问:
Random variable transformation in $ \mathbb{R}^{2}$
Let $X, Y$ be random variables with joint probability density function
$$
f_{X,Y}\,\left(x,y\right) = \frac{1}{2}\,\mathrm{e}^{-x}\,\,
\mathbb{\large I}_{\,\left[\,{x\ >\ \left\vert\,{y}\,\right\vert}\,\right]}\,\,\,,
\quad (x,y) \in \mathbb{R}^{2}
$$

Now we define $U =X$ and $V=X- |Y|$ and we would like to find joint probability density of $U,V$ and marginal distribution of $V$.

We want to apply the theorem about transformation of random variables and since there is $|Y|$, we need to separate it into two subsets.

Let $ S = \{ x>0, \ x> |y| \}$ and $G_{1} = S \cap [y>0]$, $G_{2} = S \cap [ y<0]$. Then for $$ \varphi_{1} \begin{cases} u = x \\ v=x-y \end{cases} \text{ and } \varphi_{2} \begin{cases} u = x \\ v=x+y \end{cases} $$ we that the absolute value of their Jacobian matrix determinant is one and $$ \varphi_{1} (G_{1}) = \varphi_{2} (G_{2}) = (0, \infty )^{2}$$

Also $$ \mathbb{I}_{[ x > |y| ]} = \mathbb{I}_{[u>|u-v|]} = \mathbb{I}_{[u> |v-u|]}, $$ hence $$f_{U,V} (u,v) = \frac{1}{2} e^{-u} \mathbb{I}_{[u> |v-u|]} \mathbb{I}_{[u>0]} \mathbb{I}_{[v>0]} \cdot 2 =  e^{-u} \mathbb{I}_{[2u>v>0]} $$

However, by intergrating $f_{U,V}$ over $ \mathbb{R}^{2}$ we get $2$, so there must be $ \frac{1}{2}$ lost somewhere. I would appreciate some help at this point.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{f}_{UV}\,\pars{u,v} & =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{1 \over 2}
\expo{-x}\bracks{x > \verts{y}}
\delta\pars{u - x}\ \times
\\ &
\phantom{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\quad}
\delta\pars{v - x + \verts{y}}\dd x\,\dd y
\\[5mm] & =
{1 \over 2}\expo{-u}
\int_{-\infty}^{\infty}\bracks{u > \verts{y}}
\delta\pars{v - u + \verts{y}}\dd y
\\[5mm] & =
{1 \over 2}\expo{-u}\bracks{u > 0}
\int_{-u}^{u}\delta\pars{v - u + \verts{y}}\dd y
\\[5mm] & =
\bracks{u > 0}\expo{-u}
\int_{0}^{u}\delta\pars{v - u + y}\dd y =
\\[5mm] & =
\bracks{u > 0}\expo{-u}
\int_{0}^{u}\delta\pars{y - \bracks{u - v}}\dd y
\\[5mm] & =
\bracks{u > 0}\expo{-u}\bracks{0 < u - v  v > 0}\expo{-u}} \\ &
\end{align}

Note that
$\ds{\iint_{\mathbb{R}^{2}}\on{f}_{UV}\,\pars{u,v}
\,\dd u\,\dd v = \color{red}{1}}$.

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{f}_{UV}\,\pars{u,v} & =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{1 \over 2}
\expo{-x}\bracks{x > \verts{y}}
\delta\pars{u - x}\ \times
\\ &
\phantom{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\quad}
\delta\pars{v - x + \verts{y}}\dd x\,\dd y
\\[5mm] & =
{1 \over 2}\expo{-u}
\int_{-\infty}^{\infty}\bracks{u > \verts{y}}
\delta\pars{v - u + \verts{y}}\dd y
\\[5mm] & =
{1 \over 2}\expo{-u}\bracks{u > 0}
\int_{-u}^{u}\delta\pars{v - u + \verts{y}}\dd y
\\[5mm] & =
\bracks{u > 0}\expo{-u}
\int_{0}^{u}\delta\pars{v - u + y}\dd y =
\\[5mm] & =
\bracks{u > 0}\expo{-u}
\int_{0}^{u}\delta\pars{y - \bracks{u - v}}\dd y
\\[5mm] & =
\bracks{u > 0}\expo{-u}\bracks{0 < u - v < u}
\\[5mm] & =
\bbx{\bracks{u > v > 0}\expo{-u}} \\ &
\end{align}

Note that
$\ds{\iint_{\mathbb{R}^{2}}\on{f}_{UV}\,\pars{u,v}
\,\dd u\,\dd v = \color{red}{1}}$.


问:
Specific counterexample for the non-Markovianness of the elephant random walk
Consider the Elephant Random Walk $(S_n)_{n\in\mathbb{N}}$ defined by
$S_n:=X_1+\ldots+ X_n, n\in\mathbb{N},$
whose increments $X_k:=S_k-S_{k-1}$, $k\ge 1$, are recursively defined as follows:

The distribution of $X_1$ is given by $P(X_1=+1)=q\in(0,1)$ and $ P(X_1=-1)=1-q$.

At any subsequent time $n\ge 2$, one draws randomly an integer time index $k\in\{1,\ldots, n-1\}$ with uniform probability and lets $X_n:=X_k$ with probability $p$ and $X_n:=-X_k$ with probability $1-p$.

The process was introduced in https://arxiv.org/pdf/cond-mat/0406593.pdf and was also referenced in an answer to this question: Example of a stochastic non Markov process?.

Does anyone have a specific counterexample of a trajectory of the process showing the non-Markovianness of the process? So far I've tried comparing $P(S_3=1|S_2=0)$, $P(S_4=2|S_3=1)$, $P(S_4=0|S_3=1)$, $P(S_4=0|S_3=-1)$ with the conditional probabilities of the respective possible complete trajectories and couldn't find any counterexample. I am grateful for any further ideas, especially since I don't exclude the possibility of making computation mistakes.

答:
As far as I can tell, though the elephant random walk has a very "non-Markovian description", it is actually a Markov chain - though not a time-homogeneous one, and many people talking about Markov chains do assume homogeneity. That is,
$$
   \Pr[S_n = s_n \mid S_{n-1} = s_{n-1}, S_{n-2} = s_{n-2}, \dots, S_1 = s_1] = \Pr[S_n = s_n \mid S_{n-1} = s_{n-1}]
$$
for every possible trajectory $(s_1, s_2, \dots, s_n)$. However, it's possible that for $m \ne n$, $$\Pr[S_n = x \mid S_{n-1} = y] \ne \Pr[S_m = x \mid S_{m-1} = y].$$

Here's my logic. If we want to compute $\Pr[S_{n+1} = s+1 \mid S_n = s]$ (and similarly $\Pr[S_{n+1} = s-1 \mid S_n = s]$, all we have to do is realize that in order to get to $S_n = s$ in $n$ steps, $\frac{n+s}{2}$ of the steps must have been $+1$ and $\frac{n-s}{2}$ of the steps must have been $-1$. That means when we choose a random $k \in \{1,2,\dots,n\}$, we have a $\frac{n+s}{2n}$ chance of picking a $k$ with $X_k = 1$ and a $\frac{n-s}{2n}$ chance of picking a $k$ with $X_k = -1$. Overall, there is a $$p \cdot \frac{n+s}{2n} + (1-p) \cdot \frac{n-s}{2n}$$ chance of ending up choosing $X_{n+1}=1$, and therefore getting $S_{n+1} = s+1$.

Conditioning on any other history of the Markov chain is irrelevant: it can tell us which steps were $+1$ and which were $-1$, but we already know how many of each there are. So the Markov property will in fact always hold.

However, the formula above depends on $n$, and not only on $s$. If we get to $s$ at the earliest possible time $n=|s|$, we must have taken steps that all went in the same direction, and so we have a $p$ chance of continuing in that direction. If we get to $s$ at a much later time, then $\frac{n+s}{2}$ and $\frac{n-s}{2}$ will be close to each other, and the probability of going in either direction is close to $\frac12$.

So there is no fixed probability of going from $s$ to $s+1$ (or from $s$ to $s-1$), which is what we'd want if the Markov chain were time-homogeneous.

ContentLicense: CC BY-SA 4.0


问:
Proper action in "Metric spaces of non-positive curvature"
(This question is related to Proper action and compactness)

In (I.8.2) of "Metric spaces of non-positive curvature", properly action is defined as: for each $x$, there exists a $r$ such that $\{\gamma | \gamma B(x,r) \cap B(x,r)\}$ is finite. (I think this does not implies that
$\{\gamma | \gamma B(x,r) \cap B(x,r)\}$ is finite for any $r$).

Then I find a problem about the proper action on Page 233, which is highlighted:

From the proof of proposition 6.10 (3) on page 233,

we know that for all $n$ sufficently large  $\gamma_n B(x, r+a+1) \cap B(x, r)\ne \emptyset$. Then the original proof say this contradicts to the properness of the action (see I.8.3 (1)).
But the definition of proper action is I.8.2 as above, and I.8.3(1) just explain the proper action in (I.8.2) is more restrictive than the another proper action, which is defined as: for every compact subset $K$, the set $\{\gamma | \gamma . K \cap K\}$ is finite.
Why I.8.3(1) implies a contradiction?

(In fact in Proper action and compactness
, after the answer by Seirios, there are comments by GGT, which ask the same question)

答:
I think they simply forgot to assume properness of $X$ or they forgot their definition of a proper action. Or they wrote the proof using one definition of a proper action, later change their definition but forgot to correct the proposition. The right thing to do is to write to Martin Bridson and ask him to add an erratum for this proposition.

Here is a counterexample to their claim of discreteness of the set of translation lengths (proposition 6.10(3)). Let $H$ be an infinite dimensional separable Hilbert space with an orthonormal basis $({\mathbf e}_n), n\in {\mathbb N}$. Take the sequence $a_n=1-\frac{1}{2n}$ and consider the translations
$$
T_n: {\mathbf x}\mapsto  {\mathbf x} + a_n {\mathbf e}_n, {\mathbf x}\in H.
$$
Let $G$ denote the (free abelian) subgroup of isometries of $H$ generated by the translations $T_n$, $n\in {\mathbb N}$. Then the group $G$ acts properly on $H$ (in the sense of Bridson-Haefliger which, incidentally, is equivalent to the standard notion of properness for isometric actions of discrete groups): For each point ${\mathbf x}\in H$
the set
$$
\{g\in G| g B({\mathbf x}, \frac{1}{3}) \cap B({\mathbf x}, \frac{1}{3})\ne \emptyset\}
$$
is finite.

Let $X\subset H$ denote the $G$-orbit of zero, equipped with the restriction of the standard distance function on $H$. Then $G$ acts on $X$ properly, cocompactly (since the action is transitive). But the set of translation lengths of the elements of $G$ is not a discrete subset of ${\mathbb R}$ since it contains the sequence $(a_n)$.

Here is another counterexample, of combinatorial origin: Take the countably infinite bouquet of circles $Z$; assign length $a_n= 1- \frac{1}{2n}$ to the $n$th circle. This yields a $G$-invariant path-metric on the universal covering space $X$ of $Z$, where $G$ is the group of covering transformations. The group $G$ acts discretely and transitively on the vertex set of $X$. However, the set of translation lengths includes the sequence $(a_n)$, hence, is nondiscrete.


问:
Evaluating $\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$
I need help starting with this. I can't find an example like this anywhere in my book

$$\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$$

答: So you can do this. Multiply $y$ in the numerator and the denominator so that your function becomes $\frac{x^{2} e^{y^{2}} y}{y^{6}}$ and then substitute $y^{2}=t$ to integrate.</p>

ContentLicense: CC BY-SA 3.0


问:
Lengendre symbol calculation
I'm trying to calculate the lengendre symbol of (3/383) without using the Quadratic Reciprocity Law, and with not much success.
I've thought about checking if 2^191 is congroent to 1 modulo 383 but it seems too complicated.
I'd be grateful if some could point me to the solution.
Thanks in advance.</p>

答:
You want to compute
$$
\left( \frac{2}{383} \right)
$$
you're after, $p = 383$ is a prime, so you calculate indeed
$$
2^{(p-1)/2} = 2^{191} \pmod{p},
$$
and you get $1$, so $2$ is a square modulo $p$.

If you have to do the calculation by hand, use the method of repeated squaring.

Title: Lengendre symbol calculation


问:
What is a manifold on a Euclidean space?
In this semester I study differential geometry and in this chapter we want to define what is a surface. In order to do that we first define what a manifold is on a Euclidean space, not generally what is a manifold, and Euclidean space I mean $(\mathbb{R}^{n},\left \| \cdot  \right \|)$.

That been said, if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

Can we just say it's a homomorphic function from open sets to open sets ?

Older posts had been made on this topic (but I don't think they answer well  my question, in my option) and have already been answered if there isn't  anything new to add  I will delete this.

答:
A two dimensional manifold in Euclidean space can be bent, stretched, and/or cut to make a flat surface (i.e., a subset of a plane).

There are, however, caveats as to what cuts are allowed, and it's hard to cover them while remaining "simple". When you cut the object, the region around the cut has to be bendable/stretchable to a flat surface. So if you take two pieces of paper and make a T-shape out of them, that's not a manifold, because where the two pieces of paper meet can't be bent or stretched to make a single piece of paper, and we're not allowed to fix this with cutting. Cutting is allowed in situations such a cylinder, where any small region can be bent into a flat surface, but the overall surface can't be bent flat; the pieces can't be pressed flat simultaneously.

