问:
Decomposition(or splitting) of compact operator.
I want to know why the below is.

Let $K:H\rightarrow H$ be a compact operator and $\delta>0$ be given. Then, there are two operators $T$ and $\epsilon$,
$$ K=T+\epsilon$$
where $T$ and $\epsilon$ satisfy $\dim({range(T)})<\infty$, $\|\epsilon\|_{H\rightarrow H}<\delta$ and $H$ is Hilbert space(it may serparable).

The $\| \cdot \|_{H\rightarrow H}$ is operator norm.

How can I show it ? Are $T$ and $\epsilon$ can be represented explicitly ?

答:
Let $B$ be the closed unit ball of $H$. Then $K(B)$ is pre-compact and hence has a $\delta/2$-net $y_1, \dots, y_n$. Let $P$ be the orthogonal projection onto $\operatorname{span}\{y_1, \dots, y_n\}$. Let $T = P \circ K$ and $\epsilon = K - T$. Then for $x$ with $\|x\| = 1$, pick $y_i$ such that $\|Kx - y_i\| < \delta/2$.

Notice that one has
$$\|Tx - y_i\| = \|PKx - y_i\| = \|P(Kx - y_i)\| \leq \|Kx - y_i\| < \delta/2$$

so that
$$ \|\epsilon x \| \leq \| Kx - y_i \| + \|Tx - y_i\| < \delta.$$

This shows that $\|\epsilon\| < \delta$ and $T$ clearly has finite dimensional range so we are done.


问:
Calculating a probability using the join density function
here's the problem I'm currently working on :

Let $X$ and $Y$ be two continuous random variable with join density function given by $$ f_{X,Y}(x,y)=e^{-x^2y}1_{[1,\infty)}(x)1_{[0,\infty)}(y). $$

Since $e^{x}>0, \forall x \in \mathbb{R}$, we have that  $f_{X,Y}>0, \forall (x,y) \in \mathbb{R}^2$. Also we see that

\begin{align*}
\int \int_{\mathbb{R}^2} f_{X,Y}(x,y)dxdy &=\int_{1}^{\infty} \left( \int_{0}^{\infty} e^{-x^2y} dy \right) dx\\
                     &=\int_{1}^{\infty} \left[ \frac{-1}{x^2}e^{-x^2y} \right]_{0}^{\infty}dx\\
                     &=\int_{1}^{\infty} \frac{1}{x^2}dx\\
                     &=\left[ \frac{-1}{x} \right]_{1}^{\infty}\\
                     &=1.
\end{align*}

Therefore, we know that $f_{X,Y}$ is a density function and we can move on to the next question. It is asked to evaluate the folowing probabilty : $\mathbb{P}(X^2Y>1)$.

I am unsure how to find the right boundaries for the integrals. Here's what I've done so far :

The condition $X^2Y>1$ gives us the following inequalities :

$$1 \leq x < \infty \hspace{1cm} \text{and} \hspace{1cm} \frac{1}{x^2} \leq y < \infty$$

or

$$0 \leq y < \infty \hspace{1cm} \text{and} \hspace{1cm} \frac{1}{\sqrt{y}} \leq x < \infty.$$

From that I deduce the two folowing double integrals :

$$ \mathbb{P}(X^2Y>1)=\int_{1}^{\infty} \left( \int_{x^{-2}}^{\infty} e^{-x^2y}dy \right)dx $$

and

$$ \mathbb{P}(X^2Y>1)=\int_{0}^{\infty} \left( \int_{y^{-1/2}}^{\infty} e^{-x^2y}dx \right)dy. $$

I already evaluated the first integral but the second one is more complicated. I am right ? How do you find the boundaries for such problems ?

Thanks

答:
The first integral is correct and gives you the right solution:

$$\mathbb{P}[X^2Y>1]=\frac{1}{e}$$

the second one is wrong

The correct one should be

$$\mathbb{P}[X^2Y>1]=\int_0^{1}\Bigg[\int_{y^{-\frac{1}{2}}}^{\infty}e^{-x^2y}dx   \Bigg]dy+\int_1^{\infty}\Bigg[\int_{1}^{\infty}e^{-x^2y}dx   \Bigg]dy$$

It is self evident that it is better the first choice.

This is a clear example showing that not always integration's order is neutral.

The requested probability is

$$\mathbb{P}[X^2Y>1]=\mathbb{P}[Y>\frac{1}{X^2}]$$

Thus the area to be integrated is the following purple one (the blu curve is the equation $y=\frac{1}{x^2}$)

In your second way to solve it you tried to integrate also the yellow one

ContentLicense: CC BY-SA 4.0


问:
Laplace's equation and separation of variables in spherical coordinates done rigorously
Consider a region $\Omega \subset \mathbb{R}^3$. A well-known method of solving the Laplace equation $\Delta f = 0$ on $\Omega$ (we're looking for clasical, i.e. $C^2$ solutions) is to employ the separation of variables, i.e. to look for functions $f: \mathbb{R}^3 \setminus \{ 0 \} \to \mathbb{R}$ of the form
$$f(r, \theta, \varphi) = R(r)\Theta(\theta)\Phi(\varphi)$$
and then claim that a general solution can be expressed by a series of such ansatz solutions. However, most references I found (Jackson, Griffiths and various resources found online) skip the details of this approach. I'd like to find a reference that treats this method rigorously

Precise description of the details I'm missing below.

Since the Laplace equation in spherical coordinates reads
$$\nabla^2 f = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial f}{\partial r}\right)
  + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial f}{\partial \theta}\right)
  + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2 f}{\partial \varphi^2} = 0$$

Substituting $\psi(r, \theta, \varphi) = R(r)\Theta(\theta)\Phi(\varphi)$ we get
$$\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) + \frac{1}{\Theta}\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial \Theta}{\partial\theta}\right) + \frac{1}{\Phi}\frac{1}{\sin^2\theta}\frac{\partial^2\Phi}{\partial\varphi^2} = 0$$

After multiplying by $\sin^2\theta$ we immediately get that for some constant $\lambda$
$$\frac{1}{\Phi}\frac{\partial^2\Phi}{\partial\varphi^2} = \lambda$$
Since $\Phi$ is continuous and $2\pi$-periodic, then necessarily $\lambda < 0$. At this point we're allowed to substitute $$\frac{1}{\Phi}\frac{\partial^2\Phi}{\partial\varphi^2} = -m^2$$
which yields solutions $\Phi(\varphi) = \operatorname{Re} Ae^{\pm im \varphi}$. For some reason most resources write $\Phi(\varphi) = e^{\pm im \varphi}$, but such a function is not a real function.

Similarly, we must have
$$\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = \alpha, \qquad \frac{1}{\Theta}\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial \Theta}{\partial\theta}\right) = -\alpha$$

At this point one does a substitution $\alpha = l(l+1)$. However, this substitution is only if $\alpha \geq  - \frac {1}{4}$. Moreover, the texts seem to assume that $l$ is an integer, and it's completely unclear why we can assume it. None of the texts I found really comments why this substitution is allowed.

Furthermore, $Y_{lm}$ is defined as
$$Y_{lm}(\theta, \varphi) = c_{lm} P_{lm}(\cos \theta) e^{im\varphi}$$
However, we were looking for real solutions, and this solution is a complex one, the entity are mismatched.

Finally the general solution is claimed to be of the form:
$$f(r, \theta, \varphi) = \sum_{l=0}^\infty \sum_{m=-l}^\ell a_{lm} r^l  Y_{lm} (\theta, \varphi )$$
However, all the texts I found don't justify why functions of the form $r^l  Y_{lm}$ form a complete set of functions. It's a priori unclear why functions $r^l$ form an orthogonal basis of $L^2([0, \infty))$.

答:
There are a lot of great questions here, many of which I struggled with myself when I was first learning about the physical applications of PDEs. I'll go point by point and try to give the best explanations that I can. First, some housekeeping - whenever you use spherical coordinates, it is always good to explicitly state the coordinate transformation and notation you're using for reasons I mention in this answer and also discussed on Wolfram Mathworld. So, here we go:

I will use the $\displaystyle ( r,\theta ,\phi )$ (radial, azimuthal, polar) convention for spherical coordinates, as used by Wolfram Mathworld. Explicitly, the coordinate transformation is
\begin{equation*}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =\begin{bmatrix}
r\cos \theta \sin \phi \\
r\sin \theta \sin \phi \\
r\cos \phi
\end{bmatrix}
\end{equation*}

Yes, I know the theta and phi are in the opposite order as to what you're using, but this is just the notation that I'm the most comfortable with and trying to switch it around will only make me more prone to errors. Let's get on with actually answering your questions then.

First question: Why do we have complex solutions to what should be a real valued PDE?

This is probably the easiest question to answer. Take for instance the ordinary differential equation
$$y''+y=0$$
We can solve this in the complex plane using the usual methods as
$$y(x)=c_1e^{-ix}+c_2e^{ix}$$
This is the most general solution and allows for complex valued $y$, and, critically, complex valued coefficients. Going a bit further, we can break $c_1,c_2$ and $e^{-ix},e^{ix}$ into their real and imaginary parts:
$$y(x)=(a_1+ib_1)(\cos(x)-i\sin(x))+(a_2+ib_2)(\cos(x)+i\sin(x))$$
Which now allows us to split $y$ into its real and imaginary parts:
$$y(x)=\operatorname{Re}y+\operatorname{Im}y=\bigg[(a_1+a_2)\cos(x)+(b_1-b_2)\sin(x)\bigg]+i\bigg[(a_2-a_1)\sin(x)+(b_1+b_2)\cos(x)\bigg]$$
If we make the artificial restriction that $y$ must be real valued, i.e $\operatorname{Im}y=0$, we get $a_1=a_2~;~b_1=-b_2\implies c_1=c_2^*\implies y(x)=A\sin(x)+B\cos(x)$.
In short, if we were to write
$$y''+y=0\implies y(x)=c_1 e^{-ix}+c_2e^{ix}$$
This is still valid even if we only want real valued $y$! It's just implied that $c_1=c_2^*$. Why do we do this? Sometimes, writing the solution in terms of exponentials can make expressions a little bit shorter and easier to manipulate, even if it means we have to introduce complex valued coefficients.

Second question: Why does everyone just assume $l$ is an integer? And what do we do if $\alpha<-1/4$?

The second part of this question is in some ways easier to answer than the first. the answer to the first question is, sort of, "because that's the simplest possible case", but there's more to it than that (which I'll get to after I answer the second part.)

In short, complex numbers come to save the day again. Suppose $\alpha=-2$. Then,
$$l(l+1)=-2$$
$$l^2+l+2=0$$
$$l=\frac{-1\pm\sqrt{1-4\cdot2}}{2}=\frac{-1}{2}\pm i\frac{\sqrt{7}}{2}$$
This is totally allowed! While this will no longer be a Legendre polynomial, we do have what is known as the Legendre function (of the first kind):
$$\mathscr{P}_{l,m}(z)=\frac{1}{\Gamma(1-m)}\left(\frac{1+z}{1-z}\right)^{m/2}{}_2F_1\left([-l,l];[1-m];\frac{1-z}{2}\right)$$
Which uses the (Gaussian) hypergeometric function. It can be shown (though not easily) that in the case of $m,l\in \mathbb{N}$ and $m<l$ this reduces to the normal associated Legendre polynomial.

On to the first part then: why does everyone just assume $l$ is an integer? To answer this, let's solve the simpler problem of the spherical harmonics where we assume there is no $\theta$ (azimuthal) angle dependency. That is, let's have a look at the PDE
$$\nabla^2 U(r,\phi)=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial U}{\partial r}\right)+\frac{1}{r^2\sin\phi}\frac{\partial}{\partial \phi}\left(\sin\phi \frac{\partial U}{\partial \phi}\right)$$
Assuming separability, i.e $U(r,\phi)=R(r)\Phi(\phi)$, we get (also note that I've multiplied through by $r^2$ which is fine as long as we don't care about the origin)
$$\frac{\partial}{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)=\alpha R$$
And
$$\frac{1}{\sin\phi}\frac{\partial}{\partial \phi}\left(\sin\phi \frac{\partial \Phi}{\partial \phi}\right)=-\alpha \Phi$$
A bit of an inspired substitution $\alpha=l(l+1)$. This substitution is particularly nice if $\alpha$ is a positive integer of the form $n^2+n$, so that $l$ is a positive integer, but in general we just have $l=(-1\pm \sqrt{1-4\alpha})/2$. Recognizing that the $R$ equation is a Cauchy-Euler ODE, we get
$$R(r)=Ar^l+B\frac{1}{r^{l+1}}$$
Now for the $\Phi$ equation. We have, after some manipulation
$$\Phi''+\cot(\phi)\Phi'+l(l+1)\Phi=0$$
Which, after a substitution $\mu=\cos\phi$, becomes (I'll let you work out the details)
$$(1-\mu^2)\frac{\mathrm{d}^2\Phi}{\mathrm{d}\mu^2}-2\mu\frac{\mathrm{d}\Phi}{\mathrm{d}\mu}+l(l+1)\Phi=0$$
The reduced form of Legendre's differential equation.

