问:
Laplacian of bipartite graph
It is well know that in the case of weighted graph with positive weights, the dimension of the kernel of the Laplacian is the number of connected components of the corresponding graph.

This fails when negative weights are allowed, but I was wondering if this remained true if we restrained ourselves to bipartite graph?

In other words, for bipartite graphs with possibly negative weights, is the number of connected components the dimension of the kernel of the associated Laplacian?

答:
If I haven't made any error in my code, the answer is no

Here is a counter example:

$L=\begin{bmatrix}
  1 & 0 & 0 & -1 & 0 & 0\\
  0 & 3 & 0 & -1 & -1 & -1\\
  0 & 0 & 2 & 0 & -1 & -1\\
  -1 & -1 & 0 & 2 & 0 & 0\\
  0 & -1 & -1 & 0 & 2 & 0\\
  0 & -1 & -1 & 0 & 0 & 2\\
\end{bmatrix}
$

This graph is connected, and the Laplacian has eigenvalues $\left({(5 + \sqrt{17})/2, 3, 2, 2, (5 - \sqrt{17})/2, 0}\right)$

Now take the weights:

$L_w=
\begin{bmatrix}
  1. & 0. & 0. & -1. & 0. & 0.\\
  0. & 1. & 0. & 2. & -4. & 1.\\
  0. & 0. & -3. & 0. & 4. & -1.\\
  -1. & 2. & 0. & -1. & 0. & 0.\\
  0. & -4. & 4. & 0. & 0. & 0.\\
  0. & 1. & -1. & 0. & 0. & 0.\\
\end{bmatrix}
$

this has eigenvalues $({-6.82741, 5.88255, -2.55556, 1.50042, 0, 0})$.

for completeness, here is the python code I used to find the counterexample:

```<code>    import networkx as nx
    import numpy as np
    import matplotlib.pyplot as plt
    %matplotlib inline
    #magic
    import warnings
    warnings.filterwarnings('ignore')

    import random

    np.set_printoptions(precision=3)
    import math

    def randomWeightLaplacian(A):
        #given an adjacency matrix, give random symetric weights between -10 and 10 (avoiding 0)
        #return the weighted Laplacian and the Laplacian of the unweighted graph
        n=A.shape[0] #order of the graph
        L=np.zeros((n,n)) #weighted Laplacian
        Lb=np.zeros((n,n),int) #unweighted Laplacian

        for i in range(n):
            for j in range(i+1,n):
                if A[i][j]==1:
                    w=random.choice([-1,1])*random.choice([i for i in range(1,10)])
                    L[i][j]=w
                    L[j][i]=w
                    Lb[i][j]=-1
                    Lb[j][i]=-1

        for i in range(n):         #diagonal
            L[i][i]=-sum(L[i][:])
            Lb[i][i]=-sum(Lb[i][:])
        return L,Lb

    def numberSmall(array,pres=1e-5):
        #number of small element in the array
        cc=0
        for i in array:
            #print(i)
            if math.isclose(i, 0., abs_tol=pres):
                #print(i)
                cc=cc+1
        return cc

    n,m=3,3       #size of the partition
    ccw,ccb=0,0   #multiplicity of 0 in L and Lb
    i=1
    while ccw==ccb:
        print(i)
        i=i+1
        G=nx.bipartite.random_graph(n,m, 1/3)
        A=nx.to_numpy_array(G)
        #networkx function giving the adjacency matrix of a random bipartite graph

        L,Lb=randomWeightLaplacian(A)
        G=nx.from_numpy_array(A)

        eigenval,eigenvect=np.linalg.eig(L)
        eigenvalb,eigenvectb=np.linalg.eig(Lb)

        ccw=numberSmall(eigenval)
        ccb=numberSmall(eigenvalb)

    nx.draw(G)

    print("number of connected components:",ccb," mutliplicity of 0 in the weighted graph:",ccw,"\n")
    print(L)
    print(Lb)
</code>```

Edit

my example has a diagonal element $0$ artificially (some weights cancel out) but this is not necessary. One can adapt the code adding/replacing with this piece, and still get counterexamples

```<code>L,Lb=randomWeightLaplacian(A)
flag=True
for i in range(n+m):
    if L[i][i]==0 and Lb[i][i]!=0:
        flag=False

if flag:
    eigenval,eigenvect=np.linalg.eig(L)
    eigenvalb,eigenvectb=np.linalg.eig(Lb)

    ccw=numberSmall(eigenval)
    ccb=numberSmall(eigenvalb)
</code>```

Title: Laplacian of bipartite graph

Body:
If I haven't made any error in my code, the answer is no

Here is a counter example:

$L=\begin{bmatrix}
  1 & 0 & 0 & -1 & 0 & 0\\
  0 & 3 & 0 & -1 & -1 & -1\\
  0 & 0 & 2 & 0 & -1 & -1\\
  -1 & -1 & 0 & 2 & 0 & 0\\
  0 & -1 & -1 & 0 & 2 & 0\\
  0 & -1 & -1 & 0 & 0 & 2\\
\end{bmatrix}
$

This graph is connected, and the Laplacian has eigenvalues $\left({(5 + \sqrt{17})/2, 3, 2, 2, (5 - \sqrt{17})/2, 0}\right)$

Now take the weights:

$L_w=
\begin{bmatrix}
  1. & 0. & 0. & -1. & 0. & 0.\\
  0. & 1. & 0. & 2. & -4. & 1.\\
  0. & 0. & -3. & 0. & 4. & -1.\\
  -1. & 2. & 0. & -1. & 0. & 0.\\
  0. & -4. & 4. & 0. & 0. & 0.\\
  0. & 1. & -1. & 0. & 0. & 0.\\
\end{bmatrix}
$

this has eigenvalues $({-6.82741, 5.88255, -2.55556, 1.50042, 0, 0})$.

for completeness, here is the python code I used to find the counterexample:

    import networkx as nx
    import numpy as np
    import matplotlib.pyplot as plt
    %matplotlib inline
    #magic
    import warnings
    warnings.filterwarnings('ignore')

    import random

    np.set_printoptions(precision=3)
    import math

    def randomWeightLaplacian(A):
        #given an adjacency matrix, give random symetric weights between -10 and 10 (avoiding 0)
        #return the weighted Laplacian and the Laplacian of the unweighted graph
        n=A.shape[0] #order of the graph
        L=np.zeros((n,n)) #weighted Laplacian
        Lb=np.zeros((n,n),int) #unweighted Laplacian

        for i in range(n):
            for j in range(i+1,n):
                if A[i][j]==1:
                    w=random.choice([-1,1])*random.choice([i for i in range(1,10)])
                    L[i][j]=w
                    L[j][i]=w
                    Lb[i][j]=-1
                    Lb[j][i]=-1

        for i in range(n):         #diagonal
            L[i][i]=-sum(L[i][:])
            Lb[i][i]=-sum(Lb[i][:])
        return L,Lb

    def numberSmall(array,pres=1e-5):
        #number of small element in the array
        cc=0
        for i in array:
            #print(i)
            if math.isclose(i, 0., abs_tol=pres):
                #print(i)
                cc=cc+1
        return cc

    n,m=3,3       #size of the partition
    ccw,ccb=0,0   #multiplicity of 0 in L and Lb
    i=1
    while ccw==ccb:
        print(i)
        i=i+1
        G=nx.bipartite.random_graph(n,m, 1/3)
        A=nx.to_numpy_array(G)
        #networkx function giving the adjacency matrix of a random bipartite graph

        L,Lb=randomWeightLaplacian(A)
        G=nx.from_numpy_array(A)

        eigenval,eigenvect=np.linalg.eig(L)
        eigenvalb,eigenvectb=np.linalg.eig(Lb)

        ccw=numberSmall(eigenval)
        ccb=numberSmall(eigenvalb)

    nx.draw(G)

    print("number of connected components:",ccb," mutliplicity of 0 in the weighted graph:",ccw,"\n")
    print(L)
    print(Lb)

Edit

my example has a diagonal element $0$ artificially (some weights cancel out) but this is not necessary. One can adapt the code adding/replacing with this piece, and still get counterexamples

L,Lb=randomWeightLaplacian(A)
flag=True
for i in range(n+m):
    if L[i][i]==0 and Lb[i][i]!=0:
        flag=False

if flag:
    eigenval,eigenvect=np.linalg.eig(L)
    eigenvalb,eigenvectb=np.linalg.eig(Lb)

    ccw=numberSmall(eigenval)
    ccb=numberSmall(eigenvalb)


问:
Fastest way to find derivative of the function at the given point

What is derivative of $f(x)=\left(\dfrac{\sqrt[3]{x^2+2x}}{x^2-x}\right)^3$ at the point $x=2$? $$1)-\frac34\qquad\qquad2)-\frac54\qquad\qquad3)-\frac52\qquad\qquad4)-\frac{15}4$$

This is a problem from an timed Exam, so I am looking for the fastest way to solve this. Here is my solution:

We have $f(x)=\dfrac{x^2+2x}{(x^2-x)^3}$ . by using Quotient rule, derivative at  $x=2$ is:
$$\dfrac{6\times8-(3\times3\times4)\times8}{8^2}=\frac34-\frac92=-\frac{15}4$$
Although the way I putted the values of functions instead of writing the whole derivative of function seems to be fast , it is hard to avoid algebraic mistakes and find the correct answer in the exam condition (having time pressure, stress  and so on).

After all, is there a better approach (faster) to solve this problem ?

答:
Perhaps slightly...but very slightly...easier: using that $\;(x^2-x)^3=x^3(x-1)^3\;$ and also  $\;x=2\implies (x-1)^n=1\;$

$$f(x)=\frac{x^2+2x}{(x^2-x)^3}=\frac{x+2}{x^2(x-1)^3}=\frac1{x(x-1)^3}+\frac2{x^2(x-1)^3}\implies$$

$$f'(x)=\left(\frac1{x(x-1)^3}\right)'+2\left(\frac1{x^2(x-1)^3}\right)'=$$

$$=-\frac{(x-1)+3x}{x^2(x-1)^4}-2\frac{2(x-1)+3x(x-1)}{x^3(x-1)^4}\implies$$

$$f'(2)=-\frac{1+6}{4\cdot1}-2\frac{2+6}{8\cdot1}=-\frac74-2=-\frac{15}4$$

Another Option: Write your function multiplicatively and use the product rule, which surely will be easier than before:

$$f(x)=x^{-2}(x+2)(x-1)^{-3}\implies $$

$$f'(x)=-2x^{-3}(x+2)(x-1)^{-3}+x^{-2}(x-1)^{-3}-3x^{-2}(x+2)(x-1)^{-4}\implies$$

$$f'(2)=-2\cdot\frac18\cdot4+\frac14-\frac34\cdot4=-1+\frac14-3=-4+\frac14=-\frac{15}4$$

I'd go with this last option. As thumb rule, product rule nice than quotient rule

ContentLicense: CC BY-SA 4.0


答:
Cancelling out the $x$ yields
$$\dfrac{x^2+2x}{(x^2-x)^3}=\dfrac{x^2+2x}{x^3(x-1)^3}=\dfrac{x+2}{x^2(x-1)^3}.$$
If we take the logarithm on both sides we get
$$\log f(x)=\log(x+2)-\left[2\log x+3\log(x-1)\right].$$
Now taking the derivative of this expression is trivial,
$$\frac{f'(x)}{f(x)}=\frac{1}{x+2}-\frac 2 x-\frac{3}{x-1}\Rightarrow \frac{f'(2)}{f(2)}=\frac{1}{4}-1-3=-\frac{15}{4}.$$
Last step was obtained by $f(2)=1$.

Title: Fastest way to find derivative of the function at the given point

Body:
What is derivative of $f(x)=\left(\dfrac{\sqrt[3]{x^2+2x}}{x^2-x}\right)^3$ at the point $x=2$? $$1)-\frac34\qquad\qquad2)-\frac54\qquad\qquad3)-\frac52\qquad\qquad4)-\frac{15}4$$

This is a problem from an timed Exam, so I am looking for the fastest way to solve this. Here is my solution:

We have $f(x)=\dfrac{x^2+2x}{(x^2-x)^3}$ . by using Quotient rule, derivative at  $x=2$ is:
$$\dfrac{6\times8-(3\times3\times4)\times8}{8^2}=\frac34-\frac92=-\frac{15}4$$
Although the way I putted the values of functions instead of writing the whole derivative of function seems to be fast , it is hard to avoid algebraic mistakes and find the correct answer in the exam condition (having time pressure, stress  and so on).

After all, is there a better approach (faster) to solve this problem ?


问:
Use Dedekind's Theorem to show that there does not exists $b$ such that $O_K=\mathbb{Z}[b]$
This is a past exam question that I couldn't quite nail down. Any help is much appreciated!

Let $K=\mathbb{Q}(\sqrt{d_1}, \sqrt{d_2})$ where $d_1$ and $d_2$ are both $1$ mod $3$. Moreover $d_1$ and $d_2$ are square-free. The question asks why does it not exist a $b\in O_K$ such that $O_K=\mathbb{Z}[b]$, using Dedekind's theorem on factorising $(p)$ in $O_K$ where $p$ is a rational prime (i.e., just a prime number in the normal sense).

So I factored $(3)$ into the following $(3)=(3,b_1)(3,b_2)(3,b_3)(3,b_4)$ where $b_1=\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_1d_2}, b_2=\sqrt{d_1}-\sqrt{d_2}-\sqrt{d_1d_2},b_3=-\sqrt{d_1}+\sqrt{d_2}-\sqrt{d_1d_2}, b_4=-\sqrt{d_1}-\sqrt{d_2}+\sqrt{d_1d_2}.$ Each factor is a prime ideal and I am convinced that up to this stage everything is correct.

Use Dedekind's Theorem I found that $[O_K:\mathbb{Z}[\sqrt{d_1}+\sqrt{d_2}]]$ is divisible by $3$ because otherwise the decomposition of $(3)$ will look really different from the one I wrote above.

But what next? How do I show the non-existence of such $\beta?$ I mean the fact that it can be divided by $3$ just shows that there is a subgroup which is isomorphic to $C_3$ for this $O_K:\mathbb{Z}[\sqrt{d_1}+\sqrt{d_2}]$ as an additive group but what can I deduce after that?

Many thanks in advance!!

