问:
Does the limit of proportions of objects in a sequence depend on the order of the objects in that sequence?
Motivation : Suppose A is an infinite sequence of some objects, and I want to get some sense of the proportion of objects satisfying some property P. Intuitively, I can define

proportion = $\lim \limits_{n \to \infty} \frac {\sum_{k=1}^{n} f(A_k)}{n}$, where $f(A_k)=1$ if $A_k \in P$, and $0$ otherwise.

Question : Does this limit (or even its existence) depend on the order of the objects in the sequence A?

答:
Yes, quite dramatically. Take $A = \mathbb{N}$ to be the positive integers WLOG and let $S$ be the subset having property $P$. The full group $\text{Aut}(\mathbb{N})$ of permutations of $\mathbb{N}$ acts transitively on the set of all infinite subsets of $\mathbb{N}$ whose complement is also infinite, so if $S$ is infinite with infinite complement then the limit you describe can have any value in $[0, 1]$, or be undefined, depending on the order.

For example, take $S$ to be the odd numbers. In the usual order the limit is $\frac{1}{2}$; this is called the natural density. But if we consider the order

$$1, 2, \\ 3, 5, 4, \\ 7, 9, 11, 6, \\ 13, 15, 17, 19, 8, \dots$$

in which we alternate between one even number and $k$ odd numbers for all positive integers $k$, then the limit is $1$. Variations of this idea can be used to produce any other limit or no limit.

To produce a limit $r \in [0, 1]$ take a sequence $\frac{p_k}{q_k}$ of rational approximations converging to $r$ (which can become eventually constant if $r$ is rational) and alternate between $q_k$ even numbers and $p_k$ odd numbers. To produce no limit, alternate between $2^k$ even numbers and $2^k$ odd numbers; this causes the density to fluctuate between $\frac{1}{2}$ and $\frac{1}{3}$. This is related to the fact that the "Benford subset" of positive integers whose first digit is $1$ has no natural density.

On the other hand, if $S$ is finite then the density is $0$ regardless of the order, and if $S$ is cofinite (has finite complement) then the density is $1$ regardless of the order.


答:
Yes the limit does exist.  Define an event of picking up an object and noting whether it has property P.  Then your limit function is basically the definition of the probability of an event.  Although your function f will flip-flop between 0 and 1, the impact of each new 0 or each new 1 on the total sum will become vanishingly small as n becomes increasingly large.

So if my objects are coins and P is the property (lying heads-up), then as I examine more and more coins the closer your limit will get to 0.5. By the time I've looked at 1,000,000 coins, say, the sum might be at 502,301/1,000,000.  If the next coin is a head, it will become 502,302/1,000,001.  If it is a tail, 502,301/1,000,000, but the difference is negligible.

ContentLicense: CC BY-SA 4.0

Title: Does the limit of proportions of objects in a sequence depend on the order of the objects in that sequence?

Body:
Motivation : Suppose A is an infinite sequence of some objects, and I want to get some sense of the proportion of objects satisfying some property P. Intuitively, I can define

proportion = $\lim \limits_{n \to \infty} \frac {\sum_{k=1}^{n} f(A_k)}{n}$, where $f(A_k)=1$ if $A_k \in P$, and $0$ otherwise.

Question : Does this limit (or even its existence) depend on the order of the objects in the sequence A?


问:
visualizing 1-parameter family of persistence modules by vineyard technique
In the paper "The structure and stability of persistence modules", Page 49, they use vineyard technique to visualize the 1-parameter family of persistence modules produced by three interpolation constructions described in the paper, generating the following figure:

Can someone explain to me how they come from? Thank you!

答:
I'll explain the $2$-interleaving case; the $3$-interleaving case is left to you.

The interpolating families are constructed out of the maps in the sequence
$$\begin{align}
&\color{blue}{\mathbb{I}^{[0,4)}[q-3] \oplus \mathbb{I}^{[1,6)}[p-3]} \\
&\xrightarrow{\Omega'} \color{red}{\mathbb{I}^{[0,4)}[p-1] \oplus \mathbb{I}^{[1,6)}[q-1]} \\
&\xrightarrow{\Omega} \color{green}{\mathbb{I}^{[0,4)}[q+1] \oplus \mathbb{I}^{[1,6)}[p+1]} \\
&\xrightarrow{\Omega''} \color{orange}{\mathbb{I}^{[0,4)}[p+3] \oplus \mathbb{I}^{[1,6)}[q+3]}.
\end{align}$$
Since we are using the canonical interleaving of interval modules, the entries of the maps in its standard matrix form are all $\pm 1$, if it is not necessarily zero because either the domain or codomain is zero.

Recall also that as persistence modules over the $(p,q)$-plane,
$$\begin{align}
\mathbb{I}^{[a,b)}[p-c] &= \mathbb{I}^{[a+c,b+c) \times (-\infty, \infty)} \\
\mathbb{I}^{[a,b)}[q-c] &= \mathbb{I}^{(-\infty, \infty) \times [a+c, b+c)},
\end{align}$$
so it is easy to read off the supports of these modules.

We shall visualize the interpolating families by sketching their supports.

Let us begin with $\mathrm{im}(\Omega)$.  We want the image of $$\color{red}{\mathbb{I}^{[0,4)}[p-1] \oplus \mathbb{I}^{[1,6)}[q-1]} \xrightarrow{\Omega} \color{green}{\mathbb{I}^{[0,4)}[q+1] \oplus \mathbb{I}^{[1,6)}[p+1]}$$ inside the diagonal band $$\Delta_{[-1,1]} = \{(p,q) \mid -2 \leq q-p \leq 2\}.$$

The diagonal band is bordered by the brown lines above.  The image, indicated in yellow, is the overlap between the red and green regions.  We see that on $\Delta_{-1} = \{q-p=-2\}$ we recover the interval module $\mathbb{I}^{[0,4)}$, and on $\Delta_1 = \{q-p=2\}$ we get $\mathbb{I}^{[1,6)}$, so sweeping along diagonals of slope $1$ from the bottom right to the upper left, the yellow region describes an interpolating family between these two interval modules.  In this case, we begin at $\mathbb{I}^{[0,4)}$, move linearly to $\mathbb{I}^{[0.5,5.5)}$, and finally move linearly to $\mathbb{I}^{[1,6)}$, as the vineyard confirms.

Now let's consider the kernel of $$\color{green}{\mathbb{I}^{[0,4)}[q+1] \oplus \mathbb{I}^{[1,6)}[p+1]} \xrightarrow{\Omega''} \color{orange}{\mathbb{I}^{[0,4)}[p+3] \oplus \mathbb{I}^{[1,6)}[q+3]}.$$

The kernel is the region in green that is not covered by orange, with the caveat that "double green" cancels out and is included regardless, based on the signs in the definition of $\Omega'$.  So the $\ker(\Omega'')$ is the region in yellow, which describes an interpolating family that moves linearly from $\mathbb{I}^{[0,4)}$ to $\mathbb{I}^{[-0.5,4.5)}$ to $\mathbb{I}^{[1,6)}$.

Finally, let us look at the cokernel of $$\color{blue}{\mathbb{I}^{[0,4)}[q-3] \oplus \mathbb{I}^{[1,6)}[p-3]} \xrightarrow{\Omega'} \color{red}{\mathbb{I}^{[0,4)}[p-1] \oplus \mathbb{I}^{[1,6)}[q-1]}.$$

We want the red region that is not covered by the blue region, where the red region is counted with multiplicity (but not the blue, since $\Omega'$ has rank $\leq 1$).  This is outlined in yellow.  This time, in addition to the $\mathbb{I}^{[0,4)}$ to $\mathbb{I}^{[1.5,5.5)}$ to $\mathbb{I}^{[1,6)}$ family, this is also an extra family indicated by the cross-hatching that starts at $\mathbb{I}^{[3,3)}$, grows for a bit, then dies at $\mathbb{I}^{[2,2)}$ since the red region is counted with multiplicity $2$.  This agrees with the vineyard shown.

As a word of encouragement, these calculations are not too difficult. You just need to have a clear understanding of the definitions, and exercise care with the linear algebra.


问:
What is this symmetric-like property of $n$-ary relations?
Let's consider that there is a set $S$ for which $R \subseteq S \times S \times S$ is a relation on that set. With $x,y,z \in S$, imagine that the following holds:

$$R(x, y, z) \iff R(x, z, y) \iff R(y, x, z) \iff R(y, z, x) \iff R(z, x, y) \iff R(z, y, x)$$

This 3-ary relation has a property that reminds me of symmetric relations in which the order of the elements doesn't matter. This property can hold non-trivially for arities $n>1$, and still hold for $n=0$ or $n=1$ in a trivial way, making it something rather general. What is this property called of an $n$-ary relation?

答:
The following is a published definition for a symmetric $n$-ary relation, with the qualifier of strongly to denote that all permutations must hold.

[An $n$-ary (finitary) relation $\rho$ is said to be] strongly symmetric if $(x_1, \cdots, x_n) \in \rho$ implies $(x_{\sigma(1)}, \cdots, x_{\sigma(n)}) \in \rho$ for any permutation $\sigma$ of the set $\{ 1, \cdots, n\}$.

Reference

https://www.sciencedirect.com/science/article/pii/S0195669809001589


问:
Calculate the kernel of the linear transformation $T: \mathbb{R}^6 \rightarrow \mathbb{R}^2$
Calculate the kernel of the linear transformation $T: \mathbb{R}^6 \rightarrow \mathbb{R}^2$ given by:

$$T(x_1,...,x_6) = (x_1-x_2+2x_4-3x_5+x_6,2x_1-x_2-x_3+3x_4-4x_5+4x_6)$$

By definition, this we have that $ker(T) = \{(x_1,...x_6) \in \mathbb{R}^6\ |$ $x_1-x_2+2x_4-3x_5+x_6=0 $ $\text{and}$  $2x_1-x_2-x_3+3x_4-4x_5+4x_6=0  \}$

How do I go about calculating this more explicitly?

My attempt: I see by inspection that $(1,0,1,1,1,0) \in ker(T)$, I would like to somehow extend this to a basis for the kernel.

Thank you

答: You have this system$$\left\{\begin{array}{l}x_1-x_2+2x_4-3x_5+x_6=0\\2x_1-x_2-x_3+3x_4-4x_5+4x_6=0.\end{array}\right.$$If you replace the second equation by its sum with the first one times $-2$, you get$$\left\{\begin{array}{l}x_1-x_2+2x_4-3x_5+x_6=0\\x_2-x_3-x_4+2x_5+2x_6=0.\end{array}\right.$$Now, if you replace the first equation by its sum with the second one, you get$$\left\{\begin{array}{l}x_1-x_3+x_4-x_5+3x_6=0\\x_2-x_3-x_4+2x_5+2x_6=0.\end{array}\right.$$This is equivalent to$$\left\{\begin{array}{l}x_1=x_3-x_4+x_5-3x_6\\x_2=x_3+x_4-2x_5-2x_6\end{array}\right.$$and therefore$$\ker(T)=\{(a-b+c-3d,a+b-2c-2d,a,b,c,d)\mid a,b,c,d\in\Bbb R\}.$$


问:
Convergence of $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$, where $a > \frac{1}{2} n$.
I want to show that $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$ converges, where $a > \frac{1}{2} n$. This problem is from Cambridge Math Tripos 1904. I am not sure why there is no lower bound for this integral. Anyone knows what is the default value for the empty lower bound? I can see if $n = 3$, we can use the spherical triple integrals to solve the problem. But how do we show that the general case for nth repeated integrals?

答:
Using the AG-GM  :

If $x_1,x_2,\cdots,x_n$ are positive real numbers, then

$$n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}\leq x_1^2+x_2^2+ \cdots x_n^2$$

You can write:
$$
  \int_1^\infty \cdots \int_1^\infty
  \frac{dx_1 \cdots dx_n}{(x_1^{2}+\cdots + x_n^{2})^a}
  \leq \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
  \frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}}
$$
But:
$$
\frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
\frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}} =
\frac{1}{n} \int_{1}^\infty \frac{dx_1}{x_1^{2a/n}} \cdots \int_{1}^\infty \frac{dx_n}{x_n^{2a/n}}
$$
And, since $\frac{2a}{n}> 1$,for each $i \in [1,\infty)$:
$$
\int_{1}^\infty \frac{dx_i}{x_i^{2a/n}}  = \left[ \frac{x_i^{1-2a/n}}{1-2a/n} \right]_1^\infty = \frac{1}{1-2a/n}
$$


问:
ex falso quodlibet in natural deduction (ND)?
My understanding of inference rules is that they should be intuitively acceptable (like axioms). But my only intuition for ex falso quodlibet in natural deduction is not immediate but comes from its proof, which rests on disjunctive syllogism (which apparently holds in classical propositional logic and intuitionistic logic).
I am probably missing something obvious. Is the principle of explosion really basic/intuitive and why?

The closest justification I see, based on the current answers, is that Ex Falso encodes a (metalogical) desire that the proof system be consistent. But also see the references to intuitionism in the answers.

Edit: as pointed out below, it may have been a distraction for me to raise "intuitionism" in the original question, although Ex Falso arises in classical and intuitionistic logic. This does not exclude that I would like axioms and rules to be intuitively acceptable (to me).

" There  is  surprisingly  little  agreement  about  the  exact  status  of EFQ in  the literature.  Gentzen groups the ex  falso rule alongside the other operational rules(1969,  p.  77).   Prawitz  follows  this  practice.   He  treats ⊥ as  a  ‘0-place  sentential operation’ (Prawitz 1978, p. 38), one ‘for which there is no canonical proof’ (Prawitz1977, p. 26). EFQ is then understood as the elimination rule for ⊥. Others, including Dummett, present ⊥ as an elimination rule for ¬ (e.g. Dummett 1991, p. 291).  I contend that both of these approaches are mistaken...." ref.

