问:
How to find the speed of a bob when its wire has reduced its radius?
The problem is as follows:

A brass sphere is hanging from a ceiling as it is seen in the figure
from below. Assuming that at the position $A$ the speed is
$3\frac{m}{s}$ the bob swings and hits a bolt located in point $B$. As
a result the bob keeps swinging in the trajectory indicated by the
dashed line.

Using this information find the speed of the bob when it passes
through point $C$. You may assume that the acceleration due gravity is
$g=10\frac{m}{s^2}$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{3.6 meters per second}\\
2.&\textrm{2.8 meters per second}\\
3.&\textrm{4.2 meters per second}\\
4.&\textrm{1.8 meters per second}\\
\end{array}$

So far what I've attempted to do is to use the conservation of mechanical energy on this system as follows:

Assuming that the moment when the bob is at $A$ it will have the same energy as when it passes below the point $B$. I'm assuming that the reference for the potential energy is just below point $B$. I mean where the bolt is put in the wall.

$E_{kA}+E_{uA}=E_{kB}$

Well after doing this I'm getting:

$\frac{1}{2}mv^2_{A}+mgR=\frac{1}{2}mv^2_{B}$

Cancelling the mass and replaing the known quantities I'm getting:

$v^2_{B}=9+8$

$v^{2}_{B}=17$

Now in this second segment I'm assuming that the energy on point $B$ will remain the same as when it passes through point $C$.

Therefore:

$\frac{1}{2}mv^2_{B}=\frac{1}{2}mv^2_{C}+mgR$

But since R=0.4 this will make that the speed on $C$ to be also 3 meters per second.

What could it be wrong in this analysis?. Did I overlooked something?. Can someone help me to spot what did I do wrong?.

The more I look into this question, I think that this may have to do to work with the tension of the wire.

I have not used the fact that of the angle which was given. Should this be used?.
If I were to use the angle initially I could calculate the initial tension of the wire. I mean on point $A$.

But this tension will not be the same when the bob passes through point $C$. Perhaps does it exist a way to find it?.

Can someone guide me here?. I feel lost. Please try to include a step by step explanation so I can understand.

答:
The diagram is deceptive: the bob is $ \ 1.2 \cos 60º \ = 0.6 $ m. below the pivot, which is higher than point C at $ \ 0.8 $ m. below the pivot. [The bob goes from 0.6 m. to 0.4 m. above the lowest point of the pendulum swing.]  So the change in potential energy of the bob is
$ \ -0.2 mg \ $ from point A to point C .

The speed at point C  is then given by

$ \ \frac{1}{2} \ v^{2} \ = \  \frac{1}{2} · 3^{2}  \ + \ (0.6 - 0.4)  · g  \ $ ;

this does give one of the available choices for the result.

Concerning the tension in the wire, while it is true that $ \ T \ $ throughout the swing varies, this force is always perpendicular to the circular arc of the swing.  So in a simple pendulum, the tension does no work on the bob, and so does not enter into the kinetic energy - potential energy calculation.  The tension force only serves to control the trajectory of the bob.


问:
Fun calculus problem I can't seem to solve
I've recently picked up a math book I haven't read since college (highly recommended reading by the way!). I was reviewing multi-dimentional derivatives and such, and I stumbled upon a problem I've been trying to solve for two days, and I can't get it out of my head, so please help me out! = )

Problem (from memory):

There is a rabbit that runs in a perfect circle of radius $r$ with a constant speed $v$. A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed $v$ such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?

I tried using the fact that $|x'(t)| = |y'(t)| = v$ where $x(t)$ is fox's position and $y(t)$ is rabbit's position, and that $x'(t)x''(t) = 0$ because of constant speed restriction, but I'm still failing to find a solution.

Anyone feel like attacking this one?

答:
Assuming a smart fox (and aren't they all), calculus is not required. If the fox anticipated the rabbit's path, and took the shortest path, then it has to travel a distance r. With speed v, the time is simply r/v. (I.e., the fox can travel in a straight line from the center of the circle to the point in the circle the rabbit will be when it gets there.)

Edit to add: note that this does not solve the requested problem because the fox will not always be between the rabbit and the center of the circle, as required.

ContentLicense: CC BY-SA 3.0


问:
Maximization problems - Using implicit function to avoid KKT
I am wondering if the following procedure to solve a maximization problem in - let's say - two variables with inequality and non-negative constraints works.

More specifically, let's assume something like the following concrete example:

$ \text{max } F(x,y)=x^{\frac{2}{3}} y^{\frac{1}{3}}$

$ \text{sub } G(x,y)=x^2 + y^2 \leq 2 \hspace{0.5cm}\text{and } x,y\geq 0$

I thought that, instead of using Kuhn-Tucker conditions, which would bring me to a real nightmare, it would be better to go for the following steps:

I realize that the objective function is increasing in $\mathbb{R}^3_{+}$.

For this reason the constraint has to work as an equality constraint and it is binding.

I take the implicit function of $F(x,y)$  of $y$ in terms of $x$ and I put it equal to the same thing, for the inequality constraint.

I create a system with the previous equation and the inequality constraint expressed as an equality constraint.

I focus on the positive value of $x$ and $y$ and that's it.

Does it sound reasonable?

Any feedback is welcome!

答:
Your assumptions of (1) and (2) are correct and then you can use a more formal way. To maximize $F(x,y)$ the first differential must be zero
$$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy=\frac 23 \frac{y^{1/3}}{x^{1/3}}dx+\frac 13 \frac{x^{2/3}}{y^{2/3}}dy=0$$
Due to constraint $dx$ and $dy$ are related in a way that $x$ and $y$ must always be on the curve $G(x,y)=2$. To find the relation we use the first differential
$$dG=\frac{\partial G}{\partial x}dx+\frac{\partial G}{\partial y}dy=2x\,dx+2y\,dy=0\Rightarrow dy=-\frac xy dx$$
Now we can use this relation to eliminate $dy$ in $dF$
$$dF=\frac 23 \frac{y^{1/3}}{x^{1/3}}dx-\frac 13 \frac{x^{2/3}}{y^{2/3}}\frac xy dx=0$$
$$dF=\bigg(\frac 23 \frac{y^{1/3}}{x^{1/3}}-\frac 13 \frac{x^{2/3}}{y^{2/3}}\frac xy\bigg) dx=0$$
The above equation can be satisfied for any $dx$ (or in any directions) if
$$\frac 23 \frac{y^{1/3}}{x^{1/3}}-\frac 13 \frac{x^{2/3}}{y^{2/3}}\frac xy=0$$
$$2 \frac{y^{1/3}}{x^{1/3}}=\frac{x^{5/3}}{y^{5/3}}$$
$$2 y^2=x^2\Rightarrow y=\frac x{\sqrt 2}$$
Now we satisfied that objective $F(x,y)$ is maximized along the curve $G(x,y)$. We need to satisfy one more condition that $G(x,y)=2$ by using the relation between $y$ and $x$
$$x^2+y^2=x^2+\frac{x^2}2=2\Rightarrow x=\frac 2{\sqrt 3}$$
and
$$y=\frac x{\sqrt 2}=\frac {\sqrt 2}{\sqrt 3}$$
So you neither used Lagrange multipliers nor KKT.</p>


问:
Does $(\sqrt{2x^{2}})(\sqrt{6x}) $ simplify to $2x\sqrt{3x}$, or to $2|x|\sqrt{3x}$?
I am having trouble simplifying these expressions.

$$(\sqrt{2x^{2}})(\sqrt{6x}) $$

The textbook states that for n is even

$$|a| = \sqrt[n]{a^n}$$

The answer key says that the answer is

$$ 2x\sqrt{3x}$$

But when I use the rules stated in the textbook, I get

$$ 2|x|\sqrt{3x}$$
because $|a| = \sqrt[n]{a^n}$.

This has happened a couple of times where I simplify to an absolute value variable where the answer key says it's just a variable.

Where am I missing a step?

答: Both are correct. Since the domain is $x \ge 0$ because of the term $\sqrt{x}$, your answer can be simplified to what stated in the book.

ContentLicense: CC BY-SA 4.0


问:
Difference in the interpretation of cross-correlation and convolution?
I'm confused on the interpretation between convolution and cross-correlation integral transforms. The convolution integral is defined as

$f (t) \otimes g (t) = \int f(\tau) g(t-\tau) d\tau$

and the cross-correlation is defined as

$f (t) * g(t) = \int f^{*}(\tau)g(t+\tau)$

However, when looking up the final interpretation of this result, convolution is cited as saying it is "how one function is shaped by another one" while cross-correlation is "measure of similarity", yet I am confused as to how you can extract two seemingly different interpretations from the formulas above since the difference seems quite small (complex conjugate and no reflection of function $g$).

I've taken a look at it and all I see is that convolution seems to be totally symmetric, since $f\otimes g = g\otimes f$ where as this is not necessarily the case for cross-correlation. Is convolution just a measure of the similarity of the shape only while cross-correlation encompasses both the correlation of the shape and the order at which they come by? I see on wikipedia that cross correlation is the convolution of an adjoint $f$ function but this doesn't help me.

答:
I don't know where you got these interpretations from and the context within which they are discussed. On their own, they make little sense indeed. However, they are motivated by the applications that these integrals are used or encountered.

The convolution integral naturally appears in signals and systems theory, when a signal $x(t), t \in \mathbb{R},$ passes through a "linear time invariant filter" (or "channel") with, so called, impulse response $g(t), t \in \mathbb{R}$. The convolution integral provides the output for all $t\in \mathbb{R}$. Many operations in nature such as propagation of electromagnetic signals over a medium have this mathematical representation.

The cross-correlation integral is essentially the inner product of the "vector" $x(\tau), \tau \in \mathbb{R}$ and the "vector" $g(\tau), \tau \in \mathbb{R}$, shifted by $t$. Recall from elementary linear algebra that the inner product of two vectors is a measure of their similarity, thus it makes sense to interpret cross-correlation as such.

Clearly, the two operations are equivalent in the sense that you can always treat cross-correlation as a convolution with an adjoint version of the second function, and vice versa. However, the applications these operations correspond to are so common that it turns out it is more convenient to use both terms to clearly identify what we are talking about.


问:
$G$ conex planar graph, the total number of faces
Let $G$ be a conex planar graph with n more or equal than $3$ vertices, so that each face will contain in its frontier a circuit of length at least $5$.So, the number of faces will be at most $(2n-4)$$/$$3$.
Can someone pleae help me with this problem?

答: From the given, $5f\le 2e$. Use $n+f=e+2$ to eliminate $k$.


答:
Lemma: If $G$ is a planar graph with $V$ vertices, $E$ edges, and finite girth $g$ then, $$E \leq \frac{g(V−2)}{V−2}$$

Look here: girth in planar graph

So $g=5$ and thus $3E\leq 4V-10$. Using Euler theorem: $$V-E+F =2$$ we have $$3V+3F -6 \leq 4V-10$$ so $$F\leq {V-4\over 3}$$

which is much better bound.

Title: $G$ conex planar graph, the total number of faces

Body:
Let $G$ be a conex planar graph with n more or equal than $3$ vertices, so that each face will contain in its frontier a circuit of length at least $5$.So, the number of faces will be at most $(2n-4)$$/$$3$.
Can someone pleae help me with this problem?


问:
If $n^{p}P(|X_1|>n) \to 0$ then there exist $y_n$ such that $\frac{1}{n^{1/p}}\sum_{k=1}^n X_k-y_n \to 0$ in probability
Let $p>1,(X_n)_n$ a sequence of i.i.d random variables such that $\lim_nnP(|X_1|>n^{1/p})=0.$ Let $Y_n=\sum_{k=1}^n X_k.$

Prove that there exist $y_n$ such that $\frac{1}{n^{1/p}}Y_n-y_n$ converges in probability to $0.$

Hint: Consider $Y'_n=\sum_{k=1}^n X_k1_{|X_k| \leq n^{1/p}}$ and prove that $Y_n-Y_n'$ converges in probability to $0.$

Considering $Y_n'=\sum_{k=1}^n X'_k,$ $\forall \epsilon>0,P(|Y_n-Y_n'|>\epsilon)\leq P(Y_n \neq Y_n') \leq nP(|X_1|>n^{1/p}) \to0.$ Clearly $X_1 \in L^1$ since $n^{p}P(|X_1|>n) \to 0.$ Then $\frac{1}{n^{1/p}}Y_n-y_n=\frac{1}{n^{1/p}}(Y_n-Y_n')+\frac{1}{n^{1/p}}Y_n'-y_n,$ so we must find a suitable $y_n$ such that $\frac{1}{n^{1/p}}Y_n'-y_n$ to $0$ in probability.

What is the form of $y_n$? Is it $y_n=n^{1-1/p}E[X_11_{|X_1| \leq n^{1/p}}]$? Or $y_{n}=n^{1-1/p}E[X_1]$?

