问:
Iterative calculation of mean and standard deviation
I'm looking for a formula, to iteratively calculate the mean and standard deviation of a huge list of data points.

I found some examples here (formula 15 f.) and here, but both seem to be falling for my very simple testcase [10,100].

Source 1 states:

$M_1 = x_1$

$S_1 = 0$

$M_k = M_{k-1}+(x_k-M_{k-1})/k$

as well as

$S_k = S_{k-1}+(x_k-M_{k-1})*(x_k-M_k)$

with

$\sigma = \sqrt{S_n/(n-1)}$

This leads me to $M_1 = 10, S_1 = 0$ and $M_2 = 10+(100-10)/2 = 55$ but $S_2 = 0+(100-10)*(100-55) = 4050$ and therefore with $n=2$ to $\sigma \approx 63.6396$. The correct value is $45$, which I get, when I plug in $n = 3$ in the formula for $\sigma$.

Do I understand the formula wrong?

Source 2:

$M_{n+1}=M_n+x_{n+1}$

$S_{n+1}=S_n+\frac{(n*x_{n+1}−M_n)^2}{n(n+1)}$

with the mean given by

$\bar{x}_n= \frac{M_n}{n}$

and the unbiased estimate of the variance is given by

$\sigma_n^2=\frac{S_n}{n+1}$

which leads me to

$M_1 = 10, M_2 = 110, S_1 = 0$

$S_2 = 0+\frac{(2*100-10)^{2}}{2(2+1)} = 6016.6667$

however, if I plug in $n=1$ again this is correct.
I feel, that my understanding of indexes is wrong, but why?

答:
The computationally cheapest way to do this, and also the optimal choice if you want to calculate standard deviations "on the fly" (=for each added point again), is to first calculate a running total of the values and a running total of the squared values.
Finally you can calculate the standard deviation at each point (or only for the last point) using this identity

$$\sqrt{\frac{1}{N}\sum_{i=1}^N{(x_i-\overline{x})}^2}  = \sqrt{\frac{1}{N}\sum_{i=1}^N{x_i^2}-{\Biggl(\frac{1}{N}\sum_{i=1}^N{x_i}\Biggr)^2}} $$

or in pseudo programming code:

$$  sdev = sqrt( v2sum /N - sqr(vsum/N) ) $$

where $vsum$ is the total of all values, and $v2sum$ is the total of all $N$ values squared (individually). As said, very useful if you want to know the standard deviations on the fly for each new point, as the the calculation time will stay linear (instead of growing with N², when using the standard formula).


问:
Calculating Minimum Exam Score to Pass a Course
So for a course Im taking this semester 70% of the grade is based on the exam and 30% is based on project work. In order to pass the course I need 40% combined score on the project and the exam. I got 85% on the project.

How do I now go about calculating the minimum score I need on the exam to pass? Would it be simply calculating that I have $0.85*0.3 \approx 25$% score so far so I need to score $15$% or higher on the exam? Or am I missing something?

答: Say you score $e%$ in your exam. Then you need $$85\% \times 30\%+ e \% \times 70\%\ge 40\%  $$ or $$e\% \times 0.7 \ge 0.145 $$ or $$e\ge 20.714 $$ approximately. Good luck with your exam!

ContentLicense: CC BY-SA 4.0


问:
Lagrangian Dual
As part of an applied math class for biologists I am in which is partly concerned with optimization and  duality we have been referred to a text to help us understand the concepts better.  I do not understand the solution to one of the questions.  It is question 5.13 from the pdf: https://see.stanford.edu/materials/lsocoee364a/hw5sol.pdf.
I absolutely cannot figure out how they eliminated the dual multiplier $v_i$ for this portion of the solution.  Also I do not understand what the notation $min\{0,4(c_i + a_{i}^T \mu\}$ means.  Forgive me for I am not a math student.  I am amazed I can even understand what a relaxation is and how it is used in duality.  I may have used the wrong tags for that reason as well.

I am specifically stuck here:

The step before " in order to simplify this dual..." I believe to be...

$$ \max -\frac{1}{4\mu}\Sigma_{i=1}^n (c_i + a_{i}^T \mu - v_i)^2/ v_i $$
$$s.t. \text{  } v_i \geqq 0, \mu_i \geqq 0$$

答:
Your dual problem is
$$\max -b^\top \mu - \frac{1}{4}\sum_{i=1}^n\frac{(c_i+a_i^\top\mu-\nu_i)^2}{\nu_i} \quad \mbox{s.t.} \quad \mu \geq 0, \nu \geq 0.
$$
Obviously the term $b^\top \mu$ does not depend on $\nu$.
Hence, to further investigate the role of $\nu$ it is suffcient to consider the latter part of the objctive function, which is
$$\max - \frac{1}{4}\sum_{i=1}^n\frac{(c_i+a_i^\top\mu-\nu_i)^2}{\nu_i} \quad \mbox{s.t.} \quad \mu \geq 0, \nu \geq 0.$$
Now, the idea is to focus on the single summands and to maximize those summands. The background is that if we maximize all summands, then we also maximize the whole sum. Thus, we consider (for fixed $\mu$) the summand for an arbitrary $i = 1,\ldots,n$ which leads to the mentioned problem
$$ \sup -\frac{(c_i+a_i^\top\mu-\nu_i)^2}{\nu_i} \quad \mbox{s.t.} \quad \nu_i \geq 0.$$
We find out, that the term that we want to maximize (find the supremum of) has negative sign. Also both parts of the fraction are not negative. Hence, the supremum is definitely bounded (above) by $0$.
As written in the solution we can now distinguish between two cases: If $c_i+a_i^\top\mu \geq 0$ then we can always find $\nu_i \geq 0$ such that $c_i+a_i^\top\mu - \nu_i = 0$ and hence the supremum is $0$.

If $c_i+a_i^\top\mu < 0$ then one can find out analytically (using first + second derivative) that the supremum is obtained for $\nu_i = -c_i+a_i^\top\mu$. More precisely, define
$$\begin{align}
g(\nu_i) := &-\frac{(c_i+a_i^\top\mu-\nu_i)^2}{\nu_i}\\
= &-\frac{(c_i+a_i^\top\mu)^2+\nu_i^2-2(c_i+a_i^\top\mu)\nu_i}{\nu_i}\\
= &-\frac{(c_i+a_i^\top\mu)^2}{\nu_i}-\nu_i+2(c_i+a_i^\top\mu).
\end{align}$$
Then we have
$$ \begin{align}
g^\prime(\nu_i) &= \frac{(c_i+a_i^\top\mu)^2}{\nu_i^2}-1\\
g^{\prime\prime}(\nu_i) &= -\frac{2(c_i+a_i^\top\mu)^2}{\nu_i^3}
\end{align}$$
This is all we need to find the global maximum:
$$g^\prime(\nu_i) = 0 \Leftrightarrow \nu_i^2 = (c_i+a_i^\top\mu)^2 \Rightarrow \nu_i = -(c_i+a_i^\top\mu).$$
The sign is known since we have $\nu_i \geq 0$ and $c_i+a_i^\top\mu < 0$.
Also, this choice of $\nu_i$ implies $\nu_i > 0$ and thus, $g^{\prime\prime}(\nu_i) < 0$. Hence, we have $$-\frac{(c_i+a_i^\top\mu-\nu_i)^2}{\nu_i} = -\frac{(c_i+a_i^\top\mu-(-c_i+a_i^\top\mu))^2}{-c_i+a_i^\top\mu} = 4(c_i+a_i^\top\mu) \leq 0$$
in this case.
For this reason, we can write that the the supremum equals the minimum of those two terms from the case destinction above, i.e.,
$$ \sup_{\nu_i \geq 0} -\frac{(c_i+a_i^\top\mu-\nu_i)^2}{\nu_i} = \min \{0,4(c_i+a_i^\top\mu)\}.$$
(The minimum of those two terms is $0$ if $c_i+a_i^\top\mu \geq 0$ and $c_i+a_i^\top\mu$ if $c_i+a_i^\top\mu < 0$ - exactly what we found out above.)

As $\nu$ is no longer part of that term (of the minimum above), it can be eliminated from the dual problem when inserting the minimum for the summands.
This finally leads to the dual problem
$$\max -b^\top \mu + \sum_{i=1}^n \min \{0,c_i+a_i^\top\mu\} \quad \mbox{s.t.} \quad \mu \geq 0.$$

Title: Lagrangian Dual


问:
The general way to show completeness of a theory
I have difficulty showing the completeness of a certain theory.
Namely, we know the DLO (dense linear order) without endpoints is complete, but what is the clearest way to show that? Similar variance is DLO with one endpoint, with both side endpoints, etc...

Thanks.

答:
There is no "general way" to show the completeness of an arbitrary theory. (I think this is most likely a provable statement: you should be able to reduce the halting problem to deciding the completeness of a recursively axiomatized theory, although I don't feel confident enough in recursion theory to say that with certainty.)

The usual way of proving completeness is via quantifier elimination in a language we understand (with the caveat that we can only expand the language by symbols definable in the theory). For instance, DLO has quantifier elimination in the usual language $<$. DLO with the left endpoint has q.e. once you add a constant symbol for the endpoint, etc.

Once you have quantifier elimination, completeness is typically immediate (it is enough to show that quantifier-free sentences are decided).


问:
On the composition of a function and its inverse
In The Principles of Mathematical Analysis, 3rd Ed. by Rudin, while proving Theorem 4.14 (if $f$ is a continuous function from a compact metric space $X$ to a metric space $Y$, then $f(X)$ is compact), he mentions in passing that $f\circ f^{-1}(E) \subset E$, where $E \subset Y$, and that the equality need not hold, as in, it isn't necessary that $f\circ f^{-1}(E) = E$.

Is this only because $E$ being any subset of $Y$ may include points that aren't in the range of $f$? Otherwise, the equality always holds, right?

答: Yes, $f^{-1}(E)=\{x\in X; f(x)\in E\}$, so if you take a point $a\in E\setminus Im(f)$ (if exists), then $a\notin f\circ f^{-1}(E)$ as you said.


答:
Score in exam $ = S_e$

Score in project $ =  S_p$

To pass, total score $S = 0.7 S_e + 0.3S_p \geq 0.4$

$S_p = 0.85 \, (85\%)$. Now at equality, you find min $S_e$ from the above equation. Convert into percentage.

Title: Calculating Minimum Exam Score to Pass a Course

Body:
So for a course Im taking this semester 70% of the grade is based on the exam and 30% is based on project work. In order to pass the course I need 40% combined score on the project and the exam. I got 85% on the project.

How do I now go about calculating the minimum score I need on the exam to pass? Would it be simply calculating that I have $0.85*0.3 \approx 25$% score so far so I need to score $15$% or higher on the exam? Or am I missing something?


问:
Impossibility of polynomial approximation
This is exercise 12.6 in David Ullrich's Complex Made Simple. He has discussed many ways to prove the existence of polynomial approximations to functions in the complex plane, but not how to show such approximations are impossible in certain cases, which is the point of the problem. I have an idea, but I'm unsure about its validity.

Problem: Fix $M<0$. Let $f(z)$ be defined on $\mathbb C$ so that $f(0)=1$ and $f(z)=0$ for $z\neq 0$. Show there does not exist a sequence of polynomials $\{p_n\}$ such that $\lim_{n\rightarrow \infty} p_n(z) =f(z)$ if we require $|p_n(z)|<M$ for all $z$ with $|z|\le 1$ for every $n$.

Proposed Solution: let $\gamma$ be the unit circle, traversed once. Consider the rational functions $p_n/z$. Using the dominated convergence theorem, permissible because the $f(z)$ are uniformly bounded along $\gamma$, we have

$$\lim_{n\rightarrow \infty} \int_\gamma p_n(z)/z \ dz = \int_\gamma \lim_{n\rightarrow \infty} p_n(z)/z \ dz = \int_\gamma f(z)/z = 0.$$

But the limit on the left is just the limits of the residues of $p_n(z)/z$ at $0$, which is (modulo a $2\pi i$) the constant term in the power series expansion of $p_n(z)$ at $0$, or $p_n(0)$, And we are given this goes to $1$. We get $1=0$, and the contradiction shows the impossibility of the specified approximation.

Is this correct? And is there a better or more elementary way? Ullrich makes a point of avoiding theorems requiring measure theory to prove, like the dominated convergence theorem, so I suspect there is.

答: How about this: The sequence $p_n$ is a uniformly bounded sequence of holomorphic functions on the (open) unit disc, and therefore a normal family (Montel's theorem). If they converge to $f$, they must converge uniformly on compact sets along some subsequence. But then $f$ would be continuous (indeed, holomorphic), and it isn't.</p>

ContentLicense: CC BY-SA 3.0


问:
if $y_1=\frac{\sin x}{\sqrt x}$ is solution of $x^2y''+xy'+(x^2-1/4)y=3 \sqrt x^3 \sin x$, find the other lineraly independent solutions
If $y_1=\displaystyle\frac{\sin x}{\sqrt x}$ is a solution of the differential equation $$x^2y'' + xy' + \left(x^2-\frac{1}{4}\right)y=3\sqrt x^3 \sin x,$$ find the other linearly independent solution.

答:
Set, inspired by the given solution and the form of the right side, $u(x)=\sqrt{x}y(x)$. Then its derivatives are
\begin{align}
u'(x)&=\sqrt{x}y'(x)+\frac1{2\sqrt x}y(x)\\
u''(x)&=\sqrt{x}y''(x)+\frac1{\sqrt x}y'(x)-\frac1{4\sqrt x^3}y(x)
\end{align}
so that inserted into the given equation
$$
\sqrt x^3u''(x)=x^2y''(x)+xy'(x)-\frac14y(x)=\sqrt x^3(3\sin x - u(x))
$$
Now cancelling again the common factor one is left with
$$
u''(x)+u(x)=3\sin(x)
$$
which is a standard forced harmonic oscillator (with resonance).


问:
Rate of convergence of $\int_0^c \sqrt{|\log(x)|} dx$ to zero as $c \to 0$
I have been looking for a way to quantify the rate of convergence for the function

$$ m(c) := \int_{0}^c \sqrt{ |\log(x)| } dx $$

in terms of $c$, as $c \to 0$. Of course, since $m(1) < \infty$, a dominated convergence argument shows that $m(c) \to 0$ as $c \to 0$, but is there also a function $g$ such that $\lim_{c \to 0} g(c) = 0$ and  $m(c) = O(g(c))$ for all $c$ small enough?

