问:
How many coins are still heads up?
I'm watching this documentary: https://www.youtube.com/watch?v=qVjnPvFENEE&t=2212s and around the 6 minute mark, the professor asks a question that goes like this:

"I have 1000 coins and they're all heads up. I go along and turn over every 2nd coin, so 2, 4, 6, 8 and so on. I then turn over every 3rd one, so 3, 6, 9, 12 and so on. So now coin 6 and 12 are back in the heads position. I do this all the way up to 1000, so which coins are heads up?"

I attempted it and I thought it was all the odd numbers but the coins turning back onto heads tripped me up so what do you guys think it is?

Thanks,

PS. I don't have any degree in maths, I'm just a curious kid.

答:
A useful technique in solving such problems is to scale things down--work on a smaller case, to detect patterns.

For instance, suppose we have only $6$ coins.  Put them in a row, and number them from $1$ to $6$:  $$\{1, 2, 3, 4, 5, 6\}.$$

Now look at any single coin, say coin $4$.  How many times does it get flipped when we go through this process?  Well, it gets flipped once when going through the multiples of $2$:  $\{2, 4, 6\}$.  It does not get flipped when going through the multiples of $3$, since $4$ is not a multiple of $3$.  And then it does get flipped once again when we go through the multiples of $4$.  So it get flipped a total of $2$ times.

How many times does $5$ get flipped?  Well, only once, when we go through multiples of $5$.

And how many times does $6$ get flipped?  Three times.

Clearly, this situation has something to do with the divisors of the number on the coin.  The divisors of $4$ are $1, 2, 4$.  And the divisors of $5$ are $1, 5$, and the divisors of $6$ are $1, 2, 3, 6$.  So it looks like the number of times a given coin is flipped is one less than its number of divisors.  The reason why it is one less, is because we don't start the process by flipping every coin--we start with the multiples of $2$.

Let's work out a larger example.  Say we are interested in coin number $100$.  The divisors of $100$ are:  $$1, 2, 4, 5, 10, 20, 25, 50, 100.$$  There are $9$ divisors, so we suspect that this coin would be flipped $8$ times.  Can you check that this is true?

So if there are $1000$ coins, the number of coins that are heads up is equal to the number of coins that have an odd number of divisors (including $1$, and the number itself).  Can you go through a list of the first $1000$ such numbers and see which ones have an odd number of divisors?  I will start you out with a partial list:

$$\begin{array}{c|c}
n & \sigma_0(n) \\
\hline 1 & 1 \\
 2 & 2 \\
 3 & 2 \\
 4 & 3 \\
 5 & 2 \\
 6 & 4 \\
 7 & 2 \\
 8 & 4 \\
 9 & 3 \\
 10 & 4 \\
 11 & 2 \\
 12 & 6 \\
 13 & 2 \\
 14 & 4 \\
 15 & 4 \\
 16 & 5 \\
\end{array}$$

The second column with heading $\sigma_0(n)$ represents the number of divisors of $n$.  Which of these are odd?  Do you see a pattern?  If you do, how would you go about proving this pattern?

Title: How many coins are still heads up?


问:
Questions about random variables and modulo
In my lecture notes, there is a question about the independence of the random variable.

Suppose that $X_1$ and $X_2$ are independent Bernoulli trials each with probability 0.5, and let $X_3 = (X_1 + X_2) \mod 2$.

Show that $X_1$, $X_2$ and $X_3$ are pairwise independent, but $X_3$ and $X_1 + X_2$ are not
independent.

Credit to @HallaSurvivor for explaining what $X_1+X_2$ means What does adding two random variables mean?

I would like to know why $X_3 = (X_1+X_2) \mod 2$ is not independent of $(X_1+X_2)$.

Edit: I often screw up organizing things starting from few weeks ago, apologies for that.

答:
Remember from your last question that, intuitively, random variables are independent exactly when information about one doesn't help you predict the outcome of the other.

A classic example is a dice roll and a coin toss. Knowing you rolled a $4$ doesn't tell you if you're about to flip a heads.

So saying that $X_1 + X_2$ is not independent of $X_1 + X_2 \text{ (mod 2)}$ is saying that, if I know the outcome of $X_1 + X_2$, I might be able to learn something about the outcome of $X_1 + X_2 \text{ (mod 2)}$. Let's see why:

$$
\begin{array} {|c|c|}\hline X_1 & X_2 & X_1 + X_2 & X_1 + X_2 \text{ (mod 2)} \\ \hline 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 1 & 1 \\ \hline 1 & 1 & 2 & 0 \\ \hline  \end{array}
$$

Each row happens with probability $\frac{1}{4}$ (do you see why?). But look at how similar the $X_1 + X_2$ and $X_1 + X_2 \text{ (mod 2)}$ entries are!

If you know $X_1 + X_2$, then you're guaranteed to know $X_1 + X_2 \text{ (mod 2)}$ as well! (Again: do you see why?)

Since knowledge of $X_1 + X_2$ determines the outcome of $X_1 + X_2 \text{ (mod 2)}$, the two random variables cannot be independent. The outcome of one depends on the outcome of the other.

I hope this helps ^_^

ContentLicense: CC BY-SA 4.0


问:
Algebra equation with integer solutions
If the solution for $(7-k)(8-k)x^2 - (112-15k)x + 56 = 0$ are all integer solutions, what could the value for $k$ be? (There are 4 values)

My first thought was to put in the quadraic formula and see what the solution will look like.

So put in:
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Where we'll get
$$x = \frac{112-15k\pm \sqrt{337k^2-5020k+18816}}{2(7-k)(8-k)}$$
after simplification.

However, I found that $337k^2-5020k+18816$ cannot be factorized. What should I do?

答:
You didn't solve the quadratic equation in $x$ correctly. I obtained
$$
x=\frac{7}{7-k},\quad x=\frac{8}{8-k}.
$$
Indeed, $b^2-4ac=k^2$.

Title: Algebra equation with integer solutions


问:
Quadratic gauss sum equivalence of definition
Let $r$ be a prime number and $\zeta_r$ be the $r$th rot of unity. I know quadratic gauss sum can be expressed in two ways, $g=\sum_{i=0}^{r-1}\zeta_r^{i^2}$ and $g=\sum_{i=0}^{r-1}(\frac{i}{r})\zeta_r^{i}$
$g$ being the sum. But I am not getting how to go from the first definition to the second. Can anyone help?

答:
The connection is

$$
g=\sum_{i=0}^{r-1}\zeta_r^{i^2}=\sum_{i=0}^{r-1}\left(1+\left(\frac ir\right)\right)\zeta_r^i=\sum_{i=0}^{r-1}\zeta^i+\sum_{i=0}^{r-1}\left(\frac ir\right)\zeta^i=0+\sum_{i=0}^{r-1}\left(\frac ir\right)\zeta^i=\sum_{i=0}^{r-1}\left(\frac ir\right)\zeta^i
$$

where $\sum_{i=0}^{r-1}\zeta_r^{i^2}=\sum_{i=0}^{r-1}\left(1+\left(\frac ir\right)\right)\zeta_r^i$ as, $i^2\;(mod\;r)$ is a quadratic residue, and the sum only has the quadratic residues in the exponent of $\zeta_r$, each term is multiplied by $2$ as their are two solution of $x^2-a=0\;(mod\;r)$. So, now it can be easily seen that, two expressions are equivalent.


问:
Prove that $\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$ doesn't exist without L'hopital rule
I need to prove that the following limit doesn't exist:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$$

Itried to simplify $1+\cos(2x)$ into $2\cos^2(x)$, in order to calculate both of the one-sided limits, but I am not sure how to calculate it without using L'hopital rule.

答: $\cos (3x)=4\cos^{3}x-3\cos x$ and $1+\cos (2x)=2\cos^{2}x$. So the given limit is $\lim_{x \to \pi/2} [2\cos x-\frac 3 {2\cos x}]$ which does not exist.


问:
Monomorphism of a group automorphisms
Let $\mathbb{A}$ and $\mathbb{B}$ be two algebras on the same language $\mathcal{L}.$ How can I prove that there exists a monomorphism of the group

${\rm Aut}\mathbb{A}\times{\rm Aut}\mathbb{B}$ to the group ${\rm Aut}(\mathbb{A}\times\mathbb{B})?$

答:
Let $f\in$Aut$ \mathbb{A}$ and $g\in$ Aut$\mathbb{B}.$ Let us show that $f\times g\in$ Aut($\mathbb{A}\times\mathbb{B}$), where $(f\times g)(a,b)=(f(a),g(b)).$
$f\times g$ is 1-1 because

$(f\times g)(a,b)=(f\times g)(c,d)\Rightarrow (f(a),g(b))=(f(c),g(d))\Rightarrow f(a)=f(c), g(b)=g(d).$

Because $f$ and $g$ are automorphisms, we have $a=c$ and $b=d,$ so $(a,b)=(c,d).$

$f\times g$ is sirjection because for all $(c,d)\in \mathbb{A}\times\mathbb{B}$ we know there exists $a\in\mathbb{A}$ and $b\in\mathbb{B}$ such that $f(a)=c$ and $g(b)=d$ because $f$ and $g$ are sirjections. So, $(f\times g)(a,b)=(f(a),g(b))=(c,d).$

$f\times g$ is a homomorphism because

for $c\in$ Const$_{\mathcal{L}}$ we have $(f\times g)(c^{\mathbb{A}\times\mathbb{B}})=(f\times g)(c^{\mathbb{A}},c^{\mathbb{B}})=(f(c^{\mathbb{A}}),g(c^{\mathbb{B}}))=(c^{\mathbb{A}},c^{\mathbb{B}})=c^{\mathbb{A}\times \mathbb{B}},$

for $F\in$ Fun$_{\mathcal{L}},$ $ar(F)=n$ and $(a_1,b_1),\ldots, (a_n,b_n)\in A\times B$ we have

$(f\times g)(F^{\mathbb{A}\times \mathbb{B}}((a_1,b_1),\ldots, (a_n,b_n)))=(f\times g)(F^{\mathbb{A}}(a_1,\ldots,a_n),F^{\mathbb{B}}(b_1,\ldots,b_n))=(f(F^{\mathbb{A}}(a_1,\ldots,a_n)),g(F^{\mathbb{B}}(b_1,\ldots,b_n)))=(F^{\mathbb{A}}(f(a_1),\ldots,f(a_n)), F^{\mathbb{B}}(g(b_1),\ldots, g(b_n)))=F^{\mathbb{A}\times \mathbb{B}}((f(a_1),g(b_1)),\ldots,(f(a_n),g(b_n)))=F^{\mathbb{A}\times\mathbb{B}}((f\times g)(a_1,b_1),\ldots,(f\times g)(a_n,b_n)).$

Now, the map $h:$ Aut$\mathbb{A}\times$ Aut$\mathbb{B}\rightarrow$ Aut($\mathbb{A}\times \mathbb{B}$) defined with $h(f,g)=f\times g$ is a desired monomorphism.

Title: Monomorphism of a group automorphisms


问:
Subharmonic functions
Consider:

if $\log u$ and $\log v$ are both subharmonic function on an open set $U$, then $\log(u+v)$ is also subharmonic on $U$.

How can I prove this?

答:
The assumption about the regularity is not necessary here.

Indeed, we know that we can find a sequence of smooth functions $u_n$ and $v_n$ decreasing to $u$ and $v$ respectively and such that $\log u_n, \log v_n$ are subharmonic (just use a convolution with a radial smoothing kernel).

We may then use the argument of JohnD which guarantees that $\log(u_n+v_n)$ is subharmonic. But a limit of a decreasing sequence of subharmonic functions is still subharmonic, so that $\log(u+v)$ is subharmonic.

Moreover, a direct proof (with no assumption on regularity) may be obtained once we know some basic stuff about sh functions. Namely, $f(x,y)=\log (e^x+e^y)$ defines a a convex function, non-decreasing in each variable, therefore $f(\log u, \log v)=\log(u+v)$ is subharmonic.

ContentLicense: CC BY-SA 2.5

Title: Subharmonic functions

ContentLicense: CC BY-SA 3.0


问:
Dimension problems
Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation. Prove that there exists a subspace
$S \subseteq \mathbb{R}^n$ that verifies $\ker(f^2) = \ker(f) \bigoplus S$ and $\dim S = \dim f(S) \leq \dim(\ker(f))$.

My attempt at solution: So far I extended from the $\ker(f)$ to the $\ker(f^2)$ like this: let a base of $\ker(f)$ be $B = \{v_1,\ldots,v_r\}$ extending to a bae of $\ker(f^2)$ we get $B' = \{v_1,\ldots,v_r,v_{r+1},\ldots,v_m\}$. So I know that from $1$ to $r$ are elements linearly independent of $\ker(f)$, and from $r+1$ to $m$ I have my desired $S$. This verifies part of the second thing I'm asked, which is that $\dim S = \dim f(S)$, the thing is I'm having trouble with the inequality $\dim f(s) \leq \dim \ker(f)$. I've been trying to get the inequality using the dimension theorem but I cant seem to get it.

答: You're defining $S$ to be the subspace with basis $\{v_{r+1}, \dots, v_m\}$, where $m \leq n$, right? The key property of this set is that $S \subset \ker(f^2)$ but $S$ is not in $\ker(f)$.  That means that if $w \in S$, then $f(w) \neq 0$ but $f(f(w)) = 0$.  Thus, $f(w) \in \ker(f)$.  You can conclude that $f(S) \subset \ker(f)$, which gives your inequality.</p>

Title: Dimension problems


问:
Markov Chain with 2 machines and 2 repair men
If we have 2 machines and 2 repair men. The machines each break down at different rates $\alpha_a$ for machine A and $\alpha_b$ for machine B. If we assume that the repairmen have different abilities, they each repair with exponential random times with parameters $\beta_1$ and $\beta_2$ respectively(naturally, this rate does not depend on which machine they repair).

Define a Markov chain, specify the state space and the transition matrix. Also determine the communicating classes.

My attempt: I am unsure of how to proceed for this case. Should it been machine at the same rate, and the repairmen with same abilities, I know how to solve this question. But now that the rates have been split, I am unsure how to construct a state space.

