问:
Models between a ground model and its forcing extension
Jech’s book states and proves a standard result about forcing extensions:

Suppose $B$ is a Boolean algebra and let $G$ be $B$-generic over $V$. If $M$ is a model of ZFC such that $V \subset M \subset V [G]$, then there exists a complete subalgebra $D \subset B$ such that $M = V [D \cap G]$.

My aim is to prove an analogue result about partial orders. That is,

Suppose $P$ is a p.o. and let $G$ be $P$-generic over $V$. If $M$ is a model of ZFC such that $V \subset M \subset V [G]$, then there exists a poset $Q\subset P$ such that $M = V [H]$ for some $H$ which is $Q$-generic over $V$.

Kunen uses the c.t.m. approach to introduce forcing but I didn’t find a proof of this fact in his book.

Thank you.

答:
The most straightforward analogue of the result for the Boolean algebra approach can indeed fail in the poset approach. More specifically, we have the following: there is a forcing poset $\mathbb{P}$ and a generic filter $G\subseteq\mathbb{P}$ such that there is some $a\in V[G]$ such that $V(a)$, the smallest model of ZFC containing $V$ and $\{a\}$, is not a forcing extension of the form $V[G\cap\mathbb{Q}]$ where $\mathbb{Q}$ is a complete subposet of $\mathbb{P}$. By $\mathbb{Q}$ being a complete subposet of $\mathbb{P}$ ($\mathbb{Q}\subseteq_c\mathbb{P}$), I mean

$\mathbb{Q}$ is a suborder of $\mathbb{P}$

$\mathbb{P}$ and $\mathbb{Q}$ agree on the incompatibility of finitely many elements

maximal antichains in $\mathbb{Q}$ remain maximal antichains in $\mathbb{P}$

Start with ground model $V$ and use $\mathbb{P}=\text{Col}(\omega,\kappa)$ where $\kappa$ is some cardinal bigger than $|\mathcal{P}(\mathbb{R})|$. This makes the ground model power set of the reals $(\mathcal{P}(\mathbb{R}))^V$ countable in $V[G]$, where $G$ is generic for this forcing. But now we can find a real $a$ in $V[G]$ that is random over $V$ by essentially performing random forcing $V$ (generics exist because from the point of view of $V[G]$, there are only countably many dense sets for the random forcing poset).

I claim that $V(a)$, though itself being a generic extension by the random forcing, is not of the form $V[G\cap\mathbb{Q}]$ where $\mathbb{Q}$ is a complete subposet of $\mathbb{P}$. To see this, suppose towards a contradiction that for some complete subposet $\mathbb{Q}\subseteq_c \mathbb{P}$ and $H=G\cap\mathbb{Q}$ we have $V(a)=V[H]$. Now let $f=\bigcup H$ be the canonical function added by forcing with $H$, so we have $\text{dom}(f)\subseteq\omega$ and $\text{ran}(f)\subseteq \kappa$. But since $V[H]=V(a)$ and the random forcing poset has the ccc, there is some $V$-countable set $S\in V$ that covers the range of $f$ (i.e., $\text{ran}(f)\subseteq S$).

But now $V[H]=V(a)$ can be obtained by the countable forcing $\text{Fn}(\omega,S)$, the set of finite functions from $\omega$ to $S$ ordered by reverse inclusion. Since every countable forcing poset is equivalent (i.e., has the same extension) to the Cohen poset $\text{Fn}(\omega,\omega)$, we can view $V[H]$ as a Cohen forcing extension. Now the contradiction: In a Cohen forcing extension, there is a real that is not dominated by any ground model real, whereas in a random forcing extension, every real is dominated by a ground model real (a proof of this can be found in Jech Lemma 15.30). So $V(a)$ cannot be equal to $V[H]$.

I believe this was left as an exercise in the newer Kunen.


问:
Why are groups more important than semigroups?
This is an open-ended question, as is probably obvious from the title. I understand that it may not be appreciated and I will try not to ask too many such questions. But this one has been bothering me for quite some time and I'm not entirely certain that it's completely worthless, which is why I've decided to ask it here.

Why are groups so immensely more important in mathematics and its applications than semigroups are? I know little of group theory, mathematics and its applications, so I cannot understand how much more important they are. But I know they are because I see how many more people are interested in groups than in semigroups.

I wouldn't ask this question if this fact didn't seem a little strange to me. I know that groups are associated with symmetries, or with automorphisms of structures. Cayley's Theorem tells me that every group can be seen as a set (closed with respect to taking compositions and inverses) of automorphisms of a structure with no constants, functions or relations, i.e. a plain set. I know that the automorphisms of a vector space, a module or a group form a group. Obviously, automorphisms are important.

Then we have inverse semigroups. These are less popular but this I can understand. They are associated with partial symmetries, by the Wagner-Preston Theorem. Partial functions do seem much less used than functions.

But then come semigroups. Just like in the two previous cases, there is an "embedding theorem", which says that every semigroup can be embedded in a semigroup of maps from a certain set into itself. And, analogously to the case of groups, the endomorphisms of a vector space, a module or a group form a semigroup.

This seems to say that semigroups are to endomorphisms what groups are to automorphisms. The "conclusion" would be that

$\frac{\mbox{the importance of semigroups}}{\mbox{the importance of endomorphisms}}=\frac{\mbox{the importance of groups}}{\mbox{the importance of automorphisms}}$

Assuming that endomorphisms are about as important as automorphisms, we get that semigroups are about as important as groups. My feeling is that the assumption is correct.

I understand that I must be oversimplifying something at some point. But what am I oversimplifying and where?

I realize that this is possibly a very dumb question, but my confusion is genuine.

Edit: I have realized, after reading your answers and comments, that I made the mistake of using a very vague term without even attempting to define it. The problem seems to be what importance is. Is it popularity or usefulness, or being used a lot, or interesting or being interesting, or something else, or a mixture of many traits? I'm not going to force my understanding of the word now. But if someone still wants to answer this question, perhaps it would be good idea if the answer contained an attempt at clearing this up. (Or maybe not. I'm not sure!)

Edit: A question similar to mine is dealt with here.

答:
From my point of view, the "most natural" structure is to be found in the middle of the two structures, namely the one of monoid. A monoid is just a set with an associative operation, and an identity with respect to this operation. Examples are: groups, rings with respect to multiplication, endomorphisms of an object..

The reasons why I think group theory is so successful are several:

Its theory has deep roots in Galois theory, which has shown some very advanced mathematics for its time;

It is at the midpoint of a wide variety of approaches and applications: algebraic groups, invariant theory, classifying groups in algebraic topology, K-theory in homological algebra, representation theory in linear algebra and lot more, differential geometry with Lie groups, connections with (semisimple) Lie algebras, combinatorics in finite group theory, graph theory in the study of free geoups... It's really incredible.

All such quoted theories are surprisingly well understood and shows rich but not impossible structures.

However - there is always an however in good stories - sometimes it's us human that needs additional hypothesis to better study and understand things. That's why, also in the case of finite groups, many people for example will concentrate on p groups, and so on... Maybe, that's why we concentrate on groups and not on general monoids, isn't it? But let's take a step back. What we are doing? What we want from an algebraic structure? Where do they generate?

Let me introduce you the concept of Operad. An Operad is a collection of sets $O(n) $, that you should think as corollas with $n$ leaves directly connected to a root below. Such operad comes equipped with a way of composing such corollas

$$\theta:  O(n_1) \times \ldots \times O(n_k) \times O(k) \to O(n_1 + \ldots + n_k) $$

Here you have to think that you are "plugging in" the k corollas with respectively $n_1, \ldots, n_k$ leaves into a corolla of k leaves, to get a corolla of $n_1+\ldots+n_k$ leaves. This "composition of corollas", when you plug in three layers, should be somehow associative. Doesn't matter however.

Let me give you an example of an operad. This is very important. Take an object X in a monoids category. If you are not comfortable with this, you can imagine this is a set, a vector space, a topological space... The corresponding notion of endomorphism will vary as you imagine. Monoidal here roughly means that you can do the direct product, and let's assume that it is a coincidence with our monoids for the moment.

Define the endomorphism operad $End_X$ as having $End_X(n) = Hom(X^n, X) $, and the composition of corollas is given by

$$ \theta(f_1, \ldots, f_k, g) (x_1, \ldots, x_k) = g( f_1(x_1), \ldots, f_k(x_k)) $$

Where $f_i: X^{n_i}\to X$, $x_i \in X ^{n_i}, g: X^k \to X$. This is a bit difficult to digest, but you can imagine you have plugged in your inputs $x_i$ in the corollas (which are the $f_i$ and $g$) and let them flow to the root.

Let me give you another two examples. Define $Ass$ to be the operad that has in degree $n$ the symmetric group $S_n$ and the composition is given in a weird way, but you will understand it in a moment. Note that $S_n$ is in bijective correspondence with the possible ways of ordering $n$ variables in a product. Define $Comm$ to have just one element in any degree.

A last boring definition. A morphism of operads $O \to P$ is a degreewise map $O(n) \to P(n) $ that commutes with corollas composition. Suppose now you have a morphism of operad $Comm\to End_X$ for some $X$. What does this stands for? Well, for example, the image of the unique element in $Comm(0)$ defines an element in $e \in Hom(X^0,X)\simeq X$. It also chooses an element $\mu \in Hom(X^2, X)$, by taking the image of the only element in $Comm(2) $. If you check carefully the commutativity relations of an operad morphism, you will discover that this $\mu$ should define a commutative and associative product on $X$, and that $e$ should be the neutral element wrt $\mu$. Conversely, a commutative and associative structure on $X$ with a neutral element defines such a morphism. Cool, isn't it? You can do the same trick with $Ass$, and you will discover that he is responsible for the creation of associative products.

In analogous ways, you can produce a wide variety of operads that encode several algebraic structures. Monoids, for examples, are encoded by $Ass$. There also Lie Algebras, Rings, Hopf Algebras, Little disk Algebras, Gerstenhaber Algebras, Poisson Algebras... There is also a great feature of this way of crafting algebraic structures: relatioms that holds with an equal can be relaxed to hold with an "homotopic to", and all homotopies appearing in this way can be organized into an higher structure. Infinity category deals with formalizing this stuff. That's a bit of an " adult" observation, but I swear you it can be of concrete importance (it is for example in my present work).

But... Operads can't encode groups! You can just say that are special kind of monoids, but there isn't any operad "Grps" encoding the axioms of a group. This is not my main argument, though; the main thing is that all the construction we made relies upon monoidal categories, which are fundamental to all of the factory of algebrsic structures. And yes, a symmetric monoidal category is in first approximation exactly a monoid in which we use categories to encode the axioms: is a category $C$ with a functor $ \mu: C \times C \to C$ that is associative, and with a "neutral element", i.e. an element $1 \in C$ such that $\mu(X, 1) \simeq X \simeq \mu(1, X) $. There is some relatioms that this isomorphisms should satisfy (penthagon relations I think), but this is a technicality). In Lurie treatment of Infinity categories, which is at the boundary of algebra and topology, one cannot live without symmetric monoidal categories (but could live without group theory).

Sorry for the lengthy answer. Please comment and let me know what you think! I am myself doubtful about this monoid-groups battle... I think I'll opt for polygamy in the end :)

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Complex Integration Question : Integrate $f(z) = x^2+iy^2$ when $z(t) = t^2+it^2$ and $0 \le t \le 1$
I'm trying to integrate this as a definite integral.
$f(z) = x^2+iy^2$ when $z(t) = t^2+it^2$ and $0 \le t \le 1$.

I know that:

$$\int_{z(a)}^{z(b) }f[z(t)]z'(t) \,dt= \int_C f(z) dz$$

I'm confused on how to parametrize the $x$ and $y$ terms in the $f(z)$ function and complete the integration:

$$\int_0^{(1+it)} f[z(t)]z'(t) \,dt= \int_C x^2+iy^2 dz$$

EDIT:

So this is what I have so far. From, $z(t) = t^2+it^2$ and $0 \le t \le 1$,

$x(t) = t^2$ an $y(t) = t^2$

I think we need to substitute these values in for $x^2+iy^2$

So we would have
$$f(z(t)) = [(t^2)^2+i(t^2)^2] $$
$$z'(t) = 4t^3+ 4it^3$$
$$\int_C x^2+iy^2 = \int_0^1 f(z(t))(z'(t))dt = \int_0^1 [(t^2)^2+i(t^2)^2][4t^3+ 4it^3]dt $$

Am I on the right track?

答:
Probably the confusing thing is the definition of $f$, it is the map $x+iy\mapsto x^2+iy^2$, where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$, then you have that $(f\circ z)(t)=(t^2)^2+i(t^2)^2=t^4(1+i)$, and
$$
dz(t)=d(t^2+it^2)=2tdt+2itdt=2t(1+i)dt
$$
Then, assuming that $C$ is the curve parametrized by $z$, we have that
$$
\int_{C}f(z)dz=\int_{0}^1 (f\circ z)(t)dz(t)=\int_0^{1}2t^5(1+i)^2dt=(1+i)^2\cdot \frac1{3}=i\frac2{3}
$$


问:
Finding the limit of a function?
The limit is actually easy: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t}$

One can use hopitals rule and say that ultimately the upper function will be reduced to a constant while the lower function will remain the same. Hence, the limit is zero. I was wondering if there is another way or a more mathematical representation of the answer.

答:
Alternatively, we can proceed by induction on $k\in \mathbb{N}$.

Base Case: For $k=0$, we need only apply L'Hôpital's rule once: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t}{e^t} = \lim \limits_{t\to\infty}\dfrac{1}{e^t} = 0$.

Inductive Hypothesis: Assume that $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t} =0$ holds for $k=r$.

It remains to prove that the equation holds for $k=r+1$. Then by one more application of L'Hôpital's rule and the induction hypothesis, observe that:
$$\lim_{t\to\infty}{\frac{t^{r+2}}{e^t}}=\lim_{t\to\infty}{\frac{(r+2)t^{r+1}}{e^t}}=(r+2)\lim_{t\to\infty}{\frac{t^{r+1}}{e^t}}=(r+2)(0)=0$$
as desired.

Title: Finding the limit of a function?


