问:
Is Counting Regarded as Repeated Addition?
I am uncomfortable with the following statement in the draft of my paper:

[T]he unary system of representing $n$ by $n$ contiguous dots uses a singleton alphabet and requires counting rather than addition [emphasis mine] for evaluation of its words.

I am concerned readers might object to the italicized phrase, saying: "Hey, wait a second... how can you say that, when counting implies successive additions?"

For my own part, I see counting as iterated application not of addition but of the successor function, which as we know from Peano's axioms can be regarded as the foundation of addition.

Will the quote above from my paper get me into trouble with my readers?  If so, what might an appropriate fix be?

答: Multiplication is repeated addition; that's the first indication you're wrong. In fact, addition is repeated counting. You'll know this if you've ever seen a small child who's playing a board game advance $5$ steps one at a time, rather than just moving straight from $12$ to $17$.


问:
Complex number being grater than $0$ in the definition of inner product
I was reading the definition of inner product here and they state the following:

Let $V$ be a vector space over a field $\mathbb F$ ( $\mathbb R$ or $\mathbb C$). A map $\left< \cdot , \cdot \right>: V \times V \to \mathbb F$ is called an inner product if, for all $x,y,z \in V$ and $\alpha \in \mathbb F$:

$\left< \alpha x , y \right> =  \alpha \left< x , y \right>$ and $\left< x + z , y \right> = \left< x , y \right> + \left< z , y \right>$

$\left< x , y \right> = \overline{\left< y , x \right>}$

$\left< x , x \right> > 0$, if $x \neq 0$.

My question is about property 3. Let $V$ be a vector space over $\mathbb C$, then $\left< \cdot , \cdot \right>: V \times V \to \mathbb C$, this means that for all $x \in V$, $\left< x , x \right>$ is a complex number. So what do they mean by $\left< x , x \right> > 0$? Do they mean that the inner product of $x$ with itself is allawys a real number, so if $\left< \cdot , \cdot \right>$ is an inner product then $\Im(\left< x , x \right>) = 0$? If not, what does it mean for a complex number to be greater than $0$ in this context?

答: Because of the second property, $\langle x,x \rangle = \overline{\langle x,x \rangle}$, so $\Im(\langle x,x \rangle)=-\Im(\langle x,x \rangle)=0$, i.e. $\langle x,x \rangle$ is real.

ContentLicense: CC BY-SA 4.0


答:
Because of point 2.:

$$\left< x , y \right> = \overline{\left< y , x \right>}$$ applied for $x= y$, you get

$$\left< x , x \right> = \overline{\left< x , x \right>}.$$

Therefore $\left< x , x \right>$ is a real number.

Title: Complex number being grater than $0$ in the definition of inner product

Body:
I was reading the definition of inner product here and they state the following:

Let $V$ be a vector space over a field $\mathbb F$ ( $\mathbb R$ or $\mathbb C$). A map $\left< \cdot , \cdot \right>: V \times V \to \mathbb F$ is called an inner product if, for all $x,y,z \in V$ and $\alpha \in \mathbb F$:

$\left< \alpha x , y \right> =  \alpha \left< x , y \right>$ and $\left< x + z , y \right> = \left< x , y \right> + \left< z , y \right>$

$\left< x , y \right> = \overline{\left< y , x \right>}$

$\left< x , x \right> > 0$, if $x \neq 0$.

My question is about property 3. Let $V$ be a vector space over $\mathbb C$, then $\left< \cdot , \cdot \right>: V \times V \to \mathbb C$, this means that for all $x \in V$, $\left< x , x \right>$ is a complex number. So what do they mean by $\left< x , x \right> > 0$? Do they mean that the inner product of $x$ with itself is allawys a real number, so if $\left< \cdot , \cdot \right>$ is an inner product then $\Im(\left< x , x \right>) = 0$? If not, what does it mean for a complex number to be greater than $0$ in this context?


问:
Prove that any $T_1$ topological vector space is regular

Proposition

For a $T_1$ space $(X,\mathscr T)$ the following three conditions are equivalent:

X is regular;

for any $x\in X$ and any $U$ containing it there exist an open neighborhood of $x$ such that $V\subseteq\overline V\subseteq U$;

any $x\in X$ has a local base whose element are closed.

Theorem

Suppose $K$ and $C$ are disjoint subsets of a topological vector space X such that K is compact and C is closed. so there exist a symmetric (open) neighborhood $V$ of $0$ such that
$$
(K+V)\cap(C+K)=\emptyset
$$

Corollary

Let be $X$ a topological vector space. If $\mathscr B(0)$ is a local (open) base centered at $0$ then any its element contains the closure of another one element.

So knowing these results I ask to me if any $T_1$ topological vector space is regular too and so I arranged the following arguments. If $U$ is an open set containing $x$ then $U-x$ is an open set (any raslation is an homeomorphism!) containing $0$ and so there exist an open neighborhood $V$ of $0$ whose closure is contained $U-x$ so that $V+x$ is an open set containing $x$ whose closure is contained in $U$ and thus the statement follows immediately by the proposition mentioned above. So is the argument true and are my arguments correct? Could someone help me, please?

答: Your argument is correct. You could also apply the theorem directly: if $x\in U\subseteq X$, where $U$ is open, then $\{x\}$ is compact, and $C=X\setminus U$ is closed and disjoint from $\{x\}$, so there is an open nbhd $V$ of $0$ such that $(x+V)\cap(C+V)=\varnothing$. But then $C+V$ is an open nbhd of $C$, so $F=X\setminus(C+V)$ is a closed subset of $U$, and $x+V$ is an open nbhd of $x$ such that $\operatorname{cl}(x+V)\subseteq F\subseteq U$.


问:
If $K$ is compact, then $C(K,\mathbb{R}^n)$ is a Banach space under the norm $\|f\|=\sup_{x\in K} \|f(x)\|$

Let $K$ be a topological space that is compact. Show that the space $C(K,\mathbb{R}^n)$ of all the continuous functions $f:K\to\mathbb{R}^n$ is a Banach space with the norm $\|f\|=\sup_{x\in K} \|f(x)\|$.

Below is my attempt:

$\|f\|=\sup_{x\in K} \|f(x)\|$ is a norm:

i. Suppose that $\|f\|=\sup_{x\in K} \|f(x)\|= 0$; then $f=0$. Conversely, suppose that $f=0$. Then $0=\|f\|=\sup_{x\in K} \|f(x)\|$.

ii. $\|\alpha f\|=\sup_{x\in K} \|\alpha f(x)\|=\sup_{x\in K} |\alpha|\|f(x)\|=|\alpha|\sup_{x\in K} \|f(x)\|=|\alpha|\|f\|$

iii. $\|f+g\|= \sup_{x\in K} \|f(x)+g(x)\|\leq \sup_{x\in K} \|f(x)\| + \|g(x)\|= \sup_{x\in K} \|f(x)\| + \sup_{x\in K} g(x)\|$

$C(K,\mathbb{R}^n)$ is complete (my doubt is here)

i. For all $\epsilon>0$, $\exists\, n_{0}$ such that if $m,n>n_{0}$ then $d(f_m,f_n)<\epsilon$.

We know that $|f_m-f_n|\leq \|f_m -f_n\|= \sup_{x\in K} \|f_m(x)-f_n(x)\|< \epsilon $ but how can I prove the last inequality?

ii. How to prove that $f_n$ converges to an $f$ in $C(K,\mathbb{R}^n)$?

答:
Continuity at $x\in K$ for $f$ can be seen as
 $|f(y)-f(x)|\le |f(y)-f_n(y)|+|f_n(y)-f_n(x)|+|f_n(x)-f(x)|$ and the fact $f_n$ is a Cauchy sequence in $C(K,R^n)$. Also, $f_n \rightarrow f$ can be easily seen.

ContentLicense: CC BY-SA 3.0


问:
Show that the chain rule $(G\circ F)_* = G_* \circ F_*$ holds.
Show that the chain rule $(G\circ F)_* = G_* \circ F_*$ holds.

I understand the chain rule from calculus but I am not sure how to show that it holds when also considering the pushforward.

So if we let $F: M \to N$ and $G: N \to P$ be two diffeomorphisms of smooth manifolds.

I know that I could use
$$d(g\circ f) = \sum_{i=1}^n \frac{\partial(g\circ f)}{\partial x^i}dx^i$$
and then apply the chain rule to that but I'm not sure how to do that or how it helps me.

Let $F:M \to N$ be a diffeomorphism of smooth manifolds. Then the pushforward $F_*$ of F, is defined by
$$F_* : X(M) \to X(N),     X \mapsto F_*X = dF\circ X \circ F^{-1}$$
map pushes forward vector fields on M to vector fields on N.

I am not sure how to use this knowledge to show the chain rule.
I understand that I could write
$$F_*X = dF\circ X \circ F^{-1}$$
$$G_*X = dG\circ X \circ G^{-1}$$
Therefore
$$G_*X \circ F_*X = dG\circ X \circ G^{-1} \circ dF\circ X \circ F^{-1}$$
However, I'm not sure how to manipulate this such that $(G\circ F)_* = G_* \circ F_*$.

答:
Careful! The expression $G_* X \circ F_* X$ does not make sense. A vector field takes in a point on a manifold and outputs a tangent vector (formally known as a section of the tangent bundle). But, if $p \in M$, then $G_* X$ would be a vector field "evaluated" on a tangent vector $F_* X$, which of course is non-sense. Moreover if $M \neq N$ then the expression $G_* X$ does not make sense either, since $G_*$ takes in vector fields on $N$.

What $G_* \circ F_*$ really does to vector fields $X$ on $M$ is the following:
\begin{align*}
(G_* \circ F_*)(X) &= G_* (dF \circ X \circ F^{-1}) = dG \circ dF \circ X \circ F^{-1} \circ G^{-1} \\&= d(G \circ F) \circ X \circ (G \circ F)^{-1} = (G \circ F)_* (X)
\end{align*}
Note $G_* (dF \circ X \circ F^{-1})$ is an evaluation of $G_*$ on $F_* X = dF \circ X \circ F^{-1}$, not a composition, and $dF$ sends a tangent vector of $p$ to a tangent vector of $F(p)$ (the intuition here is that the derivative of a function at a point is a linear approximation, so the derivative of a function is just a pairing of points and these linear maps).


问:
How many sub choices...
How many sub choices if there are 3 types of bread, 6 different types of meat, 8 different veggies, 4 different kinds of cheese

You must choose 1 bread.
You can choose any Meats, including none.
You can choose any veggies, including none.

You must choose 1 cheese.

I am confident in finding the number of choices regarding the bread and cheese 3(4) = 12, so my question lies in the toppings

Should I find the sum of each topping combination (Example: meats = 6C0 + 6C1 + 6C2...+ 6C6) then multiply the results with the sum of the veggies with the 12 above?

[sum of meats * sum of veggies * 12]

Or, should I treat the meat and veggie as one category? So, there would be 16 choices of toppings including having none [6+8+2(no cheese and no meat)] then calculate the sum of the combinations (16C0 + 16C1 + 16C2...+ 16C16) and multiply it by the 12 above?

[sum of all toppings * 12]

Thank you for your insight. I hope I'm somewhere in the ballpark!

答:
Your suggestion for the meat options, "6C0 + 6C1 + 6C2...+ 6C6", is correct but there is an easier way to formulate it. You are either choosing or not choosing each of the meat varieties, giving $2$ options per variety and thus $2^6$ options in the meat section.

That can be seen as a reason "why" the combinatoric identity holds:

$$ \sum_{k=0}^{n} \binom nk = 2^n $$

Title: How many sub choices...


答:
Yes you can group them together, but in this case you can directly choose a subset of the set of 8 + 6 = 14 (meats + veggies).

That gives finally $12.2^{14}$

Title: How many sub choices...

Body:
How many sub choices if there are 3 types of bread, 6 different types of meat, 8 different veggies, 4 different kinds of cheese

You must choose 1 bread.
You can choose any Meats, including none.
You can choose any veggies, including none.

You must choose 1 cheese.

I am confident in finding the number of choices regarding the bread and cheese 3(4) = 12, so my question lies in the toppings

Should I find the sum of each topping combination (Example: meats = 6C0 + 6C1 + 6C2...+ 6C6) then multiply the results with the sum of the veggies with the 12 above?

[sum of meats * sum of veggies * 12]

Or, should I treat the meat and veggie as one category? So, there would be 16 choices of toppings including having none [6+8+2(no cheese and no meat)] then calculate the sum of the combinations (16C0 + 16C1 + 16C2...+ 16C16) and multiply it by the 12 above?

[sum of all toppings * 12]

Thank you for your insight. I hope I'm somewhere in the ballpark!


问:
Local martingale that is not integrable
I have just been introduced to the notion of a local martingale, and I have asked myself the question whether the fact that $(X_{t})_{t\geq 0}$ is a local martingale with associated $\{\tau_{n}\}_{n}$ stopping times that go to infinity implies that for every $X_{t}$ is indeed integrable.

I cannot prove it and thus I believe it is false but I am struggling to find counterexamples. Could anyone give me a hint as to what such a process $X$ and the stopping times would look like so that they are indeed a local martingale however not integrable.

答:
Suppose $(B_t)_{t\ge 0}$ is a standard Brownian motion, with its natural filtration $(\mathcal F_t)_{t\ge 0}$.

Example: Let $H$ be $\mathcal F_1$ measurable, but not integrable. R.M. Dudley has shown ["Wiener functionals as Itô integrals," Ann. Probability,  5 (1977), pp. 140–141); http://www.jstor.org/stable/pdf/2242810.pdf]  that the $\mathcal F_1$ measurability of $H$ alone ensures the existence of a progressive (even predictable) process $(K_s)_{0\le s\le 1}$ such that the stochastic integral $\int_0^1 K_s\,dB_s$ exists and is equal to $H$ a.s. Define
$$
L_s:=\cases{K_s,&$0\le s<1$;\cr 0,& $s\ge 1$.\cr}
$$
The associated stochastic integral process
$$
M_t:=\int_0^t L_s\,dB_s
$$
is then a local martingale. By design,  $M_1=H$ is not integrable. In particular, $(M_t)$ is not a martingale.

