问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
The $\delta$ "function" is not really a function at all. We define $\delta$ as a sort of special notation. When we write

$\int \delta(x) f(x) dx$

This is just "syntactic sugar" (shorthand) for $f(0)$. It has many of the same properties that an ordinary function does - for example,

$\int \delta(x) (f(x) + g(x)) dx = \int \delta(x) f(x) dx + \int \delta(y) g(y) dy$

But do not mistake $\delta$ for a "real function". It is not one.


答:
No. Defining $\delta(0)=\infty$ is not at all right, because it has to equal $\infty$ in just the "right way". You can define this as a distribution, which ignoring all technicalities, just means that it is an "evaluation linear map". I.e, given an appropriate vector space $V$ of functions $\Bbb{R}^n\to\Bbb{C}$, we consider $\delta_a:V\to\Bbb{C}$ defined as $\delta_a(f):= f(a)$. In other words, $\delta_a$ is an object which eats a function as input and spits out the value of the function a the point $a$.

Afterall, this is literally what we want it to do: when we write $\int_{\Bbb{R}^n}\delta_a(x)f(x)\,dx=\int_{\Bbb{R}^n}\delta(x-a)f(x)\,dx=f(a)$, we literally mean that $\delta_a$ is an object such that whenever we plug in a function $f$, we get out its value at $a$. Of course, writing it in this way inside an integral is a priori just nonsense, because there is no function $\delta_a:\Bbb{R}^n\to\Bbb{C}$ for which the above equality can hold true.

So, in summary, the dirac delta is a function, but it's just that the domain of the dirac delta is a space $V$ of functions, and the target space for $\delta_a$ is $\Bbb{C}$. In short, it is the "evaluation at $a$ map".

As a side remark: the dirac delta is not in any way magical/esoterical once you think of it as an evaluation map. The concept of a function as being a mapping from one set to another set is (from our luxurious perspective of having hindsight) a completely standard concept. So, in this regard the dirac delta is simply a function/mapping. The only difference is that the domain is a space of functions.

Furthermore, the concept of evaluation maps is a very basic concept in linear algebra (e.g., if you study the isomorphism between a finite-dimensional vector space and its double dual you'll see exactly what I'm talking about).

Now, the "difficulty" which comes with this is the question of "how to do calculus with these new types of objects". What I mean is in the ordinary setting of discussing functions $f:\Bbb{R}\to\Bbb{R}$, we have a notion of convergence/limits (i.e a topology), we have a notion of derivative (the study of differential calculus), and we have the notion of anti-derivatives/finding primitives etc. Extending these ideas to the more general setting is where the heart of the matter lies, and to fully appreciate that one should study more closely Laurent Schwartz's theory of distributions.

ContentLicense: CC BY-SA 4.0

Title: Is it meaningful to define the Dirac delta function as infinity at zero?

Body:
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$


问:
$f: X \to Y$ is $\mathcal{A}-\mathcal{B}$-measurable if and only if $f^{-1}(B) \in \mathcal{A}$ for all $B \in \delta$
Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ measurable space. A function $f: X \to Y$ is $\mathcal{A}-\mathcal{B}$-measurable if $f^{-1}(B)\in \mathcal{A}$ for all $B\in \mathcal{B}$

Suppose that $\delta \subset \mathcal{P}(Y)$ fulfills that $\sigma(\delta)=\mathcal{B}$.

$\sigma(\delta)$ is the sigma algebra generated by $\delta$

Proof that:

$f: X \to Y$ is $\mathcal{A}-\mathcal{B}$-measurable if and only if $f^{-1}(B) \in \mathcal{A}$ for all $B \in \delta$

Let $B\in \delta$ given that $\delta \subset \sigma (\delta)=\mathcal{B}$ then $B\in \mathcal{B} $, as $ f: X \to Y$ is $\mathcal{A}-\mathcal{B}$ measurable then $f^{- 1}(B)\in \mathcal{A}$.

For the other implication, I set a $B \in \mathcal{B}$ I should show that $f^{- 1}(B) \in \mathcal {A}$ but my working hypothesis only solves the problem for me when $B\in \delta$, how can I resolve this implication?

答:
Suppose $f$ is $\mathcal{A}-\mathcal{B}$ measurable. Then clearly, we see that if $B \in \delta$, we have $B \in \sigma(\delta) = \mathcal{B}$. Then $f^{-1}(B) \in \mathcal{A}$.

Now suppose that for all $B \in \delta$, $f^{-1}(B) \in \mathcal{A}$. Define $\Sigma = \{C \subseteq Y | f^{-1}(C) \in \mathcal{A}\}$. It is easy to verify that $\Sigma$ is a $\sigma$-algebra on $B$. And we see that $\delta \subseteq \Sigma$. Therefore, $\mathcal{B} = \sigma(\delta) \subseteq \Sigma$. That is, for $B \in \mathcal{B}$, $f^{-1}(B) \in \mathcal{A}$. Then $f$ is measurable.


问:
Is the graph of this problem statement incorrect?

Here is the problem statement:

According to the parabola in the figure of the function $f(x) = ax^2 + bx + c$, it is given that $a + b + c = 0$. The points $A$ and $B$ are the x-intersects, and the point $C$ is the y-intersect of the parabola. Let the origin be the point $O$, and determine the area of the triangle $COA$.

Here are my solving steps and why I think the plot doesn't seem right.

From the figure, it is apparent that $C = 3$, which leaves me with $a + b = -3$, from which I can write $b = - a - 3$.

I find the roots of the parabola, which turn out to be $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Substituting $b = - a - 3$ and $c = 3$ the roots are $x_1 = 1$ and $x_2 = \frac{3}{a}$.

However, since the value from under the square root on step 2 above is a positive value, I can assume that $x_1$ > $x_2$, simply because I used $+$ of the $\pm$ for $x_1$, and the $-$ for $x_2$. But the problem's answer accepts 1 to be the x coordinate of the point $A$ and proceeds on calculating the area from there with $\frac{3\times1}{2}=1.5$. I'm intending to appeal for the problem, but can't make sure whether or not I'm missing something.

答:
In solving the quadratic equation, x= $\frac{(a+3)\pm\sqrt{(a+3)^2-12a}}{2a}$ =$\frac{(a+3)\pm|a-3|}{2a}$

when a$\geq$3, $x_1$=1>$x_2$=$\frac{3}{a}$;x2 is the smaller root.
In this case, the area of $\Delta$ AOC has many values pending on value of a.

when  a$\leq$3, $x_1$=1<$x_2$=$\frac{3}{a}$;1 is the small root.
In this case, the area of $\Delta$ AOC is $\frac{3}{2}$.

Hope this clarify the picture.


问:
Why does a Reuleaux triangle appear when drilling a hole with a knife
As the season commands, I was blowing out eggs for the kindergarten. In order to have large holes at both ends of the egg I first poke a small hole with a pin and then enlarge the said hole with a pointy knife (by rotating said knife in the hole in the hope of creating a round hole). To my surprise, enlarging the hole in such a fashion ended up giving me (in 7 out of 8 tries) a shape which rather looks like a Reuleaux triangle than a circle.

Now it's obvious that the resulting shape should have constant width. But my [first and vague] question is: are there reasons not to get a circle? (and is the phenomenon confirmed?)

In some sense a "Reuleaux triangle"-like shape is more probable since it has a much smaller symmetry group (so if the medium is not uniform, it might be a more "optimal" solution [to some optimisation problem]). So my [hopefully more precise] question is:

Question: What is a good model for this "drilling process" which could explain a "Reuleaux triangle"-like shape?

Just for details (but I doubt it matters) the knife was only sharp one side and both sides of the knife touch the edge of the hole at all times while the drilling occurs.

答:
I wouldn't say this counts as a "good model for the drilling process", but...

The rouleax triangle has much smaller area than the circle (and although I didnt check, it's probably also smaller than the 5-gon etc).  E.g. for diameter $1$:

$$Area(Rouleax) = \frac{\pi - \sqrt{3}}{2} \approx 0.705 < Area(Circle) = \frac{\pi}{4} \approx 0.785$$

This means it's a lesser amount of eggshell to be removed before your knife can make full rotations.

While this isn't a "mechanics" or "process" based explanation, maybe this counts as a "physics" style or "principle of least action / minimum energy" style explanation?  :)

Just for Fun postscript: There must be another constraint at work, coz my explanation cannot explain why you don't get even smaller-area (non-convex, non-constant-width) Kakeya sets.


答:
Adding to the existing answers and comments, I think a good way to argue against the slogan "$\delta(0)=\infty$" is to point out how limiting it is with respect to developing intuition for other things of the same "type" as $\delta$ itself. The $\delta$-"function" is, as others have said, not a function; rather, it's just something which can be integrated, which is not quite the same thing!

Making this rigorous leads to a lot of very interesting mathematics, to which the whole "$\delta(0)=\infty$" slogan is (in my opinion at least) a conceptual obstacle. Even if it doesn't get in the way of how one works with $\delta$ specifically, ignoring subtleties early on will just make things harder when they become central later.


问:
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$? (Singapore 1970)

I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $2^{2001}\mod{10}\equiv 4$ and $2^{2001}\pmod{10}\equiv 5$. Could you please explain to me how to solve this question? This question is multiple choice with options $1999, 2003, 4002, 6003, 2002$

答:
The answer is
$$
\overbrace{\lfloor2001\log_{10}(2)\rfloor+1}^\text{digits in $2^{2001}$}+\overbrace{\lfloor2001\log_{10}(5)\rfloor+1}^\text{digits in $5^{2001}$}
$$
However, we also have, using Iverson Brackets,
$$
\lfloor x\rfloor+\lfloor y\rfloor=\lfloor x+y\rfloor-[\{x\}+\{y\}\ge1]
$$
So we need to know $\{2001\log_{10}(2)\}+\{2001\log_{10}(5)\}$, but since $2001\log_{10}(2)+2001\log_{10}(5)=2001$, we know that the sum of their fractional parts is exactly $0$ or exactly $1$. Since the fractional parts are both positive, we must have exactly $1$.

Therefore,
$$
\begin{align}
\lfloor2001\log_{10}(2)\rfloor+1+\lfloor2001\log_{10}(5)\rfloor+1
&=\lfloor2001\log_{10}(2)+2001\log_{10}(5)\rfloor+1\\
&=2002
\end{align}
$$


问:
The graph of $x^{n}+y^{n}=r^{n}$ for sufficiently large $n$
The graph of the function $x^{n}+y^{n}=r^{n}$ for certain large values of $n$ looks suspiciously like a square.

See this page from wolframalpha. Have any results been proven regarding this observation? What do we call this figure anyway?

答:
By symmetry, you can consider the equation $y^n+x^n=r^n$ for $0 \leq x \leq r$. Rewrite as
$$
y(x) = \sqrt[n]{{r^n  - x^n }} = \sqrt[n]{{r^n  - r^n \bigg(\frac{x}{r}\bigg)^n }} = r\sqrt[n]{{1 - \bigg(\frac{x}{r}\bigg)^n }},
$$
for $0 \leq x \leq r$. This shows that $y$ is strictly decreasing from $r$ to $0$ as $x$ varies from $0$ to $r$, respectively, and that the sequence of functions $y(x) = y_n (x)$ converges pointwise, as $n \to \infty$, to the function $f$ defined by $f(x)=r$ if $0 \leq x < r$ and $f(r)=0$; moreover, the convergence to $f$ is uniform for $x \in [0,a]$, for any $0 < a <r$ (but not for $x \in [0,r]$, since $y(r)=0$). This accounts for the square shape.</p>

ContentLicense: CC BY-SA 3.0


问:
Number of pieces produced by cutting a folded ribbon
A piece of ribbon 1 metre long is folded in half so that the two ends are on top of each other. This doubled piece of ribbon is then folded in half again. The folded piece of ribbon is then cut  right through its midway point (see my drawing below).

This produces 5 pieces of ribbon of lengths: $\frac{1}{8},\frac{1}{4},\frac{1}{4}, \frac{1}{4}, \frac{1}{8}$.

My question is what happens after $n$ folds?

