问:
When does a map induce a quasi-isomorphism on associated graded complexes?
Question

I have a map $f: C \to D$ of chain complexes (of modules over some ring $R$) that I know is a quasi-isomorphism. (i.e. the induced map $f_*: H_*(C) \to H_*(D)$ is an isomorphism.)

I also have (bounded) filtrations of $C$ and $D$, and I know that $f$ respects these filtrations. (i.e. $f(\mathcal{F}_pC) \subseteq \mathcal{F}_pD$.)

Is there a necessary and sufficient third condition to ensure that $f$ induces a quasi-isomorphism on the associated graded complexes? (i.e. $(f_p)_*: H_*(\text{gr}_p(C)) \to H_*(\text{gr}_p(D))$ is an isomorphism for all $p$.)

What I've tried

A third condition is necessary because of counter-examples like $f: 0 \subseteq 4\mathbb{Z}\subseteq \mathbb{Z} \to 0 \subseteq 2\mathbb{Z} \subseteq \mathbb{Z}$ via the identity, where here I'm identifying a module $M$ with the chain complex $0 \to M \to 0$.

I believe strictness $f(\mathcal{F}_pC)=f(C)\cap\mathcal{F}_pD$ may be the correct condition, based on the Stacks Project page on filtrations here, but I can't seem to prove it myself (or come up with a counter-example).

Any help or advice would be appreciated. Thank you!

Edit: Loosening the question

Is this any easier if we let $R$ be a field? It's still not a trivial question as far as I can tell, because of the following example: let $C$ be a chain complex of $\mathbb{Q}$-vector spaces with basis $\{x,y\}$ and differential $d(x)=y$, $d(y)=0$. Let $\mathcal{F}_1$ be the filtration of $C$ given by $0 \subseteq \text{span}(y) \subseteq C$ and let $\mathcal{F}_2$ be the filtration of $C$ given by $0 \subseteq C \subseteq C$. Then the map $f: (C,\mathcal{F}_1) \to (C,\mathcal{F}_2)$ given by the identity is a quasi-isomorphism, and respects the filtrations, but does not induce a quasi-isomorphism on the associated graded complexes.

答:
I've decided there is likely no nice  (non-tautological) condition on $f$ to ensure that $f$ induces a quasi-isomorphism on associated graded complexes. Even assuming that $R$ is a field, that $f$ is strict, and that $f$ is either injective or surjective fails to imply what I want. This is demonstrated by the (embarrassingly simple) examples below.

Let $0$ be the zero chain complex, and let $(C,\mathcal{F}_1)$ be the complex defined in the edit to the question. Then the zero maps $0 \to C$ and $C \to 0$ are quasi-isomorphisms, respect the filtration, and are injective/surjective, respectively, but do not induce quasi-isomorphisms on the associated graded complexes.


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
Note that $\dfrac{(x+h)^2-x^2}{h}$ is undefined at $h=0$ and that, when $h \ne 0$,

$$\dfrac{(x+h)^2-x^2}{h} = \frac{2hx+h^2}{h} = 2x+h$$

However, the function $:x \mapsto 2x+h$ is defined, continuous, and has a value of $2x$ at $h=0$.

We also need to use

$$\lim_{h \to 0}\frac{2hx+h^2}{h}
= \lim_{h \to 0}\frac hh \; \lim_{h \to 0}\frac{2x+h}{1}
= \lim_{h \to 0} (2x+h) = 2x$$

The rest follows.

ContentLicense: CC BY-SA 4.0


问:
When can the Killing form be put in the form $\kappa_{ij}=-\kappa\delta_{ij}$, and thus be treated as an Euclidean dot product?
Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{h}$ is a Cartan subalgebra. We know when the lie algebra is semi simple then the killing form, $\kappa(X,Y) = Tr(X^{ad}Y^{ad})$, is non degenerate in $\mathfrak{g}$ and more importantly in $\mathfrak{h}$. So it is a bilinear form analogous to the metric providing an isomorphism between $\mathfrak{h}$ and $\mathfrak{h}*$ and allowing us to take the inner product across these spaces.

One thing that isn't clear fromt the literature I've read (and from my Universities lecture notes) is when we are able to to put the killing form in the form $\kappa_{ij} = -\kappa \delta_{ij}$ and so we can treat the inner product  as the normal euclidean dot product. Some notes I have read say this is only true for compact groups, others I have read say you can always do it and some don't even mention it (my lecture notes).

答:
Call a semisimple Lie algebra $\mathfrak{g}$ compact if its Killing form is negative definite. Then the simply connected Lie group $G$ with Lie algebra $\mathfrak{g}$ has its adjoint representation on $\mathfrak{g}$ preserving the Killing form; its image lies in the Lie subgroup of $\text{Aut}(\mathfrak{g})$ preserving the Killing form, which is an orthogonal group $O(\mathfrak{g})$ and in particular which is compact. It follows that the adjoint form $G/Z(G)$ of $G$ (the image of the Killing form) is compact, and hence if the center of $G$ is finite that $G$ is compact. (I think this is always true with these hypotheses but I'm not sure how to prove it.)

The semisimple Lie algebras which arise in this way are exactly the compact real forms of the semisimple complex Lie algebras, e.g. $\mathfrak{su}(n), \mathfrak{so}(n), \mathfrak{sp}(n)$ and the exceptionals.

In the noncompact case the Killing form may have indefinite signature; for example, the Killing form on $\mathfrak{sl}_2(\mathbb{R})$ has signature $(2, 1)$, and the adjoint representation induces an exceptional isomorphism $\mathfrak{sl}_2(\mathbb{R}) \cong \mathfrak{so}(2, 1)$. This is the split analogue of the exceptional isomorphism $\mathfrak{su}(2) \cong \mathfrak{so}(3)$, and both of these isomorphisms complexify to the exceptional isomorphism $\mathfrak{sl}_2(\mathbb{C}) \cong \mathfrak{so}_3(\mathbb{C})$.


问:
probability of getting a double six ($2$ dice) rolling them $24$ times
This is what I got. $\dfrac{1}{6} \cdot \dfrac{1}{6} = 2.78\% \cdot 24 = 66.72\%$

I believe that since it is a six sided dice, since you roll both of them simultaneously it would be $\dfrac{1}{6} \cdot \dfrac{1}{6}$.

So since they are rolling them $24$ times, I would just multiply it by $24$, so $2.78\% * 24$ would be $66.72\%$, which would mean I have a  $67.7\%$ chance of rolling a double six.

Do you think this is correct? Am i doing this correctly?

答:
We calculate the probability of rolling at least one double-six in $24$ rolls of two dice. The probability we roll a double-six is, as you point out, $\frac{1}{36}$.

So, on any roll, the probability of not getting a double six is $\frac{35}{36}$.

The probability of "failure" $24$ times in a row is therefore $\left(\frac{35}{36}\right)^{24}$.

So the probability of at least one "success" in $24$ rolls  is $1-\left(\frac{35}{36}\right)^{24}$.

Remark: If we interpret the question as asking for the probability of exactly one double-six, the answer is different, and more complicated. In that case, we want the probability of $1$ success and $23$ failures. This probability is $\binom{24}{1}\left(\frac{1}{36}\right)\left(\frac{35}{36}\right)^{23}$.

However, the usual interpretation is "at least one six."

This problem goes back to the the seventeenth century beginnings of probability theory. The Chevalier de Montmort was curious about what was more likely, at least one six in $4$ rolls of a die, or at least one double-six in $24$ rolls  of two dice. Each has probability about $50\%$ and the expected number of successes is the same. At the time, expectation was the primary notion.

ContentLicense: CC BY-SA 3.0


问:
Is this proof about Jordan algebra's correct?
Let $A$ be a unital commutative non-associative alternative algebra.

Let $J$ be a unital Jordan algebra.
Notice Jordan algebra's are commutative and non-associative by definition.

Conjecture :

The set of $A$ algebra's and of $J$ algebra's are identical.

Is that statement true ?

I propose the following proof :

Let $x,y$ be elements.

Alternative means ( in the context of commutative multiplication )

$$ x(xy) = x^2 y $$

The jordan identity means

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

So we are left with proving those conditions are equivalent.

Start with the first one and transform it into the second :

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

multiply both sides by $x$ on the left

$$ x^2(x y) = x(x^2 y)$$
done

In the other direction ( changing the second one in the first one )

$$x^2(xy) = x(x^2 y)  $$

divide both sides by $x$ on the left

$$  x^{-1} x^2 (xy) = x^{-1} x (x^2 y) $$

simplify

$$ x(xy) = x^2 y $$

QED

Is this proof correct ?

答:
You've correctly showed that a commutative alternative algebra is a Jordan algebra. Your proof of the converse is invalid though, and indeed there are Jordan algebras which are not alternative.

Here is on example: the spin factor $J=\mathbb{R}\oplus\mathbb{R}^n$. We think of its elements as formal sums of scalars and vectors from $\mathbb{R}^n$. The multiplication of a scalar with a scalar and of a scalar of a vector are what you think they'd be, and the product of two vectors is their inner product (thus, vector x vector = scalar).

Explicitly, $(a+\mathbf{x})(b+\mathbf{y})=ab+a\mathbf{y}+b\mathbf{x}+\mathbf{xy}=(ab+\langle\mathbf{x},\mathbf{y}\rangle)+(a\mathbf{y}+b\mathbf{x})$.

Exercise. Show the spin factor $J$ is a non-alternative Jordan algebra (for $n>1$).


问:
$ \sqrt n \log \sqrt n$ is lower than $n$ ?!
I see an example in asymptotic notation that not very sensible for me.

what is the intuitive idea to understand $ \sqrt n \log \sqrt n$ is lower than $n$ in concept of asymptotic view?

答: It's similar to $n \log n$ being less than $n^2$.


答:
$$\lim\limits_{x \to +\infty}\frac{\ln \sqrt{x}}{\sqrt{x}} = \lim\limits_{y \to +\infty}\frac{\ln y}{y} = \lim\limits_{y \to +\infty}\frac{1}{y} = 0$$

Addition:

Without using L'Hôpital's rule we can show, that for all $\boldsymbol n$ $\forall n \in \mathbb{N}$ holds $ \log _{2} n < n^{\alpha} $ for $  \alpha \in  \left[  \frac{1}{e \cdot ln2}, 1\right)  $: as it is same with $ \dfrac{\ln(\log _{2} n)}{\ln(n)} < \alpha $, then we need only to find maximum of function $ \dfrac{\ln(\log _{2} x)}{\ln(x)} $ which is in $ x = 2^{e} $. Now, knowing  $ \frac{1}{e \cdot ln2} \approx  0.530737$ we have result for $\alpha=\frac{1}{2}$.

We can obtain without L'Hôpital also $ \log _{2} n < n^{\alpha} $ for $  \alpha \in  \left(  0, 1\right)  $, but only on some subset of $\mathbb{N}$, which needs little more subtle evaluation.

Title: $ \sqrt n \log \sqrt n$ is lower than $n$ ?!

Body:
I see an example in asymptotic notation that not very sensible for me.

what is the intuitive idea to understand $ \sqrt n \log \sqrt n$ is lower than $n$ in concept of asymptotic view?


答: Intuitively, you should already see that $\log(\sqrt{n})$ is asymptotically less than $\sqrt{n}$. Therefore $\sqrt{n}\log(\sqrt{n})$ is asymptotically less than $n$.


问:
Is R is an equivalence relation?
Hello so I'm trying to make an exercise, but it makes no sense to me it goes as follows.

$Fun(X,Y)$ is the set of all functions $f:X\rightarrow Y$. We define a relation $R$ on $Fun(X,Y)$ by saying that $(f,g)\in R$ if and only if $$\{x \in X \mid f(x) \neq g(x)\}$$
is a finite set.

Prove that R is an equivalence relation?

But isn't that R is not an equivalence relation?

Because for example if I would prove that this function is reflexive I would have to say that $f(x) \neq f(x)$.

Hower I can say it is symmetric because if $f(x) \neq g(x)$ then surely $g(x) \neq f(x)$.

And at last it is not transitive because if $f(x) \neq g(x)$ and $g(x) \neq h(x)$ then $f(x) \neq h(x) $ is not implied.

Is my thoughtprocess wrong or is it just the exercise?
If you can help me please share your wisdom

Thanks in advance

答:
Your problem is that you’re not actually using the definition of $R$. For each ordered pair $\langle f,g\rangle$ of functions from $X$ to $Y$ you have to look at

$$\{x\in X:f(x)\ne g(x)\}\,,$$

the set of points in $X$ at which $f$ and $g$ disagree. For instance, if $X=Y=\Bbb N$, $f(n)=n$ for all $n\in\Bbb N$, and $g(n)=\max\{3,n\}$ for all $n\in\Bbb N$, then $f(n)=g(n)$ for all $n\ge 3$, but $f(0)=0\ne 3=g(0)$, $f(1)=1\ne 3=g(1)$, and $f(2)=2\ne 3=g(2)$:

$$\{n\in\Bbb N:f(n)\ne g(n)\}=\{0,1,2\},,$$

because $f$ and $g$ differ at the points $0,1$, and $2$ and at no others. And this set is finite, so $\langle f,g\rangle\in R$.

Change the definition of $g$ by setting $g(n)=\min\{3,n\}$ for each $n\in\Bbb N$, and you’d find that $f(n)=g(n)=n$ for $n\in\{0,1,2,3\}$, but $f(n)=n\ne 3=g(n)$ for $n>3$. Now

$$\{n\in\Bbb N:f(n)\ne g(n)\}=\{n\in\Bbb N:n\ge 4\}$$

is an infinite set, so $\langle f,g\rangle\notin R$.

It’s inconvenient to have to keep writing $\{x\in X:f(x)\ne g(x)\}$, so for $\langle f,g\rangle\in\operatorname{Fun}(X,Y)$ let

$$D(f,g)=\{x\in X:f(x)\ne g(x)\}\,;$$

this abbreviation will save a good bit of writing and reduce the visual clutter. By definition, then, $\langle f,g\rangle\in R$ if and only if $D(f,g)$ is finite.