问:
Calculate $\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$, whose indefinite integral is of non-elementary function

Calculate $$\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$$

I know an approach is using the change of variables $t=-u$, so we have the value for the integral $\frac{1}{3}$. How can I calculate that integral using other approach?

I thought about this  problem when observing this indefinite integral indefinite integral.

Body:
Let
$$I=\int_{-1}^{1}\frac{t^{2}}{1+e^t}dt=\int_{-1}^{0}\frac{t^{2}}{1+e^t}dt+\int_{0}^{1}\frac{t^{2}}{1+e^t}dt$$$$. $$
Under $t\to-t$,
$$\int_{-1}^{0}\frac{t^{2}}{1+e^t}dt=\int_0^1\frac{t^2}{1+e^{-t}}dt=\int_0^1\frac{t^2e^t}{1+et^t}dt$$
and hence
$$ I=\int_{-1}^{0}\frac{t^{2}}{1+e^t}dt+\int_{0}^{1}\frac{t^{2}}{1+e^t}dt=\int_0^1\frac{t^2+t^2e^t}{1+t^2}dt=\int_0^1t^2dt=\frac13. $$


问:
Find curve at the intersection of two level surfaces.
Let C be the curve at the intersection of two level surfaces M(x, y, z) = 5 and N(x,y,z) = 0, passing through a point P (1,1,1). Let $M(x, y, z) = 2x^2 + y^2 + 2z^2$ and $N(x, y, z) = xy-z$. Find curve C in the parametric form <x(t), y(t), z(t)>.

I am quite stuck on this one. I could let $z = xy$ and then plugging that into M I'd have $2x^2 + y^2 + 2x^2y^2 = 5$, and I have no idea what would be good ways to parametrize that.

Another way is to let $y=z/x$, in which case I end up with $2x^4 + (2z^2-5)x^2+z^2 = 0$, but again, I'm quite stuck. Any help would be really appreciated!

Or, if it is impossible to find the curve C, the question I'm working on actually asks for the tangent line to curve C at P if x'(0) = 3. If this makes things easier, how exactly is using this tangent more useful?

答:
In the equation $2x^2 + y^2 + 2 x^2 y^2 = 5$  put $x = r \cos t$, $y = \sqrt{2} \,r \sin t$, get $r = \frac{\sqrt{5} }{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } }$, and  the parametrization
$$\left( \frac{\sqrt{5} \cos t}{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } } , \frac{\sqrt{10}\sin t}{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } } \right)$$

Note that $t= \tan^{-1} (\frac{1}{\sqrt{2}})$ maps to $(1,1)$.


问:
Stochastic integral, different way?
is there a faster/easier way to compute $\int_a^bW(t)dW(t)$, $a<b$ where W(t) is the wiener process?

My method is the following:

Define $dX(t)=0dt+1dW(t)$ ($X(t)=W(t)$) and apply ito to $f(x)=x^2$.

$dW^2(t)=dt+2W(t)dW(t)$

$W(t)dW(t)=\frac12[dW^2(t)-dt]$

$\int_a^bW(t)dW(t)=\int_a^b\frac12[dW^2(t)-dt]=\frac12[W^2(b)-W^2(a)-(a-b)]$

答:
This may be faster, it's likely about the same though. Stochastic integration by parts looks like
$$\int_a^b f(t)g'(W_t)dW_t = f(t)g(W_t)|_a^b - \int_a^b f'(t)g(W_t)dt - \frac{1}{2}\int_a^b f(t)g''(W_t)dt,$$

where $f(t)$ is a deterministic function and $g(W_t)$ is a function of a Wiener process. For the integral you want we can let $f(t) = 1$ and we get

$$\int_a^b g'(W_t)dW_t = g(W_t)|_a^b - \frac{1}{2}\int_a^b g''(W_t)dt.$$

So for the integral $\int_a^b W_tdW_t$ we can let $g(x) = \frac{x^2}{2}$ and this gives

$$\int_a^b W_tdW_t = \frac{W_t^2}{2}|_a^b - \frac{1}{2}\int_a^b dt.$$

Title: Stochastic integral, different way?


问:
Principal value of complex exponential
Here you can see that Cauchy's principle value ($PV$) of the complex exponential is zero, i.e.,

$$
PV\left[\int_{-\infty}^{+\infty} e^{iax}dx\right] = 0
\tag1 $$

I'm trying to prove this a follows:

$$
PV\left[\int_{-\infty}^{+\infty} e^{iax}dx\right] = \lim_{T\rightarrow +\infty}\int_{-T}^{+T} e^{iax}dx = \frac{1}{a} \lim_{T\rightarrow +\infty} \left[ \sin(T) - \sin(-T)\right] \tag2
$$

In principle, one would write $\sin(T) - \sin(-T) = 2\sin(T)$ and would realize that the limit does not exist so the only thing I can try to prove the result in Eq. (1) is to work out both sines separatedly and assume that

$$
\lim_{T\rightarrow +\infty} \sin(T) = \lim_{T\rightarrow +\infty}\sin(-T) \tag3
$$

so finally you arrive to Eq. (1). Nevertheless, this is not very convincing from my point of view because  RHS limit is undetermined as well as LHS, so you should not assume Eq. (3) is right. Then, what's the right approach?

答:
Note that

$$e^{iax}=\cos(ax)+i\sin(ax)$$

So, the integral is equal to

$$I=2\int_0^\infty \cos ax\, dx + i \int_0^\infty \sin ax\, dx - i \int_0^\infty \sin ax\, dx$$

All three integrals are Cesaro-summable:

$$\int_0^\infty \cos ax\, dx = 0$$

$$\int_0^\infty \sin ax\, dx = 1$$

So, we have

$$I=2\cdot0+i\cdot1-i\cdot1=0$$

Title: Principal value of complex exponential


问:
How do you prove that the recurrence $T(n) = 2T(\lfloor{n/2}\rfloor) + n$ is $\Omega(n \lg n)$?
I'm working through CLRS (Introduction to Algorithms, 3rd edition) and I'm in the section about using the "substitution method" and induction to prove asymptotic bounds for recurrences. I'm trying to solve this problem (4.3-3) and am running into problems dealing with the floor function.

Show that $T(n) = 2T(\lfloor{n/2}\rfloor) + n$ is $\Omega(n \lg n)$

This was my original attempt for the substitution step, and I also see this published online in a non-official solution guide:

$$
\begin{align}
T(n) &\ge 2c\lfloor n/2 \rfloor \lg (\lfloor n/2 \rfloor) + n\\
&\ge cn \lg (n/2) + n\\
&= cn \lg n - c n \lg 2 + n\\
&= cn \lg n \quad \text{if} \space c \le 1
\end{align}
$$

But removing the floor functions directly between lines 1 and 2 seems like it could invalidate the inequality. I found another solution guide that gets around this by substituting $n/4$ for $\lfloor n/2 \rfloor$ but ends with this, which also seems to invalidate the inequality:

$$
\begin{align}
T(n) &\ge c(n/2)\lg n - cn + n\\
&\ge cn \lg n \quad \text{if} \space c \le 1
\end{align}
$$
This also doesn't seem to hold, for example, if $c = 1$. Are these approaches both just wrong or am I missing something? What's the best way to deal with the floor division in this problem?

答:
Yes, the two solutions are not so strict as you found. I'll give my solution:
$$
\begin{aligned}
 T\left( n \right) &\ge 2c\lfloor n/2 \rfloor \lg \left( \lfloor n/2 \rfloor \right) +n\\
 &\ge c\left( n-2 \right) \lg \left( \left( n-2 \right) /2 \right) +n\\
 &=cn\lg \left( \left( n-2 \right) /2 \right) -2c\lg \left( \left( n-2 \right) /2 \right) +n\\
 &=cn\lg \left( n-2 \right) -2cn\lg 2-2c\lg \left( n-2 \right) +2c\lg 2+n\\
 &=c\left( n-2 \right) \lg \left( n-2 \right) +2\lg \left( n-2 \right) -2cn\lg 2-2c\lg \left( n-2 \right) +2c\lg 2+n\\
&\geq c\left( n-2 \right) \lg \left( n-2 \right)\\
&=\Omega(n \lg n)
\end{aligned}
$$
Since
$$
\underset{n\to \infty }{\text{lim}}\frac{c (n-2) \lg (n-2)}{n \lg n} = c>0
$$

$c\left( n-2 \right) \lg \left( n-2 \right)=\Omega(n \lg n)$ is true from the definition of $\Omega$:
$$
\text{if }{\displaystyle \liminf _{n\to \infty }{\frac {f(n)}{g(n)}}>0}, \text{then } f(n)=\Omega (g(n))
$$


问:
Fundamental Solution of $\Delta^{2}$ in $\mathbb{R^{2}}$
we are to show that the fundamental solution of $\Delta^{2}$ in $\mathbb{R^{2}}$ is given by :
$$
V(x_{1},x_{2})= \frac{1}{8\pi}r^{2}\ln(r),\qquad r=|x-\xi|,
$$
and write the solution of $\Delta^{2}w=F(x_{1},x_{2})$.

My approach: Let $x=\left(x_{1}, x_{2}\right) .$ We have :
$$
V(x)=\frac{1}{8 \pi} r^{2} \log (r)
$$
In polar coordinates: (here, $V_{\theta \theta}=0$ )
$$
\begin{aligned}
\Delta V &=\frac{1}{r}\left(r V_{r}\right)_{r}=\frac{1}{r}\left(r\left(\frac{1}{8 \pi} r^{2} \log (r)\right)_{r}\right)_{r}=\frac{1}{8 \pi} \frac{1}{r}(r(2 r \log (r)+r))_{r} \\
&=\frac{1}{8 \pi} \frac{1}{r}\left(2 r^{2} \log (r)+r^{2}\right)_{r}=\frac{1}{8 \pi} \frac{1}{r}(4 r+4 r \log r) \\
&=\frac{1}{2 \pi}(1+\log r)
\end{aligned}
$$
The fundamental solution $V(x)$ for $\Delta^{2}$ is the distribution satisfying: $\triangle^{2} V(r)=\delta(r)$
$$
\begin{aligned}
\Delta^{2} V &=\Delta(\Delta V)=\Delta\left(\frac{1}{2 \pi}(1+\log r)\right)=\frac{1}{2 \pi} \Delta(1+\log r)=\frac{1}{2 \pi} \frac{1}{r}\left(r(1+\log r)_{r}\right)_{r} \\
&=\frac{1}{2 \pi} \frac{1}{r}\left(r \frac{1}{r}\right)_{r}=\frac{1}{2 \pi} \frac{1}{r}(1)_{r}=0 \quad \text { for } r \neq 0
\end{aligned}
$$
Therefore, we have that $\Delta^{2}V(r)=\delta(r)$. Hence, $V$ is the fundamental solution and the solution of:
$$
\Delta^{2} \omega=F(x)
$$
if given by:
$$
\omega(x)=\int_{\mathbb{R}^{2}} V(x-y) \Delta^{2} \omega(y) d y=\frac{1}{8 \pi} \int_{\mathbb{R}^{2}}|x-y|^{2} \log |x-y| F(y) d y
$$

Is this proof correct? if so, how could we establish an even more rigorous proof, since this proof is not rigorous enough?

答:
For a more rigorous proof (as you wish), we have that for $v \in C_{0}^{\infty}\left(\mathbb{R}^{n}\right),$ we want to show
$$
\int_{\mathbb{R}^{n}} K(|x|) \Delta^{2} v(x) d x=v(0)
$$
Suppose $v(x) \equiv 0$ for $|x| \geq R$ and let $\Omega=B_{R}(0) ;$ for small $\epsilon>0$ let
$$
\Omega_{\epsilon}=\Omega-B_{\epsilon}(0)
$$
$K(|x|)$ is biharmonic $\left(\Delta^{2} K(|x|)=0\right)$ in $\Omega_{\epsilon} .$ Consider Green's identity $\left(\partial \Omega_{\epsilon}=\partial \Omega \cup\right.$
$\left.\partial B_{\epsilon}(0)\right)$
$$
\begin{aligned}
\int_{\Omega_{\epsilon}} K(|x|) \Delta^{2} v d x &=\underbrace{\int_{\partial \Omega}\left(K(|x|) \frac{\partial \Delta v}{\partial n}-v \frac{\partial \Delta K(|x|)}{\partial n}\right) d s+\int_{\partial \Omega}\left(\Delta K(|x|) \frac{\partial v}{\partial n}-\Delta v \frac{\partial K(|x|)}{\partial n}\right) d s}_{\displaystyle=0,\;\;\;\text{since $v\equiv0$ for $x\geq R$}} \\
&+\int_{\partial B_{\epsilon}(0)}\left(K(|x|) \frac{\partial \Delta v}{\partial n}-v \frac{\partial \Delta K(|x|)}{\partial n}\right) d s+\int_{\partial B_{\epsilon}(0)}\left(\Delta K(|x|) \frac{\partial v}{\partial n}-\Delta v \frac{\partial K(|x|)}{\partial n}\right) d s \\
\lim _{\epsilon \rightarrow 0}\left[\int_{\Omega_{\epsilon}} K(|x|) \Delta^{2} v d x\right] &=\int_{\Omega} K(|x|) \Delta v^{2} d x
\end{aligned}
$$
On $\partial B_{\epsilon}(0), \quad K(|x|)=K(\epsilon) .$ Thus :\begin{aligned}
\left|\int_{\partial B_{\varepsilon}(0)} K(|x|) \frac{\partial \Delta v}{\partial n} d S\right| &=|K(\epsilon)| \int_{\partial B_{\epsilon}(0)}\left|\frac{\partial \Delta v}{\partial n}\right| d S \leq|K(\epsilon)| \omega_{n} \epsilon \max _{x \in \bar{\Omega}}|\nabla(\Delta v)| \\
&=\left|\frac{1}{8 \pi} \epsilon^{2} \log (\epsilon)\right| \omega_{n} \epsilon \max _{x \in \bar{\Omega}}|\nabla(\Delta v)| \rightarrow 0, \quad \text { as } \epsilon \rightarrow 0 \\
\int_{\partial B_{\epsilon}(0)} v(x) \frac{\partial \Delta K(|x|)}{\partial n} d S=& \int_{\partial B_{\epsilon}(0)}-\frac{1}{2 \pi \epsilon} v(x) d S \\
&=\int_{\partial B_{\epsilon}(0)}-\frac{1}{2 \pi \epsilon} v(0) d S+\int_{\partial B_{\epsilon}(0)}-\frac{1}{2 \pi t}[v(x)-v(0)] d S \\
&=-\frac{1}{2 \pi \epsilon} v(0) 2 \pi \epsilon-\underbrace{\max _{x \in \partial B_{\epsilon}(0)}|v(x)-v(0)|}_{\rightarrow 0,(v \text { is continuous})}=-v(0) .
\end{aligned}
\begin{aligned}
\left|\int_{\partial B_{\epsilon}(0)} \Delta K(|x|) \frac{\partial v}{\partial n} d S\right|=&|\Delta K(\epsilon)| \int_{\partial B_{\epsilon}(0)}\left|\frac{\partial v}{\partial n}\right| d S \leq\left|\frac{1}{2 \pi}(1+\log \epsilon)\right| 2 \pi \epsilon \max _{x \in \bar{\Omega}}|\nabla v| \rightarrow 0, \quad \text { as } \epsilon \rightarrow 0 \\
\int_{\partial B_{\epsilon}(0)} \Delta v \frac{\partial K(|x|)}{\partial n} d S &=\int_{\partial B_{\epsilon}(0)}\left(-\frac{1}{4 \pi} \epsilon \log \epsilon-\frac{1}{8 \pi} \epsilon\right) \Delta v(x) d S \\
& \leq \frac{\epsilon}{4 \pi}\left|\log \epsilon+\frac{1}{2}\right| \cdot 2 \pi \epsilon \max _{x \in \partial B_{\epsilon}(0)}|\Delta v| \rightarrow 0, \text { as } \epsilon \rightarrow 0
\end{aligned}
$$
\therefore \quad \int_{\Omega} K(|x|) \Delta^{2} v d x=\lim _{\epsilon \rightarrow 0} \int_{\Omega_{\epsilon}} K(|x|) \Delta^{2} v d x=v(0) $$