A higher dimensional manifold is harder to describe in "simple" terms; it's the same basic idea of bending, stretching, and cutting into an n-dimensional flat surface, but it's harder to visualize what those mean in higher dimensions. A 1-dimensional manifold is quite simpler, as it's simply a curve, and a 0-dimensional manifold is trivial.


问:
Category with zero morphisms implies zero object?
Let $\mathscr{A}$ be a category. Then we say that $\mathscr{A}$ is a category with zero morphisms if for every $A,A'\in\mathscr{A}$ there is a zero morphism $0_{AA'}\in\mathscr{A}(A,A')$, and the zero morphisms obey a particular commutative diagram (see wiki). Now suppose $\mathscr{A}$ has a zero object $0$. Then $\mathscr{A}$ is a category with zero morphisms, and every zero morphism factors through the zero object uniquely. So how about the converse? If $\mathscr{A}$ is a category with zero morphisms, does it necessarily have a zero object? If not, is there any simple counterexample(s)?

答:
No, a category with zero morphisms need not have a zero object. A simple counterexample is to consider a nonzero ring $R$ considered as a one-object category (even a one-object $\text{Ab}$-enriched / pre-additive category), or more generally a monoid with a zero element / absorbing element and at least one other nonzero element (but nonzero rings are nice as a common and familiar example of these).

What is true is that given a category with zero morphisms there is a unique way to adjoin a zero object to it if it doesn't already have one: it has a unique morphism to and from every other object, and every composition involving these morphisms is zero. This construction is the left adjoint of the inclusion of (categories with zero objects) into (categories with zero morphisms), where in both cases morphisms are functors that preserve zero morphisms.

Also, if a category with zero morphisms has either an initial or terminal object, that object is automatically a zero object, and a functor between two categories-with-zero-objects that preserves zero morphisms automatically preserves zero objects. I go into a bit more detail in this blog post.


问:
Plus construction of a presheave factors every sheaf-valued morphism.
I'm having some trouble understanding the correctness of some proof in Sheaves in Geometry and Logic (Mac Lane, Moerdijk). It concerns the lemma III.5.3 :

If $F$ is a sheaf and $P$ a presheaf, then any map $\phi \colon P \to F$ of
  presheaves factors uniquely through $\eta$ as $\phi = \tilde\phi \circ \eta$.

$$ \begin{matrix} P & \stackrel \eta \to & P^+ \\ & \!\!\!\!\!\!_\phi\searrow & \downarrow \small{\tilde\phi} \\ & & \!\!\!\!F\end{matrix}$$

Recall the plus construction : $P^+(C)$ is the equivalent class of matching families under the relation $(x_f)_{f\in S} \sim (y_g)_{g \in R}$ ($R,S$ covering sieves of $C$) iff there is some covering sieve $T \subseteq S \cap R$ on which $x_f = y_f$ forall $f \in T$.

Recall the morphism $\eta$ : $$\eta_C \colon P(C) \to P^+(C),\, x \to [(Pf(x))_{f \colon D \to C}].$$

I easily understand the definition of $\tilde\phi_C(\mathbf x)$ for some equivalent class $\mathbf x = (x_f)_{f\in S}$ : we push the matching family $(x_f)$ by $\phi_C$ into a matching family of $F$ ; being a sheaf, $F$ admit a unique amalgamation in $F(C)$ that we define to be $\tilde\phi_C(\mathbf x)$. The well-definition is no problem. What bother me is to check that $\tilde\phi = (\tilde\phi_C)_C$ is actually natural in C.

We want a commutative diagram
$$ \begin{matrix} P^+C & \stackrel {Ph} \to & P^+D \\
\small{\tilde\phi_C}\downarrow \ \ \ & & \ \ \ \downarrow\small{\tilde\phi_D} \\
FC & \stackrel{Fh} \to & FD \end{matrix}$$
for all $h \colon D \to C$. For an element $\mathbf x = (x_f)_{f\in S}$, and $h \in S$, the stability axiom of Grothendieck topologies assure $h^\ast S = \hom(-, D)$, and so the commutativity is immediat from the definition of $\tilde\phi_C(\mathbf x)$ as amalgamation.

But I'm stuck with the case where $h \notin S$, and (it seems to me that) it is not treated in the proof of Mac Lane. Maybe can we always find $\mathbf y = (y_g)_{g\in R}$ with $\mathbf x \sim \mathbf y$ and $h \in R$, but I can't see it.

答:
I don't know about you, but I find the whole business with equivalence classes very confusing. Looking at the question from a more category theoretical point of view should make it easier:

Let $P$ be a $\mathscr A$-valued presheaf on a site $\mathscr C$ and $X\in \mathscr C$. Then for any covering family $\mathscr U=\{U_i\rightarrow X\}$, define $P_X(\mathscr U)$ as the equalizer of the sequence
    \begin{equation}
  P_X(\mathscr U)\rightarrow \prod_i P(U_i)\rightrightarrows \prod_{i,j}P(U_i\times_XU_j).
 \end{equation}
    Further define $P^+(X)$ as the colimit
    \begin{equation}
  P^+(X)=colim_{\mathscr U}P_X(\mathscr U)
 \end{equation}
    running over all covering families $\mathscr U$ of $X$.

Then the universal property gives you immediately the morphism $\eta: P\to P^+$, and also a morphism $\tilde \phi: P^+\to F^+=F$. Being defined via the universal property, naturality should be easy to check.

To see how you get a morphism $P\to P^+$, note that the restriction map

$$P(X)\to \prod_i P(U_i)$$ induces a map $P(X)\to P_X(\mathscr U)$ via the universal property of the equalizer. Now the colimit comes with maps $P_X(\mathscr U)\to P^+(X)$, which after composition yields $P(X)\to P^+(X)$. This is independent of the choice of the covering, because for a refinement $\mathscr V=\{V_i\rightarrow X\}$ of $\mathscr U$, the diagram

$$ \begin{matrix} P(X) &  \to & P_X(\mathscr U) \\ & \searrow & \downarrow  \\ & & \!\!\!\!P_X(\mathscr V)\end{matrix}$$

commutes.


问:
Is there a closed-form or combinatorial proof for the 'multisections' of the binomial series $\sum_{k=0}^{n-1}\binom{n}{k}$?
Throughout, let $m,n\in\mathbb{N}^+$. As is well-known,
$$
\sum_{k=0}^{r-1}\binom{r}{k}=2^r-1
$$I want to understand the multisections of this sum, namely for $m=2,3,4,5\ldots, 0\le j<m$ I want to evaluate
$$
S_{m,j}(n) = \sum _{k=0}^{n-1} \binom{m n}{k m+j}
$$Numerical results suggest that $S_{m,j}(n)$ is roughly but not exactly equally distributed in $j$ for fixed $m,n$. I computed the results for $m=2,3,4,6$ (I did $5$ as well but it was much more opaque):

\begin{array}{c|cccccc}\\
m\setminus j& 0 & 1 & 2 & 3 & 4 & 5\\
\hline 2 & 2^{2 n-1}-1 & 2^{2 n-1} &  & &  &  \\
3 & \frac{2 (-1)^n}{3}+\frac{8^n}{3}-1 & \frac{1}{3} (-1)^{n+1}+\frac{8^n}{3} & \frac{1}{3} (-1)^{n+1}+\frac{8^n}{3} &  &  & \\
4 & (-1)^n 2^{2 n-1}+4^{2 n-1}-1 & 4^{2 n-1} & (-1)^{n+1} 2^{2 n-1}+4^{2 n-1} & 4^{2 n-1} & \text{} & \text{} \\
6 & \frac{1}{3} 2^{6 n-1}+(-1)^n 3^{3 n-1}-\frac{2}{3} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^n 3^{3 n-1}-\frac{1}{6} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^{n+1}
   3^{3 n-1}-\frac{1}{6} & \frac{1}{3} 2^{6 n-1}+(-1)^{n+1} 3^{3 n-1}+\frac{1}{3} & \frac{1}{3} 2^{6 n-1}+\frac{1}{2} (-1)^{n+1} 3^{3 n-1}-\frac{1}{6} & \frac{1}{3}
   2^{6 n-1}+\frac{1}{2} (-1)^n 3^{3 n-1}-\frac{1}{6} \\
\end{array}

For instance, this table says $S_{3,1}(n) = \sum _{k=0}^{n-1} \binom{3 n}{3k+1} = \frac{1}{3} (-1)^{n+1}+\frac{8^n}{3}$. Clearly we should have $\sum_{j=0}^{m-1}S_{m,j}(n)=2^{mn}-1$ but past that I'm not sure what the multisections should look like. Perhaps there is a combinatorial proof for these sums?

答:
With the adjustment mentioned in the comments (adding 1 to $S_{m,0}(n)$), we have that $S_{m,j}(n)$ is the number of ways to color $mn$ distinguishable objects black and white so that the number of black objects is congruent to $j \pmod m.$

It turns out there's no particular reason to restrict attention to the case where the number of objects is divisible by $m$, so let's talk instead about $\tilde S_{m,j}(n)$, defined to be the number of ways to color $n$ distinguishable objects black and white so that the number of black objects is congruent to $j \pmod m$. So we have $$S_{m,j}(n) = \tilde S_{m,j}(mn).$$

If we're fixing $m$ there's a nice recursive relationship between these: $$\tilde S_{m,j}(n+1) = \tilde S_{m,j}(n) + \tilde S_{m,j-1}(n).$$

That's because given a coloring of $n+1$ balls black or white with $j \pmod m$ black balls, either the last ball is white and the coloring of the first $n$ balls has $j \pmod m$ black balls, or the last ball is black and the coloring of the first $n$ balls has $j-1 \pmod m$ black balls. (Here, $j-1$ is evaluated modulo $m$, of course.)

For a fixed $m$, the set of these observations for all $j$ combines into one matrix equation:

$$
\pmatrix {\tilde S_{m,0}(n+1) \\ \tilde S_{m,1}(n+1) \\ \vdots \\ \tilde S_{m,m-1}(n+1)} =
\pmatrix {1 & 0 & 0 & \dots & 0 & 1 \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & 1 & \dots & 0 & 0 \\ \vdots \\ 0 & 0 & 0 & \dots  &1 & 1 }
\pmatrix {\tilde S_{m,0}(n) \\ \tilde S_{m,1}(n) \\ \vdots \\ \tilde S_{m,m-1}(n)}
$$

So,

$$
\pmatrix {\tilde S_{m,0}(n) \\ \tilde S_{m,1}(n) \\ \vdots \\ \tilde S_{m,m-1}(n)} =
\pmatrix {1 & 0 & 0 & \dots & 0 & 1 \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & 1 & \dots & 0 & 0 \\ \vdots \\ 0 & 0 & 0 & \dots  &1 & 1 } ^ n
\pmatrix {1 \\ 0 \\ \vdots \\ 0}
$$

So diagonalizing that matrix should give you explicit formulas for each $\tilde S_{m, j}(n)$ as a linear combination of exponential terms in $n$ where the bases of the exponent are the eigenvalues of the matrix.

Diagonalizing it is pretty easy, if $M$ is that matrix and $R$ is the rotation matrix

$$R = \pmatrix {0 & 0 & 0 & \dots & 0 & 1 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ 0 & 1 & 0 & \dots & 0 & 0 \\ \vdots \\ 0 & 0 & 0 & \dots  &1 & 0 }, $$
we have $M = I + R$ and the eigenvalues of $R$ are precisely the $m$'th roots of unity. That is, the eigenvalues of $M$ take the form $\lambda_0 = 2, \lambda_1 = 1+\zeta, \lambda_2 = 1+\zeta^2, ..., \lambda_{m-1} = 1+\zeta^{m-1}$ where $\zeta$ is a primitive $m$'th root of unity.

So, explicitly we expect for each $j$ there should be some coefficients $C_{j, k}$ for which

$$\tilde S_{m,j}(n) = C_{j,0} 2^n + C_{j, 1} (1+\zeta)^n + \dots + C_{j, m-1} (1+\zeta^{m-1})^n$$

In your case, we have

$$S_{m,j}(n) = C_{j,0} 2^{mn} + C_{j, 1} (1+\zeta)^{mn} + \dots + C_{j, m-1} (1+\zeta^{m-1})^{mn}.$$

The thing that makes things nice in the examples you're looking at ($m = 2, 3, 4, 6$) is the fact that the $m$'th power of one plus an $m$'th root of unity is nice for those things. That is for a primitive $m$'th root of unity $\zeta$ with $m$ in that set, $(1+\zeta^k)^m$ is always just an integer. This is not true for general $m$.

If you like, equivalently, the matrix $M^m$ has particularly nice eigenvalues for those values of $m$.

Uniformity

You mentioned that these are roughly uniform. That's because the eigenvector of $M$ corresponding to the largest (in absolute value) eigenvalue $2$ is the uniform vector $(1, \dots, 1)$. All the other exponential terms are relatively small.

Random walks

Another interpretation of this is that $\tilde S_{m, j}(n)$ the number of walks of length $n$ on $\mathbb Z / m \mathbb Z$ which land on $j$, where each step in the walk either leaves you where you started or moves you one step up in $\mathbb Z / m \mathbb Z$. Or, if you normalize, $\tilde S_{m, j}(n) / 2^n$ is the probability that a random walk in $\mathbb Z / m \mathbb Z$ ends up at $j$ after $n$ steps, where the steps in the random walk are chosen uniformly from those two options.

