问:
limit continuity for value K does not exist
$$f(x) := \begin{cases}
\frac{x^2+2kx+1}{x+1}  & \text{for $x$ < -1}, \\[2ex]
x^2+2 & \text{for $x$ $\ge$ -1}.
\end{cases}$$
I tried using L'Hopital, but after seeing it in desmos, the graph doesn't seem to be continuous at the point that I got.

答:
I am assuming that in the problem it is given that the function is continous in its domain

so $\lim_{x \to -1}f(x)$  must exist
$\Rightarrow (-1)^2 - 2k \ + \ 1 = 0 $

which gives $k=1$

so $\lim_{x \to -1^-} f(x) = 0 \neq f(-1) $

so $f$ is continous for no value of k


问:
Prove Direct sum of subsets of basis $V$
I have a task to prove this thing and I think I did but I'm not sure if the proof is good.

Question :
Let $B$ be basis of $V$ such that $B=B_1∪....∪B_k$ for $i=1....k$

So $U_i=\operatorname{Span}({B_i})$ --> Prove $V=U_1⊕....⊕U_k$

My answer :
(1)
$B$ is basis so the vectors in it are lineary independent which means $B_1∪....∪B_k$ are lineary independent.

so lets assume there is a Vector $v∈B_1∩....∩B_k$ which means the vector $v∈B_1,...,B_k$

so $v$ can be written as linear equation of the vectors in $B_1,...,B_k$. ----> We get that we have a common vector in $B_1$ and $B_2$... which means the basis $B$ is dependent and contradicts our assumption. ---> the intersection = ${0_V}$.

(2) Prove $V=U_1,...,U_k$ : it's trivial because we know $V=U_1+...+U_k$ by the given question.

Is that right or I proved something wrongly / assumed wrongly?
Thank you very much!!

答:
If $\dim V$ is finite. Let $\mathcal{B}=\{v_1, v_2,\ldots, v_n\}$ a basis of V, where $n=\dim V$. So, define $$U_i:=span(v_i).$$

Let $v\in V$, then $$v=c_1v_1+c_2v_2+\cdots+c_nv_n,$$ Now, note that for all $i$, $c_iv_i\in span(v_i)$, So $$V=U_1+U_2+\cdots+U_n$$

Finally, you must to show that $U_1\cap U_2\cap\cdots\cap U_n=\{0\}$ (Show that if $w\in U_1\cap U_2\cap\cdots\cap U_n$ so $w=0$). With all you must conclude that $$V=U_1⊕U_2⊕\cdots ⊕U_n.$$

ContentLicense: CC BY-SA 4.0


问:
General Inequality for Convex Functions
Suppose we have a convex function $f:[-1,1]\to \mathbb{R}$ with $f(-1)=f(1)=0$ and $\{t_i\}_{i=1}^k$ is a sequence of numbers such that $k\ge 2$ and

$$
\sum_{i=1}^k \frac{t_i}{f(t_i)+f(-t_i)}=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \       (A)
$$

Is the following necessary?

$$
\sum_{i=1}^k \frac{f(t_i)}{f(t_i)+f(-t_i)}\ge 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \       (B)
$$

I suspect $1$ can be replaced by $k/2$, in which case it works for $k=1$ and arguably $k=0$, too. I would also like to generalize this to n dimensions eventually.

PROGRESS

f(x) EVEN

(B) is true without (A) when $f$ is even since $k/2\ge 1$. Specifically, notable is when $f$ is a semicircle. I believe this implies (B) is also true when $f$ describes an (oblique) semi-ellipse.

f(x) LINEAR

If $f(x)$ is linear in a region containing all $t_i$, (A) becomes

$$
\sum_{i=1}^k\frac{t_i}{2f(0)}=0
$$

which gives (B) through

$$
\sum_{i=1}^k\frac{f(t_i)}{2f(0)}=\sum_{i=1}^k\frac{(\textrm{slope})t_i+f(0)}{2f(0)}=k/2\ge 1
$$

Maybe next is where all $t_i$ are contained on a few linear pieces, although I've run into muck that way.

INEQUALITY?

If there is a good inequality to prove (B) in the form

$$
\sum_{i=1}^k \frac{t_i}{f(t_i)+f(-t_i)}\cdot \frac{f(t_i)}{t_i}\ge 1
$$

since we know the left factors sum to $0$.

REPARAMETRIZE

Finally, I tried reparametrizing in terms of $u(t)=\frac{t}{f(t)+f(-t)}$. First, we need to show that $u$ is invertible. Write $F(t):=f(t)+f(-t)$.

$$
u^\prime(t)=\frac{F(t)-t(F^\prime(t))}{F^2(t)}
$$

The numerator is strictly negative (always $\le F(0)$) and the denominator is positive (excluding $t=\pm 1$). So, $u$ is strictly decreasing, hence invertible. Equation (B) becomes for any $\{c_i\}_{i=1}^k$ such that $\sum c_i=0$,

$$
\sum_{i=1}^k c_i\frac{f(u^{-1}(c_i))}{u^{-1}(c_i)}\ge 1
$$

Then, we would be quickly finished if we could show $|f(u^{-1}(s))/u^{-1}(s)|$ were convex on $(-\infty,0)$ and on $(0,\infty)$, as we could merge all the $c_i>0$ to one value and all the $c_i<0$ to the opposite value (no harder than $k=2$), and maintain the inequality the whole time. So, we take derivatives. Note $d/ds\  u^{-1}(s)=1/u^\prime(u^{-1}(s))$.

JUNK SCRATCH

$$
u^{\prime}(t)=\frac{F-tF^\prime}{F^2}
$$

$$
u^{\prime\prime}(t)=\frac{-tF^{\prime\prime}F-2F^\prime(F-tF^\prime)}{F^3}
$$

$$
\left.\frac{d}{ds}\frac{f(u^{-1}(s))}{u^{-1}(s)}=\frac{f^\prime(t)t-f(t)}{u^\prime(t)\cdot t^2}\right\vert_{t=u^{-1}(s)}
$$

$$
\left.\frac{d^2}{ds^2}\frac{f(u^{-1}(s))}{u^{-1}(s)}=\frac{f^{\prime\prime}(t)t\cdot u^\prime(t)t^2-(f^\prime(t)t-f(t))(u^{\prime\prime}(t)t^2+2tu^\prime(t))}{(u^\prime(t))^3\cdot t^4}\right\vert_{t=u^{-1}(s)}
$$

Recall that $u^\prime <0$. Thus (B) will follow from (A) if for all $|t|\le 1$,

$$
f^{\prime\prime}\cdot t^3\cdot u^\prime\le (f^\prime\cdot t-f)(u^{\prime\prime}\cdot t^2+2tu^\prime)
$$

or,

$$
f^{\prime\prime}\cdot t^3\cdot (F-tF^\prime)\le (f^\prime\cdot t-f)((-tF^{\prime\prime}-2F^\prime+2t(F^\prime)^2/F)\cdot t^2+2t(F-tF^\prime))
$$

$$
\frac{f^{\prime\prime}\cdot t^3}{f-tf^\prime}\le\frac{F^{\prime\prime}\cdot t^3}{F-tF^\prime}-2t\frac{F-tF^\prime}{F}
$$

If this is going to be true for all convex $f$, it needs to be true when $F=2f$, which says $0\le t(F-tF^\prime)$..... ah, the sign change at $t=0$! This is why I split left and right.

%$$
%(f^{\prime\prime}\cdot t^3-2t^2f^\prime+2tf)\cdot (F-tF^\prime)\le (f^\prime\cdot t-f)((-tF^{\prime\prime}-2F^\prime+2t(F^\prime)^2/F)\cdot t^2)
%$$

%$$
%f^{\prime\prime}\cdot t\cdot u^\prime\le (f/t)^\prime(u^\prime t^2)^\prime
%$$

答:
Let's work where $-f(t)$ is convex instead of $f(t)$.

Note that $t_i=0$ can be omitted, as it adds nothing to (A) but increases (B).

First, we will only work with $t_i>0$ and the other half will work symmetrically. Call $$c_i:=\frac{1}{f(t_i)+f(-t_i)}$$ and let $T=\sum_{t_i>0}c_it_i$ be fixed. We can show $f(t)/t$ is strictly decreasing by $$\left(\frac{f}{t}\right)^\prime=\frac{-(f-tf^\prime)}{t^2}$$ and thinking of $f-tf^\prime\ge f(0)$ as the "Legendre transform" as in the OP graph. By the same argument, $t/(f(t)+f(-t))$ is strictly increasing.

Now, we would like to bound the quantity

$$
\frac{\sum_{t_i>0} c_it_i\cdot f(t_i)/t_i}{\sum_{t_i>0} c_it_i} \ \ \ \ \ \ \ (B')
$$

which is a weighted average of values of $f(t)/t$. Since $t/(f(t)+f(-t))$ is increasing, the largest possible $t_i$ under fixed $T$ occurs when $\#\{t_i>0\}=1$, i.e. $T=c_mt_m$ for some $m$. So, (B') finds its minimum under fixed $T$ at $c_m f(t_m)$.

Similarly, the minimum over $\{t_i<0\}$ occurs when $\#\{t_i<0\}=1$. The only configurations under $k=2$ are when $t_1=-t_2$, due to invertibility of $u(t):=t/(f(t)+f(-t))$. It follows that the minimum value of (B) when there are both positive and negative $t_i$ values is $1$.

I suspect the $1$ bound cannot be replaced by $t/2$ since that would appear to be true iff $|sf(u^{-1}(s))/u^{-1}(s)|$ were convex on each side $(-1,0)$ and $(0,1)$ separately.

It's interesting to note that $$\frac{d^2}{dt^2}\left|\frac{f(t)}{t}\right|=-\text{sgn}(t)\frac{t^2f^{\prime\prime}(t)-2f^\ast(t)}{t^3}$$ is positive when $1/2 f^{\prime\prime}t^2<f^\ast(t)$, which looks like it could be formed into a physics problem. If $f(t)$ is position over time, $1/2 f^{\prime\prime}t^2$ is a displacement under constant acceleration equal to $f^{\prime\prime}$. Also, $f^\ast$ represents a displacement under no accelaration. Or, rather, $1/2 f^{\prime\prime} t^2 \le f^\ast=tf^\prime-f$ when the best quadratic approximation $y=1/2 f^{\prime\prime}(x-t)^2+f^\prime(x-t)+f$ has its $y$-intercept on the same side of the x-axis as the graph of $f$.

Title: General Inequality for Convex Functions


问:
Looking for a characterization of the image of continuous probability densities supported in $[0,1]$ via the operator $f\mapsto (f*I)f$
Let
\begin{equation*}
U = \big\{f\in C(\mathbb{R}) \mid f=f\mathbb{I}_{[0,1]}\big\},
\end{equation*}
where $\mathbb{I}_{[0,1]}$ is the indicator function of the interval $[0,1]$.
Let
\begin{equation*}
V = \bigg\{f\in U \mid \big(f\ge 0\big)\land\bigg(\int_0^1 f(t)\operatorname{d}t = 1\bigg)\bigg\}.
\end{equation*}
Let
\begin{equation*}
W = \bigg\{f\in U \mid \int_0^1 f(t)\operatorname{d}t = 0\bigg\}.
\end{equation*}
Let
\begin{equation*}
I:\mathbb{R} \to \mathbb{R}, x\mapsto x.
\end{equation*}
Define
\begin{equation*}
\Gamma\,\colon V\to W, f\mapsto (f*I)f,
\end{equation*}
where $*$ is the convolution operator, i.e. $g*h(x)=\int_{\mathbb{R}}g(t)h(x-t)\operatorname{d}t$.

I'm trying to solve the equation $\Gamma(f) = \varphi$, i.e. I'm looking for a characterization of the image of the operator $\Gamma$.

Taking Fourier transform the equation becomes
\begin{equation*}
\hat{f}'=\hat{f}'(0)\hat{f}-2\pi i\hat{\varphi}
\end{equation*}
so a viable strategy seems to solve the following linear ODE with a constraint on the derivative in $0$
\begin{equation*}
\begin{cases}
u'&=au-2\pi i\hat{\varphi} \\
u'(0) &= a
\end{cases}
\end{equation*}
then, anti-trasforming the equation $\hat{f} = u$, and finally see for which $\varphi \in W$ this procedure ends with an element in $V$.

However - especially the last step - this seems a little bit messy... is it clear when this is so? Or maybe has anyone a simpler/more elementary idea about how to tackle this problem?

答:
As mentioned in the title, $V$ is the collection of continuous pdfs of random variables (rv) supported on $[0,1]$. Let $f \in V$ be the pdf of the rv $X$. Then,

$$f\ast I(x) = \int_0^1 f(t)(x-t)\,dt = x\int_0^1 f(t)\,dt - \int_0^1 tf(t)\,dt = x - \mathbb{E}[X].$$

Let $\phi_f = \Gamma f = (f\ast I)f$. Also for any $\phi$ in the image of $\Gamma$, let $f_{\phi} = \Gamma^{-1}\phi$ whenever the inverse is unique and associate it with the rv $X_{\phi}$.

First, suppose that $f > 0$ on $(0,1)$. Then $\phi_f(t) = 0$ precisely when $t \in \{0,\mathbb{E}[X],1\}$. So if the unique zero of $\phi$ in $(0,1)$ is some $y \in (0,1)$, then we know that $\mathbb{E}[X_{\phi}] = y$. Then since $\phi(t) = (t-y)f_{\phi}(t)$, it follows that,

$$f_{\phi}(t) = \frac{\phi(t)}{t - y}.$$

So in the class of functions $\phi \in W$ such that $\phi$ has a unique zero, $y\in (0,1)$, $\phi$ is in the image of $\Gamma$ if and only if $\frac{\phi(t)}{t-y} \in V$ and this candidate solution has the right expectation. That is,

$\phi(t) \leq 0$ if $t < y$ and $\phi(t) \geq 0$ if $t > y$.

$\phi$ is differentiable at $y$. (Originally this was $\lim_{t \to y} \frac{\phi(t)}{t-y} = 0$, but that's a mistake see Edit 2)

$\int_0^1 \frac{\phi(t)}{t-y}\,dt = 1$.

$\int_0^1 \frac{t\phi(t)}{t-y}\,dt = y$.

In particular, this shows that although $W$ is closed under scalar multiplication, $\Gamma(V)$ is not.

Now we consider the general case. For each $y \in (0,1)$, define

$$f_{\phi,y}(t) = \frac{\phi(t)}{t-y}.$$

Suppose $\phi$ has multiple zeros in $(0,1)$. Then it should be easy to see that for each such zero $y$, if $f_{\phi,y}$ satisfies the four points above, then $f_{\phi,y} \in V$ and $\Gamma f_{\phi,y} = \phi$. It should also be clear by now that if there does not exist such a $y$, then there is no $f \in V$ such that $\Gamma f = \phi$. Also, if there exist $y_1 \neq y_2$ satisfying the 4 conditions, then $\Gamma f_{\phi,y_1} = \Gamma f_{\phi,y_2} = \phi$. However, I'm not sure if this ever happens or not.