[Note] : To prevent any misunderstanding, note that in $K(|x|)=K(c)$, for $|x|=\epsilon$, we have :
$$
\begin{aligned}
K(|x|) &=K(\epsilon)=\frac{1}{8 \pi} \epsilon^{2} \log \epsilon, & \Delta K=\frac{1}{2 \pi}(1+\log \epsilon) \\
\frac{\partial K(|x|)}{\partial n} &=-\frac{\partial K(\epsilon)}{\partial r}=-\frac{1}{4 \pi} \epsilon \log \epsilon-\frac{1}{8 \pi} \epsilon, & \frac{\partial \Delta K}{\partial n}=-\frac{\partial \Delta K}{\partial r}=-\frac{1}{2 \pi \epsilon}
\end{aligned}
$$


问:
how to show that $f(x)$ can be expressed uniquely as follows: $f(x)=\prod_{i=1}^k[f_i(x)]^{n_i}$
Let $f(x)\in F[x],~F$ being a field, be monic. Then how to show that $f(x)$ can be expressed uniquely as follows:

$$f(x)=\prod_{i=1}^k[f_i(x)]^{n_i}$$

for some monic irreducible polynomial $f_i(x)\in F[x]$ and $k\in \mathbb N~(n_i$'s are positive integers)

答: If $F$ is a field, $F[X]$ is a Principal Ideal Domain (take an ideal $I\neq 0$ of $F[x]$ and a polynomial of minimal degree in it, then conclude by Euclide's algorithm). Since any PID is also a Unique Factorization Domain (a proof of this fact can be found in most of abstract algebra's book, such as in Jacobson's Basic Algebra I), you're done. </p>


问:
$Aut(G) = \{id\}$ iff $|G| \leq 2$.
I'm having problems with the first implication. I already know how to prove that  $|G| \leq 2$ implies $Aut(G) = \{id\}$. But how can I prove the converse? I found it kind of simple but also a little hard to write. I did this:

Because $Aut(G) = \{id\}$ then there is only one way to rearrange all the elements, and by definition of $Aut(G)$, then $|G| \leq 2$, because you have to fix the neuter and the other element should be fixed if $|G| = 2$.

But I'm not very convinced myself. Anny help would be appreciated.

答: Note that $G/Z(G) \cong Inn(G)\leq Aut(G) = \{id\}$, so $G$ must be abelian. In an abelian group, you always have the autmorphism sending each element to its inverse. In order for this to be the trivial automorphism, each element must equal its inverse, so $G$ has only elements of order $2$, i.e. it is isomorphic to some number of copies of $\mathbb{Z}/2\mathbb{Z}$. If there is more than one copy, you can construct a nontrivial isomorphism which swaps copies of $\mathbb{Z}/2\mathbb{Z}$, which leaves two possibilities, either $G \cong \mathbb{Z}/2\mathbb{Z}$ or $G \cong \{id\}$.

Title: $Aut(G) = \{id\}$ iff $|G| \leq 2$.


问:
Prove "If $\{R, Q\} \vDash P$, then $¬P \vdash \lnot Q \lor \lnot R$".
Using the book Dirk van Dalen. "Logic and Structure (Universitext)" as reference text.

Decide whether the following statement is true or false, justifying the answer.

If $\{R, Q\} \vDash P$, then $¬P \vdash \lnot Q \lor \lnot R$.

I claim this statement is true.

Proof:

Assume $[[R]]_v=[[Q]]_v=1 \Rightarrow [[P]]_v=1$.

Let v an arbitrary valuation.

By completeness theorem,

$¬P \vDash \lnot Q \lor \lnot R \Rightarrow ¬P \vdash \lnot Q \lor \lnot R$

So, we can now try to prove $¬P \vDash \lnot Q \lor \lnot R$.

Assume $[[\lnot P]]_v=1$ (*).

Assume $\lnot ([[\lnot Q \lor \lnot R]]_v = 1)$

Assume $[[R]]_v=[[Q]]_v=1$

$[[P]]_v=1$

$[[P]]_v=0$ (From (*))

$\bot$

$\lnot([[R]]_v=[[Q]]_v=1)$

$[[\lnot Q]]_v=1 \lor [[\lnot R]]_v=1$

$[[\lnot Q \lor \lnot R]]_v = 1$

$\bot$

$[[\lnot Q \lor \lnot R]]_v = 1$

$[[¬P]]_v=1 \Rightarrow [[\lnot Q \lor \lnot R]]_v=1$

$¬P \vDash \lnot Q \lor \lnot R$

I will appreciate if someone could give some insight into the general structure of the proof and I would like to know whether my attempt is correct.

答:
The basic proof plan is correct ... but it is a little werd to use propositional logic notation for a meta-proof.

For example, you say things like

$\lnot([[R]]_v=[[Q]]_v=1)$

and

$[[\lnot Q ]]_v =1 \lor [[\lnot R]]_v = 1$

so now you are using propositional logic operators to make your meta-logical claims, which is very confusing.

So, I would suggest to just stick with English, just as you would with any mathematical proof, and instead say:

it is not the case that both $[[R]]_v =1$ and $[[Q]]_v =1$

and

either $[[\neg Q]]_v =1$ or $[[\neg R]]_v =1$


答: Let $x\in G$, then $y\mapsto xyx^{-1}$ is an automorphism of $G$ and thus is trivial. Therefore for all $y\in G$, $xyx^{-1}=y$ and $G$ is commutative. We can finally consider the automorphism $x\mapsto -x$, it is trivial and therefore $2x=0$ for all $x\in G$ and $G$ is a $\mathbb{Z}/2\mathbb{Z}$-vector field, thus there exists $I$ such that $G\simeq(\mathbb{Z}/2\mathbb{Z})^I$. If ${\rm Card}(I)\geqslant 2$, then ${\rm Aut}(G)\neq\{{\rm Id}\}$ because we can swap two distinct elements of $G$ and this gives a non-trivial automorphism, therefore ${\rm Card}(I)\leqslant 1$ and $o(G)\leqslant 2$.

Title: $Aut(G) = \{id\}$ iff $|G| \leq 2$.

Body:
I'm having problems with the first implication. I already know how to prove that  $|G| \leq 2$ implies $Aut(G) = \{id\}$. But how can I prove the converse? I found it kind of simple but also a little hard to write. I did this:

Because $Aut(G) = \{id\}$ then there is only one way to rearrange all the elements, and by definition of $Aut(G)$, then $|G| \leq 2$, because you have to fix the neuter and the other element should be fixed if $|G| = 2$.

But I'm not very convinced myself. Anny help would be appreciated.


问:
Does $1^0$, as a limiting form of the $n$ $n$-th roots of unity $1^{1/n}$, represent every complex value on the unit circle?
In the context of algebraic numbers, the $n$-th root has $n$ solutions in the complex plane. The equations look like: $1=x^n$, which gives $1^{1/n}=x$ (which can be rotated around the complex plane to generate all the solutions).  And then, if we let $n$ become large, we get many solutions. But then, when we let $n$ become arbitrarily large, we get $1^0 = x$, which only has one solution, namely $1$, because anything to the zeroth power gives $1$. This seems discontinuous to me, and it confuses me.

Does anyone know what I'm missing?

I have a feeling that perhaps $1^0$ actually is every value around the complex unit circle, which would be the natural extension of the limiting process. But, if that's true, there are tons of implications that don't seem right to me.

Maybe the following two facts are useful? I'm just spitballling on these two.

This reminds me of the fact that $a*0=0$ has infinitely many solutions.

And perhaps the following fact is useful: $2^0 = 2 ^ {100 * 0} =  (2 ^ {100})  ^ 0$

答:
I think I might have a solution to the problem, and I think the solution is that yes, the solution set to $z^0 = 1$ is $\{(x,y) | x^2 + y^2 = 1\}$

First of all, we can start with the following: $z = 1^n$, which gives $z^{1/n} = 1$.  We know that the solution set to this problem starts with one point at $(1,0)$, and then it has $n-1$ points evenly distributed throughout the rest of the unit circle.  We will call this set $S_n$

Now we take the limit as n approaches infinity.  In other words, we want $\lim_{n\to \infty} S_n$.  To get this, we are going to use the Hausdorff metric (has a wikipedia page) on the subsets of the unit circle, then we will show that we can make this metric arbitrarily small as $n\to \infty$.  The Hausdorff metric is actually a pseudometric, but it can be treated as a metric if we treat the space as the quotient space of everything which is equal under the pseudometric.  What this means is that because the unit circle of real numbers and the unit circle of algebraic numbers are indistinguishable under this pseudometric, they are the same element of the quotient space. And you can see that as $n\to \infty$, the hausdorff metric goes to zero, and thus the solution to our problem is indeed this element of the quotient space.  So this answers the question that the limit is equal to the unit circle, but we cannot tell whether it is the real unit circle or the algebraic number unit circle.


问:
Prove that the signs of a symmetric random vector are i.i.d. coin flips

Let $X=(X_1,\ldots,X_n)$ be a symmetric random vector, where symmetric means that
$$(X_1,\ldots,X_n)\overset{\mathrm{d}}{=}(a_1 X_1,\ldots,a_n X_n),$$
for all $a\in\{\pm1\}^n$. Assume that for each $j$ we have $X_j\neq0$ a.s.

Show that, conditional on $(|X_1|,\ldots,|X_n|)$, the variables $$\mathrm{sign}(X_1),\ldots,\mathrm{sign}(X_n)$$
are i.i.d. Rademacher distributed (i.e. they are independent coin flips).

Remember that: $$Z\sim\mathrm{Rademacher}\iff\mathbb{P}(Z=1)=\mathbb{P}(Z=-1)=1/2$$
I proved that $\mathrm{sign}(X_1),\ldots,\mathrm{sign}(X_n)$ are i.i.d. rademacher distributed, so it remains to show that they are independent of $(|X_1|,\ldots,|X_n|)$.

This is how I proved the first part:

Let $A=(A_1,\ldots,A_n)$ be a vector of i.i.d Rademacher variables such that $A$ is independent of $X$, $|X|$ and $ \mathrm{sign}(X)$.

I am using the notations $|X|=(|X_1|,\ldots,|X_n|)$ and $\mathrm{sign}(X)=(\mathrm{sign}(X_1),\ldots,\mathrm{sign}(X_n))$.

The symmetry property above implies that $X\overset{\mathrm{d}}{=}A\odot X$ (here $\odot$ stands for componentwise multiplication).

Simplifying $|X|$ both sides we get $$\mathrm{sign}(X)\overset{\mathrm{d}}{=}A\odot\mathrm{sign}(X).$$ Let $B=A\odot\mathrm{sign}(X)$. I showed that vector $B$ is i.i.d. rademacher distributed, thus $\mathrm{sign}(X)$ is i.i.d. rademacher distributed too. Infact, for all $b\in\{\pm1\}^n$ we have
$$\mathbb{P}(B=b)=\sum_{s\in\{\pm1\}^n}\mathbb{P}(\mathrm{sign}(X)=s)\,\mathbb{P}(A\odot s=b\mid \mathrm{sign}(X)=s)\\
=\sum_{s\in\{\pm1\}^n}\mathbb{P}(\mathrm{sign}(X)=s)\,\mathbb{P}(A\odot s=b)\\=\sum_{s\in\{\pm1\}^n}\mathbb{P}(\mathrm{sign}(X)=s)\,\mathbb{P}(A=b\odot s)=\sum_{s\in\{\pm1\}^n}\mathbb{P}(\mathrm{sign}(X)=s)\Bigl(\frac{1}{2}\Bigr)^n=\Bigl(\frac{1}{2}\Bigr)^n$$

It remains to prove that $\mathrm{sign}(X)$ is independent of $|X|$. I showed that $B$ is independent of $|X|$, but here I am stuck.

Thank you all!

