Body:
$$I=\int_{0}^{1} \frac{r^3 dr}{\sqrt{1-r^2}}$$
Let $r=\sin t \implies dr= \cos t dt$, then
$$I=\int_{0}^{\pi/2} \sin^3 t dt=\int_{0}^{\pi/2} \sin t(1-\cos^2 t) dt$$
$$I=\int_{0}^{\pi/2} (\sin t -\cos^2 t \sin t) dt$$
$$ I=-\cos t+\frac{\cos^3 t}{3}|_{0}^{\pi/2}=-(-1+1/3)=2/3.$$


答:
Or just integration by parts:

\begin{eqnarray*}\int_0^1 r^2 \frac{r}{\sqrt{1-r^2}}dr
& = & \left.-\sqrt{1-r^2}\cdot r^2\right|_0^1 + \int_0^1 (2r)\sqrt{1-r^2}\;dr \\
& = & \left.-\frac 23\left(1-r^2\right)^{\frac 32}\right|_0^1 \\
& = & \frac 23
\end{eqnarray*}

ContentLicense: CC BY-SA 4.0

Title: Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$

Body:
Evaluate

$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$

My attempt : I put  $t= \sqrt{1-r^2}$ now  $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$

$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$

$$= 2\int_{0}^{1} r^2 dt$$

$t= \sqrt{1-r^2}\implies r^2= t^2-1$

$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$


问:
are geodesic shortest path or quickest path?
I'm a bit confused with geodesics : are they the shortest path (in distance) or the quickest path (in time). For example, Let take a triangle ABC. I'm using a car. I'm in $A$ and I have to go in $B$. The path $AB$ is 2km long, but I can go at 10 km/h only, where as the path that path through C has 4 km length, but it's a free way and I can go at 100 km/h.

Clearly, the path through C is quicker, but the path AB is shorter. What is going to be the Geodesic ? The path through $C$ or the path $AB$ ?

答:
(Disclaimer: It should be a comment, but I don't have enough rep yet)

Mathematically the other answers are correct, that the "shortest path" is always a geodesic.
But you also asked about "quickest" and if you look at it physically, in GR the "time" of a path as measured by the proper time $d \tau^2 = -ds^2$ (which is the time experienced by an observer traveling in the spacetime), is maximized along a geodesic ($ds^2$ is minimized thus $-ds^2$ is maximized). So a geodesic is not the quickest but the slowest path when it comes to experienced time while traveling along the geodesic.

An easy way to see this is actually in flat Minkowski spacetime (remember that we live on a 4D manifold with a Lorentzian metric.. at least according to GR). In the twin paradox the "stationary" twin (A) is moving along a geodesic (stationary in space, but moving in time) while the twin (B) flying in a spaceship is accelerating and thus not on a geodesic. When both meet, A is older than B. A's proper time was maximized since his path was minimized.

Needless to say, this does not make sense if you consider euklidian space without time, but then again, without time you can not talk about "fastest" or "quickest".


问:
Proving that primes can't have two different decompositions
My previous questions have been on a very similar topics but I am having trouble with understanding this:

I understand that $p_1$ or $q_1$ must be greater than the other but I have no idea what is happening when he goes into m' and what that is. Is that "m prime"? I would appreciate some advice on what he is doing here?

Thank you

答:
No, $m'$ is not intended to be prime.

This is a proof by contradiction; we temporarily assume that our goal (every positive integer has a unique prime factorization) is false, and then show that it being false leads to a contradiction, therefore implying that it must have been true.

So, we assume that there exists some positive integer that has two different prime factorizations. The collection of all positive integers with two different factorizations, being non-empty, has a smallest element; we let $m$ be this smallest element. Thus, any positive integer smaller than $m$ will have a unique factorization.

We are constructing an number $m'$ that will be be a positive integer and smaller than $m$, so that it has only one prime factorization; by studying the prime factorization of $m'$, we can reach a contradiction (like we wanted to).

ContentLicense: CC BY-SA 3.0


问:
Solving a non-linear equation in terms of known functions
I have the following equation:

$r_{*}=r+\frac{r_{+}^{2}}{\left(r_{+}-r_{-}\right)} \ln \left|r-r_{+}\right|-\frac{r_{-}^{2}}{\left(r_{+}-r_{-}\right)} \ln \left|r-r_{-}\right|$

and I need to solve for $r$, I am trying to do so in terms of known functions in hope to make the solution less numerically expensive.

For example, if $r_{-}$ is zero, I can solve for $r$ as:

$r = r_{+}\left (W\left(\frac{e^{r_{*}/r_{+} - 1}}{r_{+}}\right) + 1\right) $, where $W$ is the Lambert $W$ function.

In the case where $r_{-}$ is not zero I have not been able to do something similar. Is this even possible? maybe in terms of other functions?

I would very much appreciate any input, or ideas. Thank you for your time

<img src="https://i.stack.imgur.com/yyYKA.png" alt="Plot of the function in the case of $r_{-}=0$, $r_{-}=e$, $r_{+} = 2\pi$, $2\pi<r<10\pi$" />

答:
For personal conveniency, I shall change notations : $a=r_+$, $b=r_-$ and $y=r_*$. To get rid of the absolute values, I shall use $\log(|t|)=\frac 12 \log(t^2)$ and do not make any hypothesis about the signs of $a$ and $b$ but assume $a>b$.

So, the equation write
$$y=r+\frac{a^2 }{2 (a-b)}\log \left((r-a)^2\right)-\frac{b^2 }{2(a-b)}\log \left((r-b)^2\right)$$ Since $a>b$, I would generate an estimate ignoring the last term. As you observed, if $b=0$, we have a solution which is now
$$r_0=a+\frac {a^2}{a-b}\,W(\pm \sqrt t)\quad \text{with} \quad t=\frac{(a-b)^2}{a^4}\exp\Bigg[ \frac{2 (a-b) (y-a)}{a^2}\Bigg]$$

For a simple test, I used $a=2\pi$, $b=e$ and $y=50$. This gives $r_0 \sim 20.558733$.

Now, Newton iterates
$$\left(
\begin{array}{cc}
n & r_n \\
 0 & 20.558733 \\
 1 & 24.157560 \\
 2 & 24.331464 \\
 3 & 24.331763
\end{array}
\right)$$


问:
The distribution(or just expectation) of $\min\{U_{1}, U_{2}\}\times \min\{U_{1}, U_{3}\}$?
I'm trying to find the expectation of $\min\{U_{1}, U_{2}\}\times \min\{U_{1}, U_{3}\}$, where $U_{1}, U_{2}, U_{3}$ are independent from $\mathrm{Unif}(0, 1)$, is there any idea?

答:
I can start you off, use the Law of Total Expectation and this link : Expectation of Minimum of $n$ i.i.d. uniform random variables. (for the case of two iid uniforms)

\begin{align}
E[\min\{U_1,U_2\}\min\{U_1,U_3\} ]=& \frac{1}{2}\Big(E[\min\{U_1,U_2\}\min\{U_1,U_3\} ~|~ U_1<U_2]
\\
&+E[\min\{U_1,U_2\}\min\{U_1,U_3\} ~|~ U_2<U_1]\Big)
\\
=& \frac{1}{2}\Big(E[U_1\min\{U_1,U_3\}]
\\
&+E[U_2\min\{U_1,U_3\}~|~U_2<U_1]\Big).
\end{align}

Now lets deal with $E[U_2\min\{U_1,U_3\}~|~U_2<U_1]$.

\begin{align}
E[U_2\min\{U_1,U_3\}~|~U_2<U_1]=& E[U_2\min\{U_1,U_3\}~|~U_2<U_1,U_3<U_1]
\\
&+E[U_2\min\{U_1,U_3\}~|~U_2<U_1,U_1<U_3]
\\
=& E[U_2U_3]+E[U_2U_1~|~U_2<U_1]
\end{align}

Now solve $E[U_2U_1~|~U_2<U_1]$ using a double integral.


问:
Successor of a countable ordinal is also countable.
I had this question when I was proving this theorem:

Suppose $\omega_{1} = \left\{\alpha: \alpha \textrm{ is a countable ordinal}\right\}$. Then $\bigcup\omega_{1} = \omega_{1}$.

We may assume to know that $\omega_{1}$ is an uncountable ordinal (already proved). The following formula is easy to prove using transistivity of $\in$ for ordinals:

$\bigcup\omega_{1} \subseteq \omega_{1}$,

and what is left is the inverse:

$\omega_{1} \subseteq \bigcup\omega_{1}$.

That is, suppose $\beta \in \omega_{1}$, we need to show that $\beta \in \bigcup\omega_{1}$, which is equivalent to showing that $\omega_{1}$ is a limit ordinal. Thus, we need to show that $\beta^{+} \in \omega_{1}$. If $\beta$ is finite, the proof is very obvious. If $\beta$ is countably infinite, that is $\beta \approx \mathbb{N}$, we need to prove that $\beta \cup \left\{\beta\right\}$ is also countable, or $\beta \cup \left\{\beta\right\} \approx \mathbb{N}$. I guess it is obvious that $\beta \cup \left\{\beta\right\} \approx \mathbb{N} \cup \left\{\mathbb{N}\right\}$. Then what is left is to show $\mathbb{N} \approx \mathbb{N} \cup \left\{\mathbb{N}\right\}$. I have come up with a function which may be able to establish this relation:
\begin{equation}
\begin{aligned}
f:\ &\mathbb{N} \to \mathbb{N} \cup \left\{\mathbb{N}\right\}\\
&\boldsymbol{0} \mapsto \mathbb{N}\\
&\boldsymbol{n}^{+} \mapsto \boldsymbol{n},\ \boldsymbol{0} < \boldsymbol{n}.
\end{aligned}
\end{equation}
I guess there is no problem arguing that this $f$ is a bijection. So $\beta^{+}$ is indeed countable. Thus $\omega_{1}$ is indeed a limit point, and $\omega_{1} \subseteq \bigcup \omega_{1}$. Is this process valid?

答:
Yes the procedure is fine. Just generalise it: if $\alpha$ is a countable ordinal we have a bijection $F: \Bbb N \to \alpha$.

Then define a bijection $F^+: \Bbb N \to \alpha^+= \alpha \cup \{\alpha\}$ by

$$F^+(n) = \begin{cases} \alpha & n=0\\
 F(n-1) & n \ge 1 \end{cases}$$

This takes of the successor step. You also need to note that at limit stages, we take countable unions of countable ordinals and so these limits of ordinals below $\omega_1$ are still counatble ordinals too.


问:
Modules which are both left and right modules over a noncommutative ring
Given a noncommutative ring $R$, is there a name for those (left) $R$-modules $M$ for which
$$
m.(rs) = m.(sr), ~~~ \textrm{ for all } r,s \in R, \textrm{ and } m \in M?
$$
(Note that we have denoted the action of $R$ on $M$ by a dot.)
In other words, what does one call modules which are both left and right modules.

答:
If $k \rightarrow A$ is a commutative $k$-algebra with $k$ a commutative unital ring,
you may consider the tensor product $A(d):=A^{\otimes_k d}$ of $A$ with itself over $k$ d times. There are canonical maps

$$p_i: A \rightarrow A(d)$$

defined by

$$p_i(a):=1\otimes \cdots 1 \otimes a \otimes 1 \otimes \cdots \otimes 1$$

where the $a$ is "placed in position $i$". Here $i=1,..,d$. With this definition you get $d$ commuting $A$-module structures on $A(d)$: You define for any $a\in A$ the $i$'th left multiplication on $A(d)$ as follows. Let $b:=b_1\otimes \cdots \otimes b_d \in A(d)$. define

$$  a^ib:=b_1 \otimes \cdots \otimes ab_i \otimes \cdots \otimes b_d:=p_i(a)b.$$

You get for any elements $u,v\in A$

$$ u^i(v^jb):=p_i(u)(p_j(v)b)=p_j(v)(p_i(u)b)=v^j(u^ib)$$

hence these $d$ module structures commute. Hence you do not need $A$ to be non-commutative. I do not know if there is a word reserved for this structure. Maybe Bourbaki's "Commutative algebra" book can give some answers. They study bimodules and more generally "multi modules" and I believe the above construction is referred to as a multi module.


问:
Given only a system transfer function, after changing it using state feedback, can we know if the resulting system is controllable? Observable?

Practice test question

Given

$$ G(s) = \frac{(s-2)(s-5)}{(s+1)(s-3)(s+4)} $$

Use state feedback to change it to

$$ \frac{(s-5)}{(s+1)(s+4)}$$

Is the resulting system controllable? Observable? If no, explain why not

My answer

a.) Using state feedback, change the transfer function

By putting the function into controllable canonical realization, we get

$$ \dot{x}_c = \begin{bmatrix}0&1&0\\0&0&1\\12&11&-1\end{bmatrix}x_c + \begin{bmatrix}0\\0\\1\end{bmatrix}u, \quad y_c = \begin{bmatrix}10&-7&1\end{bmatrix}x_c $$

Using state feedback (omitting the steps here) we change the transfer function to

$$ G(s) = \frac{(s-2)(s-5)}{s^3+(2+k_{c3})s^2+(-11+k_{c2})s + (-12+k_{c1})} $$

Setting the denominator equal to $(s-2)(s+1)(s+4)$ by letting $k_c = \begin{bmatrix}4&5&1\end{bmatrix}$ changes the transfer function to the desired value.

b.) Is the resulting system controllable? Observable?

It can't be both controllable and observable, as we have canceled out a term in the transfer function. However, without knowledge of the original $\mathbf{A}$ and $\mathbf{b}$ matrices, we cannot derive the transform matrix $\mathbf{T}$ to find $k = k_c\mathbf{T}$

I'm pretty confident about the first part of the question, but I'm not sure if there's more I could be doing on the second question. Is this really the best answer?