So we can now resort to the Frobenius method and try to come up with a solution $\Phi(\mu)=\sum_{n=0}^\infty a_n \mu^n$. I'll cut to the chase (you can find the details on Wolfram Mathworld) and assert that after some work we get a recurrence
$$a_{n}=\frac{\left(l+(n+1)\right)(l-n)}{(n+1)(n+2)}a_{n-2}$$
So what's the upshot here? Yes, you are correct: $l$ doesn't have to be an integer for the above to simply be a solution to our differential equation. BUT, If $l$ wasn't an integer, then the equation $a_n=0$ has no solutions since $l+n+1=0$ or $l-n=0$ is clearly impossible if $l$ is not an integer. So, as long as $a_0\neq 0$, none of the coefficients in the expansion of $\Phi(\mu)$ would zero, i.e, it would be an infinite sum with quickly growing coefficients and thus would diverge for $\mu=1$ which corresponds to $\phi=0,\pi$. Since typically we are solving for an electric potential, it doesn't make any physical sense to have a electrical potential that blows up near the poles. However if $l$ is an integer, we will at some point reach some value of $n$ such that $a_n=0$ which would lead to $a_{n+2}=a_{n+4}=...=0$. All the previous even coefficients will be nonzero and their precise form will follow from the recurrence relation. As for the odd coefficients, i.e $a_{n+2m+1}$, we can simply choose all of these to be $0$. Since the recurrence is of second order, we need only worry about every second coefficient and we can choose the odd ones to all be zero which trivially satisfies the recurrence. Which half of the coefficients we set to $0$ depends on the parity of $l$. The precise series expansions for each case are listed on Wolfram Mathworld. To illustrate the above, the expansions will look like
$$P_l(\mu)=a_0\mu^0+0\mu^1 +a_2\mu^2+0\mu^3...+0\mu^{l-1}+a_l\mu^l+0\mu^{l+1}+0\mu^{l+2}+0+...\text{  for even }l$$
$$P_l(\mu)=0\mu^0+a_1\mu^1+0\mu^2+a_3\mu^3+...0\mu^{l-1}+a_l\mu^l+0\mu^{l+1}+0\mu^{l+2}+0+...\text{  for odd }l$$
As you can see, the expansions terminate which will mean that $\Phi$ will remain bounded at the poles.

To summarize:
No, $l$ need not be an integer in order to get something that solves the ODE. But, in order to get a solution of the ODE that remains bounded, $l$ must be an integer.

Finally, onto complete function sets and orthogonal bases. It's pretty obvious that $r^l$ form an basis (but not an orthogonal basis) for $L^2[0,\infty)$ - simply consider Taylor expansions. However, the Legendre polynomials do provide an orthogonal basis for $L^2[-1,1]$. We can arrive at this by applying the Gram-Schmidt process to $\{1,x,x^2,x^3,...\}$ on the interval $[-1,1]$. (REFERENCE: https://en.wikipedia.org/wiki/Orthogonal_functions#Polynomials) If we instead want an orthogonal basis on $[0,\infty)$, we can perform a Cayley Transform which brings us to the Legendre rational functions. Finally, about the completeness of $\{r^lY_{l,m}\}$ - I don't think this is required. $\{r^l Y_{l,m}\}$ need only form a basis, not necessarily an orthogonal one.


问:
Evaluating limits; direct substitution all at once or one by one?
When evaluating limits by direct substitution, is it necessary to give the limiting value to all the variables at the same time, or is it okay to simplify smaller expressions in the overall function by substituting the variable with the value one by one?

For an (impractical) example, $\frac{(e^{(x+2)ln4})^\frac{\lfloor x+1\rfloor}{4}-16}{4^x - 16}$. This is the expression as the function approaches 2 from the left. On simplifying it, you get $\frac{2^{\frac{\lfloor x+1 \rfloor (x+2)}{2}}}{2^{2x}-16}$. One solution from the internet broke down the $\lfloor x+1\rfloor$ into 2, but kept everything else the same.

How does that work out?

答:
This is due to a property of the floor function. Note that for $x \in [1,2)$, $\lfloor x+1 \rfloor = 2$. Hence

$$\frac{2^{\frac{\lfloor x+1 \rfloor (x+2)}{2}}}{2^{2x}-16} = \frac{2^{\frac{2 (x+2)}{2}}}{2^{2x}-16}$$

Since we only care about the value of the function near $2^-$, writing

$$\lim_{x \to 2^-}\frac{2^{\frac{\lfloor x+1 \rfloor (x+2)}{2}}}{2^{2x}-16} = \lim_{x \to 2^-}\frac{2^{\frac{2 (x+2)}{2}}}{2^{2x}-16}$$

is valid. One cannot substitute the limit value of $x$ one by one, e.g.

$$4=\lim_{x \to 2} \frac {x^2-4}{x-2} \ne \lim_{x \to 2}\frac {4-4}{x-2} =0$$

but what we did to your example is not substitution. Another example would be:

$$\lim_{x \to 0^-}\frac {|x|}{x} = \lim_{x \to 0^-}\frac {-x}{x} = -1$$

since near $0^-$, $|x| = -x$.


问:
A limit involving the integer nearest to $n$-th power
Find all $x\in\mathbb{R}$ such that $$\lim_{n\to\infty}|x^n-\langle x^n\rangle|=0$$ where $\langle t\rangle$ is the integer nearest to $t$ (eg. $\langle\frac{1}{3}\rangle=0$, $\langle\frac{8}{3}\rangle=3$, $\langle k+\frac{1}{2}\rangle$ is not defined for $k\in\mathbb{Z}$).

I found this somewhere in the internet (today I searched again in IMO shortlists and didn't found, so it's probably not from there), tried to solve for a long time, but without nontrivial results (numbers $x\in\mathbb{Z}$ and $x\in(0,1)$ satisfy this, but I have no idea how to examine e.g. $x=\sqrt{2}$).

答: A Pisot number will certainly work. For instance, $\sqrt{2}+1$.


问:
Lower semicontinuous envelope of the upper semicontinuous envelope of a function.
Let $u: R^n \rightarrow R$ be a given function. The lower semicontinuous envelope of $u$ is defined as the following $$u_{*} (x) = \sup_{r>0} \inf_{ 0<|y-x|<r} u (y)$$ and the upper semicontinuous envelope $u^{*}$ of $u$ is defined as $$ u_{*} (x) = \inf_{r>0} \sup_{ 0<|y-x|<r} u (y) .$$ My question is whether the following equality holds?
$$u_{*} (x) = (u^{*})_{*} (x)$$
How to prove this?

答:
A counterexample is $u: \Bbb R^n \to \Bbb R$ defined by $u(x) = 0$ if $x$ has only rational components, and $u(x) = 1$  otherwise.

Then $u_*(x) = 0$, $u^*(x) = 1$, and $(u^*)_*(x) = 1 \ne u_*(x)$ for all $x \in \Bbb R^n$.


问:
Does $\Vert f\Vert_{L^\infty}^2\leq \Vert f\Vert_{L^2}\Vert f'\Vert_{L^2}$ hold?
I know that Sobolev's embedding implies that $H^1(\mathbb{R})\hookrightarrow C(\mathbb{R})\cap L^\infty(\mathbb{R})$, which in particular implies that $$
\Vert f\Vert_{L^\infty}\leq \Vert f\Vert_{H^1}.
$$
Now, I was wondering about the following slightly more specific (?) inequality. More precisely, I was wondering if $f\in H^1(\mathbb{R})$, then the following inequality holds:
$$
\Vert f\Vert_{L^\infty}^2\leq \Vert f\Vert_{L^2}\Vert f'\Vert_{L^2}.
$$
The problem here is to be able to put a factor only with the $L^2$ norm of $f$. Does this inequality hold?

答:
Let $f \in \mathcal{S} (\mathbb{R})$. Let $x^*$ be any point realizing the maximum of $|f|$. Then

$$2|f(x^*)| = \left|\int_{-\infty}^{x^*} f'(t) \ dt\right| + \left|\int_{x^*}^{+\infty} f'(t) \ dt\right| \leq \|f'\|_{\mathbb{L}^1}.$$

Hence, $\|f\|_{\mathbb{L}^\infty} \leq \frac{1}{2} \|f'\|_{\mathbb{L}^1}$. Applying this inequality to $f^2$ and using the Cauchy-Schwarz inequality,

$$\|f\|_{\mathbb{L}^\infty}^2 = \|f^2\|_{\mathbb{L}^\infty} \leq \frac{1}{2} \|(f^2)'\|_{\mathbb{L}^1} = \|ff'\|_{\mathbb{L}^1} \leq \|f\|_{\mathbb{L}^2} \|f'\|_{\mathbb{L}^2}.$$

Finally, $\mathcal{S} (\mathbb{R})$ is dense in $H^1 (\mathbb{R})$, and all of $f \mapsto \|f\|_{\mathbb{L}^\infty}$, $f \mapsto \|f\|_{\mathbb{L}^2}$ and $f \mapsto \|f'\|_{\mathbb{L}^2}$ are continuous in the $H^1$ norm (the later two by definition of the $H^1$ norm, the former by the Sobolev embedding). Hence the claimed inequality actually holds in $H^1 (\mathbb{R})$.


问:
Generalisation of Dominated Convergence Theorem
Wikipedia claims, if $\sigma$-finite the Dominated convergence theorem is still true when pointwise convergence is replaced by convergence in measure, does anyone know where to find a proof of this? Many thanks!

Statement of the theorem:

Let $\mu$ be $\sigma$-finite, $|f_n|\leq g$ and $f_n\rightarrow f$ in measure, then we must have

$\int f_n \rightarrow \int f$ and $\int|f_n-f| \rightarrow 0$

答:
I don't know why this question is reopened. However,  given none of the proofs really pleasing me, I'll present one.

So just by interchanging the integral signs, we see that:

$$\int |f-f_n| =  \int_{0}^{\infty} \underbrace{ \mu\left( |f_n-f|>t\right) }_{=:A_n(t)}dt \text{ (1) }$$
And under the condition of boudedness of $f_n$, we have :
$$0 \le A_n(t) \le \mu( 2g >t) := G(t)$$
Note that $$\int_{0}^{\infty} G(t)= \int 2g < \infty$$
Thus the DCT conditions for the first integrals are fulfilled,  so :

$$ \lim_{n \rightarrow + \infty} \int |f-f_n| = \lim_{n \rightarrow + \infty} \int_{0}^{\infty}  \mu\left( |f_n-f|>t\right) = \int_{0}^{\infty} \lim_{n \rightarrow + \infty}  \mu\left( |f_n-f|>t\right) = 0 $$

ContentLicense: CC BY-SA 3.0


问:
Probability of a random variable is greater than its predecessors
Suppose I have a sequence of iid random variables $(X_n)_{n\in \mathbb{N}}$ would I be correct in thinking that the following is true?

$$\mathbb{P}(X_k>X_1,\cdots, X_{k-1})=\mathbb{P}(X_k>X_1)^{n-1}$$

My thoughts for this is that $$\mathbb{P}(X_k>X_1,X_k>X_2,\cdots,X_k>X_{k-1})=\mathbb{P}(X_k>X_{k-1})\cdots\mathbb{P}(X_k>X_1)=\mathbb{P}(X_k>X_1)^{n-1}$$
where the first equality follows from independence and second equality follows from identically distributed.

However, there is also a part of me that goes against this argument, let me illustrate with an example:

Suppose we just consider $k=3$ then

$$\mathbb{P}(X_3>X_1,X_2)=\mathbb{P}(X_3>X_2>X_1)+\mathbb{P}(X_3>X_1>X_2)=2\mathbb{P}(X_3>X_1)^2.$$

where I obtained the square on $\mathbb{P}(X_3>X_1)$ is due to I splitted up $\mathbb{P}(X_3>X_2>X_1)=\mathbb{P}(X_3>X_2)\mathbb{P}(X_2>X_1).$
Then evidently two are not the same.

Where have I gone wrong in applying independence to this problem? Without further context on $X_n$ how can we reach a correct expression to the concerned probability? (which is $\mathbb{P}(X_k>X_1,\cdots, X_{k-1})$)

Many thanks!

答:
If $P_1, P_2, P_3$ are i.i.d., then

$$\mathbb{P}(X_3>X_2>X_1) \neq \mathbb{P}(X_3>X_2)\mathbb{P}(X_2>X_1)$$

because the two events on the right are not independent.

The way we usually approach this is to observe that since the variables are i.i.d., no one variable is more likely to be greatest than any other:

$$\mathbb{P}(X_1>X_2,X_3) = \mathbb{P}(X_2>X_1,X_3) = \mathbb{P}(X_3>X_1,X_2). $$

If the variables have continuous distributions then the probability that any two are exactly equal is zero, and the probability is $1$ that there will be one variable that is greater than the other two:

$$\mathbb{P}(X_1>X_2,X_3) + \mathbb{P}(X_2>X_1,X_3) + \mathbb{P}(X_3>X_1,X_2) =1. $$

From these two equations you can find $\mathbb{P}(X_3>X_1,X_2).$

If the distribution is not continuous then the calculation is typically a bit more complicated because you have to account for the non-zero probability that no variable is greater than the other two.


问:
weighted sum of exponential random variables
Suppose $X_i$ are i.i.d random variables and $X\sim\operatorname{Exp}(\lambda)$ i.e.
$Pr(X\le x)=1-e^{-\lambda x}$ for $x \ge 0$.
What is the density function of $Z=\sum_{i=1}^{N}\alpha_iX_i$ where $\alpha_i$ are constants?</p>

答: Actually, the weighted exponential random variable is still an exponential random variable. Only the mean is scaled with the same weight. Therefore, the sum $Z$ is an Hypo-exponential random variable and it is summarized in wiki's webpage Hypoexponential.</p>


问:
Optimal transport and total variation distance
I have a question regarding the following concept equating total variation distance with a particular case of optimal transport.

I don't understand why equality (6.11) holds.  We know by Kantorovich duality that the RHS is equal to $$2 \sup_{\phi \text{ Lipschitz} \\ |\phi|_{\text{Lip}} \leq 1} \int \phi d\mu - \int \phi d\nu \equiv f(\mu, \nu)$$ as a function is $c-$convex for a distance function $c = 1(x \ne y)$ if and only if it is $1-$Lipschitz.

As for the total variation, it is defined as
$$T(\mu, \nu) \equiv  \sup_{A \in \mathcal{F}} |\mu(A) - \nu(A)|$$
where $\mathcal{F}$ is our $\sigma-$algebra on whichever Polish space we're working with.  It is obvious that for $\phi(x) = 1_A(x)$, we have that $\phi$ is $1-$Lipschitz and therefore $T(\mu, \nu) \leq f(\mu, \nu)$.  I'm confused why we need the $2$ here, and how the other direction of the inequality would be shown?

Specifically, I need that for any $1-$Lipschitz function, there exists a set $A \in \mathcal{F}$ such that $|\mu(A) - \nu(A)| \ge 2 \int \phi d\mu - \int \phi d \nu$, but I have no idea how to get this right.  Any help would be massively appreciated.

