问:
Assuming the matrix A has got an inverse,prove by induction that $(A^k)^{-1}=(A^{-1})^k$
I started off with the normal procedure:

$(A^k)^{-1}=(A^{-1})^k$, let's try with a base case of k=1, this gives:
$(A^1)^{-1}=(A^{-1})^1$ which can be simplified to $A^{-1}=(A^{-1})$. So it is true for k=1.
Now I assume it is true for k=p and so I get:
$(A^p)^{-1}=(A^{-1})^p$, and if it is true for k=p then it is also true for k=p+1:
$(A^{p+1})^{-1}+(A^{-1})^p=(A^{-1})^{p+1}$ But this is clearly not true, so I have for sure made an error because if the quantities from the start are equal, LHS can't have an extra factor of $(A^{-1})^p$ So there is a big flaw in my setup.

答:
You want to show that $(A^{p+1})^{-1}=(A^{-1})^{p+1}$. In other words, $(A^{-1})^{p+1}$ is the inverse of $A^{p+1}$, or equivalently, $A^{p+1}(A^{-1})^{p+1}=I$ where $I$ denotes the identity matrix. But
$$
A^{p+1}(A^{-1})^{p+1}=A\times A^p\times(A^{-1})^p\times A^{-1}=A\times I\times A^{-1}=I,
$$
hence it is true for $k=p+1$.


问:
Likelihood of of a sum of densities
I am trying to find the likelihood, and then the log likelihood for the following function:
$f(x)= cf_1(x) + (1-c)f_2(x)$. I am stuck on how to find this likelihood because the two densities are being added. Any tips or hints to get started would be greatly appreciated.

答:
You're working with a mixture model with $c$ and $1 - c$ being the mixing proportions. I'll call them $\pi_1$ and $\pi_2$ instead. Your likelihood function is exactly what you think it would be. Assuming independent samples $x_1, \dots, x_N$, it's
$$
L = \prod_{n=1}^N (\pi_1 f_1(x_n) + \pi_2 f_2(x_n))
$$
and the log-likelihood is
$$
\log L = \sum_{n=1}^N \log(\pi_1 f_1(x_n) + \pi_2 f_2(x_n)).
$$
At this point you'd typically rely on the EM algorithm for maximum likelihood estimation of the parameters in $f_1$, $f_2$, and for $\pi_1$, $\pi_2$. For that you might look for stationary points of
$$
\sum_{n=1}^N \log(\pi_1 f_1(x_n) + \pi_2 f_2(x_n)) + \lambda(\pi_1 + \pi_2)
$$
where the Lagrange multiplier $\lambda$ is introduced to represent the constraint $\pi_1 + \pi_2 = 1$.

ContentLicense: CC BY-SA 4.0

Title: Likelihood of of a sum of densities


问:
Genus of a curve
My question is:

How to show that for all $g>0$ there exists an algebraic curve of genus $g$?
The intuition of this is simple, but I don't know how to show it.

答:
Here is a perspective from working over the field of complex numbers. I'll assume you want to talk about nonsingular curves, in which case the genus unambiguously means the geometric genus $h^0(X,\Omega^1)$, i.e. dimension of the global sections of the holomorphic cotangent bundle. Let $X$ denote a Riemann surface. It is well known that the geometric genus of $X$ defined as above equals the topological genus of the underlying real surface.

Next, using Kodaira's Embedding Theorem (or simply any other way to prove that a Riemann surface holomorphically embeds in $\Bbb{P}^n$), you can holomorphically embed $X$ as an analytic subvariety of some $\Bbb{P}^n$ and by Chow's Theorem this realizes the image of $X$ as an algebraic variety in $\Bbb{P}^n$. The genus of this variety is then $g$.

Of course, a better construction in some sense is given by using hyperelliptic curves as mentioned in Somos' answer.

Title: Genus of a curve


问:
Why is the function differentiable in the point $(0,0)$?
I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.

Considering my function: $f(x, y) = x^3 - 3xy^2$

The partial derivative of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partial derivative of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$

Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy  \end{pmatrix} $$

How can I show that my function is point-wise differentiable in $(0,0)$?

答: Since the partial derivatives exist and are continuous at every point, then the function is differentiable at every point.</p>


问:
Is there an example of a non-noetherian one-dimensional UFD?
Or in the contrapositive form

Is every one-dimensional UFD noetherian?

I know how to construct a non-noetherian UFD (polynomials in infinite number of variables over a field) and I know that it is even possible to construct a finite dimensional non-notherian UFD from here, but this example is 3-dimensional.

Since a noetherian one-dimensional UFD is a PID and vice versa, the question could be also rephrased as

Is every one-dimensional UFD a PID?

I guess  the answer is negative, since it is a nice and elegant statement that have never heard about.

答:
Every one-dimensional UFD is a PID (hence noetherian).

Let $\mathfrak p$ be a non-zero prime ideal of $R$ and $a\in\mathfrak p$, $a\neq 0$. Then $a$ can be written as a product of prime elements, so $\mathfrak p$ contains a prime element, say $p$. Then $(p)\subseteq\mathfrak p$ and since the dimension of $R$ is $1$ we get $(p)=\mathfrak p$. This shows that all prime ideals of $R$ are principal, and therefore $R$ is a PID.

ContentLicense: CC BY-SA 3.0


问:
How to write the equation of an off-center circle in terms of r (parameterization)
I've looked around online for a while and I can't seem to find a question that properly captures what I'm talking about (or answers that really help me).

For a homework problem, I have to find the area bounded by $x^{2} + y^{2} = 4$ and $(x-1)^{2} + (y-1)^2 = 1$ using double integrals in polar form. I can easily parameterize the first equation as just $r = 2$ using simple algebra.

However, for the second equation, I get stuck quickly. I can first expand the equation to get $x^{2} -2x + 1 + y^{2} -2y + 1 = 1$. I can simplify to get $x^2 + y^2 + 1 = 2(x + y)$. From here I can substitute the polar equivalents of x and y, to get

$r^2 + 1 = 2\big(rcos(\theta) + rsin(\theta)\big)$

From here I have no idea what to do. The math to get here seems simple enough, yet it's somehow wrong.

Once I have the second equation in terms of $r$, I'd imagine I could set the two equations equal to each other, and find at what angles they are equal (giving me the range of $\theta$ over which I need to integrate), then I'd have the second integral for $r$ to be over the bounds of the two equations. But then, I don't know what function I'd be integrating over that specific range.

Thanks for any help. I really am at a loss as to where to turn

答:
Introduce two different polar coordinates (let's call them $s$ and $\phi$, just so that we don't confuse them with $r$ and $\theta$), where:

$s$ is the length of a vector pointing from the center of the circle
to some point in space, and

$\phi$ is the angle that vector makes
with the positive $x$-axis.

Then the conversion between these new coordinates and the usual rectangular coordinates is:
$$ x = 1 + s \cos \phi \\ y = 1 + s\sin \phi$$
and the equation of the circle is $s = 1$.


答:
Besides from the solution proposed by @SonGohan, you can prove it by the definition.

The candidate to be the derivative is given by $L(x,y) = (3x^{2} - 3y^{2},-6xy)$.

Now you can prove that
\begin{align*}
\lim_{(x,y)\to(0,0)}\frac{|f(x,y) - f(0,0) - L(x,y)((x,y) - (0,0))^{T}|}{\|(x,y)\|} = 0
\end{align*}

Indeed, this is the case.

To begin with, notice the numerator can be written as
\begin{align*}
x^{3} - 3xy^{2} - (3x^{3} - 3xy^{2} - 6xy^{2}) = -2x^{3} - 6xy^{2}
\end{align*}

Thence we conclude that the proposed limit equals zero indeed.

Such claim results from the convenient application of the squeeze theorem.

To comprehend this, let us notice that

\begin{align*}
0 < x^{2} \leq x^{2} +y^{2} & \Rightarrow 0 < |x| \leq \sqrt{x^{2} + y^{2}}\\\\
& \Rightarrow 0 < \frac{|x|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
& \Rightarrow 0 < \frac{|x|^{3}}{\sqrt{x^{2}+y^{2}}} \leq |x|^{2}\\\\
& \Rightarrow 0 < \left|\frac{x^{3}}{\sqrt{x^{2}+y^{2}}}\right| \leq |x^{2}|
\end{align*}

Similarly, we have that
\begin{align*}
0 < y^{2} \leq x^{2} + y^{2} & \Rightarrow 0 < |y| \leq \sqrt{x^{2}+y^{2}}\\\\
& \Rightarrow 0 < \frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
& \Rightarrow 0 < \frac{|xy^{2}|}{\sqrt{x^{2}+y^{2}}} \leq |xy|\\\\
& \Rightarrow 0 < \left|\frac{xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq |xy|
\end{align*}

Finally, since we also have
\begin{align*}
0 < \left|\frac{2x^{3} + 6xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq \left|\frac{2x^{3}}{\sqrt{x^{2}+y^{2}}}\right| + \left|\frac{6xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq |2x^{2}| + |6xy|
\end{align*}

we are done.

Hopefully this helps!

Title: Why is the function differentiable in the point $(0,0)$?

Body:
I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.

Considering my function: $f(x, y) = x^3 - 3xy^2$

The partial derivative of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partial derivative of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$

Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy  \end{pmatrix} $$

How can I show that my function is point-wise differentiable in $(0,0)$?


问:
If $R$ is reduced residue system then $\exists! b\in R$ such that $a\equiv b\pmod n$

If $R$ is a reduced residue system modulo $n$ and $\gcd(a,n)=1$ then $\exists! b\in R$ such that $a\equiv b\pmod n$.

Recall:

$$R=\{a\in C \mid \gcd(a,n)=1\}$$ Where $C$ Is a complete residue system modulo $n$

I didn't quite understand the theorem even when i‘ve flipped the paper so i can see the proof of it, let’s take an example to clarify things.

Example:

Let $C=\{0,1,2,3\}\implies R=\{1,3\}$ but where is $a$ here? If $a=3,n=4$ Then $b$ here is just $3$, and it’s obvious that $3\equiv 3\pmod 4$
, so my question is what is $a$.

答:
I believe the proper formulation is the below:

If $R$ is a reduced residue system modulo $n$ and $a\in\Bbb Z$ such that $\gcd(a,n)=1$, then $\exists b\in R$ such that $a\equiv b\mod n$.

Since we have already used up the variable name $a$ in the above statement, we will write $R=\{x\in C:\gcd(x,n)=1\}$.

The proof of this statement is simple: let $a\equiv y\in C$. Then $\exists k\in \Bbb Z|a=kn+y$. Thus $1=\gcd(a,n)=\gcd(kn+y,n)=\gcd(y,n)$ which implies $y\in R$.

Edit: If $\exists!b$ means "one and only one $b$", we should add that if $a\equiv b_1\equiv b_2\in R$, then $b_1=b_2$ as no two distinct elements of $C$ are congruent.


问:
Finding general solution using Euclid's extended algorithm
I have a problem where I'm supposed to find the general equation to $242x+1870y=66$.

I used Euclid's extended algorithm to find $x=8$ and $y=-1$, but am not sure how to find all possible solutions using this information.

Is there a formula I can plug into? or can someone assist me in finishing this problem?

答:
Hint: Consider the homogenous equation ($242x + 1870y = 0$). Perhaps, the solutions to your equations are exactly the sum of your particular inhomogenous solution and a solution to the homogenous problem, since the equations are linear. (prove it!)

Now note that you can multiply both sides of the homogenous equation ad libitum to generate all solutions.


问:
Radical equation $x- 5\sqrt{x} + 4 = 0$
Solve: Radical equation $x- 5\sqrt{x} + 4 = 0$

I think I understand what I had did wrong. Instead of squaring a product of a number I was squaring the equation.

答:
Hint: The easiest method of solution is to let $x=y^2$, so you have
$$
y^2 - 5y + 4 = 0
$$
Solve for $y$, then substitute into $x=y^2$ to get your final answer.</p>

Title: Radical equation $x- 5\sqrt{x} + 4 = 0$


问:
Fourier Series: question on the period and terms
Wiki defines the Fourier coefficients as:

$a_n = \frac 2 P \int_P s(x) cos(2\pi x \frac n P) dx, b_n = \frac 2 P \int_P s(x) sin(2\pi x \frac n P) dx$

It's my understanding the leading term $\frac 1 P$ is for the average value theorem.

Question : Does anyone know why the $2$ is in the numerator? For average valueI I would expect just $\frac 1 P$.

The Fourier Series is defined:

$s_N(x) = \frac {a_0} 2 + \sum_{n=1}^N (a_n cos(\frac {2 \pi n x} P) + b_n sin(\frac {2 \pi n x} P))$

Question : similarly, why is $\frac {a_0} 2$ divided by $2$?

I ask the first question because on another site the coefficients are different:

Where $L = \frac P 2$, then

$a_n = \frac 1 L \int_{2L} s(x) cos(\pi x \frac n L) dx, b_n = \frac 1 L \int_{2L} s(x) sin(\pi x \frac n L) dx$

Substituting in $P/2$ for $L$ equals the same equation at the beginning of this post:

$a_n = \frac 2 P \int_{P} s(x) cos(2 \pi x \frac n P) dx, b_n = \frac 2 P \int_{P} s(x) sin(2 \pi x \frac n P) dx$

My suspicion is there is something going on with the period that I'm overlooking.  It appears in form where $L = P/2$ the average value form is as expected.

答:
To make things easier, I will explain it for the case $P=2\pi$, since this is the standard case. Furthermore let $\mathcal T_n=\text{span}\{\unicode{x1D7D9},\cos(\cdot),\dots,\cos(n\cdot),\sin(\cdot),\dots,\sin(n\cdot)\}$ (where $\unicode{x1D7D9}(x):=1$) the vector space of all trigonometric functions of maximum degree $n$ and let $\mathcal L^1_{2\pi}:=\{s:\mathbb R\to\mathbb R| 2\pi\,\,\text{ periodic, and absolutly integrable}\}$.
(We can clearly see that $\mathcal T_n\subset \mathcal L^1_{2\pi}$).

