问:
Nyquist diagram not encircle the origin
Given the function: $\frac{s-5}{s^2-4s +5}$

The function have one zero and two poles in the RHP. So according to principle of argument (or Nyquist diagram), the Nyquist plot encircle the origin N = |Z-P| = |1-2| = 1 time. But the Nyquist plot does not show that:

What am I missing here?

答:
It is not entirely clear what you are trying to do.

When you apply the argument principle, the contour must not pass through any of the poles or zeros.

In this case, $g(s)= {s-5 \over s^2-4s+5}$ has a zero at infinity and so the curve 'passes' through zero.

Since you are interested in the difference of right half plane poles & zeros, the contour can be modified to be $0 \to iR$, then $R e^{i\theta}$ for $\theta$ going from ${\pi \over 2}$ to $-{\pi \over 2}$ and then from $-iR$ back to zero, for some large $R$. If $R$ is sufficiently large, then the curve will pass to the right of
$0$.

Note that the winding number of the right half plane poles & zeros is $-1$ with
this curve, and the contour had a winding number of $+1$ around the origin, so the argument principle tells us that
$1 = \sum_{z} \eta(z, \gamma) -\sum_{p} \eta(p, \gamma) = \#\text{poles} - \#\text{zeroes}$.

However...

If you are trying to apply the Nyquist stability criterion to $g$, then
you are trying to determine the stability of the closed loop transfer function ${g \over 1+g}$ where $g$ is the open loop transfer function (assuming unity feedback). In this case, the system is stable if $1+g$ has no right half plane zeros.

So, either plot $1+g$ and count encirclements of zero, or plot $g$ and count encirclement of $-1$.

If this is what you are trying to go then the issue of passing through zero does not arise here.

Title: Nyquist diagram not encircle the origin


问:
For $\lambda(A)>0$, prove $\exists B\subset A\ \ni0<\lambda(B)<\lambda(A)$ where $\lambda$ is Lebesgue Measure.
Suppose $\lambda(A)=\infty$

Claim: $\lambda([a,b]\cap A)>0$ for some $a<b$

Assume on contrary that $\lambda([a,b]\cap A)=0$ for all $a<b$.

Then $[-n,n]\cap A\nearrow A$. This implies
$$
\lambda(A)=\lim\limits_{n\to\infty}\lambda( [-n,n]\cap A)=0
$$
It contradicts our assumption $\lambda(A)=\infty$.

Hence, $\lambda([a,b]\cap A)>0$ for some $a<b$.

Take $B=[a,b]\cap A\subset A$, then $0<\lambda(B)<\infty=\lambda(A)$.

Now suppose $0<\lambda(A)<\infty$

There is $x\notin A$, so $A\subset \mathbb{R}\setminus \{x\}$.

Now $[x-1/n,x+1/n]^c \cap A\nearrow \mathbb{R}\setminus \{x\}\cap A=A$

So, $\lim\limits_{n\to\infty} \lambda([x-1/n,x+1/n]^c \cap A)=\lambda(A)>0$. This implies there exists $n\in \mathbb{N}$ such that $\lambda([x-1/n,x+1/n]^c \cap A)>0$, and thus $\lambda([x-1/n,x+1/n] \cap A)<\lambda(A)$ as $\lambda([x-1/n,x+1/n]^c \cap A)+\lambda([x-1/n,x+1/n] \cap A)=\lambda(A)$.

I call $B_1=[x-1/n,x+1/n]^c\cap A$ and $B_2=[x-1/n,x+1/n]\cap A$. Then
$0<\lambda(B_1)\le \lambda(A)$ and $0\le \lambda(B_2)<\lambda(A)$.

But I am unable to get a $B\subset A$ such that $0<\lambda(B)<\lambda(A)$.

Can anyone help me find a way out from this? Thanks for help in advance.

答:
First suppose $\lambda (A) <\infty$. Let $f(x)=\lambda ((-\infty,x )\cap A)$. Verify that $|f(x)-f(y)| \leq \lambda [x,y) =y-x$ for $x <y$. Conclude that $f$ is a  continuous fucntion. Now check that $f(x) \to 0$ as $ x \to -\infty$ and $f(x) \to \lambda (A) >0$ as $x \to \infty$. Use IVP of continuous functions.

For the general case observe that there exist $N$ such that $0<\lambda ((-N,N)\cap A) <\infty$ and apply the first case to $(-N,N)\cap A$

ContentLicense: CC BY-SA 4.0


问:
Find the point on the line $y = 2x + 1$ that is closest to the point $(5, 2)$?
The question is: Find the point on the line $y = 2x + 1$ that is closest to the point $(5, 2)$.
I know the high (or middle) school method. Create a line that is perpendicular to this line, so the slope will be $-1/2$. Then the intersection is the answer.

But I meet this question from a linear algebra textbook. I want to use projection onto a line method. Projecting vector $b$ onto vector $a$ with the error $e=b-\hat{x}a$.

Thus, $a \cdot (b-\hat{x}a)=0$ or $a\cdot b-\hat{x}a\cdot a=0$.
Thus, $\hat{x}=\frac{a\cdot b}{a\cdot a}$

But the problem is that here $a$ and $b$ are pure vectors. My question is given a specific line. Then I don't know how to do with it.

答:
Let the point on the line which is closest to (5, 2) be $(x, 2x + 1)$.

Let a be a vector from (0,1) to $(x, 2x + 1)$, so a $= (x, 2x).$

Let b be a vector from (2,5) to $(x, 2x + 1)$, so b $= (x-5, 2x-1).$

a $\cdot$ b = 0$ (since a and b must be perpendicular to each other)

$x(x-5) + 2x(2x - 1) = 0$

$5x^2 - 7x = 0$

$x = 0$ or $x = \frac{7}{5}$

But $x \neq 0$ so $x = \frac{7}{5}$

Sub $x = \frac{7}{5}$ into $y = 2x + 1$ to get $y = \frac{19}{5}$.


问:
Trouble with gradient intuition
I'm currently learning about gradients, and I thought khanacademy could help me acquiring some intuition. The actual computation is clear to me, however I'm having trouble understand the intuition.

This is the video I'm talking about, and the part I have a part with is 6:11. He says the gradient vector shows the direction you have to travel in the x,y-plane in order to get a maximum slope in the z-direction.

This sounds like gibberish to me. How do you have to 'travel' it? Why do you get a maximum slope? I just have no clue.

答:
Gradient intuition.
I had problems too trying to visualize intuitively the gradient in R 3.

I want to share how I finally got it.

