问:
Classifying sharply 3-transitive actions on spheres.
It is well known that mobius transformations act sharply $3$-transitively on the Riemann sphere (that is, any three points can be mapped to any other three points in a unique way). In this question, I asked generally about (continuous) sharply k-transitive actions on spheres, and it was shown that for any such action, k is at most $3$, and $k=3$ occurs only on the 1 and 2 spheres.

My question is, are the only sharply $3$-transitive actions on spheres the mobius transformations, up to conjugation by a self-homeomorphism of the sphere? I'm also interested in the analogous question when we look at the extended real line and real mobius transformations.

答:
I am sure that the answer to the question of classifying sharply 3-transitive actions of Lie groups on the 2-sphere is: Only the standard action of $PSL(2,C)$, up to topological conjugation. However, the proof, as I see it, is by an unpleasant case-by-case analysis. Here is the proof of a similar result for group actions on $S^1$ (still unpleasant, but fewer cases to analyze).

Suppose that $G$ is a Lie group which has a 3-transitive action on $S^1$. Then $G$ is homeomorphic to the torus $T^3$ minus the big diagonal, i.e. to $Mob(S^1)$, the full Moebius group of the  circle, i.e. the full isometry group of the hyperbolic plane (since the latter also acts 3-transitively on the circle). In particular, $G$ has two connected components, is 3-dimensional and its maximal compact subgroup is isomorphic to $O(2)$. The number of 3-dimensional Lie algebras is very small, especially since you already know that the Lie algebra is "noncompact." From this, you get a classification of connected Lie groups with maximal compact subgroup $U(1)=S^1$:

$PSL(2,R)$ or its finite-sheeted covering. Concisely, the Lie algebra of $G$ is isomorphic to $sl(2,R)$.

$R^2\rtimes S^1$.

$Aff_+(R)\times S^1$.

$Nil_3/{\mathbb Z}$, where $Nil_3$ is the group of strictly upper-triangular 3-by-3 matrices and ${\mathbb Z}$ is a subgroup of the (1-dimensional) center of $Nil_3$.

Here $Aff_+(R)$ is the group of orientation-preserving affine transformations of the real line. This gives us all the possible options for $G^0$, the identity component of $G$.
Next, since $G$ acts on $S^1$ transitively, pick a point $z\in S^1$ and let  $G_z$ denote the stabilizer of $z$ in $G$. Then $G/G_z$ is homeomorphic to $S^1$. In the cases 2 and 3, the only options for such $G^0_z$ are the (semi)direct factors $R^2$ and  $Aff_+(R)$ respectively. But these subgroups act trivially by the left multiplication on the quotients $G^0/G^0_z$, from which we see that in the cases 2 and 3 the action of $G$ on $S^1$ would have to contain the subgroups  $R^2$ and  $Aff_+(R)$ respectively in the kernel of the action. Hence, a 3-transitive action is impossible in these two cases.

In the fourth case, one check that $Nil_3/{\mathbb Z}$ contains no 2-dimensional subgroups which have trivial intersection with the (central) compact subgroup $S^1$. This rules out the fourth case.

We conclude that the Lie algebra of $G$ is isomorphic to $sl(2,R)$. Let's finish the proof in the case when $G\cong Mob(S^1)$. As above, pick a point $z\in S^1$ and obtain a homeomorphism $G/G_z\cong S^1$. Thus, $G_z$ is a 2-dimensional subgroup of $G$ which has nontrivial intersection with both connected components of $G$ (since $S^1$ is connected and $G$ is not). Up to conjugation, $PSL(2,R)$ has only one 2-dimensional subgroup, namely one isomorphic to $Aff_+(R)$, the stabilizer of "infinity." Thus, $G_z$ has to be conjugate to the subgroup $Aff(R)< G$. From this, we obtain that the action of $G$ on $S^1$ is equivariantly homeomorphic to the action of $G$ on $G/Aff(R)$ by the left multiplication, which is the standard action of the full Moebius group on the circle.

It remains to exclude the cases when $G^0$ is a nontrivial covering of $PSL(2,R)$. The same analysis as above, shows that the only possible actions are given by left multiplication of $G$ on $G/G_z$, where $G^0_z$ (up to conjugation) is the identity component  of the preimage of
$Aff_+(R)< PSL(2,R)$ under the covering map $p: G^0\to PSL(2,R)$. Let us understand the corresponding actions of $G^0$ on $S^1$ assuming that $p$ is an $n$-fold covering map, $n\ge 2$.

To get some idea of what is going on, take first $n=2$, i.e. $G^0=SL(2,R)$. The subgroup $G^0_z$ then (up to conjugation)  is the group of upper-triangular matrices with positive diagonal entries.  The action of $G^0$ on $G^0/G^0_z$ then is the action of $SL(2,R)$ on the space $\Sigma$ of rays emanating from the origin in the plane. Each hyperbolic element $h$ of $SL(2,R)$ (with positive eigenvalues) has four fixed points in $\Sigma$: Two for each eigenspace. Two of these rays ($a^1, a^2$) are attractive (they correspond to the eigenvalue $>1$) and two ($r^1, r^2$) are repulsive. Now, if $g\in G$ swaps $a^1$ and $r^1$, it will have to conjugate $h$ to $h^{-1}$, hence, $g$ necessarily also swaps $a^2$ and $r^2$. Thus, $g$ cannot swap $a^1$ and $r^1$ while, say, leaving $a^2, r^2$ fixed. In particular, the action of $G$ on $\Sigma$ cannot be 3-transitive. The same argument works if you have $n>2$: Each hyperbolic element of $PSL(2,R)$ has a lift $\tilde{h}\in G^0$ which has
$n$ attractive fixed points $a^i$ and $n$ repulsive fixed points $r^j$. One cannot swap $a^1$ and $r^1$ while leaving, $a^2, r^2$ fixed.

Edit. Here is a sketch of the proof for actions on $S^2$. If $G$ admits a sharply 3-transitive action on $S^2$ then $G$ is homeomorphic to $PSL(2,C)$. In particular, the maximal compact subgroup of $G$ is isomorphic to $SO(3)$. From this, one concludes that $G$ is isomorphic to:

Either $PSL(2,C)$.

Or a semidirect product $H\rtimes SO(3)$, where $H$ is a 3-dimensional simply-connected solvable group (of which there is a continuum).

Since $SO(3)$ has only two real linear 3-dimensional representations (trivial and standard, up to isomorphism), one then sees that in Case 2 we  have

2i. Either $G\cong H\times SO(3)$.

2ii. Or $G\cong SE(3)$, the group of orientation-preserving isometries of the Euclidean 3-space $R^3$.

Then, as in the case of the analysis of actions on the circle, the point-stabilizer $G_z$ is 4-dimensional, not containing a conjugate of the maximal compact subgroup. In Case 2, all such subgroups will contain the solvable radical $H$, resulting in non-faithful action of $G$ on $G/G_z$, which is a contradiction. This leaves us only with Case 1. In this case each 4-dimensional subgroup is a conjugate of $Aff(C)$, the group of complex-affine transformations. Hence, the action on $S^2$ is standard.


问:
I have decided to study math on my own and I came across these questions. Would really appreciate if anyone could help and explain how to solve these.

I could not come up with a solution for both of the problems, so help is appreciated

ContentLicense: CC BY-SA 4.0

Body:
For the second question,

$$n^3 + 4n^2+3n+1 \equiv (-2)^3+4(-2)^2+3(-2)+1\equiv 3 \pmod{n+2}$$

so $(n+2) | 3, n+2=1, -1, 3, -3$.

(I assume $n$ is an integer.)

If you don't know modular math, then

$$n^3+4n^2+3n+1=n^3+2n^2+2n^2+4n-n+1=(n+2)(n^2+2n-1)+3$$

Again $\frac{3}{n+2}$ is an integer.


问:
Two independent uniformly distributed random variables on $A$ in $\Bbb R^2$
This is a homework question from my probability class.

Let $X$, $Y$ be two random variables. Suppose $X$ is uniformly distributed on $A$, $Y$ is uniformly distributed on $A$, $A$ $\subseteq$ $\Bbb R^2$. Further assume $X, Y$ are independent.

Prove/Disprove: $A$ is a rectangle with lines parallel to the axes.

Because $X,Y$ are uniformly distributed on $A$ I know that their PDF is equal to:

$f_X(x)=f_Y(y)=1/\lambda(A)$ where $\lambda$ is the Lebesgue measure on the plane.
because $X,Y$ are independent we get:
$f_{(X,Y)}(x,y)= 1/\lambda(A)^2$

Now I got confused...
How can I do this?

Thanks!

答:
Since $X$ and $Y$ are independent,
$ f_{XY}(x,y)=
\begin{cases}
f_X(x)f_Y(y), & (x,y)\in A \\
0, & \text{otherwise}
\end{cases}$.

Suppose that $f_{XY}(x,y)>0, \forall (x,y)\in A$. So, for any $x_0$ that $f_X(x_0)>0$, $f_{XY}(x_0,y)>0$ for all $y$ that $f_Y(y)>0$, and $f_{XY}(x_0,y)=0$ for all $y$ that $f_Y(y)=0$, i.e., the support set of non-zero probability for $Y$ remains the same for any arbitrary $x_0$ that $f_X(x_0)>0$. A similar argument applies to $Y$. Hence, $A$ is a rectangle.

The uniformity of distribution is not necessary for $A$ to be a rectangle.


问:
Can the plane be covered by open disjoint one dimensional intervals?
I remember I heard this question a while ago but have never figured it out. Let an open interval in $\mathbb{R}$ of the form $(a,b)$. Now, imaging placing it in $\mathbb{R}^2$, anywhere we want while keeping its length but perhaps rotating it if we want. Essentially this creates a line segment without its start/endpoints parametrized by

$P(1-t)+Qt$ for $0<t<1$ where $P$ and $Q$ are its start and end points and with length $||P-Q||=b-a$.

As an example, if we take the unit interval $(0,1)$, we can place it as it is in $\mathbb{R}^2$ as $(0,1)\times \{0\}$ or we can rotate it as $\{1\}\times (0,1)$ etc. I hope it is clear what I mean. My question is the following.

Can we cover the entire $\mathbb{R}^2$ with disjoint such "open intervals" or if you prefer, line segments of finite length without their start/endpoints?

I don't remember if the initial question I read was using only unit intervals, or of any length $(a,b)$ and I don't know if that plays a role in the answer but as a gut feeling, I doubt it does. My impression is that the answer is negative and we can't cover the plane in such a manner, or if it is possible it would use some weird abstract construction perhaps involving AC. Any help is appreciated!

Edit: It seems that I could have phrased my wording a bit better although that was my last paragraph in the post above that I wasn't certain if it mattered whether we need to have fixed length intervals or not. Apparently it does, it is impossible with using only fixed length intervals but it is possible if we allow intervals of different lengths. Proof of impossibility in one case given in a paper given in the answers, and a construction in the other case is given by some answers and also found in that paper itself. Thanks everyone!

答:
Yes, it can be done.  Here is a recursive construction that uses only axis-aligned segments.  Start by covering the open unit square $(0,1)\times(0,1)$ with horizontal segments.  Then, given an initial open square, make 8 more copies and arrange them around the initial square, producing a larger open square with an uncovered hash symbol or octothorpe (#).  Then cover the octothorpe with two long horizontal segments and six short vertical segments.  You’ve now tripled the size of the initial open square.  Repeating this indefinitely covers all of $\mathbb{R}^2$.

It is an interesting question whether there is a covering using segments that are all the same length.  The best I see right now is a covering using segments of length $1$ and countably many segments of length $2$.


问:
Let $A_1,A_2,\dots,A_d$ be $d\times d$ matrices that are strictly upper triangular. Then, the product of $A_1,A_2,\dots,A_d$ is the zero matrix.
SOLUTION: Let $B_i$ be given by $A_1A_2\dots A_i$. It can be shown inductively that $B_i$ is also strictly triangular, but with at least $i$ zero rows and columns. Therefore, $B_d$ will be the zero matrix.

I can't quite wrap my head around this. I would expect that in fact, the product of any two strictly triangular matrices would be the $0$ matrix. And therefore any other matrix (can be non-zero) that is multiplied with it will be the $0$ matrix. But the answer seems to say that it only has at least $i$ zero rows and columns after multiplying $i$ strictly trangiangular matrices. It might be that I am misunderstanding the entire answer in general but any simplification explanation is welcomed.

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (Problem 1.23)

答:
It is not true that the product of two upper triangular matrices is necessarily $0$. As an example, take $d = 4$ and
$$
A_1 = A_2 = A_3 = A_4 =
A = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0}.
$$
In a sense, this is the "classic" example of a strictly upper triangular matrix considered in the context of Jordan normal form. Notice what happens when we multiply these matrices:
$$
A^2 = \pmatrix{0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0}, \quad
A^3 = \pmatrix{0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \quad A^4 = 0.
$$
In a sense, what happens here is what happens in general: multiplying two upper-triangular matrices "pushes" the first non-zero "diagonal" further up. In fact, I recommend that you prove the following:

We will say that $A$ is "upper triangular of order k" if its entries are such that $a_{ij} = 0$ whenever $j < i+k$. Prove that if $A$ is upper-triangular of order $p$ and $B$ is upper-triangular of order $q$, then $AB$ is upper-triangular of order $p + q$.

This is a more precise version of the pattern observed in the solution. It is possible to prove this fairly quickly using only the definition of matrix multiplication. Once you prove this result, perhaps you'll be able to see how the full solution holds as a consequence.

In a more abstract sense, what upper-triangularity tells us about the transformation associated with the matrices $A_1,\dots,A_d$ is that there are subspaces
$$
\{0\} = V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_{n-1} \subsetneq V_n = \Bbb R^d
$$
for which $A(V_i) \subseteq V_{i-1}$ for each $i = 1,\dots, n$. That is, for every $v \in V_i$, $Av$ is an element of $V_{i-1}$.

