问:
When does a functional from a vector's span to $\mathbb{R}$ have a unique norm-preserving extension to $\mathbb{R}^n$?
Here's my question:

Consider the space $(\mathbb{R}^n, \| \cdot \|_\infty).$ For which vectors $v \in \mathbb{R}^n$ can we say that any linear functional $f: \langle v \rangle \to \mathbb{R}$ has a unique norm-preserving extension to the whole space?

My thought is that this question can be reduced to asking for which vectors we can find a sub-linear function $p$ such that $f$ is dominated by $p$ on $v$'s span. Hahn-Banach then guarantees that $f$ has an extension to the whole space, but that extension doesn't necessarily have to be unique, which I think is where this question gets tricky. At the moment, I'm trying to think about what further conditions we'd need to be guaranteed a unique extension. I'd really appreciate any help.

答:
The problem is evidently invariant under scalar multiplication so we may suppose that both $\|v\|=1$ and $\|f\|=1$, in
which case $f$ is necessarily given by $f(\lambda v)=\pm\lambda $.  We may further replace $v$ by $-v$, if necessary, so as to be able
to assume that $f(\lambda v)=\lambda $.

Next  notice that a linear functional $g$, defined on the whole ${\bf R}^n$,  has norm less than or equal to one, if and only
if the affine hyperplane
$$
  \{x\in {\bf R}^n:g(x)=1\}
  $$
has empty intersection with the open unit ball $B(0, 1)$.

In order to produce a norm-preserving extension  of our given functional $f$,  we must therefore construct an affine
hyperplane passing thru $v$,  and leaving the whole of $B(0, 1)$ to the same side.

Noticing that $B(0, 1)$ is actually the cube $(-1,1)^n$,
observe that if $v$ is one of the vertices of that cube, say
$v = (\pm 1,\pm 1,\ldots , \pm1)$, then there are many ways to build a hyperplane thru $v$ not intercepting $B(0,1)$.

The same goes for the case in which $v$ is in any edge of the cube, but of course not if $v$ is an interior point of one
of its faces.  Precisely, this means that $v$ must have one, and no more that one, coordinate with absolute value 1, so
voilà the
condition for the Hahn-Banach extension to be unique!


问:
Consider the equation $2R*x^2 - (2R\cos\frac{B-C}{2})*x+r=0$
$2R*x^2 -  (2R\cos\frac{B-C}{2})*x+r=0$

We have the triangle ABC

Where $r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

R is the circumradius of the triangle ABC.

I have the following answer options:

$cos^2\frac{B-C}{2}<\frac{2r}{R}$

$2r > R$

$\sin\frac{A}{2}$ Is a root

The equation has no real roots.

I honestly don't even know how to start here, any help would be greatly appreciated

答:
Well, here is a simple way to get some progress.

If (2) is true, then $\frac{2r}{R}>1$ and that would imply that (1) is true. Thus, proving (2) is false proves that both (1) and (2) are false.

If you take a look at Euler's theorem in geometry (https://en.wikipedia.org/wiki/Euler%27s_theorem_in_geometry), it shows that $R>2r$ proving (2) is false. This knocks out two answer choices.

Then, if you look at the discriminant, using that same property of $R>2r$, you can find a real solution by making $\cos(\frac{B-C}{2})$ equal 1. Since a real solution is possible that rules out (4).

ContentLicense: CC BY-SA 4.0


问:
Some questions about the Exercise $3.15$ in Jech Set Theory and the Axiom of Choice.
I have $3$ questions about the following exercise in Jech Set Theory (Exercise $3.15$ ($ii$)):

Show that the set of all finite one-to-one sequences in a Dedekind-finite set is Dedekind-finite.

Part of my solution:

Assume that it is Dedekind-infinite. Then, it has a countable subset. Call it $S = \{S_i \mid i<\omega\}$, where $S_i$'s are finite one-to-one sequences. Now, let $s = \bigcup_{i<\omega} \operatorname{ran}(S_{i})$. It is easy to show that $s$ is not finite, so it is infinite. My first question is: Can I now say directly that $s$ is countably infinite without the Axiom of Choice? If this is valid, then we would arrive at a contradiction since it is a subset of the original set.

In case this cannot be said without the Axiom of Choice, I have thought about the following:

For $a \in s$, let $n(a)$ be the least finite ordinal $i$ such that $a \in \operatorname{ran}(S_{i})$. Now, define the ordering $\prec$ on $s$ by:

$$(a \prec b) \ \ \ \text{iff} \ \ \ ((n(a) < n(b)) \ \ \ \text{or} \ \ \ (n(a) = n(b) \ \ \ \text{and} \ \ \ \text{$a$'s index is less than $b$'s index in the common range}))$$

Now, since this is a well-order on $s$, and $s$ is infinite, we can form a countable subset of $s$ by:

$$a_0 = \text{the least element of $s$}$$
$$a_{n+1} = \text{the least element of $s \setminus \{a_1, \ldots, a_n \}$}$$

My second question is: Is this approach correct?

Finally, my last question is a follow up to this exercise: What if the question asked the same thing but with Dedekind-finite sequences? Can we prove it in that case without the Axiom of Choice? Thank you.

答:
No, you can't say that a countable union of finite sets is countable without appealing to the axiom of choice. The second approach is indeed correct.

For the Dedekind-finite sequences, that's a very nice question, and that is more intricate. Of course you mean injective sequences, otherwise just the finite sequences are already Dedekind-infinite (fix $a$ and consider $s_n$ to be the sequence which is just $n$ times $a$).

But now comes the question, what exactly do we mean by that.

Do we mean just injective functions from a subset of $X$ to $X$? In that case, there's no reason to expect that the result is Dedekind-finite. For example, if $X$ can be mapped onto $\omega$, i.e. it has a Dedekind-infinite power set, then we can simply pick countable many subsets and consider the identity function. But note that it is certainly possible that those countably many subsets are indeed finite, which is now in contradiction with what you just proved.

Do we mean injective functions from cardinals of subsets of $X$ to $X$? That's not always a possible definition. If $X$ is amorphous, then every subset is finite or co-finite, but we cannot choose a representative for each infinite cardinal, since that would allow us to define a surjection from $X$ onto $\omega$ (take the complements, these are finite, now map each element in $X$ to the least index of a finite set in which it appears, or $0$ otherwise; easily this map has an infinite range).

So we are in a situation where "the right definition" is not obvious at all. So there's no good answer, other than "it's complicated and it depends on both $X$ and on what exactly you mean by that".


问:
Prove that there's only 3 normal subgroups in $G$.
Let $G$ be a group and $H$ a normal subgroup such that $|H| = 15$.
And also $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

The question is to show that there are only 3 normal subgroups of order 30 in $G$.

I only know some pretty easy facts from Lagrange's theorem, and don't know how it affects normal subgroups.

答:
You have misunderstood something about the question.  The conditions you've given tell us that $\vert G \vert = 100$ so that $H$ is a $5$-Sylow subgroup of $G$, which means (since you're told that it's normal) that $H$ is the unique subgroup of $G$ of order $25$.

Conversely, if $G$ has $3$ subgroups of order $25$, then they are conjugates of each other and are not normal.

Perhaps you're supposed to show that $G$ has $3$ subgroups of order $50$?

Edited to respond to corrected question

There are at least $3$ such subgroups, corresponding to the non-trivial proper subgroups of $G/H$, and those are the only three such subgroups that contain $H$.

Note also that the Sylow subgroups of $H$ are normal in $H$, which means that they're unique in $H$, which means that they're also normal in $G$.  Thus, $H$ is the unique subgroup of order $15$ in $G$.

But any order $30$ subgroup of $G$ would have to contain $H$ (because it contains the unique $3$-Sylow and $5$-Sylow subgroups of $G$), so the three subgroups you've already identified are the only three there can be.


问:
Issue with derivation of $\nabla \times\nabla{\phi} =0 $
To prove that $\nabla \times\nabla{\phi} =0 $.  The argument defines that $\nabla{\phi} = \frac{\partial}{\partial{x}} \hat{i} + \frac{\partial}{\partial{y}} \hat{j} + \frac{\partial}{\partial{z}} \hat{k}$. But shouldn't the dot product result in a scalar and not a vector? Though If it did result in a scalar, one cannot take the cross product of a scalar and a vector, so why does the dot product result in a vector here ?

答: You've conflated the gradient $\nabla\phi$ of a scalar $\phi$ with the divergence $\nabla\cdot V$ of a vector $V$. Gradients are vectors; divergences are scalars.


问:
Can these simultaneously be integers?
I am trying to figure out if there exist $k \in \mathbb{N}$ and $\omega \in \mathbb{R}$ ($\omega>0$ and  $\omega$ is not odd) such that
\begin{align}
\frac{k}{1+\omega} +\frac12  \quad \text{and}  \quad  \frac{k}{1+\tfrac{1}{\omega}} +\frac12 \\
\end{align}
are both integers  strictly greater than $1$.

I know I have a possibly strange constraint on $\omega$ but it is relevant to the problem I am trying to work on.

Assuming that $\frac{k}{1+\omega} +\frac12  = n $ for some $n \in \mathbb{N}$ and
$ \frac{k}{1+\tfrac{1}{\omega}} +\frac12 =m $ for some $m \in \mathbb{N}$

I arrive at
\begin{equation}
(2n-1)(1+\omega) = (2m-1)(1+\tfrac{1}{\omega})
\end{equation}

My suspicion is that there is no solution to this problem, and I tried to arrive at a contradiction that $\omega$ would have to be odd.

Is this the right approach? Or is my conjecture false and there is a solution?

I would greatly appreciate any help. Thank you!

答:
$$\frac{k}{1+\omega}+\frac{1}{2} = m $$
is equivalent to $$ \omega = \frac{2k}{2m-1} - 1 $$
Here $\omega$ will not be an odd integer if $k$ is not a multiple of $2m-1$.
Now
$$ \frac{k}{1+1/\omega} + \frac{1}{2} = k - m + 1 $$
so if $m$ and $k$ are integers, this is automatically an integer.  You also want $m > 1$ and $k - m + 1 > 1$, and this is easy to arrange.  For example, take $m = 3$, $k = 4$, $\omega = 3/5$.

Title: Can these simultaneously be integers?


问:
Continuous functions quotient derivative at $0$
This problem was on an exam I took (I have tried to remember it how it was but I don't have the original transcript).

Let $X$ be a metric space and let $f_1,f_2:X\rightarrow \mathbb{R}$ be two continuous differentiable functions. Suppose at some point $x_0$, $f_1(x_0)=0=f_2(x_0)$ and $f'_1(x_0)=C_1$ and $f'_2(x_0)=C_2$ for some non-zero finite constants $C_1$ and $C_2$. Prove that:

$$\lim_{x\rightarrow x_0}\frac{f_1(x)}{f_2(x)}=K$$

for some finite constant $K$.

The exam is over and I got my mark but they don't give us the answers and it has bugged me since. Noone I know seems to have been able to answer it. Could anyone help me with this?

答:
Directly applying the limit $x \rightarrow x_0$ yields the indeterminate form $\frac{0}{0}$. Thus (and since your functions are continuous and differentiable) you may then apply L'Hopital's Rule to obtain a quotient of the derivatives. Applying the limit now yields a constant (no longer an indeterminate form):

$$\lim_{x\rightarrow x_0}\frac{f_1(x)}{f_2(x)}=\lim_{x\rightarrow x_0}\frac{f'_1(x)}{f'_2(x)}=\frac{f'_1(x_0)}{f'_2(x_0)}=\frac{C_1}{C_2} \equiv K$$

where the first equality is obtained by L'Hopital's Rule.


问:
Sigma-algebra independent from itself
How can I construct a measure $\mu$ such that a $\sigma$-algebra $A$ on a non-empty set $X$ is independent from itself, that is $\mu(B \cap C)=\mu(B)\mu(C)$ for every $B,C\in A$.
Thank you

答: This can only be a 0-1-measure. Indeed, if $\mu(C)=a\neq 0$, then $\mu(C) = \mu(C\cap C) = \mu(C)^2$, which only has solutions $\mu(C)=1$ or $\mu(C)=0$. So, $\mu(B)=1$ or $\mu(B)=0$ for all $B$.

Title: Sigma-algebra independent from itself


答:
From $\mu(B)=\mu(B\cap B)=\mu(B)^2$, you get that $\mu(B)$ is $0$ or $1$.

If $B\cap C=\emptyset$, then $0=\mu(B\cap C)=\mu(B)\mu(C)$.

If $\mu(B)=1$ for a certain $B$, then $\mu(C)=1$ for all $C\supset B$.

So an example would be any Dirac measure $\delta_{x_0}$ for $x_0\in X$ and $\mathcal A=\mathcal P(X)$.

Or more generally, fix a set $B_0\subset X$, take $\mathcal A$ any $\sigma$-algebra with $B_0\in\mathcal A$ and such that no nonempty proper subset of $B_0$ is in $\mathcal A$, and define $\mu(B)=1$ if $B_0\subset B$, and $0$ otherwise.

Title: Sigma-algebra independent from itself

Body:
How can I construct a measure $\mu$ such that a $\sigma$-algebra $A$ on a non-empty set $X$ is independent from itself, that is $\mu(B \cap C)=\mu(B)\mu(C)$ for every $B,C\in A$.
Thank you


问:
Definition of Strongly Shattering
I am familiar with what it means for a collection $\mathcal{F}\subseteq 2^{[n]}$ to shatter a set $S$, $$2^{S} = \{F \cap S: F\in \mathcal{F}\}.$$

I'm reading a paper and they introduce the notion of strongly shattering a set $S$. If there exists an $I\subseteq [n]\backslash F$ such that $$2^F + I = \{H\cup I: H\subseteq F\}\subseteq \mathcal{F}.$$

This definition does not appear to follow directly from the definition of shattering a set. I would appreciate if someone could explain this definition in plain English so that I could better understand it.