Ex Falso has an ‘anomalous position [. . . ]  inside the scheme of introduction/elimination rules’, tarnishing the neat symmetry of the intuitionistic system" (Weir 1986, p. 461).

According to here: "Kolmogorov’s criterion whether to keep an axiom was whether a proposition has an “intuitive foundation” or “possesses intuitive obviousness” (van Heijenoort 1967: 421, 422) ... Kolmogorov said that, just like PEM, Ex Falso “has no intuitive foundation” (van Heijenoort 1967: 419). In particular, he says that Ex Falso is unacceptable for the reason that it asserts something about the consequences of something impossible (van Heijenoort 1967: 421)."

... "I submit that these difficulties can be overcome simply by treating⊥as a punc-tuation marker, as Tennant suggests.  Of course this means thatEFQis left withouta  logical  constant  which  it  serves  to  eliminate.   But  on  the  view  I  am  proposingno such constant is needed becauseEFQis not an operational rule at all.  Its roleis  best  accounted  for  by  according  it  the  status  of  astructural  rule.   It  simply  isthe structural rule that tells us that any sentence whatsoever follows from a con-tradiction.  Such a license isnotspecific to any logical constants, but amounts to ablanket policy.  Our reclassification ofEFQthus fits neatly with our characterizationof structural rules as global rules that assign properties to our deducibility relation." ref.

Edit: I came across this reformulation of my initial uneasiness.

"... adopting disjunctive syllogism as an axiom or as a primitive rule of inference is not an option.  Hence, it is ex  falso that is naturally treated as primitive, not the principle of disjunctive syllogism. Therefore, rather than thinking of ex  falso as the product of the mean-ings of disjunction and negation, we must treat disjunctive syllogism as a product of the ex falso rule.  This suggests that the law of disjunctive syllogism is not a con-sequence only of the meanings of the constants involved, but also hinges on one’s stance concerning the general principle expressed by EFQ." ref.

答:
Here is another possible idea behind the rule, although it is technically just presenting things Couchy said in a different way.

Because there is no specific way to prove $⊥$, for every proof we do come up with, we could have ostensibly just systematically replaced $⊥$ with $A$ to instead get a proof of $A$ (though there may be technicalities with respect to non-logical axioms). So, the elimination rule for $⊥$ is a local rule that internalizes this global property. The reason one introduces the local rule is to solidify the role of $⊥$ as classifying such degenerate proofs.

By contrast, minimal logic's $⊥$ is useful for other purposes. It can be useful to consider 'negating' with respect to an arbitrary unknown proposition that is not intended to be trivial in this way.


问:
Limit involving log

I was asking why this limit is $-\infty$ and not $+\infty\ ?$.

I thought it was $+\infty$, but Wolfram Alpha says $-\infty$.

My guess is that $\log\left(\log\left(x\right)\right) -\log\left(\log\left(x - 1\right)\right)$ goes to zero faster than $-0.5x$ goes to $-\infty$.

$$
\lim_{x \to \infty}\left\{\left(\frac{x^{2}}{2} - x\right) \left[\log\left(\log\left(x\right)\right) -
\log\left(\log\left(x - 1\right)\right)\right] -
\frac{x}{2} -\frac{x}{\log\left(x\right)}\right\}
$$

答:
We'll study the expression for $x\to\infty$:

$\Big(\dfrac{x^2}{2}-x\Big)(\log(\log x)-\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x}$, which is asymptotic to
$$\dfrac{x^2}{2}(\log(\log x)-\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x}=$$

$$=\dfrac{x^2}{2}\log(\log x)-\dfrac{x^2}{2}\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x} \sim$$

$$\sim \dfrac{x^2}{2}\log(\log x)-\dfrac{x^2}{2}\log(\log x)-\dfrac{x}{2}-\dfrac{x}{\log x}\underset{D.H.}{\sim}-\dfrac{x}{2}-\dfrac{x'}{(\log x)'}=$$

$$=-\dfrac{x}{2}-\dfrac{1}{\dfrac{1}{x}}=-\dfrac{3x}{2}\overset{x\to \infty}{\longrightarrow} -\infty.$$

Title: Limit involving log


答: If the lower bound was the origin, there would be a problem.  However assuming a small sphere around (radius $=x$) the origin is excluded, then we can change the integral to n-dimensional spherical coordinates, integrate the finite surface term and get $K\int_x^\infty \frac{r^{n-1}}{r^{2a}}dr$.  Since $2a\gt n$, the integral =$K\frac{1}{2a-n}x^{n-2a}$ is finite.

Title: Convergence of $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$, where $a > \frac{1}{2} n$.

Body: I want to show that $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$ converges, where $a > \frac{1}{2} n$. This problem is from Cambridge Math Tripos 1904. I am not sure why there is no lower bound for this integral. Anyone knows what is the default value for the empty lower bound? I can see if $n = 3$, we can use the spherical triple integrals to solve the problem. But how do we show that the general case for nth repeated integrals?


问:
When is a multivector in Cl(n) invertible?

My question

Let $\mathrm{Cl}(n)$ be the Clifford algebra over $\mathbb{R}^n$ with the usual inner product. That is, it's the quotient of the tensor algebra over $\mathbb{R}^n$ by the ideal generated by

$$\{ v^2 - v \cdot v : v \in \mathbb{R}^n \}.$$

Given an element of $\mathrm{Cl}(n)$, is there a "simple" procedure to decide whether it is invertible? For example, it's easy to check whether a matrix is invertible by checking whether its determinant is nonzero.

What I know

I know that every Clifford algebra is isomorphic to a direct sum of one or two matrix algebras, so I could use that isomorphism and then check the determinant. But I'm hoping that you know a way that doesn't involve quaternion matrix determinants and five cases depending on the dimension. :)

Maybe the determinant formulae for each case will end up identical; I haven't tried it yet.

答:
If anything else fails, you always have this tedious but straightforward method at hand.

Let $A$ be any finite-dimensional unital algebra over a field $F$, say $\dim A=n$. Then, $A$ acts on itself by multiplication from the left, providing an algebra homomorphism $\varphi:A \rightarrow \operatorname{End}(A)$ to the algebra of $F$-linear endomorphisms of $A$. Explicitly, $\varphi(a)x = ax$.

The kernel of $\varphi$ consists of elements $a\in A$ such that $\forall x\in A : ax=0$. Substitute $x=1$ to get $a=0$. Thus, $\varphi$ is injective, and $A$ is isomorphic to its image $\varphi(A) \subset \operatorname{End}(A)$. So, $a\in A$ is invertible iff $\varphi(a)$ is.

Finally, $\varphi(a)$ is a linear operator on $A$, thus an $n\times n$-matrix in some basis of $A$, and invertibility of $\varphi(a)$ can be decided by computing the determinant $\det\varphi(a)$.

Consider the case of $A = \operatorname{Cl}^{1,0}(\mathbb R)$ the Clifford algebra of a one-dimensional Euclidean space. It has a basis $\{1, e\}$ with $e^2 = 1$. An element $a + be \in A$ corresponds (in this basis) to a matrix

$$\begin{pmatrix}a & b \\ b & a\end{pmatrix}$$

with determinant $a^2-b^2$. Indeed, whenever $a^2-b^2\neq 0$, we have

$$(a+be)^{-1} = \frac{a-be}{a^2-b^2}$$

If $a^2-b^2 = 0$, then $(a+be)(a-be) = a^2 - b^2 = 0$, so $a+be$ is a zero divisor and cannot be invertible.

As another example, take $A = \operatorname{Cl}^{0,1}(\mathbb R)$ the Clifford algebra of a one-dimensional space with negative-definite form. It is known that $A \cong \mathbb C$, the complex numbers, so every nonzero element should be invertible.

Take the basis $\{1, e\}$ with $e^2=-1$. Now $a+be$ corresponds to

$$\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$$

with determinant $a^2+b^2$, which is zero iff $a=0$ and $b=0$.


问:
Density of a normed space equals density of unit ball (or of the unit sphere)
Let X be a normed space, define $\mathrm{dens} \,X$, the density character of $X$, to be the smallest cardinality of a dense subset of $X$. (E.g.: X separable iff $\mathrm{dens}\, X= \lvert \mathbf{N} \rvert$). How can I prove that $$\mathrm{dens}\, X=\mathrm{dens}\, B_X=\mathrm{dens}\, S_X,$$ where $B_X$ denote the (closed, but I think this is not relevant) unit ball of $X$ and $S_X$ the unit sphere of $X$? I know how to do this in the case in which one of this spaces is known to be separable but I can't extend the proof to this general case.

答:
For the record, the usual notation, at least in topology, is $d(X)$.

Suppose that $D$ is dense in $S_X$; Then $\Bbb Q^+D=\{qx\in\Bbb Q\times D:q>0\}$ is dense in $X$, and since $D$ is infinite, $|\Bbb Q^+D|=\aleph_0\cdot|D|=|D|$. Thus, $d(X)\le d(S_X)$.

If $D$ is dense in $X$, $\left\{\frac1{\|x\|}x:x\in D\setminus\{0\}\right\}$ is dense in $S_X$, so $d(S_X)\le D(X)$, and hence $d(S_X)=d(X)$.

If $D$ is dense in $X$, $D\cap B_X$ is dense in $B_X$, so $d(B_X)\le d(X)$. Finally, if $D$ is dense in $S_X$, $\{qx\in\Bbb Q\times D:0<q\le 1\}$ is dense in $B_X$, so $d(B_X)\le d(S_X)=d(X)$, and all three densities are equal.


问:
Why the identity matrix in first-order linear systems?
I was reviewing homogenous systems of first-order ODEs and was wondering how the identity matrix $I$ in

$$(\textbf{A}-r\textbf{I})\textbf{v}=\textbf{0}$$

came about.

Here's some exposition:

Consider the homogenous first-order linear system

$$\bf{x'=Ax}\quad(\star)$$

where

$$\textbf{x'}=\frac{d}{dt}\begin{pmatrix}x_1\\x_2\end{pmatrix}\quad\quad \bf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\quad\quad\bf{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}.$$

We look for solutions of the from

$$\textbf{x}=\textbf{v}e^{rt}\quad (\star\star )$$

where $r$ and $\textbf{v}=(v_1\;v_2)^T$ are to be determined.

Plugging $(\star\star)$ back into $(\star)$ yields

$$\textbf{Av}=r\textbf{v},$$

since $e^{rt}$ is non-zero.

Furthermore we have that

$$(\textbf{A}-r\textbf{I})\textbf{v}=\textbf{0}\quad (\star\star\star)$$

where $I$ is the $2\times2$ identity matrix.

Therefore, to solve the system $(\star)$, we must solve the homogenous algebraic system $(\star\star\star)$ for $\bf v$.

This means that for a non-trivial solution of $(\star)$,
$$\det(\textbf{A}-r\textbf{I})=\textbf{0}\quad (*)$$
must be satisfied.

Question:
Where does the identity matrix $I$ come from? I know that without the $I$ the statement $(*)$ is meaningless, but what I'm asking is why is it this precise matrix and not any other matrix?

Thanks.

答:
There are several errors in your question statement, so let me try to fix those as I answer your question.

We are given an equation $$\textbf{x}'=A\textbf{x},$$ where $\textbf{x}$ is a vector of length $n$, and $A$ is an $n\times n$ matrix.  Suppose this has a solution of the form $\textbf{x}=e^{rt}\textbf{v}$, where $\textbf{v}$ is a constant vector and $r$ is a constant scalar.  Since $(e^{rt}\textbf{v})'=re^{rt}\textbf{v}$, we obtain
$$
re^{rt}\textbf{v}=A(e^{rt}\textbf{v})
$$
and hence
$$
r(e^{rt}\textbf{v})-A(e^{rt}\textbf{v})=\textbf{0}.
$$
Here we would like to factor out $e^{rt}\textbf{v}$, but there is a problem:  Since $r$ is a scalar and $A$ a matrix, the expression $(r-A)$ makes no sense---unless we stipulate that $r$ is really $rI$, where $I$ is the $n\times n$ identity matrix.  This works because
$$
r(e^{rt}\textbf{v})-A(e^{rt}\textbf{v})=\textbf{0}
\;\;\;\Leftrightarrow\;\;\;
rI(e^{rt}\textbf{v})-A(e^{rt}\textbf{v})=\textbf{0}
$$
and hence
$$
(r-A)(e^{rt}\textbf{v})=\textbf{0}
\;\;\;\Leftrightarrow\;\;\;
(rI-A)(e^{rt}\textbf{v})=\textbf{0}.
$$


问:
Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition.
Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition.</p>

答:
Let $\mathcal{P}$ the set of the disjoint pairs $(A,B)$, where $A,B\subseteq\mathbb{R}^+$ are not empty and each one is closed under addition and multiplication by a positive rational number. Note that $\mathcal{P}\neq\emptyset$ because if we consider $X=\mathbb{Q}^+$ and $Y=\{n\sqrt{2}:n\in\mathbb{Q}^+\}$, then $(X,Y)\in\mathcal{P}$. Define $\leq$ as follows:

$(X_1,Y_1)\leq(X_2,Y_2)$ if and only if $X_1\subseteq X_2$ and $Y_1\subseteq Y_2$, for all $(X_1,Y_1),(X_2,Y_2)\in\mathcal{P}$.

Clearly, $(\mathcal{P},\leq)$ is a partially ordered set. Furthermore, it is easy to see that every chain in $(\mathcal{P},\leq)$ has an upper bound. We can now apply Zorn's lemma, so $(\mathcal{P},\leq)$ has a maximal element. Let $(A,B)\in\mathcal{P}$ a maximal element of $(\mathcal{P},\leq)$. We only have to show that $A\cup B=\mathbb{R}^+$.