Now is important that we have a skalar product on $\mathcal L^1_{2\pi}$ given by
$$\langle f,g\rangle:=\tfrac 1 \pi\int_0^{2\pi}f(x)\cdot g(x)\,dx$$
and the norm
$$||f||:=\sqrt{\langle f,f\rangle}.$$
We need that $\tfrac1 \pi$ to make thinks work good, because we want that the Fourier Sum of degree $n$ is just the orthogonal Projection from $\mathcal L^1_{2\pi}$ to $\mathcal T_n$ . That means we want define an operator $$S_n:\mathcal L^1_{2\pi}\to \mathcal T_n\\f\mapsto\sum_{k=0}^{2n}\langle f,u_k\rangle\cdot u_k$$
with an orthonormal Base $u_k$ of $\mathcal T_n$. (Maybe you have seen this before in linear algebra.)

Furthermore we want to use our Base of $\mathcal T_n$ that I wrote before. Our Base is already Orthogonal (independent of the factor $\tfrac 1 \pi$) but this factor is important that $\sin(kx)$ and $\cos(kx)$ has the norm 1 since:
$$\int_0^{2\pi}(\cos(kx))^2dx=\int_0^{2\pi} (\sin(kx))^2dx=\pi\quad\quad k\geq 1$$
We get now:
$$||\cos(k\cdot)||:=\left(\langle \cos(k\cdot),\cos(k\cdot)\rangle\right)^{\tfrac1 2}=\left(\frac 1 \pi\int_0^{2\pi}\cos(kx)^2dx\right)^{\tfrac 1 2}=\left(\frac \pi \pi \right)^{\tfrac 1 2}=1$$
(Analogously for $\sin(kx)$).
So this is the reason why we have this special factor infront of the integral. We just want that our standard Base has norm 1. Maybe you have already notice that I ignored our Base function $\unicode{x1D7D9}$ all the time, this guy has (sadly) not norm 1:
$$||\unicode{x1D7D9}||=\left(\frac 1\pi\int_0^{2\pi}\unicode{x1D7D9}(x)^2dx\right)^{\tfrac 1 2}=\left(\frac 1 \pi \int_0^{2\pi}dx\right)^{\tfrac 1 2}=\sqrt 2$$
So if we want that the last Base function has norm 1 we need to devide it by $\sqrt 2$:
$$||\tfrac {1} {\sqrt{2}} \unicode{x1D7D9}||=\left(\frac 1{\pi}\int_0^{2\pi}(\tfrac {1} {\sqrt{2}}\unicode{x1D7D9}(x))^2dx\right)^{\tfrac 1 2}=\left(\frac 1 {2\pi }\int_0^{2\pi}dx\right)^{\tfrac 1 2}=1$$
So if we want to have our Base now orthonormal we need to change the first Base Vector we now have the new Base :
$\mathcal T_n=\text{span}\{\tfrac 1 {\sqrt 2}\unicode{x1D7D9},\cos(\cdot),\dots,\cos(n\cdot),\sin(\cdot),\dots,\sin(n\cdot)\}$. We can now rewrite $\tfrac {a_0} 2$:
$$\frac {a_0} 2=\frac 1 {2\pi}\int_0^{2\pi} f(x)\underbrace{\cos(0\cdot x)}_{=1=\unicode{x1D7D9}(x)}dx=\frac 1{2\pi}\int_0^{2\pi}f(x)\unicode{x1D7D9}(x)dx=\langle f,\tfrac 1{\sqrt 2} \unicode{x1D7D9}\rangle \tfrac 1 {\sqrt 2}\unicode{x1D7D9}$$
This is the reason why we need to devide $a_0$ by 2. We now have  that the Fourier sum is the Orthogonal Projection:
$$\frac {a_0} 2 +\sum_{k=1}^{n} a_k\cdot \cos(kx)+b_k\cdot\sin(kx)=\\ \langle f,\tfrac 1{\sqrt 2} \unicode{x1D7D9}\rangle \tfrac 1 {\sqrt 2}\unicode{x1D7D9}+\sum_{k=1}^{n}\langle f,\cos(k\cdot)\rangle\, \cos(kx)+\langle f,\sin(k\cdot)\rangle\,\sin(kx)$$
That was all what we wanted, I hope that anwsers your questions.


问:
Summing the trigonometric series $\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)....$
This is a problem of Loney's "Plane trigonometry, part 2"

$$\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)...=\sum_{n=0}^{+\infty}\frac{1}{2^n}\sin[(n+1)a]$$

Here is what I've got so far:

Let $S=\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)_....\tag{1}$

$C=1+\cos(a)+\frac{1}{2}\cos(2a)+\frac{1}{2^2}\cos(3a)+\frac{1}{2^3}\cos(4a)...\tag{2}$

$C+iS=1+[\cos(a)+i\sin(a)]+\frac{1}{2}[\cos(2a)+i\sin(2a)]+\frac{1}{2^2}[\cos(3a)+i\sin(3a)]+...\tag{3}$

$=1+e^{ia}+\frac{1}{2}(e^{ia})^2+\frac{1}{2^2}(e^{ia})^3+...\tag{4}$

I am stuck here. I am trying to find the general formula for $(4)$, $(4)$ resembles a geometric series, but not quite. I have looked into $-\log(1-x)$ with $x=e^{ia}$, but this doesn't work.

If I let $e^{ia}=x$, then I have a power series

$$1+x+\frac{1}{2}x^2+\frac{1}{2^2}x^3+...$$

But I can't come up with the general formula.

The answer at the back of the book says the sum of this series is $\frac{4\sin(a)}{5-4\cos(a)}$.

My guess with this problem is that you will arrive at a formula that is the sum of $C$ and $S$, then by equating the real and imaginary part, you will be able to arrive at the sum of this series.

答:
You're almost there, but you shouldn't add $1$ in the cosine summation. You can rather normalize the power of $1/2$:
\begin{align}
S&=\dfrac{1}{2}\sin(a)+\dfrac{1}{2^2}\sin(2a)+\dots+\dfrac{1}{2^n}\sin(na)+\dotsb
\\
C&=\dfrac{1}{2}\cos(a)+\dfrac{1}{2^2}\cos(2a)+\dots+\dfrac{1}{2^n}\cos(na)+\dotsb
\end{align}
Then your series is $2S$. You have
$$
C+iS=\sum_{n\ge1}\Bigl(\frac{e^{ia}}{2}\Bigr)^{\!n}=\frac{e^{ia}}{2}\frac{1}{1-e^{ia}/2}=\frac{e^{ia}}{2-e^{ia}}
$$
You can now rationalize, so the sum is
$$
\frac{e^{ia}(2-e^{-ia})}{(2-e^{ia})(2-e^{-ia})}
$$
The denominator is $5-4\cos a$, while the numerator is
$$
2e^{ia}-1=2\cos a+2i\sin a-1
$$
so twice the imaginary part is
$$
\frac{4\sin a}{5-4\cos a}
$$


问:
Maximal element of a poset
I am currently studying Zorn's lemma and its my understanding that the definition of a maximal element is a primitive to the lemma.

From my understanding a maximal element $M$ of a partially ordered set  $(S,<)$  is an element $M \in S$ where, if $M \leq x $ for some $ x \in S$, then $x=m$.

This easilty illustrated when the set is {∅, {1}, {2}, {3}, {1, 2}}. Then the maximal elements are {3} and {1,2}.

But what if the partially ordered set i am considering is $(S,R)$.
So for example let S=P({a,b,c}) which is the power set of {a,b,c} and the relation $\subseteq$.
Now whats is the maximal element in this set?

Thank you in advance.

答:
In general posets a maximum element $M$ of $S$ is one such that $$\forall x \in S: x \le M$$

(all elements lie under $M$)
and a maximal element $M$ of $S$ is one such that

$$\forall x \in S: (M \le x) \to (x=M)$$

(nothing is properly larger than $M$).

A maximum element is always a maximal element, and is always unique.

As you see in your fist example where $\{3\}$ and $\{1,2\}$ are both maximal, in general, we can have many non-comparable maximal elements, and this is typically the kind of poset Zorn is applied in.

In $\mathscr{P}(\{1,2,3\})$ we have a maximum element $\{1,2,3\}$. This is also the only maximal element.

Title: Maximal element of a poset


问:
Prove $F_n=(-1)^{n+1}F_{-n}$ without induction
Consider the Fibonacci sequence defined by
$$F_{n+2}=F_{n+1}+F_n,~~~~~~F_1=F_2=1$$
I can prove via induction that $F_n=(-1)^{n+1}F_{-n}$, but how can it be proven without using induction?

Thank you for your help.

答: Instead define $F_{-k}:=(-1)^{k+1}F_k$ for $k>0$, then verify the recursion relation still works viz.$$F_{-k}-F_{-k-1}-F_{-k-2}=(-1)^{k+1}(F_k+F_{k+1}-F_{k+2})=0.$$


答:
Binets formula
$$F_n = \frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}},$$
where $\phi$ denotes the golden ratio, is an explicit formula for the $n$-th Fibonacci number $F_n$. We have to prove that
$$\underbrace{\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}}_{F_n}=(-1)^{n+1}\cdot\underbrace{\frac{\phi^{-n}-(-\phi)^{n}}{\sqrt{5}}}_{F_{-n}}.$$
We can rewrite $(-1)^{n+1}\phi^{-n}$ to $-(-\phi)^{n}.$ Also $(-1)^{n+1}(-\phi)^{-n}$ is equal to $-\phi^n$. Putting all of this together yields
$$(-1)^{n+1}\cdot{\frac{\phi^{-n}-(-\phi)^{n}}{\sqrt{5}}} = \frac{-(-\phi)^{-n}-[-\phi^{n}]}{\sqrt{5}}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} = F_n$$

Title: Prove $F_n=(-1)^{n+1}F_{-n}$ without induction

Body:
Consider the Fibonacci sequence defined by
$$F_{n+2}=F_{n+1}+F_n,~~~~~~F_1=F_2=1$$
I can prove via induction that $F_n=(-1)^{n+1}F_{-n}$, but how can it be proven without using induction?

Thank you for your help.


问:
Find all the cosets of $A=[a^3]$ in a cyclic group $B=[a]$ if we know that order of $B$ is $12.$
I was trying to solve the following question:

Find all the cosets of $A=[a^3]$ in a cyclic group $B=[a]$ if we know that order of $B$ is $12.$

But I am confused about $a^3$. Could you point me in the right direction for the solution?

Similarly, there is a subproblem:

Find all the cosets of $H=[3]$ in the modular additive group $\Bbb Z_{12}$

答:
The cosets are $[a^3], a[a^3], a^2[a^3], a^3[a^3]$ since $a^4[a^3]=\{a^4b\mid b\in [a^3]\}=[a^3]$ as $a^{12}=e$ (and the quotient of a cyclic group is cyclic).

For the subproblem, establish the isomorphism $\varphi$ given by $a\mapsto [1]_{12}$.


问:
Inequalities proofs in Spivak's Calculus
From Chapter 1, page 9 of Spivak's Calculus, 3rd edition:

In fact, it is convenient to consider the collection of all positive numbers, denoted by $P$, as the basic concept, and state all properties in terms of $P$:

(P10) (Trichotomy law) For every number $a$, one and only one of the following holds:
(i) $a=0$, (ii) $a$ is in the collection $P$, (iii) $-a$ is in the collection $P$.

(P11) (Closure under addition) If $a$ and $b$ are in $P$, then $a+b$ is in $P$.

(P12) (Closure under multiplication) If $a$ and $b$ are in $P$, then $a\cdot b$ is in $P$.

This is question 8 on page 15:

Although all the basic properties of inequalities were stated in terms of the collection $P$ of all positive numbers, and < was defined in terms of $P$, this procedure can be reversed. Suppose that P10-P12 are replaced by

(P'10) For any numbers $a$ and $b$ one, and only one, of the following holds: (i) $a=b$, (ii) $a<b$, (iii) $b<a$.

(P'11) For any numbers $a$, $b$, and $c$, if $a<b$ and $b<c$, then $a<c$.

(P'12) For any numbers $a$, $b$, and $c$, if $a<b$, then $a+c<b+c$.

(P'13) For any numbers $a$, $b$, and $c$, if $a<b$ and $0<c$, then $ac<bc$.

Show that P10-P12 can then be deduced as theorems.

And here's the solution as in the manual:

Two applications of P'12 show that if $a<b$ and $c<d$, then $a+c<b+c<b+d$, so $a+c<b+d$ by P'11. In particular, if $0<b$ and $0<d$, then $0<b+d$, which proves P11. It follows, in addition, that if $a<0$, then $-a>0$; for if $-a<0$ were true, then $0=a+(-a)<0$, contradicting P'10. Consequently, any number $a$ satisfies precisely one of the conditions $a=0$, $a>0$, $a<0$, the last being equivalent to $-a>0$. This proves P10. Finally, P'13 shows that if $0<a$ and $0<c$, then $0<ac$, which proves P12.

My question is, why doesn't P10 immediately follow from P'10, by substituting $0$ for $b$ ?

答:
I had the same thought about this problem. It's easy to be confused about which properties still hold.

The key point is that here, we aren't allowed to use the equivalence of $a<0$ and $-a>0$. This equivalence is actually what Spivak's proving.

See if you can follow how his proof shows this.

Once he has this, he just sticks it into 'P10 to complete the proof.

As an aside, his explanation seems to skip a possibility: he rules out $-a <0$ but not $-a=0$.

This can be ruled out by assuming it's true and then adding $-a$ to both sides of $a<0$, per 'P12. (It's the same argument he uses to rule out $-a <0$).

Finally, note that Spivak's argument for why $a< 0$ implies $-a > 0$ is perhaps needlessly circuitous.

An alternate, more direct approach:
Begin as Spivak does with $a < 0$.

Add $-a$ to both sides per 'P12.


问:
What is infinity divided by infinity?
This should be a simple question but I just want to make sure.

I know $\infty/\infty$ is undefined. However, if we have 2 equal infinities divided by each other, would it be 1?

