问:
$A\subset\mathbb{R}, t>0\Rightarrow |A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}-(-t,t))|$
I am trying to prove the following statement from Axler's MIRA book and I would appreciate an hint about how to finish my proof (NOTE: $||$ refers to outer measure):

"$A\subset\mathbb{R}, t>0\Rightarrow |A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}-(-t,t))|$"

What I have done up to now:

Let $t>0$ and $A$ be a subset of $\mathbb{R}$: then we have that

$$|A|=|A\cap\mathbb{R}|=|A\cap ((-,t,t)\cup\mathbb{R}-(-t,t))|=|(A\cap (-t,t))\cup (A\cap\mathbb{R}-(-t,t))|\leq |(A\cap (-t,t))|+|(A\cap\mathbb{R}-(-t,t))|$$ where the first equality follows from the fact that $A\subset\mathbb{R}$, the second from the fact that $\mathbb{R}=(-t,t)\cup (\mathbb{R}-(-t,t))$ and the inequality from the countable subadditivity of outer measure.

So, what I have to do now is find a way to get an inequality of the form $$|(A\cap (-t,t))|+|(A\cap\mathbb{R}-(-t,t))|\leq |A|$$ but I don't see how to do this, so I would appreciate an hint (not a solution) about how to do this.

答:
It really is immediate from the definition. First a technicality:

Lemma The definition of $|A|$ is unchanged if closed or half-open intervals are allowed as well as open intervals.

Proof: Say $|A|'$ is the outer measure including all sorts of intervals. The inf of a larger set is smaller, so $|A|'\le|A|$.

To show the other inequality: If $I$ is any interval with endpoints $a<b$ let $B(I,\epsilon)=(a-\epsilon/2, b+\epsilon/2)$. Let $\epsilon>0$. Choose intervals $I_1,\dots$ so$$A\subset\bigcup I_n$$ and $$\sum l(I_n)<|A|'+\epsilon.$$Let $I_n'=B(I_n,\epsilon/2^n)$. Since $I_n'$ is open and $A\subset\bigcup I_n'$ we have
$$|A|\le\sum l(I_n')=\sum(l(I_n)+\epsilon/2^n)=\epsilon+\sum l(I_n)\le|A|'+2\epsilon.$$
Since $\epsilon>0$ is arbitrary this shows $|A|\le|A|'$.

Now to show $|A\cap(-t,t)|+|A\setminus(-t,t)|\le|A|$: Wlog $|A|<\infty$. Let $\epsilon>0$. Choose intervals $I_1,\dots$ so $$A\subset\bigcup I_n$$ and $$\sum l(I_n)<|A|+\epsilon.$$ Let $I_n'=I_n\cap(-t,t)$, $I_n''=I_n\cap(-\infty,-t]$, and $I_n'''=I_n\cap[t,\infty)$. Then $A\cap(-t,t)\subset\bigcup I_n'$ and $A\setminus(-t,t)\subset\bigcup(I_n''\cup I_n'''), $ so $$|A\cap(-t,t)|+|A\setminus(-t,t)|\le\sum l(I_n')+\sum(l(I_n'')+l(I_n'''))=\sum l(I_n)\le|A|+\epsilon;$$as before, let $\epsilon\to0$.


问:
Are the strong and the weak operator topology well defined in abstract C* and Von Neumann algebras?
If $A$ is an abstract $C^*$ algebra, then a $A$ is isometrically isomorphic to a subalgebra of bounded operators $B(H)$ for some Hilbert space $H$. We can restrict the weak operator toplogy and the strong operator topology of $B(H)$ to $A$.

Are these definitions independent of $H$? And what about the case of $A$ being an abstract Von Neummann Algebra (a $C^*$ algebra who has a predual)?

答:
No, the weak and strong operator topologies depend strongly on the chosen representation.

For example, if you represent $C([0,1])$ on $L^2([0,1])$ by multiplication operators, then the functions $$f_n(x) = x^n$$ converge strongly to zero.  However, if you take the direct sume of that representation with the one dimensional representation corresponding to the character
$$f\mapsto f(1),$$
then the $f_n$ do not converge strongly to any $f$ in $C([0,1])$.

ContentLicense: CC BY-SA 4.0


问:
Why is $\kappa(A^{-1})= \kappa(A)$?
In my syllabus we have the alternative definition of the condition of a matrix:

$$\kappa(A)= \frac{\text{max}_{\| \vec{y} \| =1}\| A \vec{y} \|}{\text{min}_{\| \vec{y} \| =1}\| A \vec{y} \|}$$

In it, it also says that by definition of the condition of a matrix it follows that $\kappa(A^{-1})= \kappa(A)$. So there is no explanation for this. Therefore, my question is: Why is $\kappa(A^{-1})= \kappa(A)$?

答: For an invertable matrix, your numerator and denominator are the absolute values of the largest and smallest eigenvalues. The eigenvalues of the inverse matrix are the reciprocals of its eigenvalues.</p>

Title: Why is $\kappa(A^{-1})= \kappa(A)$?


问:
A result of Chebychev and an $\epsilon$ away from Bertrand's postulate
I came across the following result in a number theory book. In proving a result of Chebychev that there exist positive constants $x_0$, $c_1$ and $c_2$ such that $$ c_1 \frac{x}{\log x} \le \pi(x) \le c_2\frac{x}{\log x}$$ for all $x>x_0$, the following bounds on $\psi(x)$ were used
$$ x\log 2 + O(\log x) \le \psi(x) \le 2x\log 2 + O(\log^2(x))$$ together with the result that
$$\frac{\psi(x)}{x} = \frac{\pi(x)\log x}{x} + o(1)$$

The author says that for any $c_1 < \log 2$ and any $c_2 > 2\log 2$ the theorem holds. In an attempt to fill the details, I did the following:

starting from $$ x\log 2 + O(\log x) \le \psi(x) \le 2x\log 2 + O(\log^2(x)),$$ divide through by $x$ to obtain
$$\log 2 + o(1) \le \frac{\psi(x)}{x} \le 2\log2 + o(1)$$

Now, replace $\frac{\psi(x)}{x}$ with $\frac{\pi(x)\log x}{x} + o(1)$ so that
$$\log 2 + o(1) \le \frac{\pi(x)\log x}{x} + o(1) \le 2\log2 + o(1)$$
Here's a step I'm not sure of and need clarification: can the little $o$ term in the middle be subtracted across the board to get
$$\log 2 + o(1) \le \frac{\pi(x)\log x}{x}\le 2\log2 + o(1)$$
$$\frac{x}{\log x}(\log 2 + o(1)) \le \pi(x) \le \frac{x}{\log x}(2\log2 + o(1))$$ and hence for any $c_1 < \log2, c_2 > 2\log 2$ the theorem follows?

Secondly, the author says that since $\frac{2\log 2}{\log 2}=2,$ a corollary of this result is that for every $\epsilon > 0$, there exists a prime number in the interval $[x, (2+\epsilon)x]$ for all $x > x_0(\epsilon)$. I do not see this immediately. How can this be verified?

答:
Thanks to @GregMartin's guidance, here's how to verify the last claim that was made.

Suppose for some $\epsilon > 0$ there is no prime in the interval $[x, (2+\epsilon)x]$ for some $x$. Then,
$$\begin{align}
0 = \pi((2+\epsilon)x) - \pi(x) &\ge \frac{c_1(2+\epsilon)x}{\log((2+\epsilon)x)} - \frac{c_2x}{\log x} \\
&= \frac{c_1(2+\epsilon)x}{K + \log x} - \frac{c_2x}{\log x},  \ \ \text{ where } K=\log(2+\epsilon) \\
&= \frac{x}{\log x} \left(\frac{c_1(2+\epsilon)}{1+K/\log x} - c_2\right) \\
&= \frac{x}{\log x} \left(\frac{c_1(2+\epsilon)}{1+o(1)} - c_2\right) \ \ \text{ for } x \text{ large}\\
&= \frac{x}{\log x} \left(c_1(2+\epsilon) +o(1) - c_2\right) \\
&= \frac{x}{\log x}(c_1\epsilon + o(1)) \ \ \text{ since } 2c_1 - c_2 = o(1) \\
\end{align}$$
which is positive as $x \rightarrow \infty$, a contradiction.


问:
Example: Principal transformation groupoid with groupoid topology strictly finer than product topology

Out of curiosity, what would be an example for the following situation:

Let $G\curvearrowright X$ be a free action of a locally compact Hausdorff group $G$ on a Hausdorff space $X$.

By the freeness condition, the tranformation groupoid is algebraically principal,
$$X\times G\subseteq X\times X\text{ where }(x,g)\mapsto(x,gx).$$
Moreover one has $X\times G=X\times X$ precisely if the action is transitive.

But the groupoid topology is generally finer than the product topology induced by $X\times X$.

More precisely, convergence in the groupoid topology is given by
$$(x_n,g_n)\to (x,g)\iff x_n\to x\text{ and }g_n\to g,$$
while convergence induced by the product topology is given by
$$(x_n,g_nx_n)\to (x,gx)\iff x_n\to x\text{ and }g_nx_n\to gx,$$
whence the former is generally finer the latter.

What would be an example where the topology is strictly finer?

What I tried so far were examples along the lines $\mathbb{C}\setminus0\curvearrowright\mathbb{C}\setminus 0$,

but from here it looks one requires maybe some sort of "hyperbolic" behavior?

Parallely out of curiosity, I learned from Ian Putnam the following interesting example:

Inside the unit square, consider the diagonal plus the other two corner points,
$$\Delta\cup\{(0,1),(1,0)\}\subseteq [0,1]\times[0,1].$$
I think this example cannot arise as transformation groupoid, but I'm not sure.

So my question, is there maybe a simple argument which forbids such a representation resp. does it admit one?

答:
Consider the action of $G=\mathbb Z$ on $X=S^1$ given by
$$
  \alpha_n(z)=e^{2\pi i n\theta }x, \quad \forall n\in\mathbb Z, \quad \forall x\in S^1,
  $$
where $\theta$ is an irrational number.  Can you show that the two topologies on the transformation groupoid are very different?


答:
First you must accept that if $A^{-1}$ doesn't exist then by convention $\kappa(A)=\infty$.

Once you've postulated that $A^{-1}$ exists, you want to show

$$\max_{\| x \|=1} \| A^{-1} x \|=\left ( \min_{ \| x \|=1 } \| Ax \| \right )^{-1}.$$

To see that, take a minimizer $x^*$ on the right side. Let $b=\frac{Ax^*}{\| Ax^* \|}$; then look at $A^{-1} b=\frac{x^*}{\| Ax^* \|}$. You picked $x^*$ so that among unit vectors, $\| Ax^* \|$ is as small as possible, so $\| A^{-1} b \|=\frac{1}{\| Ax^* \|}$ is as large as possible among unit vectors $b$.

Intuitively, if $A$ maps $x^*$ to some much smaller vector $b$, then $A^{-1}$ maps $b$ back to $x^*$, which is much bigger than $b$. Juggling the normalizations obfuscates this intuition a little bit, I think.

Title: Why is $\kappa(A^{-1})= \kappa(A)$?

Body:
In my syllabus we have the alternative definition of the condition of a matrix:

$$\kappa(A)= \frac{\text{max}_{\| \vec{y} \| =1}\| A \vec{y} \|}{\text{min}_{\| \vec{y} \| =1}\| A \vec{y} \|}$$

In it, it also says that by definition of the condition of a matrix it follows that $\kappa(A^{-1})= \kappa(A)$. So there is no explanation for this. Therefore, my question is: Why is $\kappa(A^{-1})= \kappa(A)$?


问:
Difficulty in plotting polar plot for trigonometric functions
I'm having some difficulty with the following question.

Sketch the graph of $r=\tan(\theta)$,where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ and $r=\frac{\sqrt{2}}{2}\csc(\theta)$ and find the area between the two curves.

I'm given the hint that the graph has two vertical asymptotes. This is a polar plot in case that wasn't clear.

I've put this into various software, but I am not really sure how to sketch the curves. I get that in cartesian coordinates, there are vertical asymptotes at $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ but I am not sure how that can help me in plotting the polar curve. Can someone help out here?

For the $\frac{\sqrt{2}}{2} \csc(\theta)$, it's just a horizontal line but I'm not really sure how that's being done either...

I feel like I could get the area if I knew how to plot this but I can't even plot this. Can someone please help out here?

答:
It is more convenient to express the two curves in Cartesian coordinates, i.e.

$$y= \frac1{\sqrt2}, \>\>\>\>\> y= \frac{x^2}{\sqrt{1-x^2}}$$
as shown below. Note that they meet at $x=\frac1{\sqrt2}$ and the enclosed area is

$$Area = 2\int_0^{\frac1{\sqrt2} }\left( \frac1{\sqrt2} - \frac{x^2}{\sqrt{1-x^2}}\right)dx=\frac32-\frac\pi4
$$


问:
Proving by induction $4^n < 5^n$ where $n > 0$
I have done a few normal inductions (like divisibility and sums) and know a little bit about strong induction. I can't find any problem similar to this one and am struggling to understand any intuition that would make proving this statement any easier.

Here is how far I have gotten (not very much):

Base Case:

$4^1 < 5^1$

$4^2 < 5^2$

$16 < 25$

Induction:

$4^k < 5^k$

$4^{k+1} < 5^{k+1} = 4\times4^k < 5\times5^k$

I have no clue where to go or how to force the k+1 equation to somehow equate to the k equation.
Sorry if this has been asked before.
Thank you.

答:
You can relate the inequality for $k+1$ to the one for $k$ like this

$$
4 \times 4^k < 5 \times 4^k < 5\times 5^k
$$

where the first inequality follows from the fact that $4a < 4a + a = 5a$ for any positive $a$ and the second inequality is your inequality for $k$ multiplied by $5$ on both sides.


问:
Linearizing equations of motion
I have been looking at the operations of a quadrotor drone. I am reading the maths behind it and in one section it mentions:

"These equations of motion are linearized with respect to an equilibrium point" . I have some questions regarding this.

1) What exactly does it mean to linearize an equation of motion ? how exactly is this usually done?

2) If say I linearize the equations of motion for a drone or car or whatever, why is this useful, what information can it tell me?

Thank you

Here is a sort of idea of it, I cant post them all since it is a 70 page document (Don't think I can attach documents) but it mostly relates to the translational and altitude dynamics.