答:
In $\Bbb{Z}[\sqrt{d_1},\sqrt{d_2}]$ we have $4$ distinct maximal ideals above $3$:

$(3,  \sqrt{d_1}\pm 1, \sqrt{d_2}\pm 1)$ all with residue field $\Bbb{F}_3$.

The product of those 4 ideals is $(3)$.

Thus the same holds in $O_K$.

If $O_K=\Bbb{Z}[b]$ then the maximal ideals above $3$ would be of the form $(3,b-n)$ with only $3$ possible choices for $n$, not $4$.


问:
Find all basis for a set V
The set $V= \{ v_1, v_2,..., v_6 \}$ is represented by $V = \{ (1, 0, 2, 3), (4, 0, 8, 12), (4, 3, 2, 1), (4, 6, -4, -10), (1, 1, 1, 1), (2, 1, 4, 3)$. Of course the set is in $\mathbb{R^4}$. Find all the bases that are subsets of V. First, to satisfy the condition for a basis, it first needs to be linearly independent. So should I make this into a matrix, and convert it to reduced row echelon? From there, I can found out which vector is linearly independent by looking at the pivot columns, which are guaranteed to be linearly independent by definition.

Title: Find all basis for a set V

Body:
$V = \{ v_1=(1, 0, 2, 3), v_2=(4, 0, 8, 12), v_3=(4, 3, 2, 1), v_4=(4, 6, -4, -10), v_5=(1, 1, 1, 1), v_6=(2, 1, 4, 3)\}$. Here I am using $[v_1, v_2, v_3, v_4]$ as the matrix generated by the row vectors $v_1, v_2, v_3, v_4$ (it could be column vectors, without prejudice to what you want). Doing the calculations, we have

\begin{eqnarray*}
\det\,[v_1, v_2, v_3, v_4] &=& 0\\
\det\,[v_1, v_2, v_3, v_5] &=& 0\\
\det\,[v_1, v_2, v_3, v_6] &=& 0\\
\det\,[v_2, v_3, v_4, v_6] &=& -32\\
\end{eqnarray*}

Next, I am using the $\equiv$ symbol to indicate the (triangular) decomposition of the matrix.

The matrix $[v_1, v_2, v_3, v_4]$ has rank 2.

$$
[v_1, v_2, v_3, v_4] \equiv \left( \begin{array}{rrrr}
1&0&2&3\\
0&3&-6&-11\\
0&0&0&0\\
0&0&0&0\\
\end{array}\right)
$$

The matrix $[v_1, v_2, v_3, v_5]$ has rank 3.

$$
[v_1, v_2, v_3, v_5] \equiv \left( \begin{array}{rrrr}
1&0&2&3\\
0&3&-6&-11\\
0&0&1&\frac{5}{3}\\
0&0&0&0\\
\end{array}\right)
$$

The matrix $[v_1, v_2, v_3, v_6]$ has rank 3.

$$
[v_1, v_2, v_3, v_6] \equiv \left( \begin{array}{rrrr}
1&0&2&3\\
0&3&-6&-11\\
0&0&2&\frac{2}{3}\\
0&0&0&0\\
\end{array}\right)
$$

The matrix $[v_2, v_3, v_4, v_5]$ has rank 3.

$$
[v_2, v_3, v_4, v_5] \equiv \left( \begin{array}{rrrr}
4&0&8&12\\
0&3&-6&-11\\
0&0&1&\frac{5}{3}\\
0&0&0&0\\
\end{array}\right)
$$

The matrix $[v_2, v_3, v_4, v_6]$ has rank 3.

$$
[v_2, v_3, v_4, v_6] \equiv \left( \begin{array}{rrrr}
4&0&8&12\\
0&3&-6&-11\\
0&0&2&\frac{2}{3}\\
0&0&0&0\\
\end{array}\right)
$$

The matrix $[v_3, v_4,v_5, v_6]$ has rank 4.

$$
[v_2, v_3, v_4, v_6] \equiv \left( \begin{array}{rrrr}
4&3&2&1\\
0&3&-6&-11\\
0&0&1&\frac{5}{3}\\
0&0&0&-\frac{8}{3}\\
\end{array}\right)
$$

Except for the $(v_3, v_4, v_5, v_6)$ combination, the other vector combinations are linearly independent, that is, they generate the $\mathbb{R}^4$ space.


问:
Proving the equivalence of two definitions of adjoint functors without using the triangle identities
Here's a theorem from Leinster's textbook:

He proves it using triangle identities. As I mentioned in a comment to this answer, I suspect that the direct proof (that doesn't appeal to the triangle identities) would be more intuitive. So I attempted to prove this directly (see below), and I was wondering if this proof is correct (modulo the uniqueness part that I couldn't prove). And also I was wondering how to prove uniqueness.

We first prove that given an adjunction between $F$ and $G$ as in (a), there is a natural transformation $\eta$ as in (b). This is just done by defining $\eta$ to be the unit of the given adjunction, and the details are given in Lemma 2.3.5 of Leinster's textbook (triangle identities aren't used in the proof).

Now we need to prove that this correspondence is bijective. That is, if $\eta$ is as described in (b), then there exists a unique adjunction as in (a) with unit $\eta$.

Existence. We know that $\eta_A$ is initial in $(A\implies G)$, i.e., for all $\phi:A\to G(B)$, there is a unique $\phi':F(A)\to B$ such that $\phi=G(\phi')\circ \eta_A$. Define the maps $$\mathscr B(F(A),B)\cong\mathscr A(A,G(B)) \\ \phi' \text{ <--| } \phi\\\psi\mapsto G(\psi)\circ \eta_A$$

Now we need to show that (1) these two maps are inverses of each other, (2) that this bijection is natural in $A$ and $B$, (3) the unit of this adjunction is $\eta$.

(1) First composition: $\phi\mapsto \phi'\mapsto  G(\phi ')\circ \eta_A$. Need to show that $\phi=G(\phi ')\circ \eta_A$. But this is exactly the condition mentioned in (b) that we're assuming.

Second composition: $\psi\mapsto G(\psi)\circ \eta_A \mapsto (G(\psi)\circ \eta_A)'$. Need to show that $\psi = (G(\psi)\circ \eta_A)'$. By definition, $(G(\psi)\circ \eta_A)'$ is the unique arrow $F(A)\to B$ with the property that $G(\psi)\circ \eta_A=G((G(\psi)\circ \eta_A)')\circ \eta_A$. But $\psi$ is also an arrow $F(A)\to B$ that makes the same triangle commute. Thus $(G(\psi)\circ \eta_A)'=\psi$.

(2) Naturality in $B$ amounts to the commutativity of this diagram:

and it follows from the fact that $G$ is a functor that the diagram commutes.

Naturality in $A$ amounts to the commutativity of this diagram:

and this follows from the fact that $\eta$ is a natural transformation ($\eta$ being a natural transformation implies that $\eta_A\circ p=GF(p)\circ \eta_{A'}$, and now we just need to compose this with $G(\psi)$ to get the commutativity of the diagram above).

(3) To check that $\eta$ is the unit of the adjunction we defined, we need to check that $1_{F(A)}$ goes to $\eta_A$ under the adjunction map. Indeed, $1_{F(A)}\mapsto G(1_{F(A)})\circ \eta_A=1_{GF(A)}\circ \eta_A=\eta_A$.

This proves existence.

Uniqueness.

I don't know how to approach this. Any ideas? (Again, Leinster gives a proof of this using the triangle identities, but I'd like to avoid using them.)

答:
The suggested approach to uniqueness is the approach you already used on other level, namely let $X:=\{$adjunctions between $F$ and $G\}$ and $Y:=\{$natural transformations $\eta:1_{\mathcal A}\to GF$ such that...$\}$, and specify two maps $\Phi:X\to Y$ and $\Psi:Y\to X$ and verify that they are inverses of each other.

Part of this work is already done: we have $\Phi$ of an adjunction is just its unit, and in the 'existence' part you describe a nice candidate for $\Psi$, and you also have $\Phi(\Psi(\eta))=\eta$.

So, the only missing part is $\Psi(\Phi(\alpha))=\alpha$ where $\alpha$ is a natural bijection with components $\alpha_{A,B}:\mathcal B(FA,B)\to \mathcal A(A,GB)$ determining the adjunction.

Let's name these $\mathcal A^{op}\times\mathcal B\to \mathcal Set$ functors:
$$U:=\mathcal B(F-,-),\ \quad\ V:=\mathcal A(-,G-)\,,$$
so that $\alpha:U\to V$ is a natural isomorphism.

Note that if $f:A\to A',\ g:B\to B'$ then
$$U(f,g)=\underset{FA'\to B}h\,\mapsto\  \underset{FA\to B'}{g\circ h\circ Ff}\\
V(f,g)=\underset{A'\to GB}h\,\mapsto\  \underset{A\to GB'}{Gg\circ h\circ f} \,.
$$

By your definition of $\Psi$, we have $\Psi(\eta)_{A,B}=\underset{FA\to B}g\,\mapsto \ Gg\circ\eta_A$.

Now, if we spell out what the unit specifically is for the adjunction given by $\alpha$, we simply obtain
$$\eta_A=\alpha_{A,FA}(1_{FA})\,,$$
then by naturality of $\alpha$ we get for any $g:FA\to B$
$$\Psi(\Phi(\alpha))_{A,B}\,(g)\ = Gg\circ \alpha_{A,FA}(1_{FA})\ =
\ V(1_A,g)\big(\alpha_{A,FA}(1_{FA})\big)\ = \\
\ = \alpha_{A,B}\circ U(1_A,g)\,\big(1_{FA}) \ = \ \alpha_{A,B}(g)\,.$$


答:
Here is a way, not sure how much faster you'll find it, using Taylor series.
You have
$$
f(x) = \frac{x^2+2x}{(x^2-x)^3} = \frac{x+2}{x^2(x-1)^3}
$$
You want to look at things around $x=2$, so set $u:= x-2$:
$$
\frac{x+2}{x^2(x-1)^3} = \frac{u+4}{(u+2)^2(u+1)^3}
= \frac{1+\frac{u}{4}}{(1+\frac{u}{2})^2(1+u)^3}
$$
Now you want to look at things around $u=0$: use $(1+u)^a = 1+au + O(u)$ for any constant $a\in\mathbb{R}$ to get
$$
\frac{1+\frac{u}{4}}{(1+\frac{u}{2})^2(1+u)^3} = \left(1+\frac{u}{4}\right)\left(1-u + o(u)\right)\left(1-3u+ o(u)\right)
$$
Expand, only keeping the first two terms:
$$
\frac{1+\frac{u}{4}}{(1+\frac{u}{2})^2(1+u)^3} = 1+\left(\frac{1}{4}-1-3\right)u+o(u)
= 1 + \boxed{\frac{-15}{4}}u+o(u)
$$
You get the derivative $f'(2)=-\frac{15}{4}$ you wanted here. ((Also, the first term, $1$, is $f(2)$).


问:
Is multinormality required for zero covariance to imply independence?
I have two questions question which are bugging me regarding the multivariate normal/ multinormal framework. I can not find a satisfactory answer anywhere I have looked so far, perhaps it is because I don't know what to look for.

Firstly:
Can we simply assemble a multinormal distribution from two univariate normals and a covariance?

i.e can we take two univariate normals, and then express a multinormal using a covariance between them?

\begin{equation}
X \sim N(0,1) Y \sim N(0,1) \operatorname{Cov}(X, Y)=0.5 \Leftrightarrow\left(\begin{array}{l}
X \\
Y
\end{array}\right) \sim N_{2}\left(\left(\begin{array}{l}
0 \\
0
\end{array}\right),\left(\begin{array}{cc}
1 & 0.5 \\
0.5 & 1
\end{array}\right)\right)
\end{equation}

Also, regarding independence, I have been showed many times the perennial example of $Y=X^2$ where $X \sim N(0,1)$ as an example of two random variables which have zero covariance, yet a clear form of dependence between them. However, it occured to me, that Y is of course chi-squared, rather than normal, so of course we are not in a "multinormal setting".

Therefore, my second question is as follows: If we are in a truly multi-normal framework, is it fair to say that if we have zero covariance, than the two variables are truly independent? Can there exist a form of dependence expressed in a multinormal setting which is not captured by covariance?

Thanks for the insight!

答: The answer to your question is: $0$ covariance does not imply independence unless the variables are jointly normal.  Simple example: let $X$ be standard normal ($N(0,1)$) and let $Y=cX$, where $c$ (independent of $X$) is standard Bernoulli $(P(c=1)=P(c=-1)=\frac{1}{2})$.  Then $Y$ is also standard normal and $E(XY)=0$.  However they are not independent since $P(X=Y)=P(X=-Y)=\frac{1}{2}$.


问:
Formal Laurent series in more variables
The formal Laurent series in one variables is

$$\begin{equation} \mathbb{C}((t))=\mathbb{Q}(\mathbb{C}[[t]])=\mathbb{C}[[t]][\frac{1}{t}] \end{equation}$$

What is the difference between this field and

$$\begin{equation} \bigg\{\sum_{n\in\mathbb{Z}}a_{n}t^{n}\quad\text{such that}\quad \forall n \in \mathbb{Z} \ a_{n}\in\mathbb{C}\}\bigg\}\end{equation}$$

In more variables what is the difference with this case in the definition?

答:
The ring of formal Laurent series comprises formal series $\sum_{n\in \Bbb{Z}}a_nt^n$, where, for some $i \in \Bbb{Z}$, the $a_n$ with $n < i$ are all zero: so we could write the formal series as $\sum_{n=i}^\infty a_nt^n$. This reflects the fact that there is only one pair of square brackets round $\frac{1}{t}$ in $\Bbb{C}[[t]][\frac{1}{t}]$ - a notation that really means $\Bbb{C}[[t]][u]/I$ where $I$ is the ideal $(tu - 1)$. As mentioned in a comment this is the same as the field of fractions of the ring of formal power series $\Bbb{C}[[t]]$.

The element $tu - 1$ is a unit in the ring of formal power series $\Bbb{C}[[t, u]]$, implying that the quotient ring $\Bbb{C}[[t, u]]/(tu - 1) = \{0\}$. This reflects the fact that the convolution product doesn't make sense if you allow non-zero coefficients $a_n$ for negative values of $n$ of arbitrarily large absolute value.