For simplicity let us consider a function valued at x=0, y=0 that takes a value z=0 at that point. The shape of this function does not matter because we will look at the tangent plane at x=0, y=0, z=0. Let us also assume that the partial derivative with respect to x at x=0 is 1 and that the partial derivative with respect to y is also 1.
Let us now look at the tangent plane defined by the two partial derivatives.
(The tangent lines).
The partial derivative with respect to x will bisect the z x plane with a slope of 45 degrees. Similarly, the partial derivative with respect to y will bisect the z y plane with a slope of 45 degrees too.

So these two lines define our tangent plane. This tangent touches our z function at z=0.

Now imagine to project a plane, parallel to the x y plane that that intersects the z axis at z=1.

This plane parallel to the x y plane will intersect the tangent plane ( the one mentioned above defined by the two partial derivatives). The intersection is a line.
This line passes through the points  x=0, y=1, z=1 and x=1, y=0, z=1.
Now let us project this line on the x y plane. This projection is the contour line for z=1 on the x y plane.
This contour line (A) passes through the points x=1, y=0, z=0 and x=0, y=1, z=0.

Now, what is the shortest direction to take from the origin (the point where the tangent plane touches our function) to reach on the the x y plane the contour line for z=1 described above (A)?
It is obvious that the direction is 45 degrees, or the components of the the vector describing the direction are <1,1>. Same as the partial derivatives.
One can see that this vector is perpendicular to the contour line on the x y plane.

Now assume that the partial derivative with respect to y is 2, while the partial derivative with respect to x remains 1.

The tangent plane will shift. The plane parallel to the x y plane and with z= 1 will intersect the new tangent plane at points
x=1, y=0, z=1 and x=0, y=1/2, z=1.

The projection of this line on the x y will give us the new contour line for z=1.

This new contour line (B) will pass through the points
x=1, y=0, z=0 and x=0, y=1/2 z=0.

What is the direction to take now to reach from the origin this new contour line (B) on the x y plane?
The vector must be perpendicular to the new contour line (B) and so has components <1,2>, same numbers as the partial derivatives.

ContentLicense: CC BY-SA 3.0

Title: Trouble with gradient intuition


问:
Why is the integral value on complex domain?
$B_n=\{z \in C^n : |z|<1\}$ $,\quad$ $dv$ is volume measure on $B_n$

$\displaystyle \int_{B_n}z^{\alpha}\bar z^{\beta} dv(z)=0\;(?)$ when $\alpha \neq \beta$

how to prove it.

i learned multiple integral, so $\displaystyle \int \displaystyle \int_{D} f(z)dA=\displaystyle \int_{0}^{2\pi} \displaystyle \int_{0}^{1} f(z)dxdy$ style solution please...

答:
It is enough to show that the intergal w.r.t. $j-$th coordinate is $0$ where $j$ is chosen such that $\alpha _j \neq \beta_j$.  In other words the proof reduced to one-dimensaional case. So let $n=1$.

Write $z=re^{i\theta} $ with $r \geq 0, 0<\theta \leq 2\pi$ and note that the given integral becomes $\int_0^{1}\int_0^{2\pi} r^{\alpha} e^{i\alpha \theta} r^{\beta} e^{-i\beta \theta} d\theta dr$ and the inside integral is $e^{i(\alpha -\beta)\theta}|_0^{2\pi}$ which is zero if $\alpha \neq \beta$.


问:
Gaussian concentration of measure, equivalent definitions
I need some help going between two equivalent definitions.

First some notation :

$\bullet$ For $A\subseteq \mathcal{X}$ and $r>0$ define what is called the r-blowup of $A$ as \begin{equation}
A_r = \{ x \in \mathcal{X} : d(x,A)<r \}.
\end{equation}

$\bullet$ We have a metric probability space $(\mathcal{X},d,\mu)$, i.e $(\mathcal{X},d)$ is a Polish space, and $\mu$ a probability measure on its Borel sets.

Now I am going to give two definitions for  $\mu$ satisfying a Gaussian concentration of measure. Can anyone show how to go from one definition to the other (of course with a change of constants).

$\textbf{Def.1}$
$\mu$ is said to have Guassian concentration on $(\mathcal{X},d)$ if there exists $K,\kappa >0 $ such that

\begin{equation}\label{Gauss concentration}
\textit{whenever}~~\mu(A)\geq \frac{1}{2} ~~ \textit{it implies} ~~ \mu(A_r)\geq 1-K e^{-\kappa r^2}
\end{equation}

$\textbf{Def.2}$
We say $\mu$ has Gaussian concentration of measure if there exists $K,\kappa >0$ and some $r_0>0$, such that
\begin{equation}
\textit{whenever}~~\mu(A)\geq \frac{1}{2} ~~ \textit{it implies} ~~ \mu(A_r) \geq 1-Ke^{-\kappa(r-r_0)^2} ~~ \forall r\geq r_0
\end{equation}

[1, Page 103] claims that we can go between the two with a change of constants, how? Note going from $\textbf{Def.1}$  to
$\textbf{Def.2}$ is fine just replace $r$ by $r-r_0$.

[1] Raginsky, Sason. Concentration of Measure Inequalities in Information Theory, Communications and Coding. 2014

答:
As you mentioned, it is easy to go from Definition $1$ to Definition $2$.

Let $K$, $\kappa$, and $r_0$ be the constants in Definition 2. We will now find constants $K'$ and $\kappa'$ for Definition $1$. Set $K' := \max(K,1)e^{4 \kappa' r_0^2}$ and $\kappa':= \kappa/4$ to be the constants in Definition 1. We will show that these constants suffice by considering two cases:

Case 1: $r \in [0,2r_0]$. We have that
$$
K' \exp(- \kappa'r^2) = \max(K,1)\exp(\kappa'(4r^2_0 - r^2)) \geq 1, $$
and thus the desired inequality $\mu(A_r)\geq 1 - K' \exp(- r^2)$ holds trivially.

Case 2: $r \geq 2r_0$.
We have the following series of inequalities:
$$ \begin{aligned}1 - \mu(A_r) &\leq K e^{- \kappa (r - r_0)^2} \\
&\leq K e^{- \kappa r^2/4}\\
&\leq K'e^{- \kappa r^2/4} = K'e^{- \kappa' r^2},
\end{aligned}  $$
where we used that $K \leq K'$ and $|r - r_0| \geq r/2 $ for $r \geq 2r_0$.


问:
Can Strong Induction Be Used to Prove the Infinite Case?
I started real analysis and picked up a book called Understanding Analysis by Stephen Abbott. I understand why induction cannot be used prove the infinite case. From my understanding, induction proves that a statement is true for all natural numbers but not the infinite case because it only takes into consideration one case at a time (Is this correct?). Abbot uses the example $\bigcap^{n}_{i=1}A_i\ne \emptyset
$ while
$\bigcap^{\infty}_{i=1}A_i=\emptyset$ when $A_i$ is defined as a subset of $\mathbb{N}$ to show why induction would not work. Would strong induction suffice as a proof instead? What led me to this was how Abbott proved that for a set $A_n$=$\mathbb{N}\cap[n,\infty)$ and n $\in\mathbb{N}$, $\bigcap^{\infty}_{n=1}A_n=\emptyset$ using strong induction:

Suppose we had some natural number m which satisfies $m\in\bigcap^{\infty}_{n=1}A_n$. But since $m{\notin}A_{m+1}$, we have a contradiction so $\bigcap^{\infty}_{n=1}A_n=\emptyset$.

