答:
Hint: with $g(a,b,c,d,e) = a + b + c + d + e$ and
$$f(a,b,c,d,e) = \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}$$
we can consider optimizing $f$ over the set of $(a,b,c,d,e)$ for which $g(a,b,c,d,e) = 1$. The method of lagrange multipliers gives a necessary condition for the existence of a critical point on this set: $\nabla f = \lambda \nabla g$ for some $\lambda \in \mathbb{R}$.

However we know that $\nabla g$ is the column vector $(1,1,1,1,1)$, which means that
$$
\frac{1}{(1-x)^2} = \lambda
$$
for each $x = a,b,c,d,e$ individually. Notice that this condition is the same for each coordinate.

What can you conclude about the solutions to the lagrange multiplier problem, and how can you connect this information to the inequality you wish to prove?

Body:
I tried to solve this inequality:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$
with
$$a+b+c+d+e=1$$
I am stuck at this. I don't want the full solution, a hint would be enough.</p>


问:
Laplace Transformation of Differential Equation
I have a DE

$$y' + 2y = u' - u$$ and it's transfer function was given as
$$G(s) = \dfrac{(s-1)}{(s+2)}$$
Can someone elaborate on the steps in between the conversion?

(There were no initial condition $y(0)$ and $u(0)$, hence I can't seem to figure how it was transformed)

答: With the assumption $y(0)=u(0)$, taking the Laplace transform of both sides$$sY(s)-y(0)+2Y(s)=sU(s)-u(0)-U(s)\\\rightarrow G(s)=\frac{Y(s)}{U(s)}=\frac{s-1}{s+2}$$

ContentLicense: CC BY-SA 4.0


问:
Find orthogonal matrix missing values
I was solving this problem, where I need to find the value x, which is missed in the orthogonal matrix A.

$$
A =
\begin{pmatrix}
x&0.5&-0.5&-0.5\\
x&0.5&0.5&0.5\\
x&-0.5&-0.5&0.5\\
x&-0.5&0.5&-0.5\\
\end{pmatrix}
$$

One of the properties of orthogonal matrix is that the dot product of orthogonal matrix and its transposed version is the identity matrix:

$$I=A.A^T$$

$$
A^T =
\begin{pmatrix}
x&x&x&x\\
0.5&0.5&-0.5&-0.5\\
-0.5&0.5&-0.5&0.5\\
-0.5&0.5&0.5&-0.5\\
\end{pmatrix}
$$

So I figured that $$x=0.5$$

That would result in identity matrix I:

$$
I =
\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{pmatrix}
$$

As the result I had to provide the sum of missing values, which $0.5\times4 = 2$. The grader (program) returned the error. Is there something I am missing (there most likely is, I guess)?

Thank you in advance!

答: Thank you everyone, indeed there are two possible answers $$x = 1/2$$ $$x = -1/2$$, therefore there can't be a definitive answer.

Title: Find orthogonal matrix missing values


问:
Convergence of $\sum_{k = 1}^{\infty} \cos (\log k)$ and $\sum_{k = 1}^{\infty} \sin (\log k)$
I was trying to determine if

$$\sum_{k = 1}^{\infty} \cos (\log k)$$

(with $k = 1, 2, \ldots$) converges. $\log$ is the natural logarithm.

I tried the ratio test,

$$\lim_{k \to \infty} \frac{\cos \left[ \log ( k + 1 ) \right]}{\cos \left[ \log ( k ) \right]} = 1$$

but it is not useful to determine the convergence.

According to Wolfram Alpha, it diverges. How to prove it?

And what about $\displaystyle \sum_{k = 1}^{\infty} \sin (\log k)$?

What I can empirically observe is that, regardless of how big $\log k$ can become, its cosine will always be a number $x$ such that $-1 \leq x \leq 1$. But I can't immediately see that this leads to an infinite sum.

Trivially, the term test is:

$$\lim_{k \to \infty} \cos (\log k)$$

and this limit is not defined, due to the cosine. Maybe this is sufficient to state that the series diverges.

答: Suppose that $a_n =\cos (\log n )\to 0$ then of course $a_{10^k} =\cos k \to 0.$ But it is well known  that the last is not true.


问:
Proving a topological space is Tychonoff (completely regular) iff it has a base of cozero sets.
There is a particular characterization of Tychonoff spaces (also known as completely regular spaces) which uses cozero sets which I'm trying to prove. As a quick reminder,

For a space $X$, a set $A \subseteq X$ is said to be a zero set of $X$ if there exists a continuous function $f: X \rightarrow [0,1]$ such that $A = f^{-1}(\{0\})$. We say that a set $B \subseteq X$ is a cozero set of $X$ if it is the complement of a zero set; i.e., there exists a continuous function $g: X \rightarrow [0,1]$ such that $B = g^{-1}((0,1])$.

With this in mind, I'm trying to prove that a $T_1$ topological space $X$ is Tychonoff if and only if it has a base of cozero sets. This seems to be a pretty well known result, but I'm struggling to prove it (or for that matter find a proof). I've shown that if a space is Tychonoff then it has a base of cozero sets, but after a couple of weeks of failing, I can't prove the fact that if a space has a base of cozero sets, then it must be Tychonoff.

Any pointers / proofs / articles which have this proof on them?

答:
Actually, I have some alternative proof for the sufficent condition:

Let $G$ be a open set in $(X,T)$. So, $X\setminus G$ is closed in $(X,T)$, and let $x\in G$.

As $(X,T)$ is completely regular, there exists some $f:X\rightarrow I$ continuous that $f(x)=1$ and $f(X\setminus G)=0$.

Then:

$x\notin f^{-1}(0)\Rightarrow x\in X\setminus f^{-1}(0)$

$X\setminus G \subset f^{-1}(0)\Rightarrow G\supset X\setminus f^{-1}(0)$

In consequence we have that $x\in X\setminus f^{-1}(0)\subset G$. So T has a basis of cozero sets.

I think here I haven't used that $(X,T)$ is a $T_1$ space, am I right?


问:
Can I get a hint for this rates of change question?
A ball of radius $24cm$ has its volume decreasing at a rate of three times its surface area per second. What is the time taken for the volume to be reduced to one-eighth of the initial size?

Initial volume is $18432π$ and one with of this is $2304π$. I tried to integrate as follows

$$\int _{2304\pi }^{14832\pi }\:\frac{dV}{dt}$$
Where $\frac{dV}{dt}=4\pi r^2\cdot\frac{dr}{dt}$

Also from the given statement $\frac{dV}{dt}=-3\:\cdot4\pi r^2$ implying that $\frac{dr}{dt}=-3$

I couldn't solve it this way so I assume it is wrong. How do I do it?

答:
One observation. The model does not make much phyisical sense, as it is written.

The OP and the suggested solution use this :

$\frac{dV}{dt}=-3A$

But this dimensionally does not make sense. As a check, a change of units $r \rightarrow \lambda r$, would lead to:

$\frac{dV}{dt} \rightarrow \lambda^3\frac{dV}{dt} $

and

$-3A \rightarrow -3\lambda^2A $

Clearly the equation is than not a physical one because it cannot be valid for every unit system.


问:
How to characterize the net convergence in final topology?
Given a set $X$ and an indexed family $\{(Y_i,\mathscr T_i)\}_{i\in I}$ of topological spaces with functions $f_i:Y_i\to X$. Let $\tau_{\text{final}}$ be the final topology in $X$ w.r.t. $\{f_i\}_{i\in I}$, that is, it's the finest topology such that each $f_i:(Y_i,\mathscr T_i)\to (X,\tau_{\text{final}})$ is continuous.

Now consider a net $\{x_\alpha\}_{\alpha\in A}\subset X$. The question is how to characterize the convergence of the net $\{x_\alpha\}_{\alpha\in A}$ in $(X,\tau_{\text{final}})$?

It's easy to get the following characterization for net convergence in initial topology:

Given a family of functions $g_i:X\to Y_i$, let $\tau_{\text{initial}}$ be the initial topology w.r.t. $\{g_i\}_{i\in I}$, that is, it's the coarsest topology such that each $g_i:(X,\tau_{\text{initial}})\to (Y_i,\mathscr T_i)$ is continuous. Then the net $\{x_\alpha\}_{\alpha\in A}\to x$ in $(X,\tau_{\text{initial}})$ if and only if for all $i\in I$, $\{g_i(x_\alpha)\}_{\alpha\in A}\to g_i(x)$ in $(Y_i,\mathscr T_i)$.

So is there any analogous characterization for that of final topology? Any comments or hints will be appreciated!

答:
A convergence definition on a set $X$ that just respects subnets i.e. if a net is defined to converge then any subnet of that also converges by the definition, generates the finest topology in which every a-priori convergent net converges. But there might be convergent nets in that generated topology that are not among the a-priori ones. The usual textbook properties of topological convergence prevent this but a topology can be generated with less than those.

If there is a topology $\tau$ on $X$ already and the a-priori convergent nets do converge in $\tau$, then the generated topology may be the same as $\tau$ and it may be said that the convergence generates $\tau$. But, as explained above, such a convergence may not fully characterize the topological convergence of $\tau$. For the quotient topology $\pi\colon Y\rightarrow X$, the convergence definition $x_\lambda\rightarrow x$ if there is a $y_\lambda\in Y$ such that $\pi(y_\lambda)=x_\lambda$ such that $y_\lambda\rightarrow y$ and $\pi(y)=x$, generates the quotient topology but there may be nets $x_\lambda\in X$ which converge but have no such convergent lift to $Y$.

It may be useful to note that you can still work with an a-priori convergence definition when it just generates a topology because it does characterize closed sets and continuity, even though you don't necessarily get all the topologically convergent nets. For example, with the quotient topology $\pi\colon Y\rightarrow X$, $f\colon X\rightarrow Z$ is continuous if and only if $f\circ\pi$ is, from which $f$ is continuous if and only if $f(x_\lambda)\rightarrow f(x)$ for all $x_\lambda$ that do lift to $Y$. You have to be careful, though: just because some topological notions are characterized by the generating nets does not mean they all are. Some very plausible looking statements involving generating nets may be false.

For a summary that tries to be self-contained and generally accessible, you might be interested in the Appendix of https://www.aimsciences.org/article/doi/10.3934/jcd.2021003.


问:
Can you please help with the inscribed angle theorem?
I have this theorem in my textbook

The angle subtended at the centre by an arc of a circle is double the
angle which this arc subtended at any point on the remaining part of
the circle.

Now consider this diagram

Now if we go my the above theorem the yellow angle must be equal to 'x'(which is half the angle subtended at the center).
But it is not the case.Why so?

答:
"On the remaining part of the circle" means not on the arc that was mentioned in the first part of the theorem.

The radii drawn in red divide the circle into two arcs.
Arc $A$ is the short arc that sits inside the angle marked $2x$ and is the "arc of a circle" that subtends the angle $2x$ at the center of the circle.
Arc $B$ is the long arc that goes around the top of the circle from one end of the short arc to the other. Part B is "the remaining part of the circle".

From any point on arc $B,$ arc $A$ subtends an angle $x.$

On the other hand, you could take arc $B$ as the "arc of a circle"
and take arc $A$ as the "remaining part",
in which case you find that arc $B$ subtends the angle
$2\pi - 2x$ (that is, $360^\circ - 2x$)
and the yellow angle on arc $A$ is half that angle.


答:
The yellow $\bbox[black, 5px]{\color{yellow}x}$ is half of the reflex $\angle O = 360^\circ - 2x$. i.e.

$$\bbox[black, 5px]{\color{yellow}x} = 180^\circ - x$$

Imagine standing at where the blue lines meet (or anywhere on the minor arc), the angle of view of the major arc is $\bbox[black, 5px]{\color{yellow}x}$.

And imagine standing at $O$. The angle of view of the major arc is not $2x$, but the reflex angle $360^\circ - 2x$. (And you have to turn to see the full major arc) Then the theorem says

$$360^\circ - 2x = 2\ \bbox[black, 5px]{\color{yellow}x}$$

Title: Can you please help with the inscribed angle theorem?

Body:
I have this theorem in my textbook

The angle subtended at the centre by an arc of a circle is double the
angle which this arc subtended at any point on the remaining part of
the circle.

Now consider this diagram

Now if we go my the above theorem the yellow angle must be equal to 'x'(which is half the angle subtended at the center).
But it is not the case.Why so?


问:
The equation of a sphere tangent to a plane at a point whose center belongs to a plane
Determine the equation of a sphere that is tangent to the plane $x+y+z-9=0$ at the point $M(2,3,4)$ and whose center belongs to the plane:
$$ (P) : 7x-4y+5z-14=0 $$
My solution: it is clear that the point $A(2,5,4)$ belongs to the plane $(P)$. Then, let $d$ be the line
passing through $A$, $d$ orthogonal to $(P)$ (so that the director vector is equal with the normal vector of the plane, $v(7,-4,5)$ ). From the parametric equations of this line I found that the center of the sphere is of the form $C(2+7t, 5-4t, 4+5t)$. At this point, if I calculate the distance from the center of the sphere to the first plane I get ${|8t+11|\over\sqrt{3}}=r$, where $r$ is the radius of the sphere. And I'm stuck at this point because I don't really see how I can use the point $M$ to solve this problem. Any ideas? Thanks!

