问:
Questions about foundations of mathematics
It seems to me that in trying to create a foundation of mathematics, mathematicians are trying to create a formal system that models the language used in what I view as common-sense, real mathematics. Whatever the number $1$ is, I doubt it's $\{\{\}\}$, no? The set containing the empty set is merely a representation of $1$ in this formalism. So then is a foundation of mathematics merely a formalism used to model 'real' mathematics? If not, what is it?

I thought I was in on a little-known secret when I was (perhaps not intentionally) lead to believe that $1$ is the same as $\{\{\}\}$, but now I see how preposterous a claim that is, especially considering that the empty set is probably but a linguistic construct.

If the answer to my first question is 'yes', then presumably the reason for the foundation is to check it for self-consistency, for if the foundation is self-consistent, the thought process probably goes, then 'real' mathematics is self-consistent, yes? How can one know that the foundation (always) parallels 'real' mathematics? How do we know that the two discourses are isomorphic? If we can't know then why bother with a foundation in the first place? If we can know then presumably it's due to some metalanguage, but then... is the metalanguage self-consistent? How do we know? Isn't this just kicking the can down the road?

答:
I strongly empathize with your questions. However, you are unlikely to get satisfactory answers on a site like this. One reason I say this lies with the distinction you draw between "real" mathematics and studies in mathematical logic. It may be the case that what someone holds to be mathematics depends greatly upon what they have been taught under the guise of mathematics. A great deal of mathematics is taught in philosophy departments. So people who had received their initial training in advanced mathematics in such an environment will not even acknowledge the distinction you seem to be making.

I have studied the foundations of mathematics as an avocation for the last 35 years.  I received my undergraduate training in the mathematics department of the University of Chicago. Their mathematics department is a part of their Physical Sciences Collegiate Division. So, you may surmise that my understanding of "real" mathematics is not derived from philosophical methods. I had been accepted to graduate programs at schools like Princeton, Harvard, Yale, and UC Berkeley. Unfortunately, a personal tragedy ended any hope of an academic career. I never received the benefit of a graduate program.

I had been intrigued with the continuum hypothesis. I had begun with the assumption that the applicability of mathematics to the sciences indicated that "real" mathematics was largely unproblematic. So, something was probably wrong with what I had been taught in my mathematical logic courses. After all of this time, I have learned that I think that mathematics ought to be constructive.  My use of that word is very different from how those who understand the various foundational paradigms would interpret it. But, my point is that there is an aesthetic component to what someone will judge to be foundational with respect to mathematics.

You ask what a foundation might be if it is not a mere formalism. Setting aside the claims of entrenched paradigms, foundational analysis ought to be directed at discerning and exposing presuppositions assumed in the practice of mathematics. I personally discount the reduction of mathematics to linguistic forms. But, there can be no question that the practice of paraphrasing statements and using formal methods to represent mathematical assertions has enriched the field. The entrenched paradigms carry specific claims with respect to the nature of mathematics.

What you have said about the singleton with only the empty set as a member is simply untrue for a logicist committed to the views of Frege and Russell. Logicism is a metaphysical foundation. To say "mathematics is extensional" is ultimately at odds with the assertion that mathematics is formal. Frege had been critical of formalism, and, Russell challenged the formalists to give a meaningful account of numerical succession. Frege had actually done that using inclusive disjunction. A fundamental difference between Frege and Russell, however, lay with the fact that Frege's account of truth had not been classical (an aesthetic difference with historical consequences). Yet, today Frege's account of zero is seen to be an application of the principle of indiscernibility of non-existents from negative free logic. And once this is understood and recognized, one sees that this principle is ubiquitous in mathematics. Where it is applied, it establishes well-formedness conditions.

To the best that I can ascertain, this is ignored in the study of formal systems because formal systems stipulate a number of well-formedness conditions externally with respect to what modern formalists call an object language. I first became aware of this when  reading Universal Algebra by Graetzer. The development of formal systems arises from the pursuit of a logical calculus. Clearly, mathematicians communicate with proofs. So, one might easily pursue foundational studies from the perspective that logic is presupposed by mathematicians. Of course, when I work on a proof, I engage in "synthesis" when asking what follows from the asserted assumptions and "analysis" when assuming the conclusion in order to consider possible antecedents. But, the syntactic structure of the completed proof only presents the synthetic activity.  Some might hold that this is all there is to mathematics.

Rudolph Carnap is credited with the distinction between syntax, semantics, and pragmatics. Only pragmatics takes the relationship of the "language user" into account. The formal methods you have questioned concern themselves only with syntax and semantics. I would attribute this to the idea that mathematics has some profound relationship to material reality. This idea certainly influenced a great deal of foundational thought in the late nineteenth century and early twentieth century. While the beliefs of nineteenth century scholars might seems irrelevant, the apparent failure of unwarranted beliefs about geometry led to the arithmetization of mathematics.

Personally, I need a modern argument to convince me that geometry is useless in the study of foundations.  You will often find this kind of statement made by advocates of certain entrenched paradigms.  In so far as unary negation is eliminable, the 16 basic Boolean functions associated with propositional semantics relate to one another as points in a finite affine plane. Negation is then understood as a particular collineation. Along similar lines, the order-theoretic presentation of logical constants is nearly isomorphic with the order-theoretic presentation of a tetrahedron in combinatorial topology. Actually, though, the comparison should be made with a 3-dimensional projection of a tesseract. I see geometry because no one taught me to disparage geometry.

You mention metalanguages at the end of your question. If, in fact, mathematics begins with axioms, what significance does any metamathematics really have for "real" mathematics? De Morgan's work on symbolic algebras is clear with regard to meaningless symbols manipulated according to stipulated rules of operation. The only exception he makes rests with the sign of equality. Given an informative equality statement (x=y), a substitution may be performed. What justifies a substitution -- that is, its warranting -- necessarily presupposes "meaning." The law of identity provides such a meaning for reflexive equality statements. But, the necessary truth of reflexive equality statements does not follow from a semantic conception of truth. In this sense, a semantic conception of truth reaches back to Frege's account of truth.  In so far as formal systems relate to the paraphrasing of natural language, the necessary truth of reflexive equality statements enforces the uniformity of interpretation across occurrences. Of course, this conveniently coincides with the metaphysical law of identity. However, the linguistic mandate follows from what is called the analytic conception of truth.  One way of understanding formal systems is to understand the syntactic component as complying with an analytic conception of truth before a semantic conception of truth is applied to its terms. It is a well-formedness presupposition of an entrenched paradigm that is simply stipulated to be mathematics.  But, whether or not it is "real" mathematics is a matter of aesthetics.

Supposing that your "real" mathematics attributes some notion of truth to mathematical statements, the idea that meaning is conveyed by truth conditions alone will force you to consider metalanguages. And, questions about truth in the initial metalanguage will force you to accept an unending hierarchy of metalanguages. Unless you are willing to forego your attribution of truth to mathematical statements, you will be forced to concede that "common-sense" has led you to philosophical considerations you would prefer to avoid. If you think that self-consistency is the answer, you should look at what is called the coherence theory of truth. Realists have many legitimate objections to such accounts of truth.

My own aesthetic differs by virtue of a distinction that I have found only in the work of Aristotle. In his syntactic analysis (Prior Analytics) he distinguishes "demonstrations" as a subclass of "deductions." In his further analyses of logic (Posterior Analytics, Topics, and Rhetoric) he distinguishes between "demonstrative argumentation" and "dialectical argumentation." In so far as the notion of a "first principle" applies, it relates to demonstrative logic.  The experience of learning mathematics through its careful use of definition corresponds with the ordered presentation of "knowledge statements" that characterizes Aristotle's account of demonstrative argumentation (Categories).  But, the emphasis on semantics in the modern development of mathematical logic is aligned with Aristotle's dialectical argumentation.  Since my axiom systems require different quantifier rules to capture both, I see foundations as a branched system.

In addition, my aesthetic motivation leads to the idea that "foundational theories" are necessarily intensional. As noted above, one must forego one's attribution of truth to mathematical statements to avoid hierarchies of metalanguages.

Hopefully, you will derive some benefit from these remarks. It takes a great deal of effort to read and (hopefully) understand ideas with which one does not agree. Because of what economists call opportunity cost, my studies in foundations have taken away from ordinary mathematics which I would have liked to have learned. But, the questions which have led to different foundational programs are well-motivated. I have not suffered for my choices beyond the ad hominem attacks I have experienced on sites like this.


问:
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...
Question: How to show the relation
$$
J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt
=-\frac 1{64}\pi^4
$$
(using a "minimal industry" of relations, possibly remaining inside the real analysis)?

So i have found a solution to the problem, it is part of my solution for
math.stackexchange.com - questions - 3854736,
but not a satisfactory solution. "There should be more", explaining why there is a "clean result" for the integral.

Here, i am not strictly interested in a computational approach. I just want to share this with the community in these days of isolation. Any idea to attack this, or a related integral involving "three log factors" is welcome. (Well, the $\arctan$ is a sort of $\log$ in a sense that i don't want to define closer, see below.) Computations may be safely done "modulo integrals involving two or one log factor". But an illuminating, short way to show the above formula for $J$ would be wonderful.

Motivation: The above relation appeared as i tried to solve the integral posted at the above link:

Calculate $\displaystyle\int_0^{2\pi} x^2\; \cos x \cdot\operatorname{Li}_2(\cos x)\; dx$ .

After several simplifications and substitutions, it turns out that the above integral is related to integrals of the shape

$\int_0^1\log t\; R(t)\; dt$ ,  and

$\int_0^1\arctan t\cdot \log t\; R(t)\; dt$ ,  and

$\int_0^1\arctan^2 t\cdot \log t\; R(t)\; dt$ ,

and "similar" expressions.

Here $R$ is in each case a (rather simple) rational function.
(The more log and/or arctangent factors, the higher the computational complexity.)

I could compute more or less algorithmically most of the the needed integrals to solve the linked problem, all of them but the integral
$$
K=\int_0^1\arctan^2 t\cdot\log t\cdot\frac2{1-t^2}\; dt\ ,
$$
which turned out to be very hard to attack with the methods of real analysis.
Computing this integral is more or less equivalent to computing $J$, and the question wants $J$ instead, since we have a "clean formula", so that some speculation about a "clever substitution" may be accepted.

My solution (for $K$) works in complex analysis, the first step is to write
$$
\int_0^1 =\int_0^i+\int_i^1\ ,
$$
then parametrize the first integral using a linear path, the second one using a path on the unit circle.

Some comments: I will say some more words, because the situation is rich in coincidences. Since a numerical evidence is the simplest and shortes way to present (instead of showing how to show), i will use this method to at least list the coincidences. Many equalities below are "equivalent" (modulo computation of integrals of lower complexity) to the formula for $J$.

First of all, a numerical experiment using pari/gp delivers some connection between $K$ and a "cousin" of $J$:

```<code>  ? 2 * intnum( t=0, 1, atan(t)^2 * log(t) / (1-t^2) )
  %88 = -0.357038604620289042902893412499686912781214141574556097366337
  ? real(intnum( t=0, I, (pi/4 - atan(t))^2 * log(t) / (1-t^2) ))
  %89 = -0.357038604620289042902893412499686912781214141574556097366337
  ? intnum( t=0, 1, (pi/4 - atan(t))^2 * log(t) / (1-t^2) )
  %90 = -0.357038604620289042902893412499686912781214141574556097366337
</code>```

In words:
$$
\begin{aligned}
K
&=
\int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt
\\
&=
\Re
\int_0^i\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt
\\
&=
\int_0^1\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt
\ .
\end{aligned}
$$
Note the integration margins. What happens if we take the integral on $[0,i]$ instead of $[0,1]$ in the $K$-integral? Numerically:

```<code>    ? 2 * real(intnum( t=0, i, atan(t)^2 * log(t) / (1-t^2) ))
    %98 = 1.52201704740628808181938019826101736327699352613570971392919
    ? pi^4/64
    %99 = 1.52201704740628808181938019826101736327699352613570971392919
</code>```

In words:
$$
\begin{aligned}
K^*
&:=
\Re\int_0^i\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt
\\
&=\frac 1{64}\pi^4
\\
&=
-\int_0^1\left(\frac \pi 4+\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt
\\
&=-J\ .
\end{aligned}
$$

(These observations were leading to the formula for $K$ in loc. cit. .)

