问:
functional calculus on C*-algebra and majoration of $||f(A+B)-f(A)||$
In introductory books on C*-algebra, functionnal calculus is quicly presented as it is a powerful tool in the field. It is for example prooved that for an operator $a$ and a function $f$ continous on Sp$(a)$, $f(a)$ is well defined and that $||f(a)|| \leq \sup\{|f(x)|, x \in \text{ Sp}(a) \}$.

However for the moment I never find some reference that deal with functionnal inequality with two operators. For exemple it is known that for two operators a,b we have that Sp$(a+b) \subset$ Sp$(a) +$ Sp$(b)$ so personnaly for a function $f$ continuous on Sp$(a) +$ Sp$(b)$, I would expect to have an explicit inequality that bound $f(a+b)-f(a)$ and decay when $b$ decrease toward 0.

For the moment I only find one proof (see Continuity of functional calculus) that show that the it go to zero when b go to zero. But it's only a proof with abstract epsilon cutting that do not give any uniform bound. So I would like to know if there is any litterature that go into a more uniform majoration of this quantity.

答:
Bounding differences $\|f(a+b)-f(a)\|$ in terms of $\|b\|$ is much harder than the norm bound from the first paragraph, so I'll just give you some pointers to the relevant literature.

As a first example, it is known [1] that there does not exist a constant $C>0$ such that
$$
\|\,|a+b|-|a|\,\|\leq C\|b\|
$$
for all bounded self-adjoint operators $a,b$.

Continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ with the property that there exists a constant $C>0$ such that $\|f(a+b)-f(b)\|\leq C\|b\|$ for all bounded self-adjoint operators are called operator Lipschitz functions [2]. Similarly one can define operator Hölder functions, and this set is again a proper subset of Hölder functions.

To get bounds on the difference $\|f(a+b)-f(b)\|$, one typically uses the theory of double operator integrals [3,4]. If $a,b$ are bounded self-adjoint operators on $H$ with spectral measures $e,f$, then
$$
f(a+b)-f(b)=\int_{\mathbb{R}}\int_{\mathbb{R}}\phi_f(s,t)\,de(s)\,b\,df(t)
$$
with
$$
\phi_f(s,t)=\begin{cases}\frac{f(s)-f(t)}{s-t}&\text{if }s\neq t,\\ 0&\text{if }s=t.\end{cases}
$$
Then $\|f(a+b)-f(b)\|$ can be bounded by the norm of $\phi_f$ in the extended Haagerup tensor product $L^\infty(\mathbb{R},e)\otimes_{eh}L^\infty(\mathbb{R},f)$ times the norm of $b$.

[1] T. Kato: Continuity of the map $S \mapsto |S|$ for linear operators. Proc. Jpn. Acad. 49, 157–160 (1973)

[2] A. B. Aleksandrov and V. V. Peller. Operator Lipschitz functions. Russ. Math. Surv. 71 605 (2016)

[3] M. Sh. Birman and M. Solomyak. Double operator integrals in a Hilbert space. Integral Equations Operator Theory, 47(2):131–168 (2003)

[4] B. de Pagter, H. Witvliet, and F.A. Sukochev. Double Operator Integrals, Journal of Functional Analysis, 192(1):52-111 (2002)


问:
Partition Properties
I'm currently trying to understand the proof of Erdős–Rado in Jech's Set Theory book.

But since I've already problems in the first steps, I went back to the proof of:
$$2^\kappa \not\rightarrow (\omega)_{\kappa}^{2}$$

He wants to find a partition that has no homogeneous set of size 3. $S=\{0,1\}^{\kappa}$ and $F:[S]^2\rightarrow \kappa$ by $F(\{f,g\})=$ the least $\alpha < \kappa$ such that $f(\alpha)\not=g(\alpha)$.

Now comes the part I don't understand:

If $f,g,h$ distinct elements of $S$, then $F(\{f,g\})=F(\{f,h\})=F(\{g,h\})$ is impossible.

Why is it impossible? why can't there be an $\alpha$ s.t. $f(\alpha)\not= g(\alpha)\not=h(\alpha)$?

If someone also knows a nice proof of the Erdős–Rado theorem I'd be happy to hear about it!

Best, Luca

答:
Miha Habič and Asaf Karagila have already answered your main question.

The simplest proof of the Erdős-Rado theorem that I’ve seen that does not explicitly use model-theoretic tools is this one. Peter Komjáth presents it in this video clip; I’ve expanded a few of the details, but I’ve kept pretty much the same notation, in case you want to see and hear the lecture as well.

Suppose that $F:\left[(2^\kappa)^+\right]^2\to\kappa$; we’ll show that $(2^\kappa)^+$ has a homogeneous subset of power $\kappa^+$.

Let $S=\{\alpha\in(2^\kappa)^+:\operatorname{cf}\alpha=\kappa^+\}$; $S$ is a stationary subset of $(2^\kappa)^+$. (The general fact is that $\{\alpha\in\lambda:\operatorname{cf}\alpha=\mu\}$ is stationary in $\lambda$ whenever $\mu$ and $\lambda$ are regular cardinals with $\mu<\lambda$.) For each $\alpha\in S$ and $i\in\kappa$ we’ll try to build an increasing sequence $\langle x_{\alpha,i}(\xi):\xi<\eta_{\alpha,i}\rangle$ of ordinals less than $\alpha$ so that if $X_{\alpha,i}=\{x_{\alpha,i}(\xi):\xi<\eta_{\alpha,i}\}$, then $X_{\alpha,i}\cup\{\alpha\}$ is homogeneous for color $i$.

Fix $\alpha\in S$ and $i\in\kappa$. If there is no $x<\alpha$ such that $\{x,\alpha\}$ is homogeneous for color $i$, then $\eta_{\alpha,i}=0$, and $X_{\alpha,i}=\varnothing$. Otherwise, let $x_{\alpha,i}(0)=\min\{x<\alpha:\{x,\alpha\}\text{ is homogeneous for color }i\}$. Suppose that $\eta\le\kappa^+$, and we’ve already chosen $x_{\alpha,i}(\xi)$ for all $\xi<\eta$. Let $X_{\alpha,i}(\eta)=\{x_{\alpha,i}(\xi):\xi<\eta\}$. If there is an ordinal $x<\alpha$ such that $X_{\alpha,i}\cup\{x,\alpha\}$ is homogeneous for color $i$, let $x_{\alpha,i}(\eta)$ be the least such ordinal. Otherwise, $\eta_{\alpha,i}=\eta$.

If there are an $\alpha\in S$ and $i\in\kappa$ such that $\eta_{\alpha,i}=\kappa^+$, we’re done: $X_{\alpha,i}\cup\{\alpha\}$ is homogeneous for $i$ and is certainly of power $\kappa^+$. Indeed, we’ve shown more: it’s of order type $\kappa^++1$. Suppose, then, that $\eta_{\alpha,i}<\kappa^+$ for each $\alpha\in S$ and $i\in\kappa$.

For $\alpha\in S$ let $X_\alpha=\bigcup_{i\in\kappa}X_{\alpha,i}$; $|X_{\alpha,i}|\le\kappa$ for each $i\in\kappa$, so $|X_\alpha|\le\kappa$. Let $f(\alpha)=\sup X_\alpha$; $\operatorname{cf}\alpha=\kappa^+$, so $f(\alpha)<\alpha$. Thus, $f:S\to(2^\kappa)^+$ is a pressing-down (or regressive) function on the stationary set, so by the pressing-down lemma (Fodor’s lemma) there are a stationary $S'\subseteq S$ and $\gamma\in(2^\kappa)^+$ such that $f(\alpha)=\gamma$ for each $\alpha\in S'$.

For each $\alpha\in S'$, then, $\big\langle\langle x_{\alpha,i}(\xi):\xi<\eta_{\alpha,i}\rangle:i\in\kappa\big\rangle$ is a $\kappa$-sequence of $\le\kappa$-sequences in $\gamma$. Now $|\gamma|\le 2^\kappa$, so there are at most $(2^\kappa)^\kappa=2^\kappa$ $\le\kappa$-sequence in $\gamma$, and similarly there are at most $2^\kappa$ $\kappa$-sequences of these sequences. On the other hand, $S'$, being stationary, is certainly cofinal in $(2^\kappa)^+$, so $|S'|>2^\kappa$. Thus, there must be distinct $\alpha,\beta\in S'$ such that $\eta_{\alpha,i}=\eta_{\beta,i}$ for each $i\in\kappa$ and $x_{\alpha,i}(\xi)=x_{\beta,i}(\xi)$ for each $i\in\kappa$ and $\xi<\eta_{\alpha,i}$.

Without loss of generality assume that $\alpha<\beta$, let $i=F(\{\alpha,\beta\})$, and let $\eta=\eta_{\alpha,i}=\eta_{\beta,i}$; then $$X_{\beta,i}\cup\{\alpha,\beta\}=X_{\alpha,i}\cup\{\alpha,\beta\}$$ is homogeneous for $i$. But this means that $\{\alpha,\beta\}\cup X_{\beta,i}(\eta)$ is homogeneous for color $i$, where $\alpha<\beta$, and therefore in the third paragraph we would have defined the ordinal $x_{\beta,i}(\eta)$ and ensured that $\eta_{\beta,i}>\eta$. This contradiction shows that in fact there must be some $\alpha\in S'$ and $i\in\kappa$ such that $\eta_{\alpha,i}=\kappa^+$, which completes the proof.

ContentLicense: CC BY-SA 3.0

Title: Partition Properties


答:
For linear multistep methods to detect the order it is sufficient to insert $$u(x)=e^x,~~~ u_j=e^{jh},$$ at $j=0$ or $j=1$ into the step equation and to compute the defect between both sides. After cancelling the common power it remains
$$
E(h)=e^h-(1-a)e^0-ae^{-h} - \frac{h}{4}((a+3)e^h+(3a+1)e^{-h})~
$$
Now insert the power series for the exponential and collect the coefficients of equal degree.
If the defect has order $E(h)=O(h^{p+1})$, then $p$ is the order of the method.