If you are looking at formal Laurent series in more than one variable, the same considerations apply: e.g., $\Bbb{C}[[t, v]][\frac{1}{t}, \frac{1}{v}]$ means $\Bbb{C}[[t, v]][u, w]/(tu - 1, vw - 1)$ and comprises formal series $\sum_{m,n \in \Bbb{Z}}a_{mn}t^mv^n$, where at most finitely many $a_{mn}$ with $m < 0$ or $n < 0$ are non-zero.

Title: Formal Laurent series in more variables


问:
Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$
I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$

Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$

Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.

Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.

Body:
\begin{align*}J&=\int_0^1 \frac{x^2\ln(1-x^4)}{1+x^3}dx\\
&\overset{y=\frac{1-x}{1+x}}=3\ln^2 2+3\underbrace{\int_0^1 \frac{\ln y}{1+y}dy}_{=-\dfrac{\pi^2}{12}}+\underbrace{\int_0^1 \frac{\ln(1+y^2)}{1+y}dy}_{=\dfrac{3}{4}\ln^2 2-\dfrac{1}{18}\pi^2}-2\ln^2 2-6\ln 2\underbrace{\int_0^1 \frac{y}{1+3y^2}dy}_{\dfrac{1}{3}\ln 2}-\\&2\underbrace{\int_0^1 \frac{y\ln y}{1+3y^2}dy}_{z=y^2}-2\underbrace{\int_0^1 \frac{y\ln(1+y^2)}{1+3y^2}dy}_{z=y^2}+8\int_0^1 \frac{y\ln(1+y)}{1+3y^2}dy\\
&=-\frac{1}{4}\ln^2 2-\frac{5}{48}\pi^2-\frac{1}{2}\underbrace{\int_0^1 \frac{\ln z}{1+3z}dz}_{=A}-\underbrace{\int_0^1 \frac{\ln(1+ z)}{1+3z}dz}_{=B}+8\underbrace{\int_0^1 \frac{y\ln(1+y)}{1+3y^2}dy}_{=C}
\end{align*}
\begin{align*}
A&\overset{\text{IBP}}=-\frac{1}{3}\int_0^1 \frac{\ln(1+3z)}{z}dz\\
&\overset{u=\frac{1}{1+3z}}=\frac{1}{3}\int_{\frac{1}{4}}^1 \frac{\ln u}{u(1-u)}du\\
&=-\frac{2}{3}\ln^2 2+\frac{1}{3}\int_{\frac{1}{4}}^1 \frac{\ln u}{1-u}du\\
&=-\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{1}{3}\int_0^{\frac{1}{4}} \frac{\ln u}{1-u}du\\
&\overset{x=4u}=-\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{1}{12}\int_0^1 \frac{\ln\left(\frac{1}{4}x\right)}{1-\frac{1}{4}x}dx\\
&=\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{2}{3}\ln 2\ln 3-\frac{1}{12}\int_0^1 \frac{\ln x}{1-\frac{1}{4}x}dx\\
&=\boxed{\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{2}{3}\ln 2\ln 3+\frac{1}{3}\text{Li}_2\left(\frac{1}{4}\right)}\\
B&\overset{x=\frac{1}{1+3z}}=\frac{1}{3}\int_{\frac{1}{4}}^1 \frac{\ln(1+2x)}{x}dx+\frac{2}{3}\ln^2 2-\frac{2}{3}\ln 2\ln 3\\
&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln x\ln(1+2x)\Big]_{\frac{1}{4}}^1-\frac{2}{3}\int_{\frac{1}{4}}^1 \frac{\ln x}{1+2x}dx+\frac{2}{3}\ln^2 2-\frac{2}{3}\ln 2\ln 3\\
&=-\frac{2}{3}\int_{\frac{1}{4}}^1 \frac{\ln x}{1+2x}dx\\
&=-\frac{2}{3}\underbrace{\int_0^1 \frac{\ln x}{1+2x}dx}_{y=2x}+\frac{2}{3}\underbrace{\int_0^{\frac{1}{4}} \frac{\ln x}{1+2x}dx}_{y=4x}\\
&=\frac{1}{6}\int_0^1 \frac{\ln\left(\frac{y}{4}\right)}{1+\frac{y}{2}}dy-\frac{1}{3}\int_0^2 \frac{\ln\left(\frac{y}{2}\right)}{1+y}dy\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{2}{3}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^2 \frac{\ln y}{1+y}dy\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{2}{3}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^1 \frac{\ln y}{1+y}dy-\frac{1}{3}\underbrace{\int_1^2 \frac{\ln y}{1+y}dy}_{z=\frac{1}{y}}\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{2}{3}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^1 \frac{\ln y}{1+y}dy+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln z}{(1+z)z}dz\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{1}{2}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^1 \frac{\ln y}{1+y}dy-\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln z}{1+z}dz\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{1}{2}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{2}{3}\underbrace{\int_0^1 \frac{\ln y}{1+y}dy}_{=-\frac{1}{12}\pi^2}+\frac{1}{3}\underbrace{\int_0^{\frac{1}{2}} \frac{\ln z}{1+z}dz}_{x=2z}\\
&=\boxed{\frac{2}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{5}{6}\ln^2 2-\frac{2}{3}\ln 2\ln 3 +\frac{1}{18}\pi^2}\\
\end{align*}
\begin{align*}
C'&=\int_0^1 \frac{y\ln(1-y)}{1+3y^2}dy\\
C+C'&\overset{z=1-y^2}=\frac{1}{8}\int_0^1\frac{\ln z}{1-\frac{3}{4}z}dz\\
&=-\frac{1}{6}\text{Li}_2\left(\frac{3}{4}\right)\\
C^\prime-C&\overset{z=\frac{1-y}{1+y}}=\int_0^1 \frac{(1-z)\ln z}{2(1+z)(z^2-z+1)}dz\\
&\overset{\text{IBP}}=\left[\left(\frac{1}{3}\ln(1+z)-\frac{1}{6}\ln(1-z+z^2)\right)\ln z\right]_0^1-\frac{1}{3}\int_0^1 \frac{\ln(1+z)}{z}dz+\frac{1}{6}\int_0^1 \frac{\ln\left(\frac{1+z^3}{1+z}\right)}{z}dz\\
&=-\frac{1}{2}\int_0^1 \frac{\ln(1+z)}{z}dz+\frac{1}{6}\underbrace{\int_0^1 \frac{\ln\left(1+z^3\right)}{z}dz}_{u=z^3}\\
&=-\frac{4}{9}\underbrace{\int_0^1 \frac{\ln(1+z)}{z}dz}_{=\frac{1}{12}\pi^2}\\
&=\boxed{-\frac{1}{27}\pi^2}
\end{align*}
Therefore,
\begin{align*}
C&=\boxed{-\frac{1}{12}\text{Li}_2\left(\frac{3}{4}\right)+\frac{1}{54}\pi^2}\\
J&=\boxed{\ln 2\ln 3-\frac{17}{12}\ln^2 2-\frac{1}{6}\text{Li}_2\left(\frac{1}{4}\right)+\frac{7}{432}\pi^2-\frac{2}{3}\text{Li}_2\left(-\frac{1}{2}\right)-\frac{2}{3}\text{Li}_2\left(\frac{3}{4}\right)}
\end{align*}
Since,
\begin{align*}0\leq x<1,\text{Li}_2(x)+\text{Li}_2(1-x)&=\frac{\pi^2}{6}-\ln(1-x)\ln x\\
-1\leq x<1,\text{Li}_2(x)+\text{Li}_2(-x)&=\frac{1}{2}\text{Li}_2(x^2)
\end{align*}
Then,
\begin{align*}
\text{Li}_2\left(\frac{3}{4}\right)&=\frac{1}{6}\pi^2-4\ln^2 2+2\ln2 \ln 3-\text{Li}_2\left(\frac{1}{4}\right)\\
\text{Li}_2\left(-\frac{1}{2}\right)&=\frac{1}{2}\text{Li}_2\left(\frac{1}{4}\right)-\underbrace{\text{Li}_2\left(\frac{1}{2}\right)}_{=\frac{1}{12}\pi^2-\frac{1}{2}\ln^2 2}\\
\end{align*}
Thus,
$\boxed{\displaystyle J=\frac{11}{12}\ln^2 2-\frac{1}{3}\ln 2\ln 3-\frac{17}{432}\pi^2+\frac{1}{6}\text{Li}_2\left(\frac{1}{4}\right)}$

NB: I assume that,
\begin{align*}-1\leq a\leq 1,\text{Li}_2\left(a\right)&=-a\int_0^1 \frac{\ln t}{1-at}dt\\
\int_0^1 \frac{\ln x}{1+x}dx&=-\frac{1}{12}\pi^2\\
\int_0^1 \frac{\ln(1+x)}{x}dx&=\frac{1}{12}\pi^2\\
\end{align*}


问:
For $u,v,x \in \mathbb{R}^m$, let $T(u,v)(x)=x - \frac{\langle(u+v),x\rangle}{1+\langle u,v\rangle}(u+v)+2\langle u,x\rangle v$
The function in question appears in lemma $6.3$ of Milnors characteristic classes. Let $\langle \cdot, \cdot\rangle$ denote the dot product on $\mathbb{R}^m$

Let $u,v \in \mathbb{R}^m$ be unit vectors with $u \neq -v$. Let $T(u,v)$ denote the unique rotation of $\mathbb{R}^m$ which carries vector $u$ to vector $v$ and leaves everything orthogonal to $u$ and $v$ fixed. Alternatively, $T(u,v)$ can be defined by:

$$T(u,v)(x)=x - \frac{\langle(u+v),x\rangle}{1+\langle u,v\rangle}(u+v)+2\langle u,x\rangle v$$

Can somebody help me understand why this formula above works? I'm having troubles understanding why this formula gives the desired rotation. Thank you.

答: Page 77 in Milnor and Stasheff. They are required to be unit vectors.


问:
GRE topology definition
So on a certain GRE subject problem they define for a given a subset $Y$ of a topological space $X$

$$Y' = \{x \in X : x \in \text{cl}(Y \setminus \{x\})\}$$

Where $\text{cl}$ is closure. Is this just a fancy definition for the set of limit points of $Y$

Thanks!

答:
Yes.  This is known as the derived set, and is typically denoted with a 'prime'.

If the space is $T_1$, one can say that $Y'$ consists in those points $x $ for which there exists a non-trivial net in $Y $ converging to $x $.

A net is a generalization of a sequence.

Title: GRE topology definition


答:
Hints: Using $A$ for the given linear map and assuming $u$ and $v$ are unit vectors.

If $x\perp u,v$ then $A(x)=x$.

$A(u)=v$.

$A$ preserves norm and inner product in the span of $u,v$.

Deduce that $A$ is an orthogonal transformation.

Title: For $u,v,x \in \mathbb{R}^m$, let $T(u,v)(x)=x - \frac{\langle(u+v),x\rangle}{1+\langle u,v\rangle}(u+v)+2\langle u,x\rangle v$

Body:
The function in question appears in lemma $6.3$ of Milnors characteristic classes. Let $\langle \cdot, \cdot\rangle$ denote the dot product on $\mathbb{R}^m$

Let $u,v \in \mathbb{R}^m$ be unit vectors with $u \neq -v$. Let $T(u,v)$ denote the unique rotation of $\mathbb{R}^m$ which carries vector $u$ to vector $v$ and leaves everything orthogonal to $u$ and $v$ fixed. Alternatively, $T(u,v)$ can be defined by:

$$T(u,v)(x)=x - \frac{\langle(u+v),x\rangle}{1+\langle u,v\rangle}(u+v)+2\langle u,x\rangle v$$

Can somebody help me understand why this formula above works? I'm having troubles understanding why this formula gives the desired rotation. Thank you.


问:
Euler formula, polyhedron with a hole
I'd like to ask help for this question. I know that I need to have a little more concept on what kind of polyhedra that Euler's formula is valid.

Choose the false alternative:

a) F = 12

b) V = 20

c) E = 30

d) Euler's formula F + V - E = 2 applies

e) All above are true

Actually, by looking at this picture we can conclude that E is correct, since all of them seem to be true. However, shouldn't this formula be F + V - E = -2, since the genus is 2 ?

答:
See: https://en.wikipedia.org/wiki/Polygon_with_holes#In_polyhedra

$\chi =V-E+F-\text{holes}=2(1-g)$

$V=20, E=30, F=12, H=4$ (2 faces each with 2 holes)

$\chi=20-30+12-4=-2=2(1-g)
\implies g=2$ so it's $2$-holed polyhedron.

Title: Euler formula, polyhedron with a hole


问:
When $G/H$ is cyclic?
Clearly $G/H$ is cyclic if $G$ is cyclic. But sometimes $G/H$ is cyclic without $G$ necessarily cyclic, for example: $(\mathbb{Z}_6\times \mathbb{Z}_6)/\left \langle (2,3) \right \rangle \cong \mathbb{Z}_6$

In cases like the latter, how can one determine if the group is cyclic? I mean, is there something in general that lets me know?

For example

Write $(\mathbb{Z}_{20}\times \mathbb{Z}_6)/\left \langle (10,2) \right \rangle $ as an external direct product of cyclic groups of prime power order.

Is $(\mathbb{Z}_{20}\times \mathbb{Z}_6)/\left \langle (10,2) \right \rangle $ isomorphic to $\mathbb{Z}_{4}\times\mathbb{Z}_{5}$ or isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_2\times \mathbb{Z}_5$?

答:
According to the fundamental theorem of Abelian groups, you can write $G$ as a product $\mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_m}$. Now consider $f$ the morphism from $\mathbb{Z}^m$ to $G$, that consists of natural projections onto $n_i$. Let $\pi$ be the natural projection of $G$ onto $\frac{G}{H}$. Then $\frac{G}{H}$ is isomorphic to $\frac{\mathbb{Z}^m}{ker(\pi \circ f)}$.

Now we use the structure theorem of finitely generated modules over a PID. There exists a basis $(e_1 ,\cdots ,e_m)$ of $\mathbb{Z}^m$ and unique sequence of positive $d_1$ dividing $d_2$ dividing $d_3$ etcetera dividing $d_r$, such that $(d_1 e_1 , \cdots ,d_r e_r)$ is a basis of $ker(\pi \circ f)$, with $r$ being the rank of $ker(\pi \circ f)$. From this, we can simply deduce that $\frac{\mathbb{Z}^m}{ker(\pi \circ f)}$ is cyclic if and only if: $m=r$ and $d_1=\cdots = d_{m-1}=1$.