It is clear to me that if $2^F \subseteq \mathcal{F}$ then $\mathcal{F}$ shatters $F$, although this is not the only way in which $\mathcal{F}$ can shatter $F$. Is the idea that $\mathcal{F}$ shatters $F$ and more?

答:
The statement "$\mathcal F$ shatters $S$" can be stated as "For all subsets $A \subseteq S$, there is a set $B$ disjoint from $S$ such that $A \cup B \in \mathcal F$".

The statement "$\mathcal F$ strongly shatters $S$" can be stated as "There is a set $B$ disjoint from $S$ such that for all subsets $A \subseteq S$, $A \cup B \in \mathcal F$".

The only difference is that we must use the same $B$ for every $A$. This is a stronger condition: if $\mathcal F$ strongly shatters $S$, then $\mathcal F$ shatters $S$, but the reverse is not necessarily true.

For example:

to shatter $\{1,2\}$, $\mathcal F$ might contain the sets $\{3,4,5\}$ (for $\emptyset \subseteq \{1,2\}$), $\{1, 6\}$ (for $\{1\} \subseteq \{1,2\}$),  $\{2,4,8\}$ (for $\{2\} \subseteq \{1,2\}$) and $\{1,2,3\}$ (for $\{1,2\} \subseteq \{1,2\}$).

to strongly shatter $\{1,2\}$, $\mathcal F$ might contain all the sets $\{3,5\}$, $\{1,3,5\}$, $\{2,3,5\}$, and $\{1,2,3,5\}$.

Title: Definition of Strongly Shattering


Title: Can these simultaneously be integers?

Body:
I am trying to figure out if there exist $k \in \mathbb{N}$ and $\omega \in \mathbb{R}$ ($\omega>0$ and  $\omega$ is not odd) such that
\begin{align}
\frac{k}{1+\omega} +\frac12  \quad \text{and}  \quad  \frac{k}{1+\tfrac{1}{\omega}} +\frac12 \\
\end{align}
are both integers  strictly greater than $1$.

I know I have a possibly strange constraint on $\omega$ but it is relevant to the problem I am trying to work on.

Assuming that $\frac{k}{1+\omega} +\frac12  = n $ for some $n \in \mathbb{N}$ and
$ \frac{k}{1+\tfrac{1}{\omega}} +\frac12 =m $ for some $m \in \mathbb{N}$

I arrive at
\begin{equation}
(2n-1)(1+\omega) = (2m-1)(1+\tfrac{1}{\omega})
\end{equation}

My suspicion is that there is no solution to this problem, and I tried to arrive at a contradiction that $\omega$ would have to be odd.

Is this the right approach? Or is my conjecture false and there is a solution?

I would greatly appreciate any help. Thank you!

Body:
Note that
$$
(2n-1)(1+\omega) = (2m-1)(1+\tfrac{1}{\omega}) \Leftrightarrow \\
\omega(2n-1)=(2m-1) \Leftrightarrow \\
\omega =\frac{2m-1}{2n-1}
$$
set
$$
k=m+n-1
$$

Then
$$
\frac{k}{1+\omega} +\frac12 = \frac{m+n-1}{\frac{2m+2n-2}{2n-1}}+\frac{1}{2}=m \\
\frac{k}{1+\tfrac{1}{\omega}} +\frac12 =\frac{m+n-1}{\frac{2m+2n-2}{2m-1}}+\frac{1}{2}=n\\
$$


问:
Is there a space of all finite models?
Let $\mathcal{L}=\{R_i\}_{i \in I}$ be a relational language. There is a natural construction of a compact space of countable $\mathcal{L}$-structures, namely, the space $$ \prod_{R_i \in \mathcal{L}} 2^{\mathbb{N}^{n_i}}$$
where $n_i$ is the arity of the relation symbol $R_i$.

Here the intuition behind the topology is that the more the relations of two structures agree on finite fragments of the underlying set (which is chosen as $\mathbb{N}$ without loss of generality) the closer these structures are. This idea will clearly break down if one considers finite $\mathcal{L}$-structures since a finite amount of data regarding its relations will enable us to distinguish a structure from all the others, so we should end up getting the discrete topology. Indeed, we do get this if $\mathcal{L}$ is finite and we replace $\mathbb{N}$ by $\{0,1,\dots,n\}$ and take the disjoint union of these spaces as $n$ ranges over $\mathbb{N}$.

Can one construct a topological space of all finite $\mathcal{L}$-structures other than the discrete space? Just to avoid any non-trivial topologies arising from the cardinality of $\mathcal{L}$, let us assume that $\mathcal{L}$ is finite; so we are looking for a countably infinite (preferably Hausdorff) topological space.

Obviously I'd prefer to have some kind of intuition behind the notion of closeness in this space. Given how much the space of countable structures has been studied, surely someone must have tried to construct such a space.

答:
This is not an answer, but rather an addendum to your answer that is far too long for comments. To try to start getting a sense of your topology, I looked at three simple cases.

If $\mathscr{L}$ is empty, you’re just getting $\omega$ with the topology whose non-empty sets are the tails; call this space $W$.

If $\mathscr{L}$ has a single constant symbol, we can identity $\operatorname{Mod}_{<\omega}(\mathscr{L})$ with

$$\{\langle n,m\rangle\in\omega\times\omega:m\le n\}\,:$$

$\langle n,m\rangle$ corresponds to the model $\mathscr{M}(n,m)$ on $\{0,\ldots,n\}$ with the constant interpreted as $m$. $U_{\mathscr{M}(n,m)}=\{\mathscr{M}(k,m):k\ge n\}$, so $\operatorname{Mod}_{<\omega}(\mathscr{L})$ is homeomorphic to $W\times\omega$, where $\omega$ has the discrete topology.

If $\mathscr{L}$ has a single unary relation symbol $R$, $\operatorname{Mod}_n(\mathscr{L})$ is a discrete set of power $2^{n+1}$. For each $s:\{0,\ldots,n\}\to\{0,1\}$ let $\mathscr{M}_s\in\operatorname{Mod}_n(\mathscr{L})$ be the model for which $s$ is the indicator function of the interpretation of $R$. If $m,n\in\omega$, $m\le n$, $\mathscr{M}_s\in\operatorname{Mod}_m(\mathscr{L})$, and $\mathscr{M}_t\in\operatorname{Mod}_n(\mathscr{L})$, then $\mathscr{M}_t\in U_{\mathscr{M}_s}$ if and only if $s=t\upharpoonright\{0,\ldots,n\}$.

Let $T=\bigcup_{n\in\omega}{^n}2$, so that $\langle T,\subseteq\rangle$ is the full binary tree of height $\omega$, and for $t\in T$ let $U_t=\{s\in T:t\subseteq s\}$; $\{U_t:t\in T\}$ is a base for a topology $\tau$ on $T$. The previous paragraph shows that in this case $\operatorname{Mod}_{<\omega}(\mathscr{L})$ is homeomorphic to $2\times T$, where $2$ has the discrete topology. It would be homeomorphic to $T$ itself if you included the empty model.

Title: Is there a space of all finite models?


问:
Why are abelian groups of interest? What is their usefulness?
I am reading about Abelian groups

So apparently it is a set, with an associative binary operation, and identity element, an inverse operation and the binary operation must also be symmetric.

But it is not clear to me how they are useful. Trying to find why they are important it seems they arise as "additive structures" in various systems but this is too abstract for me.

Could someone give some less formal/more practical application or usages showing what is important about abelian groups (layman's terms basically)?

答:
Informally speaking, an abelian group is a place where you can sum. It's the essence of what defines the sum of ordinary 'counting numbers'.

This allows you - after some exposure - to handle very abstract objects with the same familiarity you have for the latter.

As for any abstract definition, it is not going to make sense unless you have some examples at hand. And as for most abstract definitions, the examples came first. Many, many objects come with a binary operation that is associative and commutative (with unit and inverses) so we gave it at name.

Any "number system" $R$ (technically, I mean a ring here) has a notion of sum $+$, and $(R,+)$ is an abelian group. This includes very familiar number systems such as the integers, rational, real and complex numbers.

But is also includes for example matrices over these number systems. In general, product of matrices is known to depend on the order of the factors, but not their sum. Hence, we can sum matrices 'as if they were numbers'.

Another example is the "number system" $\mathbb{Z}_{12}$, which behaves like hours in a clock. You can think of it as the numbers from $1$ to $12$, but here e.g. $11+4 = 3$. It may be uncomfortable to work with this at first, but knowing that $+$ behaves similarly to ordinary numbers helps.

The list goes on, of course, but it becomes more abstract.

Another thing which may be useful to think about is... well, non-abelian groups. As we said before, in general for matrices $A,B$ we have $AB \neq BA$. Things get non-abelian really quickly in real life too: it is not the same to put your jacket on first and then your t-shirt than doing so in the reverse order.

Commutativity is far from a 'given', hence it is important to know when it does hold. It makes some things easier to organize. But as I said before, if you are not convinced that groups are important in the first place then it may not be clear why them being abelian is a thing to care about.


问:
Integrate $\frac{\log(x^2+4)}{(x^2+1)^2}$.

Using residue calculus show that
$$\int_0^{\infty}\frac{\log(x^2+4)}{(x^2+1)^2}dx=\frac{\pi}2\log 3-\frac{\pi}6.$$

I was thinking of using some keyhole or semi-circular contour here. But the problem is apart from poles at $x=-i$ and $x=i$, the logarithm has singularities when $x=\pm 4i$.

I consider $C_R$, a semicircle contour oriented clockwise with radius $R$ centered at origin. The semicircle resides in the lower half plane, so that it encloses $x=-i$ as a pole.  I set
$$\int_{C_R} \frac{\log(2-ix)}{(x^2+1)^2}dx$$
But it seems like it leads to wrong answer.

答:
Integrating
$$
\int_\gamma\frac{\log\left(z^2+4\right)}{\left(z^2+1\right)^2}\,\mathrm{d}z\tag1
$$
along the contour

gives $2\pi i$ times the residue at $z=i$.

The integral along the curved pieces vanish as the radius of the large semi-circle grows to $\infty$.

The integral along the downward line along the right side of the imaginary axis to $2i$ (in red) is
$$
-i\int_2^\infty\frac{\log\left(x^2-4\right)+\pi i}{\left(x^2-1\right)^2}\,\mathrm{d}x\tag2
$$
The integral along the upward line along the left side of the imaginary axis from $2i$ (in green) is
$$
i\int_2^\infty\frac{\log\left(x^2-4\right)-\pi i}{\left(x^2-1\right)^2}\,\mathrm{d}x\tag3
$$
For $z=ix$ and $x\gt2$, we have one of $\log\left(z^2+4\right)=\log\left(x^2-4\right)\pm\pi i$. Integrating $\frac1{z-2i}+\frac1{z+2i}$ clockwise around $2i$ decreases $\log\left(z^2+4\right)$ by $2\pi i$. Thus, on the right side of the branch cut, we have $\log\left(x^2-4\right)+\pi i$, and on the left side, we have $\log\left(x^2-4\right)-\pi i$.

The sum of the integrals in $(2)$ and $(3)$ is
$$
2\pi\int_2^\infty\frac1{(x^2-1)^2}\,\mathrm{d}x=\frac\pi6(4-3\log(3))\tag4
$$
The integral along the real axis (in blue) is
$$
2\int_0^\infty\frac{\log(x^2+4)}{(x^2+1)^2}\,\mathrm{d}x\tag5
$$
Furthermore,
$$
2\pi i\operatorname*{Res}_{z=i}\left(\frac{\log(z^2+4)}{(z^2+1)^2}\right)=\frac\pi6(2+3\log(3))\tag6
$$
Subtracting $(4)$ from $(6)$ and dividing by $2$ gives
$$
\int_0^\infty\frac{\log(x^2+4)}{(x^2+1)^2}\,\mathrm{d}x=\frac\pi2\log(3)-\frac\pi6\tag7
$$


问:
Do solutions to the forced Stokes equations for zero-Reynolds number flows always exist?
I'm in the process of solving a forced Stokes flow problem in two dimensions, the governing equations of which are given by
$$
\nabla p = \eta \Delta \mathbf{v} + \mathbf{f}, \quad \nabla\cdot \mathbf{v} = 0.
$$
where $\mathbf{f} = f_x\mathbf{e}_x + f_y\mathbf{e}_y$ is known. I am attempting to solve the problem via a stream function: we let
$$
\mathbf{v} = \frac{\partial\psi}{\partial y} \mathbf{e}_x - \frac{\partial\psi}{\partial x} \mathbf{e}_y.
$$
By taking the curl of the Stokes equation it can be shown that $\psi$ then satisfies a bi-Poisson equation
$$
\Delta^2 \psi = \frac1\eta \left[\frac{\partial f_y}{\partial x} - \frac{\partial f_x}{\partial y}\right].
$$
My question is are there any restrictions on the forcing term $\mathbf{f}$, other than differentiability, for the problem to be well-posed? In the case I'm working on (I haven't included the details as it is a messy boundary layer approach since $\mathbf{f}$ is a bunch of Bessel functions that must be approximated) I have successfully solved for the stream function but cannot integrate it up to find the pressure: backfilling into the original Stokes equation gives me an expression for $\nabla p$ that is not the gradient of a function. Can anyone suggest why this might be?

