问:
How do I find $P(1< Z <2)$ for given problem?
Suppose  a value z of the continuous random variable Z is generated as follows:
First, a fair die is rolled.If the side of the die facing up has 1 or 2 dots, then z is drawn from a Unif(0,2) distribution;otherwise z is drawn from a Unif(1,4) distribution.
Now for such random variable Z how can you find a P(1<Z<2)??
I am just unable to comprehend what the Z random variable would even look like.</p>

答:
There's a 1/3 chance the die rolls 1 or 2 dots (called "pips," incidentally), and then a 1/2 chance the variable is between $1$ and $2$.

There's a 2/3 chance the die rolls 3, 4, 5 or 6 pips, and then a $1/3$ chance the variable is between $1$ and $2$.

So....


问:
Collecting proofs $SO(3)\cong\Bbb R P^3$
There are a few questions on SE about the well-known homeomorphism $SO(3)\cong \Bbb R P^3$, but I thought it would be fun to collect as many proofs as we can find in one spot.

Background:

$SO(3)$ is the group of orientation-preserving orthogonal transformations of $\Bbb R^3$, or equivalently the group of $3\times 3$ real matrices $A$ with $AA^\top=I$ and $\det A=1$

$\Bbb R P^3$ is the space of lines going through the origin in $\Bbb R^4$ topologized equivalently as

$\left(\Bbb R^4\setminus\{0\}\right)/\sim$, where $x\sim cx$ for all $x\in\Bbb R^4\setminus\{0\}$ and $c\in\Bbb R\setminus\{0\}$ or

$D^3/\sim$, where $p\sim -p$ for all $p\in S^2$

Both spaces are oriented $3$-manifolds with smooth structures

$SO(3)$ is diffeomorphic (and thus homeomorphic) to $\Bbb R P^3$

As varieties, $SO(3)$ is affine while $\Bbb R P^3$ is projective. Thus, they are not isomorphic as varieties.

You can find some proofs of the homeomorphism $SO(3)\cong \Bbb R P^3$ here, here, and in Hatcher $\S3.D$ (currently at the bottom of page $293$).

I will start with Hatcher's proof along with a version I've written using quaternions.

One direction I'm curious about is to think of $SO(3)$ as the Stiefel manifold $V_2(\Bbb R^3)$, and to somehow turn a pair of orthonormal vectors in $\Bbb R^3$ into a line in $\Bbb R^4$.

答:
Here is a completely non-explicit proof of a homeomorphism. We construct a homology equivalence $SO(3) \rightarrow \mathbb{R}P^3$, strengthen this to a homotopy equivalence, strengthen this to a simple homotopy equivalence, and strengthen this to a homeomorphism.

By calculating the cohomology of $SO(3)$ we can conclude there is a map $SO(3) \rightarrow \mathbb{R}P^\infty$ pulling back the generator of $H^1 (\mathbb{R}P^\infty ; \mathbb{Z}/2)$ to the generator of $H^1(SO(3); \mathbb{Z}/2)$. This necessarily factors through $\mathbb{R}P^3$ by cellular approximation. This new map has the property that it pulls back the generator of $H^3 (\mathbb{R}P^3 ; \mathbb{Z})$ to an odd multiple of the generator of $H^3 (SO(3) ; \mathbb{Z})$ since we can see that on $\mathbb{Z}/2$ cohomology, this pulls a generator back to a generator (seen by the ring structure).

$\mathbb{R}P^3$ has a group structure since it is a group quotient of $S^3$, we can now construct a degree 1 map by virtue of the fact that we have a degree 2 map $SO(3) \rightarrow \mathbb{R}P^3$ given by taking the degree 1 map to $S^3$ and postcomposing with the quotient to $\mathbb{R}P^3$. Simply use the group structure to take an appropriate combination of this degree 2 map and the odd degree map from before to get a degree 1 map.

By Poincare duality and the computation of the cohomologies of each space, this is then a homology equivalence.

It is easy to see that this induces an isomorphisms on fundamental groups from how the map is constructed (check it on first $\mathbb{Z}/2$ cohomology and use the fact that first cohomology is dual to the fundamental group). Since both of these spaces are simple (as they are groups), we deduce this homology equivalence is actually a homotopy equivalence.

In fact this homotopy equivalence must be a simple homotopy equivalence, since by calculation the Whitehead group for $\mathbb{Z}/2$ is trivial.

Finally, a simple homotopy equivalence of 3 manifolds is always homotopic to a homeomorphism by work of Turaev: "Towards the topological classification of geometric 3-manifolds".

ContentLicense: CC BY-SA 4.0


问:
Finding representing matrix to curvature tensor in orthonormal basis
Let $M^2 \subset\mathbb{R}$ be a regular surface and $ A \in M$.

Assume the gaussian curvature of M at the point A is K
and $e_1, e_2 \in T_AM$ orthonormal basis for the tanget space.

I want to find the representing matrix for the tensor $R(e_1, e_2)$ with respect to the orthonormal basis $e_1, e_2$.

Note: $R: Vec(M)^3 \to Vec(M)$
, given by, $R(X,Y)Z = \nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$
, where X, Y, Z Are vector fields on M, $\nabla$ is the Levi-Civita connection and $[X,Y]$ is the commutator of X and Y.

My initial idea was to look at the embedding of $M^2$ in $\mathbb{R}^3$, by $i :M \to \mathbb{R}^3$.
We have $g_M=i^*(g_{\mathbb{R}^3)}$, so representing $g_M$ in basis $e_1, e_2$ will result in $g_M=I$.
From here, I can easily see that all of the Christoffel ($\Gamma^k_{ij}$) symbols are vanishing at the point A.

If we look at Ricci's curvature $\mathscr{R}_{ij}=R^k_{kij}=\frac{\partial\Gamma_{ij}^k} {\partial q^k} - \frac{\partial\Gamma_{kj}^k} {\partial q^i}$ (I use the notation $\partial q^i=e_i$) and using the fact that $g^{ij }\mathscr{R}_{ij}=\mathscr{K}=2K$ (This is how we formulated theorema egregium in our class)
I ended up with the following equation :

$2K=\frac{\partial\Gamma_{22}^1} {\partial q^1}+\frac{\partial\Gamma_{11}^2} {\partial q^2}-\frac{\partial\Gamma_{21}^2} {\partial q^1}-\frac{\partial\Gamma_{12}^1} {\partial q^2}$

Now
$R(e_1,e_2)_{ij} = \begin{bmatrix}R_{121}^1&R_{121}^2\\R_{122}^1&R_{122}^2\end{bmatrix}$
And generally because the Christoffel coefficients vanishes :
$R_{12k}^l= \frac{\partial\Gamma_{2k}^l} {\partial q^1}-\frac{\partial\Gamma_{1k}^l} {\partial q^2}$.

I hoped that getting to this stage things would start cancelling out, but they don't.

I would appriciate any help completing this question!

答:
After few days of thinking I will expand Ted's comment a little in case anyone else would need it:

I will use the notation $R(X,Y,Z,W) = <R(X,Y,Z),W>$.

We have $R(X,Y,Y,X) = \frac {B(X,X)B(Y,Y)-B(X,Y)}{|X|^2|Y|^2-<X,Y>^2}$ and for the orthonormal basis $R(e_1,e_2,e_2,e_1)=<R(e_1,e_2)e_2,e_1>=K$, symmetric properties of R we get $<R(e_1,e_2)e_1,e_2>=<-R(e_2,e_1)e_1,e_2>=-K$, and $<R(e_1,e_2)e_2,e_2>=<R(e_1,e_2)e_1,e_1>=<R(e_1,e_1)e_1,e_2>=0$ and therefore

$R=\begin{bmatrix}0&K\\-K&0\end{bmatrix}$


问:
Limit of an interception point as one parameter changes.
Consider the functions
\begin{align*}
h_1(z)&=m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)\\
h_2(z)&=\frac{-M}{(a^2+z^2)^{3/2}}
\end{align*} where $M$, $m$ and $e$ are real parameters and $e<0$. I know that these two only have one interception point, $z_0$ (I checked it graphically). I also know that the limit of $z_0$ as $m$ approaches infinity is equals to $2e$ (It was also checked graphically).
BUT, if possible, I'd like to show it analytically. i.e, show that
$$\lim_{m \rightarrow \infty} z_0 = 2e$$
Is it possible to show it by a "non numerical" approach? Any tips/help will be very appreciated.

答:
It should actually be $-2e$, which assuming that $e<0$ will be positive.  We can show this by considering solutions to $h_1(z) = h_2(z)$, since a point $z = z_0$ which satisfies this equation is an intercept.  So, we have
\begin{align*}
h_1(z) &= h_{2}(z)\\[5pt]
\implies m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)&=\frac{-M}{(a^2+z^2)^{3/2}}\\[5pt]
\implies \frac{1}{z^3} + \frac{2e}{z^4} &= \frac{-M/m}{(a^2+z^2)^{3/2}}.
\end{align*}

Now, taking the limit of each side as $m\to \infty$:
$$\lim_{m\to\infty}\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg) =\lim_{m\to\infty}\left(\frac{-M/m}{(a^2+z^2)^{3/2}}\right) = 0.
$$

So, as $m\to\infty$, we have that $z_0$ solves
\begin{align*}
\frac{1}{z_0^3} + \frac{2e}{z_0^4} &= 0\\
\implies\frac{1}{z_0^3} &= -\frac{2e}{z_0^4}\\
\implies z_0 &= -2e.
\end{align*}


问:
Recommendation for Low-dimensional topology textbook
Can anyone here recommend a low-dimensional topology textbook that contains knot theory and 3,4-manifolds?Or should I look for these subjects in separate textbooks?</p>

答:
These are diverse subjects studied with different tools. You will need a strong background in algebraic topology, as well as manifold topology (PL or differentiable, most likely the latter), to do well with the material.

Here are some very good introductory texts in each.

Knot theory: Rolfsen's "Knots and links". Background: in addition to a good full course in algebraic topology, know differentiable manifolds through the isotopy extension theorem.

3-manifold topology: Hempel's book is the classic. Hatcher's short set of notes is a good substitute, though it doesn't cover as much. At some point you should read Peter Scott's paper on geometries of 3-manifolds.

The theory of 4-manifolds is too diverse to be well-discussed in one book. One should read Gompf and Stipsicz, "4-manifolds and Kirby Calculus". After that you can try to feel around for your particular taste. Some people like the book by Scorpan but I haven't read it.


问:
Definite integral $\int_0^{\pi} \sin^{n} x \ln(\sin x) dx$?
Is the integral
$$I(n) = \int_0^{\pi} \sin^{n}(x) \ln(\sin(x)) dx$$
analytically tractable for $n \in \mathbb{N}$? If not, are there good upper and lower bounds?

To clarify, although this has a fairly standard-looking form, I haven't been able to find an answer anywhere and I've tried a few of the usual tricks (although I'm a tad rusty).

I have tried integration by parts to reduce to a simpler or recursive form but this seems to make the expression more complex, and likewise with the substitution $u=\sin(x)$.

Equally, it is relatively easy to see that the expression under the integral is non-positive. Taking a derivative
$$ \frac{d}{dx} \sin^{n}(x) \ln(\sin(x)) = \sin^{n-1}(x) \cos(x) (1 + n \ln(\sin(x))) = 0$$
gives a has minimum at $-\frac{1}{en}$.

Together these give the reasonably simple bounds: $$-\frac{\pi}{en} < I(n) < 0.$$

But numerically these don't appear to be particularly tight, so I'm still hopeful of an exact solution or an improvement.

答:
Mathematica says:

$$\fbox{$\frac{\sqrt{\pi } \left(H_{\frac{n-1}{2}}-H_{\frac{n}{2}}\right) \Gamma
   \left(\frac{n+1}{2}\right)}{n \Gamma \left(\frac{n}{2}\right)}\text{ if }\Re(n)>-1$}.$$


问:
Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$
So i was given this question. Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$

Here is my solution:

Given $2n$ objects, split them into $2$ groups of $n$, $A$ and $B$. $2$-combinations can either be assembled both from $A$, both from $B$ or one from each. There are ${n \choose 2}$ from $A$ and ${n \choose 2}$ from $B$. For the mixed pair, each choice from $A$ can be coupled with $n$ choices from $B$ so the total is $n^2$.

Therefore ${2n\choose 2} = 2{n \choose 2} + n^2$

Is this correct. Is there any other combinatoric proof to solve this?