Reflexivity should now be clear: for any $f:X\to Y$, $f(x)=f(x)$ for every $x\in X$, so $D(f,f)=\varnothing$. $\varnothing$ is finite, so $\langle f,f\rangle\in R$, and $R$ is reflexive.

Symmetry is also easy. For any $f,g:X\to Y$, $D(f,g)=D(g,f)$. If $\langle f,g\rangle\in R$, then $D(f,g)$ is finite, so $D(g,f)=D(f,g)$ is finite, and therefore $\langle g,f\rangle\in R$.

It does take a little more work to prove that $R$ is transitive. Suppose that $\langle f,g\rangle,\langle g,h\rangle\in R$, so that $D(f,g)$ and $D(g,h)$ are finite subsets of $X$; we’d like to prove that $D(f,h)$ is also finite. This would certainly be true if we could show that $D(f,h)\subseteq D(f,g)\cup D(g,h)$, since the union of two finite sets is finite.

Suppose that $x\in X\setminus\big(D(f,g)\cup D(g,h)\big)$. Then $f(x)=g(x)$ (since $x\notin D(f,g)$), and $g(x)=h(x)$ (since $x\notin D(g,h)$), so $f(x)=h(x)$, and therefore $x\notin D(f,h)$, i.e., $x\in X\setminus D(f,h)$. In other words, we’ve shown that

$$X\setminus\big(D(f,g)\cup D(g,h)\big)\subseteq X\setminus D(f,h)$$

and hence that

$$D(f,h)\subseteq D(f,g)\cup D(g,h)\,,$$

as desired.

Title: Is R is an equivalence relation?


问:
How to prove that $\shortparallel T \shortparallel \leq \|T\|^{n}$?
My question. How to prove that  $\shortparallel T \shortparallel \leq \|T\|^{n}$ and, if $T$ is iniective, prove that $\|T^{-1}\|^{n}\leq \shortparallel T \shortparallel \leq \|T\|^{n}$?

First I need to state what I am using: I have two Hilbert spaces $V,W$ and $L:V\to W$ is a linear application between them. Then $\DeclareMathOperator{\dimn}{\operatorname{dim}} \DeclareMathOperator{\Sym}{\operatorname{Sym}}$

if $\dimn V\leq \dimn W$ then exist $S\in \Sym (n)$ and $O\in O(V,W)$ such that $L=O\circ S$,

If $\dimn W\leq \dimn V$  then exist $S\in \Sym (n)$ and $O\in O(V,W)$ such that $L=S\circ O^{*}$,

where $O(V,W)$ is an orthogonal injective linear transformation and $\Sym(n)$ is a linear symmetric transformation.

In this case, when $L=O\circ S$ or $L=S\circ O^{*}$ then $\DeclareMathOperator{\Det}{\operatorname{Det}}$
$$
\shortparallel L \shortparallel\, \triangleq |\Det S|.
$$
What I have tried. Let $T\in L(\mathbb{R}^{n},\mathbb{R}^{m})$ with $n\leq m$ and consider
$$
\|T\|=\sup \frac{\|T(x)\|}{\|x\|},
$$
My idea is to use the following inequality
$$
\begin{split}
\|T\|^{n} & = \sup \frac{(\|T(x)\|)^{n}}{\|x\|^{n}} = \sup \frac{(\|O\circ S(x)\|)^{n}}{\|x\|^{n}} \\
& = \sup \frac{(\| S(x)\|)^{n}}{\|x\|^{n}}\geq \frac{\|S(x)\|^{n}}{\|x\|^{n}} = \frac{\|T^{*}\circ T (x)\|^{n}}{\|x\|^{n}},
\end{split}
$$
but how to continue? Maybe could I use the fact that
$$
(\det(T^{*}\circ T))^{\frac{1}{2}}= |\det S| =  \shortparallel T \shortparallel\;? $$
Please I need some help or any hint, thank you so much.

答:
There is a way to do it without explicitly using SVD, though, SVD is super useful, and you should definitely learn about it. Anyways, let $A$ be the matrix associated with $L$. By definition:

$$ \|L\|_2 = \|A\|_2
= \sup_{x:x\neq 0} \frac{\|Ax\|_2}{\|x\|_2}
= \max_{x: \|x\| = 1} \|Ax\|_2
= \max_{x: \|x\|^2 = 1} \|Ax\|_2^2
$$

The Lagrangian of this constrained optimization problem is $\mathcal L(x, \lambda) = \|Ax\|_2^2 - \lambda (\|x\|_2^2-1)$ which yields the first order condition

$$ 0=\frac{\partial}{\partial x} \mathcal L(x, \lambda) = 2A^* Ax-2\lambda x \iff A^* A x=\lambda x $$

Thus the maximum is achieved for an eigenvector; in fact $\|A\|_2^2=\lambda_{\max}(A^* A)$. On the other hand you have already seen that $\shortparallel T \shortparallel^2 = \det(T^* T) = \prod_k \lambda_k(T^* T)$,  then$^{(*)}$, since all eigenvalues of $T^* T$ are positive it follows that

$$
\shortparallel T \shortparallel^2 = \prod_k \lambda_{k=1}^n(T^* T) \le \prod_{k=1}^n \lambda_{\max}(T^* T) =\lambda_{\max}(T^* T)^n =\|T\|_2^{2n}
$$

Note that by definition $\lambda_k (T*T) = \sigma_k(T)$ are exactly the singular values of $T$

$(*)$ if $\dim V <\dim W$, instead we have $\shortparallel T \shortparallel^2 = \det(TT^*)$ and the same argument holds


问:
Number of solutions of $ax+by+cz=n$ in nonnegative integers
Let $a,b,c,n\in\mathbb{Z}_0^+$. The number of solutions of
$$ax+by+cz=n$$
(i.e. the number of ordered triplets $(x,y,z)$ satisfying the equation) in $\mathbb{Z}_0^+$ as a function of $n$ is
$$\frac{1}{n!}\lim_{w\to 0}\frac{d^n}{dw^n}\frac{1}{(1-w^a)(1-w^b)(1-w^c)}=\frac{1}{2\pi i}\oint \frac{dw}{(1-w^a)(1-w^b)(1-w^c)w^{n+1}}$$
by the theory of generating functions and Cauchy's integral formula.

If $a,b,c$ are small, then the $n$th derivative/contour integral is amenable for computation using partial fraction decomposition. But for large $a,b,c$, the partial fraction decomposition becomes very tedious.

Are there any faster methods of computing this particular integral, or faster methods of obtaining the number of solutions in general?

答:
The topic is standardly known as the Frobenius coin problem, and
it is quite complicated in general.

In the 2D case $ax+by=n$ there is an interesting theorem, the Popovicius' theorem
which tells that the number of non-negative solutions $ p_{\left\{ {a,b} \right\}} (n)$ is given by

$$
\eqalign{
  & p_{\left\{ {a,b} \right\}} (n) = \left| {\,\left\{ \matrix{
  0 \le x,y,a,b,n \in Z \hfill \cr
  \gcd (a,b) = 1 \hfill \cr
  ax + by = n \hfill \cr}  \right.\;} \right| =   \cr
  &  = {n \over {ab}} - \left\{ {{{b^{\,\left( { - 1} \right)} n} \over a}} \right\}
 - \left\{ {{{a^{\,\left( { - 1} \right)} n} \over b}} \right\} + 1 \cr}
$$
where
$$
\left\{ x \right\} = x - \left\lfloor x \right\rfloor
 \quad b^{\,\left( { - 1} \right)} b \equiv 1\;\left( {\bmod a} \right)
\quad a^{\,\left( { - 1} \right)} a \equiv 1\;\left( {\bmod b} \right)
$$

For the 3D case, like the present, we can apply the above to $ax+by=n-cz$ and sum over $z$.
$$
\eqalign{
  & p_{\left\{ {a,b,\,c} \right\}} (n)\quad \left| {\; \gcd(a,b) = 1} \right.\quad
  = \sum\limits_{z = 0}^{\left\lfloor {n/c} \right\rfloor } {p_{\left\{ {a,b} \right\}} (n - cz)}  =   \cr
  &  = \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor } {\left( {{{n - ck} \over {ab}}
 - \left\{ {{{b^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over a}} \right\}
 - \left\{ {{{a^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over b}} \right\} + 1} \right)}  =   \cr
  &  = {n \over {ab}}\left( {\left\lfloor {{n \over c}} \right\rfloor  + 1} \right)
 - {c \over {2ab}}\left\lfloor {{n \over c}} \right\rfloor \left( {\left\lfloor {{n \over c}} \right\rfloor
  + 1} \right) + \left( {\left\lfloor {{n \over c}} \right\rfloor  + 1} \right) +   \cr
  &  - \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor } {\left( {\left\{ {{{b^{\,\left( { - 1} \right)}
 \left( {n - ck} \right)} \over a}} \right\} + \left\{ {{{a^{\,\left( { - 1} \right)}
 \left( {n - ck} \right)} \over b}} \right\}} \right)}  =   \cr
  &  = \left( {{n \over {ab}} - {c \over {2ab}}\left\lfloor {{n \over c}} \right\rfloor  + 1} \right)
\left( {\left\lfloor {{n \over c}} \right\rfloor  + 1} \right)
 - \sum\limits_{k = 0}^{\left\lfloor {n/c} \right\rfloor }
 {\left( {\left\{ {{{b^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over a}} \right\}
 + \left\{ {{{a^{\,\left( { - 1} \right)} \left( {n - ck} \right)} \over b}} \right\}} \right)}  \cr}
$$

Example:

$$n=13, \, a=3, \, b=4, \, c=5$$
the solutions are
$$
\left( {x,y,z} \right) \in
 \left\{ {\left( {0,2,1} \right),\left( {1,0,2} \right),\left( {3,1,0} \right)} \right\}
$$
The modular inverses of $a$ and $b$ are
$$
\eqalign{
  & 3 \cdot 3 \equiv 1\left( {\bmod 4} \right) \Rightarrow a^{\,\left( { - 1} \right)}  = 3  \cr
  & 1 \cdot 4 \equiv 1\left( {\bmod 3} \right) \Rightarrow b^{\,\left( { - 1} \right)}  = 1 \cr}
$$
and $p_{\left\{ {a,b,\,c} \right\}} (n)$ becomes
$$
\eqalign{
  & p_{\left\{ {3,4,5} \right\}} (13) =   \cr
  &  = \left( {{{13} \over {12}} - {5 \over {24}}\left\lfloor {{{13} \over 5}} \right\rfloor  + 1} \right)
\left( {\left\lfloor {{{13} \over 5}} \right\rfloor  + 1} \right)
 - \sum\limits_{k = 0}^2 {\left( {\left\{ {{{\left( {13 - 5k} \right)} \over 3}} \right\}
 + \left\{ {{{3\left( {13 - 5k} \right)} \over 4}} \right\}} \right)}  =   \cr
  &  = 5 - \sum\limits_{k = 0}^2 {\left( {\left\{ {{{13} \over 3}} \right\}
 + \left\{ {{{39} \over 4}} \right\} + \left\{ {{8 \over 3}} \right\} + \left\{ {{{24} \over 4}} \right\}
 + \left\{ {{3 \over 3}} \right\} + \left\{ {{9 \over 4}} \right\}} \right)}  =   \cr
  &  = 5 - \sum\limits_{k = 0}^2 {\left( {{1 \over 3} + {3 \over 4} + {2 \over 3} + {1 \over 4}} \right)}  = 3 \cr}
$$


问:
partial derivative of function with variables as powers of e

Edit:
From the @Klaus in comments, the original function is

In the above example,from Imperial College of London math for Machine Learning on Coursera, I understand partial f/partial x since it is easy to treat the e term as a constant but for partial f / partial y, if y is the only term differentiated, my guess would be that the derivative is sin(x) * e ^z^2 as y becomes 1 and everything else is left untouched.  For  same reason partial f / partial z, my guess is sin(x) * e ^2yz.

But then somewhere I remember the derivative of e to anything is just the same so that is adding to my confusion.
I think the product rule is supposed to be applied here but browsing through Khan academy and a few other online resources did not help me.  Can someone please explain to my feeble mind how this unfolds step by step?

答:
When taking partial derivatives, you treat all other variables as parameters. So, for example, if you take the partial derivative with respect to $y$, then $x$ and $z$ are parameters. After that, you can use the usual differentiation rules from one-dimensional calculus. In this case, we have to use the chain rule:
$$(g_1 \circ g_2)'(y) = g_1'(g_2(y))g_2'(y).$$
In the case at hand, $g_1(y) = \sin(x)e^y$  and $g_2(y) = yz^2$ so that $(g_1 \circ g_2)(y) = \sin(x)e^{yz^2}$. Hence, applying the chain rule, we get
$$ \frac{\partial f}{\partial y}(x,y,z) = (g_1 \circ g_2)'(y) = g_1'(g_2(y))g_2'(y) = \sin(x)e^{g_2(y)}g_2'(y) = \sin(x)e^{yz^2}z^2$$
because $g_1'(y) = \sin(x)e^y$ and $g_2'(y) = z^2$. Similarly, you can do the partial derivative with respect to $z$.


问:
Proving $\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$
For any real $a>0$:
$$\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$$
I got the hint to use Stirling's approximation but I cant seem to solve it. Any tips would be very appreciated.