答:
I don't have a yes/no answer, but it sounds like you have a vector function, $\eta\Delta v+f$, with curl zero but it is not a gradient. This can happen in  a region that is not simply-connected. The classic example is
$$ \frac{(-y,x)}{x^2+y^2}, \qquad \qquad (x,y)\ne(0,0)$$
which looks like the gradient of $\tan^{-1}(y/x)$ but isn't because the inverse tangent function is not continuous all the way around the origin. Maybe this kind of thing has happened if you are working with an external flow around an airfoil, for example.
It doesn't mean there is no solution, but it might mean there is no stream function.


答:
I'll start with the assumption that you think that the integers $\Bbb{Z}$, the rational numbers $\Bbb{Q}$, and/or the real numbers $\Bbb{R}$ are useful or interesting. All of these are examples of Abelian groups. An Abelian group is just an arithmetic system where "addition" makes sense (and is commutative, associative, etc.). It is a common idea in math to take an object of interest and to study an object one step more abstract. For instance, we formalize in the properties of an Abelian group the fundamental properties of $\Bbb{Z}$. Then, any theorem we can prove for Abelian groups $A$ will apply to any of the examples we care about, like $\Bbb{Z}$.

One might reasonably argue that proving results for $\Bbb{Z}$ on its own might be easier than proving more general results about Abelian groups since more should be true in a specific case. This is true to an extent, but sometimes in stripping down objects to their core properties the important features are no longer hidden by the less important features. This is the sort of thing you start to realize with experience, I think.

More concretely, we know that to study divisibility in the integers, modular arithmetic is essentially indispensable. While we could try to formulate theorems about $\Bbb{Z}$ and $\Bbb{Z}/n\Bbb{Z}$ (integers modulo $n$) separately, we would quickly begin to realize that we are wasting time as there are similar structures common to both of these objects. The notion of Abelian group allows us to treat $\Bbb{Z}$ and all of the $\Bbb{Z}/n\Bbb{Z}$'s on even footing.

Sometimes, it is also more conceptually simple to think of things in a more abstract manner. The polynomial rings $\Bbb{Q}[x],\Bbb{R}[x],\Bbb{C}[x]$ all have a division algorithm which gives them the structures of a Euclidean domain. When considering this, it is impossible to miss the resemblance to the division algorithm we learn in school for $\Bbb{Z}$. Viewing these objects as being members of the same family of objects allows us to prove theorems for them all at the same time and to compare and contrast their individual properties. For instance, we can compare the notions of prime factorization in these rings.

Finally, you might think about this like learning a foreign language in some sense. If you only speak your mother tongue, you will have a strong grasp of it no doubt. However, once you begin to learn other languages you can appreciate the features they have that your own language does not. In turn, this typically gives you a deeper appreciation of the structure of your own language. I would say it's the same thing: learning about Abelian groups will change the way you understand polynomials and the integers and enrich your knowledge of them.

Title: Why are abelian groups of interest? What is their usefulness?

Body:
I am reading about Abelian groups

So apparently it is a set, with an associative binary operation, and identity element, an inverse operation and the binary operation must also be symmetric.

But it is not clear to me how they are useful. Trying to find why they are important it seems they arise as "additive structures" in various systems but this is too abstract for me.

Could someone give some less formal/more practical application or usages showing what is important about abelian groups (layman's terms basically)?


问:
Separated sets proof
For sets $X, Y$ and $Z$, if $X$ and $Y$ are separated, and $X$ and $Z$ are separated, then $X$ and $Y \cup Z$ are also separated.

I am reading up on some set theorems and I am looking for a complete proof of this one.

What I know:

If $X$ and $Y$ are separated then: $X \cap \overline{Y} = \overline{X} \cap Y = \emptyset$.

In the same way $X \cap \overline{Z} = \overline{X} \cap Z = \emptyset$.

It needs to be shown that $X \cap \overline{Y \cup Z} = \overline{X} \cap (Y \cup Z) = \emptyset$.

Any assistance with a proof is much appreciated.

答:
$X \cap \overline{(Y \cup Z)} = X \cap (\overline{Y} \cup \overline{Z}) = (X \cap \overline{Y}) \cup ( X \cap \overline{Z}) = \emptyset \cup \emptyset= \emptyset$

and

$\overline{X} \cap (Y \cup Z) = (\overline{X} \cap Y) \cup (\overline{X} \cap Y) = \emptyset \cup \emptyset= \emptyset$

does the job. We only really need that closure commutes with finite unions for the first, plus the distributive properties of union and intersections.

Title: Separated sets proof


问:
Prove that map from $X$ to $RP^2\vee S^1$ is null homotopic
I am doing this UW Madison geometry qualifying exam, and I got stuck on this one :

Let $R=S^1\vee S^1$ as in the figure, and $X$ be the cellular 2-complex obtained by attaching two 2-cells to $R$ as indicated. Let $Y=RP^2\vee S^1,$ show that every map $f:X\rightarrow Y$ is nullhomotopic.

It is not hard to see using algebraic means that the induced map on $\pi_1$ is the trivial map (it has been asked before but with no answer: Map to $RP^2 \vee S^1$ nullhomotopic), however because the universal cover of $RP^2\vee S^1$ is not contractible, this is not enough. Can someone please help me finish the argument please ? Thank you !

答:
EDIT: My previous answer gave an argument which only showed that a map $X \to S^1 \vee \mathbb{R}P^3$ is nullhomotopic.  Thanks to @Jason Devito for pointing it out.  I planned to delete this incorrect answer, but I had some other thoughts which I offer as a mea culpa below.

One way to approach the problem is to note that the space $X$ is the cofiber (mapping cone) of a map $S^1 \vee S^1 \to S^1 \vee S^1$ given by "$(a^3, (ab)^5)$". Therefore, there is a long exact sequence of pointed sets $$\cdots \to [S^2 \vee S^2, Y] \xrightarrow{\text{(ii)}} [S^2 \vee S^2, Y] \to [X, Y] \to [S^1 \vee S^1, Y] \xrightarrow{\text{(i)}} [S^1 \vee S^1, Y].$$  In order to show that $[X,Y] = \{*\}$, we would need to show that the map labelled by (i) is injective and the map labelled by (ii) is surjective.

That (i) is injective follows from the algebraic manipulations alluded to in the OP.  The map (i) takes a pair $(a, b)$ of elements in $\pi_1 Y$ to the pair $(a^3, (ab)^5)$.  Since $\pi_1 Y = \mathbb{Z} * \mathbb{Z}/2$, the relation $(a^3, (ab)^5) = (1,1)$ forces $a = b = 1$, so (i) is injective.

For (ii), we have $\pi_2 Y \cong \mathbb{Z}$, but the map $\mathbb{Z}^2 \to \mathbb{Z}^2$ given by $(m,n) \mapsto (3m, 5(m+n))$ is not surjective.  So this would suggest that $[X,Y] \neq \{*\}$.


问:
Distributive property of matrix-vector multiplication?
I know that matrix multiplication is not distributive but what about matrix-vector multiplication? If $A \in \mathbb{R^{m \times n}}$ and $\vec{x}+\delta \vec{x} \in \mathbb{R^{n \times 1}}$. Then can I write $A(\vec{x}+\delta \vec{x})$ as $A(\vec{x}) + A(\delta \vec{x})$? If yes, what is this property/law called?

答:
Matrix-vector multiplication is a special case of matrix multiplication, which is distributive.

(In general, matrix multiplication is not commutative, but it is distributive.)

Your claim that $A(\vec{x}+\delta \vec{x})=A(\vec{x}) + A(\delta \vec{x})$ can also be seen as linearity.


问:
Automata | Proving that if $L$ is regular then $L'$ (word in $L'$ is a word from $L$ without the first and the last letter) is regular too.
I've see couple of approaches to this kind of questions yet I have no clue how to approach this one.

Let $L$ be a regular language and then let $L'$ be: $L' = \{w \in \Sigma^\star : awb \in L, a \in \Sigma, b \in \Sigma \}$ (word in $L'$ is a word from $L$ without the first and the last letter). Prove that if $L$ is regular then $L'$ is regular too.

I tried to make a new DFA inherited from DFA that accepts words in $L$ but I'm stuck on how to do this. Any help appreciated :)

答:
SKETCH: Start with a DFA $M$ for $L$. Let $q_0$ be the initial state of $M$, let $Q_1$ be the set of states of $M$ that can be reached from $q_0$ in one transition, and let $Q_a$ be the set of states of $M$ that have at least one transition to an acceptor state of $M$. Modify $M$ as follows to get an NFA $M'$:

Change each transition $q_0\overset{x}\longrightarrow q$ with $q\in Q_1$ to an $\epsilon$-transition $q_0\overset{\epsilon}\longrightarrow q$.

Let $Q_a$ be the set of acceptor states of $M'$.

Then show that the NFA $M'$ accepts $L'$.


问:
If $f$ is an integrable function on $[0, \infty)$, does $f$ have an antiderivative?
I am not sure about continuity of an integrable function after integration.

Let's assume, $f$ is integrable at $[0, \infty)$. Then let $F$ be the antiderivative of $f$:

$$\int f(x)dx = F(x)$$

Now the integration being defined and real, means that $F$ is differentiable at $[0, \infty)$, we are given the anti derivative

$$\frac{dF}{dx} = f(x)$$

$$dF = f(x)dx = 0 \\ \lim_{\Delta x \rightarrow 0}\Delta F(x) = 0$$

Therefore it is continuous at $[0, \infty)$, I can say for example directly that

$$\lim_{x \rightarrow 0^+}F(x) = F(0)$$

And in a side note I know that continuity implies the limit at point $x$ from the left and right ($x+0, x-0$)

答:
There is a detailed discussion in chat about this question. There was some confusion about the relationship between the antiderivative of a function, and the integrability of the function.

First of all, note that the fundamental theorem of calculus says that if $f$ is a continuous function, then the definite integral $$F(x) = \int_{0}^{x}f(t)\, dt$$
satisfies $F'(x) = f(x)$ for all $x$. We call a function $F$ with this property an antiderivative of $f$.

Not all integrable functions are continuous. The confusion in the OP was about the antiderivative of $f$ in this situation. It turns out that merely integrable functions might not have an antiderivative.

Consider, for example, the function $$f(x) = \begin{cases} 1 & x=0\\0 & x \ne 0\end{cases}$$

Then $f$ is a Riemann integrable function. In fact,

$$F(x) = \int_{0}^{x}f(t)\, dt = 0$$

for all $x$. So $F'(x) = 0$ for all $x$, meaning that the definite integral is not an antiderivative of $f$. In fact, there is no differentiable function $F$ with $F'(x) = f(x)$ for every $x$. To see this, note that if there were such an antiderivative, it would have to be constant on $(0, \infty)$ (because the derivative is 0 there). But then, because it's continuous, it has to have $F(0)$ be the same constant - meaning that $F'(0) = 0 \ne f(0)$.

To come back to the original point: If $f$ has an antiderivative $F$, then $F$ is differentiable by definition, so it is continuous too. But if $f$ is integrable, it might not have an antiderivative.


问:
What does it mean that a diagram is commutative in category theory?
'In category theory, a diagram commutes means that between any two objects in the diagram, there is a unique morphism. And the identity morphisms and compositions of morphisms might be omitted in the diagram.'

I can understand the first half of this paragraph. However, the meaning of the last half is vague. My question is very explicit, as follows:

If there is one object, which appears in the diagram twice, is there an implicit identity morphism between its twice appearance? And does it mean that all the compositions of morphisms from object $X$ to object $X$ is the identity morphism of $X$?

答:
In general I would assume that a diagram said to commute and containing
the same object twice does not have an implicit identity between those appearances of the same object.

You do get the same object twice in examples such as those shown in:
https://en.wikipedia.org/wiki/Monad_(category_theory)
with the diagrams defining a monad. In one diagram $T^2$ appears twice with no identity being intended. In the other both $T^2$ and $T$ appear twice with an explicit identity on $T$ but definitely no intention that there is one on $T^2$.


问:
My question is about operators on $L^p$ spaces
I am not sure how to proceed. This is the question:
let $1<p<+\infty$ and let $q$ be the conjugate exponent. Show that if $f\in L^q((0,\infty))$ then
$Tf(x)=x^{-1/p}\int_0^x f(t)dt\in C_0((0,+\infty))$
and furthermore
$\|Tf\|_\infty \leq C\|f\|_q.$

答:
First of all, notice that $T: L^q(\mathbb R^+_0) \to C(\mathbb R^+_0)$ is well defined by Holder's inequality since
$$|Tf(x)| = \left|x^{-1/p}\int_0^x f(t)~dt\right| \le x^{-1/p} \|\boldsymbol 1_{(0, x)}\|_p \|f\|_q = \|f\|_q. $$
Moreover, for $y \in (0, +\infty)$ and $(x_n)_n \subset (0, \infty)$ satisfying $x_n \to y$, we have $\boldsymbol 1_{(0,x_n)}(x) f(x) \to f(x)\boldsymbol 1_{(0,y)}(x)$ a.e. By dominated convergence theorem, we conclude that
$$\int_0^{x_n}f(t)~dt \to \int_0^y f(t)~dt.$$
Since the function $x \mapsto x^{-1/p}$ is continuous on $(0, +\infty)$, we deduce that $Tf(x)$ is also continuous as a product of continuous functions.