答: There is a generalization of this fact that I cannot resist sharing. Here is how I came across it: Consider the scenario where you keep additively splitting a positive integer into two positive integers as a tree diagram, until you are left with only $1$'s as the leaves. Then you take the product of the two numbers at each branching and sum all such products. Remarkably, applying this process to $n$ will always yield $\binom{n}{2}$ as the final number. Try it with $8$ and you get $28$ every time. The inductive step of the strong induction argument for this fact uses the fact that, if you split $n=a+b$ then $$\binom{a}{2}+\binom{b}{2}+ab=\binom{a+b}{2}=\binom{n}{2}.$$ A combinatorial proof double counts the number of ways of choosing two people out of $n=a+b$ people. We can either choose $2$ out of the $a$ or $2$ out of the $b$ or $1$ from each. The original post asks for a combinatorial proof of this fact when $a=b.$

ContentLicense: CC BY-SA 3.0


问:
Trouble understanding double integral
I have the following double integral that intakes two functions:

$$F_{W}\left ( w \right ) = \int_{0}^{w}\int_{0}^{\sqrt{w^{2}-h^{2}}}f_{Z}\left ( z \right )f_{X}\left ( x \right ) \,dz\,dx \ \ \mathrm{for } \ 0\leq w< H$$

Where the functions $f_{X}\left ( x \right )$ and $f_{Z}\left ( z \right )$ are given by:

$$f_{X}\left ( x \right )= \frac{1}{H}$$

$$f_{Z}\left ( z \right )= \frac{2z}{R^2}$$

Based on this paper by Sharma et al (Lemma 1 in Appendix), the resulting integral is:

$$F_{W}\left ( w \right ) = \frac{2}{3}\frac{w^3}{R^2H}  \ \ \mathrm{for } \ 0\leq w< H$$

I don't get this answer, and I have tried looking up at the "
Table of Integrals, Series, and Products" but I do not get how it was done. Any help or hints on this blackbox would be highly appreciated, cause I've been trying for days to understand this part of the paper but it's impossible.

答:
As one of the commentators has already mentioned, it seems to be that $h = x$, hence the equation would be written as:

$$F_{W}\left ( w \right ) = \int_{0}^{w}\int_{0}^{\sqrt{w^{2}-x^{2}}}f_{Z}\left ( z \right )f_{X}\left ( x \right ) \,dz\,dx \ \ \mathrm{for } \ 0\leq w< H$$

I have checked the equation with the rest of the integrals that the paper presents and it is a valid result.

Title: Trouble understanding double integral


问:
Induced map on $K_{1}$-group
Let $A$ be a unital $C^{\ast}$-algebra. Any automorphism $\alpha$ of $A$ induces a map on $K_{1}(A)$ by $\alpha_{\ast}[u]=[\alpha(u)]$.

Let the automorphism $\alpha$ be inner, does it follow that $\alpha_{\ast}$ is the identity map?

I guess the answer is negative because unitary equivalence is far from homotopy of unitaries. The answer may also depend on the type of the $C^{\ast}$-algebra. Any suggestions?

答:
If I may add a little to the above answer, $K_1(A)$ is by definition the group of homotopy classes of unitaries in the unitization  $(K \otimes A)^{\sim}$ of its stabilization $K \otimes A$ (replace $A$ by $A^{\sim}$ in the non-unital case as julien stated above.)

Polar decomposition shows that every unitary in $(K \otimes A)^{\sim}$ is homotopic to one of the form $u \oplus  (1 - 1_{M_n(A)} ) $ where $u \in M_n(A)$.

Also, for any unitary $u$ in a corner $p (K \otimes A)^{\sim} p$ and for any partial isometry $s$ with $s^*s = p$, $u \oplus (1-p)$ is homotopic to $sus^* \oplus (1-ss^*)$.

So every $K_1$ class can be represented by some unitary $u \in M_n(A)$ after adding $1$'s down the diagonal and you can move a unitary $u \in M_n(A)$ up and down the diagonal without changing its $K_1$-class. It is immediate that $K_1$ is abelian and inner-automorphisms descend to the identity map on $K_1$.

Title: Induced map on $K_{1}$-group


问:
normal distribution in 2D polar coordinates - why is the expected value not zero?
There is something I don't get. Say I have a probability density function:
$$f\left(x,y\right)=\frac{1}{2\pi\sigma^{2}}e^{-\frac{x^{2}}{2\sigma^{2}}}e^{-\frac{y^{2}}{2\sigma^{2}}}$$

Now, in polar coordinates:

$$f(r) = \frac{1}{2\pi\sigma^{2}}e^{-\frac{r^{2}}{2\sigma^{2}}}$$

But if I calculate the expected value of this probability density function, I don't get 0. Why is it? intuitively, that is the peak of the gaussian so that's what is most probable. It is this way with a one dimensional gaussian centered at the origin, but why isn't it the case here?

答:
What you have computed as $f(r)$ is incorrect, not only just for the reason pointed out in the comment by NN2, but because what you would have found if you had done the transformation correctly is that if the joint density of two variables $X$ and $Y$ is
$$f_{X,Y}(x,y) = \frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right),\tag{1}$$ then the joint density of $R =\sqrt{X^2+Y^2}$ and $\Theta = {\mathsf{atan2}}(X,Y)$ is
$$f_{R,\Theta}(r,\theta)= \frac{1}{2\pi}\cdot r\exp\left(-\frac{r^2}2\right), \quad 0 \leq r < \infty, 0\leq \theta < 2\pi,\tag{2}$$
where $$f_R(r) = r\exp\left(-\frac{r^2}2\right),\quad 0 \leq r < \infty \tag{3}$$ is a Rayleigh density, and $$f_\Theta(\theta) = \frac{1}{2\pi},\quad 0\leq \theta < 2\pi,\tag{4}$$ is a uniform density (and also showing that $R$ and $\Theta$ are independent random variables_).

It is not clear what you understand as the expected value of $f_{X,Y}(x,y)$ but if you mean $(E[X],E[Y])$ being $(0,0)$, then you need to understand that neither $E[R]$ nor $E[\Theta]$ equal $0$. Note that $f_R(r) = 0$ for $r\leq 0$ which means that all the probability mass lies to the right of the origin, implying that $E[R]>0$. Nor is $0$ the mode of the Rayleigh density.


问:
Partial derivative of differential equation

Question

I have the following ordinary differential equation (ODE)

$\frac{du(s)}{ds}=f(s) + u(s)$

which can be written in my case as

$\frac{d}{ds}u[x(s),y(s)]=f[x(s),y(s)] + u[x(s),y(s)]$.

How do I find $\frac{\partial u}{\partial y}$?

What I tried

I take the partial derivative on both sides and I exchange the order of derivatives

$\frac{d}{ds}\left\{\frac{\partial}{\partial y}u[x(s),y(s)]\right\}=\frac{\partial}{\partial y}\left\{f[x(s),y(s)] + u[x(s),y(s)]\right\}$

now, I define

$g(s)=\frac{\partial u}{\partial y}$

which implies I can find $\frac{\partial u}{\partial y}$ by solving the following ODE for $g(s)$

$\frac{dg(s)}{ds}=\frac{\partial}{\partial y}f[x(s),y(s)] + g(s)$

Could someone confirm whether this is correct?

Counter-Example

Here is an example where my above statement is wrong. I define

$x(s)=\cos(s)\\
y(s)=s^2\\
u(x,y)=2x+y^2\\
f(x,y)=4y^{3/2}-y^2-2x-\sqrt{1-x^2}$

one can verify with some algebra that it satisfies (by design)

$\frac{du(s)}{ds}=f(s) + u(s)$

in fact, both sides are equal

$\frac{du(s)}{ds}=4s^3-2\sin(s)=f(s) + u(s)$

However

$\frac{d}{ds}\left\{\frac{\partial}{\partial y}u[x(s),y(s)]\right\}=4s$

while

$\frac{\partial}{\partial y}\left\{f[x(s),y(s)] + u[x(s),y(s)]\right\}=6s$

So my original question remains: given an ODE for $u(s)$ what ODE defines $\frac{\partial u}{\partial y}$?

Solution

Based on feedback from @lord-commander, I realized that my proposed approach was completely wrong. In fact, the unknown partial derivative $\partial u/\partial y$ that I am looking for is related to the given functions by the following partial differential equation, not by an ODE.

$\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=f(s)+u(s)$

It seems to me then that my only option is to solve the original linear ODE

$\frac{du(s)}{ds}=f(s) + u(s)$

which has a general solution

$u(s) = c_1 e^s + e^s \int_1^s f(s') e^{-s'} ds'$

Then I write this as a function of $(x,y)$ and take the partial derivative.

This is the approach I was trying to avoid. In fact, my expression is complicated and the integral has to be done numerically, for many values of $s$ with $f(s)$ a computationally expensive function.

答:
I have the following differential equation

$$\frac{du(t)}{dt}=f(t) + u(t)$$

I decided to use $t$ because my eyes just didn't work with $s$, but feel free to read the last equation and change the $t$-s back to $s$.

which can be written in my case as

$$\frac{d}{dt}u(x,y)=f(x,y)+ u(x,y)$$

Both $x(t)$ and $y(t)$ depend parametrically on $t$.

$$u(x(t),y(t))=u(x,y)$$

By the chain rule:

$$\frac{du}{dt}=\frac{d}{dt}u(x,y)=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}$$

$d$ is the total derivative, $\partial$ is a partial derivative.

$$\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=f(x,y)+u(x,y)$$

$$\frac{d}{dy}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\right)=\frac{d}{dy}\left(f(x,y)+u(x,y)\right)$$

$$\frac{d\;\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}}{d y}+\frac{d\;\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}}{dy}=\frac{d}{dy}\left(f(x,y)+u(x,y)\right)$$

$$\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}\right)\cdot \frac{\partial t}{\partial y}+\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\right)\cdot\frac{\partial{t}}{\partial y}=\frac{d}{dy}\left(f(x,y)+u(x,y)\right)$$

$$\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}\right)\cdot \frac{\partial t}{\partial y}+\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\right)\cdot\frac{\partial{t}}{\partial y}=\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial y}+\frac{\partial u}{\partial y}\underbrace{\frac{\partial y}{\partial y}}_{1}\right)$$

To be completely frank i am not really sure if you can multiply out by $\frac{\partial }{\partial t}$ as the functions they are differentiating are different. I could be wrong though.


问:
Showing $\sum_{k=0}^{\infty} \sum_{i+j=k}a_i b_j = \sum_{k=0}^{\infty} a_k \sum_{k=0}^{\infty} b_k$
I want to show $\sum_{k=0}^{\infty} \sum_{i+j=k}a_i b_j = \sum_{k=0}^{\infty} a_k \sum_{k=0}^{\infty} b_k$ rigoursly.

How I can prove rigorously? (Of course suppose $\sum_{k=0}^{\infty}a_k < \infty$, $\sum_{k=0}^{\infty}b_k < \infty$ condition is required)

答:
Let $P(x) = \sum_{i=0}^\infty a_i x^i,$ and let $Q(x) = \sum_{j=0}^\infty b_j x^j.$
Then, your limit just says that

$P(1) Q(1) = (PQ)(1).$ Notice that for this to make sense, both series have to have radius of convergence $1$ (and as pointed out in the other answer, one of them has to converge absolutely at $1.$).


问:
GRE number theory question
What is the greatest integer that divides $p^4-1$ for every prime $p>5$?

So I do not even understand the solution they omitted lots of detail.

So I know that

$$p^4-1=(p-1)(p+1)(p^2+1).$$

The solution says the first two factors are divisible by $2$ and $4$, one by $2$ and the other by $4$. I do not see why, and they say the last one is divisible by $2$ as well. And so they get that $2 \cdot 4 \cdot 2 = 16$ divides $p^4-1$. Is it because $p>5$ and so subtracting $1$ we get something divisible by 4?

And furthermore, why is it that by "casework" (is what they call it) we have

$$3 \mid (p^2-1)$$

and

$$5 \mid (p^2-1).$$

Putting these together clearly I see why the answer is $3 \cdot 5 \cdot 16.$ But I do not understand the steps that got me here. At first glance I thought to use wislons theorem but that does not work here does it?
Thanks in advance, some of these GRE problems are really truly something aren't they.

答:
If $p$ is prime and $p > 5$ then it must be odd, so $p \equiv 1$ or $3 \pmod{4}$. Hence, one of $p+1$ and $p-1$ is divisible by $4$ and the other one only by $2$.

On the other hand, for any $p \neq 3$ we have $p \equiv 1,2 \pmod{3}$. It is an easy check that these are both solution to $x^2 -1 \equiv 0 \pmod{3}$.

You can do a similar thing for $5 | p^4 -1$ (it is not true $5 | p^2 -1$, try $p = 17$). A quicker way could be seeing that the group of units of $\mathbb{Z}/5 \mathbb{Z}$ is cyclic with order $4$, so the polynomial $x^4 -1$ makes the entire group vanish.

Title: GRE number theory question


答:
My book preferences are different from the other answer! My preferences are the following:

Knot theory: Lickorish "An introduction to knot theory". You don't need much background apart from singular homology/fundamental groups.

3-manifolds: Saveliev "Lectures on the topology of 3-manifolds". Some of the earlier content overlaps with Lickorish, but it's also good! Again if you know some stuff about singular homology you'll be able to read this book. It's quite selective about content though - if you're interested in things like contact topology, geometrisation etc, you'll need other sources.

4-manifolds: Scorpan "Wild world of 4-manifolds". The previous answer also mentioned this book - I like it a lot! The earlier chapters develop theory within the algebraic topology framework, but eventually we need more tools: later on Seiberg-Witten gauge theory is introduced and a bunch of wild 4-dimensional results are proven (e.g. uncountably many smooth structures on R^4)

Title: Recommendation for Low-dimensional topology textbook

Body: Can anyone here recommend a low-dimensional topology textbook that contains knot theory and 3,4-manifolds?Or should I look for these subjects in separate textbooks?</p>


问:
If $f \in L^1_{loc}(\mathbb R^n)$ and $\varphi \in C^\infty_0(\mathbb R^n)$ then $ \varphi\star f \in C^\infty(\mathbb R^n)$.
Let $f \in L^1_{loc}(\mathbb R^n)$ and $\varphi \in C^\infty_0(\mathbb R^n)$. I know that it is a very common result that $\varphi\star f$ is in $C^\infty(\mathbb R^n)$, but I am not able to prove it and I haven't found any answers on Stackexchange that proves the statement the way I want to. Perhaps one of you can redirect me to another question.