答:
Firstly, notice that form $ \lim_n n\mathsf{P}(|X_1|>n^{1/p})=0 $ it is easy to get
\begin{gather*}
 \lim_{x\to\infty}x^p\mathsf{P}(|X_1|>x)=0,\\
 \lim_{x\to\infty}\sup_{y\ge x}[y^p\mathsf{P}(|X_1|>y)]=0\qquad \tag{1}
\end{gather*}
and for $ c>0, 0<A<x $,
\begin{align*}
 &\frac1{x^c}\mathsf{E}[|X_1|^{p+c}1_{\{|X_1|\le x}\}]\\
 &\quad=\frac{p+c}{x^c}\mathsf{E}\Big[\int_0^x y^{p+c-1}1_{\{y<|X_1|\le x\}}dy\Big]\\
 &\quad\le \frac{p+c}{x^c}\Big[\int_0^A+\int_A^x\Big]y^{c-1}y^p\mathsf{P}(|X_1|>y)dy\\
 &\quad \le \frac{p+c}{x^c}\int_0^Ay^{p+c-1}\mathsf{P}(|X_1|>y)dy
      +\sup_{y\ge A}[y^p\mathsf{P}(|X_1|>y)]\frac{p+c}{c}.
\end{align*}
Letting $ x\to\infty, A\to\infty $ successively, get
\begin{gather*}
 \varlimsup_{x\to\infty}\frac1{x^c}\mathsf{E}[|X_1|^{p+c}1_{\{|X_1|\le x}\}]
 \le \sup_{y\ge A}[y^p\mathsf{P}(|X_1|>y)]\frac{p+c}{c},\\
 \lim_{x\to\infty}\frac1{x^c}\mathsf{E}[|X_1|^{p+c}1_{\{|X_1|\le x}\}]=0.\tag{2}
\end{gather*}
Taking $ c=2-p,x=n^{1/p} $ in (2),
\begin{equation*}
 \lim_{n\to\infty}\frac{n}{n^{2/p}}\mathsf{E}[|X_1|^21{\{|X_1|\le n^{1/p}\}}]=0. \tag{3}
\end{equation*}
Similarly, for $ 0<c\le p $,
\begin{gather*}
 \lim_{x\to\infty}x^c\mathsf{E}[|X_1|^{p-c}1_{\{|X_1|>x\}}]=0,\\
 \lim_{n\to\infty}n^{1-\frac1p}\mathsf{E}[|X_1|1_{\{|X_1|>n^{1/p}\}}]=0. \tag{4}
\end{gather*}
As $ n\to\infty $,
\begin{equation*}
 \mathsf{Var}\Big[\frac{Y^\prime_n}{n^{1/p}}\Big]\le
 \frac{n}{n^{2/p}}\mathsf{E}[|X_1|^21{\{|X_1|\le n^{1/p}\}}]\rightarrow 0
\end{equation*}
from (3). Hence
\begin{equation*}
 \frac{Y^\prime_n}{n^{1/p}}-\frac{n}{n^{1/p}}\mathsf{E}[X_11_{\{|X_1|\le n^{1/p}\}}]
 \stackrel{pr}{\longrightarrow}0.
\end{equation*}
Furthermore,
\begin{gather*}
 \frac{Y_n}{n^{1/p}}-\frac{n}{n^{1/p}}\mathsf{E}[X_11_{\{|X_1|\le n^{1/p}\}}]
 \stackrel{pr}{\longrightarrow}0.  \\
 \frac{Y_n}{n^{1/p}}-\frac{n}{n^{1/p}}\mathsf{E}[X_1]\stackrel{pr}{\longrightarrow}0.
 \tag{5}
\end{gather*}
from $ \mathsf{P}(Y^\prime_n\ne Y_n)\le n\mathsf{P}(|X_1|>n^{1/p}) $ and (4). For all $ y_n=n^{1-\frac1p}(\mathsf{E}[X_1]+o(1)) $,
\begin{equation*}
 \frac{Y_n}{n^{1/p}}-y_n\stackrel{pr}{\longrightarrow}0.
\end{equation*}
from (5) also.


问:
If each Ci is a closed subset in their topology, is it closed in the countably infinite product topology
I have just learned about countably infinite product topology and I am already confused by the way it is defined. The definition I was given: The product space, ∏(Xi,τi), consists of the product ∏Xi with the topology τ having as its basis the family B={∏Oi:Oi ∈τi and Oi=Xi for all but a finite number of i.

This is not what I expected the basis to be. Of course, I learned about the box topology which is what one would expect the basis to be but the text stated that the reason why the product topology is not defined that way is because the box topology lacks many good properties we want.

So if we take a familiar example like R^infinity where R is the real numbers, a basic open set would be something like R x R x (0,2) x (6,10) x R x R x (9,10) x R....with the rest all Rs. And another basic open set would be something like (0,2) x (6,10) x R x R x R....with the rest all Rs. As long as all but finite number of open sets in the product are Rs, it would be consider a basic open set right? I know there is a good reason why the countably infinite product topology is defined this way and I hope to learn the reason soon, but does that mean a set like (0,2) x (0,2) x (0,2) x ...with the rest all (0,2) would not even be considered an open set?

Now having said all that, the problem I have to work on is: For each i∈N, let Ci be a closed subset of a topological space (Xi,τi). Prove that ∏Ci is a closed subset of ∏(Xi,τi).

I am pretty this is the product topology and not the box topology. But I don't see how this statement could be true. If we go back to a familiar example R^infinity, lets take the infinite countable product of [0,4]x[0,4]x[0,4]x[0,4]x[0,4]x[0,4]x[0,4]x[0,4]....with the rest all being [0,4]. So the complement of this should be open in the product topology, and the complement of each is (-infinity,0)u(4,infinity) in R. But the text just said all basic open sets cannot be of this form, so there is no way that is open in the product topology, so I am probably misunderstanding something.

答: $C_1\times C_2\times C_3...=\bigcap_{n=1}^{\infty} (C_1\times C_2\times C_3...\times C_n\times \mathbb R\times \mathbb R\times ...)$ and intersection of closed sets is always closed.


答: Hint: show that $\prod C_i= \bigcap  π_i^{-1}(C_i)$ where $π_i: \prod X_i \to X_i $ is the i-th projection

Title: If each Ci is a closed subset in their topology, is it closed in the countably infinite product topology

Body:
I have just learned about countably infinite product topology and I am already confused by the way it is defined. The definition I was given: The product space, ∏(Xi,τi), consists of the product ∏Xi with the topology τ having as its basis the family B={∏Oi:Oi ∈τi and Oi=Xi for all but a finite number of i.

This is not what I expected the basis to be. Of course, I learned about the box topology which is what one would expect the basis to be but the text stated that the reason why the product topology is not defined that way is because the box topology lacks many good properties we want.

So if we take a familiar example like R^infinity where R is the real numbers, a basic open set would be something like R x R x (0,2) x (6,10) x R x R x (9,10) x R....with the rest all Rs. And another basic open set would be something like (0,2) x (6,10) x R x R x R....with the rest all Rs. As long as all but finite number of open sets in the product are Rs, it would be consider a basic open set right? I know there is a good reason why the countably infinite product topology is defined this way and I hope to learn the reason soon, but does that mean a set like (0,2) x (0,2) x (0,2) x ...with the rest all (0,2) would not even be considered an open set?

Now having said all that, the problem I have to work on is: For each i∈N, let Ci be a closed subset of a topological space (Xi,τi). Prove that ∏Ci is a closed subset of ∏(Xi,τi).

I am pretty this is the product topology and not the box topology. But I don't see how this statement could be true. If we go back to a familiar example R^infinity, lets take the infinite countable product of [0,4]x[0,4]x[0,4]x[0,4]x[0,4]x[0,4]x[0,4]x[0,4]....with the rest all being [0,4]. So the complement of this should be open in the product topology, and the complement of each is (-infinity,0)u(4,infinity) in R. But the text just said all basic open sets cannot be of this form, so there is no way that is open in the product topology, so I am probably misunderstanding something.


问:
If $17 \mid \frac{n^m - 1}{n-1}$ find the values of $n$ where $m$ is even but not divisible by $4$

Let $m, n \in \mathbb{Z}_+$ with $n > 2$, and let $\frac{n^m-1}{n-1}$ be divisible by $17$. Show that either $m$ is even:$ m \equiv 0 \mod 17$ and $n \equiv 1 \mod 17$. Find all possible values of $n$ in the cases when $m$ is even but not divisible by $4$, or divisible by $4$ but not divisible by $8$.

So far, I have done this: First, let's assume that $n \equiv 1 \mod 17$. Also, we know that we can write

$$\frac{n^m-1}{n-1} = \sum_{i = 0}^{m-1} n^i.$$

As $n \equiv 1 \mod 17 \implies n^i \equiv 1 \mod 17$ and so $\sum_{i = 0}^{m-1} n^i \equiv \underbrace{1 + 1 + \cdots + 1} \equiv  m$.

Also, we know that $\sum_{i = 0}^{m-1} n^i \equiv 0 \mod 17 \implies m \equiv 0 \mod 17$.

Now, let's assume that $n \not\equiv 1 \mod 17$. Then, $\frac{n^m - 1}{n - 1}$ is divisible by $17$ if $n^m - 1$ is divisible by $17$, i.e $n^m - 1 \equiv 0 \mod 17 \implies n^m \equiv 1 \mod 17$. Fermat's little theorem tells us that $n^{p-1} \equiv 1 \mod p \implies n^{16} \equiv 1 \mod 17$. From here, we have that $m \mid 16$ which tells us that $m$ must be even.

However I am now stuck on the next bit, I'm not sure how to find those values of $n$. Can someone help me please. Thank you.

答:
You have made considerable progress, indeed you are near the end. If $n\equiv 1\pmod{17}$, then by what you wrote we have $m$ must be divisible by $17$, and any $m$ divisible by $17$ will do. The condition $m$ divisible by $2$ but not by $4$ is easy to meet. We need $m\equiv{34}\pmod{68}$. Then any $n\equiv 1\pmod{17}$ will do the job.

Now let us look at the case $n\not\equiv 1\pmod{17}$. So $n-1$ and $17$ are relatively prime. So our ratio is divisible by $17$ if and only if $n^m\equiv 1\pmod{17}$.  This automatically holds if $16$ divides $m$. But our condition on $m$ is that $m$ is divisible by $2$ but not by $4$, so we are far from $16$ divides $m$.

The order of $n$ modulo $17$ divides $16$, so is a power of $2$. We want the order to be exactly $2$. This is the case if and only if $n\equiv -1\pmod{17}$.

For the second part ($m$ divisible by $4$ but not by $8$), there will be some work left to do. You need to determine which $n$ have order $4$ modulo $17$.


问:
Rock/Paper/Scissors strategy when opponent is restricted to only one choice (i.e. (0,0,1))?
Trying to help my college kid with a probability problem, but sad to say I can't figure it out! Is there a simple explanation?

What is the probability you win in a game of r/p/s.

I think this one is just 1/9, since the events are independent but didn't know for sure.

Say you and your friends are playing rock-paper-scissors. Suppose your friend knows which choice they’ll make this round (so one of P(r), P(p), P(s) = 1, and the other two are 0), and tells you this, but doesn’t tell you which choice they made. How do you choose a strategy (i.e your P(r), P(p), P(s)) to maximize your winning in the worse-case situation in this scenario?

No idea how to approach this question. I remember making some type of tables, but can't for the life of me figure this out. Wouldn't any choice simply lead to an expected value of 0? Since you don't know which  choice your opponent will be taking?

Please help me out!

答: In the second question, unless you make all your probabilities $=\frac13$, you must make at least one of them $>\frac13$. It may happen that this is the one choice losing against your opponent’s fixed choice. As we are asked about the worst case, we lose with more than $\frac13$ unless we make $P(r)=P(p)=P(s)=\frac13$.


问:
Dividing and multiplying by the derivative in an integral
$y' + y = y^{-2}$

I divided everything by $y^{-2}$ and let $v = y^3$, then solved as normal. (I relaced the $y^2$ with $y^{2}=\frac{1}{3}\frac{dv}{dy}$ after differentiating the $v$.

After $v$ substitution, $\frac{1}{3}\frac{dv}{dx} + v = 1$ and $\frac{1}{3}\int \:\frac{dv}{1-v}=\int \:dx$

Now how can I solve this if you need the derivative of $v$ in order to get a log?
Can I just multiply and divide by $\frac{dv}{dx}$? Can someone finish it off?

答:
You can proceed as

$$\dfrac{1}{3}\int \:\dfrac{dv}{1-v} = -\dfrac{1}{3}\log(1-v)$$


问:
Does this tetrahedron have a name?
Consider the tetrahedron with vertices at:

$$(0,0,0)$$
$$(2,-\sqrt2,0)$$
$$(2,\sqrt2,0)$$
$$(2,0,2)$$

This tetrahedron is not regular but does it have any notable properties? It appears to have some symmetries. For example, its faces are congruent. Its two long edges appear somehow symmetrical with respect to each other (and separate mutually perpendicular faces). Does the tetrahedron have a name?

答:
Sometimes it's called a tetragonal disphenoid,
sometimes a digonal antiprism.
However it is only an isogonal polyhedron.

--- rk

Title: Does this tetrahedron have a name?


答:
Are you just looking to solve the equation? If that is the case, this is separable.

$$\begin{split}\frac{dy}{dx}&=\frac{1-y^3}{y^2}\\
\frac{dx}{dy}&=\frac{y^2}{1-y^3}\\
\int dx&=\int\frac{y^2}{1-y^3}dy\end{split}$$

Let $u=1-y^3,du=-3y^2dy$

$$\begin{split}x&=-\frac 1 3\int\frac 1 udu\\
&=-\frac 1 3\ln|u|+c\\
x&=-\frac 1 3\ln|1-y^3|+c\end{split}$$

Equivalently, from where you left off, let $u=1-v,du=-dv$. Then

$$\begin{split}-\frac 1 3\int\frac 1 u du&=x+c\\
-\frac 1 3 \ln|1-v|&=\\
-\frac 1 3\ln|1-y^3|&=x+c\end{split}$$

Title: Dividing and multiplying by the derivative in an integral

Body:
$y' + y = y^{-2}$

I divided everything by $y^{-2}$ and let $v = y^3$, then solved as normal. (I relaced the $y^2$ with $y^{2}=\frac{1}{3}\frac{dv}{dy}$ after differentiating the $v$.

After $v$ substitution, $\frac{1}{3}\frac{dv}{dx} + v = 1$ and $\frac{1}{3}\int \:\frac{dv}{1-v}=\int \:dx$

Now how can I solve this if you need the derivative of $v$ in order to get a log?
Can I just multiply and divide by $\frac{dv}{dx}$? Can someone finish it off?


