答:
You're probably making this more complicated than it needs to be.  You got this far $$\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}=\frac{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}.$$  Note that by multiplying the all three fractions by $c/c$ we can rewrite it as as $$\frac{c + V_{12}}{c - V_{12}} = \frac{\frac{c + V_{2}}{c - V_{2}}}{\frac{c + V_{1}}{c - V_{1}}}.$$  Rewriting the right-hand side so we don't have two stacked fractions:
$$\frac{c + V_{12}}{c - V_{12}} = \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}.$$  Now we multiply the denominator of the left-hand side over and distribute:
$$c + V_{12} = c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) - V_{12}\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right).$$  Collect the $V_{12}$ terms on the left-hand side:
\begin{align}
V_{12} + V_{12}\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) &= c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) - c\\
V_{12}\left(1 + \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}\right) &= c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}} - 1\right),
\end{align}
which gives us
$$V_{12} = \frac{c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}} - 1\right)}{1 + \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}}.$$
At this point it's just a matter of expanding and simplifying the right-hand side to get the result:
\begin{align}
V_{12} &= \frac{c\left(\frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}} - 1\right)}{1 + \frac{c + V_{2}}{c - V_{2}}\cdot \frac{c - V_{1}}{c + V_{1}}}\\
&=\frac{\frac{(c + V_{2})(c - V_{1}) - (c - V_{2})(c + V_{1})}{(c - V_{2})(c + V_{1})}}{\frac{(c - V_{2})(c + V_{1}) + (c + V_{2})(c - V_{1})}{(c - V_{2})(c + V_{1})}}\\
&=\frac{(c + V_{2})(c - V_{1}) - (c - V_{2})(c + V_{1})}{(c - V_{2})(c + V_{1}) + (c + V_{2})(c - V_{1})}\\
&=\frac{2cV_{2} - 2cV_{1}}{2c^{2} - 2V_{1}V_{2}}\\
&=\frac{V_{2} - V_{1}}{1 - \frac{V_{1}V_{2}}{c^{2}}}.
\end{align}

Title: Tips for solving equations for a variable

Body:
I am trying to solve the following equation for $V_{12}$:

$$\sqrt{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}\sqrt{\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}}=\sqrt{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}$$

(this is not really relevant, but it is the derivation of the formula for the relativistic addition of velocities using the Doppler k-factor as presented in this paper).

For me as a highschool student, this isn't an easy equation to solve. Of course, I would square everything to remove the square roots. Then I thought that it might be good to have $V_{12}$ (the variable I'm trying to solve for) on one side and the two fractions with $V_1$ and $V_2$ on the oder side. This way, I got

$$\frac{1+\frac{V_{12}}{c}}{1-\frac{V_{12}}{c}}=\frac{\frac{1+\frac{V_{2}}{c}}{1-\frac{V_{2}}{c}}}{\frac{1+\frac{V_{1}}{c}}{1-\frac{V_{1}}{c}}}$$

However, I am a bit clueless what to do next. I thought of multiplying by $\left(1-\frac{V_{12}}{c}\right)$ to get rid of the fraction on the right side, but then I would again have $V_{12}$ on the right side. Whatever I tried seemed not to be helpful (for example, I had the idea to multiply out everything after squaring and then multiply by the denominators to remove the fractions, but this would only lead to overly complicated expressions with even more fractions like $\frac{V_1V_2V_{12}}{c^3}$.

The correct result should be

$$V_{12}=\frac{V_2-V_1}{1-\frac{V_1V_2}{c^2}}$$

but I cannot figure out how to get there.

So: What are gerneral reccomendations and tips to approach solving an equation like this for one variable? (Note that I'm not specifically asking for a derivation of the last mentioned formula from the first since this would probably be outside the scope of this site)

(I also tried using Wolfram Alpha to solve for $V_{12}$ and then analyse the step-by-step solution, but it didn't work)


问:
$f^n(x) = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
$f^n$(x)$ = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
Where $f^n(x)$ is the $n^{th}$ derivative of $f(x)$. And $f(x) = (7-5x)^{\frac{1}{2}}$.
So far I have made my base case of $n=1$ and showed that it equals the derivative of $f(x)$ and then I assumed n=k. I then state that: $f^{k+1} = \frac{d}{dx}(f^k)$. I know that my end goal is to get: $f^{k+1} = \frac{5-10(k+1)}{2}*(7-5x)^{-1} * f^{(k+1)-1}(x)$ which equals $f^{k+1} = \frac{-5-10k}{2}*(7-5x)^{-1} * f^{k}(x)$.
I've tried plugging the equation for $f^{k+1}$ into $f^{k+1} = \frac{d}{dx}(f^k)$ but couldn't seem to get the result I wanted, as I ended up with:
$-25*(7-5x)^{-2}*f^k(x)$
Help would be much appreciated.

ContentLicense: CC BY-SA 4.0

Body:
Hint

First of all, $f^{n+1}(x)=(f^n)'(x)$. See that both $(7-5x)^{-1}$ and $f^{n-1}(x)$ depends on $x$, so
$$f^{n+1}(x)=\frac{5-10n}{2}\left[5(7-5x)^{-2}f^{n-1}(x)+(7-5x)^{-1}f^{n}(x)\right]$$
$$f^{n+1}(x)=\left[\frac{5-10n}{2}(7-5x)^{-1}f^{n-1}(x)\right][5(7-5x)^{-1}]+\frac{5-10n}{2}[(7-5x)^{-1}f^{n}(x)]$$

by HI
$$f^n(x)=\frac{5-10n}{2}(7-5x)^{-1}f^{n-1}(x)$$

Can you finish?


问:
Second derivative of the Christoffel symbols in normal coordinates
According to wikipedia the Taylor expansion of the Christoffel symbols of a Riemannian manifold $(M,g)$ in normal coordinates is given by
$$
{\Gamma^{\lambda}}_{\mu\nu}(x)= -\frac 13 (R_{\lambda\nu\mu\tau}(0)+R_{\lambda\mu\nu\tau}(0))x^\tau+ O(|x|^2).
$$
Is there any reference for the calculation of the $O(|x|^2)$ term?
I did the computations as Yuval proposed and ended up with
$$
6\dfrac{\partial }{\partial x_v} \dfrac{\partial }{\partial x_w}{\Gamma^{n}}_{ij}(x)
\\= \frac 12 g^{nl}
\left( \nabla_i R_{jvwl} + \nabla_i R_{jwvl} +\nabla_w R_{jivl} + \nabla_v R_{jiwl} + \nabla_w R_{jvil} + \nabla_v R_{jwil} \right) +
\left( \nabla_j R_{ivwl} + \nabla_j R_{iwvl} +\nabla_w R_{ijvl} + \nabla_v R_{ijwl} + \nabla_w R_{ivjl} + \nabla_v R_{iwjl} \right) -
\left( \nabla_l R_{ivwj} + \nabla_l R_{iwvj} +\nabla_w R_{ilvj} + \nabla_v R_{ilwj} + \nabla_w R_{ivlj} + \nabla_v R_{iwlj} \right)
$$
If I apply the symmetries of the curvature tensor
$$
\left( \nabla_i R_{jvwl} + \nabla_i R_{jwvl}  + 2\nabla_w R_{jvil} + 2\nabla_v R_{jwil} \right) +
\left( \nabla_j R_{ivwl} + \nabla_j R_{iwvl}   + 2\nabla_w R_{ivjl} + 2\nabla_v R_{iwjl} \right) -
\left( \nabla_l R_{ivwj} + \nabla_l R_{iwvj}  \right)
$$
I feel like the derivatives involving derivatives in $i,j,l$ should somehow cancel out with the Bianchi identities but I don't get it to work.

答:
This needs some computations involving the Jacobi field ODE. You just need an Taylor expansion for $g_{ij}$ since $\Gamma_{ij}^k$ involves $g$ and its
first derivative. See

John Lee & Thomas Parker "Yamabe Problem", Bull. AMS. vol 17. no 1 (1987),
see page 60-61, especially equation (54).

or

Schoen & Yau "Lectures on Differential Geometry", p210-211. You can use (3.4) there to compute the $O(|x|^2)$ term.


问:
Finding the Jordan Form of matrices with mostly 0
I have been studying linear algebra and came across this question which I'm not quite sure how to solve.
The problem asks whether the matrices are similar. I wanted to find the Jordan form of 3 matrices and compare to see if they are similar:
$A=\begin{bmatrix} 0 & 1 & 1 & 0\cr 0 & 0 & 0& 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\end{bmatrix} \mbox{ , }
B=\begin{bmatrix} 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\end{bmatrix} \mbox{ , }
C=\begin{bmatrix} 0 & 0 & 1 & 1\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\end{bmatrix}$

I know that you have to find the eigenvalues from the characteristic polynomial but I'm not sure what the next step is from there. Also, since the matrices mostly contain $0$s is there a shortcut?

Any help would be greatly appreciated!

答:
The number of Jordan blocks of size $n$ for the eigenvalue $\lambda$ is $$2\dim\ker(A-\lambda I)^n-\dim\ker(A-\lambda I)^{n+1}-\dim\ker(A-\lambda I)^{n-1}$$

For the matrix $A$, we have $\dim\ker A=2$, $\dim\ker A^2=3$ and $\dim\ker A^3=4$. This determines completely its Jordan form, which is $J(0,1)\oplus J(0,3)$.

For the matrix $B$, we have $\dim\ker B=2$ and $\dim \ker B^2=4$, therefore its Jordan form is $J(0,2)\oplus J(0,2)$.

For the matrix $C$, we have $\dim \ker C=2$, $\dim \ker C^2=3$ and $\dim\ker C^3=4$, therefore it is similar to $A$.


问:
Proving $A=U_{r}S_{r}V_{r}^{T}$ for a Full SVD
Given the full SVD decomposition of a rank $r$ matrix $A\in\mathbb{R}^{m\times n}$, how can we show that $A=U_{r}S_{r}V_{r}^{T}$ where $U\in\mathbb{R}^{m\times m}$, $V\in\mathbb{R}^{n\times n}$, and $S\in\mathbb{R}^{m\times n}$?

答:
Note that :
\begin{align*}
A&=U S V^{T}=\left[U_{r} \mid U_{m-r}\right]\left[\begin{array}{c|c}
S_{r} & 0_{r, n-r} \\
\hline 0_{m-r, r} & 0_{m-r, n-r}
\end{array}\right]\left[\begin{array}{c}
V_{r}^{T} \\
\hline V_{n-r}^{T}
\end{array}\right]=\left[U_{r} \mid U_{m-r}\right]\left[\frac{S_{r} V_{r}^{T}}{0_{m-r, n}}\right]\\
&=U_{r} S_{r} V_{r}^{T}
\end{align*}


问:
Prove the following inequality using the AM-GM Mean
Let $x_0>x_1>x_2>\ldots>x_n$ be real numbers.

Prove $\displaystyle x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$.

I thought I would try and deal with the denominators by letting $a_k=x_{k-1}-x_k$

Which gives:

LHS $\displaystyle=x_0+\frac{1}{a_1}+\frac{1}{a_2}+\ldots +\frac{1}{a_n}$

I am unsure how to use the AM-GM mean from here.

答:
Required to prove:

$\displaystyle x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$

i.e., To prove that (TPT)
$$\sum a_k + \sum \frac{1}{a_k} \geq 2n$$ where $a_k=x_{k-1}-x_k$ for $k = 1, \ldots, n$

i.e., TPT $$nA + \frac{n}{H} \geq 2n$$ where $A = \frac{\sum a_k}{n}$ is the arithmetic mean and $H = \frac{n}{\sum \frac{1}{a_k}}$ is the harmonic mean

i.e., TPT $$A + \frac{1}{H} \geq 2$$

However, $$A + \frac{1}{H} \geq A + \frac{1}{A} \geq 2$$

Hence the proof.


问:
On some inequality in $C^\ast$-algebra
I am reading a proof of some properties of quasi-central approximate unit. At some point the author claims that is using the following inequality, valid for any families $(a_j)_{j\in J}$, $(b_j)_{j\in J}$, indexed by a given finite set $J$, in a given $C^\ast$-algebra $A$:
$$\left\|\sum\limits_{j\in J} a_jb_j\right\|\leq \left\|\sum\limits_{j\in J}a_ja_j^\ast\right\|^{1/2}\left\|\sum\limits_{j\in J}b_j^\ast b_j\right\|^{1/2}.$$

I do not see how exactly it is used in this specific situation, so let me briefly summarize what was done, and then I will formulate my question. First, we have an ideal $\mathcal{I}\subset A$, and a non-decreasing net $(\sigma_i)$ in the unit ball of $\mathcal{I}$. Moreover, $\sigma_i\geq 0$. It satisfied some futher conditions but they are irrelevant for what I need now. Take some $a,b\in A$. It is claimed that the above inequality immediately implies the following one:
$$\|\sigma_i^{1/2}a\sigma_i^{1/2}+(1-\sigma_i)^{1/2}b(1-\sigma_i)^{1/2}\|\leq \max\{\|a\|,\|b\|\}.$$
Why it is the case? Maybe I do not see something obvious, but I couldn't find a way how to use the previous inequality to deduce this one.

Title: On some inequality in $C^\ast$-algebra

Body:
Suppose that $\|\sigma\|\leq1$. Take $$a_1=\sigma^{1/2},\ \  a_2=(1-\sigma)^{1/2}, \ \ b_1=a\sigma^{1/2},\ \ b_2=b(1-\sigma)^{1/2}.$$

Then, using  the inequality in the first line,
\begin{align}
\|\sigma^{1/2}a\sigma^{1/2}+(1-\sigma)^{1/2}b(1-\sigma)^{1/2}\|
&\leq\|\sigma+1-\sigma\|^{1/2}\,\|\sigma^{1/2}a^*a\sigma^{1/2}+(1-\sigma)^{1/2}b^*b(1-\sigma)^{1/2}\|^{1/2}\\[0.3cm]
&=\|\sigma^{1/2}a^*a\sigma^{1/2}+(1-\sigma)^{1/2}b^*b(1-\sigma)^{1/2}\|^{1/2}\\[0.3cm]
&\leq\big\|\sigma^{1/2}\,\|a^*a\|\,\sigma^{1/2}+(1-\sigma)^{1/2}\,\|b^*b\|\,(1-\sigma)^{1/2}\big\|^{1/2}\\[0.3cm]
&=\big\|\sigma^{1/2}\,\|a\|^2\,\sigma^{1/2}+(1-\sigma)^{1/2}\,\|b\|^2\,(1-\sigma)^{1/2}\big\|^{1/2}\\[0.3cm]
&\leq\max\{\|a\|\,\|b\|\}\,\|\sigma+1-\sigma\|^{1/2}\\[0,3cm]
&=\max\{\|a\|,\|b\|\}.
\end{align}
All that's used in the remaining inequalities is that if $0\leq x\leq y$, then $\sigma x\sigma^*\leq\sigma y\sigma^*$, and $\|x\|\leq\|y\|$.