These $d_i$ can be calculated algorithmically if one knows a generating family of $ker(\pi \circ f)$. Now if we take $u_1, \cdots , u_k$ a family in $G$  generating $H$ (which is how one would typically describe a subgroup). Then if you take $v_1, \cdots , v_k \in \mathbb{Z}^m$, such that $f(v_i)=u_i$, then $v_1 , \cdots, v_m$ together with a family generating $ker(f)$ generate $ker(\pi \circ f)$.

Using that, you calculate the invariants $d_1, \cdots, d_m$  for the submodule $ker(\pi \circ f)$ and you can then conclude on whether $\frac{G}{H}$ is cyclic.

Now I am sure that for most cases you'll encounter, this method is an overkill, but it should work in general.

Title: When $G/H$ is cyclic?


问:
show $L_p(X)$ is a Banach space
Let $X$ be a Banach space and let $(\Omega, \Sigma, \mu)$ be a measure space. For simple functions
$$
f(\omega)=
\begin{cases}
x_1, \text{ if } \omega \in A_1,\\
\vdots \\
x_n, \text{ if } \omega \in A_n,\\
\end{cases},
$$
define
$$\|f\|_{L_p(X)} = \left( \sum_j \mu(A_j)\|x_j\|^p \right)^{1/p}$$
provided the $A_j$s are disjoint. In general,
$$\|f\|_{L_p(X)} = \left( \int \|f(\omega)\|^p_X \mathrm d\mu \right)^{1/p}$$
for all $f$ in the closure of simple functions. The completion is denoted by $L_p(X)$. Show that $L_p(X)$ is a Banach space.

I was able to show $\|cf\|_{L_p(X)} = |c|\|f\|_{L_p(X)}$ and $\|f\|_{L_p(X)} = 0 \implies f = 0$. But I do not know how to show the triangle inequality. Any help is appreciated. Thanks!

答:
So we need to show that

$$\|f\|_{L_p(X)} = \left( \int \|f(\omega)\|^p_X \mathrm d\mu \right)^{1/p}$$

is a norm, particularly the triangle inequality.

The triangle inequality:

$$\|f+g\|_{L_p(X)}^p = \left( \int \|f+g\|^p_X \mathrm d\mu \right)$$
$$\leq\left( \int (\|f\|_X+\|g\|_X)^p \mathrm d\mu \right)^{1/p}$$
$$=\left( \int  (\|f\|_X+\|g\|_X)(\|f\|_X+\|g\|_X)^{p-1} \mathrm d\mu \right)$$
$$=\left( \int  \|f\|_X(\|f\|_X+\|g\|_X)^{p-1} \mathrm d\mu +\int  \|g\|_X(\|f\|_X+\|g\|_X)^{p-1} \mathrm d\mu  \right)$$
$$\leq (\|f\|_{L_p(X)}+\|g\|_{L_p(X)})\|f+g\|_{L_p(X)}^{p-1}$$

Where in the last line we used Hölder, this is the Minowski inequality.

Title: show $L_p(X)$ is a Banach space


问:
Counting different weights in representations of Lie algebras
There is a well known way to compute dimensions of irreducible representations of classical Lie algebras using weights. However, there are weights that appear a few times in the representation. My question is: how can one know how many different weights appear in the representation? For example, a (1,1) representation of A2 is 8 -dimensional, but there are 2 coincident weights, and therefore we are left with just 7 different weights. So I am looking for a way to arrive to 7 and not to 8 in analogous cases, and specifically in the case of representations (1,1....1) of An.</p>

答:
If you are specifically interested in the representation $(1,1,\dots,1)$ for $A_n$ (often denoted by $\rho$ in the literature), then this has a very nice answer. The number of distinct weights is equal to the number of integral points of the "permutohedron," which is also equal to the number of forests of trees on $n$-labelled vertices. This is sequence A001858 of the OEIS.

More generally, the number of distinct weights in a given irreducible representation of any semisimple Lie algebra with highest weight $\lambda$ is the number of integral points in the weight polytope $\{W.\lambda\}$, where $W$ is the Weyl group.


问:
Is it possible to think of the number e as a fraction of pi?
$20$ years ago I did my Calculus $101$ "term paper" on this (Okay, what else would you call it? It was more of a story problem that we presented to the class, but as everyone else chose a practical applied story problem from the book, I was the only one to tackle a theoretical problem.... )

I remember yes, you can, e is just a fraction multiple of pi or something like that..., but I can't find out/remember exactly how to write it....

Thanks

答:
I think you should be more precise as "fraction" is an imprecise term. Is $e$ a rational multiple of $\pi$, or, equivalently, can you write $\frac{e} {\pi} = \frac ab$ where $a, b$ are integers sharing no common factors other than $1$? This is an open question, answer unknown. It is strongly suspected to be "no" (that $\frac{e} {\pi}$ is irrational) but without a conclusive proof one way or the other, the question remains open.

However, there is a very similar question about whether the product $e\pi$ is rational. This is also an open question. But the two quantities cannot both be rational (can you see why)?

Just for interest, there are similar open questions about $e+\pi$ and $e-\pi$, but again, both of these can't be simultaneously rational.

In fact of these four: $e+\pi, e-\pi, e\pi, \frac e{\pi}$, it is quite easy to show that at most one can be rational. Again, this is left as an exercise for you (should it interest you).


问:
Transform the squared error loss function to matrix form
I have n data, $x_i$ are the inputs, and $y_i$ are the labels. $x_i$ is $d$-dimension, and $y_i$ is $p$-dimension. We have a prediction function that is $f(x) = W^Tx$. $W$ is a matrix of dimension $d \times p$.

The squared error loss function of the prediction function is:
$$
  J(W) = \sum_{i=1}^n||y_i - W^Tx_i||_2^2
$$

I would like to write $J(W)$ in matrix form. How should I convert it to an equation involving $Y$ ($n \times p$ dimension), $W$ ($d \times p$ dimension) and $X$ ($n \times d$ dimension). Hence the loss function will be in matrix form instead of a sum?

答:
Given the set of data $D_{m}:=\left\{\left(x_{i}, y_{i}\right) \mid i=1,2, \cdots, m\right\}$ such that $x_{i} \neq x_{j}$ for $i \neq j .$ we must find the line $y=\alpha x+\beta$ that best
fits the data, i.e. find $(\alpha, \beta)$ such that for all $(a, b) \in \mathbb{R}^{2}$ we have
\begin{equation}
0 \leq \Phi(\alpha, \beta)=\sum_{i=1}^{m}\left(y_{i}-\alpha x_{i}-\beta\right)^{2} \leq \sum_{i=1}^{m}\left(y_{i}-a x_{i}-b\right)^{2}=\Phi(a, b)
\end{equation}
in matrix form, $\Phi(\alpha, \beta)=\Phi(z)=\|y-A z\|_{2}^{2}$, thus inequality $(1)$ is equivalent to the least square problem:
$$
\Phi(z)=\|y-A z\|_{2}^{2}=\min_{w \in \mathbb{R}^{n}}\|y-A w\|_{2}^{2}=\min_{w \in \mathbb{R}^{n}} \Phi(w)
$$
Since $y_{i}-a x_{i}-b=(y-A w)_{i}$ where : $w=[a ; b], z=[\alpha ; \beta], \mathbf{y}=\begin{bmatrix}y_{1} \\ y_{2} \\ y_{3} \\ \vdots \\ y_{m}\end{bmatrix}, \mathbf{A}=\begin{bmatrix}x_{1} & 1 \\ x_{2} & 1 \\ x_{3} & 1 \\ \vdots & \vdots \\ x_{m} & 1\end{bmatrix}, \mathbf{w}=\begin{bmatrix}a \\ b\end{bmatrix},$ and $\mathbf{z}=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}$.


问:
Hint to show $\frac{1}{n^2-n+1} \leq \frac{2}{n^2}$
Looking for a hint as to show the following inequality. I know it should be relatively simple but I just don't see it. I apologise as I'm aware it's rather basic, but I'm getting frustrated.

I'm reading an example concerning the proof of convergence of the following series
$$ \sum_{n=1}^\infty \frac{1}{n^2-n+1} $$
using a comparison with a series I know converges, naimley $\sum_{n=1}^\infty \frac{2}{n^2}$.

For this to work i need to show
$$ \frac{1}{n^2-n+1} \leq \frac{2}{n^2} $$

Algebraic manipulation hasn't gotten me anywhere so far. The subtraction of $n$ in the denominator is what's giving me issues I think. Many thanks.

Body:
Note that $$n^2 - 2n + 2 = (n^2 - 2n + 1) + 1 = (n-1)^2 + 1 > 0$$ so that
$$2(n^2 - n + 1) = 2n^2 - 2n + 2 > n^2.$$
In particular $n^2 - n + 1 > 0$, and you may divide both sides by $n^2(n^2 - n + 1)$ to get
$$\frac{2}{n^2} > \frac{1}{n^2 - n + 1}.$$


答:
For $n\ge 1$ the denominators are positive, so the inequality

$$\frac1{n^2-n+1}\le\frac2{n^2}$$

is equivalent to the inequality

$$n^2\le 2(n^2-n+1)\,,$$

which in turn is equivalent to $0\le n^2-2n+2$. Now think of the righthand side as $(n^2-2n+1)+1$ and rewrite it in a way that makes clear that it really is non-negative.

Title: Hint to show $\frac{1}{n^2-n+1} \leq \frac{2}{n^2}$

Body:
Looking for a hint as to show the following inequality. I know it should be relatively simple but I just don't see it. I apologise as I'm aware it's rather basic, but I'm getting frustrated.

I'm reading an example concerning the proof of convergence of the following series
$$ \sum_{n=1}^\infty \frac{1}{n^2-n+1} $$
using a comparison with a series I know converges, naimley $\sum_{n=1}^\infty \frac{2}{n^2}$.

For this to work i need to show
$$ \frac{1}{n^2-n+1} \leq \frac{2}{n^2} $$

Algebraic manipulation hasn't gotten me anywhere so far. The subtraction of $n$ in the denominator is what's giving me issues I think. Many thanks.


问:
prove that $f$ has a root in $K$ if and only if there exists a point $(b,c)$ such that $(b,c)≠ (0,0)$ and $f^h(b,c)=0$
Given the following definition:

For any univariate polynomial $f= a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ of degree $m$ in the ring $K[x]$, its homogenization in the ring $K[x,y]$ is
$f^h=a_mx^m+a_{m-1}x^{m-1}y+...+a_1xy^{m-1}+a_0y^m$

My question: How do I prove that $f$ has a root in the field $K$ if and only if there exists a point $(b,c)$ such that $(b,c)≠ (0,0)$ and $f^h(b,c)=0$

I am having trouble understanding the homogenization definition, it's just the same as the polynomial with another variable? also is the question saying if there exists a point in the homogenization that equals zero then $f$ has a root? Any help would be great.

答:
I think the homogenisation of $f$ is fairly straightforward. You basically do what it says, and get a polynomial of two variables rather than one. For example, if $f(x) = 3x^2 + 2x - 6$, then $f^h(x, y) = 3x^2 + 2xy - 6y^2$. If $f$ has degree $m$, another way to write it is:
$$f^h(x, y) = y^mf(xy^{-1}).$$
Indeed, this expression is very helpful in order to prove the result. If $f(b, c) = 0$ and $(b, c) \neq (0, 0)$, then either $b = 0$ or $c = 0$. If $c \neq 0$, then we get
$$0 = f^h(b, c) = c^m f(bc^{-1}) \implies f(bc^{-1}) = 0,$$
since $c^m \neq 0$. If $c = 0$, then we can't really use the above expression, since it features $c^{-1}$. However, we have
$$0 = f^h(b, 0) = a_m b^m  + 0 + 0 + \ldots + 0.$$
Since $f$ is of degree $m$, we have $a_m \neq 0$, and hence $b^m = 0$. As we are in a field, the only way this can happen is if $b = 0$, and hence $(b, c) = 0$, against assumption. Thus, necessarily, $c \neq 0$, and we have a root.

The converse is even easier. If $f$ has a root $x$, then $f(x) = f^h(x, 1) = 0$.


问:
martingale question for sums of N(0,1) random variables
Suppose $X_i$ are independent $N(0,1)$ distributed random variables.  Let $S_n=X_1+\cdots+X_n$ and let $Y_n=e^{S_n-\frac{n}{2}}$ and let $F^{Y_n}$ be the algebra generated by $Y_1, \ldots, Y_n$.  Show that $Y_n$ is a martingale in this algebra.

The problem basically comes down to finding the conditional expectation for each $Y_n$ but I'm now sure how to proceed.  Any help would be appreciated.

答:
$Y_n$ is integrable since it is lognormally distributed. It is also $F^{Y_n}$-measurable. Then
\begin{align*}\forall n\in\mathbb{N}^* : \quad \mathbb{E}\left[Y_{n+1}\mid F^{Y_n}\right]&=\mathbb{E}\left[\exp\left(S_n-\frac{n+1}{2}\right)\exp\left(Y_{n+1}\right)\mid F^{Y_n}\right] \\ &=\exp\left(S_n-\frac{n+1}{2}\right)\mathbb{E}\exp Y_{n+1} \\
&=Y_n\end{align*}
where I use the fact that $Y_n$ is $F^{Y_n}$-measurable and that $Y_{n+1}$ is independent of $F^{Y_n}$ in the second equality, and the fact that $\mathbb{E}e^{Y_{n+1}}=e^{1/2}$ in the third equality.


问:
Is $x+2$ invertible in given quotient ring?

Given the quotient ring $R=\mathbb{Q}[x]/(x^2-7)$. Is $x+2$ invertible in $R$?

I guess it is, but how to find it?