This gives us an alternative proof. For any vector $v \in \Bbb R^d = V_n$, $A_d v$ is an element of $V_{n-1}$. Similarly, $A_{d-1}(A_d v)$ is an element of $V_{n-2}$. Continuing in this fashion, we can conclude that $A_2 \cdots A_d v \in V_1$, so that
$$
A_1 A_2 \cdots A_d v = A_1(A_2 \cdots A_d v) \in V_0 = \{0\},
$$
which is to say that $A_1 A_2 \cdots A_d v = 0$ for every vector $v$. It follows that $A_1 A_2 \cdots A_d = 0$.


问:
Unstable formulas
A formula $\phi(x,y)$ has the order property if the are $(a_i)_{i<\omega}$ and $(b_j)_{j<\omega}$  such that $$ i< j \,\,\,\,\, \,\, \text{iff} \,\,\,\,\, \,\,  \models \phi(a_i, b_j).$$
A theory $T$ is called stable if there is no formula with the order property.

Is the theory of graphs stable?

答:
No, the theory of graphs is not stable.

Consider the graph with elements $\{a_i,b_i\mid i\in \omega\}$ and edge relation $R$, where $a_iRb_j$ if and only if $i<j$. This shows that the formula $xRy$ has the order property relative to the theory of graphs.

Title: Unstable formulas


问:
Strange inverse limit of infinite direct sums $S_n$
Let $p_j$ be the prime numbers in order, indexed by $j$. Define $S_n$ as follows:

\begin{equation}
\begin{aligned}
S_1 &:= \mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / 2 \mathbb Z &\oplus &\cdots \\
S_2 &:= \mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\cdots \\
S_3 &:= \mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\mathbb Z / (2 \cdot 3 \cdot 5 \mathbb Z) &\oplus &\mathbb Z / (2 \cdot 3 \cdot 5 \mathbb Z) &\oplus &\cdots \\
\cdots
\end{aligned}
\end{equation}
We have an obvious inverse system $\{S_n, f_n\}$, where $S_n \xrightarrow {f_n} S_{n-1}$.

What, then, is the inverse limit $\varprojlim S_n =: S$?

I am not sure how to describe it. Clearly, it is not $P := \prod_{0<i}^\infty \left( \mathbb Z / \left( \left(\prod_{0<j}^i p_j\right)\mathbb Z\right)\right)$, as $(1,1,1,\cdots)$ is an element of P, but not of $S$ (note that $S_n$ is an infinite direct sum, not an infinite direct product). Also, it is not $S' := \bigoplus_{0<i}^\infty \left( \mathbb Z / \left( \left(\prod_{0<j}^i p_j\right)\mathbb Z\right)\right)$, as $(1,2,6,30,210,...,\prod_{0<j}^n p_j,...)$ does not belong to $S'$, but it does belong to $S$, since the element's restriction to any $S_n$ is a finite sum.

So apparently it is something in-between, and I wonder whether anybody knows of a nice description.

So far I have failed to do any better than something in English, such as: "The elements of $P$ of which, for any prime $p$, only finitely many coordinates are not divisible by $p$".

答: Using the Chinese remainder theorem to split everything as a direct sum of cyclic groups of prime order, you can describe each $S_n$ as just a direct sum of countably infinitely many copies of $\mathbb{Z}/p\mathbb{Z}$ for each of the first $n$ primes $p$, with $f_n$ just being the projection that drops the last prime.  This gives a simple description of $S$ as $$\prod_{p\text{ prime}}(\mathbb{Z}/p\mathbb{Z})^{\oplus\mathbb{N}}.$$


问:
Find the limit joint distribution of the random vector uniformly distributed in the ball
Given an integer $n \geq 1$, define $\left(X_{1}^{(n)}, X_{2}^{(n)}, \ldots, X_{n}^{(n)}\right)$ as a random vector uniformly distributed in the ball
$$
 \left(X_{1}^{(n)}\right)^{2}+\left(X_{2}^{(n)}\right)^{2}+\cdots+\left(X_{n}^{(n)}\right)^{2} \leq n
 $$
Find the limit joint distribution of the random vector $\left(X_{1}^{(n)}, X_{2}^{(n)}, X_{3}^{(n)}\right)$ as $n \rightarrow \infty$.

Thanks for your help!

答:
If $(X_1^{(n)},\ldots,X_n^{(n)})$ is uniformly distributed in $\Big\{\vec{x}\in\mathbb{R}^n:||\vec{x}||^2< n\Big\}$ then the pdf of $(X_1^{(n)},\ldots,X_n^{(n)})$ is the function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ defined by $$f(x_1,\dots,x_n)=\frac{\Gamma\Big(\frac{n}{2}+1\Big)}{(n\pi)^{n/2}}$$ whenever $x_1^2+\dots+x_n^2 < n$ and $f(x_1,\dots, x_n)=0$ elsewhere. We find the joint pdf of $(X_1^{(n)},X_2^{(n)},X_3^{(n)})$ to be $$f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)=\int_{\mathbb{R}^{n-3}}f(x,y,z,x_4,\dots, x_n)dx_4\ldots dx_n$$ If $x^2+y^2+z^2\geq n$ then $f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)=0$ so assume $x^2+y^2+z^2 < n$ for the remainder of this post. We get $$f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)=\frac{\Gamma\Big(\frac{n}{2}+1\Big)}{(n\pi)^{n/2}} \int_{\{x_4^2+\dots +x_n^2 < n-x^2-y^2-z^2\}}dx_4 \dots dx_n$$ Using the fact that a ball of radius $R$ in $\mathbb{R}^{k}$ is $\frac{\pi^{k/2}}{\Gamma\Big(\frac{k}{2}+1\Big)}R^k$ we have $f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z)$ equals
the following expression $$\frac{\Gamma\Big(\frac{n}{2}+1\Big)}{\Gamma\Big(\frac{n-3}{2}+1\Big)}\Bigg(1-\frac{x^2+y^2+z^2}{n}\Bigg)^{n/2} \frac{1}{\Big(\pi(n-x^2-y^2-z^2)\Big)^{3/2}}$$ Using Stirling's approximation you will quickly see $$ f_{X_1^{(n)}X_2^{(n)}X_3^{(n)}}(x,y,z) \rightarrow \frac{1}{(2\pi)^{3/2}}e^{-\frac{1}{2}(x^2+y^2+z^2)}$$ as $n \rightarrow \infty$ i.e. the limit joint distribution of $(X_1^{(n)},X_2^{(n)},X_3^{(n)})$ is $N\Big(\vec{0},I_{3\times 3}\Big)$.


问:
Avoid Loss of significance error
I want to compute $f(x)$to avoid loss of significance$$f(x)=\frac{1-\cos x}{x^2}$$

One way is writing Taylor series approximation for $\cos x$ about $x=0$. what about this approach:

$$f(x)=\frac{2\sin^2(\frac x2)}{x^2}$$

Is this ok ?

答:
You are already mindful of rounding errors, but that is not the only issue here.

In IEEE floating point arithmetic, $0/0$ evaluates as NaN (not a number). You need a conditional statement to ensure that $f(0)$ is evaluated as $\frac{1}{2}$.

In IEEE double precision, the largest positive number is $(2-2^{-52})\cdot 2^{1023}$, so $x^2$ will certainly overflow to infinity when $|x| \ge 2^{512}$ and the default rounding mode is active. If you evaluate $f$ as you suggested, i.e., $$f(x) = 2 \frac{\sin^2\left(\frac{x}{2}\right)}{x^2},$$ and $x^2$ overflows, then the computed result of $f(x)$ is zero. You avoid this issue by evaluating $f$ as $$f(x) = 2 \left(\frac{\sin\left(\frac{x}{2}\right)}{x}\right)^2.$$

Be mindful of approximations that are more accurate than the rounding error associated with their evaluation.

In general, we seek to compute a target value $T$. We most then choose an approximation $A$ as well as an algorithm for evaluating $A$. In general, the computed value $\hat{A}$ of $A$ will be different from $A$. The total error is $$T-\hat{A} = (T - A) + (A - \hat{A}).$$ Ideally, the two contributions, i.e., the approximation error $T-A$ and the rounding error $A - \hat{A}$ should have the same size and the total error should be less than the user's tolerance $\tau$. Suppose $|T-\hat{A}| < \tau$.

If $|T-A| \ll |A-\hat{A}|$, then we are guilty of mathematical overkill and are probably wasting CPU cycles because very accurate approximations tend to be more costly to evaluate than less accurate approximations.

If $|A-\hat{A}| \ll |T-A|$, then we are guilty of algorithmic overkill and we may be able to find a faster, less accurate algorithm. It may well be enough to execute the same basic algorithm in single precision instead of double precision.

Title: Avoid Loss of significance error


问:
Fourier transform of Green's Function of Laplacian $-\frac{1}{4 \pi |x|}$
I feel that this "simple" question might have been asked before, but after some searching, I still cannot find it. We often need to know the Fourier transform of the Green's function of $\Delta$ in $\mathbb R^3,$
$$
G(x) =-\frac{1}{4\pi |x|},
$$
which can be written as
$$
\hat G(\xi) =-\int_{\mathbb R^3} \frac{e^{ix\cdot \xi}}{4\pi |x|}dx.
$$

What are some ways to evaluate this integral? What will the resulting function be like?

答:
Well, $$\Delta G=\delta$$ by definition, so taking the Fourier transform of this gives that $$\hat{G}(\xi)=-(2\pi)^{-3/2}|\xi|^{-2}\in\mathcal{S}'.$$

In fact, $$\hat{G}(\xi)=-(2\pi)^{-n/2}|\xi|^{-2}$$ on $\mathbb{R}^n,$ and this is in $\mathcal{S}'$ for all $n\geq 3$, as $|\xi|^{-2}\in L^1_{loc}$ for such $n$.

Here, I used the convention that $$\hat{u}(\xi)=(2\pi)^{-n/2}\int\limits_{\mathbb{R}^n} u(x)e^{-ix\xi}\, dx.$$


问:
$i+j-1$ tokens on row $i$ column $j$ of square grid
Consider an $n\times n$ grid, with rows and columns numbered from $1$ to $n$. On the cell in row $i$ column $j$, I place $i+j-1$ tokens. How many tokens are on the grid in total?

I have done the double sum directly:
$$\sum_{i=1}^n\sum_{j=1}^n(i+j-1)=\sum_{i=1}^n\left[n(i-1)+\frac{n(n+1)}{2}\right]=\frac{n^2(n-1)}{2}+\frac{n^2(n+1)}{2}=n^3.$$
With an answer as neat as this, I'm wondering if there's a simple (perhaps geometric) interpretation/argument?

答:
Picture for $n=6$:

Take a horizontal cut at the height of $n$ tokens. The tokens above the cut are all in the bottom right corner. Flip this shape over and it will fit snugly into the top left corner to form an $n\times n\times n$ cube of tokens.


问:
Finding the measure of an angle

Let E be a point outside of the square $ABCD$ such $\triangle ABE$ is an equilateral triangle.  What is the measure of $\angle CED$, in degrees?

I need help with this problem. I made a diagram beforehand to help me with this problem. From doing this, I find that $m(\angle CAE)$ is the same as $m(\angle DBE)$. I think these angles are $150^\circ$ because the angles of a square are each $90^\circ$, plus when we have point $E$ makes an equilateral triangle (which has three $60^\circ$ angles). So $60^\circ + 90^\circ = 150^\circ$. This is all I have so far.

答:
Draw the figure.  For definiteness, I would, like most mathematicians do, arrange the labels of the square counterclockwise around the square.  So put $A$ at top right, $B$ at top left, $C$ at bottom left, and $D$ at bottom right.
Then $E$ is somewhere above the line segment $AB$. (I am doing this to make sure that we are using the same diagram.  Yours was I think labelled differently; if so, please relabel.)

It is clear that $\angle EBC$ is $150^\circ$ ($90+60$).  So the two angles $BEC$ and $BCE$ add up to $180-150$, which is $30$.  But $BE=BC$, so the two small angles are equal.  It follows that each is $15^\circ$.

Note: I did not see your diagram, but it looks as if you were probably very close to a solution. The only thing that you likely missed is the fact that what I called $\triangle EBC$ is isosceles.

ContentLicense: CC BY-SA 3.0

Title: Finding the measure of an angle


问:
Kernel of the evaluation map on a power series ring
Let $R$ be a commutative ring with unity and $r \in R$ a nilpotent element. Is it true that if $f \in R[[\epsilon]]$ satisfies $f(r) = 0$, then $(\epsilon - r) | f$ in $R[[\epsilon]]$? I tried solving for the coefficients of $f/(\epsilon - r)$ inductively and got myself confused.</p>

答: Yes. Since $r$ is nilpotent, the map $\phi:f(\epsilon) \mapsto f(\epsilon-r)$ is an automorphism of the power series ring, and it is definitely true that if $f(0)=0$ then $\epsilon$ divides $f$.</p>


问:
Understanding Needham's derivation of equivalence between symbolic and geometric rules for complex multiplication
Tristan Needham's Visual Complex Analysis contains a proof that the symbolic (i.e. $(a+ib)(c+id) = (ac-bd) + i(bc+ad)$) and geometric (i.e. "multiply the lengths and add the angles") rules for complex multiplication are equivalent (pages 8-10). The proof goes like this:

Geometric to symbolic: The symbolic rule is determined as soon as we say that $i^2=-1$ and that the distributive property holds, so it suffices to show that these two properties hold. The former is true since $i$ has length $1$ and angle $\pi/2$, so squaring it produces length $1$ and angle $\pi$, i.e. $-1$. The latter is true since rotations and dilations preserve parallelograms.

Symbolic to geometric: $iz = i(x+iy) = -y + ix$ according to the symbolic rule, so multiplication by $i$ is equivalent to a right-angle rotation. Using the distributive property we then have $(x'+y'i)z = x'z + y'(iz) = x'z + y'(z \text{ rotated by right angle})$. Drawing the picture, the triangle formed by points $(0,0), (x',0), (x',y')$ is now similar to the triangle formed by points $(0,0), x'z, (x'+y'i)z$, which shows that the length has been multiplied by the length of $x'+y'i$ and that the angle has been rotated by the angle of $x'+y'i$.

My basic question is something like: In what sense was the above a proof? How can we formalize what must be shown, and how does the above prove that?