答:
Let $ a,x\in\left(1,+\infty\right) $, we have the following : $$ x^{a}\int_{0}^{1}{y^{a-1}\left(1-y\right)^{x-1}\,\mathrm{d}y}=\int_{0}^{x}{y^{a-1}\left(1-\frac{y}{x}\right)^{x-1}\,\mathrm{d}y}=\int_{0}^{+\infty}{y^{a-1}f_{x}\left(y\right)\mathrm{d}y} $$

Were $ f_{x} $ is the function defined on $ \mathbb{R}_{+} $ as follows : $$ \left\lbrace\begin{matrix}f_{x}\left(y\right)=\left(1-\frac{y}{x}\right)^{x-1},\ \ \ \ \ \ \ \textrm{If }y\leq x\\ f_{x}\left(y\right)=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,\ \ \ \ \ \ \ \textrm{If }y\geq x \end{matrix}\right. $$

Using the fact that $ \left(\forall y\in\mathbb{R}_{+}\right),\ f_{x}\left(y\right)\leq\mathrm{e}^{-y}=f\left(y\right)$, and that $ f_{x}\left(y\right)\underset{x\to +\infty}{\longrightarrow}f\left(y\right) $, by using the dominated convergence theorem, we have the following : $$ \int_{0}^{+\infty}{y^{a-1}f_{x}\left(y\right)\mathrm{d}y}\underset{x\to +\infty}{\longrightarrow}\int_{0}^{+\infty}{y^{a-1}f\left(y\right)\mathrm{d}y} $$

In other words : $$ x^{a}\int_{0}^{1}{y^{a-1}\left(1-y\right)^{x-1}\,\mathrm{d}x}\underset{x\to +\infty}{\longrightarrow}\int_{0}^{+\infty}{y^{a-1}\,\mathrm{e}^{-y}\,\mathrm{d}y} $$

But $ \int_{0}^{1}{y^{a-1}\left(1-y\right)^{x-1}\,\mathrm{d}x}=\beta\left(a,x\right)=\frac{\Gamma\left(a\right)\Gamma\left(x\right)}{\Gamma\left(a+x\right)} $, and $ \int_{0}^{+\infty}{y^{a-1}\,\mathrm{e}^{-y}\,\mathrm{d}y}=\Gamma\left(a\right) $, thus : $$\fbox{$\begin{array}{rcl} \displaystyle\frac{\Gamma\left(x\right)}{\Gamma\left(a+x\right)}\underset{x\to +\infty}{\sim}\frac{1}{x^{a}} \end{array}$}$$

It's another way of saying that $ \lim\limits_{x\to +\infty}{\frac{\Gamma\left(a+x\right)}{x^{a}\Gamma\left(x\right)}}=1 $.


问:
Why is $\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$ divergent?
I am stuck on this problem. I have checked on wolfram alpha, and the calculator says that this series is divergent. However, I can't find a way to prove it.

The problem:

$$\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$$

So I tried doing this problem with the direct comparison test.

I have concluded that

$$
\frac{n^2}{n^3} >\frac{n^2-1}{n^3+1}
$$
$$
b_{n}> a_{n}
$$

$b_{n}$ is a p-series, where p=1. This means that the series $b_{n}$ is divergent, making the test direct comparison test inconclusive.
$$
\frac{n^2}{n^3} = \frac{1}{n}
$$

I appreciate any help... thanks! :)

Body:
Note that

$$\frac{n^2-1}{n^3+1} \geq \frac{n^2 - \frac{n^2}{2}}{n^3+n^3} =  \frac{1}{4}\frac{n^2}{n^3}  =\frac{1}{4n}. $$


问:
Could exists a vector field on $\mathbb{S}^{2}$ with exactly $n$ zeroes?
I just started to learn index theory of tangent vector fields. I'm aware of two examples on the sphere $\mathbb{S}^{2}$ with exactly one zero, which, which are $F(x,y) = (1-x^2-y^2)\partial x$ thought on $\mathbb{D}^2$ and then indentify the boundary, I think it is

And one with two zeroes should be something like this

I was wondering, it exists a tangent vector field with $n$ zeroes on the sphere ? I thought that if I have $n$ zeroes (isolated), then thanks to the Poincarè-Hopf theorem the sum of the indices must be  $\chi(\mathbb{S}^{2}) = 2$. I thought that some combination of $-1$ and $1$ and a final zero of index $2$ could do the job, but can this vector field actually exist?
Something like defining a polynomial with $n$ distinct zeroes $p : \mathbb{C} \longmapsto \mathbb{C}$ can be extended to $\mathbb{S}^{2}$ to gain a tangent vector field on the sphere?

Thanks in advance, any help would be appreciated.

答:
There is a useful smooth vector field $\mathcal V$ to know about on the plane, satisfying the following conditions

$\mathcal V$ has one zero at $\mathcal O = (0,0)$.

Except for $\mathcal O$, every vector in the vector field $\mathcal V$ is a positive multiple of $\frac{\partial}{\partial x}$.

Outside of the ball of radius $r$ around the origin (where $r>0$ can be anything you want), the vector field $\mathcal V$ is equal to the constant vector field $\frac{\partial}{\partial x}$.

To construct $\mathcal V$, multiply $\frac{\partial}{\partial x}$ by a non-negative bump function on the plane, which has a unique zero at $\mathcal O$, and is equal to $1$ outside of the ball of radius $1$ around $\mathcal O$.

You can combine this example with known examples to get any finite number of zeroes (except zero zeroes, of course). For instance, start with your "one zero" example. Choose $k \ge 0$ points $p_i$, $1 \le i \le k$,  distinct from that zero. There exist respective neighborhoods $U_i$ of these points, with disjoint closures, and diffeomorphisms $\phi_i : U_i \to B_i \subset \mathbb R^2$, where $B_i$ is an open ball, and $\phi_i(p_i) = \mathcal O$, and such that the vector field on $U_i$ is equal to the pullback of the vector field $\mathcal V$ with respect to $\phi_i$ and a choice of radius $r_i$, equal to half the radius of $B_i$. Now replace the given vector field on $B_i$ with the pulback of $\mathcal V$, which has the effect of adding an isolated zero at $p_i$. This gives a vector field on $S^2$ with $1+k$ zeroes.


问:
using the fact that $\int_{0}^{\infty} \frac{|\sin (u)|}{u^{1+\alpha}}du<\infty$ to show the following inequallity
This is not the complete question, just the part I didn't succed so this given might be unneccesary.

Let $f \in L_{p c}^{1}(\mathbb{R}) \cap L_{p c}^{2}(\mathbb{R})$ and also for $A>0$ and $\alpha>0$
$$
|\widehat{f}(s)| \leq \frac{A}{1+|s|^{1+\alpha}}
$$
and let :$$
f(x+h)-f(x)=\int_{-\infty}^{\infty} \widehat{f}(s) e^{i s x}\left(e^{i s h}-1\right) d s
$$

show that :
$$\int_{0}^{\infty} \frac{|\sin (u)|}{u^{1+\alpha}}du<\infty$$
$$
\int_{0}^{\infty} \frac{|\sin (h s / 2)|}{|h|^{\alpha}\left(1+s^{1+\alpha}\right)} d s \leq 2^{-\alpha} \int_{0}^{\infty} \frac{|\sin (u)|}{u^{1+\alpha}}du
$$

I already succeded to show that the integral converge but I dont know how to show this inequality

答: Use the  substitution $u=\frac {|h|s} 2$ after using the fact that $1+s^{1+\alpha} \geq s^{1+\alpha}$.


问:
Counterexample: a finite-order and non-unitary operator (Hilbert space).
I need for this to be true but I can't find a suitable operator:

On a complex Hilbert space $\mathcal{H}$ let $T\in \mathcal{L}(\mathcal{H})$, which means a linear and continuous operator $T:\mathcal{H} \to \mathcal{H}$; if $T$ has finite orded, i.e., that there exists an integer $k$ such that $T^k=\mathbb{I}$ (identity), T is not necessarily unitary.

Does somebody have at least a hint?

答: Try $T = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$.


问:
Prove $\sum_{n=1}^{\infty} {X_n}$ is convergent with probability $1$ $\iff$ $P({|X_1|} =0)=1$
Let $X_1,X_2,\ldots$ be iid.

Goal: $\sum_{n=1}^{\infty} {X_n}$ is convergent with probability $1$ $\iff$ $P({|X_1|} =0)=1$.

I already proved one direction: $\sum_{n=1}^{\infty} {X_n}$ is convergent with probability $1\,\Longrightarrow\,P({|X_1|} =0)=1$ using Kolmogorov's three-series theorem.

To prove another implication i tried Kolmogorov's three-series theorem too, but i stuck in series of variance and expected value. Is this valid method?

I appreciate any help.

答:
The correct statement is:

$\sum  X_n$ is convergent with probability $1$ iff $P(|X_1|=0)=1$.

If the series converges with probability $1$ then $X_n \to 0$ with probabiltiy $1$ and hence $X_n \to 0$ in probability. Thus $P(|X_n| >\epsilon)  \to 0$. This means $P(|X_1| >\epsilon) = 0$ for every $\epsilon >0$. Can you finish?

Converse is obvious.


答:
The point of asymptotic behavior is to abstract away implementation details and look at the "big picture" of how long an algorithm takes.

An $O(n)$ algorithm has to (roughly) read (most of) the input.  A $O(\sqrt n \log \sqrt n)$ algorithm doesn't even have to read the entire input, it reads a significantly smaller amount.  That's not a small detail, so you should expect them to be different.


Title: Why is $\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$ divergent?

Body:
I am stuck on this problem. I have checked on wolfram alpha, and the calculator says that this series is divergent. However, I can't find a way to prove it.

The problem:

$$\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$$

So I tried doing this problem with the direct comparison test.

I have concluded that

$$
\frac{n^2}{n^3} >\frac{n^2-1}{n^3+1}
$$
$$
b_{n}> a_{n}
$$

$b_{n}$ is a p-series, where p=1. This means that the series $b_{n}$ is divergent, making the test direct comparison test inconclusive.
$$
\frac{n^2}{n^3} = \frac{1}{n}
$$

I appreciate any help... thanks! :)

Body:
$$
\frac{n^2-1}{n^3+1} = \frac{n^2+1 -2}{n^3+1} = \frac{n^2+1}{n^3+1} -\frac{2}{n^3+1} \geq \frac{1}{n} - \frac{2}{n^3+1} \quad \forall\ n \in \mathbb{N}.
$$

Therefore,

\begin{align}\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1} = \sum\limits_{n=1}^{\infty}\left(\frac{n^2+1}{n^3+1} -\frac{2}{n^3+1}\right)\\
\\
= \sum\limits_{n=1}^{\infty}\frac{n^2+1}{n^3+1}\ -\ \sum\limits_{n=1}^{\infty}\frac{2}{n^3+1} \\
\\
\geq \left(\sum\limits_{n=1}^{\infty} \frac{1}{n}\right)\ -\ L \\
\end{align}

for some $L\in \mathbb{R}.$


问:
How to find second-order partial derivatives?
Please, explain to me as you can how to find first-order and second-order derivatives for this function. It would be really helpful! Here is the function:

z=cot(y/x)

Seems like my solution isn't right and I can't do it right.

I started, but didn't continue

The 2 second-order derivatives seems like should be equal.

答:
For the first order partial derivatives, you have the equations

$$ \frac{\partial z}{\partial x} = \frac{-y}{\sin^2 \left(\frac{y}{x}\right) x^2 }$$

$$ \frac{\partial z}{\partial y} = \frac{-1}{\sin^2 \left(\frac{y}{x}\right) x} $$

by computation similar to what you have. Notice that you do not need the quotient rule for $\frac{y}{x}$ because either one of $y, x$ is a constant.


问:
Negative base of log if working in $\mathbb{C}$?
Would a negative base of a logarithm be allowed if one were working in the field of complex numbers, $\mathbb{C}$?

答: Yes using the base change rule $\frac{ln(c)}{ln(-b)}=\log_{-b}(c)$ and $ln(-n)=ln(n)+\pi i$ so $log_{-b}(x)=\frac{ln(x)}{ln(b)+\pi i}$


问:
Concrete and abstract numbers
Recently I heard about the concepts of concrete number = numerus numeratus and abstract number = numerus numerans. See here. These seem to be mathematical-philosophical specialist terms of medevial (European) origin with a "scholastic sound".

In

Smith, D.E. (1953). History of Mathematics. Vol. II. Dover. pp. 11–12. ISBN 0-486-20430-8.

one finds

Question: Is this distinction still in use in contempory mathematics, at least at some places? Or is it a premodern concept which has completely disappeared?

A search for "concrete number" gives 54 results in this forum, but I do not have the impression that its use has a philosophical background. Similarly "abstract number" gives 15 results.

Update:

Here are some sources which prove the origin in the medieval (European) world of ideas:

1. Joh. Micrealius, Lexicon Philosophicum (1661)

Numerus est compositarum unitatum aggregatio. Vocatur alias multitudo & quantitas discreta.

Numerus physicis alius est numerans, alius numeratus.

Numerus numerans seu formalis est, quem anima apprehendit abstractum ab omnia materia. Dicitur etiam matbematicus.

Numerus numeratus et materialis est, cujus unitates sunt res. Sic Aristoteles definivit tempus per numerum, puta numeratum, quia tempus est numerus motus per prius et posterius. Dicitur etiam physicus.

2. Time and Eternity in Augustine
(and in Medieval Scholasticism)

In short, using scholastic terminology, time is the numerus numeratus motus, but something is numerus numeratus, only if there is a numerans, although that which is the numerus numeratus can exist even if no numerans exists, but then it is not numerus numeratus, but numerus numerabilis.

3. Thomas de Aquino, Commentaria, Libros Physicorum, IV, Lectio 17

Uno modo id quod numeratur actu, vel quod est numerabile, ut puta cum dicimus decem homines aut decem equos; qui dicitur numerus 'numeratus', quia est numerus applicatus rebus numeratis. Alio modo dicitur numerus 'quo numeramus', idest ipse numerus absolute acceptus, ut duo, tria, quatuor.

答:
Past personal studies of mine of the mathematical foundations of dimensional analysis lead me to believe that the following answers are plausible ones.

Is it a premodern concept which has completely disappeared?

Quite the opposite. These concepts continued to be developed until they gave rise to a branch of mathematical physics known today as dimensional analysis. The concept of an abstract number is nowadays refereed as a dimensionless quantity and the concept of concrete number as dimensional quantity.

Is this distinction still in use in contemporary mathematics, at least at some places?