It seems as if strong induction uses all the previous cases to imply that the next case would be true so I was wondering if it would work for the infinite case. Apparently, transfinite induction extends to infinity which I learned from:

Why doesn't induction extend to infinity? (re: Fourier series)

and it seems very similar to strong induction such as the successor case which requires P($\alpha$) and sometimes P($\beta$) to be true for all $\beta\lt\alpha$ in order to prove P($\alpha$+1). This may be wrong though because I know nothing about ordinals. Here is the link: https://en.wikipedia.org/wiki/Transfinite_induction. Thank you for your time.

答:
The proof that $\bigcap_{n\ge 1}A_n=\varnothing$ when $A_n=\Bbb N\cap[n,\to)$ doesn’t use induction at all. It’s clear that the intersection is a subset of $\Bbb N$, since each of the sets $A_n$ is, so to show that the intersection is empty, we need only show that it contains no element of $\Bbb N$. And that follows immediately from the definition of intersection: for any $k\in\Bbb N$, $k\notin A_{k+1}$, so $k\notin\bigcap_{n\ge 1}A_n$.

By the way, strong induction is equivalent to ordinary induction: anything that can be proved using one can be proved using the other, though converting a proof by strong induction to a proof by ordinary induction typically makes it a little more complicated.


问:
Show that $R_{w}$ is bounded/compact using no arbitrage or non-redundancy arguments
Consider a financial market with $d+1$ assets in a one-period model. The $0-$th asset is considered the risk-free asset, the others are risky. The vector $\overline{\pi} \in \mathbb R^{d+1}$ denotes the initial price vector at $t=0$, and the random vector $\overline{S} \in \mathbb R^{d+1}$ denotes prices at $t=1$. Further we have $\pi^{0}=1,\; S^{0}=1+r$, as the $0-$th asset is risk-free.

Here a strategy $\overline{\xi}\in \mathbb R^{d+1}$ is called an arbitrage opportunity if $\overline{\xi}⋅\overline{\pi}≤0$ but $\overline{\xi}⋅\overline{S}\geq 0$ and $\mathbb P(\overline{\xi}⋅\overline{S}> 0)>0$. Further the market model is called non-redundant if
$\overline{\xi}\cdot \overline{S}=0 \; \;\mathbb P\text{-a.s.}\implies \overline{\xi}=0$.

By the Law of One price we know that in an arbitrage-free model, for any $\overline{\rho},\; \overline{\xi}$ such that $\overline{\xi}\cdot\overline{S}=\overline{\rho}\cdot\overline{S}$ a.s. it must follow that $\overline{\xi}\cdot\overline{\pi}=\overline{\rho}\cdot\overline{\pi}$.

Now as in  https://quant.stackexchange.com/q/17839/42184 we need to show that for $w>0$,

$$\mathcal{R}_{w}:=\{\overline{\xi}\in \mathbb R^{d+1}:\overline{\xi}\cdot \overline{\pi}=w,\;\overline{\xi}\cdot\overline{S}\geq 0\; \text{a.s.}\}\;\; \text{ is compact} $$

As we are in finite-dimensions, I assume that we need to use $\text{compact}\iff \text{closed and bounded}$.

Closedness is no problem at all, using convergence arguments. However, I am struggling to show boundedness. As I have not used the no arbitrage arguments or the non-redundancy condition, I assume that I will need to use them in my argument. Any ideas?

答:
This is just a re-statement of the following geometric fact about finite dimensional Hilbert spaces:

Proposition Let $V$ be a finite dimensional Hilbert space, $\bar{\pi} \in V$, and $\mathcal{S} = \{ \bar{S}_{\omega}\}_{\omega \in \Omega} \subset V$ such that:

$\bar{\pi}$ lies in the interior of the positive cone generated by $\mathcal{S}$, i.e. if $\langle \bar{\xi}, \bar{\pi} \rangle \leq 0$, then either $\langle \bar{\xi}, \bar{S}_{\omega} \rangle < 0$ for some $\omega$ or $\langle \bar{\xi}, \bar{S}_{\omega} \rangle = 0$ for all $\omega$.

$\mathcal{S}$ is separating in the sense that, if $\bar{\xi} \neq 0$, then $\langle \bar{\xi},  \bar{S}_{\omega} \rangle \neq 0$ for some $\omega$.

Then , for all $w > 0$,
$$
\mathcal{R}_w =  \{\bar{\xi}:  \langle \bar{\xi}, \bar{\pi} \rangle = w, \; \langle \bar{\xi}, \bar{S}_{\omega} \rangle \geq 0  \}
$$
is compact in $V$.

In your context, $V = \mathbb{R}^{1+d}$, $\Omega$ is the underlying probability space, Condition 1 is no-arbitrage, and Condition 2 is non-redundancy.
(Replace $\Omega$ by a set with probability $1$ if necessary. No-arbitrage and non-redundancy are really $\omega$-by-$\omega$ conditions.)

In the finite dimensional case (finitely many assets/commodities), the geometry of the statement is clear---e.g. take a basis $\bar{S}_{1}, \cdots, \bar{S}_{1+d}$ in
$\mathbb{R}^{1+d}$, and a $\bar{\pi}$ that lies in the interior of the positive cone they generate.
(Condition 2 implies that $|\Omega| \geq 1+d$, i.e. non-redundancy means there cannot be more assets than states. This means the linear span---equivalently, its closure---of $\mathcal{S}$ is $V$.)

As you said, closedness and convexity of $\mathcal{R}_w$ is clear.
$\mathcal{R}_w$ is an intersection of closed and convex sets,
$$
\mathcal{R}_w =  \bar{\pi}^{-1}( w ) \bigcap \left( \bigcap_{\bar{S} \in \mathcal{S}} \bar{S}^{-1}[0, \infty) \right)
$$
where, as a slight abuse of notation, $\bar{\pi}(\cdot)$ also denotes the linear functional/dual element $\bar{\xi} \mapsto \langle \bar{\xi},  \bar{\pi} \rangle$, same for $\bar{S}(\cdot)$.