One basic confusion I have is that the structure of the theorem at a high level looks like $A=B$  (symbolic=geometric), and yet we are having to show the equivalence in both directions. If we want to show $A=B$, we only need to show $A=B$ and not $B=A$ also. Maybe another way to state this is that it seems we just need to define complex multiplication in one way, then prove the other rule as a theorem, which seems to require only one direction. So in one sense by showing two directions the above proof seems overkill.

Okay so what are we trying to show? It seems like we have two representations of complex numbers, the algebraic/symbolic one $\mathbf C$ and the geometric one $\mathbf C'$. And we have multiplication $\cdot$ defined on $\mathbf C$, and multiplication $\cdot'$ defined on $\mathbf C'$. We want to show that there is some bijection $f : \mathbf C \to \mathbf C'$ such that $f(z\cdot w) = f(z)\cdot' f(w)$ for all $z,w\in \mathbf C$.

How does the above proof produce such an $f$? The "Geometric to symbolic" direction first shows that $f(i)^2 = f(-1)$. It then shows that $f(z)(f(w) + f(u)) = f(z)f(w) + f(z)f(u)$. If we assume $f$ interacts well with addition and multiplication by real numbers, we have $f(a+ib)f(c+id) = f((ac-bd) + i(ad+bc))$. But are we allowed to assume this? If so it sounds like we are done, and if not then I'm not sure how this direction of the proof is useful. For "Symbolic to geometric" direction I'm not sure how to translate it in terms of $f$.

答:
We have two models for the complex numbers:

In the geometric model, every complex number is a position vector $\vec{v}$ in the plane.

In the symbolic model, every complex number is an expression of the form $a + i b$ for real numbers $a$, $b$.

The additive operations in the two models are defined as follows:

In the geometric model, the sum of two complex numbers, $\vec{v}$ and $\vec{w}$, is their vector sum, $\vec{v} + \vec{w}$, which is defined by the parallelogram law.

In the symbolic model, the sum of two complex numbers, $a+ib$ and $c+id$, is defined to be $(a+c)+i(b+d)$.

The multiplicative operations in the two models are defined as follows:

In the geometric model, the product of two complex numbers, $\vec{v}$ and $\vec{w}$, is the vector $\vec{z}$ with $\lvert \vec{z} \rvert = \lvert \vec{v} \rvert \cdot \lvert \vec{w} \rvert$ and $\angle \vec{z} = \angle \vec{v} + \angle \vec{w}$.

In the symbolic model, the product of two complex numbers, $a+ib$ and $c+id$, is defined to be $(ac-bd)+i(ad+bc)$.

Now, what Needham actually shows in the first part is the following.

Consider the bijection $f \colon \mathbb{C}_{\mathsf{GEOM}} \to \{ a + ib : a, b \in \mathbb{R} \}$ given by $f(\vec{v}) = v_x + i v_y$, where $\vec{v}$ is the position vector of the point $(v_x, v_y)$, and the codomain is just the set of all expressions of the form $a+ib$ for $a, b \in \mathbb{R}$. Then, we can transport the structure of $\mathbb{C}_{\mathsf{GEOM}}$ onto the codomain by defining $(a+ib)+(c+id) = f(f^{-1}(a+ib)+f^{-1}(c+id))$ and $(a+ib) * (c+id) = f(f^{-1}(a+ib) * f^{-1}(c+id))$.

Then, Needham shows that the resultant structure on the codomain is precisely $\mathbb{C}_{\mathsf{SYMB}}$.

Similarly, in the second part Needham shows the following.

Consider the bijection $g \colon \mathbb{C}_{\mathsf{SYMB}} \to \{ \vec{v} = (v_x, v_y) : v_x, v_y \in \mathbb{R} \}$ given by $g(a + ib) = \vec{v}$, where $\vec{v}$ is the position vector of the point $(a, b)$, and the codomain is just the set of all position vectors in the plane. Then, we can transport the structure of $\mathbb{C}_{\mathsf{SYMB}}$ onto the codomain by defining $\vec{v} + \vec{w} = g(g^{-1}(\vec{v})+g^{-1}(\vec{w}))$ and $\vec{v} * \vec{w} = g(g^{-1}(\vec{v}) * g^{-1}(\vec{w}))$.

Then, Needham shows that the resultant structure on the codomain is precisely $\mathbb{C}_{\mathsf{GEOM}}$.

Consider the first part. (Let $\vec{\iota}$ be the position vector of the point $(0,1)$.) How does Needham show that the resultant structure is indeed $\mathbb{C}_{\mathsf{SYMB}}$?

Note that $f(\vec{\iota}) = i$ by the definition of $f$.

Needham notes that it suffices to check that $i^2 = -1$ and that the distributive property holds.

Since we have just transported the structure, $i^2 = -1 \iff \vec{\iota}^2 = \vec{-1}$, where $\vec{-1}$ is the position vector of the point $(-1,0)$.

Similarly, the distributive property holds for the codomain with the transported structure iff it holds for the domain $\mathbb{C}_{\mathsf{GEOM}}$.

Needham already showed how $\vec{\iota}^2 = \vec{-1}$ in an earlier section, and he shows that the distributive property holds in $\mathbb{C}_{\mathsf{GEOM}}$ to complete the proof.

Of course, one also needs to verify that the additive structure matches, which Needham leaves as an exercise. By the transport of structure, we have by definition that $(a+ib)+(c+id) = f(f^{-1}(a+ib)+f^{-1}(c+id))$, that is, $f(\vec{v}) + f(\vec{w}) = f(\vec{v}+\vec{w})$ if $f(\vec{v}) = a+ib$ and $f(\vec{w}) = c+id$, so one needs to check that $f(\vec{v}+\vec{w}) = (a + c) + i(b + d)$. And, this is true by the way the parallelogram law of vector addition is defined.

For the second part, Needham shows that the resultant structure on the codomain is $\mathbb{C}_{\mathsf{GEOM}}$ as follows (again, let $\vec{\iota}$ be the position vector of the point $(0,1)$.):

First, the verification that the additive structure matches is left as an exercise. (We have by definition that $\vec{v} + \vec{w} = g(g^{-1}(\vec{v})+g^{-1}(\vec{w}))$, and we can again complete the verification using the parallelogram law of addition.)

For the verification of the multiplicative structure, it suffices to check that for a fixed element $\vec{v}$, the map $\vec{w} \mapsto \vec{v} * \vec{w}$ rotates $\vec{w}$ by $\angle \vec{v}$ in the counter-clockwise direction and expands it by $\lvert \vec{v} \rvert$.

Needham first verifies this when $\vec{v} = \vec{\iota}$, and then uses the distributive property of $\mathbb{C}_{\mathsf{SYMB}}$ to show this for general $\vec{v}$.

Lastly, I believe the point that Needham is trying to emphasise here by transporting the structures from one model to the other and showing the equivalences (note that $g \circ f$ is the identity map on $\mathbb{C}_{\mathsf{GEOM}}$ and $f \circ g$ is the identity map on $\mathbb{C}_{\mathsf{SYMB}}$) is that neither is more fundamental than the other and that one should keep both models in mind when studying complex analysis since one viewpoint can be more useful than the other depending on the problem at hand. In fact, the entire book is based on the idea that the geometric viewpoint is far more useful than it is given credit for.


问:
Showing $X_n \rightarrow X$ and $X_n \rightarrow Y$ implies $X\overset{\text{a.s.}}{=}Y$ for four types of convergence
I need to show that if $X_n \rightarrow X$ and $X_n \rightarrow Y$, then $X\overset{\text{a.s.}}{=}Y$ for convergence in probability, convergence almost surely, as well as for convergence in mean and quadratic mean ($\mathcal L^1$ and $\mathcal L^2$ convergence).

Convergence in Probability:

For any $\epsilon>0$ and for any $n\in\mathbb N$ we have

$$\begin{align}
\mathbb P(|X-Y|\geq\epsilon)
&\leq\mathbb P(|X-X_n|+|X_n-Y|\geq\epsilon)\\\\
&\leq\mathbb P\left((|X-X_n|\geq\epsilon/2)\cup(|X_n-Y|\geq\epsilon/2)\right)\\\\
&\leq\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2)
\end{align}$$

so that

$$\mathbb P(|X-Y|\geq\epsilon)\leq\lim_{n\rightarrow\infty}\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2)=0$$

Since $$\{|X-Y|>0\}=\underbrace{\bigcup_{n=1}^\infty \underbrace{\left\{|X-Y|>\frac{1}{n}\right\}}_{=\emptyset}}_{=\emptyset}=\emptyset$$

we have that $\mathbb P\{|X-Y|>0\}=0$ and so $\mathbb P(X\ne Y)=0$. Hence $\mathbb P(X= Y)=1$ which means that $X\overset{\text{a.s.}}{=}Y$.

Convergence Almost Surely:

Since almost sure convergence implies convergence in probability, the result follows immediately from the last part. However, I'd like to show this without making use of that result. Since $X_n$ converges almost surely to both $X$ and $Y$ then $\mathbb P(\lim_{n\rightarrow\infty}X_n=X)=1$ and $\mathbb P(\lim_{n\rightarrow\infty}X_n=Y)=1$. From here it seems obvious to me that $X\overset{\text{a.s.}}{=}Y$ but I'm not sure how to show this formally.

Convergence in Mean:

$$\begin{align}
\mathbb E(|X-Y|)
&\leq\mathbb E\left(|X-X_n|+|X_n-Y|\right)\\\\
&=\mathbb E\left(|X-X_n|)+\mathbb E(|X_n-Y|\right)
\end{align}$$

so

$$\mathbb E(|X-Y|)\leq\lim_{n\rightarrow\infty}\mathbb E(|X-X_n|)+\mathbb E(|X_n-Y|)=0$$

so $X\overset{\text{a.s.}}{=}Y$

Convergence in Quadratic Mean:

I tried continuing with the same logic but it's not the case that

$$
\mathbb E(|X-Y|^2)\leq\mathbb E\left(|X-X_n|^2+|X_n-Y|^2\right)$$

so I'm not sure how to proceed.

Is my reasoning correct for the first and third? How can I proceed with the other two?

答: It follows from convergence a.s. and from convergence in $L_p$ that there is convergence in probability. So it was enough to condsider convergence in probability.


问:
Spivak Calculus Logarithms
In Spivak's calculus book, when introducing logarithms, he says that we seek to find a function $f$ such that:

$f(x+y) = f(x) f(y)$, for all $x, y$.

He then says, the condition $f(1) = 10$ implies from the previous equation that $f(x) = 10^x$ for rational $x$.

Could someone please explain how?

Thanks!

答:
For any integer $n$, we have $(f(1/n))^n = f(1) = 10$, so $f(1/n) = 10^{1/n}$.

Now for any integer $m$, we have $f(m/n) = (f(1/n))^m = (10^{1/n})^m = 10^{m/n}$.

So the equation holds for all rationals, as required.

Title: Spivak Calculus Logarithms


问:
Solving equations involving modulo operator
In computer programming languages we have an operator called % which expresses the remainder between two numbers. For example $123\%100 = 23$.

I have an equation evolving this operator, namely,

$$\frac{5}{3}(N\%36) - \frac{2}{3}(N\%6) + 2(N\%25) - (N\%5) = (2N)\%100$$

Is there some easy way to solve this equation in $N$ or at least count the solutions within some range of numbers? (I would like some techniques from number theory rather than a brute force solution, which I have already calculated myself.)