Suppose that $\mathbb{R}^+\not\subseteq A\cup B$. Therefore, there exists $x\in\mathbb{R}^+$ such that $x\not\in A\cup B$. Consider $A_x=\{kx+a:k\in\mathbb{Q}^+\cup\{0\}\textrm{ and }a\in A\}$ and $B_x=\{kx+b:k\in\mathbb{Q}^+\cup\{0\}\textrm{ and }b\in B\}$. Then $A\subseteq A_x$, $B\subseteq B_x$ and $A_x,B_x\subseteq\mathbb{R}^+$. It is easy to verify that $A_x$ and $B_x$ are closed under addition and multiplication by a positive rational number.

If either $A_x\cap B=\emptyset$ or $A\cap B_x=\emptyset$, then $(A,B)$ would not be a maximal element of $(\mathcal{P},\leq)$. Therefore, $A_x\cap B\neq\emptyset$ and $A\cap B_x\neq\emptyset$, so there is some $q_0\in\mathbb{Q}^+$ and $a\in A$ such that $q_0x+a\in B$, and there is some $q_1\in\mathbb{Q}^+$ and $b\in B$ such that $q_1x+b\in A$. Note that $q_0,q_1\neq 0$ because $A\cap B=\emptyset$. It is easy to see that $q_0q_1x+q_1a+q_0b\in A\cap B$, so $A\cap B\neq\emptyset$, a contradiction. Therefore, $A\cup B=\mathbb{R}^+$.

ContentLicense: CC BY-SA 3.0


问:
Understanding the derivation of the PDF of a two-point mixed distribution from its CDF

A random variable $X$ has the cumulative distribution function
$$\begin{cases}
0 & x<1 \\
\dfrac{x^2-2x+2}{2} & 1 \le x<2 \\
1 & x\ge 2
\end{cases}$$
Calculate $E[X]$.

The answer uses the following pdf to get $E[X]$:
$$\begin{cases}
\dfrac{1}{2} & x=1 \\
x-1 & 1 < x<2 \\
0 & \text{otherwise}
\end{cases}$$

The book I am reading doesn't have much (any!) information on two-point mixed distributions and I want to make sure that I understand how this PDF was derived from the CDF. My understanding so far is that firstly, we need to figure out from the CDF that $X$ follows a mixed distribution. Note that $F(1) = 1/2 \ne 0$, which indicates that there is a jump in the CDF of $X$ at $x=1$. Since $X$ is continuous from $1<x<2$, $X$ must follow a two-point mixed distribution. Now, the magnitude of the jump in the graph is $\frac{1}{2}-0 = \frac{1}{2}$ which gives $f(x) = \frac{1}{2}$ if $x=1$. The rest of the pdf can be obtained using routine computations. Finally, we need to compute the probabilistic "weights" to get to the CDF. Clearly, the "weight" for the discrete part is $\frac{1}{2}$, so the "weight" for the continuous part must be $1-\frac{1}{2} = \frac{1}{2}$, and we have

$$E[X] = \int_1^2 (x-1) \cdot x dx + \text{Weight for the discrete part} \cdot 1 = \int_1^2 x(x-1) dx + P[X=1] \cdot 1$$
$$= \int_1^2 x(x-1) dx + \frac{1}{2}$$

which gives the correct answer. However, I am more interested in learning whether my process for obtaining the answer (which was based mostly on deduction and intuition) is correct. Can someone please critique my post? Thanks!

答:
Your intuition is right (ignoring the measure-theoretic technicalities), and will serve in the vast majority (if not all) situations you will run into with actuarial exam problems.

An alternate solution to consider is as follows: let $X$ be a non-negative random variable (i.e., $\mathbb{P}(X \geq 0) = 1$) with CDF $F(x)$. Let $a \geq 0$ be  the largest value such that $\mathbb{P}(X \geq a) = 1$. Then assuming it exists, we have
$$\mathbb{E}[X^k] = k\left\{a^k + \int_{a}^{\infty}x^{k-1}\left[1-F(x) \right]\text{ d}x\right\}$$
This formula works whether your random variable $X$ is discrete, continuous, or mixed. It is a consequence of the "Darth Vader Rule" coined by Muldowney, Ostaszewski and Wojdows (2012).

One can observe through the CDF that $a = 1$ is the largest value such that $\mathbb{P}(X \geq 1) = 1$. Thus,
$$\begin{align}
\mathbb{E}[X^1] &= \mathbb{E}[X] \\
&= 1\left\{1^1 + \int_{1}^{\infty}x^{1-1}\left[1-F(x) \right]\text{ d}x\right\} \\
&= 1 + \int_{1}^{\infty}\left[1 - F(x)\right]\text{ d}x\text{.}
\end{align}$$
We next observe that
$$1 - F(x) = \begin{cases}
1, & x < 1 \\
1 - \dfrac{x^2 - 2x + 2}{2}, & 1 \leq x < 2 \\
0, & x \geq 2
\end{cases} = \begin{cases}
1, & x < 1 \\
\dfrac{- x^2 + 2x}{2}, & 1 \leq x < 2 \\
0, & x \geq 2
\end{cases}$$
thus
$$\int_{1}^{\infty}\left[1 - F(x)\right]\text{ d}x = \dfrac{1}{2}\int_{1}^{2}(- x^2 + 2x)\text{ d}x = \dfrac{1}{3}$$
leading to
$$\mathbb{E}[X] = 1 + \dfrac{1}{3} = \dfrac{4}{3}\text{.}$$

The formula I provided above is often written more compactly as
$$\mathbb{E}[X^k] = \int_{0}^{\infty}kx^{k-1}[1 - F(x)]\text{ d}x$$
and you can show this will yield an equivalent answer to the above. However, note this only applies to non-negative random variables.


问:
Conditional expectation of multiple random variables
We were given the following question. If N is the number of customers entering a store, and $X_{i}$ is the amount of money the $i^{th}$ customer spends and each of the $X_{i}s$ are identically distributed independent random variables, what is the expected value of T, the total amount of money spent? I am not really sure how to approach this question, except conditioning on the fact that $E[T]= E[\sum_{i=1}^{N}X_{i}]$. I know there are 2 random variables here, but where to go on from there?

答: It is assumed here that $N$ is independent of th $X_i$'s. Conditioning on $N$ we have $E[T]=\sum P(N=n)E[\sum\limits_{i=1}^{n} X_i]=\sum P(N=n)(nE[X_i])=E[X_1] \sum\limits_{i=1}^{n} nP(N=n) =E[X_1]E[N]$ since $E[X_i]=E[X_1]$ for all $i$.


答:
If $N$ is a given number,

For linearity of expectation

$$E[X_1+X_2+ \dots +X_n]=E[X_1]+E[X_2]+\dots+E[X_n]=nE[X_1]$$

If $N$ is a rv, first assume independence between $N$ and $X_i$'s then condition the expectation using

$$E(\Sigma_{i=1}^{N}X_i)=E[E(\Sigma_{i=1}^{N}X_i|N)]=E[N E(X_1)]=nE(X_1)$$

Where I set $n=E(N)$

Title: Conditional expectation of multiple random variables

Body: We were given the following question. If N is the number of customers entering a store, and $X_{i}$ is the amount of money the $i^{th}$ customer spends and each of the $X_{i}s$ are identically distributed independent random variables, what is the expected value of T, the total amount of money spent? I am not really sure how to approach this question, except conditioning on the fact that $E[T]= E[\sum_{i=1}^{N}X_{i}]$. I know there are 2 random variables here, but where to go on from there?


问:
Intersection of subspaces: Does there exist for a subspace a non trivial subspace such that that their intersection has dimension 1?
More specifically: Let $V$ be an $n$ dimensional vector space over a finite field $F_{q}$. Let $S\subset V$ be a $k$-dimensional ($k> 1$) subspace. Does there exist an $m$-dimensional subspace $W\subset V$, with $m=\dim(W)>1$ such that $\dim(S\cap W)=1$?

答: Given a vector basis of V, and considering the subbasis that spans the subspace $S$ (S is not all V) then it is enough to take the basis of V, and substract all the vectors except 1 of those that span $S$ (considering the subbase for $S$ formed by vectors of the base for $V$). Then the resulting spanned subspace is the $W$ i have been looking for. It is clear that $W$ is not unique.


问:
Calculating the integral $\int_0^\infty \frac{\cos x}{1+x^2}\, \mathrm{d}x$ without using complex analysis
Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form?
$$\int_0^\infty\frac{\cos x}{1+x^2}\,\mathrm{d}x$$

答:
I am surprised nobody has written about Schwinger's method, which can be used to evaluate loop integrals in quantum field theory, and which nicely computes this integral in combination with Glasser's master theorem. First notice that the integral can be re-written as
$$\int_0^\infty \cos(x)\int_0^\infty e^{-u(1+x^2)}\,du\,dx.$$
Now, switch the order of integration and extend the range of one of the integral to the entire real line, gaining  a factor of $1/2$:
$$\frac{1}{2}\int_0^\infty e^{-u}\int_{-\infty}^\infty e^{-x^2+ix}\,dx\,du.$$
Complete the square in the exponential, and evaluate the now Gaussian integral (use the analyticity of $e^{-z^2}$ to bring the line of integration back to the real line) to obtain
$$\sqrt{\pi}\int_0^\infty \frac{du}{2\sqrt{u}}\,\exp-\left(u-\frac{1}{4u}\right).$$
The time is ripe for a substitution $x=\sqrt{u}$, bringing us to
$$\frac{\sqrt{\pi}}{2}e^{-1}\int_{-\infty}^\infty dx\, \exp-\left(x-\frac{1}{2x}\right)^2.$$
Glasser's master theorem  now tells us how to compute this last integral, reducing it to a Gaussian, and the original integral is just
$$\frac{\pi}{2}e^{-1}.$$


问:
Finding a subspace of orthogonal projection
I would like some help solving this question

P is an orthogonal projection that is defined by:

$$P:\Bbb R^3\to \Bbb R^3$$
$$P(v)=1/180\begin{pmatrix}144&72&0\\72&36&0\\ 0&0&180\end{pmatrix}·v$$
How do I find subspace $W$ so that $$P=P_w, \quad ?$$
I've searched everywhere online but i can't seem to find a solution to it.

答:
First, I checked and, yes, the determinant of that matrix is 0 so it does map R^3 onto a proper subspace of dimension 2, or 1.
Let $v= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$ be a vector in $R^3$.

Then $Pv= \frac{1}{180}\begin{bmatrix}144 & 72 & 0 \\ 72 & 36 & 0 \\ 0 & 0 180 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \frac{1}{180}\begin{bmatrix}144x+ 72y \\ 72x+ 36y \\ 180z\end{bmatrix}$

If x= y= 0 and z= 1, that is $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ so that is one basis vector.

If x= z= 0 and y= 1, that is $\begin{bmatrix}\frac{72}{180} \\ \frac{36}{180} \\ 0 \end{bmatrix}= \begin{bmatrix}\frac{2}{5} \\ \frac{1}{5} \\ 0 \end{bmatrix}$.  That is not a multiple of $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ so it is another basis vector.

We can now say that, since the maximum dimension of the subspace was 2, these two vectors form a basis.

If you were wondering, yes, we could also take y= z= 0 and x= 1 to get
$\begin{bmatrix}\frac{144}{180} \\ \frac{72}{180} \\ 0 \end{bmatrix}= \begin{bmatrix}\frac{4}{5} \\ \frac{2}{5} \\ 0 \end{bmatrix}$.  But $\begin{bmatrix}\frac{4}{5} \\ \frac{2}{5} \\ 0 \end{bmatrix}= 2 \begin{bmatrix}\frac{2}{5} \\ \frac{1}{5} \\ 0 \end{bmatrix}$ so that is not an independent vector.


问:
Physical representation of volume to surface area
I was looking at this XKCD what-if question (the gas mileage part), and started to wonder about the concept of unit cancellation. If we have a shape and try to figure out the ratio between the volume and the surface area, the result is a length. For example, a sphere of radius 10cm has the volume of $\approx 4118 cm^3$ and an area of $\approx 1256 cm^2$. Therefore, the volume : surface area is $\approx 3.3 cm$.

My question is: what is the physical representation of length in this ratio?

答:
In this particular case, since $V = \frac43 \pi R^3$ and $A=4\pi R^2$, the ratio $\frac{V}{A}=\frac{R}{3}$.

In general, dimensional analysis (unit matching) is a tricky beast to deal with. It is usually only a valid approach to a problem if you are certain (or nearly certain) that you have identified all the variables which can possibly affect the problem being "solved". It should also be noted that any equations outputted are only valid up to numerical constants, which are hard to determine. The wikipedia page has a reasonably good (though quite dense) description of the ideas behind it:

http://en.wikipedia.org/wiki/Dimensional_analysis#A_simple_example:_period_of_a_harmonic_oscillator

A good example, commonly used, of the process working well is when it is applied to the simple pendulum. In general though, it can produce some unusual results and can't always be trusted. Since this question was inspired by xkcd, I think it's fair to link to this for an example:

http://xkcd.com/687/


问:
Extend a set to a basis for the kernel of a linear transformation
Calculate the kernel of the linear transformation $T: \mathbb{R}^6 \rightarrow \mathbb{R}^2$ given by:

$$T(x_1,...,x_6) = (x_1-x_2+2x_4-3x_5+x_6,2x_1-x_2-x_3+3x_4-4x_5+4x_6)$$

By definition, this we have that $ker(T) = \{(x_1,...x_6) \in \mathbb{R}^6\ |$ $x_1-x_2+2x_4-3x_5+x_6=0 $ $\text{and}$  $2x_1-x_2-x_3+3x_4-4x_5+4x_6=0  \}$

In this thread:
Calculate the kernel of the linear transformation $T: \mathbb{R}^6 \rightarrow \mathbb{R}^2$

Somebody helped show me to calculate that $$\ker(T)=\{(a-b+c-3d,a+b-2c-2d,a,b,c,d)\mid a,b,c,d\in\Bbb R\}.$$

My question is then this: how do I extend this set of 2 vectors: $\{(1,0,1,1,1,0),(0,-1,0,1,1,0) \}$ to a basis for $Ker(T)$?