The function $x \mapsto \varphi(x - y) f(y) \in C^\infty (\mathbb R^n)$ almost everywhere. I want to use the Leibniz rule to prove that, for $\alpha \in \mathbb N^k$,
$$\partial^\alpha(\varphi \star f) = (\partial^\alpha\varphi) \star f.$$
To this end, I am looking for an inequality that looks like the following
$$|\partial_x^\alpha(\varphi(x - y)f(y))| = |(\partial^\alpha \varphi)(x - y)f(y)| \le \|\partial^\alpha\varphi\|_\infty |f(y)|.$$
Even if the last inequality is true, it does not help to prove the statement since $f$ is not integrable in the all $\mathbb R^n$. Is it the right way to do that ?

答:
As $\varphi$ is compactly supported locally we can assume that $f\in L^1$

With $D_v$ the directional derivative in the direction $v\in \Bbb{R}^n$ then $$D_v (f\ast \varphi)=f\ast D_v\varphi$$ follows from

$$\lim_{h\to 0} \|f\ast T_{v,h}\|_\infty \le \lim_{h\to 0}\|f\|_{L^1} \|T_{v,h}\varphi\|_\infty=0$$
where $$T_{v,h}\varphi=\frac{\varphi(x+hv)-\varphi(x)}{h}-D_v\varphi(x)=\int_0^1 (D_v \varphi(x+thv)-D_v \varphi(x))dt$$
As $D_v\varphi$ is $C^\infty_c$ you can repeat to get $D_{v_1}\ldots D_{v_k}(f\ast \varphi)=f\ast D_{v_1}\ldots D_{v_k}\varphi$.


问:
How to prove $\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \frac{\pi}{2}\ln(2)$?
I was messing around with some integrals and series when I arrived at the result (and WolframAlpha agrees):

$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$

I therefore have a proof, but I would like to see how a proof would be done the other way around (i.e. starting with this series). Notice that the series could also be expressed with double factorials:

$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^2}$$

My Derivation

My derivation is quite simple. I was considering (and trying to solve) the integral

$$\int_0^1 \frac{\arcsin(x)}{x} \, dx$$

I couldn't work it out, however I found the exact same question here on math exchange and the answers show that the integral equals $\frac{\pi}{2}\ln(2)$. What I did found however, was that I could plug the Taylor series for $\arcsin(x)$ (the proof of the taylor series can be found here) into the original series to only be left with a series. Here's the Taylor series for $\arcsin(x)$:

$$
  \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}
$$

So plugging in the series above into the integral gives us:

$$\int_0^1 \frac{1}{x} \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} \, dx$$

Now here we have to be careful, but since the series works on the integral's interval $[0,1]$ and the summand is only positive, we can interchange the the integral and the sum, leaving us with:

$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1} \, \int_0^1 \frac{x^{2n+1}}{x} \, dx$$

And then solving the integral in the sum gives us the final result:

$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$

And then expanding the binomial gives the original series in the title.

答:
Consider the well-known series $$f(x)=\sum_{n=0}^{\infty}{2n\choose n}x^n=\frac{1}{\sqrt{1-4x}}$$

We will attempt to evaluate the series

$$I(x)=\sum_{n=0}^{\infty}{2n\choose n}\frac{x^n}{(2n+1)^2}$$

A simple relation between these two series exists:

$$\frac{d^2}{dt^2}[e^tI(e^{2t})]=e^t f(e^{2t})=\frac{e^t}{\sqrt{1-4e^{2t}}}$$

To obtain our goal series we just need to integrate twice with an appropriate boundary condition

$$e^t I(e^{2t})=\int_{-\infty}^tdx\int_{-\infty}^{x}dT\frac{e^T}{\sqrt{1-4e^{2T}}}$$

The inner integral is easy to do which leads us to the expression

$$I(x)=\frac{1}{2\sqrt{x}}\int_{-\infty}^{\frac{\ln x}{2}}dt ~ \arcsin(2e^t)$$

which can be further simplified to

$$I(x)=\frac{1}{2\sqrt{x}}\int_{0}^{2\sqrt{x}}\frac{\arcsin u}{u}du$$

Our goal is to evaluate this at $x=1/4$. Performing an integration by parts and afterwards the substitution $u=\cos t$ we reduce the entire thing to a well-known integral that can be calculated in various ways

$$I(1/4)=-\int_0^{\pi/2}\ln \cos t dt=\frac{\pi\ln 2}{2}$$


问:
Proof that one of three terms with 3 variables is always smaller or equal to 1
Let $x, y$ and $z$ positive real numbers with $x + y + z = 3$. Prove that at least one of the three numbers

$x (x + y − z), y (y + z − x)$ or $z (z + x − y)$

is less than or equal to $1$.

Can someone give me a hint where to start with this?

答: The three quantities are $x(3-2 z), y(3-2 x), z(3-2y).$ One of $x, y, z$ is not greater than $1.$ Suppose, without log of generality, that it is $x.$ Then, if both $y, z$ are $\geq 1,$ then the first sum is $\leq 1.$ So, let us suppose that $x, y \leq 1,$ and so $z\geq 1,$ then the first quantity is less than $1.$ The alternative is that $x, z \leq 1, y\geq 1,$ then the the third quantity is $\leq 1.$


问:
Function not integrable but absolute function integrable?
I know that the function

$$f(x) = \begin{cases}
1&\text{if } x \in \mathbb{Q}\\
0&\text{if } x \notin \mathbb{Q}
\end{cases}$$

is not integrable. Then how would I come up with a new function $g(x)$ such that $g(x)$ is not integrable but $|g(x)|$ is integrable? First, if a function is absolutely integrable, then I think the function should be integrable.

答:
What about:

$$f(x) = \begin{cases}
1&\text{if } x \in \mathbb{Q}\\
-1&\text{if } x \notin \mathbb{Q}
\end{cases}$$


问:
$\int_{0}^{\infty} \lfloor ne^{-x} \rfloor \mathrm{d}x$ where $n\in \mathbb{N} $ and $x\in \mathbb{R} ^{+} $
Find

$\int_{0}^{\infty} \lfloor ne^{-x} \rfloor \mathrm{d}x$  where $n\in \mathbb{N} $ and $x\in \mathbb{R} ^{+} $

I started thinking like this -

$f(x) =ne^{-x} $. So, $f'(x) =-ne^{-x} < 0$

So $f$  is decreasing in $0$ to $\infty$. Now let us break the interval for $x>\ln n$ then $e^x>n$ so $ne^{-x} <1$.So,we have $\lfloor ne^{-x} \rfloor =0$ for $x>\ln n$

So the integral becomes

$\int_{0}^{\infty}\lfloor ne^{-x}\rfloor \mathrm{d}x =\int_{0}^{\ln n} \lfloor ne^{-x} \rfloor \mathrm{d}x +\int_{\ln n} ^{\infty} \lfloor ne^{-x} \rfloor \mathrm{d}x$ .

The second part is zero. But how to tackle $\int_{0}^{\ln n} \lfloor ne^{-x} \rfloor \mathrm{d}x$

Can I carry on from here? If there is any independent way,that is also welcome. Thanks in advance.

答:
You have already found
$$ I_n \equiv \int \limits_0^\infty \lfloor n \mathrm{e}^{-x} \rfloor \, \mathrm{d} x = \int \limits_0^{\ln(n)} \lfloor n \mathrm{e}^{-x} \rfloor \, \mathrm{d} x $$
for $n \in \mathbb{N}$. The integrand decreases in steps and changes its value whenever $n \mathrm{e}^{-x}$ is an integer. More precisely, for $m \in \{1, \dots, n-1\}$ we have $\lfloor n \mathrm{e}^{-x}\rfloor = m \, \Leftrightarrow \ln \left(\frac{n}{m+1}\right) < x \leq \ln \left(\frac{n}{m}\right)$. Following blamethelag's suggestion we write
$$ [0, \ln(n)] = \{0\} \cup \bigcup_{m=1}^{n-1} \left(\ln\left(\frac{n}{m+1}\right), \ln \left(\frac{n}{m}\right)\right] \, .$$
The integrand is equal to $m$ on the $m$-th half-open interval. Its value $n$ on $\{0\}$ does not matter for the integral, since it is only assumed at a single point. Therefore, we obtain
\begin{align}
I_n &= \sum \limits_{m=1}^{n-1} \int \limits_{\ln\left(\frac{n}{m+1}\right)}^{\ln\left(\frac{n}{m}\right)} \lfloor n \mathrm{e}^{-x} \rfloor \, \mathrm{d} x = \sum \limits_{m=1}^{n-1} \int \limits_{\ln\left(\frac{n}{m+1}\right)}^{\ln\left(\frac{n}{m}\right)} m \, \mathrm{d} x = \sum \limits_{m=1}^{n-1} m \left[\ln\left(\frac{n}{m}\right) - \ln\left(\frac{n}{m+1}\right)\right] \\
&= \sum \limits_{m=1}^{n-1} m \left[\ln(m+1) - \ln(m)\right] = \sum \limits_{m=1}^{n-1} \left[(m+1)\ln(m+1) -  m  \ln(m) - \ln(m+1)\right] \\
&= \ln \left(\frac{n^n}{n!}\right)
\end{align}
for $n \in \mathbb{N}$ (exploiting that the first part of the final sum is telescoping). By the way, we can now use Stirling's formula to compute the limit
$$ \lim_{n \to \infty} I_n - n + \frac{1}{2} \ln(n) = - \frac{1}{2} \ln(2 \pi) \, .$$


问:
Estimate error using Taylor Series
I was told to use Taylor Series to find the error of this but I am not confident in doing this.

Estimate the error in approximating $\cos x$ by $1-(x^2/2)$ on the interval $[-0.5, 0.5]$.

Title: Estimate error using Taylor Series

Body:
A related problem. Hint:

$$
    |\sum_{k=0}^\infty(-1)^k\,a_k\,-\,\sum_{k=0}^m\,(-1)^k\,a_k|\le |a_{m+1}|. $$


问:
Question on positive semidefiniteness
Suppose we have symmetric matrices $A$ and $B$, where

$$A \succ 0, \qquad B \succeq 0, \qquad A - B \succeq 0$$

Here, $A \succeq B$ means that $x^\top Ax \geq x^\top Bx$ for all $x$.

Do we have $A^2\succeq B^2$?

If $S$ is positive definite, do we have $A S A \succeq B S B$?

It seems right, but I do not know how to prove or disprove it.

答:
Question 1

No, for a counterexample consider
$$
A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}
$$
$$
B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}
$$

Then $A-B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \geq 0$ but $A^2 - B^2 = \begin{pmatrix} 3 & 1 \\ 1 & 0\end{pmatrix}$ is not PSD.

Question 2

No, take $S$ to be the identity matrix then $S$ is positive definite but by part $1$ the inequality cannot hold.

Title: Question on positive semidefiniteness


问:
If all 2-faces of a polytope are $2n$-gons, is the edge-graph bipartite?
I wonder if I can see easily whether the edge graph of a (convex) polytope $P\subset\Bbb R^d$ is bipartite or not.

A graph is bipartite if and only if all its cycles have even length.
I thought about the following: maybe, a polytope is bipartite if and only of all its 2-dimensional faces are $2n$-gons.
But then, the 2-faces are not all the cycles of the edge graph. So this might not be true.

Question: If all 2-faces of a polytope are $2n$-gons, is the edge graph of $P$ bipartite?

It is certainly true for $d=3$, as the cycle space of the edge graph of $P$ (a planar graph) is generated by the facial cycles.

The argument must make use of convexity or the spherical topology of $P$, as one can easily find polytopal complexs for which this statement is false (e.g. see the image below, which is taken from here).

答:
H.S.M. Coxeter proves this in section 8.6 of his book Regular Polytopes.
In section 1.6, he shows "that alternate vertices of any even-faced simply-connected map can be picked out in a consistent manner (so that every edge joins two vertices of opposite types)."
He proves this by observing that any circuit of $N$ edges decomposes the map into two regions. Replacing one region by a single $N$-sided face, all faces but this one have evenly many sides, so $N$ must be even (since the number of odd faces must be even.)

Then in 8.6, he states "In fact, the argument used at the end of §1.6 may be extended to any number of dimensions, with the conclusion that alternate vertices of a simply-connected polytope or honeycomb can be selected whenever every face $\Pi_2$ has an even number of sides."

I will just quote the section:

...a circuit of edges can be shrunk to a point without leaving the manifold (i.e., without leaving the "surface" of the polytope). In other words, such a circuit is (generally in many different ways) the boundary of a two-dimensional region or "2-chain," consisting of a number of $\Pi_2$'s fitting together in such a way as to be topologically equivalent to a circle with its interior, or to the curved surface of an ordinary hemisphere.
If the circuit consists of $N$ edges, we can transform the 2-chain into a complete topological polyhedron (or two-dimensional map) by adding one $N$-sided face. Hence, by the remark at the end of §1.4, if every $\Pi_2$ has an even number of sides, the number $N$ must be even too. Finally, we select those vertices of the polytope or honeycomb which can be reached from a given vertex by proceeding along an even number of edges.