答:
The controllability of a LTI state space model does not change under state feedback. This can be shown by using the Hautus test to check for controllability

$$
\begin{bmatrix}
\lambda\,I - A + B\,K & B
\end{bmatrix} =
\begin{bmatrix}
\lambda\,I - A & B
\end{bmatrix}
\begin{bmatrix}
I & 0 \\ K & I
\end{bmatrix},
$$

since the right most matrix is square and full rank and thus the rank of the remaining matrices have to be equal to each other. So just by looking at the transfer function and noticing that there is a pole-zero cancelation one could already conclude that it would not be observable (using the Hautus test for observability one can conclude it would lose one rank).

It can also be noted that all state space models (of the same size), that have the same corresponding transfer function, are equivalent to each other. So the results regarding controllability and observability of your model should also hold for other equivalent state space realizations. So your choice for $A$ and $B$ would allow you to say something about it (assuming one has to use a minimal realization).

I do have a remark regarding the last part of the question, regarding controllability and observability. Namely, one can always start with a non-minimal state space realization, which is not controllable and not observable. Such system will always remain that way.


问:
Show that algebraic direct sum is $\sigma$-weakly dense.
Consider the abstract von Neumann algebra
$$M:= \ell^\infty-\bigoplus_{i \in I} B(H_i)$$
which consists of elements $(x_i)_i$ with $\sup_i \|x_i\| < \infty$ and $x_i \in B(H_i)$.

Let $N$ be the algebraic direct sum $\bigoplus_i B(H_i)$, thus it consists of elements $(x_i)_i$ such that only finitely many $x_i$ are non-zero. I want to show that $N$ is $\sigma$-weakly dense in $M$.

I guess my main problem is that I don't understand the $\sigma$-weak topology on $M$. By a result of Sakai, it is the unique topology on $M$ coming from a weak$^*$-topology when we realise $M$ as the dual of some other Banach space.

Maybe, we can do the following
$$M \cong \ell^\infty-\bigoplus_{i \in I} T(H_i)^* \cong \left(\ell^1-\bigoplus_{i \in I} T(H_i)\right)^*$$
But this does not seem very practical! Any insight in the matter will be appreciated!

答:
Lemma: Let $(X_i)_{i\in I}$ be a collection of Banach spaces indexed by a set. Then there exists an isometric isomorphism
$$\bigoplus_{i\in I}^{\ell^\infty}(X_i)^*\cong\bigg(\bigoplus_{i\in I}^{\ell^1}X_i\bigg)^* $$

I leave the proof to you, but we can discuss the details if there is any problem.

By Sakai's theorem the predual is unique and thus the $\sigma$-weak topology is canonical, in the sense that it is an intrinsic object that is independent of the concrete representation of the von Neumann algebra. More specifically, if $M_i=\mathcal{B(H}_i)$, then
$$\bigoplus_{i\in I}^{\ell^\infty}M_i=\bigoplus_{i\in I}^{\ell^\infty}(M_{i*})^*\cong(\bigoplus_{i\in I}^{\ell^1}M_{i*})^*$$
where $A_*$ denotes the predual of the von Neumann algebra $A$. In other words, the predual of $M$ is the same as the $\ell^1$ direct sum of the preduals of the $M_i$, as you correctly guessed. You believe that this is not practical and you are wrong about that!

Fix $x=(x_i)_{i\in I}\in M$. We wish to find a net of elements such that only finitely many coordinates are non-zero that converges ultraweakly to our point $x$. Let $\Lambda$ be the set of finite subsets of $I$ and we direct $\Lambda$ by the usual inclusion, so $\Lambda$ becomes a directed set. We denote the elements of $\Lambda$ by $\lambda$, so $\lambda$ is simply a finite subset of $I$. For $\lambda\in\Lambda$ we define $x_\lambda=(x_\lambda^{(i)})_{i\in I}$ where $x_\lambda^{(i)}=0$ if $i\in I\setminus\lambda$ and $x_\lambda^{(i)}=x_i$ if $i\in\lambda$. So the net $(x_\lambda)_{\lambda\in\Lambda}$ lies in the subspace $N$ you have specified. We will show that $x_\lambda\to x$ ultraweakly.

We fix $\varepsilon>0$ and a functional $\Phi$ in the predual of $M$ and we want to find $\lambda_0\in\Lambda$ so that for any $\lambda\geq\lambda_0$ (here $\geq$ refers to the direction we gave to $\Lambda$ of course) we have that $|\Phi(x)-\Phi(x_\lambda)|<\varepsilon$.

Edit: Wait a minute; how am I fixing a functional in the predual, how does that make any sense? Well, recall that the predual of a von Neumann algebra $A$ is identified with the set of normal linear functionals on $A$ and therefore a net $(a_\mu)\subset A$ converges ultraweakly to $a\in A$ if and only if $\phi(a_\mu)\to\phi(a)$ for all normal linear functionals $\phi$. Now that we have cleared this out, let's go on with our proof:

By the identification of the preduals that we explained in the first paragraph, $\Phi$ is identified with a sequence $(\phi_i)_{i\in I}$ where $\phi_i\in M_{i*}$ such that $\sum_{i\in I}\|\phi_i\|<\infty$. Since the generalized series converges, we can find a finite subset $F\subset I$ so that $\sum_{i\in I\setminus F}\|\varphi_i\|<\frac{\varepsilon}{\|x\|}$. Set $\lambda_0=F$ and suppose that $\lambda\geq\lambda_0$, i.e. $\lambda\supset\lambda_0$. Then

$$|\Phi(x)-\Phi(x_\lambda)|=|\sum_{i\in I}\phi_i(x_i)-\sum_{i\in I}\phi_i(x_\lambda^{(i)})|=|\sum_{i\in I\setminus\lambda}\phi_i(x_i)|\leq\sum_{i\in I\setminus\lambda}|\phi_i(x_i)|\leq\sum_{i\in I\setminus F}|\phi_i(x_i)|\leq$$ $$\leq\|x\|\cdot\sum_{i\in I\setminus F}\|\phi_i\|<\varepsilon $$

and we are done.


问:
How to fully factorise $2x^4+7x^3+4x^2-4x$?

How to fully factorise $2x^4+7x^3+4x^2-4x$?

I'm struggling to factorise polynomials like this one. I'm not sure how to best approach this problem. I've tried using the remainder and factor theorems but they seem to work only when at least one factor is known. Since I only know the polynomial itself, I'm struggling to efficiently solve these types of questions. Any help would be greatly appreciated

答:
As $x$ is the common factor:

$$2x^4+7x^3+4x^2-4x$$
$$x\left(2x^3+7x^2+4x-4\right)$$
$$x\left(2x-1\right)\left(x+2\right)\left(x+2\right)$$
$$\boxed{x\left(2x-1\right)\left(x+2\right)^2}$$


问:
Show that $[U]_\beta={[T]_\beta}^*$ for any orthonormal basis $\beta$
Let $V=F^n$ and let $A \in M_{n*n}(F)$ and $A^*$ mean conjugate transpose.

Define linear operators $T$ and $U$ on $V$ by $T(x)=Ax$ and $U(x)=A^*x$.

Show that $[U]_\beta={[T]_\beta}^*$ for any orthonormal basis $\beta$ for $V$.

Too tricky to solve!

答:
In short, the question really is: prove that $[T^*]_\beta=[T]_\beta^\ast$ for every orthonormal basis $\beta$.

Let $\beta$ be an orthonormal basis of $V$. In general, for any linear operator $S:V\longrightarrow V$, the $(i,j)$ coefficient $a_{ij}$ of $[S]_{\beta}$ is $(Se_j,e_i)$. Indeed, $Se_j=\sum_{i=1}^na_{ij}e_i$ by definition of $[S]_\beta$. So it suffices to take the inner product of the latter equation with $e_i$ to obtain the claim.

Now
$$(T^*e_j,e_i)=(e_j,Te_i)=\overline{(Te_i,e_j)}\qquad \forall 1\leq i,j\leq n$$
says precisely that the coefficients of $[T^\ast]_\beta$ and $[T]_\beta^\ast$ are the same.

Note: with the canonical basis $\beta_0$, you get that the matrix of $T$ is $A$, the matrix of $U$ is $A^*$. So $A^*=[U]_{\beta_0}=[T]_{\beta_0}^\ast=[T^\ast]_{\beta_0}$, whence $U=T^*$. So the question does amount to what I said in my first sentence.


问:
What does capital A mean in vector field notation?
This might be a dumb question but I am stuck halfway through an assignment where I am supposed to draw a bunch of vector fields because I don't understand an element of the notation.

The equation that is causing me trouble is the one below, describing a vector field in the plane, not including (0,0).

$\textbf{v}= \frac{-Ay}{r^2}\textbf{i}+\frac{Ax}{r^2}\textbf{j}; r^2=x^2+y^2$

The assignment does not include any addition on the capital A, so I assume that it has some conventional use that I am not familiar with?

答: If nothing else is said about it, I would assume $A$ is meant to be an arbitrary real number. For the drawing, I would assume $A > 0,$ since for $A=0$ you get a zero vector field and for $A < 0$ all the vectors will reverse direction.


问:
Given the equations $y+z=a$ and $y \cdot z = b$, how do I solve for $y$ and $z$?

Background

Recently, I have been factoring quadratic equation of the form
$$ x^2 + ax + b $$
into
$$ (x+y) \cdot (x+z) $$

e.g. From $x^2 + 3528x + 63180$ to $(x+3510) \cdot (x+18)$

Normally, it is possible to obtain $y$ and $z$ by looking into every factor of $b$, and find 2 of them, where there product is $b$ and their sum is $a$.

However, I find this method too tedious, especially when dealing with numbers that are hard or impossible to get its factors. So, I decided to use equations to directly obtain $y$ and $z$ given $a$ and $b$.

Question

According to the scenario,

$$ y+z=a $$
$$ y \cdot z = b $$

and so, how do I solve this system of equations, so that I get

$$ y = \frac{-\sqrt{a^2 - 4 \cdot b} + a}{2} \, \text{and} \, z = \frac{\sqrt{a^2 - 4 \cdot b} + a}{2}$$
$$ \text{or} \, y = \frac{\sqrt{a^2 - 4 \cdot b} + a}{2} \, \text{and} \, z = \frac{-\sqrt{a^2 - 4 \cdot b} + a}{2}$$
?

答:
If I understood the question, you want to know how to get the formulae for the roots of the second-degree polynomial equation. That isn't that hard. First, consider the equation written in the following way (this is the standard way):$$ax^2+bx+c=0$$ and let's assume $a\not=0$, $b\not=0$ and $c\not=0$ (otherwise, the roots are trivial!). Since $a\not=0$, we can divide by $a$ the whole equation, leading to the following expression: $$x^2+mx+n=0;\quad m=\frac{b}{a}, n=\frac{c}{n}.$$ Now add the term $\frac{m^2}{4}$ in both sides (complete the square). By doing this we get $$x^2+mx+n+\dfrac{m^2}{4}=\dfrac{m^2}{4}\Longleftrightarrow \left(x+\dfrac{m}{2}\right)^2=\dfrac{m^2}{4}-n.$$ Now, just solve for $x$ and replace $m$ and $n$ by the original values (In terms of $a$, $b$, and $c$):$$\left(x+\dfrac{m}{2}\right)^2=\dfrac{m^2}{4}-n\Longrightarrow x=-\dfrac{m}{2}\pm \sqrt{\dfrac{m^2}{4}-n}\Longrightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$Now, if we call $r_0$ and $r_1$ to each of the roots we get that $$r_0+r_1=-\dfrac{b}{a};\quad r_0+r_1=\dfrac{c}{a},$$ which is the answer to your question.

Note: One can write the expressions for $r_0$ and $r_1$ in other ways, like:$$r_0=\dfrac{2c}{-b+\sqrt{b^2-4ac}};\quad r_0=\dfrac{2c}{-b-\sqrt{b^2-4ac}}$$ or $$r_0=\dfrac{q}{a};r_1=\dfrac{c}{q};\quad q=-\dfrac{1}{2}\left(b+sign(b\sqrt{b^2-4ac})\right).$$All these three ways of writting the roots are equivalent, but from the point of view of numerical computation the last one is the best amonst the three (ie, if one of the roots is very large while the other is very small, computationally speaking is more accurate and stable).


问:
Question about norm convergence in quotient spaces
Let $X$ be a normed spaces, $U\subseteq X$ a closed subspace.

Is it true that, if $\bar{x}_n \rightarrow \bar{x}$ in $X/U$, then I can find $u_n \in U$ with $x_n+u_n\rightarrow x$?

For some context, I'm trying to show that the index is a continuous map from the multiplicative group of the Calkin algebra of $X$ (plus some other hypothesis).

My try: $\overline{x_n-x}\rightarrow 0$, so that $\inf_{u \in U}||x_n-x+u||\rightarrow _n 0$. By picking $u_n$ at each step so that $||x_n-x+u_n||$ is small enough (thanks to the infimum properties) one should be able to get the claim. Is this correct?

答:
Let's denote the quotient map by $\pi:X\to X/U$. So we have a sequence $(x_n)\subset X$ and a point $x\in X$ so that $\pi(x_n)\to\pi(x)$. Equivalently, $\pi(x_n-x)\to0$, so without loss of generality we can assume that $x=0$ and thus $\pi(x_n)\to0$. Now for any point $z\in X$ we have that
$$\|\pi(z)\|=\inf_{u\in U}\|z+u\|$$
by definition. So for each $n\geq1$, by the elementary properties of the infimum, you can choose $u_n\in U$ so that $\|x_n+u_n\|\leq\|\pi(x_n)\|+\frac{1}{n}$. But now $\|\pi(x_n)\|\to0$ since $\pi(x_n)\to0$, so the right hand side of the inequality converges to $0$, so $\|x_n+u_n\|\to0$, i.e. $x_n+u_n\to0$.