Thank you

答:
Those two vectors have two things in common, besides being elements of $\ker(T)$:

the sixth coordinate is equal to $0$;

the fourth and the fifth coordinates are equal.

So, add to them an element of $\ker(T)$ for which the first condition holds but not the second one (such as $(-1,1,0,1,0,0)$) and an element of $\ker(T)$ for which the second condition holds but not the first one (such as $(-3,-2,0,0,0,1)$) and then you will have $4$ linearly independent elements of $\ker(T)$, that is, you will have a basis of $\ker(T)$.


问:
Need help finding $E(X^2)$ for geometric distribution?
Michael plays a random song on
his iPod. He has $2,781$ songs, but only one favorite song. Let X be the number
of songs he has to play on shuffle (songs can be played more than once) in order
to hear his favorite song.

a) find $E(X)$

b) find $E(X^2)$

a is easy, just use $1/p$ where $p = 1/2781$ to find that $E(X)= 2781$

What I struggle with is b)

What is the formula for $E(X^2)$ for a geometric distribution?

Is it simply $E(X^2) = \cfrac{1}{p^2}$?

This doesn't seem right to me, but I can't find anything in my book that would give me $E(X^2)$, can someone show me what the formula is for $E(X^2)$ for a geometric distribution and the derivation if possible?

答:
$$V(X)=E(X^2)-E^2(X)$$

thus

$$E(X^2)=V(X)+E^2(X)$$

If you do not know the variance's formula you have to calculate $E(X^2)$ with its definition

$$E(X^2)=\Sigma_x x^2pq^{x-1}=\Sigma_x [x(x-1)+x]pq^{x-1}=\Sigma_x x(x-1)pq^{x-1}+E(X)=\dots=\frac{2q}{p^2}+\frac{1}{p}$$

where $q=1-p$

the proofs of the involved series are not difficult so I leave you this as an exercise


问:
In Cantor's Theorem, can the diagonal set D be empty?
Cantor's Theorem:

For any set $A, |P(A)| > |A|$. That is, the power set of any set $A$ has a strictly greater cardinality than set $A$.

Proof of not surjection:

Suppose that there is an onto $f : A → P(A)$. Let $D$ be defined as follows:

$D = \{x ∈ A\ |\ x \notin f(x)\}$.

Note that, $D ⊆ A$, hence $D ∈ P(A)$.
Since $f$ is onto, then there is an $x ∈ A$ such that $f(x) = D$. Now either $x ∈ D$ or $x \notin D$. If
$x ∈ D$, then $x \notin f(x)$ by definition of $D$. But $f(x) = D$, so $x \notin D$, a contradiction. If $x \notin D$, then $x ∈ f(x)$. But $f(x) = D$, so $x ∈ D$, a contradiction. In any case, a contradiction arises, hence the assumption that there is an $x ∈ A$ such that $f(x) = D$ is false. Hence $f$ cannot
be onto.

My Question:

What if $f(x)=\{x\}$? Then wouldn't $D$ be empty? Please tell me why this won't work.

答:
I think this is a situation where reframing the argument helps clarify it: while the diagonal argument is generally presented as a proof by contradiction, it is really a constructive proof of the following result:

For every set $A$ and function $f: A\rightarrow\mathcal{P}(A)$, there is some $X\subseteq A$ such that $X\not\in ran(f)$.

Of course this implies the usual phrasing, that there is no surjection $A\rightarrow\mathcal{P}(A)$, but in my experience it's often less mysterious to students. I think this is because it frees us from having to think about objects that we're actively trying to show don't exist; in some contexts this isn't particularly bad, but once we start looking at more abstract topics like set theory it can form a real impediment to understanding at first.

In fact, the diagonal argument describes an explicit method for getting such an $X$ from a given $f$: we just set $X=\{a\in A: a\not\in f(a)\}$. Let's call this the $\color{red}{\mbox{antidiagonal set}}$ associated to $f$.

Now let's think about things from this perspective. Fixing a set $A$, consider the "singletoning" function: $$f: a\mapsto\{a\}.$$ When we construct the corresponding antidiagonal set we get $\emptyset$, since $$\{a\in A: a\not\in f(a)\}=\{a\in A: a\not\in \{a\}\}=\emptyset.$$ But this is fine: we clearly have $\emptyset\not\in ran(f)$ as desired, since every element of $ran(f)$ is nonempty ($ran(f)$ consists exactly of the one-element subsets of $A$, and $\emptyset$ does not have one element).

So it's entirely possible that the diagonal construction, applied to a given $f$, yields the emptyset; however, that's fine, since in that case the emptyset will be one of the things $f$ misses in its range.


答:
I'm going to assume the lower bounds of these integrals are all equal to $1$.

Obviously $\int_1^{\infty}\frac{dx}{x^{2\alpha}}$ converges when $\alpha>1/2$ by your standard $p-$test, so let's assume this statement is true when $n=k>1$. Choose $\alpha>\frac{k+1}{2}$ arbitrarily. Using the trigonometric substitution $$x_{k+1}=\tan(\theta)\sqrt{x_1^2+\dots +x_k^2}$$ it follows that for any fixed $(x_1, \ldots , x_k)\in (1,\infty)^k$ that $$\int_{1}^{\infty}\frac{dx_{k+1}}{\big(x_1^2+\dots+x_k^2+x_{k+1}^2\big)^{\alpha}}=\frac{1}{\big(x_1^2+\dots+x_k^2\big)^{\alpha-\frac{1}{2}}}\int_{\arctan\Big(1/\sqrt{x_1^2 + \dots + x_k^2}\Big)}^{\pi/2}\cos^{2\alpha-2}(\theta)d\theta \leq \frac{1}{\big(x_1^2+\dots+x_k^2\big)^{\alpha-\frac{1}{2}}}$$ since $\alpha>\frac{k+1}{2}>1$. Hence $$\int_{(1,\infty)^{k+1}}\frac{dx_1 \dots dx_k dx_{k+1}}{\big(x_1^2+\dots+x_k^2+x_{k+1}^2\big)^{\alpha}}\leq \int_{(1,\infty)^k}\frac{dx_1 \dots dx_k}{\big(x_1^2+\dots+x_k^2\big)^{\alpha-\frac{1}{2}}}$$ Using the induction hypothesis the RHS converges since $\alpha-\frac{1}{2}>\frac{k}{2}$.


问:
Sorting an array to get the maximum combined sum of the differences between every two adjacent elements
This is related to this question: Stack Overflow

Problem description

We are given a sequence of n positive integers. How to sort the elements to get the maximum combined sum of the differences between every two adjacent elements.

For example, for a sequence {8, 4, 1, 2, 3} the starting difference is (8-4) + (4-1) + (2-1) + (3-2) = 9, but 14 can be reached for {4, 8, 1, 3, 2}.

The best arrangement is {4, 1, 8, 2, 3}, sum 17.

Claim

Optimal solution can always be reached by either starting with the smallest or largest element and then alternately putting currently largest/smallest elements on both sides of the sequence.

That's not all yet: for the starting sequence {1, 2, 3, 4} we would start with 4 $\longrightarrow$ 1 4 2 $\longrightarrow$ 3 1 4 2 or 1 4 2 3. So this approach is not strict enough yet.

However, if we always first put to the left and then to the right (or otherwise), we are guaranteed that the number on the left side will be smaller/bigger than the corresponding number on the right side.

I tested the solution above so I am quite sure it works.
I am interested how would you prove it.

答:
Let the given integers be $a_1,\dots,a_n$.  You can formulate the problem as a traveling salesman problem on a complete graph with $n+1$ nodes.  Besides one node for each array element, introduce a dummy node with $0$ cost to the other nodes.  Otherwise, the cost of edge $\{i,j\}$ is $-|a_i-a_j|$, and you want to find a tour of minimum cost.

If there is a counterexample to your conjectured optimal algorithm, this formulation could help you find one by solving the corresponding TSP to check optimality.


问:
Finding a certain base change matrix
Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation defined by:

$$T(a,b) = (2a+b,a-3b)$$

Let $\alpha$ be the standard ordered basis for $\mathbb{R}^2$ and let $\gamma$ be the basis $\{(1,1),(1,2)\}$. Find the matrix $A$ of $T$ relative to $\alpha$, the matrix $B$ of $T$ relative to $\gamma$, and a matrix $Q$ such that $Q^{-1}AQ = B$

My thoughts:

So to find a matrix $A$ relative to $T$, we need to find the columns of $A$. The image of each basis element of $\alpha$ under $T$ wil be the columns of $A$. The same goes for finding a matrix of $T$ relative to $\gamma$. Is this correct?

I do not, however, know how to find $Q$ and would appreciate some help. Thank you.

Title: Finding a certain base change matrix

Body: To get $Q$ you must express old basis in new one:  $$(1,0)=2(1,1)-(1,2)$$ and $$(0,1)= -(1,1)+(2,1)$$ so $$Q^{-1}=\pmatrix{2&-1\\-1&1}$$


问:
What is a good reference to read about "the degree of an algebraic subset of $\mathbb{P}^n$ is the number with a generic linear subspace"?
I read already a lot that the degree of a closed algebraic subset of $\mathbb{P}^n$ with dimension $k$ can be seen as the number of intersection points with a generic linear subspace of codimension $k$, however I have never seen a detailed development of such a statement. I'm also particularly interested in a much weaker statement: how to see that there even exists a linear subspace of codimension $k$ that intersects the closed algebriac set in a finite number of points? What would be a good reference for this, so that I can read more in detail about this fact?

答:
Let me give a proof for the weaker statement that you ask about. As with many similar statements about incidences of subvarieties in projective space, one can use a so-called incidence correspondence to linearise the problem in some sense.

In my answer I use $\mathbf G(n-k,n)$ to denote the Grassmannian of $n-k$-dimensional linear subspaces in $\mathbf P^n$.

So let $Z \subset \mathbf P^n$ be a closed algebraic subset of dimension $k$. It is enough to consider the case where $Z$ is irreducible. Consider the incidence correspondence

$$ I = \left\{ (L, z) \mid L \in \mathbf G(n-k,n), z \in Z \cap L \right\} $$

Now two subvarieties of $\mathbf P^n$ of complementary dimension must intersect (as proved e.g. in Chapter 1 of Hartshorne). So for any $L \in \mathbf G(n-k,n)$, the intersection $Z \cap L$ is nonempty. Put differently, the first projection map $\pi_1: I \rightarrow \mathbf G(n-k,n)$ is surjective.

We want to prove that a general fibre of the first projection $\pi_1: I \rightarrow \mathbf G(n-k,n)$ is finite. Since this map is surjective, it is enough to prove that $I$ is irreducible and that $I$ has the same dimension as $\mathbf G(n-k,n)$.

To see these, consider now the second projection $\pi_2: I \rightarrow Z$. For a point $z \in Z$, the fibre $\pi_2^{-1}(z)$ is just the set of $n-k$-dimensional linear subspaces  in $\mathbf P^n$ which pass through $z$. This is isomorphic to the Grassmannian $\mathbf G(n-k-1,n-1)$. In particular every fibre is irreducible of dimension $k(n-k)$. This proves two things:

(a) since $\pi_2$ is a surjective morphism onto the irreducible variety $Z$ with all fibres irreducible, the set $I$ is also irreducible. (This follows from results of Chapter of Shafarevich, if I recall correctly).

(b) the dimension of $I$ equals

$$\operatorname{dim} Z + \operatorname{dim}(\pi_2^{-1}(z))\\
= k + k(n-k)\\
= k(n-k+1) $$

On the other hand, the variety $\mathbf G(n-k,n)$ has dimension $(n-k+1)(n-(n-k))=k(n-k+1)$. So the dimensions of the source and target of $\pi_1$ are equal, as required.

In fact one can use a similar approach to give a full proof that the definition you are asking about does in fact make sense. The basic point is to show that for any finite morphism $f:Y \rightarrow X$ of projective varieties, there is a nonempty open set $U \subset X$ for which the cardinality of the fibres of $X$ is constant. Applying this with $Y=I$ and $X=\mathbf G(n-k,n)$, we get the statement you need. Again if I recall correctly, the relevant results are in Chapter 1 of Shafarevich, but I don't have a copy to hand.


问:
A slight variation on the Pythagorean theorem
Are there any solution to $$a^2+b^2=c^2+1 $$ where $a \not=0$ and $b\not=0$

This is a follow-on from a previous question For what $n$ and $m$ is this number a perfect square?, which ultimately boils down to the above. Any help would be greatly appreciated.

答:
More generally, the complete answer to,

$$x_1^2+x_2^2=y_1^2+y_2^2$$

was known way back to Euler as,

$$(ac+bd)^2 + (ad-bc)^2 = (ac-bd)^2 + (ad+bc)^2$$

Thus, your question is equivalent to equating one term to $\pm1$. The situation is very reminiscent of a Pell equation and, in fact, a subset of solutions can be given as such like,

$$(dx)^2 + (d^2y^2-1)^2 = (d^2y^2+d)^2 + 1$$

where $x,y$ satisfy the Pell equation $x^2-2(d+1)y^2 = 1$. (Avoiding certain $d$.)  This is by Gerardin. There are many other solutions, some are polynomial parameterizations. I've collected some here as Form 4.