Edit: I thought of one other thing that might help in practice. Suppose $\phi$ has multiple zeros. If $y$ and $y'$ are both zeros satisfying condition 1, then it must follow that $\phi(t) = 0$ for all $t \in [y,y']$. Thus, the candidate zeros will always fall in some interval $[a,b]\subset (0,1)$ such that $\phi|_{[a,b]} \equiv 0$. Given some $\phi$ which may or may not be in the image of $\Gamma$, there exists an interval $[a,b]$ on which condition 1 easily holds. Condition 2 trivially holds on the interval $(a,b)$ and should be simple to verify or reject on the boundaries. The integrals in conditions 3 and 4 may be viewed as equations in terms of $y$ which can be solved for on the interval $[a,b]$.

Edit 2: Looking back at this today I realize I made a mistake with condition 2. We need to verify that $f_{\phi}$ is continuous at $y$, not $0$. Condition 2 should actually be $\lim_{t \to y} \frac{\phi(t)}{t-y}$ exists which is equivalent to stating that $\phi$ is differentiable at $y$.


问:
Showing $(\Omega, \mathcal{F}, P)$ is a probability space.
Can I please get help proving the problem below? I feel like I wrote a lot of superfluous stuff. Thank you.

Let $\Omega = \mathbb{R}, \mathcal{F} = $ all subsets so that $A$ or $A^c$ is countable, $P(A) = 0$ in the first case and $ = 1$ in the second. I wish to show that $(\Omega, \mathcal{F}, P)$ is a probability space.

$\textit{Proof.}$ Observe that $\mathcal{F}$ is a sigma field, and $\emptyset \in \mathcal{F}$ as $\emptyset$ is countable by definition, $\Omega \in \mathcal{F}$ as $\Omega^c = \emptyset$ is also countable by definition. If $A\in \mathcal{F}$ then $A$ is countable as $A^c$ is countable. Now, if $A$ is countable then $A^c \in \mathcal{F}$ as $(A^c)^c$ is countable and if $A^c$ is countable then clearly $A^c \in \mathcal{F}.$ If $A_n's \in \mathcal{F}$ then if all $A_n's$ are countable then $\cup A_n's$ are countable then $\cup A_n \in \mathcal{F}.$ If there exists $N$ such that $A_N^c$ is countable then $$\left(\bigcup_{n=1}^\infty A_n\right)^c = \bigcap_{n=1}^\infty A_n^c \subseteq A_N^c$$ is countable then $\bigcap_{n=1}^\infty A_n^c$ is countable then $\cup_{n=1}^\infty A_n \in \mathcal{F}.$ Hence $\mathcal{F}$ is a sigma field.

By definition, $P(\emptyset) = 0$ as $\emptyset$ is countable and $P(\Omega) = 1$ as $\Omega^c = \emptyset$ is countable. To show that if $A_n's$ are disjoint and $A_n's \in \mathcal{F}$ then $$P\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty P(A_n)$$ if all the $A_n's$ are countable then $P(A_n) = 0$ for all $n.$ As $\cup_{n=1}^\infty A_n$ is countable then $P(\cup A_n) = \sum P(A_n) = 0.$ Now suppose there exists $N_0$ such that $A_{N_0}$ is uncountable as $A_{N_0} \in \mathcal{F}$ then $(A_{N_0})^c$ is countable and $A_n \subseteq (A_{N_0})^c$ for all $n \ne N_0$ as $A_n's$ are pairwise disjoint then $P(A_n) = 0$ for all $n=\ne N_0$ and $P(A_{N_0}) = 1.$ Now $\cup_{n=1}^\infty A_n$ is uncountable implies $$P(\cup_{n=1}^\infty A_n) = 1 = P(A_{N_0}) = \sum_{n=1}^\infty P(A_n)$$ as $P(A_n) = 0$ for all $n\ne N_0$ and $P(A_{N_0}) = 1.$ This shows $(\Omega, \mathcal{F}, P)$ is a probability space as $P(\Omega) = 1.$

答:
(Note that $\mathcal F$ is sometimes called the countable-cocountable $\sigma$-algebra.)

There are two things we must show - that $\mathcal F$ is a $\sigma$-algebra and that $\mathbb P$ is a probability measure. As for the first, clearly $\Omega^c=\varnothing$ is countable, so $\Omega\in\mathcal F$. If $E\in\mathcal F$ is countable, then $(E^c)^c=A$ is countable, so $E^c\in\mathcal F$. If $E\in\mathcal F$ such that $E^c$ is countable, then by similar reasoning $(E^c)^c$ is co-countable, and hence $E^c\in\mathcal F$. Now let $(E_n)_n$ be a sequence of sets in $\mathcal F$. If each $E_n$ is countable, then $E:=\bigcup_{n=1}^\infty E_n$ is as well, and hence $E\in\mathcal F$. If not, there exists a positive integer $m$ for which $E_m$ is co-countable. Since
$$
E^c:=\left( \bigcup_{n=1}^\infty E_n\right)^c = \bigcap_{n=1}^\infty E_n^c \subset E_n^c,
$$
it follows that $E$ is co-countable, and hence $E\in\mathcal F$.

For $\mathbb P$ to be a probability measure, we must show that it is a (positive) measure, and that $\mathbb P(\Omega)=1$. By definition, $\mathbb P(E)\in\{0,1\}$ for any $E\in\mathcal F$, and in particular $\varnothing$ is countable, so $\mathbb P(\varnothing)=0$. If $(E_n)_n$ is a sequence of pairwise disjoint sets in $\mathcal F$, then either each $E_n$ is countable, so that
$$
\mathbb P\left(\bigcup_{n=1}^\infty E_n\right) = 0 = \sum_{n=1}^\infty \mathbb P(E_n),
$$
or there exists a positive integer $m$ for which $E_m$ is co-countable. Then for each positive integer $n\ne m$, we have $E_n\subset E_m^c$, so that $E_n$ is countable and hence $\mathbb P(E_n)=0$. This implies that $\sum_{n=1}^\infty \mathbb P(E_n) = \mathbb P(E_m) = 1$. Again, as $E^c\subset E_m^c$, we see that $E$ is co-countable, so that $\mathbb P(E)=1$ - together, we have
$$
\mathbb P\left(\bigcup_{n=1}^\infty E_n\right) = 1 = \sum_{n=1}^\infty \mathbb P(E_n).
$$
(Note that these are the only two cases we need consider, as if $E$ and $F$ were  co-countable subsets of $\Omega$, then they could not be disjoint. Feel free to prove this as an exercise.)
Finally, since the complement of $\Omega=\mathbb R$ is the empty set which is clearly countable, we see that $\mathbb P(\Omega)=1$, as desired.


问:
Distance between $P$ and a line in $\mathbb{R}^3$
Find the distance between $P=(6,4,-4)$ and a line that crosses points $A=(2,1,2)$ and $B=(3,1,-4)$. Note: Using dot product only. (No cross product).

I've tried doing the proyection of $P$, but since the canonical equation of the line does not exist, the solution was wrong. How could I apply the dot product for this problem?

答:
First the equation of the line $\vec{AB}$ going through points $A$ and $B$:

Direction vector $\vec{AB} = (3 - 2, 1 - 1, -4 - 2) = (1, 0, -6)$

Points on line $AB$ will be given by $(2, 1, 2) + (1, 0, -6) t = (2 + t, 1, 2 - 6t)$

Now if a perpendicular from point $P$ to the line $AB$ meets at point $C$, the direction vector of line $\vec{PC} = (2 + t - 6, 1 - 4, 2 - 6t - (-4)) = (t - 4, -3, 6 - 6t)$

As they are perpendicular to each other $\vec{AB} \cdot \vec{PC} = 0$

So, $(t-4,-3,6-6t) \cdot (1, 0, -6) = (t - 4 - 36 + 36t) = 37t - 40 = 0$

So the point $C$ on line $AB$ for $t = \frac{40}{37}$ is $(\frac{114}{37}, 1, \frac{-166}{37})$.

Distance from point $P$ to the line $AB, PC = \sqrt{(\frac{108}{37})^2 + 3^2 + (\frac{18}{37})^2} = \sqrt{\frac{657}{37}}$


问:
Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$?
As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Comparison Test for series multiple times, I derived the following estimates:

$$\frac{1}{k+1}<\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k<\ln\left(1+\frac{1}{k}\right)<\frac{1}{2k+2}+\frac{1}{2k}$$

The lower estimate

$$\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k$$

which I derived with WolframAlpha's help, is pretty useless. $1/(k+1)<R_k$ is good, but I want to improve it. Given how clean and tight the upper estimate

$$R_k<\ln\left(1+\frac{1}{k}\right)$$

is, I figured I could find a similar lower estimate that's just as clean and tight. After thinking for a while, I came up empty handed. I can't seem to find a positive sequence $a_n$ lying between $1/(n^2+1)$ and $1/n^2$ for which $\sum_{n=1}^{k}a_n$ has a clean expression. Any ideas or hints?

Edit: I'm not trying to prove the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2}$ nor any other series.

Body:
Consider $$g(n) = \frac{1}{n-1/2} - \frac{1}{n+1/2}$$
Then
$$   g(n) - \frac{1}{n^2} =  \frac{1}{4n^4 - n^2} > 0 \ \text{for}\ n \ge 1$$

Now $1/(4n^4 - n^2)$ is a decreasing function of $n$ for $n > 1$, so
$$\eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &= \frac{1}{N+1/2} - \sum_{n=N+1}^\infty \frac{1}{4n^4-n^2}\cr & > \frac{1}{N+1/2} - \int_{N}^\infty \frac{dx}{4x^4 - x^2} \cr
&= \frac{1}{N+1/2} - \ln \left(\frac{2N+1}{2N-1}\right) + \frac{1}{N}}$$
while on the other side
$$ \eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &< \frac{1}{N+1/2} - \int_{N+1}^\infty \dfrac{dx}{4x^4 - x^2}\cr
&= \frac{1}{N+1/2} - \ln\left(\frac{2N+3}{2N+1}\right)+ \frac{1}{N+1}}$$


问:
prove: $\dfrac{2^{n+1}+(-1)^n}{3}$
I am asked to prove this notation with induction for $n\in \mathbb{N}$:

real problem is to fill the area with tilings. and for $n\in \mathbb{N}$ there are exactly so many chances to fill the area as the formula says:  $$\dfrac{2^{n+1}+(-1)^n}{3}$$

I started in this way:

Base Case: $n=1$:

$$\dfrac{ 2^2-1}{3} = 1$$

Inductive step: $n\rightarrow n+1$:

$$\dfrac{2^{n+2}+(-1)^{n+1}}{3} = help..$$

my problem is: what am i doing, i seem to have already proven it in this last step which is obviously wrong. what am i missing here in this proof?

thanks for help

ContentLicense: CC BY-SA 3.0

Title: prove: $\dfrac{2^{n+1}+(-1)^n}{3}$

Body:
Hint: $$\dfrac{2^{n+2}+(-1)^{n+1}}{3}=\dfrac{2^{n+1}+(-1)^n}{3}+\dfrac{2\cdot 2^n+2\cdot(-1)^{n-1}}{3}$$
(factorize the second term)


问:
How to prove that $\lim_{x\to\infty}\frac{(\log_2 x)^3}{x^n}=0$

I need help with proving the following $$\lim_{x\to\infty}\frac{\left({\log}_2x\right)^3}{x^n}=0\ , \quad \forall n>0.$$

I used wolframalpha and got it

and I had kind of an intuition because of the limit $\frac{\ln(x)}x$ , and still I have no idea how to formaly proof it.

Will appreciate any help!

答:
Using the hint of @Kavi Rama Murthy, note that $$f(x)\to 0 \implies f^{3}(x)\to 0$$
In your problem, you can define that $f(x):=\frac{\log_{2}(x)}{x^{n/3}}$ so, by L'Hospital's Rule you can find that $$\frac{\log_{2}(x)}{x^{n/3}} \to 0 \quad \text{as} \quad x \to \infty$$
So, by the hint and since that $f^{3}(x)=\frac{\log_{2}^{3}(x)}{x^{n}}$ so $$\frac{\log_{2}^{3}(x)}{x^{n}} \to 0 \quad \text{as} \quad x\to \infty$$
this's true for all $n>0$.


问:
Tilings of the plane and quotient spaces
A square lattice has two discrete translation symmetries, isormorphic to $\mathbb{Z}^2$. If we take the quotient of the plane by this group, $\mathbb{R}^2/\mathbb{Z}^2$, we get the square with opposite sides identified, which is the torus.

Can a similar argument be developed for a regular hexagonal lattice? I mean, the square has two independent translational symmetries; so does the hexagon; what is the difference, topologically? To put it differently, the hexagon with opposite sides identifies can be seen as the plane "quotiented" by some equivalence relation. But can it be seen as the quotient of two groups? Or at least two spaces?

Same as above but for equilateral triangle lattice.

Also, the symmetry group of the square lattice is bigger than just translations, it has rotations and reflections. Can we consider a different quotient in which in the numerator $\mathbb{R}^2$ is replaced by the whole 2D Euclidian group, and in the denominator $\mathbb{Z}^2$ is replaced by the whole symmmetry group of the square lattice? What do we get?

Where can I read an accessible introduction to these themes?

答:
the hexagon with opposite sides identifies can be seen as the plane "quotiented" by some equivalence relation. But can it be seen as the quotient of two groups?

Yes. Let $H$ be a subgroup of $\Bbb R^2$ consisting of elements $x$ such that a translation of $\Bbb R^2$ by $x$ keeps a given regular hexagonal lattice. The group $H$ is generated by three translations by one cell, whose sum is the identity map. That is, in fact, $H$ is generated by any two of these translations and $H$ is isomorphic to $\Bbb Z^2$. A quotient topological group $\Bbb R^2/H$ is homeomorphic to a hexagonal cell of the lattice with the opposite sides identified, that is a torus again.

Can we consider a different quotient in which in the numerator $\Bbb R^2$ is replaced by the whole 2D Euclidian group, and in the denominator $\Bbb Z^2$ is replaced by the whole symmmetry group of the square lattice? What do we get?

I suppose that by the whole 2D Euclidian group you mean the group $G$ of isometries of $\Bbb R^2$. Let $H$ be the group of the elements of $G$ preserving the lattice $\Bbb Z^2$. For our purpose it is convenient to consider elements of $G$ in terms of complex numbers. In the description in terms of $a$ and $\omega$, the elements of $H$ are exactly those with  $a\in\Bbb Z^2$ and $\omega\in \{1,-1,i,-i\}$.  Since $H$ contains a map $\bar z$, each coset $fH$ of $G/H$ intersects both $G_+$ and  $G\setminus G_+$, where $G_+$ consists of maps $z\mapsto a+\omega z$. So we shall study the quotient set $G_+/H_+$, where $H_+=G_+\cap H$.

