问:
$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $
I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:

$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2}  $$

Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2}  \sim   \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.

In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$

but I don't understand this solution.

Trying to solve the limit with $a^2=1+2k$:

$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot  \frac{\sin (ax)  a  x^2 - (1-\cos (ax))  2x}{x^4}}{2x} $$$$ =  \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$

答:
This part of what you wrote is incorrect:
$$\frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2}  \sim   \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}$$

You're taking the limit of the top of the fraction, but not the bottom. Instead, you need to consider under what circumstances the limit of the entire fraction might exist. Of course, this can only happen if the top tends towards zero, i.e. if
$$\cos (\pi \cdot \frac{a^2}{2}) = 0$$

So the limit might exist if $a^2 = 1 + 2k$ for some $k \in \mathbb N$. Next, assuming $a$ is one of these values, compute the limit of the entire fraction using L'Hospital's rule:

$$
\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} = \lim_{x\to 0} \frac{\frac{d}{dx} \space \cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{\frac{d}{dx} x^2} $$

$$ = \lim_{x\to 0} \frac{-\pi \left(\frac{a\sin(ax)x - 2(1 - \cos(ax))}{x^3}\right)\sin(\pi\cdot\frac{1 - \cos(ax)}{x^2})}{2x}
$$

$$
= \left( \lim_{x\to 0} \frac{-\pi\cdot(a \sin(ax) x - 2\cdot(1 - \cos(ax)))}{2x^4} \right) \left( \lim_{x\to 0} \sin\left(\pi \frac{1 - \cos(ax)}{x^2}\right) \right)
$$

You already know the second limit is $\sin\left(\pi\cdot\frac{a^2}{2}\right) = \pm 1$ (depending on the particular value of $a$). The first limit is straightforward. Just apply L'Hospital's Rule repeatedly until you have your answer.


问:
Proof by induction: base case for a limit as x goes to infinity?
With induction we always start with a base case; What would the base case for this be?
Choosing 1 seems nonsensical.
Choosing infinity seems wrong as well.

Prove, using induction, that $\lim\limits_{x\to\infty}\dfrac{(\ln x)^k}x=0$.

答:
Hint:

By L'Hospital, for $k>0$,

$$\lim_{x\to\infty}\frac{\log^k(x)}{x}=\lim_{x\to\infty}\frac{k\log^{k-1}(x)}{x}.$$

Now the induction should be obvious.

ContentLicense: CC BY-SA 4.0


问:
Does this type of equation have a name $b^x+b^y=x^b+y^b$?
I was "randomly" suggested the video "Grapesでいろんな二次グラフを描いてみた", on the Hi Shibacchi YouTube channel, of various graphs. Japanese text and frantic synth music were in the background. Eventually, it came to a set of equations of the form $$b^x+b^y=x^b+y^b$$ for some constant b. I would like to know if this type of equation has a name. I put it in Desmos and played around with the sliders and it has some neat properties.

If b is even then the equation forms two curves, one of which is closed. But if it's odd, the equation forms only one curve -which is closed. There is also symmetry along the line y=x.

Edit:
After looking at it some more I found that for some reason b=2 is unique, in that it seems to be the only even number for which the closed curve doesn't contain the origin. Also, 1 and 3 are interesting. 1 doesn't contain a closed curve; additionally, for all the other odd numbers the closed curve crosses both the x and y axes at b, except for 3, which has no real roots.

Edit2:

Implicit differentiation with respect to b yields:
$$xb^{x-1}+yb^{y-1}=ln(x)x^b+ln(y)y^b$$
Which when graphed is similar for all N > 3, it consists of two curves, starting and ending on the y and x axis, respectively. It's really cool when b = 3, or less! I'm not sure what the mathematical terminology is, but the two curves "pinch" together. At b = 2 the equation seems to be inconsistent for all x and y. And at b = 1 there is only one curve!

答:
I would call it a
mixed exponential-polynomial
equation.</p>


问:
If $T$ is injective, there exists left inverse $S:W\rightarrow V$
I apologize for this repeat question but I've been trying to figure this problem out for at least 3 hours and haven't made any progress. I've seen other Math SE posts to no avail.

Assume $W$ is finite dimensional and $T:V\rightarrow W$ is linear.
Show that if $T$ is injective, then there exists a linear map
$S:W\rightarrow V$ so that $ST=I_V$.

Like I've said, I've seen many Math SE posts and they all seem to do the same thing: The image of $T$ is a subspace of $W$;  hence,  finite dimensional. Then,  it has some basis $w_1,...,w_n$.  Moreover, since every linearly independent collection of vectors in $W$ can be extended to a basis of $W$, the basis of the image can be extended to a basis $w_1,...,w_m$ of $W$ (with $m≥n$).

Okay, so far, so good. I'm not totally sure why we're doing this but I follow what is being done.

Now, let $S:W→V$ be the linear map that sends each $w_i$ to its unique preimage if $ i≤n$ or to 0 if $i > n$.

This is where I get so confused. Why define $S$ for just the basis vectors? Moreover, why are we allowed to just set $S(w)=0$ if $i > n$? More importantly, how do we even know this is a linear mapping?

EDIT: Defining $S$ on just the basis vectors for $i \leq n$ case doesn't make much sense to me, but I can get around it by just defining it like so: $S(T(v)) = v$. But what I really don't understand is the $S(w)=0$ case.

答:
Suppose we have (finite dimensional) vector spaces $W$ and $V$ (over some field $K$) and some linear function $f \colon W \to V$.
If $(w_1, ..., w_n)$ is a basis of $W$, then $f$ is entirely determined by the values $f(w_1), ..., f(w_n)$ because any vector $w \in W$ can be written as $w = a_1 w_1 + ... + a_n w_n$ (with $a_i \in K$ appropriately chosen), and hence we have
$$
\begin{align}
f(w)
&= f(a_1 w_1 + ... + a_n w_n) \\
&= f(a_1 w_1) + ... + f(a_n w_n) \\
&= a_1 f(w_1) + ... a_n f(w_n).
\end{align}
$$
It follows that we can define any such linear map by taking a basis of $W$ and defining the images of the basis vectors (basically this is why we can use matrices to represent linear functions!).


答:
This is a classical linear algebra.

Given two vector spaces (finite dimensional or not doesn't matter) $V$ and $W$ and a basis $B$ of $V$ then every function $f:B\to W$ can be uniquely extended to a linear map $F:V\to W$. This is because given any vector $v\in V$ it can be uniquely expressed as $v=\sum_{b\in B} \lambda_b\cdot b$ for some scalars $\lambda_b$ taken from the underlying field $k$. All those scalars except finitely many are zero. With that we have $F(v):=\sum_{b\in B}\lambda_b\cdot f(b)$.

And so in order to define any linear map $V\to W$ it is enough to define its values on a basis $B$.

Note that there is no restriction on $f:B\to W$. We can setup those initial values to be anything we want, in particular there is nothing prohibiting us to define $f(w)=0$ for some elements from $B$.

but I can get around it by just defining it like so: $S(T(v)) = v$

That's not a definition. That's an equation. And yes, ultimately we want that property to be true. But we want to define $S(x)$ for $x\in W$. There are two cases: $x$ does and does not belong to the image of $T$. If it does then we can simply define $S(x):=T^{-1}(x)$. Note that $T^{-1}(x)$ is a singleton, because $T$ is injective and so this is a well defined function.

But what happens if $x$ does not belong to the image of $T$? Meaning $T^{-1}(x)=\emptyset$. That's why this naive approach fails. We fix that by considering basis of the image of $T$ and its extension to full $W$. Defininig $S(w)=0$ on those basis elements outside of the image of $T$ is arbitrary. Any $S(w)$ will work fine.

The whole idea here is similar to classical sets: given an injective function $f:X\to Y$ how do we define its partial inverse $g:Y\to X$? Well, we define $g$ to be the inverse of $f$ on the image of $f$. For elements outside of the image of $f$ we define $g$ arbitrarily.

A vector basis allows us to treat linear maps as classical functions to some degree.

Title: If $T$ is injective, there exists left inverse $S:W\rightarrow V$

Body:
I apologize for this repeat question but I've been trying to figure this problem out for at least 3 hours and haven't made any progress. I've seen other Math SE posts to no avail.

Assume $W$ is finite dimensional and $T:V\rightarrow W$ is linear.
Show that if $T$ is injective, then there exists a linear map
$S:W\rightarrow V$ so that $ST=I_V$.

Like I've said, I've seen many Math SE posts and they all seem to do the same thing: The image of $T$ is a subspace of $W$;  hence,  finite dimensional. Then,  it has some basis $w_1,...,w_n$.  Moreover, since every linearly independent collection of vectors in $W$ can be extended to a basis of $W$, the basis of the image can be extended to a basis $w_1,...,w_m$ of $W$ (with $m≥n$).

Okay, so far, so good. I'm not totally sure why we're doing this but I follow what is being done.

Now, let $S:W→V$ be the linear map that sends each $w_i$ to its unique preimage if $ i≤n$ or to 0 if $i > n$.

This is where I get so confused. Why define $S$ for just the basis vectors? Moreover, why are we allowed to just set $S(w)=0$ if $i > n$? More importantly, how do we even know this is a linear mapping?

EDIT: Defining $S$ on just the basis vectors for $i \leq n$ case doesn't make much sense to me, but I can get around it by just defining it like so: $S(T(v)) = v$. But what I really don't understand is the $S(w)=0$ case.


问:
How to prove (analytically) that $\sin(\tan^{-1}(\sqrt{2}))=\sqrt{\frac{2}{3}}$?
I calculated $\sin(\tan^{-1}(\sqrt{2}))$ using Wolfram and got it as a result $\sqrt{\frac{2}{3}}$. However, I was unable to find an identity to obtain this result.

Is there any way to calculate this value without using Wolfram?

答:
Also, the angle $t$ of right triangle satisfies

$$\sin^2 t= \dfrac{2}{3}$$
$$\cos^2 t= \dfrac{1}{3}$$
$$\tan^2 t= 2 $$


问:
Success in maths (soft question)
I would really like to hear from any professional mathematicians who didn't just sail through their university education. If one looks at the pages of many of today's mathematicians, one finds that they usually aced their university exams (often coming top of the year at prestigious universities etc). This can be disheartening (at times) for someone who hasn't followed that path. Is there actually a non-negligible hope for such people? Thank you.</p>

答:
I wanted to post the excerpt below somewhere because I think many will find it interesting and, as far as I can tell, it seems to be nowhere on the internet (see my comments here). After several minutes of searching, this question wound up being the closest fit that I was able to find.

What follows are the italicized (except for 9 words) introductory comments to Part 3. Mathematics in Isaac Asimov’s 1969 book Opus 100 (amazon.com page and Wikipedia page)—pages 89−91 in my 1969 Dell paperback version and, I think, pages 87−90 in the 1969 Houghton Mifflin hardback version. Asimov has three more personal commentaries in Part 3 (pages 94−95, 102−104, 113−115 in my paperback version), but those other commentaries veer away from the main topic of what follows.

WHEN I WAS IN GRADE SCHOOL, I had an occasional feeling that I might be a mathematician when I grew up. I loved the math classes because they seemed so easy. As soon as I got my math book at the beginning of a new school term, I raced through it from beginning to end, found it all beautifully clear and simple, and then breezed through the course without trouble.

It is, in fact, the beauty of mathematics, as opposed to almost any other branch of knowledge, that it contains so little unrelated and miscellaneous factual material one must memorize. Oh, there are a few definitions and axioms, some terminology—but everything else is deduction. And, if you have a feel for it, the deduction is all obvious, or becomes obvious as soon as it is once pointed out.

As long as this holds true, mathematics is not only a breeze, it is an exciting intellectual adventure that has few peers. But then, sooner or later (except for a few transcendent geniuses), there comes a point when the breeze turns into a cold and needle-spray storm blast. For some it comes quite early in the game: long division, fractions, proportions, something shows up which turns out to be no longer obvious no matter how carefully it is explained. You may get to understand it but only by constant concentration; it never becomes obvious.

And at that point mathematics ceases to be fun.

When there is a prolonged delay in meeting that barrier, you feel lucky, but are you? The longer the delay, the greater the trauma when you do meet the barrier and smash into it.

I went right through high school, for instance, without finding the barrier. Math was always easy, always fun, always an “A-subject” that required no studying.

To be sure, I might have had a hint there was something wrong. My high school was Boys High School of Brooklyn and in the days when I attended (1932 to 1935) it was renowned throughout the city for the skill and valor of its math team. Yet I was not a member of the math team.

I had a dim idea that the boys on the math team could do mathematics I had never heard of, and that the problems they faced and solved were far beyond me. I took care of that little bit of unpleasantness, however, by studiously refraining from giving it any thought, on the theory (very widespread among people generally) that a difficulty ignored is a difficulty resolved.