问:
The minimum of $f(x)=1+x+\cdots+x^{2n}$
For the function $$
f(x)=\sum_{k=0}^{2n}{x^k}
$$
I think this function has no zeros, and if ${f(x_0)}$ is the minimum point, then ${x_0\in \left[-1,0 \right] }$.

For $1+x+x^2+x^3+x^4$ and $1+x+x^2$, I can get the minimum by calculating $f'(x)$ and then using the root fomula. But this method doesn't work for $n\geqslant 3$ since there is no root fomula for equations higher than the fifth degree.

I have no idea for the situation where  $n\geqslant 3$, so any advice could be helpful

答:
@ClaudeLeibovici and I had a lot of fun with this one. The link to the arXiv article below provides a derivation of the minimum of this polynomial for any $n\in\Bbb N$.

Exact and approximate solutions to the minimum of $1+x+\cdots+x^{2n}$

Summary of paper:

To find the minimum of this polynomial, denoted $f_{2n}(x)$, we first derived the relationship
$$
\inf_x f_{2n}(x)=\frac{1+2n}{1+2n(1-x_{2n})},
$$
with $x_{2n}$ being the argument of the minimum. It was then shown that $x_{2n}$ satisfies
$$
x_{2n}^{2n}\left(1-x_{2n}+\tfrac{1}{2n}\right)-\tfrac{1}{2n}=0,
$$
and an exact solution of this equation was subsequently found via Lagrange inversion. For the purposes of numerical computation we also derived a faster converging perturbation series for $x_{2n}$.

Title: The minimum of $f(x)=1+x+\cdots+x^{2n}$


问:
Minimize the standard deviation in list of numbers by subtracting from each value
Suppose we have a list of values, $L = (x_1, x_2, ..., x_n)$, and another list of numbers, $Q = (q_1, q_2, ..., q_n), q_i$ being the maximum value we can subtract from $x_i$ (so from $x_i$ we can subtract any number in $[0; q_i]$. I want to find a combination of subtractions such that the standard deviation of the resulting list will be minimal. (In other words, I want to find a set of values, $R = (r_1, r_2, ..., r_n)$, $r_i \in [0; q_i]$ such that the standard deviation of $L' = (x_1-r_1, x_2-r_2, ..., x_n-r_n)$ is minimal.)

A graphical example (the blue points are the plotted $x_i$):

Any help would be appreciated. Thanks.

答:
Here's an algorithmic way to do it.

Set up $3n$ registers labelled $ij\;$ ($i=1,...,n;\;j=1,2,3$). Initially $x_i$ is placed in registers $i1$ and $i2$ while $q_i$ is placed in register $i3\,$ ($i=1,...,n$); the content of register $i1$ will be updated at each stage, and called $x_i'$, while the respective contents $x_i$ and $q_i$ of registers $i2$ and $i3$ will be kept as a constant reference. A further register holds the mean $m'=(x_1'+\cdots+x_n')/n$ of the current numbers $x_i'$.

Reduce each (original) $x_i$ by whatever allowed amount will minimize the distance of the reduced number from $m'$, and update the content of register $i1$ with this reduced number. Thus, numbers $x_i'$ less than $m'$ will remain unchanged, while a number $x_i'$ that exceeds $m'$ will be reduced either to $m'$ (if $x_i-q_i\leqslant m'$) or to $x_i-q_i$ (if $x_i-q_i>m'$). The registers $i1\;$($i=1,...,n$) are then refreshed with the corresponding newly reduced numbers $x_i'$, and the new mean of these numbers replaces the previous mean in the allocated register.

Repeat the above operation until the process converges.

To show the convergence and (at least local) optimality of the above algorithm, define $x_{ij}$ and $m_j$ recursively by
$$x_{i1}:=x_i,\qquad m_j:=\frac1n\sum_{i=1}^nx_{ij},$$
$$x_{i,\,j+1}:=\begin{cases}\quad x_{ij}&\text{if}\quad x_{ij}\leqslant m_j,\\
\quad m_j&\text{if}\quad x_i-q_i<m_j<x_i,\\
x_i-q_i&\text{if}\quad m_j\leqslant x_i-q_i,\end{cases}$$
for $i=1,...,n$ and $j=1,2,...$. Since the sequence $(x_{i1},x_{i2},...)$ is decreasing and bounded below by $x_i-q_i$, it has a limit, which we denote by $l_i$. We denote the mean $(l_1+\cdots +l_n)/n$ by $m$.

We now show that the variance of the data $l_1,...,l_n$ is minimal, in the sense that a small perturbation of any one of them increases the variance.
For simplicity, let us shift the original data so that $m=0$. Then it follows from the definition of $x_{ij}$ that

$$l_i=\begin{cases}\quad x_i&\text{if}\quad x_i\leqslant 0,\\
\quad 0&\text{if}\quad x_i-q_i<0<x_i,\\
x_i-q_i&\text{if}\quad 0\leqslant x_i-q_i.\end{cases}$$
Suppose, for convenience, that the perturbed datum is the first one ($i=1$). The unperturbed variance may be written
$$\frac1nl_1^2+\frac1n\sum_{i=2}^nl_i^2.$$
Suppose that $l_1$ becomes $l_1+n\varepsilon$, with $|\varepsilon|$ suitably small, so that the corresponding mean is perturbed to $-\varepsilon$, and the variance becomes
$$\frac1n(l_1+n\varepsilon+\varepsilon)^2+\frac1n\sum_{i=2}^n(l_i+\varepsilon)^2.$$
The increase in variance is therefore
$$\frac2n(n+1)\varepsilon l_1+\frac2n\varepsilon\sum_{i=2}^nl_1+\frac{n^2+2n+2}n\varepsilon^2$$
$$=2\varepsilon l_1+\frac2n\varepsilon\sum_{i=1}^nl_i+|O(\varepsilon^2)|$$
$$=2\varepsilon l_1+|O(\varepsilon^2)|.$$
Consider the three cases for $i=1$ in the formula for $l_i$ above. In the first case, the only allowed perturbation will be negative, namely with $\varepsilon$ negative, and $l_1$ cannot be positive either. So the first-order increase $2\varepsilon l_1$ in the variance cannot be negative. Similar reasoning shows that the variance increase is non-negative in the third case too. In the middle case, while there is a zero first-order increase in the variance, there is still a positive increase, from the second-order term.

Added note: Although the above method converges, there are extreme types of data for which the convergence is very slow. In particular, consider the case when $n=100,\,$ $q_1=\cdots=q_n=1$, $x_1=0,$ and $x_2=\cdots=x_n=1$. Then all the $x_{ij}$ converge to $0$, but after $100$ steps all but one are still about $1/\mathrm e$ from the limit. There is thus plenty of room for tweaking this algorithm to speed it up when the data are like this.


问:
Proper notation for subset/selection from a vector
I have a column X vector which is the result of some calculations. X is comprised of N+M elements, where the first N elements are electrical voltages and the last M elements are currents.
\begin{equation}
 X = \begin{bmatrix}
      V\\
      I\\
    \end{bmatrix}
\end{equation}

Following this, I wish to use only the last M elements of X in some further calculations, but I am not sure how to write it correctly. How do I denote that I want only the elements of the I vector?
I could write an intermediate step such as $$P = [X_{N+1},...,X_{N+M}]$$ but I wasn't sure if there was a sleek way of writing it all in one step.

Is there an accepted notation for this?

答:
Short answer:

I would use the notation $I$. There is no danger of confusion.

Additional remark: The reason there is no danger is that you have nicely prefaced your $X\,= [\,]$ statement with clear words explaining what you are up to.


问:
Must the action of a torus on a manifold have a fixed point if each circle subaction has a fixed point?
Let $T = (S^1)^k$ act smoothly on a compact manifold $M$ via $\rho : T \times M \to M$.
Suppose for every cocharacter/group map $\sigma : S^1 \to T$, the circle subaction $\rho \circ \sigma : S^1 \times M \to M$ has a fixed point.
Does there exist a fixed point of $\rho$ on $M$?

I'm interested in the case where $M$ is a symplectic manifold and $\rho$ is a Hamiltonian torus action.
The moment map of $\rho \circ \sigma$ will have a maximum, and this will be a fixed point of $\rho \circ \sigma$. Indeed, this is the origin of my assumption about circle subactions.

For $k=2$, I wonder whether we can show existence by contradiction.
Assume no fixed point of $\rho$ exists, and let $p_\sigma \in M$ be a fixed point of $\rho \circ \sigma$ for all $\sigma : S^1 \to T$. The set $\{ p_\sigma \}$ is an infinite set, and thus has a limit point in $M$. Can we show that the points $p_\sigma$ must be isolated (giving a contradiction)?

Any assistance or ideas would be welcome.

答:
Regarding a Hamiltonian torus action on a (compact connected) symplectic manifold, your question is answered positively by Atiyah--Guillemin--Sternberg's convexity theorem, which states that the image of the moment map is a (compact) convex polytope whose vertices have as preimages precisely the fixed point set of the torus action.

Besides the original research papers, the proof of Atiyah can be found in McDuff's and Salamon's Introduction to Symplectic Topology, whereas Guillemin's and Sternberg's proof is somewhat scattered in their book Symplectic Techniques in Physics. Atiyah's proof is more straightforward and relies on Morse-Bott theory; As you anticipated, the Hamiltonian functions associated to the 1-parameter subgroups of the torus play a key role. Guillemin's and Sternberg's proof also involves bits of Morse-Bott theory, but their proof relies on more general facts about smooth actions which allow to answer your question affirmately in the smooth setting too (at least when $k=2$):

Proposition (27.4 in their book). Let a compact group $G$ act smoothly on a compact manifold $M$. Then at most finitely many conjugacy classes of subgroups of $G$ appear as isotropy groups of points of $M$ for this action.

Here, the conjugation equivalence on subgroups of $G$ is given by $A \sim B$ iff there exists $g \in G$ such that $A = gBg^{-1}$. (Remark: This result is apparently a special case of a theorem of Mostow for continuous actions.) It follows from this proposition that when $G$ is abelian, i.e. a torus, subgroups coincide with conjugacy classes, hence only finitely many subgroups of $G$ can be isotropy groups of points of $M$ for the given torus action.

For $k > 1$, the torus $T^k$ admits infinitely many (conjugation inequivalent) circle subgroups. Your assumption on the torus actions however implies that each one of these circle subgroups fixes at least a point in $M$. In view of the proposition, most of those circle subgroups are only proper subgroups of the isotropy groups of the points they fix. Hence there are two nonequivalent circle subgroups belonging to the isotropy group of some point, implying that this isotropy group contains at least a 2-dimensional torus. When $k=2$, this proves that every smooth $T^2$-action on a compact manifold has a fixed point. I expect that pursuing this sort of argument would establish the corresponding result for any $k > 2$.


答:
Start with well-known series:
$$
f(x)=\sum_{k=0}^\infty\binom{2n}n\left(\frac x2\right)^{2n}=\frac1{\sqrt{1-x^2}}\tag1
$$
From this one obtains:
$$\begin{align}
&\sum_{k=0}^\infty\binom{2n}n\frac1{2n+1}\left(\frac x2\right)^{2n}=\frac1x\int_0^x \frac{dt}{\sqrt{1-t^2}}=\frac{\arcsin(x)}x\\
&\sum_{k=0}^\infty\binom{2n}n\frac1{(2n+1)^2}\left(\frac 12\right)^{2n}=\frac12\int_0^1 \frac{\arcsin(t)}tdt.\\
\end{align}$$
The last integral can be evaluated as:
$$\begin{align}
\int_0^1 \frac{\arcsin(t)}tdt&=\left[\arcsin(t)\log(t)\right]_0^1-
\int_0^1 \frac{\log(t)}{\sqrt{1-t^2}}dt\\
&=-\int_0^\pi \log(\sin u)du=\pi\log(2).
\end{align}$$

Title: How to prove $\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \frac{\pi}{2}\ln(2)$?

Body:
I was messing around with some integrals and series when I arrived at the result (and WolframAlpha agrees):

$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$

I therefore have a proof, but I would like to see how a proof would be done the other way around (i.e. starting with this series). Notice that the series could also be expressed with double factorials:

$$\sum_{n=0}^\infty \frac{(2n)!}{(2^nn!(2n+1))^2} = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^2}$$

My Derivation

My derivation is quite simple. I was considering (and trying to solve) the integral

$$\int_0^1 \frac{\arcsin(x)}{x} \, dx$$

I couldn't work it out, however I found the exact same question here on math exchange and the answers show that the integral equals $\frac{\pi}{2}\ln(2)$. What I did found however, was that I could plug the Taylor series for $\arcsin(x)$ (the proof of the taylor series can be found here) into the original series to only be left with a series. Here's the Taylor series for $\arcsin(x)$:

$$
  \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1}
$$

So plugging in the series above into the integral gives us:

$$\int_0^1 \frac{1}{x} \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} \, dx$$

Now here we have to be careful, but since the series works on the integral's interval $[0,1]$ and the summand is only positive, we can interchange the the integral and the sum, leaving us with:

$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1} \, \int_0^1 \frac{x^{2n+1}}{x} \, dx$$

And then solving the integral in the sum gives us the final result:

$$\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{(2n+1)^2} = \frac{\pi}{2}\ln(2)$$

And then expanding the binomial gives the original series in the title.