The group $G_+$ can be identified with the set of pairs $(a,\omega)$, where $a\in\Bbb C$, $\omega\in\Bbb T$, and $\Bbb T=\{z\in\Bbb C:|z|=1\}$. This identification imposes on $G_+$ a topology of a subset of $\Bbb C^2$. Remark that the identification imposes on $\Bbb C\times \Bbb T$ an operation $(a,\omega) (a’,\omega’)=(a’+\omega’ a, \omega\omega’)$ for any $a,a’\in\Bbb C$ and $\omega,\omega’\in\Bbb T$. Taking representatives, we see that $G_+/H_+$ is homeomorphic to a cube $[0,1]^3$ with some identifications of its boundary points. Namely, for each $x,y,z\in [0,1]$, $(0,y,z)\sim (1,y,z)$, $(x,0,z)\sim (x,1, z)$, and $(x,y,0)\sim (y,x,1)$. This is a torus, whose surface we inflated a bit, and then we put it into a high-dimensional space and glue its outer surface with the inner, swapping parallels with meridians.


问:
How do I approximate a covariance in $\langle \mathbb{R}^3, SO(3) \rangle$ as a covariance in $SE(3)$?
I am working with two different code bases. One represents the state of the system as $\langle \mathbb{R}^3, SO(3) \rangle$ while the other (correctly) represents the state of the system as $SE(3)$. I need to use information from the first source in the second source. This requires me to convert the state and its covariance from the $\langle \mathbb{R}^3, SO(3) \rangle$ to an $SE(3)$ representation. Converting the state itself seems simple; it's the covariance that is more difficult.

Pretty much everything I know about Lie groups comes from "A micro Lie theory
for state estimation in robotics". Combining the information from Example 7 of this paper with my knowledge of covariance matrices, here are my thoughts.

For a simple vector space, consider a random vector $\boldsymbol{x}$ with mean $\bar{\boldsymbol{x}}$.
The covariance of $\mathbf{x}$ is given by
\begin{align}
\mathbf{P}_{\mathbf{x}\mathbf{x}} &= \mathbb{E} \left[ \left( \mathbf{x} - \bar{\mathbf{x}} \right) \left( \mathbf{x} - \bar{\mathbf{x}} \right)^\text{T} \right] \label{eq:Pxxdefinition} .
\end{align}

By analogy with the above equation, the $SE(3)$ covariance is
\begin{align}
\mathbf{P} &= \mathbb{E} \left[ \left( \mathbf{T} \ominus \bar{\mathbf{T}} \right) \left( \mathbf{T} \ominus \bar{\mathbf{T}} \right)^\text{T} \right] \label{eq:PSe3def} .
\end{align}

And the $\langle \mathbb{R}^3, SO(3) \rangle$ covariance is
\begin{align}
\mathbf{Q} &= \mathbb{E} \left[ \left( \mathbf{T} \boxminus \bar{\mathbf{T}} \right) \left( \mathbf{T} \boxminus \bar{\mathbf{T}} \right)^\text{T} \right] \label{eq:PSO3xR3def} .
\end{align}

with the $\ominus$ and $\boxminus$ subtraction operators defined in Example 7 of the above paper (except that my $\boxminus$ here should be a \diamondminus).

Now I've attempted to expand each of these expressions to find a valid approximation. Starting with $SE(3)$, we can expand $\ominus$ operator such that
\begin{align}
\mathbf{T}_2 \ominus \mathbf{T}_1 &= \begin{bmatrix} \mathbf{V} ( \text{Log} (\mathbf{R}_1^\text{T} \mathbf{R}_2 ) )^{-1} \mathbf{R}^\text{T}_1 \left( \mathbf{p}_2 - \mathbf{p}_1 \right) \\
 \text{Log}  (\mathbf{R}^\text{T}_1 \mathbf{R}_2) \end{bmatrix}
 = \begin{bmatrix} \Delta \boldsymbol{\rho} \\ \Delta \boldsymbol{\theta} \end{bmatrix} \label{eq:SE3minusoperator}.
\end{align}

The $\langle \mathbb{R}^3, SO(3) \rangle$ minus operator can similarly be expanded as
\begin{align}
\mathbf{T}_2 \boxminus \mathbf{T}_1 &= \begin{bmatrix} \left( \mathbf{p}_2 - \mathbf{p}_1 \right) \\
 \text{Log} ( \mathbf{R}^\text{T}_1 \mathbf{R}_2 ) \end{bmatrix} =
 \begin{bmatrix} \Delta \mathbf{p} \\ \Delta \boldsymbol{\theta} \end{bmatrix} =
 \begin{bmatrix} \mathbf{R}_1 \mathbf{V} ( \Delta \boldsymbol{\theta} ) \Delta \boldsymbol{\rho} \\ \Delta \boldsymbol{\theta} \end{bmatrix} \label{eq:SO3xR3minusoperator}.
\end{align}

So while
$$
\mathbf{P} = \mathbb{E} \begin{bmatrix} \Delta \boldsymbol{\rho} \Delta \boldsymbol{\rho}^\text{T} & \Delta \boldsymbol{\rho} \Delta \boldsymbol{\theta}^\text{T} \\ \Delta \boldsymbol{\theta} \Delta \boldsymbol{\rho}^\text{T} & \Delta \boldsymbol{\theta} \Delta \boldsymbol{\theta}^\text{T} \end{bmatrix},
$$

$$
\begin{align}
\mathbf{Q} &= \mathbb{E} \begin{bmatrix} \Delta \mathbf{p} \Delta \mathbf{p}^\text{T} & \Delta \mathbf{p} \Delta \boldsymbol{\theta}^\text{T} \\ \Delta \boldsymbol{\theta} \Delta \mathbf{p}^\text{T} & \Delta \boldsymbol{\theta} \Delta \boldsymbol{\theta}^\text{T} \end{bmatrix} \\
&= \mathbb{E} \begin{bmatrix} \mathbf{R}_1 \mathbf{V} ( \Delta \boldsymbol{\theta} ) \Delta \boldsymbol{\rho} \left( \mathbf{R}_1 \mathbf{V} ( \Delta \boldsymbol{\theta} ) \Delta \boldsymbol{\rho} \right)^\text{T} & \mathbf{R}_1 \mathbf{V} ( \Delta \boldsymbol{\theta} ) \Delta \boldsymbol{\rho} \Delta \boldsymbol{\theta}^\text{T} \\ \Delta \boldsymbol{\theta} \left( \mathbf{R}_1 \mathbf{V} ( \Delta \boldsymbol{\theta} ) \Delta \boldsymbol{\rho} \right)^\text{T} & \Delta \boldsymbol{\theta} \Delta \boldsymbol{\theta}^\text{T} \end{bmatrix} \\
\end{align}
$$

I'm not sure how to simplify this to get something more useful. Ideally I'd like to find some function such that
$$
\mathbf{P} \approx convert(\mathbf{Q})
$$

Edit:
The above math suggests that an exact conversion can be given by
$$
\mathbf{P} = \mathbf{H} \mathbf{Q} \mathbf{H}^\text{T}
$$
where
$$
\mathbf{H} = \begin{bmatrix} \mathbf{V}^{-1} \mathbf{R}_1^\text{T} & \mathbf{0}_{3\times3} \\ \mathbf{0}_{3\times3} & \mathbf{I}_{3\times3} \end{bmatrix}
$$

however, I am leery of this answer because I don't think there should exist an exact conversion. I know that an $SE(3)$ covariance matrix can represent uncertainty in a banana-shaped distribution while $\langle \mathbb{R}^3, SO(3) \rangle$ cannot represent such distributions. See "The Banana Distribution is Gaussian:
A Localization Study with Exponential Coordinates" for more details.

答:
The fundamental problem in your analysis is that the $\Delta \rho$ in the $SE(3)$ case and the $\Delta \rho$ in the $(SO(3)\times\mathbb{R}^3)$ case are not the same thing.

I think it's easiest to consider how perturbing the different components of each "error" space would change the associated group.

$SE(3)$:  If I perturb the translation part of the lie algebra what happens?  Remember that the translation part is "curved" by the rotation in the screw motion.  This is why we have the $V$ term in the exponential.  We are taking that linear velocity-like term and curling it by recursively multiplying in the rotation (think about the definition of $\exp(A)$).   Therefore, if I "bump" the $\mathbf{t}_x$-component of the lie algebra, then the change in the resulting translation is going to be dependent on the rotation.  And the first-order approximation will have a dependency on $V$  to predict the final translation change. It can be helpful for me to think of the algebra of $SE(3)$ as the 6-DOF body-frame velocities that would yield some transform after 1 second of integration.

$\left(SO(3)\times\mathbb{R}^3\right)$: There is no curving motion.  If I bump the x-component of the "algebra" then it will exactly correspond to a change in the translation, no dependence on the rotation.  This would be like pinning the linear velocities in an inertial frame and moving only along that direction while also rotating.
If I want to convert from $\left(\mathfrak{so}(3)\times\mathbb{R}^3\right)$ to $\mathfrak{se}(3)$ (note the algebra) it makes sense to me that there is a conversion happening in the translation part.  They describe different things.  "How different" you ask? Well, in how much curving motion is being captured by the object.  This is encapsulated in $V$.

So, then how do I convert a covariance represented in $\mathfrak{se}(3)$ to $\left(\mathfrak{so}(3)\times\mathbb{R}^3\right)$?

It's going to depend on a couple of factors:

The "sided-ness" of the covariances.  (left or right side)

If there are also any coordinate frame transformations from one parameterization to the other.

Without knowing those items, I can't say for sure what it is, but there is definitely some (likely non-linear) function
$$
y = f\left(x\right)\\
x\in \mathfrak{se}(3),\;y\in\left(\mathfrak{so}(3)\times\mathbb{R}^3\right)
$$
that converts from one parameterization to the other.  Given this
and the respective covariance matrices
$$
\Sigma_x=E\left[xx^\top\right],
$$
and
$$
\Sigma_y=E\left[yy^\top\right].
$$
we can derive $\Sigma_y$ with a Taylor Series approximation.
$$
\begin{aligned}
\Sigma_y&=E\left[yy^\top\right] \\
        &=E\left[f(x)f(x)^\top\right] \\
        &\approx E\left[\frac{\partial f}{\partial x}x\left(\frac{\partial f}{\partial x}x\right)^\top\right] & (\mathrm{taylor\ series})\\
        &= \frac{\partial f}{\partial x} E\left[xx^\top\right]\frac{\partial f}{\partial x}^\top\\
        &= \frac{\partial f}{\partial x} \Sigma_x \frac{\partial f}{\partial x}^\top
\end{aligned}
$$

You may find these tables helpful when working out the jacobian $\frac{\partial f}{\partial x}$

So what do you lose?  The $AQA^\top$ is going to end up converting the "banana" into an ellipse.  There is no way around this, since covariance matrices can't represent anything other than ellipsoids.  The only reason that a covariance over $\mathfrak{se}(3)$ can be projected into a banana is because we are really capturing uncertainty that then gets "curled".  It's actually an ellipsoid on the manifold.  When we map it into euclidean coordinates then we see that it looks like a banana.

Where will this break down?
We need to think about how this is actually a linearization.  It's going to have problems wherever the magnitude of the rotation uncertainty is large relative to the magnitude of translation uncertainty.  Basically trying to fit an ellipsoid to a banana gets harder the more like an arc the banana is.  Luckily, you're dealing with uncertainties which should always be small.


问:
If $X^Y = Y^X$, prove that $X = Y$

$X$ and $Y$ are two sets. $X^Y$ denotes the set of all functions from $Y$ to $X$. Similarly, $Y^X$ denotes the set of all functions from $X$ to $Y$. If $X^Y = Y^X$, prove that $X = Y$.

I tried to show this by way of contradiction, but my argument seems to be lacking rigor. To begin, assume that $X\neq Y$. Then there exists $y\in Y$ such that $y \notin X$. If we think about $X^Y$, then this set contains functions $f:Y\to X$, which assign a unique image $f(y)$ to $y\in Y$. As $y\notin X$, we can obviously not find such functions in $Y^X$. So $X^Y \neq Y^X$, which is a contradiction. Hence, $X=Y$.

Are there better ways of proving this? Is my argument complete? How can I make it more rigorous?

答:
A function maps every element in the domain to some element in the co-domain.

So, pick your favorite $y \in Y$. Let $f:X\to Y$ be $f(x) = y$. But that means $f: Y \to X$ defined as $f(x) = y$ also exists. Since this is the same $f$, the domains must be the same, so $X = Y$.

Title: If $X^Y = Y^X$, prove that $X = Y$


答: If $X = Y = \emptyset$ then this is trivial. Otherwise, let $f\in X^Y = Y^X$ and let $x \in X$. Then $f(y) = x$ for some $y \in Y$, but also $f(y) = f(x_{1})$ for some $x_{1} \in X$ so that $x\in Y$. The other way is proven similarly.

Title: If $X^Y = Y^X$, prove that $X = Y$

Body:
$X$ and $Y$ are two sets. $X^Y$ denotes the set of all functions from $Y$ to $X$. Similarly, $Y^X$ denotes the set of all functions from $X$ to $Y$. If $X^Y = Y^X$, prove that $X = Y$.

I tried to show this by way of contradiction, but my argument seems to be lacking rigor. To begin, assume that $X\neq Y$. Then there exists $y\in Y$ such that $y \notin X$. If we think about $X^Y$, then this set contains functions $f:Y\to X$, which assign a unique image $f(y)$ to $y\in Y$. As $y\notin X$, we can obviously not find such functions in $Y^X$. So $X^Y \neq Y^X$, which is a contradiction. Hence, $X=Y$.

Are there better ways of proving this? Is my argument complete? How can I make it more rigorous?


答:
HINT

$$\lim_{x\rightarrow\infty}\frac{\left({\log}_2{x}\right)^3}{x^n}=\lim_{x\rightarrow\infty}\frac{t\left(\ln{x}\right)^3}{x^n}$$
where $t=\ln^32$
$$\lim_{x\rightarrow\infty}\frac{t\left(1-\frac1x\right)^3}{x^n}\le\lim_{x\rightarrow\infty}\frac{t\left(\ln x\right)^3}{x^n}\le\lim_{x\rightarrow\infty}\frac{t\left(x-1\right)^3}{x^n}$$

$$\to0\le\lim_{x\rightarrow\infty}\frac{t\left(\ln x\right)^3}{x^n}\le\to0$$

Now the limit is obvious from squeeze theorem.