答:
Consider the sphere with center at $M(2,3,4)$ and radius zero
$$(x-2)^2+(y-3)^2+(z-4)^2=0$$
A linear combination of that degenerate sphere and the tangent plane $P:x+y+z-9=0$ represents any sphere tangent to $P$ at the point $M$.

$$(x-2)^2+(y-3)^2+(z-4)^2+\lambda(x+y+z-9)=0\tag{1}$$
expland and collect
$$x^2+y^2+z^2+(\lambda -4) x+(\lambda -6) y+(\lambda -8) z+29-9 \lambda =0$$
whose center is
$$C_\lambda=\left(\frac{4-\lambda}{2},\frac{6-\lambda}{2},\frac{8-\lambda}{2}\right)$$
in order to belong to the plane $7x-4y+5z-14=0$ we must have
$$\frac{7 (4-\lambda )}{2}-2 (6-\lambda )+\frac{5 (8-\lambda )}{2}-14=0$$
which gives $\lambda=2$. Substitute this value in the equation $(1)$ to get
$$x^2+y^2+z^2-2 x-4 y-6 z+11=0$$


问:
Finding the sum of a series using method of differences
I was given this question on an assignment and was unable to get the same answer. Can someone explain this to me please. Thank you.

Using method of differences prove that $$\sum_{r=1}^n \frac{1}{r(r+2)}=\frac{n(an+b)}{4(n+1)(n+2)}$$
where $a$ and $b$ are to be found.

Hence show that
$$\sum_{r=n+1}^{2n} \frac{1}{r(r+2)}=\frac{n(4n+5)}{4(n+1)(n+2)(2n+1)}$$

Body:
$$\sum_{r=1}^{n} \frac{1}{r(r+2)}$$= $\frac{1}{2} $$\sum_{r=1}^{n} \frac{1}{r}-\frac{1}{r+2}$$
=\frac{1}{2} (1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2})
=\frac{3n^2+5n}{8n^2+12n+4}
=\frac{n(3n+5)}{4(n+1)(n+2)}$
${a}=3 , {b}=5$

$$\sum_{r=n+1}^{2n} \frac{1}{r(r+2)}$$= ($\frac{1}{2} $$\sum_{r=1}^{2n} \frac{1}{r}-\frac{1}{r+2} $) - ($\frac{1}{2} $$\sum_{r=1}^{n} \frac{1}{r}-\frac{1}{r+2}$)

$\frac{1}{2} $$\sum_{r=1}^{2n} \frac{1}{r}-\frac{1}{r+2}= \frac{1}{2} (1+\frac{1}{2}-\frac{1}{2n+1}-\frac{1}{2n+2})
=\frac{6n^2+5n}{8n^2+12n+4} $

$$\sum_{r=n+1}^{2n} \frac{1}{r(r+2)}=\frac{6n^2+5n}{8n^2+12n+4}-=\frac{n(3n+5)}{4(n+1)(n+2)}
=\frac{6n^2+5n}{4(n+1)(2n+1)}-=\frac{n(3n+5)}{4(n+1)(n+2)}
=\frac{4n^2+5n}{4(n+1)(n+2)(2n+1)}
=\frac{n(4n+5)}{4(n+1)(n+2)(2n+1)}$$


问:
What is rigorous notation for functions?
I have seen many ways to denote a function: $f(x)=x^2, y=x^2, f: x\mapsto x^2$ and so on. What is exact notation for functions? Please include lethal doses of rigor, set theory, and of course notational exactness.

Note: I am very familiar with functions in general. I just know that a lot of mathematical literature abuses notation when it comes to functions.

答:
$$f:E\to F,\quad x\mapsto f(x)$$

For example, the function $f:\mathbb R\to\mathbb R$, $x\mapsto x^2$, is not the function $g:\mathbb R_+\to\mathbb R$, $x\mapsto x^2$, but the functions $h:\{-1,0,1\}\to\mathbb R$, $x\mapsto x^2$ and $k:\{-1,0,1\}\to\mathbb R$, $x\mapsto |x|$ are equal.

ContentLicense: CC BY-SA 3.0


问:
Help With Proof by Induction on an Iterative Function
Stuck at the inductive step with this proof.

Given the four subfunctions $P_1,P_2,N_1$, and $N_2$ each with subdomain and subrange as shown above, which together make up an iterative piecewise function that we will call $f(x)$. Note that all values in the range of $f$ also appear in the domain of $f$, and vice versa, this guarantees continuous iterations under $f$.

I want to show that the trajectory of all values in the domain of $f(x)$ will converge to a single value, $1$. Part of the tree is shown below.

Base Step: Show that the initial term of each subfunction does converge to 1.

The initial term of $P_1$ is 2 which has trajectory $2→3→-4→-2→-1→1$. This composition is $N_1 ∘N_2∘N_2 ∘P_2∘P_1 (x)=\frac{9}{32}x+\frac{14}{32}$. The initial term of $P_1$ does converge to $1$, so this partial base step is true.

The initial term of $P_2$ is $1$ which has trajectory $1→-1→1$. This composition is $N_1∘P_2 (x)=\frac{3}{4} x+\frac{1}{4}$. The initial term of $P_2$ does converge to $1$, so this partial base step is also true.

The initial term of $N_1$ is $-1$ which has trajectory  $-1→1$. This composition is $N_1 (x)=\frac{-1}{2} x+\frac{1}{2}$. The initial term of $N_1$ does converge to $1$, so this partial base step is also true.

The initial term of $N_2$ is $-2$ which has trajectory $-2→-1→1$. This composition is $N_1∘N_2 (x)=\frac{-1}{4} x+\frac{1}{2}$. The initial term of $N_2$ does converge to $1$, so this partial base step is also true.

Since all the partial base steps are true, the base step is true.

The Inductive Step (updated)

Assume that an arbitrary positive integer $x_i$ will converge to $1$ under $f(x)$ after $k$ iterations, that is, there exists a composition $C(x_i)=\frac{3^m}{2^n}x_i+\frac{c}{2^n} =1$ . (This argument will also hold for a negative integer). We need to show that $x_i + 2$ will also converge.

This is as far as I can go.

On a different note however, looking at the digraph, I recognize that all the trajectories flow in one direction. This led me to wonder if once the initial terms converge, wouldn't all subsequent term converge, since iteration of $f$ on any subsequent term is an iteration on the initial term plus some integer. In other words, no number can flow in the other direction to diverge.

Any help is appreciated. Any related resource will also be appreciated.

答:
We will now show that the $(k + 1)th$ iteration of $x_i$ under $f$ will also converge to 1.

This isn't what you want to show (it's obvious: the $(k+1)th$ iteration of $x_i$ is $f(1) = -1$, which you already covered in your base case). What you've shown is that if some $x_i$ converges to $1$, then $-1$ converges to $1$, which doesn't actually tell you anything about any points other than those in your base case.

I'd also object to "randomly" - that word has a meaning, and you are not using probabilistic or stochastic methods.


答: The yellow angle does not intercept the same arc as the angle at centre. It intercepts the circle minus this arc, i.e. $2(\pi -x)R$, if $R$ is the radius of the circle.


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:
You need to ensure that $$\dfrac{1-\cos ax} {x^2}\to\dfrac{2k+1}{2}$$ so that the limit in question exists. This clearly means that $a\neq 0$ and further $$a^2=2k+1$$ and thus $k\geq 0$.

Now we can write expression in question as $$(-1)^k\cdot\dfrac {\sin\left(\dfrac{(2k+1)\pi}{2}\left(1-\dfrac{2}{2k+1}\cdot \dfrac{1-\cos ax}{x^2}\right)\right)}{x^2}$$ which can be written as $(-1)^{k}(\sin f(x)) /x^2$ where $f(x) \to 0$. And we rewrite it as $$(-1)^k\cdot \frac{\sin f(x)}{f(x)} \cdot\frac{f(x)} {x^2}$$ Thus the desired limit equals the limit of $$(-1)^k\cdot\frac{1}{x^2}\cdot\frac{(2k+1)\pi}{2}\left(1-\frac{2}{2k+1}\cdot\frac{1-\cos ax} {x^2}\right)=(-1)^k\cdot \frac{\pi} {2}\frac{a^2x^2-4\sin^2(ax/2)}{x^4}$$ Putting $ax=2t$ we can see that the above equals $$(-1)^k\frac{a^{4}\pi}{8}\frac{t^2-\sin^2t}{t^4}$$ You can easily check via factoring that last factor tends to $1/3$ and hence the desired limit is $$(-1)^k\cdot\frac{(2k+1)^2\pi}{24}$$


问:
Integral equations for research in quantum mechanics
In my research, I have reached a point where I need to find a nonnegative function which satisfies the following properties:

It's an even function:

$$f(-\theta) = f(\theta)$$

It's normalised as:

$$\int^\pi_0f(\theta)d\theta = \frac{\pi}{2}$$

For any $\phi$ in the interval $[0,\pi]$,the following property is satisfied:

$$\int^\pi_0[f(\theta)+f(\theta - \phi)-2f(\theta)f(\theta - \phi)]d\theta = \pi\sin^2\frac{\phi}{2}$$

I would be very much in your debt if you help me find such a function.

答:
There are two functions that satisfy the updated conditions.

Let us start by writing $f(\theta)$ as a Fourier series involving only cosines (which are even functions):

\begin{equation}
f(\theta)=\sum_{n=0}^{+\infty}f_n\cos(n\theta)
\end{equation}

the first condition is immediately satisfied. Let us impose the second condition:

\begin{equation}
\int_{0}^{\pi}d\theta \,\,f(\theta)=\sum_{n=0}^{+\infty}f_n\int_{0}^{\pi}\cos(n\theta)=\sum_{n=0}^{+\infty}f_n\frac{\sin{(\pi n)}}{n} \stackrel{!}{=}\frac{\pi}{2}
\end{equation}

this expression vanishes except when $n=0$, in that case
\begin{equation}
\frac{\sin{(\pi n)}}{n}=\pi
\end{equation}
or, in other words, $f_0=\frac{1}{2}$. Hence,

\begin{equation}
f(\theta)=\frac{1}{2}+\sum_{n=1}^{+\infty}f_n\cos(n\theta)
\end{equation}

Now we have to look at condition 3. We have

\begin{equation}
\int_{0}^{\pi}d\theta\,\,\cos\left(n(\theta-\phi)\right)=\frac{\sin(n\phi )}{n}[1-(-1)^n],
\end{equation}

\begin{equation}
\begin{split}
\int_0^{\pi}d\theta\,\,\left(\frac{1}{2}+\sum_{n= 1}^{+\infty}f_n\cos(n\theta)\right)\left(\frac{1}{2}+\sum_{m=1}^{+\infty}f_m\cos(m(\theta-\phi))\right) \\ \\
=\frac{\pi}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}\frac{\sin(n\phi )}{n}\left[1-(-1)^n\right]
+\sum_{n,m\neq0}f_nf_m\int_0^{\pi}d\theta\,\,\cos(n\theta)\cos(m(\theta-\phi))
\end{split}
\end{equation}

The last integral is
\begin{equation}
\int_0^{\pi}d\theta\,\,\cos(n\theta)\cos(m(\theta-\phi))=\frac{m}{m^2-n^2}[\sin(m\phi )-(-1)^n\sin(m(\phi-\pi))]
\end{equation}

Let us study the limit $n\to m$
\begin{equation}
\lim_{m\to n}\frac{m}{m^2-n^2}[\sin(m\phi )-(-1)^n\sin(m(\phi-\pi))]=\frac{\pi}{2}\cos\left(n\phi\right)
\end{equation}

Hence, imposing condition 3 means that

\begin{equation}
\frac{\pi}{2}+\frac{\pi}{2}+\sum_{n=1}^{\infty}f_n \frac{\sin(\phi n)}{n}[1-(-1)^n]-2\bigg\{\frac{\pi}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi)}{n}[1-(-1)^n]\\
+\frac{\pi}{2}\sum_{n=1}^{+\infty}f^2_n\cos(n\phi)+\sum_{n\neq m\neq 0}f_nf_m \frac{m}{m^2-n^2}\sin(m\phi )[1-(-1)^{n+m}]\bigg\}\stackrel{!}{=}
\pi\sin^2\left(\frac{\phi}{2}\right)
\end{equation}

Now the right hand side can be written as

\begin{equation}
\pi\sin^2\left(\frac{\phi}{2}\right)=\frac{\pi}{2}(1-\cos\phi)
\end{equation}

Therefore we have obtained the equation

\begin{equation}
\begin{split}
\pi\sum_{n=1}^{+\infty}f^2_n\cos(n\phi)+2\sum_{n\neq m\neq 0}f_nf_m \frac{m}{m^2-n^2}\sin(m\phi )[1-(-1)^{n+m}]=\frac{\pi}{2}\cos(\phi)
\end{split}
\end{equation}

We can satisfy this equation if $f^2_1\pi=\frac{\pi}{2}$ and if $f_{k}=0$ whenever $k>1$. This means $f_1=\pm \frac{1}{\sqrt{2}}$.

We conclude that the functions that you are looking for are

\begin{equation}
f(\theta)=\frac{1}{2}\pm\frac{1}{\sqrt{2}}\cos\theta
\end{equation}

If you want $f$ to be non negative, then just choose the plus sign.