At Columbia I took up analytical geometry and differential calculus and, while I recognized a certain unaccustomed intellectual friction heating up my mind somewhat, I still managed to get my A’s.

It was when I went on to integral calculus that the dam broke. To my horror, I found that I had to study; that I had to go over a point several times and that even then it remained unclear; that I had to sweat away over the homework problems and sometimes either had to leave them unsolved or, worse still, worked them out incorrectly. And in the end, in the second semester of the year course, I got (oh shame!) a B.

I had, in short, reached my own particular impassable barrier, and I met that situation with a most vigorous and effective course of procedure—I never took another math course.

Oh, I’ve picked up some additional facets of mathematics on my own since then, but the old glow was gone. It was never the shining gold of “Of course” anymore, only the dubiously polished pewter of “I think I see it.”

Fortunately, a barrier at integral calculus is quite a high one. There is plenty of room beneath it within which to run and jump, and I have therefore been able to write books on mathematics. I merely had to remember to keep this side of integral calculus.

In June, 1958, Austin Olney of Houghton Mifflin (whose acquaintance I had first made the year before and whose suggestion is responsible for this book you are holding) asked me to write a book on mathematics for youngsters. I presume he thought I was an accomplished mathematician and I, for my part, did not see my way clear to disabusing him. (I suppose he is disabused now, though.)

I agreed readily (with one reservation which I shall come to in due course) and proceeded to write a book called Realm of Numbers which was as far to the safe side of integral calculus as possible.

In fact, it was about elementary arithmetic, to begin with, and it was not until the second chapter that I as much as got to Arabic numerals, and not until the fourth chapter that I got to fractions.

However, by the end of the book I was talking about imaginary numbers, hyperimaginary numbers, and transfinite numbers—and that was the real purpose of the book. In going from counting to transfiinites [sic], I followed such a careful and gradual plan that it never stopped seeming easy.

Anyway, here’s part of a chapter from the book, rather early on, while I am still reveling in the simplest matters, but trying to get across the rather subtle point of the importance of zero.

ContentLicense: CC BY-SA 3.0

Title: Success in maths (soft question)


问:
Geometric-Variational idea of Sine
I like to see sin(theta) as a property of a line making clockwise angle of theta with a horizontal axis.That property would dictate how much %  the vertical component of each point in the line represents in relation the hiponetuse component of the point. Sin(80degrees) = 0,98 . That is, choosing any point on a line making 80degrees with a horizontal axis, we get that the vertical component is only 98% of the hipotenusa component.

I'm trying to go from there to the graph of a sin function.

Can anyone help me see intuitively why this % ( the values of the sine ) goes smoothly decelerating ( forming that semi-circle in the sin graph ) as we increase the angle towards 90?       Why doesn't it go linearly decelerating for example ?

I don't wanna bring circles in since sin is just one component of the circle.

Thanks a lot.

答:
Why isn't the graph of the sine function linear?

Well, think about a line of length 1 with one end at the origin, making an angle $\theta$ with the $x$-axis. As you say, $\sin\theta$ is the $y$-coordinate of the other end of this line.

Now think about what happens as this line rotates, and $\theta$ increases.

When $\theta$ is close to zero, changing $\theta$ by one degree will produce a significant change in the $y$ coordinate of the end of the line, so $\sin\theta$ will increase fairly rapidly.

Next, consider what happens when $\theta$ is around 85 degrees, say. In this case, increasing $\theta$ by one degree will make very little difference in the $y$ coordinate of the end of the line, so $\sin\theta$ will increase very slowly. By the time we get to $\theta = 90$ degrees, changing $\theta$ by a small amount won't change the $y$ coordinate at all, so we would expect the graph of the sine function to be flat (which indeed, it is).

Title: Geometric-Variational idea of Sine


问:
Algorithm for Weighted Geospatial Clustering
Given a set of values (x, y, z) = (x-coord (long), y-coord (lat), revenue), is there an algorithm similar to k-means clustering that will help find the most revenue concentrated clusterings of a fixed radius r and allowing the number of clusters to vary?

I understand that what I am laying out as the structure of this model is somewhat ambiguous and not entirely concrete, so slight alterations are welcome, but am essentially looking to find the must revenue dense areas.

答:
You can try to compute kernel density plots for spatial point patterns, see this link: density.ppp function in the R spatstat package. Your $r$ would be the radius of whatever kernel you choose and let the intensity of the kernel be dependent upon revenue. The output of this process is a surface may analyze further, it will give you some idea of where revenue is concentrated, but it will be quite dependent upon your sampling method.

Edit: You may also want to try to compute different types surfaces to gauge revenue. Inverse distance weighting, for example. Spatial interpolation like Kriging might also be useful. A small list of different methods is here;


问:
Definition of uniform convergence
I don't see how the two formulations for uniform convergence are equivalent:

$$
\lim_{n\rightarrow\infty}\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n\left(x\right)\right)\right\}=0  \\
\Leftrightarrow \\
\forall \varepsilon >0:\ \exists N\in \mathbb{N}:\ \forall x\in X:\ n\ge N\ \Longrightarrow \ d_Y\left(f\left(x\right){,}\ f_n\left(x\right)\right)<\varepsilon
$$

By the definition of limit, the above is clearly equivalent to

$$
\forall\varepsilon>0:\ \exists N\in\mathbb{N}:\ n\ge N\ \Longrightarrow\ \left|\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)\right\}-0\right|<\varepsilon,
$$
and since distance is non-negative,
$$ \left|\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)\right\}-0\right|=\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n\left(x\right)\right)\right\}<\varepsilon.$$

I am not sure however if $$
\sup_{x\in X}\left\{d_Y\left(f_n\left(x\right){,}\ f\left(x\right)\right)\right\}<\varepsilon\ \Leftrightarrow\ d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)<\varepsilon.
$$

Could someone clarify?

答:
I'm assuming you meant to write

$$\sup_{x\in X}\left\{d_Y\left(f_n\left(x\right){,}\ f\left(x\right)\right)\right\}<\varepsilon\ \Leftrightarrow\ \forall x \in X, d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)<\varepsilon,$$

since that would facilitate the proof.

The reverse implication does not hold with strict inequality, but this presents no major problem.

It should be if $d_Y(f(x), f_n(x)) < \epsilon$ for all $x \in X$, then $\sup\{d_Y(f(x), f_n(x)): x \in X\} \leqslant \epsilon$.

To prove this, suppose $\sup\{d_Y(f(x), f_n(x)): x \in X\} = \alpha > \epsilon$ and take $\beta$ such that $\epsilon < \beta < \alpha$. There must exist $x' \in X$ such that $d_Y(f(x'),f_n(x')) > \beta> \epsilon$, otherwise $\beta$ is an upper bound less than the least upper bound $\alpha$.  However, this contradicts the hypothesis that $d_Y(f(x),f_n(x)) < \epsilon$ for all $x \in X$.

The forward implication that $\sup\{d_Y(f(x), f_n(x)): x \in X\} < \epsilon$ implies  $d_Y(f(x), f_n(x)) < \epsilon$ holds as you have written and is straightforward to prove.

Title: Definition of uniform convergence


问:
$f$ nonnegative, continuous in $[a,b]$ and such that $\int_a^b f(x)dx=0$ implies $f(x)=0$ in $[a,b]$
My book proves the following statement: "Let $f$ be nonnegative and continuous in $[a,b]$ and such that
$$\int_a^b f(x)dx=0$$
then $f$ is identically $0$ in $[a,b]$." using an argument by contradiction.

However, I've tried to prove it in a direct way (which takes way more effort, but I would like to know if there is a flaw in my reasoning or not), can someone check if my work is okay please?

My try: by hypothesis $f$ is nonnegative in $[a,b]$: if $f(x)=0$, the theorem is true.

So let's suppose $f(x)>0$: let $x_0 \in [a,b]$, by hypothesis $f$ is continuous in $[a,b]$, so in particular it is continuous in $x_0$. Since $f$ is positive in this case, I can chooose $\epsilon=\frac{1}{2}f(x_0)$ in the definition of continuity to get the inequality $\frac{1}{2}f(x_0) \leq f(x) \leq \frac{3}{2}f(x_0)$ if $x_0-\delta_\epsilon < x < x_0+\delta_\epsilon$. Since by hypothesis $f$ is continuous in $[a,b]$, it is integrable in $[a,b]$ as well, so by monotonicity of integral it is
$$\int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} \frac{1}{2}f(x_0)\,dx \leq \int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} f(x)\,dx \leq \int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} \frac{3}{2} f(x_0)\,dx$$
By hypothesis $\int_a^b f(x)\,dx=0$, so in particular $\int_{x_0-\delta_\epsilon}^{x_0+\delta_\epsilon} f(x)\,dx=0$; hence from the last inequality it follows that $\delta_\epsilon f(x_0) \leq 0 \leq 3\delta_\epsilon f(x_0)$. Since $\delta_\epsilon>0$, This means that $f(x_0)$ is both $\geq0$ and $\leq 0$, the only possibility is that $f(x_0)=0$. Since $x_0 \in[a,b]$ is arbitrary, this holds for all $x\in[a,b]$ and so $f(x)=0$ for all $x\in[a,b]$.

Is this correct? If it is maybe I should've put more attention in the intervals of the kind $[a,x_0+\delta_\epsilon]$ and $[x_0-\delta_\epsilon,b]$ because since $\delta_\epsilon$ if $x_0$ is too much near the boundary I risk to "go out" of $[a,b]$, but I'm not sure if this is a problem or not since if $\delta_\epsilon$ works a lesser $\delta'_\epsilon$ would work too and so I can restrict it as much as I need.

答:
If $f(x)$ is non-negative on $[a,b]$ then that suggests that $f(x)\ge0$ on $[a,b]$ and so:
$$\int_a^bf(x)dx\ge0$$
for a function to be non-zero and the integral be zero over $[a,b]$ there must be a point of discontinuity $c\in(a,b)$ such that:
$$\int_a^cf(x)dx=-\int_c^bf(x)dx$$
but this is a contradiction because this function is discontinuous and would have to be negative at some point, and so the only solution is:
$$iff\,\,\,\int_a^bf(x)dx=0,f(x)\ge0\Rightarrow f(x)=0$$


问:
How to find the eigen vector of a system that does not have similar equations?
Suppose I am trying to solve for an eigenvector provided
my matrix
$$A=\left(
\begin{array}{cc}
 2 & 7 \\
 7 & 2 \\
\end{array}
\right)$$

eigenvalues are $9$ and $-5$

How can I get an eigenvector when I set lambda to $-5$?

$$
\begin{array}{cc|c}
 -3 & 7&0 \\
 7 & -3&0 \\
\end{array}
$$

答:
$$\left(
\begin{array}{cc}
 2 & 7 \\
 7 & 2 \\
\end{array}
\right)-(-5)I=(0,0)$$
$$
\left(
\begin{array}{cc}
 7 & 7 \\
 7 & 7 \\
\end{array}
\right)\left(
\begin{array}{c}
 x \\
 y \\
\end{array}
\right)=\left(
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right)$$
$7x+7y=0\to y=-x$ so eigenvector is $(1,-1)$

In a similar way for the other eigenvalue you get $(1,1)$


问:
What will be the count of "4cardstraight" in a Poker Game
Let's modify the  poker hand(fake):  "4cardstraight"  -- i.e. a straight, but with only 4 cards in a row instead of 5.

Rules of the game:

A hand is counted as a "4cardstraight" if it includes 4 cards in a row, but not 5 (so it's not a regular 5-card straight).

When a hand qualifies as multiple types, it should be counted as  the highest-value hand; for this purpose, the value of "4cardstraight" will be just below a straight.

In such game what will be count of "4cardstraight" ?

I found an answer for the regular poker game to count the straight at wikipedia page and explanation in this answer.
So my approach is to count all the number of sequences(having a straight of 4 and 5) and then subtract 10*4^5  from the answer to get count of "4cardstraight".

But I am not able to figure out that how can I calculate the count for all straight of size >=4.

Thanks

答:
It seems straightforward to count directly. From your count of straights, it looks as if you are allowing Aces high or low.

Viewing by bottom card, there are $11$ types of $4$-card straight. But $2$ of the types are special, the Ace low and the Ace high. For each of these $2$ types, the cards that make up the straight can be chosen in $4^4$ ways, and the odd card in $44$ ways. (if we are building a $4$-card straight of type Ace, 2, 3, 4, we will not be using any alredy chosen card, and we must avoid 5's). This gives a total of $(2)(4^4)(44)$.

For each of the $9$ other types, the odd card can have any of $40$ values. For example, if we are building a $4$-card straight that goes 8, 9, 10, J, we must avoid the 7's and the Queens.  This gives a total of $(9)(4^4)(40)$.