Trying a few more examples has led me to guess that after $n$ folds: there are $2^n + 1$ pieces of ribbon, with two strands of length $\frac{1}{2^{n+1}}$ and the rest of length $\frac{1}{2^{n}}$. I am having difficulty creating a convincing proof of this.

I came across this problem on an Oxford University Thinking Skills Admissions test ( Question 38, page 23 ). If this is a well known problem, I would also be interested in a source to read about the question in more detail.

答: Imagine the ends of the ribbon are tied together after the first fold, making it into a loop. As antkam hinted, each fold doubles the number of layers, so we end up making $2^n$ cuts, dividing the loop into $2^n$ pieces. Every piece is the same length (as it goes from the cut to the edge and back), and exactly one of these pieces consists of two equal-length parts that we tied together.


问:
Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
In Thomas's Calculus (11th edition), it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x) \, dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential)? Then, if we rearrange we get $dy/dx=f'(x)$, and so $dy/dx$ can be seen as a ratio of $dy$ and $dx$. I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, and for $dy/dx$ to be a ratio both variables need to be independent.

答:
$dy/dx$ is quite possibly the most versatile piece of notation in mathematics. It can be interpreted as

A shorthand for the limit of a quotient: $$
    \frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta
    x} \, .$$

The result of applying the derivative operator, $d/dx$, on a given expression $y$.

The ratio of two infinitesimals $dy$ and $dx$ (with this interpretation made rigorous using non-standard analysis).

The ratio of two differentials $dy$ and $dx$ acting along the tangent line to a given curve.

All of these interpretations are equally valid and useful in their own way. In interpretations (1) and (2), which are the most common, $dy/dx$ is not seen as a ratio, despite the fact that it often behaves as one. Interpretations (3) and (4) offer viable alternatives. Since Mikhail Katz has already given a good exposition of infinitesimals, let me focus the remainder of this answer on interpretation (4).

Given a curve $y=f(x)$, the equation of the line tangent to the point $(a,f(a))$ is given by
$$
g(x)=f'(a)(x-a)+f(a) \, .
$$
This tangent line provides us with a linear approximation of a function around a given point. For small $h$, the value of $f(a+h)$ is roughly equal to $g(a+h)$. Hence,
$$
f(a+h) \approx f(a) + f'(a)h
$$
The linear approximation formula has a very clear geometric meaning, where you imagine staying on the tangent line rather than the curve itself:

We can then define $dx=h$ and $dy=g(a+h)-g(a)$:

Since the tangent line has a constant gradient, we have
$$
\frac{dy}{dx}=\frac{g(a+h)-g(a)}{h}=g'(a)=f'(a) \, ,
$$
provided that $dx$ is non-zero. Here, $dy$ and $dx$ are genuine quantities that act along the tangent line to the curve, and can be manipulated like numbers becasuse they are numbers.

This notion of $dy/dx$ becomes very useful when you realise that, in a very meaningful sense, the tangent line is the best linear approximation of a function around a given point. We can turn the linear approximation formula into an exact equality by letting $r(h)$ be the remainder term:
$$
f(a+h)=g(a+h)+r(h)=f(a)+f'(a)h+r(h) \, .
$$
As $h\to0$, $r(h)\to0$. In fact, the remainder term satisifes a stronger condition:
$$
\lim_{h \to 0}\frac{r(h)}{h}=0 \, .
$$
The fact that the above limit is equal to $0$ demonstrates just how good the tangent line is at approximating the local behaviour of a function. It's not that impressive that when $h$ is small, $r(h)$ is also very small (any good approximation should have this property). What makes the tangent line unique is that when $h$ is small, $r(h)$ is orders of magnitude smaller, meaning that the 'relative error' is tiny. This relative error is showcased in the below animation, where the length of the green line is $h$, whereas the length of the blue line is $r(h)$:

Seen in this light, the statement
$$
\frac{dy}{dx}=f'(a)
$$
is an expression of equality between the quotient of two numbers $dy$ and $dx$, and the derivative $f'(a)$. It makes perfect sense to multiply both sides by $dx$, to get
$$
dy=f'(a)dx \, ,
$$
Note that $dy+r(h)=f(a+h)-f(a)$, and so
\begin{align}
f(a+h)-f(a) - r(h) &= f'(a)h \\
f(a+h) &= f(a) + f'(a)h + r(h) \, ,
\end{align}
which gives us another notion of the derivative, through the lense of linear approximation.


问:
Quadratic Rings which are UFDs but not PIDs
D=-1,-2,-3,-7 and-11 are the only five negative square free integers for which the corresponding quadratic Rings are Euclidean domains. Also D=-19,-41,-43 and-167 are the only four square free negative integers for which the  corresponding quadratic rings are PIDs but not EDs. Are there any negative square free integers <-167 for which the corresponding quadratic rings are UFDs but not PIDs? Thank you in advance for your valuable time.</p>

答:
Assuming that by "quadratic rings" you mean the integer ring of a quadratic extension of $\Bbb Q$.

The answer is then "no".

The integer ring of any number field is a Dedekind domain. It is a standard result that a ring is a PID if and only if it is a Dedekind domain and a UFD.

The proof can be found e.g. here.


问:
Concerning a result of Formanek about $k[F_1,\ldots,F_n][F_{n+1}]=k[x_1,\ldots,x_n]$
A result of E. Formanek says:
Let $k$ be a field of characteristic zero and let $R=k[x_1,\ldots,x_n]$ be the polynomial ring in $n$ variables.
Let $F_1,\ldots,F_n \in R$ have invertible Jacobian in $R$ (namely, the Jacobian is a non-zero scalar).
If there exists $F_{n+1} \in R$ such that
$k[F_1,\ldots,F_n][F_{n+1}]=R$,
then $k[F_1,\ldots,F_n]=R$.

Now let $I$ be the ideal of $R$ generated by $F_1,\ldots,F_n$:
$I=\langle F_1\ldots,F_n \rangle= RF_1+\cdots+RF_n$.

Is it possible to prove the following claim by adjusting Formanek's proof:

If there exists $F_{n+1} \in R$ such that
$k[F_1,\ldots,F_n][F_{n+1}]+I=R$,
then $k[F_1,\ldots,F_n]=R$.

Of course, if $I \subseteq k[F_1,\ldots,F_n][F_{n+1}]$, then this is just Formanek's result.

Now also asked in MO.
This is a relevant question.

Thank you very much!

答: Your condition $k[F_1,\ldots,F_n][F_{n+1}]+I=R$ is equivalent to $k[F_{n+1}]\to R/I$ is onto for a suitable $F_{n+1}\in R$. If $I=R$, there is nothing to prove (and Jacobian conjecture says this can not happen) and if not, the Jacobian condition says $R/I$ is a finite product of fields $L_1\times\cdots\times L_r$, where each $L_i$ is a finite extension of $k$. In characteristic zero, these are all simple extensions and then I will leave you to check that such an $F_{n+1}$ can be found.


问:
can we prove that is an increasing function
I have a function $f: \mathbb{R}\to [0,+\infty)$
wich is continuous, convex, even and
f(t)=0 if and only if t=0

Can we prove that f is increasing

答:
Take $$f(x)=x^2$$

$$(\forall x\in \Bbb R)\;\; f''(x)=2\ge 0$$

$$\implies f \text{ is convex at } \Bbb R$$
and
$$f(x)=0 \iff x=0$$

but $ f $ is not increasing.


问:
Inequality into Binary variable in Linear Programming
I want to solve this problem :
We have 100 cities where if there is a fire we need to call at least 1 police officier and 3 firefighters from place that are in distance of 100 km or less.
The distance from city i to j is : Di
The cost for keeping a firefighter in a city is A and the cost for the police officier is B.

$$ \sum_{i=1}^{100} A.X_i + B*Y_i $$
$$st  \sum_{i=1}^{100} X_i*Y_{ij} >= 3 $$
$$ \sum_{i=1}^{100} Y_i*Y_{ij} >= 1$$

$$ \text{where} \space X_i \space  \text{is indicator for placing firefighter in city I and}  \space Y_i \space \text{is for police officier} $$

But I cannot manage to create a binary variable for the constraint of city of distance of 100 km or less , like let $Y_{ij}$ be the indicator if the distance is 100 km or less.

答:
Your existing constraints are not linear, and you should not use both $Y_i$ and $Y_{i,j}$.

Let $E$ be the set of $(i,j)$ pairs where cities $i$ and $j$ are within distance 100 of each other.  For $(i,j)\in E$, let $F_{i,j}$ indicate whether a firefighter from city $i$ serves city $j$, and let $P_{i,j}$ indicate whether a police officer from city $i$ serves city $j$.  The desired linear constraints are
\begin{align}
\sum_{i: (i,j)\in E} F_{i,j} &\ge 3 &&\text{for all $j$} \\
\sum_{i: (i,j)\in E} P_{i,j} &\ge 1 &&\text{for all $j$} \\
F_{i,j} &\le X_i &&\text{for all $(i,j)\in E$} \\
P_{i,j} &\le Y_i &&\text{for all $(i,j)\in E$} \\
\end{align}


问:
Why is my proof of the integral of the dirac function being equal to 1 incorrect?
I know that there are questions on this website on how to prove that the integral of the Dirac function is $1$. However, I am interested in knowing why my proof is wrong and if there is a way to salvage it. So here's the question :

The Dirac delta function can be defined by the limit of a short pulse: $$\delta(t-t_0) = \lim_{\Delta \rightarrow 0}({f_{\Delta}(t)})$$ where $f_{\Delta}(t) = \frac{1}{\Delta}$ for $t_0\leq t\leq t_0 + \Delta$ and $f_{\Delta}(t) = 0$ otherwise. Convince yourself that the integral $\int_{t_1}^{t_2} \delta(t-t_0) dt = 1$ if $t_1\leq t_0 \leq t_2 $

Here's my approach:
$$\int_{t_1}^{t_2} \delta(t-t_0) dt = \int_{t_1}^{t_2} \lim_{\Delta \rightarrow 0}({f_{\Delta}(t)}) dt$$ $$=\int_{t_1}^{t_2} \lim_{\Delta \rightarrow 0}(\frac{1}{\Delta})dt$$ $$=\lim_{\Delta \rightarrow 0}(\frac{1}{\Delta})\int_{t_1}^{t_2}dt$$ $$=\lim_{\Delta \rightarrow 0}(\frac{1}{\Delta}) \cdot (t_2-t_1)$$ $$=\lim_{\Delta \rightarrow 0}(\frac{t_2-t_1}{\Delta})$$ If we let $t_1 = t_0$ and $t_2 = t_0 + \Delta$, then we get $$\lim_{\Delta \rightarrow 0}(\frac{\Delta}{\Delta})$$ $$ = 1$$

Here's why I doubt my proof.

First, in the third line of my development, I took out the limit from the integral. My reasoning behind the maneuver is that this limit does not depend on $t$ and therefore I can treat it as a constant. However, upon re-reading my proof, I am wondering if I am really "allowed" to do this. Since I am treating the limit as a constant, I am assuming that it exists and is finite, which is not the case, right? (since $\lim_{x \rightarrow 0} (1/x)$ does not exist).

Secondly, to get to the 6th line, I defined $t_2 = t_0 + \Delta$. Am I allowed to do this? I defined it that way only because it was mathematically convenient, not because it actually makes sense, which is why I doubt its validity.

I personally think that the proof is incorrect, but I'm not sure. Perhaps one could argue otherwise and I would be interested to know how.

So, can you confirm to me that my proof is incorrect, and can you please let me know if there's anything I can do to "salvage" it, or if I need to try another approach

答:
Your problem begins with the definition of Dirac delta
$$ \delta(t-t_0) = \lim_{\Delta \rightarrow 0}({f_{\Delta}(t)}) \tag{1} $$
which is not an equality of functions since the pointwise
limit does not exist at $\,t=t_0.\,$ Thus, your very first
step of writing
$$\int_{t_1}^{t_2} \delta(t-t_0) dt = \int_{t_1}^{t_2}
 \lim_{\Delta \rightarrow 0}({f_{\Delta}(t)}) dt \tag{2} $$
is problematic unless interpreted as an appropriate kind
of integral suitable for distributions.
More problematic is that, as the Wikipedia article
Dirac delta function
states:

[...] it is understood that the limit is always taken outside the integral.