答:
Your thoughts are correct, however, for the standard basis, you can deduce the matrix $A$ directly from the equations:
$$A=\begin{bmatrix}1&1\\1&-3\end{bmatrix}.$$

As to the change of basis matrix fromù  the standard basis to the basis $\gamma$, their columns are the coordinates of the vectors of $\gamma$ in the standard basis, namely
$$Q=\begin{bmatrix}1&1\\1&2\end{bmatrix}.$$
Last, there's a formula to deduce $T$ from $A$:
$$T=Q^{-1}AQ.$$

Title: Finding a certain base change matrix

Body:
Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation defined by:

$$T(a,b) = (2a+b,a-3b)$$

Let $\alpha$ be the standard ordered basis for $\mathbb{R}^2$ and let $\gamma$ be the basis $\{(1,1),(1,2)\}$. Find the matrix $A$ of $T$ relative to $\alpha$, the matrix $B$ of $T$ relative to $\gamma$, and a matrix $Q$ such that $Q^{-1}AQ = B$

My thoughts:

So to find a matrix $A$ relative to $T$, we need to find the columns of $A$. The image of each basis element of $\alpha$ under $T$ wil be the columns of $A$. The same goes for finding a matrix of $T$ relative to $\gamma$. Is this correct?

I do not, however, know how to find $Q$ and would appreciate some help. Thank you.


答:
A common trick is to use the variance to back out $E(X^2)$ as follows:
$$Var(X)=E(X^2)-(E(X))^2 \iff$$
$$E(X^2)=Var(X)+(E(X))^2$$

For a random variable $X$ that follows a geometric distribution, we have mean $E(X)=\frac{1}{p}$ and $Var(X)=\frac{1-p}{p^2}$ for some success probability p. Hence,
$$E(X^2)=\frac{1-p}{p^2}+(\frac{1}{p})^2=\frac{2-p}{p^2}$$

Title: Need help finding $E(X^2)$ for geometric distribution?

Body:
Michael plays a random song on
his iPod. He has $2,781$ songs, but only one favorite song. Let X be the number
of songs he has to play on shuffle (songs can be played more than once) in order
to hear his favorite song.

a) find $E(X)$

b) find $E(X^2)$

a is easy, just use $1/p$ where $p = 1/2781$ to find that $E(X)= 2781$

What I struggle with is b)

What is the formula for $E(X^2)$ for a geometric distribution?

Is it simply $E(X^2) = \cfrac{1}{p^2}$?

This doesn't seem right to me, but I can't find anything in my book that would give me $E(X^2)$, can someone show me what the formula is for $E(X^2)$ for a geometric distribution and the derivation if possible?


答:
Suppose that there is an infinite sequence of customers, but we only see the first $N$ of them. Thus, we can think of $X_i$ as defined for all $i$.

Let $I\{i \le N\} = 1$ if $i \le N$ and $0$ otherwise.

The expected money from the first $N$ of them is
$$
E\left[ \sum_{i = 1}^N X_i \right] = E\left[ \sum_{i = 1}^\infty I\{i \le N\} X_i \right]
$$

Since the expectation is linear, we can write this as:
$$
E\left[ \sum_{i = 1}^\infty I\{i \le N\} X_i \right] =
\sum_{i = 1}^\infty E[ I\{i \le N\} X_i ]
$$

If the $N$ and the $X_i$ are independent, then we have $E[ I\{i \le N\} X_i ] = E[I\{i \le N\}]E[X_i]$. Then:
$$
\sum_{i = 1}^\infty E[ I\{i \le N\} X_i ] =
\sum_{i = 1}^\infty E[I\{i \le N\}]E[X_i] =
E[X_i]\sum_{i = 1}^\infty E[I\{i \le N\}] =
E[X_i] E[N]
$$


问:
Study of Set theory: Book recommendations?
Can you suggest a good book for set theory?

I have just started reading about Group theory and want to learn set theory on my own.

Thanks in advance

答:
A couple of ‘entry level’ treatments that can be confidently recommended.

Herbert B. Enderton, Elements of Set Theory (Academic Press, 1997) is particularly clear in marking off the informal development of the theory of sets, cardinals, ordinals etc. (guided by the conception of sets as constructed in a cumulative hierarchy) and the formal axiomatization of ZFC. It is also particularly good and non-confusing about what is involved in (apparent) talk of classes which are too big to be sets – something that can mystify beginners. It is written with a certain lightness of touch and proofs are often presented in particularly well-signposted stages. The last couple of chapters or so perhaps do get a bit tougher, but overall this really is quite exemplary exposition.

Derek Goldrei, Classic Set Theory: For guided independent study (Chapman & Hall/CRC 1996) is written by a staff tutor at the Open University in the UK. It is as you might expect extremely clear, it is quite attractively written (as set theory books go!), and is indeed very well-structured for independent reading. The coverage is very similar to Enderton’s, and either book makes a fine introduction (but for what it is worth, I prefer Enderton ).

For more see entries on set theory in this Guide to reading on logic.


问:
How to get the maximum perimeter in a set of straight squares?
The problem is as follows:

Rudy has a toy set which consists of $11$ squares. All of them are equal
in size. The lenght of each edge is equal to $\textrm{1 cm.}$ Assuming
he's been given the instruction to make with those, plane figures that
are made up of squares joined by a complete side. If at the end of the
day he succeeds and builds one such figures whose perimeter is
maximum. What is the value of such perimeter?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{24 cm}\\
2.&\textrm{26 cm}\\
3.&\textrm{28 cm}\\
4.&\textrm{22 cm}\\
\end{array}$

I'm not very sure on this question. The situation is that the kid has been given the task to make different figures using 11 squares, given that these must share a full side or in other words two squares must be glued together not counting corners then he succeeds into making those figures. But it just happens that he makes one peculiar figure which has the largest possible perimeter.

But how to get that?. My first guess is that the only possible way to make a figure given those conditions is to put the squares one next to each other.

Given this it would follow that:

$1*11*2+2=24\,cm$

But I don't know if this should be the right answer?. Is it possible to make a larger figure with those requirements?. Is there any way to ensure that's the only way to maximize the perimeter?.

Can someone help me here?. Please try to include a figure in the answer I don't know but it seems that this may need it to justify the situation.

答:
There are $4 \times 11 = 44$ edges in total. To attach two squares uses up two edges. You must make $10$ attachments. That leaves
$44-2\times 10 = 24$ edges for the perimeter.

That will be the perimeter of any figure made from all the squares that does not circle back on itself.


问:
Matrix derivative is invertible.
I'm stuck with the next exercise.

Let $f,g:U\subseteq\mathbb{R}^{2}\to\mathbb{R}$ a $C^1$ functions over $U$. Consider $F:U\subseteq\mathbb{R}^{2}\to\mathbb{R}^{2}$ and $H:U\subseteq\mathbb{R}^{2}\to\mathbb{R}$ defined by $F(x,y)=(f(x,y),g(x,y))$ and $H(x,y)=||F(x,y)||^{2}$. Prove that for all $x\in U$, one of the next two conditions is not fulfilled.

$H$ attains a local maximum at $x$

The derivative of $F$ at $x$ is invertible, i.e., the matrix $DF(x)$ is invertible.

My idea was to proceed by contradiction, i.e., suposse that there exist $x_0\in U$ such that (1) and (2) holds. For this, (1) means that there exist $\varepsilon>0$ such that for all $z\in B(x_0,\varepsilon)\subseteq U$, $H(z)\leq H(x_0)$. Moreover, $\nabla H(x_0)=\overline{0}$. Here, $$\nabla H(x_0)=\left(2 f(x_0)\dfrac{\partial f}{\partial x}(x_0)+2g(x_0)\dfrac{\partial g}{\partial x}(x_0), 2 f(x_0)\dfrac{\partial f}{\partial y}(x_0)+2g(x_0)\dfrac{\partial g}{\partial y}(x_0) \right)=(0,0)$$By (2), $DF(x_0)$ is an invertible matrix. Then
$$DF(x_0)=\begin{pmatrix}
  \dfrac{\partial f}{\partial x}(x_0) & \dfrac{\partial f}{\partial y}(x_0)\\
  \dfrac{\partial g}{\partial x}(x_0) & \dfrac{\partial g}{\partial y}(x_0)
\end{pmatrix}$$
Beign invertible, then $\dfrac{\partial f}{\partial x}(x_0)\dfrac{\partial g}{\partial y}(x_0)-\dfrac{\partial f}{\partial y}(x_0)\dfrac{\partial g}{\partial x}(x_0)\neq 0$.
But, from here, I don't know how to continue. Maybe we can join the two conditions to derive a contradiction, but it's not clear to me. Any hint? Thanks for your help.

答:
You just need to manipulate the $\nabla H(x_0)$ equation as follows:

$$2f(x_0) \frac{\partial f(x_0)}{\partial x} = -2g(x_0)\frac{\partial g(x_0)}{\partial x}$$

$$2f(x_0) \frac{\partial f(x_0)}{\partial y} = -2g(x_0)\frac{\partial g(x_0)}{\partial y}$$

Divide the top equation by the bottom equation:

$$\frac{\frac{\partial f(x_0)}{\partial x}}{\frac{\partial f(x_0)}{\partial y}} = \frac{\frac{\partial g(x_0)}{\partial x}}{\frac{\partial g(x_0)}{\partial y}} $$

Then you get the contradiction when you cross multiply and compare to the 2nd equation.  You may need some special cases for when one of the partials is zero so that you aren't dividing by zero.

Title: Matrix derivative is invertible.


问:
Ideals of $\mathbb{Z}[\sqrt{-10}]$ containing $6$
I want to find all Ideals of $\mathbb{Z}[\sqrt{-10}]$ that contain $6$.

My guess is that the only Ideals are $\mathfrak{p}_{2}=(2,\sqrt{-10})$, $(2)$, $\mathfrak{p}_{3}=(3)$, $\mathfrak{p}_{2}\mathfrak{p}_{3}$, $2\mathfrak{p}_{3}$ and $(1)$.

First notice that this ring is the ring of integers of the number field $\mathbb{Q}(\sqrt{-10})$. And thus we have Unique prime ideal factorisation. If $\mathfrak{p}$ divides such an ideal I that contains $(6)$, then it also divides $6$. But from here I can’t complete the argument.

答:
It’s enough to find all the ideals of $R=\mathbb{Z}[\sqrt{-10}]/(6)$. But $R \cong\mathbb{Z}[X]/(6,X^2+10) \cong (\mathbb{Z}/6\mathbb{Z})[X]/(X^2+10)$, so by CRT $R \cong\mathbb{F}_2[X]/(X^2+10) \times \mathbb{F}_3[X]/(X^2+10) \cong \mathbb{F}_2[X]/(X^2) \times \mathbb{F}_3[X]/(X^2+1)$.

Now, the second factor is a field so has two ideals, and the first factor has three ideals, so there are six ideals in total, and you listed six distinct ones.


问:
Confusion on the definition of a neighbourhood
So I've just started learning metric spaces and I'm getting a bit confused on some of the definitions namely the definition of a neighbourhood.
I looked quite a bit online and the definitions always seem to vary my main confusion is the epsilon ball around the point x a subset of the neighbourhood or a proper subset?.
A formal definition of a neighbourhood in a metric space would be helpful
Thanks in advance.</p>

答:
A set $X$ is open if, for every point $x \in X$ in the set $X$, there exists some $\epsilon > 0$ such that the ball $B_{\epsilon}(x) = \{ y : d(x,y) < \epsilon \} \subseteq X$ is contained in $X$.

An open neighborhood of a point $x$ is any open set containing $x$. Some texts use this interchangeably with neighborhood. Others will define a neighborhood of $x$ as any set that contains an open neighborhood of $x$.

For almost all the concepts you'll see in an introductory textbook, both definitions will yield the results, and the distinction does not matter.


答: Think of the classic "Snake" game. As long as the snake does not have any part of its body adjacent to any other part, then it will be a figure with maximum perimeter. The straight line is just one option. If any part of its body does become adjacent to another part, you lose that part of the perimeter. And since all 11 pieces have to be connected, this is indeed the best you could do.</p>

Title: How to get the maximum perimeter in a set of straight squares?

Body:
The problem is as follows:

Rudy has a toy set which consists of $11$ squares. All of them are equal
in size. The lenght of each edge is equal to $\textrm{1 cm.}$ Assuming
he's been given the instruction to make with those, plane figures that
are made up of squares joined by a complete side. If at the end of the
day he succeeds and builds one such figures whose perimeter is
maximum. What is the value of such perimeter?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{24 cm}\\
2.&\textrm{26 cm}\\
3.&\textrm{28 cm}\\
4.&\textrm{22 cm}\\
\end{array}$

I'm not very sure on this question. The situation is that the kid has been given the task to make different figures using 11 squares, given that these must share a full side or in other words two squares must be glued together not counting corners then he succeeds into making those figures. But it just happens that he makes one peculiar figure which has the largest possible perimeter.

But how to get that?. My first guess is that the only possible way to make a figure given those conditions is to put the squares one next to each other.

Given this it would follow that:

$1*11*2+2=24\,cm$

But I don't know if this should be the right answer?. Is it possible to make a larger figure with those requirements?. Is there any way to ensure that's the only way to maximize the perimeter?.

Can someone help me here?. Please try to include a figure in the answer I don't know but it seems that this may need it to justify the situation.