Title: Local martingale that is not integrable


问:
ex falso quodlibet in natural deduction (ND)?
My understanding of inference rules is that they should be intuitively acceptable (like axioms). But my only intuition for ex falso quodlibet in natural deduction is not immediate but comes from its proof, which rests on disjunctive syllogism (which apparently holds in classical propositional logic and intuitionistic logic).
I am probably missing something obvious. Is the principle of explosion really basic/intuitive and why?

The closest justification I see, based on the current answers, is that Ex Falso encodes a (metalogical) desire that the proof system be consistent. But also see the references to intuitionism in the answers.

Edit: as pointed out below, it may have been a distraction for me to raise "intuitionism" in the original question, although Ex Falso arises in classical and intuitionistic logic. This does not exclude that I would like axioms and rules to be intuitively acceptable (to me).

" There  is  surprisingly  little  agreement  about  the  exact  status  of EFQ in  the literature.  Gentzen groups the ex  falso rule alongside the other operational rules(1969,  p.  77).   Prawitz  follows  this  practice.   He  treats ⊥ as  a  ‘0-place  sentential operation’ (Prawitz 1978, p. 38), one ‘for which there is no canonical proof’ (Prawitz1977, p. 26). EFQ is then understood as the elimination rule for ⊥. Others, including Dummett, present ⊥ as an elimination rule for ¬ (e.g. Dummett 1991, p. 291).  I contend that both of these approaches are mistaken...." ref.

Ex Falso has an ‘anomalous position [. . . ]  inside the scheme of introduction/elimination rules’, tarnishing the neat symmetry of the intuitionistic system" (Weir 1986, p. 461).

According to here: "Kolmogorov’s criterion whether to keep an axiom was whether a proposition has an “intuitive foundation” or “possesses intuitive obviousness” (van Heijenoort 1967: 421, 422) ... Kolmogorov said that, just like PEM, Ex Falso “has no intuitive foundation” (van Heijenoort 1967: 419). In particular, he says that Ex Falso is unacceptable for the reason that it asserts something about the consequences of something impossible (van Heijenoort 1967: 421)."

... "I submit that these difficulties can be overcome simply by treating⊥as a punc-tuation marker, as Tennant suggests.  Of course this means thatEFQis left withouta  logical  constant  which  it  serves  to  eliminate.   But  on  the  view  I  am  proposingno such constant is needed becauseEFQis not an operational rule at all.  Its roleis  best  accounted  for  by  according  it  the  status  of  astructural  rule.   It  simply  isthe structural rule that tells us that any sentence whatsoever follows from a con-tradiction.  Such a license isnotspecific to any logical constants, but amounts to ablanket policy.  Our reclassification ofEFQthus fits neatly with our characterizationof structural rules as global rules that assign properties to our deducibility relation." ref.

Edit: I came across this reformulation of my initial uneasiness.

"... adopting disjunctive syllogism as an axiom or as a primitive rule of inference is not an option.  Hence, it is ex  falso that is naturally treated as primitive, not the principle of disjunctive syllogism. Therefore, rather than thinking of ex  falso as the product of the mean-ings of disjunction and negation, we must treat disjunctive syllogism as a product of the ex falso rule.  This suggests that the law of disjunctive syllogism is not a con-sequence only of the meanings of the constants involved, but also hinges on one’s stance concerning the general principle expressed by EFQ." ref.

答:
The reason Brouwer used the term "intuitionist logic" is that a proof in intuitionistic logic necessarily corresponds to a construction. In particular, the rules of intuitionistic logic come in pairs: each symbol $\wedge,\vee,\to$ has an introduction rule and elimination rule (except for your observation that $\bot$ has no introduction rule), whereas excluded middle ($\vdash \phi\vee\neg\phi$) has no counterpart.

So the term "intuitionistic" is incidental.

The best explanation I can give for the elimination rule for $\bot$ (EFQ) is that its intended meaning is contradiction. Namely, the rule says "if a contradiction is derived, then anything is derivable." However, since the presence of a contradiction implies that the system is inconsistent, this can be reworded to mean "anything is derivable in an inconsistent system".

Being able to derive anything in a logical system would make the system useless (or trivial). So EFQ essentially guarantees that if a contradiction is present in the system, then the system is useless.

So to prove that a logical system is useful, or non-trivial (or consistent), we must show that no contradiction is derivable. Note that the absence of an introduction rule for $\bot$ does not guarantee that it cannot be derived. This is because of modus ponens, (or the cut rule in sequent calculus), which is why we need the cut elimination theorem to prove consistency results.

Mauro ALLEGRANZA pointed out that minimal logic (more generally paraconsistent logic) excludes EFQ. To quote wikipedia,

The characteristic or defining feature of a paraconsistent logic is that it rejects the principle of explosion. As a result, paraconsistent logics, unlike classical and other logics, can be used to formalize inconsistent but non-trivial theories.


问:
Let $k$ be an integer not divisible by $5$. Show that either $k^2-1$ or $k^2+1$ is divisible by $5$.

Let $k$ be an integer not divisible by $5$. Show that either $k^2-1$ or $k^2+1$ is divisible by $5$.

Looking at this mod $5$ it seems that $k$ can be one of the form $k \equiv 1,2,3,4 \pmod{5}$. This would imply that $k^2 \equiv 1,4,9,16 \equiv 1,4,4,1 \pmod{5} \Longrightarrow k^2-1 \equiv0,3,3,0 \pmod{5}$ so $k^2-1$ is divisible by $5$? Similar reasoning could be used for $k^2+1$, but that wouldn't give any zero remainders so $5$ doesn't divide $k^2+1$ right?

I feel like I'm missing something since it seems that $k^2-1$ could also give a remainder $3$?

答:
We must have $k \equiv 1,2,3,4 \pmod 5$. Working$\pmod5$ one finds the following:

$k \equiv 1 \implies k^{2}-1 \equiv 0$

$k \equiv 2 \implies k^{2}+1\equiv5\equiv0$

$k \equiv 3 \implies k^{2} + 1 \equiv 10 \equiv 0$

$k \equiv 4 \implies k^{2} - 1 \equiv 15 \equiv 0$.


问:
How can evaluate $\lim_{x\to0}\frac{\sin(3x^2)}{\tan(x)\sin(x)}$
I know this:

$$\lim_{x\to0}\frac{\sin(3x^2)}{\tan(x)\sin(x)}$$

But I have no idea how make a result different of: $$\lim_{x\to0}\frac{3(x)}{\tan(x)}$$
Any suggestions?

Body:
\begin{align}
\ \lim_{x\to 0}\frac{\sin(3x^2)}{\tan(x) \sin^2(x)}=\lim_{x\to 0}\cos(x) \lim_{x\to 0} \frac{\sin(3x^2)}{\sin^2(x)}\\
&=\lim_{x\to 0}\frac{\sin(3x^2)}{\sin^2(x)}\\
&=\Biggl(\lim_{x\to 0}\frac{\sin(3x^2)}{3x^2}3x^2\Biggl)\Biggl(\lim_{x\to 0}\frac{x^2}{\sin^2(x)}\frac{1}{x^2}\Biggl)\\
&=\Biggl(\lim_{x\to 0}\frac{\sin(3x^2)}{3x^2}\Biggl) \Biggl(\lim_{x\to 0}3x^2\Biggl)\Biggl(\lim_{x\to 0}\frac{x^2}{\sin^2(x)}\Biggl)\Biggl(\lim_{x\to 0}\frac{1}{x^2}\Biggl)\\
&=\lim_{x\to 0}3x^2 \Biggl(\lim_{x\to 0}\frac{1}{x^2}\Biggl)=\lim_{x\to 0}\frac{3x^2}{x^2}\\
&=3
\end{align}


问:
Analytic solution to $\alpha , \beta \in \mathbb{R}$ such that $\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)$?
Are there any $\alpha , \beta \in \mathbb{R}$ such that $$\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)?$$

The trivial solutions are $(\alpha, \beta)=(0, \dfrac{\pi}{2})$. But are there more? I dont see an obvious way of tackling this problem but I keep coming back to it because it seems interesting to me (look at the graph of the solutions in desmos: https://www.desmos.com/calculator/ms8ad8cxqt).

I tried simplifying matters by trying out the case where $\alpha = \beta$ where we get that $$\cos \alpha \cdot \sin \alpha = \cos \bigl(\sin (\alpha ^{2})\bigr)$$ and using the fact that $2\sin \alpha \cos \alpha = \sin 2\alpha$ we really want to find $\alpha \in \mathbb{R}$ such that $$\dfrac{1}{2}\sin 2\alpha = \cos \bigl(\sin (\alpha ^{2}) \bigr),$$ though I have to admit that this does not make the problem easier (it seems to me at least).

Has anyone looked at this problem before and have a solution or maybe some hints or suggestions on how to go further in solving the problem?

答:
The solution set $S$  consists of disjoint  flowers $\gamma_{jk}$.  These flowers are more or less circular "distorted Lissajous figures" and become more complex further out in the plane. It will be difficult to analyze the individual $\gamma_{jk}$, but we can  put each of them in a nice small $(x,y)$-square with center $(0,0)$.

To this end we draw $\xi$- and $\eta$-axes with $\pm45^\circ$ slopes, intersecting at  the point ${\bf c}_{00}=\bigl(0,{\pi\over2}\bigr)$.  The units on these axes are chosen such that the centers ${\bf c}_{jk}$ of the flowers obtain new coordinates $${\bf c}_{jk}=[j\pi,k\pi]\qquad\bigl((j,k)\in{\mathbb Z}^2\bigr)\ ,$$ whereby we use square brackets for such coordinates. This setup leads to the transformation formulas
$$\alpha=\xi+\eta,\qquad \beta=-\xi+\eta+{\pi\over2}\ .\tag{1}$$
We now partition the plane into squares, aligned with the new axes, by putting
$$\xi=j\pi+x,\quad \eta=k\pi+y\qquad\bigl(\ [x,y]\in\left[-{\pi\over2},{\pi\over2}\right]^2\ \bigr)\ .$$
In this way each  ${\bf c}_{jk}$ is the center of its square of (new) side length $\pi$. The formulas  $(1)$ now appear as
$$\alpha=(j+k)\pi+x+y,\qquad\beta=(-j+k)\pi-x+y+{\pi\over2}\ .\tag{2}$$
This implies
$$\cos\alpha\sin\beta=(-1)^{k+j}\cos(x+y)(-1)^{k-j}\cos(y-x)
=1-\sin^2 x-\sin^2 y\ ,\tag{3}$$
where the RHS is independent of $j$ and $k$.
For points of  $S$ we want $$\cos\alpha\sin\beta=\cos\bigl(\sin(\alpha\cdot\beta)\bigr)\geq\cos1>{1\over2}\ .$$
From $(3)$ it then follows that, independently of $j$ and $k$, we must have $$\sin^2 x+\sin^2 y<{1\over2}\qquad\bigl(\> [j\pi+x,k\pi+y]\in S\>\bigr)\ ;\tag{4}$$
in particular $$|x|<{\pi\over4}, \quad |y|<{\pi\over4}\qquad \bigl([x,y]\in\gamma_{jk}\bigr)\ .$$ Formula $(4)$ is the decisive equation of the present answer: It shows that the formal procedure has lead to the right separation of the different flowers $\gamma_{jk}$.

To sum it up: When $(j,k)\in{\mathbb Z}^2$ is given and fixed, by $(3)$ and $(2)$  the flower $\gamma_{jk}$ has an $[x,y]$-equation of the form
$$1-\sin^2 x-\sin^2 y=
 \cos\circ\sin\left(\bigl((j+k)\pi+x+y\bigr)\bigl((-j+k)\pi-x+y+{\pi\over2}\bigr)\right)\ .$$
I don't see an obvious simplification of the $\bigl(\ldots\bigr)$ part here.


问:
Permutation combination problem
This is how Edward’s Lotteries work. First, 9 different numbers are selected. Tickets with exactly 6
of the 9 numbers randomly selected are printed such that no two tickets have the same set of numbers.
Finally, the winning ticket is the one containing the 6 numbers drawn from the 9 randomly. There is
exactly one winning ticket in the lottery system. How many tickets can the lottery system print?</p>

答:
The number of tickets is the number of ways to choose $6$ numbers from $9$.

How do we calculate that that number? It is simplest if we know a formula for the binomial coefficient $\binom{9}{6}$ aka $C(9,6)$ aka $C^9_6$. One expression for this is $\frac{9!}{6!3!}$. Now we compute. There is a lot of cancellation.

Or we can figure out the number from first principles. How many ways are there to choose $6$ numbers (the ones that will appear on tickets) from $9$? Just as many as there are ways to choose the $3$ numbers that don't appear. (Yhis "feels" easier, since $3$ is smaller than $6$.)

There are $(9)(8)(7)$ three number strings made up of different numbers. The $6$ strings $abc,acb, bac,bca,cab,cba$ all represent the same ticket, the same choice of rejected numbers. So the total number of tickets we can print is $\frac{(9)(8)(7)}{6}$.

Title: Permutation combination problem


问:
$\lim_{n\to\infty} \int_0^1e^{2\pi inp(x)}=0$
Let $p$ non constant real polynomial, I want to prove that $\lim_n \int_0^1e^{2\pi inp(x)}=0$

I am not sure even how to begin with this, whether to use measure theory, fourier, complex analysis. It reminds me somewhat of Riemann-Lebesgue lemma from Fourier, but I am not really sure how to proceed with that.

My intuition would be to try and use measure theory/convergence theorems to prove $\lim_n\int_0^1cos(2\pi n x^m)=0$ for every m and similarly for sine and then somehow generalize for general polynomials. Take the simple case for $m=2$ for example. Taking $u=2\pi nx^2$ we get $\int_0^1cos(2\pi n x^m)=\int_0^{2 \pi n} \frac{\sqrt{2\pi n}}{4\pi n}\frac{1}{\sqrt{u}}cos(u)=\int_{\mathbb R} \frac{\sqrt{2\pi n}}{4\pi n}\frac{1}{\sqrt{u}}cos(u)\chi_{[0,2\pi n]}$ but this doesn't seem to help much as I can't use any of the convergence theorems.

Any hints as to at least what would be a good strategy to approach this problem would be appreciated.