Add up the two numbers obtained.


问:
Finding the solution to a third-order linear homogeneous differential equation
The equation to solve is: $(4D^3+27D^2+70D+39)y=0$

My first step is to identify the characteristic equation :
$4D^3+27D^2+70D+39$

Then, I find the roots of the characteristic equation. There is only one root: -3/4.

Finally, I find the complimentary solution: $y = C_{1}e^{-3x/4}$

I have also verified that placing this complimentary solution into the original differential equation does indeed equal 0.

Can I consider that I have solved this differential equation? I am confused as to whether or not I need to find a particular solution for this problem.

Thank you.

答:
Next you extract the factor $4D+3$ to get the deflated quadratic polynomial,
$$\begin{align}
4D^3+27D^2+70D+39&=(4D+3)D^2+24D^2+70D+39\\
&=(4D+3)(D^2+6D)+52D+39\\
&=(4D+3)(D^2+6D+13)
\end{align}$$
Then find the roots of the quadratic equation $0=D^2+6D+13=(D+3)^2+4$ etc.


问:
How to define my moments of a fake dice?
Say I have a dice.

Say I suspect that the dice is unfair with probability 75 % with probabilities $p_1, ..., p_6$ that are NOT all equal to 1/6.

However, with 25 % probability, the dice is actually fair and the probabilities are indeed 1/6.

I have two questions:

What is the mean and variance of this dice's outcomes?

Are there different ways of defining what I meant by "fake with 75 % probability" and do those different ways lead to different answers? For example, maybe the die is EITHER fake OR real, and I don't know which one it is .... that is one way to define it. OR maybe the die is BOTH fake AND real, and it switches between these two states every time I throw it. This definition is different, but does it give the same answer? Are there other definitions that make sense?

答:
Regarding your second question, you can get different answers depending on what you mean by "unfair with 75% probability". In most cases, you will have dice with fixed characteristics that are either fair or unfair, with some fixed probability distribution over the faces. If the die is either fake or real, the expected value of the die is either 3.5 (if it's real) or whatever the expected value of the biased die is. Regardless of how much you believe the die is fair or biased, the die's actual expected value must be one of two values, and not the average of both.

If you did have a die that could change characteristics, the expected value would be the weighted average of the fair and biased expected values. The expected value of the die in this case is based on real, physical characteristics of the die, and not your own beliefs about the die's characteristics.

You could alternatively look at this as a process in which you randomly select a die from 3 biased and 1 fair die, and then roll it. In this case, the expected value of the process is again the weighted average of the fair/unfair expected values. But if you only have one die with immutable characteristics, its expected value is determined only by those characteristics and not how likely you think it is that the die is fair. The die actually is fair or not, and has some fixed, true expected value which is unaffected by how strongly you believe the die is fair.


问:
A detail on the proof of Theorem 6.9 of Steel's note on iterated ultrapowers
I am reading Steel's note on iterated ultrapowers, and I am filling the gaps in the proof of Theorem 6.9. However, I am stuck on how to proceed with it.

I am checking the last part of the proof (page 31 of the linked note), which claims $P_{n+1}\in V^Q_{\operatorname{lh}E}=V^{P_n}_{\operatorname{lh}E}$. However, I do not see why the last equality holds. I tried to prove it from the equality $V^{N_n}_{\operatorname{lh}F_n}=V^{N_{n+1}}_{\operatorname{lh}F_n}=V^{\operatorname{Ult}(N_k,F_n)}_{\operatorname{lh}F_n}$. While $\pi_n(V^{N_n}_{\operatorname{lh}F_n})=V_{\operatorname{lh}E}^{P_n}$ seems obviously true, I have no idea why $\pi_n(V^{\operatorname{Ult}(N_k,F_n)}_{\operatorname{lh}F_n})=V^Q_{\operatorname{lh}E}$ holds.

I do not see how iterations over a tree work, so I must be missing something on very simple facts. I would appreciate any help.

答:
I get how to proceed. Note that $k$ is the least number such that $\operatorname{crit} F_n<\operatorname{lh} F_k$. Moreover, if we take $\tau_k=\sup \sigma^" \operatorname{lh} F_k$, then $P_k\underset{\tau_k}{\backsim} P_n$ by the inductive assumption.

Fix $\xi$ such that $\operatorname{crit} F_n<\xi<\operatorname{lh} F_k$. Then $$\operatorname{crit}E=\pi_n(\operatorname{crit}F_n)<\pi_n(\xi)=\sigma(\xi)<\tau_k,$$ where the equality follows from Exercise 34. By Lemma 6.2, we have $\operatorname{Ult}(P_k,E)\backsim_{i_E(\operatorname{crit}E)+1} \operatorname{Ult}(P_n,E)$.
Since $\operatorname{lh}E\le i_E(\operatorname{crit}(E))$ and $P_n\models \text{$E$ is nice}$, we have $V^Q_{\operatorname{lh}E}=V^{\operatorname{Ult}(P_k,E)}_{\operatorname{lh}E}=V^{\operatorname{Ult}(P_n,E)}_{\operatorname{lh}E}=V^{P_n}_{\operatorname{lh}E}$.


问:
Compact set in lp space
$1\leq p< \infty$
$$\ell_p:=\{x=(x_n):\sum_{i=1}^{\infty}|{x_i}|^p<\infty\}$$ sequance space
$$\| x\|_p:=(\sum_{i=1}^{\infty}|{x_i}|^p)^{\frac{1}{p}}$$and p norm
$$ $$
I know $ (\ell_p,\|.\|_p)$ Banach space
$$A:=\{x \in \ell_p: \forall n \in \Bbb{N}; \ |x_n| \leq n^{\frac{-2}{p}}\}$$
My question is show A is a compact subset of $\ell_p$ how can I do that I  try to show sequantial compactnes of A  but I can't please help .

答:
Consider the topological space $$X=\prod_{n=1}^\infty D_n$$
where $D_n=\{z\in\mathbb{C}: |z|\leq n^{-2/p}\}$, where $X$ is endowed with the product of the standard topology that disks have. Since this is a product of closed and bounded subsets of the plane (which are compact subsets), it is a compact space, by Tychonoff's theorem.

Define $f:X\to\ell^p$ by $f(x_n)=(x_n)$. The image of $f$ is precisely set $A$. But $f$ is continuous: Let $(x_\lambda)$ be a net converging at $x\in X$, say $x_\lambda=(x_n^\lambda)$ and $x=(x_n)$. Let $\varepsilon>0$ and we take an integer $M\geq1$ so that $\sum_{n=M+1}^\infty\frac{1}{n^2}<\frac{\varepsilon}{2^{p+1}}$, which is possible because the series converges. By the definition of the product topology, we can find $\lambda_0$ far out in the net so that for all $\lambda\geq\lambda_0$ we have that $|x_n^\lambda-x_n|<(\frac{\varepsilon}{2M})^{1/p}$ for all $n=1,\dots, M$. Therefore, for such $\lambda$ we have that
$$\|f(x_\lambda)-f(x)\|^p=\sum_n|x_n^\lambda-x_n|^p=\sum_{n=1}^M|x_n^\lambda-x_n|^p+\sum_{n=M+1}^\infty|x_n^\lambda-x_n|^p\leq$$ $$\leq\frac{\varepsilon}{2}+\sum_{n=M+1}^\infty(\frac{1}{n^{2/p}}+\frac{1}{n^{2/p}})^p\leq\frac{\varepsilon}{2}+2^p\cdot\sum_{n=M+1}^\infty\frac{1}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$
where we used the triangle inequality together with the fact that $x^\lambda_n, x_n\in D_n$.

But a continuous function maps compact sets to compact sets, so since $X$ is compact we have that $A$ (which is equal to $f(X)$) is compact.

Title: Compact set in lp space


问:
Differential equation and solution involving light passing through a glass.
I been stumped on this question for a few days and can't seem find an elegant solution ($x = 8.66cm$).   I have found $k = 0.0811$.

Any help appreciated.

The intensity I of light diminishes as it passes through glass, the reduction in intensity being proportional to the thickness x of glass.  The intensity is 15 as it enters a block of glass 5cm thick, and is 10 when it leaves, find the minimum thickness of glass required to reduce the intensity to zero for the same light source.

I have been using the differential equation $\frac{dI}{dx} = -kI.$

Thanks in advance

Postscript:  Thanks to A-Level Student the DE to find the solution is

$\frac{dI}{dx} = -kx.$  This reduces the problem to $I = \frac{x^2}{5} + 15.$  Then setting $I = 0$ gives $x = 8.66cm.$

答:
The equation that you use is incorrect: $$dI=-k\ dx$$
The solution of this is $$I_0-I(x)=kx$$
So first you calculate $k$:
$$k=\frac{I_0-I(5)}{5}=1$$
Then you calculate $x$ for $I(x)=0$
$$I_0-0=kx\\15-0=x\\x=15$$


问:
Could someone give a detailed (yet elementary) proof for Jensen's inequality?
I want to prove that

Suppose there is a function $f:[a,b] \to \mathbb R$, and there are $x_i \in [a,b], w_i \gt 0 $ for $i=1,\dots,n$ such that $\sum_{i=1}^nw_i=1$, then if the function is convex, the following inequality holds $$\sum_{i=1}^nw_if(x_i) \ge f(\sum_{i=1}^n w_ix_i)$$
  and if the function is concave, flip the sign of inequality.

Please include (very) detailed explanation, because I do not have much mathematical maturity, and perhaps may not be able to understand what the equation defining convex means if the explanation is too concise.

答:
Lemma: If you have some weighted points scattered in a plane, then their centre of mass lies in their convex hull. To understand convex hull, see the pictures on page-768 of the paper

Proof:

Suppose the point was outside and let's say we marked that in red, then we can draw a line through that point (one in blue), now if we were to draw vectors from that point to vertices where the particles are, it is clear that the vector sum can't be zero because the vector to each particle has some component along the normal. Now, recall the definition of the center of mass,

$$ \sum_{i=0}^n m_i \vec{r_i} =(\sum_{i=0}^n m_i)\vec{r_{com} }$$

Or,

$$ \sum_{i=0}^n m_i( \vec{r_i} - \frac{\vec{r_{com} } }{n} )= 0$$

So, the centre of mass can be though of the point from which the vectors to the mass particles add to zero, but here it is clear that this doesn't happen. Hence, it is necessary that centre of mass lies inside the convex hull.

I'll do the proof for a convex function $(f'' \geq 0)$ and I'll leave it to you to prove it for a concave one. A convex function is something which looks like this:

Now, suppose I placed a mass of $m$ on three points on the curve:

It is easy to see the centre of mass of these points must have coordinates $$ ( \frac{x_1 + x_2 + x_3}{3} , \frac{ f(x_1) + f(x_2) + f(x_3)}{3})$$ (purple dot)

And, we can that the point on curve at same x coordinate (green dot) is given as: $$( \frac{x_1 + x_2 + x_3}{3} ,  f( \frac{x_1 + x_2 +x_3}{3})) $$

And since the purple dot is above the green dot, we can write the following inequality:

$$ f( \frac{x_1 + x_2 +x_3}{3}) \leq \frac{ f(x_1) + f(x_2) + f(x_3)}{3}$$

And of course, this argument would still if you took more points / had different amount of mass on each point. Hence, we have found a simple visual understanding of Jensen's inequality.


答: I don't think you can reduce it to zero unless you account for the quantum nature of light.   Proportion means it will  scale the intensity down.   But if k and x are greater than zero , then k*x will always be nonzero.  The exception is when you reduce the intensity to less than a photon.  The intensity can be made as close to zero as you want, however.</p>

Title: Differential equation and solution involving light passing through a glass.

Body:
I been stumped on this question for a few days and can't seem find an elegant solution ($x = 8.66cm$).   I have found $k = 0.0811$.

Any help appreciated.

The intensity I of light diminishes as it passes through glass, the reduction in intensity being proportional to the thickness x of glass.  The intensity is 15 as it enters a block of glass 5cm thick, and is 10 when it leaves, find the minimum thickness of glass required to reduce the intensity to zero for the same light source.

I have been using the differential equation $\frac{dI}{dx} = -kI.$

Thanks in advance

Postscript:  Thanks to A-Level Student the DE to find the solution is

$\frac{dI}{dx} = -kx.$  This reduces the problem to $I = \frac{x^2}{5} + 15.$  Then setting $I = 0$ gives $x = 8.66cm.$


问:
$X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, $Y$ is $\mathcal{G}$-measurable, then $f(X,Y)$ is independent of $\mathcal{G}$?
Let $(\Omega, \mathcal{F}, P)$ be a probability space,  $\mathcal{G}\subset \mathcal{F}$ be a sigma algebra, $X, Y : \Omega \to \mathbb{R}^n$ be random vectors, and $f$ be a measurable function on $\mathbb{R}^{2n}$.
Suppose that $X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, and $Y$ is $\mathcal{G}$-measurable. Is it true that $f(X,Y)$ is independent of $\mathcal{G}$?

