问:
Two confusions on Great Picard Theorem
I am learning complex variables again with a greater depth, and I learnt the Great Picard Theorem, which is stated as follows:

If $z_{0}$ is an essential singularity of $f$, the image of any punctured neighborhood of $z_{0}$ via $f$ is the whole complex plane, except possibly for one point, infinitely often.

What does it mean by "infinitely often" here? What I know about i.o. is in the sense of probability: say we have $\{A_{n}\}_{n=1}^{\infty}$ a infinite sequence of events, then $$\{A_{i}\ \text{i.o}\}=\{\omega\in\Omega:\omega\in A_{i}\ \ \text{for infinitely many of}\ \ i\in \{1,2,3,\cdots\}\}.$$ I also know the related Borel-Cantelli.

However, what does i.o. imply here? Like, there are infinitely many points in any punctured neighborhood of $z_{0}$ that can be mapped by $f$ to a point in the complex plane? What does this even mean?

My second confusion is from this post: Entire function whose square or composition with itself is a polynomial

In this post, we have $f$ entire such that $f^{2}$ is a polynomial. The post says that as $f$ cannot achieves any values in $\mathbb{C}$ infinitely many times, it must be a polynomial by Great Picard Theorem.

I don't follow. It is indeed correct that as $f$ is entire, there is no singularity, so there is no essential singularity. But great picard theorem does not give a necessary consequence of the non-existence of singularity, like, for instance, no essential singularity implies no value can be achieved i.o.

I understand this theorem is really deep and really powerful, and it has many extension. But for now I am quite confused about it..

Thanks in advanced for any help!

Edit 1: Answers

Thanks to everybody that participated in this discussion, I know understand both the meaning of i.o. and the proof of the linked post.

Firstly,

i.o. means that for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ where $\omega_{0}$ possibly exists, $f^{-1}(\omega)$ is an infinite set. In other words, $f(z)=\omega$ has infinitely many solutions, for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$  and for any $z$ that is not the essential singularity (because any punctured neighborhood can be mapped to the whole $\mathbb{C}\setminus\{\omega_{0}\}$).

Second, if $f^{2}(z)$ is a polynomial, we suppose $f(z)$ is not a polynomial, as $f(z)$ is entire, it means that $f(z)$ has an essential singularity at $\infty$. By Great Picard Theorem, any punctured neighborhood of $z_{0}:=\infty$ then will be mapped to the whole $\mathbb{C}\setminus\{\omega_{0}\}$ for some possibly existing $\omega_{0}$, and $f(z)=\omega$ has infinitely many solutions for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ and $z\neq\infty$. It then implies that $$f^{2}(z)=\omega^{2}$$ has infinitely many solutions for any $\omega\in\mathbb{C}\setminus\{\omega_{0}\}$ but this is not possible as $f^{2}(z)$ is a polynomial, a contradiction.

I am really grateful for everyone who helped clarify this theorem. Now i can see its power. :)

答:
When we say that the image of $f$ “is the whole complex plane, except possibly for one point, infinitely often”, what that means is that, for some $w_0\in\Bbb C$, whenever $w\in\Bbb C\setminus\{w_0\}$, then $f^{-1}(\{w\})$ is an infinite set.

On the other hand, a corollary of the Great Picard Theorem states that if $f$ is an entire non-polynomial function, then, for some $w_0\in\Bbb C$, the equation $f(z)=w$ has infinitely many solutions for any $w\in\Bbb C\setminus\{w_0\}$. Sometimes, people call “Great Picard Theorem” to this corollary.

Title: Two confusions on Great Picard Theorem


问:
Rotating the graph of a function.
The problem is to rotate a function f(x) by 90 degrees and obtain a new rotated function g(x). I've tried to use the rotation matrix and the formula g(x,y) =(f(xcos(a) - y sin(o)) - xsin(o) ) / cos(o) and they seem to work, but I don't really end up with a form of g(x).
Is there even a way to do that, because on the internet I only read that the rotated graph isn't a function anymore?

In the attached pictures the problem is visualized. In the first figure the original function f(x) and in the second picture the wished rotated function is shown.

original function f(x)

rotated function g(x)

答:
Let $L$ be the inverse image of the $y$-axis under your rotation. If any line parallel to $L$ cuts the graph in more than one point, then the resulted rotated points cannot be a function of $x$ anymore after rotation.

Under some circumstances you may get lucky, as you did in the example you posted.

The graph of your function is the set of points $(x,f(x))$, and you can always rotate that with the linear transformation $(x,f(x))\mapsto (\cos(\theta)x+\sin(\theta)f(x), -\sin(\theta)x+\cos(\theta)f(x))$, but the resulting set of points is no longer guaranteed to be the graph of a function of $x$.

For example, when you rotate the graph of $f(x)=x^2$ by $\pi/4$ clockwise, you get this graph:

You can see there on the $y$-axis it cannot be expressed as a function of $x$ alone. You can however express it as a function of a parameter $t$ which outputs ordered pairs.

ContentLicense: CC BY-SA 4.0

Title: Rotating the graph of a function.


问:
Textbooks for learning: Lie algebra, topology, and differential geometry?
I'm a PhD student trying to learn the math required to understand Gauge theory. From some reading over the past few weeks, I think I need to learn the following:

Lie algebra / Lie groups

Algebraic and geometric Topology

Differential geometry

I only need understand the material enough to be literate (my research hopes only to use some results from these fields and not to contribute to them).
Does anyone have any good recommendations for textbooks that assume little mathematical knowledge beyond some undergrad. courses in linear algebra, vector calc., probability, and PDEs?

In particular, I'm hoping to find a textbook that assumes little prior knowledge while being somewhat brief its exposition (in my reading so far I'm finding myself getting lost in the volume of definitions).

Thanks for reading, I appreciate any suggestions!

edit: hardmath asked where I feel I've been getting lost in definitions. I've started reading "A Course in Modern Mathematical Physics" by Peter Szekeres (suggested by a friend). This seems like a great book and I've learnt a lot in the first few chapters but I'm struggling with the amount of terminology. For example, by chapter 3 my notes contained about 70 definitions.

答:
Here are my picks for best intro books:

Lie Algebras: https://www.amazon.com/Introduction-Algebras-Springer-Undergraduate-Mathematics/dp/1846280400

Differential Geomtry: https://www.amazon.com/First-Steps-Differential-Geometry-Undergraduate/dp/146147731X

Geometric Topology:

https://www.amazon.com/Introduction-Theory-Translations-Mathematical-Monographs/dp/0821810227

Algebraic Topology:

Fundamental group / covering spaces:
https://www.amazon.com/Introduction-Topological-Manifolds-Graduate-Mathematics/dp/1441979395

Homology/Cohomlogy:
https://www.amazon.com/Elements-Algebraic-Topology-James-Munkres/dp/0201627280/ref=sr_1_2?dchild=1&keywords=munkres+algebraic+topology&qid=1607026571&s=books&sr=1-2


问:
Are neither open nor closed sets in the Euclidean space locally compact?
It is known that the Euclidean space $\mathbb{R}^n$ is locally compact, and that all open or closed subsets of a locally compact Hausdorff space are locally compact. What can we say about neither open nor closed subsets in the Euclidean space? Are they locally compact?

答:
Claim: If $X$ is locally compact Hausdorff, then a subset $A\subseteq X$ is locally compact in the subspace topology if and only if it is locally closed.

Here a subset $A$ is locally closed if it is open in its closure $\overline A$. Equivalently if $A$ is the intersection $U\cap C$ of an open subset $U\subseteq X$ and a closed subset $C\subseteq X$.

Proof of claim: $\Rightarrow$ Suppose $A$ is locally compact. To show that it is locally closed in $X$ it will suffice to show that each point $x\in A$ has an open neighbourhood $U\subseteq X$ such that $U\cap\overline A\subseteq A$. Proceed as follows. Since $A$ is locally compact there is a compact $K\subseteq A$ and an open $U\subseteq X$ such that $x\in U\cap A\subset K\subset A$. Then $K$ is also compact in $X$, and hence closed there too, since $X$ is Hausdorf. Thus $\overline{U\cap A}\subset K\subset A$. Now notice that $U\cap\overline A\subseteq \overline{U\cap A}$. For if $y\in U\cap\overline A$ and $V$ is any neighbourhood of $y$, then $U\cap V$ is a a neighbourhood of $y$, which therefore meets $A$, so $V\cap(U\cap A)=(U\cap V)\cap A\neq\emptyset$ and hence $x\in \overline{U\cap A}$. Thus we have $x\in U\cap \overline A\subset \overline{U\cap A}\subseteq A$, as required.

$\Leftarrow$ The subset $\overline A$ is locally compact as it is closed in the locally compact space $X$. Thus if $A$ is an open subset of $\overline A$, then $A$ is also locally compact. $\square$

Remarks: The first part of the proof establishes that any locally compact subspace of a Hausdorff space $X$ is locally closed (regarless of whether $X$ itself is locally compact). The second part of the proof establishes that any locally closed subset of a locally compact space is locally compact (regardless of any separation assumptions. Here the correct definition of local compactness is that each point have a base of compact neighbourhoods).

Examples: 1. Any half-open interval $[a,b)$ or $(a,b]$ is a locally compact subspace of $\mathbb{R}$ which is neither open nor closed. Since finite products of locally compact spaces are locally compact, $\prod^{n}_{i=1}[a_i,b_i)$ and $\prod^{n}_{i=1}(c_i,d_i]$ and mixed products thereof are locally compact subspaces of $\mathbb{R}^n$, and in particular locally closed there. 2. Neither $\mathbb{Q}$ nor its complement $\mathbb{P}=\mathbb{R}\setminus\mathbb{Q}$ is locally compact, since neither is locally closed in $\mathbb{R}$. Products like $\mathbb{Q}^n$ and $\mathbb{P}^n$ are not locally closed in $\mathbb{R}^n$ and are not locally compact.


问:
The relationship of ${\frak m+m=m}$ to AC
Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.)

Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$?

If not, what is the relationship of this statement for all cardinals ${\frak m}$ to AC?

The weaker principle $\aleph_0\le{\frak m\Rightarrow m}+1={\frak m}$ can be derived in ZF, and it is a well-known result of Tarski that $\forall{\frak m}\ge\aleph_0,{\frak m\times m=m}$ implies the axiom of choice, but I am not sure about the intermediate result with ${\frak m+m=m}$.

Edit: A related question: is it consistent with ZF that there exists a cardinal ${\frak m}\ge\aleph_0$ such that ${\frak m=n+p}$ with ${\frak n<m}$ and ${\frak p<m}$? It's hard to say whether ${\frak m+m=m}$ precludes this scenario.

答:
To add slightly on Andres' answer, it is not only that $\frak m+m=m$ does not imply $\sf AC_\omega$, but in the other direction we can have $\sf DC_\kappa$ (for an arbitrary $\kappa$) but $\frak m+m=m$ fails.

We say that $A$ is a $\kappa$-amorphous set if every subset of $A$ has cardinality $<\kappa$, or its complement has such cardinality - but not both.

Note that if $A$ is $\lambda$-amorphous, then for $\kappa>\lambda$, $A$ is $\kappa$-amorphous as well. We say that $A$ is a properly $\kappa$-amorphous set if it is $\kappa$-amorphous, and $\aleph(A)=\kappa$. In other words, if every ordinal smaller than $\kappa$ can be injected into $A$.

For every regular $\kappa$ it is consistent with $\sf ZF+DC_{<\kappa}$ that there exists a [properly] $\kappa$-amorphous set.

Now, if $A$ is a $\kappa$-amorphous set, then $|A|<|A|+|A|$. This is a trivial observation since $A\times2$ can be written as the union of two sets, neither has complement of size $<\kappa$. So there cannot be an injection from $A\times2$ into $A$, but there is an obvious injection in the other direction.

Therefore $\sf ZF+DC_\kappa$ cannot prove $\frak\forall m. m+m=m$.

(On a slightly more confusing note, I am switching between $\sf DC_\kappa$ and $\sf DC_{<\kappa}$, but the former can be thought as $\sf DC_{<\kappa^+}$.)

To the added question, by the way, this is a simple result. If I recall correctly it is by Tarski.

The axiom of choice is equivalent to the assertion that whenever $\frak p,m,n$ are cardinals and $\frak p+m=n$ then $\frak p=n$ or $\frak m=n$.

The proof is simple, let $\frak a$ be a cardinal, and let $\kappa=\aleph(\frak a)$ the Hartogs number of a set of size $\frak a$. Then $\frak a+\kappa=a$ or $\frak a+\kappa=\kappa$. But $\kappa\nleq\frak a$ so the first option is impossible, therefore $\frak a\leq\kappa$, and can be well-ordered. $\square$

In particular if the axiom of choice fails, let $\frak a$ be a non-well ordered cardinal and $\kappa$ its Hartogs number, then $\frak a+\kappa$ is a set which can be decomposed into two strictly smaller cardinals.

It is also consistent with $\sf ZF$ that every cardinal which is not well-orderdable can be written in such way:

Monro, G.P. Decomposable Cardinals. Fund. Math. vol. 80 (1973), no. 2, 101–104.

ContentLicense: CC BY-SA 3.0


问:
Proving the logarithm rule for the matrix exponential
I am currently doing some work out of Undergraduate Analysis by Serge Lang and I have run into a bit of trouble with this question on the matrix logarithm.