For the inequality, refer this

Here's a convincing graph:

Title: How to prove that $\lim_{x\to\infty}\frac{(\log_2 x)^3}{x^n}=0$

Body:
I need help with proving the following $$\lim_{x\to\infty}\frac{\left({\log}_2x\right)^3}{x^n}=0\ , \quad \forall n>0.$$

I used wolframalpha and got it

and I had kind of an intuition because of the limit $\frac{\ln(x)}x$ , and still I have no idea how to formaly proof it.

Will appreciate any help!


问:
Is it wrong to use Binary Vector data in Cosine Similarity?
I am doing Information Retrieval using Cosine Similarity.

My data is a binary vector.

Since most of the references I read were using non-binary vector (non-binary matrix) data, I am wondering if it is wrong to use binary vector data in the cosine similarity function.

答:
In a similar senario with binary vectors (recommender systems in my case which can be see as a special case of information retrieval) I had very good success with cosine similarity.

When measuring Rankscore I got much better rate on my retrievals when I used Cosine similarity than with the Tanimoto coefficient, but this might be very application specific. You should probably experiment with different similarity measurements and see what yields the best result in you application.

I compare the result of the effects from different similarity measures between my binary vectors here. As mentioned above I have the greatest success with cosine similarity.


问:
functional equation in the set of positive integers.

Let $k$ be a fixed positive integer. Find all functions $f :\mathbb{N}\to\mathbb{N}$ such that for any distinct positive integers $a_1,a_2,\ldots,a_k$, there exists a permutation $(b_1,b_2,\ldots,b_k)$ of $\{a_1,a_2,\ldots,a_k\}$ such that
$f(a_1)/b_1+ f(a_2)/b_2+\ldots+ f(a_k)/b_k$
is a positive integer.

Please help, I'm really stuck here, I guess $f$ must be constant or linear, I tried induction on $k$, but the permutation mess everything, the case $k=1$ is trivial, idk how to deal with $k>1$.

答:
Lemma: If $p$ is a prime number, $A, B, M \in \mathbb{N}$, $p \nmid B$, $M$ and $p$ are coprime, then the sum $\frac{A}{M} + \frac{B}{p}$ cannot be a positive integer.

Proof of the lemma is straightforward: $\frac{A}{M} + \frac{B}{p}=N \Rightarrow pA + MB=NMp \Rightarrow p | MB$, a contradiction.

Now we prove $\forall n \in \mathbb N$, we must have $n|f(n)$. We construct a sequence $a_1, a_2, \ldots, a_k$ in the following fashion: $a_1=n$, and $\forall 2\leqslant j \leqslant k$,

$$
a_j > \max\{a_m, f(a_m) , \forall 1\leqslant m<j\}, \text{ and } a_j \text{ is a prime number}.
$$

Then there's a permutation $\{ b_j, 1\leqslant j \leqslant k \}$ of $\{ a_j, 1\leqslant j \leqslant k \}$ such that $S \equiv \sum_{j=1}^k f(a_j)/b_j$ is a positive integer. Assume $b_i = a_k$ for some index $i$, we have

$$
S = \frac{A}{\prod_{j=1}^{k-1} a_j} + \frac{f(a_i)}{a_k}.
$$

From the lemma, since $\gcd(\prod_{j=1}^{k-1} a_j, a_k)=1$, we know $a_k | f(a_i)$. By construction $a_k > f(a_j), \forall j<k$, therefore $i=k, a_k | f(a_k)$.

It follows that $\sum_{j=1}^{k-1} f(a_j)/b_j$ is also a positive integer. By the same reasoning we have $b_{k-1} = a_{k-1}, a_{k-1} | f(a_{k-1})$. We repeat this process until finally we have $f(n)/n$ as a positive integer.

On the other hand if $n|f(n), \forall n\in \mathbb{N}$, the identity permutation guarantees that $\sum f(a_j)/a_j$ is a positive integer. $\blacksquare$


问:
Where is the empty set in a venn diagram?
"Show that the empty set is a subset of every set."  This is the problem I'm working on. It seems standard, but the nuance is that if two sets are disjoint, and the empty set is a subset of both, how do we draw the empty set in a Venn Diagram?

I searched and found this article: venn diagram of power set and empty set, but the person's response is that the empty set is in the intersection of all the other sets... my point is that those sets could be disjoint, so there would be no intersection.  In that case, how do we draw it?

I recognize that the answer might just be that venn diagrams break down and don't illustrate the empty set.  If so, are there other visualization techniques that are general enough? I know enough category theory to briefly think about it from that perspective.

Thanks in advance

答:
You can draw a dot (circle with area zero) anywhere in the Venn Diagram and that can sort of serve as an intuition for the empty set. It contains nothing and consequently, has no area.

In fact, you can draw many dots spread out all over the Venn Diagram like you spilled glitter on your diagram. It doesn't matter how many you draw or where you draw them, as long as they are area zero.

You'll see that the empty set is, indeed, a subset of every set $A$, because the statement: "For every element $e$ in the empty set, $e \in A$" is vacuously true.


问:
Suppose we have 44 books on Math, 55 books on English and 66 books on History, In how many ways you can put them on your bookshelf?
Suppose we have $4$ books on Math, $5$ books on English and $6$ books on History. In how many ways you can put them on your bookshelf if you want :-

$1)$ The first book is a math book.

$2)$ All math books are at the beginning.

$3)$ Math and English books will stay together.

$4)$ The first book and the last book are both Math books.

(I'm completely lost with this problem. If someone can give me an explanation that would be great.)

I know that we have $3$ subjects so there are $3$! Or $6$ possibilities and for $4$ math books we have $4!$ or $24$ ways to order. When it is $5$ English books we have $5!$ or $120$ ways to order. If it is $6$ history books we have $6!$ Or $720$ ways to order. So I think the answer is $6 \times 24 \times 120 \times 720=12,441,600$ ways to order the books .

答:
These are $4$ separate homework problems, right? In total we have $15$ books.

$1)$ $4 * 14!$ (First choose any math book, then for each choice, we can order the remaining 14 however we want)

$2)$ $4! * 11!$ (First, order the math books 4! number of ways, then for each ordering, we can order the remaining 11 however we want)

$3)$ Unclear/left as an exercise for the reader.

$4)$ $4 * 3 * 13!$ (First, choose the first book to be a math book, then choose the last book to be a math book, then order the remaining 13 in the middle however we want)


问:
Hamiltonians, PDE, Density of a Fluid.
I'm struggling with Hamiltonians and PDE. My question is loosely based on the PDE of this paper by Figalli, Gangbo and Yolcu.

Before asking a question, let me spell out some prior knowledge, coming from the Wikipedia page on the Hamiltonian:

In Hamiltonian mechanics a system is described by canonical coordinates $(q,p)$: for simplicity lets think of $q$ as position in $\mathbb{R}^d$ and $p$ the momentum (of a particle).

The Hamiltonian of the system captures the change in time of these coordinates. Hamilton's equations are
\begin{align}
\frac{dp}{dt}=&-\frac{\partial H}{\partial q}
\\
\frac{dq}{dt}=&\frac{\partial H}{\partial p}
\end{align}
If $H=\frac{p^2}{2m}+V(q)$ where ($m$=mass and $V$ is some energy potential) then Hamilton's equations say that the time derivative of position is velocity (i.e Newton's 2nd Law), and the change in momentum is equal to the negative gradient of the potential energy (i.e. the system tries to minimize energy).

Now in the paper by Figalli et al referenced above, they study the following  PDE
$$
\partial_t \rho_t(x)=-\text{div}\Big(\rho_t(x)\nabla_p H\big(x,-\rho_t^{-1}(x)\nabla P(\rho_t(x))  \big)\Big),
$$
where $\rho_t$ is interpreted as the density of a fluid at time $t$, and  they call $P$ its pressure.

Moreover, the Hamiltonian is customarily chosen as depending only on its second argument, i.e.
$$
H(x,p)=H(p)=\frac{1}{2}\|p\|^2,
$$
(note in this case if we pick $P(a)=a$ then we get the heat equation $\partial_t \rho_t(x)=\Delta\rho_t(x)$).

Question:

My confusion is why they call $H$ the Hamiltonian: how does it relate to the Hamiltonian I described above? Generally how to write a PDE for a density?

Moreover, since $\rho_t$ only acts on $x$, why does the Hamiltonian depend on the second coordinate?

答:
The Euler equations for an inviscid fluid are an example of an Hamiltonian system if all the forces involved are conservative and the following relationship exists between pressure $P$ and fluid density $\rho_t$ (for smooth, real-valued functions $\rho_t\in C^k(\mathbb{R}\times\mathbb{R}^3)$ and $P$ defined on $\mathbb{R}^3$ and with $d$ the exterior derivative)
\begin{equation}
\begin{aligned}
d \rho_t\wedge dP &= 0 \\
\iff(\partial_y\rho\partial_z P - \partial_z \rho\partial_y P) dydz +(\partial_z\rho\partial_x P - \partial_x \rho\partial_z P)dzdx + (\partial_x\rho\partial_y P - \partial_y\rho\partial_x P)dxdy&=0 \\
\iff (\nabla\rho_t \times \nabla P)\cdot dS&=0,
\end{aligned}
\end{equation}
where in the second line,  $\rho_t:= \rho$ and $\partial/\partial x := \partial_x$ (similarly for $y$ and $z$), while $dS:=(dydz,dzdx,dxdy)$ is an infinitesimal surface element dotted into the cross product. Euler fluids satisfy a conservation equation
$$\partial_t\rho_t + \mathrm{div}(\rho_tV_t)=0, $$
where $V_t$ is identified as the velocity field of the fluid; in many cases, this is a smooth, non-singular vector field on some simply connected region $\Omega\subset\mathbb{R}^3$.

In the paper you cite, found below equation (1.1), they explicitly define the product of fluid density and the velocity field as the argument you have inside the divergence. Assuming $d\rho_t\wedge dP=0$ (an isothermal-like condition), the Euler equations $$\underbrace{\partial_t V_t+(V_t\cdot\nabla)V_t}_{:=D_t V_t} - F = \frac{1}{\rho_t}\nabla P,$$ where $F$ is a another smooth, conservative vector field on the same region, can then be written as $$D_t V_t = -\nabla (U+\alpha),$$
since there exists a smooth map we call potential $U:\Omega\rightarrow \mathbb{R}$ satisfying $F=-\nabla U$. Since we also require $d\rho_t\wedge dP=0$, we may take $\alpha$ to be understood as the Volterra-type integral $$\alpha:=\int_{(0,0)}^{(t,x)}\frac{dP}{\rho_t}\Rightarrow d\alpha = \frac{dP}{\rho_t}.$$

The Hamiltonian is then defined as the sum of kinetic and potential energies $$ H = \frac{1}{2}(V_t\cdot V_t) + (U+\alpha),$$ where only the first term has velocity-dependence and only the second term $(U+\alpha)$ has positional dependence (explicitly, $V_t\in\ker\partial_q$ and $(U+\alpha)\in\ker\partial_p$). That is, we consider $q\in \Omega$ as the position vector at which the velocity field $V_t$ is defined for time $t$ so that component-wise $\partial_t q = V_t = \partial H/\partial V_t$. For the sake of notation, write the velocity field $V_t := p$ so that we can also write Hamilton's equations $$\partial_t p = -\dfrac{\partial H}{\partial q}\hspace{0.5cm}\text{and}\hspace{0.5cm}\partial_t q = \dfrac{\partial H}{\partial p}.$$

The first of the equations explains why one may replace the velocity field $V_t$ with the gradient ($\nabla_p$) of a Hamiltonian. As far as why the second coordinate of the Hamiltonian is of the form $-\rho_t^{-1}\nabla[P(\rho_t)]$, one may identify this by raising the 1-form $$d\alpha=\frac{dP}{\rho_t}=\rho_t^{-1}(\partial_x P dx + \partial_y Pdy + \partial_z Pdz)$$ to $\rho_t^{-1}\nabla P$.

Title: Hamiltonians, PDE, Density of a Fluid.


问:
Proving $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$
This question is the last part of a problem leading to proof of Stirling's approximation. I've already proved that $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists and that $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$.

Hence, the question asks to assume $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists, and then asks to use $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$ to show $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$.

My attempt goes like this:

Since $\sqrt{x}$ is continuous, we can use $\sqrt{\underset{n\to \infty }{\text{lim}}f(n)}=\underset{n\to \
\infty }{\text{lim}}\sqrt{f(n)}$ to get $\underset{n\to \infty }{\text{lim}}\frac{2^{2 n} (n!)^2}{(2 n)! \
\sqrt{2 n+1}}=\sqrt{\frac{\pi }{2}}$.

Then we can eliminate $2^{2 n}n!$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{\sqrt{1+2 n} \left(n-\frac{1}{2}\right) \left(n-\frac{3}{2}\right) \cdots  \
\frac{3}{2}\frac{1}{2}}$$

Then factor out $n^n$ and adjust $\sqrt{1+2n}$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{n^{n+\frac{1}{2}} \sqrt{2+\frac{1}{n}} \left(1-\frac{1}{2n}\right) \left(1-\frac{3}{2n}\right) \cdots  \
\frac{3}{2n}\frac{1}{2n}}$$

The $\sqrt{2+\frac{1}{n}}$ factor will give a $\sqrt{2}$, so the remaining $\prod _k^n \left(1-\frac{2 k-1}{2 n}\right)$ must somehow relate to $\sqrt{2} e^{-n}$.

However, I'm not sure how to do this.

答:
I didn't use the given limit, but Stirling's approximation can solve this in one shot. You can rewrite the given limit as:
$$L = \lim_{n \to \infty}\frac{n!}{\left( \frac ne\right)^n \sqrt n}$$
Then, by stirling's approximation $n! = \sqrt{2\pi n}\ \left( \frac ne \right)^n + O(\frac{1}{n})$, the limit becomes
$$L = \frac{\sqrt{2\pi} \cdot n!}{n!} = \sqrt{2\pi}$$


问:
$(n + i)(n + i + 2)$ is divisible by at most $N$ primes
Let, $\Pi_{n,m}= n(n+1) \cdots (n+m-1)$ and $P (m)$ denotes
greatest prime divisor of a positive integer $m$. For fixed $m$ and $t$, we will in fact consider the problem of classifying those positive
integers $n$ for which

$(2.1) \; \; P(\Pi_{n,m)} ≤ p_t$

where $p_t$ is the $t^{th}$ prime.

we will argue as follows. For
any given $m ≥ 3$, we will split $4 [m/4] $(or $ m − 1,$ if $m ≡ −1 \pmod 4$) of the indices
$0, · · · , m − 1$ into disjoint pairs of the form $(i,  i + 2)$, where $i ≡ 0, 1 \pmod 4$. There
are

$[\frac m4]+ [\frac {m+1}{4}]$ such pairs. Choose $t_0$ such that

$π(m − 1) ≤ t_0 ≤ t.$

If $m ≥ 3$, we can thus find an $i \in  [0, m − 3]$ such that the product $(n + i)(n + i + 2)$
is divisible by at most

$(2.2)\; \;  N= [\frac{t-t_0}{[\frac m4]+ [\frac {m+1}{4}]}]$

of the primes in

$(2.3) \; \;  \{ p_{t_0+1}, · · · ,p_t\}$.

