问:
Frobenius-Schur indicator: why doesn't the Weyl unitary inner product qualify?
As best as I understand, the Frobenius Schur indicator tells us for a finite group $G$ and an irreducible representation $\rho: G \rightarrow \operatorname{Aut}(V)$ (where $V$ is a vector is a vector space over the complex numbers):

what $G$ invariant bilinear forms does $V$ possess?

I am extremely confused; Why doesn't the standard inner product:

$$
\langle\cdot,\cdot\rangle: V \otimes V \rightarrow \mathbb R;
\langle v, w \rangle \equiv \sum_{g \in G} [\rho(g)(v), \rho(g)(w)]
$$
for some inner product $[\cdot,\cdot]: V \otimes V \rightarrow \mathbb R$ qualify?

I also know that separately, the space of $G$ invariant bilinear forms must be 1-dimensional. This is because a $G$ invariant bilinear form carries the same data as a G-invariant mapping $f: V \mapsto V^\star$. However, since in the finite dimensional case, $V \simeq V^*$, we have by Schur's lemma that $f = \lambda I$ for some $I$.

Combining the two "facts", we have that

$[\cdot, \cdot]$ is a $G$ invariant symmetric bilinear form on $V$

The space of $G$-invariant symmetric bilinear forms is one dimensional

Thus, all $G-$invariant bilinear forms are 1-dimensional and symmetric

This is clearly absurd, since the entire point of the Frobenius Schur indicator is to tell us if the space of bilinear forms is symmetric or antisymmetric.

What am I missing?

答:
You said yourself the indicator applies to complex vector spaces, then proceed to talk about invariant inner products on real vector spaces. See what the issue is?

For complex inner product spaces, the inner product is sesquilinear, not symmetric bilinear.


问:
Complete square values of quadratic formula
For Which values of $x \in [0,n]$ the polynomial $P(x)=ax^2+bx+c$  $\ $ gives a complete square value.
For instance the polynomial $P(x)=3x^2+5x+7$ on the interval $x \in [0,100]$ has only two values for $x=3  \rightarrow P(3)=49 $ and for $x=54  \rightarrow P(54)=9025 $.
I was wondering if there is a method to find such values of $x$ directly.

答:
So I will interpret this question as asking which integers $x \in \mathbb{Z}$ satisfy
$$ax^2 + bx + c = y^2$$
for some $y \in \mathbb{Z}$.

In general there may be no such $x, y$. Consider for example $a = -1$, $b=0$ $c = -1$. Then the LHS is always negative, hence has no solutions, even in $\mathbb{R}$.

The name for this problem is determining whether a quadratic form has integer solutions, and has motivated a lot of modern number theory.

In particular, consider the following situation, when $b=0$ we are often concerned with finding integers $c$ represented by forms
$$y^2- ax^2$$
(i.e., can be written in this was $c = y^2 - ax^2$). Even answering this question with $c = 1$ yields the problem of Pell's equation, and otherwise is the Generalised Pell's equation (you can read the wikipedia page for an overview).

You may be interested to know that to find rational solutions to this equation, it suffices to find a single solution - in which case one can "project away" from the point to find a rational parametrisation for this curve (viewing it in the $(x,y)$-plane).

ContentLicense: CC BY-SA 4.0


问:
How do I prove the set of the equivalence classes of $R$ has the same cardinality as the set of all finite sets of primes?
I have been stuck on this for a while now.

Given the equivalence relation $R$ over $\mathbb{Z^+}: aRb \leftrightarrow \exists q \in \mathbb{Q}(\frac{a}{b} = q^2) $ how does one prove that the set of its equivalence classes $B$ has the same cardinality as the set of all finite sets of primes $C$.

What I am looking for, is some kind of proof that $B$ is a countable set.

I tried to come up with a mapping function between the two, compare them both to something else but everything was without any meaningful result. Any help is greatly appretiated.

答:
Let $m$ and $n$ be two positive square-free integers (i.e., $p^2$ doesn't divide $n$ or $m$, for any prime $p$), and suppose that $mRn$.

If $mRn$, then there exists a rational $q$ such that $\frac{m}{n}=q^2$.

Now, we may assume $q>0$, and this rational number can be written as $q = \frac{r}{s}$, where $r,s$ are positive integers and $\gcd(r,s)=1$.

So we have
$$
r = p_1^{\alpha_1}\cdots p_k^{\alpha_k}
\quad\text{and}\quad
s = p_1^{\beta_1}\cdots p_k^{\beta_k},
$$
where $\alpha_i\beta_i=0$, that is, for each $I$, we can't have both $\alpha_i\neq0$ and $\beta_i\neq0$.

It follows that
$$ms^2=nr^2,$$
and so $r^2|m$ and $s^2|n$ (because $\gcd(r,s)=1$).
But then $p_i^{2\alpha_i}|m$ and $p_i^{2\beta_i}|n$.

Since $m$ and $n$ are square-free, it must be $\alpha_i = \beta_i =0$, for all $i$.
But that means $q=r=s=1$ and $m=n$.

So, if $m \neq n$ then they represent different classes.

Given that they are square-free, each number if fully determined by the (finite) set of primes that divide it.

Edit.
The reasoning above shows that each square-free positive integer is in a different class.

To finish the desired correspondence, one still has to prove that there are no other classes, i.e., that every $n \in \mathbb Z^+$ is related to some square-free one.

This follows from the fact that $nRnp^2$, for every prime $p$ and integer $n$.

Indeed, there exists $q \in \mathbb Q$ such that $\frac{np^2}{n}=q^2$: just take $q=p$.


问:
estimate for a solution in some norm
Consider the following problem

$-(x^2+1)u''-2xu'=\sin(2\pi x), \;x \in (0,1)$

$u(0)=0$, $u(1)=0$.
I want to show that there exists a unique solution $u \in H^1_0(0,1)$ and find an estimate for

$\|u'\|_{L^2(0,1)}$

Here is the solution:

I have a few questions: What is $a_{ij}$ ? Is it $-(x^2+1)$? and if so how does that form a matrix?
Is it possible to prove uniqueness by using Lax-Milgram?

I don't understand the integrating by parts. What is our $u$ and $dv$?

Why are we inner producting with u? Shouldn't we use a test function but then again it is stated that u is in $H^1_0(0,1)$. I'd still like an answer on this, if possible

Finally the lines after the Cauchy Schwarz inequality are a bit mysterious.
Please shed some light on it.

And it looks like the problem was written incorrectly. Is should be find an estimate $||u'||_{H^1_0(0,1)}$ but I could be wrong

If you have a different solution that you would like to share, please feel free to do so

答:
Suppose $u_1$ and $u_2$ on $[0, 1]$  both satisfy
$$0 = (x^2 + 1)u''(x) + 2xu'(x) + \sin(2\pi x)$$
on $(0, 1)$ with the boundary conditions $u(0) = u(1) = 0$.
Then $w := u_1 - u_2$ satisfies
\begin{align}
0 &= (x^2 + 1)w''(x) + 2xw'(x) \\
&= ((x^2 + 1)w'(x))'
\end{align}
with $w(0) = w(1) = 0$. This shows $(x^2 + 1) w'(x)$ is a constant, say $C$, and that
$$ w(1) = w(0) + C \int_0^1 (x^2 + 1)^{-1} dx.$$
You know that $w(0) = 0$, that the integrand is positive on $(0, 1)$, so the only possible way you'll get $w(1) = 0$ is if $C = 0$.

You might consider more generally the ODE $u''(x) + b(x)u'(x) = f(x)$, where $b$ and $f$ are continuous on $[0, 1]$, same boundary conditions. With $u_1$ and $u_2$ also satisfying these, again set $w := u_1 - u_2$. Using the integrating factor $I(x) := \exp(\int_0^x b(\xi)d\xi)$ you can write
\begin{align}
0 &= w''(x) + b(x)w'(x) \\
  &= (I(x)w'(x))'/I(x)
\end{align}
and use the same trick to show $w = 0$.

All that being said, I understand it doesn't really answer the question about weak solutions, in this case in $H^1_0(0, 1)$, being unique.

Your original ODE can be written in the form $au'' + a'u' = f$, and for any $v \in H^1_0(0,1)$ you have
$(au'' + a'u')v = fv$. Integrating over $(0, 1)$ you get
\begin{align}
\int_0^1 fv &= \int_0^1 (au'' + a'u')v \\
&= \int_0^1 (au')'v \\
&= \int_0^1 (au'v)' - \int_0^1 au'v' \\
&= -\int_0^1 au'v'.
\end{align}
which provides a weak formulation.
Doing the $w := u_1 - u_2$ trick again you find $w$ has to satisfy
$$ 0 = \int_0^1 a w'v'.$$
The specific choice $v = w$ shows $\int_0^1 a (w')^2 = 0$.

Title: estimate for a solution in some norm


问:
$f : M \longmapsto N \hspace{0.1cm} C^{\infty}$ if dim $M > $dim $N$ then $f$ can't be injective
Is the statement true ? Let's suppose $M,N$ be smooth-manifolds without boundary for the moment, I think I got the proof if I'm able to find $y \in \text{RegVal}(f) \cap \text{Im}(f)$ arguing as follows : by the sake of contradiction let's $f$ be injective, then $f^{-1}(y) = \left\lbrace x\right\rbrace$, but for a known theorem $f^{-1} (y)$ is an $(m-n)$ manifold.

Since $m > n$, we have that this can't happens since $\left\lbrace x\right\rbrace$ is a $0-$manifold.

This seems correct, but I can always find some $y \in \text{RegVal}(f) \cap \text{Im}(f) ?$ And what if boundaries are allowed ?

Any help would be appreciated

答:
No, you most certainly cannot always find a regular value in the image of $f$. Consider the map $\Bbb R^m\to\Bbb R^n$ (with $m>n$) given by
$$f(x_1,\dots,x_n,x_{n+1},\dots,x_m) = (x_1,\dots,x_{n-1},0).$$
That said, this map has constant rank and certainly will not be injective.
In fact, even though $f$ may not be a submersion, in a neighborhood of a point of maximum rank the rank of $f$ will be constant and the map will, in appropriate coordinates, look like such a mapping. Thus, it fails to be injective in a neighborhood of any point of maximum rank. Where the rank drops, things get much more difficult. But this is still enough to answer your question.


问:
How to show that we can't use the negative root in $e^x$ = $y\pm\sqrt{y^2+1}$?
We have that $e^x$ = $y\pm\sqrt{y^2+1}$ but we know that $e^x$ > 0 so we must use $e^x$ = $y+\sqrt{y^2+1}$; how do I show that $y-\sqrt{y^2+1}$ will always be $\le 0$?

Body: First show that $y^2 < y^2+1$.  Then $y < \sqrt{y^2+1}$, so $y-\sqrt{y^2+1} < 0$.


Title: How to show that we can't use the negative root in $e^x$ = $y\pm\sqrt{y^2+1}$?

Body: We have that $e^x$ = $y\pm\sqrt{y^2+1}$ but we know that $e^x$ > 0 so we must use $e^x$ = $y+\sqrt{y^2+1}$; how do I show that $y-\sqrt{y^2+1}$ will always be $\le 0$?

Body: $ y^2<y^2+1\implies  y<\sqrt{y^2+1}\implies y-\sqrt{y^2+1}<0$


问:
Find $u\in\mathbb{R}$ such that $\mathbb{Q}(u) = \mathbb{Q}(2^{1/2}, 5^{1/3})$.
I am having trouble finding such a $u$.  My instincts at first told me to do the obvious thing and let $u = 2^{1/2}5^{1/3}$ but $u^{2} = \left(2^{1/2}5^{1/3}\right)^{2} = 2\cdot5^{2/3}$ but we want $\frac{1}{2}u^{2} = 5^{1/3}$ right??  Any hints on how to approach the exercise?

Then the next part ask to describe how you would find all $w\in\mathbb{Q}(2^{1/2}, 5^{1/3})$ such that $\mathbb{Q}(w) = \mathbb{Q}(2^{1/2}, 5^{1/3})$ I have no approach on this one except to possible look at linear combinations of the basis terms but dont know how to go about it.

答:
we want $\frac{1}{2} u^2 = 5^{1/3}$ right?

We just want $5^{1/3}$ to be in this field somewhere. You've shown it contains $5^{2/3}$: try squaring that.

ContentLicense: CC BY-SA 3.0


问:
Does zero vector in the span
This is seems to be an trivial question, I just want to know does zero vector always in the span since we could produce zero vector by multiply zero in each component of the basis set.</p>

答:
If our set $S$ is non-empty than it's pretty straight-forward that $0\in S$. If $S=\emptyset$, then the question gets trickier. The answer is still "yes",but some thought needs to be done. Normally you write:

$$\langle S \rangle=\langle v_1,...,v_n\rangle=\left\{\sum_{i=1}^n a_iv_i:a_i\in F\right\}$$

A more general notation:
$$\langle S \rangle=\langle v_1,...,v_n\rangle=\left\{\sum_{i\in \{1,...,n\}} a_iv_i:a_i\in F\right\}$$
So if our set is empty:
$$\langle S \rangle=\langle \emptyset \rangle=\left\{\sum_{i\in \emptyset} a_iv_i:a_i\in F\right\}$$
So our sum has no summands!(it's an empty sum)
What's the value of an empty sum?
Well obviously if the sum is empty we would like that:
$$1+\sum_{i\in \emptyset} \alpha_i=1$$
So the only meaningful definition is:
$$\sum_{i\in \emptyset} \alpha_i=0$$
And this means that:
$$\langle S \rangle=\langle \emptyset \rangle=\{0:a_i\in F\}=\{0\}$$
It's always an interesting logic exercise to reason on this kind of subtleties. If you want a little challenge, what should be the value of an empty product?
$$\prod_{i \in \emptyset} \alpha_i=?$$
(However this is more than a mental exercise, the fact that $\langle \emptyset \rangle=\{0\}$, guarantees the existence of a basis of the null vector space)

Title: Does zero vector in the span


问:
Prove that $\sin^2{x}+\sin{x^2}$ isn't periodic by using uniform continuity
Before the problem is a proof that says a periodic function whose domain is $\mathbb{R}$ is uniformly continuous.So actually the problem is to prove $\sin^2{x}+\sin{x^2}$ isn't uniformly continuous.I hope to fellow the problem.Thanks for the zealous!</p>

答:
We proceed from the definition of uniform continuity. (One can get the result more simply by considering the derivative of $\sin(x^2)$.)