I showed that on the space of $n\times n$ real matrices, if $|I-C|<1$ we can define the matrix logarithm as follows: $$\log(C)=\sum (-1)^{n+1}\frac{(C-I)^n}{n}$$
As well as the standard power series definition of the matrix exponential.

The author then goes on to say "Show that if $A,B$ commute, then $$\exp(A+B)=\exp(A)\exp(B)$$ State and prove the similar theorem for the log."
Proving the identity for the exponent was relatively easy, only requiring some use of the binomial theorem and some rearranging, however, I am not sure how to proceed with proving the identity for the matrix logarithm. I believe we want, if $A$ and $B$ commute,we get $$\log(AB)=\log(A)+\log(B)$$ We then get $$\log(AB)=\sum (-1)^{n+1} \frac{(AB-I)^n}{n}=\sum_{n} \frac{(-1)^{n+1}}{n}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}A^kB^k$$ but I am unsure how to proceed from here. Anyone have any suggestions? So far we have learned how to deal with series in complete normed vector spaces, continuous maps between arbitrary complete normed vector spaces, continuous linear maps between normed vector spaces and the shrinking lemma.

答:
Using the fact that $\log(\exp(A))=\exp(\log(A))=A$, we have $$\exp(\log(A)+\log(B))=\exp(\log(A))\exp(\log(B))$$ which reduces to $$\exp(\log(A)+\log(B))=AB.$$ If we take the Log of both sides we have $$\log(\exp(\log(A)+\log(B)))=\log(AB)$$
and hence $$\log(A)+\log(B)=\log(AB)$$


问:
Followup to question concerning $N_G(H) / C_G(H) \cong B \leq \mathrm{Aut}(H)$

Prove $H \leq G$ implies $N_G(H) / C_G(H) \cong B \leq \mathrm{Aut}(H)$ where $B$ is some subgroup of $\mathrm{Aut}(H).$

I did see this question:

For $H \leq G$, showing that $N_G(H)/C_G(H) \leq \text{Aut}(H)$

which has been helpful. My question now is with the Proposition 13 that the question makes reference to:

$\textbf{Proposition 13: }$ Let $H$ be a normal subgroup of the group $G$. Then $G$ acts by conjugation on $H$ as automorphisms of $H$. More specifically, the action of $G$ on $H$ by conjugation is defined for each $g \in G$ by $$h\mapsto ghg^{-1} \text{ for each $h \in H$.}$$ For each $g \in G$, conjugation by $g$ is an automorphism of $H$. The permutation representation afforded by this action is a homomorphism of $G$ into $\mathrm{Aut}(H)$ with kernel $C_G(H)$. In particular, $G/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$.

Specifically, my question is what do they mean by

"The permutation representation afforded by this action..."?

I see that the conjugation map stated in the proposition above is an automorphism of $H$ and so it just permutes the elements of $H$ but I don't know what the permutation representation is or means.

My goal is, I think at least for now, to prove this proposition and the referred question above states that the result I need to show follows from there.

答:
$\textbf{Proof of Proposition 13}$: As $H$ is normal in $G$, $ghg^{-1}\in gHg^{-1}=H$ and so each $ghg^{-1}\in H$. If $h,k \in H$ then
$$f_g(hk)=g(hk)g^{-1} \text{ and } f_g(h)f_g(k)=(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$$
so $f_g(hk)=f_g(h)f_g(k)$ so $f$ is a homomorphism. If $h \in H=gHg^{-1}$ then $h=gh'g^{-1}$ for some $h' \in H$ and so
$$f_g(h')=gh'g^{-1}=h$$
so $f_g$ is a surjection. We have that
\begin{align*}
\mathrm{Ker}(f_g) &= \{ h \in H: f_g(h)=e_H \} \\
                &= \{ h \in H: ghg^{-1}=e_H \} \\
                &= \{ h \in H: gh=e_Hg \} \\
                &= \{ h \in H: gh=ge_H \} \\
                &= \{ h \in H: h=e_H \} \\
                &= \{ e_H \}
\end{align*}
so $f_g$ is injective. We conclude that $f_g$ is an isomorphism and as $f_g:H \rightarrow H$, we have $f_g$ is an automorphism.

Let $f$ be as above. If $g,k \in G$ then
$$f(gk)=f_{gk} \text{ and } f(g) \circ f(k) = f_g \circ f_k.$$
For any $h \in H$, $f_{gk}(h)=(gk)h(gk)^{-1}$ and
$$(f_g \circ f_k)(h)=f_g( f_k(h) ) = f_g(khk^{-1})=g(khk^{-1})g^{-1}=(gk)h(gk)^{-1}=f_{gk}(h)$$
so $f_{gk} = f_g \circ f_k$ meaning
$$f(gk)=f(g) \circ f(k).$$
Therefore $f$ is a homomorphism. We have then
\begin{align*}
\mathrm{Ker}(f) &= \{ g \in G : f(g)=\mathrm{id}_H \} \\
                &= \{ g \in G: f_g = \mathrm{id}_H \}
\end{align*}
and so if $g \in \mathrm{Ker}(f)$ then for all $h \in H$, $f_g(h)=ghg^{-1}=h$ and so $gh=hg$ meaning that $g \in C_G(H)$. Further if $g \in C_G(H)$ then $f_g(h)=ghg^{-1}=hgg^{-1}=h=\mathrm{id}_H$ so $g \in \mathrm{Ker}(f)$. We have shown then
$$\mathrm{Ker}(f)=C_G(H).$$
Therefore, we can take the function $j: G \rightarrow \mathrm{Im}(f)$ by $j(g)=f(g)$. This function $j$ is a surjective homomorphism and so by the first isomorphism theorem we have that since $\mathrm{Ker}(f)=\mathrm{Ker}(j)$ that
$$G/C_G(H) \cong \mathrm{Im}(f)$$
and we know that $\mathrm{Im}(f) \leq \mathrm{Aut}(H)$ proving what was needed.

$\textbf{Proof of desired thing: }$ We have that $H$ is a normal subgroup of $N_G(H)$ and so from above  $N_G(H)/C_{N_G(H)} \cong B \leq \mathrm{Aut}(H)$ and further $C_{N_G(H)}(H)=C_G(H)$ proving the statement.


问:
Trying to find out what is wrong with my methodology for this Laplace Transform Example.
I am just starting to learn convolutions and Laplace Transforms and am doing an example in my book and I don't understand why my methodology is incorrect so I will give the book's methodology first and then my methodology and I would appreciate if someone can tell me where I am misunderstanding something.

Find the PS to the following IDE:

$x'(t)+2x(t)-4\int_0^te^{(t-\tau)}x(\tau)d\tau = e^t$

where $x(0) = 0$

Book's Methodology:

let X(s) = $\scr{L}(x(t))$

Apply LaPlace Transform to both sides of the equation to get

$(sX(s)-x(0))+2X(s)-4\scr{L}(e^t\circledast x(t))=\scr{L}(e^t)$

We Know $ \scr{L}(e^t\circledast x(t)) = \scr{L}(e^t)\scr{L}(x(t))$

So let's also apply the IC's and rewrite the equation as

$sX(s)+2X(s)-4\cfrac{1}{s-1}X(s)=\scr{L}(e^t)$

Solving for X(s) and performing partial fraction, we obtain

$X(s) = \cfrac{1}{5}(\cfrac{1}{s-2}+\cfrac{4}{s+3})\scr{L}(e^t) $

Then applying inverse LT, we get

$x(t) = \cfrac{1}{5}\scr{L}^{-1}((\cfrac{1}{s-2}+\cfrac{4}{s+3})\scr{L}(e^t)) = \cfrac{1}{5}(e^{2t}+4e^{-3t}) \circledast e^t$

Using $e^{\alpha t} \circledast e^{\beta t} =\cfrac{e^{\alpha t}-e^{\beta t}}{\alpha - \beta}$

we get

$x(t) = \cfrac{1}{5}(\cfrac{e^{2t}-e^{t}}{2 - 1} + 4\cfrac{e^{-3t}-e^{t}}{-3 - 1}) = \cfrac{1}{20}(-3e^t+4e^{2t}-e^{-3t})$

Now What is wrong with my methodology?

My Methodology:

Apply LaPlace Transform to both sides of the equation to get

$(sX(s)-x(0))+2X(s)-4\scr{L}(e^t\circledast x(t))=\scr{L}(e^t)$

We Know $ \scr{L}(e^t\circledast x(t)) = \scr{L}(e^t)\scr{L}(x(t))$

So let's also apply the IC's and rewrite the equation as

$sX(s)+2X(s)-4\cfrac{1}{s-1}X(s)=\cfrac{1}{s-1}$ (Here I use $\cfrac {1}{s-1}$ instead on the RHS but I don't think there should be an issue here? Am I mistaken?)

$X(s)(\cfrac{s^2+s-6}{s-1}) = \cfrac{1}{s-1}$

$X(s) = \cfrac{(s-1)}{(s-1)(s^2+s-6)} = \cfrac{1}{(s^2+s-6)} = \cfrac{1}{5}(\cfrac{1}{s-2}) -\cfrac{1}{5}(\cfrac{1}{s+3})$ (I used partial fraction decomposition)

Now apply inverse laplace Transform

$x(t) = \cfrac{1}{5}\scr{L}^{-1}(\cfrac{1}{s-2})-\cfrac{1}{5}\scr{L}^{-1}(\cfrac{1}{s+3}) = \cfrac{1}{5}(e^{2t}-e^{-3t})$ which does not match the book's answer
What am I doing wrong?
Thank you for any help

答:
They made a mistake in the book after this line:
$$x(t) = \cfrac{1}{5}\left(\cfrac{e^{2t}-e^{t}}{2 - 1} + 4\cfrac{e^{-3t}-e^{t}}{-3 - 1} \right) $$
It should give:
$$x(t) = \cfrac{1}{5}({e^{2t}\color{red}{-e^{t}}} -{e^{-3t} \color{red}{+e^{t}}})$$
This gives the same answer as yours:
$$x(t) = \dfrac{1}{5}\left(e^{2t} -e^{-3t}\right)$$


问:
Is this explanation of normal subgroups and quotient groups correct?
I apologize for the long post, but I'm currently a student finishing up his first semester in group theory. My introduction was pretty definition-heavy so I've found I can internalize concepts (such as quotient groups, normal subgroups, etc.) myself by forming my own way of motivating and teaching them intuitively. I'd like to know if my current presentation/understanding is correct.

I think after learning about subgroups and Lagrange's theorem, a natural question is then if we can break down a group $G$ to better understand its parts and hopefully the whole $G$ (as is tradition in any analytical endeavor). But if we want to pull back any useful understanding of $G$ from this smaller group, it ought to preserve some structure of $G$. That structure is exactly how the operation acts on elements of $G$ (since groups are just elements with an operation relating them).

So for the sake of exploration, we pretend to have the magic function $\phi : G \rightarrow H$ that does exactly this for us–maps $(G, *_G)$ to some smaller part $(H, *_H)$, then ask what we can say about $\phi$. Our original goal was for $\phi$ to preserve the operation, i.e. for all $a, b \in G$ that $\phi(a *_G b) = \phi(a) *_H \phi(b)$.

The next thing I would observe is that since $H$ has smaller order, $\phi$ necessarily maps a multiple elements, say $a, b$, in $G$ to the same element in $H$. In this sense, $a$ and $b$ are "equivalent" under $\phi$. Given that we have "willed" $\phi$ to operation-preserving, we can see that a natural way this arises by letting $ak = b$ for some $k \in G$:

$$ak = b \implies \phi(b) = \phi(a *_G k) = \phi(a) *_H \phi(k).$$

If we want $\phi(b) = \phi(a) *_H \phi(k) = \phi(a)$ then $\phi(k) = e_H$. I think leads naturally to the definition of the kernel: it's a set of elements that maps to the identity, and makes $a$ equivalent to $b$ mod $\ker \phi$. And in fact, as I've learned from this answer, we naturally get equivalence classes of elements that partition the group into cosets analogous to modular arithmetic. So, $\phi$ takes elements and puts them neatly into these equivalence classes (abstracting away some of the details in $G$ that look "the same" in $H$, leading–at least for me–directly to the First Isomorphism Theorem). Then it makes sense to propose the map $\phi : g \mapsto g \ker \phi$.

The next question becomes what the operation of this $H$ looks like. We've established that elements of $H$ are cosets (and equivalence classes), so for two elements in cosets $g_1 \ker\phi$ and $g_2 \ker\phi$, once combined by $*_H$ we'd want for the result to be in $g_1g_2 \ker\phi$ (modular arithmetic analogy works here as well). Set-wise, we might write

$$g_1\ker\phi \cdot g_2\ker\phi = g_1g_2\ker\phi.$$

But does this come for free? For an element $g_1k_1g_2k_2 \in g_1\ker\phi \cdot g_2\ker\phi$ to look like $g_1g_2k$ for some $k$, it must be that $k_1g_2 = g_2k_3$ for some $k_3$. Set-wise this can be written as $g\ker\phi = \ker\phi g$, i.e. left-costs = right-cosets and it turns out it indeed satisfies this condition and we are safe to proceed.