Next, your step equating the integral to
$$ =\lim_{\Delta \rightarrow 0}(\frac{1}{\Delta})\int_{t_1}^{t_2}dt \tag{3} $$
is flawed because you should not be integrating the constant function
$1$ but instead the function $\,\Delta f_\Delta(t).\,$
As the definition states:

$f_{\Delta}(t) = \frac{1}{\Delta}$ for $t_0\leq t\leq t_0 + \Delta$ and $f_{\Delta}(t) = 0$ otherwise.

What you are integrating here has the value $1$ for $t_0\leq t\leq t_0 + \Delta$
and $0$ otherwise. Thus the integral is not $\,(t_2-t_1)\,$ as you have it in your next step.


问:
Why if the columns of $Q$ are an orthonormal basis for $F^n$ then $Q$ is unitary?
Why if the columns of $Q$ are an orthonormal basis for $F^n$ then $Q$ is unitary?

There exist any available theorems to prove it?

答:
Here's one way to see it. If the columns form an orthonormal basis, you can directly check that $QQ^\ast = Q^\ast Q = I$. So $Q^{-1} = Q^\ast$. Then we can compute what $Q$ does to norms using the inner product: $$\begin{align*} \|a\|^2 &= (a,a) \\
&= (Ia, a) \\
&= (Q^{-1}Qa,a) \\
&= (Q^\ast Qa,a) \\
&= (Qa, Qa) \\
&= \|Qa\|^2 \end{align*}$$ and so $Q$ preserves norms and is bijective since it has an inverse.</p>


问:
Solutions of $u_{xx} + 2 \mathrm i u_{xy} + u_{yy} = 0$
A problem in A. Friedman, Partial Differential Equations, 1969, chapter 1, section 14, is concerned with the elliptic, but not strongly elliptic PDE on the open unit disk $\mathbb D = \{z= x+ \mathrm iy \mid |z| < 1\}$ given by
\begin{equation*}
u_{xx} + 2\mathrm i u_{xy} + u_{yy} = 0, \qquad u(z) = 0, \forall z\in \partial \mathbb D.
\end{equation*}
Friedman claims that for any complex analytic function $f(z)$ defined on the closed unit disk $\bar{ \mathbb D}$, the function $u(z) = (1-|z|^2) f(z)$ solves the Dirichlet problem.

However, taking $f = 1$ already shows that this is not true. Can you explain that? Can you give a better example?

答:
There appears to be a typo in the equation Friedman treats. Whenever $f(z)$ is analytic,
the function $u(z)=(1-|z|^2)f(z)$ does satisfy
$$ u_{xx}+2{\rm i}u_{xy} - u_{yy}=0.$$
The symbol of this equation is
$$
a(\xi)=\xi_1^2+2{\rm i}\xi_1\xi_2-\xi_2^2 = (\xi_1+{\rm i}\xi_2)^2,
$$
which does not vanish for any real vector $\xi=(\xi_1,\xi_2)$ except for $(0,0)$.
This makes the PDE elliptic in the sense that I believe Friedman's book uses (which I take from a 1955 paper of Nirenberg titled Remarks on strongly elliptic partial differential equations).
But it is not ``strongly elliptic,'' since it is false that
$$\Re a(\xi)=\xi_1^2-\xi_2^2 \ge c(\xi_1^2+\xi_2^2)$$
for some postive constant $c$.


问:
Can we construct a function that is continuous at each rational point and discontinuous at uncountably many irrational points?
Some functions on $\mathbb{R}$ are continuous at every irrational and discontinuous at every rational, but there aren't any functions that are continuous at every rational and discontinuous at every irrational. Additionally, it is easy to come up with a function that is continuous at every rational and discontinuous at countably many irrationals. Now, I am wondering if there are any functions from $\mathbb{R}$ to $\mathbb{R}$ that are continuous at every rational point, but discontinuous at uncountably many irrational points. If so, could you give me some examples?

答:
I’m turning my comment into an answer. We know that, as $\mathbb{Q}$ has measure zero, there is an open subset $U \subset \mathbb{R}$ with arbitrary small measure containing $\mathbb{Q}$. We just assume that the Lebesgue measure of $U$ is finite.

Let $f$ be the indicator of $U$. Clearly, $f$ is continuous at any point in $U$. Since $U$ is dense, $f$ cannot be continuous at any point outside $U$, so the set of discontinuities of $f$ is $\mathbb{R} \backslash U$ m, which contains only irrationals and has infinite Lebesgue measure (and in particular is uncountable).


问:
Correct interpretation of noise-based wind field generation by Khorloo Oyundolgor
I am interstring in the creation 3D wind field for my falling snow scene using DirectX.
After learning different approaches, I came to a conclusion to use the following method because of the detailed description and good computational complexity as well. The study is quite big, but actually, looking upon sections 2.3, 2.4, 3.5, and 5.4 may be enough.

The bottom line is to construct wind power spectrums for each wind direction and apply inverse Fourier transform based on the dimension of the simulated wind field for each wind component: diagram.

Implementation steps:

Step 1: Determine characteristics of wind fields by defining wind parameters such as mean wind speed $U(z)$ at the given height $z$ and the terrain roughness $z_0$.

Step 2: Calculate friction force using Equation 3.2.

Step 3: Calculate parameter values of $A_u$, $A_v$, and $A_w$, as defined in Chapter 5.4.2, for each wind velocity component.

Step 4: Select parameter $\beta$ as $5.0/3.0$ according to Kolmogorov’s law applied in the along-mean-wind direction.

Step 5: Construct wind power spectrums $S_u$, $S_v$, $S_w$ for each wind direction according to Equations 5.6, 5.8, and 5.9.

Step 6: Apply inverse Fourier transform based on the dimension of thesimulated wind field for each wind component: $IFFTnD(S_u)$, $IFFTnD(S_v)$, $IFFTnD(S_w)$

Despite the in-depth description of the model, I am not sure that understand some fundamental ideas:

Question 1

According to diagram, we got a 3D wind field for the giving height ($z$, $U(z)$). Does it mean that I can choose some base height $z_{base}$ and construct the frozen-wind model once only for my scene?

Question 2

What is the output of the $IFFT$ operation? Am I right that $IFFTnD(S_u)$ gives a function that returns a velocity value along the $u$ axis for point coordinate?

Question 3

In this paper, $1/f^{\beta}$ noise is generated using "Fourier filtering". In fairness, I have not learned this part of study properly yet, so answer me just if it's obvious for you:

How the $1/f^{\beta}$ function will look like?

Is it possible to use another implementation of $1/f^{\beta}$, for example, based on Simplex Noise?

Thanks a lot in advance for your response.

答:
Ok, I've learned the issue and going to try to answer myself. Still glad to get your comments.

Question 1

You (me) totally missed the key idea.  Equations 5.9 is evaluating for each point in the N-space. Eventually, you got an nD float array which is input for inverse fast Fourier transform operation.

Question 2

No. The iFFT's input/output is the discrete sets. And yes, every set's value describes turbulence speed components.

Question 3

Actually, no difference in how to implement fBm noise generator.


问:
Grammatical standard for ordinals of complex expressions
In English, ordinals are usually created by appending a “-th” or “-eth” sound to the end. In mathematics, this extends to variables, hence the word “$n$th”. But the numbers 1, 2, 3, and 5 are irregular—first, second, third, and fifth instead of *oneth, *twoth, *threeth, and *fiveth, however those would be pronounced. 5 is the odd one out here because the difference is only apparent if you spell it out (“fifth”) or pronounce it aloud (/fɪfθ/) rather than using the digit (“5th”) as is more common in mathematics. This applies not only to these irregular numbers themselves, but to all numbers that end in those same digits: twenty-first, twenty-second, hundred-and-first, etc.

What does this mean for mathematical writing when you have an expression that ends one of these numbers? What's the ordinal for $n+1$? $n+1$st? $n+1$th? $(n+1)$th? $n$th$+1$? $n$th$+1$st? In informal contexts I have seen all of these examples.

Alterantively, are these all proscribed, and if so, what alternative is there? Do you introduce a new variable on the spot (e.g. “$m$th where $m=n+1$”)?

(Not sure how to tag this question, by the way; edit me if I’m wrong there.)

答:
The standard is to append the -th ending to all integer-denoting expressions other than those ending in the digits $1$, $2$, or $3$. If the expression contains typographic spacing (as e.g. surrounds the plus sign in $n+1$), then the expression must be enclosed in parentheses or brackets, to prevent the -th from visually attaching specifically to the elements of the expression following the last space. Thus we might have the $[m+\frac12n(n+1)]$th term in a sequence, but not the $m+\frac12n(n+1)$th term. The only use of -st, -nd, or -rd will then occur in such “words” as $4371$st, $22$nd, or $50$,$003$rd.

An exceptional case arises if we use a thin space rather than an unspaced comma to parse the groups of three digits in a large number. For example, $29\,403\,002$nd would be more natural than $(29\,403\,002)$th.


问:
Decomposition of a sum of cubes
I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing  the squares. Can you give me a hint?

Title: Decomposition of a sum of cubes

Body: Hint:  Use $a^2+b^2-ab=(a-b)^2+ab$


Title: Decomposition of a sum of cubes

Body:
I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing  the squares. Can you give me a hint?

Body:
Compare $S = (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

with $T = (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.$

Adding: $S + T = 2(a^3 + 3ab^2).$


问:
Defining the order topology on the natural numbers
How might one go about defining the order topology on the natural numbers with natural order (<)? I'm working on constructing the basis for the topology, and I realize that the intervals [1,b) and (a,b) must be present in the basis, since 1 is the least element of the naturals, but I'm not sure how to finish it off and properly describe the topology</p>

答:
$[1,2) = \{1\}$ in $(\Bbb N, <)$, so $\{1\}$ is basic open, as you mentioned.

For all $n >1$, $(n-1,n+1) = \{n\}$ is basic open (as an open interval).

But if all singletons are in the base for the order topology, all subsets of $\Bbb N$ are open, being unions of those. So we get the discrete topology. Nothing new but a valid order topology on such a "discrete" order.


问:
What is the probability of a random normal point falling inside an intersection of two half-spaces?
Given two fixed vectors $x_i, x_j \in \mathbb{R}^n$, what is the probability that a random point $y\sim \mathcal{N}(0, \mathbf{Id})$ in $\mathbb{R}^n$ falls inside the intersection of their upper half-spaces i.e. $y \in \{ z \mid z^T x_i > 0 \mbox{ and } z^Tx_j > 0\}$.

I suspect $\frac{\pi - \angle(x_i,x_j)}{2\pi}$ by pure intuition (from a 2D case and the angle of the dual-cone) but any brute-force solution (using say generalized spherical coordinate) seems messy. Is there an easy way to see why this is true? Any pointer is greatly appreciated. Thanks!

答:
Your problem in fact reduces to the 2D case: The spherical normal distribution is invariant under rotations, so we may assume without loss of generality that $x_1 = (1,0,0,\ldots,0)$ and $x_2 = (a_1,a_2,0,\ldots,0)$. Now, your above reasoning applies.

To elaborate a bit further:

First, we may assume that $\|x_1\|_2 = \|x_2\|_2 = 1$ because $z^\top x > 0$ if and only if $z^\top \tilde x > 0$, where $\tilde x = x\,/\,\|x\|$. We also assume $x_1$ and $x_2$ are linearly independent; otherwise, the answer is easy to deduce.

Now, take $U$ to be an orthogonal matrix such that $U x_1 = (1,0,0,\ldots,0)$ and $U x_2 = (a_1, a_2, 0, \ldots, 0)$. (To construct such a $U$, we can take its first row to be $x_1$ and its second row to be the normalization of $x_2 - \langle x_1, x_2\rangle\cdot x_2$; the remaining rows can be chosen arbitrarily to complete an orthonormal basis.) Note that $(Uz)^\top (Ux_k) = z^\top x_k$. Furthermore, $Uz\sim\mathcal N(0, I)$. Hence, it suffices to answer: What is the probability that $\hat z^\top \hat x_k > 0$ for both $k\in\{1,2\}$, where $\hat z = Uz\sim\mathcal N(0, I)$ and $\hat x_k = Ux_k$? In particular, only the first two coordinates of each $\hat x_k$ are non-zero. Hence, we may "forget" about the remaining coordinates and focus only on the first two coordinates of $\hat x_1$ and $\hat x_2$.