Using a computer algebra system (like the CAS Magma this can be done automatically and gives

```<code>> A<a>:=FunctionField(Integers());
> P<h>:=PowerSeriesRing(A);
> Exp(h)-(1-a)-a*Exp(-h)-h/4*((a+3)*Exp(h)+(3*a+1)*Exp(-h));

(-a - 1)/3*h^3 + (a - 1)/24*h^4 + (-a - 1)/30*h^5
    + (a - 1)/360*h^6 + (-a - 1)/840*h^7 + ...
</code>```

So order $p=2$ for $a\ne -1$, order $p=3$ for $a=-1$.

ContentLicense: CC BY-SA 4.0

Title: What is the order of linear multi step method $u_{j+1}=(1-a)u_j+au_{j-1}+h/4\{(a+3)u'_{j+1}+(3a+1)u'_{j-1}\}$ for solving $u'=f(x,u)~?$

Body:
Problem: The order of linear multi step method $~u_{j+1}=(1-a)u_j+au_{j-1}+h/4\{(a+3)u'_{j+1}+(3a+1)u'_{j-1}\}~$ for solving $~u'=f(x,u)~$ is

$(1)~~~2~~~~$ if $~~a=-1$

$(2)~~~2~~~~$ if $~~a=-2$

$(3)~~~3~~~~$ if $~~a=-1$

$(4)~~~3~~~~$ if $~~a=-2$

My work: The general multi-step (k-step) method for the solution of $u'=f(x,u)~,~~u(x_0)=u_0~$ may be written as
$$u_{j+1}=a_1u_j+a_2u_{j-1}+\cdots+a_ku_{j-k+1}+h\left(b_0u'_{j+1}+b_1u'_j+\cdots+b_ku'_{j-k+1}\right)$$ and $~T_{j+1}~$ which is the local truncation error is written as
$$T_{j+1}=c_0u(x_j)+c_1hu'(x_j)+c_2h^2u''(x_j)+\cdots+c_ph^pu^{(p)}(x_j)+T_{p+1}$$where $~c_0=1-\sum_{i=1}^ka_i~,$

$c_q=\frac{1}{q!}\left[1-\sum_{i=1}^ka_i(1-i)^q\right]-\frac{1}{(q-1)!}\sum_{i=0}^kb_i(1-i)^{q-1}~,~~~q=1,2,\cdots,p$

The linear multi-step method is said to be of order $~p~$ if $~c_0=c_1=c_2=\cdots=c_p=0~$ and $~c_{p+1}\ne0~.$

Given that $~u_{j+1}=(1-a)u_j+au_{j-1}+h/4\{(a+3)u'_{j+1}+(3a+1)u'_{j-1}\}~$

Comparing $~a_1=1-a~,~~a_2=a~,~~b_0=(a+3)/4~,~~b_1=0~,~~b_2=(3a+1)/2~$ and $~k=2~.$

So $~c_0=1-\sum_{i=1}^2a_i=1-a_1-a_2=1-1+a-a=0~$

$c_1=\frac{1}{1!}\left[1-\sum_{i=1}^2a_i(1-i)^1\right]-\frac{1}{(0)!}\sum_{i=0}^2b_i(1-i)^{0}=1+a_2-b_0-b_2=0$

Similarly, $~c_2=0~,~~c_3=-1/3(1+a)$

if $~a=-1~,~$ then $~c_3=0~$

if $~a=-2~,~$ then $~c_3=1/3\ne0~$

Hence if $~a=-2~,~$ then $~c_0=c_1=c_2=0~$ but $~c_3\ne0~.$
Therefore the order is $~2~$ for $~a=-2~$  and then option $\bf (2)$ is correct.

Also by the similar process, $~a_4=-1/24(1-a)~$
if $~a=-1~,~c_4=-1/12\ne0~.$
 Therefore the order is $~3~$ for $~a=-1~$ and then option $\bf (3)$ is correct.

My Query: Please anyone help me to check  if there be any wrong. Or let me inform, if there be any alternative process to solve it. Thank you.

Body:
For linear multistep methods to detect the order it is sufficient to insert $$u(x)=e^x,~~~ u_j=e^{jh},$$ at $j=0$ or $j=1$ into the step equation and to compute the defect between both sides. After cancelling the common power it remains
$$
E(h)=e^h-(1-a)e^0-ae^{-h} - \frac{h}{4}((a+3)e^h+(3a+1)e^{-h})~
$$
Now insert the power series for the exponential and collect the coefficients of equal degree.
If the defect has order $E(h)=O(h^{p+1})$, then $p$ is the order of the method.

Using a computer algebra system (like the CAS Magma this can be done automatically and gives

> A<a>:=FunctionField(Integers());
> P<h>:=PowerSeriesRing(A);
> Exp(h)-(1-a)-a*Exp(-h)-h/4*((a+3)*Exp(h)+(3*a+1)*Exp(-h));

(-a - 1)/3*h^3 + (a - 1)/24*h^4 + (-a - 1)/30*h^5
    + (a - 1)/360*h^6 + (-a - 1)/840*h^7 + ...

So order $p=2$ for $a\ne -1$, order $p=3$ for $a=-1$.


问:
Proof of separating hyperplane theorem using minimax theorem for zero sum games
Let $x_1,\ldots,x_M\in\mathbb{R}^N$. Let $P \equiv conv \{x_1,\ldots,x_M\}$ denote the convex hull of these points. Using the Minimax Theorem prove that for all $y\in\mathbb{R}^N \setminus P$ there exists a vector $v\in[-1,1]^N$ such that $v\cdot(x-y)>0$ for all $x\in{}P$.

(Hint: Find a way of identifying $v$ and $x$ with mixed
strategies in a zero-sum game.)

My attempt: I have been able to prove the other direction (that too with help from online lecture notes) - i.e. how to use the separating hyperplane theorem to prove the minmax theorem. But I'm unable to find a way to proceed with the direction required in the problem. Any help is appreciated.

答:
Using LinAlg's hint, here is my attempt:

Consider the two-player zero-sum game where player 1 (the maximizer) chooses $g\in\{-1,1\}^N$ and player 2 (the minimizer) chooses one of the points $j\in\{1,\ldots,M\}$, and player 1's payoff is $\sum_{k=1}^Ng_k\cdot(p^j_k-q_k)$. By the Minimax Theorem, this game has a value $v$. Note that we can identify mixed strategies of player 1 with elements of $[-1,1]^N$, and mixed strategies of player 2 can be identified with elements of $P$ (by taking expectations). So the expected utility of player 1 from a mixed strategy profile $(\gamma,p)$ is $\sum_{k=1}^N\gamma_k(p_k-q_k)$.

Let $\gamma^*\in[-1,1]^N$ be a maxminimizer and let $p^*\in P$ be a minmaximizer. Then $v=\sum_{k=1}^N\gamma^*_k\cdot(p^*_k-q_k)$. Note that $\gamma=p^*/\lVert p^*\rVert\in[-1,1]^N$, so $v\geq\sum_{k=1}^{N}\gamma_k\cdot(p^*_k-q_k)=\lVert p^*-q\rVert>0$ (since $q\notin P$, and hence $q\neq p^*$). Thus, the value is positive, and so for any mixed strategy $p$, we have $\sum_{k=1}^{N}\gamma^*_k(p_k-q_k)>0$, as desired.


问:
Differentiability of a function of several variables at $0$
The following is a problem from Spivak's Calculus On Manifolds

Q) Let $ f:\mathbb R^n \to \mathbb R $ be a function such that $|f(x)| \leq |x|^2 $. Show that $f$ is a differentiable at $0$.

The solutions given on the solution manuals is as follows.

$$ |f(0)| \leq ||0||^2 \implies f(0)=0 $$

$$ Since \  \lim_{x\rightarrow 0}\frac{|f(x)-f(0)|}{||x||}= \lim_{x\rightarrow 0}\frac{|f(x)|}{||x||}\leq \lim_{x\rightarrow 0}||x||=0$$ $$\implies Df(0)(x)=0$$
My question is how did we come to the conclusion that $Df(0)(x)=0$ ? Did we just guess the value of the linear transformation $Df(0)(x)$ so that the function $f$ could be differentiable at $0$. Aren't we supposed to check for differentiability ? I understand that if the function $f$ is differentiable at $0$ then the derivative of the function at the point $0$ is a linear transformation $Df(0):\mathbb R^n \to \mathbb R$ such that
$$ \lim_{x\rightarrow 0}\frac{||f(x)-f(0)-Df(0)(x)||}{||x||}=0$$ How can we prove differentibility using this definition ?

答:
Let $e_1,\ldots,e_n$ be the canonical base of $\mathbb{R}^n$. Then, using that $|e_i|=1$ for all $i\in\{1,\ldots,n\}$ we get:

$$\dfrac{|f(t\cdot e_i)-f(0)|}{|t|}=\dfrac{|f(t\cdot e_i)|}{|t|·||e_i||}=\dfrac{|f(t·e_i)|}{||t\cdot e_i||}\leq ||t\cdot e_i||=|t|\cdot ||e_i||=|t|$$

Taking limit $t\to 0$ we get that: $$\lim_{t\to 0} \dfrac{|f(t\cdot e_i)-f(0)|}{|t|}=0 \implies D_i f(0):=\lim_{t\to 0}\dfrac{f(t\cdot e_i)-f(0)}{t}=0$$


问:
What does the notation "$\;(f·g)(x)\;$" mean?
Suppose I have two functions $f(x)$ and $g(x)$. What does it mean to have $(f·g)(x)$?

This notation can be found in a paper on Partial Differential Equations and Image Processing:

What we call a multiscale analysis is a family of transforms $(T_t)_{t>0}$ of an image $u^o(x)$, such that $u(t,x) = (T_t \cdot u^o)(x)$ can be seen as the image $u^o$ at a coarser scale $t$.

答:
When working with functions there are several standard notations that you should have always present:

The $\cdot$ is used to describe the usual product (sometimes called pointwise product):$$f(x)\cdot g(x)=f(x)g(x).$$
You can also compose both functions using $\circ$ in the following way:$$f\circ g = f(x)\circ g(x)=f(g(x)).$$
Finally, the notation used in the paper you mentioned, has to do with the convolution. The symbol $*$ is used to express the convolution of $f$ with $g$ (it is read this way), and it is defined in the following way:$$f*g(x)=(f*g)(x)=\int_{-\infty}^{\infty}f(\xi)g(x-\xi)d\xi.$$ This operation is closely related to the notion of Fourier transform and very popular in the real analysis, signal processing and differential equations fields.


问:
Winding number of a curve (not complex analysis)
I am asked to calculate the winding number of an ellipse (it's clearly 1 but I need to calculate it)

I tried two different aproaches but none seems to work.

I would like to know why none of them work (I believe it is because these formulas only work if I have a curve parametrized by arc lenght).

Approach 1:

A valid parametrization : $\gamma=(a\cos t,b\sin t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$

$\dot{\gamma}(t)=(-a\sin t,b\cos t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$

$\ddot{\gamma}(t)=(-a\cos t,-b\sin t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$

$\det(\dot{\gamma}(t)|\ddot{\gamma}(t)) = \renewcommand\arraystretch{1.2}\begin{vmatrix}
-a\sin t & -a\cos t \\ b\cos t & -b\sin t \end{vmatrix}=ab \sin^2 t+ab \cos^2 t=ab$

$||\dot{\gamma}(t)||^3=(\displaystyle\sqrt{(-a\sin t)^2+(b\cos t)^2})^3=(\displaystyle\sqrt{a^2\sin^2 t+b^2\cos^2 t})^3=a^3b^3$

$\kappa(t)=\displaystyle\frac{ab}{a^3b^3}=\displaystyle\frac{1}{a^2b^2}$

$\mathcal{K}_\gamma = \displaystyle\int_{0}^{2\pi} \displaystyle\frac{1}{a^2b^2} \ dt= \displaystyle\frac{2\pi}{a^2b^2}$, $\mathcal{K}_\gamma$ is the total curvature of the curve.

$i_\gamma=\displaystyle\frac{\displaystyle\frac{2\pi}{a^2b^2}}{2\pi}=\displaystyle\frac{1}{a^2b^2}$...which is not necessarily 1.

Approach 2:

Winding # = $\displaystyle\frac{1}{2\pi}\displaystyle\int_{\gamma}\displaystyle\frac{-y}{x^2+y^2}\>dx+\displaystyle\frac{x}{x^2+y^2}\>dy$

That gives us $\displaystyle\frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\left( \displaystyle\frac{-b\sin t}{a^2\cos^2 t+b^2\sin^2 t}(-a\sin t)+\displaystyle\frac{a\cos t}{a^2\cos^2 t+b^2\sin^2 t}(b\cos t) \right)\>dt$

$\displaystyle\frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\left( \displaystyle\frac{ab}{a^2\cos^2 t+b^2\sin^2 t }\right)\>dt$, which I computed and cannot be calculated.

Clearly the second approach is valid if we are dealing with a circumference of radius 1. We can generalize for the elipsee using Green's Theorem. I would also like if someone could show me this way as well.

Thank you

答:
The last integral indeed has a closed form to evaluate you can exploit the symmetry about $\pi$ to get the integral as $$ \begin{align}
 \pi I &= ab \int_{0}^{\pi} \dfrac{1}{a^2 \cos^2(t) + b^2 \sin^2(t)}dt\\
&= 2ab\int_{0}^{\pi/2} \dfrac{\sec^2(t)}{a^2+b^2 \tan^2(t)} dt \end{align}$$
Now substituite $\tan(t) = u$
$$ \dfrac{\pi}{2ab} I = \int_{0}^{\infty}\dfrac{1}{a^2+b^2x^2}dx = \dfrac{\pi}{2ab} $$
thus giving $I =1$


问:
Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $f(mn)=f(m)+f(n)+3f(m)f(n)$ for $m,n \geqslant1.$

Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $$f(mn)=f(m)+f(n)+3f(m)f(n)$$ for $m,n \geqslant1.$

This was a problem from a Putnam practice book and I couldn't seem to figure out how to show this. Initially, I started to wonder that if the function were bijective then it has to be injective as well as surjective. So trying to determine if its injective let $a,b \in \mathbb{N}$ then if it were injective $$f(mn)=f(ab) \implies f(m)+f(n)+3f(m)f(n) =f(a)+f(b)+3f(a)f(b)$$ but this doesn't seem to help since I don't have explicitly $f$. How should I look at this problem?

答:
1.Show that the function is well defined
2.Prove that there exist a kernel.
3. Prove that identity maps to element of the kernel
If you can't prove 1 and 2 the function or mapping is not bijective.
In other words, you can prove that the function is both injective and surjective( i.e if the inverse of the function maps back to every element in the domain) as required.</p>


问:
Hard improper integral $\int_0^\infty \frac{(1+\frac{1}{x})^{-x}}{x^2}dx $
I have been stuck on this problem for a while:
$$\int_0^\infty \frac{(1+\frac{1}{x})^{-x}}{x^2}dx $$

I have toyed around with substitution and integration by parts with no luck. I have also searched for this on the internet but nothing has come up. Any help would be appreciated

答:
Note that $f:(0, 1] \to \Bbb R$ defined by

$$f(x) = \left(1 + \frac1x\right)^{-x}$$
is a decreasing function and thus, $f(x) \ge f(1) = 0.5$.

As a result, your integrand exceeds $0.5x^{-2}$ on $(0, 1]$. However, the improper integral of $x \mapsto x^{-2}$ diverges on $(0, 1]$. By comparison, so does your integral.


问:
Can this Diophantine equation have infinitely many solutions?
I came across this as a intermediate step while solving another problem in analysis.

Suppose $m$, and $n$ are integers satisying,

$$\frac{1}{n} - \frac{1}{m} = \alpha$$ where $\alpha$ can be any non zero real.

I essentially need to show that only finitely many $(n,m)$ pair solutions are possible to complete my proof. How can I proceed? Clearly, we only need to check when $\alpha$ is rational.

答: Write the equation as $\frac{1}{n} - \frac{1}{m} = \frac{r}{s}$ with $r,s\in\mathbb Z$. Clear denominators to get $rmn-sm+sn=0$. The standard procedure to solve this diophantine equations is to multiply by $r$ and factor it as $(rm+s)(rn-s)=-s^2$. This gives you only a finite possible values for $rm+s$ and $rn-s$ since they are divisors of $-s^2$.


问:
Finding the values of $z$ s.t. $\sum_{n=0}^{\infty} \left( \frac{z}{1+z} \right)^n$ is convergent
I manipulated the series to $\sum_{n=0}^{\infty} \left( \frac{1}{1-(-1/z)} \right)^n$, which converges for $|-1/z|<1$ by geometric series. Then solving for $z$, I obtained $z>(1/\bar{z})$.

Is this correct? I was expecting some numerical values, so I'm not sure about my answer.

答:
Let me suggest another approach. Put $w=\dfrac{z}{z+1}$. Where will $$\sum_{n=0}^\infty w^n$$ converge? What does this tell you about the possible domain of $z$?

ADD

I am getting $$2\Re z+1>0$$ Note you cannot really use complex numbers in an inequality, since they are not ordered in any way. However, from $|z|<|1+z|$ you get the equivalent $$z\bar z<(1+z)(1+\bar z)$$ since everything is real. This gives $$z\bar z <1+(z+\bar z)+z\bar z$$ which is the same as $$0<1+2\Re (z)$$


问:
Analytic method for ODE problem
I am studying on a drag force ODE. My question is:

Is there any analytic method to solve $$\frac{dv}{dt}+\alpha v^n=g\\ n \in(1,2]$$ It is somehow look like Bernoulli Differential Equations $y' + p\left( x \right)y = q\left( x \right){y^n}$ but not exactly. It is originated from falling body equation with air drag force.$mg-kv^n=ma$ or $mg-kv^n=m\frac{dv}{dt}$ that divide both sides by $m$ and rewrite as $\frac{dv}{dt}+\frac{k}{m}v^n=g$,$g=9.8\ \frac{m}{s^2}$ for the case of $n=1 $ it is easy to solve $v=g\frac mk (1-e^{-\frac kmt})$ but how to do it for the case of $n=\frac 32 \ $ or $n=2 $

Realy I got to struggle with special two cases... can someone help me? I want to solve to at least for $n=2 $ to compare with the solution that I got by numerical solutions.

Thanks in advance.

答:
$$\frac{dv}{dt}+\alpha v^n=g$$
$$\frac{dt}{dv}=\frac{1}{g-\alpha v^n}$$
$$t=\int \frac{dv}{g-\alpha v^n}+constant$$
$$t=\frac{v}{g}\:_2F_1\left(1\:,\:\frac{1}{n}\:;\:1+\frac{1}{n}\:;\:\frac{\alpha}{n}v^n \right)+c$$
$\:_2F_1\:$ is the Gauss hypergeometric function. https://mathworld.wolfram.com/HypergeometricFunction.html

$v(t)$ is the inverse function, generally not nice.

In some particular values of $n$ the solution is simpler.

Case $n=1 \quad$ linear ODE $\qquad\qquad v(t)=\frac{g}{\alpha}+c\:e^{-\alpha\;t}$

Case $n=2 \quad$ Riccati ODE $\qquad\qquad v(t)=\sqrt{\frac{g}{\alpha}}\tanh\left(\sqrt{\alpha g}\:(t-c) \right)$

Case $n=\frac12 \quad$ the solution involves the LambertW function.

Title: Analytic method for ODE problem


问:
How to write 'and' in mathematical language?
How to write 'and' in mathematical language?

I'm not from the field of math, but I needed to write two very simple equations in the same small line and I'm afraid it'll be confusing since they're very close to each other. So I found it best to put a symbol which means "and", and it'll separate them. Does such a symbol exist?

答: Yes: $\wedge$. For instance, you can write: let $x$ and $y$ be real numbers such that$$x^2+y=2=1\wedge y=2x.$$However, it is more common to write it as$$\left\{\begin{array}{l}x^2+y^2=1\\y=2x.\end{array}\right.$$And there is nothing wrong in writing$$x^2+y=2=1\text{ and }y=2x.$$


答:
This is a perfectly good example:

Consider the equations

$$F(x,y)=0 \quad \text{and}\quad G(x,y)=0.$$

Body:
How to write 'and' in mathematical language?

I'm not from the field of math, but I needed to write two very simple equations in the same small line and I'm afraid it'll be confusing since they're very close to each other. So I found it best to put a symbol which means "and", and it'll separate them. Does such a symbol exist?


问:
Expansion of $(\sinh(x))^{\frac15}$ around 0 for x > 0
I'm aware of the series expansion of the hyperbolic functions, but how does one expand a fractional power of sinus hyperbolicus, i.e. e.g. $(\sinh(x))^{\frac15}$?

答:
The way I usually approach these kind of expansions is by looking at the power series expansion of the inner and outer function. For this problem, the series expansion for $\sinh(x)$ (inner function) is given by