Boundedness can be shown via the very nice argument given by @E-A. (When $\Omega$ is finite, you can also argue directly using

$$
\bar{\xi} \in \mathcal{R}_w \; \Rightarrow \; 0 \leq \sum_{k=1}^{K} q_k \langle \bar{\xi},  \bar{S}_{\omega_k} \rangle = (1+r) w,
$$
where $q_k$'s form an equivalent martingale measure.
)

Compactness of $\mathcal{R}_w$ does not hold in general for infinite dimensional Hilbert spaces. $\mathcal{R}_w$ remains closed but is no longer bounded in general.
For example, take $\mathcal{S} = \{ e_k, k \geq 1 \}$ to be an orthonormal basis of a separable Hilbert space $V$, and $\bar{\pi} = \sum\limits_{k \geq 1} \frac{1}{k^2} e_k$. Then both Conditions 1 and 2
hold. However, $\mathcal{R}_w$ is not bounded---e.g. consider
$$
\bar{\xi}_n = (w - \frac{1}{n}) e_1 + n e_n, \;\; n \geq 2.
$$
in $\mathcal{R}_w$.

Where the boundedness argument breaks down is that the unit sphere in $V$ is only weakly compact when $V$ is finite dimensional.
A sequence of unit vectors $\frac{\bar{\xi}_n}{\|\bar{\xi}_n\|}$ need not have a non-zero weak limit---which would then be an arbitrage portfolio if $\limsup\limits_n \|\bar{\xi}_n\| = \infty$.


问:
Can random variables be something else other then discrete or continuous?
My textbook asks me the following question as a concept check:

Do you think that there are random variables which are neither discrete nor continuous? If yes, try to
construct a simple example. If no, then discuss why not.

I wrote No, how else can outcomes be measured if they aren't finite or countably infinite (which falls under discrete random variables) or if they can't be measured using a range of values (which falls under continuous random variables)? So the answer is no.

My question is, am I right? Can a random variable be described as something other then discrete or continuous? I don't think so based off my reasoning above, but I could be wrong and I'd appreciate it if someone could tell me whether I have this concept check right.

答:
You can have bijective mixed function in just complete space.

f(x) = ⌊x⌋ + 1/2 + sgn-parity(x - ⌊x⌋) × (x - ⌊x⌋) /2

The sgn-parity gives positive or negative identity dependent on parity of the significancy of x. Rational x gives unequivocal parity and discrete points but irrational x is a superposition state.

The example show how random distribution could be useful as mixed continuity.


问:
Evaluating $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$
I am trying to evaluate $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$. Factoring and using partial fraction decomposition, I have found that the indefinite integral is:
$$\ln{|x-5|}-\ln{|x-3|} + C$$

But when evaluating the improper integral, I don't know how to deal with the difference between two diverging limits, or even whether or not it even converges. My intuition tells me that since the denominator of the original integral is a second degree polynomial, it converges, but I'm not entirely sure.

$$\lim_{b\to\infty}{(\ln{(b-5)}-\ln(b-3))}$$

How do you evaluate this limit?

答: Use that $$\log x-\log y=\log\dfrac xy$$</p>


问:
If $(f_n)_{n\in\mathbb N}$ converges in $L^p$, does the sequence of normalized functions converge as well?
Let $(E,\mathcal E,\mu)$ be a measure space and $p\ge1$. If $(f_n)_{n\in\mathbb N}\subseteq L^p(\mu)$ and $f\in L^p(\mu)$ with $$\left\|f_n-f\right\|_{L^p(\mu)}\xrightarrow{n\to\infty}0\tag1,$$ can we infer that $\left(\frac{f_n}{\left\|f_n\right\|_{L^p(\mu)}}\right)_{n\in\mathbb N}$ is $L^p(\mu)$-convergent as well?

Note that clearly $\left\|f_n\right\|_{L^p(\mu)}\xrightarrow{n\to\infty}\left\|f\right\|_{L^p(\mu)}$. So, the answer should be positive.

答:
If the measure space is finite and $f_n$ is the constant function $(-1)^{n} /n$ then $f_n \to 0$ in any $L^{p}$ but $\frac {f_n} {\|f_n\|_p}$ does not converge.

The conlusion holds if $\|f\|_p \neq 0$ and this is general fact valid in any normed linear space:

Let $(x_n)$ be sequence in a normed linear space $X$ such that $\|x_n-x\| \to 0$. If $x \neq 0$ then $\frac {x_n} {\|x_n\|} \to \frac x {\|x\|}$

Proof: $|\|x_n\|-\|x\|| \leq \|x_n-x\| \to 0$ so $\|x_n\| \to \|x\|$. It is quite easy to see that $\|x_n-x\| \to 0$ together with $a_n \to a, (a_n,a \in [0,\infty)) $ implies $a_nx_n \to ax$. (Use triangle inequality). Now take $a_n=\frac 1 {\|f_n\|_p}, a=\frac 1 {\|f\|_p}$.


问:
Local Homeomorphism with constant fiber is a covering map
Let $f:X\to Y$ be a local homeomorphism between Hausdorff spaces such that $f$ is surjective and $f^{-1}(y)$ is compact $\forall y\in Y$. If $|f^{-1}(y_1)|=|f^{-1}(y_2)|,\forall y_1,y_2\in Y$ then show that $f$ is a covering map.

Attempt: I know $f^{-1}(y)$ is finite as it has a finite cover $\cup_{i=1}^nU_i\supseteq f^{-1}(y)$ with each $U_i$ open such that $f_{U_i}$ is a homeomorphism. I considered $V_i\subseteq U_i$ with $V_i\cap V_j=\emptyset$ but the problem is the preimage of $\cap_{i=1}^nf(V_i)$ is not necessarily in $\cup_{i=1}^nU_i$ otherwise I would have found an evenly covered neighborhood of $y$. I need to show that $f$ is closed so I can use the set $(\cap_{i=1}^nf(V_i))\backslash (f(X\backslash \cup_{i=1}^nU_i))$ instead or consider another approach by using the constant preimage assumption which I do not see how it fits there...

答:
You don't need to show that $f$ is closed.

Let $y \in Y$. Since the fiber $f^{-1}(y)$ is compact, then it is closed because $X$ is Hausdorff, also, since $f^{-1}(y)$ is finite, then it is discrete.

Say $f^{-1}(y) = \{x_1, x_2, \dots, x_n\}$, then by the Hausdorff property of $X$, we can find open sets $\{U_1, U_2, \dots, U_n\}$ such that $x_i \in U_i$ and $U_i \cap U_j = \phi$ for each $i \neq j$.

Since $f$ is a local homeomorphism, we can find for each $x_i$ an open neighborhood $V_i$ such that $f(V_i)$ is an open neighborhood of $y$ and that the restriction $f|_{V_i}$ is a homeomorphism.