答:
By the stationary phase method, the integral is approximately given by
$$ \int_0^1 e^{inp(x)}dx= \sum_{x_0} \sqrt{\frac{2\pi}{np''(x_0)}}e^{inp(x_0)+\mathrm{sign}(p''(x_0))i\pi/4}+o(1/\sqrt{n}),$$
where the sum is over the stationary points of $p(x)$ in the integration interval (roots of $p'(x)$).

So it decays like $\sim 1/\sqrt{n}$ for large $n$, provided there are no double roots of $p'(x)$. If there are double roots, it will decay slower, but will still decay.


问:
Residue theorem. Why is my solution wrong?
Question

Calculate the integral:
$$I(a) = \int_0^\infty \frac{x^a}{x^2 + 1} dx$$
for -1 < a < 1.

In my solution, I use the domain $\Omega = \{\epsilon <|z|< R, z \notin \mathbb{R}_+ \}$, where $\epsilon$ is small and $R$ is big. Then I divide $\partial \Omega$ into $\gamma_R$, $\gamma_{\epsilon}$, $l_+$ and $l_-$ with orientation, as shown in the picture (this might not be valid, so if it looks suspicious to you, please skip to the given solution).\

I. Integration: By integrating
$$f(z) = \frac{z^a}{z^2 + 1}$$
on the lines and curves, I get for the curves
$$\int_{\gamma_\epsilon}f(z)dz \quad \text{and} \quad \int_{\gamma_R}f(z)dz \to 0$$
and for the lines
$$\int_{l_+} f(z) dz \to \int_0^\infty \frac{x^a}{x^2 + 1} dx$$
$$\int_{l_-} f(z) dz \to -\int_0^\infty \frac{x^ae^{2a\pi i}}{x^2 + 1} dx$$
as $\epsilon \to 0$ and $R \to \infty$, which means
$$\int_{\partial \Omega}f(z)dz \to (1 - e^{2a\pi i})\int_0^\infty \frac{x^a}{x^2 + 1} dx$$
as $\epsilon \to 0$ and $R \to \infty$.

II.Residue Theorem: Since there are two singularities $i$ and $-i$ with order 1,
$$\int_{\partial \Omega}f(z)dz = 2\pi i (Res(f,i) + Res(f,-i)) = \pi(e^{a\pi i/2} - e^{-a\pi i/2})$$

Overall, we have
$$I(a) = \frac{\pi(e^{a\pi i/2} - e^{-a\pi i/2})}{1 - e^{2a\pi i}} = -\pi e^{-a\pi i}\frac{\sin{a\pi/2}}{\sin{a\pi}}$$
which is not even real.

Given Solution

In the solution given, the domain is the upper half disc of radius R instead. The boundary is divided into the arc, the positive real-axis and the negative real-axis. Integration on the boundary gives
$$\int_{\partial \Omega}f(z)dz = (1 + e^{a\pi i})I(a)$$
Only $i$ is included in the domain as a point for residue calculation, which gives
$$\int_{\partial \Omega}f(z)dz = 2\pi i \frac{e^{a\pi i/2}}{2i} = \pi e^{a\pi i/2}$$
Overall,
$$I(a) = \frac{\pi}{e^{-a\pi i/2} + e^{a\pi i/2}} = \frac{\pi}{2\cos(a\pi/2)}$$

Calculator has proven the above result right. So what went wrong in my solution?

Thank you all in advance!

答:
The function $z^a$ (and hence $\frac{z^a}{z^2 + 1}$) has a branch cut from $0$ to $\infty$.  You have to put the cut on a path somewhere.  The solution you give forces the cut to run along the positive real axis because you want to use Cauchy's results, so the function has to have only point singularities inside your path.  Since the only region you leave outside your path runs along the positive real axis, the integral you write cannot possibly represent the integral you want -- the integral you want is the integral along the branch cut, so depends sensitively on which points of the positive real axis are on this side or that side of the cut.

You want the branch cut to be outside of your path.  The provided solution leaves the entire lower half-plane for the branch cut to run through.  This allows the branch cut to be nowhere near the positive real axis.

Your path is very unlikely to give you the result you want, in general.  Without a branch cut, the lines $\ell_+$ and $\ell_-$ should cancel out, leaving a contribution of $0$ from the positive real axis.  With the branch cut, you are getting two contributions which don't cancel, only one of which is part of the integral you actually want and the other is junk that you have no leverage to remove.

So you want a path that (1) has a component that runs alon the positive $x$ axis (in either direction, but the positive direction is convenient), (2) that encloses one or both of the poles at $\pm \mathrm{i}$ (so, hopefully, you can assign all of the contribution by the residue to the integral of interest), (3) has tiny radius and huge radius circular arcs that can be shown to $\rightarrow 0$ as tiny $\rightarrow 0$ and huge $\rightarrow \infty$, and (4) ends before forcing the branch cut to land on the interval of integration in the original integral and ends on an integral you can do, that reduces to some multiple of the original integral, or has some other (perhaps symmetry) property that allows you to integrate along this path.  This means that you really have no choice on the $\ell_+$, $\gamma_\varepsilon$, and $\gamma_R$ portions of the path except "at what angle do I want this region to stop to satisfy part (4) above -- and it can't be $2\pi$ because then the branch cut stomps on your desired integral.


问:
Example of subdirectly irreducible algebra that contains a proper sub-algebra
I read that an algebra $\textbf{A}$ is subdirectly irreducible if whenever $\textbf{A}$ is isomorphic to a sub-algebra of a direct product $\Pi_{i \in I}\textbf{C}_i$, it has to be the case that $\textbf{A}$ is isomorphic to $\textbf{C}_j$ for some $j \in I$.

I would like to ask, is it possible for there to be a subdirectly irreducible algebra that contains proper sub-algebras ? Is there any example in the case of groups ?

答:
Your definition of subdirectly irreducible is  not quite right: the correct conclusion is that $\mathbf{A}$ is isomorphic to a subalgebra of $\mathbf{C}_j$ for some $j\in I$, and that the isomorphism is induced by the embedding into the direct product.

In fact, the usual definition is the following (e.g., Jacobson's Basic Algebra II):

Definition. An $\Omega$-algebra $\mathbf{A}$ is subdirectly irreducible if and only if the intersection of all nontrivial congruences of $\mathbf{A}$ is nontrivial.

For instance, in the case of groups this would become:

Definition. A group $G$ is subdirectly irreducible if and only if it contains a minimum nontrivial normal subgroup $N$ (i.e., a subgroup $N$ such that $N\triangleleft G$ and if $K\triangleleft G$, then either $K=\{e\}$ or $N\leq K$).

And for rings,

Definition. A ring $R$ is subdirectly irreducible if and only if it contains a a minimum nontrivial ideal.

And for modules,

Definition. An $R$-module $M$ is subdirectly irreducible if and only if the intersection of all nontrivial submodules of $M$ is nontrivial.

In the case of groups, for instance, we would have:

Theorem. A group $G$ is subdirectly irreducible if and only if for every $f\colon G\to \prod_{i\in I}G_i$, if $f$ is one-to-one then there exists $j\in I$ such that $\pi_j\circ f\colon G\to C_j$ is one-to-one. In particular, $G$ is isomorphic to a subalgebra of $C_j$.

Proof. Assume first that $G$ has a minimal nontrivial normal subgroup $N$, and that $f\colon G\hookrightarrow \prod_{i\in I}G_i$ is an embedding into a direct product. Consider the maps $\pi_i\circ f\colon G\to G_i$. The kernel of $\pi_i\circ f$ is either trivial or contains $N$. Since $f$ is one-to-one, the intersection of all kernels must be trivial. Thus, at least one kernel $\mathrm{ker}(\pi_j\circ f)$ does not contain $N$, which means that it is trivial. Hence, $\pi_j\circ f$ is an embedding and $G$ is isomorphic to a subgroup of $G_j$.

Conversely, the family of maps $\pi_K\colon G\to G/K$ as $K$ ranges over all nontrivial normal subgroups of $G$ induces a morphism $f\colon G\to \prod_{\{e\}\neq K\triangleleft G} G/K$. The kernel of this map is the intersection of all nontrivial normal subgroups of $G$. Since none of the maps $\pi_K\colon G\to G/K$ are embeddings, the morphism $f$ is not an embedding, hence the intersection of all nontrivial normal subgroups of $G$ is nontrivial, proving that $G$ has a minimum nontrivial normal subgroup, as claimed. $\Box$

Examples.

The groups $\mathbf{C}_{p^n}$, $p$ a prime, $n\geq 1$, are subdirectly irreducible. When $n\gt 1$, they all have proper nontrivial subgroups.

The Prüfer $p$-group $\mathbf{C}_{p^{\infty}}$ is subdirectly irreducible.

The groups $S_n$, $n\geq 2$, are subdirectly irreducible. For $n\neq 4$, the minimum normal subgroup is $A_n$; for $n=4$ it is the Klein $4$-subgroup $\{e, (12)(34), (13)(24), (14)(23)\}$.

Fields are subdirectly irreducible; any non-prime field contains a proper subfield.


答:
I may suppose you are doing everything allright, but you have to be careful when identifying poles. As far as I understand you are integrating (circle radius R) counter clockwise; you function should be continuous and single-valued in all plane, except for the cut along the positive part of the axis. It means that the first pole is $e^{\pi i /2}$ (you've made a quarter of full turn counter clockwise), while the second pole is $e^{\pi i3 /2}$ (you've made 3/4 a full turn counter clockwise, i.e. 3/4 of $2\pi$).

Hopefully this will be helpful.

Title: Residue theorem. Why is my solution wrong?

Body:
Question

Calculate the integral:
$$I(a) = \int_0^\infty \frac{x^a}{x^2 + 1} dx$$
for -1 < a < 1.

In my solution, I use the domain $\Omega = \{\epsilon <|z|< R, z \notin \mathbb{R}_+ \}$, where $\epsilon$ is small and $R$ is big. Then I divide $\partial \Omega$ into $\gamma_R$, $\gamma_{\epsilon}$, $l_+$ and $l_-$ with orientation, as shown in the picture (this might not be valid, so if it looks suspicious to you, please skip to the given solution).\

I. Integration: By integrating
$$f(z) = \frac{z^a}{z^2 + 1}$$
on the lines and curves, I get for the curves
$$\int_{\gamma_\epsilon}f(z)dz \quad \text{and} \quad \int_{\gamma_R}f(z)dz \to 0$$
and for the lines
$$\int_{l_+} f(z) dz \to \int_0^\infty \frac{x^a}{x^2 + 1} dx$$
$$\int_{l_-} f(z) dz \to -\int_0^\infty \frac{x^ae^{2a\pi i}}{x^2 + 1} dx$$
as $\epsilon \to 0$ and $R \to \infty$, which means
$$\int_{\partial \Omega}f(z)dz \to (1 - e^{2a\pi i})\int_0^\infty \frac{x^a}{x^2 + 1} dx$$
as $\epsilon \to 0$ and $R \to \infty$.

II.Residue Theorem: Since there are two singularities $i$ and $-i$ with order 1,
$$\int_{\partial \Omega}f(z)dz = 2\pi i (Res(f,i) + Res(f,-i)) = \pi(e^{a\pi i/2} - e^{-a\pi i/2})$$

Overall, we have
$$I(a) = \frac{\pi(e^{a\pi i/2} - e^{-a\pi i/2})}{1 - e^{2a\pi i}} = -\pi e^{-a\pi i}\frac{\sin{a\pi/2}}{\sin{a\pi}}$$
which is not even real.

Given Solution

In the solution given, the domain is the upper half disc of radius R instead. The boundary is divided into the arc, the positive real-axis and the negative real-axis. Integration on the boundary gives
$$\int_{\partial \Omega}f(z)dz = (1 + e^{a\pi i})I(a)$$
Only $i$ is included in the domain as a point for residue calculation, which gives
$$\int_{\partial \Omega}f(z)dz = 2\pi i \frac{e^{a\pi i/2}}{2i} = \pi e^{a\pi i/2}$$
Overall,
$$I(a) = \frac{\pi}{e^{-a\pi i/2} + e^{a\pi i/2}} = \frac{\pi}{2\cos(a\pi/2)}$$

Calculator has proven the above result right. So what went wrong in my solution?

Thank you all in advance!


答:
Split $[0,1]=\bigcup_j [a_j,b_j]$ where $p$ is injective on $(a_j,b_j)$ (ie. $p'$ doesn't vanish) then $$\int_{a_j}^{b_j} e^{2i\pi np(x)}dx=\int_{p(a_j)}^{p(b_j)} \frac{e^{2i\pi n y}}{p'(p^{-1}(y))}dy$$

Since $$\int_{p(a_j)}^{p(b_j)} |\frac{1}{p'(p^{-1}(y))}|dy= |b_j-a_j|$$ then$\frac1{p'(p^{-1}(y))}$ is $L^1[p(a_j),p(b_j)]$ and we can apply Riemann-Lebesgue lemma.

Title: $\lim_{n\to\infty} \int_0^1e^{2\pi inp(x)}=0$

Body:
Let $p$ non constant real polynomial, I want to prove that $\lim_n \int_0^1e^{2\pi inp(x)}=0$

I am not sure even how to begin with this, whether to use measure theory, fourier, complex analysis. It reminds me somewhat of Riemann-Lebesgue lemma from Fourier, but I am not really sure how to proceed with that.

My intuition would be to try and use measure theory/convergence theorems to prove $\lim_n\int_0^1cos(2\pi n x^m)=0$ for every m and similarly for sine and then somehow generalize for general polynomials. Take the simple case for $m=2$ for example. Taking $u=2\pi nx^2$ we get $\int_0^1cos(2\pi n x^m)=\int_0^{2 \pi n} \frac{\sqrt{2\pi n}}{4\pi n}\frac{1}{\sqrt{u}}cos(u)=\int_{\mathbb R} \frac{\sqrt{2\pi n}}{4\pi n}\frac{1}{\sqrt{u}}cos(u)\chi_{[0,2\pi n]}$ but this doesn't seem to help much as I can't use any of the convergence theorems.

Any hints as to at least what would be a good strategy to approach this problem would be appreciated.