$$\sinh(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}=x+\frac{x^3}{6}+\frac{x^5}{120}+\mathcal{O}(x^7)$$
and for the outer function we have
$$(1+x)^{\frac{1}{5}}=1+\frac{x}{5}-\frac{2x^2}{25}+\mathcal{O}(x^3).$$
Therefore, only including the highest order, the expansion reads:
$$\sinh^{\frac{1}{5}}(x)=\Big(x+\frac{x^3}{6}+\mathcal{O}(x^5)\Big)^{\frac{1}{5}}=x^{\frac{1}{5}}\Big(1+\frac{x^2}{6}+\mathcal{O}(x^4)\Big)^{\frac{1}{5}}=x^{\frac{1}{5}}\Big(1+\frac{x^2}{30}+\mathcal{O}(x^4)\Big).$$
The next highest order term would have the form
\begin{align}
\sinh^{\frac{1}{5}}(x)&=\Big(x+\frac{x^3}{6}+\frac{x^5}{120}+\mathcal{O}(x^7)\Big)^{\frac{1}{5}}\\&=x^{\frac{1}{5}}\Big(1+\frac{x^2}{6}+\frac{x^4}{120}+\mathcal{O}(x^6)\Big)^{\frac{1}{5}}\\&=x^{\frac{1}{5}}\Big(1+\frac{1}{5}\Big(\frac{x^2}{6}+\frac{x^4}{120}\Big)-\frac{2}{25}\Big(\frac{x^2}{6}+\frac{x^4}{120}\Big)^2+\mathcal{O}(x^6)\Big)\\&=x^{\frac{1}{5}}\Big(1+\frac{x^2}{30}-\frac{x^4}{1800} +\mathcal{O}(x^6)\Big)
\end{align}
where in the last line because of the precision up to order $x^6$ the higher order terms from the expansion of $\big(\frac{x^2}{6}+\frac{x^4}{120}\big)^2$ were discarded. This procedure then can be iterated futher and further for higher orders in the expansion.

Bonus: For the more generalized version of your problem, the expansion up to the second highest order has the from
$$\sinh^\alpha(x)=x^\alpha\bigg(1+\frac{\alpha}{6}x^2+\frac{\alpha(5\alpha-2)}{360}x^4+\mathcal{O}(x^6)\bigg).$$


问:
How to denote a sequence indexed by another sequence?
Lets say I have two corresponding finite sequences $A = (3, 4, 6, 1)$ and $B = (x_3, x_4, x_6, x_1)$, and a space $C$ such that $x_i \in C$ for any $i \in \mathbb{N}$.

How can I denote $B$ in terms of $A$? The expression
$$
B = (x_i)_{i \in A}
$$
would be convenient, but, presumably incorrect, since $\in$ isn't used for sequences.

答:
We can use the projection map
\begin{align*}
&\pi_k:X^n\to X\qquad\qquad\qquad 1\leq k\leq n\\
&(x_1,\ldots,x_n)\to x_k\tag{1}
\end{align*}
to extract the $k$-th component of an $n$-tuple $(x_j)_{1\leq j\leq n}$.

Using (1) we can write OPs example as
\begin{align*}
B=\left(x_{\pi_{k}(A)}\right)_{1\leq k\leq |A|}
\end{align*}

Note: With respect to ease readability it might be more convenient to simply write:

Let $A=(a_j)_{1\leq j\leq n}$. We consider $B=\left(x_{a_j}\right)_{1\leq j\leq n}$ ...


问:
Concurrence of three chords
The center $A$ of a circle lies on another circle which cuts the former in $B,C$. $AD$ is a chord of the latter circle meeting $BC$ in $E$. From $D$ are drawn $DF$ and $DG$ to touch the former circle. Prove that $G,E$ and $F$ lie on a straight line.
I tried inversion, but could not prove the result. Any help would be appreciated.

答:
Let the circle $(A)$ be $C_1$ and $(ABC)$ be $C_2$. Note $A,F,D,G$ lie on a circle. Call it $C_3$.

Now $BC$ is radical axis of $C_1,C_2$. Since $C_2,C_3$ intersect in $A,D$, their radical axis is $AD$. Since $C_1,C_3$ intersect in $F,G$, their radical axis is $FG$.

Since none of these three circles are concentric, by radical axis theorem, $BC$, $AD$, $FG$ concur at the $E$. Hence $E$ is collinear with $F,G$.

Title: Concurrence of three chords


问:
Integral over null set is zero
I am sorry for this elementary question, but i could not figure out a rigorous proof for why the Lebesgue integral of any function over a null set is zero.

Thanks for helping!

答:
Hello this was asked over 7 years ago so I am sure that this is most definitely no longer relevant for you in particular but I thought I would answer this for people in the future who might was a slightly different answer. Infact that statement holds for any measure not just Lebesgue.

Firstly a little about the notation I will use:

$\mathbb{1}_A$ is the "indicator function" of the set $A$ you may have instead seen this as $\chi_A$ or $\mathcal{1}_A$ or something else. It simply takes the value $1$ on $A$ and $0$ otherwise.

This proof will require:

Theorem 1:

If $f = g$ $ \mu$ almost everywhere then $\int f d\mu = \int g d\mu$

(A proof of this theorem can be found at the bottom of this answer but this is a fairly simple result in most measure theory courses.)

Proof of main result

We wish to show that $\int_A f d \mu = 0 $ whenever $\mu(A) = 0 $

Firstly $\int_A f d \mu = \int f \mathbb{1}_A d\mu $

Secondly the function $g := f \mathbb{1}_A $ is $0$ almost everywhere as by the definition of indicator functions it is non zero only on $A$ which again has measure zero from the question.

More formally let us define $ N := \{ x \in X | g(x) \not = 0 \} $ any $x \in N $ must be in $A$ because $g(x)$ for $x \not \in A = 0 $ hence $N \subseteq A$ Hence $\mu(N) = 0 $ (monotonicity of measure)

Hence $g = 0 $ almost everywhere and so by Theorem 1:

$\int_A f d\mu = \int f \mathbb{1}_A d \mu = \int g d\mu = $(Theorem 1)$ = \int 0 d\mu = 0 $

Title: Integral over null set is zero


问:
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$
It is visible that the result is 0, but I can't calculate it.
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$

It occurred to me to express it as a product and take a ratio test, but I'm not sure

Body:
Note that $$\frac{3^n}{(1+3)(1+3^2)\dots(1+3^n)}\leq\frac{3^n}{(1+3^{n-1})(1+3^n)}$$
and $$\lim_{n\to\infty}\frac{3^n}{(1+3^{n-1})(1+3^n)}=\lim_{n\to\infty}\frac{1}{1+3^{n-1}}\lim_{n\to\infty}\frac{3^n}{1+3^n}=0\cdot 1=0$$
Hence by the squeeze theorem our original limit is $0$.


Body:
It is visible that the result is 0, but I can't calculate it.
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$

It occurred to me to express it as a product and take a ratio test, but I'm not sure

Body:
Hint

$$\begin{align}0≤\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}≤\lim_{n\to\infty} \frac{3^n}{3×3^2\ldots 3^n}=\lim_{n\to\infty}\dfrac {3^n}{3^{\frac {n(n+1)}{2}}}=0\end{align}$$


问:
Differentiating $ f(x)= \frac{x + \sin x}{x - \cos x}$
Can someone help me? I'm having some trouble with this:
how can I differentiate $$ f(x)= \frac{x  + \sin x}{x - \cos x} \quad ?$$

P.S. Is there any trick or something to derive this kind of limit?
Thanks!

答:
You can use the product rule for this on the function $f(x)=(x+\sin{x})(x-\cos{x})^{-1}$.

The product rule states that for $f(x)=u\cdot v$, $f'(x)=u\cdot v'+ v\cdot u'$.

So we have: $$f'(x)=(x+\sin{x})(-1)(1+\sin{x})(x-\cos{x})^{-2}+(x-\cos{x})^{-1}(1+\cos{x})$$

Alternatively, you can use the quotient rule. The quotient rule states that for $f(x)=u/v$, $f'(x)=\frac{u'v-uv'}{v^2}$

So you have:

$$f'(x)=\frac{(1+\cos{x})(x-\cos{x})-(x+\sin{x})(1+\sin{x})}{(x-\cos{x})^2}$$

These two expressions are equivalent.


问:
Optimization problem solution (water-filling)
Let $x\in\mathbb{R}^n$ be an optimization variable and $\alpha\in\mathbb{R}^n$ be an n-dimensional vector. The standard water-filling problem is formulated as

\begin{equation}
\begin{array}{ll}
\operatorname{minimize} & -\displaystyle\sum_{i=1}^{n} \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1
\end{array}
\end{equation}

and has a well known solution (see Boyd & Vandenberghe pag. 245).

My question is about a similar problem:

\begin{equation}
\begin{array}{ll}
\operatorname{minimize} & \max_{i}- \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1
\end{array}
\end{equation}

The problem is still convex since the maximum of convex function is convex. How the solution would chance for this slightly modified optimization problem? It would be interesting to understand, also in terms of a wireless communication interpretation. Thanks!

答:
You can solve this problem by using a greedy algorithm.  To see intuitively why this is the case, imagine that $\ \alpha_i\ $ is the initial amount of water in the $\ i^\text{th}\ $ of $\ n\ $ buckets, to any selection of which you are required to add water amounting to a total of $\ 1\ $ unit.  If $\ w_i\ $ is the amount of water in the $\ i^\text{th}\ $ bucket after you have finished topping them up, and $\ w_\min=\min_\limits{1\le i\le n}w_i\ $ is the smallest amount of water in any of the buckets, you receive a payoff of $\ \log\big(w_\min\big)\ $, which you wish to naximise.  Letting $\ x_i\ $ be the amount of water you add to the $\ i^\text{th}\ $ bucket, then (because $\ \log\ $ is a strictly increasing function) your problem is to

\begin{equation}
\begin{array}{ll}
\operatorname{maximise} & \min_\limits{1\le i\le n} \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1\ ,
\end{array}
\end{equation}
which is equivalent to the problem posed:

\begin{equation}
\begin{array}{ll}
\operatorname{minimize} & \max_{i}- \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1\ .
\end{array}
\end{equation}

I shall assume here—as in the problem cited from Boyd and Vandenberghe—that $\ \alpha_i>0\ $ (otherwise there are elements of the feasible set at which the payoff function is not well-defined), and, without loss of generality, that $\ \alpha_i\ $ have been labelled so as to be in non-deceasing order:
$$
0<\alpha_1\le\alpha_2\le\dots\le\alpha_n\ .
$$
If $\ \alpha_1<\alpha_2\ $, then when you start adding water to the buckets, there is no point adding it to any other bucket than the first, because as long as that bucket has received no additional water, your payoff will stay at $\ \log\big(\alpha_1\big)\ $, whereas, by adding $\ x_1\ $ of water to that bucket, your payoff will increase to $\ \log\big(\alpha_1+x_1\big)\ $ as long as $\ x_1\le\alpha_2-\alpha_1\ $.  Once $\ x_1\ $ reaches  $\ \alpha_2-\alpha_1\ $, there is no point adding more water to the first bucket unless you add some to the second as well, since your payoff would otherwise remain stuck at $\ \log\big(\alpha_2\big)\ $. The best solution would thus appear to be to top up as many of the emptiest buckets as you can to the same level, until your reservoir has been used up.

Suppose that your reservoir gets used up once you've raised all of the buckets $\ 1,2,\dots,m\ $ to a level $\ \ell\ge\alpha_i\ $ for $\ i=1,2,\dots\,m\ $.  Then you must have $\ \ell\le\alpha_{m+1}\ $—otherwise you would have had to include bucket $\ m+1\ $ among those you've added water to—and $\ x_i=\ell-\alpha_i\ $, so $\ 1=\sum_\limits{i=1}^m\big(\ell-\alpha_i\big)=m\ell-\sum_\limits{i=1}^m\alpha_i\ $, or
$$
\ell=\frac{1}{m}\left(1+\sum_\limits{i=1}^m\alpha_i\right)\ .
$$
This leads us to define
\begin{align}
m^*&=\max\left\{m\in\mathbb{N}\,\bigg|\,1+\sum_\limits{j=1}^m\alpha_j\ge m\alpha_i\ \ \text{ for }\ 1\le i\le m\right\}\\
\ell^*&=\frac{1}{m^*}\left(1+\sum_\limits{j=1}^{m^*}\alpha_j\right)\\
x_i^*&=\cases{\ell^*-\alpha_i&if $\ 1\le i\le m^*$\\
0&otherwise.}
\end{align}
With these definitions, we have
\begin{align}
\ell^*&=\frac{1}{m^*}\left(1+\sum_\limits{j=1}^{m^*}\alpha_j\right)\\
&\ge\alpha_i\ \ \text{ for }\ 1\le i\le m^*\ \text{, and}\\
1+\sum_\limits{j=1}^{m^*+1}\alpha_j&< \big(m^*+1)\alpha_{m^*+1}\ ,
\end{align}
from which it follows that $\ x_i^*\ge0\ $ and $\ \ell^*\le\alpha_k\ $ for $\ m^*+1\le k\le n\ $. We also have $\ \sum_\limits{i=1}^nx_i^*=\sum_\limits{i=1}^{m^*}\big(\ell^*-\alpha_i\big)=1\ $, from the definition of $\ \ell^*\ $.  So $\ x^*\ $ is a feasible solution to the problem with objective value
$$
\min_\limits{1\le i\le n} \log\big(\alpha_i+x_i^*\big)=\log\big(\ell^*\big)\ ,
$$
because $\ \alpha_i+x_i^*=\ell^*\ $ for $\ 1\le i\le m^*\ $, and $\ \alpha_j+x_j^*=\alpha_j\ge\ell^*\ $ for $\ m^*+1\le j\le n\ $.

It is easy to show that $\ x^*\ $ is optimal.  If $\ x\ $ is any feasible solution, then since $\ \sum_\limits{i=1}^{m^*}x_i\le1=\sum_\limits{i=1}^{m^*}x_i^*\ $, there must exist at least one $\ k=1,2,\dots,m^*\ $ for which $\ x_k\le x_k^*\ $.  Therefore
\begin{align}
\min_\limits{1\le i\le n}\log\big(\alpha_i+x_i\big)&\le
\log\big(\alpha_k+x_k\big)\\
&\le\log\big(\alpha_k+x_k^*\big)\\
&=\log\big(\ell^*\big)\ .
\end{align}
That is, the objective value for $\ x\ $ cannot exceed its value $\ \log\big(\ell^*\big)\ $ attained by $\ x^*\ $.


问:
Geometric interpretation of the 9 order 3 points of Elliptic curves
Silverman in their book Rational Points on Elliptic
Curves have a theorem;

Theorem 2.1: Let $C$ be a non-singular cubic curve
$$C : y^2 = f (x) = x^3 + ax + bx + c.$$

(c) A point $P = (x, y) \neq \mathcal{O}$ on $C$ has order three if and only if $x$ is a root of the polynomial
$$\psi_3 (x) = 3x^4 + 4ax^3 + 6bx^2 + 12c^x + 4ac − b^2.$$

(d) The curve $C$ has exactly nine points of order dividing three. These nine points form a group that is a product of two cyclic groups of order three.

We say that a point $P$ has order 3 if $[3]P = \mathcal{O}$, where $\mathcal{O}$ is the identity element, not necessarily the point at infinity. Then we have $[2]P = -P$ and looking at the $x$ coordinates $$x([2]P) = x(-P)=x(P)$$ then we can conclude that $x([2]P)=x(P)$ if $P \neq \mathcal{O}$

When we consider a curve  like $y^2 +y = x^3$

If the two intersection points of the curve with the $y$-axis are inflection points then we will have the 6 points.

Where are all nine points on the curve geometrically?

答:
Based on the clarification in the comments I am going to prove the following result:

Lemma Let $E/\mathbb{R}$ be an elliptic curve. Further suppose that $E$ is given in simplified Weierstrass form
$$E : y^2 = x^3 + ax + b$$
with $a,b \in \mathbb{R}$, and $4a^3 + 27b^2 \neq 0$. Then we have that $E(\mathbb{R})[3]$ is either trivial or $ \mathbb{Z}/3\mathbb{Z}$

Proof: The $x$-coordinates of the $3$-torsion points of $E$ are given by the roots of the $3$-division polynomial
$$\psi_3(x) = 3x^4 + 6ax^2 + 12bx - a^2$$

Now consider the splitting field $K$ of $\psi_3$. Since we are working over $\mathbb{R}$, we have that either $K = \mathbb{R}$ or $\mathbb{C}$.

Now by general field theory $K$ contains $\mathbb{R}(\sqrt{\delta})$ where $\delta$ is the discriminant of the polynomial $\psi_3(x)$. But we can just compute that
$$\delta = -2^8 3^3 (4a^3 + 27b^2)^2$$

But we may remove square factors, hence $K$ contains $\mathbb{R}(\sqrt{-3}) = \mathbb{C}$. In particular $\psi_3$ cannot split over $\mathbb{R}$ (indeed, an analysis of the subgroup lattice of $S_4$ shows that $\psi_3$ can have at most one root by Galois theory over the function field $\mathbb{R}(a,b)$).

But even without this remark we have proved that $E(\mathbb{R})[3]$ does not have order $9$ (since at least one of the $x$-coordinates cannot be contained in $\mathbb{R}$). It is easy to come up with examples of curves with both of the other cases.

This is the low key way of proving that $K(E[N])$ contains the $N^{th}$ roots of unity, which can in general be seen from the Weil pairing.


答:
To find $f'(x)$, use the quotient rule:

$$f'(x) = \frac{(x-\cos(x))(1+\cos(x)) - (x + \sin(x))(1+\sin(x))}{(x-\cos(x))^{2}}$$

Can you simplify the algebra from here?

Title: Differentiating $ f(x)= \frac{x + \sin x}{x - \cos x}$

Body:
Can someone help me? I'm having some trouble with this:
how can I differentiate $$ f(x)= \frac{x  + \sin x}{x - \cos x} \quad ?$$

P.S. Is there any trick or something to derive this kind of limit?
Thanks!


问:
Using Runge-Kutta 4th order to solve a complex matrix differential equation
I'm trying to solve a differential equation of the form

\begin{equation}
\frac{dQ(t)}{dt} = - i H(t)Q(t)
\end{equation}

$Q$ and $H$ are both complex matrices. I'm using RK4. I implemented it as if I was treating a single variables so I'll define $k1,k2,k3,k4$ as matrices and do the normal RK4 calculations. I'm using armadillo library with C++ so it should be able to handle complex matrices without any problem. I'm not getting the results I'm expecting so I was wondering if the Runge-Kutta should be changed for this case?

答:
Take $L=\max_{t\in[a,b]}\|H(t)\|_{op}$, this is the global Lipschitz constant (constant of the Lipschitz condition) of the ODE over the integration interval.

Evaluating the stability region of the RK4 method gives that one needs $Lh<2.5$ for any hope of a non-chaotic result, $Lh<1.5$ for a qualitatively reasonable result, $Lh=0.1$ for a quick graphical check and $Lh\sim 0.001$ for the optimal result in double precision 64bit floating point numbers.

The topic of Runge-Kutta-like methods on manifolds like the Lie groups is a well-researched one. For example, a modification/correction of RK4 is derived and tested in Munthe-Kaas (1998): "Runge-Kutta methods on Lie groups".


问:
Find permutations of variable length
I want to generate a random string consists of lower- upper-case letters and digits. In addition, the length of this string is variable ranging from 4 to 6.
For example, the first word is 3rT6, the second is DGJeH, third is LLG5us.

What is the permutations formula to calculate it.

Is it correct to calculate it using the following:

```<code>w!(n-w)!/n!
</code>```

W is the number of characters and n is the number of the character set,

thanks

答: If you are asking for the formula that counts the number of strings with $w=4$, $w=5$, and $w=6$ symbols from your lower-upper-digit set, it is just $$n^w,$$ since there are no additional restrictions. If you would like your word to have no repeated symbols, or no vowels together (for example), then the answer will be different.

Title: Find permutations of variable length


问:
find the range of following function : $f(x)=\frac {x^2 +x -2}{x^2 -x -2}$
I am trying to find the range of following function: $f(x)=\frac {x^2 +x -2}{x^2 -x -2}$, when I try it, I sketch the graph and it was obvious that its range is $\Bbb R$, but I cannot find its range algebraically.

Note: I tried polynomial division and obtained $1 + \frac{2x}{x^2 -x -2}$, and I take derivatives to find critical points for global max or min but it did not work to obtain $\Bbb R$.

My question is how to find its range without using graphs?

答: Since $f(1)=0$, $0$ belongs to the range. Since $\lim_{x\to2^-}f(x)=-\infty$, it follows from this and from the intermediate value thorem that the range contains $(-\infty,0]$. And, since $\lim_{x\to-1^+}f(x)=\infty$ and still because $f(1)=0$, the range contains $[0,\infty)$. Therefore, the range is $\Bbb R$.


Body: Hint: $$f'\left(\dfrac{u}{v}\right)=\frac{u'v-uv'}{v^2}$$


Title: $\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$

Body:
$$
f(n)=\frac{3^n}{(1+3)(1+3^2)\cdots(1+3^n)}
=\frac{\frac{3^n}{3^1 3^2 \cdots 3^n}}{\frac{1+3^1}{3^1}\frac{1+3^2}{3^2}\cdots\frac{1+3^n}{3^n}}\\
=\frac{\frac{3^n}{3^{\frac{n(n+1)}2}}}{(1+3^{-1})(1+3^{-2})\cdots(1+3^{-n})}
=\frac{2\cdot 3^{-\frac{n(n-1)}{2}}}{\left(-1;\frac{1}{3}\right)_{n+1}}
$$
in terms of a $q$-Pochhammer symbol in the denominator.
For large $n$ this denominator approaches a constant,
$$
\lim_{n\to\infty}\left(-1;\frac{1}{3}\right)_{n+1}
= \left(-1;\frac{1}{3}\right)_{\infty}
\approx 3.129868037134023\ldots,
$$
in Mathematica,

    QPochhammer[-1, 1/3] // N

and therefore $\lim_{n\to\infty}f(n)=0$.


答:
HINT

$$f(x)=\frac {x^2 +x -2}{x^2 -x -2}=t$$

$$\implies x^2(1-t)+x(t+1)+(2t-2)=0$$

$$\implies \Delta = (t+1)^2-4(1-t)(2t-2)≥0$$

So, you can solve last inequality by using elementary algebra.

Title: find the range of following function : $f(x)=\frac {x^2 +x -2}{x^2 -x -2}$

Body:
I am trying to find the range of following function: $f(x)=\frac {x^2 +x -2}{x^2 -x -2}$, when I try it, I sketch the graph and it was obvious that its range is $\Bbb R$, but I cannot find its range algebraically.

Note: I tried polynomial division and obtained $1 + \frac{2x}{x^2 -x -2}$, and I take derivatives to find critical points for global max or min but it did not work to obtain $\Bbb R$.

My question is how to find its range without using graphs?