This implies that $W_i = U_i\cap V_i$ is an open neighborhood of $x_i$ for all $i$ and that $f(W_i)$ is an open neighborhood of $y$ from the fact that $f|_{V_i}$ is a homeomorphism, and thus, maps open sets in the subspace $V_i$ (each $W_i$ is an open set in the subspace $V_i$) to open sets in the subspace $f(V_i)$ which are also open sets in $Y$. Also, each $f|_{W_i}$ is a homeomorphism.

Therefore, $B = \bigcap\limits_{i}^{n}f(W_{i})$ is an open neighborhood of $y$.

Moreover, let $A_i = f^{-1}|_{W_i}(B) = W_i \cap f^{-1}(B)$. Since $f|_{W_i}$ is a homeomorphism onto $f(W_i)$, then it is also a surjection and an injection, and from the fact that $A_i \subseteq W_i$, $f(A_i) = f|_{W_i}(A_i) = B$. So each $A_i$ is mapped homeomoprhically to $B$.

Also, $x_i \in A_i$ for all $i$, and thus, $f^{-1}(y) \subseteq \bigcup\limits_{i}^{n}A_i$. Hence, $\{y\} = f(f^{-1}(y)) \subseteq f(\bigcup\limits_{i}^{n}A_i) = \bigcup\limits_{i}^{n}f(A_i) = B$. In other words, $B$ is a neighborhood of $y$.

The problem now is that we are not sure whether
$f^{-1}(B)$ is contained in $\bigcup\limits_{i}^{n}A_i$.

Let $x' \in f^{-1}(B)$, then $f(x') = y' \in B$. Therefore, $y' \in f(W_i)$ for all $i$. Hence, we can find $x'_i$ in each $W_i$ such that $f(x'_i) = y'$. Using the fact that $|f^{-1}(y_1)| = |f^{-1}(y_2)|$ for all $y_1 \neq y_2$ and that $|f^{-1}(y)| = n$, we conclude that $f^{-1}(y')=\{x'_1, x'_2, \dots, x'_n\} \subseteq \bigcup\limits_{i}^{n}A_i$. But $x'$ must be one of $\{x'_1, x'_2, \dots, x'_n\}$, and thus, $x' \in \bigcup\limits_{i}^{n}A_i$ which completes the proof.


问:
Can someone explain why the probability of getting a royal flush is only based on the choosing the suit?
So $|S| = 52C5, |E| = 4C1$ therefore $P = \frac{1}{649740}$. If you only select the suit, how do you guarantee that the cards OF the suit will be 10, J, Q, K and Ace? It seems so random that only $4C1$ is required and I don't understand. I know that the cards would be "fixed" since you MUST have the 10 at least. But even so, I am questioning it.

答:
$$\text{Probability} = \frac{\text{N(umerator)}}{\text{D(enominator)}}$$

where $N =$ # of successful 5 card combinations

and $D =$ total # of 5 card combinations.

$$D = \binom{52}{5}$$

and

$$N = 4 ~~\text{(i.e. one combination of AKQJ10 for each suit)}.$$


答:
Of course it can, by a simple coin toss.

For example, let $X$ be a discrete distribution and $Y$ be a continuous distribution. Define the random variable $Z$ to be as follows : let $H$ be an independent (of $X$ and $Y$) Bernoulli-$\frac 12$ random variable, and define $Z = X$ if $H=0$ and $Z= Y$ if $H=1$.

Clearly $Z$ is not discrete or continuous (can be easily proven, since there are points which alone have non-zero probability by themselves, but not every point has this property as well), you may describe it as half of each. That gives rise to what is called as a mixed distribution.

The sentence "No, how can outcomes be measured ... variables)?" in your second paragraph misses the fact that an outcome , or event, need not just be a range or a set of points, but could be a mixture of both (for example, the speed of my car is either $10, 20$ or between $30$ and $40$) and hence be measured by breaking into these parts.

If one thinks of continuous random variables as those having an associated "density function" (PDF/CDF) as is usually the case, then funnily enough there is an entire class of distributions which aren't discrete, continuous, or even a mix of these. These are the singular distributions, the most well-known example being the Cantor distribution. This is not discrete (in terms of points having weights) or continuous (in terms of having a PDF/CDF) but is still a probability distribution. The buck stops here, though : every probability distribution is a mixture of a singular, continuous, and a discrete distribution.

Title: Can random variables be something else other then discrete or continuous?

Body:
My textbook asks me the following question as a concept check:

Do you think that there are random variables which are neither discrete nor continuous? If yes, try to
construct a simple example. If no, then discuss why not.

I wrote No, how else can outcomes be measured if they aren't finite or countably infinite (which falls under discrete random variables) or if they can't be measured using a range of values (which falls under continuous random variables)? So the answer is no.

My question is, am I right? Can a random variable be described as something other then discrete or continuous? I don't think so based off my reasoning above, but I could be wrong and I'd appreciate it if someone could tell me whether I have this concept check right.


问:
Solution verification: showing $|x|^2 = x^2$ and $|x|<|y| \iff x^2 < y^2$
I have two propositions to prove and below are my proofs.

Any helps or comments would be appreciated!

Proposition $1$: For all $x∈ \Bbb R$, $|x|^2 = x^2$.

Proof: Let $x∈ \Bbb R$ and suppose $x≥0$.

By the definition of the absolute value, $|x|$ is defined to be $x$ if $x≥0$.

Then $|x| = x$, and $|x|^2 = |x| \cdot |x| = x \cdot |x| = x \cdot x = x^2$.

Now, suppose $x<0.$ By the definition of the absolute value, $|x|$ is defined to be $–x$ if $x<0$.

Then $|x| = -x$ and $|x|^2 = |x| \cdot |x| = -x \cdot |x| = -x \cdot -x = x^2$.

Hence $|x|^2 = x^2$.

Proposition $2$: Let $x,y∈ \Bbb R$. Then $|x|<|y|$ if and only if $x^2 < y^2$.

Proof:

Let $x,y ∈  \Bbb R$ and suppose $|x|<|y|$. Then $|x| \cdot |x| < |x| \cdot |y| < |y| \cdot |y|$ and therefore, $|x|^2 < |y|^2$.

Then, by proposition $1$, $|x|^2 < |y|^2 \implies x^2 < y^2$.

Now, suppose $x^2 < y^2$.

Then, by proposition $1$, $x^2 < y^2 \implies |x|^2 < |y|^2$ and since $|x|,|y|≥0$,  $|x| < |y|$.

答:
You're pretty much right.

I would say your phrasing for the second proof could be a bit better to show, properly, each implication. Here's how I'd approach things...