问:
'Proof' that $f''(x)=\frac{f'(x)}{x}$
Consider the following:

$$f''(x)=\lim_{h\to0}\frac{f'(x+h)-f'(x)}{h}$$
Now using L'Hopital's rule (as this is a case of an inderterminate) we have
$$f''(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{hx}$$
But this is
$$\frac{1}{x}\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
which is $\frac{1}{x}f'(x)$.

So it would seem that
$$f''(x)=\frac{f'(x)}{x}$$ which is quite obviously false. Where is my error?

Thanks in advance.

答: L'Hôpital's rule actually implies$$\lim_{h\to0}\frac{f(x+h)-f(x)}{hx}=\lim_{h\to0}\frac{f^\prime(x+h)}{x},$$not$$\lim_{h\to0}\frac{f(x+h)-f(x)}{hx}=\lim_{h\to0}\frac{f^\prime(x+h)-f^\prime(x)}{h}.$$As @Brainstorming notes, your derivative was with respect to the wrong variable.

Title: 'Proof' that $f''(x)=\frac{f'(x)}{x}$


问:
Is this a characteristic function? Why (not)?
I have recieved a home assignment and the question is if $(1-|t|)e^{-|t|}$ is a characteristic function. If it is a yes, then what distribution does it correspond to?

I was wondering if that is a combination of convex characteristic functions, the Polya theorem sadly does not work. I would appreciate any ideas.

答: Since this is a homework problem, I will not give a full answer. The candidate characteristic function is integrable on the real line. Therefore, the candidate distribution has a density that can be computed as the inverse Fourier transform. One gets an expression for a potential density; if it qualifies, then we will also be able to determinate the distribution. If not, we will know that the aforementioned function is not a characteristic function.</p>


问:
two asymptotic comparison on sparse graph and misunderstanding point
I read in my notes:

if we use Dijkstra $|V|$ times ($|V|$ number of vertexes) for finding all pairs shortest paths in graph $G$, We get time complexity for Dijkstra algorithm as $O(VE+ V^2 log V)$ and if we run bellman ford algorithm $|V|$ times we get time $O(V^2E)$.

The above details is not important.

I read too, in sparse graph Dijkstra works better (i.e: $O(VE+ V^2 log V)$ is better asymptotic respect to $O(V^2E)$ in sparse graph).

I think sparse graph is $ |E| \in O(V)$  or $E=o(V^2/\log V)$. Infact I have two misunderestanding:

$1)$ which one is used to show a graph is sparse commonly?

$2)$ according to $(1)$ how we can intuitively understand $O(VE+ V^2 log V)$ is better asymptotic respect to $O(V^2E)$ at least I think the reverse is true.

答:
As well said by wikipedia The distinction between sparse and dense graphs is rather vague, and depends on the context.

The accepted answer in this StackOverflow question, also state

Dense graph is a graph in which the number of edges is close to the maximal number of edges. Sparse graph is a graph in which the number of edges is close to the minimal number of edges.

Sparse or Dense is not really a proper property of a graph. it will indeed depend on the context. Your two "definitions" are perfectly both acceptable.

For the second point, if you suppose that $E=O(V)$, then

Dijkstra works in $O(VE+V^2\log V)=O(V^2+V^2\log V)=o(V^3)$

bellman ford works in $O(V^2E)=O(V^3)$

if you suppose that $E=o(V^2/\log V)$, then

Dijkstra works in $O(VE+V^2\log V)=o(V^3/\log V)$

bellman ford works in $O(V^2E)=o(V^4/\log V)$

So that Dijkstra is more efficient in both cases.


问:
Intuitive explanation of order of distribution
Definition. We say that $T\in \mathcal{D}'(\Omega)$ has $\textbf{order}$ $m$ if $m$ is the minimum integer for which there exists a constant $C>0$ such that $$|\langle T,f\rangle_{\mathcal{D}'\leftrightarrow\mathcal{D}}|\leq C\left(\sum_{|\alpha|\leq m}\|D^\alpha f\|_{\sup(\Omega)}\right),\quad\forall f\in\mathcal{D}(\Omega)$$ If a distribution does not have order $m$ for any $m\in\mathbb{N}$, then it has infinite order.

I understand the definition, however, not sure what this means in more intuitive sense. I see the right-hand side is somewhat semi-norm on $\mathcal{D}(\Omega)$, so when the order of the distribution is higher, then it seems like the distribution is "wilder" in some sense. I will appreciate it if someone can help me to understand the motivation behind of defining the order of distribution.

答:
Intuitive, but imprecise:

The order of a distribution $u$ is the highest order of derivative of the test function $\varphi$ that the value $\langle u, \varphi \rangle$ depends on.

An easy case: the order of $\partial^\alpha \delta$ is $|\alpha|$.


问:
Curl of cross product of two vectors: $\operatorname{curl}(r\times a)$
I cannot find mistake in any method
hand written is mine, can someone point out the mistake
book image
my solution</p>

答:
Your error is that $ \nabla \cdot a \neq 0$ but $\nabla \cdot a = a_1 \partial_1 + a_2 \partial_2 + a_3 \partial_3$. Therefore, the first term reads

$(\nabla \cdot \vec{a}) \vec{r} = a_1 \cdot \partial_1(x) \cdot \vec{i} + a_2 \cdot \partial_2(y) \cdot\vec{j} + a_3 \cdot \partial_3(z) \cdot\vec{k} = \vec{a}$,

as (due to my notation) $\partial_1(x) := \partial_x(x) = 1$ and so on.

Therefore, as the calculation of the second term was correct, we obtain $(\nabla \cdot \vec{a})\vec{r} - (\nabla \cdot \vec{r})\vec{a} = \vec{a} - 3 \vec{a} = -2 \vec{a}$


问:
Size of the image (rank) in $T_n$
For $\alpha, \beta \in T_n$ (full transformation semigroup/monoid - set of all maps from $\{1,2,\ldots, n\}$ to itself), show that $|\text{Im}(\alpha\beta)| \leqslant |\text{Im}(\alpha)|$ and $|\text{Im}(\alpha\beta)| \leqslant |\text{Im}(\beta)|$.

How do you show this (apparently it's easy)?

答:
If $k \in \operatorname{Im}(\alpha\beta) \subseteq \{1, \dots, n\}$, then there is some $i \in \{1, \dots, n\}$ such that $\alpha(\beta(i)) = k$.

This equation shows that $\alpha$ maps $\beta(i) \mapsto k$.  In other words, $k \in \operatorname{Im}(\alpha)$, and so $\operatorname{Im}(\alpha\beta) \subseteq \operatorname{Im}(\alpha)$.  Note that we only have equality $\operatorname{Im}(\alpha\beta) = \operatorname{Im}(\alpha)$ if $\beta$ is onto (hence bijective, as $\{1, \dots, n\}$ is finite).

For the second inequality, the correspondence $\operatorname{Im}(\alpha\beta) \to \operatorname{Im}(\beta), k \mapsto \beta(i)$, which involves arbitrarily choosing a preimage $i$, is injective.  Why?  Let $k, k' \in \operatorname{Im}(\alpha\beta)$ and choose $i, i' \in \{1, \dots, n\}$ such that $\alpha(\beta(i)) = k$ and $\alpha(\beta(i')) = k'$.  If $\beta(i) = \beta(i')$, then by application of $\alpha$, it follows that $k = k'$.  Note that we only have the bijection $\operatorname{Im}(\alpha\beta) \cong \operatorname{Im}(\beta)$ if $\alpha$ is one-to-one (hence bijective, as $\{1, \dots, n\}$ is finite).

ContentLicense: CC BY-SA 3.0

Title: Size of the image (rank) in $T_n$


答:
Thanks PTDS.
I saw your solution and had a light bulb moment.
Recall that the AM-GM mean for two(2) numbers can give us: $\displaystyle a+\frac{1}{a}\ge2$

Restate the original problem as:

{$\displaystyle x_0-x_n+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\ldots +\frac{1}{x_{n-1}-x_n}\ge2n$}

Now let $a_k=x_{k-1}-x_k$

So, $x_0 - x_n = a_1 + a_2 + \ldots + a_n$

Thus,

LHS $=\displaystyle a_1+a_2+\ldots+a_n+\frac{1}{a_1}+\frac{1}{a_2}+\ldots +\frac{1}{a_n}$

$\displaystyle =a_1+\frac{1}{a_1}+a_2+\frac{1}{a_2}+\ldots a_n+\frac{1}{a_n}$

$\ge 2+2+\ldots +2$ (n times)

$=2n$

$=$RHS

Title: Prove the following inequality using the AM-GM Mean

Body:
Let $x_0>x_1>x_2>\ldots>x_n$ be real numbers.

Prove $\displaystyle x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$.

I thought I would try and deal with the denominators by letting $a_k=x_{k-1}-x_k$

Which gives:

LHS $\displaystyle=x_0+\frac{1}{a_1}+\frac{1}{a_2}+\ldots +\frac{1}{a_n}$

I am unsure how to use the AM-GM mean from here.


问:
Is all of mathematics set theory in disguise?
Is all of mathematics, just set theory in disguise? We know the ZFC axioms allow us to encode mathematical objects as sets of some kind. So, even theorems about, say, Hilbert spaces can be written down, tediously, as formulas of ZFC. So, does this mean all of mathematics is set theory in disguise?</p>

答:
No.

Say you've identified a mathematical theory $T$ with a model in $\mathsf{ZFC}$. It has multiple models in $\mathsf{ZFC}$, each with different this-object-is-that-set identifications. Not only are these identifications different by model, they also have at-the-set-level consequences that don't mean anything about $T$ itself. For example, some but not all models of $\Bbb N$ in $\mathsf{ZFC}$ satisfy $m\le n\implies m\subseteq n$, where the first statement is about naturals but the second is about sets with which they're identified when specifying a model. It would be silly to write $2\subseteq 3$, or $2\in 3$ (these are examples of what @NoahSchweber's answer calls junk theorems), because these identifications don't tell us what $T$'s objects "really are". The only reason models are interesting is that, as long as you can find one, $\mathsf{ZFC}$ (or whatever baseline we picked) implies $T$ is consistent.

You could just as easily use something else as a "base" for mathematics, e.g. category theory.


问:
Interest rate that generates the maximum profit
Story Problem:

A savings and loan association estimates that the amount of money on
deposit will be 1 million times the percentage rate of interest. For
instance, a 4% interest rate will generate $4 million in deposits. If
the savings and loan association can loan all the money it takes in at
10% interest, what interest rate on deposits generates the greatest
profit?

This is a problem that is part of a chapter where derivatives are used to solve business and economic problems.

The problem says to find the greatest profit. The profit function and associated functions/vars are

Profit: P(x) = R(x) - C(x)

number of items a business has sold = x

Revenue: R(x) = x*p

price of an item sold = p

Cost: C(x) = cost of selling x

I'm having a hard time converting the info in the story problem to the profit formula.

It seems like:

Revenue = 1million * interest rate

Cost = ?

I'm not sure what the cost is.  I don't see any indication of what the cost of a deposit is.

Profit = (1,000,000 * i) - ?Cost?

Thank you for your help!

答:
Looking at revenue and cost for a bank is confusing because it's often backwards from how people or companies think of it.  The bank loses money by paying interest on deposits.  Why does it take deposits then?  So that it  can turn around and loan that money out to other people at a higher interest rate.

The bank makes money by collecting interest on loans.  The higher the interest on the loan, the more revenue.  So the revenue will be the total amount of money the bank is able to loan out times the loan interest rate. The bank cannot loan out more money than it has received in deposits.

number of items a business has sold = dollars loaned

price of an item  sold = loan interest rate

The bank loses money by paying interest on deposits.  That's how it convinces customers to use its bank.  So the cost is the total amount of money on deposit times the deposit interest rate.

cost of selling 1 item = cost of loaning one dollar out = cost of convincing one person to deposit one dollar in the bank = deposit interest rate

This problem is extra confusing because both revenue and cost are (total money)*(interest rate).  You have to pay attention to whether it is the interest rate on loans the bank is offering, or the interest rate on deposits.


问:
Solving the system with $\tan x+\tan y=1-\tan x\tan y$ and $ \sin2y-\sqrt 2 \sin x=1$
I have came across this system of trigonometry equations.

$$\tan x + \tan y=1-\tan x\tan y$$
$$ \sin2y-\sqrt 2 \sin x=1$$

I have tried with expressing $\tan$ as $\frac{\sin}{\cos}$ and using the formula for $\sin2\alpha$. But I haven't gone any further than the start.

答:
HINT

The first equation can be rewritten as
\begin{align*}
\tan(x) + \tan(y) = 1 - \tan(x)\tan(y) & \Longleftrightarrow \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)} = 1\\\\
& \Longleftrightarrow \tan(x+y) = \tan\left(\frac{\pi}{4}\right)\\\\
& \Longleftrightarrow x + y = \frac{\pi}{4} + k\pi
\end{align*}
where $k\in\mathbb{Z}$. Substituting such relation into the second equation, one has that
\begin{align*}
\sin(2y) - \sqrt{2}\sin(x) = 1 & \Longleftrightarrow \sin\left(\frac{\pi}{2} - 2x\right) - \sqrt{2}\sin(x) = 1\\\\
& \Longleftrightarrow \cos(2x) - \sqrt{2}\sin(x) = 1\\\\
& \Longleftrightarrow 1 - 2\sin^{2}(x) - \sqrt{2}\sin(x) = 1\\\\
& \Longleftrightarrow \sin(x)(\sqrt{2}\sin(x) + 1) = 0
\end{align*}

Can you take it from here?

