问:
Prove that $f(z)$ is a onto group homomorphism?
Given  $G= \{ z \in \mathbb{C} |  \exists \ n \in \mathbb{Z}^{+} \text{such that} \ z^n=1\}$.
Define a map $f : G \to G $ by  $$f(z)=z^k$$

where $k>1$ is fixed and $k \in \mathbb{Z}^+$

My question :  Prove that  $f(z)$ is  a onto group homomorphism ?

My attempt  : No , take  $z^{n/k} \in G$  now $f(z^{n/k})=z^n=1$  for all  $z^{n/k} \in G$

$\implies$  $f$ become  a  constant i,e  $f=1$

we know that constant function never give onto

therefore  $f$ is not a onto homomorphism

答:
The problem comes from the fact that $n$ is not fixed ! For $z \in G$, a priori, the "corresponding $n$" could be anything.

A good idea to start would be to understand what the elements of $G$ look like. In fact, you can try to show (or may be you already know) that if $z^n = 1$, then you can write $z = e^{\frac{2ir\pi}n}$ for some $0 \leq r < n$.

This means that $G = \{e^{\frac{2ir\pi}n}\;|\; n \in \mathbb N^\ast , 0\leq r < n\}$. Now you can work out more precisely how your function acts on this set !


问:
Number of long and short diagonals for an n-sided regular polygon
In reference to a question I got in a test, I asked a teacher how many long and short diagonals were in a 15-sided regular polygon. Considering a long diagonal to be any diagonal with a minimum of 2 vertices in between the vertices joined to make the diagonal, and a short diagonal made such that only 1 vertex lies in between, he replied 15 for both.

My question is can we generalise this result? n long and n short diagonals for an n sided regular polygon? It definitely doesn't work for smaller values of n, but I guess the picture seems to blur at bigger values. If not, what's the correct answer? Please do let me know.

答:
From Finding the diagonals of an n-sided polygon,
we know that an $n$-sided regular polygon ($n \geq 3$) will have $\frac13 n(n-3)$ diagonals altogether.

If $n$ is even, then $n\geq 4$ and from any given vertex, there is one and only one "longest" diagonal that connects to that vertex. We can identify the following subsets of vertices: the $2$ vertices at each end of the diagonal, $\frac12(n-2)$ vertices on one side of the diagonal, and $\frac12(n-2)$ vertices on the other side of the diagonal.

We can then partition the vertices into $\frac n2$ sets of two vertices connected by longest diagonals. There is exactly one longest diagonal for each of these sets of two vertices. Hence there are exactly $\frac n2$ longest diagonals.

For every other diagonal in any regular polygon, there will be more vertices on one side of the diagonal than on the other. Choose $k$ such that $0 < k < \frac12(n-2)$;
then $k < n - 2 - k$ and from any given vertex there are two diagonals that each have $k$ vertices on one side and $n - 2 - k$ on the other.

Each vertex is connected to two such diagonals but each diagonal connects two vertices, so the number of diagonals is the same as the number of vertices.
If you're not familiar with that kind of reasoning,
starting any given vertex traverse the perimeter of the polygon counterclockwise through $k$ other vertices and then stop at the next vertex; construct a diagonal through the starting vertex and the stopping vertex, and this diagonal has exactly $k$ vertices on one side. Each vertex gives you exactly one such diagonal, and every diagonal is produced by exactly one such vertex (you get a different diagonal if you start at the other vertex), so there are exactly as many such diagonals as there are vertices.

For $n \geq 5$ we have $n$ diagonals with one vertex between the ends of the diagonal. (We need $n \geq 5$ because in the case $n = 4$ there are only two such diagonals and for $n < 4$ there are none.)
For $n \geq 7$ we have $n$ diagonals with exactly two vertices between the ends of the diagonal. For $n \geq 9$ we have $n$ diagonals with exactly three vertices between the ends, and so forth.

So, in summary, in a regular $n$-sided polygon, we have the following diagonals:

Altogether there are $\frac12 n(n-3)$ diagonals of all lengths.

If $n$ is even, there are $\frac n2$ longest diagonals.

For any diagonal that is not a longest diagonal of polygon with an even number of sides, there are $n$ diagonals of that length
(where the "length" can be identified by counting the number of vertices on one side of the polygon, choosing the side with the smaller number of vertices).

ContentLicense: CC BY-SA 4.0


问:
Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$
Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?

答:
There is actually a much, much neater method. Following your solutions, we subtract the equations (or substitute $20$) to get,
$$(u-v)+(v^2-u^2)=0$$
Using the familiar identity $a^2-b^2$ $=(a-b)(a+b)$,
$$-(v-u)+(v-u)(v+u)=0$$
which, on factoring, implies
$$(v-u)(-1+v+u)=0$$
which implies
$$v=u\text{ or }v=1-u$$
All that remains is mere substitutions into our original equations.

Hope this helps. Have a wonderful day. Ask anything if not clear :)


答:
You have the following problem: $$\begin{bmatrix}x^2+y^4=20\\ x^4+y^2=20\end{bmatrix}$$

In the first equation, you can isolate $x$ to get $x=\sqrt{20-y^4}$ and $\:x=-\sqrt{20-y^4}$. You then substitute the $x$ values into the second equation to get:
$$\left(\sqrt{20-y^4} \right)^4 +y^2=20, \left(-\sqrt{20-y^4} \right)^4 +y^2=20$$

Simplifying the equations will result in $y=2, y=-2$. You can then substitute the solution into the first equation and you will get $x=2,-2$ depending on which value of $y$ you substitute. Therefore, the final solutions are:

$$\begin{pmatrix}x=2,\:&y=2\\ x=-2,\:&y=2\\ x=2,\:&y=-2\\ x=-2,\:&y=-2\end{pmatrix}$$

Title: Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$

Body: Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2+v+380=0$$ which I can't solve because we haven't studied any methods for solving equations of fourth degree (except $ax^4+bx^2+c=0$). Any other methods for solving the system?


问:
Is node connectivity simply the smallest node degree in the network?
I was reading Network Science with Python and NetworkX, which has the following definition:

The node connectivity is the smallest min-cut over all node pairs. The edge connectivity is defined similarly.

It then proceeds to show that Zachary's Karate Club, as well as several other examples, has a node connectivity of 1. Does this simply means that there exists a node with a single edge (i.e. a degree of 1)?

In general, could the node connectivity ever exceed the smallest degree node in the network? Will it ever be less than that?

答:
I'm going to have a stab at a proof too, please feel free to leave any feedback on this.

Suppose that you had a graph $G = (V, E)$ that is $\kappa$-connected. Let $\delta(G)$ be the minimum degree.

Suppose that $\kappa(G) > \delta(G)$. Now select $v \in V$ that has degree $\delta(G)$. Now remove $\delta(G)$ edges from $v$. By selection, this vertex must now be disconnected, which would mean that $G$ had been disconnected by removing fewer than $k$ edges, which contradicts the claim that $\kappa(G) > \delta(G)$.

To prove that $\kappa(G) \le \delta(G)$, consider the following examples.

First, consider a graph with two nodes with a single undirected edge connecting them. In this case, both nodes have a degree of 1, and the graph would become disconnected by removing the single edge. In this case, clearly $\kappa(G) = \delta(G) = 1$.

Secondly, consider the following graph:

In this case, $\delta(G) = 2$ and $\kappa(G) = 1$, so $\kappa(G) < \delta(G)$. Therefore, it is possible that either $\kappa(G) = \delta(G)$ or $\kappa(G) < \delta(G)$, but it is not possible that $\kappa(G) > \delta(G)$. Therefore, $\kappa(G) \le \delta(G)$.


问:
$\int_\Omega F(u_n)\to0$ implies $\int_\Omega F(au_n)\to 0$ for $a\in [0,\infty)$?
Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $F:[0,\infty)\to [0,\infty)$ be a convex, strictly increasing and continuous function satisfying $F(0)=0$. Suppose that $u_n\in L^\infty(\Omega)$ (Lebesgue measure) is a sequence of non-negative functions satisfying $$\int_\Omega F(u_n)\to0$$

Can I conclude that for every $a\in [0,\infty)$ the following convergence is true: $$\int_\Omega F(au_n)\to 0$$

Remark 1: Convexity and $F(0)=0$ implies that $F(ax)\leq aF(x)$ for $a\leq 1$, hence the only case that needs to be treat is the case $a>1$.

Update 1: In the circumstances described abode, $u_n\to 0$ in measure. Maybe this can help...

Thank you

答:
You cannot conclude that without the extra assumption that $F(a x) / F(x)$ is bounded as $x \to \infty$.

Suppose there is a sequence $\{x_n\}$ such that
$$
\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{F(a x_n)}{F(x_n)} = \infty.
$$
Define $u_n(t) = x_n\chi(t)$ where $\chi$ is the characteristic function of a set of measure $1/F(a x_n)$. Then
$$
\int_\Omega F(u_n) = \frac{F(x_n)}{F(a x_n)} \to 0
$$
but
$$
\int_\Omega F(a u_n) = \frac{F(a x_n)}{F(a x_n)} = 1.
$$
So, as long as $F$ is bounded by a polynomial it will work, but for e.g. $F(x) = e^x - 1$ it fails.

ContentLicense: CC BY-SA 3.0


问:
How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $?
How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta  $?

I have tried substitution of cosine and sine, dividing and multiplying by Sine and cosine, converting everything to tan. It can probably be solved using beta/gamma function but I can't figure it out.

答: Idea: use $\sin^2\theta = 1- \cos^2 \theta$ and work out the polynomial in $\cos \theta$, and apply techniques from here, e.g., or here on this site in a more modern look.


问:
Interpretation of $f(x)(1 + o(1))$
What is an exact interpretation of the expression $f(x)(1 + o(1))$?

$o(1)$ - 'o' here is referring to little-oh notation. I understand that any function $g(x)$ with $o(1)$ refers to the fact that function $g(x)$ asymptotically reaches zero, i.e $lim_{x\to\infty}g(x) \to 0$. But when the expression is embedded in such expression above, I am not sure what to make of it.

How does $f(x)$ gets distributed to the expression inside? Is it $f(x) + f(x)o(1)$ or something else?

What does it really mean when you multiply $o(1)$ to some function? Back to my original question.

答:
First the notation only makes sense if you add where the Landau notation $o$ is taken. Assumes that it is at $x_0 \in D$ where $D$ stands for the domain of $f$. Then it means that it exists a map $\epsilon: D \to \mathbb R$ such that $\lim\limits_{x \to 0} \epsilon(x)=0$ where $g(x) = f(x)(1 + o(1))$ is equal to

$$g(x) = f(x) (1+\epsilon(x-x_0)).$$

And if you're considering a neighborhood of $\infty$, then you have $g(x) =f(x)(1+\epsilon(x))$ where $\lim\limits_{x \to \infty} \epsilon(x)=0$.

Title: Interpretation of $f(x)(1 + o(1))$


问:
Solving differential equations with trigonometric functions using laplace
Can this DE be solved using Laplace transform?

$$
L\;\frac{d^2\theta}{dt^2} + A\cos(\theta) +g\sin(\theta) = 0
$$

where g , A , L are constants

答:
No, the Laplace transform works for linear equations.

You can handle your equation as
$$\dot\theta\ddot\theta+\dot\theta(a\cos\theta+b\sin\theta)=0$$
and by integration
$$\dot\theta^2=2a\sin\theta-2b\cos\theta+c.$$

This separable equation leads you to an elliptic integral, then to a Jacobi function, something not too exotic. (Note that by adding a suitable constant to $\theta$, you can let one of the sine or cosine term vanish.)


问:
Verify that integral satisfies boundary conditions
Consider $$u(x_0,y_0,z_0)=\frac{z_0}{2\pi}\int_{\mathbb R^2}\frac{f(x,y)}{\left[(x-x_0)^2+(y-y_0)^2+z_0^2\right]^{3/2}}\,dx\,dy.$$
Show that
(i) $u\to f(x_0,y_0)$ in the limit as $z_0\to0$ and (ii) $u\to0$ in the limit as $x_0^2+y_0^2\to\infty$.

Some motivation: this form of $u$ is the solution to the Laplace equation $\nabla^2u=0$ on the upper half-space $z>0$, subject to the boundary conditions $u(x,y,0)=f(x,y)$ and $u\to0$ as $x^2+y^2\to\infty$. This can be obtained e.g. by the reflection method. I want to verify directly that this form of $u$ does indeed satisfy those boundary conditions, but am struggling with the computation.

I have been given as a hint the substitution $z_0^2t^2=(x-x_0)^2+(y-y_0)^2$, but I'm not really sure how to use this?

答:
The statement (i) can be proven following Evan's book approach (Partial Differential Equations - page 38). I will assume that the function $f$ is continuous and bounded in $\mathbb{R}^2$.

