{-\varepsilon}^{\varepsilon}x\delta'(x)dx=\left[x\delta(x)\right]_{-\varepsilon}^{\varepsilon}-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx=0-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx=-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx$$

I see the right and left sides are equal under this integration, but how does that tell me anything about their value at 0? Couldn't the integral be equal around 0 in  "as small interval as we wish" yet still the original functions get different values at zero?


问:
Can someone explain the math behind tessellation?
Tessellation is fascinating to me, and I've always been amazed by the drawings of M.C.Escher, particularly interesting to me, is how he would've gone about calculating tessellating shapes.

In my spare time, I'm playing a lot with a series of patterns which use a hexagonal grid, and a tessellating design I saw on a chair in my dentist's surgery.

Here's some examples of patterns I've worked up...

Because I don't really understand the math, I've had to create the basic unit shape by hand, I can see the 60/120 degree angles well enough, but I really want to figure out the proportions of the shape, which appear (from just looking) to be tightly interrelated.

A few observations I've made during this process (aside from it being a bit more tricky than I first imagined.)

There appears to be a equilateral triangle which contains the key dimensions, and of course the regular hexagon is important, since this essentially forms 2 interwoven hexagonal grids.

Beyond this I'm unable to express mathematically what is going on. Key things I'd like to learn are...

How to calculate the relative positions of each shape to build the grid programmatically.

How to construct the shape unit mathematically.

I know that this doesn't help too much in understanding less rigid tessellation designs, such as Escher's reptiles, which I assume MCE made intuitively (I read that he wasn't a mathematician, which I was surprised by.)

Anyway, I hope this question isn't too sprawling, and thank you for looking.

Additional notes on the origin of this pattern.

Previously I'd said, (I believe the pattern is pretty old and of Chinese origin.).

However, this statement was pure speculation based on the appearance of similar patterns in Chinese art, also using the p31m wallpaper tiling method, this one below for example..

Well, it turns out that this specific pattern:

Appears originally in Islamic / Arabic / Moorish art. Within the confines of my limited research, it seems that geometric designs in Islamic art hold no specific religious significance (I expect this is a highly controversial statement.) and instead provide an opportunity for craftsmen to demonstrate the skill and subtlety of their workmanship, and to dazzle and intrigue the viewer with their complexity.

It seems that using circle construction of a grid of equilateral triangles, this pattern can be discovered, along with numerous variations, eg:

After working with this and other Islamic patterns I find that this not only appeals  to the viewer but as much to the artist. Every time I begin working with these patterns I'm constantly amazed by them.

Apparently Keith Critchlow an artist, lecturer, author, and professor of architecture, has made extensive studies in Islamic architecture, arts and decoration, and I've ordered a couple of books written by him, Order in Space and Islamic Patterns: An Analytical and Cosmological Approach, to further my own education on this topic. (I should note that Critchlow received high praise from Buckminster Fuller, who commented on his genius. Personally, as a long time acolyte of Fuller, I'm very intrigued to see Professor Critchlow's books.)

New York Metropolitan Museum of Art provides this PDF book for free download...

Animated Construction

Hexagonal Tessellation Construction (draft) from ocodo on Vimeo.

Click the image to watch...

Update

So I discovered through pencil/compass construction that the base unit is 1/8th of the radius of the initial construction circle (i.e. the length of one side of the hexagon.)

The width of the struts are 1/4 (or 2 base units)

I'm pushed for time to get this done tonight so I'm just posting progress as I work things out... (and noting where I make mistakes)

Construction:

Here is the construction mapped to the the 31m wallpaper group...

(handily iPhone4 wallpaper sized... well, the link through is.)

Gallery.

I've added a Flickr set to post various designs on the hexagonal lattice theme, take a look.

Interactive pattern generators

I've made some processing sketches which you might like to see, they generate patterns like this...

and this...

(They have some keyboard controls to tame the randomness a bit.)

3D / Blender experiments.

I have these shapes being 3D printed, I'll post a report when I get them in the mail (will be about 10 days) (I used Shapeways.com who print 3D objects in plastic, metal and other assorted materials.)

The intention with this set of 7 is to assemble them, and frame them nicely. I also get to test the precision of the 3D print material.

3D prints arrive.

Due to my inexperience with solid manufacture, I completely overlooked the need to allow for manufacturing/material tolerance. After a brief study on this aspect I'm remodelling my figures to add this in. However I need to fabricate a few sets to test optimal tolerance.

Anyway, this is what I've got back from shapeways, (nb: they use UPS shipping, included in the cost and the items are packaged very well, there is no risk of breakage.)

When I get the tolerance right I'll post again.

By the way, I found a very cool feature in CARMetal, (the software I've used to do the geometric construction.) I can add controls like sliders and text inputs that affect dimensions. For example, I create a slider t (tolerance) and can then apply t to the radius of my tolerance circles, which I use subsequently to construct the line segments which make up the adjusted shape, these all stay in sync with the slider value.

It's probably easier to see it, the tolerance is marked green (way too much in this example.)

Hand sanded

Clay tiles...

Well, not actually clay, I've used Sculpey, and pressed the tiles out using these tile cutters, which were 3D printed by Shapeways.

Additional notes...

(These are mostly for my reference, I'm summarising, but please feel free to correct me where I've gone wrong, it would be a big help.)

In addition to the p31m symmetry, there's an interesting relationship within the shape dimensions. The struts are of course all of equal width, and obviously all the angles are multiples of $60^\circ$. I've noted before, that the strut width is one quarter of the construction radius ( $w = \frac{r}{4}$ or as shown below, $2u$.) To calculate all required measurements we need to use the base unit $u = \frac{r}{8}$. The distance between the centres of each shape is $9u$. The length of a edge of the exterior arrow head (and of course it's corresponding interior edge) is $5u$. The remaining dimensions are the interior
lengths of the arrow head, which are both $2u$ (of course, these interlock with each other.)

Another minor update

I found some interesting information in this article (and page 2) on Islamic art.

I also took a photo of the chair in my dentists surgery, which began this little study of mine.

答: I recommend reading "Symmetry" by Hermann Weyl.  He first discusses the history of symmetry and then classifies the point symmetry groups, the frieze groups and finally the wallpaper groups.  Note that spherical tilings are just as interesting (and as storied) as Euclidean tilings.  Hyperbolic tilings are more recent -- they are a less fundamental yet somewhat deeper subject. </p>


答:
Your equation is linear in $u=y^2$. You can consider this as a case of a Bernoulli equation
$$
u'=u\cos x+2\sin(2x).
$$

Title: Please help me to solve the differential equation $y' = (y/2) \cos x + (1/y) \sin (2x)$

Body: Please help me to solve this differential equation: $y' = \frac{y}{2}\cos (x) + \frac{1}{y}\sin (2x)$?


问:
Identity and substitution in Intuitionistic Logic
I am a beginner in mathematical logic. I have a basic question about identity propositions in intuitionistic logic. For example, from $(*)$ and $(**)$:
$$\Gamma\vdash a=b\quad\quad(*)\quad\quad\quad\quad\quad\quad\Gamma\vdash P(a)\quad\quad(**)$$
we shall deduce that $\Gamma\vdash P(b)$ also holds. But what if we have $(***)$?
$$\Gamma\vdash \neg\neg(a=b)\quad\quad(***).$$
My question is: from $(**)$ and $(***)$, shall we conclude $\Gamma\vdash P(b)$ or $\Gamma\vdash\neg\neg P(b)$?

I don't know how to handle identity propositions because I have not met these in my textbook (von Plato, 2014). I would appreciate it a lot if anyone can answer my question and if possible provide some useful references for handling identity propositions. Thanks in advance!

答:
I. From $\Gamma \vdash \neg\neg (a = b)$ and $\Gamma \vdash P(a)$, can we conclude $\Gamma \vdash P(b)$?

No. In the general case, $\Gamma \vdash P(b)$ does not follow from $\Gamma \vdash \neg\neg(a = b)$ and $\Gamma \vdash P(a)$.

Assume that we could conclude $\Gamma \vdash P(b)$ from $\Gamma \vdash \neg\neg(a = b)$ and $\Gamma \vdash P(a)$ for any context $\Gamma$ and formula $P$. Then in particular this would work for $P(x) \leftrightarrow a = x$. We'd have $\Gamma \vdash P(a)$ by the reflexivity of equality, so we would be able to conclude $\Gamma \vdash P(b)$, that is, $\Gamma \vdash a = b$. This would mean that equality is always stable under double-negation, i.e. that $\Gamma \vdash \neg\neg (a = b) \rightarrow a = b$ is provable in every context.

However, we can construct theories (models, topoi) where equality is not $\neg\neg$-stable, e.g. any model of Smooth Infinitesimal Analysis.

II. From $\Gamma \vdash \neg\neg (a = b)$ and $\Gamma \vdash P(a)$, can we conclude $\Gamma \vdash \neg\neg P(b)$?

We can always do this. The trick is to realize that double negation distributes over implication, even in intuitionistic logic.

Since $\Gamma \vdash a = b \rightarrow (P(a) \rightarrow P(b))$ is provable, we can introduce a double negation to get $\Gamma \vdash \neg\neg (a = b \rightarrow P(a) \rightarrow P(b))$, and distribute the double negations to get $\Gamma \vdash \neg\neg (a = b) \rightarrow \neg\neg P(a) \rightarrow \neg\neg P(b)$. Since we already know $\Gamma \vdash \neg\neg (a = b)$, we can conclude that $\Gamma \vdash \neg\neg P(a) \rightarrow \neg\neg P(b)$. Similarly, we already know $\Gamma \vdash P(a)$, from which it follows that $\Gamma \vdash \neg\neg P(a)$. Putting everything together we get $\Gamma \vdash \neg\neg P(b)$, which was to be shown.


问:
How to show that $T^n = A^n = A$ for an idempotent matrix $A$
suppose $A$ is a $2\times 2$ matrix given as

$$\pmatrix {3 & −6\cr 1 & -2}$$

Find the linear map $T : {\bf R}^2 \to{\bf R}^2$ associated to $A$ and find $T^{5001}$ based on its associated matrix.

I know that $A$ is an idempotent matrix but how do I solve this question with a general proof?

答:
I suppose what you mean by "general proof" is to use induction to solve this.

So maybe first write down the statement you want to show:
In this case it is $P_n:A^n=A$ for $n\in\mathbb{N}$.

So first you show that the base case is true, that is when $n = 1$. This is trivial.

Then you show that assuming $P_k$ is true for some $k\in \mathbb{N}$, i.e. $A^k=A$, then $P_{k+1}$ is true $(A^{k+1}=A)$.

Therefore $A^n=A$ $\forall n\in \mathbb{N}$.