It is enough to show that $\sin(x^2)$ is not uniformly continuous. For if $\sin^2 x+\sin(x^2)$ were uniformly continuous, then since $\sin^2 x$ is uniformly continuous, the difference  $\sin^2 x+\sin(x^2)-\sin^2 x$ would be uniformly continuous.

Let $\epsilon=\frac{1}{2}$. We show that there does not exist a $\delta\gt 0$ such that for all $x$, if $|x-y|\lt \delta$, then $|\sin(x^2)-\sin(y^2)|\lt \epsilon$.

Let $\delta$ be a fixed positive quantity. Let $x=\sqrt{2n\pi}$, where $n$ is a large integer to be chosen later. Then $\sin(x^2)=0$. Let $n$ be such that $2\delta\sqrt{2n\pi}\gt \frac{\pi}{2}$. Note that
$$(x+\delta)^2=x^2+2x\delta+\delta^2\gt x^2+2x\delta\gt 2n\pi +2\delta\sqrt{2n\pi}\gt 2n\pi+\frac{\pi}{2}.$$

It follows that there is a $y$ such that $x\lt y\lt x+\delta$ and $\sin(y^2)=1$. In particular, $|\sin(x^2)-\sin(y^2)|\gt \frac{1}{2}$. This completes the proof.


问:
Action of a Lie group on a smooth manifold
Suppose a Lie group acts smoothly (but not necessarily properly or freely)
on a smooth manifold $M$. Show that each orbit is an immersed submanifold
of $M$; which is embedded if the action is proper.

This is problem 15 of chapter 21 of Lee's smooth manifolds book I was searching to solve the above problem and I'm now a little confused writing an easy argument for this. I found the following argument but this is not easy enough for me to understand:

For each $p\in M$, the isotropy group $G_p = \{g\in G: g\centerdot p=p\}$ is a closed subgroup of $G$.

The quotient space $G/G_p$ has a unique smooth manifold structure, and $G$ acts transitively on this smooth manifold by left multiplication.

The orbit map $\mathscr O_p\colon G\to M$ given by $\mathscr O_p(g) = g\centerdot p$ descends smoothly to the quotient $G/G_p$ to give a smooth map $F_p\colon G/G_p\to M$ that is a bijection from $G/G_p$ onto the orbit $G\centerdot p$.

The map $F_p$ is equivariant with respect to the left actions of $G$ on $G/G_p$ and $M$, and therefore it has constant rank.

Because it is a constant-rank injection, $F_p$ is a smooth immersion, and therefore its image $G\centerdot p$ is a smoothly immersed submanifold of $M$.

Could anyone help me prove this?

答:
So I am going to do this step by step. So I belive it's clear that $G_p$ is a closed subgroup of $G$, and we can consider the proper free action $G\times G_p\rightarrow G_p$ by left translation , and since $G_p$ is a closed subgroup it also has a manifold structure and we can consider the space $G/G_p$. Now we get a bijection from $G/G_p$ to the orbit $g.p$ of $p$ since we are doing the quotient by the points that would fix the orbit. Then you have that $F_p$ is an equivariant map because $g.F_p(hG_p)=g(hp)=(gh)p=(gh)G_p p=F_p(ghG_p)$. Using the fact that the action of $G$ in $G_p$ is transitive and the equivariant rank theorem you have that $F_p$ must have constant rank and then using the fact that locally for suitable coordinates $F_p$ looks like $(x_1,...,x_r)\rightarrow (x_1,...,x_k,0,...,0)$ where $k$ is the rank of $F_p$, but since $F_p$ is injective this forces $k=r$ and so you get that it's an immersion,  and so since the image of $F_p$ is the orbit of $p$ and that $F$ is an injective immersion you get the desired result.

I don't know what much could be added , is there anything that needs clarification ?


问:
Ideal of a group
The ideal is defined in the ring theory;

In ring theory, a branch of abstract algebra, an ideal of a ring is a special subset of its elements

In the answer to this question What is the exponent of a group?, the term ideal is used for groups as;

The exponent of a group $G$ is the non-negative generator of the ideal $\{z \in \mathbb{Z} : \forall g \in G (g^z=1)\}$.

Is this usage of the term ideal correct? If not, what is the correct term?

答: The ideal $\{z \in \mathbb{Z} : \forall g \in G (g^z = 1)\}$ does not live in $G$, it lives in $\mathbb{Z}$. Because $\mathbb{Z}$ is a PID whose units are $\{1, -1\}$, we may speak of "the non-negative generator" of any ideal.

Title: Ideal of a group


问:
Propriety of a spectral family on a Hilbert space
Let $\{E(\lambda)\}_{\lambda \in \mathbb{R}}$ be a spectral family on a Hilbert space $\mathcal{H}$, meaning a collection of orthogonal projectors satisfying, for any $f \in \mathcal{H}$

$ \quad (i) \ \lim \limits_{\lambda \to -\infty} E(\lambda)f = 0 \quad $ and $\quad \lim \limits_{\lambda \to +\infty} E(\lambda)f = f$

$ \quad (ii) \ \ (E(\lambda)f,f) \le (E(\mu)f,f) \ \ $ for $\lambda \le \mu$

$ \quad (iii) \ \lim \limits_{\epsilon \to 0^+} E(\lambda + \epsilon)f = E(\lambda)f$

Then, $E(\lambda)E(\mu) = E(\mu)E(\lambda) = E(\lambda) \ \ $ for $\lambda \le \mu$

My attempt

Since $E(\lambda)$ is a projector, $f \in \mathrm{Ran} \ E(\lambda) \iff f = E(\lambda)f$. $\\$ Also, from $(ii)$, one gets $\mathrm{Ran} \ E(\lambda) \subseteq \mathrm{Ran} \ E(\mu)$. Using these two: since obviously $E(\lambda)f \in \mathrm{Ran} \ E(\lambda) \subseteq \mathrm{Ran} \ E(\mu)$, it follows $E(\mu)(E(\lambda)f) = E(\lambda)f$, so $E(\mu)E(\lambda) = E(\lambda)$.

I am however having trouble proving the other half of the statement, $E(\lambda)E(\mu) = E(\lambda)$. Maybe I am missing something obvious

答:
Upon request, here is the answer contained in my comment above:

Once the OP has already proven that $E(\mu)E(\lambda) = E(\lambda)$, and observing that orthogonal projectors are self adjoint, we have
$$
E(\lambda)E(\mu) =
\big(E(\mu)E(\lambda)\big) ^* = E(\lambda)^* = E(\lambda).$$


问:
keeping track of proper classes vs sets when showing ZFC has a countable model (assuming ZFC is consistent)
I've heard the claim that ZFC has a countable model in ZFC if ZFC is consistent. As far as I can tell, this is a straightforward consequence of Gödel's completeness theorem and the Löwenheim–Skolem theorem.

I'm curious, though, how careful one needs to be about bookkeeping details related to sets and proper classes when formulating the argument ... or if the issue just doesn't come up for an obvious reason that I'm missing.

So, my question is twofold

Does the completeness theorem give us a model over a set anyway?

If we pick a model over a proper class initially, can we use Löwenheim–Skolem?

When I nonstandardly say a model over X, I mean a model whose universe is X. I nonstandardly say a model in X when the fundamental set theory used to construct it is the set theory X.

ZFC is a first order theory with one relation symbol $\in$, by Gödel's completeness theorem it has a model. I'm not sure whether this theorem promises us the existence of a model over a set or if the promised model could be a model over a proper class instead.

Also, ZFC does not have any finite models, since $\emptyset, P(\emptyset), P(P(\emptyset)), \cdots$ are all distinct, where $P$ denotes the powerset.

Suppose for the sake of argument that we conclude that ZFC has a model over a proper class (using ZFC as the raw material for building the model). In that case, we interpret $\in$ as $\in$ in ZFC. So, we've built a trivial model, let's call it $M$. The universe of $M$ is the class of all sets in ZFC. It's possible that by applying the completeness theorem, we could have gotten a model over a set, which would make this problem go away. I'm not sure. Suppose we decided to pick $M$ as our model even if it's an inconvenient choice.

The second prong of the argument, as far as I can tell, is an application of the (downward) Löwenheim-Skolem theorem. The downward Löwenheim-Skolem theorem does not address the [proper class]-set boundary explicitly, so I'm wondering how the argument works. This question invokes Löwenheim-Skolem as an explanation for the existence of countable models.

In the proof of the Löwenheim-Skolem theorem sketched in the Wikipedia article, we use the axiom of choice repeatedly to select elements of the universe that have to be included in the new universe for the model we're constructing.

A strict reading of the axiom of choice promises nothing about the ability to make arbitrary choices over proper classes (and a choice function cannot have a proper class as a domain), so I'm not sure how to get from my trivial self-model of ZFC $M$ to any model whose universe is a set, let alone a countable set.

答:
Does the completeness theorem give us a model over a set anyway?

Yes.  The completeness theorem says that every consistent theory has a model, and in this context, "model" is defined to mean a model over a set.

If we pick a model over a proper class initially, can we use Löwenheim–Skolem?

This is a subtle issue--what does it even mean to have a model over a proper class?  The issue is that the usual recursive definition of what it means for a structure to satisfy a formula does not work when your structure is a proper class.  So, if you have a structure $(M,\epsilon)$ where $M$ is a proper class, you cannot even state $(M,\epsilon)\models ZFC$ in the language of ZFC.

However, for any individual sentence $\varphi$, you can state $(M,\epsilon)\models \varphi$, by carrying out the recursive definition "by hand" in the specific case of $\varphi$.  So typically, when people talk about "class models", what they mean instead is that there is a theorem scheme which says that for each axiom $\varphi$ of ZFC, you can prove that $(M,\epsilon)\models\varphi$.  Alternatively, you could assume that your class model has the special property that you actually can define a formula satisfaction relation for it, but this is rarely actually true in practice.

As for carrying out a Löwenheim–Skolem argument for class models, you can do this as long as you can define satisfaction with respect to all the formulas you care about.  (This means you can carry it out as usual if your class model has a satisfaction relation, or if it doesn't, you can only carry it out to get a submodel that satisfies finitely many particular axioms of ZFC, rather than all of ZFC.)  The issue with the axiom of choice that you point out can be avoided using Scott's trick.  For instance, every time the proof of Löwenheim–Skolem would have you make a choice of an element with some property, instead of choosing just one element, you can pick all the elements with the desired property that have minimal rank.  There may be uncountably many such elements so this may not give you a countable model, but it will at least get you down to a set-sized model.  Then, you can use ordinary Löwenheim–Skolem to get down to a countable model.

Note that in this weak (theorem scheme) sense of "class model", you do not even need to assume that ZFC is consistent in order to get a class model of ZFC: $(V,\in)$ is a class model of ZFC.  The argument above then shows that for any finite subset of the axioms of ZFC, you can prove there is a (set) model of those axioms.  This is known as the reflection principle. (Crucially, though, this is a separate theorem for each finite set of axioms, rather than a single theorem with a quantified variable representing an arbitrary finite set of axioms.  As a result, you cannot combine it with compactness to conclude that ZFC is consistent.)


问:
Proving or disproving: $a^b<b^a$?
So, I have two real numbers, and I need to prove or disprove the statement: $a^b<b^a$.

Note: The question contains real numbers and not parameters. I just don't want you to solve it for me, so I don't reveal the numbers... This is a calculus I question.

I have thinking about this question for a while, with no clue on what to do.

I thought of defining some functions, and use some theorems we learned...

Any hints on how to deal with those kind of questions will be much appreciated!

Thanks a lot!

Title: Proving or disproving: $a^b<b^a$?

Body:
As shown in the other answers,
the hard case is when
$a < e < b$.

Many years ago,
I proved the following
result in this case:

If
$1 < x < e$
and $e < y < e^2/x$
(i.e., $xy < e^2$)
then
$y^{1/y} > x^{1/x}
$.

I don't know about
$y > e^2/x$.

I've reconstructed my proof
and here it is.

Let
$f(x)
=\dfrac{x^{1/x}}{(e^2/x)^{x/e^2}}
$
with
$1 < x < e$.

The proof shows that
$\ln(f(x))
\lt 0$
for $1 < x < e$.
This reduces to
showing that
$h(z)=(1-z)\cosh(z)-e^{-z}
\lt 0$
for $0 < z < 1$.

As usual,
I'm sure there are
simpler proofs than mine,
which shows that
the coefficients of
$h(z)$
are all negative.

Show
$f(x) <1 $.