答:
We note that $N$ solves your equation if and only if $900+N$ solves your equation.  Then, some quick code (e.g. the following)

```<code>   Select[Range[900], 5 Mod[#, 36]/3 - 2 Mod[#, 6]/3 + 2 Mod[#, 25] - Mod[#, 5]
       == Mod[2 #, 100] &]
</code>```

gives the following solutions in $[1,900]$:

$$\{1, 2, 3, 4, 30, 31, 32, 33, 34, 360, 361, 362, 363, 364, $$
$$390, 391, 392, 393, 394, 720, 721, 722, 723, 724, 900\}$$

We see that each solution occurs in a block of $4$, so we can think of our solutions as elements of
$$\{0,30,360,390,720\},$$
shifted by an element of
$$\{i+900j : i \in [0,4],\, j \in \mathbb{Z}\}.$$

Body:
We note that $N$ solves your equation if and only if $900+N$ solves your equation.  Then, some quick code (e.g. the following)

   Select[Range[900], 5 Mod[#, 36]/3 - 2 Mod[#, 6]/3 + 2 Mod[#, 25] - Mod[#, 5]
       == Mod[2 #, 100] &]

gives the following solutions in $[1,900]$:

$$\{1, 2, 3, 4, 30, 31, 32, 33, 34, 360, 361, 362, 363, 364, $$
$$390, 391, 392, 393, 394, 720, 721, 722, 723, 724, 900\}$$

We see that each solution occurs in a block of $4$, so we can think of our solutions as elements of
$$\{0,30,360,390,720\},$$
shifted by an element of
$$\{i+900j : i \in [0,4],\, j \in \mathbb{Z}\}.$$


答:
$f(x)=f(x/2 + x/2) = f(x)^2 \geqslant 0$ and if $f(x)=0$ then $\forall y,f(y)=0$ and this is not possible since $f(1)=10\neq 0$ So $g:x\mapsto \ln f(x)$ is well defined.

Then $g(x+y)=g(x)+g(y)$ for all $x,y$.

We easily get $g(n) = n g(1)$ by induction for all $n\in \mathbb{N}$, and then for $\mathbb{Z}$.

If $r=p/q$ is rational then $g(p)=g(qr)=qg(r)$ so $g(r)=p/q$.

So for all rational $x$, $g(x)=x g(1)$.

So for all rational $x$, $f(x) = \exp (g(x)) = e^{g(1)x}$.

The condition $f(1) = 10$ implies $g(1)=\ln 10$ so $$\forall x, f(x) = e^{\ln 10 x}=10^x$$

Title: Spivak Calculus Logarithms

Body:
In Spivak's calculus book, when introducing logarithms, he says that we seek to find a function $f$ such that:

$f(x+y) = f(x) f(y)$, for all $x, y$.

He then says, the condition $f(1) = 10$ implies from the previous equation that $f(x) = 10^x$ for rational $x$.

Could someone please explain how?

Thanks!


问:
Computing a determinant with modulo $2$
\begin{vmatrix} \lambda & 1 & 1\\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix}

If we want to see that the determinant is non zero we can do modulo 2, and we get if $\lambda$ is even:
$$ \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$

The determinant is in this case not zero.

But... if we input $\lambda = -2$, then the determinant is zero. Shouldn't the determinant be non zero for every even number in the diagonal since it's modulo 2 determinant is not zero.

EDIT: We want to see if determinant is non zero in integers and used modulo 2 only as a tool for quicker computation. Like shown here in the first answer Is the following matrix invertible?

Title: Computing a determinant with modulo $2$

Body: $$ \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}=2\equiv0\pmod2$$


问:
Prove that the linear combination of two vectors is a plane containing the two vectors
How do you prove that any vector from the linear combination of two vectors forms a plane containing them for two vectors of any dimension?
Can you take the cross product of two vectors and find an equation for their plane and then plug an arbitrary linear combination of the two vector back into that equation to see if it works?</p>

答:
Nick is correct. How to prove this depends on how you define "plane" in this context. Since Nick has discussed the vector space meaning, I'll approach it from the geometric side. The first thing that must be done is to establish the vector space - an algebraic object - as a geometric object. For our purposes, this needs a couple definitions:

point = "vector".

Given any two distinct vectors $v, w$, the line through them is the set $\{tv + (1-t)w\mid t \in \Bbb R\}$.

Now we add the geometric definition of a plane:

A set $A$ of points is affine if for any two points in $A$, the entire line through them is a subset of $A$. (Note that the intersection of a collection of affine sets must also be affine.)

The affine span of a set $B$ is the intersection of all affine sets containing it. (Since the entire space is affine, there is at least one.)

A plane is the affine span of three non-collinear points.

And a final definition to recognize that the concept of "vector" is actually more detailed than that of a "point", more closely represented by line segments from a common endpoint:

The plane through two non-parallel vectors $v, w$ is the affine span of $0, v, w$.

Because the vectors are not parallel, the line through $v, w$ cannot contain $0$ (why?). So $0, v, w$ are not collinear, and their span is a plane by the previous definition. Since we are recognizing the "vector-ness" of $v, w$ in this definition, the plane should not just contain the points $v, w$, but the entire line segments. Since it is affine, this only requires that the other endpoint, $0$, also be included.

So for non-parallel $v, w$, let $V = \{ av + bw \mid a,b \in \Bbb R\}$ be their span, and $A$ be their affine span, that is, the plane through $v, w$.

Note that $0 = 0v + 0w \in V$, and if $u_1 = a_1v + b_1w, u_2 = a_2v + b_2w$ are two vectors in $V$, then for any $t \in \Bbb R$,
$$tu_1 + (1-t)u_2 = [ta_1 + (1-t)a_2]v + [tb_1 + (1-t)b_2]w \in V$$
So the entire line through $u_1, u_2$ is also in $V$. Therefore $V$ is an affine set containing $0, v, w$, and so $A \subseteq V$.

Conversely, if $u = av + bw$ with $a+b \ne 0$, then letting $t = \frac a{a+b}$, we see that $u=(a+b)x$ where $x = tv + (1-t)w$. Since $v, w \in A$, so is $x$. Since $0, x \in A$, so is $u$. Also because $0, w \in A$, we have $-w \in A$ as well. Therefore $\frac 12v + \frac12(-w)$ must also be in $A$, and so must be $t(v-w) + (1-t)0$ for any $t$. Since any $u$ as above with $a+b = 0$ has form $a(v-w)$, this means that for all $u \in V, u \in A$.

So $A = V$. The set $V$ of all linear combinations of the two vectors $v, w$ is the same as the plane $A$ through $v$ and $w$.


问:
find the last 4 dgits of $2^{2016}$
I am trying to find the last 4 digits of $2^{2016}$.

Here is where I am up to:

2^2016 = x mod 10000

2^2016 = x mod 625 - I used Eulers function to find that 2^16 = 1 mod 625 = 536

2^2016 = x mod 16

Now I am stuck. The second equation can not use Euler so how do I continue?

Thanks

Title: find the last 4 dgits of $2^{2016}$

Body: $2^{2016} = 16 \times 2^{2012} \equiv 0 \mod{16}$.


问:
Should the unbiased estimator of the variance of the sample proportion have (n-1) in the denominator?
I know that the variance of the sample mean is:

$$\frac{\sigma^2}{n}$$

And that the unbiased estimator for that expression is:

$$\frac{\sigma^2}{n-1}$$

The variance of the sample proportion is:

$$\frac{p(1-p)}{n}$$

Does an unbiased estimator of variance of the sample proportion also need to have (n-1) in the denominator, like the unbiased estimator of the variance of the sample mean?

I feel the resoning for the denominator should be applicable for both situations, but I haven't seen it mentioned anywhere...

答:
If you have a random variable $X$ with a binomial distribution with parameters $n$ and $p$, then $\mathbb E\left[\frac{X}{n} \right]=p$, so the sample proportion $Y=\frac{X}{n}$ is an unbiased estimator of $p$.

Meanwhile $\mathrm{Var}\left(Y \right) = \mathrm{Var}\left(\frac{X}{n} \right) = \frac{p(1-p)}{n}$ is the variance of the sample proportion,

implying $\mathbb E[Y^2] = \frac{p(1-p)}{n}+\frac{p^2}{n^2}$,

so $\mathbb E\left[Y(1-Y) \right] = p - \frac{p(1-p)}{n}-\frac{p^2}{n^2} = p(1-p)\frac{n-1}{n} $ and thus   $\mathbb E\left[\frac{Y(1-Y)}{n-1} \right] = \frac{p(1-p)}{n}$,

showing $\frac{Y(1-Y)}{n-1} $ is an unbiased estimator of the variance of the sample proportion $\frac{p(1-p)}{n}$


答:
We will show that $f(x)=10^x$ first for $x\in\mathbb N$, then we will expand to $x\in\mathbb Z$ and, finally, to $x\in\mathbb Q$.

Case $x\in\mathbb N$

First, let's find $f(0)$. Note that $10=f(1)=f(1+0)=f(1)f(0)=10f(0)$. By canceling $10$ on both sides, we conclude that $f(0)=1$.

Next, let's prove that, for $n\in\mathbb N$, $f(nx)=f(x)^n$. This goes by induction:

$n=0$: true as we have just proved that $f(0)=1=f(x)^0$

$n\to n+1$: $f((n+1)x)=f(nx)\cdot f(x)=f(x)^n\cdot f(x)=f(x)^{n+1}$

From this, putting $x=1$ we see that $f(n)=f(1)^n=10^n$.

Case $x\in\mathbb Z$

Next, let's assume $n\in\mathbb Z$: $1=f(0)=f(n+(-n))=f(n)\cdot f(-n)$, so $f(-n)=\frac{1}{f(n)}$. In particular, as we know $f(n)=10^n$ for $n\in\mathbb N$, then $f(-n)=\frac{1}{10^n}=10^{-n}$.

Case $x\in\mathbb Q$

Finally, let's take $x\in\mathbb Q$. Say $x=p/q$ with $p\in\mathbb Z, q\in\mathbb N\setminus\{0\}$. We have already proven that $f(qx)=f(x)^q$. However, this means that $10^p=f(p)=f(qx)=f(x)^q$, so $f(x)=(10^p)^{1/q}=10^{p/q}=10^x$

Thus, the formula $f(x)=10^x$ is valid for each $x\in\mathbb Q$.

Title: Spivak Calculus Logarithms


问:
Graph automorphisms and the diamond graph
I thought I understood graph automorphisms, but looking at a specific example is tripping me up. Consider the diamond graph:

Apparently, the Klein 4-group is the automorphism group of this graph. The Klein 4-group is
$$\{(), (12)(34), (13)(24), (14)(23)\}$$

So, if we take $\pi=(12)(34)$ as an example, the definition of graph isomorphism says that $ij$ is an edge in $G$ if and only if $\pi(i)\pi(j)$ is an edge in $G$. But if we look at nodes $2,4$, then $24$ is an edge in $G$ but $\pi(2)\pi(4) = 13$ is not an edge in $G$. So why is $(12)(34)$ an automorphism for this graph?

What am I doing wrong?
What is the best way to visualize the action of a permutation of the vertices of a graph?

Body: Hint: Consider $\{(), (24), (13), (13)(24)\}$.


答:
The Klein $4$-group is the name given to any group with $4$ elements with each element being self-inverse.  The graph automorphism group of a graph on $4$ vertices is, as you wrote, a subgroup of the symmetric group on $4$ elements.  The graph automorphism group of the diamond graph you drew is
$$
\{(), (13), (24), (13)(24)\}
$$

This group is a Klein $4$-group, just not the one that you wrote.

Title: Graph automorphisms and the diamond graph

Body:
I thought I understood graph automorphisms, but looking at a specific example is tripping me up. Consider the diamond graph:

Apparently, the Klein 4-group is the automorphism group of this graph. The Klein 4-group is
$$\{(), (12)(34), (13)(24), (14)(23)\}$$

So, if we take $\pi=(12)(34)$ as an example, the definition of graph isomorphism says that $ij$ is an edge in $G$ if and only if $\pi(i)\pi(j)$ is an edge in $G$. But if we look at nodes $2,4$, then $24$ is an edge in $G$ but $\pi(2)\pi(4) = 13$ is not an edge in $G$. So why is $(12)(34)$ an automorphism for this graph?

What am I doing wrong?
What is the best way to visualize the action of a permutation of the vertices of a graph?


问:
Relation between LCM and GCD
I am trying to prove relation between G.C.D and L.C.M for three integers.
Is the following relation even true. If yes, then I need hints please.
let a, b and c be any integers then,

$$lcm(a,gcd(b,c))=gcd(lcm(a,b),lcm(a,c))$$
Or in notations,
$$[a,(b,c)]=([a,b],[a,c])$$

答:
Hint:

Try using the fundamental theorem of arithmetic, but write with all prime numbers (some with power $0$) rather than just the ones in the factorization.

Then if you look at each prime number, you will find the problem becomes essentially $max(A, min(B,C))=min(max(A,B), max(A,C))$ which I will leave you to prove.

Title: Relation between LCM and GCD


问:
Can a function be differentiable at a jump discontinuity?
I learnt in spivak's calculus that if a function is differentiable at a point then it is continuous at that point however I am confused about this function for example
$$
f(x)=\begin{cases}
-2x & x<4, \\
8 & x=4.
\end{cases}
$$
Is this function differentiable at $x=4$ since this point is an isolated point of the functions range the function is continuous at that point.
However why can we not calculate the derivative at $x=4$ the right hand limit need not exsist as the function is not even defined for values of x greater than 4 so why is the derivative not defined ?

答:
As pointed out in the comments, $f$ is not continuous at $4$, since $\lim_{x\rightarrow 4^{-}}f(x)=-8\neq f(4)$. Recall that $x\in D(f)$ is an isolated point of the domain if there exists $\delta> 0$ such that $[(x-\delta,x+\delta)\setminus \{x\}]\cap D(f)=\emptyset$. Check that this is not the case.

Now, differentiability $\Rightarrow$ continuity still holds for left-differentiability and left-continuity. As the function is not left-continuous, it cannot be left-differentiable at $x=4$.


问:
Complex Integral Without Using Residue Theorems or Cauchy's Theorem $\int_{\gamma} \frac{1}{(z-a)(z-b)} dz$
I am trying to compute the integral $\int_{\gamma} \frac{1}{(z-a)(z-b)} dz$ without any residue theorems or Cauchy's theorem, where $\gamma$ is a circled center at $0$ of radius $r$ with $|a| < r < |b|$. Using PFD, the integral can be rewritten as

$$\frac{1}{a-b} \left[ \int_{\gamma} \frac{1}{z-a} dz - \int_{\gamma} \frac{1}{z-b} dz \right].$$

If $\gamma$ is parametrized as $\gamma(t) = re^{it}$, then the first integral becomes $$\int_{0}^{2 \pi} \frac{ i r e^{i t}}{re^{it} - a} dt.$$

I'm not sure how to deal with that integral. It looks like the integral of $u'/u$, but that would involve logs and one has to be careful with branch cuts....

答:
For $a\in[0,\infty)$ and all $t\in (0,2\pi)$, $re^{it}-a\in\mathbb C\setminus[0,\infty)$, so we can take the branch cut to be at the positive real axis and the integral is $\lim_{t\to 2\pi^-}\log(re^{it}-a)-\lim_{t\to 0^+}\log(re^{it}-a)=2\pi i$.
The general case is similar, simply that the branch cut is along a different line.


答: It doesn't make sense to ask whether $f$ is continuous at 4 because differentiability is only defined for interior points of the domain, but 4 lies on the boundary.

Title: Can a function be differentiable at a jump discontinuity?

Body:
I learnt in spivak's calculus that if a function is differentiable at a point then it is continuous at that point however I am confused about this function for example
$$
f(x)=\begin{cases}
-2x & x<4, \\
8 & x=4.
\end{cases}
$$
Is this function differentiable at $x=4$ since this point is an isolated point of the functions range the function is continuous at that point.
However why can we not calculate the derivative at $x=4$ the right hand limit need not exsist as the function is not even defined for values of x greater than 4 so why is the derivative not defined ?


问:
Quotient Map from R to S1
Hi please could someone check my understanding here please? I've been asked to find a quotient map from $\mathbb R$ to $S^1$.

Is the quotient map the surjective function from $\mathbb R$ to $\mathbb R/\sim$ where an open set in $\mathbb R/\sim$ is open iff it's pre-image is open in $\mathbb R$, or is it the surjective function from $\mathbb R$ to $S^1$ where an open set in $S^1$ is open iff it's pre-image is open in $\mathbb R$?

I've been using my lecture notes and the book by W A Sutherland but I'm struggling to follow the phrasing for questions such as this.