In this post, it is shown that the Poisson kernel in the Half-space given by
$$ K(x,y) = \frac{2x_n}{n \alpha(n)} \frac{1}{|x-y|^n} \hspace{1 cm} (x \in \mathbb{R}^{n}_{+}, \hspace{0.3 cm} y \in \partial \mathbb{R}^{n}_{+})$$
(where $n\alpha(n)$ is the surface area of the unit sphere in $\mathbb{R}^n$) satisfies the relation
$$ \int_{\partial \mathbb{R}^{n}_{+}} K(x,y) dy = 1.$$
This can be used to prove the desired result. Indeed,  if you fix $(x_0,y_0)$ in $\mathbb{R}^2$ and $\epsilon >0$, we may choose $\delta >0$ so that
\begin{equation}
\label{2}|f(x,y)-f(x_0,y_0)|< \epsilon \hspace{0.5 cm} \text{if} \hspace{0.5 cm} |(x,y)-(x_0,y_0)|< \delta. \hspace{1 cm } (1)
\end{equation}

If we take $|(x,y) - (x_0,y_0)|< \frac{\delta}{2}$, then
\begin{eqnarray*}
|u(x,y,z) - f(x_0,y_0)| &=& |\int_{\mathbb{R}^2} K(x,y,z,w)f(w) dw -f(x_0,y_0)  | \\
&\leq& \int_{\mathbb{R}^2} K(x,y,z,w) |f(w)-f(x_0,y_0)| dw \\
&=& \int_{\mathbb{R}^2 \cap B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw \\
&+& \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw. \\
\end{eqnarray*}
Now, by (1) we have
$$ \int_{\mathbb{R}^2 \cap B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw \leq \epsilon.$$
On the other hand, if $|(x,y)-(x_0,y_0)| < \frac{\delta}{2}$ and $|w-(x_0,y_0)|> \delta$,
we have
\begin{eqnarray}
|w-(x_0,y_0)| &\leq& |w-(x,y)|+|(x,y)-(x_0,y_0)|\\
&\leq &  |w-(x,y)| + \frac{\delta}{2}\\
&<& |w-(x,y)|+ \frac{1}{2}|w-(x_0,y_0)|
\end{eqnarray}
Thus, we obtain $|w-(x,y)|> \frac{1}{2}|w-(x_0,y_0)|$ and therefore

\begin{eqnarray}
\int_{\mathbb{R}^2 - B((x_0,y_0),\delta))}K(x,y,z,w) |f(w)-f(x_0,y_0)| dw &\leq& 2||f||_{\infty} \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))}K(x,y,z,w) dw \\
&\leq & C\frac{ z}{2 \alpha(2)} ||f||_{\infty} \int_{\mathbb{R}^2 - B((x_0,y_0),\delta))} \frac{dw}{|w-(x_0,y_0)|^{2}}
\end{eqnarray}
and this last integral tends to zero when $z \longrightarrow 0$, which shows that $u \to f(x_0,y_0)$ for any $(x_0,y_0) \in \mathbb{R}^2$.

I think that statement (ii) is not true. You may take, for instance, $f(x,y)=1$ and then,
$$u(x,y,z) = \int_{\mathbb{R}^2} K(x,y,z,w) f(w) = \int_{\mathbb{R}^2} K(x,y,z,w) = 1$$
and this function doesn't satisfy the required condition at the limit.


问:
Solve $\sqrt{3}\tan\theta=2\sin\theta$
I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.

$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$

$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$

I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.

Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.

Where am I going wrong please?

答: Why square the equation? $\cos \theta = \frac{\sqrt 3}{2}$ is easier to solve. Also, include the solutions of $\sin \theta = 0$ before canceling $\sin \theta$.</p>

Title: Solve $\sqrt{3}\tan\theta=2\sin\theta$


问:
Upper and Lower Triangular Matrices
Given the matrix A=$ \left( \begin{array}{ccc}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8\\
1 & -1 & 2 & 3 \\
2 & 1 & 1 &2\end{array} \right) $, write it in the $L_{4 \times 4}U_{4 \times 4}$, where L is the lower triangular matrix and U is the upper triangular matrix.

To be honest, I don't even understand what the question is asking of me, however I do know what upper and lower triangular matrices are.

Thanks in advance for your help.

答:
As @Julien pointed out, every square matrix admits a $PLU$ decomposition, where $P$ is a permutation matrix. We have: $A = P \cdot L \cdot U$, such that:

$A=\begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\0 & 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 & 0\\1 & 1 & 0 & 0\\2 & 1 & 1 & 0\\5 & \dfrac{4}{3} & \dfrac{5}{3} & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 & 4\\0 & -3 & -1 & -1\\0 & 0 & -4 & -5\\0 & 0 & 0 & -\dfrac{7}{3} \end{bmatrix}$

You could try manually cranking this one to find its $LU$ factorization. We want:

$L \cdot U = \begin{bmatrix} 1 & 0 & 0 & 0\\l_{21} & 1 & 0 & 0\\l_{31} & l_{32} & 1 & 0\\l_{41} & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} u_{11} & u_{12} & u_{13} & u_{14}\\0 & u_{22} &u_{23} & u_{24}\\0 & 0 & u_{33} & u_{34}\\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}$

We start off by solving the first row, so we get:

$$u_{11} = 1, u_{12} = 2, u_{13} = 3, u_{14} = 4$$

The portion of the multiplication that determines the remaining entries in the first column of $A$ yields:

$$l_{21}u_{11} = 5 \rightarrow l_{21} = 5$$

$$l_{31}u_{11} = 1 \rightarrow l_{31} = 1$$

$$l_{11}u_{11} = 2 \rightarrow l_{41} = 2$$

At this point rewrite all the variables you solved for and then continue the process and see if you can solve the remaining variables. Of course it is easy to check the result if you can solve all of the equations.

So, we currently have:

$L \cdot U = \begin{bmatrix} 1 & 0 & 0 & 0\\5 & 1 & 0 & 0\\1 & l_{32} & 1 & 0\\2 & l_{42} & l_{43} & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 & 4\\0 & u_{22} &u_{23} & u_{24}\\0 & 0 & u_{33} & u_{34}\\0 & 0 & 0 & u_{44} \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 & 4 \\5 & 6 & 7 & 8\\1 & -1 & 2 & 3 \\2 & 1 & 1 &2 \end{bmatrix}$

Try solving for $u_{22},u_{23}, u_{24}$, and then $l_{32}, l_{42}$ and continue this process.

Title: Upper and Lower Triangular Matrices


问:
Even Numbers with Set Notation
I am thinking about the many ways to write even numbers for set notation.

I found one that I havent been able to find any confirmations online.

$$\{x\vert x/2 \in\mathbb{Z}\}$$ Is this too simple of an answer?

I did find the other ways to solve but tried simplifying it.

答:
Is this too simple of an answer?

It's not that it's too simple, but it's a little unnatural: For instance, if I extended division by $2$ to apply to some shapes so that, say $\triangle/2=-17$, then it would seem that $\triangle$ should be in your set $\left\{x\mid x/2\in\mathbb Z\right\}$. So your notation requires an underlying assumption that there aren't any weird definitions floating around.

It's a better habit to clearly write where each variable lives, as in $\left\{2k\mid k\in\mathbb Z\right\}$ (common) or $\left\{x\in\mathbb Q\mid x/2\in\mathbb Z\right\}$ (not common, but avoids the problem I mentioned above).

You can see a couple of different-looking ways the set of even naturals might be written in set builder notation at the Examples subsection of the English Wikipedia page for "set-builder notation".

Title: Even Numbers with Set Notation


问:
Is there a closed form for this recursion equation?

Given
$$f(a)=\frac{1}{6} \sum _{i=1}^6 \left[1-(1-f (a-1)) \left(\frac{i}{6}\right)^a\right]$$
with $f(1)=0$, $a\in\mathbb{N}$ and $f:\mathbb{N}\to\mathbb{Q},$

is there a way to write a closed form solution for the above recursion function. I am conducting a research and obtained the above result, but was interested if a function of $f(a)$ only in terms of $a$ exists, for arbitrary $a$. This would save a lot of computational work for my research. This in fact, is a small result which will be used in more complex calculations, therefore a closed form would be optimal. Thank you.

答:
We have to study the recursion

$$f(a) = \frac{1}{6} \sum_{i=1}^{6}\left( 1-\left( 1-f(a-1) \right)(\frac{i}{6})^a  \right)\tag{0}$$

Letting

$$g(a) = \frac{1}{6}\sum_{i=1}^{6} (\frac{i}{6})^a\tag{1}$$

the recursion reads

$$f(a) = 1-g(a) + g(a) f(a-1), f(1) = 0, a=2, 3, ...\tag{2}$$

Is is a linear inhomogenous recursion relation which can be solved by standard methods.

This means that the general solution is gven by the sum of the general solution to the homogeneous equation and one special solution to the inhomogeneous equation.

Now the homogenous equation

$$f(a) = g(a) f(a-1)\tag{3}$$

is easily solved generally by

$$f_h(a) = c \prod_{m=1}^{a-1} g(1+m)\tag{4}$$

here $c$ is an arbitrary constant.

We can find a special solution to the inhomogeneous equation $(2)$ using a trick called variation of the constant ($c$) (which can be easily found on the internet) applied to the expression $(4)$

Doing this carefully the solution is finally found to be

$$f_{s}(a) = \left(\prod _{m=0}^{a-1} g(m+1)\right) \left(\sum _{n=-1}^{a-1} \frac{1-g(n+1)}{\prod _{m=0}^n g(m+1)}-\frac{5}{7}\right)\tag{5}$$

Shifting indices and noticing that $g(0)=1$ this can be written more elegantly as

$$f_{s}(a) = \left(\prod _{m=1}^{a} g(m)\right) \left(\sum _{n=1}^{a} \frac{1-g(n)}{\prod _{m=1}^{n} g(m)}-\frac{5}{7}\right)\tag{5a}$$

Expression $(5a)$ is similar to that of Raffaele.

To derive $(5)$ we have used that

$$g(0) = 1, g(1) = \frac{7}{12}\tag{6}$$

Checking $f_{s}(1)=0$:

The second factor in $f_{s}(a)$ vanishes because

$$\sum _{n=-1}^{a-1} \frac{1-g(n+1)}{\prod _{m=0}^n g(m+1)}=\left(\frac{1-g(n+1)}{\prod _{m=0}^n g(m+1)}\text{/.}\, n\to -1\right)+\left(\frac{1-g(n+1)}{\prod _{m=0}^n g(m+1)}\text{/.}\, n\to 0\right)=0+\frac{5}{7}$$

Checking $f_{s}(2) = \frac{125}{216}$: OK

In fact from the original recusion $(0)$ we get

$$f(2)=\frac{1}{6} \left(-\frac{1}{36} 41 (1-f(1))+\frac{7}{18} (f(1)-1)+f(1)+5\right)\text{/.}\, f(1)\to 0=\frac{125}{216}$$

which conincides with $f_{s}(2)$.

Discussion

As noticed in the solution of Jean-Claude Arbaut my formula $(5a)$ can be transformed into this much simpler expression:

$$f(a) = 1 - \prod_{k=2}^{a} g(k)\tag{5b}$$

Noticing also that we can write

$$g(k) = \frac{1}{6^{k+1}} H_6^{(-k)}\tag{5c}$$

where $H_n^{(m)}$ is the harmonic number of order $m$ we get from $(5b)$ that

$$f(a) = 1-6^{\frac{1}{2} \left(-a^2-3 a+4\right)} \prod _{k=2}^a H_6^{(-k)}\tag{5d}$$

so that $f$ is expressed in terms of known (but higher trancendental) functions.


答:
From Numbra answer

Motive : to show  $f$ is onto

$G = \{e^{\frac{2ir\pi}n}\;|\; n \in \mathbb N^\ast , 0\leq r < n\}$.

Now let $f $ is given by $e^{\frac{2ir\pi}n} \in G \implies a=e^{\frac{2ir\pi}{kn}}\in G$

according to question  $f(a)=a^k=e^{\frac{2ir\pi}n}$

so  $f$ is onto

Title: Prove that $f(z)$ is a onto group homomorphism?

Body:
Given  $G= \{ z \in \mathbb{C} |  \exists \ n \in \mathbb{Z}^{+} \text{such that} \ z^n=1\}$.
Define a map $f : G \to G $ by  $$f(z)=z^k$$

where $k>1$ is fixed and $k \in \mathbb{Z}^+$

My question :  Prove that  $f(z)$ is  a onto group homomorphism ?

My attempt  : No , take  $z^{n/k} \in G$  now $f(z^{n/k})=z^n=1$  for all  $z^{n/k} \in G$

$\implies$  $f$ become  a  constant i,e  $f=1$

we know that constant function never give onto

therefore  $f$ is not a onto homomorphism


答:
Let $\theta =2t $ and use the identities $1+\cos2t=2 \cos^2 t,\>
\sin2t = 2\sin t \cos t$
to rewrite the integral as
$$I=\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4d\theta = 128\int_0^{\pi/2} (\cos^{10}t - \cos^{12} t )dt$$
Then, apply the recursive result
\begin{align}
K_n= \int_0^{\pi/2} \cos^ntdt = \frac{n-1}n K_{n-2},\>\>\>\>\>K_0=\frac\pi2
\end{align}
to arrive at

$$I=128 (K_{10}- K_{12} )=\frac{32}3K_{10}=\frac{32}3\frac9{10}\frac78\frac56\frac34\frac12K_0=\frac{21}{16}\pi$$

Body:
How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta  $?

I have tried substitution of cosine and sine, dividing and multiplying by Sine and cosine, converting everything to tan. It can probably be solved using beta/gamma function but I can't figure it out.