答:
$\, y:=x\!+\!2,\,\ 0 = x^2\!-7 = (y\!-\!2)^2\!-7 = y(y\!-\!4)-3\,\Rightarrow\,  1/y = (y\!-\!4)/3 = (x\!-\!2)/3\ \ $ QED

Similarly, over a field,  one can use a shift to transform a nonzero polynomial with $\,r\,$ as a root, into a polynomial having $\,s = r+a\,$ as a root. Any such polynomial yields the inverse, since

$$\ 0 = f(s) = s g(s) - f_0\ \Rightarrow\ s g(s) = f_0\ \Rightarrow \dfrac{1}s = \dfrac{g(s)}{f_0}$$

Note that we can always assume $\, f_0 = f(0)\ne 0\,$ by cancelling all factors of $\,x\,$ from $\,f(x).\ $ This can be considered a generalization of rationalizing denominators (as in Zev's post), which uses the fact that an algebraic number divides its norm (a special case of the above, when $\, f_0 =$ norm).

ContentLicense: CC BY-SA 3.0


问:
Is there a sense in which the Chi-squared distribution is an inner product?
I have been self-studying statistics recently, and the apparent similarities between linear algebra (especially Hilbert spaces) and statistics have been popping out to me. Linear independence gets replaced with statistical independence. The mean is like an origin to which random variables can be translated to change from an affine space into a vector space. The covariance is basically just an inner product.

At first I thought that these similarities might be coincidental, but after a lot of research, it became clear that they are not. There appears to be quite a lot of literature on this topic, but I have not yet studied this relationship very closely and am in the process of finding a textbook to read.

That said, based on the research that I have done so far, there is still one connection to vectors that has not been discussed anywhere I've looked: the Chi-Squared distribution.

Given the analogies described above, it seems obvious that the analogy for independent, standard normal random variables of the form

$$X=\frac{x-\mu_x}{\sigma_x}, \ \ Y=\frac{x-\mu_y}{\sigma_y}, \text{ etc.}
$$

would be orthogonal unit vectors.

With this understanding, one might construct a "random variable vector" ( $\chi$ ) of the form

$${\large \chi}(a,b,c) = aX+bY+cZ= \pmatrix{a \\ b \\ c}
$$

where $a$, $b$, and $c$ are constant, real-valued, and non-random. This vector is itself a normally distributed random variable with mean $\mu_\chi=a\mu_x+b\mu_y+c\mu_z=0$ and variance $\sigma_\chi^2=a^2\sigma_x^2+b^2\sigma_y^2+c^2\sigma_z^2=a^2+b^2+c^2$. So far, it is obeying all of the analogical rules established above.

That would lead naturally to the Chi-squared distribution

$${\large \chi^2} =\pmatrix{1 & 1 & 1} \pmatrix{1 \\ 1 \\ 1}=X^2+Y^2+Z^2
$$

which is not a "vector" (as expected). However, it is not clear to me how to interpret its meaning - especially its mean (in this case, $\mu_{\chi^2}=3$) or variance (in this case, $\sigma_{\chi^2}^2=6$).

So I have 3 inter-related questions:

1) Does it make sense to interpret independent, standard normal random variables as being akin to "orthogonal unit vectors"?

2) If it does, is there any meaning or utility in constructing a "random variable vector" as described above? If there is not, is there another way to construct a random variable vector that is meaningful?

3) Finally, is there any sense in which the Chi-squared distribution can be seen as an inner product between a "random variable vector" and itself? Can that inner product be generalized to be between different vectors? Can it be used to construct a "norm" on this space (say, with the Chi distribution)? If so, what exactly does that mean?

Note: I know that this is a question from ignorance built on possibly tenuous analogies. Since I am self-studying, I can't help but draw connections between this new material and the sort of math that I am already comfortable with. If there is any way that I can improve this question, please let me know.

答:
Interesting question!

First of all, yes, it does make sense to define independent random variables as different components of a vector.  For example, the position of a particle in three dimensions might well be determined by $x,y,$ and $z$ each being a standard normal random variable.  So it can make literal geometric sense to do as you suggest.

People make random vectors all the time.  They see common use in linear regression with multiple variables, for example.  Observations of several variables at once are often listed forming a vector.  Many of the regression formulas are written with matrices and vectors.

You can call something a vector if the collection of things like it satisfies the rules of a vector space. Identity, inverse, commutativity of addition, multiplication by a constant, and so forth. I see no reason why you couldn't make your vectors of normal random variables into a vector space.  However, you need to decide if you want to use just normal random variables, or allow any random variables.  The sum of two normal random variables is normal, but that isn't true for many other distributions.  Also you have to be careful what you are using as scalars, if you want to multiply components by them!

The problem comes in when you sum them.  If you add three random variables together with coefficients, you get, as you pointed out, one random variable.  The closest geometric analogy would be using the Manhattan metric in three dimensions, where your sum might be the distance between two points moving only along a rectangular grid.  But three components of a vector do not sum up to one of anything in particular, generally.

The $\chi^2$ random variable for positive integer degrees of freedom can be considered as a sum of squared normal variables, so it does look rather like a dot product, doesn't it?  However, as I pointed out above, you need to know what your scalars are.  If they are regular numbers, then a chi-square random variable is not a number and so cannot be a dot product result.  If you wanted to use chi square random variables, then you have the problem of multiplying a $Z$ by a $\chi^2$ and not getting a $Z$ back again, so that you wouldn't have a vector any more.

If you were to make a very general space, where components can be any random variable at all, and the scalars are also random variables, then you might be able to make it work.  Covariance might not makes sense any more as a dot product in that case, however.


问:
Let $\alpha$ be algebraic over $F$. Suppose the degree of $\alpha$ over $F$ is odd. Show that $F(\alpha)=F(\alpha^2)$?
I am trying to solve the following exercise:

Let $\alpha$ be algebraic over $F$. Suppose the degree of $\alpha$ over $F$ is odd. Show that $F(\alpha)=F(\alpha^2)$

It's not clear to me how to solve this. I've been thinking and I guess that $F(\alpha^2)\subset F(\alpha)$ but I don't know about the other direction. Can you help me?

答: $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$ since $[F(\alpha):F]$ is odd, we deduce that $[F(\alpha):F(\alpha^2)]$ is odd and is $1$ since the fact $\alpha$ is a solution of $X^2-\alpha^2$ implies that $[F(\alpha):F(\alpha^2)]\leq 2$.


问:
A simple problem of linear algebra in infinite dimension
Let $E$ be a vector space (of infinite dimension) and $u : E \rightarrow E$. Suppose that $E/u(E)$ has finite dimension. Is it true then that this dimension is equal to $\dim \ker u$ ?

In finite dimension the rank theorem says yes. But here $E$ is typically infinite-dimensional. I realize it is still true that there is an isomorphism $\overline{u} : E/\ker u \rightarrow u(E)$, so I guess if we could prove that $E/(E/\ker u)) \simeq \ker u$ it would be enough.

EDIT : I'm not sure if this is relevant, but in my situation there is a (Banach) topology on $E$ and $u$ is continuous. Maybe this is important if we need to have direct spaces. However the question is purely algebraic, so I don't really know...

答:
No, this isn't necessarily true. One of the main novel aspects of infinite-dimensional vector spaces is that they can be isomorphic to proper subspaces, i.e. $u$ may map $E$ onto $D\subsetneq E$ with no kernel. This is essentially no different that the set theoretic fact that infinite sets are in bijection with proper subsets.

For a concrete example, consider the vector space $\Bbb{R}[X]$ of real polynomials and have $u(p)=pX$. $u$ has no kernel, but the one-dimensional subspace of constant polynomials is missing from the image. It doesn't matter that your $E$ is a Banach space, as similar examples are then forthcoming.


问:
Fatou's lemma - Royden's proof
Royden states that to prove Fatou's lemma it is ncessary and sufficent to show that if $h$ is any bounded measurable function of finite support for which  $0\leq h\leq f$ on $E$, then
$$
\int
_E h\leq \lim \inf \int_E f_n$$

Why did Royden choose to construct such a function instead of working directly with $f$?

答: By definition, $\int f$ is the supremum of $\int h$ for $0 \leq h \leq f$ bounded measurable of finite support. So at the very end, you just need to take the sup over such $h$ to get $\int f \leq \liminf \int f_n$. He uses such $h$ is order to apply the bounded dominated convergence theorem, which he proves earlier. This allows him to prove Fatou in terms of dominated convergence instead of giving a completely independent proof (and potentially repeating a bunch of work).

Title: Fatou's lemma - Royden's proof


问:
From the given information, how do we use the Remainder Theorem to reach the answer?
Why does the sequence for the polynomial alternate when it equals $-30$, and not alternate when it equals $10$?  Why is the denominator when the sum of the geometric series is equal to $30$, $1+r$, and not $1-r$?  I am unable to correctly relate the Remainder Theorem to its use in the given answer.

The question:

When a polynomial of degree $3$ is written in descending powers of $x$ the coefficients of its four terms form a geometric sequence.

When the polynomial is divided by $x+1$ it s remainder is $-30$.  Then it is divided by $x-1$ its remainder is $10$.

The answer:

Use the Remainder Theorem.  We have

$-a+ar-ar^2+ar^3=-30$

and

$a+ar+ar^2+ar^3=10$

Solving simultaneously:

$\frac{a(1-r^4)}{1+r}=30$

and

$\frac{a(1-r^4)}{1-r}=10$

Giving $r=-\frac{1}{2}$ and $a=16$

The polynomial is therefore $16x^3-8x^2+4x-2$

答:
The polynomial is a cubic whose coefficients are a geometric sequence (in the order of descending powers, i.e. starting with $x^3$ and ending with the constant term), meaning that our polynomial must take the form
$$p(x) = ax^3 + arx^2 + ar^2x + ar^3.$$
Remainder theorem states that the remainder of $p(x)$ when divided by $x - \alpha$, is $p(\alpha)$.

So, when dividing $p(x)$ by $x - 1$ (i.e. $\alpha = 1$), the remainder is $p(1)$. Thus,
$$10 = p(1) = a1^3 + ar1^2 + ar^21 + ar^3 = a + ar + ar^2 + ar^3.$$

Similarly, when dividing $p(x)$ by $x + 1$ (i.e. $\alpha = -1$), the remainder is $p(-1)$. Thus,
$$-30 = p(-1) = a(-1)^3 + ar(-1)^2 + ar^2(-1) + ar^3 = -a + ar - ar^2 + ar^3.$$

This should hopefully explain the alternating signs. The next observation you need to make is that both of the above expressions are geometric series. Each have $4$ terms. The first has an initial term of $a$ and a common ratio of $r$, while the second has an initial term of $-a$ and a common ratio of $-r$. The first one is precisely the usual formula for geoemetric series:
$$10 = a + ar + ar^2 + ar^3 = a\frac{1 - r^4}{1 - r}.$$

To get a formula for the second one, we simply replace $r$ with $-r$ and $a$ with $-a$, to get
$$-30 = -a + ar - ar^2 + ar^3 = (-a)\frac{1 - (-r)^4}{1 - (-r)} = -a\frac{1 - r^4}{1 + r}.$$
To get the equation from the solution, simply multiply both sides by $-1$.


问:
powers of linear operator
Let $T \in L(V)$, where $V$ is a finite dimensional vector space over a field and $L(V)$ is the space of all linear operator. In Sheldon Axler’s textbook, I am asked to verify:

$T^mT^n = T^{m+n}$, and $(T^m)^n = T^{mn}$.

However, I am not sure what the question is asking; is there even anything to verify? It seems to me that there is nothing to be proved here.

Thanks in advance.

答:
By definition, $T^m$ is $T$ composed with itself $m$ times, and $T^n$ is $T$ composed with itself $n$ times. So if you compose $T^m$ with $T^n$, this is $T$ composed with itself $m+n$ times. So by definition, $T^mT^n=T^{m+n}$.

Similarly, if we compose $T^m$ with itself $n$ times, this is the same as $T$ composed with itself $mn$ times, thus by definition we have $(T^m)^n=T^{mn}$.

Title: powers of linear operator


问:
Finding a paper
Sometimes I have trouble finding particular papers.  Does anyone have any general  suggestions?

I find myself in this situation frequently:  An author references a paper.  I find it on MathSciNet and read the abstract.  I then realize I want to read the paper.  Then I cannot find the paper.

Is there any better approach besides just using Google Scholar, Google, the author's webpage, JStor and MathSciNet?  I am at a University, and have some access to catalogues.

Specifically I wanted to read this but I can't seem to do it online.

Thanks a lot!

Update:  I ended up finding the paper by talking to a Professor I know who had a PDF on hand.  None of the links below lead to the correct paper, which is why I have not accepted an answer.

答: It may be worth considering ResearchGate.</p>

Title: Finding a paper


问:
DFCC (discrete finite chain condition )
A space $X$ has the DFCC (discrete  finite chain condition ) provided every discrete family of nonempty open sets is finite. (see A survey on star covering propertiesMV Matveev - Topology Atlas, preprint, 1998)

Does the discrete family of nonempty open sets mean that every point $x \in X $ has a nbhd that intersects at most one set of the given family?

However, in the proof of the Theorem 14 in the charpter 2, why he said if the sequence $S=\{U_n:n\in \omega\}$ is not DFCC, then there exists $x \in X$ is an accumulation point for $S$, each nbhd of $x$ meets infinitely many $U_n$'s? Why is "infinitely"? This is puzzling me.

Thanks for any help:)

答:
Does the discrete family of nonempty open sets mean that every point $x\in X$ has a nbhd that intersects at most one set of the given family?

Accordingly to Engelking’s “General topology”, yes.

Why is "infinitely"?

It seems that there is an error in the proof exposition.  The contradiction is already implied from the fact that a neighborhood $V$ intersects two different $U_n$-s. Therefore no such neighborhood of $x$ can exist and the sequence $s$ should be discrete.

See, for instance, the proof of Theorem 2.1.7 (that each DFCC space is 2-starcompact) in "Star-covering properties" by E.K. van Douwen, G.M. Reed, A.W. Roscoe, and I.J. Tree.