TIA

答:
Let's consider the map $\exp_I:I\to S^1$ such that $t\mapsto(\cos(2\pi t),\sin (2\pi t))=e^{2\pi it}$.

First we want to show that $\exp_I$ is an identification. So, in particular we have to prove that the topology on $S^1$ is the same topology induced by $\exp_I$.

Firts we can observe that $\exp_I$ is continuous because $\exp:\mathbb R\to \mathbb R^2$ is continuous.

We call $\mathcal T_d$ the topology on $S^1$ induced by the euclidean distance and we'll show that $\mathcal T_d=\mathcal T_{\exp_I}$.

The open saturated sets in $[0,1]$ are like $[0,a)\cup(b,1]$ or $(c,d)$, with $0<a<b<1$ and $0<c<d<1$. Notice that $\exp_I^{-1}(\exp_I([0,a)))=[0,a)\cup \{1\}$, which is not open.

The open sets in $(S^1,\exp_I)$ are given by the images of the sets in $\{[0,a)\cup(b,1],(c,d)\}\overset{a,b,c,d\in \mathbb R}{\underset{0<c<d<1}{_{0<a<b<1}}}$, which are arches on $S^1$. Since open sets of $\mathcal T_d$ are given by the intersection of $\epsilon$-balls in $\mathbb R^2$ with the circle $S^1$, we have shown that $\mathcal T_{\exp_I}<\mathcal T_d$.

Conversely is quite easy because open sets for $\mathcal T_d$ are the sets $B_{\epsilon}(p)\cap S^1$ that are open for $\mathcal T_{\exp_I}$ and this shows that $\mathcal T_d<\mathcal T_{\exp_I}\implies \mathcal T_d=\mathcal T_{\exp_I}\implies\exp_I$ is an identification and we can use the universal property of quotient.

We put on $I$ the topology induced by the euclidean topology on $\mathbb R$, so $A\subseteq I$ is open if $\exists U \subseteq \mathbb R$ open such that $A=U\cap I$.

On $S^1$ we consider the topology induced by the map $\exp_I$, so a set $A$ is open for $\mathcal T_{\exp_I}\iff \exp^{-1}(\exp_I(A))=A$.

We now define an equivalence relation $\sim$ such that $$t\sim t'\iff\begin{cases}t=t'\\t,t'\in \{0,1\}\end{cases}\iff\exp_I(t)=\exp_I(t').$$
Let $\pi:I\to I/\sim$ be the quotient map, whose topology on $I/\{0,1\}$ is the topology induced by the map $\pi$.

Since $\exp_I$ is an identification and respect the property $[t]_{\sim}=[t']_{\sim}\implies \exp_I(t)=\exp_I(t')$, we say that $\bar{\exp_I}:I/\{0,1\}\to S^1$ is well-defined, hence injective. Furthermore $\bar{\exp_I}$ is surjective because $\exp_I$ is surjective and it's also continuous for the universal property of quotient $\implies\bar{\exp_I}$ is an homeomorphism whose inverse is $\bar \pi:S^1\to I/\{0,1\}$.

We can conclude that $(I/\{0,1\},\mathcal T_{\pi})\cong(S^1,\mathcal T_d)$.

Title: Quotient Map from R to S1


问:
In order to induce group sparsity, can we use $\left \| A \right \|_{1,1}$ instead of $\left \| A \right \|_{1,2}$?
Let's define $\left \| A \right \|_{p,q}$ as follows:
$$\left \| A \right \|_{p,q} = \sum_{i=1}^n \left \| \alpha^i \right \|_q^p$$.

Where $\alpha^i$ is the i-th row of the matrix A.

The above norm is used for example in the following paper for inducing group sparsity with p = 1 and q = 2.

[Mairal, Julien, et al. "Non-local sparse models for image restoration." 2009 IEEE 12th international conference on computer vision. IEEE, 2009.]

With p = 1 and q = 2, this function essentially first computes $l_2$ norm on the rows of A. So, the matrix become a vector. Then. it computes the $l_1$ norm of that vector.

My question is what are the consequences of using $\left \| A \right \|_{1,1}$ instead of $\left \| A \right \|_{1,2}$? For example, if we want to solve a problem like this:
$$(U,A) = \text{argmin}_{U,A} \left \| Y - UA \right \| + \lambda\left \| A \right \|_{1,2}$$

Does it make the optimization more difficult?

答:
There shouldn't be any difference from the optimization perspective, as both $L_{1,2}, L_{1,1}$ are convex, sub-differentiable functions. One different might be, if you're using a proximal algorithm, is the prox-map, but I think that both have explicit formulas.

However the question to ask is why use an $L_{1,1}$ regularizer. $L_{1,2}$ encourages group sparsity, that is, all elements in the group are likely to be set to zero or all will be non-zero. $L_{1,1}$ however doesn't have this property. It's as if you impose an $l_1$ norm on a matrix viewed as a long vector, hence it will encourage "individual" sparsity, without considering groups.


答: This is just a question of definition. If you look at the definition of left-differentiability (here, for instance), you will see that according to that definition your function is not left-differentiable at $x=4$. And that's that.


问:
What is the integrand of a path integral?
I have been reading the Deligne et. al notes "Quantum Fields and Strings: A Course For Mathematicians" and I am trying to reconcile what I learned with what physicists are doing. I am confused about what physicists mean by expressions of the form $$\langle 0|\phi(x_1)...\phi(x_n)|0\rangle= \int \mathcal{D}\phi \; \phi(x_1)...\phi(x_n)e^{iS[\phi]}$$
I am not asking about the measure $\mathcal{D}\phi$. This integral is supposed to be the "$n$-point correlation function of the field $\phi$" and should return a scalar, but the $\phi(x_i)$ are all supposed to be operators on a Hilbert space, so it is not clear to me how this happens.

This notation also suggests to me that this correlation function is independent of the field $\phi$ (since $\phi$ is also used for the variable of integration), but this seems unreasonable. Would it be correct to say $$\langle 0|\psi(x_1)...\psi(x_n)|0\rangle= \int \mathcal{D}\phi \; \psi(x_1)...\psi(x_n)e^{iS[\phi]}$$ instead?

答:
The integrand is the functional $F[\phi]:= \phi(x_1)\cdots\phi(x_n)\exp(iS[\phi])$. Your confusion probably stems from the fact that physicists like to jump around between the Hamiltonian picture where your field is some kind of operator-valued distribution, and the Feynman picture where you're working with path integrals and $\phi$ is just a function (or distribution). The link between them is precisely that the vacuum expectation value of the operator $\phi(x_1)\cdots\phi(x_n)$, i.e.
$$
\langle 0|\phi(x_1)\cdots\phi(x_n)|0\rangle,
$$ can be calculated as the path integral of the integrand I wrote above.


问:
$f(z)$ is constant on a path $\rightarrow f'(z) = 0$?
$f(z): D\longrightarrow \mathbb C$ is in analytic in $D$ and constant on a diffrentiable path $\Gamma$.

Is necessarily $f'(z) =0$ for every $z \in Γ$? proove or give an example against.
Tried to solve it without success.

I will be happy for help.
thanks :)

答:
If there exists $c\in\mathbb{C}$ such that $f=c$ on $\Gamma$ then the zeros of $f-c$ are not isolated because for any $\varepsilon>0$ there exists $\delta>0$ such that for all $u,v\in\mathbb{R}$, if $|u-v|<\delta$ then $|\Gamma(u)-\Gamma(v)|<\varepsilon$. Therefore if $u\in\mathbb{R}$, there exists $v\in\mathbb{R}$ such that $\Gamma(v)\in D(\Gamma(u),\varepsilon)$ and there is no isolated points in $\Gamma$. According to the isolated zeros principle, $f$ is constant on the connected part of $D$ containing $\Gamma$, so in particular $f'(z)=0$ for all $z\in\Gamma$.

EDIT : Here is another solution : since $f$ and $\Gamma$ are differentiable, then $f\circ\Gamma$ is also differentiable. $f\circ\Gamma$ being constant, we have $\Gamma'(t)f'(\Gamma(t))=0$ for all $t$. I suppose that $\Gamma$ is never constant (that means there is no $a<b$ such that $\Gamma_{|(a,b)}$ is constant), otherwise of course if $\Gamma$ is a constant path it does not work, and if $\Gamma_{|(a,b)}$ is constant then you can get rid of $\Gamma(t)$ for $t\in (a,b)$ by juxtaposing $\Gamma(a)$ with $\Gamma(b)$. Therefore if $\Gamma'(t)\neq 0$ then $f'(\Gamma(t))=0$ and if $\Gamma(t)=0$, there exists a sequence $(t_n)$ converging toward $t$ and such that $\Gamma'(t_n)\neq 0$ for all $n$ (this is because $\Gamma$ is never constant). Hence $f'(\Gamma(t_n))=0$ for all $n$ and taking as limit as $n\rightarrow +\infty$ gives $f'(\Gamma(t))=0$.


答:
The field $\phi$ has slightly different meanings inside the path integral and outside of it, as the Feynman path integral is relating two different interpretations.

Inside the path integral, imagine that we're summing an action of classical fields over a measure of all possible values for those classical fields. As a variable of integration, $\phi$ is thus a classical field and each value $\phi(x_i)$ is a scalar value.

Outside the path integral is where $\phi$ is a Hilbert space operator.

Title: What is the integrand of a path integral?

Body:
I have been reading the Deligne et. al notes "Quantum Fields and Strings: A Course For Mathematicians" and I am trying to reconcile what I learned with what physicists are doing. I am confused about what physicists mean by expressions of the form $$\langle 0|\phi(x_1)...\phi(x_n)|0\rangle= \int \mathcal{D}\phi \; \phi(x_1)...\phi(x_n)e^{iS[\phi]}$$
I am not asking about the measure $\mathcal{D}\phi$. This integral is supposed to be the "$n$-point correlation function of the field $\phi$" and should return a scalar, but the $\phi(x_i)$ are all supposed to be operators on a Hilbert space, so it is not clear to me how this happens.

This notation also suggests to me that this correlation function is independent of the field $\phi$ (since $\phi$ is also used for the variable of integration), but this seems unreasonable. Would it be correct to say $$\langle 0|\psi(x_1)...\psi(x_n)|0\rangle= \int \mathcal{D}\phi \; \psi(x_1)...\psi(x_n)e^{iS[\phi]}$$ instead?


答:
The (one-sided) derivative at $4$ would be

$$ \lim_{x\to 4} \frac{f(x) - f(4)}{x-4} = \lim_{x\to 4} \frac{-2x - 8}{x-4} = \lim_{x\to 4}\left(-2- \frac{16}{x-4} \right)$$

which doesn't exist. So $f$ is not differentiable at $4$, nor is it continuous at $4$: $$\lim_{x\to 4} f(x) = -8 \neq f(4).$$

In order to define a meaningful notion of "the limit of $f(x)$ as $x$ approaches $a$" you need $a$ to be a limit/cluster point of the domain of $f$. So in particular if $a$ is an interior point then you can define the two-sided limit that you're used to. If $a$ isn't an interior point, but still a cluster point, then you can still talk about the limit as $x \to a$ but just be careful that some of the theorems you have seen might no longer apply if $a$ is not an interior point.

In this case, however, it is still true that if $f$ is differentiable at $a$ (with $a$ a cluster point) then $f$ is continuous at $a$.


问:
Can the plane be covered by open disjoint one dimensional intervals?
I remember I heard this question a while ago but have never figured it out. Let an open interval in $\mathbb{R}$ of the form $(a,b)$. Now, imaging placing it in $\mathbb{R}^2$, anywhere we want while keeping its length but perhaps rotating it if we want. Essentially this creates a line segment without its start/endpoints parametrized by

$P(1-t)+Qt$ for $0<t<1$ where $P$ and $Q$ are its start and end points and with length $||P-Q||=b-a$.

As an example, if we take the unit interval $(0,1)$, we can place it as it is in $\mathbb{R}^2$ as $(0,1)\times \{0\}$ or we can rotate it as $\{1\}\times (0,1)$ etc. I hope it is clear what I mean. My question is the following.

Can we cover the entire $\mathbb{R}^2$ with disjoint such "open intervals" or if you prefer, line segments of finite length without their start/endpoints?

I don't remember if the initial question I read was using only unit intervals, or of any length $(a,b)$ and I don't know if that plays a role in the answer but as a gut feeling, I doubt it does. My impression is that the answer is negative and we can't cover the plane in such a manner, or if it is possible it would use some weird abstract construction perhaps involving AC. Any help is appreciated!

Edit: It seems that I could have phrased my wording a bit better although that was my last paragraph in the post above that I wasn't certain if it mattered whether we need to have fixed length intervals or not. Apparently it does, it is impossible with using only fixed length intervals but it is possible if we allow intervals of different lengths. Proof of impossibility in one case given in a paper given in the answers, and a construction in the other case is given by some answers and also found in that paper itself. Thanks everyone!

答:
Assuming that the intervals can be of different lengths, I think it can be done. I sketched something on a paper and it seems to me like this is a solution that only uses 2 different lengths for the intervals, namely $1$ and $\frac 12$:

First, create a fabric-looking structure from horizontal intervals of the form $(m, m+1) \times \{k\}$ and $\left(m+\frac 12, m+\frac 32\right) \times \left\{k+\frac 12\right\}$ and vertical ones of the form $\{k\} \times \left(m+\frac 12, m+\frac 32\right)$ and $\left\{k+\frac 12\right\}\times(m,m+1)$, $\forall k,m\in\mathbb{Z}$. Then just complete it with parallel segments of length $\frac 12$, for example, with all-vertical segments: $\{k+r\}\times\left(m,m+\frac 12\right), \{k+r\}\times\left(m+\frac 12, m+1\right)$, $\forall k,m \in\mathbb{Z}, r\in(0,1)\setminus\left\{\frac 12\right\}$.

I would have added an illustration, but all I have is a pen-on-paper sketch and I'm using a phone. I can add a photo of that if requested.


问:
Evaluate $\lim\limits_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}.$
Suppose $\alpha>1$, and $a_n$ be the largest natrual number $k$ such that $k!\le \alpha^n$.Evaluate $$\lim\limits_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}.$$

According to the assumption, we obtain

$$(a_n)!\le \alpha^n<(a_n+1)!=(a_n+1)(a_n)!.$$
Therefore
$$ (a_1)!(a_2)!\cdots(a_n)!\le \alpha^{1+2+\cdots+n}<(a_1+1)(a_2+1)\cdots(a_n+1)(a_1)!(a_2)!\cdots(a_n)!.$$
But how to go on?

答:
I guess answer is 0.5!

Here's some lemma I could prove:

Lemma 1: $\{a_n\}_{n=1}^{\infty}$ is increasing. (obviously)

Let $a_{t-1} \leq \alpha \leq a_t $.

Lemma 2: for $i\geq t, a_{i+1} = \ a_i$ or $a_{i+1}= a_i + 1$.

Proof: if $a_i = a_{i+1}=...=a_{i+r}\leq a_{i+r+1}$ then

$$a_{i+r}!\leq \alpha ^{i+r-1} \leq (a_{i+r}+1)!$$
$$a_{i+r+1} !\leq \alpha ^{i+r}\leq (a_{i+r+1}+1)!$$
so $\alpha\ a_{i+r}!\leq \alpha ^{i+r} \leq \alpha\ (a_{i+r}+1)!$ then $a_{i+r+1}!\leq \alpha\ (a_{i+r}+1)! \leq a_i \ (a_{i+r}+1)!$. Now if $a_{i+r+1} >a_{i+r}+1$ we come to a contradiction by last inequality!!! So $a_i\leq a_{i+1}\leq a_i +1$ for $i\geq t.$

Lemma 3: by Stirling's approximation $a_n!\approx \sqrt{2\pi a_n}(\frac{a_n}{e})^{a_n}\leq \alpha^n\Rightarrow a_n \ln a_n \leq n \ln \alpha\Rightarrow \frac{\ln a_n}{\ln \alpha} \leq \frac{n}{a_n}\Rightarrow \lim_{n\to \infty}\frac{n}{a_n}=\infty\Rightarrow a_n\leq n$

Lemma 4: let $A_n = \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n} $ then for large enough $n, \lim_{n\to\infty} A_n < \infty.$

Proof:
$$\lim_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}\ =\ \lim_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_{t-1}a_t \cdots a_n}}{a_n}=
 \lim_{n \to \infty} \frac{\sqrt[n]{a_ta_t\cdots a_t\ (a_t+1)(a_t+2) \cdots (a_t+n-t)}}{a_n}< \lim_{n \to \infty} \frac{a_t+a_t+\cdots +a_t+ (a_t+1)+(a_t+2)+ \cdots +(a_t+n-t)}{na_n}=\lim_{n \to \infty} \frac{na_t+1+2+\cdots (n-t)}{na_n}=\lim_{n \to \infty} \frac{a_t}{a_n}+\frac{(n-t+1)(n-t)}{2na_n}=0+\lim_{n \to \infty} \frac{n}{a_n}=\infty$$

Finally by running a programm in matlab:\