Because $\hat x_1$ and $\hat x_2$ both are of unit norm, the probability that $\hat z^\top \hat x_k > 0$ for both $k = 1, 2$ is, as you computed above,
$$\frac 12 - \frac{\arccos(\langle \hat x_1, \hat x_2\rangle)}{2\pi} = \frac 12 - \frac{\arccos(\langle x_1, x_2\rangle)}{2\pi}$$
(where $\arccos$ returns a value in $[0, \pi)$). Note that $\langle \hat x_1, \hat x_2\rangle = \langle Ux_1, U x_2\rangle = \langle x_1, x_2\rangle$ because $U$ is orthogonal. Hence we recover an expression for the probability in terms of $x_1$ and $x_2$.


问:
Where can I find the proof of this theorem about the smoothness of the initial data?
What to follow is a reference from the text Differential forms by Victor Guillemin and Peter Haine

So clearly the theorems $2.2.4$ and $2.2.5$ follow directely by the Cauchy theorem for ordinary differential equation provided that the vector field if locally lipschitz or of class $C^r$ for $r\ge 1$ since any function of calss $C^1$ is locally lipschitz. However how prove the last theorem? In particular I know that if the vector field $\vec v$ is of class $C^r$ for $r\ge 1$ then the solution of the correspondent system of differential equations is of class $C^r$ too but unfortunately this did not help me to prove the theorem. So could someone idicates where I can find the proof of the last theorem, please? I point out I did NOT study lebesgue integration and measure theory so that I ask courteously to not use them, thanks.

MY PROOF ATTEMPT

So by the regularity theorem we know that the function $\gamma_p$ is of class $C^\infty$ so that if $A_p$ is an open neighborhood of any $p\in V$ contained in $V$ then the set $J:=\gamma^{-1}_p[A_p]$ is open an open interval containing $a\in I$ and so if $\pi$ is the projection of $V\times I$ onto $I$ the statement follows proving that
$$
h(q,t)=[\gamma_p\circ\pi](q,t)
$$
for any $(q,t)\in A_p\times J$ because the set $A_p\times J$ is open and the function $\gamma_p\circ\pi$ is of class $C^\infty$ being a composition of such functions. So how prove the last equality? Is effectively it hold?

答:
There are quite a few alternative proofs of Theorem 2.2.6 and of many of its variants.

Incidentally, Theorem 2.2.6 is just a very particular case of the general situation (most notably, in general it is quite restrictive to assume that $I$ is the same for all $p$ as in Theorem 2.2.6 but it is fine in case we just care about a small neighborhood).

Unfortunately, there are no simple proofs (read short if you want since all proofs are classical), although in the present case most tend to be slightly simple since it is much harder to prove that $C^k$ implies $C^k$ for some finite $k$ then for $k=\infty$. The reason is that for a finite $k$ it is usually necessary a different argument to obtain say $C^k$ and not only $C^{k-1}$ (I am really simplifying this a lot since the theory is quite vast). Still all these proofs for $k=\infty$ are quite long.

Overall it is thus better to give a pointer to a canonical reference, such as Ordinary Differential Equations, by Jack K. Hale that unfortunately is not with us anymore. Have a look at Theorem 3.3 in page 21 (a bit too concise but one of the best places).


Title: Decomposition of a sum of cubes

Body:
hint

put
$$z=2y$$

and use

$$a^3+b^3=(a+b)(a^2-2ab+b^2+ab)$$
$$=(a+b)\Bigl((a-b)^2+ab\Bigr)$$

where
$$a=x+z \text{ and } b=x-z$$


问:
n roots in the finite field
I'm trying to prove the following statement:

$x^n-1$ has $n$ roots in the finite field $\mathbb{F}_q$ if and only if $q = 1\pmod n$

For the forwards implication, I know that $x^n-1$ is separable but then I'm not sure where to go from here.

For the backwards implication, I used $q = 1\pmod n)$ to get that $n \mid q-1$, and then I've worked out the formal derivative $nx^{n-1}$ because I want to somehow show that they don't share a common factor, to then conclude that that $x^n-1$ is indeed separable. But I'm not sure how to reach that so I'm stuck on this bit too.

Any help would be greatly appreciated!

答:
A finite field of order $q$ has cyclic multiplicative group of order $q^n-1$.

If $a\in\mathbb{F}_q$ is a root of $x^n-1$, then $a^n=1$; in particular, the multiplicative order of $a$ divides $n$. Conversely, if the multiplicative order of $a$ divides $n$, then $a^n=1$, hence $a$ is a root of $x^n-1$.

Thus, the roots of $x^n-1$ are precisely the elements of multiplicative order dividing $a$.

If $n|q-1$, then the multiplicative group of $\mathbb{F}_q$ has a unique subgroup of order $n$, and its elements are precisely the elements with multiplicative order dividing $n$; there are exactly $n$ of them, so $x^n-1$ has $n$ distinct roots.

Conversely, since the elements of multiplicative order dividing $n$ form a subgroup of the multiplicative group, if there are $n$ of them, then by Lagrange’s theorem you have that $n$ divides $q-1$.

Title: n roots in the finite field


问:
Finding the rational values of constant for which these constants are roots of equation
Problem :

Determine all rational values for which $a,b,c$ are the roots of $x^3+ax^2+bx+c=0$

Solution :

Sum of the roots $a+b+c = -a$ ........(i) ( Since , as per question $a,b,c$ are roots of equation and we have to find values )

$\sum_{a,b,c} ab = b ........(ii)$

$abc= -c \Rightarrow ab = -1 \Rightarrow a =\frac{-1}{b} .....(iii)$

Solving (i),(ii) and (iii) simultaneously we get an equation which is biquadratic in "a" which is :

$2a^4-2a^2-a+1=0$ Here we can find that $a = 1$ is a factor of this equation and by long division of polynomial we can divide $a-1$ with the given biquadratic we get cubic polynomial in a which is $2a^3+2a^2-1$ ...

Please guide further and suggest whether this is correct or not.. Thanks...

答:
Neither $-1$ nor $1$ is a solution to your final cubic, so this calls for the use of the cubic formula.

$$a = 2, b = 2, c = 0, d = -1$$
$$\Delta_0 = 2^2 - 3(2)(0) = 4$$
$$\Delta_1 = 2(2^3) - 9(2)(2)(0) + 27(2^2)(-1) = 16 - 108 = -92$$
$$C = \sqrt[3]{\frac{-92 + \sqrt{(-92)^2 - 4(4^3)}}{2}} = \sqrt[3]{\frac{-92 + \sqrt{8208}}{2}} = \sqrt[3]{\frac{12\sqrt{57} - 92}{2}} = \sqrt[3]{6\sqrt{57} - 46}$$

We can divide $8208$ by $-27a^2$ (or $-108$) to get the equation's discriminant, $-76$. This means the equation has one real root and two complex roots. That real root is given by:

$$-\frac{1}{6}\left(2 + \sqrt[3]{6\sqrt{57} - 46} + \frac{4}{\sqrt[3]{6\sqrt{57} - 46}}\right)$$

A number which is definitely not rational. So you have found the only rational root, 1.


问:
expected value of a log-normal distribution

Let $X$ be a continuous random variable with log-normal distribution, i.e.
$$
f_X(x)=\begin{cases}
    \frac{1}{x\sqrt{2\pi}} e^{-\frac{1}{2}(\log x)^2}, & \text{for x>0} \\
    0, & \text{otherwise}. \\
\end{cases}
$$
Determine the value of $A=\sqrt{2\pi}\mathbb{E}[|\log X|]$.

I know that generally, $\mathbb{E}[X]=e^{\mu+\frac{1}{2}\sigma^2}$. And in this case, $\mu=0$ and $\sigma^2=1$. My question is how do I apply this to $\mathbb{E}[|\log X|]$?

答:
By the Law of the Unconscious Statistician, you want to compute
$$
A = \sqrt{2\pi}\mathbb{E}[|\log X|] = \sqrt{2\pi}\int_{0}^\infty |\log x| f_X(x)dx
= \int_{0}^\infty \frac{|\log x|}{x} e^{-\frac{1}{2}\log^2 x}dx
$$
(first step: you should have checked the integral converges. Shouldn't be more than a formality.)

Now, to get rid of the absolute value, you can rewrite this as
$$\begin{align}
A &= \int_{0}^1 \frac{\log\frac{1}{x}}{x} e^{-\frac{1}{2}\log^2\frac{1}{x}}dx
+ \int_{1}^\infty \frac{\log x}{x} e^{-\frac{1}{2}\log^2 x}dx\\
&= 2\int_{1}^\infty \frac{\log x}{x} e^{-\frac{1}{2}\log^2 x}dx
\end{align}$$
the second equality by the change of variables $u=\frac{1}{x}$ in the first integral. Now, do the change of variables $t=\log x$ to get
$$\begin{align}
A &= 2\int_{1}^\infty \frac{\log x}{x} e^{-\frac{1}{2}\log^2 x}dx
&= 2\int_{0}^\infty t e^{-\frac{1}{2}t^2}dt
\end{align}$$
since $dt = \frac{dx}{x}$. But now, we know how to solve this! Either you recognize $\sqrt{2\pi}\mathbb{E}[|Y|]$ for $Y\sim\mathcal{N}(0,1)$; or you just notice that the integrand is a derivative, and you get
$$\begin{align}
A &= 2\left[-e^{-t^2/2}\right]_0^\infty = \boxed{2}
\end{align}$$


问:
Little $o$ notation problem
Determine the values of $a$ and $b$ if

$$ \sqrt{x^2 +1}=ax + b +o(x) $$

when $x\to 0$ and prove the result,
where $o(x)$ corresponds to the notation little $o$.

My first attempt was to rewrite the equation as

$$ \sqrt{x^2 +1} -ax - b = o(x) $$

and use the definition of little $o$, then I would need to show that the next lim is equal to zero

$$ \lim_{x \to 0} \frac{\sqrt{x^2 +1} -ax - b }{x} = 0$$

in order to use L'Hopital I need $0/0$ in the direct evaluation. So if $f(x) = \sqrt{x^2 + 1} -ax -b$ then I need $0= f(0) = 1-b $ or equivalent $b = 1$.

Using L'Hopital we need

$$  \frac{  \lim_{x \to 0} \frac{x-a}{\sqrt{x^2+1}} }{\lim_{x \to 0}  1} = 0$$

Substituting x,

$$ \lim_{x \to 0} \frac{ -a }{1} = 0$$

So $a=0$. But I'm not sure I'm susing correctly the implications in L'Hopital Theorem cause I known the limit exists and I would like L'Hopital to be true but I feel I'm forcing the hyptesis. Is there anoher way t find $a$ and $b$?

答:
There is simpler, set $f(x)=\sqrt{x^2+1}$ which is derivable in $0$.

Notice $f(0)=1$ therefore $\dfrac{f(x)-f(0)}{x}\to f'(0)=0$

Which is nothing more than $\sqrt{1+x^2}=1+o(x)$

Title: Little $o$ notation problem


Title: Decomposition of a sum of cubes

Body:
Use $(a+b)+(a-b)=2a$

and $(a+b)^2+(a-b)^2=2a^2+2b^2$

and $(a-b)(a+b)=a^2-b^2$

Then
$$
\underbrace{(\color{#C00}{(x+2y)}+\color{#090}{(x-2y)})}_{2x}\left(\!\vphantom{2^2}\right.\underbrace{\color{#C00}{(x+2y)}^2+\color{#090}{(x-2y)}^2}_{2x^2+8y^2}-\underbrace{\color{#C00}{(x+2y)}\color{#090}{(x-2y)}}_{x^2-4y^2}\left.\!\vphantom{2^2}\right)
$$


问:
Define a relation $R$ on the set $\mathbb{Q}$ by $aRb$ if and only if $|a - b| \leq 1$. Is $R$ reflexive, symmetric, and/or transitive?
My process so far.