答:
To prove sign($X$) is independent of $|X|$, it's enough to show
$$P(\operatorname{sign}(X)=a, |X|\le c)=P(\operatorname{sign}(X)=a)P(|X|\le c)\tag1$$
for every $a\in\{\pm 1\}^n$ and every $c:=(c_1,\ldots,c_n)\ge 0$, where the notation $\{|X|\le c\}$ stands for the event
$$\{|X_1|\le c_1, |X_2|\le c_2, \ldots, |X_n|\le c_n\}.$$
To do this, argue by symmetry that the LHS of (1) equals
$$P(\operatorname{sign}(X)=a, |X|\le c)=P(0\le X_1\le c_1, \ldots, 0\le X_n\le c_n)=:P(0\le X\le c).\tag2$$
Since (2) holds for all $a$ and there are $2^n$ such $a$, conclude
$$P(|X|\le c)=\sum_a P(\operatorname{sign}(X)=a, |X|\le c)=2^nP(0\le X\le c).\tag3
$$
Now solve (3) for $P(0\le X\le c)$ and plug back into (2):
$$P(\operatorname{sign}(X)=a, |X|\le c)=\frac1{2^n}P(|X|\le c)\tag4$$
Let $c\to\infty$ to deduce $P(\operatorname{sign}(X)=a)=\frac1{2^n}$, and you're done. (This argument gets you the distribution of sign($X$) for free.)


问:
Can you split the product of two functions of two variables into the sum of two others?
I'm trying to design a set of functions to demonstrate a control algorithm I'm working on.

Does there exist a non-trivial $(v(x), f(y), w(x), g(y))$ such that $v\cdot w=f+g \; \forall (x,y)$? $x$ and $y$ are independent variables. By non-trivial, I mean that the four functions can't just be constants. Each of the functions maps real scalars to real scalars.

答:
If there is $b$ so that $w(b)=0$, then
$f(x)+g(b)=v(x)w(b)=0$, this implies $f(x)=-g(b)$ is a constant: that is a "trivial" solution.

So assume $w(y)\neq 0$ for all $y$ and similarly, $v(x)\neq 0$ for all $x$.
Now
$$
\frac{f(x_1)+g(y)}{f(x_2)+g(y)}=\frac{v(x_1)w(y)}{v(x_2)w(y)}
=\frac{v(x_1)}{v(x_2)},
$$
If $f(x_1)\neq f(x_2)$, this equation implies $g(y)$ is a constant. So either $f$ or $g$ is a constant. So there is only "trivial" solutions.


问:
Expansion of trigonometric function squared (with two different method)
A series expansion of $\cos(x)$ is

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

I want to get the series expansion of $\cos^2(x)$, but using two different method I get different result

First Method : Using Trigonometric Identities

$$
\cos^2(x)=\frac{1+\cos(2x)}{2}
$$

Then subtituting the series
$$
\cos(2x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(2x)^{2n} \\
\cos^2(x)=\frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}2^{2n}\frac{(-1)^n}{(2n)!}x^{2n}
$$

Second Method : Squaring The Series

By directly squaring the series
$$
\cos^2(x)=[\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}]^2
$$

I get

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

Which is different result, the only way I can think off to make this the same result as the first method above is to somehow show that

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

and

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

Can I use the second method ? If so how to show the double summation lead back to result in the first method (especially the second one) ?

答:
In your second method, make a partition according to the value of $m+n$, you'll get
$$ \cos^2(x)=\sum_{k=0}^{+\infty}\sum_{\substack{n,m\geqslant 0 \\ n+m=k}}\frac{(-1)^k}{(2n)!(2m)!}x^{2k}=\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k}}{(2k)!}\sum_{n=0}^k\binom{2k}{2n}$$
But $$\sum_{n=0}^k\binom{2k}{2n}=\sum_{n=0}^{2k}\frac{1+(-1)^n}{2}\binom{2k}{n}=\frac{2^{2k}+0^{2k}}{2}$$
Therefore the above sum is $2^{2k-1}$ when $k\geqslant 1$ and $1$ when $k=0$, which means that
$$ \cos^2(x)=1+\frac{1}{2}\sum_{k=1}^{+\infty}2^{2k}\frac{(-1)^k}{(2k)!}x^{2k} $$
and you get back your first sum !


问:
How to solve a nonhomogeneous Cauchy-Euler differential equation without using Wronskians
I need to solve the non-homogenous Cauchy-Euler differential equation

$x^2y''-4xy'+6y=3x$

for class, but we haven't learned anything about Wronskians and that's mentioned in every explaination I see for how to do these problems. Is there a way to solve this without using Wronskians?

答:
$$x^2y''-4xy'+6y=3x$$
For the inhomogeneous part, try:
$$y_p=Ax$$
Plug this in the differential equation and find the constant A. Then the solution is:
$$y=y_h+y_p$$
Where $y_h$ is the solution of the homogeneous equation:
$$x^2y''-4xy'+6y=0$$


问:
Convergence in Distribution ($Y \mid N = n)$ is $\chi^2_{2n}$ and $N$ is Poisson($\theta$))
Suppose I have a hierarchical model like the following:

$Y\mid N=n$ is governed by a chi-square distribution with $2n$ degrees of freedom, and

$\theta$ is governed by a Poisson distribution.

I want to show that as $\theta \rightarrow \infty$, $\frac{Y - E[Y]}{\sqrt{\operatorname{Var}(Y)}} \rightarrow$ standard normal in distribution.

I know that to show convergence in distribution, I have to show that the pdf (or cdf)
$$\lim_{n \rightarrow \infty} f_{X_n}(x) = f_X(x),$$
but I am unsure how to go about this. How could I show that $\frac{Y - E[Y]}{\sqrt{\operatorname{Var}(Y)}} \rightarrow$ standard normal in distribution?

答:
For clarity I will leave out routine algebra and computations.
Let $\theta$ be fixed with $N$ being Poisson with mean $\theta$. You can first compute the MGF of $Y$ by conditioning on $N$ and using the chi-squared MGF:
$$M_{Y}(t) = E[\exp(tY)] = E[E[\exp(tY) | N]].$$
You can also compute $\mu = E[Y]$ and $\sigma^2 = Var(Y)$ by conditioning on $N$.
Now let
$$X = \frac{Y - \mu}{\sigma}.$$
Simple algebra reveals that the MGF of $X$ is easily obtained from the MGF of $Y$:
$$M_{X}(t) = \exp\left(-\frac{t\mu}{\sigma}\right)M_{Y}\left(\frac{t}{\sigma}\right).$$
After computing $M_X(t)$ you can take the limit as $\theta \to \infty$ and you will find that
$$\lim_{\theta \to \infty}M_{X}(t) = \exp\left(\frac{t^2}{2}\right),$$
the standard normal MGF.


问:
Does there exist $g$ s.t $g'=f$?
I have the following homework question:

Let G be the bounded open set shown in gray in this picture, whose
  boundary consists of eight line segments. The endpoints of those
  segments are, as shown, the points $-2,-1,-1+4i,1+4i,1,2,2+5i,-2+5i$.

Let $f:\, G\to\mathbb{C}$ be an arbitrary function which is
  holomorphic in $G$. Does there a function $g:\, G\to\mathbb{C}$ which
  satisfies $g'(z)=f(z)$ for all $z\in G$ ? one tool that might be
  useful here is the Identity Theorem.

What I did:

I believe that I can construct such a function, but I am unsure if
my construction is correct:

I would take some sequence of points $\{z_{i}\}_{i=1}^{\infty}$,
s.t there exist $\{r_{i}\}_{i=1}^{\infty}$ s.t $D(z_{i},r_{i})\subseteq G$
and s.t $$\cup_{i=1}^{\infty}D(z_{i},r_{i})=G$$

I believe that such a sequence of points can be obtained by choosing
all the points $x+iy\in G$ s.t $x,y\in\mathbb{Q}$.

Moreover, I think that the points can be arranged s.t $$D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$

We first consider $D(z_{0},r_{0})$ - there is some $g_{0}:\, D(z_{0},r_{0})\to\mathbb{\mathbb{C}}$ s.t
 $g_{0}'=f$ in $D(z_{0},r_{0})$, since $f$ is holomorphic.

We continue with $z_{1}$and get $\widetilde{g_{1}}$, since $g_{0}'=\widetilde{g_{1}}'$
for all points in $D(z_{0},r_{0})\cap D(z_{1},r_{1})$ then $g_{0}-\widetilde{g_{1}}$
is a constant $c$, and we can pick $g_{1}=\widetilde{g_{1}}+c$.
We take $g(z)$ will be eventually defined for every $z\in G$ as
one of the $g_{i}$'s.

This construction looks a bit fishy to me, I am not sure about the
"moreover" part and even if so, I am not totally convinced I can
arrange the constants to fit to get a holomorphic function $g$.

ADDED:

Instead of $$D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$
  we can relax the condition to be $$\cup_{i=1}^{r}D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$

Is my construction correct ?

I would also appreciate to see another approach (maybe one that uses
the Identity Theorem as suggested)

答:
You don't need so many disks. The region $G$ is the union of three rectangles $R_i$ intersecting in two squares $Q_1=R_1\cap R_2$ and $Q_2=R_2\cap R_3$. In each of the $R_i$ the given function $f$ has a primitive $F_i$, by the most elementary version of Cauchy's theorem. In $Q_1$ the function $F_2-F_1$ has derivative zero, so it is a constant $c_1$, and similarly in $Q_2$ the function $F_3-F_2$ is a constant $c_2$. It follows that the function
$$F(z):=\cases{F_1(z)+c_1\quad&$(z\in R_1)$ \cr
F_2(z)\quad&$(z\in R_2)$ \cr
F_3(z)-c_2\quad&$(z\in R_3)$ \cr}$$
is a well defined primitive of $f$ in $G$.

Of course observing that $G$ is simply connected makes all of the above superfluous. It is a standard theorem of complex analysis that a function $f$ holomorphic in a simply connected (i.e., without holes) domain $G$ has a primitive on $G$.

Title: Does there exist $g$ s.t $g'=f$?


问:
Geodesic distance of two quaternions
Let $q_0$ and $q_1$ be unit quaternions, and $d(q_0, q_1)$ the angular geodesic distance between the two.

I gather that the angular distance can be calculated with the formula:

$$d(q_0, q_1) = \lVert \ln{(q_0^{-1}q_1)}\rVert,$$

however, I cannot find derivation nor source for this. Where can I find a reference for this formulation?

答:
The geodesics on a sphere are arcs, which are pieces of great circles. Great circles on a sphere may be parametrized by $\cos\theta$ and $\sin\theta$ in two coordinates and $0$ in others for some choice of orthonormal coordinates. Equivalently, by $(\cos\theta)\mathbf{a}+(\sin\theta)\mathbf{b}$ for perpendicular vectors $\mathbf{a}$ and $\mathbf{b}$.

For $S^3$ specifically, if $\mathbf{u}$ is a unit vector and $p$ is any versor (unit quaternion) then $p$ and $p\mathbf{u}$ are orthogonal with respect to the inner product, so $(\cos\theta)p+(\sin\theta)p\mathbf{u}=p\exp(\theta\mathbf{u})$ parametrizes a great circle. If we restrict the domain of $\theta$ to an interval $[0,\phi]$ we get an arc.

The angle between $q_0=p$ and $q_1=p\exp(\theta\mathbf{u})$ is $\theta$. Note we can write

$$ \exp(\theta\mathbf{u}) = q_0^{-1}q_1 $$

and solve $\theta=\|\ln(q_0^{-1}q_1)\|$ (assuming we define $\ln$ by the rule $\ln\exp(\mathbf{v})=\mathbf{v}$ for $\|\mathbf{v}\|<\pi$).

Note that given $q_0=p$ and $q_1$ arbitrary, it is always possible to write $q_1=p\exp(\theta\mathbf{u})$, as that is just a matter of expressing $q_0^{-1}q_1$ in polar form with a convex angle... unless of course $\theta=\pi$, so $q_0$ and $q_1$ are antipodal, in which case $\ln(-1)$ isn't really defined although you could argue $\|\ln(-1)\|$ should be defined as $\pi$ still.

I think this is basic enough to not need a source in a research article, only a swift explanation. Is there another reason you want a source for this fact?

Title: Geodesic distance of two quaternions


问:
Does the union of two subgroups being a subgroup imply that they are subsets?
I saw this exercise in a book, not sure if is as trivial as I'm imagining:

Let $G$ be a group. Let $H$ and $K$ be subgroups of $G$. Now, if $H \cup K$ is a subgroup, does this imply that $H \subset K$ or $K \subset H$.

I believe the answer is yes, however is it this trivial:

Since $H$ and $K$ are subgroups of $G$ they must both contain the unique identity element, $e$, which means $e \in H$ and $e \in K$. Therefore it is immediate that $H \subset K$ or $K \subset H$?

答:
What you've written isn't a full proof -- to show that $H\subset K$ or $K\subset H$, you'd need to show that every element of $H$ is inside $K$ (or that every element of $K$ is inside $H$), not just that one particular element of $H$ is inside $K$.

For a hint on how to prove this, assume the contrary -- i.e. that there exists $x\in H$ that's not in $K$ and $y\in K$ that's not in $H$. See if you can construct something that would be in $H\cup K$ (if it is a subgroup), but can't be in $H$ or $K$.


问:
Mandelbrot-like sets for functions other than $f(z)=z^2+c$?
Are there any well-studied analogs to the Mandelbrot set using functions other than $f(z)= z^2+c$ in $\mathbb{C}$?</p>

答:
Ultrafractal has a particularly impressive collection of thousands most of which have configurable parameters as well as julia variations. Some of these parameters are shared among many such as starting position and bailout whereas others are specific to individual fractals, most notably I would say:

Mandlebrot

Phoenix

Newton

Magnet & Magnet 2

Nova

Burning Ship

Lambda

Barnsley 1, Barnsley 2, Barnsley 3

Tricorn

Manowar

Druid

There are other things to explore as well, as just one example take coloring, the inside of sets for example is usually coloured solid black but Ultrafractal alone contains hundreds of other possibilities such as triangular inequality average.

http://www.incubism.com/dan/novaFractal/img/novaInter_cloudyInteriour_mandelBollock.png


问:
$\lim_{y \to \infty}\int_{R}f(x-t)\frac{t}{t^2 +y^2}dt=0?$ for $f\in L^{p}$, $p \in [1,\infty)$
For $f\in L^{p}$, $p \in [1,\infty)$
we want to prove:
$$\lim_{y \to \infty}\int_{R}f(x-t)\frac{t}{t^2 +y^2}dt=0$$

I'm not sure whether we can exchange the limit and the integral, cuz I cannot find the integral function $g(t)$ such that $|f(x-t)\frac{t}{t^2 +y^2}| \leq |g(t)|$.