问:
$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = +\infty \implies \lim_{x \to a^+}\frac{f'(x)}{g'(x)} \neq -\infty $
Let $a \in \mathbb R $ and $f,g:]a$,+∞[$\to \mathbb R$ two differentiable functions such that $g'$ is never equal to $0$. I need to prove that $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = +\infty \implies \lim_{x \to a^+}\frac{f'(x)}{g'(x)} \neq -\infty $

I tried to use mean value theorem but it does not lead me anywhere. I am also not very familiar with proving things with $\neq$...

I am not allowed to use L'Hopital.

答:
According to Darboux's theorem, as $g^\prime$ never vanishes it is always positive or always negative. $\lim_{x \to a^+} g(x) = +\infty$ implies that $g^\prime(x) \lt 0$ for $x \in (a, \infty)$.

$\lim_{x \to a^+}\frac{f^\prime(x)}{g^\prime(x)} = -\infty $ would imply the existence of $\delta \gt 0$ such that $f^\prime(x) \gt 0$  for $x \in (a, a+ \delta)$. A contradiction with $\lim_{x \to a^+} f(x) = +\infty$.


问:
Formal proof for detection of intersections for constrained segments
They told me it was off-topic at stackoverflow.
So I am trying my luck here. Yes, it's a homework, but I'm looking for some guidance (or related literature) instead of complete solutions.

Please see the drawing that accompanies the question:

I need to come up with a formal proof for the following statement:

Given an arbitrary count of line segments (S1 -- S3 in the drawing) constrained by two vertical lines (M & N) an answer whether there is at least one intersection of the segments (in the whole picture) can be given by checking each of the two segments that have their left endpoints adjacent on M in descending order (in respect to their position on line M).

For example, in drawing checking could be done in such order:

S1 and S2 (no intersection)

S2 and S3 (intersection detected)

While it seems very obvious, there shall be a formal proof for that (most likely, a simple one). Any ideas?

答: If you define a function which is the difference in height of the two segments an intersection is represented by the function going through $0$ between $M$ and $N$.  For $S1, S2$ it is positive at both ends.  For $S2, S3$ it is positive on one end and negative on the other.</p>


问:
Pigeon holes principle
Let $P$ be a group that it's elements are 257 sentences in which only atomic sentences from $A,B,C$ exist (i.e. $A \iff B,\space\space A \wedge B \wedge C, \space\space...$) Show that there exists two different $p_1, p_2 \in P$ so that the sentence $p_1 \iff p_2$ is a tautology.

Pigeons are the $257$ sentences but I can't think of a way to prove the asked.

答: Consider the truth tables.  How many different possible truth tables are there for sentences with three variables?  What can you say about sentences that have the same truth table?</p>

Title: Pigeon holes principle


问:
If conic is represented by $21x^2 -6xy +29y^2 +6x-58y-151=0$, then find the centre, length of axes and eccentricity
Using partial derivatives, I found centre of the conic as $(0,1)$ and I think the conic represents an ellipse. But I am not able to find the rest of the answers.

答:
Let
$$f(x,y)= 21x^2 -6xy +29y^2 +6x-58y-151 $$
$$f_x’(x,y)= 42x-6y+6$$
$$f_y’(x,y)= 6x+58y-58$$
The center $(0,1)$ is obtained via $f’_x=f’_y=0$
and by matching the normal vectors at  vertexes, i.e.
$$f’_x:f’_y=(x-0):(y-1)$$
the equations for the major and minor axes are obtained
$$3y-x-3=0,\>\>\>\>\>y+3x-1=0$$
Then, substitute  them into $f(x,y)= 0$ to get the major vertexes
$(\pm\frac9{\sqrt{10}},1\pm\frac3{\sqrt{10}})$ and the minor vertexes
$(\mp\sqrt{\frac35},1\pm3\sqrt{\frac35})$, and in turn their respective lengths $2a=6$ and $2b=2\sqrt6$. Thus, the eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}= \frac1{\sqrt3}$.


问:
What's the limit of the function $f(x)=\ln x-ax$ $(a>0)$, when $x$ approach $0$ and infinity?
I'm a beginner in calculus.

I know when $x \to 0$, $\ln x \to -\infty$ and $-ax \to 0,$ then $\ln x-ax \to -\infty$.

But when $x$ approaches infinity, $\ln x \to \infty$ and $-ax \to -\infty$.

Which one changes faster?

Thanks a lot.

答:
$$f(x) = \ln x - a x.$$
$$e^{f(x)} = x e^{-ax} \to 0, \quad \text{ as } x\to \infty.$$
$$\lim_{x\to\infty}f(x)=-\infty.$$


答:
If you are not allowed to use integrals which the standard method of invoking Abel's Theorem requires, then pretty much all you can do is get creative. Denote:

$$S_n = \sum_{k=0}^n\left(\frac{1}{6k+1} + \frac{1}{6k+3} + \frac{1}{6k+5}\right)$$
and
$$T_n = \sum_{k=0}^n\left(\frac{1}{6k+2} + \frac{1}{6k+4} + \frac{1}{6k+6}\right)$$
for starters. Then, you can see that $S_n+T_n = H_{6n+6}$, the usual Harmonic sum. Note that you have to consider these finite truncates, for otherwise each of $S_n$ and $T_n$ diverge.

On the other hand, consider the remaining term in your original sum:

$$R_n = \sum_{k=1}^n\left(\frac{1}{8k} + \frac{1}{8k+2} + \frac{1}{8k+4} + \frac{1}{8k+6}\right)$$

$$= \frac{1}{2}\sum_{k=1}^n\left(\frac{1}{4k}+\frac{1}{4k+1}+\frac{1}{4k+2}+\frac{1}{4k+3}\right) = \dfrac{H_{4n+3} -1-\frac 12-\frac 13 }{2} = \dfrac{H_{4n+3}}{2} - \frac{11}{12}.$$

Now, we just need to "evaluate" $T_n$ nicely:

$$T_n = \frac{1}{2}\sum_{k=0}^n\left(\frac{1}{3k+1} + \frac{1}{3k+2} + \frac{1}{3k+3}\right) = \dfrac{H_{3n+3}}{2}.$$
Therefore, the $n$ -th partial sum of your limit is then:

$$S_n-R_n - \left(1+\frac 13+\frac 15\right) = H_{6n+6} - \dfrac{H_{3n+3}}{2} - \frac{23}{15}-\dfrac{H_{4n+3}}{2}+\frac{11}{12} = $$
$$ \sim \ln(6n+6) - \ln(3n+3)/2-\ln(4n+3)/2-\dfrac{37}{60} = \ln\sqrt{\dfrac{12n+12}{4n+3}}-\dfrac{37}{60}$$
so the answer is:
$$\lim\limits_{n\to\infty}\left(\ln\sqrt{\dfrac{12n+12}{4n+3}}-\dfrac{37}{60}\right) = \frac{1}{2}\ln 3 - \frac{37}{60}$$

Title: Evaluate $ \sum_{k=1}^\infty ( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6})$

Body:
I need to evaluate the sum given by:

$$\displaystyle \sum_{k=1}^\infty \left( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6} \right)$$

I know that:

for $k = 1$ I get:
$\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}\mathbf{-\frac{1}{14}}$

for $k = 2$ I get:
$\frac{1}{13}+\frac{1}{15}+\frac{1}{17}-\frac{1}{16}-\frac{1}{18} \mathbf{-\frac{1}{20}-\frac{1}{22}}$

for $k = 3$ I get:
$\frac{1}{19}+\frac{1}{21}+\frac{1}{23}-\frac{1}{24} \mathbf{-\frac{1}{26}-\frac{1}{28}-\frac{1}{30}}$

for $k = 4$ I get:
$\frac{1}{25}+\frac{1}{27}+\frac{1}{29} \mathbf{ -\frac{1}{32}-\frac{1}{34}-\frac{1}{36}-\frac{1}{38}}$

I can write that sum for $k = 4$ as:
$$ \frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15}-\frac{1}{16}+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}+\frac{1}{21}-\frac{1}{22}+\frac{1}{23}-\frac{1}{24}+\frac{1}{25}-\frac{1}{26}+\frac{1}{27}-\frac{1}{28}+\frac{1}{29} -\frac{1}{30} \mathbf{ -\frac{1}{32}-\frac{1}{34}-\frac{1}{36}-\frac{1}{38}}$$

And I can write that sum for $k = 8$ as:
$$ \frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+\frac{1}{13} -\frac{1}{14}+\frac{1}{15}-\frac{1}{16}+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}+\frac{1}{21}-\frac{1}{22}+\frac{1}{23}-\frac{1}{24}+\frac{1}{25}-\frac{1}{26}+\frac{1}{27}-\frac{1}{28}+\frac{1}{29} -\frac{1}{30} + \frac{1}{31}-\frac{1}{32}+\frac{1}{33}-\frac{1}{43}+\frac{1}{35}-\frac{1}{36}+\frac{1}{37} -\frac{1}{38}+\frac{1}{39}-\frac{1}{40}+\frac{1}{41}-\frac{1}{42}+\frac{1}{43}-\frac{1}{44}+\frac{1}{45}-\frac{1}{46}+\frac{1}{47}-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}+\frac{1}{51}-\frac{1}{52}+\frac{1}{53} -\frac{1}{54} \mathbf{ -\frac{1}{56}-\frac{1}{58} -\frac{1}{60}-\frac{1}{62}-\frac{1}{64}-\frac{1}{66}-\frac{1}{68} -\frac{1}{70}}$$

I see that for every k I get $1$ extra negative element at the end of the sum. I had an idea to rewrite it that way:
$$\displaystyle \sum_{k=1}^\infty \left( \frac{1}{6k+1}+\frac{1}{6k+3}+\frac{1}{6k+5}-\frac{1}{8k}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6} \right) = $$
$$\displaystyle \sum_{k=1}^\infty \left( (-1)^{n+1}\frac{1}{n} \right) -1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6} - \displaystyle \sum_{k=1}^\infty X$$

That way I know that the first sum converges (Dirichlet), but still I don't know how to evaluate that expression. I don't know how to evaluate $\displaystyle \sum_{k=1}^\infty ( (-1)^{n+1}\frac{1}{n}$ ) and I don't know how to include those negative elements in my sum (those are marked as X).


问:
sum of weigted multiplicative function
Let $f$ is mulitplicative function, and $N$ is given integer,

then,
$$\sum_{n=1}^N \sum_{l=1}^{n} \lfloor \frac{n}{l} \rfloor f(l) $$

How compute this nicely?

Is there no way to get this with O(N) if $f$ is preprocessed?

答:
We have a double sum, which (if implemented naively) would require $O(N^2)$ integer operations, additionally to the cost of computing the $f$-values. A standard technique in such cases is to change the order of summation. The sum is over all pairs $(n,l)$ satisfying $1\le l\le n\le N$, so we can sum over $n$, first:
$$\sum_{n=1}^N \sum_{l=1}^{n} \lfloor n/l \rfloor f(l)=\sum^N_{l=1}f(l)\sum^N_{n=l}\lfloor n/l \rfloor.$$
Now $\lfloor n/l \rfloor$ is constant between consecutive multiples of $l$, so we have some blocks of $l$ equal values plus some rest of a period where $\lfloor n/l \rfloor=\lfloor N/l \rfloor$, meaning
$$s(N,l)=\sum^N_{n=l}\lfloor n/l \rfloor=l\,\frac{\lfloor N/l \rfloor(\lfloor N/l \rfloor-1)}2+(N \bmod l+1)\,\lfloor N/l \rfloor.$$ Using $N \bmod l=N-l\,\lfloor N/l \rfloor$, we can write this as $$s(N,l)=(N+1)\,\lfloor N/l \rfloor-l\,\frac{\lfloor N/l \rfloor(\lfloor N/l \rfloor+1)}2.$$ Depending on the shape of $f$, the resulting sum $$\sum_{n=1}^N \sum_{l=1}^{n} \lfloor n/l \rfloor f(l)=\sum^N_{l=1}f(l)\,s(N,l)$$ can often be computed still more efficiently, using the fact that $\lfloor N/l \rfloor$ takes only $O(N^{1/2})$ different values.

Title: sum of weigted multiplicative function


答:
The fact that there is an $xy$ term means that the axes of the conic are rotated from the coordinate axes.  So the first thing I would do is rotate to new $x'y'$ at some angle $\theta$:
$$x= x'\cos(\theta)- y'\sin(\theta), \quad y= x'\sin(\theta)+ y'\cos(\theta)$$

\begin{align}
21x^2&= 21(x'^2\cos^2(\theta)- 2x'y'\sin(\theta)\cos(\theta)+ y'^2\sin^2(\theta))\\
-6xy&= -6(x'^2\cos(\theta)\sin(\theta)+ x'y'(\cos^2(\theta)- \sin^2(\theta))- y'^2\sin(\theta)\cos(\theta))\\
29y^2&= 29(x'^2\sin^2(\theta)+ 2x'y'\sin(\theta)\cos(\theta)+y'^2\cos^2(\theta)\\\\
21x^2- 6xy+ 29y^2&= (21\cos^2(\theta)- 6\cos(\theta)\sin(\theta)+ 29\sin^2(\theta))x'^2\\&\qquad+ (-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta))x'y'\\
&\qquad+ (21\sin^2(\theta)+ 6\sin(\theta)\cos(\theta)+ \cos^2(\theta)y'^2
\end{align}

The point is to eliminate the coefficient of $x'y'$ so we must have $$-6\cos^2(\theta)+ 6\sin^2(\theta)+ 21\sin(\theta)\cos(\theta)+ 58\sin(\theta)\cos(\theta)= 0$$ or
$$6\sin^2(\theta)+ 69\sin(\theta)\cos(\theta)- 6\cos^2(\theta)= 0$$

We can think of that as the quadratic equation $6X^2+ 69XY- 6Y^2= 0$.  By the quadratic formula $$X= \frac{-69Y\pm\sqrt{4761Y^2+ 144Y^2}}{12}= \frac{-69Y\pm Y\sqrt{4905}}{12}= \frac{1.03}{12}Y.$$
So $\sin(\theta)= \dfrac{1.03}{12}cos(\theta)$ and $\tan(\theta)= \dfrac{1.03}{12}$.