We know that $aRa$ means that $|a - a| \leq 1$. Which is true, thus, $R$ is reflexive.

Now onto symmetric. $aRb \implies bRa$. This means we have $|a - b|\leq 1$. Now observe:
$$|a - b| \leq 1$$
$$-1 \leq a - b \leq 1$$
$$1 \geq b - a \geq 1\quad\quad \text{times }(-1)$$
Thus, $R$ is symmetric.

Now we need to show whether or not $R$ is transitive. This is where I am not really sure where to proceed.

Any help to point me in the right direction is great! Thanks!

答:
Well, consider $|a- b| \le 1$ and $|b-c| \le 1$.

$|a-b| \le 1$

$-1 \le a-b \le 1$ a bit of manipulation to focus on the range of $a$ in terms of $b$

$b-1 \le a \le b+1$

And to do the same for $|b-c| \le 1$

$-1 \le b-c \le 1$

$1 \ge c-b \le -1$

$b-1 \le c \le b+1$

If we have $a$ as low as possible, say $a = b-1$ and $c$ as high as possible, say $c= b+1$ then we have $|a-c| = |(b-1) - (b+1)| = 2$.

So not transitive.

Are you familiar with the triangle inequality?

It states that $|m| + |n | \ge |m+n|$ (try and play and see why this is so...try to guess why it is called the triangle inequality.)

Thus if $|a-b|\le 1$ and $|b-c| \le 2$ then $|a-b| + |b-c| \ge |(a-b)+(b-c)| = |a-c|$. so we have $2 \ge |a-b| +|b-c| \ge |a-c|$.

This implies we might be able to have $2 \ge |a-c| > 1$ be choosing $a$ and $b$ and $b$ and $c$ far enough apart.


问:
You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. Find angle $\angle MCN$
You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. If $KL$ is the diameter, $KL\parallel AB$, $KM\perp AB$ and $LN\perp AB$ Find angle $\angle MCN$.

Measuring this angle it works out to be $45^o$, but I don't know how to work it out. I attempted to utilize $KLNM$ being a rectangle (something which is proved since $KM\perp AB$ and $LN\perp AB$ which means $KM\parallel LN$ and also $KL\parallel MN$ hence $KLMN \#$ and since $KM\perp MN$ then $KMNL$ is a rectangle), but this didn't work out for me. Could you please explain to me how to solve the question?

答: It's apparent that $|OM|=|ON|=|OC|(=\sqrt 2\cdot|OK|)$ and that $\angle MON =90^\circ$. Then by the inscribed angle theorem, $\angle MCN =45^\circ$.


问:
Evaluating $\int_0^1 (1-x^2)^n dx$
In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.

b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:

$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$

and so in part (b) i got:

$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$

I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

答: Evaluate $f_n(x):=\int_0^x(1-t^2)^ndt$ with integration by parts viz. $u=(1-t^2)^n,\,v=t$ so$$f_n=x(1-x^2)^n+2n\int_0^xt^2(1-t^2)^{n-1}dt=x(1-x^2)^n+2n(f_{n-1}-f_{n+1}).$$Definite integration on $[0,\,1]$ simplifies this to $f_{n+1}=f_{n-1}-f_n/(2n)$, making b) tractable by induction.

Title: Evaluating $\int_0^1 (1-x^2)^n dx$


问:
Integrating with longitudinal strips to find the surface area of a shell
Consider a thin spherical shell. Lets look at 1 small element of the shell, which is like a longitude of the shell, and let it subtend an angle of $d\theta$ at the centre of the shell.

It has a surface area $dA=2\pi$r*rd$\theta$.

So we can consider many such small elements, each like a longitude of the shell, and each passing through the same 2 poles of the sphere.

Thus we can integrate this expression for surface area, taking limits as 0 to $\pi$, and this will cover the entire area of the sphere. So,

$$A=2\pi r^2\int_0^\pi\,d\theta$$

i.e. $A=2\pi^2$r$^2$

Where have I gone wrong?

Here in the image, the green part represents one element, not red.

答:
Consider the violet element of area in your picture, comprised between $\theta$, $\theta+d\theta$ in longitude, and between $\phi$, $\phi+d\phi$ in latitude. Its area is
$$
dA'=r\cos\phi\,d\theta\cdot r\,d\phi.
$$
Integrate this on $\phi$ for $-\pi/2\le\phi\le\pi/2$ to get the area of the whole slice as $dA=2r^2\,d\theta$.

Integrating again on $\theta$ for $0\le\theta\le2\pi$ one gets the area of the sphere.


问:
What is the result of multiplying two disjunctive probabilities
I have seen this as a part of an equation numerous times -> $p(1-p)$ where $p$ is a probability that an event is going to happen. My question is what is the event that the result of this equation gives us.

答:
The term $p(1-p)$ often appears in the context of Bernoulli trials with success probability $p$. Specifically, it is the probability that the first of two Bernoulli trials is successful and the second is not. It is also the probability that the second of two Bernoulli trials is successful and the first is not.

Note that by the above observations, the probability of exactly one success in two trials is $2p(1-p)$ since there are two ways we can choose the successful trial (either as the first or as the second). This is a special case of the binomial distribution.


问:
An inequality for $\lfloor x \rfloor- \lfloor y \rfloor$
In wikipedia, I come across this identity for the floor function:

In my question, I am interested in $x,y>0$.
\begin{equation}
 \lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x + y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor + 1.
\end{equation}

Now, making some substitutions I arrive at
\begin{equation}
 \lfloor x -  y \rfloor  \leq  \lfloor x \rfloor - \lfloor y \rfloor \leq \lfloor x -  y \rfloor + 1.
\end{equation}

However, I think that because of my positivity restriction, it is even stronger than that.
\begin{equation}
 \lfloor x -  y \rfloor  \leq  \lfloor x \rfloor - \lfloor y \rfloor < \lfloor x -  y \rfloor + 1.
\end{equation}
Is this true?

Or is there a case where we have equality
\begin{equation}\label{key}
 \lfloor x \rfloor - \lfloor y \rfloor = \lfloor x -  y \rfloor + 1,
\end{equation}
I suspect this is not possible, since the only possible `edge' cases are half integers. However, equality does not hold for such cases.

答:
If $x = 2, y = 1.5$, then
$$
\lfloor x \rfloor - \lfloor y \rfloor = 2 - 1 = 0 + 1 = \lfloor x - y \rfloor + 1.
$$


答:
So $A$ is an open set if every $a\in A$ is an interior point.

$a$ is an interior point if there is an $r> 0$ so $N_r(a) = \{n\in \mathbb N: d(n,a)= |n-a| < r\}\subset A$.

$N_r(a) = \{n\in \mathbb N: |n-a|< r\} =\{n\in \mathbb N: a-r < n < a+r\}$.  If $r < \frac 12$ then $N_{\frac 12}(a) = \{n\in \mathbb N: a-\frac 12 < n < a+ \frac 12\} = \{a\}$.

And as $a \in A$ then $\{a\} \subset A$ so $a$ is always an interior point of any set it belongs to.

So... all sets are open.  And that's the discrete topology

Title: Defining the order topology on the natural numbers

Body: How might one go about defining the order topology on the natural numbers with natural order (<)? I'm working on constructing the basis for the topology, and I realize that the intervals [1,b) and (a,b) must be present in the basis, since 1 is the least element of the naturals, but I'm not sure how to finish it off and properly describe the topology</p>


问:
Convolution of regulated functions
In his Analysis 1, Konrad Königsberger defines a convolution of two $2\pi$ periodic regulated functions (that have left and right limits at every point ) $f$ and $g$ as:

$$
(f*g)(x):=\frac{1}{2\pi}\int_{\mathbb{T}}f(t)g(x-t)dt
$$

where $\mathbb{T}$ is an arbitrary $2\pi$ long interval. Right after this, he states that $f*g$ is clearly $2\pi$ periodic and regulated. To me, it is not obvious that it's regulated.

I've managed to prove that if the functions are continuous, then their convolution is also continuous, but I can't use the same argument to prove that it's regulated.

What seemed as a promising attempt was to prove that the convolution of two step (piecewise-continuous) functions is a piecewise linear function. The problem is that I cant find a nice way to show this. Knowing that this holds, one could approximate $f(t)g(x-t)$ with $\varphi(t)\psi(x-t)$ such that $|f(t)g(x-t)-\varphi(t)\psi(x-t)|<\frac{\varepsilon}{2(b-a)}$ for all $t$, and $\varphi*\psi$ with a step function $\rho$ such that $||\varphi*\psi-\rho|| < \frac{\varepsilon}{2}$.

Edit:

I managed to prove it using piecewise linear and step functions: First we must prove that the convolution of two step functions with at most one point of discontinuities is a piecewise linear function. To do this, we can let:

\begin{equation*}
\begin{split}
&\varphi(t)=
\begin{cases}
\alpha & x\in[0,a)\\
\beta & x\in[a,2\pi)
\end{cases}\\
&\psi(t)=
\begin{cases}
\gamma & x\in[0,b)\\
\delta & x\in[b,2\pi)
\end{cases}
\end{split}
\end{equation*}

They're both extended $2\pi$ periodically. We can assume that $a\leq b$. Let $x\in[0,2\pi)$

\begin{equation*}
\begin{split}
&2\pi(\varphi*\psi)(x) = \int_0^{2\pi}\varphi(t)\psi(x-t)dt = \alpha\int_0^a\psi(x-t)dt + \beta\int_a^{2\pi}\psi(x-t)dt
&
\end{split}
\end{equation*}

Now you simply look at cases $x\in[0,a);x\in[a,b);x\in[b,2\pi)$ and notice, after a few substitutions, that the function is linear on all of them.

When $\varphi$ and $\psi$ are arbitrary step functions they can be written as sums of step functions with one point of discontinuity. Using the bilinearity of the convolution and the fact that the sum of piecewise linear functions is piecewise linear, we get that convolution of any two step functions is piecewise linear, thus obviously regulated.

Finally, if we let $f,g$ be $2\pi$ periodic regulated functions, we pick step functions $\varphi,\psi,\rho$ such that:

\begin{equation*}
\begin{split}
&1.||g-\psi|| < \frac{\varepsilon}{3||f||}\\
&2.||f-\varphi|| < \frac{\varepsilon}{3||\psi||}\\
&3.||\varphi*\psi-\rho||<\frac{\varepsilon}{3}
\end{split}
\end{equation*}

From here, we get that for all $x\in[0,2\pi)$

\begin{equation*}
\begin{split}
|(f*g)(x)-\rho(x)|\leq|(f*g)(x)-(f*\psi)(x)|+|(f*\psi)(x)-(\varphi*\psi)(x)| +
|(\varphi*\psi(x)-\rho(x))|<\varepsilon
\end{split}
\end{equation*}

答:
Let $\varphi(x)=1_{[a,b]}(x)$ be the indicator function of the interval $[a,b]$. Then $f*\varphi$ is continuous by the fundamental theorem of calculus (in Königsberger it is proven for integrands which are regulated functions).

If $\varphi$ is a step function, i.e. a linear combination of indicator functions of singletons and intervals as above, then $f*\varphi$ is still continuous.

Any regulated function on a compact interval is bounded. So $\|f\|_{[-\pi,\pi]}$ exists.

Let $x_0 \in \mathbb{R}$ and $\varepsilon>0$.

Since $g$ is regulated and periodic, there exists a step function $\varphi$ (actually it is the periodic extension of a step function) such that $\|g-\varphi\|_{\mathbb{R}} < \varepsilon$.

Choose $\delta>0$ such that $|(f*\varphi)(x) - (f*\varphi)(x_0)|<\varepsilon$ for all $x \in (x_0-\delta,x_0+\delta)$. Then

$$
\begin{align*}
|(f*g)(x) - (f*g)(x_0)| &\leq |(f*g)(x) - (f*\varphi)(x)| + |(f*\varphi)(x) - (f*\varphi)(x_0)| \\
&\quad + |(f*\varphi)(x_0) - (f*g)(x_0)| \\
&\leq 2\varepsilon\|f\|_{[-\pi,\pi]} + \varepsilon
\end{align*}
$$

for all $x \in (x_0-\delta,x_0+\delta)$. This shows that $f*g$ is continuous on $\mathbb{R}$. Of course, any continuous function is regulated.