How could I argue this? Appreciate with any hints!

答:
If $p>1$, use Hölder's inequality. Then write
$$\int_{\Bbb R}\left|\frac{t}{t^2+y^2}\right|^qdt\leqslant \int_{\{|t|\geqslant R\}}\frac 1{|t|^q}dt+(2R)^{Q+1}y^{-2q}.$$

If $p=1$, use dominated convergence theorem, where $|f(x-t)|\frac{|t|}{t^2+1}$ is a dominating function ($y>1$).


问:
What is the formal proof that the reciprocal of a number approaches 0 as the number increases without bound?
In Calculus textbooks it seems assumed that
$$ \lim_{n\to\infty} \frac{1}{n} = 0 ,$$
meaning that "the limit of the reciprocal of $n$ as $n$ increases without bound is zero".

This makes intuitive sense: if $a/b$ gives the (potentially fractional) number of times $b$ fits into $a$, then as $b$ increases without bound, $a/b$ should get increasingly small.

What would be the formal proof that $$\lim_{n\to\infty} \frac{1}{n} = 0?$$

More generally, this is usually expressed as
$$ \lim_{n\to\infty} \frac{a}{n} = 0 ,$$
though if $a$ does not contain $n$ it gives the same result,
$$ \lim_{n\to\infty} \frac{a}{n} = a \lim_{n\to\infty} \frac{1}{n} = a(0) = 0 .$$

答:
To show that $\lim_{n\to\infty} \frac 1n=0$, we need to show that, for every $\epsilon$, there exists a threshold $n_0$ for which, if $n>n_0$, then $\left|\frac 1n\right|\leq \epsilon.$

Take any $n_0$ greater than $1/\epsilon$. Then, for any $n>n_0$,
$$0\leq \frac 1n<\frac{1}{n_0}<\epsilon,$$
and we are done.

(This last chain of inequalities corresponds to $1/\epsilon<n_0<n$.)


问:
How do we find $P(Z > -1.40)$?
Can we simply do $P(Z \le 1.4)$ ?

I'm confused because in cases like
$P(-1.4 < Z < 1.63)  = P(Z < 1.63) + P( Z < 1.4) - 1$

Why do we subtract one in the second case?

This is the case for $Z \sim (0, 1)$

答:
If $Z\sim Normal(0,1)$ so we can see by symmetry $$\mathbb{P}[Z>-1.40]=\mathbb{P}[Z\leq 1.40]$$
or $$\mathbb{P}[Z>-1.40]=1-\mathbb{P}[Z\leq -1.40]$$then you can use the table or
the definition.

Note that $$\mathbb{P}[Z\leq 1.40]=\int_{-\infty}^{1.40}f(x)dx=\int_{-\infty}^{1.40}\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{x^{2}}{2} \right)dx=0.919243.$$

Also, you can use the online-calculator.

Title: How do we find $P(Z > -1.40)$?


答:
Hint

You are actually referring to the limit of a sequence

$$1, (1/2), (1/3), \cdots $$

rather than the limit of a function.

Given $\epsilon > 0,~$ take $M \in \mathbb{Z^+}$ such that $$\frac{1}{\epsilon} < M.$$

Title: What is the formal proof that the reciprocal of a number approaches 0 as the number increases without bound?

Body:
In Calculus textbooks it seems assumed that
$$ \lim_{n\to\infty} \frac{1}{n} = 0 ,$$
meaning that "the limit of the reciprocal of $n$ as $n$ increases without bound is zero".

This makes intuitive sense: if $a/b$ gives the (potentially fractional) number of times $b$ fits into $a$, then as $b$ increases without bound, $a/b$ should get increasingly small.

What would be the formal proof that $$\lim_{n\to\infty} \frac{1}{n} = 0?$$

More generally, this is usually expressed as
$$ \lim_{n\to\infty} \frac{a}{n} = 0 ,$$
though if $a$ does not contain $n$ it gives the same result,
$$ \lim_{n\to\infty} \frac{a}{n} = a \lim_{n\to\infty} \frac{1}{n} = a(0) = 0 .$$


答:
We make the concept of "approach" in limits rigorous through epsilon-delta proofs, which you can look up.

Here is the proof for your limit: limit as $x$ approaches infinity of $\frac{1}{x}$


答:
Consider the set $A = \{1/n\mid n\in\mathbb{N}_{>0}\}$.

Clearly, it is bounded below by $0$ and it is not empty. Consequently, it admits an infimum $L\geq 0$.

Suppose that $L > 0$. Hence $2L > L$, which means there exists an $n_{0}$ such that $2L > 1/n_{0}$.

In other words, we have just proven that $L > 1/2n_{0}$, which contradicts our assumption.

Thus we can conclude that $\inf(A) = 0$.

Now we are ready to prove the proposed claim. More precisely, we want to prove the following proposition:
\begin{align*}
(\forall\varepsilon > 0)(\exists n_{0}\in\mathbb{N}_{>0})\:\text{s.t.}\left(n\geq n_{0}\Rightarrow \frac{1}{n} < \varepsilon\right)
\end{align*}

Indeed, this is the case.

Let us consider an arbitrary positive real number $\varepsilon > 0$.

Given that $\varepsilon > \inf(A)$, there exists an $n_{0}$ such that $\varepsilon > 1/n_{0}$.

Consequently, for every $\varepsilon  > 0$, there corresponds an $n_{0}\in\mathbb{N}_{>0}$ such that
\begin{align*}
n\geq n_{0} \Rightarrow \frac{1}{n} \leq \frac{1}{n_{0}} < \varepsilon
\end{align*}

Hopefully this helps!


问:
ODE describing insect population
The rate of increase of the population of insects is proportional to the number, , of insects present at any time  days after the beginning of the experiment. Given that the population of insects at the beginning is , and after 10 days is 2. Find in terms of ,
a) The population of insects after 5 days.
b) The days taken for the population of insects to be 3 times the original</p>

答:
Remember that the derivative is the rate of change. The problem states that the rate of increase of the population is proportional to the population size. In a formula, this can be written as
\begin{equation}
P'(t) = kP(t).
\end{equation}
This equation has solution $P(t) = P_0 e^{kt}$. So we need to compute $k$. We know that $P_0 = S$, since $P(0)=S=P_0 e^{0}$. We also know $P(10) = S_2 = S  e^{10k}$. We can solve for $k$ and obtain $k = \frac{1}{10}\ln(\frac{S_2}{S}) = \ln((\frac{S_2}{S})^{\frac{1}{10}})$. From here it is smooth sailing.

Title: ODE describing insect population


问:
Homeomorphisms that lift must lift to homeomorphisms
I am trying to understand the Burau representation of the braid group, and I am stuck on a particular assertion regarding lifts of homeomorphisms. This braid group/burau context may or may not be necessary: (paraphrasing)

Let $D_n$ be the $n$-times punctured disk. We have a homomorphism $\phi: D_n \to \mathbb{Z}$ sending a curve to its winding number. Let $p:\tilde{D}_n \to D_n$ be the covering corresponding to the kernel of $\phi$. Let $h$ be some homeomorphism of $D_n$ fixing the boundary pointwise (a braid). For any loop $\gamma$, we know $\phi(h\gamma) = \phi \gamma$, so $h$ lifts to a homeomorphism $\tilde{h}$ of $\tilde{D}_n$.

I (think I) understand everything but the last statement. I am interpreting the $\phi(h\gamma) = \phi \gamma$ condition to be satisfying the lifting criterion---it is saying in particular that $(h \circ p)_\ast (\pi_1(\tilde{D}_n)) \subseteq p_\ast (\pi_1(\tilde{D}_n))$. I believe that one can obtain the lift $\tilde{h}: \tilde{D}_n \to  \tilde{D}_n$, but how do I know it is a homomorphism? If you can prove the inverse lifts, you can use this answer, but I don't know how to do this.

Edit: After typing this out and reading it, I see that the $\phi(h\gamma) = \phi \gamma$ condition also guarantees the inverse lifts. However, think my question may still be interesting in the general setting.

Say I have a cover $p:\tilde{X} \to X$ and a homeomorphism $h:X \to X$, with the property that $(h \circ p)_\ast (\pi_1(\tilde{X},\tilde{x}_0)) \subseteq p_\ast (\pi_1(\tilde{X},\tilde{x}_0))$, so that $h \circ p$ lifts to a map $\tilde{h}:\tilde{X} \to \tilde{X}$. Must $\tilde{h}$ be a homeomorphism?

My suspicion is that the answer is no, since we are not guaranteed that $f^{-1}$ lifts, but it is not clear to me whether this fact is necessary.

答:
Indeed, $\tilde{h}$ need not be an isomorphism in general.  Writing $G=\pi_1(X,x_0)$ and $H=\pi_1(\tilde{X},\tilde{x}_0)$, we may identify $H$ as a subgroup of $G$ via $p$, and our assumption is that $h_*(H)\subseteq H$.  From the commutative square $$\require{AMScd}
\begin{CD}
\tilde{X} @>{\tilde{h}}>> \tilde{X}\\
@V{p}VV @VV{p}V \\
X @>{h}>> X
\end{CD}$$
we see that $\tilde{h}_*:H\to H$ is just the restriction of $h_*:G\to G$.  So, if $h_*$ maps $H$ to a proper subgroup of itself, $\tilde{h}_*$ will not be surjective, so $\tilde{h}$ is not a homeomorphism.

For an explicit example, you could take $X=B\mathbb{Q}$ and $h:X\to X$ the homeomorphism induced by the multiplication by $2$ map $\mathbb{Q}\to\mathbb{Q}$.  Then $h_*$ maps the subgroup $\mathbb{Z}\subset\mathbb{Q}$ to itself, so $h$ lifts to the associated covering space, but the lift is not a homeomorphism since multiplication by $2$ is not surjective on $\mathbb{Z}$.


问:
What is the geometric interpretation of the rowspace?
I am missing something essential in my understanding of matrices as linear mappings (or matrices in general). To explain, I think I have a decent intuition for columnspace of a matrix. Basically, the columnspace can be thought of as the set of points other than the origin which can be mapped by the matrix of a linear mapping (I like to think of matrices in general as linear mappings).

What I have trouble with is the significance and geometric interpretation of the rowspace of a matrix. By the definition, I can see that the rowspace and columnspace are analagous; the rowspace is merely the columnspace of the transpose of the matrix of the linear mapping. However, I cannot explain this anymore intuitively than above. To quote Albert Einstein, "if you don't understand it simply, you don't understand it well enough". I am looking for a simple explanation (or geometric interpretation that will contribute to my imagination of linear mappings).

How would you explain this concept?

To give you some context, I ran into problems when my course started to discuss the following:

For an $n \times n$ matrix $P$, the following are equivalent:

(1) The columns of $P$ form an orthonormal basis for $\mathbb{R}^n$

(2) $P^T = P^{-1}$

(3) The rows of $P$ form an orthonormal basis for $\mathbb{R}^n$.

I understand the algebra perfectly, but I seek a level of intuition so that I would not need to fiddle around with algebra to discover these facts.

How would you explain the above concepts intuitively or geometrically, without the need for algebra to make a case?

Many thanks in advance.

答:
This is in a sense trivial but also maybe not something you'd immediately think of.

Consider a square matrix $A$ (though it doesn't have to be).

The columns are a basis for the output (column) space.  The image of the cartesian grid in the input space is an affine, probably skewed (unless the matrix is orthogonal) grid in the output space.

The grid lines are parallel to these basis vectors (the columns of $A$).

Consider the image of the unit sphere in the input space.  This will be some ellipsoid in the output space.

Choose a particular cartesian axis in the output space, say the z (3rd) axis.

What point on the ellipsoid image "sticks out" furthest along the z axis? i.e., what point has the greatest z coordinate?

That point will lie on the span of the vector that is the image under $A$ of the 3rd row of $A$ used as an input to $A$. i.e., the span of $A\vec{v}$ where $\vec{v}^T$ is the 3rd row of $A$.

So if the 3rd row of $A$ is $\left[a\ b\ c\right]$, then the vector in the output space whose tip is at the $\left[a\ b\ c\right]$ point of the (skewed) output space grid will run through that point of the ellipse.

One way to explain this is that the "Jacobian" of $\vec{f}(\vec{x}) = A\vec{x}$, of $\vec{f}$ with respect to $\vec{x}$, is $A$.  So the direction in the input space of greatest increase of the mth output dimension (coordinate) is the mth row.

Looking at the image of the unit sphere of the input space corresponds to limiting yourself to a fixed "step size" to take along the gradient.


问:
Antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$
Is it possible to explicitly find the antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$? It is rather straight-forward to find it when $a=0$, or $b=0$ or $c=0$. This is also easy in the case where $a=b$, or $b=c$, or $a=c$. But I have been struggling to do this in the general case. Any help is appreciated. Thanks!

答:
Hint:

$\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}=\sqrt{(x+a)(x-a)(x+b)(x-b)(x+c)(x-c)}$

consult with Lauricella function


问:
An issue with approximations of a recurrence sequence
By trying to give an approximation to a given recurrence sequence I encountered a problem.

To be more precise I have a method but it fails if the right condition is not met and I wonder how I should proceed then.

An example will clarity.

General Question:

Let $a_0=c>0$ , $a_n=f(a_{n-1})$ and the sequence is strictly decreasing and has limit equal to $0$. Also for every finite $n$ we have $a_n >0$.

Specific Example:

Let $f$ be the function $log(1+x)$.

THE NECC CONDITION :
$\dfrac{x^m}{f(x)}$ is analytic at $x=0$ for some integer $m$.