One idea is to use partial integration in $J$ or $K$. Well, we have for $K$:
$$
\begin{aligned}
K
&=
\int_0^1\arctan^2 t\cdot\log t\;\left(-\log\frac {1-t}{1+t}\right)'\; dt
\\
&=
\underbrace{\int_0^1\arctan^2 t\cdot\frac 1t\cdot \log\frac {1-t}{1+t}\; dt}_{=2K\text{ (why?)}}
\\
&\qquad\qquad+
\underbrace{
\int_0^1
2\arctan t\cdot\frac 1{1+t^2}\cdot \log t\cdot \log\frac {1-t}{1+t}\; dt
}_{=-K\text{ (why?)}}
\ .
\end{aligned}
$$

Note that $\arctan$ is related to the logarithm (over $\Bbb C$), we have the relation
(around $0$)
$$
\arctan t=\frac 1{2i}\log\frac {1+it}{1-it}\ .
$$
The substitution $t=\frac{1-u}{1+u}$ and the formula for $\tan(\arctan 1-\arctan u)$ are giving:
$$
\begin{aligned}
K
&=
\int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt
\\
&=\int_0^1
\left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {1-u}{1+u}\cdot \frac {du}u\ .
\\
&=\int_1^\infty
\left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {u-1}{1+u}\cdot \frac {du}u\ .
\end{aligned}
$$
(Write $\log t=\frac 12\log t^2$ to have the same expression under the integral on $(0,1)$ and on $(1,\infty)$.)

Note the fact that the factor $\frac 2{1-t^2}$ is not "random". It is the right one to make things feasible. It is the derivative of $\displaystyle -\log\frac{1-t}{1+t}$, and plugging in $t=iu$ into $\displaystyle \log\frac{1-t}{1+t}$ leads to an expression related to $\arctan u$. And conversely, $\arctan(iu)$ is related to such a logarithmic expression in $u$.

答:
This not an answer

Considering that
$$\big[\tan ^{-1}(t)\big]^2=\sum_{n=1}^\infty (-1)^{n+1}\,a_n\,t^{2n}$$ where
$$a_n=\frac 1n\sum_{k=1}^n\frac 1 {2k-1}=\frac{H_{n-\frac{1}{2}}+2 \log (2)}{2 n}$$and using the fact that
$$\int_0^1\frac {t^{2n}}{1-t^2}\log(t)=-\frac{1}{4} \zeta \left(2,\frac{2n+1}{2}\right)$$
$$K=2\int_0^1\big[\tan ^{-1}(t)\big]^2\,\frac{\log (t)}{1-t^2}\,dt$$
$$K=\frac 14 \sum_{n=1}^\infty (-1)^n \frac{\zeta \left(2,n+\frac{1}{2}\right) \left(H_{n-\frac{1}{2}}+2 \log
   (2)\right)}{n}$$ which converges extremely slowly.

ContentLicense: CC BY-SA 4.0


问:
Projective scheme over a ring
I’m reading Qing Liu’s Algebraic Geometry and Arithmetic Curves. I’m confusing  my double definitions of projective morphism over a ring $A$.
In definition 2.3.42, projective scheme over $A$ is an $A$-scheme that is isomorphic to a closed subscheme of $\operatorname{Proj} A[x_0,...,x_n]$.
In definition 3.1.11, $X \rightarrow \operatorname{Spec A}$ is projective if it factors in to a closed immersion $X \rightarrow \operatorname{Proj} A[x_0,...,x_n]$ followed by the canonical morphism $\operatorname{Proj} A[x_0,...,x_n] \rightarrow \operatorname{Spec} A$.
So I think A scheme $X$ can be said projective in definition 3.1.11 sense.
So I think this two definitions should be coincide.

I can understand if $X$ is projective in the sense of definition 3.1.11, $X$ is projective in the sense of 2.3.42.
But I can’t understand the converse, when $X$ is projective in the sense of 2.3.42, why does $f:X \rightarrow \operatorname{Spec} A$ have a decomposition two morphisms.

答: Given a scheme $S$, an $S$-scheme is a scheme $X$ together with a morphism $X \to S$ (often referred to as the structure map), and a morphism of $S$-schemes is a morphism of schemes $X \to Y$ such that the structure map $X \to S$ factors as $X \to Y \to S$. So, phrased in this language, the second definition literally just says that $X$ can be realized as a closed $A$-subscheme of some projective space over $A$. The factorization of $X \to \operatorname{Spec} A$ as a composition $X \to \operatorname{Proj} A[x_0, \dots, x_n] \to \operatorname{Spec} A$ is encoded in the fact that a projective scheme over $A$ is (by definition) an $A$-scheme, and the isomorphism to a closed subscheme of projective space is a morphism of $A$-schemes.

Title: Projective scheme over a ring


问:
"composition" of two characteristic functions
Suppose $\varphi_Y$ is a characteristic function. Let $p \in (0,1),\ q = 1-p$.

Show $\bar{\varphi}(t) = \frac{p}{1-q \varphi_Y(t)}$ is a characteristic function.

My thoughts:

Let X ~ Ge(p).

Then $\varphi_X(t) = \frac{p}{1-q \exp(it)}$

$\varphi_X(\frac{\ln(\varphi_Y(t))}{i}) = \frac{p}{1-q \varphi_Y(t)} = \bar{\varphi}(t)$

$\bar{\varphi}(t) = \varphi_X(\frac{\ln(\varphi_Y(t))}{i}) = E(\exp( i(\frac{\ln(\varphi_Y(t))}{i})X)) = E(\varphi_Y(t)X) = E(X) \varphi_Y(t) = \frac{q\varphi_Y(t)}{p}$

I don't think this can be right, but I also don't know what I did wrong. Can someone point out what I did wrong? Thanks!

答:
First of all logarithms of characteristic functions are not well defined. Logarithms in the complex plane are quite complicated. Even if we assume  that $\phi_Y(t)$ is positive for each $t$ your calculations are wrong.

$e^{(\ln (\phi_Y(t)) (X)}=((\phi_Y(t))^{X}$, not $(\phi_Y(t)) X$.


问:
On normals and parabolas
Question

Let $P(p^2, 2p)$, where $p > 0$, be a variable point on the curve whose equation is $y^2 = 4x$. The normal at P meets the curve again at the point $Q(q^2, 2q)$.

(i) Express $q$ in terms of $p$.

(ii) Prove that the length, $L$, of the line segment $PQ$ is given by $$L^2 = \frac {k(1 + p^2)^3} {p^4}\ ,$$ where $k$ is a constant to be determined.

My answer

Since $P$ is a variable point, it could be any point right? In particular, it can be $Q$, so $q = p$. However, this is clearly wrong, because I am unable to proceed and prove (ii).

Any hints/suggestions as to why my answer to (i) is wrong and what the correct answer is would be greatly appreciated :)

Edit

Following some help, I know that $q = -\sqrt {p^2 + 4}$.

\begin{align}
L^2 & =
(2p - 2q)^2 + (p^2 - q^2)^2
\\[5 mm] & =
4p^2 - 8pq + 4q^2 + 16
\\[5 mm] & =
8p^2 + 32 - 8pq
\\[5 mm] & =
8p^2 + 32 + 8p\sqrt {p^2 + 4}
\end{align}

I am not sure how to proceed from here to prove (ii). In particular, I notice that the required form does not have a square root, so how should I get rid of the square root in my answer?

答:
First, let us find the implicit derivative of the curve $y^{2} = 4x$. Differentiating both sides, we have:

$$2yy' = 4$$

$$y' = \frac{2}{y}$$

Thus, the slope of the tangent line at the point $(p^{2}, 2p)$ is $\frac{1}{p}$, which means the normal is $-p$. Then, using point-slope form to find the equation of this normal line:

$$y - 2p = -p(x - p^{2})$$

$$y = -px + p^{3} + 2p$$

Now, we can substitute into the given equation for the curve to find the points of intersection:

$$(-px +p^{3} + 2p)^{2} = 4x$$

$$p^{2}x^{2} - (2p^{4} + 4p^{2} + 4)x +(p^{3} + 2p)^{2}= 0$$

Dividing both sides by $p^{2}$:

$$x^{2} - \frac{2p^{4} + 4p^{2} + 4}{p^{2}}x + (p^{2} + 2)^{2}$$

Because the normal intersects the curve a $(p^{2}, 2p)$, we know that $p^{2}$ must be a solution. Using this and the fact that the product of the roots is $(p^{2} + 2)^{2}$, the other solution is $x = \big(\frac{p^{2} + 2}{p}\big)^{2}$.

Now, since it is given that $p > 0$, we note that the $y$-coordinate of the second intersection point must be negative. (Think of this as a line intersecting both branches of a sideways parabola, one branch above the $x$-axis and one part below). Then, we must have:

$$\boxed{(q^{2}, 2q) = \bigg(\bigg(\frac{p^{2} + 2}{p}\bigg)^{2}, -\frac{2p^{2} + 4}{p}\bigg)\text{ and }q = -\frac{2p^{2} + 4}{p}}$$

Now, using the distance formula:

$$PQ = L = \sqrt{\bigg(p^{2} - \bigg(\frac{p^{2} + 2}{p}\bigg)^{2}\bigg)^{2} + \bigg(2p + \frac{2p^{2} + 4}{p}\bigg)^{2}}$$

$$L^{2} = \bigg(p^{2} - \bigg(\frac{p^{2} + 2}{p}\bigg)^{2}\bigg)^{2} + \bigg(2p + \frac{2p^{2} + 4}{p}\bigg)^{2}$$

Multiplying out by $p^{4}$ to clear denominators:

$$p^{4}L^{2} = (p^{4} - (p^{2} + 2)^{2})^{2} + (2p^{3} + 2p^{3} + 4p)^{2}$$

$$p^{4}L^{2} = (-4p^{2} - 4)^{2} + (4p^{3} + 4p)^{2}$$

$$p^{4}L^{2} = (-4)^{2}(p^{2} + 1)^{2} + (4p)^{2}(p^{2} + 1)^{2}$$

$$p^{4}L^{2} = (16 + 16p^{2})(p^{2} + 1)^{2}$$

$$p^{4}L^{2} = 16(p^{2}+ 1)^{3}$$

$$\boxed{L^{2} = \frac{16(p^{2} + 1)^{3}}{p^{4}}, k = 16}$$

Title: On normals and parabolas


问:
Evaluate $\lim_{n\to\infty} \prod_{k=1}^n \frac{2k}{2k-1}\int_{-1}^{\infty} \frac{{\left(\cos{x}\right)}^{2n}}{2^x} \; dx$
Problem 9 in the JHMT 2013 Calculus Test asks to evaluate
$$\lim_{n\to\infty} \prod_{k=1}^n \frac{2k}{2k-1}\int_{-1}^{\infty} \frac{{\left(\cos{x}\right)}^{2n}}{2^x} \; dx$$
The answer is $\pi\cdot 2^\pi /(2^{\pi}-1)$. How can I show this?  I know that the infinite product diverges and the limit cannot be moved into the integral, but I don't know what to do.  Maybe I can represent the integral as a summation?

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n \to\infty}\,\,
\prod_{k = 1}^{n}{2k \over 2k - 1}
\int_{-1}^{\infty}{\cos^{2n}\pars{x} \over 2^x}
\,\dd x = {2^\pi \over 2^{\pi} - 1}\,\pi} \approx
3.5431:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\prod_{k = 1}^{n}{2k \over 2k - 1}} =
\prod_{k = 1}^{n}{k \over k - 1/2} =
{n! \over \pars{1/2}^{\overline{n}}}
\\[5mm] = &\
{n! \over \Gamma\pars{1/2 + n}/\Gamma\pars{1/2}} =
\root{\pi}{n! \over \pars{n - 1/2}!}
\\[5mm] &\
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\root{\pi}{\root{2\pi}n^{n + 1/2}\,\,\expo{-n} \over \root{2\pi}\pars{n - 1/2}^{n}\expo{-n + 1/2}}
\\[5mm] = &\
\root{\pi}{n^{n + 1/2} \over
n^{n}\,\bracks{1 - \pars{1/2}/n}^{\,n}\expo{1/2}}
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\bbx{\root{\pi}n^{1/2}}\label{1}\tag{1} \\ &
\end{align}

\begin{align}
&\bbox[5px,#ffd]{
\int_{-1}^{\infty}{\cos^{2n}\pars{x} \over 2^x}
\,\dd x}
\\[5mm] = &\
\int_{-1}^{0}2^{-x}\cos^{2n}\pars{x}\,\dd x +
\sum_{k = 0}^{\infty}\,\,\int_{k\pi}^{k\pi + \pi}2^{-x}
\cos^{2n}\pars{x}\,\dd x
\\[5mm] = &\
\int_{0}^{1}2^{x}\cos^{2n}\pars{x}\,\dd x +
\sum_{k = 0}^{\infty}\,\,2^{-k\pi}
\int_{0}^{\pi}2^{-x}\cos^{2n}\pars{x}\,\dd x
\\[5mm] = &\
\underbrace{\int_{0}^{1}2^{x}\cos^{2n}\pars{x}\,\dd x}
_{\ds{\cal I}}\ +\
{1 \over 1 - 2^{-\pi}}\
\underbrace{\int_{0}^{\pi}2^{-x}\cos^{2n}\pars{x}
\,\dd x}_{\ds{\cal J}}
\end{align}