问:
Find line orthogonal to another line passing through point p in R³
So in R³, given a line $G$ in parameter form and a point Q, how do I compute the line passing through Q that is orthogonal to $G$?

In order for two vectors to be orthogonal, the dot product has to be zero. Also, for a point to be located on a line, inserting the point into the line equation must result in equality, right? Not quite sure how to use this for the task.

A hint to get me on the right track would suffice

答:
For $P \in G$, compute $\vec{PQ}$ (which depends on the parameter $\lambda$). Then $\vec{PQ} \cdot \vec{G}=0$ (where $\vec{G}$ is the direction of $G$) is a linear equation in $\lambda$.

If $G=\{\lambda (g_x, g_y, g_z) + (p_x, p_y, p_z)\}$ and $Q=(q_x, q_y,q_z)$, then $\vec PQ=(\lambda g_x + p_x -q_x, \lambda g_y + p_y -q_y, \lambda g_z+ p_z -q_z)$ and $\vec PQ \perp G$ iff $g_x(\lambda g_x + p_x -q_x)+g_y(\lambda g_y + p_y -q_y+ g_z(\lambda g_z+ p_z -q_z)$.
That is, iff $\lambda (g_x^2 + g_y^2 + g_z^2) = g_x(p_x-q_x)+g_y(p_y-q_y)+g_z(p_z-q_z)$.


问:
Finding all ordered quadruples $(a, b, c, d)$ of primes
Find all ordered quadruples $(a, b, c, d)$ of primes such that
\begin{align*}
13a^2 + b^2 + 13c^2 &= 2b(3a + 2c) \\
a^3 −(bc)^3 &= (b+c−a)d+2021.
\end{align*}

A quick WolframAlpha search yielded that the answer to this was no pairs, but I wasn't quite sure how to derive it.

答:
The first equation, considered as quadratic equation in $c$, has no real solution in $c$, except for the case $3b=13a$.

Then we have $a=3$ and $b=13$, so that the first equation gives $c=2$. This indeed solves the first equation. However, then the second equation is $6d + 9785=0$, so that $d$ is not an integer. Contradiction.


问:
Is the space of $\sigma$-finite measures measurable?
Let $(X,\mathcal{F})$ be a measurable space. There is a natural $\sigma$-algebra on the space of all measures on $\mathcal{F}$, defined as being generated by all maps $\mu\mapsto\mu(A)$ with $A\in\mathcal{F}$. It is obvious that, for instance, the set of probability measures and the set of all finite measures on $\mathcal{F}$ are both elements of this $\sigma$-algebra. Is the same true of the $\sigma$-finite measures? I suspect not but don't know how to prove it.

答:
I'm fairly certain that if $\mathcal{F}$ is big enough, then the set of $\sigma$-finite measures shouldn't be in this $\sigma$-algebra. However, I also don't know how to prove it. Maybe an example will give someone some inspiration, or maybe I'll come up with some real progress later.

If $X = \mathbb{N}$ and $\mathcal{F} = \mathcal{P}(\mathbb{N})$, then every measure $\mu$ on $(X, \mathcal{F})$ is completely determined by its values on each $\{n\}$, for $n \in \mathbb{N}$. In particular, a measure $\mu$ here is $\sigma$-finite if and only if $\mu(\{n\}) < \infty$ for all $n \in \mathbb{N}$. Thus, using the notation in my comment to the question,
$$
\{\mu : \mu \text{ is } \sigma \text{-finite}\}
= \bigcap_{n \in \mathbb{N}} f_{\{n\}}^{-1}([0, \infty)),
$$
which is a member of our $\sigma$-algebra since it's a countable intersection of sets in our $\sigma$-algebra.


问:
How union and intersection implies complement?
Defining the family of Borel sets to be the smallest σ–algebra that contains all the closed sets in a metric space X, I want to show that it is equivalent to the following definition :

The only part remained to show is that if conditions 1. and 2. of the Theorem 3.3 holds then for any set $E\in B$, also $X\backslash E\in B$. I know being closed under union and complement implies being closed under intersection, but how to prove being closed under union and intersection implies being closed under complement?

Other way to finish my proof is to show that in a metric space every open set is an $F_σ$ ; this is easy for $\mathbb{R}$ but how to prove that for a general metric space?

答:
This is actually a rather clever little proof.

Consider the set $C = \{S \in B : X \setminus S \in B\}$. I claim that this set satisfies the properties

If $E_1, E_2, E_3, ... $ belong to $C$ then $\bigcup\limits_{i = 1}^\infty E_i \in C$

If $E_1, E_2, E_3, ... $ belong to $C$ then $\bigcap\limits_{i = 1}^\infty E_i \in C$

Every closed set is in $C$.

Let's begin with 1. Suppose $E_1, E_2, E_3, ...$ belong to $C$. Then $E_1, E_2, E_3, ...$ belong to $B$. Then $\bigcup\limits_{i = 1}^\infty E_i \in B$. Now consider $X \setminus \bigcup\limits_{i = 1}^\infty E_i = \bigcap\limits_{i = 1}^\infty (X \setminus E_i)$. Since each $X \setminus E_i$ is in $B$, so too is $\bigcap\limits_{i = 1}^\infty (X \setminus E_i)$.

Showing 2 is almost identical to showing 1, so we will skip it for brevity.

Now we must show that every closed set is in $C$. It suffices to show that every open set is in $B$. Consider an open set $U$, and consider the sets $K_n$, indexed by $n \in \mathbb{N}$, defined by $K_n = \{x | \forall y \in X \setminus U, d(x, y) \geq 1/n\} = \bigcap\limits_{y \in X \setminus U} \{x | d(x, y) \geq 1/n\}$. Then clearly, $K_n$ is a closed set for all $n$. Then for all $n$, we have $K_n \in B$. Then in particular, we have $\bigcup\limits_{n = 1}^\infty K_n \in B$.

I claim that $U = \bigcup\limits_{n = 1}^\infty K_n \in B$. For suppose that $x \in U$. Then take $\delta$ such that $B_\delta(x) \subseteq U$. Take $n$ such that $1/n < \delta$. Then $x \in K_n$.

And suppose that $x \in K_n$. Suppose that $x$ is not in $U$. Then we have $d(x, x) = 0 < 1/n$; contradiction. So $x \in U$.

Since $C$ satisfies 1, 2, and 3, we see that $B \subseteq C$. Then $B$ is closed under complementation.


问:
Generating a 'path' of uniformly distributed continuous random variables
Let $X$, $Y$ be independent continuous random variables both uniformly distributed on $[0,1]$. Is there a function $F(t;X,Y):[0,1]^{\times 3}\rightarrow [0,1]$ satisfying:

$F$ is continuous in $t$, with $F(0)=X$ and $F(1)=Y$;

For any $t_0\in [0,1]$, $F(t_0;X,Y)$ is uniformly distributed?

In a sense, this is a path in the space of uniformly distributed measurable functions. Note, each individual condition is rather easily achievable:

Example 1. Let $F(t;X,Y)=(1-t) X+tY$. It is obvious that $F$ is continuous in $t$, with $F(0)=X$ and $F(1)=Y$. However, for $t_0\neq 0,1$, $F(t_0;X,Y)$ is not uniformly distributed, but rather trapezoidally distributed on $[0,1]$.

Example 2. Let $F(t;X,Y)=\max((1-t),t)^{-1}((1-t) X+tY)\ {\rm mod}\ \max(1-t,t)$. For any $t_0\in [0,1]$, $F(t_0;X,Y)$ is uniformly distributed on $[0,1]$; heuristically, note that the probability density function of $(1-t) X+tY$ is trapezoidal, and the modulus in essence combines triangular portions of the density complementarily to obtain a uniform density function. However, this is typically not continuous in $t$ due to the modulus function.

Motivation: Here $F=F(t)$ is a parameter to a Monte Carlo simulation, and $t$ is a deterministic input chosen by the user. It is known that the parameter $F$ should be uniformly distributed at each input $t$, and that it is independent for $t$ far enough away from one another, but it is not clear how values should be correlated otherwise. Also, one would expect $F$ to be continuous (or at least $\epsilon$-close to a continuous function for some reasonably small $\epsilon$).The question posed to me is whether there is a sensible way to use draws for $F(0)$ and $F(1)$ to deterministically populate $F(t)$ for other choices of $t$ so that it represents the correct distribution, and it seemed to me an obvious and interesting question with no easy obvious answer, but I am not an expert in probability theory.

答:
This is possible!

I start with a Gaussian process (GP). This is simpler since those are uniquely determined by covariance function and mean function.

The GP is a function $G:\Omega\times [0,1]\rightarrow\mathbb{R},\, (t,\omega)\mapsto G_t(\omega)$ such that for every every choice ${t_1}, \ldots {t_n}$ the random variables $G_{t_1}, \ldots G_{t_n}$ have a joint Gaussian distribution. Choose the mean function of $G$ as zero and define the covariance as
$$ \text{Cov}[G_t,G_s]=\max\left(0, 1-2 \left\lvert t -s\right\lvert\right).$$

Note that $Cov[G_0,G_1]=0$, i.e. they are uncorrelated. Since $G_0,G_1$ are Gaussian they are also independent. Each sample path $G(\omega)$ is a continuous function this can be seen from correlation function (see the Wikipedia article).

Finally, this Gaussian process must be converted to one with uniform marginals. This is performed by the Gaussian cumulative distribution function $\Phi$. Simply set $U_t=\Phi(G_t).$ Then each margin is uniform and continuous. The independence remains as well.


问:
Families of Morse functions
I don't have a clue with this problem. Thank you very much for your help & guidance.

(a) Suppose $F(x,t): X\times I \rightarrow R$ is a homotopy of Morse functions. That is, $f_t: X \rightarrow R$ is Morse for every $t$. Show that the set $C = \{(x,t)\in X\times I : d(f_t)_x = 0\}$ forms a closed, smooth submanifold of dimension one of $X\times I$. Assume the homotopy is constant near the ends of I and use an open interval.

(b) Let $\pi: X \times I \rightarrow I$. Show that $d(\pi|_C)_{(x,t)}: TC_{(x,t)} \rightarrow TI_t$ is surjective.

(c) Show that if $X$ is compact, there is no homotopy of Morse functions between two Morse functions with different numbers of critical points.

答:
(a) Whenever you are asked to show that something defined by an equation is a smooth manifold, the Implicit Function theorem should come to mind. Let $n$ be the dimension of manifold $X$. Since being a submanifold is a local property, we can work with a coordinate patch: that is, consider $X$ as being $\mathbb R^n$. Then $d(f_t)$ is a smooth map into $\mathbb R^n$, let's denote it by $F:\mathbb R^n\times (0,1)\to \mathbb R^n$. We must prove that the derivative of $F$ has maximal rank at the points where $F=0$. This is where you will use the Morse condition.

(b) Between one-dimensional manifolds, "surjective derivative" is the same as "nonzero derivative". The implicit function theorem can tell you why it's nonzero.

(c) Use parts (a) and (b) to show that for every integer $k$, the set of values $t$ for which $f_t$ has exactly $k$ critical points, is open.

Title: Families of Morse functions


问:
Prove that an element in the quotient field is actually in the ring itself
Let $R$ be a noetherian integral domain and $K$ the quotient field of $R.$ Suppose $f \in K.$ Suppose for each maximal ideal $M$ of $R,$ we can find $h,k\in R$ so that $f = h/k$ and $k \not\in M$. Show $f \in R.$

Note: $R$ is not assumed a UFD!

My attempt:
I tried to prove that the summation of ideals generated by all the k that are not in some maximal ideals is the whole ring. But I dont know how to use that to move on. Also, I do not understand how to show an element is in the ring. I tried to prove that k has to divide h, but since R is not a UFD, I am not sure to achieve that.

答:
Define $I=\{a\in R: af\in R\}\subseteq R$. This is clearly an ideal of $R$. If we can show that $I=R$ then it means $1\in I$ and so $f\in R$, as we want. So suppose $I\ne R$. Then $I$ is contained in some maximal ideal $M$. By assumption, there are $h,k\in R$ such that $f=\frac{h}{k}$ and $k\notin M$. But then note that:

$kf=k\frac{h}{k}=h\in R$

And so $k\in I\subseteq M$, a contradiction.

Note that you don't even need $R$ to be Noetherian.


答:
All you need is for $R$ to be a domain. We'll prove
$\newcommand{\m}{\mathfrak{m}}$

Proposition. $R=\bigcap_{\m}R_\m$.

In the above, the intersection runs over all maximal ideals of $R$, and we identify $R_\m$ with the subring $\{\frac rs\mid r\in R, s\in R\setminus\m\}$ of $K$. One inclusion is obvious: the canonical maps $R\hookrightarrow R_\m\hookrightarrow K$ show that $R$ is a subset of the intersection $\bigcap_{\m}R_\m$.