问:
Methods for Finding Raw Moments of the Normal Distribution
I'm having some trouble with finding raw moments for the normal distribution. Right now I am trying to find the 4th raw moment on my own. So far, I know of two methods:

I can take the 4th derivative of the moment generating function for the normal distribution and evaluate it at 0.

I can use the fact that $E(x^4)$ is an expectation of a function of x to write  $$E({X}^{4})=\int_{Sx}^{} {x}^{4} f(x) dx=\int_{-\infty}^{\infty} {x}^{4}\frac{{e}^{\frac{{(x-\mu )}^{2}}{2{\sigma }^{2}}}}{\sqrt{2\pi }\sigma } dx$$

I'm wondering if there's a 3rd method. We haven't covered integrating the normal pdf in class, and taking the 4th derivative of ${e}^{\frac{{t}^{2}{\sigma }^{2}}{2}+t\mu }$ seems really messy/inelegant, so I'm wondering if there is some conceptual piece about moment generating functions I am missing. Thanks in advance!

答:
There is a nice recurrence for the raw moments of a normal distribution with mean $\mu$ and variance $\sigma^2$:

$$\mathbb E\left[X^{n+1}\right] = \mu \mathbb E\left[X^{n}\right] + n \sigma^2 \mathbb E\left[X^{n-1}\right] $$ starting at $\mathbb E\left[X^{0}\right]=1$ and $\mathbb E\left[X^{1}\right]=\mu$. So you get:

$$\begin{matrix}
\mathbb E\left[X^{0}\right]=&1 \\
\mathbb E\left[X^{1}\right]=&\mu \\
\mathbb E\left[X^{2}\right]=&\mu^2 +\sigma^2 \\
\mathbb E\left[X^{3}\right]=&\mu^3 +3\mu\sigma^2 \\
\mathbb E\left[X^{4}\right]=&\mu^4 +6\mu^2\sigma^2 +3\sigma^4\\
\mathbb E\left[X^{5}\right]=&\mu^5 +10\mu^3\sigma^2 +15\mu\sigma^4 \\
\mathbb E\left[X^{6}\right]=&\mu^6 +15\mu^4\sigma^2 +45\mu^2\sigma^4 +15\sigma^6
\end{matrix}$$

and so on


问:
Time derivative of polar coordinates
For the equation:
$r = (x^2 + y^2)^{\frac{1}{2}}$, $r \cos{\theta} = x$ and $r \sin{\theta} = y$

My hand written notes indicate:

$$\begin{align}
\ddot{x} &= (\ddot{r} - r \dot{\theta}^2) \cos{\theta} - (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \sin{\theta} \tag{A}\\
\ddot{y} &= (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \cos{\theta} + (\ddot{r} - r \dot{\theta}^2) \sin{\theta} \tag{B}
\end{align}$$

I cannot remember where i got it form and am not comfortable in fully deriving it.  Can someone confrim if the above is correct or link somewhere i can find it.

答:
from $x(t)=r(t)\cos\theta(t)$

you derive once wrt $t$
$$x'(t)=r'(t) \cos (\theta (t))-r(t) \theta '(t) \sin (\theta (t))$$
then derive again
$$x''(t)=\cos (\theta (t)) \left(r''(t)-r(t) \theta '(t)^2\right)-\sin (\theta (t)) \left(2 r'(t) \theta '(t)+r(t) \theta ''(t)\right)$$

same for $y(t)=r(t)\sin\theta(t)$
$$y'(t)=r'(t) \sin (\theta (t))+r(t) \theta '(t) \cos (\theta (t))$$
and
$$y''(t)=(2 r'(t) \theta '(t) +r(t) \theta ''(t)) \cos (\theta (t))+\sin (\theta (t)) \left(r''(t)-r(t) \theta '(t)^2\right)$$

Title: Time derivative of polar coordinates


问:
Substitution in ordinary differential equation gone wrong
Given

$$
y(t) = \left( \int u(t) \; \text{dt} \right)^2,
$$

the derivative of $y$ over $t$ is

$$
\dot{y} = 2 u(t) \int u(t) \; \text{dt}.
$$

Substituting $\int u$ in the above equation yields

$$
\dot{y} = 2u(t) \sqrt{y}.
$$

But integrating this differential equation with $y(0) = 0$ and $u(t) \equiv 1$ yields $y \equiv 0$ for all $t$, whereas $y(t) = t^2$ in the original equation.

Why is that?

答:
Your approach to the question is wrong, from the beginning itself. After integration, $y$ is no longer a function of $t$. Assuming that you mean $\int_0^x$ by the indefinite integral. We write
\begin{equation}
y(x) = \left(\int\limits_0^x u(t)dt\right)^2
\end{equation}
Differentiating using the Newton-Leibniz rule, we get,
\begin{equation}
y'(x) = 2u(x)\int\limits_0^x u(t)dt
\end{equation}
Assuming that $\int\limits_0^x u(t)dt>0$ and using the first equation in the above equation, we get,
\begin{equation}
\dfrac{dy}{dx} = 2u(x)\sqrt y
\end{equation}
Putting $u(x)\equiv1$, we get,
\begin{align}
&\dfrac{dy}{dx}=2\sqrt y\\
\implies &\dfrac{dy}{2\sqrt y} = dx\\
\implies &\sqrt y = x+C&&(\text{By integrating both sides})\\
\implies &y = x^2&&(\text{Using the initial condition})
\end{align}
This is what you desired. Also note that we can divide both sides by $\sqrt y$ only by assuming $\sqrt y\ne0$. $y\equiv0$ is a trivial solution to the differential equation.


答:
Note that by Vieta's formulas, $y$ and $z$ satisfy $x^2-ax+b$, where you can use then the quadratic formula. However, many quadratics can be factored efficiently using a substitution.

You can use the AC method in reverse. In your example, divide by the largest perfect square divisor of $63180$, which is $324 = 18^2$ to get:

$$\frac{x^2}{18^2} + \frac{98}{9}x + 195=0$$
$$\Rightarrow (\frac{x}{18})^2 + 98 \cdot 2 \cdot\frac{x}{18} + 195 = 0$$
$$\Rightarrow X = \frac{x}{18}: X^2+196X+195 = 0$$

which factorises as $(X+195)(X+1)$. In general, $x^2 + (k+1)x + k$ also factorises as $(x+k)(x+1)$. To find the roots of the original quadratic, find $x$ by using $X = x/18$ again.

Title: Given the equations $y+z=a$ and $y \cdot z = b$, how do I solve for $y$ and $z$?

Body:
Background

Recently, I have been factoring quadratic equation of the form
$$ x^2 + ax + b $$
into
$$ (x+y) \cdot (x+z) $$

e.g. From $x^2 + 3528x + 63180$ to $(x+3510) \cdot (x+18)$

Normally, it is possible to obtain $y$ and $z$ by looking into every factor of $b$, and find 2 of them, where there product is $b$ and their sum is $a$.

However, I find this method too tedious, especially when dealing with numbers that are hard or impossible to get its factors. So, I decided to use equations to directly obtain $y$ and $z$ given $a$ and $b$.

Question

According to the scenario,

$$ y+z=a $$
$$ y \cdot z = b $$

and so, how do I solve this system of equations, so that I get

$$ y = \frac{-\sqrt{a^2 - 4 \cdot b} + a}{2} \, \text{and} \, z = \frac{\sqrt{a^2 - 4 \cdot b} + a}{2}$$
$$ \text{or} \, y = \frac{\sqrt{a^2 - 4 \cdot b} + a}{2} \, \text{and} \, z = \frac{-\sqrt{a^2 - 4 \cdot b} + a}{2}$$
?


问:
Natural number solutions to $\frac{xy}{x+y}=n$ (equivalent to $\frac 1x+\frac 1y=\frac 1n$)
I have a question about the following problem from a Putnam review:

Let $n\in \mathbb{N}$. Find how many pairs of natural numbers $(x, y)\in \mathbb{N}\times \mathbb{N}$ solve
$$
\frac{xy}{x+y}=n.
$$

I have found some solutions for all $n$, such as $(2n, 2n)$ and $(n+1, n(n+1))$, but I feel as though this is the wrong approach, as the question is only asking for the number of solutions to the equation.

I don't really want a complete solution, but any hints would be greatly appreciated.

答:
Let $d=\gcd (x,y)$ and $x=dp, y=dq$ and $\gcd (p,q)=1$. Then we have $$\frac{dpq}{p+q}=n$$ Now $\gcd(pq,p+q)=1$ implies $(p+q)|d$. So $d=k(p+q)$ where $k\in\mathbb{N}$. Hence all the solutions are given by, $$x=kp(p+q), y=kq(p+q)$$
 where $k,p,q\in\mathbb{N}$, $\gcd(p,q)=1$ and $n=kpq$. </p>


问:
Do the slopes of supporting lines tend to zero at a minimum point?
Let  $F:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$. Suppose also that $F$ is convex on $[1-\epsilon,1]$, for some $\epsilon>0$.

Let $y_n \in [1-\epsilon,1)$, $y_n \to 1$, and suppose $m_n$ is a slope of a supporting line to $F|_{[1-\epsilon,1]}$ at $y_n$, i.e.
$$
F(x) \ge F(y_n)+m_n(x-y_n) \, \, \, \text{ for every } \, \, x \in [1-\epsilon,1].
$$
Is it true that $\lim_{n \to \infty}m_n=0$?

By plugging $x=1$, we get $0=F(1) \ge F(y_n)+m_n(1-y_n)$ so $m_n<0$.

If $F$ is differentiable at $y_n$, then this forces $m_n=F'(y_n)$, so if $F$ is $C^1$ near $x=1$, we have $m_n=F'(y_n) \to F'(1)=0$. However, I don't want to assume that $F$ is differentiable.

答: No.  Consider $F(x)=\lvert x-1\rvert+(x-1)^2$, which is strictly convex on $\mathbb{R}$ and differentiable except at $x=1$.  We have obviously $\lim_{n\to\infty} m_n=-1$ coming from the $\lvert x-1\rvert$ for $x<1$.


答:
Not under the hypotheses given.  Consider $F(x) = |x-1|$.  In this case, the unique supporting line for $y_n \in [1-\varepsilon, 1]$ is 1, so $m_n$ is a constant sequence.

More generally, I think you're after a sequential lower hemicontinuity condition for the subdifferential of $F\vert_{[1-\varepsilon, 1]}$.  As a correspondence the subdifferential will be (cyclically) monotone, hence every selection from it will be nondecreasing, but absent stronger conditions not every value of $\partial F\vert_{[1-\varepsilon, 1]}(0)$ will be obtainable as a limit (from the left) of a selection.

Title: Do the slopes of supporting lines tend to zero at a minimum point?

Body:
Let  $F:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$. Suppose also that $F$ is convex on $[1-\epsilon,1]$, for some $\epsilon>0$.

Let $y_n \in [1-\epsilon,1)$, $y_n \to 1$, and suppose $m_n$ is a slope of a supporting line to $F|_{[1-\epsilon,1]}$ at $y_n$, i.e.
$$
F(x) \ge F(y_n)+m_n(x-y_n) \, \, \, \text{ for every } \, \, x \in [1-\epsilon,1].
$$
Is it true that $\lim_{n \to \infty}m_n=0$?

By plugging $x=1$, we get $0=F(1) \ge F(y_n)+m_n(1-y_n)$ so $m_n<0$.

If $F$ is differentiable at $y_n$, then this forces $m_n=F'(y_n)$, so if $F$ is $C^1$ near $x=1$, we have $m_n=F'(y_n) \to F'(1)=0$. However, I don't want to assume that $F$ is differentiable.


问:
Has someone ever tried to extend the complex number to $i^x=-1$?
I recently watched a series of videos about the history of complex numbers, and as I got to the part about their geometrical representation, I started to get more curious about the square in $i^2=-1$.

Indeed, we all know that the complex plane is made of two orthogonal spaces (Real and imaginary numbers), and I find it almost too neat that multiplying by i is essentially applying a rotation by $90°$ (this gives that the "angle" between the two spaces has to be $90°$)

My question is, has someone ever tried to see what would happen, and what kind of mathematical theories we could end up with, by analysing another set of numbers defined by $i^x=-1$? (x would be a real number, positive or negative). I'm guessing then the two spaces would not be orthogonal anymore, but there could also be a lot of other unforeseen consequences.

答:
Yes. But not in this sense. Quaternions are usually termed as the extension of the complex numbers which was first discovered by William Hamilton. You can have a look here:
https://en.wikipedia.org/wiki/Quaternion

Also you can see some videos on YouTube for the same. Usually like complex numbers are intuitively used to solve signal or circuit problems where current and voltage are some phase apart, quaternions are used to solve many problems in the Robotics domain specially the POSE problem (position + orientation). Roll Pitch and Yaw are the $3$ rotations about X,Y and Z axes which can be combinedly representated as a single quaternion.
Hope this helps...


问:
logic formalizing: The multiplication of every two positive numbers is also positive
EDITED THE QUESTION WITH THE RIGHT SIGNS

I have to formalize the expression: "The multiplication of every two positive numbers is also positive" which I think is:

$$\forall a \forall b(ab>0 \to \left((a>0) \land (b>0)\right))$$

but I have to choose between two options:

$\forall a \forall b(ab \le 0  \to \left((a>0) \to (b
\le 0))\right)$

$\forall a \forall b (\left((a>0) \land (b<0)\right) \land ab<0))$

which one is more correct?

答:
This “∀∀(>0→((>0)∨(>0)))” formalizes that if the product of two numbers a, b is positive then one of them is positive. This is simply false (as long as you include negative numbers in your universe)

I don’t see how 2. or 3. formalizes that the product of two positive numbers is positive. The easiest way to do it is $\forall a \forall b (a > 0 \land b > 0 \implies ab > 0)$


答:
The slopes $m_n$ increase with increasing $y_n$ so that $\lim_{n \to \infty}m_n$ exists. But
$$
 F_-'(y_n) \le m_n \le F_+'(y_n) \le F_-'(1)
$$
so that limit is strictly negative if the left derivative $F_-'(1)$ is negative.