答:
The perimeter will be

$$(11 \times 4) - 2j$$

where $j$ is the number of joins.  That is, each time one square is joined with another, 2 of the sides become internal and thus not part of the perimeter.  Since you start with 11 squares, there must be at least 10 joins.

Addendum

Responding to the OP's comment/question:

I do agree with the OP's solution, since the minimum value for
$j$ is $10$, and since $(11 \times 4) - (2 \times 10) = 24.$  I actually just made up the formula from intuition.  Suppose that two squares are joined.  Then the perimeter has changed from $8$ to $6$, because the (two) joined edges have become internal.

Imagine using 9 squares to form a $3 \times 3$ larger square.

The perimeter, which is equal to $(4 \times 3),$ can be computed by noticing
that in each row there are $2$ joins and that there are $3$ joins between both
the 1st and 2nd row and between the 2nd and 3rd row.  Therefore, the  perimeter
will equal $(9 \times 4) - \{2 \times [(3 \times 2) + (2 \times 3)]\} = 12.$

The underlying idea is merely that each join causes exactly two sides to become internal.


问:
How to solve this logarithmic equation for x
How can the following equation solved for x?

$\ln(x) - \ln(1-x) = \ln(1-y) - \ln(1-z) - \ln(v)$

I assume it simplifies to:

$\frac{x}{1-x} = \frac{1-y}{(1-z)*v} $

I have tried to solve this by factorizing the terms, but it does not lead anywhere. Any hints?

答:
\begin{align*}
\frac{x}{1-x} &= \frac{x\overbrace{-1+1}^{0}}{1-x}  \\
    &= \frac{x-1}{1-x} + \frac{1}{1-x}  \\
    &= -1 + \frac{1}{1-x}
\end{align*}
So you have
\begin{align*}
-1 + \frac{1}{1-x} &= \frac{1-y}{(1-z)v}  \\
\frac{1}{1-x} &= 1 + \frac{1-y}{(1-z)v}  \\
\frac{1-x}{1} &= \frac{1}{1 + \frac{1-y}{(1-z)v}  }  \\
x-1 &= \frac{-1}{1 + \frac{1-y}{(1-z)v}  }  \\
x &= 1 + \frac{-1}{1 + \frac{1-y}{(1-z)v}  }  \\
\end{align*}

Notice that your original equation has the constraints $x > 0$, $1-x > 0$, $1-y > 0$, $1-z > 0$, and $v > 0$.  The above introduces the constraint $1 + \frac{1-y}{(1-z)v} \neq 0$ (although this follows as a consequence of the others).


答:
The "one such figure" with maximum perimeter should obviously use all squares - otherwise you can add a square somewhere and increase the perimeter.

And you should expose as much of each square's edges as possible on that perimeter. With two joined squares, you don't have an option; both squares show three of their edges of the figure perimeter for a total of $6$ cm. However whenever you add another square, you always lose at least the side that you connect it with and one of the edges in the smaller figure. So every additional square can only gain you $2$ cm of perimeter. (and you can gain less than $2$ cm, or even lose perimeter length, if you connect a new block to more than one side of the smaller figure, such as when you "fill a gap" in some way, e.g. to make a $2\times2$ block or form a loop).

So for three squares you can get $8$ cm of perimeter; for four squares $10$ cm, etc. but not more. Most straightforwardly you can do this with the squares attached in a straight line, but this is not required - any figure that doesn't close a loop or form a $2\times2$ block should work.


问:
How can I solve this Laws of Sines problem?
This is a homework question that was set by my teacher, but it's to see the topic our class should go over in revision, etc.

I have calculated $AB$ to be 5.26m for part (a). I simply used the law of cosines and plugged in the numbers.

part (b) is the question I have been trying for quite a while. I tried to do the law of sines, to no avail. To calculate $BC$ I need the angle opposite that which I do not have (or know how to work out.) The triangle $BDC$ has a right-angle, but this does not work as $sine(90) = 1$

A step in the right direction would be more beneficial than a full answer.

答:
steps to solve this problem

1)

use cosine formula in $\Delta ABD$ to find side $AB$.

$AB^2=88.3^2+91.2^2-2\times88.3\times 91.2\times \cos2.8^\circ$

$AB=5.257$m

2)

find $\angle ABD$ in above triangle using same cosine formula.

$\cos \angle ABD=\dfrac{5.257^2+88.3^2-91.2^2}{2\times 5.257\times 88.3}$

$\angle ABD=122.069^\circ$

3)

Now find out $\angle BDC=\angle ABD-90^\circ=32.069^\circ$

4)

Now we can use simple trigonometry to solve

$\dfrac{BC}{DC}=\tan 32.069^\circ\implies DC=\dfrac{BC}{\tan 32.069^\circ}$

and in $\Delta ACD$

$\dfrac{AB+BC}{DC}=\tan (32.069^\circ+2.8^\circ)$

$\dfrac{5.257+BC}{DC}=\tan 34.869^\circ$

${5.257+BC}={DC}\tan 34.869^\circ$

${5.257+BC}=\dfrac{BC}{\tan 32.069^\circ}\tan 34.869^\circ$

$BC=46.879$m and $AB=5.257$m


问:
Prove that the image of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .
let $G$ and $H$ be groups and let $\varphi : G \to H$ be  a homomorphism.Prove that the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ .

My attempt : To  show the image  of $\varphi$ , $\varphi(G)$ is a subgroup of $H $

It is given that $\varphi :G \to H$ is  a homomorphism and  $\varphi (G)$ denote the image of $\varphi$.

Now we will used the lagrange theorem's  because  $\varphi$ is homomorphism

let $ y\in \varphi(G)$ and  $g\in G$.Then  $y^{-1} = \varphi(g)^{-1}=\varphi(g^{-1})\in \varphi (G)$

$y^{-1}= \varphi(g^{-1}) $ This implies that  $\varphi(G)$ is satisfying  the lagrange theorem

Therefore $\varphi (G)$ is a  subgroup of $H$

答:
How about using the subgroup criterion: $\varphi(G)\ne\emptyset$ and $x,y\in \varphi(G)\implies xy^{-1}\in\varphi(G)$?

Or, you could use the first isomorphism theorem, which tells us that the image is isomorphic to $G/\rm{ker}\varphi$.  $\ker\varphi$ is a normal subgroup of $G$, so the quotient is indeed a group.


答:
$$\frac{x}{1-x} = \frac{1-y}{(1-z)v} \implies \frac{x}{1-y}=\frac{1-x}{(1-z)v} = \frac{x+1-x}{1-y+(1-z)v}=\frac{1}{1-y+(1-z)v}\\
\implies x = \frac{1-y}{1-y+(1-z)v}$$

where we used the fact that if $\frac ab = \frac cd$ then
$$\frac ab = \frac cd = \frac{a+c}{b+d}\tag1$$

Title: How to solve this logarithmic equation for x

Body:
How can the following equation solved for x?

$\ln(x) - \ln(1-x) = \ln(1-y) - \ln(1-z) - \ln(v)$

I assume it simplifies to:

$\frac{x}{1-x} = \frac{1-y}{(1-z)*v} $

I have tried to solve this by factorizing the terms, but it does not lead anywhere. Any hints?


问:
The left cosets of $H$ in $G$ partition $G$

Let $G$ be a group and $H$ a subgroup. Then the left cosets of $H$ in $G$
partition $G$. In particular, $(1)$ each $a$ ∈ G is in exactly one left coset, namely $aH$, and $(2)$ if $a, b \in G$, then either $aH = bH$ or $aH \cap bH = \emptyset $.

The part $(2)$ is done. My problem is in part $(1)$, I tried this but not really sure:

Let $a\in G$, we have that $e\in H$, so $a\in aH$, since $a=ae$. This shows that $a$ lies in some left coset, namely $aH$.

Now if $a\in aH$ and $a\in bH$, we have that $a=ae=abh$, so $bh=e$ and thus $a$ lies in exactly one left coset.

Am I right?

答:
Assuming that you proved (2) I proceed:

$\mathbf{Theorem  1:}$ For $a,b \in G$ prove that $aH=bH$ iff $a^{-1}b \in H$.

$\mathbf{Theorem  2:}$ For $a,b \in G$ prove that $b \in aH$ iff $a^{-1}b \in H$

Then the following conditions are equivalent:
$$b \in aH \equiv a^{-1}b \in H \equiv aH=bH$$ Since $e \in H, a=ae \in aH$.
Let $a \in bH$. Then $aH=bH$. Thus $a$ belongs to exactly one left coset.


问:
Evaluating $\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$
I am trying to evaluate the following integral:

$$\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$$

I thought about using the substitution $\beta \doteq \tfrac{(t-a)}{C}$ so the integral would become:

$$\int_{0}^{\infty} \beta^2 \cdot \exp (- \beta^2 ) \cdot \exp(i w (\beta C - a)) \ C \ d\beta$$

However, I don't know how to evaluate this either.

Any idea on how to do this?

EDIT: As Caran-d'Ache pointed out, I did in fact lose a minus! I'm still just learning Mathematica, how did you enter the integral?

答:
So one has the following integral:$$\mathrm{I}=\int_{0}^{\infty} \beta^2e^{- \beta^2} e^{i w (\beta C - a)} \ C \ d\beta$$
$$\mathrm{I}=Ce^{-i w a}\int_{0}^{\infty} \beta^2 e^{- \beta^2}   e^{i w \beta C } d\beta$$
Let's for simplicity set $Ce^{-i w a}=\alpha$, $w C=s$, so:
$$\mathrm{I}=\alpha\int_{0}^{\infty} \beta^2 e^{- \beta^2}   e^{i \beta s } d\beta$$
Joining the exponent terms and completing the square in the power of the resulting exponent:
$$\mathrm{I}=\alpha\int_{0}^{\infty} \beta^2 e^{- (\beta-\frac{i s}{2})^2}   e^{-\frac{s^2}{4}} d\beta$$
Again $\tilde{\alpha}=\alpha \ e^{-\frac{s^2}{4}}$ and changing the variable (and the limits of integration) from $\beta$ to $t=\beta-\frac{i s}{2}$ one can obtain:
$$\mathrm{I}=\tilde{\alpha}\int_{-\frac{i s}{2}}^{\infty-\frac{i s}{2}}\left(t+\frac{i s}{2}\right)^2 e^{- t^2}   dt$$
Expanding the terms in the brackets:
$$\mathrm{I}=\tilde{\alpha}\int_{-\frac{i s}{2}}^{\infty-\frac{i s}{2}}\left(t^2+i t s-\frac{s^2}{4}\right) e^{- t^2}=\frac{\tilde{\alpha}}{8} i \left(2 s-\sqrt{\pi } e^{-\frac{s^2}{4}} \left(s^2-2\right) \left(\text{erfi}\left(\frac{s}{2}\right)-i\right)\right)$$
where $ \text{erfi}$ - is the imaginary error function. After that one should assume all the substitutions what were made.</p>


问:
If conic is represented by $21x^2 -6xy +29y^2 +6x-58y-151=0$, then find the centre, length of axes and eccentricity
Using partial derivatives, I found centre of the conic as $(0,1)$ and I think the conic represents an ellipse. But I am not able to find the rest of the answers.

答:
Since Quanto has given an answer using calculus, I'll give an answer using linear algebra alone. First, we may express this conic in terms of matrix multiplication:

$$21x^2 -6xy +29y^2 +6x-58y-151=v^\top M v=0$$
where $v=(x,y,1)^\top$ and

$$M=M^\top=\begin{pmatrix} 21 & -3 & 3 \\ -3 & 29 & -29 \\ 3 & -29 & -151\end{pmatrix}.$$

The presence of nonzero off-diagonal entries in the last row and last column signals that it's not centered, since these lead to the linear terms in the equation. These may be eliminated by adding the second row to the third and then the second column to the third. These are jointly implemented by a similarity transformation with the appropriate upper-triangular matrix $U$:

$$M'=U^\top M U = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix} \begin{pmatrix} 21 & -3 & 3 \\ -3 & 29 & -29 \\ 3 & -29 & -151\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 21 & -3 & 0 \\ -3 & 29 & 0 \\ 0 & 0 & -180\end{pmatrix}.$$

We can now diagonalize $M'$ by hand. By inspection, it has an eigenvalue $-180$. The remaining $2$-by-$2$ block has determinant $600$ and trace $50$, and thus we deduce that the remaining eigenvalues are $30,20$. Computing the eigenvectors by inspection, we deduce that $M''$ is diagonalized as $M'=R \Lambda R^{-1}$ where $\Lambda=\operatorname{diag}(30,20,-180)$ and
$$R=\begin{pmatrix} \frac{1}{\sqrt{10}}  & \frac{3}{\sqrt{10}}&0 \\ -\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0 \\ 0 & 0 & 1\end{pmatrix}.$$
Note that the columns of $R$ are the orthonormal eigenvectors of $M'$ and thus $R^\top=R^{-1}$. Putting these together, we have
\begin{align}
v^\top M v
&=v^\top (U^{-1})^\top M' U^{-1} v\\
&=(U^{-1}v)^\top  R  \Lambda R^{-1} U^{-1} v\\
&=(R^{-1}U^{-1}v)^\top \Lambda (R^{-1} U^{-1})v\\
&=u^\top \Lambda u
\end{align}
where $u=R^{-1}U^{-1} v=(UR)^{-1}v$ gives the transformation to a coordinate system in which the ellipse is aligned with the axes. Explicitly, we have
$$u=R^{-1}U^{-1}v
=\begin{pmatrix} \frac{1}{\sqrt{10}}  & -\frac{3}{\sqrt{10}}&0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix} x \\ y \\ 1\end{pmatrix}
=\begin{pmatrix} \frac{1}{\sqrt{10}}(x-3y+3) \\ \frac{1}{\sqrt{10}}(3x+y-1) \\ 1\end{pmatrix}.$$
We may thus write $u=(x',y',1)^\top$. In terms of the new coordinates, we have $$30x'^2+20y'^2=180\implies\frac{x'^2}{6}+\frac{y'^2}{9}=1.$$ The lengths of the axes are evidently $\sqrt{6}$ and $3$ respectively, and the eccentricity is $\sqrt{1-6/9}=1/\sqrt{3}$. Since the new coordinates are obtained by first translating and then rotating the system, this transformation doesn't change the dimensions and therefore these values also describe the original ellipse. In addition, we may note that the point $(x,y)=(0,1)$ is mapped to $(x',y')=(0,0)$ which we recognize as the center of the ellipse in these coordinates. Hence $(0,1)$ is indeed the center of the original ellipse.