Motivation: This question arises from a randomized algorithm. In this algorithm, each iteration receives some randomness that is independent of the previous iterations. In my question, $\mathcal{G}$ indeed represents the randomness of the algorithm up to iteration $k$, $Y$ is a vector generated by iteration $k$, $X$ is some new randomness injected into the algorithm at iteration $k+1$, and $f(X,Y)$ is a quantity computed from $X$ and $Y$. It turns out that $f(X,Y)$ is statistically independent of $Y$. I would like to prove that $f(X,Y)$ is independent of the first $k$ iterations, which will be interesting and convenient.

Any comments or criticism will be appreciated. Thank you.

A related question: Uniform distribution on the unit sphere rotated by a random orthogonal matrix.

答:
The answer is yes.  To see this, I claim that if $A \in \mathcal{G}$ and $M > 0$, then
\begin{equation*}
\mathbb{E}(f(X,Y) : A, |f(X,Y)| \leq M) = \mathbb{E}(f(X,Y) : |f(X,Y)| \leq M) \mathbb{P}(A).
\end{equation*}

First, notice that the independence of $X$ from $\mathcal{G}$ implies that
\begin{equation*}
\mathbb{E}(f(X,Y) 1_{\{|f(X,Y)| \leq M\}} \mid \mathcal{G}) = \mathbb{E}(f(X,Y) 1_{\{|f(X,Y)| \leq M\}} \mid Y).
\end{equation*}

The only way I can see to prove this (in general) is to approximate $f(X,Y) 1_{\{|f(X,Y)| \leq M\}}$ by a sequence of (sums of) random variables of the form $g_{n}(X) h_{n}(Y)$, which is possible since, for instance, continuous functions with compact support can be approximated in this way.  (We can derive $\mathbb{E}(g_{n}(X) h_{n}(Y) \mid \mathcal{G}) = \mathbb{E}(g_{n}(X)) \mathbb{E}(h_{n}(Y) \mid Y)$ immediately by independence and $\mathcal{G}$-measurability.)

Next, notice that since $f(X,Y)$ is independent of $Y$, it follows that $\mathbb{E}(f(X,Y) 1_{\{|f(X,Y)| \leq M\}} \mid Y) = \mathbb{E}(f(X,Y) :|f(X,Y)| \leq M)$ almost surely.

Putting the two insights together, we deduce that $\mathbb{E}(f(X,Y) 1_{\{|f(X,Y)| \leq M\}} \mid \mathcal{G}) = \mathbb{E}(f(X,Y) : |f(X,Y)| \leq M)$ almost surely.  Therefore, by definition of conditional expectation,
\begin{equation*}
\mathbb{E}(f(X,Y) : A, |f(X,Y)| \leq M) = \mathbb{E}(f(X,Y) : |f(X,Y)| \leq M) \mathbb{P}(A).
\end{equation*}


问:
Minimum DFA's number of states accepting strings whose length is not a multiple of 9 based on an alphabeth composed by 4 symbols
How can I calculate the minimum DFA's number of states accepting strings whose length is not multiple of 9 based on an alphabet composed by 4 symbols?

My attempt is:

$L = \{w \in \{a,b,c,d\}^* : \text{9 does not divide |w|}\}$

that means there is not a whole number $z$ such that $|w| = 9\cdot z$

therefore $|w| \in \mathbb{N} - \{9, 18, 27, ...\}$

How can I continue?

答: The number of characters in the alphabet is irrelevant, since you’re interested only in the length of the word. In order to recognize when the length is a multiple of $9$, you need $9$ states, one for each possible remainder. The best that you can do is ring of $9$ states, all of them acceptor states except the initial state. If you’re not sure, start with that DFA and apply the algorithm to minimize it.


问:
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas
I have this identity:

$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$

If I write this like as:

$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$

I can use Werner and prostaferesis formulas and I find the identity $(1)$.

But if we suppose of not to use these formulas I have done a try writing:

$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$

$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$

$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$

Then I have abandoned because I should do long computations and I think that is not the right way.

Is there any trick without to use the prostapheresis or Werner's formulas?

答:
hint

Put $$a=2\alpha$$
and prove simply that

$$\cos^2(2a)=\cos^2(a)-\sin(3a)\sin(a)$$

using

$$2\sin(3a)\sin(a)=$$
$$\cos(3a-a)-\cos(3a+a)$$


问:
Injective or surjective linear transformations
Let $f: \mathbb{R}^3 \to \mathbb{R}^4$ be an arbitrary linear transformation. Does an $f$ exist, which is injective/surjective.

I feel like the answer to both should be yes, because although the matrix $A$ corresponding to that transformation is not a square, we only need to find a matrix $A^{-1} \in \mathbb{R}^{3 x 4}$ that is the left-inverse of $A$. Surjectivity would be ensured that way, because given a vector $w$, we only need to calculate $A^{-1}w$. With injectivity I am not quite sure how to argue my case.

答: Suppose such linear transformation can be surjective and try applying the rank-nullity theorem, that should give you a contradiction...</p>


问:
$E^*(\mathbb{C}P^{\infty})=\bigoplus_{n\in\mathbb{Z}}E^n(\mathbb{C}P^{\infty})$ or $\prod_{n\in\mathbb{Z}}E^n(\mathbb{C}P^{\infty})$?
PRELIMINARY DEFINITIONS:

Let $E^*$ be a multiplicative generalized cohomology theory. By the suspension isomorphism we have:
$$
\tilde{E^2}(S^2)\cong\tilde{E^0}(S^0)=E^0(pt)
$$
So there is a special element $t\in\tilde{E^2}(S^2)$ which corresponds to $1_E\in E^0(pt)$ under the above isomorphism.

$E^*$ is said to be complex orientable if the inclusion map $i:S^2=\mathbb{C}P^1\to\mathbb{C}P^{\infty}$ induces a sujective morphism $i^*:\tilde{E^2}(\mathbb{C}P^{\infty})\to\tilde{E^2}(S^2)$. A pair $(E^*,x_E)$ of a complex orientable cohomology theory $E^*$ and a choice of an element $x_E\in\tilde{E^2}(\mathbb{C}P^{\infty})$ such that $i^*(x_E)=t$ is called an oriented cohomology theory.

It is well known that singular cohomology, complex K-theory and complex cobordism are examples of orientable cohomology theories.

QUESTION:

According to various textbooks, for example [1, Theorem 3.10], we have the following classical result: for any complex oriented cohomology theory $(E^*,x_E)$, there is an isomorphism
$$
E^*(\mathbb{C}P^{\infty})\cong E^*(pt)[[x_E]]
$$

Now my question is: $E^*(\mathbb{C}P^{\infty})=\bigoplus_{n\in\mathbb{Z}}E^n(\mathbb{C}P^{\infty})$ or $E^*(\mathbb{C}P^{\infty})=\prod_{n\in\mathbb{Z}}E^n(\mathbb{C}P^{\infty})$? More generally, how is $E^*(X)$ defined for a topological space $X$?

Indeed, if I assume that $E^*(X)$ is the direct sum (as in [1, page 26]), then I found a contraddiction taking as oriented cohomology the singular cohomology with the first Chern class:
$$
\mathbb{Z}[c_1]=H^*(\mathbb{C}P^{\infty},\mathbb{Z})\neq \mathbb{Z}[[c_1]]
$$
Taking $E^*(X)$ to be the cartesian product I do not found any contraddiction, but it is a bit messy if compared with the standard notation in algebraic topology, where usually the cohomology ring is defined to be the direct sum. For example, if we consider K-theory, we know that $K^{2i+1}(pt)=0$ and $K^{2i}(pt)=\mathbb{Z}\beta^i$, where $\beta\in K^2(pt)$ is the Bott element. So, according to this definition $K^*(pt)=\mathbb{Z}[[\beta,\beta^{-1}]]$ and thus
$$
K^*(\mathbb{C}P^{\infty})=\mathbb{Z}[[\beta,\beta^{-1}]][[x_K]]
$$
where $x_K$ is the first Chern class in K-theory. This result seems a bit strange to me, since I used to know that $K^*(\mathbb{C}P^{\infty})=\mathbb{Z}[\beta,\beta^{-1}][[x_K]]$.

However, the definition of $E^*(X)$ via the cartesian productseems to be the right choice, as suggested in this previous question Is $H^*(\mathbf{C} P^\infty)=R[X]$ or $R[[X]]$?.  Someone can help? What am I missing?

REFERENCES:

[1] A.Kono, D.Tamaki-Generalized cohomology

[2] J.Lurie-Chromatic homotopy theory course http://people.math.harvard.edu/~lurie/252x.html

[3] M.Hopkins-Complex oriented chomology theories and the language of staks https://people.math.rochester.edu/faculty/doug/otherpapers/coctalos.pdf

答:
The correct definition is to use neither products nor direct sums, but to instead define graded rings as sequences.  That is, define a $\mathbb{Z}$-graded ring $A$ to be a sequence of abelian groups $(A_n)_{n\in\mathbb{Z}}$ together with multiplication maps $A_n\times A_m\to A_{m+n}$ satisfying the usual axioms.  If $A$ is a $\mathbb{Z}$-graded ring, then you can define a power series graded ring $A[[x]]$ (where $x$ has degree $d$) for which $A[[x]]_n$ is the set of formal expressions $\sum_{k=0}^\infty a_kx^k$ where $a_k\in A_{n-dk}$ for each $k$, with multiplication defined in the obvious way.  The statement $$E^*(\mathbb{C}P^{\infty})\cong E^*(pt)[[x_E]]$$ is then an isomorphism of graded rings in this sense.

As you have observed, this statement is not true in general if you define both sides as rings by taking direct sums.  It also does not make sense in general if you take direct products, though: $E^*(pt)$ is not even a ring in general if you take direct products!  If $E$ is connective then it is a ring and you can define multiplication in the obvious way, but if $E^*(pt)$ is nontrivial in both positive and negative degrees that are arbitrarily large then there may be products that do not make sense.  For instance, for $K$-theory, if you tried to define $K^*(pt)=\mathbb{Z}[[\beta,\beta^{-1}]]$ as a direct product, then how would you compute $$\left(\sum_{n\in\mathbb{Z}} \beta^n\right)^2?$$  There would be infinitely many terms that contribute to the same degree.

(I suppose you could get a statement that is always correct by defining $E^*(X)=\prod_{n\geq 0} E^n(X)\oplus \bigoplus_{n<0}E^n(X)$; i.e., you take a sort of restricted product where you can have infinitely many nonzero coordinates only in the positive direction.  I don't know of any benefit to doing this instead of just leaving your graded rings as sequences, though.)


问:
Is $\textbf{NC}_n$ a ranked lattice?
A set partition $\prod_n$ is a $\textit{noncrossing partition}$ (NCP) if its associated equivalence relation $\cong$ satifies the following condition: $$\text{for all } i<j<k<l \text{ if } i \cong k \text{ and } j \cong l \text{ then } i \cong j \cong k \cong l$$

The set of NCPs of order $n$ is denoted $\textbf{NC}_n$. Ordering by reverse refinement makes  $\textbf{NC}_n$ into a subposet of the partition lattice $\prod_n$

I'm interesting in showing that $\textbf{NC}_n$ is a ranked lattice, and was also pondering the natural question as to if it is a sublattice of $\prod_n$?

I"ve been relentlessly trying to find a counter example, as my gut is telling me there is no way that $\textbf{NC}_n$ can be ranked... However, such a counter example has not been forthcoming. I've been fiddling around with this for too long and would appreciate if someone could help me escape my tunnel vision... Thanks!

答:
In this lattice, a covering relation merges two parts (under the condition that these two parts are neighbors in the appropriate sense), just as in the lattice of all set partitions.

So the number of parts provides the rank function.

EDIT:

To understand the rank function, it is enough to look for a function on the elements of the posets that is increased by one along every edge of the Hasse diagram, i.e. along every cover relation.

So you need first to understand the cover relations.

The partial order relation between two noncrossing partitions is given by inclusion of each block of one in some block of the other. An increasing relation is therefore obtained by merging some parts together. Cover relations turn out to be obtained if and only if exactly two parts are merged, and for this to be possible, the two parts must see each other, namely not be separated by another part.

Once one understands the cover relations, it is easy to check that the function "minus number of parts" does the job, indeed increases by one along cover relations.

Title: Is $\textbf{NC}_n$ a ranked lattice?


问:
Why don't we extend the naming of powers into higher dimensions?
This is probably a stupid question.

If 2^2 is known as "Two squared" and 2^3 is "Two cubed", then why do we stop at 2^3? Why do we not call 2^4 "Two tesseracted", and so on?