(The excerpt is from Villani (2009))

答:
I think there is some difference in definition. Look the lecture notes Probability in High Dimensions by Van-Handel. In example 4.14 the author writes:

$$
||\mu - \nu||_{TV} = \inf_{M\in\mathcal C(\mu,\nu)}M(X\neq Y)
$$

And he then goes on to prove this.

What might be happening is a different definition of the T.V metric.

Indeed, we can prove that using your definition of TV, the equality
$$||\mu - \nu||_{TV} = \sup_A|\mu(A) - \nu(A)| = 2\inf P[X\neq Y]$$

Would be inconsistent. Note:

$$\mu(A) - \nu(A) = P[X \in A] - P[Y \in A] = $$
$$=
P[X \in A, X=Y] - P[X \in A,X\neq Y]+
P[Y \in A,X=Y] - P[Y \in A,X\neq Y] =
$$
$$
= P[X \in A, X\neq Y] - P[Y \in A, X \neq Y] \leq P[X\neq Y]
$$
Therefore,
$$\sup_A|\mu(A) - \nu(A)| \leq P[X\neq Y]$$
Hence, $\sup_A|\mu(A) - \nu(A)|>0 \implies 2P[X\neq Y]> \sup_A|\mu(A)-\nu(A)|$


问:
Proving the solution of an ode converges as $t$ goes to infinity given its norm is bounded by a convergent function
Question

Let $f \in \mathcal C(\mathbb R_+ \times \mathbb R^n, \mathbb R^n)$ and $g \in \mathcal C(\mathbb R_+ \times \mathbb R, \mathbb R_+)$. Assume that

$g(t,u)$ is non-decreasing in $u$ for each fixed $t$ and
$\vert f(t,x)\vert \lt g(t,\vert x \vert), (t,x)\in \mathbb R_+ \times \mathbb R^n$

Every solution $u(t) = u(t,t_0,u_0)$ of the IVP $u^\prime(t) = g(t,u),
u(t_0)=u_0$ is bounded.

Show that every solution of the IVP $x^\prime(t) = f(t,x), x(t_0) = x_0$ exists on $[t_0,\infty)$ and converges to a constant vector as $t \to \infty$.

My Attempt

I have completed the first part of the question (show every solution of the IVP $x' = f(t,x), x(t_0) = x_0$  exists on $[t_0,\infty)$) using following corollary.

Corollary: Let $f \in \mathcal C(J \times \mathbb R^n, \mathbb R^n)$  where $J=[a,b)$ and $\vert f(t,x) \vert \lt g(t,\vert x \vert)$ for $(t,x)\in J \times R^n$ and $g \in \mathcal C(\mathbb R_+ \times \mathbb R, \mathbb R_+)$. If $\gamma(t)$ is the maximal solution of $u^\prime = g(t,u), u(t_0)=u_0$ existing on $J$, then the solution $x(t) = x(t,t_0,x_0)$ of $x^\prime = f(t,x), x(t_0) = x_0$ where $\vert x_0 \vert \leq u_0$ exists on $J$ and $\vert x(t) \vert \leq \gamma(t)$ for all $t\in J$.

Simply let $\gamma(t)$ be the maximal solution of  $u^\prime = g(t,u), u(t_0)=\vert x_0 \vert$. Since we know all solutions of this IVP are bounded (by (2)), we know that $\gamma(t)$ must exist on $J = [t_0,\infty)$. By the corollary, $x(t)$ also exists on $[t_0,\infty)$. Moreover, $\vert x(t) \vert<\gamma(t)\forall t\in[t_0,\infty)$.

Having said that, I do not know how to prove that $x(t)$ must converge. I have the following results:

Since $g(t,u)$ is always positive, we know that $\gamma(t)$ is an increasing bounded continuous function of $t$ (by (2)). Thus, $\gamma(t)$ must converge. Hence, $\vert x(t) \vert$ is bounded from above by a convergent function.

$g(t,u)$ is a non-decreasing function of $u$ for every fixed $t$, thus $\vert x^\prime(t) \vert \leq \vert f(t,x) \vert \leq \vert g(t,\vert x(t) \vert) \vert \leq \vert (g(t,\gamma(t)) $

but I'm not sure what they imply about the convergence of $x(t)$. Any help would be greatly appreciated.

答:
We have for all $t \ge0$

$$\vert x^\prime(t) \vert = \vert f(t,x(t)) \vert \leq g(t,\vert x(t) \vert)\leq g(t,\gamma(t))=\gamma^\prime(t)$$

Hence for $ 0 \lt u \lt v$

$$0 \le \left\vert x(v) - x(u) \right\vert = \left\vert\int_u^v x^\prime(t)  \ dt\right\vert \le \int_u^v \vert x^\prime(t) \vert \ dt \le \gamma(v) - \gamma(u)$$

As $\gamma$ has a limit at $\infty$, $\gamma(v) - \gamma(u)$ can be as small as desired for $u,v$ large enough. Which implies that it is also the case for $\vert x(v) - x(u) \vert$ and according to Cauchy criterion that $\lim\limits_{x \to \infty} x(t)$ exists.


问:
For $a,b>0$ and $c\ge0$ prove that $\frac{2(a+b)}{3a+6b+9c}+\frac{6(b+c)}{5a+2b+3c}+\frac{3(a+c)}{2a+8b+6c}\ge\frac{3}{2}$
For $a,b>0$ and $c\ge0$ prove that $\frac{2(a+b)}{3a+6b+9c}+\frac{6(b+c)}{5a+2b+3c}+\frac{3(a+c)}{2a+8b+6c}\ge\frac{3}{2}$

This question is easily solved if you say: $x=3a+6b+9c$, $y=5a+2b+3c$, $z=2a+8b+6c$.

From this we have that:

$2(a+b)=\frac{-x+y+z}{2}$

$6(b+c)=\frac{x-y+z}{2}$

$3(c+a)=\frac{x+y-z}{2}$

Hence it is enough to prove that $\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\ge6$ which holds true.

I couldn't solve this question, as I couldn't think of the substitution. Could you please explain to me intuitively why I should have thought of doing the particular $x,y,z$ substitutions?

答:
It is not too the intuitive for me. The more intuitive substituion for me is

$$
\begin{align}
x&=2(a+b)\\
y&=6(b+c)\\
z&=3(a+c)
\end{align}
$$

which will give us

$$
\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\geq\frac{3}{2}
$$

which is a well known inequality


问:
A problem on polynomial
Let $p(x)$ be polynomial of degree $10$. It is given that $p(0)=1$ and $p(x)=x$ for $x=1$ to $10$. What is the value of $p(11)$. </p>

<p>I am able to show that it is certainly greater than or equal to $10$, but unable to find the exact value.  </p>

答: Consider the polynomial $q(x):=p(x)-x$. This will likewise be of degree $10$ (why?), and it can readily be seen that $1,2,...,10$ are roots of this polynomial. They are in fact the only roots of the polynomial, and each has multiplicity $1$ (why?). Hence, $$q(x)=\alpha(x-1)(x-2)\cdots(x-10)$$ for some nonzero constant $\alpha$. Use the fact that $q(0)=1$ to find $\alpha$. Finding $q(11)$ will then be straightforward, and from there we find $p(11)$.

Title: A problem on polynomial


问:
Example of functions where limits don't commute
Can we have a function $f(h,k)$ such that $$\lim_{h\to 0}\lim_{k\to 0}f(h,k)\neq \lim_{k\to 0}\lim_{h\to 0}f(h,k)?$$

答:
Moore-Osgood theorem deals with interchange iterated limits like this, See here and here

Simple counter example from the wiki:

$$ \lim_{x,y \to 0,0} f(x,y) = \lim_{x \to 0 } \lim_{ y \to 0} \frac{x^2}{x^2 +y^2}$$

More discussion here detailed discussion about this here


答: Say $$f(x,y)=\frac {x-y}{x+y}\quad(x+y\neq0).$$You can extend the domain to $\Bbb R^2$ by defining $f(x,y)=0$ when $x+y=0$.

Title: Example of functions where limits don't commute

Body: Can we have a function $f(h,k)$ such that $$\lim_{h\to 0}\lim_{k\to 0}f(h,k)\neq \lim_{k\to 0}\lim_{h\to 0}f(h,k)?$$


问:
$V_{\mathbb{A}^n}(\sum_{\imath}\mathfrak{a}_{\imath}) = \bigcap_{\imath} V_{\mathbb{A}^n}(\mathfrak{a}_\imath)$
Let
$$
V_{\mathbb{A}^n}(\mathfrak{a}) := \{x \in \mathbb{A}^n \; \vert \; f(x) = 0 \; \text{for all} \; f \in \mathfrak{a}\},
$$
for some affine space $\mathbb{A}^n$, a polynomial $f$ and an ideal $\mathfrak{a}$. I want to proof that
$$
V_{\mathbb{A}^n}(\sum_{\imath}\mathfrak{a}_{\imath}) = \bigcap_{\imath} V_{\mathbb{A}^n}(\mathfrak{a}_\imath),
$$
and that
$$
V_{\mathbb{A}^n}(\bigcap_{i=1}^{n} \mathfrak{a}_i) = \bigcup_{i=1}^{n}V(\mathfrak{a}_i).
$$

How can I prove above equations?
And why is the finite case in the second one so important?

Thoughts on the first case: The sum of two ideals is again an ideal as an abelian group can be subdivided into subgroups or joint under the group operation to a larger abelian subgroup. Multiplication with an element of the ring preserves still the group structure, thus yields again an element of the abelian group.

答:
This is Hartshorne Lemma II.2.1(b), but I appreciate that Hartshorne's proof is a little sparse in terms of details.

As a set $V_{\mathbb{A}^n}(\mathfrak{a})$ is the set of all prime ideals containing $\mathfrak{a}$, so $V_{\mathbb{A}^n}(\sum \mathfrak{a}_i)$ is the set of all prime ideals containing $\sum \mathfrak{a}_i$. Recall that $\sum \mathfrak{a}_i$ is the smallest ideal containing every $\mathfrak{a}_i$, so an ideal $\mathfrak{p}$ contains $\sum\mathfrak{a}_i$ if and only if $\mathfrak{p}$ contains each $\mathfrak{a}_i$. Thus the set $V_{\mathbb{A}^n}(\sum_i \mathfrak{a}_i)$ of prime ideals containing $\sum_i \mathfrak{a}_i$ is the same as the set $\bigcap_i V_{\mathbb{A}^n}(\mathfrak{a}_i)$ of prime ideals contatining each $\mathfrak{a}_i$.

The second one is similar and I encourage you to try and prove it - once you have the $n = 2$ case you're done by induction, so try to understand why $V(\mathfrak{ab}) = V(\mathfrak{a}) \cup V(\mathfrak{b})$; if you have a prime ideal containing $\mathfrak{ab}$ can you show that it contains either $\mathfrak{a}$ or $\mathfrak{b}$ (think about why we used a prime ideal here).

The finite case is important because of the definition of topological spaces; the set of opens is closed under arbitrary union but closed under only finite intersections. This statement is saying that the $V_{\mathbb{A}^n}(\mathfrak{a})$ form a base for a topology (the Zariski topology).

edit: as per the reccomendation of LetGBeTheGraph in the comments, let's see the example of why the union of every prime ideal of $\mathbb{Z}$ is not closed in this Zariski topology.

Recall that $n \in (k)$ iff $n$ is a multiple of $k$. Thus for example the intersection $(2) \cap (3) = (6)$ because if $n \in \mathbb{Z}$ is a multiple of two and three then it's a multiple of six. Thus any $a \in \bigcap_{p\text{ prime}} (p)$ must simultaneously be a multiple of every prime number, no such integer satisfies this, so there is no prime ideal containing $\bigcap_{p\text{ prime}} (p)$, in our terminology; $V_{\mathbb{Z}}(\bigcap_{p\text{ prime}} (p)) = \emptyset$.

We compare this to $\bigcup_{p\text{ prime}} V_{\mathbb{Z}}((p))$. Given any prime $p$ we have that $(p)$ is a prime ideal containing $(p)$, i.e. $(p) \in V_{\mathbb{Z}}((p))$, so at very least we have that $(p) \in \bigcup_{p\text{ prime}} V_{\mathbb{Z}}((p))$. Thus we have that $\bigcup_{p\text{ prime}} V_{\mathbb{Z}}((p)) \neq V_{\mathbb{Z}}(\bigcap_{p\text{ prime}} (p))$.


问:
Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$
How should we find two rational numbers $\alpha$ and $\beta$ such that

$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$

The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you

答:
Apply the detesting formula
$$\sqrt[3]{a+b \sqrt R}=\frac12\sqrt[3]{3bs-a}\left(1+\frac1s \sqrt R\right)\tag1$$
where $s$ is the solution of
$$s^3-\frac{3a}b s^2+3R s-\frac{a}b R =0$$

So, for $\sqrt[3]{7+5\sqrt{2}}$, solve for $s$ in
$$s^3-\frac{21}5s^2+6s-\frac{14}5=\frac15(s-1)(5s^2-16s+14)=0$$
which yields $s=1$. Plug into (1) to obtain the denestation
$$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$
or, $\alpha=\beta=1$.


问:
What is the meaning of "independent events " and how can we logically conclude independence of two events in probability?

What is the meaning of  "independent events " in probability

For eg: Two events (say A and B)are independent , what I understand is the occurrence of A is not affected by occurrence of B .But I am not comfortable with this understanding , thinking this way every events I meet upon are independent!, does there exist a more mathematical definition( I don't want the formula associated with it)

Another thing I want to know in a real case or a problem how do we
  understand(logically) if two events are independent?

That is without verifying that $P(A\cap B) = P(A)P(B) \Large\color{blue}{\star}$  how do we conclude that two events $A$ and $B$ are independent or dependent?.

Take for eg an example by "Tim"

Suppose you throw a die. The probability you throw a six(Event $A$) is $\frac16$ and the
  probability you throw an even number(Event $B$) is $\frac12$.