In fact one has $\lim_{n \to \infty}m_n = F_-'(1)$. If not, then $m_n \le a < F_-'(1)$ for all $n$. Then
$$
 F(x) \ge F(y_n)+m_n(x-y_n) \ge F(y_n) + a(x-y_n)
$$
for $1-\epsilon < x < y_n$. It follows that $F(x) \ge F(1) + a(x-1)$ for $1-\epsilon < x < 1$ and therefore $F_-'(1) \le a$, contrary to the assumption.


问:
Probability on product spaces
I am having some trouble, more of an argument with someone else, about a simple question regarding product spaces.

Let $X_1,X_2,\dots,X_n$ a set of independent and identically distributed random variables from a population $P\in\mathcal{P}$, where $\mathcal{P}$ is a family of probability measures (non explicitly parameterized). The generic random variable $X$ is a measurable function from a probability space $\left(\Omega,\mathcal{F},P'\right)$ to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

In order to build the random vector containing the elements of the sample, $\tilde{X}=[X_i]_{n\times 1}$, and be able to calculate probability measures of events like $\{X_1<X_2\}$ or $\{X_1=X_2\}$, I decided to build a product space $(\Omega^n=\Omega\times\Omega\dots\times\Omega,\sigma(\mathcal{F}^ n)=\sigma(\mathcal{F}\times\mathcal{F} \dots,\times\mathcal{F}),P=P'\times P'\times\dots\times P')$, and let the vector function $\tilde{X}$ go from this product space to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$.

Therefore, the events

$\{X_1=X_2\}=\{(w_1,w_2,\dots,w_n)\in\Omega^n:X_1(\omega_1)=X_2(\omega_2)\}$
and

$\{X_1<X_2\}=\{(w_1,w_2,\dots,w_n)\in\Omega^n:X_1(\omega_1)<X_2(\omega_2)\}$

have a clear meaning.

However, the person I am having the argument with argues that

$\tilde{X}$ goes from the original probability space $\left(\Omega,\mathcal{F},P'\right)$ to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$, and

If I keep doing it the way of product spaces it is a lot harder to calculate the probabilities I want.

If my friend's argument is true, I am having trouble finding the meaning of, and calculating the probabilities of the events above.

What am I missing?

Best regards,

JM

答:
That's a good question. The point is that if you have a random variable
$$
 X:(\Omega,\mathscr F,P)\to (\Bbb R,\mathscr B(\Bbb R))
$$
and you need to consider, say, two copies of this variable $X_1$ and $X_2$, a direct way would be to define the vector $\tilde X = (X_1,X_2)$ on a product space $(\Omega,\mathscr F,P)\otimes (\Omega,\mathscr F,P)$. This is intuitive, this way always works, and IMHO it's easier to compute the probabilities you've mentioned over a product space - you have clear image of the diagonal in your mind when dealing with $\{X_1 = X_2\}$ and of a subdiagonal triangle when dealing with $\{X_2\leq X_1\}$. The latter make easier computations of the correspondent double integrals.

On the other hand, formally speaking, you don't have to construct a product space in most of the practial cases. That is, most of the probability spaces we're dealing with are standard. For example, since $X$ is a real-valued random variable, you can always take
$$
  (\Omega,\mathscr F,P) = ([0,1],\mathscr B([0,1]),\lambda)
$$
where $\lambda$ is the Lebesgue measure. As a result, the product space is isomorphic to the original space and hence any random vector defined over the product space can be defined over the original space. However, I wouldn't suggest going that way due to the following reasons:

It does not always work: if $\Omega$ has just two elements and $\mathscr F$ is its powerset, then you can't defined $\tilde X$ over the original space.

I disagree with your friend that it is easier to compute probabilities when defining $\tilde X$ over the original state space, rather than over the product space.

It is less intuitive, more technically involved and unnecessary.

Please, tell me whether the answer is clear to you.

Title: Probability on product spaces


问:
Evaluate $\int_0^\infty f(x)dx$ when ,$f(x)=e^{-e^{x}}e^{x}\sin x$ by residues or otherwise
Evaluate improper integral $\int_0^\infty f(x)dx$ .
$$f(x)=e^{-e^{x}}e^{x}\sin x$$

My try-
Can I have a semi circular path from $0$ to $R$ and counterclockwise semi-circle?

答:
Your integral is
\begin{align*}
\int_0^{ + \infty } {e^{ - e^x } e^x \sin xdx} & = \Im \int_0^{ + \infty } {e^{ - e^x } e^x e^{ix} dx}  = \Im \int_0^{ + \infty } {e^{ - e^x } e^{(1 + i)x} dx} \\ & \mathop  = \limits^{t = e^x } \Im \int_1^{ + \infty } {e^{ - t} t^i dt}  = \Im \Gamma (1 + i,1) = 0.1866485915\ldots,
\end{align*}
where $\Gamma(a,z)$ is the (upper) incomplete gamma function.


问:
Trigonometry - worm thread related word problem

The Question

The machine tool diagram shows a symmetric worm
thread, in which a circular roller of diameter 1.5 inches sits.
Find the amount d that the top of the roller rises above the
top of the thread, given the information in the diagram. (Hint:
Extend the slanted sides of the thread until they meet at a point.)

[

My Understanding

I found w and h using basic trig

For w

$$\sin15°= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{0.75}{w}$$   $$w = \frac{0.75}{\sin15°}$$ $$\sin15°= \frac{\sqrt{6} - \sqrt{2}}{4}$$
$$\Longrightarrow w = 2.897$$

For h $$\tan15° =  \frac{0.85}{h}$$ $$h = 3.172$$

$$h-w = 0.2744$$

I've been trying to find angles I could use, but nothing comes to mind, so I'll be grateful if you give me some hints. English is not my first language, so sorry for possible mistakes.


问:
Graded Integral domain
A property of a $\mathbb{Z}$-graded integral domain arose in lecture, and I for the life of me cannot see why this is the case.

Consider $$
R=\bigoplus_{i\in \mathbb{Z}}R_i
$$
Here $R$ is a $\mathbb{Z}$-graded integral domain. We considered the subring $R_0$. (Of course showing that $R_0$ is not trivial, but I was able to manage that). It was stated that $R_0$ is algebraically closed in $R$. After looking through definitions, it is still not clear to me as to why this is the case. Here being algebraically closed means that $R_0$ is equal to its algebraic closure, and the algebraic closure of $R_0$ in $R$ is the set of elements $a \in R$ such that $a$ is algebraic over $R_0$. Lastly the notion of $a \in R$ being algebraic over $R_0$ means that there exists a polynomial, say $g$, with coefficients in $R_0$ such that $g(a) = 0$. Any help would be appreciated.

答:
Basically the idea is that if $a$ has some non-zero part in a degree other than $0$, then taking powers will shift this away from degree $0$ and so a non-zero polynomial cannot have $a$ as a zero. More precisely:

Let $0\ne g\in R_0[X]$ be a polynomial with coefficients in $R_0$ and write $g=b_nX^n+b_{n-1}X^{n-1}+\dots+b_0$ with $b_n\ne0$. Let $a\in R$ and $a_i$ be the $i$-th homogeneous component of $a$ (so that $a=\sum_{i\in \Bbb Z}a_i$). Assume that $a\notin R_0$. Assume that $a$ has some non-zero component in positive degree and let $m$ be the largest positive integers for which $a_m\ne0$ (if $a$ has no non-zero component in positive degree we take $m$ to be the minimal integer with $a_m\ne0$). Now if we plug $a$ into $g$ and write out all the homogeneous terms, we see that in the expression $g(a)$ there is exactly one term of degree $nm$, namely $b_na_m^n$. As the other non-zero $a_i$ have lower degree there is no other term of degree $nm$. By assumption $R$ is an integral domain, so we have $b_na_m^n\ne0$. In particular this means that $g(a)$ cannot be zero as it isn't zero in the $nm$-th component. We showed that no $a\in R\setminus R_0$ can satisfy a non-trivial polynomial equation over $R_0$, i.e. $R_0$ is algebraically closed in $R$.

In fact, we see that we can weaken the integral domain hypothesis: we only need that $R$ is reduced and torsionfree as an $R_0$-module.

Title: Graded Integral domain


问:
find matrix gradient
$f: \Omega \rightarrow \mathbb{R}$
defined by

$\Omega=\left\{X \in \mathbb{R}^{m \times n}: X^{\top} A X+B^{\top} X+X^{\top} B+C \succ 0\right\}$

with
$A \in \mathbb{S}^{m}, B \in \mathbb{R}^{m \times n}, C \in \mathbb{S}^{n}$
arbitrary;

$\qquad f(X)=\log \operatorname{det}\left(X^{\top} A X+B^{\top} X+X^{\top} B+C\right)$

Find the gradient.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\B{\big}\def\L{\left}\def\R{\right}\def\o{\operatorname}$Let's
use a colon to denote the trace/Frobenius product, i.e.
$$\eqalign{
A:B &= \o{Tr}\L(A^TB\R) \\
A:A &= \big\|A\big\|_F^2 \\
}$$
And for typing convenience, define the symmetric matrix
$$\eqalign{
M &= X^TAX + B^TX + C \\
dM &= 2\operatorname{Sym}\L(X^TA\,dX\R) + B\,dX \\
}$$
where $\;\operatorname{Sym}\L(X\R)\doteq\tfrac 12\L(X+X^T\R)$

Write the function using the notation above.
Then calculate its differential and gradient.
$$\eqalign{
f &= \log\det(M) \\
df&= d\operatorname{Tr}(\log M) \\
  &= M^{-1}:dM \\
  &= M^{-1}:\big(dX^TAX + X^TA\,dX + B^TdX\big) \\
  &= M^{-1}X^TA:dX^T + AXM^{-1}:dX \,+\, BM^{-1}:dX \\
  &= \B(2AX+B\B)M^{-1}:dX \\
  &= \B(2AX+B\B)\,\B(X^TAX + B^TX + C\B)^{-1}:dX \\
\p{f}{X}
  &= \B(2AX+B\B)\,\B(X^TAX + B^TX + C\B)^{-1} \\\\
}$$

The first step in the derivation makes use of Jacobi's formula.

The next step uses a formula from the Matrix Cookbook
$$\frac{d}{dX}\o{Tr}(f(X)) = f'(X)^{T}$$
And subsequent steps make use of the fact that the terms in a Frobenius product can be rearranged in many equivalent ways, due to the properties of the underlying trace function, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB \\
}$$

Title: find matrix gradient


问:
Velocity of the shock in traffic flow problem
Hello so my teacher went over an example in class, and I just wanted to verify I understand it

The example was
$$u_t + (1-2u) u_x = 0$$
such that
$u(x,0 )=
\begin{cases}
      0 & x\leq 0 \\
      0.5 & \geq 1 \\
      x/2 & \text{otherwise}
   \end{cases}
$
and we wanted to compute the velocity of the shock for the initial conditions. We said in class that we will denote it $u_s$ and it is defined as follows

$u_s = \frac{F^+ - F^-}{u^+ - u^-}$, where $F^+$ is the right flux , where the Flux is defined as $F = v(u) u$ and we assume $v(u) = 1 - u$. So in this case

$u_s = \frac{F^+ - F^-}{u^+ - u^-} = \frac{0.5*0.5 - 0*1}{0.5 - 0}= 0.5$

My question is two fold,

Is it normal to assume $v$ which corresponds to the velocity to be equal to $1- u$ and what is the physical interpretation of this in terms of the traffic flow application? My guess is that we are saying as there is more density of cars the speed of the cars is slowing down.

When we computed the speed of the shock we took into account only $0$ and $.5$, but I think if we draw the characteristics we have characteristics from the middle region corresponding to $x/2$ also intersecting at the point (1,1), so shouldn't the shock speed involve them as well?

Thank you for reading this.

答:
The macroscopic traffic-flow model by Lighthill-Whitham-Richards (LWR) is an hydrodynamic model for traffic flow on a single infinite road. It consists of a scalar hyperbolic conservation law, which represents the conservation of cars (continuity equation):
$$
u_t + F (u)_x = 0 \, , \qquad F(u) = v(u)\, u
$$
where the flux $F(u)$ depends only on the density of cars $u \in [0,u_\max]$. Introduced by Greenshields, the simplest expression for the car velocity $v(u)$ reads
$$
v(u) = v_\max \left(1 - \frac{u}{u_\max}\right) ,
$$
which stipulates that the car velocity is decreasing linearly with the car density. With the numerical values $v_\max = 1$ m/s and $u_\max=1$ car/m, the car flux $F(u) = u\, (1-u)$ vanishes when the road is either empty or full ($u \in \lbrace 0,1\rbrace$), and the maximum flux is reached for a half-empty road $u = 0.5$. Thus, the present situation represents a road that is empty for $x \leq 0$ ($u=0$), with a continuous transition towards a road that is at optimal equilibrium for $x\geq 1$ ($u=0.5$). In the transition part $0\leq x\leq 1$, the car density $u$ and car flux $F(u)$ both increase with $x$, but the car velocity decreases.

This part refers to the picture in this post. The method of characteristics is no longer valid when characteristic curves intersect. Here, all those lines coming from $0\leq x\leq 1$ intersect at $(x,t) = (1,1)$. Hence they should be stopped there. However, the other lines coming from $x< 0$ and $x> 1$ can be continued. Beyond formation of the shock wave at $(1,1^+)$, we have $u_- = 0$ on the left side of the shock and $u_+=0.5$ on the right side. The shock speed $u_s$ is deduced from the Rankine-Hugoniot condition
$$
u_s = \frac{F(u_+) - F(u_-)}{u_+ - u_-} = \frac{0.25 - 0}{0.5 - 0} = \frac12 .
$$


问:
Circular argument for infinitude of primes from Euler product?
Many textbooks will explain how the Euler product formula can be used as a basis for proving the infinite of primes:

$$\sum_{n}\frac{1}{n^{s}}=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{-1}$$

My concern is that this is a circular argument because the derivation of the product formula uses the fact there are infinite primes.