$\begin{array}\\
g(x)
&=\ln(f(x))\\
&=(1/x)\ln(x)-(x/e^2)\ln(e^2/x)\\
&=(1/x)\ln(x)-(x/e^2)(\ln(e^2)-\ln(x))\\
&=(1/x)\ln(x)-(2x/e^2)+(x/e^2)\ln(x)\\
&=(1/x+x/e^2)\ln(x)-(2x/e^2)\\
&=(e^{-y}+e^y/e^2)y-(2e^y/e^2)
\qquad x = e^y, 0 < y < 1\\
&=(e^{-y}+e^{y-2})y-2e^{y-2}\\
&=e^{-1}(e^{-y+1}+e^{y-1})y-2e^{y-2}\\
&=\dfrac{2y}{e}\dfrac{e^{-y+1}+e^{y-1}}{2}-2e^{y-2}\\
&=\dfrac{2y}{e}\cosh(-y+1)-\dfrac{2}{e}e^{y-1}\\
&=\dfrac{2}{e}(y\cosh(-y+1)-e^{y-1})\\
&=\dfrac{2}{e}((1-z)\cosh(z)-e^{-z})
\qquad z = 1-y, y = 1-z, 0 < z < 1\\
&=\dfrac{2}{e}h(z)
\qquad h(z)=(1-z)\cosh(z)-e^{-z}\\
&=\dfrac{2}{e}\left((1-z)\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}-\sum_{n=0}^{\infty} \dfrac{(-1)^nz^n}{n!}\right)\\
&=\dfrac{2}{e}\left(\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}-z\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}-\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}+\sum_{n=0}^{\infty} \dfrac{z^{2n+1}}{(2n+1)!}\right)\\
&=\dfrac{2}{e}\left(-\sum_{n=0}^{\infty} \dfrac{z^{2n+1}}{(2n)!}+\sum_{n=0}^{\infty} \dfrac{z^{2n+1}}{(2n+1)!}\right)\\
&=\dfrac{2}{e}\left(\sum_{n=0}^{\infty} \dfrac{z^{2n+1}(1-(2n+1))}{(2n+1)!}\right)\\
&=-\dfrac{2}{e}\sum_{n=0}^{\infty} \dfrac{2nz^{2n+1}}{(2n+1)!}\\
&\lt 0
\qquad\text{for } z > 0\\
\end{array}
$

Therefore
$g(x) < 0$
so $f(x) < 1$.

Therefore,
if $r(x) = x^{1/x}$,
then,
if $1 < x < e$,
$r(x)
\lt r(e^2/x)
$.

Since $e^2/x > e$,
if
$e < y <  e^2/x$,
$r(y)
\gt r(e^2/x)
\gt r(x)
$.

Therefore,
if $1 < x < e < y < e^2/x$,
or $1 < x < e$
and $xy < e^2$
then
$r(y) > r(x)
$.


问:
$\dim V_\lambda=\dim V_{w(\lambda)}$ when $V$ is an integrable module over a Kac-Moody algebra

Let $V$ be an integrable module over a Kac-Moody algebra. Then $\dim V_\lambda=\dim V_{w(\lambda)}$ for each $\lambda\in\mathfrak{h}^*$ and $w\in W$ (the Weyl group).

It's a proposition stated in Kac's book Infinite dimensional Lie algebras, however he just states it follows form the two following lemmas:

$\textrm{ad}\ e_i$ and $\textrm{ad}\ f_i$ are locally nilpotent over a Kac-Moody algebra.

If $\lambda$ is a weight of $V$, then $\lambda-\lambda(\alpha_i^\vee)\alpha_i$ is also a weight of the same multiplicity.

However I don't see why this is the case?

Also the first lemma is pretty clear to me, but I don't see why the second lemma holds?

I'm also happy to see, if there is another way of proving this. Maybe it can somehow be reduced to the finite-dimensional case like when showing that "The module $L(\Lambda)=M(\Lambda)/N(\Lambda)$ (the quotient of a Verma module) over a Kac-Moody algebra is integrable if and only if $\lambda(\alpha_i^\vee)\geq 0$ for all $i=1,\cdots,n$"?

It should also be noted, that it've been earlier shown that "If $\alpha\in\Delta, w\in W$ then $w\alpha\in W$ and $\textrm{mult}(\alpha)=\textrm{mult}(w\alpha)$."
Here $\Delta$ denotes the union of the positive and negative roots.

I've managed to prove that $V$ decomposes into a direct sum of finite-dimensional irreducible $\mathfrak{sl}_2$-modules.

Now in order to finish the proof, we need to justify that given a module $N$ in the $\mathfrak{sl_2}$-decomposition then $\dim N_\lambda=\dim N_{r_i\lambda}$. I think this i exactly the second lemma stated by Kac above. However we also need to justify that the weights have the form Joppy mentions. We also need to justify that $\lambda(\alpha_i^\vee)\in\mathbb{Z}$. This is where I'm still in doubt.

答:
When $\mathfrak{g}$ is a Kac-Moody algebra with Cartan indexing set $I$, the vector space $\mathfrak{g}_{(i)} := \mathbb{C}\{e_i, \alpha_i^\vee, f_i\}$ is a Lie subalgebra isomorphic to $\mathfrak{sl}_2$ for each $i \in I$. An $\mathfrak{h}$-diagonalisable $\mathfrak{g}$-module $V$ is integrable if it is a locally finite $\mathfrak{g}_{(i)}$-module for all $i \in I$. (Recall that to be locally finite means that $V$ is a union of finite-dimensional modules). This means that an integrable module $V$ decomposes as a (possibly infinite) direct sum of finite-dimensional $\mathfrak{g}_{(i)}$-modules, or in other words a direct sum of $\mathfrak{sl}_2$-modules.

The weights of finite-dimensional $\mathfrak{sl}_2$-modules are symmetric around zero, with the reflection $s: \lambda \mapsto - \lambda$ giving the symmetry. Similarly, the weights of $\mathfrak{h}$-invariant finite-dimensional $\mathfrak{g}_{(i)}$-modules are symmetric around the hyperplane $\ker \langle \alpha_i^\vee, - \rangle \subseteq \mathfrak{h}^*$, with the simple reflection $r_i(\lambda) = \lambda - \langle \alpha_i^\vee, \lambda \rangle \alpha$ giving the symmetry. If we were to zoom in on a single finite-dimensional $\mathfrak{g}_{(i)}$-submodule in $V$, with highest weight vector $v_{\Lambda}$ say, then we would find that its weights were
$$ \Lambda, \Lambda - \alpha, \Lambda - 2\alpha, \cdots, r_i(\Lambda) + \alpha, r_i(\Lambda). $$
What I picture in my head is that I have the large representation $V$, and when I only look at a single $i \in I$, it breaks down into finite $\mathfrak{sl}_2$-strings. On this finite-dimensional $\mathfrak{g}_{(i)}$-submodule $U \subseteq V$ we have $\dim U_\lambda = \dim U_{r_i \lambda}$ for all $\lambda$, therefore adding this up across all the submodules we get $\dim V_\lambda = \dim V_{r_i \lambda}$. Since we can do this for any $i \in I$, we get $\dim V_\lambda = \dim V_{w \lambda}$ for any element $w$ of the Weyl group.


答:
Here's some partial progress. Maybe someone will have further ideas.

By homogeneity, we can set $a_1 + \cdots + a_n = 1$ throughout. First dispense with boundary behavior. Fix $1 \leq k \leq n$ and note that
$$\frac{\prod_{i=1}^k (a_i+a_{i+1})}{\prod_{i=1}^n a_i} = \prod_{i=1}^{k-1} \frac{a_i + a_{i-1}}{a_i} \cdot \prod_{i=k}^{n-1} \frac{a_i + a_{i+1}}{a_{i+1}} \cdot \frac{1}{a_k} \geq \frac{1}{a_k}.$$

Hence $F_n(a_1, \ldots, a_n) \geq \frac{1}{\min_i a_i}$, so $F_n \to \infty$ uniformly as we tend to the boundary.

Let $G_n(a_1, \ldots, a_n) = \log F_n(a_1, \ldots, a_n)$, which we may as well minimize. Recalling that $a_1+\cdots+a_n=1$ throughout, consider when $\nabla G_n = 0$. We find
$$\begin{align*}a_1 \partial_1 G_n &= \frac{a_1}{a_1+a_2} + a_1 - 1 \\
a_i \partial_i G_n &= \frac{a_i}{a_{i-1} + a_i} + \frac{a_i}{a_i + a_{i+1}} + a_i - 1 \qquad \text{for }1 < i < n \tag{*}\label{*}\\
a_n \partial_n G_n &= \frac{a_n}{a_{n-1}+a_n} + a_n - 1.
\end{align*}$$
In particular we have the relation
$$\sum_{i=1}^n a_i \partial_i G_n = 0$$
which seems important, though I haven't found a use for it.

Using $\eqref{*}$, we can rewrite the condition $\nabla G_n = 0$ as either
$$\begin{align*}
\frac{a_2}{a_1+a_2} &= a_1 \\
\frac{a_3}{a_2+a_3} &= a_1 + a_2 \\
&\vdots \\
\frac{a_n}{a_{n-1}+a_n} &= a_1 + a_2 + \cdots + a_{n-1} \\
1 &= a_1+\cdots+a_n
\end{align*}$$
or
$$\begin{align*}
\frac{a_2}{a_1+a_2} &= a_1 \\
\frac{a_3}{a_2+a_3} &= \frac{a_2}{a_1+a_2} + a_2 \\
&\vdots \\
\frac{a_n}{a_{n-1}+a_n} &= \frac{a_{n-1}}{a_{n-2}+a_{n-1}} + a_{n-1} \\
a_1+\cdots+a_n &= 1.
\end{align*}$$

Either of these forms is suitable for computer algebra systems. I threw the first 12 into Mathematica; here's the output. Using the first $n-1$ relations, we may iteratively solve for $a_2, a_3, \ldots, a_n$ as rational functions of $a_1$, and then we may impose the final condition to solve for $a_1$. This shows that $a_1$ (and hence all the others) are algebraic, so the algebraic minimum math110 suggests in a comment above is at least plausible on its face. Edit: indeed, I checked through $n=12$ that the values of $F_n$ when $\nabla G_n = 0$ are precisely the Galois conjugates of math110's guess of $\left(\sin\left(\frac{2\pi}{n+2}\right)/\sin\left(\frac{\pi}{n+2}\right)\right)^{n+2}$.

Based on that data, it seems that when $n$ is odd, $a_1$ satisfies $p_n(t) := \sum_{k=0}^{(n+1)/2} \binom{n+1-k}{k} (-t)^k = 0$. Intriguingly, this family of polynomials appears to be tending pointwise to 0 on $[0, 1]$ as $n \to \infty$, and they seem to be real-rooted. Moreover, it appears that the unique value of $a_1$ solving the above system subject to $a_i > 0$ and $a_1 + \cdots + a_n = 1$ is the smallest root of $p_n(t)$.

I get the sense that I'm missing some sort of physical meaning in this problem. I also can't see why $\eqref{*}$ implies the symmetry $a_i = a_{n-i+1}$, though it should. There's a bizarre amount of structure throughout this problem that I'm at a loss to explain!

Title: How find this nice minmum of this value $\frac{\prod_{i=1}^{n-1}(a_{i}+a_{i+1})\sum a_{i}}{\prod_{i=1}^{n}a_{i}}$

Body:
Let $n$ be give postive integer number. For any $a_{i}>0 (i=1,2,\cdots,n)$, find the minimum of the value
$$F_{n}(a_{1},a_{2},\cdots,a_{n})=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})\cdots (a_{n-1}+a_{n})(a_{1}+a_{2}+\cdots+a_{n})}{a_{1}a_{2}a_{3}\cdots a_{n}}.$$(by wang yong xi)

I try when $n=2$,then
$$F_{2}(a_{1},a_{2})=\dfrac{(a_{1}+a_{2})(a_{1}+a_{2})}{a_{1}a_{2}}=\dfrac{(a_{1}+a_{2})^2}{a_{1}a_{2}}\ge 4$$when $a_{1}=a_{2}$  is minimum

(2):when $n=3$,
$$F_{3}=\dfrac{(a_{1}+a_{2})(a_{2}+a_{3})(a_{1}+a_{2}+a_{3})}{a_{1}a_{2}a_{3}}$$WLOG $a_{3}=1$,so
$$F=\dfrac{(a_{1}+a_{2})(a_{2}+1)(a_{1}+a_{2}+1)}{a_{1}a_{2}}$$
I use this find this minimum is $$(F_{3})_{min}=\dfrac{1}{2}(11+5\sqrt{5})$$
when $a_{1}=1,a_{2}=\dfrac{\sqrt{5}-1}{2}$ seelinks


问:
How to find the shaded area of a triangle formed by the tangent of a circle?

Hey guys,

The last question of the chapter has stumped me once again! I need to find the area of part of this triangle. I’m not sure how to attack the question considering that I’m not given the radius of the circle and I’m not sure how the measurement of 4cm will help me?

Thanks,
Scratch Cat

答:
One way is to find the difference in areas of a right-angled triangle and a circular sector.

Finding both areas requires finding the radius of the circle, which may be found through the triangle

$$\begin{align*}
\frac{r}{r+4} &= \sin 35^\circ\\
r &= (r+4)\sin 35^\circ\\
(1-\sin 35^\circ)r &= 4\sin 35^\circ\\
r &= \frac{4\sin 35^\circ}{1-\sin 35^\circ}
\end{align*}$$


问:
Existential Instantiation Intuition
I am a little confused with the Existential Instantiation. If the domain = {a,b}, and the hypothesis is ∃xP(x) for some predicate P, why can we claim that both P(a) and P(b) are true via P(a)∧(a is in the domain) and WLOG for the case with b. What is the intuition behind this? My interpretation is that "there exists an x" corresponds to "at least one", so why would this be true if we aren't using the universal quantifier?

Edit: Here is some context:

I am trying to show the following argument is invalid over the domain {a,b}:

$$h_{1}:\forall x (P(x) \rightarrow Q(x))$$
$$h_{2}:\exists x \neg P(x) $$
$$ \therefore  \neg Q(x)$$

The source says that $h_{1}$ is true because P(x) is false for both inputs. I am not sure how one can deduce that P(a) and P(b) are false for $\exists x \neg P(x)$

Edit 2: I misunderstood the exercise - I was supposed to give values to each predicate so that the argument holds invalid. This clarifies a lot here.

答:
Existential instantiation says that you can remove an outermost existential quantifier and replace the variable that it introduced with a fresh constant that doesn't appear earlier in the proof.

A fresh constant is not the same thing as an element of your domain. It is a new constant symbol that stands for a specific element element of your domain (but an arbitrary one).