Solution for the Specific Example:

Step 1. Since
$$0 < a_{n+1} = \log (1+a_n) < a_n,$$
it is a monotone decreasing sequence which is bounded. Thus it must converge to some limit, say $\alpha$. Then $\alpha = \log(1 + \alpha)$, which is true precisely when $\alpha = 0$. Therefore it follows that
$$a_n = o(1). \tag{1}$$

Before going to the next step, we make a simple observation: it is easy to observe that the function
$$\frac{x}{\log(1+x)}$$
is of class $C^3$. In particular, whenever $|x| \leq \frac{1}{2}$, we have
$$ \frac{x}{\log (1+x)} = 1+\frac{x}{2}-\frac{x^2}{12}+O(x^3). $$
This can be rephrased as
$$ \frac{1}{\log(1+x)} = \frac{1}{x}+\frac{1}{2}-\frac{x}{12}+O(x^2). \tag{2}$$
Here, we note that the bound, say $C > 0$, for the Big-Oh notation does not depend on $x$ whenever $|x| \leq \frac{1}{2}$.

Step 2.
By noting $(1)$, we fix a positive integer $N$ such that whenever $n \geq N$, we have  $|a_n| \leq \frac{1}{2}$. Then by $(2)$,
$$ \frac{1}{a_{n+1}} - \frac{1}{a_n} = \frac{1}{2} + O(a_n), $$
where the bound for Big-Oh notation depends only on $N$. Indeed, we may explicitly choose a bounding constant as
$$C'=\frac{1}{12} + \frac{1}{2}C,$$
where $C$ is the same as in $(2)$. Thus if $n > m > N$, we then have
$$ \begin{align*}
\frac{1}{a_n}
&= \frac{1}{a_{m}} + \sum_{k=m}^{n-1} \left( \frac{1}{a_{k+1}} - \frac{1}{a_k} \right) \\
&= \frac{1}{a_{m}} + \sum_{k=m}^{n-1} \left( \frac{1}{2} + O(a_k) \right) \\
&= \frac{1}{a_{m}} + \frac{n-m}{2} + O((n-m)a_m).
\end{align*} $$
Thus we have
$$ \left|\frac{1}{n a_n} - \frac{1}{2}\right| \leq \frac{1}{n}\left(\frac{1}{a_m} + \frac{m}{2} + C'm a_m \right) + C' a_m.$$
Taking limsup as $n\to\infty$, we have
$$ \limsup_{n\to\infty}\left|\frac{1}{n a_n} - \frac{1}{2}\right| \leq C' a_m. $$
Since now $m$ is arbitrary, the right-hand side can be made as small as we wish. Thus the left-hand side must vanish, yielding
$$ \frac{1}{n a_n} = \frac{1}{2} + o(1),$$
or equivalently
$$ n a_n = 2 + o(1). \tag{3} $$

However notice how much that method depended on the NECC CONDITION , SINCE we needed a taylor series with nonzero radius when expanded from $0$.

So my question is what should I do if that NECC condition is not met for $f$ ?

MORE SPECIFIC How do I proceed for an $f(x)$ that is not meromorphic at $0$ but is meromorphic on the halfopen interval $)0,1/2)$ ?

I considered using the taylor expansion of $f$ at $\frac{1}{q}$ and taking the limit for large positive $q$ going to $\infty$. But Im not sure how to do that. Or even if its a good idea.

Im also unsure if it matters what type of singularty we have at $0$.

But I believe there must be a way!

ONE MORE THING

I assumed that $f$ has no natural boundary. Not sure if it matters or not.
Such as if $f(x)$ is defined for $x<0$ matters or not.

To end this question maybe a more specific question : ADD to the above question the condtion that $f(f(x)-x) - (f(x)-x) = g(x)$ where $g$ is the FUNCTIONAL INVERSE of an entire function.

Maybe that makes the question easier because it is more specific.

答:
Hence the question [is] how to handle it.

"It" corresponds to $a_0=c$ with $c\gt0$ and $a_{n+1}=f(a_n)$ for every $n\geqslant0$, where
$$
f(x)=\frac1{\frac1x+\log\frac1x}.
$$
Such a function cries out for conjugacy... Thus, let us consider the sequence defined by $b_n=1/a_n$, then $b_{n+1}=g(b_n)$ for every $n\geqslant0$, where
$$
g(x)=x+\log x.
$$
If $b_0\gt1$, that is, if $c\lt1$, then $(b_n)$ is increasing and $b_n\to+\infty$ hence $a_n\to0$. If $b_0=1$, that is, if $c=1$, then $b_n=a_n=1$ for every $n$. If $b_0\lt1$, that is, if $c\gt1$, then $b_n\lt0$ after a while hence $(b_n)$ is not defined.

Thus, the case of interest is $c\lt1$. Associated to $b_{n+1}=g(b_n)$ is the ordinary differential equation $x'(t)=g(x(t))-x(t)=\log x(t)$, solved by
$$
t=\int_{x(0)}^{x(t)}\frac{\mathrm du}{\log u}=\mathrm{li}(x(t))-\mathrm{li}(x(0)),
$$
where $\mathrm{li}$ denotes the logarithmic integral. Thus, $\mathrm{li}(x)\sim x/\log x$ and $x(t)\sim t\log t$ when $t\to+\infty$. This suggests, and indeed one can show, that $b_n\sim n\log n$ when $n\to\infty$.

More precisely, $\log t\geqslant\log b_n$ for every $t$ in $(b_n,b_{n+1})$ hence
$$
\mathrm{li}(b_{n+1})-\mathrm{li}(b_n)=\int_{b_n}^{b_{n+1}}\frac{\mathrm dt}{\log t}\leqslant\frac{b_{n+1}-b_n}{\log b_n}=1,
$$
that is,
$$
\mathrm{li}(b_n)\leqslant\mathrm{li}(b_0)+n.
$$
Likewise, $b_{n+1}-\log b_{n+1}\leqslant b_n$ hence $\log(t-\log t)\leqslant\log b_n$ for every $t$ in $(b_n,b_{n+1})$ hence
$$
\int_{b_n}^{b_{n+1}}\frac{\mathrm dt}{\log( t-\log t)}\geqslant\frac{b_{n+1}-b_n}{\log b_n}=1,
$$
that is,
$$
\int_{b_0}^{b_{n}}\frac{\mathrm dt}{\log( t-\log t)}\geqslant n.
$$
Since $\log( t-\log t)\sim\log t$ when $t\to\infty$, all this yields $\mathrm{li}(b_n)\sim n\sim b_n/\log b_n$, and finally,
$$
\lim_{n\to\infty}n(\log n)a_n=1.
$$


问:
Matrix with entities 0 and 1
The number of $3\times 3$ matrix A whose entities are either 0 or 1 and for which the system $A\left[ {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
0\\
0
\end{array}} \right]$ has exactly two distinct solution is

(A) 0

(B) $2^9-1$

(C) 168

(D) 2

How do we proceed this type of problem I have asked a similar type of problem Symmetrical Matrices with entities $0$ and $1$ but how do we do this problem as the matrix may or may not be symmetric.

答:
Obviously the correct choice is (A)

Three situations can only occur while solving a system of three linear equations in three variables:

(i) No solution(the three planes form a triangular prism i.e line of intersection of the two planes is parallel to the third plane or at least two of the three planes are parallel)

(ii) Exactly one solution(the three planes intersect in a point)

(iii) Infinite number of solutions(the three planes intersect in a line or the three equations represent one plane only) .

So there cannot be any matrix $A$ which creates system of linear equations having exactly two solutions

Title: Matrix with entities 0 and 1


问:
Proof of lifting the exponent lemma
I recently read about the lifting the exponent lemma in a handout I understood most part of it but couldnt understand the proof from where n(exponent of the numbers) was assumed as $p^ab$ p and b coprime with p (prime) any help to make me understand it would be great.Thanks in advance.
https://brilliant.org/wiki/lifting-the-exponent/
Proof of lemma step 2

答:
Given $n$ and $p$, we can define $k$ to be $v_p(n)$. Then $p^k$ divides $n$, and so $n$ can be written as $n=p^ka$ for some integer $a$. We know
$$k=v_p(n)=v_p(p^ka)=v_p(p^k)+v_p(a)=k+v_p(a),$$
so $v_p(a)=0$, and so $p$ cannot divide $a$.

Title: Proof of lifting the exponent lemma


问:
Find the intersection point and verifying its direction
The tangent to the curve $y = e^x$  drawn at the point $(c,e^c)$ intersects the line joining the points $(c – 1, e^{c-1})$ and $(c+1, e^{c+1})$

(A) on the left of x = c

(B) on the right of x = c

(C) at no point

(D) at all point

My approach is as follow

$y = {e^x},y' = {e^x} \Rightarrow y' = {e^c}$

$y = {e^c}x + {c_{in}}$

Where $c_{in}$ represent the y intercept

${c_{in}} = {e^c}\left( {1 - c} \right)y = {e^c}x + {e^c}\left( {1 - c} \right)$

The points are $\left( {c - 1,{e^{c - 1}}} \right)\& \left( {c + 1,{e^{c + 1}}} \right) \Rightarrow m = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}$

$y = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}x + c{'_{in}}$...eq(i)

${e^{c + 1}} = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)\left( {c + 1} \right)}}{2} + c{'_{in}} \Rightarrow {e^c}\left( {e - \frac{{\left( {e - {e^{ - 1}}} \right)\left( {c + 1} \right)}}{2}} \right) = c{'_{in}} \Rightarrow c{'_{in}} = {e^c}\left( {\frac{{ - e\left( {c - 1} \right) + {e^{ - 1}}\left( {c + 1} \right)}}{2}} \right)$

$y = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}x + {e^c}\left( {\frac{{ - e\left( {c - 1} \right) + {e^{ - 1}}\left( {c + 1} \right)}}{2}} \right)$...eq(ii)

How we will proceed

Definitely we need to find the value of intersection x and see whether it is left or right of c

答:
Detailed solution can be found on Page 6

https://drive.google.com/file/d/0BwEaIp7y7BJTMU5jd3JvQl9rTEE/view


问:
Suppose $\gcd(a,b,c)=1$. Then is $a^n+b^n+c^n$ divisible by $a+b+c$?
I tried some examples and they all worked, but I have no idea how to factorize $a^n+b^n+c^n$ into the product of $a+b+c$ and something else.

Could anyone tell me whether this result is true or not? And, if it is true, how to prove it?

答:
There’s a well-known [classical] recursion:

$a^n+b^n+c^n = (a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+ac+bc)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3}).$

I don’t believe there’s any factorization with only $a+b+c$.


问:
Find general solution of ODE
I have this equation $xy''+y'= -1$ with y(0)=1 and y(1)=0.

So far I tried to solve the homogeneous equation:

$y(x)=c_1ln(x)+c_2$

but when I try the boundary conditions I'm not getting anywhere.

答:
Observe $-x$ is a solution (not satisfying the boundary condition);
add a suitable homogeneous solution, we see $y=1-x$ is the solution.

Title: Find general solution of ODE


问:
Convex hull of the union of compact and convex sets
Let X be a normed space and $A,B \subset X$.

I need to prove that if $A$ and $B$ are compact and convex then $conv(A \cup B)$ is compact. (Here $conv(A \cup B)$ is the convex hull of $A \cup B$)

If $U_{i \in I}$ is a finite cover for $A$ and $V_{j \in J}$ is a finite cover for $B$, how do I cover the elements of $z \in conv(A \cup B)$ such that $z \notin A$ and $z \notin B$?

I tought to pick one element of each $U_{i}$ and $V_{j}$ and for each pair $(u_{i},v_{j})$, the number of pairs $(u_{i},v_{j})$ is finite, create a line segment. Then at each midpoint create an open ball that covers the entire line. But I can't prove that this will cover all that points in the convex hull and I think that this will not be the right path.

答: A more general version of this claim is proved in Aliprantis-Border (2006, Lemma 5.29, page 183, Infinite Dimensional Analysis) for an arbitrary topological vector space.</p>


问:
Find the Number of Group Homomorphisms from $\mathbb{Z}/10\mathbb{Z}$ to $A_4$.
Is this reasoning correct?

$\mathbb{Z}/10\mathbb{Z}$ is isomorphic to $\mathbb{Z}_{10}$ which is generated by $1$, hence we may look at the number of homomorphisms between $\mathbb{Z}_{10}$ and $A_4$ for the answer to the original question. A homomorphism $\phi$ must map the element $1$ of $\mathbb{Z}_{10}$ to an element of $A_4$ such that its order divides both $10$ (by properties of homomorphisms) and $12$ (by Lagrange's Theorem). Thus, the only possible candidates for the mapping of $1$ must have order $1$ or $2$. This in turn yields $4$ possible elements in $A_4$, namely, $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$. All of these mappings preserve the group operation, thus the answer to the original question is $4$.

答: Yes, your reasoning is correct. Yet another way to tackle the problem is by the fundamental homomorphism theorem. Here we are looking for the (number of) possible subgroups of $A_4$ isomorphic to quotients of $\Bbb Z_{10}$; the (normal) subgroups of $\Bbb Z_{10}$ are: $\{0\}$, $\{0,5\}\cong \Bbb Z_2$, $H:=\{0,2,4,6,8\}\cong \Bbb Z_5$ and $\Bbb Z_{10}$; for order reasons (Lagrange theorem), the only options are $\Bbb Z_{10}/\Bbb Z_{10}\cong\Bbb Z_1$ and $\Bbb Z_{10}/H\cong\Bbb Z_2$. Therefore, if we call $\phi$ such a homomorphism, we can have either $\phi(\Bbb Z_{10})=\{()\}$ (the trivial case) or any among $\phi(\Bbb Z_{10})=\{(),(12)(34)\}$, $\phi(\Bbb Z_{10})=\{(),(13)(24)\}$, $\phi(\Bbb Z_{10})=\{(),(14)(23)\}$, so $4$ altogether.


问:
Proof that $\int_0^\infty\frac{\ln x}{x^3 - 1} \, dx = \frac{4 \pi^2}{27}$
I realise this question was asked here, but I'm not able to work with any of the answers. The hint given by my professor is

Integrate around the boundary of an indented sector of aperture $\frac{2 \pi}{3}$

but when I try that I can't figure out how to deal with the (divergent) integral along the radial line at angle $2 \pi / 3$. My issue with the accepted answer is that it uses the residue theorem where it doesn't apply, at least as we've learned it, since
$$z \mapsto \frac{\log^2z}{z^3 - 1}$$
has non-isolated singularities on the closed region bounded by the proposed contour (due to branch cuts), and I am not sure how to relate the integral along the real axis to one over a contour modified to avoid the branch cut.