I am a little bit surprised that I ended up showing that $f*g$ is continuous, but in light of the answer given by nullUser it might not be out of this world because the regulated function $g$ equals an (everywhere) left-continuous or (everywhere) right-continuous function in all except countably many points.

Title: Convolution of regulated functions


问:
Truth tables with $3$ rows of variables

$$\begin{array}{cccc}
\textbf{X} &\textbf{Y} &\textbf{Z} & \textbf{A}\\
 T & T & T & T\\
 T & T & F & T\\
 T& F& T& F\\
 T& F& F& F\\
 F& T& T& F\\
 F& T& F& T\\
 F& F& T& F\\
 F& F& F& T\\
\end{array}$$

I'm fairly new to truth tables so please bear with me. I was working on some truth table values and I was doing fine until I came across a truth table with $3$ rows to consider. Can you please help me figure out what expression will give the $A$ values?

答:
Use a 3-input (4x2) Karnaugh map:

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

With (X,Y) indexing the rows, and Z indexing the columns.

You should end up with two clusters of two $T$s.

Each cluster allows you to treat one of the 3 inputs as a "don't care" for that cluster (e.g. $Z$ might be a "don't care").  The AND of the two other two input variables in a cluster (e.g. maybe $X$ and $Y$), gives you the expression for the cluster, e.g. $X\cdot Y$

The OR of all of the cluster expressions gives you the overall expression.

You should end up with
$$ A = X \cdot Y + X' \cdot Z' $$
or something equivalent.

Title: Truth tables with $3$ rows of variables


Title: Evaluating $\int_0^1 (1-x^2)^n dx$

Body:
In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.

b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:

$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$

and so in part (b) i got:

$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$

I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

Body:
Part b) can be obtained just by induction. The result is immediate for $n=1$ and
\begin{align*}
\int_0^1 (1-x^2)^{n+1}dx = & \int_0^1 (1-x^2)(1-x^2)^ndx = \int_0^1(1-x^2)^n dx-\int_0^1 x^2 (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!} + \frac{1}{2} \int_0^1 x (-2x) (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!}-\frac{1}{2n+2}\int_0^1(1-x^2)^{n+1} dx
\end{align*}

This implies that
$$
\int_0^1 (1-x^2)^{n+1}dx = \frac{2n+2}{2n+3} \frac{2^{2n}(n!)^2}{(2n+1)!}=\frac{(2n+2)^2 2^{2n} (n!)^2}{(2n+1)!(2n+2)(2n+3)}=\dfrac{2^{2n+2}((n+1)!)^2}{(2n+3)!}.
$$

\begin{align*}
\int_0^1 x (-2x) (1-x^2)^n dx = &\left[x \dfrac{(1-x^2)^{n+1}}{n+1} \right]_0^1 - \frac{1}{n+1}\int_0^1 (1-x^2)^{n+1}
\end{align*}


问:
What does the dot product of two vectors represent?
I know how to calculate the dot product of two vectors alright. However, it is not clear to me what, exactly, does the dot product represent.

The product of two numbers, $2$ and $3$, we say that it is $2$ added to itself $3$ times or something like that.

But when it comes to vectors $\vec{a} \cdot \vec{b}$, I'm not sure what to say. "It is $\vec{a}$ added to itself $\vec{b}$ times" which doesn't make much sense to me.

答:
When one calculates A.B, two measurements happen: measurement of how small the angle between them is, and how long A and B are. A.B basically means projection length of A on B, with this length then scaled by the absolute length of B.

One way to think about the interpretation of the dot product is to think how would one maximise or minimise the dot product between two vectors. Let's assume we are trying to maximise the dot product between two vectors that we can modify:

The dot product will be grow larger as the angle between two vector decreases. The dot product A.B will also grow larger as the absolute lengths of A and B increase. This is because as A gets larger, its projected length will be longer, and as B's length gets larger, the scaling of A's projection will grow larger, given that B's absolute length will act as a scaler of A's projection length.


答:
Let $Y=\log(X)=g(X)$ Then

$$f_Y(y)=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}=\frac{f_X(e^y)}{e^{-y}}=\frac{\frac{1}{e^y\sqrt{2\pi}} e^{-\frac{1}{2}(y)^2}}{e^{-y}}=\frac{1}{\sqrt{2\pi}} e^{-\frac12 y^2}$$

which is a standard normal. (That's precisely why $X$ is lognormal!).

Then $$E[|Y|]= 2 \frac{1}{\sqrt{2\pi}} \int_0^\infty y e^{-\frac12 y^2} dy =2 \frac{1}{\sqrt{2\pi}} \int_0^\infty e^{-u} du = 2 \frac{1}{\sqrt{2\pi}}$$

Hence $$\sqrt{2\pi}\mathbb{E}[|\log X|]=2$$

Title: expected value of a log-normal distribution

Body:
Let $X$ be a continuous random variable with log-normal distribution, i.e.
$$
f_X(x)=\begin{cases}
    \frac{1}{x\sqrt{2\pi}} e^{-\frac{1}{2}(\log x)^2}, & \text{for x>0} \\
    0, & \text{otherwise}. \\
\end{cases}
$$
Determine the value of $A=\sqrt{2\pi}\mathbb{E}[|\log X|]$.

I know that generally, $\mathbb{E}[X]=e^{\mu+\frac{1}{2}\sigma^2}$. And in this case, $\mu=0$ and $\sigma^2=1$. My question is how do I apply this to $\mathbb{E}[|\log X|]$?


问:
What is the use of the Dot Product of two vectors?
Suppose you have two vectors a and b that you want to take the dot product of, now this is done quite simply by taking each corresponding coordinate of each vector, multiplying them and then adding the result together. At the end of performing our operation we are left with a constant number.

My question therefore is what can we do with this number,why do we calculate it so to speak? I mean it seems almost useless to me compared with the cross product of two vectors (where you end up with an actual vector).

答:
When one calculates A.B, two measurements happen: measurement of how small the angle between them is, and how long A and B are. A.B basically means projection length of A on B, with this length then scaled by the absolute length of B.

One way to think about the interpretation of the dot product is to think how would one maximise or minimise the dot product between two vectors. Let's assume we are trying to maximise the dot product between two vectors that we can modify:

The dot product will be grow larger as the angle between two vector decreases. The dot product A.B will also grow larger as the absolute lengths of A and B increase. This is because as A gets larger, its projected length will be longer, and as B's length gets larger, the scaling of A's projection will grow larger, given that B's absolute length will act as a scaler of A's projection length.

Hence in problems where is desirable to maximise or minimise the size of vectors and minimise the deviation or angle between them, quantifying it using the dot product can be useful.


问:
Difference between complete and closed set
What is the difference between a complete metric space and a closed set?

Can a set be closed but not complete?

答:
An interesting thing about you question need to be noticed. A complete metric space $Y$ is a metric space $(Y,d_Y)$ such that every Cauchy sequence determined by the metric $d_Y$ is convergent for some point of $Y$. A closed subset $Y$ of a metric space $X$ is a set such that every convergent sequence of points in $Y$ converges to a point of $Y$ - with the metric $(X,d).$

This raise a funny observation. Let us prove that every complete subspace $Y$ is closed in $X$; that is, if $Y$ is endowed with the same metric as X and is complete relative to its metric, then $Y$ is closed in $X.$

Proof: Let $d_Y: Y \times Y \rightarrow \mathbb{R}$ be the induced metric of $d: X \times X \rightarrow \mathbb{R}.$ Then if $y \in X$ is limit of a convergent sequence of points in $Y$ with the metric $d$, say $\left(y^{k}\right)_{n\in\mathbb{N}}$, then for each $\epsilon>0$ there exist a $N_{\epsilon}$ such that $d(y_m,y_n)<\epsilon$ for all $n,m>N_{\epsilon}.$ Now, note that we can evaluate the function $d_Y$ at $(y_m,y_n),$ since each coordinate belongs to $Y$. Then for each  $\epsilon>0$ there exist a $N_{\epsilon}$ such that $d_{Y}(y_m,y_n)<\epsilon$ for all $n,m>N_{\epsilon}.$ Hence $\left(y^{k}\right)_{n\in\mathbb{N}}$ is Cauchy and there exist a point $\bar{y} \in Y$ such that $d_Y(y^{k}, \bar{y}) \rightarrow 0$. Since $d$ is equal to $d_Y$ for every point of $Y$, then $d(\bar{y},y) \leq d(\bar{y},y^{k})+d(y^{k},y)\rightarrow 0$. Finally, we can conclude that $\bar{y}=y$ with $\bar{y} \in Y$ and $y$ belongs to $Y.$

This is a detailed proof. I found it quite funny to prove it without using a circular argument.


问:
Five digit password whose digits are all among $\{0,2,3,5,7\}$
A password of computer used five digits where they are from 0 and 9. What is the probability that the digits of the password  are all among $\{0,2,3,5,7\}$?

Edit:

I tried this way but not sure is it correct or not.

Number available 0 to 9 .

total = 10 numbers .

The code is of 5 numbers .

first number can be selected in 10c1 ways .

Since the numbers are not unique and repetition is allowed we can select the second number in 10c1 ways and so on .

So total possible ways in which this code can be made is (10c1)(10c1)(10c1)(10c1)(10c1).

Now the possibility of finding the code with only prime numbers and zero
.

So total numbers to select from 0,2,3,5,7 .

Total numbers 5 .

The first number can be selected in 5c1 ways again repetition allowed so second number can be selected in 5c1 ways and so on .

So numbers of ways in which the code with odd numbers is selected is (5c1)(5c1)(5c1)(5c1)(5c1) .

Probability = (number of required outcomes )/ (total possible outcomes)

= (5c1)(5c1)(5c1)(5c1)(5c1) / (10c1)(10c1)(10c1)(10c1)(10c1)

= 1/32

答:
That way works, but if you think a bit you'll see combinations were unnecessary here, and that's not where I think Cameron was leading you. Combinations only become important when you start sampling without replacement (hence my request for clarification whether repetition was allowed).

You want every digit in the password to be in the set $\{0,2,3,5,7\}$. All possible passwords must have digits in $\{0,1,2,3,4,5,6,7,8,9\}$. Therefore you have $5$ possibilities for a success out of $10$ total in each of the $5$ (independent) digits of the password, or
$$\left(\frac{5}{10}\right)^5 = \left(\frac{1}{2}\right)^5=\frac{1}{32}.$$

A more pressing issue, however, is, who chooses a password like this? :)


Title: Evaluating $\int_0^1 (1-x^2)^n dx$

Body:
I assume integration is done between $0$ and $1$.

For $n\in\mathbb{N}$, let's have $I_n=\int_{0}^1(1-x^2)^{n}\,dx$.

Then for any $n\in\mathbb{N}$,

$$I_{n+1}=\int_{0}^1 (1-x^2)(1-x^2)^{n}\,dx=I_n-\int_{0}^1 x^2(1-x^2)^{n}\,dx$$

Integratin par parts :

$$\int_{0}^1 x^2(1-x^2)^{n}\,dx=\int_{0}^1 x\times x(1-x^2)^{n}\,dx=[x\times\frac{-(1-x^2)^{n+1}}{2n+2}]_0^1-\int_{0}^{1}\frac{-1}{2n+2}(1-x^2)^{n+1}dx$$

Therefore $$\int_{0}^1 x^2(1-x^2)^{n}\,dx=\frac{1}{2(n+1)}I_{n+1}$$

Going back to the first equality :

$$I_{n+1}=I_n-\frac{1}{2n+2}I_{n+1}$$

Reordering everything :

$$I_{n+1}=\frac{2(n+1)}{2n+3}I_n$$

Since $I_0=1$, the former equality implies $I_n\neq 0$ for all $n\in\mathbb{N}$. Let $n$ be a non-negative integer.