答:
Control is the question of designing an input[s] $u(t)$ for a system to bring it state to a desired point. In your application, you are finding a set of input forces $u(t)$ for a general nonlinear system

$$\dot{x}(t) = f(x(t), u(t))\quad (1)$$

so that the solution $x(t)$ tends towards a desired setpoint $x_0.$ This would be the desired configuration (position and orientation) of the quadcopter. In your case, $f(x,u)$ would be the general equations of motion (Newton-Euler equations).

However, it turns out it is much easier to control linear differential systems that take the form

$$\dot{x}(t) = A\,x(t) + B\,u(t)\quad (2)$$

where $x$ is a vector and $A, B$ are constant, real matrices than it is to control a general nonlinear system (1). So the first question we ask is

Can we approximate equation (1) with (2) for some constant $A, B$ under a reasonable assumption?

The answer is yes if there exists a constant applied input $u_0$ so that the desired setpoint $x_0$ is an equilibrium point, i.e. when $f(x_0, u_0) = 0.$ You can think of $u_0$ as the applied force needed to keep a quadcopter at the desired configuration once it actually gets there. For example, you must maintain a non-zero applied thrust if you want to keep the quadcopter at a positive altitude to compensate gravity.

Once you have computed $u_0$ to keep your quadcopter in a desired setpoint $x_0,$ you can linearize the dynamics by evaluating the jacobian (first derivative matrix) of $f$ at the equilibrium point $(x_0, u_0).$ In particular you would define

$$A := \left.\frac{d f}{d x} \right|_{x=x_0, u=u_0}, \quad B := \left.\frac{d f}{d u} \right|_{x=x_0, u=u_0}$$

and then define the perturbations from the equilibrium point

$$\delta x := x - x_0, \quad \delta u := u - u_0,$$

and it will turn out that the dynamics of $\delta x$ are approximated by

$$\delta \dot{x} = A\,\delta x + B\,\delta u.$$

At this point, a control engineer can do some math (linear control theory) to answer the question

Does there exist a rule for $\delta u$ that ensures $\delta x \to 0$ ?

There are many many many textbooks on how to do this. For example, you might implement state feedback $\delta u = -K \delta x$ and choose $K$ cleverly so that the resulting dynamics

$$\delta\dot{x} = (A - B K)\, \delta x$$

satisfy the property. Upon doing so, one can solve for $u(t)$ using the definition of $\delta u$ and the resulting control law will work, for states close enough to the desired configuration, to bring the configuration $x$ towards $x_0.$ Improving the response of the system can be done by tuning the feedback law designed for $\delta u$ however linearization constrains your performance guarantees to a small region near the equilibrium point.

A good start to all of this would be a textbook on Linear Control Systems. Some cover a bit of modelling and linearization theory. I believe "Modern Control Systems" by Dorf & Bishop does but a few others do as well.

Title: Linearizing equations of motion


问:
Lattices in vector space over local field contained in a common lattice
Let $A=\mathbb{F}_q[t]$, let $v_\infty$ be the usual valuation on $k=\mathbb{F}_q(t)$ such that $ v_\infty(1/t)=1$, and let $k_\infty=\mathbb{F}_q((t^{-1}))$ the completion of $k$ with respect to this valuation. Let $V$ be a finite dimensional $k_\infty$- vector space. I found the following claim which I don't know how to prove:

Take a $\mathbb{F}_q[t]$-lattice $\Lambda_1\subset V$ (since $A$ is PID, $\Lambda_1 $is of the form $\Lambda_1= v_1A+...+v_rA$, where $r=\dim_{k_\infty} V$). Then the set of lattices $\Lambda_2$ such that there is a lattice $\Lambda$ containing both $\Lambda_1$ and $\Lambda_2$ is dense in the set of all latices $\mathcal{L}(V)$.

Any help towards proving this will be greatly appreciated

答:
What is a $\mathbb{F}_q[t]$-lattice in $k_\infty{}^n$, a free discrete $\mathbb{F}_q[t]$-submodule such that $k_\infty{}^n/\Lambda$ is compact ?

If so then (being a PID) any finitely generated $\mathbb{F}_q[t]$-submodule is free, and $k_\infty{}^n/\Lambda$ is compact iff $k_\infty \Lambda=k_\infty^n$.

(if $k_\infty \Lambda$ is smaller than $k_\infty^n$ then the quotient contains a copy of $k_\infty$, it is not compact, if $k_\infty \Lambda=k_\infty^n$ then up to a $k_\infty$-linear transformation it reduces to a quotient of $k_\infty{}^n / \Bbb{F}_q[t]^n$ which is $\cong t^{-1} \Bbb{F}_q[[t^{-1}]]^n$ as topological groups)

So it remains to check the discreteness.

Let $\Lambda_1$ be a lattice which is of rank $n$ as a $\Bbb{F}_q[t]$-module. With a $k_\infty$-linear transformation we can assume that $\Lambda_1=\Bbb{F}_q[t]^n$.

Then $\Lambda_1+ \Bbb{F}_q[t] u$ is discrete iff $u\in \Bbb{F}_q(t)^n$: look at the image of $\Bbb{F}_q[t] u$ in the compact set $k_\infty{}^n/\Lambda_1$, for it to have no accumulation point we need that the image of $\{ t^au, a\ge 0\}$ is a finite set whence $(t^a-t^b) u \in \Lambda_1$ for some $a>b$.

Conversely if $u\in \Bbb{F}_q(t)^n$ then the image of $\Bbb{F}_q[t] u$ in $k_\infty{}^n/\Lambda_1$ is a $\Bbb{F}_q[t]/(f(t))$ module, where $f$ is the denominator of $u$, so it is a finite set.

Therefore take any lattice $\Lambda_2$ and approximate it by a lattice $\Lambda_2'$ in $\Bbb{F}_q(t)^n$ to obtain that $\Lambda_1+\Lambda_2'$ is a lattice.


问:
Problem with operations on exponents
Okay, this is a very basic doubt, and I would like to know where exactly the mistake is.

Let's take the equality $1^2=1$

Now I do it this way Step 1: $1^2 = 1^\frac{4}{2}$

Step2: $\implies (1^4)^\frac{1}{2}$

Step3: $\implies 1^\frac{1}{2} = \pm 1$

The answer should be only one but I am also getting minus one.
I know it obviously is a mistake, but I would like to know which exact step and in which particular operation the mistake is....

答: Given a positive number $a$, $a^\frac12$ is defined to be the positive number $b$ such that $b^2=a$. In particular, $1^\frac12=1,$ even though $(-1)^2=1,$ too.

ContentLicense: CC BY-SA 3.0

Title: Problem with operations on exponents


问:
Let $f$ differentiable on $(a, \infty)$ such that $\lim_{x\to\infty}\frac{f(x)}{x}=0$. Show that $\lim\inf_{x\to\infty}|f'(x)|=0$

Let $f$ differentiable on $(a, \infty)$ such that $\lim_{x\to\infty}\frac{f(x)}{x}=0$. Show that $\lim\inf_{x\to\infty}|f'(x)|=0$. I would like to know if my proof holds, please and have a feedback.

As $\lim_{x\to\infty}\frac{f(x)}{x}=0$, we have by definition that: $\forall \epsilon>0 \ \exists x_0 \ \forall x\ge x_0$: $|\frac{f(x)}{x}|<\epsilon$.

Let $k \in (a,\infty)$. We have then $\forall \epsilon>0, \ \forall x\ge x_0$:

$|\inf|f'(x)||=\inf|f'(x)|\le|f'(x)|=\Big|\lim_{x\to k}\frac{f(x)-f(k)}{x-k}\Big|\le \lim_{x\to k}\Big|\frac{f(x)-f(k)}{x-k}\Big|< \lim_{x\to k}\epsilon=\epsilon$

So $\lim\inf_{x\to\infty}|f'(x)|=0$ as $k$ is arbitrary.

I'm not sure at all of this proof. The last inequaility seems to me false, but I didn't find a counter example. Another approach was to distinguish cases: $f$ converges or $f$ diverges (then apply L'Hôpital).

答:
Prove by contradiction. Suppose the contrary that $\liminf_{x\rightarrow\infty}|f'(x)|=l>0$.
Note that $\lim_{x\rightarrow\infty}\inf_{y\geq x}|f'(y)|=\liminf_{x\rightarrow\infty}|f'(x)|>l/2$,
so there exists $x_{0}$ such that $\inf_{y\geq x}|f'(y)|>l/2$ whenever
$x>x_{0}$. In particular, $|f'(y)|>l/2$ for all $y>x_{0}$.

Fix $x_{1}=x_{0}+1$. Let $x>x_{1}$ be arbitrary. By mean-value theorem, there
exists $\xi_{x}\in(x_{1},x)$ such that $|\frac{f(x)-f(x_{1})}{x-x_{1}}|=|f'(\xi_{x})|>l/2$.
On the other hand,
\begin{eqnarray*}
 &  & \frac{f(x)-f(x_{1})}{x-x_{1}}\\
 & = & \frac{\frac{f(x)}{x}-\frac{f(x_{1})}{x}}{1-\frac{x_{1}}{x}}.
\end{eqnarray*}
Note that as $x\rightarrow\infty$, $\frac{f(x_{1})}{x}\rightarrow0$,
$\frac{f(x)}{x}\rightarrow0$, and $\frac{x_{1}}{x}\rightarrow0$.
Therefore $\lim_{x\rightarrow\infty}\frac{f(x)-f(x_{1})}{x-x_{1}}=0$.
This contradicts to $|\frac{f(x)-f(x_{1})}{x-x_{1}}|>l/2$ whenever
$x>x_{1}$.


问:
Game-like probability problem
I'm quite new to an intermediate study of probability and i would like to ask how this problem can be modeled.

Let's suppose that i have a set of states $(X_{1},X_{2}...X_{n})$ with n given and I know that to reach $X_{2}$ from $X_{1}$ i have a probability $p=0.9$, to reach $X_{3}$ from $X_{2}$  i have a probability $P_{2}=0.4$ and so on until i reach $X_{n}$ from $X_{n-1}$ with some probability $p_{n}$.

If I'm on the state $X_{i}$ and the event "Reach $X_{i+1}$" fails, i have to stay on $X_{i}$ until it succeeds.

Let's suppose that i know $n$ and the transition probabilities of each state, how can i model this problem? markov chain? join of geometric distributions?
What can i say about the final probability to reach $X_{n}$ with some confidence?

(The question may be stupid but i'm interested into sharpening my skills and learning more :) ).

Thanks.

答:
Let $p_i$ be the probability of moving from state $i$ to state $i+1$. The additional time, $T_i$, to reach state $i+1$ given that at a fixed time it is in state $i$ has a Geometric distribution. $P[T_i=t]=(1-p_i)^{t-1}p_i$ because the first $t-1$ units of time must be failures to move to the different state followed by a success. The mean and variance are $\frac{1}{p_i}$ and $\frac{1-p_i}{p_i^2}$. The total time to reach state $n$ is the sum of these independent Geometric random variables.

For example, the density for $T_1+T_2+T_3$ is
$$P[T_1+T_2+T_3=t] \\ =\sum_{n_1=1}^{t-2}\sum_{n_1=1}^{t-n_1-2}P[T_1=n_1]P[T_2=n_2]P[T_3=t-n_1-n_2]$$

$$=\frac{p_1 p_2 p_3 }{(p_2-p_1)(p_1-p_3)(p_2-p_3)}\left((p_2-p_1)(1-p_3)^{t-1}+(p_1-p_3)(1-p_2)^{t-1}-(p_2-p_3)(1-p_1)^{t-1} \right)$$
for $t=3,4,...$.

The sum has mean $\sum_{i=1}^{n-1}\frac{1}{p_i}$ and variance $\sum_{i=1}^{n-1}\frac{1-p_i}{p_i^2}$. The sum is approximately normal. You can use that to find an approximation for the density or cumulative distribution function. For example, the probability the sum is equal to $t$ is approximately $P[t-\frac{1}2<X<t+\frac{1}2]$ where $X$ is normal with the mean and variance above.

Title: Game-like probability problem


问:
How to solve $\int \csc^{4} \frac{3}{2} x \,dx$ using integration by parts

How to solve $\int \csc^{4} \frac{3}{2} x \,dx$

This is what I have so far:

$$\int \csc^{4} \frac{3}{2} x \,dx=\frac{3}{2} \int \csc^{2} x \csc^{2}x\,dx.$$

Let $u=\csc^{2}$ and $dv=\csc^{2} \,dx$. Then we have $du=-2\csc^{2}x\cot x\,dx$ and $v=-\cot x$.

\begin{align}
\frac{3}{2} \int \csc^{2} x \csc^{2} x \,dx&=\frac{3}{2}\left(-\csc^{2}x\cot x-\int-\cot x(-2\csc^{2}x\cot x\,dx)\right)\\
&= \frac{3}{2}\left(-\csc^{2}x\cot x-2\int \csc^{2}x\cot^{2}x\,dx\right)
\end{align}

At this point, I am uncertain about how to continue. Do I find the integral in the right hand side again? Or perhaps I missed something? Thank you.

答:
Just to avoid the ambiguity pointed out in the comments, I'll carry out the integral of $\csc^4(x)$. (If $\frac32$ is a constant outside the argument to $\csc$, then just multiply this result by $\frac32$. If it belongs inside the argument, just make the appropriate substitution of $x$ with $\frac{3x}2$, keeping in mind the chain rule.)

For the remaining integral, recall that

$$\cot^2(x)=\csc^2(x)-1$$

Then

$$\int\csc^2(x)\cot^2(x)\,\mathrm dx=\int\csc^4(x)\,\mathrm dx-\int\csc^2(x)\,\mathrm dx$$

which gives

$$\color{red}{\int\csc^4(x)\,\mathrm dx}=-\csc^2(x)\cot(x)-2\left(\color{red}{\int\csc^4(x)\,\mathrm dx}-\int\csc^2(x)\,\mathrm dx\right)$$

i.e. a simple linear equation whose solution is the antiderivative you seek.


答:
For the base case, directly checking the inequality
$$
4^1 < 5^1
$$
is enough.

Now, if for some positive integer $k$ we have
$$
4^{k} < 5^{k}, \quad \mbox{(*)}
$$
we have to infer from it
$$
\mbox{the desired inequality } 4^{k+1} < 5^{k+1}.
$$

To make this inference, write
$$
\begin{array}{llll}
& 4^{k+1} & \\
= & 4^{k} & 4^1 \\
< & 4^{k} & 5^1 \quad \mbox{(by the base case)}\\
< & 5^{k} & 5^1 \quad \mbox{(by (*))}\\
= & 5^{k+1}.
\end{array}
$$

(So, basically, the task is to construct a chain of inequalities that leads to the desired result and where each inequality follows from something that's already been proved.)