So, in the end, we have designed $H$, a broken-down version of $G$. And how did we do it? By "dividing out" or "quotienting out" the information that looks the same under $\phi$ in $H$–$\ker \phi$. Thus we write $H$ as $G/\ker\phi$, aptly called a quotient group.

Although, you could flip this presentation, and instead of viewing from the kernel perspective, suppose $K$ is some arbitrary group. Then it must satisfy the condition of left-cosets = right-cosets (which we name normality because it is a nontrivial property that gives us a usable quotient) for $G/K$ to be a group, as $\ker\phi$ already does, and through satisfying normality automatically becomes the kernel of some homomorphism (namely the natural, which I've presented).

My questions are:

Is this presentation correct (on an intuitive level, I know there are lots of places for concrete proofs)? It feels right to me, but I also feel like I may have gotten definitions vs. implications mixed up.

If so, does any textbook follows this approach that I can dig into?

I think homomorphisms also can fit into this framework, given I suggest $\phi$ pretty early on, but how would non-surjective homomorphisms be explained?

答:
I know you've gotten one satisfactory answer, but let me weigh in here a bit.

I'll first mention that what you present is generally correct, and a valid way to approach this. The idea of "breaking down a (finite) group into smaller pieces" is in fact behind the idea of classifying finite simple groups (groups that cannot be broken down), together with the theory of group extensions (trying to understand what a group $G$ "is" if you have a normal subgroup $N\triangleleft G$, and you understand both $N$ and $G/N$).

But let me offer you a different perspective and a different way into the isomorphism theorems...

After you learn about groups, and subgroups and Lagrange's Theorem, maybe Cauchy's Theorem, we come to a crossroads in how to try to better understand a given group.

One way to try to learn things about a given group $G$ is to just stare at it until you notice some interesting things about $G$. However, generally speaking, a much more fruitful approach in algebra is to take a less static approach and to consider two things: what the group $G$ "can do", and how it interacts with other groups.

What a group "can do" is in fact historically how groups were originally understood. The original notion of a group was a "group of permutations": a collection of operations acting on a set in specific ways. Even as late as the turn of the 20th Century, Burnside's book on groups still defines a group as a collection of "operators" acting on "some objects". It was Cayley who introduced the abstract definition of a group as a "set with a binary associative operations satisfying certain conditions", and then immediately went on to prove that this did not change the objects of study, as any "group of permutations" was a group under his new proposed definition, and any object that satisfied this new proposed definition could be understood as a "group of permutations". This is the notion behind Cayley's Theorem, and why it is, in my opinion, more important historically than practically today. But this already introduces the notion of functions: what does "can be understood as a group of permutations" mean? It means you can biject it with such a group in a way that respects the operation.

This also leads us to functions. To justify why we want to think about functions, let's consider two areas where functions play a major role: the real numbers/calculus, and linear algebra.

The key property of the real numbers was that they were "continuous": they have no 'holes'. Rather than just stare at real numbers and see if we can say interesting things about them, it turns out to be much more fruitful and interesting to consider functions from $\mathbb{R}$ to itself that respect this "continuity". And so we get the notion of continuous functions, and the study of continuous functions, as a way to shed light on the nature of the real numbers themselves.

Similarly, with Linear Algebra, staring at vector spaces only takes you so far; the real power of vector spaces only emerges when you start considering linear transformations.

In both cases, you don't just want any old function; you want functions that "preserve" whatever it is that makes your objects interesting. For real numbers, continuity; for vector spaces, the addition and scalar product.

So with groups. A group is characterized by three things (bear with me): a binary operation $G\times G\to G$, that assigns to any pair of elements $g_1,g_2\in G$ their "product" $g_1g_2$. A distinguished element $e_G\in G$ with the property that $ge_G=e_Gg=g$ for all $g\in G$. And a function $G\to G$ that assigns to every element $g\in G$ its "inverse", $g^{-1}$, which has the property that $gg^{-1}=g^{-1}g=e_G$.

So if we have two groups $G$ and $H$, then a "function that preserves this structure" would be a function $f\colon G\to H$, such that

Respects products: if $g_1,g_2\in G$, then $f(g_1g_2) = f(g_1)f(g_2)$.

Respects the identity: $f(e_G) = e_H$.

Respects inverses: if $g\in G$, then $f(g^{-1}) = (f(g))^{-1}$.

It turns out that two of these conditions are superfluous, but that is how we want to start. As you know, if 1 holds for a function between groups, then 2 and 3 will automatically hold as well. One can then define a group homomorphism as simply a function that satisfies 1 and prove 2 and 3; I prefer to define it as a function that satisfies 1, 2, and 3, and then prove that if it satisfies 1, then it must satisfy 2 and 3. The reason I prefer is that I think it makes the definition more natural.

Okay, so these are the functions that will play the role of "linear transformations" and "continuous functions". We call them, as I mentioned above, "group homomorphisms." They also are the type of functions needed in Cayley's argument that any group "can be seen" as a group of permutations, because that corresponds to a one-to-one function $f\colon G\to S_X$ (for some set $X$), that satisfies 1, 2, and 3. So that $f(G)$ is "essentially the same" (as far as the group structure is concerned) as $G$, but now it consists of permutations on a set $X$.

Now, given a function $f\colon G\to H$ (in fact, any function between two sets $X$ and $Y$), there is a natural equivalence relation that we can define on $G$. Let us say that two elements $x,y\in G$ are "$f$-equivalent", $x\sim_f y$,  if and only if $f(x)=f(y)$. This is easily verified to be an equivalence relation, and so it partitions $G$ into equivalence classes.

But because $f$ is a group homomorphism, we have the following consequences: if $x\sim_f y$ and $z\sim_f w$, then $xz\sim_f yw$, and $x^{-1}\sim_f y^{-1}$. So we can make the set of equivalence classes, $G/\sim_f$ into a group! Let $[x]_f$ be the equivalence class of $x$. Then we can define $[x]_f*[y]_f = [xy]_f$, $e_{G/\sim_f} = [e_G]_f$, and $([x]_f)^{-1} = [x^{-1}]_f$. It is then an easy exercise to show that this is indeed a group.

What relation does this group have with $G$ and with $f$? Well, that's the first isomorphism theorem: there is a bijective group homomorphism between the group $G/\sim_f$ and the image group $f(G)$, given by sending $[x]_f$ to $f(x)$.

How is this related to normal subgroups? Ah, well, these equivalence classes have an interesting property: because of the property that $x\sim_f y$ implies $x^{-1}\sim_f y^{-1}$, and if $x\sim_f y$ and $z\sim_f w$ then $xz\sim_f yw$, we have
$$x\sim_f y \iff xy^{-1}\sim_f e_G.$$
That is: we can completely determine the equivalence relation by just knowing $[e_G]_f$. Moreover, this collection is a subgroup of $G$!

Will any subgroup work? No, it turns out it doesn't. If $N$ is a subgroup and we try to define an equivalence relation $x\sim y$ if and only if $xy^{-1}\in N$, we do get an equivalence relation, but we do not get an equivalence relation that lets you define a group structure on $G/\sim$. The condition that lets you do that is precisely that $N$ must be a normal subgroup. I go into much more detail about this in this answer.

So, "good" equivalence relations, those coming from functions (they are called "congruences"), correspond to normal subgroups. In fact, as with Cayley's Theorem before (which gave a separate definition of "group" and then showed it was really the same as the old one), so it is with "good" equivalence relations:

Theorem. A subgroup $N$ of a group $G$ is normal in $G$ if and only if there exists a group $H$ and a homomorphism $f\colon G\to H$ such that $N=[e_G]_f$.

This then leads to the usual First Isomorphism Theorem, which says: this construction of taking quotients of a group is "essentially the same" as looking at the image of $G$ under a group homomorphism, in that given any homomorphism $f\colon G\to H$, if $N=[e_G]_f$, then $G/N = G/\sim_f$ is "essentially the same" as $f(G)$: there is a bijective group homomorphism between them.

The Third Isomorphism Theorem corresponds to compositions of morphisms: if $f\colon G\to H$ and $g\colon H\to K$, then $f(G)/\sim_g$ is essentially the same as $g\circ f(G)/\sim_{g\circ f}$. That is, if $N\triangleleft G$, $K\triangleleft G$, $N\subseteq G$, then $K/N \triangleleft G/N$ and $(G/N)/(K/N)\cong G/K$.

The Fourth, or lattice, Isomorphism Theorem establishes a correspondence between the subgroups of $f(G)$ and the subgroups of $G$ that contain $[e]_f$. One then asks... okay, and what about other subgroups of $G$? That's what the Second Isomorphism Theorem gives you: if $f\colon G\to H$ is a homomorphism, and $K$ is an arbitrary subgroup of $G$, then $f(K)$ corresponds to $K/(K\cap N)$ (where $N=[e]_f$). And this image is "the same" as the image of $KN$. That is,
$$\frac{K}{K\cap N} \cong \frac{KN}{N}.$$

So in summary:

First Isomorphism Theorem tells you that images of groups correspond to quotients and vice-versa.

Third Isomorphism Theorem tells you that this correspondence plays well with composition.

Fourth Isomorphism Theorem tells you that there is a very nice correspondence between the subgroups of $f(G)$ and the subgroup of $G$ that contain $[e]_f$.

And the Second Isomorphism Theorem tells you how the rest of the subgroups of $G$ behave under the homomorphism $f$.

Thus, the importance of normal subgroups corresponds simply to the importance of homomorphisms. Images of a group are like "shadows" of the group, and so will hopefully sometimes be easier to understand. Simple groups are the ones that we cannot simplify this way: we'll just have to stare at them intently until we understand them. And if we can understand simple groups, and we can understand how to put groups together (group extensions) from $N$ and $G/N$, then perhaps we can leverage our (hypothetical) understanding of simple groups into a (even more hypothetical) understanding of all groups. Turns out this is too naïve a hope, unfortunately, but perhaps it can help justify why we care about morphisms, normal subgroups, quotients, etc.


问:
Constructively embedding $\mathbb{Q}^\mathbb{N}$ into $\mathbb{R}$
Using the axiom of choice it is provable that $\mathbb{R}$ is isomorphic to $\mathbb{Q}^\mathbb{N}$ as a vector space over $\mathbb{Q}$. (Assuming AC, both spaces have a Hamel basis over $\mathbb{Q}$ of the same cardinality and are thus isomorphic.)

So my question is whether such an isomorphism between $\mathbb{R}$ and $\mathbb{Q}^\mathbb{N}$ can be constructed without AC or, at least, whether we can embed $\mathbb{Q}^\mathbb{N}$ into $\mathbb{R}$ without AC. (By embedding I mean constructing an injective $\mathbb{Q}$-linear map from one space into the other.)

The latter is equivalent to asking whether we can construct a subspace of $\mathbb{R}$ that has a schauder-basis over $\mathbb{Q}$, as such a subspace should automatically be isomorphic to $\mathbb{Q}^\mathbb{N}$.

Thanks for the help!

答:
In fact it's consistent with ZF that there are no nontrivial homomorphisms $\mathbb{R} \to \mathbb{Q}$. Quoting from a previous answer where this came up:

There's a model of ZF constructed by Shelah in which every set of real numbers has the Baire property. This implies, if I understand correctly, that there are no nonzero homomorphisms from $\mathbb{R}$ to any countable abelian group (since any countable abelian group with the discrete topology is a Polish group, so in this model any homomorphism from $\mathbb{R}$ to such a group is automatically measurable and so automatically continuous). So $\mathbb{R}$, and $SO(2)$, have no subgroups of countable index in this model.

This doesn't rule out the possibility of an explicit embedding $\mathbb{Q}^{\mathbb{N}} \to \mathbb{R}$; I'm not sure one way or the other whether such a thing exists but off the top of my head I'd bet it doesn't. I'd bet it's consistent with ZF that every linear map $\mathbb{Q}^{\mathbb{N}} \to \mathbb{R}$ factors through the projection to some finite subset of its factors.


问:
maximum and minimum of the function $ y=2 \sin3x+\sin 6x$
I have to find maximum and minimum of the function
$$ y=2 \sin3x+\sin 6x$$

$y'= (\cos3x+\cos6x)= 6 \cos 3x (1+2 \sin 3x)=0 \Rightarrow \cos3x=0  \lor \sin3x=- \frac{1}{2}  \Rightarrow 3x= \frac{\pi}{2} \lor 3x= \frac{3\pi}{2} \lor 3x= \frac{7\pi}{6}\lor 3x= \frac{11\pi}{6}$.
Analyzing the sign of the derivative we can say that in $[0,2\pi]$:

$y'>0 \Rightarrow 0< 3x< \frac{\pi}{2} \lor \frac{7}{6} \pi<3x < \frac{3}{2}\pi \lor \frac{11}{6} \pi<3x <2 \pi $.

For $3x= \frac{\pi}{2}$ and $3x= \frac{3}{2} \pi$ there are two maxima.

For $3x= \frac{7}{6}\pi$ and $3x=  \frac{11}{6}\pi$ there are two maxima.

The first two positions can be summarized in $3x= \frac{\pi}{2}+k\pi \Rightarrow x= \frac{\pi}{6}+\frac{k \pi}{3}= \frac{\pi+2k \pi}{6}= \frac {1+2k}{6} \pi$.

The other two disequalities can be summarized in $3x= \frac{\pi}{2}+ \frac{2 \pi}{3} k \Rightarrow x=  \frac{\pi}{6}+ \frac{2 \pi}{9} k $.