答:
When $\beta_1 = \beta_2$, this would be a classic 4-state (AB, !AB, A!B, !A!B) case. For $\beta_1 \ne \beta_2$, you would have to expand the cases for the availability of the repairmen. This would mean however that the failure transitions have to be split on two state variants where either repairman 1 or 2 is available to repair. But if you need to adjust $\alpha_a$ and $\alpha_b$ for this, you might as well adjust the effective repair rates of the 4-state case by an effective repair rate weighed by the repairmen availability:
$$\beta_{eff} = {{\beta_1 P\{1\;available\}+\beta_2 P\{2\;available\}}\over {P\{1\;available\}+P\{2\;available\}}}$$
If the repairmen availability is proportional to their repair rate, i.e. $P\{i\;available\}\propto \beta_i $:
$$\beta_{eff} = {{\beta_1^2+\beta_2^2}\over {\beta_1+\beta_2}}$$


问:
How to solve this trigonomectric equation $2^{\lvert\sin x\rvert} + 2^{\lvert\cos x\rvert} = 3$?
I thought its solutions are $(\sin x;\cos x) = \{(\pm1;0),(0;\pm 1)\}$

But how can I solve it?

Please help :(

答:
For simplicity we have to solve for $x\ge 0$ $$f(x)=2^x+2^{\sqrt{1-x^2}}=3$$.

Note that $$f'(x)=2^x\log 2-\frac{2^{\sqrt{1-x^2}}x\log 2}{\sqrt{1-x^2}}$$

if   $x\le \frac{1}{\sqrt{2}}$ then $x\le \sqrt{1-x^2}$
then we have

Lemma: $$\frac{2^x}{x}\ge \frac{2^{\sqrt{1-x^2}}}{\sqrt{1-x^2}}$$

proof: show that $g(x)=\frac{2^x}{x}$ is decreasing in given interval.

From this lemma it follows that $f'(x)\ge 0$.Thus for all $x\in (0,\frac{1}{\sqrt{2}}]$ :$f(x)>f(0)=3$

Similarly prove that $f(x)$ is decreasing for all $x\in [\frac{1}{\sqrt{2}},1]$ hence deduce that $f(x)>f(1)=3$ for all $x\in [\frac{1}{\sqrt{2}},1)$.

The conclusion is now obvious:i.e $x=0,x=1$ are the only solutions.


问:
Derivative for $4x^2+\frac{216}{x}$ .
$4x^2+\frac{216}{x}$ .

Steps that I used was to differentiate the the first term using formula .
2nd step was that 216 is a constant and we don’t differentiate values which are present in denominator.
Therefore 1/x.

It’s differentiation came to be equal to this by me.

I got is $8x + \frac{1}{x}$

But the answer is $8x-\frac{216}{x^2}$.

How come the constant is not differentiated and x value got squared ?

答:
If $c$ is a constant, then $(cf(x))'=c(f(x))'$; you get the square because $\frac{1}{x}=x^{-1}$ and $(x^{-1})'=(-1)\cdot x^{-2}=-\frac{1}{x^2}$.
To obtain the correct result you have to use these two rules.

Title: Derivative for $4x^2+\frac{216}{x}$ .


Title: Derivative for $4x^2+\frac{216}{x}$ .

Body:
$4x^2+\frac{216}{x}$ .

Steps that I used was to differentiate the the first term using formula .
2nd step was that 216 is a constant and we don’t differentiate values which are present in denominator.
Therefore 1/x.

It’s differentiation came to be equal to this by me.

I got is $8x + \frac{1}{x}$

But the answer is $8x-\frac{216}{x^2}$.

How come the constant is not differentiated and x value got squared ?

Body:
$\frac{1}{x}=x^{-1}$, hence

$$\frac{d}{dx}\frac{1}{x}=(-1)x^{-2}=-\frac{1}{x^2}$$


答:
Starting from your last line:
$$L=\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}=\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{2\cos^2 (x)}$$
$$L=\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{2\sin^2 (\pi/2-x)}$$
Change variable $x=\dfrac {\pi}2
-u$:
$$L=\lim_\limits{u\to 0}\frac{\cos(3\pi /2-3u)}{2\sin^2 (u)}=\lim_\limits{u\to 0}\frac{\sin(-3u)}{2\sin^2 (u)}$$
$$L=-\dfrac 12\lim_\limits{u\to 0}\dfrac{\sin (3u)}{3u}\dfrac{u^2}{\sin^2 (u)}\frac 3 u$$
Then use: $$\lim_\limits{u\to 0}\frac{\sin u}{u}=1$$

Title: Prove that $\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$ doesn't exist without L'hopital rule

Body:
I need to prove that the following limit doesn't exist:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$$

Itried to simplify $1+\cos(2x)$ into $2\cos^2(x)$, in order to calculate both of the one-sided limits, but I am not sure how to calculate it without using L'hopital rule.


问:
Changing model function in non linear curve fitting
I have some $(t,S)$ data and this four-parameters stretched exponential model (from diffusion NMR signals):

$S(t) = S_0\cdot e^{-D\cdot t^a}+c$ .

Here $D$, $S_0$, $a$, $c$,  are parameters to estimate. They must be all positive except for $c$ that is a little noise-term and can be negative.
I'm using standard scipy.optimize library for fitting but this is not relevant. Sometimes it does not converge at all.

So i'm trying to use less parameters by dividing my $S$-data array for the first value $S(t_0)$ so it starts from $1$, using this three-parameters function:

$S(t)/S_0 =  e^{-D\cdot t^a}+\tilde{c}$ ,

with $\tilde{c}=c/S_0$.

I know that i'm assuming that $S(t_0) \simeq S_0$ but i think i'm doing something wrong.
Should i use this function instead?

$S(t)/S_0 =  e^{-D\cdot (t-t_0)^a}+\tilde{c}$,

答:
You are trying to fit data with a very highly nonlinar model. The main reason for difficulties is, much more than often, the lack of "reasonable" (or at least consistent) guesses of the parameters.

I shall change your notation and let $k=-D$.

Let us use you idea of
$$\frac S{S_0}=e^{k\, (t-t_0)^a}+\tilde{c}$$ and for the time being "ignore" the $\tilde{c}$ or give it a fixed value (I shall come back later to this point). Now, take logarithms
$$\log\left(\frac S{S_0}-\tilde{c} \right)=k(t-t_0)^a$$Logarithms again
$$\log\left(\log\left(\frac S{S_0}-\tilde{c} \right) \right)=\log(k)+a \log(t-t_0)=b+a \log(t-t_0)$$
Considering that you have $n$ data points $(t_i,S_i)$, for a given value of $\tilde{c}$ define new variables, namely
$$x_i=t_i-t_0 \qquad \text{and} \qquad y_i=\log\left(\log\left(\frac {S_i}{S_0}-\tilde{c} \right) \right)$$ and your model is just
$$y_i=b+a t_i$$ which is more than simple. From the parameters, recompute
$y_i^{calc}$ and from that the corresponding $\left(\frac {S_i^{calc}}{S_0}-\tilde{c}\right)$ and the corresponding sum of squares $(SSQ)$. Now, play a little bit with the value assigned to $\tilde{c}$ until you see a minimum value for $SSQ$. When this will be done, you have the reasonable estimates for $k$ and $a$ and you can start the nonlinear regression.


问:
Are There Any Symbols for Contradictions?
Perhaps, this question has been answered already but I am not aware of any existing answer. Is there any international icon or symbol for showing Contradiction or reaching a contradiction in Mathematical contexts? The same story can be seen for showing that someone reached to the end of the proof of a theorem (i.e. as shown the tombstone symbol ∎, Halmos).

答: I'm 8 years late, but I've seen that this answer is not on the list. In Romania we use a symbol similar to a pair of scissors, that I have learned from both my teachers I've studied with. Unfortunately, I was unable to find such a symbol in MathJax, but I will try reproducing it: $\require{HTML} \style{display: inline-block; transform: rotate(30deg) scaleX(-1)}{\flat} \style{display: inline-block; transform: rotate(-30deg) scaleX(1)}{\flat}$. Yes, something like that, but with smaller "ears".

OwnerDisplayName: Babak Sorouh


问:
Defining the exponential map without explicit affine connection
Given a smooth manifold $M$, a smooth vector field $V$, and some $x_0 \in M$, we can define the exponential map through the map $\phi$, where $\phi$ is defined on some open interval of $\mathbb{R}$, $\phi(t_0) = x_0$, and $x = \phi(t)$ satisfies the differential equation
$$\frac{dx}{dt} = V(x),$$
and we let $\phi(t) = \exp(t\cdot V, x_0)$, so $\exp(v, x) = \phi(1)$.

This is taken from Theorem 3.1 of J-M Souriau's Structure of Dynamical Systems. At this point of the book, he hasn't introduced any concept of an affine connection, only assuming a Hausdorff manifold, but here we have somehow defined (and shown the existence of a unique) exponential map. Is there some natural affine connection he has assumed, or can we find the connection corresponding to this given definition of an exponential map? I saw this related question: Exponential maps depends on Riemannian metric?, but I'd like to pinpoint where exactly the metric or associated connection would show up.

I know that the exponential map can be defined in terms of geodesics, and I guess the ODE $dx/dt = V(x)$ is the geodesic equation, which can be solved componentwise with respect to some a suitable coordinate basis on the tangent bundle (e.g., Prop 6.1.2 here). But what metric or connection does this geodesic equation correspond to?

答:
The author seems to refer to exponential map for any "time 1 of a flow". This is a sort of generalization of:

the exponential map of a Riemannian manifold, which is in fact the time 1 map of the geodesic flow (defined on the unit tangent bundle)

the exponential map of a Lie group, which is the time 1 map of a natural flow of left invariant vector fields

These notions are the same in some particular cases - when a Lie group is equipped with a bi-invariant metric. In general, they differ, but they share some common properties, for example, $\psi_t\psi_s = \psi_{t+s}$ and $\psi_0=\mathrm{id}$, which is a very-well known property of the complex exponential function $\exp : \mathbb{C}\to \mathbb{C}^*$.

There is no reason for a time 1 map of a flow to be induced by some geodesic flow of a particular riemannian metric, so there may be no connextion inducing this flow and giving sense to the term "exponential map" as understood as a riemannian notion.


问:
Circle intersections algorithm with $O((n+s)\log(n))$ complexity
I have found a few different posts about finding the intersections of circles in $O((n+s)\log(n))$ time with $n$ the amount of circles and $s$ the amount of intersections. The problem is that none of  the answers are giving a clear and steady answer.

I found this answer that looked very interesting, but I still have my doubts about it. Snetfel explained that we can use a sweep-line algorithm. Here is a brief summary:

We make a priority queue containing all $x$ points of left $(x-r)$ and right $ (x+r)$ extremas of the circles. We
then use a binary tree to respresent the sweep-line status.

Whenever we encounter a left extrema of a circle, we add it to the status and check for intersections with every other active circle in the status.

Whenever we encounter a right extrema of a circle, we remove it from the status.

Here you can find a video I made so you can have a better understanding. But does this method do the job in $O((n+s)\log(n))$ time as we are still checking each added circle to all other circles in the status?

I tried another method which failed miserably.

My failed attempt

What I tried is the following.

We again make a priority queue containing all $x$ points of left $(x-r)$ and right $(x+r)$ extremas of the circles.

Start of circle $c$

Insert the circle with its center y value into a binary tree. (so you sort the binary tree by its $y$ value)

Check for intersection between circle $c$ and its predecessor.

Check for intersection between circle $c$ and its successor.

End of circle $c$

Check for intersection between the predecessor and successor of circle $c$.

Remove the circle from the binary tree.

But I noticed that this doesn't find all intersection points on a circle. For example, a circle with a high $y$ value and a huge radius, will not be compared to a circle with a very low $y$ value, eventhough they have an intersection.

Here you can find a sketch of my idea.

I found on different sources online that you can divide the circle in two semicircles. But by what value do you key your tree? How do you sort it so you can find the right neighbours?

答:
But does this method do the job in $O((n+s)\log(n))$ time as we are still checking each added circle to all other circles in the status?

This is not true.  We only need to check for intersection with the newly added circle’s predecessor and successor – that’s why we need the (balanced) binary search tree (e.g., an AVL tree), as it can perform this operation in $O(\log n)$ time.

When performing the line sweep, we have three types of events that we process with the priority queue:

Start of circle $c$

Insert $c$ into the binary search tree

Check for intersection between $c$ and its successor

Check for intersection between $c$ and its predecessor

End of circle $c$

Check for intersection between the predecessor of $c$ and the successor of $c$

Remove $c$ from the binary search tree

Intersection between $c_1$ and $c_2$

Note the intersection

Swap $c_1$ and $c_2$ within the binary search tree

Check for intersection between $c_1$ and its new neighbor

Check for intersection between $c_2$ and its new neighbor

Because there are $O(n + s)$ events and both the operations on the priority queue and the tree take $O(\log n)$ time, the final complexity is $O((n + s) \log n)$.

By the way, a tricky point is that we don’t have a well-defined ordering of the circles along the sweep line.  However, we can rectify this by using the top and bottom semicircles instead.

EDIT: To clarify, when you insert a new (semi)circle into the binary search tree, you use its current $y$-coordinate and compare that to the current $y$-coordinates of the other (semi)circles to determine where on tree the new one should go.  By “current” I mean relative to the $x$-coordinate of the sweep line, e.g.,
$$
y(x) = y_0 \pm \sqrt{r^2 - (x - x_0)^2}\text.
$$


问:
Power series $\sum\limits_{n=0}^\infty \frac{x^n}{(n!)^2}$
It is well known that $\sum\limits_{n=0}^\infty \frac{x^n}{(n!)} = e^x$.

What is $\sum\limits_{n=0}^\infty \frac{x^n}{(n!)^2}$ ?

答: $$\sum\limits_{n=0}^\infty \frac{x^n}{(n!)^2}=I_0\left(2 \sqrt{x}\right)$$ where appears the modified Bessel function of the first kind.


问:
Linear programming problem neither max nor min
Heres the actual question:

television provider broadcasts two movie channels, A and B. Channel A broadcasts 1 romantic
  movie, 3 action movies and 3 comedies per month at a cost of 50 Euro. Channel B broadcasts
  3 romantic movies, 4 action movies and 1 comedy per month at a cost of 25 Euro. Suppose that
  you like to see 8 romantic movies, 19 action movies and 7 comedies at minimal cost in the coming
  months. For how many months should you request both channels?

a. Formulate this problem as an LP problem

Now my way of thinking was that I would first write it down more clearly, A and B are the television channels.

$A: 1R,3A,3C => 50 euro$

$B: 3R, 4A, 1C => 25 euro$

And I would define a new variable $x_i :=$ cost of taking a channel one month.

But then I stumble upon the problem of defining the hard limits they have set, the 8 R, 19 A, and  7 C.