In order to evaluate the asymptotic integrals; I'll use, hereafter, the Laplace's Method:
\begin{align}
{\cal I} & \equiv
\int_{0}^{1}2^{x}\cos^{2n}\pars{x}\,\dd x =
\int_{0}^{1}2^{x}\exp\pars{2n\ln\pars{\cos\pars{x}}}
\,\dd x
\\[5mm] &\
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\int_{0}^{\infty}\exp\pars{-nx^{2}}\dd x =
{\root{\pi} \over 2}\,n^{-1/2}
\end{align}
Similarly,
\begin{align}
{\cal J} & \equiv
\int_{0}^{\pi}2^{-x}\cos^{2n}\pars{x}\,\dd x =
\int_{-\pi/2}^{\pi/2}2^{-x - \pi/2}\,\,
\sin^{2n}\pars{x}\,\dd x
\\[5mm] = &\
2^{-\pi/2}\int_{0}^{\pi/2}\pars{2^{-x} + 2^{x}}
\sin^{2n}\pars{x}\,\dd x
\\[5mm] = &\
2^{-\pi/2}\int_{0}^{\pi/2}\pars{2^{-\pi/2 + x}\ +\ 2^{\pi/2 - x}}
\cos^{2n}\pars{x}\,\dd x
\\[5mm] &\
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\pars{2^{-\pi} + 1}\int_{0}^{\infty}
\exp\pars{-nx^{2}}\dd x
\\[5mm] = &\
\pars{2^{-\pi} + 1}{\root{\pi} \over 2}\,n^{-1/2}
\end{align}
Then,
\begin{align}
&\bbox[5px,#ffd]{
\int_{-1}^{\infty}{\cos^{2n}\pars{x} \over 2^x}
\,\dd x}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
\overbrace{{\root{\pi} \over 2}\,n^{-1/2}}
^{\ds{\sim {\cal I}\ \mrm{as}\ n\ \to\ \infty}}
\\[2mm] + &\
{1 \over 1 - 2^{-\pi}}\
\underbrace{\pars{2^{-\pi} + 1}\,
{\root{\pi} \over 2}\,n^{-1/2}}
_{\ds{\sim {\cal J}\ \mrm{as}\ n\ \to\ \infty}}
\\[5mm] = &\
\bbx{{\root{\pi} \over n^{1/2}}
\,{2^{\pi} \over 2^{\pi} - 1}} \label{2}\tag{2} \\ &
\end{align}

With (\ref{1}) and (\ref{2}):
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to\infty}\,\,
\prod_{k = 1}^{n}{2k \over 2k - 1}
\int_{-1}^{\infty}{\cos^{2n}\pars{x} \over 2^x}
\,\dd x} =
\bbx{{2^\pi \over 2^{\pi} - 1}\,\pi}
\\[5mm] \approx &\ 3.5431
\end{align}


问:
When defining a function, is it more correct to use $f(x)=x^2$ or $y=x^2$?
In school I had a teacher that would use the following notation (seemingly) interchangeably:

$f(x)$ and $y$ and y$(x)$. What are their differences? From what I put together, $f(x)$ is a function named "f" that takes the argument x. The notation $y$ is by convention used to refer to the dependent variable. The notation $y(x)$ is the same as $y$ except it makes clear $y$ depends on elements from the set $x$. So if you're defining a function, wouldn't it be most correct to use $f(x)=x^2$ instead of $y=x^2$ ? Also of what use is arrow notation if $f(x)=x^2$ is the same as $f: x \mapsto x^2$ ?

I took a calculus course where, as a side comment, the teacher said sometimes mathematicians omit parts and instead of writing $\int f(x) dx$ they just write $\int f$ because the variable can be inferred from context. In a sense notation isn't important, it's always the concept being conveyed that is. Am I looking at it correctly?

答:
None of the notations you mentioned is really more correct. They're different ways of doing it, sometimes with certain advantages, as for instance shorthand.

For instance, there's Leibniz's notation for the derivative, $\dfrac{\rm dy}{\rm dx}$, suggestive as it is for cancellation in the chain rule, among other things, I guess; as compared with Newton's $f'(x)$.

The arrow convention is somewhat suggestive, of course.

A lot of these things can be a matter of preference.  Remember, as it said outside Robert D. Edwards' office at UCLA: Mathematics: the Ultimate Art.


问:
Is it possible to define maximum elements in strict partial orders without using equality?
This is similar to a question I asked a few months ago. It is possible to define maximum elements in reflexive partial orders $\leq$, without using equality: $(\forall x)x \leq c$. I wonder if it is possible to do so in strict partial orders. To make my question precise, suppose we are given a strict partial order $<$, and a constant $c$. Is it possible to define a first-order formula that does not mention equality, that holds precisely when $c$ is the maximum element? My intuition tells me it is not.

答:
As a quick point of clarification: I'll write "the greatest" instead of "the maximal" since maximality usually refers to the weaker notion of having nothing strictly greater. That notion is of course trivially $\mathsf{FOL_{w/o=}}(<)$-definable.

Your intuition is correct (but see below). We can prove this using a technique which has been used to answer other questions of yours, including the one linked to in the OP: namely, that in $\mathsf{FOL_{w/o=}}$ surjections which preserve and reflect atomic formulas preserve arbitrary-complexity formulas.$^*$

This is also true of $\mathsf{FOL}$ of course, but in full $\mathsf{FOL}$ we have the atomic formula "$x=y$" reflection of which forces injectivity. So in $\mathsf{FOL}$ this reduces to the statement that isomorphisms are elementary embeddings; it's only when we drop equality that we get to consider more interesting maps than bijections.

Specifically, let $D_1,D_2$ be the $1$- and $2$-element discrete strict partial orders respectively, and let $f$ be the unique map $D_2\rightarrow D_1$. In the sense of $\mathsf{FOL_{w/o=}}(<)$, the map $f$ is a surjection which preserves and reflects all atomic formulas, so for every $\overline{a}\in D_2$ and every $\mathsf{FOL_{w/o=}}(<)$-formula $\varphi$ we have $$D_2\models\varphi(\overline{a})\quad\iff\quad D_1\models\varphi(f(\overline{a})).$$ Since $D_1$ has a greatest element but $D_2$ doesn't, this means that the property of being a greatest element cannot be expressed by a $\mathsf{FOL_{w/o=}}(<)$-formula.

OK, what if we restrict attention to strict partial orders which do have a greatest element; can we now find that greatest element in $\mathsf{FOL_{w/o=}}$?

It turns out the answer is yes. Consider the formula $$\gamma(x)\equiv\forall y,z(y<z\rightarrow y<x).$$ It's easy to see that if $D$ is a strict partial order with a greatest element $a$ then $\gamma^D=\{a\}$.

However, note that this $\gamma$ behaves weirdly in strict partial orders without greatest elements. For example, let $L_{1,2}$ be the partial order with three elements $a,b,c$ such that $b<c$ and no other comparison holds. Then $\gamma^{L_{1,2}}=\{c\}$ even though $c$ is not the greatest element of $L_{1,2}$. Basically, in an arbitrary strict partial order $\gamma$ merely pins down the set of elements which are strictly above all non-maximal elements!

So the more nuanced answer is:

There is an $\mathsf{FOL_{w/o=}}(<)$-formula which identifies the greatest element of a strict partial order when such an element exists; however, there is no $\mathsf{FOL_{w/o=}}(<)$-sentence  which holds in exactly the strict partial orders which have a greatest element, and so consequently the aforementioned formula must behave weirdly on some strict partial orders without a greatest element.

$^*$Actually, the "right" statement of this fact is a bit different.

Suppose $\mathcal{A},\mathcal{B}$ are structures in the same language with domains $A,B$, and $\mathbb{F}\subseteq A\times B$ is such that:

$\mathbb{F}$ is total: for all $a\in A$ there is some $b\in B$ such that $a\mathbb{F}b$.

$\mathbb{F}$ is surjective: for all $b\in B$ there is some $a\in A$ such that $a\mathbb{F}b$.

$\mathbb{F}$ preserves and reflects atomic formulas: for every atomic formula $\varphi(x_1,...,x_n)$ and every $a_1,...,a_n\in A$ and $b_1,...,b_n\in B$ with $a_i\mathbb{F}b_i$ for all $1\le i\le n$, we have $$\mathcal{A}\models\varphi(a_1,...,a_n)\quad\iff\quad\mathcal{B}\models\varphi(b_1,...,b_n).$$

Then $\mathbb{F}$ preserves and reflects all $\mathsf{FOL_{w/o=}}$-formulas: if $\psi(x_1,...,x_n)$ is an arbitrary $\mathsf{FOL_{w/o=}}$-formula, $a_1,...,a_n\in A$, and $b_1,...,b_n\in B$ with $a_i\mathbb{F}b_i$ for all $1\le i\le n$, we have $$\mathcal{A}\models\psi(a_1,...,a_n)\quad\iff\quad\mathcal{B}\models\psi(b_1,...,b_n).$$ In particular, if such an $\mathbb{F}$ exists then $\mathcal{A}\equiv_\mathsf{w/o=}\mathcal{B}$.

Note that this version of the statement is symmetric and compositional, and arguably such relations give the right notion of "equality-free isomorphism"; that said, it's a bit harder to think about at first.


问:
Inconsistent System of Linear Equations
Let $A ∈ M_{n\times n}(F)$. Suppose that the system of linear equations $AX = B$
has more than one solution. Prove that there is a column $C ∈ F^n$ such that
the system of linear equations $AX = C$ is inconsistent.

I thought it'd be clever to use this (underlined by my pencil):

答:
Hints: What does the premise imply about the kernel of $A$? Also:

$$\dim(\text{Im } A)+\dim(\ker A)=n.$$

ContentLicense: CC BY-SA 3.0

Title: Inconsistent System of Linear Equations


答:
In mathematics exists something called 'abuse of notation', where you omit something when you know everyone reading it knows what you are meaning (or it is just impossible to have a crystal clear notation as in many diagramms in homological algebra), or we use the same notation for different objects. Again normally everyone knows what is meant, but it can be inconvenient for students new in the subject.

At the question with the function notation.
Basically you see it correctly.

There are different notations which are used.
For a function $f:X\to Y$ you can write $x\mapsto f(x)$, or $f(x)=y$ and this would mean exactly the same.

When first introduced to functions one would note it like $y=x$ or $y=x^2$. This notation has it flaws.

First of all the f(x)-notation can contain more information.
When we have $f(x)=x^2$ then $f(-1)=1$ and $f(1)=1$. So this notation gives us the point we are talking about (1|1) or (-1|1) while the $y=\dotso$ notation does not remember the point $x$, and only its value.
When first learning functions we create these value tables, where we keep track of $x$ and $y$.

The notation $y=\dotso$ should be viewed from a purley didactic point of view.
Making clear how to get an y-value and so on.
That these value tables are important, and how to scetch a graph.
As many problems at this level have to be solved by putting a given point in the equation.
Like calculate $m$ and $b$ in $y=mx+b$, when you have points given.
From the coordinate names it is clear where to put them.

Later when the student already is familiar with this, we can move on to a more fitting notation, which might be difficult for younger people.
When we care about other questions. What is the derivative, how to get extremum and so on.
However, students that learn the f(x) notation often do not really understand it.
They know what to do, but they would talk about the function f(x) not understanding that this is supposed to note a value on the y-axis.

Also do not treat this equation correctly.
One would see calculations like

$$f(x)=2x^2+4x+2 |:2$$

$$f(x)=x^2+2x+1 $$

when correct would be $\frac{f(x)}{2}=x^2+2x+1$, what comes from that most of the time the questions these equations have to answer is when $f(x)=0$, so it does not matter.

Just my two cents.

Title: When defining a function, is it more correct to use $f(x)=x^2$ or $y=x^2$?

Body:
In school I had a teacher that would use the following notation (seemingly) interchangeably:

$f(x)$ and $y$ and y$(x)$. What are their differences? From what I put together, $f(x)$ is a function named "f" that takes the argument x. The notation $y$ is by convention used to refer to the dependent variable. The notation $y(x)$ is the same as $y$ except it makes clear $y$ depends on elements from the set $x$. So if you're defining a function, wouldn't it be most correct to use $f(x)=x^2$ instead of $y=x^2$ ? Also of what use is arrow notation if $f(x)=x^2$ is the same as $f: x \mapsto x^2$ ?

I took a calculus course where, as a side comment, the teacher said sometimes mathematicians omit parts and instead of writing $\int f(x) dx$ they just write $\int f$ because the variable can be inferred from context. In a sense notation isn't important, it's always the concept being conveyed that is. Am I looking at it correctly?


问:
What is the Probable Number of Occurrences when the Odds Change at a Set Rate over Time?
Odd Saturday night question:

At time = 0s, the probability of an event is .25.

At time = 100s, the probability of an event is 1.

The probability increase is linear, so the equation for the probability as a function of time is .0075*t + .25

Each second a roll is made to see if there has been a success or not. Successes are noted, and time marches forward.

I would like to determine the probable number of successes between time 0 to 100s, but I can't figure out how to do so without calculating the odds for each event and doing some kind of an average of the sum. If that's even the right way to go.

Any help would be appreciated!

Edit: Maybe, because the growth is linear, I could say that the average probability of success would be the average of the probability at time 100 and time 0? So, in this case .75/2?

If that's correct, then the number of success expected would be 101 * .75/2 = 37.875?

Or am I really off?