The above excerpt is taken from "PRIME FACTORS OF CONSECUTIVE INTEGERS by MARK BAUER AND MICHAEL A. BENNETT".

QUESTION:
I am having trouble to understand $(2.2)$.

How the product $(n + i)(n + i + 2)$ is divisible by at most $ N=[ \frac{t-t_0}{[\frac m4]+ [\frac {m+1}{4}]}]$? Please write down the detail derivation. Thanks.

答:
Let us define $S$ as
$$S:=\bigg\{(n+i)(n+i+2)\ |\ \text{$i$ is such that $i\in[0,m-3]$ and $i\equiv 0,1\pmod 4$}\bigg\}$$

Then, we have
$$|S|=\bigg[\frac m4\bigg]+\bigg[\frac {m+1}{4}\bigg]$$

We want to prove the following claim :

Claim 1 : If $m ≥ 3$, we can find an $i \in  [0, m − 3]$ such that the product $(n + i)(n + i + 2)$ is divisible by at most $N=\bigg[\frac{t-t_0}{|S|}\bigg]$ of the primes in $\{ p_{t_0+1}, · · · ,p_t\}$.

In order to prove Claim 1, it is sufficient to prove the following claim :

Claim 2 : If $m ≥ 3$, we can find an $i$ satisfying $i \in  [0, m − 3]$ and $i\equiv 0,1\pmod 4$ such that the product $(n + i)(n + i + 2)$ is divisible by at most $N=\bigg[\frac{t-t_0}{|S|}\bigg]$ of the primes in $\{ p_{t_0+1}, · · · ,p_t\}$.

So, in the following, let us prove Claim 2.

Proof of Claim 2 :

Assume to the contrary that for every $i$ satisfying $i \in  [0, m − 3]$ and $i\equiv 0,1\pmod 4$, the product $(n + i)(n + i + 2)$ is divisible by at least $\bigg(\bigg[\frac{t-t_0}{|S|}\bigg]+1\bigg)$ of the primes in $\{ p_{t_0+1}, · · · ,p_t\}$.

Now, let us use the following lemma (a proof of the lemma is written at the end of this answer) :

Lemma : For every prime $P$ in $\{ p_{t_0+1}, · · · ,p_t\}$, the number of the elements of $S$ divisible by $P$ is at most $1$.

Now, consider the product of all the elements of $S$.

It follows from lemma 1 that the product of all the elements of $S$ is divisible at least by$$|S|\bigg(\bigg[\frac{t-t_0}{|S|}\bigg]+1\bigg)\tag1$$
of the primes in $\{ p_{t_0+1}, · · · ,p_t\}$.

Here, $(1)$ is bigger than $t-t_0$ since, using $[x]\gt x-1$, we have
$$|S|\bigg(\bigg[\frac{t-t_0}{|S|}\bigg]+1\bigg)\gt |S|\bigg(\frac{t-t_0}{|S|}\bigg)=t-t_0$$

This implies that there have to exist more than $t-t_0$ primes in $\{ p_{t_0+1}, · · · ,p_t\}$.

This contradicts that there are $t-t_0$ primes in $\{ p_{t_0+1}, · · · ,p_t\}$.$\quad\blacksquare$

Finally, let us prove the lemma :

Lemma : For every prime $P$ in $\{ p_{t_0+1}, · · · ,p_t\}$, the number of the elements of $S$ divisible by $P$ is at most $1$.

Proof of Lemma : Assume to the contrary that there are two non-negative integers $i,j$ such that $i,j\equiv 0,1\pmod 4$ and $0\le i\lt j\le m-3$ and $0\equiv (n+i)(n+i+2)\equiv (n+j)(n+j+2)\pmod P$.

If $n+i\equiv n+j\equiv 0\pmod P$, then we get $j-i\equiv 0\pmod P$ which is impossible since $1\le j-i\le m-3\lt m-1\lt P$.

If $n+i\equiv n+j+2\equiv 0\pmod P$, then we get $j+2-i\equiv 0\pmod P$ which is impossible since $3\le j+2-i\le m-1\lt P$.

If $n+i+2\equiv n+j+2\equiv 0\pmod P$, then we get $j-i\equiv 0\pmod P$ which is impossible since $1\le j-i\le m-3\lt m-1\lt P$.

If $n+i+2\equiv n+j\equiv 0\pmod P$, then we get $j-i-2\equiv 0\pmod P$. Since $-1\le j-i-2\lt m-1\lt P$, we have to have $j=i+2$. Since $i\equiv 0,1\pmod 4$, we have $j\equiv 2,3\pmod 4$ which contradicts that $j\equiv 0,1\pmod 4$.$\quad\blacksquare$


问:
How to solve this linear recurrence relation
how to solve following recurrence relation :

$f(n) = 3 * f(n - 1) + 4$

i've got that recurrence relation from following sequence, where f(n) is nth value of the following sequence.

$7, 25, 79, 241, 727, 2185$, and so on.

So $f(0) = 7$, $f(1) = 25$. etc.

答:
The hint from Neat Math suggests $f(n)=3f(n-1)+4\iff f(n)+2=3(f(n-1)+2)$

or $g(n)=3g(n-1)$ where $g(n)=f(n)+2$, so $g(n)=3^{n}g(0)$, with $g(0)=9$,

so $g(n)=3^{n+2}$, so $f(n)=3^{n+2}-2$.

A more pedantic solution would be the following:

$f(n)-3f(n-1)=f(n-1)-3f(n-2)$, so $f(n)=4f(n-1)-3f(n-2)$.

The roots of the characteristic equation $r^2=4r-3$ are $r=1,3$,

so the solution is $f(n)=A\cdot3^n+B\cdot1^n$.

Solve for $A$ and $B$ given $f(0)=7$ and $f(1)=25$.


问:
How would I determine the spanning set for the kernel of a map that sends each polynomial to its value at 0?
I am given a linear transformation, T, such that T: P --> R (where P is the vector space of polynomials and R is all real numbers), and that p|-->p(0).

I know that the kernel of a transformation, T, is the set of all u in V such that T(u) = 0

So I'm looking for the spanning set of all p for which T(p)= p(0) = 0, but I'm not sure what 0 (the zero vector) looks like in this example, and I also don't know how to notate a spanning set with a polynomial.

I think of p(0) as the polynomial given by a(0)^0 + b(0) + c(0)^2 + ... + z(0)^n, but other than that, I feel pretty lost with finding a spanning set of the kernel of T.

Any guidance would be greatly appreciated!

答:
The zero element of this vector space is the zero polynomial. A spanning set for the kernel  is $\{x,x^{2},x^{3},...\}$.

[The kernel consists of all polynomials with no constant term].


Title: prove: $\dfrac{2^{n+1}+(-1)^n}{3}$

Body:
I am asked to prove this notation with induction for $n\in \mathbb{N}$:

real problem is to fill the area with tilings. and for $n\in \mathbb{N}$ there are exactly so many chances to fill the area as the formula says:  $$\dfrac{2^{n+1}+(-1)^n}{3}$$

I started in this way:

Base Case: $n=1$:

$$\dfrac{ 2^2-1}{3} = 1$$

Inductive step: $n\rightarrow n+1$:

$$\dfrac{2^{n+2}+(-1)^{n+1}}{3} = help..$$

my problem is: what am i doing, i seem to have already proven it in this last step which is obviously wrong. what am i missing here in this proof?

thanks for help

Body:
Looks like you would need the n - 1 case.

$2^{n+1} + (-1)^n = (2^{n+1} + (-1)^n)(2 + -1) = 2^{n+2} + 2(-1)^n + -2^{n+1} + (-1)^{n+1} = (2^{n+2} + (-1)^{n+1}) + 2(2^n + (-1)^{n-1}(-1)^2)$


问:
If all action profiles in a payoff matrix are Pareto optimal does this automatically mean that the game is a constant-sum?
My guess is no, because constant sum means that the payoff is constant in every action profile, but I'm not quite sure.</p>

答:
Having all the action profiles in a payoff matrix Pareto optimal does not imply that the game is constant sum. A counter example is given by

\begin{array}{|c|c|c|}
\hline
& L & M & R \\
\hline
H & (3, 4) & (4, 3) & (5, 2) \\ \hline
T & (6, 0) & (2, 5) & (1, 6) \\ \hline
\end{array}

As we can see they are all Pareto optimal. This game is not constant sum though because of $(T, L)$.


答:
Here is the solution for the $3$rd part .

Assume the $4$ Math and $5$ English books to be in a line, as in each arrangement they have to be consecutively placed. Take this to be a block of $9$ books.

In this block the number of ways to arrange the $4$ Math and $5$ English books would be :- $$\frac{9!}{4!*5!} = 126$$

Now this whole block can be placed with $6$ History books in this number of ways :-
$$\frac{7!}{6! * 1!} = 7$$

Add both of these up to get total number of ways in which Math and English books stay together :- $126 + 7 = 133$ .

Title: Suppose we have 44 books on Math, 55 books on English and 66 books on History, In how many ways you can put them on your bookshelf?

Body:
Suppose we have $4$ books on Math, $5$ books on English and $6$ books on History. In how many ways you can put them on your bookshelf if you want :-

$1)$ The first book is a math book.

$2)$ All math books are at the beginning.

$3)$ Math and English books will stay together.

$4)$ The first book and the last book are both Math books.

(I'm completely lost with this problem. If someone can give me an explanation that would be great.)

I know that we have $3$ subjects so there are $3$! Or $6$ possibilities and for $4$ math books we have $4!$ or $24$ ways to order. When it is $5$ English books we have $5!$ or $120$ ways to order. If it is $6$ history books we have $6!$ Or $720$ ways to order. So I think the answer is $6 \times 24 \times 120 \times 720=12,441,600$ ways to order the books .


答:
Might this relate to the fact that we can take that product out as a limit and notice that
${\underset{n\to \infty }{\text{lim}} (1-\frac 1{n})^n}=e^{-1}$. Then if we can separate the limits and the product will be $e^{-n}$ Then maybe there is some relation to the multiples of two in the denominator and the $\sqrt {2}$? I am not sure if there are some binomial tricks (considering we can make $n$ as small as we like) to bound this changing numerator $2k+1$ when $n \to \infty$.

Title: Proving $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$

Body:
This question is the last part of a problem leading to proof of Stirling's approximation. I've already proved that $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists and that $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$.

Hence, the question asks to assume $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists, and then asks to use $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$ to show $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$.

My attempt goes like this:

Since $\sqrt{x}$ is continuous, we can use $\sqrt{\underset{n\to \infty }{\text{lim}}f(n)}=\underset{n\to \
\infty }{\text{lim}}\sqrt{f(n)}$ to get $\underset{n\to \infty }{\text{lim}}\frac{2^{2 n} (n!)^2}{(2 n)! \
\sqrt{2 n+1}}=\sqrt{\frac{\pi }{2}}$.

Then we can eliminate $2^{2 n}n!$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{\sqrt{1+2 n} \left(n-\frac{1}{2}\right) \left(n-\frac{3}{2}\right) \cdots  \
\frac{3}{2}\frac{1}{2}}$$

Then factor out $n^n$ and adjust $\sqrt{1+2n}$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{n^{n+\frac{1}{2}} \sqrt{2+\frac{1}{n}} \left(1-\frac{1}{2n}\right) \left(1-\frac{3}{2n}\right) \cdots  \
\frac{3}{2n}\frac{1}{2n}}$$

The $\sqrt{2+\frac{1}{n}}$ factor will give a $\sqrt{2}$, so the remaining $\prod _k^n \left(1-\frac{2 k-1}{2 n}\right)$ must somehow relate to $\sqrt{2} e^{-n}$.

However, I'm not sure how to do this.


问:
What to study after Miranda's "Algebraic curves and Riemann surfaces"?
I am supervising a small reading group on Riemann surfaces. We are following Rick Miranda's book "Algebraic curves and Riemann surfaces". We will probably be done at the end of the year, and we would like to continue the seminar. What would be the next best thing to study ?

The students are undergrad, so they know topology, algebra, complex analysis and multivariable calculus. We will also, roughly, be familiar with most of the book. (One of the students really wants to study sheaf theory, so something with some sheaf theory would be nice.) They don't know algebraic geometry (other than what is in Miranda).

I have some ideas of course, in particular "chapter on algebraic surfaces" by Miles Reid, and "Hodge theory and complex geometry I" by Claire Voisin. But that might be too hard just after Miranda, so I am interested in other propositions. If possible, avoid suggestions like reading Hartshorne (it's a lot of heavy machinery, and for example most of the applications of chapter 4 can be obtained by elementary methods over $\mathbb C$, like in Miranda's book.)

答:
Huybrechts' book titled Complex Geometry and Griffiths Harris 'Principles of algebraic geometry" chapters 0 and 1

Source:

What are some alternatives to Griffiths Harris 'Principles of algebraic geometry" chapters 0 and 1?


答:
Elaborating on Daniel Schepler's comment: if $L = \lim_{n \to \infty} \frac{n!}{n^{n+1/2} e^{-n}}$ then
$$\sqrt{\pi/2}
= \lim_{n\to \infty} \frac{2^{2n} (n!)^2}{(2n)!\sqrt{2n+1}}
= \lim_{n \to \infty} \frac{2^{2n}n^{2n+1} e^{-2n} L^2}{(2n)^{2n+1/2} e^{-2n} L \sqrt{2n+1}}
= L\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{2}\sqrt{2n+1}}
= \frac{L}{2}.$$

For what it's worth, I tried to continue your approach by approximating $\log \left(\prod_{k=1}^n (1 - \frac{2k-1}{2n})\right)$ with the integral $n \int_0^{1-1/(2n)} \log(1-x) \, dx$ but it was terribly messy and didn't seem tight enough to get the exact equivalence with $\sqrt{2} e^{-n}$. But I think the above approach is the intended approach.


问:
The differential equation of the change of the amount of the nutrition inside of the incubator.
The problem statement is as below.

We will handle the incubator with the microbes inside of it.

The microbe increases consuming a nutrition.

The amount of the nutrition is never increased.

$10^3*P(t):=$ The number of the microbes inside of the incubator at time $t$[h].

$f(t):=$ The amount of the nutrition inside of the incubator.

$P(t):=P_0*$exp$(k*t);$

$k:=0.1386;$

$P_0:=$The return value of $P(t)$ as time $t=0$;

$c\Delta t:=$The amount of the nutrition which is consumed at micro time $\Delta t[h]$ by the $50*10^3$ microbes.