P.S. I can't resist doing a little bit of calculus. Writing $f(x,y)=v^\top M v$ as above, we may write $f_x=2 v^\top M e_1$ and $f_y=2v^\top Me_2$. The condition that $f_x=f_y=0$, then, corresponds to choosing $v=(x,y,1)^\top$ such that it's orthogonal to both $Me_1=(21,-3,3)^\top$ and $Me_2=(-3,29,-29)$. This is readily accomplished by taking the cross product of these last two vectors, obtaining $$(Me_1)\times (Me_2) = (0,600,600).$$ From this we deduce that the vector $(0,1,1)$ will suffice and therefore the center is at $(0,1)$.


问:
$f:\Bbb R\mapsto\Bbb R$ such that $f(x+y)f(x-y)=2x+f(x^2-y^2)$

Find all functions $f:\Bbb R\mapsto \Bbb R$ satisfying $$f(x+y)f(x-y)=2x+f(x^2-y^2)$$

I let $a=x+y$ and $b=x-y$ and noticed that $f(a)f(b)=a+b+f(ab)$. It's obvious now that $f(n)=n+1$ satisfies, for all real $n$.

How do I show this more rigorously, and that no other functions satisfy?

Below is my slightly more rigorous approach.

I let $y=0$ so $f(x)^2=2x+f(x^2)$ and let $f(x)=p+q$ for real $p$ and $q$. We see that $$2pq+p^2+q^2=2x+f(x^2).$$ Since $2pq$ and $2x$ have the same integer coefficient, letting $p=kx$ and $q=\frac 1k$ for some $k$ implies $pq=x$ so the $2x$ terms cancel out, thus $f(x)=kx+\frac 1k$ or $$k^2x^2+\frac 1{k^2}=kx^2+\frac 1k$$ for all $x$. Since this must be valid for all $x$, it must be valid for $x=1$, so substituting that neatly removes the $x$ variable. Trivially $k\neq 0$ and upon solving for $k$, since $f:\Bbb R\mapsto\Bbb R$ then $k$ must be real, and we can thus show that $k=1$ can only satisfy, yielding $f(x)=x+1$.

However, although this approach is more rigorous, it assumes $pq=x$ out of elegant consequence. So I tried a different approach:

Let $x=y=0$, then we can see $f(0)^2=f(0)$ thus $f(0)\in\{0,1\}$. Now let $x=y$ and we have $$f(2x)f(0) = 2x+f(0).$$ This must be true for all $x$ but plugging $f(0)=0$ wouldn't make the statement true for all $x$, so $f(0)=1$ and therefore $f(2x)=2x+1$ or, through $2x\mapsto x$, we have  $f(x)=x+1$.

This approach looks good to me, but I don't know how to show no other functions exist to satisfy the problem. Since $f(x+m)-f(x)=m$ which is constant for any $x$, I thought of employing this property to attempt showing that $f$ is linear. But I think I could only pull this off if $f:\Bbb Z\mapsto \Bbb Z$ instead.

Any thoughts?

Thanks.

答: This is much simpler. Put $x=y=0$ to see that $f(0)^{2}=f(0)$. Hence $f(0)=0$ or $1$. But if $f(0)=0$ we get a contradiction by putting $y=x$ so we must have $f(0)=1$. Now put $x=y$  to get $f(2x)=2x+1$ for all $x$. Hence $f(x)\equiv x+1$.


问:
Show that the number of Sylow $p$-subgroups of $G$ is either $1$ or $p+1$.
Let $p$ be a prime and $G$ a group with $|G|=(p-1)p(p+1)$. Show that the number of Sylow $p$-subgroups of $G$ is either $1$ or $p+1$.

Hint: Write $n_p(G)=kp+1$ for some $k \in \mathbb{Z}$ and show that $p^2-1=(kp+1)l$ for some $l \in \mathbb{Z}$. Conclude that $p$ divides $l+1$ and use this to prove the wanted conclusion.

答:
Since $|G|=p(p-1)(p+1)$ and $gcd(p,(p-1)(p+1))=1$. The theorem of Sylow implies that $n_p(G)$ divides $(p-1)(p+1)$.

We have $(p-1)(p+1)=p^2-1=(kp+1)l$ implies that $p^2=klp+l+1$, we deduce that $p$ divides $klp+l+1$ and $p$ divides $l+1$. $l+1=ap, l=ap-1$.
Since $l$ and $n_p(G)$ are positive, $k,\geq 0, a>0$. Suppose that $k\geq 2$,
$p^2-1=(p-1)(p+1)\geq 2p(ap-1)$. Impossible since $ap-1\geq p-1$ and $2p> p+1$.

If $k=1$, $n_p(G)=p+1, k=0, n_p(G)=1$.


问:
Regular singular points of a Fuchsian equation
I was reading the book Painleve transcendents: The Riemann-Hilbert approach and it said the following:

Suppose that the rational matrix function $A(\lambda)$ has only simple poles. Then all singular points of equation
$$
\frac{d\Phi(\lambda)}{d\lambda} = A(\lambda)\Phi(\lambda)
$$
are Fuchsian, and the equation itself is called Fuchsian equation. Assuming that $a_1,\dots,a_m$ are all the singular points and $a_m = \infty$, then a Fuchsian equation with $m$ regular singular points can be written as
$$
\frac{d\Phi(\lambda)}{d\lambda} =\sum_{k=1}^{m-1} \frac{A_k}{\lambda-a_k}\Phi(\lambda)
$$

The book then said that

Observe that if $\sum A_k = 0$, then the infinity is actually a regular point and the equation has in fact $m-1$ singular points located on the affline part of $\mathbb{C}P^1$, the one-dimensional complex projective space(isomorphic to the Riemann sphere $\overline{\mathbb{C}}$).

and I am not sure how to observe that.

答:
To study the ODE you described
near $\lambda=\infty$, it is instructive to change variables to $t=1/\lambda$ and consider the result near $t=0$.

Writing $\Phi(\lambda)=\Psi(t)$, and noticing that
$$ \frac{d\Psi(t)}{dt}=-\,\lambda^2\frac{d\Phi(\lambda)}{d\lambda},$$
the ODE becomes
$$ \frac{d\Psi(t)}{dt}=-\,\frac{1}{t}\sum_{k=1}^{m-1}\frac{A_k}{1-a_k\,t}\Psi(t).$$
Expand the coefficient function on the right side of this equation around $t=0$ to see that the first term in the expansion of the coefficient is
$$-\,\frac{1}{t}\sum_{k=1}^{m-1}A_k.$$
So it is singular at $t=0$, unless the sum $\sum_{k=1}^{m-1}A_k$ vanishes.

If this sum does indeed vanish, then $\lambda=\infty$ is not singular, that is, the ODE is regular at infinity (in $\lambda$). (Note that infinity is not always regarded as a point, but it can be regarded as one in complex projective space $\mathbb C\mathbb P^1$. If you’re unfamiliar with such spaces, see https://en.wikipedia.org/wiki/Complex_projective_space .)


问:
Formal proof for detection of intersections for constrained segments
They told me it was off-topic at stackoverflow.
So I am trying my luck here. Yes, it's a homework, but I'm looking for some guidance (or related literature) instead of complete solutions.

Please see the drawing that accompanies the question:

I need to come up with a formal proof for the following statement:

Given an arbitrary count of line segments (S1 -- S3 in the drawing) constrained by two vertical lines (M & N) an answer whether there is at least one intersection of the segments (in the whole picture) can be given by checking each of the two segments that have their left endpoints adjacent on M in descending order (in respect to their position on line M).

For example, in drawing checking could be done in such order:

S1 and S2 (no intersection)

S2 and S3 (intersection detected)

While it seems very obvious, there shall be a formal proof for that (most likely, a simple one). Any ideas?

答: Once the segments are numerated try to use induction on $n$ which is the number of the segment, checking whether it is intersecting the next one. Good luck</p>


问:
Why is left multiplication on a group bijective?
I noticed the fact "left multiplication on a group is always bijective" is a very common argument (for instance to prove Sylow's theorem)

I don't see why left (and right) multiplication is bijective in general.
All I can see is that left multiplication is an action of a group on itself, so let $G$ be a group:
$$\cdot: G \times G \longrightarrow G$$
$$(g,g') \longmapsto g \cdot g'=gg'$$
Properties of group actions hold. I tried to prove this by contradiction but I'm not able to find any useful argument.
Could anyone enlighten me?

答:
The map $\ell_g$ defined by $\ell_g(h)=gh$ is a bijection $G\to G$. Indeed, note that the map $\ell_{g^{-1}}=\ell_g^{-1}$ because
$$ (\ell_{g^{-1}}\circ\ell_g)(h)=g^{-1}gh=h$$
and similarly
$$ (\ell_g\circ \ell_{g^{-1}})(h)=gg^{-1}h=h.$$
The analogous argument shows that the right multiplication operator is a bijection. Note that none of these operators is a group homomorphism except for example $\ell_e=\Bbb{1}_G.$


问:
Convergence of $\sum_{n=0}^{\infty}\frac{\cos(n \sqrt{2})}{\sqrt{n}}$. Is my thinking correct?
I need to examine convergence of that sum:

$$\sum_{n=0}^{\infty}\frac{\cos(n \sqrt{2})}{\sqrt{n}}$$

I think that it can be done like that:

$$ - \infty\leq \sum_{n=0}^{\infty}\frac{-1}{\sqrt{n}} \leq \sum_{n=0}^{\infty}\frac{\cos(n \sqrt{2})}{\sqrt{n}} \leq \sum_{n=0}^{\infty}\frac{1}{\sqrt{n}} \leq \infty$$

Sums on both hand sides converge to $0$.

However, I am nervous since I got $n \sqrt{2}$ under $cos$ and I not sure if my solution is correct (if I shouldn't find something more "advanced" / creative).

答:
The analysis is not correct.  First, $\sum_{n=1}^\infty \frac1{\sqrt{n}}$ diverges.  So, the bounds in the OP are useless.

We can use Dirichlet's test.  Here we note the $\frac1{\sqrt n}$ monotonically decreases to $0$ and that the sum $\sum_{n=1}^N \cos(n\sqrt 2)$ is uniformly bounded (i. e.  . there exists a number $M$ such that the absolute value of the sum is less than $M$ for all $N$).  The theorem gurantees that the series of interest converges.


问:
(Co)tangent Map of the d^th Veronese Embedding
This question is regarding Vakil 19.1 D, self-study.

What does the (co)tangent map of the $d^{th}$ Veronese embedding over an algebraically closed field $k$

$$\mathbb P^n_k \to \mathbb P^N_k$$

look like on closed points? Specifically, I would like to see the tangent map is injective (or the cotangent map is surjective). If we choose a closed point in $\mathbb P^N_k$ and look at it inside an affine patch, we get a residue field that looks like (say in $U_0$)

$$k[y_1,...,y_N]/(y_1-a_1,...,y_N-a_N)$$

where we think of the $y_i$ as degree $d$ monomials in the $x_i$ divided by $x_0^d$. This maps to $k[x_1,...,x_n]/(x_1-a_1,...,x_n-a_n)$ by (I think) sending each $y_i$ to the corresponding monomial in the $x_i$. The cotangent map appears to be

$$(y_1-a_1,...,y_N-a_N)/(y_1-a_1,...,y_N-a_N)^2 \to (x_1-a_1,...,x_n-a_n)/(x_1-a_1,...,x_n-a_n)^2$$

sending the $y_i$ to the same monomials as above, and keeping the constants fixed. How is this onto? What maps to $x_i-a_i$ for a fixed $i$, for example?

答:
As in your previous question I continue to think it is more intuitive to look at the tangent map. The Veronese embedding can be written in a coordinate-free way as

$$\mathbb{P}(V) \ni [v] \mapsto [v^n] \in \mathbb{P}(S^n(V))$$

where $V$ is a finite-dimensional vector space and $S^n(V)$ is its $n^{th}$ symmetric power. The Zariski tangent space at a point $[v] \in \mathbb{P}(V)$ can naturally be identified with the quotient $V/v$ of $V$ by the subspace spanned by $v$. Writing a tangent vector at $[v]$ in the form $[v + \epsilon w]$ where $\epsilon^2 = 0$, the differential of the Veronese embedding is

$$[v + \epsilon] \mapsto [(v + \epsilon w)^n] = \left[ v^n + n \epsilon wv^{n-1} \right] \in \mathbb{P}(S^n(V))$$

so to show injectivity it suffices to show that if $nw v^{n-1} = 0$ in the tangent space $S^n(V)/v^n$ to $v^n$, then $w = 0$ in $V/v$. If the characteristic of $k$ does not divide $n$, so that we can divide by $n$, this follows from the fact that the symmetric algebra $S(V) = \sum_{n \ge 0} S^n(V)$, which is a polynomial algebra, is an integral domain, so if $wv^{n-1} = c v^n$ then $v^{n-1}$ can be canceled from both sides to give $w = cv$.