Title: Proving by induction $4^n < 5^n$ where $n > 0$

Body:
I have done a few normal inductions (like divisibility and sums) and know a little bit about strong induction. I can't find any problem similar to this one and am struggling to understand any intuition that would make proving this statement any easier.

Here is how far I have gotten (not very much):

Base Case:

$4^1 < 5^1$

$4^2 < 5^2$

$16 < 25$

Induction:

$4^k < 5^k$

$4^{k+1} < 5^{k+1} = 4\times4^k < 5\times5^k$

I have no clue where to go or how to force the k+1 equation to somehow equate to the k equation.
Sorry if this has been asked before.
Thank you.


问:
A nice combinatorial identity: $\sum_{k=1}^{n-1}\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk}=1$
While answering a recent question I came across the following nice-looking identity:
$$
\sum_{k=1}^{n-1}\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk}=1
$$
valid for all integer $n\ge2$.

Is there a simple algebraic proof of this identity?

答:
Write this as
$$
\sum_{k=1}^{n-1}\frac{\binom{k}{n-k-1}\left(\frac{2k+n-1}{k}+1\right)}{\binom{n}k}=\frac{\binom{k}{n-k-1}(3k+n-1)}{\binom{n}kk}
$$
Now this is written as a hypergeometric series. There are several ways to do this, but I have carefully chosen the way so that no division by zero occurs in the range $1\le k\le n-1$.

We can now use Zeilberger's algorithm to find a recurrence this sum satisfies. Doing so, we find that when
\begin{align}
F(n,k)&=\frac{\binom{k}{n-k-1}(3k+n-1)}{\binom{n}kk},
\\
R(n,k)&=\frac{(2k+1-n)(2k-n)}{(n-k)(3k+1-n)},
\\
G(n,k) &= F(n,k)R(n,k),
\end{align}
we have
$$
F(n,k)=G(n,k+1)-G(n,k)
$$
This is actually better than what Zeilberger's usually gives us. We have found an anti-difference for $F(n,k)$, allowing us to compute $\sum_{k=a}^bF(n,k)$. In particular,
$$
\sum_{k=1}^{n-1}F(n,k)=G(n,n)-G(n,1)
$$
It is not hard to see that $G(n,1)=0$, either because of the factors $(2k+1-n)(2k-n)$ when $n\in\{2,3\}$, or because of the $\binom{k}{n-k-1}$ factor when $n\ge 4$. However, $G(n,n)$ is an indeterminate form $0/0$. Let us look carefully at $G(n,k)$:
$$
G(n,k)=\frac{\binom{k}{n-k-1}(3k+n-1)}{\binom{n}kk}\cdot \frac{(2k+1-n)(2k-n)}{(n-k)(3k+1-n)}\\=\frac{\color{red}{\binom{k}{n-k-1}}(2k+1-n)(2k-n)}{\color{red}{(n-k)}\binom{n}kk}
$$
The problem factors are in red. However, we can rewrite $\binom{k}{n-k-1}\frac{1}{n-k}=\binom{k+1}{n-k}\frac1{k+1}$, and the problem disappears. Doing so, we find $G(n,n)=1$, and we are done.

Note that you can avoid all of these division by zero considerations if you write everything in terms of Gamma functions, and take limits if the singularities of $\Gamma$ get involved.


问:
How to solve for the probability of this problem?
Suppose Boyet, a basketball player, has an 85% chance of making free throw. Over the whole game, he attempts 5 free throws. What is the probability that he will miss at least one of them?

So I'm literally having a hard time understanding what how to get the independent probability on this problem. How would I know what's the two events that intersected in this situation? How can I use the independent formula in this one?

答:
Total probability is equal to the chance the desired situation happens plus the chance that it does not happen.

P(missing no shots) + P(missing 1 or more shots) = 1

P(missing 1 or more shots) = 1 - P(missing no shots)

Each free throw is independent of each other with the same probability

P(missing no shots) = P(making one shot)^(attempted shots)

As the probability of making one shot is given as 85% and the number of attempted shots is given as 5, we can evaluate the this equation.

P(missing no shots) = (0.85)^5

P(missing 1 or more shots) = 1-0.4437053125

P(missing 1 or more shots) = 0.5562946875 = 55.62946875%


问:
Calculating combinations of a password
I want to find the possible amount of combinations of a password that is 5 characters long, and each character could either be a lowercase letter or a digit. Also, each password should start with a lower case letter and contain at least two digits. The answer I got was 26 * 10 * 10 * 36 * 36.

Am I correct?

答:
Given there are only $5$ characters you can do this directly by working individual cases.

Case $1$: There are exactly two digits out of last four characters -

$\displaystyle 26 \cdot {4 \choose 2} \cdot 10^2 \cdot 26^2$

Case $2$: There are exactly three digits out of last four characters -

$\displaystyle 26 \cdot {4 \choose 3} \cdot 10^3 \cdot 26$

Case $3$: Last four characters are all digits  -

$\displaystyle 26 \cdot 10^4$

Add all of the arrangements from the three cases.

You can also calculate by subtracting cases where there are no digits or only one of the last four characters is a digit, which is given by -

$\displaystyle 26 \cdot (36^4 - 26^4 - 4 \cdot 10 \cdot 26^3)$

Title: Calculating combinations of a password


问:
Factoring $x^4 + 12x^3 + 46x^2 + 59x + 18$

How do I factorize the following?
$$x^4 + 12x^3 + 46x^2 + 59x + 18$$

I've tried looking for a root by trial and error to no avail.

The answer is
$$(x^2 + 5x + 2)(x^2 + 7x + 9)$$

答:
There are quiet a few ways to do it.

First of all you can find the zeros of this polynomial by Ferrari's method. As you find it, you can surely factorize the polynomial.

Another way  is to use Descartes' rule of sign which gives you a hint where the roots of the equation $f(x)=0$ lie.

Also if $f(\alpha).f(\beta)<0$ then you are quiet sure that at least one root of $f(x)=0$ lies between $\alpha$ and $\beta$. By trial and error method, you can find that root.


问:
derivative of a numerical expression
I have a function $$g(y) = \int{f(y, t)dt}$$ and I am currently able to evaluate the integral numerically (doesn't seem like it is possible to find an analytical solution...), which I have also plotted for various values of $g(y)$. All good so far. However, I am also interested in the quantity $$\frac{dg(y)}{dy}$$ Is there a way to obtain this derivative?</p>

答:
You can use finite differences as below
$$g(y_0+h)\approx g(y_0)+g'(y_0)\cdot h+O(h^2)$$
$$g(y_0-h)\approx g(y_0)-g'(y_0)\cdot h+O(h^2)$$
By substracting second from first
$$g(y_0+h)-g(y_0-h)\approx 2g'(y_0)\cdot h$$
$$\Rightarrow g'(y_0)\approx\frac{g(y_0+h)-g(y_0-h)}{2h}$$
Or in discrete notation
$$\Rightarrow g'(y_i)\approx\frac{g(y_{i+1})-g(y_{i-1})}{2h}$$</p>

Title: derivative of a numerical expression


答:
Hint:

An elementary rule for inequalities, seen in middle school, asserts that, for positive numbers,
$$a<b\enspace\text{ and }\enspace a'<b'\implies aa'<bb'.$$


答:
Let $t=\frac32x$ and integrate by parts as follows

\begin{align}
 \int \csc^{4} \frac{3}{2} x \,dx
&= \frac23\int \csc^{4} t \,dt=- \frac29\int \sec^{2} t \>d(\cot^3 t)\\
&= -\frac29\csc^2 t\cot t+\frac49 \int\csc^2 t dt\\
 &= -\frac29\csc^2 t\cot t-\frac49 \cot t +C
\end{align}

Title: How to solve $\int \csc^{4} \frac{3}{2} x \,dx$ using integration by parts

Body:
How to solve $\int \csc^{4} \frac{3}{2} x \,dx$

This is what I have so far:

$$\int \csc^{4} \frac{3}{2} x \,dx=\frac{3}{2} \int \csc^{2} x \csc^{2}x\,dx.$$

Let $u=\csc^{2}$ and $dv=\csc^{2} \,dx$. Then we have $du=-2\csc^{2}x\cot x\,dx$ and $v=-\cot x$.

\begin{align}
\frac{3}{2} \int \csc^{2} x \csc^{2} x \,dx&=\frac{3}{2}\left(-\csc^{2}x\cot x-\int-\cot x(-2\csc^{2}x\cot x\,dx)\right)\\
&= \frac{3}{2}\left(-\csc^{2}x\cot x-2\int \csc^{2}x\cot^{2}x\,dx\right)
\end{align}

At this point, I am uncertain about how to continue. Do I find the integral in the right hand side again? Or perhaps I missed something? Thank you.


问:
Simulation of orbiting bodies
I am writing a computer program to simulate orbiting bodies such as planets and stars.

I wish to have a starting point in which a number of bodies are randomly scattered around a central heavy body. Each body (except the central one) is sent off at the correct orbiting velocity but in a random direction.
I am using an x,y,z coordinate system.
For each orbiting body, I can establish the starting position and orbiting velocity.

I want each orbiting body to start moving in a random direction (but on the right plane) for which I need to calculate the initial x, y and z velocity components. This bit I cannot figure out how to do.
Does anyone have any suggestions?

答:
The task posed is to create the initial velocity as if the body under consideration was alone in the system with the fixed center of mass $M$ and has a circular orbit in a random plane. Of course this initial order will be beautifully destroyed in the first steps of the simulation.

For the randomly generated vector $\vec x_0=r\hat s$, $\|\hat s\|=1$, generate a likewise random, but normalized orthogonal vector $\hat n$. Then the operator $(\hat n\times)$ acts like a complex unit $J$, $J^2=-I$, in the plane orthogonal to $\hat n$. The circular orbit can thus be written as
$$
\vec x(t)=e^{J\omega t}\vec x_0
$$
using the matrix exponential. Now insert into the gravity equation
$$
-ω^2\vec x(t)=\ddot{\vec x}(t)=-GM\frac{\vec x(t)}{\|\vec x(t)\|^3}
\\~\\
\implies ω=\sqrt{\frac{GM}{r^3}}.
$$
This determination now can be translated into the initial velocity
$$
\dot{\vec x}(t)=\omega J\vec x(t)\implies
\color{blue}{\vec v_0=\sqrt{\frac{GM}{r^3}}(\hat n\times\vec x_0)}=\sqrt{\frac{GM}{r}}(\hat n\times\hat s)
$$
The initial speed can be read off as $\|\vec v_0\|=ωr=\sqrt{\frac{GM}{r}}$.

Title: Simulation of orbiting bodies


问:
What does this definition of a function mean?
I've never seen before the following function definition.

Image taken from Ch. 2 of the book "Principles of Eventual Consistency"

What does it try to convey in this specific form and how does it differ from the commonly used notation?

(The book is freely accesible from https://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/final-printversion-10-5-14.pdf)

答:
I’ve not seen the notation before, but it appears that:

$(A\to B)$ is the set of all functions from $A$ to $B$;

$(A\rightharpoonup B)$ is the set of partial functions from $A$ to $B$, i.e., the set of all functions whose domains are subsets of $A$ and whose ranges are subsets of $B$; and

$(A\times B)$ is the set of all relations from $A$ to $B$, so that it is simply $\wp(A\times B)$.

Thus, $(A\to B)\subseteq(A\rightharpoonup B)\subseteq(A\times B)$: every function from $A$ to $B$ is a partial function, and every partial function is a relation.

Finally, if $f:A\rightharpoonup B$ is a partial function (and I’m including the possibility that $f$ is a total function, i.e., with domain all of $A$), $a\in A$, and $b\in B$, we can define a new function $f[a\mapsto b]$ whose domain is $\{a\}\cup\operatorname{dom}(f)$:

$$f[a\mapsto b](x)=\begin{cases}
b,&\text{if }a=b\\
f(x),&\text{otherwise.}
\end{cases}$$

This function sends $a$ to $b$ but otherwise acts exactly as $f$ does.


答:
There is an algorithm for factorizing quartic polynomials with rational coefficients.

The first step is to use the rational roots theorem to determine whether or not the polynomial has rational roots. Yours has none.

The second step is to replace your quartic by its reduced version: in your case it's$$q(x)=p(x-3)=x^4-8x^2-x+12.$$Now, let $r(x)$ be the resolvent cubic of $q(x)$, which is $y^3-16y^2+16y-1$. Does it have a rational root which happens to be the square of another rational number? Yes: $1(=1^2)$. Then it follows from what is described here that$$q(x)=\left(x^2-x-4\right) \left(x^2+x-3\right)$$and that therefore$$p(x)=r(x+3)=\left(x^2+5 x+2\right) \left(x^2+7 x+9\right)$$indeed.

Title: Factoring $x^4 + 12x^3 + 46x^2 + 59x + 18$

Body:
How do I factorize the following?
$$x^4 + 12x^3 + 46x^2 + 59x + 18$$

I've tried looking for a root by trial and error to no avail.

The answer is
$$(x^2 + 5x + 2)(x^2 + 7x + 9)$$


问:
Proof Explanation: Nonempty perfect sets of $(\Bbb R, |\cdot|)$ are uncountable
The proof I have some trouble with:

Suppose $\varnothing\ne A\subset\Bbb R$ is perfect. $A$ is perfect, so $A$ is closed and every point in $A$ is a limit point of $A$. Hence, $A$ cannot be finite. $A$ is infinite. We want to show that $A$ is uncountable. Suppose $A$ is countable. Then, $A = \{x_1,x_2,...\}$ is an enumeration of $A$. Consider $x_1$, and an interval $[a_1,b_1]$ s.t. $x_1\in (a_1,b_1)$. Since $x_1$ is a limit point of $A$, $(a_1,b_1)\setminus\{x_1\}\cap A\ne \varnothing$ (in fact, intersects at infinitely many points). Now, choose $x_{n_1}\in A$, s.t. $x_{n_1}\ne x_1,x_2$ and consider $(a_2,b_2) \ni x_{n_1}$ s.t. $(a_2,b_2)\subset (a_1,b_1)$ and $\displaystyle b_2-a_2 \le \frac{b_1-a_1}{2}$ and $x_1,x_2\notin [a_2,b_2]$.

Q1. Why can we find such a subset? i.e. why can we find $(a_2,b_2)$ with the above properties?