```<code>figure;
for j=0:1:10
alpha = 1+j;
A = [];
B = [];
for i=1:100
    k = 1;
    while (factorial(k) <= power(alpha, i))
        k = k + 1;
    end
    A = [A, k-1];
    B = [B, nthroot(prod(A),i)/A(end)];
end

%display(A);
%display(B);
%figure;
%plot(A);
%hold on;
%figure;
plot(B);
hold on;
end
</code>```

limit of A_n

limit of a_n

Body:
I guess answer is 0.5!

Here's some lemma I could prove:

Lemma 1: $\{a_n\}_{n=1}^{\infty}$ is increasing. (obviously)

Let $a_{t-1} \leq \alpha \leq a_t $.

Lemma 2: for $i\geq t, a_{i+1} = \ a_i$ or $a_{i+1}= a_i + 1$.

Proof: if $a_i = a_{i+1}=...=a_{i+r}\leq a_{i+r+1}$ then

$$a_{i+r}!\leq \alpha ^{i+r-1} \leq (a_{i+r}+1)!$$
$$a_{i+r+1} !\leq \alpha ^{i+r}\leq (a_{i+r+1}+1)!$$
so $\alpha\ a_{i+r}!\leq \alpha ^{i+r} \leq \alpha\ (a_{i+r}+1)!$ then $a_{i+r+1}!\leq \alpha\ (a_{i+r}+1)! \leq a_i \ (a_{i+r}+1)!$. Now if $a_{i+r+1} >a_{i+r}+1$ we come to a contradiction by last inequality!!! So $a_i\leq a_{i+1}\leq a_i +1$ for $i\geq t.$

Lemma 3: by Stirling's approximation $a_n!\approx \sqrt{2\pi a_n}(\frac{a_n}{e})^{a_n}\leq \alpha^n\Rightarrow a_n \ln a_n \leq n \ln \alpha\Rightarrow \frac{\ln a_n}{\ln \alpha} \leq \frac{n}{a_n}\Rightarrow \lim_{n\to \infty}\frac{n}{a_n}=\infty\Rightarrow a_n\leq n$

Lemma 4: let $A_n = \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n} $ then for large enough $n, \lim_{n\to\infty} A_n < \infty.$

Proof:
$$\lim_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}\ =\ \lim_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_{t-1}a_t \cdots a_n}}{a_n}=
 \lim_{n \to \infty} \frac{\sqrt[n]{a_ta_t\cdots a_t\ (a_t+1)(a_t+2) \cdots (a_t+n-t)}}{a_n}< \lim_{n \to \infty} \frac{a_t+a_t+\cdots +a_t+ (a_t+1)+(a_t+2)+ \cdots +(a_t+n-t)}{na_n}=\lim_{n \to \infty} \frac{na_t+1+2+\cdots (n-t)}{na_n}=\lim_{n \to \infty} \frac{a_t}{a_n}+\frac{(n-t+1)(n-t)}{2na_n}=0+\lim_{n \to \infty} \frac{n}{a_n}=\infty$$

Finally by running a programm in matlab:\

figure;
for j=0:1:10
alpha = 1+j;
A = [];
B = [];
for i=1:100
    k = 1;
    while (factorial(k) <= power(alpha, i))
        k = k + 1;
    end
    A = [A, k-1];
    B = [B, nthroot(prod(A),i)/A(end)];
end

%display(A);
%display(B);
%figure;
%plot(A);
%hold on;
%figure;
plot(B);
hold on;
end

limit of A_n

limit of a_n


问:
Prove that if k is odd and G is a k-regular (k-1)-edge-connected graph, then G has a perfect matching.
I understand that if k is odd and G is k-regular, then each vertex must have odd degree. I am having trouble understanding how to incorporate the edge-connectivity.

(This is my first undergraduate graph theory class, so a simple explanation would be helpful.)

Thanks!

答:
In fact, $k$ could be even and odd and theorem holds.

We can use Tutte's theorem to prove the result.

Let $S \subset V(G)$, and $odd(G/S)$ the number of odd components in $G-S$.

Then the number of edges out of $S$ is at most $|S|*k$. Now consider an odd component $O$ in $G-S$, we have:
$$ k*|O|=\sum_{v \in V(O)}deg(v) = 2|In|+|Out| $$
Here $|Out|$ denotes the number of edges with only one end in $O$ and $In$ denotes the number of edges with both end in $O$. Now because $|O|$ is odd so $k$ has the same parity as $|Out|$.

Edit: Now $|Out| \geq k-1$, so $|Out| \geq k$(here we use the fact that $|Out|$ and $k$ has the same parity).

to Now $|Out| \geq k-1$, so $|Out| \geq k$(here we use the fact that $|Out|$ and $k$ has the same parity). So thanks to the comment due to @Misha, we realized that $|Out| \geq k-1$ is true if $k$ is odd. Otherwise $|Out|$ could be even and hence equals to $0$ and then the argument breaks.

Now the number of edges with one end in an odd component and one end in $S$ is at least $k|odd(G-S)|$ which is also less than or equals to $|S|*k$. So $|S| \geq |odd(G-S)|$. So by Tutte's theorem the graph has a perfect matching.


问:
Mapping between a set and its power set
For a set $X$, let $P(X)$ be the set of all subsets of $X$ and $\Omega(X)$ be the set of all functions from $f:X \to \{0,1\}$. Then

$(1)$ If $X$ is finite , then $P(X)$ is finite

$(2)$ if $X$ and $Y$ are finite  and if there is a one-one correspondence between $P(X)$ and $P(Y)$, then there is a one-one correspondence between $X$ and $Y$

$(3)$ There is no one-one correspondence between $X$ and $P(X)$

$(4)$ There is a one_one correspondence between $\Omega(X)$ and $P(X)$

My attempt:-

$(1)$ is true.

$(3)$ True by Cantor's Theorem

I know there are answers to $(4)$ on this site but I am specially interested in $(2)$ which I know to be true.

Let $f:P(X) \to P(Y)$ be a one-one correspondence. Since $X$ and $Y$ are finite we have $|P(X)|=|P(Y)|\Rightarrow |X|=|Y|$ otherwise $|P(X)|\neq P(Y)|$. Since X and Y are finite sets with same no. of elements , there is a bijection . Is this a good enough proof? Can this be  generalized to two infinite sets having a one-one correspondence between their power sets.

Thanks for your thoughts and time.

答:
I think this answer will help you a lot.

Given that $~~P(X):$  set of all subsets of $X$
 $~~~~~~~~~~~~~~~~~~~Ω(X):$ set of all functions $~f:X→\{0,1\}$

For option $\bf(1)$, if $X$ is finite, let $|X|=n$, then $|P(X)|=2^n$ which is again finite. Hence option $(1)$ is true.

For option $\bf(2)$, let $X$ and $Y$ are finite and there is one-one correspondence between $P(X)$ and $P(Y)$.$$\implies |P(X)|=|P(Y)|\implies 2^{|X|}=2^{|Y|}\implies |X|=|Y|$$Hence there exists one-one correspondence between $X$ and $Y$. Thus option $(2)$ is true.

For option $\bf(3)$, since $~|P(X)|=2^{|X|}\gt|X|~$
Thus there is no one-one correspondence between $X$ and $P(X)$. Therefore option $(3)$ is true.

For option $\bf(4)$, since $~|P(X)|=2^{|X|}=|\Omega(X)|~$
Therefore there exists a one-one correspondence between $\Omega(X)$ and $P(X)$. Therefore option $(4)$ is true.

Title: Mapping between a set and its power set


问:
Uniform convergence of power series on open interval
Let be $S:=\sum\limits_k^{\infty}a_kx^k$ a real valued power series with convergence radius of $R>0$. Then, we know that the power series converges uniformly on every compact subset of $(-R,R)$.

Why do we require a compact set instead of an open set?

We could simply choose an $r$ with $0<r<R$ and then define an interval $M:=(-r,r)$. If I go through the proof which shows that $S$ is pointwise convergent on $(-R,R)$ I don't see where it should fail if I choose $r$ as an upper bound for all $x\in(-r,r)$? So I would conclude that $S$ is uniformly convergent on each open subset $(-r,r)$ with $0<r<R$ of $(-R,R)$.

答:
We require a compact set because we have to exclude $(-R,R)$ itself. For any open subset you mention, the closure is then compact and also contained in $(-R,R)$, so the power series converges uniformly there, hence also on the open set.

Generally, compact convergence is quite common because as you mentioned power series converge compactly, a lot of other stuff also converges compactly (for example continuous functions on $\mathbb{R}^n$ converging pointwise to another continuous function). And they have the property that you can switch limits: if $x_n\to x$ and $f_n\to f$ compactly, then $$\lim_{n\to\infty}\lim_{m\to\infty} f_n(x_m) = \lim_{m\to\infty}\lim_{n\to\infty} f_n(x_m),$$ so that you can for example exchange integration or differentiation with compact limits.

In a more abstract setting, if $X$ is a locally compact topological space, the compact topology is the natural topology to give to $C(X,Y)$ (the space of continuous functions to another space $Y$), and the functor $C(X,-)$ is the right-adjoint for the functor $-\times X$ (topological tensor-hom adjunction).


问:
Lines on a singular cubic surface
How many lines the cubic surfaces $xyz=w^3 \in \mathbb P^3$ has? I found only three: $x=w=0$, $y=w=0$ and $z=w=0$. How to prove that there are no other lines? Also, this surface is singular, is it possible to calculate these lines with some natural multiplicities, summing up to 27?</p>

答:
Here is a direct way of showing the cubic surface $S=\{xyz=w^3\}$ with three cusp singularities has only three lines: $x=w=0$, $y=w=0$ and $x=w=0$.

Suppose there is another line $L$ in $S$ different from the above, then $L$ must contain a point $p$ of the form $$p=(x_0,y_0,z_0,1),$$

and $L$ lies in the tangent hyperplane $T_pS$. The condition $p\in S$ requires that $x_0y_0z_0=1$, so $T_pS$ has equation

$$\frac{x}{x_0}+\frac{y}{y_0}+x_0y_0z-3w=0.$$

The intersection $C=T_pS\cap S$ is a cubic plane curve and $C$ contains $L$ as an irreducible component, therefore $C$ is reducible. However, by isolating $w$ in the linear equation and canceling $w$ in the cubic surface equation, one finds the equation of $C$ is
$$xyz=\frac{1}{27}(\frac{x}{x_0}+\frac{y}{y_0}+x_0y_0z)^3$$

in the projective plane with coordinate $x,y,z$. Of course, $C$ is singular at $p=(x_0,y_0,\frac{1}{x_0y_0})$ since it is tangency point. However, a direct computation of partial derivatives shows that $p$ is the only singularity of $C$ and the Hessian at $p$ is always nondegenerate, so it is an ordinary double point. Therefore $C$ is irreducible, which is a contradiction.

Title: Lines on a singular cubic surface


问:
Understanding the definition the norm of a principal ideal of a number field
Hi I'm a grad student and I was re reading the lecture notes my professor from a previous semester posted for algebraic number theory.

The lecture notes state: "If $O_K$ is a number field and $I$ is an ideal of $O_K$, then $N(I)$ is defined as $N(I) = |O_K : I|$."

I took this definition to mean, if $N(I) = n$ then there are $n$ elements $a_1, a_2, \cdots a_n \in O_K$ such that every element of $a \in O_k$ is of the form $i + a_j$ where $i \in I$, and $1 \leq j \leq n$.