EDIT

Notice that $\tan(x)\tan(y) = 1$ iff
\begin{align*}
\tan(x)\tan(y) = 1 \Longleftrightarrow
\begin{cases}
\sin(x)\sin(y) = \cos(x)\cos(y)\\\\
\cos(x)\cos(y) \neq 0
\end{cases} & \Longleftrightarrow
\begin{cases}
\cos(x+y) \neq 0\\\\
\cos(x)\cos(y) \neq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
\displaystyle x + y \neq \frac{\pi}{2} + m\pi\\\\
\displaystyle x \neq \frac{\pi}{2} + n\pi\\\\
\displaystyle y \neq \frac{\pi}{2} + p\pi
\end{cases}
\end{align*}

where $m,n,p\in\mathbb{Z}$. As it can be checked, the proposed solutions satisfy such restrictions.


问:
Solving an equation with least residues and roots modulo $n$
I am facing the problem of proving $z^2 \equiv 1\pmod 8$ for all odd $z\in\mathbb{N}$.

What I know is that for each $z \in \mathbb{N}$, $z = m^2-1$, for $m\in\mathbb{N}$.

What I also know is that in a least residue system there are the following solutions for all $z\in\mathbb{N}$.

The solutions are:
$$n \equiv 1,3,5,7$$

Which lead to solutions of the initial problem:
$$n = 8 \cdot k + 1\\
n = 8 \cdot k + 3 \\
n = 8 \cdot k + 5 \\
n = 8 \cdot k + 7$$
where $k\in\mathbb{N}$.

My question now is how I can transfer my knowledge about the solutions for all $n\in\mathbb{N}$ to the key problem where $n$ must be an odd number and the proofs needs to be done for all odd $n$.

I also know that $8 = 2^3$ which leads to $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod 8$$ $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod {2^3}$$ but I do not know what this tells me right now.

答: $\bmod 8$ there are only eight numbers to check.  All the odd $z\in \Bbb N$ are in one and only one of the residue classes $\bar1,\bar3,\bar5$ or $\bar7$.  For each of these we can verify the condition is satisfied.  Thus it is true for all odd naturals.


问:
Formula for determinant
For $n \geq 2$, let $_$ be the $n \times n$ matrix defined by

$(G_n)_{ij} = 1 \text{ if } i = j$

$(G_n)_{ij} = 2 \text{ if } j = i + 1$ (mod n)

$(G_n)_{ij} = 0$ otherwise

I am told there is a general formula for det$(G_n)$ for $n \geq 2$. This is what I'd like to find.

I find that $G_2$ = $\begin{bmatrix} 1 & 2  \\ 2 & 1 \end{bmatrix}$ with det($G_2$) = -3.
$G_3$ = $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2  \\ 2 & 0 & 1\end{bmatrix}$ with det(G_3) = 9

$G_4$ = $\begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 2 & 0 & 0 & 1\end{bmatrix}$ with det($G_4$) = -15

I can guess the general formula det$(G_n) = (-1)^{n+1}(2^n-1)$. How might I show that this is true for all n? As hint I was told to use the permutation definition of the determinant.

Any help would be appreciated.

edit:

Attempted proof:

I will use the fact that (1) If $\pi \in S_n$ is a permutation for which either $\pi(i) = i$ or $\pi(i) = i + 1 $(mod $n)$ holds for all i, then we either have $\pi(i) = i$ for all $i$ or $\pi(i) = i + 1 $(mod $n$) for all $i$.

Let det$(G_n) = \sum_{\pi \in S_n}sgn(\pi)\prod_{i=1}^nG_{i,\pi(i)}$.

Since $\pi(i) = i$ is only true for the identity map. The other n! - 1 permamutations must be of the form $\pi(i) = i + 1$.

We get det$(G_n) = sgn(id)(1)^n$ + $\sum_{\pi \in S_n - id }sgn(\pi)2^n = 1 - 2^n$. Note that the sum will be negative since we took out an even permutation and there will be one more odd permutation than even.

What have I messed up here?. This formula is correct for $G_2$ and $G_4$ but not $G_3$

答:
If we want to use the permutation formula for the determinant, then it would suffice to prove the following.

Claim: If $\pi \in S_n$ is a permutation for which either $\pi(i) = i$ or $\pi(i) = i+1 \pmod n$ holds for all $i$, then we either have $\pi(i) = i$ for all $i$ or $\pi(i) = i+1 \pmod n$ for all $i$.

Proof: Suppose for the purpose of contradiction that $\pi(i_1) = i_1$ and $\pi(i_2) = i_2 + 1 \pmod n$.

Suppose without loss of generality that $i_1 > i_2$. It cannot hold that $i_1 = i_2 + 1$, or we'd have $\pi(i_1) = \pi(i_2)$. Let $S = \{i : i_2 < i < i_1\}$; note that $S$ is non-empty. Since $\pi$ satisfies the hypothesis, we must have $\pi(S) \subset S \cup \{i_1\}$. However, $\pi$ is injective, which means that
$$
\pi(S) \subset [S \cup \{i_1\}] \setminus \{\pi(i_1),\pi(i_2)\} = S \setminus \{\pi(i_2)\}.
$$
This means that $\pi(S)$ is a smaller set than $S$, which is impossible by the pigeonhole principle.

Regarding your edit:

I will use the fact that (1) If $\pi \in S_n$ is a permutation for which either $\pi(i) = i$ or $\pi(i) = i + 1 $(mod $n)$ holds for all i, then we either have $\pi(i) = i$ for all $i$ or $\pi(i) = i + 1 $(mod $n$) for all $i$.

Let det$(G_n) = \sum_{\pi \in S_n}sgn(\pi)\prod_{i=1}^nG_{i,\pi(i)}$.

So far, so good.

Since $\pi(i) = i$ is only true for the identity map. The other n! - 1 permutations must be of the form $\pi(i) = i + 1$.

This is incorrect. There is exactly one permutation $\pi$ for which $\pi(i) = i$ holds for $i = 1,\dots,n$, and there is also exactly one permutation $\pi$ for which we have $\pi(i) = i + 1 \bmod n$ for $i = 1,\dots,n$. The key fact here is that for the remaining $n! - 2$ permutations, the product $\prod_{i=1}^n G_{i,\pi(i)}$ will be zero.

With that said, the remainder of your proof is wrong since it builds on this incorrect foundation. Let $\pi_1 \in S_n$ denote the identity map, and let $\pi_2 \in S_n$ denote the map with the formula $\pi_2(i) = i + 1 \bmod n$. With the observations in the previous paragraph, we have
$$
\begin{align}
\det(G) &= \sum_{\pi \in S_n}\operatorname{sgn}(\pi) \prod_{i=1}^n G_{i,\pi(i)}
\\ & = \operatorname{sgn}(\pi_1) \prod_{i=1}^n G_{i,\pi_1(i)}
+ \operatorname{sgn}(\pi_2) \prod_{i=1}^n G_{i,\pi_2(i)}
\\ & = 1 \cdot 1^n + (-1)^{n-1} \cdot 2^n = 1 + (-1)^{n-1} 2^n.
\end{align}
$$
To justify this, I still need to argue that $\operatorname{sgn}(\pi_2)$ is an even permutation for odd $n$ and an odd permutation for even $n$.  To see that this is the case, it suffices to note that $\pi_2$ can be implemented as a succession of the following $n-1$ transpositions:
$$
\pi_2 = \tau_{1,2} \circ \cdots \circ \tau_{n-1,n}
$$
where $\tau_{i,j}$ is the permutation that switches the $i$th and $j$th elements, but otherwise acts as the identity permutation.

An interesting observation from the alternative approach: if we use the formula for the determinant of a circulant matrix provided here, we end up with the following. Take $f$ to be the function $f(x) = 1 + 2x$, and $\omega = e^{2 \pi i/n}$. We have
$$
\begin{align}
\det(G) = \det(G^T) = \prod_{j=0}^{n-1} f(\omega^j).
\end{align}
$$
If $n$ is odd, write this as
$$
\begin{align}
\det(G) &= \prod_{j=0}^{n-1} f(\omega^j)
= f(1) \cdot \prod_{j=1}^{(n-1)/2} f(\omega^j)f(\omega^{-j})
\\ & = 3 \cdot \prod_{j=1}^{(n-1)/2}(1 + 2\omega^j)\overline{(1 + 2 \omega^j)}
\\ & = 3 \cdot \prod_{j=1}^{(n-1)/2} |1 + 2\omega^j|^2
\\ & = 3 \cdot \prod_{j=1}^{(n-1)/2} [(1 + 2\cos(2 \pi j/n))^2 + 4 \sin^2(2 \pi j/n)]
\\ & = 3 \cdot \prod_{j=1}^{(n-1)/2} [5 + 4 \cos(2 \pi j/n)].
\end{align}
$$
Apparently, this should be equal to $1 + 2^n$.

Title: Formula for determinant


问:
Counting Problem - N unique balls in K unique buckets w/o duplication $\mid$ at least one bucket remains empty and all balls are used
I am trying to figure out how many ways one can distribute $N$ unique balls in $K$ unique buckets without duplication such that all of the balls are used and at least one bucket remains empty in each distribution?

Easy, I thought. I'll just hold a bucket in reserve, distribute the balls, and place the empty bucket. I get:

$ K\cdot N! / (N-K-1)! $

Even were I sure this handles the no duplicates condition, what if $K \geq N$?

Then I get a negative factorial in the denominator. Is the solution correct and/or is there a more general solution?

Thanks!

答:
A slightly different approach using the twelvefold way.

If $K>N$ then it doesn't matter how you distribute the balls since at least one bucket will always be empty. In this case we are simply counting functions from a $N$ element set to a $K$ element set. Therefore the number of distributions is $K^N$.

If $K=N$ then the only bad assignments are the ones in which every bucket contains precisely one ball. This happens in precisely $N!$ ways, so just subtract out these cases for a total of $N^N - N!$ distributions.

If $K < N$, we first choose a number of buckets which cannot be filled and then we fill the remaining buckets. If we choose $m$ buckets to remain empty, then the remaining $K-m$ buckets must be filled surjectively. The number of surjections for each $m$ is
$$(K-m)!{N\brace K-m}$$
where the braced term is a Stirling number of the second kind. Summing over $m$ gives the required result
$$\sum_{m=1}^{K-1}(K-m)!{N\brace K-m}\binom{K}{m}$$
I am not sure if this simplifies or not. In summary, if we let $f(N,K)$ denote the number of distributions, then
$$f(N,K) = \begin{cases}K^N & K > N\\
N! & K=N\\
\sum_{m=1}^{K-1}(K-m)!{N\brace K-m}\binom{K}{m} & K < N\end{cases}$$


问:
Prove that KDE is convolution of EDF and Kernel
This is equation 6.23 in Elements of Statistical Learning, page 209.

It concerns KDE (kernel density estimation). They let $\phi_h$ be the kernel function, the Gaussian distribution pdf with mean $0$ and standard deviation $h$.

Their statement is as such: $\hat{f}_X(x) = \frac{1}{N} \sum_{i = 1}^N \phi_h (x - x_i) = (\hat{F} * \phi_h)(x)$, where $\hat{F}$ is the empirical distribution function, $*$ is the convolution operator.

They say this without proof, and I was trying to recreate the results for a paper I'm writing. However, I get something different. The $1_{[\varphi(x)]}$ here are indicator functions. $\Phi_h$ is the normal cdf with mean $0$ and standard deviation $h$.

$$[\hat{F}*\phi_h](x) = \int \hat{F}(t) \phi_h(x - t) \, dt = \int \frac{\# \{ x_i \leq t \}}{N} \phi_h(x - t) \, dt = \frac{1}{N} \int \# \{x_i \leq t \} \phi_h(x - t) \, dt $$
$$= \frac{1}{N} \int \sum_{i = 1}^N {1}_{[x_i \leq t]} \phi_h(x - t) dt =\frac{1}{N} \sum_{i = 1}^N \int {1}_{[x_i \leq t]} \phi_h(x - t) dt = \frac{1}{N} \sum_{i = 1}^N \int_{x_i}^{\infty} \phi_h(x - t)  $$
$$ = \frac{1}{N} \sum_{i = 1}^N  \left[ -\Phi(x - t)  \right]_{t = x_i}^{\infty} =  \frac{1}{N} \sum_{i = 1}^N  \Phi_h(x - x_i)$$

So essentially I get that $\hat{f}_X(x) = \frac{d}{dx} \left[(\hat{F} * \phi_h)(x) \right]$, which is still a pleasant result. Is my derivation correct? Is this an error in the book? It has not yet been listed on the list of errors.

答:
I think the authors glossed over some technical details in order to keep the book flowing.
To fully justify this on a technical level I would make use of Lebesgue integration theory, but combining convolution with the Dirac delta also shows what is going on.
If you are comfortable extending the convolution
$$
(f*g)(x) := \int_{\mathbb{R}} f(x-y)g(y)dy
$$
from $L^1$ functions to also work with the Dirac delta $\delta_a$ (see: https://en.wikipedia.org/wiki/Dirac_delta_function), then you can directly write
\begin{align}
\widehat{f}_n(x) &= \frac{1}{n} \sum_{i=1}^n k_h(x-x_i) = \frac{1}{n} \sum_{i=1}^n (k_h*\delta_{x_i})(x) = \frac{1}{n} \sum_{i=1}^n \int_{\mathbb{R}} k_h(x-y)\delta_{x_i}(y) dy\\
&= \int_{\mathbb{R}} k_h(x-y) \frac{1}{n} \sum_{i=1}^n \delta_{x_i}(y) dy\\
&= \left( k_h * \frac{1}{n} \sum_{i=1}^n \delta_{x_i} \right)(x)
\end{align}
where I used $(f*\delta_a)(x) = \int_{\mathbb{R}} f(x-y)\delta_{a}(y) dy = f(x-a)$ and the linearity of the integral.
What this shows is that the kernel density estimator is not the convolution with the empirical distribution function, but rather with something resembling an empirical density.