Maths history texts will suggest that Euler's derivation of the product formula was done by a process of "sieving the zeta function", outlined next:

We start with the zeta function, having proven (elsewhere) $s>1$ for convergence.

$$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\ldots$$

We can divide this series by $2^{s}$.

$$\frac{1}{2^{s}}\zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\frac{1}{12^{s}}\ldots$$

These denominators are multiples of $2^{s}$. By subtracting these terms from $\zeta(s)$, we sieve out terms with these multiples of $2^{s}$.

$$(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\ldots$$

Repeating for $3^s$ gives us:

$$(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

And again for $5^s$:

$$(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

Repeating the process, we can only remove multiples of successive primes. We can't remove multiple of a number C if it is non-prime, C=AB, because we would have already sieved them when we removed multiples of its factors A or B.

In the limit of this process (having removed multiples of an infinitude of primes), only 1 remains on the RHS, and on the left we have an infinite product over all primes:

$$\ldots\cdot(1-\frac{1}{11^{s}})\cdot(1-\frac{1}{7^{s}})(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1$$

Back to my question.

This derivation assumes there are an infinitude of primes in order to attain the single 1 on the RHS. Therefore this Euler product can't be used as basis to say that there is an infinitude of primes.

Where is the flaw in my logic?

Comment: I am aware of the other derivation which uses the FTA and doesn't assume an infinitude of primes, but this is the method used by Euler, and he used it to demonstrate the infinitude of primes, for the first time since Euclid, according to the maths history books.

I am not a university trained mathematician so would appreciate replies with minimal technical terminology.

答:
The proofs given below for the Euler product of $\zeta(2)$ and the fact that $\zeta(2)=\frac{\pi^2}{6}$ don't use that there are infinitely many primes.

Now assume that there are only finitely many primes. Then the Euler product for $\zeta(2)$ is a rational number, so that $\frac{\pi^2}{6}$ is a rational number - contradiction.

References:

Euler Product formula for Riemann zeta function proof

Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)


问:
Dedekind infinite set definition
I was reading a book about elementary set theory, and I've found this defenition:

I have two questions:

Question 1:

Why one-to-one correspondence were assumed to be equal to injection?

Question 2:

Why we say $f:X\to X$ is an injection ? It cannot be bijection?

答:
It is common in English literature to use one-to-one as synonymous with injective. Anyways, even if it meant that $f:X\to Y$ is bijective, the hypothesis $Y\subsetneq X$ would make it equivalent to $f:X\to X$ being injective and not surjective nevertheless.

Since $f(X)\subsetneq X$ is a hypothesis, bijectivity of $f$ is impossible.

Title: Dedekind infinite set definition


答:
Yes, a one-to-one correspondence is an injective function. And no, it cannot be a bijection, because while a bijection is an injection, the requirement from $f$ is that $f(X)$ is a proper subset of $X$, or in other words: $f$ is not a surjection.

For example, $\Bbb N$ is a Dedekind-infinite set, since $f(x)=x+1$ is an injection which is not a surjection. In contrast, $\{0,1\}$ is not Dedekind-infinite, since there are only two injective functions $f(0)=0, f(1)=1$ and $g(0)=1,g(1)=0$ and they are both surjective.

Title: Dedekind infinite set definition

Body:
I was reading a book about elementary set theory, and I've found this defenition:

I have two questions:

Question 1:

Why one-to-one correspondence were assumed to be equal to injection?

Question 2:

Why we say $f:X\to X$ is an injection ? It cannot be bijection?


问:
Modules which are both left and right modules over a noncommutative ring
Given a noncommutative ring $R$, is there a name for those (left) $R$-modules $M$ for which
$$
m.(rs) = m.(sr), ~~~ \textrm{ for all } r,s \in R, \textrm{ and } m \in M?
$$
(Note that we have denoted the action of $R$ on $M$ by a dot.)
In other words, what does one call modules which are both left and right modules.

答:
Basically, your condition amounts to the fact that the action of $R$ on $M$ factors through the "abelianization" of $R$ (not sure if this is standard terminology for rings).

Specifically, for any ring $R$, you can form the ideal $I$ generated by all elements of the form $xy-yx$, and $R^{ab} = R/I$ is then the minimal commutative quotient of $R$.

Then a left $R$-module $M$ is also a right module iff the action of $R$ factors through $R^{ab}$. So studying "left-and-right" modules over $R$ is the same thing as studying modules over the commutative ring $R^{ab}$.


问:
What is computed if you maximise likelihood of an exponential distribution without resorting to logs?
I would like to get some explanation or intuition behind the process of finding the maximum likelihood of an exponential distribution.

Given the following likelihood function:

$L\left( \theta \right) = \theta ^{4}e^{-8.65\theta }$

I would like to find $\theta$ for which the likelihood is maximised. I know I can achieve this by taking the derivative and solving the equation:

$\dfrac{dL\left( \theta \right) }{d\theta }=0$

Having calculated the derivative I obtained:

$\dfrac{dL\left( \theta \right) }{d\theta } = \theta ^{3}\left( 4-8.65\theta \right) \cdot e^{-8.65\theta }$

Solving it for $0$ yields ~ $0.46$.

When one takes a look at the plot of this function, one can see that this function doesn't have the maximum in that place. Since it is $e^{-x}$, the max value goes to the inf.

What am I actually determining, if I say that likelihood is maximised at 0.46?

答:
It has a maximum at $\theta\approx 0.4624$ even if it is just $L(0.46)\approx 0.0008$

$$...$$


答:
One (probably the most straighforward) correct logic expression can be: $$\forall a \forall b(((a\gt 0)\land (b\gt 0))\rightarrow ab \gt0) \tag 1$$
We shall compare others with this one to check their validity.

The expression $\forall a \forall b(ab\gt 0\rightarrow((a\gt 0)\land(b\gt 0)))$ means that if the product of two numbers is positive, then both of them are positive. This is the converse of what was asked. So, it's incorrect.

In option $2$, $\forall a \forall b (\left((a>0) \land (b>0)\right) \land ab<0))$ means that the all two numbers are positive and their product is negative. This is clearly incorrect. Thus, it doesn't generalize the statement either.

In option $1$,  $\forall a \forall b(ab\le 0  \to \left((a\gt0) \to (b\le 0))\right)$ is correct. It can be modified into $\forall a \forall b(ab\le 0  \to \left((a\le0) \lor (b\le 0))\right)$ which the contrapositive of $(1)$.

Thus, option $1$ is correct.

Title: logic formalizing: The multiplication of every two positive numbers is also positive

Body:
EDITED THE QUESTION WITH THE RIGHT SIGNS

I have to formalize the expression: "The multiplication of every two positive numbers is also positive" which I think is:

$$\forall a \forall b(ab>0 \to \left((a>0) \land (b>0)\right))$$

but I have to choose between two options:

$\forall a \forall b(ab \le 0  \to \left((a>0) \to (b
\le 0))\right)$

$\forall a \forall b (\left((a>0) \land (b<0)\right) \land ab<0))$

which one is more correct?


问:
Degree of dual variety of complete intersection
Let $X$ be a complete intersection in $\mathbb P^n$, does anyone know how to calculate the degree of a dual variety of $X$?

Some preliminaries can be found in 3264 and all that, chapter 10. For instance, such dual varieties are always codimension one (except when $X$ is linear); and when $X$ is a hypersurface, the answer is well know.

In particular, I'd like to know the answer when $X$ is the curve as a complete intersection by a cubic and quadric in $\mathbb P^3$.

答: I think the $h^n$ in the explanation should be $h^{n-k}$. E.g. in the case $k=n$ the expression in the parentheses must be equal to $1$.


问:
$ \sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$
Let $ f: \mathbb N  \to \mathbb N$ be a bijective function such that:

$$ \sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$$

Now my question is does any such $f$ exists?

答:
Let $f$ be any given bijection from $\Bbb N\to\Bbb N$. Consider the partial sum $S_n=\sum_{k=1}^n\frac{f(k)}{k^2}$. Suppose $\{t_i\}_{i=1}^n$ is a strictly increasing sequence such that $t_i\in f(\{1,2,...,n\})$. Clearly, $t_i\ge i$. Then, from the rearrangement inequality$$S_n\ge\sum_{i=1}^n\frac{t_i}{i^2}\ge\sum_{i=1}^n\frac i{i^2}$$

Thus, $\lim_{n\to\infty}S_n\ge\lim_{n\to\infty}\sum_{i=1}^n\frac i{i^2}\to\infty$.


问:
Local linearity definition of derivative of multivariable function

This is an alternate definition of a derivative of a function from R^2 to R. How do we go from this definition to the old definition or show they are equivalent?

Also, is this actually a better definition of the derivative? If so, why don't we do this for single variable calculus?

答:
The "old definition" does not work for functions from $\mathbb{R}^2\to\mathbb{R}$. Because the Newton quotient
$$
\frac{f(x+h)-f(x)}{h}
$$
does not make sense when the numerator $h$ is a vector in $\mathbb{R}^2$.

An equivalent way to write (3.2) in your post is
$$
f(x+h)=f(x)+\ell h+r(h)
$$
where
$$
\lim_{h\to 0}\frac{|r(h)|}{\|h\|}=0\;.
$$


答:
Define $$g(m)=\sum_{n=1}^m\dfrac{f(n)}{n^2}$$
Now note that if the $f$-image of the set $[m]=\{1,2,\dots, m\}$ is $[m]$ itself, then the minimum value of $g(m)$ is taken when $f(k)=k\:\forall\:1\le k\le m$ (by rearrangement inequality).

Otherwise there exists a subset $S$ of $[m]$ such that $f(S)\subset[m]$ and $f([m]\setminus S)\cap[m]=\emptyset$. In this case, we may note that $g(m)$ is term-wise more than or equal to the partial sum obtained by replacing $f([m]\setminus S)$ by $[m]\setminus f(S)$ and if we do that, the lower bound obtained above using rearrangement inequality again applies. Therefore, we have,
\begin{equation}
\sum_{n=1}^\infty\dfrac{f(n)}{n^2} =\lim_{m\to\infty}g(m)\ge\lim_{m\to\infty}\sum_{i=1}^m\dfrac1n=\infty
\end{equation}
Hence, this series is divergent.

Body:
Let $ f: \mathbb N  \to \mathbb N$ be a bijective function such that:

$$ \sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$$

Now my question is does any such $f$ exists?


问:
Regarding Picard's iteration its relation to numerical methods, such as Euler's method.
I've been told numerical methods in solving ODEs, such as Euler's method and Runge-Kutta, are all in some way approximations to Picard's iteration, and I'm trying to understand how.

Suppose we have a differential equation on an interval $[x_0,x_L]$:

$$\frac{dy}{dx}=f(x,y)$$
with initial condition
$y(x_0)=y_0$

I would like to numerically solve the equation on a set of points $\{x_0<x_1<\dots<x_n\}$, i.e. obtain approximations $y_i$ to the true solution $y(x_i)$ for each $x_i$.

Picard's iteration works as follows:

$$y_{0,0}=y_0$$

$$y_{0,k}(x_1)=y_0+\int_{x_0}^{x_1} f(x,y_{0,k-1}(x))dx \;\; \mathrm{for} \;\; k \geq 1$$

suppose we stop for $k=m$, then take $y_1=y_{0,m}(x_1)$

We then repeat the process for $i \geq 1$.

$$y_{i,0}=y_i$$

$$y_{i,k}(x_{i+1})=y_i+\int_{x_i}^{x_{i+1}} f(x,y_{i,k-1}(x))dx \;\; \mathrm{for} \;\; k \geq 1$$

$$y_{i+1}=y_{i,m}(x_{i+1})$$

So is the idea of a numerical method (e.g. Euler's method) to replace the integral $\int_{x_i}^{x_{i+1}} f(x,y_{i,k-1}(x))dx$ with an approximation such as $(x_{i+1}-x_{i})f(x_i,y_i)$ (Euler's method)? What I don't understand is why numerical methods only iterate once for each point $x_i$ (in other words, $m=1$) but Picard's iteration suggests you should iterate multiple (potentially many) times for each $x_i$?

答:
Euler method stops at just one iteration, but Runge-Kutta goes for higher iterates.  For example, the RK2 (midpoint method):
$$
y_{n+1}=y_n+hf(x_{n+\frac12}, y_n+\frac12hf(x_n,y_n))
$$
is the result of the second Picard iteration: the first use the constant $y^0(x)=y_n$ for $\int_0^{th}f(x_n+\xi,y^{0}(x_n+\xi))\,\mathrm{d}\xi$ with Euler and get $y^1(x_n+th)=y_n+thf(x_n,y_n)$ for $t\in[0,1]$ and then the second iteration
$$
\int_0^{h}f(x_n+\xi,y^{1}(x_n+\xi))\,\mathrm{d}\xi
$$
we use the mid-point rule.

Similarly, you have predictor-corrector methods which you can view the corrector method as approximating the repeated Picard iterations as many times as you want.


问:
Construction of the addition function
I am reading a book called Analysis I by Herbert Amann and Joachim Escher. I am currently stuck on page 33 where they construct the addition operator using functions. One property the addition operator should fulfil is the following
$$n+ v(m) = v(n+m)$$

where $v$ is the succesor function. In order for the $+$ operator to obtain this property, the authors propose the following set:

$$ N:= \{n \in \mathbb{N}; \exists \ \varphi_n: \mathbb{N} \to \mathbb{N} \ \ \text{with } \  \varphi_n(0) = v(n) \ \ \text{and } \ \varphi_n(v(m)) = v(\varphi_n(m)) \ \forall \ m \in \mathbb{N}\} $$

I somewhat do see why they propose this set. It makes sense later on once you define the binary operator $+$. But before they ultimately define the addition operator they first prove that $N = \mathbb{N}$. As always, they prove it with Peano's induction Axiom. First they define $\varphi_0 := v$. It is clear to me that $0 \in N$ since $\varphi_0(0)= v(0)$ and $\varphi_0(v(m)) = v(\varphi_0(m))$. By Peano's axiom which states that there exists a successor function these two conditions are true statements hence $\varphi_0$ exists. From this point on I am not able to follow their thinking. The next step is to show that if $n \in \mathbb{N}$, then $v(n) \in N$. I don't really understand how they prove that if $n \in N $ then $v(n) \in N$.