For instance, suppose $ \exists x \mathop. \varphi$ is true  and the constant symbol $c$ has not been used so far in the proof. You can then conclude the following.

$$ \varphi[x := c] $$

This means $\varphi$, but with free occurences of $x$ replaced with $c$. For example

$$ \exists x \mathop. P(x) \to Q(x) $$
$$ P(c) \to Q(c) $$

Quoting Wikipedia briefly,

The rule has the restrictions that the constant $c$ introduced by the
rule must be a new term that has not occurred earlier in the proof,
and it also must not occur in the conclusion of the proof.

The reason for freshness constraint is to prevent us from concluding that $\exists x \mathop. \varphi$ and $\exists x \mathop. \psi$ implies $\exists x \mathop. \varphi \land \psi$. The reason to prohibit $c$ from appearing in the conclusion is its arbitrariness. We don't know anything about $c$ other than that it's an element of the domain of discourse and it satisfies the properties specified in the statement it's taken from ($\varphi$ and $P(x) \to Q(x)$ in the above examples).

Title: Existential Instantiation Intuition


问:
Show a sequence is convergent, and find it's limit.
I need to show that the sequence (n^2+1)/(3n^3-2n^2) for n=1,2,... is convergent and find the limit.
Can someone cofirm i've done this correctly.

(n^2+1)/(3n^3-2n^2)
dividing top and bottom by n^3 we get, (1/n +1/n^3)/(3-2/n)
As 1/n has limit 0 we get (0+0)/3-0)=0/3 = 0
So this is convergent with limit0

Thanks
Kara

答:
You can get
an explicit upper bound
by noting that,
for $n \ge 2$,
$\dfrac{1/n +1/n^3}{3-2/n}
\le \dfrac{2/n}{3-1}
=\dfrac1{n}
$.


问:
How to determine which probability distribution this distribution is closest to?
I have run a simulation and obtained a large number of values which I have used to plot a discrete probability distribution. How do I determine which probability  distribution this distribution is closest to? I originally thought it was a binomial distribution but was told I was incorrect.

Distribution image:

答:
I will illustrate how to fit a dataset with some distribution and then establish whether such a fit is reasonable.

Suppose we're given the following hypothetical grouped dataset: $$\text{Class}:0,1,3,8,9,11$$ $$\text{Observed Frequency}:4,5,4,3,1,1$$ If we seek to fit this grouped data with a $\text{Poisson}(\lambda)$ distribution we will need to use our dataset to estimate $\lambda$. It is well$-$known that the MLE for $\lambda$ is the sample mean $\frac{\sum xf_x}{\sum f_x}=\frac{61}{18}$. To see if this assumption is reasonable we can perform the hypothesis test $$H_0: \text{We're sampling from a Poisson}\Big(\frac{61}{18}\Big) \text{ distribution}$$ $$H_a: \text{We're not sampling from a Poisson}\Big(\frac{61}{18}\Big) \text{ distribution}$$ Assuming $H_0$ is true the expected frequency for the class $0$ is $$P(X=0)\cdot \sum f_x \approx 0.607$$ where $X\sim \text{Poisson}\Big(\frac{61}{18}\Big)$. Continuing in this manner we get a list of "expected counts" shown below: $$\text{Class}:0,1,3,8,9,11$$ $$\text{Expected Frequency}:0.607,2.059,3.940,0.262,0.099,0.010$$ The test statsitic for this hypothesis test is $\sum \frac{(O-E)^2}{E}\approx 158$ which possesses a chi$-$squared distribution having $5$ degrees of freedom. The $p-$value is $$P(\chi^2_5\geq 158)\approx 0$$ So we'll reject $H_0$ and try to find some other distribution to fit our dataset.


问:
A question about the complex conjugate bundle
If a complex vector bundle is constructed by the complexification of a real vector bundle, say $E=F\otimes \mathbb{C}$, then there's a conclusion that $E$ is isomorphic to its conjugate bundle by setting  $f: x+iy \to x-iy$. However, usually a complex bundle is not isomorphic to its conjugate bundle. So what's the intrinsic difference between them? I'm reading the section on Pontrjagin classes in Milnor and Stasheff's "Characteristic Classes" and I am a little confused.

答:
A complex conjugation on a complex vector space $V$ is an antilinear map $\sigma : V \to V$ such that $\sigma\circ\sigma = \operatorname{id}_V$; note that $\sigma$ induces an isomorphism $V \to \overline{V}$. Choosing a complex conjugation on $V$ is equivalent to choosing an isomorphism between $V$ and the complexification of a real vector space; see this answer. Every finite-dimensional complex vector space $V$ admits a complex conjugation because $V \cong \mathbb{C}^n$ for some $n$ and $\mathbb{C}^n\cong\mathbb{R}^n\otimes_{\mathbb{R}}\mathbb{C}$.

By the same argument as above, a complex vector bundle admits a complex conjugation (and hence is isomorphic to its conjugate bundle) if and only if it is the complexification of a real vector bundle. However, not every rank $n$ complex vector bundle is isomorphic to the trivial rank $n$ complex vector bundle (which is isomorphic to the complexification of the trivial rank $n$ real vector bundle). So a non-trivial bundle need not admit a complex conjugation map.

A natural question to ask is whether the existence of an isomorphism $E\to \overline{E}$ implies the existence of a complex conjugation. The answer is no! That is, there are complex vector bundles $E$ which do not admit a complex conjugation but nonetheless are isomorphic to $\overline{E}$.

Example: For any four-dimensional CW complex $X$, there is a bijection between isomorphism classes of principal $SU(2)$-bundles (or equivalently, rank two complex vector bundles with first Chern class zero) and $H^4(X; \mathbb{Z})$ given by $E \mapsto c_2(E)$; see this question.

Let $E \to S^4$ be a complex rank two vector bundle; note that $c_1(E) = 0$. As $c_1(\overline{E}) = 0$ and $c_2(\overline{E}) = c_2(E)$, we see that $E$ is isomorphic to $\overline{E}$. Suppose now that $E$ admits a complex conjugation. Then $E$ is isomorphic to $E_0\otimes_{\mathbb{R}}\mathbb{C}$ for some real rank two vector bundle $E_0$. As $w_1(E_0) = 0$, the bundle $E_0$ is determined up to isomorphism by its Euler class which is necessarily zero, so $E_0$ is trivial. Therefore, if $E$ admits a complex conjugation, then $E$ is trivial.

So for any $\alpha \in H^4(S^4, \mathbb{Z}) \cong \mathbb{Z}$ with $\alpha \neq 0$, the complex rank two vector bundle $E \to S^4$ with $c_2(E) = \alpha$ satisfies $E \cong \overline{E}$ but does not admit a complex conjugation.


问:
Gradient of trace of Kronecker product
Consider the following equation

$$
f(X)=\operatorname{tr}[(X \otimes I_N)^{-1}B],
$$
where $X$ is positive-definite with dimension $L$, and $I_N$ is an identity matrix of dimension $N$. $B$ is also positive-definite.

The goal is to compute the gradient of $f$ over $X$, but I have not found a way to do that yet.

My first trial was to reformulate the term to get rid of the $\otimes$ operation, however, this could not be achieved by combining the properties of the Kronecker product and trace.

答:
First, decompose $B$ into a sum of the form
$$
B = \sum_{i,j = 1}^N B_{ij} \otimes E_{ij},
$$
where each $B_{ij}$ has size $L \times L$ and $E_{ij}$ denotes the $N \times N$ matrix with a $1$ as its $i,j$ entry and zeros elsewhere.  Then, rewrite the function as
$$
\begin{align}
f(X) &= \operatorname{tr}[(X^{-1} \otimes I_N) B] =
\sum_{i,j = 1}^N \operatorname{tr}[(X^{-1} \otimes I_N)(B_{ij} \otimes E_{ij})]
\\ & = \sum_{i,j=1}^N \operatorname{tr}[(X^{-1}B_{ij}) \otimes E_{ij}]
= \sum_{i,j = 1}^N \operatorname{tr}(X^{-1}B_{ij}) \operatorname{tr}(E_{ij})
\\ & = \sum_{i=1}^N \operatorname{tr}(X^{-1}B_{ii}) = \operatorname{tr}\left(X^{-1}\sum_{i=1}^N B_{ii}\right)
\\ & = \operatorname{tr}(X^{-1} \operatorname{tr}_2(B)),
\end{align}
$$
where $\operatorname{tr}_2$ denotes a partial trace. From there, you have a matrix whose derivative can be looked up in a standard reference; see for instance this wikipedia table. The "numerator layout" derivative in this case will be
$$
\frac{\partial f}{\partial X} = -X^{-1}\operatorname{tr}_2(B) X^{-1}.
$$
The "gradient" of your function, as the term is typically used in this context (i.e. the direction of steepest ascent), will be the transpose of the above expression.

Title: Gradient of trace of Kronecker product


答:
For the induction the following will be useful to show that the difference is a multiple of 11:

$10^0 - (-1)^0 = 1-1 = 11 \times 0$

$10^{n+1} - (-1)^{n+1} =  10\times (10^n -(-1)^n) -11 \times (-1)^n $, i.e. the difference between (by the inductive hypothesis that $10^n -(-1)^n$ is a multiple of 11*) a multiple of 11 and another multiple of 11.

An alternative method is

$$10^{n+1} - (-1)^{n+1} = 11 \times \sum_{i=0}^{n} 10^i(-1)^{n-i}$$

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


答:
Step 1

Step 1: Using our knowledge that $∠BAC = ∠DAE = 35^o$ and that $∠ADE = 90^o$ (tangentiality), we can calculate section $AD = AE/cos(35) \approx 4.88$

Step 2

Step 2: Using Thales' Theorem, we know that $∠ADF=90^o$.

Thus, we can simply calculate $AF=AD/cos(35) \approx 5.96$

Then, using simple trigonometry again we obtain $AB = AF/cos(35)\approx7.28 $

Step 3

Step 3: Invoking Thales' theorem again, we obtain that $∠ABC = 90^o$.

We then calculate $AC=AB/cos(35)\approx8.88$

Similarly, $FB=AF*tan(35)\approx4.17$

Step 4: The area of ABC is then = $0.5*FB*AC = 18.54$

The radius of the large circle = $8.88-4 \approx 4.88$

The area of the large circle = $\pi*r^2 \approx 74.91$

Step 5: However, we only need to know the part within the triangle. As $∠ABC = 90^o$ and $∠BAC=35^o$, $∠BCA = 55^o$.

Thus, the relevant area $ \approx 55/360*74.91 = 11.44$

Answer: The answer is then $18.54-11.44 = 7.10cm^2$

Title: How to find the shaded area of a triangle formed by the tangent of a circle?

Body:
Hey guys,

The last question of the chapter has stumped me once again! I need to find the area of part of this triangle. I’m not sure how to attack the question considering that I’m not given the radius of the circle and I’m not sure how the measurement of 4cm will help me?

Thanks,
Scratch Cat


问:
Basic 3-joint inverse kinematics question

In the above diagram, I am trying to solve for the 3 angles, x, y, and z (hope they are clearly labeled). The circles next to the angles just denote where the joints are; this is for a robot arm. Any solution of these angles will work, with the one caveat that none of the joints can go "above" the hypotenuse c. Distances a, b, and c are all known, as are the lengths of the three arm parts, h1, h2, and h3. h3 > h2 = h1 in length.

One thing that may help is that I can "fix" part h3 at a certain angle if need be, like saying that it will always be 5 degrees away from c. I think that may help, but I am not sure how. I think, if possible, it would actually be preferable to fix part h3. How can I solve for x, y, and z in terms of the other knowns? Thanks in advance for your help, I am a trigonometry/inverse kinematics noob, and Google has not led to a solution so far.

答:
In most cases there is no unique solution. For example, say $c/3<h_1=h_2=h_3<c$. I can show 3 trivial solutions:

$h_1$ is along $c$, and $h_2$, $h_3$ form an isosceles triangle with the base $c-h_1$.

$h_3$ is along $c$, and $h_1$, $h_2$ form an isosceles triangle with the base $c-h_3$.

$h_2$ is parallel to $c$, and you form an isosceles trapeze

But there are infinitely many possibilities, so you need more constraints.

EDIT
For case 3, you know that the long base of the trapezoid is $c$, the short base is $h=h_1=h_2=h_3$. If you draw perpendiculars from the $y$ and $z$ joints onto $c$ you form a rectangle and two right angle triangles. The hypotenuse if $h$ and one of the sides is $$\frac{c-h}2$$Then the angle between $c$ and $h_1$ is $$\alpha=\arccos\frac{c-h}{2h}$$
The angle between $h_1$ and $h_2$, and the angle between $h_2$ and $h_3$ will be $$90^\circ+(90^\circ-\alpha)=180^\circ-\alpha$$

EDIT2
New request: $h_1=h_2=h<h_3$. We are still going to try to create a trapezoid, $h_2||c$, but it won't be isosceles anymore. In your sketch, draw a parallel to $h_3$ through the joint between $h_1$ and $h_2$. You will have a parallelogram with lengths $h_3$ and $h$, and a triangle with lengths $h$, $h_3$, and $c-h$. You can use Law of cosines to calculate the angles in this triangle.


问:
Fourier transform of hyperbolic function $\frac{d}{dx}\log({\sinh{x}})$
I am trying to calculate the FT of $\frac{d}{dx}\log({\sinh{x}})$.
I did not calculate it directly but I used the result of its derivative.
This is a tentative: I found the following result in the literature but they did not mention how they get the result except adding a small positive $\epsilon$ which I believe is considered as the imaginary part of $x$ :
$\hat{F}_k\Biggl\{\frac{d^2}{dx^2}\log({\sinh{x}})\Biggr\}=\frac{k}{1-e^{-\pi k}}$
then
$\hat{F}_k\Biggl\{\frac{d}{dx}\log({\sinh{x}})\Biggr\}=\frac{1}{ik}\hat{F}_k\Biggl\{\frac{d^2}{dx^2}\log({\sinh{x}})\Biggr\}=\frac{i}{e^{-\pi k}-1}$
First, is my reasoning correct ?  I want to understand how they get the result by adding $ i \epsilon$ and if it is possible how do I calculate the FT of $\frac{d}{dx}\log({\sinh{x}})$ without using FT properties.