Title: If conic is represented by $21x^2 -6xy +29y^2 +6x-58y-151=0$, then find the centre, length of axes and eccentricity

Body: Using partial derivatives, I found centre of the conic as $(0,1)$ and I think the conic represents an ellipse. But I am not able to find the rest of the answers.


问:
Given $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$, find the value of $f(1009, 2019)$

Let $f:(\mathbb{N}\ \cup \{0\})^2$ $\rightarrow\mathbb{N}\ \cup\{0\}$ be such that $f(0, x) = f(x, 0) = 1$ for all x $\in$ $\mathbb{N}\ \cup$ {$0$}.
Also for all $i, j \in\ \mathbb{N}$ we have $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$. Find $f(1009, 2019)$.

What I've tried:

I took two cases: $i>=j$ and $i<j$.

As $j$ is reduced by $1$ in both $f(i-1,j-1)$ and $f(i,j-1)$, if $j<=i$, then $j$ will reach $0$ first irrespective of $i$. So the value of $f(i,j)$ is $2^{j}$. Here's a branch diagram to illustrate what I meant

So $f(3,2)=2^2=4$

However I'm stuck at finding the pattern for the other case ($j>i$). Example for $f(i,j)$ where $i<j$:

$f(2,3)=7$

I do recognize that I can stop evaluating once I reach a point where $i=j$ but I can't find at which places and how many times they reach that point. It would be highly helpful If you would help me find the pattern. Also, am I going in the right direction? Is there any other way to solve this problem(a formula or something)? Help would be highly appreciable!

Body:
I programmed the sequence to see if there is any pattern.
Below is the $13 \times 13$ matrix including enough initial values to notice the pattern to be proven by induction after all:

[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
[1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79]
[1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299]
[1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794]
[1, 2, 4, 8, 16, 32, 63, 120, 219, 382, 638, 1024, 1586]
[1, 2, 4, 8, 16, 32, 64, 127, 247, 466, 848, 1486, 2510]
[1, 2, 4, 8, 16, 32, 64, 128, 255, 502, 968, 1816, 3302]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1013, 1981, 3797]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2036, 4017]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4083]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095]
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096]

So, the pattern I noticed is as follows:
$$f(m,n) = \begin{cases} \displaystyle\sum\limits_{k=0}^{m} {n \choose k},\ \  m < n \\[1mm] 2^n, \qquad \ \ \ \ ~ m\ge n \end{cases}$$
Next, we should prove it by induction.

The case $m < n$ (which is more interesting since $1009 < 2019$):

$\square$ We have enough information for the base case (see the table above). Now assume that
$$f(m-1,n-1) = \sum_{k=0}^{m-1}{n-1 \choose k} $$ and $$ f(m,n-1) = \sum_{k=0}^{m}{n-1 \choose k}$$ are true.

We should show that $f(m-1,n-1) + f(m,n-1) = f(m,n)$ is found by the same formula; that is, we are to prove
$$\sum_{k=0}^{m-1}{n-1 \choose k} + \sum_{k=0}^{m}{n-1 \choose k} = \sum_{k=0}^{m} {n \choose k}$$

Indeed,
$$
\begin{align}
\sum_{k=0}^{m-1}{n-1 \choose k} + \sum_{k=0}^{m}{n-1 \choose k}
&= \sum_{k=1}^{m}{n-1 \choose k-1} + {n-1\choose 0} + \sum_{k=1}^{m}{n-1 \choose k} \\[1mm]
&=  {n-1\choose 0} +  \sum_{k=1}^{m}\left[{n-1 \choose k-1} +{n-1 \choose k}\right] \\[1mm]
&\overset*= {n\choose 0} + \sum_{k=1}^{m}{n\choose k} \\[1mm]
&= \sum_{k=0}^m {n\choose k}
\end{align}
$$
where (*) denotes the Pascal's formula. $\blacksquare$

The case $m \ge n$:

$\square$ The base case and the induction hypothesis are analogous and the induction step is trivially proven by noting that
$$\underbrace{2^{n-1}}_{f(m-1,n-1)}+\underbrace{2^{n-1}}_{f(m,n-1)} = \underbrace{2^n}_{f(m,n)}. \ \ \blacksquare$$

Finally, we note that $$\boxed{f(1009, 2019) = \sum_{k=0}^{1009}{2019 \choose k}}$$


问:
for which $a$ is $f$ integratble
for which $a$, with $a>0$ is this function integratble $f: [0,1] \times [0,1] \to [0,\infty]: \mathcal{f(x,y)} = \left\{ \begin{matrix} \frac{xy}{(x+y)^{a}} & (x,y-\neq(0,0) \\ \infty & (x,y)=(0,0) \end{matrix}\right.$

My intuition says that $a$ needs to be $< 1$. But I have no idea how I can prove it correctly.

First I tried to solve the integral. Because $f$ is a positive continuous (so measurable) function, it's enough to see for which $a$ the integral is finite. But this didn't work.

Second, I tried by saying $\frac{xy}{(x+y)^{a}}$<$\frac{y}{(x+y)^{a}}$. I tried with change of variables (x,y)=(u-y,y) an then we can say that $\frac{y}{(u)^{a}}$= $\theta(u^{a})$. I thought this meant that the function is only integrateable if $a< 1$.

EDIT: a good suggestion $\frac{xy}{(x+y)^{a}}$<$\frac{1}{(x+y)^{a}}$

答:
The function is integrable when $a<4$.

Due to the fact that $f(x,y)$ is always positive, Fubini's Theorem allows us to work with the iterated integrals. As the only possible problem occurs at $0$, we can also integrate on a smaller rectangle.

If $a\geq 4$, then
$$
\int_0^{1/2}\int_0^{1/2}\frac{xy}{(x+y)^a}\,dx\,dy
\geq\int_0^{1/2}\int_0^{1/2}\frac{xy}{(x+y)^4}\,dx\,dy=\int_0^{1/2}\frac{3y+1}{6y(y+1)^3}\,dy=\infty.
$$

And when $a<4$, I don't see a nice argument, but it can be done by brute force. We may assume that $a$ is not an integer, because on $[0,1/2]\times[0,1/2]$ we have $(x+y)^{-a}\leq (x+y)^{-a-\delta}$ for any $\delta>0$. We may also assume $a>0$, because the case $a\leq 0$ is trivial.

It turns out that the antiderivatives are not that hard to calculate, using parts and a bit of patience.

$$
\int_c^1\int_0^1 \frac{xy}{(x+y)^a}\,dx\,dy=b_0+\frac{b_1 c}{(c+1)^a}+\frac{b_2 c^{a+3}}{(c+1)^a}+b_3 \,c^{4-a}+\frac{b_4 c^{a+1}}{(c+1)^a}+\frac{b_5c^{a+2}}{(c+1)^a}
$$
for appropriate constants $b_0,\ldots,b_5$, where in particular
$$
b_3=\frac{1}{(a-4)(a-2)(a-1)},
$$
so nonzero. As we need to take the limit as $c\to0$, the limit exists as long as $a<4$ to guarantee that the $b_3$ term converges.

Title: for which $a$ is $f$ integratble


问:
How can I solve this Laws of Sines problem?
This is a homework question that was set by my teacher, but it's to see the topic our class should go over in revision, etc.

I have calculated $AB$ to be 5.26m for part (a). I simply used the law of cosines and plugged in the numbers.

part (b) is the question I have been trying for quite a while. I tried to do the law of sines, to no avail. To calculate $BC$ I need the angle opposite that which I do not have (or know how to work out.) The triangle $BDC$ has a right-angle, but this does not work as $sine(90) = 1$

A step in the right direction would be more beneficial than a full answer.

答:
To compute directly an angle or a length, you must have at least 2 legnth and an angle or 2angle and a length, and then use the law of Sines. For the moment you have 1 angle and 1 length

The idea is to use the other triangle (on which you have or may have full information) to gain the missing information. You could compute the angle ABD within the first triangle, and then deduce the angle CBD on the second one. And then you have the necessary information to use the law of Sines.


问:
Probability of presence of at least one color blind person in a group
The frequency of red-green color vision defects among populations of Northern European origin
is around 8% of males and 0.5% of females.

Deeb S. The molecular basis of variation in human color vision: Variation in human color vision. Clinical Genetics.
2005;67(5):369-377. doi:10/cn5s4g

One souce I found for the probability of at least one color blind person in a group of three men is (https://jfly.uni-koeln.de/color/): $0.22 = 1-(1-0.92)^3$

Does that source do the correct calculation?
I think it should be $0.22= 1 -(1-0.08)^3$

How could I calculate the chance of presence of at least one color blind person in a group of two women and two men?

Would $0.16 = 1 - (0.92^2 * 0.995^2)$ be correct?

Thank you for your help.

答: Yes, they're both correct. The source is calculating the probability of the opposite event (which is easier). Note $P(Event) = 1 - P(Opposite \space event)$. There you have it


问:
Bounds on the diagonal elements of inverse of negative definite matrix
Let $X$ be an $n\times n$ negative definite matrix such that $X_{ii}<0$ for all $i$ and $X_{ij}\geq 0$ for $i\neq j$. Denote by $Y=X^{-1}$ the inverse of $X$. I would like to find bounds on the diagonal elements of $Y$. Can we find $\underline{Y}$ and $\overline{Y}$ such that $\underline{Y}\leq Y_{ii} \leq \overline{Y}$ for all $i$?

Note that $-X$ satisfies the definition of an $M$-matrix and so its inverse is such that all its elements are nonnegative. This implies that all the elements of $Y$ are nonpositive and so one potential upper bound is $\overline{Y}=0$ but I would like to find a tighter bound if possible.

Also, in the special in which $X$ is diagonal, it is easy to show that
$$
\left(\max_i X_{ii}\right)^{-1}\leq Y_{ii} \leq \left(\min_i X_{ii}\right)^{-1}.
$$
I am hoping that a result like this extends to nondiagonal $X$'s.

答:
You can use the Rayleigh quotient. For any hermitian matrix $A$ and nonzero vector $x$,
$$\lambda_{\min}(A) \leq \frac{x^TAx}{x^Tx} \leq \lambda_{\max}(A).$$
By taking $x=e_i$, the $i^{th}$ standard basis vector (with a $1$ at position $i$ and zeros everywhere else), you obtain:
$$\lambda_{\min}(X^{-1}) \leq (X^{-1})_{ii} \leq \lambda_{\max}(X^{-1}).$$
You have $\lambda_{\min}(X^{-1}) = 1/\lambda_{\max}(X)$ and $\lambda_{\max}(X^{-1}) = 1/\lambda_{\min}(X)$ since all eigenvalues are negative, and since if $\lambda$ is an eigenvalue of $X$ then $1/\lambda$ is an eigenvalue of $X^{-1}$ (if $Xv = \lambda v$ then $X^{-1}v = (1/\lambda)v$).


问:
Bracing a polygon without triangles
The following Laman graph braces a square without triangles. Stated another way, this is a unit-distance rigid graph without 3-cycles. It seems to be the smallest example of a triangle-free braced polygon.  This happens to be a subgraph of the unit sticks cube graph.

What other regular polygons have triangle-free rigid constructions?

答: As an addendum to Parcly Taxel's answer, his hexagon bracings are a subset of a whole 2DOF family of hexagonal bracings.  Here are two especially symmetric members of this family. (The dotted lines indicate unit-separations that are not included as edges.)

Title: Bracing a polygon without triangles


答: As far as I can see, your error is just being inconsistent about the branch of $z^a$ that you're using. In the first part of the calculation, you used a branch that gives the real value of $z^a$ along $l_+$ and gives that real value times $e^{2\pi ia}$ along $l_-$. Then you have to use the same branch when calculating the residues.  You have the right residue at $i$, but the residue at $-i$ should not involve $e^{-a\pi i/2}$ but rather $e^{3a\pi i/2}$. If my (hurried) arithmetic is right, that correction will eliminate the $-e^{-a\pi i}$ factor and will leave you with an answer that agrees with the given solution (thanks to the double-angle formula for the sine function).


问:
Choice of postulate 1b in Kleene's Introduction to Metamathematics
In Introduction to Metamathematics, Kleene introduces a formal system where the first three postulates in the group for propositional calculus are:

$$
1a. A \to (B \to A)\\
1b. (A \to B) \to ((A \to (B \to C)) \to (A \to C))\\
2. {A, A \to B \over B}
$$

As far as I understand, $1a$ and $2$ are typical for Hilbert-style deductive systems, but $1b$ is not. A more traditional choice, serving pretty much the same purpose (e.g. proving $A \to A$ to start with) would have been:
$$
(A \to (B \to C)) \to ((A \to B) \to (A \to C))
$$

What is the rationale for the unique choice made for $1b$ in Introduction to Metamathematics?

References:

S. C. Kleene, 1952, Introduction to Metamathematics (Ishi Press: 2009 reprint), page 82

https://en.wikipedia.org/wiki/Hilbert_system

https://en.wikipedia.org/wiki/List_of_Hilbert_systems

答:
Let's write:
\begin{align*}
(1b) &\quad (A\to B)\to ((A\to (B\to C))\to (A\to C))\\
(1b') &\quad (A\to (B\to C))\to ((A\to B)\to (A\to C))
\end{align*}

Intuitively, $(1b)$ and $(1b')$ have the same meaning: If you know $(A\to B)$ and you know $(A\to (B\to C))$, then you can conclude $(A\to C)$, by using Modus Ponens $(2)$ twice. The order in which the hypotheses are stated doesn't matter.

The role of $(1b)/(1b')$ in a Hilbert system is to "internalize" Modus Ponens in order to prove the deduction theorem. Both $(1b)$ and $(1b
)$ are adequate to prove the deduction theorem, and once we have the deduction theorem in place, our particular choice of axioms for $\to$ doesn't matter.


问:
Is there a standard or commonly used symbol for the set of recursive (aka computable) real numbers?
The sets of naturals, signed integers, rationals, reals are denoted by the standard symbols $\mathbf{N},\mathbf{Z},\mathbf{Q},\mathbf{R}$ or $\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R}$. Is there a similar symbol commonly used to denote the recursive (aka computable) reals?