In short: how do I define a constraint that is hard (e.g. not a maximum)

答:
Normally I'd just give you hints but I actually have a couple of problems with the problem formulation. First of all, the proper model here is not a linear program (LP), but an integer linear program (IP). I'm assuming that you cannot subscribe to fractional months, after all, since that is standard practice in the cable industry. This is the model I'm thinking of:
$$\begin{array}{ll}
\text{minimize}   & 50 A + 25 B \\
\text{subject to} & A + 3 B \geq 8 \\
                  & 3 A + 4 B \geq 19 \\
                  & 3 A + B \geq 7 \\
                  & A, B \geq 0 \\
                  & A, B ~ \text{integer}
\end{array}$$
Now of course, it's possible they're permitting you to subscribe to fractional months. But you'd better ask the person requesting the model if that's the case. If they do allow fractional months, then indeed you can go ahead and drop the integer constraint, and you have an LP. (LATE EDIT: actually, fractional months really don't make sense, because then you'd get fractional movies!)

But even this is not enough, because this only tells you the total number of months to subscribe to A and B, not the number of months you must subscribe to both channels. To answer that question, you need more information than is provided. Specifically, you need to know the maximum number of months you are allowed to wait to complete your movie watching task.

For instance, suppose the model yields $A=3$, $B=2$. If you're allowed to take 5 months to watch the movies on your list, then you don't need to subscribe to both channels at the same time ever. Just subscribe to A for three months, then switch to B.

On the other hand, if you're given an additional rule that you must subscribe for a minimum number of total months, then of course the total number of months required is $\max\{A,B\}$, and the number of months subscribed to both channels is $\min\{A,B\}$. But again, without additional information, the answer cannot be given---and neither answer is obtained directly from the LP or IP model, but rather from post processing the model output.

So the task "Formulate this problem as an LP problem" is, actually, ill-posed. Not only does it likely require an IP, but it requires additional modeling work after the optimization model is solved. "Solve this problem using an LP" would be OK, if they make it clear that fractional months are permitted.


问:
Computing operator norm $\ell_p\ni(x_n)\longrightarrow T(x_n)=(a_nx_n)\in\ell_q$, $q<p$
I want to compute the norm of the operator $T:\ell_p\longrightarrow\ell_q$ with $1\leqslant q<p<\infty$, defined by $T(x_n)=(a_nx_n)$. By hypothesis $T$ is a bounded linear operator. My initial doubt is what conditions must satisfy $a=(a_n)\in\mathbb{C}^\mathbb{N}$ for that to be true. I think I can prove that $T$ is a bounded linear operator iff $(a_n)\in\ell_s$ where $s=pq/(p-q)$ using Hölder's inequality. If my reasoning is correct I know that $\|T\|\leqslant\|a\|_s$. But I have no idea how to go on from there. Any hint?

Apologies in advance for my English...

答:
Hint: $T$ is  abounded operator iff $(a_n) \in \ell_{s}$ where $s=\frac {pq} {p-q}$ and the operator norm is $\|T\|=(\sum |a_n|^{s})^{1/s}$. This can be proved using Holder's inequality and condition for equality in Holder's inequality: choose $x_n$ to be of the form $\pm |a_n|^{\alpha}$ for  a suitable $\alpha$ to see that the norm is exactly $\|(a_n)\|_s$.

Recall that $\|T\| \geq \frac {\|Tx\|} {\|x\|}$ for any non-zero $x$, in particular for the suggested sequence $(x_n)$.


问:
Find the possible values of $p$ for which the equation has coincident roots.
Find the possible values of $p$ for which the equation $(2p+3)x^2+(4p-14)x+16p+1=0$ has coincident roots.

Coincident roots means 'equal roots'. For equal roots, we should use: $b^2-4ac=0$

$(4p-14)^2-4(2p+3)(16p+1)=0$

After solving this, I get: $-112p^2-536p+184=0$

I solved the above equation by this formula: $\frac{-b\pm \sqrt {b^2-4ac}}{2a}$

Is this right? Maybe there is some mistake but I can't find it. My answers are wrong. I got $-5.11$, $0.32$, my book says: $\frac{1}{2}$, $-3\frac{2}{7}$. Help. :(

Body:
If it has coincident roots then $b^2-4ac = 0$
$$(4p-1)^2 - 4(2p+3)(16p+1) = 0$$
Expand and you get
$$14p^2 + 39p - 23 = 0$$
$$(2p-1)(7p+23) = 0$$
$$P = 1/2\quad\text{or}\quad P = -23/7$$


问:
Bernoulli's beam equation approximative solution
I'm trying to solve a problem of approximate solution of a Bernoulli's beam equation by numerical integration in Python for which I already posted a question in the stackoverflow (https://stackoverflow.com/questions/65017578/python-solving-bernoullis-beam-equation-with-scipy).
I however suspect that I have a problem in the mathematical understanding of the problem. Runge-Kutta's method of numerical integration only accepts the first order differential equation, so I have to use a new variable to transform the original equation.

Bernoulli's beam equation for a cantilever beam with a point load P at the free end:
$$
E*I*y''−M(x)=0
$$
$$
y'' = -(P/(E*I))*x
$$
Boundary conditions are
$$
y(x=L) = 0
$$
$$
y'(x=L) = 0
$$
$$y'''(x=0) = F
$$
If I understand this correctly, this is a linear, second order, non-homogeneous differential equation with constant coefficients, which has a neutral form of;
$$
y'' + a1(x)*y' + a2(x)*y = f(x)
$$
and the coefficients a1 and a2 are equal to zero.

Then I need to implement a new variable h to reduce the order:
$$
y' = h
$$
$$
dh/dt = x
$$
Does this means that I'm implementing a function h and a variable t. I just can't wrap my head around this. Does this perhaps mean that instead of second order differential equation, I'm getting a set of 2 first order differential equation?
Also, when I define the y' as h, this means that I have to integrate y'' to define h? When I do this using non-definitive integrals, I get an unknown constant C. How do I implement this numerically?

I searched for explanations of this, but I only found them for order reduction of ordinary diff equations of second order, while the beam equation above is not ordinary?

答:
If you only have
$$y''=f(x)$$
Integrate
$$y'=\int f(x)\,dx+ c_1=F(x)+c_1$$ Integrate again
$$y=\int F(x)\,dx+ c_1x+c_2$$


问:
$ \iint_{R} \cos((3-y)^2) dx dy$ over the region $R=\{(x,y): |x-1|+x \le y \le x+1)\}$
Not sure how to start this question, since $\cos((3-y)^2)$ cannot be easily integrated. I believe there is a change of variable that needs to take place but I'm not sure what values to use.

答:
Note that you need to calculate $$\iint_{\mathscr{R}:\{(x,y):|x-1|-x\leq y\leq x+1\}}\cos((3-y)^{2})dxdy$$
Now, by definition we need to evaluate
$$\int_{a}^{b}\int_{g_{1}(y)}^{g_{2}(y)}\cos((3-y)^{2})dxdy$$
Since that region of integration is $$\mathscr{R}:\{(x,y):|x-1|-x\leq y\leq x+1\}$$
so, we get $$g_{1}(y)=y-1 \quad \text{and} \quad g_{2}(y)=\frac{y+1}{2}$$
and $$a=1 \quad b=3$$
Therefore,
\begin{eqnarray*}
\iint_{\mathscr{R}:\{(x,y):|x-1|-x\leq y\leq x+1\}}\cos((3-y)^{2})dxdy&=&\int_{1}^{3}\int_{y-1}^{\frac{y+1}{2}}\cos((3-y)^{2})dxdy\\
&=&\int_{1}^{3}\cos((3-y)^{2})\left[\frac{y+1}{2}-(y-1) \right]dy\\
&=&\int_{1}^{3}\cos((3-y)^{2})\frac{y+1}{2}dy-\int_{1}^{3}\cos((3-y)^{2})(y-1)dy\\
&=&\left(\sqrt{2\pi}C\left[2\sqrt{\frac{2}{\pi}} \right]-\frac{\sin(4)}{4}\right)-\left(\sqrt{2\pi}C\left[2\sqrt{\frac{2}{\pi}} \right]-\frac{\sin(4)}{2}\right)\\
&=&\boxed{\frac{\sin(4)}{2}-\frac{\sin(4)}{4}}
\end{eqnarray*}
where $C(x)$ is theFresnel $C$ integral.


问:
Distance between point and arc
Knowns:

$c$ = center of rotation for arc

$\theta$ = arc angle

$r$ = radius of arc

$R$ = some constant threshold

2 dimensions

Arc always swept out from the origin.

$P$ = some arbitrary point.

I need to calculate the shortest distance from some arbitrary point with respect to two arcs : $ (r - R) \theta$ and $(r + R) \theta$, such that the distance returned is the closest distance of the point from either of these arcs.

Excuse the poor drawing, I'm not well versed in the digital tools to illustrate the problem, but my preview seems clear.

I can think of a way involving discretizing the two arcs into points, and taking the distance of the point to each of those discrete points, taking the minimum of those distances, and then taking the minimum of the two minimum distances of the point from the two arcs, but this is an approximate solution, and I am looking for an exact solution.

What is the general exact solution to this problem?

答:
Let

$\phi$ be the angle of $P$ from $c$, where $0$ is straight down and with the same orientation as $\theta$, so we can compare them.

$r_p = |P-c|$ (that is, the dist of $P$ from $c$),

$E = (r\sin\theta,r(1-\cos\theta))$ be the endpoint of the arc,

Then I believe

$\quad\phi < \theta \quad\Rightarrow\quad d = \min\ \big(|r_p - r + R|,|r_p - r-R|\big)$ (as Adrian Keister answered),

$\quad$else
$\quad\quad\quad\quad \ d = \min\ \big(|P-(0,0)|,\ |P-E|\big)$

Title: Distance between point and arc


问:
Why we use subtract to represent distance
I am confused about that:
|x-y|= the distance between x and y

why do we use '-' to represent distance? Where does the principle come from?
I know that if y>0, x-y is exactly the distance; if y<0, x-y means you add the distance from the origin to y and x-y is the distance. But why is that? Somebody set this principle that we should use '+' present positive direction and '-' represent negative direction, and x-y represent reversing the negative to the positive then we can get the distance?
Maybe I am confused about '-' itself. Hoping get your answer! Thank you!

答:
We can think of one-dimensional space as a number line.

The  Ruler Postulate states that

The points on a line can be put into a one-to-one correspondence (paired) with the real numbers. The distance between any two points is represented by the absolute value of the difference between the numbers.

On a number line, the number that corresponds to a point is called the ordinate. Suppose, for example, that point $P$ has ordinate $5$ and point $Q$ has ordinate $-2$. It's an abuse of notation, but we often write this as $P=5$ and $Q=-2$. (A point is not "equal to" a number, it "corresponds to" a number.).

The distance between points $P$ and $Q$ can be written as $d(P,Q)$. So the ruler postulate tells us that

$$d(P,Q) = |5 - (-2)| = |(-2) - 5| = 7$$

In general, this can be written as $$d(P,Q) = |P-Q|$$

If $x$ is a real number, then $|x| = \sqrt{x^2}$. So we can write the above as

$$d(P,Q) = \sqrt{(P-Q)^2}$$

We can think of two-dimensional space as a coordinate plane.

How do we find the distance between two points $P$ and $Q$ on a plane? Here is a derivation of the formula

$$d(P,Q) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$

for the distance between the points $P$ and $Q$ with coordinates $P = (x_1, y_1)$ and $Q=(x_2,y_2)$.

We can write

$$P - Q = (x_1, y_1) - (x_2, y_2) = (x_1 - x_2, y_1 - y_2)$$

It is (well, used to be) common notation to express the length of the vector
$V=(x,y)$ as $\|V\| = \sqrt{x^2+y^2}$. Then the length of the vector $P-Q$ is
$$\|P-Q\| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$

It follows that $$d(P,Q) = \|P-Q\|$$

You should be able to see how this can be generalized to $n-$dimensional space.

By the way, the use of the double lines, $\|$, to distinguish between the number line and vector spaces in general is not required. $|P-Q|$ seems to be used just as often as
$\|P - Q\|$.


问:
Prove that $\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$ doesn't exist without L'hopital rule
I need to prove that the following limit doesn't exist:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$$

Itried to simplify $1+\cos(2x)$ into $2\cos^2(x)$, in order to calculate both of the one-sided limits, but I am not sure how to calculate it without using L'hopital rule.

答:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}=\lim_\limits{y\to 0}\frac{\sin(3y)}{1-\cos(2y)}$$ Using Taylor series and long division
$$\frac{\sin(3y)}{1-\cos(2y)}=\frac{3}{2 y}-\frac{7 y}{4}+O\left(y^3\right)$$


问:
Which function is represented in this contour plot?
This is a kind of unusual question, but I am pretty sure the function in this contour plot has an analytic form. I just don't remember what it was.

I found it in Stephen Boyd & Lieven Vandenberghes book (figure 3.2 page 113), but there seems no formula given.

Does anybody know what function this is? I would also help to know the formula for a function similar to this.

答:
The paper Determining an Aesthetic Inscribed Curve describes how to generate smooth convex curves from convex polygons (Voronoi cells in that paper). It describes both implicit and parametric formulations.
Here is a sample image from that paper:


问:
Harmonic conjugate of $u(r,\theta)=r^3\cos(3\theta)$

Verify that the following function $u$ are harmonic, and in each case give a conjugate harmonic function $v$ (i.e $v$ such that $u+iv$ is analytic)

$$u(r,\theta)=r^3\cos(3\theta)$$

I need to find the harmonic conjugate of the equation but I can't seem to prove it is harmonic. Does anybody got any idea? Thank you so much.

答:
$z^3$ is holomorphic, hence $\Re(z^3)$ and $\Im(z^3)$ are harmonic conjugates; writing

$z = re^{i\theta}, \tag 1$

we see that

$z^3 = (re^{i\theta})^3 = r^3(e^{i\theta})^3 = r^3(\cos 3\theta + i \sin 3 \theta) =r^3\cos 3\theta +  ir^3 \sin 3 \theta); \tag 2$

in follows that

$\Re(z^3) = r^3\cos 3\theta  \tag 3$

and

$\Im(z^3) = r^3\sin 3\theta, \tag 4$

are both harmonic functions, with $r^3\sin 3\theta$ conjugate to $r^3\cos 3\theta$.


问:
Possible arrangements of the word ARRANGEMENT
In how many the alphabets of the word ARRANGEMENT can be arranged such that :

Exactly two pairs of consecutive identical alphabets appear in the arrangement?