Later on the lecture notes state:

"Let $\gamma \in O_K$, and $\{\beta_1, \cdots \beta_n \}$ be an integral basis of $O_K$. Let $$\gamma\beta_j = \sum_{k = 1}^n a_{j,k}\beta_k, \, \, \, \, \, j = 1, 2, \cdots, n,$$
then \begin{equation*} \begin{bmatrix}
\gamma \beta_1 \\
\vdots \\
\gamma \beta_n
\end{bmatrix} = A\begin{bmatrix}
 \beta_1 \\
\vdots \\
\beta_n
\end{bmatrix}, \, \, \, \, \, A = [a_{j,k}] \in \mathbb{Z}^{n \times n} \end{equation*}

Let $I=\langle \gamma \rangle$.
Now $N(I) = |O_K :I| = |det(A)|$."

So I'm having trouble seeing why $|O_K :I| = |det(A)|$?

答:
First, consider the case where $A$ is diagonal with positive diagonal entries $d_1,\dots,d_n$. Note that if
$$
\pmatrix{\gamma \beta_1 \\ \vdots \\ \gamma \beta_n} = \pmatrix{d_1 \beta_1\\ \vdots \\ d_n \beta_n},
$$
Then the elements of the form $a= \sum_i c_i \beta_i$ with $c_i \in \{0,1,\dots,d_i-1\}$ are such that every element of $O_K$ is of the form $m + a$ for some $m \in \langle \gamma\rangle$ and some such element $a$. There are $d_1d_2 \cdots d_n$ such elements, so $|O_K:I| = |\det(A)|$ holds.

For the general case, note that $A$ has a smith normal form decomposition $A = UDV$ with $U,V$ unimodular and $D$ diagonal with positive diagonal entries. Define $\delta_1,\dots,\delta_n$ such that
$$
\pmatrix{\delta_1\\ \vdots \\ \delta_n} = V \pmatrix{\beta_1\\ \vdots \\ \beta_n}.
$$
Note that $\delta_1,\dots,\delta_n$ is another integral basis of $O_K$. We have
$$
\pmatrix{\gamma\beta_1\\ \vdots \\ \gamma \beta_n} =  U\pmatrix{d_1 \delta_1 \\ \vdots \\ d_n \delta _n}.
$$

Note that for $c = (c_1,\dots,c_n) \in \Bbb Z^n$, $\gamma \sum_i c_i \beta_i$ can be computed as
$$
\pmatrix{c_1 & \cdots & c_n}\pmatrix{\gamma\beta_1\\ \vdots \\ \gamma \beta_n}
= c^TA \pmatrix{\beta_1 \\ \vdots \\ \beta_n}.
$$
Thus, each element of $\langle \gamma \rangle$ can be expressed in the form
$$
c^TU\pmatrix{d_1 \delta_1 \\ \vdots \\ d_n \delta _n} = (U^Tc)^T \pmatrix{d_1 \delta_1 \\ \vdots \\ d_n \delta _n},
$$
which is to say that the set $\{d_i \delta_i\}$ forms an integral basis for $\langle \gamma \rangle$. So, the elements of the form $a= \sum_i c_i \delta_i$ with $c_i \in \{0,1,\dots,d_i-1\}$ are such that every element of $O_K$ is of the form $m + a$ for some $m \in \langle \gamma\rangle$ and some such element $a$. There are $d_1d_2 \cdots d_n$ such elements, so $|O_K:I| = \det(D) = |\det(A)|$ holds, which was what we wanted.


问:
$i+j-1$ tokens on row $i$ column $j$ of square grid
Consider an $n\times n$ grid, with rows and columns numbered from $1$ to $n$. On the cell in row $i$ column $j$, I place $i+j-1$ tokens. How many tokens are on the grid in total?

I have done the double sum directly:
$$\sum_{i=1}^n\sum_{j=1}^n(i+j-1)=\sum_{i=1}^n\left[n(i-1)+\frac{n(n+1)}{2}\right]=\frac{n^2(n-1)}{2}+\frac{n^2(n+1)}{2}=n^3.$$
With an answer as neat as this, I'm wondering if there's a simple (perhaps geometric) interpretation/argument?

答:
Perhaps the following is an overkill, but explains how to backtrack arguments with combinatorials. Call $[n]=\{1,2,\cdots ,n\}.$ You can backtrack the argument. Suppose you want to count the number of functions from $[3]$ to $[n].$ this is clearly $n^3$ now, fix the image of $1$ so
$$n^3=n\cdot n^2=n\cdot \left (2\binom{n}{2}+\binom{n}{1}\right ),$$
where the second equality is given by either you choose two images or one for $2,3.$

Now, one copy of $\binom{n}{2}$ you combined it with $\binom{n}{1}$ getting $n^3=n\left (\binom{n}{2}+\binom{n+1}{2}\right)$ using the recursion of pascal interpretation(basically if $n+1$ belongs to the subset is because you were fixing just one image. The function is constant in $2,3$). Now you use the hockey stick interpretation(this is getting the biggest element in the subset and then the other ones.) to get
$$n^3=\sum _{j=1}^{n-1}\left (n\cdot j+\binom{n+1}{2}\right )=\sum _{j=1}^{n}\left (n\cdot (j-1)+\binom{n+1}{2}\right ),$$
again Hockeystick to get
$$n^3=\sum _{j=1}^{n}\sum _{i=1}^nj-1+i,$$
so the whole interpretation becomes as follows:

Suppose you have a function $f:[3]\longrightarrow [n]$ then either $f(2)>f(3)$ or $f(2)\leq f(3).$ If it is the first call $i$ the image of $1$ and call $j$ the image of $f(2)$ then you have  $j-1$ options for $f(3)$ giving you $\sum _{i,j}j-1.$ If it is the other case, call $j$ the image $f(1)$ and call $i+1$ the image of $3,$ then you have $i$ options for $f(2).$ If $i+1=n+1,$ then your function will be constant on $2,3$ and you have $n$ options to pick this number giving you $\sum _{i,j}i.$ Because this options are disjoint, you add them.


问:
Let $A,B$ be self-adjoint operators and $A\leq B$. Is $e^A \leq e^B$ true?
If $A,B$ commute, that $e^A\leq e^B$ follows from functional calculus.

Is this still true when $A,B$ do not commute?

Also I wonder if $A^{2n+1}\leq B^{2n+1}$ is true for natural number $n$.

答:
Example.

$$
A = \begin{bmatrix}
1/2&1\cr
1&1
\end{bmatrix}
\\
B = \begin{bmatrix}
1&0\cr
0&3
\end{bmatrix}
$$
Then $A \le B$ is true, but $A^3 \le B^3$ is false and
$e^A \le e^B$ is false.

How did I find it?

For a $2 \times 2$ matrix
$$
\begin{bmatrix}
a&b\cr
c&d
\end{bmatrix}
$$
the conditions for positive semidefinite are
$$
b=c,\quad a \ge 0,\quad ad-bc \ge 0 .
$$
I know that a self-adjoint matrix can be diagonalized, and I normalized an entry to $1$.  Using
$$
A = \begin{bmatrix}
a&b\cr
c&d
\end{bmatrix},\qquad
B = \begin{bmatrix}
1&0\cr
0&f
\end{bmatrix}
$$
the conditions for $A \le B$ are
$$
b=c,\quad a\le 1,\quad ad-af-cb - a + f \ge 0 .
\tag1$$
The conditions for $A^3 \le B^3$ are
$$
-{a}^{2}b-abd-{b}^{2}c-b{d}^{2}=-{a}^{2}c-acd-b{c}^{2}-c{d}^{
2},
\\
a^3+2abc+bcd \le 1
\\
{a}^{3}{d}^{3}-{a}^{3}{f}^{3}-3\,{a}^{2}b
c{d}^{2}+3\,a{b}^{2}{c}^{2}d-2abc{f}^{3}-{b}^{3}{c}^{3}-bcd{f}^{3}-a
bc-2bcd-{d}^{3}+{f}^{3} \ge 0 .
\tag2$$
Then we have to find values that satisfy $(1)$ but not $(2)$.  Mostly I did it with equality for the inequalities, except for one strict inequality forward in $(1)$ (choosing $a=1/2$) and then one backward in $(2)$ (choosing $d=1$).


问:
What does "attain values" mean for an analytic function here?
Here's a question in complex analysis that I don't get, neither do I understand the first part of its answer given in the book. I would really appreciate some help.

Question:

Show that if $f(z)$ is an entire function, and there is a nonempty disk
such that $f(z)$ does not attain any values in the disk, then $f(z)$ is
constant.

Answer:

If $f$ does not attain values in the disk $|w - c| < \epsilon$, then
$1/(f - c)$ is bounded, hence constant by Liouville's theorem, and $f$
is constant.

Entire function: a function that is analytic on the entire complex plane.

Liouville's theorem: every bounded entire function is a constant.

Source of the question: Gamelin, Complex Analysis.

Happy New Year!

答:
It means that for some disc $B(a,r)$ we have that

$$ f(\mathbb{C}) \cap B(a,r) = \emptyset$$

Then, for all $z \in \mathbb{C}$,

$$ | f(z) - a | \geq r$$

and you can consider its multiplicative inverse and apply Liouville.


问:
Is there a continuous $f(x,y)$ which is not of the form $f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$
For a continuous function $f : [0,1]^2 \to \mathbb{R}$, let us say $f$ is a sum of products (SOP) if there exist an integer $n > 0$ and continuous functions $g_1, \dots, g_n, h_1, \dots, h_n : [0,1] \to \mathbb{R}$ such that $$f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$$
for all $x,y \in [0,1]$.

How can I show there exists a continuous $f$ which is not an SOP?

I think this should be easy, but for some reason I don't see how to proceed.

Note that the Stone-Weierstrass theorem says that every continuous function on $[0,1]^2$ is a uniform limit of SOPs.  I want to see that you can't drop the limit.

Is the same true if we replace $[0,1]$ by an arbitrary infinite compact Hausdorff space?

答:
We have a writing $f(x,y) =\sum_{i=1}^n g_i(x) h_i(y)$, equivalently
$$f(\cdot, y) = \sum_{i=1}^n h_i(y) g_i(\cdot)$$
means that the system of function   $(f(\cdot, y))_y$ is  contained in the span of n functions $g_i(\cdot)$, $i=1, \ldots, n$, equivalently,
this system of functions $(f(\cdot, y))_y$ has rank at most $n$ ( remember: functions are "vectors" ).

But for a general function $f(x,y)$ ( analytic, not a polynomial) the system $(f(\cdot, y))_y$ is infinite dimensional ($e^{x y}$ (@Daniel Fischer: answer), $\frac{1}{x+y}$, $\sqrt{x+y}$, $\frac{1}{1-x y}$ $\&$ c.)

It's worth looking at $f(x,y)$ as an infinite matrix indexed by $x$, $y$. There exists such a writing if and only if the matrix has rank $\le n$, that is, if and only if every minor
$$\det (f(x_i, y_j))_{i,j=1, n+1}=0$$


问:
Marginalization of Gaussian canonical form
I'm struggling with deriving the marginalization of Gaussian canonical form.

Suppose a joint Gaussian $x = [x_1 \  x_2]^T$ in the moment form
\begin{align}
p(x;\mu,\Sigma) = \frac{1}{\sqrt{(2\pi)^d|\Sigma|}} \exp \left( -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu) \right),
\end{align}
where
\begin{align}
\mu=\begin{bmatrix}
\mu_1 \\
\mu_2
\end{bmatrix},
\end{align}
\begin{align}
\Sigma=\begin{bmatrix}
\Sigma_{11} & \Sigma_{12} \\
\Sigma_{21} & \Sigma_{22}
\end{bmatrix}.
\end{align}
Its canonical form is
\begin{align}
p(x;\eta,\Lambda) = \frac{|\Lambda|}{\sqrt{(2\pi)^d}} \exp \left( -\frac{1}{2}x^T \Lambda x + \eta^T x - \frac{1}{2} (\eta^T \Lambda \eta) \right),
\end{align}
where
\begin{align}
\eta = \Sigma^{-1} \mu = \begin{bmatrix}
\eta_1 \\
\eta_2
\end{bmatrix},
\end{align}
\begin{align}
\Lambda = \Sigma^{-1} = \begin{bmatrix}
\Lambda_{11} & \Lambda_{12} \\
\Lambda_{21} & \Lambda_{22}
\end{bmatrix}.
\end{align}
We can use elementary transformation to derive
\begin{align}
\Lambda = \begin{bmatrix}
\Sigma_{11} & \Sigma_{12} \\
\Sigma_{21} & \Sigma_{22}
\end{bmatrix}^{-1}
= \begin{bmatrix}
\Sigma_{11}^{-1}+\Sigma_{11}^{-1}\Sigma_{12} (\Sigma/\Sigma_{11}) \Sigma_{21}\Sigma_{11}^{-1} & -\Sigma_{11}^{-1}\Sigma_{12}(\Sigma/\Sigma_{11}) \\
-(\Sigma/\Sigma_{11})\Sigma_{21}\Sigma_{11} & (\Sigma/\Sigma_{11})
\end{bmatrix},
\end{align}
where $(\Sigma/\Sigma_{11})=\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}$ is the Schur complement of $\Sigma$ w.r.t. $\Sigma_{11}$.