Then the normalized vector

$$ 2^{-n} \pmatrix {\tilde S_{m,0}(n) \\ \tilde S_{m,1}(n) \\ \vdots \\ \tilde S_{m,m-1}(n)} $$

just keeps track of the probability distribution of your location in $\mathbb Z / m \mathbb Z$ after $n$ steps of this random walk, and the normalized matrix $M/2$ is the transition matrix for this random walk.

This should also make it fairly clear that the distribution should approach a uniform one, since there's nothing in the random walk that tends to keep you in near one element of $\mathbb Z / m \mathbb Z$ over another.

More combinatorial approach

For the nice values of $m$ in the question, $M^m$ diagonalizes over the rational numbers. With enough effort, it should be possible to make this observation into a more combinatorial argument for the identities that you're talking about. (There's lots of things you could mean by "combinatorial", e.g. I consider the above to already be fairly combinatorial, especially in terms of the "random walk" interpretation above.) You can be explicit about the diagonalization, and write down some combinatorial interpretations of the individual terms in the diagonalization. You might need to do messy things like putting negative terms on one side of an equation and positive terms on another, and construct bijections recursively and hope that those bijections turn out to be something nice. I don't think the outcome of this process is likely to be worth the effort. It might be easier to work with $M^{2m}$ instead, since then the eigenvalues will all be positive.

The combinatorial things you put together this way can only exist and be meaningful for these special values of $m$, so I don't think they'll give that much insight over the more abstract approach. I think the more abstract approach (with a concrete interpretation in terms of random walks) is "the right way to think about it".

Edit: One exciting form of argument that might exist here: For those values $m$, it might be relevant that $\mathbb Q[\zeta]$ is a degree 2 extension of $\mathbb Q$, and that there's a corresponding rank 2 lattice of algebraic integers. Maybe the walks in $\mathbb Z / m \mathbb Z$ have some interesting combinatorial relationship to this lattice (and walks in this lattice?).


问:
Show $\int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x} = \frac{\pi z}{sin(\pi z)}$
I need to solve the following integral:
$$
I = \int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x}.
$$

Wolfram Alpha gives the answer as $ \frac{\pi z}{sin(\pi z)}$, or equivalently, $\pi z csc(\pi z)$

My ultimate goal is to demonstrate that $z!(-z)! = \frac{\pi z}{sin(\pi z)}$

So far, I arrived at this integral by gamma and beta functions:

$$
z!(-z)! = \Gamma(z+1)\Gamma(-z+1) = \Gamma(m)\Gamma(n)\\
= B(z+1,-z+1)\Gamma(m+n) = B(z+1,-z+1)\Gamma(2)\\
\\
=B(z+1,-z+1) = \int_{0}^{\infty} \frac{x^{-z+1-1}}{(1 + x)^{2}} ~ \mathrm{d}{x}\\
= \int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x}
$$

This integral is effectively the answer. Can you solve this with a contour integral? Also, is there a special name for this integral?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{x^{-z} \over \pars{1 + x}^{2}}\,\dd x} =
\int_{0}^{\infty}
x^{-z}\bracks{\sum_{k = 0}^{\infty}{-2 \choose k}x^{k}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
x^{-z}\bracks{\sum_{k = 0}^{\infty}{k + 1 \choose k}
\pars{-1}^{k}x^{k}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
x^{\pars{\color{red}{1 - z}} - 1}\bracks{\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{2 + k}}
{\pars{-x}^{k} \over k!}}\,\dd x
\\[5mm] = &\
\Gamma\pars{\color{red}{1 - z}}
\Gamma\pars{2 - \bracks{\color{red}{1 - z}}} =
\Gamma\pars{1 - z}\Gamma\pars{z}z\quad
\pars{\substack{\mbox{Ramanujan's}\\[0.5mm] \mbox{Master}
\\[1mm] \mbox{Theorem}}}
\\[5mm] = &\
\bbx{\pi z \over \sin\pars{\pi z}} \\ &
\end{align}


问:
Affinely independent Points
In my lecture notes I got this definition. But I don't see how, this last sum can be found.

The points $x_1,\ldots,x_n\in\Bbb R^d$ are said to be affinely independent if the relations $$\sum_{i=1}^n\lambda_ix_i=0,\quad\sum_{i=1}^n\lambda_i=0$$ imply that $\lambda_1=\ldots=\lambda_n=0$.  For any $x_r$, $r\in\{1,\ldots,n\}$ and $\sum_{i=1}^n\lambda_i=0$ it follows that $$0=\sum_{i=1}^n\lambda_ix_i=\sum_{i=1\\i\neq r}^n\lambda_i(x_i-x_r).$$

答:
If $\sum\limits_{i=1}^{n} \lambda_i=0$ then $\sum\limits_{i=1}^{n} \lambda_i x_r=0$

If also $\sum\limits_{i=1}^{n} \lambda_i x_i=0$ then by subtraction $\sum\limits_{i=1}^{n} \lambda_i(x_i- x_r)=0$

Clearly $\lambda_r(x_r-x_r)=0$ too, so these together imply $\sum\limits_{\substack{i=1\\i \not = r}}^{n} \lambda_i(x_i- x_r)=0$

Title: Affinely independent Points


答:
If $\sum_{i=1}^n\lambda_ix_i=0$ and $\sum_{i=1}^n\lambda_i=0$, then

$$\begin{align*}
\sum_{\substack{i=1\\i\ne r}}^n\lambda_i(x_i-x_r)&=\sum_{\substack{i=1\\i\ne r}}^n\lambda_ix_i-x_r\sum_{\substack{i=1\\i\ne r}}^n\lambda_i\\
&=\left(\sum_{i=1}^n\lambda_ix_i-\lambda_rx_r\right)-x_r\left(\sum_{i=1}^n\lambda_i-\lambda_r\right)\\
&=(0-\lambda_rx_r)-x_r(0-\lambda_r)\\
&=0\,.
\end{align*}$$

Title: Affinely independent Points

Body:
In my lecture notes I got this definition. But I don't see how, this last sum can be found.

The points $x_1,\ldots,x_n\in\Bbb R^d$ are said to be affinely independent if the relations $$\sum_{i=1}^n\lambda_ix_i=0,\quad\sum_{i=1}^n\lambda_i=0$$ imply that $\lambda_1=\ldots=\lambda_n=0$.  For any $x_r$, $r\in\{1,\ldots,n\}$ and $\sum_{i=1}^n\lambda_i=0$ it follows that $$0=\sum_{i=1}^n\lambda_ix_i=\sum_{i=1\\i\neq r}^n\lambda_i(x_i-x_r).$$


问:
How there can be $64$ triples $(x, y,z)$, made out of $4$ elements?
It is stated in the book titled: Abstract Algebra, by Pierre Antoine Grillet; on pg. 5.

$$\begin{array}{c|ccccc}
Example\, \, 1.7\, & a & b & c & d \\\hline
a     & a & b & c & b\\
b     & b & c & a & c\\
c     & c & a & b & a\\
d     & b & c & a & c
\end{array}$$

Example 1.7 is commutative. Associativity, however, is a different kettle of beans: the $4$ elements of Example 1.7 beget $64$ triples $(x, y,z)$, each with two products $(xy)z$ and $x (yz)$ to compare.

I cannot get how there can be $64$ triples $(x, y,z)$, made out of $4$ elements.

The choice of $3$ elements out of four can be done in $4$ ways:

$abc$,
 $abd$,
 $acd$,
 But, from the rest only one choice is left: $b,c,d$.

So, such choices are: $4P3=4$.

Now, there need be ordering among these $3$ elements, i.e. $xyz, xzy, yxz, yzx, zxy, zyx$, leading to $6$ elements for each choice of $3$ elements.

So, the number of combinations is: $4*6=24$.

I searched for errata, but could not find anywhere for the book. So, kindly help me in

答: If there are four choices for each of the three entries (i.e., repetitions are allowed), then there are $4^3=64$ possibilities.  (When you calculated $4\times3\times2=24$, you assumed that the three entries had to be distinct.)


问:
Why does this weird iteration converge to the square root ??
Let $1 < x < 4$ , $a_1 = x$ and $b_1 = 0$.

Now consider the (conditional) iterations

if $a_n > b_n$ then

$a_{n+1} = 4(a_n - b_n - 1)$

$b_{n+1} = 2(b_n + 2)$

else ( $a_n = b_n$ or $a_n < b_n$ )

$a_{n+1} = 4 a_n$

$b_{n+1} = 2 b_n$

Now consider

$c_n = \frac{b_n}{2^n}$

Now define

$f(x) = \lim_{n \to \infty} c_n $

Now apparently $ f(x)  = \sqrt x$

Despite being a simple limit of a conditional iteration, I do not get it.

I am even surprised limits of conditional iterations can be analytic functions with respect to the starting value.

I have never seen these type of iterations during my education nor in a book.
Not even in books about numerical methods.

Sure the algorithm converges slowly but it has benefits too and it does not use complicated functions.

Im clearly missing the big picture here.

Conditional iterations (or recursions) is a topic I know nothing about it seems.

How and why does this algorithm give the square root ?

What is the bigger picture ? How to generalize this ?

I could be wrong, but is this related to the mediant when $x$ is rational ?

答:
Claim: This process calculates the square root in binary, by calculating the next digit.

The described process is how one would calculate square roots in binary using the long division method. I'm guessing that you can find articles about the process.

Fill you get stuck, add in details of what you've been working on and why you're stuck.

Observations:

$c_n = \frac{b_n}{2^n}$ suggests we want to look at $b_n$ in binary.

$b_{n+1} = 2(b_n + 2) $ or $ 2b_n \Rightarrow c_{n+1} = c_n + \frac{1}{2^n} $ or $ c_n$. In fact, ignore $b_n$ and just work with $c_n$.

The idea is that if you know $ c_n ^2 \leq x \leq (c_n + \frac{1}{2^{n-1}})^2$, then to figure out the next digit you look at $ x - ( c_n^2 + \frac{1}{2^n} )^2$ and see if it's positive or negative.


问:
Ideal over a Banach algebra

Let $ I $ be a closed ideal of a commutative unital Banach algebra such that $\mathcal{A}/I\ncong \mathbb{C}$.

Let $[a]\in \mathcal{A}/I$ such that $[a]\notin \mathbb{C}[1]$ and let $\lambda\in \sigma ([a])$. Then, $a-\lambda1\notin I$ and $I+\mathcal{A}(a-\lambda1)$ is an ideal in $\mathcal{A}$ such that $I+\mathcal{A}(a-\lambda1)\not = \mathcal{A}.$

Therefore, $ I $ is not an maximal ideal.

I have already proven that $ a- \lambda1 \notin I $ and that $ I + \mathcal{A} (a- \lambda1) $ is an ideal. But I have not been able to prove that $ I + \mathcal{A} (a- \lambda1) \not = \mathcal{A} $, an idea is to show that $ 1 \notin I + \mathcal {A} (a- \lambda1) $. How can I do this? How can I conclude that $ I $ is not a maximal ideal? Thanks so much for reading.

答:
Assume that $I+\mathcal A(a-\lambda 1) = \mathcal A$. Then, there are $x\in I$ and $y\in A$ such that:
$$1 = x+(a-\lambda 1)y$$
Reducing this modulo $I$ yields:
$$[1] = ([a]-\lambda[1])[y]$$
Therefore, $[a]-\lambda[1]\in\mathcal A/I$ is invertible, i. e. $\lambda\in\sigma([a])$. That is a contradiction to the choice of $\lambda$.

Thus, $I+\mathcal A(a-\lambda 1) \neq \mathcal A$. Since $[a]\notin\mathbb C[1]$, $[a]-\lambda[1]\neq [0]$, i. e. $a-\lambda\notin I$.

This implies that $I$ is not a maximal ideal because of the following proper inclusions of ideals:
$$I\subsetneq I+\mathcal A(a-\lambda 1)\subsetneq\mathcal A$$

Title: Ideal over a Banach algebra


问:
Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$
Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.

My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.

Any help would be appreciated.

答:
Another method, because it wasn't mentioned, is using Lagrange's trigonometric identities (which should not be difficult to prove using complex numbers, like here)
$$\sum _{k=1}^{n}\sin(k\theta )=
{\frac {1}{2}}\cot {\frac {\theta }{2}}-{\frac {\cos \left(\left(n+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac {\theta }{2}}\right)}}=...$$
with $\theta=\frac{1}{n^2}$. Given $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$$...={\frac {1}{2}}\cot {\frac {\theta }{2}}-\frac{1}{2}\left(\cos\left(n\theta\right)\cot\frac{\theta}{2}-\sin(n\theta)\right)=\\
\frac{1}{2}\cdot\left(\sin(n\theta)+\cot\frac{\theta}{2}\cdot\left(1-\cos\left(n\theta\right)\right)\right)=\\
\frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cot\frac{1}{2n^2}\cdot\left(1-\cos\frac{1}{n}\right)\right)=...$$
considering that $1-\cos{a}=2\sin^2\frac{a}{2}$
$$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cot\frac{1}{2n^2}\cdot \sin^2\frac{1}{2n}\right)=...$$
also considering $\color{red}{\lim\limits_{x\to0}\frac{\sin x}{x}=1}$
$$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cos\frac{1}{2n^2}\cdot \frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\cdot \frac{\sin^2\frac{1}{2n}}{\frac{1}{4n^2}}\cdot\frac{\frac{1}{4n^2}}{\frac{1}{2n^2}}\right)=\\
\frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cos\frac{1}{2n^2}\cdot \color{red}{\left(\frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\right)}\cdot \color{red}{\left(\frac{\sin\frac{1}{2n}}{\frac{1}{2n}}\right)^2}\right)\to \frac{1}{2}, n\to\infty$$


问:
The irreducibility of $f(X,Y)$
This is an exercise from the book Introduction to algebra by А. И. КОСТРИКИН.