In fact, you can in pretty much the same way write $\widehat{f}_n$ as a Lebesgue integral of $k_h$ with respect to the empirical distribution function (more correctly with the measure associated to the empirical distribution function) which incidentally places the mass $1/n$ at each $x_i$, i.e., write
\begin{align}
\widehat{f}_n(x) = \int_{\mathbb{R}} k_h(x-y) d\widehat{F}_n(y),
\end{align}
where $\widehat{F}_n(x) = \frac{1}{n} \sum_{i=1}^n \mathbf{1}_{(-\infty,x]}(x_i)$ is the empirical distribution function.


问:
$100$th derivative of $y=\ln(2x-x^2)$ at $x=1$

What is $100$th derivative of $y=\ln(2x-x^2)$ at $x=1$?

$a)2\times99!$

$b)-2\times99!$

$c)2\times101!$

$d)-2\times101!$

I tried to find a pattern by calculating first derivatives of $y$:

$$y'=\cfrac{2-2x}{2x-x^2} ,\quad y'(1)=0$$

$$y''=\cfrac{-2(2x-x^2)-(2-2x)^2}{(2x-x^2)^2},\quad y''(1)=-2$$

And $y'''$is going to get really ugly. so I couldn't find any pattern.

答:
You split them first:

$$y = \ln(2-x) + \ln(x)\implies y' = \frac{1}{x-2}+\frac 1x\dots{}{} $$


问:
If $x\in L$ lies in exactly one Cartan subalgebra, then $x$ is regular

(Question 15.3 in Humphreys's book Introduction to Lie Algebras) Let $L$ be semisimple (char $\mathbb F = 0$), $x\in L$ semisimple. Prove that if $x$ lies in exactly one CSA (Cartan subalgebra), then $x$ is regular.

Definitions. An element $x$ is said regular if $C_L(x) = \{y\in L: [y,x] = 0\}$ is a maximal toral subalgebra. In particular, when $L$ is semisimple, being maximal toral is equivalent to being a CSA which is equivalent to being a minimal Engel subalgebra.
An Engel subalgebra is one of the form $L_0(\mbox{ad }x) = \{y \in L: (\mbox{ad }x)^ky =0 \mbox{ for some integer } k \}, x\in L$.

Remark. This question have been asked here Cartan Subalgebra and regular elements and here $x$ regular $\Leftrightarrow$ $x$ is in exactly one CSA. Both of them lack of an answer. Moreover, the suggestion given in the latter is in my opinion wrong, since taking "$y\in C_L(x)\setminus H$ such that $L_0(\mbox{ad }y)$ is minimal" (I'm assuming he meant minimal among all the $y\in C_L(x)\setminus H$) does not ensure that it is minimal among all Engel subalgebras. I've tried to work a little bit more with this argument as follows. Of course, if $L_0(\mbox{ad }y)$ is not minimal, it will contain a minimal Engel subalgebra, say $L_0(\mbox{ad }z)$ ($L$ has finite dimension). But I can't see how to argue that $x\in L_0(\mbox{ad } z)$ to obtain the absurd of $x$ lying both in $H$ and $L_0(\mbox{ad } z)$.

Some ideas. Suppose $x$ is not regular. On one hand, being semisimple implies that $C_L(x) = L_0(\mbox{ad } x)$. Furthermore, there exists a maximal toral $H\subseteq C_L(x),$ where $H$ is the subalgebra spanned by the semisimple elements of $L$. Since $H = C_L(H)$, it follows that $x\in H$. On the other hand, $x$ being irregular implies that $L_0(\mbox{ad }x) = C_L(x)$ is not miminal, so let us take $L_0(\mbox{ad }y)\subseteq C_L(x)$ properly and minimal. In particular, $y\in C_L(x)$ gives us that $x\in L_0(\mbox{ad }y).$  Both $H$ and $L_0(\mbox{ad }y)$ are CSA, so it remains to prove that they are not the same to obtain a contradiction. $H$ and $L_0(\mbox{ad }y)$ are too generic, I can't see a way to argue that. I'm stuck and out of ideas.

Any insight or help is very much appreciated.

答:
Say $x\in\mathfrak{a}$ is the Cartan subalgebra. Note that we can decompose $L=L_0\oplus\bigoplus_{\lambda\in\Omega}L_\lambda$ for some finite set $\Omega\subseteq\mathfrak{a}^*$ that does not contain $0$ ($L_\lambda$ are the eigenspaces of the pairwise commuting operators $ad(h), h\in\mathfrak{a}$). But, $L_0=\mathfrak{a}$ since Cartan subalgebras are self-centralizing (I think this fact is proved in Chapter 8). I claim that $\lambda(x)\neq 0$ for $\lambda\in\Omega$. Indeed, if $\lambda(x)=0$, then $ad(x)$ annihilates $L_\lambda$<s> so $L_\lambda+\mathbb{F}x$ contains an abelian subalgebra of dimension at least $2$ and contains $x$ (Take an element $y\in L_\lambda$, form $\mathbb{F}x+\mathbb{F}y$).</s> so there exists $0\neq y\in L_\lambda$ such that $ad(x)(y)=\lambda(x)y=0$. Pick $h\in\mathfrak{a}$, $z\in L_{-\lambda}$ such that $y,h,z$ is an $\mathfrak{sl}_2$ triple, i.e. $$
\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}\mapsto y,\quad\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\mapsto h,\quad\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix}\mapsto z
$$ is an isomorphism between $\mathfrak{sl}_2$ and $\mathbb{F}y+\mathbb{F}h+\mathbb{F}z$. Note that $y+z$ is semisimple since its preimage in $\mathfrak{sl}_2$ is semisimple. Moreover, $x$ commutes with $y+z$, so $\mathbb{F}x+\mathbb{F}(y+z)$ is abelian and contains semisimple elements. This contradicts the fact that $x$ is contained in a unique Cartan algebra and we deduce that $\lambda(x)\neq 0$.

This finishes the proof: $L_\lambda$ are also eigenspaces of $ad(x)$ and we have shown that $\lambda(x)$ is never $0$. Thus, the nullspace of $ad(x)$ is exactly $L_0=\mathfrak{a}.$


问:
Interior of sum of discrete set and a set with non-empty interior
I'm trying to prove the following:

Let $Z=\{z_1, \cdots, z_m\}$ and $K$ subsets of $\mathbb{R}^n$ with $\operatorname{int}(K) \not= \emptyset$, then $\operatorname{int}(Z+K)=Z+\operatorname{int}(K)$

I know it is true for just a traslation and it is easy to see the $\supset$-inclusion, but I'm having troubles proving the other one.

Any ideas?

答:
No. Here is a counterexample. Let $n=1$, $Z=\{0,1\}$, $K=[0,1]$.
$$\operatorname{int}(Z+K)=\operatorname{int}([0,1]\cup[1,2])=\operatorname{int}([0,2])=(0,2)$$
$$Z+\operatorname{int}(K)=\{0,1\}+(0,1)=(0,1)\cup(1,2)$$
So $\operatorname{int}(Z+K)\not\subseteq Z+\operatorname{int}(K)$, since $1\in\operatorname{int}(Z+K)$ but $1\notin Z+\operatorname{int}(K)$.


问:
When is the cone $C(X)$ on a space locally compact?
There are a few questions in this forum dealing with special cases of this question, for example this for $X = \mathbb Z$ and this for $X = \mathbb R$. My question is

What are necessary and sufficient conditions on $X$ which assure the local compactness of the cone $C(X) = (X \times I)/(X \times \{1\})$?

The answer may depend on the precise definition of local compactness. Here are two variants:

$X$ is locally compact if each $x \in X$ has a compact neigborhood.

$X$ is locally compact if each $x \in X$ has a neigborhood basis consting of compact sets.

Clearly 2. is stronger than 1. If "compact" is assumed to include "Hausdorff", then 1. and 2. are equivalent. The same is true for Hausdorff $X$ independent of the interpretation of "compact".

The reader is encouraged to use his favorite interpretation.

An obvious sufficient condition is this:

If $X$ is compact, then $C(X)$ is compact and thus locally compact in the sense of 1. If $X$ is compact Hausdorff, then $C(X)$ is compact Hausdorff and thus locally compact in the sense of 2.

Similarly an obvious necessary condition is this:

If $C(X)$ is locally compact, then $X$ is locally compact.

In fact, $X$ is homeomorphic to the base $X \times \{0\}$ of $C(X)$ which is closed in $C(X)$, thus locally compact.

If $X$ is locally compact, then clearly the open subspace $C(X) \setminus \{*\} \approx X \times [0,1)$ is locally compact, where $*$ is the tip of $C(X)$, i.e the common equivalence class of the points in $X \times \{1\}$.

It seems to me that a non-compact $X$ cannot have a locally compact cone. The reason is that if $C(X)$ is locally compact, then $*$ must have a compact neigborhood. I can prove a partial result (see my answer to my own question). But I am interested whether there is a more general theorem.

答:
Here is a partial answer.

Let $X$ be a normal (including Hausdorff) countably paracompact space. Then the following are eqiuvalent:

$X$ is compact.

$C(X)$ is compact.

$C(X)$ is locally compact.

This applies to all paracompact Hausdorff spaces $X$, in particular to all metrizable $X$.

The equivalence of 1. and 2. is obvious, and 2. implies 3. It remains to show that 3. implies 1. Our strategy is to embed $X$ as a closed subset of a compact neigborhood of the tip $*$ of $C(X)$. This will be done by shifting the base $X \times \{0\}$ of $C(X)$ towards $*$.

Let $U$ be an open neigborhood of $*$ in $C(X)$ with compact closure $K \subset C(X)$. If $p : X \times I \to C(X)$ denotes the quotient map, then $V = p^{-1}(U)$ is an open neigborhood of $X \times \{1\}$ in  $X \times I$. For each $x \in X$ let $f(x) = \inf\{ t \in I \mid \{x \} \times [t,1] \subset V \}$. Clearly $0 \le f(x) < 1$ because $V$ is open. Moreover $\{x \} \times (f(x),1] \subset V$. The function $f$ is upper semicontinuous: Let $f(x) < r$. Pick $t$ such that $f(x) < t < r$. Then  $\{x \} \times [t,1] \subset V$ and thus there exists an open neigborhood $W_x$ of $x$ in $X$ such that $W_x \times [t,1] \subset V$. Then $f(y) \le t < r$ for $y \in W_x$. Since $f(x) < 1$ for all $x$ and the constant function $1$ is lower semicontinuous, a theorem which was proved independently by Dowker (see "On countably paracompact spaces." Canadian Journal of Mathematics 3 (1951): 219-224 / Theorem 4) and by Katetov (see "On real-valued functions in topological spaces." Fund. Math. 38 (1951): 85-91 / Theorem 2) says that there exists a continuous $h : X \to \mathbb R$ such $f(x) < h(x) < 1$ for all $x$. Define $H : X \to C(X), H(x) = p(x,h(x))$. This is an embedding: In fact, the restriction $\bar p : X \times [0,1) \stackrel{p}{\to} C(X)$ is an embedding and $\bar h : X \to  X \times [0,1), \bar h(x) = (x,h(x))$, is an embedding. Moreover, $H(X)$ is closed in $C(X)$ and $\bar h(X) \subset V$, thus $H(X) \subset U \subset K$. We conclude that $H(X)$ is compact. Therefore $X$ is compact.

Update:

The above theorem says that a normal (including Hausdorff) countably paracompact space $X$ which is not compact cannot have a locally compact cone.

In the special case of a $\sigma$-compact locally compact Hausdorff $X$ we can give an alternative proof which does not use the above "sandwich theorem" for upper and lower semicontinuous functions.

So let $C(X)$ be locally compact,  $U$ be an open neigborhood of $*$ in $C(X)$ with compact closure $K \subset C(X)$ and $V = p^{-1}(U)$ which is an open neigborhood of $X \times \{1\}$ in $X \times I$.

We have $X = \bigcup_{n=1}^\infty K_n$ with compact $K_n \subset X$ such that $K_n \subset \operatorname{int}K_{n+1}$. There exists open $W_n \subset X$ and $t_n \in (0,1)$ such that $K_n \times \{1\} \subset W_n \times (t_n,1] \subset V$. W.l.o.g. we may assume that the sequence $(t_n)$ is non-decreasing. Note that $s_n = (1+t_n)/2$ is contained in $(t_n,1)$. Let $B_n = \operatorname{bd} K_n$ which is compact (but possibly empty; in that case $K_n$ is clopen). The sets $C_n = K_n \setminus \operatorname{int}K_{n-1}$ are compact and contain the disjoint set $B_n$ and $B_{n-1}$ (formally we set $K_0 = \emptyset$). We inductively construct continuous $f_n : C_n \to I$ as follows: For $n=1$ let $f_1(x) = s_2$. Given $f_1,\ldots, f_n$ such that $f_i(x) = s_i$ for $x \in B_{i-1}$, $f_i(x) = s_{i+1}$ for $x \in B_i$ and $f_i(x) \in [s_i,s_{i+1}]$ for all $x \in C_i$ we use the Urysohn theorem to find $f_{n+1} : C_{n+1} \to I$ such that $f_{n+1}(x) = s_{n+1}$ for $x \in B_n$, $f_{n+1}(x) = s_{n+2}$ for $x \in B_{n+1}$ and $f_{n+1}(x) \in [s_{n+1},s_{n+2}]$ for all $x \in C_{n+1}$. The collection of all these $f_n$, $n \in \mathbb N$, can be pasted to a continuous $f : X \to I$ having the property that $(x,f(x)) \in V \setminus X \times \{1\}$. In fact, for $x \in C_n$ we have $f(x) = f_n(x) \in [s_n,s_{n+1}] \subset (t_n,1)$ and thus $(x,f(x)) \in C_n \times (t_n,1) \subset W_n \setminus X \times \{1\} \subset V \setminus X \times \{1\}$. By construction $X' = \{(x,f(x)) \mid x \in X \}$ is a closed subset of $C(X)$ which is homeomorphic to $X$ and, being a closed subset of $K$, compact.