And event $C$ such that $A$ and $B$ both happen would mean $A$ should happen(As here $A$ is a subset of $B$) hence it's too $\frac16$

My thought:

The above example suggest something like this "Suppose I define two events $A$ and $B$ and if one of the even is a subset of the other then the events are not independent". Of course such an argument is not a sufficient condition for dependency , for eg: consider an event $A$ throwing a dice and getting an odd number and event $B$ getting 6. They aren't independent ($\Large\color{blue}{\star}$ isn't satisfied) . So again I improve my suggestion "Suppose I define two events $A$ and $B$ and if one of them is a subset of the other or their intersection is a null-set then the events are not independent $\Large\color{red}{\star}$"

So at last is $\Large\color{red}{\star}$ a sufficient condition?. Or does there exist a suffcient condition (other than satisfying the formulas $\Large\color{blue}{\star}$? And what is the proof, I can't prove my statement as the idea for me is not that much mathematical .

答:
I think you are asking why is the notion of independent events when there is no relation between them!

Notion of independent events help to calculate the probability of occurring two independent events on like p(A)Up(B),P(A) intersection P(B) on contrary to calculating dependent events.

Independent events--> their probabilities of occurring is not dependent on one another
Dependent events ---> Calculation of probabilities dependent.


问:
Application of Hölder to prove $L^q \subset L^p$ for $1\leq p\leq q<\infty$ with Lebesgue measure
I'm working on the following problem and got stuck. Any help would be really appreaciated.
Let $a,b\in\mathbb{R}$ and $1\leq p \leq q < \infty$. Show that for any $f\in L^q([a,b])$ the following holds:

$$\frac{\lvert\lvert f\rvert\rvert_p}{(b-a)^{1/p}} \leq \frac{\lvert\lvert f\rvert\rvert_q}{(b-a)^{1/q}}$$

Thus $L^q([a,b])\subset L^p([a,b]).$

So I noticed that we could maybe use Hölder inequality with $g=\chi_{(a,b)}$, then $g$ is measurable and integrable, in particular:

$$g\in L^q,L^p$$
$$\lvert\lvert g\rvert\rvert_p = \left(\int\lvert g\rvert^p\right)^{1/p}= (b-a)^{1/p}$$
Then, if we suppose $f\in L^p$. By Hölder:

$$\lvert\lvert f \cdot g\rvert\rvert_1=\left(\int\lvert f\cdot g\rvert \right) \leq \lvert\lvert f\rvert\rvert_q\cdot \lvert\lvert g\rvert\rvert_p = \lvert\lvert f\rvert\rvert_q \cdot (b-a)^{1/p} $$

If we use Hölder again:

$$\frac{\lvert\lvert f \cdot g\rvert\rvert_1}{(b-a)^{1/p}} \leq \frac{\lvert\lvert f\rvert\rvert_p}{(b-a)^{1/p}}\cdot\lvert\lvert g\rvert\rvert_q = \frac{\lvert\lvert f\rvert\rvert_p}{(b-a)^{1/p}}\cdot (b-a)^{1/q} \leq \lvert\lvert f\rvert\rvert_q$$

Which would give us the inequality. But this would only be true if:

$\frac{1}{q}+\frac{1}{p}=1$

We don't know if $f\in L^p$

What am I missing or doing wrong?
Thanks for the help.

答:
Your last inequality seems obscure.

Anyway, perhaps you were hindered by the usual setting of Holder, so let me rephrase it for you :
$$ \left( \int |h| \right)^m \left( \int |g| \right)^n \ge \left( \int |h|^{\frac{m}{m+n}}.|g|^{ \frac{n}{m+n}} \right)^{m+n}$$

for all measurable functions $f,g$ and positive real number $m,n$.

Your desired equality is the special case of Holder with :
$$ h:= |f|^q ; g:= \mathbb{1}_{[a,b]} ; m=1$$
and $n>0$ satisies $ p =q\frac{m}{m+n}$ (note that $p<q$)


问:
Given a linear transformation $T$ find a basis $B$ such that $[T]_B$ is a diagonal matrix
Given

$ A = \begin{bmatrix} 1 & 3 \\ 0 & 6 \end{bmatrix} $

and the linear transformation $T:M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$ which is defined as

$T(v) = Av$

find a basis $B$ of $M_2(\mathbb{R})$ such that $[T]_B$ is a diagonal matrix.

I have no idea how to approach this problem, any help would be appreciated.

答:
A matrix $M = \begin{pmatrix} x & y \\ z & t \end{pmatrix}\neq 0$ is an eigen vector of $T$ for eigenvalue $\lambda$ if and only if
\begin{align}
\begin{pmatrix} x + 3z & y + 3t \\ 6z & 6t \end{pmatrix} = \begin{pmatrix} \lambda x & \lambda y \\ \lambda z & \lambda t \end{pmatrix}
\end{align}
and then, if and only if
\begin{align}
(1-\lambda)x + z &= 0 \\
(1-\lambda )y+  t&= 0 \\
(6-\lambda)z &=0\\
(6-\lambda)t&=0
\end{align}
So now, the problem reduces to determine for which value of $\lambda$ the above system have a non-zero solution. If you can determine $4$ independant matrices $M_1,M_2,M_3$ and $M_4$ that are solution (maybe they are eigenvectors with the same eigenvalue $\lambda$!), you can diagonalize $T$ in the basis $(M_1,M_2,M_3,M_4)$.

A trick is to notice that $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ is an eigenvector of $M$ with eigenvalue $1$, so $M_1=\begin{pmatrix} 1 & 0\\ 0 & 0  \end{pmatrix}$ and $M_2 = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ are eigenvectors of $T$ with eigenvalue $1$. You already have two independant eigenvectors for $T$, so just focus on finding two others.

In fact, if $V_1$ and $V_2$ are two linearly independant eigenvectors of $M$, then the $2\times 2$ matrices $(V_1,0)$, $(0,V_1)$, $(V_2,0)$ and $(0,V_2)$ will be independant eigenvectors of $T$. One can see this by looking at the above system: the $x$ and $z$ coordinates are coupled, independantly of the $y$ and $t$ coordinates (in fact, they are not so independant because they have to have the same eigenvalue!). So $(x,z)$ and $(y,t)$ have to be eigenvectors of $M$ with the same eigenvalue $\lambda$.


问:
Bijection with constraints between sets of polynomial
How can I show the existence (or even better: construct explicitly) of a bijection (if one exists) between the set of all polynomial that have integer coefficients and a integer root and the set of all polynomials that have positive integer coefficients and positive integer roots ?

If I drop the additional "root" constraint, the problem is very simple to solve. But with the root constraint I have no idea how to solve it, since my knowledge of algebra is rather limited. For example if I use the function $h:\mathbb{Z} \rightarrow \mathbb{N}, \ h:=\begin{cases}
2t & ,t\geqslant0\\
-2t-1 & ,t<0\end{cases} $ to map the integer coefficients bijectively to positive integer coefficients, I can find counterexample that have no integer roots whatsoever.

答: A polynomial with positive coefficients can't have a positive root - just think about what happens when you plug a positive number into such a polynomial. </p>


问:
Number of Rectangular prisms
How many rectangular prisms are possible with a volume of 100 cubic cm and sides of integral dimensions?</p>