答:
For each $n\in\Bbb N$ let $C_n=[0,4]$, and let $C=\prod_{n\in\Bbb N}C_n$. Then

$$\Bbb R^\infty\setminus C\ne\prod_{n\in\Bbb N}(\Bbb R\setminus C_n)\,,$$

so the fact that $\prod_{n\in\Bbb N}(\Bbb R\setminus C_n)$ is not open tells you nothing about whether $C$ is closed or not.

Suppose that $x=\langle x_n:n\in\Bbb N\rangle\in\Bbb R^\infty$; when is $x$ not in $C$? In order for $x$ to be in $C$, we must have $x_n\in[0,4]$ for every $n\in\Bbb N$. Thus, $x\in\Bbb R^\infty\setminus C$ exactly when there is at least one $n\in\Bbb N$ such that $x_n\notin[0,4]$. If for each $k\in\Bbb N$ we set

$$U_k=\left\{\langle x_n:n\in\Bbb N\rangle\in\Bbb R^\infty:x_k\notin[0,4]\right\}\,,$$

then $\Bbb R^\infty\setminus C=\bigcup_{k\in\Bbb N}U_k$. If each of the sets $U_k$ is open, then so is their union, and therefore $C$ is closed.

And in fact each $U_k$ is even a basic open set in the product topology on $\Bbb R^\infty$; can you see how to write $U_k$ in the form $\prod_{n\in\Bbb N}V_n$, where each $V_n$ is open in $\Bbb R$, and all but finitely many of the $V_n$ are equal to $\Bbb R$?


问:
How is formula for p value derived?
Suppose that we want to test the mean of a normal population. Let $X_1,\dots,X_n\overset{i.i.d.}\sim N(\mu,1)$ (for simplicity, I'm assuming $\sigma^2 = 1$. Suppose that we are testing $H_0:\mu = \mu_0$ against the alternative $H_1:\mu\neq \mu_0$. We are looking for a rejection region of the form $|\overline{X}-\mu_0|>c$.

Therefore
\begin{eqnarray*}
P_{\mu_0}(|\overline{X}-\mu_0|>c) &=& \alpha\\
\text{i.e.,}\quad P_{\mu_0}(\sqrt{n}(|\overline{X}-\mu_0|)>c\sqrt{n}) &=& \alpha
\end{eqnarray*}
This implies that $c\sqrt{n} = z_{{\alpha}/{2}}$.

Thus we arrive at a test that calls for rejection of the null-hypothesis if $\sqrt{n}(|\overline{X}-\mu_0|) >z_{{\alpha}/{2}}$.

In this test, why is the p-value
\begin{equation}
\label{p-val}
P(|Z|>\text{TS}),
\end{equation}
where $\text{TS} = \sqrt{n}(|\overline{X}-\mu_0|)$?

Although it looks intuitively alright, I am just wondering if there is a mathematical rigorous sequential way of obtaining the above formula?

PS: I asked this question in statistics stack exchange, where I couldn't get a convincing answer.

答:
Although it looks intuitively alright, I am just wondering if there is a mathematical rigorous sequential way of obtaining the above formula?

Yes, there is!

First way, analytical

Using the Generalized Likelihood Ratio you get that

$$\sup_{\theta \in \Theta_0}L\propto\exp\left\{\frac{1}{2}\Sigma_i(x_i-\mu_0)^2 \right\} $$

$$\sup_{\theta \in \Theta}L\propto\exp\left\{\frac{1}{2}\Sigma_i(x_i-\overline{x}_n)^2 \right\} $$

Do the ratio, manipulate is algebraically (try because it is not difficult) and get the desired solution

Second way, using a property of Gaussian model

If the model is gaussian, in the two tails hypothesis test there is a well known method known as "Confidence interval method" by which you calculate the confidence interval of $\mu\in [a;b]$ with the desired confidence level $(1-\alpha)100\%$

You use the following decision rule: Reject $H_0$ if and only if $\mu_0 \notin [a;b]$

Following this method you get exactly the solution you want to prove

Finally observe that this test IS NOT a UMP test.

Title: How is formula for p value derived?


问:
Confused about interpretation and satisfiability in first order logic.
In my textbook("Introduction to mathmatical logic" by Elliot Mendelson , there is a definition of satisfiability:-

If $\mathscr B$ is an atomic wf $A^k_n(t_1 , ... , t_n)$ and $(A^n_k)^M$ is the corresponding n-place relation of the interpretation , then a sequence $s = (s_1 , s_2 , ...)$ satisfies $\mathscr B$ if and only if $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$.

Shouldn't it say "iff $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$ is true" or "iff $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$ holds" ? maybe it is saying implicitly $(A^n_k)^M(s^*(t_1) , ... , s^*(t_n))$ is a theorem.But where are the axioms and rules from which it can be even deduced? I am confused.
 For me it is kind of like saying "the painting is good iff $3$"

答:
The logical form of the definition  of satisfiability is: (the sentence) “Plato is a Philosopher “ is true iff Plato is a philosopher.

Thus, the locution “is satisfiable”, that like “is true” applies to a linguistic expression, is on the left side of the definition: formula ... is satisfiable iff..., while on the right side we have a fact, to which “it holds” applies, but it is redundant.

See A.Tarski.


问:
Difference between Propositional and Judgmental Equality
I'm reading the  HoTT  book and actually I'm a bit confused on core difference between
propositional equality (noation: $a=_A b$ where $a, b:A$ and judgmental equality
$ a \equiv b $.

The propositional equality is just a proposition and therefore a type $a=_A b$ and
checking if $a=_A b$ is true is equivalent to find a term $t: a=_A b$.

In contrast $ a \equiv b $ is a judgement, so statement about the theory and
not in. In the book one involved reason why we need also the judgmental equality
is to be able to make reasonable definitions like declaring a function
$f: A \to B$ to be $x^2$: $f(x) \equiv x^2$.

Intuitively this differs from propositional equality because it's a definition
we did and the theory not 'knows it'. So it seems senseless to try to prove
in the theory if $f(x) = x^2$, because we defined it so while the theory
we want to talk about is stiff, informally speaking 'we cannot define something
within the theory'.

At least, that's how I understood the difference between propositional equality
and judgmental equality. First of all, is this 'intuition' correct or
'misleading' and(or) wrong? The reason why I think that it might be shortcoming is the remark
on page 19 that 'As type theory becomes more complicated, judgmental equality
can get more subtle than this.' So I doubt that what I write above hits
to essential nature of the difference between propositional
and judgmental equality.

Question: Can the difference between propositional and judgmental equality made
be precise and does it coinside with my understandings of it sketched above
or did I miss the point?

答:
Your intuition is correct, but is a simplified understanding of the difference between propositional and judgemental equality. That's exactly the meaning of the sentence in the HoTT book that you are quoting in chapter 1.1: this chapter is meant to be an introduction to help understand type theory, and does not pretend to be exact and exhaustive in the details.

When you try to formalize a (informal) theory, you end up with a syntaxical representation of the objects of the theory, and syntaxic rules to manipulate these objects. With a simple theory, each object has a unique syntaxical representation: then there is no need of definitional (or judgmental) equality.

But most of the time, it is impossible to have a single syntaxical representation of each object; then you need to add additional rules, the conversion rules, that specify when 2 different syntaxical expressions represent the same object in the theory. 2 syntaxical expressions are definitionaly equal when they are different representations of the same object in the theory.

For instance, you usually want to ignore the differences in the naming of the variables: you need an $\alpha$-conversion rule to specify that two expressions differing only by the names encode the same object of theory.

Then you have the computation rules, that identify the syntaxical representation $f(a)$ with the result of the computation of $f$ applied to $a$ ($\beta$-conversion). You also want to identify an expression using notations with the expression obtained by replacing the notation by its definition.

The exact border between definitional and propositional equality is not always easy to draw (e.g. this note). It may be desirable, e.g. to simplify the formal proofs, that two objects that are proved to be propositionally equal are made definitionally equal with a new specific conversion rule. This can possibly be done with an equality reflection rule; however, such a rule has an impact on the meta-theoretical properties of the theory, such as the decidability of type checking.

It is desirable that, in the formalization of a theory, objects have a unique syntaxical representation, or, at least, can be converted into a unique syntaxical representation (strong normalization property). For a theory dealing with simple objects, propositional equality is then trivial and reduces to definitional equality. But for a theory dealing with more complex objects such as spaces or structures, admitting possibly different equivalent definitions inside the theory, the desired notion of equality of objects (space or structure) should be an equivalence of objects (space or structure), and therefore cannot be reduced to the definitional equality. Homotopy type theory and its univalence axiom is meant to achieve such a goal.


问:
arc length of $\sqrt{x}$ from $0$ to $a$.
Find the arc length of $\sqrt{x}$ from $0$ to $a\ (a\in \mathbb{R})$.

I found that the arc length above is the same arc length as $x^2$ from $0$ to $\sqrt{a}$. Then I used the formula to calculate arc length, I obtained that  $$s(x)=\int_{0}^{\sqrt a} \sqrt{1+4x^2}\ dx$$ and made the substitute $2x= \tan(t)$
but then I got stuck.

Any help would be great.

答:
$$2x = \tan \theta$$
$$dx = \frac12 \sec^2 \theta \: d\theta$$
$$\displaystyle\int_0^{\sqrt{a}} \sqrt{1 + 4x^2} \: dx = \frac12 \int_0^{\arctan \sqrt{a}} \sec^3 \theta \: d\theta$$
Use integration by parts with $u = \sec \theta$ and $dv = \sec^2 \theta \: d\theta$. Then $du = \sec \theta \tan \theta \: d \theta$ and $v = \tan \theta$. This leads to
$$\frac12 \int_0^{\arctan \sqrt{a}} \sec^3 \theta \: d\theta = \frac12 \sec \theta \tan \theta \biggr|_0^{\arctan \sqrt a} - \int_0^{\arctan \sqrt{a}} \tan \theta \sec^2 \theta \: d\theta$$
You have to apply the identity $\sec^2 \theta = 1 + \tan^2 \theta$. You will end up with the original integral on the RHS, which you then add to both sides, then divide by 2.


问:
Fastest Algorithm for NLP with linear constraints
I have an minimization problem of the following form:

(Im not a mathematician, i come from the programming side, so excuse me if i have not the perfect standard of writing the formulas)

$Z(x) = \sqrt[2]{x_1^2 \cdot a_1+...+x_n^2 \cdot a_n} + x_1 \cdot b_1+...+x_n \cdot b_n$

with the constraints:

(1) $\sum_{i=1}^n{x_i} =1$

(2) $0<x_i<=c_i \ \text{with} \ c = \{c_1,..,c_n\}$

Im interested in an algorithm that finds a local minimum as fast as possible, since i want to implement it.

I have already read a lot of stuff with different approaches for different type of constraints, but they mostly didnt fit or where hard to understand, so i could not be sure if they fit.

I understand that this is a NLP with linear constraints, but since im pretty new to NLP i hope for someone with more experience who can help me.
Thank you!

答:
First, a couple minor points. I'm assuming $a_i > 0$ so the problem is convex, and $\sum_{i=1}^n c_i \ge 1$, so the problem is feasible. Also, there is no guarantee that there will be a minimum with $x_i$ strictly positive, so the constraints should be $0 \le x_i \le c_i$. If it's unacceptable in your application for the $x_i$ to be zero, then you should specify some minimum nonzero value for them.

While I'm not sure what the best algorithm for this problem would be, here's at least a partial solution. For this particular problem, you can almost find a closed form solution. That is, you can reduce it to searching over 2 parameters.

Let $C$ be the feasible region (intersection of a box and a plane):
$$
C := \{ \mathbf{x} \in \mathbb{R}^n \ \mid \ 0 \le x_i \le c_i, \ \sum_{i=1}^n x_i =1 \}
$$
In this particular case, it's relatively easy to project onto $C$. By introducing another parameter, you can see that the optimal $\mathbf{x}$ will be a (scaled) projection onto $C$. This uses the standard trick that $\sqrt{w} = \min_{s>0} \frac{s}{2} + \frac{w}{2s}$.
$$
\begin{aligned}
\min_{\mathbf{x} \in C} \sqrt{ \sum_{i=1}^n a_i x_i^2} + \sum_{i=1}^n b_i x_i
&= \min_{\mathbf{x} \in C} \min_{s > 0} \frac{s}{2}+ \frac{1}{2s}\sum_{i=1}^n a_i x_i^2 + \sum_{i=1}^n b_i x_i\\
&= \min_{s > 0} \frac{s}{2} +\frac{1}{2s} \left(\min_{\mathbf{x} \in C}\sum_{i=1}^n a_i x_i^2 +  2 s b_i x_i \right) \\
&= \min_{s > 0} \frac{s}{2}\left(1 - \sum_{i=1}^n b_i^2 a_i^{-1} \right) +\frac{1}{2s} \left(\min_{\mathbf{x} \in C}\sum_{i=1}^n a_i (x_i +   s b_i a_i^{-1})^2 \right)
\end{aligned}
$$
I'll skip the derivation of the formula for the scaled projection onto $C$, but it ends up being the following. Define $\mathbf{y}(s,t)$ by:
$$
y_i(s,t) := \max(0,\min(c_i,(t-b_i s)/a_i))
$$
Then the optimal solution to the problem is $\mathbf{x}^* = \mathbf{y}(s^*,t^*)$, where the optimal parameters $s^*$ and $t^*$ are chosen to satisfy:
$$
\begin{gathered}
\sum_{i=1}^n y_i(s^*,t^*) = 1\\
\sqrt{\sum_{i=1}^n a_i y_i(s^*,t^*)^2} = s^*
\end{gathered}
$$
So then the problem reduces to finding $s^*$ and $t^*$. For fixed $s$, you can solve for $t$ which satisfies $\sum_{i=1}^n y_i(s,t) = 1$ in $O(n)$ time. This paper has the case for the unit simplex, see the references on the linked page for the general case.