问:
Identity with Bernoulli numbers: $\sum\limits_{k=1}^{n}k^p=\frac{1}{p+1}\sum\limits_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j}$
How I can prove that
$$\sum_{k=1}^{n}k^p=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j},$$
where $B_j$ is the $j$th Bernoulli number?

I hope to find the answer. Thanks for help.

答:
This is a consequence of the use of the Bernoulli polynomials. We can define them by $$B_0(x)=1$$ $$B'_{n}(x)=nB_{n-1}(x)$$and $$\int_0^1 B(x)dx=0$$

They have the particular property that $$B_{p+1}(x+1)-B_{p+1}(x)=(p+1)x^{p}$$ which is why we use them to evaluate such a sum, also $B_{n}(0)=B_{n}$, the $n$-th Bernoulli number.

You can find any $B_n(x)$ with the above, and thus evaluate any sum $$\sum_{k=1}^nk^p$$

Give it a try! Set $x=k$ and sum away, to obtain that $$\sum\limits_{k = 1}^n {{k^p} = } \frac{{{B_{p + 1}}\left( {n + 1} \right) - {B_{p + 1}}\left( 0 \right)}}{{p + 1}}$$

In fact, to get your sum to look like in your question you will have to show that $$B_n(x)=\sum_{j=0}^n \binom{n}{j}B_jx^{n-j}$$ where $B_j:=B_j(0)$. Induction should be enough.

ADD This is Faulhaber's Formula.


问:
Assume the distribution is normal. Use the area of the normal curve to answer the question. Round to the nearest whole percent.
The average size of the fish in a lake is 11.4 inches, with a standard deviation of 3.2 inches. Find the probability of catching a fish longer than 17 inches.

I posted another question on standard deviation earlier, but I'm unable to find clear results on how to solve this problem specifically either.

I know the equation for standard deviation but the part I cannot find any information online is how to solve "longer than 17 inches". I keep finding similar question that will ask to find the probability in between two numbers but I can't find much else.

答:
You have $X \sim\mathsf{Norm}(\mu=11.4,\sigma=3.2).$ You seek $P(X > 17) - 1 - P(X \le 17) = 0.0401,$ to four places.

Here is the exact numerical answer from R, where pnorm is a normal CDF.[If you're using a statistical calculator you should be able to use it to get the same answer.]

```<code>1 - pnorm(17, 11.4, 3.2)
[1] 0.04005916
</code>```

Because $17 - 11.4 = 5.6$ and $5.6/3.2 = 1.75$ you're asking for the probability of a fish with length more than 1.75 standard deviations above the mean. By the Empirical Rule the probability in a normal distribution more than $2\sigma$ above the mean is only about 2.5%.
So, if you're familiar with this rule, you should not be surprised that the numerical answer is so small.

If you are supposed to get the answer by standardizing and using a printed CDF table of the standard normal distribution, you could begin as follows:

$$P(X \le 17) = P\left(\frac{X=\mu}{\sigma}
\le \frac{17 - 11.4}{3.2}\right) = P(Z \le 1.75),$$
where $Z$ has a standard normal distribution.

I will leave it to you to figure out how to get the
answer from a printed table. Various textbooks have
different styles of normal tables, so you will have
to figure out how to use the table in your book.
You will start by finding 1.75 in the margins of the table and getting a relevant probability from the body of the table.

In the figure below, the left-hand panel shows the
problem on a scale of inches and the right-hand panel
shows the 'standardized' version of the problem compatible with printed tables. In both cases the total area under the density curve is $1,$ and the final numerical answer is the area to the right of the dotted vertical line.

R code to make the figure (in case anyone wants to see it):

```<code>par(mfrow=c(1,2))
 curve(dnorm(x, 11.4, 3.2), 0, 25, lwd=2, ylab="Density",
       xlab="Inches", main="NORM(11.4, 3.2)")
  abline(h=0, col="green2")
  abline(v=17, col="red", lwd=2, lty="dotted")
 curve(dnorm(x, 0, 1), -3.5, 3.5, lwd=2, ylab="Density",
       xlab="z", main="NORM(0, 1)")
  abline(h=0, col="green2")
  abline(v=0, col="green2")
  abline(v=1.75, col="red", lwd=2, lty="dotted")
par(mfrow=c(1,1))
</code>```

Body:
You have $X \sim\mathsf{Norm}(\mu=11.4,\sigma=3.2).$ You seek $P(X > 17) - 1 - P(X \le 17) = 0.0401,$ to four places.

Here is the exact numerical answer from R, where pnorm is a normal CDF.[If you're using a statistical calculator you should be able to use it to get the same answer.]

1 - pnorm(17, 11.4, 3.2)
[1] 0.04005916

Because $17 - 11.4 = 5.6$ and $5.6/3.2 = 1.75$ you're asking for the probability of a fish with length more than 1.75 standard deviations above the mean. By the Empirical Rule the probability in a normal distribution more than $2\sigma$ above the mean is only about 2.5%.
So, if you're familiar with this rule, you should not be surprised that the numerical answer is so small.

If you are supposed to get the answer by standardizing and using a printed CDF table of the standard normal distribution, you could begin as follows:

$$P(X \le 17) = P\left(\frac{X=\mu}{\sigma}
\le \frac{17 - 11.4}{3.2}\right) = P(Z \le 1.75),$$
where $Z$ has a standard normal distribution.

I will leave it to you to figure out how to get the
answer from a printed table. Various textbooks have
different styles of normal tables, so you will have
to figure out how to use the table in your book.
You will start by finding 1.75 in the margins of the table and getting a relevant probability from the body of the table.

In the figure below, the left-hand panel shows the
problem on a scale of inches and the right-hand panel
shows the 'standardized' version of the problem compatible with printed tables. In both cases the total area under the density curve is $1,$ and the final numerical answer is the area to the right of the dotted vertical line.

R code to make the figure (in case anyone wants to see it):

par(mfrow=c(1,2))
 curve(dnorm(x, 11.4, 3.2), 0, 25, lwd=2, ylab="Density",
       xlab="Inches", main="NORM(11.4, 3.2)")
  abline(h=0, col="green2")
  abline(v=17, col="red", lwd=2, lty="dotted")
 curve(dnorm(x, 0, 1), -3.5, 3.5, lwd=2, ylab="Density",
       xlab="z", main="NORM(0, 1)")
  abline(h=0, col="green2")
  abline(v=0, col="green2")
  abline(v=1.75, col="red", lwd=2, lty="dotted")
par(mfrow=c(1,1))


问:
${\left(\frac{\partial}{\partial x_j}\right)}_p = {\left(\frac{\partial}{\partial y_j}\right)}_p$ if $x_j = y_j$?
Let $M$ be a (real) manifold and let $(U , \varphi)$ and $(U , \psi)$ be two charts on $M$ with $\varphi = (x_1 , \ldots , x_n)$ and $\psi = (y_1 , \ldots , y_n)$. If $j \in \{1 , \ldots , n\}$, $p \in U$ and $x_j = y_j$, is ${\left(\frac{\partial}{\partial x_j}\right)}_p = {\left(\frac{\partial}{\partial y_j}\right)}_p$ true? I think it should not be necessarily true, but I do not know how to show it.

答:
No.  Partial derivative $\frac{\partial f}{\partial x_j}$ depends not only on $x_j$ but the other variables as well.

Example.

Let's do change of variables from $(x_1,x_2)$ to $(y_1,y_2)$ like this:
$$
x_1 = y_1+y_2,
\\
x_2 = y_2,
$$
or in reverse
$$
y_1 = x_1-x_2,
\\
y_2 = x_2.
$$
So if your function is $f(x_1,x_2) = x_1$, which is the same function as $f(y_1,y_2) = y_1+y_2$,
then we get
$$
\frac{\partial f}{\partial x_2} = 0,
\\
\frac{\partial f}{\partial y_2} = 1.
$$
despite the fact that $x_2 = y_2$.


答:
We can be a bit clever instead. Notice that

$$\ln(2x-x^2) = \ln(1-1+2x-x^2) = \ln(1-(x-1)^2)$$

$\ln(1-x)$ has a known Taylor series

$$-\sum_{n=1}^\infty \frac{x^n}{n}$$

which means the Taylor series of $\ln(1-(x-1)^2)$ is

$$-\sum_{n=1}^\infty \frac{(x-1)^{2n}}{n}$$

Comparing terms to Taylor's theorem will give you that the answer is b.

Body:
What is $100$th derivative of $y=\ln(2x-x^2)$ at $x=1$?

$a)2\times99!$

$b)-2\times99!$

$c)2\times101!$

$d)-2\times101!$

I tried to find a pattern by calculating first derivatives of $y$:

$$y'=\cfrac{2-2x}{2x-x^2} ,\quad y'(1)=0$$

$$y''=\cfrac{-2(2x-x^2)-(2-2x)^2}{(2x-x^2)^2},\quad y''(1)=-2$$

And $y'''$is going to get really ugly. so I couldn't find any pattern.


问:
Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$

For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$

This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.

Is my reasoning fully acceptable throughout my solution?

答:
If $\arg(f(z))=\pi /2$, that means that $\Re(f(z))=0$ and as well $0 < \Im (f(z))$: that's a more "sure" way to go.

The steps in detail are
$$
\eqalign{
  & \arg \left( {{{z - 8} \over {z - 2}}} \right) = {\pi  \over 2}\quad  \Rightarrow \quad
 \left\{ \matrix{
  {\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right) = 0 \hfill \cr
  0 < {\mathop{\Im}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right) \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ \matrix{
  z = x + i\,y \hfill \cr
  {{z - 8} \over {z - 2}}\quad \left| {\;z \ne 2} \right.\quad  =  \hfill \cr
   = {{x - 8 + i\,y} \over {x - 2 + i\,y}}
 = {{\left( {x - 8 + i\,y} \right)\left( {x - 2 - i\,y} \right)}
 \over {\left( {x - 2} \right)^{\,2}  + \,y^{\,2} }} =  \hfill \cr
   = {{\left( {\left( {x - 8} \right)\left( {x - 2} \right) + \,y^{\,2} } \right) + i\,y\left( {x - 2 - x + 8} \right)}
 \over {\left( {x - 2} \right)^{\,2}  + \,y^{\,2} }}\quad \left| \matrix{
  \;x \ne 2 \hfill \cr
  \;y \ne 0 \hfill \cr}  \right. \hfill \cr
  0 = {\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right)
 = \left( {x - 8} \right)\left( {x - 2} \right) + \,y^{\,2} \quad \left| \matrix{
  \;x \ne 2 \hfill \cr
  \;y \ne 0 \hfill \cr}  \right. \hfill \cr
  0 < {\mathop{\Im}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right)
 = 6y\quad \left| \matrix{
  \;x \ne 2 \hfill \cr
  \;y \ne 0 \hfill \cr}  \right. \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ \matrix{
  z = x + i\,y \hfill \cr
  0 = \left( {x - 5 - 3} \right)\left( {x - 5 + 3} \right) + \,y^{\,2}  =  \hfill \cr
  \left( {x - 5} \right)^{\,2}  + \,y^{\,2}  - 9\quad \left| {\;0 < y} \right. \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ \matrix{
  z = x + i\,y \hfill \cr
  (x,y) \in circle(center\,(5,0),\,radius\,3)\; \wedge \;0 < y \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left\{ \matrix{
  \left| {z - 5} \right| = 3 \hfill \cr
  0 < {\mathop{\Im}\nolimits} z = {\mathop{\Im}\nolimits} \left( {z - 5} \right) \hfill \cr}  \right. \cr}
$$

The circle is the same as you found, but (specially in the complex field) you must be careful in justifying every step that you do


答:
It depends what you mean by "is."

There is certainly a sense in which we can "embed" all of mathematics inside $\mathsf{ZFC}$. Consequently, if we imagine a person who for whatever reason is only ever willing to talk about actual formal proofs from $\mathsf{ZFC}$, they would still be able to prove not-explicitly-set-theoretic theorems by proving the appropriate set-theoretic translation.

However, $\mathsf{ZFC}$ - and more generally the framework of set theory - is not unique in this sense. Personally I find it the most natural framework, but this is absolutely a subjective position and there are good arguments against it; the one I find most compelling is the "non-structural" nature of $\mathsf{ZFC}$, mentioned in J.G.'s answer, namely that when we implement a piece of non-set-theoretic mathematics in $\mathsf{ZFC}$ we wind up with "junk theorems" in addition to the theorems which are translations of meaningful results in the original context.

So is mathematics set theory? Well, it can be if you want it to - but that says more about your choice of style than about mathematics.

Title: Is all of mathematics set theory in disguise?

Body: Is all of mathematics, just set theory in disguise? We know the ZFC axioms allow us to encode mathematical objects as sets of some kind. So, even theorems about, say, Hilbert spaces can be written down, tediously, as formulas of ZFC. So, does this mean all of mathematics is set theory in disguise?</p>


问:
Equation to track the change in a definite integral
Given the following equations

$$
f(x) = \sqrt{x}\\
\int^x_0f(x)dx
$$

Would it be possible to calculate an equation to calculate the change in the definite integral over x? For example, the change between f(0) and f(2) is 0.667 and 1.219, $f(1) - f(0)$ and $f(2) - f(1)$ respectively. Is there an equation to determine these numbers without utilizing the definite integral for each number from $\int^x_{x-1}f(x)dx$

答: Well, the rate at which the area under the curve changes is simply the function itself.  See this</p>


问:
How to deal with polynomial quotient rings
The question is quite general and looks to explore the properties of quotient rings of the form $$\mathbb{Z}_{m}[X] / (f(x)) \quad \text{and} \quad \mathbb{R}[X]/(f(x))$$
where $m \in \mathbb{N}$

Classic examples of how one can treat such rings is to  find relationships like $$\mathbb{Z}[x]/(1-x,p) \cong \mathbb{Z_{p}}$$ for prime $p$.