Yes. Whenever in mathematical modeling we feel the presence of dimensional physical quantities in a determined model is making its analysis harder, in the light of Buckingham $\pi$ theorem we might try to remove them through a technique known as nondimensionalization. Roughly, we try to algebraically combine dimensional quantities (concrete numbers) into dimensionless quantities (abstract numbers) and then rewrite the model in terms of these dimensionless quantities. A proof of Buckingham's $\pi$ theorem as well as the mathematical foundations of dimensional analysis can be found at Jan-David Hardtke (2019). On Buckingham's Π-Theorem. arXiv:1912.08744 [math-ph].

An example of nondimensionalization in the branch of the qualitative analysis of differential equations can be found at section 1.2 of J.D. Murray (2011). Mathematical Biology: I. An Introduction, Third Edition., and in the branch of statistics can be found at Pearson's correlation coefficient which is essentially the nondimensionalization of the covariance between two random variables.

In case you don't get any answer you find helpful then I suggest emailing Jan-David Hardtke whose e-mail can be found at his above linked article.

Title: Concrete and abstract numbers


问:
Prove that $\{x \in X \mid f(x) < g(x)\}$ is an open set in $X$.
Let $Y$ be an ordered set in the order topology. Let $f,g:X \rightarrow Y$ be continuous functions form a topological space $X$.

Prove that $\{x \in X \mid f(x) < g(x)\}$ is an open set in $X$.

$\textbf{Proof:}$ Since $Y$ is an ordered set in the Order Topology, then $Y$ is Hausdorff.

So, $\forall a \neq b \in Y \,\, \exists$open sets $U,V \subset Y$ such that $a \in U$, $b \in V$, and $U \cap V = \emptyset$.

Pick open subsets $U,V \subset Y$ with $a \in U$ and $b \in V$.

Since $f,g$ are continuous functions, then $f^{-1}(U)$, $g^{-1}(V) \subset X$ are open.

Let $x \in f^{-1}(U) \cap g^{-1}(V)$ which is also open. Then, $x \in f^{-1}(U)$ and $x \in g^{-1}(U)$.

So, $f(x) \in U \subset Y$ and $g(x) \in g^{-1}(V) \subset Y$. Since $Y$ is Hausdorff, then $U \cap V = \emptyset$ and $f(x) \neq g(x)$.

So, either $f(x)<g(x)$ or $f(x)>g(x)$.
Without loss of generality, suppose $f(x)<g(x)$.

Then, $x \in \{x \in X \mid f(x)<g(x)\}$

$\therefore x \in f^{-1}(U) \cap g^{-1}(V) \subset \{x \in X \mid f(x)<g(x)\} \subset X$.

$\therefore \{x \in X \mid f(x)<g(x)\}$ is open in $X$. $\blacksquare$

I wasn't given any feedback on my proof, so I am trying to see if this is well written. I would also like to know if the WLOG makes sense to use here.

答:
I’m afraid that it does not work: it’s true that $f(x)\ne g(x)$ whenever $x\in f^{-1}[U]\cap g^{-1}[V]$, but it’s entirely possible that there are $x_0,x_1\in f^{-1}[U]\cap g^{-1}[V]$ such that $f(x_0)<g(x_0)$, but $f(x_1)>g(x_1)$. (It would be a good exercise to construct an example of this phenomenon.)

There is also a notational problem: when you write

$$x\in\{x\in X:f(x)<g(x)\}\,,$$

you’re using $x$ to mean two completely different things in the same expression. The first $x$ refers to a specific point of $X$; the instances of $x$ between the curly braces are instances of a dummy variable used to define the set. It should be

$$x\in\{u\in X:f(u)<g(u)\}$$

or the like.

Let $A=\{x\in X:f(x)<g(x)\}$. The most straightforward way to show that $A$ is open is to let $x\in A$ be arbitrary and find an open nbhd of $x$ contained in $A$. To do this we will use disjoint open nbhds of $f(x)$ and $g(x)$, but that isn’t enough: we have to choose them a bit carefully.

If there is a $u\in Y$ such that $f(x)<u<g(x)$, let $U=(\leftarrow,u)$ and $V=(u,\to)$, the open rays to the left and to the right of $u$; clearly $f(x)\in U$, $g(x)\in V$, and $U\cap V=\varnothing$. If there is no such $u$, then $g(x)$ is the immediate successor of $f(x)$ in $Y$, and we let $U=\big(\leftarrow,g(x)\big)$ and $V=\big(f(x),\to\big)$; again we have $f(x)\in U$, $g(x)\in V$, and $U\cap V=\varnothing$. Note that in both cases $y_0<y_1$ whenever $y_0\in U$ and $y_1\in V$: this is the fact that makes the argument work and that was missing in your construction.

Let $W=f^{-1}[U]\cap g^{-1}[V]$; clearly $W$ is an open nbhd of $x$. If $z\in W$, then $f(z)\in U$ and $g(z)\in V$, so $f(z)<g(z)$, and therefore $z\in A$. Thus, $W\subseteq A$, and $A$ is open in $X$.


答:
Complex polynomials won't work on their own, but a generalization can be made which allows one to produce an explicit formula (in stereographic coordinates) for any combination of indices which sum to $2$.

We can cover the Riemann sphere with two sets of opposite stereographic coordinates in the standard way, so that the transition function between them is $\tau(z)=1/z$, thinking of the tangent spaces $T_z\mathbb{C}$ as copies of $\mathbb{C}$ in the standard way, the differential of the transition function $d_z\tau:T_z\mathbb{C}\to T_{1/z}\mathbb{C}$ is just multiplication by $-1/z^2$.

We may treat a function $F:\mathbb{C}\to\mathbb{C}$ as a vector field in stereographic coordinates, and to see if it extends smoothly to the Riemann sphere, we can write its representative in the opposite coordinates, centered at infinity:
$$
\widetilde{F}(z)=-z^2F(1/z)
$$
We see that a complex polynomial of degree $>2$ will have a nonremovable singularity at infinity, but for degree $\le 2$ things behave as expected: each root has index equal to its algebreic multiplicity, and each polynomial of degree $d\in\{0,1\}$ has an additional zero of index $2-d$ at infinity. To get the other cases, we have to allow for negative indices; one way to do this is by complex conjugation.

Choose $a_1,\cdots,a_{n+2},b_1,\cdots,b_{n}\in\mathbb{C}\setminus\{0\}$, and define
$$
F(z)=\frac{(z-a_1)\cdots(z-a_{n+2})(\bar{z}-b_1)\cdots(\bar{z}-b_n)}{1+|z|^{2n}}
$$
This function is smooth on $\mathbb{C}$, and applying the transition function and doing a bit of algebra we obtain
$$
\widetilde{F}(z)=-a_1\cdots a_{n+2}b_1\cdots b_n\frac{(z-1/a_1)\cdots(z-1/a_{n+2})(\bar{z}-1/b_1)\cdots(\bar{z}-1/b_n)}{1+|z|^{2n}}
$$
And so $F$ extends smoothly to the Riemann sphere, and does not vanish at infinity. Additionally, the indices of the zeros of $F$ can be easily computed: for any zero $z_0\in\mathbb{C}$, the index of $z_0$ is the number of $a_i$ equal to $z_0$ minus the number of $b_i$ equal to $z_0$. This can be seen by first showing that the index of the vector field $V(z)=z^p\bar{z^q}$ is $p-q$ using polar decomposition, and looking at $F(z_0+z)$ in a small neighborhood around the origin.

Using this functional form, to obtain zeros of indices $I_1,I_2,\cdots,I_m\in\mathbb{Z}$ at distinct coordinates $z_1,\cdots,z_m\in\mathbb{C}$ we can choose positive integers $r_1,\cdots,r_m,s_1,\cdots s_n\in\mathbb{N}$ such that $r_i-s_i=I_i$ and then choose $a_1,\cdots,a_{n+2}$ to contain $r_i$ copies of each $z_i$ and $b_1,\cdots,b_n$ to contain $s_i$ copies of each $z_i$.

Title: Could exists a vector field on $\mathbb{S}^{2}$ with exactly $n$ zeroes?

Body:
I just started to learn index theory of tangent vector fields. I'm aware of two examples on the sphere $\mathbb{S}^{2}$ with exactly one zero, which, which are $F(x,y) = (1-x^2-y^2)\partial x$ thought on $\mathbb{D}^2$ and then indentify the boundary, I think it is

And one with two zeroes should be something like this

I was wondering, it exists a tangent vector field with $n$ zeroes on the sphere ? I thought that if I have $n$ zeroes (isolated), then thanks to the Poincarè-Hopf theorem the sum of the indices must be  $\chi(\mathbb{S}^{2}) = 2$. I thought that some combination of $-1$ and $1$ and a final zero of index $2$ could do the job, but can this vector field actually exist?
Something like defining a polynomial with $n$ distinct zeroes $p : \mathbb{C} \longmapsto \mathbb{C}$ can be extended to $\mathbb{S}^{2}$ to gain a tangent vector field on the sphere?

Thanks in advance, any help would be appreciated.


问:
$\lim_{x\to 0} \frac{\sin(t\sqrt{x^2-k^2})}{\sqrt{x^2-k^2}}=?$
$\lim_{x\to 0} \frac{\sin(t\sqrt{x^2-k^2})}{\sqrt{x^2-k^2}}=?$

Anyone can help, please.
Does is equal to $t$?

Body: $$\lim_{x\to 0} \frac{\sin(t\sqrt{x^2-k^2})}{\sqrt{x^2-k^2}}=\frac{\sin(t\sqrt{-k^2})}{\sqrt{-k^2}}$$


问:
Show that $R_{n}=\left\{z \in \mathbb{C} \mid z^{n}=1\right\}$ is a subgroup of $S=\{z \in \mathbb{C} : |z|=1\}$.
Show that for each $n \in \mathbb{N}, n \geq 2$ the set of roots $n$ n-th of the unity: $R_{n}=\left\{z \in \mathbb{C} \mid z^{n}=1\right\}$ is a subgroup of $S=\{z \in \mathbb{C}  : |z|=1\}$.

I try:

$1 \in R_n $ because $1^n=1$ and $1*z^n=1*1=1$ for all $z \in R_n$

If $z,x \in R_n$ then $z^n*x^n=(z*x)^n=1*1=1 $ then $z^n*x^n \in R_n$

and if $z\in R_n$ then $z^{-1} \in R_n$ because $z^{-1}=\frac{1}{z^n}=\frac{1}{1}=1=z*z^{-1}$

is right?

答:
I will use the one-step subgroup test.

Fix $n\in\Bbb N$, $n\ge 2$.

Clearly $1\in R_n\neq\varnothing $ and $R_n\subseteq S$. Indeed,

$$\begin{align}
z^n=1&\implies|z^n|=1\\
&\implies|z|^n=1\\
&\implies|z|=\pm1\\
&\implies|z|=1.
\end{align}$$

Let $y,z\in R_n$. Then $y=e^{i2k\pi/n}, z=e^{i2\ell\pi/n}$ for $k,\ell\in \Bbb Z$. So

$$\begin{align}
yz^{-1}&=e^{i2k\pi/n}e^{-i2\ell\pi/n}\\
&=e^{i2(k-\ell)\pi/n}\\
&\in R_n
\end{align}$$

since $k-\ell\in\Bbb Z$.

Hence $R_n\le S$.


问:
Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$
Show that for $r \in (-1,1)$

$$ \int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt = 0$$

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$
The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.

Body:
hint

$$1-2r\cos(t)+r^2=$$
$$(r-\cos(t))^2+\sin^2(t)$$


问:
Isomorphim between $\mathbb F_{2^k}$ with different reduction polynomials
Let $P_0(x)$ and $P_1(x)$ be two given distinct irreducible binary polynomials of same degree $k$. For both $P_i(x)$, the set $\mathbb F_{2^k}$ of binary polynomial of degree less than $d$, together with polynomial addition, and polynomial multiplication followed by reduction modulo $P_i$, form a field. It's well-known these fields are isomorphic (see e.g. this question).

Is that isomorphism unique? How can we express it (or one) using $P_0(x)$ and $P_1(x)$?

That is, is there a unique permutation $g$ on the set $\mathbb F_{2^k}$ such that $$\begin{align}\forall A(x)\in\mathbb F_{2^k},\,\forall B(x)\in\mathbb F_{2^k},\
&g\big(A(x)+ B(x)\big)=g\big(A(x)\big)+ g\big(B(x)\big)\\
&g\big(A(x)\cdot B(x)\bmod P_0(x)\big)=g\big(A(x)\big)\cdot g\big(B(x)\big)\bmod P_1(x)
\end{align}$$
and how do we express it (or one)?

I see that $g(0)=0$, $g(1)=1$, and it's enough to know $g(x)$ to fully define $g$. But I fail to prove that $g$ is unique, or find what $g(x)$ can be. I've tried to suppose there are two distinct $g$ and $g'$, define $h=g'\circ g^{-1}$, state there is a polynomial $D(x)$ with $h(D(x))\ne D(x)$ and come to a contradiction, but nothing happens.

答: $P_1(T)\in \Bbb{F}_2[T]$ has a root $a$ in the field $\Bbb{F}_2[x]/(P_0(x))$, the other ones are $a^{p^m}$, and the isomorphisms $$\Bbb{F}_2[y]/(P_1(y))\to \Bbb{F}_2[x]/(P_0(x))$$ send $y$ to $a^{p^m}$ for some $m$ ($k$ possible choices).


问:
Problem getting to and understanding the answer of $\lim \limits_{x \to 0^+} \frac{(\sin{x})^x}{x}$
$$\lim \limits_{x \to 0^+} \frac{(\sin{x})^x}{x}$$

Somehow right limit ($0^+$) is $+\infty$ but left ($0^-$) one is undefined. I wrote limit like $ \lim \limits_{x \to 0^+} \frac{\sin{x}}{x} * \lim \limits_{x \to 0^+} (\sin{x})^{x-1}$ and the first one is 1 so it leaves us with just $\lim \limits_{x \to 0^+} (\sin{x})^{x-1}$. So if we put the values of the limit in: $\sin{x}$ when $x$ is converging to $0$ is $0$ and we get $0^{0-1}$ or $\frac{1}{0}$. So if the right limit is $\frac{1}{0^+} = +\infty$ why isn't $\frac{1}{0^-} = -\infty$?