Is it correct or I am missing something?

I don't understand my book shows a different solution.

答:
Your function and its derivative are periodic, with period $\frac{2\pi}3$, so it  suffices to see what happens on an interval of length $\frac{2\pi}3$. Further more, the function is odd, and its derivative is even, so the interval $\bigl[0,\frac{\pi}3\bigr]$ will do.

Now, since $f'(x)=6(\cos 3x+\cos 6x)=6(2\cos^2 6x+\cos 6x -1)$, set $u=\cos 6x$. We have  $f'(x)=2u^2+u-1$,  and the roots of this quadratic polynomial are $u=-1,\frac12$. As $u$ is a cosine, we see  that, on the chosen interval, taking into account that $\cos 3x$ is decreasing,
\begin{align}
f'(x)=2u^2+u-1\ge 0\iff \frac12\le u\le 1&\iff \cos \frac\pi3\le\cos 3x\le \cos 0\\&\iff 0\le x\le\frac\pi 9\\
f'(x)=2u^2+u-1\le 0\iff -1\le u\le \frac12&\iff \cos\pi\le\cos 3x\le \cos \frac\pi3\\&\iff\frac\pi 9\le x\le \frac\pi 3.
\end{align}
Therefore the maxima are attained at $x\equiv\frac\pi 9\mod\frac{2\pi}3$.


问:
Expected number of a pattern in a random sequence stopped after appearance of another symbol for the first time
Let $X\sim p_1\delta_1+p_2\delta_2+p_3\delta_3$. So the variable $X$ takes the values 1,2,3 with different probabilities. If I have a sequence of iid random variables $(X_i)_{i\geq 1}$ with $X_i\sim X$, what is then the expected value of the number of times i can observe the pattern "13" before seeing 2 for the first time?

Let $N_2=\inf\{n:X_n=2\}$ and $T_{13}$ the variable that counts the number of times "13" appears in the sequence. Then

$$T_{13}=\sum_{i=1}^{N_2-2}1_{X_i=1,X_i+1=3}$$

I tried using the Wald-Identity, but that gave me
$$E(T_{13})
=\left(\frac1{p_2}-2\right)\cdot\frac{p_1p_3}{p_1+p_3}
%=\left(\frac{1-p_2}{p_2}\right)\cdot\frac{p_1p_3}{p_1+p_3}=\frac{p_1p_3}{p_2}$$

Now, for $p_i=1/3$ that implies $E(T_{13})=\frac{1/9}{2/3}=1/6$. But simulations show that in this case the probability should be around 1/3.

Also, the formula gives negative results for $p_2>1/2$, which is another contradiction.

Any suggestions on how to approach such a problem in a more rigorous way?

EDIT: Via a Markov Chain Pattern approach i got $$E(T_{13})=\frac{p_1p_3}{p_2}$$ (got the initial conditions wrong.)

Surprisingly, using
$$T_{13}=\sum_{i=1}^{N_2-1}1_{X_i=1,X_i+1=3}$$
gives the correct value using the wald-identity...

答:
Here are a couple of approaches.

In either case, create a 4 state markov chain with states

$\big\{1: (1), 2:(1,3), 3: (\text{"something other than 1"},3), 4: (2)\big\}$

with some care, any state except 1 may be interpreted as a 'cold start' for the process here but it it can be easy to confuse this.  In the below 'approach 1' the natural interpretation is to have state 3 as the 'cold start' for the process and in 'approach 2' state 4:(2) is the sensible 'cold start' state.  The transitions are hopefully obvious, but here is an example, with $p_1 = 0.10$, $p_2 = 0.05$, $p_3 =0.85$.

$A:=\displaystyle \left[\begin{matrix}0.1 & 0.85 & 0 & 0.05\\0.1 & 0 & 0.85 & 0.05\\0.1 & 0 & 0.85 & 0.05\\0.1 & 0 & 0.85 & 0.05\end{matrix}\right]$

approach 1

The above chain is shown in recurrent form.  The first approach is to make seeing a '2' an absorbing state.  The transient chain, then is given by deleting row and column 4 and calling the resulting matrix $B$.  This gives a fundamental matrix

$N^{-1} = \big(I_3-B\big)$

and

$N=\big(I_3-B\big)^{-1} = \displaystyle \left[\begin{matrix}3.0 & 2.55 & 14.45\\2.0 & 2.7 & 15.3\\2.0 & 1.7 & 16.3\end{matrix}\right]$

and we can simply read off the number of times we visit the pattern (1,3), represented by state 2, given our 'cold start' of state 3.  We visit this $1.7$ times.  The above was done with explicit numbers for concreteness but you can work with $N^{-1}$ symbolically and then get $N$ via the adjugate, etc.

approach 2

viewing this as a Renewal Reward Process:

Define an epoch by starting in state 4:(2) in the original markov chain, and ending in state 4:(2).  I.e. each visit to state 4 marks a 'renewal' and a new epoch.  Designate a reward of 1 for each time you see the pattern (1,3) during an epoch.  The renewal reward limit theorem tells us

$\lim_{t\to \infty}\frac{\mathbb E[R(t)]}{t}= \frac{\mathbb E[R_1]}{\mathbb E[T_1]}$

i.) $\mathbb E[T_1]=p_2^{-1}$ as the distribution of times for a given epoch is geometric

ii.)  $\lim_{t\to \infty}\frac{\mathbb E[R(t)]}{t}=p_1\cdot p_3$

This can be found by solving the standard problem of runs, or computing $A_\infty = \lim_{t\to \infty}A^t$

and looking at component (2,2) of said matrix.  A slightly better interpretation: you can solve for $\pi^T A = \pi^T$  and read off the steady state probability of visiting state 2:(1,3).

putting this all together

$\mathbb E[R_1] =  \lim_{t\to \infty}\frac{\mathbb E[R(t)]}{t}\cdot \mathbb E[T_1] = p_1\cdot p_3\cdot p_2^{-1}$

so the expected reward (= expected visits) in the first epoch is $p_1\cdot p_3\cdot p_2^{-1}$ which is also the expected number of visits during an arbitrary epoch since this is a proper renewal process.


问:
Show that the Mobius transformations form a group.
I'm wondering if using the 4 statements below to help prove that Mobius Transformations form a group is enough. If so, how can I use M(z)=(az+b)/(cz+d) to do that?

Thank you in advance, and please feel free to fix any formatting as I'm not great with that. :)

答:
For example, we can show closure as follows. Let $M_1,M_2$ be the Mobius transformations
$$
M_1(z) = \frac{a_1 z + b_1}{c_1 z + d_1}, \quad
M_2(z) = \frac{a_2 z + b_2}{c_2 z + d_2}.
$$
To show that condition 1 holds, show that the function $M_1 \circ M_2$ is a Mobius transformation.
$$
M_1(M_2(z)) = \frac{a_1 \frac{a_2 z + b_2}{c_2 z + d_2} + b_1}{c_1 \frac{a_2 z + b_2}{c_2 z + d_2} + d_1} = \cdots
= \frac{[a_1a_2 + b_1 c_2]z + [a_1b_2 + b_1 d_2]}
{[c_1 a_2 + d_1 c_2]z + [c_1 b_2 + d_1 d_2]}.
$$
We can see that this could be a Mobius transformation because it can be expressed in the form
$$
M(z) = \frac{az + b}{cz + d}
$$
for some complex numbers $a,b,c,d$. In fact, to properly show that $M$ is a Mobius function, we should also argue that $ad \neq bc$, which is to say that there is no cancelation and $M$ is not a constant function.

One way to argue that $M$ is not a constant function is to come back once you've proven point 4 (the existence of inverse). If $M(z)$ where a constant function, then the function $M_1^{-1}(M(z)) = M_2(z)$ would also be a constant function. That can't be true because, by the definition of the Mobius group, $M_2(z)$ was not a constant function (i.e. $a_2 d_2 \neq b_2 c_2$).

For 2, note that for two Mobius functions $M_1,M_2$: by the definition of your group, if $M_1(z) = M_2(z)$ for all complex numbers $z$, then $M_1 = M_2$. With that in mind, we note that for any $z \in \Bbb C$, we have
$$
[(M_1 \circ M_2)\circ M_3](z) = (M_1 \circ M_2)(M_3(z)) = M_1(M_2(M_3(z))),\\
[M_1 \circ (M_2 \circ M_3)](z) = M_1((M_2 \circ M_3)(z))
= M_1(M_2(M_3(z))).
$$
That is, we have
$$
[(M_1 \circ M_2)\circ M_3](z) = [M_1 \circ (M_2 \circ M_3)](z)
$$
for all complex numbers $z$, so that $(M_1 \circ M_2)\circ M_3 = M_1 \circ (M_2 \circ M_3)$. In other words, the defined multiplication is associative.

For 3, let $I$ denote the Mobius function
$$
I(z) = z = \frac{1\cdot z + 0}{0\cdot z + 1}.
$$
Verify that for any Mobius function $M$, we have $M \circ I = I \circ M = M$.


问:
Why can I estimate the variance if I already estimated the mean in this way?
In my lectures it's discussed the estimation of $Var(X)$ in the case I have already estimated $E(X)$, but there is something I don't know if it's a problem or not.

Suppose $Var(X) = f(E(X))$ , for example in the case of a bernoulli r.v. we have $Var(X) = E(X)(1-E(X))$ , besides I have estimated $ E(X) $ to be $ \hat{\theta} $ then I could use as a estimate of $Var(X)$ the function $f(\hat{\theta})$ .

However, in this reasoning I think there is something not so clear about the error in the estimate because if the estimate of $E(X)$ is not exact (which is always the case) then the estimate of $Var(X)$ could be much more inaccurate depending on the actual function $f$.

Is this method used because it's assumed that the estimate $\hat{\theta}$ is equal to the actual value of $E(X)$? I don't have much knowledge in statistical inference so I don't know if this is a real problem. Is this a true problem?

答:
If you have an estimator $\hat\theta$ for a parameter $\theta$ and you are interested in estimating $h(\theta)$ for some function $h$, then it is legitimate to propose $h(\hat\theta)$ as an estimator for $h(\theta)$. You are correct that if there is uncertainty in $\hat\theta$, then that uncertainty will propagate to $h(\hat\theta)$. A simple way to assess the uncertainty is to expand $h(\hat\theta)$ in a one-term Taylor approximation around $\theta$:
$$
h(\hat\theta) \approx h(\theta) + h'(\theta)(\hat\theta-\theta).\tag1$$
This gives a rough measure of the variance for $h(\hat\theta)$, namely:
$$
\operatorname{Var}(h(\hat\theta))\approx [h'(\theta)]^2 \operatorname{Var}(\hat\theta).\tag2
$$
You can interpret (2) as an approximate relation between the variance of $\hat\theta$ and the variance of $h(\hat\theta)$ .

Taking expectations in (1) also gives a rough measure of the expectation of $h(\hat\theta)$. This first-order approximation to the mean and variance of $h(\hat\theta)$ is often called the delta method. An example of its use, applied to estimation of $p(1-p)$ in the Bernoulli case, is here.


问:
Basis of a subspace starting from a given subspace.
I was given the following subspaces:
$$E=\mathrm{Span}\{(1,0,1), (2,0,-1),(1,0,0)\}\\F=\mathrm{Span}\{(1,0,2),(1,1,1),(1,-1,3)\}$$
I found that a basis for $E\cap F$ is $\{(\frac{1}{2},0,1)\}$. Now, I am asked to expand this basis to get a basis for $E$, $F$ and $E+F$. I do not know how to proceed know, can somebody help me?

答:
You can observe that $\text{dim}(E)=\text{dim}(F)=2$, and
$$E=\langle(1,0,1),(2,0,-1)\rangle=\langle u,v\rangle \text{ and }F=\langle(1,0,2),(1,1,1)\rangle=\langle s,t \rangle$$ (with gaussian elimination you find that the matrices whose columns are the vectors that span $E$ and $F$ have two non-zero pivot, so their dimension is two).

We want now to find a basis for $E\cap F$: imposing $\alpha u+\beta v=\gamma t+\delta s$ and solving the matrix associated to the system $\alpha u+\beta v-\gamma t-\delta v=0$, we obtain $(\alpha,\beta,\gamma,\delta)=(-5t,t,-3t,0)$, $t\in\mathbb R$.
One possible vector that span $E\cap F$ is $(1,0,2)$.

We know that $\text{dim}(E)+\text{dim}(F)-\text{dim}(E\cap F)=\text{dim}(E+F)=3=\text{dim}(\mathbb R^3)$.

To extend the basis of the intersection to $E$, $F$ and $E+F$ we only have to add vectors linearly independent with $(1,0,2)$. For example, we can generate $E+F$ with $(1,0,2),(1,0,0),(0,1,0)$ and we can easily add the vector $u$ and $t$ to generate respectively $E$ and $F$.

Rmk: If $W$ is a vector subspace of dimension $k$ and $w_1,\dots,w_k$ are linearly independent, than $\{w_1,\dots,w_k\}$ is a basis for $W$.