答:
They use induction to prove that $N=\mathbb N$.

By definition above, $N \subseteq \mathbb N$; thus, it is enough to prove that $N$ includes all naturals.

How to do this? By induction.

The base step: $0 \in N$, is Ok.

The next step is to show that if $n \in N$, then also $\nu(n) \in N$.

Thus, assume that $n \in N,$ i.e. there is a function $\varphi_n$ that satisfies the condition specified by the definition of $N$, and construct $\varphi_{\nu(n)}$.

They introduce an "auxiliary" function $\psi$ defined by:

$\psi(m)=\nu(\varphi_n(m))$, for every $m$.

And we have to check that is satisfies the defining condition.

Firstly, $\psi(0)=\nu(\varphi_n(0))$ and $\varphi_n$ is already defined in such a way that $\varphi_n(0)=\nu(n)$.

Thus, $\psi(0)=\nu[\nu(n)]$ and this "match" with our expectation that $\varphi_{\nu(n)}(0)=\nu[\nu(n)]$ (read it as: $\varphi_{n+1}(0)=\nu(n)=n+1$, because $0+(n+1)=n+1$).

Next, $\psi(\nu(m))=\nu (\varphi_n(\nu(m)))$.

Again, $\varphi_n$ is already defined; thus we know its value for every argument, $\nu(m)$ included.

But $\varphi_n$ satisfies the defining condition: $\varphi_n(\nu(m))=\nu(\varphi_n(m))$, and thus we can "commute" it to get: $\psi(\nu(m))=\nu (\nu(\varphi_n(m))$.

But $\psi(m)$ has been defined as: $\nu(\varphi_n(m))$, and thus we get:

$\psi(\nu(m))=\nu(\psi(m))$.

Thus, we have checked that the specified condition is fully satisfied, and we can use $\psi$ and $\varphi_{\nu(n)}$.

Conclusion: we have showed that $\varphi_0$ exists, and we have showed that, if $\varphi_n$ exists, also $\varphi_{\nu(n))$ exists.

Applying axiom (N1) we conclude that every $n \in \mathbb N$ is in $N$, i.e. that $N = \mathbb N$.

The gist of the construction is to define "sum" through an infinite sequence of functions: $\{ \varphi_0, \varphi_1, \ldots, \varphi_n, \ldots \}$ where $\varphi_0$ is "add $0$" to every number $m$ [i.e. $\varphi_0(m)=m+0$] and $\varphi_{\nu(n)}$ is "add $\nu(n)$" to every number $m$ [i.e. $\varphi_{\nu(n)}(m)=m+\nu(n)=m+(n+1)$].

The result is that:

$\varphi_n(m)=m+n$, for every $n,m$.

Title: Construction of the addition function


问:
Product of signs of two juxtaposition of two disjoint permutations
Edited: Suppose we are given two disjoint permutations $(a_1,\dots,a_m)$ and $(b_1,\dots,b_n)$. I want to prove that (if it is correct)

$$\text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)\times \text{sgn}(b_1,\dots,b_n, a_1,\dots,a_m)=(-1)^{\sum_{i=1}^m (a_i-1)+\sum_{i=1}^n (b_i-1)}.$$

Where $\text{sgn}$ denotes the sign of permutation which is $+1$ or $-1$ depending on whether the permutation is even or
odd.
<s> I know that parity of a $k$-cycle is $k-1$. I think $(a_1,\dots,a_m, b_1,\dots,b_n)$ is same as $(a_1,\dots,a_m)(b_1,\dots,b_n)$ because they are disjoint cycles, so $\text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)=\text{sgn}(a_1,\dots,a_m)\cdot\text{sgn}(b_1,\dots,b_n)$. Right? And $\text{sgn}(a_1,\dots,a_m)=(-1)^{m-1}$ and $\text{sgn}(b_1,\dots,b_n)=(-1)^{n-1}$ But these seems do not imply the above claim.</s>

Any suggestion?

答:
$\text{sgn}((a_1,\cdots,a_m)) = (-1)^{m-1}$ indeed.

Also, by disjointness $(a_1,\cdots,a_m)(b_1,\cdots,b_n) = (b_1,\cdots,b_n) (a_1,\cdots,a_m)$. So the product of their signs is $(-1)^{m-1} (-1)^{n-1} = (-1)^{m+n-2}$.

I don't know what you mean by $\text{sgn}(a_1,\dots,a_m, b_1,\dots,b_n)$ though. Is that the combined cycle? But this is not the same as the product.. In either order this new cycle has sign $(-1)^{m+n-1}$, being of length $m+n$..


问:
Non-linear second order equation with the derivative term
I am trying to solve the following equation:

$y''(x) - \beta H(x) y'(x)=- (2 p - 2) y' ^ {2}(x)/y(x)$

where p and $\beta$ are constant and H is a function of x. Please help me with this equation.

答:
Dividing by $y'$ we get $$\frac{y''}{y'} -\beta H =(2-2p)\frac{y'}{y}$$

integrating with respect to $x$ we obtain

$$\ln |y'| +(2p-2)\ln |y| =\beta \int H dx $$

$$\ln \left|\left(\frac{y^{2p-1}}{2p-1}\right)' \right|=\beta \int H dx $$
$$\left(\frac{y^{2p-1}}{2p-1}\right)' =e^{\beta \int H dx }$$
$$\frac{y^{2p-1}}{2p-1}=\int \left(e^{\beta \int H dx}\right)dx$$


问:
A question on logic - where intuition can fail
Suppose I have two predicates $P(x)$ and $Q(x)$, such that $\overline{P(x)\wedge Q(x)}$ holds for all $x$.

Now, if $\displaystyle \bigwedge_{x\in A}P(x)$ for a set $A$, it must be certainly true, that $\displaystyle \overline{\bigwedge_{x\in A}Q(x)}$ by intuition. But suppose, that $A = \{\;\;\}$, then obviously both $\displaystyle \bigwedge_{x\in A}P(x)$ and $\displaystyle\bigwedge_{x\in A}Q(x)$.

Where is my mistake? I guess it is in the assumption that
$$\bigwedge_x\overline{P(x)\wedge Q(x)} \rightarrow \left(\bigwedge_{x\in A}P(x) \rightarrow \overline{\bigwedge_{x\in A}Q(x)}\right)?$$

*Note: $\displaystyle \overline{P(x)\wedge Q(x)} \equiv \lnot(P(x)\land Q(x))$

$\displaystyle \quad\quad\quad\quad \overline{\bigwedge_{x\in A}Q(x)} \equiv \lnot \left(\bigwedge_{x\in A}Q(x)\right)$

答:
Your intuition is based on the set $A$ having some nontrivial size.  But as you note, the intuition is wrong if $A$ is empty.

You only need to change your intuition slightly.  If $P$ holds for all $x \in A$, then we know that $Q$ cannot hold for any $x \in A$.  That is, $\bigwedge_{x\in A} \lnot Q(x)$.

The mistaken intuition is in transforming
$\bigwedge_{x\in A} \lnot Q(x)
\ \ \ \rightarrow \ \ \
\lnot \bigwedge_{x\in A} Q(x)$.
This step is only correct if $A$ is nonempty, but even then it is a weak step (for large $A$ the left hand side is a strong statement while the right hand side is relatively weak), so you would do best to forget about this sort of transformation — try to erase it from your intuitive repertoire!

The correct transformation would be that of De Morgan:
$\bigwedge_{x\in A} \lnot Q(x)
\ \ \ \rightarrow \ \ \
\lnot \bigvee_{x\in A} Q(x)$.


问:
Proving $f(x) := \sin(ax) - \sin(bx)$ has absolute value greater or equal to 1 for some $x>0$
I am trying to prove that for any choice of reals $a$ and $b$  such that
$a \neq b $
the function $f(x) := \sin(ax) - \sin(bx)$ has $|f(x)| \geq 1$ for some $x>0$

(This is not an exercise question but something I am trying to prove for myself. Plots with Wolfram Alpha seem to indicate it might be true, and so counterexamples are very much welcome)

Here is my attempt.
For this I wrote $f$ of 3 variables $a,b,x$. We find that value of $x_0$
for which $f$ attains a maximum and minimum and then show that $|f(x_0)|$
greater than 1.

$g(a,b,x)=f(x)$ and then zeroing the gradient. (pending the 2nd derivative test)

$f_a = x \cos(ax)$

$f_b = -x \cos(bx)$

$f_x = a\cos(ax) - b\cos(bx)$

Since the point of minima/maxima are conjectured $x>0$ , this means $\cos(ax)$ and $\cos(bx)$ are both zero
and the third equation is redundant.

If $a$ and $b$ are rational then obviously we can find such an $x$.
But when one of them is irrational, I don't know how I would prove this.

Another attempt was using the $\sin(u)-\sin(v)$ product formula, but that
led nowhere too.

Any pointers?

答:
Hint :

For $x_k=\pi/2a + 2*k\pi/a $, $\sin(ax_k) =1$ You "just" have to prove that for a certain $x_k$ $\sin(bx_k)$ is inferior than 0.

By this method, you could also prove a greater result : There exists $x$ where $f(x)>A$ for every $A<2$


问:
1D Lax-Friedrichs scheme leading error term

The Lax-Friedrichs scheme for the 1D linear advection equation $$\frac{}{}+ _\frac{}{} = 0$$ with a
constant flow speed of $_$ is
$$(+Δ,)=\frac{1}{2}[(,+Δ)+(,−Δ)]−\frac{Δ_}{2Δ}[(,+Δ)−(,−Δ)]$$
where $Δ$, $Δ$ and $_$ are the time step, the cell width of a 1D uniform mesh and the flow
speed respectively. Applying Taylor series expansions of $(+Δ,)$, $(,+Δ)$ and
$(,−Δ)$ about $(,)$, show that the leading error term (...) of the Lax-Friedrichs
scheme is
$$...=\frac{1}{2}\left[\frac{Δ^2}{Δ}− Δ^2_\right]\frac{^2}{^2}$$

Could someone please explain how to do this as I can't find out how to work out the Taylor series

答:
Let us use the notation $U_j^n$ for approximations of $U(x_j, t_n)$ with $x_j = j\Delta x$ and $t_n = n\Delta t$. Using Taylor series, one shows that for smooth solutions
\begin{aligned}
\frac{U_j^{n+1} - \frac12( U_{j+1}^n + U_{j-1}^n )}{\Delta t} &= (U_t)_j^n + \frac{\Delta t}{2}(U_{tt})_j^n - \frac{\Delta x^2}{2\Delta t} (U_{xx})_j^n + O(\Delta t^2, \Delta x^2)
\\
\frac{U_{j+1}^{n} - U_{j-1}^n}{2\Delta x} &= (U_x)_j^n + O(\Delta x^2) .
\end{aligned}
Now let us compute the difference between the PDE and its approximation deduced from the Lax-Friedrichs method. Up to higher-order terms, the scheme produces
$$
\begin{aligned}
(U_t + v_x U_x)_j^n
& - \frac{U_j^{n+1} - \frac12( U_{j+1}^n + U_{j-1}^n )}{\Delta t} - v_x \frac{U_{j+1}^{n} - U_{j-1}^n}{2\Delta x} \\
&= \underbrace{-\frac{\Delta t}{2}(U_{tt})_j^n + \frac{\Delta x^2}{2\Delta t} (U_{xx})_j^n}_\text{l.e.t.}
\end{aligned} $$
Then, the transport equation yields the relationships $U_{tx} = -v_x U_{xx}$ and $U_{tt} = -v_x U_{xt}$ between temporal and spatial derivatives,
which lead to $U_{tt} = v_x^2 U_{xx}$. Finally,
$$
\text{l.e.t} = \frac{1}{2}\left(\frac{\Delta x^2}{\Delta t} - \Delta t\, v_x^2\right) (U_{xx})_j^n \, ,
$$
where $j$ and $n$ can be removed.


问:
Why is this a topology on $Y$?
Let $f$ be a continuous map from $X$ onto $Y$, and $f$ is a quotient map. Why is the family $\tau_q $ of all $f(U)$, where $U$ is open in $X$, a topology on $Y$?</p>

答:
It is not, in fact, a topology. For a simple counterexample, let $X=\{1,2,3,4\}$ with the topology $\{ \{1,2\}, \{3,4\}, X, \emptyset\}$. Let $Y=\{1,2,3\}$ with the indiscrete topology. Define $f$ as $f(1)=1, f(2)=f(3)=2, f(4)=3$. Now,
$$ \tau_q = \{ Y, \emptyset, \{1,2\}, \{2,3\}\}$$ which is not closed under intersection.

Edit: $f$ is also a quotient map.

Title: Why is this a topology on $Y$?


问:
Finding a method to calculate the line integral of a certain function with a certain domain
For example, let us calculate the line integral $\int_{\partial} \frac{1}{(z-1)} $ where the domain is $D(0,|2|)$.

I first generalize the whole reasoning into: for integral $\int\frac{1}{(z-a)} $, if the domain contains a, then its value is 2$\pi i$, on the other hand, if

the $\int\frac{1}{(z-a)^n} $ with n equal or great than 2

for integral $\int\frac{1}{(z-a)} $ the point a does not contain in the domain

The integral goes directly to 0.

Is it valid?