答:
We could, but don't for several reasons.

Squares and cubes calculate areas and volumes; in everyday life we
never encounter the geometric generalizations.

You can't keep inventing a new word for each dimension. There isn't
one for hypercubes in dimensions greater than four. So it's not clear
what your "and so on" should be.

In higher dimensions mathematicians want the name of the calculation
to tell you the dimension, so you use the dimension itself, not a
made up word. So "two to the $17$th" for $2^{17}$.

"Tesseracted" is ugly.


问:
Can we show that $(x_n)$ is relatively compact if and only if $(\langle x_n,\;\cdot\;\rangle)$ restricted to a closed ball is relatively compact?
Let $E$ be a Banach space, $\iota:E\to E''$ denote the canonical injection of $E$ into $E''$, i.e. $$\iota x:=\langle x,\;\cdot\;\rangle\;\;\;\text{for }x\in E,$$ $\sigma_c(E',E)$ denote the topology of compact convergence on $E$ and $(x_n)_{n\in\mathbb N}$.

Are we able to show that $(x_n)_{n\in\mathbb N}$ is relatively compact if and only if there is a $r>0$ such that $(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is relatively compact with respect to the topology of uniform convergence on $C(B^{E'}_r(0),\sigma_c(E',E))$?

The claim can be found in the proof of Linde's book Probability in Banach Spaces - Stable and Infinitely Distributions, Lemma 2.3.6:$^1$

He also seems to conclude (most probably using the Arzelà-Ascoli theorem) that "$(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is relatively compact with respect to the topology of uniform convergence on $C(B^{E'}_r(0),\sigma_c(E',E))$" follows from "$(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is $\sigma_c(E',E)$-equicontinuous". But this is only true, if $(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is pointwise bounded, which should be the case if (and only if) $(x_n)_{n\in\mathbb N}$ is bounded. Why is that the case?

$^1$ $V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$ is endowed with the restriction of the topology $\sigma_c(E',E)$ and $\delta_x$ denotes the Dirac measure concentrated at $x$.

答:
Given a set $E$, denote by $M(E)$ the set of all bounded functions on $E$ with the supremum norm topology.  We will consider $M(V_{\delta})$, where $V_{\delta} = \{\psi \in E' \, \mid \, \|\psi\| \leq \delta\}$.

Claim: $(x_{n})_{n \in \mathbb{N}}$ is relatively compact in $E$ if and only if, for some $\delta > 0$, $(i x_{n})_{n \in \mathbb{N}}$ is relatively compact in $M(V_{\delta})$.

The claim follows from the following equivalence: given any subsequence $(n_{j})_{j \in \mathbb{N}}$ and any $\delta > 0$, $(x_{n_{j}})_{j \in \mathbb{N}}$ is a Cauchy sequence in $E$ if and only if $(ix_{n_{j}})_{j \in \mathbb{N}}$ is a Cauchy sequence in $M(V_{\delta})$.

Indeed, notice that, given $\delta > 0$, we have
\begin{align*}
\sup \left\{ |ix_{n_{j}}(\psi) - ix_{n_{k}}(\psi)| \, \mid \, \psi \in V_{\delta} \right\} &= \sup \left\{ |\psi(x_{n_{j}}) - \psi(x_{n_{k}})| \, \mid \, \psi \in V_{\delta} \right\} \\
&= \delta \|x_{n_{j}} - x_{n_{k}}\|.
\end{align*}

Thus, by the definition of Cauchy sequence, $(x_{n_{j}})_{j \in \mathbb{N}}$ is Cauchy for the norm metric in $E$ if and only if it is Cauchy for the supremum norm metric in $M(V_{\delta})$.


问:
Several graph theory proofs

Let $G=(V,E)$ be a graph with $|V|=n$. Prove that $G$ is connected if $d(v)\geq \frac{n-1}{2}$ for all $v\in V$.

Let $G=(V,E)$ be a graph with $|V|=n$ and $|E|=m$. Prove that $\min\limits_{u\in V} d(u) \leq 2\dfrac{m}{n}\leq \max\limits_{v\in V}d(v)$.

My attempts:

$|E|\leq n(n-1)$ since every node can have at most $n-1$ edges, and there are $n$        nodes in total. Hence making use of the formula $\sum\limits_{v\in V}d(v)=2|E|$, we obtain $\sum\limits_{v\in V}d(v)\leq 2n(n-1)$. I don't know how to proceed from here.

All I can come up with is that $\max\limits_{v\in V}d(v)=n-1$. I don't think this helps me though.

Could anyone please provide any additional hints to these problems? Thank you in advance

答:
Actually, $|E|\leq n(n-1)/2$; you are counting each edge twice, when it goes from a node, and when it goes to a node. Take two arbitrary nodes $x,y$. If they are not connected, then there is not an edge between them, so all the outgoing edges from $x$ must go to at least $(n-1)/2$ nodes other than $x$. Similarly all the outgoing edges from $y$ must go to at least $(n-1)/2$ nodes, all distinct from both $x$ and the neighbors of $x$, or otherwise $G$ would be connected. But wait, how many nodes have we used already...?

Use the formula $\sum_v d(v)=2m$. Suppose $\min_v d(v)\leq 2m/n$ were false, then $d(v)>2m/n$ for all $v$. Insert into the formula. Do the same thing for the inequality $2m/n\leq \max_v d(v)$.

Title: Several graph theory proofs


问:
Simple probability problem with negation

A survey in a high school conducted that from 417 students 186 is doing sports, 136 music and 74 music and sports. What is the probability that randomly chosen student doesn't play sports and doesn't play music?

Isn't this just $\frac{21}{417} \cdot 100 = 5.03597122$ since the number of people that doesn't do anything is $417-(186+136+74)=21?$ I keep questioning my self right now...

答:
It's easiest in these sorts of problems to separate the students into disjoint groups. Let $w$ be the number of students who do neither music nor sports, $x$ the number of students who do music only, $y$ the number of students who do sports only, and $z$ the number who do both music and sports. These four groups are mutually exclusive, so we have

$$ w + x + y + z = 417 \Rightarrow w = 417 - (x + y + z) . $$

We also know from the given information that

$$ x + z = 136 , $$

$$ y + z = 186 , $$

$$ z = 74 . $$

Combining these three equations, we have

$$ (x + z) + (y + z) - z = 136 + 186 - 74 $$
$$ \Rightarrow x + y + z = 136 + 186 - 74 = 248 $$

Then, $w = 417 - (x + y + z) = 417 - 248 = 169 $.

As a percentage, we have

$$\frac{169}{417} \approx 40.5\% .$$


问:
Find two matrices A,B$\in {{M}_{n\times n}}$ with the propertie that there not exists such a continuous function $f:[0,1]\to M$ with $f(0)=A,f(1)=B$
If $M_{n\times n}$ is the set of invertible matrices with real entries. Find two matrices $A,B\in M_{n \times n}$ with the propriety that there not exists  such a continuous function

$$f:[0,1]\to M, \quad f(0)=A, f(1)=B $$

the only way i was thinking was is the  inverse function such as $f^{-1}(A)=0, \quad f^{-1}(B)=1,$
but this doesnt seem to get me anywhere.

答: If $\det(A)>0>\det(B)$, then there is no such function, because otherwise the range of the map $\det\circ f$ would contain $\det(A)$ and $\det(B)$, but not $0$.


问:
Fixed Point Derivative Question
Let $f,g: [0,1] \to \mathbb{R}$, $x_0$ a fixed point in $(0,1)$. Assume $f(x_0) = g(x_0) = 0$, $f$ and $g$ are differentiable at $x_0$ and $g'(x_0) \neq 0$.

Prove: $$ \lim_{t \to x_0} \frac{f(t)}{g(t)} = \frac{f'(x_0)}{g'(x_0)}$$

Hints preferred rather than being given the whole solution. Thanks!

答:
Remember that a derivative is itself a limit

$$
\frac{f'(x_0)}{g'(x_0)} =  \frac{ \lim_{t \to x_0}\frac{{f(t) - f(x_0)}}{t-x_0}}{ \lim_{t \to x_0}\frac{g(t)-g(x_0)}{t-x_0}} =\lim_{t \to x_0}
\frac{ \frac{{f(t) - f(x_0)}}{t-x_0}}{\frac{g(t)-g(x_0)}{t-x_0}} =
 \lim_{t \to x_0} \frac{f(t)-f(x_0)}{g(t)-g(x_0)} = \cdots
$$
Do you see how to proceed? Make sure to justify why each step is valid.

Title: Fixed Point Derivative Question


问:
Is it possible to calculate how much percent the function $\cos ax-\cos 2ax$ is positive on its periodic domain?
We know that $\cos ax$ on its period $0<x<\frac{2\pi}{a}$ is $50$ % positive and $50$ % negative.

My question is that are we able to conclude such a result for the following function?
$$\cos ax-\cos 2ax$$
I mean can we say something about how much percent this function is positive on its periodic domain? According to the plots, I see that this percent does not depend on the parameter $a$; is that correct?

Here, $a$ and $x$ are both reals and positive.

答: Since the function is periodic we need only find the proportion of the curve which is positive over the first period. To do this, we can integrate between $0$ and the first root, of the function. This will give the positive area. Do the same to find the area of the curve which is negative, and then combine the two to find the percentage you require.


问:
Book recommendation for Introductory Differential Geometry, with lots of examples (calculations)
Could you give a recommendation for differential geometry or curves and surfaces with lots of examples and calculations?

I need lots of examples to understand a topic therefore it would great if there are some book with good calculations.

答:
The ones I used and liked were:

Differential Geometry of Curves and Surfaces - Manfredo Do Carmo

Modern Differential Geometry of Curves and Surfaces with Mathematica - Alfred Gray


问:
differential equation of a population growth and change
I want to formulate a system of equations and initial conditions of the following data:
Each year the population1 grows by 4% and population2 by 2%.
Also each year 3% of population1 leaves it and go to population2
and 1% vice versa, leaves population2 and goes to population1.

The initial populations are: population1 = 20000000, population2 = 5000000.

I thought of the next formalism of this:
$$x(t+1)=1.04*x(t)-0.03*x(t)+0.01*y(t)$$
$$y(t+1)=1.02*y(t)-0.01*y(t)+0.03*x(t)$$

where $x(t)$ is the size of population1 and $y(t)$ is the size of population2.
Is this is correct in those term or maybe I should reformulate it in another way ?

I think that I need some how to get something like this (ignore the concrete number and function):
$$\overline{X'}=\begin{pmatrix}1 & 1\\-2 & -2\end{pmatrix}\overline{X}+\begin{pmatrix}e^{-t}\\-2e^{-t}\end{pmatrix}$$

with an initial conditions like that:
$$\overline{X}(0)=\begin{pmatrix}-1\\0 \end{pmatrix}$$

but I don't know how to do it, so will be grateful for some help.

An idea I had is:
$$x'\left(t\right)=\frac{x\left(t+1\right)−x\left(t\right)}{\left(t+1−t\right)}=\left(1.04−0.03−1\right)x\left(t\right)+0.01y\left(t\right)$$

is it correct ?

答:
For population  denoted by $x$, you get $$x'=.04x-.03x+.01y$$ and for the second population denoted by $y$ you get $$ y'=.02y+.03x-.01y$$

That simplifies to $$x'=.01x+.01y$$ and $$y'=.03x+.01y$$

Now you write it in matrix form.


问:
Does the integral of a nonconstant monomial over the unit disk vanish?
I've been thinking recently about integrals over fixed domains and what they do to monomials.

Let $[\varphi]$ be an Iverson bracket, sending true to 1 and false to 0. Let $a \equiv_{(n)} b$ mean that $a$ is equivalent to $b$ mod $n$.

The unit interval has a nice formula.

$$ \int_0^1 x^n \text{d}x = \frac{1}{n+1} $$

The interval $[-1, 1]$ also has a nice formula, in particular the odd terms drop out, leaving only the even terms.

$$ \int_{-1}^1 x^n \text{d}x = [n \equiv_{(2)} 1] \frac{2}{n+1} $$

However, the interval $[-1,1]$ is kind of a cross section of the complex unit disk. That got me thinking about what non-constant monomials look like on the unit disk ... and I think the graph of $\text{Re}(z^n)$ can always be rotated a little bit to look like the graph of $\text{Re}(-z^n)$ for positive $n$, but I'm not sure.

I think we can make this argument rigorous by "pairing up" every $z$ with $u_{2n}z$ with $u_{2n}$ being the root of unity with least positive $\theta$ in the polar representation. $(u_{2n})^nz^n$ would then equal $-1z^n$.