答:
Ave probability is 1/2 (1 + 0.25) = .625

expected # successes is $101 \times 0.625.$


问:
How can I solve this matrix?
I have a set of RGB colour values detected by a camera $C_{i_{RGB}}$ which are to be described by the following:

$C_{i_{RGB}} = X F_{i_{rgb}}$

where $F_{i_{rgb}}$ is the component incident light at the detector, where the components $rgb$ are fixed at narrow bands and $X$, an $(n\times 3)$ matrix, is a description of the colour band overlap (cross talk if you will) for specific frequencies.  In a simplified RGB scheme, $X$ would be described:

$\begin{align}
\mathsf X &= \begin{bmatrix}
  1 & R_{GR} & R_{BR} \\
  R_{RG} & 1 & R_{BG} \\
  R_{RB} & R_{G} & 1
\end{bmatrix}\end{align}$

That is, the diagonal entries of $X$ are $1$'s, and the off-diagonals are rational.

How can I solve for the values of $X$ only knowing $C_{i_{RGB}}$? Eventually, given $C_i$ I need to find the corresponding $F_i$ using the constant $X$.

Eventually added colour bands (band passes) will be added which will turn the above relation into something like this (and I fear the case where I have 5, 6, and 7 band passes to deal with):

$\begin{bmatrix}
  C_{i_R} \\
  C_{i_G} \\
  C_{i_B}
\end{bmatrix} = \begin{align}\begin{bmatrix}
  X_{aR} & X_{bR} & X_{cR} & X_{dR} & X_{eR} & X_{fR} \\
  X_{aG} & X_{bG} & X_{cG} & X_{dG} & X_{eG} & X_{fG} \\
  X_{aB} & X_{bB} & X_{cB} & X_{dB} & X_{eB} & X_{fB}
\end{bmatrix}\end{align}\begin{bmatrix}
  F_{i_a} \\
  F_{i_b} \\
  F_{i_c} \\
  F_{i_d} \\
  F_{i_e} \\
  F_{i_f}
\end{bmatrix}$

This is for a graphics programming application and I lament that my first year algebra was 20 years ago.

答: If you don't know both $X$ and $F_i$ then you have an infinite number of solutions.

ContentLicense: CC BY-SA 2.5

Title: How can I solve this matrix?


问:
"Hilbert foresaw the possibility of negative solutions to some mathematical problems"
In My Collaboration with JULIA ROBINSON it is said

Hilbert foresaw the possibility of negative solutions to some mathematical problems

What evidence is there that Hilbert knew (around the time when he set the problems) or suspected that some decision problems (like resolution of Diophantine equations) can be impossible?

答:
See Hilbert's famous 1900 lecture:

http://aleph0.clarku.edu/~djoyce/hilbert/problems.html

Occasionally it happens that we seek the solution under insufficient hypotheses or in an incorrect sense, and for this reason do not succeed. The problem then arises: to show the impossibility of the solution under the given hypotheses, or in the sense contemplated. Such proofs of impossibility were effected by the ancients, for instance when they showed that the ratio of the hypotenuse to the side of an isosceles right triangle is irrational. In later mathematics, the question as to the impossibility of certain solutions plays a preeminent part, and we perceive in this way that old and difficult problems, such as the proof of the axiom of parallels, the squaring of the circle, or the solution of equations of the fifth degree by radicals have finally found fully satisfactory and rigorous solutions, although in another sense than that originally intended. It is probably this important fact along with other philosophical reasons that gives rise to the conviction (which every mathematician shares, but which no one has as yet supported by a proof) that every definite mathematical problem must necessarily be susceptible of an exact settlement, either in the form of an actual answer to the question asked, or by the proof of the impossibility of its solution and therewith the necessary failure of all attempts.

I don't think it gets more explicit than that.

That being said, I think he believed that his own problems (the Tenth problem, the Entscheidungsproblem...) can be solved - at least, I have never seen evidence to the contrary (not that I know much about these things...)


问:
Is iteration process through equal sign possible?
Is iteration process through equal sign possible?

for an example: repeatedly pressing on the equal button of calculator

Repeatedly pressing on the equal button of calculator until full converging the Answer.

this is (Ahmad Reza Estakhr's) symbol of iteration of equal sign $\Re_{i=1}^{\infty}=$ which means repeatedly pressing on equal sign.

答:
The answer is: yes, probably, subject to a few caveats. Here are two I can think of:

it must be running an algorithm that will actually converge (it's sometimes rather hard to prove that an algorithm converges with a given input, e.g. Newton-Raphson with a given polynomial and 'starting value'),

it must be performing an algorithm that isn't very sensitive to rounding errors (because calculators will make them!).

But there are probably more.


问:
Scope in $\forall x P(x) \rightarrow Q(x)$
For $ \exists x P(x) \wedge Q(\color{blue}x) $

The scope of the quantifier, is just the predicate $P(x)$, thus the occurrence of x in blue is free.

I am unsure exactly the scope of the quantifier in this formula:

$ \forall x P(x) \rightarrow Q(\color{red}x) $

Based off the previous formula, I think the occurrence of the x in red is free but I'm not sure.

Is the occurrence of $\color{red}{x}$ free or bound? Would this vary based off convention?

答:
For the first one you can think of it as the same as $\exists \color{red}x(P(\color{red}x) \land Q(y))$ because they're under different domains.

The second one I am guessing that they are both under different domains. What makes me say this is that quantifiers are usually before a set of parentheses. $$(1) \space \forall \color{red}x(P(\color{red}x) \rightarrow Q(\color{red}x)) \text{ vs. } (2)\space \forall \color{red}x(P(\color{red}x) \rightarrow Q(\color{blue}y))$$

I think $(2)$ is your answer.


问:
Continuous-time Markov Chain for wrong seats on a plane
I found this variation of the wrong seats on a plane problem.

A passenger plane with $N$ numbered seats is about to take off; $N-1$
seats have already been taken, and now the last passenger enters the
cabin. The first $N-1$ passengers were advised by the crew, rather
imprudently, to take their seats completely at random, but the last
passenger is determined to sit in the place indicated on his ticket.
If his place is free, he takes it, and the plane is ready to fly.
However, if his seat is taken, he insists that the occupier vacates
it. In this case the occupier decides to follow the same rule: if the
free seat is his, he takes it, otherwise he insists on his place being
vacated. The same policy is then adopted by the next unfortunate
passenger, and so on. Each move takes a random time which is
exponentially distributed with mean $\mu^{-1}$. What is the expected
duration of the plane delay caused by these displacements?

What is the state space for this CTMC? In the classic problem of 100 seats and only the first passenger being unruly, how I imagined it is a "linked list" which closed on itself when a person finally sits in the correct seat. I am unsure of how to extend this idea to this CTMC problem. Would appreciate some insight!

答:
We say that a passenger is "active" between the time they were requested to move until the time they got to their seat, (and then: either a new passenger is active or none is active, according to whether our passenger's seat was taken or not).

Let ${\bf X}=(X_t:t\ge 0)$ count the number of passengers that were activated by time $t$ if there is an active passenger at time $t$, and zero otherwise. Since no more than $N-1$ passengers can be activated, the state space is $0,\dots, N-1$.

Also, at time $0$, the chain is either equal to $0$ with probability $1/N$ (see calculation below) of equal to $1$ with probability $(N-1)/N$.

Now for the dynamics of the chain.

Clearly, if $X_t=0$, then $X_{t+s}=0$ for all $s\ge 0$ (none is active now then none will ever be activated).  That is $0$ is absorbing.  If $X_{t}=j$ for $j=1,\dots,N-1$,  then at time $t$ we are looking at the $j$-th customer to be asked to move. Besides this customer, we have:

the last customer to board: already seated in their place.

$j-1$ customers previously activated, also seated in their places.

$N-1-j$ customers, uniformly seated in the $N-j$ remaining seats.

There are $(N-j) \times \cdots\times  2$ ways to seat the last group of $N-1-j$ customers in the $N-j$ seats. There are $(N-j-1)!$ ways to seat them all in seats excluding the one belonging to our currently active customer. Therefore, the probability that the currently active customer's seat is available is $1/(N-j)$.

Since we assumed exponential rate $\mu$ from every state, the transitions (other states) from $j$ are to:

$0$ with exponential rate $\mu / (N-j)$; or

$j+1$ with rate $\mu * (N-j-1)/(N-j)$

Now for the expected delay. Call it $T$. It is simply the hitting time of state $0$ by the chain ${\bf X}$.

Let $M$ be the number of passengers to be ever activated. Observe that  $P(M\ge  1) = \frac{N-1}{N}$ and $P(M\ge j+1| M\ge j)= \frac{N-j-1}{N-j}$. Therefore by induction,
$P(M \ge j+1) = \frac{N-j-1}{N}$, $j=0,1,2,\dots,N-1$.

In other words, $M$ is uniform on $\{0,\dots,N-1\}$. As a result $E[M] =\frac{N-1}{2}$.

Since the time until absorption is the sum of $M$ independent exponential RVs with rate $\mu$, it follows that the expected time until absorption is $\mu^{-1} E[M]$, that is

$$E[T]  = \mu^{-1}\frac{N-1}{2}.$$


问:
Free group of a free group.
Let $G=\frac{F}{R}$ be the presentation of a group $G$. Clearly here $F$ is free group. Can we obtain a free group $\mathbb{F}$ of the group $F$. Is $\mathbb{F}\cong F?$ What will be the presentation of $\mathbb{F}.$ actually I am studying the presentation of direct product of groups Group Presentation of the Direct Product.. Here free product of two groups is used before, that's why I am thinking about this.

In simple words my question is to obtain a free presentation of  a free group $F$.

答:
Yes, you can make a free group $\mathbb{F}$ from any set, including the underlying set of a (possibly free) group $F$.

That said, $\mathbb{F} \not\cong F$. For example, suppose $F$ was the free group on a set $\{a, b\}$, then $a$ and $a^{-1}$ are distinct elements of $F$, which happen to multiply to give the identity. But, when generating $\Bbb{F}$, each distinct element of $F$ becomes an unrelated generator of $\Bbb{F}$. So, in $\Bbb{F}$, $a$ and $a^{-1}$ are interpreted as strings of length $1$, with the two different symbols $a$ and $a^{-1}$. Making the free group $\Bbb{F}$ ignores the former relationship between these elements of $F$, and so $aa^{-1}$ becomes an irreducible word in $\Bbb{F}$.

In a similar manner, $a^2$ and $a$ are also distinct elements of $F$, and as such, $a^2a$, $aa^2$, and $a^3$ are three distinct elements of $\Bbb{F}$.

Title: Free group of a free group.


问:
Complex analysis for real valued integrals.
I'm asked to compute

$$\int_{-\infty}^{\infty}\frac{\sin(ax)}{x(1+x^4)}dx$$

To do this, I consider the function

$$f(z)=\frac{e^{aiz}}{z(1+z^4)}$$

and the contour composed by the segment on the real positive line going from $\epsilon$ to $R$, the half-upper circumference, the segment from  $-R$ to $-\epsilon$ and the half-lower circumference around zero.
When the radius of the small circumference tends to zero, the border of the contour contains a pole of the function, $0$ in particular. What should I do in this case?

Can you please show me how to compute this integral?
Thanks in advance.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{p_{1} \equiv \expo{\pi\ic/4}}$ and
$\ds{p_{2} \equiv \expo{3\pi\ic/4}}$ are the integrand poles in the complex plane upper half. Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}
{\sin\pars{ax} \over x\pars{1 + x^{4}}}\,\dd x} =
\on{sgn}\pars{a}\int_{-\infty}^{\infty}
{\sin\pars{\verts{a}x} \over x\pars{1 + x^{4}}}\,\dd x
\\[5mm] = &\
\on{sgn}\pars{a}\,\Im\int_{-\infty}^{\infty}
{\expo{\ic\verts{a}x} - 1 \over x\pars{1 + x^{4}}}
\,\dd x
\\[5mm] = &\
\on{sgn}\pars{a}\,\Im\braces{%
2\pi\ic\sum_{n = 1}^{2}
\on{Res}\pars{{\expo{\ic\verts{a}x} - 1 \over x\bracks{1 + x^{4}}}}_{\ x\ =\ p_{k}}\,\,}
\\[5mm] = &\
2\pi\on{sgn}\pars{a}\,\Re\bracks{%
\sum_{n = 1}^{2}
{\expo{\ic\verts{a}p_{k}} - 1 \over p_{k}
\pars{4p_{k}^{3}}}}
\\[5mm] = &\
-\,{\pi \over 2}\on{sgn}\pars{a}\,\Re
\sum_{n = 1}^{2}
\pars{\expo{\ic\verts{a}p_{k}} - 1}
\\[5mm] = &\
{\pi \over 2}\on{sgn}\pars{a}
\left[\vphantom{\Large A}2\right.
\\[2mm] &\ \!\!\!\!\!\left.-\Re\pars{%
\expo{\ic\verts{a}\root{2}/2 - \verts{a}\root{2}/2} +
\expo{-\ic\verts{a}\root{2}/2 - \verts{a}\root{2}/2}
\,\,\,}\right]
\\[5mm] = &\
{\pi \over 2}\on{sgn}\pars{a}\ \times
\\[2mm] &\ \bracks{
2 - 2\exp\pars{-\,{\root{2}\verts{a} \over 2}}
\cos\pars{\root{2}\verts{a} \over 2}}
\\[5mm] = &\
\pi\on{sgn}\pars{a}\ \times
\\[2mm] &\
\bracks{
1 - \exp\pars{-\,{\root{2}\verts{a} \over 2}}
\cos\pars{\root{2}a \over 2}}
\end{align}


问:
Why does the series diverge?
Why does the series $\sum_{n=1}^\infty(-1)^n n$ diverge?