Title: Is there a closed form for this recursion equation?

Body:
Given
$$f(a)=\frac{1}{6} \sum _{i=1}^6 \left[1-(1-f (a-1)) \left(\frac{i}{6}\right)^a\right]$$
with $f(1)=0$, $a\in\mathbb{N}$ and $f:\mathbb{N}\to\mathbb{Q},$

is there a way to write a closed form solution for the above recursion function. I am conducting a research and obtained the above result, but was interested if a function of $f(a)$ only in terms of $a$ exists, for arbitrary $a$. This would save a lot of computational work for my research. This in fact, is a small result which will be used in more complex calculations, therefore a closed form would be optimal. Thank you.

Body:
Let's call $\varphi(k)=6^{-k-2} \left(1+2^{k+1}+2^{2 k+2}+3^{k+1}+5^{k+1}+6^{k+1}\right)$

and $\gamma(h)=6^{-h-2} \left(1+2^{h+1}+2^{2 h+2}+3^{h+1}+5^{h+1}-5\ 6^{h+1}\right)$

Then we have

$$f(a)= \left(\prod _{k=1}^{a-1} \varphi(k)\right) \sum _{h=0}^{a-1} \left(-\frac{\gamma(h)}{\prod _{k=1}^h \varphi(k)}\right)-\frac{5}{12} \prod _{k=1}^{a-1} \varphi(k)$$

first values are
$$\left(0,\frac{125}{216},\frac{26645}{31104},\frac{231720479}{241864704},\frac{3720223313533}{3761479876608},\ldots\right)$$


问:
Why does the condition $|a_{n+1}-a_n|\le1/n$ not imply the sequence is Cauchy?
While going through this condition I noticed that the sequence $a_n=\sum_{n=1}^{\infty}1/n$ satisfies
$|a_{n+1}-a_n|\le1/n$ $\forall n\in$ $\mathbb{N}$. However $a_n$ is not a Cauchy sequence, and with this I had established a counter-example for the condition. But ultimately I failed to understand why such a thing happened even though the difference of the terms were bounded by $1/n$, and $1/n$ is a sequence whose terms condense to a point as $n$ tends to $+\infty$(essentially the idea behind Cauchy Sequences).
Also how can I check if the following conditions imply "Cauchyness":
1.$|a_{n+1}-a_n|\le(n+1)^n/n^{3/2+n}$,
2.$|a_{n+1}-a_n|\le (n\log\ n)/e^n$

答:
Cauchy sequences are about $|a_m-a_n|$ being arbitrarily small if $m$ and $n$ are both large enough

But in this case you have  $a_{2n}-a_n =\sum\limits_{j=n+1}^{2n}\frac1j \ge n \frac1{2n} = \frac12$, which does not get arbitrary small.


问:
Proving the Integral is holomorphic
It is an integral from a note.

We concerend conformal map on $\mathbb{C}_+=\left\{ z:\text{Im}z>0 \right\} $ which satisfies $$
\text{Im}f\left( z \right) >0,\text{Im}f'\left( z \right) \ne 0\qquad on~\mathbb{C}_+
$$
A such example is given that
$$
f_{\alpha}\left( z \right) =\int_0^{\infty}{\left( z+t \right) ^{\frac{1}{2}}\left( 1+t \right) ^{-\alpha -2}}dt
$$
for $\alpha>-1/2$.

And I want to show this integral is meaningful. So we should prove two things.One is that it is convergent,and the other is holormorphic.The latter one is easy by Fubini theorem and Morera theorem.However,I don't know how to show this integral is convergent,for it is hard for me to find an estimation.And I want to split $\mathbb{C}_+$ but failed. Besides,I think that $\sqrt{z+t}$ has no bounds,and the integral seems not to be convergent?

答:
Assuming that $z$ doesn't belong to the real interval $(-\infty,0)$, the map $g_{\alpha,z}(t) = \left( z+t \right) ^{\frac{1}{2}}\left( 1+t \right) ^{-\alpha -2}$ is continuous on $[0, \infty)$ and

$$\vert g_{\alpha,z}(t) \vert \le \left(\vert z \vert+t \right) ^{\frac{1}{2}}\left( 1+t \right) ^{-\alpha -2} = \left(\frac{\vert z \vert+t}{(1+t)}\right)^{1/2}\left( 1+t \right) ^{-\alpha -3/2}$$

As $\lim\limits_{t \to \infty} \left(\frac{\vert z \vert+t}{(1+t)}\right)^{1/2} = 1$ and $\alpha +3/2 \gt 1$, the integral is convergent.

Title: Proving the Integral is holomorphic


问:
First derivative of a quadratic function
I have the following question about the derivative.

Find out the quadratic function, which takes the value $41$ at $x=-2$ and the value $20$ at $x=5$ and is minimized at $x=2$ . Calculate the minimum value of this function.

If the function is minimized at $x=2$, then $f'(2) = 0$.

My question is: the derivative of the quadratic is a line tangent to the function or the slope of the function for a given $x$? Then the derivative equals $0$ because it is minimized at the vertex and at the vertex there is no slope?

This is not really about the question above but i just want to be sure if that is the case.

答:
Given

$f(x) = Ax^2 + Bx + C.$

$f(-2) = 41.$

$f(5) = 20.$

$f(x)$ achieves a minimum at $x = 2.$

To do:

Determine $A,B,C.$

Determine $f(2).$

$f'(x) = 2Ax + B$ and $f''(x) = 2A.$

From the constraint that $f(x)$ has a minimum at $x=2,$ you conclude that $f'(2) = 0$ and $f''(2) > 0.$

This implies that

$$A > 0 ~\text{and that}~ B = (-4A).\tag1$$

At this point it is well to take a step back and realize that the constraint re the location (i.e value of $x$) that generates the minimum value of $f(x)$ is absolutely independent of $C$.  The only effect of the term $C$ is to translate the graph of the function $f(x) ~:~C$  units (either up or down).  Therefore, from an intuitive viewpoint, equation (1) above exhausts the info re $f$ has a minimum at $x=2$.

Therefore, there is no choice re continuing the problem.  Coefficient $B$ has been solved (re $B = -2A$) and you are forced to use the other premises (i.e. Given's) to solve for $A$ and $C$.  Once this is done, then $f$ will be completely determined, so the computation of $f(2)$ will then be straightforward.

$$A(5^2) + (-4A)(5) + C = 20 \implies 5A + C = 20.\tag2$$

$$A([-2]^2) + (-4A)(-2) + C = 41 \implies 12A + C = 41.\tag3$$

Subtracting equation (2) from equation (3) above, you have that

$$A = 3 \implies \{C = 5 ~\text{and}~ B = (-12)\}.$$


答:
There is another solution using the "reverse" of multiple angle formulae. In your case
$$(1-\cos^2(t))(1+\cos(t))^4=$$ $$\frac{3 \cos (t)}{2}-\frac{15 \cos (2 t)}{32}-\frac{5\cos (3 t)}{4} -\frac{13\cos (4 t)}{16}
   -\frac{\cos (5 t)}{4} -\frac{\cos (6 t)}{32} +\frac{21}{16}$$

Have a look here.

Title: How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $?


问:
Prove by mathematical induction: $x^n - y^n$ is divisible by $x - y$, for all positive integers $n$.
Here is what I have so far:

Base: When $n=1$, we have $x^1 - y^1 = x - y$. Hence, P(1) is true.

Inductive hypothesis: We assume that P(k) is true: $x^k - y^k$ is divisible by  $x - y$. That is: $x^k-y^k=(x - y)z$, for some integer $z$

Inductive step: We show that P(k+1) is true. That is:

$x^{k+1} - y^{k+1} = (x - y)z$

I don't quite know where to go from there.

答:
The induction hypothesis should be “$x-y$ divides $x^k-y^k$”, not that they're equal.

Hint: $x^{k+1}-y^{k+1}=x^{k+1}-xy^{k}+xy^k-y^{k+1}$

Alternate hint: from the induction hypothesis you have $x^k-y^k=(x-y)z$, so $y^k=x^k-(x-y)z$ and therefore
$$
x^{k+1}-y^{k+1}=x^{k+1}-x^ky+(x-y)yz
$$


问:
Evaluate the following integral involving $\sin \pi x$

Let $F: \Bbb{R} \to \Bbb{R}$ be defined by $$F(s)=\begin{cases}1, & \text{if }s\ge \dfrac12 \\[0.2cm]0, & \text{if }s< \dfrac12 \end{cases}$$
  I need to evaluate $$\int^{1}_{0} F(\sin \pi x) dx\,$$

I noticed that $\sin (\pi x)$ is greater than $\frac{1}{2}$ for $\frac{1}{6}\le x \le \frac{5}{6}$ thus integral reduces to $\int^{\frac{5}{6}}_{\frac{1}{6}} 1 dx=\frac{2}{3}$. Is it okay?

答:
Let's write the function $F(\sin\pi x)$ like this
$$F(\sin\pi x)=
\begin{cases}1, & \text{if }\sin\pi x\ge \dfrac12 \\[0.2cm]0, & \text{if }\sin\pi x< \dfrac12 \end{cases}$$
Now we solve the inequality $\sin\pi x \ge \frac12$. We only care about the solutions in the range $[0;1]$. You already did it.  It holds for $x \in [\frac16;\frac56]$. So we can rewrite the function $F(\sin\pi x)$ like this:
$$F(\sin\pi x)=
\begin{cases}1, & \text{if } x \in \big[\frac16;\frac56\big] \\[0.2cm]0, & \text{if } x \in \big[0;\frac16\big)\cup\big(\frac56;1\big] \end{cases}$$
Now we can split the integral to two parts, one for each case of this function:
$$
\int_0^1 F(\sin\pi x)dx = \int_{[0;1]} F(\sin\pi x)dx =\\= \int_{\big[\frac16;\frac56\big]}F(\sin\pi x)dx + \int_{\big[0;\frac16\big)\cup\big(\frac56;1\big]}F(\sin\pi x)dx = \int_{\big[\frac16;\frac56\big]}1dx + \int_{\big[0;\frac16\big)\cup\big(\frac56;1\big]}0dx
$$
If you aren't familiar with this notation, it is the same as:
$$
\int_\frac16^\frac561dx+\left(\int_0^\frac160dx+\int_\frac56^10dx\right)
$$
The second term is zero, so only the first term remains. And you see that your result was correct.


问:
Derivative of Associated Legendre polynomials at $x = \pm 1$
I'm creating meshes for spherical harmonics, and I need a normal at a given point. Whenever I'm at the poles, $\cos{\theta} = \pm 1$, and I do not know how to find the derivative there. All the formulas I have found to describe the derivative have an $1 - x^2$ in the denominator, and I get an indeterminate form. For reference, the one I'm using is:
$$(P_\ell^m)^\prime(x) = \frac{\sqrt{1-x^2} P_\ell^{m+1}(x) + mx P_\ell^m (x)}{x^2 - 1}$$

I found the derivatives for some cases, and it seems that $m = \pm 1$ results in $\pm \infty$, $m = 0$ yields triangular numbers, and $|m| \ge 3$ makes the derivative $0$. But I can't find an overarching pattern or algorithm I can use to produce these. Is there a nice way?

答:
Could you try using the formula from Wikipedia ?

\[  P_m^l (x) = \frac{(-1)^m}{2^\ell \ell!} (1-x^2)^{m/2} \frac{d^{\ell + m}}{dx^{\ell + m}}(1 - x^2)^{\ell} \]

Let $y = 1-x \approx 0$,

\[  P_m^l (y) = \frac{(-1)^m}{2^\ell \ell!} y^{m/2}(2-y)^{m/2} \frac{d^{\ell + m}}{dy^{\ell + m}} y^{\ell}(2-y)^\ell \propto y^{m/2} \approx 0\]

I am guessing the derivative is always zero unless $m = 0$, the only interesting case.

Or plug in $x = \cos \theta$ and take the limit $\theta \to 0$:

\[  (P_\ell^m)^\prime(\cos \theta) = \frac{\sin \theta \, P_\ell^{m+1}(\cos \theta) + m \cos \theta \,  P_\ell^m (\cos \theta )}{\sin^2 \theta} \]

Not sure how to evaluate this... spherical harmonics are rotationally symmetric, so maybe they shouldn't have non-zero derivatives at the poles?


问:
Determine if $1.001^n$ will grow faster than $n^{10}$.
We know that $n^k \leq k^n$, where $n^k$ is a polynomial and $k^n$ is an exponential function. However, what if we have a case such as $n^{10}$ and $1.001^n$. How can I determine which function will grow faster?

Background - this is a problem from my complexity class and specifically the topic about the Big O-Notation.

答:
We wish to determine if:

$$\lim_{n\to\infty}\frac{n^{10}}{1.001^{n}} < 1$$

Using l'Hopital's rule once:

$$\lim_{n\to\infty}\frac{n^{10}}{1.001^{n}} = \lim_{n\to\infty}\frac{10n^{9}}{\ln(1.001)1.001^{n}}$$

Twice:

$$=\lim_{n\to\infty}\frac{10n^{9}}{\ln(1.001)1.001^{n}} = \lim_{n\to\infty}\frac{10\cdot 9n^{8}}{\ln(1.001)^{2}1.001^{n}}$$

All the way until ten times:

$$ = \lim_{n\to\infty}\frac{10\cdot 9\cdot 8\cdots 2\cdot 1}{\ln(1.001)^{10}1.001^{n}} = \lim_{n\to\infty}\frac{\text{constant}}{1.001^{n}} = 0$$

Because the exponential grows without bound. Thus, $\boxed{1.001^{n}\text{ grows faster than }n^{10}.}$ Of course, this can be generalized to say that any function showing exponential growth will eventually be larger than any polynomial.