For a fixed $\varepsilon > 0$, and for any $\delta \in (0, 1 - \varepsilon)$, we could let $\log_{-\delta / 2}$ be the branch of the logarithmic function with a cut along the ray $\operatorname{arg}z = -\delta / 2$ and define a contour which goes along the positive real axis from $\varepsilon$ to $1 - \delta$, a semicircle in the upper half plane of radius $\delta$ around $1$, the positive real axis from $1 + \delta$ to $2$, an arc of radius $2$ around $0$ with central angle $2 \pi - \delta$, the ray $\operatorname{arg}z = 2 \pi - \delta$ from $r = 2$ to $r = \varepsilon$, and finally an arc of radius $\varepsilon$ around $0$ back to $\varepsilon$. But then, for example, I don't know how to calculate the limit of integral along the arc of radius $\varepsilon$
$$\lim_{\delta \to 0}\int_0^{2 \pi - \delta}\frac{\log_{-\delta / 2}^2(\varepsilon e^{i \theta})}{\varepsilon^3 e^{3 i \theta} - 1} \varepsilon i e^{i \theta} \, d\theta.$$

If I instead try to first use the substitution $x = e^u$ on the real integral and then compute a contour integral, I still get a divergent integral that I don't know how to handle, this time along the top of an indented rectangle.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\cal C}$ is a key-hole contour which "takes care" of the $\ds{\ln}$-branch cut with $\ds{0 < \arg\pars{z} < 2\pi}$.
Integrand poles are $\ds{\expo{2\pi\ic/3}}$ and
$\ds{\expo{4\pi\ic/3}}$.

Residue Theorem yields:
\begin{align}
\oint_{\cal C}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z& =
2\pi\ic\bracks{%
\pars{{2\pi \over 3}\,\ic}^{2}\,{\expo{2\pi\ic/3} \over 3} +
\pars{{4\pi \over 3}\,\ic}^{2}\,{\expo{4\pi\ic/3}  \over 3}}
\\[5mm] & = -\,{4 \over 27}\pars{3\root{3} - 5\ic}\pi^{3}
\end{align}

Note that the singularity at $\ds{z = 1}$ lies on the
$\ds{\ln}$-branch cut. Hereafter we properly deals with this fact ( see below ).

Namely,
\begin{align}
&\bbox[5px,#ffd]{\oint_{\cal C}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z}
\\[5mm] = &\
\int_{0}^{\infty}{\ln^{2}\pars{x} \over
\pars{x - 1 + \ic 0^{+}}\pars{x^{2} + x + 1}}\,\dd x
\\[2mm] + &\
\int_{\infty}^{0}{\bracks{\ln\pars{x} + 2\pi\ic}^{2} \over
\pars{x - 1 - \ic 0^{+}}\pars{x^{2} + x + 1}}\,\dd x
\\[5mm] = &\
\mrm{P.V.}\int_{0}^{\infty}{\ln^{2}\pars{x} \over
\pars{x - 1}\pars{x^{2} + x + 1}}\,\dd x
\\[2mm] - &\
\mrm{P.V.}\int_{0}^{\infty}{\ln^{2}\pars{x} + 4\pi\ic\ln\pars{x} - 4\pi^{2} \over
\pars{x - 1}\pars{x^{2} + x + 1}}\,\dd x
+ {4\pi^{3} \over 3}\,\ic
\\[5mm] = &\
-4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}\,\dd x +
4\pi^{2}\
\underbrace{\mrm{P.V.}\int_{0}^{\infty}{\dd x \over x^{3} - 1}
\,\dd x}_{\ds{-\,{\root{3} \over 9}\,\pi}}\  +\
{4\pi^{3} \over 3}\,\ic
\end{align}

\begin{align}
&\int_{0}^{\infty}{\ln\pars{x} \over x^{3} - 1}\,\dd x
\\[2mm] = &\
{-\,\pars{4/27}\pars{3\root{3} - 5\ic}\pi^{3} -
4\pi^{2}\pars{-\root{3}\pi/9} - 4\pi^{3}\ic/3 \over -4\pi\ic}
\\[2mm] & =
\bbx{4\pi^{2} \over 27} \approx 1.4622 \\ &
\end{align}


问:
Question regarding second derivatives and maximums. Justifying a specific part
Note: My question is regarding the bolded parts. Read those and my explanation below to see what I am asking.

Let $f$ be a real-valued function on an open subset $U$ of $R$ that is twice differentiable at $x_0 \in U$. Show that if $f'(x_0) = 0$ and $f''(x_0) < 0$, then the restriction of $f$ to some open ball of center $x_0$ attains a maximum at $x_0$.

Here is my proof

To show there is a maximum, we want to show that

$f(x) < f(x_0) + f'(x_0)(x - x_0) \ \forall x \neq x_0, x_0 \in U$

Case 1.) Assume $x < x_0$. Applying the Mean Value Theorem on the interval $[x,x_0], \exists c \ where \ x < c < x_0 \ s.t.$

$f(x_0) - f(x) = f'(c)(x_0 - x)$

$f(x) = f(x_0) - f'(c)(x_0 - x)$

$f(x) = f(x_0) + f'(c)(x - x_0)$

Since $f''(x) < 0 \forall x \in U, f'(x)$ must be decreasing.

Since we know from the MVT that $c < x_0$, we can put these results together to have.

$f'(x_0) < f'(c)$

Multiplying this by $x - x_0$ (which is negative, so flip the inequality),

$f'(c)(x - x_0) < f'(x_0)(x - x_0)$

Adding $f(x_0)$ to both sides,

$f(x_0) + f'(c)(x - x_0) < f(x_0) + f'(x)(x - x_0)$

The left side is $f(x)$

$f(x) < f(x_0) + f'(x)(x - x_0)$ as desired.

Case 2.) Assume $x_0 < x$. Using the MVT on the interval $[x_0,x], \exists d \ where \ x_0 < d < x_0 \ s.t.$

$f(x) - f(x_0) = f'(c)(x - x_0)$

$f(x) = f(x_0) + f'(c)(x - x_0)$

Similar to before, $f''(x) < 0$ means $f'(x)$ is decreasing. We know from this application of MVT that $x_0 < c$

Therefore,

$f'(c) < f'(x_0)$

Multiplying by $x - x_0$ (positive this time)

$f'(c)(x - x_0)) < f'(x_0)(x - x_0)$

Adding $f(x_0)$,

$f(x_0) + f'(c)(x - x_0) < f(x_0) + f'(x_0)(x - x_0)$

The left side is $f(x)$

$f(x) < f(x_0) + f'(x_0)(x - x_0)$ still holds.

So we have shown that $\forall x \in U \ s.t. \ x \neq x_0$, the function will be concave down. Therefore, it has a maximum at $x_0$

I bolded the parts that were unclear. My professor told me that this is not necessarily true. The function will admit a negative second derivative at $x_0$, but not in the entire neighborhood.

He said in this case, what I did is correct, but he said it is not obvious as to why. He said I have to say something about why I can make that claim. Can someone help me figure out what I can say to make that part clear? I'm not seeing it right now.
Edit: I am thinking it maybe due to the fact that I'm claiming this happens on all of R. So like, the neighborhood wouldn't really matter? But then again, I feel like that would be trivial, so I don't think he would ask the question. I think I just defined it that way. I'm really not sure. Please, any help is appreciated!
Thanks in advance!

答:
Your professor is correct.

What we are dealing here with is called a local maximum.

Let $f$ be a real valued function defined on some open subset $U$ of $\mathbb {R}$. A point $x_0\in U$ is said to be a point of local maximum of $f$ if there is an open interval $I$ such that $x_0\in I, I\subseteq U$ and $f(x_0)\geq f(x) $ for all $x\in I$.

Here is the result you seek:

Theorem: Let $U$ be an open subset of $\mathbb {R} $ and let $f:U\to\mathbb {R} $ be a function. If $x_0\in U$ is such that $f'(x_0)=0,f''(x_0)<0$ then $x_0$ is a point of local maximum of $f$.

Observe that we don't have any information about existence of second derivative at any point except at $x_0$. But the existence of second derivative $f''$ at $x_0$ implies the existence of first derivative $f'$ in some neighborhood of $x_0$.

Next we need to analyze the meaning of $f''(x_0)<0$. Since $$f''(x_0)=\lim_{x\to x_0}\frac{f'(x)-f'(x_0)}{x-x_0}$$ the quotient on the right hand side in above equation must be negative for all $x$ in some deleted neighborhood $I$ of $x_0$. This means that $f'(x) - f'(x_0)=f'(x)$ has the sign opposite to that of $x-x_0$ in $I$. It follows that if $x\in I, x>x_0$ then $f'(x_0)<0$ and if $x\in I, x<x_0$ then $f'(x) >0$.

What this means is that $f$ is strictly increasing on the left side of $x_0$ in $I$ and strictly decreasing on the right side of $x_0$ in $I$. This clearly means that $f(x_0)\geq f(x)$ for all $x\in I$.

Your approach has issues at two steps. First is the definition of local maximum. The definition of local maximum does not need any idea of derivative. I don't see why you write $$f(x) <f(x_0)+f'(x_0)(x-x_0)$$ Next you are using values of $f''$ at points in some neighborhood of $x_0$. This is not given. The fact that $f''(x_0)<0$ does not imply that $f'$ is strictly decreasing in some neighborhood of $I$. This is the part in bold where you have some confusion. You have somehow assumed $f''$ to be negative in some neighborhood of $x_0$ and then deduce the conclusion that $f' $ is strictly decreasing in that neighborhood.

You can see from the argument in my answer that we don't need to know whether $f'$ is decreasing or not. We just need to know that derivative $f'$ is positive to the left of $x_0$ and negative to the right of $x_0$ in some neighborhood. This is sufficient to get the desired conclusion.


问:
Suppose $gH = Hg, \forall g \in G$. does this imply that $g^{-1}Hg \subseteq H, \forall g \in G$?
Let $G$ be a group and $H \leq G$. Suppose $gH = Hg,  \forall g \in G$. We know that this is equivalent to $gHg^{-1} \subseteq H,  \forall g \in G$ and $gHg^{-1} = H,  \forall g \in G$. However I haven't seen any textbook write what was stated in the title. Is it false?

答: It's obviously true.  If $gH = Hg$, multiply both sides by $g^{-1}$ to get $gHg^{-1} = H$, and hence $gHg^{-1} \subseteq H$.  Since $g$ can be anything, replace $g$ by $g^{-1}$ to get $g^{-1}Hg \subseteq H$.


答:
$y''+\dfrac1xy'+\dfrac1x=0$

$(y'x)'=y''x+\dfrac1xy'x=(y''+\dfrac1xy')x$

$\dfrac1x (y'x)'+\dfrac1x =0\implies (y'x)'=-1\implies y'x=-c$

$y=-c\log(x)+c'$, but $\log(0)$ is undefined, which means that $y=c'=1$.

Let $u(x)=ax^2+bx+c, u'(x)=2ax+b,u''(x)=2a\implies u''+\dfrac1xu'+\dfrac1x=4a+\dfrac1x(b+1)$, setting it to be the zero function gives $a=0, b=-1$, so $y=1-x$ is the solution

Title: Find general solution of ODE

Body:
I have this equation $xy''+y'= -1$ with y(0)=1 and y(1)=0.

So far I tried to solve the homogeneous equation:

$y(x)=c_1ln(x)+c_2$

but when I try the boundary conditions I'm not getting anywhere.


问:
Proof of Banach-Alaoglu theorem by Douglas
I am struggling to understand a step in the proof of the Banach-Alaoglu theorem (1.23) given in Douglas - Banach Algebra Techniques in Operator Theory. Here is the relevant section of the book:

I am having trouble understanding why the highlighted portions are true. This feels like the usual "linear maps are determined by their images on the basis," except there is no basis here. I imagine the fact that $\lVert\varphi_1\rVert,\lVert\varphi_2\rVert\le1$ must be relevant, but I do not see how. I would appreciate any help.

答:
If two linear functionals $f$ and $g$ are equal on the closed unit ball they are equal everywhere: For any $x$ with $\|x\| >1$ we have $f(x)=\|x\|f(\frac x {\|x\|})=\|x\|g(\frac x {\|x\|})=g(x)$ by linearity of $f$ and $g$.

The product topology is the topology of convergence of each coordinate  and weak* convergence is the topology of convergence at each point $x$.


问:
When is the derivative of $f(g(x))$ equal to $g(f'(x))$?
By the chain rule, we know that the derivative of $f(g(x))$ is $f'(g(x))g'(x)$.

Question: When is the derivative of $f(g(x))$ equal to $g(f'(x))$?

Trivial solutions include the following:

Let $f$ be any differentiable function and $g$ be the identity function ($g(x)=x$) or the zero function ($g(x)=0$).

Let $f$ be any constant function and $g$ be any differentiable function that fixes zero.

On the other hand,  $f(x)=x^2$ and $g(x)=\sin x$ form a nontrivial solution (by the double angle formula for sine).

If $g(x)=x+a$, where $a$ is a constant, then $f'(x+a)=f'(x)+a$, so $f(x)=\frac{1}{2}x^2+bx+c$ would work for any two constants $b$ and $c$.

Finally, if $f(x)=mx+b$, where $m$ and $b$ are constants, then $mg'(x)=g(m)$, so $g'(x)=\frac{g(m)}{m}$ and $g(x)=\frac{g(m)}{m}x+c$ for some constant $c$. But this must in particular be true for $x=m$, so $c$ must be zero and $g(x)=nx$ would then work for any constant $n$.

But I do not know any other solutions. Can anyone help find one?

Note that $f$ may be replaced by any other function with the same derivative (i.e. differing from $f$ by a constant) without changing the validity of the equation, so we may assume without lost of generality that $f$ fixes zero (assuming, of course, that zero is in the domain of $f$).

答:
Not an answer, but too long for a comment: Let $F(x) = f'(x)$. Then you are trying to solve $$F(g(x))g'(x) = g(F(x))$$

Let $g(x)$ be any differentiable function. This becomes a functional equation for $F(x)$. For $g(x) = x^n$, where $n > 0$, it is $$F\left(x^n\right)nx^{n-1} = F(x)^n$$

The solution (found using Mathematica) is $$F(x) = n^{\frac{1}{n-1}}x^{c - \frac{(n-1)\ln(\ln(x))}{n\ln(n)}}$$

$f(x)$ would then be the integral of $F(x)$.