答:
If I have a real function of real variable how this:

$$\psi(x)=f(x)^{g(x)}$$

your domain is $$\text{dom}(\psi(x))=\{x\in\Bbb R \mid f(x)>0\,\wedge x \in \operatorname{dom}(g(x))\} \tag 1$$ Considering the $(1)$, the function

$$h(x)=\frac{(\sin{x})^x}{x}$$
have a domain, $D$ with $x\neq 0 \ \wedge \ \sin(x)>0$. If you solved this $0^-\notin D$. Hence:

$$\lim_{x \to 0^-} \frac{(\sin{x})^x}{x}=\nexists$$

After, being $0<\sin(x)\leq 1$, and $x\to 0^+\implies$

$$\lim_{x \to 0^+} \sin^{x}{(x)} = \lim_{x \to 0^+} e^{\ln (\sin^{x} x)}=e^{\lim_{x \to 0^+} \frac{\ln{\sin x}}{\frac{1}{x}}} \tag 2$$
where in the ratio you have an indeterminal form $(\infty/\infty)$. You use the l'Hopital's rule. After you will have an indeterminal form $(0/0)$. Using another times the l'Hopital's rule you can try that you will have

$$e^{- \lim_{x \to 0^+} \frac{2 x}{\tan^{2}{x} + 1}} = e^{- \lim_{x \to 0^+} 2 x \cos^{2}{x}}=1$$

Therefore

$$\lim_{x \to 0^+} \sin^{x}{x}=1$$ and

$$\lim_{x \to 0^+} \frac{(\sin{x})^x}{x}=+\infty$$


问:
How far would Rebecca walk if she walks 100 feet in a straight line and turns $165^{\circ}$ clockwise, repeating this pattern until...
If Rebecca walks 100 feet in a straight line. She turns $165^{\circ}$ clockwise and walks another 100 feet, and turns $165^{\circ}$ clockwise again. If she continues this pattern until she reaches the point where she started. How far did she walk?

I can't see how she would ever get back to her origin.
She turns $\frac{11}{24}$ of full rotation each time, this will not form a regular polygon.
It only makes sense she gets back to start if she turns for example $20^{\circ}$ - convenient factor of $360$, here I can use the fact the exterior angles of polygon add up $360^{\circ}$ to find the number of angles formed and derive an answer.

答:
The shape that Rebecca walks is an example of a
regular star polygon.
Unlike the sides of an ordinary regular polygon,
the sides of a regular star polygon intersect each other.

You can show that after walking a certain number of $100$-foot lengths and performing the $165$-degree turn at the end of each $100$ feet, Rebecca will have made several full turns plus an additional $180^\circ.$
That is, she will be facing exactly opposite the direction in which she first started walking. The next $100$ feet will cancel the first $100$ feet she walked, then she will cancel the next $100$ feet, and so forth until she has canceled all of the "certain number of" $100$-foot lengths and is back where she started.

I leave it as an exercise to figure out how many times Rebecca has to turn before she is facing opposite her original direction for the first time.

Note that the vertices of a regular star polygon are also vertices of an ordinary regular polygon. If you can figure out how many turns Rebecca makes to get back where she started, you can figure out the number of vertices of the star polygon.
You can then try to figure out how to inscribe the star polygon in a regular polygon with that number of vertices in case you are really curious to see what her path is like and exactly how she gets back to her starting point.
Technically, however, you don't need to draw the polygon in order to find the answer to the question.
You just need to know how many sides it has.

As a hint to how to figure out the number of vertices without having to use the internal-angle formula from the web page (which is suspect anyway, since it only accounts for the $p$ in the notation $\{p/q\}$, whereas $\{7/3\}$ clearly has a different angle than $\{7/2\}$),
consider that after turning $165^\circ$ clockwise at every vertex,
Rebecca must be facing in the same direction she started.
That is, she must have turned through a whole multiple of $360^\circ.$
Also consider that if there are $n$ vertices, Rebecca would have turned
$n \times 165^\circ$ clockwise

So try to see what is the least multiple of $165^\circ$ that is also a multiple of $360^\circ.$
As a further hint, notice that $165^\circ = 180^\circ - 15^\circ.$

A postscript:

I would also like to note that it is possible for a star polygon to have an odd number of sides (the five-pointed star is an obvious example), in which case none of the walks along the sides is canceled by another side in a $180$-degrees opposite direction. But the final displacement is the vector sum of all the displacements along the individual sides, which combine by adding the vectors; and the vector sum is zero. One way to see this is to rearrange the sequence of sides to that they form a regular convex polygon.


问:
Area for multivariable function
I have to find the area of the graph of the function $z(x,y)= \arctan \frac{y}{x} $  Defined on $\{(x,y) | x^2 + y^2 \leq 1 \text{ and }|x|\leq y \}$

So, I have to use integration. But how to deal with the domain? I guess I have to reformulate my domain as integration bounds, right?

答:
By drawing $D = \{ (x,y) | x^2 + y^2 \le 1 ~~ \text{and} ~~ -y \le x \le y \} = \{(x,y) | x^2 + y^2 \le 1 ~~ \text{and} ~~ -1 \le \frac{x}{y} \le 1 \}$

we can actually see that it is basically a slice of the unit circle, a quarter to be exact:

$$ |x| \le y \rightarrow  -y \le x \le y$$

Is basically the function $|x|$ and all the area 'above' the function.

(I cannot post images because the lack of reputation here, and so I highly recommend you to look it up on sites such as Desmos here: https://www.desmos.com/calculator)

We can do a substitution of variables as so:

$$\begin{aligned} \left\{\begin{matrix}
u = x^2 + y^2\\ v = \frac{x}{y}
\end{matrix}\right. \end{aligned} $$

And so our new set $D$ will be:

$$ D = \{(u,v) | u \le 1 , -1 \le v \le 1 \} $$

Jacobian:

$$J = \frac{D(x,y)}{D(u,v)} = \frac{1}{ \frac{D(u,v)}{D(x,y)}} = \frac{1}{\begin{vmatrix}
u_x & u_y\\
v_x & v_y
\end{vmatrix}} = \frac{1}{\begin{vmatrix}
2x & 2y\\
\frac{1}{y} & \frac{-x}{y^2}
\end{vmatrix}} = \frac{1}{-2 \frac{x^2}{y^2} -2} = \frac{1}{-2(v^2 + 1)}$$

Back to the integral(s)

$$ I = \int \int_{D_{uv}} \arctan( \frac{1}{v} ) \cdot \frac{1}{-2(v^2 + 1)} dudv \\ -\frac{1}{2} \left ( \int_{-1}^{1} \frac{\arctan{ \frac{1}{v}}}{1 + v^2} dv \right ) \cdot \left ( \int_{0}^{1} du \right )$$

The first integral seems intimidating, however we can notice it is an odd function meaning $f(v) = -f(-v)$ on a symmetric interval $(-1,1)$ and thus the area is $0$.

Title: Area for multivariable function


答:
Whether for the number or its base the same rule

For $(x,y ) >0$
$$\log_y (x)=\dfrac{\log x}{\log y}$$

for $ x>0, y <0$
$$\log_y (x)=\dfrac{\log x}{\log y+  (2k-1) \pi i}$$

for $ x<0, y <0$
$$\log_y (x)=\dfrac{\log x+  (2n-1) \pi i}{\log y+  (2k-1) \pi i}.$$

Title: Negative base of log if working in $\mathbb{C}$?

Body: Would a negative base of a logarithm be allowed if one were working in the field of complex numbers, $\mathbb{C}$?


问:
Find a value of joint density for transformed random variables.
I am working on prep questions for the Qualifying exam and stuck with the following one. I tried to find the joint density using the transformation by the Jacobian, but it did not work out because I could not figure out the inverse of $U$ and $V$. Any helps would be appreciated!

Let $X$ and $Y$ be independent Exponential random variables with mean 1. Let
$$U = \exp(X) + 2\exp(Y), V = 2\exp(X^2) + \exp(Y^2)$$
and let $g(u,v)$ be a joint density for $U$ and $V$ which is as continuous as possible. Find the value of $g(3e,3e)$.

答: Define $(u,v)=\big(e^x+2e^y,2e^{x^2}+e^{y^2}\big)$ where $(x,y)\in (0,\infty)^2$. Observe how $$(u,v)=(3e,3e) \iff (x,y)=(1,1)$$ Therefore $$g(3e,3e)=f_{XY}(1,1)\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|_{(u,v)=(3e,3e)}\Bigg|=\frac{f_{XY}(1,1)}{\Bigg|\frac{\partial(u,v)}{\partial(x,y)}\Bigg|_{(x,y)=(1,1)}\Bigg|}$$ Since $X,Y\sim \exp(1)$ are independent we have $f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)=e^{-(x+y)}$ which is supported $(0,\infty)^2$. Also $\frac{\partial(u,v)}{\partial(x,y)}=2ye^xe^{y^2}-8xe^{x^2}e^y$ so that $g(3e,3e)=\frac{e^{-2}}{\big|-6e^2\big|}=\frac{1}{6e^4}$.


问:
How to construct abstractly and symmetrically the map $^{n+k}M⊗^kN→^nM$ given a pairing $M⊗N→R$?
Let $R$ be a commutative ring and let $M,N$ be $R$-modules.

Let $\langle-,-\rangle$ be a pairing $M⊗N→R$. We can construct a pairing $\Lambda^kM⊗\Lambda^kN→R$ by sending $(a_1∧⋯∧a_k)⊗(b_1∧⋯∧b_k)$ to $\det(\langle a_i, b_j \rangle)_{ij}$.
We can more generally construct a map $\Lambda^{n+k} M ⊗ \Lambda^n N → \Lambda^k M$ that can be defined recursively by
$$ \langle a_1∧⋯∧a_{n+k}, b_1∧⋯∧b_k \rangle = \sum_{i=1}^{n+k} (-1)^{i-1} \langle a_i,b_1\rangle \langle a_1∧⋯∧\widehat{a_i}∧⋯∧a_{n+k}, b_2∧⋯∧b_k \rangle $$
where $\widehat{a_i}$ means that $a_i$ is omitted. The result is a sum over all injections $f : [\![1,k]\!] → [\![1,n+k]\!]$ of $$± \bigg(\prod_{i=1}^k \langle a_{f(i)}, b_i \rangle \bigg) \bigwedge_{j \not\in \operatorname{image}(f)} a_j.$$

Bourbaki gives an abstract construction of these maps as follows, but it is not symmetrical and I don't know how to prove the symmetry of the definition without referring to the formula. Here is the construction of Bourbaki (Algèbre III):

We start by defining a coalgebra structure on $\Lambda M$ by applying $\Lambda$ to the coalgebra structure on $M$ given by the diagonal $M→M⊕M$ and the zero map $M→0$. We get a comultiplication $\Lambda M → \Lambda(M⊕M) = \Lambda M ⊗ \Lambda M$ and a counit $\Lambda M→0=R$. Dualizing degreewise this structure, we get an algebra structure on the graded dual
$$(\Lambda M)' = \bigoplus_{k=0}^∞ (\Lambda^k M)^*.$$ This is an anticommutative alternating graded algebra (which can be deduced from the fact that $\Lambda M → \Lambda (M⊕M)$ is a morphism of algebras, without computation), so by the universal property of $\Lambda (M^*)$, the morphism $M^* → (\Lambda M)'$ gives an algebra morphism $\Lambda (M^*) → (\Lambda M)'$ and thus an action on $\Lambda M$.

So this construction gives the pairing $\Lambda^k M ⊗ \Lambda^k M^* → R$ and if we have a morphism $N→M^*$, we can deduce the pairing $\Lambda^k M ⊗ \Lambda^k N → \Lambda^k M ⊗ \Lambda^k M^* → R$.

Is there an other abstract construction that is more symmetrical than Bourbaki's?

PS: I include the tag "category theory" because perhaps people watching this tag know an answer.

答:
Note that  $\wedge M $ is an $R$ algebra with the following universal property: we have the map $m \mapsto m \in \wedge^1(M) \simeq M$, and $m^2=0$ in $\wedge M$. Moreover, whenever we have an $R$ algebra $A$ and an $R$ linear map from $M$ to $A$ such that any $m$ maps to an element of square $0$, then there exists a map... ( a particular case of Clifford algebra, when the quadratic form is $0$). Now, you need to know how to take the graded tensor products of graded commutative algebras. Once you know that, you know what is $\wedge M \otimes \wedge M$, another graded commutative algebra. Now consider the $R$ linear map from $M$ to $\wedge M \otimes \wedge M$, $m \mapsto m\otimes 1 + 1\otimes m$. Note that we have
$$( m\otimes 1 + 1\otimes m)( m\otimes 1 + 1\otimes m)= m\wedge m \otimes 1 + m \otimes m - m \otimes m + 1 \otimes m \wedge m = 0$$
We conclude there exists a morphism of $R$ algebras
$$\wedge M \to (\wedge M )\otimes (\wedge M)\\
m \mapsto m \otimes 1 + 1 \otimes m$$

Let's calculate for instance the image of $m_1 \wedge m_2 \wedge m_3$. We have
$$m_1 \wedge m_2 \wedge m_3\mapsto ( m_1 \otimes 1 + 1 \otimes m_1)( m_2 \otimes 1 + 1 \otimes m_2)(m_3 \otimes 1 + 1 \otimes m_3) = \\=m_1 \wedge m_2 \wedge m_3\otimes 1  +  (m_1 \wedge m_2 \otimes m_3 + m_2 \wedge m_3 \otimes m_1  +  m_3 \wedge m_1 \otimes m_2) + \\+(\cdots )+ 1 \otimes m_1 \wedge m_2 \wedge m_3$$

We can describe the image of an element $m_1 \wedge \cdots \wedge m_l$. The rule of signs is like the one for the Laplace expansion. The important thing is that globally the permutation across the $\otimes$ sign is even. I hope it is clear enough.