Title: DFCC (discrete finite chain condition )


问:
For $u,v,x \in \mathbb{R}^m$, let $T(u,v)(x)=x - \frac{\langle(u+v),x\rangle}{1+\langle u,v\rangle}(u+v)+2\langle u,x\rangle v$
The function in question appears in lemma $6.3$ of Milnors characteristic classes. Let $\langle \cdot, \cdot\rangle$ denote the dot product on $\mathbb{R}^m$

Let $u,v \in \mathbb{R}^m$ be unit vectors with $u \neq -v$. Let $T(u,v)$ denote the unique rotation of $\mathbb{R}^m$ which carries vector $u$ to vector $v$ and leaves everything orthogonal to $u$ and $v$ fixed. Alternatively, $T(u,v)$ can be defined by:

$$T(u,v)(x)=x - \frac{\langle(u+v),x\rangle}{1+\langle u,v\rangle}(u+v)+2\langle u,x\rangle v$$

Can somebody help me understand why this formula above works? I'm having troubles understanding why this formula gives the desired rotation. Thank you.

答:
Note $\|u+v\|^2=2(1+\langle u,v\rangle)$. If we write $\displaystyle w=\frac{u+v}{\|u+v\|}$, the given formula is

$$ R = \mathrm{Id}-2ww^T+2vu^T. $$

Where does this formula come from?

One clue is that $\mathrm{Id}-2ww^T$ is the Householder matrix for reflection across $w$'s orthogonal complement. Indeed, a good geometry exercise is that the product of two hyperplane reflections is a plane rotation by $2\theta$, in the plane spanned by the hyperplanes' two normals, where $\theta$ is the angle between the hyperplanes. One can argue it suffices to do this in 2D, and then it becomes a synthetic geometry exercise.

Note that $w$ is halfway between $u$ and $v$, so the angle $\angle uw$ is half the angle $\angle uv$. Therefore we may take the product of the two Householder reflections associated with $u,w$'s complements:

$$ (\mathrm{Id}-2ww^T)(\mathrm{Id}-2uu^T). $$

If you FOIL this out, then simplify with $w=u+v$, you get $R$ (exercise).


问:
If $(N_t)_{t\geq 0}$ is a Poisson Process w/ parameter $\lambda$, then $(N_{t+s} - N_s)_{t\geq 0}$ is also PP$(\lambda)$
All of the notes I've seen on Poisson Processes just hand wave this proof and say that it just immediately follows from the memory-less property of Exponential RVs. Could someone walk though how one could show this? Thanks.</p>

答:
Let $N(t)$ be a Poisson process. Then

Then $N(0)=0$.

The increments are independent, e.g. $N(t_1,t_1+h_1),N(t_2,t_2+h_2)$ are independent as long as they don't overlap.

$N(h,h+t)$ follows $Poi(t\lambda)$.

Now consider $N^*(t)=N(t+s)-N(s)=N(s,s+t)$.

First $$N^*(0)=N(t-0)-N(t)=0$$.

Second for any $i$ when the intervals don't overlap $$N^*(t_i,t_i+h_i)=N(t_i+s,t_i+h_i+s),$$

so that the increments are independent by 2.

Finally $N^*(h,h+t)=N(s+t)-N(s+h)=N(s+h,s+h+t)$, which follows $Poi(\lambda t)$ by 3.


问:
Understanding the Basic Mathematics of a Kalman Filter

In Introduction to Linear Algebra, Gilbert Strang, it says that for a Kalman filter
$$
\hat{x}_1=\hat{x}_0+K_1(b_1-A_1\hat{x}_0)
$$
where the Kalman gain matrix $K_1=W_1A_1^TV_1^{-1}$ and covariance of errors in $\hat{x}_1$ is $W_1^{-1}=W_0^{-1}+A_1^{T}V_1^{-1}A_1$.

I think I understand what Kalman filter is trying to do, when new data is coming instead of computing the least square solution for the whole data we are trying to make use of the least square solution for old data to find that for the whole data. And I think here we are assuming the data error has a normal distribution and that is why the covariance matrix comes into picture.

I also understand weighted least square solution for $Ax=b$ is $\hat{x}=(A^TV^{-1}A)^{-1}A^TV^{-1}b$, and how we arrive at this expression.

But how do we arrive at the term $K_1(b_1-A_1\hat{x}_0)$ ?

I am having difficulty following the mathematical form of the equation as it is ?

Reference: Page 560,Chapter 12-Linear Algebra in Probability & Statistics, Introdcution to Linear Algebra, Gilbert Strang

答:
A quick note before answering your question.

The Kalman filter (KF) is most often used if the parameters being estimated are changing in time (i.e., dynamic parameters); however, the KF can also be used if the parameters being estimated are not changing in time (i.e., static parameters). In such cases, the KF is equivalent to weighted least squares. For example, if you look at the KF equations as described by Wikipedia, the equations look different; however, I can ensure you that the equations are identical with some algebra (except for the value covariance matrix used in the gain, which I'll mention in a moment as I suspect this is a typo in the reference, but I am not completely sure.).

Answer:

The estimate $\hat{\mathbf{x}}_0$ given $\mathbf{b}_0$ is given by (15) as follows:
$$ \hat{\mathbf{x}}_0 = (A_0^T V_0^{-1} A_0)^{-1} A_0^T V_0^{-1} \mathbf{b}_0 $$

If we expand (16), then we have the following:
$$
\begin{align}
     A_0^T V_0^{-1} A_0 \hat{\mathbf{x}}_1 + A_1^T V_1^{-1} A_1 \hat{\mathbf{x}}_1 = A_0^T V_0^{-1} \mathbf{b}_0 + A_1^T V_1^{-1} \mathbf{b}_1,
\end{align}
$$
and simplifying and rearranging, we have
$$
\begin{align}
A_0^T V_0^{-1} A_0 \hat{\mathbf{x}}_1 &= A_0^T V_0^{-1} \mathbf{b}_0 + A_1^T V_1^{-1} \mathbf{b}_1 - A_1^T V_1^{-1} A_1 \hat{\mathbf{x}}_1 \\
&= A_0^T V_0^{-1} \mathbf{b}_0 + A_1^T V_1^{-1} (\mathbf{b}_1 - A_1 \hat{\mathbf{x}}_1)
\end{align}
$$
then, left multiplying by $(A_0^T V_0^{-1} A_0)^{-1}$ gives the following:
$$
\hat{\mathbf{x}}_1  = \underbrace{(A_0^T V_0^{-1} A_0)^{-1}A_0^T V_0^{-1} \mathbf{b}_0}_{\hat{\mathbf{x}}_0} + \underbrace{(A_0^T V_0^{-1} A_0)^{-1} A_1^T V_1^{-1}}_{K_1} (\mathbf{b}_1 - A_1 \hat{\mathbf{x}}_1)
$$

However, notice in the equation above that $K_1 = W_0 A_1^T V_1^{-1}$ instead of $K_1 = W_1 A_1^T V_1^{-1}$, which is the equation provided in the reference. In the KF (assuming static parameters), the updated covariance is given as follows:
$$ W_k^{-1} = W_{k-1}^{-1} + A_k^T V_k^{-1} A_k, $$
which is consistent with the reference. However, the covariance matrix that should be used in the Kalman gain is the prior covariance matrix, but in the reference, the posterior covariance matrix is used instead. The posterior covariance matrix represents the uncertainty in the estimate after performing the update, so the posterior covariance should be not used in the gain. I believe this is a typo in the reference.


答:
Another counterexample but this time: $dim ker u > dim(E/u(E))$.

Consider $E$ to be teh space of all polynomials with real or complex coefficients and consider $u$ to be the derivative operator. The image is the entire space while kernel is non-empty.

$$ u(a_{0} + \cdots + a_{n}x^{n}) = a_{1} + 2a_{2}x + \cdots + na_{n}x^{n-1}$$

Title: A simple problem of linear algebra in infinite dimension

Body:
Let $E$ be a vector space (of infinite dimension) and $u : E \rightarrow E$. Suppose that $E/u(E)$ has finite dimension. Is it true then that this dimension is equal to $\dim \ker u$ ?

In finite dimension the rank theorem says yes. But here $E$ is typically infinite-dimensional. I realize it is still true that there is an isomorphism $\overline{u} : E/\ker u \rightarrow u(E)$, so I guess if we could prove that $E/(E/\ker u)) \simeq \ker u$ it would be enough.

EDIT : I'm not sure if this is relevant, but in my situation there is a (Banach) topology on $E$ and $u$ is continuous. Maybe this is important if we need to have direct spaces. However the question is purely algebraic, so I don't really know...


问:
About the curvature of a curve
Let $\alpha : I \rightarrow R^2$ be a smooth curve. For each $t \in I$ consider $N(t)$ the normal unit vector at the point $\alpha(t)$. Fix $\lambda > 0 $ a constant and define the parallel curve $\gamma(t) = \alpha (t) + N(t) \lambda$. Someone can give me a help to calculate the curvature of $\gamma$ with respect to the curvature of $\alpha$? My intuition says that will be something like:

$$K_{\gamma}(t) = \frac{K_{\alpha} (t)}{ 1 + \lambda K_{\alpha(t)}}$$

where $K_{\gamma}$ and $K_{\alpha}$ are the curvatures of $\gamma$ and $\alpha$ respectively. Someone can give me a help ? I am not good with these things ...

Thanks in advance

答:
The unit speed condition is not essential, the only required condition is $\alpha$ to be regular. If $\alpha$ is unit speed computations are simpler. Anyway, if $\alpha$ is regular the curve $\gamma(t)$ is regular whenever $\lambda k_{s}  \neq 1$, where $k_{s}$ is the signed curvature of $\alpha$. To see this, let $s$ be the arc-length parameter of $\alpha$ and $\mathbf{t}$ be the unit tangent vector of $\alpha$, then
$$
\frac{d \gamma}{dt} = \frac{d \alpha}{dt} + \lambda \frac{d\mathbf{n}}{dt} = \frac{ds}{dt}\mathbf{t} + \lambda \frac{d\mathbf{n}}{ds} \frac{ds}{dt} ,
$$
but it is known that $\frac{d\mathbf{n}}{ds} = -k_{s} \mathbf{t}$, so
$$
\frac{d\gamma}{dt} = \frac{ds}{dt} \mathbf{t} - \lambda k_{s} \frac{ds}{dt} \mathbf{t} = (1 - \lambda k_{s}) \frac{ds}{dt} \mathbf{t}. \qquad (1)
$$
Therefore, $\gamma$ is regular whenever $1 - \lambda k_{s} \neq 0$, as desired. An additional fact is that the signed curvature of $\gamma$ is
$$
k_{s}^{\lambda} = \frac{k_{s}}{\lvert 1 -
\lambda k_{s} \rvert }.
$$
To show this, let $s^{\lambda}$ be the arc-length parameter of $\gamma$, then by equation (1) we have
$$
\frac{d s^{\lambda}}{dt} = \lvert 1 - \lambda k_{s} \rvert \frac{ds}{dt}.
$$
Let the $\mathbf{t}^{\lambda}$ be the unit tangent vector of $\gamma$, then
$$
\mathbf{t}^{\lambda} = \frac{d \gamma}{d s^{\lambda }} = \frac{d\gamma}{dt}\frac{dt}{ds^{\lambda}} = \frac{d\gamma}{dt}/\frac{ds^{\lambda}}{dt} = (1- \lambda k_{s}) \frac{ds}{dt} \mathbf{t}\left/\lvert 1 - \lambda k_{s} \rvert \frac{ds}{dt}\right. = \pm \mathbf{t} = \epsilon \mathbf{t}.
$$
Then, the unit normal vector of $\gamma$ is ${n}^{\lambda} = \epsilon \mathbf{n}$. By definition, the signed curvature $k_{s}^{\lambda} $of $\gamma$ is given by
$$
k_{s}^{\lambda} \mathbf{n}^{\lambda} = \frac{d \mathbf{t}^{\lambda}}{d s^{\lambda}} = \frac{d \mathbf{t}^{\lambda}}{dt}\left/ \frac{ds^{\lambda}}{dt}\right. = \epsilon \frac{d \mathbf{t}}{dt}\left/ \lvert 1 - \lambda k_{s} \rvert\frac{ds}{dt}\right.
$$
that is to say,

$$
k_{s}^{\lambda} \mathbf{n}^{\lambda} = \epsilon \lvert 1 - \lambda k_{s} \rvert ^{-1} \frac{dt}{ds} \frac{d \mathbf{t}}{dt} = \epsilon  \lvert 1 - \lambda k_{s} \rvert ^{-1} \frac{d \mathbf{t}}{ds} = \epsilon \lvert  1- \lambda k_{s} \rvert ^{-1} k _{s} \mathbf{n} =k_{s}\lvert 1 - \lambda k_{s} \rvert^{-1} (\epsilon \mathbf{n}) = k_{s} \lvert 1 - \lambda k_{s} \rvert^{-1} \mathbf{n}^{\lambda},
$$
which implies
$$
k_{s}^{\lambda} = \frac{k_{s}}{ \lvert 1 - \lambda k_{s}\lvert.}
$$

The curve $\gamma$ is parallel to $\alpha$ in the sense that the distance bewteen $\gamma$ and $\alpha $ is $\lVert \gamma(t) - \alpha (t) \rVert = \lambda$, for all $t$, and $(\gamma(t) - \alpha(t))\cdot \mathbf{t}(t) = 0 $, for all $t$; however this is not the case for $\lambda$ too large, as is shown in the next example

where $\alpha(t) = (2 \cos t, \sin (t))$ and $\gamma = \alpha + \lambda \mathbf{n}$.

Title: About the curvature of a curve


问:
Smooth Coordinates for which a Vector Field is a Coordinate Vector Field
Given a vector field $W = x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}$, I want to find smooth coordinates around $(1, 0)$ such that this vector field is a coordinate vector field. I know that the flow for this field is given by $\theta_t(x,y)=(xe^t,ye^{2t})$.

I think that I need to find a smooth coordinate representation of $W$ such that mapping $(1,0)\mapsto \frac{\partial}{\partial x^i}\big|_{(1,0)}$ gives us our original field, but I have no clue explicitly how to calculate it. I think I need to calculate flows for the representation (probably parameterized by other variables), but I'm not sure. If it helps, this is problem 9-10 from Lee's Intro to Manifolds.

答:
In Lee's book, take a close look at Theorem 9.22, which has the label "Canonical Form Near a Regular Point" (it is on page 220 in the edition I'm using). It is the theorem which states that what you are trying to do is possible (i.e. find coordinates so that the chosen vector field is a coordinate vector field). This is also often called the "straightening theorem" (e.g. this Wikipedia page)

If you read the proof, it tells you how to construct such coordinates.