答:
Quick and dirty and dumb MATLAB code:

```<code>for h = 1 : 100,
    for w = 1 : h,
        for l = 1 : w,
            if w*h*l == 100,
                [h,w,l]
            end
        end
    end
end
</code>```

Confirms 8 solutions.

Title: Number of Rectangular prisms

Body:
Quick and dirty and dumb MATLAB code:

for h = 1 : 100,
    for w = 1 : h,
        for l = 1 : w,
            if w*h*l == 100,
                [h,w,l]
            end
        end
    end
end

Confirms 8 solutions.


问:
cardinality of the union of 2 sets with at least two elements and cardinality of cartesian product
Suppose X,Y are sets with at least 2 elements. Show that
$X\cup Y\le X\times Y$

So my first thought was that cardinality $|X|\ge 2$ and the same for $|Y|\ge 2$ but by the inclusion-exclusion principle we have $|X\cup Y|=|X|+|Y|-|X\cap Y|$ but the problem does not say if they are disjoint or not.
If we assume that they are disjoint then $|X\cup Y|\ge 4$ right?
But from the definition of the cardinality for the cartesian product we have similar
$|X\times Y|\ge 4$. But I have to find an injective function (including the number of elements) s.t $f:X\cup Y\to X\times Y$ and thus conclude the result.
How should I do it?

答:
Suppose $x_1$ and $x_2$ are distinct elements of $X$, $y_1$ and $y_2$ are distinct elements of $Y$.

If $X$ and $Y$ are disjoint, you can define $f\colon X\cup Y\to X\times Y$ by
$$
\begin{cases}
f(x)=(x,y_1) & \text{if $x\in X$, $x\ne x_1$}\\
f(x_1)=(x_1,y_2) \\
f(y)=(x_2,y) & \text{if $y\in Y$, $y\ne y_1$}\\
f(y_1)=(x_1,y_1)
\end{cases}
$$
This function is injective. Two distinct elements in $X$ have distinct images and two distinct elements in $Y$ have distinct images. So we can check whether it's possible that an element in $X$ and one in $Y$ have the same image: suppose $f(x)=f(y)$; if $x=x_1$, we have $f(x)=(x_1,y_2)$, but $f(y)=(x_2,y)$ or $f(y)=(x_1,y_1)$ and equality is impossible. So $x\ne x_1$ and $f(x)=(x,y_1)$. But, again $f(y)=(x_2,y)$ or $f(y)=(x_1,y_1)$; the second case is impossible, so we should have $y\ne y_1)$ and so $f(y)=(x_2,y)\ne(x,y_1)=f(x)$.

Suppose now $X$ and $Y$ are not disjoint and write $X\cup Y=X\cup(Y\setminus X)$.

If $Y\setminus X$ has at least two elements, we get an injective map
$X\cup Y\to X\times(Y\setminus X)\subseteq X\times Y$ by the argument above. Otherwise $Y\setminus X$ is empty or it has just one element.

If $Y\setminus X=\emptyset$, we have $Y\subseteq X$ and we can define
$f\colon X\to X\times Y$ by $f(x)=(x,y_1)$ which is injective.

If $Y\setminus X=\{y_0\}$ we can define $f\colon X\cup\{y_0\}\to X\times Y$ by $f(x)=(x,y_0)$ for $x\in X$ and $f(y_0)=(x_1,y_1)$ if $y_0\ne y_1$ or $f(y_0)=(x_1,y_2)$ if $y_0=y_1$.


问:
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved
How are can the above recurrence relation be solved?

I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$

And then here:

$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$

And we are given that these: $T(0) = 2, T(1) = 3$

So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.

答:
Rewrite the expression as
$$
T_{n+2}=T_{n+1}+T_{n}+3n+7
$$
Now postmultiply by $z^k$ and sum over $k$ to get the generating function $G_{T}(z)=\sum_{k=0}^{\infty}T_k z^k$. Here
$$
\sum_{k=0}^{\infty}T_{k+2}z^k = \frac{1}{z^2}\bigg(G(z)-T_0-T_1 z\bigg)\\
\sum_{k=0}^{\infty}T_{k+1}z^k=\frac{1}{z}\bigg(G(z)-T_0 \bigg)
$$
Now do the algebra and find the expression with $G(z)$ on the LHS of the expression and $\sum_{k \geq 0}\varphi_k z^k$. The $\varphi_n$ term corresponding to $z^n$ will be the solution.  </p>


问:
Finding the independent term of an expansion
The question and the answer are provided below; however, I have no idea what strategy was implemented to achieve that answer. I would like to know what is happening to solve this question and what each of the lines serves to do. In other words, Just walk me through it, thanks!

答:
What they used (twice) is $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
Here $a$ and $b$ are some simple powers of $x$ multiplied by some numbers. In the first expression $a=1\cdot x^{-2}$ and $b=(-3)\cdot x^1$. Then they did the same thing for the second term, with $c=(1/2)x^{-1}$ and $d=1\cdot x^2$.
You can now figure out the powers of $x$ in the expanded from of each term. So for the first expression, in the order that I wrote the first equation, the powers of $x$ are $-2\cdot3,-2\cdot2+1,-2+1\cdot2, 1\cdot3$ or $-6, -3, 0, 3$. You can do the same for the second expression. When you multiply, you make all possible combinations with one power from the first expression, added with a power in the second one. Choose only the terms where the sum is $0$.


问:
Derivative of trace involving the Hadamard product
I am trying to find a matrix product of a co-occurrence matrix

$$\mathbf{Y} \approx \mathbf{U} \mathbf{V}^\intercal,$$ where $\mathbf{Y} \in \mathbb{R}^{I \times J}$, $\mathbf{U} \in \mathbb{R}^{I \times K}$ and $\mathbf{X} \in \mathbb{V}^{J \times K}$.

To find this, I am going to minimise the error (note I have added regularisation terms to prevent overfitting)

$$\xi(\mathbf{U},\mathbf{V}) =  ||\mathbf{Y} - \mathbf{U} \mathbf{V}^\intercal ||^2_F + \lambda_u || \mathbf{U} \mathbf{U}^\intercal ||^2_F + \lambda_v || \mathbf{V} \mathbf{V}^\intercal ||^2_F. $$

Then to minimise this I will find the derivatives with respect to both $\mathbf{U}$ and $ \mathbf{V} $:

$$\nabla_\mathbf{U} = -2\mathbf{V} \left( \mathbf{Y} - \mathbf{U}\mathbf{V}^{ \intercal} \right) + 2\lambda_u \mathbf{U}$$
$$\nabla_\mathbf{V} = -2\mathbf{U}\left( \mathbf{Y} - \mathbf{U}\mathbf{V}^\intercal \right) + 2\lambda_v \mathbf{V}.$$

These have been calculated using the identities:

$$ ||\mathbf{A}||^2_F = \text{tr}(\mathbf{A}\mathbf{A}^\intercal) $$ and $$ \frac{\partial}{\partial \mathbf{Y}} \text{tr}((\mathbf{A}\mathbf{Y} + \mathbf{C})(\mathbf{A}\mathbf{Y} + \mathbf{C})^\intercal) = \mathbf{A^\intercal(\mathbf{A}\mathbf{Y} + \mathbf{C})}.$$

I have no issue with these derivations, the problem is I am now trying to solve the same problem but with an added $\textit{masking matrix}$ (for reasons I won't go into here), which is a randomly populated binary matrix $\mathbf{M} \in \mathbb{R}^{I \times J}$. This changes the error function to

$$\xi(\mathbf{U},\mathbf{V}) =  ||\mathbf{M}\odot(\mathbf{Y} - \mathbf{U} \mathbf{V}^\intercal) ||^2_F + \lambda_u || \mathbf{U} \mathbf{U}^\intercal ||^2_F + \lambda_v || \mathbf{V} \mathbf{V}^\intercal ||^2_F. $$

I'm unsure how I would find the derivative of this new error function as the Hadamard product complicates things - if anyone could shed some light on this it would be much appreciated!

Thank you.

答:
$\def\p{\partial}$
The binary matrix $M$ has the property that $\;M\odot M = M$

Let a colon denote the trace/Frobenius product, i.e.
$$B:C={\rm Tr}(B^TC)={\rm Tr}(C^TB)=C:B$$
The Hadamard and Frobenius products commute in the following senses
$$\eqalign{
A:B &= B:A \\
A\odot B &= B\odot A \\
C:(A\odot B) &= (C\odot A):B \\
}$$
For typing convenience define the matrices
$$\eqalign{
X &\doteq UV^T-Y \\
W &\doteq M\odot X \\
}$$
Notice that
$$\eqalign{
M\odot W &= M\odot M\odot X \\&= M\odot X \\&= W
}$$
Write the error function in terms of these and calculate its differential.
$$\eqalign{
{\cal E} &= W:W + \lambda_uU:U + \lambda_vV:V \\
\\
d{\cal E}
 &= 2W:dW + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2W:(M\odot dX) + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2W:(dU\,V^T+U\,dV^T) + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2WV:dU + 2U^TW:dV^T + 2\lambda_uU:dU + 2\lambda_vV:dV \\
 &= 2(WV+\lambda_uU):dU
  \;+\; 2(W^TU+\lambda_vV):dV \\
}$$
From which the gradients can be identified as
$$\eqalign{
\frac{\p{\cal E}}{\p U} &= 2(M\odot X)V + 2\lambda_uU \\
\frac{\p{\cal E}}{\p V} &= 2(M\odot X)^TU + 2\lambda_vV \\
}$$
Note that setting $M=J\,$ (i.e. the all-ones matrix) recovers the gradients in the original (non-Hadamard) case, since for any matrix $A$ one has $\;J\odot A = A$.


问:
eigen vectors and linear independence
I know that it is the case that if we have eigenvectors $v_1....v_r$ that correspond to distinct eigen values $λ_1....λ_r$ of an $n$ x $n$ matrix, then {$v_1....v_r$} is linearly independent.

But is is it possible to have $a_1$ linearly independent eigen vectors corresponding to $λ_1$, $a_2$ linearly independent eigen vectors corresponding to $λ_2$... $a_p$ linearly independent eigen vectors corresponding to $λ_n$, and the set of all $r$ eigenvectors where $r=a_1+a_2+...+a_p$ be linearly dependent?

For example: consider a $3x3$ matrix. Imagine we have one vector(call it $v_1$) forming a basis for the eigenspace of a one eigen value, but two vectors(call them $v_2$ and $v_3$) forming a basis for different eigen value. are we guaranteed that {$v_1,v_2,v_3$} are linearly independent?

答:
Suppose  is any field. For example, it could be ℝ or ℂ.
Suppose $M$ is an n×n -matrix.
Suppose $\lambda_1, \dots, \lambda_r$ is a list of distinct eigenvalues of $M$.

Suppose:

$v_{1, 1}, \dots, v_{1, k_1}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_1$.

$v_{2, 1}, \dots, v_{2, k_2}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_2$.

$\vdots$

$v_{r, 1}, \dots, v_{r, k_r}$ is a linearly independent list of eigenvectors of $M$ corresponding to $\lambda_r$.

Suppose $\alpha_{1, 1}, \dots, \alpha_{1, k_1}, \alpha_{2, 1}, \dots, \alpha_{2, k_2}, \dots, \alpha_{r,1}, \dots, \alpha_{r, k_r} \in \mathbb{F}$ are scalars such that
$$(\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}) +
  (\alpha_{2, 1} v_{2,1} + \dots + \alpha_{2, k_2} v_{2, k_2}) + \dots +
(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r}) = 0.$$

Define $u_1 = (\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}), \dots,
u_r=(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r})$.

Clearly, $u_1$ is either the zero vector or a $\lambda_1$-eigenvector of $M$. Analagously, $u_2$ is either the zero vector or a $\lambda_2$-eigenvector of $M$. And so on for $u_3, \dots, u_r$.

By the theorem you say you know, it must be the case that for each $i \in \{1, \dots, r\}$, $u_i$ is actually the zero vector.
But then for each $i \in \{1, \dots, r\}$, all the coefficients $\alpha_{i,1}, \dots, \alpha_{i, k_i}$ must be zero, because the vectors $v_{i,1}, \dots, v_{i, k_i}$ are linearly independent.

We have shown that any linear combination of $v_{1, 1}, \dots, v_{1, k_1}, \dots, v_{r, 1}, \dots, v_{r, k_r}$ equal to zero must actually be the trivial combination (i.e., with all coefficients zero). ∎

Title: eigen vectors and linear independence


问:
Possible ways to do stability analysis of non-linear, three-dimensional Differential Equations
For example Lorenz system,
$$
\frac{d}{dt}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
-\sigma & \sigma & 0\\
\rho & -1 & -x\\
y & 0 & -\beta
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}
$$

The article on Wiki describes the system of differential equations to have stable equilibrium for a specific value of $\rho$.

Given some three dimensional system of differential equations, I would like to know how to do the stability analysis? or explain how stability analysis of Lorenz system is done ...

答:
Consider the Lorenz equations:

$$x' = f(x,y,z) = \sigma(y-x)$$

$$y' = g(x,y,z) = \rho x - y -xz$$

$$z' = h(x,y,z)= -\beta z + xy$$

where $\sigma$ and $\beta$ are viewed as fixed positive constants and $\rho$ is a parameter.

Here is a guide to work through given that the Wiki describes what happens at each of the critical points. The outline of how to derive those is as follows:

$(1)$ Find the critical points of the system, that is, where you simultaneously have $x' = y' = z' = 0$.

Clearly, a critical point is given by $(x, y, z) = (0,0,0)$.

$(2)$ Find the Jacobian matrix of the system of equations.

The Jacobian matrix (using partial derivatives) is given by:

$$\tag 2 \displaystyle J = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}& \frac{\partial f}{\partial z} \\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}& \frac{\partial g}{\partial z}\\ \frac{\partial h}{\partial x} & \frac{\partial h}{\partial y}& \frac{\partial h}{\partial z}\end{bmatrix} = \begin{bmatrix}-\sigma & \sigma & 0\\\rho - z & -1 & -x\\ y & x & -\beta\end{bmatrix}$$

$(3)$ Lastly, evaluate the eigenvalues of the Jacobian evaluated at each of the critical points for stability analysis. So, evaluating $(2)$ at the CP $(0,0,0)$, yields the matrix:

$$\displaystyle J_{0,0,0} = \begin{bmatrix}-\sigma & \sigma & 0\\\rho & -1 & 0\\ 0 & 0 & -\beta\end{bmatrix}$$

To find the eigenvalues, we set-up and solve for the characteristic polynomial using:

$$| A - \lambda I| = 0 \rightarrow -\beta \lambda^2-\beta \lambda+\beta r \sigma-\beta \lambda \sigma-\beta \sigma-\lambda^3-\lambda^2+\lambda \rho \sigma-\lambda^2 \sigma-\lambda \sigma = 0$$

This leads to the three eigenvalues (note, we could have written these straight off given the special form of the diagonal matrix above) and their corresponding eigenvectors as:

$\displaystyle \lambda_1 = -\beta, v_1 = (0, 0, 1)$

$\displaystyle \lambda_2 = \frac{1}{2} \left(-\sqrt{4 \rho \sigma+\sigma^2-2 \sigma+1)}-\sigma-1\right), v_2 = -\frac{-1+\sigma+\sqrt{1-2 \sigma+4 \rho \sigma+\sigma^2}}{2 \rho}, 1, 0)$

$\displaystyle \lambda_3 = \frac{1}{2}\left(\sqrt{4 \rho \sigma+\sigma^2-2 \sigma+1)}-\sigma-1\right), v_3 = -\frac{-1+\sigma-\sqrt{1-2 \sigma+4 \rho \sigma+\sigma^2}}{2 \rho}, 1, 0)$

Of course those $\lambda$ are the eigenvalues we use to do a stability analysis.

To study the stability of this, we need some complex center manifold theory, so I am going to stay away from that. It leads to a pitchfork bifurcation at the origin.

It is worth noting that there are two other pairs of fixed points at:

$x = y = \pm \sqrt{\beta(\rho - 1)}, z = \rho -1$

To find this, notice that the first equation gives us $y=x$, substituting this into the second equation gives $x(\rho - 1) - xz = 0 \rightarrow z = \rho-1$, and we substitute these previous two values into the third and have $x^2 = \beta z \rightarrow x =  \pm \sqrt{\beta(\rho - 1)}$.

To simplify matters, lets linearize the system by removing the nonlinear terms and analyze the simpler system.

The linearization (eliminate the $xy$ and $xz$ nonlinear terms from the system) is given by:

$$x' = \sigma(y-x)$$

$$y' = \rho x - y$$

$$z' = -\beta z$$

If you look at the equation for $z'$, you notice that it is decoupled from the other two equations and it can be solved outright as $z(t) = c_1 e^{-\beta t}$ and $z(t) \rightarrow 0$ exponentially fast. The other two directions are governed by the system:

$$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}-\sigma & \sigma\\\rho & -1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$

You should be able to analyze this system from all the data provided now.

As you can see, these systems are very complex and include a rich set of mathematics to deal with, so it takes time to get there.

Here is another guide with some of the details for you to work out.


问:
Problem on temperate distribution theory

Problem: Give an example of an infinitely differentiable function f on $\mathbf{R}$ such that (1): there does not exist a polynomial P on $\mathbf{R}$ such that $|f(x)| \le |P(x)|, x \in \mathbf{R}$ and (2): the mapping $\mathbf{S}(\mathbf{R}) \ni \varphi \rightarrow \int f(x) \varphi(x)dx$ defines a temperate distribution.

Denote by $\mathbf{S}$ the Schwartz space of infinitely differentiable functions $\phi(x)$ such that $|\frac{\partial^k \varphi(x)}{\partial x^k}| \le C_{m,k}(1+|x|)^{-m}$ for any $k$ and any positive integer $m$. Introduce the norms $|[\varphi]|_{m,S}=\textbf{sup}_{x \in \mathcal{R}^n} (1+|x|)^m \sum_{|k|=0}^m |\frac{\partial^k \varphi(x)}{\partial x^k}|$. We say that $\varphi_n \rightarrow \varphi in S$ if $|[\varphi_n-\varphi]|_{m,S} \rightarrow 0$ as $n \rightarrow \infty$ for all $m=1,2,...,$

Definition: Linear continuous functional on S is called a tempered distribution. The linear space of tempered distributions is denoted by $S'$.

Theorem 9.1: For any tmepered distrubition f, there exist an integer m and a constant C such that $|f(\varphi)| \le C |[\varphi]|_{m,S} \forall \varphi \in S$.

Theorem 9.2: Any $f \in S'$ can be represented in the following form: $f=\sum_{|k|=0}^{m_1} \frac{\partial^k \varphi(x)}{\partial x^k} f_k$, where $f_k$ are regular functions in $S'$ corresponding to continuous functions $f_k(x)$ satisfying the estimates $|f_k(x)| \le C_k(1+|x|)^{m_2}$. Here $m_1,m_2$ are integers.

Question: I have no clue how to approach this problem. Any help is appreciated.

答:
Pick your favorite bump function $\Psi(x)$ then define
$$