The irreducible factors of the homogeneous polynomial function
$$
f(X, Y)=a_{0} X^{n}+a_{1} X^{n-1} Y+\cdots+a_{n-1} X Y^{n-1}+a_{n} Y^{n} \in \mathbb{Q}[X, Y]
$$
are also homogeneous, and $f(X,Y)$ is irreducible if and only if $f(X,1)=a_0X^n+a_1X^{n-1}+\cdots+a_{n-1}X+a_n\in \mathbb Q[X]$ is irreducible.

My question is, how to prove these two statements? I can intuitively know the correctness of the statements, but I cannot prove them strictly.

答:
Let's look at the first claim first. Say first to set the stage that we're dealing with polynomials with coefficients in one variable. Then we would want to show that the factors of a monomial must also be monomials. If they weren't, then we'd have some factorization $f=gh$, where $f$ is a monomial but one of the factors, say $g$, is not. To show that this means that $gh$ is not a monomial, what we can do is take the largest-degree and smallest-degree terms from $g$ and $h$ and multiply them. Since $\mathbb{Q}$ is a domain, the products of these terms will be nonzero. The product of the largest-degree terms will have a power different from the other pairwise products of monomials in $gh$, as will the product of the smallest-degree terms. That means that these terms can't be cancelled out, so they are both in $f$, so $f$ is not a monomial.

When we look at $f$ in 2 variables, we kind of have to extend our definition of "largest-degree." We'll call our extended definition the extended degree of a monomial. Say that the extended degree of a larger-degree monomial is larger than that of a smaller-degree polynomial, as you would expect, but say also that given two monomials of the same degree, the one with a higher power of $X$ has a higher extended degree. Then we do the same thing as before. Consider some product $gh$. Take their terms of largest and smallest extended degrees and multiply them. Again, nothing can cancel those two terms out, so that they are both present in $f$. And if $g,h$ are not homogenous, then these terms will have different degrees so $f$ is not homogenous.

I'm not so sure about the second claim. The function $XY$ is definitely not irreducible, but the function $X$ definitely is. But I think the intention of this problem is that if we have a factorization in one, then we have a factorization in the other. If we have a factorization of $f(X,1)$, then we can definitely multiply powers of $Y$ by the various terms so that each factor is homogenous, and then multiplying by some power of $Y$ at the end we will have a factorization of $f(X,Y)$. The contrapositive of this is that if $f(X,Y)$ is irreducible then $f(X,1)$ is irreducible.

Now like I said the first direction isn't totally correct as-is. But it's right in spirit: if we have a factorization of $f(X,Y)$, then we can plug in 1 for $Y$ and we will have a factorization of $f(X,1)$. The only problem is that it's possible that some of these factors become trivial: in particular, any factor which is a power of $Y$ will simply get mapped to 1. Thus, if we add the stipulation that $f(X,Y)$ has a factorization where at least two factors are not powers of $Y$, then this is correct. In particular, if we only consider $f(X,Y)$ where $a_0\neq0$ so that $Y$ is not a factor, then $f(X,1)$ being irreducible will imply that $f(X,Y)$ is irreducible.

Title: The irreducibility of $f(X,Y)$


问:
Basis and dimension for $\mathbb{R}^2$?
For a vector space $\mathbb{R}^2$ under usual addition and usual scalar multiplication over $\mathbb{R}$

Why is that a basis for $\mathbb{R}^2$ is the set $\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix}\right\}$ and not just $\begin{bmatrix}1\\1\end{bmatrix}$?

答:
The vector $(1,1)$ is not a basis for $\Bbb R^2$ for the simple reason that not every vector in $\Bbb R^2$ can be expressed as a scalar multiple of it.

More generally: a set $\beta \subseteq V$ with elements $\beta_i$ is a basis for a vector space $V$ over a field $F$ if two conditions are met, one of which is that $\rm{span}(\beta) = V$. Or, that is, for every $v \in V$, we can choose $\beta_i$ from $\beta$ such that

$$v = \sum_{i=1}^n \alpha_i \beta_i$$

where $\alpha_i \in F$ are scalars. If you can always find scalars $\alpha_i \in F$ such that $v$ is a sum of multiples of those $\beta_i \in \beta$, in other words, then $\beta$ is a vector space for $V$.

However, in the case of $V = \Bbb R^2$ and $\beta = \left\{ (1,1) \right\}$, it is easy to see that this condition is not met. Namely, how would you write $(0,1)$ or $(1,0)$, each elements of $\Bbb R^2$, solely in terms of scalar multiples of elements of $\beta$?

In fact, as it happens: for $V$ a finite-dimensional vector space with dimension $\dim V = n$, it happens that any basis $\beta$ of $V$ satisfies $|\beta| = n = \dim V$. You need one vector in the basis to account for each of the dimensions of the vector space itself, loosely speaking. This is another means by which we see $\left\{ (1,1) \right\}$ is not a basis for $\Bbb R^2$, since it has cardinality $1$, where $\Bbb R^2$ has dimension $2$ over $\Bbb R$. (Or infinite dimensionality over the smaller subfields of $\Bbb R$ like $\Bbb Q$, but that's another matter altogether.)

To reiterate, to perhaps clear up some confusion you have in your comments: your dimension is always with respect to the field $F$ your vector space $V$ is over, usually denoted with a subscript. For instance:

$$\dim \limits_{\Bbb R} \Bbb C = 2 \text{ whereas } \dim \limits_{\Bbb C} \Bbb C = 1$$

Why is this the case? Well, this goes back to the fact that linear combinations of your spanning set take scalars from the field $F$.

If we're only allowed real scalars, then $\left\{1,i\right\}$ are a spanning set for $\Bbb C$. After all, every $z \in \Bbb C$ may be expressed as $z = x+iy$ for $x,y \in \Bbb R$. You can't have a dimension of $1$ because then your claimed spanning set would only span a line in $\Bbb C$ and not the entirety of $\Bbb C$ itself.

On the other hand, if we're allowed complex scalars, then of course $\dim_\Bbb C \Bbb C = 1$. That is because $z = z \cdot 1$ for every $z \in \Bbb C$: that is, we treat $z$ as the scalar in the case on the right-hand side. Of course, that also means the standard basis is $\left\{1\right\}$ in that case.

(And of course, more generally: $F$ is always able to be regarded as a vector space over itself for any field $F$, with basis $\left\{1\right\}$. Then $\dim_F F = 1$.)

Title: Basis and dimension for $\mathbb{R}^2$?


问:
dot product between vector and matrix
In my book on fluid mechanics there is an expression
$$
\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij}
$$
where
$\boldsymbol{\tau}_{ij}$ is a rank-2 tensor (=matrix). Given that $\boldsymbol\nabla=(\partial_x, \partial_y, \partial_z)$, a vector, what do I get when I dot it with a matrix?

If I was to write $\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij}$ in Einstein notation, then how would it look?

The tensor $\boldsymbol{\tau}_{ij}$ is given by
$$
 \begin{pmatrix}
  \tau_{xx} & \tau_{yx} & \tau_{zx} \\
  \tau_{xy} & \tau_{yy} & \tau_{zy} \\
  \tau_{xz} & \tau_{yz} & \tau_{zz}
 \end{pmatrix}
$$
and the dot-product yields (by comparison with later expressions in the chapter)

$$
\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij} = \mathbf{i}(\partial_x \tau_{xx} + \partial_y \tau_{yx} + \partial_z \tau_{zx})+\mathbf{j}(\partial_x \tau_{xy} + \partial_y \tau_{yy} + \partial_z \tau_{zy}) + \mathbf{k}(\partial_x \tau_{xz} + \partial_y \tau_{yz} + \partial_z \tau_{zz})
$$

However, I don't see how this last expression comes about.

答:
Let $n$ be any $(1,0)$ - (contravariant) tensor and $\tau$ be $(0,2)$ - (covariant) tensor. Then, $n = n^i e_i$ and $\tau = \tau_{ij}e^i \otimes e^j $ and therefore, we have
$$
\begin{align}
n\cdot\tau
&= \left<n^i e_i,~ \tau_{jk} e^j \otimes e^k \right>\\
&= n^i \tau_{jk} (e_i \cdot e_j) \otimes e^k\\
&= g_{ij} n^i \tau_{jk} e^k
\end{align}
$$
where $\left<\cdot,\cdot\right>$ denote the inner product, $e_i$ is a basis vector, $\otimes$ is a tensor product, and $g_{ij} = e_i \cdot e_j$ is a metric tensor, which is a Kronecker delta $\delta_{ij}$ in the Cartesian coordinate system. That is,
$$
\begin{align}
n\cdot\tau
&= \delta_{ij} n^i \tau_{jk} e^k\\
&= n^i \tau_{ik} e^k
\end{align}
$$
which is corresponding to your problem. You have to be familiar with tensor algebra to follow this.

Title: dot product between vector and matrix


问:
Characterization of Hilbert spaces using orthogonal decomposition
Using the closest point property of Hilbert spaces we can prove that for any closed subspace $A \subset H$ we can decompose H as
$$H = A \oplus A^{\bot}.$$

Is this property characterizing Hilbert spaces? Namely, if we have an inner product space H such that for every closed subspace $A \subset H$ we can write
$$H = A \oplus A^{\bot}$$
then space is Hilbert? I've seen some specific cases, where the space was not complete and $H \neq A \oplus A^{\bot}$ for some closed $A$; for example for $H = c_{00}$ (space of sequences with finite support) with inner product inherited from $\ell^2$ and $A = \{ x \in c_{00}: \sum_{n=1}^{\infty} \frac{x_n}{n}=0\}$ so I wonder if we can generalize example like this and always find a closed subspace $A$ of an incomplete inner product space $H$, such that $H \neq A \oplus A^{\bot}$? Or if there is a different way of approaching this question perhaps.

答:
In fact $H$ is complete iff for every closed subspace $K\subseteq H$, one has that $H=K\oplus K^\perp$.  Actually we will prove
something a bit stronger:

Theorem.  Let $H$ be a separable, infinite dimensional inner-product space such that  $K^\perp\neq \{0\}$, for every proper closed subspace $K\subseteq H$.  Then
$H$ is complete.

Proof. Find an orthonormal basis $\{e_n\}_n$ for $H$ (the existence of bases
relies only on Graham-Schmidt and hence works  on any inner-product space).

As usual we may then use $\{e_n\}_n$ to view  $H$ as  a subspace of
$\ell ^2$ containing the compactly supported sequences.

If $H$ is not complete, then $H\subsetneq \ell ^2$, so we may pick a vector $\xi \in \ell ^2\setminus H$.   It is then obvious that
$$
  K:= H\cap \{\xi \}^\perp
  $$
is a proper closed subspace of $H$, and I claim that the othogonal complement of $K$ within $H$ coincides with $\{0\}$.
In other words we claim that
$$
  K^\perp\cap H=\{0\}.
  $$

To see this, suppose by contradiction that $\eta $ is a nonzero vector in $K^\perp\cap H$.

Fix any $m$ such that $\langle e_m, \xi \rangle \not=0$, and for every $n$ consider the vector
$$
  u_n = e_n - \frac{\langle e_n, \xi \rangle }{\langle e_m, \xi \rangle }e_m.
  $$
Observing that $\langle u_n, \xi \rangle =0$, we see that $u_n\in K$, so
$$
  0 = \langle u_n, \eta \rangle  =
  \langle e_n, \eta \rangle  - \frac{\langle e_n, \xi \rangle }{\langle e_m, \xi \rangle }\langle e_m, \eta \rangle ,
  $$
whence
$$
  \langle e_n, \eta \rangle  = \frac{\langle e_n, \xi \rangle }{\langle e_m, \xi \rangle }\langle e_m, \eta \rangle  = \langle e_n, \xi \rangle \frac{\langle e_m, \eta \rangle }{\langle e_m, \xi \rangle }.
  $$
Since $m$ is fixed we see that the coefficients of $\eta $ and $\xi $ are proportional, meaning that also  $\eta $ and $\xi $
themselves are proportional, but this is a contradiction because $\eta $ lies in $K$ and $\xi $ does not.


问:
What exactly are axioms?
The title says it all. What exactly are axioms? I mean, I know that is a statement which do not need to be proven. But what are the requirements to be an axiom? For instance, may I state something that seems to be  false as an axiom? May I state an axiom that is already a conclusion of another theorem or conclusion?

Are there any books on how to build axiomatic systems and explains this "philosophic" part of mathematics? Is Russell's a nice option for this?