问:
Kernel of morphism out of direct sum
Let $\mathcal{C}$ be an abelian category with objects $A,B,C \in \mathcal{C}$ for which we have a surjection $f:A\rightarrow C$ and an arbitrary morphism $g:B\rightarrow C$. Consider
\begin{align*}
h:A\oplus B \stackrel{f+g}{\longrightarrow} C
\end{align*}
with $\ker(h)= \{(a,b) \in A \oplus B \mid f(a)=-g(b)\}$.
Intuitively I thought $\ker(h)=\ker(f)\oplus B$.

As $f$ ist surjective $f^{-1}(g(b))\neq \emptyset$ for every $b\in B$. Hence we have a surjection
\begin{align}
\ker(h) \stackrel{\phi}{\rightarrow} B \\
(a,b) \mapsto b
\end{align}
with $\ker(\phi)=\ker(f) \oplus (0)\cong \ker(f)$. Therefore we have a short exact sequence
\begin{equation*}
0 \rightarrow \ker(f) \rightarrow \ker(h) \rightarrow B \rightarrow 0.
\end{equation*}
But so far I couldn't find any general splitting for this sequence. Is my intuition wrong?

答: $A= \mathbb{Z}, B = \mathbb{Z}/2\mathbb{Z}. C = \mathbb{Z}/4\mathbb{Z}, f(n) = n\,(\mathrm{mod}\,4), g(m) = 2m$ gives a counterexample: the kernel of $h = f \oplus g$ is $\{(2k, k\,\mathrm{mod}\,2)\,|\, k\in \mathbb{Z}\}\simeq \mathbb{Z}$.

Title: Kernel of morphism out of direct sum


问:
Interesting Logic
At a basketball fair, everyone is a fan of just one team (Celtics, Lakers, or 76ers). Celtics fans always lie to Lakers fans, Lakers fans always lie to 76ers Fans, and 76ers Fans always lie to Celtics fan. Other than this, everyone always tells the truth. Everyone else has been going to this fair for 20 years now so they all know which team others support. This is my first time at the fair so I don't know anybody's affiliations, however, because I am wearing a Celtics shirt, everybody else knows that I am a Celtics fan.

What question should I ask someone to determine if they are a Lakers fan?

Let's say you overhear a conversation between two people. Does a single statement exist such that upon overhearing it, I can determine whether the two people support the same basketball team?

答:
For the first question, you can ask whether the person in question is a Laker-or-76er fan. If it is a Celtic fan, he answers no and you know that he isn't a Laker-or-76er fan. If it is a 76er fan, he will lie and tell that he isn't a Laker-or-76er fan, so he tells you he's a Celtic fan essentially. If you ask the question to a Laker, he'll answer positive, and hence you know for sure that he is a Laker.

As for the second question, there does not exist a single statement such that we always know whether the two are fans of the same club or not. If two lakers fans are chatting and we overhear some statement $\varphi$ such that we immediately know they are fans of the same club, then this does have nothing to do with the two actually being fans of the same club, but only with our ability to do some logical deduction, hence we will always deduce the same conclusion. Or, as summarized in Lulu's comment; every sentence has only one interpretation.

Title: Interesting Logic


答:
Let's go step by step.  We want an equation of the following form:

$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}\star&0&0&0\\\star&\star&0&0\\\star&\star&\star&0\\\star&\star&\star&\star\end{pmatrix}\begin{pmatrix}\star&\star&\star&\star\\0&\star&\star&\star\\0&0&\star&\star\\0&0&0&\star\end{pmatrix}$$

From the first column and first row of our known matrix, it's not too hard to see that we can start with this:

$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&\star&0&0\\1&\star&\star&0\\2&\star&\star&\star\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&\star&\star&\star\\0&0&\star&\star\\0&0&0&\star\end{pmatrix}$$

Next, we can choose the diagonal elements of our upper triangular matrix to be $1$, and fill in the lower triangular matrix column by column:

$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&-4&0&0\\1&-3&\star&0\\2&-3&\star&\star\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&1&\star&\star\\0&0&1&\star\\0&0&0&1\end{pmatrix}$$

$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&-4&0&0\\1&-3&5&0\\2&-3&1&\star\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&1&2&\star\\0&0&1&\star\\0&0&0&1\end{pmatrix}$$

$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\1&-1&2&3\\2&1&1&2\end{pmatrix}=\begin{pmatrix}1&0&0&0\\5&-4&0&0\\1&-3&5&0\\2&-3&1&7/5\end{pmatrix}\begin{pmatrix}1&2&3&4\\0&1&2&3\\0&0&1&8/5\\0&0&0&1\end{pmatrix}$$

Title: Upper and Lower Triangular Matrices

Body:
Given the matrix A=$ \left( \begin{array}{ccc}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8\\
1 & -1 & 2 & 3 \\
2 & 1 & 1 &2\end{array} \right) $, write it in the $L_{4 \times 4}U_{4 \times 4}$, where L is the lower triangular matrix and U is the upper triangular matrix.

To be honest, I don't even understand what the question is asking of me, however I do know what upper and lower triangular matrices are.

Thanks in advance for your help.


问:
if $f(a_n)=(-1)^n \cdot n$ then $f'(x)$ returns any real value
Let $a_n$ be a positive sequence monotonic decreasing so $a_n\to 0$. Let $f$ has  a derivative function in $(0,\infty)$ so for every $n\in\mathbb{N}$ (without zero) we get $f(a_n)=(-1)^n \cdot n$. Prove or disprove: $f'(x)$ returns any real value.

I tried to build $f(x)$. If $a_n=1/n$ then $a_n$ be a positive sequence monotonic decreasing so $a_n\to 0$. So we want $f(1/n)=(-1)^n\cdot n$ so $f(n)=(-1)^{1/n}\cdot \frac{1}{n}$ so we get $f(x)=(-1)^{\frac{1}{x}}\cdot \frac{1}{x}$. But then $f$ does not have a derivative function in some points of $(0,\infty)$. How to solve it?

As I understand, to prove that $f'(x)$ gets any real number, I need to show $\lim_{x\to\infty} f'(x)=\infty$ and $\lim_{x\to-\infty} f'(x)=-\infty$.

答: You can use the mean value theorem and Darboux's theorem. The first one shows that there exists $c_n \in (a_{n+1}, a_{n})$ where $|f'(c_n)|$ can be arbitrarily large with alternating signs as $n$ grows, and the second theorem says that intermediate value property holds for $f'(x)$ for any differentiable (not necessary $C^{1}$ function $f(x)$.


答:
Your recurrence can be written, with $v(a)=1-f(a)$,

$$v(a)=v(a-1)\frac16\sum_{i=1}^6\left(\frac{i}{6}\right)^a$$

And $v(1)=1$.

This leads immediately to:

$$f(a)=1-\frac{1}{6^{a-1}}\prod_{k=2}^{a}\left[\left(\frac{1}{6}\right)^{k}
+\left(\frac{2}{6}\right)^{k}+\left(\frac{3}{6}\right)^{k}+\left(\frac{4}{6}\right)^{k}+\left(\frac{5}{6}\right)^{k}+\left(\frac{6}{6}\right)^{k}
\right]$$

Not sure it qualifies as a "closed form", but I suspect you can't do much better. If the problem is performance, notice that you can improve it very much by keeping all preceding values of $f(a)$ in memory. Then computing the next one simply requires applying your recurrence, which is not difficult to compute. Notice also that $f(a)$ converges to $1$ linearly.


问:
How's the 'integer lattice' or the 'direct sum' of $\mathbb{Z} \oplus \mathbb{Z}$ not a Free group?
My question is about the direct sum $\mathbb{Z} \oplus \mathbb{Z}$ which is a Free Abelian group and not a free group.

The the integer lattice, or what I think is the direct sum $\mathbb{Z} \oplus \mathbb{Z}$, is known as a Free Abelian group which can have basis such as $e_{1} = (1,0), e_{2} = (0,1)$. You can check exactly what I've read here: https://en.wikipedia.org/wiki/Free_abelian_group#Integers_and_lattices

The integers $\mathbb{Z}$ are a free group with one generator and thus are a free Abelian group, yet groups that comprise of two generators are not Abelian. And again you can check what I've read here https://en.wikipedia.org/wiki/Free_group#Examples

Why is the integer lattice not also a free group with two generators $\{(1,0), (0,1)\}$ that is Abelian and with elements of the type $e_{1}^{4}e_{2}^{5}$ that equal elements such as $4e_{1}+5e_{2}$ of a Free Abelian group?

答: Not a free group due to commutativity.</p>

Body: Not a free group due to commutativity.


答:
In a group with a single generator, it is easy to see that this is abelian. A relation on such a group would be $a^n=1$ where $1$ is the identity, and such a relation renders the group finite.

In the case of two or more generators, the relation $xy=yx$ does not necessarily hold. If we wish to force it to hold we have to specify it as a relation in the definition of the group, and since it is not a trivial relation, the group which arises cannot be free - a free group is subject to no non-trivial relations.

A free abelian group is one whose only relations ensure that the generators commute (and hence that all the elements commute). The word "free" is still used, but it is a different idea. A free abelian group is "torsion free" because it has no elements except the identity of finite order.

Title: How's the 'integer lattice' or the 'direct sum' of $\mathbb{Z} \oplus \mathbb{Z}$ not a Free group?

Body:
My question is about the direct sum $\mathbb{Z} \oplus \mathbb{Z}$ which is a Free Abelian group and not a free group.

The the integer lattice, or what I think is the direct sum $\mathbb{Z} \oplus \mathbb{Z}$, is known as a Free Abelian group which can have basis such as $e_{1} = (1,0), e_{2} = (0,1)$. You can check exactly what I've read here: https://en.wikipedia.org/wiki/Free_abelian_group#Integers_and_lattices

The integers $\mathbb{Z}$ are a free group with one generator and thus are a free Abelian group, yet groups that comprise of two generators are not Abelian. And again you can check what I've read here https://en.wikipedia.org/wiki/Free_group#Examples

Why is the integer lattice not also a free group with two generators $\{(1,0), (0,1)\}$ that is Abelian and with elements of the type $e_{1}^{4}e_{2}^{5}$ that equal elements such as $4e_{1}+5e_{2}$ of a Free Abelian group?


问:
Show that the sequence $(s_n) = ((-1)^n n)$ is unbounded.
I need to prove that $(s_n)$ diverges. Since every convergent sequence is bounded, all I have to do is prove that $(s_n)$ is unbounded. But how?

答:
The first step would be to write down a more tractable definition of unbounded. Negating the definition of bounded means that for each $r >  0$ we should be able to find some $n$ for which $|s_n| \geq r$.

Note that, by archimedianity, if we can do this for $r \in \mathbb N$ we can do it for any $r$ (by doing it for a larger number which is natural). Think about this for a while if it is not clear at first.

So, given $r \in \mathbb N$, can you find $n \geq 1$ such that $|s_n| \geq r$?


问:
Differential Equations Reference Request
Currently I'm taking the Differential Equations course at college, however the problem is the book used. I'll try to make my point clear, but sorry if this question is silly or anything like that: the textbook used (William Boyce's book) seems to assume that the reader doesn't familiarity with abstract math, so it lacks that structure of presenting motivations, then definitions, then theorems and corolaries as we see in books like Spivak's Calculus or Apostol's Calculus.

I've already seem a question like this on Math Overflow, however unfortunatelly some people felt offended somehow and said: "are you saying that Boyce is easy? It doesn't matter if you know how to prove things, you must learn to compute", and my point is not that: Apostol and Spivak teaches how to compute also, however since their books are aimed to mathematicians they take care to build everything very fine and their main preocupation is indeed in the theoretical aspects.

I really don't like the approach of: well, in some strange way we found that this equation works, so memorize it, know how to compute things and everything is fine. I really want to understand what's going on, and until now I didn't find this possible with Boyce's book (certainly there are people that find this possible, but I'm used to books like Spivak's and Apostol's, so I don't really do well with books like Boyce's).

I've already seem Arnold's book on Differential Equations, but the prerequisites for reading it are bigger: he uses diffeomorphisms many times and although I'm currently also studying differential geometry, I don't feel yet comfortable on reading a book like this one.

Can someone recommend a book that covers ordinary differential equations, systems of differential equations, partial differential equations, and so on, but that can be read without much prerequisite, and that still has the structure of a book of mathematics? And when I say "structure of book of mathematics" is being like Spivak's and Apostol's books: not mixing up definitions, theorems and examples inside histories of how the theory developed. Since I'm a student of Mathematical Physics, of course motivations and examples from Physics are welcome, but not all mixed up in the text.

In truth I don't really believe that a book like the one I described exists (if a book on this topic is good in my point of view, I believe it'll have a lot of prerequisites). Anyway I hope that I don't get misunderstood in what's my doubt, and really sorry if this question is silly in someway.