OwnerDisplayName: Arthur

OwnerDisplayName: Ruben Campos Delgado


问:
Radius of convergence of $f(z)=\frac{z^4e^z}{(6-z)(2i-z)} $
The Question is:

Compute the radius of convergence for the expansion at the point
$z=4+3i$ of the function $$f(z)=\frac{z^4e^z}{(6-z)(2i-z)}. $$

In my text book the solution is written like this:
$\sqrt{(4-6)^2+3^2}$,  but I don't understand how they calculate it. Can someone please explain it?

Thanks!

答:
The function $f$ is holomorphic on $\mathbb{C} \setminus \{6, 2i\}$. The Radius of Convergence is given by the minimum of the distances from $4+3i$ to $z = 6$ or $z = 2i$. That's due to the fact that we can write $f$ as a power series on $D_R(4+3i)$ (meaning the open disc of radius $R$ around $4+3i$). You also might fancy a look at Radius of Convergence.

The distance from $4 + 3i$ to $z = 2i$ is given by $\sqrt{4^2 + 1^2}$. The distance from $4 +3i$ to $z = 6$ is given by $\sqrt{2^2 + 3^2}$. This will immediately become clear if you draw a figure and use Pythagoras.

Since
\begin{align}
\sqrt{17} = \sqrt{4^2 + 1^2} > \sqrt{2^2 + 3^2} = \sqrt{13}
\end{align}
and $R = \operatorname{min}\{\sqrt{4^2 + 1^2}, \sqrt{2^2 + 3^2}\}$ we find $R = \sqrt{2^2 + 3^2}$ or as the text book has written it: $R = \sqrt{(4-6)^2 + 3^2}$.


问:
Why can we use eigendecomposition in a funciton in this way
On this Wikipedia page, it says that

$$f(A) = Qf(\Lambda)Q^{-1}$$

where $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal, and $Q$ is the eigenvector matrix. In other words, they the eigendecomposition of $A$, such that $A=Q\Lambda Q^{-1}$. Why is equation above true? Why are we able to pull out the eigenvector matrix this way?

Edit: I have tried to find proof. The closest thing I could find is a definition in a book, which states it as if it is a fact. It is not an assignment, only something I fail to see why it works. I genuinely would like to understand. Please don't downvote the question to oblivion.

答:
Inspired by this answer to another question, here is my solution. Please correct me if I am wrong.

The basic idea is to use Taylor expansion and the properties of $Q$.

Using Taylor series, we can expand the matrix function $f(A)$ as

$$
f(A) = f(0) + \frac{f'(0)}{1!}A + \frac{f''(0)}{2!}A^2 + \frac{f'''(0)}{3!}A^3 + .....
$$

In a more compact form,

$$
f(A) = \sum_{n=0}^\infty C_n A^n,
$$

where coefficient $C_n$ is the $k_{\text{th}}$ derivative evaluated at $0$ divided by $n!$.
Each component of the Taylor expansion is linearly independent to each other. For each element, we can show that

$$
A^n = Q(\Lambda^n)Q^{-1}
$$

Therefore we have

$$
f(A) = \sum_{n=0}^\infty C_n Q(\Lambda^n)Q^{-1} = Q \big(\sum_{n=0}^\infty C_n\Lambda^n\big) Q^{-1} = Q f(\Lambda) Q^{-1}.
$$


问:
Why are most ordered pairs (ind, dep) but we use (r, theta) for polar functions?
Most of high school mathematics works on the assumption that the abscissa is the independent variable and the ordinate is the dependent variable and the ordered pair is written as such (abs, ord) = (ind, dep). Yet when we consider polar coordinates, r is written as a function of theta (r = f(theta)). So theta is independent and the radius dependent. Yet when we write the ordered pair we reverse the order - (r, theta) = (dep, ind).
Why is this?
When considering ordered pairs and their definition, what is the best way to prepare students for college-level (and higher) mathematics?</p>

答:
I think your question presumes an order of dependence between the coordinates that is not always the case.  Yes, it is a convention that we often plot functions of the form $y = f(x)$ in the order that you describe, where the horizontal axis is the independent variable and the vertical the dependent.  And similarly, in polar coordinates, it is a convention to conceptualize the radius $r$ as a function of angle $\theta$.  But this is not always true; for example, we could plot a relation  $$x^2 + y^2 = 1$$ that describes a circle in the coordinate plane.  And in polar coordinates, that same circle would have the equation $$r = 1.$$

Moreover, in a polar coordinate system, we could write more complex curves in which neither $r$ nor $\theta$ can be expressed as an explicit function of the other.  Mathematics is rife with examples like this in which we write things according to convention, rather than some underlying consistent rule.


问:
Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)

I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).

Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$

Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
 | 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$

答:
Second solution of the problem

We prove by absurdity. Suppose
$$\not\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta \tag{1}$$

We have $(1)$ is equivalent to
$$\forall\delta>0, \exists x \in ]0,\pi[, \forall n\in \Bbb N, |\sin(xk^n)|< \delta \tag{2}$$

Construct a sequence $\delta_p$ such that $\delta_p \xrightarrow{p\rightarrow+\infty}0$. Then there exists a sequence $x_p \in ]0,\pi[$ such that
$$\forall n\in \Bbb N, |\sin(x_pk^n)|< \delta_p$$

Take $n=0$, we have $|\sin(x_p)| \xrightarrow{p\rightarrow+\infty}0$. Because the function $|\sin(x)|$ continue and have only 2 roots in $[0,\pi]$, the sequence $x_p$ must then converge to the set of two values $\{\{0\},\{\pi\} \}$ (we remind that $0<x_p<\pi$ for all $p$). We can extract from the sequence $\{x_p \}$ a sub sequence $\{x_q \}$ such that  $x_q \xrightarrow{q\rightarrow+\infty} r \in \{\{0\},\{\pi\} \}$. Without loss of generality, we suppose $r=0$.

As $\delta_q \xrightarrow{q\rightarrow+\infty}0$ and $x_q \xrightarrow{q\rightarrow+\infty} 0$, , we can choose an $q$ such that
$$\delta_q < \sin(\frac{\pi}{k^2})$$ and $$x_q<\frac{\pi}{k^2} \tag{3}$$

From $(2)$, we have
$$\forall n\in \Bbb N, |\sin(x_qk^n)|< \delta_q < \sin(\frac{\pi}{k^2}) \tag{4}$$
We will prove that (4) can't hold true by seeking an $n$ such that
\begin{align}
(5):\frac{1}{k^2}\pi<x_qk^n<\frac{k^2-1}{k^2}\pi &\iff \ln(\frac{1}{k^2}\pi)<\ln(x_q)+n\ln(k)<\ln(\frac{k^2-1}{k^2}\pi) \\
&\iff \frac{\ln(\frac{1}{k^2})}{\ln(k)}<\frac{\ln(\frac{x_q}{\pi})}{\ln(k)}+n<\frac{\ln(\frac{k^2-1}{k^2})}{\ln(k)} \\
&\iff \frac{\ln(\frac{1}{k^2})}{\ln(k)}-\frac{\ln(\frac{x_q}{\pi})}{\ln(k)}<n<\frac{\ln(\frac{k^2-1}{k^2})}{\ln(k)} -\frac{\ln(\frac{x_q}{\pi})}{\ln(k)}\\
\end{align}

This interval has the length of $\frac{\ln(\frac{k^2-1}{k^2})-\ln(\frac{1}{k^2})}{\ln(k)} = \frac{\ln(k^2-1)}{\ln(k)} >1$. Besides, from (3), the left term $\frac{\ln(\frac{1}{k^2})}{\ln(k)}-\frac{\ln(\frac{x_q}{\pi})}{\ln(k)}$ is always postive. Then, there always exists $n_0 \in \Bbb N^*$ satisfying (5).

And we notice that, from (5), we have $$|\sin(x_0k^{n_0})|>\sin(\frac{\pi}{k^2}) \tag{6}$$

From (4) and (6), we have a contradiction. Hence, (1) can't hold true. Consequently, we can conclude that
$$\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$$


问:
Understanding the proof that countable product of second countable space is second countable under the product topology
This is a very short proof but I dont understand the reason the proof was done this way and why we have a countable basis for the product topology.

"A countable product of second countable spaces is second countable."

Suppose that for each $n$ ,$X_n$ is second countable, say $\mathcal{B}_n$ is a countable basis. Then $$\{\prod_n B_n : B_n = X_n \text{ for all but finitely many } n; \text{ if } B_n \neq X_n, \text{ then } B_n \in \mathcal{B}_n \}$$ is a countable basis for $X = \prod_n X_n$, so  $X$ is second countable.

答:
The proof consists of a few things:

Note that all members of $\mathcal{B}$ are product open sets (of the basic kind: infinite cubes where all but finitely many factors (“sides”) are the whole space in that factor space $X_n$).

The collection $\mathcal{B}$ is countable: to construct an element we make finitely many independent choices from countable sets: first a finite subset of $\Bbb N$: the coordinates that are allowed to be non-trivial (there are countably many finite subsets for a countable set), and next for such a subset $F$ a member $B_n$ from the countable set $\mathcal{B}_n$ for all $n \in F$, and then our member from $\mathcal{B}$ is determined. Underlying this is theorems like the countable union of countable sets is countable and finite products of countable sets is countable. It’s basically set theory.

The collection $\mathcal{B}$ is a base for the product topology. For this it suffices to show that for each product open $O$ and $x \in O$ we can find a member of the base containing $x$ sitting inside $O$. To this end we first find a standard basic open subset $U:=\prod_{n \in \Bbb N} U_n$ such that $x \in U \subseteq O$ and being basic open means there is a finite subset $F$ such that for $n \notin F$, we have $U_n = X_n$ and for $n \in F$ we have that $U_n \subseteq X_n$ is open. So for those $n \in F$: as $x_n \in U_n$ and $\mathcal{B}_n$ is a base for $X_n$ we find $B_n \in \mathcal{B}_n$ such that $x_n \in B_n \subseteq U_n$, and then we define $B = \prod_n B_n$ by also defining $B_n = X_n$ for $n \notin F$. Then $B \in \mathcal{B}$, and $$x \in B =\prod_n B_n \subseteq \prod_n U_n = U \subseteq O$$ as required.


问:
Asymptotic solution to the differential equation $y''+(\pi^2+\varepsilon)y+y^2=0$ with $y'(0)=y'(1)=0$
In studying perturbation methods, I am stuck with the following problem.

Let $\varepsilon>0$ be a small parameter. Find the asymptotic approximation to the non-constant solution of the differential equation
$$
y''+(\pi^2+\varepsilon)y+y^2=0,\\
y'(0)=y'(1)=0.
$$

I tried to solving it by perturbation method. Assume that we have the expansion
$$
y(x)=\varepsilon^\alpha y_1(x)+\varepsilon^\beta y_2(x)+\cdots, \quad (*)$$
where $0<\alpha<\beta$. Substituting it into the original differential equation, at leading order $O(\varepsilon^\alpha)$, we obtain
$$
y''_1+\pi^2 y_1=0,\\
y'_1(0)=y'_1(1)=0,
$$
which yields $y_1(x)=A\cos(\pi x)$, where $A$ is a constant. To find the value of $A$, we need to look at the next order, which is
$$
\varepsilon^\beta y''_2+\varepsilon^\beta y_2+\varepsilon^{1+\alpha }y_1+\varepsilon^{2\alpha} y_1^2=0,\\
y'_2(0)=y'_2(0)=0$$
I tried to balance the terms by choosing $\beta=1+\alpha$ or $\beta=2\alpha$ or $1+\alpha=2\alpha$ or $\beta=1+\alpha=2\alpha$. However, in each of the cases,  I ended up with $A=0$ when substituting the solution of $y_2$ into the boundary conditions.

On the other hand, integrating the original differential equation once, we have
$$
y'^2+(\pi^2+\varepsilon)y^2+\frac{2}{3}y^3=C,$$
where $C$ is a constant. Using the boundary conditions $y'(0)=y'(1)=0$, we can find the values of $y(0)$ and $y(1)$ once $C$ is specified. By separating variables, we see that
$$
\int_{y(0}^{y(1)}\frac{dy}{\sqrt{C-(\pi+\varepsilon^2)y^2-\frac{2}{3}y^3}}=\int_{0}^1 dx=1,$$
which can be used to find the constant $C$. Numerical calculations show that for small $\varepsilon>0$ , $C$ exists, so the differential equation has non-constant solutions for small $\varepsilon>0$. For instance, when $\varepsilon=0.01$, $C\approx 1.1699$ and the graph of a non-constant solution plotted by Mathematica is

I think that problem is that the asymptotic expansion of the non-constant solution may not be of form (*), but I don't know how what the correct form is.

答:
Equivalent forms of the equation

With the substitution $y=Au(Bt)$ scaling both the domain and the range, the equation reduces to
$$
u''+\frac{\pi^2+ε}{B^2}u+\frac{A}{B^2}u^2=0.
$$
The scales can now be chosen in a way that produces the unperturbed equation with $ε=0$ using $A=B^2=1+\frac{ε}{\pi^2}$, or a form closer to the standard harmonic oscillator
$$
x''+x+δx^2=0
$$
with $B^2=\pi^2+ε$, $A=δB^2$.