Thanks

EDIT

Thanks for such many responses. I am asking this because recently, I have read an "article" in which the author tries to use the non-aggression principle as an axiom, like mathematicians and scientists do, to prove some things like mathematicians do: Establish the axioms and prove conclusions. But this principle, in my view, seems not be true, since aggression does exist.

I guess this question should be redirected to some exchange on Philosophy...

答:
See Schlimm - Axioms in Mathematical Practice

This paper has a nice philosophical discussion of axioms.

Title: What exactly are axioms?


问:
Why can we use a weaker metatheory for consistency results?
I'm not entirely sure why we can get away with using weaker metatheories for consistency results.

For instance consider the following, $ZFC \not \vdash Con(ZFC) \rightarrow Con(ZFC+I)$. This is because $ZFC + I \vdash Con(ZFC)$, and then if $ZFC \vdash Con(ZFC) \rightarrow Con(ZFC + I)$, we'd get $ZFC + I \vdash Con(ZFC + I)$, contradicting Godel's incompleteness theorem. So $ZFC$ can't be a metatheory for proving $Con(ZFC) \implies Con(ZFC + I)$. But $ZFC$ is far stronger than $PA$, so if it can't prove the consistency statement, why can $PA$? For instance, if $PA \vdash Con(ZFC) \implies Con(ZFC + I)$, then obviously $ZFC + I \vdash Con(PA)$, so why doesn't the same contradiction as above occur? Is it that $ZFC + I$ doesn't necessarily prove $Con(ZFC) \rightarrow Con(ZFC + I)$ even if $PA \vdash Con(ZFC) \rightarrow Con(ZFC+I)$? Then why did it work in the contradiction above? Was it because $ZFC \subseteq ZFC + I$?

This is just a specific example, but my question in general is: it seems that we can't prove consistency results in fairly strong theories, yet I often hear people say that $PA$ or even a finite fragment of $PA$ is enough for most consistency proofs.

Another example: using $ZFC + Con(ZFC) + \neg Con(ZFC + $"Inaccessibles exist"$)$ as our metatheory we definitely can't prove $Con(ZFC) \rightarrow Con(ZFC + $"inaccessibles exist"$)$. But the latter is a fairly easy consistency proof which we say that can be done in fairly weak metatheories. $ZFC + Con(ZFC) + \neg Con(ZFC + $"inacessibles exist"$)$ seems pretty strong to me.

Overall I'm interested in how the metatheory comes into play when describing consistency statements; it seems to me that the metatheory is very much important and yet people usually don't even mention what they're using.

答:
I think this question is based on an error, namely the claim that $Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$ is considered a theorem of the "standard metatheory" of mathematics. This is not the case. In particular, when people (correctly!) say that most relative consistency theorems can be proved in $\mathsf{PA}$ or even weaker systems, there's no tension here since $Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$ is not a relative consistency theorem in the sense meant.

Specifically, here's how I would phrase the heuristic claim above:

In practice so far, whenever $T,S$ are "naturally occurring" theories containing $\mathsf{ZF}$ with $\mathsf{ZF}\vdash Con(T)\rightarrow Con(S)$, we have $\mathsf{PA}\vdash Con(T)\rightarrow Con(S)$.

(And in fact even less than $\mathsf{PA}$ is needed here; I'm not an expert here, but my understanding is that we can go down to $\mathsf{I\Sigma_1}$ without serious effort and even further if we take more care. Hajek/Pudlak's book is a good resource here if you're interested.)

Note that the clause "containing $\mathsf{ZF}$" there is crucial, since if we go below $\mathsf{ZF}$ things trivialize: e.g. we boringly have $$\mathsf{ZF}\vdash Con(\emptyset)\rightarrow Con(\mathsf{PA})$$ since $\mathsf{ZF}\vdash Con(\mathsf{PA})$ outright, but of course $$\mathsf{PA}\not\vdash Con(\emptyset)\rightarrow Con(\mathsf{PA}).$$

The point then is that $T=\mathsf{ZFC}, S=\mathsf{ZFC+I}$ does not constitute a counterexample to this heuristic since - as you observe - we don't have $\mathsf{ZF}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$ (unless $\mathsf{ZF}$ is inconsistent that is).

As to justifying stronger systems from below, Godel's second incompleteness theorem - as you observe in the OP - shows that we cannot reach $\mathsf{ZFC+I}$ from $\mathsf{ZFC}$ in the same way that we can reach $\mathsf{ZFC}$ from $\mathsf{ZF}$ (or $\mathsf{ZFC+GCH}$ from $\mathsf{ZF}$, or similar). The relevant term here is consistency strength: $\mathsf{ZFC+I}$ has strictly greater consistency strength than $\mathsf{ZFC}$ (unless the latter is inconsistent of course).


问:
Showing if two events determined by a die toss are independent or not.
I'm having trouble setting up a problem with random variables to determine if two events A and B are independent or not. The problem is as follows:

A fair die is repeatedly toss until the first even number apears. The random variable $X$ is defined as the number of toss in which the first even number appears. Define the following events: $A:=X$ is even. $B:=X$ is a multiple of 3.
Determine if A and B are independent.

This is my approach:

We know that two events A and B are independent iff $\,\mathbb{P}(A\cap{B})=\mathbb{P}(A)\mathbb{P}(B)$. I will find the probability of A, B and their intersection and then show they are not equal. Hence, proving that A and B are not independent.

Define $\Omega:=\{(E),(O,E),(O,O,E),(O,O,O,E),...\}$. Each $\omega$ in $\Omega$ is the event in which $n$ die tosses are required to get an even number. Now, our random variable $X$ can be defined as:

$$X(\omega)=|\omega|$$

The number of elements in each $\omega$, i.e., the number of die tosses. Now, we can compute the probability of A, B and $A\cap{B}$ as follows:

$$\mathbb{P}(A)=\mathbb{P}(X=2n)\,n\in\mathbb{Z}^{+}$$

Which we can write as:

$$\mathbb{P}(\{\omega\in\Omega:X(\omega)=2n\})=\mathbb{P}\{(O,E),(O,O,O,E),(O,O,O,O,O,E),...\}$$

Since these events are disjoint, we can put the probability as:

$$\mathbb{P}(\{\omega\in\Omega:X(\omega)=2n\})=\mathbb{P}\{(O,E)\}+\mathbb{P}\{(O,O,O,E)\}+\mathbb{P}\{(O,O,O,O,O,E)\}+...$$

The probability of obtaining an even or an odd face in the die is $\frac{1}{2}$. Since each toss is independent and the die is fair, each case can be seen as:

$$\mathbb{P}\{(O,E)\}+\mathbb{P}\{(O,O,O,E)\}+\mathbb{P}\{(O,O,O,O,O,E)\}+...=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{6}+...=\sum_{n=1}^{\infty}{\left(\frac{1}{2}\right)}^{2n}$$

Which can be simplified to:

$$\mathbb{P}(A)=\sum_{n=1}^{\infty}{\left(\frac{1}{4}\right)}^{n}=\frac{1}{3}$$

By a similar procedure it can be shown that:

$$\mathbb{P}(B)=\sum_{n=1}^{\infty}{\left(\frac{1}{2}\right)}^{3n}=\frac{1}{7}$$

And:

$$\mathbb{P}(A\cap{B})=\sum_{n=1}^{\infty}{\left(\frac{1}{2}\right)}^{6n}=\frac{1}{63}$$

Which shows that $\mathbb{P}(A\cap{B})\neq\mathbb{P}(A)\mathbb{P}(B)$

Then, the events are not independent.

答:
I'm having trouble setting up a problem with random variables to determine if two events A and B are independent or not.

No you are not.  You have done it.

You have basically shown that $X$ is a geometrically distributed random variable with success-rate parameter $1/2$, so:  $$\mathsf P(X{=}n)=2^{-n}\,\mathbf 1_{n\in\Bbb N^+}$$

Then you have evaluated the probability of $X$ being divisible by a natural-number using the closed form of the geometric series.
$$\begin{align}\mathsf P(X\in\{kn:n\in\Bbb N^+\}) &=\mathsf P(\bigcup_{n=1}^\infty \{X{=}kn\})\\[1ex]&=\sum_{n=1}^\infty (2^{-k})^n&(\text{provided }k\in\Bbb N^+)\\[2ex]&=\dfrac{1}{2^k-1}\\[3ex]\mathsf P(A)&=\dfrac{1}{2^2-1}\\\mathsf P(B)&=\dfrac 1{2^3-1}\\\mathsf P(A\cap B)&=\dfrac{1}{2^6-1}\end{align}$$

Thus proving that the events, $A$ and $B$, are not independent.


问:
How do you enter a variable>variable inequality constraint into a Simplex method calculator?
In 2-variable linear programming problems, constraints can take the form of either $aX+bY < C$ or $mX > nY$. Both lines graph to form linear bounds so the graphical solution applies. But in 3+ variable problems, you need to use the Simplex method, and the calculators for the Simplex method apparently don't accept constraints of the form $mX > nY$.

I tried converting them algebraically to $mX-nY > 0$, but in a 2 variable problem the Simplex calculator gives me "no solution" when the graphical method finds a solution. Is there another way to enter these sorts of constraints in Simplex calculators?

This the problem I tried:

Maximize: $Z=5x+7y$

$x+y≤30$

$2x+y≤50$

$4x+3y≥60$

$2x≥y$

$x≥0,y≥0$

The graphical method finds that the solution is $x=10, y=20, Z=190$.

I entered that into the online calculator here as

$F(x)=5X1+7X2$

$1X1+1X2≤30$

$2X1+1X2≤50$

$4X1+3X2≤60$

$2X1+-1X2≥0$

And it returns:

"Answer:

The system has no solutions or has many solutions."

答:
You changed $\ge 60$ to $\le 60$, and you also need to specify the $\ge 0$ bounds.  But even with these changes, that online calculator yields the wrong conclusion. The online calculator provides the complete steps for the big-M method. After the first step you notice that it has a basis where $x_{10}=-30$, which is not feasible. Apparently it makes an error with the ratio test. The same website has a dual simplex solver which gives the correct output.

However, I verified your solution graphically here.


答:
Axioms are nothing more than the premises of your specific problem that are justified by the meanings of their predicates. The axioms should not conflict with the other axioms you may assume. Therefore, if they are false in your axiomatic system, they cannot be added as axioms (premises) to your problem. The propositions that are conclusions of the other theorems are theorems, but not axioms.

Refer to 2011 - Barker-Plummer - Language, Proof, and Logic.

Title: What exactly are axioms?

Body:
The title says it all. What exactly are axioms? I mean, I know that is a statement which do not need to be proven. But what are the requirements to be an axiom? For instance, may I state something that seems to be  false as an axiom? May I state an axiom that is already a conclusion of another theorem or conclusion?

Are there any books on how to build axiomatic systems and explains this "philosophic" part of mathematics? Is Russell's a nice option for this?

Thanks

EDIT

Thanks for such many responses. I am asking this because recently, I have read an "article" in which the author tries to use the non-aggression principle as an axiom, like mathematicians and scientists do, to prove some things like mathematicians do: Establish the axioms and prove conclusions. But this principle, in my view, seems not be true, since aggression does exist.

I guess this question should be redirected to some exchange on Philosophy...


问:
Quotients of algebras
Let $A$ be a finite dimensional algebra over an algebraically closed field $k$.

Can $A$ admit infinitely many non-isomorphic quotient algebras?

答:
Yes.  For instance, let $A=k[x,y]/(x,y)^5$.  For any $c\in k\setminus\{0,1\}$, consider the algebra $B_c=A/(xy(x+y)(cx+y))$.  Write $V_c=m/m^2$ where $m=(x,y)\subset B_c$ is the unique maximal ideal.  We can identify $V_c$ with the set of homogeneous linear polynomials in $x$ and $y$, and thus identify its projectivization with $\mathbb{P}^1_k$.  Now note that there is a unique set of four elements of $\mathbb{P}^1_k$ which, if lifted to elements of $V_c$ and then to elements of $B_c$, have a product of $0$, namely the elements corresponding to $x,y,x+y,$ and $cx+y$.  Identifying $\mathbb{P}^1_k$ with $k\cup\{\infty\}$ in the usual way, these four points correspond to $\infty,0,1,$ and $c$.

Now, any isomorphism $B_c\to B_d$ would induce a linear isomorphism $V_c\to V_d$; projectivizing this linear isomorphism would give an isomorphism $\mathbb{P}^1_k\to\mathbb{P}^1_k$ which maps $\infty,0,1,$ and $c$ to $\infty,0,1,$ and $d$ in some order.  Taking cross-ratios, this means $d$ must be the cross-ratio of $\infty,0,1,$ and $c$ in some order.  There are only finitely many such orders, and thus only finitely many values of $d$ such that $B_c$ and $B_d$ are isomorphic.  Since there are infinitely many different values $c$ can take, this means the quotients $B_c$ of $A$ must have infinitely many different isomorphism types.

Title: Quotients of algebras


问:
Nilpotent matrices proof

Prove that there does not exist a $7 \times 7$ matrix $B$ with real entries such that $B^2 + B + I$ is nilpotent.

Since I need to show this statement for any $7 \times 7$ matrix $B$, I know that it isn't enough to give a counterexample, so I need to proxy it generally. I know the definition of nilpotent: the existence of $k \geq 0$ such that $(B^2 + B + I)^k = 0$. But I cannot fiugre out how to start a rigorous argument, other than that it may be appropriate to proceed by contradiction. I would appreciate any help on this problem.