答:
As a consequence of the analyticity of holomorphic functions you get the formula:
\begin{equation}
\frac{f^{n)}(z_{0})}{n!} = \frac{1}{2 \pi i} \int_{\partial}\frac{f(w)}{(w-z_{0})^{n+1}}dw
\end{equation}

$\partial$ is oriented counterclockwise, $z_{0}$ is inside $\partial$ and $f(w)$ is analytic on and inside $\partial$.

You can see a proof here (only for circles): https://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions

You can also see this proof (not only for circles but the proof is not complete):
http://mathonline.wikidot.com/cauchy-s-integral-formula-for-derivatives


问:
de Rham and Hodge cohomologies become locally free after localization?
This is a follow-up question to these two: 1 2.

I am studying a paper about the degeneration of the de Rham-Hodge spectral Sequence. The following data are given: A field $k$ and a smooth proper $k$-scheme $X\longrightarrow\operatorname{Spec}(k)$, a finitely generated $\mathbb{Z}$-algebra $A$ and a smooth proper scheme $\mathfrak{X}\longrightarrow A$ so that $\mathfrak{X}_{k}\cong X$. The author claims:

After possibly localizing $A$, we can assume that the coherent $A$-modules $H^{n}_{\operatorname{dR}}(\mathfrak{X}/A)$ and $H^{n}_{\operatorname{Hodge}}(\mathfrak{X}/A)$ are locally free.

Why?

I am by no means an expert on the topic, so please be as détailed as possible in your answers; a reference might also help.

答:
this is a general property of coherent modules. First it is a very good idea to look at the paper of Illusie:Frobenius and Hodge degeneration.which gives much more detail than the paper you are reading now.(and written by the person who prove the theorem!)

about your question, look at the generic point $\eta$ of $A$(the point associated to the prime ideal 0) local ring of $A$ at this point is just your fraction field. now assume that your coherent sheaf is given by the module $M$,then $M_\eta$ is finite free over $A_\eta=Frac(A)$.let $(m_1/a_1,m_2/a_2,...,m_n/a_n)$ be a basis for the module $M_\eta$. consider generators for $M$,$(k_1,...,k_t)$ and write $$k_i=\Sigma \frac{b_{ij}}{c_{ij}}\frac{m_j}{a_j}$$. now if invert all the $a_i,c_{ij}$ in $A$ and call the ring $A_S$. now you have a surjective map $A_S^n\to M_S$ by sending $e_i\to \frac{m_i}{a_i}$.call the kernel of this map $N$, so you have $$0\to N\to A_S^n\to M_S\to 0$$
but you know that $A_\eta^n\to M_\eta$ is an isomorphism and $A_S\to A_\eta$ is flat,so from the above exact sequence we have $N_\eta=0$. because $N$ is finitely generated(by coherence) there should be an element $a\in A$ such that $aN=0$. so in summary after inverting $a$ the map between $A^n\to M$ become an isomorphism and hence $M$ would be free.

now back to your problem you know that $R^if\Omega_{X/A}$ is coherent and so after going to a localisation $A_S$,${(R^if\Omega_{X/A})}_S$ is free. but this is ${R^if\Omega_{X_S/A_S}}$(because R^if is always quasi cohehrent) so we are done.


问:
Natural number solutions to $\frac{xy}{x+y}=n$ (equivalent to $\frac 1x+\frac 1y=\frac 1n$)
I have a question about the following problem from a Putnam review:

Let $n\in \mathbb{N}$. Find how many pairs of natural numbers $(x, y)\in \mathbb{N}\times \mathbb{N}$ solve
$$
\frac{xy}{x+y}=n.
$$

I have found some solutions for all $n$, such as $(2n, 2n)$ and $(n+1, n(n+1))$, but I feel as though this is the wrong approach, as the question is only asking for the number of solutions to the equation.

I don't really want a complete solution, but any hints would be greatly appreciated.

答:
This can be rewritten as

$$\frac 1x+\frac 1y=\frac 1n$$

$$\frac 1y=\frac 1n-\frac 1x$$
$$y=\frac{nx}{x-n}$$

Changing variables to $k=x-n$:

$$y=\frac{n(n+k)}{k}$$

$$y=\frac{n^2}k+n$$

The left side is an integer. What does that tell you about the allowed values of $k$, and the corresponding number of valid $x$ values?


问:
If a set is "not open", what is it?
Physics background here, no formal pure maths education so please try to reply accordingly.

What are the possibilities for a set which is "not open"? Are they:

1)

closed

both open and closed

neither open nor closed

or

2)

closed

neither open nor closed?

In other words if a set is said to be "not open" does that mean that it could be both open and closed?

Accordingly for "not closed".

Or am I entirely confused?

答:
A very common mistake is thinking that a not open set is closed. This is false, e.g. $[2,5) \subset \mathbb{R}\ $ is not open and it is also not closed. This shows that: "Not open does not imply closed".

What is true is that the complement of an open set is closed.

So always be aware of this.

As to your question, Xander Henderson answered correctly in the comments: it can't be (1) because a not open set cannot be both open and closed because it cannot be open.

Title: If a set is "not open", what is it?


问:
Conditional expectation of sum of numbers on 2 throws of a dice given their difference
Let$  X$ and $Y$ be the numbers obtained on first and second throws of a fair die .
Calculate $E(X+Y|(X-Y)^2=1)$

I am doing basically $$E(X|(X-Y)^2=1)=E(Y|(X-Y)^2=1)$$
So $E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
$E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
implies $E(X|(X-Y)^2=1)=1/6(1+2+3+4+5+6)$

So I am getting it $7$ please help

答:
The result is correct.

To "see" the solution observe that throwing 2 fair dice the events that correspond to

$$(X-Y)^2=1$$

are 10

$$\{(1,2);(2,1);(2,3);(3,2);(3,4);(4,3);(4,5);(5,4);(5,6);(6,5)\}$$

given this sample space, the sum $X+Y$ is a rv taking values in

$$ Z=
\begin{cases}
\frac{1}{5},  & \text{if $z=3$}  \\
\frac{1}{5},  & \text{if $z=5$}  \\
\frac{1}{5},  & \text{if $z=7$}  \\
\frac{1}{5},  & \text{if $z=9$}  \\
\frac{1}{5},  & \text{if $z=11$}
\end{cases}$$

Thus

$$\mathbb{E}[Z]=7$$


问:
Finding the range of a matrix $A$
Consider the linear transformation define by

$$u(x) = \begin{bmatrix}
x_1 - \bar{x} \\
x_2 - \bar{x} \\
\vdots \\
x_m - \bar{x}
\end{bmatrix},
$$
for all $x \in R^m$, where $\bar{x} = (1/m) \sum x_i$.

Find the matrix $A$ for which $u(x) = Ax$, and then determine the dimension of the range and null spaces of $A$.

I found $A$ to be:

$$
\begin{bmatrix}
(1-\frac{1}{m}) - \frac{1}{m} \cdots -\frac{1}{m}\\
-\frac{1}{m} + (1-\frac{1}{m}) \cdots - \frac{1}{m}\\
-\frac{1}{m} - \frac{1}{m} + (1-\frac{1}{m})
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\vdots\\
x_m
\end{bmatrix}
=
I_m -\frac{1}{m}1_m 1'_m
$$

I am stuck at how to find the dimension of the range because the book has the answer has $\dim(R) = m-1$ but I'm not seeing where they get that from matrix $A$.

答:
Hint: The easiest approach is to start by finding the nullspace, which is the space of solutions to the equation $u(x) = 0$. From there, the rank-nullity theorem says that we have
$$
\dim(\mathcal R(u)) = n - \dim(\mathcal N(u)).
$$

Title: Finding the range of a matrix $A$


问:
$\epsilon,\delta$-proof for the limit of $\log \sinh{(x^2)}-x^2$ for $x \rightarrow \infty$
I want to use a $\epsilon,\delta$-proof for the existence and value for the limit of $$\log \sinh{(x^2)}-x^2$$
for $x \rightarrow \infty$.

Now, I know the definition for such proof to be $\forall \epsilon > 0 \exists c \forall x > c: \left | f(x)-L \right |<\epsilon$.

I am struggeling to approach this problem as I am not familiar with the method for  such proofs. What if the limit as $x \rightarrow \infty$ is $\infty$ then this definition would crumble, right?

答: The limit is $\log\left(\frac12\right)$. In order to see why, note that\begin{align}\log\left(\sinh\left(x^2\right)\right)-x^2&=\log\left(e^{x^2}-e^{-x^2}\right)-\log(2)-\log\left(e^{x^2}\right)\\&=\log\left(1-e^{-2x^2}\right)+\log\left(\frac12\right)\end{align}and that therefore$$\log\left(\sinh\left(x^2\right)\right)-x^2-\log\left(\frac12\right)=\log\left(1-e^{-2x^2}\right).$$So, since $1-e^{-2x^2}<1$ for any real number $x$, we have$$\left|\log\left(\sinh\left(x^2\right)\right)-x^2-\log\left(\frac12\right)\right|=-\log\left(1-e^{-2x^2}\right).$$So, given any $\varepsilon>0$, take $M=\sqrt{-\frac{\log\left(1-e^{-\varepsilon}\right)}2}$ and then\begin{align}x>M&\iff x>\sqrt{-\frac{\log\left(1-e^{-\varepsilon}\right)}2}\\&\iff2x^2>-\log\left(1-e^{-\varepsilon}\right)\\&\iff-2x^2<\log\left(1-e^{-\varepsilon}\right)\\&\iff e^{-2x^2}<1-e^{-\varepsilon}\\&\iff1-e^{-2x^2}>e^{-\varepsilon}\\&\iff\log\left(1-e^{-2x^2}\right)>-\varepsilon\\&\iff-\log\left(1-e^{-2x^2}\right)<\varepsilon.\end{align}


答:
The question is somewhat like asking if to compute $1+1$ one necessarily has to apply residual calculus. That said, it is more the fundamental theorem than the Picard iteration that guides the constructions of some basic numerical methods
$$
y_{n+1}-y_n=\int_{x_n}^{x_{n+1}}f(x,y(x))\,dx
$$
What one can reasonably motivate with this formula and related iteration schemes are one-step methods up to order 2 and Adams-Bashford-Moulton multistep formulas.

The origin of Runge-Kutta methods is indeed from attempts to elevate known quadrature methods to correspondingly higher order ODE integration methods. Runge (1895) started from the Simpson formula to get a 4-stage 3rd order method. Heun (1900), while computing his methods via matching coefficients in Taylor expansions, still extensively discussed methods based on the 4th order Gauß quadrature.

However, Kutta (1901) in introducing what is now known as general explicit Runge-Kutta methods, computed the order conditions directly from the Taylor expansions and fully solved the $s$-stage order-$s$ methods for $s=2,3,4$. Only in examples did he then take known quadrature methods as design decisions to reduce the number of free variables. This approach, find the order equations, establish some simplifying design decisions, solve the algebraic system, has remained valid for the construction all of the modern higher order methods.

Title: Regarding Picard's iteration its relation to numerical methods, such as Euler's method.

Body:
I've been told numerical methods in solving ODEs, such as Euler's method and Runge-Kutta, are all in some way approximations to Picard's iteration, and I'm trying to understand how.

Suppose we have a differential equation on an interval $[x_0,x_L]$:

$$\frac{dy}{dx}=f(x,y)$$
with initial condition
$y(x_0)=y_0$

I would like to numerically solve the equation on a set of points $\{x_0<x_1<\dots<x_n\}$, i.e. obtain approximations $y_i$ to the true solution $y(x_i)$ for each $x_i$.

Picard's iteration works as follows:

$$y_{0,0}=y_0$$

$$y_{0,k}(x_1)=y_0+\int_{x_0}^{x_1} f(x,y_{0,k-1}(x))dx \;\; \mathrm{for} \;\; k \geq 1$$

suppose we stop for $k=m$, then take $y_1=y_{0,m}(x_1)$

We then repeat the process for $i \geq 1$.

$$y_{i,0}=y_i$$

$$y_{i,k}(x_{i+1})=y_i+\int_{x_i}^{x_{i+1}} f(x,y_{i,k-1}(x))dx \;\; \mathrm{for} \;\; k \geq 1$$

$$y_{i+1}=y_{i,m}(x_{i+1})$$

So is the idea of a numerical method (e.g. Euler's method) to replace the integral $\int_{x_i}^{x_{i+1}} f(x,y_{i,k-1}(x))dx$ with an approximation such as $(x_{i+1}-x_{i})f(x_i,y_i)$ (Euler's method)? What I don't understand is why numerical methods only iterate once for each point $x_i$ (in other words, $m=1$) but Picard's iteration suggests you should iterate multiple (potentially many) times for each $x_i$?


问:
prove the inequality $x + y = 2$. Prove that $x^2y^2 (x^2 + y^2)\leq2$
Positive numbers  $x + y = 2$. Prove that $x^2y^2 (x^2 + y^2)\leq2$

答:
Setting $y=2-x$ and substituting this into the inequality we have $$ x^2y^2(x^2+y^2)-2\le 0 \Longleftrightarrow
(x^4 - 4x^3 + 5x^2 - 2x - 1)(x - 1)\le 0.
$$
Since $y=2-x\ge 0$ we have $0<x<2$. Now show that the claimed inequality is true for all such $x$.


问:
Example of a continuous point on function with no RHS and LHS derivative?
I am reading smirnov's "A Course of Higher Mathematics" and Struggling with one of the example given below

screenshot on the example

There is no equation given for this curve but just the figure itself. Can anyone please explain this example in a more detailed manner ?

I understand that f(x) can be continuous for x = c but at the mean time not differentiable. But I just don't get this example here.

答:
The question asks about a depiction of an example of a continuous function which is claimed not to have left or right hand derivatives at a point $c$. The graph of $f$ is drawn as oscillating back and forth between two lines of different slopes that intersect at $c$ (where the value of $f(c)$ is the common value of the two lines) and one is to infer from the surrounding text that this oscillation continues indefinitely with the graph of $f$ as $x$ approaches $c$.