Represent the differential equation of the change of the value of $f(t)$ as parameter $t$ changes with using $c$  ,$P(t)$ .

答:
Here is how I look at it -

Number of microbes at time $t = 10^3 P(t)$

$f(t)$ is the amount of the nutrition at time $t$.

$c \Delta t$ is the amount of nutrition consumed in time $\Delta t$ by $50 \times 10^3$ microbes.

So amount of nutrition consumed in $\Delta t$ at time $t = \displaystyle \frac{c \Delta t}{50 \times 10^3} \times 10^3 \times P(t)$

$\Delta f(t) = \displaystyle \frac{c}{50} P(t) \Delta t$

lead to $\displaystyle \frac{df(t)}{dt} = \frac{c}{50} P(t) \, \, $ (in terms of $c, P(t)$)

If you have to find in terms of $P_0$ replace $P(t)$ with its formula in terms of $P_0$ and $t$.


问:
Prove that the norm on the space of real-valued Lipschitz functions is a norm
Let $\mathcal{L}$ be the space of real-valued Lipschitz functions of order 1 defined on $[0,1]$. That is, the class of functions $f$ such that
$$
\sup\limits_{(x,y)\in[0,1]\times[0,1], x\neq y}
\dfrac{\vert f(x)-f(y)\vert}{\vert x-y\vert}
=K(f)<\infty.
$$
Let $\Vert f\Vert_1=\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +K(f)=\Vert f\Vert + K(f)$.
Show that $\Vert\cdot\Vert_1$ is a norm on $\mathcal{L}$.

To show $\Vert\cdot\Vert_1$ is a norm on $\mathcal{L}$, I use 3 axioms of definition of norm.

Take $f,g\in \mathcal{L}$ and $a\in \mathbb{R}$.

(1)
\begin{align*}
\Vert f+g\Vert_1=
\sup\limits_{0\leq t\leq 1}\vert f(t)+g(t)\vert +K(f+g)
&\leq \sup\limits_{0\leq t\leq 1}\vert f(t)\vert +\sup\limits_{0\leq t\leq 1}\vert g(t)\vert +K(f)+K(g)\\
&= \left(\sup\limits_{0\leq t\leq 1}\vert f(t)\vert + K(f)\right) + \left(\sup\limits_{0\leq t\leq 1}\vert g(t)\vert + K(g)\right)\\
&=\Vert f\Vert_1+\Vert g\Vert_1
\end{align*}

(2)
\begin{align*}
\Vert af\Vert_1=
\sup\limits_{0\leq t\leq 1}\vert af(t)\vert +K(af)
&= a\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +aK(f)\\
&\leq \vert a\vert \sup\limits_{0\leq t\leq 1}\vert f(t)\vert +\vert a\vert K(f)\\
&= \vert a\vert \left(\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +K(f)\right)\\
&= \vert a\vert \Vert f\Vert_1\\
\end{align*}

(3)
If
\begin{align*}
\Vert f\Vert_1=
\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +K(f) = 0
\end{align*}
then we will prove $f=0$.

\begin{align*}
\sup\limits_{0\leq t\leq 1}\vert f(t)\vert  &= -K(f)\\
&= -\sup\limits_{(x,y)\in[0,1]\times[0,1], x\neq y}
\dfrac{\vert f(x)-f(y)\vert}{\vert x-y\vert}.
\end{align*}

In this case, I don't know how to get $f=0$.

My question

Does my proof in (1) and (2) correct? In (3), how to get $f=0$?

答: Your proof of 1) and 2) are correct. For 3) you are over-thinking. Sum of two non-negative numbers cannot be $0$ unless they are both zero. Hence $\|f\|_1=0$ implies $\sup|f(t)|=0$ so $f(t)=0$ for all $t$.


问:
Likelihood function of $n$ for $X\sim \text{Bin}(n,0.5)$
I have to find the likelihood function for $X\sim \text{Bin}(n,0.5)$ where we know that $x=14$ and so $n \ge 14$. Several trials were conducted counting successes (14 of which there were) but the number of trials was forgotten, so we need to estimate $n$.

I know that the pmf of a Binomial random variable is $$P(X=x)=\binom{n}{x}\cdot p^x \cdot(1-p)^{n-x}$$ and in this case we have $$P(X=14)=\binom{n}{14}\cdot 0.5^{14} \cdot(1-0.5)^{n-14}=\binom{n}{14}\cdot0.5^{n}\,,$$ however I'm not sure this would be useful since for $n \ge 14$, the function seems to increase at an exponential rate that doesn't appear to have a maximum.

So I'm not sure using the pmf is appropriate - does anyone have any other ideas? Any help would be greatly appreciated.

答:
I have to find the likelihood function ...(omissis)... Several trials were conducted counting successes (14 of which there were) but the number of trials was forgotten, so we need to estimate n.

No, if you read well your problem you have only to find a way to estimate the $n$ trials with 14 known successes among them

First Idea

using the hint of the text, I would approximate

$$B\Big(n;\frac{1}{2}\Big)\approx N\Big(\frac{n}{2}; \frac{n}{4} \Big)$$

Thus I would maximize

$$f(n)\propto n^{-\frac{1}{2}}e^{-\frac{1}{2n}(2x-n)^2  }$$

Finding

$$\hat{n}=\Bigg\lceil\frac{-1+\sqrt{1+16 x^2}}{2}\Bigg\rceil_{x=14}=28$$

Second idea,  Method of Moments

Using a Negative Binomial, in average, how many trials are  neeeded to get 14 successes?

$$\hat{n}=\mathbb{E}[N]=\frac{14}{\frac{1}{2}}=28$$

As you can see, the result is 28 in both cases


问:
Prove the following Lemma in the polynomial rings.
Let $R$ be a ring. Then, the natural inclusion
$R \to R[x]$
which just sends an element $r \in R$ to the constant polynomial $r$, is a ring homomorphism.

Attempts:
Let $r \in R$ and define $g : R \to R[x]$ as $g(r) = f(x)$ where $f(x) = r$ for all $x \in R$.

Then,
\begin{align*}
g(r_1 + r_2) &= r_1 + r_2 \\
&= g(r_1) + g(r_2)
\end{align*}
and
\begin{align*}
g(r_1 r_2) &= r_1 r_2 \\
&= g(r_1) g(r_2)
\end{align*}

Hence, proved.

Is above true?

答:
Your proof looks fine. Depending on your definition of group homomorphism, you may also want to verify that $g(0)=0$ and $g(1)=1$.

A nice general fact is that if $R$ is a subring of a ring $S$, then the inclusion $R\ \hookrightarrow\ S$ is a group homomorphism.


问:
Approach the value through derivative
We have the function $y=50-x^2$.
We are asked to calculate approximately the change that $y$ will undergo when $x$ increases from $2$ to $4$. We are mainly asked to use the expression dy=y'(x)*dx.

What I did was to calculate $y'(x) =-2x$ thus $y'(2)=-4$ and then:
$dy=y'(2)(4-2)=-4\cdot(4-2)=-8$

So the answer is that through the differential of the function we found that finally $y$ decreases by $8$, when $x$ increases from $2$ to $4$. Is my way of thinking correct?

I would greatly appreciate your guidance.

Thank you very much in advance.

答: Your way of thinking is correct. But the approximation becomes better (in this case it is even exact) if you use $y'$ for the midpoint of the change rather than the starting point. Which is to say, $y'(3)$ instead of $y'(2)$.

Title: Approach the value through derivative


问:
Irreducible polynomials over the field $GF(2)$ corresponds to prime numbers. Is this a known theorem?
Conjecture: Consider the field $GF(2) = {(0, 1)}$. An irreducible polynomial over this field corresponds to a prime number.

For example: $x^4 + x^0$ is irreducible and corresponds to $2^4 + 2^0 = 16 + 1 = 17$, which is prime.

I'll take a prime number $11 = 2^3 + 2^1 + 2^0$. Converting to binary: 1011. The corresponding polynomial over the field $GF(2)$ is $x^3 + x^1 + x^0$. This polynomial is irreducible.

I suspect there is a theorem for this that could be more general. If there is a theorem for this, what is it called?

Please note: I am not a Mathematics Major.

答:
The conjecture is unfortunately false; we can see that
$5 = 2^{2}+1$ is prime, but the corresponding polynomial $(x^2 + 1) = (x+1)(x+1)$ over $\mathbb{F}_{2}$ so is not irreducible; conversely, $x^{4}+x^{3}+1$ is irreducible over $\mathbb{F}_{2}$, but $2^{4}+2^{3}+1 = 16+8+1=25$ is not prime.

I think you have likely found a coincidence for some small numbers, since the numbers corresponding to an irreducible polynomial will always be odd, and most of the odd numbers less than 32 (so corresponding to polynomials of degree 4 or less) are prime. I would not expect there to be anything interesting to discover here.

In fact, if you consider the irreducible polynomials of degree 8 or less, roughly half (36 of 71) correspond to primes and half correspond to composites. Polynomials of degree 15 or less, about 1/4 correspond to primes (1156 of 4720). Once we consider polynomials of degree 20 or less, we are down to about 16% corresponding to primes (17772 out of 111013). I would guess that as the degree of the polynomial gets large, the percentage of irreducible polynomials corresponding to prime numbers approaches 0.

(On the other side of this, there are 82025 primes less than $2^{20}$ only about 22% of which will give an irreducible polynomial.)


问:
prove: $\dfrac{2^{n+1}+(-1)^n}{3}$
I am asked to prove this notation with induction for $n\in \mathbb{N}$:

real problem is to fill the area with tilings. and for $n\in \mathbb{N}$ there are exactly so many chances to fill the area as the formula says:  $$\dfrac{2^{n+1}+(-1)^n}{3}$$

I started in this way:

Base Case: $n=1$:

$$\dfrac{ 2^2-1}{3} = 1$$

Inductive step: $n\rightarrow n+1$:

$$\dfrac{2^{n+2}+(-1)^{n+1}}{3} = help..$$

my problem is: what am i doing, i seem to have already proven it in this last step which is obviously wrong. what am i missing here in this proof?

thanks for help

Title: prove: $\dfrac{2^{n+1}+(-1)^n}{3}$

Body:
$$\dfrac{2^{n+2}+(-1)^{n+1}}{3}$$
$$\dfrac{2^{n+1}\cdot2+(-1)^{n}\cdot(-1)}{3}$$
$$\dfrac{2^{n+1}\cdot2-(-1)^{n}}{3}$$
$$\dfrac{2^{n+1}\cdot(3-1)-(-1)^{n}}{3}$$
$$\dfrac{2^{n+1}\cdot(3)-2^{n+1}-(-1)^{n}}{3}$$
$$\dfrac{2^{n+1}\cdot(3)-(2^{n+1}+(-1)^{n})}{3}$$
$$\dfrac{2^{n+1}\cdot(3)}{3}-\dfrac{(2^{n+1}+(-1)^{n})}{3}$$

Proven.


问:
General formula for $\int_a^b(x-a)^m(x-b)^ndx$
Recently, I stumbled upon this:

$$\int_a^b (x-a)^3(x-b)^4dx$$

I did this in many ways - using IBP multiple times and by even expanding both terms like a madman.

The result was: $$\frac{(a-b)^8}{280}$$
I was astonished by this cute little result. So I started wondering about evaluating the following:

$$\int_a^b(x-a)^m(x-b)^ndx$$

I wanted to use reduction to obtain a result similar to Wallis' formula for sine and cosine

Any other method is always welcome

Body:
Start using IBP:

$$I_{(m,n)}=\int_a^b(x-a)^m(x-b)^ndx=0-\int_a^b m(x-a)^{m-1}\frac{(x-b)^{n+1}}{n+1}dx=-\frac{m}{n+1}I_{(m-1,n+1)}$$

Applying IBP multiple times, we obtain $$I_{(m,n)}=(-1)^{m+1}\frac{m!.n!}{(n+m)!}I_{(0,n+m)}$$

$$\boxed{I_{(m,n)}=(-1)^{m+1}\frac{m!.n!}{(n+m+1)!}(a-b)^{n+m+1}}$$


问:
How to deduce the expression of Sz from the raising and lowering operators?
I know that we can define the $S_x$ and $S_y$ operators as follows:
$$
S_x=\frac{1}{2}(S^++S^-),\\
S_y=\frac{1}{2i}(S^+-S^-).
$$
The question is, how to deduce the expression of $S_z$ from the raising $S^+$ and lowering $S^-$ operators?

答:
Write $$S^+=S_x+iS_y\\S^-=S_x-iS_y$$
Then $$S^+S^-=S_x^2+iS_yS_x-iS_xS_y+S_y^2=(S_x^2+S_y^2)-i(S_xS_y-S_yS_x)$$
Using $$S^2=S_x^2+S_y^2+S_z^2$$and $$[S_x,S_y]=i\hbar S_z$$
you get $$S^+S^-=S^2-S_z^2+\hbar S_z$$
Similarly $$S^-S^+=S^2-S_z^2-\hbar S_z$$
Subtracting the two, you get the commutator for $S^+$ and $S^-$:
$$[S^+,S^-]=2\hbar S_z$$


问:
Find the distribution of X and the conditional distribution of Y given X. Are these distributions known? Which?
I have this problem, I am supposed to obtain the distribution of X, for this I know that I must integrate the joint density, but there are many parameters that are not familiar to me, until now by definition I have to:

$F_X(x)=\int_0^x \frac{\lambda} {\Gamma (\alpha)}(\frac{\lambda}{x})^\alpha y^{\alpha-1}e^{-\lambda (u+\frac{y}{u})} du $

But I don't know how to continue.

The exercise says:

Given $\alpha > 0$ and $\lambda> 0$ let $(X, Y)$ be an absolutely continuous random vector with joint density:

$$f_{XY}(x,y) = \frac{\lambda} {\Gamma (\alpha)}(\frac{\lambda}{x})^\alpha y^{\alpha-1}e^{-\lambda (x+\frac{y}{x})}$$
$$0<x,y<\infty$$

a) Find the distribution of $X$ and the conditional distribution of $Y$ given $X$. Are these distributions.

b) Prove that $\frac{y}{x}$ has distribution ${\Gamma (\alpha,\lambda)}$ and is independent of $X$ without appealing to the theorem of change of variables.
known? Which?

I appreciate your ideas.