Updated attempt based on a helpful answer:

Seeking a contradiction, suppose that $B^2 + B + I$ is nilpotent, so there is a $k > 0$ such that $(B^2 + B + I)^k = 0$. Let $B$ be a $7 \times 7$ matrix with real entries. Define its characteristic polynomial
\begin{align*}
q_B (\lambda) = \det(\lambda I - B).
\end{align*}
This is a monic polynomial of degree $7$, and every odd polynomial has a real root, so there exists some $\lambda \in \mathbb{R}$ such that $\det(\lambda I - B) = 0 $. This $\lambda$ is an eigenvalue of $B$, so there exists a non-zero eigenvector $v$ such that $Bv = \lambda v$.

Here is my thought as to how to proceed. I plugged in values of $k$ and multiplied out $(B^2 + B + I)^k$ and ended up with a polynomial in $\lambda$. For example:
\begin{align*}
(B^2 + B + I)v & = B^2 v + Bv + Iv = \lambda^2 v + \lambda v + v \\
(B^2 + B + I)^2 v & = (B^4 + 2B^3 + 3B^2 + 2B + I)v = B^4 v + 2B^3 v + 3B^2 v + 2B + v.
\end{align*}
It must be the case that $(B^2 + B + I)^k v \neq 0$ because $v \neq 0$ and we're always adding $v$ at the end. But I can't prove rigorously that the terms don't "cancel out."

答:
Hint: Since $B$ is a real $7 \times 7$ matrix, the characteristic polynomial $p(\lambda) = \det(\lambda I-B)$  is a real polynomial with degree $7$. Hence, $p(\lambda)$ has at least one real root, and thus, $B$ has a real eigenvalue.

Now, let $\lambda$ be this real eigenvalue and $v \neq 0$ be the corresponding eigenvector, i.e. $Bv = \lambda v$. Can you express $(B^2+B+I)^kv$ in terms of $\lambda$, $k$, and $v$? Is it possible for $(B^2+B+I)^kv = 0$ for some integer $k \ge 0$?

Title: Nilpotent matrices proof


问:
Single predictor soft thresholding
I have found in this book the following problem
$$\underset{\beta}{\mathrm{minimize}}\left\{\frac{1}{2N}\sum\limits_{i=1}^N(y_i-z_i\beta)^2+\lambda|\beta|\right\}$$
for parameter $\beta>0$ known real sequences $y_i$s, $z_i$s and known $N\in\mathbb{N}$. The non-differentiability at zero and the fact that this cannot be decomposed into $N$ minimizations since $\beta$ is common for all measurements are my problems here. So (by inspection) they propose the following optimal solution for $\beta$:
$$
\hat{\beta}=
\left\{
\begin{array}{ll}
\frac{1}{N}\langle\mathbf{z},\mathbf{y}\rangle-\lambda,&\quad\frac{1}{N}\langle{\mathbf{z}},\mathbf{y}\rangle>\lambda\\
0,&\quad\left|\frac{1}{N}\langle{\mathbf{z}},\mathbf{y}\rangle\right|\leq\lambda\\
\frac{1}{N}\langle\mathbf{z},\mathbf{y}\rangle+\lambda,&\quad\frac{1}{N}\langle{\mathbf{z}},\mathbf{y}\rangle<-\lambda
\end{array}
\right.
$$
But the proof of this is not clear to me. Can you give any hints?</p>

答:
This boils down to minimizing the function
$$
 f(x)
 =\frac12(x-y)^2+\lambda|x|
$$
as a function of $x\in\mathbb R$, where $y\in\mathbb R$ and $\lambda>0$. It seems that this can be done by inspection, but I would suggest to use the concept of subdifferentials and look for the subdifferential that contains $0$ as its element.

Denote the subdifferential of $f$ at $x$ by $\partial f(x)$. Then
$$
 \partial f(x)=
 \begin{cases}
 \{x-y-\lambda\}&\text{if}\ x<0;\\
 [-y-\lambda,-y+\lambda]&\text{if}\ x=0;\\
 \{x-y+\lambda\}&\text{if}\ x>0.
 \end{cases}
$$
Hence,
$$
 x^\star=
 \begin{cases}
 y+\lambda&\text{if}\ y<-\lambda;\\
 0&\text{if}\ |y|\le\lambda;\\
 y-\lambda&\text{if}\ y>\lambda.
 \end{cases}
$$
Equivalently,
$$
x^\star
 =\mathcal S_\lambda(y)
 =\operatorname{sign}(y)(|y|-\lambda)_+,
$$
where $\mathcal S_\lambda:\mathbb R\to\mathbb R$ is the soft-thresholding operator with $\lambda\ge0$.

We have that
\begin{align*}
 &\frac1{2N}\sum_{i=1}^N(y_i-z_i\beta)^2+\lambda|\beta|=\\
 &=\frac1{2N}\sum_{i=1}^N(y_i^2-2y_iz_i\beta+z_i^2\beta^2)+\lambda|\beta|\\
 &=\frac12\Bigl[N^{-1}\sum_{i=1}^Ny_i^2-2N^{-1}\sum_{i=1}^Ny_iz_i\beta+N^{-1}\sum_{i=1}^Nz_i^2\beta^2\Bigr]+\lambda|\beta|\\
 &=\frac12[N^{-1}\|y\|_2^2-N^{-2}(y^{\operatorname T}z)^2]+\frac12[\beta^2-2\beta N^{-1}y^{\operatorname T}z+N^{-2}(y^{\operatorname T}z)^2]+\lambda|\beta|\\
 &=\frac12[N^{-1}\|y\|_2^2-N^{-2}(y^{\operatorname T}z)^2]+\frac12(\beta-N^{-1}y^{\operatorname T}z)^2+\lambda|\beta|,
\end{align*}
where we used the assumption that $N^{-1}\sum_{i=1}^Nz_i^2=1$. It follows that
$$
\hat\beta=\mathcal S_\lambda(N^{-1}y^{\operatorname T}z)
$$
or, equivalently,
$$
 \hat\beta=
 \begin{cases}
 N^{-1}y^{\operatorname T}z+\lambda&\text{if}\ N^{-1}y^{\operatorname T}z<-\lambda;\\
 0&\text{if}\ |N^{-1}y^{\operatorname T}z|\le\lambda;\\
 N^{-1}y^{\operatorname T}z-\lambda&\text{if}\ N^{-1}y^{\operatorname T}z>\lambda.
 \end{cases}
$$

I hope this helps.

Title: Single predictor soft thresholding


问:
How do I write another vector as a linear combination without row reducing?

I'm not sure how I can do this problem without row reduction. I typically just put the 4 vectors in an augmented matrix (with the (1,2,3) being my fourth column in my augmented matrix) and then go to reduced echelon form. How else can I solve this without row reducing?

Here's the answer just for reference; I have no clue how they got that without row reducing:

答:
You want $\begin{bmatrix}3&-1&-1\\1&2&-4\\1&1&7\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$.

You could solve for $\begin{bmatrix}a\\b\\c\end{bmatrix}$ by multiplying both sides by $\begin{bmatrix}3&-1&-1\\1&2&-4\\1&1&7\end{bmatrix}^{-1}$.


问:
Changing the Form of an Exponential Function
I am having trouble understanding how to change the form of an exponential function. Can someone explain the process in which the function below

$$T(t) = e^{-0.0407409t+3.89467} +26$$

is changed into this form.

$$T(t) = 49.1398 e^{-0.0407409 t} + 26$$

Thanks

答: Starting with $$T(t) = e^{-0.0407409t+3.89467} +26$$ we can use rules of exponents to rewrite the first term as $$e^{-0.0407409t}e^{3.89467},$$ and since $$e^{3.89467} \approx 49.1398$$ we can say that $$T(t) \approx 49.1398e^{-0.0407409t} +26.$$  Note that you really shouldn't say that $$T(t) = 49.1398e^{-0.0407409t}$$ because some rounding was done to go from $e^{3.89467}$ to $49.1398$.


问:
Integration of a rational function from +/- infinity
I am trying to calculate the integral

$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$

where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

答:
Why don't you want to simplify it by putting the term under the integral in the following manner:
$$\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}=\frac{\alpha +x}{\beta  (c (x-a)^2+1)}-\frac{\gamma }{\delta +x}$$
with $$\alpha =\frac{4 a^2 c+a b^2 c+1}{b^2 c}$$
$$\beta =\frac{4 a^2 c+4 a b^2 c+b^4 c+1}{b^2 c}$$
$$\gamma =\frac{b^2}{4 a^2 c+4 a b^2 c+b^4 c+1}$$
$$\delta =a+b^2$$
which are just constants.
So
$$
\begin{eqnarray*}
\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx=\frac{\alpha}{\beta c}\int_{-\infty}^{\infty} \frac{1}{ (x-a)^2+\frac{1}{c}}dx&+&\\+\frac{1}{\beta c}\int_{-\infty}^{\infty} \frac{x}{ (x-a)^2+\frac{1}{c}}dx &-&\\-\gamma\int_{-\infty}^{\infty} \frac{1}{\delta +x}dx
\end{eqnarray*}
$$</p>


问:
What is the meaning of group extension?
I frequently see the notation

$1\to N\to G\to Q\to 1$.

Wikipedia explains that this is the notation to describe situation where there is an injective map from $N$ to $G$ called $i$ and $G/i(N)\cong Q$. It says that if $N\subset Z(G)$ then it is called central extension and if $N$ is cyclic, then this is called successive cyclic extension.

Why is there $1$ on the end in this notation and why is this notation important? I was just learning about supersolvable group and nilpotent group, and this notation seems out of context. Can you explain a little bit of context with simple terms without using Galois extension?

答:
The first $1$ indicates that the map $i:N\to G$ is injective.  The last that the map from $G$ to $Q$ is surjective.  This is because the sequence is exact.  That's the kernel of each map is the image of the previous one.

The sequence is called a short exact sequence, and an extension is a super-group, or a group that contains another as a subgroup.

We get $G/i(N)\cong Q$ by the first isomorphism theorem.

Title: What is the meaning of group extension?


问:
Is this probability inequality still true if we conditioned the term on a variable?
When I read Part III of this scientific paper in telecommunication, I see that the author use the following result from the Jensen Inequality

$E\left[ {\max \left( {X,Y} \right)} \right] \geqslant \max \left[ {E\left( X \right)E\left( Y \right)} \right]$

Where $E\left[ . \right]$ is the expectation operator.

Does the inequality still true if we conditioned on a variable $T$ like this:

${E_{\left\{ T \right\}}}\left[ {\left. {\max \left( {X,Y} \right)} \right|T = t} \right] \geqslant \max \left[ {{E_{\left\{ T \right\}}}\left( {\left. X \right|T = t} \right){E_{\left\{ T \right\}}}\left( {\left. Y \right|T = t} \right)} \right]$

Thank you for your enthusiasm !

答:
I'm not sure what you mean by $E_{T}$ But normally, you see:
$$ \mathbb{E}_{T}[max(X,Y)| T=t] \ge \mathbb{E}_{T}[X| T=t]$$
and $$ \mathbb{E}_{T}[max(X,Y)| T=t] \ge \mathbb{E}_{T}[Y| T=t]$$
Hence, your conclusion


问:
Extension of uniform convergence to the end point points : Series of continuous functions.
Let $\left(f_{n}\right)_{n \geq 1}$ be a sequence of continuous functions $f_{n}: \mathbb{R} \rightarrow[0,+\infty) .$ Suppose that the series $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly in an interval $(a, b)$ and $\sum_{n=1}^{\infty} f_{n}(y)<\infty$ for $y \in\{a, b\} .$ Prove that $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly on $[a, b]$

Can I deduce that  $\sum_{n=1}^{\infty}f_n(a),~\sum_{n=1}^{\infty}f_n(b),$ are uniformely Cauchy by using the boundedness, continuity and the uniformity in convergence on $(a,b)$? How to apply Weierstrass M test here?

答:
Your first idea is correct.

Since convergence is uniform for $y \in (a,b)$, for any $\epsilon > 0$ there exists $N(\epsilon) \in \mathbb{N}$ such that for all $m > n > N(\epsilon)$ and $y \in (a,b)$, we have

$$\tag{*}\left|\sum_{k=n+1}^m f_k(y) \right| < \epsilon$$

Pointwise convergence of the series at the endpoints also guarantees the existence of $N_a(\epsilon), N_b(\epsilon)$ such that (*) holds for all $m > n > \max(N_a(\epsilon),N_b(\epsilon))$ and $y \in \{a,b\}$.

Thus, (*) holds for all $m > n > \max(N(\epsilon),N_a(\epsilon),N_b(\epsilon))$ and $y \in [a,b]$ and the series converges uniformly on $[a,b]$ by the uniform Cauchy criterion.


问:
set builder mathematical notation in clustering algorithm
I came across a clustering algorithm, which uses the set builder notation to specify the clusters (or elements in the cluster). The algorithm is here.