Rice, in the paper where he proves they are a field, Recursive real numbers (1954), denotes their field with "$\mathscr{E}$".

答:
The notation I've most commonly seen in this context is "$\mathsf{REC}$." This is also used variously to denote the set of computable binary sequences (= computable points in Cantor space) and the set of computable sequences of natural numbers (= computable points in Baire space), so the context needs to be clear enough to indicate which meaning is meant. When multiple versions are considered at once, in my experience the "context" is added as a subscript, as $\mathsf{REC}_\mathbb{R}$ vs. $\mathsf{REC}_{2^\omega}$ vs. $\mathsf{REC}_{\omega^\omega}$ respectively.

For whatever reason, I've not seen "$\mathsf{COM}$" or "$\mathsf{COMP}$" used analogously with anywhere near the same frequency.


问:
Stuck on proving convergence of infinite series over the quotient of two values of a convergent sequence
in preparation for my real calculus exam, I found an exercise problem in my text book which I can't seem to solve. Given is a sequence $(a_n)$, which is positive, monotone and has an upper bound. The task is to proof, that
$$\sum_{k=1}^\infty \left(\frac{a_{k+1}}{a_k} - 1\right)$$
converges, but I don't seem to make any progress.

Obviously, $(a_n)$ converges. I thought about using Cauchy's convergence test, but I didn't make much progress with that because I don't see a way to find a $n_0(\epsilon) \in \mathbb{N}$.

Another approach I thought of would be to directly prove that $$\lim_{n\to\infty} \sum_{k=0}^n\left(\frac{a_{k+1}}{a_k} - 1\right)$$ exists. This sequence of partial sums if monotone, because $\frac{a_{k+1}}{a_k}$ is positive, so it would be sufficient to find an upper bound. Because I couldn't think of a reasonable one I thought about a proof by contradiction, assuming that $$\forall b \in \mathbb{R} \exists n\in\mathbb{N} : \sum_{k=1}^n \left(\frac{a_{k+1}}{a_k}-1\right) > b$$
This would imply that $$\sum_{k=1}^n\left(\frac{a_k}{a_{k+1}}\right) > b + n$$ but I don't see a way to find a contradiction from there.

As you can tell, I'm really stuck on that problem and would greatly appreciate any help!

答:
Assume $a_k$ is non-decreasing and bounded from above by $M$.

Let $\epsilon_k = \frac{a_{k+1}}{a_k} - 1 \ge 0$, we have
$$\sum_{k=1}^p \epsilon_k \le \prod_{k=1}^p (1+\epsilon_k) - 1
= \prod_{k=1}^p \frac{a_{k+1}}{a_k} - 1 = \frac{a_{p+1}}{a_1} - 1 \le \frac{M}{a_1} - 1 < \infty $$
So the partial sums of $\epsilon_k$ is bounded from above. Since $\epsilon_k$ is non-negative, the partial sums is monotonic increasing. As a result, the partial sums converges as a sequence. This is equivalent to the convergence of the sum/series:
$$\sum_{k=1}^\infty \epsilon_k = \sum_{k=1}^\infty\left(\frac{a_{k+1}}{a_k}-1\right)$$


问:
Compute the indefinite integral of a function whose primite is arccos
While I was studying Kepler's first law, I encountered the following indefinite integral:
$$\int \frac{\lvert b \rvert}{x^2 \sqrt{-1+\frac{2a}{x}-\frac{b^2}{x^2}}}\,dx.$$
($a,b\in \Bbb{R}$, $\lvert b\rvert=\sqrt{b^2}$, $a^2-b^2>0$ and may assume $x>0$.)
It is claimed that, up to a constant (clearly), $\arccos\Bigl(\frac{b^2-ax}{x\sqrt{a^2-b^2}}\Bigr)$ is the primitive of the integrand.
An easy derivation confirms the assertion.
But I was not able to find the primitive by myself.
Can you help me to understand how such a computation should be handled, please?

答:
I'm not sure if this goes through rigorously (feel free to yell at me in the comments) but this was my process:

First, complete the square in the denominator:
$$\sqrt{-1 + \dfrac{2a}{x} - \dfrac{b^2}{x^2}} = \sqrt{\dfrac{a^2}{b^2}-1 - \dfrac{b^2}{x^2} + \dfrac{2a}{x} - \dfrac{a^2}{b^2}} = \sqrt{\dfrac{a^2-b^2}{b^2}-\left(\dfrac{b^2}{x^2}-2\dfrac{b\cdot a}{x \cdot b} + \dfrac{a^2}{b^2}\right)} = \sqrt{\dfrac{a^2-b^2}{b^2}-\left(\dfrac{b^2-ax}{bx}\right)^2}=\dfrac{\sqrt{a^2-b^2}}{|b|}\sqrt{1-\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)^2}$$

Plugging this into our integrand:
$$\int \dfrac{|b|}{x^2 \sqrt{-1 + \dfrac{2a}{x} - \dfrac{b^2}{x^2}}} dx = \int\dfrac{|b|^2}{x^2\sqrt{a^2-b^2}}\cdot \dfrac{1}{\sqrt{1-\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)^2}}dx$$

Notice that by the linearity of absolute values, $|b|^2 = |b||b| = |b^2|$, and because $b^2$ is always nonnegative, $|b^2| = b^2$.

Now, introduce the trigonometric substitution:
$$\cos{\theta}=\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}$$

Differentiating on both sides:
$$-\sin{\theta} d\theta = \dfrac{x\sqrt{a^2-b^2}(-a) - (b^2-ax)\sqrt{a^2-b^2}}{x^2(a^2-b^2)}dx = \dfrac{-b^2}{x^2\sqrt{a^2-b^2}}dx$$

Notice the similarity between this and our integrand's first fraction. Therefore:
$$\int\dfrac{b^2}{x^2\sqrt{a^2-b^2}}\cdot \dfrac{1}{\sqrt{1-\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)^2}}dx = \int \dfrac{-1}{\sqrt{1-\cos^2{\theta}}} \cdot -\sin{\theta} d\theta$$

Now, supposing that $\sin\theta > 0$:
$$\int \dfrac{-1}{\sqrt{1-\cos^2{\theta}}} \cdot -\sin{\theta} d\theta = \int \dfrac{\sin\theta}{|\sin\theta|} d\theta = \int d\theta = \theta + C$$

By definition, $\cos\theta = \dfrac{b^2-ax}{x\sqrt{a^2-b^2}}$, so $\theta = \arccos\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right)$.

$$\int \dfrac{|b|}{x^2 \sqrt{-1 + \dfrac{2a}{x} - \dfrac{b^2}{x^2}}} dx = \arccos\left(\dfrac{b^2-ax}{x\sqrt{a^2-b^2}}\right) + C$$

I believe this works, I am a little unsure about the finer details of assuming $\sin\theta > 0$ and whether the use of $\arccos$ is appropriate, so there may be some details left which need to be filled in, but I would say that this is at least a solid motivation for the given primitive.


答:
To begin, I'll copy some ideas from my earlier post here.

Let $s$ be equal to the absolute value of the smallest (most negative) diagonal entry of $X$.  We can rewrite $X$ as
$X = A - sI$,
where $A$ is a non-negative matrix.

By the Perron-Frobenius theorem, $A$ must have a positive eigenvalue equal to $\rho(A)$.

Furthermore, for any eigenvalue $\lambda$ of $A$, $\lambda - s$ is an eigenvalue of $X$.  Thus, $\rho(A) - s$ is an eigenvalue of $A$.  Since the eigenvalues of $X$ all have negative real part, we note that
$$
\text{Re}(\rho(A) - s) > 0 \implies \rho(A) < s
$$
So, we have $\rho(A) < s$.  Denote $B = \frac 1s A$.
We note that $\rho(B) = \rho(A)/s < 1$. Let $X' = \frac 1s X = B - I$.  Since $\rho(B) < 1$, we can show that the infinite series $-\sum_{n \geq 0} B^n$ converges to $(B-I)^{-1}$.  With that established, we have
$$
X^{-1} = [sX']^{-1} = -s^{-1} \sum_{n=0}^\infty B^n = -s^{-1} \sum_{n=0}^\infty \left(I + \frac 1s X\right)^n.
$$
From there, one could apply the ideas from this recent post (with the same asker as this one). In particular, let $v$ denote a Perron eigenvector of $B$ (i.e. $v > 0$ and $Bv = \rho(B)v$). Let $D = \operatorname{diag}(v)$. We see that $D^{-1}BD$ is a matrix with constant row-sum, and that
$$
X^{-1} = -s^{-1} D\left[\sum_{n=0}^\infty (D^{-1}BD)^n\right]D^{-1}.
$$

Title: Bounds on the diagonal elements of inverse of negative definite matrix

Body:
Let $X$ be an $n\times n$ negative definite matrix such that $X_{ii}<0$ for all $i$ and $X_{ij}\geq 0$ for $i\neq j$. Denote by $Y=X^{-1}$ the inverse of $X$. I would like to find bounds on the diagonal elements of $Y$. Can we find $\underline{Y}$ and $\overline{Y}$ such that $\underline{Y}\leq Y_{ii} \leq \overline{Y}$ for all $i$?

Note that $-X$ satisfies the definition of an $M$-matrix and so its inverse is such that all its elements are nonnegative. This implies that all the elements of $Y$ are nonpositive and so one potential upper bound is $\overline{Y}=0$ but I would like to find a tighter bound if possible.

Also, in the special in which $X$ is diagonal, it is easy to show that
$$
\left(\max_i X_{ii}\right)^{-1}\leq Y_{ii} \leq \left(\min_i X_{ii}\right)^{-1}.
$$
I am hoping that a result like this extends to nondiagonal $X$'s.


问:
binomial-coefficients question $(\frac{x\sqrt[3]{x}}{2}+\frac{1}{\sqrt[15]{x^28}})^n$
I need to find the The coefficient of the free element (when the power is 0) for

$\left(\frac{x\sqrt[3]{x}}{2}+\frac{1}{\sqrt[15]{x^{28}}}\right)^n$

such that the sum of first coefficient and the coefficient second from the end is 13.

I used binom newton thoerm

where $\sum_{i}^{n}\binom{n}{i}(\frac{1}{2})^{i}*1^{n-i}*x^{\frac{3}{4}(i)-\frac{28}{15}(n-i)}$

I think the first coefficient is when $i=0$  and second from the end is $i=n-1$

I tried to substitute the $i's$ and equaling the sum of coefficients to 13 but I dont think that's the way..

答:
The "sum of coefficients" likely refers to the sum of just the binomial coefficients $\binom ni$. Otherwise the sum of the first and penultimate coefficient is $\binom n0+\binom n{n-1}\cdot2^{1-n}=1+n/2^{n-1}\ne13$ as $n\le2^{n-1}$ for $n\in\Bbb Z$.

Another reason to believe that $13$ is the sum of the binomial coefficients only is that the sum of the first and penultimate coefficient is different when we consider $\left(\frac{x^{4/3}}2+\frac1{x^{28/15}}\right)^n$ and $\left(\frac1{x^{28/15}}+\frac{x^{4/3}}2\right)^n$.

So we have $\binom n0+\binom n{n-1}=13$ giving $n=12$. For the constant term we require the powers of $x$ to cancel out, and thus $\color{red}{\frac43}i=\frac{28}{15}(12-i)$ giving $i=7$.


问:
When is $f(x,y)=|x|^a |y|^b$ on the unit disk integrable?
When is $f(x,y)=|x|^a |y|^b$ (for $x,y \neq 0; 0$ otherwise) on the unit disk  in $\mathbb{R}^2$ integrable?

答:
Since you tagged measure theory, I assume you mean Lebesgue integrable. If $a,b>-1$ then $f$ is integrable because
\begin{align}
\int_Df\, dm_2 \leq \int_{[-1,1]^2}f\, dm_2 = 4\int_{[0,1]^2}f\, dm_2 = 4 \left(\int_0^1x^a\, dx\right)\left(\int_0^1y^b\, dy\right)<\infty.
\end{align}

Conversely, if $a\leq -1$ or $b\leq -1$, then $f$ is not integrable, because the square $[-\epsilon,\epsilon]^2$ is contained in the unit disk $D$ if $\epsilon>0$ is small enough, so
\begin{align}
\int_Df\,dm_2 \geq \int_{[-\epsilon,\epsilon]^2}f\, dm_2 = 4 \left(\int_0^{\epsilon}x^a\,dx\right)\left(\int_0^{\epsilon}y^b\,dy\right) = \infty,
\end{align}
where the last line is because $a\leq -1$ or $b\leq -1$ implies one of the bracketed terms is infinite. In other words, $f$ is integrable on $D$ if and only if $a,b>-1$.


问:
$ e^{At}$ for $A = B^{-1} \lvert \cdots \rvert B $
For a homework problem, I have to compute $ e^{At}$ for

$$ A = B^{-1}  \begin{pmatrix}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \end{pmatrix} B$$

I know how to compute the result for $2 \times 2$ matrices where I can calculate the eigenvalues, but this is $3 \times 3$, and I cannot compute eigenvalues, so is there any identity or something which allows computing such exponentials?
Thanks!

Body: Joseph is right and  $ D^n=\begin{pmatrix}(-1)^n&0&0\\0&2^n&0\\0&0&3^n \end{pmatrix}so (Dt)^n=\begin{pmatrix}(-t)^n&0&0\\0&(2t)^n &0\\0&0&(3t)^n\end{pmatrix}$ so $\left(I+Dt+\frac{1}{2!}(Dt)^2+\frac{1}{3!}(Dt)^3+\cdots\right)=\begin{pmatrix}e^{-t}&0&0\\0&e^{2t}&0\\0&0&e^{3t}\end{pmatrix}$


问:
How to calculate $\lim\limits_{n \rightarrow +\infty} \int_{0}^{+\infty} \frac{1}{(1+\frac{x}{n})^n x^{1/n} } dx$ using Beppo Levi theorem?
I'm trying to find  $\lim\limits_{n \rightarrow +\infty} \int\limits_{0}^{+\infty} \frac{1}{(1+\frac{x}{n})^n x^{1/n} } dx$. My numerical test with large numbers instead of the infinity shows that result is 1 (I suppose). Also I tried to apply some facts from real analysis. For example,  Beppo Levi theorem.