答:
$$ a \cdot b = |a| |b| \cos \theta$$

$$ a \times b = |a| |b| \sin \theta$$

Square the two equations and add:

$$ (a \cdot b)^2 + (a \times b)^2 = |a|^2 |b|^2$$

Now you should be able to finish :-)

Title: using the $sin^2x=1-cos^2x$,show that |axb| = $\sqrt{a^2.b^2−(a.b)^2}$

Body:
using $sin^2x=1-cos^2x$ in $|v|=|a|.|b|.sinx$, where  $v=a x b$ (vector product)

show that  |axb| = $\sqrt{a^2.b^2−(a.b)^2}$

ı dont understand what ı have to do exactly.
according to equation, must be Zero.

,so,Do I need to find the vectors what makes equation 0 ?
if I do,where ı have to begin the  solution ?


问:
5 cars, 4 parking places. Derangements and permutations with fixed points
I found an exercise in combinatorics:

In the parking of a building, there a re five parking spots, with their owner cars assigned
  to them. One day only four cars arrived. In how many ways can they park so that not one cars
  parks on their corresponding spot?

So I think the following way: consider the missing car. It may arrive and park on its own parking spot, there are $5\cdot D_4$ ways of doing that, where $D_n$ is the number of derangements of $n$-element set, or it may arrive and find that his place is already taken. Then it takes someone other's place, thus making a complete derangement of 5 element set. So the total number is $5\cdot D_4+D_5$. Am I thinking correctly?

The real problem I have with this is that I found it in some early pop-quiz some teacher gave during the introduction to probability class. Isn't there an easier way?

答: The four employee cars can either leave the chief's place free and park falsely in $D_4=9$ ways, or they use the chief's place, and then the chief coming later would also be in a false place. This can happen in $D_5=44$ ways. It follows that there are $53$ ways for the four cars to park falsely.


问:
Intersection with boundary
Let $X$ be a metric space. If $A \subset X$ is a closed set, and $x \notin A$ with $b_r(x) \cap A \neq \emptyset$ where $b_r(x)$ denotes the open ball centered at $x$ of radius $r$, is it true that $b_r(x) \cap \delta A \neq \emptyset$? ($\delta A$ denotes the boundary of $A$). This seems intuitively true to me but I'm not quite sure how to show that.

答: Since $\delta A$ contains the limit points of $A$, one may pick $y_1, y_2, ... \in b_r(x) \cap A$ such that $(y_i)$ converge to some point $y\in \delta A$, which is to say $y\in b_r(x) \cap \delta A$.

Title: Intersection with boundary


问:
Geometric sum converges to 1/2
$$\sum_{n=0}^\infty\frac{1}{3^{2^n}-3^{-2^n}}=\frac{1}{2}$$

I assumed this is a geometric series, but im not 100% sure about it. I checked if its convergent anyway, and it is, but other than that I cant prove it converges to $\frac{1}{2}$. Limit when n approaches infinity of $x_n$ is 0.

答:
To see that this sum is telescopic, rewrite the general term as

$$\frac {3^{2^n}} {3^{2^{n+1}} - 1} = \frac 1 {3^{2^n} - 1} - \frac 1 {3^{2^{n+1}} - 1}$$

Another way is to simply observe the partial sums

\begin{alignat}{2}
\frac38 &= \ \ \ \frac38 & & = \frac12-\frac18\\
\frac38 + \frac 9 {80} &= \ \ \frac {39}{80} & &= \frac12-\frac1{80}\\
\frac38 + \frac 9 {80} + \frac {81} {6560} &= \frac {3279}{6560}& &= \frac12-\frac1{6560}
\end{alignat}

and try to prove this observation by induction.

Title: Geometric sum converges to 1/2


问:
Proof that $\sin x=4^{-x/\pi}\sum_{n=0}^\infty (-1)^n\binom{4x/\pi}{2n+1}$ for $x\ge 0$
Is there any proof of the identity
$$\sin x=4^{-x/\pi}\sum_{n=0}^\infty (-1)^n\binom{4x/\pi}{2n+1}$$
where $x\ge 0$?

If we replace $4x/\pi$ by $x$ and write the binomial in terms of the falling factorial $(x)_n$, we get
$$\sum_{n=0}^\infty (-1)^n \frac{(x)_{2n+1}}{(2n+1)!},$$
which is remarkably similar to the power series for $\sin x$ centered at $0$:
$$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}.$$

答:
Start with:
$$
(1+z)^A = \sum_{n=0}^{\infty}\binom{A}{n}z^n
$$Split this into even and odd pieces:
$$
(1+z)^A =  \sum_{n=0}^{\infty}\binom{A}{2n+1}z^{2n+1}+ \sum_{n=0}^{\infty}\binom{A}{2n}z^{2n}
$$Playing around with the even-only piece (exercise!), we have
$$
\sum_{n=0}^{\infty}\binom{A}{2n}z^{2n}=\frac{1}{2} \left((1-z)^A+(1+z)^A\right)
$$So,
$$
 \sum_{n=0}^{\infty}\binom{A}{2n+1}z^{2n} =\frac{(1+z)^A-(1-z)^A}{2 z};
$$put $z=i$, $A=4x/\pi$ and try and simplify it using Euler's Formula.


问:
$ 2\cot2x = 4$ from $0$ to $\pi$
$$ 2\cot2x = 4$$
I need to solve this equation from $0$ to $\pi$.

I have used many trig identities and gotten down to

$$2\cos^2x - 1 = 4\sin x\cos x$$

yet I have no idea how to get it into one trig function.

答:
We use relation:

$$cotan(2x)=\frac{1-tan^2 (x)}{2 tan(x)}$$

So we have following equation:

$$tan^2(x)+4tan(x)-1=0$$

Which gives :

$$tan x=2 ±\sqrt 5 $$

$tan x=2+\sqrt 5 $ gives $x≈77^o$

$2 cotan(2\times 77)≈-4$

which is not acceptable.

$tan x= 2-\sqrt 5≈ -0.236$ gives $x=-13$ and $x≈180-13=167^o$

But $2cotan (-26)≈-4$

So $x=13$ can be acceptable because :

$2 cotan(26^o)≈4$

Hence solution is $x=2k \pi +\frac{13}{180}\pi$

and $x=(2k+1)\pi+\frac{13}{180}\pi$

Title: $ 2\cot2x = 4$ from $0$ to $\pi$


问:
Prove that $T^2 = T \iff \text {rank}\ (T) + \text {rank}\ (I - T) = \dim V.$

Let $V$ be a finite dimensional vector space over $\Bbb C$ or $\Bbb R$ and let $T$ be a linear operator on $V.$ Prove that $T^2 = T$ if and only if $$\text {rank}\ (T) + \text {rank}\ (I - T) = \dim V.$$

"$\implies$" part $:$ By rank nullity theorem it follows that $$\text {rank}\ (T) + \text {nullity}\ (T) = \dim V.$$ So If we can show that $\text {nullity}\ (T) = \text {rank}\ (I - T)$ then we are through. Now since $T^2 = T$ it follows that $T(I - T) = 0$ which in turn implies that $\text {Im}\ (I - T) \subseteq \text {Ker}\ (T).$ Hence we have $\text {rank}\ (I - T) \leq \text {nullity}\ (T).$ Also for $v \in \text {Ker}\ (T)$ we have $(I - T) (v) = v$ i.e. $v \in \text {Im} (I - T).$ So $\text {Ker}\ (T) \subseteq \text {Im}\ (I - T)$ and hence we have $\text {nullity}\ (T) \leq \text {rank}\ (I - T).$ Combining this with the previous inequality we find that $\text {rank}\ (I - T) = \text {nullity}\ (T),$ as required.

Now how do I prove "$\impliedby$" part?

Any help will be highly appreciated. Thanks in advance.

答:
If $\operatorname{rank}(T) + \operatorname{rank}(I - T) = \dim V$, then using the rank-nullity theorem on both $T$ and $I - T$,
$$\operatorname{null}(T) + \operatorname{null}(I - T) = \dim V.$$
Note that $v \in \operatorname{ker}(T) \cap \operatorname{ker}(I - T)$ implies that $Tv = 0$ and $0 = v - Tv = v$, hence the two kernels sum directly to $V$. Hence, we can adjoin bases for each kernel together to make a full basis $B$ for $V$.

For the vectors $v \in B$ from $\operatorname{ker}(T)$, we note that $(T - T^2)v = (I - T)Tv = (I - T)0 = 0$.

For the vectors $v \in B$ from $\operatorname{ker}(I - T)$, we note that $(T - T^2)v = T(I - T)v = T0 = 0$.

Therefore, $T - T^2$ maps the basis $B$ to $0$, and thus by linearity, maps every vector to $0$. Therefore, $T = T^2$.


问:
Probability of $X$ out of $N$ dice landing on $M$
The problem is as follows:

We have $N$ dice and we throw them on a table. What is the probability that $M$ will fall $X$ times?

Specific example:
We have $10$ dice and we throw them on a table. What is the probability that $3$ will occur $7$ times (so there are seven dice out of $10$ which have $3$ on top)?

I am really bad at probability. I've checked a few sites about how I could solve this but it seems too hard figure it out.

答:
From the N dice choose X and ensure that those X show you required no. but others do not.
So it is just (N choose X)(1/6)^X(5/6)^(N-X)


问:
Confused with the notation used in proof of monotone class theorem and the logic behind the proof
We were introduced to the monotone class theorem which stated that (in my own understanding) the minimal sigma field generated by the field $\mathcal{C}$ (denoted $\sigma[\mathcal{C}]$) equals to the minimal monotone class generated by that same field $\mathcal{C}$(denoted $\mathcal{M[C]}$). i.e. $$m[C]=\sigma[C] \ \ \ \text{when } C \text{ is a field}$$

I was introduced to this theorem by the textbook Probability for Statisticians 2nd edition by Galen R.Shorack first. I was able to follow the proof in the textbook by checking all the statements are true, but I didn't quite get the idea behind the proof. I revisited the concept and the proof today using the Youtube source: https://youtu.be/4XhMgu_ekeY?t=606.

I find myself confused with the notation of the proof: I think if $E\subset \mathcal{M[Q]}$ (In this series of video, $ \mathcal{Q}$ is an algebra), then $E$ must be a class of sets. However, in $g(E)=\{F\in\mathcal{M[Q]}:E\backslash F,E\cap F,F\backslash E \in\
\mathcal{M(Q)}\}$ the symbol $E$ seems to represent the element of class, which is set. Also, in the claim, $E\in\mathcal{Q}$ also suggests $E$ is an element in the class.

The lecturer constructs $g(E)$ and show it contains $Q$ and is a monotone class. It seems to me This step means for any given $E\in \mathcal{M(Q)}$, by the function $g$, we will get a corresponding class of sets $g(E)$ which is a monotone class. Does that mean we are mapping a set to a class of set? In fact, in the textbook that I used for studying, the notation is introduced as: For each $A \in \mathcal{M}, \operatorname{let} \mathcal{M}_{A} \equiv\left\{B \in \mathcal{M}: A B, A^{c} B, A B^{c} \in \mathcal{M}\right\} .$ And we were shown that $\mathcal{M}_{A}$ is a monotone class. It didn't occur to me we can think of the construction as some sort of function.