答: "Theory of Ordinary Differential equations" by Coddington. If you find the previous one advanced, there is "An Introduction to Ordinary Differential Equations" by Coddington too. The first one is very complete and have a lot of things that I have never seen in other books. The second one is very basic (there is almost no qualitative theory), but proves the validity of some techniques usually used to solve ordinary differential equations. They also gives some explicit formulas for solution in special cases.Carmen Chicone's book "Ordinary Differential Equations with Applications" is good too if you want to study better the qualitative theory.</p>


问:
How Can the Vector Space $\mathbb{R}^n$ be identified with the Column space
Hi i am reading An introduction to manifolds by Loring and have some doubts in section 2.1 and 2.3. First i am writing some statements from the sections from which i have doubts and then i'll state my doubts.So the first statement that i have doubt in is:

By the correspondence between arrows and column vectors, the space $\mathbb {R}^n$ can be identified with this column space.

The second statement that i have doubt in is:

We usually denote the standard basis for $\mathbb {R}^n$ or $T_p(\mathbb R^n)$ by $e_1,...e_n$.

The third statement i have doubt in is:

This proves that $\mathcal D=\mathcal D_v$ for $v=\langle \mathcal Dx^1,...\mathcal xD^n \rangle$.This theorem shows that one may identify the tangent vectors at $p$ with the derivations at $p$. Under the vector space isomorphism $T_p\mathbb R^n\simeq\mathcal D_p\mathbb R^n$ the standard basis $e_1,...e_n$ for $T_p(\mathbb R^n)$ corresponds to the set $\left.\frac{\partial}{\partial x^1}\right|_p,...,\left.\frac{\partial}{\partial x^n}\right|_p$.

My doubts are as follows:

How can we identify the space $\mathbb R^n$ with the column space? Is this an isomorphism ?

What does it mean that we have identified the space $R^n$ with column space? Does this mean we can generate the space $R^n$ as linear combinations of the vectors from the column space? Also it is written in the book that "we visualize the tangent space $T_p \mathbb R^n$ at $p$ in $R^n$ as the vector  space of all vectors emanating from $p$". So will this vector space include all the possible vectors that have $p$ as starting point,even the normal vectors? Or only the vectos that pass tangentially through $p$?

In the second statement(doubt) why we are denoting the standard basis $e_1,...,e_n$ both for $R^n$ and $T_p \mathbb R^n$? Aren't they different vector spaces? How can they have the same basis. For example, in $\mathbb R^3$ we have $i,j,k$ as basis vectors.Are these basis vectors in $R^3$ different from $\left.\frac{\partial}{\partial x^1}\right|_p,\left.\frac{\partial}{\partial x^2}\right|_p,\left.\frac{\partial}{\partial x^3}\right|_p$?

In the third(doubt) how does $\mathcal D=\mathcal D_v$ for $v=\langle \mathcal Dx^1,...\mathcal xD^n \rangle$ mean allow us to identify the tangent vectors at $p$ with the derivations at $p$? I mean i know that the tangent vector is $v$ , $\mathcal D$ is the derivation at $p$ and $\mathcal D_v=v^i \frac{\partial}{\partial x^i}$ is the directional derivative at $p$.

For reference i am attaching the screenshots of the statement where i have highlighted different statements from the book.

答:
In linear algebra, by default, the elements of $\Bbb R^n$ are considered as column vectors, so that the linear maps can be written as $x\mapsto Ax$ with a matrix (after choosing coordinate systems), corresponding to the way we write functions in analysis: $x\mapsto A(x)$.

(While if we chose row vectors to work with, we need to consider the action of functions on the right: $x\mapsto xA$.)

On the other hand, to distinguish notation, the points of the space are written as row vectors in the text.

Geometrically, the tangent plane of a plane (embedded in the space) at any point is the plane itself. Likewise, the tangent space $T_p(\Bbb R^n)$ at any point $p$ of the full space $\Bbb R^n$ is the full space, and there is no surface or anything to talk about normal vectors, all vectors are tangent vectors of the space.

You can visualize it by drawing the vector $v\in T_p(\Bbb R^n)$ that corresponds to $\tilde v\in\Bbb R^n$ from point $p$ to point $p+\tilde v$.

But be careful here, the tangent vectors at $p$ are probably defined abstractly: as the equivalence classes of smooth curves through $p$, with the equivalence given by having identical derivative at $p$.

The correspondence to $\Bbb R^n$ is simply given by taking the derivative (velocity vector) of the curve at $p$, all done in the standard coordinates.

Well, when we write 'identifying' two kinds of objects, then afterwards they get identified in the text.

By (the above) definition, we primarily consider tangent vectors as equivalence classes of curves. Then we introduce the directional differentiation operators that map real numbers to real valued smooth functions and satisfy the Leibniz rule at $p$, and these can also be identified with the tangent vectors: for a curve $\gamma:(-\varepsilon,\varepsilon)\to M$ with $\gamma(0)=p$, the directional derivative is $f\mapsto (f\circ\gamma)'(0)$.

In the specific case when $M=\Bbb R^n$ (or locally, using a chart [=local coordinate system] of the $n$ dimensional manifold), the tangent vectors at a point $p$ can further be identified with the actual elements of $\Bbb R^n$, because they form a vector space with basis given by the curves $t\mapsto p+t\cdot e_i$ or by the corresponding directional differentiations $\frac\partial{\partial x^i}$.

So, the column vector $v=\pmatrix{v_1\\v_2\\ \vdots \\ v_n}$ is the correspondent of the directional differentiation $\partial_v=\sum_i v_i\frac\partial{\partial x^i}$ and also of the equivalence class of the curve $t\mapsto p+tv$.


问:
Homotopical factorization of maps that are zero in homology
Suppose $f: M \rightarrow N$ is a cellular map between CW-complexes of dimension $m$ and $n$ respectively such that the induced maps in homology are zero after a while (i.e. $H_i(f) = 0$ for some $k$, $n\geq k>0$).

Is it true that, up to homotopy, $f$ factors through a subcomplex $X \subseteq N$ of dimension $k$?

If not, is there a way to classify "geometrically" these maps?

In particular, I was trying to study the set of maps up to pointed homotopy $\left[T^n, S^2\right]_*$ ($T^n$ is the $n$-fold torus, i.e. $S^1\times S^1 \dots \times S^1$), so I would love to have some results for this very specific case if the question is too broad.

Are there good references on the topic?

答:
Given a map $f:M\rightarrow N$ between finite-dimensional CW complexes $M,N$ you can always take $k=\dim N$ to get a facorisation of $f$ though a $k$-dimensional subcomplex of $N$. In general $f$ can be homotoped to a cellular map, so if $\dim M<\dim N$ then this improves after letting $k=\dim M$ and factoring $f$ through the $k$-skeleton of $N$.

However while these trivial factorisations always exists, there is in general no way to minimise the value of $k$ with respect to the action of $f$ on homology.

For instance, if $n>1$, then there are infinitely many integers $m>n$ for which there is a non-homotopically trivial map $f:S^m\rightarrow S^n$. For dimensional reasons this map induces $0$ on $\widetilde H_*$. The fact that $f$ is not null-homotopic, however, means that it does no factor through the zero-skeleton of $S^n$, or in fact the $k$-skeleton for any $k<n$.

For an explicit example of such a map we can take the Hopf map $\eta:S^3\rightarrow S^2$. Moreover we can make this relevant to the problem at hand by considering the map $f:T^3\rightarrow S^2$ which is given by the composition
$$f:T^3\rightarrow S^3\xrightarrow\eta S^2,$$
where the first map collapses the $2$-skeleton of $T^3$ to a point. It is a consequence of my answer here that $f$ is not null-homotopic. Yet $f$ induces $0$ on $H_2T^3$ because $\eta$ does. This example generalises to any $n\geq 3$ for which there is a non-trivial map $S^n\rightarrow S^2$.


答:
Free as in free groups means there are no relations. So a free group on two generators $a,b$ contains arbitrary words with the letters $a, b, a^{-1}, b^{-1}$ that do not contain $aa^{-1}, a^{-1}a, bb^{-1}, b^{-1}b$.

A free abelian group on $a,b$, however, always has the relation $ab=ba$, which is a relation not occurring in a free group. The two notions free and free abelian only coincide for free (abelian) groups with a single generator.

The concept of freeness comes from somewhere deeper: groups form a category. You can assign to a group its underlying bare set, that’s the forgetful functor. By definition, a free functor is a left adjoint to the forgetful functor, which for groups turns out to be the above construction. For abelian groups, however, the left adjoins to the (same) forgetful functor looks different; it’s precisely the free abelian group.


问:
Proof, Factor theorem.
Question is :

If $f(x)$ is a polynomial with integral coefficients and, suppose that $f(1)$ and
$f(2)$ both are odd, then prove that there exists no integer n for which $f(n) = 0.$

My approach :

I tried writing the function in form of quotient, remainder and divisor.
(I have assumed the remainder to be linear.)

$f(x) = Q(x)D(x) + ax + b$

$f(x) = K(x-1)(x-2) + ax + b  \longrightarrow (1)$

$f(x) = K_1(x-1) + (2n_1 +1)  \longrightarrow (2)$

$f(x) = K_2(x-1) + (2n_2 +1)  \longrightarrow (3)$

I tried the above, but it isn't proving to be of much help.

On the other hand, if I think reverse, the possible way I can think of when a polynomial with integral coefficients has no integer n for which $f(n)=0$  is probably when it has odd coefficients, because in that case, the roots aren't rational. I don't know how to relate this with the question, however.

Please help me with a quick method for this problem. Kindly verify if I have done any mistakes. Thanks in advance!

答:
Suppose, there is an integer $\ n\ $ with $\ f(n)=0\ $

Then, $\ x-n\ $ is a factor of $\ f(x)\ $ which is even for either $\ x=1\ $ or $\ x=2\ $ , hence at least one of $\ f(1)\ $ and $\ f(2)\ $ must be even.

Title: Proof, Factor theorem.


问:
Interpretation of $f(x)(1 + o(1))$
What is an exact interpretation of the expression $f(x)(1 + o(1))$?

$o(1)$ - 'o' here is referring to little-oh notation. I understand that any function $g(x)$ with $o(1)$ refers to the fact that function $g(x)$ asymptotically reaches zero, i.e $lim_{x\to\infty}g(x) \to 0$. But when the expression is embedded in such expression above, I am not sure what to make of it.

How does $f(x)$ gets distributed to the expression inside? Is it $f(x) + f(x)o(1)$ or something else?

What does it really mean when you multiply $o(1)$ to some function? Back to my original question.

答:
For clarity, I will assume that all the asymptotic notations are at $+ \infty$, and will note e.g. $o(1)$ instead of the more correct $o_{+\infty}(1)$. All what follows holds anyway even when we are not "close to $+\infty$ but close to a specific point $x_0$", but it is simply easier to write and to make proofs.

You can simply think of $o(1)$ as some function which converges to $0$. More generally, $o(f)$ can be thought of as a function $g$ such that $\frac g f \to 0$.

You can more formal definitions in the other answers. However, one cool thing is that if you think of them as functions, you get a good intuition for some properties that are effectively true !

For example, if you take two functions $f$ and $g$, then $f \times o(g) = o(fg)$

In your case, you get $f(1+o(1)) = f + o(f)$.

Intuitively, you multiply $f$ by something that is roughly equal to $1$. Hence, the result is $f$ but some error, and the "order of magnitude" of this error has been multiplied by $f$. So to answer your two questions:

You can distribute as usual, and whenever you encounter a product of the form $f \times o(g)$, you can slide the $f$ inside the $o$ to get $o(fg)$. This is also why you will never really encounter the notation $f(x)o(1)$, because it is not really practical and does not give more information than $o(f)$. Because we tend to think of $o(f)$ as "error terms", we "add" them to the rest of our equalities.

When you multiply $o(1)$ by some function $f$, you are multiplying $f$ by something which tends to $0$. You don't know at which rate, but you know that it goes to $0$ as you get closer to $+\infty$. Hence, the only thing that you know about $f \times o(1)$ is that it is negligible compared to $f$, because it is the product of $f$ and something converging to $0$. This is exactly what the notation $o(f)$ means (and as I already mentioned, it is in fact correct to write $f \times o(1) = o(f)$).

Don't worry, the Landau notations take a while to get used to, but you will eventually get a better grasp on what they really mean and what you are allowed to do with them.

Title: Interpretation of $f(x)(1 + o(1))$


答:
Let $f(n)=1.001^n$ and $g(n)=n^{10}$

For $n \approx 116735$ we have $f(n)=g(n)\approx 4.699\times 10^{50}$

and for $n>116735$ we have $f'(n)=\log_{e}(1.001)f(n) \approx 0.001 f(n)$ while  $g'(n) = \frac{10}{n} g(n) \lt 0.0001 g(n) $

so for $n>116735$ we have $f(n) > g(n)$ and $f'(n) > g'(n)$

so past this point $1.001^n$ is larger than $n^{10}$ and is growing faster

Title: Determine if $1.001^n$ will grow faster than $n^{10}$.

Body:
We know that $n^k \leq k^n$, where $n^k$ is a polynomial and $k^n$ is an exponential function. However, what if we have a case such as $n^{10}$ and $1.001^n$. How can I determine which function will grow faster?