问:
For $1 \le p \lt q \le \infty$ show that neither $L^p(\Bbb{R},\lambda)$ or $L^q(\Bbb{R},\lambda)$ contain the other
Let $1 \le p \lt q \le \infty$, and $\lambda$ denote the Lebesgue measure. Show that $L^p(\Bbb{R},\lambda) \not\subset L^q(\Bbb{R},\lambda)$ and $L^q(\Bbb{R},\lambda) \not\subset L^p(\Bbb{R},\lambda)$ by constructing functions where each belongs to one but not the other.

Hint: You will likely want to handle $q = \infty$ separately.

My idea:

After some searching online, I came across this suggestion:

For the first non-inclusion, consider $f(x) = \begin{cases} x^{\frac{1}{p}} &\text{if } 0 \le x \le 1 \\ 0 &\text{o.w.} \end{cases}$

Here is my attempt to use this function:

Then $\int_{\Bbb{R}}(x^{\frac{1}{p}}d\lambda(x))^p = \int_0^1xd\lambda(x) \lt \infty$

And$\int_{\Bbb{R}}(x^{\frac{1}{p}}d\lambda(x))^q = \int_0^1x^{\frac{q}{p}}d\lambda(x)$ $?= \infty$

I'm not quite sure how to show this. If anyone can help that would be great!

答:
Here is something stronger (adapted from Exercise 6.6 in Folland, 1999, p. 187):

For any given $p\in(0,\infty]$, there exists some $f:(0,\infty)\to\mathbb R$ such that $f\in L^p$ but $f\notin L^q$ for any $q\in(0,\infty]$ other than $p$.

Consider functions of the form $$f(x)\equiv\frac{|\ln x|^b}{x^a}\chi_I(x)+\frac{|\ln x|^d}{x^c}\chi_J(x)\quad\text{for $x\in(0,\infty)$}$$ with suitably chosen $a,b,c,d\in\mathbb R$ and intervals $I,J\subseteq(0,\infty)$.


问:
Understanding what $H^{-1}$ means when $H$ is a subset and when $H$ is a group.

A non-empty subset $H$ of a group $G$ will be a subgroup iff $HH^{-1}=H$

In this theorem, I don't understand what $H^{-1}$ means in both cases when $H$ is a subset and when $H$ is a group

Suppose $H$ is any subset. What does the notation $H^{-1}$ means?
please explain with an example

Suppose $H$ is a group. What does the notation $H^{-1}$ means?
please explain with an example

答:
$H^{-1}$ is the set of all inverses of elements in $H$ i.e. $H^{-1}=\{g^{-1}| g\in H\}$ of course, this notation only makes sense when $H\subset G$ for some group $G$. Say we take $G=\mathbb{Z}$ and $H=\{1,2,3\}$ then $H^{-1}=\{-1,-2,-3\}$

When $H$ is a group $H^{-1}$ is again all of the inverses of elements in $H$. So if we again take $G=\mathbb{Z}$, the additive group over the integers. and $H=2\mathbb{Z}=\{2n| n\in \mathbb{Z}\}$ then we have $H^{-1}=\{-2n| n\in\mathbb{Z}\}=H$


答:
$MM^{-1}(z)={a{dz-b\over -cz+a}+b\over c{dz-b\over -cz+a}+d}$ = ${{adz-ab\over -cz+a}+b}\over{{-cdz+cb\over -cz+a}+a}$

$MM^{-1}(z)=\frac{(adz - ab) + b(-cz + a)}{(-cdz+cb) + a(-cz + a)} = \frac{(ad - bc)z}{ad -bc}$

Would this be okay for proving the inverse? :)

Title: Show that the Mobius transformations form a group.

Body:
I'm wondering if using the 4 statements below to help prove that Mobius Transformations form a group is enough. If so, how can I use M(z)=(az+b)/(cz+d) to do that?

Thank you in advance, and please feel free to fix any formatting as I'm not great with that. :)


问:
Interpretation of costates in an optimal control problem
Consider the optimal control problem
\begin{gather}
\max_{u}\int_{t_{0}}^{t_{1}} f(t,x_{1},x_{2},u)\, dt \nonumber \\
\begin{aligned}[t]\label{sys:bkgrd_OC_problem}
&\text{subject to:}\quad x_{1}'(t) = g_{1}(t,x_{1},x_{2},u),\quad x_{1}(t_{0}) = c_{1} \\
&\phantom{\text{subject to:}\quad} x_{2}'(t) = g_{2}(t,x_{1},x_{2},u),\quad x_{2}(t_{0}) = c_{2}
\end{aligned}
\end{gather}

where $f,g$ are sufficiently smooth.  The typical solution via Pontryagin's Maximum Principle involves defining a Hamiltonian $$H = f(t,x_{1},x_{2},u) + \lambda_{1}g_{1}(t,x_{1},x_{2},u) + \lambda_{2}g_{2}(t,x_{1},x_{2},u)$$
where the costates $\lambda_{1}$ and $\lambda_{2}$ satisfy
\begin{align*}
\lambda_{1}'(t) &= -\frac{\partial H}{\partial x_{1}},\quad \lambda_{1}(t_{1}) = 0\\
\lambda_{2}'(t) &= -\frac{\partial H}{\partial x_{2}},\quad \lambda_{2}(t_{1}) = 0.
\end{align*}

A common interpretation of $\lambda_{1}(t)$ and $\lambda_{2}(t)$ are that they represent the marginal values of the states $x_{1}$ and $x_{2}$ at time $t$, respectively.

So, with that in mind, here is my question:

If we modify the problem so that we replace $f(t,x_{1},x_{2},u)$ with $f(t,x_{1},u)$, i.e. remove the explicit dependence of the objective functional on $x_{2}$, but leave everything else the same, does this change how we interpret $\lambda_{1}(t)$ and $\lambda_{2}(t)$?  If so, what is the new interpretation?  I haven't been able to find a reference that addresses this sort of interpretation question for anything but the simplest one state/one control types of optimal control problems.

答:
Thinking in terms of units:
$$[\text{unit of } \lambda_i] = \dfrac{[\text{unit of } f] \times [\text{unit of } t]}{[\text{unit of } x_i]} = \dfrac{[\text{unit of Cost}] }{[\text{unit of } x_i]}, \quad \quad i = 1,2$$

The unit of $f$ does not depend on its inputs. Therefore, changing the number of arguments will not change the interpretation of $\lambda$ in terms of its units.

However, two linearly independent Lagrange terms, $f_a(t, x_1, x_2, u)$ and $f_b(t, x_1, u)$, will always yield different solutions of state, control and costate, even if their units are the same.


问:
Inverse Laplace Transform, am stuck simplifying
I want to find the inverse Laplace transform of $F(s)= \frac{4s}{(s^2+5)^2-4s^2}$, but I'm having trouble simplifying this fraction so I'm able to take the Laplace transform.

I simplified the denominator and got $s^4+6s^2+25$, but I'm not sure how that helps me.

I also tried looking at $4s^2$ and $(2s)^2$, but still am stuck.

Any suggestions?

Body:
Hint

$$\frac{4 s}{\left(s^2+5\right)^2-4 s^2}=\frac{4s}{(s^2+5+2s)(s^2+5-2s)}=\frac{As+B}{s^2+5+2s}+\frac{Cs+D}{s^2+5-2s}$$


问:
Is $\langle f, h \rangle_{\omega}$ real on a Kaehler manifold with Kaehler metric $\omega$?
Let $X$ be a Kaehler manifold equipper with a metric $\omega$. Suppose $f, h$ are smooth real functions on $X$, then $\int_{X} h \Delta_{\omega} f \frac{\omega^{n}}{n!} = -\int_{X} g^{i \bar j} h_{i} f_{\bar j} \frac{\omega_0^n}{n!} = -\int_{X} \langle h, f \rangle_{\omega} \frac{\omega^n}{n!}$ using integraion by parts. $\Delta_{\omega} f$ is defined to be $g^{i \bar j} \partial_{i} \partial_{\bar j} f$ which should be real(unchanged under complex conjugation). This is suggesting that $\langle h, f \rangle_{\omega}$ should be real as well. But why is that the case?

I computed $\overline{\langle h, f \rangle_{\omega}} = \overline{ g^{i \bar j} h_{i} f_{\bar{j}}} = g^{\bar i j} h_{\bar i} f_{j}$ which does not seem to be equal to $\langle h, f \rangle_{\omega}$.

答:
There are several issues here.

First, what do you mean by $\langle h,f\rangle_{\omega}$? When $f$ and $h$ are scalar functions, the pointwise inner product $\langle h,f\rangle_{\omega}$ usually just means their product $hf$ (or, if they're complex-valued, $h\overline f$). But that is not what should appear in this integration-by-parts formula. Instead, it should involve first derivatives of $h$ and $f$ (see below).

Next, in order for a formula like this to hold, you need to assume either that $X$ is compact (without boundary) or that $f$ and $h$ are compactly supported.

There are actually three different formulas like this on a Kähler manifold, corresponding to the fact that the three Laplacians $\Delta = d^*d$, $\Delta_{\partial} = \partial^* \partial$, and $\Delta_{\overline\partial} = \overline\partial{}^*\overline\partial$ are all equal up to constant multiples:
$$
\Delta = 2 \Delta_{\partial} =  2\Delta_{\overline\partial} .
$$
Correspondingly, there are three integration-by-parts formulas:
\begin{align*}
\int_X h \Delta f\, dV_\omega &= \int_X \langle dh,df\rangle_\omega dV_\omega,\tag{1}\\
\int_X h \Delta_\partial f\, dV_\omega &= \int_X \langle \partial h,\partial f\rangle_\omega dV_\omega,\tag{2}\\
\int_X h \Delta_{\overline\partial} f\, dV_\omega &= \int_X \langle {\overline\partial} h,{\overline\partial}f\rangle_\omega dV_\omega.\tag{3}
\end{align*}
(Here I've written $dV_\omega = \omega^n/n!$. Note also that these Laplacians differ by a sign from the one you denoted by $\Delta_\omega$, which accounts for the lack of negative signs on the right-hand sides.) When $f$ and $h$ are real, these right-hand sides can be written in holomorphic coordinates as follows:
\begin{align*}
 \langle dh,df\rangle_\omega &= g^{i\bar j} \left(h_i f_{\bar j} + h_{\bar j} f_i\right),\\
\langle \partial h,\partial f\rangle_\omega &=
g^{i\bar j}  h_i f_{\bar j},\\
\langle {\overline\partial} h,{\overline\partial}f\rangle_\omega &=
g^{i\bar j}  h_{\bar j} f_i.
\end{align*}
The first of these expressions is real, but the other two need not be.

Importantly, the fact that the right-hand side of an integral formula like (2) or (3) is real does not imply that the integrand is real. Just to emphasize the point, consider the real-valued functions on the torus $\mathbb C/\mathbb Z^2$ given by $h(x+iy) = \sin 2\pi x$ and $f(x+iy) = \sin 2\pi y$. Then $h\Delta_{\partial} f = \tfrac12 h\Delta f$ is certainly real, but using the Euclidean metric, for which $g^{1\bar 1} = 2$,
$$
g^{i\bar j} h_i f_{\bar j} =
g^{1\bar 1} \partial_z h \,\partial _{\bar z} f =
2\pi^2 i \cos(2\pi x)\cos(2\pi y),
$$
which is not real.


问:
Prove a necessary condition for the irreducibility of a polynomial over F_p
Let $K=\mathbb{F}_p$, with $p$ prime, and $f(x)\in K[x]$ be a polynomial of degree $n>0$. Show that if $f(x)$ is irreducible then P has no roots in $\mathbb{F}_{p^d}$ with $d\le \frac{n}{2}$.

答:
As $p$ is supposed to be irreducible, and $a$ is a root of $p$ in $\mathbb F_d$ we have

$$d=[\mathbb F_{p^d}: \mathbb F_p]=[\mathbb F_{p^d}:\mathbb F_p(a)][\mathbb F_p(a):\mathbb F_p]\ge n.$$

Hence $p$ can’t have a root in $\mathbb F_{p^d}$ if $d \le n/2$.


问:
Calculating the matrix exponential in this case?
I am wondering how in general one can calculate the matrix exponential. For example, I am working in the Lie algebra $\mathfrak{su}(3)$. I have two matrices $$ m_1 = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ and $$m_4 = \begin{pmatrix} 0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$ Their linear combination $\alpha m_1 + \beta m_4$ can be written as $$ X := \alpha m_1 + \beta m_4 = \begin{pmatrix} 0 & - \overline{z} & 0 \\ z & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ where $z = \alpha + i \beta$.

Now I wish to obtain a closed formula (if possible) for the matrix exponential $e^{tX}$ (where $t$ is some real number), perhaps in terms of sines and cosines? How does one in general calculate the matrix exponential like this?