Now, the algebra $\wedge M \otimes \wedge M$ is bi-graded. So we have maps from
$$(\wedge M \otimes \wedge M)^{n+k} \to \wedge^k M \otimes \wedge^n M$$   This gets us an $R$ linear  map
$$\wedge^{n+k} M \to \wedge^k M \otimes \wedge^n M$$
From the above it should be clear what this map is. We used the graded tensor product of $\wedge M$ with itself  to justify that the map is OK, and the correct form. Otherwise, one can just take it as a definition. Anyways, this linear map gets us a linear map
$$\wedge^{n+k} M \otimes \wedge^k N \to \wedge^k M \otimes \wedge^n M\otimes \wedge^k N$$

But we have a linear map
$$\wedge^k M \otimes \wedge^k N\to R\\
(m_1 \wedge\ldots\wedge m_k) \otimes (n_1 \wedge\ldots\wedge n_k) \mapsto \det( \langle m_i, n_j\rangle)$$

Now compose and we get our desired map.

We should also mention the following important fact that shows that one can define the interior product inductively. Say we have $\alpha \in \wedge^{n+k_1+ k_n}$, $\beta_i\in \wedge^{k_i}N$, $i=1,2$. Denote the pairing
$ \wedge^{n+k}M \otimes \wedge^k N \to \wedge^n M$ by
$(\alpha, \beta) \mapsto \alpha\Finv \beta$. Then we have
$$\alpha\Finv(\beta_1\wedge \beta_2) = (\alpha\Finv \beta_1) \Finv\beta_2$$
This follows from the Laplace expansion of determinants.

$\bf{Added:}$ We can see that $\Finv \beta$ is the adjoint of $\beta \wedge \cdot $  on the $N$ side.


问:
Interpreting the logarithm as a sum of simple poles along the negative real axis
I've heard it remarked that you can basically consider $\log z$ to be a function which has simple poles everywhere on the negative real axis (with a constant "residue density" at each pole).  This would be something like
$$ \log z = \int_0^\infty \frac{dx}{z + x} $$
But of course the integral on the right-hand side actually diverges.  We actually get
$$ \int_0^\infty \frac{dx}{z + x} = \lim_{b \to \infty} \int_0^b \frac{dx}{z + x} = \lim_{b \to \infty} \left( \log(z + b) - \log(z) \right) = \infty $$

In physics, there are a variety of methods for subtracting out the divergent part of such a limit to get a finite answer (various flavors of regularization and renormalization).  I'm wondering whether there is a standard approach here so that something similar can be done to "rescue" the first equation above from the divergent part of the integral.

Another way of phrasing the problem above is that I showed that the Stieltjes transform of a constant on the interval $(-\infty, 0]$ does not exist.  But perhaps there is another density function $\rho$ so that $\log z$ is the Stieltjes transform of $\rho$.
$$ \log z = \int_0^\infty \frac{\rho(x)}{z + x} dx $$
What is $\rho$?  Well, the Stieltjes inversion formula says that it should be given by
$$ \begin{align} \rho(x) &= \lim_{\epsilon \to 0} \frac{\log(x+i\epsilon) - \log(x-i\epsilon)}{2\pi i} \\ &= \frac{(\log |x| + \pi i) - (\log |x| - \pi i)}{2\pi i} \\ &= 1 \end{align} $$

But this gets me exactly back to the integral that I started with, which is divergent!  Hopefully I am just missing something obvious.

Edit: Alternate statement of question

There has been a lot of confusion in the comments below about what I am looking for, so let me restate it in a very narrow way.  I would be satisfied with either of the following:

A sequence of meromorphic functions $f_n(z)$ with simple poles along the negative real axis with the following properties:

a. The poles become dense in the limit $n \to \infty$.

b. $\lim_{n \to \infty} f_n(z) = \log z$

A proof that there is no such sequence.

答:
The correct formula is
$$\log z = \lim_{T\to \infty} \log(z)-\log(1+\frac{z-1}{T+1})$$ $$=\lim_{T\to \infty} (\log(T+1)-\log(1))-(\log(T+z)-\log(z))$$ $$= \lim_{T\to \infty} \int_0^T (\frac1{1+x}-\frac{1}{z + x}) dx=\int_0^\infty (\frac1{1+x}-\frac{1}{z + x}) dx$$

There is no distribution such that $\log (x+iy) = \frac1{x+iy}\ast \rho$ because this distribution would have to be $C+1_{x>0}$ yielding a divergent integral.


答:
You seem to confuse the definition of $R_n$ being about elements $z$ so that $z^n$ has a property, with believing the elements of $S$ are the $z^n$.  Any you seem to confusing your task of discovering things about elements $x,z \in R_n$ with discovering things about $x^n,z^n$.

In proving that $1$ is an identity of $R_n$ you must prove for all $z \in R_n$ that $1*z = z$.  You seem to think we want to prove that $1\cdot z^n = z^n$ (or $1\cdot z^n = 1$.... your argument is not clear.

Anyway as $1$ is the identity element for $\mathbb C$, $1$ will inherit its identity function for any subset containing $1$ (whether or not the subset is a group or not).  So all we need to do is show $1 \in \mathbb R_n$.  Which it is as $1^n = 1$.

And in proving muliplication is closed within elements of $R_n$, we must show if $x,z\in R_n$ that $xz\in R_n$.  Start to show this correctly but then you conclude with the wrong conclusion $x^nz^n \in R_n$ (for which your argument isn't valid).

What we wish to show is that $x,z\in R_n$ will imply $xz \in R_n$.  And, as multiplication is commutative $(xz)^n = x^nz^n$ and as $x,z\in R_n$ we hae $x^nz^n=1\cdot 1 =1$ and so $xz\in R_n$.

Your third argument is incomprehensible.

We must show that if $z\in R_n$ that $z^{-1}\in R_n$.  That is if $z^n =1$ then $(z^{-1})^n = 1$.  You could do it the way it seemed like you were attempting by noting as $z^n =1$ then $z^{-1} = z^{-1}z^n = z^{n-1}$ and $(z^{n-1})^n = z^{n(n-1)} = (z^n)^{n-1} = 1^{n-1} =1$ so $z^{-1}\in R_n$ but that seems a bit convoluted.  There is a more straightforward way:  $(z^{-1})^n = (z^{-1})^n\cdot 1 = (z^{-1})^n\cdot z^n = 1$.

....

But before you do any of this you must actually prove $R_n$ is a subset of $S$.  That is if $z^n = 1$ you must prove $|z| = 1$.

Title: Show that $R_{n}=\left\{z \in \mathbb{C} \mid z^{n}=1\right\}$ is a subgroup of $S=\{z \in \mathbb{C} : |z|=1\}$.

Body:
Show that for each $n \in \mathbb{N}, n \geq 2$ the set of roots $n$ n-th of the unity: $R_{n}=\left\{z \in \mathbb{C} \mid z^{n}=1\right\}$ is a subgroup of $S=\{z \in \mathbb{C}  : |z|=1\}$.

I try:

$1 \in R_n $ because $1^n=1$ and $1*z^n=1*1=1$ for all $z \in R_n$

If $z,x \in R_n$ then $z^n*x^n=(z*x)^n=1*1=1 $ then $z^n*x^n \in R_n$

and if $z\in R_n$ then $z^{-1} \in R_n$ because $z^{-1}=\frac{1}{z^n}=\frac{1}{1}=1=z*z^{-1}$

is right?


问:
Given a group $G$ and given any element $g_i \in G$ does $g_i^n \in G$?
Given a finite group $G$ and assuming $G = \{g_0,g_1,g_2,\dots,g_m\}$ for a positive $m$ value.

If we choose a $g_i$ where $g_i \in G$.
Does the sequence $\{g_i,g_i^2,g_i^3 \dots \ \} \subset G$ due to the fact it is closed under binary operation? How do I guarantee that no element of the sequence will be bigger than $g_m?$ It must have some kind of 'circularity' to make things hold.

Can someone enlight me on how this statement holds(if holds) and why?

答:
This is part of the definition of a binary operation and a group.  The product of any two elements in a group is again an element of that group.

As for being "bigger than $g_m$", that doesn't make any sense.  There is no inherent notion of size in a group, so there is no bigger to speak of.  The indexes you've chosen to use to label your group elements are completely arbitrary and have nothing to do with the size of those elements.

It may help to work out a few examples of small finite groups.  Write down all their elements and work out how to compute products of those elements.


问:
Could exists a vector field on $\mathbb{S}^{2}$ with exactly $n$ zeroes?
I just started to learn index theory of tangent vector fields. I'm aware of two examples on the sphere $\mathbb{S}^{2}$ with exactly one zero, which, which are $F(x,y) = (1-x^2-y^2)\partial x$ thought on $\mathbb{D}^2$ and then indentify the boundary, I think it is

And one with two zeroes should be something like this

I was wondering, it exists a tangent vector field with $n$ zeroes on the sphere ? I thought that if I have $n$ zeroes (isolated), then thanks to the Poincarè-Hopf theorem the sum of the indices must be  $\chi(\mathbb{S}^{2}) = 2$. I thought that some combination of $-1$ and $1$ and a final zero of index $2$ could do the job, but can this vector field actually exist?
Something like defining a polynomial with $n$ distinct zeroes $p : \mathbb{C} \longmapsto \mathbb{C}$ can be extended to $\mathbb{S}^{2}$ to gain a tangent vector field on the sphere?

Thanks in advance, any help would be appreciated.

答:
Here's a more topological way to look at it, in terms of height functions and Morse theory. When $n$ is even, take a sphere that's reshaped to have $n/2$ local maxima, one global minimum, and $n/2-1$ saddle points (so it looks like a camel with $n/2$ humps, closed up at the bottom). (Then we have $\chi(S^2) = n/2-(n/2-1)+1 = 2$, as it should.) We can't make this work when $n$ is odd, since twice the number of saddle points must equal $n-2$.

At any rate, you get the vector field on the sphere by taking the gradient of the height function, i.e., by projecting a vertical unit vector (up or down doesn't matter) onto the tangent plane of the sphere at each point.


问:
Is the image of a subspace a subspace?
I'll clarify this question a bit. Let $f: E \rightarrow F$ be a linear map between these two vector spaces. Let $H1 \subset E$.

Now, I know that $Im(E) \subset F$. But is $Im(H1) \subset F$ true? My initial assumption is that this isn't always true, since, for example:

$f: R \rightarrow R-{0}$ ;  $x \mapsto x-2x$. Now the image of the subsapce which contains every $2x$ isn't a subspace since it would be equal to $0$, and therefore isn't defined in $R-0$.

Assuming this is correct (correct me if I got something wrong, please!), then in what cases would the image of $H1$ be a subspace of $F$? When it's bijective, for example?

答:
Your mistake is that the map you give from $f: \mathbb{R} \to \mathbb{R}\setminus 0 $ by
$$
x \mapsto x - 2x
$$
Doesn't actually map to $\mathbb{R}\setminus 0 $ since $f(0) = 0 \not\in \mathbb{R}\setminus 0$ (note also that $\mathbb{R}\setminus 0$ does not form a vector space under addition). In the general case, yes you can always restrict the domain of a linear transformation from vector spaces, and the image of the restriction will always be a subspace of the target space. Further given a linear transformation $f:V\to W$ with $V$ and $W$ being vector spaces, if $S$ is a subspace of $V$, then the image of $S$ is a subspace of the image of $V$.

Title: Is the image of a subspace a subspace?


问:
Deduce that Bd$(A)=\overline{A}-\text{Int}(A)$.
My textbook defines the boundary of $A$ as Bd$(A)=\overline{A} \cap \overline{A^c}$.

Deduce that Bd$(A)=\overline{A}-\text{Int}(A)$.

$\textbf{Proof:}$ $(\implies)$ We know that Bd$(A) \subset \overline{A}$, int$(A) \subset A \subset \overline{A}$, and Bd$(A) \cap$ int$(A) = \emptyset$.

Thus, let $x \in$ Bd$(A)$. Then $x \in \overline{A} -$ int$(A)$.

$(\impliedby)$ Let $a \in \overline{A}-$int$(A) \implies a \not \in$ int$(A)$.

So, $a \in G \subset A$ where $G$ is open and $\exists a \neq b$ where $b \in G$ and $b \not \in A$.

$\therefore a \in$ Bd$(A)$.

$\therefore$ Bd$(A)=\overline{A}-$int$(A)$. $\blacksquare$

Some remarks that were left on my proof were:

Why is Bd$(A)-$int$(A)= \emptyset$?

When you "Let $a \in \overline{A}-$int$(A)$," this makes $a$ specific, so you can't say $\exists a$.

Finally, "$a \in G \subset A$ where $G$ is open," but $a \not \in$int$(A)$?

Any help with the remarks and the proof will be very helpful.

答: It is heard to follow your second direction of the proof. I will write my proof here. $C(A)=\mathbb R\setminus A$ let $a\in \overline A-\text{int(A)}$ then $a\in \overline A$ and $a\notin \text{int}A$. So, $a\notin\text{int}A$ implies that  for all open set $G$ such that $a\in G$ we have $a\in G\not\subset A$ . This means $G\cap C(A)\neq\emptyset$. So, $a\in\overline {C(A)}.$ Hence. $a\in\text{Bd}(A)).$ Another way to see it. By using the very important result which is $$\text{int(A)}=C(\overline{(C(A)})$$ Your lemma, we will follow immediately from  that.


答:
"due to the fact it is closed under binary operation? "

yes, is binary operation so $g_i^2\in g$ and $g*g_i^2 = g^3\in G$ and so by induction for all powers.

"How do I guarantee that no element of the sequence will be bigger than gm?"