答:
Try writing it as

\begin{gather*}
y=\frac{x^{2} +x-2}{x^{2} -x-2}\\
y\left( x^{2} -x-2\right) =x^{2} +x-2\\
( y-1) x^{2} -( y+1) x+( 2-2y) =0\\
\\
\\
\end{gather*}
As x belongs to R, the discriminant of the above equation must be greater than or equal to zero.
Find the discriminant, solve the inequality and you get the range of y.


问:
How to prove that the intersection of two planes is a line (using parameteric equations)?
I'm reading the explanation to why the intersection between two planes is a line in the textbook. This seems reasonable enough, but I don't understand the last part of the proof. This indicates that my understandingn of parametric equations may be flawed, which might come back to haunt me later.

According to the book we start with the equations of two planes, which we write as augmented matrices, so:

\begin{array}{rrr|r}
    A_1 & B_1 & C_1 & -D_1 \\
    A_2 & B_2 & C_2 & -D_2
  \end{array}

Through gaussian elimination (omitting some steps) this can be reduced to:

\begin{array}{rrr|r}
    0 & 1 & a & b \\
    1 & 0 & f & g
  \end{array}

Which would mean that in the intersection of the two planes, the following would apply:

$$y+az=b$$
$$x+fz=g$$

The text then goes on to claim that

(1) z can be treated as a paramater (even though it should be a variable, I've never encountered such paramaters, what would that even mean?) and
(2) that because of this, we could write that:

$$x=g-ft$$
$$y=b-at$$
$$z=t$$

It then goes on to clame (3) that because of this, the solution is a three dimensional line.

Can someone explain these three steps in a little more detail? Or maybe point me to a helpful guide somewhere?

答: As I understand it, you can't actually solve a system of 3 variables with only two equations, therefore you set one variable to be a parameter (in this case z) and solve for x and y. As you'll see, there are infinite parametric equations for each line, with each of them a different fixed t ∈ ℝ, but they all describe the same line.</p>


问:
On gaps in the dimension of subspaces
Let $P\subseteq\Bbb N\cup\{\infty\}$, with $0\in P$. Does there exist a "nice" space $X_P$ such that $X_P$ has a subspace of dimension $n$ iff $n\in P$? By nice I mean nice enough that there's no need to worry about what "dimension" means, ideally a Polish space, but separable metrizable spaces would also work.

I know that surprisingly for $P=\{0,\infty\}$ there is even a compact Polish example, but I'm not sure whether there can be such gaps in the dimension of subspaces for finite dimensional spaces, so an easier question for which I'd also like to know the answer is the following: Let $X$ be a Polish space with $\dim X=n<\infty$. Must $X$ have subspaces of all dimensions between $0$ and $n$?

答:
No, there is such an $\{n,0\}$ example in Polish spaces for every $n$. So of dimension $n$ but a non-empty subspace can only have dimension $n$ or $0$.
See van Mill's book on infinite-dimensional topology (his first) for a proof and references.

Title: On gaps in the dimension of subspaces


问:
Find the maximal interval where this IVP has unique solution
Consider the IVP

$t\dfrac{dy}{dt} = t + \left|y\right|$, where $y(-1) = 1$, I need to find the largest interval where this has a unique solution.

Here, if I consider $f(t,y) = 1 + \dfrac{\left|y\right|}{t}$ then  there exist a rectangle
$\left|t+1\right| \le a$, $\left|y-1\right| \le b$ where this $f$ is continuous.

Also, $f$ is Lipchitz Continuous, in this given rectangle, so now by Uniqueness and Existence theorem we get an interval

$\left|t + 1\right|\le h $ where $h = \text{min}(a, b/M)$, $M = \text{max f}$ in the given domain.

Here, I am stuck as I don't get how to compute the maximum of this function on this domain.

The answer provided is $(-\infty,0)$ but I am clueless how to reach this interval ?

Thank You.

答:
On every domain $(t,y)\in[-b,-a]\times \Bbb R$ with $0<a<1<b<\infty$ you have $L=\frac1a$ as constant in the Lipschitz condition, which holds globally on that domain. This implies that the IVP has a solution on the interval $[-b,-a]$. By the uniqueness part of the theorem, this solution can be uniquely extended to any larger interval of this type. The union of intervals of this type is $(-\infty,0)$.

Determining the maximum intervall of existence

Picard-Lindelöf theorem variations

Globally Lipschitz implies solutions exist for all time

Questions about the Picard–Lindelöf theorem for an ODE

Showing Global Lipschitz $\implies$ Global Solution ODE


答:
Basically, getting the curve of intersection of two planes is equivalent to solving the two equations below:

\begin{gather*}
A_{1} x+B_{1} y+C_{1} z=D_{1}\\
A_{2} x+B_{2} y+C_{2} z=D2\\
\end{gather*}
Now, as I'm sure you know, to get the DEFINITE and UNIQUE solution to a n-variable linear system, you need at least n equations.
Therefore, it is impossible to get a definite and unique solution to these equations.
Hence we use a PARAMETER, to find the set of values of x,y and z that satisfy both these equations.
All three of x, y and z will be defined by linear relationships to the parameter (henceforth called t).
As the parameter changes, the values of x, y and z would change in such a way that they satisfy both equations 1 and 2 at all times.

Now, we know that x, y and z are defined by linear relationships with a parameter t.
Well, there's a theorem in analytical geometry, that if x, y and z are related by linear relationships to the same parameter, they all lie on a line.

Hope this answers your question!

Title: How to prove that the intersection of two planes is a line (using parameteric equations)?

Body:
I'm reading the explanation to why the intersection between two planes is a line in the textbook. This seems reasonable enough, but I don't understand the last part of the proof. This indicates that my understandingn of parametric equations may be flawed, which might come back to haunt me later.

According to the book we start with the equations of two planes, which we write as augmented matrices, so:

\begin{array}{rrr|r}
    A_1 & B_1 & C_1 & -D_1 \\
    A_2 & B_2 & C_2 & -D_2
  \end{array}

Through gaussian elimination (omitting some steps) this can be reduced to:

\begin{array}{rrr|r}
    0 & 1 & a & b \\
    1 & 0 & f & g
  \end{array}

Which would mean that in the intersection of the two planes, the following would apply:

$$y+az=b$$
$$x+fz=g$$

The text then goes on to claim that

(1) z can be treated as a paramater (even though it should be a variable, I've never encountered such paramaters, what would that even mean?) and
(2) that because of this, we could write that:

$$x=g-ft$$
$$y=b-at$$
$$z=t$$

It then goes on to clame (3) that because of this, the solution is a three dimensional line.

Can someone explain these three steps in a little more detail? Or maybe point me to a helpful guide somewhere?


问:
Find the values for which $\frac{x}{1-x^2} > 0$
It's been quite long since high school and I forgot some stuff.
I'm trying to find the values for which $\frac{x}{1-x^2} > 0$.

Now, obviously this occurs when $x < -1$ and $x>0$, yet if you just solve the inequality systematically, you would just get $x > 0$ which is not completely true:

$$\frac{x}{1-x^2} > 0$$
$$(1-x^2)\frac{x}{1-x^2} > 0(1-x^2)$$
$$x > 0$$

Would appreciate someone explaining why this doesn't seem to work, thanks in advance.

答:
You can multiply both sides of the inequality (and keep it the same) if the function you're multiplying with has the ame sign as the inequality sign.
For example, you can safely multiply (x^2+1), because it is positive everywhere.
But is (1-x^2) positive everywhere?</p>


问:
$|\phi(G):\phi(H)|$ divides $|G:H|$

Let $\phi$ be a homomorphism defined on a finite group $G$, and let $H\subseteq G$. Show that $|\phi(G):\phi(H)|$ divides $|G:H|$.

Not quite sure where to start on this one. We have a theorem saying that $\phi(H)$ is a subgroup of $\phi(G)$, but what is $|\phi(H)|$? If $H$ contains the kernel $N$ of $\phi$, then I think everything behaves nicely and $|\phi(G):\phi(H)|=|G:H|$. But if $N\not\subseteq H$, I don't know what I can conclude.

答:
In one answer it is claimed that $G$ must be finite, in one answer it is claimed that we only need that $|G:H|$ is finite. In fact, no finiteness assumptions are necessary (on the other hand, in the infinite case, the statement is rather boring due to cardinal arithmetic).

Here is an alternative bijective proof, which is perhaps more conceptual and works without any finiteness assumptions: The map $G/H \to \phi(G)/\phi(H), [g] \mapsto [\phi(g)]$ is a well-defined surjective homomorphism of $G$-sets. Thus it suffices to prove:

Lemma. If $f : X \to Y$ is a surjective homomorphism of $G$-sets and $X$ is transitive, then $|Y|$ divides $|X|$. Namely, if $y_0 \in Y$, then $|X|=|Y| \cdot |f^{-1}(y_0)|$.

Proof: Since $f^{-1}(gy)=g f^{-1}(y)$, every fiber of $f$ has the same cardinality $N$. Then $X = \coprod\limits_{y \in Y} f^{-1}(y)$ (as a set) witnesses $|X| = |Y| \cdot N$. $\square$

By the way, applying the Lemma to $G/K \twoheadrightarrow G/H$ for subgroups $K \subseteq H \subseteq G$, we get Lagrange's Theorem $|G:K|=|G:H| \cdot |H:K|$. It is also useful in other contexts.

Title: $|\phi(G):\phi(H)|$ divides $|G:H|$


答: The problem is that when you multiply by $(1-x^2)$, you assume this expression is positive. When you multiply both sides of an inequality by a negative number the inequality sign should be reversed.

Title: Find the values for which $\frac{x}{1-x^2} > 0$

Body:
It's been quite long since high school and I forgot some stuff.
I'm trying to find the values for which $\frac{x}{1-x^2} > 0$.

Now, obviously this occurs when $x < -1$ and $x>0$, yet if you just solve the inequality systematically, you would just get $x > 0$ which is not completely true:

$$\frac{x}{1-x^2} > 0$$
$$(1-x^2)\frac{x}{1-x^2} > 0(1-x^2)$$
$$x > 0$$

Would appreciate someone explaining why this doesn't seem to work, thanks in advance.


Body:
Two cases:

$x>0$ and $1-x^2>0$, which means $x>0$ and $x^2<1$, i.e.,$0<x<1$.

$x<0$ and $1-x^2<0$, which means $x<0$ and $x^2>1$, i.e., $x<-1$.

Thus the domain is $(-\infty,-1)\cup (0,1)$.


问:
Collecting math websites
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.

答:
Great question. It would probably also be interesting to think about what could be useful but is not yet out there. Here are a few picks that I can think of:

A useful website for ring theorists is the Database of Ring Theory. It is actually managed by rschwieb, a frequent user of MSE. You can look for examples or counterexamples of your favourite properties of rings and modules in the database.

In a similar vein, here is a Database for Number Fields.

Super useful for number theorists is also LMFDB. It's another database with which you can look for number-theoretic objects with certain desired properties.

For category lovers, the equivalent of the Stacks Project is Kerodon on Higher Category Theory. I think this one is still growing.

And of course, obvious ones like Wolfram Alpha and Wikipedia!

Maybe I should also mention certain social media meme groups in which you can apply your knowledge to understand the most recent homological algebra memes. But maybe this is not quite the right place for that... ;-)

Title: Collecting math websites


问:
How to find position vector of circle?
so I have a question on solving the following:

"A particle moves around the circle $x^2+y^2=25$ at constant speed, making one revolution in 2s. Find its acceleration?"

In the solution it says, that it's easy to see that the position vector is given by $$\textbf{r}=5cos(\omega t)\textbf{i}+5sin(\omega t)\textbf{j}$$
but unfortunately for me is not so easy to see, why can I say that this is the vector position? Does the formula $\textbf{r}=acos(\omega t)\textbf{i}+bsin(\omega t)\textbf{j}$ only hold for circles (does this formula has a name?)?

and the last thing I want to ask is with one revolution, do they mean that the particle has travelled one cicle?

If you know the answer and would like to share it, it would be very much appreciated

Thanks in advance!

答:
Hint: what is the paramtric form of a point on the circle?

This corresponds to the position of the rider at any time t.

For your second question, yes, this means the particle has travelled one full circle, from starting point to starting point.

Title: How to find position vector of circle?


答:
I have found these extremely useful, from time to time:

Online Encyclopedia of Integer Sequences

NIST Digital Library of Mathematical Functions

They both provide facts of a specialized nature, rather than understandings or arguments.  That is, they are thin on explanations and proofs.  The OEIS does have very thorough references, so this is not a serious problem.  The DLMF also has many references, which I then skim with MathSciNet.

I have mixed feelings about Wikipedia's mathematics articles.  Individual articles are often imprecise, more interested in giving the gist of a topic  than in giving precise details; this is especially bad when it comes to precise statements of theorems or (worse) of proofs.  There is usually  only a loose connection between the cited references and the contents of the article: which (say) of the six cited sources contains details of this or that statement in the article?  And there is poor coordination between collections of articles, which often use differing terminology and notation.  On the other hand, it is a good source for understanding (say) current usage of mathematical terminology.  I typically find 75 percent of MSE questions use some technical term in a way I do not understand, and Wikipedia often gives an indication of what the questions' posers had in mind.

Title: Collecting math websites

Body:
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.


答:
You need the intersection of sets $\{x| x>0\} $ and $\{x| (1-x^2)>0\}$ which happens for $0<x<1$.

Another possibility is to get the intersection of $\{x| x<0\} $ and $\{x| (1-x^2)<0\}$ which happens for $x<-1$.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
I absolutely was shocked when I learned about the exact formula for the number of partitions of an arbitrary natural number.  This formula is amazing for so many reasons, including not only the simple fact that it exists at all, but also that it is so intimidatingly complicated, in the typical style of a result of Ramanujan's.