There are many proofs in the textbook involving constructing a set of this kind and prove the constructed sets satisfies the certain property. Whenever I checked the properties of the set as stated, those properties always hold as if by magic. I fancy by thinking those constructions as function may be insightful and illuminating but I lack the mathematical background to make my mind clearer. I apologize for the lengthy explanation of my confusion which may be something very basic and simple. Could someone please explain to me how to think the proof from the perspective of a function and explain the intuition behind the proof? If any fundamental concepts (for example, in topology, functional analysis, set theory) were needed to better understand the idea behind the proof I am more than happy the direction could be pointed.

答:
$\newcommand{\C}{\mathscr{C}}$
$\newcommand{\G}{\mathscr{G}}$

I think your confusion makes perfect sense and I agree that the notation used in the video is at least very loose, if not wrong. Below is a clearer proof.

As you may have already familiarized, to finish the proof, you need to justify two directions:

$\sigma(\C) \subset m(\C).$

$m(\C) \subset \sigma(\C).$

Since (by definition) $\sigma(\C)$ is the smallest sigma-field that contains $\C$ and $m(\C)$ is the smallest monotone class that contains $\C$, to show $1$, it would be sufficient to show $m(\C)$ is a sigma-field (as we already know $m(\C)$ contains $\C$); similarly, to show 2, it suffices to show $\sigma(\C)$ is a monotone class.

2 is immediate, as a sigma-field is of course a monotone class. In fact, if $A_1, A_2, \ldots \in \sigma(\C)$ and $A_n \uparrow A$, then $A = \cup_{n = 1}^\infty A_n \in \sigma(\C)$, since any sigma-field is closed under the formation of countable unions. Similarly you can check the other condition in the monotone class is also satisfied.

1 is trickier. First note that if $m(\C)$ is a field, then it must be a sigma-field (left you as an exercise, if you have difficulty, I can edit the answer later), thus it suffices to show $m(\C)$ is a field. To this end, first consider the class
\begin{align*}
\G = \{A: A^c \in m(\C)\}.
\end{align*}
Since $m(\C)$ is monotone, so is $\G$. Since $\C$ is a field, $\C \subset \G$, and so $m(\C) \subset \G$ (again, use the minimality). Hence $m(\C)$ is closed under complementation.

Next define the class
\begin{align*}
\G_1 = \{A: A \cup B \in m(\C) \text{ for all } B \in \C\}.
\end{align*}
Then it is straightforward to check (use definition) that $\G_1$ is a monotone class and $\C \subset \G_1$; from the minimality of $m(\C)$ follows $m(\C) \subset
\G_1$. Now define
\begin{align*}
\G_2 = \{B: A \cup B \in m(\C) \text{ for all } A \in m(\C)\}.
\end{align*}
Then $\G_2$ is a monotone class, and from $m(\C) \subset \G_1$ it follows that $A \in m(\C)$ and $B \in \C$ together imply that $A \cup B \in m(\C)$; in other words, $B \in \C$ implies that $B \in \G_2$. Thus $\C \subset \G_2$; by minimality, $m(\C) \subset \G_2$. Therefore, $A, B \in m(\C)$ implies that $A \cup B \in m(\C)$.

This completes showing $m(\C)$ is a field.

To summarize, I hope the route of reaching the conclusion is now clear to you: that is,

Why we would think of defining classes $\G, \G_1, \G_2$ and how they help us to achieve our ultimate goal. I think this corresponds to your statement "Whenever I checked the properties of the set as stated, those properties always hold as if by magic. I fancy by thinking those constructions as function may be insightful and illuminating but I lack the mathematical background to make my mind clearer." At the first glance, the creation of $\G, \G_1, \G_2$ does look like a magic coming out from nowhere, but what if you compare them with the definition of field, which is the destination that we want to arrive? Did you see any similarity? In fact, $\G, \G_1, \G_2$ are so defined that all properties of the field can be checked through them.

How "minimality" are repeatedly are used in the whole argument.

This trick is extensively used in measure theory, to get more practice of implementing it, you may consider reviewing the proof of the Dynkin's $\pi$-$\lambda$ theorem and trying to prove this exercise by yourself.


问:
Symbolic definition for $i$?
Please help!

I am trying to find a definition for $i$ that doesn't work for $-i$.

let $j$ be either $i$ or $-i$.

saying $j^2=-1$ doesn't help since $(i)^2=-1$ and $(-i)^2 = -1$

saying $j=\sqrt{-1}$ doesn't halp since $(-1)^{\frac{1}{2}}$ is multivalued $i$ and $-i$

saying $j=e^{i\pi/2}$ doesnt help since $e^{i\pi/2}=i$ but $e^{(-i)\pi/2}=-i$ too (yeah I know I said j=$e^{i\pi/2}$ and not $j=e^{j\pi/2}$ I did that to hopefully make it more intuitive)

saying $j=ln(i)/(\pi/2)$ doesn't work since $i=ln(i)/(\pi/2)$ and $-i=ln(-i)/(pi/2)$ (check with a phase plotter, ln is multivalued but includes i and -i in their respective functions shown above http://davidbau.com/conformal/#log(z)%2F(pi%2F2)-z)

saying $Im(j) > 0$ doesn't work because $Im(z)=Re(z/i)$ and $Re(i/i)>0$ but $Re((-i)/(-i))>0$

Is this parity some sort of law? And yet $i\neq-i$ since $i=(-1)\cdot(-i)$. Unless...

Post Scriptum: Irionically, it is very easy to symbolically represent $1$ vs $-1$ even though $(1)^2=1$ and $(-1)^2=1$, we can just say $x$ such that $x=x^2$

EDIT: can quaternions help? Is $i$ a set of 2 numbers? Are they always equal?

答:
An idea: using the "identification" (in fact isomorphism, but making it a simple identification) of $\;\Bbb C\;$ with $\;\Bbb R^2\;$, we have that $\;z=x+iy \stackrel{\text{Ident.}}\sim (x,y)\;$, and thus

$$i\sim(0,1)\;,\;\;\text{whereas}\;\;-i\sim (0,-1)$$

and that's one way (a rather simple and algebraic one) to distinguish completely $\;i\;$ from $\;-i\;$ .

Title: Symbolic definition for $i$?


问:
Conceptually(not algebraically), why do I get a quadratic equation when I subtract a linear equation form a quadratic equation.
While simultaneously solving a quadratic equation and a linear equation, I noticed that their intercepts are actually equal to the x intercepts of another quadratic function. I find it hard to grasp that subtracting a linear equation from a quadratic equation results in a symmetric curve(parabola).

I understand that the result of the subtraction will give me a polynomial of degree 2 which forms a parabola, but why?