The following is copied directly from the source, where the optimization function is this:

$$\mathit{E}(\mathbf{Z}) = \sum_{j=1}^k \frac{\mathbf{z}^⊤_j \mathbf{Az}_j}{f_j(\mathbf{Z})}$$

where:

$\mathbf{A} \in \mathbb{R}^{n×n}$ is the affinity matrix of n data points

$\mathbf{Z} = [\mathbf{z_1},···,\mathbf{z_k}] \in \mathcal{Z}, \mathcal{Z} = \{\mathbf{Z} \in \{0, 1\}^{n×k} | \mathbf{Z1} = \mathbf{1}\}$ is an assignment matrix, in which k is the number of clusters and 1 is the all-one vector

The optimization function and the affinity matrix $\mathbf{A}$ are logical, however, when you come to the notation of the clusters Z is where the problem lies. I'm not sure what the following notations mean:

The equation for $\mathcal{Z}$ is based on the predicate $\mathbf{Z1} = \mathbf{1}$, what does that mean? And why do the clusters $\mathbf{Z}$ belong to $\mathcal{Z}$ while $\mathcal{Z}$ is constructed by $\mathbf{Z}$ that belong to the $\{0, 1\}^{n×k}$ set?

The "all-one" vector is what? Presumably a row or column vectors of all 1s?

Thank you.

答: $\mathbf{Z}$ is an $n \times k$ matrix, and the first $\mathbf{1}$ is a $k \times 1$ matrix (column vector) of 1s (often written as $\mathbf{e}$).  The second $\mathbf{1}$ is an $n \times 1$ matrix (column vector) of 1s, and $\mathbf{Z 1} = \mathbf{1}$ means $$\sum_{j=1}^k Z_{i,j} = 1$$ for $i \in \{1,\dots,n\}$.  Because $Z_{i,j} \in \{0,1\}$, this implies that each row $i$ contains exactly one 1 and exactly $k-1$ 0s.


答:
Short exact sequences

That's what's called a "short exact sequence", and the notation only really makes sense in light of exact sequences more generally. We say a sequence of groups and homomorphisms between them like

$$ \dots \to G_{-1} \to G_{0} \to G_{1} \to G_{2} \to \dots $$

is exact if the kernel of the map from $G_k$ to $G_{k+1}$ is the image of the map from $G_{k-1}$ to $G_k$ for each $k$. This turns out to be a useful concept, but explaining why it's useful is, I think, out of scope.

A special case of this is an exact sequence like the one in your question

$$ 1 \to N \xrightarrow{f} G \xrightarrow {g} Q \to 1.$$

The claim that this sequence is exact tells you three things:

(1) The image of the map $1 \to N$ is the kernel of the map $N \xrightarrow{f} G$. Since the image is trivial that means $f$ is injective.

(2) The image of the map $N \xrightarrow{f} G$ is the kernel of the map $G \xrightarrow {g} Q.$ That is, $\operatorname{im} f$ (which is a copy of $N$ in $G$) inside $G$ is $\ker g$.

(3) The image of the map $G \xrightarrow {g} Q$ is the kernel of the map $Q \to 1$. Since the kernel is all of $Q$, that means $g$ is surjective.

So this can be summarized as follows: The sequence

$$ 1 \to N \xrightarrow{f} G \xrightarrow {g} Q \to 1.$$

is said to be a short exact sequence if $f$ is injective, $g$ is surjective, and $\operatorname{im} f = \ker g$. The only special role that the $1$'s play in this is that they are convenient ways to indicate that $f$ is injective and that $g$ is surjective, given that we are talking in terms of exact sequences anyway. The fact that these claims fit nicely into the framework of exact sequences is part of what makes the framework useful, but it's not useful in an obvious way until you've worked with it a bit, especially if you're only touching short exact sequences.

Special examples of this:

If $g:G \to H$ is any surjective map, then $1 \to \ker g \xrightarrow{f} G \xrightarrow {g} H \to 1$ is a short exact sequence, where $f: \ker g \to G$ is the inclusion map.

If $N$ is a normal subgroup of $G$, then $1 \to N \xrightarrow{f} G \xrightarrow {g} G/N \to 1$ is a short exact sequence, where $f$ is the inclusion map and $g$ is the obvious projection map.

But you don't have to use short exact sequences to talk about group extensions.

Group extensions

A group extension of a group $H$ is a group $G$ together with a surjective map $\pi: G \to H$. The extension is said to be central if $\ker \pi \subseteq Z(G)$. It's said to be cyclic if $\ker \pi$ is cyclic, etc.

The basic idea is that elements of $G$ are elements of $H$, but "with some extra information attached". If you started with $G$, you could reconstruct $H$ as a quotient: $H \cong G/\ker \pi.$

The nature of the extra information is, of course, related to what information you forget when you map elements of $G$ down to elements of $H$, i.e. information about the kernel $\ker \pi$. That's why a lot of data about the extension is stated in terms of the kernel: roughly the messier the kernel (and its relationship to $G$), the messier the information you're attaching. The simpler the kernel (and its relationship to $G$), the simpler the information you're attaching.

Group extensions are a basic way of building more complicated groups out of simpler ones. In that sense they're a bit like products of groups, but messier and richer.

Title: What is the meaning of group extension?

Body:
I frequently see the notation

$1\to N\to G\to Q\to 1$.

Wikipedia explains that this is the notation to describe situation where there is an injective map from $N$ to $G$ called $i$ and $G/i(N)\cong Q$. It says that if $N\subset Z(G)$ then it is called central extension and if $N$ is cyclic, then this is called successive cyclic extension.

Why is there $1$ on the end in this notation and why is this notation important? I was just learning about supersolvable group and nilpotent group, and this notation seems out of context. Can you explain a little bit of context with simple terms without using Galois extension?


问:
Are root functions continuous given any domain?
For instance take $f(x)=\sqrt{4-x}$. For all cases, we can say that the function is continuous at $x=a$ if $\lim\limits_{x\to a}f(x)=f(a)$. However, if a limit exists it is unique. Thus, the limit of $f(x)$ does not exist for any $x=a$ because $\sqrt{a}=\pm b$ (where $b$ is a positive real number), and thus the graph function is discontinuous. From what I have seen people tend to disregard the $-b$. In that case, the $\lim\limits_{x\to a}f(x)=f(a)$ for any $a>0$. Thus the graph would be continuous everywhere (given the proper domain) but at $x=0$.

I'm just a little confused. So if someone could explain how this works I would greatly appreciate it.

答:
The radical sign represents the principal or nonnegative square root. The $-b$ isn't being ignored: it's just negative (if $b$ is a positive real number, as you say) and thus it isn't $\sqrt{a}$ but is instead $-\sqrt{a}$.

As a result, when you have the function $f(x)=\sqrt{4-x}$, your outputs are going to be nonnegative because the radical sign represents the principal square root of $4-x$.


问:
finding the population in Statistics
I have the following question:

Netflix would like to estimate the proportion of their current subscribers who would pay extra for a premium membership including access to more movies and TV shows. To do this they plan to calculate a 95% confidence interval to estimate the proportion. They would like a margin of error of 0.05. How many subscribers must they sample to obtain this interval?

I know the answer is 385 but I keep getting an answer of 36.8 when I use this formula: n = ((z*/m)^2)p(1-p*) and it doesn't work when I use this formula either: n = ((z* * sigma)/m)^2.
Can anyone tell me where I'm going wrong?

答:
You will find your answer here

Reference

Title: finding the population in Statistics


答:
We can check if every $n$th root function is continuous in a slightly sneaky way by making use of a version of the inverse function theorem, which states the following:

If $I\subseteq\mathbb{R}$ is an interval and $f:I\to\mathbb{R}$ is injective and continuous on $I$, then the function $f^{-1}:I\to f(I)$ is  continuous.

Consider the function $f:[0,\infty)\to [0,\infty)$ defined as $f(x)=x^n$ for some $n\in\mathbb{N}$. It is not too difficult to show that all polynomial functions are continuous. It is also good to note that $f$ is bijective on the specified domain and codomain, so $f$ satisfies the hypothesis of the version of the inverse function theorem stated above. This implies that $f^{-1}:[0,\infty)\to[0,\infty)$ is continuous, that is $f^{-1}(x)=\sqrt[n]{x}$ is continuous on $[0,\infty)$.

Title: Are root functions continuous given any domain?

Body:
For instance take $f(x)=\sqrt{4-x}$. For all cases, we can say that the function is continuous at $x=a$ if $\lim\limits_{x\to a}f(x)=f(a)$. However, if a limit exists it is unique. Thus, the limit of $f(x)$ does not exist for any $x=a$ because $\sqrt{a}=\pm b$ (where $b$ is a positive real number), and thus the graph function is discontinuous. From what I have seen people tend to disregard the $-b$. In that case, the $\lim\limits_{x\to a}f(x)=f(a)$ for any $a>0$. Thus the graph would be continuous everywhere (given the proper domain) but at $x=0$.

I'm just a little confused. So if someone could explain how this works I would greatly appreciate it.


问:
Matrix, complex vector spaces
Suppose $M$ is a real 2 by 2 matrix with the property that $M^2+I=0$, and the eigenvalues of $M$ is $i$ and $-i$.

Prove that there exists a complex matrix P with the property that $$P^{-1}(aI+bM)P=\begin{bmatrix}
a & -b\\ b
 &a
\end{bmatrix}$$ for all $a,b\in\mathbb{R}$

I have some progress of this question, so I am not sure if this is correct $$P^{-1}(aI+bM)P=aP^{-1}IP+bP^{-1}MP=aI+bP^{-1}MP$$ ???? how do you show this?well, if that is correct, I some how want to show that $$P^{-1}MP$$ is the matrix \begin{bmatrix}
0 &-1 \\1
 &0
\end{bmatrix}
then something about the two matrix are similar, I am thinking really hard, all of this now looks like a mess in my head, can someone please help me? Thank you.

答:
As you have correctly established, what we're looking for is $P$ such that $PMP^{-1} = J$, where
$$
J = \pmatrix{0&-1\\1&0}.
$$
One standard approach to this is to use diagonalization over the complex numbers: because both $M$ and $J$ are diagonalizable with eigenvalues $\pm i$, they are similar to the same complex matrix and therefore similar to each other. This tells us that there is an invertible matrix $P$ with possibly complex entries for which $PMP^{-1} = J$.

Here is an approach which guarantees that $P$ is real. The operation $M\mapsto PMP^{-1}$ is a change of basis, so what we want to show is that there is a basis $\mathcal B = \{v_1,v_2\}$ such that the matrix of $M$ relative to $\mathcal B$ is $J$.

With that in mind, take $v_1$ to be any non-zero vector. I claim that $Mv_1$ and $v_1$ are linearly independent. Otherwise, $Mv_1$ would be a real vector that is a multiple of $v_1$, which would mean that $v_1$ is an eigenvector of $M$ corresponding to a real eigenvalue, which is impossible. Define $v_2 = Mv_1$. Note that
$$
Mv_2 = M(Mv_1) = M^2v_1 = (-I)v_1 = -v_1.
$$
Thus, we have $Mv_1 = 0\cdot v_1 + 1 \cdot v_2$, and $Mv_2 = -1 \cdot v_1 + 0 \cdot v_2$. It follows that the matrix $[M]_{\mathcal B}$ of $M$ with respect to $\mathcal B$ is $J$. Equivalently, if we take $P$ to be the matrix whose columns are $v_1,v_2,$ then $PMP^{-1} = J$, which was what we wanted.

Title: Matrix, complex vector spaces


答:
The solution to the problem is very simple as long as you bear in mind the important fact the previous answer highlights, namely that any real polynomial (by which I mean a polynomial in the ring $\mathbb{R}[X]$) of odd degree has at least one real root together with the following general:

Lemma. Let $K$ be an arbitrary commutative field, $n \in \mathbb{N}^{\times}=\mathbb{N} \setminus \{0\}$ a nonzero natural, $f \in K[X]$ an arbitrary polynomial and $M \in \mathscr{M}_n(K)$ a square matrix of order $n$ over $K$. If $\lambda \in K$ is an eigenvalue of $M$ then $f(\lambda)$ is an eigenvalue of $f(M)$.

In your particular instance, since $B^2+B+\mathrm{I}_7=(X^2+X+1)(B)$ is required to be nilpotent, all its eigenvalues must be $0$. Fixing $\alpha \in \mathbb{R}$ to be a certain eigenvalue of $B$ -- which exists by virtue of the order $7$ being odd -- the lemma above would entail that $\alpha$ is also a root of the polynomial $X^2+X+1$. However, this polynomial is irreducible over $\mathbb{R}$ and has no real roots.

The general claim that any eigenvalue of a nilpotent matrix $M \in \mathscr{M}_n(K)$ must be $0_K$ -- where $K$ is once again an arbitrary commutative field -- is easily proved by remarking that relation $MX=\lambda X$ -- where $\lambda$ is an arbitrary eigenvalue of $M$ and $X \in \mathscr{M}_{n, 1}(K)$ a corresponding eigenvector -- entails $M^r=\lambda^rX$ for any $r \in \mathbb{N}$. If in particular $M^r=\mathrm{O}_n$ for a certain $r \neq 0$, it then follows that $\lambda^r=0_K$ or $X=\mathrm{O}_{n1}$. Since eigenvectors are by definition non-zero vectors, it can only be that $\lambda^r=0_K$ which further entails $\lambda=0_K$, as $r \neq 0$ and since in any field (be it even noncommutative) a product is zero only if one of the factors is zero.

It can be furthermore shown that the characteristic polynomial of a nilpotent matrix, square of order $n$ over a commutative field $K$ is $X^n \in K[X]$, however your particular problem does not require a result of this depth.