答:
I guess you should make use of the identity $$\coth(x)=\frac{1}{x} + \sum_{n=1}^\infty \frac{2x}{x^2+n^2\pi^2} \, .$$
The Fourier-Transform $${\cal F}\left[\coth(x)\right](k)=PV\int_{-\infty}^\infty {\rm d}x \, e^{-ikx} \coth(x)$$
(where $PV$ stands for the principal-value at $x=0$) is not convergent due to the behaviour of the cotangens hyperbolicus at $\pm \infty$. The integral can be regularized by introducing a dampening factor $e^{-\epsilon |x|}$ and eventually taking the limit $\epsilon \rightarrow 0^+$. This gives rise to the regularized Fourier-Transform
$$ {\cal F}\left[\coth(x)\right](k) = \lim_{\epsilon \rightarrow 0} PV \int_{-\infty}^\infty {\rm d}x \, e^{-ikx-\epsilon|x|} \coth(x) = I_1(k) + I_2(k)$$
where
$$I_1(k) = \lim_{\epsilon \rightarrow 0} PV\int_{-\infty}^\infty \frac{e^{-ikx-\epsilon|x|}}{x} \, {\rm d}x = PV\int_{-\infty}^\infty \frac{e^{-ikx}}{x} \, {\rm d}x = -i\pi \, {\rm sign}(k) \\
I_2(k) = \lim_{\epsilon \rightarrow 0}\int_{-\infty}^\infty {\rm d}x \sum_{n=1}^\infty \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx-\epsilon|x|} = \lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty \int_{-\infty}^\infty {\rm d}x \, \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx-\epsilon|x|} \\
= \sum_{n=1}^\infty \int_{-\infty}^\infty {\rm d}x \, \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx} = \sum_{n=1}^\infty -2\pi i  \,{\rm sign}(k) \, e^{-\pi|k|n} = \frac{-2\pi i \, {\rm sign}(k)}{e^{\pi|k|}-1} \, .$$
The integral $I_1(k)$ converges even for $\epsilon=0$ so that the limit can be taken immediately. For $I_2(k)$ the principal-value is redundant as the integrand is well-behaved at $x=0$. Summation and integration can be interchanged by dom. convergence, since the RHS of $$\int_{-\infty}^\infty {\rm d}x \, \underbrace{\left|\sum_{n=1}^N \frac{2x}{x^2+n^2\pi^2}\right|}_{\leq 1 \,\forall N\in{\mathbb N}} \, e^{-\epsilon|x|} \leq \int_{-\infty}^\infty {\rm d}x \, e^{-\epsilon|x|}$$
converges for $\epsilon>0$. Since the integral and sum converge even for $\epsilon=0$, the limit can be taken immediately. The resulting integral can be calculated by means of the residue theorem for the 2 cases $k>0$ and $k<0$. ${\rm sign}(k)$ is the signum function.

Finally you can easily check that $$I_1(k)+I_2(k)=-i\pi \, {\rm sign}(k) \left(1+\frac{2}{e^{\pi|k|}-1}\right) = -i\pi \coth\left(\frac{\pi k}{2}\right) \, .$$


问:
A question about direct sums of subspaces...

Let $W_1$ be a subspace of a finite dimensional vector space $V$. Prove that there exists a subspace $W_2\subset V$ such that $V=W_1\oplus W_2$.

EDIT$^1$:

This may be of use here:

What does
$$ W_1 \cap W_2 = \{0\}$$
even mean? The zero vector is the only thing shared in common by the two subset?

答:
Here are some explicit examples for you to play around with:

Let $V$ be 2-dimensional, say $\langle x, y\rangle$, and let $W_1 = \langle x\rangle$. Can you see that $W_2 = \langle y\rangle$ works? Can you see that $\langle x+y\rangle$ also works, and that $\langle 3x - 5y \rangle$ also works, but that $\langle x \rangle$ doesn't, that $\{0\}$ doesn't, and that $\langle x+y, x-y\rangle$ doesn't? (Draw pictures.)

Let $V = \langle x_1, \dots, x_n\rangle$ (n large), $W_1 = \langle x_1, x_2, x_3\rangle$, and show that $W_2 = \langle x_4, \dots, x_n\rangle$ works.

Let $V = \langle x, y, z\rangle$, and let $W_1 = \langle x\rangle$. Can you pick the obvious space that works as $W_2$? Can you pick a non-obvious space that works? Can you show why $\langle y\rangle$ and $\langle 3x+5y, x-y\rangle$ don't work?


问:
How to find the angle of a rotation 2 by 2 matrix?
A 2 by 2 matrix is given with irrational elements and we can derive that the top left element is equal to $\cos(θ)$, and the bottom left element is equal to $\sin(θ)$.

How could we find the value of $θ$ and therefore the angle that the rotation does. I think by using the signs of the trigonometric functions in each quadrant.

答:
As shown in this answer, a nice formula for getting $\theta$ from $\sin(\theta)$ and $\cos(\theta)$ that takes care of quadrants is
$$
\theta=2\tan^{-1}\left(\frac{\sin(\theta)}{1+\cos(\theta)}\right)\tag1
$$
The only ambiguity might be when $\cos(\theta)=-1$, where we get $2\tan^{-1}\left(\frac00\right)$. In this case, $\theta$ could be  $\pm\pi$.


问:
Calculator similar to Desmos but for $3$D
Is there a calculator with functionality similar to Desmos but in $3$ dimensions?  I am looking to learn about families of quadric surfaces so I am looking for a $3$D calculator with sliders.

答:
Just about all of the 3D plotters I've seen online fail in a few major areas:

Asymptotes. For example, $f(x,y) = \tan(x)$ shouldn't have vertical planes in its graph.

Plot boundaries. Functions with restricted domains, i.e. $f(x,y)=\sqrt{1-x^2-y^2}$, should not have jagged triangles where the domain ends (in this case the unit circle on the $xy$ plane).

Discontinuities. Related to #1, but also applies to step functions and the like. Discontinuities should not be visible as an abrupt vertical plane.

Awhile ago I attempted to address these issues in Koval's 3D Grapher. Now at the time I wasn't nearly as clever as I really needed to be, so it still struggles with some things (notably $f(x,y)=\tan(x)+\tan(y)$ and other "2D asymptotes"), but in general it does pretty well:

Also supports spherical and cylindrical coordinates and some 1D and 2D parameterized graphs.


问:
A subset of the vector space of all real sequences with finitely-many non-zero elements. Boundedness, closedness, compactness and completeness.
Let $X$ be the space of all real sequences with finitely many nonzero elements, equipped with the supremum norm. That is, for each $x= (x_k)_{k \in \mathbb N}$ we have:
$$ \|x \| = \sup \{ |x_k| k\in \mathbb N \}.$$
Consider the subset $V$ of $X$ defined by:
$$ V:= \{ x= (x_k)_{k \in \mathbb N} \in X | \sum_{k=1}^\infty |x_k| \leq 1 \}.$$

I wish to answer the following practice exam questions:

Is $V$ bounded in $X$?

Is $V$ closed in $X$?

Is $V$ compact in $X$?

Is $V$ complete in $X$?

Consider $x=(x_n)_{n \in \mathbb N}\in V$, we then have that for each $n\in \mathbb N$ we may write:
$$ |x_n| \leq \sum_{k=1}^\infty|x_k| \leq 1 $$
Each term is nonnegative and therefore the entire sum is certainly more than a single term. We thus find that $1$ is an upper bound for each element of the sequence hence certainly it is larger than or equal to the least upper bound:
$$ \| x\| \leq 1.  $$
We conclude that $V$ is bounded.

I think this is not true as there are sequences in $V$ that leave the space (and $X$). Consider $x^{(n)}$ given by:
$$x^1= (\frac{1}{2}, 0, 0, \dots) $$
$$x^2=(\frac{1}{2}, \frac{1}{4}, 0, \dots) $$
$$x^3=(\frac{1}{2}, \frac{1}{4}, \frac{1}{8},0, \dots) $$
Which might be described as $(x^n_k)_{k \in \mathbb N}$ with $x^n_k=\frac{1}{2^{k}}$ for $k\leq n$ and $0$ otherwise ($0 \not \in \mathbb N$). Each of the elements of this sequence lives in $X$ and since the geometric series converges to $1$ also in $V$. So $$ \sum_{k=1}^\infty |x^k_n| \leq 1 .$$ However, we run into a problem since  its limit is $x=(x_k)_{k\in \mathbb N}$ with $x_n=\frac{1}{2^n}$, which does not have finitely many nonzero elements hence it is not in $X$ and then certainly not in $V$.

Compact normed vector spaces are closed, since the space is not closed, it cannot be compact.

The space cannot be complete as the example in (2) is a Cauchy-sequence, but the sequence does not converge to a limit in $V$. In a complete space every Cauchy sequence is convergent (with limit in the space).

Did I do this okay?

答:
(1) Your solution looks correct.

(2) To show that $V$ is not closed in $X$ you need to find a sequence of elements in $V$ that converge to an element in $X$. Your sequence contains only elements of $V$ but it does not converge to an element in $X$. For that reason it is not a valid proof.

(3) Your proof relies on (2) which is not proven. You could use (4) to prove this (Hint: Is it possible for a space that is not Cauchy complete to be compact?)

(4) This proof is correct since being Cauchy complete is a property inherent to the space and does not depend on the ambient space $X$ (this is in contrast to the property of being closed which does depend on $X$).


答:
Choose a basis for $W_1$.  This basis can be extended to a basis for $V$ (well-known result).  Now let $W_2$ be the span of the newly added vectors.  Then
$W_1 \cap W_2 = \{0\}$ by linear independence and $W_1 + W_2 = V$.</p>

Title: A question about direct sums of subspaces...

Body:
Let $W_1$ be a subspace of a finite dimensional vector space $V$. Prove that there exists a subspace $W_2\subset V$ such that $V=W_1\oplus W_2$.

EDIT$^1$:

This may be of use here:

What does
$$ W_1 \cap W_2 = \{0\}$$
even mean? The zero vector is the only thing shared in common by the two subset?


问:
optimization Determine the coordinates of point
Determine the coordinates of point P of the line passing through points (a; 0) and (0; b), where
a and b are positive, which maximize the area of the rectangle shown in the diagram.

diagram

if someone can give me some explanation it would be appreciated.

答:
You can solve the problem in many different ways.

The equation of the straight line is $\frac{x}{a} + \frac{y}{b} = 1$

Let the co-ordinates of $P$ be $(t, b(1-\frac{t}{a}))$

Area of the rectangle $f(t) = bt(1-\frac{t}{a})$

You need to find the value of $t$ that maximizes $f(t)$

Method 1: Algebra

Since $f(t) = \frac{b}{a} \left(\frac{a^2}{4} - (t-\frac{a}{2})^2 \right)$

it is maximum when $t = \frac{a}{2}$, i.e., the co-ordinates of $P$ are $(\frac{a}{2}, \frac{b}{2})$

Method 2: Calculus

Set $f'(t) = 0 \implies t = \frac{a}{2}$ and $f''(t) < 0$ means this is indeed a maxima

Method 3: Lagrange multiplier

Let the co-ordinates of $P$ be $(x, y)$

Consider $L = xy + \lambda (\frac{x}{a} + \frac{y}{b} - 1)$

Set $\frac{\partial L}{\partial x} = 0$, $\frac{\partial L}{\partial y} = 0$, $\frac{\partial L}{\partial \lambda} = 0$ and solve to obtain $x = \frac{a}{2}$ and $y = \frac{b}{2}$


问:
If $M$ is a finitely generated module, $S$ is a submodule of $M$, $M/S$ is free, and $M\cong S \oplus M/S$, why must $S$ be finitely generated?
I am working through Rotman's Theorem 9.3 in Advanced Modern Algebra. If $M$ is finitely generated, $S$ is a submodule of $M$, and $M/S$ is free, we have that $M\cong S\oplus M/S$. My professor told me that because $S$ is a direct summand of a finitely generated module, then $S$ itself has to be finitely generated. I am wondering why he chose to word it like this instead of saying that since $S$ is a submodule of $M$ and $M$ is finitely generated, then $S$ must be finitely generated, too. I do not see explicitly why his statement is true (even though it makes sense intuitively).

答:
The fact that $M/S$ is free is completely irrelevant. Any homomorphic image of a finitely generated module is finitely generated as well.

More precisely, let $f\colon M\to N$ be a surjective homomorphism. If $\{x_1,\dots,x_n\}$ is a finite set of generators for $M$, then $\{f(x_1),\dots,f(x_n)\}$ is a set of generators for $N$. Can you prove it? Of course $f$ being surjective is the key.

A direct summand is a homomorphic image.

Your argument fails: a submodule of a finitely generated module need not be finitely generated. Standard example: take the ring $R=k[x_0,x_1,\dotsc]$ of polynomials over the field $k$ in infinitely many variables. Then the ideal generated by the variables is not a finitely generated $R$-module, whereas $R$ is obviously finitely generated as an $R$-module.

A module such that every submodule thereof is finitely generated indeed has a special name: it is called a Noetherian module. If the ring is Noetherian (as a module over itself), then every finitely generated module is Noetherian. The ring in the above example is not Noetherian.


问:
Quotient space form by the action of the discrete Heisenberg group on the Heisenberg group
Though I am a beginner to differential topology, pardon me for something very basic. Here is my attempt!

