问:
How to prove $\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1})= -\infty$?
I tried to prove by using definition for divergence:

We say that a sequence $a_n$ diverges to $-\infty$ if for every $M>0$, there is an integer $n_0$ such that $a_n < -M$ whenever $n>n_0$

So I start:
$$\frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
Because $|cos x|\leq 1:$
$$-\frac{n}{2} \leq -\frac{n}{1+\sqrt{n}} \leq \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
$$-\frac{n}{2} < -M$$
$$n>2M$$

Therefore $n_0$ exists and we can say (for example): $n_0=2M+1$.

Is this correct? Is there some other way we could prove this?

答:
hint

Sorry, your first inequality is not correct.

The only way is to write

$$\cos(\frac{\pi n}{n+1})=\cos(\pi-\frac{\pi}{n+1})$$
$$=-\cos(\frac{\pi}{n+1})$$

And

$$\frac{n}{1+\sqrt{n}}\ge \frac{n}{2\sqrt{n}}$$

But, for $ n $ great enough $(n\ge N_1)$, we have
$$-\cos(\frac{\pi}{n+1})<-\frac 12$$
since it goes to $ -1$.
thus

$$n>N_1\;\implies \; a_n<-\frac{\sqrt{n}}{4}$$
Given $ M>0 $

You just need to find $ N_2 $ such that
$$n> N_2 \;\implies \; \frac{\sqrt{n}}{4}>M$$

You can take $ N_2=16M^2$.

Finally, if you take $ N=\max(N_1,N_2) $, then

$$n>N\;\;\implies \;\;a_n<-M$$


问:
Decompose $\frac{1}{a \sin x +b}$ into sum of trigonometric functions
I am wondering if it is possible to write

$$f(x)=\frac{1}{a \sin x +b}$$

as a sum of functions that don't have trigonometric functions in the denominator? Or just pure trigonometric functions, without a constant added. So maybe something like this:

$$\frac{1}{a \sin x +b}=A\sin^3 x \cos^2x + B\sin x \cos ^4 x + C\frac{1}{\cos x}+ D\frac{\cos^2 x }{\sin x}+\dots $$

The reason I'm asking is that I need to integrate products of $f(x)$ with powers $\cos^n x$, and it would be really neat to get rid of the constant in the denominator, as integrals of the form $\int\frac{\cos^n x}{a \sin x +b}dx$ are extremely hard to solve, whereas integral tables contain solutions to $\int \sin^p x \cos^q x dx$ and similar integrals.

One potential starting point could be

$$\frac{1}{a \sin x +b}=\frac{a \sin x -b}{(a \sin x +b)(a \sin x -b)}=\frac{a \sin x -b}{(a^2 \sin^2 x -b^2)}=...$$
with the goal to eventually make the replacement $\sin^2 x +\cos^2 x =1$. Can such a method be formalized? Or is the decomposition I have in mind provable not possible? Thanks for your help!

答:
Assume $b>a$ such that $b+a\sin(x)\ne0$. Then, per the Fourier series

$$\frac{\sqrt{b^2-a^2}}{b+a\sin(x)}=1+2\sum_{n=1}^{\infty}\left(\frac{\sqrt{b^2-a^2}-b}{a}\right)^n[\sin\frac{n\pi}2\sin{(nx )}+ \cos\frac{n\pi}2\cos{(nx )} ]$$

ContentLicense: CC BY-SA 4.0


问:
About the infinite solutions of a Diophantine equation
Consider the following problem:
$$\sum_{k=1}^N k^2=q^2$$
where q is an integer number. This can be written as:
$$\frac{1}{3}N^3+\frac{1}{2}N^2+\frac{1}{6}N=q^2$$
In the same way we can write:
$$\sum_{k=1}^N k^3=q^3$$
that means:
$$\frac{1}{4}N^4+\frac{1}{2}N^3+\frac{1}{4}N^2=q^3$$
In general, we can consider the following equation:
$$\sum_{k=1}^N k^r=q^r$$
with $r$ integer number.
The question is: does this equation has infinite solutions for every $r$ integer?
Thanks.</p>

答: Not always,when $r=2$, the equation $x(x+1)(2x+1)=6y^2$ has only these solutions:$(x,y)=(0,0)(-1,0)(1,±1)(24,±70).$</p>

ContentLicense: CC BY-SA 3.0


问:
Is there an efficient algorithm to generate a symmetry group from a finite set of symmetry operation?
While thinking about the generation of a lot of similar yet distinct pictures from a simple set of primitives I ended up trying to find an efficient algorithm to generate the elements of a symmetry group generated from a finite set of operations.

For example, in 2D, given the two operations $F = \text{"reflection through an axis A"}$ and $R = \text{"clockwise 90° rotation"}$, I can generate the symmetry group of the square (I think it is called $D_4$).

In my head I could come to the conclusion that this group has exactly 8 elements, written in terms of F and R, (I being the identity): $\{I, R, R^2, R^3, AR, A, RA, R^2A\}$.

However, I cannot find either a formal proof that there are no more elements, or an algorithm  that would produce those elements without trial and error.

Is there such an algorithm that could be applicable to other set of generator (say 60° rotation and one reflection) ?

PS: When I say efficient algorithm I really mean deterministic, provably terminating algorithm, the complexity is probably terrible anyway if such an algorithm exists
PPS: I think there is a way to formulate this in term of permutation, but I don't have the required background and cannot understand how it works, let alone the answer to a similar question in an isomorphic space.

答: If you're in the group $D_k$ for some $k \geq 3$ (with $r$ as the rotation and $t$ as the reflection), you always have the identity $rt = tr^{k-1}$ hold, so you only need to consider elements of the form $r^x$ or $tr^x$, which adds up to $2k$ elements total that you can simply write down one by one.


问:
Non-standard form of Sobolev's embedding: Does $\Vert f\Vert_{L^2}^2\leq C \Vert f\Vert_{L^\infty}\Vert f\Vert_{H^1}$ hold??
I am wondering exactly the question of the title. Is it true that there exists a constant $C>0$ such that for all $f\in H^1(\mathbb{R})$ it holds $$
\qquad \qquad \Vert f\Vert_{L^2}^2\leq C \Vert f\Vert_{L^\infty}\Vert f\Vert_{H^1} \quad ?? \qquad (*)
$$
I know that by Sobolev's embedding we have that $$
\Vert f\Vert_{L^\infty}\leq C \Vert f\Vert_{H^1},
$$
and of course we have the trivial bound $\Vert f\Vert_{L^2}\leq \Vert f\Vert_{H^1}$. However, it is not clear to my if something like the inequality in $(*)$ holds. Does anyone has any thoughts?

答:
No, this inequality cannot hold for any $f\in H^1(\mathbb{R})$.

When you wonder if such a functional inequality can hold, a standard test is the scaling argument. Here is how it goes.

Let $f\in H^1(\mathbb{R})$ be non-identically $0$ and define $f_\lambda(x) = \sqrt{\lambda}f(\lambda x)$ with $\lambda>0$. Then one easily computes $\|f_\lambda\|_{L^2} = \|f\|_{L^2}$, $\|f_\lambda\|_{L^\infty} = \sqrt{\lambda}\|f\|_{L^\infty}$ and $\|f_\lambda'\|_{L^2} = \lambda\|f'\|_{L^2}$. By contradiction assume that $(\ast)$ hold. Then since $f_\lambda\in H^1(\mathbb{R})$, it must also satisfy $(\ast)$. Therefore
$$
\|f\|_{L^2}^2 \leq C\,\sqrt{\lambda}\|f\|_{L^\infty}(\|f\|_{L^2}^2 + \lambda^2\|f'\|_{L^2}^2)^{1/2}
$$
By letting $\lambda\to 0$ we obtain a contradiction.


问:
Commutator $[A, -A]$ not vanishing
I think this might be trivial but I have a question about the commutator. I want $A$ to be an operator (might be non-linear) on a Hilbert space (e.g. $L^2$). Is
$$ [A,-A] =0$$
always true?
I think this is wrong, my attempt: By definition $$[A, -A] = A(-A)- (-AA) = A(-A) +A^2$$ and $A(-A) \neq  -A^2$ might be the case because $A$ might be non-linear. Is that correct or did I miss something there?

答:
Your commutator vanishes only if $A(-f) = -A(f)$. For a counter example, consider $A(f) = \text{abs} f = |f|$: the two terms in the commutator,

\begin{align}
\text{abs}( -\text{abs}(f)) &= +\text{abs}(f) \\
-\text{abs}( \text{abs}(f)) &= -\text{abs}(f)
\end{align}

are not identical, so the commutator does not vanish.

OwnerDisplayName: kade

Title: Commutator $[A, -A]$ not vanishing


问:
Chebyshev's inequality for sum of indicator variables
If $N = \sum_{i=0}^{n} I_i$ where $I_i$ is the indicator variable of event $A_i$, with $P(A_i)=p_i$, how can one show that

$$
P(N=0) \leq \frac{\operatorname{Var}(N)}{E(N)^2}
$$

using Chebyshev's inequality or otherwise?

I don't know how to apply Chebyshev's inequality here – it doesn't seem like there's anything that has the right form.

Edit:
The first part of the question is to calculate the expectation and variance of $N$ in terms of $p_i$ and $p_{ij} = P(A_i \cap A_i)$ - perhaps this could be useful for the second part?

Edit:
Full question:

答:
I think I have an answer to this question but would appreciate any feedback! :)

Firstly note that
$$
P(N=0) \leq P(|N-E(N)| = E(N))
$$

We know that

$$
P(|N-E(N)| \geq E(N)) \leq \frac{Var(N)}{E(N)^2}
$$
using Chebyshev.

We want to connect these two expressions. We can do this by noting that
$$
P(|N-E(N)| \geq E(N)) = P(|N-E(N)| = E(N)) + P(|N-E(N)| > E(N)) \geq P(|N-E(N)| = E(N))
$$

Therefore, putting this all together, we have

$$
P(N = 0) \leq P(|N-E(N)| = E(N)) \leq P(|N-E(N)| \geq E(N))\leq \frac{Var(N)}{E(N)^2}
$$


问:
How many pillars do we need to surround a triangular area?
Suppose, I have a triangular garden with sides of $25$ feet, $40$ feet and $55$ feet. If I want to surround my garden with pillars set $5$ feet apart, how many pillars will I need?

I calculated the triangle's perimeter and divided it by $5$. I got 24.

Since it is a enclosed area, I will need one less pillar. So I will need a total of $23$ pillars.

But someone suggested me that I actually need $22$ pillars, but he didn't explain why. So am I miscalculating something?

答:
Your reasoning for $23$ pillars seems to be flawed, as explained in another answer. This is a variation of the classic "fencepost" problem.

So if the requirement is that the pillars must be five feet apart measured along the perimeter, you need $24$ pillars, and $23$ are not sufficient.

But if the requirement is merely that each pillar must be no more than $5$ feet from two others, then there is a trick answer.

First put a pillar at each vertex of the triangle and at intervals of $5$ feet along each side. This takes $24$ pillars.

Next, let's look at the vertex between the sides of length $25$ and $55$.
The angle between these sides is slightly less than $42$ degrees.
If we look at the pillars $5$ feet away from the vertex on each side, those two pillars are less than $3.6$ feet apart. So you can take away the pillar at the vertex and still have every pillar no more than $5$ feet from two others.

Now let's look at the vertex between the sides of length $40$ and $55.$
The angle between these sides is slightly less than $25$ degrees.
If we look at the pillars $10$ feet away from the vertex on each side, those two pillars are less than $4.3$ feet apart. So you can take away the pillar at the vertex and the two pillars $5$ feet from the vertex, and still have every pillar no more than $5$ feet from two others.

According to this interpretation, then, you can remove $4$ pillars from the original $24$ (one at one vertex, three at or near the other vertex) and enclose your garden with $20$ pillars spaced $5$ feet apart.

I don't imagine that this is the intended solution; it is just a play on the ambiguous statement of the question.

Indeed, one might ask how a set of pillars can "enclose" a garden when they are $5$ feet apart and one can simply walk in or out between two pillars.
To "enclose" a garden one might imagine using the pillars as fence posts
and stringing wires or rails or some other material between them.
In that case, if you only use $20$ pillars as suggested in my "trick" interpretation, the wires or rails will cut off two corners of the garden. To really enclose a garden by a fence using fence posts connected by straight line segments, you truly do need $24$ fence posts. If the sides had not been chosen to be multiples of $5$ but had been some other three random numbers adding to $120,$ for example $26, 40, 54,$ you would need more than $24$ fence posts in order to have one at every vertex.


答: Observe that $$x_{n+1}^2=x_n^2+2+\frac1{x_n^2}$$ Thus, you might notice that $$x_{n+2}^2=x_{n+1}^2+2+\frac1{x_{n+1}^2}=\left(x_{n}^2+2+\frac1{x_n^2}\right)+2+\frac1{x_{n+1}^2}>x_{n}^2+4$$ In fact, if you keep descending and use similar estimations, you will reach to the following conclusion $$x_{n+2}^2>x_1^2+2(n+1)=2n+3$$ You may use induction to prove this more rigorously. We are now done, since this yields $$x_{100}^2>2\cdot 98+3=199>196=14^2$$

Title: For $X_{n+1}=X_n+1/X_n$ with $ X_1=1 $, prove that $X_{100}>14$.

Body:
Given the sequence:
$$X_{n+1}=X_n+1/X_n \qquad X_1=1, $$ prove that  $X_{100}>14$


问:
Gaps between prime numbers
Counting up from zero, if you find a prime number, is there a certain intervall you can skip knowing for sure that sequent prime number will not be skipped?

For reference, I am writing a program which searchs for prime numbers by checking every number up from 2. The problem is that once the number is big enough it takes very much time to find the next number. For example, if the program detects a 10-digit prime number, can I skip a certain intervall knowing that the next one won't be left out?