问:
Variety of Nilpotent Matrices
Let $k$ be an algebraically closed field and view $M_n(k)$ as $\mathbb{A}^{n^2}$.  $A\in M_n(k)$ is nilpotent if and only if $A^n=0$.  Since the equation $A^n=0$ is given by $n^2$ polynomial equations, it defines a variety in $\mathbb{A}^{n^2}$.  What is the dimension of this variety?  Is it irreducible?  If not, what are its irreducible components?</p>

答:
It's dimension is $n(n-1)$ and it is indeed irreducible.

The variety you are describing is the nullcone of the Lie algebra $\mathfrak{gl}_n$.  It's known (with maybe some mild hypotheses that $\mathfrak{gl}_n$ satisfies) that the nullcone of a Lie algebra is irreducible and its dimension is twice the dimension of a maximal unipotent subalgebra.  In the case of $\mathfrak{gl}_n$ this will be the subalgebra $\mathfrak{u}_n$ of upper triangular matrices with zero's on the diagonal, which has dimension $\frac{1}{2}n(n-1)$.

Both these facts are hard, I can't explain the proof to you here.  If you want a reference, see Jantzens section of the Birkhauser Lie Theory book.

ContentLicense: CC BY-SA 3.0

Title: Variety of Nilpotent Matrices


问:
Probability of being in each transient state in an absorbing Markov chain, given that you are not in an absorbing state?
I have tried to derive a generalized answer to this question, but don't know how to check my work. To clarify, I am asking for the probability of being in each transient state of an absorbent Markov chain after an unknown, finite amount of time has passed, given that you are not in an absorbing state.

I'll start with an example to show my approach.

Start with a Markov chain with N states, where the $n^{th}$ state is absorbing ($p_{nn} = 1, p_{ni} = 0$ $\forall$ $i \not= n$). We will call the transition matrix for this Markov chain P, where each element $p_{ij}$ is the probability of transitioning from state i to state j. We want to know the probability of being in each state i given that we are not in state n after some finite, unknown amount of time has passed. Note that we are not solving for the steady state distribution, which gives you the probabilities of being in each state after an infinite amount of time has passed.

My approach from here was to construct a new transition matrix $Q$, which contains the probability of transitioning between all the transient states of this MC. To do so, we remove the $n^{th}$ row and column from $P$ to create a new matrix, say, $P(A)$. Now, we must rescale the values of $P(A)$ such that the row elements all add up to 1 for all rows. One way to do so is:

take the dot product of $P(A)$ and a vector of ones with N-1 dimensions, and call the resulting N-1 vector $\vec S$

diagonalize this vector into a matrix called D, so that $D_{ii} = S_i$ and $D_{ij} = 0$ $ \forall i \not = j$.

Now, $Q = D^{-1}P(A)$

If I am not mistaken, the elements of $Q$, called $q_{ij}$, should represent the probability of transitioning from state i to state j, given that i does not transition to state n (the absorbing state). I (quite possibly mistakenly) believe that the steady state vector of Q gives the answer to my original question. Can anyone provide any insight here? I am struggling to find any resources that I can use to confirm my approach.

答:
Your approach will not work.

Suppose we have the following $3$-state Markov chain. States $1$ transitions to itself with probability $0.99$, and to state $2$ with probability $0.01$. State $2$ always goes to state $3$, and state $3$ is absorbing. We start in state $1$.

Then at any time $t>0$, the probability of being in state $2$, given that you're not in state $3$, is $0.01$ exactly. Being told that you're not in state $3$ tells us that we stayed put in state $1$ for the first $t-2$ transitions, and in particular that we were in state $1$ at time $t-1$ (if we went to state $2$ earlier, we'd be in state $3$ by now). Then we have a $0.99$ chance of being in state $1$ at time $t$, and a $0.01$ chance of being in state $2$.

On the other hand, the approach of deleting state $n$ (state $3$ in this case) and rescaling will make state $2$ absorbing, which means the probability of ending up in state $2$ will only grow.

Instead, to solve the problem for a known time $t$, compute the row vector $\vec w = \vec v P^t$, where $\vec v$ is the row vector of the initial probabilities. In other words, find the probability vector at time $t$. Then, set $w_n$ to $0$ and rescale the rest of $\vec w$ to add up to $1$. That's what conditioning on an event is.

You cannot solve this problem for a totally unknown time $t$. Even if you can't say what $t$ is for certain, you should have a probability distribution on $t$; in this case, compute the expected value $\vec w = \mathbb E[\vec v P^t]$, and use that instead.


问:
Why are pull backs always defined for 1 forms but not push forwards?
So I know that given a differentiable function $\mathcal{F}: \mathcal{M} \rightarrow \mathcal{N}$, we can define the push forward map:
$$
\mathcal{F}_*: T_p(\mathcal{M}) \rightarrow T_{\mathcal{F}(p)}(\mathcal{N})
$$

However, the pull back map for vectors would require the inverse function $\mathcal{F}^{-1}$ which may not necessarily exist and this is why we cannot define a pull back for vector fields.

Now coming to forms, I do not understand why the pull back is always defined while the push forward is not? That is, given that I know $\langle \omega, v \rangle$, shouldn't I be able to define $\langle \mathcal{F}_* (\omega), \mathcal{F}_*(v)\rangle$? Why is this not possible?

答:
I just want to add some intuition here for you. You should think of pullback of forms in terms of substitution (as you did with substitutions in integrals in your second calculus course). For example, if $f\colon\Bbb R\to \Bbb R^2$, say $f(t)=(t\cos t, t^2)$, and $\omega$ is a $1$-form on $\Bbb R^2$, say $\omega = P\,dx + Q\,dy$, then
$$f^*\omega = f^*(P\,dx + Q\,dy) = P(t\cos t,t^2)d(t\cos t) + Q(t\cos t,t^2) d(t^2)$$
is what you get merely by substituting $x=t\cos t$, $y=t^2$, and using $dg=g'(t)\,dt$ as you did in substitutions in integrals. Indeed, the whole premise of substitution is that you're pulling back and doing the integral in a new coordinate system.


问:
counting numbers with specific Quality
What is the count of 10 digit numbers with the exact sequences 72 and 081 in them with no repetition .
Is there any fast way to calculate this because of the 081 the places of 72 are limited and not coherent. For example placing 081 nearly at the end with 1 digit next to it potentially removes the ability to place 72 there</p>

答:
As the number must contain $081$ and $72$, we have to count how many numbers are there containing "pieces": $081$, $72$, $3$, $4$, $5$, $6$, $9$ with no repetitions. There are $7$ pieces, so if we allowed the number to start with $081$, the pieces could be arranged in $7!$ ways. However, as the number cannot start with $081$, we can place any of the $6$ remaining pieces in the first position, any of the $6$ pieces (the remaining ones and $081$) in the second position, any of the $5$ pieces in the third position and so on.

We conclude that there are therefore $6\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ such numbers.

Title: counting numbers with specific Quality


问:
Show $\mu_*$ is a measure

Suppose $\mu$ is a finitely additive set function on Borel $\sigma$ algebra on $\mathbb R$ with $\mu(\mathbb R)<\infty$. Define $\mu_*(A)=\sup \{\mu(K)|K\subset A,K$ compact$\}$. Show that $\mu_*$ is a measure on Borel sets.

I cannot even show $\mu_*$ is finitely additive. More specifically, I need help in the following direction. If A,B are disjoint Borel then $\mu_*(A\cup B)\leq \mu_*(A)+\mu_*(B)$. The issue is, if I take any compact K subset of $A\cup B$ then I cannot split K to get compact sets K1 and K2 subsets of A and B respectively.

答:
There is a not-mentioned hypothesis, which I think is needed, which is that $\mu\geq0$.

Note first that $\mu$ preserves order, since for $A\setminus B$ we have
$$
\mu(A)\leq\mu(A)+\mu(B\setminus A)=\mu(A\cup(B\setminus A))=\mu(B).
$$
And $\mu$ is subadditive, since $$
\mu(A\cup B)=\mu(A\cup(B\setminus A))=\mu(A)+\mu(B\setminus A)\leq\mu(A)+\mu(B).
$$

Let $A,B$ be Borel sets and $K\subset A\cup B$ be compact. Then $A\cap K$ and $B\cap K$ are Borel sets, so
$$
\mu(K)\leq\mu((A\cap K)\cup (B\cap K))\leq\mu(A)+\mu(B).
$$
Then $\mu_*(A\cup B)\leq\mu(A)+\mu(B)$.

If $\mu_*(A)=\infty$ or $\mu_*(B)=\infty$, the inequality holds. Otherwise, Fix $\varepsilon>0$ and choose $K_1\subset A, K_2\subset B$ compact with $\mu(A)<\mu(K_1)+\varepsilon$, $\mu(B)<\mu(K_2)+\varepsilon$. Then
$$
\mu_*(A\cup B)\leq \mu(A)+\mu(B)\leq \mu(K_1)+\mu(K_2)+2\varepsilon\leq\mu_*(A)+\mu_*(B)+2\varepsilon.
$$
As $\varepsilon>0$ was arbitrary, $\mu_*(A\cup B)\leq\mu_*(A)+\mu_*(B)$.

Title: Show $\mu_*$ is a measure


问:
For which values of $ \alpha$ ,$ e^{\alpha x} - (1+\alpha) \cdot x^2$ is convex in $\mathbb{R}$
I have to say for which values of $ \alpha$ the following function is convex in  $\mathbb{R}$:
$$ e^{\alpha x} - (1+\alpha) \cdot x^2$$

The second derivative is $$ \alpha^2 e^{\alpha x} - 2(1+\alpha) >0 $$

But I don't know how to proceed.

答:
HINT: Use the fact that a differentiable function $f$ is convex if and only if
$$f(x) \ge f(y) + f'(y) (x - y).$$


问:
Estimate the scale of the power series with Poisson pdf-like terms
Sorry to bother you, but I guess that this question is not appropriate for MO, so I repost it here hoping that someone could give me a clue.

I would like to have an estimate for the series
$$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m,$$
where $e$ is the base of natural logarithm, $k!$ is the factorial of the integer $k$, $t$ represents the time and $t>0$, $m$ is a positive integer and $m>1$ (Obviously, $P(t)=1$ when $m=1$ since $P(t)$ is exactly the cdf of Poisson distribution with associated parameter $t$ in this case.).

I am interested in showing that the order of the scale could look something like
$$P(t) = O(t^{-\alpha m}),\alpha > 0.$$

But I find it difficult for me to extend their method to the cases when $m>2$.

 I also tried to calculate $P(t)$ in Mathematica and it gave me the result like
$$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m = e^{-mt}\text{HypergeometricPFQ}[...,t^m]$$
It seems that the decreasing speed of $P(t)$ is much slower than that of $e^{-mt}$ as $t$ increases. But I could not find the closed-form bound for $P(t)$ when $m>2$ just like $O(1/\sqrt t)$ when $m=2$.

Does anyone know of a scale or a bound of $P(t)$ in the literature? Any comments and answers would be highly appreciated. Many thanks!

答:
The function you have, and this is also the closed form that Mathematica gave you, is related to the hypergeometric function,
$$ P(t) = e^{-m t}\,\,{}_0F_{m-1}(;1,1,\ldots,1; t^m). $$
To calculate its behaviour as $t\to\infty$, at least in this case, is relatively easy. In particular the DLMF gives the formulas necessary, 16.11.9 and 16.11.1-4.

The result is
$$ P(t) = \frac{1}{(2\pi)^{\frac{m-1}{2}}\sqrt{m}}\frac{1}{t^{\frac{m-1}{2}}} + O(t^{\frac{-1-m}{2}}). $$

You can also check this for small values of $m$ using Mathematica's function Series, which understands how to calculate asymptotic behaviour.


问:
The Baire space is homogeneous,
it's kwown that the Cantor space $2^{\omega}$ is a homogeneous topological space (because it is a topological group). Does anyone have any idea why $\omega^{\omega}$ is a homogeneous topological space?
Thanks.

答:
If $x,y\in\omega^\omega$, for each $n\in\omega$ let $f_n:\omega\to\omega$ be the map the interchanges $x_n$ and $y_n$ and leaves everything else fixed. Let

$$h:\omega^\omega\to\omega^\omega:z\mapsto\langle f_n(z_n):n\in\omega\rangle\,;$$

then $h(x)=y$, and $h$ is an autohomeomorphism of $\omega^\omega$.

Another approach is to note that $\omega^\omega$ is homeomorphic to $\Bbb Z^\omega$, and that the product topology on $\Bbb Z^\omega$ is the same as the order topology induced by the lexicographic order $\preceq$. Now let $\sigma,\tau\in\mathbb{Z}^\omega$, and define $\delta\in\mathbb{Z}^\omega$ by $\delta(n)=\tau(n)-\sigma(n)$. Then the translation $$s:\mathbb{Z}^\omega\to\mathbb{Z}^\omega:\varphi\mapsto\varphi+\delta=\langle\varphi(n)+\delta(n):n\in\omega\rangle$$ is an order-automorphism of $\langle\mathbb{Z}^\omega,\preceq\rangle$ taking $\sigma$ to $\tau$ and thus also an autohomeomorphism of $\Bbb Z^\omega$ with the order topology induced by $\preceq$.

Title: The Baire space is homogeneous,


答: It's also a topological group, just view it as $\Bbb Z^\omega$ in the obvious coordinatewise group operation. $\Bbb Z \simeq \omega$ topologically after all.

Title: The Baire space is homogeneous,

Body:
it's kwown that the Cantor space $2^{\omega}$ is a homogeneous topological space (because it is a topological group). Does anyone have any idea why $\omega^{\omega}$ is a homogeneous topological space?
Thanks.