With the above notation, we can write the marginal and conditional in the moment form as follows.
\begin{align}
\mu_1^\text{Marg} &= \mu_1 \\
\Sigma_1^\text{Marg} &= \Sigma_{11}
\end{align}
\begin{align}
\mu_{2|1}^\text{Cond} &= \mu_2 + \Sigma_{21}\Sigma_{11}^{-1}(x_1-\mu_1)\\
\Sigma_{2|1}^\text{Cond} &= (\Sigma/\Sigma_{11}) = \Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}.
\end{align}

I understand that with the moment form, this can be done by manipulating the quadratic polynomial in the exponential into $p(X_1)$ and $p(X_2 \mid X_1)$. But I'm stuck at deriving the marginal and conditional in the canonical form
\begin{align}
\Lambda_{2|1}^\text{Cond} &= \Lambda_{22} \\
\eta_{2|1}^\text{Cond} &= \eta_2 - \Lambda_{21} x_1
\end{align}
\begin{align}
\Lambda_{1}^\text{Marg} &= \Lambda_{11} - \Lambda_{12} \Lambda_{22}^{-1} \Lambda_{21} \\
\eta_{1}^\text{Marg} &= \eta_1 - \Lambda_{12} \Lambda_{22}^{-1} \eta_2.
\end{align}

I guess I should also begin with the joint distribution and do some manipulation to obtain $p(X_1)$ and $p(X_2 | X_1)$
\begin{align}
p(x;\eta,\Lambda) = \frac{|\Lambda|}{\sqrt{(2\pi)^d}} \exp \left( -\frac{1}{2} \begin{bmatrix}x_1 \\ x_2\end{bmatrix}^T  \begin{bmatrix}
\Lambda_{11} & \Lambda_{12} \\
\Lambda_{21} & \Lambda_{22}
\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} + \begin{bmatrix}\eta_1 \\ \eta_2\end{bmatrix}^T \begin{bmatrix}x_1 \\ x_2\end{bmatrix} - \frac{1}{2} (\eta^T \Lambda \eta) \right)
\end{align}

But I cannot split this thing into the correct marginal and conditional. Any help would be appreciated!

Reference:
https://people.eecs.berkeley.edu/~jordan/courses/260-spring10/other-readings/chapter13.pdf

答:
Since the vector $X=[X_1,...,X_d]^T$ is normal, its elements are jointly normal, i.e., any linear combination of them is normal. Let $a=[1, 0, ...,0]^T$, then $X_1 = a^T X$ is normal with mean and variance as follows.

$\mu_{X_1}=E\{X_1\}=a^T E\{X\} = a^T \mu = \mu_1$, and

$\sigma_{X_1}^2=E\{(X_1-\mu_1)(X_1-\mu_1)^T\} = a^TE\{(X-\mu)^T(X-\mu)\}a=a^T\Sigma a=\Sigma_{11}$.

Choosing $a\in \{0,1\}^d$ accordingly, you may derive all marginal distribution. As for the conditional distributions, applying $a$ results in the omission of the rows and columns of $\Sigma$ corresponding to the non-zero elements of $a$. Since the determinant of product is the product of determinants, the normalizing factor also checks out.


问:
Find continuous $f$ such that $f(x+1)=f(x)+f(\frac{1}{x})$

Find all real continuous functions that verifies :
$$f(x+1)=f(x)+f\left(\frac{1}{x}\right)    \ \ \ \ \ \  (x\neq 0)    $$

I found this result $\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$ and I tried to study the behaviour of the function $g$ defined as $g(x)=\frac{x}{x-1}$ and compare it with $x$ in order to use fixed point theorem but it won't work.

I need a hint and thanks.

答:
Some partial results:

Letting $x+1= {1\over x} $ we get $$x_{1,2}= {-1\pm \sqrt{5}\over 2}$$ are zeroes for $f$.

If $x\to {1\over x}$ we get $$f(1+{1\over x}) = f({1\over x})+f(x)=f(1+x)$$ so $$\boxed{f(1+{1\over x}) = f(1+x);\;\;\;\forall x}$$

So taking $x\to 0$ in this equation we get $$f(1) = \lim _{x\to
   \pm \infty}f(x)$$ and on the other hand we have, from starting equation $$f(1)-f(0) =\lim _{x\to
   \pm \infty}f(x)$$
so $f(0)=0$ and we have $x_3=0$.

From boxed equation we get $x\to x-1$ we get, for $\forall x\ne 1$: $$f(x) = f\Big({x\over x-1}\Big)$$

 Since last formula is not defined for $x=1$ it is tempting to check whether $$\boxed{g(x) = {x^2+x-1\over x-1}}$$ satisfies the condition, which does, but unfortunately, it is no defined on whole $\mathbb{R}$.

Finally, if we set $V$ as a set of all solution to this equation, we see that $V$ is a vector space over $\mathbb{R}$.


问:
How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$
How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$ since $e^0 = 0^0$, and we know the right hand side is undefined?</p>

答:
Actually, there are many arguments to define $0^0 := 1$. This is one of the many compelling reasons.

Arguably the most persuasive is the fact that if $f$ and $g$ are analytic functions with $f(a)=g(a)=0$, not identically zero in a neighborhood of $a$, then
$$
\lim_{x\to a} f(x)^{g(x)} = 1.
$$


问:
Finding all pairs $(m,n)$ of positive integers such that $m(n + 1) + n(m − 1) = 2013$

Find all pairs $(m, n)$ of positive integers such that
$$m(n + 1) + n(m − 1) = 2013$$

What I've done: I tried to make the problem simpler by $2mn+m-n$ and factoring it equals $(2m-1)(2n+1)$

Where do I go now?

Body:
If you've gotten $m(n+1) + n(m-1)= 2mn + m -n$

And $(2m-1)(2n+1) = 4mn +2m - 2n - 1$ so your factoring is not correct.

However if $m(n+1) + n(m-1) = 2mn + m -n = 2013$ then

$(2m-1)(2n+1) = 4mn + 2m -2n - 1 = 2(2mn + m -n) - 1 = 2*2013 - 1 = 4025$.

So now it's a matter of prime factoring $4025$.

Clearly $25|4025$ so $4025 =  161*5^2$ and $161=7*23$ so $4025 = 5^2*7*23$ and so

$(2m-1)(2n + 1) = 5^2 *7*23$ so you have the following problems to solve.

If you break $4025= 5^2*7*23$ into two factors, $4025 = A*B$ you will be get several ($24$) sets of equations to solve and several ($24$) pairs of solutions..

$2m-1 = A$ and $2n+1 = B$ so $m = \frac {A+1}2$ and $n = \frac {B-1}2$.

Where $(A,B) = $

$(1, 4025)$

$(5, 805)$

$(7, 575)$

$(23, 175)$

$(25, 161)$

$(35, 115)$,

..... etc..... (Don't forget the reverse pairs $A = 4025$ and $B=1$ and $A=805$ and $B = 5$ as well as $A=1; B=4025$ and $A=5;B=805$)....


问:
Let $\phi$ a homomorphism from $G$ to $H$. Let $K \leq G$ contain the kernel of $\phi$. Show that $\phi ^{-1}\left( \phi\left( K \right) \right) = K.$

Let $\phi$ a homomorphism from $G$ to $H$. Let $K \leq G$ contain the kernel of $\phi$. Show that $\phi ^{-1}\left( \phi\left( K \right) \right) = K.$

I have been asked to prove this statement. It looks quite simple to prove, but for some reason I am very stuck trying to prove it. I have taken the approach of trying to show that the sets on each side of the equality are subsets of each other. I have that the set on the left side of the equality is essentially
$$
\left\{ g \in G: \phi\left( g \right)=\phi \left( k \right) \text{ for some }k\in K \right\}
$$
I also see that since the kernel of $\phi$ is in $K$, then we have that
$$
\phi\left( g \right)=e \implies g\in K
$$
I am using the approach of trying to prove that if $ \phi\left( g \right)=\phi \left( k \right) \text{ for some }k\in K $, then $ g \in K $. Obviously the other containment $\left(( K \subseteq \phi ^{-1}\left( \phi\left( K \right) \right) \right)$ is quite trivial, but I cannot see is how to prove the harder containment. I know this is probably very easy, but for some reason I can't work it out. Thanks for any help.

答: Hint: $\phi$ is a homomorphism, so if $\phi(g)=\phi(k)$, then $\phi(gk^{-1})=?$


问:
Group algebra of Lie group
Let $G$ be a compact Lie group over a field $K$, say $\mathbb{R}$ or $\mathbb{C}$. Can we define the group algebra $KG$ of $G$? I assume that since $G$ is continuous, we might have to look at integrals, so a general element $a \in KG$ would look something like $\int_G \lambda_x x$. This is well-defined since $G$ is compact. Does something like this exist? If so, are there any interesting results/literature on them?

答:
Sure you can define the group algebra $K[G]$ as usual. Problem is a bit that the word group algebra is quite overloaded, and esp. when talking about topological groups or Lie groups you could also be talking about a lot of other things.

Other objects also called group algebra include:

If $G$ is topological: $C(G)$ (continuous functions on $G$ w/ convolution), $\mathcal{M}(G)$ (regular signed/complex measures on $G$ w/ convolution)

If $G$ is topological and locally compact: $L^1(G)$ (integrable functions w.r.t. some fixed Haar measure w/ convolution), $C_c(G)$ (compactly supported continuous functions), $\mathcal{M}_c(M)$ (compactly supported signed/complex measures)

If $G$ is Lie: $C^\infty(G)$ and $C_c^\infty(G)$ (smooth (compactly supported) functions w/ convolution), $\mathcal{D}(G)$ and $\mathcal{D}_c(G)$ ((compactly supported) distributions w/ convolution)

and so on. Note that most of them don't have a 1 (exceptions: measures and distributions), and that in the case of $G$ finite they all reduce more or less to the group algebra that we know. Given a representation of $G$, you can also build a representation of most of these algebras (but you probably have to worry about finiteness of the integrals, so better stick to the compactly supported ones... Or to compact groups), which is where/how I first heard about those other group algebras: https://encyclopediaofmath.org/wiki/Infinite-dimensional_representation

Title: Group algebra of Lie group


问:
Are there infinitely-many numbers that are both square and triangular?
I just started to read "Friendly Introduction to Number Theory" but I am getting stuck in the first exercises.

1.1. The first two numbers that are both squares and triangles are 1 and 36. Find the next one and, if possible, the one after that. Can
  you figure out an efficient way to find triangular–square numbers? Do
  you think that there are infinitely many?

https://www.math.brown.edu/~jhs/frintch1ch6.pdf

I found how to find out the number which is both square and triangle. (don't know if this is effective way)

https://github.com/y-zono/friendly-introduction-number-theory/blob/master/01/1-1/main.go

However how can I answer "Do you think that there are infinitely many?"? I think I need to find the formula but no idea yet. Can you please help me?

答:
Assume that $$\frac{n(n+1)}{2}=m^2\iff n(n+1)=2m^2$$ Since $\gcd(n,n+1)=1$ we need to factor $m^2$ as $m^2=a^2b^2$ with $\gcd(a,b)=1$. Assume without loss of generality that $a^2\leq b^2$, then there are 3 cases to consider given that now $$n(n+1)=2a^2b^2$$

$n=a^2, n+1=2b^2$. Then $$1=2b^2-a^2\geq b^2$$ so $a=b=\pm 1$.

$n=2a^2, n+1=b^2$. Then $$1=b^2-2a^2$$ One solution to this is $(a,b)=(2,3)$, and we can always generate a new solution by $$(a',b')=(2ab, b^2+2a^2)$$ The motivation for this is to consider the equation $1=b^2-2a^2$ over the ring $\Bbb Z[\sqrt 2]$, with norm $N(a+b\sqrt 2)=a^2-2b^2$. Then $N(\alpha\beta)=N(\alpha)N(\beta)$, in particular if $N(\alpha)=1$ then $N(\alpha^n)=1$.

$n=b^2, n+1=2a^2$. Then $$1=2a^2-b^2$$ multiply by $2$ to make the equation into $$2=(2a)^2-2b^2=(2a+\sqrt 2 b)(2a-\sqrt 2 b)\in\Bbb Z[\sqrt 2]$$ one solution to this equation is $(a,b)=(1,1)$. To generate more solutions we may multiply by any unit $u$, for instance $u=3+2\sqrt 2$. $(2a+\sqrt 2 b)(3+2\sqrt 2)=6a+4b+(4a+3b)\sqrt 2$. $2=(10+7\sqrt 2)(10-7\sqrt 2)$, so $$1=2\cdot 5^2-7^2$$ Therefore $$(a',b')=(3a+2b, 4a+3b)$$ and so there are also infinitely many solutions in this case also.


问:
If in a Triangle ABC b , c , B ( where A ,B ,C denotes angles and a , b ,c denotes sides of the triangle ABC ) are given and b < c . Prove that
If in  a Triangle ABC    b , c ,  B  ( where A ,B ,C  denotes angles  and  a,b,c denotes sides of the triangle ABC ) are given and b < c prove that $$ \ sin \frac  {(A_1-A_2)} {2}\ = \frac { (a_1-a_2)} {(2b)}   $$

MY ATTEMPT :

taking cosine rule $$ cos B = \frac{\ a^2 + c^2 - b^2} {2ac} $$  and  rearranging the terms we get $${a^2 -2ac cosB +(c^2-b^2)} = 0 $$    taking $ a_1 $ and $a_2$ as roots and solving $ a_1 $ + $a_2$ = $ 2c$ $cos B$ and  $a_1$ $a_2$ = $ c^2$ - $b^2$   we get $$  ( a_1 -  a_2  )^2  =  4b^2 -4c^2 cos^2 B $$

I am  struck here please help me to solve this question

答:
As shown in the picture, $BC=a_1, BC'=a_2, \theta = {A_1 - A_2\over 2}$.

Therefore $\sin({A_1 - A_2\over 2}) = {{a_1-a_2\over 2}\over b}$ by definition of sine function in right triangle.


问:
Two identity element?
On $\Bbb N=\{0,1,2,...\}$ we define the operation $\otimes$ by $m\otimes n= |m-n|$. Are there any identity element?

I came up with two identity elements. $e= 0$ and $e=2m$ for an element $m$ in the set, but how is that possible? I thought on a operation $\otimes$ that it would always be one identity element?

答:
It's not possible. If a set with binary operation has an identity element then it must be unique. No matter if associativity, commutativity or other things hold.

If $1, 1'$ are identity elements then $1\cdot 1' = 1 = 1'$ from the properties of identity elements.

Identity element can't depend on other elements. There is however a different but similar notion of local identity element, which depends on an element chosen. We say that $e$ is a local identity element for $x$ if $e\cdot x = x\cdot e = x$. We may denote  $e$ by $e(x)$ to indicate that $e$ depends on $x$.

I saw this one used in some texts about theory of quasigroups, without going into details of what a quasigroup is.

Here, $0$ and $2m$ are all possible local identities for $m$, but only $0$ is the identity element.

Title: Two identity element?