The reverse inclusion is equivalent to
$$
K \setminus R
\subset
K\setminus \bigcap_{\m}R_\m
\quad\text{i.e.}\quad
K \setminus R
\subset
\bigcup_{\m}~(K\setminus R_\m)
$$
So consider $x\in K\setminus R$ and put $J=\{r\in R\mid rx\in R\}$. Then $J$ is a proper ideal of $R$: it's obviously an ideal of $R$ and not all of $R$ since $1\notin J$. By Zorn's lemma, $J$ is contained in some maximal ideal $\m_0$. The claim is now that $x\in K\setminus R_{\m_0}$. Otherwise we could write $x=\frac rs$ with $r\in R$, $s\in R\setminus\m_0$, but then $sx=r\in R$ and $s\in J\subset\m_0$, which is a contradiction. Therefore $x\in K\setminus R_{\m_0}$ and we are done.

Title: Prove that an element in the quotient field is actually in the ring itself

Body:
Let $R$ be a noetherian integral domain and $K$ the quotient field of $R.$ Suppose $f \in K.$ Suppose for each maximal ideal $M$ of $R,$ we can find $h,k\in R$ so that $f = h/k$ and $k \not\in M$. Show $f \in R.$

Note: $R$ is not assumed a UFD!

My attempt:
I tried to prove that the summation of ideals generated by all the k that are not in some maximal ideals is the whole ring. But I dont know how to use that to move on. Also, I do not understand how to show an element is in the ring. I tried to prove that k has to divide h, but since R is not a UFD, I am not sure to achieve that.


问:
Why can't a set have two elements of the same value?
Suppose I have two sets, $A$ and $B$:

$$A = \{1, 2, 3, 4, 5\} \\ B = \{1, 1, 2, 3, 4\}$$

Set $A$ is valid, but set $B$ isn't because not all of its elements are unique. My question is, why can't sets contain duplicate elements?

答:
It stated that members of a set should have a common character so that we can easily identify the non-members of a given set.
Supposing we have:
S={ 2,4,6,8,10...}
The comm characteristics that exist among the members of set S is that. We always get the same number when we subtract on member from the other member at the immediate next.
I. e 4-2=2
6-4= 2
8-6=2. The two(2)s that appears is a unique feature the sets have.

But if say A= { 1,1,2, 1,3 4...}
It will be difficult to get the unique features the exist between the of the set. Hence, set A may be said to be not well defined set.


问:
Existence of an inequality
Let $f: [a,b] \to \mathbb{R}$ be a differentiable function with continuous derivative. Prove the existence of $>0$ such that for all $x, y\in[a, b]$ the inequality |()−()|≤ |−| is true.

I was thinking of proving this using the Mean Value Theorem but I am not really sure how to proceed. Any help is appreciated.

答: The derivative is continuous on the compact set $[a,b]$ and hence bounded by some M.  Now use MVT

Title: Existence of an inequality


问:
$p_1,p_2$ roots of $x^2+3x+5=0$, then the value of the expression $M=\frac{p_1^2+5p_1+7}{p_1^2+7p_1+5}+\frac{p_2^2+5p_2+7}{p_2^2+7p_2+5}$
If $p_1,p_2$ are the roots of the equation $x^2+3x+5=0$, then the value of the expression $M=\frac{p_1^2+5p_1+7}{p_1^2+7p_1+5}+\frac{p_2^2+5p_2+7}{p_2^2+7p_2+5}$

I solved the question as follows:

Initially I tried to solve it by finding what the roots are but since $b^2-4ac=9-20<0$, a real solution doesn't exist. Henceforth I attempted to solve it using Vieta's formulas, by saying that $p_1+p_2=-\frac{a_{n-1}}{a_n}=-\frac{3}{1}=-3$ and similarly $p_1*p_2=5$ and then $M=\frac{(p_1^2+5p_1+7)(p_2^2+7p_2+5)+(p_2^2+5p_2+7)(p_1^2+7p_1+5)}{(p_1^2+7p_1+5)(p_2^2+7p_2+5)}$

and eventually after a lot more calculations:

$M=\frac{7}{10}$

I find this method extremely laborious and I believe that there must exist a faster and easier method of solving this question. Could you please explain an easier and faster method?

Body:
Since $p_1,p_2$ are roots of $x^2+3x+5=0$, then $p_1^2+3p_1+5=p_2^2+3p2+5=0$ and we can simplify the expression to
$$
\frac{2p_1+2}{4p_1}+\frac{2p_2+2}{4p_2}=\frac{p_1+1}{2p_1}+\frac{p_2+1}{2p_2}=\frac{p_1p_2+p_2+p_1p_2+p_1}{2p_1p_2}=1+\frac{p_1+p_2}{2p_1p_2}
$$
Now $p_1+p_2=-3$ and $p_1p_2=5$ so $M=\frac{7}{10}$


答:
hint

$ f $ is continuous and differentiable at $ [a,b] \implies$

$$(\forall x,y\in[a,b])\;(\exists c\in(x,y))\;:$$
$$\; f(x)-f(y)=(x-y)f'(c)$$

but $ f' $ is continuous at the compact $ [a,b] $
$$\implies f' \text{ is bounded}$$

$$\implies (\exists M>0)\;:$$
$$\;(\forall t\in [a,b])\;|f'(t)|\le M$$

Title: Existence of an inequality

Body:
Let $f: [a,b] \to \mathbb{R}$ be a differentiable function with continuous derivative. Prove the existence of $>0$ such that for all $x, y\in[a, b]$ the inequality |()−()|≤ |−| is true.

I was thinking of proving this using the Mean Value Theorem but I am not really sure how to proceed. Any help is appreciated.


问:
Deriving the basis-dependent definition of the Hodge star operator
Let $M$ be a $n$-dimensional manifold and let

$$\star : \Lambda^p(M) \longrightarrow \Lambda^{n-p} : \beta \longmapsto \star \beta$$

be the Hodge star operator, such that

$$ ( \alpha, \beta ) = \int_M \alpha \wedge \star \beta$$

for all $\alpha \in \Lambda^p$. Here, $\big( \boldsymbol{\cdot}, \boldsymbol{\cdot} \big)$ denotes the inner product, defined as

$$ (\alpha, \beta) = \frac{1}{p!} \int_M \alpha_{i_1 \ldots i_p} \; \beta_{j_1 \ldots j_p} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; dV$$

Show that

$$\star \big( dx^{i_1} \wedge \ldots \wedge dx^{i_p} \big) = \frac{\sqrt{\det g}}{(n-p)!} \cdot g^{i_1 j_1} \cdots g^{i_p j_p} \cdot \varepsilon_{j_1 \ldots j_n} \cdot dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n}$$

Now, I suppose this is a pretty standard problem, when dealing with differential geometry -- however, I neither managed to solve it myself nor did I find a complete proof of that proposition. Most of the proofs I found were either missing some vital step, or did not bother to expand the proof as detailed as I would need it to understand.

Can anyone help me by providing a source for a proof or can contribute a proof itself? I would really appreciate it.

P.S.: I found the answer to How to derive the coordinate expression of the Hodge dual? useful, but I don't understand some of the steps. I copied the proof and marked the steps I don't understand with a question mark (?)

Since both sides are linear in $\omega$, it suffices to check on a basis. Let $\omega := \mathrm{d}x^{1}\wedge...\wedge \mathrm{d}x^{k}$. Then, one immediately notices that $\eta \overset{?}{=} \varepsilon^{i_1 ... i_k} \omega$. Observe that
\begin{align*}
⟨\omega,\eta⟩ \ vol \enspace &= \enspace \varepsilon^{i_1 ... i_k} \ ⟨\omega,\omega⟩ \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&\overset{?}{=} \enspace \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ g^{1 j_1} ... g^{k j_k} \ \varepsilon_{j_1 ... j_k} \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&= \enspace \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&= \enspace \sqrt{g} \ \Big( \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \Big) \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&\overset{?}{=} \enspace \mathrm{d} x^1 \wedge ... \wedge \mathrm{d} x^k \wedge \Big( \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \Big) \ \mathrm{d} x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\\
&= \enspace \omega \wedge \widetilde{\eta}
\end{align*}

答:
I now figured out something, but I am not sure whether my calculations are correct, so please tell me if my approach is acceptable.

Let $\alpha \in \Lambda^p$ such that $\alpha = \frac{1}{p!} \; \alpha_{k_1 \ldots k_p} \; dx^{k_1} \wedge \ldots dx^{k_p}$ and let $\beta = dx^{j_1} \wedge \ldots \wedge dx^{j_p}$. Then by assumption

\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha \wedge\Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big) \\
&= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!}  \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha_{k_1 \ldots k_p} \; \Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge \Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big)
\end{align}

By renaming the indices, one obtains

\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \underbrace{\Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge\Big(dx^{k_{p+1}} \wedge \ldots \wedge dx^{k_n} \Big)}_{= \; \varepsilon^{k_1 \ldots k_n} \; dx^1 \wedge \ldots \wedge dx^n} \\
&= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \; \varepsilon^{k_1 \ldots k_n} \; \Big( dx^1 \wedge \ldots \wedge dx^n \Big)
\end{align}

Using $dV =\sqrt{\det g} \cdot dx^1 \wedge \ldots \wedge dx^n$, one gets

\begin{align}
 \alpha \wedge \star \beta \enspace &= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \varepsilon^{k_1 \ldots k_n} \; \alpha_{k_1 \ldots k_p}\\
&= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; (n-p)! \; \delta^{k_1 \ldots k_p}_{j_1 \ldots j_p} \; \alpha_{k_1 \ldots k_p} \\
&= \enspace \frac{dV}{p!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; p! \; \alpha_{[j_1 \ldots j_p]} \\
&= \enspace dV \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \alpha_{j_1 \ldots j_p}
\end{align}

which coincides with the inner product. Here, the antiymmetry of $\alpha$ was used. Note that the factor of $\tfrac{1}{p!}$ within the inner product arises from a $p$-form $\gamma = \tfrac{1}{p!} \, \gamma_{m_1 \ldots m_p} \, dx^{m_1} \wedge \ldots \wedge dx^{m_p}$ being inserted into the inner product, i.e. $(\alpha, \gamma)$.

Please tell me what you think.


问:
Compact set proof
Let $D \subseteq \mathbb{R}^2$ be a non-empty subset that is compact in the Euclidian metric space $(\mathbb{R}^2, d)$ and let $E = \{x^2 + y^2: (x, y) \in D\} \subseteq \mathbb{R}$ be another subset. Prove that $E$ is a compact subset of $(\mathbb{R}, | \cdot |)$.

Note: Such a set is called compact if it is closed and bounded.

My attempt:

Bounded:

Since $D$ is bounded, there is some $M$ so that $d(x, y) < M$ for all $x, y \in D$. Suppose $x = (a_1, a_2)$ and $y = (b_1, b_2)$ then

$$d(x, y) = d((a_1, a_2), (b_1, b_2)) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2} < M$$

From this we need to show that the usual metric on $\mathbb{R}$, denoted by $d_1(x, y) = |x - y|$ is bounded too.

Not sure how to proceed from here.

(Does bounded here mean bounded above and below? Or does what I've done suffice?)

Closed:

A set is closed if either all its limit points are contained in the space or if its complement is open. Finding the limit points of $E$ doesn't seem doable. More so, looking at $E$, I can't seem to figure out its complement.

Another approach could be to show that $E$ is complete. Then, $E$ would be closed but I can't think of how to show that all Cauchy sequences converge in $E$.

Not sure how to do this one.

Any assistance is much appreciated.

答:
The function $f(x,y)=x^2+y^2$ is continuous and $f(D)=E$.  Since D is compact, and since f is continuous, E must be compact.  Here we are using the fact that the continuous image of a compact set is compact.

Edit:  I'll try to give another proof without using topological machinery.

Bounded:  If $D$ is bounded then it is contained in some disk $||(x,y)|| \leq R$.  Since $x^2+y^2=||(x,y)||^2$ we have $f(x,y) \leq R^2$ for each $(x,y) \in D$ (here we used $f(D)=E$).

Closed:  This follows from the reverse triangle inequality $| ||x||-||y|| | \leq ||x-y||$ where $||x-y||=d(x,y)$

Title: Compact set proof


问:
Is every additive monofunctor between abelian categories left exact?
Is there an additive functor between abelian categories, which preserves monomorphisms, but is not left exact?</p>

答:
Consider an integral domain $R$ which is not a field, and choose $a \in R \setminus (\{0\} \cup R^*)$. Consider the functor $\mathsf{Mod}(R) \to \mathsf{Mod}(R)$ which maps an $R$-module $M$ to its submodule $aM$. Clearly this preserves monomorphisms, but it is not left exact: The exact sequence $0 \to R \to Q(R) \to Q(R)/R$ becomes $0 \to aR \to Q(R) \to Q(R)/R$ after applying the functor.

Alternatively, it suffices to give an additive functor between abelian categories which preserves epimorphisms, but is not right exact. Take an integral domain $R$, which is not a field, and consider the functor $\mathsf{Mod}(R) \to \mathsf{Mod}(R)$ which takes an $R$-module to its maximal torsionfree quotient, i.e. which mods out the torsion submodule. This functor preserves epimorphisms, but is not right exact: The exact sequence $R \to Q(R) \to Q(R)/R \to 0$ becomes $R \to Q(R) \to 0 \to 0$ after applying the functor.