Assume $|x|<|y|$. Then

\begin{align*}
x^2 &= |x|^2 \tag{proposition 1} \\
&= |x||x| \tag{def. of square} \\
&< |y||y| \tag{$|y|>|x|\ge 0$} \\
&=|y|^2 \tag{def. of square} \\
&=y^2 \tag{proposition 1}
\end{align*}

Thus $|x|<|y| \implies x^2 < y^2$. Now, conversely, suppose $x^2 < y^2$. Then

\begin{align*}
|x|^2
&= x^2 \tag{proposition 1} \\
&< y^2 \tag{assumption} \\
&= |y|^2 \tag{proposition 1}
\end{align*}

Thus, $x^2 < y^2 \implies |x|^2 < |y|^2$. Since $\sqrt z$ is monotone increasing on $\Bbb R_{\ge 0}$, it follows that $|x|^2 < |y|^2 \implies |x| < |y|$, the desired conclusion.

Alternatively, go for contrapositive: show that $|x| \ge |y| \implies |x|^2 \ge |y|^2$. I think this might be preferable to avoid invoking things that might not be "kosher" for the material at the time, however obvious. This proof is pretty straightforward:

\begin{align*}
|x|^2
&= |x||x| \tag{def. of square} \\
&< |y||y| \tag{$|y| \ge |x| \ge 0$} \\
&= |y|^2 \tag{def. of square}
\end{align*}

Thus, $x^2 < y^2 \implies |x|^2 <|y|^2 \implies |x|<|y|$, completing the proof.

That said, overall, I think you have the spirit of that proof right, and your first proof seems pretty good, so good job.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


答: It doesn't converge in the traditional sense, as it is unbounded at $ x = 5 $ and $ x = 3 $ (and is unbounded in such a way that the integral does not converge). It does have a finite Cauchy Principal Value though. </p>

Title: Evaluating $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$

Body:
I am trying to evaluate $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$. Factoring and using partial fraction decomposition, I have found that the indefinite integral is:
$$\ln{|x-5|}-\ln{|x-3|} + C$$

But when evaluating the improper integral, I don't know how to deal with the difference between two diverging limits, or even whether or not it even converges. My intuition tells me that since the denominator of the original integral is a second degree polynomial, it converges, but I'm not entirely sure.

$$\lim_{b\to\infty}{(\ln{(b-5)}-\ln(b-3))}$$

How do you evaluate this limit?


问:
$\cap_{n=1}^{\infty}A_n$ and infinity
One question:

If the definition of $\cap_{n=1}^{\infty}A_n=\{x\in A_i\forall n\in N\}$ and it is nonempty, then does it mean that its elements belong to the infinite intersection of $A_n$ or any finite intersection of $A_n$ for all natural numbers?

To further elaborate, I would like to show how I feel towards this confusing notation $\cap_{n=1}^{\infty}A_n$.

Understanding Analysis Steven Abbott

Example 1.2.2 in which it defines $A_i = \{x\in N: x\geq i\}$. By induction, it is nonempty for each finite intersections. But a proof by contradiction can show that when it goes to infinite case, which uses the notation $\cap_{n=1}^{\infty}A_i$, it is a null set. In other words, in this example, this notation is used for infinite intersection.

Theorem 1.4.1 in which it proves the nested interval property. $I_n = \{x\in R: a_n\leq x\leq b_n\}$. Here, it doesn't specify whether this is infinite intersection or not. Instead, it said, $\exists x\forall n\in N x\in I_n$. Hence, that $x\in\cap_{n=1}^{\infty}A_n$. In other words, in this example, this notation is used for every finite natural number

Theorem 1.5.8 says If $A_n$ is a countable set for each $n\in N$, then $\cup_{n=1}^{\infty}A_n$ is countable. In other words, in this example, this notation is used for infinite intersection.

I am confused by this notation in a sense that the notation includes the infinity sign but its definition means every natural number. Hence, whenever I see it, I just don't know which one to apply.

Say if I go for the direction in which it is applicable $\forall n\in N$, then induction should work because induction is doing exactly the same thing! Though, this post suggests otherwise by saying the notation is about infinity.

Fine, I switch direction in which it is about infinite intersection. But then in some cases, for example, the one I listed above, somehow if something is applicable for all natural numbers, it is fine to be part of this notation.

So in short, I feel that this notation has 2 conflicting meanings

$\forall n\in N$

Infinity

I have done researches and asked questions before but I still don't understand. So I guess I got something every wrong and confused in some definitions.

答:
$\bigcap_{n=1}^\infty A_n$ is a set. What set? The set of all things that belong to every one of the sets $A_n$ for $n\in\Bbb Z^+$. Let $\mathscr{A}=\{A_n:n\in\Bbb Z^+\}$; then $\bigcap\mathscr{A}$ means exactly the same thing. $\bigcap_{n=1}^\infty A_n$ is simply a customary notation that means neither more nor less than $\bigcap_{n\ge 1}A_n$, $\bigcap\mathscr{A}$, and $\bigcap\{A_n:n\in\Bbb Z^+\}$. There is no $A_\infty$: the $\infty$ is just a signal that the index $n$ is to assume all positive integer values.

Suppose that for each positive real number $x$ I let $I_x$ be the open interval $(-x,x)$. Then $\bigcap_{x\in\Bbb R^+}I_x$ is the set of all real numbers that belong to every one of these open intervals. If $\mathscr{I}=\{I_x:x\in\Bbb R^+\}$, then

$$\bigcap\mathscr{I}=\bigcap_{x\in\Bbb R^+}I_x=\bigcap_{x\in\Bbb R^+}(-x,x)=\{0\}\,.$$

How do I know? If $y\in\Bbb R\setminus\{0\}$, then $y\notin(-|y|,|y|)=I_{|y|}$, so there is at least one member of $\mathscr{I}$ that does not contain $y$, and therefore by definition $y$ is not in the intersection of the sets in the family $\mathscr{I}$. On the other hand, $0\in(-x,x)=I_x$ for every $x\in\Bbb R^+$, so $0$ is in the intersection $\bigcap\mathscr{I}$.

In neither case have we used induction anywhere. In the case of the sets $A_n$ we might be able to use induction on $n$ to show that each of the sets $A_n$ has some property $P$, but we could not extend that induction to show that $\bigcap\mathscr{A}$ has $P$. We might somehow be able to use the fact that each $A_n$ has property $P$ to show that $\bigcap\mathscr{A}$ also has $P$, but that would require a separate argument; it would not be part of the induction. The induction argument in that case would prove that

$$\forall n\in\Bbb Z^+(A_n\text{ has property }P)\,;$$

the separate argument would then show, using that result and other facts, that the single set $\bigcap\mathscr{A}$ has property $P$. You could call this set $A_\infty$ if you wished to do so, but that would just be a label; you could equally well call it $A$, or $X$, or even $A_{-1}$, though offhand I can’t imagine why you’d want to use that last label.