Again, $(a_2,b_2)\setminus \{x_{n_1}\}\cap A$ has infinitely many points. Choose $x_{n_2} \ne x_1,x_2,x_3$. Consider $(a_3,b_3)\ni x_{n_2}$ s.t. $(a_3,b_3)\subset (a_2,b_2)$ and $\displaystyle b_3 - a_3 \le \frac{b_2-a_2}{2} \le \frac{b_1-a_1}{2^2}$.

The same question again, I do not understand the construction of these intervals!

Continuing this way, one obtains a nested sequence of closed and bounded intervals $\{[a_k,b_k]\}_{k=1}^\infty$ s.t. $\displaystyle b_k - a_k \le \frac{b_1-a_1}{2^{k-1}}$ and $x_{n_{k-1}} \ne x_1,x_2,\ldots,x_k$ with $x_{n_{k-1}}\in (a_k,b_k)$ s.t. $x_1,\ldots,x_k \notin [a_k,b_k]$.

By the Nested Interval Theorem, we can find $x\in \bigcap_{k=1}^\infty [a_k,b_k]$ s.t. $x\in A$. Observe $x\ne x_n, n\ge 1$ by the construction above. This is not possible as $A = \{x_n\}_{n=1}^\infty$. Therefore, $A$ is uncountable.

Q2. How is the last step concluded? Where is the contradiction?

It'd be very helpful if someone could explain the proof to me in detail, I am not able to follow especially the construction of the intervals $[a_k,b_k]$ and what the contradiction is. Other than that, I was able to understand several intermediate steps. Thank you!

答:
Answer to Q1: Consider $x_1$ which being a limit point has the property that every open neighborhood of it will have infinitely many points of $A$.In particular, you can choose $a_1,b_1$ and denote $I_1=[a_1,b_1]$ such that $I_1\cap A$ is non empty. Note that $I_1$ contains infinitely many points of $A$ so at least one of the halves of $I_1$ will have infinitely many points of $A$. WLOG let left half of $I_1$ have infinitely many points of $A$. Let's call it $I_1'$. Choose $x_{n_1}\in I_1'-\{x_1,x_2\}$. As before, choose $a_2,b_2$ such that $x_{n_1}\in [a_2,b_2]$ and $I_2=[a_2,b_2]\subset I_1'\subset I_1$. So length of $I_2=b_2-a_2\lt (b_1-a_1)/2$.

Answer to Q2: Note that $I_1\supset I_2\supset\cdots$. Now define $J_n= I_n\cap A$. Now verify that $J_n$'s are also nested closed and bounded sets. By nested intervals property, $\exists x\in A\cap (\bigcap_{k=1}^\infty [a_k,b_k])\implies x\in A$. $\exists m: x_m=x$. But $x_m\notin [a_m,b_m]=I_m $. Contradiction.

Alternatively, you may prove it like this also:

Suppose on the contrary that $A$ is countable hence $A=\{x_1,x_2,... \}$. Nested closed and bounded non empty sets in $\mathbb R$ have non empty intersection. Let's make use of that property.

Take any $r\gt 0$ and define $N_1=\{t\in\mathbb R: |t-x_1|\lt r\}$. Define $\overline N_1=\{t\in\mathbb R: |t-x_1|\le r\}$. Note that $N_1\cap P$ is non empty.

Consider an open neighborhood $N_n$ of $x_n$. $N_n$ has infinitely many points of $A$. Define $N_{n+1}\subset N_n$ such that $x_n\notin \overline N_{n+1}$ and $N_{n+1}$ contains infinitely many points of $A$. In this way construct sequence $\{\overline N_n\}$.

Define $S_n=\overline N_n\cap A$. Note that $S_n$ is closed and bounded.

The way we have constructed the sequence $N_n$ implies that $x_k\in S_k \implies x_k\notin S_{k+1}\implies $ $\cap_{n=1}S_n$ is empty which is a contradiction as nested closed and bounded non-empty sets must have non empty intersection.


问:
Iterative injective change of variable formula application question
Let $g_1:\mathbb{R}^{d_1} \rightarrow \mathbb{R}^{d_2}$ and $g_2:\mathbb{R}^{d_2} \rightarrow \mathbb{R}^{d_3}$ be injective and differentiable with $d_1 < d_2 < d_3$. Let $g = g_2 \circ g_1$ and $A \in \mathbb{R}^{d_1}$. I am interested in computing the volume of $A$:
\begin{equation}
\displaystyle \int_A dx
\end{equation}
using the change of variable formula, and derived this in two different ways, arriving at two different results which I do not think match in general. I would appreciate any help pointing out what I am doing wrong.

First way: By the change of variable formula and then applying the chain rule,
\begin{align}
\displaystyle \int_A dx & = \int_{g(A)}|\det J_g^\top(g^{-1}(z))J_g(g^{-1}(z))|^{-1/2}dz\\
& = \int_{g(A)} |\det J_{g_1}^\top(g^{-1}(z))J_{g_2}^\top(g_2^{-1}(z))J_{g_2}(g_2^{-1}(z))J_{g_1}(g^{-1}(z))|^{-1/2}dz
\end{align}
where $J_g(x)$ denotes the Jacobian matrix of $g$ evaluated at $x$, and $g^{-1}$ is well defined over $g(A)$.

Second way: Again, by the change of variable formula, but this time applying it twice,
\begin{align}
\displaystyle \int_A dx & = \int_{g_1(A)}|\det J_{g_1}^\top(g_1^{-1}(y))J_{g_1}(g_1^{-1}(y))|^{-1/2}dy \\
& = \int_{g_2(g_1(A))} |\det J_{g_1}^\top(g_1^{-1}(g_2^{-1}(z)))J_{g_1}(g_1^{-1}(g_2^{-1}(z)))|^{-1/2} |\det J_{g_2}^\top(g_2^{-1}(z))J_{g_2}(g_2^{-1}(z))|^{-1/2}dz\\
& = \int_{g(A)} |\det J_{g_1}^\top(g^{-1}(z))J_{g_1}(g^{-1}(z))|^{-1/2} |\det J_{g_2}^\top(g_2^{-1}(z))J_{g_2}(g_2^{-1}(z))|^{-1/2}dz\\
\end{align}

Now, because this holds for any $A$, it would imply the integrands are equal almost surely and thus:
\begin{equation}
|\det J_1^\top J_2^\top J_2 J_1| = |\det J_1^\top J_1||\det J_2^\top J_2|
\end{equation}
where I simplified notation for clarity. I don't think the above equation is true for arbitrary $J_1 \in \mathbb{R}^{d_2 \times d_1}$ and $J_2 \in \mathbb{R}^{d_3 \times d_2}$ (or even if $J_1$ and $J_2$ are full rank as would be the case due to injectivity). Any help is very appreciated!

答:
The problem is in the "second way", the second equal sign. The term $\text{det}J_2^TJ_2$ contains too much information, e.g. how $d_2$-dimensional volume is being deformed by map $g_2$ from ${\mathbb R}^{d_2}$ to $ {\mathbb R}^{d_3}$. But in reality we only need to check how the
$d_1$-dimensional volume in the direction of $g_1(A)$, which is a $d_1$-dimensional submanifold of $ {\mathbb R}^{d_2}$, is being deformed by $g_2$.

$\text{det}J_2^TJ_2$ brings in information like, among other things, how the geometry in the direction perpendicular to $g_1(A)$, is being deformed by $g_2$; this information is totally irrelevant for our purpose.


问:
"Surface Area" of $k$ simplex in $\mathbb{R}^{k}$?
Consider the $k+1$ vertices $(x_1,\ldots,x_{k+1})$ with $x_i\in\mathbb{R}^k,i=1,\ldots,k+1$. I know that the "volume" of the $k$-dimensional simplex  formed by these vertices is proportional to

$$\left|\det\left( \begin{array}{ccc}
1 & \ldots & 1 \\
x_1^{\top} & \ldots & x_{k+1}^{\top} \end{array} \right)\right|$$

My question is: what is the formula to compute surface "area" of the simplex formed by  $(x_1,\ldots,x_{k})$  in terms of determinant of these vertices?

Edit:

An example to make this more intuitive.

If suppose $k=3$, then I am looking for the area of the triangle with vertices $(x_1,x_2,x_3)$ (which is one of the 4 faces of the tetrahedron with vertices $(x_1,x_2,x_3,x_4)$ whose volume is proportional to:

$$\left|\det\left( \begin{array}{ccc}
1 & 1 & 1 & 1 \\
x_1^{\top} & x_2^{\top} & x_3^{\top} & x_{4}^{\top} \end{array} \right)\right|$$

)

答: The problem with the formula you're using is that it can only compute the volume of the $k$-simplex in $k$-dimensional space. There are other, more appropriate options, for example the Cayley-Menger determinant. It gives the volume of the simplex in terms of its edge lengths and thus is independent of the dimension of the space. Simply use that for computing the "volumes" of the sides of your simplex.


问:
Is Calkin algebra $B(H)/\mathcal K(H)$ isomorphic to some subalgebra of $B(H)$?
$B(H)$ is the operator algebra on separable Hilbert space. $\mathcal K(H)$ is the compact operator algebra.

I have no idea on this. Could someone please give me a hint?

答:
It is easy to show that the Calking algebra has an uncountable family of pairwise orthogonal projections, so it cannot embed in $B(H)$.

The way to show this is to construct an uncountable family$\{A_t\}$ of infinite subsets of $\mathbb N$ such that, for all $s\ne t$, the set $A_t\cap A_s$ is finite. Then, given a countable family $\{E_n\}$ of pairwise orthogonal rank-one projections in $B(H)$, one forms projections
$$
P_t=\sum_{n\in A_t}E_n.
$$
Then
$
\displaystyle P_tP_s=\sum_{n\in A_t\cap A_s}E_n
$ is finite-rank, so in the Calkin
$$
\pi(P_t)\pi(P_s)=\pi(P_tP_s)=0.
$$