I'm not sure exactly which complex integral to use, and contour integrals seem to apply to the wrong kind of object, so I'm expressing it as two independent real integrals over the real and pure imaginary axes.

$$ \int_{-1}^{1} \int_{-1}^{1} [a^2+b^2 < 1] (a+bi)^n \;\text{d}a\text{d}b = 0 \;\; \text{when $n \ge 1$} $$

Is this true?

答:
$$
\iint_D (a+bi)^n dadb
 = \int_0^{2\pi} \int_0^1 \left(re^{it}\right)^n rdrdt
 = \int_0^{2\pi} e^{itn}dt \int_0^1 r^{n+1} dr
$$
and the left integral is zero since $e^{2\pi i} = 1$.

This also would follow from the Cauchy Theorem since $z^n$ is analytic in the disk.


问:
How to evaluate $\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx$
I want to evaluate: $$\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx,$$ but I don't see how can I achieve so.

My attempts so far have been rewriting the integral using algebraic identities such as: $$ab=\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{2}\left(a-b\right)^2=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2,$$ that yield other integrals like:
$$\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(1-x\right)}{1+x}\:dx,\:\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)}{1+x}\:dx.$$Normally the beta function and expanding terms into series is used to deal with these kind of integrals yet neither can be used because of the upper bound.

What else can be done in order to compute the main integral? Thanks.

答:
An idea by Cornel Ioan Valean

Exploit the result
$$f(a)=\int_0^{1/2} \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x$$
$$\small =\frac{1}{2 a}\log ^3(2)-\frac{3 }{4 a}\zeta (3)-\frac{\log ^2(2) }{2 a}\log \left(1-\frac{a}{2}\right)-\frac{\log ^2(2) }{2 a}\log \left(\frac{a-2}{a-1}\right)+\frac{\log(2)}{a}\text{Li}_2\left(\frac{a}{2}\right)$$
$$\small+\frac{\log (2) }{a}\text{Li}_2\left(\frac{a}{2 (a-1)}\right)+\frac{1}{a}\text{Li}_3\left(\frac{a}{2}\right)+\frac{1}{a}\text{Li}_3\left(\frac{a}{2 (a-1)}\right)-\frac{1}{a}\text{Li}_3\left(\frac{a}{a-1}\right)-\frac{\text{Li}_3(a-1)}{a}.$$

This is a modified form of Lemma 2 in the paper  A special way of extracting the real part of the Trilogarithm,  $\displaystyle  \operatorname{Li}_3\left(\frac{1\pm i}{2}\right)$ easily obtained by the means presented in the paper https://www.researchgate.net/publication/337868999_A_special_way_of_extracting_the_real_part_of_the_Trilogarithm_Li_31i2

A nice fact: maybe good to mention also this integral variant

$$\int_0^{1/2}\frac{\log(1-x)\log(x)\log(1+x)}{x}\textrm{d}x=\int_{-1}^0 f(a) da.$$


答:
As @binary_string100101 says:

$$\int\limits_{x=0}^{2 \pi a} [\cos (a x) - \cos (2 a x)]^+  dx = \frac{3 \sqrt{3}}{2}$$

where $[\cdot]^+$ is non-zero only if the argument is non-negative.

The same holds for the negative portion.

So $50\%$ above, and $50\%$ below.

Title: Is it possible to calculate how much percent the function $\cos ax-\cos 2ax$ is positive on its periodic domain?

Body:
We know that $\cos ax$ on its period $0<x<\frac{2\pi}{a}$ is $50$ % positive and $50$ % negative.

My question is that are we able to conclude such a result for the following function?
$$\cos ax-\cos 2ax$$
I mean can we say something about how much percent this function is positive on its periodic domain? According to the plots, I see that this percent does not depend on the parameter $a$; is that correct?

Here, $a$ and $x$ are both reals and positive.


答:
Barrett O'Neill's Elementary Differential Geometry. A classic, you can't go wrong with this one..

Download link

Title: Book recommendation for Introductory Differential Geometry, with lots of examples (calculations)

Body:
Could you give a recommendation for differential geometry or curves and surfaces with lots of examples and calculations?

I need lots of examples to understand a topic therefore it would great if there are some book with good calculations.


答: Let A be any matrix with positive determinant, B any matrix with negative determinant. If there is such a $f$ , then $det(f(t))$ is a continuous function of $t\in[0,1]$ (it is a polinomial), with $det(f(0))=det(A)>0$ and $det(f(1))=det(B)<0$. Then, by the intermediate value theorem there must be a $t\in[0,1]$ such that $det(f(t))=0$ giving a contradiction.

Title: Find two matrices A,B$\in {{M}_{n\times n}}$ with the propertie that there not exists such a continuous function $f:[0,1]\to M$ with $f(0)=A,f(1)=B$

Body:
If $M_{n\times n}$ is the set of invertible matrices with real entries. Find two matrices $A,B\in M_{n \times n}$ with the propriety that there not exists  such a continuous function

$$f:[0,1]\to M, \quad f(0)=A, f(1)=B $$

the only way i was thinking was is the  inverse function such as $f^{-1}(A)=0, \quad f^{-1}(B)=1,$
but this doesnt seem to get me anywhere.


问:
Calculate the mean of the number of balls taken out of the box
A box contains 10 balls that are numbered $1,\ldots, 10$. We roll a dice and then take out balls one after the other with repetition until the number of the ball that came out is less than or equal to the result of the roll of the cube. Let $N$ be the number of balls taken out of the box. Calculate $E(N)$.

As I understand, the result should be $E(N)=\frac{49}{12}$. But I can't seem to figure it out. Is it possible to explain how did they got this result?

答:
This will not be a complete answer but hopefully it helps you get started.

Say we roll the die and get an $i$ where $i \in \{ 1,\dots,6\}$. And let $N :=$ number of balls drawn to get a ball less than or equal to $i$.

Let's analyze some probabilities of $N$:

$P(N=1):$ for this we would need to draw a ball numbered in the range $1, \ldots i$. Now under the assumption we may choose any ball uniformly at random as these sorts of problems are usually assumed to be we get

$\displaystyle P(N=1) = \frac{i}{10}$ since we have $i$-many balls numbered less than or equal to $i$.

For instance, if $i = 4$ then for $N = 1$ we would need to draw one of the following balls numbered $1,2,3$ or $4$ on the first try. Each has a $\frac{1}{10}$ chance of being drawn and so we sum them up for $\frac{4}{10} = \frac{i}{10}$.

Now let's consider the case $P(N=2)$, this event occurs if we have drawn a number (strictly) greater than $i$ on the first try and then a ball numbered less than or equal to $i$ on the second try. If we have $i$ as our roll then there are $10-i$  many balls (strictly) greater than $i$. Additionally, after we draw this ball there will now be $9$ balls remaining. If we put this together we get:

$\displaystyle P(N=2) = P(\text{ 1st ball is > $i$ and 2nd ball is } \le i)$

We may continue in this fashion.

We will need to consider what happens as $i$ changes which hints that we should condition on the outcome of our roll of the die.


问:
Given that $y = (1+\ln(4x))^{3/2}$ find $y'$ at $x = 4e^{-3}$
I've been trying to attempt this question but I keep getting the wrong answer.

For $y'$, I get

$$y' = \frac{3}{2}(1+\ln(4x))^{1/2}\times\frac{1}{x}$$

which I think can be rewritten as:

$$y' = \frac{3}{2x}(1 + \ln(4x))^{1/2}$$

I then sub the value of $x$ in and I get $3/0.5e^3$ and I think that's $6e^-3$.

But the correct answer is $12e^{-3}$.

Any help or pointing out any mistakes I've made will be super helpful.

Thank you.

Body:
See MathJax to format the question.

Did you evaluate $(1+\ln(4x))^{\frac{1}{2}}$ at $x=\frac{1}{4}{e^3}$?

Since $$\frac{dy}{dx}=\frac{3}{2x}(1+\ln(4x))^{\frac{1}{2}}$$

At $x=\frac{1}{4}{e^3}$ we have

$$\frac{3}{2(\frac{1}{4}{e^3})}(1+\ln(e^{3}))^{\frac{1}{2}}=\frac{3}{0.5\cdot e^{3}}\cdot2=12e^{-3}.$$


问:
Explicit form of recursive formula
So I was asked to give some background on the problem so here it goes. I was working on creating a problem with functions and series and at one point I ended up creating a bunch of these recurrence relation problems. So this one in particular seemed to me it was solvable, and I was pretty sure I saw something very similar in the past, however standard techniques failed to me and I could not even guess the answer to use induction, so that's why I posted the question, and the answers were extremely helpful to be honest, so thanks :).

What is the explicit form of the following recursive sequence:

$a_0 = 0 \\
a_1=2 \\
a_{n+1}=2a_n-a_{n-1}+2(n+1)
$

How can one solve this without guessing the answer, which btw I could not.

答:
Hint

Call $b_n=a_{n}-a_{n-1}$ and get
$$b_{n+1}-b_n=2(n+1)$$

Now use a telescoping sum.

Title: Explicit form of recursive formula


问:
How to distinguish between exclusive "or" and inclusive "or" when the truth value of p and q cannot be simultaneously true
I just started learning Discrete Mathematics using Kenneth H. Rosen's "Discrete Mathematics and Its Application", and I'm having trouble understanding the solution for example 9 in section 1.1.2.

Since the disjunction of p and q (or the inclusive "or") is the proposition that states that either p is true, or q is true, or both p and q are true, if the "or" in the statement is an inclusive "or", then if p and q are both true, the truth value of the statement has to be Truth. Therefore, what the explanation is trying to do (or at least what I think it's trying to do) is to show that when both p and q is true then the truth value for the statement is False which means the "or" must be an exclusive "or" . However, if p and q cannot be simultaneously true, then how can there be a truth value for the statement when p and q are both true? The truth table for the statement, whether with the inclusive "or" or the exclusive "or", will only have 2 cases: p is T and q is F, and p is F and q is T.

I guess what I want to ask is how to distinguish between exclusive "or" and inclusive "or" when the truth value of p and q cannot be simultaneously true?

答:
You have r=I spend all my money on Europe or a car, p=I spend all my money on Europe, and q=I spend all my money on a car. The truth table is

```<code>p q r (p or q) (p exclusive or q)
T F T T        T
F T T T        T
F F F F        F
</code>```

Therefore either of those two translations matches with the values of p, q, and r that we have been given in the problem. So either is valid, because they give the same truthedness.

However say you were given a different problem such as the following (in particular, p and q can both be true, and r has a corresponding truthedness corresponding to it). Coming up with an example in words is another story altogether, so I leave that out of the discussion.

```<code>p q r (p or q) (p exclusive or q)
T T F T        F
T F F T        T
F T T T        T
F F F F        F
</code>```

Then neither of these formulae would work because r has been defined in such a way that neither p or q nor p exclusive or q are an accurate translation.

Conclusion

Either the regular or or exclusive or will provide a correct translation in this case because p and q cannot simultaneously happen. Your task is to determine the truthity of r corresponding to each combination of p and q that can happen, and come up with a formula that matches those truth results.

Body:
You have r=I spend all my money on Europe or a car, p=I spend all my money on Europe, and q=I spend all my money on a car. The truth table is

p q r (p or q) (p exclusive or q)
T F T T        T
F T T T        T
F F F F        F

Therefore either of those two translations matches with the values of p, q, and r that we have been given in the problem. So either is valid, because they give the same truthedness.

However say you were given a different problem such as the following (in particular, p and q can both be true, and r has a corresponding truthedness corresponding to it). Coming up with an example in words is another story altogether, so I leave that out of the discussion.

p q r (p or q) (p exclusive or q)
T T F T        F
T F F T        T
F T T T        T
F F F F        F

Then neither of these formulae would work because r has been defined in such a way that neither p or q nor p exclusive or q are an accurate translation.

Conclusion

Either the regular or or exclusive or will provide a correct translation in this case because p and q cannot simultaneously happen. Your task is to determine the truthity of r corresponding to each combination of p and q that can happen, and come up with a formula that matches those truth results.


答:
The other answers to this question are responding to how much of the area between the curve and the $x$-axis is positive as a proportion of the total area. I read the question as being what proportion of the domain is the function positive over. If that's the case, the solution is below:

You need to know where the function is positive and where it's negative. The easiest way to do that is to find the roots:

$$ \cos(ax) - \cos(2ax) = 0$$

$$ \Rightarrow \cos(ax) - \left( 2\cos^2 (ax) - 1 \right) = 0$$

$$ \Rightarrow 2\cos^2 (ax) - \cos(ax) - 1 = 0$$

Note that this equation is quadratic in $\cos(ax)$, so we can use the quadratic formula:

$$ \cos(ax) = \frac{1 \pm \sqrt{1-4(2)(-1)}}{4} $$

$$ \Rightarrow \cos(ax) = 1, -\frac{1}{2} $$

For the period between $x = 0$ and $x = \frac{2\pi}{a} $, the solutions to these equations are:

$$ x = 0, \frac{2\pi}{3a}, \frac{4\pi}{3a}, \frac{2\pi}{a} $$

Then, the function is nonnegative on the intervals $[0,\frac{2\pi}{3a}]$ and $[\frac{4\pi}{3a}, \frac{2\pi}{a}]$. The ratio of the measure of the union of these intervals to the interval for the period is then

$$ \frac{(\frac{2\pi}{a} - \frac{4\pi}{3a}) + (\frac{2\pi}{3a} - 0)}{\frac{2\pi}{a} - 0} $$

$$ = \frac{\frac{4\pi}{3a}}{\frac{2\pi}{a}} $$

$$ = \frac{2}{3} .$$

So, $\cos(ax) - \cos(2ax)$ is positive for $\frac{2}{3}$ of its interval.