The behavior is analogous to the function $f$ given by $f(x) = x \sin(1/x)$ for nonzero $x$ and $f(0) = 0$.  (This function is continuous at $0$ because $|f(x)| \leq |x|$ holds for all $x$ and $f(0) = 0$.)  If you look at the graph of $f$ you can see that it oscillates between the graphs of the lines $y = x$ and $y = -x$ and it is possible to prove via formulas that there are sequences of values of $h$ going to $0$ from either side for which the limit of the difference quotient
$$
\frac{f(h) - f(0)}{h}
$$
is $1$ and $-1$, respectively, and hence the limit from either side does not exist.  (And note of course that $1$ and $-1$ are the actual slopes of the two lines $y = x$ and $y = -x$.)

Consider for example sequences $c_n$ and $d_n$ going to $\infty$ (or $-\infty$) for which $\sin(c_n) = 1$ and $\sin(d_n) = -1$ hold for all $n$, respectively (as can easily be constructed using the periodicity of the sine function; it's integer multiples of $2\pi$ plus either $\pi/2$ or $3\pi/2$) and then consider the values $h = 1/c_n$ or $1/d_n$ as desired.


问:
Regarding Picard's iteration its relation to numerical methods, such as Euler's method.
I've been told numerical methods in solving ODEs, such as Euler's method and Runge-Kutta, are all in some way approximations to Picard's iteration, and I'm trying to understand how.

Suppose we have a differential equation on an interval $[x_0,x_L]$:

$$\frac{dy}{dx}=f(x,y)$$
with initial condition
$y(x_0)=y_0$

I would like to numerically solve the equation on a set of points $\{x_0<x_1<\dots<x_n\}$, i.e. obtain approximations $y_i$ to the true solution $y(x_i)$ for each $x_i$.

Picard's iteration works as follows:

$$y_{0,0}=y_0$$

$$y_{0,k}(x_1)=y_0+\int_{x_0}^{x_1} f(x,y_{0,k-1}(x))dx \;\; \mathrm{for} \;\; k \geq 1$$

suppose we stop for $k=m$, then take $y_1=y_{0,m}(x_1)$

We then repeat the process for $i \geq 1$.

$$y_{i,0}=y_i$$

$$y_{i,k}(x_{i+1})=y_i+\int_{x_i}^{x_{i+1}} f(x,y_{i,k-1}(x))dx \;\; \mathrm{for} \;\; k \geq 1$$

$$y_{i+1}=y_{i,m}(x_{i+1})$$

So is the idea of a numerical method (e.g. Euler's method) to replace the integral $\int_{x_i}^{x_{i+1}} f(x,y_{i,k-1}(x))dx$ with an approximation such as $(x_{i+1}-x_{i})f(x_i,y_i)$ (Euler's method)? What I don't understand is why numerical methods only iterate once for each point $x_i$ (in other words, $m=1$) but Picard's iteration suggests you should iterate multiple (potentially many) times for each $x_i$?

答:
All these methods do approximate the solution of the ODE. Picard's method is more a theoretical tool. It can be used to obtain approximate analytical solutions, or to prove theorems. As far as I know, it is never used with a numerical integration method, as that would be quite inefficient.

The basic numerical method is indeed Euler's, and Runge-Kutta are improvements that gives better convergence order for smooth functions. (The same way that the rectangle rule is the basic numerical integration method for explicit functions while Newton-Cotes are improvements.)


答:
Alternatively,

$x^2 y^2 (x^2+y^2) - 2 \leq 0\iff x^2y^2(4-2xy) - 2 \leq 0 \iff -2(xy-1)(x^2y^2 - xy -1) \leq 0$

edit.

This is actually what @Albus Dumbledore suggested in the comment section.

Title: prove the inequality $x + y = 2$. Prove that $x^2y^2 (x^2 + y^2)\leq2$

Body: Positive numbers  $x + y = 2$. Prove that $x^2y^2 (x^2 + y^2)\leq2$


问:
When a proposition is satisfiable and valid?
The statement of the problem is confusing me. I don't understand whether it is asking for the definition of satisfiable proposition and valid proposition separately, or it is asking when a proposition is both (satisfiable and valid). Following is the approach of mine for later case.

A statement is satisfiable if it contains a model having truth value
$T$ and a statement is valid if the truth values of all of its
models are $T$. So the the a statement is satisfiable and valid if it
is a tautology.

What should I do? Whether I consider the definitions separately or tautology as the answer to the problem?

答:
There are two definitions:

satisfiable: there is a model (or: interpretation) such that the statement is true in it.

Example: $p \lor q$ is a formula of propositional logic that is satisfied (evaluated to true) by a valuation (or: truth assignment) $v$ such that $v(p)= \text T$.

valid: a statement is valid when it is true in every model.

Example: $p \lor \lnot p$ is a formula of propositional logic that is valid by every valuation.

Obviously, every valid statement (formula) is satisfiable, but not vice versa. Thus

a valid formula is satisfiable and valid.

A tautology (in propositional logic) is a formula that is true in every valuation; thus it is a valid formula.


问:
Integral of product of Gamma, Erfc and Exponential
I can confirm that:
$$\int_{-\infty }^{\infty } \frac{e^{-\frac{K^2}{2}} 2^{-n-\frac{1}{2}} n \Gamma \left(\frac{1}{2},\frac{K^2}{2}\right) \text{erfc}\left(\frac{-K}{\sqrt{2}}\right)^{n-1}}{\pi } \, dK = \frac{1-2^{-n}}{n+1}$$
numerically, but I can't work out the derivation symbolically. Is there a symbolic derivation?

答: Substitute $x=\operatorname{erfc}(-K/\sqrt2)$; then $dx/dK=\sqrt{2/\pi}e^{-K^2/2}$, and $\color{blue}{\Gamma(1/2,z)=\sqrt\pi\operatorname{erfc}\sqrt{z}}$ for $z\geqslant 0$, thus $\Gamma(1/2,K^2/2)=\sqrt\pi\operatorname{erfc}(|K|/\sqrt2)=\sqrt\pi(1-|1-x|)$. The given integral is $$2^{-n-1/2}\frac{n}{\pi}\int_0^2\sqrt\pi(1-|1-x|)x^{n-1}\sqrt{\pi/2}\,dx=\frac{n}{2^{n+1}}\int_0^2 x^{n-1}(1-|1-x|)\,dx.$$ This is evaluated easily (say, by splitting $\int_0^2=\int_0^1+\int_1^2$), and the result is as expected.


问:
Choosing the right probability space for a sample space that contains functions
If I want to make some probability measurements for a sample space that would be, for example, all the continuous functions on the interval $ [0,1] $. Is there a "natural" sigma algebra and probability measure to choose for this specific sample space?

I'll give an example to what I call "natural"

for a finite set sample space, a natural sigma algebra would be the power set, the full sigma algebra, and I guess the natural probability measure would be the uniform measure.

Im intresed in question such "what is the probability of a continuois function to.be differentiable" and Im looking for the suitable probability soace that would make the question well defined.

Thanks in advance.

答:
The space $C([0,1])$ of continuous functions on $[0,1]$ can be equipped with a natural norm, which is the supremum norm (inducing the topology of uniform convergence). That makes $C([0,1])$ into a Banach space (and in fact also a Polish space). Hence I would say that the most "natural" $\sigma$-algebra on $C([0,1])$ is its Borel $\sigma$-algebra $\mathcal{B}_{C([0,1])}$, generated by all the open sets.

However, in the example you gave with differentiability it is not clear whether the set of differentiable function would be a Borel set. (I actually suspect it might not be; similar to the fact that the set of continuous functions is not measurable with respect to the product $\sigma$-algebra $\mathcal{B}_{\mathbb{R}}^{\otimes [0,1]}$ on $\mathbb{R}^{[0,1]}$).

I am also not sure what you would consider to be a "natural" probability measure on $C([0,1])$, at least there probably can't be such a thing as a "uniform" distribution  -- I mean, even if you look at $\mathbb{R}$, is not possible as there is no "uniform distribution" on $\mathbb{R}$...

Edit: I found this interesting answer, which deals with the question about the probability of a function being differentiable anywhere


问:
Help proof by induction
I just got my first induction assignment in a new course. They want me to prove by induction that:
$$\sum_{s=1}^k s*s! = (K+1)!-1.$$

The way I understand induction is that I test for the first value. Then for n and then n+1, to show the given expression is true.
I've done it for 1 and n, however I'm stuck at n+1.
$$\sum_{s=1}^{n+1} s*s!=(n+1)!-1+(n+1)*(n+1)!$$
Using maple I can see the expression is $-1+(n+2)!$ (which is true) however i dont know how to reduce/rewrite $(n+1)!-1+(n+1)(n+1)!$ to $-1+(n+2)!.$$

I've asked my friends and a older student, but to no avail. I'm hoping you guys can help.

答:
Just factor out the common factor $(n+1)!$
$$\begin{align}
\color{blue}{(n+1)!}-1+(n+1)\cdot\color{blue}{(n+1)!} &= \color{blue}{(n+1)!}(1+(n+1))-1 \\
&= \color{blue}{(n+1)!}(n+2) - 1 \\
&= (n+2)!-1\end{align}$$

For the sake of another direct proof, see this answer.

Title: Help proof by induction


问:
Connected sum of a link and a knot
Let $L=K_1 \cup K_2$ be a two-components link in a copy of $S^3$ and let $K$ be a knot, thought in a different copy of $S^3$. In other words, we have two couples $(S^3, L)$ and $(S^3, K)$. Let us define a notion of "connected sum" between objects of this kind: we choose a component of $L$ (for example $K_1$) and two arcs $S \subset K_1$ and $S' \subset K$. We define the connected sum of the couples $(S^3, L)$ and $(S^3, K)$ along $K_1$ as the object $(S^3 \# S^3 \simeq S^3, K \# K_1 \cup K_2)$, where the connected sum of the two copies of $S^3$ is performed along disks which intersect the links in the chosen intervals $S$ and $S'$.

I am not sure that this operation is well defined; it would be nice if it was but I wouldn't bet on that. What I would like to prove is that if $L$ is a split link, then the result of this operation is again a split link. The problems seems to be the fact that the bars used to perform the connected sum of knots could intersect the other component ($K_2$); I think that it could be possible to perform the connected sum of knots "inside" the disks used to perform the connected sum of the two copies of $S^3$. Is it correct? Would it be enough to state the the resulting link is split, under the hypothesis that $L$ is split? Thanks in advance.

P.S. I found this definition of connected sum of links in "Three-dimensional link theory and Invariants of Plane Curve singularities" by Eisenbud & Neumann. This definition can be in fact easily generalized to the connected sums of two links, but things may turn to be more complicated in that case.

答:
As stated this is not a correct definition of the connected sum of two knots/links. This is called a 'band sum'. The band sum is not well-defined: you can obtain many different knots as a result of the band sum of two given knots. (Try to picture this by imagining your band winding around one of the knots a few times before actually connecting up with it.)

The correct definition demands that there be a regular planar projection of the knots so that the projection of this disc only intersects the projection of the knots in the arcs you have already specified.

That is, take two knot diagrams, and do your connect-sum by attaching a band which only intersects the knot diagrams in those arcs. This prevents the "winding" I did above.

I believe but have not checked that the definition you get from this is well-defined (after specifying the component of L, of course); that this is well-defined for knots ought to be in any half-decent knot theory textbook (say Rolfsen). And it should be easy to see that the resulting link is indeed split (choose a diagram so that L is split by a line in the plane, then take K to be on one side of that line and your band to be on one side of that line).

Title: Connected sum of a link and a knot


问:
Inverse Laplace transformation using reside method of transfer function that contains time delay
I'm having a problem trying to inverse laplace transform the following equation

$$
  h_0 = K_p * \frac{1 - T s}{1 + T s} e ^ { - \tau s}
$$

I've tried to solve this equation using the residue method and got the following.

$$
  y(t) = 2 K_p  e ^ {- \tau s} e ^ {-t/T}
$$

$$
  y(t) = 2 K_p e ^ {\frac{\tau}{T}} e ^ {-t/T}
$$

And that is clearly wrong.

Is it so that you can't use the residue method on functions that contains time delay, or is it possible to do a "work around" or something to get to the right answer?

答:
First do polynomial division to simplify the fraction:

$$\frac{1-Ts}{1+Ts}=-1+\frac{2}{1+Ts}$$

Now expand $h_0$:

$$h_0=-K_pe^{-\tau{s}}+2K_p\frac{1}{Ts+1}e^{-\tau{s}}$$

Recall the time-domain shift property:

$$\scr{L}(f(t-\tau))=f(s)e^{-\tau{s}}$$

$$\scr{L}^{-1}h_0=-k_p\delta{(t-\tau)}+2k_p g(t-\tau)$$

Where $g(t)=\scr{L}^{-1}\frac{1}{ts+1}$.

To take the inverse Laplace transform of this term, recall the frequency domain shift property:

$$\scr{L}^{-1}f(s-a)=f(t)e^{at}$$

$$\frac{1}{Ts+1}=\frac{1}{T(s+\frac{1}{T})}=\frac{1}{T}\frac{1}{s+\frac{1}{T}}$$
Therefore the inverse Laplace transform is:

$$\frac{1}{T}e^{-\frac{t}{T}}$$

Finally, putting all of it together, the full inverse Laplace transform of the original expression is:

$$-K_p\delta(t-\tau)+\frac{2K_p}{T}e^{-\frac{t-\tau}{T}}$$


Body:
Let $x=1+k$ and $y=1-k$, for some $k\in(-1,1)$. Then
\begin{align}
x^2y^2(x^2+y^2)&=(1-k^2)^2\times2(1+k^2)\\
&=2(1-k^2)(1-k^4)\\
&\le 2
\end{align}
Further, equality holds iff $k=0$, i.e. $x=y=1$.