At least two pairs of consecutive identical alphabets appear in the arrangement?

What I've tried so far:

The alphabets that can be repeated should appear more than once in the given word, these are $$\text{AA,EE,NN,RR}$$

The answer to the first part is given by :
$$\binom{4}{2}\left[\frac{9!}{\left(2!\right)^{2}}-\binom{2}{1}\frac{8!}{\left(2!\right)^{1}}+\binom{2}{2}\frac{7!}{\left(2!\right)^{0}}\right]$$

For the second question it's needed to subtract the cases where less than two pair of consecutive alphabets appear in the arrangement from the total arrangements of the given word:

$$\frac{11!}{\left(2!\right)^{4}}$$
$$-\left[\underbrace{\frac{11!}{\left(2!\right)^{4}}-\binom{4}{1}\frac{10!}{\left(2!\right)^{3}}+\binom{4}{2}\frac{9!}{\left(2!\right)^{2}}-\binom{4}{3}\frac{8!}{\left(2!\right)^{1}}+\binom{4}{4}\frac{7!}{\left(2!\right)^{0}}}_{\text{Arrangements with no pair of consecutive identical alphabets}}\right]$$
$$-\binom{4}{1}\left[\underbrace{\frac{10!}{\left(2!\right)^{3}}-\binom{3}{1}\frac{9!}{\left(2!\right)^{2}}+\binom{3}{2}\frac{8!}{\left(2!\right)^{1}}-\binom{3}{3}\frac{7!}{\left(2!\right)^{0}}}_{\text{Arrangements with only one pair of consecutive identical alphabets}}\right]$$
$$=\color{green}{398160}$$

But the other way I came with was to compute the number of arrangements with two pairs of consecutive identical alphabets which is given by $$\binom{4}{2}\frac{9!}{\left(2!\right)^{2}}$$

I thought that these are the arrangements that at least two such pairs will appear but I don't understand why this way gives the wrong answer?

答:
On your second approach for at least two pairs of consecutive identical letters,

Yes $\binom{4}{2} \frac{9!}{\left(2!\right)^{2}}$ contains all arrangements of at least two doubles. The problem is that it overcounts them.

For any two given pairs (say $AA, EE$), number of arrangements is $\frac{9!}{(2!)^{2}}$ and this contains all arrangements of triples of $(AA, EE, NN)$ and $(AA, EE, RR)$ - two possible triples for the given two pairs. But when your given two pairs are (say $EE, NN$), your arrangements again include triples of $(AA, EE, RR)$.

In short there are only $\binom{4}{3}$ ways of choosing at least $3$ pairs of two identical letters but you are counting them $\binom{4}{2} \times 2 \,$ times. So you need to subtract those extra $8$ times.

Similarly for arrangements of $4$ pairs, there is only one way to choose them. But each of the $\binom{4}{2} = 6 \,$ given two pairs counted them once. Then when we subtracted $8$ times of at least three pairs, we subtracted them $8$ times and so we are net $(-2)$ on arrangements of $4$ pairs. We need to add $3$ arrangements of $4$ pairs to make sure we are counting all $4$ pairs once. Hence our answer should be -

$\displaystyle \binom{4}{2} \frac{9!}{\left(2!\right)^{2}} - 2 \times \binom{4}{3} \frac{8!}{\left(2!\right)^{1}} + 3 \times \binom {4}{4} \frac{7!}{\left(2!\right)^{0}}$


问:
What is the intuition behind R squared of linear regression? from the perspective of vector projection

I copied this photo from Anupam Kundu's answer on Quora.

As explained on Quora, linear regression can be imagined as a projection from the dependent variable on the hypo plane defined by the columns of $X$ matrix.

Now,  if go one step further, what is the geometric meaning of $R^{2}$ in such projection? Can anyone share some ideas?

答:
I don't think you can see the $R^2$ in this illustration since it is a a so-called "dimensionless" scalar
$$
R^2 = \frac{Y^T(H-\frac{1}{n}J)Y}{Y^T(1-J\frac{1}{n})Y} =
 1 - \frac{Y^T(I-H)Y}
{Y^T(1-J\frac{1}{n})Y},
$$
where $H = X(X^TX)^{-1}X^T$ is the projection matrix onto the column space of $X$, and $J = \mathrm{1}\mathrm{1}^T$.


问:
Express your answer using factorial notation
If $ f(x) = \frac{x^{46} + x^{45} + 2}{1 + x} $, calculate $ f^{(46)}(3) $. (Express your answer using factorial notation.)

I simplificated the function :  $(3^{45}+ \frac{1}{2} )^{46} $  and i don't know how to express this answer using factorial notation give me a hint. I think that I must use BINOMIAL formula but show me the next step

Body: Following @Matti.P's explanation that $f^{(n)}(x)$ means $\tfrac{d^nf}{dx^n}$, note $\frac{x^{46}+x^{45}+2}{1+x}=x^{45}+\frac{2}{1+x}$, so$$f^{(46)}(x)=2\tfrac{d^{46}}{dx^{46}}\tfrac{1}{1+x}=\tfrac{46!2}{(1+x)^{47}}$$(you may need to prove by induction that $\tfrac{d^n}{dx^n}\tfrac{1}{1+x}=\tfrac{n!(-1)^n}{(1+x)^{n+1}}$ if the result is unfamiliar). Hence $f^{(46)}(3)=\tfrac{46!}{2^{93}}$.


答:
To prove that the limit doesnot exist we proceed as following:
$$\lim_\limits{x\to \frac{\pi}{2}}\frac{\cos(3x)}{1+\cos(2x)}$$
Put $x = \frac{\pi}{2}$ to get
$$ \frac{\cos(\frac{3\pi}{2})}{1+\cos(2\frac{\pi}{2})} = \frac{0}{0} $$ which is the indeterminate form. Now to prove that the limit doesnot exist we consider the Left Hand Limit and the Right Hand Limit for the function.

Left hand limit:
Let $x$ approach $\frac{\pi}{2}$ from a value slightly lower than it i.e. $x \lt \frac{\pi}{2}$. This is the intuition for LHL. Formally,
$$\lim_\limits{x\to {\frac{\pi}{2}}^-}\frac{\cos(3x)}{1+\cos(2x)}$$
Put, $$x = \frac{\pi}{2} - h$$ where $h$ is an infinitely small positive number. Now as $$x = \frac{\pi}{2}-h$$ $x$ tends to $\frac{\pi}{2}$,  $h$ tends to $0$ and the limit becomes:

$$\lim_\limits{x\to {\frac{\pi}{2}^-}}\frac{\cos(3x)}{1+\cos(2x)} = \lim_\limits{h \to 0}\frac{\cos3(\frac{\pi}{2}-h)}{1+\cos2(\frac{\pi}{2}-h)}$$
$$=\lim_\limits{h \to 0}\frac{-\sin3h}{1-\cos2h}$$
Solve it further to get,
$$\lim_\limits{h \to 0}\frac{-\sin3h}{1-\cos2h} = \lim_\limits{h \to 0}\frac{-2\sin(\frac{3h}{2})\cos(\frac{3h}{2})}{2\sin^2{\frac{h}{2}}} = \lim_\limits{h \to 0}\frac{-3}{\sin\frac{h}{2}} = \frac{-3}{0} = -\infty$$
PS: Take care of the negative sign.

Proceed the same for the RHL:

Right hand limit:
Let $x$ approach $\frac{\pi}{2}$ from a value slightly above than it i.e. $x \gt \frac{\pi}{2}$ $$\lim_\limits{x\to {\frac{\pi}{2}}^+}\frac{\cos(3x)}{1+\cos(2x)}$$
Put, $$x = \frac{\pi}{2} + h$$ where $h$ is an infinitely small positive number. Now as $$x = \frac{\pi}{2}+h$$ $x$ tends to $\frac{\pi}{2}$,  $h$ tends to $0$ and the limit becomes:
$$\lim_\limits{x\to {\frac{\pi}{2}^+}}\frac{\cos(3x)}{1+\cos(2x)} = \lim_\limits{h \to 0}\frac{\cos3(\frac{\pi}{2}+h)}{1+\cos2(\frac{\pi}{2}+h)}$$
$$=\lim_\limits{h \to 0}\frac{\sin3h}{1-\cos2h}$$
Solve it further to get,
$$\lim_\limits{h \to 0}\frac{\sin3h}{1-\cos2h} = \lim_\limits{h \to 0}\frac{2\sin(\frac{3h}{2})\cos(\frac{3h}{2})}{2\sin^2{\frac{h}{2}}} = \lim_\limits{h \to 0}\frac{3}{\sin\frac{h}{2}} = \frac{3}{0} = \infty$$

You can see from the above analysis that,
$$LHL = -\infty$$
$$RHL = \infty$$
$$LHL \neq RHL$$
So the limit doesnot exist. Hope this helps...


问:
Find triple functions $ (g_0,g_1,g_2)$ such that $g_0+g_1'+g_2'' = \delta_0-\delta_1$
I want to find a triple of compactly supported continuous functions $ (g_0,g_1,g_2)$ on $\mathbb{R}$ such that $$g_0+g_1'+g_2'' = \delta_0-\delta_1$$

This is seemingly not so hard but ive broken my over this,
tried combinations of $x\cdot \mathbb{I}_{[0,1]} $ and other functions but I just dont get it..

Suggestions as to what might work are greatly appreciated.

答:
Let's start with finding a continuous $f$ with $f'' = \delta_0 - \delta_1$. For some $a \in \mathbb R$, we have
$$ f'(x) = a + \mathbb 1_{[0,1]}(x)  = \begin{cases} a & x \not\in [0,1] \\ a + 1 & x \in [0,1] \end{cases} $$
(denoting by $\mathbb 1_E$ the indicator function of a set $E \subseteq \mathbb R$)
and hence by integrating once more for some $b \in \mathbb R$
$$ f(x) = ax + x\mathbb 1_{[0,1]}(x) + \mathbb 1_{(1,\infty)}(x) + b $$
lets choose $a = b = 0$, leaving us with
$$ f(x) = x\mathbb 1_{[0,1]}(x) + \mathbb 1_{(1,\infty)}(x) $$
Now let $\eta \in C^\infty_c(\mathbb R)$ be a test function with $\eta|_{[-1, 2]} = 1$ and $g_2 = f\eta$, then
$$ g_2'' = f''\eta + 2f'\eta' + f\eta'' $$
Now $f''\eta = f''$, as $\eta$ equals 1 in an open neighbourhood of the support of $f''$. If we let now $g_1 = -2f\eta'$, we have $g_1' = -2f'\eta' - 2f\eta''$, giving
$$ g_2'' + g_1' = f'' - f\eta'' $$
now let $g_0 = f\eta''$ and get
$$ g_2'' + g_1' + g_0 = f'' = \delta_0 - \delta_1. $$</p>


问:
Does a cone in $\mathbb{R}^3$ admit a warped geometry?
Let

$$C = \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 \leq r^2 z^2 \text{ and } z > 0 \}$$

be a certain convex cone in $\mathbb{R}^3$, where $r > 0$.

Let $D$ the the closed unit disk in $\mathbb{R}^2$. Is it possible to find a Riemannian metric $g$ on $D \times (0, \infty)$ of the form $g = f(t)^2 dx^2 + dt^2$ ($dx^2$ is the flat metric on $D$ and $dt^2$ is the usual metric on $\mathbb{R}$) such that $C$ is isometric to $(D \times (0, \infty), g)$?

答:
Here is an answer to the title "does the cone admit a warped product metric" and the answer is yes:

One natural wraped product on $C$ will be induced by a spherical section. In $\mathbb{R}^3$, the polar decomposition gives a natural metric on $\mathbb{R}^3 \setminus \{0\}$ which is
$$
\mathrm{d}r^2 + r^2 \overset{\circ}{g}
$$
where $\overset{\circ}{g}$ is the round metric of the unit sphere. It gives an isometry between $(0,\infty)\times \mathbb{S}^2$ and $\mathbb{R}^3\setminus\{0\}$.

As $C$ is a natural riemannian submanifold of $\mathbb{R}^3 \setminus \{0\}$, in polar coordinates, it is isometric to $(0,\infty) \times ({C\cap\mathbb{S}^2})$. Note that $C\cap \mathbb{S}^2$ is a "round" disk. Thus, one can find a diffeomorphism $f:D \to C\cap \mathbb{S}^2$ and pull-back the round metric on $D$.

One then has an isometry $(0,\infty) \times D \to C$, in which metric is the warped metric
$$
\mathrm{d}r^2 + r^2 f^*\left(\overset{\circ}{g}|_{C\cap \mathbb{S}^2} \right)
$$
So one has a wraped metric on $(0,\infty)\times D$. But indeed, the metric on $D$ here is not flat (it has sectionnal curvature $1$).

Now, here is a remark, more than an answer, to the question "can we give a product warped metric that involves the flat metric on the cone":

Note that if $\mathrm{d}r^2 + f(r)g_D$ is a warped product of the cone involving the flat metric, in these coordinates $(r,p) \in (0,\infty)\times D$, the level surfaces $\{r_0\}\times D$ are orthogonal to the geodesics $(0,\infty)\times\{p_0\}$. Then, the $"r"$ coordinate will not be the distance to the origin in $\mathbb{R}^3$. Otherwise, these radial geodesics in $\mathbb{R}^3$ will be orthogonal to a same flat disk, which is not true. This shows that there exists in the cone a flat disk orthogonal to a familly geodesics in $\mathbb{R}^3$, and the cone would be foliated by a family of geodesic rays all orthogonal to the same flat disk. This gives me the intuition that the answer will be "no", but I do not have a formal proof for this right now, and I may be wrong.


问:
Find kernel of a homomorphism $f$
I have $f: \mathbb{Z}^2 \rightarrow \mathbb{Z}^2$ , $f$ is homomorphism such that $f(1,0) = (2,-1)$ and $f(0,1) = (-6,5)$. Find kernel of $f$. My progress:

$\ker(f)=\{(a,b)\in \mathbb{Z}^2: f(a,b)=(0,0) \}$

$f(a,b)=f(a,0)+f(0,b)=af(1,0)+bf(0,1)=a(2,-1)+b(-6,5)=(2a-6b,-a+5b) = (0,0)$

Now I get $2a-6b=-a+5b \Rightarrow b = 3/11a$. However, this seems to be wrong:

$b = 3/11a \Rightarrow (-4/11a,4/11a)=(0,0)$. If $a=11$,    $(−4/11a,4/11a)\neq (0,0)$.