\Psi_{p,w,h}(x)=h\Psi\left(\frac{x-p}{w}\right)\ .
$$
Here $p$ allows to pick the position of where you want it translated. $h$ allows you to adjust the height and $w$ controls the width.
You can construct $f$ in the form
$$
f(x)=\sum_{n=0}^{\infty}\Psi_{n,w_n,h_n}(x)
$$
for suitable sequences $(w_n)$ and $(h_n)$.
First pick the $h_n$ so they grow faster than any polynomial, e.g., $h_n=e^n$.
Then pick the widths small enough so that 1) the bumps do not touch and the areas of the bumps are so small so the total area under $f$ is finite. Then $f(x)dx$ is a bounded measure and thus a temperate measure and it will define a temperate distribution.


问:
How to know if the solution for this angle ϕ is in units of radians or degrees?
I am studying out of interest a modelling approximation for a plectrum plucking a string (eg. guitar, piano, harpsichord) as given by these articles:

http://recherche.ircam.fr/pub/dafx11/Papers/24_e.pdf (relevant part on page 3 of pdf)

https://stacks.stanford.edu/file/druid:wp454hs7976/Jack%20Perng%20Thesis-augmented.pdf (relevant part on page 75 of pdf)

The model works roughly like this:

The string is represented here as the yellow circle in cross-section. The red part on the left is the component the plectrum is attached to that moves (moving upwards along the y-axis here). The black bendy part is obviously the plectrum.

$F_p$ (the force from the plectrum) can be broken down in terms of $x$ and $y$ vector components as per:

Where E is the Young's Modulus, I is moment of inertia, and L' is how far along the length of the plectrum the string is contacting.

The only trouble I'm having in understanding this model is concerning the units of $\phi$. Is this angle solution going to be in radians or degrees? In C++, sin and cos take radians by default and you have to convert degrees to radians first if working from degrees. So I need to know which one I'm getting from this equation for $\phi$.

Is $\phi$ here in radians or degrees?

Thanks for any guidance.

答:
Oddly enough, the angle should usually be given in radians. However, looking at the first referenced paper, I am 99% sure that they are using angles in degrees. The authors speak of ( after (6) )

For small-angle approximations ($\phi<<1$), this reduces [...].

If this would be in radians, this would mean roughly 57° which is not really a "small-angle approximation".
Also, they say that

when $\theta=90°$, there is no bending [...].

Which also indicates the use of angles in degrees. I would suggest contacting the authors. Usually, authors are very happy if someone is interested in their work and are willing to help. You could also try to replicate the given results since the authors seem to have provided all necessary data.


问:
Bolzano-Weierstrass on $\mathbb{R}^n$ with different Norms
I have been dealing with Topology recently and was told that the Bolzano-Weierstrass-Theorem still applies for every bounded sequence in $\mathbb{R}^n$, no matter the metirc, if it is induced by a norm. My question is how we can just jump from $\mathbb{R}^n$ to $\mathbb{R}^m$ dimensions while the Bolzano-Weierstrass-Theorem still holds true and why it's validity doesn't change with the Norm ($\mathbb{R}^n$, $\lVert x\rVert_2$) versus ($\mathbb{R}^n$, $\lVert x\rVert_\infty$), for example.

答:
All norms are equivalent on finite dimensional vector spaces.

That is, there are positive $c,C$ such that $$c\|x\|_1\leq \|x\|_2\leq C \|x\|_1$$ for all vectors $x$. In other words, they generate the same topology.


问:
About the empty-set and rules about it
Is $\emptyset$ a set of ordred triplets, pairs? is it a set of any property we will name (Maybe "the set of elephants" - with no elephants )?

I get that not all adjectives / properties can describe it, such as "the class with the number 1 in it" - not the empty set for sure, but what about the properties I wrote above?

I think that it isn't - because it doesn't have any elements - so it does not have any ordered pairs or triplets, but maybe it is because it has nothing then we cannot decide - so the answer is yes?

I am really confused by this question, I would appreciate your help!

答:
You can certainly not characterize the empty set as the set of all ordered pairs or triplets, or $n$-tuples of any kind. However, you can characterize the empty set in several ways. One way that is sometimes used in courses in introductory set theory is the following: If you have some universe $\mathcal{U}$, you can use the seperation axiom to construct the empty set like this

$$ \emptyset = \{ x \in \mathcal{U} \, | x \neq x \}.$$

However, this is just a "trick" that I think for most people brings no intuition or deeper understanding. The empty set is simply the set which has no elements. You could also construct it as follows, assuming the same universe $\mathcal{U}$:

$$\forall x \in \mathcal{U} \quad x \notin \emptyset.$$

This is however less rigorous, and if possible, even less didactic, so forget about it now that you have seen it. Finally, addressing your question as to whether you can say that the empty set is a set of "something", you could actualy say that about anything since the empty set is a subset of every set, i.e.

$$\forall A \quad \emptyset \subseteq A,$$

(do you see why?).

Title: About the empty-set and rules about it


问:
Both quasiconvex and quasiconcave function
The function $$f(x) = x^2$$ is both quasiconcave and quasiconvex. How can I show this?

答:
Any convex function is quasi-convex. Thus, $f=x^2$ is quasi-convex. Your claim of quasi-concavity is not correct. Function $f$ is quasi-concave if
$$f(\lambda x +(1-\lambda )y)\ge min \{f(x), f(y)\},$$
which is apparently wrong for $f=x^2$.


问:
Prove the $\lim _{z \rightarrow 1}\frac{\overline{z}-1}{z-1}$ doesn't exist.
Prove the $\lim _{z \rightarrow 1}\frac{\overline{z}-1}{z-1}$ doesn't exist.

We see that $\frac{\overline{z}-1}{z-1}=\frac{x-iy-1}{x+iy-1}$, but when we take $z=1+iy$ we have $\lim _{z \rightarrow 1} \frac{1-iy-1}{1+iy-1}=-1$. But I don't see other points such that the limit is diffferent to $-1$. Could you help me with that? or can I take $z=1+x+i0$ such that $x \rightarrow 0$?

答: If $z=1+h$ with $h\in(0,\infty)$, then$$\frac{\overline z-1}{z-1}=\frac hh=1.$$This, together what you did, shows that the limit doesn't exist.


问:
If $f(x) > 0$ and Riemann Integrable on $[a,b]$ is $\frac{1}{f(x)}$ Riemann integrable?
Problem

If $f(x) > 0$ and Riemann Integrable on $[a,b]$ is $\frac{1}{f(x)}$ Riemann integrable on $[a,b]$?

My attempt

One has to show that $$S(P) - s(P) < \epsilon$$
$$S(P) = \sum_{j=1}^NM_j(x_j - x_{j-1})$$
$$S(P) = \sum_{j=1}^Nm_j(x_j - x_{j-1})$$
Now I consider the difference between these two:
$$S(P) - s(p) = \sum_{j=1}^N(M_j-m_{j-1})(x_j - x_{j-1})$$

How do I continue from here?

答:
Here's a way to view the problem. Denote $M_j^f=\sup_{[x_{j-1},x_j]}f(x)$ and $m_j^f=\inf_{[x_{j-1},x_j]}f(x)$. Define $g(x):=\frac{1}{f(x)}$. Then it follows that $M_j^g=\frac{1}{m_j^f}$ and $m_j^g=\frac{1}{M_j^f}$ since $f(x)>0$. That is, assuming there exists $\alpha>0$ such that $m_j^f\geq \alpha >0$ for all $j$. In other words, if $f(x)\geq \alpha >0$ for all $x\in[a,b]$.

Then given $\epsilon>0$, there exists a partition  $P$ such that $U(P,f)-L(P,f)<\alpha^2 \epsilon$. For that same partition, we have
\begin{align}
U(P,g)-L(P,g)&=\sum_{j=1}^n(M_j^g-m_j^g)(x_j-x_{j-1}) \\
&=\sum_{j=1}^n\left(\frac{1}{m_j^f}-\frac{1}{M_j^f}\right)(x_j-x_{j-1}) \\
&=\sum_{j=1}^n\left(\frac{M_j^f-m_j^f}{M_j^f m_j^f}\right)(x_j-x_{j-1}) \\
&\leq\sum_{j=1}^n\left(\frac{M_j^f-m_j^f}{\alpha^2}\right)(x_j-x_{j-1}) \\
&=\frac{1}{\alpha^2}(U(P,f)-L(P,f)) \\
&<\epsilon.
\end{align}
Unfortunately, we cannot assume $f(x)\geq \alpha>0$ so any such counterexample would have to violate this.


答:
The answer to your question is in two steps. I assume you know the Bolzano-Weierstrass in $(\mathbb{R},|\cdot|)$, stating that a closed interval $[a,b]$ is compact.

First step The metric space $\left(\mathbb{R}^n,\|\cdot\|_{\infty}\right)$ is the product metric space $\prod_{i=1}^n \left(\mathbb{R},|\cdot|\right)$ with the supremum distance. Moreover, a finite product of compact topological spaces is compact, so any $\left(\prod_{i=1}^n [a_i,b_i]\right)\subset \mathbb{R}^n$ is compact for the topology of $\|\cdot\|_{\infty}$. Thus, the Bolzano-Weierstrass theorem is true for that particular norm.

Second step On $\mathbb{R}^n$, all norms are equivalent, that is for $N_1$ and $N_2$ two norms, there exists two constants $C_1$ and $C_1$ such that
\begin{align}
\forall x \in \mathbb{R}^n,C_1 N_1(x) \leqslant N_2(x) \leqslant C_2N_1(x)
\end{align}

So every norm is equivalent to $(\|\cdot\|_{\infty})$. The topologies induced are the same, and the compact subsets are the same. The particular case of step 1 show the generalized Bolzano-Weierstrass theorem.

Edit Another way to think this proof is this one: let $(p_k)$ be a bounded sequence in $\left(\mathbb{R}^n,\|\cdot\|\right)$, say $p_j = (x_1(k),x_2(k),\ldots,x_n(k))$. As all norms are equivalent, $(p_k)$ is also bounded for the $\|\cdot\|_{\infty}$ norm. Consequently, each coordinate sequence $(x_i(k))_k$ is bounded in $\mathbb{R}$.
The sequence $(x_1(k))_k$ is bounded. By the one dimensional Bolzano-Weierstrass theorem, there exists $\phi_1 : \mathbb{N} \to \mathbb{N}$ increasing such that $(x_1(\phi_1(k))_k$ is convergent to a number $x_1$.

The sequence $x_2(\phi_1(k))$ is bounded as a subsequence of $(x_2(k))$. Then there exists $\phi_2$ inscreasing such that $x_2(\phi_1(\phi_2(k)))$ converges to a number $x_2$. And so on.
Define $\phi = \phi_1\circ \phi_2\circ \cdots \circ \phi_n$. Then $p_{\phi(k)}$ has a limit $(x_1,\ldots,x_n)$ in $\mathbb{R}^n$ for the $\|\cdot\|_{\infty}$ metric. Use again the equivalence of norms to show that this is also the limit of $p_{\phi(k)}$ for the previous norm.

Title: Bolzano-Weierstrass on $\mathbb{R}^n$ with different Norms

Body: I have been dealing with Topology recently and was told that the Bolzano-Weierstrass-Theorem still applies for every bounded sequence in $\mathbb{R}^n$, no matter the metirc, if it is induced by a norm. My question is how we can just jump from $\mathbb{R}^n$ to $\mathbb{R}^m$ dimensions while the Bolzano-Weierstrass-Theorem still holds true and why it's validity doesn't change with the Norm ($\mathbb{R}^n$, $\lVert x\rVert_2$) versus ($\mathbb{R}^n$, $\lVert x\rVert_\infty$), for example.


答:
How about, approach along a diagonal?
$$  \lim_{x \rightarrow 0} \frac{(1+x) - \mathrm{i}x - 1}{(1+x) + \mathrm{i}x - 1} = -\mathrm{i}  \text{.}  $$

Body:
Prove the $\lim _{z \rightarrow 1}\frac{\overline{z}-1}{z-1}$ doesn't exist.

We see that $\frac{\overline{z}-1}{z-1}=\frac{x-iy-1}{x+iy-1}$, but when we take $z=1+iy$ we have $\lim _{z \rightarrow 1} \frac{1-iy-1}{1+iy-1}=-1$. But I don't see other points such that the limit is diffferent to $-1$. Could you help me with that? or can I take $z=1+x+i0$ such that $x \rightarrow 0$?


答:
Consider $z = 1 + \varepsilon i$, where $\varepsilon > 0$. Then $\frac{\overline z - 1}{z - 1} = -1.$

Now consider $z = 1 + \varepsilon$, where $\varepsilon > 0$. Then $\frac{\overline z - 1}{z-1} = 1$.

Clearly, $1 \neq -1$.

Title: Prove the $\lim _{z \rightarrow 1}\frac{\overline{z}-1}{z-1}$ doesn't exist.


问:
Explain the formula for the size of an orbit...
Explain the formula for the size of an orbit and show that this always is a divisor of the group order $|G|$. (You may use Lagrange's theorem!)

So I would like to know how i can go about answering this. I know there are a few question fairly similar to this but all answers involve using cosets, and proving bijections. I am not to sure whether i would need to do that or whether there is a simpler way?

答:
I think you can easily find more about this basic and important point in any permutation book. But,let $G$ is a group which acts on the set $\Omega$ transitively. It means that $$\forall\omega_1,\omega_2\in\Omega; ~\exists g\in G,~~~\omega_1^g=\omega_2$$ Set $\Omega^*=\{G_{\omega}x\mid x\in G\}$ wherein $G_{\omega}$ is the stabilizer of $\omega\in\Omega$. In fact, you are considering the right coset of $G_{\omega}$ in the group. Define the following map: $$f:\Omega^*\to\Omega\\\ f(G_{\omega}x)=\omega^x$$ It can be easily verified that $f$ is well-defined and one-one function:
$$G_{\omega}x=G_{\omega}y\longleftrightarrow xy^{-1}\in G_{\omega}\longleftrightarrow\omega^{xy^{-1}}=\omega\longleftrightarrow \omega^x=\omega^y$$ it is onto also (Why?) Therefore, $$|\Omega|=|\Omega^*|=[G:G_{\omega}]$$ Now think that the action is not transitive. What will happen? Indeed, we should replace $|\Omega|$ by $|\omega^G|$ the orbit corresponding to $\omega$. In fact, $$|G|=|\omega^G||G_{\omega}|$$</p>


问:
Isomorphism between $D_{2n}/Z(D_{2n})$ and $D_n$
I need some help with the following task:

Show that  for all $n\geq2$ is $D_{2n}/Z(D_{2n})\cong D_n$

$D_n$ is the dihedrial group with $2n$ elements and looks like $D_n=\{id,r,...,r^{n-1},s,sr,...,sr^{n-1}\}$

Sadly I have no idea how $D_{2n}/Z(D_n)$  looks like but I guess I need to know that to show that it is a homomorphism and bijectiv.

Thanks for helping me

答:
Hint: By the first isomorphism theorem, it is equivalent to find a homomorphism $f:D_{2n} \to D_n$ with $\ker(f) = Z(D_{2n})$.

Let $s_n,r_n$ denote the generators of $D_{n}$ and let $s_{2n},r_{2n}$ denote the generators of $D_{2n}$. Define the map $f:D_{2n} \to D_n$ by
$$
f(s_{2n}^j r_{2n}^k) = s_{n}^jr_n^k.
$$
Verify that $f$ is a homomorphism and that $\ker(f) = Z(D_{2n})$. In verifying that $f$ is a homomorphism, you might find the identity
$
sr^ks = r^{-k}
$
to be useful.


答:
The reason why Bolzano-Weierstrass holds in every $\mathbb{R^n}$ and its independent of the norm that you have on $\mathbb{R^n}$ is based on the following facts (which are consequences of the finite dimensionality of the space).

Proposition 1 A normed linear space $(X,||.||)$ (for example $\mathbb{R^n})$ is finite dimensional if and only if its closed unit ball $B_X=\{x\in X: ||x||\leq 1\}$ is compact.

Now Proposition 1 answers to your first question, i.e. to why the Bolzano Weierstrass holds in every $\mathbb{R^n}$ indepedently of $n$. (And actually holds in every finite dimensional normed space)

Too see this, assuming we know Proposition 1 and we have a bounded sequence $(x_m)$ in $\mathbb{R^n}$ then by the boundedness of $(x_m)$ there would be a ball with center $0$ and radius $M>0$ (the bound of $x_m$) such that $x_m\in B_X(M)=\{x\in \mathbb{R^n}: ||x||\leq M\}$. By proposition 1, $B_X$ is compact, and therefore $B_X(M)$ is also compact. Hence, $x_m$ must have a convergent subsequence.

An important corollary of the Bolzano-Weierstrass is the following characterization of compactness in the context of finite dimensional normed spaces:

Corollary 1 In a finite dimensional linear space $(X,||.||)$, every subset $K$ of $X$ is compact if and only if it is closed and bounded.

Now, to your second question, to why is independently of the norm that you have in $\mathbb{R^n}$ is because of the following

Proposition 2 In every finite dimensional linear space, every two norms $||.||_1,||.||_2$ are equivalent. Meaning, that there exists two constants $C_1,C_2>0$ such that
$$C_1||x||_1\leq ||x||_2\leq C_2||x||_1$$

Now its easy to check that if a sequence converges with the usual norm $||.||_2$ in $\mathbb{R^n}$ it will also converges with any other norm.

I hope that this helps!


答:
If you define a vector $\mathbf{U}_{n}$ by
$$
\mathbf{U}_{n}=\left(
\begin{array}
[c]{c}%
T\left(  n\right)  \\
T\left(  n-1\right)  \\
n\\
1
\end{array}
\right)  \text{   }n\geq1
$$
then your recurence relation can be written
$$
\mathbf{U}_{n+1}=\mathbf{AU}_{n}
$$
where $\mathbf{A}$ is the constant 4x4 matrix
$$
\mathbf{A}=\left(
\begin{array}
[c]{cccc}%
1 & 1 & 3 & 4\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1
\end{array}
\right)
$$
So the problem is equivalent to finding an expression for the matrix powers
$\mathbf{A}^{m}$.  This should be straightforward via the eigenvalues of $\mathbf{A}$, which are $\lambda$=1 (twice), and $\frac{1\pm\sqrt{5}}{2}$.</p>

Title: How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved

Body:
How are can the above recurrence relation be solved?

I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$

And then here:

$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$

And we are given that these: $T(0) = 2, T(1) = 3$

So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.


问:
Transcendental equation solutions
I'm working on solving a 3d particle in a well for quantum mechanics.

I have the transcendental equation
$$ - \cot(z) = \sqrt{(z_0/z)^2-1} $$

A graph of this looks like