问:
Combinatorics: precomputed all combination without repetition, need fast indexing
I've asked a similar (but not identical) question on SO here https://stackoverflow.com/questions/5984102 but the answers aren't that helpful for I need very fast indexing (and maybe that my question wasn't clear enough).

I precomputed a lot of values that took days to compute (with the load spread on several computers).  Now I've got a huge lookup table that I need to frequently access and that access has to be as fast as possible.

Basically, I've got a problem like this: I precomputed C(38,6) values (i.e. 2 760 681 values)  that are now store in a big lookup table.  Memory ain't an issue.  Precomputing ain't an issue.  The "order" ain't an issue and I can re-order in any way I like.

Also, C(38,6) is just an example, I could be needing C(42,7) as well, etc.

When I did the precomputing, I basically looped like this (the actual code is more complicated for the load needed to be spread across several machines and for the actual computation aren't here but here's the meat of it):

```<code>for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
        for (int k = j + 1; k < n; k++) {
            for (int l = k + 1; l < n; l++) {
                for (int m = l + 1; m < n; m++) {
                    for (int o = m + 1; o < n; o++) {
                       ...
                    }
                }
            }
        }
    }
}
</code>```

The only thing that matters is to have the fastest lookup time possible.

I've got i,j,k,l,m and n all in [0..37] (or [1..38], doesn't matter) and I need to access the corresponding precomputed element in my huge lookup array as fast as possible.

Is there a closed formula for this that is fast to compute?

What I've got now is a binary search: my i shall always be lower than j, j lower than k etc.  I then created an arbitrary index doing:

```<code>(i + 1) * (n exp 5) + (j + 1) * (n exp 4) + (k + 1) * (n exp 3) + (l + 1) * (n exp 2) + (m + 1) * n + o
</code>```

This is probably overkill but guarantees no collisions. I then do a binary search on that index and find where the precomputed value is stored.

This is kinda fast but I was wondering if there was a faster way to "reach" my precomputed index?

Basically because I don't know enumerative combinatorics well enough I'm precomputing my own index and using that index to do a binary search.  This is fast but I'm looking for something faster, like a closed formula (I doubt anything doing "enumerations" and i++ or things like that would be faster than those few multiplication to compute that index + the binary search).

I think the complexity of my method is log(n) but I'm just not sure.

答:
It is pretty absurd to store this at all!
There are various ways of writing down functions $f:[0,\binom nk]\to\mathcal{P}(\{1,2,\dots,n\}$ which are bijections, and computable orders of magnitude faster than any look-up involving I/O (you can surely do it without even getting a caché miss…)</p>


问:
Spin group without Clifford algebras
I have to build the spin group $Spin(n)$ without use Clifford algebras. Can I find a complete description of spin group with a topological method? How can I build $Spin(n)$ as the double covering of $SO(n)$?</p>

答: Define $Spin(n)$ as the universal cover of $SO(n)$. The universal cover of a Lie group is a Lie group (e.g. Theorem 2.5 on p. 10 of these notes). In particular, when $n\geq 3$, the homotopy exact sequence gives $\pi_1(SO(n)) = \mathbb{Z}_2$, so $Spin(n)$ is a double covering of $SO(n)$.</p>

Title: Spin group without Clifford algebras


问:
Convolving the relation $o(x)$
I'll get right to the point.  For sequences $\{a_n\}$ and $\{b_n\}$ in $\mathbf C$, let $c_n = \sum_{d \mid n} a_db_{n/d}$. Set $A(x) = \sum_{n \leq x} a_n$, $B(x) = \sum_{n \leq x} b_n$, and $C(x) = \sum_{n \leq x} c_n$. If $A(x) = o(x)$ and $B(x) = o(x)$, must it be true that $C(x) = o(x)$, or is there a (simple) counterexample? And if there is a counterexample, is the conclusion valid if we also assume $A(x)$ or $B(x)$ is $O_\delta(x^{1-\delta})$ for some $\delta$ in $(0,1)$ but introduce no further assumptions?

This question came up when editing some course notes. When proving various relations are equivalent to the prime number theorem, the proof that $\sum_{n \leq x} \mu(n) = o(x)$ implies $\psi(x) \sim x$ appears to be basically the same everywhere I have looked and I'd prefer if there were a different argument. In the standard proof (whose basic structure goes back to Landau (see Section 1)), you reduce the problem to checking that
$$
\sum_{dd' \leq x} \mu(d)(\log(d') - \tau(d') + 2\gamma) = o(x),
$$
where $\tau(n) = \sum_{d \mid n} d$ and $\gamma$ is Euler's constant. The proof of this $o(x)$ estimate uses the following three properties:

$\sum_{n \leq x} \mu(n) = o(x)$,

$\sum_{n \leq x} (\log(n) - \tau(n) + 2\gamma) = O(\sqrt{x})$,

$|\mu(n)| \leq 1$ for all $a$.

Abstracting what makes the proof work, we can say for my question at the start that $A(x) = o(x)$ and $B(x) = o(x)$ implies $C(x) = o(x)$ if we add the conditions that $\{a_n\}$ is bounded (think $a_n = \mu(n)$) and $B(x) = O_\delta(x^{1-\delta})$ for some $\delta$ in $(0,1)$ (think $b_n = \log(n) - \tau(n) + 2\gamma$ and $\delta = 1/2$). I would prefer a proof of $C(x) = o(x)$ that is more symmetric in the assumptions on $\{a_n\}$ and $\{b_n\}$ while still being applicable to the intended example above ($a_n = \mu(n)$ and $b_n = \log n - \tau(n) + 2\gamma$).

I have not seen a result of the form "$A(x) = o(x)$ and $B(x) = o(x)$ implies $C(x) = o(x)$" anywhere, so I suspect it might not be true, but I have not seen (or created) a counterexample either.

答:
For the simple counterexample:

Take $f$ increasing to $\infty$ such that $\tau(n)>f(n)$ for a positive proportion of integers.

($f(n)=\log\log(n+2)$ should work).

Let $$a(n)=b(n)= \frac1{\sqrt{f(n)}},\qquad a(d)b(n/d)\ge a(n)b(n)=\frac1{f(n)}$$

$a(n)=b(n)=o(1)$ so $A(n)=B(n)=o(n)$ but $$c(n)\ge \frac{\tau(n)}{f(n)}\ge 1$$ for a positive proportion of integers, therefore $C(n)\ne o(n)$.

Title: Convolving the relation $o(x)$


问:
Use Schur to prove that for any norm $\lim_{n \rightarrow \infty} ||A^n|| = 0 \text{ iff } \rho(A) < 1$ where $\rho$ denotes spectral radius.
I know I need to prove both directions of the biconditional, but am having some trouble in both ways.

First, looking left to right, If we assume the limit equals zero, it means that the $A$ matrix keeps getting smaller each time we apply $A$. Thus, since eigenvalue represent the effect of the matrix the maximum would have to be less than one but this doesn't seem very rigorous.

Then, I'm not sure how to proceed by assuming the spectral radius is less than 1 except for doing the backwards reasoning of what I talked about above. Any help is appreciated.

答:
$\rho(A)=\lim\sup ||A^{n}\|^{1/n}$. If $\rho(A) <1$ and $r =\frac {1+\rho (A)} 2$ then $\|A\|^{n}<r^{n}$ for $n$ sufficiently large and $r^{n} \to 0$ since $r<1$.

Suppose $\|A^{n}\| \to 0$. There exists $k$ such that  $\|A^{k}\|< \frac 1  2$. If $\lambda $ is an eigen value of $A$ then $\lambda ^{k}$ is an eigen value of $A^{k}$ so $|\lambda^{k} |<1$ which implies $|\lambda|<1$. Hence, $\rho(A) <1$.


问:
Linear independence of solutions to Floquet equation
I have a question about solutions to the Floquet equation, i.e.:
$$
\dot{x} = A(t) x
$$
where $x$ is an $N$-dimensional time-dependent column vector and $A(t)$ is a $T$-periodic $N \times N$ matrix.

To solve the equation I first find the eigenvalues $q_\beta$ and eigenvectors $\phi_\beta(t)$ of the equation
$$
( A(t) - \frac{d}{dt} ) \phi_\beta = q_\beta \phi_\beta
$$
where the $\phi_\beta$ live in the space of $T$-periodic column vectors.
With the $q_\beta$ and $\phi_\beta$ we form the Floquet equation solutions $\psi_\beta(t)$ by
$$
\psi_\beta(t) = \phi_\beta(t) e^{q_\beta t}
$$
As I understand it, there can only be $N$ linearly independent solutions $\psi_\beta$ out of the infinite number of $q_\beta, \phi_\beta$ pairs.
This is because the Floquet equation is an ordinary linear differential equation and so possesses an $N$-dimensional space of solutions.
The redundancy of solutions (again, as I understand it) is explained by the fact that for every periodic eigenvector $\phi$ with eigenvalue $q$ we can generate another $\phi'(t) = \phi(t) e^{i 2 \pi n \frac{t}{T}}$ with eigenvalue $q' = q - i \frac{2 \pi n}{T}$.

My question is: if $N$ eigenvectors $\phi_\beta$ are selected so that their associated Floquet solutions $\psi_\beta$ are linearly independent, under what conditions are the solutions orthogonal? For instance, the matrix $A(t)$ that I currently working on is real, normal ($AA^T=A^TA$), and has the property that $A + A^T$ is constant.

答:
Ok, I've been looking a little more thoroughly at what you're doing. ;-) To keep things simple, let's assume the matrix $A$ is constant. Then what you're looking at are the ODEs $\dot x = (A-q)x$ and you want to find solutions $x$ that are $T$-periodic. Now, the solution to this equation is $x(t) = e^{t(A-q)}a = e^{-tq}e^{tA}a$, where $a$ is some vector. Periodicity means that $x(0) = x(T)$, i.e., $a = e^{-Tq}e^{TA}a$ or $e^{TA}a = e^{Tq}a$. This is an eigenvalue equation for $A$ which can only be satisfied if $e^{Tq}$ is an eigenvalue for $A$. There are $N$ of these pairs $(q_k,a_k)$. In fact, you may choose the eigenvalues $q_k$ and eigenvectors $a_k$ for $A$ itself because $Aa_k = q_ka_k$ implies $e^{TA}a_k = e^{Tq_k}a_k$.

Now, if $(q_1,a_1)$ and $(q_2,a_2)$ are such that $q_1\neq q_2$, then $a_1$ and $a_2$ are indeed orthogonal because $A$ is normal. So, for the solutions $x_1$ and $x_2$ of the Floquet equation you get $x_j(t) = e^{-tq_j}e^{tA}a_j = e^{-tq_j}e^{tq_j}a_j = a_j$, which is constant. So,
$$
\langle x_1(t),x_2(t)\rangle = \langle a_1,a_2\rangle = 0.
$$
This of course has to be modified for non-constant $A$, where the $x_j(t)$ are surely not constant.


问:
ultrafilter is principal if and only if it contains a finite set
Prove that an ultrafilter is principal if and only if it contains a finite set.

First we will show that an ultrafilter is principal if it contains a finite set.

Define a principal ultrafilter
$$\mathscr{U}_{x}=\{U \subseteq I | x \in U\}$$
Note that $\{x\}$ is always an element of  $\mathscr{U}_{x}$, thus every principal ultrafilter contains a finite set.

Now we want to show that if an ultrafilter contains a finite set then it is principal.

Let $\mathscr{U}$ be an ultrafilter on $I$ and let U be a finite subset of $\mathscr{U}$.

I am stuck here. I know I need to show that there is an $x \in U$ such that $\{x\} \in \mathscr{U}$ which will imply that $\mathscr{U} = \mathscr{U}_x$. I am not sure how to find\define such an x. Can I just let an $x \in U$ and go from there??

答:
Prove the lemma: if $F \cup G \in \mathcal{U}$ then $F \in \mathcal{U}$ or $G \in \mathcal{U}$ when $\mathcal{U}$ is an ultrafilter. By induction, this extends to finite unions as well. Apply that to a finite set, split into singletons.

The lemma follows from the following fact on ultrafilters $\mathcal{U}$ on $X$:

$(\ast)$ For every $A\subseteq X$ either $A \in \mathcal{U}$ or $X\setminus A \in \mathcal{U}$.

Suppose we know this fact and $F \cup G \in \mathcal{U}$ and $F \notin \mathcal{U}$ and $G \notin \mathcal{U}$ then by this fact $F^c \in \mathcal{U}$ and $G^c \in \mathcal{U}$ and so $F^c \cap G^c \in \mathcal{U}$. But $(F \cup G) \cap (F^c \cap G^c)=\emptyset$, so we'd have a contradiction. So $F \in \mathcal{U}$ or $G \in \mathcal{U}$ must hold.

The proof of the fact $(\ast)$ follows from the maximality of $\mathcal{U}$: if $A \not \mathcal{U}$ this means there exists some $U \in \mathcal{U}$ such that $A \cap U = \emptyset$. Why? If this were not the case, $A$ would have non-empty intersection with all $U \in \mathcal{U}$ but then $\mathcal{U}':= \{B \subseteq X\mid \exists U \in \mathcal{U}: U \cap A \subseteq B\}$ would be a strictly larger (as $A \in \mathcal{U}'\setminus \mathcal{U}$) filter on $X$, which cannnot be!; (check that this is indeed a filter under these circumstances)
Having such a $U$ we then have that $A \cap U = \emptyset$ so $U \subseteq A^c$ and as $U \in \mathcal{U}$ and filters are closed under supersets, $A^c \in \mathcal{U}$ and we have shown that whenever $A \notin \mathcal{U}$ then $A^c \in \mathcal{U}$, which is what we wanted.