答:
One way of looking at it is through finite differences. A quadratic function has the property that successive differences follow a linear progression. For example, with $f(n) = n^2 + n + 1$:

```<code>n  f(n)  difference
-------------------
0    1
          2
1    3
          4
2    7
          6
3   13
          8
   ...
</code>```

A linear function has the property that successive differences are constant. Thus, adding a linear function to a quadratic function will shift the differences of the latter without disturbing their linearity.

The argument in the previous paragraph relies on an understanding that "constant + linear = linear", so in fact you can see that you can use the same reasoning inductively for adding polynomials of any degree.

Body:
One way of looking at it is through finite differences. A quadratic function has the property that successive differences follow a linear progression. For example, with $f(n) = n^2 + n + 1$:

n  f(n)  difference
-------------------
0    1
          2
1    3
          4
2    7
          6
3   13
          8
   ...

A linear function has the property that successive differences are constant. Thus, adding a linear function to a quadratic function will shift the differences of the latter without disturbing their linearity.

The argument in the previous paragraph relies on an understanding that "constant + linear = linear", so in fact you can see that you can use the same reasoning inductively for adding polynomials of any degree.


问:
If the equation $3x+5y=k$ has exactly $4$ positive integral solutions and $k$ is the largest number for which this is possible, then what is $k$?

If the equation $$3x+5y=k$$ has exactly $4$ positive integral solutions and $k$ is the largest number for which this is possible, what is the value of $k$?

I can't wrap my head around the fact that why the values of $y$ will be limited to only $1$,$2$ and $3$. Why can not we go beyond that?

答:
$ax+by=c$ if $a,b,c,x,y$ are positive integers and $gcd(a,b)=1$, then the number of positive integer pairs are given by $N+1$ if integer part of $c/(ab)$ is $N$ and the remainder ($r$) when $c$ is divided by $(ab)$ has a solution for $r=ax_0+by_0$ where $x_0,y_0$ are positive integers.

So if $3x+5y=59$, then $N=3$ and $r=14$, check that $14=3x+5y$ has asolution $(3,1)$.
So the starting eqn will have 4 solutions.

But if $k=60$, the given equation will have $60/15=4$ number of solution and no more becaues $3x+5y=0$ cannot have solution as positive integer pair, as $0$ is not a positive integer.

Finally, $k_{max}=60$.


答:
So if $$ 2\cot 2x = 4 \Leftrightarrow \cot 2x = 2 $$
then $$ 2x = \cot^{-1}2 + k\pi\Leftrightarrow x = \frac{\cot^{-1}2}{2} + \frac{k\pi}{2} $$
approximately we have
$$ x = 0.2318 + \frac{k\pi}{2} $$
and in range $(0,\pi)$:
$$ x \in \{0.2318, 1.8026\} $$

Title: $ 2\cot2x = 4$ from $0$ to $\pi$

Body:
$$ 2\cot2x = 4$$
I need to solve this equation from $0$ to $\pi$.

I have used many trig identities and gotten down to

$$2\cos^2x - 1 = 4\sin x\cos x$$

yet I have no idea how to get it into one trig function.


问:
Combinatorics: Distributing tasks in class
Suppose you sit in a class room with $n$ students. There are $k$ question and every student can pick $5$ of them. How many different ways are there such that no question is left out?

In total we have $5\cdot n$ decisions. This is a really hard question, I tried simulating some cases but I don't see a common pattern.

答:
There are $\binom{k}5$ different sets of $5$ questions, so there are $\binom{k}5^n$ different ways for each of $n$ students to choose a set of $5$ questions. From this figure we need to subtract the number of ways that leave out at least one question. For any question $Q$ there are $\binom{k-1}5$ sets of $5$ questions that do not include $Q$, so $\binom{k-1}5^n$ of the $\binom{k}5^n$ ways leave out $Q$. Since there are $k$ questions, this might suggest that the answer is $$\binom{k}5^n-k\binom{k-1}5^n\,,\tag{1}$$ but that fails to take into account the fact that every combination of choices that leaves out two questions has been subtracted twice, once for each of the omitted questions, and is therefore counted $-1$ times in $(1)$. There are $\binom{k}2$ pairs of questions, and there are $\binom{k-2}5^n$ ways for the class to leave out a specific pair of questions, so in $(1)$ we’ve subtracted $\binom{k}2\binom{k-2}5^n$ too much; restoring it gives us

$$\binom{k}5^n-k\binom{k-1}5^n+\binom{k}2\binom{k-2}5^n\,.\tag{2}$$

We’re not done, since any set of class choices that leaves out three questions is counted $1-3+3=1$ time in $(2)$, and it shouldn’t be counted at all. Can you complete the inclusion-exclusion calculation from here?


问:
Distribution of $Y=e^{-X}$ for Gamma-distributed $X$
Suppose we have a random variable $X \sim \text{Gamma}(\alpha, \,\beta)$ with shape $\alpha>0$, rate $\beta > 0$, and pdf
\begin{align}
f_X(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x}, \quad \text{for } x \in (0,\,\infty).
\end{align}
I'm curious about the properties of $Y=g(X)=e^{-X}$ and if it is another well-known distribution. Since $g(X)$ is strictly decreasing in $X$, and thus has inverse $g^{-1}(X) = -\log(X) = \log\left(\frac1X\right)$, we can find
\begin{align}
f_Y(y) = f_X\left(g^{-1}(y)\right)\left|\frac{d}{dy} g^{-1}(y)\right| = \frac{\beta^\alpha}{\Gamma(\alpha)} \left[\log\left(\frac1y\right)\right]^{\alpha - 1} y^{\beta -1}, \quad \text{for } y \in (0,\,1).
\end{align}
However, "neatly" obtaining the cdf $F_Y(y)$, mean $\text{E}[Y]$, variance $\text{Var}[Y]$, MGF $M_Y(t)$ etc. from this pdf seem rather difficult to do, as trying to integrate $f_Y(y)$ likely involves the incomplete gamma function. Is there some insight I'm missing (perhaps a change of variable or something similar) that would allow one to compute these integrals by hand?

EDIT: I actually realized after posting this that the moments of $Y$ are not too difficult to compute if one makes the substitution $\log\left(\frac1y\right) = u$. A similar thing can be done for the higher-order moments.

答:
Using the notation on the wiki page, for the CDF, you can proceed by noting that
$$
F_Y(y) = P(Y \le y) = P(e^{-X} \le y) = P (X> -\ln y) = 1-F_X(-\ln y),
$$
and since $F_X(x) = \frac{1}{\Gamma(\alpha)} \gamma(\alpha,\beta x)$, we have
$$
F_Y(y) = 1- \frac{1}{\Gamma(\alpha)} \gamma(\alpha,-\beta \ln (y))
$$
For the expectation, we can recall that the MGF of a gamma random variable is
$$
m_X(t) = E(e^{tX}) = \left ( 1- \frac{t}{\beta} \right)^{-\alpha}, \quad t < \beta,
$$
Then note that
$$
E(Y) = E(e^{-X}) = m_X(-1) = \left ( 1+ \frac{1}{\beta} \right)^{-\alpha}
$$
Similarly,
$$
E(Y^2) = E(e^{-2X}) = m_X(-2),
$$
and so you can use this to compute $V(Y) = E(X^2) - [E(X)]^2$


问:
Find the limit of the sequence $ \frac{x^n}{n^k}$ as $n \to \infty$ for all values of$ x > $0 and $k = 1, 2,\cdots$
I have tried using the ratio lemma to tackle this question and also the fact $(n+1)^k \geq 1 + nk$ and I haven't reached an answer. How should I go about solving this problem?

答:
HINT

$x\left (\frac{ n^{k}}{n+1^{k}} \right )= x\left ( \frac{n}{n+1} \right )^{k}=x\left ( \frac{n+1}{n} \right )^{-k}= x\left ( 1+\frac{1}{n} \right )^{-k}$


问:
Module isomorphism, simple modules, and quotients
I'm reading R.S. Pierce's Associative Algebras. While proving a preliminary lemma to Nakayema's Lemma, the following is mentioned:

Let $M$, $N$, be two $A$-modules where $N$ is a submodule of $M$ and $M/N$ is simple. Let $u\in M$, $u\notin N$. Then $uA+N=M$.

This must be really elementary but I can't see why. I'm assuming it's related to some isomoprhism theorems and probably Schur's lemma. Or maybe it follows directly from the definitions of simple modules or quotient rings? Anyway, I don't think the assumption $u\notin N$ is necessary.

Any help would be much appreciated.

答: Because of the isomorphism theorem for modules there is an inclusion-preserving bijection between the submodules of $M/N$ and the submodules of $M$ containing $N$. Thus, because $uA+N$ properly contains $N$ and $M/N$ is simple it has to be equal to $M$.</p>


问:
Help needed in Unit 5 Section 1 Algebra by Hungerford
I am self studying Field Theory from Textbook Hungerford and couldn't solve this problem on page 241 .

For (a)  X would be composite of LM if X = intersection of all subfields P such that  $L\bigcap M\subseteq P$.

So, I tried using this defination to prove  $LM \subseteq  L(S) $ and $L(S) \subseteq LM$.

I tried proving $LM \subseteq L(S) $but I got confused by inclusion relations as K is subset of $ L \bigcap M $ and then M= K(S) for some S subset of M.

It's a very hard nut to crack due to it and I am unable to.

I have done (b).

For (c) I have no clue. (a) can't be used in C as $E_i$ can't be necessarily subset of each other,in my opinion.

So, can you please tell how to solve (a) and Give hint for (c).

答:
Part a), your definition of the composition $LM$ seems to be wrong. The definition of $LM$ is the the smallest subfield containing both $L$ and $M$, not the intersection $L\cap M$.

First note that $L(S) \supset K(S) =M$ (because $L\supset K$). Clearly $L\subset L(S)$. So $LM \subset L(S)$. On the other hand, $L(S) \subset (LM)(S) = LM$ since $L\subset LM$ and $S\subset M$.

Part c), you can prove inductively on $n$. (Try prove the case for $n=2$ first.)


问:
Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots

Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.

Obviously as soon as you factor this equation to $$(4x^2+8x+a^2)(x^2+ax+1)=0$$ then you have finished.
However, how am I supposed to think of factoring the equation in such a manor? That you are supposed to factor it, is obvious, but how can you find out which are its factors, aside from blind luck?

答:
Rewrite it as a polynomial in $a$:

$$ a^3 x + a^2 (x^2 + 1) + a ( 4x^3 + 8x^2 ) + (4x^4 + 8x^3 + 4x^2 + 8x) = 0.$$

The factorization almost immediately jumps out by observation:

$$ (a^2 + 4x^2 + 8x ) ( ax + x^2 + 1) = 0 $$

The idea of changing the variable is a common trick. It can be helpful when factoring (e.g. applying the Reminder Factor Theorem), or finding real roots (e.g. apply the quadratic discriminant to another variable).

As an example, try to factorize $ a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 - 2c^2a^2$.

There are numerous approaches that one can use, so in the spirit of this question, consider it as a quadratic in $a^2$ first.


问:
Distance from vectors in $\mathbb{Z}^d$ to the cube $[-1/2,1/2]^d$
Let $\|\;\|_2$ be the Eucliden norm on $\mathbb{R}^d$.

Problem: Suppose $x\in Q:=\big[-\tfrac12,\tfrac12\big]^d$. Is
$$
\|x-k\|_2\geq \frac{1}{2\sqrt{d}}\|k\|_2
$$
for all $k\in\mathbb{Z}^d$?

Notice that for all $x\in Q$,  $\|x-k\|_2\geq d(k,Q):=\inf_{x\in Q}\|x-k\|_2$. So it is to enough to show that
$$d(k,Q)\geq \frac{\|k\|_2}{2\sqrt{d}},\quad k\in\mathbb{Z}^d$$

I think this holds but I can't reproduce a proof at the moment. The relevance of this simple geometric result is that it provides a criteria for absolute and uniform convergence of the Poisson summation