Is it because the expanded series:

$ -1 + 2 -3 + 4 -5 + 6... $

is the difference between the infinite sum of the even and odd numbers which would be infinity?

答:
A series $\sum_{n=1}^\infty a_n$ converges if the sequence of partial sums, i.e. $\sum_{k=1}^n a_k$, i.e.
\begin{align*}
&a_1 \\
&a_1 + a_2 \\
&a_1 + a_2 + a_3 \\
&a_1 + a_2 + a_3 + a_4 \\
&\vdots
\end{align*}
converges in the sense of sequences. In the case of the given series, the partial sums come to:
\begin{align*}
-1 &= -1 \\
1 &= -1 + 2 \\
-2 &= -1 + 2 - 3 \\
2 &= -1 + 2 - 3 + 4 \\
-3 &= -1 + 2 - 3 + 4 - 5 \\
3 &= -1 + 2 - 3 + 4 - 5 + 6 \\
&\vdots
\end{align*}
One can prove the pattern emerging with induction if they wish (or indeed by simply grouping and summing pairs of consecutive terms), but what is clear is that the sequence of partial sums is oscillating, positive and negative, while getting larger and larger. This prevents the partial sums from converging, and hence prevents the series from converging.

Title: Why does the series diverge?


Title: Why does the series diverge?

Body:
Why does the series $\sum_{n=1}^\infty(-1)^n n$ diverge?

Is it because the expanded series:

$ -1 + 2 -3 + 4 -5 + 6... $

is the difference between the infinite sum of the even and odd numbers which would be infinity?

Body:
Method 1

Try grouping 2 terms at a time
$-1+2-3+4-5+...$

$=(-1+2)+(-3+4)+(-5+6)+...$

$=1+1+1+...$

$=∞$

Method 2

$-1=-1$

$-1+2=1$

$-1+2-3=-2$

$-1+2-3+4=2$

$-1+2-3+4-5=-3$

$-1+2-3+4-5+6=3$

As we can observe the sequence doesn't seem to(and will not) converge at a particular points on the number line


问:
Internal Bisector of a triangle
In $\Delta ABC$, the bisector of the angle A meets the side BC at D and circumscribed circle at E, then DE equals to

(A) $\frac{{{a^2}\cos \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(B) $\frac{{{a^2}\sec \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(C) $\frac{{{a^2}\sin \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(D) $\frac{{{a^2}\cos ec\frac{A}{2}}}{{2\left( {b + c} \right)}}$

My approach is as follow

Internal Bisector property

$\frac{{AB}}{{AC}} = \frac{{BC}}{{DC}} \Rightarrow \frac{c}{b} = \frac{t}{{a - c}}$

How do I proceed in order to find DE

答:
By Sine Theorem
Since $\angle ABC=\angle AEC$ and $\angle ACB=\angle AEB$ (angles are equal subtended same segment)

$$\frac{a}{\sin\angle A}=\frac{b}{\sin\angle AEC}=\frac{c}{\sin\angle AEB}=\frac{b+c}{\sin\angle AEC + \sin \angle AEC}$$

$$\sin\angle AEC + \sin \angle AEC=\frac{(b+c)\sin\angle A}{a}$$

$\angle BCE=\angle CBE=\angle BAE=\angle CAE=\frac{A}{2}$
(angles are equal subtended same segment and $AE$ bisects $\angle A$)

$$\frac{DE}{\sin\angle \frac {A}{2}}=\frac{DC}{\sin\angle AEC}=\frac{BD}{\sin\angle AEB}=\frac{BD+DC}{\sin\angle AEC + \sin \angle AEC}=\frac{a}{\sin\angle AEC + \sin \angle AEC}$$

$$\sin\angle AEC + \sin \angle AEC=\frac{a\sin\angle \frac {A}{2}}{DE}$$

$$\frac{(b+c)\sin\angle A}{a}=\frac{a\sin\angle \frac {A}{2}}{DE}$$

$$DE=\frac{a^2\sin\angle \frac {A}{2}}{2(b+c)\sin\angle \frac {A}{2}\cos\angle \frac {A}{2}}=\frac{a^2}{2(b+c)\cos\angle \frac {A}{2}}$$

Title: Internal Bisector of a triangle


问:
If $X$ is a random variable and $a$ is a positive number then does $Y = \max (a,X)$ is a legitimate random variable?
If $X$ is random variable with a given PDF and $a$ is a positive number then could $Y = \max (a,X)$ be considered as a legitimate random variable ?

Mathematica result look strange !

For example, when $Y = \max (3,X)$ and $X$ follow the exponential distribution with parameter $\lambda$, could $Y$ be considered as a random variable ?

If so then how can I interpret the result attach in the picture ?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Basically $\ds{\pars{~\mbox{besides notation details}~}}$, it's something like
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\on{P}\pars{x}
\,\delta\pars{y - \max\braces{a,x}}\dd x}
\\[5mm] = &\
\int_{-\infty}^{a}\on{P}\pars{x}
\,\delta\pars{y - a}\dd x +
\int_{a}^{\infty}\on{P}\pars{x}
\,\delta\pars{y - x}\dd x
\\[5mm] = &\
\delta\pars{y - a}\int_{-\infty}^{a}\on{P}\pars{x}
\,\dd x +
\on{P}\pars{y}
\bracks{y > a}
\end{align}

For instance: With $\ds{\on{P}\pars{x} = \bracks{x > 0}\lambda\expo{-\lambda x}}$ and $\ds{\lambda, a > 0}$; the result is given by
\begin{align}
&\delta\pars{y - a}\pars{1 - \expo{-\lambda a}} +
\bracks{y > 0}\lambda\expo{-\lambda y}
\bracks{y > a}
\\[5mm] = &\
\bbx{\delta\pars{y - a}
\pars{1 - \expo{-\lambda a}} +
\bracks{y > a}\lambda\expo{-\lambda y}}
\\ &
\end{align}


问:
Generating primes from other primes
For a natural number $n$ let $M$ be an $n$ by $n$ matrix w/$0$'s on diagonal and natural numbers off diagonal and let $p_1, p_2, \dots, p_n$ be a set of prime numbers.

Note then, that

$$p=\sum_{i=1}^n (\pm) \prod_{j=1}^n p_i^{m_{ij}}$$

is not divisible by any of the primes $p_1, p_2, \dots, p_n$.

Questions: Is there a finite set of primes that would generate
 (a)   all co-primes to the set,
 (b)   all other primes
in this fashion (varying $M$ and signs).

Example: $p_1 = 2$, $p_2=3$,

$$ 2+3 = -2^2 + 3^2 =5, $$
$$ 2^2+3 = 2^4 - 3^2 = 7, $$
$$ 2^3+3 = -2^4+ 3^3 = 11, $$
$$ 2^2+ 3^2 = 13,$$
$$\dots,$$
$$ ? =37. $$

What restriction on $M$ and signs would produce only primes for a set $p_1, p_2, \dots, p_n$?

答:
Intuitively, the answer to the first two questions is no and the last is that there is no restriction on $M$ and signs that will produce only primes.  The fundamental reason is that powers get very far apart when the exponent is large.

Pick your set $\{p_1,p_2,p_3,\ldots p_n\}$.  It is finite and has a maximum.  Now pick a prime and a composite that are about $10^{100^n}$ times the largest one.  It is quite unlikely that you can express them this way.  We could calculate how many powers of the primes are less that this and find it is small compared to the number of numbers to express.  This will not be rigorous-there could be some cancellation between larger powers of the primes that brings it back into range.

For the last, note that $2^5+3=35$.  You just need to find one prime that $2$ is a primitive root of and you will be able to find a multiple of that prime.

Title: Generating primes from other primes


问:
Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$
I am supposed to prove the following inequality. One standard method that can be put to use is considering a function $f(x)=x^2-(1+x)(\ln(1+x))^2$ and take its derivative to comment on the behaviour of its increase and arrive at concluding the inequality using the fact that $f(0)=0$.

$$x^2\gt (1+x)(\ln(1+x))^2,~~ \forall x\gt 0$$

I was wondering if there was a neater way to come up with the proof, especially as this is a contest-practice problem. Any hints are appreciated. Thanks.

答:
let $x+1=e^t$ then $t>0$.

to prove $(1)$ is:we will prove $$F(t)=e^t+\frac{1}{e^t}-2-x^2>0$$
notice that $$F''(t)\ge 0 \tag{$F'$ is increasing}$$ and $F''(t)=0$ only when $t=0$
$$F'(0)=F(0)=0$$
which means  $F(t)>0$ for $t>0$


问:
How can this sum sign be expanded?
How can this sum sign

$$\sum_{i=1}^{\log (n)} 2^i$$

be expanded to

$$2^{\log (n)+1}-2$$

Which rules do I need to look at to understand this?

答:
I disagree with the other responses.  In my opinion, absent any agreed convention on how to perform the summation when the upper bound is not an integer, the summation is gibberish.

Suppose that you regard $f(i) = 2^i.$  In order to evaluate the summation, first you have to clearly express it as the sum of terms.  For example, $f(1) + f(2) + \cdots + f(k)$ or $f(1) + f(2) + f(3) + \cdots$ [re when the upper bound is $\infty$].

Only after the summation is clearly expressed as the sum of terms can you then try to simplify the expression.  To try to simplify the expression without a clear summation is simply putting the cart before the horse so that you can ignore that the summation is gibberish.

It may well be that there exists an already agreed upon convention that would allow the summation to be evaluated as other responses have.  However, prior to this query, I (for one) have never seen such a convention anywhere.

My total ignorance of such a convention should have some relevance.  How will math students react?  Is the OP's puzzlement really so surprising?

Title: How can this sum sign be expanded?


问:
Extend usual properties of 1D random variables to multidimensional random variables
It is well-known that for two linearly independent, mean 0 $\mathbb{R}$-valued random variables $X$ and $Y$ that $$\text{Var}(aX+bY) = a^2\text{Var}(X) + b^2\text{Var}(Y)$$ for any $a,b \in \mathbb{R}$, where $\text{Var}(X) := \int_{\mathbb R} x^2 f(x) dx$ if the law of $X$ (with mean 0) is $f$. I am not sure whether a multidimensional version of the analogous statement if still true. Specifically, if two linearly independent, mean $\bf{0}$ (here $\bf{0}$ denotes the dimensional zero vector) $\mathbb{R}^d$-valued random variables $X$ and $Y$, do we still have something like  $$\text{Var}(aX+bY) = a^2\text{Var}(X) + b^2\text{Var}(Y)$$ for any $a,b \in \mathbb{R}$, but now with the "variance" defined via $\text{Var}(X) := \int_{\mathbb R^d} |{\bf x}|^2 f({\bf x}) d{\bf x}$ if the law of $X$ (with mean $\bf{0}$) is $f$. I am not sure whether the dimension $d$ plays a role in such generalizations, so I appreciate any help or references.

答:
In the first line you wrote 'linearly independent  for '(stochastically) independent'. Linear independnece does not yield any of these equations.

In order to extend this to higher dimensions you need orthogonality :$E \langle X, Y \rangle=0$ This is again implied by indepedence: $E \langle X, Y \rangle=\sum E(X_iY_i) =0$ . Hence we get $E|aX+bY|^{2}=a^{2}E|X|^{2}+b^{2}E|Y|^{2}$ under independence.


问:
A condition for a Toeplitz operator to be Fredholm and find its index
I have a question about showing that a Toeplitz operator is Fredholm:

Show that the operator $T_{e^{inx}}:L_2^+\to L_2^+$ acting on $L_2^+=\sum_{k\ge
0}a_ke^{ikx}$ with $\sum_{k}|a_k|^2<\infty$ is Fredholm for every
$n\in\mathbb{Z}$ and find its index.