Thank you in advance.

答:
For $c < 1$, the substitution $x = e^{-u}$ transforms the integral into
$$\int_{-\log c}^{\infty} \sqrt{u} e^{-u}\,du = \Gamma(3/2, -\log c)\,.$$
This is an upper incomplete Gamma function, and we have the known asymptotic behaviour
$$\lim_{x \to +\infty} \frac{\Gamma(s,x)}{x^{s-1}e^{-x}} = 1$$
which is easily derived via integration by parts. Repeated integration by parts yields more terms of the asymptotic expansion.

Thus for $m(c)$ we obtain
$$m(c) \sim c\sqrt{\lvert \log c\rvert}\,.$$
(We could also get that from direct integration by parts in the definition of $m(c)$ writing the integrand as $1 \cdot \sqrt{\lvert \log x\rvert}$, but writing it as a known function first has its advantages.)


答:
so I need to score 15% or higher on the exam? Or am I missing something?

Your answer is not "totally" to waste.

first you did not take 25% in the project score but you took $0.3\times0.85=25.5\%$

given the fact that you need more a minimum of $14.5\%$ you can easy calculate the minimum exam's score in the following way

$$\frac{0.145}{0.70}\approx 20.71\%$$

as showed by all other guys who answered


问:
How many solutions to $a^n-b^n=2^{100}$
Are there a distinct number of solutions to:

$a^n-b^n=2^{100}$

If a, b and n are positive integers greater than 1

I have tried to solve it with code but it takes to long.

```<code> for (<span class="math-container">$a=3; $</span>a -lt 5000; <span class="math-container">$a++) {
    for ($b=2; $b -lt $a; $b++){
        for ($n=2; $n -lt 200; $n++) {
            $x=[bigint]::Pow($a,$n)-[bigint]::Pow($b,$n)
            if ( [bigint]::Pow(2,100) -eq $x) {write-host $a,$b,$</span>n}
            }
        }
   }
</code>```

答:
It has exactly $49$ solutions for $n=2$ given by $a = 2^{99-k}+2^{k-1}$, $b = 2^{99-k}-2^{k-1}$ with $k = 1, 2, \ldots, 49$.

For $n>2$ it has no solutions.

Hints for the proof:
Reduce to the case where $n$ is an odd prime or a power of $2$.

Title: How many solutions to $a^n-b^n=2^{100}$


问:
Mean reverting Ornstein-Uhlenbeck SDE
A mean reverting Ornstein-Uhlenbeck SDE is given by
$$=(−)+;_0=,$$
where m and  are positive constants and  is a standard Brownian motion in 1 dimension. I have obtained the solution of this equation, $$X_t = xe^{-t} + m(1-e^{-t}) + \sigma \int_0^t e^{-(t-s)} dW_s.$$ My questions are:
Is my solution correct?
What is the long time behaviour of the solution?
How to obtain the equation of the second moment?

答:
Hint

I didn't check into detail the solution you found, but at least, the solution is something of this form. Now, one can prove that if $f\in L^2[0,t]$, then $$\int_0^tf(s)\,\mathrm d W_s\sim \mathcal N\left(0,\int_0^t f(s)^2\,\mathrm d s\right).$$

Title: Mean reverting Ornstein-Uhlenbeck SDE


问:
Are the graphs of these two functions equal to each other?
The functions are: $y=\frac{x^2-4}{x+2}$ and $(x+2)y=x^2-4$.

I've seen this problem some time ago, and the official answer was that they are not.
My question is: Is that really true?

The functions obviously misbehave when $x = -2$, but aren't both of them indeterminate forms at that point? Why are they different?

答:
Forget functions for a moment.  An equation is a sentence, and the graph of an equation (in two variables) is the set of pairs $(x,y)$ which satisfy the sentence.  So $(1, -\frac{1}{3})$ is in the graph of the equation $y = \frac{x^2 -4}{x+2}$ and $(x+2)y = x^2-4$.  You can see this by plugging in $x = 1$ and $y = -1/3$ -- each sentence is true.

The point $(-2, 0)$ lies in the graph of the second equation, but not in the first, because $\frac{0}{0}$ is a nonsense phrase (i.e. "undefined").  It doesn't make sense to call nonsense like $y = \frac{0}{0}$ true or false.  Moreover, $(-2, 3)$ and $(-2, 4)$ also satisfy the second sentence.  In fact, ANY pair $(-2, y)$ satisfies the second equation, so all those points lie on the graph of the second equation.

Now let's return to functions.  The first equation is the graph of a function.  The second equation is not, because it fails the vertical line test.  To be really nit-picky, a function is a rule which assigns to each element of the domain a unique element of the range.  So an equation is not a function.

Finally, "indeterminate forms" concern limits, which don't seem immediately relevant here.


问:
Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$
Question: Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$

Attempt: I want to use Euclid's Algorithm because it seemed to be easier than what my book was doing which was manually finding the linear combination.

Euclid's Algorithm states that we let $a,b \in N $. By applying the Division Algorithm repeatedly...then $\gcd(a,b) = r_j$ will be the last non-zero remainder. By the Well Ordering Principle, there is a smallest element of the set of natural numbers less than or equal to $r_j$.

I have used long division, but since I can't get it to show up here, I will type what I've done.

Starting at $\gcd(2n+1,9n+4)$,

$\frac{2n+1}{9n+4}$

I can multiply $2n+1$ four times and I would have a remainder of $n$ because $(9n+4)-(8n+4) = 9n+4-8n-4=n$

so we have $4 \cdot (2n+1)+n$ if I apply Euclid's Algorithm.

For $\gcd(2n+1, n)$,

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

I can multiply $n$ 2 times and I will have only 1 as the remainder because $(2n+1)-(2n+0) = 2n+1-2n+0=1$

Therefore, we have $2 \cdot (n) +1$

and  $\gcd(n,1)$  which is $1$

Since the end result is $1, \gcd(2n+1,9n+4)=1$

I followed an example from this link

http://cms.math.ca/crux/v33/n5/public_page274-276.pdf

Am  I doing this correctly?

答:
There is also the systematic matrix approach:

$$
\begin{pmatrix} a \\ b \end{pmatrix}
=
\begin{pmatrix} 2 & 1 \\ 9 & 4 \end{pmatrix}
\begin{pmatrix} n \\ 1 \end{pmatrix}
\implies
\begin{pmatrix} n \\ 1 \end{pmatrix}
=
\begin{pmatrix} -4 & \hphantom-1 \\ \hphantom-9 & -2 \end{pmatrix}
\begin{pmatrix} a \\ b \end{pmatrix}
$$
In particular, $1=9a-2b$, which implies $\gcd(a,b)=1$.

The key point here is that the first matrix has determinant $-1$ and so is invertible over the integers.


问:
Rigorous definition of a type
What is a type in type theory? I tried to find a rigorous definition without luck. And that makes me wonder.. maybe there isn't any rigorous definition?

My aim is to see how homotopy type theory provides a foundation for mathematics. An attempt is given here. Despite the nice resources given there, however, I found that my problem was more basic than knowing what a homotopy type theory is. Rather, it's about type theory: I'm in search of a rigorous definition of type theory.

Now, since this question is close to the fundamentals.. I reckon that I might have to specify what I really need. First of, I know how the basics of type theory works. I know that a type is roughly a collection of things, called terms, whatever these means. I know there might be rules to tell if a given term is of a given type. I know there are types derived from predefined types.. etc. If you ask me to follow the rules and play the game, I think I won't have a problem. What bothers me is that I don't really know what a type is, whenever addressed.

So, really, what is a type? What is a term? What is a function? Are they fundamentally undefined?

My feeling is.. it's necessary to introduce the rules of the game we're playing if one has to more rigorously refine type. And to do that, deductive system and lambda calculus seem to be a must. To define a function, it seems that certain realistic or imaginary machine should be involded.. for example Turing machine. And are there other systems to be introduced, before defining rigorously what a type is? Also, a natural question then arises: can all these systems be rigorously defined?

Remark

I have to admit at the end that I might not fully understand what I'm asking. I know I'm in search of a full and rigorous definition of type.. but you might ask what I mean by "rigorous"? Well.. I don't know. How can I define "rigorous"? How can I define "define"? What is "what"? And what is "what is what"? ..? It might sound like I'm making fun of words, but I'm serious.

EDIT: Zhen Lin asked a great question in the comment, which helped me express my expectation through set theory.

@ZhenLin I'm much comfortable with set theory because I sort of know what are undefined. We start with a set of undefined symbols (characters) and make up strings. There are some symbols that are "special": " ", "(", ")", "$\exists$", "$\forall$, "$\wedge$, "$\to$... etc. There is a list of pre-defined rules, that tell us how to reduce one to another. There is a list of pre-defined axioms that tell us where we could start. There is a list of translation rules that translate what's in it to English. And then we obtain meanings from that..

Related

What do logicians mean by "type"?

答:
Your question is very typical for a small class of individuals. Unfortunately, reduction to "first principles" is something you need to look for in the papers reprinted in sourcebooks, modern commentary written mostly by philosophers, and the arduous task of sifting through bibliographies. If you do so, you will find that different authors will use the same words differently. You will then be faced with deciding upon reconciliations that work for you personally.

With that said, almost all foundational literature begins with "motivation." Rarely does the technical development return to what serves as motivation to recover the heuristic under the formal. In my personal opinion, the HOTT book is utterly flawed for this reason. If you claim a foundation, its explanation does not belong in appendixes.

In so far as formalist foundations are associated with Hilbert, it essentially invokes the paraphrase of natural language fragments. There is no meaningful notion of definition. Its relationship to "language signatures" is closely bound to the study of symbolic algebras pursued in the tradition of British mathematics (de Morgan and Peacock). And, this had been motivated by the desire to do for logic what Newton had done for analytical geometry.

De Morgan speaks of "peculiar symbols," "meanings," and "rules of operation." Formalism is bound to peculiar symbols and rules of operation without meanings. But, de Morgan observes that the sign of equality is a  exception. Substitutivity in a calculus requires a warrant.

This leads to the difference between Frege's logicism and Russell's logicism. With respect to intentionality, Frege's logic had been non-classical. Russell's interpretation of definite descriptions as a form of quantifier recovers classical ambivalence.  But,  this effectively undermines the generally understood role of definition with respect to first principles.

Russell's account of definite descriptions introduces the notion of "self-identity" into the jargon of logic and its defense of the law of identity. It also accounts for Russell's stipulation that "term" is to be identified with "individual." A term can clearly be stipulated to mean something else. But, this lies at the heart of first-order logic. Russell's recovery of classical bivalence depends upon the fact that collections are composed of individuals.

One influence on Russell had been Peano's distinction between membership and inclusion. Earlier authors had not made this distinction. It lies at the heart of extensionality and semantics.

However, although Russell had been committed to a logicist interpretation for the sign of equality, he considered other views. He has a chapter on "quantity and magnitude" in "Principles of Mathematics" in which he considers alternatives. His "relative equality" does not assume the necessity of reflexive equality statements, although they are provable. In a parenthetical remark that follows, he says that reflexive equality statements maybe assumed as a consequence to a "type" stipulation (If A be a number, then A=A).

Backtracking this even further, Aristotle's analysis of logic speaks of genera, species, in individuals. All of it involves "names." Individuals are associated with proper names and are referred to as "primary substances." Other names are "secondary substances." These would be comparable with "types."

You can get from Russell to HOTT by reading Martin-Lof.

Russell's account of foundations is bound to his logical atomism. One need not take that view.  Category theory does not. Lawvere makes an oblique comment about how philosophy detoured away from mathematics because of how the logicists deprecated inclusion. Since HOTT uses a categorical notion of set, this is relevant to your question.

Title: Rigorous definition of a type


问:
Vector geometry proof (triangle)
Consider the triangle $ABC$, with point $D$ midpoint of side $AC$, and $E$ midpoint of segment $BD$. Show that $\vec{CE}=\frac{1}{4}\vec{CA}-\frac{1}{2}\vec{BC}.$

What I have tried: since $\vec{CE}=\vec{CD}+\vec{DE}$ I tried to to reformulate $\vec{CD}$ and $\vec{DE}$ in terms of $\vec{CA}$ and $\vec{BC}$. Now, $\vec{CD}=\frac{1}{2}\vec{CA}$ but for $\vec{DE}$ apart from the fact that it is half of $\vec{DB}$ I don't see how to connect it to $\vec{BC}$. I would appreciate an hint about how to proceed with this approach or how I could attack this problem in another way altogether.

Thanks.

答:
Hint 1:  If the origin of the plane where the triangle lies is $O$, then $\vec {OX}$ is the position vector, p.v. of point $X$ in the plane, and like you have used, the  triangle law of vector addition  gives for two points $A,B$, we have $$\vec{OA} + \vec{AB} = \vec{OB} \\ \implies \vec{AB} = p.v(B) - p.v(A)$$

Hint 2: The midpoint of two points $A,B$ identified by their position vectors $(p.v(A)=\vec{OA}, p.v(B)=\vec{OB})$ is given by $C$, such that $$p.v(C) = \dfrac{p.v(A)+p.v(B)}2$$

Note that $p.v(A)=\vec{OA}=a_x \hat{i} + a_y \hat{j}$ (for $2$ dimensions) is just another vector.

Title: Vector geometry proof (triangle)


问:
Proving whether an argument is valid or not when the truth value of a proposition just can't be found.
The question I must solve is the following:

Given Problem: Use the rules of inference, show that the given argument is valid or not. The domain is all persons.

“You are listening, and I am speaking. If you are listening, then it is not silence. If music system is off then it is
silence. If music system is not off then we are in happy mood, imply the conclusion that we are in happy mood and
we are good friends.”.
Give a reason for each step of proof.

My attempt at a Solution:
I begin solving it by first creating some propositions as follows:

p = "You are listening"

q = "I am speaking"

r = "It is silence"

s = "Music system is off"

t = "We are in a happy mood"

u = "We are friends"

Now I rewrite the given arguments as follows:

$p \land q$

$p \rightarrow \lnot r$

$s \rightarrow r$

$(\lnot s \rightarrow t) \rightarrow (t \land u)$

Using these premises I do:

$$p\land q \tag{Premise}\\$$
$$p \tag{Simplification}$$
$$p \rightarrow \lnot r \tag{Premise}$$
$$\lnot r \tag{Modus Ponens}$$
$$s\rightarrow r \tag{Premise}$$
$$\lnot s \tag{Modus Tollens}$$
$$\lnot s \rightarrow t \tag{Premise}$$
$$t \tag{Modus Ponens}$$
$$t \land u \tag{Premise}$$

Now I don't know the truth value of u, if I did I could derive it and I think that would prove that the argument is valid.

Question: Is the argument invalid since there is no way for me to derive anything further? Or is there a flaw in my reasoning here?

答:
It is obvious that you cannot derive that 'we are good friends', and so this argument is invalid.

But, you cannot show the argument to be invalud using rules of inference. Using rules of inference you can only show arguments to be valid.

So .. I have no idea how you are supposed to do what you are asked to do.

What a terrible problem!  Please let me know what book this is from so I can be sure never to get it.


问:
ODE with variable coefficients
Suppose I have the following ODE:
$$Au(x)- Bxu'(x) - Cu''(x)=0,$$
where $A,B,C>0$ are constants. If $u'(x)$ would not be multiplied by the $x$ factor, it would have been easy to obtain the general solution, but now I have no clue how to solve this ODE. What is here the best  strategy to obtain the general solution?

答:
Since $C>0$ you can rewrite your equation:
$$
u''(x) = au(x)-bxu'(x)
$$
where $a=\frac{A}{C}$ and $b=\frac{B}{C}$. Introduce the phase coordinates: $(v(x),w(x))=(u(x),u'(x))$ and you obtain a system of non autonomous first order ODE:
$$
\left\{\begin{array}{cl}v'(x) &= w(x) \\ w'(x) &=av(x)-bxw(x)
\end{array}\right.
$$
which in matrix notations is:
$$
\begin{pmatrix}v'(x)\\w'(x)\end{pmatrix} = \begin{pmatrix}0&1\\a&-bx\end{pmatrix}\begin{pmatrix}v(x)\\w(x)\end{pmatrix}
$$

I now assume starting conditions $(v(0),w(0))=(v_0,w_0)$. The general solution of this equation is given by,

$$
\begin{pmatrix}v(x)\\w(x)\end{pmatrix}=M(x)\begin{pmatrix}v_0\\w_0\end{pmatrix}
$$

where $M(x)$ is the resolvant of the system, i.e. the matrix solution of
$$
M'(x)=\begin{pmatrix}0&1\\a&-bx\end{pmatrix}M(x), \quad M(0)=Id.
$$

Remark: If $b=0$ then it is an autonomous ODE (as you observe) and then the resolvant is given by the matrix exponential: $M(x)=\exp\begin{pmatrix}0&t\\at&0\end{pmatrix}$

Title: ODE with variable coefficients


问:
Prove that the image of $F$ is a closed set in $\mathbb R^n$
Being $F : \mathbb R^n\to \mathbb R^n$ a continuous function such that for every $x,y\in \mathbb R^n$,
$$\|x − y\| \leq\|F(x) − F(y)\|\,.$$
I'd like to prove that the image of $F$ is a closed set in $\mathbb R^n$.

So far I have tried to use that if $x_n$ is a sequence in $\mathbb R^n$ such that $F(x_n)$ is a Cauchy sequence, $x_n$ must be Cauchy as well, in addition to the fact that any closed set has to contain all the limits of its sequences, but with this approach I haven't been able of find a proof.

Any other ideas on how to solve this proof?

答:
Let $\{y_n\}_{n\in \mathbb{N}}$ be a convergent sequence in $Im(F)$, since $\mathbb{R}^n$ is a complete space, $y_n$ is a Cauchy sequence, so, for a given $\epsilon >0;~\exists n_0\in \mathbb{N}$ such that $\lVert y_n-y_m\rVert <\epsilon ~\forall n,m\geq n_0$.

Now, construct a sequence $\{x_n\}_{n\in \mathbb{N}}$ putting $x_n$ as any element of $f^{-1}(y_n)$. With this you have that for $n,m\geq n_0; \lVert x_n-x_m\rVert \leq\lVert y_n-y_m\rVert <\epsilon$. That is, $x_n$ is Cauchy, so is convergent.

Let $x=lim(xn)$, using that $F$ is continuous we have that $lim(y_n)=lim(F(x_n))=F(lim(x_n))=F(x)$. That is, $y_n$ converge to a point in $Im(F)$, so, this is a closed set.


答:
Let's consider point $B$ as reference. Then,

$\vec{CE} = \vec{BE} - \vec{BC} = \frac12 \vec{BD} - \vec{BC}$.

Now (considering point $C$ as reference)

$\vec{BD} = \vec{CD} - \vec{CB} = \vec{CD}+\vec{BC} = \frac12\vec{CA} +\vec{BC}$.

So,

$\vec{CE} = \frac14\vec{CA} + \frac12\vec{BC} - \vec{BC} = \frac14\vec{CA}-\frac12\vec{BC}$

Title: Vector geometry proof (triangle)

Body:
Consider the triangle $ABC$, with point $D$ midpoint of side $AC$, and $E$ midpoint of segment $BD$. Show that $\vec{CE}=\frac{1}{4}\vec{CA}-\frac{1}{2}\vec{BC}.$

What I have tried: since $\vec{CE}=\vec{CD}+\vec{DE}$ I tried to to reformulate $\vec{CD}$ and $\vec{DE}$ in terms of $\vec{CA}$ and $\vec{BC}$. Now, $\vec{CD}=\frac{1}{2}\vec{CA}$ but for $\vec{DE}$ apart from the fact that it is half of $\vec{DB}$ I don't see how to connect it to $\vec{BC}$. I would appreciate an hint about how to proceed with this approach or how I could attack this problem in another way altogether.

Thanks.


问:
Are rational homology 3-spheres atoroidal?
I can't seem to pin down whether rational homology 3-spheres are atoroidal. For a 3-manifold to be atoroidal then there must be no $\pi_{1}$-injective tori. That is, there is no injection $\mathbb{Z}\oplus\mathbb{Z}\hookrightarrow \pi_{1}(M)$.

Of course, we cannot simply abelianize since there are 3-manifolds where there is an injection $\mathbb{Z}\oplus\mathbb{Z}\hookrightarrow \pi_{1}(M)$, but $\mathbb{Z}\oplus\mathbb{Z}\to H_{1}(M)$ is not injective. (At least, I'm not certain of this, but I could believe it. For example, the group $G = \langle a,b | a^{2} = b^{2} = 1\rangle$ has a subgroup generated by $ab$ that is isomorphic to $\mathbb{Z}$, but the abelianization $Ab(G)$ of $G$ has no injection $\mathbb{Z}\to Ab(G)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$).

Moreover, every nontrivial class in $H_{2}(M;\mathbb{Z})$ can be realized by a closed essential orientable surface. So if $M$ is a rational homology sphere, this doesn't rule out the fact that $M$ could have closed orientable essential tori. However, if $M$ contains an embedded incompressible torus $T$, then wouldn't the embedding $i:T\to M$ induce an injection $i_{*}:H_{1}(T)\to H_{1}(M)$?

I could be horribly on the wrong track, so any help would be appreciated!

答:
No, this is quite false. Even integer homology 3-spheres can contain incompressible tori. Let $K$ be a (smooth or polygonal) knot in $S^3$, the exterior $E(K)$ of $K$ is the complement in $S^3$ to an open tubular neighborhood of $K$. Thus, $\partial E(K)$ is homeomorphic to the 2-dimensional torus $T^2$. Furthermore, $H_*(E(K))\cong H_*(S^1)$ and the homomorphism induced by the inclusion $H_1(\partial E(K))\to H_1(E(K))$ is surjective. In particular, one of the simple loops on the boundary torus generates $H_1(E(K))$. (Proving these statements is a nice exercise in algebraic topology.) A less trivial fact is that if $K$ is a nontrivial knot (i.e. is not an unknot) then $\partial E(K)$ is incompressible in $E(K)$. One can derive this, for instance, from the Loop Theorem.

Now,  let $M_1, M_2$ be exteriors of some nontrivial knots in $S^3$.
Next, glue $M_1, M_2$ by a homeomorphism $h$ of their boundaries such that $h$ sends a loop generating  $H_1(M_1)$ to a null-homologous loop on the boundary of $M_2$ and vice-versa. The result will be a closed 3-manifold $M$ which is integer homology sphere containing an incompressible torus, namely, the common image  of $\partial M_1, \partial M_2$ in $M$. (The first claim is an exercise in application of the Mayer-Vietoris sequence; the second is    an exercise in application of the Seifert- Van Kampen Theorem.)


问:
Is this elementary number theory proof correct?
Let $A(n)$ be the number of primes less than $n$, divided by $n$ (so for example, $A(n) \leq 1$, as there cannot be more primes less than $n$ as there are integers less than $n$). Suppose that $n$ is a positive multiple of the positive integer $q$. Show that $$A(n) \leq \frac{q+(n-q)\frac{\phi({q})}{q}}{n}$$

This is what I have done.

First note that the above is equivalent to showing that (no. of primes $<n$) $\leq {q+(n-q)\frac{\phi({q})}{q}}$

Let $n=kq$

Therefore (no. of primes $<n$) $=$ (no. of primes $<kq$) $\leq \phi(kq)$ (remember $1$ is not prime)

$\phi(kq)=\phi(k)\phi(q) \leq (k-1)\phi(q)$ as $\phi(k) \leq (k-1)$.

$(k-1)\phi(q) < q+(k-1)\phi(q)=q+(qk-q)\frac{\phi(q)}{q}=q+(n-q)\frac{\phi(q)}{q}$ as required.

The reason I'm worried about this is that I have actually established a stricter inequality than in the question. However, I can't see anything wrong with this. Any reassurance or criticism would be much appreciated.

答:
You do not know that $(n/q,q)=(k,q)=1$, so you don't know that $\phi(qk)=\phi(q)\phi(k)$. Let me shine some light on a line of reasoning that involves just counting numbers.

Split the set $\{1,\cdots,kq\}$ into $k$ strings of consecutive integers,

$I_0=\{1,\cdots,q\}$

$I_1=\{q+1,\cdots,2q\}$

$I_2=\{2q+1,\cdots,3q\}$

$\cdots$

$I_{k-1}=\{(k-1)q+1,\cdots,kq\}$

Note that each interval contains exactly $q$ numbers and is a full system of representatives for the residues mod $q$. Anyway, let $A\subseteq\{1,\cdots,kq\}$ containing all of $I_0$, and for $i>0$, $A$ contains all those elements of $I_i$ that are coprime to $q$. Now, $iq+r$ is coprime to $n$ iff $r$ is coprime to $n$, so in fact $\#(I_i\cap A)$ is equal to the number of elements in $I_0$ coprime to $q$, which is $\phi(q)$. Hence

$$|A|=|I_0|+\sum_{i=1}^{k-1}|I_i\cap A|=q+(k-1)\phi(q).$$

Observe that $k-1=(n-q)/q$ and the set of all primes contained in the interval $\{1,\cdots,kq\}$ is a subset of $A$, so the $\#$ of primes is $\le |A|$. One motivation behind treating $I_0$ differently than $I_i$ for positive $i$ is that $q$ may be prime, but $q$ is not coprime to $q$, so $q\in I_0$ but $q$ is not in the subset of $I_0$ of elements coprime to $I_0$, suggesting $I_0$ requires special treatment.

You are right in thinking that this is a crude bound, though. (In fact, this bound grows linearly with $n$, as it is of the form $a+bn$, whereas $\pi(n)\sim n/\log n$ by the prime number theorem.)


问:
proving a formula related with ζ
Could someone show me how to prove the following formula?

prove that
when $a_n=\frac{ζ(2n)}{Π^{2n}}$  (n=1.2.3……)

$$(n+\frac{1}{2})a_n=\sum_{k=0}^{n-1}a_{n-k}{a_k}$$ (n≥2)  and find out the value of $a_2,a_3,a_4,a_5,a_6$

zeta-function is soooo difficult and i almost give up

I need your help and thank you very much

答:
The generating function for $\zeta(2n)/\pi^{2n}$ is

$$\frac{1}{2}-\frac{\pi x}{2}\cot(x)=\sum_{n=1}^\infty\frac{\zeta(2n)}{\pi^{2n}}x^{2n}.$$

Combining this with the fact that $\cot(x)$ satisfies the differential equation

$$f’(x)=-1-(f(x))^2$$

and using the Cauchy product gives the desired  recurrence relation.

Title: proving a formula related with ζ


问:
Help with Taylor polynomial
I need to find the Taylor polynomial of order $2n$ of the function
$$f(x,y)=\frac{1}{1+xy}$$ on  $(x_0,y_0)=(0,0)$.
Can anyone give me a hand please?

Sorry, i make a mistake, their should say "polynomial" instead of "series, i have corrected now.

Title: Help with Taylor polynomial

Body:
Hint, for $|u|<1$,
$$\frac{1}{1-u}=1+u+u^2+u^3+\ldots.$$


问:
How to approximate a double integral numerically using the Simpson's Rule
.I'm trying to approximate the integral :

$$\int_{0}^{1} \int_{0}^{1} \ln(e^x+e^y+x+y)\, dx\,dy$$ with an error $\leq 10^{-4}$.
I tried to use the Simpson's rule, and trapezoidal rule.

But I had difficulty with this, since in these rules I have to find $n$ where $h=\frac{b-a}{n}$.
In these two rules we have a formula for the error, that includes a derivative of $f$ in a point (that belongs to $[a,b]$). But I could manage to how calculate $n$.

It is the first time for me to calculate this, so I apologize if I was not accurate.

答:
The double integral Composite Simpson's error term is given by

$$E = \dfrac{-(d-c)(b-a)}{180}\left[h^4 ~\mbox{max} \left(\dfrac{\partial^4f}{\partial x^4}\right) + k^4 ~\mbox{max} \left(\dfrac{\partial^4f}{\partial y^4}\right) \right]$$

For this particular problem, we will assume that we use an equal step size, that is $h = k = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}$, which makes the error

$$E = \dfrac{-(d-c)(b-a)}{180 ~n^4}\left[~\mbox{max} \left(\dfrac{\partial^4f}{\partial x^4}\right) +  ~\mbox{max} \left(\dfrac{\partial^4f}{\partial y^4}\right) \right]$$

Our function is $f(x, y) = \ln(e^x+e^y+x+y)$, so we have

$\dfrac{\partial^4f}{\partial x^4} = -\dfrac{6 \left(e^x+1\right)^4}{\left(x+e^x+e^y+y\right)^4}+\dfrac{12 e^x \left(e^x+1\right)^2}{\left(x+e^x+e^y+y\right)^3}-\dfrac{4 e^x \left(e^x+1\right)}{\left(x+e^x+e^y+y\right)^2}+\dfrac{e^x}{x+e^x+e^y+y}-\dfrac{3 e^{2 x}}{\left(x+e^x+e^y+y\right)^2}$

$\dfrac{\partial^4f}{\partial y^4} = -\dfrac{6 \left(e^y+1\right)^4}{\left(x+e^x+e^y+y\right)^4}+\dfrac{12 e^y \left(e^y+1\right)^2}{\left(x+e^x+e^y+y\right)^3}-\dfrac{4 e^y \left(e^y+1\right)}{\left(x+e^x+e^y+y\right)^2}+\dfrac{e^y}{x+e^x+e^y+y}-\dfrac{3 e^{2 y}}{\left(x+e^x+e^y+y\right)^2}$

Using a numerical solver, the max of $\dfrac{\partial^4f}{\partial x^4}$ and $\dfrac{\partial^4f}{\partial y^4}$ over the region $0 \le x \le 1, 0 \le y \le 1$ is $-0.0186659$.

We want an $E \le 10^{-4}$, and have

$$E = -\dfrac{1}{180~ n^4}\left(-0.0186659 -0.0186659\right) = \frac{0.000207399}{n^4} \le 10^{-4}$$

This results in $n \ge 1.2$, and because we need an integer $n$, we choose $n = 2$, but this is not very satisfying as maybe we need to choose $n = 3$ or more.

Let's check this result, using Composite Simpson with $n = 2$, we get an error of $$|1.4562429599654196 - 1.4573986597429938| = 0.0011557 = 1.1157 \times 10^{-3}$$

That was pretty close, but if we had chosen $n = 3$, we have $E = 4.63677 \times 10^{-4}$, which aligns with the ask much better.

I think what this question is asking is for you to use $n = 2, 3, \ldots$ until you get a result that has a smaller error than required.

Update

This is the 2D-Simpson Formula I used above for the case $n = m = 2$

$h=\frac{b-a}{2 m}$

$k=\frac{b-a}{2 n}$

$X_{\text{i$\_$}}=a+h i$

$Y_{\text{j$\_$}}=a+j k$

$\int_{a}^{b} \int_{c}^{d} f(x, y)\, dx\,dy = \frac{1}{9} h k \left(4 \sum _{j=1}^n f\left(a,Y_{2 j-1}\right)+2 \sum _{j=1}^{n-1} f\left(a,Y_{2 j}\right)+4 \sum _{j=1}^n f\left(b,Y_{2 j-1}\right)+2 \sum _{j=1}^{n-1} f\left(b,Y_{2 j}\right)+4 \sum _{i=1}^m f\left(X_{2 i-1},c\right)+2 \sum _{i=1}^{m-1} f\left(X_{2 i},c\right)+4 \sum _{i=1}^m f\left(X_{2 i-1},d\right)+2 \sum _{i=1}^{m-1} f\left(X_{2 i},d\right)+16 \sum _{j=1}^n \left(\sum _{i=1}^m f\left(X_{2 i-1},Y_{2 j-1}\right)\right)+8 \sum _{j=1}^{n-1} \left(\sum _{i=1}^m f\left(X_{2 i-1},Y_{2 j}\right)\right)+8 \sum _{j=1}^n \left(\sum _{i=1}^{m-1} f\left(X_{2 i},Y_{2 j-1}\right)\right)+4 \sum _{j=1}^{n-1} \left(\sum _{i=1}^{m-1} f\left(X_{2 i},Y_{2 j}\right)\right)+f(a,c)+f(a,d)+f(b,c)+f(b,d)\right).$


问:
Two men and four women line up at a checkout counter in a store
Two men and four women line up at a checkout counter in a store.

$a)$ In how many ways can they line up?

$b)$ In how many ways can they line up if the first person line is a woman, and then the line changes by gender $(w, m, w, w, m, w)$?

答:
a) it is $6! = 720$ as you mentioned

b) If the order is $(w, m, w, w, m, w)$, you can permute women and men in $4!$ and $2!$ ways so number of arrangements $= 4! \times 2! = 48$


问:
Using method of separation of variables to find special and typically solutions of $\frac{dy}{dx}=|y|^{\frac13}$
So I've got the special solution which is $y=0$.
In terms of separating the variables and integrating, I know it involves sgn but I'm struggling to complete it.

答:
$$\frac{dy}{dx}=\pm y^{\frac13}$$

$$\int y^{-\frac13} dy=\pm \int dx$$

Can you finish it?


问:
Unclear integration by parts of function $f\in L^2$
I came across this mathematical statement while investigating on Fourier analysis and $L^2$ spaces. I need some clarification on why the first term isn't taken into consideration. Originating from this article http://www-users.math.umn.edu/~garrett/m/fun/uncertainty.pdf, Paul Garett. Firstly, it states that through integration by parts the following statement can be said for a function $f\in L^2$.
$$\int{|f(x)|^2dx}=-\int{x(|f(x)|^2)'dx}$$

Logically this originates from
$$\int{|f(x)|^2dx}=x|f(x)|^2-\int{x(|f(x)|^2)'dx}$$

However, I don't understand why the first term was taken into consideration, thanks.

答:
I am not fully convinced that the author isn't cheating somewhere, though as I was writing I think I've got his point. Notice that if $f\in L^2$, then $\liminf_{\lvert x\rvert\to\infty} \lvert x\rvert\lvert f(x)\rvert^2=0$, therefore there are sequences $x_n\to \infty$ and $y_n\to-\infty$ such that $x_k\lvert f(x_k)\rvert^2\to 0$ and $y_k\lvert f(y_k)\rvert^2\to 0$. Thus, \begin{align}&\int_{\Bbb R}\lvert f(x)\rvert^2\,dx=\lim_{k\to\infty} \int_{y_k}^{x_k} \lvert f(x)\rvert^2\,dx=\\&=\lim_{k\to\infty} x_k\lvert f(x_k)\rvert^2-y_k\lvert f(y_k)\rvert^2-\int_{y_k}^{x_k}x(\lvert f(x)\rvert^2)'\,dx=\lim_{k\to\infty}-\int_{y_k}^{x_k}x(\lvert f(x)\rvert^2)'\,dx\end{align}

Therefore, if $x(\lvert f(x)\rvert^2)'$ is in $L^1$, which is a prerequisite to write $\int_{\Bbb R}$, then we have the identity $$\int_{\Bbb R}\lvert f\rvert^2\,dx=\lim_{k\to\infty}-\int_{y_k}^{x_k}x(\lvert f(x)\rvert^2)'\,dx=-\int_{\Bbb R} x(\lvert f(x)\rvert^2)'\,dx$$

Notice also that, in principle, under the weaker hypothesis that $\lim_{t\to\infty}\int_0^t x(\lvert f(x)\rvert^2)'\,dx$ and $\lim_{t\to-\infty}\int_t^0 x(\lvert f(x)\rvert^2)'\,dx$ exist, the identity $$\int_{\Bbb R}\lvert f\rvert^2\,dx=-\int_{-\infty}^\infty x(\lvert f(x)\rvert^2)'\,dx$$ holds as well.


问:
$1 - \beta = P(U > 10.91) + P(U < 1.91)$?
A power supply manufacturer is interested in the variability of the output voltage, a variable known to follow a normal law. He tried $12$ selected random blocks and obtained the following results, the standard deviation of which is $0.322$: $5.34$, $5.00$, $5.07$, $5.25$, $5.65$, $5.55$, $5.35$, $5.35$, $4.76$, $5.54$, $5.44$, $4.61$.

If the true value of $\sigma^2$ is $1.0$, what is the probability that the hypothesis to be rejected with $\alpha = 0.05$.

So the hypothesis is $H_0: \sigma^2 = 0.5$ and $H_1: \sigma^2 \not= 0.5$.

$$\beta = P(\chi_{1-\frac{\alpha}{2}; 11} \leq U_0 \leq \chi_{\frac{\alpha}{2};11} | \sigma_1^2 = 1)$$
$$=P(3.82 \leq U_0 \leq 21.82 | \sigma_1^2 = 1)$$
$$=P(3.82 \leq \frac{(n-1) S^2}{\sigma_0^2} \leq 21.82 | \sigma_1^2 = 1)$$
$$=P(\frac{3.82\sigma_0^2}{\sigma_1^2} \leq \frac{(n-1) S^2}{\sigma_1^2} \leq \frac{21.82 \sigma_0^2}{\sigma_1^2})$$
$$P(1.91 \leq U \leq 10.91)\text{, where } U \sim \chi^2_{11}$$
So $$1 - \beta = P(U > 10.91) + P(U < 1.91)$$

The answer is $0.45$, but I obtained $0.55$. What am I doing wrong?

答:
Setting $U=\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(11)}^2$, you have

$$1-\alpha=0.95=\mathbb{P}[5.124\leq U\leq 21.920]$$

thus the "Acceptance region"  for $H_0$ is

$$0.2329\leq S^2\leq 0.9964$$

Given $\sigma^2=1$ you can calculate the same region by

$$\mathbb{P}[2.56\leq U\leq 10.96]=0.995-0.447\approx0.55$$

leading to the desired result of

$$1-0.55=0.45$$


答:
Assume the triple $\,(A,B,C)\,$ as a barycentric coordinate system of the plane.

We have, using barycentric coordinates,

$$D=\frac12(A+C) \qquad \text{and}\qquad E=\frac12(B+D)=\frac12 B+\frac14(A+C). $$

Therefore

\begin{align} E-C &= \frac12(B-C)+\frac14(A-C) + \frac14(C-C)=\\[2ex]
 &=\frac14(A-C)-\frac12(C-B),
\end{align}

i.e. our thesis.

Title: Vector geometry proof (triangle)


问:
$n\log_2(n)\in\Omega\left(10^3n\right)$
I am struggling with a new topic in my discrete math class.
I must show that

$$n\log_2(n)\in\Omega\left(10^3n\right)$$

A detailed explanation will be awesome!

答: Hint: $\Omega(10^3n) = \Omega(n)$.

Title: $n\log_2(n)\in\Omega\left(10^3n\right)$

Body: Hint: $\Omega(10^3n) = \Omega(n)$.


问:
Multinomial theorem for 0's and 1's