答: It looks like your integral limits are wrong. You have to integrate with respect to $y$, not $u$. For every $x,y$ ranges from $0\to\infty$ so the integral is$$\begin{align*}F_X(x)&=\int_0^\color{red}\infty \frac{\lambda} {\Gamma (\alpha)}\left(\frac{\lambda}{x}\right)^\alpha y^{\alpha-1}e^{-\lambda (x+y/x)} dy\\&=\frac{\lambda e^{-\lambda x}}{\Gamma(\alpha)}\int_0^\infty\left(\frac {\lambda y}x\right)^{\alpha-1}e^{-\lambda (y/x)} d(\lambda y/x)\end{align*}$$since $x$ is constant when you integrate with respect to $y$. Now keep $m=\lambda y/x$, then $m$ ranges from $0\to\infty$ since $x,\lambda>0$. The integral becomes$$\begin{align*}F_X(x)&=\frac{\lambda e^{-\lambda x}}{\Gamma(\alpha)}\int_0^\infty m^{\alpha-1}e^{-m} dm\\&=\lambda e^{-\lambda x}\end{align*}$$which is the PDF of exponential distribution.


问:
Are there any Combinatoric proofs of Bertrand's postulate?
I feel like there must exist a combinatoric proof of a theorem like: There is a prime between $n$ and $2n$, or $p$ and $p^2$ or anything similar to this stronger than there is a prime between $p$ and $(\prod_p p) + 1$ (Euclid's theorem).