Unfortunately, I'm not sure what the best approach is for finding both $s^*$ and $t^*$, though I'd expect somewhere in the CS literature there's an algorithm with this problem as a special case which is runs in $O(n)$ or $O(n^2)$ time. If you just want to use a simple bisection approach, you can use the following bounds to get your starting iterations:
$$
\begin{gathered}
\min_i \sqrt{a_i} \le s^* \le \max_i \sqrt{a_i} \\
\min_i b_i s^* \le t^* \le \max_i b_i s^* + c_i
\end{gathered}
$$


问:
Taylor series alternative without using derivatives
The Taylor series approximates a function as a polynomial, given that you know it's value at point $a$. The coefficients of Taylor series polynomial for the function $f(x)$ are defined as:

$p[n]=\frac{f^{(n)}(a)}{n!}$

where $f^{(n)}(a)$ denotes the nth derivative of f evaluated at the point a. (The derivative of order zero of f is defined to be f itself). For example, the first 5 Taylor coefficients of $e^x$ are:
$$\langle1,\frac{1}{2},\frac{1}{6},\frac{1}{24},\frac{1}{120}\rangle$$

However, I need a power series like Taylor series that does not require computing derivatives, should work on a wide variety of functions, and must use ONLY real numbers. It can use everything else (integral, infinite sum etc.), and may not be a polynomial. I have tried Fourier series and applying polynomial regression, but it doesn't work for most functions. Laurent series is not fitting my criteria since it requires complex numbers. Are there any others?

答:
Just like the Fourier series, you can use integration to find approximate functions. The advantage here is that you can always do integration numerically so this method lends itself well to numerical methods.

Another difference from a Taylor series expansion is that it can produce the best fit curve within a range of values instead of only around one value. Of course, Taylor is guaranteed to have zero error at this value, but with the integration methods, zero error will happen at an unknown position.

Example, approximate $y(x) = \sin( \pi x)$ using a polynomial between $x=0 \ldots 1$. Lets try $$y(x) \approx c_1 x + c_3 x^3 + c_5 x^5$$

First order integral $$ \int_0^1 y(x)\,{\rm d}x = \int_0^1 (c_1 x + c_3 x^3 + c_5 x^5)\,{\rm d}x $$ $$ \frac{2}{\pi} = \frac{c_1}{2} + \frac{c_3}{4} + \frac{c_5}{6} \tag{1} $$

Second-order integral. Choose a simple function $f(x)$ that mimics $y(x)$. Use it on both sides of the integral. In our case $y(x)$ is simple enough to make $f(x)=\sin(\pi x)$ also  $$ \int_0^1 y(x) \sin(\pi x)\,{\rm d}x = \int_0^1 (c_1 x + c_3 x^3 + c_5 x^5) \sin(\pi x)\,{\rm d}x $$ $$ \frac{1}{2} = \frac{c_1}{\pi} + \left( \frac{1}{\pi} - \frac{6}{\pi^3} \right) c_3 + \left( \frac{1}{\pi} - \frac{20}{\pi^3} + \frac{120}{\pi^5} \right) c_5 \tag{2}$$

Third-order integral. Similar to the above, but with a higher-order basis function. Three terms is all we need since we have 3 coefficients in our approximation $$ \int_0^1 y(x) \sin(2 \pi x)\,{\rm d}x = \int_0^1 (c_1 x + c_3 x^3 + c_5 x^5) \sin(2 \pi x)\,{\rm d}x $$ $$ 0 = -\frac{c_1}{2 \pi} + \left( \frac{3}{4\pi^3} - \frac{1}{2\pi} \right) c_3 - \left( \frac{1}{2\pi} - \frac{5}{2 \pi^3} + \frac{15}{4\pi^4} \right) c_5 \tag{3} $$

Take equations (1), (2), and (3) and solve for the coefficients $c_1$, $c_3$ and $c_5$.

$$ \boxed{ \sin(\pi x) \approx 3.1154891857 x - 4.8785662612 x^3 + 1.7911004686 x^5 } $$

You can plot the above in Wolfram Alpha to get

As you can see it is a pretty good approximation between 0 and 1, but also between -1 and 0 since it is an odd function. Below is a plot of the error

which is maximum near and beyond the boundary of $x=1$, and zero at four places between 0 and 1. The compromise here is that the slope at $x=0$ is different from $\pi$ so the function deviates faster near zero than $\pi x$ which is the first term of the Taylor series.


问:
closure of A in the product space
For each $\alpha\in I=[0,1]$, let $X_{\alpha} = \{0,1\}$ be the discrete topology. For every $\Delta\subseteq I$, define $f_\Delta=(x_\alpha)\in\prod_{\alpha\in I}X_\alpha$, where $x_\alpha=1$ if $\alpha\in\Delta$, and $x_\alpha=0$ if $\alpha\notin\Delta$. Let $A=\{f_\Delta:\Delta\text{ is a finite subset of }I\}$. Show that $f\in\operatorname{cl}A$, where $\operatorname{cl}A$ is the closure of $A$ in the product space $\prod_{\alpha\in I}X_\alpha$.

答:
Let $f$ be an arbitrary function in $\prod_{\alpha \in I} X_\alpha$, and $O$ be an arbitrary open neighbourhood of $f$. By the definition of the product topology (via its standard base) there is a basic open set $f \in B:=\prod_{\alpha \in I} O_\alpha \subseteq O$, which is such that there is a finite subset $F$ of $I$ such that $O_\alpha = X_\alpha $ for all $\alpha \in I\setminus F$ and such that all $O_\alpha$ are open in $X_\alpha$.

Let $\Delta = f^{-1}[\{1\}] \cap F$, which is a finite subset of $I$ too, so $f_\Delta \in A$, and $f_\Delta \in B$ by the definitions: if $\alpha \notin F$, then $f_\Delta(\alpha) \in X_\alpha = O_\alpha$ trivially, and if $\alpha \in F$ we have $f_\Delta(\alpha) = f(\alpha) \in O_\alpha$ as well. So $O \cap A\neq  \emptyset$ and as $O$ was arbitrary , $f \in \operatorname{cl} A$, as required.

Note that we don't need the $X_\alpha$ to be discrete spaces.

Title: closure of A in the product space


问:
Derivation help for $E[x^{\top}Ax] = tr(AP)$: missing the trace operator, using the moment generating function for Gaussian $x$ and gradient
Following notations are used:

$x$: vector-valued random variable $x$ (with its mean $\bar{x}$ = 0).

$P$: covariance of $x$

$tr(\cdot)$: trace operator

$E[\cdot]$: expectation

$M_x(s)$: Moment generating function of $x$

$\nabla_s$: gradient with respect to $s$

This is the statement from my reading:

"~ for a Guassian random vector $x$, the identity $E[x^{\top}Ax] = tr(AP)$  can be derived using the moment generating function $M_x(s) = E[e^{s^{\top}x}] = e^{\frac{1}{2}s^{\top}Ps + s^{\top}\bar{x}}$ and the gradient operator $\nabla_s$".

I follow the partially shown derivation:
\begin{align}
E[x^{\top}Ax] &= E[(\nabla_s e^{s^{\top}x})^{\top}Ax]|_{s = 0} \qquad \text{(since $x = \nabla_s e^{s^{\top}x}|_{s=0}$)} \\
              &= E[\nabla^{\top}_sAx e^{s^{\top}x}]|_{s=0} \quad\qquad \text{(since $e^{s^{\top}x} $ is a scalar)} \\
              &= \nabla^{\top}_sAE[xe^{s^{\top}x}]|_{s=0} \quad\qquad \text{(A is constant and $E[\cdot]$ and $\nabla_s$ are linear)} \\
              &= \nabla^{\top}_sAE[\nabla_s e^{s^{\top}x}]|_{s=0} \quad\qquad \text{($xe^{s^{\top}x} = \nabla_s e^{s^{\top}x}$)} \\
              &= \nabla^{\top}_sA\nabla_sE[ e^{s^{\top}x}]|_{s=0} \quad\qquad \text{($E[\cdot]$ and $\nabla_s$ linear)} \\
              &= \nabla^{\top}_sA\nabla_sM_x(s)|_{s=0} \quad\qquad \text{(definition of $M_x(s)$)} \\
\end{align}

But when evaluating the gradient from the last line, I run into a problem of missing the trace operator somewhere below:

\begin{align}
E[x^{\top}Ax] &= \nabla^{\top}_sA\nabla_sM_x(s)|_{s=0} \quad\qquad  \\
              &= \nabla^{\top}_s A \nabla_s \left(e^{\frac{1}{2}s^{\top}Ps} \right)|_{s=0}\quad\qquad \qquad\text{(since $\bar{x} = 0$ )} \\
              &= \nabla^{\top}_s A(Ps)e^{\frac{1}{2}s^{\top}Ps}|_{s=0}\qquad\qquad\qquad \text{(chain rule and $P$ symmetric)} \\
              &=?\quad  A\left[Pe^{\frac{1}{2}s^{\top}Ps} + (Ps)(Ps)^{\top} e^{\frac{1}{2}s^{\top}Ps}\right]|_{s=0}\quad\qquad \text{(product rule)} \\
              &=?\quad  AP \qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{(evaluating at $s = 0$)} \\
\end{align}

I am wrong but don't know where I am wrong. I suspect the second from the last line. If so, how? Where should a trace operator pop up?

Based on the answer, a complete derivation would be:

\begin{align}
E[x^{\top}Ax] &= \nabla^{\top}_sA\nabla_sM_x(s)|_{s=0} \quad\qquad  \\
              &= \nabla^{\top}_s A \nabla_s \left(e^{\frac{1}{2}s^{\top}Ps} \right)|_{s=0}\quad\qquad \qquad\text{(since $\bar{x} = 0$ )} \\
              &= \nabla^{\top}_s A(Ps)e^{\frac{1}{2}s^{\top}Ps}|_{s=0}\qquad\qquad\qquad \text{(chain rule and $P$ symmetric)} \\
              &= tr(AP)e^{\frac{1}{2}s^{\top}Ps} + (Ps)^{\top}(APs)e^{\frac{1}{2}s^{\top}Ps}|_{s=0}\quad\qquad \text{(product rule)} \\
              &= tr(AP) \qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{(evaluating at $s = 0$)} \\
\end{align}

答:
to cut down on the symbol manipulation, I'd suggest a different path:

$\mathbb E\Big[\mathbf x^T A \mathbf x\Big]$

$=\mathbb E\Big[\text{trace}\big(\mathbf x^T A \mathbf x\big)\Big]$

$=\mathbb E\Big[\text{trace}\big( A \mathbf x\mathbf x^T\big)\Big]$

$=\text{trace}\Big(\mathbb E\big[ A \mathbf x\mathbf x^T\big]\Big)$

$=\text{trace}\Big(A\mathbb E\big[  \mathbf x\mathbf x^T\big]\Big)$

$=\text{trace}\Big(AP\Big)$


答:
Let $B$ be a matrix. We have
\begin{equation}
\nabla_s^\top B s = \sum_i\frac{\partial}{\partial s_i}\sum_j b_{i j} s_j
=\sum_i\sum_j b_{i j}\frac{\partial s_j}{\partial s_i} =
\sum_i\sum_j b_{i j}\delta_i^j=\text{tr}(B)
\end{equation}

Title: Derivation help for $E[x^{\top}Ax] = tr(AP)$: missing the trace operator, using the moment generating function for Gaussian $x$ and gradient

Body:
Following notations are used:

$x$: vector-valued random variable $x$ (with its mean $\bar{x}$ = 0).

$P$: covariance of $x$

$tr(\cdot)$: trace operator

$E[\cdot]$: expectation

$M_x(s)$: Moment generating function of $x$

$\nabla_s$: gradient with respect to $s$

This is the statement from my reading:

"~ for a Guassian random vector $x$, the identity $E[x^{\top}Ax] = tr(AP)$  can be derived using the moment generating function $M_x(s) = E[e^{s^{\top}x}] = e^{\frac{1}{2}s^{\top}Ps + s^{\top}\bar{x}}$ and the gradient operator $\nabla_s$".

I follow the partially shown derivation:
\begin{align}
E[x^{\top}Ax] &= E[(\nabla_s e^{s^{\top}x})^{\top}Ax]|_{s = 0} \qquad \text{(since $x = \nabla_s e^{s^{\top}x}|_{s=0}$)} \\
              &= E[\nabla^{\top}_sAx e^{s^{\top}x}]|_{s=0} \quad\qquad \text{(since $e^{s^{\top}x} $ is a scalar)} \\
              &= \nabla^{\top}_sAE[xe^{s^{\top}x}]|_{s=0} \quad\qquad \text{(A is constant and $E[\cdot]$ and $\nabla_s$ are linear)} \\
              &= \nabla^{\top}_sAE[\nabla_s e^{s^{\top}x}]|_{s=0} \quad\qquad \text{($xe^{s^{\top}x} = \nabla_s e^{s^{\top}x}$)} \\
              &= \nabla^{\top}_sA\nabla_sE[ e^{s^{\top}x}]|_{s=0} \quad\qquad \text{($E[\cdot]$ and $\nabla_s$ linear)} \\
              &= \nabla^{\top}_sA\nabla_sM_x(s)|_{s=0} \quad\qquad \text{(definition of $M_x(s)$)} \\
\end{align}

But when evaluating the gradient from the last line, I run into a problem of missing the trace operator somewhere below:

\begin{align}
E[x^{\top}Ax] &= \nabla^{\top}_sA\nabla_sM_x(s)|_{s=0} \quad\qquad  \\
              &= \nabla^{\top}_s A \nabla_s \left(e^{\frac{1}{2}s^{\top}Ps} \right)|_{s=0}\quad\qquad \qquad\text{(since $\bar{x} = 0$ )} \\
              &= \nabla^{\top}_s A(Ps)e^{\frac{1}{2}s^{\top}Ps}|_{s=0}\qquad\qquad\qquad \text{(chain rule and $P$ symmetric)} \\
              &=?\quad  A\left[Pe^{\frac{1}{2}s^{\top}Ps} + (Ps)(Ps)^{\top} e^{\frac{1}{2}s^{\top}Ps}\right]|_{s=0}\quad\qquad \text{(product rule)} \\
              &=?\quad  AP \qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{(evaluating at $s = 0$)} \\
\end{align}

I am wrong but don't know where I am wrong. I suspect the second from the last line. If so, how? Where should a trace operator pop up?