So, can I just swap the integral and limit by this theorem?

Thank you for help!

答:
Hint: (DCT)

For all $n \geqslant 2$, we have $\displaystyle\left(1 + \frac{x}{n} \right)^{-n}x^{-1/n} \leqslant \begin{cases}\left(1 + \frac{x}{2} \right)^{-2}, &x \geqslant 1\\x^{-1/2}, &0 < x < 1 \end{cases}$


问:
Showing that the gamma function is integrable for $x \in \mathbb{C}_{Re>0}$
Let $\mathbb{C}_{Re>0}=\{x \in \mathbb{C} \space | \space Re(x)>0 \}$.

For all $x \in \mathbb{C}_{Re>0}$ the function $\gamma_x: \mathbb{R}_+ \to \mathbb{C}, \gamma_x(t)=t^{x-1}\exp(-t)$ is integrable.

How do I show this?

答:
Note that $\forall x,a,b\in\mathbb{R}:x^{a+ib}=x^ae^{ib\ln(x)}$. Hence, for any $x=a+bi$ with $a>0$ and $b\in\mathbb{R}$:
$$\int_{\mathbb{R}^+}|t^{x-1}e^{-t}|dt = \int_{\mathbb{R}^+}|t^{a-1}e^{-t}|dt=\int_{\mathbb{R}^+}t^{a-1}e^{-t}dt.$$

We know that $\lim_{t\to 0}t^{a-1}e^{-t} = 0$, so there are no problems around zero. Let's look at the asymptotic behavior of $t^{a-1}e^{-t}$ at infinity. First of all, note that $t^{a-1}=O(t^{a-1})$ as $t\to \infty$. Furthermore, it is true that $$e^{-t} = o\left(\frac{1}{t^b}\right)\hspace{0.4cm}\text{as}\hspace{0.4cm}t\to\infty,$$
for all $b>0$. Take $b=2a$, then $$t^{a-1}e^{-t} = O\left(\dfrac{1}{t^{a+1}}\right)\hspace{0.4cm}\text{as}\hspace{0.4cm}t\to\infty.$$
Since $a>0$, this means that $t^{a-1}e^{-t}$ is integrable on $\mathbb{R}^+$.


问:
Initial Value Problem with wrong solution, I can't find my mistake
Currently I'm working on the following exercise:

$$
4y''+ 4y'+ 5y = 0\,,\qquad y\left(\pi\right) = 1
\,,\quad y'\left(\pi\right) = 0
$$

My solution to this exercise is
$$
{\rm e}^{-x/2}\left[\,{-\cos\left(x\right)/{\rm e}^{-\pi/2} - \sin\left(x\right)/{\rm e}^{-\pi/2}}\,\,\right],
$$
but I'm not sure about this answer since Symbolab (https://www.symbolab.com/solver/simplify-calculator/4y''%2B4y'%2B5y%3D0%2C%20y%5Cleft(%5Cpi%5Cright)%3D1%2C%20y'%5Cleft(%5Cpi%20%5Cright)%3D0) tells me something else.

Can someone either correct my answer or confirm the correct answer?

Thanks!

答:
The general solution is given by

$$y(x) = c_1 e^{-x/2} \sin(x) + c_2 e^{-x/2} \cos(x)$$

The derivative is given by

$$y'(x) = \dfrac{1}{2} e^{-x/2} (c_1 (2 \cos(x) - \sin(x)) - c_2 (2 \sin(x) + \cos(x)))$$

Evaluating $y(x)$ at $x = \pi$

$$y(\pi) = -c_2 e^{-\pi/2} = 1$$

Evaluating $y'(x)$ at $x = \pi$

$$y'(\pi) = \dfrac{1}{2} e^{-\pi/2} (c_2 - 2 c_1) = 0$$

Can you solve those two for $c_1$ and $c_2$?


问:
$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. Am I correct?
I have to find this limit:

\begin{align} \lim_{n \rightarrow \infty}
 \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

My attempt:

Since we know that $-1\leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$ we have:

\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot(-1)}{n+1}\underbrace{\leq}_{\text{Is this fine?}}&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{(1)^{n}\cdot\sqrt{n}\cdot(1)}{n+1}\\ \\ \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{n}}{n+1}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n+1}
\end{align}

My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have

\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\\ \\  \Rightarrow \ \ \ 0 \leq &\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq 0 \\ \\ \therefore& \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1}=0
\end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!

答:
The first inequality is not true in general: if $a\leqslant b$ and $c<0$, then $ac\color{red}{\geqslant}bc$.

But note that$$\left|\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}\right|\leqslant\frac{\sqrt n}{n+1}\to0,$$and therefore your limit is $0$ indeed.


答:
You're either right or very nearly right: in either case I would be more comfortable putting some absolute value signs on the function to get rid of the $(-1)^n$ terms.

Here's why: let's say we have some $f(x)$ such that $\lim_{x \to a} |f(x)| = 0$. Notice by the linearity of limits that $\lim_{x \to a} -|f(x)| = -\lim_{x \to a} |f(x)| = 0$.

So, because for any $f(x)$ we have $-|f(x)| \leq f(x) \leq |f(x)|$, by the squeeze theorem we have that $\lim_{x \to a} f(x) = 0$.

So, the $(-1)^n$ terms aren't an issue, but I would state this explicitly in your work.

Title: $\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. Am I correct?

Body:
I have to find this limit:

\begin{align} \lim_{n \rightarrow \infty}
 \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

My attempt:

Since we know that $-1\leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$ we have:

\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot(-1)}{n+1}\underbrace{\leq}_{\text{Is this fine?}}&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{(1)^{n}\cdot\sqrt{n}\cdot(1)}{n+1}\\ \\ \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{n}}{n+1}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n+1}
\end{align}

My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have

\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\\ \\  \Rightarrow \ \ \ 0 \leq &\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq 0 \\ \\ \therefore& \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1}=0
\end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!


问:
Pigeon holes principle
Let $P$ be a group that it's elements are 257 sentences in which only atomic sentences from $A,B,C$ exist (i.e. $A \iff B,\space\space A \wedge B \wedge C, \space\space...$) Show that there exists two different $p_1, p_2 \in P$ so that the sentence $p_1 \iff p_2$ is a tautology.

Pigeons are the $257$ sentences but I can't think of a way to prove the asked.

答: HINT: Notice that $257=2^8+1$, so you might guess that there will be $2^8$ pigeonholes. You have three atomic sentences. They can have $2^3=8$ different combinations of truth values. How many different truth tables can you build from these $8$ combinations of truth values? Now notice that if $p_1\iff p_2$ is a tautology, then $p_1$ and $p_2$ have the same truth table. (Why?)

Title: Pigeon holes principle


问:
Proving two Topologies to be equal
Let $(X,\mathcal T)$ and $(Y,\mathcal V)$ be topological spaces, let $f$ be a function that maps $X$ onto $Y$, and let $\mathcal U$ be the quotient topology on $Y$ induced by $f$. Proved that if $f$ is continuous and  closed then $\mathcal U= \mathcal V$.

All I can deduce with the information given in the question is that $f$ is quotient map.

Can you please tell how should I proceed with this problem. I am not good in problems related to quotient topology and I am still learning.Just telling the main elements of proof will be sufficient.I will fill details myself.

Thanks!

答:
Since $\mathcal U = \{U \in \mathcal P(Y) : f^{-1}(U) \in \mathcal T\}$, we see that $\mathcal U$ is the biggest topology on $Y$ that makes $f$ continuous, that is, if $\mathcal T'$ is a topology on $Y$ such that $f : (X,\mathcal T) \to (Y,\mathcal T')$ is continuous, then $\mathcal T' \subseteq \mathcal U$. Why? Well, if $U \in \mathcal T'$, then $f^{-1}(U) \in \mathcal T$ (this is just the continuity of $f$), and this means precisely that $U \in \mathcal U$, so, $\mathcal T' \subseteq \mathcal U$.

Once observed this, if $f : (X,\mathcal T) \to (Y,\mathcal V)$ is continuous, $\mathcal V \subseteq \mathcal U$. On the other hand, if $U \in \mathcal U$, then $f^{-1}(U) \in \mathcal T$ (by definition) and $f(X \setminus f^{-1}(U))$ is closed in $(Y,\mathcal V)$ (since $f$ is closed). But,
$$f(X \setminus f^{-1}(U)) = f(f^{-1}(Y \setminus U)) \stackrel{!}= Y \setminus U$$ meaning that $U \in \mathcal V$, and so, $\mathcal U \subseteq \mathcal V$ (what happens in $(!)$?).

By the way, if you know that those conditions implies that $f$ is a quotient map, then $\mathcal U = \mathcal V$ by merely definition of quotient map, in other words, this problem is more elementary.

Title: Proving two Topologies to be equal


答:
If $(a_n)$ is a bounded sequence and $(b_n)$ is an infinitesimal sequence, then the product sequence $(a_n \cdot b_n)$ is also infinitesimal.

Starting from your sequence,

$$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$$

Being $\forall n\in \mathbb{N}$ then $-1 \leq (-1)^{n} \le 1$
and $-1\leq \sin (n^n)\leq 1$  two bounded sequences then $$\lim_{n \rightarrow \infty} \frac{(-1)^{n}}{n+1}=\lim_{n \rightarrow \infty} \frac{\sin (n^n)}{n+1}=0$$

$$\lim_{n \rightarrow \infty} \underbrace{[(-1)^{n}\sin(n^{n})]}_{\text{bounded sequences}}\,\frac{\sqrt{n}}{n+1}=\lim_{n \rightarrow \infty} [(-1)^{n}\sin(n^{n})]\cdot \underbrace{\sqrt{\frac{n}{(n+1)^2}}}_{\text{infinitesimal sequence}}=0$$


问:
Are the irreducible rational polynomials of given degree dense?
If $f\in \mathbb{Q}[X]$ is a polynomial of degree $n\geq 2$ and $\varepsilon>0$, is there always an irreducible $g\in \mathbb{Q}[X]$ of degree $n$ with $$\sum_{k=0}^n |f_k-g_k| <\varepsilon\quad\text{ (where $f_k$, $g_k$ are the coefficients of $f,g$)}$$
?

edit : My idea so far: Change the coefficients so that the denominators are large integers, which leaves more room for changing the numerators and apply a irreducibility criterion for polynomials in $\mathbb{Z}$ ?

答:
Following @Jakobian's comment:

For $f\in\mathbb{Q}[X]$ choose $N\in\mathbb{N}$ such that $N\cdot f\in\mathbb{Z}[X]$. For large $l\in\mathbb{N}$ consider the polynomials $F:=2^l N f$ and $G$ with $G_k:=F_k; 1\leq k\leq n-1$; $G_0=F_0+2$; $G_n=F_n+1$. By Eisenstein criterion $G$ and $g:=\frac{G}{2^l}$ are irreducible. For suitable large $l$ we get
$$\sum_{k=0}^n |f_k-g_k| <\varepsilon$$


答:
\begin{align} \lim_{n \rightarrow \infty}
 \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}

This is one of those problems where you are supposed to take a step back and make things easy on yourself.

Let

$$f(n) = (-1)^{n}\sin(n^{n})$$

and let

$$g(n) = \frac{\sqrt{n}}{n+1} < \frac{1}{\sqrt{n}}.$$

You want

$$\lim_{n\to \infty} [f(n) \times g(n)].$$

As $n \to \infty, f(n)$ is a bounded function.

That is $~-1 \leq f(n) \leq +1~$ for all $n$.

Also, clearly, as $n \to \infty, g(n) \to 0.$

Therefore, as $n \to \infty, \{f(n) \times g(n)\} \to 0.$

Edit

Adding some formality:

For all $n, |f(n) \times g(n)| \leq |g(n)|$ which goes go zero.

Therefore, since $|f(n) \times g(n)| \to 0,~ \{f(n) \times g(n)\} \to 0.$


答:
Fact 1.  If $a_n\leq b_n$ for large $n$ then $\displaystyle\lim_{n\to\infty}a_n\leq\lim_{n\to\infty}b_n$, provided the limits exist.

Fact 2.  If $\displaystyle\lim_{n\to\infty}|a_n|=0$ then $\displaystyle\lim_{n\to\infty}a_n=0$.

From fact 1 we have
$$
0\leq\lim_{n\to\infty}\left|\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}\right|
\leq\lim_{n\to\infty}\frac{\sqrt{n}}{n+1}
=0
$$
and so we can apply fact 2 to obtain
$$
\lim_{n\to\infty}\frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}
=0.
$$


答:
How many different truth-functions are expressible using just three propositional variables? (Hint: how many lines are there in a truth-table in three variables? how many truth-possibilities are there for each line? So how many different truth-functions in three variables in total?)

If there are $x$ different truth-functions in three variables and $y$ different wffs built up from three variables, and $x < y$, how might the pigeon hole principle be applied?

Title: Pigeon holes principle

Body:
Let $P$ be a group that it's elements are 257 sentences in which only atomic sentences from $A,B,C$ exist (i.e. $A \iff B,\space\space A \wedge B \wedge C, \space\space...$) Show that there exists two different $p_1, p_2 \in P$ so that the sentence $p_1 \iff p_2$ is a tautology.

Pigeons are the $257$ sentences but I can't think of a way to prove the asked.


Title: How to calculate $\lim\limits_{n \rightarrow +\infty} \int_{0}^{+\infty} \frac{1}{(1+\frac{x}{n})^n x^{1/n} } dx$ using Beppo Levi theorem?

Body:
I'm trying to find  $\lim\limits_{n \rightarrow +\infty} \int\limits_{0}^{+\infty} \frac{1}{(1+\frac{x}{n})^n x^{1/n} } dx$. My numerical test with large numbers instead of the infinity shows that result is 1 (I suppose). Also I tried to apply some facts from real analysis. For example,  Beppo Levi theorem.

So, can I just swap the integral and limit by this theorem?

Thank you for help!