And if we have an infinity divided by another half-as-big infinity, would we get 2? For example $\frac{1+1+1+\ldots}{2+2+2+\ldots}=\frac12$?

答:
To elaborate a bit on the comment by sos440, there are at least two approaches to the issue of infinity/infinity in calculus:

(1) $\frac \infty\infty$ as an indeterminate form.  In this approach, one is interested in the asymptotic behavior of the ratio of two expressions, which are both "increasing without bound" as their common parameter "tends" to its limiting values;

(2) in an enriched number system containing both infinite numbers and infinitesimals, such as the hyperreals, one can avoid discussing things like indeterminate forms and tending, and treat the question purely algebraically: for example, if $H$ and $K$ are both infinite numbers, then the ratio $\frac H K$ can be infinitesimal, infinite, or finite appreciable, depending on the relative size of $H$ and $K$.

One advantage of approach (2) is that it allows one to discuss indeterminate forms in concrete fashion and distinguish several cases depending on the nature of numerator and denominator: infinitesimal, infinite, or appreciable finite, before discussing the technical notion of limit which tends to be confusing to beginners.

Note 1 (in response to user Xitcod13): Here an infinitesimal number, in a number system $E$ extending $\mathbb{R}$, is a number smaller than every positive real $r\in\mathbb{R}$.  An appreciable number is a number bigger in absolute value than some positive real.  A number is finite if it is smaller in absolute value than some positive real.

Title: What is infinity divided by infinity?


问:
Proving that $\frac{\text{d}\vec{v}}{\text{dt}}=\frac{\text{d}|\vec{v}|}{\text{dt}}\hat{v}+\frac{\text{d}\hat{v}}{\text{dt}}|\vec{v}|$.
We were taught the following equation on a physics lecture:

$$\frac{\text{d}\vec{v}}{\text{d}t}=\frac{\text{d}|\vec{v}|}{\text{d}t}\hat{v}+\frac{\text{d}\hat{v}}{\text{d}t}|\vec{v}|$$

where $\vec{v}$ is a velocity vector dependent on time $t$. I was wondering how the equation could be proven.

答:
Just write

$$
\vec{v} = |\vec{v}| \hat{v} \tag{1}
$$

that is to say: vector $\vec{v}$ is of magnitude $|\vec{v}|$ and pointing in the direction $\hat{u}$. After that, apply the product rule on Eq. (1)


答:
If $\hat{v}$ is the unit vector with the same direction as $v$, then this should be true by the fact that
$$
\vec{v}=|\vec{v}|\hat{v}
$$
and by using the product rule for derivative.

Title: Proving that $\frac{\text{d}\vec{v}}{\text{dt}}=\frac{\text{d}|\vec{v}|}{\text{dt}}\hat{v}+\frac{\text{d}\hat{v}}{\text{dt}}|\vec{v}|$.

Body:
We were taught the following equation on a physics lecture:

$$\frac{\text{d}\vec{v}}{\text{d}t}=\frac{\text{d}|\vec{v}|}{\text{d}t}\hat{v}+\frac{\text{d}\hat{v}}{\text{d}t}|\vec{v}|$$

where $\vec{v}$ is a velocity vector dependent on time $t$. I was wondering how the equation could be proven.


答:
Set $y^2=x$ so that the equation becomes $y^2-5y+4=0$.

This has solutions $y=1, y=4$ by any standard method.

So $x=y^2=1 \text{ or } 16$.

Title: Radical equation $x- 5\sqrt{x} + 4 = 0$

Body:
Solve: Radical equation $x- 5\sqrt{x} + 4 = 0$

I think I understand what I had did wrong. Instead of squaring a product of a number I was squaring the equation.


答: For scalar $\phi$ and vector $A$,$$\frac{\text{d}}{\text{d}t}(\phi A_i)=\frac{\text{d}\phi}{\text{d}t}A_i+\frac{\text{d}A_i}{\text{d}t}\phi\implies\frac{\text{d}}{\text{d}t}(\phi A)=\frac{\text{d}\phi}{\text{d}t}A+\frac{\text{d}A}{\text{d}t}\phi,$$by contraction with the standard basis $e^i$. Now take $\phi:=|v|,\,A:=\hat{v}$.


问:
Bound on Truncated Alternating Harmonic Series
I'm trying to prove the following inequality

$$\sum _{n=1}^{2N}\frac{(-1)^n}{n}+\log (2)<\frac{1}{4N+1}.$$

Most of the traditional inequalities I've seen for harmonic numbers aren't tight enough to prove this, unless I'm doing something wrong.

Body:
As the taylor series of $\ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}x^n}{n}$, we have

$$\ln(2)=\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n}$$

then
\begin{align}
\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n} + \ln(2) &= \sum_{n=2N+1}^{+\infty} \frac{(-1)^{n-1}}{n} \\
&= \sum_{k=N}^{+\infty} \left( \frac{(-1)^{2k}}{2k+1}+\frac{(-1)^{2k+1}}{2k+2} \right) \\
&= \sum_{k=N}^{+\infty} \left( \frac{1}{2k+1}-\frac{1}{2k+2} \right) \\
&= \sum_{k=N}^{+\infty} \frac{1}{(2k+1)(2k+2)} \\
&= \sum_{k=N}^{+\infty} \frac{4}{(4k+2)(4k+4)} \\
&< \sum_{k=N}^{+\infty} \frac{4}{(4k+1)(4k+5)} =\sum_{k=N}^{+\infty} \left( \frac{1}{4k+1}-\frac{1}{4(k+1)+1}  \right) = \frac{1}{4N+1}\\
\end{align}


问:
Prove using the definition of θ.
How I can prove that:
$$
((n^2) − 3*n + 2) ∈ θ(n^2)
$$
using the definition of $θ$ ?

θ : an asymptotic notation in data structure algorithm.

答:
One way is to find, on $n \in [1,\infty)$, the minimum and maximum of
$$  \frac{n^2 -3n + 2}{n^2}  \text{.}  $$
Let $\ell$ be the global minimum and $u$ be the global maximum.  Then you know
$$  \ell n^2 \leq n^2 - 3n + 2 \leq u n^2  \text{,}  $$
where $\ell$ and $u$ are constants (independent of $n$).  This establishes the result you want.

In this particular case, there is a critical point giving the lower bound, $-1/8$, and then you can take a limit to get the upper bound, $1$.

Of course, since $\Theta$ allows the bounds to hold on any interval of the form $[n_0, \infty)$.  You can choose to optimize on a smaller half-line.  If, say, you optimize on $[5,\infty)$, the rational function above is strictly monotonically increasing, so you can evaluate the ratio at $5$ to get a lower bound and take the limit as $n \rightarrow \infty$ (which is very, very easy for rational functions and is taught prior to calculus when graphing rational functions) to get an upper bound.

Title: Prove using the definition of θ.


问:
Limit with a set proof
Let $x_n = n^{-1}$ and let $X \subset \mathbb{N}$ so that $\sum_{n \in X} x_n$ is convergent. Define $p(m) = 2^m$ and let $Y = \{x \in X | p(m) \leq x < p(m+1)\}$. Prove that $\liminf_{m\to\infty} \frac{cardinal(Y)}{p(m)}$ converges to $0$.

This is what I have so far:

Assume $\sum_{n \in X} x_n$ converges. Then $\lim_{n\to\infty} x_n = 0$.By definition of the limit, for all $\epsilon > 0$, there is a $N \in \mathbb{N}$ such that $|x_n| = |n^{-1}| = |\frac{1}{n}| < \epsilon$ for $n \geq N$

We need that $\liminf_{m\to\infty} \frac{cardinal(Y)}{p(m)} = 0$. We require an $N_2 \in \mathbb{N}$ so that $|\frac{cardinal(Y)}{p(m)}| = \frac{cardinal(Y)}{p(m)} < \epsilon_2$ for $n \geq N_2$.

I have no idea how to solve this problem particularly because it spans and requires arguments from two topics: sets and limits. Any assistance is appreciated.

Edit: I tried using the comparison test as an argument but to no avail.

答:
Reach into your bag of tricks of what looks familiar.

We know $\frac 11 + \frac 12 +\frac 13 +.... = \sum_{n\in \mathbb N} x_n$ diverges.

And we prove $\sum_{n\in \mathbb N}x_n$ diverges because $\frac 1{2^m} + \frac 1{2^m+1} + ..... + \frac 1{2^{m+1} - 1} > \frac 1{2^{m+1}} + \frac 1{2^{m+1}} + ..... + \frac 1{2^{m+1}} = 2^m\times \frac 1{2^{m+1}} = \frac 12$ so

$\sum_{k=1}^{\infty} \frac 1k = \sum_{m=0}^{\infty} (\sum_{k=2^{m}}^{2^{m+1}-1} \frac 1k )> \sum_{m=0}^\infty \frac 12 = \infty$.

Now to use the notation of the problem If we let $W(m) = \{2^m,2^m + 1,..... 2^{m+1}-1\}=\{n\in \mathbb N|2^n\le n< 2^{m+1}\}= \{w\in \mathbb N: p(m) \le w < p(m)\}\supseteq \{x\in X|p(m) \le x < p(m)\} = Y_m$

So we have for each $n\in Y(m)$ then $2^m = p(m)\le n < p(m+1)=2^{m+1}$ then $x_n = \frac 1n \ge \frac 1{p(m+1)}$ and $\sum_{n\in Y(m)} x_n > \frac 1{p(m+1)}\times |Y_m|$

And $\sum_{w\in X}x_w=\sum_{m=0}^{\infty}(\sum_{w\in Y(m)} x_w > \sum_{m=0} \frac {|Y_m|}{p(m+1)}$

But this converges so $\lim_{m\to \infty} \frac {|Y_m|}{p(m+1)}= \lim_{m\to \infty} \frac {|Y_m|}{2p(m)}= 0$.  So $\lim_{m\to \infty}\frac {|Y_m|}{p(m)} = 0$. and $0\le \lim \inf_{m\to\infty}\frac {|Y_m|}{p(m)} \le \lim_{m\to \infty} \frac {|Y_m|}{p(m)} = 0$

Title: Limit with a set proof


问:
When is a locally Borel set Borel?
Let $X$ be a LCH space. Call $E\subseteq X$ locally Borel if $E\cap K$ is Borel for all compact $K\subseteq X$. Evidently, locally Borel sets form a $\sigma$-algebra and every Borel set is locally Borel.

My question is the following. Suppose $X$ is a group, or more generally there is a Radon measure $\mu$ on $X$ (using, say, the definition in Folland: finite on compact sets, outer regular on all Borel sets, and inner regular on all open sets) such that $\mathrm{supp}\,\mu=X$. Does it follow that every locally Borel set is Borel?

Note some assumption on $X$ like this is necessary: for example, the first uncountable ordinal has subsets which are locally Borel but not Borel. On the other hand, if $X$ is $\sigma$-compact, a set is Borel if and only if it is locally Borel.

答:
Following a now-deleted comment which was on the right track, I'm now pretty certain the following groups are counterexamples:

$\mathbb{R}$ times an uncountable discrete group,

An uncountable restricted direct product of copies of $\mathbb{Q}_p$.

In particular, they each decompose as an uncountable coproduct (i.e. topological disjoint union) of pieces $X_i$ with Borel subsets $E_i$ such that the rank of the sets $E_i$ in the Borel hierarchy grows without countable bound. The union of the sets $E_i$ is then locally Borel but not Borel.

It still may be interesting to figure out what the "correct" assumptions on $X$ are, but I'll consider this particular question answered unless I am wrong in calling these counterexamples.

Title: When is a locally Borel set Borel?


问:
Proving continuity from local uniform convergence

Prove that if $(f_n (x))$ converges locally uniformally to $f$ and all $f_n$ are continuous, then $f$ is continuous.

I know how to prove that uniform convergence implies continuity of $f$, but not local uniform convergence, unless the only difference is, in proving continuity of $f$ at $x$, we restrict ourselves to a neighborhood at which the convergence is uniform and then use the same argument.

Any help would be appreciated.

答: Continuity is a local property. So, if $x_0$ belongs to the domain $D$ of your functions, take a neighborhood $V$ of $x_0$ in $D$ such that $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$ on $V$. The, since uniform convergence preserves continuity, $f|_V$ is continuous. But $V$ is a neighborhood of $x_0$. So, $f$ is continuous at $x_0$.


问:
A polynomial function that is non-constant is injective iff it contains one root that is non-zero, or no roots at all.
I was messing around the other day in desmos, and I began to graph functions which have certain roots, (for instance $x^2-x-1$), and noticed that these functions are not injective. In this case, $\phi$ and $\frac{1-\sqrt5}{2}$, are both $0$ once plugged into this function. So I started thinking, is this true generally? And I came up with this claim:

Claim: A polynomial function that is not constant is injective iff it contains one root that is non-zero (such as $\phi\neq 0$), or has no roots at all.

I do not yet have a proof, but I am working on it. Would this claim be correct? If not, provide a counterexample please. :)

答:
(EDITED)
A nonconstant polynomial with real coefficients is injective on the real numbers  if its derivative has no real roots.

A polynomial with no real roots can never be injective on the real numbers.


问:
About the definition of vertex
Following is the definition of a vertex from $\textit{Elementary Differential Geometry}$ by Pressley

Definition.
A vertex of a curve $\gamma(t)$ in $\mathbb{R^2}$ is a point where its signed curvature $\kappa_s$ has a stationary point, i.e., where $d\kappa_s/dt=0$.

And here is an exercise and solution from the book about how to find number of vertices of a parameterised curve.

Show that the limacon $\gamma(t)=((1+2\cos t)\cos t,(1+2\cos t)\sin t)$ has only two vertices.

Solution.