答:
Hint.

The existence of the function $f$ means that there exists a path between the two matrices $A$ and $B$, which implies that $A$ and $B$ are in the same connected component of $M:=M_{n\times n}$.

On the contrary, the continuous map $\det:M\to\mathbb{R}$ tells you that $M$ is not connected (since $\det(M)$ is not).

Now, find two elements of $M$ that are not in the same component of $M$.


问:
Catalan numbers using the recurrence.
A gambler tosses a coin, winning 1\$ on heads and losing 1\$ on tails. We know they got $t$ tails and $k+t$ heads, thus winning $k$\$. The total ways in which this could have happened is ${2t+k \choose t}$ and the number of ways such that they reached $k$\$ for the first time on the last toss they made (call this "the Catalan constraint") is given by the Catalan numbers: ${2t+k-1 \choose t}-{2t+k-1 \choose t-1}$. See here: Probability that random walk will reach state $k$ for the first time on step $n$.

Now, let $n(k,t)$ be the number of ways in which they get to $k$ for the first time on toss $2t+k$ (which we know from other means is the generalized Catalan numbers). Then, conditioning on the first toss, we get the recurrence:

$$n(k,t)=n(k-1,t)+n(k+1,t-1) \tag{1}$$

Both the total number of paths (total ways of getting to $k$\$ in $2t+k$ tosses), as well as the paths with the Catalan constraint (ways of getting to $k$\$ for the first time after $2t+k$ tosses) satisfy the recurrence above. The only difference is in the initial conditions. For both series, we have:

$$n(0,0)=1$$

For paths with the Catalan constraint;
$$n(0,t)=0 \;\;\forall \;\; t>0$$

While for un-constrained paths;

$$n(0,t)={2t \choose t}$$

There must be a way to go from the recurrence in equation (1) to the solutions for the counts of the two kinds of paths (depending on which starting conditions we substitute), using generating functions or other techniques. Is there a way to do this?

Note that @robjohn's answer in the question linked above does do this, but he uses the interpretation that these are paths on the grid. I was hoping for a solution that doesn't depend on this intuition and only equation (1) with the starting conditions.

Specifying the initial conditions explicitly. In addition to the recurrence given by equation (1) which both series satisfy, we have the following initial conditions:

Case-1:
Paths that satisfy the Catalan constraint.
$$n(0,0)=1$$
$$n(0,t)=0 \forall t>0$$

We know the solution of the recurrence in equation (1) with these initial conditions is:

$$n(k,t) = {2t+k-1 \choose t}-{2t+k-1 \choose t-1}$$

Case-2:
All paths (no constraint).
$$n(0,0)=1$$
$$n(0,t)={2t \choose t} \forall t>0$$

We know the solution of the recurrence in equation (1) with these initial conditions is:

$$n(k,t) = {2t+k \choose t}$$

答:
We are trying to solve the functional equation $$f(n, k) = f(n-1, k) + f(n+1, k-1)$$ with different boundary conditions.

Case $1$

$$f(0, k) = \begin{cases}
1 & k = 0\\
0 &\text{otherwise}
\end{cases}$$

Case $2$

$$f(0, k) = \begin{cases}
1 & k = 0\\
\binom{2k}{k} &\text{otherwise}
\end{cases}$$

Define the generating function $$F_n(x) = \sum_{k=0}^\infty f(n, k) x^k$$

Multiply the original equation by $x^k$ and sum over $k \geq 1$ to obtain

$$\sum_{k \geq 1} f(n, k) x^k = \sum_{k \geq 1} f(n-1, k) x^k + \sum_{k \geq 1} f(n+1, k-1) x^k$$

Or, $$F_n(x) - 1 = F_{n-1}(x) - 1 + x F_{n+1}(x)$$

Or, $$x F_{n+1}(x) - F_n(x) + F_{n-1}(x) = 0$$

The characteristic equation of the above recurrence relation for the generating function $F_n(x)$ is given by
$$xm^2 - m + 1 = 0 \implies m = \frac{1 \mp \sqrt{1-4x}}{2x}$$

Hence the solution is $$F_n(x) = c_1 m_1^n + c_2 m_2^n$$ where
$m_1 = \frac{1 - \sqrt{1-4x}}{2x} = \frac{2}{1+ \sqrt{1-4x}}$
and
$m_2 = \frac{1 + \sqrt{1-4x}}{2x} = \frac{2}{1- \sqrt{1-4x}}$

Note that $m_2$ is not defined for $x = 0$

We assume that $F_n(x)$ is defined for $x = 0$ and hence we set $c_2 = 0$

Finally $$F_n(x) = c_1 \left(\frac{2}{1+ \sqrt{1-4x}}\right)^n$$

Since the coefficient of $x^k$ in $F_n(x)$ gives the value of $f(n, k)$, we need to calculate that value in order to find $f(n, k)$

Case $1$: In this case, $$F_n(0) = 1 \implies c_1 = 1 \implies F_n(x) = \left(\frac{2}{1+ \sqrt{1-4x}}\right)^n$$

The coefficient of $x^k$ in $F_n(x)$ or $$f(n, k) = \binom{n+2k-1}{k} - \binom{n+2k-1}{k-1}$$

Case $2$: In this case, $$F_n(0) = \sum_{k = 0}^\infty \binom{2k}{k} x^k = \frac{1}{\sqrt{1-4x}}$$

$$\implies c_1 = \frac{1}{\sqrt{1-4x}} \implies F_n(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{2}{1+ \sqrt{1-4x}}\right)^n$$

The coefficient of $x^k$ in $F_n(x)$ or $$f(n, k) = \binom{n+2k}{k}$$

Title: Catalan numbers using the recurrence.


答:
Arnaldo has suggested a very nice elementary approach. It can also be done using generating functions. The recurrence

$$a_n=2a_{n-1}-a_{n-2}+2n$$

is valid for all $n\in\Bbb Z$ if we assume that $a_n=0$ for $n<0$. The generating function for the sequence $\langle a_n:n\ge 0\rangle$ is $g(x)=\sum_na_nx^n$, so multiply through by $x^n$ and sum over $n$:

$$\begin{align*}
g(x)&=2\sum_na_{n-1}x^n-\sum_na_{n-2}x^n+2\sum_nnx^n\\
&=2x\sum_na_{n-1}x^{n-1}-x^2\sum_na_{n-2}x^{n-2}+2\sum_nnx^n\\
&=2xg(x)-x^2g(x)+2x\sum_nnx^{n-1}\\
&=\left(2x-x^2\right)g(x)+2x\left(\frac1{1-x}\right)'\\
&=\left(2x-x^2\right)g(x)+\frac{2x}{(1-x)^2}\,,
\end{align*}$$

so

$$g(x)=\frac{2x}{(1-x)^2(1-2x+x^2)}=\frac{2x}{(1-x)^4}\,.$$

Now it helps to know some basic power series, in this case that

$$\frac1{(1-x)^4}=\sum_n\binom{n+3}3x^n\,,$$

so that

$$\begin{align*}
\frac{2x}{(1-x)^4}&=2x\sum_n\binom{n+3}3x^n\\
&=2\sum_n\binom{n+3}3x^{n+1}\\
&=\sum_n2\binom{n+2}3x^n\,,
\end{align*}$$

and

$$a_n=2\binom{n+2}3\,.$$

Title: Explicit form of recursive formula

Body:
So I was asked to give some background on the problem so here it goes. I was working on creating a problem with functions and series and at one point I ended up creating a bunch of these recurrence relation problems. So this one in particular seemed to me it was solvable, and I was pretty sure I saw something very similar in the past, however standard techniques failed to me and I could not even guess the answer to use induction, so that's why I posted the question, and the answers were extremely helpful to be honest, so thanks :).

What is the explicit form of the following recursive sequence:

$a_0 = 0 \\
a_1=2 \\
a_{n+1}=2a_n-a_{n-1}+2(n+1)
$

How can one solve this without guessing the answer, which btw I could not.


答:
For the geometry of surfaces I recommend the book Differential Geometry of Curves and Surfaces by Manfredo Do Carmo,  which is clear, self-contained and with a lot of examples and exercises.

If you want to go further and learn modern differential geometry (manifolds etc) an excellent book is An introduction to manifolds by Loring Tu, which contains a lot of examples and explicit calculations. Moreover it is self-contained, and has a lot of appendices in the case you have to strengthen you background.

Another very good book about manifolds is Introduction to smooth manifolds by Lee. You can read some of its chapters as a complement of Tu’s book, indeed it contains some important topics that you can’t find in the book of Loring Tu.


问:
Derivation of the Probability Mass Function of Bernoulli distribution
The Probability Mass Function of Bernoulli distribution looks like this:

\begin{equation}
p : \mathcal{X} \to [0,1], x\mapsto p(x) := \mu^{x}(1 - \mu)^{1-x} \mbox{ for } \mu \in [0,1].
\end{equation}

I am interested in it's derivation. How did someone determine it to have exactly this form, are there any mathematical steps, which lead to this function?

Body:
The expression $\mu^x(1-\mu)^{1-x}$ is equal to $\mu$ when $x = 1$ and $1-\mu$ when $x = 0$. It's just a more compact way of writing $P[x=1] = \mu$ and $P[x=0] = 1-\mu$:

$$ P[x=0] = \mu^0(1-\mu)^{1-0} = 1-\mu , $$

$$ P[x=1] = \mu^1(1-\mu)^{1-1} = \mu . $$


答:
Your observation is correct: if we know $P$ and $Q$ can't both be true, then the expressions $P\lor Q$ and $P\oplus Q$ are equivalent (in that they have the same truth value). We could write this as the following theorem:

$$\lnot(P\land Q)\implies(P\lor Q\iff P\oplus Q).$$

Your question "if p and q cannot be simultaneously true, then how can there be a truth value for the statement when p and q are both true?" is a little confusing. The truth functions corresponding to the logical operators are defined for all inputs even if some of the function values are irrelevant in a given situation.

Edit: I looked at the exercise you linked to in the comments, and the answer given in the book is questionable. They ask you to "translate" a sentence of the form "P or Q", and expect you to get $P\oplus Q$ by noticing and incorporating additional information (that P and Q can't both be true). It makes sense that one would want to do this, because saying $P\oplus Q$ is "stronger" or "more informative" than saying $P\lor Q$, but I'd argue that this is more than just a translation from English into logical symbols. If the exercise writer wanted $P\lor Q$ to be an incorrect answer, they should've said "write a logical expression that exactly describes the scenarios that are possible" or something, not just "translate".

Title: How to distinguish between exclusive "or" and inclusive "or" when the truth value of p and q cannot be simultaneously true

Body:
I just started learning Discrete Mathematics using Kenneth H. Rosen's "Discrete Mathematics and Its Application", and I'm having trouble understanding the solution for example 9 in section 1.1.2.

Since the disjunction of p and q (or the inclusive "or") is the proposition that states that either p is true, or q is true, or both p and q are true, if the "or" in the statement is an inclusive "or", then if p and q are both true, the truth value of the statement has to be Truth. Therefore, what the explanation is trying to do (or at least what I think it's trying to do) is to show that when both p and q is true then the truth value for the statement is False which means the "or" must be an exclusive "or" . However, if p and q cannot be simultaneously true, then how can there be a truth value for the statement when p and q are both true? The truth table for the statement, whether with the inclusive "or" or the exclusive "or", will only have 2 cases: p is T and q is F, and p is F and q is T.

I guess what I want to ask is how to distinguish between exclusive "or" and inclusive "or" when the truth value of p and q cannot be simultaneously true?


问:
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas
I have this identity:

$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$

If I write this like as:

$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$

I can use Werner and prostaferesis formulas and I find the identity $(1)$.

But if we suppose of not to use these formulas I have done a try writing:

$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$

$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$

$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$

Then I have abandoned because I should do long computations and I think that is not the right way.

Is there any trick without to use the prostapheresis or Werner's formulas?