Then

$$\frac{I_n}{I_0}=\prod_{d=0}^{n-1}\frac{I_{d+1}}{I_d}=\prod_{k=0}^{n-1}\frac{2d+2}{2d+3}=\prod_{k=0}^{n-1}\frac{2d+2}{2d+3}\frac{2d+2}{2d+2}=\frac{2^{2n}(n-1+1)!^2}{(2(n-1)+3)!}=\frac{2^{2n}n!^2}{(2n+1)!}$$


问:
Central Difference Approximation with sin(y^2)
Consider the Boundry Value ODE:

$(d^2\ T)/(dy^2\ )=\sin(y^2\ )$

Derive the equation for approximating the interior points using the central difference method.

I'm kinda stuck on this one. I was able to derive the equation down to this form:

$T_{i-1}[\frac{1}{∆y^2}]+T_i [\frac{-2}{∆y^2}]+T_{i+1} [\frac{1}{∆y^2} ]=sin(y_i)^2$

With the forcing function on the right, but I feel like I am missing something or doing something wrong. i.e. Am i supposed to do something with the sin at the end?

答:
As you said, you get the linear equations
$$
\dfrac{T_{i-1}-2T_i + T_{i+1}}{\Delta y^2} = \sin (y_i^2), \quad i=1, \cdots, n
$$

Now you just need to solve this (tridiagonal) linear system, getting the values of $T_0$ and $T_{n+1}$ from the boundary conditions. The $\sin (y_i^2)$ terms are know and correspond to the rhs of the linear system.


问:
Order topology related question
Say I have two topological spaces $X,Y$ s.t $Y$ has the order topology, and $f,g$ are two continuous functions $g,f:X\rightarrow Y$. I want to show that the set $O=\{x:f(x)> g(x)\}$ is open in $X$. So my attempt was defining $h=f-g$ and looking at $O=\{x:h(x)>0\}$ and proving it's open in $X$, but I'm not even sure if I'm allowed to use the concept of $0$ since I don't know anything about $Y$.

Say I can use this concept, I want to show that there exists such open $U\subseteq Y$ s.t $h^{-1}(U)=O$, and by continuity I'm done. So I thought about defining $U=Im(h)\cap V$ where $V=\{y\in Y:y>0\}$ which is simply the ray $(0,\infty)$ in $Y$ and obviously open. So $U$ is open in the subspace topology of $Im(h)$ and therefore $O$ is open in $X$.

Is my approach correct?

答:
Fix $x\in O$. We will find a neighborhood $W$ of $x$ with $W\subseteq O$.

Case 1. There is some $z\in(g(x),f(x))$. Since $g(x)\in(\leftarrow,z)$
(and $(\leftarrow,z)$ is open and $g$ is continuous),
there is a neighborhood $U$ of $x$
such that $g(U)\subseteq(\leftarrow,z)$.
Similarly there is a neighborhood $V$ of $x$
such that $f(V)\subseteq(z,\rightarrow)$. Let $W=U\cap V$.
If $y\in W$ then $g(y)<z<f(y)$ hence $y\in O$ and $W\subseteq O.$

(If you prefer to work with preimages, then fix $x\in O$ and in Case 1 fix
$z\in(g(x),f(x))$ as before, and
let $W=\big(g^{-1}(\leftarrow,z)\big)\cap\big(f^{-1}(z,\rightarrow)\big)$.
Then $W$ is open and $x\in W\subseteq O$.)

Case 2. $(g(x),f(x))=\varnothing.$
In this case $(\leftarrow,f(x))=(\leftarrow,g(x)]$ and
$(g(x),\rightarrow)=[f(x),\rightarrow)$, open sets.
Let $W=\big(g^{-1}(\leftarrow,g(x)]\big)\cap\big(f^{-1}[f(x),\rightarrow)\big)$.
Then $W$ is open and $x\in W\subseteq O$. (Indeed, if $y\in W$ then $g(y)\le g(x)<f(x)\le f(y).$)

Title: Order topology related question


问:
What is the minimum weighting trials to be made with a two pan scale to find the lighter marble?
The puzzle is as follows:

Over a table there are three bags with marbles, on the first one there are 100 blue marbles, on the second one there are 100 red marbles and on the third one there are 100 green marbles. All the marbles inside the bags are in equal weight, except two of them whose weight is slightly less from the others and identical, which are on different bags. Considering all of this, How many weighing trials at minimum must be made by using a two pan scale to find with certainty those two mysterious lighter marbles?

The choices given are as follows:

$\begin{array}{ll}
1.&\textrm{11 trials}\\
2.&\textrm{8 trials}\\
3.&\textrm{9 trials}\\
4.&\textrm{10 trials}\\
\end{array}$

I'm not sure exactly how to solve this problem, my best effort goes as this:

Assuming that those bags fit in the two pan scale (hence marbles are sufficiently small) then you put in the left pan one bag and in the other the other bag.

The best thing that it can happen is that right off the bat one bag is heavier than the other because it contains the heavier marbles. The worst thing is that they are equal in weight because they both contain those mysterious lighter marbles. But either one way or another. Assuming we know beforehand there are two marbles whose weight is less than the others and those two are in two separate bags then this means that if those two bags weight the same, then those two bags contains the marbles we're looking for. While the other bag has the heavier marbles.

All of this account for one weighing trial.

Now that we discriminated those bags, we know our two bags which contains the lighter marbles we split them in groups of 50. This accounts for four groups.

Assuming that we do not mix those, we put those two groups in one pan and the other two in the other pan.

The worst that can happen is that they are even, and we still cannot tell on which side there are the ligther msrbles.

Thus a new weighing trial is needed.

This accounts for a second trial.

In the next trial, we switch the groups by fixing two groups. After doing this, the scale will reveal which side is the lighter one. Thus the heavier marbles can now be returned back to the group of the marbles which do weight more.

This accounts for our third trial.

We repeat the cycle again. But now we have 100 marbles remaining, as the other 100 went back with their set of heavier marbles. Thus, we split those in 4 groups of 25.

Assuming that it happens the same as before, we can't tell the different first, so a new weighing trial is needed.

This accounts for our fourth trial.

We repeat the weighing trial, switching the groups, and we know for sure 50 marbles are heavier and the other 50 must contain the lighter marbles.

This accounts for the fifth trial.

We now have 50 marbles whose weight must contain those two marbles. We again repeat the cycle. In this case we cannot divide 25 by 4. Remember that all the time we are keeping track on which side belongs to which bag, so we have took care not to mix the groups. So we use one of these groups of 25 (and save the other for later) from which we know it contains a ligther marble. But because 25 cannot be split in two. We take out from that group a marble, and we have two groups of 12.

We perform a new weighing trial. The best thing it can happen is both sides are even and we got our lighter marble. The worst thing is that we need a new weighing trial.

This accounts for a sixth trial.

We collect the heavier marbles as this was revealed by the scale plus the marble what we took out first and we end up with 12 ones from which we don't know where is the mysterios marble and perform a new weighing trial. This time splitting in six and six.

Again we assume that the worst thing that can happen is that we are still unable to know where is the mysterious marble.

From those we know that six are heavier as the scale revealed this to us by tilting it to one side.

This account for a seventh trial.

We now have six from which we don't know where is the ligther marble. We split them in groups of 3 and 3 and perform a new weighing trial, from it we do know the heavier ones.

This accounts for the eight trial.

Now it comes finally our last 3 marbles. From these we must take out one. In any case it will be revealed which is ligther. If the scale is even, then the one which we took out is the lighter. If one side is heavier the other resting in the other pan is the ligther one, which belongs to one bag.

Remember that we saved one group from before performing the sixth trial. Thus as we required 3 steps to discover which marble was the ligther one, this adds to the already weighing trials that we accounted.

This means $8+3=11$

Thus I conclude that at minimum we do need $11$ weighing trials for that, or choice 1. But of course, I could be overlooking something.

Can someone please check my logic?. I am not sure if it can done in less steps or in a different way, but to my understanding this is the most logical way to do it.

Thus can someone help me here?.

答:
Here is a way to do it in $10$ total weighings, which as saulspatz pointed out, is optimal.

First weighing: all of one color vs all of another color.  From this you know the colors of the two lightweight or "bad" marbles.  (BTW this weighing reduces the no. of possibilities from $3 \times 10^4$ to $10^4$ which is the best you can do.)

For the rest, lets assume the two colors with bad marbles are Red and Green.

At this point, what saulspatz's "oracle" argument shows is that if you never mix the two colors in a subsequent weighing, you cannot do better than $5$ more per color, or $11$ total.  So the key is indeed to having further weighing(s) with both colors.

Let $R$ be a fixed subset of $81$ red marbles, and $R'$ be the remaining $19$ red marbles.  Similarly let $G$ be a fixed subset of $81$ green marbles, and $G'$ be the remaining $19$ green marbles.

Second weighing: $R$ vs $G'$ plus $81-19 = 62$ blue ("good") marbles, so that there are $81$ on each side.  We have $3$ possibilities:

The most common case is $R < G' + B$.  In this case we know:

The bad red marble is in $R$, which we can find out using $4$ more weighings  because $3^4 = 81$.

The bad green marble is in $G$, which we can also find out using $4$ more weighings.

So the total no. of weighings $= 1+1+4+4 = 10$.

The least common, and also easiest case is $R > G' + B$.  In this case we know:

The bad red marble is in $R'$, and we can find it using $3$ more weighings coz $3^3 > 19$.

The bad green marble is in $G'$, another $3$ weighings.

So the total no. of weighings $= 8$.

The last case, conceptually the most complicated, is $R = G' + B$.  In this case, the sets containing the bad marbles can be (i) $R$ and $G'$, or, (ii) $R'$ and $G$.

Luckily, we have enough leeway to "waste" the third weighing to find out which of (i) or (ii) is true.

Third weighing (for $R = G' + B$ case):  Simply weight $R$ vs $81$ blue marbles to see if $R$ contains a bad marble.  Since the blues are all good, there can be only two outcomes:

If $R$ weighs less, then we have case (i) where we know the bad marbles are in $R$ and $G'$.  The one in $R$ can be found by $4$ more weighings, and the one in $G'$ can be found by $3$ more, for a total of $1 + 1 + 1 + 4 + 3 = 10$ weighings.

If $R$ weighs equal with the blues, then we have case (ii) which is symmetric.

BTW, visualizing the $100 \times 100$ possibilities (after the first weighing) as a $100\times 100$ grid helped me solve this problem.


问:
A function with the following properties is always bounded, uniformly continuous, and if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant.
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function with the following property: Given $\epsilon > 0$, there exists a positive real number $r > 0$ and a set $D \subset \mathbb{R}$ such that

Any interval $I$ with length $r >0$, is such that $D \cap I \ne \emptyset$

$|f(x+t) - f(x)| < \epsilon$ for any $t \in D$ and $x\in \mathbb{R}$

Show that a function with those properties is always bounded, uniformly continuous, and if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant.

I've been stuck trying to prove this for a while. I think that any $f$ is a periodic function, like $\sin$, and that would imply that it is bounded and that if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant. But I don't know how to show that $f$ is periodic, nor if it is the best approach for this proof.

答:
To show that $f$ is bounded, take $\epsilon = 1$ (or any other number), and get the associated $r$ and $D$. Since $f$ is continuous, it is bounded on the interval $[0, r]$ -- call a possible bound $M$. Then for any $x \in \mathbb R$, we can find $t \in D \cap [-x, -x + r]$, so that
$$
|f(x)| = |f(x) + f(x+t) - f(x+t)| \leq |f(x + t)| + |f(x) - f(x+t)| \leq M + 1.
$$
Thus $f$ is bounded by $M+1$.

To show that $f$ is uniformly continuous, let $\epsilon$ be arbitrary. Now get $r, D$ associated to $\epsilon/4$. Since $f$ is continuous, it is uniformly continuous on $[0, r + 1]$, and we can find a $\delta < 1$ such that for any $x, y \in [0, r + 1] $ we have that $|x-y| < \delta \implies |f(x) - f(y)| < \epsilon/4$. We claim that this $\delta$ works to show uniform continuity for $f$ for our original $\epsilon$. To this end, take $x, y \in \mathbb R$ with $|x - y| < \delta$. Again take $t \in D \cap [-x, -x + r]$, so that
$$
|f(x) - f(y)| \leq |f(x) - f(x+t)| + |f(x + t) - f(y + t)| + |f(y + t) - f(y)| \leq \frac\epsilon4 + \frac\epsilon4 + \frac\epsilon4 < \epsilon.
$$
Together with Crostul's partial answer, this answers the question.