答:
No, but you can do something similar.  The first 3 primes are $2, 3, 5$ and $2\cdot 3\cdot 5 = 30$.  If a number is greater than $30$ and prime then it must be congruent to $1, 7, 11, 13, 17, 19, 23,$ or $29$ modulo $30$.

So for every block of 30 numbers, you need only check 8 of them.

You can expand this method by using more than the first 3 primes.

Title: Gaps between prime numbers


答:
Chebyshev said it, I'll say it again:

There's always a prime between $n$ and $2n$.

Title: Gaps between prime numbers

Body:
Counting up from zero, if you find a prime number, is there a certain intervall you can skip knowing for sure that sequent prime number will not be skipped?

For reference, I am writing a program which searchs for prime numbers by checking every number up from 2. The problem is that once the number is big enough it takes very much time to find the next number. For example, if the program detects a 10-digit prime number, can I skip a certain intervall knowing that the next one won't be left out?


问:
Trouble in understanding Predicate logic semantics
I am studying predicate logic and I have a problem understanding its semantics. I am reading "Logic for computer scientists" by Uwe Schoning and the passage I have a problem with is this:

If $F$ has the form $F=P(t_1,\dots,t_k)$ where $t_1\dots t_k$ are terms and $P$ is a predicate symbol of arity $k$, then its semantics are defined by
$$\mathcal{A}(F) =\begin{cases}1& \text{ if } (\mathcal{A}(t_1),\dots,\mathcal{A}(t_k)\in P^{\mathcal{A}}\\ 0& \text{ otherwise}\end{cases}$$

Here $\mathcal{A}$ is a structure and $P^{\mathcal{A}} = I_{\mathcal{A}(P)}$ is described as below:
Predicate logic structure pair

I think I kinda get what structure is about, my problem is about the first picture. I didn't get what does it mean. Does it mean that if we had a predicate symbol, the value of that predicate symbol is true if all its terms are defined by that structure? If so then if for example we have $P^2(x,f(x))$ and both $x$ and $f(x)$ are defined then $P$ is true regardless? That doesn't seem right!

I tried googling and reading other documents but it confused me more. I'll be happy if someone explains the predicate logic semantics to me in a clear and simple way with a few examples. Thanks in advance

答:
As per the linked definitions, a structure $\mathcal A$ is made of a domain $U$ of "objects" (like e.g. numbers) and an interpretation $I$, i.e. a function that maps terms and predicates  of the languages to objects and subsets of the domain.

Intuitively, terms are "names" for objects.

An unary predicate $P$ (intuitively: a property of objects) is mapped by the interpretation to a subset of the domain, i.e. to the subset of the domain of all and only those objects for which $P$ holds.

Finally, formulas are evaluated true or false according to the interpretation of terms and predicates occuring in them.

This is the gist of the above definition:

$\mathcal A(F) = 1 \text { iff  } (\mathcal A(t_1),\ldots,A(t_k)) \in P^{\mathcal A}$.

Consider a simple example with domain the set $\mathbb N$ of natural numbers.

Let $t$ a term, $P$ an unary predicate and consider a structure $\mathcal A =(\mathbb N, I)$ such that: $t^{\mathcal A}=2 \in \mathbb N$ and $P^{\mathcal A} = \{ n \mid n \text { is even } \} \subseteq \mathbb N$.

Thus $P(t)^{\mathcal A}=1$, because $2$ is even, and $(\forall n P(n))^{\mathcal A}=0$.

What happens when terms and formulas have free variables?

Here we need the third clause of the definition of interpretation: it provides a reference to every variable.

Thus, according to the previous example, let $I$ such that $x^I=2$ and consider the term $t := x+1$.

Now $t^{\mathcal A}$ will be $3$ and thus $P(t)^{\mathcal A}=0$.


问:
Cyclic shifts when multiplied by $2$.
I was trying to solve the following problem:

Find a number in base $10$, which when multiplied by $2$, results in a number which is a cyclic shift of the original number, such that the last digit (least significant digit) becomes the first digit.

I believe one such number is $105263157894736842$

This I was able to get by assuming the last digit and figuring out the others, till there was a repeat which gave a valid number.

I noticed that each time (with different guesses for the last digit), I got an $18$ digit number which was a cyclic shift of the one I got above.

Is there any mathematical explanation for the different answers being cyclic shifts of each other?

答:
Suppose an $N$ digit number satisfies your condition, write it as $N= 10a + b$, where $b$ is the last digit. Then, your condition implies that

$$ 2 (10a + b) = b\times 10^{N-1} + a $$

Or that $b \times (10^{N-1} -1 ) = 19 a$.

Clearly, $b$ is not a multiple of 19, so we must have $10^{N-1} -1$ to be a multiple of 19. You should be able to verify (modular arithmetic), that this happens if and only if $N\equiv 0 \pmod{19} $. This gives us that $N = 19k$, and the numbers have the form

$$\frac{ 10^{19k} + 9 } {19} b  = 10 \lfloor \frac{10^{19k-1} } {19} b \rfloor$$

[The equality occurs because $\frac{1}{19} < 1$.]

I now refer you to "When is $\frac{1}{p}$ obtained as a cyclic shift", for you to reach your conclusion.

There is one exception, where we need the leading digit to be considered as 0 (not the typical definition, so I though I'd point it out). We can get more solutions, obtained by concatenations of your number with itself, and several cyclic shifts (and again accounting for a possible leading digit of 0).

Title: Cyclic shifts when multiplied by $2$.


答:
As suggested in the comments, by induction you may prove that:
$$ (X_n)^2 \geq 2n-1 $$

Base case $n=1$:
$$1^2 \geq 2-1=1.$$
(IH) Then suppose  for some $k\in \mathbb N_{\geq 1}$ we have that:
$$ (X_k)^2 \geq 2k-1$$
We will now make the inductive step. Observe that by definition of the sequence we have:
$$ (X_{k+1})^2 =(X_k + 1/X_k)^2 =(X_k)^2 + 2 \frac{X_k}{X_k} + \frac{1}{(X_k)^2} \geq (X_k)^2 + 2 .$$
By the induction hypothesis we have:
$$ (X_{k+1})^2 \geq (X_k)^2 + 2 \geq (2k-1)+2 = 2(k+1)-1. $$
Then by the principle of mathematical induction the estimate holds for all $n \in $$\mathbb N_{\geq 1}$, in particular for $n=100$ we have that:
$$(X_{100})^2  \geq 200-1=199 > 196=14^2$$
Hence $X_{100}>14$.


问:
Infinite sum of inverse of products of successive primes
If we iteratively remove the multiples of the succesive prime numbers from the natural numbers, starting from 2, i.e.,

$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,  ...  \rightarrow$

$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21,  ...  \rightarrow$

$1, 5, 7, 11, 13, 17, 19, 23, 25,  ...  \rightarrow$

we are consecutively removing fractions

$$\frac{1}{2}, \; \frac{1}{2·3}, \; \frac{1}{2·3·5}, \; \frac{1}{2·3·5·7},\;  ... $$

from the total amount of natural numbers. This suggests that the infinite sum

$$\frac{1}{2}+ \frac{1}{2·3}+ \frac{1}{2·3·5}+ \frac{1}{2·3·5·7}+ ...$$

is equal to 1, but I couldn't figure out a proof for this result. Some references about this particular sum would be appreciated too.
Thank you in advance

答:
This is not the case:
$$\frac12+\frac1{2\cdot 3}+\frac1{2\cdot 3\cdot 5}+\frac1{2\cdot 3\cdot 5\cdot 7}+\frac1{2\cdot 3\cdot 5\cdot 7\cdot11}+\cdots \le  \frac12\left(1+\frac1{3}+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\cdots \right)=\frac34$$

It seems that your series converges to
$$0.70523017179180096514743168288824851374357763910915432819226791381\ldots$$
and does so way faster than the geometric series used in the above comparison.
After all, the $n$th summand is $$\le \frac12\cdot\frac{2^n\cdot n!}{(2n)!}\approx \frac{2^nn^ne^{-n}\sqrt{2\pi n}}{2\cdot(2n)^{2n}e^{-2n}\sqrt{4\pi n} }=\frac1{2\sqrt 2}\cdot \left(\frac e{2n}\right)^n.$$


问:
Differentiation of $v(a)=18^2 \cdot a$
$$\frac{d(v)}{da}=  \frac{d}{da} \left( 18^2 a+ 12a^2 -144a \right).$$

I get the answer as $0$ but it is given $18^2$ as answer for only the first term I am talking about.

Not getting why the constant is not differentiated?

答:
Use the product rule:

$\dfrac {d (u v)} {d a} = u \dfrac {dv} {da} + v \dfrac {du} {da}$

In this case, $u = 18^2$ and $v = a$.

So you get $18^2 \times \dfrac {da} {da} + a \dfrac {d (18^2)} {da}$.

This of course works out to $18^2 \times 1 + a \times 0$.

Title: Differentiation of $v(a)=18^2 \cdot a$


问:
Probability of sequence
The question is as follow: Let $(X_{n})_{n}$ be a sequence of Random variable that is independent with the probability

$P(X_{n}=1)=1-P(X_{n}=0)=\frac{1}{n}$

Show that $P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =1$

I tried to use the Borel-Cantelli Theorem to prove this, however I obtain that

$P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =0$

${ F }_{ n }:=\left\{ { X }_{ n }=0 \right\} $

$\sum _{ n=1 }^{ \infty  }{ P\left( { F }_{ n }^{ C } \right)  } =\sum _{ n=1 }^{ \infty  }{ 1-P\left( X_{ n }=0 \right)  } =\sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ n }  } =\infty$

By Borel-Cantelli, we know that $P\left( { \underset { n\rightarrow \infty  }\limsup  }{ F }_{ n }^{ C } \right) =1$

Since ${ \left( { \underset { n\rightarrow \infty  }{ \liminf }  }{ F }_{ n } \right)  }^{ C }=\left( { \underset { n\rightarrow \infty  }{\limsup}  }\quad { F }_{ n }^{ C } \right) $,

(ie) $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { F }_{ n } \right) =1-P\left( { \underset { n\rightarrow \infty  }{ \limsup }  }\quad { F }_{ n }^{ C } \right)$,

I obtain that $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { X }_{ n }=0 \right) =0$

Since this is the opposite of what I want to prove and the statement is said to be true, I can't see where I went wrong in my argument.

答:
You have a mistake in the last line of your argument
$$ \liminf_{n \to \infty} F_n = \liminf_{n \to \infty} \lbrace X_n = 0 \rbrace \neq \lbrace \liminf_{n \to \infty} X_n = 0 \rbrace$$

The event $\liminf F_n$ is the event that eventually $F_n$ will be true. That is from some moment $k$ for every $n > k$, $F_n$ will happen. In this case this is equivalent to say that the sequence will be constant and equal to zero.

Independent Borel-Cantelli will assure you that both $\limsup F_n$ and $\limsup F_n^c$ will be true, since both probability series diverge. Remember that the interpretation of limsups of events is that the event will happen infinitely often (i.o). That means that both $\lbrace X_n = 0 \rbrace$ and $\lbrace X_n = 1 \rbrace$ will happen i.o. which implies that almost surely $\liminf X_n = 0$ and $\limsup X_n = 1$

Title: Probability of sequence


问:
For what values of $c$ does the curve $ y = cx^{3} + e^{x} $ have inflection points?
For what values of $c$ does the curve $ y = cx^{3} + e^{x} $ have inflection points?

at first I found first derivative $ f^{'}(x) = 3cx^2 + e^{x} $

then second derivative $ f^{''} (x) = 6cx + e^{x} $

now the second derivative should  go from positive to negative or from negative to positive yes?
and I wrote this system
$$ -e^{x} < 6cx < 0 $$

so give me a hint i don't know that to do

答:
You can have an inflection where the second derivative cancels, or

$$6cx+e^x=0.$$

This transcendental equation cannot be solved for $x$, but it can for $c$. If you "decide" an inflection point at $x$, then

$$c=-\frac{e^x}{6x}$$ and it suffices to check the range of $c$. The computation is quite tractable.

Note that an extra condition is that the third derivative

$$6c+e^x$$ does not vanish at the same time. This is possible with $x=1$ and $c=-\dfrac e6.$


问:
Mathematical Formulation - No number less than Supremum M can be an upper bound
I was reading up on supremum and infimum and came across a line whose mathematical formulation has confused me. It says, "No number less than the supremum, M of a set S can be an upper bound of S." This line is very clear. Since M is itself the smallest member of the set of all upper bounds of S, this line is only obvious. But I am confused with the mathematical formulation that follows -

For any positive number , epsilon, however small, there exists a y belonging to S such that
y > M - epsilon

Can someone help me with this?

Attaching an image of it as well.

答: If there were no $y$ in $S$ such that $y>M-\varepsilon$, y.e. if $y\le M-\varepsilon$ for all $y\in S$, $M-\varepsilon$ would be a smaller upper bound.


问:
Does equality of sets follow not only from what they contain but also from what they are contained by?
The Axiom of extensionality states that two sets are equal if they contain the same elements:

$\forall A \forall B [\forall x (x \in A \Leftrightarrow x \in B) \Rightarrow A = B]$

Can it be replaced by the statement that two sets are equal if they are contained by the same elements?

$\forall A \forall B [\forall x (A \in x \Leftrightarrow B \in x) \Rightarrow A = B]$

答:
The Axiom of Extensionality cannot be replaced by the statement that two sets are equal if they are contained by the same sets (this latter principle is a form of Leibniz's law, and I will refer to it by that name).

First, let me prove that, unlike Extensionality,  $\forall A,B. (\forall x. A \in x \leftrightarrow B \in x) \rightarrow B = A$ follows from the other axioms of set theory, without invoking either Extensionality or Leibniz's law in the proof.