If the characteristic of $k$ does divide $n$ then this definition of the Veronese embedding differs from the one using monomials. I assume that whatever source this came from is implicitly working in characteristic $0$. It may be possible to rescue this version of the Veronese embedding using divided powers.

Edit: Here is some detail on the identification of the Zariski tangent space at $v$ with $V/v$, as requested. Working slightly informally (every manipulation I'm doing with infinitesimals can be justified by defining the Zariski tangent space using $k[\epsilon]/\epsilon^2$ but I have not done so here), consider the quotient map

$$V \setminus \{ 0 \} \ni v \mapsto [v] \in \mathbb{P}(V).$$

Since $V$ is a vector space the Zariski tangent space at every point can be canonically identified with $V$. The differential of the quotient map at $v \in V$ sends a tangent vector $v + \epsilon w$ to $[v + \epsilon w]$. The kernel of the differential consists of $\epsilon w$ such that $[v + \epsilon w] = [v]$, meaning $v + \epsilon w$ is a scalar multiple of $v$, meaning $w$ is a scalar multiple of $v$. And the differential is surjective (I don't know what the cleanest way to see this is but it follows from working in an affine chart of $\mathbb{P}(V)$). So it identifies the tangent space at $[v]$ with $V/v$ as desired.

Edit #2: As Georges says in the comments, this identification of the Zariski tangent space is not quite right; it should be $\text{Hom}(v, V/v)$ where $v$ is shorthand for the line spanned by $v$. I neglected to take into account the effect of scaling $v$ properly. However, if we pick a specific $v \in V$ lifting $[v] \in \mathbb{P}(V)$ we can ignore this which is implicitly what I've done above so the calculations above are still fine (as far as I can tell).


答:
Suppose $gx=gy$.  Then $g^{-1} gx=g^{-1} gy \implies x=y$.  That proves injectivity.

Suppose $g'\in G$.  Then $g (g^{-1}g')=g'$ implies surjectivity.

In addition, in terms of actions, left multiplication by an element of a group is transitive and faithful. These are somewhat similar notions.

Title: Why is left multiplication on a group bijective?

Body:
I noticed the fact "left multiplication on a group is always bijective" is a very common argument (for instance to prove Sylow's theorem)

I don't see why left (and right) multiplication is bijective in general.
All I can see is that left multiplication is an action of a group on itself, so let $G$ be a group:
$$\cdot: G \times G \longrightarrow G$$
$$(g,g') \longmapsto g \cdot g'=gg'$$
Properties of group actions hold. I tried to prove this by contradiction but I'm not able to find any useful argument.
Could anyone enlighten me?


问:
Adjoint group of a reductive group
Let $G$ be a (not necessarily quasi-split) reductive group over $\mathbb{Q}$. In order to use results from semi-simple Lie groups for the reductive group $G(\mathbb{R})$, it seems common to implicitly use the adjoint group of $G$. If I understand correclty, it is defined by the exact sequence of algebraic groups
$$
1 \rightarrow Z_G \rightarrow G \rightarrow G^{\operatorname{ad}} \rightarrow 1
$$
where $Z_G$ denotes the center of $G$. My questions are the following:

What is the relation between $G^{\operatorname{ad}}(\mathbb{R})$ and $\operatorname{Ad}(G(\mathbb{R}))$, the image of $G(\mathbb{R})$ under the Adjoint representation? From what I understand, the latter is isomorphic to $G(\mathbb{R}) / C_{G(\mathbb{R})}\,(G(\mathbb{R})^\circ)$, the quotient modulo the centralizer of the connected component of $G(\mathbb{R})$. (see Wikipedia on the Adjoint representation)

Is $G^{\operatorname{ad}}$ an adjoint semisimple group? (Def: $G$ is called adjoint, if for any connected group $H$, any isogeny $G \rightarrow H$ is an isomorphism. This implies for example that $G^{\operatorname{ad}}$ is a direct product of its simple factors).

To compare group decompositions of both groups (like Iwasawa decomposition), one needs to fix maximal compact subgroups. So let $K \subset G(\mathbb{R})$ and $\tilde{K} \subset G^{\operatorname{ad}}(\mathbb{R})$ be maximal compact subgroups. Is there any way to get a relation between $K$ and $\tilde{K}$? Since $\operatorname{Ad}$ is continuous, the image $\operatorname{Ad}(K)$ should be compact in $\operatorname{Ad}(G(\mathbb{R}))$, but I see no reason for it to be maximal.

A toy example would maybe be $GL_2$ with maximal compact subgroup $O(2) \subset GL_2(\mathbb{R})$. The Adjoint group should be $PGL_2$ with maximal compact subgroup $PO(2) = O(2) / Z(O(2))$ and $Z(O(2)) = Z_{GL_2} \, ((\mathbb{R})) \cap O(2) = \{ \pm 1 \}$.

I would be grateful for any answer. Thank you!

答:
Preamble:
Before we begin, let us give some very brief background that might be helpful.

So, let $G$ be a reductive group over $\mathbb{R}$. Then, we have defined $G^\mathrm{ad}:=G/Z(G)$ where this quotient is taken in the sense of algebraic groups (i.e. the fppf/etale sheafification of the functor on $\mathbb{R}$-algebras which takes such an algebra $R$ to $G(R)/Z(G)(R)$).

More concretely, one can think of

$$G^\mathrm{ad}(\mathbb{R})=(G(\mathbb{C})/Z(G)(\mathbb{C}))^\sigma$$

where $\sigma$ is the non-trivial element of $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ acting on $G(\mathbb{C})/Z(G)(\mathbb{C})$— so in particular, an element of $G^\mathrm{ad}(\mathbb{R})$ needn’t come from an element of $G(\mathbb{R})$ since these are elements of $G(\mathbb{C})$ which are fixed by $\sigma$, whereas elements of $G^\mathrm{ad}(\mathbb{R})$ are represented by elements of $G(\mathbb{C})$ which needn’t be fixed by $\sigma$ on the nose but, instead, fixed up to multiplication by an element of $Z(G)(\mathbb{C})$.

In particular, one has an exact sequence of groups

$$1\to Z(G)(\mathbb{R})\to G(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})\to H^1(\mathbb{R},Z(G))$$

where this latter group is $H^1(\mathbb{R},G):=H^1(\mathrm{Gal}(\mathbb{C}/\mathbb{R}),G(\mathbb{C})$ the first Galois cohomology of $Z(G)(\mathbb{C})$ with the obvious $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ action. This is particularly nice since $H^1(\mathbb{R},Z(G))$ is a finite abelian group which can be explicitly calculated (e.g. see [1, Corollary 25.29] for an explicit description). In particular, if $Z(G)$ is a split torus then Hilbert’s theorem 90 implies that $H^1(\mathbb{R},Z(G))=0$ and so $G^\mathrm{ad}(\mathbb{R})=G(\mathbb{R})/Z(G)(\mathbb{R})$.

Question 1: Ok, so let $\mathrm{Ad}:G\to \mathrm{GL}(\mathrm{Lie}(G))$ be the adjoint representation. Then, we have that $\ker(\mathrm{Ad})=Z(G)$—for example see [1, Proposition 21.7] and note that while he assumes that $G$ is split, the desired equality can be checked after base change to $\mathbb{C}$, so it’s OK.

So then, we see that

$$\ker(\mathrm{Ad}_{\mathbb{R}})=\ker(\mathrm{Ad})(\mathbb{R})=Z(G)(\mathbb{R})$$

where I am using $\mathrm{Ad}_{\mathbb{R}}$ for the $\mathbb{R}$-points of the map $\mathrm{Ad}$. In particular, we see that

$$\mathrm{Ad}_\mathbb{R}(G(\mathbb{R}))\cong G(\mathbb{R})/Z(G)(\mathbb{R})$$

and, in fact, $Z(G)(\mathbb{R})=Z(G(\mathbb{R}))$ (e.g. this follows easily from the fact that $G(\mathbb{R})$ is Zariski dense in $G$ by unirationality—e.g. see [1, Theorem 17.93]).

So, you want to see the difference between $G^\mathrm{ad}(\mathbb{R})$ and $G(\mathbb{R})/Z(G)(\mathbb{R})$. But, not much can be said beyond what we said above. Namely, from our discussion above the difference is precisely measured by $H^1(\mathbb{R},G)$. One consequence of this though, is that the map

$$\mathrm{Ad}_\mathbb{R}(G(\mathbb{R}))=G(\mathbb{R})/Z(G)(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})$$

is an isomorphism on connected components. Namely, let us denote by $H(\mathbb{R})^+$ the connected component of $H(\mathbb{R})$ in the real topology (i.e. not the Zariski topology). Then, by the above we have that

$$(G(\mathbb{R})/Z(G)(\mathbb{R}))^+\to G^\mathrm{ad}(\mathbb{R})^+$$

is an injection, which must be a surjection since it has finite cokernel (since $(G(\mathbb{R})/Z(G)(\mathbb{R}))^+$ has finite index in $G(\mathbb{R})/Z(G)(\mathbb{R})$ by [1, Corollary 25.55] and $G(\mathbb{R})/Z(G)(\mathbb{R})$ has finite cokernel in $G^\mathrm{ad}(\mathbb{R})$, namely $H^1(\mathbb{R},G)$).

One special case of this is the following. If $G(\mathbb{R})$ is connected then the map

$$(G(\mathbb{R})/Z(G)(\mathbb{R}))\to G^\mathrm{ad}(\mathbb{R})$$

is an isomorphism onto $G^\mathrm{ad}(\mathbb{R})^+$. This connectivity hypothesis is, shockingly, satisfied if $G$ is simply connected as an algebraic group (i.e. that $G(\mathbb{C})$ is a simply connected complex Lie group)—this follows form [1, Theorem 25.54]).

Question 2: Yes. This is true. In fact, it’s not hard to see that a semisimple group is $H$ adjoint (which to me means connected) if and only if it has trivial center. Indeed, if $H$ has non-trivial center then $H\to H/Z(H)$ is a non-trivial isogeny (note $Z(H)$ is finite since $H$ is semisimple). Conversely, if $f:H\to H’$ is a non-trivial isogeny then $\ker(f)$ is a non-trivial subgroup of $H$ and, in fact, it’s contained in $Z(H)$. The reason is that $\ker(f)$ is finite and so the conjugation action $H\to \mathrm{Aut}(\ker(f))$ has trivial image since $\mathrm{Aut}(\ker(f))$ is a finite group (scheme) and $H$ is connected.

So, from this, we see that your question becomes: is $Z(G^\mathrm{ad})=0$? The answer to this is yes, although the proof in complete generality is more difficult than one might first imagine. See [1, Corllary 17.62(e)].

Question 3: Let’s try to split our understanding into two parts. Namely, let us denote by $K’$ the image of $K$ under the map $G(\mathbb{R})\to G^\mathrm{ad}(\mathbb{R})$. We first try to understand the difference between $\widetilde{K}$ and $K’$, then $K’$ and $K$. For simplicitly, I’m going to assume that $G$ is semisimple. If you have a burning desire to know what happens in the general case, I can return later.

First, $K’$ is, up to conjugacy, contained in $\widetilde{K}$ (since $K’$ is contained in some maximal compact subgroup, and all such subgroups are conjugate—e.g. see [2, Theorem D.2.8]). It is also going to be finite index. Indeed, note that $\widetilde{K}^+$ is finite index in $\widetilde{K}$ and $K^+$ is finite index in $K$ and the map $G(\mathbb{R})^+\to G^\mathrm{ad}(\mathbb{R})^+$ is surjective (by prior discussion) from where the claim follows. But, in general, they won’t be equal. Indeed, note that if $G(\mathbb{R})$ is connected, then so is $K$ (e.g. see [2, Theorem D.2.8]) and thus so is $\widetilde{K}$. But, if $G^\mathrm{ad}(\mathbb{R})$ is disconnected then so is $\widetilde{K}$ (again by [2, Theorem D.2.8]) and so you’re in big trouble. This happens with $G=\mathrm{SL}_2$.

Finally, to understand the difference between $K’$ and $K$ we note that since $Z(G)(\mathbb{R})$ is compact it’s contained in every $K$ and so $K’=K/Z(G)(\mathbb{R})$.

EDIT: I wrote the below quickly, so I would double-check it for correctness.

To handle the case when $G$ is not semisimple, note that $G^\mathrm{der}\subseteq G$ is a semisimple group and the composition $G^\mathrm{der}\to G\to G^\mathrm{ad}$ realizes $G^\mathrm{ad}$ as the adjoint group of $G^\mathrm{der}$ as well (e.g. this follows from ). So then, note that $K\cap G^\mathrm{der}(\mathbb{R})$ is a compact subgroup, which must be maximal since if $L$ is a compact subgroup of $G^\mathrm{der}(\mathbb{R})$ properly containing $K\cap G^\mathrm{der}$ then $LK$ is a compact subgroup of $G(\mathbb{R})$ properly containing $K$. Note that the image of $K\cap G^\mathrm{der}(\mathbb{R})$ in $G^\mathrm{ad}(\mathbb{R})$ is contained in $K’$ up to finite index, because the product map $Z(G)(\mathbb{R})\to G^\text{der}(\mathbb{R})\to G(\mathbb{R})$ has finite cokernel and kernel (with cokernel equal to $H^1(\mathbb{R},Z(G^\text{der}))$). So, you can deduce from this that, roughly, the same argument in the semisimple case applies— $K’$ and $\widetilde{K}$ differ by finite index.

References:

[1] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[2] Conrad, B., 2014. Reductive group schemes. Autour des schémas en groupes, 1, pp.93-444.

Title: Adjoint group of a reductive group