In the case of the sets $I_x$ there is no possibility of using induction to show that each $I_x$ has some property: these sets cannot be listed as $I_1,I_2,I_3$, and so on, because there are uncountably many of them. We can still prove things about the set $\bigcap\mathscr{I}$, however. And we could give it any convenient label. $\bigcap\mathscr{I}$ is informative but perhaps a little inconvenient; I might choose to give it the handier label $I$.

In the case of $\mathscr{A}$ there happens to be a customary notation that uses the symbol $\infty$, but that is simply a consequence of the fact that the sets $A_n$ are indexed by integers. We’re doing exactly the same sort of thing in the example with $\mathscr{I}$, but in that case there is no possibility of using a limit of $\infty$ on the intersection, because there is no way to index the uncountably many sets $I_x$ by integers.

Title: $\cap_{n=1}^{\infty}A_n$ and infinity


问:
Probability of 4 of a kind in 7 cards, if first dealt 2 different cards?
Question: In Texas Hold 'Em, you choose your hand from among seven cards: two that are dealt to you, plus five more that are available to all players.

Suppose the two cards dealt to you are of different denominations. What is the probability that you'll be able to make four of a kind out of all seven cards available to you?

I'm trying to use Bayes conditional to solve this:

Prob (4 of a kind in 7 cards after first being dealt 2 different denominations ) = ( Prob(Being dealt 2 different denominations first (in 2 cards), given 4 of a kind in 7 cards) * Prob(4 of a kind in 7 cards) ) / Prob (Being dealt 2 different denominations first (in 2 cards))

I'm looking for clarity on how to think about this problem (and, I believe, how I should avoid overthinking it as I've spent too much time pondering "what ifs").

I can't solve "Prob (Being dealt 2 different denominations first (in 2 cards), given 4 of a kind in 7 cards)", so I'm not sure if I'm stating the possibilities properly. What if I'm dealt Queen, King, and 3 more Kings in the 5 cards. Or what if there are 4 Jacks in the 5 cards and the 4-of-a-kind is made completely in the 5? The location of the 4 of a kind (if linked with one of the initial 2 cards or not), along with how to separate the probabilities of the initial 2 cards from the 5 is confusing me.

答:
The idea that the probability cant be counted is a bit incorrect, tho not as clean.

C1 = $\frac{52}{52}$ or $P(1)$ out of 52 possible choices

C2 = $\frac{48}{48}$ or $P(1)$ out of 52 - 4 (so that the C1 and C2 don't match)

C3 = $\frac{6}{50}$ or $P(\frac{3}{25})$ since there are 3 cards to match c1 and 3 cards to match c2

C4 = $\frac{2}{49}$ or $P(\frac{2}{49})$ since there are only 2 cards in the deck to match c3

C5 = $\frac{1}{48}$ or $P(\frac{1}{48})$ since there are only 1 card in the deck to match c4

C6 = $\frac{47}{47}$ or $P(1)$ This card don't matter and since it can be anything, a probability of 1 makes things easier

C7 = $\frac{46}{46}$ or $P(1)$ This card don't matter and since it can be anything, a probability of 1 makes things easier

Thus: $\frac{52}{52}*\frac{48}{48}*\frac{3}{25}*\frac{2}{49}*\frac{1}{48}*\frac{47}{47}*\frac{46}{46}$ $=$ $1*1*\frac{3}{25}*\frac{2}{49}*\frac{1}{48}*1*1$ $=$ $\frac{3}{25}*\frac{2}{49}*\frac{1}{48}$ = $\frac {1}{9800}$ = .000102

HOWEVER

This outcome can happen with $\binom {5}{2}$ or 10 different card locations/positions (combinations)!

So:  $\frac {1}{9800}*\frac {10}{1} = \frac {1}{980}$ = .0010204...

This obviously under-counts something (by -.000239) but this is a good thing as the probability of getting 4 of a kind with out any help from the first 2 pocket cards equals????? ~.000239.

C1 = $\frac{52}{52}$ or $P(1)$ out of 52 possible choices (this card can not match the river cards)

C2 = $\frac{51}{51}$ or $P(1)$ out of 51 cards left (this card can not match the river card but could match c1)

C3 = $\frac{4}{50}$ or $P(\frac{4}{50})$ since there are 4 cards available to be used for 4 of a kind

C4 = $\frac{3}{49}$ or $P(\frac{3}{49})$ since there are only 3 cards in the deck to match c3

C5 = $\frac{2}{48}$ or $P(\frac{2}{48})$ since there are only 2 card in the deck to match c3, c4

C6 = $\frac{1}{47}$ or $P(\frac{1}{47})$ since there is only 1 card in the deck to match c3, c4, c5

C7 = $\frac{46}{46}$ or $P(1)$ This card don't matter and since it can be anything, a probability of 1 makes things easier

Thus: $\frac{52}{52}*\frac{51}{51}*\frac{4}{50}*\frac{3}{49}*\frac{2}{48}*\frac{1}{47}*\frac{46}{46}$ $=$ $1*1*\frac{4}{50}*\frac{3}{49}*\frac{2}{48}*\frac{1}{47}*1$ $=$ $\frac{4}{50}*\frac{3}{49}*\frac{2}{48}*\frac{1}{47}$ = $\frac {1}{230300}$ = .000004

Now take .000004 and multiply it by $\binom {5}{4}$ or 5 and multiply by the renaming possible denominations for the 5th card in the river.

$\frac {1}{230300}$ * 5 * 11 =  $\frac {55}{230300}$ = $\frac {11}{46060}$ = .000239

Next add P(of 3 cards in the river that match one card in the pocket) + P(4 of a kind in the river) = .0010204 + .000239 = .001259.

BOOM! If it's not fun, it's not math :D

NN + APL


问:
In Homotopy Type Theory, how do the continuous notions of spaces and paths, match the discrete notions of constructible terms and proofs?
Context: I'm starting with being interested in type theory as a framework within which to do mathematics (e.g. practically, using proof assistants based on type theory), and my understanding is that HoTT is motivated partially/primarily from that perspective. I and am not interested in homotopy theory per se, or  in using type theory to understand homotopy theory.

My understanding of homotopy type theory is limited, but I know that in HoTT we think of a type $T$ as a space, of instances $x,y:T$ of the type as points in that space, and of proofs $p:x=_Ty$ of equality as paths in the space.

How I can understand "proofs of equality as paths"

My interpretation of proofs of equality is that they are computational paths between terms. If we think of $x$ and $y$ as actual constructed terms in e.g. a proof assistant such as Coq, then the way to prove $x=_Ty$ is to show a sequence of computational steps (or at least, a program that would produce one, such as in an inductive proof) from $x$ to $y$. E.g. with the standard inductive definition of the natural numbers, it could be a transformation from $(1+1)+1$ (i.e. $x$) to $3$ (i.e. $y$), by repeatedly applying the addition rules to get $s (s (s (0)))$ and then applying the definition of $3$. i.e. proofs of $x=_Ty$ are computationally valid paths between terms, using the underlying computational system's (lambda calculus I'd presume) valid computation rules (e.g. beta-reduction, delta-reduction of lambda-terms).