$$\sum_{k\in\mathbb{Z}^d}f(x+k)=\sum_{k\in\mathbb{Z}^d}\widehat{f}(k)e^{2\pi ik\cdot x}$$
in the $\mathbb{T}^d$ torus, where  $f\in L_1(\mathbb{R}^d)$. If $f$   can be bounded poitwise by an integrable decreasing radial function, i.e. $|f(x)|\leq \phi_0(\|x\|_2)$, where $\phi_0$ is monotone non increasing and $\phi_0\circ\|\;\|_2\in L_1(\mathbb{R}^d)$

A proof or a good hint will be appreciated.

答:
After giving it more thought, I came up with the following  solution below. I am still  interested  in accepting a more elegant solution which may make  clever use of the law of cosines and/or the triangle inequality without resorting to induction.

My solution:

As stated, $Q=\big[-\tfrac12,\tfrac12\big]^d$. For any $\mathbf{k}=(k_1,\ldots,k_d)\in\mathbb{Z}^d$, define
$$\Delta(\mathbf{k})=\{1\leq j\leq d: k_j\neq0\}$$
By compactness, there is $\mathbf{q}\in\partial Q$ such that
$$d(\mathbf{q},\mathbf{k})=d(\mathbf{k},Q)$$
Since $0<\big|m-\operatorname{sign}(m)\frac12\big|\leq |m-t|$ for all integer $m\neq0$ and $|t|\leq\frac12$
$$
d(\mathbf{k},Q) =\sum_{j\in\Delta(\mathbf{k})}\big(|k_j|-\tfrac12\big)^2
$$
That is, $\mathbf{q}=(q_1,\ldots,1_d)$ where $q_j=\operatorname{sign}(k_j)\frac12$ when $j\in\Delta(\mathbf{k})$ and $q_j=0$ otherwise.

We proceed to show

$$\begin{align}
d(\mathbf{k},Q)\geq\frac{\|\mathbf{k}\|_2}{2\sqrt{d}}\tag{1}\label{one}
\end{align}
$$
or equivalently
$$\begin{align}
4d\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2\geq\sum_{j\in\Delta(\mathbf{k})}k^2_j\tag{1'}\label{onep}
\end{align}$$
by induction on the dimension $D$ of the space.

For $D=1$, \eqref{onep} holds trivially for $k=0$, and also for $k\neq0$ since
$2\big||k|-\tfrac12\big|=2|k|-1\geq|k|$ since $|k|\geq1$.

Asume \eqref{onep} holds for $D=1,\ldots,d-1$, where $d-1\geq1$. Let $\mathbf{k}\in\mathbb{Z}^d$ and $\alpha_\mathbf{k}=\#\Delta(\mathbf{k})$.
If $\alpha_\mathbf{k}<d$, then
$$
\begin{align}
4d\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2\geq4\alpha\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2\geq \sum_{j\in\Delta(\mathbf{k})}k^2_j
\end{align}
$$
by the indiction hypothesis. If $\alpha=d$, then $|k_j|>0$ for all $1\leq j\leq d$ and so,
\begin{align}
4d\sum_{j\in\Delta(\mathbf{k})}(|k_j|-\tfrac12)^2 &=4d\sum^{d-1}_{j=1}(|k_j|-\tfrac12)^2 + 4d(|k_d|-\tfrac12)^2\\
&\geq  4(d-1)\sum^{d-1}_{j=1}(|k_j|-\tfrac12)^2 + 4(|k_d|-\tfrac12)^2\\
&\geq \sum^{d-1}_{j=1}k^2_j + k^2_d=\|\mathbf{k}\|^2_2
\end{align}

This completes the induction argument.


问:
Does there exist a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$ such that $\mathbf{A}^{2}=-\mathbf{I}$?
Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,

$$\mathbf{A}^2=-\mathbf{I}$$

I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?

答:
When $A$ is an $n\times n$ real matrix, it is possible if and only if $n$ is even.

If $n$ is odd, then $\det(A^2)=(\det A)^2\ge 0$ but $\det(-I)=-1$, so $A^2=-I$  is impossible.

If $n$ is even, you can construct $A$ from the case $n=2$ easily.


问:
How do we solve the equation $x^{2} - 4x + 4 - i = 0$?
I have the following problem:

Find all the roots of $x^2 − 4x + 4 − i$ in standard form.

I have no clue where to start. I tried separating the real and imaginary parts, getting $x^2 − 4x + 4 = 0$, but obviously $-i$ doesn't equal 0.

I went along with that just to see and set $x=2-i$ after solving the real part $x^2 − 4x + 4 = 0$ and adding the $-i$ just to see where it would take me, and that didn't work.

I'm left with trying to solve $x^2 − 4x + 4 − i$ by brute force, and can't seem to find a solution. Any help is much appreciated.

答:
HINT

To begin with, notice that

\begin{align*}
x^{2} - 4x + 4 - i = 0 \Longleftrightarrow (x-2)^{2} = i = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)
\end{align*}

Then you can find the solutions through the relation
\begin{align*}
z^{n} = \cos(\theta) + i\sin(\theta) \Longleftrightarrow z = \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)
\end{align*}
where $0\leq k\leq n-1$ and $k\in\mathbb{N}$. At your case, $\theta = \pi/2$ and $n = 2$. Can you take it from here?

BONUS

You can also apply the Bhaskara's formula where $a = 1$, $b = -4$ and $c = 4-i$. Thus we have that
\begin{align*}
z_{1,2} = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 16 + 4i}}{2} = \frac{4 \pm 2\sqrt{i}}{2} = 2\pm\sqrt{i}
\end{align*}


问:
Showing a sequence space is totally bounded
I am trying to show that $X= \{(x_n)\in \mathbb{R}^\mathbb{N}:x_n\in [0,1]\space\forall \space n\}$ with the metric $d(x,y) = \sum^\infty_{k=1} \frac{|x_k−y_k|}{2^k}$ is totally bounded.

To show that it is totally bounded, I think that for any $\epsilon >0$ (where WLOG I can consider consider $\epsilon=2^{-n}$), I have to find finitely many sequences $\{ x_1,...,x_{n_\epsilon}\}$ such $\{ B(x_i,\epsilon): i\in \{1,...,n_\epsilon\}\}$ covers X, but I haven't made much progress with this.

Any help would be much appreciated.

答:
Let $\varepsilon>0$ arbitrary. Then there exists an $n\in\mathbb N$, such that
$$
\varepsilon>\frac{1}{2^n}.
$$
Consider now $S$ the finite set of all the elements $\boldsymbol{x}\in X$, of the form $\boldsymbol{x}=(x_1,\ldots,x_n,x_{n+1},0,0,\ldots)$, where
$$
x_j\in\left\{
\frac{k}{2^{n+1}}: k=0,1,2,\ldots,2^{n+1}
\right\}.
$$
Then, for every $\boldsymbol{y}=(y_1,\ldots,y_n,\ldots)\in X$, there exists a  $\boldsymbol{x}\in S$, such that
$$
|x_j-y_j|<\frac{1}{2^{n+1}}, \quad j=1,\ldots,n+1,
$$
and hence
$$
d(\boldsymbol{x},\boldsymbol{y})<\sum_{j=1}^{n+1}\frac{1}{2^{n+1}}\cdot\frac{1}{2^j}+\sum_{j=n+2}^{\infty}
\frac{|y_j|}{2^j}<\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}=\frac{1}{2^n}<\varepsilon
\quad\Longrightarrow\quad \boldsymbol{y}\in B(\boldsymbol{x},\varepsilon).
$$
Hence
$$
X\subset \bigcup_{x\in S}B(x,\varepsilon).
$$


答:
HINT: Notice that for any $n\in\Bbb N$ and $x,y\in X$,

$$\sum_{k>n}\frac{|x_k-y_k|}{2^k}\le\sum_{k>n}\frac1{2^k}=\frac1{2^n}\,,$$

so

$$d(x,y)\le\frac1{2^n}+\sum_{k=1}^n\frac{|x_k-y_k|}{2^k}\,.$$

If you knew that $$\frac{|x_k-y_k|}{2^k}<\frac1{n2^n}\tag{1}$$ for $k=1,\ldots,n$, you could conclude that $$d(x,y)<\frac1{2^{n-1}}$$ no matter what $x_k$ and $y_k$ are when $k>n$. Find a finite subset $F$ of $[0,1]^n$ such that for each $\langle x_1,\ldots,x_n\rangle\in[0,1]^n$ there is a $\langle y_1,\ldots,y_n\rangle\in F$ such that $(1)$ holds, and use it to get what you need for $\epsilon=\frac1{2^{n-1}}$.

Title: Showing a sequence space is totally bounded

Body:
I am trying to show that $X= \{(x_n)\in \mathbb{R}^\mathbb{N}:x_n\in [0,1]\space\forall \space n\}$ with the metric $d(x,y) = \sum^\infty_{k=1} \frac{|x_k−y_k|}{2^k}$ is totally bounded.

To show that it is totally bounded, I think that for any $\epsilon >0$ (where WLOG I can consider consider $\epsilon=2^{-n}$), I have to find finitely many sequences $\{ x_1,...,x_{n_\epsilon}\}$ such $\{ B(x_i,\epsilon): i\in \{1,...,n_\epsilon\}\}$ covers X, but I haven't made much progress with this.

Any help would be much appreciated.


答:
I doubt this is the intended solution but here it is anyway.

The main tool is the discriminant. Wikipedia says:

The discriminant [of a quartic] is zero if and only if two or more roots are equal. If the coefficients are real numbers and the discriminant is negative, then there are two real roots and two complex conjugate roots. Likewise, if the discriminant is positive, then the roots are either all real or all non-real.

The discriminant of the quartic in question is $-16 (a - 2)^6 (a + 2)^2 (5 a^2 + 12 a + 20)^2 \le 0$. Therefore, that quartic has only real roots iff its discriminant is zero.

Factoring the discriminant is the hard part here, harder than factoring the quartic in the first place.

Title: Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots

Body:
Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.

Obviously as soon as you factor this equation to $$(4x^2+8x+a^2)(x^2+ax+1)=0$$ then you have finished.
However, how am I supposed to think of factoring the equation in such a manor? That you are supposed to factor it, is obvious, but how can you find out which are its factors, aside from blind luck?


问:
Is verify the same as proof?
I am solving a matrix problem where I have to verify whether the given matrix is orthogonal. But the matrix is not orthogonal. Is the question wrong or does it have a different interpretation of verify? What's the difference between "verify" and "proof"?

答:
It depends on the subtleties of the wording. Compare: "Verify that the matrix is orthogonal" (it is, and you must confirm it) to "Verify whether the matrix is orthogonal [or not]" (maybe it is it, maybe it isn't).

The word "verify" comes from Latin veritas, "truth"; it means to show that something is true. With that in mind, if you need to find out whether or not a certain property holds, it might be better to say, "Determine whether the matrix is orthogonal or not."

Title: Is verify the same as proof?


答:
The answer is no in $\mathbb{R}^{3\times 3}$. Recall that $\det AB=\det A\det B$ and so if $\det AA=\det -I$ then $(\det A)^2=-1$ but $\det A$ is a real number for all $A\in\mathbb{R}^{3\times 3}$ and so no such $A$ can exist.

This extends easily to all odd $n$ for $\mathbb{R}^{n\times n}$ and also suggest that solutions exist in $\mathbb{C}^{n\times n}$. Most notably, the matrix $A=\text{diag}(i,\ldots, i)$ has the property that $A^2=-I$.

Title: Does there exist a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$ such that $\mathbf{A}^{2}=-\mathbf{I}$?

Body:
Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,

$$\mathbf{A}^2=-\mathbf{I}$$

I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?


答:
It sounds like you have picked up on the fact that conjugation $$f: a+bi \mapsto a-bi$$ is a ring isomorphism where $i$ and $-i$ correspond to each other.

Addition preserving: $f((a+bi)+(c+di))=f((a+c)+(b+d)i)=(a+c)-(b+d)i=(a-bi)+(c-di)=f(a+bi)+f(c+di)$

Multiplication preserving: $f((a+bi)(c+di))=f((ac-bd)+(ad+bc)i)=(ac-bd)-(ad+bc)i=(a-bi)(c-di)=f(a+bi)f(c+di)$

$f(1)=1$

Conjugation is clearly bijective too.

Title: Symbolic definition for $i$?

Body:
Please help!

I am trying to find a definition for $i$ that doesn't work for $-i$.

let $j$ be either $i$ or $-i$.

saying $j^2=-1$ doesn't help since $(i)^2=-1$ and $(-i)^2 = -1$

saying $j=\sqrt{-1}$ doesn't halp since $(-1)^{\frac{1}{2}}$ is multivalued $i$ and $-i$

saying $j=e^{i\pi/2}$ doesnt help since $e^{i\pi/2}=i$ but $e^{(-i)\pi/2}=-i$ too (yeah I know I said j=$e^{i\pi/2}$ and not $j=e^{j\pi/2}$ I did that to hopefully make it more intuitive)

saying $j=ln(i)/(\pi/2)$ doesn't work since $i=ln(i)/(\pi/2)$ and $-i=ln(-i)/(pi/2)$ (check with a phase plotter, ln is multivalued but includes i and -i in their respective functions shown above http://davidbau.com/conformal/#log(z)%2F(pi%2F2)-z)

saying $Im(j) > 0$ doesn't work because $Im(z)=Re(z/i)$ and $Re(i/i)>0$ but $Re((-i)/(-i))>0$

Is this parity some sort of law? And yet $i\neq-i$ since $i=(-1)\cdot(-i)$. Unless...

Post Scriptum: Irionically, it is very easy to symbolically represent $1$ vs $-1$ even though $(1)^2=1$ and $(-1)^2=1$, we can just say $x$ such that $x=x^2$