问:
What is the limit of this sequence defined below?
Suppose that in a nxn grid, we choose a lattice point and took the sum of the distances from that point to every other lattice point in the grid (including the edge points) and then divide the sum by n^2. We then do the same thing so that the "center" is at another point and do this for all points in the grid  and take the average. Let a_n be defined as the average for the nxn grid multiplied by 2/n. I ran a python script and from the first 75 terms in this sequence, it looks like it's converging to 1.043 or it's just diverging very slowly. Does the sequence converge and is there a way to find an explicit formula for a_n?</p>
```<code>def center_Distance(n,m, x, y):
total = 0
for i in range(n):
    for j in range(m):
        total += ((x-i)**2 + (y-j)**2)**0.5
return total/(n*m)

li = []
mi = []
M = 75
for i in range(1, M):
#li will contain the sum of the distances for each center

li = []
    for k in range(i):
        for m in range(i):
            li.append(center_Distance(i, i, k, m))

    # find average of li and multiply by 2/i
    total = 2*sum(li)/(len(li)*i)
    print(total)
</code>```

答: This is not really an answer (especially since it doesn't contain any work of my own), but too long and unwieldy for a comment: This is essentially the average distance of two points in an $n\times n$-grid. Why it's multiplied by $2/n$ in the end, I can't say, multiplying by $1/n$ (giving a value of about $0.521$ in the limit) would reduce it to the average distance of two points in a unit square with a discrete uniform distribution. As $n\to\infty$, we can expect it to converge to the average distance of two points independently uniformly distributed in the unit square, and that is known: Average distance between two randomly chosen points in unit square (without calculus)


问:
Find the generating elements of the cyclic group
I have the following question:

Let $ (\mathbb {Z} _4, +) $ be a cyclic group, such that $ \mathbb {Z}_{4}= \{0,1,2,3\}$. Find the generating elements of $ \mathbb {Z}_4 $.

The feedback says that the answer is 1 and 3, but I don't understand how I should proceed to find that result. I thought it was just using the powers ($1 ^ 0, 1 ^ 1, 1 ^ 2$, etc.), but I can't go on from there.

答:
The group is small enough to test each element by taking successive powers; since the group is additive, this amounts to taking multiples of the elements.

In general, though, as a hint, look at the greatest common divisor of the order of the cyclic group and the representative of the equivalence class that is each element of the group.


问:
Gauss map of a minimal surface converse
It can be shown that the Gauss map of a minimal surface is conformal quite easily, but does the converse of this statement hold? Specifically, what surfaces have a conformal Gauss map?</p>

答:
HINT: The characteristic polynomial of the Weingarten (derivative of Gauss) map at a point $p$ is, of course, $t^2-2H(p)t+K(p)$. Apply the Cayley-Hamilton Theorem. (You might first observe that the Gauss map is always conformal at an umbilic point.)

Even easier, write the Weingarten map in diagonal form. When is it conformal?

Title: Gauss map of a minimal surface converse


问:
Prove: $\sum\limits_{j=1}^{2^n}\frac{1}{j} \geq 1 + \frac{n}{2}$ for all positive integers $n$.
Given the following proof.

Proof: by induction on n

Base Case: $n = 1$ is trivial. Done

Inductive Step: suppose for $n \geq 1$ we have  $\sum\limits_{j=1}^{2^n}\frac{1}{j} \geq 1 + \frac{n}{2}$. We show for $n + 1$. Observe
$$
\begin{equation} \begin{split}
(1)& \quad\sum_{j=1}^{2^{n+1}}\frac{1}{j} &= \sum_{j=1}^{2^n} \frac{1}{j} + \sum_{j=2^n + 1}^{2^{n+1}}\frac{1}{j} \\
(2)&&\geq 1 + \frac{n}{2} + \left(2^{n+1} - 2^n\right)\frac{1}{2^{n+1}} \\
&&= 1 + \frac{n}{2} + 1 - \frac{1}{2} \\
&&= 1 + \frac{n+1}{2}
\end{split} \end{equation}
$$
Done.

Two questions: How does the procedure of separating the summation of the LHS in (1) to the two summands on the RHS in (1) work? In addition, how does one evaluate $\sum\limits_{j=2^n + 1}^{2^{n+1}} \frac{1}{j}$ to be the third summand in (2)?

NOTE: The summations limits in second summand of the RHS of (1) may be wrong. I could not remember fully, what it was.

In addition, I am NOT asking why we separate in (1) to prove. I am asking how the algebraic manipulations works in going from LHS of (1) to RHS of (1).

答:
As the number of summations is finite, you can break the summation from any point you desire. Hence, it is straightforward to reach RHS (1) from LHS (1).

For the second part of your question it is easy to replace $\frac{1}{j}$ with minimum value in the range of summation $\frac{1}{2^{n+1}}$. Therefore,

$$\sum_{j = 2^n + 1}^{2^{n+1}}\frac{1}{j} > \frac{1}{2^{n+1}}\sum_{j = 2^n + 1}^{2^{n+1}}1 = (2^{n+1} - (2^n + 1) + 1)\frac{1}{2^{n+1}} = (2^{n+1} - 2^n)\frac{1}{2^{n+1}}$$.


问:
Norm bounded sets in $C^*$-algebras/von Neumann algebras
I was reading about the weak operator topology in wiki, and I saw that: Norm-bounded sets in $B(H)$ are pre-compact in WOT.

I was wondering: Is there an analogous statement for von Neumann algebras? $C^*$-algebras?

答:
The statement would have been that norm bounded sets are pre-compact, as being  bounded does not guarantee that it is closed.

Other than that, yes. You would usually see a von Neumann algebra $M$ as embedded in $B(H)$. So a wot-closed bounded set is a closed subset of a compact set (a closed ball in $B(H)$) and thus compact.

In the case of a C$^*$-algebra, you would have to embed it in some $B(H)$ to even have a wot topolgy. In that case, bounded sets would be pre-compact.


问:
Linear Dependence / Independence of Matrices
If $\{A_1, A_2, ..., A_k\}$ is a linearly independent subset of $M_{n \times n}(F)$, is $\{A_1^n, A_2^n, ..., A_k^n\}$ also linearly independent for $n > 1, n \in  \mathbb{Z^+}$?

Thanks!

答: No, let $A_1, ..., A_k$ be $k$ many independent matrices with zero entries on and below diagonal... Then all $A_j^n=0$.


问:
Prove that if $G$ is a finite simple group containing a subgroup $H$ of order $27$, then $|G : H| \ge 9.$

Prove that if $G$ is a finite simple group containing a subgroup $H$ of order $27$, then $|G : H| \ge 9.$

I have so far been solving this by looking at $|G:N|=1, 2, 3...$ etc., all the way up to $8$ and showing that each case is not possible. This is a method that I believe will work in solving the problem if I continue using it (I have shown it does not work for $1, 2$, and $3$).

However, I was wondering if there was a more efficient way to go about a problem like this.

答: $G$ acts by left multiplication on the left quotient $G/H$. If $G$ is simple, then $G$ embeds into $S_{[G:H]}$ and hence $|G|$ divides $[G:H]!$. But $|G|=27[G:H]$, so $|G|$ divides $[G:H]!$ if and only if $3^3 \mid ([G:H]-1)!$ if and only if $[G:H]\ge 10$. Therefore, a stronger condition holds, actually.


答:
In common cases it is possible to swap the integral and the derivative so that:
$$\dfrac{\partial g(y)}{\partial y} = \int \dfrac{\partial f(y, t)}{\partial y}dt$$

For this equality to be true you need an inversion result, a sufficient condition is that the partial derivative of $f$ is continuous w.r.t. both $y$ and $t$ and bounded uniformly w.r.t. $y$ on the domain of integration. The last condition means that there exist a function $h$ on the domain of $t$ such that for all $x, t$: $\|\dfrac{\partial f(y,t)}{\partial y}\| \leq h(t)$.
See http://en.wikipedia.org/wiki/Leibniz_integral_rule.

Title: derivative of a numerical expression

Body: I have a function $$g(y) = \int{f(y, t)dt}$$ and I am currently able to evaluate the integral numerically (doesn't seem like it is possible to find an analytical solution...), which I have also plotted for various values of $g(y)$. All good so far. However, I am also interested in the quantity $$\frac{dg(y)}{dy}$$ Is there a way to obtain this derivative?</p>


问:
Roots of a certain sixth order polynomial
I am looking for the roots (or basically any information regarding them) of the sixth order polynomial
$$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$
for positive, real constants $a,b$.
Since $p(0)=-b^2<0$ and $\lim_{x\to\pm\infty}p(x)=+\infty$ we have at least two real roots $x_0,x_1$. Moreover, plotting the function for several parameters suggests, that the remaining four roots are non-real (i.e. two pairs of complex conjugate solutions).

Unfortunately this is as far as I got. I am especially interested in the complex solutions and their behavior for large parameters $a,b$.
Any idea on how to tackle this problem is very welcome!

答:
$\color{brown}{\textbf{Parametrization.}}$

Searching the given polynomial's representation in the form of
$$P(a,b,x) =(a+2p)(x^6+t^6)+p(x^2+t^2)^3-3px^2(x^2+t^2)^2 + 2bx^3,\tag1$$
easily to get
$$ax^6 -3pt^2 x^4 + 2b x^3 + (3p+a) t^6=ax^6+(a+1)x^4+2bx^3-b^2,$$

$$t^2=-\dfrac{a+1}{3p},\tag2$$
$$\quad (3p+a)(a+1)^3=27b^2p^3,\quad p^3=\dfrac{c^2}4(3p+a),$$
where
$$c= \dfrac2b\sqrt{\dfrac{(a+1)^3}{27}}.\tag3$$

Then
$$4\left(\dfrac pc\right)^3-3\,\dfrac pc = \dfrac ac.$$
Taking in account identities
$$4s^3-3s = \cos(3\arccos s),$$
one can obtain
$$p=c\cos\left(\dfrac23k\pi+\dfrac13\arccos\dfrac ac\right),\tag4$$
where $\;k\in\{-1,0,1\}\;$ provides the real positive result.

$\color{brown}{\textbf{Substitution.}}$

Assume
$$y=x+\dfrac {t^2}x,\tag5$$
then
$$x^3+\dfrac{t^6}{x^3}=\left(x+\dfrac {t^2}x\right)^3-3t^2\left(x+\dfrac {t^2}x\right)=y^3-3t^2y,$$
Therefore, the given polynomial $(1)$ presented in the form of
$$P(a,b,x) = x^3Q\left(a,b,x-\dfrac{a+1}{3px}\right),\tag6$$
where
$$Q(a,b,y) = (a+3p)y^3-3py^2+(a+2p)\dfrac{a+1}p y+2b.\tag7$$
Therefore, the given polynomial is reduced the cubic polynomial of the general form with the known constant cofficients.

$\color{brown}{\textbf{Example.}}$

Let $\;a=7,b=5,\;$ then $\;c\approx1.741\,859\,372\,646,\;
p\approx2.172\,466\,644\,575.\;$

Then WA gives $\;y\approx-0.227\,483\,364\,391,\;$ with the right root
$\;x_1\approx 1.000\,000\,000\,000.\;$

$\color{brown}{\textbf{Conclusions.}}$

From $(2)-(7)$ should the next:

The roots of the given polynomial are solvable in the closed form via elementary functions;

Identity $\;P\left(a,b,-\dfrac{a+1}{3px}\right)=-\dfrac{(a+1)^6}{(3px)^6}P\left(a,b,x\right)\;$ defines the correspondence between the values on the positive and negative arguments.


问:
Boolean Algebra: How does $\bar A\bar BC+\bar A\bar C\bar D+A\bar CD+\bar AB\bar C$ become $\bar A\bar BC + \bar A\bar C\bar D+A\bar CD+B\bar CD$?
I'm trying to understand one of the steps taken during the process of getting a cnf in Boolean algebra but I just cant understand what is happening here.

$$\bar A \bar B C + \bar A \bar C \bar D +  A \bar C  D + \bar A  B  \bar C$$
$$\bar A \bar B C + \bar A \bar C \bar D +  A \bar C  D + B  \bar C D$$

It seems like they just exchange the !A for D , but I cannot understand which of the Boolean algebra laws they used.

Could someone help me understand it ?

答:
$$
\bar{A}B\bar{C} = \bar{A}B\bar{C}D + \bar{A}B\bar{C}\bar{D}
$$

the second piece $\bar{A}B\bar{C}\bar{D}$ is already a subset of $\bar{A}\bar{C}\bar{D}$ so you get

$$
\begin{align}
\bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + \bar{A}{B}\bar{C} &= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + (\bar{A}B\bar{C}D + \bar{A}B\bar{C}\bar{D}) \\
&= \bar{A}\bar{B}{C} + (\bar{A}\bar{C}\bar{D} + \bar{A}B\bar{C}\bar{D}) + {A}\bar{C}{D} + \bar{A}B\bar{C}D \\
&= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + \bar{A}B\bar{C}D
\end{align}
$$

Now, ${A}B\bar{C}D$ is a subset of ${A}\bar{C}D$ so you can split it out:

$$
\begin{align}
\bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + \bar{A}{B}\bar{C} &= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + (\bar{A}B\bar{C}D + \bar{A}B\bar{C}\bar{D}) \\
&= \bar{A}\bar{B}{C} + (\bar{A}\bar{C}\bar{D} + \bar{A}B\bar{C}\bar{D}) + {A}\bar{C}{D} + \bar{A}B\bar{C}D \\
&= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + \bar{A}B\bar{C}D \\
&= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + ({A}\bar{C}{D} + {A}B\bar{C}D) + \bar{A}B\bar{C}D \\
&= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + ({A}B\bar{C}D + \bar{A}B\bar{C}D) \\
&= \bar{A}\bar{B}{C} + \bar{A}\bar{C}\bar{D} + {A}\bar{C}{D} + B\bar{C}D
\end{align}
$$


问:
Can someone explain this integration trick for log-sine integrals?
I was working on this rather challenging log-sine integral:

$$
\int_{0}^{2\pi}x^{2}\ln^{2}\left(2\sin\left(x \over 2\right)\right)\,{\rm d}x = {13\pi^{5} \over 45}
$$

The upper limit is a waiver from the norm of $\frac{\pi}{2}$. Anyway, when integrating log-sin integrals one can often times use the famous identity

$$\displaystyle -\ln(2\sin(x/2))=\sum_{k=1}^{\infty}\frac{\cos(kx)}{k}$$