Denote $S^1$ to be the unit circle. The space $L_2^+$ is space of functions $f(z)$ on $S^1$ with $z=e^{i\theta}$ $$f(z)=\sum_{n\ge 0}a_nz^n$$ with norm $$\frac{1}{2\pi}\int_{0}^{2\pi}|f(e^{i\theta})|^2d\theta=\sum_{n\ge 0}|a_n|^2$$

$P^+$ is the projection from $L_2(\mathbb{Z})$ onto $L_2^+$, $\sum_{k\in\mathbb{Z}}a_ke^{ikx}\mapsto\sum_{k\ge
0}a_ke^{ikx}$. If $h(e^{i\theta})$ is a bounded function on $S^1$, $T_h:L_2^+\to L_2^+$ is defined as $$T_h(f)=P^+hf$$
and $\|T_h\|\leq\sup_{\theta}|h(\theta)|$

I'm trying to show that this Toeplitz operator is Fredholm and find its index. My attempt is to show the following: if $h$ is a continuous mapping from $S^1$ to $\mathbb{C}$ and $h(z)\neq 0$ on $S^1$, then $T_h$ is Fredholm. Once I have this, how can apply this to $T_{e^{inx}}$? Added: This works.

For the index, I have read from various sources that $\text{ind}(T_h)=-w(h)$, the winding number of $h$. Is it true for my case? Added: It is $-n$ by computing the winding number of $e^{inx}$ directly.

Thank you.

Added: if $h$ is a continuous mapping from $S^1$ to $\mathbb{C}$ and $h(z)\neq 0$ on $S^1$, then $T_h$ is Fredholm.
The theorem I used can be proved by:

If $h_1$ and $h_2$ are continuous, $T_{h_1}T_{h_2}-T_{h_1h_2}$ is compact.

答:
It is true that for $h\in C(S^1)$, the Toeplitz operator $T_h$ is Fredholm if and only if $0$ is not in the range of $h$, and that $\operatorname{ind}(T_h)=-\operatorname{wind}(h)$.  However, this using this result to compute $\operatorname{ind}(T_{e^{inx}})$ is both unnecessary (the index can be computed directly from the definition) and circular (to prove the above index theorem, one reduces to the case that $h=e^{inx}$ for some $n$, then computes $\operatorname{ind}(T_{e^{inx}})=-n$).

In any case, one only needs to compute the index of $T_{e^{ix}}$.  Note that this operator is an isometry, so $\ker(T_{e^{inx}})=\{0\}$.  Also, $T_{e^{-ix}}=(T_{e^{ix}})^*$ has a one-dimensional kernel spanned by the constant functions, and therefore $$\operatorname{ind}(T_{e^{ix}})=\dim\ker(T_{e^{ix}})-\dim\ker(T_{e^{-ix}})=-1.$$


答:
You can use the nth term test for divergence.

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

If $\lim_{n \to \infty}a_n \ne 0$, then the series diverges.

Title: Why does the series diverge?


问:
why does integral of exponent becomes 1 / something?
Could anyone explain to me why at step 3, after integration it becomes $1/(a-jw)$?

答:
Let $z=b+ic,b >0$
$$\int_{0}^{\infty} e^{-zt} dt=-\left .\frac{e^{-zt}}{z}\right|_{0}^{\infty}=\frac{1}{z}=\frac{1}{b+ic}.$$
As $e^{-zt}$ vanishes at $t=\infty$ if only $\Re(z)=b>0$$


问:
How to prove AC = BD?
Triangle ABC has $\widehat{A} = 130^\circ;\, \widehat{B} = 30^\circ.$ D and E are the intersection between the mid-perpendicular of $AC$ and $BC, AB,$ respectively.

a) Prove ADE is an isosceles triangle

b) AC = BD

a) is an easy one, but b) is in a different level (at least with me)

How can I prove AC= BD?

Please help me!

答:
Hint:

Draw perpendicular bisector of BD( a perpendicular passing the midpoint M of DB) . Draw a line from D parallel with AC to intersect the drawn bisector at F.We have:

$\angle ADB=\angle CAD=20^o$

$\angle ADB=\angle FDB=\angle FBD=20^o$

Since $\angle ADF=20^o$ F must be on circle center on D and passing A, therefore $AD=DF$. Also $AD=DC$ and $DF=FB$

Hence triangles ACD and FDB for SAS (two sides and angles they make are equal)are equal so AC=BD.

Title: How to prove AC = BD?


问:
Limit as N goes to Infinity
Consider this limit: $$\lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n^2} = x$$

I thought the way to solve this for $x$ was to reduce it using the fact that as $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$:
$$\therefore \lim_{n\rightarrow\infty}(1+0)^{n^2} = x$$

Apparently this is wrong! Why is it wrong?

答:
But as $n\rightarrow \infty $ that exponent gets ever larger.  A smaller argument, yes, but raised to an ever larger power.  Can't assume that the whole thing is going to go to one.  Try a few examples on your calculator to see what happens.

For a little more detail, if $n$ is some big number, when you multiply everything out, the first two terms (look up the binomial expansion) will be

$1+ n^2 * \displaystyle\frac{1}{n} = 1+n \rightarrow \infty$

None of the remaining terms will be negative, so you know this is a bottom limit on what you are going to get.

Title: Limit as N goes to Infinity


问:
Clarifying expectation with respect to a function, and the Radon-Nikodym derivative
I am trying to understand induced measures in the context of expectation. I have commonly seen formulas such as KL-divergence written like so:

$$\text{KL}(p||q) = \int\log\dfrac{p(x)}{q(x)}dp = \int p(x) \log\dfrac{p(x)}{q(x)}dx = \mathbb{E}_{p(x)}[\log\dfrac{p(x)}{q(x)}]$$

which reads as the expectation of the formula $\log\dfrac{p(x)}{q(x)}$ with respect to the measure $p(x)$. But over what set is this measure being taken?

Is it over the entire real line: $\int_{-\infty}^\infty p(x) \log\dfrac{p(x)}{q(x)}dx$?

Or is it over the support of $p$: $\int_{x:p(x) >0} p(x) \log\dfrac{p(x)}{q(x)}dx$?

For the KL-divergence formula these turn out to be the same, but consider the expectation $\mathbb{E}_{p(x)}[\dfrac{q(x)}{p(x)}] = \int p(x)\dfrac{q(x)}{p(x)}dx = \int q(x)dx$.

In the case of the real line, we have $\int_{-\infty}^\infty q(x)dx = 1$.

In the case of the support of $p$, we have $\int_{x:p(x)>0} q(x)dx$, which is obviously very different.

I believe the second is correct, but I was unable to prove this in a way that satisfied me intellectually. My work is as follows:

Interpreting $p$ as the Radon-Nikodym derivative of an induced measure $\nu$, and letting $S$ represent the support $S = \{x: p(x) > 0\}$, we can calculate $\int \log\dfrac{q(x)}{p(x)}dp = \int_S \dfrac{q(x)}{p(x)}dp + \int_{S^C} \dfrac{q(x)}{p(x)}dp$. We have $\nu(S^C) = \int_{S^C} p(x)dx  = 0$, thus the second integral $\int_{S^C} \dfrac{q(x)}{p(x)}dp = 0$ and that leaves us with the integral over the support as desired.

To me, this proof seems valid but it also seems like cheating. I just basically look at the set of measure $0$ and declare that I can ignore it even though the value being integrated over there ($\dfrac{q(x)}{p(x)}$) is infinite. This post on a similar question stated I could do that, but I don't understand why.

答:
$E_{X \sim p} \log \frac{p(X)}{q(X)}$ is actually shorthand for $E_{X \sim p} f(X)$ where $f(x) = \begin{cases} \log \frac{p(x)}{q(x)} & p(x) > 0 \\ 0 & p(x) = 0\end{cases}$. The assumption of absolute continuity ($q(x)=0$ implies $p(x)=0$) ensures that $f$ is well-defined (no division by zero). This needs to be explicitly baked into the definition of the KL divergence, since "$\log \frac{0}{0}$" and "$0 \log \frac{0}{0}$" are ambiguous.

Similarly, if you write $E_{X \sim p} \frac{q(X)}{p(X)}$, you need to be explicit in how you define the function. You could say $E_{X \sim p} g(X)$ where $g(x) = \begin{cases} \frac{q(x)}{p(x)} & p(x) > 0 \\ 0 & p(x) = 0\end{cases}$ if you wish, but this is something that needs to be stated; $E_{X \sim p} \frac{q(X)}{p(X)}$ alone is ambiguous, since you get a $0 \cdot \infty$ situation.


问:
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano

Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$

Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.

Also, can we prove that there is only one real solution without using the discriminant?

Body:
Let cubic equation $f=ax+bx^2+cx^3$, then for equation $x^3 - 9 x^2 - 15 x - 6 = 0$ we have $a=-15,b=-9,c=1,f=6$.

Non-Cardano cubic formula:

$X=\dfrac{-(a b+9 c f)+\sqrt{(a b+9 c f)^2-4 (b^2-3 a c) (a^2+3 b f)}}{2 (b^2-3 a c)}=-1/2$

$A=a+2 b X+3 c X^2=-21/4$

$F=a X+b X^2+c X^3-f=-7/8$

$G=A^3-27 c F^2=-1323/8$

$Q=\Big\{G^{1/3},-(-1)^{1/3} G^{1/3},(-1)^{2/3} G^{1/3}\Big\}=\Big\{(3/2) (-1)^{1/3} 7^{2/3}, -(3/2) (-7)^{2/3}, -(3/2) 7^{2/3}\Big\}$

$x=X+\dfrac{3 F}{Q-A}=\Big\{-(1/2)\Big(1+\dfrac{7}{7 + 2 (-1)^{1/3} 7^{2/3}}\Big), \dfrac{7 - (-7)^{2/3}}{-7 + 2 (-7)^{2/3}},\dfrac{7 - 7^{2/3}}{-7 + 2\cdot 7^{2/3}}\Big\}=\\
\Big\{3 - 7^{1/3} - 7^{2/3}/2 - i \sqrt{3} (-7^{1/3} + 7^{2/3}/2), 3 - 2 (-7)^{1/3} + (-7)^{2/3}, 3 + 2\cdot 7^{1/3} + 7^{2/3}\Big\}=\\
\Big\{-0.742584 + 0.144242\cdot i, -0.742584 - 0.144242\cdot i, 10.4852\Big\}$

And note: $z^3=w \implies z=\Big\{w^{1/3},-(-1)^{1/3}w^{1/3},(-1)^{2/3}w^{1/3}\Big\}$,

where $\Big\{-(-1)^{1/3},(-1)^{2/3}\Big\}=\Big\{e^{-2\pi\cdot i/3},e^{2\pi\cdot i/3}\Big\}=\Big\{-0.5 - 0.866025\cdot i, -0.5 + 0.866025\cdot i\Big\}$


答:
If your method isn't wrong so we have also

$$\lim_{n\to\infty}\frac{1}{n}\times n=\lim_{n\to\infty}0\times n=0???$$

Title: Limit as N goes to Infinity

Body:
Consider this limit: $$\lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n^2} = x$$

I thought the way to solve this for $x$ was to reduce it using the fact that as $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$:
$$\therefore \lim_{n\rightarrow\infty}(1+0)^{n^2} = x$$

Apparently this is wrong! Why is it wrong?


答:
as $A,B,C,E$ are concyclic.$$ED.DA=BD.DC$$
using this Formula:$$AD=\frac{2\sqrt{bcs(s-a)}}{b+c}$$
Aslo by angle bisector rule$$BD=\frac{ca}{b+c},CD=\frac{ba}{b+c}$$

Can you finosh it?

Title: Internal Bisector of a triangle

Body:
In $\Delta ABC$, the bisector of the angle A meets the side BC at D and circumscribed circle at E, then DE equals to

(A) $\frac{{{a^2}\cos \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(B) $\frac{{{a^2}\sec \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(C) $\frac{{{a^2}\sin \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(D) $\frac{{{a^2}\cos ec\frac{A}{2}}}{{2\left( {b + c} \right)}}$

My approach is as follow

Internal Bisector property

$\frac{{AB}}{{AC}} = \frac{{BC}}{{DC}} \Rightarrow \frac{c}{b} = \frac{t}{{a - c}}$

How do I proceed in order to find DE


答:
$[ABC] = \frac{ab \sin c}2$. Let $D$ be the feet of the internal $A$ bisector.

$[ADB] + [ADC] = [ABC] \iff cd\sin(\frac{\angle A}2) + bd \sin(\frac{\angle A}2) = bc \sin(\angle A) \iff$

$ AD = \frac{2bc \cos (\frac{\angle A}2)}{b+c}$

Now $DE$ seems a bit trickier, how about:

$ce\sin(\frac{\angle A}2) + be \sin(\frac{\angle A}2) = bc \sin(\angle A) + \frac{a^2}2 \tan (\frac{\angle A}2)$ ?

$e = \frac1{b+c} (2bc \cos (\frac{\angle A}2) + \frac{a^2}{2 \cos (\frac{\angle A}2)}) = \frac{\sec (\frac{\angle A}2)}{b+c}(\frac{a^2}2 + bc(\cos (\angle A)+1))$

Title: Internal Bisector of a triangle


答:
Integral of exponent does not become 1/something. The basic rule is

$$ \int  e^{ax} dx =  \frac{e^{ax}}{a} +c   ......->1$$

In the integrations you perform above,you integrate, for example, $e^{(a-j\omega)t}$ with respect to $dt$. Put $a-j\omega$ as $a$ above in 1. You just get the coefficient of the variable (wrt which you are performing integration) as the denominator. It happens in logarithmic and trigonometric integrations too. I think learning the basics of integration from any good book will help you.