答:
I think that we can establish the identity without too much work; it certainly isn't short though. Your statement is equivalent to $$\cos^2 4x-\cos^2 2x=-\sin 6x\sin2x$$We'll start with left hand side; my strategy will be to make everything in terms of $\cos 2x$. I will make use of the following identities:
$$\begin{align}
\sin^2\theta&\equiv1-\cos^2\theta\\
\cos2\theta&\equiv2\cos^2\theta-1\\
\sin3\theta&\equiv3\sin\theta-4\sin^3\theta
\end{align}$$
Here we go:
$$\begin{align}
\cos^2 4x-\cos^2 2x&\equiv(2\cos^2 2x-1)^2-\cos^22x\\
&\equiv4\cos^42x-5\cos^22x+1
\end{align}$$
Now for the right hand side:
$$\begin{align}
-\sin6x\sin2x&\equiv-\sin2x(3\sin2x-4\sin^3 2x)\\
&\equiv4\sin^42x-3\sin^2x\\
&\equiv4(1-\cos^22x)^2-3(1-\cos^22x)\\
&\equiv4(\cos^4 2x-2\cos^2x+1)-3+3\cos^22x\\
&\equiv4\cos^42x-5\cos^22x+1
\end{align}$$
Hence we have $$\cos^2 4x-\cos^2 2x\equiv-\sin 6x\sin2x$$ as required.

I personally would also be very interested in any shortcut or trick that could be used here instead; if you find one please let me know.


问:
Find an expression to $ \operatorname{Res}(g,\frac{z_0}{a})$ with $\operatorname{Res}(f,z_0)$

Given $z_0$ singular point of $f(z)$ and let $a>0$ real number, setting $g(z)=f(az)$ find an expression to $\operatorname{Res}(g,\frac{z_0}{a})$ with $\operatorname{Res}(f,z_0)$

My attempt

$Res(g,\frac{z_0}{a})=\frac{1}{2i\pi}\int_{|\xi-\frac{z_0}{a}|=\epsilon}g(\xi)d\xi=\frac{1}{2i\pi}\int_{|\xi-\frac{z_0}{a}|=\epsilon}f(a\xi)d\xi=\frac{1}{2i\pi}\int_{|\xi-z_0|=\epsilon}\frac{1}{a}f(\xi)d\xi=\frac{1}{a}Res(f,z_0)$

Not sure if this is the right answer

答:
Expanding on Desperado's comment, assume that the function $f(z)$ has a Laurent expansion at $z=z_{0}$.

Then   $$ f(z) = \sum_{n=-\infty}^{\infty} b_{n} (z-z_{0})^{n}.$$

Replacing $z$ with $az$, $a >0$, we get $$g(z) = f(az) = \sum_{n=-\infty}^{\infty} b_{n} (az-z_{0})^{n}  = \sum_{n = -\infty}^{\infty} a^{n} b_{n} \left(z-\frac{z_{0}}{a} \right)^{n}.$$  Therefore, $$\operatorname{Res}\left[ g(z), \frac{z_{0}}{a}\right] = a^{-1}b_{-1} = \frac{1}{a} \, \operatorname{Res} \left[f(z), z_{0} \right].$$


问:
Definable vs Interpretable Order
This is probably a trivial question but I'll ask it nonetheless.

What is the difference between an order being definable, and an order being interpretable? All texts I've read make a distinction between the two, but they are advanced enough to not provide definitions.

I know that a linear/partial order $<$ on $M^n$ is definable if there exists $\phi(\bar x,\bar y)$ such that $M \models \phi(\bar a,\bar b)$ if and only if $\bar a< \bar b$. What would it mean for such an order to be interpretable?

答:
Briefly, "interpretable" means "definable on a quotient of a definable set by a definable equivalence relation" or, equivalently, "definable in $M^{eq}$". In the case of orders, an interpretable partial order is the same thing as a definable pre-order.

Recall the following four axioms for a binary  relation $\leq$ on $X$:

(Reflexivity) For all $x\in X$, $x\leq x$.

(Transitivity)  For all $x,y,z\in X$, if $x\leq y$ and $y\leq z$, then $x\leq z$.

(Anti-symmetry) For all $x,y\in X$,  if $x\leq  y$ and $y\leq x$, then $x = y$.

(Linearity) For all $x,y\in X$, $x\leq y$ or $y\leq x$.

The following types of order are defined by combinations of these axioms:

Pre-order: 1 and 2

Partial order: 1, 2, and 3

Linear pre-order: 1, 2, and 4

Linear order: 1, 2, 3, and 4

For any pre-order $\leq$ on $X$, there is an equivalence relation $\sim$ defined by $x\sim y$ if and only if $x\leq y$ and $y\leq x$. Then the relation $\preceq$ defined by  $[x]_\sim \preceq [y]_\sim$ if and only if $x\leq y$ is a well-defined partial order on $X/{\sim}$. If $\leq$ is a linear pre-order, then $\preceq$ is a linear order.

Now if the pre-order $\leq$ is definable (on the definable set $X$), then $\sim$ is a definable equivalence relation on $X$, and the induced partial order $\preceq$ on $X/{\sim}$ is interpretable.

Conversely, if $\preceq$ is an intepretable partial order (on the quotient $X/E$ of a definable set $X$ by a definable equivalence relation $E$), then we can "pull back" $\preceq$ to a definable binary relation on $X$ by $x\leq y$ if and only if $[x]_E \preceq [y]_E$. And this relation $\leq$ is a definable pre-order on $X$.

Regarding the paper by Pierre Simon that you link to: The statement that SOP is equivalent to "admits a definable partial order with infinite chains" is not quite right. Either "partial order" should be changed to "preorder", or "definable" should be changed to "interpretable".

So the correct statement is that an NIP unstable theory always has an interpretable partial order with infinite chains, and the goal of the paper is to improve this to an interpretable infinite linear order. The main result gives something weaker than interpretable: namely definable on a quotient by a $\bigvee$-definable (but not definable) equivalence relation. In the $\aleph_0$-categorical case, we do actually get an interpretable infinite linear order. But the quotient is essential: We can pull back this interpretable linear order to a definable linear preorder, but not a definable linear order.

Title: Definable vs Interpretable Order


问:
Subrings and homomorphisms of unitary rings
Let $(R,+,\cdot)$ be a ring (in the definition i use multiplication is associative operation and it's not assumed that there is unity in the ring).

I've seen two definitons of subring.

1) non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring

2) Let $(R,+,\cdot)$ be unitary ring with unity $e$. Non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring and $e \in S$.

I'm looking for an example of such ring R and its subset S that $(S,+,\cdot)$ is a ring but not a unitary ring.

I will be also very grateful for an example of such UNITARY rings $(R_1,+_1,\cdot_1)$, $(R_2,+_2,\cdot_2)$ and function $f: ~~ R_1 \longrightarrow R_2$ that

$(1) \forall a,b \in R_1 ~~~ f(a+_1 b) = f(a) +_2 f(b) $

$(2) \forall a,b \in R_1 ~~~ f(a\cdot_1 b) = f(a) \cdot_2 f(b)$

$(3) f(e_1) \neq e_2$

where $e_1$ is unity in $R_1$; $e_2$ - in $R_2$.

Thanks in advance.

答:
$R = \mathbb Z$ and $S = 2\mathbb Z$

$R_1 = R_2 = \mathbb Z$ and $f(a) = 0$.

More generally let $R$ be a unitary ring. Any proper ideal $S$ provides an example for 1. For arbitrary unitary rings $R_1$ and $R_2\neq\{0\}$, the zero map is always an example for 2.


答:
There's a bit of ambiguity in the way things are stated, but it seems you have something like this in mind.

For a year you're modeling the
overall change in the first population, $x(t + 1) - x(t)$, as being the result of:

Some sort of inherent growth, adding $0.04 \times x(t)$.

A loss to population 2, removing $0.03\times x(t)$.

New additions coming from population 2, adding $0.01 \times y(t)$.

That gives you
\begin{equation}
x(t + 1) - x(t) = (0.04 - 0.03)\,x(t) + 0.01\,y(t).
\end{equation}

Same sort of thing for population 2:
\begin{equation}
y(t + 1) - y(t) = 0.03\,x(t) + (0.02 - 0.01 )\,y(t).
\end{equation}
Maybe easier to write
\begin{align}
x(t + \Delta t) - x(t)) &= (A\Delta t)x(t) + (B\Delta t)y(t) \\
y(t + \Delta t) - y(t)) &= (C\Delta t)x(t) + (D\Delta t)y(t)
\end{align}
where $A$, $B$, $C$, $D$ are rates since it looks like you might have an eye on shifting to a continuous model ($\Delta t \rightarrow 0$). In that case you get the system
\begin{align}
x'(t) &= A x(t) + B y(t) \\
y'(t) &= C x(t) + D y(t),
\end{align}
and typically you'd have initial conditions.

But, if that's not what you're after, the original problem would end up looking like
\begin{equation}
\begin{bmatrix}
x(t + 1)\\
y(t + 1)
\end{bmatrix}
=
\begin{bmatrix}
0.01 & 0.01 \\
0.03 & 0.01
\end{bmatrix}
\begin{bmatrix}
x(t) \\
y(t)
\end{bmatrix},
\end{equation}
and if you're working from initial conditions you end up with
\begin{equation}
\begin{bmatrix}
x(n)\\
y(n)
\end{bmatrix}
=
\begin{bmatrix}
0.01 & 0.01 \\
0.03 & 0.01
\end{bmatrix}^n
\begin{bmatrix}
x(0) \\
y(0)
\end{bmatrix}.
\end{equation}

Title: differential equation of a population growth and change

Body:
I want to formulate a system of equations and initial conditions of the following data:
Each year the population1 grows by 4% and population2 by 2%.
Also each year 3% of population1 leaves it and go to population2
and 1% vice versa, leaves population2 and goes to population1.

The initial populations are: population1 = 20000000, population2 = 5000000.

I thought of the next formalism of this:
$$x(t+1)=1.04*x(t)-0.03*x(t)+0.01*y(t)$$
$$y(t+1)=1.02*y(t)-0.01*y(t)+0.03*x(t)$$

where $x(t)$ is the size of population1 and $y(t)$ is the size of population2.
Is this is correct in those term or maybe I should reformulate it in another way ?

I think that I need some how to get something like this (ignore the concrete number and function):
$$\overline{X'}=\begin{pmatrix}1 & 1\\-2 & -2\end{pmatrix}\overline{X}+\begin{pmatrix}e^{-t}\\-2e^{-t}\end{pmatrix}$$

with an initial conditions like that:
$$\overline{X}(0)=\begin{pmatrix}-1\\0 \end{pmatrix}$$

but I don't know how to do it, so will be grateful for some help.

An idea I had is:
$$x'\left(t\right)=\frac{x\left(t+1\right)−x\left(t\right)}{\left(t+1−t\right)}=\left(1.04−0.03−1\right)x\left(t\right)+0.01y\left(t\right)$$

is it correct ?


问:
Any good upper bounds for $\sum_{i = 0}^{m} 2^i {k+i \choose i}$?
I am trying to solve a recursion (I faced it when I wanted to analyze the running time of an algorithm precisely). I reduced it to compute a "good" upper bound for $\sum_{i = 0}^{m}2^i {k + i \choose i}$. We have these assumptions that $1 < m \leq Ck$, for some $C\in \mathbb{N}$.
I have tried many approximations but the bounds are not what I want. They are very loose. On the other hand, it doesn't look like a very sophisticated summation but I have so much difficulty solving it!
Does anyone have any ideas?

答:
Define $\{a_m\}$ by
$$a_m = \sum_{i=0}^m 2^i \binom{k+i}{i}$$
Then we can show
$$a_m = \frac{2^{m+1} m^k}{k!} \left( 1 + O \left(\frac{1}{m} \right) \right)$$
by application of Note IV.26, page 256, in Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick.

If you should happen to consult the book, be aware that Note IV.26 contains a typographical error: the factor $\alpha_1^{-n+r}$ should be $\alpha_1^{-n-r}$.

Note IV.26, corrected and slightly paraphrased, reads:

Let $f(z)$ be a rational generating function with a single dominant
pole at $z=\alpha_1$ with multiplicity $r$.  Then  $$f_n =
 \frac{C}{(r-1)!} \alpha_1^{-n-r} n^{r-1} \left( 1 + O \left(
 \frac{1}{n} \right) \right)$$ with $$C = \lim_{z \to \alpha_1}
 (z-\alpha_1)^r f(z)$$
[end of Note]

A "dominant pole" is a pole nearest the origin.

The generating function for $a_m$ is
$$f(z) = \frac{1}{(1-z)(1-2z)^{k+1}}$$
which has a dominant pole at $z=1/2$ of multiplicity $r=k+1$.
Applying Note IV.26 we find $C= (1/2)^k$, yielding the asymptotic relation stated above.