So, what is wrong here?

Title: Find kernel of a homomorphism $f$

Body:
$$f(a,b) = (0,0)\iff  (2a-6b, 5b-a) = (0,0)$$
$$\iff a = 3b, a = 5b$$
$$\iff a = b = 0$$

Hence, $\ker f = \{(0,0)\}$ and $f$ is injective.


答:
First set $$f(x) = \left\{\begin{array} 0 0 &:if\quad x\leq 0 \\ x &:if\quad 0\leq x\leq 1 \\ 1 &:if\quad 1\leq x\end{array}\right.$$

Then $f''(x) = \delta_0 - \delta_1$

Now suppose I can find  some twice differentiable function $g$ with $g(x) = 0$ for sufficiently small $x$ and $g(x) = 1$ for sufficiently large $x$.

Then
$$\begin{array}{rcl}g_1(x)&=&0 \\ g_2(x) &=& g'(x) \\ g_3(x) &=& f(x) - g(x) \end{array}$$
Will do nicely.

I assume you won't have much trouble finding a $g$.

Title: Find triple functions $ (g_0,g_1,g_2)$ such that $g_0+g_1'+g_2'' = \delta_0-\delta_1$

Body:
I want to find a triple of compactly supported continuous functions $ (g_0,g_1,g_2)$ on $\mathbb{R}$ such that $$g_0+g_1'+g_2'' = \delta_0-\delta_1$$

This is seemingly not so hard but ive broken my over this,
tried combinations of $x\cdot \mathbb{I}_{[0,1]} $ and other functions but I just dont get it..

Suggestions as to what might work are greatly appreciated.


问:
Cayley's theorem and unique groups of particular order?
We have groups of prime order that are unique for that order as well as groups of order $pq$ where $p,q$ are prime and $p \nmid (q-1)$ that are unique. However, Cayley's theorem shows that there's a symmetric group that is isomorphic to that group.

So shouldn't there be at least 2 groups rather than 1 unique group for those specific orders (the group itself and another $S_n$ group)?

答: Cayley's theorem shows that there is a subgroup of a symmetric group that is isomorphic to that group, and in your cases that subgroup happens to be cyclic.


答:
The homomorphism $f$ is injective because
$$\det f=\begin{vmatrix}2&-6\\-1&5\end{vmatrix}=4\ne 0.$$

More generally, over a commutative ring $R$, an endomorphism of a free $R$-module is injective if and only if $\det f$ is a non-zero divisor – which means nonzero if $R$ is an integral domain.

Title: Find kernel of a homomorphism $f$

Body:
I have $f: \mathbb{Z}^2 \rightarrow \mathbb{Z}^2$ , $f$ is homomorphism such that $f(1,0) = (2,-1)$ and $f(0,1) = (-6,5)$. Find kernel of $f$. My progress:

$\ker(f)=\{(a,b)\in \mathbb{Z}^2: f(a,b)=(0,0) \}$

$f(a,b)=f(a,0)+f(0,b)=af(1,0)+bf(0,1)=a(2,-1)+b(-6,5)=(2a-6b,-a+5b) = (0,0)$

Now I get $2a-6b=-a+5b \Rightarrow b = 3/11a$. However, this seems to be wrong:

$b = 3/11a \Rightarrow (-4/11a,4/11a)=(0,0)$. If $a=11$,    $(−4/11a,4/11a)\neq (0,0)$.

So, what is wrong here?


问:
$\underset{i\rightarrow\infty}{\lim}p_i=p$; $f$ is continuous if and only if $\underset{i\rightarrow\infty}{\lim}f(p_i)=f(p)$
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$. Take a sequence $\{p_i\}_{i\ge1}\in\mathbb{R}^n$ such that $\underset{i\rightarrow\infty}{\lim}p_i=p$. Then. $f$ is continuous if and only if $\underset{i\rightarrow\infty}{\lim}f(p_i)=f(p)$.

Proof.

Only if part:

$f$ is continuous $\Rightarrow f$ is continious at $p$ $\Rightarrow\forall\varepsilon>0~\exists\delta_{\varepsilon,p}>0$ such that $\forall y\in\mathbb{R}^n$, if $\|y-p\|<\delta_{\varepsilon,p}$ then $\|f(y)-f(p)\|<\varepsilon$.

Also, $\underset{i\rightarrow\infty}{\lim}p_i=p\Rightarrow\forall\varepsilon>0~\exists N_\varepsilon\in\mathbb{N}$ such that $\forall n\ge N_\varepsilon$, $\|p_i-p\|<\varepsilon$.

Fix $\varepsilon>0$ arbitrarily. Then, for $\delta_{\varepsilon,p}~\exists N\in\mathbb{N}$ such that $\forall n\ge N$, $\|p_i-p\|<\delta_{\varepsilon,p}$$\Rightarrow$$\|f(p_i)-f(p)\|<\varepsilon$.

$\because\varepsilon>0$ was arbitrarily fixed,

$\therefore\forall\varepsilon>0~\exists N\in\mathbb{N}$ such that $\forall n\ge N$, $\|f(p_i)-f(p)\|<\varepsilon\Rightarrow\underset{i\rightarrow\infty}{\lim}f(p_i)=f(p)$.

The "only if part" only seems to require the fact that $f$ is continuous at $p$; everywhere continuity is redundant. Besides, if we are only given information for a given point $p$, it would be weird for it to affect continuity or discontinuity everywhere so, I think the statement needs to be modified to include $f$ is continuous at $p$. Also, I am not sure how to prove the "if part".

答:
The statement is incorrect (at the very least, it is extremely misleading) in the form it is currently phrased. Here are the correct statements:

Let $f:\Bbb{R}^n\to \Bbb{R}^m$ be a function, and fix a point $p\in\Bbb{R}^n$. Then, $f$ is continuous at $p$ if and only if for every sequence $\{p_i\}_{i=1}^{\infty}$ in $\Bbb{R}^n$, $\lim\limits_{i\to \infty}p_i=p$ implies $\lim\limits_{i\to \infty}f(p_i)=f(p)$.

This is the pointwise version. The "global" version (which follows immediately from the pointwise version) is

Let $f:\Bbb{R}^n\to \Bbb{R}^m$ be a function. Then, $f$ is continuous on $\Bbb{R}^n$ if and only if for every point $p\in\Bbb{R}^n$ and every sequence $\{p_i\}_{i=1}^{\infty}$ in $\Bbb{R}^n$, $\lim\limits_{i\to \infty}p_i=p$ implies $\lim\limits_{i\to \infty}f(p_i)=f(p)$.


问:
Function equal its integral implies zero function
Let $f:\mathbb{R}\to \mathbb{R}$ be continuous such that
$$f(x) = \int_0^xf(s)ds $$
and $f(1) = 0$. Can we prove that $f=0$ without resorting to ODE solving techniques?

答:
There is a straightforward way to get your result. You will have
$$
f(1)=\int_0^1f(s)ds=0.
$$
This identity can be solved taking for example $f(s)=Ce^s$ because the exponential has the property to be equal to its integral. Putting this in the above equation will yield $C=0$ and you are done.


问:
How to construct a symmetric and idempotent matrix from given two non-colinear vectors?
Given a vector $x\in \Bbb R^n$ it is easy to see that $A:=\dfrac{xx^t}{x^tx}$ is symmetric and idempotent. i.e. $A^t=\dfrac{(xx^t)^t}{x^tx}=\dfrac{((x^t)^t(x)^t)}{x^tx}=A$ and $A^2=\dfrac{xx^txx^t}{x^txx^tx}=\dfrac{xx^t}{x^tx}=A$. and this matrix is same as the matrix of $z:=\alpha x$ for all $\alpha\neq 0$.

I want to see the 2-dimensional case. e.g. Given two non-colinear vectors $x,y\in \Bbb R^n$, the matrix $B:=\dfrac{xx^t}{x^tx}+\dfrac{yy^t}{y^ty}$ is clearly symmetric. But is it also idempotent and independent of the vectors $x,y\in \text{plane}(x, y)$? (I think it is not because of the sum or subtraction of two idempotent matrices is not idempotent most of times.) i.e. $B^2=B$ and $B=\dfrac{ww^t}{w^tw}+\dfrac{zz^t}{z^tz}$ for all non-colinear vectors $w, z\in \text{plane}(x, y)\subset \Bbb R^n$.

My try: $B^2=B+\dfrac{xx^tyy^t}{x^txy^ty}+\dfrac{yy^txx^t}{x^txy^ty}$ that seems it has some extra terms. So my construction is not working.

How to construct a symmetric and idempotent matrix from given two non-colinear vectors $x,y\in \Bbb R^n$?

答:
When $x$ and $y$ are linearly independent, $u=\frac{x}{\|x\|}+\frac{y}{\|y\|}$ and $v=\frac{x}{\|x\|}-\frac{y}{\|y\|}$ form an orthogonal basis of the linear span of $x$ and $y$. Therefore $B=\frac{uu^T}{u^Tu}+\frac{vv^T}{v^Tv}$ is an orthogonal projection onto the linear span of $x$ and $y$. Unfortunately, this method does not generalise to three or more vectors.

For a truly general construction, use Moore-Penrose pseudoinverse. For any complex matrix $A$, $AA^+$ is the orthogonal projection onto the column space of $A$. In particular, when $A$ is a real matrix of full column rank, we have $AA^+=A(A^TA)^{-1}A^T$.


问:
On proving $\ker(TT^*+T^*T)=\ker(T^*)\cap \ker(T)$
Let $T:V\to V$ be a linear transformation of finite dimensional inner product space.

I want to show that  $$\ker(TT^*+T^*T)=\ker(T^*)\cap \ker(T).$$

I showed already that $$\ker(T^*)\cap \ker(T)\subseteq \ker(TT^*+T^*T).$$

How do I showing the other inclusion?

($T^*$ is adjoint of $T$)

Thank you!

答:
Let $x \in \ker(TT^* + T^*T)$. Then $TT^*x = - T^*Tx$, hence
\begin{align*}
  \|Tx\|^2 &= (Tx,Tx)\\
           &= (x, T^*Tx)\\
           &= -(x, TT^*x)\\
           &= -(T^*x, T^*x)\\
           &= -\|T^*x\|^2
\end{align*}
As both $\|Tx\|^2$ and $\|T^*x\|^2$ are non-negative, this is only possible for $Tx = T^*x = 0$.</p>


问:
Prove that there are positive real solutions to $x+y=xyz \land xy=x+z \land y+z=yz$
For $x,y,z \in \mathbb{R}_{> 0}$, how can I prove that there are solutions to this set of equations,

$$x+y=xyz \land xy=x+z \land y+z=yz$$
in a legitimate way, not by feeding it to WolframAlpha?

I have found out that

$x = \frac{y}{yz-1} = \frac{z}{y-1}$

$y = \frac{x}{xz-1} = \frac{x+z}{x} = \frac{z}{z-1}$

$z = \frac{x+y}{xy} = xy-x = \frac{y}{y-1}$

but I am not sure how to progress from there.

答:
The solutions can be parametrized by
$$
x=y - 2, \quad z=y^2 - 3y + 2,
$$
where $y$ has to satisfy the equation
$$
y^3 - 4y^2 + 4y - 2=0.
$$
Every cubic polynomial has a real root, and this one has a positive one satisfying $y>2$. So we are done, since $x,z>0$ for any $y>2$.


问:
Proving a convex function is of bounded variation
Let $f: I \rightarrow \mathbb{R}$ be a convex function, where $I$ denotes a closed interval. If $f$ is convex, how would I prove that $f$ is of bounded variation?

答:
A small variation (!) of Jonathan's answer: Let $f: [a, b] \to \Bbb R$ be convex. Let
$$
 c = \sup \{ x \in [a, b] \mid \text{$f$ is decreasing on $[a, x]$} \} \, .
$$
Then $f$ is decreasing on $[a, c]$ and increasing on $[c, b]$, and therefore
$$
V_a^b(f) = V_a^c(f) + V_c^b(f) = (f(a)-f(c)) + (f(b) - f(c)) = f(a) - 2f(c) + f(b) < \infty \, .
$$

In other words: a convex function is piecewise monotone and therefore of bounded variation on a compact interval.


问:
Order of the stabiliser of a vector
$\overrightarrow{e}_1=(1\space 0 \space 0\space 0\space 0\space 0)^T$, $\overrightarrow{e}_2=(0\space 1 \space 0\space 0\space 0\space 0)^T$,... $\overrightarrow{e}_6=(0\space 0 \space 0\space 0\space 0\space 1)^T$ is the canonical base of the vector space $\mathbb{R}^6$.

Define an action of the symmetric group $S_6$ on $\mathbb{R}^6$ by "permutation of the coordinates" as follows: For a permutation $\sigma$ in $S_6$ and a vector $\lambda_1\overrightarrow{e}_1 + ... \lambda_6\overrightarrow{e}_6$ in $\mathbb{R}^6$ we set $\sigma \cdot (\lambda_1\overrightarrow{e}_1 + ... \lambda_6\overrightarrow{e}_6)= \lambda_1\overrightarrow{e}_{\sigma(1)} + ... \lambda_6\overrightarrow{e}_{\sigma(6)}$

For this action, what is the order of the stabiliser of the vector $(1\space 2\space2\space 3\space 3\space3 )^T$ ?

What I know:

I think the group has order $6!$ and the stabiliser of the vector is a $6 \times 6$ matrix because the stabiliser $\cdot$ the vector has to give the vector. But I don't know how to compute the order of a matrix.

答:
We are looking for the permutations $\sigma$ such that
$$\sigma \cdot (1\space 2\space2\space 3\space 3\space3 )^T = (1\space 2\space2\space 3\space 3\space3 )^T$$
that is
$$1\overrightarrow{e}_{\sigma(1)} + 2\overrightarrow{e}_{\sigma(2)} +2\overrightarrow{e}_{\sigma(3)} + 3\overrightarrow{e}_{\sigma(4)}+ 3\overrightarrow{e}_{\sigma(5)}+ 3\overrightarrow{e}_{\sigma(6)} = (1\space 2\space2\space 3\space 3\space3 )^T$$
So, we can look at our options now. We must have $\sigma(1) =1$ since $\overrightarrow{e}_{\sigma(1)}$ is the only term with coefficient $1$. Similarly, we can have $\sigma(2) = 2 \text{ or }3$ and $\sigma(3) = 2 \text{ or }3$. And finally, we can have $\sigma(4) = 4 \text{ or } 5 \text{ or } 6$, $\sigma(5) = 4 \text{ or } 5 \text{ or } 6$ and $\sigma(6) = 4 \text{ or } 5 \text{ or } 6$ (Of course since $\sigma$ is a permutation, the choices for these are not independent from each other). Can you find the corresponding stabilizer elements in $S_6$ now?