问:
Area under a cycloid
If the length of the cycloid is $4$ times the diameter of a rotating circle, then the area under the arch traced out by that cycloid is how many times the area of the rotating circle?

I tried using the parametric equations for a cycloid: $x = a(t-\sin{t})$, and $y = a(1-\cos{t})$.

The answer is three. Can someone please explain this? I am preferably looking for an answer that does not involve calculus.

答:
To use the parametric equation:

Render the area as $\int y dx$ between the appropriate limits, which are clearly from $t=0$ to $t=2\pi$.  Plug in the parametric equations to get

$a^2\int_0^{2\pi}(1-\cos t)d(t-\sin t)$

To get an integral with just $dt$ differentiate the $t-\sin t$ function and combine the derivative with the preexisting factor $1-\cos t$.  Thus

$a^2\int_0^{2\pi}(1-\cos t)^2dt$

This may be simplified by putting in $(1-\cos t)^2=1-2\cos t+\cos^2t$ and then $\cos^2 t=(1+\cos 2t)/2$.  Then the cosine integrate to zero over full periods, allowing those terms to be canceled out and leaving just a constant ($3/2$) remaining in the inte-grand.  Thus the final answer $3\pi a^2$ where the parametric equations had defined $a$ as the radius.

Title: Area under a cycloid


问:
Question about a proof of Cauchy $\implies $ Convergent in $\mathbb R^k$

Suppose $\{x_n\}$ is bounded. Then the range of $\{x_n\}$ is in some $k$-cell $I$ which is compact. Thus $\{x_n\}$ converges in $I$.

We are done if we can prove $\{x_n\}$ is bounded. Let $E_N$ be the range of the subsequence $x_N, x_{N +1}, x_{N + 2}, \ldots.$ Then $\displaystyle{\lim_{N \to \infty} \operatorname{diam} E_N = 0}$. $\color{red}{\text{Thus there's $N \in \mathbb N$ with $d(\operatorname{diam}E_N, 0) < 1$}},$ $ \color{blue}{\text{so the distance between any two points of $E_N$ is less than $1$}}$, so $E_N$ is bounded. Because $\{x_1, x_2, \ldots, x_{N - 1}\}$ is finite, we can bound it by the maximum distance between two of its points. Note, $\{x_1, x_2, \ldots, x_{N - 1}\} \cup E_N$ which is the range of $\{x_n\}$ is a union of two bounded sets and so $\{x_n\}$ is bounded.

My question is about the parts in $\color{red}{\text{red}}$ and $\color{blue}{\text{blue}}$ of the quote above. If $\displaystyle{\lim_{N \to \infty} \operatorname{diam} E_N = 0}$, then for any $\epsilon > 0$, there's some $K \in \mathbb N$ s.t. $N \ge K \implies d(\operatorname{diam} E_N, 0) < \epsilon.$ Since $d(\operatorname{diam} E_N, 0) < \epsilon$ is true for any nonnegative $\epsilon$, in particular we have $d(\operatorname{diam} E_N, 0) < 1 \iff |\operatorname{diam}E_N| < 1$ which means the distance between any two points of $E_N$ is less than $1$. But this is true only for $N \ge K.$ That means all of $E_N, E_{N+1}, E_{N+2},\ldots, E_{K -1}$ have a diameter greater or equal to $1$. Shouldn't we have to have $|\operatorname{diam}E_N| < 1$ true for all $N$ for the $\color{blue}{\text{conclusion in blue}}$ above to hold? Thanks.

答:
In the sentence you've highlighted red and blue, the proof is choosing a particular value of $N$. Before the red part, $N$ could be any natural number, and by the time we reach the blue part, $N$ is to mean a certain number where we do know that $|\mathrm{diam}\,E_N| < 1$.

You are right that the proof would be clearer if it didn't "reuse" $N$ in this way, instead saying

Thus there's $K \in \mathbb{N}$ with $d(\mathrm{diam}\, E_K,0) < 1$, so the distance between any two points of $E_K$ is less than $1$, so $E_K$ is bounded.

and so on, using $K$ instead of $N$ for the remainder of the text.


问:
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$

The solution goes as follows:

$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+y}$. We can see that:

$ab+bc+ac+2abc=1$

It's enough if: $(a+b+c)^2\ge 4-14abc$

$(a+b+c)^2\ge 4(ab+bc+ab+2abc)-14abc$

From Schur it is enough if: $6abc\ge\frac{9abc}{a+b+c}$ which is true from Nesbit.

Could you please provide a more intuitive and easier approach?

Body:
Version of 01.04.2021.

Since the task is homogenious, one may denote
$$s=x+y+z=3,\quad r=xy+yz+zx,\quad p=xyz,\tag1$$
wherein
$$r\le\dfrac13s^2=3,\quad p\le\dfrac1{27}s^3=1. \tag2$$
At the same time,
$$p=(3-u)v,\quad r=(3-u)u+v,\tag3$$
where
$$u=x+y\in[0,3],\quad v=xy\in\left[0,\dfrac{u^2}4\right].\tag4$$

The given inequality takes the forms of
$$4\big(x(s-y)(s-z)+y(s-z)(s-x)+z(s-x)(s-y)\big)^2$$
$$ \ge \big(16(s-x)(s-y)(s-z)-56xyz\big)(s-x)(s-y)(s-z)$$
$$ = \big(4(s-x)(s-y)(s-z)-7xyz\big)^2 -49(xyz)^2,$$
or
$$4(s^3-2sr+3p)^2+49p^2\ge (4sr-11p)^2,\tag5$$
wherein, taking in accouint $(1)-(4),$

$$4(s^3-2sr+3p)^2+49p^2-(4sr-11p)^2$$
$$=4(27-6r+3p)^2+49p^2-(12r-11p)^2$$
$$=12(-108r+10pr-3p^2+54p+243)=12f(u,v),$$
$$f(u,v)=243-324u+108u^2+54v+36uv-60u^2v+10u^3v+3v^2+8uv^2-3u^2v^2,$$
$$f'_v(u,v)=54+36u-60u^2+10u^3+6v+16uv-6u^2v.$$
Then at the bounds by $\;v\;$

$f(u,0)=27(3-2u)^2\ge0,$

$f\left(u,\dfrac{u^2}4\right)=\dfrac3{16}(1296-1728u+648u^2+48u^3-79u^4+16u^5-u^6)$
$=\dfrac3{16}(9-u^2)(u^2-8u+12)^2\ge0.$

At the stationary points
$$2f(u,v)=2f(u,v)-v\,f'_v(u,v)=g(u,v),$$
where

$g(u,v)=486-648u+216u^2+(54+36u-60u^2+10u^3)v,$

$g(u,0)=54(3-2u)^2\ge0,$

$g\left(u,\dfrac{u^2}4\right)=\dfrac12(972-1296u+459u^2+18u^3-30u^4+5u^5)$
$=\dfrac52\left(u+\dfrac{58}{15}\right)\left(u-\dfrac85\right)^2\left(u-\dfrac{10}3\right)^2 +\dfrac{253}{15}\left(u+3\right)\left(u-\dfrac53\right)^2$
$+\dfrac1{1350}\left(95150-119630u+38207u^2\right)\ge0.$

Since $\;g(u,v)\;$ is a linear function by $\;v,\;$ it achives the least value at the edges of the area.

Proved!


问:
Explicitly exhibit that vector fields over the 2-sphere is a projective module
We know by the Serre Swan theorem that smooth vector fields over a smooth manifold form a projective module over the ring of smooth functions. We also know that the hairy ball theorem implies that the module of vector spaces on the sphere $\mathfrak X(S^2)$ is not a free module. Hence, it must be a "real" projective module, not a free module.  So,  there must exist some non-trivial module $F$ over $C^\infty(S^2)$ such that $\mathfrak X(S^2) \oplus F \simeq \oplus_i C^\infty(S^2)$.

Do we know what $F$ is, explicitly? Do we know what the right hand side of the equality $\mathfrak X(S^2) \oplus F \simeq \oplus_i C^\infty(S^2)$ should be explicitly? (what is the rank of the free module?)

Can we compute $F$ "in general" for a given (nice) smooth manifold $M$?

答: The normal bundle of $S^2$ embedded in $\mathbb{R}^3$ is trivial (see this question), so the sections of this bundle give a rank $1$ free $C^{\infty}S^2$ module $F$ so that $\mathfrak X(S^2) \oplus F= \oplus_{i=1}^3 C^{\infty}S^2$.


问:
What are the cluster points of this filter?
Let X be a topological space, $A\subset X$ and $\mathcal{F}=\{F\subset X|A\subset  F\}$. Then $\mathcal{F}$ is a filter on X. I would like to know what are the cluster points of this filter.

I am preparing for an exam and I need help. Thank you in advance.

答: HINT: The cluster points of any filter $\mathscr{F}$ on a space $X$ are the points of the set $\bigcap_{F\in\mathscr{F}}\operatorname{cl}F$. What is $\bigcap_{F\in\mathscr{F}}\operatorname{cl}F$ for your particular filter?</p>


问:
Combination problem
There are N advertisement boards of which M consecutive boards should have at least K advertisements. How to find number of ways in which this is possible keeping cost minimum.
Eg: N=6,M=3,K=2 which has 6 possible ways: {1 1 0 1 1 0}, {0 1 1 0 1 1}, {0 1 1 1 1 0}, {1 0 1 1 1 0}, {0 1 1 1 0 1}, {1 0 1 1 0 1} [1=Ad,0=No Ad]
and when N=3,M=2,K=1 has only 1 possible way: {0,1,0} [since cost should be minimum(use least no of ads)]

N>=M>=K>=1

答:
First, let's deal with the case $N\leq M$. Clearly $K$ advertisements placed in any way will suffice as the minimum cost. There are ${N+K-1 \choose K}$ ways to arrange them in any way.

Now, let's deal with $N > M$.

Hint: Show that you need at least $\lfloor \frac{N}{M} \rfloor K$ advertisements.

Hint: Construct an explicit example to show that  $\lfloor \frac{N}{M} \rfloor K$ advertisements is sufficient.

This shows that $\lfloor \frac{N}{M} \rfloor K$ advertisements is the minimum cost.

Let $N_i$ denote the number of advertisements on the $i$th board.

Hint: Show that $N_i = N_{i+M}$, where the indices make sense.

Hint: If $N = mM + d$, then show that $N_1 = N_2 = \ldots N_d = 0 $.

This means that we must place $K$ advertisements in the remaining $M-d$ spots.

Hint: Show that any arrangement of $K$ advertisements into these $M-d$ spots will work.

Hint: Show that there are thus ${M-d+K -1 \choose K}$ ways to arrange them.

With $N=3, M=2, K=1$, we get $d=1$, and the answer of ${2-1+1-1 \choose 1} = 1$.

Title: Combination problem


问:
Prime ideal after modding out by homogenizing variable implies original ideal prime?
Let $R:=\mathbb{C}[x_1,\ldots,x_d]$ be a polynomial ring. Let $I=(f_1,\ldots,f_n)$ be an $R$-ideal. Let $f^*_i$ be the homogenization of $f_i$ in $R[t]$.

Question: If $(\overline{f_1^*},\ldots,\overline{f_n^*})$ is prime, where $\overline{f_i^*}$ denotes the image of $f_i^*$ in $R$ under the map $t\mapsto 0$, is $I$ prime?

Edit: I changed the question to be specifically about homogenizing. I also changed notation to use $t$ to hopefully clarify the homogenization.

I want to rephrase the question using the filtration language:

First we define an orderly $\mathbb{Z}$-filtration $\mathcal{F}$. This means $\mathcal{F}$ satisfies the four filtration axioms:

$F_g\subset F_h$ for $g<h$

$\bigcup_{g\in\mathbb{Z}} F_g R = R$

$(F_g R)(F_h R)\subseteq F_{g+h}R$ for all $g,h\in\mathbb{Z}$

$1\in F_0R\setminus F_{<0}R$ (this is set minus),

where $F_{<g}:=\bigcup_{h<g}F_hR$,

along with the orderly axiom:

For all nonzero $a\in R$ there exists a smallest element $g\in\mathbb{Z}$ such that $a\in F_g R$. The smallest element $g$ is then called the order of $a$. Note that an orderly $\mathbb{Z}$-filtration is separated.

Now we define a degree filtration by letting $\alpha=(\alpha_1,\ldots,\alpha_d)$ be a vector of positive integers and assign $R$ the structure of a positively graded $\mathbb{C}$-algebra by setting $\deg_\alpha x_i=\alpha_i$ for $1\leq i\leq d$. Define $\mathcal{F}^\alpha=\{F_i^\alpha\}$ by setting
$$F_i^\alpha R=\bigoplus_{j\leq i}R_j.$$
Here $R_j$ denotes the $j^{th}$ homogeneous component with respect to the $\alpha$-graduation.

Define the associated graded ring
$$\text{gr}_{\mathcal{F}^a}(R)=\bigoplus_{g\in\mathbb{Z}}F_g R/F_{<g}R.$$
Define the initial form of $a\in R$ of order $g$ to be the residue class $\text{in}_{\mathcal{F}^\alpha}(a)=a+F_{<g}R$.

The question now becomes: If $(\text{in}_{\mathcal{F}^\alpha}(f_1),\ldots,\text{in}_{\mathcal{F}^\alpha}(f_n))$ is prime, then is $(f_1,\ldots,f_n)$ prime?