Take any two sets $A,B$, and assume that $\forall x. A \in x \leftrightarrow B \in x$. Define the set $\overline{A} = \{ y \in \mathcal{P}(A)  \:|\: y = A \}$ using the Powerset axiom and the Separation schema. Clearly $A \in \overline{A}$, hence by our assumption $B \in \overline{A}$ as well. But by the defining property of $\overline{A}$, the following holds: $\forall y. y \in \overline{A} \leftrightarrow y \in \mathcal{P}(A) \wedge y = A$. In particular, since $B \in \overline{A}$, we can conclude $B = A$.

This proves Leibniz's law in set theory, and since the proof never invokes Extensionality, this implication holds even in set theory without Extensionality.

So all that remains to be shown is that the axiom of Extensionality is not a redudant axiom, i.e. that Extensionality, unlike Leibniz's law, does not follow from the other axioms of set theory. So we have successfully reduced your question to a previous Math.SE question.


问:
Permutations of Number sequence with Minimum cost
There are numbers from 1 to N and our task is to create a permutation such that the cost of the permutation is minimum. For each number, there is a left and right cost. To put number p (i<=p<=N) at the  ith (1-based index) index, it costs Lp*(i-1) + Rp*(N-i) where L[] (left array) and R[] (right array) cost is given. (size of Left[] and Right are same = N)

We do not need to find that permutation but we need to find the total minimum possible cost of the permutation.

Example:- N = 3
L[] (left array) = {1,2,6}
R[] (right array)  = {4,7,2}

Answer = 13..because possible permutation is {3,1,2} and cost = (0 * 6 + 2 * 2) + (1 * 1 + 1 * 4) + (2 * 2 + 0 * 4) = 13 which is minimum possible cost.