Title: why does integral of exponent becomes 1 / something?

Body: Could anyone explain to me why at step 3, after integration it becomes $1/(a-jw)$?


问:
Conditional probability (cards question).
Given a piece of cards with $52$ total cards, $13$  of each one from the $4$ options ($\heartsuit, \spadesuit, \diamondsuit, \clubsuit$) . You and your friend getting randomly $7$ cards each,

what is the probability that your friend have one heart card given that you have three heart cards?

*note that I mean to the amount of cards from type hearts  not the number 1$\heartsuit$.

My attempt

Let $A$ and $B$ be two events :

$A$- my friend have $1$ heart cards

$B$- I have $3$ heart cards

we need to find $$P(A|B)$$
I want to use the formula $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

now I am stuck how to calculate $P(B)$ and $P(A\cap B)$

In the book there is 4 options $$1. \ \frac{12!}{3!\cdot9!}\cdot\frac{32!}{5!\cdot27!}\cdot\frac{8!\cdot36!}{44!}$$
$$2.\ \frac{12!}{3!\cdot9!}\cdot\frac{33!}{4!\cdot29!}\cdot\frac{7!\cdot38!}{45!}$$ $$3.\ \frac{10!}{1!\cdot9!}\cdot\frac{34!}{7!\cdot27!}\cdot\frac{8!\cdot36!}{44!}$$ $$4.\ \frac{10!}{1!\cdot9!}\cdot\frac{35!}{6!\cdot29!}\cdot\frac{7!\cdot38!}{45!}$$

I did not understand this pattern with the factorial, how did they build it?

答:
I agree with Robert Shore's answer.  I further agree with his comment that for this particular problem, it is most readily solved without using Bayes Theorem.  However, I still think that it is important to understand how to set up the conditional probability mathematics.

Let $S$ denote the event that your 7 card hand has exactly 3 hearts.

Let $T$ denote the event that your friend's 7 card hand has exactly 1 heart.

Then $$p(T|S) ~=~ \frac{p(S ~\text{and} ~T ~\text{both occur})}{p(S)}.$$

$${p(S ~\text{and} ~T ~\text{both occur})}
~=~ \frac{\binom{13}{3}\binom{39}{4}}{\binom{52}{7}} \times \frac{\binom{10}{1}\binom{35}{6}}{\binom{45}{7}}.$$

$${p(S)}
~=~ \frac{\binom{13}{3}\binom{39}{4}}{\binom{52}{7}}.$$


问:
Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$
The function

$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$

Has the following poles of order 2:

$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$

$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$

$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$

$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$

$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$

$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$

$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$

$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$

$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$

We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{6}\over \pars{x^{4} + a^{4}}^{2}}\,\dd x}
\\[5mm] \stackrel{x\ =\ \verts{a}t^{1/4}}{=}\,\,\,&\
{1 \over 4\verts{a}}\int_{0}^{\infty}{t^{\color{red}{7/4} - 1} \over \pars{1 + t}^{2}}\,\dd t
\end{align}
Note that
\begin{align}
{1 \over \pars{1 + t}^{2}} & =
\sum_{k = 0}^{\infty}{-2 \choose k}t^{k} =
\sum_{k = 0}^{\infty}{k + 1 \choose k}
\pars{-1}^{k}\,t^{k}
\\[5mm] & =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{2 + k}}
{\pars{-t}^{k} \over k!}
\end{align}
With the
Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{6}\over \pars{x^{4} + a^{4}}^{2}}\,\dd x} =
{1 \over 4\verts{a}}\Gamma\pars{7 \over 4}
\Gamma\pars{2 - {7 \over 4}}
\\[5mm] = &\
{1 \over 4\verts{a}}{3 \over 4}\Gamma\pars{3 \over 4}
\Gamma\pars{1 \over 4} =
{3 \over 16\verts{a}}\,{\pi \over \sin\pars{\pi/4}}
\\[5mm] = &\
\bbx{{3\root{2}\pi \over 16}\,{1 \over \verts{a}}}
\\ &
\end{align}


答: You have an indeterminate form:  $1^\infty$, so need some finer analysis than just plugging in.  We know $\displaystyle \lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n} = e$ so we might expect yours to diverge as $e^n$</p>

Title: Limit as N goes to Infinity


问:
Solve the differential equation $\frac{dx}{dy} -\frac{x \log x}{1+\log x} =\frac{e^y}{1+\log x}$, if $y(1)=0$
I don’t know what form this. It’s probably a reducible to linear form equation, but I don’t know which term to substitute or how to rearrange. Can I get a hint?</p>

答:
We need to solve
\begin{eqnarray*}
(IVP): \left\{\begin{aligned} \frac{dx}{dy}-\frac{x\ln(x)}{1+\ln(x)}=\frac{e^{y}}{1+\ln(x)}\\
y(1)=0 \\ \end{aligned}\right.\\
\end{eqnarray*}
Now, rewrite the ODE, you have $$e^{y}+x\ln(x)-(\ln(x)+1)\frac{dx}{dy}=0$$
(If you have difficulty getting to this part, tell me to put the details).
Let $P(y,x)=e^{y}+x\ln(x)$ and $Q(y,x)=-\ln(x)-1$, so we can see this ODE is not an exact equation, because $P_{x}=\ln(x)+1\not=0=Q_{y}$.

Now, find an integrating factor $\mu(y)$ such that $$\mu(y)R+\frac{dx}{dy}\mu(y)Q=0$$ is exact.
Solving this part, we can see that $\mu(y)=e^{-y}$, and multiply both sides of $$e^{y}+x\ln(x)+(-1)(1+\ln(x))\frac{dx}{dy}=0$$ by $\mu(y)$ we have a exact equation.

Now, using the method for the method to solve exact equations (if you don't know this method, please tell me to fill in the details), we have that exists a function $f$ such that $$f_{y}=P \quad f_{x}=Q$$ and $$f(x,y)=c$$
Solving this, we have $$f(y,x)=y-x\ln(x)e^{-y}$$
so, the solution for the ODE is $$\boxed{y-x\ln(x)e^{-y}=c}$$

Now, just use the initial condition $y(1)=0$ so, you have the solution for IVP. Solving this last part we have $c=0$ and the solution for IVP is $$\boxed{y-x\ln(x)e^{-y}=0}$$


问:
Find all possible digits such that $\overline{ab}\mid\overline{a0b}$
My question is : how to find all possibilities for the $\overline{ab}\mid\overline{a0b}$ if we rewrite as $$10a+b\mid100a+b$$ it can easy to show for $b=0$ it is has solution like $$10\mid100\\20\mid200\\...$$ and for $$a=4,b=5 \to 45\mid405$$ but I am for stuck to solve it for more possible numbers. Can someone help me?

答:
Writing it as$$100a+b = (10a+b)k \\
10(10-k)a = (k - 1)b$$

Now we only need to see $5$ (quite) easy cases:

Since $a > 0$:

$a=1 $: $$10(10-k) = (k-1)b \implies b = \cfrac{10(10-k)}{k-1}$$
Since $b<10$, we must have $$\frac{10-k}{k-1}<1 \implies 5 < k \le 10$$
Checking for these cases, we get $k = 6,7,10$

$a=2 $: $$10(10-k)2 = (k-1)b \implies b = \cfrac{ 10\cdot2(10-k)}{k-1}$$
Since $b<10$, we must have $$\frac{10-k}{k-1}<\frac 12 \implies 7 < k \le 10$$
Checking for $k=8, 9$, we do not get an integer result. $k=10$ is valid.

$a=3 $: $$10(10-k)3 = (k-1)b \implies b = \cfrac{ 10\cdot3(10-k)}{k-1}$$
Since $b<10$, we must have $$\frac{10-k}{k-1}<\frac 13 \implies 8 < k \le 10$$
Checking for $k=9$, we do not get an integer result. $k=10$ is valid.

$a=4:$ $$10(10-k)3 = (k-1)b \implies b = \cfrac{ 10\cdot4(10-k)}{k-1}$$
Since $b<10$, we must have $$\frac{10-k}{k-1}<\frac 14 \implies 8 < k \le 10$$
Checking for $k=9, 10$, we see these are valid $k$.

For $a\ge 5$, it is impossible to get $b<10$ by the same argument.

So, now you have all the cases done.


问:
a new method to solve some congruence problems
I'm new to number theory and I've newly learnt congruence and there are a special kind of questions in my book which I usually face problems to solve an example of them is as follows:

If $18a\equiv12b\ (\mathrm{mod}\ 9)$ then which of the following statments is wrong?
$$(1)\ a\equiv0\ (\mathrm{mod}\  2)$$
$$(2)\ b\equiv0\ (\mathrm{mod}\ 3)$$
$$(3)\ 3a\equiv b\ (\mathrm{mod}\ 3)$$
$$(4)\ 3a\equiv2b\ (\mathrm{mod}\ 3)$$

The point is that I can solve them but with a long and hard effort so I was thinking of a shorter way and I remember the short way my teacher told me for division problems which is as follows:

If $d|2n+5, d|9n+2$,then $d|\det\begin{bmatrix}
2 & 5 \\
9 & 2 \\
\end{bmatrix}$ but he also mentioned that we have to check the determinant for paradoxes.

Anyway I don't know where this formula has come from and I wonder to know two things.

Where does this method come from and can I use it for the problems I mentioned first(since you can write every congruence as a division)?

Actually I have some ideas when they don't change to $\mathrm{mod}$ in question choices but I don't know what to do when there's a change in the base.

Any help is appreciated!

答:
Three tips:

You can divide both sides of a congruence by an integer if you divide the modulus by that same integer; so for instance, $18a\equiv12b \pmod 9$ is exactly the same as $6a\equiv4b \pmod 3$.

We might as well reduce all integers that appear modulo the modulus; so for instance, since $6\equiv0\pmod3$ and $4\equiv1\pmod3$, we know that $6a\equiv4b \pmod 3$ is exactly the same as $0a\equiv1b \pmod 3$, or $b\equiv0\pmod3$.

Finally, if two moduli are relatively prime, then whether the congruences hold or not are completely independent of each other. For example, since $\gcd(9,2)=1$, whether $18a\equiv12b \pmod 9$ is completely independent of whether $a\equiv0\pmod 2$ (they might be both true, both false, or one of each either way).


问:
Calculating the convex hull of a set.
I have a question about a answer in this post: convex hull specifically from the answer given by @Per Manne.

I know that $$B=\left\{ \begin{pmatrix} x \\ y \end{pmatrix}\in \mathbb{R}^{2}: y>0\right\}$$ is convex set, I could easily verify using the definition of convexity (in fact geometrically it is very intuitive), but how can I prove that $B$ is convex hull of $$A=\left\{\begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^{2}: y \geq \frac{1}{1+x^{2}}\right\}?$$

I think that intuitively it is true, since it is known that by extensivity $A \subseteq conv(A):=B$.

But, by definition how can I prove that fact?

答: Let $(x,y) \in B$. Let $n$ be a positive integer such that $ n >|x|$ and $1+n^{2} >\frac 1 y $. Let $t=\frac {n-x} {2n}$. Then $0<t<1$. Now $(x,y)=t(-n,y)+(1-t)(n,y)$. Note that $(-n,y)$ and $(n,y)$ both belong to $A$. Hence any element of $B$ is a convex combination of two points of $A$.

Title: Calculating the convex hull of a set.


问:
Find the generating function. Put stamps in order and for a specific value.
We have stamps of 3-cent, 4-cent and 20-cent and it is required to put stamps on an envelope of total cost n cents. The order in which we put the stamps does matter. For example for n = 40 the order [3][3][4][3][4][20][3] is different from [3][4][3][3][4][20].</p>

答:
Hint:

You know that $(x^3 + x^4 + x^2)^k$ is the generating function whose $n$th coefficient tells you how many ways there are to make $n$ with exactly $k$ stamps. So if you want to make $n$ with any number of stamps, then that's going to be

(the number of ways to make $n$ with 0 stamps) plus
(the number of ways to make $n$ with 1 stamp) plus
(the number of ways to make $n$ with 2 stamps) plus
(the number of ways to make $n$ with 3 stamps) plus ...

So if $a_{n,k}$ is the coefficient of $x^n$ in $(x^3 + x^4 + x^2)^k$, then you want
the coefficient of $x^n$ in the solution, to be $a_n = \sum_k a_{n,k}$.

Do you see how to find this generating function using the $(x^3 + x^4 + x^2)^k$ that you already have?

I hope this helps ^_^


问:
How do I answer this question on counting?
I know that this question is pretty simple to answer, but I am just not able to.

A class has 100 students.

40 are good in calculus.

70 are good in trigonometry.

21 are good in neither calculus nor trigonometry.

How many are good at both?

I know that I need to apply some variation of the formula
$|P ∪ Q| = |P| ∪ |Q| - |P ∩ Q|$

Where $P$ is the number of students good in calculus, and $Q$ is the number of students good in trigonometry.