Also note that stabilizer is not a $6 \times 6$ matrix since the action here is defined as $S_6 \times \mathbb{R}^6 \to \mathbb{R}^6$ so stabilizer of an element in $\mathbb{R}^6$ must be a subgroup of $S_6$.

Title: Order of the stabiliser of a vector


问:
Probability of sequence
The question is as follow: Let $(X_{n})_{n}$ be a sequence of Random variable that is independent with the probability

$P(X_{n}=1)=1-P(X_{n}=0)=\frac{1}{n}$

Show that $P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =1$

I tried to use the Borel-Cantelli Theorem to prove this, however I obtain that

$P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =0$

${ F }_{ n }:=\left\{ { X }_{ n }=0 \right\} $

$\sum _{ n=1 }^{ \infty  }{ P\left( { F }_{ n }^{ C } \right)  } =\sum _{ n=1 }^{ \infty  }{ 1-P\left( X_{ n }=0 \right)  } =\sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ n }  } =\infty$

By Borel-Cantelli, we know that $P\left( { \underset { n\rightarrow \infty  }\limsup  }{ F }_{ n }^{ C } \right) =1$

Since ${ \left( { \underset { n\rightarrow \infty  }{ \liminf }  }{ F }_{ n } \right)  }^{ C }=\left( { \underset { n\rightarrow \infty  }{\limsup}  }\quad { F }_{ n }^{ C } \right) $,

(ie) $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { F }_{ n } \right) =1-P\left( { \underset { n\rightarrow \infty  }{ \limsup }  }\quad { F }_{ n }^{ C } \right)$,

I obtain that $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { X }_{ n }=0 \right) =0$

Since this is the opposite of what I want to prove and the statement is said to be true, I can't see where I went wrong in my argument.

答:
Are you certain about the statement of the problem? I can't see what is wrong with your proof,  and using Borel 0-1 law (in French, but this is basically a statement similar to, while slightly easier to use than, the Borel-Canetelli lemma; see also Pro. 2.2 from these lecture notes with $F_n$ (independent events, with $\mathbb{P} F_n=\frac{1}{n}$), one also gets that
$$
 \mathbb{P}\{ \limsup_{n\to+\infty} F_n \} = 1
$$
since $\sum_{n=1}^\infty \mathbb{P} F_n = +\infty$. In other terms, infinitely many of the $F_n$'s occur, i.e. $X_n=0$ for infinitely many values of $n$. But then, one cannot have $X_n=0$ ultimately (that is, $\forall n \geq N$ for some $N$) w.p. $1$, which is the result you are apparently asked to show.

Edit: what I wrote is wrong (from the beginning), in the sense that it is not the right quantity that is considered. See Did's answer.

Title: Probability of sequence


答:
Note that the differential equation you derived implied that $$ \frac{\partial}{\partial y}\left( f(x+y)f(x-y)\right)=f'(x+y)f(x-y)-f(x+y)f'(x-y) =0.$$
And hence that $$f(x+y)f(x-y) = g(x).$$
We can specify the function $g$ by letting $y = 0$ to get
$$f(x)^2 = g(x)$$

and hence

$$f(x+y)f(x-y) = f(x)^2.$$

now sub $x=y$ and note that $f(0) = 0$ to get that

$$f(x)^2 =0$$

Title: Function equal its integral implies zero function

Body:
Let $f:\mathbb{R}\to \mathbb{R}$ be continuous such that
$$f(x) = \int_0^xf(s)ds $$
and $f(1) = 0$. Can we prove that $f=0$ without resorting to ODE solving techniques?


问:
Prove that the norm function $R \to \mathbb{Z}$ defined by $N(a+\sqrt db) = a^2 - db^2$ is a multiplicative integer-valued function

Let $R = \mathbb{Z}[\sqrt d]$ where $d$ is a square-free integer.
Prove that the norm function $R \to \mathbb{Z}$ defined by $N(a+\sqrt
db) = a^2 - db^2$ is a multiplicative integer-valued function, with $N(\alpha) = 0 $ if
only if $\alpha = 0$.

For this problem, should I simply show that for $\alpha, \beta \in R, N(\alpha\beta) = N(\alpha)N(\beta)$?  Is that what the question is asking for?

答:
$\begin{align}
&\text{Let }\;R=\mathbb{Z}[\sqrt d]\;\text{where }\;d\;\text{ is a square-free integer.}\\
&\text{Prove that the norm function }R\to\mathbb{Z}\text{ defined by }\\
&N(a+\sqrt db)=a^2-db^2\text{ is a multiplicative integer-}\\
&\text{valued function, with }N(\alpha)=0\text{ if only if }\alpha= 0\;.
\end{align}$

For all $\alpha,\beta\in R$, it results that

$N(\alpha\beta)=N\big((a_1+\sqrt db_1)(a_2+\sqrt db_2)\big)=$

$=N\big(a_1a_2+db_1b_2+\sqrt d(a_1b_2+a_2b_1)\big)=$

$=(a_1a_2+db_1b_2)^2-d(a_1b_2+a_2b_1)^2=$

$=a_1^2a_2^2+d^2b_1^2b_2^2+\color{blue}{2da_1a_2b_1b_2}-da_1^2b_2^2-da_2^2b_1^2-\color{blue}{2da_1a_2b_1b_2}=$

$=\color{brown}{a_1^2a_2^2}+\color{green}{d^2b_1^2b_2^2}-\color{green}{da_1^2b_2^2}-\color{brown}{da_2^2b_1^2}=$

$=a_2^2(a_1^2-db_1^2)-db_2^2(a_1^2-db_1^2)=$

$=(a_1^2-db_1^2)(a_2^2-db_2^2)=$

$=N(a_1+\sqrt db_1)N(a_2+\sqrt db_2)=$

$=N(\alpha)N(\beta)\;.$

Moreover,

$N(\alpha)=0\iff N(a+\sqrt db)=0\iff a^2-db^2=0\;.$

Now we are going to prove by contradiction that $\;b=0\;.$

Let $\;p\;$ be a prime factor of $\;d\;,\;$ then

$d=pd_1\;$ and $\;p\nmid d_1\;$ (indeed $\;d\;$ is square-free),

$b=p^{n}b_1\;$ where $\;n\in\mathbb{N}\cup\left\{0\right\}\;$ and $\;p\nmid b_1\;.$

If $\;b\ne0\;,\;$ then $\;db^2=pd_1p^{2n}b_1^2=p^{2n+1}d_1b_1^2\;,\;$ consequently

$a^2=db^2=p^{2n+1}d_1b_1^2\;,$

but it is impossible for the unique-prime-factorization theorem, indeed $\;a^2\;$ has the prime factor $\;p\;$ with an even exponent, while $\;p^{2n+1}d_1b_1^2\;$ has the prime factor $\;p\;$ with an odd exponent.

Hence $\;b=0\;.$

Since $\;a^2=db^2\;$ and $\;b=0\;,\;$ it follows that $\;a=0\;.$

Consequently, $\;\alpha=a+\sqrt db=0\;.$

So far we have proved that $\;N(\alpha)=0\implies\alpha=0\;.$

Conversely,

if $\;\alpha=0\;,\;$ then $\;N(\alpha)=N(0+\sqrt d\cdot0)=0^2-d\cdot0^2=0\;.$

Hence, $\;\alpha=0\implies N(\alpha)=0\;.$


问:
Fundamental theorem of finite Abelian group
In Gallian's Contemporary Abstract Algebra, he mentioned about this Greedy Algorithm for an Abelian group. It's basically a procedure of expressing a finite Abelian group as an internal direct product which is a result from fundamental theorem of finite Abelian group. But the validity of the algorithm is not clear for me.

The algorithm is given as follows:

Greedy Algorithm for an Abelian Group of Order $p^n$ ($p$ is prime)

Compute the orders of the elements of the group $G$.

Select an element $a_1$ of maximum order and define $G_1=\langle a_1\rangle$.

If $|G_1|=|G|$, stop. Otherwise, replace $i$ by $i+1$.

Select an element $a_i$ of maximum order $p^k$ such that $p^k\leq|G|/|G_{i-1}|$ and none of $a_i$, $a_i^p$, $a_i^{p^2}$,..., $a_i^{p^{k-1}}$ is in $G_{i-1}$ and define $G_i=G_{i-1}\times\langle a_i\rangle$.

Return to Step 3.

Internal direct product requires that none of the elements of $\langle a_i\rangle$ except identity should be in $G_{i-1}$ so why the condition in step 4 is weaker?

答:
As written, the algorithm is... poorly defined. I guess that you meant:

Let $i=1$, and $G_0= \emptyset$.

Compute the orders of the elements of the group G.

Select an element $a_i$ of maximum order $p^k$ such that $p^k\leq|G|/|G_{i-1}|$ (not sure this is necessary to state) and such that none of $a_i, a_i^{p}, \dots , a_i^{p^{k-1}}$ is in $G_{i-1}$ and define $G_i=G_{i-1} \times ⟨a_i⟩$.

If $|G_i|=|G|$, stop. Otherwise, replace $i$ by $i+1$. and repeat step 2.

That being said, you are asking why it is enough to restrict attention to powers of $a_i$ that have a power of $p$ as exponent. The answer is that $a_i^p$ and $a_i^{pq}$ whenever $q$ does not divide $p$ generate the same subgroup. In fact, $G$ is a $p$-group and all elements have order $p^m$ for some $m \in \mathbb{N}$, so $a_i^{pq}$ has to have the same order as $a_i^p$. So this seemingly weaker hypothesis is actually equivalent to the standard requirement.


问:
Formula for the differential of a left-invariant 1-form
I am currently working through the first volume of spivak's differential geometry and am currently on the chapter on Lie groups. After showing that the differential of a left invariant form is itself left invariant, he then does the following

$d\omega(\tilde{X}, \tilde{Y})=\tilde{X}(\omega(\tilde{Y}))-\tilde{Y}(\omega(\tilde{X}))-\omega([\tilde{X},\tilde{Y}])$
$=-\omega([\tilde{X},\tilde{Y}])$

to demonstrate a simplified formula for the differential of a 1-form.

Here, $\omega$ is left-invariant 1-form and $\tilde{X}$, $\tilde{Y}$ are left-invariant vector fields.

While I can understand the first equality (it is simply the coordinate-free formula for a 1-form), I can't understand what motivates the second equality.

Is there a reason the two left terms should be equal or is it that they both go to zero for left-invariant vector fields?

答:
Suppose $X$ and $Y$ are left invariant vector fields, and that $\omega$ is a left invariant $1$-form. Then
\begin{align}
\forall g \in G,~ X(g) &= g\cdot X(e) \\
Y(g)& = g\cdot Y(e) \\
\omega_g(X(g))&= \omega_e(g^{-1}X(g)) = \omega_e(g^{-1}g\cdot X_e) = \omega_e(X_e)\\
\omega_g(Y(g))&= \omega_e(g^{-1}Y(g)) = \omega_e(g^{-1}g\cdot Y_e) = \omega_e(Y_e)
\end{align}

Then, $g \mapsto \omega(X) (g)$ and $g\mapsto \omega(Y)(g)$ are constant functions, and thus, $X\cdot \omega(Y) = Y\cdot\omega(X) = 0$. Hence
$$
\mathrm{d}\omega(X,Y) = X\cdot \omega(Y) - Y\cdot \omega(X) - \omega\left([X,Y]\right) = - \omega([X,Y])
$$


问:
How to construct a self orthogonal Latin square of order 5
I have been unable to find an elegant method of constructing self orthogonal Latin squares. However, I came across this question: construct a self orthogonal Latin square of order 5 using the fact that the set of elements on the main diagonal of a self orthogonal Latin from a transversal. This seems to imply that you can use this fact alone to construct self orthogonal Latin squares or at-least ones of order five.  </p>

<p>I have been unable to figure out how I can do this question and would appreciate any help. </p>

答: Use the elements of $\;\mathbb{Z}_v$ as the names of the rows and columns of your Latin square. Let $\boldsymbol{A}=(a_{ij})$ such that $a_{ij}=2i-j\in \mathbb{Z}_v$. This forms a self-orthogonal Latin square always when the $\gcd(v,6)=1$, so it works for the order $5$ case as well.


问:
Cooley Tukey DFT splitting doubt (should be simple)
I can't understand the basic principle on which the Cooley Tukey algorithm is based, the algorithm says I can split in two parts the DFT computation like in the following

$$\begin{matrix} X_k= \underbrace{\sum \limits_{m=0}^{N/2-1} x_{2m}   e^{-\frac{2\pi i}{N/2} mk}}_{\mathrm{DFT\;of\;even-indexed\;part\;of\;} x_m} {} +  e^{-\frac{2\pi i}{N}k} \underbrace{\sum \limits_{m=0}^{N/2-1} x_{2m+1} e^{-\frac{2\pi i}{N/2} mk}}_{\mathrm{DFT\;of\;odd-indexed\;part\;of\;} x_m} =  E_k + e^{-\frac{2\pi i}{N}k} O_k.\end{matrix}$$
Formula 1.0

The algorithm then says that I may avoid the computation process of half the data because it is valid

$$\begin{matrix} X_k & =& \left\{\begin{matrix}E_k + e^{-\frac{2\pi i}{N}k} O_k & \mbox{if } k < N/2 \\ \\E_{k-N/2} - e^{-\frac{2\pi i}{N} (k-N/2)} O_{k-N/2} & \mbox{if }k \geq N/2. \end{matrix} \right. \end{matrix}$$ Formula 1.1

But I can't understand this.

I tried with an example: I took a simple vector

x=[0 1]

to transform with a DFT with Cooley Tukey, here's what I did: following the basic DFT formula

$$X_k = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} nk}$$

I'm getting

$$X_0 = 0 + 1$$

$$X_1 = 0 + e^{-\pi i}$$

And this should be the classic approach. Now if I split the x vector in two elements, I'll have two parts: one (the 0 element) with even indices and the other (the 1 element) with odd indices. Applying the 1.0 formula everything is going well, but the 1.1 formula isn't working..  I can't obtain X0 and X1 by using that

Where am I getting wrong?