Because that would be logically impossible.  As $G$ has only $m$ elements and all $g_i^k\in G$ there can only be at most $m$ (or fewer) of these $g_i^k$.

"It must have some kind of 'circularity' to make things hold."

Yes it must. And it does.

There must be infinitely many $g_i^k = g_i^l =g_i^w=.....$.

You will discover later that if $G$ is finite that there will be some value of $k$ where $g_i^k =e$, the identity element, and that therefore $g_i^{nk + r} = g_i^r$ and thus the infinite repeats.  The smallest value of $k$ so that $g_i^k =e$ is called the order of $g_i$.

[I haven't proven this is so.  But by the fact that there are infinitely many $n \in \mathbb N$ but only finitely many elements in $G$ there must be some $i,j\in \mathbb N, i < j$ but $g^i = g^j$.  From there we can prove that $e=(g^{-1})^i g^{i} = (g^{-1})^ig^{j} = g^{j-i}$.  So there is a $k:= j-i$ where $g^k = e$.]

......

But note.  If $G$ is infinite then all the $g_i^k$ can be distinct.

And BTW, please to not get confused be the subindexes.  They have nothing to do with anything.

If I had my way I'd have said the group, $G$,  has $m$ elements and not bother giving them labels until needed.

and the statement about powers of elements, I'd say as this: for any $g\in G$ we have $g^k \in G$.  (Which does mean that the set of $\{g,g^2,g^3,.....\}$ must be finite and the there must be (infinitely many) $i,k \in \mathbb N; i\ne k$ but where $g^i = g^k$.  (In other words "It must have some kind of 'circularity' )

I hope things are clear.

......

points to the OP for the phrase ""It must have some kind of 'circularity' " which is absolutely true.

And thanks to LSpice for telling me not to label the elements of $G$ as $\{e,a,b,c,....,z\}$ which just trades one confusion for another.

Title: Given a group $G$ and given any element $g_i \in G$ does $g_i^n \in G$?

Body:
Given a finite group $G$ and assuming $G = \{g_0,g_1,g_2,\dots,g_m\}$ for a positive $m$ value.

If we choose a $g_i$ where $g_i \in G$.
Does the sequence $\{g_i,g_i^2,g_i^3 \dots \ \} \subset G$ due to the fact it is closed under binary operation? How do I guarantee that no element of the sequence will be bigger than $g_m?$ It must have some kind of 'circularity' to make things hold.

Can someone enlight me on how this statement holds(if holds) and why?


答:
I used the naive method and from your insight $f(r,t)=\log(1-e^{it}r)+\log(1-e^{-it}r)$. Then we have, by substituting the inner integrand to be $u$ for each integral,

\begin{equation}
\begin{split}
\int_0^\pi\log(1-e^{it}r)dt+\int_0^\pi \log(1-e^{-it}r) dt&=\int_{1-r}^{1+r}\log u\dfrac{du}{i(u-1)}\\ &+\int_{1-r}^{1+r}\log u\left(-\dfrac{du}{i(u-1)}\right)=0
\end{split}
\end{equation}
as desired.

Title: Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$

Body:
Show that for $r \in (-1,1)$

$$ \int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt = 0$$

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$
The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.


问:
Is it possible to make a regular 3-polygon by selecting $3$ points in $S$
I would appreciate if somebody could help me with the following problem:

Let $\mathbb{Z}$ be the set of all integers and let $ S = \mathbb{Z} \times \mathbb{Z} $.

Question:

1). Is it possible to make a regular 3-polygon by selecting $3$ points in $S$ ?

2). Is it possible to make a regular 5-polygon by selecting $3$ points in $S$ ?

3). Is it possible to make a regular $n$-polygon by selecting $n$ points in $S$ ?

答:
Hints:

A regular $\,3$-gon is just an equilateral triangle. To simplify things assume one of the vertices of this triangle is $\,(0,0)\,$ and one of the sides is on the positive $\,x$-axis. For a second side we need a straight line through the origin forming an angle of $\,60^\circ\,$ with the positive $\,x$-axis, i.e. the line

$$y=\sqrt 3\,x$$

If an element of $\,(n,m)\in S\,$ is on the above line, then

$$m=\sqrt 3\,n\,\,.\;\text{ Do you think this is possible for}\;\; \,m,n\in \Bbb Z\;?\ldots$$

Do something similar to the above for the regular pentagon: we need then a line with slope equal to $\,\tan 108^\circ\,$ and etc.


问:
Residues of $z^{\alpha}\Re(z)$ and $z^{2\alpha + 1}\Re(z^{2})$
For the function $z^{\alpha}\Re(z)$ where $\Re(z)$ is a rational function with a zero of at least two at infinity, at most a simple pole at the origin and $0<\alpha<1$.

How do we show that the residues of $z^{\alpha}\Re(z)$ in the whole plane is the same as the residues of $z^{2\alpha + 1}\Re(z^{2})$ in the upper half plane?

This is based on an extract of a paragraph from Ahlfors Complex Analysis

答:
If $\phi$ is biholomorphic $V\to U$ and $a\in U$ is a simple pole of $h$ meromorphic on $U$ then for $z$ close to $a$ $$h(z)=\frac{Res(h(w),a)}{z-a}+O(1)$$ For $s$ close to $b=\phi^{-1}(a)$ $$h(\phi(s))=\frac{Res(h(w),a)}{\phi(s)-a}+O(1)=\frac{Res(h(w),a)}{\phi'(b)(s-b)+O(s-b)^2}+O(1)$$ $$
=\frac{Res(h(w),a)}{\phi'(b)(s-b)}+O(1)$$
thus $$Res(h(\phi(s)),b)=\frac{Res(h(z),a)}{\phi'(b)}$$
It fails if $a$ is a double pole, try with $\phi(s)=1/s, h(z)=1/(z-1)^2$, $h(\phi(s))=1+\frac{2s-1}{(s-1)^2}$ has a non-zero residue at $1$.


问:
$f: B^2 \rightarrow B^2$ be a continuous map such that $f(x) = x$ for every $x \in S^1$, show $f$ is surjective
$B^2 = \{x \in \mathbb{R}^2; ||x|| \leq 1\}$ is the unit disk inthe plane, and let $f: B^2 \rightarrow B^2$ be a continuous map such that $f(x) = x$ for every $x \in S^1 = \{x \in \mathbb{R}^2; ||x|| = 1\}$. Show that $f$ is surjective.

This my following attempt at a solution:

If $f$ isn't surjective, then there exists a point $y \in B^2$ such that $f(x) - y$ is never $0$ for all $x \in B^2$. Thus $\frac{f(x) - y}{|f(x) - y|}$ is well defined and exists on $S^1$.

however, I'm sure this is a dead end. Some hints would be greatly appreciated!

答:
Note that since $y \notin S^1$, $ty \notin S^1$ for all $t \in [0,1]$. This implies that $H(x,t)= \frac{x-ty}{|x-ty|}$ is a homotopy between $1_{S^1}$ and the map $g$ that you defined composed with $\iota : S^1 \rightarrow B^2$. Furthermore, this homotopy fixes the point $y/|y| \in S^1$. Thus, we have that
$$S^1 \xrightarrow{\iota} B^2 \xrightarrow{g}S^1$$ is (pointed) homotopic to $1_{S^1}$. Now apllying we the fundamental group we have that the identity

$$1_\mathbb{Z}: \mathbb{Z} \rightarrow 0 \rightarrow\mathbb{Z}$$
factors through the zero group (since homotopic maps induce the same group morphism), a contradiction.


问:
Quadratic Functions
Consider the strictly convex quadratic function $f(x) = \frac{1}{2}x^tPx - q^tx + r,$ where $P \in \mathbb{R}^{n \times n}$ is a positive definite matrix, $q \in \mathbb{R}^n$ and $r \in \mathbb{R}.$ Let $\mathcal{H} := \{H: H \text{ is a }k- \text{dimensional subspace in } \mathbb{R}^n\}.$ Clearly, the restriction of $f$ to any $H \in \mathcal{H}$ is again a strictly convex function. For any $H \in \mathcal{H},$ we will use $x_H$ to denote the unique optimal point of the following problem

\begin{equation*}
\underset{x \in H}{\text{min.}} \;  f(x).
\end{equation*}

Now consider the map, $\psi(H) = x_H.$

Prove / Disprove: The map $\psi$ is bijective.

Remark: It is assumed that $P$ is invertible and $q \neq \mathbf{0}.$

答:
Edit: Whether $\psi$ is surjective depends on the definition of its codomain. For example, if $n=2$ and $k=1$, the image of $\psi$ is a curve on $\mathbb{R}^2$. So, if the codomain of $\psi$ is $\mathbb{R}^2$, $\psi$ is certainly not surjective.

At any rate, $\psi$ may not be injective even if the origin is not a global minimum. Consider $a=(1,0,0)^T$ and $f(x)=\|x-a\|^2$. Let $\{e_1,e_2,e_3\}$ be the canonical basis of $\mathbb{R}^3$. Then $x_H=a$ when $H=\operatorname{span}\{e_1,e_2\}$ or $\operatorname{span}\{e_1,e_3\}$.

Title: Quadratic Functions


问:
How to prove this function is $L^{r}(E)$
Let $p,q\in[1,\infty)$, $E\subseteq\mathbb{R}$ measurable with finite measure. Let $u\in L^{p}(E)\cap L^{q}(E)$, fix $\alpha\in[0,1]$ and define:
$$\frac{1}{r}=\frac{1-\alpha}{p}+\frac{\alpha}{q}$$
Show $u\in L^{r}(E)$ and: $$\|u\|_{L^{r}(E)}\leq\|u\|^{1-\alpha}_{L^{p}(E)}+\|u\|^{\alpha}_{L^{q}(E)}$$

答: See the interpolation section in https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality.</p>


答:
Yes, and it is possible to choose the indices of the zeros freely as long as the sum is $2$. The first part of this answer is a slight variation of Kajelad's.

Let $n_1,\dots,n_k > 0$ with $n_1+\cdots+n_k \geq 2$ be the positive multiplicities. The polynomial $(z-1)^{n_1}\cdots(z-k)^{n_k}$ gives a vector field on $\mathbb{C}$ that looks like $z^{n_1+\cdots+n_k}$ when $z$ is big. Let's compute what this field looks like around $∞$ on the Riemann sphere. We apply the change of charts given by $z \mapsto 1/z$, so we precompose our field by this function and divide by $(1/z)' = -1/z^2$. This gives $(1/z)^{n_1+\cdots+n_k-2}$. The $2$ of the Euler characteristic appears here. There is a pole at $0$ but we can transform it to a zero instead by inverting the norm of the field, with a smooth transition. It gives $\overline{z}^{n_1+\cdots+n_k-2}$. The $\overline{z}$ means that the multiplicity is negative instead of positive but except from that, we are in the same situation as before and we can apply the same technique to separate this zero into as much zeroes of negative multiplicities as we want.

We can add zeroes of index $0$ by multiplication with a real field vanishing at the desired points, as in the answer of Elliot Yu.

Here is an example of a field with four zeroes of multiplicities $1$ (we only see three of them) and two zeroes of multiplicities $-1$ (we only see one of them).

More generally, if you have a finite set of closed loops with self-intersections on the sphere (here, two perpendicular circles separating the sphere into four quarters), then you can build a field by placing one zero of multiplicity $1$ into each of the delimited region and a zero of multiplicity $-1$ at each intersection point of the loops. Each zero of multiplicity $1$ has a "typhoon" going around it as in the drawing above and the directions of these typhoons alternate as we change of region. To see that it always works, we can do by recursion on the number of self-intersections: remove an intersection point by merging two opposite regions.

For instance, a loop drawing an 8 will produce three zeroes of multiplicities $1$ and one zero of multiplicity $-1$.

Look also at this old video of Vi Hart: https://www.youtube.com/watch/heKK95DAKms
When she colors the regions, the black ones are those oriented in one direction and the white ones are those oriented in the other direction.

I don't know how this reverses nor the general theory behind all of this, so if someone knows the keywords, I'm interested.


问:
Evaluating $\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$

Evaluate the series $$S=\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$$

I have tried many values of $(a,b,c)$ and plugged into Wolframalpha, it always converges. I know that for particular values of $a,b,c$, we solve it by forming a telescoping series by using the fact that $\displaystyle \arctan x-\arctan y=\arctan\left(\dfrac{x-y}{1+xy}\right)$ and converting it into a form $f(r+1)-f(r)$.

But I think that we cannot convert all types into this form. Even if this was possible, what is the way for us to know what $f$ to use? In particuar, I was evaluating $\displaystyle \sum_{r=1}^{\infty} \cot^{-1}\left(3r^2-r-\frac13\right)$, but couldn't convert it into telescoping series. So, how then, do we solve this? and for what values of $(a,b,c)$ is the sum convergent?

答:
Given quadratic $ar^2 +br+c$, suppose that we can find some function $f(r)$ such that
$$
(\dagger)\quad\quad\frac{1+f(r+1)f(r)}{f(r+1)-f(r)} = ar^2 + br + c.
$$
Then as $$\cot^{-1}(ar^2+br+c)=\arctan\frac{1}{ar^2+br+c}\\
=\arctan \frac{f(r+1)-f(r)}{1+f(r+1)f(r)}=\arctan(f(r+1))-\arctan(f(r)),
$$
we can proceed to find $\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)$ telescopically as mentioned. Indeed we would have $$\sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r)),$$
provide convergence.

Now to find $f(r)$, we propose to seek it in the form $$f(r)=\frac{Ar+B}{Cr+D},$$ a linear fraction. If so, then the LHS of the $(\dagger)$ equation will be a quadratic in $r$. Indeed, substituting this linear fractional into the LHS yields

$$\frac{A^2+C^2}{AD-BC}r^2 + \frac{A^2+C^2+2AB+2CD}{AD-BC} r + \frac{B^2+D^2+AB+CD}{AD-BC}.$$

So the task becomes to find $A,B,C,D$ such that
$$ \begin{eqnarray}a &=& \frac{A^2+C^2}{AD-BC} \\
b &=& \frac{A^2+C^2+2AB+2CD}{AD-BC}\\
c &=& \frac{B^2+D^2+AB+CD}{AD-BC}\end{eqnarray}$$

Since here we overparametrize $a,b,c$ with $A,B,C,D$, we can in principle (see remark below) find these values. For instance you could set $A=0$ to simplify your search. Also note that as $\displaystyle \lim_{r\to\infty}f(r) = \frac AC$, we will have $ \displaystyle \lim_{r\to\infty}\arctan(f(r)) = \arctan\left(\frac AC\right)$.