Based on the answer, a complete derivation would be:

\begin{align}
E[x^{\top}Ax] &= \nabla^{\top}_sA\nabla_sM_x(s)|_{s=0} \quad\qquad  \\
              &= \nabla^{\top}_s A \nabla_s \left(e^{\frac{1}{2}s^{\top}Ps} \right)|_{s=0}\quad\qquad \qquad\text{(since $\bar{x} = 0$ )} \\
              &= \nabla^{\top}_s A(Ps)e^{\frac{1}{2}s^{\top}Ps}|_{s=0}\qquad\qquad\qquad \text{(chain rule and $P$ symmetric)} \\
              &= tr(AP)e^{\frac{1}{2}s^{\top}Ps} + (Ps)^{\top}(APs)e^{\frac{1}{2}s^{\top}Ps}|_{s=0}\quad\qquad \text{(product rule)} \\
              &= tr(AP) \qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{(evaluating at $s = 0$)} \\
\end{align}


问:
A query regarding sets from Quantum computing since Democritus
I was reading the book "Quantum Computing Since Democritus".

"The set of ordinal numbers has the important property of being well
ordered,which means that every subset has a minimum element. This is
unlike the integers or the positive real numbers, where any element
has another that comes before it."

Unlike integers? Let's consider a set $\{1,2,3\}$ This has a minimum element.

Do you get what does the author wants to say here?

答: The integers mean $\Bbb Z$ here, so there we have sets like $\{-1,-2,-3,-4,\ldots\}$ (or $\Bbb Z$ itself) which do not have a minumum element and every $n$ has an element $n-1$ before it. So there are not well-ordered. The positive integers, i.e. $\Bbb N$, are.


问:
Local minimal about x=0
What should be the value of a for wich the function f(x)=$ (x^2-2ax+2a)/(x-a) $  have a local mínimum in x=0?

Derivating (i don´t include the calculus) one a twince i arrive to the answer that this funtion doesn´t has a minimun.

If a=0, the the funtion is reduced to f(x)=$x^2/x$ that is f(x)= x for all x not equal to zero, and indefined if x=0, so, i think, is imposible that f has a local minimun here due to, in x=0 the function is not defined.

The other options that I have are: a=1 and a=2, and i think nethier of them are correct.

Is all correct in my treatment of the problema?

答:
We can get a pretty fair idea of the function's behavior by writing its expression as

$ \ f(x) \ = \ \frac{x^2 \ - \ 2ax \ + \ 2a}{x \ - \ a} \  = \ \frac{x^2-2ax+a^2}{x-a} \ + \frac{2a \ - \ a^2}{x-a} \ = \ \frac{(x \ - \ a)^2}{x-a} \ + \frac{a \ (2 \ - \ a)}{x-a} $ $  = \ (x \ - \ a) \ + \frac{a \ (2 \ - \ a)}{x-a} \ , \ x \neq a \ . $

We see immediately from this that the graph of $ \ f(x) \ $ for  $ \ a \ = \ 0 \ $ is the line  $ \ y \ = \ x \ $ with a "hole" at the origin, so the function is undefined at  $ \ x \ = \ 0 \ $ , as you observed, and possesses no minimum value.  A similar situation holds for $ \ a \ = \ 2 \ $ , for which the graph is the line $ \ y \ = \ x \ - \ 2 \ $ with a "hole" at  $ \ x \ = \ 2 \ $ .

For all other values of  $ \ a \ $ , the function expression is the sum $ \ (x \ - \ a) \ + \frac{k}{x-a} \ $ , $ \ k \ $ a constant, which has a graph resulting from the sum of a line with a slope of $ \ 1 \ $ and a "diagonal" hyperbola.

In the cases with $ \ a \ < \ 0 \ $ and $ \ a \ > \ 2 \ $ , we have $ \ (x \ - \ a) \ - \frac{k}{x-a} \ , \ k \ > 0 \ $ , so the "branch" of the hyperbola to the "left" of the vertical asymptote is "concave upward" (curves upward) and the branch to the "right" is "concave downward". Alternatively, we can say that both terms is the function expression are always increasing, so $ \ f(x) \ $ has only the critical point at $ \ x \ = \ a \ $ where $ \ f'(x) \ $ is undefined, and so has neither a maximum nor a minimum value.

For $ \ 0 \ < \ a \ < \ 2 \ $  , the expression is $ \ (x \ - \ a) \ + \frac{k}{x-a} \ , \ k \ > 0 \ $ , so the  hyperbola is oriented with  the "left" branch  "concave downward", and the "right" one "upward". But this causes the function to "run toward negative infinity" as $ \ x \ \rightarrow \ a^{-} \ $ and "toward positive infinity" as $ \ x \ \rightarrow \ a^{+} \ $ . We now have critical points, but the one to the "left" of the vertical asymptote is a relative maximum .  For $ \ a \ = \ 1 \ $ , there is such a relative maximum at $ \ x \ = \ 0 \ $ . So we never have a relative minimum at  $ \ x \ = \ 0 \ $ .

If we take the trouble to calculate the first derivative, we obtain $ \ f'(0)  \  = \ \frac{2a \ · \ (a-1)}{a^2} \ $ , which confirms that we can only have a critical point at $ \ x \ = \ 0 \ $ for $ \ a \ = \ 1 \ $ (and that we have a meaningless result for $ \ a \ = \ 0 \ $ ) .

The "geometrical" analysis above is somewhat easier to deal with than computing the second derivative.  If we do so, though, we find that  $ \ f''(0)  \  = \ \frac{2 \ · \ (a-2)}{a^2} \ $ ; this is negative for  $ \ a \ = \ 1 \ $ , confirming a relative maximum.  So your surmise that the function you present has no relative minimum at  $ \ x \ = \ 0 \ $ is correct.

Title: Local minimal about x=0


问:
The Fundamental Theorem of Lebesgue Calculus

Let $[a, b]$ be an interval of $\mathbf{R}$. If $\varphi:[a, b] \rightarrow \mathbf{R}$ is continuous, then the function $F:[a, b] \rightarrow \mathbf{R}$ defined by $F(x)=\int_{[a, x]} \varphi(x) d \lambda(x)$, for $x \in[a, b]$, is differentiable on $[a, b],$ and $F^{\prime}=\varphi$.

For completeness, I'll type the full proof:

Let $x \in[a, b],$ and let $h \neq 0$ such that $x+h \in[a, b] .$ Assuming for example that $h>0,$ we have
$$
\frac{F(x+h)-F(x)}{h}=\frac{1}{h} \int_{(x, x+h]} \varphi(y) d \lambda(y)=\int_{(0,1]} \varphi(x+h u) d \lambda(u)
$$
where (this is the part that I don't understand) the last equality follows from the change of variable $y=x+h u .$ Since $\varphi$ is continuous, when $h \rightarrow 0,$ the function $u \mapsto \varphi(x+h u)$ converges pointwise to $\varphi(x)$ on $[a, b]$. Moreover all these functions are bounded by $|\varphi|$, which is a bounded, hence integrable function on $[a, b] .$ By dominated convergence, we thus have
$$
\int_{(0,1]} \varphi(x+h u) d \lambda(u) \underset{h \rightarrow 0}{\longrightarrow} \varphi(x)
$$

The question is, how do I justify the above change of variables and the integrals that follow? This is the form of change of variables that I know. I am unable to use this to conclude the above (I suspect $f(\alpha)= x+h\alpha$).

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and $(\mathcal{Y},\mathcal{B})$ be a measurable space. $f: (\Omega, \mathcal{F}, \mu) \to (\mathcal{Y},\mathcal{B})$ is a measurable map. Let $f_* \mu (A):= \mu(\{ f^{-1}(A)\})$ for $A \in \mathcal{B}$ be a pushed forward measure. Let $\phi: (\mathcal{Y},\mathcal{B}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be measurable. If $\phi$ is non-negative or integrable, then $\int_\Omega \phi \circ f d \mu= \int_\mathcal{Y} \phi d (f_* \mu)$.

答:
In your case, $\ \Omega=(0,1],$$\ \mathcal{Y}=(x,x+h],$$\ \mathcal{F}=\{A\cap(0,1]\,|\,A\in\mathcal{L}\}\ $ and $\mathcal{B}=\{B\cap(x,x+h]\,|\,B\in\mathcal{L}\}\ $, where $\ \mathcal{L}\ $ is the set of Lebesgue measurable subsets of $\ \mathbf{R}\ $, and $\ \mu=\lambda\ $, Lebesgue measure on $\ (\Omega, \mathcal{F})\ $.  With $\ f(u)=x+hu\ $, the measure $\ f_*\mu\ $ is simply
\begin{align}
f_*\lambda(A)&=\lambda\left(f^{-1}(A)\right)\\
&=\lambda\left(\frac{A-x}{h}\right)\\
&=\frac{\lambda(A)}{h}\ .
\end{align}
That is, $\ f_*\mu\ $ is just Lebesgue measure on $\ (\mathcal{Y},\mathcal{B})\ $, scaled down by a factor of $\ h\ $.  Thus, your change of variable formula gives you
\begin{align}
\int_{(0,1]}\phi(x+hu)d\lambda(u)&=\int_\mathcal{X}\phi\circ f d\mu\\
&=\int_\mathcal{Y}\phi(y)df_*\mu(y)\\
&=\int_{(x,x+h]}\phi(y)\frac{d\lambda(y)}{h}\ .
\end{align}


问:
how do you find the $x$ value for $-\sin x+\cos x=0$

Find the sationary points of the curve and their nature for the equation $y=e^x\cos x$ for $0\le x\le\pi/2$.

I derived it and got $e^x(-\sin x+\cos x)=0$.

$e^x$ has no solution but I don't know how to find the $x$ such that $-\sin x+\cos x=0$

答:
Since you are asking for $0\le x \le \pi/2$ it's quite trivial that $x = \pi/4$ is the only solution of $\sin x = \cos x$

Edit:

Reasoning in $0\le x \le \pi/2$: notice that $\cos x = 0$ only if $x =\pi/2$ and for $x = \pi/2$ we have $\sin \pi/2 = 1 \ne 0 = \cos \pi /2$.

So we can say for sure that $x = \pi/2$ is not a solution of the equation $\sin x = \cos x$.

Divide by $\cos(x)$ both sides of  $\sin x = \cos x$ and you get $\tan x = 1$ which is satiesfied in $0\le x \le \pi/2$ only for $x = \pi/4$


问:
Computing limit involving integration (application of FTC)
The question asks:

Let $f(x)$ be continuous with domain $\mathbb{R}$. Compute the following limit:
$$\lim_{x \to 0} \int_{0}^{x} \frac{(x-t)f(t)}{x^2} \,dt$$

I sense I have to use Fundamental Theorem of Calculus somehow but I don't know how. Could anyone help? Thank you so much.

答: The given  limit is just $\lim_{x \to 0} \frac 1 x \int_0^{x} f(t)dt-\lim_{x \to 0} \frac {\int_0^{x} tf(t)dt} {x^{2}}$. The first limit is $f(0)$ by FTC. Now write $$\lim_{x \to 0} \frac {\int_0^{x} tf(t)dt} {x^{2}}$$ $$=\lim_{x \to 0} \frac {\int_0^{x} t[f(t)-f(0)]dt} {x^{2}}-\lim_{x \to 0} \frac {\int_0^{x} tf(0)dt} {x^{2}}.$$ Write down the last limit by direct calculation. For showing that $\lim_{x \to 0} \frac {\int_0^{x} t[f(t)-f(0)]dt} {x^{2}}=0$ use the fact that $|f(t)-f(0)| <\epsilon$ for all $t$ between $0$ and $x$ provided $|x|$ is suffciently small.


答:
Hint:

$$-\sin x+\cos x=0$$
If and only if

$$\tan x=1$$

Can you get it from here?

Title: how do you find the $x$ value for $-\sin x+\cos x=0$

Body:
Find the sationary points of the curve and their nature for the equation $y=e^x\cos x$ for $0\le x\le\pi/2$.

I derived it and got $e^x(-\sin x+\cos x)=0$.

$e^x$ has no solution but I don't know how to find the $x$ such that $-\sin x+\cos x=0$


问:
Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?

Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin

First, I get the line which passes through point of contact of tangents from origin using result here which is :

$$ -12x-16y-2 \cdot 75 = 0$$

Or,

$$ 6x + 8y -75 =0 $$

Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:

$$ C(0,0) C(x,y) =  (6x+8y-75)^2$$

This leads to:

$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$

$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$

For applying the result in this answer, then the formula

$$ a= 75 - 36,  b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$

Or,

$$ a = 39, b= 11 , h=-48$$

$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$

However, the intended answer was:

$$ \tan \theta = \frac{1}{\sqrt{3} } $$

Where have I gone wrong?