by switching sum and integral, integrating, then evaluating the sums. However, this is only valid when $x\neq \pm \pi,\pm2\pi,\ldots$. So, just for the heck of it, I decided to do something I thought would not be viable but have done it anyway. I integrated $$\displaystyle \int_{0}^{2\pi}\frac{x^{2}\cos^{2}(kx)}{k^{2}},$$ evaluated the resulting sums and arrived at $\displaystyle \frac{41{\pi}^{5}}{180}$.

Now, take the integral $$\displaystyle \int_{0}^{\pi}x^{2}\ln^{2}(2\cos(x/2))dx=\frac{11{\pi}^{5}}{180}.$$ This one can be done by using Cauchy's cosine formula, differentiating, and so on. Anyway, note that when I add the two results, the correct result for the integral at hand is obtained.

$$\displaystyle \frac{41{\pi}^{5}}{180}+\frac{11{\pi}^{5}}{180}=\frac{13{\pi}^{5}}{45}.$$

By quandary/query is why does this happen to work?. A fluke? Does that cosine sum represent something that when evaluated and added to the above log-cos integral happens to be equivalent to the log-sine integral in question?.

See what I am trying to explain?. I done something I knew I was not supposed to, but it happened to work out. If viable, what would $\displaystyle \int_{0}^{2\pi}\frac{x^{2}\cos^{2}(kx)}{k^{2}}$ represent such that when it is added to $\displaystyle \int_{0}^{\pi}x^{2}\ln^{2}(2\cos(x/2))dx=\frac{11{\pi}^{5}}{180}$ equals $\displaystyle \int_{0}^{2\pi}x^{2}\ln^{2}(2\sin(x/2))dx=\frac{13{\pi}^{5}}{45}$.

I knew I could not just square the cosine and multiply by $x^2$. But,sometimes it can be done due to Parseval, i.e., $\displaystyle \ln^{2}(2\sin(x/2))=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(nx)\cos(kx)}{nk}$. Then, because $\displaystyle \int_{0}^{\pi}\cos(nx)\cos(kx)dx=0, \;\ n\neq k$, one can integrate $\displaystyle \int_{0}^{\pi}\frac{\cos^{2}(kx)}{k^{2}}$, then sum.

This works for $2\pi$ as well, but that $x^2$ term tends to throw a wrench in things But I do not think I can do this in my case because of the $x^2$ term.  Or can I?. I was surprised when I saw this. I thought perhaps I stumbled onto a cool way to evaluate this, but I am not entirely sure what I have.

Or, if anyone has their own clever method?.

EDIT:  I managed to evaluate this integral by considering the identity

$$\ln\left(2\sin\left(\frac{x}{2}\right)\right)=\ln(1-e^{ix})+\frac{i}{2}(\pi -x)$$

答:
Preliminaries

Looking at the real part of $\log\left(1-e^{ix}\right)$, we get
$$
\log\left(2\sin\left(\frac x2\right)\right)=-\sum_{k=1}^\infty\frac{\cos(kx)}{k}\tag1
$$
Integrating by parts twice, we get
$$
\int_0^{2\pi}x^2\cos(kx)\,\mathrm{d}x=\left\{\begin{array}{}\frac{8\pi^3}3&\text{if }k=0\\\frac{4\pi}{k^2}&\text{if }k\ne0\end{array}\right.\tag2
$$

In the derivations below, we use the values of $\zeta(2)$ and $\zeta(4)$ computed at the end of this answer.

Start to evaluate one sum, then get another:
$$
\begin{align}
\color{#C00}{\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{(k+j)^2kj}}
&=\sum_{j=1}^\infty\sum_{k=1}^\infty\left(\frac1{kj}-\frac1{k(k+j)}\right)\left(\frac1{kj}-\frac1{j(k+j)}\right)\tag{3a}\\[3pt]
&=\color{#00F}{\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\frac1{j^2}}-2\color{#090}{\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{k^2(k+j)j}}+\color{#C00}{\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{(k+j)^2kj}}\tag{3b}\\
\color{#090}{\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{k^2(k+j)j}}
&=\frac12\color{#00F}{\zeta(2)^2}\tag{3c}\\[3pt]
&=\frac{\pi^4}{72}\tag{3d}
\end{align}
$$
Explanation:

$\text{(3a)}$: $\frac1{kj}-\frac1{k(k+j)}=\frac1{j(j+k)}$ and $\frac1{kj}-\frac1{j(k+j)}=\frac1{k(j+k)}$

$\text{(3b)}$: expand the products and separate the sums

$\phantom{\text{(3b):}}$ note that $\sum\limits_{j=1}^\infty\sum\limits_{k=1}^\infty\frac1{k^2(k+j)j}=\sum\limits_{j=1}^\infty\sum\limits_{k=1}^\infty\frac1{k(k+j)j^2}$

$\text{(3c)}$: cancel the red sums, move the green sum to the left side, and divide by $2$

$\text{(3d)}$: simplify

Evaluate the sum we started before (using the other sum we got before):
$$
\begin{align}
\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{(k+j)^2kj}
&=\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac1{k^2(k-j)j}\tag{4a}\\
&=\sum_{k=1}^\infty\sum_{j=1}^{k-1}\frac1{k^2(k-j)j}\tag{4b}\\
&=\sum_{k=1}^\infty\sum_{j=1}^{k-1}\frac1{k^3}\left(\frac1{k-j}+\frac1j\right)\tag{4c}\\
&=2\sum_{k=1}^\infty\left(\frac{H_k}{k^3}-\frac1{k^4}\right)\tag{4d}\\
&=2\sum_{k=1}^\infty\sum_{j=1}^\infty\frac1{k^3}\left(\frac1j-\frac1{j+k}\right)-2\zeta(4)\tag{4e}\\
&=2\sum_{k=1}^\infty\sum_{j=1}^\infty\frac1{k^2(k+j)j}-2\zeta(4)\tag{4f}\\
&=\frac{\pi^4}{36}-\frac{\pi^4}{45}\tag{4g}\\[6pt]
&=\frac{\pi^4}{180}\tag{4h}
\end{align}
$$
Explanation:

$\text{(4a)}$: substitute $k\mapsto k-j$

$\text{(4b)}$: swap order of summation

$\phantom{\text{(4b):}}$ (we include $k=1$ since the inner sum is then $0$)

$\text{(4c)}$: partial fractions

$\text{(4d)}$: $\sum\limits_{j=1}^{k-1}\frac1{k-j}=\sum\limits_{j=1}^{k-1}\frac1j=H_k-\frac1k$

$\text{(4e)}$: $H_k=\sum\limits_{j=1}^\infty\left(\frac1j-\frac1{j+k}\right)$

$\text{(4f)}$: simplify the summand

$\text{(4g)}$: apply $(3)$

$\text{(4h)}$: simplify

Putting The Preliminaries To Work
$$
\begin{align}
&\int_0^{2\pi}x^2\log\!\left(2\sin\left(\frac{x}2\right)\right)^2\,\mathrm{d}x\\
&=\int_0^{2\pi}x^2\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{\cos(kx)}{k}\frac{\cos(jx)}{j}\,\mathrm{d}x\tag{5a}\\
&=\int_0^{2\pi}x^2\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{\color{#C00}{\cos((k+j)x)}+\color{#090}{\cos((k-j)x)}}{2kj}\,\mathrm{d}x\tag{5b}\\
&=\color{#C00}{2\pi\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{(k+j)^2kj}}+\color{#090}{4\pi\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac1{(k-j)^2kj}+\frac{4\pi^3}3\sum_{k=1}^\infty\frac1{k^2}}\tag{5c}\\
&=2\pi\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{(k+j)^2kj}+4\pi\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{k^2(k+j)j}+\frac{4\pi^3}3\zeta(2)\tag{5d}\\
&=\frac{\pi^5}{90}+\frac{\pi^5}{18}+\frac{2\pi^5}9\tag{5e}\\[6pt]
&=\frac{13\pi^5}{45}\tag{5f}
\end{align}
$$
Explanation:

$\text{(5a)}$: apply $(1)$

$\text{(5b)}$: $\cos(a)\cos(b)=\frac{\cos(a+b)+\cos(a-b)}2$

$\text{(5c)}$: apply $(2)$

$\text{(5d)}$: substitute $k\mapsto k+j$ in the middle sum

$\text{(5e)}$: apply $(3)$ and $(4)$

$\text{(5f)}$: simplify


问:
Question on The multiplicative magic square
Consider this magic square:

$$\begin{array} {|r|r|}\hline a & b & c \\ \hline d & e & f \\ \hline j & h & i \\ \hline  \end{array} $$
Where $a,b,c,d,e,f,j,h,i\in \mathbb N^*$ and the $\gcd$ of all nine elements is $1$.
This magic square is A multiplicative one such that the product in each row, column and diagonal is equal to $P$

The first question: Prove that $e^3=P$

The second question: construct a multiplicative magic square with the divisors of $100$

My attempt:

It’s not actually an attempt, it is just an observation, since the $\gcd(a,b,c,d,e,f,j,h,i)=1$ And $$aei=cej=beh=def=P$$
$e\in\mathbb N^* \implies e\geq1$, so we can divde by it:
$$ai=cj=bh=df$$
Wich mean that $a\mid cj,bh,df$ And $i\mid cj,bh,df$ and so on..., but what’s next?

答:
$$e^3=\frac{aei\cdot beh\cdot ceg}{abc\cdot ghi}=\frac{P^3}{P^2}=P$$

The knight move strategy seems to work just as in the additive case:

$$\begin{matrix}50 &1&20\\4&10& 25\\5&100 &2\end{matrix}$$


问:
How to prove the non-existence of solution of a contour problem with ordinary differential equation methods?
Let
$$\begin{cases}
-y^{''}(x)-y(x)= \sin(x), \forall x \in [0, \pi] \\
y(0)=y(\pi)=0
\end{cases}\label{eq1}\tag{1}$$

Prove that it has no solution.

It is a non-homogeneous linear ordinary differential equation of the second order, so the general solution can be written as: $ y (x) = y_ {h} (x) + y_ {p} (x) $ where

$y_{h} (x) $ is the solution of the homogeneous equation $ -y^{''} - y = 0 $, which is $ y_{h} (x) = c_{1} \cos (x) + c_{2} \sin (x) $.

$ y_{p} (x) $ is the particular solution that satisfies the non-homogeneous equation $ -y^{''}(x) -y(x) = \sin (x) $, then we assume it as $y_{p} (x) = A \cos (x) + B \sin (x) $, deriving and substituting we get that: $ -y_{p}^{''} (x) -y_{p}(x) = \sin(x) \longrightarrow A \cos(x) + B \sin(x) - A \cos(x) - B \sin (x) = \sin(x) $,

i.e. $ 0 = \sin (x) $
but this is not true for all $ x \in [0, \pi] $.

Would this prove that $ (1) $ has no solution?

And if not, can you tell me how to test it?

答:
The method of undetermined coefficients demands to add a factor $x$ in the case of resonance,
$$
y_p(x)=x(A\cos(x)+B\sin(x))
$$
with
$$
-y_p''(x)-y_p(x)=2A\sin(x)-2B\cos(x)\implies A=\frac12, ~~B=0.
$$
In total with the left side boundary condition
$$
y(x)=\frac12x\cos(x)+C\sin(x),
$$
but at the right boundary $0=y(\pi)=\frac{\pi}2\cos(\pi)=-\frac{\pi}2$ is impossible to satisfy.


问:
How do you find the maximum value of $r$ in a polar function?
I have $\, r=\cos\alpha +\sin2\alpha,\quad 0\le\alpha\le\frac{\pi}{2}.$

Do you then find $\dfrac{dr}{d\alpha}$ and let that $=0$ ?

I am after just a few set of instructions.

答:
We have to investigate
$$r(\alpha):=\cos\alpha+\sin(2\alpha)\qquad(0\leq\alpha\leq{\pi\over2})\ .$$
We note in passing that $r(0)=1$, $\ r({\pi\over2})=0$ and compute
$$r'(\alpha)=-\sin\alpha +2\cos(2\alpha)=-(4u^2+u-2)\ ,$$
where we have put $\sin\alpha=:u$. Solving $r'(\alpha)=0$ gives
$$u_\pm={-1\pm\sqrt{33}\over 8}\ ,$$
and since a usable $u$ has to lie in the interval $[0,1]$ we are left with
$$\sin\alpha_0=u_+={\sqrt{33}-1\over 8}\doteq0.593\ ,$$
which leads to $$\cos\alpha_0=\sqrt{1-u_+^2}={\sqrt{30+2\sqrt{33}}\over8}\ .$$
In this way we obtain
$$r(\alpha_0)=\cos\alpha_0(1+2\sin\alpha_0)={\sqrt{30+2\sqrt{33}}\over8}{\sqrt{33}+3\over 4}\doteq1.760\ .$$
Since $r(\alpha_0)$ is greater than both $r(0)$ and $r({\pi\over2})$ it follows that the obtained value $r(\alpha_0)\doteq1.760$ is the maximal value for $r$ in the given interval.

It is not necessary to compute second derivatives here: The extreme values of $r$ are among $r(0)$, $\>r({\pi\over2})$, and the values of $r$ at the zeros of $r'$ in the interval $\ \bigl]0,{\pi\over2}\bigr[\ $.


问:
Showing $1+\cos(\theta)+\dots+\cos (n\theta)=\frac12+\frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$

Suppose that $n \geq 1$ is a natural number.

(i) Show that for $z$ in $\mathbb{C}$, provided $z \neq 1$,

$$ \frac{1-z^{n+1}}{1-z} = 1+z+...+z^n.$$

(ii) Using de Moivre’s Theorem, part (b)(i) and standard trigonometric identities, show that for $\theta$ in $(0,2\pi)$,

$$ 1+\cos(\theta) + \dots + \cos (n\theta) = \frac{1}{2} + \frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$$

I have set $z=a+bi$ and subbed this into the top equation but dont seem to be getting anwhere with this problem. Any guidance will be great!

答:
From part (i), for $z\in\mathbb C\setminus\{1\},$
$$1+z+...+z^n=\frac{1-z^{n+1}}{1-z}.$$

Let $z=e^{i\theta}$ and $\theta\in(0,2\pi).$ Then
$$1+e^{i\theta}+...+e^{in\theta} = \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}\\
=\frac{e^{i\frac{n+1}2\theta}\left[e^{-i\frac{n+1}2\theta}-e^{i\frac{n+1}2\theta}\right]}{e^{i\frac\theta2}\left[e^{-i\frac\theta2}-e^{i\frac\theta2}\right]}\\
\color{#C00}{\overset1=}e^{i\frac{n}2\theta}\frac{-2i\sin\left(\frac{n+1}2\theta\right)}{-2i\sin\left(\frac\theta2\right)}\\
=\frac{\sin\left(\frac{n+1}2\theta\right)}{\sin\left(\frac\theta2\right)}e^{i\frac{n}2\theta}.$$

By Euler's formula and equating real parts,
$$ 1+\cos(\theta) + \dots + \cos (n\theta) = \frac{\sin\left(\frac{n+1}2\theta\right)}{\sin\left(\frac\theta2\right)}\cos\left(\frac{n}2\theta\right) \\
\color{#C00}{\overset2=} \frac1{\sin\left(\frac\theta2\right)}{\frac12\left[\sin\left(\left(n+\frac12\right)\theta\right)+\sin\left(\frac\theta2\right)\right]} \\
= \frac{1}{2} + \frac{\sin\left(\left(n+\frac12\right)\theta\right)}{2\sin\left(\frac\theta2\right)}.\\$$

1 using $e^{i\alpha}-e^{-i\alpha}=2i\sin(\alpha)$

2 using one of the product-to-sum trigonometric identities


问:
Solving $x^5 - 2x^3y^2 + xy^4 = y^2$

Find all integer solutions to the equation $$x^5 - 2x^3y^2 + xy^4 = y^2.$$

I wasn't quite sure how to start on this problem, as factoring out $y^2$ would likely give me a headache and mean that I would have to deal with nasty fractions. I also tried subtracting, but that didn't get me anywhere either.

答:
$$x^5 - 2x^3y^2 + xy^4 = x(x^2-y^2)^2 = y^2$$

As pointed out in the comments, $x^2 - y^2$ must be a factor of $y$. Hence both $x \pm y$ are factors of $y$. Write

$$y = m(x-y), y = n(x-y)$$

where $m,n$ are integers. Then if $y \ne 0$,

$$ny = mn(x-y), my = mn(x+y)$$

$$(m-n)y = 2mny \leadsto -1 = 4mn - 2m + 2n- 1 = (2m + 1)(2n-1)$$

Thus $\{2m+1, 2n-1\} = \{\pm1\}$, i.e. $m=n=0$ or $m=-1, n=1$.

The first possibility gives $y=0$, which leads to $x=0$.

The second possibility gives $x=0$, which leads to $y=0$.

Thus $(0,0)$ is the only solution.

Title: Solving $x^5 - 2x^3y^2 + xy^4 = y^2$


问:
About the connection of $L^2$-convergence and convergence in distribution.
Let $(T_{1,n})_{n\in\mathbb{N}}$ and $(T_{2,n})_{n\in\mathbb{N}}$ two sequences of real valued random variables in $L^2(\Omega,\mathbb{A},P)$. Suppose that