问:
Determine the exact point(s) of intersection between f(x)=x^2-x-13 and it’s reciprocal function
does anybody know how to solve this? I’m not sure what the reciprocal function means.</p>

答: The reciprocal of a function $f(x)$ is simply $\frac{1}{f(x)}$. So let's denote it by $r(x)$ for convenience. Then, to find the point(s) of intersection, you want to find all $x$ that would satisfy $f(x) = r(x)$, or, written explicitly, $$x^2-x-13 = \frac{1}{x^2-x-13}.$$ Let's analyse this expression. When could it be that $a = 1/a$? Multiplying by $a$ we get $a^2=1$ which results in $a=\pm \sqrt{1}=\pm 1$. So the points of intersection of $f(x)$ and $r(x)$ are only when $f(x)=r(x) = \pm 1$. The goal now is to find the corresponding $x$-coordinates. In other words, you need to solve the two equations $$x^2-x-14 = 0, \\ x^2-x-12 = 0.$$ I trust you can take it from here?


问:
Good abstract algebra books for self study
Last semester I picked up an algebra course at my university, which unfortunately was scheduled during my exams of my major (I'm a computer science major). So I had to self study the material, however, the self written syllabus was not self study friendly (good syllabus overall though).

The course was split up into 3 parts, group theory, ring theory and field theory. As a computer science major we only had to study the first 2.

Now that I passed the exam for this course I want to study the field theory part ( which covers Galois theory, etc).

So, now I want to ask whether any of you know any good books on abstract algebra, which lift off at basic ring theory and continue to more advanced ring theory and to finite fields, Galois theory, ...

Please keep in mind that I am not a math major, and that I would like books which are suited for self study (thus a lot of examples and intuition).

Thanks in advance!

答: My favorite is Introduction to Abstract to Algebra by Keith Nicholson. There is a course in YouTube, by James Cook base on that book.</p>


问:
Estimate error using Taylor Series
I was told to use Taylor Series to find the error of this but I am not confident in doing this.

Estimate the error in approximating $\cos x$ by $1-(x^2/2)$ on the interval $[-0.5, 0.5]$.

答:
The Lagrange error bound states that the gap between $f(x)$ and its Taylor expansion in $x_0$ of order $k$ is
$$
R_k(x) = \frac{f^{(k+1)}\big(\xi(x)\big)}{(k+1)!} (x-x_0)^{k+1}
$$
where $\xi(x)$ is a point lying in the interval $[x_0,x]$ or $[x,x_0]$ (depending whether $x_0\leq x$ or $x_0\geq x$). In your case,
$$
1-\frac{x^2}2
$$
is the 3rd order approximation of $\cos x$ in $x_0=0$ (since $\cos'''(0)=\sin(0)=0$), so
$$
R_3(x)
~=~
\cos x - \left(1 - \frac{x^2}2\right)
~=~
\frac{\cos{\big(\xi(x)\big)}}{24}x^4
$$
If $x$ ranges in $[-0.5,0.5]$, then so does $\xi(x)$. Therefore
$$
\max_{[0.5,0.5]}{\big|R_3(x)\big|}
~\leq~
\frac{\max_{[0.5,0.5]}{\big|\cos(x)\big|}}{24}
\max_{[0.5,0.5]}{\big|x^4\big|}
~=~
\frac{1}{24}\cdot\frac{1}{2^4}
~=~
\frac{1}{384}
~\approx~
0.0026\ldots
$$</p>

Title: Estimate error using Taylor Series


问:
Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005)
Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.

Source : Vietnamese TST 2005

I'm sure that the only solutions are the zero function, the identity, and the negative of the identity.  First, when $x=y=z=0$ we have $f(0)=3f(0)^3$ so $f(0)=0$ is the only integer solution.  Next if $y=-x$ and $z=0$ then $f(-x)=-f(x)$ so $f$ is odd.  Then $x=y=1$ and $z=-1$ give $f(1)^3=f(1)$ so $f(1)=0,1,-1$.

Since $f$ is a soln iff $-f$ is, we can take $f(1)=0,1$.  Then $f(2)=f(1^3+1^3+0^3)=0,2$.  Then $f(3)=f(1^3+1^3+1^3)=0,3$.  Also $f(4^3-3^3-3^3)=f(10)=f(1^3+1^3+2^3)$, so $f(4)=0,4$.  Then $f(5^3-4^3-4^3)=f(-3)$ so $f(5)=0,5$.  Then $f(6)=f(2^3-1^3-1^3)=0,6$.  Then, $f(7)=f(2^3-1^3+0^3)=0,7$.  Then $f(8)=f(2^3+0^3+0^3)=0,8$.  Then $f(9)=f(2^3+1^3+0^3)=0,9$ and $f(10)=f(2^3+1^3+1^3)=0,10$.  Etc...

There should be a smart way to make induction, but I haven't found that.  May be a nice identity.  This is the closest to the claim that $n^3$ is the sum of five cubes of integers with absolute values $<|n|$:
$$n^3=6^3+(n-2)^3+(n-3)^3-(n-6)^3-(n-11)^3+(n-12)^3.$$
At least the identity above gives me $f(11)=0,11$ and $f(12)=0,12$.

答:
As requested I'm posting the above comment as an answer here.

In order to show that every $n\geq 8$ can be written as sum of five cubes with absolute value less than $n$:

we can write for odd numbers $n=2k+1$ with $k\geq 4$
$$(2k+1)^3 = (2k-1)^3 + (k+4)^3 - (k-4)^3 - 5^3 - 1^3$$

for even numbers $n\geq 8$, we can write $n=2^a m$ where $m$ is $4$, $6$, or an odd number $\geq 5$. Then $m$ can be decomposed as $m=m_1^3+m_2^3+m_3^3+m_4^3+m_5^3$ so that
$$(2^a m)^3 = (2^a m_1)^3 + (2^a m_2)^3 + (2^a m_3)^3 + (2^a m_4)^3 + (2^a m_5)^3$$


答:
In order to prove that $E$ is bounded take $z\in E$. Then $z=x^2+y^2$ for some $(x,y)\in D$. Since $D$ is bounded, there is some $M\in\Bbb R_+$ such that the distance from each element of $D$ to $(0,0)$ is smaller than $M$. But then\begin{align}z&=x^2+y^2\\&=(\text{distance from $(x,y)$ to $(0,0)$})^2\\&<M^2.\end{align}

I shall prove now that it is closed. Let $(e_n)_{n\in\Bbb N}$ be a sequence of elements of $E$ which converges to some $l\in\Bbb R$. For each $n\in\Bbb N$, let $(x_n,y_n)\in D$ be such that $x_n^{\,2}+y_n^{\,2}=e_n$. The sequence $\bigl((x_n,y_n)\bigr)_{n\in\Bbb N}$ has a subsequence $\bigl((x_{n_k},y_{n_k})\bigr)_{k\in\Bbb N}$ which converges to some $(x_0,y_0)\in D$. But then, since the map$$\begin{array}{ccc}D&\longrightarrow&E\\(x,y)&\mapsto&x^2+y^2\end{array}$$is continuous,$$l=\lim_{k\to\infty}e_{n_k}=\lim_{k\to\infty}x_{n_k}^{\,2}+y_{n_k}^{\,2}=x_0^{\,2}+y_0^{\,2}\in E.$$Since every convergent sequence of elements of $E$ converges to an element of $E$, $E$ is closed.

Title: Compact set proof

Body:
Let $D \subseteq \mathbb{R}^2$ be a non-empty subset that is compact in the Euclidian metric space $(\mathbb{R}^2, d)$ and let $E = \{x^2 + y^2: (x, y) \in D\} \subseteq \mathbb{R}$ be another subset. Prove that $E$ is a compact subset of $(\mathbb{R}, | \cdot |)$.

Note: Such a set is called compact if it is closed and bounded.

My attempt:

Bounded:

Since $D$ is bounded, there is some $M$ so that $d(x, y) < M$ for all $x, y \in D$. Suppose $x = (a_1, a_2)$ and $y = (b_1, b_2)$ then

$$d(x, y) = d((a_1, a_2), (b_1, b_2)) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2} < M$$

From this we need to show that the usual metric on $\mathbb{R}$, denoted by $d_1(x, y) = |x - y|$ is bounded too.

Not sure how to proceed from here.

(Does bounded here mean bounded above and below? Or does what I've done suffice?)

Closed:

A set is closed if either all its limit points are contained in the space or if its complement is open. Finding the limit points of $E$ doesn't seem doable. More so, looking at $E$, I can't seem to figure out its complement.

Another approach could be to show that $E$ is complete. Then, $E$ would be closed but I can't think of how to show that all Cauchy sequences converge in $E$.

Not sure how to do this one.

Any assistance is much appreciated.


问:
What information do I get about the linear relations between a Matrix' lines from its rank?
So, let's say I have a 4x2 matrix with a positive rank.

The maximum rank of the matrix is 2, and if the rank of the matrix is 1 I then know that one of the lines in the matrix is a scalar times the other line of the matrix.

Now, let's say I have a 6x3 matrix with a positive rank.
The maximum rank of the matrix is 3, but I struggle to understand the difference between when the rank is 1 and when the rank is 2. I understand that a matrix' rank is the number of non-zero rows in its row echelon matrix, but I struggle to understand what it tells me.

Let's say I have rank 2. Does it mean that I get a linear combination of 2 lines that provides the 3rd? If so, what does rank 1 get you?

I really hope the question is coherent.
Thank you and have a blessed day.

答:
A 6x3 matrix has rank 3 if and only if all three rows (I would not say "lines") are inependent.

If it has rank 2 that means that two of the rows are independent and the third is a linear combination of those two.

If it has rank 1 that means that each row is a scalar multiple of one row.


答:
If you think of a matrix as a linear transformation, then the rank is the dimension of the range.  This is the definition that I find most illuminating, as well as most useful.

In the spirit of your question however, the rank is the maximum number of linearly independent rows (or columns).  So you are correct in your example about the meaning of rank $2$.  If the rank were $1$, it would mean that there is some row, such that each of the other rows is a scalar multiple of it.  We can choose any nonzero row as the distinguished row.  In general, if there rank is $r$ there are $r$ rows, such that every row of the matrix is a linear combination of those $r$ rows, and this isn't true for any number less than $r$.

Title: What information do I get about the linear relations between a Matrix' lines from its rank?

Body:
So, let's say I have a 4x2 matrix with a positive rank.

The maximum rank of the matrix is 2, and if the rank of the matrix is 1 I then know that one of the lines in the matrix is a scalar times the other line of the matrix.

Now, let's say I have a 6x3 matrix with a positive rank.
The maximum rank of the matrix is 3, but I struggle to understand the difference between when the rank is 1 and when the rank is 2. I understand that a matrix' rank is the number of non-zero rows in its row echelon matrix, but I struggle to understand what it tells me.

Let's say I have rank 2. Does it mean that I get a linear combination of 2 lines that provides the 3rd? If so, what does rank 1 get you?

I really hope the question is coherent.
Thank you and have a blessed day.


问:
How to calculate the ILT of $e^{-\sqrt{s}}/s^{3/2}$?
Mathematica can calculate $e^{-\sqrt{s}}/s^{3/2}$, and the result is
$$-1 + \frac{2\sqrt{t} e^{-\frac{1}{4t}}}{\sqrt{\pi}} + \mbox{Erf}\left[\frac{1}{2\sqrt{t}}\right]$$.

However, when I used the residue theorem to calculate it, the integral along the small circle around zero is divergent, $\frac{1}{2\pi i} \int \frac{e^{-\sqrt{s}}}{s^{3/2}} e^{st} ds,\ s=\epsilon e^{i\theta},\ \theta \in(\pi,-\pi) $.

Would you like to help me with this problem?
Thanks in advance.

答:
To find the inverse LT, we may use the definition, which requires integration in the complex plane:

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-\sqrt{s}} e^{s t} $$

where $C$ is the following contour:

(NB I have used this contour for years and lots of people have relied on it in turn.  But please keep in mind that this contour is really an approximation, one that works but is a simplification of the actual contour.)

We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis.  There are $6$ pieces to this contour, $C_k$, $k \in \{1,2,3,4,5,6\}$, as follows.

$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.

$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.

$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.

$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.

$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.

$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.

We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.

On $C_2$, the real part of the argument of the exponential is $R t \cos{\theta} + \sqrt{R} \cos{(\theta/2)}$, where $\theta \in [\pi/2,\pi)$.  Clearly, $\cos{\theta} < 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$ as:

$$R \int_{\pi/2}^{\pi} d\theta R^{-3/2} e^{R t \cos{\theta}+\sqrt{R} \cos{(\theta/2)}} \le R^{-1/2} \int_0^{\pi/2} d\theta \, e^{-R t \sin{\theta}+\sqrt{R} \sin{\left (\frac{\pi}{4}-\frac{\theta}{2} \right )}} \le \frac{\pi e^{-\sqrt{R}/2}}{2 R^{1/2}(t R - \sqrt{R}/2)} $$

(NB it is assumed in that computation that $R \gt \frac1{4 t^2}$.)

On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$.  This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.