Background - this is a problem from my complexity class and specifically the topic about the Big O-Notation.


问:
John Wallis and calculating ratios
I was asked the following question:

Using Wallis's method, calculate the values $n=0,\frac12,1,\frac32,2,\frac52$ in row $p=\frac32$ of his ratio table.

The Wallis rereferred to here is John Wallis of England (1616-1703). The ratio table looks like this: $$\begin{matrix}p/n&0&1&2&3&4&5&6&7&...\\0&1&1&1&1&1&1&1&1&...\\1&1&2&3&4&5&6&7&8&...\\2&1&3&6&10&15&21&28&36&...\\3&1&4&10&20&35&56&84&120&...\\...&...&...&...&...&...&...&...&...&...&\\\end{matrix}$$

I don't understand how to read this table or how to actually go about solving the question. The table's goal is to find the ratio of the area of the unit square to the area aenclosed in the first quadrant by the curve $y=(1-x^{\frac1p})^n$.

答:
The ratio table represents the ratio of the unit square over the area under the stated curve, $y = (1-x^{1/p})^n$.  We shall assume $p, n \geqslant 0$ and adopt the convention that the area is $1$ when $p=0$.The area of the unit square is always $1$, while the area under the curve is
\begin{align}
A_{p,n} = \int_0^1 (1-x^{1/p})^n ~ dx
\end{align}
and the relevant Wallis number in the table is simply the reciprocal, $W_{p,n} = 1 / A_{p,n}$.  It would seem by convention the area $A_{p,n}$ is taken to be $1$ when $p=0$.

To evaluate the integral, you can begin by making the substitution $x = \sin^{2p} \theta$ for $0 \leqslant \theta \leqslant \pi/2$, so that $dx = 2p\sin^{2p-1}\theta \cos \theta ~d\theta$
\begin{align}
A_{p,n} &= \int_0^{\pi/2} (1-\sin^2\theta)^n \cdot 2p \sin^{2p-1}\theta \cos \theta d\theta \\
&= 2p\int_0^{\pi/2} \cos^{2n+1} \theta \sin^{2p-1}\theta ~d\theta.
\end{align}
In this case you are only interested in a particular instance, $p=3/2, n=0,\frac{1}{2},1, \cdots$, giving
$$A_{p,n} = 2p \int_0^{\pi/2}\sin^2\theta \cos^{2n+1}\theta ~d\theta$$ where $m=2n+1=1,2,3,\cdots$.  Integrate once by parts,
\begin{align}
A_{p,n} = 2p \int_0^{\pi/2} \cos^{2n+1} \theta - \cos^{2n+3}\theta  ~ d\theta .
\end{align}
These integrals are standard, see the Wikipedia entry for Wallis's integral, giving,
\begin{align}
\begin{array}
~A_{3/2,0} = 1 & A_{3/2,1/2} = \frac{3\pi}{16} \\
A_{3/2,1} = \frac{6}{15} & A_{3/2,3/2} = \frac{3\pi}{32} \\
\cdots & \cdots\\
n=k, k=0,1,\cdots & n=k+\frac{1}{2},k=0,1,\cdots\\
A_{3/2,n}=3\frac{2^{2k}k!^2}{(2k+1)!}\frac{1}{2k+3} &
A_{3/2,n}=3\frac{(2k+2)!}{2^{2k+3}(k+1)!^2}\frac{1}{k+2} \frac{\pi}{2}
\end{array}
\end{align}
and the corresponding Wallis ratios are the reciprocals of these numbers.

More generally, you can evaluate $A_{p,n}$ for $p$ and $n$ when either is a multiple of $\frac{1}{2}$ by considering the integral,
$$I_{r,s} = \int_0^{\pi/2} \sin^r\theta \cos^s \theta~d\theta.$$
in which $r, s$ are non-negative integers;  first notice $I_{r,s}=I_{s,r}$.  Evaluate the special cases when $r=0,1, s \geqslant 1$ and $s=0,1, r \geqslant 1$.  Then use integration by parts to obtain the recurrence relation,
$$I_{r,s}=\frac{s-1}{r+1} I_{r+2,s-2}$$
where by symmetry we can take $r \geqslant s \geqslant 2$.  Using the special cases this should give all values for $I_{r,s}$.  Then $A_{p,n} = 2p I_{2p-1,2n+1}$.

Title: John Wallis and calculating ratios


问:
Negative continued fraction of Euler Mascheroni constant
$\gamma$, the Euler-Mascheroni constant, has the following simple regular continued fraction:

$$\gamma=[0; 1, 1, 2, 1, 2, 1,\dots]=0+
\cfrac{1}{
  1+\cfrac{1}{
    1+\cfrac{1}{
      2+\cfrac{1}{
        1+\ddots
      }
    }
  }
}$$

I wondered which would be its negative continued fraction, but after searching I have not found anything about it in the literature I have checked.

Just using rough approximation methods, I have derived that it seems that

$$\gamma=1-
\cfrac{1}{
  3-\cfrac{1}{
    2-\cfrac{1}{
      3-\cfrac{1}{
        2-\ddots
      }
    }
  }
}$$

(although there is an initial pattern in the alternating $2$ and $3$, this pattern ceases).

I would like to know if there is some way to transform a regular continued fraction into a negative continued fraction, to see if that transformation could give some valuable info about $\gamma$. Also, if the negative continued fraction of $\gamma$ is already known, it would be great if you could provide where I can consult it.

Thanks in advance!

答:
I found this interesting research that I share with you in case you are interested in the relationship between the negative and positive continued fractions:

https://scholarship.claremont.edu/cgi/viewcontent.cgi?filename=2&article=1183&context=hmc_theses&type=additional

According to Definition 8, the first terms of the negative continued fraction of the Euler Mascheroni constant would be:

$$\gamma=[1,3,2,3,2,3,2,2,2,5,2,2,2,2,2,2,2,2,2,2,2,2,7,3,2,2,2,2,2,2,2,3,2,...]$$


问:
Help to know what kind of differential equation is this
I've to solve this differential equation:

$
(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0
$

But I even don't know what kind of differential equation is it.

Please, any help?

Thanks in advance.

答:
$$(2y^3-2y^2+3x)dx+(3xy^2-2xy)dy=0$$
Rewrite it as:
$$2(y^3-y^2)dx+3xdx+xd(y^3-y^2)=0$$
$$2wdx+3xdx+xdw=0$$
Where $w=y^3-y^2$. It's easier to solve now. The integrating factor is obvious:
$$\mu(x)=x$$
$$(wdx^2+x^2dw)+3x^2dx=0$$
$$dwx^2+3x^2dx=0$$


答:
The presentation for $\mathbb{Z}\oplus\mathbb{Z}$ is $$\langle a,b\ |\ aba^{-1}b^{-1}\rangle,$$
where the $aba^{-1}b^{-1}$ term is there to guarantee commutativity. (It is, however, a free abelian group on two generators. But that's a definition thing.)


问:
Derivative of trace involving hadamard product and product of inverse matrices
I need to find the derivative with respect to $\mathbf{\Omega}$ of

$$
 Tr\left(\left(\left(\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}\right)\circ\mathbf{I}\right)\mathbf{S}\right)
$$

In the above, $\mathbf{\Omega}^{-1}$ is symmetric, $\mathbf{C}$  is symmetric, $\mathbf{I}$ is the identity matrix and $\mathbf{S}$ is symmetric.

I understand that writing this using $:$ notation might help, which if I'm correct allows me to write

$$
\begin{align}
\phi&=\left(\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}\right)\circ\mathbf{I}:\mathbf{S}\\
&=\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}:\mathbf{I}\circ\mathbf{S}
\end{align}
$$

However from there I'm unsure - do I need a version of the product rule to deal with the right hand side - in which case how is this written when using $:$ notation? Or using the rule that $d\mathbf{\Omega^{-1}}=-\mathbf{\Omega^{-1}}d\mathbf{\Omega}\,\,\mathbf{\Omega^{-1}}$?

答:
You're on the right track, that is the correct product rule.

Here is a useful relationship when $A,B$ are symmetric matrices
$$\eqalign{
{\rm Sym}(X) &\doteq \tfrac 12\left(X+X^T\right) \\
d(BAB) &= dB\,AB + BA\,dB \\
 &= dB\,AB + (dB\,AB)^T \\
 &= 2\;{\rm Sym}(dB\,AB) \\
}$$

Putting the pieces together yields
$$\eqalign{
\phi &= (\Omega^{-1}C\Omega^{-1}\circ I):S \\
 &= (I\circ S):\Omega^{-1}C\Omega^{-1} \\\\
d\phi
 &= (I\circ S):d\left(\Omega^{-1}C\Omega^{-1}\right)
    &\big({\rm sym\,rule}\big) \\
 &= (I\circ S):2\;{\rm Sym}\left(d\Omega^{-1}C\Omega^{-1}\right) \\
 &= 2\;{\rm Sym}(I\circ S):d\Omega^{-1}C\Omega^{-1} \\
 &= 2\;(I\circ S):d\Omega^{-1}C\Omega^{-1} \\
 &= 2\;(I\circ S)\Omega^{-1}C:d\Omega^{-1}
    &\big({\rm product\,rule}\big) \\
 &= -2\;(I\circ S)\Omega^{-1}C:\Omega^{-1}d\Omega\,\Omega^{-1} \\
 &= -2\;\Omega^{-1}(I\circ S)\Omega^{-1}C\Omega^{-1}:d\Omega \\\\
\frac{\partial\phi}{\partial\Omega}
 &= -2\;\Omega^{-1}(I\circ S)\Omega^{-1}C\Omega^{-1}
    &\big({\rm gradient\,matrix}\big) \\\\
}$$
Here is another identity which was used above
$$\eqalign{
X:{\rm Sym}(Y)
 &= X:\tfrac 12(Y+Y^T) \\
 &= \tfrac 12X:Y + \tfrac 12X:Y^T \\
 &= \tfrac 12X:Y + \tfrac 12X^T:Y \\
 &= \tfrac 12(X+X^T):Y \\
 &= {\rm Sym}(X):Y \\
}$$
A similar identity exists for the function
$${\rm Skew}(X) = \tfrac 12\left(X-X^T\right)$$


答: $|s_{n}| = n$. Imagine $s_{n}$ was bounded above and below by $M_{1}, M_{2}$ respectively, then $|s_{n}|< max(|M_{1}|, |M_{2}|)$, however choosing $n= \ceiling{max(|M_{1}|, |M_{2}|)}$ we get a contradiction. Thus $s_{n}$ is unbounded.

Title: Show that the sequence $(s_n) = ((-1)^n n)$ is unbounded.

Body: I need to prove that $(s_n)$ diverges. Since every convergent sequence is bounded, all I have to do is prove that $(s_n)$ is unbounded. But how?


问:
How use density argument to extend operator?
Let $(X, \|\cdot\|_{X})$ be Banach space (e.g, $X=L^p$) and let  $\mathcal {S}$ is dense in $X$.

Assume that $T$ is map  such that
$$\|Tf\|_{X} \leq \|f\|_{X}\quad \forall f\in \mathcal{S}.$$

My Question is: Can we expect to use density argument to prove that
$$\|Tf\|_{X} \leq \|f\|_{X}\quad \forall f\in X?$$ If so, what is natural way to define $T$ on $X$?

My attempt:
Let $f\in X\setminus \mathcal {S}.$ Since $\mathcal{S}$ is dense in $X,$ there exists $(f_n)\subset X$ such that $f_n\to f$ in $X.$
Define $Tf:= \lim f_n$.
My questions: Can we say the definition of $Tf$ is independent of $f_n$ (as there might exist $(g_n)$ such that $g_n\to f$ in $X$).

$\|Tf\|_{X}= \|\lim Tf_n\|_{X} \leq \lim \|Tf_n\| $ (Can we do this? Can we take limit out side the norm? (I guess norm function is continuous so it make sense? correct?)). And then we get $\|Tf\|_{X}\leq \lim \|f\|_{X}= \|f\|_{X}.$

答:
You're pretty close, I would say!

To make this more formal and to avoid abuses of notation which at this stage are not yet justified and can also be confusing and/or misleading, let's define a new operator $\widetilde{T}:X\to X$ by the following: if $f\in S$, then $\widetilde{T}f = Tf$ and for $f\in X\setminus S$, $\widetilde{T}^{(f_n)}f = \lim_n T f_n$ where $f_n\to f$ (density of $S$ is needed at this stage). Note that $\widetilde{T}^{(f_n)}f$ to this point may depend on the choice of $f_n$ - we will show it does not. I'm adding the superscript sequence notation just to make the issue clear. You've already thought about it this way, but it helps to write it out as a new operator (which it technically is).

We'll first show that the operator norm condition extends to $\widetilde{T}$. As you note, $\|\widetilde{T}^{(f_n)}f\| = \|\lim_n Tf_n\| = \lim_n \|Tf_n\| \le \lim_n \|f_n\|$ as the norm is continuous (almost by definition since limits in $X$ come from the norm). Since $f_n\to f$ (meaning $\lim_n \|f_n-f\| = 0$), $\lim_n \|f_n\| = \|f\|$ and so $\|\widetilde{T}^{(f_n)}f\| \le \|f\|$.