$\dot\gamma=(\color{red}{-\sin t-2\sin2t},\color{blue}{\cos t+2\cos 2t})$ and $\Vert\dot\gamma\Vert=\color{green}{\sqrt{5+4\cos t}}$, so the angle $\varphi$ between $\dot\gamma$ and the $x$-axis is given by $$\cos\varphi=\frac{\color{red}{-\sin t-2\sin 2t}}{\color{green}{\sqrt{5+4\cos t}}},\sin\varphi=\frac{\color{blue}{\cos t+2\cos 2t}}{\color{green}{\sqrt{5+4\cos t}}}$$
Differentiating the second equation gives
$$\dot\varphi\cos\varphi=\frac{-\sin t(24\cos^2 t+42\cos t+9)}{(5+4\cos t)^{3/2}}\text{ so}$$
$$\dot\varphi=\frac{\sin t(24\cos^2 t+42\cos t+9)}{(5+4\cos t)(\sin t+2 \sin 2t)}=\frac{9+6\cos t}{5+4\cos t}$$
Hence, if $s$ is the arc-length of $\gamma$, $\kappa_s=\frac{d\varphi/dt}{ds/dt}=\frac{9+6\cos t}{(5+4\cos t)^{3/2}}$ , so $\dot\kappa_s=\frac{12\sin t(2+\cos t)}{(5+4\cos t)^{5/2}}$. This vanishes at only two points of the curve, where $t=0$ and $t=\pi$.

Since for signed curvature $\kappa_s$, have $\kappa_s=\vert\kappa\vert$, that $d\kappa_s/dt=0$ iff $d\kappa/dt=0$, is it really necessary to find $\kappa_s$ every time; In general, can we just use $\kappa$, as it's much easy to calculate? Like suggested in this post (defined vertex as where $d\kappa/dt=0$).

For example in above exercise we can just compute $\kappa=\frac{9+6\cos t}{(5+4\cos t)^{3/2}}=\kappa_s$, and find where $d\kappa/dt$ vanishes that gives the same result, which seems to be a much shorter approach than compute the sign curvature. (However, all the examples from the book are calculating $\kappa_s$ instead.) So I'm not sure, based on definition of this book, is this approach correct?

答:
The reason for the signed in the definition is that you want to rule out a point like the origin on $y=x^3$. The signed curvature of this curve is $\kappa_s(x) = \dfrac{6x}{(1+9x^4)^{3/2}}$, and it has no critical point at $x=0$. However, the unsigned curvature $\kappa = |\kappa_s|$ has a minimum at $x=0$ (although it is not differentiable there).

Vertices should correspond to extreme points of curvature, not just any old critical points.

If you do not like the lack of differentiability of $\kappa$ in the example, change to $y=x^5$ instead. The same phenomenon persists, but now $\kappa$ will be differentiable at $x=0$.

Title: About the definition of vertex


问:
If $R$ is reduced residue system then $\exists! b\in R$ such that $a\equiv b\pmod n$

If $R$ is a reduced residue system modulo $n$ and $\gcd(a,n)=1$ then $\exists! b\in R$ such that $a\equiv b\pmod n$.

Recall:

$$R=\{a\in C \mid \gcd(a,n)=1\}$$ Where $C$ Is a complete residue system modulo $n$

I didn't quite understand the theorem even when i‘ve flipped the paper so i can see the proof of it, let’s take an example to clarify things.

Example:

Let $C=\{0,1,2,3\}\implies R=\{1,3\}$ but where is $a$ here? If $a=3,n=4$ Then $b$ here is just $3$, and it’s obvious that $3\equiv 3\pmod 4$
, so my question is what is $a$.

答:
well, as $R \subset C$ and, by definition of complete residue system, then for any $a\in \mathbb Z$ there is a unique $b \in C$ so that $a\equiv b \pmod n$.

So this statement is equivalent to proving, if $a \equiv b \pmod n$ then $\gcd(a,n) = 1 \iff \gcd(b,n) = 1$

And that follows fairly straightforwardly.

$\gcd(a,n) = \gcd(a+kn, n)$ for all integers $k$.

And $a \equiv b \pmod n$ is defined as $n|b-a$ which means that there is an integer $k$ so that $a = b+kn$.   So $\gcd(b,n) = \gcd(a + kn, n) =\gcd(a,n)$.

....

Perhaps a stronger and more useful statement would be:

Lemma:  If $a\equiv b \pmod n$ then $\gcd(a,n) =\gcd(b,n)$.


问:
Can you use "limits" in geometry proofs?
Let's say that I wanted to prove power of a point for a tangent and secant. The diagram would look something like this. Let's say that I am allowed to use the power of a point for a a secant and a secant. Would this proof be valid?

Now, I take the "limit" as $B$ approaches $A$. It would look something like this.

So, in the final diagram, points $A$ , $B$, are in the same place, let's call this place $P$. Since the power of point for secant-secant remained through through this whole transformation until this very exact point, we still have that $(EB)(EA) = (ED)(EC)$ because even a very small nudge away makes this true, so this is still basically true even though the points are in the same place. However, in this case, $EB$ & $EA$ have become the single segment $PE$. Thus our original equation becomes $(PE)^2 = (ED)(EC)$, thus proving power of a point for a secant and a tangent.

答:
With a little bit of rigor, you can do this of course.

You need to define what means $B$ approaches $A$. For example, you can take the angle $\theta=\angle DEB$ and say $\theta\to\theta_{\tau}$, where $\theta_\tau$ is the angle between $CD$ and the tangent from $E$ to the circle (let's call the point $A_\tau=B_\tau=T$).

Next, you want to show that indeed $A\to T$, when $\theta\to \theta_\tau$. With a little bit of geometry involved, you can show that $AT(\theta_\tau-\theta)<k\sqrt{\theta_\tau-\theta}$, where $k$ is some constant. Since $AT\to 0$, then $A\to T$.

Finally, since the distance is a well-behaving function of two points (triangle inequality), $A\to T$ implies $EA\to ET$.

Repeat the same with $B$ and you are done.

Of course, if you are describing your proof to someone informally, you can just say what you said “as A approaches B, secant becomes tangent and bla-bla”. Meaning, the listener can easily fill in the gaps listed above.


问:
Proving that $\exists t \in \mathbb R: ||t u|| = d(0,K)$
Let's consider $\mathbb R^n$ with the usual inner product $\cdot$, norm $||\cdot||$ and distance $d(\cdot,\cdot)$ induced by that inner product.

If $X \subseteq \mathbb R^n$ and $a \in \mathbb R^n$ we can define $d(a,X) := \inf \{d(a,x) : x \in X\}$.

Let $K$ be a Hyperplane, meaning that $\exists p \in \mathbb R^n$, and $\exists S \leq \mathbb R^n$ with $\dim S = n - 1$ such that:

$$K = \{p\}  + S$$

Let $u \in \mathbb R^n$ such that $u$ spans $S^\perp$, so $\{u\}$ is a basis of $S^\perp$.

Then:

$$\exists t \in \mathbb R: ||t u|| = d(0,K)$$

I came up with this by observing some specific cases in $\Bbb R^2$ and $\Bbb R^3$ and this was my attempt of generalizing. It's easy to intuitively understand why this is true in $\Bbb R^2$ and $\Bbb R^3$ but I'm not being able to prove this for any $\Bbb R^n$. So my questions are:

Is this even true for $\Bbb R^n$?

If so, how can I prove this?

答:
Well, $d(0,K) \le \|p\|$ since certainly $p \in K$. Therefore we can take the infimum over $K \cap \overline{B}(0,p)$ without changing anything:
$$d(0,K) = \inf\{\|x\|:x \in K\} = \inf\{\|x\|:x \in K \cap \overline{B}(0,p)\}$$
and the latter is an infimum of a continuous function over a compact set so it is attained. Hence let $x_0 \in K \cap \overline{B}(0,p)$ be such that $d(0,K) = \|x_0\|$.

Now if we define $t := \frac{\|x_0\|}{\|u\|} \in \Bbb{R}$ then
$$\|tu\| = t\|u\| = \frac{\|x_0\|}{\|u\|}\|u\| = \|x_0\| = d(0,K).$$


问:
Is a continuous locally square-integrable martingale uniquely (in distribution) determined by its angle bracket?
Question: Given a real-valued continuous increasing predictable process $A$, is there a real-valued continuous locally square-integrable martingale $M$, such that $\langle M\rangle = A$ ? If so, is this $M$ unique in distribution, that is, if there is an $N$ fulfill the assumption, then $M$ and $N$ has the same law in the path space ?

The angle bracket $\langle M\rangle$ denotes the predictable quadratic variation of $M$, that is, the unique continuous increasing predictable process starting at zero such that $M^2 − \langle M\rangle$ is a local martingale.

Motivation: This question arises when I was reading the book of J. Jacod & A. N. Shiryaev. They define the characteristics of semimartingales in Section II.2. Given a truncation function $h$, the characteristics of a semimartingale $X$ associated with $h$ is a unique triplet $(B,C,\nu)$. So a natural question is: can a triplet $(B,C,\nu)$ for a fixed truncation function $h$ determine a semimartingale uniquely in the sense of distribution? The begining question is a special case and part of this question.

Part of the answer: If the increasing process $A=\{A_t\}_{t\ge0}$ in the question is an absolutely continuous function of $t$ a.s., then the martingale representation theorem asserts that there is a stochastic integral with respect to a Brownian motion such that the angle bracket of this stochastic integral is $A$.

答:
I think you can let $B$ be a Brownian motion and set $M_t = B_{A_t}$.  You need to be a bit careful about the filtrations to make that work, but that can be handled if you look at the proofs about continuous martingales as time changed Brownian motion.  I'm not sure whether or not this is unique in distribution, though.

Edit: This $M$ is not unique.  Let $B_t$ be a Brownian motion, set $\tau := \inf\{t : B_t = -1\}$, and let $A_t := t \wedge \tau$.  The processes $M_t := B_{t \wedge \tau}$ and $N_t := - B_{t \wedge \tau}$ are both continuous locally square integrable martingales with $\langle M \rangle_t = \langle N \rangle_t = t \wedge \tau = A_t$, but $\mathbb{P}(M_t < -1) = 0$ and $\mathbb{P}(N_t < -1) > 0$.


问:
Inductively prove that this sequence of integrals is bounded.
EDIT: I have an attempted solution to this in a post below, it is very long, but still incomplete.

EDIT:Alright, I've pretty much almost finished my solution, but my biggest problem is the 2nd integral, rightmost inequality. I cannot find a way to evaluate it and compare it to the n=1 case for the upper integral, which should give me what I need.

EDIT: still stuck!

EDIT: 10 days later, I still need help with this..

Can someone please help me with this? I'm not even sure where to go from here.

Here is the problem:

Prove that

$$\frac{1}{2}<\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,$$

for any integer $n\ge1$,

And that

$$\frac{1}{2}<\int_{0}^{1}\frac{dx}{\sqrt{4-x+x^{3}}}\le0.52359$$

Now I can do this for the $n=1$ case, as obviously this simply integrates to $\arcsin{\frac{1}{2}}$ giving $\frac{\pi}{6}$, but I'm completely stumped on any doing any further for this question.

All help very appreciated, thank you for your time.

答:
Alright, here is my attempt at a full solution, can anyone tell me if it is correct?

EDIT: Somewhere along the way I realise I have went wrong, which is pointed out in this post.

For the first integral:

Trivially:

$$\frac{1}{\sqrt{1-x^{2n}}}\ge1,    \forall x \in \mathbb{R}, \forall n \in \mathbb{N}$$

From here, we use

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\ge\int_0^\frac{1}{2}dx=\frac{1}{2}$$

$$\implies \frac{1}{2}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}} \forall n\in\mathbb{N}$$

[Note, I have proven the $\le$ relationship here, but I don't know how to also confirm the $<$ relationship between the integral and $\frac{1}{2}$, any help here?]

Now, for $x \in[0,\frac{1}{2}],x^{2(n+1)}\le x^{2n}$

$$\implies \frac{1}{\sqrt{1-x^{2(n+1)}}}\le \frac{1}{\sqrt{1-x^{2n}}}\forall n\in\mathbb{N},x\in[0,\frac{1}{2}]$$

$$\implies\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2(n+1)}}}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\forall n\in\mathbb{N}$$

With this information, we can now prove the statement:

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,\forall n\in\mathbb{N}$$

For $n=1$,

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2(1)}}}=\left[\arcsin{x}\right]_0^\frac{1}{2}=\frac{\pi}{6}\approx0.52359$$

Assume statement true for $n=k$

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2k}}}\le0.52359$$

Now, for $n=k+1$, we know:

$$\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2(k+1)}}}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2k}}}\le0.52359$$

Therefore

$$\frac{1}{2}\le\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,\forall n\in\mathbb{N}$$

Again, I still can't prove the strict inequality on the left, unless I already have and just cannot see it.

Now for the second integral:

For $x\in[0,1]$,$$4\ge 4-x+x^3\ge3$$

As $x\ge x^3$

So we know that $$\sqrt{4}=2\ge\sqrt{4-x+x^3}$$

$$\implies \frac{1}{2}\le\frac{1}{\sqrt{4-x+x^3}}$$

$$\implies\frac{1}{2}=\int_0^\frac{1}{2}\frac{1}{2}dx\le\int_0^1\frac{dx}{\sqrt{4-x+x^3}}$$

Again, I cannot prove strict inequality here, and would greatly appreciate help for that.

Now for the rightmost inequality.

Statement:

$$\frac{1}{2}\left[\int_0^1\frac{dx}{\sqrt{1-x^2}}-\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}\right]=\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}$$

The proof for this is very trivial. The first integral equates to $\frac{\pi}{2}$, the second to $\frac{\pi}{6}$

Moving on, we know that $$\forall x\in[0,1], \sqrt{4-x+x^3}>\sqrt{1-x^2}, always.$$

$$\implies \frac{1}{\sqrt{4-x+x^3}}<\frac{1}{\sqrt{1-x^2}}$$

In typing this, I have just realised I made a huge mistake, and the rest of my solution to this is now invalid.

I was going to go on to say that $$\int_0^1\frac{dx}{\sqrt{4-x+x^3}}\le\frac{1}{2}\left[\int_0^1\frac{1}{\sqrt{4-x+x^3}}-\int_0^\frac{1}{2}\frac{1}{\sqrt{4-x+x^3}}\right]\le\frac{1}{2}\left[\int_0^1\frac{dx}{\sqrt{1-x^2}}-\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}\right]=\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}$$

But this is clearly incorrect. Any ideas, anyone?