Title: Nilpotent matrices proof

Body:
Prove that there does not exist a $7 \times 7$ matrix $B$ with real entries such that $B^2 + B + I$ is nilpotent.

Since I need to show this statement for any $7 \times 7$ matrix $B$, I know that it isn't enough to give a counterexample, so I need to proxy it generally. I know the definition of nilpotent: the existence of $k \geq 0$ such that $(B^2 + B + I)^k = 0$. But I cannot fiugre out how to start a rigorous argument, other than that it may be appropriate to proceed by contradiction. I would appreciate any help on this problem.

Updated attempt based on a helpful answer:

Seeking a contradiction, suppose that $B^2 + B + I$ is nilpotent, so there is a $k > 0$ such that $(B^2 + B + I)^k = 0$. Let $B$ be a $7 \times 7$ matrix with real entries. Define its characteristic polynomial
\begin{align*}
q_B (\lambda) = \det(\lambda I - B).
\end{align*}
This is a monic polynomial of degree $7$, and every odd polynomial has a real root, so there exists some $\lambda \in \mathbb{R}$ such that $\det(\lambda I - B) = 0 $. This $\lambda$ is an eigenvalue of $B$, so there exists a non-zero eigenvector $v$ such that $Bv = \lambda v$.

Here is my thought as to how to proceed. I plugged in values of $k$ and multiplied out $(B^2 + B + I)^k$ and ended up with a polynomial in $\lambda$. For example:
\begin{align*}
(B^2 + B + I)v & = B^2 v + Bv + Iv = \lambda^2 v + \lambda v + v \\
(B^2 + B + I)^2 v & = (B^4 + 2B^3 + 3B^2 + 2B + I)v = B^4 v + 2B^3 v + 3B^2 v + 2B + v.
\end{align*}
It must be the case that $(B^2 + B + I)^k v \neq 0$ because $v \neq 0$ and we're always adding $v$ at the end. But I can't prove rigorously that the terms don't "cancel out."


问:
Does there exist a smooth function with following properties?
Let $f \in C^\infty(\mathbb{R}^d; \mathbb{C})$.
Does there exist $g \in C^\infty(\mathbb{R}^d; \mathbb{C})$ such that $g(x)f(x) = 0$ and $|g(x)|=1$ for all $x \in \mathbb{R}^d$ ?

答:
It may not exist. Consider a smooth function $\phi\colon \mathbb{R}\to \mathbb{R}$, $\phi(x) = \exp (-\frac{1}{x})$ for $x > 0$ and $\phi(x) = 0$ for $x\le 0$. Now take $f(x) = (\phi(x), \phi(-x))$.

$\bf{Added:}$

In the case where the domain is $\mathbb{R}$ we need the zero of $f$ to be of infinite order.

However, if the domain is $\mathbb{R}^2 = \mathbb{C}$, we have the example of $f(z) = z$.


问:
Prove that $\{a_n\}$ converges if $|a_n|\le 2$ and $|a_{n+2}-a_{n+1}|\le \frac{1}{8} |a_{n+1}^2-a_n^2|$
$\{a_n\}$ is a sequence such that $|a_n|\le 2$ and $$|a_{n+2}-a_{n+1}|\le \frac{1}{8} |a_{n+1}^2-a_n^2|$$
How to prove that $\{a_n\}$ converges using only inequalities, Cauchy-sequences and the squeeze theorem?

I have no clue how to start with the proof. I have tried some nonsensical algebra but I think I'm stuck. Could it help if I knew what is the limit of the sequence?
I would like some direction.

Body:
Hint:

\begin{align}
|a_{n+2}-a_{n+1}|&\leq \frac{1}{8} |a_{n+1}^2-a_n^2| \\
&=\frac{1}{8}|a_{n+1}+a_n| |a_{n+1}-a_n| \\
&\leq \frac 12 |a_{n+1}-a_n|
\end{align}

Hence, $|a_{n+k}-a_n|\leq ...$?


问:
Is this characterization of a curve on a surface using lines of curvature correct?
Let $\alpha$ be a curve on a surface with Gaussian curvature $K=-1$ and let $N$ be the surface normal along $\alpha$. Assume that
\begin{equation}
\langle\alpha'(t),N(t)\rangle=0\quad\text{and}\quad\langle\alpha'(t),N'(t)\rangle\neq 0
\end{equation}
for all $t$. I came across a document where the author says that if this holds then

$\alpha'(t)$ and $N'(t)$ are everywhere or nowhere parallel, i.e., $\alpha$ is everywhere or nowhere tangent to a principal direction.

I don't seem to know how to use the two given assumptions to prove that $\alpha'(t)$ and $N'(t)$ are everywhere or nowhere parallel, nor that $\alpha$ is everywhere or nowhere tangent to a principal direction. Actually, it is not clear to me why these two conclusions are equivalent: if for each $t$ we have that $\alpha$ is tangent to a principal direction, then $\alpha$ is a line of curvature. This means, using Rodrigues' curvature formula, that $N'(t)+k_i(t)\,\alpha'(t)=0$, where $k_i(t)$ is the principal curvature associated with the direction to which $\alpha$ is tangent. Then, N'(t) is everywhere parallel to $\alpha'(t)$ provided the $k_i(t)$ are constant, which is not the case in general.

Any idea regarding this proof of why the staments are not correct is really appreciated.

答: This seems to be false. What is correct is that $\alpha$ is never tangent to an asymptotic direction. The curve can certainly vary from a line of curvature without ever being tangent to an asymptotic direction. (The first condition is totally redundant.)


问:
Complex number question?
Are there only finitely many complex numbers $z$ such that $z^{634}=1$?

I think there are only finitely many; is that not so?

答:
Yes, there are only finitely many $634^{th}$ complex roots of unity.

This is easily seen with polar notation.  Let $z=re^{i\theta}$ and $z^{634}=1$.

Then $r=1$ and $\theta=\dfrac{2\pi n}{634}$.

There are only finitely many values of $n$ (viz., $634$) such that $0\le\dfrac{2\pi n}{634}<2\pi$.

Title: Complex number question?


问:
How would you find the least number of popular votes to win an election given it is split in electoral districts?
I was given a problem to find the least number of popular votes to win an election through an electoral college. They gave me $N$ number of counties with $K$ required electoral votes to win. Each county had a population and a number of electoral votes given to the winner if he won the county. If you won about half of the county's popular votes you would get all electoral votes. How would you go about this sort of problem? Would you compare vote ratios or something else?

答: You can solve the problem via integer linear programming as follows.  For each county $i\in\{1,\dots,N\}$, let $p_i$ be the population and let $e_i$ be the number of electoral votes. Let binary variable $x_i$ indicate whether county $i$ is won.  The problem is to minimize $$\sum_{i=1}^N \left(\left\lfloor\frac{p_i}{2}\right\rfloor +1\right) x_i$$ subject to $$\sum_{i=1}^N e_i x_i \ge K.$$


问:
Complex number $|z|$=1
Are there only finitely many complex numbers $z$ with $|z|=1$?

I think there are infinitely many such numbers.

答: Yes, there are infinitely many complex numbers $z$ with $|z|=1$, including $e^{2 \pi i/n}$ for all $n\in \Bbb N$.

Title: Complex number $|z|$=1


答: Note that $$Q^{-1} \underbrace{\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}}_{:=J} Q = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \quad \textrm{where } Q := \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}$$ and then, since there exists a complex matrix $R$ with $$R^{-1}MR = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}$$ we have $$QR^{-1}MRQ^{-1} = Q \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} Q^{-1} = J.$$ Thus, for $P := RQ^{-1}$, $$P^{-1}(aI+bM)P = aI+bP^{-1}MP = aI + bJ = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}.$$

Title: Matrix, complex vector spaces

Body:
Suppose $M$ is a real 2 by 2 matrix with the property that $M^2+I=0$, and the eigenvalues of $M$ is $i$ and $-i$.

Prove that there exists a complex matrix P with the property that $$P^{-1}(aI+bM)P=\begin{bmatrix}
a & -b\\ b
 &a
\end{bmatrix}$$ for all $a,b\in\mathbb{R}$

I have some progress of this question, so I am not sure if this is correct $$P^{-1}(aI+bM)P=aP^{-1}IP+bP^{-1}MP=aI+bP^{-1}MP$$ ???? how do you show this?well, if that is correct, I some how want to show that $$P^{-1}MP$$ is the matrix \begin{bmatrix}
0 &-1 \\1
 &0
\end{bmatrix}
then something about the two matrix are similar, I am thinking really hard, all of this now looks like a mess in my head, can someone please help me? Thank you.


问:
Prove or disprove statements about the greatest common divisor
Help with prove or disproving either of these statements would be really appreciated, one or the other is fine, I just need a start or a solution to one and I'm sure I could probably figure the other out.

(a) For all $a, b, c \in \mathbb{Z}$, $\gcd(ac, bc) = c \gcd(a, b)$

(b) For all $a, b, c \in \mathbb{Z}$, $\gcd(ab, c) = \gcd(a, b) \gcd(b, c)$

答:
Concerning (b), perhaps it's worth mentioning that $\gcd$ and $\operatorname{lcm}$ distribute each with respect to the other, so you have
$$
\begin{align}
&\gcd(\operatorname{lcm}(a, b), c) = \operatorname{lcm}(\gcd(a, c), \gcd(b, c)),
\\&
\operatorname{lcm}(\gcd(a, b), c) = \gcd(\operatorname{lcm}(a, c), \operatorname{lcm}(b, c)).
\end{align}
$$
The first of these equalities is the correct form of (b).</p>


答: Indeed, all complex numbers on the circle $S^1$: namely all numbers of the form $e^{i\theta}$ for $\theta\in \mathbb{R}$

Title: Complex number $|z|$=1

Body:
Are there only finitely many complex numbers $z$ with $|z|=1$?

I think there are infinitely many such numbers.


问:
Variance and Mean of a random variable
Recall that a fair coin is one where we have equal probability of flipping a heads or a tails; that is, $P(H) = P(T) = 1/2$.

Let $D_n$ be a random variable counting the difference in the number of heads and number of tails after n flips of a fair coin. That is, $D_n = H_n − T_n$ where $H_n$ is the random variable counting the number of heads after n flips of a fair coin, and $T_n$ is defined similarly for tails.

What is the expected value of $D_n$?

What is the variance of $D_n$? The following identity may be useful: $V(aX+b) = a^{2}V(X).$

I have found the expected value of $D_n$ to be 0 but am having some trouble calculating the variance of $D_n$ and was wondering if I could get some help with that.

答:
If there are $n$ flips, then $H_n + T_n = n$ for any possible outcomes of $H_n$ and $T_n$.  For example, if $n = 10$ flips, and you observe $H_{10} = 7$, then you already know $T_{10} = 3$ because you can only get heads or tails on any given coin toss.

So if $H_n$ and $T_n$ add up to $n$ always, then you can express $D_n$ in terms of only one of these two random variables; e.g., $$D_n = H_n - (n - H_n) = 2H_n - n.$$  So to use the example, we have $D_n = 7 - 3 = 4$, or we can also calculate it as $D_n = 2(7) - 10 = 14 - 10 = 4.$

Now how do you compute the expectation and variance of $2H_n - n$?

Title: Variance and Mean of a random variable


Title: Complex number $|z|$=1

Body:
$$|z|=1$$
$$\sqrt{x^2+y^2}=1$$
$${x^2+y^2}=1.$$


问:
Covering space of Bouquet of Two circles
Given a bouquet of 2 circles $B_2$, is there a way to find a covering space of $B_2$ such that the deck transformation group of our covering space is isomorphic $F_2 / <R>$. i.e. the group $G = \langle x,y |R_1,R_2,.. \rangle$?

In general I want to find a covering space of $B_2$ whose deck transformation group is the alternating group $A_4$. I know that the fundamental group of $B_2$ is the free group on two generators and I know I can write $A_4$ as such $A_4 = \langle x,y | x^2 = y^3 = (xy)^3 = 1 \rangle$. But I'm not really sure how to go about finding a covering space that satisfies this. Could I perhaps relate this to the Cayley graph of $A_4$?

I've been searching everywhere to find some sort of connection between the deck transformations and the relations of the free group but I've come up empty. I've been using Hatcher's algebraic topology.

Is this fact true in general that given a group $F_n / <R>$ we can find a covering space of $B_n$ whose deck transformation group is $F_n / <R>$?

答:
Yes, this is true in general.  Suppose that $X$ is a wedge of $k$ circles so that its fundamental group $F_k$ is free of rank $k$.  Orient each loop. These loops, $x_1, \dots, x_k$, define a basis.  Suppose we are given a group $G$ (such as $A_4$) and a surjective homomorphism $f:F_k \to G$.  This defines a generating set for $G$, namely $f(x_1), \dots, f(x_k)$.  Let $C$ be the Cayley graph of $G$ with respect to this generating set.  Thus, the vertices are $G$ and there is an oriented edge $(g,gs)$ for each $g \in G$ and generator $s$.  The edge is labeled $s$.  The group $G$ acts on $C$ on the left via $h.(g,gs) = (hg,hgs)$.

Prove this is a covering space action.  This is essentially Exercise 17 in Hatcher.  Proposition 1.40 relates the deck group to $G$: they are isomorphic

The picture for $A_4$ is pleasing.  Try to visualize the deck maps defined by the generators in the presentation you give above.