```<code>Plot[{Sqrt[(5*\[Pi]/z)^2 - 1], -Cot[z]}, {z, 0, 5*\[Pi]}]
</code>```

A solution manual for Griffiths QM book states that the solutions ONLY occur at multiples of pi, and supplies the following graph

why are those other non-intervals of pi intersections not considered?

答: If that's what the manual says, it is wrong.  Multiples of $\pi$ are not solutions, as $\cot$ is not defined there.  Rather, there is one solution in each interval between two multiples of $\pi$.

Title: Transcendental equation solutions


问:
Curious minimization problem
Well, this problem may look easy, but I'm not able to get the equations to minimize.

We have a hole in the floor and need to make a lid for it so that:

i) It can never fall into the hole, that is, the lid cannot pass through the constructed hole, in no position. We understand that the cap is slightly larger than the hole
(with the same shape) and that at the bottom of the lid there is a small edge, very close to the edge of the lid, to fit the gap.

ii) We have some flat rectangular metal plates, a little less than 70 centimeters inwide by three meters long. These plates must be able to pass through the hole.

iii) They should use the least amount of material possible, to minimize costs, but respecting
the above requirements.

We are asked to get the exact area of the lid.

Any help to translate it to a math language and how to solve it, would be really nice.

答: Make the hole just large enough to pass the plate through, something like 1 cm by 70 cm.  The problem is that you can orient the lid with the 70 cm longways.  You need a thick part of the lid that is more than 1 cm.  It doesn't need to be much.  A 1.25 by 1.25 cm part on the end of the lid will be enough.  There is no absolute minimum for the lid, as it needs to be wider than 1 cm but can be any amount wider.</p>

Title: Curious minimization problem


问:
Can I know the Fourier coefficients of a function that uniformly converges? if so why is it?
For example the function

$$f(x) = \sum\frac{1}{2^n}\sin(2^nx)$$
we can see that the function uniformly converges

Can I know that the Fourier coefficients are $$\hat{f}(x)=\frac{1}{2^n}$$

if so why is that possible?

答:
Denote $s_n(x) = \frac{1}{2^n}\sin(2^nx)$. $f$ is the continuous uniform limit of the series $\sum s_n$.

Therefore the Fourier coefficients of $f$ exist as $\int f(x) \sin nx, \int f(x) \cos nx$ exist and $a_n = 0$, $b_{2^p} = \frac{1}{2^p}$ and $b_n = 0$ for $n \neq 2^p$ for all $n \in \mathbb N$. This is by definition of those coefficients.

And $f$ is the sum of its Fourier sum also due to its definition:

$$f(x) = \sum_{n=0}^\infty\frac{1}{2^n}\sin(2^nx)$$


问:
Which of the two methods is correct and why?

Problem

A deck of card is shuffled then divided into two halves of 26 cards
each. A card is drawn from one of the halves, it turns out to be an
ace. The ace is then placed in the second half-deck. The half is then
shuffled, and a card is drawn from it. Compute the probability that
this drawn card is an ace.

my attempt:-

let A = number of aces in the second deck before an ace added to it

let B = number of aces in the first deck before an ace is removed from it

let D = event that an ace is drawn from the second deck after an ace is added to it

Now, what we are asked to find is: $$\sum_{i=0}^3P(A=i|B>0)\cdot P(D) = \sum_{i=0}^3P(A=i|B>0)\cdot \frac{i+1}{27} $$

now, there are two ways of calculating $P(A=i|B>0)$

(Protocol A)restrict the sample space way:-
since B has at least one ace, we restrict the sample space to 51 cards out of which 3 are aces. $$P(A=i|B>0) = \frac{{{3}\choose{i}}{{48}\choose{26-i}}}{{{51}\choose{26}}}$$

(Protocol B)$P(A|B) = \frac{P(A\cap B)}{P(B)}$:-
$$P(A=i|B>0) = \frac{P(A=i\cap B>0)}{P(B>0)} = \frac{{{4}\choose{i}}{{48}\choose{26-i}}\bigg{/}{{52}\choose{26}}}{\bigg{[}{{52}\choose{26}}-{{4}\choose{0}}{{48}\choose{26}}\bigg{]}\bigg{/}{{52}\choose{26}}}$$

My question is : (out of Protocols A and B) which method is the correct one? and Why?

Protocol A has already been used here to answer this question correctly. So, protocol A is obviously correct. So, I guess my question becomes: Why is protocol B incorrect?

答:
Not an answer to your question but an alternative that might interest you.

The ace that was drawn from the first half has probability $\frac1{27}$ to become the last drawn card.

All other cards have equal probability to become the last drawn card, so if $p$ denotes this probability then:$$51p+\frac1{27}=1$$This makes clear that:$$p=\frac1{51}\frac{26}{27}$$

Three of those cards are aces so the probability that the last drawn card is an ace equals:$$\frac1{27}+3p=\frac1{27}+\frac3{51}\frac{26}{27}=\frac{43}{459}$$


问:
How to project a 3d-line (represented in Plücker coordinates) into 2d image
I have a fully calibrated camera setup (that means $K$ and $P = [R|t]$ are known) and want to project a 3d line into the camera image.

The 3d line is defined via world coordinates $A$ and $B$ (as $4x1$ homogeneous vectors) and represented with Plücker coordinates:

$$
L = A\cdot B^T - B\cdot A^T\\
L_{coord} = \{l_{12},l_{13},l_{14},l_{23},l_{42},l_{34}\}
$$

How is than the projection performed? I assume as the projection matrix for points is $3x4$ the new projection matrix should something like $3x6$. I could not find a good resource about this topic.

答:
You could have simply projected the points A and B to image points a and b

$$a=PA$$
$$b=PB$$

Then compute the 2D line from the 2D image points with cross product.


问:
Prove by smallest counterexample $3 | (5^n - 2^n)$
I have it up to here:

For $k - 1$ (which is true) $3 | (5^{k-1} - 2^{k-1})$

$5^k - 2^k = 3n$, where $n\in Z$

So $5(5^{k-1}) - 2(2^{k-1})$

But now what? Is there some algebra trick I'm missing because I am stuck here

答: What you should be saying is assume there is an $n$ such that $3 \not \mid 5^n-2^n$  Based on that assumption, show that $3 \not \mid 5^{n-1}-2^{n-1}$ and the failure continues.  It would force $3 \not \mid 5^1-2^1$, which is false.  This means the original assumption is wrong and $3 \mid 5^n-2^n$ for all $n$


答:
Let $k$ be the smallest positive integer for which $3\not|~5^k - 2^k$. Of course $k\neq 1$, so $k-1$ is a positive integer, therefore

$$3~|~5^{k-1} - 2^{k-1}$$

so

$$3~|~2(5^{k-1} - 2^{k-1}) + 3\cdot 5^{k-1} = 5^k - 2^k$$

which is a contradiction.

Title: Prove by smallest counterexample $3 | (5^n - 2^n)$

Body:
I have it up to here:

For $k - 1$ (which is true) $3 | (5^{k-1} - 2^{k-1})$

$5^k - 2^k = 3n$, where $n\in Z$

So $5(5^{k-1}) - 2(2^{k-1})$

But now what? Is there some algebra trick I'm missing because I am stuck here


问:
Limit as n approaches infinity involving roots
$$\lim_{n\to\infty}\frac{n}{1+2\sqrt{n}}$$

Given my understanding of how to solve these problems, I need to take the highest power of $n$ in the denominator and then divide both the numerator and denominator by that power. But the highest power in the denominator is lower than the one in the numerator, so I will still be left with $n^{1/2}$ in the numerator, which gives $\frac{\infty}{2}=\infty$. Is that the correct answer or am I missing something? I would have thought that since the power of the denominator is smaller than the numerator that you could factor it by long division, but I'm not sure I know how to do that.

答:
It's correct, but I'd avoid writing things like $\dfrac{\infty}{2}=\infty$ for anything other than intuition.

You could use long division if you really wanted to:
$$\dfrac{n}{1+2\sqrt{n}} = \dfrac{1}{2}\sqrt{n} + \dfrac{1}{4} - \frac{1}{4} \cdot \dfrac{1}{1+2\sqrt{n}}$$
This clearly gives you the same answer.


问:
Derangement and $1/e$ - how to make this proof rigorous?
It is well known that the number of derangements (permutations $\sigma \in S_n$ with no fixed points) is given as
$$
D_n =  n!\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}.
$$
From this, we can show that the probability of a given permutation to be a derangement converges to $1/e$ as $n$ goes to infinity. Now, here's a variation of the problem: for given $k$, what is the limit probability that the permutation of a sequence $(1, ..., 1, 2, ..., 2, ..., n, ..., n)$ (each $i \in [n]$ shows exactly $k$ times) is derangement as $n\to \infty$? I heard that one can find the number of such derangements in terms of Bernoulli polynomials and the answer is $1/e^k$, but I wonder if there's more easier way to show this.

In fact, here's my "proof": for $k=1$, for each position, the probability where $\sigma(i) \neq i$ will be $\frac{n-1}{n}$ as $n$ is large enough, so that the probablity will be the limit of $(\frac{n-1}{n})^{n}$, which is $1/e$. Similar arguments also works for general $k$. I think this "proof" is correct at least intuitively, but I wonder if this can be made rigorous.

答: Calculate the probability where $\sigma(i)\neq i$ in your sequence to be $\frac{(nk-k)(nk-k-1)...(nk-2k)((nk-k)!)}{nk!}$ then pair things up as $\frac{nk-k}{nk}\frac{nk-k-1}{nk-1}...\frac{nk-k-k}{nk-k}$ then by substitute $n+\frac{j}{k}$ for $n$ to get $\frac{nk-k}{nk}$ so the limit goes to $(\frac{nk-k}{nk})^k$, which when you raise to the $n$ power to include all $i$ gives the desired result is reached.


问:
Length of evolute for general regular $C^2$ curve (is $\kappa'$ continuous?)
Let $c(t)$ be a regular $C^2([a,b])$ curve parameterized by arc length. It's curvature satisfies $\kappa(t)\ne 0$ and $\kappa'(t)\ne 0$.

I want to show that the length of the evolute $e(t) = c(t)+\frac{1}{\kappa(t)}N(t)$ is given by
$$L(e) = \left\lvert\frac1{\kappa(b)}-\frac1{\kappa(a)}\right\rvert$$
I have already found the tangent vector of $e$, and am stuck on this particular integral:
$$\int_a^b \frac{\lvert\kappa'(t)\rvert}{\kappa^2(t)}dt$$
The absolute value is giving me all the trouble.

As $c$ is only $C^2$, by the Frenet-Equations for planar curves, $\kappa'$ could be non-continuous as $c'''$ could be non-continuous. So I can't guarantee that $\kappa'$ won't change signs. Or can I?

Is $\kappa'$, in fact, continuous for some reason? Is there some other way to crack that absolute value?

答:
Here is a trick. The Darboux's theorem state that if $f$ is a differentiable function on an interval, then $f'(I)$ is an interval $J\subset \mathbb{R}$. Thus, if $k$ is differentiable and $k'$ does not vanish, then $k'$ must have constant sign.

This does not, by the way, guarantee that $k'$ is locally integrable. I think you should look closely to the assumptions: the curvature $\kappa$ is defined for a $C^2$ parametrized curve but assuming $k$ is differentiable is closely related to assuming that the curve is three-time differentiable.

I would say that the good assumptions for your exercice are that your curve is $C^3$, or that it is $C^2$, three times differentiable, with $k'$ locally integrable.