Right now I'm trying to toy with the relationship$$\frac{1}{3}\int_0^1\frac{dx}{\sqrt{1-x^2}}=\int_0^\frac{1}{2}\frac{dx}{\sqrt{1-x^2}}$$


问:
Associativity of addition in integers modulo $n$.
My proof that addition in $G =\mathbb{Z}/n\mathbb{Z}$ is associative is as follows:

Proof：Before taking (mod $n$), we add members of $G$ in the same way as members of $H=(\mathbb{Z}, +)$ where we know that addition in $H$ is associative. Since taking (mod n) does not affect this addition, the addition in $G$ is also associative.

Does this proof seem like enough? I think I'm right but I often risk falling short in my proofs at times.

答:
As pointed out by @ArturoMagidin, your attempt assumes too much.

Here is a proof.

Let $[a]_n, [b]_n, [c]_n\in \Bbb Z/n\Bbb Z$. Then, by definition,

$$[a]_n=\{x\in \Bbb Z: n\mid (a-x)\}\quad\text{and}\quad [a]_n+_n[b]_n=[a+b]_n.$$

(I'll leave it to you to show that $+_n$ is well-defined.)

Now

$$\begin{align}
[a]_n+_n([b_n]+_n[c]_n)&=[a]_n+_n[b+c]_n\text{ (by definition of }+_n\text{)}\\
&=[a+(b+c)]_n\text{ (by definition of }+_n\text{)}\\
&=[(a+b)+c]_n\text{ (by associativity of }+\text{)}\\
&=[a+b]_n+_n[c]_n\text{ (by definition of }+_n\text{)}\\
&=([a]_n+_n[b]_n)+[c]_n\text{ (by definition of }+_n\text{)}.
\end{align}$$


问:
Why is $\operatorname{Aut}(\mathbb{Z}_p)\cong \mathbb{Z}_p^*$?
Let $\mathbb{Z}_p$ denote the $p$-adic integers. Why is $\operatorname{Aut}(\mathbb{Z}_p)\cong \mathbb{Z}_p^*$?

I know that $\operatorname{Aut}(\mathbb{Z}/p^n\mathbb{Z})\cong(\mathbb{Z}/p^n\mathbb{Z})^*$, but I'm not sure which result from group theory to apply.

答:
There is an injective map
$$
\mathbb{Z}_p^\times \to \text{Aut}_{\text{ab. gp.}}(\mathbb{Z}_p)
$$
given by taking $\alpha$ to left-multiplication by $\alpha$ (this is true for any ring).

You wish to know this is surjective. Say $\phi$ is an automorphism of $\mathbb{Z}_p$. Then $\phi$ has to preserve the subgroup $p^n\mathbb{Z}_p$, and therefore $\phi$ has to induce an automorphism of $\mathbb{Z}/p^n\mathbb{Z}$.

This automorphism must be multiplication by a unit $\beta_n$ in $\mathbb{Z}/p^n\mathbb{Z}$, since all automorphisms of this abelian group arise in this way (you seem to already believe this, but proof: take the fact above that was "true for any ring" and observe that injective functions from finite sets to themselves are bijective).

So we have a whole bunch of $\beta_n$. Since for $n > m$ the diagram

$\require{AMScd}$
\begin{CD}
\mathbb{Z}_p / p^n \mathbb{Z}_p  @>\phi>> \mathbb{Z}_p / p^n \mathbb{Z}_p \\
@V  V V @VV  V\\
\mathbb{Z}_p / p^m \mathbb{Z}_p  @>\phi>> \mathbb{Z}_p / p^m \mathbb{Z}_p
\end{CD}
commutes, where the vertical arrows are the natural reduction and the horizontal are the maps induced by $\phi$, we must have that $\beta_n$ reduces to $\beta_m$ mod $p^n$.

So the $\beta_n$ form an element $\beta$ of
$$
\varprojlim{(\mathbb{Z}_p/p^n\mathbb{Z}_p})^\times = \mathbb{Z}_p^\times,
$$
and our automorphism $\phi$ must be multiplication by this $\beta$.


问:
$P(Y<X<Z)$ for independent Poisson distributions
Suppose $X, Y, Z$ all follow a Poisson distribution each with a distinct mean parameter $\lambda_x, \lambda_y, \lambda_z$ respectively. Additionally assume distribution is independent of the others.

How would I go about calculating $P(Y < X < Z)$? My first guess would be something like

$$\sum_{k=1}^\infty \sum_{j=1}^\infty \sum_{i=1}^j f(i, j,k)$$

But I'm not sure what function to be summing over. Would it just be the product of their PMF's?

答:
Let us use the convenient symbols $x,y,z$ as the bound variables for the series.

Due to the independence, the joint probability mass function will be the product of their probability mass functions.  So the term for the series is: $$\def\pois{\operatorname{\cal Pois}}\pois(x;\lambda_{\small X})\pois(y;\lambda_{\small Y})\pois(z;\lambda_{\small Z})$$

The series needs to measure over the event of $Y{<}X{<}Z$, so the domain is: $$\{\langle x,y,z\rangle{\in}\Bbb N^3: 1{\leqslant}x, 0{\leqslant}y{\leqslant}x{-}1, x{+}1{\leqslant}z\}$$

Or such.

$$\displaystyle\mathsf P(Y{<}X{<}Z) ~=~ \sum_{x=1}^\infty\pois(x;\lambda_{\small X})\left(\sum_{y=0}^{x-1}\pois(y;\lambda_{\small Y})\right)\left(\sum_{z=x+1}^{\infty}\pois(z;\lambda_{\small Z})\right)$$


问:
Convolution of probability mass functions (3 non-parametric distributions)
I am familiar with the convolution of probability mass function when it involves two random variables, but I get a little confused when there's a third one.

I have to find the probability mass function of Z = W + X + Y where W, X and Y are independent random variables and with the given table:

I am not sure about the general formula to solve this. Thank you in advance for your help!

答:
Unless this is an exercise in theoretical methods of getting
the distributions of sums of discrete random variables, I would
do this by simulation. It seems that the sum of your three random
variables is reasonably well-approximated by a normal distribution.
[One analytic method might be to multiply the three moment generating
functions, where the distribution of the sum can be found
by collecting similar terms.]

Simulation in R. With a million iterations you can
expect about 2 or 3 significant digits of accuracy.

```<code>set.seed(2021)
m = 10^6
x1 = sample(0:4, m, rep=T, p=0:4)
summary(x1); var(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.000   2.000   3.000   2.999   4.000   4.000
[1] 1.001328  # aprx Var(x1)

x2 = sample(0:4, m, rep=T, p=c(2,3,3,2,0))
summary(x2); var(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
    0.0     1.0     1.0     1.5     2.0     3.0
[1] 1.049731

x3 = sample(0:4, m, rep=T, p=4:0)
summary(x3); var(x3)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   0.000   1.000   1.001   2.000   3.000
[1] 1.001272

s = x1 + x2 + x3
summary(s);  sd(s);  var(s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
    1.0     4.0     6.0     5.5     7.0    10.0
[1] 1.748697   # aprx SD(S)
[1] 3.057941   # aprx Var(S)
</code>```

Simulated distribution table:

```<code>table(s)/m
s
       1        2        3        4        5
0.008076 0.034235 0.084840 0.161267 0.211413
       6        7        8        9       10
0.211874 0.161167 0.084805 0.034200 0.008123
</code>```