答: Geometry hints at a counterexample: the union of the two lines $y = x$ and $y = 1$ is not irreducible, but its intersection with the $x$-axis is a single point. So, translating into commutative algebra, consider the ideal $\langle (y - x)(y - 1) \rangle \subset \mathbb{C}[x, y]$, which is not prime. Its image under the projection map is the prime ideal $\langle x \rangle \subset \mathbb{C}[x]$.


问:
Computing the surface area 2-form for a paraboloid
Suppose I want to find a 2-form in ${\bf R}^3$ which gives the surface area to the paraboloid $D$ in some bounded region,

$$\phi(s,t)=(s,t,s^2+t^2)$$

I know that a general 2-form in ${\bf R}^3$ has the form,

$$\omega=f_1dx\wedge dy + f_2dy\wedge dz+f_3dx\wedge dz$$

Then this becomes,

$$\int_D\omega=\int_{\phi^{-1}(D)}\Big[f_1(\phi(s,t))-f_2(\phi(s,t))2s+f_3(\phi(s,t))2t\Big]ds\wedge dt$$
$$$$

If $J$ is the Jacobian for $\phi$, then since I know that the surface area element is given by $\sqrt{\det(J^TJ)}=\sqrt{1+4s^2+4t^2}$, then that means one possible solution would be,

$$f_1=f_2=f_3=\frac{\sqrt{1+4x^2+4y^2}}{1-2x+2y}$$

More generally, for surfaces in ${\bf R}^3$, it seems like one can always determine a (the?) surface area 2-form by computing,

$$f_1=f_2=f_3=\frac{\sqrt{(\det J_{xy})^2 + (\det J_{yz})^2 + (\det J_{xz})^2}}{\det J_{xy} + \det J_{yz} + \det J_{xz}}\circ\phi^{-1}$$

Clearly any time a parametrization can be formed by simply projecting down onto the xy-plane, computing this is trivial, however I imagine that computing $\phi^{-1}$ in general may be intractable. Nevertheless, $\phi$ is injective, so should it not always exist in theory?

Moreover, what exactly is this object? When can it be computed? does it always in theory exist? And is this generalizable to any dimension.

答:
In general, the surface area $2$-form will be $$\omega = n_1\,dy\wedge dz + n_2\,dz\wedge dx + n_3\,dx\wedge dy,$$
where $\vec n$ is the unit outward-pointing normal. I've derived this in other posts on here.


问:
For which values of $ \alpha$ ,$ e^{\alpha x} - (1+\alpha) \cdot x^2$ is convex in $\mathbb{R}$
I have to say for which values of $ \alpha$ the following function is convex in  $\mathbb{R}$:
$$ e^{\alpha x} - (1+\alpha) \cdot x^2$$

The second derivative is $$ \alpha^2 e^{\alpha x} - 2(1+\alpha) >0 $$

But I don't know how to proceed.

答: Clearly, if $\alpha < -1$ so that $(1+\alpha) <0$. Then the second derivative is always strictly positive, whence $f$ is convex. You can argue that $\alpha > -1$ won't work as well. Basically, if $\alpha > -1$ so that $(1+\alpha) > 0$, then the term $\alpha^2\cdot\mathrm{e}^{\alpha x}$ will go to zero as either $x \to \infty$ or $x \to -\infty$, so the expression for the second derivative of $f$ cannot be positive in this case.


问:
Prove that for a given $r>0$ there exists $\theta >0$ so that for all $a \leq x \leq b$, $\theta |f(x)|\leq r$
Let $f(x)$ be a given continuous function on a closed interval $[a,b]$. Prove that for a given $r>0$ there exists $\theta > 0$ so that for all $a \leq x \leq b$, $\theta |f(x)| \leq r$.

Hello, can anyone help me with this question? From what I understand, we should use the extreme value theorem?

答: Let $r>0$ be arbitrary. By the extreme value theorem, there exist $\overline x,\underline x\in[a,b]$ such that $f(\overline x)\geq f(x)$ and $f(\underline x)\leq f(x)$ for all $x\in[a,b]$. Since $c:=\max\{\vert f(\overline x)\vert, \vert f(\underline x)\vert\}$ is a real number, there exists a $\theta>0$ such that $\theta c\leq r$. Pick e.g. $\theta = \frac{r}{c+1}$. Finally, the statement to be proven follows from the fact that $\vert f(x)\vert \leq c$ for all $x\in[a,b]$.


问:
Probability of choosing ace of spades before any club
From a deck of $52$ cards, cards are picked one by one, randomly and without replacement. What is the probability that no club is extracted before the ace of spades?

I think using total probability for solve this

$$P(B)=P(A_1)P(B\mid A_1)+\ldots+P(A_n)P(B\mid A_n)$$

But I am not sure how to solve this. Can someone help me?

答:
There is a nice answer to this with an "Aha" moment. We need to understand that the only cards that matter are the ace of spades and the 13 clubs.

The rest of the cards might as well be pieces of candy or pebbles. They don't have a role to play in the probability in the question.

The answer is (ace of spades) /(ace of spades+13 clubs) =1/14.
We need to get it in our heads that picking a card other than the ones mentioned above have no impact in the overall outcome. Assume you take out an ace of hearts from the deck. This act favors neither of the two outcomes: The ace of spades coming up first OR a club coming up first.


问:
Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$

Calculate the sum of all irrational roots of
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$

I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}- xy+ y^{2} \right ),$ but have had no meaningful results.

答:
If we assume "cube root" is the real-valued root, then we have

$$\begin{cases}4\sqrt[3]{8x-3}-8x^3-3=0 \\ x\in\mathbb R  \end{cases} \Longleftrightarrow  512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0$$

$$\Longleftrightarrow (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0$$

Then using standard calculus tools (but, only algebraic ways are also possible ) we get

$$64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73 > 0$$

for all $x\in\mathbb R$.

So, we have only $2$ irrational roots:

$$x_1+x_2=\frac{-2}{4}=- \frac 12$$

Then, answer is equal to $- \dfrac 12.$

If we use the principal root instead of "real-valued cube root",  then we have

$$4\sqrt[3]{8x-3}-8x^3-3=0 \Longleftrightarrow \begin{cases} 512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0 \\ 8x-3≥0  \end{cases} $$

$$\Longleftrightarrow \begin{cases} (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0 \\ x≥\frac 38\end{cases}$$

The last result implies, we have only $1$ irrational root:

$$x=\frac {-1+\sqrt {13}}{4}$$

So, answer is equal to $\dfrac {\sqrt {13}-1}{4}.$


问:
Find the spectrum of $Ax(t)=x(t+s)$
Let be $\mathcal{C}_\mathbb{R}$ the set of the continuous and bounded function in $\mathbb{R}$ with norm $ \|x\|=\underset{t\in\mathbb{R}}{sup}|x(t)|$ Consider the operator $A$ such that $Ax(t)=x(t+s)$ for some $s\in\mathbb{R}$. Show that the spectrum of A is the unit sphere $=\{\lambda\in\mathbb{C}: |\lambda|=1\} $.

Please, I don't know how to aboard this exersice. I find the norm of $A$, $\|A\|=1$ so $\sigma(A)\subseteq \bar{D}(0,1)$. Then I was trying to show that if $|\lambda|<1$ then $A-\lambda Id$ is invertible, so I began from show that it's injective but I only find that $|x(t+s)|<|x(t)|$ for all $ t\in\mathbb{R}$ and I don't know why this implies that $x=0$.
Or maybe this is not te correct way to make this exercise please could you give me some hints?

答:
$\|Ax\|=\|x\|$ and $A$ is surjective: Given $y$ note that $Ax=y$ where $x(t)=y(t-s)$. Since $\|A\| =1$ we see that $\sigma (A) \subseteq \{\lambda: |\lambda | \leq 1 \}$. Using the fact that $\|A^{-1}\| \leq 1$ we see that  $\sigma (A) \subseteq \{\lambda: |\lambda | \geq 1 \}$. To finish the proof note that $|\lambda|=1$ implies that thre exists a continuous function $x$ with $Ax=\lambda x$.

[ Suppose $s>0$. Let $x$ be any continuous function on $[0,s]$ with $x(0)=x(s)=0$, say $x(t)=s(t-s)$. Define $x$ on $[ns,(n+1)s]$ by $x(t)=\lambda ^{n} x(t-ns)$. You can check that this function is continuous, bounded and $Ax=\lambda x$].

Title: Find the spectrum of $Ax(t)=x(t+s)$


问:
What is Inverse Fourier Transform of $2^2/(^2−^2)$?
What is the inverse Fourier transform of  $F(w) = \frac{4}{4+(j2\pi f)^2} $?
 I have two suggested solutions:

Assume $j2\pi f = \omega$ and use the standard transform
$e^{-\alpha |\tau |} = \frac{2\alpha}{\alpha ^2 + \omega ^2}$ to obtain $\frac{1}{4}e^{-2|\tau|}$

Calculate $f(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\alpha}{\alpha^2-\omega^2}e^{i\omega\tau}d\omega$

For (2), how can I solve the integral in order to get the inverse Fourier transform $f(\tau)$?

答:
$\newcommand{\eqd}{\triangleq}$
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$\newcommand{\thme}[1]{#1}$
$\newcommand{\opFT}{\mathrm{F}}$
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$\newcommand{\Real}{\mathrm{Re}}$

I'll be honest with you---it looks like there is no Inverse Fourier Transform for $2\alpha^2/(\alpha^2-\omega^2)$.
I'd be happy to be proven wrong...but at least consider the following...

(1) Let $s\eqd j\omega$. Then the Fourier Transform $\int_{-\infty}^{\infty} \fg(\tau)e^{-j\omega\tau}\dtau$ of $\fg(\tau)$
becomes the (Two-Sided) Laplace Transform $\int_{-\infty}^{\infty} \fg(\tau)e^{-s\tau}\dS$ of $\fg(\tau)$.

(2) Note that what we want is a function $\fg(\tau)$ with Laplace Transform $\frac{2\alpha}{\alpha^2+s^2}$. Why?
Because then when we substitute back in $s=j\omega$ we would get
$$\brlr{\frac{2\alpha}{\alpha^2+s^2}}_{s=j\omega} = \frac{2\alpha}{\alpha^2-\omega^2}$$
and the world would be a happier place.

(3) If only there was such a function $\fg(\tau)$ ... hmmmmm ... But wait! There is a function like that! In fact, if you happened to guess that the function $\fg(\tau)$ is
$$\fg(\tau)\eqd 2\sin(\omega\tau)\step(\tau)$$
where $\step(\tau)=1$ for $\tau\geq1$ and $0$ otherwise (the step function), then you would be right! And you can check that here (where the One-Sided Laplace is used and so $\step(\tau)$ is implied).

(4) So the Inverse Laplace Transform of $\fG(s)\eqd\frac{2\alpha}{\alpha^2+s^2}$ is $\fg(\tau)\eqd 2\sin(\omega\tau)\step(\tau)$. So we just set $s=j\omega$ and we're all done here, right? Well not quite. That Laplace Transform $\fG(s)$ has a Region of Convergence of $\Re(s)>0$, where $\Re(s)$ is the real component of the complex $s$. The problem here is that the Fourier Transform exists along the imaginary axis of $s$---that is, at $s=j\omega$, which means $\Re(s)=0$ ... which is not in the Region of Convergence of $\fG(s)$ ... and so rather, it diverges or "goes to infinity" or "blows up" at $s=j\omega$.

And so, while the Laplace Transform of $\fg(\tau)$ does exist (for $\Re(s)>0$), it does not exist (it diverges) for $\Re(s)\leq0$. And so by extension, it looks like the Fourier Transform $\fG(\omega)=\frac{2\alpha}{\alpha^2-\omega^2}$ does not exist (because it is not in the Region of Convergence of $\fG(s)$).


问:
How to prove the following integral inequality?

Let $f:[0,1]\to \Bbb R$ be differentiable function with $f(0)=0$ and $0\leqslant f(x)'\leqslant 1$. Then
  $$3\left(\int_0^1f(x)^2dx\right)^3\leqslant \int_0^1f(x)^8dx.$$

My attempt:

$0\leqslant f(x)'\leqslant 1$ implies $0\leqslant f(x)\leqslant x$. then $0\leqslant f(x)^2\leqslant xf(x)$ and $0\leqslant f(x)^2\leqslant x^2$.

$$0\leqslant \int_0^1f(x)^2\leqslant 1/3$$

But these inequality can't relate $f(x)^2$ and $f(x)^8$. Any idea?

答: Your inequality does not seem to be true in general. Take $f(x)=ax$ with $a<1$: for this function we have that $0<f'<1$  and $f(0)=0$, so it is a legit function according to your requests. By using your inequality,  a direct calculation gives $(a^6)/9<(a^8)/9$, namely $1<a^2$. This is a contradiction since we assumed $a<1$.


问:
Find MGF for $f(x) = \frac{\theta^x e^{-\theta}}{x!}, x=0,1,2,3,.. $
Given the following probability distribution $$f(x) = \frac{\theta^x e^{-\theta}}{x!}, x=0,1,2,3,.. $$
Find the moment generating function, MGF, then find mean of random variable $Y=(X+1)^3$

To my knowledge $$M_x(t) = \sum{e^{tx}f(x)}$$

$$\sum{\frac{\theta^x e^{tx-\theta}}{x!}}, x=0,1,2,..$$

How to solve this summation ? then I am supposed to get MGF for $Y$, but I don't know what to do with power ?