问:
Incompleteness Theorems for Limit-Computable Formal Systems
Godel’s First and Second Inconpleteness Theorems are about Peano Arithmetic, but their punchlines are respectively that “Any sufficiently expressive computable formal system cannot be complete and consistent” and that “Given a computable emulation of the proof theory of such a system inside itself, no proof consistency can be given, if the system is consistent”. In mathematics we like certainty (or something like it, for those of you with less objective dispositions), so this is a problem. However, practical considerations do not always need certainty. In this spirit, we might not require a computable formal system, but merely a limit-computable one. By this I mean that there is a theoretical machine $m$ which takes pairs of sentences $s$ of the system and natural numbers $n$ as inputs and outputs $0$ or $1$ assuming it halts. Provability of $s$ in the system is that
$$\lim_{n\to\infty}m(s,n)=1$$
which just says that, for all but finitely many $n$, $m$ outputs $1$ on input $(s,n)$. In this way, if we accept the system, we can at least be sure that it will eventually tell us the right answer. Are there analogues of the Incompleteness Theorems for such systems? Or similar results?

答:
I'm going to focus on the first incompleteness theorem below, except for the final paragraph.

There are two problems with the theory you describe. The first is that, as you observe, it is not actually computable - and as finite entities, we can't really make that much use of limit computability. The second issue is that it is unsound: it is complete and consistent, but proves false sentences.

When we take soundness into account we see that there is indeed a version of the incompleteness theorem which "relativizes" to limit computation and beyond. Specifically, for each $n>0$ we have:

($\mathsf{GIT_n}$):  Suppose $T$ is a consistent theory of arithmetic which contains Robinson arithmetic, is ${\bf 0^{(n-1)}}$-computably axiomatizable, and is $\Pi^0_n$-sound. Then $T$ is incomplete.

(As usual this can be generalized substantially, but let's focus on this version for simplicity.)

A couple quick comments:

"${\bf 0^{(n-1)}}$" is the $n-1$th iterate of the Turing jump. The case $n=1$ corresponds to computably axiomatizable theories (= the original GIT).

"$\Pi^0_n$" is a particular class of sentences in the language of arithmetic - see here. "$\Pi^0_n$-soundness" is the property of only proving true $\Pi^0_n$ sentences. Note that when $n=1$ this is a redundant hypothesis, since every theory containing Robinson arithmetic proves all true $\Sigma^0_1$ sentences and so cannot prove any false $\Pi^0_1$ sentences without being inconsistent; this is why this additional hypothesis doesn't appear in the usual statement of the first incompleteness theorem.

The two notions above are related to each other and to (a generalization of) limit computability: see Post's theorem and Shoenfield's limit lemma. In particular, we have "computable = $\Delta^0_1$ = ${\bf 0^{(0)}}$-computable" and "limit computable = $\Delta^0_2$ = "${\bf 0^{(1)}}$-computable."

The $\mathsf{GIT_n}$s above can be proved by relativizing any of the computability-theoretic proofs of the usual incompleteness theorem; a useful trick here is that if $T$ satisfies the hypotheses of $\mathsf{GIT_n}$ then so does $T$ + the set of true $\Sigma^0_n$ sentences (this is a good exercise). Proof-theoretic arguments can also be used, but their relativizations are in my opinion less immediate.

[The second incompleteness theorem poses an interesting new difficulty, on the other hand; see an earlier edit of this question in conjunction with Andreas Blass' comment below.]


问:
If $f^2(b) - f^2(a) = b^2 - a^2$, then the equation $f'(x)f(x) = x$ has at least one root in (a, b).
Suppose $f$ is continuous on [a, b] and differentiable on the open
interval (a, b). How to  that if $f^2(b) - f^2(a) = b^2 - a^2$,

then the equation

$f'(x)f(x) = x$

has at least one root in (a, b).

答:
Consider
$$
F(x)=f^2(x)-x^2
$$
and apply Rolle's theorem</p>


问:
Find the values of $k$ , for which the system is stable.
I have a multiple choice question , that says , if $y(t) = (k^2 -3k -4)\log(x) + \sin(x)$ , find the values for which the system is stable ( I guess it means BIBO stable) .

$$a. 1
 \ \ \ \ b. 3 \ \ \ \ c.-1 \ \ \ \ d. 5 \quad e. 4
\quad f. 2$$

We are allowed to pick as many choices as we want.
I circled every single choice, am I missing something?

Let's suppose $ \exists M \in R $ such that: $|x(t)| < M  \quad\forall t \in \mathbb{R}$

$ |y(t)| = |(k-4)(k+1)\log(x) + \sin(x)| \leq |(k-4)(k+1)\log(x)| + 1 $

since $\log(x)$ is defined, then $x(t)>0 \forall t \in \mathbb{R} \rightarrow |x(t)| = x(t) < M$

and since $\log_{10}(x) $ is an increasing function $\rightarrow x(t) < M \rightarrow \log(x(t)) < \log M$

$ |y(t)|  \leq |(k-4)(k+1)\log(M)| + 1 $

For every given value of $k$,  $|(k-4)(k+1)\log(M)| \in [0,+\infty]$ so I can't see why not , every value is possible

答:
Okay after some rethinkig i ended ip with the conclusion:

If $(k^2 -3k 4) = (k-4)(k+1) = 0 \rightarrow k \in {-1,4}$ then $|y|=|sinx|<2$ it is BIBO stable

but if $c=(k^2 -3k 4) \neq 0$ then $|y| = |c||logx(t)| + |sinx|<=|c||logx(t)| +1$

Suppose $x(t) = \frac{1}{10^|t|}<=1 \forall t $ but then ...

$lim_{|y| \rightarrow \infty} = \infty$ it is not BIBO -BOUNDED


问:
Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$
How should we find two rational numbers $\alpha$ and $\beta$ such that

$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$

The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you

答:
By Gauss' lemma, if $\alpha$ and $\beta$ are rational, then they are integers. Simply cube both sides to get
$$7+5\sqrt{2}=(\alpha+\beta\sqrt{2})^3.$$
Expanding the right hand side and comparing coefficients shows that
\begin{eqnarray*}
7&=&\alpha^3+6\alpha\beta^2&=&\alpha(\alpha^2+6\beta^2),\\
5&=&3\alpha^2\beta+2\beta^3&=&\beta(3\alpha^2+2\beta^2),
\end{eqnarray*}
so $\alpha$ divides $7$ and $\beta$ divides $5$. This leaves very few options to check.


答:
Yes, it is correct. Another way to demonstrate it is to multiply the numerator and denominator by the conjugate:

$$\frac{n}{1+\sqrt{n}}=\frac{n}{1+\sqrt{n}}\frac{1-\sqrt{n}}{1-\sqrt{n}}$$

Doing the multiplication, the ratio becomes

$$\frac{n-n\sqrt{n}}{1-n}=\frac{1-\sqrt{n}}{-1+\frac{1}{n}}$$

The denominator converges to -1, while the numerator diverges to $\infty$.

Title: Limit as n approaches infinity involving roots

Body:
$$\lim_{n\to\infty}\frac{n}{1+2\sqrt{n}}$$

Given my understanding of how to solve these problems, I need to take the highest power of $n$ in the denominator and then divide both the numerator and denominator by that power. But the highest power in the denominator is lower than the one in the numerator, so I will still be left with $n^{1/2}$ in the numerator, which gives $\frac{\infty}{2}=\infty$. Is that the correct answer or am I missing something? I would have thought that since the power of the denominator is smaller than the numerator that you could factor it by long division, but I'm not sure I know how to do that.


问:
Question: Rudin exercise 3.8
I know there has been many posts about this question here already but I'm not sure how to resolve my dilemma.

***If $ \sum{a_n} $ converges and $\{b_n\}$ is bounded and monotonic, prove that $\sum{a_nb_n}$ converges. ***

This is my work thus far:
I have $\sum_na_n$ is convergent which implies for all $\varepsilon > 0$ there exists $N$ s.t.

$\left\vert \sum_{k=n}^ma_k\right\vert =|s_m-s_n|\leq \varepsilon$ for $m \geq n\geq N$ where $\{s_n\}$ is a Cauchy sequence.

$\{b_n\}$ is bounded so there exists $M\in \mathbb{N}$ s.t. $\vert b_n \vert \leq M.$

Then we have $\vert a_nb_n\vert=|a_n|b_n|\leq |a_n|M.$

I know $\left\vert \sum_{n}a_k\right\vert\leq \sum_n|a_n|$ in general but I don't have that $\sum_na_n$ converges absolutely. I haven't learned Dirichlet's Test yet.

What I want to do is show that $|a_nb_n|\leq c_n, n\leq N_0\in \mathbb{N}$ and $\sum c_n$ converges. For now, I think my $c_n=|a_n|M$ but I don't see how I can show $\sum_n |a_n|M=M\sum_n |a_n|$ converges. Any tips please?

答:
I reworked it; please let me know if this makes sense!!

$\{b_n\}$ is bounded so $|b_n|\leq M$.

$\sum_n a_n$ is convergent so $\forall \varepsilon >0, \exists N$ s.t. $\vert \sum_n^m a_k|\leq \varepsilon/M,\,\,m\leq n \leq N.$

$\vert \sum_{k=n}^m a_kb_k\vert\leq \vert \sum_{k=n}^m a_k M|=|M\sum_{k=n}^m a_k\vert=M\vert\sum_{k=n}^m a_k\vert\leq M\cdot \varepsilon/M=\varepsilon.$

Title: Question: Rudin exercise 3.8


答:
Hint::

$n\ge 1$;

Consider the binomial expansion of

$5^n=(3+2)^n; $

All terms except the last term $2^n$ have a factor $3$.

What can you say about $5^n-2^n$ ?


问:
Degenerate Pi-notation
can i say that a degenerate productory like

$$
\prod_{t=1}^{0} (1+r_t)
$$

is equal to one? I cant seem to find a precise answer about this anywhere.

答: Yes – if you don't multiply by anything at all, it's the same as if you had multiplied by $1.$ That's why $x^0=1$ and why $0!=1.$

Title: Degenerate Pi-notation


问:
Probability of first to draw red ball from urn without replacement
There are R red and G green balls in a box. Alice and Bob pick one ball from the box in turns without replacement. First one to draw a red ball wins. What is the probability that Alice wins if she draws first?

Here is my attempt - Could someone please advise where I went wrong?

For any draw, the probability of picking a red ball is equal to picking $k$ green balls before that and a red ball on $k+1$ draw.

$\displaystyle P(i=k)=\frac{1}{(k+1)}\frac{\dbinom{G}{k}\dbinom{R}{1}}{\dbinom{N}{k+1}}=\displaystyle \frac{R}{(k+1)} \frac{G!}{k!(G-k)!}\frac{(N-k-1)!(k+1)!}{N!}$

$\displaystyle=\frac{G!}{N!} \frac{(R+G-k-1)!}{(G-k)!}.R\frac{(R-1)!}{(R-1)!}=\displaystyle\frac{1}{\binom{R+G}{G}}\dbinom{R+G-k-1}{G-k}$

Let $P(A)$=Alice wins, $P(B)$=Bob wins. Let $P(x)=P(A) - P(B)$. Therefore $P(A) =\displaystyle\frac{1}{2}(1+P(x)) $

$\displaystyle P(x)=\frac{1}{\dbinom{R+G}{G}} \sum_{k=0}^{G} (-1)^{k} \dbinom{R+G-k-1}{G-k}$

If $G$ is even, $-1^k=-1^{G-k}$; using binomial identity

$\dbinom{n}{k} = -1^k \dbinom{k-n-1}{k}$, we get,

$\displaystyle P(x)=\frac{1}{\dbinom{R+G}{G}} \sum_{k=0}^{G} \dbinom{-R}{G-k}$

$\displaystyle P(x)=\frac{1}{2^N\dbinom{R+G}{G}} \sum_{k=0}^N \dbinom{N}{k} \sum_{k=0}^G \dbinom{-R}{G-k}$

$\displaystyle P(x)=\frac{1}{2^N\dbinom{R+G}{G}} \sum_{k=0}^N \sum_{k=0}^G \dbinom{G+R}{k} \dbinom{-R}{G-k}$

Using Chu-Vandermonde identity, we get,

$\displaystyle P(x)=\frac{1}{2^N\dbinom{R+G}{G}} \sum_{k=0}^N \dbinom{G}{G}$

$\displaystyle P(x)=\frac{N+1}{2^{N}\dbinom{R+G}{G}}$ - which seems wrong e.g. set $G=0$

答:
Let random variable $R_1$ be the number of draws to get the first red ball.  For $k\in\{1,\dots,G+1\}$, we have
\begin{align}
P(R_1=k)
&= \left(\prod_{d=0}^{k-2} \frac{G-d}{R+G-d}\right)\frac{R}{R+G-(k-1)} \\
&= \frac{\binom{G}{k-1}}{\binom{R+G}{k-1}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-(k-1)}{G-(k-1)}}{\binom{R+G}{G}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-(k-1)}{R}}{\binom{R+G}{G}}\frac{R}{R+G-(k-1)}\\
&= \frac{\binom{R+G-k}{R-1}}{\binom{R+G}{G}}.
\end{align}
So the probability that Alice wins is
\begin{align}
\sum_{k=0}^{\lfloor G/2 \rfloor} P(R_1=2k+1)
&=\sum_{k=0}^{\lfloor G/2 \rfloor} \frac{\binom{R+G-(2k+1)}{R-1}}{\binom{R+G}{G}} \\
&=\frac{1}{\binom{R+G}{G}}\sum_{k=0}^{\lfloor G/2 \rfloor}\binom{R-1+G-2k}{R-1} \\
&=\frac{1}{\binom{R+G}{G}} \sum_{k=0}^G\frac{1+(-1)^k}{2}\binom{R-1+G-k}{R-1} \\
&=\frac{1}{2\binom{R+G}{G}}\left(
\sum_{k=0}^G\binom{R-1+G-k}{R-1}
+\sum_{k=0}^G(-1)^k\binom{R-1+G-k}{R-1}
\right)\\
&=\frac{1}{2\binom{R+G}{G}}\left(
\binom{R+G}{R}
+(-1)^G \sum_{k=0}^G\binom{-(G-k)-1}{R-1}
\right)\\
&=\frac{1}{2}+\frac{(-1)^G}{2\binom{R+G}{R}}
\sum_{k=0}^G\binom{-(G-k)-1}{R-1}
\\
\end{align}