I was trying to prove one by the Sieve on this grid

```<code>    1     2     3     4 ... p
  p+1   p+2   p+3   p+4 ... 2p
 2p+1  2p+2  2p+3  2p+4 ... 3p
...........................
...........................
........................... p^2
p^2+1 p^2+2 p^2+3 p^2+4 ...
</code>```

but it did not work.

Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes?

答:
Here is an elementary argument which shows that there are "lots" of primes. Consider the number of numbers in the interval $[1, n]$. On the one hand, this is $n$. On the other hand, by unique factorization every such number can be uniquely written in the form $\prod_{i=1}^{\pi(n)} p_i^{e_i}$. It follows that $n$ is equal to the number of non-negative integer solutions to

$$e_1 \log p_1 + ... + e_{\pi(n)} \log p_{\pi(n)} \le \log n.$$

This is easily seen to be at most $\left( \log_2 n \right)^{\pi(n)}$, hence

$$n \le (\log_2 n)^{\pi(n)}$$

from which it follows that

$$\pi(n) \ge \frac{\log n}{\log \log_2 n}.$$

It might be possible to extract a weak form of Bertrand's postulate from this argument.


问:
Existence of a self-homeomorphism $\phi$ of a connected Hausdorff manifold satisfying $\phi(x)=y$
Suppose $M$ is a connected Hausdorff manifold, then show that for any $x,y \in M$ there exist a homeomorphism $\phi:M \to M$ such that $\phi(x)=y$

Using the definitions, I concluded that there exists open neighborhood $U,V$ of $x,y$ respectively and open sets $R_1 \subset \Bbb R^m$ and $R_2 \subset \Bbb R^m$ such that $U$ and $R_1$ , $V$ and $R_2$ are homeomorphic.

But I am unable to proceed from here.

答:
Lemma: Let $x,y\in \text{int}\left(\Bbb D^n\right):=\{z\in\Bbb R^n:||z||< 1\}$.
Then, there is a homeomorphism $\varphi:\Bbb D^n\to \Bbb D^n$ such
that $\varphi(w)=w$ for $||w||=1$ and $\varphi(x)=y$.

Consider the homeomorphism $\psi:\text{int}(\Bbb D^n)\to \Bbb R^n$ given by $$\psi(z)=\frac{z}{1-||z||}.$$ Let $T:\Bbb R^n\to \Bbb R^n$ be the translation given by $T(z)=z-\psi(x)+\psi(y)$. Then, consider $\psi^{-1}\circ T\circ \psi:\text{int}(\Bbb D^n)\to \text{int}(\Bbb D^n)$ and extend it to whole $\Bbb D^n:=\{z\in\Bbb R^n:||z||\leq 1\}$.

Why such an extension is possible? Note that for $||z||<1$, write $z=rv$ for some $v\in\Bbb S^{n-1}$ and some $r\in[0,1)$. Then, $\psi(z)=\frac{r}{1-r}v$ and for any $R\in [0,\infty)$ and any $w\in \Bbb S^{n-1}$ we have $\psi^{-1}(Rw)=\frac{R}{1+R}w$. So, for any $r\in [0,1)$ and any $v\in \Bbb S^{n-1}$ we have
$$\psi^{-1}\circ T\circ \psi(rv)$$$$=\frac{\left|\left|\frac{r}{1-r}v-\psi(x)+\psi(y)\right|\right|}{1+\left|\left|\frac{r}{1-r}v-\psi(x)+\psi(y)\right|\right|}\cdot\frac{\frac{r}{1-r}v-\psi(x)+\psi(y)}{\left|\left|\frac{r}{1-r}v-\psi(x)+\psi(y)\right|\right|}$$$$=\frac{\left|\left|rv-(1-r)\psi(x)+(1-r)\psi(y)\right|\right|}{(1-r)+\left|\left|rv-(1-r)\psi(x)+(1-r)\psi(y)\right|\right|}\cdot\frac{rv-(1-r)\psi(x)+(1-r)\psi(y)}{\left|\left|rv-(1-r)\psi(x)+(1-r)\psi(y)\right|\right|}.$$

Every connected topological manifold is homogeneous i.e. for a connected manifold $M$ and any two points $a,b\in  M$ there is a homeomorphism
$\Phi:M\to M$ such that $\Phi(a)=b$.

To prove this, consider the non-empty set $$S:=\{x\in M|\text{ there is a homeomorphism }f:M\to M\text{ with }f(x)=b\}.$$ Now, consider $y\in S$ with a homeomorphism $g:M\to M$ such that $g(y)=b$. Let $\psi:U\big(\subseteq_{\text{closed}}M\big)\to \Bbb D^n$ be a homeomorphism with $y\in \text{int}(U)$. Now, for any $x\in\text{int}(U)$, choose $\varphi:\Bbb D^n\to\Bbb D^n$ such that $\varphi\left(\psi(x)\right)=\psi(y)$ and $\varphi(w)=w$ for $||w||=1$. So, define a homeomorphism $f:M\to M$ as $$f(z)=\begin{cases}g\big(\psi^{-1}\circ \varphi\circ \psi(z)\big) &\text{ if }z\in\text{int}(U),\\g(z)&\text{ if }z\in M\backslash \text{int}(U). \end{cases}$$ In other words, $\text{int}(U)\subseteq S$. That is $S$ is open in $M$. Similarly, prove that $M\backslash S$ is open. Now, $M$ is connected implies the result.

Some extra fact: Every connected manifold $M$ is $2$-homogeneous, i.e. given
$\{a_1,a_2\}\cup\{b_1,b_2\}\subseteq M$, we have a homeomorphism
$\psi:M\to  M$ such that $\psi(a_k)=b_k$ for each $k=1,2$.

Let $$T:=\big\{(x_1,x_2)\in M\times M\big|\text{there is a homeomorphism }$$$$f:M\to M\text{ with }f(x_1)=b_1,f(x_2)=b_2\big\}.$$ Now, consider $(y_1,y_2)\in T$ with a homeomorphism $g:M\to M$ such that $g(y_1)=b,g(y_2)=b_2$. Let $\psi_k:U_k\big(\subseteq_{\text{closed}}M\big)\to \Bbb D^n$ be a homeomorphism with $y_k\in \text{int}(U_k)$ for $k=1,2$ with $U_1\cap U_2=\emptyset$. Now, for any $x_k\in\text{int}(U_k)$, choose $\varphi_k:\Bbb D^n\to\Bbb D^n$ such that $\varphi_k\left(\psi_k(x_k)\right)=\psi_k(y_k)$ and $\varphi_k(w)=w$ for $||w||=1$ where $k=1,2$. So, define a homeomorphism $f:M\to M$ as $$f(z)=\begin{cases}g\big(\psi_1^{-1}\circ \varphi_1\circ \psi_1(z)\big) &\text{ if }z\in\text{int}(U_1),\\g\big(\psi_2^{-1}\circ \varphi_2\circ \psi_2(z)\big) &\text{ if }z\in\text{int}(U_2),\\g(z)&\text{ if }z\in M\backslash\big( \text{int}(U_1)\cup \text{int}(U_2)\big). \end{cases}$$ In other words, $\text{int}(U_1)\times \text{int}(U_2)\subseteq T$. That is $T$ is open in $M\times M$. Similarly, $\big(M\times M\big)\backslash T$ is open. Now, $M\times M$ is connected implies the result.

Similarly, one can show, every connected topological manifold is $k$-homogeneous for every
$k\in\Bbb N$.


答:
Let $u=(x-a)/(b-a)$ so that
\begin{align}
\int_a^b(x-a)^m(x-b)^ndx=(-1)^m(b-a)^{m+n+1}\int_0^1u^m\left(1-u\right)^ndx
\end{align}
and latter integral is a well-known quantity.

Title: General formula for $\int_a^b(x-a)^m(x-b)^ndx$

Body:
Recently, I stumbled upon this:

$$\int_a^b (x-a)^3(x-b)^4dx$$

I did this in many ways - using IBP multiple times and by even expanding both terms like a madman.

The result was: $$\frac{(a-b)^8}{280}$$
I was astonished by this cute little result. So I started wondering about evaluating the following:

$$\int_a^b(x-a)^m(x-b)^ndx$$

I wanted to use reduction to obtain a result similar to Wallis' formula for sine and cosine

Any other method is always welcome


问:
Hamilton quaternion ring over $\mathbb{Z}$ is a domain
assuming that $a,b\in H_\mathbb{Z}$ such that $a=a_0+a_1i+a_2j+a_3k$ and $b=b_0+b_1i+b_2j+b_3k$, $a_i ,b_i \in \mathbb{Z}$
I wanna prove that $H_\mathbb{Z}$ is a domain.

my attempt: I first wanted to use the definition of domain and prove that $H_\mathbb{Z}$ is a domain. so if $ab=0$ then I have to show that $a=0$ or $b=0$. if $a=0$ then it is proved. so assuming $a$ is nonzero how do I show that $b$ should be zero?

PS:I want to prove this directly using the definition of domain.
thanks in advance!

答:
Consider the quaternion $2$. It cannot be inverted inside $H_{\Bbb Z}$, because $$2(a_1+a_2i+a_3j+a_4k)$$ must have all coefficients even, and thus it cannot equal $1$.

Edit: $H_{\Bbb Z}$ has no zero divisors.

Take two quaternions, and multiply them:
$$
(a_0 + a_1i + a_2j + a_3k)(b_0 + b_1i+b_2j+b_3k)\\
\begin{align}= &a_0b_0 - a_1b_1-a_2b_2-a_3b_3\\
& +i(a_0b_1 + a_1b_0 + a_2b_3 - a_3b_2)\\
& +j(a_0b_2 + a_2b_0 + a_3b_1 - a_1b_3)\\
& +k(a_0b_3+a_3b_0 + a_2b_1 - a_1b_2)
\end{align}
$$
We want to prove that if this is zero, one of the original quaternions must have been zero. This product is zero iff all the four coefficients above are zero. And they are all zero iff the sum of their squares is zero. So we calculate the sum of their squares.
$$
\begin{align}&(a_0b_0 - a_1b_1-a_2b_2-a_3b_3)^2\\
& +(a_0b_1 + a_1b_0 + a_2b_3 - a_3b_2)^2\\
& +(a_0b_2 + a_2b_0 + a_3b_1 - a_1b_3)^2\\
& +(a_0b_3+a_3b_0 + a_2b_3 - a_3b_2)^2
\end{align}
$$
After a bit of checking, we see that all terms of the form $a_i^2b_j^2$ survive, with one term for each possibility for $i$ and $j$. And we see that all the other terms cancel (for instance, $-2a_0b_0a_1b_1$ from the first square cancels against $2a_0b_1a_1b_0$ from the second square).

So what we are left with is that the sum of squares above reduces to
$$
a_0^2b_0^2 + a_0^2b_1^2 + a_0^2b_2^2 + a_0^2b_3^2\\
+a_1^2b_0^2 + a_1^2b_1^2 + a_1^2b_2^2 + a_1^2b_3^2\\
+a_2^2b_0^2 + a_2^2b_1^2 + a_2^2b_2^2 + a_2^2b_3^2\\
+a_3^2b_0^2 + a_3^2b_1^2 + a_3^2b_2^2 + a_3^2b_3^2
$$
But this is equal to $(a_0^2+a_1^2+a_2^2+a_3^2)(b_0^2+b_1^2+b_2^2+b_3^2)$, which can only be $0$ if one of the original two quaternions is equal to $0$.

Thus, if a product of (integral) quaternions is zero, one of the factors must be zero.


问:
Prove by induction $n^{n}>2^{n}\times n!$
Prove by induction $n^{n}>2^{n}\times n!$ for all $n\geq6$.

This is what I've got at the moment. Still not sure whether is the right way to solve it by induction:

Basic Step: Show that $S(1)$ is true for  $n=6$:
$$6^{6}>2^{6}\text{·}6!$$
$$46656>46080$$

...

Induction step: Show that $S\left(n+1\right)$ is true, that is:
$$(n+1)^{(n+1)}>2^{n+1}\times(n+1)!$$
Adding $(n+1)$ to both sides of $n^{n}>2^{n}\times n!$ we have
$$\left(n+1\right)n^{n}>2^{n}\times\left(n+1\right)n!$$

$$n^{n+1}+n^{n}>2^{n}\times\left(n+1\right)!$$

答:
The inequality you would like to prove is rather

$$n^n < \color{blue}{3}^n\cdot n!$$

A proof by induction can be done using the fact that

$$\left(1+\frac 1n\right)^n < 3 \text{ for all } n \in \mathbb{N}$$

The induction step could then look as follows:

Under the assumption that $n^n < 3^n\cdot n!$ is true, to show is
$$(n+1)^{n+1} < 3^{n+1}\cdot (n+1)!$$
or equivalently
$$(n+1)^n < 3\cdot 3^n\cdot n!$$

But this is true because

$$(n+1)^n = \frac{(n+1)^n}{n^n}\cdot n^n$$ $$ = \left(1+\frac 1n\right)^n\cdot n^n < 3\cdot n^n \stackrel{n^n < 3^n\cdot n!}{<} 3\cdot 3^n\cdot n!$$


问:
Are these two binary operations commutative? $x*y = (x+y)/(xy)$ or $x\&y=(x+1)/(y+1)$?

Are these two binary operations commutative?$$x*y = \frac{x+y}{xy}\\x\&y=\frac{x+1}{y+1}$$

Can you help me to identify if both of these are communtative or not?

Thank you very much in advance.

答:
First binary operation is commutative.

\begin{align}
x*y&=\dfrac{x+y}{xy}\\
&=\dfrac{y+x}{yx}\\
&= y*x
\end{align}

Second binary operation is not commutative.

\begin{align}
x\&y&=\dfrac{x+1}{y+1}\\
&\neq\dfrac{y+1}{x+1}\text{, except } (y+1)^2=(x+1)^2\\
&= y\&x.
\end{align}


问:
How to prove continuity of a map whose image is a clopen set
I read that a map $f:X \rightarrow Y$ is continuous, if for any open set $O \subseteq Y$, the preimage $f^{-1}(O)$ is open in $X$. Alternatively, if $O$ is closed in $Y$, the preimage is closed in $X$.

However, suppose that $Y$ is equipped with the discrete topology where all subsets of $Y$ are clopen. Is there an instance where a continuous function maps an open set in $Y$ to an open set in $X$, but does not do so for closed sets ?

答:
If $f:X\to Y$ is continuous, then $f^{-1}[U]$ is open in $X$ for every open $U\subseteq Y$, and $f^{-1}[F]$ is closed in $X$ for every closed $F\subseteq Y$. It doesn’t matter what the topologies are: all that matters is that $f$ is continuous.

If $Y$ has the discrete topology, however, every subset of $Y$ is clopen, so $f^{-1}[A]$ is clopen in $X$ for every $A\subseteq Y$.


问:
tensor product and matrix multiplication distributive properties
I am trying to find partial trace of some matrix of the form

$M = (A \otimes B)\times (A^{T*} \otimes B^{T*})$

in which $\otimes$ is tensor product, $\times$ is matrix multiplication, $T*$ is conjugate transpose of the matrix. $A$ and $B$ are vectors (but can be matrices too).

I was wondering if I could use the following relation?

$(A \otimes B)\times (A^{T*} \otimes B^{T*}) = (A \times A^{T*})\otimes (B \times B^{T*})$

It checks out for $A=\left(\begin{matrix} a\\b \end{matrix}\right)$ and $B=\left(\begin{matrix} c\\d \end{matrix}\right)$.

if this relation is general, what about this one (for an arbitrary vectors (or matrices) C and D)

$(A \otimes B)\times (C \otimes D) = (A \times C)\otimes (B \times D)$

although I don't know how to trace this matrix. I would appreciate any help. thank you in advance

答:
Consider matrices $A,B,C,D$ of sizes such that the products $AC$ and $AD$ can be formed. We can use block matrix multiplication to show that $(A\otimes B)\,(C\otimes D)=(AC)\otimes(BD)$.

We will use the notation $A\otimes B = (a_{ij} B)_{ij}$ to denote block matrices, where indices are always supposed to range approriately. Then
\begin{align*}
(A\otimes B)\,(C\otimes D) &= (a_{ij} B)_{ij}\, (c_{ij} D)_{ij} \\
&= \left(\sum_k (a_{ik} B)(c_{kj} D)\right)_{ij} \\
&= \left( \left(\sum_k a_{ik} c_{kj}\right) BD\right)_{ij.}
\end{align*}
Note that $\sum_k a_{ik} c_{kj}$ is the $i,j$-th entry of $AC$ so the result is indeed equal to $(AC)\otimes (BD)$.

Since traces of Kronecker products are given as $\operatorname{Tr}(A\otimes B)=\operatorname{Tr}(A) \operatorname{Tr}(B)$, this yields
$$
\operatorname{Tr}\left((A\otimes B)\,(C\otimes D)\right)
= \operatorname{Tr}(AC) \operatorname{Tr}(BD).
$$
In your case that gives
$$
\operatorname{Tr}\left((A\otimes B)\,(\overline{A}^T\otimes \overline{B}^T)\right) = \operatorname{Tr}(A\overline{A}^T) \operatorname{Tr}(B\overline{B}^T) = \|A\|_F^2\, \|B\|_F^2,
$$
where $\|\cdot\|_F$ denotes the Frobenius norm.


问:
How to prove that $\lim_{x\to\infty}\frac{(\log_2 x)^3}{x^n}=0$

I need help with proving the following $$\lim_{x\to\infty}\frac{\left({\log}_2x\right)^3}{x^n}=0\ , \quad \forall n>0.$$

I used wolframalpha and got it

and I had kind of an intuition because of the limit $\frac{\ln(x)}x$ , and still I have no idea how to formaly proof it.

Will appreciate any help!

答:
Here is a direct way without L'Hospital:

Substituting $x=e^t$ and using $\log_2 x= \frac{\ln x}{\ln 2}$ you have

$$\lim_{x\to \infty} \frac{(\log_2 x)^3}{x^n} =\frac 1{\ln^3 2}\lim_{t\to \infty}\frac{t^3}{e^{nt}}$$

Now, since $e^u = \sum_{k=0}^{\infty}\frac{u^k}{k!}$, you have $e^u > \frac{u^4}{4!}$ for any $u>0$. Hence,

$$0\leq \frac{t^3}{e^{nt}} < \frac{t^3}{\frac{(nt)^4}{4!}}=\frac{4!}{n^4}\cdot \frac 1t \stackrel{t\to\infty}{\longrightarrow}0$$


问:
On the image of an embedding of $\mathbb{R}$ in $\mathbb{R}^3$
The function $f:\mathbb{R}\to\mathbb{R}^3$ given by $t\mapsto(t,t^2,t^3)$ is clearly an injection and also an immersion. Also, using Heine Borel theorem one can show that it is a proper map, and thus is an embedding. Thus, the image $Z$ of this map is a submanifold of $\mathbb{R}^3$, of codimension $2$. Thus there exist two real functions $g_1$ and $g_2$ on $\mathbb{R}^3$ which cut out $Z$ and are locally independent, i.e., $Z$ is the set of all common zeros of $g_1$ and $g_2$ and the differentials of $g_1$ and $g_2$ are linearly independent on an open neighbourhood of $Z$. I want to find out two such functions explicitly. Any help will be appreciated.

Body: $g_1(x)=x_2-x_1^2$ and $g_2(x)=x_3-x_1^3$ do that.


问:
Show $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is a Banach spaces
Let $\mathcal{L}$ be the space of real-valued Lipschitz functions of order 1 defined on $[0,1]$. That is, the class of functions $f$ such that
$$
\sup\limits_{(x,y)\in[0,1]\times[0,1], x\neq y}
\dfrac{\vert f(x)-f(y)\vert}{\vert x-y\vert}
=K(f)<\infty.
$$
Given $\Vert f\Vert_1=\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +K(f)=\Vert f\Vert + K(f)$ is a norm on $\mathcal{L}$.

Show that $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is a Banach spaces.

To show $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is a Banach spaces, I use the definition of Banach spaces, i.e.
$\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is complete.

To show $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is complete, I will prove every Cauchy sequence on $\mathcal{L}$ is converge, i.e.

Let $\{f_i\}_{i\in \mathbb{N}}$ be the sequence in $\mathcal{L}$.
For all $\varepsilon>0$ there exist $N\in \mathbb{N}$ such that for all natural number $m,n>N$
$$\Vert f_m-f_n\Vert_1<\varepsilon.$$

Proof.

Let $\varepsilon>0$.

\begin{align*}
\Vert f_m-f_n\Vert_1 &=\sup\limits_{0\leq t\leq 1}\vert f_m(t)-f_n(t)\vert +K(f_m-f_n)\\
&\leq \sup\limits_{0\leq t\leq 1}\vert f_m(t)\vert + \sup\limits_{0\leq t\leq 1}\vert f_n(t)\vert +K(f_m-f_n)\\
\end{align*}

Now, I don't know (don't have an idea) how to conclude less than $\varepsilon$. Any hint to prove it?

答: Let $\epsilon >0$. There exists $N$ such that $\sup _t |f_n(t)-f_m(t)| +K(f_n-f_n)<\epsilon$ for all $n,m >N$. This implies that $\sup _t |f_n(t)-f_m(t)| <\epsilon$ for all $n,m >N$ and $K(f_n-f_m)<\epsilon$ for all $n,m >N$. $(f_n(t))$ is a Cauchy sequence so $f(t)=\lim f_n(t)$ exists for each $t$. Letting $m \to \infty$ in the inequality $|f_n(t)-f_m(t)| <\epsilon$ we get $|f_n(t)-f(t)| \leq \epsilon$ for $n >N$. Similarly, we can let $m \to \infty$ in $K(f_n-f_m) <\epsilon$ to get $k(f_n-f) \leq \epsilon$ for $ n>N$. Rest of the argument should now be clear.


答:
I liked your method the best.But if you are fine with using wallis formula,here is a hint;

Hint use the substituition $$x=a+({b-a})\sin^2 \theta$$
$$dx=2(b-a)\sin \theta \cos \theta d\theta$$

can you take it from here?


问:
Using matrices for a recurrence to compute $a_n$ in $O(\log n)$
How would I compute $a_n$ in $O(\log n)$ given they have a linear recurrence, such as $a_i=3b_{i-1}+2b_{i-2}+6^i, b_i=14a_{i-1}-b_{i-2}-6^i$ with matrices?

I know how to deal with recurrences like  $a_i=b_0a_{i-1}+b_1a_{i-2}+\cdots+b_ka_{i-k-1}$ in $O(k^3 \log n)$

Dealing with constants isn’t hard; for $a_i=3a_{i-1}+2a_{i-2}+1$, I can let $b_i=a_i-\frac 14$ to clear the constant term.

I don't know how to deal with an exponential term and multiple sequences.

答:
Hint:

Look at new sequences $A_i = \frac{a_i}{6^i}$ and $B_i = \frac{b_i}{6^i}$. By dividing your original recurrences by $6^i$, we see that these must satisfy the following recurrences:

$$
A_i = \frac{3 B_{i-1}}{6} + \frac{2 B_{i-2}}{6^2} + 1
\quad \quad
B_i = \frac{14 A_{i-1}}{6} - \frac{B_{i-2}}{6^2} -1
$$

From here, standard matrix theoretic techniques will work. After finding the $A_i$ and $B_i$, you can recover $a_i = 6^i A_i$ and $b_i = 6^i B_i$.

As an aside, though -- $a_n$ will not be $O(\log(n))$. In fact, $a_i$ seems to grow extremely quickly (proof by sage):

Edit:

I realize I misunderstood your question - you don't want to show that $a_n \in O(\log(n))$. Instead you want to compute $a_n$ in $O(\log(n))$ time by using the repeated squaring trick.

While you could still use the trick of $A_i$ and $B_i$ to turn our recurrence into something more manageable, we can also do the whole problem in one step:

We want to work with $a_i$ and $b_i$, and to compute them we need to keep track of $a_{i-1}$, $b_{i-1}$, and $b_{i-2}$.

So we package this data into a vector, and want to find a matrix $M$ so that the following holds:

$$
\begin{pmatrix} a_i\\ b_i\\ a_{i-1}\\ b_{i-1}\\ 6^i \end{pmatrix}
=
M
\begin{pmatrix} a_{i-1}\\ b_{i-1}\\ a_{i-2}\\ b_{i-2}\\ 6^{i-1} \end{pmatrix}
$$

Notice we keep a $6^i$ in our vector because we need it in order to compute the future $a_i$.

If we find such a matrix $M$, then notice $a_n$ is just the first entry of $M^n v$ for some vector of initial conditions $v$ (why?). We can then use the repeated squaring trick to compute $M^n$ (and thus $a_n$) efficiently.

Do you see how to construct such a matrix $M$? Hint: The $a_i$ and $b_i$ are given as linear combinations of earlier $a$s and $b$s, as well as a $6^i$, all of which we're keeping track of in the vector!

I hope this helps ^_^


问:
Product of two algebraic varieties is affine... are the two varieties affine?
Let $X_1$ and $X_2$ two algebraic varieties such that their product $X_1\times X_2$ is affine. Are $X_1$ and $X_2$ affine then?

If this is not true, could you give a counterexample?

答: If say $X_1$ is projective then $X=X_1\times X_2$ could not be affine, as every function on $X$ would be of the form $f=(c,f_2)$, for $c$ some constant (since functions on projective varieties are necessarily constant). Certainly if $X$ was affine then there would be more functions on $X$, namely functions which are non-constant in the first factor. </p>


问:
prove that f is a surjection
Let C be a partition of a given set A.

Prove that there exists a set B and $ f : A \rightarrow B$ such that

$ \quad $ (i) f is surjective

$\quad $ (ii) $ C = \{\{x \in A : f(x)=y\} : y \in B \}.  $

Let $ C = A_1 \cup A_2 ... \cup A_n$, n = infinity if A is infinite set.

Let f be a function s.t. $ \forall x,y \in A, \exists i \in \{1,2,....n\} $ where $ x,y \in A_i \iff f(x) = f(y) = i. $

$ B = \{1,2,...n\}. $

Take any $ y \in B,  f(x) = y, \forall x  \in A_y $. Therefore, f is surjective

I am not sure if this proof is correct?

答: If $C$ is an arbitrary partition then it looks like one can take $B = \{c_{1},...c_{n},...\}$ where $c_{i}$ corresponds to the first element in the $i$th partition subset of $A$. If you have a uncountable partition set and/or a "first element" is not defined then one can choose one using a random function. However, I don't know how to make this rigourous because it completely depends on what the partitions are. If the partitions are intervals then of course you can take all midpoints. If it is a partition of a tricky nowhere continuous function then you might need to be a bit more creative.

Title: prove that f is a surjection


问:
Is there an algebraic way to solve this equation?
Is there an algebraic trick using some change of variable or properties of exponent or logarithms to solve the following equation algebraically for $x\in \mathbb{R}$?

$\log_{\frac{1}{4}}(x)=(\frac{1}{4})^x$

答: Put $y = 4^{-x}$, then you want to solve $4^{-y} = y$ (and it follows that $y=x$). A unique solution exists because $4^{-y}-y$ has negative derivative and is positive at $0$, negative at $1$ (so the solution is between $0, 1$). One can see that $y=1/2$ is a solution, so $x = 1/2$.


问:
When does $A$ iff $B$, imply $A=B$?
I understand that it goes in the other direction, that is, if $A=B$, then $A$ if and only if $B$ is also true.

(Just do a proof by contradiction, given $A=B$ assume $\lnot(A\leftrightarrow B)$, then $\lnot(A\leftrightarrow A)$, so the assumption was wrong and $A=B$ implies $A\leftrightarrow B$.)

However I want to know under what conditions does the implication go the other way? Is there any general set of requirements I can look for, to know that it will hold in the reverse case? Under what special circumstances can I conclude $A=B$ when I already know $A\leftrightarrow B$?

You are free to interpret $A$ and $B$ as whatever object you wish. If it helps, they can be restricted to a certain type to answer the question.

答:
I found an answer to the question by looking into the propositional extensionality that @Daniel Schepler linked in a comment. The trick is to take $A$ and $B$ as propositions in the case of $A\leftrightarrow B$, and as names for propositions (denoted as $'A'$ and $'B'$) in the case of $'A'='B'$. Then the implication, if $A\leftrightarrow B$ then $'A'='B'$ holds.

Reference:
https://ncatlab.org/nlab/show/propositional+extensionality

Title: When does $A$ iff $B$, imply $A=B$?