Citing a previously performed perturbation analysis

This last form, after switching the sign in $x$, was explored in a previous topic. Fixing the amplitude $R$, the solution close to the circle of radius $R$ is given by the parametrization
$$
x\sqrt{1+\frac23δx}=-R\cos\phi, ~~ \dot x=R\sin\phi,\\
\implies x=-R\cos\phi-δ\frac{R^2}3\cos^2\phi+O(δ^2),\\
\left[\small\begin{aligned}
\implies 1&=(1+\frac{2δR}3\cos\phi+...)\dot\phi,\\
t&=\phi+\frac{2δR}3\sin\phi+...\iff \phi=t-\frac{2δR}3\sin t+...\\
x(t)&=-R\left[\cos t+\sin t\left(\frac{2δR}3\sin t\right)+...\right]-δ\frac{R^2}3\cos^2t+O(δ^2)\\
&=-R\cos t-δ\frac{R^2}3[\cos^2t+2\sin^2 t]+O(δ^2)
\end{aligned}\right]
$$
for the half-period of interest $\phi\in[0,\pi]$. Taking the derivative of the first equation one can derive a first-order ODE for the angle. Doing some series expansion one gets an expression for the half-period $\phi(T)=\pi$ as
$$
T=\pi\left(1+\frac{5}{12}(δR)^2+\frac{385}{576}(δR)^4+O(δ^6)\right).
$$
Per the parametrization we want $B=T$, which in the leading order gives
$$
(\piδR)^2=\frac65ε.
$$
This suggests $δ=\sqrtε$, $R=\frac{\sqrt{6/5}}{\pi}$.

Fall back to perturbation series for numerics

Now set $x=-R\cos t+δx_1$ to result in the equation
$$
x_1''+x_1+x^2=0,
$$
which in first order has the initial condition $x_1(0)=-\frac{R^2}3$ and the solution
$$
x_1=\frac{R^2}6(\cos(2t)-3).
$$
This can be used in an approximation $x=x_0+δx_1+δ^2v$ or just as initial guess in the boundary value problem for $v$ in $x=x_0+δv$.