$p(n)=\frac{1}{\pi \sqrt{2}} \sum_{k=1}^\infty \sqrt{k}\, A_k(n)\,
\frac{d}{dn} \left(
    \frac {1} {\sqrt{n-\frac{1}{24}}}
    \sinh \left[ \frac{\pi}{k}
    \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right]
\right)
$

where

$A_k(n) = \sum_{0 \,\le\, m \,<\, k; \; (m,\, k) \,=\, 1}
e^{ \pi i \left[ s(m,\, k) \;-\; \frac{1}{k} 2 nm \right] }.$

LastEditorDisplayName: anon


Body:
Hint:
$$f(x)=\frac{g(x)}{h(x)}$$

$$f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$


答:
Function $f = \frac{x^2+x-2}{x^2-x-2}$ has horizontal asymptote $y= 1$ which is given by the quotient of the leading coefficients.

Moreover $x^2-x-2 = (x+1)(x-2)$ and so $f$ has poles (and thus vertical asymptotes) $x_1=-1$ and $x_2=2$.

Behavior at the borders:

$\lim_{x\rightarrow-\infty} = \lim_{x\rightarrow-\infty} = 1$ as above.

$\lim_{x\rightarrow-1^-} = -\infty$ and $\lim_{x\rightarrow-1^+} = +\infty$ (the pole has multiplicity 1 and so there is a sign change).

$\lim_{x\rightarrow 2^-} = -\infty$ and $\lim_{x\rightarrow 2^+} = +\infty$ (the pole has multiplicity 1 and so there is a sign change).

The whole analysis shows that the domain is $\Bbb R\setminus\{-1,2\}$ and the range is $\Bbb R$ since $f(0)=1$ is attained (despite the fact that $y=1$ is an asymptote).


答:
The problem is, you seem to believe that you can multiply an inequality by any number, just like in equations. However, note that $5>3$ is correct. Multiply by $-1$ to get $-5<-3$, which is obviously false. You must note here, that

An inequality can be multiplied by any positive number, and will still hold true. But multiplying by a negative number, we need to reverse the sign.

This is why, in inequalities, often to get rid of the denominator, we multiply by the square of the denominator (it is always positive), and not the denominator. So, for your case,
$$x(1-x^2)>(1-x^2)^2\implies x-x^3>1-2x^2+x^4$$
which is not very nice. Instead of this, you can use the simple fact that
$$\frac ab > 0 \iff a,b>0 \text{ or }a,b<0$$
And divide the problem in two cases and solve separately.

Hope this helps. Ask anything if not clear :)


问:
Trigonometry - word problem

The Question

The machine tool diagram shows a symmetric die punch.
In this view, the rounded tip is part of a circle of radius r, and the slanted
sides are tangent to that circle and form an angle of 54◦. The top and
bottom sides of the die punch are horizontal. Use the information in the
diagram to find the radius r.

The given diagram

My attempts to solve

ABNM is a rectangle, therefore
$$AB = MN$$
$$AC = \frac {AB}{2} = ML$$
$$Θ = \frac {54°}{2} = 27°$$
In △LNK
$$ \tan27° = \frac{17/16}{h} $$
$$ h = \frac{17/16}{\tan27°} $$

$$w = h - 1\frac{1}{2}$$
$$\tan27° = \frac{x}{w}$$
$$x = w\tan27°$$

I don't think I've come any close to the solution, so I'll be grateful if you give me some hints or corrections.

答:
You are closer than you think.

Drop a perpendicular from $O$ to $KN$. Since the circle is tangent to $KN$, we have:

$$\frac r{r+w} = \sin 27^\circ$$

or

$$r = \frac {w\sin 27^\circ}{1-\sin 27^\circ}$$

Title: Trigonometry - word problem


问:
Prove that sequence diverges by definition
Can you help me to prove that $ \lim_{n\to\infty} (n-2n^2)= -\infty $ given definition $\forall \varepsilon>0  \  \exists N  \in  \mathbb{N}: n \geq N \Rightarrow  \left | x_n -a \right |< \varepsilon$. As far as I know, I have to use necessity and sufficienty, but I don't know how to construct the proof.

答:
Hint : If $n>1$, $$n-2n^2<-n^2.$$

If $M<0$, taking $N>\sqrt{-M}$ make the job.

Notice that $\lim_{n\to \infty }f(n)=-\infty $ if $$\forall M<0,\exists N\in\mathbb N: \forall n\in\mathbb N, n\geq N\implies f(n)<M.$$


问:
Formal Proving Discrete maths
Prove using logical equivalence rules/laws that

$$~[ ∧ ~ ∧ ( ∨ )] ∨ (~ ∧ ) ∧ ~( ∨ ~ ∨ ~) ≡ ~ ∨ b$$

答:
I'll prove the left to right half using a semantic argument, have your truth tables ready to check my reasoning:

A basic principle of disjunction (POD):

$$\Gamma, \phi \lor \psi \vDash \chi \;\;\; \text{iff} \;\;\; \Gamma,\phi \vDash \chi \,\,\, \text{and} \,\,\, \Gamma,\psi \vDash \chi$$

A basic principle of conjunction (POC):

$$\Gamma \vDash \phi \land \psi \;\;\; \text{iff} \;\;\; \Gamma \vDash \phi \,\,\, \text{and} \,\,\, \Gamma \vDash \psi$$

Deduction Teorem (DT):

$$\Gamma \vDash \phi \to \psi \;\;\; \text{iff} \;\;\; \Gamma,\phi\vDash\psi$$

Also: POD implies this introduction rule:

$$\phi \vDash \phi \lor \psi$$

And obviously assumption (Ass):
$$ \phi \vDash \phi$$

First consider the Assumption: $\Gamma,[a \land b \land (a \lor c) ]\vDash [(a \land b) \land (a \lor c) ]$; by POC we get: $$\Gamma,[a \land b \land (a \lor c) ]\vDash a\land b$$ and with another application of POC we obtain: $$\Gamma,[(a \land b) \land (a \lor c) ]\vDash a$$

Therefore, using the introduction rule: $$\Gamma,[(a \land b) \land (a \lor c) ]\vDash a \lor b.$$

Now consider (Ass): $\Gamma,[( a \land b) \land (a \lor b \lor c)] \vDash( a \land b) \land (a \lor b \lor c)$; with a simple use of POC we obtain: $$\Gamma, [( a \land b) \land (a \lor b \lor c)]\vDash a\land b$$

And thus with an equivalent argument we deduce: $$\Gamma,[( a \land b) \land (a \lor b \lor c)]\vDash a \lor b.$$

Now we use POD to obtain the formula:

$$\Gamma, \, [a \land b \land (a \lor c) ] \, \lor \, [( a \land b) \land (a \lor b \lor c)] \vDash a \lor b  $$

And finally we use the deduction theorem to get the first result we want:

$$\Gamma \vDash \{[a \land b \land (a \lor c) ] \, \lor \, [( a \land b) \land (a \lor b \lor c)]\} \to a \lor b$$

Now you must start from $a \lor b$ and deduce $[a \land b \land (a \lor c) ] \, \lor \, [( a \land b) \land (a \lor b \lor c)]$.

Title: Formal Proving Discrete maths


答:
For rational functions (ratios of polynomials), sign changes can only occur where a factor of the numerator or denominator is zero. At such a point, the function itself will be either undefined (if the denominator vanishes) or zero (if the numerator vanishes).

So the best approach is to make a sign chart for each factor. For each factor draw a number line, label the point where the factor vanishes, and label the sign on each side of the factor.

Then combine the sign charts of all the factors multiplicatively to arrive at a sign chart for your function, taking care to mark the points with "$0$" where the numerator factors vanish and as "$*$" where the denominator factors vanish.

Finally, look at the inequality you need to solve. Since it is of the form "$f(x)>0$", you will include only the parts of the number line where the overall sign is positive (not zero!). If instead it were "$f(x)\geq 0$", you would also have to include the points where the overall sign equal zero (typically transition points where the numerator factors vanish), and so forth.

Example:
Solve $\dfrac{x+1}{x-2}\leq 0$.