My approach:-

```<code>As Lp*(i-1) + Rp*(N-i) is a Summation from i = 1 to N, tried to figure out a common formula for this as

Lp*(1-1) + Rp*(N-1) + Lp*(2-1) + Rp*(N-2) + Lp*(3-1) + Rp*(N-3) ...... Lp*(N-1) + Rp*(N-N)

=> Lp*(0) + Rp*(N-1) + Lp*(1) + Rp*(N-2) + Lp*(3) + Rp*(N-3) ..... Lp*(N-1) + Rp*(0)

=> Lp*(((N-1) * N)/2) + Rp*(((N-1) * N)/2) OR (((N-1) * N)/2)*(Lp + Rp)
</code>```

After that, I put Lp and Rp values from the Left array and right array respectively to get minimum cost, but this is not working.

Can any please help me to solve these questions? (Or please tell me if I am mistaking somewhere or breaking maths rules in my derivation).

答: If you rearrange your cost, it becomes $(L_p-R_p)*i+(R_p*N-L_p)$. Note that the only part in the expression which is multiplied by $i$ is $(L_p-R_p)$ and the rest is constant, i.e, no matter how you try to arrange your numbers, this part is always going to add up into the answer. Therefore an optimal strategy would be to sort our array on the basis $(L_p-R_p)$ in decreasing manner and then take the sum of $i*(L_p-R_p)$ over all $1<=i<=N$.


问:
Constructing a most powerful a discrete distribution, if probabilities for some points are given
Let $X_{1}, \ldots, X_{n}$ be a sample from a discrete distribution. Let the null hypothesis $H_{0}$ state that the distribution is uniform on $\{0,1,2\},$ i.e., $P\left(X_{i}=0\right)=P\left(X_{i}=1\right)=P\left(X_{i}=\right.$
2)$=\frac{1}{3}$.

Let the alternative $H_{1}$ be that the distribution is given by $P\left(X_{i}=0\right)=\frac{1}{5}=P\left(X_{i}=1\right), P\left(X_{i}=2\right)=\frac{3}{5}$. Consider testing $H_{0}$ vs $H_{1}$.

I want to construct a Neyman-Pearson most powerful test among tests with level at most $\alpha .$

But I have no idea how to construct such a test from given information. Please help me with this. Thankyou.

答:
Let $m$ denote the number of $2$’s in the sample.  The Neyman–Pearson lemma tells us that we should reject the null hypothesis whenever
$$
\frac{L_1(m)}{L_0(m)} \ge k_\alpha\text,
$$
where $L_0$ is the likelihood under $H_0$, $L_1$ the likelihood under $H_1$, and $k_\alpha$ is chosen to have the appropriate power.  The likelihood ratio here is of the form
$$
\frac{\left(\frac15\right)^{n - m} \left(\frac35\right)^m}{\left(\frac13\right)^n} = \frac{3^{n + m}}{5^n}\text.
$$

You should therefore reject the null hypothesis whenever $m$ is high enough, i.e.,
$$
\frac{3^{n + m}}{5^n} \ge k_\alpha \iff m \ge m_\alpha\text.
$$

Intuitively, this makes sense because the $2$’s are more likely under $H_1$, so if we have many of them, we should prefer $H_1$!  How do we choose $m_\alpha$?  For any given critical value $m_c$,
$$
\alpha = \left(\frac{1}{3}\right)^n \sum_{j = m_c}^{n} \binom{n}{j}\text.
$$

Therefore, you can try different $m_c$ from $0$ until $n$ and see which one gives you the required power $\alpha$.


问:
Finding the probability of an unknown constant
I am trying to find the value of the unknown constant number and the probability of the length less than it.

The question is "A unit length, thin rod breaks in two places, each break independently uniformly distributed on [0,1]. Let a be a small number. Show that the probability that one of the three pieces into which the rod breaks has length less than a is approximately ka as a -> $0$ where k is a constant you should determine.

I know I can use Taylor expansion to find the approximation of a but it would be impossible to draft out an equation if a is unknown. How can I find the value of a? And how would it be possible to find the probability from knowing the value of a?

Update: Thx for everyone's help. So what I've got now is that k must be smaller than 1 in order for the rod to be shorter than a. But I'm kinda stuck in here now as I'm not sure how does it related to finding the probability?

答: Let the positions of the two breaks be $x$ and $y$.  The sample space is the unit square.  Having one piece less than $a$ in length occurs when $x \lt a, y \lt a, x \gt 1-a, y \gt 1-a, -a \lt x-y \lt a$.  A figure is below.  Compute the area of this region, the outer frame and the main diagonal, in terms of $a$.  There will be a term proportional to $a$ with a subtraction proportional to $a^2$ for the overlaps in the corners.  You are supposed to neglect the term proportional to $a^2$.


问:
Formulation weak form of PDE for solving in FeniCS
I'm trying to derive the weak form for the following Fokker-Planck equation to use in a finite element package. I have

$$\frac{\partial P}{\partial L}(L,\eta) = \frac{1}{\theta}\bigg(2\eta\frac{\partial P}{\partial \eta} +(\eta^2+1)\frac{\partial^2 P}{\partial \eta^2}\bigg), \quad \eta >1,$$

With initial condition $P(L=0) = \delta(\eta-1)$. The Dirichlet boundary conditions can be set later. I require the problem to be in the weak form (page 11) for it to be solved by the FeniCS finite element package. I choose to discretize the time derivative via

$$\bigg(\frac{\partial P}{\partial L}\bigg)^{n+1} \approx \frac{P^{n+1}-p^n}{\Delta L}$$

so

$$P^{n+1}-\frac{\Delta L}{\theta}\bigg(2\eta\frac{\partial P^{n+1}}{\partial\eta}+(\eta^2+1)\frac{\partial^2 P^{n+1}}{\partial \eta^2}\bigg) - p^n = 0.$$

The weak form of the problem I want to write like

$$a(P,v) = L_{n+1}(v)$$

Where $v$ is a test function. I think then I can write

$$\begin{align}
a(P,v) &= \int_{\Omega}P^{n+1}v-\frac{\Delta L}{\theta}\bigg(2\eta\frac{\partial P^{n+1}}{\partial\eta}+(\eta^2+1)\frac{\partial^2 P^{n+1}}{\partial \eta^2}\bigg)\,dx, \\
L_{n+1}(v) &= \int_{\Omega}p^{n}v\,dx.
\end{align}$$
I'm not sure whether this is correct, and if I should do something with the second derivative in $a(P,v).$

答:
Actually, the boundary conditions may change the expression of the weak form. Here I am going to assume that you have homogeneous Dirichlet boundary conditions at $\eta=1$ as well as at $\eta_\max>1$ which is where your are going to truncate your (at the moment unbounded, so untractable numerically) domain in $\eta$.

Observe that your equation rewrites
$$
\frac{\partial P}{\partial L}(L,\eta) = \frac{1}{\theta}\frac{\partial}{\partial \eta}\bigg( (\eta^2+1)\frac{\partial P}{\partial \eta}\bigg)
$$

So after your backward Euler discretization in time, it yields
$$
P^{n+1} - \frac{\Delta L}{\theta}\frac{\partial}{\partial \eta}\bigg( (\eta^2+1)\frac{\partial P^{n+1}}{\partial \eta}\bigg) = P^n
$$
in strong form.

To obtain the weak form you can formally integrate the equation against a test function $v$ that vanishes at the endpoints of your interval. It yields after one integration by parts in the second term of the left-hand side

$$
\int_1^{\eta_\max}P^{n+1}(\eta)v(\eta)d\eta + \int_1^{\eta_\max}\frac{\Delta L}{\theta}(\eta^2+1)\frac{\partial P^{n+1}}{\partial \eta}(\eta)\frac{\partial v}{\partial  \eta}(\eta)d\eta = \int_1^{\eta_\max}P^n(\eta)v(\eta)d\eta
$$

From there you have your bilinear form
$$
a(P,v)=\int_1^{\eta_\max}P(\eta)v(\eta)d\eta + \int_1^{\eta_\max}\frac{\Delta L}{\theta}(\eta^2+1)\frac{\partial P}{\partial \eta}(\eta)\frac{\partial v}{\partial  \eta}(\eta)d\eta
$$
and your linear form
$$
L_n(v)=\int_1^{\eta_\max}P^n(\eta)v(\eta)d\eta
$$


问:
$dT/dx=0$ always true?
Disclaimer:

I am posting it here because I am searching for mathematical arguments supporting the validity of the given differential equation

In a Classical Mechanics book I found the assumption that for an arbitrary particle with constant mass in the Real line $dT/dx=0$, with $T$ the Kinetic Energy i.e. $T=(m⋅\dot x^2)/2$

My hypothesis is that the author used the following 'identity'
$$d\dot x^2/dx=0$$

But solving the differential equation (correct me if I am wrong please) I get to $\dot x=f(t)$

Which I think could be wrong because it could be $\dot x=f(x,t)$ couldn't it?

答:
Of course, it is not always true. As you supposed, let $\dot{x} = f(t,x)$ with $x(0) = x_0$. Then,
$$
\dfrac{dT}{dx}(t,x)= \dfrac{\partial T}{\partial t} \dfrac{dt}{dx}+\dfrac{\partial T}{\partial x}
$$
which is not guaranteed to be identically zero. Think of a situation where a mass traverses the floor while losing its kinetic energy. Clearly in this case, $dT/dx \neq 0$.

Furthermore, as you know, the solution of ODE is always a function of only $t$. But the time $t$ can also be expressed as a function of $x$ (i.e., the inverse of $x(t)$) for some restricted domain on which $x$ is injective. Probably, you are confused concerning the partial derivative and total derivative.

Title: $dT/dx=0$ always true?


问:
Cubic bezier get $y$ as a function of $x$ (not $t$)
I'm trying to write a program to display curves.

I have 4 control points $ A, B, C $ and $ D $ and a value for $x$. From that, I want to find the value for $y$.

Where I've gotten so far

We can use De Casteljau's formula to describe a cubic bezier:

$$
P = (1-t)^3A+3(1-t)^2tB + 3(1-t)t^2C+t^2D
$$

Every point is a vector so

$$
x = (1-t)^3ax+3(1-t)^2tbx + 3(1-t)t^2cx+t^2dx
$$

If I solve this for $t$ I should be able to then plug that $t$ value into
$$
y = (1-t)^3ay+3(1-t)^2tby + 3(1-t)t^2cy+t^2dy
$$
To get my desired y value.

The problem is that I haven't found a way to solve this equation for t. I plugged it into wolfram alpha but it seems to be only an approximation.

Is there a more precise / simpler way to achieve what I want?

EDIT: https://github.com/gre/bezier-easing

答:
I"ll point you to another poster who seems to have started down the same road here, who seems not-so-happy with their decision.

I also found a good search term for what you want (where $x$ is time) is "easing curves", and the MDN documentation about the CSS easing-functions has some good details. The key thing I see there is that you can apply some easy-to-verify restrictions on your control points and you can be guaranteed to not have the "$x$ moving back and forth" problem.  If you could get your hands on that code you'd be golden.

There seems to be a Javascript implementation here, and they talk about having to do a lot of clever work for the $x \leftrightarrow t$ problem that you are facing. They also claim that their code is used by React (!!) and Apple (!!!), so I'll suggest that this is one of those problems where your time is better spent getting someone else's code into your project than re-implementing it yourself. The underlying math question is still interesting, and follow that up if you're curious, but I wouldn't bother if I just wanted my project to move forward.


问:
Do we know if there are more primes with even leading digits or odd leading digits?
I was just wondering, out of curiosity, do we know if there are more primes with even leading digits or odd leading digits? For example, primes with even leading digits would be $23$ or $29$ and primes with odd leading digits would be $31$ or $97$. If we haven't figured it out then how do we do something like that?

Thanks

答:
The Prime Number Theorem can be used to find the supremum and infimum of the relative densities. In particular for primes starting with odd digits the upper bound is 7/9 = 0.777... (if you sample numbers starting 1999...) and the lower bound is 41/81 = 0.506... (if you sample numbers starting 8999...).

The supremum of a set is its least upper bound. For a finite set, a supremum is its maximum. For infinite sets, though, the supremum may be ("slightly") greater than all the members of the set. For example, the set $\{x:x<3\}$ has supremum 3 even though all members are less than 3. You can see that there is no maximum for this set, since if you choose some $y$ in the set it is less than $(3+y)/2$ which is less than 3, and so also in the set.

Similarly, the infimum is the greatest lower bound of the set, the generalization of the minimum.

What you might like in a situation like this would be the relative density of the primes starting with odd digits in the set of primes. This would be
$$
\lim_{x\to\infty}\frac{\text{# of primes}\le x\text{ starting with an odd digit}}{\text{# of primes}\le x}
$$

But what happens if this limit does not exist? For example, it might go up to 70%, then back down to 55%, then back up to 70%, and repeat ad infinitum. In fact this is precisely what happens!

So instead of looking for the limit, which does not exist (just like the maximum of $\{x:x<3\}$ does not exist), we look for the limit supremum and limit infimum. These always exist, and they are equal exactly when the limit exists (in which case they are equal to the limit).

But what is a limit supremum (lim sup for short)? In this case, it is
$$
\lim_{y\to\infty}\sup_{x\ge y}\frac{\text{# of primes}\le x\text{ starting with an odd digit}}{\text{# of primes}\le x}
$$
or in other words, the largest value that the function becomes arbitrarily close to infinitely often. (If it happened only finitely often, then it happens for the last time at some $y_0$, and the limit goes beyond $y_0$ to $\infty.$)

Of course the limit inferior (lim inf) is defined similarly with $\inf$ in place of $\sup.$

So much for the preliminaries. Now it's time to look at the Prime Number Theorem. It says that there are $(1+o(1))(x/\log x)$ primes below $x.$ The meaning of $o(1)$ is technical, but basically it's just some number that becomes arbitrarily small as $x$ increases. Now for some constant $0<\alpha<1$ the number of primes between $\alpha x$ and $x$ is $(1+o(1))(x/\log x) - (1+o(1))(\alpha x/\log(\alpha x))=(1+o(1))(1-\alpha)x/\log x.$ (The different $o(1)$ are actually different numbers, so they don't cancel the way you'd expect. The notation is funny, but that's the way it's usually written.)

So basically there are just the number of primes you'd expect between $\alpha x$ and $x$ since the logarithm in the density grows so slowly.

Now we can use this to derive the densities we need. First, let's look at the relative density of the odd-starting primes between $10^{n-1}$ and $10^n.$ 5/9 start with an odd digit and 4/9 start with an even digit, since by the above use of the Prime Number Theorem the densities are what we'd expect. Now you can do the same with the primes between $10^{n-2}$ and $10^{n-1}$, etc., so for the primes below $10^n$ you have 5/9 starting with an odd digit.