Start by finding a curve through the point $(1,0)$ which is not tangent to the vector field near the point. For example, we can take the vertical line $x=1$. We can parameterize this line using the $y$-coordinate. For each point on this vertical line, we can apply the flow of $W$ for time $t$ (for $t \in (-\varepsilon, \varepsilon)$), and this will generate a small neighborhood around $(1,0)$.

In other words, our new coordinate system is $(t,v)$, which is the point obtained by starting at $(1,v)$ (i.e. a point on the vertical line, in the original $x,y$ coordinates) and flowing for time $t$, using the $\theta_t$ you mentioned. This gives the coordinate transformation

$$ (x,y) = (e^t, v \, e^{2t}) $$

You can confirm (using the chain rule) that $\frac{\partial}{\partial t}$ is the desired vector field:

$$ \frac{\partial}{\partial t} = \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial}{\partial y} = e^t \, \frac{\partial}{\partial x} + 2v e^{2t} \, \frac{\partial}{\partial y} = x \, \frac{\partial}{\partial x} + 2y \, \frac{\partial}{\partial y} $$


问:
Doob-like equality with conditional expectation
Suppose $X_t$ is a positive, continuous martingale w.r.t. a filtration $\mathcal{F}_t$ such that $X_t \to 0$ a.s. when $t \to \infty$. Given $X^* = \sup_{t \in \mathbb{R}^+} X_t$, I want to prove that, for a given $x \in \mathbb{R}^+$
\begin{equation*}
\mathbb{P}(X* \geq x \mid \mathcal{F}_0) = 1 \land \frac{X_0}{x}
\end{equation*}

Where $a \land b = \min{(a,b)}$

This is very similar to the Doob martingale inequality, and, to get rid of the conditional expectation, I have tried to work with the sequence $I_{B} X_n$ with $B \in \mathcal{F}_0$ and prove it like Rosenthal;s A First Look at Rigorous Probability Theory proof of Doob's inequality (Theorem 14.3.1):

Let $A_k = \{ X_k \geq x , X_i < x \;, \; \forall \; 0< i <j\}$. Note that $A_i \cap A_j = \emptyset$ . So

$$ A = \bigcup_{k=0}^\infty A_k = \{ X* \geq x\}$$

Then

\begin{split}
x \; \mathbb{P}(A\mid B) &= x \sum_{k=0}^\infty \mathbb{E}[{I_A I_B}] \\
&=  \sum_{k=0}^\infty \mathbb{E}[x {I_A I_B}] \\
\end{split}

But I keep stuck there. ¿Is this the right path?

答:
Disclaimer: This is only an incomplete answer.

If $x\leq X_0$ the result is immediate. Now, let $x>X_0\geq 0$ and $\tau_x:=\inf\left\{n\in\mathbb{N}: X_n\geq x\right\}$. For all $k\geq 0$, the stopped process $\left(X_{n\land k \land \tau_x}\right)_{n\geq 0}$ is a bounded martingale, so the optional stopping theorem implies
$$\forall k,n\in\mathbb{N} : \quad \mathbb{E}X_{n\land k\land \tau_x}=\mathbb{E}X_0=X_0$$
Applying the dominated convergence theorem:
$$\lim_{n\uparrow\infty}\mathbb{E}X_{n\land k\land\tau_x}=\mathbb{E}\lim_{n\uparrow\infty}X_{n\land k\land\tau_x}=\mathbb{E}\left[\mathbf{1}_{\left\{\tau_x>k\right\}}X_k\right]+\mathbb{E}\left[\mathbf{1}_{\left\{\tau_x\leq k\right\}}X_{\tau_x}\right]$$
Letting $k\uparrow\infty$ we have by a combination of the monotone convergence theorem and the dominated convergence theorem
$$\mathbb{E}\left[\mathbf{1}_{\left\{\tau_x<\infty\right\}}X_{\tau_x}\right]=X_0$$
This shows that $\mathbb{P}(X^\star\geq x)\leq X_0/x$. For the other inequality, I think that we would need some continuity argument...


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
A proof which, unlike all other suggestions, doesn't require power as a prerequisite.

This is what your son noticed by himself by looking at the first occurences in the sequence:

1 × 100 + 2 × 10 + 8 = 128

Now, let him remark that he can multiply everything by 2 without affecting the equality:

2 × (1 × 100 + 2 × 10 + 8) = 2 × 128

Assuming that he understands that the doubling distributes over the sum, he will obtain:

2 × 1 × 100 + 2 × 2 × 10 + 2 × 8 = 2 × 128

Then remark that when you double any pick you obtain of course the next pick in the sequence:

2 × 100 + 4 × 10 + 16 = 256

Repeat. Multiply both sides by two, distribute and double every pick again, and you obtain one more equality:

4 × 100 + 8 × 10 +  32 = 512

Repeat.

Now he should be pretty much convinced that his statement holds for any pick.
(Of course it could be a good opportunity to introduce recurrence!)


问:
Understanding why the definition $\cos(\widehat{vw})=\frac{v\cdot w}{\|v\| \|w\|} $ works so well.
After perusing several geometry / vector space posts on the forum to better understand formal notions of angles and rotations, while avoiding circularity, I came across a definition that I really like from How do you formally prove that rotation is a linear transformation? :

$$\cos(\widehat{vw})=\frac{v\cdot w}{\|v\| \|w\|} $$

After playing around with this definition, I noticed that one of the properties this definition (obviously quite purposefully) possesses is that holding $w$ constant, one can vary the length (but not 'direction') of $v$ without changing the value of the expression (which adheres to our intuitive notion of an 'angle' between two vectors).

I am going through the Chapter 4 appendices of Spivak's Calculus, so I have not yet really even transitioned into any notion of derivative/rates of change, etc. I was wondering if an analytical justification for my observation can be stated without using calculus.

More or less, I am wondering if I can gain insight into why the value of the expression is independent of the length of $v$.

Perhaps the very nature of this question cannot be answered without a nod to calculus, in which case, my apologies for the poorly posed question.

Edit 1:

It seems like the idea here is that, treating the dot product as a function, $\cdot(v,w)$ and $\sqrt{\cdot(v,v)}\sqrt{\cdot (w,w)}$ change equally while varying the length of $v$.

Edit 2:

After some thinking, one can reduce my argument to:

Prove $\cos \big ( \widehat{(\alpha v)w})=\cos(\widehat{vw})$ for any scalar $\alpha$.

Applying the definition: $\cos \big ( \widehat{(\alpha v)w})=\dfrac{(\alpha v) \cdot w}{\lVert \alpha v \rVert \lVert w \rVert}$

From properties of norms and dot products, we know the following:

$(\alpha v) \cdot w=\alpha(v \cdot w)$

$\lVert \alpha v \rVert = \alpha \lVert v \rVert$

Applying these two properties, we get:

$\dfrac{(\alpha v) \cdot w}{\lVert \alpha v \rVert \lVert w \rVert} = \dfrac{\alpha (v \cdot w)}{\alpha \lVert v \rVert \lVert w \rVert}$

Canceling out the $\alpha$'s, the answer follows.

Using the same approach, we can further generalize this to:

$\cos \big ( \widehat{(\alpha v)(\beta w})=\cos(\widehat{vw})$ for any scalars $\alpha$ and $\beta$.

答:
If you are willing to believe that two vectors are orthogonal to each other iff their dot product is $0,$ then consider this:

By computing the dot product, one can see that

$\dfrac{u\cdot v}{\|v\|\|v\|} v$ and $u - \dfrac{u\cdot v}{\|v\|\|v\|} v$ are orthogonal to each other, and

the first of the two vectors above is in the direction of $v,$ and

the sum of those two vectors is $u.$

Thus we have a right triangle in which the adjacent side is $\dfrac{u\cdot v}{\|v\|\|v\|} v$ and the hypotenuse is $u$. The cosine is the length of the adjacent side divide by the length of the hypotenuse.


问:
Conditional Independence and probability
There are two coins each having probability of heads as $0.9$ and $0.1$ respectively. And probability of choosing a coin for experiment is $0.5$ ( a coin once picked at beginning is used for every toss ).
P(Heads in $11$th toss) = $0.5$

P(Heads in $11$th toss | Heads in first $10$ tosses ) = $0.9$ (approx) as we are almost certain that $1$st coin is used therefore value obtained is $0.9$.
I tried to prove it mathematically

P(Heads in $11$th toss | Heads in first $10$ tosses ) = $$\frac{P(Heads\ in \ 11th \ toss \ ∩ Heads \ in \ first \ 10 \ tosses )}{ P(Heads \ in \ first \ 10 \ tosses)}$$

P(Heads in $11$th toss ∩ Heads in first $10$ tosses ) = $0.5 * ( (0.9)^{11}) + 0.5 * ( (0.1)^{11}) $

P(Heads in first $10$ tosses) =  $0.5 * ( (0.9)^{10}) + 0.5 * ( (0.1)^{10})$

P(Heads in $11$th toss | Heads in first $10$ tosses ) = $( 0.9^{11} + 0.1^{11})/(0.9^{10}+0.1^{10}) = 0.8999$

But there is another way to write P(Heads in $11$th toss ∩ Heads in first $10$ tosses ) as P(Heads in first $10$ tosses|Heads in $11$th toss  ) * P(Heads in $11$th toss)

I am not able to figure out what would be the value of  P(Heads in first $10$ tosses|Heads on $11$th toss ) and how to find it?

If there were no biases in the coins would that make the above events independent?

答:
If I understand your question correctly, I think that you should also consider the chance that a certain coin was chosen if you got 10 heads in a row.

P(0.9 coin | first 10 heads) = P(0.9 coin and first 10 heads)/P(first 10 heads) = $\frac{0.5*0.9^{10}}{0.5*0.9^{10} + 0.5*0.1^{10}} = 0.9999999997132$ (according to Wolfram Alpha). The chance of 11th coin heads is $0.9999999997132*0.9 + (1 - 0.9999999997132)*0.1$ which is super close to 0.9. Hope this helps!


问:
Quadratic equation and inequality of solution
I have a quadratic equation, $bx^2 + 2ax + b =0$, with, $a>b>0$.
I can solve this as followings,
$x_{+} = \frac{- a +\sqrt{a^2 - b^2}}{b},x_{-} = \frac{- a -\sqrt{a^2 - b^2}}{b},$

In my text book, an inequality equation, $x_{+}^{2} < 1 < x_{-}^2$, is written with the solution, but I cannot derive the equation. If you can, please teach me the derivation.
Best regards,

Body:
Given $x_{+} = \frac{- a +\sqrt{a^2 - b^2}}{b},x_{-} = \frac{- a -\sqrt{a^2 - b^2}}{b}$,  the following is true with a>b>0 :

|$x_{-}| = |\frac{ a +\sqrt{a^2 - b^2}}{b}| >|\frac{ a +\sqrt{b^2 - b^2}}{a}=1$
, |$x_{+}| = |\frac{- a +\sqrt{a^2 - b^2}}{b}|<|\frac{- b +\sqrt{a^2 - a^2}}{b}|=1$,

$\to$|$x_{+}|^{2} < 1 < |x_{-}|^2$,   $x_{+}^{2} < 1 < x_{-}^2$


答: You can also let $g_n=inf_{i \geq n} \{f_n\}$ and note that $\int g_n \leq \int f_n$ for each n.  Moreover $g_n$ increases monotonically to lim inf $f_n$ . Hence by monotone convergence theorem $\int g_n = \int lim inf f_n$.  Since $\int g_n \leq \int f_n$ we have $\int lim inf f_n \leq lim inf \int f_n$.  The final step is to note that pointwise convergence implies lim inf $f_n$ = f

Title: Fatou's lemma - Royden's proof

Body:
Royden states that to prove Fatou's lemma it is ncessary and sufficent to show that if $h$ is any bounded measurable function of finite support for which  $0\leq h\leq f$ on $E$, then
$$
\int
_E h\leq \lim \inf \int_E f_n$$

Why did Royden choose to construct such a function instead of working directly with $f$?


问:
Quadratic expression in $x$ with roots $\frac gh$ and $-\frac hg$
What quadratic expression in $x$ has roots $$\frac{g}{h}\qquad\text{and}\qquad-\frac{h}{g}?$$ I know that this can be factored as $$\left ( x-\frac{g}{h} \right )\left ( x+\frac{h}{g} \right )=0$$ But I'm really confused with the choices. The choices are as follows:

$(a)\hspace{5 mm}g^2 h^2 x^2 -\frac{g^2}{h^2}$

$(b)\hspace{5 mm}hgx^2+(g^2-h^2)x-hg$

$(c)\hspace{5 mm}hgx^2+(h^2-g^2)x+hg$

$(d)\hspace{5 mm}hgx^2+(h^2-g^2)x-hg$

Is the answer (d)?

答:
Given:

$$\boxed{\left ( x-\frac{g}{h} \right )\left ( x+\frac{h}{g} \right )=0}$$
Start by expanding the factors using the distributive property, sometimes referred to as FOIL:

$$x^2+\dfrac{xh}{g}-\dfrac{xg}{h}-\dfrac{hg}{hg}=0$$

We shall next multiply both sides of the equation by the common denominator $hg$. The purpose of this is to remove the denominator making the equation easier to manage:

$$hg\bigg(x^2+\dfrac{xh}{g}-\dfrac{xg}{h}-\dfrac{hg}{hg}=0\bigg)$$

$$hg\bigg(hgx^2+xh^2-{xg^2}-{hg}=0\bigg)$$

As we can see, the equation is quite different from the original now and free of fractions:

$$hgx^2+x(xh^2-xg^2)-{hg}=0$$

Factoring out $x$ from $xh^2-xg^2$

$$\boxed{hgx^2+x(h^2-g^2)-{hg}=0}$$

Which gives us our final answer of: D


问:
Commutator group of symmetries of the real plane is the group containing rotations.
Let $A$ be the group of symmetries of $\mathbb{R}^2$ fixing the origin and $B$ the subgroup of all rotations.

I want to show that the commutator subgroup, $[A, A] = B$.

I was able to conclude that $[A, A] \subset B$ but I have trouble proving $B \subset [A, A]$.

Any ideas on how to begin?

答:
Hint: Let $r$ be a rotation of $\mathbb R^2$ about the origin, and let $s$ be the reflection of $\mathbb R^2$ about the $x$-axis. What is $srs^{-1}$? So what is $[s, r]$?