H(The set of $3 \times 3$ unipotent matrices over $\mathbb{R}$, Heisenberg group) is homeomorphic to $\mathbb{R}^3$. It suffice to show that $\mathbb{R}^3/\mathbb{Z}^3$ is homeomophic to $T^3=\mathbb{S}^1\times\mathbb{S}^1\times\mathbb{S}^1$ Now define a map $f:\mathbb{R}^3/\mathbb{Z}^3\to S^1\times S^1\times S^1$ as follows:

$f([(x,y,z)])=q\circ p^{-1}([(x,y,z)])=q(x+n,y+n,z+n)=(e^{2\pi i(x+n)},e^{2\pi i(y+n)},e^{2\pi i(z+n)})=(e^{2\pi iz},e^{2\pi iy},e^{2\pi iz})$(As cosine and sine are $2\pi$ periodic functions), where p is the canonical quotient map from $\mathbb{R}^3$ to $\mathbb{R}^3/\mathbb{Z}^3$ and q is defined as $q(x,y,z)=(e^{2\pi ix},e^{2\pi iy}, e^{2\pi iz})$. Clearly $f$ is well defined. Now the continuity of $f$ follows by continuity of $p,q$ and openness of $p$(as $p$ is projection under a group action $\mathbb{Z}$). Surjectivity follows as given a point $a\in T^3$, $\exists b \in \mathbb{R}^3$ such that $q(b)=a$(q is surjective) and then $p(b)\in\mathbb{R}^3/\mathbb{Z}^3$. Clearly $f$ is one-one by definition. Hence, by universal mapping property of quotient map, $q$ induces a homeomorphism $f$. Therefore, $H/\Gamma\simeq \mathbb{R}^3/\mathbb{Z}^3\simeq T^3$($\Gamma$ is the discrete Heisenberg group).

But even I am really not sure whether this really works or not! As the action is defined as:
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix}
\begin{bmatrix}
m\\
n\\
o
\end{bmatrix}
=\begin{bmatrix}
x+m\\
n+ao+b\\
c+o
\end{bmatrix}

The second slot is not just the usual translation so the action may be different then the usual action of $\mathbb{Z}^3$ on $\mathbb{R}^3$. Any help is appreciated!

答:
Even though this is very late, I happened to stumble across a problem regarding this group, and I could really relate to the struggle of the OP trying to find something "elementary" regarding this particular 3-manifold. Since I've managed to figure it out, I thought it could be useful to leave this as a future reference. In particular, I'll prove that the given quotient space is a compact differentiable manifold of dimension 3.

Let $H=\left\{ \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} \in \operatorname{Mat}_{3\times 3}(\mathbb{R}) : a,b,c \in \mathbb{R} \right\}$ be the real Heisenberg group with the Euclidean topology induced by $\mathbb{R}^9$ and let $\Gamma$ be the discrete Heisenberg group, i.e. the set of all matrices in $H$ with integer entries. Consider on $H$ the equivalence relation $\sim$ defined by $$A'\sim A \iff \text{there exists } \gamma\in \Gamma \text{ satisfying } A'=\gamma A.$$ We want to prove that the quotient space $M=H/\Gamma$ is a topological 3-manifold (i.e. that it is second-countable, Hausdorff and locally 3-Euclidean) which admits a differential structure.

$\textbf{M is second-countable:}$ Let's first prove that $\sim$ is an open relation, i.e. that for any open set $U\subset H$, also $[U]=\bigcup_{A\in U}[A]$ is open in $H$. Let $\mu_{\gamma}\colon H\to H$ be the left multiplication by $\gamma\in \Gamma$, i.e. $\mu_{\gamma}(A)=\gamma A$ for any $A\in H$. Since $\Gamma$ is a subgroup of $H$ and both $\mu_{\gamma}$ and its inverse $\mu_{\gamma}^{-1}=\mu_{\gamma^{-1}}$ are continuous by definition of matrix multiplication, we have that $\mu_{\gamma}$ is a homeomorphism. Thus, for any open set $U\subset H$, $$[U]=\bigcup_{A\in U}[A] = \bigcup_{A\in U}\{\gamma A: \gamma\in \Gamma\}=\bigcup_{A\in U} \bigcup_{\gamma\in \Gamma}\{\gamma A\}=\bigcup_{\gamma \in \Gamma} \mu_{\gamma}(U)$$ is open as it is the union of open sets (since $\mu_{\gamma}$ is a homeomorphism and homeomorphism are open maps). This proves that $\sim$ is an open relation, thus the canonical map $\pi\colon H\to M$, $A\mapsto [A]$ is also open. Since $H$ has a numerable basis $\mathcal{B}=\{B_i : i\in I\}$ inherited as a subspace of the Euclidean space $\mathbb{R}^9$, we have that $\mathcal{B}'=\{\pi(B_i): i\in I\}$ is a numerable basis for $M$, i.e. $M$ is second-countable.

$\textbf{M is Hausdorff:}$ Since $\sim$ is an open relation, in order to prove that $M$ is Hausdorff it suffices to prove that $R=\{(A,A'): A\sim A'\}$ is closed in $H\times H$. Let $\iota\colon \mathbb{R}^3\to H$ be the "inclusion map" $$(a,b,c)\mapsto \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$ (which can be easily checked to be a homeomorphism) and let $f\colon H\times H \to H$, $(A,A')\mapsto A'A^{-1}$. Since $$A'\sim A \iff A'=\gamma A \iff A'A^{-1}=\gamma\in \Gamma \iff f(A,A')\in \Gamma$$ we have that $R=f^{-1}(\Gamma)$. Since $f$ is continuous and $\Gamma=\iota(\mathbb{Z}^3)\subset H$ is closed as $\mathbb{Z}^3\subset \mathbb{R}^3$ is closed, we conclude that $R\subset H\times H$ is closed, thus the quotient space $M$ is Hausdorff.

$\textbf{M is locally 3-Euclidean:}$ The key claim is that for any $A\in H$ there exists an open neighborhood $U_A \subset H$ of $A$ which intersects each orbit of the action of $\Gamma$ on $H$ at most once. To prove this, let's start by examining what each orbit looks like. Suppose that $B'\sim B\Rightarrow B'=\gamma B$ for some matrices $$ B=\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}, \ B'=\begin{pmatrix} 1 & a' & c' \\ 0 & 1 & b' \\ 0 & 0 & 1 \end{pmatrix}, \ \gamma=\begin{pmatrix} 1 & p & r \\ 0 & 1 & q \\ 0 & 0 & 1 \end{pmatrix}.$$ We can check that this is equivalent to the system of equations $$\begin{cases} a'= a+p \\ b'=b+q \\ c'= c+r+bp\end{cases} \ \ (\star)$$ where $p,q,r\in \mathbb{Z}$. Thus, letting $x=\iota^{-1}(A)\in \mathbb{R}^3$, we have that $U_A=\iota(D(x,1/2))$ has the required properties, where $D(x,1/2)=\{x'\in \mathbb{R}^3: |x'-x|< 1/2\}$. Indeed, it's clear that $U_A\subset H$ is an open neighbourhood of $\iota(x)=A$. For the sake of contradiction, suppose there existed $B,B'\in U_A$ such that $B'\sim B$, and let $y=\iota^{-1}(B),$ $z=\iota^{-1}(B')\in \mathbb{R}^3$. Since $$|y-z|\leq |y-x|+|x-z|<1/2 + 1/2 = 1,$$ looking at $(\star)$ we must also have that $|p|=|a'-a|<1$ and $|q|=|b'-b|<1$, which forces $p=q=0$ as $p$ and $q$ are integers. Plugging this back into the third equation of $(\star)$, for the same reason $c'=c+r\Rightarrow |r|=|c'-c|<1$ forces $r=0$. Since $(a',b',c')=(a,b,c)$, this proves that $B'=B$, i.e. $U_A$ contains at most one element from each orbit of the action of $\Gamma$ on $H$.

Why is this useful? Consider the canonical map $\pi\colon H\to M$. We already know from before that $\pi$ is a continuous open map; however, what we just proved shows that the restriction $\pi_{|_{U_A}}\colon U_A\to M$ is also injective, thus it is a homeomorphism onto its image. Pick an arbitrary point $v=\pi(A)\in M$, and let $V=\pi(U_A)$ where $U_A$ is defined as above. Then, $\varphi_V=\iota^{-1}\circ \pi_{|_{U_A}}^{-1}\colon V\to D(x,1/2)$ is a homeomorphism (as it is a composition of homeomorphisms), thus $V\subset M$ is an open neighbourhood of $v$ which is homeomorphic to an open set of $\mathbb{R}^3$, i.e. $M$ is locally $3$-Euclidean.

$\textbf{M has a differential structure:}$ Let $\mathcal{A}$ be the set of all charts $(V,\varphi_V)$ defined as above. We claim that $\mathcal{A}$ is an atlas for $M$. Clearly, the union of all $V$'s is $M$, thus it suffices to show that the charts are differentiably compatible. Let $(V,\varphi_V)$, $(V',\varphi_{V'})\in \mathcal{A}$ and suppose that $V\cap V'\neq \emptyset$ (otherwise it's trivial). Then, for any $x\in \varphi_V(V\cap V')$ we have that $$\varphi_{V'}\circ \varphi_V^{-1}(x)=(\iota^{-1}\circ \pi_{|_{V'}}^{-1})\circ (\iota^{-1}\circ \pi_{|_{V}}^{-1})^{-1}(x) =\iota^{-1}\circ \pi_{|_{V'}}^{-1} \circ \pi_{|_{V}} \circ \iota(x) = x $$ is obviously smooth and similarly $\varphi_{V}\circ \varphi_{V'}^{-1}(x)=x$ for any $x\in \varphi_{V'}(V\cap V')$, thus $\mathcal{A}$ is an atlas and $M$ is a differentiable $3$-manifold.

$\textbf{M is compact:}$ This follows from the fact that any $A=\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\in H$ is equivalent to $$A'=\begin{pmatrix} 1 & \{a\} & \{c-b\lfloor a\rfloor\} \\ 0 & 1 & \{b\} \\ 0 & 0 & 1 \end{pmatrix}$$ where $\lfloor \cdot \rfloor$ and $\{\cdot \}$ denote integer and fractional part, respectively. The proof of this fact follows from plugging in the values in $(\star)$, which shows that $p,q$ and $r$ are actually integers, thus $A'\sim A$. Since all entries of $A'$ are real numbers in $[0,1]$, we see that we can restrict $\pi$ to the compact "unit cube" $\iota([0,1]^3)\subset H$ and its image is still all $M$, thus $M$ is compact (as $\pi$ is continuous and surjective). $\square$


问:
Integration by parts question: why does the constant of integration get swallowed up?
To integrate $e^x\cos x$ we can use integration by parts twice:
\begin{align}
\int e^x \cos x \, dx &= e^x\sin x - \int e^x\sin x \, dx \\
&= e^x\sin x - \left(-e^x\cos x + \int e^x\cos x \, dx\right) \\
&= e^x\sin x + e^x\cos x - \int e^x \cos x \, dx \\
2 \int e^x\cos x \, dx &= e^x(\sin x + \cos x) \\
\int e^x \cos x \, dx &= \frac{e^x(\sin x + \cos x)}{2}
\end{align}
But this solution is incomplete: there should be a constant of integration at the end. Why does this happen? I think it might have something to do that you are adding a family of functions to itself, but I can’t quite pin down what the problem is.

答:
Strictly speaking, you have something more like $F(x)=e^x \sin(x) - G(x)$ then $F(x)=e^x \sin(x) + e^x \cos(x) - H(x)$, where $F$ and $H$ are both some antiderivatives of $e^x \cos(x)$ (not necessarily the same ones) and $G$ is some antiderivative of $e^x \sin(x)$. Now $F$ and $H$ differ only by a constant, so $2F(x)=e^x \sin(x) + e^x \cos(x) + C$ and then you divide.

In practice if you want to take derivations seriously, you're better off with definite integrals with variable limits, i.e. working with $\int_a^x f(y) dy$ instead of $\int f(x) dx$.


答:
Whenever you have an indefinite integral,
there is a constant of integration.

When you make the limits definite,
the constant is subtracted away
since it occurs in
both the lower and upper limit.

Title: Integration by parts question: why does the constant of integration get swallowed up?

Body:
To integrate $e^x\cos x$ we can use integration by parts twice:
\begin{align}
\int e^x \cos x \, dx &= e^x\sin x - \int e^x\sin x \, dx \\
&= e^x\sin x - \left(-e^x\cos x + \int e^x\cos x \, dx\right) \\
&= e^x\sin x + e^x\cos x - \int e^x \cos x \, dx \\
2 \int e^x\cos x \, dx &= e^x(\sin x + \cos x) \\
\int e^x \cos x \, dx &= \frac{e^x(\sin x + \cos x)}{2}
\end{align}
But this solution is incomplete: there should be a constant of integration at the end. Why does this happen? I think it might have something to do that you are adding a family of functions to itself, but I can’t quite pin down what the problem is.


答:
To complement Pedro Amaral’s answer, I’ll show that $V$ is closed in $X$.

Suppose that $x=\langle x_n:n\in\Bbb Z^+\rangle\in X\setminus V$, so that $\sum_{n\ge 1}|x_n|>1$. Let

$$\epsilon=\sum_{n\ge 1}|x_n|-1>0\,.$$

If $y\in X$, and $\|y-x\|<\epsilon$, then $|y_n-x_n|<\epsilon$ for each $n\in\Bbb Z^+$, and therefore $|y_n|>|x_n|-\epsilon=1$ for each $n\in\Bbb Z^+$. Therefore $y\notin V$, and the $\epsilon$-ball centred at $x$ is disjoint from $V$. This shows that $V$ is closed in $X$.

Title: A subset of the vector space of all real sequences with finitely-many non-zero elements. Boundedness, closedness, compactness and completeness.

Body:
Let $X$ be the space of all real sequences with finitely many nonzero elements, equipped with the supremum norm. That is, for each $x= (x_k)_{k \in \mathbb N}$ we have:
$$ \|x \| = \sup \{ |x_k| k\in \mathbb N \}.$$
Consider the subset $V$ of $X$ defined by:
$$ V:= \{ x= (x_k)_{k \in \mathbb N} \in X | \sum_{k=1}^\infty |x_k| \leq 1 \}.$$

I wish to answer the following practice exam questions:

Is $V$ bounded in $X$?

Is $V$ closed in $X$?