答:
For an even more clear approach, lets work with $2$-by-$2$ matrices. Let $$Y =\begin{pmatrix} 0 & - \overline{z}  \\ z & 0  \end{pmatrix}$$

Notice that $Y^2 = -|z|^2I$ so $Y^{2k} = (Y^2)^k = (-|z|^2)^kI$ and we find
\begin{align*}
\exp(tY) &= \sum_{n=0}^{\infty}\frac{(tY)^n}{n!}\\
&= \sum_{k=0}^{\infty}\frac{(tY)^{2k}}{(2k)!} + \sum_{k=0}^{\infty}\frac{(tY)^{2k+1}}{(2k+1)!}\\
&= \sum_{k=0}^{\infty}\frac{t^{2k}(-|z|^2)^k}{(2k)!}I + \sum_{k=0}^{\infty}\frac{t^{2k+1}(-|z|^2)^k}{(2k+1)!}Y\\
&= \sum_{k=0}^{\infty}\frac{(-1)^k(t|z|)^{2k}}{(2k)!}I + \sum_{k=0}^{\infty}\frac{(-1)^k(t|z|)^{2k+1}}{(2k+1)!}\frac{Y}{|z|}\\
&= \cos(t|z|)I + \frac{\sin(t|z|)}{|z|}Y\\
&=\begin{pmatrix} \cos(t|z|) & -\frac{\overline{z}}{|z|}\sin(t|z|)  \\ \frac{z}{|z|}\sin(t|z|) & \cos(t|z|)  \end{pmatrix}
\end{align*}
We can easily extend this to the matrix $X$ that you gave:

$$\exp(tX) =\begin{pmatrix} \cos(t|z|) & -\frac{\overline{z}}{|z|}\sin(t|z|)& 0  \\ \frac{z}{|z|}\sin(t|z|) & \cos(t|z|)& 0 \\ 0 & 0 & 1 \end{pmatrix}$$


问:
Question about proving existence of a function $f$ such that $f \circ f = g$ for an odd function $g$
I have been reading a book on elementary mathematics and have come to a problem where I don't understand where the solution comes from. The problem goes :

"Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be an odd function, such that $g(x) > 0$
whenever $x > 0$. Show that there exists a function $f : \mathbb{R} \rightarrow \mathbb{R}$
for which $g = f \circ f$."

In the answer key they just say :

"Take $f(x) = -g(x)$ for $x \leq 0$ and $f(x) = -x$ for $x \geq 0$."

I am wondering what the reasoning is behind this answer. The truth is that I just want to see a more detailed solution to the problem. Can anyone help with this ?

答:
Interesting question, as an example consider the identity function for $g$, then $f$ is either trivially the identity function or more interestingly the negative the identity function (or more generally Babbage's functional equation, but that takes us off topic). What about $f(x)=x^3$? Well it's just as easy to solve the general case.

Now consider $f\circ f(x)=g(x)$ the naive attempt would be to let $f(x)=g(x)$ but then things wouldn't workout, so to fix the issue we need to define $f(x)=x$ sometimes, and $f(x)=g(x)$ other times. If we let $f(x)=x$ if $x>0$ then we get stuck because $f\circ f(x)=x\neq g(x)$ if $x>0$, so to avoid getting stuck we let $f(x)=-x$ if $x>0$, and once we do that we know that $f(x)$ must equal $g(-x)$ if $x<0$. so that $f\circ f(x)=g(-(-x))=g(x)$ if $x>0$, or $f\circ f(x)=-g(-x)=g(x)$ if $x<0$.(Since $g$ is odd)

Notice the solution is not unique, since we could have defined $f(x)=-x$ if $x<0$ then we would define $f(x)=-g(x)$ if $x<0$. Are there more?


问:
Construct a matrix with the specified characteristics
I am taking a signals processing class online and we are covering some basic linear algebra. Here is a question that I'm struggling with from the class.

Find a matrix $A$ which satisfies all of the following characteristics:

$\text{dim }\text{col}(A) = 2$

$\begin{bmatrix} 1 & 1 & 2 & -1\end{bmatrix}^T \in \text{null}(A)$

$\begin{bmatrix} 3 & 3 & 6 \end{bmatrix}^T \in \text{col}(A)$

$\begin{bmatrix} 3 & 0 & 0 & 3\end{bmatrix}^T \in \text{col}(A^T)$

Attempt

So far, I've started by making the first two columns of my matrix $A$ independent.

$$\begin{bmatrix}
1 & 0 & \alpha_1 & \beta_1 \\
1 & 0 & \alpha_2 & \beta_2 \\
0 & 1 & \alpha_3 & \beta_3
\end{bmatrix}$$

From this I can clearly fulfill the requirement that $\begin{bmatrix} 3 & 3 & 6 \end{bmatrix}^T \in \text{col}(A)$. Then I start filling in the $\alpha_i$ and $\beta_i$ values to fulfill the other characteristics.

The problem is, whenever I manage to satisfy a couple of the characteristics, I break the third one. I know how to individually satisfy these characteristics, but not how to satisfy them together. How can I go about constructing a matrix that satisfies all these characteristics?

答:
The last condition says that the matrix has a row
$$
[3 \; 0\; 0 \; 3].
$$
Notice that this row's dot product with the generator of your nullspace, i.e. with the vector
$$
\nu = [1 \; 1 \; 2 \; −1]
$$
is zero.  Furthermore, the same is true of a row that is either a scalar multiple of the above row, or has the form
$$
[3\; \alpha\; \beta \; 3].
$$
where $[\alpha \; \beta] \cdot [1 \; 2] = 0$.

Furthermore, the requirement that the rank of the matrix is 2 can be met by making two of the three rows linearly dependent.  So, let's make the rows of our matrix:
$$
[3 \; 0\; 0 \; 3],
$$
$$
[3 \; \alpha\; \beta \; 3],
$$
and, finally, the row that equals twice the first row:
$$
[6 \; 0\; 0 \; 6].
$$
All that is left to do is choose suitable nonzero values for $\alpha, \beta$.

Each of these rows, when dotted with our desired nullspace generator $\nu$, gives zero, hence $\nu$ is indeed in the nullspace, as required.  The rank of the resulting matrix is 2 as long as $\alpha, \beta$ are not both zero.  And the first (and also the last) column satisfies the 3rd bulleted requirement.


答:
For a measure space $(X, \mu)$ and $1\leq p<q\leq \infty$, we have $L^p(X, \mu)\not\subset L^q(X, \mu)$ if and only if $(X, \mu)$ has sets of arbitrarily small, positive measure and $L^q(X, \mu) \not\subset L^p(X, \mu)$ if and only if $(X, \mu)$ has sets of arbitrarily large finite measure.

I would recommend to not focus on $(\mathbb{R}, \lambda)$ and prove the above paragraph instead, especially since it isn't that much more difficult.

Title: For $1 \le p \lt q \le \infty$ show that neither $L^p(\Bbb{R},\lambda)$ or $L^q(\Bbb{R},\lambda)$ contain the other

Body:
Let $1 \le p \lt q \le \infty$, and $\lambda$ denote the Lebesgue measure. Show that $L^p(\Bbb{R},\lambda) \not\subset L^q(\Bbb{R},\lambda)$ and $L^q(\Bbb{R},\lambda) \not\subset L^p(\Bbb{R},\lambda)$ by constructing functions where each belongs to one but not the other.

Hint: You will likely want to handle $q = \infty$ separately.

My idea:

After some searching online, I came across this suggestion:

For the first non-inclusion, consider $f(x) = \begin{cases} x^{\frac{1}{p}} &\text{if } 0 \le x \le 1 \\ 0 &\text{o.w.} \end{cases}$

Here is my attempt to use this function:

Then $\int_{\Bbb{R}}(x^{\frac{1}{p}}d\lambda(x))^p = \int_0^1xd\lambda(x) \lt \infty$

And$\int_{\Bbb{R}}(x^{\frac{1}{p}}d\lambda(x))^q = \int_0^1x^{\frac{q}{p}}d\lambda(x)$ $?= \infty$

I'm not quite sure how to show this. If anyone can help that would be great!


问:
What is the discriminant of a quadratic extension over a number field?
Let $K$ be a number field and $d \in \mathcal{O}_K \setminus \mathcal{O}_K^2$. What is the discriminant of the extension $K[\sqrt{d}]/K$ ? Do we know its ring of integers and which primes are split or not ? (I know that these questions are solved when $K=\mathbb{Q}$, but I can't find any result otherwise).</p>

答:
At least when working theoretically, or by hand, it is probably easiest to solve this sort of question locally.  (I don't have any feeling for how computationally implemented algorithms for this sort of question work, and so won't comment any more on them.)

In short, fix a prime $v$ of $K$, and consider the completion $K_v$ and the local
extension $K_v[\sqrt{d}]/K_v$.  The discriminant of $K[\sqrt{d}]/K$ will be a product of the local discriminants, and so we are reduced to computing these.

How do we analyze $K_v[\sqrt{d}]/K_v$?  Well, you could first try to compute
the finite index subgroup $(K_v^{\times})^2$ of $K_v^{\times}$.  Note that
$K_v^{\times} = \mathcal O_{K_v}^{\times} \times \pi^{\mathbb Z}$, if $\pi$
is some choice of uniformizer, and so $(K_v^{\times})^2 =
(\mathcal O_{K_v}^{\times})^2\times \pi^{2\mathbb Z}$.

Computing $(\mathcal O_{K_v}^{\times})^2\subset \mathcal O_{K_v}^{\times}$
is straightforward.  E.g. if $v$ has odd residue characteristic, then
any element in $1 + \pi \mathcal O_{K_v}$ is a square, and so it is just
a question of computing the subgroup of squares in the residue field at $v$.
If $v$ has even residue characteristic, then it is still pretty easy in any
particular case to compute the square units, say using the fact that
any unit congruent to $1 \bmod 4\pi$ is a square (as one sees using
the binomial expansion for $(1+x)^{1/2}$).

Once you've done this step, you can easily check if $d$ is a square in $K_v$,
which tells you whether or not $K[\sqrt{d}]/K$ is split at $v$.

To compute the discriminant is not that much harder.  If $v$ has odd residue characteristic,
then we may divide $d$ through by even powers of $\pi$ so that it is either
a unit, or else is exactly divisible by $\pi$.  In the first case, $K_v[\sqrt{d}]/K_v$ will be unramified at $v$, and hence the contribution to
the discriminant from $v$ will be $1$.  In the second case, $K_v[\sqrt{d}]/K_v$ will be tamely ramified at $v$, and the contribution to the discriminant from $v$ will be one power of $v$.

If $v$ is of even residue characteristic, then the computations are more involved, because even if $d$ (perhaps after dividing through by even powers of
$\pi$) is a unit, it can still happen that $K_v[\sqrt{d}]/K_v$ is ramified
(consider the case $\mathbb Q_2(i)/\mathbb Q_2$), although it need not be
(consider the case $\mathbb Q_2(\sqrt{5})/\mathbb Q_2$).

If you would like more details, I can provide them.


问:
Vector Field, Scalar Field. Which is meaningful and not?
Let a, b, c be vectors, f(x, y, z) be a scalar ﬁeld, F(x,y,z) be a vector ﬁeld. Which of the
following expressions are meaningful?

I. (a×b)×(c×b)

II. |a|(b· c) +|a|(b+c)

III. ∇ ×(f F)

IV. (∇ ×F)·(f ×F)

I know that (I) makes sense. Vector x Vector.
(II) is not meaningful

I am confused about grad and curls. I know that div and curl apply to vector fields, grad to
scalar fields.

So. curl(f F). is (f F) scalar and is (f x F) vector?

答:
Let's call a vector $v$ and a scalar $s$. Then we have:

$v\times v\quad\checkmark$

$s\cdot s + s\cdot v\quad \Rightarrow\text{nonsense}$

$v\times (s\cdot v)\quad\checkmark$

$(v\times v)\cdot (s\times v)\quad \Rightarrow\text{nonsense}$


问:
Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

Find the intersection of surfaces $^2 + ^2 = 2$ and $^2 + ^2 + ^2 = 8$. What kind of curve is it?

What I supposed to do? In 1) I need to find some transformation, which I don't know. In second I just tried to substitute equation variables one into another? but still not sure.

答:
$x²+y²-4x+3=0$ =>
$x²-4x+4+y²=0=(x-2)²+y²=1$;
it is equation of circle which centre in O(2;0; 0) and radius R=1. After rotation around y-axis we will have sphere (it's obvious, we will have a GLP that equidistant from O, what is the sphere by the definition); radius of sphere R=1; centre of sphere O. So we can write the equation of this sphere:
$(x-2)²+y²+z²=1²$.

We can find set of points of intersection just by solving the system of equations (SOE):
$x²+y²=2z$; $x²+y²+z²=8=2z+z²=8$ =>
$z²+2z-8=(z+4)(z-2)=0$; so $z=-4$ or $z=2$.
What about $z=-4$ we can see that $x²+y²=2z=-8$ impossible, so it doesn't satisfy SOE;
If $z=2$ then $x²+y²=2z=4$ => $x²+y²=4$; it's the only set of points of intersection and it's the equation of circle with centre O(0; 0; 4), radius R=2 and located in the plane $z=4$.


答:
$$
\begin{cases}
x^2+y^2=2 z\\
x^2+y^2+z^2=8\\
\end{cases}
$$
is the equation of the circle with center $(0,0,2)$ and radius $2$ in the plane $z=2$, intersection of the paraboloid $2z=x^2+y^2$ and the sphere $x^2+y^2+z^2=8$.

Indeed subtracting the two equation we get
$$z^2-2z-8=0\to z_1=2;\;z_2=-4$$
$z_2=-4$ must be discarded because $z\ge0$. So we have

$$
\begin{cases}
x^2+y^2=4\\
z=2\\
\end{cases}
$$
The curve is shown in the picture below

Body:
Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

Find the intersection of surfaces $^2 + ^2 = 2$ and $^2 + ^2 + ^2 = 8$. What kind of curve is it?

What I supposed to do? In 1) I need to find some transformation, which I don't know. In second I just tried to substitute equation variables one into another? but still not sure.