Notice in particular, $P(S=1) = P(X_1=1, X_2=0, X_3=0)$
$=(.1)(.2)(.8) = 0.008.$

```<code>hist(s, prob=T, ylim=c(0,.3), br=(-.5:12.5), col="skyblue2")
 curve(dnorm(x, mean(s), sd(s)), add=T, col="orange", lwd=2)
</code>```

Body:
Unless this is an exercise in theoretical methods of getting
the distributions of sums of discrete random variables, I would
do this by simulation. It seems that the sum of your three random
variables is reasonably well-approximated by a normal distribution.
[One analytic method might be to multiply the three moment generating
functions, where the distribution of the sum can be found
by collecting similar terms.]

Simulation in R. With a million iterations you can
expect about 2 or 3 significant digits of accuracy.

set.seed(2021)
m = 10^6
x1 = sample(0:4, m, rep=T, p=0:4)
summary(x1); var(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.000   2.000   3.000   2.999   4.000   4.000
[1] 1.001328  # aprx Var(x1)

x2 = sample(0:4, m, rep=T, p=c(2,3,3,2,0))
summary(x2); var(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
    0.0     1.0     1.0     1.5     2.0     3.0
[1] 1.049731

x3 = sample(0:4, m, rep=T, p=4:0)
summary(x3); var(x3)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   0.000   1.000   1.001   2.000   3.000
[1] 1.001272

s = x1 + x2 + x3
summary(s);  sd(s);  var(s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
    1.0     4.0     6.0     5.5     7.0    10.0
[1] 1.748697   # aprx SD(S)
[1] 3.057941   # aprx Var(S)

Simulated distribution table:

table(s)/m
s
       1        2        3        4        5
0.008076 0.034235 0.084840 0.161267 0.211413
       6        7        8        9       10
0.211874 0.161167 0.084805 0.034200 0.008123

Notice in particular, $P(S=1) = P(X_1=1, X_2=0, X_3=0)$
$=(.1)(.2)(.8) = 0.008.$

hist(s, prob=T, ylim=c(0,.3), br=(-.5:12.5), col="skyblue2")
 curve(dnorm(x, mean(s), sd(s)), add=T, col="orange", lwd=2)


问:
Hopf algebra structure on universal enveloping algebra?
Let $\mathfrak g$ be a Lie algebra. Show that on $U(\mathfrak g)$ (universal enveloping algebra) there is a natural Hopf algebra structure induced by the Hopf algebra structure on the tensor algebra $T(\mathfrak g)$.

So I know what the Hopf algebra structure on $T(\mathfrak g)$ is (the delta and epsilon functions, the antipode map, etc.). I can see how these maps induce maps on $U(\mathfrak g)$ since $U(\mathfrak g)$ is just a quotient on T(\mathfrak g), but my professor said we have to show they're well-defined. This is where I get stuck. Take the multiplication by tensors for example.

How would you show this is well-defined?

答:
The enveloping algebra is a quotient of $T(\mathfrak{g})$ by the 2-sided ideal generated by $[x,y]-x\otimes y+y\otimes x$, $x,y\in\mathfrak{g}$. Since $\Delta$ is an algebra homomorphism, it is enough to show that $\Delta([x,y]-x\otimes y+y\otimes x)$ belongs to that ideal. Equivalently, you could show that $\Delta([x,y]) = \Delta(x\otimes y-y\otimes x)$ in $U(\mathfrak{g})$.

Well, since $[x,y]\in\mathfrak{g}$, $\Delta([x,y])=[x,y]\otimes 1 + 1\otimes [x,y]$. The harder computation is
$$
\Delta(x\otimes y-y\otimes x)=\Delta(x)\otimes \Delta(y)-\Delta(y)\otimes \Delta(x).
$$
Write it out and use the relations in $U(\mathfrak{g})$ (namely, $\cdots\otimes(x\otimes y-y\otimes x)\otimes\cdots=\cdots\otimes[x,y]\otimes\cdots$) to show that it equals $\Delta([x,y])$ as written above.


问:
Prove that $$ is injective and $$ is surjective
This is the question I got and I'm not sure how to approach it.

Let $f: A → B$ and $g: B → A$ be two functions, such that $g ∘ f(x) = x$ for all $a ∈ A$. Prove that $f$ is injective and $g$ is surjective.

答:
$f$ injective: If $f(x_1)=f(x_2)$, then, applying $g$ yields $x_1=g(f(x_1))=g(f(x_2))=x_2$.

$g$ surjective: If $y\in A$, $g(f(y))=y$.


问:
Is there a power series that converges to the function $f(x)= \lvert x\rvert$ for all $x$?
Is there a power series that converges to the function $f(x)= \lvert x\rvert$ for all $x$?

I am pretty lost on how to even start this.

答:
No. This function is not even differentiable at $0$, while power series give us functions that are infinitely differentiable within their interval of convergence.

– Andres Caicedo, Apr 24 at 0:55


问:
Characterization of separable spaces
I was curious whether the following property for separable spaces characterizes them.

Let $X$ be a separable Banach space, and let $B^*$ be the closed unit ball of $X^*$. Then there exists a countable set $D \subset B^*$ such that for all $x\in X$:

$$ ||x|| = \sup_{x^*\in D} |\langle x,x^*\rangle|$$

If $\{x_i\} \subset X$ is a countable dense subset of $X$ and $x_i^*\in B^*$ are such that $\langle x,x^*\rangle=||x_i||$ (note that $x_i^*$ exists by Hahn-Banach), then $D=\{x_i^*\}$ is such a set.

My question is if such a property characterizes separable spaces. This is, if there exists a countable set $D\subset B^*$ such that for all $x\in X$:

$$ ||x|| = \sup_{x^*\in D} |\langle x,x^*\rangle|$$
then
must $X$ be separable?

答: The answer is NO. Let $X=\ell^{\infty}$ and $x^{*}_n(x)=x_n$. Then $x_n^{*}$ is a continuous linear functioinal on $X$ and $\|x\|=\sup | \langle x, x_n^{*} \rangle|$ for all $x \in X$ but $X$ is nor separable.

Title: Characterization of separable spaces


答: To show $g$ surjective, pick any $a \in A$ and since $g(f(a)) = a$, let $b = f(a)\in B$ then $g(b) = a$ showing $g$ surjective. To show $f$ injective, if $f(x) = f(y)\implies x=g(f(x)) = g(f(y))=y\implies x = y$, hence $f$ is injective.

Title: Prove that $$ is injective and $$ is surjective

Body:
This is the question I got and I'm not sure how to approach it.

Let $f: A → B$ and $g: B → A$ be two functions, such that $g ∘ f(x) = x$ for all $a ∈ A$. Prove that $f$ is injective and $g$ is surjective.


问:
What volume is enclosed by $k$ evenly-spaced, overlapping American footballs whose axes are diameters of a unit sphere?
Take $k \in \Bbb N$ intersecting American footballs and configure them inside a unit sphere such that each football touches two opposite ends of the sphere. Each of the shapes are spaced evenly apart.

In the case $k\to \infty$ the volume should be equal to the volume of a sphere.

Related Problem (intersection of $n$ congruent cylinders).

I want to tackle the case $k=2$ first. So I'm using "Cavalieri's Lemon" which is the surface of revolution of a parabolic arc, to model a football.

The equation for the parabolic arc to be revolved about the $x-$axis is: $f(x)=x(1-x).$

What is the volume in the region where the (two) footballs overlap?

I've calculated that the volume of one football is $\frac{\pi}{30}.$ If the interior shape is a well-known shape I can proabably search for the volume of that shape, but I'm having trouble visualizing the interior shape with the enclosed volume.

答:
Let's assume the axes of the footballs are perpendicular to each other, e.g., $$(y^2 + z^2) = (1/4 - x^2)^2, \\ (x^2 + z^2) = (1/4 - y^2)^2.$$ Then we are looking for the volume common to their intersection.  We are searching for the value of $x = y = c > 0$ when $z = 0$ corresponding to the intersection of the curves $y = 1/4 - x^2$, $x = 1/4 - y^2$, which is of course at $(x,y) = (c,c)$ where $c = \frac{\sqrt{2}-1}{2}$.  Hence the volume corresponds to the  integral
$$\begin{align}
V &= 16 \int_{x=0}^c \int_{y=x}^{1/4-x^2} \sqrt{(1/4 - x^2)^2 - y^2} \, dy \, dx \\
&= \frac{1}{4} \int_{x=0}^c \left(1-4 x^2\right)^2\left( \pi - 2 \tan ^{-1}\frac{4
   x}{\sqrt{16 x^4-24 x^2+1}}\right) - 8 x \sqrt{16 x^4-24 x^2+1}  \, dx \\
&= \frac{2\pi + 16 \log 2 - 13}{60} \\
&\approx 0.0728923.
\end{align}$$


问:
Series for $\pi$ which correspond to apollonian gaskets or hyperbolic tilings of the unit disk
Consider the two partitions of the unit disk in $\mathbb{R}^{2}$, the first an Apollonian gasket and the second is the $\{7,3\}$ hyperbolic tiling:

Since the unit disk has radius $1$, both of these have area $\pi$, and for each partition, there is a series representation of $\pi$, so:

Given an Apollonian gasket in the disk of unit radius, is there an straightforward way of getting the sequence of areas of disks which comprise it?

Given a regular hyperbolic tiling, is there a straightfoward way of obtaining the Euclidean area of the (distorted) polygons which comprise it?

答:
Although you could find such sequences, they would most likely themselves make heavy use of $\pi$. For this reason, the best you could hope for would be some sequence from which you could factor out $\pi$ such that the result adds up to $1$.

Apollonian gasket

You can use Descartes' theorem to compute the (inverse) circle radii, starting from four given ones. For the symmetric configuration your picture suggests, these will be non-integral (see Wikipedia on almost $D_3$ symmetry in this context). More precisely, according to this section of Erich's packing center the radii of the thee largest circles will be $r=\frac{3}{2\sqrt3 + 3}$. So you can compute subsequent radii from these, given enough patience. To turn all these radii back into areas you'd square them and multiply them by $\pi$. So if you learn that the squared radii sum up to $1$, this does not tell you one bit about the value of $\pi$.

Hyperbolic tiling

It seems you're using the Poincaré disk model of hyperbolic geometry. Which means your heptagon boundaries are not really straight lines but instead circle arcs, which makes the Euclidean area computation both painfully complicated to write and very intimately related to circles. Therefore I doubt you can get the area expressed without massive use of $\pi$. So the whole result would most likely appear like a circular argument, pardon the pun. One alternative might be using hyperbolic corners but connecting them with Euclidean straight lines. Another alternative might be the Beltrami-Klein model. Neither does look particularly suited for an easy sequence description of $\pi$, though.


问:
Is the Whitney topology finer or coincide with uniform topology for codomain is $\mathbb{R}$
Let $C^{r}(U,\mathbb{R})$ be the space of $C^{r}$ functions from open subset $U\subseteq\mathbb{R}^{n}$ to $\mathbb{R}$, $0\leq r<\infty$. This space can equipped with the Whitney topology(or strong topology), is this topology finer than the uniform topology(supremum norm as a metric), or actually coincide?

Edit
It seems that there is no way to compare this two topology in the above setting. If we only focus on subspace $C^{r}_{b}(U,\mathbb{R})$, the space of $C^{r}$ and bounded function, so that supremum norm can be defined. Can we compare their topology?

答:
This question is discussed in detail in section 1.7 of the book:

Eldering, Jaap. Normally hyperbolic invariant manifolds: the noncompact case. Vol. 2. Atlantis Press, 2013.

Here is a preprint, and here is the final paragraph of section 1.7:

We conclude that the $C^k_b$-topologies induced by our uniform norms are not equivalent to either the weak or strong Whitney topology, because the weak topology
allows arbitrary behavior of functions outside compact sets, while the strong
topology completely restricts that behavior. Our norms allow moderate variations at infinity. In general, ‘moderate behavior’ is not well-defined on a general
noncompact manifold, as it depends on the choice of charts. In the setting of
bounded geometry, though, the uniform, metric structure makes this behavior
unambiguous; we can restrict to normal coordinate charts and consider ‘moderate
behavior’ with respect to these. Note that these topologies are equivalent on
compact domains.


问:
Centralizer : center subgroup ~ normalizer : normal subgroup ~ ?: quotient group
Centralizer subgroup:
$$C_G(A)=\{g\in G\mid gag^{-1}=a,\forall a\in A\}$$
Center subgroup:
$$Z(G)=\{g\in G\mid ga=ag,\forall a\in G\}$$

Centralizer and center are subgroups of $G$. The centralizer takes a subset $A\in G$ and the center always uses the entire group $G$. So the centralizer becomes center when $A=G$
$$
C_G(G)= Z(G).
$$

My questions are that:

Under what precise mathematical conditions that the normalizer $N_G(A)$ equal to a normal subgroup $N$ of $G$? When is $N_G(A)$ not equal to a normal subgroup $N$? See for instances.

Since the quotient group is the $G/N$, are there analogous concepts of quotient (not a subgroup) for $G/N_G(A)$, important for what properties or theorems?

In short, I am exhausting the concepts of

Centralizer : center subgroup ~ normalizer : normal subgroup  ~ ?: quotient group

答:
The normalizer works "the other way around" to what you may think it does. The normalizer of $A$ in $G$ is the largest subgroup of $G$ that contains $A$ and in which $A$ is normal.

That is, $N_G(A) =\{g\in G:gA=Ag\}$. By its very definition, $A$ is normal in $N_G(A)$. However, $N_G(A)$ may not be a normal subgroup of $G$, it seems you already have examples of this. I am not sure you should expect any "reasonably general" condition that guarantees a normalizer is normal, but rather, the other way around.

One way in which normalizers and centralizers interact with quotients is as follows: each element $g\in N_G(A)$ defines an automorphism of $A$ by conjugation, and hence there is a map $N_G(A) \longrightarrow \mathrm{Aut}(A)$ that assigns $g$ to the automorphism $a\longmapsto {}^ga$. By definition, the kernel of this map is the centralizer of $A$ in $G$, so one obtains the "NC lemma" that $C_G(A)$ is normal in $N_G(A)$ and that $N_G(A)/C_G(A)$ is isomorphic to a subgroup of $\mathrm{Aut}(A)$.


问:
How do I find the tangent line to this curve?
I have a two-part question I am having trouble understanding what the second part is asking me and how to do it.

PART (a): Use the definition of derivative to calculate $f'(1)$ for $f(x) = x^2 - 3x$

PART (b): Use the result in (a), find the equation of the tangent line to the curve of $f(x)= x^2 - 3x$ at the point $(1, −2)$.

How do I find the equation of the tangent line to the curve? could I get help with the steps so I know how to do the next similar problems?