I'm looking for a special case of multinomial theorem where the indeterminates $x_1,\dots,x_m$ only take values $0$ and $1$. The power $n$ can be assumed to be greater than he number of arguments $m$, although the general case together with $n\leq m$ would be even better.

Edit: more specifically, I'm looking for a formula that is expressed as a sum of all possible monomials with a maximal degree of an indeterminate equals $1$ (e.g. $x_2, ~x_3x_4x_6, ~x_1x_6$) with some coefficients.

答:
If I understand correctly, what you want is a formula of the type:

$$(x_1+\ldots+x_m)^n=\sum_{I\subseteq [m]} \alpha_I \prod_{i\in I} x_i$$

modulo $x_i^2-x_i=0$ for $i=1,\ldots,n$. Note that from the multinomial theorem,

$$\alpha_I=\sum_{k_i\ge 1,\sum_{i\in I} k_i=n} \frac{n!}{\prod_{i\in I} k_i!}.$$

You can check that:

$$\alpha_I=|I|!{n\brace |I|}$$

where ${n\brace k}$ are Stirling numbers of the second kind.

Title: Multinomial theorem for 0's and 1's


问:
Limit Involving a Tetration of the Variable
We wish to evaluate this challenging limit
$$
\lim_{x\to 1}\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}
$$

My Observation : As an initial guess, I shall consider $f_{n}(x)=x^{x^{x{^{.^{.^{.}}}}}}$ with $n$ being the number of $x$. To simplify this problem I considered the case of $n=1$, it is easy to see that the limit is $1$, to be sure of my assertion I further considered the case for $n=2$, we receive :
$$
\left(\frac{x^{x}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}
$$
We thus simplify this expression :
\begin{align*}
\left(\frac{x^{x}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{\exp(x\ln(x))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{\exp(x(x-1+o(x-1)))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{1+x(x-1)+o(x(x-1))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(x+o(x)\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}(x-1+o(x-1))\right) \\
&=\exp(\sqrt{x-1}+o(\sqrt{x-1}))
\end{align*}
Where :$$\displaystyle\lim_{x\to 1}\exp(\sqrt{x-1}+o(\sqrt{x-1}))=1$$

Now the big question is how can we extend this for the general case that is for $n\geq3$? Any help would be much appreciated.

答:
Comment : I would like to thank Claude and Ninad for your suggestions and answers. I truly believe that this problem needs not the Lambert W function yet I shall accept Claude's answer since he put an effort in helping me with this problem. After spending several hours on this problem I have managed to prove it as follow.

My Proof : For $n\geq3$ we have that :
\begin{align*}
\exp(f_{n-2}(x)\ln(x))\ln(x)&=(1+f_{n-2}(x)\ln(x)+o(\ln(x)))(x-1+o(x-1)) \\
&=x-1+(x-1)f_{n-2}(x)\ln(x)+o((x-1)\ln(x)) \\
\end{align*}
$\implies \exp(\exp(f_{n-2}(x)\ln(x))\ln(x))-1=(x-1+(x-1)f_{n-2}(x)\ln(x))+o((x-1)\ln(x))$

And Since :
\begin{align*}
\exp(\exp(f_{n-2}(x)\ln(x))\ln(x))-1&=\exp\left(x^{f_{n-2}(x)}\ln(x)\right)-1 \\
&=\exp(f_{n-1}(x)\ln(x))-1\\
&=x^{f_{n-1}(x)}-1\\
&=f_{n}(x)-1
\end{align*}
Thus, we have that :
\begin{align*}
\frac{f_{n}(x)-1}{x-1}&=\frac{(x-1+(x-1)f_{n-2}(x)\ln(x))+o((x-1)\ln(x))}{x-1}\\
&=1+f_{n-2}(x)\ln(x)+o(\ln(x))\\
\implies& \ln\left(\frac{f_{n}(x)-1}{x-1}\right)=f_{n-2}(x)\ln(x)+o(\ln(x)) \\
\implies& \frac{1}{\sqrt{x-1}}\ln\left(\frac{f_{n}(x)-1}{x-1}\right)=\frac{f_{n-2}(x)\ln(x)}{\sqrt{x-1}}+o\left(\frac{\ln(x)}{\sqrt{x-1}}\right)\xrightarrow{x\to 1} 0 \\
\implies& \lim_{x\to 1}\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{f_{n}(x)-1}{x-1}\right)\right)=1
\end{align*}
Hence :
$$
\lim_{x\to 1}\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}=1
$$

Indeed a fascinating problem and once more thanks for Claude and Ninad for your help


问:
The theory which axiomizes infinity.
I met a strange question which gives me a Language L which only has one unary Relation R.(Set R) And it asks me to find an L-Theory which axiomizes the class of L-structure A where both R and A\R are infinite.

Is there really a way to describe infinite by those basic ->, =, etc + this R?

答:
Hint: design a sentence $\phi_n$ for $n = 1, 2, \ldots$ that says that $R$ contains at least $n$ distinct elements. E.g.,
$$
\phi_3 = \exists x_1, x_2, x_3(x_1 \neq x_2 \land x_1\neq x_3 \land x_2 \neq x_3 \land R(x_1) \land R(x_2) \land R(x_3))
$$

Similarly, design a sentence $\psi_n$ for $n = 1, 2, \ldots$ that says that $A \setminus R$ contains at least $n$ elements.

Now consider the theory $T=\{\phi_1, \psi_1, \phi_2, \psi_2, \ldots\}$. What can you say about $R$ and $A \setminus R$ in a model of $T$.

Title: The theory which axiomizes infinity.


问:
How is the simulation done of the black scholes model?
This post contains additional questions from:

How do I implement the Euler scheme for this SDE?

I think it is more appropriate to start a new post.

In an assignment the following SDE needs to be simulated:

\begin{array}{l}
d S_{t}=\alpha S_{t} d t+\sigma_{i} S_{t} d w_{t}^{1} \\
d \sigma_{t}=-\left(\sigma_{t}-\xi_{t}\right) d t+p \sigma_{t} d w_{t}^{2} \\
d \xi_{t}=\frac{1}{\alpha}\left(\sigma_{z}-\xi_{i}\right) d t
\end{array}with $S_{0}=€ 50, \quad \sigma_{0}=0.20, \quad \xi_{0}=0.20 \quad$ and $\alpha=0.10$

For $p=0($ and $\alpha \neq 0)$ the well-known Black-Scholes model is obtained:
$$
d S_{t}=\alpha S_{i} d t+\sigma_{0} S_{i} d w_{i}^{1}
$$

On page 17 and 18 of the document the same SDE is solved:
https://beta.vu.nl/nl/Images/werkstuk-dmouj_tcm235-91341.pdf

In the previous post as well in the document the stockprise the following conversion is done:

$S_t=\exp(X_t)$ or $X_t=\ln(S_t)$.

My first question is: Why is this done?

The solution of the SDE is:

$S_{t}=S_{t-1} e^{\left(\mu-\frac{1}{2} \sigma^{2}\right) d t+\sigma \varepsilon \sqrt{d t}}$

In the previous post the following solutions were given

$$\newcommand{\D}{\mathit{\Delta}}
\begin{align}
X_{t+\D t}&=X_t+(α-\tfrac12σ_t^2)\D t+σ_tz^1_t\sqrt{\D t}\\
σ_{t+\D t}&=σ_t−(σ_t−ξ_t)\D t+pσ_tz^2_t\sqrt{\D t}\\
ξ_{t+\D t}&=ξ_t+\frac1α(σ_t−ξ_t)\D t
\end{align}
$$

I'm a bit confused on how to plot the stockprice vs the time now. It's more that I'm not really sure about the approach. I want to do this in Python. I would first discretize my time interval in N intervals of $dt$. Then I would fill an array with ξ for each timestep. I would do the same for σ. At last I would calculate $X$ for each timestep. Since I know Xt, but want to plot $S_t$ v.s the time. I need to do a $Exp()$ converserion on the array.

Or is it better to only calculate ξ and σ for each timestep and then fill it in the exact formula of $S_t$. I think the result will be the same(apart from the randomness), but I'm not really sure about that

Questions:

Why is the conversion $X_t$ from $S_t$ done and why specifically $ln()$?

It the outline of simulation I described above OK?

Tim

答:
If you discretize the equation for $S_t$ directly, then there is a non-zero chance that you get negative values during the integration, especially for larger time steps. You would need some mechanism to catch that in a way that does not change the statistics of the process (too much).

Parametrizing $S_t=e^{X_t}$ forces $S$ to be positive, that the equations additionally simplify under the transformation is welcome but not the primary reason. Now the discretization of the equation for $X_t$ allows an unconstrained integration.

The formula
$$
S_{t_{k+1}}=S_{t_k} e^{\left(\mu-\frac{1}{2} \sigma^{2}\right)\,d t_k+\sigma \zeta_k \,\sqrt{d t_k}}
$$
just translates the effect of the $X$ discretization back into an evolution formula for $S$, it makes not much difference which formulations of the two you use.

You could carry out the integration of $(σ_t,ξ_t)$ first, as they do not contain $S_t$ or $X_t$, but I do not see why that would be helpful, update $(X_t,σ_t,ξ_t)$ simultaneously in one time loop.


问:
Difficulties in obtaining the Fourier series for an integral operator
I'm currently studying a paper that derives a numerical scheme to solve via Fourier Transform a certain type of PDE's, Neural-Field Equations.

The equation in study has the following integral operator:

\begin{equation}
A(\mathbf{x},t)=\int_{\Omega} d^2\mathbf{y}\int_0^{\tau_{max}} L(\mathbf{x-y},\tau)S[V(\mathbf{y},t-\tau)] d\tau,
\end{equation}
with $\mathbf{x},\mathbf{y}\in\Omega=[-\frac{l}{2},\frac{l}{2}]\times[-\frac{l}{2},\frac{l}{2}]$

Note: we have two convolutions here, one in space (2-dim) and other in time.

The authors derive the Fourier series of $A$ by stating:

\begin{equation}
A(x,y,t) = \sum_{m,n=-\infty}^{\infty}\left(e^{i(k_mx+k_ny)}\int_0^{\tau_{max}}\tilde{L}_{mn}(T)\widetilde{SV}_{mn}(t-T)\,\,dT\right),
\end{equation}
with $\tilde{L}_{mn}(T)$ and $\widetilde{SV}_{mn}(t-T)$ being the spatial Fourier transform of $L(\mathbf{x},t)$ and the image of the nonlinear functional $S[V(\mathbf{x},t)]$, respectively.

They do not do this demonstration and I really need to demonstrate this passage.

I know the spatial Fourier series of $A$ can be written as:

\begin{equation}
A(x,y,t) = \frac{1}{l^2} \sum_{m,n=-\infty}^{\infty} A_{mn}(t)e^{i(k_mx+k_ny)},
\end{equation}
with $A_{mn}$ being:
\begin{equation}
A_{mn} = \frac{1}{l^2} \int_{\Omega} A(x,y,t)e^{-i(k_mx+k_ny)}.
\end{equation}

But, if I just play around, naively, with these expressions I can't get what the authors state.
Any help?

Best regards,
Tiago

答:
Let's assign some new functions so that we can have a better view of our integral.

Let :

$L^{t}( \mathbf{x}):=L( \mathbf{x},t)$

$H^t( \mathbf{x}) := S[V(\mathbf{x},t)]$

$A^t(\mathbf{x}):= A( \mathbf{x},t)$

So the expression of $A$ is now changed to (you made a small typo in the integral domain of $d^y$):

$$A^t(\mathbf{x})=\int_{0}^{\tau_{max}} \int_{\Omega} L^{\tau}( \mathbf{x-y})H^{t-\tau}(\mathbf{y})d^2\mathbf{y}d\tau$$
From which you can realize, as you know, immediately the emergence of convolutives terms, which help us even reduce even further the formula of $A$, that is:
$$A^t(\mathbf{x}) = \int_{0}^{\tau_{max}}  \left( L^{\tau}*H^{t-\tau}\right)(\mathbf{x})d\tau$$
From this point, I guess you are clear of what to do next, that is to use a common identity we have in Fourier analysis : $ \widehat{ g*h}= \hat{g}\hat{h}$ (up to a multiplicative constant, depending on the definition of convolution).
$$ \tilde{ A^t}(\mathbf{y})\underbrace{=}_{\text{def.}} \int_{0}^{\tau_{max}} \widetilde{ \left( L^{\tau}*H^{t-\tau}\right)} (\mathbf{y}) d\tau =l\int_{0}^{\tau_{max}} \widetilde{  L^{\tau}}(\mathbf{y}) \widetilde{H^{t-\tau}} (\mathbf{y}) d\tau$$
The rest is yours.

Remark Upon finishing writing this reply, I realize your Fourier formula is
a bit different than mine ( in particular at the multiplicative constant), so while the perspective stays the same, I think you should adapt my reply to your demonstration with suitable adjustment in the multiplicative constants.


问:
How to determine if any point is in the acute or the obtuse angle between 2 planes
Show that the point $(3, 2, -1)$ lies inside the acute angle formed by the planes
 $5x-y+z+3=0$ and $4x-3y+2z+5=0$.

I have tried this by calculating the angles between the plane, passing through the line of intersection of these two planes, and these planes. But the angles can't be calculated without calculator. So I need to get the solution in an easier way without calculator.

I also tried to get a method similar as the case of origin. To get if a origin is in the acute or the obtuse angle b/w 2 planes we check the sign of $a_1a_2+b_1b_2+c_1c_2$ if equations of the planes are $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2$ provided both $d_1$ and $d_2$ are positive.

I want to get this type of solution for this case also.

答:
Every euclidian plane in 3D may be defined by a single point and an orthogonal vector to the plane. Both are not unique, but one may choose:
$$5x-y+z+3=0 \Leftrightarrow (5, -1, 1) \cdot [(x, y, z) - (0, 0, -3)] = 0$$
$$4x-3y+2z+5=0 \Leftrightarrow (4, -3, 2)  \cdot [(x, y, z) - (1, 3, 0)] = 0$$
Then, let us transpose everyone so $(3,2,-1)$ becomes the origin. We add $(-3, -2, 1)$ to the reference point in each plane.
First plane becomes:
$$(5, -1, 1) \cdot [(x, y, z) - (-3, -3, -2)] = 0 \Leftrightarrow 5x-y+z+14=0$$
And the second one:
$$ (4, -3, 2)  \cdot [(x, y, z) - (-2, 1, 1)] = 0 \Leftrightarrow 4x-3y+2z+9=0$$
Finally, apply the test:
$$(5, -1, 1) \cdot (4,-3,2) = 25 >0$$
So, if I did not make any mistake, this point lies in the obtuse angle, since the product was positive.


问:
Attaching two manifolds along their boundary
I have a question about a proof in John Lee's Introduction to Topological Manifolds.

Suppose $M$ and $N$ are two topological $n$-manifolds with nonempty boundary (for reference, the definition I am using is: $M$ and $N$ are second countable Hausdorff spaces such that each point has an open neighborhood homeomorphic to either an open set in $\mathbf{R}^n$ or an open set in $\mathbf{H}^n = \{ (x_1,\ldots,x_n) \in \mathbf{R}^n: x_n \geq 0\}$). If $\partial M$ and $\partial N$ denote the manifold boundaries of $M$ and $N$, suppose $h: \partial N \to \partial M$ is a homeomorphism. We may attach $M$ and $N$ along their boundaries by forming the adjuction space

$$ M \cup_h N = \frac{M \sqcup N}{\sim}$$

where $\sim$ is the relation generated by identifying $x\in \partial N$ with $h(x)$. The theorem is:

With $M$, $N$, and $h$ as above, $M \cup_h N$ is an $n$-manifold (without boundary).

The proof first proceeds to show that each point of the adjuction space has a euclidean neighborhood. If $q: M \sqcup N \to M \cup_h N$ denotes the quotient map, then Lee puts $S=q(\operatorname{Int} M \cup \operatorname{Int} N)$ (where $\operatorname{Int}$ denotes the manifold interior) and shows that each point of $M \cup_h N -S$ has a euclidean neighborhood. I am OK with this argument.

Now, I quote:

Suppose $s\in S$, and let $y \in \partial N$ and $x=h(y) \in \partial M$ be two points in the fiber $q^{-1}(s)$. We can choose coordinate charts $(U,\phi)$ for $M$ and $(V, \psi)$ for $N$ such that $x\in U$ and $y\in V$, and let $\hat{U}=\phi(U)$, $\hat{V} = \psi(V) \subseteq \mathbf{H}^n$. It is useful in this proof to identify $\mathbf{H}^n$ with $\mathbf{R}^{n-1}\times [0,\infty)$ and $\mathbf{R}^n$ with $\mathbf{R}^{n-1} \times \mathbf{R}$. By shrinking $U$ and $V$ if necessary, we may assume $h(V \cap \partial N) = U \cap \partial M$, and that $\hat{U} = U_0 \times [0,\epsilon)$ and $\hat{V} = V_0 \times [0,\epsilon)$ for some $\epsilon>0$ and some open subsets $U_0,V_0 \subseteq \mathbf{R}^{n-1}$.

My question: How does one choose the coordinate domains $U$ and $V$ to satisfy these two conditions simultaneously? I know how to choose $U$ and $V$ such that $h(V \cap \partial N) = U \cap \partial M$, and I know how to choose $U$ and $V$ such that their images under the coordinate maps are equal to sets of the form $U_0 \times [0,\epsilon)$ and $V_0 \times [0,\epsilon)$, but I don't know how to pick $U$ and $V$ such that both of these conditions are satisfied at the same time.

EDIT: Let me be a bit more specific. We are to choose the coordinate domains (by shrinking if necessary) to satsify the following two conditions simultaneously:

(a) $h(V \cap \partial N) = U \cap \partial M$,

(b) $\hat{U} = U_0 \times [0,\epsilon)$ and $\hat{V} = V_0 \times [0,\epsilon)$.

Now, assuming that I have adjusted $U$ and $V$ to satisfy (a), then I may adjust them further to satisfy (b). But then how do I know that $U$ and $V$ still satisfy (a)? I can be more specific by what I mean by "adjust" if neeeded.

Am I misreading the proof, or am I missing something glaringly obvious?

答:
I take it you're worried about having the $\epsilon$'s be the same? You can rescale by a linear map in the $x_n$ coordinate.

OK, if the first condition no longer holds, shrink either $U_0$ or $V_0$ to make it hold.


问:
Do you know this finitely presented group on two generators?
I computed using Sage the fundamental group of some topological space and got the infinite group
$$\langle a, b\mid aba^{-1}ba\rangle.$$
By the change of variables $x=b^{-1}$ and $y=a$, it can also be written as
$$\langle x, y \mid xy=y^2x^{-1}\rangle.$$
Do you know if this group has a name ? Do you know if there exists a table of finitely presented groups of small "complexity" like this one ?

答:
Note that the abelianized presentation is $\langle a,b\mid ab^2\rangle$. This suggest changing generators so that $ab^2$ is a generator. Define $x=ab^2$, $t=ab$, so that $a=tx^{-1}t$, $b=t^{-1}x$.  This yields the presentation
$$G=\langle t,x\mid x(t^{-2}xt^2)(t^{-1}xt)^{-1}\rangle.$$
Write $y=t^{-1}xt$. Then this yields:
$$G=\langle t,x,y\mid t^{-1}yt=x^{-1}y,\; t^{-1}xt=y\rangle.$$