Numerical experiments

```<code># global constants and invariant functions
R2 = 1.2/np.pi**2
R = R2**0.5

def x0(t): return -R*np.cos(t)
def dx0(t): return R*np.sin(t)
def x1(t): return R2/6*(np.cos(2*t)-3)
def dx1(t): return -R2/3*np.sin(2*t)

# Variables depending on the perturbation parameter
eps = 1e-8
dta = eps**0.5
B2 = np.pi**2+eps
B, A = B2**0.5, dta*B2
dvb = -dx0(B)/dta

# Boundary Value Problem
def ode_v(t,v): xt= x0(t)+dta*v[0]; return [v[1], -v[0]-xt**2]
def bc_v(v0,vb): return [v0[1],vb[1]-dvb]
# initial guess
t = np.linspace(0,B,5)
v = [ x1(t), dx1(t) ]
# solve and print (error) message
res = solve_bvp(ode_v,bc_v,t,v, tol=1e-4)
print(res.message,res.y[0,0], res.y[0,-1],-R2/3)
</code>```

This converges well and reliably. Plotting after scaling back for the original problem gives in the phase plane

Body:
Equivalent forms of the equation

With the substitution $y=Au(Bt)$ scaling both the domain and the range, the equation reduces to
$$
u''+\frac{\pi^2+ε}{B^2}u+\frac{A}{B^2}u^2=0.
$$
The scales can now be chosen in a way that produces the unperturbed equation with $ε=0$ using $A=B^2=1+\frac{ε}{\pi^2}$, or a form closer to the standard harmonic oscillator
$$
x''+x+δx^2=0
$$
with $B^2=\pi^2+ε$, $A=δB^2$.

Citing a previously performed perturbation analysis

This last form, after switching the sign in $x$, was explored in a previous topic. Fixing the amplitude $R$, the solution close to the circle of radius $R$ is given by the parametrization
$$
x\sqrt{1+\frac23δx}=-R\cos\phi, ~~ \dot x=R\sin\phi,\\
\implies x=-R\cos\phi-δ\frac{R^2}3\cos^2\phi+O(δ^2),\\
\left[\small\begin{aligned}
\implies 1&=(1+\frac{2δR}3\cos\phi+...)\dot\phi,\\
t&=\phi+\frac{2δR}3\sin\phi+...\iff \phi=t-\frac{2δR}3\sin t+...\\
x(t)&=-R\left[\cos t+\sin t\left(\frac{2δR}3\sin t\right)+...\right]-δ\frac{R^2}3\cos^2t+O(δ^2)\\
&=-R\cos t-δ\frac{R^2}3[\cos^2t+2\sin^2 t]+O(δ^2)
\end{aligned}\right]
$$
for the half-period of interest $\phi\in[0,\pi]$. Taking the derivative of the first equation one can derive a first-order ODE for the angle. Doing some series expansion one gets an expression for the half-period $\phi(T)=\pi$ as
$$
T=\pi\left(1+\frac{5}{12}(δR)^2+\frac{385}{576}(δR)^4+O(δ^6)\right).
$$
Per the parametrization we want $B=T$, which in the leading order gives
$$
(\piδR)^2=\frac65ε.
$$
This suggests $δ=\sqrtε$, $R=\frac{\sqrt{6/5}}{\pi}$.

Fall back to perturbation series for numerics

Now set $x=-R\cos t+δx_1$ to result in the equation
$$
x_1''+x_1+x^2=0,
$$
which in first order has the initial condition $x_1(0)=-\frac{R^2}3$ and the solution
$$
x_1=\frac{R^2}6(\cos(2t)-3).
$$
This can be used in an approximation $x=x_0+δx_1+δ^2v$ or just as initial guess in the boundary value problem for $v$ in $x=x_0+δv$.

Numerical experiments

# global constants and invariant functions
R2 = 1.2/np.pi**2
R = R2**0.5

def x0(t): return -R*np.cos(t)
def dx0(t): return R*np.sin(t)
def x1(t): return R2/6*(np.cos(2*t)-3)
def dx1(t): return -R2/3*np.sin(2*t)

# Variables depending on the perturbation parameter
eps = 1e-8
dta = eps**0.5
B2 = np.pi**2+eps
B, A = B2**0.5, dta*B2
dvb = -dx0(B)/dta

# Boundary Value Problem
def ode_v(t,v): xt= x0(t)+dta*v[0]; return [v[1], -v[0]-xt**2]
def bc_v(v0,vb): return [v0[1],vb[1]-dvb]
# initial guess
t = np.linspace(0,B,5)
v = [ x1(t), dx1(t) ]
# solve and print (error) message
res = solve_bvp(ode_v,bc_v,t,v, tol=1e-4)
print(res.message,res.y[0,0], res.y[0,-1],-R2/3)

This converges well and reliably. Plotting after scaling back for the original problem gives in the phase plane


问:
(Proof Verification) Show that the collection of open disks is a basis for a topology on the plane
Here's what I did.

Let $\Sigma$ be a collection of open sets for a topological space $(X,\Omega)$. By definition of a basis for the plane, $\Sigma^2$ consists of all possible open disks for the topology $\Omega$. Furthermore, since open sets in a topology $\Omega$ are defined to be the union of open disks, the collection of open disks $\Sigma^2$ must be a basis for a topology on $\mathbb{R}^2$.

Am I wrong? If so, how can I improve this?
Thanks.

答: Hint:  It suffices to show that the set of all arbitrary unions of disks is closed under unions and finite intersections (and contains the plane and the empty set).


问:
How to find $dT$? Thermodynamics (equation of state)
I recently had my first lesson thermodynamics, and there is an exercises that I don't quite understand how you have to get the answer. It goes as follows:

$$PV^{2}=nRe^{3T}$$

we are given that n en R are both constants. The question is what is $dT$?

I don't think I have the right mathematical background to solve this question. I really would love to know where I could find the name of such kind problems. I now how to derive (and partial derivation), but I'm used to working with $\frac{dy}{dx}$, but know it's only $dy$ or in my case $dT$.

If someone could help, solve this problem and where I can find theory to practice such mathematical problems that would be very helpful

Thanks in advance

答:
The problem is asking you to consider the differential form of the equation of state:
\begin{align}
PV^{2}&=nRe^{3T}\tag{1}\\
\implies 2PV\,dV + V^{2}\,dP &=3nRe^{3T}\,dT
\end{align}
where here we are thinking of $dV, dP, dT$ as infinitesimal changes in the quantities.  So, we get $$dT = \frac{2PV\,dV + V^{2}\,dP}{3nRe^{3T}}.$$  To remove the dependence on $T$ on the right hand side above, we can use $(1)$ to obtain $$dT = \frac{2PV\,dV + V^{2}\,dP}{3PV^{2}} = \frac{2P\,dV + V\,dP}{3PV}.$$


问:
Points contained in the diagonal of the product of schemes
Let $X,Y$ schemes over $S$, and $f,g$ two $S$-morphisms of schemes, $h:X \to Y\times_{S} Y$ the morphism obtained from $f$ and $g$ and $\Delta:Y \to Y \times_{S} Y$ the diagonal morphism.

I tried to prove that if $x \in X$ s.t. $f(x)=g(x)$ , then $h(x) \in $ image of $\Delta$, using the universal property of the product, but I didn't manage. Is it true ? it seems that it should be

答:
I'm happy you didn't manage to prove your statement because it is false!

Here is a counter-example:

Take $X=Y=Spec (\mathbb C)=\{\omega\}$ and $S=Spec(\mathbb R)$, so that  $Y\times_SY=Spec(\mathbb C\otimes_\mathbb R \mathbb C)$.

And take for $f,g$ the morphisms $X\to Y$ corresponding to the identity and the conjugation $\mathbb C\to \mathbb C$.

Then the morphism $h:X\to Y\times_SY$ corresponds to the $\mathbb R$-algebra morphism $\mathbb C\otimes_\mathbb R \mathbb C\to \mathbb C:z\otimes w\mapsto z\bar w  $ .

Hence  the image $h(\omega)$ of $h:X \to Y\times_{S} Y$ is the point corresponding to the  prime ideal $\mathfrak p\subset \mathbb C\otimes_\mathbb R \mathbb C$ equal to the real vector space generated by the two vectors $1\otimes 1-i\otimes i,i\otimes 1+1\otimes i\in \mathbb C\otimes_\mathbb R \mathbb C$ i.e.  $$  \mathfrak p=\mathbb C(1\otimes 1-i\otimes i)\oplus \mathbb C(i\otimes 1+1\otimes i)    \in Spec(\mathbb C\otimes_\mathbb R \mathbb C)               $$

Whereas the morphism $\Delta :Y\to Y\times_SY$ corresponds to the $\mathbb R$-algebra morphism $\mathbb C\otimes_\mathbb R \mathbb C\to \mathbb C:z\otimes w\mapsto zw  $ .

Hence the image $\Delta (\omega)$ of the diagonal morphism corresponds to  the prime ideal $\mathfrak q\subset \mathbb C\otimes_\mathbb R \mathbb C$ equal to the real vector space generated by the two vectors $1\otimes 1+i\otimes i        , i\otimes 1-\otimes i     \in \mathbb C\otimes_\mathbb R \mathbb C$ i.e. $$  \mathfrak q=\mathbb C(1\otimes 1+i\otimes i)\oplus \mathbb C(i\otimes 1-1\otimes i)       \in Spec(\mathbb C\otimes_\mathbb R \mathbb C)                $$

Conclusion:

The unique point $\omega$ of $X$ has an image not contained in the diagonal of the product:
$$ \omega \in X \;\text {satisfies}\; f(\omega)=g(\omega)\in Y \; \text {but} \;h(\omega)= \mathfrak p \notin Im(\Delta)=\{\Delta (\omega)\}=\{\mathfrak q\}                     $$


问:
Number of binary search trees on $n$ nodes of height up to $h$
How can I find the number of binary search trees up to a given height $h$, not including BSTs with height greater than $h$ for a given set of unique numbers $\{1, 2, 3, \ldots, n\}$?

For example, if the number of nodes is $4$, then the total number of BSTs is $14$.
Out of these $14$ BSTs I have to find out the number with height not more than $3$.

答:
Let us assume that the number of nodes are $n$. Also, consider that $a_{n,h}$ represents the number of different binary trees with $n$ nodes having height $h$.
We construct the solution recursively.

Consider that $m$ is the current root, then for a binary tree of height $h$ we consider two cases:

The left subtree has the height of $h-1$. Also, we know that the left subtree contains $m-1$ nodes. So, the number of binary trees with  $m-1$ nodes and height are given by $a_{m-1,h-1}$. Now, the right subtree can have any height in $[0, h-1]$, so the number of ways of forming right subtree is given by $\sum\limits_{i=1}^{h-1}a_{n-m,i} $. Therefore, total ways in this case are:  $ a_{m-1,h-1}\sum\limits_{i=1}^{h-1}a_{n-m,i} $

Now, consider that left subtree has a height less than $h-1$. Then, the right subtree(consists of $n-m$ nodes) must have a height equal to $h-1$ which is given by $a_{n-m, h-1}$. Now, the left subtree can have any height in $[0, h-2]$, so the number of ways of forming left subtree is given by $\sum\limits_{i=1}^{h-2}a_{m-1,i} $. Therefore, total ways in this case are:  $ a_{n-m,h-1}\sum\limits_{i=1}^{h-2}a_{m-1,i} $

As we have $n$ different ways to choose the node, we have the number of different binary trees of height $h$ and $n$ nodes given by the following formula:

$a_{n,h}$ = $\sum\limits_{m=1}^{n} (a_{m-1,h-1}\sum\limits_{i=1}^{h-1}a_{n-m,i}+a_{n-m,h-1}\sum\limits_{i=1}^{h-2}a_{m-1,i})$

This recurrence relation can be solved using the base case as:

$a_{0,0}=1$

$a_{i,0}=a_{0,i}=0$,  for all $i>0$

Hence the final answer to the original question becomes:

$A_{n,h}=\sum\limits_{i=1}^{h} a_{n,i}$


问:
Can someone help me identify this math symbol and its meaning?
From my university notes:

Comment to slide 9

By virtue of the result shown in this slide, we can talk about the least element of a set $D$, if one exists, and we denote it with $\perp$, pronounced "bottom". As an example, given the set $S=\{1,2,3\}$, the pair $(\mathcal{P}(S),\subseteq)$ (where $\mathcal{P}(S)$ denotes the power set of $S$) is a poset and the empty set is its bottom.

Having introduced an ordering on the partial functions, we can now summarize the require-ments for $\mathrm{FIX}\,F$:

$\mathrm{FIX}\,F$ is a fixed point of $F$, that is $F(\mathrm{FIX}\,F)=\mathrm{FIX}\,F$ and

$\mathrm{FIX}\,F$ is a least fixed point of $F$, that is if $Fg=g$ then $\mathrm{FIX}\,\color{red}{F\sqsubseteq g}$.

The next task is to ensure that all functionals $F$ that may arise in ${\tt While}$ do indeed have least fixed points.

There is this symbol which looks like a square version of an inclusion symbol but there is also an actual inclusion symbol in this same it of text. Here is another example of it's usage:
Fixed point theory

Definition A partially ordered set is a pair $(D,\sqsubseteq_D)$, where $D$ is a set and $\sqsubseteq_D$ is a reflexive, transitive, anti-symmetric relation on $D$.

Example Defined the ordering $\sqsubseteq$ on partial functions of $\mathbf{State}\hookrightarrow\mathbf{State}$:
$$g_1\sqsubseteq g_2\ \text{whenever}$$
$$\text{if}\ g_1s=s'\ \text{then}\ g_2s=s'\ \text{for all choices of}\ s\ \text{and}\ s'$$
then $(\mathbf{State}\hookrightarrow\mathbf{State},\sqsubseteq)$ is a poset.

As you can see the symbol can also contain a subscript (in this case $D$). What does this symbol mean? when is is used and what does the subscript stand for?

Both of these examples are in context to Denotational semantics. I tried to google the meaning but couldn't find anything. copy-pasting the symbol in the search bar resulted in pasting a v character

答: Detexify identifies it as \sqsubseteq. As discussed here, its meaning is highly context-dependent. But @MauroALLEGRANZA has identified the meaning applicable here.


答:
I think the topic you want to review is the Chain Rule for partial derivatives, which seems simple but can get very tricky.  First, solve your equation of state for T.  With n and R constant you have T(P,V). So dT = ∂T/∂P dP + ∂T/∂V dV.  Take the partial derivatives of your T(P,V) and plug them in.

I get T = $\frac{1}{3}(ln P + 2 ln V - ln(nR)$
So I get ∂T/∂P = $\frac{1}{3P}$ and $∂T/∂V = \frac{2}{3V}$
If you have a very hard time with differentials in particular, just divide all of them by ds or dt (small t) or some other variable not involved.  Then they are in effect derivatives.

Title: How to find $dT$? Thermodynamics (equation of state)

Body:
I recently had my first lesson thermodynamics, and there is an exercises that I don't quite understand how you have to get the answer. It goes as follows:

$$PV^{2}=nRe^{3T}$$

we are given that n en R are both constants. The question is what is $dT$?

I don't think I have the right mathematical background to solve this question. I really would love to know where I could find the name of such kind problems. I now how to derive (and partial derivation), but I'm used to working with $\frac{dy}{dx}$, but know it's only $dy$ or in my case $dT$.

If someone could help, solve this problem and where I can find theory to practice such mathematical problems that would be very helpful

Thanks in advance


问:
Question involving Taylor's theorem
Let $f \colon \Bbb R \to \Bbb R$ be a $C^2$ function. Prove:

$$\lim_{n\to\infty}\left(n^2\int_0^1f(x)\,\mathrm dx-n\sum_{k=1}^nf\left(\frac{2k-1}{2n}\right)\right)=\frac{f'(1)-f'(0)}{24}$$

Our professor gave us this question and said that it involves the use of Taylor's theorem. However I don't even know how to start this. Kindly help.

答:
We can write
$$
n^2 \int_0^1 {f(x)dx}  = n^2 \sum\limits_{k = 1}^n {\int_{(k - 1)/n}^{k/n} {f(x)dx} }  = n\sum\limits_{k = 1}^n {\int_{k - 1}^k {f\left( {\frac{t}{n}} \right)dt} } .
$$
For $k - 1 < t < k$, the Taylor formula tells us that
\begin{align*}
f\left( {\frac{t}{n}} \right) = f\left( {\frac{{2t}}{{2n}}} \right) = f\left( {\frac{{2k - 1}}{{2n}}} \right) & + f'\!\left( {\frac{{2k - 1}}{{2n}}} \right)\left( {\frac{{2t - 2k + 1}}{{2n}}} \right) \\ & + \frac{1}{2}f''\!\left( {\frac{{k - \delta _k (t)}}{n}} \right)\left( {\frac{{2t - 2k + 1}}{{2n}}} \right)^2
\end{align*}
with some $0 < \delta _k (t) < 1$. Thus, by the mean value theorem for integration
$$
n^2 \int_0^1 {f(x)dx}  = n\sum\limits_{k = 1}^n {f\left( {\frac{{2k - 1}}{{2n}}} \right)}  + \frac{1}{{24n}}\sum\limits_{k = 1}^n {f''\!\left( {\frac{{k - \zeta _k }}{n}} \right)}
$$
with some $0<\zeta_k<1$. Finally,