Body: You don't even need such an exchange. With $x=n\tan^{2}t$,$$\begin{align}\int_0^\infty\frac{x^{-1/n}dx}{\left(1+\frac{x}{n}\right)^{n}}&=\int_0^{\pi/2}2n^{1-1/n}\sin^{1-2/n}t\cos^{2n+2/n-3}tdt\\&=n^{1-1/n}\text{B}\left(1-\frac{1}{n},\,n+\frac{1}{n}-1\right)\\&=n^{1-1/n}\frac{\Gamma\left(1-\frac{1}{n}\right)\Gamma\left(n+\frac{1}{n}-1\right)}{\Gamma\left(n\right)}\\&\stackrel{n\to\infty}{\sim}1.\end{align}$$


问:
Any Difference between tensor product over integer numbers and rational numbers?
Are the following two rings isomorphic: $\mathbb{R}\otimes_\mathbb{Q}\mathbb{R}$  and $\mathbb{R}\otimes_\mathbb{Z}\mathbb{R}$

I think they are isomorphic as $\mathbb{Z}$-modules. But as rings I have no idea how to aproach the problem. So any hint or comment would be helpful.

答: A homomorphism of rings is an isomorphism of rings iff it's an isomorphism of abelian groups iff it's an isomorphism of sets. If you know that they're isomorphic as $\mathbb{Z}$-modules probably what you mean is that the natural map $\mathbb{R} \otimes_{\mathbb{Z}} \mathbb{R} \to \mathbb{R} \otimes_{\mathbb{Q}} \mathbb{R}$ is an isomorphism of $\mathbb{Z}$-modules (this is true more generally for any tensor product of $\mathbb{Q}$-vector spaces, and in particular $\mathbb{Q}$-algebras). This natural map is a ring homomorphism and hence a ring isomorphism.


问:
Count the number of sets in $S$ so $S\subseteq P(A)$
In one of previous exams there was the following question:

Let $A=\{1,2,3,4,5,6,7,8\}$ and $S\subseteq P(A)$.

Given that each set in $S$ has $4$ members, and each member of $A$ belongs to $3$ sets of $S$. How many sets are there in $S$?

There are $8$ elements in $A$. Each one of them should be a member of a set of three sets. What does it mean? Does it mean that for each $a\in A$ there exist $B\subseteq S$ so $a\in B$? What $3$ means? Also, how do I count the number of sets in $S$?

答:
$P(A)$ is the set of all subsets of $A$ and hence $S$ is a set of some subsets of $A$, say $S=\{S_1,S_1,...,S_n\}$ where $|S_i|=4$.

The question says that each $a\in A$ is contained in exactly three sets in $S$, say $S_l,S_m,S_n$. If you were to write down the elements of all $S_i$, you would have written each element of $A$ thrice, i.e. $8\times3=24$ characters. The total number of characters written is also $\sum_i|S_i|=4n$. Thus $4n=24\iff n=6$.


问:
Algebra Word Problem (Linear Equations)
I attempted this problem however I highly suspect that my answer is
incorrect, any help on how to approach it efficiently will be
appreciated. It proceeds as follows:

I bought a box of chocolates for myself last week. However, by the time
I got home I had eaten $\frac{7}{8}$ of the chocolates. As I was
putting the groceries away, I ate $\frac{2}{3}$ of what was left.
There are now 22 chocolates left in the box. How many chocolates were in
the box in the beginning?

My Attempt:

Let the original number of chocolates be $x$,

I ate $\frac{7}{8}x$, the number remaining = $\frac{1}{8}x$.

Then I ate $\frac{2}{3}(\frac{1}{8}x)=\frac{1}{12}x$, the number
remaining = $\frac{11}{12}x$, it follows that $\frac{11}{12}x=22$,
hence $x=24.$

Title: Algebra Word Problem (Linear Equations)

Body:
You eat $7/8x$ then $1/8x $ is left.

Then you eat $2/3$ of what is left that is $2/3 (1/8x)=1/12x$

Then
$$\frac{7}{8}x+\frac{1}{12}x+22=x$$
and finally $x=528$

Check: $\frac{7}{8}\,528=462$ and $528-462=66$ then $\frac{2}{3}\,66=44$ and $66-44=22$.

Notice that you ate $506$ chocolates!


问:
if $|\mathcal{H}_1 \cap \mathcal{H}_2|> n+1$ then $\mathcal{H}_1 =\mathcal{H}_2$ for hyperovals of an even order plane
Let $\Pi$ be a projective plane of even order $2n$. Let $\mathcal{H}_1,\mathcal{H}_2$ be two hyperovals in $\Pi$. Show that if $|\mathcal{H}_1 \cap \mathcal{H}_2|> n+1$ then $\mathcal{H}_1 =\mathcal{H}_2$.

We know that in a hyperoval no three points are collinear, and there are no tangent lines, namely any line intersects a hyperoval in $2$ points or it is disjoint from it. Also, a hyperoval contains $2n+2$ points. Suppose $|\mathcal{H}_1 \cap \mathcal{H}_2|> n+1$. We look at $\mathcal{H}_1 \setminus(\mathcal{H}_1 \cap \mathcal{H}_2)$. Denote  $|\mathcal{H}_1 \setminus(\mathcal{H}_1 \cap \mathcal{H}_2)|=x$. Then we have
$$2n+2 =|\mathcal{H}_1|=|\mathcal{H}_1 \setminus(\mathcal{H}_1 \cap \mathcal{H}_2)|+ |\mathcal{H}_1 \cap \mathcal{H}_2|> x + n+1$$ so $x < n+1$. I am stuck at this point, any ideas?

答:
Consider the points that are not in $\mathcal{H}_{1} \cap \mathcal{H}_{2}$. If such a point $P$ is in $\mathcal{H}_{1} \cup \mathcal{H}_{2}$, then any line on $P$ meets at most one point of $\mathcal{H}_{1} \cap \mathcal{H}_{2}$; on the other hand, if $P \not\in \mathcal{H}_{1} \cup \mathcal{H}_{2}$, then $P$ lies on $n+1$ secant lines to $\mathcal{H}_{1}$, and so at least one of these secant lines must meet $\mathcal{H}_{1} \cap \mathcal{H}_{2}$ in two points (since $P$ must be joined to every point of $\mathcal{H}_{1} \cap \mathcal{H}_{2}$, which has more than $n+1$ points, by an $\mathcal{H}_{1}$-secant). By a similar argument, this same property holds for $\mathcal{H}_{2}$.

Thus if we put
$$\mathcal{S} = \{ P \ : \ P \in (\mathcal{H}_{1} \cap \mathcal{H}_{2})^{c} \mid \mbox{ every line on $P$ meets $\mathcal{H}_{1} \cap \mathcal{H}_{2}$ in at most one point}\}$$
then we can see that
$$\mathcal{H}_{1} = \mathcal{H}_{2} = (\mathcal{H}_{1} \cap \mathcal{H}_{2}) \cup \mathcal{S}.$$


问:
Understanding the derivation of the PDF of a two-point mixed distribution from its CDF

A random variable $X$ has the cumulative distribution function
$$\begin{cases}
0 & x<1 \\
\dfrac{x^2-2x+2}{2} & 1 \le x<2 \\
1 & x\ge 2
\end{cases}$$
Calculate $E[X]$.

The answer uses the following pdf to get $E[X]$:
$$\begin{cases}
\dfrac{1}{2} & x=1 \\
x-1 & 1 < x<2 \\
0 & \text{otherwise}
\end{cases}$$

The book I am reading doesn't have much (any!) information on two-point mixed distributions and I want to make sure that I understand how this PDF was derived from the CDF. My understanding so far is that firstly, we need to figure out from the CDF that $X$ follows a mixed distribution. Note that $F(1) = 1/2 \ne 0$, which indicates that there is a jump in the CDF of $X$ at $x=1$. Since $X$ is continuous from $1<x<2$, $X$ must follow a two-point mixed distribution. Now, the magnitude of the jump in the graph is $\frac{1}{2}-0 = \frac{1}{2}$ which gives $f(x) = \frac{1}{2}$ if $x=1$. The rest of the pdf can be obtained using routine computations. Finally, we need to compute the probabilistic "weights" to get to the CDF. Clearly, the "weight" for the discrete part is $\frac{1}{2}$, so the "weight" for the continuous part must be $1-\frac{1}{2} = \frac{1}{2}$, and we have

$$E[X] = \int_1^2 (x-1) \cdot x dx + \text{Weight for the discrete part} \cdot 1 = \int_1^2 x(x-1) dx + P[X=1] \cdot 1$$
$$= \int_1^2 x(x-1) dx + \frac{1}{2}$$

which gives the correct answer. However, I am more interested in learning whether my process for obtaining the answer (which was based mostly on deduction and intuition) is correct. Can someone please critique my post? Thanks!

答:
What you wrote there isn't a PDF, since it doesn't have integral $1$. There isn't really a PDF in the usual sense at all, because there is a discrete part.

A PDF for this distribution is $f(x)=\frac{1}{2} \delta(x-1) + (x-1) 1_{(1,2)}(x)$. Here $\delta$ denotes the Dirac delta "function", which is not really a function in the same way you are used to. This funny $1$ notation is defined by

$$1_A(x)=\begin{cases} 1 & x \in A \\
0 & x \not \in A \end{cases}$$

One can evaluate $\int_{-\infty}^\infty x f(x) dx = \int_{-\infty}^\infty x \left ( \frac{1}{2} \delta(x-1) + (x-1) 1_{(1,2)}(x) \right ) dx$ to get the expectation. To do that, the delta function term winds up giving you $\left. \frac{1}{2} x \right |_{x=1}=\frac{1}{2}$, while the other term gives you $\int_1^2 x(x-1) dx$.

Note that the PDF on $(1,2)$ is in fact $x-1$: there is no need for an extra factor of $1/2$ because $\int_1^2 (x-1) dx$ is already $1/2$.


问:
Why all irreducible representations appear in the regular representation?
Let $G$ be a finite group and $R$ the regular representation. That is, as a vector space $R = F(G)$ is the free vector space with basis $G$. If the basis is $\{e_g : g \in G\}$ the action is defined by

$$g \cdot e_{g'}=e_{gg'}$$

and extended by linearity.

Now, in the book I'm studying the author states the following corolary:

Corollary 2.18: Any irreducible representation $V$ of $G$ appears in the regular representation $\dim V$ times.

The "proof" for this is a little argument before the statement:

We know tha the character of $R$ is simply

$$\chi_R(g)=\begin{cases}0, & g\neq e, \\ |G|, & g= e\end{cases}$$

Thus, we see first of all that $R$ is not irreducible if $G\neq \{e\}$. In fact, if we set $R = \bigoplus V_i^{\oplus a_i}$, with $V_i$ distinct irreducibles, then:

$$a_i = (\chi_{V_i},\chi_R)=\dfrac{1}{|G|}\chi_{V_i}(e)|G|=\dim V_i.$$

All I get from that is: if we decompose $R$ into a direct sum of irreducible representations, the multiplicites are the dimensions.

But what guarantees that any irreducible representation of $G$ appears in that decomposition of $R$? Why all irreducible representations of $G$ appear in the direct sum decomposition of the regular representation?

答:
There is a theorem that states:

Let $V$ be a linear representation of $G$ s.t. $V = W_1 \oplus \dots\oplus W_k $ (all of these are irreducible representation of $G$) with $\phi$ being the character. If $W$ is the irreducible representation with character $\chi$, then the number of $W_i$ which is isomorphic to $W$ is the scalar product of $(\phi|\chi)$.

[Serre-Linear representation of finite groups-Ch-2-theorem 4]

So, when it says that "Any irreducible representation $V$ of G appears in the regular representation", it takes into consideration the isomorphs, as we normally do in algebra.


问:
Find all planes which are tangent to a surface
I'm given the surface $z=1-x^2-y^2$ and must find all planes tangent to the surface and contain the line passing through the points $(1, 0, 2)$ and $(0, 2, 2).$ I know how to calculate tangent planes to the surface given one point, but how would I do so given two points? Any push in the right direction would be appreciated. Thank you.</p>

答:
Given $p_1,p_2$ and $G(p) = 0$ with $p = (x,y,z)$, determine a plane $\Pi\to (p-p_1)\cdot\vec n=0$ with $\{p_1,p_2\}\in \Pi$ and such that $\Pi$ is tangent to $G$ at $p^*$

Representing for $\nabla G(p^*)$ the normal to the the surface $G(p)=0$ at point $p^*$ we have the conditions.

$$
\cases{
\nabla G(p^*) = \lambda\vec n\\
(p_2-p_1)\cdot \vec n = 0\\
G(p^*) = 0\\
(p^*-p_1)\cdot\vec n = 0\\
\|\vec n\| = 1
}
$$

seven equations and seven $(p^*,\vec n,\lambda)$ unknowns.


答: Everything looks good up until you said you had $\frac{11}{12}x$ remaining.  You are correct that as you were putting groceries away you ate $\frac{1}{12}x$, but you need to subtract that from the number of chocolates you had when you got home instead of the total number you started with, so it should be $\frac{1}{8}x - \frac{1}{12}x = \frac{1}{24}x$ instead of $x - \frac{1}{12}x = \frac{11}{12}x$.  After that you set $$\frac{1}{24}x = 22$$and solve as you did.  That will give you $x = 22\cdot 24 = 528.$  Another way to do it would be to say that when you got home you had $\frac{1}{8}$ of the original amount left, and then after eating more you had $\frac{1}{3}$ of the new amount left, which gives you $$\frac{1}{3}\left(\frac{1}{8} x\right) = 22,$$ which gets you to the same place.

Title: Algebra Word Problem (Linear Equations)

Body:
I attempted this problem however I highly suspect that my answer is
incorrect, any help on how to approach it efficiently will be
appreciated. It proceeds as follows:

I bought a box of chocolates for myself last week. However, by the time
I got home I had eaten $\frac{7}{8}$ of the chocolates. As I was
putting the groceries away, I ate $\frac{2}{3}$ of what was left.
There are now 22 chocolates left in the box. How many chocolates were in
the box in the beginning?

My Attempt:

Let the original number of chocolates be $x$,

I ate $\frac{7}{8}x$, the number remaining = $\frac{1}{8}x$.

Then I ate $\frac{2}{3}(\frac{1}{8}x)=\frac{1}{12}x$, the number
remaining = $\frac{11}{12}x$, it follows that $\frac{11}{12}x=22$,
hence $x=24.$