$\|T_{1,n}\|_{L^2(\Omega)}\xrightarrow{n\rightarrow\infty} 1$,

$\|T_{2,n}\|_{L^2(\Omega)}\xrightarrow{n\rightarrow\infty} 1$ and

$\|T_{1,n}-T_{2,n}\|_{L^2(\Omega)}\xrightarrow{n\rightarrow\infty} 0$.

Suppose further that $T\in L^2(\Omega,\mathbb{A},P)$ with continuous distribution function and $T_{1,n}\xrightarrow{d} T$ (convergence in distribution).

Does it hold that $T_{2,n}\xrightarrow{d} T$?

答:
We actually just need that $T_{1,n}-T_{2,n}$ converges to $0$ in probability. We use portmanteau theorem.

Let $f\colon\Bbb R\to\Bbb R$ be a uniformly continuous bounded function. Then:
$$|E(f(T_{2,n}))-E(f(T))|\leqslant |E(f(T_{2,n}))-E(f(T_{1,n}))|+|E(f(T_{1,n}))-E(f(T))|,$$
and if $\varepsilon$ is fixed, and $\delta$ the corresponding number in the definition of uniform continuity,
\begin{align}|E(f(T_{2,n}))-E(f(T_{1,n}))|&\leqslant\int_{\{|T_{1,n}-T_{2,n}|>\delta\}}|E(f(T_{2,n}))-E(f(T_{1,n}))|d\mu\\
&+\int_{\{|T_{1,n}-T_{2,n}|\leqslant\delta\}}|E(f(T_{2,n}))-E(f(T_{1,n}))|d\mu\\
&\leqslant 2\sup |f|\cdot \mu\{|T_{1,n}-T_{2,n}|>\delta\}+\varepsilon.
\end{align}


问:
Limits of $f: E \to \mathbb R$
Let $f ∶ E \subset \mathbb R^n$ and $x_0 \in \mathbb R^n$. Show that $$\lim_{x \to x_0}f(x) = l$$
$\iff$ $\exists R > 0$ and a function $g ∶ ]0, R[ \to \mathbb R^+$ such that $\lim_{r \to 0^+}g(r)=0$ and for all $x \in B(x_0,R) \cap E/ \{x_0\}$, $|f(x) − l| \leq g(\lVert x − x_0
\rVert)$.

For the $\Longrightarrow$ implication, I have $\forall \epsilon > 0, \exists \delta > 0$ such that $0 < \lVert x- x_0\rVert < \delta \implies \lvert f(x)-l \rvert < \epsilon$.
I took $r$ to be defined as $r:= \lVert x-x_0 \rVert$ so that when $r$ goes to $0$. But then I don't really see what can be my function $g$ (I was given the hint that it should involve a sup so I thought of $g(r) := (\sup {\lvert f(x) -l\rvert}, x \in B(x_0,r) \cap E/\{x_0\})$ but I don't really understand why then what would be R ?

For the $\Longleftarrow$ implication, I take $r:= \lVert x-x_0 \rVert$ and I have $\forall \epsilon > 0, \exists \delta > 0$ such that $0 < r < \delta \implies \lvert f(x)-l \rvert < g(r) <\epsilon$ then $\lim_{x \to x_0}f(x) = l$.

Can someone help me for the first implication ?

答:
I am thinking that since $E$ is an open neighborhood around $x_0$ (not necessarily all of $\mathbb R$), take $R$ to be smaller than the radius of any open ball around $x_0$ contained in $E$ to guarantee $g$ is well-defined. This should ensure $f$ is bounded on the ball.

If you take $R$ to be too large, $\sup$ may go to infinity. For example consider $f(x)=1/x, x_0=1, R=2$ and see what happens with your $g$.

Title: Limits of $f: E \to \mathbb R$


问:
simplex volume in terms of facet areas
It is known since ancient times how the volume of triangle can be expressed in terms of its edge lengths (Heron's formula). Is there a generalization for simplices that specifies the volume of an $n$-dimensional simplex in terms of the areas of its facets?

(I'm not looking for the Cayley-Menger determinant which gives the $n$-volume in terms of all edge lengths.)

答:
Promoting comments to an answer, as requested:

Even in three dimensions, there's a problem: a tetrahedron admits six degrees of freedom, but faces account for only four. My personal "hedronometric" research introduces the notion of three pseudo-faces to make up the deficit. (There's a relation between face and pseudo-face areas that reduces the degrees of freedom to six.) With these, volume can be uniquely determined.

See my answer to the question "Does knowing the surface area of all faces uniquely determine a tetrahedron?".

I'll echo here that my note "Heron-like Results for Tetrahedral Volume" (PDF link via daylateanddollarshort.com) considers this topic at length. You don't have to take my own word for it, though; as mentioned in the note (as Theorem 7), Gerber proved in 1975 that a tetrahedron whose opposite edges are orthogonal maximizes volume for a given set of face-areas. We wouldn't need that result if face-areas determined volume uniquely.

Title: simplex volume in terms of facet areas


问:
The "best" simple and accurate approximation for $x - \cos(x) = w$

the given equation is $x-\cos(x) = w$

given some $w$ values $-100, -1.9827359827356, 50, 75.2982735, 100\ldots$ maybe many more between, say $-10000\ldots 10000$

Please find some

a) simple !!
b) enough accurate

approximation to find $x$

I tried

$a_1  \sin{(z_1  w + z_2)} + a_2  \sin{(z_3  w + z_4)} + a_3 \sin{(z_5  w + z_6)} ... + w + constant$

and found some good but it was something like at least $9 \sin$ things to get a good approximation.

Can you find more simple still somewhat accurate approx equation for this ?

答:
For convenience, I am treating $y=x-\sin(x)$, which is just a translate of the original equation.

This function is odd and can be repeated by translation, so it is enough to solve in $[-\frac\pi2,\frac\pi2]$. Inspired by its Taylor development, we look at the graph of

$$\sqrt[3]{x-\sin(x)}$$ which is very close to a straight line. Hence the linear approximation

$$\sqrt[3]y\approx\frac2\pi\sqrt[3]{\frac\pi2-1}\,x.$$


问:
How to interpret the domain and codomain of the function?
Suppose, I take for $n≥1$ we consider $\cos(nx)$ : $[0,1] \to \Bbb R $

I understand that domain is restricted between $0$ and $1$. Codomain being  $\Bbb R$.

So, is it valid to say $|\cos(nx)| \leq 1 = -1≤\cos(nx)≤1$ . I think this should be right as the range $\in$ codomain which is not restricted, so $-1≤y≤1$. Another way is to visualize a semicircle going through I and IV Quadrant (vertically)

or the domain is restricted $[0,1]$. So, $0≤\cos(nx)≤1$.

also, So every $x$ value from the domain is giving a $y$ from the codomain ( which has to between $-1$ and $+1$ unit ciricle understanding)?

I am still a bit confused. Please if you guys could clarify

答: Whatever value $nx$ may take, $\cos(nx)$ always lies between $-1$ and $1$. In your question, the domain of definition is restricted which is $[0,1]$. But your $f(x)$ can go negative for suitable choice of $n$ and $x$, though the left end of the range interval depends on the value of $n$. for every $n\ge 4$, the range set is $[-1,1]$ because for such $n$, you can always find $x=\pi/n$ which is less than $1$. Also $[-1,1]\subset\Bbb{R}$ and this is an into mapping.


问:
Geometric transformations in Olympiad problems - Putnam 2001, A-4
I have been reading Pasumarty's article A Fine Use of Transformations in Mathematical Reflections 4 (2016).

The first part of the article deals with question A-4 on Putnam 2001:

Triangle $ABC$ has an area $1$. Points $E$, $F$, $G$ lie, respectively, on $BC$, $CA$,
$AB$ such that $AE$ bisects $BF$ at point $R$, $BF$
bisects $CG$ at point $S$, and $CG$ bisects $AE$ at point $T$. Find the area of the
triangle $RST$.

I have already read the answer to this question, but I don't understand the following sentence in Pasumarty's article:

By symmetry in the equilateral case, we can see that $\dfrac{AG}{AB} = \dfrac{BE}{BC} = \dfrac{CF}{CA}  = r$.

How can I formally prove the equality of these ratios if $ABC$ is an equilateral triangle? I tried to consider the clockwise (or counterclockwise) rotation of $120^{\circ}$ about the center of the triangle, that maps $ABC$ to itself, but I can't prove that the image of $AG$ is $BE$, the image of $BE$ is $CF$ and the image of $CF$ is $AG$, or something similar.

答:
First, look at just one pair of lines with one condition: $G$ is a point on line $AB$, and $BF$ bisects $CG$ at point $S$.

If $\frac{AG}{AB} = r$ and $\frac{CF}{CA} = s$, we can find a relationship between $r$ and $s$. For example, using areas: since $S$ bisects $CG$, $[BFG] = [BFC]$. We have $[BFC] = s \cdot [ABC]$, and $$[BFG] = (1-r) \cdot [BFA] = (1-r) \cdot (1-s) \cdot [ABC],$$ so $s = (1-r)(1-s)$, which is easiest to write as $r = 1- \frac{s}{1-s} = \frac{1-2s}{1-s}$.

The actual algebraic relationship is not super important to this step, but I derive it to give a rigorous proof of an intuitive fact: as $s$ increases, $r$ decreases, and vice versa.

The same relationship must hold between any two of the three ratios, by an identical proof.

Now suppose that (for a fixed equilateral $\triangle ABC$) there are multiple possible ways to draw $AE, BF, CG$ that satisfy the condition. Choose one with the maximum possible value of $\frac{AG}{AB}$. Then it must have the minimum possible value of $\frac{CF}{CA}$. Then it must have the maximum possible value of $\frac{BE}{BC}$, from which we deduce that actually it has the minimum possible value of $\frac{AG}{AB}$. In other words, only one value of $\frac{AG}{AB}$, and similarly only one value of $\frac{CF}{CA}$ and $\frac{BE}{BC}$, is possible.

Having shown uniqueness, we are finally ready to look at what happens when we apply the $120^\circ$ rotation that takes $A$ to $B$, $B$ to $C$, and $C$ to $A$. In general, this takes the points $E,F,G$ to points $E', F', G'$ such that segments $A'E'$, $B'F'$, and $C'G'$ satisfy the condition in the problem; in other words, $BE'$, $CF'$, and $AG'$ have the property we want. But because the solution is unique, we must have $E'=F$, $F'=G$, and $G'=E$. From this, $AG=CF=BE$ in the equilateral case follows.


问:
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

答:
I came across very similar problems while doing Maths Olympiad in high school and  had only one approach to it then.

Observe the property for $a,b,c \in \Bbb K$ $$ \text {if } a+b+c=0,\ \text{ then  }\  a^3+b^3+c^3=3abc \quad (\ast)$$

This can easily be proved by expanding $(a+b+c)^3$
$$\begin{array}{rll}(a+b+c)^3&=&a^3+b^3+c^3+\underbrace{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b}_X+6abc=0\\X&=&3ab\underbrace{(a+b)}_{=-c}+3bc\underbrace{(b+c)}_{=-a}+3ca\underbrace{(c+a)}_{=-b}=-9abc \\ (a+b+c)^3&=&a^3+b^3+c^3-9abc+6abc\ =\ 0 \ \implies\  (\ast)\end{array}$$

In the problem above you have an expression of the form

$$\sqrt[3]{x+\sqrt y}+\sqrt[3]{x-\sqrt y}=z$$which can also translate to $$\underbrace{\sqrt[3]{x+\sqrt y}}_a+\underbrace{\sqrt[3]{x-\sqrt y}}_b +\underbrace{(-z)}_c=0$$
Applying the property abover
$$\begin{array}{rll}\underbrace{\left(\sqrt[3]{x+\sqrt y}\right)^3+\left(\sqrt[3]{x-\sqrt y}\right)^3}_{2x} +(-z)^3&=&3\underbrace{\left(\sqrt[3]{x+\sqrt y}\right)\left(\sqrt[3]{x-\sqrt y}\right)}_{\text{conjugates}}(-z) \\ \implies \quad 2x-z^3&=&-3z\cdot \sqrt[3]{x^2-y}\end{array}$$
In conclusion the problem resolves to finding the roots
$$z^3-3z\cdot \sqrt[3]{x^2-y} -2x = 0$$

I have also been curious about this, I asked this sometimes ago


问:
Prove: $\sum\limits_{j=1}^{2^n}\frac{1}{j} \geq 1 + \frac{n}{2}$ for all positive integers $n$.
Given the following proof.

Proof: by induction on n

Base Case: $n = 1$ is trivial. Done

Inductive Step: suppose for $n \geq 1$ we have  $\sum\limits_{j=1}^{2^n}\frac{1}{j} \geq 1 + \frac{n}{2}$. We show for $n + 1$. Observe
$$
\begin{equation} \begin{split}
(1)& \quad\sum_{j=1}^{2^{n+1}}\frac{1}{j} &= \sum_{j=1}^{2^n} \frac{1}{j} + \sum_{j=2^n + 1}^{2^{n+1}}\frac{1}{j} \\
(2)&&\geq 1 + \frac{n}{2} + \left(2^{n+1} - 2^n\right)\frac{1}{2^{n+1}} \\
&&= 1 + \frac{n}{2} + 1 - \frac{1}{2} \\
&&= 1 + \frac{n+1}{2}
\end{split} \end{equation}
$$
Done.

Two questions: How does the procedure of separating the summation of the LHS in (1) to the two summands on the RHS in (1) work? In addition, how does one evaluate $\sum\limits_{j=2^n + 1}^{2^{n+1}} \frac{1}{j}$ to be the third summand in (2)?

NOTE: The summations limits in second summand of the RHS of (1) may be wrong. I could not remember fully, what it was.

In addition, I am NOT asking why we separate in (1) to prove. I am asking how the algebraic manipulations works in going from LHS of (1) to RHS of (1).

Body:
First question: Notice:  $\sum_{k=1}^{M} a_k= a_1 + a_2 + ........   + a_{M-1}+a_M =$

$\underbrace{a_1 + a_2 + ...+ a_W} + \underbrace{a_{W+1}+.....   + a_{M-1}+a_M}=$

$\sum_{k=1}^{W} a_k + \sum_{k=W+1}^{M} a_k$.

We can split any sum anywhere that way.

Second question:

If $2^{n}+1 \le j \le 2^{n+1}$ then

$\frac 1{2^{n} + 1} \ge \frac 1j \ge \frac 1{2^{n+1}}$ so

$\frac 1j \ge \frac 1{2^{n+1}}$

SO $\sum_{j=2^n+1}^{2^{n+1}} \frac 1j \ge \sum_{j=2^n+1}^{2^{n+1}} \frac 1{2^{n+1}}=$

$\underbrace{\frac 1{2^{n+1}}+\frac 1{2^{n+1}}+\frac 1{2^{n+1}}+ .... + \frac 1{2^{n+1}}}_{(2^{n+1} - 2^n)\text{ times}}=$

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

$\frac {2^{n+1}}{2^{n+1}} - \frac {2^{n}}{2^{n+1}} = 1-\frac 12 = \frac 12$.


问:
Cardinal of an infinite Cartesian Product of finite increasing sets
Let $X_n=\{1,2,...,n\}$. I want to compute the cardinal of $\prod\limits_{n\in\mathbb{N}}X_n:=\{f:\mathbb{N}\rightarrow \bigcup\limits_{n\in\mathbb{N}}X_n: f(n)\in X_n\}$.
I believe that its cardinal is $2^{\aleph_0}$.

I have been able to show that
\begin{align}
\left|\prod\limits_{n\in\mathbb{N}}X_n\right|\geq 2^{\aleph_0}.
\end{align}