However, I often struggle to intuitively understand what the elements of such rings are, and they to compute using them.

For example, what do the elements of the set $\mathbb{Z}_{2}[X] / (x^4+1)$ actually look like. Can this set be described in a clearer way to help understand the way the rings work.

Is there some general way of describing elements of such rings so that isomorphisms and computations become easier to handle?

Apologies if this question is not sufficiently clear.

答:
Often you can use the universal properties (of quotients and polynomial algebras) in order to simplify the rings. For example,

$\mathbb{F}_2[x]/(x^4+1) = \mathbb{F}_2[x]/((x+1)^4) \cong \mathbb{F}_2[y]/(y^4)$

and this cannot be simplified anymore. You just "cut" the polynomials in $y$ above $y^4$. Of course $y^5$ is still there, but it equals zero. And $1+y$ is a unit, since $1+y+y^2+y^3$ is inverse to it (more generally, unit + nilpotent = unit, this is the geometric series).

Often the Chinese Remainder Theorem helps to simplify the rings. It shows that if $f \in K[x]$ decomposes as $f_1^{k_1} \cdot \dotsc \cdot f_n^{k_n}$ with irreducible $f_i$, then $K[x]/(f) \cong \prod_{i=1}^{n} K[x]/(f_i^{k_i})$.

For example, $\mathbb{R}[x]/(x^2-1) = \mathbb{R}[x]/((x+1)(x-1)) \cong \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x-1) \cong \mathbb{R} \times \mathbb{R}$. Another interesting example is $\mathbb{Q}[x]/(x^n-1) \cong \prod_{d|n} \mathbb{Q}(\zeta_d)$.

Since the CRT holds for arbitrary rings, we can also apply it to $\mathbb{Z}/m[x]$. But the problem is that this ring is not a PID anymore (when $m$ is not a prime), so that we have no nice factorizations and even when they exist the corresponding ideals are not coprime and therefore CRT doesn't apply. For example, I don't think that we can really simplify $\mathbb{Z}/4[x]/(2x)$. The elements have the form $a_0+a_1 x+ a_2 x^2 + \dotsc$ with integers $a_i$, where we can calculate $a_0 \bmod 4$ and $a_i \bmod 2$ for $i>0$.


问:
Laplace Transform of $f(t)=c$: Evaluating the limit of a product of functions, when one function has a complex number?
When evaluating the Laplace Transform of $f(t)=c$, assuming $s$ is some complex number $s=\sigma+i\omega$

$F(s)=\frac cs(1-\lim_{T\to \infty}e^{-\sigma T}(\cos\omega T-i\sin\omega T)$

I can see that $e^{-\sigma T}$ will approach zero as long as $\sigma>0$ but $\cos\omega T-i\sin\omega T$ doesn't approach anything.

I know that the limit of $f(x)\cdot g(x)$ is the limit of $f(x)$ multiplied by the limit of $g(x)$, provided that both functions have a limit.

But in this case, only one of them has a limit (equal to zero), while the other is undefined.

My textbook says that $\lim_{T\to \infty}e^{-\sigma T}(\cos\omega T-i\sin\omega T)=0$  provided that $\sigma >0$

I don't have experience evaluating limits like this, what am I missing?

答:
Since $|e^{xi}| = 1$ for all $x \in \mathbb R$ and $|e^a| = e^a$ for all $a \in \mathbb R$,
$$|e^{-\sigma T}e^{-\omega T i}| = |e^{-\sigma T}||e^{-\omega T i}| = e^{-\sigma T}$$ and this shows that the distance from $e^{-\sigma T}e^{-\omega T i}$ to the origin can be make arbitrarily small taking $T$ sufficiently large.


问:
Absolutely continuous measures
We have two measures with

$\mu_F(]-\infty,x])=F(x)$

$\mu_G(]-\infty,x])=G(x)$

$G(x)=F(x)^2$

Prove that $\mu_G$ is absolutely continuous with respect to $\mu_F$.

So we have to prove that $\mu_F(A)=0\Rightarrow\mu_G(A)=0$. I was able to prove this when A is an interval. But how can I argue that it is true for A that are not intervals as well?

答: Let $\mu_F(E)=0$. By the Approximation Theorem of Measure Theory (See Theorem D, p. 56 of Measure Theory by Halmos) we can find a finite disjoint union on half-closed intervals, say $H=\bigcup_{k=1}^{n} (a_k,b_k]$ such that $\mu (E\Delta H) <\epsilon$ where $\mu =\mu_F+\mu_G$. Now $$\mu_G(E)$$ $$ \leq \mu_G(H)+\mu_G(E \setminus H)$$ $$ < \sum\limits_{k=1}^{n} (G(b_k)-G(a_k))+\epsilon.$$ But $G(b)-G(a) \leq 2(F(b)-F(a))$ so we get $\mu_G(E) \leq 2\mu_F(H)+\epsilon <\mu_F(E)+2\epsilon=2\epsilon$ Since $\epsilon$ is arbitary we are done.

Title: Absolutely continuous measures


问:
Quadratic Equation with "0" coefficients
Let's say I have two objects $x$ and $y$ whose position at time $t$ is given by:
$$
x = a_xt^2+b_xt+c_x \\
y= a_yt^2+b_yt+c_y
$$

And I want to find which (if any) values of $t$ cause $x$ to equal $y$. That is
$$
(a_x-a_y)t^2+(b_x-b_y)t+(c_x-c_y) = 0
$$

This can easily be solved with quadratic equation. But what about the case when $a_x = a_y$? Now obviously this makes the problem much simpler to solve by hand! But if I were writing, lets say a simulation, where $a_x$ didn't necessarily equal $a_y$, but it could, how would I go about solving the equation? Preferably I'm looking for some sort of algorithm/equation that I can use to avoid having to write separate logic for the cases where $a_x = a_y$ and $a_x \neq a_y$.

Quick background: I have knowledge of math up to Linear Algebra (I feel a very simple answer lies here, but I can't quite work it out) and ~1/2 a course in Differential Equations (though I think that doesn't really apply here), but I'm more than willing to learn something new if it provides an easy way of solving my problem. Thanks!

答:
You can write a simulation to handle both cases if you resort to numerical solutions as opposed to analytical.

1). You try using Newton-Raphson method, $x_{n+1} = x_{n}-f(x_{n})/f'(x_{n}) $. This method does not care about what the order of f is, and will solve an approximate solution regardless.

2). you do a brute force way of putting values for t, until you get zero, or as close to zero as you can at a reasonable resolution and time. (not recommended)


问:
$A\rightarrow B$ and $\Pi x:A.b(x)\in B$
I have a question about the following two formulas (where both $A,B:\textsf{Prop}$ and $x$ is not a free variable in $B$):
$$A\rightarrow B\quad\quad\quad\Pi x:A.b(x)\in B$$
where $b(x)\in B$ is a membership type (equivalent to $\textsf{Id}(b(x),b(x),B)$ using the notation of identity type) indicating that $b(x)$ is an inhabitant of $B$. It appears to me that they are equivalent. From $\Pi x:A.b(x)\in B$, by the elimination rule of $\in$, it is easy to deduce $\Pi x:A.B$, which is in fact equivalent to $A\rightarrow B$. So we have the following:
$$(\Pi x:A.b(x)\in B)\rightarrow(A\rightarrow B)$$
But what about the other direction? Thanks in advance!

答:
Let me expand a bit on Dan Doel's point. In Martin-Löf type theory, to make a judgement that $\operatorname{Id}(b(x),b(x),B)$ is a type, you need to have already judged that $b(x)$ is of type $B$. In particular, for $$\prod_{x:A}\operatorname{Id}(b(x),b(x),B)$$ to be a type, you need to know that we can deduce $b(x):B$ from $x:A$. This means you must assume that $\lambda x.b(x)$ is a term of type $A\rightarrow B$ for your question to even make sense. If we do assume this – ie, if we assume that we have already judged $\lambda x.b(x):(A\rightarrow B)$ – then we can solve your problem, but it's perhaps a slightly unsatisfying solution:

In Martin-Löf type theory, we have the "reflexive" constructor for identity types, which I will denote $\operatorname{refl}$. The introduction rule for $\operatorname{refl}$ is that, if $C$ is a type and we have judged $c:C$, then we can judge $\operatorname{refl}_c:\operatorname{Id}(c,c,C)$. If you have this construction in your type theory, then $\prod_{x:A}\operatorname{Id}(b(x),b(x),B)$ in fact always holds, with a proof given by the term $\lambda x.\operatorname{refl}_{b(x)}$. Since we know $b(x):B$ for any $x:A$, it is clear that this will be well-typed. If you wish, we can then construct a proof of $$(A\rightarrow B)\rightarrow\prod_{x:A}\operatorname{Id}(b(x),b(x),B),$$ simply by considering the term $\lambda f.\lambda x.\operatorname{refl}_{b(x)}$.  Conversely, we can construct a proof of  $$\prod_{x:A}\operatorname{Id}(b(x),b(x),B)\rightarrow (A\rightarrow B)$$ by the term $\lambda g.\lambda x.b(x)$; since we already know $\lambda x.b(x):(A\rightarrow B)$, this will be well-typed.

If this is the answer you're looking for, great! However, there's not much "moral" content in it, as the terms we construct don't ever use their first arguments. This is why I'm not sure if we're understanding your question correctly.


问:
Solving an equation with least residues and roots modulo $n$
I am facing the problem of proving $z^2 \equiv 1\pmod 8$ for all odd $z\in\mathbb{N}$.

What I know is that for each $z \in \mathbb{N}$, $z = m^2-1$, for $m\in\mathbb{N}$.

What I also know is that in a least residue system there are the following solutions for all $z\in\mathbb{N}$.

The solutions are:
$$n \equiv 1,3,5,7$$

Which lead to solutions of the initial problem:
$$n = 8 \cdot k + 1\\
n = 8 \cdot k + 3 \\
n = 8 \cdot k + 5 \\
n = 8 \cdot k + 7$$
where $k\in\mathbb{N}$.

My question now is how I can transfer my knowledge about the solutions for all $n\in\mathbb{N}$ to the key problem where $n$ must be an odd number and the proofs needs to be done for all odd $n$.

I also know that $8 = 2^3$ which leads to $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod 8$$ $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod {2^3}$$ but I do not know what this tells me right now.

答:
If $z$ is odd, then $z-1$ and $z+1$ are two consecutive even numbers;

both of them are divisible by $2$, and one of them is divisible by $4$.

Therefore, $(z+1)(z-1)=z^2-1$ is divisible by $8$; i.e., $z^2\equiv1\pmod8$.


问:
Show that Polar of a set is convex
Polar of a set is defined as:

$C^0 = \{ y\in \mathbb{R}^n | y^Tx \leq 1, \forall x \in C\}$

Now from wikipedia, the intuitive idea of polar of a cone is easily understandable. But how can it be shown that $C^0$ is convex, even when $C$ is not?
 Any hint or explanation is highly appreciated. Thanks.

答:
The half-space proof by daw is quick and elegant; here is also a direct proof:

Let $\alpha\in\left]0,1\right[$, let $x\in C$, and let $y_1,y_2\in C^0$. Then by definition of $C^0$, $y_1^\top x\leq 1$ and $y_2^\top x\leq 1$. Since multiplication by positive numbers preserves inequality, we know $\alpha y_1^\top x\leq \alpha$ and $(1-\alpha)y_2^\top x\leq (1-\alpha)$. Adding both inequalities reveals
$$\alpha y_1^\top x+(1-\alpha)y_2^\top x=\big(\alpha y_1+(1-\alpha)y_2\big)^\top x\leq \alpha+(1-\alpha)=1,$$
so $\alpha y_1+(1-\alpha)y_2\in C^0$ which completes the proof.

Title: Show that Polar of a set is convex


问:
Probability and the Symmetric Group
Let $p_{n,k}$ be the probability that a random permutation from $S_n$, the symmetric group of order $n!$, has exactly $k$ fixed points.  I am trying to compute $\lim _{n\to \infty}p_{n,k}$.  After playing with it a bit, I am relatively confident the limit is $\frac{1}{k!}$, although I have yet to come up with a proof.

Here's what I've tried so far.  If $a_{n,k}$ is the number of elements of $S_n$ with $k$ fixed points, then we have that

$$
a_{n,k}=\binom{n}{k}a_{n-k,0}
$$

That is, there are $\binom{n}{k}$ ways to choose $k$ fixed points, and after those are chosen, there are $a_{n-k,0}$ to permute the remaining $n-k$ elements with $0$ fixed points.  After computing $a_{n,0}$ by hand for small values of $n$, I was able to look up the recursion formula $a_n=na_{n-1}+(-1)^n$ from oeis.org.  Using this, I was able to come up with my "guess" of $\frac{1}{k!}$, but this direction doesn't seem to be leading me towards finding a proof.

Any ideas how to proceed?  Group theory is not my strong point, so I would not be surprised if there is a result that makes this problem almost trivial.  A pointer to any such results would be excellent.

答:
The number of fixed points is roughly Poisson, so the correct probability is $e^{-1}/k!$. You should be able to derive this using an appropriate inclusion-exclusion formula.

First go over all $k$-sets and sum the probabilities of having fixed point there - in the limit this part contributes $1/k!$. Every $(k+1)$-set was counted $k+1$ times instead of $0$, so remove $k+1$ times the contribution of all $(k+1)$-sets - this part contributes $(k+1) \cdot -1/(k+1)! = -1/k!$ in the limit. Every $(k+2)$-set is now counted $\binom{k+2}{2} - (k+2)\cdot (k+1) = -\binom{k+2}{2}$ times instead of $0$, so add $\binom{k+2}{2}$ times the contribution of all $(k+2)$-sets - this part contributes $\binom{k+2}{2} \cdot 1/(k+2)! = 1/2k!$ in the limit. Continuing the pattern, we get
$$ \frac{1}{k!} \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \cdots \right) = \frac{e^{-1}}{k!}. $$
I'll let you work out the details.

Title: Probability and the Symmetric Group


问:
Proving convergence of a sequence for all integers
Having difficulty understanding how to approach the following sequence.