But we don't want to restrict ourselves to looking just at powers of 10. What if we looked at numbers below $2\cdot10^n$? Then half would be below $10^n$ of which 5/9 would start with an odd digit, and the half between $10^n$ and $2\cdot10^n$ all start with odd digits. Thus (5/9+1)/2 = 7/9 start with odd digits. It's not hard to see that this is the best you can do.

Similarly, on the other side, if you look at $9\cdot10^n$ you have nine groups: the ones below $10^n$ have 5/9 odd first digits, the $n-$digit ones starting with 1, 3, 5, or 7 are all odd-initial, and the $n-$digit ones starting 2, 4, 6, 8 are all even-initial. Overall that's (5/9+4)/9 = 41/81. And we're done!


问:
How to prove $\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1})= -\infty$?
I tried to prove by using definition for divergence:

We say that a sequence $a_n$ diverges to $-\infty$ if for every $M>0$, there is an integer $n_0$ such that $a_n < -M$ whenever $n>n_0$

So I start:
$$\frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
Because $|cos x|\leq 1:$
$$-\frac{n}{2} \leq -\frac{n}{1+\sqrt{n}} \leq \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
$$-\frac{n}{2} < -M$$
$$n>2M$$

Therefore $n_0$ exists and we can say (for example): $n_0=2M+1$.

Is this correct? Is there some other way we could prove this?

答:
Hi I have like no idea about this being proof enough for you but we usually do this:
$
 \lim_{n \to \infty} \frac{n}{1+\sqrt{n}}\cos\left(\frac{\pi n}{n+1}\right)= -\infty
$

$
1. \frac{\pi n}{n+1} = \frac{\pi }{1+\frac{1}{n} }
$ // divide by n

for $n\to
 \infty ,  \frac{1}{n} = 0
$

$
\frac{\pi }{1+\frac{1}{n} } = \frac{\pi }{1+0}
$
so $\cos(\pi) = -1$  // so to get to $-\infty$  we only need to show  now that:

$
2. \frac{n}{1+\sqrt{n}} \to \infty
$ // divide by n again

$
\frac{n}{1+\sqrt{n}} = \frac{n}{n\cdot\left( \frac{1}{n} + \frac{\sqrt{n}}{n}\right) }
$

for
$n->  \infty, \frac{1}{n} = 0
$
and
$  \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}=0
$

so we have
$
 \frac{1}{1\cdot( 0 + 0) }  = \frac{1}{0}
$ that goes to infinity

so we end with

$
 \infty \cdot -1  = -\infty$

Take care! I did not write the lim in every row because of fatal lazyness


问:
Finding the base in an exponential equation
I'm trying to solve for $b$ in what seems like a very simple exponential equation:
$b^{4.89}=1 182 795 699$

I know that $y=\log_b x$ is equivalent to $b^y=x$ but I don't know how that helps me to isolate $b$. I looked for the solution in WolframAlpha and it gives me $b=71.68$, but it won't show a step-by-step solution.

I would appreciate a lot if you could tell me how to solve this equation. Thanks :)

Body:
$$b^{4.89}=1 182 795 699$$
It is also equal to
$$b^{489/100}=1182795699$$
Hence,
$$b=1182795699^{100/489}$$


问:
Hilbert space Grassmannians as quotients of unitary groups
Let $\mathcal{H}$ be a complex seperable Hilbert space. By Kuiper's theorem $U(\mathcal{H})$ is contractible. In particular, it has a different homotopy type from $U(\infty) := \operatorname{colim} U(n)$.
The complex Grassmannian of $m$-dimensional subspaces in $\mathbb{C}^n$ can be defined as a quotient
\begin{equation}
Gr_m(\mathbb{C}^n) = \frac{U(n+m)}{U(n) \times U(m)}.
\end{equation}
Then the Grassmannian of $m$-dimensional subspaces in $\mathbb{C}^\infty$ is $Gr_m(\mathbb{C}^\infty) = \operatorname{colim}_n Gr_m(\mathbb{C}^n)$.

Question: Let $V \subseteq \mathcal{H}$ be a $m$-dimensional linear subspace. Does the space
\begin{equation}
\frac{U(\mathcal{H})}{U(V) \times U(V^\perp)}
\end{equation}
have the same homotopy type as $Gr_m(\mathbb{C}^\infty)$?

答:
As a classifying space, the Grassmannian $\mathrm{Gr}_m(\mathbb{C}^\infty)$ can be modelled as the quotient of a contractible space with free $U(m)$-action by the group $U(m)$.  Therefore, in order to show that $$\frac{U(\mathcal{H})}{U(V) \times U(V^\perp)}$$ is a Grassmannian, it suffices to show that the Stiefel manifold $\mathrm{St}_m(\mathcal{H}) := U(\mathcal{H}) / U(V^\perp)$ is contractible with a free $U(m)$-action.  Freed's notes proves the contractibility of $\mathrm{St}_m(\mathcal{H})$ by appealing to Kuiper's theorem or the contractibility of the unit sphere in $\mathcal{H}$, and it is not difficult to see that $U(m)$ acts freely on $\mathrm{St}_m(\mathcal{H})$.  Therefore, the quotient is a classifying space for $U(m)$.

In the comments, you mention the difference between fixing $V$ or letting $V$ vary.  But Freed's notes is not really about varying $V$: he is just being agnostic about which subspace to take.  When it comes down to forming the quotient homogeneous space $\mathrm{St}_m(\mathcal{H})$, you take the quotient by the isotropy group of some fixed subspace; it just doesn't really matter which subspace $V$ you choose since the isotropy groups are conjugate.

What may be perhaps a little unsatisfying with this argument is that we use that classifying spaces are "unique".  But I think this only extends to showing that they have the same weak homotopy type, so you need a little extra (e.g., CW structures) to show that they have the same homotopy type.  In particular, this argument does not immediately provide maps between $U(\mathcal{H}) / (U(V) \times U(V^\perp))$ and the colimit description of the infinite Grassmannian.  I believe that this can be done directly, if less slickly.


问:
Does the statement: $\exists z\in\mathbb{Z}\; \mbox{ such that }\; \forall b\in\mathbb{N},\;\exists\; a\in\mathbb{Q} \; :\; 3z-2b + a=0$ hold true?
Consider this statement:
$$\exists z\in\mathbb{Z}\; \mbox{ such that }\; \forall b\in\mathbb{N},\;\exists\; a\in\mathbb{Q} \; :\; 3z-2b + a=0.$$

I am trying to understand if it is true or false and, if it is false, I would like to understand how deny the statement.

About me, it is true because $3z-2b +a=0\iff 2b\equiv a \ \mbox{mod } 3$, but I am not sure. Could anyone please help me to understand?

Thank you in advance!

答:
This is true.

Take $z=0$ then the end of the logical sentence is $\forall b \in \mathbb{N}$, $\exists a \in \mathbb{Q}$ : $a=2b$ which is true.
Remark: it is also true for every other choice of $z$


答: Hint: Make your life as easy as possible and see what happens for $z = 0$ (even though every integer is fine). Then, given a natural number $b$, you need to find a rational number $a$ such that $-2b + a = 0$. Do you now see whether the statement is true or false?

Title: Does the statement: $\exists z\in\mathbb{Z}\; \mbox{ such that }\; \forall b\in\mathbb{N},\;\exists\; a\in\mathbb{Q} \; :\; 3z-2b + a=0$ hold true?

Body:
Consider this statement:
$$\exists z\in\mathbb{Z}\; \mbox{ such that }\; \forall b\in\mathbb{N},\;\exists\; a\in\mathbb{Q} \; :\; 3z-2b + a=0.$$

I am trying to understand if it is true or false and, if it is false, I would like to understand how deny the statement.

About me, it is true because $3z-2b +a=0\iff 2b\equiv a \ \mbox{mod } 3$, but I am not sure. Could anyone please help me to understand?

Thank you in advance!


问:
Stuck at solving differential equation by using integrating factor
This is the text:

Solve the following differential eqution: $$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$$

This particular differential equation has to be solved by finding an integrating factor. I first tried making this equation into a Darboux differential eqaution, and didn't find any luck. These are the integrating factors that I tried: $u(x,y)=u(x)$ , $u(x,y)=x^my^n$ , $u(x,y)=u(y)$ .

I don't have any idea what else to try. You don't have to answer it fully, just give me a hint or something. Thank you in advance.

答: Have you tried using $(x+y)^{-2}~$?

Body: Have you tried using $(x+y)^{-2}~$?


问:
If $\operatorname{sp}(A) \cup \operatorname{sp}(B)=\operatorname{sp}(A\cup B) \Rightarrow A\cup B$ is linearly dependent
$\newcommand{\sp}{\operatorname{sp}}$

Let $V$ be a vector space over $F$ field, and let $A,B$ be two different, disjoint, non empty sets of vectors from $V$.

If $\sp(A) \cup \sp(B)=\sp(A\cup B) \Rightarrow A\cup B$ is linearly dependent

I've started by saying that if $\sp(A) \cup \sp(B)=\sp(A\cup B)$
then $\sp(A) \subseteq \sp(B)$ or $\sp(B)\subseteq \sp(A)$.

Thus if we assume, WLOG, that $\sp(A) \subseteq \sp(B)$ and we take $v_1 \in A$ and multiply it by scalars $\alpha_1,\ldots,\alpha_k$ we'll get a vector from $B$.

Thus $ A\cup B$ is linearly dependent.

I feel like this proof is not good enough, where is it failing?

答:
Here is a direct proof:

To prove that $A \cup B$ is linearly dependent I must show that there exists a finite number of distinct vectors $w_1, \ldots,w_n$ in $A \cup B$ and scalars $c_1,\ldots,c_n$ in $F$, not all zero, such that

$$ c_1w_1 + \dots + c_nw_n = 0.$$

As already mentioned, span$(A) \cup$ span$(B) =$ span$(A \cup B)$ implies span$(A) \subset$ span$(B)$ or span$(B) \subset$ span$(A)$. I'll assume the former.

Let $x \neq 0 \in$ span$(A)$. Therefore, we can write $x = a_1v_1 + \dots + a_sv_s$ for some $v_1,\dots,v_s$ in $A$ and for scalars $a_1,\dots, a_s$ in $F$. By the inclusion above, we have that $x \in$ span$(B)$ and similarly, $x = b_1u_1 + \dots + b_tu_t$ for $u_1,\ldots,u_t \in B$ and $b_1,\ldots, b_t \in F$. Note then that we have

$$0 = x - x = (a_1v_1 + \dots + a_sv_s) - (b_1u_1 + \dots + b_tu_t)$$

Which is a linear combination of distinct vectors (as $A$ and $B$ were disjoint) of $A \cup B$ and the scalars $a_1,\ldots,a_s, b_1, \ldots, b_t$ are not all zero as $x$ was non-zero. Therefore, $A \cup B$ is linearly dependent.


问:
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$

I couldn't do this question and hence I looked at the solution which goes as follows:

if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$

I state that $y=lx$. Hence the system becomes:

$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.

Hence the possible solutions are:

$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.

My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?

Body:
$\begin{align}
x^2+2xy & =  5 \\
y^2-3xy & = -2
\end{align}$

$\begin{align}
y = & \frac{5-x^2}{2x} \\
x = & \frac{2+y^2}{3y}
\end{align}$

$\begin{align}
y = & \frac{5-\left( \frac{2+y^2}{3y} \right)^2}{2 \left(\frac{2+y^2}{3y}\right)} \\
\end{align}$

$\begin{align}
y = & \frac{5 \cdot(9y^2)-(2+y^2)^2}{2 \cdot 3y (2+y^2)} \\
\end{align}$

$\begin{align}
{2 \cdot 3y^2 (2+y^2)} = & {5 \cdot(9y^2)-(2+y^2)^2} \\
\end{align}$

$\begin{align}
7y^4-29y^2+4=0
\end{align}$

$\begin{align}
y \in \{2,-2,\frac{1}{\sqrt{7}},\frac{-1}{\sqrt{7}}\}
\end{align}$

Plugging it into $x = \frac{2+y^2}{3y} $ we finish the problem:

$(x,y) \in \left\{(1, 2), (-1, -2),
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right),
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}$


问:
if $f(0) = 0$ is it true that $\forall x \in \mathbb{R}: [ f(x)<x \implies f'(x)\leq 1]$
This is a question from my calculus book in high school.

$f(x)$ is a function and $f(0) = 0$

Is it true that $\forall x: f(x) \le x \iff f'(x) \le 1$

----edit----

what can we say about two sides of the equivalency?

1)$\forall x: f(x) \le x \implies f'(x) \le 1$

2)$\forall x: f'(x) \le 1 \implies f(x) \le x $

答:
Try $f(x)=\ln (x+1)$.
It can be shown that $f(x)\le x$ but $f'(-0.5)=2>1$


问:
Partition Theorem Conditional Probability
given pdf fx(x) = $\frac{2x}{9}$ when 0<x<3 and its 0 otherwise and

pdf for y given x is

fY|X(y|x) = $\frac{3y^2}{x^3}$ for 0<y<x

How do i find let's say P(Y<2)

I'm guessing i have to integrate as such

$$\int_{0} ^{3}P(Y<2|X=x)\frac{2x}{9}dx$$

But how do i find what P(Y<2|X=x) equals to for my integration? How do i go about doing this or generally just these type of questions?

答:
The joint density of $(X,Y)$ is the product of the conditional density of $Y$ given $X$ and the marginal density of $X$:
\begin{align}
f_{X,Y}(x,y) &= f_{Y\mid X=x}(y\mid x) f_X(x)\\
&= \left(\frac{3y^2}{x^3}\cdot \mathsf 1_{(0,x)}(y) \right)\left(\frac{2x}9\cdot \mathsf 1_{(0,3)}(x) \right)\\
&= \frac23\left(\frac yx\right)^2\cdot\mathsf 1_{(0,3)\times(0,x)}(x,y).
\end{align}
To find the marginal density of $Y$, we integrate the joint density over the range of $X$:
\begin{align}
f_Y(y) &= \int_{\mathbb R} f_{X,Y}(x,y)\ \mathsf dx\\
&= \int_y^3 \frac23\left(\frac yx\right)^2\ \mathsf dx\\
&= \frac29y(3-y),\quad 0<y<3.
\end{align}
It follows then that for $y\in(0,3)$ we have
\begin{align}
F_Y(y) &:= \mathbb P(Y\leqslant y)\\
&= \int_0^y f_Y(t)\ \mathsf dt\\
&= \int_0^y \frac29t(3-t)\ \mathsf dt\\
&= \frac{1}{27} (9-2 y) y^2.
\end{align}
When $y=2$, this quantity is $\frac{20}{27}$.


问:
Approximating $\ln(2)$ with error of $5*10^{-3}$?
How many terms of $ln(2)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}$ are needed to approximate it with an error of less than $5*10^{-3}$?

答:
There is a formula to calculate the error.

https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder

$$\text{Error}_n=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$$

In your case, $f(x)=\ln(x+1)$, $x=1$, $a=0$.


问:
Describing a differentiable bijective function
Describe a differentiable bijective function from $$\{(x, y) \in |\mathbb{R}^2\mid x^2 + y^2 < 1\}$$ to $$\{(x, y, z) \in \mathbb{R}^3\mid x^2 + y^2 + z^2 = 1, z < 1\}.$$

I'm not sure how the question wants me to lay this out or what it really wants with 'describe'. Can anyone provide any hints to help me start this?

答:
This is the answer to your question. Jacobs explicitly found a diffeomorphism between the two sets, hence a differentiable bijection.

https://math.stackexchange.com/a/2348911/334795

Your question is a very famous example that appears in topological manifold theory. We say that, from your sets, the latter is homeomorphic to the former. In fact, we can also say that it's diffeomorphic to the former in this case.


答:
Suppose, we can compute $\ln\frac{3^2}{2^3}$, $\ln\frac{3^3}{5^2}$, and  $\ln\frac{2^7}{5^3}$ with errors $<\epsilon$.
Then
$$\tag1\ln 2=9\ln\frac{3^2}{2^3} -6\ln\frac{3^3}{5^2}+4\ln\frac{2^7}{5^3}$$
with error $< 19\epsilon$.
To compute $\ln\frac 98$ etc., we can us use
$$ \ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}kx^{k}$$
and very naively stop when the next summands of this alternating series would become $<2\cdot 10^{-4}$:
$$\begin{align}\ln\frac98&\approx  \frac18-\frac1{2\cdot 8^2}+{\frac1{3\cdot 8^3}}&\frac1{4\cdot 8^4}<0.0001\\
\ln\frac{27}{25}&\approx \frac2{25}-\frac{2^2}{2\cdot 25^2}&{\frac{2^3}{3\cdot 25^3}}{<0.0002}\\
\ln\frac{128}{125}&\approx \frac 3{125}&\frac{3^2}{2\cdot 125^2}<0.0003
\end{align}$$
So using these in total six terms in $(1)$, our error is certainly below $$9\cdot0.0001+6\cdot0.0002+4\cdot 0.0003 =0.0033<5\cdot 10^{-3}.$$

If we use the same stopping criterion directly with the alternating series
$$ \ln 2= \sum_{k=1}^\infty\frac{(-1)^{k+1}}{k},$$
we need to continue until the next summand is $<5\cdot 10^{-3}$, that is until $k=200$.

Title: Approximating $\ln(2)$ with error of $5*10^{-3}$?

Body: How many terms of $ln(2)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}$ are needed to approximate it with an error of less than $5*10^{-3}$?


问:
Self-orthogonal Latin squares
A Latin square $A$ is called self-orthogonal if $A$ and $A^{T}$ are orthogonal Latin squares.

Use the elements of $\;\mathbb{Z}_v$ as the names of the rows and columns of your Latin square. Let $\boldsymbol{A}=(a_{ij})$ such that $a_{ij}=2i-j\in \mathbb{Z}_v$. Prove that this forms a self-orthogonal Latin square if the $\gcd(v,6)=1$. I have just tried to see that this works with small orders like $v=5$:
$$\boldsymbol{A}=\begin{pmatrix}
0 & 4 & 3 & 2 & 1\\
2 & 1 & 0 & 4 & 3\\
4 & 3 & 2 & 1 & 0\\
1 & 0 & 4 & 3 & 2\\
3 & 2 & 1 & 0 & 4\\
\end{pmatrix}\implies\boldsymbol{A}^{T}=\begin{pmatrix}
0 & 2 & 4 & 1 & 3\\
4 & 1 & 3 & 0 & 2\\
3 & 0 & 2 & 4 & 1\\
2 & 4 & 1 & 3 & 0\\
1 & 3 & 0 & 2 & 4\\
\end{pmatrix}$$
So it seems to work and I could probably sketch a general construction, but I don't know how the assumption $\gcd(v,6)=1$ makes this all work. Can someone explain?

答:
Let's assume $\gcd(v, 6)=1$ and work in the ring $\Bbb{Z}_v$ except where otherwise stated (so I can write $i = i'$ rather than $v \mid i - i'$ or $i \equiv i' \,(\mathrm{mod}\,v)$).

We have $a_{ij} = a_{ij'}$ iff $2i -j = 2i - j'$ iff $j = j'$. Also we have $a_{ij} = a_{i'j}$ iff $2i - j = 2i' - j$ iff $2(i - i') = 0$ iff $i = i'$, since $2$ is invertible in $\Bbb{Z}_v$ as $\gcd(v, 6) \neq  1$. Hence $A = (a_{ij})_{i, j \in \Bbb{Z}{_v}}$ is a Latin square.

$A$ is self-orthogonal if the pairs $(A_{ij}, A^T_{ij}) = (a_{ij}, a_{ji})$ are all distinct. but $(a_{ij}, a_{ji}) = (a_{i'j'}, a_{j'i'})$ iff the following equations hold:
$$
\begin{align*}
2i - j &= 2i' - j' \\
2j - i &= 2j' - i'
\end{align*}
$$
Since $2$ is invertible as $\gcd(v, 6) = 1$, we can rewrite these equations as:
$$
\begin{align*}
i - i' &= \frac{1}{2}(j - j') \\
j - j' &= \frac{1}{2}(i - i')
\end{align*}
$$
whence
$$
\begin{align*}
i - i' &= \frac{1}{4}(i - i') \\
j - j' &= \frac{1}{4}(j - j')
\end{align*}
$$

which gives us
$$
\begin{align*}
4(i - i') &= i - i' \\
4(j - j') &= j - j'
\end{align*}
$$

but if $4x = x$ in $\Bbb{Z}_v$, we must have that $v \mid 3x$ in $\Bbb{Z}$ implying that $x =0$ in $\Bbb{Z}_v$, since $v$ and $3$ are coprime by our assumption that $\gcd(v, 6) = 1$. Hence $i = i'$ and $j = j'$ and $A$ is indeed self-orthogonal.

Title: Self-orthogonal Latin squares


Title: Stuck at solving differential equation by using integrating factor

Body:
This is the text:

Solve the following differential eqution: $$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$$

This particular differential equation has to be solved by finding an integrating factor. I first tried making this equation into a Darboux differential eqaution, and didn't find any luck. These are the integrating factors that I tried: $u(x,y)=u(x)$ , $u(x,y)=x^my^n$ , $u(x,y)=u(y)$ .

I don't have any idea what else to try. You don't have to answer it fully, just give me a hint or something. Thank you in advance.

Body:
$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$

$\dfrac{dy}{dx}=-\dfrac{2x^3+3x^2y+y^2-y^3}{2y^3+3x^2y+x^2-x^3}$

Let $y=xu$ ,

Then $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

$\therefore x\dfrac{du}{dx}+u=-\dfrac{2x^3+3x^3u+x^2u^2-x^3u^3}{2x^3u^3+3x^3u+x^2-x^3}$

$x\dfrac{du}{dx}=\dfrac{(u^3-3u-2)x-u^2}{(2u^3+3u-1)x+1}-u$

$x\dfrac{du}{dx}=\dfrac{(-2u^4+u^3-3u^2-2u-2)x-u^2-u}{(2u^3+3u-1)x+1}$

$((2u^4-u^3+3u^2+2u+2)x+u^2+u)\dfrac{dx}{du}=-(2u^3+3u-1)x^2-x$

Let $x=\dfrac{1}{v}$ ,

Then $\dfrac{dx}{du}=-\dfrac{1}{v^2}\dfrac{dv}{du}$

$\therefore-\left(\dfrac{2u^4-u^3+3u^2+2u+2}{v}+u^2+u\right)\dfrac{1}{v^2}\dfrac{dv}{du}=-\dfrac{2u^3+3u-1}{v^2}-\dfrac{1}{v}$

$((u^2+u)v+2u^4-u^3+3u^2+2u+2)\dfrac{dv}{du}=v^2+(2u^3+3u-1)v$

This belongs to an Abel equation of the second kind.


问:
Difference between limit at infinity and negative infinity
I know that

\begin{align}
\lim\limits_{x\to\infty}(\sqrt {x^2 + x} - \sqrt {x^2 - 1}) & =
\lim\limits_{x\to\infty}\frac {x + 1} {\sqrt {x^2 + x} + \sqrt {x^2 - 1}}
\\[5 mm] & =
\lim\limits_{x\to\infty}\frac {1 + \frac 1 x} {\sqrt {1 + \frac 1 x} + \sqrt {1 - \frac 1 {x^2}}}
\\[5 mm] & =
\frac 1 2\ .
\end{align}

However, why is it that $\lim\limits_{x\to-\infty}(\sqrt {x^2 + x} - \sqrt {x^2 - 1}) = -\frac 1 2\ $? Regardless of whether $x \rightarrow \infty$ or $-\infty\ $, doesn't $\frac 1 x \rightarrow 0\ $? Any intuitions and explanations behind the difference in results will be greatly appreciated :)

Edit

Many thanks to everyone for their input! I know the difference now :)

Body: If $x<0$, $\sqrt {x^2 + x}=-x\sqrt{1+\frac{1}{x}}$.


问:
Solve $ \int_{0}^{y}{f(x)g(x,y)}dx=\int_{0}^{y}{h(x)}dx $
Here we get $$ \int_{0}^{y}{f(x)g(x,y)}dx=\int_{0}^{y}{h(x)}dx $$ where $ f(x),g(x,y),h(x) $ are all real functions. Let's suppose that they are all smooth functions.

I'd like to solve for $ f(x) $, and naturally, I take derivatives:
$$ G(x,y)=\int{f(x)g(x,y)}dx $$
$$ \implies G(y,y)-G(0,y)=\int_{0}^{y}{h(x)}dx
\implies \frac{d(G(y,y)-G(0,y))}{dy}=h(y)$$
and we get $$ \frac{\partial G}{\partial x}(x,y)\bigg|_{x=y}+\frac{\partial G}{\partial y}(x,y)\bigg|_{x=0}^{x=y}=h(y) $$ $$ \implies f(y)g(y,y)+\int_{0}^{y}{f(x)\frac{\partial g(x,y)}{\partial y}}dx=h(y) $$
And I'm stuck here. My $ f(x) $ seems to be unsolvable for me.

Any attention or suggestions would be greatly appreciated!

答:
Provided $g(0,0) \ne 0$, this is solvable, but that doesn't mean there is necessarily some nice expression for the solution.

To see that it is solvable, at least over a significant space of functions, consider the case where "smooth" = "analytic". That is,
$$f(x) = \sum_{i=0}^\infty a_ix^i, \quad g(x,y) = \sum_{j,k = 0}^\infty b_{jk}x^jy^k, \quad h(y) = \sum_{n=0}^\infty c_ny^n$$
Integrating your equation gives
$$\sum_{i,j,k} \frac{a_ib_{jk}}{i+j+1}y^{i+j+k+1} = \sum_n \frac{c_n}{n+1}y^{n+1}$$
which gives for each $n$
$$\sum_{i+j+k = n}\frac{a_ib_{jk}}{i+j+1} = \frac{c_n}{n+1}$$
Now $g$ and $h$ are known, and thus so are $b_{jk}$ and $c_n$, and we want to find $a_i$.  Since $i, j, k, n \ge 0$, we get the following equations:

$n = 0:\quad a_0b_{00} = c_0$, which solves to $a_0 = \dfrac{c_0}{b_{00}}$

$n = 1:\quad \dfrac {a_1b_{00}}2 + \dfrac{a_0b_{10}}2 + \dfrac{a_0b_{01}}1 = \dfrac{c_1}2$. Since we know $a_0$, this is easily solved for $a_1$.

$n = 2$ will provide a much longer sum. But there will be only one term in $a_2$, while all the other terms involve the already known values of $a_0$ and $a_1$. So again $a_2$ is uniquely determined. Similar remarks hold for all higher values of $n$.

Assuming convergence (note that $a_i$ is majorized by a certain sum of $c_n$ and various $b_{jk}$), this uniquely defines $f$.


问:
Given a series onverges does the series of the power of former series converge?
given   $[ \sum_{n=1}^{\infty} x ]$ converges, does the series of its power converge (given $x^n$ > 0)? $[\sum_{n=1}^{\infty} x^n]$ ?

So far, I know that $[ \sum_{n=1}^{\infty} 1/x]$ diverges and $[ \sum_{n=1}^{\infty} 1/x^k]$ converges with k > 1.

It would be great if you help me solve this. Thanks !

答:
No. For example $$\sum \frac{(-1)^n}{\sqrt{n}}$$ converges, but when you square the general term, you obtain $$\sum \frac{1}{n}$$ which diverges.

If the general term $x_n$ is non-negative, however, this would be true. Indeed, if $\sum x_n$ converges, then $(x_n)$ tends to $0$, and so for all $k > 1$, you have $x_n^k = o(x_n)$, so the series $\sum x_n^k$ also converges.


答:
What you have done to find the limit doesn't work if $ x<0 $.

Let's do this again : \begin{aligned}\lim_{\left|x\right|\to +\infty}{\left(\sqrt{x^{2}+x}-\sqrt{x^{2}-1}\right)}&=\lim_{\left|x\right|\to +\infty}{\frac{x+1}{\sqrt{x^{2}+x}+\sqrt{x^{2}-1}}}\\ &=\lim_{\left|x\right|\to +\infty}{\frac{x+1}{\sqrt{x^{2}\left(1+\frac{1}{x}\right)}+\sqrt{x^{2}\left(1-\frac{1}{x}\right)}}}\\ &=\lim_{\left|x\right|\to +\infty}{\frac{x+1}{\sqrt{x^{2}}\sqrt{1+\frac{1}{x}}+\sqrt{x^{2}}\sqrt{1-\frac{1}{x}}}}\\ &=\lim_{\left|x\right|\to +\infty}\frac{\frac{x}{\left|x\right|}+\frac{1}{\left|x\right|}}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}\\&=\lim_{\left|x\right|\to +\infty}\frac{\mathrm{sgn}\,{x}+\frac{1}{\left|x\right|}}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}\end{aligned}

Thus, the limit as $ x\to +\infty $ would be $ \frac{1}{2} $, and the limit as $ x\to -\infty $ would be $ -\frac{1}{2} \cdot $

Title: Difference between limit at infinity and negative infinity

Body:
I know that

\begin{align}
\lim\limits_{x\to\infty}(\sqrt {x^2 + x} - \sqrt {x^2 - 1}) & =
\lim\limits_{x\to\infty}\frac {x + 1} {\sqrt {x^2 + x} + \sqrt {x^2 - 1}}
\\[5 mm] & =
\lim\limits_{x\to\infty}\frac {1 + \frac 1 x} {\sqrt {1 + \frac 1 x} + \sqrt {1 - \frac 1 {x^2}}}
\\[5 mm] & =
\frac 1 2\ .
\end{align}

However, why is it that $\lim\limits_{x\to-\infty}(\sqrt {x^2 + x} - \sqrt {x^2 - 1}) = -\frac 1 2\ $? Regardless of whether $x \rightarrow \infty$ or $-\infty\ $, doesn't $\frac 1 x \rightarrow 0\ $? Any intuitions and explanations behind the difference in results will be greatly appreciated :)

Edit

Many thanks to everyone for their input! I know the difference now :)


问:
Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$
Solve $x^2+5x+6 \equiv 187 \mod 187$

Solution

$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$

$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17  $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equiv 10 \mod 11 ) \wedge ( 2y \equiv 1 \mod 17  \vee 2y \equiv 13 \mod 17)  $$
$$ ( y \equiv 6 \mod 11 \vee y \equiv 5 \mod 11 ) \wedge ( y \equiv 9 \mod 17  \vee y \equiv 15 \mod 17)  $$
Combining that from CRT I got:
$$ y \in \left\{49, 60,83,94 \right\} $$
and for example:
$$ x+\frac{5}{2}  \equiv 94 \mod 187$$
$$ 2x  \equiv 183 \mod 187$$
some calculus and get...
$$x \equiv 185 $$
And the same thing for each other case.

Question

Is there any faster (or smarter) way to solve equations like that?

答:
Solve $x^2+5x+6 \equiv (x+2)(x+3) \pmod {187}$.

Besides the two 'in your face' solutions, $x \equiv -2 \pmod{187}$ and $x \equiv -3 \pmod{187}$, we can buttress the argument given by José Carlos Santos to find all four solutions.

We want to find integers $x, k, j$ satifying

$\;\text{L1:}\quad x + 2 = 11k$

$\;\text{L2:}\quad x + 3 = 17j$

Subtracting $\text{L1}$ from $\text{L2}$ we write

$\tag 1 1  = 17j - 11k$

Bézout's identity gives us

$\quad 1 = 17(2) - 11(3)$

So $k = 3$ and, plugging into $\text{L1}$,  $x \equiv 31 \pmod{187}$ is a solution.

For the last solution,

$\;\text{L1:}\quad x + 2 = 17j$

$\;\text{L2:}\quad x + 3 = 11k$

Subtracting $\text{L1}$ from $\text{L2}$ we write

$\tag 2 1  = -17j + 11k$

Bézout's identity gives us

$\quad 1 = 17(2) - 11(3)$

So $k = -3$ and, plugging into $\text{L2}$,  $x = -36 \equiv 151 \pmod{187}$  is a solution.

It can also be argued that there are exactly four solutions.


问:
Question about bipartie graphs.
I have a fairly basic inquiry but i would sleep better at night if i saw a proof of it.

Q: i know that if i take a connected subgraph with at least 2 vertices of any simple bipartite graph G that it has to be bipartite.

how would one go about proving that this is the case for any simple graph G.

i think that if G had vertices of all degree 2 i could prove it easily but above that i am not sure.

答:
Hint: If $G$ is bipartite, then its vertices can be split $V=V_1 \cup V_2$ so that the edges are subset of $V_1 \times V_2$.

Let $H$ be any subgraph of $G$ and let $V' \subset V$ be its vertices. Then
you can split the vertices of $V'$

$$V'= (V'\cap V_1) \cup (V' \cap V_2)$$

and it is easy to see that this yields $H$ bi-partite.

P.S. if you are familiar with the fact that bipartite means two colorable
, if you color the vertices of $G$ with two colors, that is a good coloring for any subgraph...

Title: Question about bipartie graphs.


问:
If $ f (x) - f (y) $ is an integrable function then $ f $ is integrable.
Let $ f: [0, 1] \rightarrow [−\infty, \infty] $ be a finite and measurable function a.e. If $ f (x) - f (y) $ is an integrable function then $ f $ is integrable.

I'm trying to do this exercise showing that it is non-negative and measurable. However, this path has been cumbersome. I would like to know if there is a shorter or simplified way to do this exercise. I'm not sure if this feature is actually positive. Some help is considered.
I think it could be proved using Fubini's theorem but I don't have good ideas for this yet

答:
Fix a $y$ such that $|f(y)|<\infty,$ and let $h_y(x)=f(x)-f(y)$.
Now note that $$\int |f(x)|dx\le \int |h_y(x)|dx +\int |f(y)|dx.$$

The first term is finite becase $h_y$ is integrable, and the second term is finite because $f(y)$ is a constant and the Lebesgue measure on $[0, 1]$ is finite.


问:
Bound on tail probability for a random variable with bounded moment and density

Can we say that  $$\sum_{n=1}^{+\infty}\sup_{X}\mathbb P\left( |X| > n \right)<+\infty $$
where each sup is taken over any $X$ zero mean random variable with $\mathbb E[|X|]=1$, and
with distribution whose density function
is bounded by a constant $C>0$?

I tried to estimate $\mathbb P\left( |X| > n \right)$.
Markov's estimation gives $1/n$ as a bound, but it is clearly not optimal, since the bound is attained by atomic distributions, so the condition of the bound on the density is not utilized.