Solution: The factors are $(x+1)$ and $(x-2)$, which vanish at $-1$ and $2$ respectively. Sign charts:

```<code>number line:    _________-1______________2___________
sign of (x+1):  ----------0++++++++++++++++++++++++++
sign of (x-2):  -------------------------0+++++++++++
combined:       ++++++++++0--------------*+++++++++++
                          ^              ^
                         zero         undefined
</code>```

The combined sign is determined by first marking points where the function is zero "$0$" or undefined "$*$"; the sign on the interval in between such points is constant, and will be negative if there is an odd number of negative factors in the charts above it, and positive otherwise.

Note that after combining factors by division, the function is zero at $-1$ and undefined at $2$.

For our solution we want points on the number line where the function is negative or zero (since we are solving an inequality of the form $f(x)\leq 0$).

The solution is $[-1,2)$, which includes the left endpoint but not the right endpoint.

If the inequality were $f(x)>0$ instead, we would only want points where the combined sign is positive (not zero), so the solution would be $(-\infty,-1)\cup (2,\infty)$.

Now you should be able to do something similar. You will have three sign charts to combine.

(BTW, your solution doesn’t work because you have ignored the sign contribution of the denominator.)

Body:
For rational functions (ratios of polynomials), sign changes can only occur where a factor of the numerator or denominator is zero. At such a point, the function itself will be either undefined (if the denominator vanishes) or zero (if the numerator vanishes).

So the best approach is to make a sign chart for each factor. For each factor draw a number line, label the point where the factor vanishes, and label the sign on each side of the factor.

Then combine the sign charts of all the factors multiplicatively to arrive at a sign chart for your function, taking care to mark the points with "$0$" where the numerator factors vanish and as "$*$" where the denominator factors vanish.

Finally, look at the inequality you need to solve. Since it is of the form "$f(x)>0$", you will include only the parts of the number line where the overall sign is positive (not zero!). If instead it were "$f(x)\geq 0$", you would also have to include the points where the overall sign equal zero (typically transition points where the numerator factors vanish), and so forth.

Example:
Solve $\dfrac{x+1}{x-2}\leq 0$.

Solution: The factors are $(x+1)$ and $(x-2)$, which vanish at $-1$ and $2$ respectively. Sign charts:

number line:    _________-1______________2___________
sign of (x+1):  ----------0++++++++++++++++++++++++++
sign of (x-2):  -------------------------0+++++++++++
combined:       ++++++++++0--------------*+++++++++++
                          ^              ^
                         zero         undefined

The combined sign is determined by first marking points where the function is zero "$0$" or undefined "$*$"; the sign on the interval in between such points is constant, and will be negative if there is an odd number of negative factors in the charts above it, and positive otherwise.

Note that after combining factors by division, the function is zero at $-1$ and undefined at $2$.

For our solution we want points on the number line where the function is negative or zero (since we are solving an inequality of the form $f(x)\leq 0$).

The solution is $[-1,2)$, which includes the left endpoint but not the right endpoint.

If the inequality were $f(x)>0$ instead, we would only want points where the combined sign is positive (not zero), so the solution would be $(-\infty,-1)\cup (2,\infty)$.

Now you should be able to do something similar. You will have three sign charts to combine.

(BTW, your solution doesn’t work because you have ignored the sign contribution of the denominator.)


问:
What does $\sum_{n=0}^\infty z^{n(n+1)/2}$, $|z|<1$ converge to?
Does anyone know what the series
$$
S(z) = \sum_{n=0}^{\infty} z^\frac{n(n+1)}{2}
$$
converges to for $|z|<1$? This came up in an application where $z$ is a probability and $S(z)$ an expected value. The exponent comes from the fact that $\sum_{k=0}^nk=\frac{n(n+1)}{2}$. I think $S(z)$ should definitely converge because the coefficients are nonnegative and form a subsequence of the well-known $\sum_{n=0}^\infty z^n=\frac{1}{1-z}$. Is there a closed-form representation of $S(z)$?

答:
The Wikipedia article Ramanujan theta function
states

$\psi(q) = f(q,q^3) = \sum_{n=0}^\infty q^{n(n+1)/2}
= (q^2;q^2)_\infty(-q;q)_\infty $

and, yes, this converges only for $\,|q|<1.$
Note that his is as close to closed
form as you can get. A more complicated
expression for this is in terms of a Jacobi
theta function.


问:
What is $\lim_{n\to\infty} 1+\frac{i}{n}$

What is $$\lim_{n\to\infty} 1+\frac{i}{n}\;\;\;\;,i=\sqrt{-1}$$

Can we say that this limit is $1$? Can a limit in the complex world be a real number? If not what is the limit(if it exists) ?

答:
Can a limit in the complex world be a real number?

Yes as the real coordinate axis (X-axis where $y=0$) dwells inside the complex plane $\mathbb C$. Moreover, a complex sequence may approach to its limit point not just only from left or right (as in real axis) but via any direction (spiral, oblique, etc).


答:
There is a website devoted to counterexamples:
http://www.mathcounterexamples.net/?fbclid=IwAR0dZKurrSUV1ft8AeW-Gh189x58BdjGZ6M2KzeeyiHJQ6grdCuCgLdGUgw

The Encyclopedia of Triangle Centers:
https://faculty.evansville.edu/ck6/encyclopedia/ETC.html

and also the Periodic Table of Finite Elements:
http://femtable.org/

For curiosities, there is a website that teaches (with audio files) the pronunciations of many mathematician's names:
http://pronouncemath.blogspot.com/

Mathematical pictures:
https://images.math.cnrs.fr/?lang=fr

Finally, there is the nice Halmos' album of pictures of mathematicians:
https://www.maa.org/press/periodicals/convergence/whos-that-mathematician-images-from-the-paul-r-halmos-photograph-collection

Title: Collecting math websites


问:
Nielsen-Schreier and the Axiom of Choice
The Nielse-Schreier (NS) Theorem says that every subgroup of a free group is free. The proof uses the Axiom of Choice, and Läuchli showed in 1962 that the negation of NS is consistent with ZFA (ZF with atoms). By a result of Jech-Sochor, this result can be transferred to ZF. So some form of Choice is needed to prove NS.

In 1985, Paul Howard showed that the Axiom of Choice for sets of finite sets follows from NS.

Is the NS theorem known to be equivalent to the Axiom of Choice or some weak form of Choice?

答:
As I wrote at the end of this answer, the paper by Paul Howard doesn't appear to be cited anywhere (see this Google Scholar search).

So I doubt there has been any progress since Howard's original paper, and that there is any particular "choicey" principle that the NS theorem is equivalent to.

It should also be remarked that "weak form of choice" is a very arbitrary thing, one can define it simply as a statement provable from $\sf ZFC$ but not from $\sf ZF$, but one can also define it as a statement which explicitly says "Given a family with property X, we can choose from it". In the case of the former, NS itself appears to be a weak form of choice; but in the latter case it is not clear if such principle even exists (for example "every set can be linearly ordered" is not equivalent, to the best of my knowledge, to such principle).


问:
Does this series converge or diverge by the comparison test?
I have the following series:

$\frac{1}{2}+\frac{2}{7}+\frac{3}{11} . . . \frac{n}{4n-1}$

I rearranged it like-so:

$\frac{n}{4n-1} < \frac{n}{3n-1} < \frac{n}{2n-1} < \frac{2n}{2n-1}$

Although, it's comparatively similar to:

$\frac{1}{2} ...\frac{1}{n}$

So I've made the assumption the series diverges, however, I'm not sure if my reasoning can imply this.

Body:
$\frac{1}{2}+\frac{2}{7}......+\frac{n}{4n-1}=\frac{1} {2} +\sum\limits_{2}^{n} \frac{k}{4k-1}$

$\sum\limits_{2}^{n} \frac{k}{4k-1}=\sum\limits_{2}^{n} \frac{1}{4-\frac{1}{k}} $

$\frac{7}{2} \leq  4-\frac{1 }{k} \leq 4-\frac{1}{n}<4$

$\Rightarrow $$ \frac{1} {4}\leq\frac{1}{4-\frac{1}{k}}\leq \frac{2}{7}$

$\Rightarrow $$\sum\limits_{2}^{n} \frac{k}{4k-1}\geq \frac{n-1}{4}$

So

$ \lim_{n\to +\infty} \sum\limits_{2}^{n} \frac{k}{4k-1}=+\infty$

This series is divergent


答:
Here's an easy way to handle this:

You have $\frac {x} {1-x^2} > 0 \implies \frac {x} {(1+x)(1-x)} > 0$

Multiply both sides by $(1+x)^2(1-x)^2$. You don't have to worry about the direction of the inequality as that expression is non-negative for real $x$.

So, after cancellation, you get $x(1+x)(1-x) >0$

It's now just a matter of sketching an already fully factorised cubic (so you know the $x$ intercepts). Note that the lead coefficient will be negative so you know $y \to \pm \infty$ respectively as $x \to \mp \infty$. That allows you to complete the sketch, which immediately gives you the solution set $(x<-1) \cup (0<x<1)$.

Your mistake was in multiplying by an expression $(1-x^2)$ whose sign you cannot assume for all reals. That problem does not exist in my solution.


问:
question on a multi valued function.
Let $f(r,\theta) = (r\cos\theta, r \sin\theta)$, where $(r,\theta) \in \mathbb R^2, r\neq 0$. Then how to disprove the following?

for any $(r,\theta) $ such that $r\neq 0$,  $f$ is one-one on a neighbourhood of $(r,\theta) $.

$Df(r,\theta) = r^2I$ for any $(r,\theta) $ such that $r\neq 0$.

My attempt:- I found out that $Df(r,\theta) = \begin{pmatrix}
                                               \cos\theta & -r\sin\theta\\
                                               \sin\theta & r\cos\theta
                                              \end{pmatrix}$.
So by substituting any value of $(r,\theta)$ the matrix is not equal to $r^2I$. But I do not know how to disprove the first statement. Please help me.

答:
Let $f(r,\theta)=(r\cos\theta,r\sin\theta)$, with $r\in\mathbb R^+$.

$f\in\mathcal C^1(\mathbb R^+\times\mathbb R)$ and the Jacobian matrix of the function is
$$J_f(x,y)=\begin{pmatrix}\cos\theta&&-r\sin\theta\\\sin\theta&&r\cos\theta \end{pmatrix}.$$
The determinant is given by $\det(J)=r\cos^2\theta+r\sin^2\theta=r> 0\implies\exists U(p_0)$ and $V(f(p_0))$ open neighbourhoods $\forall p_0=(r,\theta)\in\mathbb R\setminus\{0\}\times\mathbb R$ such that $f|_U:U\to V$ is invertible.

Since the restriction is locally bijective, the function will be one-one in every neighborhood of a point in its domain.

Title: question on a multi valued function.


问:
Number of homomorphism
Recently, I was reading on group operations and came across a question like determine the number of group homomorphism for two groups. I don't specifically recall the example but how can one determine the number of homomorphism? Wouldn't that depend on the order of the group? and bigger the order, more would be the no. of homomorphism. The example that I have in mind is rather simple, determine the number of homomorphism from $\mathbb{Z_{8}}\rightarrow \mathbb{Z_{10}}$?</p>

答:
order of ${Z_8}=8$,  order of ${Z_{10}}=10$. Both are finite cyclic groups. Hence, to check number of Homomorphism between them is gcd of their orders, $\text{gcd}(8,10) = 2$, Hence the number of homomorphism function is $2$. This is not coincident.

One of The property of homomorphism states ${\phi{(g^n)= (\phi(g))^n}}$, such a homomorphism is completely specified by the image of 1. That is, if, 1 maps to a then x maps to xa.

Another property states that if ${|g|}$ is finite, then ${|\phi(g)|}$ divides ${|g|}$, so ${|a|} ={1,2}$ as then ${a={0,5}}$, i.e., ${1 \mapsto 0}$ or ${1 \mapsto 5}$ .These are the only two homomorphism is possible.

Title: Number of homomorphism


问:
Weak law and strong law of large numbers and modes of convergence
The strong law implies the weak law (although there seem to be variations on exactly what hypothesis are assumed for the strong law vs the weak law which probably make this not true depending on your definition of the strong law and weak law). The weak law states convergence in probability whereas the strong law states convergence almost surely.

Why does the strong law imply the weak law? In real analysis convergence in measure and convergence almost everywhere don't imply each other. Can someone explain what are the subtle meanings behind the difference convergences and what they mean for the law of large numbers.

答:
@Davide Giraudo's answer works quite well with the probability space, which has finite measure. And I am going to show a more general result, i.e.

$\textbf{Claim:}$ if the measurable space $(\Omega,\mathcal{F},\mu)$ has finite measure($\mu(\Omega)<\infty$), then the a.e. convergence implies convergence in measure.

which is suggested by @Ashok. Also, I will introduce another mode of convergence to help people understand the relationship between these modes of convergence(since I think it is hard to help the question owner after 8 years...).

Key observation is that

$X_n$ converges a.e. to $X$ is equivalent to(by definition)
$$\mu(\mathop{\cap}\limits_{m=1}^{\infty}\mathop{\cup}\limits_{n=m}^{\infty}\{|X_n-X|\geq\epsilon\})=0,$$

while converges in measure is equivalent to(by definition as well)
$$\lim\limits_{n\rightarrow\infty}\mu(\{|X_n-X|\geq \epsilon\})=0.$$

They are both true under the following statement:
$$\lim\limits_{m\rightarrow\infty}\mu(\mathop{\cup}\limits_{n=m}^{\infty}\{|X_n-X|\geq \epsilon\})=0,$$

in all measurable space. And this kind of convergence is called $\textbf{almost uniform convergence}$, or a.u. convergence in short, since you can read from the above formula that for any $\epsilon>0$, there is some measurable set $A\in\mathcal{F}$ such that $\mu(A)<\epsilon$ and $X_n$ converges uniformly on $A$.

(Well, it may not be direct to "read" from it, but can be proved with moderate effort, and is a very good exercise for everyone :) Or you can go to Egorov's theorem for the proof--we only need a part of it to show the equivalence. However, I still recommend you to try on yourself.

Now, you can find: if the measure space is finite, then a.e. convergence is equivalent to a.u. convergence, since you can now pull the intersection out to be a limit, then implies convergence in measure. Q.E.D.

In general measure spaces, a.e. convergence and convergence in measure can be totally irrelevant:

Let $X_n(x)=1(|x|>n)$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, then it converges a.e. but not in measure(Lebesgue measure);

Let $X_n(x)=1(x\in[\frac{i}{2^k},\frac{i+1}{2^k}))$ be the famous "typewritter" function, where $n=2^k+i$, then it converges in measure but not a.e.


问:
How to know whether a set of points can be rotated to lie in positive orthant?
I have a set of points in N-dimensional space. I want to find out whether they could, in principle, be rotated to lie solely in the positive orthant of space. Is there a property of these points that would guarantee this rotation exists?

If this question proves too difficult, perhaps another related one will be easier. I have the linear dynamical system and starting point that generated these points. Is there a property of linear dynamical systems which, if satisfied, ensure that under some rotation the trajectory will lie solely in the positive orthant?