EDIT: can quaternions help? Is $i$ a set of 2 numbers? Are they always equal?


问:
Is the $S^1$ cone $D^2$?
Let $X$ be a topological space, let $Y=X \times [0,1]$ and $A=X \times \left \{ 0 \right \} \subset Y$. the cone on X is the space $\left \{ X \times [0,1] \right \}/\left \{ X \times \left \{ 0 \right \}  \right \}$.

Now let $X=S^1$, my goal is to prove that the cone on $X$ is homeomorphic to $D^2$, where $D^2$ is the unitary closed disk in $\Bbb R^2$.
Basically, this cone is obtained from a cylinder by contraction of the lower base to a point.

Using the polar coordinates we can establish a map $f: \left \{ X \times [0,1] \right \}/\left \{ X \times \left \{ 0 \right \}  \right \} \rightarrow D^2$ defined as $f(\theta, \rho)= \rho(\cos \theta, \sin \theta)$.

This map is continuos, injective and surjective, but how to prove that is a homeomorphism?

答:
To show $f$ is a homeomorphism without using compactness, you can explicitly construct the inverse of $f$, namely
$$
g(x, y) = [t(x, y), \sqrt{x^2 + y^2}]
$$
where brackets denote equivalence class, and
$$
t(x, y) = \begin{cases}
0 & (x, y) = (0,0) \\
\frac{\pi}{2} & x = 0, y > 0\\
\frac{3\pi}{2} & x = 0, y < 0\\
\arctan(y/x) & x > 0 \\
\pi + \arctan(y/x) & x < 0
\end{cases}
$$
where each of these values is to be considered a number (mod $2\pi$), hence an element of $S^1$.

The continuity of $g$ is pretty clear everywhere except along the $y$-axis. For those, you'll need to show that $g^{-1}(U)$ is open whenever $U$ is open in the domain, which will mean writing down a way to describe open sets in the quotient, which is a pain in the neck, but you're welcome to do it.

Title: Is the $S^1$ cone $D^2$?


问:
Factorizing $12x^3 - 8x^2 - 3x + 2$
How should I factorize this cubic expression?

$$12x^3 - 8x^2 - 3x + 2$$

Title: Factorizing $12x^3 - 8x^2 - 3x + 2$

Body:
We observe that $12:8=3:2$

So

$\begin{array}
112x^3 - 8x^2 -3x + 2  \\
&=&4x^2(3x-2) - (3x-2) \\
&=&(3x-2)(4x^2 - 1)\\
&=&(3x-2)((2x)^2 - 1^2)\\
&=&(3x-2)(2x+1)(2x-1)  \\
\end{array}$


答: Take the determinant of both sides. You find that $(\det A)^2=-1$ so that $\det A=\pm i$ which is impossible for real $A$. (But trivial for complex $A$.)</p>


问:
Trouble with variable transformation over a double integral

Consider the double integral over a region $D$
$$\iint_D \cos\left(\frac{x-y}{x+y}\right) \mathrm{dA}$$
where $D$ is the triangular region in the $xy$-plane with vertices $\{(0, 0), (0, 1),
(1, 0)\}$. Sketch the image of the region $D$ in the $uv$-plane under the variable transformation $u = x - y, v = x + y$ and evaluate the double integral using the new variables $u$ and $v$.

I found the new region $S$ to be a triangular region which represented an isosceles triangle with vertices at $(0,0), (0,2) \text{ and } (1,1)$. The variable transformation also meant that $ x = \frac{1}{2}(v+u)$ and $y = \frac{1}{2}(v-u)$. I know that $$\int\int f(x,y)  \text{dydx}$$ over a region $R$ also equals to $$\int\int f(x(u,v), y(u,v))\begin{vmatrix}
    \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \\\end{vmatrix}\text{dvdu}$$ over a region $S$.

By substituting in all the equations, I got:
$$\int _0 ^1 \int _u ^{-u+2} \frac{1}{2}\cos\left(\frac{u}v{}\right)\text{dvdu}$$
I changed the order of integration and got:
$$\int _0 ^2 \int _v ^{2-v} \frac{1}{2}\cos\left(\frac{u}v{}\right)\text{dudv}$$
which equals:
$$\frac{1}{2} \times \int _0 ^2 v(\sin \left(\frac{2-v}{v}\right) - \sin (1))\text{dv}$$
which I don't know how to simplify any further but I'm sure there is a rational answer. Can anyone help me with this problem?

答:
Your bounds of integration are incorrect. Using the transformation, we have $(0,0)\to (0,0)$, $(0,1)\to (-1,1)$, $(1,0)\to (1,1)$. This is the region $-1\le u\le 1,|u|\le v\le 1$; since your substitution is linear, it is determined by three points. The Jacobian determinant is correct, as is the new integrand. So we have
$$
I=\int_{-1}^{1} \int_{|u|}^1\cos\left(\frac{u}{v}\right) \cdot \frac{1}{2}\;dvdu
$$Now change the order of integration and integrate:
$$
I=\frac{1}{2}\int_{0}^{1} \int_{-v}^v\cos\left(\frac{u}{v}\right) \;dudv
=\int_{0}^{1} v\sin\left(1\right) \;dv=\frac{\sin(1)}{2}
$$


问:
How to find "n" by using weak law of large number?
Use the weak law of large numbers to find a number $n$ such that if one tosses a (fair) coin $n$ times the proportion of "heads" will lie in the range $[0.49, 0.51]$ with probability at least $0.95$.

I really have no idea on how to start working on this question as the value of variance is not provided. Can anyone help me for it?

答:
Using Cebishev's inequality you have

$$\mathbb{P}[|\overline {X}_n-\mu|\leq \epsilon]\geq 1-\frac{V(\overline {X}_n)}{\epsilon ^2}$$

You  have only to substitute and set the probability at 0.95

The variance of the coin flip is 0.25 thus the variance $V(\overline {X}_n)=\frac{0.25}{n}$


问:
Notation for change-of-coordinate matrix
I'm currently working through David Lay's Linear Algebra and Its Applications and I'm a bit confused by the notation used for the change-of-coordinate matrix.

If $\mathscr{B}$ and $\mathscr{C}$ are bases of a vector space $V$, then the change-of-coordinate matrix from $\mathscr{B}$ to $\mathscr{C}$ and $\mathscr{C}$ to $\mathscr{B}$ are denoted by
$P_{\mathscr{C}\leftarrow\mathscr{B}}$ and $P_{\mathscr{B}\leftarrow\mathscr{C}}$ respectively.

What I'm wondering is if there is a reason that the arrow points from right to left as opposed to common limit notation in calculus that is read from left to right?

答: Intuitively, $P_{\mathscr{C}\leftarrow\mathscr{B}}$ acts on vectors from the left, taking a vector $[v]_{\mathscr{B}}$ and producing a vector $P_{\mathscr{C}\leftarrow\mathscr{B}} [v]_{\mathscr{B}} = [v]_{\mathscr{C}}$, so you can see how $\mathscr{B}$ is being sucked into $P$ to produce $\mathscr{C}$.


答:
No.  For $3\times 3$ matrices, the determinant of $-\mathbf{I}$ is $-1$, and $\mathrm{det}(\mathbf{AA}) = \mathrm{det}(\mathbf{A})^2$ is always nonnegative.

Of course, if you allow complex numbers then the answer is yes, e.g. if $\mathbf{A} = i\mathbf{I}$.


问:
Simple group of order 10000

Prove that no group of order 10,000 is simple.

Here is my attempt.

For contradiction, suppose $G$ of order $10000$ is simple. Notice that
$$10,000 = 5^4 \cdot 16.$$
By the Sylow theorems, $n_5 = 1 + 5k$ for $k \geq 0$ and $n_5 \mid 16$. Going through the cases:
\begin{align*}
k = 0 \implies n_5 = 1 & & \text{divides $16$} \\
k = 1 \implies n_5 = 6 & & \text{doesn't divide $16$} \\
k = 2 \implies n_5 = 11 & & \text{doesn't divide $16$} \\
k = 3 \implies n_5 = 16 & & \text{divides $16$} \\
k = 4 \implies n_5 = 20 & & \text{doesn't divide $16$}
\end{align*}
So $n_5 = 1$ or $n_5 = 16$, but since we're taking $G$ to be simple, we have $n_5 \neq 1$, so $n_5 = 16$.

At this point I know that I need to construct an isomorphism between $G$ and some subgroup of $S_{16}$, using conjugation, and then use Lagrange's theorem. but I cannot figure out how to do this.

答:
If using the action on $G$ by conjugation is slightly difficult for you to grasp, here might be another lengthier method to go about solving this question. (I hope its right)!

Let $n_5=16$. Let $P$ and $Q$ be two sylow-5 subgroups of order $625$ each. Now we know
$$|P \cap Q|=\frac{|P|.|Q|}{|PQ|} \geq \frac{|P|.|Q|}{|G|} \sim 39.06$$
But we know that $|P \cap Q|$ divides 625 as $P \cap Q \subset P$ and so $|P \cap Q|=125$.
This implies $|PQ|=3125$

Now as $[P:P \cap Q]=[Q: P \cap Q]=5$, which is the smallest prime divisor of 625, $P \cap Q \triangleleft P$ and $P \cap Q \triangleleft Q$.

Let $N=N_G(P \cap Q)$. Then, as $P \cap Q$ is normal in both P and Q, we have $P \subset N$ and $Q \subset N$. Now,

$p(P \cap Q)p^{-1}=P \cap Q \ \ \ \forall p \in P$ (as $P \cap Q \triangleleft P$)

and

$q(P \cap Q)q^{-1}=P \cap Q \ \ \ \forall q \in Q$ (as $P \cap Q \triangleleft Q$).

Therefore, for any $pq \in PQ$, we have
$$pq(P \cap Q)(pq)^{-1}=pq(P \cap Q)q^{-1}p^{-1}=p(P \cap Q)p^{-1}=P \cap Q$$
and so we have $PQ \subset N$ and $|PQ| \leq |N|$.

So $|N| \geq 3125$, and it must also be a divisor of $10,000$ and divisible by $625$ (as $P \subset N$ and $Q \subset N$). The only possibilities then are $5000$ and $10000$.

If $|N|=5000$, then its index in $G$ must be 2 and hence it is normal. If $|N|=10000$, then $P \cap Q$ is normal in $G$. Therefore $G$ cannot be simple.

Title: Simple group of order 10000


问:
Please help me to solve the differential equation $y' = (y/2) \cos x + (1/y) \sin (2x)$
Please help me to solve this differential equation: $y' = \frac{y}{2}\cos (x) + \frac{1}{y}\sin (2x)$?

Body:
$$y' = \frac{y}{2}\;\cos (x) + \frac{1}{y}\sin (2x)$$
$$\cos x \dfrac {dy}{d\sin x} = \frac{y}{2}\cos (x) + \frac{2}{y}\sin x \cos x$$
$$\dfrac {dy}{dv} = \frac{y}{2} + \frac{2v}{y}$$
$$\dfrac 12\dfrac {dy^2}{dv} = \frac{y^2}{2} +{2v}$$
$$\dfrac {du}{dv} = {u} +{4v}$$
$$u'-u=4v$$
Where $u=y^2$ and $v= \sin x$
$$(ue^{-v})'=4ve^{-v}$$
Integrate.


问:
Proof Paths are Preserved
I am trying to prove that for graphs $G$ and $H$ with an isomorphism existing from $G$ to $H$, $p$ is a path in $G$ from $u$ to $v$ if, and only if, $f(p)$ is a path in $H$ from $f(u)$ to $f(v)$. My textbooks proof is as follows:

Let $P_n$ be the graph denoted by $V(P_n):=\{1,2,...,n\}$ and $E(P_n):=\{e_1,e_2,...,e_n\}$ where for each $i \in \mathbb{N} , e_i = \{i,i+1\}$.

We have that:

p is a path in $G  [1]$
$\iff p \cong P_n [2]$
$\iff f(p) \cong f(P_n) [3]$
$\iff f(p) \cong P_n [4]$
$\iff f(p)$ is a path in $H [5]$

I’m not sure I understand how to get from step [2] to [3], and from step [3] to [4] in either direction. Could someone please elaborate on why this is the case?

答:
Having a path in $G$ - a subgraph isomorphic to $P_n$ for some $n$ - means that there is an injective function $p: V(P_n) \to V(G)$ such that for all edges $e = \{i,i+1\} \in E(P_n)$, $p(e) = \{p(i), p(i+1)\} \in E(G)$. If you like (and apparently your textbook does like), we can define this function $p$ to be the path.

Then $f \circ p$ is a function $V(P_n) \to V(H)$ with the same property (because $f$ preserves edges) and therefore $f \circ p$ is a path in $H$.

Conversely, because $f$ also preserves non-edges, if $f \circ p$ is a path in $H$, then $p$ must be a path in $G$.

On the other hand, I really can't get behind some of the notational choices in your textbook's proof:

If $p$ is a function, then $p \cong P_n$ doesn't make any sense, because $P_n$ is a graph and $p$ is not a graph.

Both $f(p)$ and $f(P_n)$ don't make any sense, because $f$ is a function on vertices. I would accept applying $f$ to a set of vertices, elementwise, but a graph is more than that.

Title: Proof Paths are Preserved


问:
Dividing a rectangle in 3 equal parts half at time?
Here's an interesting puzzle that has got me thinking:

You need to divide a rectangle into 3 equal parts along it's length. However you can only divide things into half. You don't have any other tools at your disposal - a saw/cutter and a scale that can accurately tell you the half of a full length. How can this be done?

Clarification:

Assume a piece of wood:

```<code>Length: (16x)      ----------------
Step 1: (2 halves) --------|--------
Step 2: (4 quarts) ----|----|----|---- (1/3rd is somewhere between the first and second cut)
</code>```

And so on till you are close to $1/3$ with some $\pm\epsilon$. How can I mathematically frame and solve this problem and say after how many divisions will I be close to the 1/3 mark?

答:
If we aren't concerned with exactitude and are willing to accept some $\pm\epsilon$ then it is merely a question of "how much" error we can tolerate.

One can mark down halves successively to get lengths of $\frac{1}{2^n}$

Once you get that measurement as an acceptable unit, you can count $x$ units off to get to $\frac{1}{3}$ and $\frac{2}{3}$

Let me walk you through a simple example:

Say the length of the block is 100.

Then we want to cut it off at 33.33 and 66.67.

Now successively measure halves (half of half of half of...) 4 times. This gives us a unit of $\frac{1}{16}*100 = 6.25$

Now we can count off 5 units as 31.25, and 6 units as 75, which are "pretty close" to 33.33 and 66.67 respectively.

If your error tolerance is low, you can just keep making the unit smaller till you fit inside your $\pm\epsilon$ range.

In the above example, if you had successively measured halves 8 times, instead of 4, the unit would have been $\frac{1}{256}*100 = 0.390625$

Counting off 85 units = 33.20 (yup, really close to 33.33)

Counting off 171 units = 66.79 (reasonably close to 67.67)

In fact you can do one better. After counting off 85 units and cutting that piece off, you are left with 66.80. Now just chop that off into half, and you'll end up with 3 pieces: 33.20, 33.40, 33.40

The benefit of using the length as 100 is this: all the numbers above can actually be expressed as %.

Now if we had to issue a blanket statement, as to how many successive measures of half it takes to arrive at the right unit, you could say
$max(\frac{1}{2^n} l) < 2 \epsilon $


问:
How to determine the order of an element of Special Linear Group
Let,

$S=\begin{pmatrix}  0  &-1 \\ 1  & 0    \end{pmatrix}$
and

$T=\begin{pmatrix}  1  &1 \\ 0  & 1    \end{pmatrix}$ .

The matrices $S$ and $T$ generate $SL_2(\mathbb Z)$. The matrix $S$ has order $4 \: (S^2 = −I_2)$, while $T$ has infinite order.

Now consider we have an arbitrary element $\beta \in SL_2(\mathbb Z)$.

What is the process/ algorithm to determine/detect whether $\beta$ is an element of finite or infinite order?

答:
A matrix with integer entries has finite order if and only if its minimal polynomial can be expressed as a product of distinct cyclotomic polynomials.

In the $2 \times 2$ case, things are easy. One method would be to follow the following procedure. Begin with the matrix $\beta \in SL(\Bbb Z)$.

If $\beta$ is either $I$ or $-I$, then it has finite order. Otherwise, proceed

Find the characteristic polynomial of $\beta$, $p(x) = x^2 - \operatorname{tr}(\beta)x + 1$.  If we obtain one of the polynomials
$$
\Phi_3(x) = x^2 + x + 1, \quad \Phi_4(x) = x^2 + 1, \quad \Phi_6(x) = x^2 - x + 1,
$$
then $\beta$ has finite order (in particular, order $2,3,4,$ or $6$ respectively). Otherwise, $\beta$ has infinite order.

To summarize: a matrix $\beta \in SL(\Bbb Z)$ has finite order if and only if it is equal to $I$ or $-I$ or its trace satisfies $\operatorname{tr}(\beta) \in \{-1,0,1\}$.


问:
Solving $(x\cos y - y\sin y)dy - (x\sin y + y\cos y)dx =0$.
$$(x\cos y - y\sin y)dy - (x\sin y + y\cos y)dx =0.$$

I know that this is not an exact differential equation. So I have tried various integrating factors that I know. How can I solve this using an exact differential equation? In fact, can it even be solved by this method? If not then can someone please tell me how it can be done?

答:
$$A(x,y)dx+B(x,y)dy$$
is exact if $$\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$$
The given differential is
$$- (x\sin y + y\cos y)dx +(x\cos y - y\sin y)dy $$
and
$$\frac{\partial}{\partial y}(-(x \sin y+y \cos y))=-x \cos y+y \sin y-\cos y$$
$$\frac{\partial}{\partial x}(x \cos y-y \sin y)=\cos y$$
$$\frac{\partial A}{\partial y}\ne\frac{\partial B}{\partial x}$$
The differential is not exact.

A method to solve it anyway can be found here.


问:
Equivalent definition of Lebesgue integral
We define $\int f d \mu=\sup \{\mathcal{L}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$ where $\mathcal{L}(f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \inf _{A_{i}} f$ for an $\mathcal{S}$-partition $P=A_{1}, \ldots, A_{n}$.

I am looking to show that when $\mu(X)< \infty$ and $f$ is bounded then this definition is equivalent to using the upper sum $\mathcal{U}(f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \sup _{A_{i}} f$ with $\int f d \mu=\inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$.

It was easy to prove that $\int f d \mu \leq \inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$ by using the fact that $\mathcal{L}(f, P) \leq \mathcal{U}(f, P)$.

I am now left to prove $\int f d \mu \geq \inf \{\mathcal{U}(f, P): P \text { is an } \mathcal{S} \text { -partition }\}$.
I think this can be done by first proving it for when $f$ is simple and then generalizing but I am struggling. Any help would be much appreciated. Thanks in advance

答:
First of all, the definition of the Lebesgue integral above won't work in general for negative (measurable) functions. Consider $f\colon \mathbb{R} \to \mathbb{R}; f(x) = -e^{-x^2}$. We have
$$ - \infty < \int_\mathbb{R} f d\lambda\,,$$
where $\int f d \lambda$ denotes the usual Lebesgue integral wrt. to the Lebesgue measure. (It's usualy defined by approximating measurable $f$ by simple functions. I'd suggest looking this up, since it is rather poorly explained in the book you mentioned.) On the other hand
$$
\sup \{\mathcal{L}(f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\} = - \infty
$$
since $\mathcal{L}(f, P) = -\infty$ for all Partitions $P$.

Now, in order to use the definition 3.3 from Axler for $\textit{measurable non-negative}$ functions and show the result for measurable bounded $f \colon X \to [0, \infty]$ with $\mu(X) < \infty$, we first use linearity of the integral
$$
\int_X f d \mu = -\int_X (-f) d \mu = - \int_X \|f\|_\infty - \|f\|_\infty -f d\mu = - \int_X \|f\|_\infty -f d\mu + \|f\|_\infty \mu(X)\,,
$$
where $\|f\|_\infty$ denotes the sup norm. Since $\|f\|_\infty -f \geq 0$ we get by definition 3.3
$$
\int_X f d \mu = -\sup \{\mathcal{L}(\|f\|_\infty -f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\,.
$$
Observe that $\mathcal{L}(-f, P)=\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \inf_{A_{i}} -f = -\sum_{i=1}^{n} \mu\left(A_{i}\right) \cdot \sup_{A_{i}} f = -\, \mathcal{U}(f,P)$, hence
$$
\begin{align}
\int_X f d \mu &= - \sup \{-\, \mathcal{U}(f - \|f\|_\infty, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\\
&=\inf \{\, \mathcal{U}(f- \|f\|_\infty, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\,.\\
\end{align}
$$
Since $\sup_{A_{i}} (f- \|f\|_\infty) = \sup_{A_{i}} (f)- \|f\|_\infty$ in the defintion of $\mathcal{U}$ it follows that
$$
\begin{align}
\int_X f d \mu &=\inf \{\, \mathcal{U}(f, P) - \|f\|_\infty \mu(X) \mid P \text { is an } \mathcal{S} \text { -partition}\}+ \|f\|_\infty \mu(X)\\
&=\inf \{\, \mathcal{U}(f, P) \mid P \text { is an } \mathcal{S} \text { -partition}\}\,.\
\end{align}
$$


问:
Why does intrinsic euler angle rotation not equal extrinsic rotation?
I am trying to implement a Rotation matrix showing that intrinsic rotations equal extrinsic rotations, with a simple reordering of matrix arguments and angles.