答:
A much more sane and simpler way:

Observe the right triangle created by the line segment from origin to center, and origin to the point which is tangent to circle. Since, tangent and radius are perpendicular, we can apply good ol' trignometery easily:

$$ \sin \theta = \frac{r}{OC}= \frac{5}{10}= \frac{1}{2}$$

Hence, the $ x= \frac{\pi}{6}$ , which makes total angle between tangent as $ x = \frac{\pi}{3}$

However, all the other answer are beautiful in demonstrating the different way of approaching the same problem.


问:
Proof that $dx/|x|$ is a Haar measure on non-zero reals?
Most importantly, what is the meaning of this notation $\lambda = dx/|x|$?  How do I compute say $\lambda(0,1)$ for example?</p>

答:
This is the Haar measure on the multiplicative group ${\bf R}^\times$ (or the group of positive reals under multiplication too, but then there's no reason to use the absolute value symbol). Note that the "$dx$" part actually corresponds to the additive Haar measure. In symbols,

$$\mu^\times(A):=\int_Ad^\times x=\int_A\frac{d^+x}{|x|}.$$

(The $d^+x$ means we perform the integral as usual in calculus, because our measure there is in fact the additive Haar measure.)

For example, let's look at an interval $I=(a,b)$ with $0<a0$ is some multiplicative scaling factor,

$$\mu^\times(cI)=\log\left(\frac{cb}{ca}\right)=\log\left(\frac{b}{a}\right)=\mu^\times(I).$$

This is an example of how the multiplicative Haar measure is translation-invariant, where translation in the multiplicative group amounts to uniform multiplication by a fixed scalar.

More generally, if $A\subset(0,\infty)$, we have

$$\mu^\times(cA)=\int_{cA}d^\times x=\int_{x\in cA}\frac{dx}{|x|}=\int_{u\in A}\frac{cdu}{|cu|}=\int_A\frac{du}{|u|}=\mu^\times(A).$$

This follows from the simple substitution $x=cu$, $dx=cdu$, $x=cu\in cA\iff u\in A$.

Body:
This is the Haar measure on the multiplicative group ${\bf R}^\times$ (or the group of positive reals under multiplication too, but then there's no reason to use the absolute value symbol). Note that the "$dx$" part actually corresponds to the additive Haar measure. In symbols,

$$\mu^\times(A):=\int_Ad^\times x=\int_A\frac{d^+x}{|x|}.$$

(The $d^+x$ means we perform the integral as usual in calculus, because our measure there is in fact the additive Haar measure.)

For example, let's look at an interval $I=(a,b)$ with $0<a<b$. Then

$$\mu^\times(I)=\int_{(a,b)}d^\times x=\int_a^b\frac{dx}{|x|}=\log b-\log a=\log\left(\frac{b}{a}\right).$$

Notice, then, that if $cI=(ca,cb)$ where $c>0$ is some multiplicative scaling factor,

$$\mu^\times(cI)=\log\left(\frac{cb}{ca}\right)=\log\left(\frac{b}{a}\right)=\mu^\times(I).$$

This is an example of how the multiplicative Haar measure is translation-invariant, where translation in the multiplicative group amounts to uniform multiplication by a fixed scalar.

More generally, if $A\subset(0,\infty)$, we have

$$\mu^\times(cA)=\int_{cA}d^\times x=\int_{x\in cA}\frac{dx}{|x|}=\int_{u\in A}\frac{cdu}{|cu|}=\int_A\frac{du}{|u|}=\mu^\times(A).$$

This follows from the simple substitution $x=cu$, $dx=cdu$, $x=cu\in cA\iff u\in A$.


答:
$$
\int_{0}^{x} \frac{(x-t)f(t)}{x^2} \,dt = \int_{0}^{x} \frac{f(t)}{x} \,dt - \int_{0}^{x} \frac{tf(t)}{x^2} \,dt = \frac 1x \int_{0}^{x} f(t) \,dt - \frac{1}{x^2}\int_{0}^{x} tf(t) \, dt
$$

Time to use FTC and the fact that antiderivative $F$ is not just continuous, but continuously differentiable for continuous $f$. Then it is possible to apply L'Hopital's rule. But firstly, it's better to integrate by parts:
$$
\int_{0}^{x} tf(t) \,dt \,dt = F(x)x - \int_{0}^{x} F(t) \,dt.
$$
So we get (assuming $F(0) = 0$)
$$
\lim_{x \to 0} \left[\frac{F(x)}{x} - \frac{xF(x)}{x^2} + \frac{1}{x^2}\int_{0}^{x} F(t) \,dt \right] = \lim_{x \to 0} \frac{1}{x^2}\int_{0}^{x} F(t) \,dt = \lim_{x \to 0} \frac{F(x)}{2x} = \lim_{x \to 0}\frac{f(x)}{2} = \frac{f(0)}{2},
$$
where L'Hopital's rule was used twice.

Title: Computing limit involving integration (application of FTC)

Body:
The question asks:

Let $f(x)$ be continuous with domain $\mathbb{R}$. Compute the following limit:
$$\lim_{x \to 0} \int_{0}^{x} \frac{(x-t)f(t)}{x^2} \,dt$$

I sense I have to use Fundamental Theorem of Calculus somehow but I don't know how. Could anyone help? Thank you so much.


问:
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4$
I've been trying to help my son with what seems an innocent enough equation, but I have to admit, I'm struggling and walking in circles - the question is:

If $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4$, what is $abc$?

First I multiply both sides with the denominators:

$a(a+c)(a+b)+b(b+c)(a+b)+c(b+c)(a+c)=4(a+b)(a+b)(b+c)$

$\Updownarrow$

$a^3+b^3+c^3=3(abc+a^2b+ab^2+b^2c+a^2c+ac^2+bc^2)$

From here I begin to run in circles - substituting $x=abc$ and expressing $a$, $b$ and $c$ in terms of $x$, for example, ends up after a while back at the beginning. So what is the trick - or is there no simple solution?

(Sorry about the tag - I couldn't find anything more suitable for such an elementary question)

Edit

The suggested duplicate is not a full answer, I think. My question concerns the same equation, but the question is different, and the answers are only partial. I will accept the answer given by Rhys Hughes, since it answers my question and gives a good explanation of how he reached it.

答:
$abc$ is not fixed in general. You can pick $a=b=\alpha$ and arrive at $$\frac{2\alpha}{\alpha+c}+\frac{c}{2\alpha}=4\implies c=\alpha\bigg(\frac{7\pm\sqrt{65}}{2}\bigg)$$Yielding $abc=\alpha^3\bigg(\frac{7\pm\sqrt{65}}{2}\bigg)$.

In this way we may have the value of $abc$ span $\Bbb R$ since for each $y\in \Bbb R$; $\alpha=\bigg(\frac{2y}{7\pm\sqrt {65}}\bigg)^\frac 13\implies abc=y$


问:
When does a system of polynomial equations have infinitely many solutions?
It seems intuitively correct to say that a system of polynomial equations has finitely many solutions if there are as many equations as there are variables in the system. However, how would you prove that a system of polynomial equations with fewer equations than variables has infinitely many solutions (corresponding to the "under-determined" case in linear systems)? How would you characterize when a subset of equations "coincide" (you can combine several to get another, like linear dependence in linear systems)?

For context, I'm in the middle of an abstract algebra course, and we've just finished Chapter 7 of Dummit & Foote (though this question isn't yet directly related to the course, that's just my background knowledge).

答:
This question turns out to be deeper than it seems, since it has a much richer theory than the linear case. In particular, it's one of the questions leading to elimination theory.

To illustrate why this problem is more complex than it seems, let me give an example of system of polynomials that can have either zero, finitely many or infinitely many solutions depending on what kind of solutions you are looking for:

$$ \{ x^2-2, y^2 + z^2 \}$$

First, note that this system has no rational solutions, since there is no $x \in \mathbb{Q}$ such that $x^2 = 2$. Next, over the real numbers, the system has exactly two solutions, $x = \pm\sqrt{2}, y = z = 0$, since for any $y, z \in \mathbb{R}$ where $y \neq 0$ or $z \neq 0$, $y^2+z^2 > 0$. Finally, over the complex numbers, this system has infinitely many solutions, namely those where $x = \sqrt{2}$ and $y = i z$ or $y = -i z$.

Over the complex numbers (or more generally, over an algebraically closed field), one can say that we have $n$ variables and $k$ polynomials, where $k < n$, there are either no solutions or infinitely many solutions, using a more general notion of dimension (a key ingredient in this argument is Krull's Principal Ideal Theorem). Over other fields, these questions turn out to be significantly more complex.

Algorithmically, one would use Buchberger's algorithm to transform the set of polynomials into a Gröbner Basis (using a lexicographic term ordering), which generalizes row-echelon form. If we have $n$ variables $x_1, \ldots, x_n$ and the Gröbner basis contains a polynomial containing only variables $x_i, \ldots, x_n$ for each $i = 1, \ldots, n$, we know that the system can have only finitely many solutions. Conversely, if the system does not contain such polynomials for one of the $i$, and the field we're interested in is algebraically closed, we know that there must be infinitely many solutions.

Edit: $x^2 = 2$ has two solutions, not one. Thanks @lisyarus!


答:
When considering an ordinal $\alpha$ (von neumann ordinals), not all its subsets will have the $0$ as the minimum element. I might be mistaken, but I got the impression that perhaps OP might be assuming that to be the case.

So if one considers $\alpha \subseteq \alpha$, its minimum element will be $0$. In other words $\min(\alpha)=0$.

Now say $\alpha=\omega^3$. We might consider a subset $S_1$ defined as: $S_1=\{\, \omega \cdot (1+x) \,|\, x \in \mathrm{Ord} \, \, \wedge x<\omega^2 \, \}$. Similarly we might define $S_2=\{\,\omega^2 \cdot (1+x) \,|\, x \in \mathrm{Ord} \, \, \wedge x<\omega \, \}$. We can also write $S_2=\{\, \omega^2,\, \omega^2 \cdot 2,\, \omega^2 \cdot 3,.... \,\}$. Both $S_1$ and $S_2$ are well-defined subsets of $\alpha=\omega^3$. However, we have $\min(S_1)=\omega$ and $\min(S_2)=\omega^2$.

Edit: An easier example similar to two above. Define $S_0=\{\, 1+x \,|\, x \in \mathrm{Ord} \, \, \wedge x<\omega^3 \, \}$. We get $S_0 \subset \omega^3$ and $\min(S_0)=1$.

Title: A query regarding sets from Quantum computing since Democritus

Body:
I was reading the book "Quantum Computing Since Democritus".

"The set of ordinal numbers has the important property of being well
ordered,which means that every subset has a minimum element. This is
unlike the integers or the positive real numbers, where any element
has another that comes before it."

Unlike integers? Let's consider a set $\{1,2,3\}$ This has a minimum element.

Do you get what does the author wants to say here?


问:
Why should we care about sequence in real analysis?
I'm self-learning Real analysis through the book Real Analysis by Terence Tao. There is a concept of sequence which I don't fully understand its intuition and motivation by constructing one. It is not quite natural for me.

I see that the concept of sequence arise for the first time when constructing real number using rational number. And then it come back again and again when dealing with continuity, derivative,...

My question is:
When first think about sequence (of object in general) I can think of many useless sequences which behave arbitrarily.
For example, the sequence $1, 3, -1, 2, 2, -1...$ doesn't mean anything (i choose these number randomly). It seems that the number of useless sequences is much more than that of useful sequence.

So, does the concept of sequence naturally arise in real analysis ? Why is the concept of sequence extensively while there are way more useless sequences than the useful ones ?

Thank you very much for your help!

答:
When first think about machine I can think of many useless machines which behave arbitrarily. For example, a random collection of levers, wheels, and pulleys connected by ropes doesn't do anything (i connect these pieces randomly). It seems that the number of useless machines is much more than that of useful machine.

So, does the concept of machine naturally arise in engineering ? Why is the concept of machine extensively while there are way more useless machines than the useful ones ?


问:
Prove the equivalence of presentation of a free group with a free product
I want to prove that the following presentation of a free group (generators and relations):

$$\left(\begin{array}{c|c}
x_0,a_0&a_0a_1x_2=x_0a_1\\
x_1,a_1&a_1a_2x_0=x_1a_0\\
x_2,a_2&a_2a_0x_1=x_2a_2
\end{array}\right)$$

is equivalent to the free product

$$(a_0,a_1)\star(x_2,A:A^2x_2=x_2A), \quad A=a_2a_0a_1.$$

I have written the presentation that way in case it is suggestive of the solution. This is an example (#11) from A Quick Trip Through Knot Theory by R. Fox, which is a chapter in Topology of 3-Manifolds, Ed. M.K. Fort.

Mostly I am just overwhelmed by the idea of trying to show that any combinbation of the six generators can always be written as alternating products of the generators of the two free groups in the product. I can show some things (for instance, I can easily show the relation $A^2x_2=x_2A$ holds), but how does one really prove things like this?

This is not a homework but since I am asking it rather like a homework, I will take hints first if that's what some people would like to offer.

答:
Having the answer makes this calculation less nasty than it might otherwise be.  Since the desired presentation doesn't involve $x_0$ or $x_1$, it's reasonable to use the first and third of the given relations to eliminate these, expressing them in terms of $x_2$ and the three $a$'s.  So your group is generated by $\{a_0,a_1,a_2,x_2\}$ subject to just the one remaining relation, the second on the original list, which becomes, after you plug in the expressions for $x_0$ and $x_1$,
$$
a_1a_2a_0a_1x_2a_1^{-1}=a_0^{-1}a_2^{-1}x_2a_2a_0.
$$
Transpose all the factors of the form $a_i^{-1}$ to the other sides, and you get $A^2x_2=x_2A$.  Finally, since $a_2$ can be expressed in terms of the other two $a$'s and $A$, namely as $a_2=Aa_1^{-1}a_0^{-1}$, we can introduce $A$ as a new generator and eliminate $a_2$.  The result is that the group is generated by $\{a_0,a_1,A,x_2\}$ subject to $A^2x_2=x_2A$, as desired. </p>


问:
In how many different ways can we make a flag of $5$ stripes if we have have $2$ red stripes, $2$ white stripes and $3$ green stripes?
The main problem that I am encountering is the fact that there is an unequal number of stripes. It can be done schematically but is quite exhausting.