答:
a) If I understood you correctly you could figure out Part a). The derivative of $f(x)=x^2-3x$ is $2x-3$. If evaluated at the point $x=1$, $f'(1)=-1$.

b) This is an example for a tangent line to a curve:

As you can see, the tangent line measures the slope of the curve at the point where they're intersecting. Do you remember how we can find the slope of a function?


问:
tensoring coordinate ring.
Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ (called the coordinate ring of the unit circle).\

I want to prove that $\mathbb C \otimes_{\mathbb R} B$ is a UFD. I know that we are tensoring up with scalars but how this will help me in proving that it is a UFD.

I know that: a unique factorization domain is an atomic integral domain in which factorization to irreducibles is unique (up to associates). And a commutative ring is atomic if each $r \in R$ is a finite product of irreducibles in R.

Any help is appreciated.

答:
Ok, here's the plan:

Show that $\mathbb{C} \otimes_{\mathbb{R}} B \cong \mathbb{C}[x,y]/(x^2+y^2-1)$. One direction of this isomorphism is determined by $z\otimes p \mapsto zp$ when $z \in \mathbb{C}$ and $p \in B$. Show that this extends to a ring homomorphism $\mathbb{C} \otimes_{\mathbb{R}} B \to \mathbb{C}[x,y]/(x^2+y^2-1)$ and find an inverse!

Show that $\mathbb{C}[x,y]/(x^2+y^2-1) \cong \mathbb{C}[t,t^{-1}]$. To do this, we need to construct homomorphisms $\mathbb{C}[x,y]/(x^2+y^2-1) \to \mathbb{C}[t,t^{-1}]$ and $\mathbb{C}[t,t^{-1}] \to \mathbb{C}[x,y]/(x^2+y^2-1)$ which compose to give the identities. The only (good) way to build maps between these $\mathbb{C}$-algebras is to use their universal properties. A $\mathbb{C}$-algebra homomorphism from $\mathbb{C}[x,y]/(x^2+y^2-1)$ is uniquely determined by the images of $x$ and $y$ (let's call these $a$ and $b$, respectively), with the restriction that $a^2 + b^2 = 1$. A $\mathbb{C}$-algebra homomorphism from $\mathbb{C}[t,t^{-1}]$ is uniquely determined by the image of $t$, with the restriction that the image of $t$ must be a unit.

Here are some hints: in order for the map $\mathbb{C}[t,t^{-1}] \to \mathbb{C}[x,y]/(x^2+y^2-1)$ to be surjective, the image of $t$ must generate the codomain as a $\mathbb{C}$-algebra. In order for the map $\mathbb{C}[x,y]/(x^2+y^2-1) \to \mathbb{C}[t,t^{-1}]$ to be surjective, the image must contain $t$ and $t^{-1}$. In particular, you need $t$ to correspond to a unit in $\mathbb{C}[x,y]/(x^2+y^2-1)$ -- think about the factorization of $x^2 + y^2$ over $\mathbb{C}$ to find this unit!

Show that $\mathbb{C}[t,t^{-1}]$ is a UFD, because it's a localization of a UFD.

Hope this helps! Happy to fill in more details if you get stuck somewhere.

Title: tensoring coordinate ring.


问:
Formal definition of a wave
I have never seen a formal definition of a wave-like function. By wave-like function, I mean, intuitively, something similar to the sine function. The closest definition I have gotten is "finite linear combination of sine functions". But this still leaves a lot of stuff out. Is there a text where there is a good definition of wave-like functions?</p>

答:
The closest I have found for the definition of a “wave” in physics is “a disturbance that propagates through a medium, or through a field of some sort”. (The “field” refers to a vector field, or a more general tensor field). This is, of course, not a precise definition but it is the best I can find, at this level of generality.

I have forgotten where I read this definition, some physics textbook.

Title: Formal definition of a wave


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: It seems weirdly arbitrary to me that you can comb a hairy n-sphere if n is odd, but that this is impossible when n is even.

LastEditorDisplayName: anon


问:
Why is lim sup necessary
I have hard time understanding why the concept of lim superior is necessary, for example the ratio test. There are two ways, it seems of phrasing this:

(Abbott, Understanding Analysis p.78) Given $\sum a_n$ with $a_n \neq 0$, if $\lim \lvert{\frac{a_{n+1}}{a_n}} \rvert = r < 1$, then the series converges absolutely

(Rudin, Principles of Mathematical Analysis p. 66) The series $\sum a_n$ converges if $\lim \sup _{n \to \infty} \lvert{\frac{a_{n+1}}{a_n}} \rvert < 1$

Why is the addition of supremum necessary, if the first one seems just fine? What exactly is the difference between THE limit and the supremum of the set of subsequential limits?

答:
Take the sequence $(a_n)_n$ defined by $a_1=1$ and
$$
a_{n+1} = \frac{2+(-1)^n}{4}\cdot a_n
$$
Then, obviously,
$$
0 \leq \frac{a_{n+1}}{a_n} = \frac{2+(-1)^n}{4}
$$
You can comnpute its limsup and apply the second theorem. The limit, however, doesn't exist, so you cannot apply the first. So the second theorem is strictly more general than the first.

(Note, of course, that if the limit exists, then it is equal to the limsup and both theorems apply.)

Title: Why is lim sup necessary


问:
Algebraically closed proof.
Let $B = \bigoplus_{i \in \mathbb Z}B_i$ be a $\mathbb Z$-graded integral domain. Given $f \in B$ define $\deg: B \rightarrow \mathbb Z \cup \{-\infty\}$ by
$$\deg(f) = \mathrm{maxSupp}{(f)} \text{ if } f \neq 0 \text{ and } \deg(0) = -\infty $$

Note that elements of $B_i$ are homogeneous and each $f \in B$ has a unique decomposition $f = \sum_{j \in \mathbb Z} f_j, f_j \in B_j.$

A function $f$ is homogeneous if it can be written as sum of monomials of different degrees.

I am trying to prove that $B_0$ is algebraically closed in $B.$

I want to show that $B_0$ is algebraically closed in $B.$ Which means that we want to show that $B_0$ is equal to its algebraic closure $\bar{B_0}$, where the algebraic closure of $B_0$ in $B$ is the subring $\bar{B_0}$ consisting of all $b\in B$ that are algebraic over $B_{0}.$ And $b\in B$ is algebraic over $B_0$ means that there exists a polynomial, say $f(b) \in B_0[x]$, i.e. with coefficients in $B_0$ such that $f(b) = 0$.

Now, assume by contradiction that $b\in B\setminus B_0$ such that $b \neq 0$ and $\deg b \neq 0$ and $b$ is algebraic over $B_0.$

Since $b$ is algebraic over $B_0,$ then there exists  $f(b) \in B_0[x]$ such that $f(b) = 0$ i.e., there exists
$c_0, c_1, \dots , c_t \in B_0$ with $c_t \neq 0$ such that  $$f(b) = \sum_{j=0}^t c_jb^j = 0 \quad (1)$$

But since $b\in B$ then $$b=\sum_{i\in \Bbb Z}b_i \quad (2)$$ i.e. $b_i$ is the $i$-the homogeneous component of $b.$  Since $b$ has some non-zero component in a positive degree by assumption, let $m$ be the largest positive integer for which $b_m\ne0$ (if $b$ has no non-zero component in positive degree we take $m$ to be the minimum integer with $b_m\ne0$)

Now, substituting from $(2)$ into $(1)$ we get $$f(b) = \sum_{n=0}^t c_n(\sum_{i\in \Bbb Z}b_i)^n = 0 \quad (3)$$

But then the term $(b_m)^n$ has degree $mn$ and there is exactly one of it because other nonzero $b_i$ have lower degree (note that degrees of $c_n$'s are $0$ because they are in $B_0$). Now, since $B$ is an integral domain, then $c_n(b_m)^n \neq 0 $ because both of $c_n,(b_m)^n$ not equal $0.$ In particular, this means $f(b)$ can not be zero as it is not zero in the $mn$-th component which is a contradiction.

But I got a hint that I have to show that the values of $\deg(b^n), n \geq 0,$ are distinct.

So I modified the proof above (which I understand with the help of a proof given by @Leoli) to the following:

$\deg \sum_{i=0}^{n-1} c_ib^i \leq \max_{0 \leq i \leq (n-1)} \deg(c_i b^i) \leq \max_{0 \leq i \leq (n-1)} \deg(b^i) = \max_{0 \leq i \leq (n-1)} \{im \} = (n-1)m$

Where the first inequality by property $3$ of $\deg$ we proved before and the second inequality because  $\deg c_i  = 0$ and the equality before last because $\operatorname {maxSupp}b$ is $m.$

But $\deg c_nb^n = nm \neq 0$ because $B$ is an integral domain and $c_n \neq 0$ and $(b_m)^n \neq 0.$

Therefore, the maxSupp term in $\sum_{i=0}^{n-1} c_ib^i$ does not cancel the maxSupp term in $ c_nb^n$

But $\sum_{i=0}^{n} c_ib^i = 0,$ then $ mn = \deg c_nb^n = \deg \sum_{i=0}^{n} c_ib^i = \deg (0) = -\infty$

A contradiction.

I feel like my proof is not ordered well, could someone criticize it and give me a more elegant proof.

答:
Your idea of using the degree function is perfectly good, so here I just offer you a way of writing your argument in a shorter way.

Consider an element $b\in B$ and write it in the form $b=t+s$ where $t\neq 0$ is homogeneous of degree $N$ and $s$ is a sum of homogeneous terms of lower degree. Let us call $t$ the leading term of $b$. Let us assume that $\deg t>0$.

Then for any $k$, we see that the leading term of $b^k$ is $t^k$, that is, $b^k = t^k+s'$ where $s'$ is a sum of elements of lower degree: this follows from the binomial expansion.

If $f\in B_0[x]$ is a polynomial of degree $d$ with coefficients of degree zero, then it follows from the above that the leading term of $f(b)$ is $c_dt^d$, so that if
$f(b) =0$, then we must in fact have that $c_dt^d=0$: this follows by using the direct sum decomposition you wrote in your post.

Since $B$ is a domain and since $c_d\neq 0$ and $t\neq 0$, we can conclude that $c_dt^d\neq 0$, so that no element with leading term which is homogeneous of positive degree can be algebraic over $B_0$, like you claimed.

Title: Algebraically closed proof.


问:
Quaternion sandwich simplification gives me a term too many
Context:
I'm reading Foundations of Game Engine Development by Eric Lengyel. On the topic of rotations with quaternions it is explained that a vector v can be rotated by being "sandwiched" by quaternions: v' = $qvq^{-1}$. When the quaternion is a unit-quaternion, the conjugate of q (denoted q*) is equal to $q^{-1})$ , such that v' = qvq*

To demonstrate how the sandwich product shown in this equation performs a rotation, the quaternion is split into a vector and a scalar part: q = b + c.

v is treated like a pure quaternion so doesn't have a scalar part.

The book gives the following formula:
$$\mathbf{q}\mathbf{v}\mathbf{q^*} = (\mathbf{b} \times \mathbf{v} + c\mathbf{v} - \mathbf{b} \cdot \mathbf{v})(-\mathbf{b}  + c)$$

Problem:

According to the author, the right side can be simplified to:
$$=(c^2 - b^2)\mathbf{v} + 2(\mathbf{v} \cdot \mathbf{b})\mathbf{b} + 2c(\mathbf{b} \times \mathbf{v})$$

But when I try to work this out myself, I have a term more, and I don't know how he got rid of it.
For me it looks like this:
$$ = (\color{red}{\mathbf{b} \times \mathbf{v} \times \mathbf{-b})} -c(\mathbf{v} \times \mathbf{b}) + \mathbf{b}(\mathbf{b} \cdot \mathbf{v}) + c(\mathbf{b} \times \mathbf{v}) + c^{2}\mathbf{v} + c(\mathbf{b} \cdot \mathbf{v}) $$

$$ = \color{red}{\mathbf{v}(\mathbf{b} \cdot \mathbf{-b}) + \mathbf{b}(\mathbf{b} \cdot \mathbf{v})}\color{blue}{ -c(\mathbf{v} \times \mathbf{b})} + \mathbf{b}(\mathbf{b} \cdot \mathbf{v})\color{blue}{ + c(\mathbf{b} \times \mathbf{v})} + c^{2}\mathbf{v} + c(\mathbf{b} \cdot \mathbf{v})  $$

$$ = \color{red}{-b^2\mathbf{v} + \mathbf{b}(\mathbf{b} \cdot \mathbf{v})} + \color{blue}{2c(\mathbf{b} \times \mathbf{v})} + \mathbf{b}(\mathbf{b} \cdot \mathbf{v}) + c^{2}\mathbf{v} + c(\mathbf{b} \cdot \mathbf{v}) $$

$$ = \color{green}{-b^2\mathbf{v}} \color{brown}{+ \mathbf{b}(\mathbf{b} \cdot \mathbf{v})} + 2c(\mathbf{b} \times \mathbf{v})  \color{brown}{+ \mathbf{b}(\mathbf{b} \cdot \mathbf{v})} \color{green}{+ c^{2}\mathbf{v}} + c(\mathbf{b} \cdot \mathbf{v}) $$

$$ = \color{green}{(c^2-b^2)\mathbf{v}} \color{brown}{+ 2\mathbf{b}(\mathbf{b} \cdot \mathbf{v})} + 2c(\mathbf{b} \times \mathbf{v})+ c(\mathbf{b} \cdot \mathbf{v}) $$

I hope the colours make it clear how I simplified. My simplification looks the same except that I end up with an extra term $c(\mathbf{b} \cdot \mathbf{v})$.

Did I make a mistake? Or does the last term equate to zero somehow in this context?

Some help would be greatly appreciated.

答:
Your mistake comes at the first step.  Remember quaternion multiplication is $(a+\mathbf{b})(c+\mathbf{d})=(ac-\mathbf{b}\cdot\mathbf{d})+(c\mathbf{b}+a\mathbf{d}+\mathbf{b}\times\mathbf{d})$ so you should get
\begin{align*}\require{cancel}
&(\mathbf{b} \times \mathbf{v} + c\mathbf{v} - \mathbf{b} \cdot \mathbf{v})(-\mathbf{b}  + c)\\
&=[(- \mathbf{b} \cdot \mathbf{v})(c)-((\mathbf{b} \times \mathbf{v} + {\color{blue}{c\mathbf{v}}})\cdot(-\mathbf{b})]\\
&\quad+[(- \mathbf{b} \cdot \mathbf{v})(-\mathbf{b})+c(\mathbf{b}\times\mathbf{v}+c\mathbf{v})+(\mathbf{b}\times\mathbf{v}+c\mathbf{v})\times(- \mathbf{b} \cdot \mathbf{v})(-\mathbf{b})]\\
\end{align*}
you missed my blue term there.

Also, since $\times$ is not associative, $\mathbf{b}\times\mathbf{v}\times(-\mathbf{b})$ is slightly ambiguous.


问:
An associated Stirling number identity related to the second-order Eulerian numbers.
A similar Stirling number identity representing the second-order Eulerian numbers
can be found at this question.

We denote the associated Stirling cycle numbers as $\left[\!\left[ n\atop  k\right] \! \right] $
A008306, A106828
and the associated Stirling set numbers as $\left\{\!\! \left\{ n\atop  k\right\} \!\! \right\} $ A008299, A137375.
Second-order Eulerian numbers as we use them are defined in A340556. They differ slightly from the second-order Eulerian numbers as defined by Graham, Knuth, and Patashnik in CMath.

In the range $0 \le k \le n$ the following identity holds:

$$ \sum_{j=0}^{k} (-1)^{k-j} \binom{n-j}{k-j}
\left\{ \!\! \left\{ n+j\atop j\right\} \!\! \right\} =
\sum_{j=0}^{n-k+1} (-1)^{n-k-j+1} \binom{n-j}{k-1}
\left[\! \left[ n+j\atop j\right] \! \right] $$

Maybe someone enjoys showing us the proof?

答:
We seek to show that with $0\le k\le n$ the following identity holds:

$$ \sum_{j=0}^{k} (-1)^{k-j} {n-j \choose k-j}
\left\{ \!\! \left\{ n+j\atop j\right\} \!\! \right\} =
\sum_{j=0}^{n-k+1} (-1)^{n-k-j+1} {n-j\choose k-1}
\left[\! \left[ n+j\atop j\right] \! \right] $$