In order to show that $\left|\prod\limits_{n\in\mathbb{N}}X_n\right|\leq 2^{\aleph_0}$, I am trying to find a one to one map from the set $\{f:\mathbb{N}\rightarrow\bigcup\limits_{n\in\mathbb{N}}X_n: f(n)\in X_n\}$ to the set $\{g:\mathbb{N}\rightarrow \{0,1\}\}$ but I am not able to find it. Is it even possible?

Any hint to find such a map?

Is it correct that $\prod\limits_{n\in\mathbb{N}}X_n$ has cardinality $2^{\aleph_0}$?

答:
Define

$$\varphi:\prod_{n\in\Bbb Z^+}X_n\to\wp(\Bbb Z^+):f\mapsto\left\{2^n3^{f(n)}:n\in\Bbb Z^+\right\}\,.$$

That should be good enough for your purposes, but if you really want a map to ${^{\Bbb Z^+}}\{0,1\}$, you can use $f\mapsto\chi_{\varphi(f)}$, mapping $f$ to the indicator (characteristic) function of $\varphi(f)$. The set $\varphi(f)$ is a fairly standard way of encoding the ordered pairs making up $f$ as single positive integers. (I have used $\Bbb Z^+$ here instead of $\Bbb N$, because for me the latter includes $0$.)


问:
Selecting a cubic root using a convention - is there a better algorithm than my naive one?
I am trying to calculate cubic roots of a complex number according to the following convention:

The convention comes from the following paper:
Ting Zhao, Dongming Wang, Hoon Hong. Solution Formulas for Cubic Equations Without or With Constraints. Journal of Symbolic Computation, Elsevier, 2011, 46 (8), pp.904-918.

What is the best way to go about calculating $\sqrt[3]{1+i}$ using this convention?

Specifically, I am implementing this in a computer program using double precision floating point numbers. (64 bit floats, 128 bit complex numbers) I would like to avoid round-off error as much as possible. Speed is secondary to precision, but still important.

The "naive" algorithm I have developed on my own is this:

Given that $x$ is $1+i$, $arg(x)=0.7853981633974483$

Therefore, we need to use the third rule from the convention... so we know that $arg(\sqrt[3]{x}) = -\frac{1}{3}arg(x)$

Compute the right hand side: $arg(\sqrt[3]{1+i}) = 0.2617993877991494$

$\sqrt[3]{1+i}$ has three roots. They are: $[(-0.7937005259840997+0.7937005259840998i),
 (-0.2905145555072513-1.0842150814913512i),
 (1.0842150814913512+0.2905145555072518i)]$

Now let's take $arg(x)$ of each root: [2.356194490192345, -1.832595714594046, 0.26179938779914974]

We see that $arg(x)$ of the 3rd root is equal to what we calculated for the right hand side in step (3) (0.26...) . Therefore, the correct root according to the convention is $1.0842150814913512+0.2905145555072518i$

Is there a better method? I am an engineer, not a mathematician, and I'm worried that perhaps I'm missing a much simpler (or at least more analytical) solution because I'm ignorant of some mathematical identity that would simplify everything.

Notes: I am calculating $arg(x)$ using Python's cmath.phase() function. This question is related to my previous question found here.

答:
The Ting et al. paper that you cite promises, using the convention on p. 906, that if you use the argument  $\pi/4$, then taking $\sqrt[3]{1+\mathrm{i}}$ to have argument  $(1/3) \cdot \pi/4 = \pi/12$ (and, of course, magnitude $\sqrt[3]{\sqrt{1^2+1^2}} = \sqrt[6]{2}$, using the real cube and sixth root functions) selects the correct root for Legendre's formulas.  Their convention works if the coefficients of the original cubic are all real.  If the coefficients of the original cubic are not all real, the best method I've seen is to try all three roots in the solutions and see which choice of cube root gives an actual root of the polynomial -- expecting a formula for this is expecting too much.  (You are testing at most three potential values.  This is trivial computer time.)

As a numerical fact: it is too much to hope that a finite precision approximation to a root will give exactly zero (or even zero to a specified precision) when passed as an argument to the polynomial.  You should either engage in sensitivity analysis to find out how (im-)precision in your approximate root leads to (im-)precision in the value of the polynomial evaluated at that approximate root or you should establish a small tolerance for the difference between the value of the polynomial and the expected value, zero.


问:
Integer programming formulation for the entire region of a triangle except the region of a rectangle inside it
Suppose we have a triangle-shaped region such as:

$x-y\ge0$

$x+y\le8$

$x,y\ge0$ and integers

How to write an IP formulation to describe all real points inside this triangle except the points inside the rectangle (Only the red-shaded region) as depicted in the figure below:

答:
You want to enforce
$$\lnot(3 < x < 5 \land 1 < y < 2),$$
equivalently,
$$x \le 3 \lor x \ge 5 \lor y \le 1 \lor y \le 2.$$
Introduce binary variables $z_i$ for $i \in \{1,2,3,4\}$ and linear constraints:
\begin{align}
x - 3 &\le (8-3)(1-z_1) \tag1\\
5 - x &\le (5-0)(1-z_2) \tag2\\
y - 1 &\le (4-1)(1-z_3) \tag3\\
2 - y &\le (2-0)(1-z_4) \tag4\\
\sum_{i=1}^4 z_i &\ge 1 \tag5
\end{align}
Constraint $(1)$ enforces $z_1=1 \implies x \le 3$.
Constraint $(2)$ enforces $z_2=1 \implies x \ge 5$.
Constraint $(3)$ enforces $z_3=1 \implies y \le 1$.
Constraint $(4)$ enforces $z_4=1 \implies y \ge 2$.
Constraint $(5)$ enforces $\lor_i (z_i = 1)$.


问:
Regarding a Paper by Paul A. Clement on Tridiagonal Matrices
I've asked this question at MathOverflow and was told it'd be better suited for here.

In Paul A. Clement's (1959) paper,

A Class of Triple-Diagonal Matrices for Test Purposes, SIAM Review, Vol. 1, No. 1 (Jan., 1959), pp. 50-52

He makes the claim that the eigenvalues of:

$ \begin{pmatrix}
0 & y_{1} & 0 & ... & 0 \\\
x_{1} & 0 & y_{2} &  & ... \\\
0 & x_{2} & 0 & ... & 0 \\\
... &  & ... &  & y_{n} \\\
0 & ... & 0 & x_{n} & 0
\end{pmatrix} $

are $\pm (n), \pm (n-2), ..., (\pm1 \; or \; 0)$ for $x_{k} = k$ and $y_{k} = n-k+1$.

Specifically, and I quote, "then a theorem of Sylvester establishes that the eigenvalues of this An+, are the
numbers".

I can't for the life of me figure out what theorem and/or how it follows from them. I am familiar with Sylvester's formula for matrices in terms of their eigenvalues, but to get Frobenius covariants of a matrix A one needs to know the eigenvalues to start with.

Am I overlooking something trivial here?

答:
EDIT: OP has changed the question, so what follows is no longer relevant, and can safely be ignored. </p>

<p>If $n=3$, the matrix is $$\pmatrix{0&3&0\cr1&0&2\cr0&2&0\cr}$$ which has characteristic polynomial $t^3-7t$ and eigenvalues $\pm\sqrt7$ and $0$, so something is wrong. </p>

Body:
EDIT: OP has changed the question, so what follows is no longer relevant, and can safely be ignored.

If $n=3$, the matrix is $$\pmatrix{0&3&0\cr1&0&2\cr0&2&0\cr}$$ which has characteristic polynomial $t^3-7t$ and eigenvalues $\pm\sqrt7$ and $0$, so something is wrong.


问:
How do I show that a domain is not simply connected?

Let $U$ be a star-shaped domain in $\mathbb{C}$. Prove that the subset obtained from $U$ by removing a finite number of points is not simply connected.

Usually when we want to show something is simply connected we show that any two curves with fixed endpoints are homotopic by finding a suitable homotopy. However, what can you do if you need to show that the domain is not simply connected? Can we somehow show that there doesn't exist a homotopy?

答:
With no technology available whatsoever this is an unreasonable question to ask a new student. It is exactly as difficult as the following theorem.

The punctured plane $\Bbb C \setminus \{0\}$ is not simply connected. In fact, the loop $\gamma_r(\theta) = (r\cos \theta, r \sin \theta)$ is not null-homotopic for any $r>0$.

If you know this fact, you can use it to prove your result.

If not, you are not ready yet to prove the stated claim. This fact is non-trivial and the proof is non-obvious.


问:
Integration by parts on the Riemann sphere
I have two ways of integrating the area of the $2$-sphere, reaching contradictory results. The first method is the usual one:

$A=\int_{0}^{2\pi}d\phi\int_0^{\pi}d\theta \,\,\,\sin\theta=4\pi $

Lets now use stereographic co-ordinates, projecting from the south pole, i.e. $z=e^{i\phi}\tan\frac{\theta}{2}$. Getting the Jacobian factors correct we find that our integral becomes:

$A=4\int d^2z \frac{1}{(1+|z|^2)^2}$

Now I can write the integrand as double derivative:

$\partial_{z}\partial_{\bar{z}}\ln(1+|z|^2)=\partial_z\frac{z}{1+|z|^2}=\frac{1}{1+|z|^2}-\frac{|z|^2}{(1+|z|^2)^2}=\frac{1}{(1+|z|^2)^2}$

So I can write the integral as:

$A=\int d^2z \partial_z\partial_{\bar{z}}\ln(1+|z|^2)$

Now using integration by parts/Stokes theorem, the integral just evaluates to its value at the boundary of the integration manifold. Because we are on the $2$-sphere/Riemann sphere and not just the complex plane, the manifold over which we are integrating does not have a boundary and hence the integral evaluates to zero:

$A=0$

Which is a contradiction. Where is the incorrect logic?

答:
The problem is that $z$ takes values on $\mathbb{C}$, and not on the sphere $\mathbb{S}^2$.

The stereographic projection relates the sphere to the plane plus infinity. The area of the sphere is
$$ A = \int\limits_{x_1^2+x_2^2+x_3^2=1} \, 1 \; . $$
This integral is the same if you integrate over the whole sphere minus one point (set of measure zero). This area is equal to the area of the sphere without the North pole
$$ A = \int\limits_{\mathbb{S}^2-\mathrm{N}} \, 1 \ $$
and the stereographic projection gives a map to the complex plane. With the usual parameterization
$$  A = 4\int\limits_{\mathbb{C}} d^2z \, \partial_{z}\partial_{\bar{z}} \, \ln (1+\vert z \vert^2)  $$
and $d^2z := d\mathrm{Re}z  \wedge d\mathrm{Im}{z} = \frac{1}{2i} \, d\bar{z} \wedge dz  \, $.

Explicitly calculating the integral

Indeed, the integral is not hard to calculate. Integrate over a disc of radius $R$ and take the limit $R\to \infty$,
$$ A=  \frac{4}{2i} \lim_{R\to \infty}\;\int\limits_{\vert z \vert < R} d\bar{z} \wedge dz\, \partial_{z}\partial_{\bar{z}} \, \ln (1+\vert z \vert^2) $$
Appling Stokes theorem
\begin{align} A &= \frac{4}{2i} \lim_{R\to \infty}\;\int\limits_{\vert z \vert = R} d\bar{z} \, \partial_{\bar{z}} \, \ln (1+\vert z \vert^2) = \frac{4}{2i} \lim_{R\to \infty}\;\int\limits_{\vert z \vert = R} d\bar{z} \, \frac{z}{1+\vert z \vert^2} \; \\
&= \frac{4}{2i} \lim_{R\to \infty}\;\int\limits_{\vert z \vert = R} \frac{d\bar{z}}{\bar{z}} \, \frac{\vert z \vert^2 }{1+\vert z \vert^2} = \frac{4}{2i}  \lim_{R\to \infty} \, 2 \pi i \; \frac{R^2}{1+R^2} \\
&= 4 \pi \; . \end{align}


问:
$\sum_{n=1}^{\infty}a_n b_n$ Convergence
If $\lim_{n \to \infty} a_n=0$ and $\lim_{n \to \infty} b_n = 0$, then does $\sum_{n=1}^{\infty}a_n b_n$ converge? I think this is false - I don't know about both $a_n$ and $b_n$ but I know that given $a_n=\frac{1}{x}$ the sum $\sum_{n=1}^{\infty}a_n$ does not converge.

答: So try $a_n=b_n=\frac1{\sqrt n}$.

Body: So try $a_n=b_n=\frac1{\sqrt n}$.


答:
With  n $\geq 4$ and  $n <4$ cases, we can get the answers directly.

Case I: n $\geq 4$  range $\to$ $[-1,1]$

To ensure your range to be $[-1,1]$,  need $nx$ to cover at least $\pi$, $nx \geq \pi$, or $n \geq \frac{\pi}{x}$. Since $x$ max is $1$, so as long as  $n \geq \pi$ (or n $\geq 4$), you will have a full range of $[-1,1]$.

Case II: $n < 4$ range $\to$  $[\cos n,1]$

For $n < 4$ (or $n= 1,2,3$), you can get the range by calculating $\cos 1 (~0.54), \cos 2 (~ -0.42)$ and $\cos 3(~=0.99)$  and replace $\cos n$ with these values directly: $[\cos n,1] $

Title: How to interpret the domain and codomain of the function?

Body:
Suppose, I take for $n≥1$ we consider $\cos(nx)$ : $[0,1] \to \Bbb R $

I understand that domain is restricted between $0$ and $1$. Codomain being  $\Bbb R$.

So, is it valid to say $|\cos(nx)| \leq 1 = -1≤\cos(nx)≤1$ . I think this should be right as the range $\in$ codomain which is not restricted, so $-1≤y≤1$. Another way is to visualize a semicircle going through I and IV Quadrant (vertically)

or the domain is restricted $[0,1]$. So, $0≤\cos(nx)≤1$.

also, So every $x$ value from the domain is giving a $y$ from the codomain ( which has to between $-1$ and $+1$ unit ciricle understanding)?

I am still a bit confused. Please if you guys could clarify


问:
Can anyone solve this E[Xi X-bar]?
I am trying to solve this expectation and I don't seem to find a logical way to solve it?

Here:

$\mathrm{E}[X_i*\bar{X}]$

$X_i$ is iid.

答:
I know this is an old question, but I think there may be an easier way to do this.

Assuming $(x_1, \cdots, x_n)$ are i.i.d. with mean $\mu$ and variance $\sigma^2$.

Given $$\bar{X} = \frac{\sum_{i=1}^{n}{X_i}}{n},$$ and $$n\bar{X}=\sum_{i=1}^{n}{X_i},$$

we can observe:

$$\mathbf{E}\left[\sum_{i=1}^{n}{X_i \bar{X}}\right] = \mathbf{E}\left[\bar{X}
 n\bar{X}\right] = n\mathbf{E}\left[\bar{X}^2\right],$$ then using the fact that $$\mathbf{E}\left[\bar{X}^2\right]=\frac{\sigma^2}{n}+\mu^2,$$ the final answer is $$n\left(\frac{\sigma^2}{n}+\mu^2\right)$$.

Title: Can anyone solve this E[Xi X-bar]?