We have $3$ options for the first stripe. Still $3$ for the second, but we can have $2$ or $3$ for the third stripe, and thus a problem.

Any ideas?

答:
Hint. There is at least a green stripe G, so we distinguish 3 cases:

If we have one G then the other stripes are RRWW. In this case, the number of different flags is $\frac{5!}{1!2!2!}$.

If we have two Gs then the other stripes are RWW or RRW. In this case, the number of different flags is ??

If we have three Gs then the other stripes are RR or WW or RW. In this case, the number of different flags is ?? + ??

Can you take it from here?


问:
Find the locus in the “circle point midpoint theorem”.
This is a simplified version of the “circle point midpoint theorem”.

Q is a point external to the circle $C : x^2 + y^2 = r^2$. P is a point on C. M is the midpoint of PQ. Find the locus of M as P moves along C.

I managed to find the answer by:-

(1) Get A and B, the points of intersections of QO extended with C.

(2) Find $M_1$ and $M_2$, the midpoints of QA and QB respectively.

(3) Knowing that the locus of M is a circle with $M_1$ and $M_2$ as its extremities, I form its equation. It happens to be the equation of the locus of M.

My work depends heavily on the assumption that the locus is a circle which is not supposed to be known beforehand. So, I am wondering is there another way to find the required locus.

答:
$Q$ is a point with coordinates $(a,0)$ with $a\le -r$. A generic point in your circle has coordinates $(r\cos t, r\sin t)$. The midpoint has coordinates
$$
(\frac{a+r\cos t}{2},\frac{r\sin t}{2})
$$
Excluding $t$ we clearly have $(2x-a)^2+(2y)^2=r^2$ or
$$
(x-a/2)^2+y^2=(r/2)^2
$$
which is a circle of radius $r/2$ centered at $(a/2,0)$.


问:
What is wrong in my attempt to expand the definite integral $\int_0^1\ln(1+x^3)dx$ about $x = 1$ through second degree terms?
What is the correct way to expand the definite integral $\int_0^1\ln(1+x^3)dx$ about $x = 1$ through second degree terms?

I am getting contradicting answers through different approaches.

First approach:

Let $f(x) = \int_x^1\ln(1+x^3)dx = \int_{b(x)}^{a(x)}\ln(1+\zeta^2)d\zeta$ with $g(x, \zeta) = \ln(1+\zeta^3)$, $b(x) = 1$, $a(x) = x$.

By Leibniz's rule,
\begin{align}
f'(x) & = g(x, b(x))\cdot \require{cancel}\cancelto{0}{b'(x)} - g(x, a(x))\cdot a'(x) + \int_{a(x)}^{b(x)}\require{cancel}\cancelto{0}{\frac{\partial}{\partial x}g(x,\zeta)}d\zeta \\
      & = -g(x, a(x))\cdot a'(x) \\
      & = -\ln(1 + x^3).
\end{align}

Now,
$$ f''(x) = -\frac{3x^2}{1+x^3}.$$

Then, using Taylor's expansion around $x = 1$ up to second degree:

$$ \int_x^1\ln(1+x^3)dx \approx  -\ln(2) (x-1)-\frac{3}{4}(x-1)^2 .$$

Since now I have an expansion, I thought I could just substitute $x = 0$ to obtain the expansion of the integral as below

$$\int_{x=0}^1\ln(1+x^3)dx \approx \ln(2)-\frac{3}{4}.$$

Second approach:
To verify if the answer above is correct, I tried to just expand from the beginning in the following way.

Let $f(x) = \int_0^1\ln(1+x^3)dx = \int_{b(x)}^{a(x)}\ln(1+\zeta^2)d\zeta$ with $g(x, \zeta) = \ln(1+\zeta^3)$, $b(x) = 1$, $a(x) = 0$.

Again, by Leibniz's rule,
\begin{align}
f'(x) & = g(x, b(x))\cdot \require{cancel}\cancelto{0}{b'(x)} - g(x, a(x))\cdot \require{cancel}\cancelto{0}{a'(x)} + \int_{a(x)}^{b(x)}\require{cancel}\cancelto{0}{\frac{\partial}{\partial x}g(x,\zeta)}d\zeta \\
      & = 0 \\
      & \implies f''(x) = 0..
\end{align}

But then I get a different expansion:

$$ \int_0^1\ln(1+x^3)dx \approx  0.$$

I am more confused because the integral can be evaluated to a constant:

$$ \int_0^1\ln(1+x^3)dx = -3 + \frac{\pi}{\sqrt{3}} + \ln(4) \approx 0.20009$$

which is different from both results.

答:
You are falling victim to your own imprecise notation.  After all, what exactly do you mean by the statement $$f(x) = \int_0^1 \log(1+x^3) \, dx?$$  The RHS is a constant.  It uses $x$ as the variable of integration, yet you are using the same variable on the left as if somehow you are expecting the RHS to vary with $x$.  That is why, in your second approach, you get $f'(x) = 0$, because you just found an elaborate way to show that the derivative of a constant is zero.  You don't have this problem in your first approach because you wrote $$f(x) = \int_{\zeta = x}^1 \log(1 + \zeta^3) \, d\zeta.$$  And now the RHS is a function of $x$ because $x$ is the lower limit of integration and you have used a different variable of integration $\zeta$.  Well, you didn't really write it exactly this way, instead you wrote the abomination $$f(x) = \int_x^1 \log (1 + x^3) \, dx,$$ but my version is what you intended.

Ultimately, were you to use more terms in your series, you would know which approach is correct:  We have $$f(x) \approx -(x-1) \log (2) -\frac{3}{4} (x-1)^2  -\frac{1}{8} (x-1)^3 + \frac{5}{32} (x-1)^4 + O((x-1)^5)$$ where upon evaluating at $x = 0$ yields $$f(0) \approx \log 2 - \frac{15}{32} \approx 0.224397$$ which is reasonably close to the exact value, with even more terms giving better convergence.


问:
Take a set $S$ which has 6 vectors, and $S$ is a subset of $\mathbb{R}^5$. Why is $S$ linearly dependent?
I know that for a vector space $\mathbb{R}^5$, $\dim(\mathbb{R}^5)=5$ and thus a basis (b) (linearly independent and $span(b) = \mathbb{R}^5$) of $\mathbb{R}^5$ should have 5 vectors. I do not know how to link ideas together to understand why $S$ is linearly dependent? Thank you! :)

答:
Let $v_1,v_2,v_3,v_4,v_5$ be a basis for $R^5$ and $u \in R^5$

If  $v_1,v_2,v_3,v_4,v_5$ forms a basis than there exists real numbers $c_i$ such that

$u=c_1v_1+c_2v_2+c_3v_3+c_4v_4+c_5v_5$

$u=c_1v_1+c_2v_2+c_3v_3+c_4v_4+c_5v_5 \Rightarrow u-(c_1v_1+c_2v_2+c_3v_3+c_4v_4+c_5v_5)=0$

As it can be seen, $v,v_1,v_2,v_3,v_4,v_5$ are linearly dependent because coefficient of $v$ is not $0$


问:
How to calculate the polar arc length of the entire cardioid $r=a(1-\cos\theta)$
I'm having a bit of an issue calculating the arc length of $r = a(1-\cos\theta)$.

I'll begin by listing the steps I made in my attempt to solve this exercise.

We know that the arc length formula for polar coordinates is $$L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \,d\theta$$

So with $r = a(1-\cos\theta)$ the arc length formula becomes $$L = \int_{0}^{2\pi} \sqrt{[a(1-\cos\theta)]^2 + (a\sin\theta)} \, d\theta.$$

Simplifying this further, we obtain $$L = a\sqrt{2}\int_{0}^{2\pi} \sqrt{1-\cos\theta} \, d\theta.$$

I'm stuck on solving the resultant integral. I've attempted u-substitution and integration by parts but to no avail. Is my resultant integral incorrect? How do I proceed?

答:
$$\cos \theta = 1 - 2 \sin^2 \displaystyle\frac{\theta}{2}$$
so $$1 - \cos \theta = 2 \sin^2 \displaystyle\frac{\theta}{2}$$
You need to be careful taking the square root of $2 \sin^2 \displaystyle\frac{\theta}{2}$, since the sign changes over your interval of integration. So, double the integral from $\theta = 0$ to $\theta = \pi$ and you should get the correct answer.


答:
Elementary counting should work here. We know at least one of the stripes has to be green ($G$). So we have to choose $4$ stripes from $2$ G, $2$ R and $2$ W colors.

Keeping one $G$ aside,

i) If we pick two $G$, we can then pick $RR, WW, RW$.

ii) If we pick one $G$, we can then pick $RRW, WWR$.

iii) If we pick no $G$, we then have only choice of $RRWW$.

So for (i), you can now arrange them to make $2 \cdot \frac{5!}{3! 2!} + \frac{5!}{3!} = 40 \ $ flags.

Similarly you can find for others.

Title: In how many different ways can we make a flag of $5$ stripes if we have have $2$ red stripes, $2$ white stripes and $3$ green stripes?

Body:
The main problem that I am encountering is the fact that there is an unequal number of stripes. It can be done schematically but is quite exhausting.

We have $3$ options for the first stripe. Still $3$ for the second, but we can have $2$ or $3$ for the third stripe, and thus a problem.

Any ideas?


答:
We can first simplyfy the equation to have;
$$\sin{x}=\cos{x}$$
$$\implies\sin^2{x}=\cos^2{x}.$$

Adding $\cos^2{x}$ to both sides gives;
$$\sin^2{x}+\cos^2{x}=\cos^2{x}+\cos^2{x}=2\cos^2{x}.$$

Note that $\sin^2{x}+\cos^2{x}=1,$ so we know that;
$$2\cos^2{x}=1$$
$$\implies\cos^2{x}=\frac{1}{2}$$
$$\implies\cos{x}=±\frac{\sqrt{2}}{2}.$$

Easily, the only solutions to this are $x=45^\circ, 135^\circ, 225^\circ,$ and $315.$ But since we also need $\sin{x}=\cos{x},$ the only solutions are $x=45^\circ$ and $225^\circ,$ or $x=\frac{\pi}{4}$ and $\frac{5\pi}{4}.$ But since $0\leq x\leq\frac{\pi}{2},$ $x=\frac{\pi}{4}$ is the only solution.


问:
How can there be multiple irreducible representations of a group each having distinct dimension?
The matrix elements for the $(2l+1)$-dimensional irreducible representation of SO(3) are given by:

$D_{m',m}^l(\phi_1,\Phi,\phi_2)=[i^{m'-m}\sqrt{(l+m')!(l-m')!(l+m)!(l-m)!}$
$\times(\frac{1+\cos\Phi}{2})^l (\frac{1-\cos\Phi}{1+\cos\Phi})^{(m'-m)/2}$
$\times\sum\limits_k \frac{(-1)^k}{(l+m-k)!(l-m'-k)!(m'-m+k)!k!}$
$\times(\frac{1-\cos\Phi}{1+\cos\Phi})^k] e^{im'\phi_2}e^{im\phi_1}$

My question is, how can there be multiple irreducible representations of SO(3) each of which has a different dimension?  I thought that by definition an irreducible representation is one which has no non-trivial invariant subspace.  I know that SO(3) can be represented by the group of 3x3 symmetric matrices of determinant = 1 (I thought this had something to do with the fact that it takes 3 parameters to describe a 3D rotation).  If I have a larger matrix (e.g. 5x5) doesn't it have to have a non-trivial invariant subspace (e.g. of dimension 2), and therefore be reducible?

The application here is the expansion of functions using generalized spherical harmonics, which is accomplished (per the Peter-Weyl theorem) using an infinite linear combination of the matrix elements of the irreducible representationS (emphasis on the S) of SO(3).

答:
Dear okj, in the $SO(3)$ case, and perhaps at least morally even more generally, you may imagine the higher-dimensional irreducible representations as tensors. In particular, the 5-dimensional representation is a symmetric traceless tensor $T$. It satisfies
$$T_{ij} = T_{ji}, \quad \sum_{i=1}^3 T_{ii} = 0 $$
The first condition reduces 9 elements of the matrix to 6 independent ones; the vanishing of the trace reduces 6 to 5. Now, if an $SO(3)$ matrix $M$ acts on vectors via
$$ \vec v \to M \vec v ,$$
then it acts on $T$ via
$$ T \to M T M^{T}.$$
Note that it preserves the symmetry of $T$ as well as its vanishing trace. However, it's equally clear that there is no linear combination of the five entries in $T$ that would be invariant under all $SO(3)$ transformations. So there is no singlet, and because $3+1+1$ and $1+1+1+1+1$ are the only possible ways how to decompose a 5-dimensional representation to representations you know, it follows that none of them works and there is no triplet (vector), either.

More general representation of $SO(3)$ may also be written as tensors with some constraints. The constraints are needed to make them irreducible.

For more general groups, both finite and Lie groups, the more complicated representations are not that easily written as tensors, but in principle, it's always possible to find any irrep in the tensor product of "fundamental representations".

Here is an important point that may be confusing you. One may build more complicated representations of a group as
$$ V_1\oplus V_2$$
which is manifestly reducible but one may also build them as
$$ V_1\otimes V_2, $$
the tensor product. This is like creating object  with many indices. There is no simple link between the direct sum and the tensor product. The tensor product typically contains completely new, higher-dimensional ireducible representations than $V_1$ and $V_2$.