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{24n}}\sum\limits_{k = 1}^n {f''\!\left( {\frac{{k - \zeta _k }}{n}} \right)}  = \frac{1}{{24}}\int_0^1 {f''(x)dx}  = \frac{{f'(1) - f'(0)}}{{24}}.
$$
I leave you the details.

Title: Question involving Taylor's theorem


问:
Intuition behind universality of uniform distribution
Assuming F is the CDF of R.V. X and is strictly increasing, the following is true: F(X) ~ Unif(0,1).

I was able to prove this as follows,

Let Y = F(X), Y is an R.V. with sample space Y = [0,1]. Mathematically,

Y = F(X) = P(X $\leq$ X),

P(Y $\leq$ y) = P(F(X) $\leq$ y) = P(X $\leq$ $F^{-1}$(y)) = F($F^{-1}$(y)) = y

I am confused by this result. Intuitively, the probability of the event (X <= X) would be 1 (once X is realized), right? Even before going with the proof, I was guessing the pdf of Y would be 1 at Y = 1 and 0 everywhere else. Can someone please help understand the gap in my understanding?

Side question: while thinking about the universality of uniform distribution and working through the above proof, I was trying to come up with an intuition for the inverse of a CDF. But, the inverse is not making a lot of sense to me. If someone has interpretations for this, I would really appreciate it.

答:
Lets remember if we have a probability space $(\Omega,\mathcal{F},\mathcal{P})$ i.e a (set, sigma algebra, probability measure), and $X$ is a (say real valued) random variable w.r.t to this guy, then $X:\Omega \to \mathbb{R}$, and so how you have defined $Y$ we similarly have $Y:\Omega \to \mathbb{R}$.

Now your proof is fine (as long as you state for $y\in[0,1]$) so I will address your concerns. Firstly you saying the pdf of $Y$ should put all the mass into $1$, i.e because $X\leq X$ is always true $Y$ should always take the value $1$. The problem here is your reading of $Y=F(X)$, this is a random variable so once we give it an $\tilde{\omega} \in \Omega$ we get

$$
Y(\tilde{\omega})=\mathbb{P}\Big(X\leq X (\tilde{\omega})\Big)=\mathbb{P}\Big(\Big\{ \omega\in \Omega ~:~X(\omega)\leq X (\tilde{\omega})\Big\}\Big),
$$

so you can see where your confusion came from.

The uniform random variable is fundamental for instance in programming / simulations, take a general random variable if you have the inverse of its CDF you can numerically draw from its distribution using the uniform random variable.


问:
Is every fundamental group of a mapping torus of a surface homeomorphism a CAT(0) group?
TLDR; Is every fundamental group of a mapping torus of a surface homeomorphism a CAT(0) group?

Let $S$ be a compact surface of negative Euler characteristic and let $f :S\to S$ be a homeomorphism. Write $M_f$ to denote the mapping torus of $f$,  $$M_f = \frac{S \times [0,1]}{(0,f(x)) \sim (1, x)}.$$

Thurston shows that:

If $f$ is pseudo-Anosov then $M_f$ admits a complete hyperbolic structure of finite volume.

If $f$ is finite order then $M_f$ admits a complete $\mathbb{H}^2 \times \mathbb{R}$ structure.

If $f$ is reducible then $M_f$ admits an embedded incompressible torus, not parallel to the boundary.

If $S$ has empty boundary, then in cases 1 and 2 we can conclude that $M_f$ acts properly and cocompactly on $\mathbb{H}^{3}$ and $\mathbb{H}^2 \times \mathbb{R}$, respectively, and thus that $\pi_1(M_f)$ is CAT(0).

My question: What can be said in the case when $S$ (and thus $M_f$) has non-empty boundary? What can be said in case 3?

答:
In Cases 1 and 2 with nonempty boundary, the complete metric on $M_f$ can be truncated to give a metric on a compact space, by "cutting off horoballs". In Case 1 you cut off a $\pi_1 M_f$-invariant, pairwise disjoint family of open, 3-dimensional hyperbolic horoballs in $\mathbb H^3$, leaving $\mathbb R^2$ boundaries, and then mod out by the group, leaving torus boundaries. In Case 2 you cut off a $\pi_1(S)$-invariant, pairwise disjoint family of open 2-dimensional hyperbolic horodiscs in the $\mathbb H^2$ factor, leaving $\mathbb R$ boundaries in that factor, and then cross with the $\mathbb R$ factor, leaving $\mathbb R^2$ boundaries; and then again you mod out by the group leaving torus boundaries. In either case, the restricted path metric is a CAT(0) metric (this is probably explained in the book of Bridson and Haefliger).

In Case 3, there's some literature you can consult. I can't find a complete description, and the totality of what I can find canot be squeezed into an answer, but there's several results in this paper of Bernard Leeb as a start. For example, in the case of empty boundary, Theorem 3.3 of that paper, combined with Thurston's geometrization theorem, implies that if at least one component of the Thurston decomposition of $f$ is pseudo-Anosov then $M_f$ admits a CAT(0) geometry.

Case 3 get more complicated in the subcase that every component of the Thurston decomposition of $f$ is finite order, and so every component of the torus decomposiiton of $M_f$ is Seifert fibered. Dropping for a moment the hypothesis that $M$ is a mapping torus, Leeb's paper describes an examples of a closed, irreducible manifold $M$ whose torus decomposition has only  Seifert fibered pieces, and where $M$ itself has no locally CAT(0) metric, and yet $\pi_1(M)$ is a CAT(0) group. But I do not know whether such examples exist where $M$ is a mapping torus, and I do not know whether any mapping torus examples exist where the group is not CAT(0).


问:
Why is it that this part of the vector has to be manipulated before we can let t=0
I'm sure I missed some important part in calculus but I don't know where to look to find the answer.

A question that I'm doing is asking me to find the limit of a vector function. When I look up the answer to the problem, it leaves the $\mathbf{i}$ and $\mathbf{k}$ components alone but differentiates the $\mathbf{j}$ component, why is this? Why does the $\mathbf{k}$ component have to be manipulated when the others don't?

https://i.stack.imgur.com/myD36.png
$$ \lim_{t \to 0} \left( e^{-3t} \mathbf{i} + \frac{t^2}{\sin^2 t} \mathbf{j} + \cos 2t \mathbf{k}\right)$$

Edit:
https://i.stack.imgur.com/TD1sN.png
It didn't click with me that maybe the solution I was talking about would be helpful to post

答: My guess is that $\sin^2t=0$ when $t=0$ so it looks like you're dividing by $0$. To get around this one could apply l'Hopital's rule  to verify that $\lim_{t \to 0}\frac{t^2}{\sin^2 t} = 1$. The other coefficients are obviously continuous at $t=0$, so no calculation is needed.


问:
Is $(\Bbb Q[x])_x/(\Bbb Q[x])$ an Artinian $(\Bbb Q[x])$-module?
Consider the ring $R = \Bbb Q[x]$ and the multiplicatively closed set $S = \{1, x, x^2, \ldots\}$.

Let $R_x = S^{-1}R$. Consider the module $M = R_x/R$. Determine whether $R_x$ and $M$ are Noetherian/Artinian $R$-modules.

Edit: Note that $R$ is an $R$-submodule of $R_x$ and the quotient is of $R$-modules.

Attempt.

Note that $R_x$ is an $R$-algebra (and hence, $R$-module) via the map $p(x) \mapsto \frac{p(x)}{1}$. Since $R$ is a domain, this map is injective and thus, we may regard $R$ as an $R$-submodule of $R_x$.

Now, $R$ is not an Artinian $R$-module and hence, neither is $R_x$.

On the other hand, we have the chain

$$\langle \overline{x^{-1}} \rangle \subsetneq \langle \overline{x^{-2}} \rangle \subsetneq \langle \overline{x^{-3}} \rangle \subsetneq \cdots$$

of $R$-submodules in $R_x/R$ and hence, $R_x/R$ is not a Noetherian $R$-module. In turn, this implies that $R_x$ is not Noetherian.

Thus, only the question of showing whether or not $M$ is an Artinian $R$-module remains, which I am unable to do.

答: By Bezout, if $\frac{p(x)}{x^k} \in R_x$ with $p(0) \neq 0$, there is a polynomial $q$ such that $x^k|pq-1$. In particular, in $R_x/R$, $q(x)\frac{p(x)}{x^k}=\frac{1}{x^k}$. It follows that the lattice of sub-$R$-modules of $R_x/R$ is isomorphic to $\mathbb{N} \cup \{\infty\}$ so it doesn’t have any strictly decreasing infinite sequence. So $R_x/R$ is Artinian over $R$.


问:
Prove area relations of triangle with giving vector equation

In a triangle $ABC$ with point $I$ inside it such that $\overrightarrow{IA}+2\overrightarrow{IB}+3\overrightarrow{IC}=0$. If $S$ is the area of $\triangle ABC$ and $T$ is the area of $\triangle AIC$ then prove that $S=3T$

By the given we could get:
\begin{equation}
-\overrightarrow{AI}+2\overrightarrow{IB}+3\overrightarrow{IC}=0\\-6\overrightarrow{AI}+2\overrightarrow{AB}+3\overrightarrow{AC}=0\\\Rightarrow 6\overrightarrow{IA}+2\overrightarrow{AB}+3\overrightarrow{AC}=0
\end{equation}
This is all I could get from the given, moreover I don't know if this problem is in 2D or 3D at all but I think maybe it is in 2D. The problem ask to prove about area so we should find some relations of the cross product that I couldn't. And for the picture I don't know how to draw it too. Please help

答: Notice that $$\overrightarrow{IA}+2\overrightarrow{IB}+3\overrightarrow{IC}=0\iff 3\cdot \frac{\overrightarrow{IA}+2\overrightarrow{IB}}{3}+3\overrightarrow{IC}=\overrightarrow 0$$ Thus, consider the point $P$ on the plane such that $$\vec{p}-\vec{i}=\overrightarrow{IP}=\frac{\overrightarrow{IA}+2\overrightarrow{IB}}{3}=\frac{\left(\vec{a}-\vec{i}\right)+2\cdot \left(\vec{b}-\vec{i}\right)}{3}=\frac{\vec{a}+2\vec{b}}{3}-\vec{i}\iff \vec{p}=\frac{\vec{a}+2\vec{b}}3$$ Hence, $P$ is the point on $AB$ such that $AP=2PB$. Go back to the original equation and substitute in $$3\overrightarrow{IP}+3\overrightarrow{IC}=0\iff \overrightarrow{IP}=\overrightarrow{CI}$$ Thus, $I$ is the midpoint of $CP$. Finally, we have $$\fbox{$\frac{S}{T}=\frac{S}{\frac12\triangle PAC}=\frac{S}{\frac12\cdot \frac23 S}=3$}$$


答: In this CS context, I think $\sqsubseteq$ would be called the refinement relation (see, for example Morgan's Programming from Specifications). The subscript $D$ is just a reminder of the set $\sqsubseteq$ is defined on. Here one partial function $g_1$ is refined by another function $g_2$,  if they agree wherever $g_1$ is defined (allowing for the possibility that the domain of $g_2$ strictly includes $g_1$, so that $g_2$ is "more defined" than $g_1$). Typically refinement is about resolving indeterminacy (a.k.a., strengthening the post-condition) and expanding the domain of definition (a.k.a., weakening the pre-condition). Here the relations being refined are deterministic (i.e., partial functions), so only weakening the pre-condition applies.

Title: Can someone help me identify this math symbol and its meaning?

Body:
From my university notes:

Comment to slide 9

By virtue of the result shown in this slide, we can talk about the least element of a set $D$, if one exists, and we denote it with $\perp$, pronounced "bottom". As an example, given the set $S=\{1,2,3\}$, the pair $(\mathcal{P}(S),\subseteq)$ (where $\mathcal{P}(S)$ denotes the power set of $S$) is a poset and the empty set is its bottom.

Having introduced an ordering on the partial functions, we can now summarize the require-ments for $\mathrm{FIX}\,F$:

$\mathrm{FIX}\,F$ is a fixed point of $F$, that is $F(\mathrm{FIX}\,F)=\mathrm{FIX}\,F$ and

$\mathrm{FIX}\,F$ is a least fixed point of $F$, that is if $Fg=g$ then $\mathrm{FIX}\,\color{red}{F\sqsubseteq g}$.

The next task is to ensure that all functionals $F$ that may arise in ${\tt While}$ do indeed have least fixed points.

There is this symbol which looks like a square version of an inclusion symbol but there is also an actual inclusion symbol in this same it of text. Here is another example of it's usage:
Fixed point theory

Definition A partially ordered set is a pair $(D,\sqsubseteq_D)$, where $D$ is a set and $\sqsubseteq_D$ is a reflexive, transitive, anti-symmetric relation on $D$.

Example Defined the ordering $\sqsubseteq$ on partial functions of $\mathbf{State}\hookrightarrow\mathbf{State}$:
$$g_1\sqsubseteq g_2\ \text{whenever}$$
$$\text{if}\ g_1s=s'\ \text{then}\ g_2s=s'\ \text{for all choices of}\ s\ \text{and}\ s'$$
then $(\mathbf{State}\hookrightarrow\mathbf{State},\sqsubseteq)$ is a poset.

As you can see the symbol can also contain a subscript (in this case $D$). What does this symbol mean? when is is used and what does the subscript stand for?

Both of these examples are in context to Denotational semantics. I tried to google the meaning but couldn't find anything. copy-pasting the symbol in the search bar resulted in pasting a v character


答:
First, yes, it is planar, because a linear combination of $\overrightarrow{IA}, \overrightarrow{IB},$ and $\overrightarrow{IC} = 0$, which means that you can express any combination of them with only two, hence two dimensional.
You made a good start.  Remember we are doing a comparison.  The area of $\Delta ABC$ is $\frac12 \lvert \overrightarrow{AC} \times  \overrightarrow{AB}\rvert$ and the area of $\Delta AIC$ is $\frac12 \lvert \overrightarrow{AC} \times \overrightarrow{AI}\rvert$, so we want to rewrite one in terms of the other somehow.  Solve your expressions for $\overrightarrow{AI} = \frac13 \overrightarrow{AB} + \frac12\overrightarrow{AC}$.  Then $\overrightarrow{AC} \times \overrightarrow{AB} = \overrightarrow{AC} \times\left(\frac13\overrightarrow{AB} + \frac12 \overrightarrow{AC}\right)$ and $\overrightarrow{AC} \times \overrightarrow{AC} = \overrightarrow{0}$. The result follows.

Title: Prove area relations of triangle with giving vector equation

Body:
In a triangle $ABC$ with point $I$ inside it such that $\overrightarrow{IA}+2\overrightarrow{IB}+3\overrightarrow{IC}=0$. If $S$ is the area of $\triangle ABC$ and $T$ is the area of $\triangle AIC$ then prove that $S=3T$

By the given we could get:
\begin{equation}
-\overrightarrow{AI}+2\overrightarrow{IB}+3\overrightarrow{IC}=0\\-6\overrightarrow{AI}+2\overrightarrow{AB}+3\overrightarrow{AC}=0\\\Rightarrow 6\overrightarrow{IA}+2\overrightarrow{AB}+3\overrightarrow{AC}=0
\end{equation}
This is all I could get from the given, moreover I don't know if this problem is in 2D or 3D at all but I think maybe it is in 2D. The problem ask to prove about area so we should find some relations of the cross product that I couldn't. And for the picture I don't know how to draw it too. Please help


问:
Necessary equation for envelopes
Consider a family of curves parameterised by $t$, where each member of the family is described by
$$F(t,x,y) = 0.$$
Define an envelope of the family to be a curve $E$ such that every point on $E$ is tangent to some member of the family. When do we have that
$$\frac{\partial F}{\partial t}(t,x,y) = 0$$
for all $(x,y) \in E$, and in those cases, why?</p>

答:
Yes, the envelope $E$ is precisely the solution set of the system
\begin{align*}
F(t,x,y) &= 0 \\
\frac{\partial F}{\partial t}(t,x,y) &= 0\,,
\end{align*}