Since $(x^{-1}y,y)$ is a basis of the free group on $x,y$, we identify a semidirect product of the free group $F(x,y)$ on $x,y$ by a cyclic group $\langle t\rangle$ acting on $F(x,y)$ through powers of the automorphism $(x,y)\mapsto (x^{-1}y,y)$. (In particular, $G$ is torsion-free.)

Note that this automorphism is not inner, but its square is $(x,y)\mapsto ((x^{-1}y)^{-1}y,y)=(y^{-1}xy,y)$, which is an inner automorphism (right conjugation by $y$). Hence the unique subgroup of index 2 in $G$ is a direct product $F(x,y)\times\langle t^2y^{-1}\rangle$. Note that the element $t^2y^{-1}$, whose centralizer is this subgroup of index 2, is just equal to $a$.


问:
Finding number of permutation of a word given conditions
Find the number of permutations of the word "ATTITUDE" such that no two "T" occur together Using inclusion exclusion principle only.

I found the total number of permutations which is $8!/3!$ then i took both T as a unit "a" thus the number of permutations with 2T's together is 7!.I subtracted them but could not get the right answer.I think that I have overcounted some cases but i dont know which one.Any help would be appreciated.Thanks in advance.

答:
Unrestricted number of ways to arrange them $ \displaystyle = \frac{8!}{3!}$

Now we need to subtract all the permutations where we have $2$ $T's$ together, that is $7!$ ways

But this also subtracts arrangements where 3rd $T$ comes either before or after $TT$
($ \, \_TT\_ \,$). So we removed $TTT$ twice instead of removing only once.

So number of desired permutations $\displaystyle  = \frac{8!}{3!} - 7! + 6! = 2400$


问:
Superfluous assumptions on Vietoris-Begle mapping theorem in Iversen's Cohomology of Sheaves
I am trying to prove the Vietoris-Begle mapping theorem as indicated in Iversen's Cohomology of Sheaves on page 203. The statement is the following:

Let $f: X\rightarrow Y$ be a proper map between locally compact topological spaces with acyclic fibres  and let $G$ be the constant sheaf on Y (with values in an abelian group G, by abuse of notation). Then the induced map
$f^{*}: H^{\cdot}(Y,G)\rightarrow H^{\cdot}(X,f^{-1}G)$
is an isomorphism.

As opposed to this statement let me propose a stronger statement:

Let $f:X\rightarrow Y$ be a closed morphism between topological spaces, and let for all $y\in Y$ the canonical map $G\rightarrow \Gamma(f^{-1}(y);f^{-1}G)$ be an isomorphism.
Then the induced map $f^{*}: H^{\cdot}(Y,G)\rightarrow H^{\cdot}(X,G)$
is an isomorphism.

Now I was wondering why Iversen would impose these additional assumptions. Is there something wrong about my approach?

Proof (of the second statement):

The crucial part is step (4), I have however added steps (1) - (3) in order to increase the readability.

(1) Firstly, since $f$ is closed, so we can deduce that for every open set $V$ of $X$ containing $f^{-1}(y)$ there exists an open set $U$ in $X$ containing $y$, such that $f^{-1}(y)\subseteq f^{-1}(U)\subseteq V$, for example if we set $U:=Y\backslash f(X\backslash V)$.

(2) Now the canonical map $\phi_{y}:(f_{*}f^{-1}G)_{y}\rightarrow \Gamma(f^{-1}(y);f^{-1}G)$ is an isomorphism:

Picking some $s\in (f_{*}f^{-1}G)_{y}$, we can choose an open neighbourhood $U$ of $y$ and a section $t\in\Gamma(U;f_{*}f^{-1}G)$ such that $t_{y}=s$. The map $\phi$ sends $s$ to the equivalence class of $t$ in $colim_{f^{-1}(y)\subseteq V\subseteq X \text{ open neighbourhood}} \Gamma(V,f^{-1}G)$. Assume it is sent to zero. This means that there is an open set $V\subseteq X$ containing $f^{-1}(y)$ such that $t_{\vert V}=0$, but we can actually choose some open neighbourhood $W\subseteq Y$ of $y$ such that $f^{-1}(W)\subseteq V$ and $t_{\vert W}=0$, so injectivity is shown.

Surjectivity is similar.

(3) Now we can prove that the unit map $G\rightarrow f_{*}f^{-1}G$ is an isomorphism: pick a point $y$ of $Y$ and observe that the composition
$G_{y}\rightarrow (f_{*}f^{-1}G)_{y}\rightarrow \Gamma(f^{-1}(y);f^{-1}G)$ is an isomorphism by assumption.

(4) This is where my approach is differing substantially from Iversen's, so if something is wrong about my proof, it will most likely be in this part:

Let for any topological space $a_{Z}:Z\rightarrow *$ be defined. The global section functor can be written as $\Gamma(Z,_)=(a_{Z})_{*}$. The unit map $G\rightarrow f_{*}f^{-1}G$ hence induces a natural isomorphism
$\Gamma(Y;G)=(a_{Y})_{*}G\rightarrow (a_{Y})_{*}f_{*}f^{-1} G=(a_{X})_{*}f^{-1}G=\Gamma(X,f^{-1}G)$.
This natural transformation is the one which induces the map $f^{*}H^{\cdot}(Y;G)= R\Gamma(Y;G)\rightarrow R\Gamma(X;f^{-1}G)$ by the universal property of the derived functor, and again by the universal property, $f^{*}$ is an isomorphism q.e.d.

I also welcome counterexamples.

答:
Thanks Roland! And here is what is wrong about my reasoning: What I have somehow hidden in the last sentence of my proof with the words "by the universal property" is the following:

Let $Sh(Y)$ the category of sheaves on $Y$ and $Ab$ the category of abelian groups. Both the functors $(a_{Y})_{*}$ and $(a_{Y})_{*}f_{*}f^{-1}$ induce functors $K^{+}(Sh(Y))\rightarrow K(Ab)$ on the category of complexes modulo homotopy (bounded below) as usual. Let $P_{Sh(Y)}:K^{+}(Sh(Y))\rightarrow D^{+}(Sh(Y))$ and $P_{Ab}:K(Ab)\rightarrow D(Ab)$ be the localisation functors. $R(a_{Y})_{*}$ comes equipped with a natural transformation $\varepsilon: P_{Ab}(a_{Y})_{*}\Rightarrow R((a_{Y})_{*})P_{Sh(Y)}$ and $R((a_{Y})_{*}f_{*}f^{-1})$ comes equipped with a natural transformation $\zeta: P_{Ab}(a_{Y})_{*}f_{*}f^{-1}\Rightarrow R((a_{Y})_{*}f_{*}f^{-1}) P_{Sh(Y)}$.

If we denote by $\tau$ the natural transformation $(a_{Y})_{*}\Rightarrow(a_{Y})_{*}f_{*}f^{−1}$, the universal property of the right derived functor yields a natural transformation $R\tau: R((a_{Y})_{*})\rightarrow R((a_{Y})_{*}f_{*}f^{−1})$ such that $\zeta\circ(P_{Ab}\tau) =(R\tau P_{Sh(Y)})\circ\varepsilon$. Of course, $R\tau$ is just $f^{*}$.

EDIT: The rest of this post contains mistakes. For a better understanding of the context of the accompanying comments I will leave the post as it is.

So far so good. At this point I have incorrecty assumed that the construction $\tau\mapsto R\tau$ is functorial. If it were so, then the existence of $\tau^{-1}$ would yield that $R(\tau^{-1})=(R\tau)^{-1}$, hence we could deduce my claim. Roland's counterexample therefore proves that the construction $\tau\mapsto R\tau$ is not functorial in general.


问:
Fluid Force on semicircular gate partially submerged

What is the best way to approach this problem?

Does the problem change since the gate is only partially submerged?

答:
There are at least a couple ways to set this up, but a convenient one is to start by calling the water level $ \ y = 0 \ $, measuring $ \ y \ $ "downward" from there as the "depth" variable.  The center of the circle, of which the semicircle is part, is then located at $ \ x = 0 , y = 1 \ , $ making the equation for the circle is then  $x^2 + (y-1)^2 \ = \ 2^2$.

The hydrostatic pressure at depth $y$ is given by $P(y) = \rho g y \ $.  The infinitesimal amount of force applied to the gate at that depth is then $ \ dF = P(y) \ dA = P(y) \cdot w(y) \ dy \ ,  $ where $w(y)$ is the "width function" for the gate.  We find this function by solving the equation of the circle for  $x$, and doubling that distance to the circular edge of the gate from its symmetry axis.  Thus,

$$x^2 \ = \ 4 \ - \ (y-1)^2 \ \Rightarrow \ w(y) \ = \ 2  \cdot  \sqrt{4 - (y-1)^2} $$

$$dF \ = \ \rho g y \ \cdot \ 2  \cdot  \sqrt{4 - (y-1)^2} \ dy \ ,$$

which we would then integrate "downward" in depth from $y = 0$ to $y = 1$ .

The set-up does change depending upon the water level, but we would still measure from the surface of the water as $y = 0$ in this approach.


问:
Some confusion about the proof for Darboux's Theorem
I'm having some confusion in the proof of Darboux's Theorem. It appears similar questions have been asked before, but I'm still confused by the replies, so I thought I would ask my own.

Here is my proof.
Let $g(x) = f(x) - \gamma x$

Assume $f'(a) < f'(b)$ w.l.o.g.

We know
$f'(a) < \gamma < f'(b)$ by hypothesis.

So,
$f'(a) - \gamma = g'(a) < 0$ and $f'(b) - \gamma = g'(b) > 0$

Since
$g'(a) < 0$ and $g'(b) > 0$, (opposite signs) we know $\exists c$ such that $g'(c) = 0$

That step right there is my confusion. I am basically using the IVT to claim there is a value in between. However, to use the IVT, the function has to be continuous. That is not an assumption in the problem, only that $f$ is continuous. I've found this question asked a couple of times, but the common reply seems to be that the derivative need not be continuous to have the intermediate value property because of Darboux's Theorem. But I am trying to prove Darboux's Theorem! So while I believe that fact, I can't use the theorem within its proof. I cannot seem to justify that step in the event that $g'$ is discontinuous.

I have been told there is another version of the proof combining the MVT and IVT. However, I've found it online in a few places, and I'm having a hard time following it. So I am trying to figure out how to do it this way since I don't understand the other way. Can someone explain to me why I can use the IVT without the derivative being continnuous?

答:
Since $g'(a)<0$, $g(x)<g(a)$ when $x>a$ and $x$ is close enough to $a$. And, since $g'(b)>0$, $g(x)<g(b)$ when $x<b$ and $x$ is close enough to $b$. So, $g$ has a minimum on $[a,b]$ which is attained at some $x_0\in(a,b)$. And so $g'(x_0)=0$, since:

if $x>x_0$, then $\frac{g(x)-g(x_0)}{x-x_0}\geqslant0$, and therefore $g'(x_0)=\lim_{x\to x_0^{\,+}}\frac{g(x)-g(x_0)}{x-x_0}\geqslant0$;

if $x<x_0$, then $\frac{g(x)-g(x_0)}{x-x_0}\leqslant0$, and therefore $g'(x_0)=\lim_{x\to x_0^{\,-}}\frac{g(x)-g(x_0)}{x-x_0}\leqslant0$.

And $g'(x_0)=0\iff f'(x_0)=\gamma$.


问:
Second Order Non-Linear ODE involving Bessel Functions
I'm trying to solve this but I'm getting nowhere. Does anyone know step-by-step solution? or at least the general techniques to use? I do know that the solution involves the Bessel functions.

$y'' + (1/x)y' - Ay = B$

Where y is differentiated with respect to  x.

答:
Well,this seems to be a nonhomogeneous Bessel equation.The solution space will be spanned by the fundamental solution of the homogeneous Bessel equation plus an solution to the nonhomogeneous one.
First consider the homogeneous equation:

                       $x^2y''+xy'-Ax^2y=0$

As we all know the fundamental solution to this very Bessel equation is:

$J_v=\sum^{+\infty}_{k=0}\frac{(-1)^k(1/2x)^{2k}}{\Gamma(k+1)k!}$ where $\Gamma$ is the standard gamma function.

For the detailed process you can just google'first Bessel kind function',but as you may notice there is difference on the coefficients here between the canonical one,I suggest a dialation trick such as let $Y=ay$ where a is to be chosen later to get the canonical equation.

The next step would be searching for a solution to the nonhomogeneous one,which is piece of cake,just let $y=-A/B$.

So the general solution to this problem is:$y=cJ_v-A/B$,where $c$ is constant.


问:
Let $\alpha$ and $\beta$ be any two distinct complex numbers,then $|\alpha-\sqrt{\alpha^2-\beta^2}|+|\alpha+\sqrt{\alpha^2-\beta^2}|=$
Let $ \alpha $    and    $\beta  $ be any two distinct complex numbers,then $|\alpha-\sqrt{\alpha^2-\beta^2}|+|\alpha+\sqrt{\alpha^2-\beta^2}|=$

My Attempt

Let $z_1=\alpha-\sqrt{\alpha^2-\beta^2}$,

$z_2=\alpha+\sqrt{\alpha^2-\beta^2}$

$z_1$ and $z_2$ are the roots of the complex valued equation

$z^2-2\alpha z+{\beta}^2=0$

So

$z_1+z_2=2\alpha$

and

$z_1z_2={\beta}^2$

now since alpha and beta are complex we cannot assume the roots to be conjugate and hence

$\overline{z_1}=z_2$ does not hold and consequently I am unable to find the necessary equations to solve for the value of

$|z_1|+|z_2|$

Please help me out!

Thanks in advance.

答:
In order to simplify the calculation, and to avoid confusion with the two possible values of the complex square root, it is useful to define $w$ as a square root of $\alpha^2 - \beta^2$ (is does not matter which one), and then work only with the property that
$w^2 = \alpha^2 - \beta^2$.

Using the parallelogram law twice one gets
$$
\begin{align}
 \bigl( |\alpha -w | + |\alpha+w| \bigr)^2 &= |\alpha -w|^2 + |\alpha+w|^2 + 2|\alpha^2-w^2| \\
 &= 2 |\alpha|^2 + 2 |w|^2 + 2|\beta|^2 \\
 &= |\alpha  + \beta|^2 + 2 |\alpha^2 - \beta^2| + |\alpha-\beta|^2 \\
 &= \bigr(|\alpha  + \beta| + |\alpha-\beta| \bigr)^2
\end{align}
$$
and therefore
$$
|\alpha -w | + |\alpha+w| = |\alpha  + \beta| + |\alpha-\beta| \, .
$$


问:
Uncountability of a ring defined as a subset of cartesian product

For a contradiction, I tried to use the cantor diagonal argument but could not conclude. Any help would be appreciated.

答:
You can embedd the power set of the natural numbers into $\Bbb Z_p$ in the following way. Recall that $\mathcal{P}(\Bbb N) \cong \{0,1\}^{\Bbb N}$.

Any sequence $(x_n)_n\in \{0,1\}^{\Bbb N}$ defines a unique sequence of natural numbers $(q_n)_n$, where $q_n = \sum \limits_{i=0}^{n-1} x_i p^i$. Obviously $\pi_n(q_{n+1} + p^{n+1}\Bbb Z) = q_n + p^n\Bbb Z$, hence the sequence $(q_n)_n$ is indeed an element of $\Bbb Z_p$. This yields an injective map $\{0,1\}^{\Bbb N} \rightarrow \Bbb Z_p$, as different subsets will be distinct in some $n$, so the numbers $q_{n+1}$ will be different.


问:
Does a cokernel in a exact sequence induce a monomorphism?
In an abelian category we have the following exact sequence:
$$0\rightarrow A^0 \xrightarrow{a^0} A^1\xrightarrow{a^1} A^2 \rightarrow \ldots$$
As part of a bigger proof I consider the cokernel of $a^0$ and its arrow to $A^2$:
$$A^1\xrightarrow{c} \text{coker}(a^0)\xrightarrow{\alpha} A^2 $$
I want to prove that the morphism $\alpha:\text{coker}(a^0)\rightarrow A^2$ is a monomorphism.

I know that $\text{im}(a^0)=\ker(\text{coker}(a^0))$ by definition of image and $\text{im}(a^0)=\ker(a^1)$ by exactness, therefore I have $\ker(\text{coker}(a^0))=\ker(a^1)$.
In most categories, that would mean that the cokernel of $a^0$ would be a subobject of $A^2$, but I don't know how to formalize that intuition.

答: Every epimorphism is the cokernel of its kernel in an abelian category, so $\operatorname{coker}(a^0)=\operatorname{coker(\ker(a^1))}$. Thus $\alpha$ is in fact the (co-)image of $a^1$, and thus a monomorphism.


问:
Norm of weakly convergent series
$H$ is an infinite dimensional Hilbert space with associated norm $\lVert \cdot \rVert$, and $(x_n)_{n≥1}$ is a sequence in $H$ and $x\in H$.
I have tried with choosing the orthonormal basis $e_n$ and showing that this converges weakly to $0$, but that the norm doesn't. Can someone help me with how to show this.

答:
It does not follow ! We have

$$(1) \quad x_n \to x$$

weakly. Now assume that

$$(2) \quad ||x_n|| \to ||x||.$$

Then (1) and (2) imply that $||x_n -x|| \to 0,$ a contradiction.

Title: Norm of weakly convergent series


问:
possible error while proving $\sum_{n=1}^{\infty}(x^{n} - x^{n-1})$ uniformly convergent in $[0,1)$ when its isn't
possible duplicate of Why is $\sum_{n=1}^{\infty} x^n-x^{n+1} $ not uniformly convergent in $[0,1)$?

I'm trying to understand where exactly I went wrong trying prove this:

"Prove/refute: $\sum_{n=1}^{\infty}(x^{n} - x^{n-1})$ uniromly convergent over $[0,1)$"

So by Cauchy criterion a series is uniformly convergent iff for every $\epsilon>0$ there exist $N$ such thart for every $m>n>N$ and for every $x\in I$ takes place that $|(\sum_{k=n+1}^{m}f_{k}(x)|<\epsilon$

So I did this:
Let $0<\epsilon$. There exists $N \in \mathbb{N}$ such that for every $N<n<m$ and for every $x \in [0,1)$:

$|\sum_{k=1}^{m}(x^{k} - x^{k+1}) - \sum_{k=1}^{n}(x^{k} - x^{k+1})|$

$<|\sum_{k=n+1}^{m}x^{k}(1-x)| <_{0\leq x<1} |\sum_{k=n+1}^{m}(1-x)|$

$ < |(m-n)(1-x)| = |m-n+ x(n-m)| $

$<_{0\leq x<1} |m-n+1(n-m)|$

$ = m-n+n-m = 0 < \epsilon$

So on one hand it seems like it is correct while here and also in my study books its not but I couldn't find where exactly I gone wrong.

答: You cannot possibly prove that $|m-n+x(n-m)|<0$, since no absolute value is negative. Your error lies in the assumption that $|m-n+x(n-m)|$ is smaller than the value that it takes when $x=1$.


问:
Python Lax-Friedrichs and central scheme programming
Good day :)

I found this exercise:

Consider $q_t + aq_x = 0$ with a constant velocity $a$.

Write a computer program that implements this equation using both the central scheme and also the Lax-Friedrichs scheme.

Use periodic boundary conditions. The initial data is a box followed by a Gaussian. Check the instabiliy of the central scheme for varaious CFL numbers. Check the dissipation of the Lax-Friedrichs scheme for large times.

Tryed to do it with Python and this is my code:

```<code>import numpy as np
import matplotlib.pyplot as plt

N = 100
t_min = 0
t_max = 1
k = (t_max-t_min)/N
x_min = 0
x_max = 1
a = 1
h = (x_max-x_min)/N
alpha = a*k/(2*h)
x = np.arange(x_min, x_max+2*h, h)
u_0 = np.exp(-225*(x-0.3)**2)
u_0[np.where((x>=0.6) & (x <= 0.8))] = 1.0
u_lf = u_0.copy()
u_cs = u_0.copy()
u_num1 = u_0.copy()
u_num2 = u_0.copy()
t = 10000

plt.clf()

for j in range(N):
    u_num1[j] = (u_lf[j+1]+u_lf[j-1])/2 - alpha*(u_lf[j+1]-u_lf[j-1])
    u_num2[j] = u_cs[j] - alpha*(u_cs[j+1]-u_cs[j-1])
u_lf = u_num1.copy()
u_cs = u_num2.copy()

u_ex = np.exp(-225*(x-0.3-(a*t)%t)**2)
u_ex[np.where((x>=0.6) & (x <= 0.8))] = 1.0

u_ex=u_ex%len(u_ex)
u_lf=u_lf%len(u_lf)
u_cs=u_cs%len(u_cs)

plt.plot(x, u_ex, 'r-', label="Exact solution", fillstyle='none')
plt.plot(x, u_lf, 'o', label="Lax-Friedrichs", fillstyle='none')
plt.plot(x, u_cs, '^', label="Central scheme", fillstyle='none')
plt.axis((0, 1, -.5, 1.5))
plt.legend(loc='lower right')
plt.suptitle("t = %1d" % (t))

plt.show()
</code>```

I think there are some errors which I don't see, since the central scheme is perfectly stable and the LF scheme is not dissipating for any time.

答:
With the above choice of parameters, we have $a k/h = 1$, which is a very particular case for which some numerical schemes provide the exact solution. This property is sometimes called the unit CFL condition, see Exercise 4.2 p. 85 of the monograph by LeVeque (2002). Besides this feature, OP uses the same number of points $N$ both in space and time (there is no reason to do so in practice). Moreover, OP forgot to implement the periodic boundary conditions $$
\begin{aligned}
Q_{0}^{n} &= Q_{N-1}^{n} \\
Q_{N}^{n} &= Q_{1}^{n}
\end{aligned}
$$
see this post for instance, and other related posts on this site.


问:
property of a subgroup generated by two subgroups
Let $A, B, C$ be subgroups of a finite group $G$. Is it true that the following holds:

$$\langle A,B\rangle\cap C=\langle A\cap C, B\cap C\rangle$$

答:
No. Consider $A=\Bbb Z/n\Bbb Z \oplus 0, B = 0 \oplus \Bbb Z/n \Bbb Z$ and $C = \langle (1,1) \rangle$ as subgroups of $(\Bbb Z/n\Bbb Z)^2$. We have
$$C = (\Bbb Z/n\Bbb Z)^2 \cap C = \langle A, B\rangle \cap C$$
but
$$ 0 = \langle 0,0\rangle = \langle A \cap C, B\cap C\rangle.$$


答: The answer is no. We can take, for example, $G=S_5$, $A=A_5$, $B=\langle(12)\rangle$ and $C=\langle (13)\rangle$. Then $\langle A,B\rangle=S_5$ and $\langle A,B\rangle\cap C=C$. On the other hand, $A\cap C=B\cap C=1$ and so $\langle A\cap C, B\cap C\rangle=1\neq C=\langle A,B\rangle\cap C$.

Title: property of a subgroup generated by two subgroups

Body:
Let $A, B, C$ be subgroups of a finite group $G$. Is it true that the following holds:

$$\langle A,B\rangle\cap C=\langle A\cap C, B\cap C\rangle$$


答:
The function $g\colon [a,b]\to\Bbb R$ is continuous, hence attains its minimum at some $x_0\in [a,b]$.

We have $g'(x_0)\ge 0$: This is clear if $x_0=b$, so assume $x_0<b$. Then for all sufficiently small $h>0$, we have $x_0+h\le b$ and $g(x_0+h)\ge g(x_0)$, hence $\frac{g(x_0+h)-g(x_0)}{h}\ge 0$. By taking the limit as $h\to 0^+$, we find $g'(x_0)\ge 0$.

We have $g'(x_0)\le 0$: This is clear if $x_0=a$, so assume $x_0>a$. Then for all sufficiently small $h>0$, we have $x_0-h\ge h$ and $g(x_0-h)\ge g(x_0)$, hence $\frac{g(x_0-h)-g(x_0)}{-h}\le 0$. By taking the limit as $h\to 0^+$, we find $g'(x_0)\le 0$.

Therefore $g'(x_0)=0$.

Title: Some confusion about the proof for Darboux's Theorem

Body:
I'm having some confusion in the proof of Darboux's Theorem. It appears similar questions have been asked before, but I'm still confused by the replies, so I thought I would ask my own.

Here is my proof.
Let $g(x) = f(x) - \gamma x$

Assume $f'(a) < f'(b)$ w.l.o.g.

We know
$f'(a) < \gamma < f'(b)$ by hypothesis.

So,
$f'(a) - \gamma = g'(a) < 0$ and $f'(b) - \gamma = g'(b) > 0$

Since
$g'(a) < 0$ and $g'(b) > 0$, (opposite signs) we know $\exists c$ such that $g'(c) = 0$

That step right there is my confusion. I am basically using the IVT to claim there is a value in between. However, to use the IVT, the function has to be continuous. That is not an assumption in the problem, only that $f$ is continuous. I've found this question asked a couple of times, but the common reply seems to be that the derivative need not be continuous to have the intermediate value property because of Darboux's Theorem. But I am trying to prove Darboux's Theorem! So while I believe that fact, I can't use the theorem within its proof. I cannot seem to justify that step in the event that $g'$ is discontinuous.

I have been told there is another version of the proof combining the MVT and IVT. However, I've found it online in a few places, and I'm having a hard time following it. So I am trying to figure out how to do it this way since I don't understand the other way. Can someone explain to me why I can use the IVT without the derivative being continnuous?


问:
Find the number of quadratic polynomials ax² + bx + c

Find the number of quadratic
  polynomials ax² + bx + c such that:

a)
  a, b, c are distinct.

b) a, b, c ε {1,
   2, 3, …2008}

c) x + 1 divides ax² + bx
  + c

Reasoning->b=a+c  b ranges from 3 to
  2008 x1+x2=n has n-1C1 solns. Using th
  elogic we get 2+3+..2007 solutions. A
  total of 2015027
  solutions.here,condition 1 comes into
  picture telling us that a,b,c are
  distinct. Therefore the even values of
  b ie.4,6,8,..2008 will have one
  solution less ie.from the given
  solution subtract 1003 thus giving us
  20104024

答:
Your question and answers to your question can be found here. For the answer see: abhitsian's Comment in this link:

http://quantologic.wordpress.com/2009/07/27/problem-of-the-day-27-07-09-2/#comments


问:
Gamma properties
How can i prove that \begin{align*}
\Gamma\left(\frac{n}{2}\right) &= \frac{\sqrt{\pi} \left(n-1\right)!}{2^{n-1} \left(\frac{n-1}{2}\right)! }
\end{align*}

I try this
\begin{align*}
\Gamma\left(\frac{n}{2}\right) &= \int\limits_0^\infty x^{\frac{n}{2}-1}e^{-x} \mathrm{d}x\qquad\qquad\mid\text{integration by parts }\int fg^\prime= fg-\int f^\prime g\\
&=-x^{\frac{n}{2}-1}e^{-x}\mathop{\big|}\limits_0^\infty+\int\limits_0^\infty \left(\frac{n}{2}-1\right) x^{\frac{n}{2}-2}e^{-x}\mathrm{d}x\\
&=-\lim\limits_{x\rightarrow\infty}x^{\frac{n}{2}-1}e^{-x}+0+\int\limits_0^\infty \left(\frac{n}{2}-1\right) x^{\frac{n}{2}-2}e^{-x}\mathrm{d}x\qquad\qquad\mid\lim\limits_{x\rightarrow\infty}x^{\frac{n}{2}-1}e^{-x}\rightarrow 0\\
&=\left(\frac{n}{2}-1\right)\int\limits_0^\infty  x^{\frac{n}{2}-2}e^{-x}\mathrm{d}x\qquad\qquad\mid\text{integration by parts }\\
&=\left(\frac{n}{2}-1\right)\left(-x^{\frac{n}{2}-2}e^{-x}\mathop{\big|}\limits_0^\infty+\int\limits_0^\infty \left(\frac{n}{2}-2\right) x^{\frac{n}{2}-3}e^{-x}\mathrm{d}x\right)\\
&=\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\int\limits_0^\infty  x^{\frac{n}{2}-3}e^{-x}\mathrm{d}x\\
&=\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\Gamma\left(\frac{n}{2}-2\right)\\
&=\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\left(\frac{n}{2}-3\right)\Gamma\left(\frac{n}{2}-3\right)\\
\end{align*}
but I don't know what to do now. Any ideas?