I think it may be possible to show that given "sufficiently nice" $g$, an $F$ exists that solves the functional equation. I'm not sure how to go about this though.


问:
Solving surface integral using divergence theorem
Here is the problem:

Use divergence theorem to compute

$S$ is boundary of solid bounded above cone $z = \sqrt{x^2+y^2}$ below sphere $x^2+y^2+z^2 = 1$.

Here is my solving process:

When setting up the integral, the bounds I chose are

$ 0 <r < \frac{\sqrt{2}}{2}$ , upper bound from 2 solid's intersection

$ 0< \theta <2\pi$

$ 0 < z <\sqrt{1-r^2}$, upper bound comes from equation of sphere

Then the $\nabla \cdot F = x^2+y^2+z^2$, and $dV = rdzdrd\theta$.

Here is my confusion:

The result I've got is incorrect, and the only problem I can think of is the lower bound of $z$. It might be $r$ (coming from the equation of cone) instead of $0$ but why? The bounded solid E starts from origin, isn't it? Hope someone will answer my doubt.

Plus, are there any other problems in my solving process? Thanks!

答:
Your divergence is correct. The problem as you mentioned is with the lower bound of $z$. In cylindrical coordinates, the integral should be set up as -

$\displaystyle \int_0^{2\pi} \int_0^{1/ \sqrt 2}\int_r^{\sqrt{1-r^2}} r(r^2+z^2) \, dz \, dr \, d\theta$

On your question as to why the lower bound should be $r$ and not $0$, first deal with the cone with base as $z = \frac{1}{\sqrt2} \,$ and vertex at the origin, without the spherical cap on top. For a given $r$, can $z$ freely vary between $0$ and $1 / \sqrt2$? That will give you the volume of the cylinder instead of the cone. In a cone, $z$ can vary from the base to the points on its rays, ($r \leq z \leq \frac{1}{\sqrt2}$). Now the upper bound of $z$ changes because we are just adding the spherical cap on top.

In spherical coordinates which is easier in this case,

$\displaystyle \int_0^{2\pi} \int_0^{\pi/4}\int_0^1 r^4 \sin \phi \, dr \, d\phi \, d\theta$


问:
If $a$ is an eigenvalue for $A$ and $b$ is an eigenvalue for $B$, what is an eigenvalue for $AB$?
Let both $A$ and $B$ be two $n \times n$ matrices such that $a$ is an eigenvalue for $A$ and $b$ is an eigenvalue for $B$. How can I show that $ab$ is an eigenvalue for $AB$?

I began my solution by stating:
$$Av = av$$ $$Bw = bw$$ $$\implies ABvw = abvw$$ $$\implies ABvw = abvw$$

But I'm not sure what my last step should be to complete the proof. Any guidance is greatly appreciated!

答:
$ab$ need not to be an eigenvalue of $AB$ in general, except the case, when the eigenvectors of $a,b$ are same.

If eigenvector of $a,b$ be same , $v$, namely.

Then , $ABv=A(Bv)=A(bv)=b(Av)=b(av)=abv$.

For counter example whenever it doesn't happen,

Take , $A=
\begin{pmatrix}
    0 & 1 \\
    1 & 0 \\
    \end{pmatrix}
$,
$B=\begin{pmatrix}
    0 & -1 \\
    -1 & 0 \\
    \end{pmatrix}
$
Now, $1$ is an eigenvalue of both $A$ and $B$, but $AB$ has the only eigenvalue $-1$.


答:
The following statements are equivalent:

(i) $gH = Hg$ for all $g \in G$;

(ii) $gHg^{-1} = H$ for all $g \in G$;

(iii) $gHg^{-1} \subseteq H$ for all $g \in G$.

Proof.

Suppose (i) and take $g \in G$. By assumption we have $gH \subseteq Hg$, which means that for every $h \in H$ there is $h' \in H$ such that $gh = h'g$, so $ghg^{-1} = h'$ and then $gHg^{-1} \subseteq H$. Also, by assumption we have $Hg \subseteq gH$, and this tells us that for every $h \in H$ there is $h' \in H$ such that $hg = gh'$, so $h = gh'g^{-1}$ and then $H \subseteq gHg^{-1}$.

Now suppose (iii) and we will prove (i). For that, take $g \in G$ and observe that $gHg^{-1} \subseteq H$ means that for all $h \in H$ there exists $h' \in H$ with $ghg^{-1} = h'$, so $gh = h'g$ and this proves $gH \subseteq Hg$. Next, since we are also assuming that $g^{-1}Hg = g^{-1}H(g^{-1})^{-1} \subseteq H$, this means that for all $h \in H$ there exists $h' \in H$ with $g^{-1}hg = h'$, so $hg = gh'$ and this proves that $Hg \subseteq gH$.

Finally, it is obvious that (ii) implies (iii).

Title: Suppose $gH = Hg, \forall g \in G$. does this imply that $g^{-1}Hg \subseteq H, \forall g \in G$?

Body: Let $G$ be a group and $H \leq G$. Suppose $gH = Hg,  \forall g \in G$. We know that this is equivalent to $gHg^{-1} \subseteq H,  \forall g \in G$ and $gHg^{-1} = H,  \forall g \in G$. However I haven't seen any textbook write what was stated in the title. Is it false?


问:
Basic question about the image and kernel of tensor products
I have a very basic question about the interactions of the image/kernel of module maps with tensor products.

For a ring $A$, $A$-modules $M, M'$ and $N$, and a map of $A$-modules $f: M\mapsto M'$, do the following hold?

$$\operatorname{Im}(f \otimes_A Id_N) = \operatorname{Im}(f) \otimes_A N $$
$$\operatorname{Ker}(f) \otimes_A N \subseteq  \operatorname{Ker}(f \otimes_A Id_N)$$

答:
Yes. For your first question, we first show the inclusion $\subseteq$. Let $x\in M\otimes_A N$ be arbitrary; then there are $m_i\in M$ and $n_i\in N$ such that $x=\sum_{i=1}^km_i\otimes n_i$. Then we have $$(f\otimes 1_N)(x)=\sum_{i=1}^k(f\otimes 1_N)(m_i\otimes n_i)=\sum_{i=1}^kf(m_i)\otimes n_i.$$ Each $f(m_i)\otimes n_i\in \operatorname{Im}{(f)}\otimes_A N$, so $(f\otimes 1_N)(x)$ is also, as desired.

Now, to see the inclusion $\supseteq$, since $\operatorname{Im}(f)\otimes_A N$ is generated as an abelian group by simple tensors of the form $f(m)\otimes n$, it suffices to show that each $f(m)\otimes n\in\operatorname{Im}(f\otimes 1_N)$. But this is immediate, since $$f(m)\otimes n=(f\otimes 1_N)(m\otimes n).$$

For your second question, since $\operatorname{Ker}(f)\otimes_A N$ is generated as an abelian group by simple tensors of the form $m\otimes n$, where $m\in\operatorname{Ker}(f)$, it suffices to show that all such elements are in the kernel of $f\otimes 1_N$. But this holds immediately: $$(f\otimes 1_N)(m\otimes n)=f(m)\otimes n=0\otimes n=0\otimes 0,$$ as desired.

Note, the reverse inclusion does not hold in general. Indeed, let $A$, $M$, and $M'$ all equal $\mathbb{Z}$, take $N=\mathbb{Z}\big/2\mathbb{Z}$, and consider the map $f:\mathbb{Z}\rightarrow\mathbb{Z}$ given by $x\mapsto 2x$. Now, the induced map $$f\otimes 1:\mathbb{Z}\otimes(\mathbb{Z}\big/2)\rightarrow\mathbb{Z}\otimes(\mathbb{Z}\big/2)$$ sends $x\otimes\overline{y}$ to $2x\otimes\overline{y}=x\otimes2\overline{y}=x\otimes \overline{0}=0\otimes \overline{0}$, and so the kernel of $f\otimes 1$ is the whole space. On the other hand, $\operatorname{Ker}(f)=\{0\}$, so $\operatorname{Ker}(f)\otimes(\mathbb{Z}\big/2)$ is the trivial submodule.


问:
Informal question about probability concerning infitie events and their limits
Suppose you have infite countable events which depend on an integer $n$, i.e. $A_1(n),...,A_k(n),....$
And you know for each event $\lim_{n\rightarrow\infty }Pr(A_k(n))=1$ for every $k\in\mathbb{N}$. What is the probability that all events happen simulaneously i.e. $\lim_{n\rightarrow\infty}\Pi_{k=1}^{n}Pr(A_k(n))$ ?

The context of my question is this one here:

Assume that for a probabilityfunktion $p(n)$ and random graphs $G(n,p(n))$ and a first order statement $A$ $\lim_{n\rightarrow\infty}Pr(G(n,p(n))\models A)=1$.

Then for every $A$, $\lim_{m,n\rightarrow\infty}Pr[(G(n,p(n)\models A)\iff H(m,p(m))\models A)] = 1$.

I now want to know the probability that $G(n,p(n))$ is elementary equivalent to $H(m,p(m))$. I.e. that $G(n,p(n))\models A \iff H(m,p(m))\models A$. For a single $A$ the probabilty approaches $1$, but can we say the same thing for all $A$s at the same time? There are infinite many countable $A$s and that is the background of my question.

Ultimately I want to show that $\lim_{m,n\rightarrow\infty}Pr[\text{Duplicator wins EHR(G(n,p(n)),H(m,p(m)),t)}]=1, \forall t$

I have to show this statement above in order to use the lemma which states:

$\text{Duplicator wins every EHR(G,H,t) forall t}\iff \forall A\in FO, G\models A\iff H\models A$

答:
It can be a lot of different things.

Let $p_k(n)$ be the probability of $A_k(n)$. If $p_k(n)=1$ for all $k$ and $n$, then clearly the product is $1$. However, say that $p_k(n)$ is $0$ if $n<k$ and $1$ otherwise. For each $k$,
$$\lim_{n\to\infty} p_k(n)=1,$$
since $p_k(n)$ is eventually $1$. However, for each $n$,
$$\prod_{k=1}^\infty p_k(n)=0,$$
since all but finitely many terms are $1$. So, the limit of the product is $0$.

More generally, if you set $p_k(n)=1$ whenever $n\neq k$ and $p_k(n)=q$ when $k=n$, then the product at each $n$ is $q$ and so the limit is always $q$. So, you can make the limit be anything between $0$ and $1$ (inclusive).


问:
simple and uniform convergence of $ \sum ( x+\frac{1}{n})^{n+\frac{x}{n}} $
It's my first post in this forum ! Hello ! (I have a very bad english sorry in advance...) I come with an exercice that I can not solve. I hope you can help me...
Q1 : simple and uniform convergence of $ \sum ( x+\frac{1}{n})^{n+\frac{x}{n}} $

Here my guess and what I have done the convergence is |x|< 1 (that I didn't proved just guessed)
I tried limited development it didn't worked.

Thank you for reading !

答:
One possible solution is to apply the $n$-root test.

To have powers well defined, assume $x\geq0$.
Let $a_n(x)=\Big(x+\frac{1}{n}\Big)^{n+\tfrac{x}{n}}=x^{n}x^{x/n}\Big(1+\frac{1}{xn}\Big)^n\Big(1+\frac{1}{xn}\Big)^{x/n}$. Then
$$
|a_n(x)|^{1/n}=|x||x|^{x/n^2}\Big|1+\frac{1}{xn}\Big|\Big|1+\frac{1}{nx}\Big|^{x/n^2}\xrightarrow{n\rightarrow\infty}|x|
$$

Hence, as you suspected, convergence holds for $0<x<1$ and uniform convergence occurs in compact subsets of $(0,1)$. Some details are left for the OP.


问:
System of distinct representatives and chessboards
I encountered the following problem, which was presented in the context of the topic of SDRs (system of distinct representatives) - I am able to solve the problem, but I make no use of a SDR, and I am wondering whether someone sees a solution which applies an SDR:

Suppose we are given an $m \times n$ chessboard, with $m$ even. Provided we have $n \geq 2$, prove that if we delete one white square and one black square, we can still tile the remaining board with dominos.

I have a reasonably straightforward way of solving the problem (basically it is sufficient to assume W.L.O.G. that the deleted cells $\textit{define}$ the chessboard, by acting as the top-left and bottom-right corners, and showing that this resulting grid can be tiled) by doing a little case-analysis, but this makes no use of an SDR (which is the context in which this question was asked). I don't know whether or not there is a solution which involves an SDR, but if there is, I cannot find it. In particular, I am not even sure how an SDR could be defined in this context. So my question is, can anyone see how to solve this problem by making use of the existence of some SDR? Again, I am aware of a solution in general, just not one which involves an SDR. Any results regarding SDRs are allowed (e.g. Hall's condition). Let me know if you have any thoughts regarding how an SDR could be defined here!

答: For each black square, consider the set of white squares which neighbor it. An SDR for this collection of sets corresponds to a domino tiling; choosing a white square $W$ from the set for a black square, $B$, means that $B$ and $W$ are covered by a domino. You then need to prove the Hall condition holds for this collection of sets, which means that every set of $k$ black squares shares a border with at least $k$ distinct white squares. I am not sure if there is any way to do this which does not at some point use the idea in your solution.


问:
Limits of integration in iterated integrals
We were doing order statistics in our probability class. And were asked to compute the following integral:

$$\iiint_{0 < x_{1} < x_{2} < x_{3} < 1} dx_{1}\, dx_{2}\, dx_{3}$$

I am unsure of how the limits of integration are being computed. Our professor told us that the answer would be $\int_{0}^{1} \int_{0}^{x_{3}} \int_{0}^{x_{2}}$. I just don't see why this has to be the case. Could someone please help me out a little?

答: $x_1$ ranges from $0$ to $x_2$. $x_2$ ranges from $x_1$ to $x_3$ but $x_1$ itself goes all the way down to $0$. So $x_2$ takes all values from $0$ to $x_3$. $x_3$  takes all values from $0$ to $1$. I hope this explanation helps.