Solution:

 Observe that $srs^{-1} = r^{-1}$, and therefore, $[s, r] = srs^{-1}r^{-1} = r^{-2}$. Thus, every rotation that can be written in the form $r^{-2}$ for some rotation $r$, is a commutator. But all rotations of the plane (about the origin) can be written in this form — to write a rotation $\rho$ as $r^{-2}$, simply take $r$ to be the rotation in the opposite direction of $\rho$, by half the angle of $\rho$. Therefore, all rotations of $\mathbb R^2$ about the origin are commutators.


问:
Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis?
Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.

Basis step: The statement is true for $n=1$ since: $y^{1} + \frac{1}{y^{1}} = y^{n} + \frac{1}{y^{n}} \in \mathbb{Z}$.

Induction hypothesis: Assume that $y^{k} + \frac{1}{y^{k}}$ and $y^{k-1} + \frac{1}{y^{k-1}}$ are integers for all $n \geq 1$.

Induction step: We show that $y^{k+1} + \frac{1}{y^{k+1}}$ is true.

$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (\frac{y^{k}}{y} + \frac{y}{y^{k}})$

$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (y^{k-1} + \frac{1}{y^{k-1}})$

Since it is given that $y + \frac{1}{y} \in \mathbb{Z}$ and by the induction hypothesis, $y^{k} + \frac{1}{y^{k}}$, $y^{k-1} + \frac{1}{y^{k-1}} \in \mathbb{Z}$:

$\Rightarrow (\mathbb{Z})(\mathbb{Z})-\mathbb{Z}=\mathbb{Z}$ (Closure property for integers)

Therefore, $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.

This is the only way I was able to do the proof. However, I want to know if I am allowed to assume that $P(k-1)$ is also true in the induction hypothesis since as far as I know we can only assume that $P(k)$ is true. Would also appreciate feedback and alternative solutions for my proof!

答:
Yes, that is "strong induction", and it follows from induction.

Strong induction:

$$P(0)\land(\forall n: (\forall k: k\leq n \Rightarrow P(k))\Rightarrow P(n+1))\Rightarrow\forall n :P(n)$$


答:
For $n=2$ and $n=1$ the statement is correct.

Because, $$\left(y^2+\frac{1}{y^2}\right)=\left(y+\frac 1y\right)^2-2$$

Then ,suppose that $n=k$ and $n=k-1$ are also correct.....

Title: Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis?

Body:
Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.

Basis step: The statement is true for $n=1$ since: $y^{1} + \frac{1}{y^{1}} = y^{n} + \frac{1}{y^{n}} \in \mathbb{Z}$.

Induction hypothesis: Assume that $y^{k} + \frac{1}{y^{k}}$ and $y^{k-1} + \frac{1}{y^{k-1}}$ are integers for all $n \geq 1$.

Induction step: We show that $y^{k+1} + \frac{1}{y^{k+1}}$ is true.

$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (\frac{y^{k}}{y} + \frac{y}{y^{k}})$

$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (y^{k-1} + \frac{1}{y^{k-1}})$

Since it is given that $y + \frac{1}{y} \in \mathbb{Z}$ and by the induction hypothesis, $y^{k} + \frac{1}{y^{k}}$, $y^{k-1} + \frac{1}{y^{k-1}} \in \mathbb{Z}$:

$\Rightarrow (\mathbb{Z})(\mathbb{Z})-\mathbb{Z}=\mathbb{Z}$ (Closure property for integers)

Therefore, $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.

This is the only way I was able to do the proof. However, I want to know if I am allowed to assume that $P(k-1)$ is also true in the induction hypothesis since as far as I know we can only assume that $P(k)$ is true. Would also appreciate feedback and alternative solutions for my proof!


问:
Gradient of squared distance function $\operatorname{dist}^2(z) = \frac{1}{2}\|z - \mathrm{proj}_{\mathcal{C}}\left( z \right) \|_2^2 .$
I am looking for a proof of the gradient of a squared distance function to a set $\mathcal{C}$, where the function can be shown as
$$f(z) := \operatorname{dist}^2\left(z\right) = \frac{1}{2}\|z -  \operatorname{proj}_{\mathcal{C}}\left( z \right) \|_2^2 .$$

What will be the gradient of $\operatorname{proj}_{\mathcal{C}}\left( z \right)$ with respect to $z$ in general?

Background: I am trying to prove the proximal operator of a squared distance function as attempted herein Proximal operator of squared Euclidean distance to a set, i.e., $\operatorname{dist}\left(\right)$, where I am unsure about the gradient of such a squared distance function.

ADD: I found same question Gradient of $\mbox{dist}\left(x, D \right)^2:= \left\| x - P_{D}(x)\right\|_2^2$, where $P_{D}(x)$ is a projection operator, but the proofs are bit too heavy (at least) for me.

答:
Let $\,p=\operatorname{proj_{\cal C}}(z).\;$ Then what the linked answer (and especially the comments) are saying is that $p$ is not differentiable, which leaves a simple quadratic problem, i.e.
$$\eqalign{
f &= \frac 12(z-p):(z-p) \\
df &= (z-p):dz \\
\frac{\partial f}{\partial z} &= (z-p) \\
}$$
As a concrete example, suppose you are projecting into the positive orthant of ${\mathbb R}^{2}$, then every $z$ in the negative orthant is projected to $\,p=(0\;\;0)^T,\,$ i.e. the origin.


问:
Using chain rule for derivative of multivariable function
I've been trying to learn more about the chain rule of a multivariable function, but I'm a bit confused on a special case I've encountered. I'm given a single valued function

$\displaystyle f(x_1,x_2,x_3)$

evaluated at $x_3 = g(x_1,x_2)$, which is also a single valved function. Note that $x_n$ are all scalars. I think I would summarize the functions as $f:\mathbb{R}^3\rightarrow \mathbb{R}$ and $g:\mathbb{R}^2\rightarrow \mathbb{R}$. I'm going to represent the function composition as

$\displaystyle h(x_1,x_2) = f(x_1,x_2,g(x_1,x_2)) \equiv (f \circ g)(x_1,x_2)$

(EDIT: The equivalence above is wrong...see answer/comments)

in which $h:\mathbb{R}^2\rightarrow \mathbb{R}$. I'm told to find the derivative of $h$ with respect to $x_2$. However, I want to write down the general expression for the derivative. Following the link above (and material within), I can use the so called derivative operator $\textbf{D}$ and write

$\displaystyle \textbf{D}h = \textbf{D}f|_{x_3=g}\textbf{D}g$

where I've introduced the notation $|_{x_3=g}$ to denote "is evaluated at $x_3 = g(x_1,x_2)$" and dropped the other function variables for clarity. What I don't follow is how to write this in terms of vector/matrix products. In some of the linked material, it says the derivative of a scalar valued function is expressed as a $1 \times n$ row vector. Therefore,

$\displaystyle \textbf{D}h = \left[ \frac{\partial h}{\partial x_1} \quad \frac{\partial h}{\partial x_2}\right]$

$\displaystyle \textbf{D}f|_{x_3=g} = \left[ \frac{\partial f}{\partial x_1}|_{x_3=g} \quad \frac{\partial f}{\partial x_2}|_{x_3=g} \quad \frac{\partial f}{\partial x_3}|_{x_3=g} \right]$

$\displaystyle \textbf{D}g = \left[ \frac{\partial g}{\partial x_1} \quad \frac{\partial g}{\partial x_2}\right]$

I know the answer, specific to the question: "find the derivative of $h$ with respect to $x_2$" is

$\displaystyle \frac{\partial h}{\partial x_2} = \frac{\partial f}{\partial x_2}|_{x_3=g} + \frac{\partial f}{\partial x_3}|_{x_3=g} \frac{\partial g}{\partial x_2}$

so, my question is not about the answer. I'm trying to see how I would arrive at that using vector/matrix products? Specifically, the form "$\displaystyle \textbf{D}h = \textbf{D}f|_{x_3=g}\textbf{D}g$" is easy to remember (from single variable days), and if I can understand how to use it for all special cases, that would be helpful (using the case presented here to showcase).

答:
Define $G(x,y)=(x,y,g(x,y))$ and consider $f \circ G=f(x,y,g(x,y))$.  Now we have

$DG=
\begin{bmatrix}
1 & 0  \\
0 & 1  \\
g_x & g_y
\end{bmatrix}
$
and $Df=[f_x, f_y, f_z]$.  Now $Df \circ DG = [f_x + f_z g_x , f_y+f_z g_y]$ and we get what you have, $h_y=f_y+f_z g_y$

The problem was that you didn't set up the $G$ correctly.


问:
Transform the squared error loss function to matrix form
I have n data, $x_i$ are the inputs, and $y_i$ are the labels. $x_i$ is $d$-dimension, and $y_i$ is $p$-dimension. We have a prediction function that is $f(x) = W^Tx$. $W$ is a matrix of dimension $d \times p$.

The squared error loss function of the prediction function is:
$$
  J(W) = \sum_{i=1}^n||y_i - W^Tx_i||_2^2
$$

I would like to write $J(W)$ in matrix form. How should I convert it to an equation involving $Y$ ($n \times p$ dimension), $W$ ($d \times p$ dimension) and $X$ ($n \times d$ dimension). Hence the loss function will be in matrix form instead of a sum?

答:
With $X = \begin{bmatrix}x_1 & \dots & x_n\end{bmatrix}$ and $Y = \begin{bmatrix}y_1 & \dots & y_n\end{bmatrix}$, you simply obtain:
$$j(W) = ||Y-W^TX||_{2,1},$$
where the $L_{2,1}$ norm is the sum of the Euclidean norms of the columns of the matrix $Y-W^TX$.

Note however the dimensions: $X \in \mathbb{R}^{d \times n}$ and $Y \in \mathbb{R}^{p \times n}$.


答: Here's a way you can see it.  Suppose $n$ is a counterexample, so that $y^n+\frac{1}{y^n} \notin \Bbb Z$ for some $n$.  Then because every non-empty set of natural numbers has a least element, there must be some least $n$ that works as a counterexample.  Therefore, for each $k \lt n, y^k+\frac{1}{y^k} \in \Bbb Z$.  Moreover, $n \neq 1, 2$.  From here, use your proof.


问:
Horn and spindle tori
I was trying to prove that the horn torus and the spindle torus are not manifolds by definition(locally diffeomorphic to some Euclidean space.). I have no idea how to do this, but I attempted it in the following way:

I failed to show that a "slice" of manifold is a manifold itself. By a slice I mean, if the manifold $X\in \mathbf R^n$, then we set the coordinates $x_i=0$ for some $i$ where $0\leq i\leq n$. I feel this would work but I have no idea how to prove it.

Then I look at the cross section for the tori I mentioned(this is equivalent to taking a slice.). For the horn torus, you have two circles touching each other. And the fore spindle torus, you have two circles intersecting each other. Since a circle is locally diffeomorphic to $\mathbf R^1$, the two circles better have to be diffeomorphic to $R^1$. Otherwise we get the result we want.

I tried to proof that a neighbourhood around the point they touch(or intersect) cannot be locally diffeomorphic to $\mathbf R^1$. I tried to do it by contradiction. However I'm stuck on finding a contradiction...

Any thoughts?

P.S. Under the request of Sam Lisi, here are the definitions or horn and spindle tori:

A torus is the set of points in $\mathbf R^3$ at a distance $b$ from the circle of radius a in the $xy$ plane. This is like you put a circle with radius $b$ in the $yz$ plane centred at $(a,0,0)$. Then you make it orbit around the origin and you get a torus.

A horn torus is when $a=b$. If you take a cross-section, you'll find that it's two circles touching each other at one point. When $a<b$, it's called a spindle torus. The cross-section would look like two circles intersecting with each other at two pionts.

There are pictures in this wikipedia article that might help you visualise the horn and spindle tori: http://en.wikipedia.org/wiki/Torus

答:
Often the best way to show a $k$-dimensional subset $M$ of $\mathbb R^n$ is not a submanifold is this: Show that there's some point $p\in M$ so that no neighborhood is a graph of a smooth function over any of the standard coordinate $k$-planes.

In the case of your horn torus, at that central pinch point, the vertical line test will fail for any choice of the 3 standard coordinate ($2$-)planes ($xy$, $yz$, $xz$). Indeed, by symmetry, it fails for every plane, but you don't need to show this. (Intuitively, if the surface were to have a tangent plane at this point, by rotational symmetry, it would have infinitely many.)

Title: Horn and spindle tori


问:
How to integrate $\frac{1}{x^2+x+2}$?
I know the result is going to be in the inverse tangent form and i know you must complete the squares of the denominator to obtain this. I also know that you must use u-substitution method. However i am confused about the steps in between. Could anyone help?

step 1 : you complete the squares in the denominator (ans: $(x+0.5)^2 + 7/4$ )

step 2 : you find the derivative of the function $x^2+x+2 = 2x + 1$

step 3 : you use $u$-substitution

That's all I know, i don't know exactly what you should substitute in $u$ for.

The answer is : $$\frac{2\arctan(\frac{2x+1}{\sqrt 7})}{\sqrt 7}  +C$$

答:
OK, you've shown some work, let me try to guide you through the rest.

First of all, completion of the square wasn't done correctly.

It should be $x^2 + x + 2 = x^2 + x + (\frac 12)^2 - \frac 14 + 2 = (x+\frac 12)^2 + \frac 74$

So now you're left with integrating $\frac 1{(x+\frac 12)^2 + \frac 74}$

I'm sure you've seen that the form $\frac 1{1+x^2}$ can be integrated to give $\arctan x + c$. The question is, can we make what we have look like this special form?

What are we lacking? Well, we have a fraction ($\frac 74$) where $1$ should be, for starters. Let's remedy that shall we?

$$\frac 1{(x+\frac 12)^2 + \frac 74} \\ = \frac {1}{\frac 74 (\frac 47(x+\frac 12)^2 + 1) } \\ = \frac 47 \cdot \frac {1}{1+(\frac 2{\sqrt 7}(x+\frac 12))^2}$$

Doesn't that look a lot like what you want?

If you now substitute $u = \frac 2{\sqrt 7}(x+\frac 12)$, what do you get? Can you finish?

Title: How to integrate $\frac{1}{x^2+x+2}$?