答: Formula 1.1 seems to be working fine. We have $E_0=x_0=0$ and $O_0=x_1=1$. Since $N/2=1$, the upper row of 1.1 applies for $k=0$ and the lower row applies for $k=1$. Plugging $k=0$ into the upper row yields $X_0=E_0+O_0=x_0+x_1=0+1$, and plugging $k=1$ into the lower row yields $x_1=E_0-O_0=x_0-x_1=0-1$, which is the same as what you wrote, since $\mathrm e^{-\pi i}=-1$.</p>


答:
To prove the other inclusion : for $x \in \ker(TT^*+T^*T)$,
\begin{eqnarray}
0 &=& (TT^*x + T^*Tx,x) \\
&=& (TT^*x,x) + (T^*Tx,x) \\
&=& (T^*x,T^*x) + (Tx,Tx) \\
&=& \lVert T^*x \rVert^2 + \lVert Tx \rVert^2
\end{eqnarray}
And thus $Tx = T^*x = 0$.</p>

Title: On proving $\ker(TT^*+T^*T)=\ker(T^*)\cap \ker(T)$

Body:
Let $T:V\to V$ be a linear transformation of finite dimensional inner product space.

I want to show that  $$\ker(TT^*+T^*T)=\ker(T^*)\cap \ker(T).$$

I showed already that $$\ker(T^*)\cap \ker(T)\subseteq \ker(TT^*+T^*T).$$

How do I showing the other inclusion?

($T^*$ is adjoint of $T$)

Thank you!


问:
Factorizing Polynomial in a Quotient Ring
I'm trying to solve this question:
Let $F=\mathbb{F}_2[y]/(y^3+y+1)$.

Show that $F$ is a field

Factorize $x^3+x+1$ and $x^3+x^2+1$ into irreducible monic polynomials in $F[X]$.

I managed to prove the first part using $\mathbb{F}_2$ is a field $\implies \mathbb{F}_2[y]$ is a PID $\implies$ an ideal is maximal iff it is generated by an irreducible element and then showing that $y^3+y+1$ is irreducible. However, I'm stuck on the second part. Any help is appreciated. Is there a general method for tackling this sort of questions?

答:
We have $F=\mathbb F_2[\alpha]$, where $\alpha^3+\alpha+1=0$. Moreover, $F$ is a field with $8$ elements, all of the form $u\alpha^2+v\alpha+w$, with $u,v,w\in\mathbb F_2$. To factor those polynomials, just find which elements are roots of each.

For instance, $0=(\alpha^3+\alpha+1)^2=(\alpha^2)^3+\alpha^2+1$ and so $\alpha^2$ is another root of $x^3+x+1$. Divide $x^3+x+1$ by $x-\alpha$ and by $x-\alpha^2$ to find the third root.


答:
Don't think of the $S_{6}$ as a group of matrices, think of it as a group of permutations; so for example the permutation $(i,j) \in S_{6}$ acts by sending $e_{i} \mapsto e_{j}$, $e_{j} \mapsto e_{i}$, and $e_{k} \mapsto e_{k}$ for $k \neq i,j$.

Then it is clear that the stabilizer group of the vector $(1,2,2,3,3,3)$ contains the permutation $(2,3)$, and also contains all permutations on $\{4,5,6\}$; this gives a permutation group $S_{\{4,5,6\}} \simeq S_{3}$, which is generated by the permutations $(4,5)$, $(4,5,6)$.

So you need to compute the order of $\langle (2,3), (4,5), (4,5,6) \rangle$. The easiest way to do this is to recognize that it is isomorphic to $\langle (1,2) \rangle \times S_{\{4,5,6\}} \simeq \mathbb{Z}_{2} \times S_{3}$ which has order 12.

Title: Order of the stabiliser of a vector

Body:
$\overrightarrow{e}_1=(1\space 0 \space 0\space 0\space 0\space 0)^T$, $\overrightarrow{e}_2=(0\space 1 \space 0\space 0\space 0\space 0)^T$,... $\overrightarrow{e}_6=(0\space 0 \space 0\space 0\space 0\space 1)^T$ is the canonical base of the vector space $\mathbb{R}^6$.

Define an action of the symmetric group $S_6$ on $\mathbb{R}^6$ by "permutation of the coordinates" as follows: For a permutation $\sigma$ in $S_6$ and a vector $\lambda_1\overrightarrow{e}_1 + ... \lambda_6\overrightarrow{e}_6$ in $\mathbb{R}^6$ we set $\sigma \cdot (\lambda_1\overrightarrow{e}_1 + ... \lambda_6\overrightarrow{e}_6)= \lambda_1\overrightarrow{e}_{\sigma(1)} + ... \lambda_6\overrightarrow{e}_{\sigma(6)}$

For this action, what is the order of the stabiliser of the vector $(1\space 2\space2\space 3\space 3\space3 )^T$ ?

What I know:

I think the group has order $6!$ and the stabiliser of the vector is a $6 \times 6$ matrix because the stabiliser $\cdot$ the vector has to give the vector. But I don't know how to compute the order of a matrix.


问:
How to solve $(1-t)^2dy+(ty-y+3)dt=0$
I want to solve this differential equation: $(1-t)^2dy+(ty-y+3)dt=0$

I tried to use separation method:

$$(1-t)^2dy+y(t-1)dt+3dt=0$$

$$\frac{dy}{y}+\frac{dt}{t-1}+\frac{3}{y(t-1)^2}dt=0$$

But in the last fraction I still have $t$ and $y$. I don't know how to separate it.

答:
Hint

With all these $(1-t)$'s, let $y=\frac z{1-t}$ and it will be very simple.

Title: How to solve $(1-t)^2dy+(ty-y+3)dt=0$


问:
Entire function bounded on vertical lines $\Re(z) = \pm1$, but not between them.
Is there an entire function $f : \mathbb{C} \to \mathbb{C}$ such that $f$ is bounded on the vertical lines $\Re(z) = \pm1$, but is unbounded on the region between them i.e.
$$
\sup_{|\Re(z)| = 1} |f(z)| < \infty
\qquad\text{and}\qquad
\sup_{|\Re(z)| < 1} |f(z)| = \infty
$$

What approach should be taken to prove/disprove that such a function exists? If it exists can I define it explicitly?

答:
The simplest example is $f(z)=e^{e^{\pi iz/2}}$. Note that for $\Re z =\pm 1, z=\pm 1+iy$ one has $e^{\pi iz/2}=\pm ie^{-\pi y/2}$, so $|f(z)|=1$ there but for $\Re z=0, z=iy$, we have $f(z)=e^{e^{-\pi y/2}}$ which is highly unbounded as $y \to -\infty$

By Phragmen Lindelof, one needs an entire function of infinite order.


问:
Confidence Intervals under different conditions of Variance??
What's the difference in CI under the situation in part (a) and (b)??

A sample of 20 random streamflow observations produced a mean of 145m3/sec and a variance of 30 (m3/sec)2. What are the 95% confidence intervals on the mean assuming a normal distribution if (a) the true variance is estimated to be 30 (m3/sec)2; (b) the true variance is 30 (m3/sec)2. Discuss the reason why the confidence intervals computed for parts (a) and (b) are different.

答: Hint: In (a) we have an estimate of the true variance and is thus a random variable. Therefore, in part (a), the confidence interval for the mean will have to account for the additional variability in the estimated variance.


答:
$$(1-t)^2dy+(ty-y+3)dt=0$$
$$(1-t)^2dy+y(t-1)dt+3dt=0$$
$$u^2dy+yudu+3du=0$$
$$u(udy+ydu)+3du=0$$
$$d(yu)+\dfrac 3udu=0$$
Where $u=t-1$.
The differential equation is now exact.  Integrate:
$$yu+ 3\ln  |u|=C$$
$$\boxed {y(t-1)+ 3\ln  |t-1|=C}$$

Title: How to solve $(1-t)^2dy+(ty-y+3)dt=0$

Body:
I want to solve this differential equation: $(1-t)^2dy+(ty-y+3)dt=0$

I tried to use separation method:

$$(1-t)^2dy+y(t-1)dt+3dt=0$$

$$\frac{dy}{y}+\frac{dt}{t-1}+\frac{3}{y(t-1)^2}dt=0$$

But in the last fraction I still have $t$ and $y$. I don't know how to separate it.


问:
How many pillars do we need to surround a triangular area?
Suppose, I have a triangular garden with sides of $25$ feet, $40$ feet and $55$ feet. If I want to surround my garden with pillars set $5$ feet apart, how many pillars will I need?

I calculated the triangle's perimeter and divided it by $5$. I got 24.

Since it is a enclosed area, I will need one less pillar. So I will need a total of $23$ pillars.

But someone suggested me that I actually need $22$ pillars, but he didn't explain why. So am I miscalculating something?

答:
Small example first to see what is going on. If we have to cover a perimeter of $15$ feet, we would need $3+1$ pillars if it would be a straight line. Notice $15/5=3$, which corresponds to the number of gaps between the poles. It would look like:
$$ \underbrace{\Big| \text{5 feet}\Big| \text{5 feet}\Big|\text{5 feet}\Big|}_{15 \text{ feet }} $$
We see that there is one more pillar than there are gaps.

However, we have $3$ pillars if the endpoints were to be connected (which is the case for our triangle). Since when we get back to the starting point we see that we have already placed a pillar there.

Indeed $25+40+55=120$ then $120/5=24$. If we were to place them along a line we would have to place $25$ pillars, but for our triangle the endpoints overlap so we only need $24$ pillars.


答:
Note that your ODE is a linear equation. So we have
$$(1-x)^{2}dy+(xy-y+3)dx=0 \iff \frac{dy}{dx}+\frac{y}{x-1}=-\frac{3}{(x-1)^{2}}$$
Let the integrant factor $\displaystyle \mu(x)=\exp\left[ \int \frac{1}{x-1}dx \right]=x-1$.
Now, multiply both sides by $\mu$ we have $$(x-1)\frac{dy}{dx}+y=-\frac{3}{x-1}$$
Solving this part $$ \int \frac{d}{dx}((x-1)y)dx=-\int \frac{3}{x-1}dx\implies (x-1)y=-2\ln|x-1|+c $$
So, we have $$\boxed{y=\frac{-3\ln|x-1|+c}{x-1}}$$

Note: If you have a ODE of the form $$y'+P(x)y=Q(x)$$
So, the integrant factor is the form $$\mu(x)=\exp\left[\int P(x)dx \right]$$
Then, you can multiply both sides of the ODE and you can get $$\frac{d}{dx}(y\mu)=Q(x)\mu(x)dx$$
So, we have $$y\mu(x)=\int Q(x)\mu(x)dx+C$$
So, the general solution for the ODE linear equation of first order is $$\boxed{y(x)=\frac{1}{\mu(x)}\left[\int Q(x)\mu(x)dx+c \right]}$$

Title: How to solve $(1-t)^2dy+(ty-y+3)dt=0$


问:
If $v_1,..,v_n$ are linearly indpendent (LI) in $(V,\mathbb{F}')$, then they are LI in $(V,\mathbb{F})$ also.
Say we have a fields $\mathbb{F}\subset
\mathbb{F}'$.

If $v_1,..,v_n$ are linearly indpendent (LI) in
$(V,\mathbb{F}')$, then they are LI in
$(V,\mathbb{F})$ also.

Converse does not hold.

Proof: Write:
$$a_1v_1+a_2v_2+...+a_nv_n=0$$ where
$a_1,a_2,...a_n  \in\mathbb{F}$, so they are in $\mathbb{F}'$
Thus we have trivial linear combination in $(V,\mathbb{F}')$ where vectors are LI, so all $a_i$ are $0$ and thus given vectors are LI in $(V,\mathbb{F})$ also.

For converse we see that $(1,0)$ and $(\pi, 0)$ are LI in $(\mathbb{R}^2,\mathbb{Q})$ but in $(\mathbb{R}^2,\mathbb{R})$ are not.

Now I wonder if

the proof is correct and

is $(\mathbb{R}^2,\mathbb{Q})$ vector subspace of $(\mathbb{R}^2,\mathbb{R})$?

答:
The proof is correct yes. A simpler example would be $1$ and $\pi$: they are linearly independent if we see $\Bbb R$ as a vector space over $\Bbb Q$, but not of you see it as a vector space over $\Bbb R$.

And, not $\Bbb R^2$ as a vector space over $\Bbb Q$ is not a subspace of $\Bbb R^2$ as a vector space over $\Bbb R$. That's so because the concept of subspace is defined only for vector spaces over the same field.


问:
Non-standard form of Sobolev's embedding: Does $\Vert f\Vert_{L^2}^2\leq C \Vert f\Vert_{L^\infty}\Vert f\Vert_{H^1}$ hold??
I am wondering exactly the question of the title. Is it true that there exists a constant $C>0$ such that for all $f\in H^1(\mathbb{R})$ it holds $$
\qquad \qquad \Vert f\Vert_{L^2}^2\leq C \Vert f\Vert_{L^\infty}\Vert f\Vert_{H^1} \quad ?? \qquad (*)
$$
I know that by Sobolev's embedding we have that $$
\Vert f\Vert_{L^\infty}\leq C \Vert f\Vert_{H^1},
$$
and of course we have the trivial bound $\Vert f\Vert_{L^2}\leq \Vert f\Vert_{H^1}$. However, it is not clear to my if something like the inequality in $(*)$ holds. Does anyone has any thoughts?

答:
Such an inequality can't hold for scaling reasons: Let, for $a>0$, $f_a(x):= a^{-1/2}f(x/a)$. Then $\| f_a\|_{L^2}=\| f\|_{L^2}$, $\| f_a\|_{L^\infty}= a^{-1/2}\| f\|_{L^\infty}$, and $\| f_a\|_{H^1}= \| f_a\|_{L^2} + \| (f_a)'\|_{L^2}= \| f\|_{L^2} + a^{-1}\| f'\|_{L^2}$. Letting $a\to \infty$ we reach a contradiction if such an inequality held.

In fact this argument shows that if you want to replace the $L^\infty$ norm by an $L^q$ norm, then necessarily $q\leq 2$.


问:
$C^\infty$ extension operator
Let $B_r$ denotes the ball $B(0,r) \subset \mathbb{R}^3$ and $h \in ]0,1]$.

I am looking for the following extension operator :

$$P:C^\infty(B_h) \rightarrow C^\infty(B_2)$$

such that for all $u \in C^\infty(B_h)$, $P(u) \in C^\infty(B_2)$ verifies the following properties, for a fixed $s>3$

$P(u)=u$ in $B_h$

$||\nabla P(u)||_{(L^s(B_2))^3} \leq C_1 ||\nabla u||_{L^s(B_h)}$, with $C_1$ a positive constant independant of $h$ and $u$

My guess is that it might comes from a Sobolev operator extension. Any help or thoughts are welcomed !