Is $V$ compact in $X$?

Is $V$ complete in $X$?

Consider $x=(x_n)_{n \in \mathbb N}\in V$, we then have that for each $n\in \mathbb N$ we may write:
$$ |x_n| \leq \sum_{k=1}^\infty|x_k| \leq 1 $$
Each term is nonnegative and therefore the entire sum is certainly more than a single term. We thus find that $1$ is an upper bound for each element of the sequence hence certainly it is larger than or equal to the least upper bound:
$$ \| x\| \leq 1.  $$
We conclude that $V$ is bounded.

I think this is not true as there are sequences in $V$ that leave the space (and $X$). Consider $x^{(n)}$ given by:
$$x^1= (\frac{1}{2}, 0, 0, \dots) $$
$$x^2=(\frac{1}{2}, \frac{1}{4}, 0, \dots) $$
$$x^3=(\frac{1}{2}, \frac{1}{4}, \frac{1}{8},0, \dots) $$
Which might be described as $(x^n_k)_{k \in \mathbb N}$ with $x^n_k=\frac{1}{2^{k}}$ for $k\leq n$ and $0$ otherwise ($0 \not \in \mathbb N$). Each of the elements of this sequence lives in $X$ and since the geometric series converges to $1$ also in $V$. So $$ \sum_{k=1}^\infty |x^k_n| \leq 1 .$$ However, we run into a problem since  its limit is $x=(x_k)_{k\in \mathbb N}$ with $x_n=\frac{1}{2^n}$, which does not have finitely many nonzero elements hence it is not in $X$ and then certainly not in $V$.

Compact normed vector spaces are closed, since the space is not closed, it cannot be compact.

The space cannot be complete as the example in (2) is a Cauchy-sequence, but the sequence does not converge to a limit in $V$. In a complete space every Cauchy sequence is convergent (with limit in the space).

Did I do this okay?


问:
Proving that there are infinitely many prime numbers of the form $4k+3$

Anyone wanna help me solve this one? Been at it for a little bit but haven't really gotten anywhere..

答:
Suppose that there were finitely many prime numbers of the form $4k+3$; call them $p_1,\ldots,p_m$. Let
$$N=4p_1\cdots p_m-1.$$
Then $N\equiv 3\bmod 4$. What can you say about the prime divisors of a number that is $3\bmod 4$?</p>


问:
what are all the continuous functions $f(x)$ that is has a domain $x>1$ and $f(ab)=f(a^2)+f(b^2)$
what are all the continuous functions $f(x)$ that is has a domain $x>1$ and $f(ab)=f(a^2)+f(b^2)$

the reason $f(1)$ is undefined is that $f(1)= 2\times f(1)$ and I don't want $f(1)=0$

答: Put $a=b$ to get $f(a^{2})=2f(a^{2})$. Hence, $f(a^{2})=0$ for all $a >1$ which implies $f(x)=0$ for all $x >1$.


问:
vector multiply question to prove
in the beginning I want to know some answers to this question: if $u, v \in \mathbb{R^{n}}$, show that if $|u+v|^{2}=|u|^{2}+|v|^{2}$, then $u\bullet v=0$. I think this one is showing how u or v is =0. I made the first step to be let u=0. Then I can prove this for u. Then I do v=0, and then prove it same as u.

答:
Recall that $\vert x+y\vert^2 = (x+y)\bullet(x+y)$. Now use the fact that $+$ is distributive over $\bullet$: $$\vert x + y\vert^2 = (x+y)\bullet(x+y) = x\bullet x + 2x\bullet y + y\bullet y = \vert x\vert^2 + \vert y\vert^2 + 2x\bullet y.$$
The only chance that this equation holds is iff $x\bullet y = 0$.

Title: vector multiply question to prove


问:
Is multiplication $\sigma$-weakly continuous on bounded subsets?
Let $M \subseteq B(H)$ be a von Neumann algebra. Let $S\subseteq M$ be a bounded subset. Is the multiplication map
$$S \times S \to M: (x,y) \mapsto xy$$
$\sigma$-weakly (= weak$^*$) continuous?

Attempt: Assume $x_i \to x$ and $y_i \to y$ in the $\sigma$-weak topology where $(x_i)_i$ and $(y_i)_i$ are bounded nets. We must show that
$x_iy_i \to xy$ in the $\sigma$-weak topology. Let $z \in L^1(H)$ be a trace class operator. We need to show that
$$|Tr(z(xy-x_iy_i)| \to 0$$

I tried
$$|Tr(zxy-x_iy_i)| = |Tr(zxy-zxy_i + zxy_i - zx_i y_i)|$$
$$\le |Tr(z(xy-xy_i)| + |Tr(z(xy_i-x_iy_i))|$$

Now I would like to use something like
$$|Tr(zx)| \leq |Tr(z)|\|x\|$$
but I'm not sure that holds.

答:
Well, I guess  Martin has already aptly (as always) answered this question but still let me expand on my previous
comment, as suggested  by the OP.

The first thing to notice is that the weak operator topology is not something intrinsic of a von Neumann algebra, as it depends on the underlying Hilbert space.

Some may say that von Neumann algebras are inextricably linked to the Hilbert space where they were born, and this is why people
don't
care at all  about representing  von Neumann algebras elsewhere, the prevailing attitude being to keep the algebra
within its craddle $B(H)$.

Nevertheless,  there is a very limited situation in which it might be useful to consider just one more  representation,
in addition to the defining one.

Given a von Neumann algebra $M\subseteq B(H)$, consider the representation of $M$ on $H\otimes \ell ^2$
given by
$$
  T\in M\mapsto T\otimes 1\in B(H\otimes \ell ^2).
  $$
Viewing  $H\otimes \ell ^2$ as an infinite direct sum of copies of $H$, one has that $T\otimes 1$ is the (block) diagonal operator, with
$T$ repeated down the diagonal infinitely many times.

$M\otimes 1$ is still weakly closed in $B(H\otimes \ell ^2)$,  hence a von Neumann algebra, while  the passage
from $M$ to $M\otimes 1$ preserves  all of the intrinsic properties of $M$.  However,  the weak operator topology may no longer be the same!

The reason is very simple.  A vector $\xi $ in $H\otimes \ell ^2$ is actually a sequence $\xi =(\xi _n)_n$ of vectors,  with each $\xi _n$ lying
in $H$,  and such that $\sum_n\|\xi _n\|^2<\infty $.

Observe that  the vector state $\varphi _\xi $ defined by a vector $\xi $, as above, may be expressed, for each $T$ in $M$, by
$$
  \varphi _\xi (T) =
  \big \langle (T\otimes 1) \xi , \xi \big \rangle  = \sum_n\big \langle T \xi _n, \xi _n\big \rangle .
  $$

Defining $u_n=\xi _n/\|\xi _n\|$,  and $\alpha _n = \|\xi _n\|^2$,  we then have that
$$
  \varphi _\xi (T) =
  \sum_n\alpha _n\big \langle T u_n, u_n\big \rangle  =
  \sum_n\alpha _n\text{tr}(TS_n) =  \text{tr}(TS),
  $$
where $S_n$ is the rank-one,  trace-class operator
$$
  S_n(\xi ) = \big \langle \xi , u_n\big \rangle u_n,
  $$
and $S$ is the absolutely converging sum of the $\alpha _nS_n$.

Following along these lines one may prove that  the vector states on $M\otimes 1$ in fact correspond to the normal states on $M$.

Summarizing, the passage from $M$ to $M\otimes 1$ turns the $\sigma $-weak operator topology into the weak operator topology.


问:
How can I generate cyclic whole number?
My goal was to create a function that would generate number from 1->N->1. 1 2 3 4 3 2 1.

Easiest approach I was going for was create a function where using cosine i'd generate integers, and where x -> 2N.

So if N is 4, then x would go upto 2x4 = 8.

1 2 3 4 5 6 7 8

then output of ƒ would be

1 2 3 4 4 3 2 1

or

1 2 3 4 3 2 1

either would have been great.

答:
If $n$ is the length of a series of $1$s, then
$\space(111\cdots)^2=(123\cdots n\cdots 321)\space$ for $n$ ones.

$$1^2=1\quad11^2=121\quad 111^2=12321\quad 1111^2=1234321\quad \cdots$$

This will work for any $n$ provided the base is at least $n+1$.

Title: How can I generate cyclic whole number?


问:
Poisson conditional probability question
So I have a question that asks the following:

The probability that $n$ electrons are emitted by a thermionic cathode in a time interval $t$ is given by a Poisson distribution with mean $\lambda$. What is the probability that $n$ electrons are emitted in a time interval $t$  in which at least 2 electrons are emitted?

I've tried to use conditional probability with the Poisson pmf given that $n$ is more than or equal to 2 but I can't seem to get anywhere and at this point any help would be really appreciated. Thanks in advance!

答:
First of all, $X\sim \mathcal{P}(\lambda t)$, i.e., $\Pr\{X=n\}=\frac{(\lambda t)^n}{n!}e^{-\lambda t}$. The probability that $n$ electrons are emitted, given at least 2 are emitted is:
$$\Pr\{X=n|X\geq 2\} = \frac{\Pr\{X=n, X\geq 2\}}{\Pr\{X\geq2 \}}$$
Note that
$$\Pr\{X=n, X\geq2\} = \begin{cases}
\dfrac{(\lambda t)^n}{n!}e^{-\lambda t}, & n\geq 2\\
0, & \text{otherwise}
\end{cases}$$
and
$$\Pr\{X\geq2\}=\sum\limits_{k=2}^{+\infty}\frac{(\lambda t)^k}{k!}e^{-\lambda t}
=1-(1+\lambda t)e^{-\lambda t}$$
So,
$$\Pr\{X=n|X\geq2\}=\begin{cases}
\dfrac{(\lambda t)^n}{n!}\dfrac{e^{-\lambda t}}{1-(1+\lambda t)e^{-\lambda t}}, & n\geq 2\\
0, & \text{otherwise}
\end{cases}$$


问:
Second order derivative with imaginary root r
I am doing assigned practice questions, and encountered this:

$y''(x)=-y(x),\,y'(0)=1,\,y(0)=0$

Here's what I tried:

Let $y=e^{rx}$, $y'=re^{rx}$, $y''=r^{2}e^{rx}$

Rearrange the equation ===> $y''(x)+y(x)=0$

Factor out $e^{rx}$, and I have $e^{rx}(r^{2}+1)=0$

Which makes $r=i$, and $y=Ce^{ix}$ for some constant C

I am stuck and need some help for this question.

Am I on the right track? Or did I messed up from the beginning?

答:
You are on the right track.  $r^2+1=0\implies r=\pm i$, so $y=C_1e^{ix}+C_2e^{-ix}$.

Now   use the initial conditions to find $C_1$ and $C_2$.

Can you solve for $C_1$ and $C_2$ and show that $y=\sin x$?


问:
Sign of the first derivative
I am confused by the statement that a read, that says $f'(x_0)>0$ does not imply that $f$ is increasing in an open interval around $x_0$. But the book also mentions that if $f(x)$ is differentiable at $x_0$ and $f'(x_0)>0$ then there is $h>0$ such that for all $x_1,x_2\in(x_0-h,x_0+h)$ if $x_1<x_0<x_2$ then $f(x_1)<f(x_0)<f(x_2)$. I feel the statement contradicts one another. Any explanations will be appreciated.

答: In the second statement you cannot take $x_1$ and $x_2$ on the same side of $x_0$. Hence there is no contradiction.

Title: Sign of the first derivative


问:
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}<a+b<-2$

Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$

I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.

Please give some idea/hint for the other part.

答:
Let $f(a,b)=a^3+b^3-6ab+11$. At the extrema, the two curves $f(a,b)=0$ and $k=a+b$ are tangential to each other, i.e.

$$\frac{f_a’}{f_b‘}=\frac{3a^2- 6b}{3b^2-6a}=1\implies (a-b)(a+b+2)=0
$$
which leads to the tangential points $a=b$ and $a+b=-2$. It is straightforward to verify the maximum $a+b<-2$ and the minimum is determined by $f(a,a)=0$, or $2a^3-6a^2+11=0$, whose sole real root is given by the Cardano’s formula and is greater than $-\frac76$.


答:
In order to find $\int f(x)\,dx$ (on an interval $I$), all one needs is to find one differentiable function $F:I\to\mathbb{R}$ such that $F'(x)=f(x)$ for every $x\in I$. Then
$$
\int f(x)\, dx=F(x)+C
$$

This is the definition of indefinite integrals. This definition is based on the observations that

If $F:I\to\mathbb{R}$ is such that $F'=f$ on I, then for any constant $K$, $(F+K)'=f$ on $I$;

If $G:I\to\mathbb{R}$ and $F:I\to\mathbb{R}$ are such that $G'=F'$ on $I$, then there exists a constant $K$ with $G=F+K$. (This is why it is important that $I$ is "connected", i.e., an interval on $\mathbb{R}$.)

Now go back to your integral where $f(x)= \int e^x\cos x$, $I=\mathbb{R}$. The formal$\dagger$ calculation just gives you one $F$ such that $F'=f$ on $I$, which give you the answer:
$$
\int e^x\cos x\,dx=\frac{1}{2}e^x(\sin x+\cos x)+C.
$$

$\dagger$ The reason it is "formal" is that by definition,
$$
\int f(x)\,dx
$$
means a family (set) of functions (on $I$); but in the calculation, one uses only one "instance" of the set. This is OK since one can eventually add the "arbitrary constant" $C$ to get the answer.