问:
Textbooks for learning: Lie algebra, topology, and differential geometry?
I'm a PhD student trying to learn the math required to understand Gauge theory. From some reading over the past few weeks, I think I need to learn the following:

Lie algebra / Lie groups

Algebraic and geometric Topology

Differential geometry

I only need understand the material enough to be literate (my research hopes only to use some results from these fields and not to contribute to them).
Does anyone have any good recommendations for textbooks that assume little mathematical knowledge beyond some undergrad. courses in linear algebra, vector calc., probability, and PDEs?

In particular, I'm hoping to find a textbook that assumes little prior knowledge while being somewhat brief its exposition (in my reading so far I'm finding myself getting lost in the volume of definitions).

Thanks for reading, I appreciate any suggestions!

edit: hardmath asked where I feel I've been getting lost in definitions. I've started reading "A Course in Modern Mathematical Physics" by Peter Szekeres (suggested by a friend). This seems like a great book and I've learnt a lot in the first few chapters but I'm struggling with the amount of terminology. For example, by chapter 3 my notes contained about 70 definitions.

答:
As I suggested in comments, "An Introduction to Manifolds" by Loring Tu is a good place to start. Points in favour:

From the preface the author wrote it with the rationale of providing "a readable but rigorous introduction that gets the reader quickly up to speed, to the point where for example he or she can compute the de Rham cohomology of simple spaces." He goes on to explain that there is a "self-imposed absence of point-set topology in the prerequisites... [His] solution is to make the first four sections of the book independent of point-set topology and to place the necessary point-set topology in an appendix." The intention is that the reader study the first four sections and Appendix A simultaneously.

The book is the first volume of a series, so if you need more than what is in this book then you might find it in another book by the same author. It is the prequel to the famous "Differential Forms in Algebraic Topology" by Bott and Tu, and there is also "Differential Geometry" by Tu (ostensibly the third book but written with only the contents of the "Introduction" as prerequisites).

It's a Springer Universitext book, which at the time of writing makes it pretty cheap as far as textbooks go.

Point against: a perhaps-more-standard set of references for similar material is the series of books by John M Lee. Lee's books make better references in part because they are written to be more encyclopaedic, which might mean that Tu's books are slightly more approachable to someone learning on their own.


问:
A bijection for counting commuting pairs in finite group
Let $G$ be a finite group and $C_G$ the set of conjugacy classes in $G$. Using the orbit-stabilizer theorem for the conjugation action, it can be found that the number of commuting pairs of elements in $G$, i.e. the size of the set
$$A_G=\{(x,y)\in G\times G, xy=yx\}$$
equals the size of $G$ times the size of $C_G$,
$$|A_G|=|G||C_G|.$$

This is surprisingly simple, and of course the right hand side counts the number of pairs containing one element from $G$ and one from $C_G$, i.e. the size of the set
$$B_G=\{(a,b), a\in G, b\in C_G\}.$$

Is there a nice bijection between sets $A_G$ and $B_G$ that works for all $G$?

答:
The answer is no in the following sense. Both the set of commuting pairs $\text{Hom}(\mathbb{Z}^2, G)$ and the set $G \times C_G$ are naturally acted on by the automorphism group of $G$, and in general it's not possible to find a bijection which is $\text{Aut}(G)$-equivariant.

For example, take $G = S_3$, the smallest nonabelian group. Then $\text{Aut}(G) \cong S_3$ acting by conjugation. $G$ has $3$ conjugacy classes, all of which are fixed by $\text{Aut}(G)$, so $G \times C_G$ splits up under the action of $\text{Aut}(G)$ into $3$ sets of $3$ orbits, of sizes $1, 1, 1, 2, 2, 2, 3, 3, 3$.

On the other hand, there's only one commuting pair of elements which is fixed by $\text{Aut}(G)$, namely the identity and the identity. So the two don't have the same fixed points.


问:
Integral of $\int \frac{\sin(x)dx}{3-\cos(x)}$
I am trying to solve this integral and I need your suggestions.

I don't know if its OK to set $3-\cos(x)$ as $t$ $\rightarrow dt = \sin(x)dx$ or just take $\cos(x)$ and set it as $t$
$$\int \frac{\sin(x)dx}{3-\cos(x)}$$
Thanks

答: Yes, making a substitution is the way to go, if you let $t=3-\text{cos}(x)$ then we have $dt=\text{sin(x)}dx$ which implies we have$$\int\frac{\text{sin}(x)}{3-\text{cos}(x)}dx=\int\frac{\text{sin}(x)}{t}\frac{dt}{\text{sin}(x)}=\int\frac{1}{t}dt$$I think you can continue from here.</p>


问:
Optimization Question Finding closest point
Find the point on the graph of the function that is closest to the given point.

The function: $f(x)= \sqrt {x-8}$; the point: $(15,0)$

答: Well, actually you can deal with this problem even without differentiating. You just expected to find the distance from any point on the graph of f(x) to (15; 0), minimize it and find out where the minimum is reached. (It can be done without using of derivative).</p>


问:
Deviation from average
Say I have a finite list of numbers of some statistical experiment, what is the best way to represent the typical distance from the mean of the numbers based on this data? I have found the standard deviation, but there is also average absolute deviation. How do they differ?

Thanks in advance

答:
Suppose a population is normally distributed with unknown mean $\mu$ and unknown variance $\sigma^2.$ Also suppose you have a random sample of size $n$ from this population. Then the 'best' estimator of the population mean $\mu$ is the sample mean $\bar X = \frac 1n \sum_{i=1}^n X_i.$ Also, the usual estimator of the population variance $\sigma^2$ is the sample variance $S^2 \ \frac{1}{n-1}\sum_{i=1}^n (X_i = \bar X)^2.$

By 'best' above, I mean that these estimators are unbiased: $E(\bar X) = \mu$ and $E(X^2) = \sigma^2.$ Moreover, among unbiased estimators, each of these has the smallest possible variance. There are other criteria for 'goodness' of estimators, and they lead to other estimators of $\sigma^2.$ But $\bar X$ and $S^2$ are the most commonly used ones.

In situations where the population distribution is not normal or is unknown in shape, other estimators of the center or dispersion may be used. For example, if the population is Laplace (double-exponential), then it may be best to estimate the
population mean (which, by symmetry, is also the population median) may be the sample median. Moreover, there are situations in which the mean absolute deviation (from the mean) $\frac 1n \sum_{i=1}^n |X_i - \bar X|$ may be a better way to measure the dispersion of the population distribution. (Other way to describe the dispersion of a population or sample are also in use.)

In any discussion of alternative estimators, there must be a clear definition of what is meant by 'best' before meaningful choices can be made among estimators.

Title: Deviation from average


问:
Zero divisors in crossed group rings
It is not difficult to see that a group ring $K[G]$ ($K$ a domain) has non-trivial zero-divisors whenever there exists a non-trivial torsion element $g\in G$. [In fact, in this case, $1-g$ is such a non-trivial zero-divisor]

My question is the following: is it possible that a crossed product $K*G$ is a domain even if $G$ has non-trivial torsion (e.g., what happens when $G$ is finite)?

My guess is that $K*G$ may very well be a domain. In fact, if $G'$ is a group such that $K*G'$ is a domain and $G=G'/H$ is quotient of $G'$ (I'm thinking for example to the case $G'=\mathbb Z$ and $G=\mathbb Z(p)$), then $D=K*H$ is a domain and so $K*G'=D*(G'/H)=D*G$ cannot have non-trivial zero divisors. It seems that this example answers my question but in a very indirect way... do you have any straightforward reason for which this phenomenon may happen? Thank you in advance.

答:
well, it is not really an answer but an illuminating example: the ring of complex numbers $\mathbb C$ can be viewed as a crossed product ring $\mathbb R*C_2$. In fact, the elements of $\mathbb C$ can be viewed as pairs $(r,s)$, with $r,s\in \mathbb R$, with the sum defined component-wise and multiplication as follows:
$$(r_1,s_1)(r_2,s_2)=(r_1r_2-s_1s_2, r_1s_2+s_1r_2)\,.$$
The unique difference with the multiplication in the group ring $\mathbb R[C_2]$ consist in a $-1$, in fact, the multiplication in this second ring is
$$(r_1,s_1)(r_2,s_2)=(r_1r_2+s_1s_2, r_1s_2+s_1r_2)\,.$$
Anyway, from the point of view of the zero divisors this makes an enormous difference, $\mathbb R[C_2]$ has (infinitely many) zero divisors, while $\mathbb C$ is a field:
$$\text{In }\mathbb R[C_2]:\ \ \ \ (1,1)(1,-1)=(1-1,1-1)=(0,0)\,,$$
$$\text{In }\mathbb C:\ \ \ (1,1)(1,-1)=(1+1,1-1)=(2,0)\neq 0\,.$$</p>

Title: Zero divisors in crossed group rings


问:
Show that $\int_0^{c} \frac{x^{k-1}}{1-x^m}=\sum_{n=0}^{\infty}\frac{c^{nm+k}}{nm+k}$
$$\int_0^{c} \frac{x^{k-1}}{1-x^m}=\sum_{n=0}^{\infty}\frac{c^{nm+k}}{nm+k}$$

where $k,m \in \mathbb{N}^{\times}:= \mathbb{N} \setminus \mbox {{0}}, c\in(0,1) $

Can I maybe use here a speciel series? I stil don't know at which point should I start.

Body:
Hint:
$$ \frac{1}{1-x} = \sum_{i=0}^{\infty} x^i$$

$$ x \to x^m$$

$$ \frac{1}{1-x^m} = \sum_{i=0}^{\infty} x^{im}$$

Multiply this by $x^{k-1}$,

$$ \frac{x^{k-1} }{1-x^m}  = \sum_{i=0}^{\infty} x^{k+im-1}$$


问:
Divergent sequence with only one accumulation point
I'm trying to  find some examples of divergent sequences that have a unique accumulation point (in the context of $R^n$, with the usual topology). I know that if a sequence converges, then it has a unique accumulation point, but the inverse sentence is not true. I found one using $n$ and $1/n$, but I can't find another example.

答: Consider the sequence $f(n) = 1$ for odd $n$ and $f(n) = n $ for even $n$.


问:
How can I find the generators of $\mathbb C[t]$ as a $\mathbb C[t+t^2]$ module
For every $f(t)=a_nt^n+...+a_1t+a_0$ how can i write this as $r_1g_1(t)+...+r_mg_m(t)$ for $r_i\in\mathbb C[t]$ and $g_i\in\mathbb C[t+t^2]$? I’m finding it impossible to find any $r_i$ and $g_i$ for a given polynomial like $f(t)=3t^5-2t^4+t^3-6$, so I’m not sure how to find $r_i$ such that they work for every polynomial. Is there even a way?

I’m trying to solve a larger problem which is to show that $\mathbb C[t]$ is integral over $\mathbb C[t+t^2]$, which our lecturer has shown how to do for other examples, and firstly finds the generators

答:
$1,t$ suffices for a basis of $\Bbb C[t]$ as a $\Bbb C[t+t^2]$ module. You can prove this by induction on the degree of a polynomial: the base case of degree zero and degree one are obvious, so let $d>1$ and assume that you've shown that every polynomial of degree $d-1$ or less can be written as a $f(t^2+t)+g(t^2+t)t$. If $p$ is of degree $d$ with top term $at^d$, then $p-at^{d-2}(t^2+t)$ and $at^{d-2}$ are of lower degree and can be written appropriately by the induction hypothesis. So $at^{d-2}(t^2+t)$ can as well, and therefore $p=(p-at^{d-2}(t^2+t))+at^{d-2}(t^2+t)$ can be written as a $\Bbb C[t^2+t]$-linear combination of $1$ and $t$.

Alternatively, one can check that $\Bbb C[t]=\Bbb C[t^2+t][t]$, and $t$ satisfies the degree-two monic polynomial $X^2+X-(t^2+t)$, which is a direct proof of integrality.


问:
Prove that if $A$ is a symmetric orthogonal matrix, then $1$ and $-1$ are the only eigenvalues.
Since $A$ is symmetric , it is orthogonally diagonalizable $\implies $ $A$ is similar to some diagonal matrix $D$. $\implies$ $\det(A) = \det(D)$

Since $A$ is orthogonal $\implies \det(A) = \pm1 \implies \det(D) = \pm1$

I know that $A$ and $D$ have the same eigenvalues and the same trace.

I am not sure what the next step is.

答:
$A$ is symmetric orthogonal matrix
so we can use one of the following :
$A^{T}=A$ and $A^{T}=A^{-1}$ or $A^{T}A=AA^{T}=I$.

Now let $k$ be an eigenvalue of $A$. Then
$Ax=kx$ with $x$ is different from zero.

$x = Ix = (A^{-1}A)x = A^{-1}(Ax)=A^{-1}(kx) =k(A^{-1}x) = k(Ax) = k^{2}x$

Then $x=k^{2}x \implies (k^2 -1)x=0 \implies k^2 -1=0 \implies k= \pm 1$.
proved :)


答:
Here's the reasoning: Since $g$ is odd, we really only need to know how it maps positive numbers to know all of $g$. We set up $f$ so that the first application gets the correct $g$ value, if you ignore the sign, and the second application fixes the sign. At least that's what it does for negative numbers. For positive numbers, it does roughly the same two steps, but in the other order.