Let us show that $\widetilde{T}$ is well-defined, i.e. if there exists another sequence $g_m\to f$, then $\widetilde{T}^{(f_n)}f = \lim_n Tf_n = \lim_m Tg_m = \widetilde{T}^{(g_m)} f$ as well. The sketch below is the basis for the argument which I will leave you to fully formalize:

\begin{align}
\|Tf_n - Tg_m\| &= \|T(f_n - g_m)\| \\
& \le \|f_n - g_m\| \\
&= \|f_n - f + f - g_m\| \\
& \le \|f_n - f\| + \|f - g_m\| \\
& \to 0
\end{align}

Thus as $n$ and $m$ tend to infinity, $Tf_n$ and $Tg_m$ converge to the same element since limits are unique in Banach spaces. Therefore we can drop the $\widetilde{T}^{(f_n)}$ notation in favor of a singular $\widetilde{T}$ notation.


问:
The relation between rational forms and Jordan forms.
Is there any algorithm / method that allows one to determine the Jordan form of a matrix after determining its rational form?</p>

答:
I explain the relationship between Jordan and rational forms in these recently written notes.  Here are the main points:

Jordan form is only available when a transformation is split, i.e., when its minimal polynomial (equivalently, its characteristic polynomial) is a product of linear factors over the ground field.

In this case, the rational canonical form takes the form of a direct sum of companion matrices for polynomials $(t-\lambda)^a$.  In the lingo of the notes, we have expressed $T$ as a direct sum of cylic, primary invariant subspaces.  On each such subspace (or "square block" of the corresponding matrix) the transformation $T-\lambda 1_V$ is nilpotent, so its rational canonical form is that of the companion matrix of the polynomial $t^a$: it has all zeros except $1$'s along the main diagonal.  Adding back $\lambda  1_V$, we get a matrix which has $\lambda$'s on the main diagonal, $1$'s on the subdiagonal and otherwise zeros.  This is a Jordan block.

Or possibly your convention on Jordan blocks is that the $1$'s should appear on the "superdiagonal" rather than the "subdiagonal".  This is an inessential difference: if you just reverse the order of the basis elements you will go from my version of Jordan normal form to yours.  I prefer mine I suppose because it is that much more closely related to rational canonical form and thus seems more natural.

Here is another way to say this: the rational normal form is more general: it exists for any transformation.  Knowing the rational normal form is equivalent to knowing a decomposition of $V$ into a direct sum of cyclic "primary" subspaces -- the latter means that the minimal polynomial is a prime power.  The minimal polynomials of these cyclic primary subspaces are precisely the elementary divisors $p_1,\ldots,p_n$  When you have the elementary divisors, the rational canonical form is a direct sum of companion matrices $C(p_1),\ldots,C(p_n)$.  When the minimal polynomial is split, each elementary divisor is of the simple form $(t-\lambda_i)^{a_i}$, and from this data you build the Jordan blocks as well, in a way which can be seen in terms of rational canonical forms as described in 2) above.  But fundamentally, both forms are obtained from the elementary divisors and the algorithmic work of finding either form is plainly equivalent to that of finding the elementary divisors.


问:
Calculate the interquartile range of probability density function. With more information attached

I have been trying to figure out the last part of this question and have tried I will add the work below but I can’t seem to figure it out I am self studying so I have no teacher to ask. Any help would be much appreciated . Thank you!

答:
The quartiles must satisfy $0 \le x \le 5$.

The lower quartile is $\frac{10-\sqrt{100-25}}{2}\approx 0.69$ (this appears in your calculations but is not the value you chose)

while the upper quartile would be $\frac{10-\sqrt{100-75}}{2}= 2.5$ (you divided by $8$ rather than $4$)


问:
$(0,1)$ without union of intervals centered at rational numbers
Let $(a_n)$ be a sequence of all rational numbers in $(0,1)$ and let
$$\mathscr{I}_t:=\bigcup_i\left(a_i-\frac{t}{2^{i+1}},a_i+\frac{t}{2^{i+1}}\right)\bigcap\,(0,1)\text{.}$$
Then $\lambda(\mathscr{I}_t)\leq t$ because $\displaystyle\sum_n\frac{t}{2^n}=t$ where $\lambda$ is the Lebesgue measure. That means $\mathscr{I}_t\subsetneq(0,1)$ for $t<1$.

I'm trying to understand what $(0,1)\setminus\mathscr{I}_t$ looks like and in which way it depends on the enumeration of the rationals.

My thoughts: Obviously it would only contain irrational numbers. Any $x\in(0,1)\setminus\mathscr{I}_t$ has to satisfy $\mid x-a_i\mid\geq\frac{t}{2^{i+1}}\forall i$. But I'm trying to build an intuitive understanding and don't know how such an $x$ would look like.

答:
What is your exact question? I'm assuming you already know the Cantor set but if not maybe that's a place to start. It's not quite the same, as the Cantor set still contains rational numbers.

It might also help to start with the "simple" enumeration 1/2, 1/3, 2/3, 1/4, 3/4, 1/5...

One thing that is interesting to notice is that your set can have a measure arbitrarily close to one but never exactly one. One can go even further and prove that any closed subset of the irrational numbers on the unit interval can't have measure 1(when using the Lebesgue measure), due to the fact, that open sets that aren't the empty set have measure greater 0.


问:
Query regarding $k$-critical graphs
I am studying coloring of graphs and having some basic doubts in the theorems regarding $k$-critical graph.

$\mathbf {Definition}$ - A graph $G$ is said to be $k$-critical if $ \chi (G) = k$ and $ \chi(H) < k$ for any proper subgraph $H$ of $G$.

Now all the theorems regarding $k$-critical graphs use the fact that – if $\chi(G) =k$ and $\chi(G \setminus v) <k$ for any vertex $v$ of $G$ , then $G$ is $k$-critical.

But I am not understanding how the above statement is equivalent to the definition. Because there may exist a proper subgraph $H$ of $G$ that has all the vertices of $G$. This may be very trivial but I am really struggling with this. Someone please help me to understand this.

Thank You.

答:
You are right and the people stating those theorems are (a little bit) wrong.

Unfortunately, many people talk about $k$-critical graphs when they really mean "$k$-vertex-critical". A $k$-vertex-critical graph is exactly one for which deleting any vertex reduces the chromatic number. Equivalently, any proper induced subgraph has a smaller chromatic number.

Just to make sure what you are saying is always clear, I would avoid using the term "$k$-critical graph" and only ever talk about "$k$-vertex-critical" and "$k$-edge-critical" graphs. Then it's clear that the second kind of graph really is sensitive to the removal of a single edge.

To reassure you that there really is a difference, consider the following graph, taken from the paper Efficient algorithms for finding critical subgraphs by Desrosiers, Galinier, and Hertz:

This graph has chromatic number $4$, and will continue to have chromatic number $4$ if the edge $v_1v_2$ is deleted, but if you delete any vertex, the result has chromatic number $3$.

Title: Query regarding $k$-critical graphs


问:
How to factor numbers that are the product of two primes
What are techniques to factor numbers that are the product of two prime numbers? For example, how would we factor $262417$ to get $397\cdot 661$? Would we have to guess that factorization or is there an easier way?</p>

答:
At the risk of being a necromancer ...

... factorising a number we know to be the product of two primes should be easier than factorising a number where we don't know that.

Has anyone done an attack based on working backwards through the number?

For example, as we know 262417 is the product of two primes, then these primes must end with 1,7 or 3,9.  Our solution is therefore abcde1 x fghij7 or klmno3 x pqrst9 where the letters need to be determined.

Now work with the last pair of digits in each potential solution (e1 x j7 and o3 x t9) and eliminate all those digits for e, j, o and t which do not produce a 1 as the fifth digit.  We have the complication of dealing with possible carries.

Obviously the tree will expand rather quickly until it begins to contract again when investigating the frontmost digits.


问:
Determine if $1.001^n$ will grow faster than $n^{10}$.
We know that $n^k \leq k^n$, where $n^k$ is a polynomial and $k^n$ is an exponential function. However, what if we have a case such as $n^{10}$ and $1.001^n$. How can I determine which function will grow faster?

Background - this is a problem from my complexity class and specifically the topic about the Big O-Notation.

答:
As other answers gave a way to actually prove the result, I would like to add something to "feel" why this is true, or at least a way to remember it without having to do the proof each time this question shows up in your life !

Surely, if one of the two functions grows faster than the other, the same can be said about their logarithms.

We will compare $\ln(1.001^n) = n \ln(1.001)$ and $\ln(n^{10}) = 10 \ln(n)$.

The graph of the first one is a straight line. It does not grow very fast, but still.

The graph of the second one is a "zoomed in" version of the graph of the $\ln$ function.

"Surely", the straight line will eventually cross the log curve.

(I know that this is very informal. However, this is an attempt at making things more "intuitive" rather than having to rely on computation and abstract rules. If it does not help, feel free to say so !)


问:
Exponential of Squared Brownian Motion
Long time lurker, first time posting! Have a problem, that looks familiar but I can't put my finger on it.

Need to calculate
$\mathbb{E} [\exp(aW_T^2)|F_t]$ where $W_t$ is an $F_t$ adapted standard Brownian motion and $t \leq T$. Any help on exponentials of squared Brownian motion is very appreciated!

答:
I think the previous answer is false, as the expectation of the product is the product of the expectations if and only if both random variables are independent, which is obviously not the case.

However the first part is true, so I will only try to correct the second part:

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = \mathbb{E} \left[ e^{u(W_T - W_t)^2} e^{2 u W_t (W_T - W_t)} \mid \mathcal{F}_t \right] e^{u(W_t)^2}$

As $(W_T - W_t) \perp \!\!\! \perp \mathcal{F}_t$ and $(W_T - W_t) \sim \mathcal{N}(0, T-t)$ we have :

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = e^{u(W_t)^2} \int_\mathbb{R} e^{u(T - t) x^2 + 2 u W_t \sqrt{(T - t)} x} \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx$

You then get this (I leave you the intermediate calculations) :

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = \frac{\exp \left( u(W_t)^2 + \frac{2 u^2  (W_t)^2 (T - t)}{1 - 2 u (T - t)} \right)}{\sqrt{2 \pi}} \int_\mathbb{R} e^{-\frac{1}{2} \left(  \sqrt{1 - 2 u (T - t)}x - \frac{2 u  W_t \sqrt{(T - t)} }{\sqrt{1 - 2 u (T - t)}} \right)^2}dx$

And then after the right (obvious here) change of variables and a simplification in the exponential on the left, you end up with :

$\mathbb{E} \left[ e^{u(W_T)^2} \mid \mathcal{F}_t \right] = \frac{ \exp \left( \frac{u(W_t)^2}{1 - 2 u (T - t)} \right)}{\sqrt{1 - 2 u (T - t)}}$

If you require further explanations, I will be glad to give it.

Title: Exponential of Squared Brownian Motion


问:
Compact Metric Problem
while defining Compact Metric Space in one of book of HL ROYDEN's REAL ANALYSIS statement says:

If T is collection of open subsets of a metric space X, then the collection F of complements of sets in T is collection of closed sets. Moreover, T is cover if and only if F has empty intersection. Thus, "By de Morgan identities a metric space X is compact if and only if every collection of closed sets with a non empty intersection has a finite sub-collection whose intersection also is nonempty"

firstly, this statement says T is cover if and only if intersection of F is empty, then it goes on saying that X is compact if and only if every collection of closed sets with nonempty intersection has finite sub-collection whose intersection also is nonempty.

Please explain it?.

答:
The easiest way to explain is with de Morgans laws:

$complement(A\cup B) =complement(A)\cap complement(B) $

So if $T$ is a cover of $X$, then

$$ \empty = complement(X) = complement(\cup_{t\in T}A_{t}) = \cap_{t\in T} complement(A_{t})$$

$X$ is compact if an only if every open cover has a finite subcover, thus $X$ is compact if and only if given any such $T$ as above there exists $T'\subset T$ with $T'$ finite so that:

$$ \empty = complement(X) = complement(\cup_{t\in T'}A_{t}) = \cap_{t\in T'} complement(A_{t})$$

Title: Compact Metric Problem


问:
First-order formula implying the existence of an infinite branch in a tree
I want to find a formula which would imply the existence of an infinite branch in a tree. I know beforehand that my tree has countably many vertices (in fact, each vertex can possibly have a countable number of sons, so we cannot directly apply the Kőnig's lemma).

My question is: quantifying only over vertices (and edges, for that matter), is it possible to write a logical formula showing that the tree has an infinite branch (i.e. a tree satisfies the formula iff it has an infinite branch) ?
It is not hard to find one implying the existence of branches of arbitrary length, but this is not what I am looking for - although I am not even sure such a property can be expressed in first-order logic.

If this is impossible, are there some other assumptions that would need to be made in order to be able to write such a formula ? (For example, if the tree had a finite branching factor - which can be expressed by a first-order formula - then we could apply the Kőnig's lemma. As I already said, this is not the case here, but does there exist other hypothesis that would achieve the same result ?)

答:
As it stands out, there is no easy way to answer my question. Indeed, one can find
here some sets of trees in term that are placed in the analytic hierarchy - rather than the arithmetic one, and I do not have any way of "discarding" those trees with the elements given in my question, meaning that the general problem is much harder that what can be dealt with FO-logic.