where we have associated Stirling numbers of the first and second kind.

Now from the combinatorial meaning of these numbers (cancel fixed
points resp. singleton sets) we have that

$$\left[\! \left[ n\atop k\right] \! \right] =
\sum_{q=0}^k (-1)^q {n\choose q} {n-q\brack k-q}$$

and

$$\left\{ \!\! \left\{ n\atop k\right\} \!\! \right\} =
\sum_{q=0}^k (-1)^q {n\choose q} {n-q\brace k-q}.$$

Consult    OEIS    A008306    and    OEIS
A008299 for more information.  We will only
use  the second  of  these but  we  show the  pair  to illustrate  the
similarity in their construction  (PIE). The combinatorial classes for
these      are     $\def\textsc#1{\dosc#1\csod}\def\dosc#1#2\csod{{\rm
#1{\small     #2}}}    \textsc{SET}(\mathcal{U}\times\textsc{CYC}_{\ge
2}(\mathcal{Z}))$ and $\textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge
2}(\mathcal{Z})).$

We start with the LHS and obtain

$$\sum_{j=0}^{k} (-1)^{k-j} {n-j \choose k-j}
\sum_{q=0}^j (-1)^q {n+j\choose q} {n+j-q\brace j-q}.$$

With $n\ge 1$ this is

$$(-1)^k \sum_{j=1}^{k} {n-j \choose k-j}
\sum_{q=1}^j (-1)^q {n+j\choose j-q} {n+q\brace q}.$$

Recall e.g. from Concrete Mathematics chapter 6.2. that

$$\bbox[5px,border:2px solid #00A000]{
{n\brace m} = (-1)^{n-m} \frac{(n-1)!}{(m-1)!}
[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}.}$$

We find for the LHS

$$(-1)^k \sum_{j=1}^{k} {n-j \choose k-j}
\sum_{q=1}^j (-1)^q {n+j\choose j-q}
(-1)^n \frac{(n+q-1)!}{(q-1)!} [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-q}
\\ = (-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
\sum_{j=1}^{k} {n-j \choose k-j}
\\ \times
\sum_{q=1}^j (-1)^{q-1} {n+j\choose j-q}
{n+q-1\choose q-1}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-q+1}
\\ = (-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
\sum_{j=1}^{k} {n-j \choose k-j}
\\ \times [w^{j-1}] (1+w)^{n+j}
\sum_{q=1}^j (-1)^{q-1} w^{q-1}
{n+q-1\choose q-1}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-q+1}.$$

Now the coefficient extractor enforces the upper limit of the inner sum
and we may extend $q$ to infinity, getting

$$(-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
\sum_{j=1}^{k} {n-j \choose k-j}
\\ \times [w^{j-1}] (1+w)^{n+j}
\frac{1}{(1+w/(\frac{1}{z}\log\frac{1}{1-z}))^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=1}^{k} {n-j \choose k-j}
[w^{j-1}] (1+w)^{n+j}
\frac{1}{(\frac{1}{z}\log\frac{1}{1-z}+w)^{n+1}}.$$

The inner term is
$$[w^{j-1}] (1+w)^{n+j}
\frac{1}{(\frac{1}{z}(\log\frac{1}{1-z}-z)+1+w)^{n+1}}
\\ = [w^{j-1}] (1+w)^{j-1}
\frac{1}{(1+\frac{1}{1+w} \frac{1}{z}(\log\frac{1}{1-z}-z))^{n+1}}.$$

Re-expanding the series,

$$(-1)^{n-k+1} n! [z^n]
\sum_{j=1}^{k} {n-j \choose k-j}
[w^{j-1}] (1+w)^{j-1} \\ \times
\sum_{q=j}^n {n+q\choose n} (-1)^q \frac{1}{(1+w)^q}
\left(\frac{1}{z}(\log\frac{1}{1-z}-z)\right)^q.$$

The upper limit on the inner sum results from $[z^n]$ because
$\frac{1}{z}  (\log\frac{1}{1-z}-z) = \frac{1}{2} z + \cdots$ and the
lower one from the fact that $[w^{j-1}] (1+w)^{j-1-q} = 0$ when $1\le
q\le j-1$; $q=0$ produces a constant. Continuing,

$$(-1)^{n-k+1} n! [z^n]
\sum_{j=1}^{k} {n-j \choose k-j}
[w^{j-1}] \\ \times
\sum_{q=j}^n {n+q\choose n} (-1)^q \frac{1}{(1+w)^{q-(j-1)}}
\left(\frac{1}{z}(\log\frac{1}{1-z}-z)\right)^q
\\ = (-1)^{n-k} n! [z^n]
\sum_{j=1}^{k} {n-j \choose k-j}
\sum_{q=j}^n {n+q\choose n} (-1)^{q-j} {q-1\choose q-j}
\left(\frac{1}{z}(\log\frac{1}{1-z}-z)\right)^q
\\ = (-1)^{n-k} n!
\sum_{j=1}^{k} {n-j \choose k-j}
\sum_{q=j}^n {n+q\choose n} (-1)^{q-j} {q-1\choose q-j}
[z^{n+q}] \left(\log\frac{1}{1-z}-z\right)^q
\\ = (-1)^{n-k} n!
\sum_{j=1}^{k} {n-j \choose k-j}
\sum_{q=j}^n {n+q\choose n} (-1)^{q-j} {q-1\choose q-j}
\\ \times \frac{q!}{(n+q)!}
\times (n+q)! [z^{n+q}] \frac{1}{q!}
\left(\log\frac{1}{1-z}-z\right)^q
\\ = (-1)^{n-k}
\sum_{j=1}^{k} {n-j \choose k-j}
\sum_{q=j}^n (-1)^{q-j} {q-1\choose q-j}
\left[\! \left[ n+q\atop q\right] \! \right].$$

It remains to simplify the binomial coefficients:

$$(-1)^{n-k}
\sum_{j=1}^{k} [u^k] u^j (1+u)^{n-j}
\sum_{q=j}^n (-1)^{q-j} {q-1\choose q-j}
\left[\! \left[ n+q\atop q\right] \! \right].$$

We see that we may raise $j$ to $n$ owing to $[u^k]$:

$$(-1)^{n-k}
\sum_{j=1}^{n} [u^k] u^j (1+u)^{n-j}
\sum_{q=j}^n (-1)^{q-j} {q-1\choose q-j}
\left[\! \left[ n+q\atop q\right] \! \right]
\\ = (-1)^{n-k}
\sum_{q=1}^n \left[\! \left[ n+q\atop q\right] \! \right]
[u^k] (1+u)^n \sum_{j=1}^q {q-1\choose q-j} (-1)^{q-j}
u^j (1+u)^{-j}.$$

The inner term is
$$[u^k] (1+u)^n \frac{u^q}{(1+u)^q}
\sum_{j=0}^{q-1} {q-1\choose j} (-1)^{j}
\frac{(1+u)^j}{u^j}
\\ = [u^k] (1+u)^{n-q} u^q
\left(1-\frac{1+u}{u}\right)^{q-1}
\\ = [u^k] (1+u)^{n-q} u  (-1)^{q-1}
= (-1)^{q-1} [u^{k-1}] (1+u)^{n-q}.$$

This yields

$$\sum_{q=0}^{n-k+1} (-1)^{n-k-q+1} {n-q\choose k-1}
\left[\! \left[ n+q\atop q\right] \! \right]$$

which is  the claim. (Here we  must have $n-q\ge k-1$  or $n-k+1\ge q$
else  the binomial  coefficient vanishes  and we  may lower  the upper
limit from $n$ to $n-k+1.$)


问:
Prove that $|\sin x|\le|x|$ without using the continuity property
Without using any property about continuity of sine function, Prove that $|\sin x|\le|x|$.
First come to my mind is that using taylor series to approximate the sine function, where we know that $\sin x=\sum_{i=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$, but i am not quite see how to prove the inequality.</p>

答: Without words... Well, nearly without words...


问:
Faster way to calculate $\frac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$
What is the value of  $\cfrac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ ?
$$1)1+\sqrt3\quad\quad\quad\quad\quad\quad2)-1+\sqrt2\quad\quad\quad\quad\quad\quad3)1-\sqrt2\quad\quad\quad\quad\quad\quad4)\sqrt2-2\sqrt3$$

It was one of the questions from timed exam (for example supposed to solve any question in average one minute).  but I have difficulty to calculate this expression fast. my approach is:

$$\left(\frac{2\sqrt2+3\sqrt3}{5-\sqrt6}\times\frac{5+\sqrt6}{5+\sqrt6}\right)-\frac2{\sqrt3-1}$$
By rationalizing the second fraction it is not hard to see it is $\sqrt3+1$. but I had difficulty to accurately calculate the first one in short time. Is it possible to evaluate it quicker?

答:
If you are very clever, it might occur to you that

$$\sqrt8+\sqrt{27}=(\sqrt2)^3+(\sqrt3)^3=(\sqrt2+\sqrt3)((\sqrt2)^2-\sqrt2\sqrt3+(\sqrt3)^2)=(\sqrt2+\sqrt3)(2-\sqrt6+3)$$

which lets you cancel out the $5-\sqrt6$ from the denominator. But that only occurred to me after the fact.


Title: Faster way to calculate $\frac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$

Body:
What is the value of  $\cfrac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ ?
$$1)1+\sqrt3\quad\quad\quad\quad\quad\quad2)-1+\sqrt2\quad\quad\quad\quad\quad\quad3)1-\sqrt2\quad\quad\quad\quad\quad\quad4)\sqrt2-2\sqrt3$$

It was one of the questions from timed exam (for example supposed to solve any question in average one minute).  but I have difficulty to calculate this expression fast. my approach is:

$$\left(\frac{2\sqrt2+3\sqrt3}{5-\sqrt6}\times\frac{5+\sqrt6}{5+\sqrt6}\right)-\frac2{\sqrt3-1}$$
By rationalizing the second fraction it is not hard to see it is $\sqrt3+1$. but I had difficulty to accurately calculate the first one in short time. Is it possible to evaluate it quicker?

Body:
$$\begin{align}
\frac{\sqrt{8}+\sqrt{27}}{5-\sqrt{6}}
&= \frac{(\sqrt{8} + \sqrt{27})(5 + \sqrt{6})}{25-6} \\
&= \frac{5 \sqrt{8} + \sqrt{48} + 5\sqrt{27} + \sqrt{81(2)}}{19} \\
&= \frac{10\sqrt{2} + 4\sqrt{3} + 15\sqrt{3} + 9\sqrt{2}}{19} \\
&= \sqrt{2} + \sqrt{3}.
\end{align}$$


问:
Homogenizing an inequality with a condition
I am stuck on the following problem, proving this with the condition that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$:
$$(a+1)(b+1)(c+1)\geq 64$$
I started by expanding the lhs, $abc+ab+bc+ac+a+b+c+1$, but now I do not know what to do. I have also found that $\frac{ab+bc+ac}{abc}=1\Rightarrow abc=ab+bc+ac$, but I don't think this helps much. Would anyone be kind enough to provide some hints as to how I can homogenize this thing?

P.S. I am aware of some of the other "easier" solutions, like the one involving the AM-GM-HM inequality, but here I really just want to practice on how to homogenize inequalities.

答:
$(a+1)(b+1)(c+1)\ge 64\Leftrightarrow 2(ab+bc+ca)+(a+b+c)\cdot \frac{abc}{ab+bc+ca}\ge 64\cdot \frac{a^2b^2c^2}{(ab+bc+ca)^2}$

This is what you want.


问:
Is $f(x) = (x,x)$ not continuous in this topology?
Let $S$ be the unit circle with the usual topology and let $\mathbb{R}$ be equipped with the discrete topology, then consider $S \times \mathbb{R}$ with the product topology, and consider the map $f(x) = (x,x)$ from $S$ to $S \times \mathbb{R}$, where here we are identifying $S$ with, say, $[0,1)$.

Is this map $f$ continuous?

The preimage of any $U \times R$ under $f$ for $U$ open in $S$ and $R$ open in $\mathbb{R}$ (any subset) is just $U$ or $\emptyset$, which is open in $S$, so it seems that $f$ should be continuous.

On the other hand...

Consider a non-closed subset $B$ of $S$. We have that $B = f^{-1}(f(B))$, but $f(B)$ is closed since any converging sequence in $S \times \mathbb{R}$ must be eventually constant. So, if $f$ were continuous we would have $B$ is closed since it is the inverse image of a closed set under a continuous map, a contradiction.

Clearly there is a mistake somewhere, but where have I gone wrong? Is the map $f$ continuous or not?

答: To see that $f$ is not continuous, observe that $f[S]$ is a discrete subset of $S\times\Bbb R$. To see this, let $x\in S$, where I identify $S$ with $[0,1)$; then $f(x)=\langle x,x\rangle$, so $S\times\{x\}$ is an open nbhd of $f(x)$ that contains no other point of $f[S]$. (In fact the discrete subsets of $S\times\Bbb R$ are precisely those $A\subseteq S\times\Bbb R$ with the property that $A\cap(S\times\{x\})$ is finite for each $x\in\Bbb R$.) Since $S$ is not a discrete space, $f$ cannot be continuous.


答:
The sine function is an odd function, so it suffices to prove the inequality for nonnegative $x$.

For $0\le x\le 1$, rewrite $\sin x$ in two different ways:
\begin{align*}
\sin x &=\sum_{n=0}^\infty\left(\frac{x^{4n+1}}{(4n+1)!} - \frac{x^{4n+3}}{(4n+3)!}\right)\tag{1}\\
&=x - \sum_{n=0}^\infty\left(\frac{x^{4n+3}}{(4n+3)!} - \frac{x^{4n+5}}{(4n+5)!}\right).\tag{2}
\end{align*}
Since each bracket term in the two infinite series is nonnegative, $(1)$ implies that $\sin x\ge0$ and $(2)$ implies that $\sin x\le x$. Therefore $|\sin x|\le |x|$.

For $x>1$, note that $\sin x = |\operatorname{imag}(e^{ix})| \le |e^{ix}|$. If we can show that $|e^{ix}|\le1$, we are done. One way to prove this is to show that $\overline{e^{ix}}=e^{-ix}$ (easy) and $e^{ix}e^{-ix}=1$ (theorems about multiplication of absolutely convergent power series, i.e. convergence of Cauchy product, is needed here).

Title: Prove that $|\sin x|\le|x|$ without using the continuity property

Body:
Without using any property about continuity of sine function, Prove that $|\sin x|\le|x|$.
First come to my mind is that using taylor series to approximate the sine function, where we know that $\sin x=\sum_{i=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$, but i am not quite see how to prove the inequality.</p>


问:
$16$ subspaces of $F_2^3$
Given the finite field, $F_2$, consider the 3-dimensional vector space $V$ which contains all of the $3$-tuples consisting of elements in $F_2$.

In my textbook, it says that $V$ has $16$ subspaces, but I can't figure out what they are. Can someone please help me? here is what I have so far:

Subsets of dimension $1$: $(0,0,0)$

Subsets of dimension $2$: $(0,0,0)$ mixed with any other element $\rightarrow 7$ possibilities

Subsets of dimension $3$: $\{ (0,0,0), (0,0,1), (1,1,0) \}, \{(0,0,0), (0,1,0), (1,0,1) \}, \{ (0,0,0), (1,0,0), (0,1,1) \} \rightarrow 3$ possibilities

There is the entire $V$.

So far, I have these $12$ subspaces. What are the last $4$?

答:
There is one $0$-dimensional subspace: $\{0 \}$.

There are $7$ $1$-dimensional subspaces: the span of any nonzero element, consisting of $0$ and that element.

A $2$-dimensional subspace with basis $\{b_1, b_2\}$ consists of $0, b_1, b_2, b_1+b_2$.
Note that any two distinct nonzero elements are linearly independent.  Each $2$-dimensional subspace thus has $3$ (unordered) bases, and there are ${7 \choose 2} = 21$ possible bases in all, so $21/3=7$ $2$-dimensional subspaces.

There is one $3$-dimensional subspace, namely the whole space.  That's a total of $16$ subspaces.

Title: $16$ subspaces of $F_2^3$


问:
Understanding why Roots of Chebyshev Polynomial Give best Interpolation
So I know that if we choose the nodes of the interpolating polynomial to be the roots of the Chebyshev polynomial then the error is minimized. My textbook doesn't fully prove this, they just give a sketch of the proof.

In the sketch they mention $\int_{-1} ^{1} |(x-x_0) \cdot\cdot\cdot (x-x_n)| dx =2^{-n} $ and give no hint as to how they got here.

I know that we get this by integrating both sides of the usual error statement for interpolating polynomials. I think the rest of the proof mainly makes sense but this step has me stumped.

I tried to use induction to prove this to myself but it wasn't working. Maybe I made a silly mistake, but I can't seem to find my error and I cannot find anything on this on the internet. Can someone please give me a hint on how to prove this?

Thanks!

答:
There is a missing absolute sign and it should be max not integral, i.e.
$$
\max_{x\in[-1,1]}\lvert (x-x_0)\dots(x-x_n)\rvert=\max_{x\in[-1,1]}\lvert 2^{-n}T_{n+1}(x)\rvert=2^{-n}.
$$

Edit: correcting not integral.
The formula for error of polynomial interpolation with nodes $x_i$, $i=0,1,\dots,n$ is
$$
f(x)-P_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n (x-x_i)
$$
so the maximum error is never more than $\sup\frac1{(n+1)!}\lvert f^{(n+1)}\rvert$ when using Chebyshev nodes.