On $C_4$, The integral behaves as

$$i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \epsilon^{-3/2} e^{-i 3 \phi/2} e^{-\epsilon^{1/2} e^{i \phi/2}}e^{\epsilon e^{i \phi} t} = -i 4 \epsilon^{-1/2} +  i 2 \pi + O \left ( \epsilon^{1/2} \right )$$

While this is singular as $\epsilon \to 0$, this divergence will cancel as follows.  On $C_3$, we have $z=e^{i \pi} x$ and the integral there is

$$e^{i \pi} \int_R^{\epsilon} dx \, x^{-3/2} e^{-i 3 \pi/2} e^{-i \sqrt{x}} e^{-x t} $$

On $C_5$, we have $z=e^{-i \pi} x$ and the integral there is

$$e^{-i \pi} \int_R^{\epsilon} dx \, x^{-3/2} e^{i 3 \pi/2} e^{i \sqrt{x}} e^{-x t} $$

Combining the two integrals, we get, integrating by parts and as $R \to \infty$,

$$\begin{align}i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \, \cos{\sqrt{x}} \, e^{-x t} &= i 4 \epsilon^{-1/2} \cos{\sqrt{\epsilon}} \, e^{-\epsilon t} - i 4 \int_{\epsilon}^{\infty} dx \, x^{-1/2} \left (\frac{\sin{\sqrt{x}}}{2 \sqrt{x}} + t \cos{\sqrt{x}} \right ) e^{-x t} \end{align}$$

As stated, the divergences cancel and the contour integral is equal to, as $\epsilon \to 0$:

$$\int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{s t}  - i 4 \int_{\epsilon}^{\infty} dx \, x^{-1/2} \left (\frac{\sin{\sqrt{x}}}{2 \sqrt{x}} + t \cos{\sqrt{x}} \right ) e^{-x t} + i 2 \pi$$

Note that the 2nd integral indeed converges.  By Cauchy's theorem, the contour integral is zero.  Thus, the ILT is

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{s t} = -1 + \frac{2}{\pi} \int_{0}^{\infty} dx \, x^{-1/2} \left (\frac{\sin{\sqrt{x}}}{2 \sqrt{x}} + t \cos{\sqrt{x}} \right ) e^{-x t}$$

The integral on the right is not too bad to evaluate; NB it is easy to show using Parseval's Theorem that

$$\frac{2}{\pi} \int_{0}^{\infty} dx \, x^{-1/2} \frac{\sin{\sqrt{x}}}{2 \sqrt{x}}  e^{-x t} = 2 \operatorname{erf}{\left ( \frac1{2 \sqrt{t}} \right )} $$

$$\frac{2}{\pi} \int_{0}^{\infty} dx \, x^{-1/2} t \cos{\sqrt{x}} e^{-x t} = 2 \sqrt{\frac{t}{\pi}} e^{-\frac1{4 t}} $$

Finally, we have the ILT in closed form:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, s^{-3/2} e^{-\sqrt{s}} e^{s t} = \frac{2}{\sqrt{\pi}} t^{1/2} e^{-\frac1{4 t}} - \operatorname{erfc}{\left ( \frac1{2 \sqrt{t}} \right )} $$


问:
How to represent math iteration and mapping issue
I want to translate a function into math notations.

I have a number X = 1000 and I have a list of numbers A = [200,300,50,50,400,500,700] and list of letters B = [AA,BB,CC,DD,EE,FF,GG] which is map for list A

I iterate over the A and compare if the first element of list A is larger than X, then in second iteration, I compare if the sum of first and second element are larger than X and repeat the process unti the sum of n element is larger than X. I took the number of iteration (index) where the condition was meet and return the corresponding element from list B.

pseudo function

```<code>function F(X,A,B) {
  for index in A; do:
    if sum(A[index0:index]) > X:
      return B[index]
}
</code>```

in this example, F(X,A,B) will return FF

Can someone please help me to put this function in mathematical notations or guide me on where I can learn how to do it.

答:
Using typical conventions like subscripts for the mathematics equivalent of array indexing, you can write the desired value of $\mathtt{index}$, which I'll call $i_1$, as follows:

$$
i_1 = \min_n \left(\sum_{i = i_0}^n A_i > X\right)
$$

(where I've written $i_0$ for your $\mathtt{index0}$ - which I presume is $0$ if you are coding in python). In words the above says $i_0$ is the least $n$ such that the sum of the $A_i$ for $i$ from $i_0$ to $n$ exceeds $X$. The final result is then $B_{i_1}$.

By the way, I think the result is actually $\mathtt{FF}$ in your example, because $200 + 300 + 50 + 50 + 400 = 1000$, so you have to carry on to $n = 5$ to exceed $1000$.

As to where to learn more, Knuth's Art of Computer Programming is the bible, for many people, but there are lots of other good books that may be more approachable. E.g. Aho and Ullman's Data Structures and Algorithms.


问:
Partition and Riemann-Stieltjes integration

Let $f$ and $g$ be bounded on $[a , b]$ and $g \in \mathcal{R}(\alpha)$ on $[a ,b]$. Also $P$ is an arbitrary partition. If $f\le g$ and $\alpha$ is increasing on $[a , b]$ determine whether the following propositions are true or false:

$1.$ $L(P,f,\alpha) \le \int_a^b gd\alpha$

$2.$ $U(P,f,\alpha) \le \int_a^b gd\alpha$

My try: I think the first one is true since $$f\le g \implies \int_a^b fd\alpha \le \int_a^b gd\alpha \\ L(P,f,\alpha)\le \int_a^b fd\alpha \implies L(P,f,\alpha)\le \int_a^b gd\alpha$$
The second proposition is wrong. Let $f=g =x$ and $a = 0,b =2, P = \{0 ,1,2\}$. Then $U(P,f,\alpha) = 3$ and $\int_a^b gd\alpha = 2$. Clearly $3 \not \le 2$, so this is false.

Are my answers correct and precise?

答:
For (1) your approach is not adequate, as it is possible that $f \leqslant g$ but $f \not\in \mathcal{R}(\alpha)$. This occurs when $f$ and $\alpha$ are both discontinuous at a point in $[a,b]$ from the right or left.

A simple argument is

$$L(P,f,\alpha) = \sum_{j=1}^n \inf_{x \in [x_{j-1},x_j]}f(x) \cdot (\alpha(x_j) - \alpha(x_{j-1})) \leqslant \sum_{j=1}^n \inf_{x \in [x_{j-1},x_j]}g(x) \cdot (\alpha(x_j) - \alpha(x_{j-1}))\\ \leqslant \int_a^b g \, d\alpha $$


问:
Showing that $a^{12}\equiv 1\pmod{210}$ if $a$ is coprime to $210$
I ran into this next question:

Show that if $a$ is coprime to $210$:

$a^{12}\equiv1 \pmod{210}$.

This is not a homework question.

Thank you very much in advance,

Yaron

答:
Hint: Since $210$ is the product of the pairwise relatively prime numbers $2$, $3$, $5$, and $7$, it is enough to show that (i) $a^{12}\equiv 1\pmod{2}$, and (ii)  $a^{12}\equiv 1\pmod{3}$, and (iii) $a^{12}\equiv 1\pmod {5}$, and (iv) $a^{12}\equiv 1\pmod{7}$.

Each of these four tasks can be done using Fermat's Theorem.

Each of the tasks can also be done using somewhat less machinery, but then writing out the details takes quite a bit longer.

Detail: We show for example that $a^{12}\equiv 1\pmod{7}$. We have that $a$ and $7$ are coprime. Fermat's Theorem says that if $p$ is prime, and $a$ and $p$ are coprime, then $a^{p-1}\equiv 1\pmod{p}$. Taking $p=7$, we find that $a^6\equiv 1\pmod{7}$. Squaring both sides of this congruence, we find that $a^{12}\equiv 1\pmod{7}$.

The argument for $5$ is almost the same. We have by Fermat's Theorem that $a^{4}\equiv 1\pmod{5}$. Since $4$ divides $12$, it follows that $a^{12}\equiv 1\pmod{5}$. The same idea works for $3$ and for $2$. In the case of $2$ at least, it is silly to use Fermat's Theorem. For $a$ and $2$ coprime just says that $a$ is odd. And if $a$ is odd, so is $a^{12}$, which says that $a^{12}\equiv 1\pmod{2}$.


问:
Showing the expectation of the record of iid random variables is infinity
Take a sequence of iid RVs: $(X_n, n \geq 1)$. I want to show $E[\inf\{n>1:X_n>X_1\}]=\infty$. I know how this would work for geometric processes, but this statement seems to be more general. Also, I think there could be a  way to write it as a markov process like the famous Gambler's ruin problem. But, I want a simpler solution.

I think I should use the fact that ($N = \inf\{n>1:X_n>X_1\}$) $E[N] = \sum_{n\geq1}P(N \geq n)$ , but I'm stuck and don't know how to proceed from here.

答:
Conditional on $X_1=x,$ the expected value of $N$ is $$E(N\mid X_1=x) = 1 + \frac{1}{P(X>x)} $$ since we are simply waiting for an event with probability $P(X>x)$ to happen. Thus, law of total expectation gives $$ E(N) - 1  = \int_{-\infty}^\infty \frac{f_X(x)}{P(X>x)}dx = \int_{-\infty}^\infty \frac{f_X(x)}{1-F_X(x)}dx = \int_0^1\frac{1}{1-F}dF = \infty$$ where $f_X$ is the PDF of $X,$ $F_X$ is the CDF, and we used the variable trasformation $F = F_X(x)$ in the second to last equality.

(On a side note I don't think this result actually requires the PDF to exist, or any assumptions about $X$ at all. So, consider the approach I took as a heuristic stand-in for something more rigorous that I don't know/remember the details of :))

Edit Now that I think about it, the approach you're starting with works. We have $$P(N\ge n+1)= P(X_n,X_{n-1},\ldots, X_2 \le X_1) $$ and by symmetry, $P(X_n,X_{n-1},\ldots, X_2 \le X_1) \ge 1/n,$ so  $\sum_n P(N\ge n)= \infty.$


问:
Step missing in the proof of $\mathbb{R}$ being uncountable.
Today I saw the proof of the uncountability of $\mathbb{R}$, where given  a list of all elements of $\mathbb{R}$, we produce an element not in the list by requiring that its $n$-th digit is different from the $n$-th digit of the $n$-th element of the list. Since it has different decimal expansion from every element in the list, it can't be on the list. I think this last statement is actually wrong since in $\mathbb{R}$ one number can have more than one decimal expansion. So even if the decimal expansion of the newly generated element differs from those of the elements on the list, this doesn't mean that the element is not in the list.

Am I right? and if so, how can the argument be fixed?

答: You are right. The argument can be fixed; it usually is fixed. The only numbers with more than one decimal expansion have exactly two of them; they are the numbers like, say, $1.26$, which is equal to $1.2599999999\ldots$ In order to fix the argument, we choose the $n$th digit such that it is different from the $n$th digit of the $n$th number and also different from $9$. And we also choose the decimal representation of the $n$th number so that it doesn't end with infinitely many $9$'s.


问:
Weaker version of inverse semigroup
Is there a weaker version of inverse semigroup where the existence of reflexive inverse is dropped, only the uniqueness remains.

This was motivated by, and resembles the weaker version of divisible: cancellable -- the former requires both existence and uniqueness, while the latter only requires uniqueness.

More formally, a "weakly inverse semigroup" is a semigroup $S$ such that $\forall a, p, q \in S\ (apa=a \wedge pap=p \wedge aqa=a \wedge qaq=q \rightarrow p=q)$.

This concept will also be a weaker version of cancellative semigroup. (I think)

Is there such a concept being used anywhere? Or if not, is there any reason why not (such as having no interesting property, etc.).

答:
According to your definition, a semigroup is weakly inverse if each of its elements has at most one inverse. Now, an element has at least one inverse if and only if it is regular.
Thus a semigroup is weakly inverse if and only if every regular element has exactly one inverse.

This class of semigroups has been widely studied, at least in the finite case, under the name of block groups, because of the very peculiar structure of their regular $\cal D$-classes, represented in the picture below, where $G$ is a maximal group of the $\cal D$-class and a star indicates the presence of an idempotent in the corresponding $\cal H$-class:

$\hskip 100pt$

Block-groups admit several characterizations in terms of Green relations.

Theorem. Let M be a finite monoid. The following conditions are equivalent:

M is a block group;

for every regular $\cal D$-class $D$ of $M$, $D^0$ is a Brandt semigroup;

for all idempotent $e, f$ of $M$, $e \mathrel{\cal R} f$ implies $e=f$ and $e \mathrel{\cal L} f$ implies $e=f$;

for all idempotent $e, f$ of $M$, $efe=e$ implies $ef = e = fe$;

the submonoid generated by the idempotents of $M$ is $\cal J$-trivial.

Finite block-trivial monoids are closed under taking submonoids, morphic images (i.e. quotient monoids) and finite direct products
and hence form a variety of finite monoids.

If $G$ is a group, then ${\cal P}(G)$, the monoid of subsets of $G$, under the operation defined by
$$XY = \{xy \mid x \in X, y \in Y\}$$
is a block-group.
A (difficult) theorem states that the variety of finite monoids PG generated by the monoids of the form ${\cal P}(G)$, where $G$ is a group, is equal to the variety BG of all finite block-groups. See [1] for a survey and further results.

[1] J.-É. Pin, PG = BG, a success story, in NATO Advanced Study Institute, Semigroups, Formal Languages and Groups, J. Fountain (ed.), Kluwer academic publishers (1995) 33-47.

Title: Weaker version of inverse semigroup