Let $f$ be differentiable on $\mathbb {R}$ with $a := \sup\{ |f'(x)| : x\in \mathbb {R}\} <1$. Select $s_0 \in \mathbb {R}$ and let
$s_1 = f(s_0)$, $s_2 = f(s_1)$ , ..., and, in general,
$s_n = f(s_{n-1})$ for $n = 1, 2, \ldots $.

Prove that $(s_n)_{n\in \mathbb {N}}$ is a convergent sequence.

答: Hint: use the fact that $|f(x)-f(y)|=|f'(c)(x-y)|\le a|x-y|$ to prove that your sequence is Cauchy.</p>


问:
Can two elements of a set be equal but not identical
I have some confusion regarding the equality produced by ordering of sets. My question is this: is it possible for a set to contain two elements which (when compared with some ordering relation) are equal, and nonetheless are distinct elements of the set.

An example of this might be a set of polynomials $\{x^2, x+2, x^3\}$ where they are ordered by evaluating them at $x=2$. Clearly the first two polynomials are "equal" on this ordering, but not the same element. Is this simply a forbidden ordering?

答:
$\newcommand{\p}{\preccurlyeq}$Definitions: A partial order relation $\p$ on a set $S$ mandates the following properties:

Reflexivity: $s \p s$ for all $s \in S$

Antisymmetry: Whenever $s \p t$ and $t \p s$, it holds that $t=s$

Transitivity: If $s \p t$ and $t \p u$, then $s \p u$

A poset $(S,\p)$ becomes a total (or linear) order with the following property:

Comparability/Connexity: For every $s,t \in S$, either $s \p t$ or $t \p s$

Let's formalize your order relation a bit more clearly. Let $S := \{x^2,x+2,x^3\}$. Define $\p$ on $S$ by

$$f \p g \iff f(2) \le g(2)$$

where $\le$ is the usual "less than or equal to" order relation on $\Bbb R$.

In this case, antisymmetry does not hold. This is because while

\begin{align*}
x+2 \p x^2 &\text{ because } 4 = x+2 \bigg|_{x=2} \le x^2 \bigg|_{x=2} = 4\\
x^2 \p x+2 &\text{ because } 4 = x^2 \bigg|_{x=2} \le x+2 \bigg|_{x=2} = 4\\
\end{align*}

it is obvious that $x^2 \ne x+2$ (at least in $\Bbb R[x]$, the space of polynomials over $\Bbb R$) despite $x^2 \p x+2$ and $x+2 \p x^2$ both holding.

Thus, we wouldn't even say $\p$ is an order relation on $S$.

Suppose for the sake of contradiction that there exist distinct $a,b$ in $S$, a set with partial order $\preccurlyeq$. Suppose, further, $a \approx b$, where $\approx$ means that $a \p b$ and $b \p a$ under $\p$. If $\p$ is indeed an order relation, however, antisymmetry requires $a=b$ in the usual sense of equality, i.e. they are identical elements.


答:
Here is perhaps a better approach. Note $\arg\left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ implies

$$\frac{z-8}{z-2}=re^{i \frac{\pi}{2}}\implies
 \frac{z-8}{z-2}= - \frac{\bar z-8}{\bar z-2}\implies |z|^2 -5(z+\bar z )+16=0
$$
which is $|z-5|^2=9$, or $|z-5|=3$, i.e. a circle of center $(5,0)$ and radius $3$.

Note that $z=\frac{8-i 2r }{1-i r} =\frac{(8+2r^2)+i 6r}{1+r^2}$, indicating a positive imaginary and the upper half circle.

Title: Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$

Body:
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$

This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.

Is my reasoning fully acceptable throughout my solution?


问:
Is all of mathematics set theory in disguise?
Is all of mathematics, just set theory in disguise? We know the ZFC axioms allow us to encode mathematical objects as sets of some kind. So, even theorems about, say, Hilbert spaces can be written down, tediously, as formulas of ZFC. So, does this mean all of mathematics is set theory in disguise?</p>

答: I would say that it's the other way round: most mathematical topics can be disguised in set theory, but their true nature is only revealed when the disguise is cast off. For a simple example, the set theoretic view of the permutation groups $S_n$ as a set of functions each function being represented as a set of pairs doesn't help much in elementary group theory.


问:
Any metric space is isometric to a subspace of some complete metric space
Let $S$ be any nonvoid set and let $B(S)$ denote the set of all complex valued functions $f$ on $S$ such that $$|| f||_u = \sup \lbrace |f(x)|: x\in S \rbrace < \infty.$$

The metric if $B(S)$ is $||f-g||_u$ for $f,g\in B(S)$. Now let $S$ be any metric space with metric $\rho$. Let $p\in S$ be fixed. For $s\in S$ define $f_s$ on $S$ by $$f_s(x) = (\rho(x,s)-\rho(x,p)).$$ Prove that me mapping $s\rightarrow f_s$ is an isometry of $S$ into $B(S)$; i.e. $$||f_s - f_t||_u = \rho(s,t)$$ for all $s,t\in S$.

I can obtain in a more or less straightforward manner that $||f_s-f_t||_u\leq \rho(s,t)$. My questions is, is it valid to say that, since $t$ is given, I can set $\tilde{p}=t$ and $$\rho(s,t) \leq \sup |\rho(x,s)-\rho(x,t)|?$$ (It seems to me that everything can be reduced to a quadrilateral in $\mathbb{R}^2$ with a "moving vertex" $x$.)

答: Note that $\rho(s,t)$ is the same as $|\rho(x,s)-\rho(x,t)|$ when $x=s$, so $\rho(s,t)$ is an element of $\{|\rho(x,s)-\rho(x,t)| : x \in S\}$ and then $$\rho(s,t) \leq \sup\{|\rho(x,s)-\rho(x,t)| : x \in S\} = \|f_s-f_t\|_u.$$


问:
Continuity of vector space operations in a normed space
Here's problem 4 immediately following section 2.3 in Erwine Kryszeg's book, Introductory Functional Analysis With Applications:

Show that in a normed space $X$, vector addition and scalar multiplication are continuous operations with respect to the norm; that is, the mappings defined by $(x,y) \mapsto x+y$ and $(\alpha,x) \mapsto \alpha x$ are continuous.

Now the map $(x,y) \mapsto x+y$ is a map from $X \times X$ to $X$; so we can consider $X\times X$ under the norm defined as follows:
$$|| (x,y) ||_{X\times X} \colon= ||x||_X + ||y||_X$$
for all $(x,y) \in X \times X$. With this norm, we can easily prove the vector addition map to be continuous.

But what about the norm on $K \times X$ for the continuity of the scalar multiplication map? Here $K$ (either $\mathbb{R}$ or $\mathbb{C}$) denotes the field of scalars.

答:
There are many forms of product norm. See, for example, P153 of Folland "Real Analysis". Instead of the additive form of product norm $\|(x,y)\|=\|x\|+\|y\|$, here I use max form $\|(x,y)\|=\max(\|x\|,\|y\|)$ to prove continuity of scale multiplication $(\alpha,x)\mapsto \alpha x$. Note that the max form is equivalent to additive form, so they induce the same metric topology. Specifically, $\|(\alpha,x)\|_{K\times X}=\max(|\alpha|,\|x\|)$. From the definition of max form of product norm, it is easy to see $|\alpha|,\|x\|\le\max(|\alpha|,\|x\|)=\|(\alpha,x)\|_{K\times X}$. Denote the induced metric in $K\times X$ as $d$.

Now fix $\alpha_0$ and $x_0$. If $x_0=0$, $(\alpha,x)\mapsto \alpha x$ is a constant function, which is trivially continuous. In the following, we assume $x_0\ne0$.

Let
$$\begin{equation*}
A=\begin{cases}
\alpha_0+1 &\text{if $\alpha_0\ge0$,}\\
-(\alpha_0-1) &\text{otherwise.}
\end{cases}
\end{equation*}
$$
For any $\epsilon>0$, let $$\delta=\min(1,\frac{\epsilon}{2\|x_0\|},A,\frac{\epsilon}{2A}).$$
Then as long as $d\bigl((\alpha,x),(\alpha_0,x_0)\bigr)<\delta$, we have $|\alpha x-\alpha_0x_0|<\epsilon$, establishing the continuity of scale multiplication. This is because, when it is the case,
$$\eqalign{d\bigl((\alpha,x),(\alpha_0,x_0)\bigr)
&=\|(\alpha,x)-(\alpha_0,x_0)\|_{K\times X}\cr
&=\|(\alpha-\alpha_0,x-x_0)\|_{K\times X}\cr
&=\max(|\alpha-\alpha_0|,\|x-x_0\|)\cr
&<\delta,}$$
so $|\alpha-\alpha_0|<1$ which forces $|\alpha|<A$. Likewise, we have $|\alpha-\alpha_0|<\frac{\epsilon}{2\|x_0\|}$ and $\|x-x_0\|<\frac{\epsilon}{2A}$. As a result,
$$\eqalign{\|\alpha x - \alpha_0 x_0\|
  & = \|\alpha x - \alpha x_0 + \alpha x_0 - \alpha_0 x_0\|\cr
  &\le\|\alpha x - \alpha x_0\| + \|\alpha x_0 - \alpha_0 x_0\|\cr
  & = |\alpha|\|x-x_0\| + |\alpha-\alpha_0|\|x_0\|\cr
  &<A\frac{\epsilon}{2A} + \frac{\epsilon}{2\|x_0\|}\|x_0\|\cr
  & = \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.}$$


问:
Can such a matrix be singular?
Let $A$ be an $n\times n$ matrix that satisfies

i. All diagonal entries of $A$ are positive, even integers,

ii. All non-diagonal entries of $A$ are positive, odd integers,

iii. $A$ is symmetric: $A_{ij}=A_{ji}$.

iv. If $i\neq j$, then $A_{ij}<A_{ii}$ and $A_{ij}<A_{jj}$.

Question: can such $A$ be singular (i.e. $|A|=0$)?

I can prove the answer is NO, when $n$ is even, and without using condition iv: reduce $A$ mod $2$ we get a matrix with zero diagonal and all other entries $1$; then we can compute directly that the determinant of such matrix is odd, in particular not zero. When $n$ is odd this argument says $|A|$ is even, I don't know if it can be zero, and it seems condition iv will be important here.

I encountered it when I was trying to solve a question in this post.

答:
I found:
$$
\begin{pmatrix}
15 & 0 \\
 0 & 15 \\
11 & 11
\end{pmatrix}
\cdot
\begin{pmatrix}
14 &  1 & 11 \\
 1 & 14 & 11 \\
\end{pmatrix}
=
\begin{pmatrix}
210 &  15 & 165 \\
 15 & 210 & 165 \\
165 & 165 & 242
\end{pmatrix}
$$
The general idea is to start with a product of rank-$(n-1)$ matrices and to get the configuration of odd and even numbers. This looks like the following, assuming that $1$ represents odd and $0$ represents even:
$$
\begin{pmatrix}
1 & 0 & \cdots & 0 & 0 \\
0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & 0 \\
0 & 0 & \cdots & 0 & 1 \\
1 & 1 & \cdots & 1 & 1 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
0 & 1 & \cdots & 1 & 1 & 1 \\
1 & 0 & \cdots & 1 & 1 & 1\\
\vdots & \vdots & \ddots & \vdots & \vdots  & \vdots \\
1 & 1 & \cdots & 0 & 1 & 1\\
1 & 1 & \cdots & 1 & 0 & 1
\end{pmatrix}
$$
Then we observe that the row vectors of the first matrix should point approximately in the same direction as the column vectors of the second matrix, all the vectors should have approximately the same length, and two row vectors of the first matrix (two column vectors of the second matrix) should enclose an angle that is as close as possible to the right angle - in order to keep the non-diagonal elements as small as possible.

In the end, we have to ensure that the resulting matrix is symmetric.

So we can use the following

$$
\begin{pmatrix}
a+n-2 & 0 & \cdots & 0 & 0 \\
0 & a+n-2 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & a+n-2 & 0 \\
0 & 0 & \cdots & 0 & a+n-2 \\
b & b & \cdots & b & b \\
\end{pmatrix}
\cdot
\begin{pmatrix}
a & 1 & \cdots & 1 & 1 & b \\
1 & a & \cdots & 1 & 1 & b\\
\vdots & \vdots & \ddots & \vdots & \vdots  & \vdots \\
1 & 1 & \cdots & a & 1 & b\\
1 & 1 & \cdots & 1 & a & b
\end{pmatrix}
$$
and adjust $a$ and $b$ with an even $a$ and an odd $b$ such that all conditions are fulfilled.

Title: Can such a matrix be singular?


问:
$n$-skeleton in the definition of CW complexes
I try to understand the abstract definition of CW complexes in Hatcher's Algebraic Topology.

Specifically, I refer to the following definition in the appendix of Hatcher's book.

A CW complex is a pace $X$ constructed in the following way:

(1) Start with a discrete set $X^0$, the $0$-cells of $X$.

(2) Inductively, form the $n$-skeleton $X^n$ from $X^{n-1}$ by attaching $n$-cells $e_\alpha^n$ via maps $\varphi_\alpha:S^{n-1}\to X^{n-1}$. This means that $X^n$ is the quotient space $X^{n-1}\sqcup_\alpha D_\alpha^n$ under the identification $x\sim \varphi_\alpha(x)$ for $x\in\partial D_\alpha^n$. The cell $e_\alpha^n$ is the homeomorphic image of $D_\alpha \setminus \partial D_\alpha^n$ under the quotient map.

(3) $X=\cup_n X^n$ with the weak topology.

I do not understand the inductive step (2) above:

What are the maps $\varphi_\alpha$ in the definition? Are they from certain assumptions?

What are the $\alpha$'s?

Could one give an example showing what exactly is the quotient space in the definition?

答:
I'll focus on stripping the induction down to its basics (for examples, there's lots in Hatcher).

To start the induction, the discrete topological space $X^0$ is given.