if we have a sufficient condition given by the Implicit Function Theorem. That is, in order for these equations to be guaranteed to define $(x,y)$ locally as a $C^1$ function of $t$, we need the $2\times 2$ matrix
$$\begin{bmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\
\frac{\partial^2 F}{\partial x\partial t} & \frac{\partial^2 F}{\partial y\partial t}
\end{bmatrix}$$
to be nonsingular.

Here's a hint to explain where the condition $\frac{\partial F}{\partial t}=0$ comes from. At the point $(x(t),y(t))\in E$ the tangent vector to $E$ should be tangent to the curve
$$F(x,y,t)=0\tag{$\star$}$$
(with $t$ fixed), hence orthogonal to the gradient vector $\big(\frac{\partial F}{\partial x}(x(t),y(t)),\frac{\partial F}{\partial y}(x(t),y(t))\big)$. Now differentiate equation $(\star)$ with $t$ varying and $x=x(t)$, $y=y(t)$.

Title: Necessary equation for envelopes


问:
Deducing a topology given its continuous functions.
Given a topology $\tau$ for a set $S$, one can easily check if some functions are continuous between some subsets $A,B \subseteq S$, inheriting a topology from $\tau$ to both. My question is if/how I can do the other way around. Choosing the kind of function $f: A \rightarrow B$ that I want to be continuous between $(A, \tau|_A)$ and $(B, \tau|_B)$, how can i construct such $\tau$?

To help understand what I mean, I'll give an example. The continous deformations (not creating holes or glueing point) can be checked to be continuous under the standard topology in $\mathbb{R}$. But suppose I don't know the existence of the standard topology, can I deduce it just by deciding that I want the continuous deformations to be my continuous functions between topological spaces with such topology? If so, there is a "standard method" to deducing such topology?

答: For whatever map $f$, the trivial topology is a topology that will make $f$ continuous for any $A,B \subseteq S$. So I don't think that you can recover a specific topology. The trivial topology will always satisfy your requirement.


问:
Compact and weakly compact
In a Banach space,

is every compact set weakly compact set?

I guess its true just want to make sure.

Thank you!!

答:
The weak topology is coarser than the strong topology, therefore strong-compact subsets are weak-compact.

This is a purely topological fact: every cover by open sets from the coarser topology is a cover by open sets from the finer topology, and therefore a finite subcover exists a fortiori.

Title: Compact and weakly compact


问:
What is the difference between a set of vectors and a span
I was trying to understand spans and it got me even more confused.

My notes about span came about like this:

Let  $ S = \{ u_1, u_2... u_k \} $ be a set of vectors in $ \mathbb R^n$.

The set of all linear combination of $ u_1, u_2, .. u_k $

$$ \{ c_1u_1 + c_2u_2 + ... + c_ku_k  | c_1, c_2, ... c_k \} $$

is called a linear span of S (or the linear span of $ u_1, u_2, ... u_k $) and is denoted by $ span(S) $  (or $ span( u_1, u_2, ... , u_k) $

So what is the difference when $ S = \{u_1, u_2... u_k\} $ vs $ span( u_1, u_2... u_k)  $ or $span(S)$ in terms of linear combinations - Isnt $S$ also a set of all linear combinations?

答:
The span is always on a set of vectors and not on ‘a single’ vector. By the end of this response, it will be clear why span is defined on a set of vectors and not on a single vector.

Let me try to explain with an analogy from simple mathematics.
Addition of 2 numbers can modelled with an equation ‘A + B = C’.
where A,B,C can be any number between ‘-infinity to +infinity’(including zero).
Lets try calculate addition for various values of A & B and capture the result in a table as shown below.

In the above table, all possible values of C(highlighted in yellow colour) is the span for all possible values of numbers A, B.
We can define ‘Span’ for the addition of 2 numbers as all possible values of C that can be obtained for all possible values of A + B where A,B,C can be any values ranging between -infinity and +infinity.

Lets extend the above analogy for Vectors.

You are buying cheese and butter from you local super market.
Cheese costs and weighs 5 dollars and 2 pounds per box.
Butter costs and weighs 8 dollars and 3 pounds per box.
You can buy any number of cheese and butter boxes. As an individual you might buy lesser quantities whereas if you are purchasing for a restaurant you might buy in large quantities.

The Vector equations for the above scenario are as follows
Cheese Vector = ( 5$ per box, 2 lbs per box) = C( 5, 2)
Butter Vector = ( 8$ per box, 3 lbs per box) = B(8, 3)
Resultant Vector( when you buy both) = Cheese Vector + Butter Vector = ( price of Cheese per box + price of Butter per box , Weight of Cheese = Weight of Butter) = R ( 5$ + 8 $, 2lbs + 3 lbs) = R(13, 5)

So the vector equations are
C(5,2) + B(8,3) = R(13,5).
Where C,B,R are the coefficients or scalars for the number of boxes purchased.

Depending your lucky day(Black Friday or special promotion etc.), you might get one as free if you buy the other one. So B or C can be represented as negative
Lets try calculate addition for various values of C & B and capture the result in a table as shown below.

In the above table, all possible values of Resultant Vector-R (highlighted in yellow colour) is the span of Cheese and Butter vectors.

Since we are interested only in 2 dimensions(price and weight), the span(possible values of R) is the entire 2-D plane where price and weight form each of the dimensions or axes. i.e. R vector can be calculated or added or placed as a coordinate points with in the 2D plane.

If we were interested in 3 dimensions(price, weight, tax) then the span(possible values of R) is the entire 3-D plane where price, weight and taxes form each of the dimensions or axes.

For B,C we can very well have decimal values as well if your local super market allows i.e. buy one and half box of cheese, two and half box of butter etc.
That's all to it!!!

So the ‘Span’ for buying of cheese and butter(read as Vector space of cheese and butter vectors) is all the possible values of R that can be obtained for all possible values of C, B where B,C can be any values ranging between -infinity and +infinity.
As it is evident, span is on both cheese and butter vectors and not on only one. if it is only one, the usual number line arithmetic itself is sufficient.
Hope this is clear.


问:
What's the difference between 'for any' and 'for all'?
There are lots of expressions like, for all x, for any x, for some x, etc.

It may be because of the fact that I am not a native English speaker, but I can't tell the difference between 'for all' and 'for any'.

(I think 'for some x' means 'there exists at least one point', right?)

Could you explain it? thank you!

答:
Your interpretation of "for some" is correct.

There is no logical difference between "for all" and "for any", they both mean $\forall$.
For any number n in the integers, there is always a number bigger than it.
For all numbers in the integers, there is always a number bigger than it.

The difference language-wise is that your sentence must stick to the plural form of nouns when using "for all", while you may use singular nouns when using "for any".


问:
Random variables with zero mean always equal zero. But do random variables with infinite mean never equal zero?
If I have a random variable $X \in \mathbb{N}$, with arbitrarily large mean, is it trivial that $P(X=0)=0$?

I am able to prove that when its mean is zero, it takes the value zero with probability one, using e.g. the Markov inequality, but not the other way around.

I can provide exact expressions for its moments as a function of some parameter $n$, but in the limit $n \to \infty$, these moments all diverge.

As such, the variance is infinite, so it typically varies "infinitely" from its infinite mean. But can it vary all the way to $0$?

答:
By $X \in \mathbb N$, I am assuming you mean $X$ is a discrete random variable with support among the natural numbers.

No, it is not necessary for $P(X = 0)$ even if $X$ has infinite mean. Here is a counterexample: Define the probability mass function $p$ of $X$ as:
$$
p(x) = \begin{cases} \frac{1}{2}, &\text{ if } x = 0 \\
\frac{3}{\pi^2 x^2}, &\text{ if } x \in \mathbb Z_+
\end{cases}
$$ and $0$ everywhere else of course. Obviously $P(X = 0) = p(0) = 1/2 \neq 0$ by definition. But is it a valid pmf? Yes:
$$
\sum_{x = 0}^\infty p(x) = \frac{1}{2} + \frac{3}{\pi^2} \sum_{x = 1}^\infty \frac{1}{x^2} = \frac{1}{2} + \frac{3}{\pi^2} \cdot \frac{\pi^2}{6} = \frac{1}{2} + \frac{1}{2} = 1
$$ And, its mean is infinite:
$$
E[X] = \sum_{x = 0}^\infty xp(x) = \frac{3}{\pi^2} \sum_{x = 1}^\infty x \cdot \frac{1}{x^2} = \frac{3}{\pi^2} \sum_{x = 1}^\infty \frac{1}{x} = \infty
$$ because the harmonic series diverges to $\infty$.


问:
Integrate by change of order
$$ \int_{0}^{1}\int_{x}^{\sqrt{2-x^2}} \frac{x}{\sqrt{x^2 + y^2}}  \,dydx $$
tried by splitting the region in 2 where the line $$y = x $$cuts the curve $$y = \sqrt{2-x^2}$$
unable to find the limits

答:
Hint.

Try polar coordinates. The region

$$
D=\{(r,\theta): 0\le r\le \sqrt{2}, \pi/4\le \theta\le \pi/2\}
$$

So the integral is
$$
\int_{\pi/4}^{\pi/2}\int_0^{\sqrt{2}}(\cos\theta)\cdot r\ drd\theta
$$

If you only want to change the order of the integrals, your integral is

$$
\int_0^1\int_0^yf(x,y)dxdy+\int_{1}^{\sqrt{2}}\int_0^{\sqrt{2-y^2}}f(x,y)dxdy
$$

Basically, you split the domain into two pieces. The lower piece gives you the first integral and the upper piece gives you the second one.

Title: Integrate by change of order


答:
I mean a really trivial thing could be to choose the discrete topology.

But let me describe a  more common scenario : imagine we have a map $f:A\to B$ (note to define a map all we need is the sets $A,B$). Further imagine we know that we have a topology on $B$, call it $\tau_B$, then

$$
A \overset{f}{\longrightarrow} (B,\tau_B),
$$

now how does one construct a topology on $A$, call it $\tau_A$ so that $f$ is continuous? Well its common to construct it exactly like that i.e

$$
\tau_A:=\{ M\subset A~:~f^{-1}(N)\in \tau_B \},
$$
then for sure $f$ is continuous, i.e this is the coarsest topology one needs for continuity of $f$. $\textbf{Of course}$ one still needs to check $\tau_A$ is a topology, it might not always be!

Title: Deducing a topology given its continuous functions.

Body:
Given a topology $\tau$ for a set $S$, one can easily check if some functions are continuous between some subsets $A,B \subseteq S$, inheriting a topology from $\tau$ to both. My question is if/how I can do the other way around. Choosing the kind of function $f: A \rightarrow B$ that I want to be continuous between $(A, \tau|_A)$ and $(B, \tau|_B)$, how can i construct such $\tau$?

To help understand what I mean, I'll give an example. The continous deformations (not creating holes or glueing point) can be checked to be continuous under the standard topology in $\mathbb{R}$. But suppose I don't know the existence of the standard topology, can I deduce it just by deciding that I want the continuous deformations to be my continuous functions between topological spaces with such topology? If so, there is a "standard method" to deducing such topology?


问:
Reference for Lemma 2 on Mean Value Theorem Wiki
I'm looking for a reference for Lemma 2 on the Mean Value Theorem Wikipedia page. I've also stated it below:

Lemma 2

Let $v:[a,b]\rightarrow\mathbb{R}^m$ be a continuous function on $[a,b]$, then we have that
$$ \left\Vert\int_a^b v(t)\,dt\right\Vert\leq\int_a^b\Vert v(t)\Vert dt. $$

答:
Edit: In my first answer, I overlooked the multi-dimensional character of your question. It still seems quite standard.

One possible reference: Rudin, Principles of Mathematical Analysis, Theorem 6.25


问:
A unique circle with 3 points proof
I have prove the theorem: There is only one circle passing through three given non-collinear points in both geometrical and algebraic ways.
THere is one question that I just have no idea with.
'the accuracy and limitations of this technique"

I dont know what to write...! thank you

答:
In the geometrical way, you can use the fact that in a circle, if you draw a line between any two points in the circle and you bisect that line, the bisected line will go through the center of the circle. So if you have three points, you can draw two bisects and they will intersect at a unique point, which will be the center of your circle. To determine its radius, just compute the distance from your center to any of your three points.

In the algebraic way, you have three points $(x_1, y_1), (x_2, y_2), (x_3,y_3)$ such that
$$
(x_i - a)^2 + (y_i - b)^2 = c^2, \qquad i=1,2,3
$$
which gives you three equations in three unknowns, namely $a,b,c$ (the coordinates of the center $(a,b)$ and the radius $c$). You can use one of the equations to reduce the other two to linear equations and solve the linear system to get the coordinates $a$ and $b$. Once you have those two, any of the three equations will give you $c$.

Hope that helps,

Title: A unique circle with 3 points proof


问:
Generalized max function that chooses N largest items.
Let $X$ be a list of real numbers. The length of $X$ is finite, i.e., $|X|=m<\infty$.

The largest element of $X$ can be denoted by $\max X$.

My question is

Is there any mathematical notation that represents $n\leq m$ largest elements of $X$? or is there any mathematical way to represent the sublist composed of the $n$ largest elements of $X$?

答: Yes. If you use a subscript in parentheses, that indicates order.  For example, if $X_1 = 5, X_2 = 7, X_3 = 1, X_4 = 6$, then $X_{(1)} = 1, X_{(2)} = 5, X_{(3)} = 6,$ and $X_{(4)} = 7$. The notation is used for order statistics in probability theory. So the largest k items in a list of N times would be $X_{(n-k+1)}, X_{(n-k+2)}, ..., X_{(n)}.$


问:
Green function of the Laplace-Dirichlet operator in a bounded domain
Let $\Omega \subset \mathbb{R}^d$, $d = 1,2,3$, be a bounded domain with $\partial \Omega$ smooth. Let $$A = -\Delta \colon H_0^1(\Omega) \cap H^2(\Omega) \to L^2(\Omega)$$be the Laplace-Dirichlet operator. I am looking for a reference that assures the existence of a Green operator for $A$, namely $G \colon L^2(\Omega) \to H_0^1(\Omega) \cap H^2(\Omega)$ such that $$v = Gf \Longleftrightarrow Av = f.$$

In addition, what do we know about it? I am particularly interested in compactness of this Green operator

答:
Consider the problem of find $u\in H_0^1$ such that $$\tag{1}\int_\Omega\nabla u\nabla v=\int_\Omega fv,\ \forall\ v\in H_0^1$$

There is a lot of ways to solve this problems, to wit, one can use Lax-Milgram theorem or Calculus of Variation or Monotone Methods operators and so on. Let's try the second one (because I really like it more than others).