It's pretty clever -- it's as if $f$ uses the sign to both encode and decide whether or not $g$ has applied yet.

Title: Question about proving existence of a function $f$ such that $f \circ f = g$ for an odd function $g$

Body:
I have been reading a book on elementary mathematics and have come to a problem where I don't understand where the solution comes from. The problem goes :

"Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be an odd function, such that $g(x) > 0$
whenever $x > 0$. Show that there exists a function $f : \mathbb{R} \rightarrow \mathbb{R}$
for which $g = f \circ f$."

In the answer key they just say :

"Take $f(x) = -g(x)$ for $x \leq 0$ and $f(x) = -x$ for $x \geq 0$."

I am wondering what the reasoning is behind this answer. The truth is that I just want to see a more detailed solution to the problem. Can anyone help with this ?


问:
Evaluating $I(n) = \int^{\infty}_{0} \frac{\ln(x)}{x^n(1+x)}\, dx$ for real $n$
I am not sure how to handle the additional parameter $n$. I first need to find out for which real values of $n$ will the integral converge. Based on intuition and checking with mathematica, I believe it will converge only for $0 < n < 1$, although I am not sure and I would like to see exactly how to determine those values. For this integral, I would normally try using a branch cut, but again I am not sure how to apply it given my additional parameter $n$.
$$I(n) = \int^{\infty}_{0} \dfrac{\ln(x)}{x^n(1+x)}\, dx$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{I}\pars{n} & \equiv
\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n}\pars{1 + x}}\,\dd x\,\right\vert_{\,\Re\pars{n}\ <\ 1}} =
\left.\partiald{}{\nu}\int_{0}^{\infty}
{x^{\color{red}{\nu - n + 1} - 1} \over
1 + x}\,\dd x\,\right\vert_{\,\nu\ =\ 0}
\end{align}
Note that
$\ds{{1 \over 1 + x} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}}
\,{\pars{-x}^{k} \over k!}}$.

Then,
\begin{align}
\on{I}\pars{n} & \equiv
\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n}\pars{1 + x}}\,\dd x\,\right\vert_{\,\Re\pars{n}\ <\ 1}}
\\[5mm] = &\
\left.\partiald{}{\nu}\,
\Gamma\pars{\color{red}{\nu - n + 1}}
\Gamma\pars{1 - \bracks{\color{red}{\nu - n + 1}}}
\,\right\vert_{\,\nu\ =\ 0}\quad
\pars{\substack{\ds{Ramanujan's}\\[0.5mm]\ds{Master}
\\[0.5mm]\ds{Theorem}}}
\\[5mm] = &\
\left.\partiald{}{\nu}\,
{\pi \over \sin\pars{\pi\bracks{\nu - n + 1}}}
\,\right\vert_{\,\nu\ =\ 0} =
\left.-\pi\,\partiald{\csc\pars{\pi\bracks{\nu - n}}}{\nu}
\,\right\vert_{\,\nu\ =\ 0}
\\[5mm] = &\
\left.\pi^{2}\csc\pars{\pi\bracks{\nu - n}}
\cot\pars{\pi\bracks{\nu - n}}
\,\right\vert_{\,\nu\ =\ 0}
\\[5mm] = &\
\bbx{\pi^{2}\csc\pars{n\pi}\cot\pars{n\pi}} \\ &
\end{align}


问:
Asymptotics of convolution of a series given the asymptotics of the series
so I have a series of real numbers $(q_n)_{n\in\mathbb N}$ ($q_i \in [0,1]$) depending on some $\alpha \in (0,1/2)$ and what I know is the following, there exists constants $c_1, c_2$ so that:

$q_i \sim c_1 i^{\alpha -1}$ for $i\rightarrow \infty$

and

$\sum_{i=0}^k q_i \sim c_2 k^{\alpha}$ for $k\rightarrow \infty$

holds.

I would be interested if there is something like a Tauberian Theorem that could give one something like

$\sum_{i\geq 0} q_i q_{i+j} \sim c_3j^{2 \alpha -1} $ for $j\rightarrow \infty$

for some constant $c_3$ or something like that? It would be even better to get asymptotic results for
$\sum_{i=0}^N q_i q_{i+j}$?  Maybe there's also something?

Would be really glad if someone has a hint or an idea!

All the best!

Body:
If $q_k \sim C k^{-r}$ with $r\in (1/2,1),C\ne 0$ then $$\sum_{k\ge 1}q_k q_{k+j} =
O(\sum_{k=1}^{\log j} k^{-r} (k+j)^{-r})+
\sum_{k>\log j} C^2 k^{-r} (k+j)^{-r}(1+o(1))$$
$$= O( (\log j)^{1-r}j^{-r} )+(C^2+o(1))\sum_{k>\log j} k^{-r} (k+j)^{-r}$$
Then $\sum_{k>\log j} k^{-r} (k+j)^{-r}>\sum_{k>\log j}^{j} k^{-r} (2j)^{-r}\sim (2j)^{-r}\frac{ j^{1-r}}{1-r}$ so

$$\sum_{k\ge 1}q_k q_{k+j} = (C^2+o(1))\sum_{k>\log j}  k^{-r} (k+j)^{-r} \sim C^2\sum_{k>1} k^{-r} (k+j)^{-r}$$
I don't know how to find the asymptotic of the latter series, but it is close to $j^{1-2r}$.


问:
How to solve for $x$ and $y$ given these two multivariable functions?
I am trying to find the critical points of this function $(x-y)(16-xy)$

I found that the partial derivatives are $f_x = 16-2xy+y^2 = 0$ and $f_y=-x^2-16+2xy = 0$. However, I have tried everything but cannot solve for $x$ and $y$ with these two equations. How do you solve for $x$ and $y$?

答:
If you solve $f_x = 0$ for e.g. x you get
$x = \frac{16 + y^2}{2 y}$. Insert this into $f_y$ you get to:
$-8 - \frac{64}{y^2} + \frac{3 y^2}{4} = 0$
You can solve this e.g. with multiplying with $y^2$, substituting $y^2$ then and get two real solutions:
$y_1 = -4$, $y_2 = 4$. Inserting this back into your $f_x$ solution you get the corresponding $x$ solutions. Assuming you are just looking for real solutions.


问:
Basic question on Fubini's theorem, and solution of $\int \int_A 1 \ dx dy, \quad$ where $A = [3,3]\times\mathbb{R}^2$
Consider:
$\int \int_A 1 \ dx dy, \quad$ where $A = [3,3]\times\mathbb{R}^2$.

I know this is probably an extremely basic question, but I've forgotten everything from my multivariable calculus class many years ago, and I'm having trouble with some conceptual questions regarding the above problem.

First question: How does one evaluate this integral? Intuitively, I feel like it should be equal to 0 since you're basically taking the area of a line. And indeed:

$\int^\infty_{-\infty} \int^3_3 1 \ dxdy = \int^\infty_{-\infty} 0 dy = 0$.

But when I switched the order of integration, I obtained:

$\int^3_3 \int^\infty_{-\infty}  1 \ dy dx, \quad$ which does not converge.

Here I thought that this can't be right, and I know that in the one variable case, an infinite integral can be expressed as a limit. So I wrote the above again as:

$\lim_{a \rightarrow \infty} \lim_{b \rightarrow -\infty}\int^a_{b} \int^3_3 1 \ dxdy = \lim_{a \rightarrow \infty}\lim_{b \rightarrow -\infty} \int^3_3 \int^a_{b} 1 \ dydx = 0$.

But I'm not entirely convinced whether I can use this one-variable concept of limits in the multi-variable case, and furthermore that I can pull this limit out of the integral. What is the best and most rigorous way to evaluate this (or show that it does not exist)?

Second question: In a more general sense, I understand Fubini's theorem to imply that if

$\int \int_A f(x,y) \ dx dy, \quad$ exists, then assuming that you have the correct bounds for integration,

$\int \int f(x,y) \ dx \ dy = \int \int  f(x,y) \ dy \ dx$

If my understanding of Fubini's theorem is true (and if not please correct me), does the converse hold? i.e. if:

$\int \int f(x,y) \ dx \ dy = \int \int  f(x,y) \ dy \ dx$

then does

$\int \int_A f(x,y) \ dx dy, \quad$ exists?

I hope those questions made sense, and thank you in advance!

Note: I haven't yet learned Lebesgue integration, nor any measure theory, so if it possible to answer the question using only Riemann integrals, and no measure theory, that would be great!

答:
For your first question: over in Riemann-land, we define integrals over unbounded domains as the limit of integrals over bounded domains(this is not the case when working from a measure-theoretic standpoint). Thus, for any $C,D > 0$, let's consider $A_C^D = [3,3] \times [-C,D]$ (I'm assuming the $\mathbb{R}^2$ is a typo, as you seem to be using $\mathbb{R}$ in its place after the first line).

Then we have $$\int_{A}1 = \lim\limits_{C\to\infty}\lim\limits_{D\to\infty}\int_{A_C^D}1,$$ by definition.

So, let's evaluate the integral inside the limits. Since $1$ is continuous, we need only exhibit one sequence of Riemann sums that we can compute the limit of, and we can be sure that all such sequences converge to the same limit.

So, keeping life easy for ourselves, let's just use the constant sequence $A_n = \{A_C^D\}$. We obviously have $A_C^D = \bigcup_{B \in A_C^D}B$ for all $n$, and $\Delta A_C^D = 0$ for all $n$, so in particular, $\Delta A_n \to 0$ as $n \to \infty$.

Thus, we have $$\int_{A_C^D}1 = \lim\limits_{n\to\infty}1\left(\frac{C+D}{2}\right)\Delta A_C^D = \lim_{n\to\infty}0 = 0,$$ (where $1$ is the constant function and the brackets denote function evaluation) and hence $$\int_A1 = \lim_{C\to \infty}\lim_{D \to \infty}0 = 0.$$


问:
Find the rotation of the object?

Image 1: The object's original height and width are 9 and 4. Here the object is not tilted, so the bounding box would cover the same dimension.

Image 2: If we tilt the object, the bounding box, as you can observe from the image, has the height and width of 10 and 8.

From this information, can we find out how much degree is the object rotated in the second image?

答:
As you can see, $\triangle ABC \sim \triangle BED$

So we have

$\frac{BC}{AB} = \frac{DE}{BE}$

$BC = \sqrt{16 - x^2}$

$\displaystyle \frac{\sqrt{16 - x^2}}{4} = \frac{8-x}{9}$

Squaring and solving the quadratic, we get $x \approx 3.4516$

Angle by which the object has been rotated is

$\angle BAC = \arccos \frac{3.4516}{4} \approx 30.6^0$

EDIT:

On the comment that the height does not add up to $10$, it will not. When the box width is $8$ due to rotation, the required height will be $\approx 9.78$ and so if the box height is $10$, it will have some room on the top. Please see below diagram that explains when the required height will be maximum.

The maximum required height will be the length of the diagonal and will be the case when the diagonal $AE$ is vertical.

So the maximum required height = $\sqrt{9^2 + 4^2} \approx 9.85$

This happens when we rotate by angle $23.95^0 (= \arctan \frac{4}{9})$. The required width at this point is $\approx 7.31$.

Please also consider that as we keep rotating, there will be a point $(90^0 - 23.95^0)$ when the required width is $9.85$ and the required height is $7.31$. Given the dimension of the object, you will never need box to go beyond $9.85$ on either side (width or height).

Here is a plot of how width and height of the box will change with the angle of rotation between $(0 \leq \theta \leq \frac{\pi}{2}) \,$ which is taking it from the current vertical position to the horizontal position. As you can see, at angle of rotation of $45^0$, the box requires the same width and height of $\approx 9.2$.

Title: Find the rotation of the object?


答:
ANSWER: NO REAL ANGLES SATISFY ALL CONSTRAINTS, PROF GAVE YOU BAD NUMBERS

The picture, due to symmetry there are 4 unknowns: a,b and x,y
Where a and b are the sides of the big tirangle and x,y are the sides of the little triangles.

We then have five equations:
$$x+b=10$$
$$a+y=8$$
$$a^2 + b^2 = 9^2$$
$$x^2 +y^2 = 4^2$$
$$ \frac{a}{x} = \frac{b}{y}$$

This is five equations and four unknowns, you can solve this and... you get that there are no solutions!?? WUT!?
https://www.wolframalpha.com/input/?i=Solve%5B%7Bx%2Bb%3D%3D10%2Ca%2By%3D%3D8%2Ca%5E2%2Bb%5E2%3D%3D9%5E2%2Cx%5E2%2By%5E2%3D%3D16%2Cx%2Fa%3Dy%2Fb%7D%2C%7Bx%2Cy%2Ca%2Cb%7D%5D

So I went digging to figure out with values do satisfy this, it looks like you have some bad numbers, $\sqrt{9^2 + 4^2} \approx 9.8 < 10$, so no real rotation angle will allow it cover a distance of 10. Tell your prof to do their own HW before they assign it!! haha

Title: Find the rotation of the object?

Body:
Image 1: The object's original height and width are 9 and 4. Here the object is not tilted, so the bounding box would cover the same dimension.

Image 2: If we tilt the object, the bounding box, as you can observe from the image, has the height and width of 10 and 8.

From this information, can we find out how much degree is the object rotated in the second image?