答:
Hint:

Based on the data and your interpretation of it, we have that those who are "good at neither calculus nor trigonometry" is $(P\cup Q)^c$. Given this, what is $P\cup Q$?


答: Let's try to answer this without using complex formulas. The first thing to know is how many are good at calculus and/or trigonometry. We can substract those who are not good at both from the total student (good at either= 100-21 = 79). Now the 79 people must consists of people who are only good at calculus, only good at trigonometry and good at both.Let the number of people good at both be $x$, and I believe there is enpugh information to solve this.

Title: How do I answer this question on counting?

Body:
I know that this question is pretty simple to answer, but I am just not able to.

A class has 100 students.

40 are good in calculus.

70 are good in trigonometry.

21 are good in neither calculus nor trigonometry.

How many are good at both?

I know that I need to apply some variation of the formula
$|P ∪ Q| = |P| ∪ |Q| - |P ∩ Q|$

Where $P$ is the number of students good in calculus, and $Q$ is the number of students good in trigonometry.


问:
Computing integral with floor function
Question

Find the exact value of $$\int_{1}^{2016}\frac {\lfloor \ln x \rfloor} {x}\ \mathrm{d}x\ .$$

My working

My intuition is to treat the integral without the floor function and integrate, then "put back" the floor function after, so $$\int_{1}^{2016}\frac {\lfloor \ln x \rfloor} {x}\ \mathrm{d}x = \frac 1 2 [(\lfloor \ln x \rfloor)^2]^{x = 2016}_{x = 1}\ .$$

May I know if my intuition is correct? If not, what would be the right way to do this? This is my first time encountering an integral with the floor function. Any help/suggestions would be greatly appreciated :)

答:
This is a plot of floor(ln(x)) along x axis.Find these points of discontinuity and break your integral at these points. The the numerator in all of these sub parts will be a constant and you will have just $\frac{some-constant}{x}$ as the integrand.

You can find these points of discontinuity using desmos.com or a scientific calculator (which is easily available on smartphones or computers these days).

Title: Computing integral with floor function


Title: Solve the differential equation $\frac{dx}{dy} -\frac{x \log x}{1+\log x} =\frac{e^y}{1+\log x}$, if $y(1)=0$

Body: I don’t know what form this. It’s probably a reducible to linear form equation, but I don’t know which term to substitute or how to rearrange. Can I get a hint?</p>

Body:
Hint; Put $x\log x=t$
we have $$\frac{dt}{dy}-t=e^y$$
$$e^{-y}dt-te^{-y}dy=1$$
$$d(e^{-y}t)=1$$


问:
Finding the gamma integral for a sample of $iid$ $Bin(1, )$ random variables with $Θ = [0, 1]$, and a prior of $$ distributed as $U[0, 1]$
I am given $X_1,…, X_n$ as a sample of $iid$ $Bin(1, )$ random variables
where $Θ = [0, 1]$, and we are assuming that the prior distribution of  is $U[0, 1]$

Therefore, the likelihood function for the observed sample is

$L(\theta) = \prod_{i=1}^n \theta^{x_i}(1-\theta)^{1-x_i} = \theta^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}$

Therefore, the posterior is written as follows:

$f(\theta|\textbf{x}) = \frac{\theta^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}}{\int_0^1 \theta^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}d\theta} = \frac{\theta^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}}{\frac{\Gamma(1 + \sum_{i=1}^nx_i)\Gamma(n-1 + \sum_{i=1}^nx_i)}{\Gamma(n+2)}}$

My question is how did we know that the $ \int_0^1 \theta^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}d\theta = \frac{\Gamma(1 + \sum_{i=1}^nx_i)\Gamma(n-1 + \sum_{i=1}^nx_i)}{\Gamma(n+2)}$?

Is there a formula or a general pattern for this?

答:
The posterior can be easily found remembering, as usual,  that

$$ \bbox[5px,border:2px solid red]
{
\pi(\theta|\mathbf{x})\propto \pi(\theta)\times p(\mathbf{x}|\theta)
\
}
$$

Thus, being $\pi(\theta)=1$, you have

$$\pi(\theta|\mathbf{x})\propto \theta^{\Sigma x}(1-\theta)^{n-\Sigma x}\sim Beta(\Sigma x+1;n-\Sigma x+1)$$

Now, remembering how the analytical expression of a Beta density is, the complete analytical expression of your posterior is

$$ \bbox[5px,border:2px solid red]
{
Beta(a,b)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\theta^{a-1}(1-\theta)^{b-1}
\qquad (1)
}
$$

Substituting (1) in your exercise you get immediately

$$\pi(\theta|\mathbf{x})=\frac{\Gamma(n+2)}{\Gamma(\Sigma x+1)\Gamma(n-\Sigma x+1)}\theta^{\Sigma x}(1-\theta)^{n-\Sigma x}$$

The result of your integral is wrong. At a certain point you wrote $\Gamma(n-1+\Sigma x)$ while the correct expression is $\Gamma(n-\Sigma x+1)$


问:
Codimension 1 subset of a normal variety is locally principal
I am trying to prove this result:

Let $X$ be a normal variety, and suppose $W$ is an irreducible closed subvariety of $X$ of codimension $1$. Then there exists an affine open $U$ $\subseteq$ $X$ such that the ideal of  $W \cap U$ in $k[U]$ is principal.

Here is what I got:

Consider the local ring of the subvariety $W$ in $X$. It is a DVR, and suppose its maximal ideal is $M$. Let $\langle U,f\rangle$ be an uniformizing parameter of it. We can very well choose $U$ to be an affine open set. Now consider the ideal of $W \cap U$ in $k[U]$, say it is generated by $g_1,g_2,...,g_n$.

Then from the very definition, we have $\langle U,g_i\rangle$ will be an element of $M$, so there exists $\langle U',f'\rangle$ in $M$ such that $\langle U,f\rangle\langle U',f'\rangle = \langle U,g\rangle$.

But this just means that $g$ is a multiple of $f$ on $U \cap U'$.

How to get that $f|g$ on whole of $U$?

Any help will be appreciated.

Please see: I have knowledge of only the first 5 sections of chapter 1 of Hartshorne, so an elementary proof will be appreciated.

答: You're on the right track. The correct thing to do here is to shrink $U$: since all the $g_i$ are multiples of $f$ in the local ring $\mathcal{O}_{X,W}$, there exist functions $h_i\in\mathcal{O}_{X,W}$ so that $g_i=fh_i$. Considering these as elements of the function field of $X$, which is exactly the fraction field of $k[U]$, we can write all the $h_i$ as fractions where the denominator doesn't vanish along $W$ by the DVR property. Now take the non-vanishing set in $U$ of the product of all of these denominators: this is affine, and each $g_i$ retricted to this set is an honest multiple of $f$, giving a principal ideal.


问:
Use an infinite series to approximate to four decimal places: integral from 0 to 1 sin(7x)/x (finding number of terms without calculator)
So I managed to solve this problem by simply taking the Maclaurin series of sin(7x), which was

$$\sum_{n=0}^∞  \frac{(-1)^n(7x)^{2n+1}}{(2n+1)!}$$

and divided the summation by x, which gave me

$$\sum_{n=0}^∞  \frac{(-1)^n(7x)^{2n}}{(2n+1)!}$$

I integrated this and got $$\sum_{n=0}^∞  \frac{(-1)^n(7x)^{2n+1}}{(2n+1)(2n+1)!}$$ and set the bounds from 0 to 1. I plugged in 1 to get

$$\sum_{n=0}^∞  \frac{(-1)^n(7)^{2n+1}}{(2n+1)(2n+1)!}$$

Now, this is where my problem came in. I managed to use a calculator to figure out what the series converges to. However, I want to know how to solve this by finding the number of terms needed to find the sum. How can one find the number of terms necessary to get this sum?

答:
There is a formula which will give you in advance the numbers to be added before computing anything.

Let us make the problem more general
$$\sum_{n=0}^\infty  (-1)^n\frac{a^{2n+1}}{(2n+1)(2n+1)!}=$$ $$\sum_{n=0}^p  (-1)^n\frac{a^{2n+1}}{(2n+1)(2n+1)!}+\sum_{n=p+1}^\infty  (-1)^n\frac{a^{2n+1}}{(2n+1)(2n+1)!}$$ and then, you want to know $p$ such that the first neglected term be smaller or equal to $10^{-k}$; that is to say
$$R_p=\frac{a^{2p+3}}{(2p+3)(2p+3)!} \leq 10^{-k}\implies(2p+3)(2p+3)! \geq a^{2p+3}\,10^{k}$$

Taking logarithms, solve for $p$ the equation
$$\log(2p+3)+\log\big[ (2p+3)!\big] - (2p+3)\log(a)-k\log(10)=0$$ and later, wa shall use $\lceil p\rceil$.

For the time being, let $m=2p+3$; using Stirling approximation and continuing with Taylor series, the above write
$$m (\log (m)-\log (e a))+\frac{3 }{2}\log (m)+ \log (\sqrt{2 \pi} )-k \log(10)+O\left(\frac{1}{m}\right)=0$$ which can be solved only if we neglect the $\frac{3 }{2}\log (m)$ term (this means that the solution would be a slight overestimate of the solution. So, the approximation
$$m=2p+3 \sim \frac{2 k \log (10)-\log (2 \pi )}{2 W\left(\frac{2 k \log (10)-\log (2 \pi )}{2 e   a}\right)}$$ where $W(.)$ is Lambert function.

Applied to your case $a=7$ and $k=4$, this gives $p=11.5664$; if the $\frac{3 }{2}\log (m)$ term was kept, the solution would have been $p=9.68912$.

Checking with your numbers
$$R_{9}=5.21 \times 10^{-4} \qquad R_{10}=4.60 \times 10^{-5} \qquad R_{11}=3.46 \times 10^{-6}$$

But, since we had to enter the beautiful world of Lambert function, we could do much better assuming $(2p+3)(2p+3)! \sim (2p+4)!$ and write the equation as
$$(2p+4)!=a^{2p+4} \frac {10^k} a$$ Now, have a look at this question of mine and admire the approximation proposed by @robjohn. Adapted to your problem, this would give
$$m=2p+4 \sim e\,a \, \exp\Bigg[W\left(\frac{2 k \log (10)-\log \left(2 \pi  a^3\right)}{2 e a}\right) \Bigg]-\frac 12$$

Applied to your case, this would give $p=9.66906$ then  $\lceil p\rceil=10$.

Suppose now that you want $100$ decimal places; this gives immediately $\lceil p\rceil=59$.

Checking
$$R_{58}=5.56 \times 10^{-99} > 10^{-100} \qquad R_{59}=1.84 \times 10^{-101} < 10^{-100}$$


问:
$f(yf(x) + y) = xy + f(y)$ for all $x, y.$ Prove $f$ is surjective
I'm stucking with this problem for days. f is a function from $\mathbb{R}$ to $\mathbb{R}$ satisfies:

$$f(yf(x) + y) = xy + f(y) \qquad \forall x, y \in \mathbb{R}$$

Prove $f$ is surjective, and hence find all the functions satisfy this equation.

I think fixing a $x$ and $y$ as a constant and study the function might be a good idea, but so far I've been only playing with $x=1,0$ and $y=1,0,$ and they didn't seem to help in proving surjectivity.

Any help is appreciated. Thanks!

(Proving injectivity is easy)

答:
First, notice that by defining$ g(x,y)=yf(x)+y$ , and using the injectivity you have found, then $f \circ g(0,y) $, you can find out that f(0)=0.

Now , find (1,y) s.t $f \circ g (1,y)$ isnt0 ,say C.

Suppose negatively that there is real number $M_0$ s.t $ M_0 $ isn't in Im(f).

$ f \circ g (1,y) = f(y) +y $

Notice that in this case (by the given equality) , for any $ y \in \mathbb{R} $, $M_0 +y $ isn't in Im(f)- for contradiction, take $y=c-M_0$


问:
Showing continuity for trigonometric function
I came across a question involving continuous extension of a trigonometric function:
$$ \frac{\sin^2(\frac{1}{x})\sin^3x}{x} $$  if $ x \lt 0$
and $$x(x-1)(x-2)(x-3)(x-4)(x-5)$$
if $x \ge 0 $. How do we go about showing that it is continuous at 0? I know you have to show that $\lim_{x\to 0^-}$ = $\lim_{x\to 0^+}$, but not really sure how to find the answer for $\lim_{x\to 0^-}\frac{\sin^2(\frac{1}{x})\sin^3x}{x}$. Is it possible to explain step by step?

答:
Fact 1 $\lim_{x\to0}\frac{\sin x}{x}=1$ See this question

Fact 2 $\lim_{x\to\infty}\frac{\sin x}{x}=0$ (since $|\frac{\sin x}{x}|\le 1/|x|\to0$).

Equivalently, $\lim_{x\to0}x\sin\frac{1}{x}=0$.

Then $$\frac{\sin^2(\frac{1}{x})\sin^3x}{x}=\left(x\sin\frac{1}{x}\right)^2\left(\frac{\sin x}{x}\right)^3\to0^2\times1^3=0$$