For example, I tried analyzing distribution with density of the kind
$$
f(x) = \begin{cases}
0 & |x|\le a\\
M|x|^{-k} & |x|>a
\end{cases}
$$

and I found that  for any $M,a,k$ satisfying the conditions, and for any $n$,
$$
\mathbb P\left( |X| > n \right)\le \frac 1e n^{-1 -\epsilon(C)}
$$
where $\epsilon = C-1 - \sqrt{C^2-2C} > 0.$
(In case you're wondering, there are no such variables with $C\le 2$)

答:
Here's a counterexample:

Let $X_n$ be a random variable with density
$$
f_n(x) = \begin{cases}
C & |x|\le a\\
C & 2n<|x|\le 2n +b\\
0 & \text{otherwise}
\end{cases}
$$
In this case we have
$$
1 = \int f_n(x) = 2C(a+b) \implies a = \frac 1{2C} -b, 0\le b<\frac 1{2C}
$$
$$
1 = \mathbb E[|X_n|] = \int |x|f_n(x) = 2C\int_0^a |x| + 2C\int_{2n}^{2n+b}|x|\\
=Ca^2 + C((2n+b)^2-(2n)^2) = C(a^2 + 4nb + b^2  )\\
=  \frac 1{4C} + (4Cn-1)b + 2Cb^2\\
\implies b = \frac 1{4C} \left[ 1-4Cn \pm \sqrt{(4Cn-1)^2 + 8C -2  }  \right]
$$
If $C>1/4$, then, for any $n$ bigger than a constant depending only on C,
$$
b = \frac {4Cn-1}{4C} \left[ -1 + \sqrt{1 + \frac{8C -2}{(4Cn-1)^2}  }  \right]\\
\ge \frac {4Cn-1}{4C} \frac{4C -1}{(4Cn-1)^2} \frac 12 = \frac{s}{4Cn-1}
$$
where $s$ is a positive constant depending only on $C$. Since
$$\sup_X\mathbb P(|X|>n) \ge \mathbb P(|X_n|>n) = bC \ge \frac{sC}{4Cn-1}$$
definitively in $n$, we conclude that the series diverges (at least as $\Omega(\log(n))$).

Update: Notice that for any $X$ with $\mathbb E[|X|]=1$, then the bound on the density cannot be less than $1/4$. If $C>1/4$, then the above counterexample works.

Surprisingly, if $C=1/4$, there's exactly one distribution for which $\mathbb E[|X|]=1$, that is
$$
f(x) = \begin{cases}
1/4 & |x|\le 2\\
0 & \text{otherwise}
\end{cases}
$$
and since it has compact support, the series converges.


问:
Inverse limit of $\left(\mathbb{Z}/p^n\mathbb{Z}\right)_{n \in \mathbb{N}}$
In the wikipedia article about the inverse limit it is stated that for a prime number $p$
$$\varprojlim_{n \in \mathbb{N}} \mathbb{Z}/p^n\mathbb{Z} = \mathbb{R}/\mathbb{Z},$$
where the arrow between the groups $\mathbb{Z}/p^n\mathbb{Z} \longrightarrow \mathbb{Z}/p^{n+1}\mathbb{Z}$ is given by multiplication by $p$.

I don't quite understand that: I interpret this "multiplication by $p$" as $[x]_{p^n} \mapsto [px]_{p^{n+1}}.$ But shouldn't the arrows point in the other direction from $\mathbb{Z}/p^{n+1}\mathbb{Z}$ to $\mathbb{Z}/p^n\mathbb{Z}$ for the inverse limit to make sense? On the other hand if one interprets it as a direct limit, then the arrows become inclusions and the limit is simply the group of all roots of unity of order $p^n$ for every $n \in \mathbb{N}$. I also find it a bit odd that one can get a "continous" object from "discrete" objects without invoking any topological notions at all.

How can this be reconciled, is this direct limit correct?

答:
This is false (I'll correct it on Wikipedia). The correct statement is that the direct limit / filtered colimit of $\mathbb{Z}/p^n \mathbb{Z}$ with these maps is the Prufer $p$-group $\mathbb{Z} \left[ \frac{1}{p} \right]/\mathbb{Z}$, which is the subgroup $\mu_{p^{\infty}}$ of $\mathbb{R}/\mathbb{Z}$ consisting of $p$-power roots of unity. The inverse limit / cofiltered limit (with a different set of maps going the other way, namely the quotient maps $\mathbb{Z}/p^{n+1}\mathbb{Z} \to \mathbb{Z}/p^n\mathbb{Z}$) is the $p$-adic integers $\mathbb{Z}_p$.

There's a reason people don't take inverse limits over diagrams that are "increasing to the right" like this diagram is:

$$\mathbb{Z}/p^0\mathbb{Z} \xrightarrow{p} \mathbb{Z}/p\mathbb{Z} \xrightarrow{p} \mathbb{Z}/p^2\mathbb{Z} \xrightarrow{p} \dots$$

and it's that the inverse limit over any such diagram is its first term.

The solenoid example is also wrong (and I will also correct it). $\mathbb{R}/p^n \mathbb{Z}$ is not a ring because $p^n \mathbb{Z}$ is not an ideal. You can in fact prove that every compact Hausdorff ring must be totally disconnected, or equivalently must be profinite, so rings like the $p$-adic integers are all you get.


问:
Closed set written as an union of images of curves implies that at least one of those images is closed?
Let $A\in \mathbb{R}^2$ be a closed set. Suppose $A=\bigcup_{i\in\Lambda}U_i$ as a disjoint union, where $U_i$ is the image of some $C^1$ injective curve $\gamma_i:\mathbb{R}\to\mathbb{R}^2$. There is $j\in\Lambda$ such that $U_j$ is closed?

答:
I am told to add at least 30 characters but isn't a picture  worth a thousand words?

EDIT.  Two curves are enough!


答:
Note that for $x<0$ when dividing both the numerator and the denominator to $\sqrt{x^2}$, you are actually dividing the numerator by $\sqrt{x^2}=-x$ rather than $x$.

An intuition that why the limits are equal to a real number, is that rather than both the square roots tending to $\infty$, they become closer and closer until their distance tends to a fixed value.


问:
How do I formally write down a Euclidean space with symbols?
A space is an ordered tuple, where the first element is a set and the following elements are describing the added structure, e.g. $(X, m)$ for a metric space, $(X, \tau)$ for a topological space. What are the following elements for a Euclidean space?

As far as I understand we need

$X=\mathbb R^n$ is the set of all n-tuples of real numbers (with $n\in\mathbb N$)

we need the elements of $X$ to be vectors - so linearly combine-able with the scalar multiplication $\times$, the field $F$ and addition $+$.

a dot product $\cdot$ between the elements of $X$.

a norm for the elements of $X$. Is this inherently included in the dot product or do I need to state it explicitly to be precise? Don't I need an additional "$-$"? http://faculty.cord.edu/ahendric/2008Fall210/subsub.pdf suggests that this is also included in the "$+$".

completeness of $X$ (is this inherently included in the fact that $X=\mathbb R^n$?)

a metric (I think this is also inherently included in the norm and the fact that the elements of $X$ are vectors, right?)

From that I infer, that a Euclidean space is $(\mathbb R^n, \cdot, +, F, \times)$. Possibly I also need a "$-$".

So: How do I formally write down a Euclidean space with symbols?

答:
You already wrote down a Euclidean space in your question: $\mathbb{R}$.

The only other thing I can think of that you might want to include is your metric. Say $(\mathbb{R},d)$ is a metric space and define d, which is the distance of any two points.

There are some axioms to remember for metrics:

$d(x,x)=0$

$d(x,y)>0$

$d(x,y)=d(y,x)$

$d(x,z)\leq d(x,y)+d(y,z)$ (called the triangle inequality; think of a right triangle, and you walk in a diagonal line to get where you need to go)

There are many metrics we could define for a space like $\mathbb{R^2}$, the real plane; the most common being $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

EDIT:

You would need to learn some topology I suppose. The Cartesian product is just one example of a more general concept which is product spaces. In topology we discuss continuity and open sets (they are not all defined the same). Say $X,Y$ are topological spaces, and the set, $U_{X_i}$ and $V_{Y_i}$ are open in their respective topologies.

We define the topology on the product space $X\,\,x\,\, V$ by just saying it "inherits" the topology of the other two spaces. A subset of $X\,\,x\,\, V$ is open if an only if $U\subset X$ and $V\subset Y$ are both open. This applies exactly the same way to our standard metric spaces, but instead the product space will inherit the metric, which can be thought of as giving us an idea of what "open" is as well!


问:
Is the logarithm of a martingale still a martingale?
For example $S_t = S_0\cdot \exp(\sigma W^*_t - 0.5\sigma^2t)$ is martingale (it's a part of Black-Scholes).

Is $\ln S_t$ a martingale?

答:
$$ln(S_t)=ln(S_0)-0.5\sigma^2 t + \sigma W^{*}_t$$

When you look at the above, you see the process $ln(S_t)$ has a drift $-0.5\sigma^2 t$, so it cannot possibly be a martingale. Just taking the expectation we see that:

$$\mathbb{E}[ln(S_t)]=ln(S_0)-0.5\sigma^2 t$$

On the other hand, when we exponentiate, the expectation is centered on the initial value (because of the mean property of lognormally distributed random variable):

$$\mathbb{E}[S_t]=\mathbb{E}[S_0e^{-0.5\sigma^2 t + \sigma W^{*}_t}]=S_0$$

So the above looks like a much more plausible Martingale candidate!

In general, intuitively, if you see a process with a (deterministic) drift, it cannot possibly be a Martingale, because it's future expected values will not be centered on the "current" value.


问:
For what values of $c$ does the curve $ y = cx^{3} + e^{x} $ have inflection points?
For what values of $c$ does the curve $ y = cx^{3} + e^{x} $ have inflection points?

at first I found first derivative $ f^{'}(x) = 3cx^2 + e^{x} $

then second derivative $ f^{''} (x) = 6cx + e^{x} $

now the second derivative should  go from positive to negative or from negative to positive yes?
and I wrote this system
$$ -e^{x} < 6cx < 0 $$

so give me a hint i don't know that to do

答:
Write the equation $y''=6cx+e^x=0$ as a system
$$\begin{cases}
y=e^x\\
y=-6cx\\
\end{cases}
$$

we see that for $c>0$ there is one and only one solution: one inflection point.

The tangent passing through the origin $(0,0)$ has equation $y=ex$, thus

for negative $c$ we have no solution when $-6c<e$ that is $-\frac{e}{6}<c<0$

for $c<-\frac{e}{6}$ we have two intersection that is two inflection points.

For $c=-\frac{e}{6}$ second derivative is $y''=e^x-ex$ which is zero at $x=1$. This is not an inflection point since third derivative $y^{(3)}=e^x-e$ is zero too at $x=1$.

It happens like in $y=x^4$ where second derivative $y''=12x^2$ is zero at $x=0$ but there is no inflection point since $y^{(3)}(0)=0$.


问:
Triple integral over a sphere

I did the first integral in a), but im not sure about b). K is the sphere x^2+y^2+z^2 ≤ 1. Why can I use symmetry in b)? The solution is (1/3) of the answer in a). I want to understand how the symmetry works out in this case.

答:
Because you can divide the integral in a into three parts and since you are integrating over a sphere of radius one centered at the origin then x, y, z have the same values and give you the same result.

So you can replace y and z with x so you got 3 times the integral in b.

Title: Triple integral over a sphere


问:
What would be the immediate implications of a formula for prime numbers?
What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer, however, I would like to know some of these implications - like the breaking of many security systems based on the primes. Moreover, there are examples of practical Math's implication, not just theoretical, of a possible prime number formula discovery? There is for example in Physics, Chemistry, Geography or Astronomy any field which would be very improved with a so great and dreamed Eureka?

答:
There are dozens, probably hundreds, of formulas for prime numbers. It's a very well-studied problem. Guy's Unsolved Problems in Number Theory has a section devoted to this, and Ribenboim's books cover this in some depth. Many formulas have been published in mathematical journals, and I've seen at least one survey paper in a journal dedicated to giving an overview of the various types of formulas for primes.

Finding a new formula for the primes might be interesting enough to be published, but not in a first-rate journal unless there's something special about it.

As for security concerns, what would be much more important would be a way to solve integer factorization quickly -- say, in polynomial time. (At the risk of verbosity, this means the time needed to factor $n$ is less than $(\log n)^k$ for some fixed $k$.) Proving that a number is prime can already be done in polynomial time, see the famous AKS algorithm (or the more practical ECPP).


问:
Analytical solution to 2D diffusion equation with a drift term
I have been trying to solve the 2D PDE
$$\frac{\partial p(x,y,t)}{\partial t}=-\frac{\partial }{\partial x}p(x,y,t)+\frac{\partial ^2}{\partial x^2}p(x,y,t)+\frac{\partial ^2}{\partial y^2}p(x,y,t)$$

This is close to the diffusion equation $$-\frac{1}{D}\frac{\partial p(x,y,t)}{\partial t}+\frac{\partial^2}{\partial x^2}p(x,y,t)+\frac{\partial ^2}{\partial y^2}p(x,y,t)=0$$ where D is a constant.
This has a solution of
$$p(x,y,t)=\frac{1}{4\pi D t}\exp\frac{-(x+y)^2}{(4Dt)}$$
(Source: http://rpdata.caltech.edu/courses/aph162/2006/Protocols/diffusion.pdf)

However, there is an additional drift term. I am hoping someone can help me out with the analytical solution to the first equation. Thank you :)

答:
The solution of the diffusion equation that you are referring to here is called the fundamental solution, i.e. the solution starting at $t=0$ at a Dirac centered at $(0,0)$. When a PDE is linear with constant coefficients the fundamental solution may be computed thanks to the Fourier transform.

In your case, if we take the Fourier transform of the equation, we get
$$
\frac{d}{dt} \hat{p}(t,\xi_x,\xi_y) = (-i\xi_x-\xi_x^2-\xi_y^2)\hat{p}(t,\xi_x,\xi_y),
$$
where $\hat{p}(t,\xi_x,\xi_y) = \int_{\mathbb{R}^2}e^{-i\xi_xx-i\xi_yy}p(t,x,y)d\xi_x d\xi_y$.

The fundamental solution corresponds to the solution for which the initial data in Fourier is identically $1$, which corresponds to a Dirac at $0$ in the physical space. By solving the ODE on $\hat{p}$ we get
$$
\hat{p}(t,\xi_x,\xi_y) = e^{-t(i\xi_x+\xi_x^2+\xi_y^2)}.
$$

The solution that you are seeking is the inverse Fourier transform of that, namely
$$
p(t,x,y) = \frac{1}{(2\pi)^2}\int_{\mathbb{R}^2} e^{i\xi_xx+i\xi_yy-t(i\xi_x+\xi_x^2+\xi_y^2)}d\xi_x d\xi_y.
$$
A change of variable and Fubini give you
$$
p(t,\sqrt{t}x+t,\sqrt{t}y) = \frac{1}{4\pi^2t}f(x)f(y)
$$
where $f(x) =\int_{\mathbb R}e^{i\xi x-\xi^2}d\xi$. Differentiation under the integral sign and integrations by parts will show you that $f'(x)= -f(x)x/2$. Since $f(0) = \sqrt{\pi}$, one has $f(x) = \sqrt{\pi}e^{-x^2/4}$ and putting all together
$$
p(t,x,y) = \frac{1}{4\pi t}\exp\left(\frac{-(x-t)^2-y^2}{4t}\right).
$$


问:
Finiteness of support of a module as $S$- and $R$- module