答:
This cannot work for just $C^\infty$-functions. Take $u(x):=\frac1{h^2-\|x\|^2}$.
It is infinitely often differentiable and unbounded on $B_h$. So it cannot be extended to $B_2$.

Title: $C^\infty$ extension operator


问:
problem requiring proof by definition
I am solving real analysis where the author has directly used the below :-

If $$\lim_{n\to\infty}{x_n} = l \qquad , x_n > 0 \; \forall \;n \in \mathbb{N} $$
then,
$$\lim_{n\to\infty}{\log {x_n} = log \;l}$$

I am trying to prove the above by using the standard definition of limits .Here's my approach -

$$since , \;\: x_n > 0 \;\;\forall \;n \in \;\mathbb{N} \Rightarrow l \; \geqslant 0 $$
$$by\;definition\;of\;limits\;,|x_n-l| < \epsilon \;\forall\; n \in \mathbb{N}$$
We need to prove -
$$|\log x_n - \log l| \lt \epsilon \;\forall\; n\;\in \;\mathbb{N} $$
$$\equiv \qquad\qquad|log \frac{x_n}{l}|\; \lt \epsilon \; \forall \;n \; \in \;\mathbb{N}$$
$$\equiv \qquad\qquad -\epsilon\lt log \frac{x_n}{l} \lt \epsilon \;\forall\;n\in \mathbb{N}$$
$$\equiv \qquad\qquad e^{-\epsilon}\lt \frac{x_n}{l}\lt e^{\epsilon} \; \forall n\in \mathbb{N}$$
$$\equiv \qquad\qquad le^{-\epsilon}\lt x_n\lt le^{\epsilon} \; \forall n\in \mathbb{N}$$

$$since , x_n \lt l + \epsilon \;and \;l + \epsilon \lt le^{\epsilon}\; by \;expamsion \;of \;exponential\;series$$

$$Hence, x_n \lt le^{\epsilon}$$

But i am unable to prove the left inequality i.e.
$$le^{-\epsilon}\lt x_n$$

Any help in this regard or any other approaches is appreciated.

答:
I think your definition of a limit of a sequence is slightly off.

The definition says $\forall \epsilon>0$ there exists a natural number N, such that if $n \ge N$ then $|x_n-l|<\epsilon$

So as you showed in the preliminary work... if we want,

$|\log x_n - \log l| \lt \epsilon$,

then that means

$$le^{-\epsilon}\lt x_n\lt le^{\epsilon}$$

$$\iff le^{-\epsilon}-l\lt x_n-l\lt le^{\epsilon}-l$$

Note that $le^{-\epsilon}-l<0$ and $le^{\epsilon}-l>0$

So we choose an $\epsilon_2 = \min\{l-le^{-\epsilon},le^{\epsilon}-l\}$.

Now you can do the actual proof. Given an $\epsilon>0$ you need to show there exists a natural number N such that if $n \ge N$, $|\log x_n - \log l| \lt \epsilon$.

We let $\epsilon_2 = \min\{l-le^{-\epsilon},le^{\epsilon}-l\}$.

Since we are given that $\lim_\limits{n\to\infty}x_n = l$, we know there's a natural number N such that if $n\ge N$, $|x_n - l|\lt \epsilon_2$

So we need to prove from $|x_n-l|<\epsilon_2$ that $|\log x_n - \log l| \lt \epsilon$. This is essentially going backwards in your derivation.

$|x_n-l|<\epsilon_2 \implies$

$|x_n-l|<l-le^{-\epsilon}$ and $|x_n-l|<le^{\epsilon}-l$

So

$le^{-\epsilon}-l<x_n-l<l-le^{-\epsilon}$

and

$l-le^{\epsilon}<x_n-l<le^{\epsilon}-l$

From these two, we can recover:

$le^{-\epsilon}-l\lt x_n-l\lt le^{\epsilon}-l \implies$

$le^{-\epsilon}\lt x_n\lt le^{\epsilon}$

And so on... keep going backwards to recover $|\log x_n - \log l| \lt \epsilon$

Title: problem requiring proof by definition


问:
Raising non-diagonalizable Markov transition matrix to the power of $n$
Consider an aperiodic and irreproducible Markov process with absorbing states. For each non-absorbing state, how can I numerically compute the likelihood of ending up in each absorbing state?

Assume we have four non-absorbing states, and two absorbing states, with the following transition matrix:

$$A = \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1
\end{pmatrix}$$

By raising $A$ to the power of $n$, I may calculate the transition probability in $n$ steps between any two states. By letting $n$ converge to infinity, I will get the likelihood of ending up in each absorbing state after infinite time. In the above example I would get that the first non-absorbing state would transition into the first absorbing state, and all other non-absorbing states in the second absorbing state.

As the matrix $A$ is non-diagonalizable, I wonder how I could numerically compute $A^n$, with $n$ approaching infinity, in a robust and efficient way.

The above example of $A$ is a simplified version of my typical transition matrix that I provide here only for illustration. Typically, my transition matrix would be much larger, e.g. involving between 1,000 and 10,000 states, with typically however only a few links between states. (I have a cell grid, allowing for movements between adjacent cells, with each cell representing a state).

Is my only possibility of numerically computing $A^n$ in a robust way by iterative multiplication? If so, what would be a good criterion for termination, i.e., what measure could I use to assess the similarity between two iterations?

答:
$A$ can be written as $J=SAS^{-1}$ such that $J=D+N$ where $D$ is diagonal and $N$ is nilpotent, such decomposition is called Dunford decomposition and is computed using the Jordan normal form $J$. In this case:

$$A^n=SJ^nS^{-1}$$

Once $S,D$ and $N$ are known, you will be left to compute $N^n$ up to the order of the nilpotent matrix $N$, denoted $k_N$, then for $n<k_N$ and by commutativity of diagonal matrices :

$$J^n=(D+N)^n=\sum_{l=0}^n \binom{n}{l} N^lD^{n-l}.$$

For $n\ge k_N$:

$$J^n=\sum_{l=0}^{k_N} \binom{n}{l} N^lD^{n-l}+D^n$$

Indeed, $N^l=0$ as soon as $l\ge k_N$.

The efficiency of such procedure depends on how far you need to compute $n$ and how much evaluations you want to make. If only a few evaluations are required, using a recursive multiplication algorithm should be faster (but more greedy in terms of memory).


问:
Find upper sum to prove not Riemann integrability
Let $f: [0,1] \to \mathbb{R}$ be defined by $f(x)=0$ if $x$ is irrational and $f(x) = x+1$ if $x$ is rational. I want to prove using an argument by upper and lower sums that $f$ is not Riemann integrable. I can easily see that $L(f) = 0$, but I'm not sure how to find $U(f)$.

Any hint is appreciated.

答:
If you split the interval $[0,1]$ into $n$ (equal) subintervals, the maximum value of $f(x)$ on the $k$-th interval is $f\bigl(\frac kn\bigr)=\frac kn + 1$, so
$$U_n(f)=\frac1n\sum_{k=1}^n\Bigl(\frac kn+1\Bigr)=\frac1n\biggl(\frac{1+2+\dots+n}n +n\biggr).$$
Can you take it from there?


问:
Probability of sequence
The question is as follow: Let $(X_{n})_{n}$ be a sequence of Random variable that is independent with the probability

$P(X_{n}=1)=1-P(X_{n}=0)=\frac{1}{n}$

Show that $P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =1$

I tried to use the Borel-Cantelli Theorem to prove this, however I obtain that

$P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =0$

${ F }_{ n }:=\left\{ { X }_{ n }=0 \right\} $

$\sum _{ n=1 }^{ \infty  }{ P\left( { F }_{ n }^{ C } \right)  } =\sum _{ n=1 }^{ \infty  }{ 1-P\left( X_{ n }=0 \right)  } =\sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ n }  } =\infty$

By Borel-Cantelli, we know that $P\left( { \underset { n\rightarrow \infty  }\limsup  }{ F }_{ n }^{ C } \right) =1$

Since ${ \left( { \underset { n\rightarrow \infty  }{ \liminf }  }{ F }_{ n } \right)  }^{ C }=\left( { \underset { n\rightarrow \infty  }{\limsup}  }\quad { F }_{ n }^{ C } \right) $,

(ie) $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { F }_{ n } \right) =1-P\left( { \underset { n\rightarrow \infty  }{ \limsup }  }\quad { F }_{ n }^{ C } \right)$,

I obtain that $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { X }_{ n }=0 \right) =0$

Since this is the opposite of what I want to prove and the statement is said to be true, I can't see where I went wrong in my argument.

答:
In a nutshell, for $\{0,1\}$-valued random variables $(X_n)$,

$$[\liminf X_n=0]=\limsup\ [X_n=0]$$
  $$[\limsup X_n=0]=\liminf\ [X_n=0]\quad(=[\lim X_n=0])$$

The event $A=[\liminf X_n=0]$ is $A=\limsup A_n$ with $A_n=[X_n=0]$. Now, $\limsup A_n=\bigcap\limits_nB_n$ with $B_n=\bigcup\limits_{k\geqslant n}A_k$. For every $n$ and $k\geqslant n$, $A_k\subseteq B_n$ and $P[A_k]=1-1/k$ hence $P[B_n]\geqslant1-1/k$ hence $P[B_n]=1$. Thus, $P[A]=1$.

This does not use the independence property, only the hypothesis that $P[X_n\ne0]\to0$.

Regarding your proof, note that you are asked to compute the probability of $[X=0]$ with $X=\liminf X_n$, and that you compute the probability of $\liminf [X_n=0]$ instead. In other words, your problem is that you assumed (wrongly) that the events
$[\liminf X_n=0]$ and $\liminf [X_n=0]$ coincide.

Title: Probability of sequence


问:
Function equal its integral implies zero function
Let $f:\mathbb{R}\to \mathbb{R}$ be continuous such that
$$f(x) = \int_0^xf(s)ds $$
and $f(1) = 0$. Can we prove that $f=0$ without resorting to ODE solving techniques?

答:
Put $g(x) = |f(x)|$. Note that for $x \ge 0,$ we have

$$g(x) = |f(x)| = \left|\int_0^x f(s) {\rm d}s\right| \le \int_0^x |f(s)| {\rm d}s = \int_0^xg(s) {\rm d}s.$$

(The reason for assuming $x \ge 0$ is to have the integration in the "correct" order and not worry about extra modulus signs.)

Thus, we have
$$g(x) \le \int_0^x g(s) {\rm d}s.$$

Using Grönwall's inequality, we directly get the answer. I give a proof anyway.

Putting $G(x) := \displaystyle\int_0^x g(s) {\rm d}s,$ we see that $G'(x) \le G(x)$ or $G'(x) - G(x) \le 0.$ Also, note that $g \ge 0$ and thus, $G(x) \ge 0$ for $x \ge 0$.

Multiplying with $\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right),$ we see that
$$(G'(x) - G(x))\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right) \le 0$$
or that
$$\left(G(x)\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right)\right)' \le 0.$$

Thus, $$G(x)\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right)$$ is a decreasing function. However, note that the above is a nonnegative function which is $0$ at $x = 0$. Thus, we must have $G(x) = 0$ for all $x \ge 0$ and thus, $g = G' \equiv 0$ on $[0, \infty)$. A similar argument shows the same for $x < 0$ as well. Thus, we get $g \equiv 0$ or that $f \equiv 0$.


问:
Existence of left invariant $n-$forms in a lie Group
I am trying to show that the space of left invariant $n$-forms for an $n-$dimensional Lie group $G$ has dimension $1$. Now I was able to see that if we have a left-invariant $1-$form then it's value depends only at the values of $\omega_e$ has so the maximum dimension of this space will be $1$. Now I am trying to prove the existence of such forms. For this take an element $\omega_e \in \Lambda^n(T_eG)$ and define $\omega_g(X_1,...,X_n)=\omega_e((dL_{g^{-1}})_g X_1,...,(dL_{g^{-1}})_gX_n)$, and it's easy to see that is left-invariant. Now my problem is checking that in fact we have smoothness. For this I tried seeing that $\omega$ defines a multilinear alernating map from $\mathfrak{X}(M)\times ... \times \mathfrak{X}(M)\rightarrow C^{\infty}(M)$. Now the problem is that if I have vector fields $X_1,...,X_n$ I can't seem to argument why the function $\omega(X_1,...,X_n)(g)=\omega_e((dL_{g^{-1}})_g X_1(g),...,(dL_{g^{-1}})_g X_n(g) )$ is smooth , since we have that the $L_g^{-1}$ are changing . Any help with this last part is aprecciated . Thanks in advance.

答:
You already proved that a left invariant $n$-form would be unique up to a constant. Now, to prove that there indeed exist such a left invariant $n$-form, let $X_1,\ldots,X_n$ be a basis of $T_eG$. Let $\theta^1,\ldots,\theta^n$ be its dual basis in $T_eG^*$. Then $\omega_e=\theta^1\wedge\cdots\wedge \theta^n$ is a non-zero differential form of degree $n$ in $T_eG$ because $\omega_e(X_1\wedge\cdots\wedge X_n) = 1$. Now, one can define a $n$-form $\omega$ on $G$ by requiring it to be left invariant, that is
$$
\omega_g = \omega_e\left((\mathrm{d}L_{g^{-1}})\cdot,\ldots,\left(\mathrm{d}L_{g^{-1}}\right)\cdot \right)
$$
It is a smooth $n$-form, and it is left invariant by construction. The smoothness follows from the fact that $g \mapsto \mathrm{d}L_{g^{-1}}$ is smooth because $G$ is a Lie group! Then, if $X$ is a vector field, in the left invariant frame $X_1,\ldots,X_n$ created from the basis on $T_eG$, one can write
$$
Y(g) =\sum_{i=1}^n Y^i(g)\left(\mathrm{d}L_gX_i\right)
$$
and then
$$
\omega_g(Y_1(g),\ldots,Y_n(g)) = \omega_g\left(\sum_{i}Y_1^j(g)X_i(g),\cdots,\sum_{i}Y_n^j(g)X_i(g) \right)
$$
And using linearity and left-invariance of $\omega_g$ and $X_j(g)$, the smoothness follows from the smoothness of the function $g\mapsto Y_j^i(g)$. Hence, $\omega$ is smooth.