问:
Why is the function $f(t)=(t,t,t,t,t,...) $continuous in the product topology but not the box topology?
The function $f:\mathbb R \to \mathbb R^\infty$ given by $f(t)=(t,t,t,t,t\dots)$ is said to be not continuous in the box topology but continuous in the product topology.

The example given is $U=(-1,1)\times(-1/2,1/2)\times (-1/3,1/3)\times \dots $. Why does this set show the box topology is not continuous by $f$? We can just take the preimage of $U$ and it would just be $U$ correct? I can see that $(-1/n,1/n)$ converges to $0$ which is closed in $\mathbb R$, but for any specific $\mathbb R_i$ in $\mathbb R^\infty$, we actually do have an open interval even if it is extremely small.

答: $f^{-1}(U)=\{t: |t| <\frac 1 n \, \forall n\}=\{0\}$ which  is not open.


答: It's conceptually simpler to begin with the arbitrary sequence notion and then specify further as needed (e.g. monotone sequence, decreasing sequence, recursive sequence, integer sequence, positive sequence, monotone decreasing sequence, rational sequence, alternating sequence, convergent sequence, etc.) than it is to start with one of these as the "base notion" and then add or subtract properties as needed. For example, suppose "sequence" means what we now call a monotone decreasing sequence. What name would you propose that we give to what we now call an alternating sequence?</p>

Title: Why should we care about sequence in real analysis?

Body:
I'm self-learning Real analysis through the book Real Analysis by Terence Tao. There is a concept of sequence which I don't fully understand its intuition and motivation by constructing one. It is not quite natural for me.

I see that the concept of sequence arise for the first time when constructing real number using rational number. And then it come back again and again when dealing with continuity, derivative,...

My question is:
When first think about sequence (of object in general) I can think of many useless sequences which behave arbitrarily.
For example, the sequence $1, 3, -1, 2, 2, -1...$ doesn't mean anything (i choose these number randomly). It seems that the number of useless sequences is much more than that of useful sequence.

So, does the concept of sequence naturally arise in real analysis ? Why is the concept of sequence extensively while there are way more useless sequences than the useful ones ?

Thank you very much for your help!


答:
There are various ways to interpret the question. Firstly, it is certainly the case that the amount of useless sequences far exceeds that of useful sequences. However, this will almost always be the case with any definition. The price you pay for putting lots of different interesting things under the same umbrella is that with them you also get lots of boring things under your umbrella. However, that is not a bug but rather feature of good definitions. The point is not to throw away the baby with the bath water. Simply allow all these useless exemplars co-exist with the ones that actually interest you. More importantly, just because we defined sequences and say things like "let $s$ be a sequence" rather than "let $s$ be an interesting sequence" does not mean we have any particular interest in all sequences. There will be toy examples that we'll look at just to clarify concepts but in practice we will always actually be looking at interesting sequences. Just don't try to visualise all sequences.

The above should answer the second half of your question. As for the first half, from an applications perspective sequences arise all the time. For instance, if you perform a measurement in physics the result you get is hardly ever precise. Too many factors affect your measurement so what you really always get is some rational number as the result of performing the measurement. Now, you can, and should if you are a good physicist, never accept the first measurement result as definitive. So, you improve your testing conditions and measure again. And again. And again. You do so a few dozen cases. What you obtain is a finite sequence. Of these values, which is the correct one? That is hard to tell, but we can idealise. If you had enough time you'd perform the measurement repeatedly forever, each time learning from the previous measurements and improving. This ideal situation leads to a genuine sequence. What is the true result of the measurement? Well, still hard to say but if the sequences converges, then there is some weight to the claim that the limit of the sequence is the true measurement result.

So, the above explains why the practicing user of analysis (e.g., physicists, engineers, chemists, etc.) are interested in sequences. They produce finite ones all the time and if only given enough time they would produce actual sequences. So it is a useful abstraction. (Incidentally, this touches upon a general principle in analysis whereby infinity is considered a very good approximation for large numbers.) But one may wonder whether sequences arise naturally also from an intrinsically mathematical perspective. The answer is yes, to some extent. It's true that the real numbers can be constructed by using sequences of rationals, but that is secondary (the reals can be constructed in many ways that do not mention sequences at all). Regardless of how the reals are constructed, or if they are constructed at all, the topology of the reals is strongly related to sequences. Convergence with respect to the natural topology of $\mathbb R$ is fully and faithfully captured by the convergence of sequences. This, however, is not true in all spaces that are directly of interest in analysis.


问:
If $A$ is an $m \times n$ matrix, show that if $A \neq 0$ then $\exists x \in \Bbb R^n$ such that $Ax \neq 0$.
Let $A$ be an $m \times n$ matrix. Show that if $A \neq 0$ then $\exists x \in  \Bbb R^n$ such that $Ax \neq 0$.

答:
$A \neq 0 \Rightarrow $ there exists a i-th column $A_i$ in $A$  that is not equal $\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0\end{pmatrix}$

$A\begin{pmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0\end{pmatrix}=A_i \neq 0$


问:
What is the meaning of "prove $\{a,b\}=\{c,d\}$" in this question?
There is a math problem:

If $a+b=c+d$ and $a^2+b^2=c^2+d^2$ prove $\{a,b\}=\{c,d\}$

So by the $\{a,b\}=\{c,d\}$,Which statement should I prove?

First:  ($a=c$ and $b=d$) or ($a=d$ and $b=c)$

Second: $a=c$ and $b=d$

I think I should prove the first statement because of properties of the sets but I am not sure.

答:
This means that you need to prove that the sets $\{a,b\}$ and $\{c,d\}$ are equal.

The first statement is correct since order is irrevelant in sets.


问:
Closed-form solution to the recurrence
Context

In a game players are divided into two groups, A and B. Players in A know identities of all players whereas players in B only know the identity of himself. In the first round Group A eliminates one player from Group B, then the eliminated choose one of the remaining players at random to eliminate. If the new victim is in B, the new victim would do the same until a player in A is eliminated, and a new round starts. The group that first have all players eliminated loses the game.

Suppose there are $n$ players in total and $x$ players in Group B. Let $p(n,x)$ denote the probability that Group B wins.

Question

What is the closed-form solution to the recurrence $$p(n+1,x+1)=\frac{n-x}{n}p(n-1,x)+\frac{x}{n}p(n,x)$$ with conditions $x\le n$ and $x,n \in\Bbb N$, and boundary conditions $p(0,0)=0$, $p(n,0)=0$ and $p(n,n)=1$ for $n\ne 0$?

答:
At least a partial answer so far. Let $a_{n,k}=p(n,n-k)$ for $0\leqslant k\leqslant n$.

Then $a_{n,n}=0$ for $n\geqslant 0$, $a_{n,0}=1$ for $n>0$, and $$na_{n+1,k}=ka_{n-1,k-1}+(n-k)a_{n,k}$$ for $n>1$ and $1\leqslant k\leqslant n$. Let $g_k(x)=\sum_{n=k}^\infty a_{n,k}x^n$; then $\sum_{n=k}^\infty[\ldots]x^{n-1}$ yields $$\big(g_k(x)/x\big)'=kg_{k-1}(x)+g_k'(x)-kg_k(x)/x$$ with $g_0(x)=x/(1-x)$; solved for $g_k(x)$, this gives the recurrence $$g_k(x)=kx(1-x)^{k-1}\int_0^x\frac{g_{k-1}(t)}{(1-t)^k}\,dt.$$

Fortunately, this recurrence has a closed-form (!) solution: $$g_k(x)=\frac{x}{1-x}\big(x+(1-x)\log(1-x)\big)^k.$$ As $g_k(x)=\sum_{n=k}^\infty p(n,n-k)x^n$ (actually the terms are nonzero only for $n>2k$), this gives a way to get a closed form of $p(n,n-k)$ for each fixed $k$, in terms of Stirling numbers of the first kind.

But it remains to find out whether there is a closed form (with no summations) for arbitrary $k$.

Title: Closed-form solution to the recurrence


问:
Proving a positive integer
Suppose that a positive integer $n$ ends with $k$ $9\text{'s}$. How can we prove that the integer $3n^4+4n^3$ ends with at least $2k$ $9\text{'s}$?

答:
Firstly note that saying a number ends with $k$ $9$s is equivalent to saying that it is congruent $-1\ (\text{mod }10^k)$, as for example $99$ ends with two $9$s and $99\equiv -1\ (\text{mod }100)$.

So write $n=A\cdot 10^k -1$, for an integer $A$. Then note:
\begin{align}
n^3&\equiv 3A\cdot10^k-1\ \quad&(\text{mod }10^{2k})\\
n^4&\equiv -4A\cdot10^k+1 \quad&(\text{mod }10^{2k})
\end{align}

As all the other terms include $10$ to a $\geq2k$ integer power. So adding:
\begin{align}
3n^4+4n^3 &\equiv (-12A\cdot10^k+3) + (12A\cdot10^k-4)\quad&(\text{mod }10^{2k})\\
&\equiv -1 \quad&(\text{mod }10^{2k})
\end{align}

Which is the required answer.

Title: Proving a positive integer


答: The first statement is intended; the second would use $()$ instead of $\{\}$. If $a=d\ne c=b$ then $a+b=c+d,\,a^2+b^2=c^2+d^2$, so the second interpretation is false.

Title: What is the meaning of "prove $\{a,b\}=\{c,d\}$" in this question?

Body:
There is a math problem:

If $a+b=c+d$ and $a^2+b^2=c^2+d^2$ prove $\{a,b\}=\{c,d\}$

So by the $\{a,b\}=\{c,d\}$,Which statement should I prove?

First:  ($a=c$ and $b=d$) or ($a=d$ and $b=c)$

Second: $a=c$ and $b=d$

I think I should prove the first statement because of properties of the sets but I am not sure.


问:
How many ways to combine two sets so that order of each set is preserved.
If we have sets $\lbrace1,2,...k\rbrace$  and $\lbrace  k+1,..., n\rbrace$, in how many ways can these sets be combined so that the original two sets respect the original order and 1 is always in front?

For example if k=3, n=5,
12345,

12453,

14523,

12435,

14235,

14253.

There are 3*2 of them.

I don't think the answer is $k*(n-k)$ in general since it doesn't hold when n=4, k=2.

答: You have $k$ slots (including the one at the end)  in the first set to put the $n-k$ numbers in where each slot can hold as many numbers as you like.  This is a weak composition of $n-k$ into $k$ parts.  There are ${ n-1 \choose k-1}$ of them</p>


问:
how do you find the $x$ value for $-\sin x+\cos x=0$

Find the sationary points of the curve and their nature for the equation $y=e^x\cos x$ for $0\le x\le\pi/2$.

I derived it and got $e^x(-\sin x+\cos x)=0$.

$e^x$ has no solution but I don't know how to find the $x$ such that $-\sin x+\cos x=0$

答:
For all $x$, $\cos(x+\pi/2)=-\sin(x)$ hence the equation is equivalent to
$$\cos(x)+\cos(x+\pi/2)=0$$

But we also have the identity

$$\cos p+\cos q=2\cos\frac{p+q}{2}\cos\frac{p-q}2$$

So your equation is also equivalent to

$$2\cos(x+\frac{\pi}4)\cos\frac{\pi}4=0$$

That is, $\cos(x+\frac\pi4)=0$.

But $\cos t=0$ iff $t=\frac{\pi}{2}+k\pi$, $k\in\Bbb Z$. Therefore, your equation is also equivalent to

$$x=\frac\pi4+k\pi$$

And the only value of $x$ in $[0,\pi/2]$ occurs for $k=0$.


答:
Graphically we see immediately that $x=\frac {\pi}{4}$. But how about deriving this analytically?

Consider $\cos(2x)$ and use the formula for the double argument $\cos(2x) = \cos(x)^2 - \sin(x)^2$

Hence if $\sin(x) = \cos(x)$ we find that

$$\cos(2x)=0$$

Since $\cos(\alpha)$ vanishes if and only if the argument is of the form  $\alpha=\pi(\frac{1}{2}+k), k=0,\pm1,\pm 2, ...$ $ we have

$$x = \frac{\pi}{2}(\frac{1}{2}+k)|_{k\to 0}\to \frac{\pi}{4}$$

in the requested range.

We can also find the value of $a=\sin(x)=\cos(x)\gt0$: from the relation $\sin(x)^2+\cos(x)^2=1$ we get $2 a^2 = 1$ and hence $a=\frac{1}{\sqrt{2}}$.


问:
Any finite subset of a linearly independent set is linearly independent

Prove that if each finite subset of a set $S$ of vectors is linearly independent, then $S$ is also linearly independent.

My attempt thus far:

Suppose that each finite subset of $S$ is linearly independent. That is, for each such subset $S_n$ of $S$, with vectors $x_1,...,x_n \in S_n$, we know that

$$a_1x_1 + ... + a_nx_n = 0 \implies a_1 = ... a_n = 0$$

for scalars $a_1, ..., a_n$.

At this point I'm stuck. I'm not sure how to proceed with this known information and reach the conclusion that $S$ itself is also linearly independent. I know that this question has been asked elsewhere (such as for example here: Prove that any finite subset of a linearly independent set is linearly independent), but I am not understanding the answers and explanations that have been provided on these other pages and would like some additional guidance on completing this proof.

答:
Linear dependence/independence can be defined in  countably infinite vector spaces for any family of its elements:

The family $u_i, i \in I$ is linearly dependent if there exists a non-empty finite subset $J \subset I$ and a family of $a_i$ coefficients from corresponding field, all non-zero, such that $\sum\limits_{i \in J}a_iu_i=0\quad (1)$.

Now if we assume, that each finite subset of a set $S$ of vectors is linearly independent, but $S$ itself is linearly dependent, then from above definition we obtain  non-empty finite subset with $(1)$ property, which contradicts assumption.