$\DeclareMathOperator\nn{\mathfrak{n}}\DeclareMathOperator\mm{\mathfrak{m}}\DeclareMathOperator\Supp{Supp}\newcommand\card[1]{\lvert#1\rvert}$Let $f:R\to S$ be a homomorphism  of commutative Noetherian  rings with identity that makes
$S$ a finite $R$-module.

Let $M$ be a (not necessarily finite) $S$-module. So $M$ is also an $R$-module.

If $\card{\Supp_S M}\lt \infty$, is $\card{\Supp_R M}\lt \infty$?

If $\card{\Supp_R M}\lt \infty$, is $\card{\Supp_S\ M}\lt \infty$?

What if:

1- $f$ is epimorphism?

or

2- $f$ is a monomorphism?

答:
$|\operatorname{Supp}_R M| \lt \infty \iff |\operatorname{Supp}_S M| \lt \infty$.

Combine the following:

If $Q \in \operatorname{Supp}_S M$ then $P=Q \cap R \in \operatorname{Supp}_R M$, because $M_Q$ is a localization of $M_P$ (as $S$-modules). (This statement doesn't require $f$ finite.)

If $P \in \operatorname{Supp}_R M$ then there is at least one $Q \in \operatorname{Supp}_S M$ which lies over $P$ (i.e. $Q \cap R = P$):  The support of a sum of modules is the union of their supports, so the statement reduces to $S$-modules generated by one element i.e. $M=S/I$. Then $P \in \operatorname{Supp}_R S/I \implies P \supset I \cap R \implies \exists Q \supset I$ such that $Q \cap R = P$ (by "lying over" for the finite ring injection $R/(I \cap R) \subset S/I)$, and such $Q \in \operatorname{Supp}_S S/I$.

For each $P \subset R$ the set of $Q \subset S$ lying over $P$ is finite, because the fiber $S_P/PS$ is an Artinian ring (finite over the field $R_P/P$).

Notes:

It is not required that rings $R, S$ are Noetherian.

A finite epimorphism of commutative rings is surjective, so $\operatorname{Supp}_R M = \operatorname{Supp}_S M$ in this case.


问:
Powers of polylogarithms
I would like to take powers of arbitrary order to polylogarithm functions. For instance, given
$$
\text{Li}_\alpha(z) = \sum_{k=1}^\infty \frac{z^k}{k^\alpha}
$$
I am interested in
$$
[\text{Li}_\alpha(z)]^m = \left(\sum_{k=1}^\infty \frac{z^k}{k^\alpha}\right) ^m
$$

Of particular interest is how the integral representation

$$
\text{Li}_\alpha(z) = \frac{z}{\Gamma(\alpha)}\int^\infty_0dt \frac{t^{\alpha-1}}{e^t-z}
$$
would behave in this case. Does anyone know of research in this area ?

答:
well:
$$\ln(1+z)=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{k}x^k$$
so:
$$\ln(1-z)=-\sum_{k=0}^\infty\frac{x^k}{k}$$
$$\therefore \operatorname{Li}_1^m(z)=(-1)^m\ln^m(1-z)$$

If we first try and start with a function of the form:
$$\frac{d}{dx}\operatorname{Li}_a(x)=\sum_{k=0}^\infty x^{k^a-1}=\frac1x\sum_{k=0}^\infty x^{k^a}$$
maybe there is a way to find a function that is represent by this? :)

Title: Powers of polylogarithms


问:
find the largest integer less than a number
I'm trying to figure out this problem for few hours now. please help.

Define $$\text{PCOM}(a,b,c) := \{ax + by + cz : x, y, z ≥ 0\}.$$

Given integers $t > c > b > a > 1,$ I need help to find the largest integer $w < t$ such that $w\notin\text{PCOM}(a,b,c).$

Example: $\text{PCOM}(3,5,7) =\{0,3,5,6,7,8,9,10,…\}.$  The largest integer less than $9$ that is not in $\text{PCOM}(3,5,7)$ is $4$.

答:
You can easily sovle this problem with dynprog, if it is nessesary. It is $O(t)$. There are for sure much better algorithms, but your question has a tag "dynamic-programming"...
http://pastebin.com/QN2BJL17</p>


问:
if $f(0) = 0$ is it true that $\forall x \in \mathbb{R}: [ f(x)<x \implies f'(x)\leq 1]$
This is a question from my calculus book in high school.

$f(x)$ is a function and $f(0) = 0$

Is it true that $\forall x: f(x) \le x \iff f'(x) \le 1$

----edit----

what can we say about two sides of the equivalency?

1)$\forall x: f(x) \le x \implies f'(x) \le 1$

2)$\forall x: f'(x) \le 1 \implies f(x) \le x $

答:
The first implication is false.

$$(\forall x\in \Bbb R)\; \; f(x)-x\le 0$$

implies that the function $ g:x\mapsto f(x)-x $ is negative but not necessarily decreasing $(g'(x)=f'(x)-1\le 0 $
Take for example $$f(x)=x-x^2$$
then
$$x>1 \implies f'(x)-1<-2$$

The second is semi true :
In fact, For $ x\ge 0$,

$$(\forall t\in[0,x])\; f'(t)\le 1\implies$$
$$\int_0^xf'(t)dt\le \int_0^xdt$$
$$\implies f(x)-f(0)\le x$$
$$\implies f(x)\le x$$

and for $ x\le 0$
$$\int_x^0f'(t)dt\le \int_x^0dt$$
$$\implies f(x)\ge x$$


问:
Correspondence between representation of $SL(2,\mathbb{C})$ and of $\mathfrak{sl}(2,\mathbb{C})$
For the Lie-Group $SL(2,\mathbb{C})$ consider the represenations definied by

$\begin{equation}\Pi_1:SL(2,\mathbb{C})\to GL(2,\mathbb{C}), A\mapsto A\end{equation}$

and

$\begin{equation}\Pi_2:SL(2,\mathbb{C})\to GL(2,\mathbb{C}), A\mapsto A^*\end{equation}$

Where $A^*$ is the elementwise complex conjugated of $A$. As far as I know those are both irreducible and not isomorphic.

On the other hand both representation correspond to a unique lie-algbera representation of $\mathfrak{sl}(2,\mathbb{C})$ of dimension 2, $\pi_1,\pi_2$. Now there is only one lie-algbera representation of $\mathfrak{sl}(2,\mathbb{C})$ of dimension 2, up to isomorphism.

Hence $\pi_1\cong \pi_2$. Another theorem says that $\pi_1$ and $\pi_2$ are isomoprhic if and only if $\Pi_1$ and $\Pi_2$ are isomorphic.

But $\Pi_1 \not\cong \Pi_2$.

My question is where exactly the mistake is. Am I missing something obvious? Which step is wrong and why?

答: If you actually differentiate $\Pi_2$ you will find straightforwardly that you don't get a complex representation of $\mathfrak{sl}_2(\mathbb{C})$, so the theorem you quote classifying such representations does not apply. As Moishe says in the comments, the fix is to only consider holomorphic representations, which $\Pi_2$ is not.


问:
Does $\lim_{n\to\infty} \frac{(-1)^{n}}{n}$ converge to $0$?
I was solving a problem in real analysis where the author used the below to support his argument.

$$\lim_{n\to\infty} \frac{(-1)^{n}}{n} = 0$$

Is the above limit valid ?

It struck me that, as $1^\infty$ is an indeterminate form, can we say that the above limit holds?
Possible known values to which $1^\infty$ can go are $1$, $e$, and $\infty$ . Can we substantiate and say that the above limit holds?

答:
The author is correct.  It is sandwiched
between $\dfrac{-1}n$ and $\dfrac 1n$.


问:
Do we know if there are more primes with even leading digits or odd leading digits?
I was just wondering, out of curiosity, do we know if there are more primes with even leading digits or odd leading digits? For example, primes with even leading digits would be $23$ or $29$ and primes with odd leading digits would be $31$ or $97$. If we haven't figured it out then how do we do something like that?

Thanks

答:
One could expect that the Benford's law applies here (as Calvin Lin comments also noted), though I seems quite difficult (if possible) to prove it. Assuming that, primes starting with an odd digit would be have a relative population of 60.9%

Update: here's a relevant paper


问:
Trouble in understanding Predicate logic semantics
I am studying predicate logic and I have a problem understanding its semantics. I am reading "Logic for computer scientists" by Uwe Schoning and the passage I have a problem with is this:

If $F$ has the form $F=P(t_1,\dots,t_k)$ where $t_1\dots t_k$ are terms and $P$ is a predicate symbol of arity $k$, then its semantics are defined by
$$\mathcal{A}(F) =\begin{cases}1& \text{ if } (\mathcal{A}(t_1),\dots,\mathcal{A}(t_k)\in P^{\mathcal{A}}\\ 0& \text{ otherwise}\end{cases}$$

Here $\mathcal{A}$ is a structure and $P^{\mathcal{A}} = I_{\mathcal{A}(P)}$ is described as below:
Predicate logic structure pair

I think I kinda get what structure is about, my problem is about the first picture. I didn't get what does it mean. Does it mean that if we had a predicate symbol, the value of that predicate symbol is true if all its terms are defined by that structure? If so then if for example we have $P^2(x,f(x))$ and both $x$ and $f(x)$ are defined then $P$ is true regardless? That doesn't seem right!

I tried googling and reading other documents but it confused me more. I'll be happy if someone explains the predicate logic semantics to me in a clear and simple way with a few examples. Thanks in advance

答:
$\newcommand\A{\mathcal{A}}$
Admittedly, the usage of the symbol $\mathcal{A}$ both standing alone as a structure and in function notation to denote the value of an expression in that structure is confusing.

$\A(\Box)$ (where $\Box$ is some expression -- a term or a formula) does not mean that $\Box$ is defined in $\A$ at all (it has to be in order to be evaluate the expression at all), but it stands for the value of the expression $\Box$ in $\A$.

$$\mathcal{A}(F) =\begin{cases}1& \text{ if } (\mathcal{A}(t_1),\dots,\mathcal{A}(t_k)) \in P^{\mathcal{A}}\\ 0& \text{ otherwise}\end{cases}$$

means that the value of the atomic formula $F := P(t_1, \ldots, t_n)$ is $1$ iff the tuple consisting of the values of the terms $t_1$ to $t_k$ in the structure $\A$ is an element of the interpretation of the predicate $P$.

Those values $\A(\Box)$ are what is defined inductively in the clauses 1. - 2. + 1. - 6. on p. 46-47.

The value of a term is an object in the universe, and the interpretation of an $n$-place predicate symbol is a set of $n$-tuples of objects from the universe. So checking the value of an atomic formula comes down to checking if the tuple of objects that the terms denote is among the tuples of objects that $P$ is defined to be true of.

A $1$-tuple can simply be identified with its only object: $(a) = a$, so the interpretation of a $1$-place predicate like $\mathrm{prime}$ can be said to be a set of individual objects rather than a set of "$1$-tuples".

Take the example on p. 45, and restrict yourself  subformula $Q(g(a,z))$ for the moment:

By a stepwise evaluation recursing down the definitions of $\A(\Box)$, we get that

$
\A(Q(g(a,z)) = 1\\
\Leftrightarrow \underbrace{(\A(g(a,z))) \in Q^\A}_{\text{def. } \A(P(t_1, \ldots, t_k))}\\
\Leftrightarrow \underbrace{g^\A(\A(a), \A(z))}_{\text{def. } \A(f(t_1, \ldots, t_k))} \in Q^\A\\
\Leftrightarrow g^\A(\underbrace{a^\A}_{\text{def. } \A(f(t_1, \ldots, t_k))}, \underbrace{z^\A}_{\text{def. } \A(x)}) \in Q^\A\\
\Leftrightarrow \underbrace{+}_{\text{def. } I_\A(g)} (\underbrace{2}_{\text{def. } I_\A(a)}, \underbrace{3}_{\text{def. } I_\A(z)}) \in \underbrace{\{n \in U_\A : n \text{ is prime }\}}_{\text{def. } I_\A(Q)}\\
\Leftrightarrow \underbrace{5}_{\text{def. } + } \in \{n \in \underbrace{\mathbb{N}}_{\text{def. } U_\A} : n \text{ is prime }\}\\
\text{Since 5 is prime, } \A(Q(g(a,z)) = 1.
$


答:
In general if your integrand $f(x,y,z)$ is such that $f(x,y,z)=f(-x,y,z)$ and the domain of integration is symmetric respect to the $yz$ plane you can split the integral as a sum.

In your case the sphere $K:=\{(x,y,z)\in \mathbb{R}^3 :x^2+y^2+z^2\le1\}$ is symmetric respect to the plane $yz$.
So, if we now consider the restriction $K^+:=\{(x,y,z)\in K:x>0\}$ and an even function $f$ (respect to the variable $x$) then we would  have
$$\iiint_Kf=2\iiint_{K^+}f$$ which can be easy solved using a change of variables.

Title: Triple integral over a sphere

Body: I did the first integral in a), but im not sure about b). K is the sphere x^2+y^2+z^2 ≤ 1. Why can I use symmetry in b)? The solution is (1/3) of the answer in a). I want to understand how the symmetry works out in this case.


问:
count of matrix of 0,1 in which each row and each column have at least one 1.
I know that the number of matrices with $0,1$ entries, in which each row and each column have at least one $1$, is equal to

$$\sum_{r=0}^n \sum_{s=0}^n (-1)^{r+s} C(n,r) C(n,s) 2^{rs}.$$

I want to prove it combinatorially using Inclusion–exclusion principle.

答:
Let's say $S_{t,u}$ is the number of binary $n$ by $n$ matrices with $t$ rows of zeroes and $u$ columns of zeroes. The rows can be chosen in $C(n,t)$ ways and the columns can be chosen in $C(n,u)$ ways. The remaining portion of the matrix has $(n-t)(n-u)$ binary elements. So
$$S_{t,u} = C(n,t)C(n,u)2^{(n-t)(n-u)}$$
(Note that this formula works even when $t=u=0$.)

By inclusion/exclusion, the number of matrices with no row of zeroes and no column of zeroes is
$$\begin{align}
N_0 &= \sum_{t=0}^n \sum_{u=0}^n (-1)^{t+u} S_{t,u} \\
 &= \sum_{t=0}^n \sum_{u=0}^n (-1)^{t+u} C(n,t)C(n,u)2^{(n-t)(n-u)}
\end{align}$$
Now make a change of indices to $r=n-t$ and $s=n-u$.  The result is
$$\begin{align}
N_0 &= \sum_{r=0}^n \sum_{s=0}^n (-1)^{2n-r-s} C(n,n-r)C(n,n-s) 2^{rs} \\
&= \sum_{r=0}^n \sum_{s=0}^n (-1)^{r+s} C(n,r)C(n,s) 2^{rs}
\end{align}$$


问:
$\lim_{n \to \infty} \int_1^a \frac{n}{1+x^n}dx$ solution with term-by-term integration of series
Term-by-term integration was given as one of the solutions:

https://math.stackexchange.com/a/339717/851293

From the theory, the power series can be integrated term-by-term on an interval lying inside the interval of convergence, but how to prove this solution is allowed for this particular case since the function also depends on $n$?

答:
it can be shown that:
$$\int_1^\infty\frac{1}{1+x^n}dx$$
converges for all $n>1$ so the integral part will converge. One nice way you could do it would be:
$$\frac{1}{1+x^n}<\frac 1{x^n}\,\,\,\,x\ge1$$
so:
$$\int_1^a\frac{n}{1+x^n}dx<\int_1^a\frac{n}{x^n}dx=\frac{(a^n-a)n}{a^n(n-1)}=a\frac{n}{n-1}\frac{a^{n-1}-1}{a^n}=\frac{n}{n-1}\frac{a^{n-1}-1}{a^{n-1}}=\frac{n}{n-1}\left(1-a^{1-n}\right)=L_n$$
which when we look at very large values of $n$ it is clear that:
$$\frac{n}{n-1}\to1$$
$$1-a^{1-n}\to1$$
also since the function is always positive and $a>1$ then the integral must also be positive, and so:
$$0\le I<1$$


问:
Proof that a sequence of random variables have finite expectation
Let $X_n$ be iid non-negatives random variables. Prove that $\mathbb{E}[X_1] < \infty$ iff $P(X_n \ge n\text{ i.o.}) = 0$

I thought I would start like this for one direction

$\infty > \mathbb{E}[X_1] \ge \sum\limits_{n=1}^\infty P( X_n > n)$ then the first Borel Cantelli Theorem states that

$$P(X_n \ge n\text{ i.o.}) = 0$$

but how would I go forward with the other direction ?

答: You can use Borel's Zero-One Law (prop. 2.2 here), using $$\mathbb{E} X_1 \leqslant \sum_{n=0}^\infty \mathbb{P}\{X_1 > n\} = \sum_{n=0}^\infty \mathbb{P}\{X_n > n\}$$ with $A_n\stackrel{\rm{}def}{=}\{X_n > n\}$.</p>