Body:
$$M_X(t) = \mathbb{E}[e^{tX}] = \sum_{x=0}^{+\infty} e^{tx} \frac{\theta^x e^{-\theta}}{x!} = e^{-\theta}\sum_{x=0}^{+\infty}  \frac{\theta^x e^{tx}}{x!} =  e^{-\theta}\sum_{x=0}^{+\infty}  \frac{(\theta e^{t})^x}{x!}.$$

As suggested by @Imaosome:

$$\sum_{k=0}^\infty\frac{a^k}{k!} = e^a.$$

Hence:

$$M_X(t) = e^{-\theta}e^{\theta e^{t}} = e^{\theta(e^t - 1)}.$$

For the mean of $Y = (X+1)^3$....

$$\mathbb{E}[Y] = \mathbb{E}[(X+1)^3] = \mathbb{E}[X^3 + 3X^2 + 3X + 1] = \mathbb{E}[X^3] + 3\mathbb{E}[X^2] + 3\mathbb{E}[X] + \mathbb{E}[1] = \\
= \left(\frac{d^3}{dt^3}M_X(t)\right)_{t=0} + 3\left(\frac{d^2}{dt^2}M_X(t)\right)_{t=0} + 3\left(\frac{d}{dt}M_X(t)\right)_{t=0} + 1 = ...\\$$


问:
Simple Calculus question (Substitution)
My question is "To show
$ \int_{1}^\alpha {1\over1+x^2} dx = \int_{1\over\alpha}^1 {1\over 1+u^2} du$

such that
$u = \frac{1}{x}$ and $\alpha $ > 1,

I got to

$ \int_{1}^\alpha {1\over1+x^2} dx = -\int_{1}^{1\over\alpha} {1\over 1+{1\over u}^2} {du\over u^2}  $

but i don't know how to swap the upper bound and lower bound around, could someone please explain it clearly to me?

答: It follows from the elementary property of integrals that $\int_{a}^{b}f(x)dx = -\int_{b}^{a} f(x)dx$, and the convention that integrals are typically written $\int_{a}^{b}f(x)dx$ where $a \leq b$. Since $\alpha > 1, \frac{1}{\alpha} < 1$. </p>

Title: Simple Calculus question (Substitution)


问:
Proving input-to-state stability for a set of equilibria
Consider the following nonlinear system:
\begin{align}
\dot{x} &= -axy \\
\dot{y} &= axy - by
\end{align}
It's easy to check that this system has a continuum of equilibria (of the form $(x,0)$). I'm trying to prove the perturbed version of this system
\begin{align}
\dot{x} &= -axy +u_1(t)\\
\dot{y} &= axy - by+u_2(t)
\end{align}
is input-to-state stable, i.e, there exist $\beta \in KL$ and $\gamma \in K$ such that

$$
|x(t)|\leq \beta (|x_{0}|,t)+\gamma (\|u\|_{\infty }).
$$

I'm not sure how to do that since the system doesn't have an isolated equilibria and most of the results for input-to-state stability seem to be concerned for that particular case. Moreover, I'm not sure how to find a Lyapunov function for the system.

答:
I would suggest to look at the literature on ecological and epidemiological systems as many of those systems have this structure. In fact, this model is the model of a reaction network with mass-action kinetics. So, you may also look at the literature on that topic.

Also, if you are actually working on that, you may need to explicitly consider the positivity of the variables. For instance, you may take the Lyapunov function $V(x,y)=x+y$. In this case, you have $\dot{V}(x,y)=-by$ and you can prove the asymptotic stability of the continuum of equilibrium points. Perhaps, you can use a variation of this Lyapunov function for your purpose.


问:
Help required with limit
In finding the asymptotic value of a certain  quantity I ended up with the following:

$$f(n)=(n-1)B_{1/2}(n,n+2)$$
$$g(n)=(n+1)B_{1/2}(n+2,n)$$
$$ h(n)=4^n \left( f(n)-g(n)\right)$$
Numerical simulations lead me to believe that  h(n) approaches 1 as n approaches infinity.However ,i am not able to prove it analytically
Can somebody kindly help me with the evaluation of $ \lim_{n \to \infty} h(n)$?
If somebody could also plug it in Mathematica, I would be highly obliged. Thanks for any help in advance.

P.S.: In the above the notation $B_z(a,b)$ stands for the incomplete beta function defined by:
$$B_z(a,b)=\int\limits_0^z u^{a-1}(1-u)^{b-1} \mathrm{d}u.$$

答:
Mathematica cannot evaluate the limit.

Use these equalities:

$$B_z(a,b)=B(a,b)-B_{1-z}(b,a)$$
$$B_z(a,b)=\frac{1}{a}((1-z)^b z^a+(a+b)B_z(a+1,b))$$
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
$$B_{1/2}(a,a)=\frac{\Gamma(a)^2}{2\Gamma(2a)}$$

and you can show $$h(n)=4^n \left(2 \left(\left(\frac{1}2 \right)^{2n + 2} +
      \left(\left(\frac{1}2 \right)^{2n + 2} + \frac{(2n +3) \Gamma(n + 2)^2}{ \Gamma(2n + 2 )} \right) \right)\\-(n+1) \frac{\Gamma(n+2)\Gamma(n)}{\Gamma(2n+2)}
   \right)$$
$$=1-\frac{\sqrt{\pi}}{2n^{1/2}}-\frac{5 \sqrt{\pi}}{16 n^{3/2}}+...$$

OwnerDisplayName: sajjad veeri

Title: Help required with limit

OwnerDisplayName: John L


答:
Exploit the symmetry

$$B(1/2;n,n+2) + B(1/2;n+2,n) = B(1;n,n+2) = B(n,n+2) = \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(2n+2)}$$

(the Beta function) to express this in terms of $B(1/2;n,n+2)$ and $B(n,n+2).$  Integrate by parts to express $B(1/2;n,n+2)$ in terms of $B(1/2;n+1,n+1) = B(n+1,n+1)/2$ (by the symmetry of its integrand) to obtain

$$f(n) - g(n) = 2^{-2n} + (n+1)(B(n+1,n+1) - B(n,n+2)).$$

Simple algebra then reduces this to

$$4^{n}(f(n) - g(n)) = 1 - \frac{n+1}{n(2n+1)}\,\frac{2^{2n}}{\binom{2n}{n}}.$$

It is well-known (and readily deducible from, say, Stirling's asymptotic approximation to the $\Gamma$ function) that $2^{-2n}\binom{2n}{n} = O(n^{-1/2})$ and the limit follows.

(Why did I express the result in this way?  Because $\binom{2n}{n}/2^{2n}$ is a familiar quantity: it expresses the proportion of row $2n$ in Pascal's Triangle occupied by the middle value.  To statisticians, this is the chance of obtaining exactly $n$ heads in $2n$ independent flips of a fair coin. The Normal approximation to the Binomial$(2n,1/2)$ distribution (which has mean $n$ and variance $n/2$) estimates this proportion as

$$2^{-2n}\binom{2n}{n} \approx \Phi\left(\frac{1/2}{\sqrt{n/2}}\right) - \Phi\left(-\frac{1/2}{\sqrt{n/2}}\right) \approx \phi(0)\left(\frac{\sqrt 2}{\sqrt{n}}\right)=\sqrt{\frac{1}{n\pi}}.$$

The approximation uses the first order expansion of the standard Normal cdf $\Phi$ at $0.$)

Title: Help required with limit

Body:
In finding the asymptotic value of a certain  quantity I ended up with the following:

$$f(n)=(n-1)B_{1/2}(n,n+2)$$
$$g(n)=(n+1)B_{1/2}(n+2,n)$$
$$ h(n)=4^n \left( f(n)-g(n)\right)$$
Numerical simulations lead me to believe that  h(n) approaches 1 as n approaches infinity.However ,i am not able to prove it analytically
Can somebody kindly help me with the evaluation of $ \lim_{n \to \infty} h(n)$?
If somebody could also plug it in Mathematica, I would be highly obliged. Thanks for any help in advance.

P.S.: In the above the notation $B_z(a,b)$ stands for the incomplete beta function defined by:
$$B_z(a,b)=\int\limits_0^z u^{a-1}(1-u)^{b-1} \mathrm{d}u.$$

OwnerDisplayName: whuber


问:
covering map $\mathbb{R}^2 - $2 points $\rightarrow \mathbb{R}^2 - $ 1 points?
Is there a covering map $\mathbb{R}^2 - $2 points $\rightarrow \mathbb{R}^2 - $ 1 points?

wanted to double check if my reasoning below is correct.

$\mathbb{R}^2 - $2 points deformation retracts to a surface with genus $2$ while $\rightarrow \mathbb{R}^2 - $ 1 deformation retracts to a surface of genus $1$. Thus, there is no such covering map.

答:
You have a good idea, but I think there are still some gaps. You have to clarify why a potential covering map restricted to your specific retract of $B = \mathbb R^2 \setminus \{point\}$ gives your specific retract of $E = \mathbb R^2 \setminus \{two \phantom{x} points\}$.

I suggest to proceed as follows. Assume there is covering map $p : E \to B$. We know that $p_* : \pi_1(E) \to \pi_1(B)$ is injective. But we have $B \simeq S^1$ and $E \simeq S^1 \vee S^1$, thus $\pi_1(B) = \mathbb Z$ and $\pi_1(E) = \mathbb Z * \mathbb Z$. This prevents the existence of an injective homomorphism as required.


问:
What is the name of the matrix operation "$\;\overrightarrow{y}\;$"?
I was doing some stuff for my computer science class, and it was talking about the vectorized form of unregularized logistic regression and I'm pretty bad at matrices and when I saw this symbol I was completely stumped. (I'm sorry for not pasting the text it wasn't coming out right so I am using an image)

Here is an image for what I am asking. The notation is above the y:

sorry for having not many specifics.
Thanks for any help you could give me!

答:
$\large \vec y$ denotes a vector; for example, $\large\vec y=(y_1,y_2,...,y_n)$ is an $n$-tuple in $\Bbb R^n$,

where $y_1,y_2,...,y_n\in\Bbb R$.  Such a vector could be multiplied by a scalar and/or a matrix.


问:
Intuition/references for why Möbius transformations are uniquely defined by three points?
Assume that I have a Möbius transformation

$$M(z)=\frac{az+b}{cz+d}$$

with real coefficients $a$, $b$, $c$, and $d$, and complex argument $z$. I have been interested in gaining intuition on why this transformation is uniquely defined by three points $(z_1,z_2,z_3)$ and their images $(w_1,w_2,w_3)$. I combed the StackExchange and found this answer by user Lehs, which suggests the reason is that the equation above can be rewritten as:

$$M(z)=\frac{a}{c}\frac{z+\frac{b}{a}}{z+\frac{d}{c}}=A\frac{z+B}{z+C}$$

where

$$A=\frac{a}{c}$$
$$B=\frac{b}{a}$$
$$C=\frac{d}{c}$$

This sort of explains how the four coefficients $a$, $b$, $c$, and $d$ can be reduced to three coefficients $A$, $B$, and $C$, but leaves me with two new questions:

Why is Möbius transformation usually expressed in terms of $a$, $b$, $c$, and $d$ if there is a seemingly more succinct way of representing this transformation?

An earlier comment in the same question also mentions 'symmetry', and suggests that while the transformation is uniquely determined, the coefficients $a$, $b$, $c$, and $d$ are not. I can see that multiplying the four coefficients with the same non-zero factor would yield the same $A$, $B$, and $C$. Does this mean that $A$, $B$, and $C$ uniquely define a transformation?

If you have any other comments which may aid my intuition, I would greatly appreciate it. Also, if you have any references you could suggest with regards to intuition about the Möbius transformation, or which detail any of the results I collected here, that would be amazing.

答: If $T, S$ have same images at 3 unique points, then $T\circ S^{-1}$ is identity map because Mobius transform has at most 2 fixed points (quadratic equation from $\frac{az+b}{cz+d}= z$, has at most 2 solutions). Then its uniquely defined.


问:
How to interpret "let" in mathematics?
Before explaining my issue, I wanted to first explain what things do make sense to me.

So, statements (used at the beginning of proofs) like "Suppose $x$ is an integer" or "Assume $x$ is an integer" make sense to me. The way I read them is like: "Let's just pretend that the symbol $x$ happens to represent an integer".

Statement (also used in proofs) like "Let $x$ be 3" or "Let $x$ equal 3" also make sense to me. The way I read them is like: "Let's just temporarily name 3 with the symbol $x$.

My issue comes with statements like "Let $x$ be an integer" or "Let $x$ $\in$ $\mathbb{Z}$". I really do not know how I should intuitively interpret such a statement. I can't interpret it in the same way as like "Let $x$ be 3" because there is not a specific object being assigned like 3. Should I interpret it like how I interpret "Suppose $x$ is an integer"?

EDIT:

So, from what I gathered from the responses, I think I understand now how I should think of "Let $x$ be an integer".

I could think of it as "Assume a newly created symbol x happens to represent an integer". However, this can cause issues as then "Let $x$ be an element of the empty set" is also completely valid.

Instead, I should not think of "Let $x$ be an integer" as an assumption, but an assignment/declaration, just like "Let $x$ be 3". One tangible way to think about this is to imagine myself assigning the newly created symbol $x$ to an integer chosen in secret by a friend. With this mindset, $x$ is not assumed to be an integer - it is an integer. It is just unknown to me what integer it is.

I hope this made sense to anybody with similar questions. If anyone thinks I have made an incorrect finding, please feel free to correct me.

答: You should take "Let $x$ be an integer" as a synonym for "Assume that $x$ is an integer" or "Consider an integer $x$".

Title: How to interpret "let" in mathematics?