Title: Evaluating $\int_0^1 (1-x^2)^n dx$

Body:
An idea: substitution $\;t=\sin x\,,\,\,dt=\cos x\,dx\;$ , so

$$\color{red}{I_n}:=\int_0^1(1-t^2)^ndt=\int_0^{\pi/2}(\cos x)^{2n+1}dx=\int_0^1\left(1-\sin^2x\right)(\cos x)^{2n-1} dx=$$

$$=I_{n-1}-\int_0^{\pi/2}\sin^2x(\cos x)^{2n-1}dx\stackrel{\text{by parts:}\\u=\sin x,\,v'=\sin x(\cos x)^{2n-1}}=\color{red}{I_{n-1}}+\overbrace{\left.\frac{\sin x(\cos x)^{2n}}{2n}\right|_0^1}^{=0}-\color{red}{\frac1{2n}I_n}$$

$$\implies I_n=\frac{2n}{2n+1}I_{n-1}$$

For $\;n=0\;$ we get

$$I_0=\int_0^1(1-t^2)^0dt=\int_0^1dt=1=\frac{2^0(0!)^2}{1!}\;\;\checkmark$$

Now induction...and the result follows at once:

$$I_n=\frac{2n}{2n+1}I_{n-1}\stackrel{\text{Ind. Hyp.}}=\frac{(2n)^2\cdot2^{2n-2}((n-1)!)^2}{(2n-1)!\cdot2n\cdot(2n+1)}=\frac{2^{2n}(n!)^2}{(2n+1)!}$$


Title: Evaluating $\int_0^1 (1-x^2)^n dx$

Body:
$$I(n)=\int_0^1(1-x^2)^ndx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}=\frac{du}{2u^{1/2}}$ and so:
$$I(n)=\frac12\int_0^1(1-u)^nu^{-1/2}du=\frac12 B\left(\frac12,n+1\right)=\frac{\Gamma\left(\frac12\right)\Gamma(n+1)}{2\Gamma\left(n+\frac32\right)}$$
Now we can use the identity that:
$$\Gamma\left(\frac12+(n+1)\right)=\frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt{\pi}$$
$$\Gamma\left(\frac12\right)=\sqrt{\pi}$$
and so:
$$\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma(n+3/2)}=\frac{4^{n+1}(n+1)!n!}{2(2n+2)!}=\frac{2^{2n+1}(n+1)(n!)^2}{(2(n+1))!}$$
now we can use the fact that:
$$(2n+2)!=(2n+2)(2n+1)!=2(n+1)(2n+1)!$$
and so we arrive at:
$$\frac{2^{2n}(n!)^2}{(2n+1)!}$$


问:
Given that  and  are positive integers with $\mid$, how would one prove that $\gcd(, ) = $?
I'm having trouble showing that if  and  are positive integers with |, then (, ) = a.

I'm aware that the greatest common divisor of two numbers is the largest number that is a divisor of both numbers. Ex) $\gcd(4,2) = 2 $. But in this case $a=4$ and $b=2$ so the expression above would not work. I suppose what I'm asking is if variable $a$ is always the smallest of the two numbers given.

Note

I now realize that I should have given the problem some more thought before posting on here. I had been overlooking the | portion of the expression. Thank you for the help!

答:
$\begin{array}{lll}\gcd(a,b) &=\min\{as+bt>0\mid (s,t)\in\mathbb Z^2\}\\
&=\min\{as+ant>0\mid (s,t)\in\mathbb Z^2\} & \text{since }a\mid b\iff b=na\\
&=a\times\min\{s+nt>0\mid (s,t)\in\mathbb Z^2\} & \text{use }\min(ax,ay)=a\min(x,y)\\
&=a\times \gcd(1,n) & \text{choosing }s=1-n,t=1\text{ gives }\gcd(1,n)=1\\
&=a\times 1=a\end{array}$


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
The Szemeredi-Trotter incidence theorem gives an upper bound on the total number of incidences between a finite set of points and a finite set of lines in the plane.  An incidence is a pair $(\ell, x)$ consisting of a line $\ell$ and a point $x \in \mathbb{R}^2$ such that $x \in \ell$.

The original proof of the theorem consisted of a tricky cell-decomposition trick which abound in additive combinatorics.  The newer slick proof comes from an argument of Szekely which uses elementary graph theory techniques.  The details are summarized very nicely by Terry tao here and here.  Ultimately the proof relied on Euler's formula $f -e + v = 2$, the basic inequalities one sees when studying graphs for the first time, such as $e \leq 3v - 6$, and an upper bound on the crossing number of a specific subgraph.  The newer proof is much easier to grasp and can certainly be taught to undergraduates.

An overview of the older formulation can be found on Terry Taos'b log (specifically here and here).


答:
My idea here is to use the Beta function:

$$ \int_0^1 t^{m}(1 - t)^n \;dt = \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}. $$

Where the Gamma function satisfies $\Gamma(n + 1) = n!$ if $n$ is an integer, and also $\Gamma$ satisfies the recurrence $\Gamma(n + 1) = n\Gamma(n)$.

So if we substitute $t = x^2$, $dt = 2x\; dx \iff \frac{1}{2\sqrt{t}}\;dt=dx$ then

\begin{align}
\int_0^1 (1 - x^2)^n \; dx &= \frac{1}{2} \int_0^1 t^{-1/2} (1 - t)^n \;dt \\
&= \frac{1}{2} \frac{\Gamma(\frac12)n!}{\Gamma(n+\frac32)}. \tag{1}
\end{align}

And now using the recurrence $\Gamma(n + 1) = n\Gamma(n)$, we find that

\begin{align}
\Gamma(n+\tfrac32) &= (n+\tfrac12)\Gamma(n + \tfrac12) \\
&= (n+\tfrac12)(n - \tfrac12) \Gamma(n - \tfrac12) \\
&\hspace{2.5mm}\vdots \\
&= (n+\tfrac12)(n - \tfrac12) \cdots \tfrac{5}{2} \cdot \tfrac{3}{2} \cdot \tfrac12 \cdot \Gamma(\tfrac12) \\
&= \frac{1}{2^n} (2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 \cdot \Gamma(\tfrac12) \tag{2}
\end{align}

The product $(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1$ is also known as the double factorial $(2n + 1)!!$. It is well-known that we can rewrite it in terms of $(2n + 1)!$ as

\begin{align}
(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 &= \frac{(2n + 1)(2n)(2n - 1)(2n - 2) \cdots 3 \cdot 2 \cdot 1}{(2n)(2n-2) \cdots 2} \\
&= \frac{(2n+1)!}{2^n \cdot n!}. \tag{3}
\end{align}

So now finally, if you combine $(1), (2), (3)$, you get the result you want.

Title: Evaluating $\int_0^1 (1-x^2)^n dx$


问:
VC Dimension of o-Minimal Families
I'm not a model-theoretic person, I'm looking at Laskowski (1992), "VC Classes of Definable Sets". Have I understood correctly that the following true: If A is an o-minimal subset of $\mathbb{R}^n \times \mathbb{R}^m$, then the collection
$\{ A_y \mid y \in \mathbb{R}^n \}$, where $A_y = \{ x \in \mathbb{R}^m \mid (y,x) \in A \}$ is the fibre of A at y, has finite VC dimension?

Excert from Laskowski (1992)

答:
"$A$ is an o-minimal subset" doesn't make any sense. You mean "$A\subseteq \mathbb{R}^{n+m}$ is a definable subset for an o-minimal structure on $\mathbb{R}$".

But yes, your reading of Laskowski is correct. More generally, a complete first-order theory $T$ is called NIP (not independence property) if for every model $M$, every family of fibers for a definable set in $M$ has finite VC dimension. And o-minimal theories are primary examples of NIP theories.

Title: VC Dimension of o-Minimal Families


问:
Rewrite piece-wise function as an integral with given condition
I have the following piece wise function. Suppose that $f : \!R \rightarrow \!R$ has derivatives of all orders.

Also, $f(0)=0$
$$
g(t)=
\begin{cases}
 f(t)/t \quad t\neq 0 \\
 f'(0)\quad t=0\\
\end{cases}
$$

Prove that $g(t) = \frac{f(t)}{t} = \int_{0}^{1} f'(tx)dx$

Below is what I have tried:

Using FTC, $\int_{0}^{1} f'(tx)dx = f(t \cdot 1) - f(t \cdot 0) = f(t) - f(0)$

Since it is provided that $f(0) = 0$, now I have just $f(t)$

And the proof becomes $g(t) = \frac{f(t)}{t} = f(t)$, for this to be true, $t$ need to be $1$. But I cannot find a reason to justify that.

Am I on the right track? Or did I mess up from the beginning?

答:
Ok I figured it out. I need to substitute $u=tx$ as a function of $x$

Now I have $\frac{du}{t}=dx$, put it back into the integral $\int_{0}^{1}f'(tx)dx$. This gives $$\int_{0}^{1} \frac{f'(u)}{t}du$$

Evaluate this integral gives $\frac{f(t)}{t} - \frac{f(0)}{t}$

And since $f(0) = 0$, then $g(t) = \frac{f(t)}{t}$


问:
how to prove using pigeonhole principle there exist two of them, $a_i$ and $a_j$, such that $$ divides $a_i − a_j$.
Let $a_1,a_2,\cdots, a_{n+1}$ be $ + 1$ distinct positive integers, show that there exist two of them, $a_i$ and $a_j$, such that $$ divides $a_i − a_j$.How to solve this using pigeon hole principle?

答:
We prove the stronger: Namely this statement but for $n$ integers than $n$+$1$.

Assume all of the integers are distinct (mod $n$) so that we don't have simple trivial cases such as $a_i$=$0$(mod $n$). Then there are $n$-$1$ remainders (excluding $0$). So we have by the pigeonhole principle, there must be two integers with the same remainder, say $a_i$ and $a_j$. In other words, $a_i=a_j$(mod $n$), which would imply the given result.

Since it holds for $n$ numbers, it must hold for $n$+$1$ numbers.


问:
Show directly from the $\epsilon$-$\delta$ definition that $\lim_{x\to0}\frac{\sin(3x)}{3x} = 1$ given $\lim_{x\to0}\frac{\sin(x)}{x} = 1$
We have the definition of what it means to say that $f(x)$ tends to $M \in \mathbb{R}$

It is: $\forall \epsilon > 0, \exists \space \delta > 0$ such that $x \in (a,b)$ and $0 < |x-c| <\delta \implies |f(x) - M| < \epsilon$.

Now, I understand that we want to show the case is true for $f(3x)$ but directly showing that from the definition is escaping me.

So far I have tried:

Assuming the result true for $y$ and then letting $y=3x$ but I'm not sure that's showing the result directly from the definition.

And I've considered using uniqueness of limits, but the question strictly states no results about limits other than the definition above can be used.

I know that $\sin(3x)$ is bounded between $-1$ and $1$ but I'm not sure how this will help.

In response to a comment:

We know: $0 < |x| < \delta' \implies |\frac{\sin{x}}{x} - 1| < \epsilon'$

and we want to show $0 < |x| < \delta \implies |\frac{\sin{3x}}{3x} - 1| < \epsilon$

答:
We have $f(x) = \frac{\sin{x}}{x}$

We know, for some $\delta > 0$ $$0 < |x| < \delta \implies |f(x) - 1| < \epsilon$$

We want to show, $\exists \space \delta' > 0$ such that $$0 < |x| < \delta' \implies |f(3x) - 1| < \epsilon$$

If we let $\delta'= \frac{\delta}{3}$, we have $$0 < |x| < \frac{\delta}{3} \iff 0 < |3x| < \delta \implies |f(3x) - 1| < \epsilon.$$

And we are done.