问:
Evaluation of $\int\sqrt[4]{\tan x}dx$
Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$

$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$

So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$

So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$

Now How can I solve after that

Help me

Thanks

Title: Evaluation of $\int\sqrt[4]{\tan x}dx$

Body:
Let $t=\sqrt[4]{\tan x}$ ,

Then $x=\tan^{-1}t^4$

$dx=\dfrac{4t^3}{t^8+1}~dt$

$\therefore\int\sqrt[4]{\tan x}~dx$

$=\int\dfrac{4t^4}{t^8+1}~dt$

$=\int\dfrac{4t^4}{(t^4-\sqrt2t^2+1)(t^4+\sqrt2t^2+1)}~dt$

$=\int\dfrac{\sqrt2t^2}{t^4-\sqrt2t^2+1}~dt-\int\dfrac{\sqrt2t^2}{t^4+\sqrt2t^2+1}~dt$ (with reference to https://www.wolframalpha.com/input/?i=4x%5E2%2F%28x%5E4%2B1%29)

$=\int\dfrac{\sqrt2}{t^2-\sqrt2+\dfrac{1}{t^2}}~dt-\int\dfrac{\sqrt2}{t^2+\sqrt2+\dfrac{1}{t^2}}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-\sqrt2}~dt+\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}+\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}+\sqrt2}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2-\sqrt2}~dt+\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2-\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2+\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2+\sqrt2}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{d\left(t-\dfrac{1}{t}\right)}{\left(t-\dfrac{1}{t}\right)^2+2-\sqrt2}+\dfrac{1}{\sqrt2}\int\dfrac{d\left(t+\dfrac{1}{t}\right)}{\left(t+\dfrac{1}{t}\right)^2-2-\sqrt2}-\dfrac{1}{\sqrt2}\int\dfrac{d\left(t-\dfrac{1}{t}\right)}{\left(t-\dfrac{1}{t}\right)^2+2+\sqrt2}-\dfrac{1}{\sqrt2}\int\dfrac{d\left(t+\dfrac{1}{t}\right)}{\left(t+\dfrac{1}{t}\right)^2-2+\sqrt2}$

$=\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tan^{-1}\dfrac{t-\dfrac{1}{t}}{2-\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tanh^{-1}\dfrac{t+\dfrac{1}{t}}{2+\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tan^{-1}\dfrac{t-\dfrac{1}{t}}{2+\sqrt2}+\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tanh^{-1}\dfrac{t+\dfrac{1}{t}}{2-\sqrt2}+C$

$=\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tan^{-1}\dfrac{\sqrt[4]{\tan x}-\dfrac{1}{\sqrt[4]{\tan x}}}{2-\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tanh^{-1}\dfrac{\sqrt[4]{\tan x}+\dfrac{1}{\sqrt[4]{\tan x}}}{2+\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tan^{-1}\dfrac{\sqrt[4]{\tan x}-\dfrac{1}{\sqrt[4]{\tan x}}}{2+\sqrt2}+\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tanh^{-1}\dfrac{\sqrt[4]{\tan x}+\dfrac{1}{\sqrt[4]{\tan x}}}{2-\sqrt2}+C$


问:
If $\tau$ and $\tau'$ define the same borel sets, then $\tau=\tau'$
I have a group $G$ with two topologies $\tau, \tau'$ on $G$ that makes it a Polish group (a completely metrizable and separable topological group). I need to show that if $\mathcal{B}(\tau)=\mathcal{B}(\tau') $ , that is they define the same borel sets, then we have $\tau=\tau'$.

I just tried the obvious, I suppose a $U\in \tau$ then $U\in \mathcal{B}(\tau)=\mathcal{B}(\tau')$ but I can't go anywhere from here...

Could you help me?

答:
It suffices to assume that they have the same sets with the Baire property.

It is enough to show that if $e\in U$ is $\tau$-open, then it contains a $\tau'$-open $V\ni e$.

Since $G$ is a $\tau$-topological group, there is a $\tau$-open $W\ni e$ with $W=W^{-1}$ and $WW\subseteq U$. But $W$ is $\tau'$-Borel, so $W$ has the Baire property with respect to $\tau'$, and it is not $\tau'$-meagre (since countably many translates cover $G$), so we can find a nonempty $\tau'$-open $V_0$ such that $W\cap V_0$ is comeagre in $V_0$, and we may assume without loss of generality that $V_0$ is symmetric (replacing it with $V_0\cup V_0^{-1}$ if necessary).

Then you can show that $V_0V_0\subseteq WW\subseteq U$. Since $V_0$ is symmetric, $e\in V_0V_0$ and we are done.
Source: Automatic continuity of group homomorphisms.


答:
The first statement is correct: see twosigma's comment.

Adding to Kavi's answer, the second statement says that $f$ is strictly increasing at the point $x_0$, but it does not say that $f$ is increasing in an open interval around $x_0$ (which it isn't necessarily).

Title: Sign of the first derivative

Body: I am confused by the statement that a read, that says $f'(x_0)>0$ does not imply that $f$ is increasing in an open interval around $x_0$. But the book also mentions that if $f(x)$ is differentiable at $x_0$ and $f'(x_0)>0$ then there is $h>0$ such that for all $x_1,x_2\in(x_0-h,x_0+h)$ if $x_1<x_0<x_2$ then $f(x_1)<f(x_0)<f(x_2)$. I feel the statement contradicts one another. Any explanations will be appreciated.


问:
Fixed points of orthogonal complement of regular representation
Let $G$ be a finite group and $N$ a normal subgroup of $G$. Set $Q:=G/N$, the quotient group. Denote by $\rho_G$ the real regular representation of $G,$ ie $\rho_G = \mathbb{R}[G].$ Now consider $\rho_Q \subseteq \rho_G.$ Then we have $\rho_G= \rho_Q \oplus \rho_Q^\perp.$

What are the subgroups $H$ of $G$, such that the fixed point set $(\rho_Q^\perp)^H\neq 0$?

In particular, for $G= C_{p^n},$ cyclic group of order $p^n,$ and $N=C_{p^k}$, (k < n) what are the subgroups of $G$ satisfying above property?

I really appreciate any help you can provide.

答:
The Artin-Wedderburn theorem says $\mathbb{C}[G]\cong\bigoplus\mathrm{End}(U)$ as $U$ ranges over all irreps of $G$.

If $N\trianglelefteq G$ is a normal subgroup then any irrep $G/N\to GL(V)$ can be turned into an irrep $G\to GL(V)$ by simply composing $G\to G/N\to GL(V)$. Then AW applies to $\mathbb{C}[G/N]$ too - the summands in $\mathbb{C}[G/N]$'s AW decomposition are precisely those in $\mathbb{C}[G]$'s decomposition corresponding to those $G$-irreps $U$ which "came from" being irreps of $G/N$ first.

Define $\rho_{G/N}$ to be this subalgebra of $\mathbb{C}[G]$. I assume by $\rho_{G/N}^\perp$ you mean the orthogonal complement with respect to the obvious inner product (in which $G$ is an orthonormal basis). Since $G$ acts on $\mathbb{C}[G]$ by left-multiplication unitarily, $\rho_{G/N}^\perp$ must be a subrep of $\rho_G=\mathbb{C}[G]$ complementary to $\rho_{G/N}$. This forces $\rho_{G/N}^\perp$ to be a direct sum of the $\mathrm{End}(U)$s corresponding to irreps $U$ that do not come from irreps of $G/N$.

The condition $(\rho_{G/M}^\perp)^H\ne0$ may be rewritten as $\dim(\rho_{G/N})^H<\dim(\rho_G)^H$.

Exercise. Show how to turn a transversal for $G/H$ into a basis for $\mathbb{C}[G]^H$.

Thus, we may further rewrite the condition as $[G:HN]<[G:H]$, or simply $N\not\subseteq H$. You can apply this to your particular situation of prime-power cyclic groups if you want.


问:
Cardinality of Set and its Power Set Singleton

I know that I should prove that there is a bijection between the two sides but I do not know how to do it.

答:
HINT: Here is an outline of a proof that does not assume any knowledge of cardinal arithmetic.

Show that there is a bijection between $\wp(A)$ and $\wp(A)\times\{\wp(A)\}$. Then show that $\wp(A)$ and $\wp(A)\times\{\wp(A)\}$ are disjoint and use this fact to define a bijection between $\wp(\wp(A))\times\wp(\wp(A))$ and $\wp\big(\wp(A)\cup\big(\wp(A)\times\{\wp(A)\}\big)\big)$.

The next step is to define a bijection between

$$\wp\left(\wp(A)\cup(\wp(A)\times\{\wp(A)\}\big)\right)$$

and $\wp\big(\wp(A\cup\{A\})\big)$.

Finally, define a bijection between $\wp\big(\wp(A\cup\{A\})\big)$ and $\wp(\wp(A))$ and put the pieces together to get the desired bijection between $\wp(\wp(A))$ and $\wp(\wp(A))\times\wp(\wp(A))$.


问:
A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$?
I am currently trying to show that $\int_{-\infty}^\infty \cos(x^2) \, \mathrm dx = \sqrt{\frac{\pi}{2}}$ and the last integral I have to evaluate is
$$\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx.$$
Now of course I'm familiar with wolframalpha, however the way it solves this integral seems very awkward and also not elegant to me, even though the function to me looks quite simple. So, is there a simpler way to solve this integral or is the way described on wolframalpha already (one of) the simplest approach(es)? I ask this because often wolframalpha doesn't see tricks (occurred to me when I wanted to find a formula for the n-th derivative of some function) which a human eye might see.

Thanks for any answers in advance.


问:
Prove that $ | \mathcal {P} (\mathbb N) | \le | \mathbb R | $.
Define $\Phi: \mathcal {P} (\mathbb N) \to \mathbb R$ by

$$
\Phi (S) = \sum_ {k \in S} 10 ^ {- k}.
$$

How to prove that $\Phi$ is injective?

I assumed that $\Phi(S) = \sum_ {k \in S} 10 ^ {-k} = \sum_ {k \in S'} 10 ^ {- k} = \Phi(S')$ and tried deriving that $S = S'$, but could not make this approach work.

答:
For every $S \in \mathcal{P}(\mathbb{N})$, $\Phi(S)$ has a unique decimal representation consisting of $0$s and $1$s.

Suppose $k \in S$ and $k \notin S'$. Then $\Phi(S)$ has a $1$ at the $k$th position to the right of the decimal point while $\Phi(S')$ has a $0$. Therefore, $\Phi(S) \ne \Phi(S')$.

Similarly if $k \in S'$ and $k \notin S$.

We conclude that $\Phi$ is injective.


问:
Sums of powers being powers of the sum
I'm looking for literature on solving problems of the form
$$
n_1^\alpha+\cdots+n_k^\alpha=(n_1+\cdots+n_k)^\beta
$$
for positive integers $n_1,\ldots,n_k$ and fixed parameters $k$ and $\alpha\ne\beta.$ Any ideas?

This is perhaps similar to multigrade equations or "equal sums of like powers" but not quite similar enough to use those results (as far as I can see). But surely this is too simple not to have been studied?

答:
I do not know about the literature, but it is clear that for each $k,$ there is only finitely many solutions. Clearly, $\alpha>\beta.$ Assuming that $n_1\le n_2...\le n_k$ we can estimate, $(kn_k)^{\beta}\ge RHS=LHS\ge n_k^{\alpha},$ hence
$n_k^{\alpha-\beta}\le k^{\beta}.$ </p>

Title: Sums of powers being powers of the sum


答:
If the two sets differ, then they differ at a least element $n$.

Suppose $n$ is in $S$ but not in $S'$. Then $\Phi(S) - \Phi(S') = 10^{-n} - (\sum_ {n < k \in S'} 10 ^ {- k} - \sum_ {n < k \in S} 10 ^ {- k})$.

Can you bound $\sum_ {n < k \in S'} 10 ^ {- k} - \sum_ {n < k \in S} 10 ^ {- k}$ and show that it is less than $10^{-n}$? This proves that the two outputs are different.

Title: Prove that $ | \mathcal {P} (\mathbb N) | \le | \mathbb R | $.

Body:
Define $\Phi: \mathcal {P} (\mathbb N) \to \mathbb R$ by

$$
\Phi (S) = \sum_ {k \in S} 10 ^ {- k}.
$$

How to prove that $\Phi$ is injective?

I assumed that $\Phi(S) = \sum_ {k \in S} 10 ^ {-k} = \sum_ {k \in S'} 10 ^ {- k} = \Phi(S')$ and tried deriving that $S = S'$, but could not make this approach work.


问:
What's wrong with my "proof" of the existence of the intersection of the void set?
I had to prove the existence of the intersection of a nonempty set S, and my first approach was the following:

Let $S$ be an arbitrary set. By the Axiom of Union, the set $ \cup S$ exists; we can then apply the Axiom of Extensionality and to talk about the set $I=\{x \in \cup S: (\forall T \in S)  x \in T \}$.

This set $I$ seemed to be the good candidate for being the intersection, but then I realized I never
used the hypothesis of $S$ being non-vacuous ! Moreover, with this "proof" I'll have that $\cap  \emptyset$ will be a subset of $\emptyset$, then equal to
the empty set, but I know very well that's false.

I already figured out a right proof of my initial problem:

let $S$ be a non-empty set; let then A be an element of S. We can then apply the Axiom of Extensionality and to talk about the set  $I=\{x \in A: (\forall T \in S) x \in T \}$, and this set is what we wanted to be something called "the intersection of the non-vacuous set $S$ ".

but now I'm curious about knowing what's wrong with my first tricky argument (just please don't be mean with me, I'm aware it can be a silly mistake).

答:
We understand intuitively that a set B is the intersection of a family $S$ iff for all $x$ we have that
$$
x \in B \iff (\forall X \in S)x \in X.
$$
It's obvious that we can draw $(\forall X \in S)x \in X$ from $x \in \{x \in \bigcup S : (\forall X \in S)x \in X\}$.

If $(\forall X \in S)x \in X$ and $S \neq \varnothing$, then there is a $B \in S$. So, $(\exists X \in S)x \in X$ and thus $x \in \bigcup S$. We conclude that $x \in \{x \in \bigcup S : (\forall X \in S)x \in X\}$.

What if $S = \varnothing$? Then the last reasoning now will be broken because we can't choose a concrete element of $S$.

Technically your first definition is valid, but the second one is simpler and more transparent with the use of $S \neq \varnothing$.