问:
$X^\emptyset$ versus $\emptyset^X$
I recently read that there is a difference between $X^\emptyset$ and $\emptyset^X$. I was hoping that someone could explain why this is the case.

I have interpreted $B^A$ as, "The set of all functions that have a domain of $A$ and a range of some subset of $B$". More formally, I thought that:

$f \in B^A \leftrightarrow \exists Z ( Z \in P(B)\  \land f \subseteq A \times Z) $ <-- Edit: $\color{red}{ \text{Here is the mistake} }$

Additionally, we know that $\emptyset \times X = X \times \emptyset=\emptyset \times \emptyset = \emptyset$

It is vacuously true that $\forall x [\emptyset \subseteq x]$

Also, $P(\emptyset)=\{\emptyset\}$, which means that $\emptyset \in P(\emptyset)$ and, more generally, $\forall x [\emptyset \in P(x)]$

Using this information, consider the following two cases:

Case 1: $X^\emptyset$

Case 2: $\emptyset^X$

Case 1: $X^\emptyset$

Consider if $f$ is the empty set function.

Does a $Z$ satisfy the following formula? $Z \in P(X)\  \land \emptyset \subseteq \emptyset \times Z$

Sure! Let $Z=\emptyset$.

$\emptyset$ is an element of $P(X)$ and $\emptyset$ is a subset of $\emptyset \times \emptyset = \emptyset$.

Therefore, $X^\emptyset = \{ \emptyset \}$

Case 2: $\emptyset^X$

Consider if $f$ is the empty set function.

Does a $Z$ satisfy the following formula? $Z \in P(\emptyset)\  \land \emptyset \subseteq X\times Z$

Yup. If $Z=\emptyset$, then $\emptyset \in P(\emptyset)$ and $\emptyset \subseteq X \times \emptyset=\emptyset$

Therefore, $\emptyset^X=\{\emptyset\}$

So where exactly did I go wrong?

答:
The empty set function isn´t a function from $X$ to $\varnothing$, because its domain is $\varnothing$, not $X$. So the problem is in case 2.

Moreover, your formula $f \in B^A \leftrightarrow \exists Z ( Z \in P(B)\  \land f \subseteq A \times Z)$ is not right, because any subset $f$ of $A\times B$ satisfies $\exists Z ( Z \in P(B)\  \land f \subseteq A \times Z)$ (putting $Z=B$).

A correct formula would be:

$f\in B^A \leftrightarrow (f\subseteq A\times B)\land(\forall a\exists b \;\langle a,b\rangle\in f)\land (\forall a\forall b\forall c\neg(b\neq c\land\langle a,b\rangle\in f\land \langle a,c\rangle\in f))$.

Where the first parenthesis says the domain and codomain are $A$ and $B$ respectively, the second one says that every element of $A$ has an image and the third one says no element of $A$ can have two different images.

Title: $X^\emptyset$ versus $\emptyset^X$


问:
Finding planar curve $\beta(s)$ given curvature $k(s) = \frac{1}{s}$, $\beta(1) = (0, 0)$, $\beta'(1) = (1, 0)$
Find the curve $\beta(s) = (x(s),y(s))$ with $|\beta'(s)|= 1$ for all $s >0$ such that $\beta(1) = (0,0)$, $\beta'(1) = (1,0)$ , and curvature $k(s) =\frac{1}{s}$ for all $s >0$.

I am guessing some sort of integral is required since we are given 'initial' conditions, but I do not see how to begin.

答:
Once you have a continuous function $k:(a,b)\rightarrow\mathbb{R}$ there exists always a planar curve $\beta:(a,b)\rightarrow\mathbb{R}^2$ parametrized by arc-lenght and such that $k_\beta(s)=k(s)$, where $k_\beta(s)$ is the curvature of such $\beta$.

In your case since $|\beta'(s)|=1$, then $k_\beta(s)=\theta'(s)$ where $\beta'(s)=T(s)=(cos\theta(s); sin\theta(s))$. In particular \begin{equation*}\theta(s)=\int k_\beta(s)ds
\end{equation*}is determined up to a constant (such constant will depend on the initial condition). In your case you have to integrate the function $\frac{1}{s}$ and once you have found $\theta(s)$ put it in the definition of $T(s)$ then integrate both coordinates to find $x(s), y(s)$ which are the components of the planar curve.

It's a useful exercise to do if your are approaching the first time to these topics.

EDIT Once you mandaged to solve the integral defining $\theta(s)$ you'll have an equation of the form $\theta(s)=f(s)+c$, where $f(s)$ is the function integrated and $c\in\mathbb{R}$. Put this (explicit) expression in $T(s)=\beta'(s)$ and use the hypothesis $\beta'(1)=(1,0)$, you'll find the constant $c$. The same will happen when you integrate $T(s)$ to find $x(s),y(s)$, you'll get another constant $\bar{c}$ and use the hypothesis $\beta(1)=(0,0)$


问:
Combinatorics Problem on proving that a particular sum is 0
I'm having some issues with proving that the following sum is $0$ for any value of $n \geq 2$:

$$
\sum_{j=1}^{n} \frac{1}{\prod_{i=1,i\neq j}^{n}(a_{j}-a_{i})}
$$

where the $a_i$ are non-zero and positive. Just wondering if anyone had any pointers, the first few cases are easy to verify, i.e $n=2,3$.

So for $n=2$ the sum reads:

$$
\frac{1}{a_2 -a_1}+\frac{1}{a_1-a_2} = 0
$$

and for $n=3$ it would be
$$
\frac{1}{(a_{3}-a_{1})(a_{2}-a_{1})}+\frac{1}{(a_1-a_2)(a_3-a_2)}+\frac{1}{(a_1-a_3)(a_2-a_3)} = 0,
$$
it's just the general case I'm struggling to find an argument for. Thanks!

答:
Not an answer... but maybe useful

Let $p$ be the polynomial $p(x) = (x-a_1) \dots (x-a_n)$. Then for $1 \le j \le n$

$$p^\prime(a_j)= \prod_{\substack{i=1 \\ i \neq j}}^{n}(a_{j}-a_{i})$$ and the result to be proven is equivalent to

$$\sum_{j=1}^n \frac{1}{p^\prime(a_j)} = 0.$$ However, I don't know how to prove that for the time being...

One can also notice that

$$\lim\limits_{x \to a_j} \frac{x-a_j}{p(x)} = \frac{1}{p^\prime(a_j)}.$$


问:
Are absolute extrema only in continuous functions?
The Extreme Value Theorem says that if $f(x)$ is continuous on the interval $[a,b]$ then there are two numbers, $a≤c$ and $d≤b$, so that $f(c)$ is an absolute maximum for the function and $f(d)$ is an absolute minimum for the function.

So, if we have a continuous function on $[a,b]$ we're guaranteed to have both absolute maximum and absolute minimum, but functions that aren't continuous can still have either an absolute min or max?

For example, the function $f(x)=\frac{1}{x^2}$ on $[-1,1]$ isn't continuous at $x=0$ since the function is approaching infinity, so this function doesn't have an absolute maximum.
Another example: suppose a graph is on a closed interval and there is a jump discontinuity at a point $x=c$, and this point is the absolute minimum.

The extreme value theorem requires continuity in order for absolute extrema to exist, so why can there be extrema where the function isn't continuous?

答:
This is a difference between necessary and sufficient conditions, see here.

The extreme value theorem states that continuity on a closed interval is sufficient to ensure that the function attains a maximum and minimum. However, this condition is not necessary. Consider
$$
f(x) = \begin{cases}
1, \text{ for $x=0$} \\
0, \text{ elsewhere}.
\end{cases}
$$
Clearly, $\max(f)=1$, $\min(f)=0$, but $f$ is discontinuous at $x=0$.


问:
why $[F(a):F(a^2)]=2?$
Prove that  if $[F(a):F]$ is odd then  $F(a)=F(a^2)$

My attempt : I got answer here

But i have some confusion in the given answer

My confusion is given below and marked in red

Problem $\mathbf{1.(a)}$. State true or false with justitication. For fields $F\subseteq K$, $\alpha\in K$, if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$.

Proof. Suppose that $F(\alpha)\ne F(\alpha^2)$. Cleary this implies $\alpha\notin F(\alpha^2)$. We can also conclude that the minimal polynomial of $\alpha$ over $F(\alpha^2)$ is $x^2-\alpha^2$. Hence $\color{red}{[F(\alpha):F(\alpha^2)]=2.}$ But we know that $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$. Hence we must have $2\mid[F(\alpha):F]$ and thus we arrive at a contradiction. So $F(\alpha)=F(\alpha^2)$ and the given stament is true.$\square$

My confusion : why $[F(a):F(a^2)]=2?$

My thinking : Here $f(x)=x^2-a^2$  is reducible in $F(a^2) \implies  a\in F(a^2)$

Put $x= a$ , then $f(a)= a^2 -a^2 =0$

So $[F(a):F(a^2)]=1$

答:
The polynomial $f(x)=x^2-a^2$ is not reducible as $a\notin F(a^2)$ by assumption (otherwise $F(a)\subseteq F(a^2)$ and hence $F(a)=F(a^2)$ as trivially $F(a^2)\subseteq F(a)$). In particular, you cannot put $x=a$ as this is not an element of your base field $F(a^2)$.

Hence, indeed $\min_{F(a^2)}(a)=x^2-a^2$ and consequently $[F(a):F(a^2)]=\deg(f)=2$.

Title: why $[F(a):F(a^2)]=2?$


问:
Number of smooth numbers less than x
A $p$-smooth number is defined as an integer whose prime factors are all less than or equal to $p$

In the wiki article about smooth numbers it states:

Let $\displaystyle \Psi (x,y)$ denote the number of $y$-smooth integers less than or equal to $x$ (the de Bruijn function). If the smoothness bound $B$ is fixed and small, there is a good estimate for $\displaystyle \Psi (x,B)$:
$$\displaystyle \Psi (x,B)\sim \frac {1}{\pi (B)!}\prod _{p\leq B}{\frac {\log x}{\log p}}$$
where $\displaystyle \pi (B)$ denotes the number of primes less than or equal to $B$.

How was this derived? Are there any other good bounds on $\Psi(x,y)$?

答: If we let $y=5$, for example, each $5-$smooth number $N$ can be written as $N=2^a3^b5^c$.  We then have $\log N=a\log 2 + b\log 3 +c\log 5$.  You can think of a lattice, three dimensional because we have three primes, with each $5-$smooth number identified with the point $(a,b,c)$ in the lattice.  The lattice spacing is $\log 2, \log 3, \log 5$ in the three dimensions.  $\Psi(x,5)$ is then the number of lattice points nearer the origin than a plane that goes through the point $\log N$ on each axis.  There are $\frac {\log N} {\log 2}$ points along the $2$ axis and similarly along the other axes.  This means the volume of a lattice cell is $\prod_{p \le 5} \log p$, which gets us the product in your expression.  The fact that we have a tetrahedron instead of a rectangular block give a factor $\frac 16$, which corresponds to the $\frac 1{\pi(B)!}$ in your expression.

Title: Number of smooth numbers less than x


问:
Differential Equations Reference Request
Currently I'm taking the Differential Equations course at college, however the problem is the book used. I'll try to make my point clear, but sorry if this question is silly or anything like that: the textbook used (William Boyce's book) seems to assume that the reader doesn't familiarity with abstract math, so it lacks that structure of presenting motivations, then definitions, then theorems and corolaries as we see in books like Spivak's Calculus or Apostol's Calculus.

I've already seem a question like this on Math Overflow, however unfortunatelly some people felt offended somehow and said: "are you saying that Boyce is easy? It doesn't matter if you know how to prove things, you must learn to compute", and my point is not that: Apostol and Spivak teaches how to compute also, however since their books are aimed to mathematicians they take care to build everything very fine and their main preocupation is indeed in the theoretical aspects.

I really don't like the approach of: well, in some strange way we found that this equation works, so memorize it, know how to compute things and everything is fine. I really want to understand what's going on, and until now I didn't find this possible with Boyce's book (certainly there are people that find this possible, but I'm used to books like Spivak's and Apostol's, so I don't really do well with books like Boyce's).

I've already seem Arnold's book on Differential Equations, but the prerequisites for reading it are bigger: he uses diffeomorphisms many times and although I'm currently also studying differential geometry, I don't feel yet comfortable on reading a book like this one.

Can someone recommend a book that covers ordinary differential equations, systems of differential equations, partial differential equations, and so on, but that can be read without much prerequisite, and that still has the structure of a book of mathematics? And when I say "structure of book of mathematics" is being like Spivak's and Apostol's books: not mixing up definitions, theorems and examples inside histories of how the theory developed. Since I'm a student of Mathematical Physics, of course motivations and examples from Physics are welcome, but not all mixed up in the text.

In truth I don't really believe that a book like the one I described exists (if a book on this topic is good in my point of view, I believe it'll have a lot of prerequisites). Anyway I hope that I don't get misunderstood in what's my doubt, and really sorry if this question is silly in someway.

答: I would recommend Differential Equations, Mechanics, and Computation You can read first pages to figure out whether it works for you. The book is 1) introductory, 2) rigorous, 3) with emphasis to physics. </p>