答:
First prove that $\Gamma\Big(\frac{1}{2}\big)=\sqrt{\pi}$

Proof:

Given the known result

$$\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}$$

set $\frac{x^2}{2}=t$ and get

$$\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}dx=2\int_0^{\infty}\frac{\sqrt{2}}{2}t^{-\frac{1}{2}}e^{-t}dt=\sqrt{2}\int_0^{\infty}t^{\frac{1}{2}-1}e^{-t}dt=\sqrt{2}\Gamma\Big(\frac{1}{2}\Big)\rightarrow \Gamma\Big(\frac{1}{2}\Big)=\sqrt{\pi}$$

Second prove that

$\Gamma(n)=(n-1)\Gamma(n-1)$

Proof:

simply integrating by parts you get

$$\Gamma(n)=\int_0^{\infty}x^{n-1}e^{-x}dx=(n-1)\int_0^{\infty}x^{n-2}e^{-x}dx=(n-1)\Gamma(n-1)$$

Finally: put the two result together and get you proof recursively

Title: Gamma properties


问:
Probability of random sample of $50$ students from a class of $300$.

A simple random sample of $50$ students is taken from a class of $300$ students. In the class, the average midterm score is $67$ and the standard deviation(SD) is $12$, there are $72$ women. Let $W$ be the number of women in the sample, and let $S$ be the average midterm score of the sampled students.

$N=300;
n=50;
μ=E[S]=67;
σ=SD=12;
W=72;$

Find $E(W)$?=Expected value of Women?
E(W)=n*p=50*72/300=12

Find $SE(W)$?=Standard error of Women?

and Find $E(S)$?= Expected value of student tast score? $E(S) =μ=E[S]=67 $

Find $SE(S)$?=Standard Error of Student midterm test Score?

$SE(S)={[σ/sqrt(n)]*[$\sqrt((N-n)/(N-1))]}$

$SE(S)={[12/sqrt(50)]*[(300-50)/(300-1)]}=1.55178$

答:
Note that here $W \sim Binomial(n=50,p=\dfrac{72}{300})$

So $E(W)=np=50 \cdot \dfrac{72}{300}$ and $SE(W)=\sqrt{np(1-p)}=\sqrt{50 \cdot \dfrac{72}{300}\left(1-\dfrac{72}{300}\right)}$.

Again $E(S)=\bar x =67$ and $SE(S)=\sqrt{\dfrac{\sigma ^2}{n}}=\sqrt{\dfrac{12^2}{50}}$


问:
Portmonteau's Theorem works if we demand only compactness, if $X_n$ are dense
When introducing Portmanteau's Theorem my Probability Theory course gave the following exercise for the reader.

Assume $X, X_n$ are real random variables such that $(X_n)_n$ is tight. Further, assume
$$\limsup_{n\rightarrow\infty}P(X_n\in K)\le P(X\in K)\tag{1}$$
holds for all compact of $K\subset\mathbb R$. Show that $X_n\rightarrow X$ in distribution. Why can't we drop the tightness assumption?

The first statement would follow directly from the assumption by Portmanteau, if we replace "for all compact" with "for all closed". Is there some way to get from compact sets to closed sets using tightness?

EDIT an answer to the second question is given here.

答:
Fix $\varepsilon >0$. Since $(X_n)_n$ is tight, there exists $a>0$ such that
$$\sup_n\mathbb{P}(|X_n|>a) < \varepsilon.$$
Now let $F \subset \mathbb{R}$ be a closed subset. We have
$$\mathbb{P}(X_n \in F) \leq \mathbb{P}(X_n \in F\cap [-a,a])+\mathbb{P}(|X_n| >a )\leq \mathbb{P}(X_n \in F\cap [-a,a])+\varepsilon.$$
Since $F\cap [-a,a]$ is compact, we get
$$\limsup_n \mathbb{P}(X_n \in F)\leq \mathbb{P}(X\in F\cap [-a,a]) + \varepsilon \leq \mathbb{P}(X\in F) + \varepsilon.$$
This holds for every $\varepsilon >0$, thus letting $\varepsilon \to 0$ we deduce that for every closed subset $F \subset \mathbb{R}$
$$\limsup_n \mathbb{P}(X_n \in F)\leq \mathbb{P}(X\in F).$$
It follows from the Portmanteau theorem that $X_n$ converges to $X$ in distribution.


问:
How to transfer a function of two variables into one variable by change of variables?
If $f(x,y)=y\, f(\frac{x}{y},1)=y\, g(\bar{x})=u(\bar{x})$ with $\bar{x}=\frac{x}{y}$,
then
\begin{align}
\frac{\partial f(x,y)}{\partial x}=& \frac{\partial [y\, f(\frac{x}{y},1)]}{\partial x} \tag 1\\
=&  \frac{\partial [y\, g(\bar{x})]}{\partial x} \tag 2\\
=&\frac{\partial [y\, g(\bar{x})]}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial x} \tag 3\\
=&\frac{\partial u(\bar{x})}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial x}. \tag 4
\end{align}
How about $\frac{\partial f(x,y)}{\partial y} $?
I'm a little confused.
Does it follow the derivation bellow?
\begin{align}
\frac{\partial f(x,y)}{\partial y}=& \frac{\partial [y\, f(\frac{x}{y},1)]}{\partial y} \tag 5\\
=&  \frac{\partial [y\, g(\bar{x})]}{\partial y} \tag 6\\
=&y\, \frac{\partial g(\bar{x})}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial y}+g(\bar{x}) \tag 7\\
=&\frac{\partial u(\bar{x})}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial y}.\tag 8
\end{align}

答:
Answer:

$$ \frac{\partial f}{\partial x} = g_{\bar x} (\bar x)$$

$$ \frac{\partial f}{\partial y} = g(\bar x) - \bar x g_{\bar x}(\bar x) $$

Derivation:

We are given that $ f(x,y) = y f(x/y, 1) = y g(\bar x)$, where $\bar x = x/y$. Using the chain rule:

$$ \frac{\partial f}{\partial x} = y \cdot g_{\bar x} (\bar x) \cdot \frac{\partial \bar x}{\partial x}  = y \cdot g_{\bar x} (\bar x) \cdot \frac{1}{y} = g_{\bar x} (\bar x)$$

and

$$ \frac{\partial f}{\partial y} = g(\bar x) + y \cdot g_{\bar x} (\bar x) \cdot \frac{\partial \bar x}{\partial y}  = g(\bar x) + y \cdot g_{\bar x} (\bar x) \cdot \frac{-x}{y^2} = g (\bar x) - \frac{x}{y}g_{\bar x}(\bar x) = g(\bar x) - \bar x g_{\bar x}(\bar x) $$


问:
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$

$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$

So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$

now i did not understand how can i solve it

Help me

Thanks

答:
The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick.

As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so
$$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$
The sneaky trick is to substitute $u=x+\frac1x$, which gives
$$I=2\int_2^\infty\frac1{\sqrt{(u^2-4)(u^2+u-1)}}\,du$$
This changes all the denominator quartic's roots from complex to real, which will make the sequel easier. Furthermore, the roots are very simple: $-2,-\varphi,1/\varphi,2$ where $\varphi$ is the golden ratio. We can now apply Byrd and Friedman 258.00 directly, which says that if $d<c<b<a<y$,
$$\int_a^y\frac1{\sqrt{(t-a)(t-b)(t-c)(t-d)}}\,dt=gF(\psi,m)\\
\text{where }g=\frac2{\sqrt{(a-c)(b-d)}},\psi=\sin^{-1}\sqrt{\frac{(b-d)(y-a)}{(a-d)(y-b)}},m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$$
($m=k^2$ is the parameter, as used by Mathematica and mpmath.) We have $d=-2,c=-\varphi,b=1/\varphi,a=2$, but since $y=\infty$ we have to take the limit of the expression for $\psi$ as $y\to\infty$, which therefore becomes
$$\psi=\sin^{-1}\sqrt{\frac{b-d}{a-d}}=\sin^{-1}\sqrt{\frac{1/\varphi+2}4}=\sin^{-1}\frac\varphi2=\frac{3\pi}{10}$$
The other key numbers follow:
$$g=\frac2{\sqrt{(2+\varphi)(1/\varphi+2)}}=\frac2{\sqrt{4\varphi+3}}$$
$$m=\frac{4(1/\varphi+\varphi)}{(2+\varphi)(1/\varphi+2)}=\frac{4(2\varphi-1)}{4\varphi+3}=8\varphi-12$$
Finally we have the answer in terms of elliptic integrals:
$$\boxed{I=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},8\varphi-12\right)}$$

```<code>#!/usr/bin/env python3
from mpmath import *
mp.dps = 150

print(quad(lambda t: 1/sqrt(polyval((1,1,1,1,1),t)), [0, inf]))

g = 4/sqrt(4*phi+3)
m = 8*phi-12
print(g*ellipf(3*pi/10, m))
</code>```

P.S. A certain Dr. Sonnhard Graubner gave an answer here, but it's incorrect.

Body:
The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick.

As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so
$$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$
The sneaky trick is to substitute $u=x+\frac1x$, which gives
$$I=2\int_2^\infty\frac1{\sqrt{(u^2-4)(u^2+u-1)}}\,du$$
This changes all the denominator quartic's roots from complex to real, which will make the sequel easier. Furthermore, the roots are very simple: $-2,-\varphi,1/\varphi,2$ where $\varphi$ is the golden ratio. We can now apply Byrd and Friedman 258.00 directly, which says that if $d<c<b<a<y$,
$$\int_a^y\frac1{\sqrt{(t-a)(t-b)(t-c)(t-d)}}\,dt=gF(\psi,m)\\
\text{where }g=\frac2{\sqrt{(a-c)(b-d)}},\psi=\sin^{-1}\sqrt{\frac{(b-d)(y-a)}{(a-d)(y-b)}},m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$$
($m=k^2$ is the parameter, as used by Mathematica and mpmath.) We have $d=-2,c=-\varphi,b=1/\varphi,a=2$, but since $y=\infty$ we have to take the limit of the expression for $\psi$ as $y\to\infty$, which therefore becomes
$$\psi=\sin^{-1}\sqrt{\frac{b-d}{a-d}}=\sin^{-1}\sqrt{\frac{1/\varphi+2}4}=\sin^{-1}\frac\varphi2=\frac{3\pi}{10}$$
The other key numbers follow:
$$g=\frac2{\sqrt{(2+\varphi)(1/\varphi+2)}}=\frac2{\sqrt{4\varphi+3}}$$
$$m=\frac{4(1/\varphi+\varphi)}{(2+\varphi)(1/\varphi+2)}=\frac{4(2\varphi-1)}{4\varphi+3}=8\varphi-12$$
Finally we have the answer in terms of elliptic integrals:
$$\boxed{I=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},8\varphi-12\right)}$$

#!/usr/bin/env python3
from mpmath import *
mp.dps = 150

print(quad(lambda t: 1/sqrt(polyval((1,1,1,1,1),t)), [0, inf]))

g = 4/sqrt(4*phi+3)
m = 8*phi-12
print(g*ellipf(3*pi/10, m))

P.S. A certain Dr. Sonnhard Graubner gave an answer here, but it's incorrect.


问:
Fourier transform of $\nabla \circ \operatorname{div}$
Thanks everyone for your attention to this question.

Let $F$ a function from Schwartz space of smooth functions on $\mathbb{R^n}$, which derivatives is rapidly decreasing. So $F$ is locally integrable over $\mathbb{R^n}$ and I can easily compute that:

$$ \mathscr{F}(\operatorname{div} \circ \space \nabla F)(\xi) = \mathscr{F}(\Delta F)(\xi) = -|\xi|^2\mathscr{F}(F). $$

Now I need to calculate the Fourier transform of this composition in reverse order, namely:

$$\mathscr{F}(\nabla \circ \operatorname{div}F).$$

I have a next hypothesis:

$$\mathscr{F}(\nabla \circ \operatorname{div}F) = -\xi\xi^\mathrm{T}\mathscr{F}(F),$$

but my calculations are not entirely accurate, in my opinion, and I am not very sure of the given answer.

Thanks you a lot for your attention!

答:
If you are dealing with Schwartz space then everything should be okay because in Schwartz space, we have the following identities:
$$ \mathscr{F}( \text{div} F )= i\xi^T  \mathscr{F}(F)$$
for $F: \mathbb{R}^n \rightarrow \mathbb{R}^{n \times m}$ and
$$ \mathscr{F}( \nabla G )= i\xi  \mathscr{F}(G)$$
for $G: \mathbb{R}^n \rightarrow \mathbb{R}^{1 \times m}$

Thus,
$$\mathscr{F}( \text{div} \circ \nabla F)= i\xi^T \mathscr{F}( \nabla F)= - \xi^T \xi F $$
for all $F: \mathbb{R^n} \rightarrow \mathbb{R}^{1 \times m} $

and
$$ \mathscr{F}( \nabla \circ \text{div} F)= i\xi \mathscr{F}( \text{div} F)= - \xi \xi^T F $$
for all $F : \mathbb{R}^n \rightarrow \mathbb{R}^n$


答:
An example: let $H=\ell^2$ and $(u_n)$ the usual orthonormal basis of $H$.

Then show that $u_n \to 0$ weakly. But $ ||u_n||=1 \to 1.$

Title: Norm of weakly convergent series

Body:
$H$ is an infinite dimensional Hilbert space with associated norm $\lVert \cdot \rVert$, and $(x_n)_{n≥1}$ is a sequence in $H$ and $x\in H$.
I have tried with choosing the orthonormal basis $e_n$ and showing that this converges weakly to $0$, but that the norm doesn't. Can someone help me with how to show this.