From wikipedia, "Any extrinsic rotation is equivalent to an intrinsic rotation by the same angles but with inverted order of elemental rotations, and vice versa."

Let $Rx$,$Ry$,$Rz$ be the common Rotation matrices given in the literature and wiki link.

Let $\alpha$, $\beta$,$\gamma$ be the angles by which the Rotations are rotated about the axes.

Suppose I use a Z-X-Z extrinsic convention.
Then I have $R$ = $Rz(\gamma$)$Rx(\beta$)$Rz(\alpha$). Physically, this means I Rotate an object in the order,starting from the rightmost angle to the left and get a resulting Rotation matrix, $R$.

If I then use a z-x'-z'' intrinsic convention, I would swap angles and get $R$ =$Rz(\alpha$)$Rx(\beta$)$Rz(\gamma$). Meaning, on the intrinsic (movable) axis, I start rotating by the $\gamma$ angle about the $Zaxis$ and continue to the left.

My understanding is both Rotations should be mathematically equivalent.

Yet when I try to use concrete values like ($\alpha$,$\beta$,$\gamma$) = (45,30,15) degrees, respectively, I get two different answers depending on the method.

Running the models in R I get for
Extrinsic:

```<code>> round(Rz(D2R(angles[3]))%*%Rx(D2R(angles[2]))%*%Rz(D2R(angles[1])),4)
>       [,1]    [,2]    [,3]
> [1,] 0.5245 -0.8415  0.1294
> [2,] 0.7745  0.4085 -0.4830
> [3,] 0.3536  0.3536  0.8660
</code>```

Intrinsic:

```<code>> round(Rz(D2R(angles[1]))%*%Rx(D2R(angles[2]))%*%Rz(D2R(angles[3])),4)
>       [,1]    [,2]    [,3]
> [1,] 0.5245 -0.7745  0.3536
> [2,] 0.8415  0.4085 -0.3536
> [3,] 0.1294  0.4830  0.8660
</code>```

Note that the Rotations are not equivalent, but similar. If I transpose the Intrinsic result, they are even closer -- the absolute values are exact.

```<code>> round(t(Rz(D2R(angles[1]))%*%Rx(D2R(angles[2]))%*%Rz(D2R(angles[3]))),4)
         [,1]    [,2]   [,3]
  [1,]  0.5245  0.8415 0.1294
  [2,] -0.7745  0.4085 0.4830
  [3,]  0.3536 -0.3536 0.8660
</code>```

Even though the values are correct, now the signs are wrong.
If I lastly convert the angles to negative values, the answers line up perfectly.

```<code>> round(t(Rz(D2R(-angles[1]))%*%Rx(D2R(-angles[2]))%*%Rz(D2R(-angles[3]))),4)
        [,1]    [,2]    [,3]
 [1,] 0.5245 -0.8415  0.1294
 [2,] 0.7745  0.4085 -0.4830
 [3,] 0.3536  0.3536  0.8660
</code>```

So, concretely, the Rotation matrices will be equal if I equate

$Rz(\gamma$)$Rx(\beta$)$Rz(\alpha$) = ($Rz(-\alpha$)$Rx(-\beta$)$Rz(-\gamma$))$^T$

Extrinsic being the left side, and intrinsic being the right.
I've looked high and low in the literature and questions here, but i still don't understand why this is the case, and have not found any concrete examples that show this.
I have seen some examples showing to invert or transpose one frame to get the other, but nothing about negative angles.
Can someone explain why this is true?

*D2R in my code is just a function to convert degrees to radians.

edit. I should also add that I"m aware the negative angles alone imply clockwise (lefthand) rotation. I still don't get the equivalence of intrinsic to extrinsic though.

答:
I think the reason you are having difficulty is because you do not know how to represent the $x'$ or $z''$ rotations in matrix notation.

A single rotation matrix takes a set of $(x,y,z)$ coordinates for one point $P$ in a particular fixed coordinate system and gives you a set of $(x,y,z)$ coordinates for a (usually different) point $P'$ in the same fixed coordinate system, namely the point $P'$ to which the rotation would move an object at point $P$.
In some applications the fixed coordinate system is called the "world" coordinates.

The rotation matrices you can obtain by plugging in various angles $\theta$ and $\phi$ in $R_x(\theta)$ and $R_z(\phi)$ will always represent rotations around the "world" $x$ and $z$ axes, never around the "body" axes $x'$ or $z''$ except in the cases where those axes happen to coincide with the world axes
(which happens when the previous intrinsic rotation angles are zero).

In order to perform a rotation around the $x'$ axis in general you need a way to perform a rotation around an arbitrary axis in the $x,y$ plane.
One way to do this is to rotate the arbitrary axis back onto the world $x$ axis (which you can do via an $R_z$ matrix), rotate things around the world $x$ axis (via an $R_x$ matrix), and then undo the first rotation (via another $R_z$ matrix).
You can multiply these matrices together to get a single matrix that represents the rotation around the desired axis, although the form of that matrix is (typically) not equal to any $R_x,$ $R_y,$ or $R_z$ matrix.

Likewise to rotate around the $z''$ axis, you can rotate the $z''$ axis back to the $z$ axis, perform your rotation there, and then undo the first rotation.

A more detailed explanation of this, showing how the complete matrix multiplication for the intrinsic rotations produces the same composed rotation matrix as the matrix multiplication for the extrinsic rotations, is given in this answer.

By the way, while you can get the inverse of a world-axis rotation by changing the sign of the angle (for example, from $R_z(\alpha)$ to $R_z(-\alpha)$),
you can also get an inverse of a rotation by transposing the rotation matrix.
That is,
$$ R_z(-\alpha) = R_z(\alpha)^T. $$

Now observe that
\begin{align}
(R_z(-\alpha)R_x(-\beta)R_z(-\gamma))^T
&= (R_z(-\gamma))^T (R_x(-\beta))^T (R_z(-\alpha))^T \\
&= R_z(\gamma)R_x(\beta)R_z(\alpha),
\end{align}

which is the identity formula you confirmed at the end of your question.


问:
Volume of polygon-based pyramid
I read in a book a few months ago that the volume of a (solid) pyramid with a base that is ANY polygon (I'm not sure if it mentioned it being regular or not) is equal to
$$\frac{1}{3}\times A\times h$$
where $A$ is the area of the base (i.e. of the polygon) and $h$ is the height of the pyramid.

This seems to be true in many cases, such as when the base is a square, a triangle and when the base is a circle (ie the pyramid becomes a cone).

My question is, how can we prove this? I simply have no idea.

Thank you for your help.

答:
Any polygon can be decomposed into triangles, giving $A = A_1 + \cdots + A_n$.

Thus,
$$\begin{align}
V &= V_1 + \cdots + V_n\\
&=\frac13A_1h + \cdots + \frac13A_nh\\
&=\frac13(A_1 + \cdots + A_n)h\\
&=\frac13Ah.
\end{align}$$

For the base case, you can decompose a prism like so. To see that the three pyramids have the same volume, note that the left and middle pyramids share a blue base and have the same height; similarly, the middle and right pyramids share a red base and have the same height.

Title: Volume of polygon-based pyramid


问:
Oxford Entrance Exam MAT 1998: Solution verifcaition
I need confirmation for the solution for part (d) of the following question from the 1998 MAT Oxford Exam:

My formula is $W_n= 2 r_{n-3} +r_{n-2}$. Is this correct for part (d)? If not please just say no as I want to go over my work and try to figure it out for myself.

答:
seems right

enumerate   $BW[\cdots]W, \;\; W[\cdots]WB, \;\; W[\cdots]W$


问:
What is the meaning of the parentheses in $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$?
I am studying homomorphisms is groups and i saw a theorem saying:

For $g$ in a group $G$, the cosets $gH$ and $Hg$ are the same, and collapsed onto the single element $\phi(g)$ by $\phi$. That is, $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$.

I am confused about the parentheses here. I think $\{\phi(g)\}$ denotes the set, formed by the elements of the mapping, but i am not sure what $[\ ]$ represent. Can anyone tell how should i interpret those parentheses? Or explain the theorem in other words?

Thank you

答:
$$Φ^{-1}\left[\{Φ(g)\}\right]=gH=Hg\tag{1}$$

$(1)$ simply denotes the preimage (the inverse mapping) of the image of the "set": $\{\Phi(g)\} \subseteq G/H$ under $\Phi$, assuming in this case we have that $H$ is normal. Since the "object" being mapped by the homomorphism $\Phi$ is technically a set, instead of enclosing the set in parentheses, as we would for an "element", it's using square brackets, but it would mean the same to write $Φ^{-1}\left(\{Φ(g)\}\right)=gH=Hg$

Added: I agree with Ted Shifrin that your text is being excessively pedantic (pedantic: overly concerned with formal rules and trivial points of learning!)

We more typically refer $\Phi (g)$ as an element in $G/H$, and could express the preimage of an element $\Phi(g) \in H/G$ by simply using:  $$Φ^{-1}[Φ(g)]=gH=Hg \;\;\;\text{ or}\;\;\; \Phi^{-1}\left(\Phi(g)\right) = gH = Hg\tag{2}$$


问:
The set of all subsequential limits of a sequence is closed.

Let $X$ be a metric space and $\{x_n\}$ be a sequence in $X$. Let $E$ be the set of all subsequential limits of $\{x_n\}$. Then $E$ is closed.

I have a idea of this proof.
Assume $E$ is a infinite set.
Let $p\in X-E$. Then $\exists r>0$ such that $B(p,r)\cap E=\emptyset$. The reason is if $p\in \bar E$, then there is some subsequential limit $y$ close to $p$, and there exists a subsequence $x_{n_k}$ converges to $y$. So, $d(x_{n_k},p)<d(x_{n_k},y)+d(y,p)<\varepsilon$ which is a contradiction.

So, $B(p,r)\subset X-E$ and $E$ is closed.

答:
Your idea is very nice, but perhaps your proof could be a little more precise. Note, however, that there is no reason to assume that $E$ is an infinite set - the proof also works for finite $E$.

Here is a suggestion.

It should be clear that a point $\xi \in X$ is the limit of some subsequence of $(x_n)$ if and only if each open neigbhorhood of $\xi$ contains infinitely many $x_n$.

Consider $p \in \overline E$ and let $U$ be an open neigbhorhood of $p$. Then $U$ contains some $\xi \in E$. But $U$ is an open neigbhorhood of $\xi$ and thus contains infinitely many $x_n$.