Next, the set of attaching maps $S^0 \to X^0$ is given, one for each 1-cell. Using those attaching maps, the 1-skeleton $X^1$ is constructed as a quotient. Formally an index set is also given, which I'll denote $I_1$, and the attaching maps are denoted $\phi_{\alpha} : S^0 \to X^0$ for each $\alpha \in I_1$.

Next, the set of attaching maps $S^1 \to X^1$ is given, one for each 2-cell. Using those attaching maps, the 2-skeleton $X^2$ is constructed as a quotient. Formally an index set is also given, which I'll denote $I_2$, and the attaching maps are denoted $\phi_{\alpha} : S^1 \to X^1$ for each $\alpha \in I_2$.

Continuing by induction assuming $X^n$ has been constructed.

Next, the next the set of attaching maps $S^n \to X^n$ is given, one for each $n+1$ cell. Using those attaching maps, the $n+1$-skeleton $X^{n+1}$ is constructed as a quotient. Formally an index set is also given, which I'll denote $I_{n+1}$, and the attaching maps are denoted $\phi_{\alpha} : S^n \to X^n$ for each $\alpha \in I_{n+1}$.

and so on ...


问:
Having Trouble Understanding How One Set Maps On To Another
This is almost verbatim from my textbook :

"Our  previous  examples  of  vector spaces, $R^n$ and  $R^\infty$, are special cases of the vector space $R^S$ (where S is any non-empty set) because a list of length $n$ of numbers in $R$ can be thought of as a function from $\{1,2,...,n\}$ to $R$, and a sequence of numbers in $R$ can be thought of as a function from the set of positive integers to $R$."

Something I do understand :

Ok, let's take the vector space $R^{[0,1]}$ which would be the set of all Real-Valued functions $f$ such that $f : [0,1] \rightarrow R$.  So, for example, $f(x)=5x \in R^{[0,1]}$.  That I get.

But, how can "a list of length $n$ of numbers in $R$ can be thought of as a function from $\{1,2,...,n\}$ to $R$"?

Right, so, if we're in $R^2$ then we've got $f:\{1,2\} \rightarrow R$.  But how is some ordered pair in $R^2$, say (7, -3) related to the set $\{1,2\}$ and what exactly is the thing in $R$ that is being mapped to?

答: The ordered pair $x = (x_1,x_2) \in \mathbb R^2$ can be identified with the function $x : \{1,2\} \to \mathbb R$ such that $x(1) = x_1$ and $x(2) = x_2$. Also, when you write "$f(x)=5x \in \mathbb R^{[0,1]}$" this is wrong. If $f : [0,1] \to \mathbb R$ is given by the rule $f(x)=5x$, then $f$ is an element of $\mathbb R^{[0,1]}$.


答:
A list of $n$ real numbers is something like $$x_1, x_2, \dots, x_n$$

We can view this as a function $f:\{1,2,\dots,n\}\to\mathbb{R}$ defined by $$f(1)=x_1\\f(2)=x_2\\\vdots\\f(n)=x_n$$

Title: Having Trouble Understanding How One Set Maps On To Another

Body:
This is almost verbatim from my textbook :

"Our  previous  examples  of  vector spaces, $R^n$ and  $R^\infty$, are special cases of the vector space $R^S$ (where S is any non-empty set) because a list of length $n$ of numbers in $R$ can be thought of as a function from $\{1,2,...,n\}$ to $R$, and a sequence of numbers in $R$ can be thought of as a function from the set of positive integers to $R$."

Something I do understand :

Ok, let's take the vector space $R^{[0,1]}$ which would be the set of all Real-Valued functions $f$ such that $f : [0,1] \rightarrow R$.  So, for example, $f(x)=5x \in R^{[0,1]}$.  That I get.

But, how can "a list of length $n$ of numbers in $R$ can be thought of as a function from $\{1,2,...,n\}$ to $R$"?

Right, so, if we're in $R^2$ then we've got $f:\{1,2\} \rightarrow R$.  But how is some ordered pair in $R^2$, say (7, -3) related to the set $\{1,2\}$ and what exactly is the thing in $R$ that is being mapped to?


问:
Complex Fourier series coefficients of a periodic function

I have tried integrating to find a relationship between $C_n^{(m)}$ and $C_n^{(m-1)}$, but I don't know if I'm right, as the integral is very confusing.

Also, even if I am wrong on the relationship, I attempted section b) and I did not know how to approach the change of variable.

Any help would be amazing. Thank you.

答:
Observe that

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

Now use (or rederive, using integration by parts) the fact that

$$\hat{f''}(n)=in\hat{f'}(n)=-n^2\hat{f}(n).$$

Take the Fourier-series transform of the first equation, specifically the left and right sides, to get:

$$-n^2c^{(m)}_n=(2m)(2m-1)c^{(m-1)}_n.$$


问:
How to calculate rotation of arrow to point along a vector in 3D?
I have a 3D vector Starting at some point A and ending at some point B. I have an arrow sitting at point A. I need to rotate the arrow such that it is pointing along the vector from point A to point B. The arrow starts pointing along the y axis (take a look at the image to see this better). I'm really not much of a mathematician and I'm struggling with this. Any help would be much appreciated.

答:
So you need three things:

normal vector using cross product

angle between vectors

a matrix that rotates the arrow

So the first step is simply cross product.
2nd step is angle between vectors using
$$ a=\frac{acos(a_1 \cdot a_2)}{len(a_1)*len(a_2)} $$, where $len(a_0) = \sqrt{a_x^2+a_y^2+a_z^2}$

3rd step is a matrix:
\begin{bmatrix}
cos(a)+(1-cos(a))*x^2 & (1-cos(a))*x*y-sin(a)*z & (1-cos(a))*x*z+sin(a)*y & 0 \\
(1-cos(a))*y*x+sin(a)*z & cos(a)+(1-cos(a))*y^2 & (1-cos(a))*y*z-sin(a)*x & 0 \\
(1-cos(a))*z*x-sin(a)*y & (1-cos(a))*z*y+sin(a)*x & cos(a)+(1-cos(a))*z^2 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
There $(x,y,z) = normalize(cross(a_1,a_2))$.

Note that you can only apply the matrix when the vectors start from origo, so you need translate the system from point A to origo using standard translate matrix. and then after applying the matrix, translate it back to point A.


答:
The functional analysis notation $Y^X$ means the space of all functions $f:X\to Y$.  This is a vector space in a natural way when $Y$ is a vector space.

So here we are just employing that notation.  That is, each element of $\Bbb R^n$, say, is an $n$-tuple $(x_1,x_2,\dots,x_n)$.  And that in turn corresponds to a unique function $f:\{1,2,\dots,n\}\to\Bbb R$, given simply by $f(i)=x_i$.

The extension to infinite sequences is straight forward.  See this.


问:
Can we efficiently modify the largest singular value of a matrix without computing the full SVD?
Say we've got $A\in\mathbb{R}^{N\times N}$, and $A=U\textrm{diag}(\sigma_1,\sigma_2,\ldots,\sigma_N) V^\top$ is its singular value decomposition. For a function $f\colon\mathbb{R}_{+}\to\mathbb{R}_{+}$, I am curious if there is a method for efficiently evaluating

$$\hat{A}=U\textrm{diag}(f(\sigma_1),\sigma_2,\ldots,\sigma_n)V^\top,$$
without having to compute the full SVD of $A$ (which can be quite expensive).

My motivation in thinking this might be possible is because for eigenvalues, one can compute the largest eigenvalue-eigenvector pair much faster (numerically speaking) than performing a full eigendecomposition. However, I feel like despite this, the answer is probably still no?

答:
I have not tried implementing this, but it is at least in theory possible to find the top singular vectors (left and right) by power method (just google "SVD power method" and such), and some others -- see here to start and follow the links.

Of course, once you know the top singular vectors and the top singular value it solves the problem in your question.

Indeed, $A=UDV^T$ means precisely that $A=\sum_i \sigma_i u_iv_i^T$ ($A$ takes the unit vector $v_i$ to $\sigma_i u_i$). To replace $\sigma_1$ by $f(\sigma_1)$ just add $(f(\sigma_1)-\sigma_1 )u_1v_1^T$ to $A$.


问:
Limit of a cube root
Given $\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1}-\alpha x-\beta)=0$ for some $\alpha, \beta \in \mathbb{R}$. I have tried equating it as $\lim_{x \to +\infty}\alpha x+\beta=\lim_{x \to +\infty} (\sqrt[3]{x^3+x^2+x+1})$. Cubing both sides would give$$(\alpha x)^3+\beta^3+3\alpha^2x^2\beta+3\beta^2\alpha x=x^3+x^2+x+1$$
Hence,
$$\alpha^3=1; \beta^3=1;3\beta^2\alpha=1;3\alpha^2\beta=1$$
However, I am unsure if I am doing it right.

答:
For large $x$, it is guaranteed that $\left|\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right| < 1$, and so we may use a Binomial expansion :

$$ \left(1 + \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right)\right)^\frac13 = 1 + \frac13 \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}\right) + O\left( \frac{1}{x^2} \right) = 1 + \frac{1}{3x} + O\left( \frac{1}{x^2} \right).$$

Therefore,

$$\sqrt[3]{x^3+x^2+x+1} = x \ \sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\alpha x-\beta$$

$$ = x\left(1 + \frac{1}{3x} + O\left( \frac{1}{x^2} \right)\right) -\alpha x  -\beta = x(1 - \alpha) + (\frac{1}{3}  - \beta) + O\left( \frac{1}{x} \right) ,$$

which tends to $0$ as $x \to \infty \iff \alpha = 1$ and $\beta = \frac13.$

$$$$

Addendum:

To answer your question for general $f(x), g(x)$ (assuming they both $\to \infty$ as $x \to \infty$): if $f(x)$ and $g(x)$ are very close for large values of $x$, such that $ f(x)-g(x) \to 0$, then it is true that, for all $ \varepsilon > 0,$ no matter how small, $\exists N$ such that $|f(x) - g(x)| < \varepsilon$ for all $x > N$.

However, this does not imply that $\left|f^3(x) - g^3(x)\right| <  \varepsilon.$ In fact,

$$\left|f^3(x) - g^3(x)\right| = \left|f(x) - g(x)\right|\left(f^2(x) + f(x)g(x) + g^2(x)\right)$$

$$ = \left|f(x) - g(x))\right|\left((f(x) - g(x))^2 + 3f(x)g(x)\right) < \varepsilon\left(\varepsilon^2 + 3f(x)g(x)\right),$$

but $3f(x)g(x)$ is not necessarily bound by $\varepsilon.$ If, for example, $3f(x)g(x) > \frac{1}{\varepsilon},$ then there may be a significant/noticable difference between $f^3(x)$ and $g^3(x)$.

Using numbers as an example, consider $10^6$ and $10^6 + \frac{1}{10^3}.$ You might think that their cubes are very close, but in fact they differ by a whooping $ \approx 3 \times 10^9$ (!)

Title: Limit of a cube root


问:
find recursive solution $T(n)=2T(n/2)+n-1$
I want to solve this:

$$T(n) = 2 T\left(\frac{n}{2}\right) + n - 1 $$

I try :

\begin{align*}
       n &= 2^m \\
       T(2^m) &= 2T(2^{m-1}) + 2^m -1 \\
       2 ^ m &= B \\
       T(B) &= 2T(B-1) + B -1 && (1) \\
       r - 2 &= 0  \implies r = 2  \implies  T (B) = C*2^B && (h)
\end{align*}

particular :
$$
       T(B) = AB + D \quad  \text{(A and D are const)}
$$
1:

\begin{align*}
       AB + DB &= 2(AB + DB) + B -1 \\
       A &= -1 \text{ and } D = -1 \\
       T(B) &= c \cdot 2^B -1 -B
\end{align*}
and I can't continue solving currectly
please help me

答:
As I said in my comment, you have an error fairly early on, so I am not going to address what all went wrong in your proof.

Here is how I would do it: Let $ k = \log_2 n $ and $ g(k) := 2^k $. Hence $$ T(g(k)) = 2T(g(k - 1)) + 2^k - 1 $$ Let $ f:= T \circ g $. Substituting this in: $$ f(k) = 2f(k - 1) + 2^k - 1 $$ Next, let, $ h(x) := 2^{-x}f(x) $. Hence, $$ h(k) = h(k-1) + 1 - 2^{-k} $$ From here, it is easy to determine a function for the case where the domain of $ h $ is the natural numbers: $$ h(k) = h(0) + \sum_{n = 1}^k 1 - 2^{-n} = h(0) + k + 1 - 2^{-k}$$ Because $ h(x) = 2^{-x} \left(T \circ g\right)(x) $, $ T(x) = 2^x h\left(g^{-1}(x)\right) = 2^{\log_2 x} \left(h(0) + \log_2 x + 1 - \frac{1}{x}\right)$, which reduces to $$ T(x) = Cx + x + x \cdot \log_2 x - 1 $$ where $ C = T(1) $.

The hard part is proving that this is the unique definition across the real numbers. While it is easy to verify that it works on such a domain, that doesn't necessarily imply uniquness.


问:
What is the maximum value of $|V|$
If $G=(V,E)$ is a connected graph with $|E|=17$ and $deg(v)>2$ for all vertices of graph $G$, what is the maximum value for $|V|=v$?

$deg(v)\geq 3$ for all $v\in V$, $\sum_{v\in V} deg(v)=2|E|=2\cdot 17=34$, thus $34\geq 3v$ and from there $v \leq \frac{34}{3} \approx 11.33$, $v\leq 11$. $v_{\text{max}}=11$.

Those were my thoughts, but it wasn't enough, apparently I only gave an upper bound.

Am I missing something?

答:
I think your reasoning is correct and one can show that there exists a connected graph with $11$ vertices and $17$ edges:

Note that I constructed this graph by using your arguments actually, since degree sum was $34$ and $v_{max} = 11$, I consider $10$ vertices of degree $3$ and $1$ vertex with degree $4$. Easiest way to construct such a graph is to consider an $11$-cycle first, then first make a vertex with degree $4$ (this also makes two of the vertices degree $3$ so we are left with $8$ vertices of degree $2$), then try to pair vertices so that they all have degree $3$.

Title: What is the maximum value of $|V|$