An example, find $\displaystyle \sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac 13\right)$. We seek $A,B,C,D$ as above that works. Take $A=0$, we have by wolfram alpha $B = 1,C = -3 ,D = 2$ (among many other possible solutions). So $f(r) = \dfrac{1}{-3r+2}$ and
$$
\sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac13\right)\\
=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r))\\
=-\arctan(-1)= \arctan(1) = \frac{\pi}4.
$$

Remark. There is a limitation to this, for instance say $a\neq 0 $ for this $f(r)$ to take this form. Indeed, if $a =0$, then we see that $A=C=0$, which will give a contradiction if $b\neq 0 $. So if $f(r) = \dfrac{Ar + B}{Cr+D}$, then $(a,b,c)$ needs to be in the range of the function
$$(A,B,C,D)\mapsto \left(\frac{A^2+C^2}{AD-BC},\frac{A^2+C^2+2AB+2CD}{AD-BC},\frac{B^2+D^2+AB+CD}{AD-BC}\right).$$

Remark 2. Despite this limitation, you can use this the other way: Pick your favorite four numbers $A,B,C,D$ and write down $f(r) = \dfrac{Ar + B}{Cr+D}$. This generates a quadratic $ar^2 + br + c$, and with this you will have the value of $\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c) = -\arctan\left(\dfrac{A + B}{C+D}\right)+\arctan\left(\dfrac{A}{C}\right)$.


问:
Log expression gives two answers based on how you think about it
when you have the expression $\log( -x)$ base $3$, isn't that undefined, because you can't raise 3 to any power which will give you a negative value?

I am supposed to graph this expression and from the transformation of functions, this makes sense though because you just flip the graph of $log x$ base $3$ over the y-axis.

I would like to know why you get two different answers?

Thanks, Amruth

答:
Since $\log(x)$ is only defined on $(0,\infty)$ we get that $\log(-x)$ is only defined on $(-\infty, 0)$ Morever, given $f(x) = \log(x)$, we get that $f(-x) = \log(-x)$ is the reflection about the y-axis.

Hence, we do not get two answers from the same input of the given function. That is to say : $\log$ is a well-defined function.

Also see here : What is $\text{log}(-x)$?


问:
Allowance distribution problem
When I say allowance, I mean like the kind you earned as a child. Here is the problem I am trying to solve.
A parent has $20 they want to give to the child for doing all of their chores. If the child does all of their chores, they get 100% of their allowance. However is thy only do some of their chores they only get a percentage of their allowance. Here is my kicker. Not all chores are weighed the same.

The chores are weighted:

Big chores = 10

Medium chores = 5

Small chores = 1

If the parent setups a set of chores for the week:

2 big chores(paint house, change oil)

4 medium chores(clean room)

7 small chores(brush teeth, wash dishes)

At the end of the week, the child only does:

1 big chores(paint house)

2 medium chores(clean room)

3 small chores(brush teeth)

I am unsure how to calculate how much allowance the child gets and what the distributions per-chore would be for the $20. Could someone point me in the right directions? The reason I have this question is because I am building an IOS app to help parents and children manage household duties. Please forgive me if I tagged this question wrong, as I am not sure what tag this falls under.

答:
Total the number of points the parent assigns for the week; in this case, $47$: twenty for big chores, twenty for medium chores, and seven for small chores.

Compute the number of points the child achieved. In this case, $23$: ten for the big chore, ten for the two medium chores, and three for the small chores.

Since he accomplished $\frac{23}{47}$ths of the points, he should get $\frac{23}{47}$ths of the original offer:
$$\left(\frac{23}{47}\right)\times \$20 \approx \$9.79.$$

Note that this matches reasonably well: the child did only a little under half the assigned chores, so he's getting only a little under half the money.

In general, total the number of points achieved, divide by the total number of points assigned, and multiply by the reward offered:
$$\left(\frac{\text{points accomplished by child}}{\text{points originally assigned by parent}}\right)\times\left(\text{full allowance amount}\right) = \text{allowance earned.}$$
With this formula, if the child were to do more than was assigned, he would get a comparative "bonus" as well. If you want to prevent bonuses, then you should replace the computation on the left with the minimum between that computation and the full allowance amount.

Title: Allowance distribution problem


问:
Find the values of a and b for which the series $\sum n^a \cdot (ln(n))^b$ converges absolutely, and for which it diverges
Find the values of $a$ and $b$ for which the series converges absolutely, and for which it diverges.

$$\sum_{n=2}^\infty\frac{1}{( n^a\cdot(ln(n))^b)}$$

I have compared $$\frac{1}{(n^a\cdot ln(n)^b)} > \frac{1}{(n^a \cdot n^b)}$$

What I see here is that if $a+b>1$ series converges By the $p$ test.
But I need a full range of values for $a$ and $b$ such that it converges and diverges.

答:
We can think about it$$\int_{2}^{+\infty}\frac{1}{x^{a}}dx$$
When $a>1$,let$$a=1+2\varepsilon(\varepsilon>0)$$
and$$\lim_{x\to+\infty}\frac{\dfrac{1}{x^{1+2\varepsilon}\ln^bx}}{\frac{1}{x^{1+\varepsilon}}}=\lim_{x\to+\infty}\dfrac{1}{x^\varepsilon \ln^bx}=0$$
So$$a>1\implies \sum_{n=2}^{\infty}\frac{1}{n^a\ln^bn}\text{converges} $$
$$a<1\implies \sum_{n=2}^{\infty}\frac{1}{n^a\ln^bn}\text{diverges}$$
When $a=1$,we think about it$$\int_{2}^{+\infty}\frac{1}{x\ln^b x}dx=\int_{\ln2}^{+\infty}\frac{1}{x^b}dx$$
There are similar ones $$b>1\implies \sum_{n=2}^{\infty}\frac{1}{n\ln^bn}\text{converges} $$
$$b\leqslant1\implies \sum_{n=2}^{\infty}\frac{1}{n\ln^bn}\text{diverges}$$


问:
Prove $f$ is continuous if $f(\overline{A}) \subset \overline{f(A)}$ for all subsets $A$.
Let $X$ and $Y$ be topological spaces. Suppose that $f: X \rightarrow Y$ is a function satisfying: For every $A \subset X$, $f(\overline{A}) \subset \overline{f(A)}$.

Prove $f$ is continuous.

$\textbf{Proof:}$ Let $X$ and $Y$ be topological spaces. Suppose $f: X \rightarrow Y$ is a function satisfying $\forall A \subset X$, $f(\overline{A}) \subset \overline{f(A)}$.

Let $a \in \overline{A}$ and then $f(a) \in \overline{f(a)}$.

Let $V \subset Y$ be open with $f(a) \in V$.

By way of contradiction, suppose $\forall \, \text{open} \, U \subset X$ with $a \in U$, $f(U) \not \subset V$.

So, $\forall a \in U$, $b \in X \backslash f^{-1}(V)$ is also $b \in U$.

Then, $a \in \overline{X \backslash f^{-1}(V)} \subset \overline{Y \backslash V}$ $\rightarrow \leftarrow$ (contradiction!) because $V \cap Y \backslash V = \emptyset$.

$\therefore f$ is continuous. $\,\, \blacksquare$

答:
Your idea basically amounts to showing that $f$ is continuous at each point of $X$, but to carry it out properly, you really should start with an arbitrary $x\in X$. Now you need an $A\subseteq X$ such that $x\in\operatorname{cl}A$. That’s easy: let $A=\operatorname{cl}\{x\}$. (Of course if $X$ is $T_1$, $A$ will just be the singleton $\{x\}$, but we don’t really care what it is as long as it contains $x$.)

Now suppose that $f$ is not continuous at $x$. Then there is an open nbhd $U$ of $f(x)$ such that $f[V]\nsubseteq U$ for every open nbhd $V$ of $x$. Thus, for each open nbhd $V$ of $x$ there is a point $x_V\in V$ such that $f(x_V)\notin U$. Let $S=f^{-1}[Y\setminus U]$. Then for each open nbhd $V$ of $x$ we have $x_V\in V\cap S$, so $x\in\operatorname{cl}S$, and therefore

$$f(x)\in\operatorname{cl}f[S]\subseteq Y\setminus U\,,$$

which is absurd, since $f(x)\in U$. Thus, $f$ is continuous at $x$ and therefore is continuous.

And when we get to this point, we see that it was really $S$ that we wanted: we never actually used $A$ at all.


问:
Compute $\tilde{E}\left(B_t - \int_0^t B_s \,ds\right)$
We have that $B$ is a Standard Brownian Motion with $B_0 = 0$ under the prob. measure $P$ and $\tilde{B} = B_t - \int_0^t B_s ds$ for $t\in [0,T]$, $T>0$. I want to calculate
$$
 \tilde{E}\left(B_t - \int_0^t B_s \,ds\right)
$$
where $\tilde{E}$ is the expectation under the probability measure given by Girsanov Theorem... My question is: Since $\tilde{B}$ is a SBM under $\tilde{P}$, is it not zero? Otherwise, how could I do it explicitly?
Thanks in advance

答:
Given
$$\bar B_t = B_t - \int_0^t B_s ds, \tag{1}$$
the corresponding Radon–Nikodym density process of $Q$ with respect to $P$ is
$$\rho_t:=\left. \frac{dQ}{dP} \right|_{\mathcal F_t}=e^{\int_0^t B_s dB_s - \frac{1}{2} \int_0^t B_s^2ds}, \quad 0 \leq t \leq T,$$
which solves the SDE
$$ \rho_t = \rho_t B_t dB_t, \quad \rho_0 = 1. \tag{2}$$
Now, apply the integration by parts formula (Ito product rule) using (1) and (2) to show that $d(\bar B_t \rho_t)=A_t \cdot dB_t$ (for some process $A$). This means that we don't have a drift in the SDE of $B_t \rho_t.$ Thus, $\{B_t \rho_t, 0 \leq t \leq T\}$ is a $P$-local martingale. Then check that $E_P[\int_0^t A^2_s ds] < \infty$ to conclude that $\{B_t \rho_t, 0 \leq t \leq T\}$ is indeed a $P$-martingale. Therefore,
$$E_Q[\bar B_t] = E_P[\bar B_t \rho_t]= E_P[\bar B_0 \rho_0]=E_P[\bar B_0]=E_P[B_0]=0.$$


问:
How do we solve the equation $3\sec^{2}(x) - 4 = 0$?
I am confused about the answer to this question:

$$3\sec^{2}(x) - 4 = 0$$

I got the answer

$$x = \frac{\pi}{6} + 2n\pi\quad\text{or}\quad x = \frac{5\pi}{6} + 2n\pi$$

However, the book's answer is $\pi/6 + n\pi, 5\pi/6 + n\pi$

Isn't the period of a secant function $2\pi$? If so, why is the answer saying $\pi$?

答: Sometimes a plot can really help:

Body: Sometimes a plot can really help:


答:
The proposed equation is equivalent to
\begin{align*}
3\sec^{2}(x) - 4 = 0 & \Longleftrightarrow 4\cos^{2}(x) - 3 = 0\\\\
& \Longleftrightarrow \cos(x) = \pm\frac{\sqrt{3}}{2}\\\\
& \Longleftrightarrow \cos(x) = \pm\cos\left(\frac{\pi}{6}\right)
\end{align*}

Based on it, we can conclude that
\begin{align*}
\cos(x) = \cos\left(\frac{\pi}{6}\right) \Longleftrightarrow x = \pm\frac{\pi}{6} + 2m\pi
\end{align*}
as well as
\begin{align*}
\cos(x) = -\cos\left(\frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \Longleftrightarrow x = \pm\frac{5\pi}{6} + 2n\pi
\end{align*}

Gathering all solutions, we obtain the following solution set:
\begin{align*}
S = \left\{x\in\mathbb{R} \mid \left(x = \frac{\pi}{6} + m\pi\right)\vee\left(x = \frac{5\pi}{6} + n\pi\right)\right\}
\end{align*}

In order to understand it properly, notice that
\begin{align*}
\begin{cases}
-\dfrac{5\pi}{6} = \dfrac{\pi}{6} - \pi\\\\
-\dfrac{\pi}{6} = \dfrac{5\pi}{6} - \pi
\end{cases}
\end{align*}

Based on it, we can express the solution set as proposed in the book.

EDIT

Considering the comment of @HansEngler, we can also solve the proposed equation as follows:
\begin{align*}
3\sec^{2}(x) - 4 = 0 & \Longleftrightarrow 4\cos^{2}(x) - 3 = 0\\\\
& \Longleftrightarrow 2\cos(2x) - 1 = 0\\\\\
& \Longleftrightarrow \cos(2x) = \cos\left(\frac{\pi}{3}\right)\\\\
& \Longleftrightarrow 2x = \pm\frac{\pi}{3} + 2k\pi\\\\
& \Longleftrightarrow x = \pm\frac{\pi}{6} + k\pi
\end{align*}

Title: How do we solve the equation $3\sec^{2}(x) - 4 = 0$?

Body:
I am confused about the answer to this question:

$$3\sec^{2}(x) - 4 = 0$$

I got the answer

$$x = \frac{\pi}{6} + 2n\pi\quad\text{or}\quad x = \frac{5\pi}{6} + 2n\pi$$

However, the book's answer is $\pi/6 + n\pi, 5\pi/6 + n\pi$

Isn't the period of a secant function $2\pi$? If so, why is the answer saying $\pi$?