To give an idea of the kind of thing I'm looking for (if it exists!): in 2D if the dot products between the set of points are all positive you can guarantee that there's some rotation that could rotate the set of points into the positive quadrant, and if a dot product is negative you know there is no such rotation. This is a nice easy test I can apply. However, this approach doesn't generalise (in 3D points can have all positive dot products, but cannot be rotated into the positive octant), can you think of one that will generalise?

Later edit - a few things I've been considering:

People study positive linear systems: linear dynamical systems, $\dot{\boldsymbol{x}}(t) = \boldsymbol{Ax}(t)$, in which the trajectory, $\boldsymbol{x}(t)$, never leaves the positive orthant. It turns out if the matrix $\boldsymbol{A}$ is Metzler, meaning all off-diagonal elements are non-negative, then a trajectory starting in the positive orthant will stay there. This is related to the Perron-Frobenius Theorem which I don't know so much about.

So it seems another way of framing my question would be: you have your linear dynamical system in an arbitrary basis. Can you tell whether the dynamics matrix $\boldsymbol{A}$ is Metzler under some rotation? The Perron-Frobenius theorem talks about rotation invariant things like eigenvalue spectra which gives me hope that this could be a constructive route, but I haven't been able to fit things together yet... Maybe the clever people on this site will be able to!

(Which is similar to this old question: Converse of Perron Frobenius Theorem: Necessary and Sufficient Conditions for positivity (or non negativity))

答:
Not a solution but some simple comments rather.

First off, one can normalize vectors $x_k$ i.e. multiply them by a positive scalar so that we can assume that $||x_k||$ = 1.
Then we look for a unitary operator, say $U$ such that $U(x_k)$ has non-negative coordinates  for any $k \in \{1,2,....m\}$.
If such $U$ can be found then for any subset $S \subset \{1,2,...,m\}$ we would have
$$||\sum_{k \in S} U(x_k)|| >= (\sum_{k \in S} ||U(x_k)||)^{\frac{1}{2}} $$
I mean, $||.||$ here is assumed to be Euclidean i.e. $l_2$ norm and for such norms it is not difficult to see that the above inequality holds (again $U(x_k)$ has nonnegative coordinates). Since $U$ is unitary we can now formulate necessary condition for the set of (normalized) points to be rotated into positive orthant:
$$||\sum_{k \in S} x_k|| >= \sqrt |S| $$
for any subset $S \subset \{1,2,...,m\}$

It is easy to see that this conditions is also sufficient in $R^{1}$ and $R^{2}$.

For n=1 it means that all vectors point in the same direction, and for n=2 we just take two vectors with the biggest angle between them and the condition above insures that this angle is not bigger than $\frac{\pi}{2}$.


问:
Extracting the coefficient from a generating function
Consider (for fixed $r$) the following function:
$$f(z) = \frac{1}{1-z-z^2-\cdots-z^r} = \frac{1}{1-z\frac{1-z^r}{1-z}}=\sum_{j=0}^\infty\left(z\frac{1-z^r}{1-z}\right)^j$$
(Assume everything is ok with regards to convergence.)

The text I am reading claims that if we were to write $f(z)$ in the form $\sum_{n=0}^\infty A_n z^n$, that
$$A_n = \sum_{j,k} (-1)^k \binom{j}{k}\binom{n-rk-1}{j-1}$$

I have no idea where this comes from. Could anyone point me in the right direction? Thanks!

Edit: Also, the text states without explanation that in the case $r=2$, the coefficients are the Fibonnaci numbers:

$$\frac{1}{1-z-z^2} = 1 + z + 2z^2 + 3z^3 + 5z^4 + 8 z^5 + 13 z^6 + \cdots$$

Is this a non-trivial result, or am I just not seeing something?

答:
$$f(z) = \sum_{j=0}^\infty \left( z \frac{1 - z^r}{1-z} \right)^j = \sum_{j=0}^{\infty}z^j \left( \sum_{k=0}^j {j\choose k} (-z^r)^k \right) \left( \sum_{m=0}^{\infty} {m+j-1 \choose j-1}z^m \right)$$

Change the order of summation. Note that in order to get a term of $z^n$, we need $j + rk + m = n$, or that $m= n - rk - j$.

The coefficient of $z^n$ is thus

$$ \sum_{j,k} (-1)^k {j\choose k} {n-rk-1 \choose j-1}$$

Let $g(x)$ be the generating function of the Fibonacci numbers. Then,

$\begin{array}{l l l l l l}
g(x) &= 1 &+ x & + 2 x^2 & + 3x^3 & + \ldots \\
xg(x) & =  & + x & + x^2 & + 2x^2 & + \ldots \\
x^2 g(x) & = & & + x^2 & + x^3 & + \ldots \\
\hline
(1-x-x^2)g(x)  & = 1 \\
\end{array}$

Hence, $g(x) = \frac{1}{1-x-x^2}$.

Note: This is with $F_1=1, F_2=2$. Depending on how you want to start your Fibonacci sequence, some people use $g(x) = \frac{x}{1-x-x^2}$ as the generating function instead.


问:
should the function with the following property vanish at some point?
Suppose $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ smooth such that $f(x+1,-y) = -f(x,y)$, should the function with this property vanish at some point?

I've tried direct inspection, tried to override one of the coordinates to see if the result was an odd function, but nothing seems to work. Is there any other way to address this problem?

答: If $f \equiv 0$, then we are done. Otherwise, there exists some $(x,y) \in \mathbb{R}^2$ such that $f(x,y) \neq 0$. In particular, $f(x,y)$ and $f(x+1,-y)$ have opposite signs. Since $f$ is continuous and $\mathbb{R}^2$ is connected, $\operatorname{im}{f}$ is connected (this is the general version of the IVT) and therefore $0 \in \operatorname{im}{f}$.


问:
On proposition I.3.2 of 'Quantum groups' by Kassel.
I am reading the book Quantum groups by Kassel. In proposition I.3.2 at the very beginning the reader is asked to show that under the identifications made, the maps $\Delta,\varepsilon$ and $S$ correspond to the maps $+,0$ and $-$.

However, $+$ is a map from $A^2$ to $A$, and I'm not sure how we can use the identifications to see that $\Delta:k[x]\rightarrow k[x',x'']$ corresponds to $+$.

It seems to me, all you can get from the identifications is that the map $\Delta$ corresponds to the element $x'+x''\in k[x',x'']$. Similarly, $\varepsilon$ corresponds to the element $0\in k$ and $S$ corresponds to the element $-x\in k[x]$.

So what exactly is meant in this proposition? I'm fairly sure this is a stupid question, but it's one that should be well-understood before proceeding any further in this theory.

Thank you in advance.

答:
Recently, I read the book by Kassel. What disturbed me about his presentation that it was not clear how to come up with the maps $\Delta, \epsilon$ and $S$ starting from the addition maps $+ : A^2 \to A$ from your algebras. However, once you think about this, the Yoneda lemma comes to the rescue. Indeed, given a commutative algebra $A$, we can consider the map

$$\text{Hom}_{\text{Alg}}(k[x',x''],A) \cong A^2 \stackrel{+}\longrightarrow A \cong \text{Hom}_{\text{Alg}}(k[x],A)$$

This gives a natural transformation between the functors
$$\text{Hom}_{\text{Alg}}(k[x',x''],-) \to \text{Hom}_{\text{Alg}}(k[x],-)$$
which by the Yoneda lemma gives an algebra morphism
$$\Delta: k[x]\to k[x',x'']\cong k[x]\otimes k[x]$$

One then calculates that $\Delta$ is the desired algebra morphism, i.e. that $\Delta(x) = x' + x''$. Under the canonical identification $k[x',x'']\cong k[x]\otimes k[x]$, we then obtain
$$\Delta(x) = x \otimes 1 + 1 \otimes x.$$


问:
How to derive differential gradient descent from cost function using matrix calculus?
I am new to matrix calculus and I hope you can help me with this question!

I’m trying to derive the update law for gradient descent to minimize a cost function
$$j(\mathbf{X}) = \frac{1}{2} \mathbf{e}^T(\mathbf{X}) \mathbf{e}(\mathbf{X})$$
with $s \in \mathbb{R}$ and $\mathbf{b}, \mathbf{e} \in \mathbb{R}^n$ and $\mathbf{A}, \mathbf{X} \in \mathbb{R}^{n\times m}$. First I calculate the gradient of $j$ w.r.t. a vectorized $\mathbf{x} := \operatorname{vec} \mathbf{X} \in \mathbb{R}^{nk}$
$$\nabla_\mathbf{x} j(\mathbf{X}) = \left(\frac{\partial j}{\partial \mathbf{x}}\right)^T = \left(\frac{\partial \mathbf{e}(\mathbf{X})}{\partial \mathbf{x}}\right)^T \mathbf{e}(\mathbf{X}).$$
For $\mathbf{e}$ it is known that
$$\frac{\mathrm{d}\mathbf{e}(\mathbf{X}(s), s)}{\mathrm{d}s} = \mathbf{A} \mathbf{X}^T(s) \mathbf{b}.$$
Now I try to somehow derive the gradient using the chain rule
$$\nabla_\mathbf{x} j(\mathbf{X}) = \left(\frac{\mathrm{d} \mathbf{e}(\mathbf{X}(s), s)}{\mathrm{d}s} \frac{\mathrm{d} s}{\mathrm{d} \mathbf{x}(s)}\right)^T \mathbf{e}
= \left(\mathbf{A} \mathbf{X}^T(s) \mathbf{b} \frac{\mathrm{d} s}{\mathrm{d} \mathbf{x}(s)}\right)^T \mathbf{e}
= \left(\frac{\mathrm{d} s}{\mathrm{d} \mathbf{x}(s)}\right)^T \operatorname{Tr}\left(\mathbf{X}^T(s) \mathbf{b} \mathbf{e}^T \mathbf{A}\right)
= \left(\frac{\mathrm{d} s}{\mathrm{d} \mathbf{x}(s)}\right)^T \mathbf{x}^T(s) \operatorname{vec}\left(\mathbf{b} \mathbf{e}^T \mathbf{A}\right)
=\ ...?$$
The update law for $\mathbf{X}$ is stated as
$$\frac{\mathrm{d} \mathbf{X}}{\mathrm{d} s} = - \mathbf{\Gamma} \mathbf{b} \mathbf{e}^T \mathbf{A}$$
where $\mathbf{\Gamma}$ is a diagonal $\mathbb{R}^{n \times n}$ matrix of update rates.

I don’t understand how the update law (last equation) can be derived from the gradient (second to last equation). Could anyone please help me? Thank you so much in advance!

答:
$\def\B{\Big}\def\L{\left}\def\R{\right}\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\P#1#2{\frac{d #1}{d #2}}$Using
the given expression for the error function, calculate its differential
$$\eqalign{
e &= e_0 + A\L(\small\int_{X_0}^{X} dX\R)^Tb
\quad\implies\quad
\color{red}{de = A\,dX^Tb} \\
}$$
Calculate the differential of the cost function and its gradient.
$$\eqalign{
j &= \tfrac 12e:e \\
dj &= e:\color{red}{de} \\
 &= e:\color{red}{A\,dX^Tb} \\
 &= e^T:b^TdX\,A^T \\
 &= be^TA:dX \\
\p{j}{X}
 &= be^TA  &\quad\big({\rm gradient}\big) \\
}$$
Now use an iterative gradient-descent method to calculate the optimal $X$
$$\eqalign{
X_{k+1} &= X_k - \lambda_k\Big(be^T_kA\Big) \\
}$$
where the scalar $\lambda_k>0$ is the step-length which minimizes the cost function for the $k^{th}$ iteration. The initial value (at $k=0$) of $X$ is likely $\,X_0=X(s_0),\,$ or it could be a random matrix.

Apparently, your textbook proposes to replace the standard scalar step-length with a matrix
$$\lambda_kI \to \Gamma_k \\\\$$

In the preceeding, a colon denotes the trace/Frobenius product, i.e.
$$\eqalign{
A:B = {\rm Tr}(A^TB) \\
}$$
The properties of the trace function allow the terms in such a product to be rearranged in various ways, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB \\
}$$


问:
Find $c\in\mathbb{R}$, and $f$ such that $\int_c^xt\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\frac{x^3}3$

Find a number and a function that satisfies:
$$\displaystyle\int_{c}^{x}t\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\dfrac{x^3}{3}$$

I tried taking derivative both sides so I got:

$$\left(\displaystyle\int_{0}^{x}t\cdot f(t)\,\mathrm{d}t-\displaystyle\int_{0}^{c}t \cdot f(t)\,\mathrm{d}t\right)^\prime=x \cos(x)+x^2 $$

It's clear that the right part of the integral will be $0$ because it's constant:

$$\left( \displaystyle\int_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right)^{\prime}=x \cos(x)+x^2$$

But I didn't do rest. Because I don't know how to derive the integral with two functions. I know with one function like $\int_{0}^{x} f(t)\,\mathrm{d}t$. Thanks!

答:
As said in answers, there is no way to find the zero of function $$f(c)=3c\sin(c)+3\cos(c)+c^3$$ but we can make nice approximations.

By inspection, the solution is "close" to $c=-\frac \pi 2$. So, we can do one single iteration of Householder method and obtain, by hand, the estimate
$$c \sim -\frac{1872 \pi +552 \pi ^3-5 \pi ^5}{1152+1104 \pi ^2+8 \pi ^4}$$ which is $-1.673493$ while the "exact" solution is $-1.673463$.


问:
How to use ratio test for the series $\frac{n}{2^n}$
given this series:

$$\frac{1}{2} + \frac{2}{2^2}...\frac{n}{2^n}$$

How to test for convergence or divergence through the ratio test?

I rearranged it likeso:

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

To show that the series diverges.

However, how do I show that:

$$\frac{a_{n+1}}{a_n}=l$$

where $l$ < 1 = convergence, and $l$ > 1 = divergence.

How do I get it into the fractional form $\frac{a_{n+1}}{a_n}$?

答:
$$\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \frac12\frac{n+1}{n}$$

which is between $\frac12$ and $\frac34$ when $n\ge 2$ so it converges.

In fact the sum is $2-\dfrac{n+2}{2^n}$ and so it converges to $2$


Title: How to use ratio test for the series $\frac{n}{2^n}$

Body:
given this series:

$$\frac{1}{2} + \frac{2}{2^2}...\frac{n}{2^n}$$

How to test for convergence or divergence through the ratio test?

I rearranged it likeso:

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

To show that the series diverges.

However, how do I show that:

$$\frac{a_{n+1}}{a_n}=l$$

where $l$ < 1 = convergence, and $l$ > 1 = divergence.

How do I get it into the fractional form $\frac{a_{n+1}}{a_n}$?

Body: $a_n=\dfrac n{2^n}\implies \dfrac{a_{n+1}}{a_n}=\dfrac{\dfrac{n+1}{2^{n+1}}}{\dfrac n{2^n}}=\dfrac{n+1}{2n}\to\dfrac12$.


答:
We know that
$$a_n :=\frac{n}{2^n}, \quad a_{n+1} = \frac{n+1}{2^{n+1}}.$$
Now we have to calculate
$$L=\lim_{n\to\infty}\cfrac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}=\lim_{n\to\infty}\frac{(n+1)2^n}{n2^{n+1}}=\frac{1}{2}\lim_{n\to\infty}\frac{n+1}{n}.$$
This limit converges to $1$. That means that $L=1/2$. Since $L<1$, the sequence converges absolutely.


问:
"Opposite" of Lebesgue's Dominated Convergence Theorem
Lebesgue's Dominated Convergence Theorem states:

Assume $g: X \to \overline{\mathbb{R}}$ is a nonnegative, integrable function and that $(f_n)$ is a sequence of measurable functions converging pointwise to f. If $|f_n|≤g$ for all $n$, then $$\lim_{n\to\infty}\int f_n d\mu=\int fd\mu$$

Is the "opposite" true, that is if $\lim_{n\to\infty}\int f_n d\mu\neq\int f d\mu$ and $|f_n|≤g$ is $g$ not integrable?

答: If $g: X \to \overline{\mathbb{R}}$ is a nonnegative, integrable function and that $(f_n)$ is a sequence of measurable functions converging pointwise to $f$ and if $|f_n|≤g$ for all $n$, then yes since your statement is just the contrapositive, hence equivalent.


问:
Question on a function $f$ such that $f: \mathbb R^n \leftarrow \mathbb R^n$,
Let  $f(x) = x||x||^2$ where $x \in \mathbb R^n$ . Then which is correct?

$(Df)(0) = 0$

$(Df)(x) = 0$ for all x

f is one to one

f has inverse.

My attempt:- I found out that $\begin{pmatrix}
                               3x_{1}^2+x_{1}^2  & 2x_{1}x_{2} \\
                               3x_{2}^2+x_{1}^2  & 2x_{1}x_{2}
                               \end{pmatrix}$
is the derivative matrix of the given function. Then option 2 is rejected. I do not know how to proceed.

答:
The Jacobian is not exactly right, you just have to switch the order of appearance of the entries in the bottom row; anyway you correctly discarded option 2.

The function describes the vectors in $\mathbb R^n$ times the square of their magnitude, so suppose you have two vectors $\mathbf v$ and $\mathbf u$ such that $\mathbf f(\mathbf v)=\mathbf f(\mathbf u)$, so $||\mathbf v||^2 \, \mathbf v = ||\mathbf u||^2 \, \mathbf u $; meaning they are parallel, so continue: do they share direction and magnitude?...

Given $\mathbf f(\mathbf x)$ can you think of a way to get back at $\mathbf x$?