Define $I:H_0^1\to\mathbb{R}$ by $$I(u)=\frac{1}{2}\int_\Omega |\nabla u|^2-\int_\Omega fu$$

Note that $I$ is the energy functional associated to the problem $(1)$, i.e. $$\langle I'(u),v\rangle = \int_\Omega\nabla u\nabla v-\int_\Omega fv, \forall\ u,v\in H_0^1$$

where $\langle \cdot,\cdot\rangle$ denotes duality between $H_0^1$ and $H^{-1}$. Hence, to solve $(1)$ is equivalently to find $u\in H_0^1$ such that $$\langle I'(u),v\rangle =0,\ \forall\ v\in H_0^1$$

The last equality is saying that $u$ is a critical point of $I$. Hence, let's try to find a critical point of $I$ (in fact a minimum point).

Note that (by using Holder and Poincare inequality) there exist a constant $C>0$ such that $$\tag{2}I(u)\geq \frac{1}{2}\|u\|_{1,2}^2-C\|u\|_{1,2}$$

From $(2)$ we conclude that $I$ is coercive ($I(u)\to\infty$ if $\|u\|_{1,2}\to\infty$). Now let's prove that $I$ is lower semicontinuous in the weak topology. Let $u_n\to u$ in the weak topology. On one hand we have that $\|\cdot\|_{1,2}$ is lower semicontinuous in the weak topology. On other hand, note that by the compact immersion of $H_0^1$ in $L^2$, we can suppose that (up to a subsequence) $u_n\to u$ strongly in $L^2$, hence, by using Lebesgue theorem, we can conclude that $\int_\Omega fu_n\to \int_\Omega fu$. Therefore $I$ is weak lower semicontinuous.

By using the fact that $I$ is coercive and weak lower semicontinuous, we can conclude that there exist $u\in H_0^1$ such that $I(u)\leq I(v)$ for every $v\in H_0^1$. Because $I$ is strictly convex, we also have that $u$ is unique, hence we can associate to $(1)$ the Solution operator or Green operator $G$, i.e. if $G(f)=u$ then $u$ satisfies $(1)$. Note that by regularizing the solution (see Brezis chapter 9) we can suppose that $u\in H_0^1\cap H^2$, therefore $$G:L^2\to H_0^1\cap H^2$$

The first thinkg we can cnote about $G$ is it's continuity: indeed, by taking $v=u=G(f)$ in $(1)$ we have by Holder and Poincare inequality that $$\|G(f)\|_{1,2}\leq C\|f\|_2$$

On the other hand, because $H_0^1\cap H^2$ is compactly embedded in $L^2$, we can conclude that $$G:L^2\to L^2$$

is a compact operator.

Remark: You can also show that $G$ is positive and symmetric.

Remark 1: All that I have said here can be found for example in: Brezis chapter 9, Evans chapter 6 and so on...


问:
How do I understand $T_p(\mathbb{R^n})\cong D_p(\mathbb{R}^n)$ intuitively?
This might be a stupid question, but I'm reading Tu's smooth manifold chapter 8 and still couldn't figure out what the tangent vector means and what the above isomorphism tells us intuitively in Euclidean space (so that I can also apply this intuition in the case of manifolds). For me, tangent space at some point $p$ in some object(manifold) is like a tangential line or a plane at that point that intersects trivially. So I want to know, is the "tangent vector" in this context something we can actually calculate for a given manifold? For example, if I'm given a manifold $M$ and the charts, or the equation for $n$ dimensional Euclidean object, we can concretely get n-1 dimensional (please correct me if I'm wrong) tangent "vectors"? And how can I understand these vectors are isomorphic to some differential operators that take some function and spits out real value?

Is it the right way to understand?

答:
For any finite dimensional vector space we have $T_p(V) \cong V$. To see this, for $v\in V$ and $f\in C^\infty (V)$ define $$D_v(f):=\left.\frac{d}{dt}f(p+tv)\right|_{t=0},$$ which is just directional derivative of $f$ at point $p$ in direction $v$.

You can check that $D_v$ is linear functional on $C^\infty(V)$ since by the chain rule we have $$\left.\frac{d}{dt}f(p+tv)\right|_{t=0} = \nabla f(p)\cdot v = \sum v^i \frac{\partial f}{\partial x^i}(p),$$ where $v^i$ are the coordinates of $v$ in standard basis.

Also, $$D_v(fg) = D_v(f)g(p) + f(p)D_v(g),$$ so $D_v$ is a derivation at point $p$, that is $D_v\in T_p(V)$.

It turns out that $v\mapsto D_v\colon V\to T_p(V)$ is isomorphism of vector spaces.
To check injectivity, assume $D_v$ is $0$ and act on coordinate function $x^j\colon V \to \mathbb R$ to get $0 = D_v(x^j) = v^j,$ for all $j = 1,\ldots,n$. So, $v = 0$. Since the dimensions coincide, the map is isomorphism.

Hopefully, this gives enough intuition why the abstract definition of tangent vectors  is a good generalization of familiar definitions in $\mathbb R^n$ (or any real finite dimensional vector space).


问:
Expressing $(5.8\sqrt{40} + 56.4) - (5.8\sqrt{10} + 56.4)$ in simplified radical form
I am trying to complete the question below, but I am not sure how to simplify the radical. What I have so far is

$$(5.8\sqrt{40} + 56.4) - (5.8\sqrt{10} + 56.4) \;=\; \text{the difference}$$

How does one simplify this expression without a calculator?

The formula $E = 5.8\sqrt{x} + 56.4$ models the projected number of elderly Americans ages $65$-$84$, $E$, in millions, $x$ years after $2020$.

a. Use the formula to find the projected increase in number of Americans ages $65.84$, in millions from $2030$ to $2060$. Express this difference in simplified radical form.

b. Use a calculator and write your answer in part (a) to the nearest tenth.

答:
Answer from the comments.

Since $\sqrt{40} = \sqrt{4} \sqrt{10} = 2 \sqrt{10}$, the difference is:

$$5.8 \cdot 2 \sqrt{10} - 5.8 \sqrt{10} = 5.8\sqrt{10} (2-1) = 5.8\sqrt{10}$$

which is $18.3$ million, rounding down to the nearest tenth (of a million).

This answer is the right order of magnitude as there are around $330$ million Americans, so the change in this age group should be an order of magnitude less.


问:
1985 USAMO #4 (Graph Theory)
There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\lfloor n/2\rfloor -1$ of them, each of whom knows both or else knows neither of the two. Assume that "know" is a symmetrical relation; $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.

I saw a solution: Consider the number of pairs $(X, \{Y, Z\})$, where $X, Y, Z$ are distinct points such that $X$ is joined to just one of $Y, Z$. If $X$ is joined to just $k$ points, then there are just $k(n - 1 - k) \leq (n - 1)^2/4$ such pairs $(X, \{Y, Z\})$. Hence in total there are at most $n(n - 1)^2/4$ such pairs. But there are $n(n - 1)/2$ possible $\{Y, Z\}$. So we must be able to find one of them $\{A, B\}$ which belongs to at most $[ (n - 1)/2 ]$ such pairs. Hence there are at least $n - 2 - [ (n - 1)/2 ] = [n/2] - 1$ points $X$ which are joined to both of $A$ and $B$ or to neither of $A$ and $B$. [If confused by the $[ ]$, consider $n = 2m$ and $n = 2m+1$ separately. But I don't understand the following inequality $k(n - 1 - k) \leq (n - 1)^2/4$. Someone can help me ou have other solution, please?

Thanks a lot!

答: Since $X$ is joined to $k$ points, to choose the set $\{Y,Z\}$, we must choose one point $X$ is joined to, and one point $X$ is not joined to. There are $k$ choices for the first point and $n-1-k$ choices for the second. Hence the number of pairs $\{Y,Z\}$ is equal to $k(n-1-k)$, which by the inequality $a(2b-a)\le b^2$, cannot exceed $(n-1)^2/4$.

Title: 1985 USAMO #4 (Graph Theory)


问:
Properties of mollifiers $L^p_{loc}$ convergence proof (Evan's)
I have a question on the proof of Evan's PDE Appendix C.5.Theorem 7(Properties of mollifiers), showing $f^{\epsilon} \to f$ in  $L_{loc}^p(U).$ The proof starts by choosing open sets compactly contained in each so that $V \subset\subset W \subset\subset U$ and showing $\|f^{\epsilon}\|_{L^p(V)} \leq \|f\|_{L^p(W)}.$

After showing $|f^{\epsilon}(x)|^p \leq \int_{B(x,\epsilon)} \eta_{\epsilon}(x-y)|f(y)|^p dy$ on $V$ which I attached in the link lllll , Evan's proceeds as
$$ \begin{align*} \int_{V} |f^{\epsilon}(x)|^p dx &\leq \int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon} (x-y)|f(y)|^p dy\right) dx\\
& \leq \int_W |f(y)|^p \left(\int_{B(y,\epsilon)} \eta_{\epsilon} (x-y) dx\right) dy = \int_W |f(y)|^p dy,
\end{align*}$$
provided $\epsilon > 0 $ is sufficiently small.

The first inequality comes from the part $|f^{\epsilon}(x)|^p \leq \int_{B(x,\epsilon)} \eta_{\epsilon}(x-y)|f(y)|^p dy$ on $V,$ and $\int_{B(y,\epsilon)} \eta_{\epsilon} (x-y) dx = 1,$ so the last inequality follows, but

Firstly, is my understanding correct about the second inequality? Here $f$ is locally integrable (i.e. $f \in L^p_{loc}(U)$), so we are simply interchanging the order of integrals by Fubini's theorem, and then the inequality comes in from expanding the space $V$ to $W,$ right? Then why would we want to work on $W$ in this proof? The later proof uses continuous approximation of $f$ on $W$ to obtain $\lim \sup_{\epsilon \to 0} \|f^{\epsilon} - f\|_{L^p(V)}$ is arbitrarily small, which concludes the proof, but doesn't seem to need $W$ if it weren't for using the inequality $\|f^{\epsilon}\|_{L^p(V)} \leq \|f\|_{L^p(W)}.$

Further, where do we need $\epsilon$ to be sufficiently small?

答:
In the first place, you are totally right about the second inequality; if we consider the integral
$$ \int_{V} \left( \int_{B(x, \epsilon)} \eta_{\epsilon}(x-y)|f(y)|^{p} dy\right) dx,$$
we can identify that the integrand is a non-negative function and since we are dealing with $\sigma$-finite measures (Lebesgue measure on $\mathbb{R}^{n}$) all the hypotheses of Fubini's theorem are fulfilled and then we can change the order of integration. Moreover, since $V \subset W$ we also have
$$  \int_{V} \left( \int_{B(x, \epsilon)} \eta_{\epsilon}(x-y)|f(y)|^{p} dy\right) dx \leq  \int_{W} \left( \int_{B(x, \epsilon)} \eta_{\epsilon}(x-y)|f(y)|^{p} dy\right) dx,$$
so, combining this with Fubini's theorem the second inequality is obtained.

Now, I think that Evans introduce the open set $V \subset \subset W$ just because it gives us an edge in the last part of the proof. In the book Evans presents the following estimate: He first fixes $\delta >0$ and then chooses $g \in C(V)$ such that $\|f-g \|_{L^{p}(V)} < \delta$  (which is possible since $C(V)$ is dense in $L^{p}(V)$) then he states that
$$ \|f^{\epsilon} - f\|_{L^{p}(V)} \leq \color{blue}{{\|f^{\epsilon} - g^{\epsilon}\|_{L^{p}(V)}}} +  \|g^{\epsilon} - g\|_{L^{p}(V)}+\|g - f\|_{L^{p}(V)}. $$
Observe that we have "control" over the second and third term of this expression: $g^{\epsilon}$ converges uniformly to $g$ on $V$
and  $ \|g-f \|_{L^{p}(V)} < \delta.$ However, we don't know a priori if the term $\|f^{\epsilon}-g^{\epsilon} \|_{L^{p}(V)}$ will tend to zero. Nevertheless, if we use the previous estimate,
$$\| f^{\epsilon}-g^{\epsilon} \|_{L^{p}(V)} \leq \| f-g \|_{L^{p}(V)}   $$
Then it becomes clear that we also have control over the first term. I consider that the use of the set $W$ is merely to give a convenient way to estimate the last inequality.

Finally, we do need to choose $\epsilon$ small enough to be sure that each convolution is well defined. Recall that when we are defining the convolution of a function $f: U \longrightarrow \mathbb{R}$ we actually define it on the set $U^{\epsilon} = \{ x \in U \hspace{0.1 cm} | \hspace{0.1 cm} dist(x,\partial U)> \epsilon\}$ the thing is that
$$f^{\epsilon}(x) = \int_{U} \eta_{\epsilon}(x-y)f(y)dy = \int_{B(0,\epsilon)} \eta_{\epsilon}(y)\color{red}{f(x-y)}dy \hspace{1 cm} (x \in U^{\epsilon})$$
and we need to be sure that the term in red is well defined, i.e.  that $x-y \in U$ for any $y \in B(0,\epsilon)$. In the specific context of the proof, you may have something like this:

So if $\epsilon$ is not small enough, you can not even guarantee that the convolution is well defined.


问:
Question about the proof of $e$ irrationality
I found an article with the following elementary proof of $e$ irrationality
https://people.ohio.edu/diao/papers/Irrationality%20of%20e.pdf.
But I am having trouble understanding one step.

It states that, from:

\begin{equation}
 \frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}
\end{equation}

follows:

\begin{equation}
 q_n \geq q_{n+1}
\end{equation}

I can not understand why this is so. Can anyone explain in more detail?

答: Both fractions$$\frac{p_{n+1}}{q_{n+1}}\quad\text{and}\quad\frac{np_n-q_n}{q_n}$$are equal to the same rational number, but the first one is an irreducible fraction. Therefore, and since both numerators and both denominators are natural numbers, $np_n-q_n\geqslant p_{n+1}$ and $q_n\geqslant q_{n+1}$. Actually, not only we have $q_n\geqslant q_{n+1}$ as, in fact, $q_n=Nq_{n+1}$ for some $N\in\Bbb N$.


问:
Question involving relation between directional derivative and the gradient of a function
Prove or disprove the following statement:
Suppose a function has all its directional derivatives $$D_\hat{p}f$$ exist at a point $$(a,b)$$, then $$D_\hat{p}f(a, b) = (\nabla f)(a,b)\cdot \hat{p}$$

I think that this is false and hence need to provide a counterexample. Could you give an example such that
the directional derivative exist but the gradient doesn't?

答:
The gradient consists of directional derivatives. Hence, there is no example for what you are asking.

However, what you are actually looking for is a function which has all directional derivatives, such that the formula
\begin{equation}\tag{$\star$}D_{\hat p}f (a,b)=\nabla f(a,b)\cdot\hat p\end{equation}
is not satisfied.

A possible example is the function $$ f(x,y)=\begin{cases}\dfrac{xy^2}{x^2+y^4} & \mbox{for}\ (x,y)\ne(0,0)\\ 0 &\mbox{for}\ (x,y)=(0,0).\end{cases}$$
One can easily check the following properties:

The partial derivative of $f$ exist and vanish at $(0,0)$, hence $\nabla f(0,0)=0$.

All directional derivatives at $(0,0)$ exist. If $\hat p=(p_x,p_y)\ne (0,1)$ we have
$$ D_{\hat p}(0,0)=\frac{p_y^2}{p_x}. $$

Hence, $(\star)$ can't hold, since $\nabla f(0,0)\cdot \hat{p} = 0$ for any $\hat p$.

Edit: The statements follow directly from below calculation:

 For $\hat p = (p_x,p_y)$ we have $$ D_{\hat p} f(0,0)=\lim_{\epsilon\to 0} \frac{f(\epsilon p_x,\epsilon p_y)-f(0,0)}{\epsilon} = \lim_{\epsilon\to 0}\frac{p_xp_y^2}{p_x^2+\epsilon^2p_y^4}.$$


答:
If (as in the paper you cite) you have:

$$
 \frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}
$$

where $p_{n+1}$ and $q_{n+1}$ are coprime, you must have
$$
q_{n+1} = \frac{q_n}{\gcd(np_n - q_n, q_n)} \le q_n
$$

Title: Question about the proof of $e$ irrationality

Body:
I found an article with the following elementary proof of $e$ irrationality
https://people.ohio.edu/diao/papers/Irrationality%20of%20e.pdf.
But I am having trouble understanding one step.

It states that, from:

\begin{equation}
 \frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}
\end{equation}

follows:

\begin{equation}
 q_n \geq q_{n+1}
\end{equation}

I can not understand why this is so. Can anyone explain in more detail?