So the way I think of it is: two terms may be different syntactically (e.g. $1+1$ and $2$), and are therefore different points in the "space belonging to their type" (e.g. $\mathbb N$). But they are semantically equal to each other, so that there is a computational path between them.

proofs of equality in homotopy type theory

In explanations of homotopy type theory that I've seen, the explanation usually asks you to think of types as spaces, where usually they are pictured as topological spaces (circles, donuts, etc). Then the point is made that two proofs $p,q:x=_Ty$ are fundamentally different, if there is no homotopy between them, and some image like this is shown to suggest that two proofs are not homotopic:

But afaik, a homotopy has to be continuous, meaning that the space has to have some kind of topological structure. However, the computational steps that we make in reasoning e.g. that $(1+1)+1 = 3$, are finite. My understanding is that it's fundamental to intuitionistic type theory that all the terms of a type are actually constructible terms, i.e. strings of symbols that we can write down, and hence that the "space" belonging to a type $T$ is discrete. But the only reasonable topology that I can think of on such a space would be the topology generated by the base set consisting of equivalence classes of terms based on $=_T$, which would I think imply that all paths $p,q:x=_Ty$ are homotopy equivalent (since any path consisting of points inside such a base set is continuous).

This makes me confused about homotopy type theory itself: It seems to me that in order to formalize the idea of proofs being fundamentally different, we'd have to think of a discrete object, probably a finite graph (where nodes are terms and links are basic computational steps), and concepts like homotopy and continuous functions and spaces are out of place.

Where is the mistake in my thinking?

答:
It's quite common to describe "nice" topological spaces (up to homotopy) by cell complexes. The idea is to specify your space by combinatorial data, and we can then use combinatorial reasoning to prove things about our spaces.

We have $0$-cells, or points. Then we have $1$-cells, which are edges connecting the points. If we have a bunch of edges which enclose a region (a polygon, say) then we can "fill" the region with a $2$-cell.

As an example, we can build the circle $S^1$ by considering one $0$-cell $\mathtt{base}$, and one $1$-cell $\mathtt{loop}$ connecting $\mathtt{base}$ to itself. If we then add a $2$-cell to this, we would get a disk $D^2$. If we add two $2$-cells, we get a sphere $S^2$. Here the original $S^1$ is the equator, and the two $2$-cells become the hemispheres. As a (fun?) exercise, you might try to build a cellular complex which describes a Torus.

Notice this is an entirely combinatorial description. We have $n$-cells in various dimensions, and we have incidence relations telling us how these cells are glued together. We're not reasoning about open sets, or actually thinking of our edges as copies of $(0,1)$. We really do have a discrete description of these topological objects, and this description is frequently more user-friendly than the classical one when it comes to explicit computations. Of course, we can cash out this description and talk about a honest topological space by putting the weak topology on the collection of cells.

Topologists already cared about cell complexes, because they let us efficiently read off topological invariants. By studying fudamental groups in this way, for instance, we are led naturally to combinatorial group theory. We can also read off more complex invariants using cellular homology. All this to say that these cellular decompositions have been of interest to topologists for a long time.

What, then, is type theory?

Given a type $A$, we can construct terms $a : A$ that inhabit $A$. Of course, some terms are syntactically different, but describe "the same thing", and we might want to consider them equal.
So we postulate an Identity Type whose terms are "proofs" that $a_1 = a_2$.

Now let's say we have a type $A$. Let's say it has one constructor
$\mathtt{base} : A$. But let's give it a higher order constructor as well. We'll give it a nontrivial proof $\mathtt{loop} : \mathtt{base} = \mathtt{base}$. If you think the data of this type sounds a lot like the data of a cellular decomposition of $S^1$, you're exactly right!

This, then, is the way that HoTT works. Using Higher Inductive Types we can consider types which have exactly the same data as the cellular decomposition of topological spaces. Then by using type theoretic arguments (for instance, path induction), we can prove things about topological spaces!

Rather excitingly, the correspondence goes both ways. Type Theories correspond to the internal logics of categories, and it turns out these logics can be understood geometrically by reasoning with cell complexes. When we draw a commutative diagram, we have objects ($0$-cells) which we connect by arrows ($1$-cells). Then saying the diagram commutes is akin to saying we can "fill it in" by a $2$-cell. But what if we have multiple ways for the diagram to commute? For instance, if we have a "pullback cube" (where we pull back an entire pullback square along an arrow $f$). Then there are two natural transformations which witness a certain commutativity, and these natural transformations are actually isomorphic too. We say that this is a $3$-cell which "fills the cube". Obviously you can continue on in this way, and if you've heard people talk about "Infinity Categories" this is roughly what they're studying.

This is why you see so many higher category theorists also excited about algebraic topology -- the two fields are deeply connected in precisely this way. The study of these higher categories is entirely analogous to the study of cell complexes, and there's a lot of very pretty interplay between the fields.

I hope this helps ^_^


答:
Calculus solution:

Let d be the distance. the point on line is where it is tangent to a circle with radius d and center (5, 2) and we can write:

$d=\sqrt{(x-5)^2+(y-2)^2}=\sqrt{(x-5)^2+(2x+1-2)^2}=\sqrt{5x^2-14x+26}$

Taking derivative and equating to 0 we get:

$x=\frac 7 5$

$y=\frac {19}5$

Title: Find the point on the line $y = 2x + 1$ that is closest to the point $(5, 2)$?

Body:
The question is: Find the point on the line $y = 2x + 1$ that is closest to the point $(5, 2)$.
I know the high (or middle) school method. Create a line that is perpendicular to this line, so the slope will be $-1/2$. Then the intersection is the answer.

But I meet this question from a linear algebra textbook. I want to use projection onto a line method. Projecting vector $b$ onto vector $a$ with the error $e=b-\hat{x}a$.

Thus, $a \cdot (b-\hat{x}a)=0$ or $a\cdot b-\hat{x}a\cdot a=0$.
Thus, $\hat{x}=\frac{a\cdot b}{a\cdot a}$

But the problem is that here $a$ and $b$ are pure vectors. My question is given a specific line. Then I don't know how to do with it.


问:
How to do integral $\int\frac1{x^3+x+1}dx$?
I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$

答:
My favored way with polynomials in denominator (for any degree with or without real roots)
$$\int\frac{dx}{x^3+x+1}=\int\frac{dx}{(x-a)(x-b)(x-c) }$$ where $(a,b,c)$ are the roots. Now, using partial fraction decomposition, the integrand is
$$\frac{1}{(a-b)
   (a-c) (x-a)}+\frac{1}{(b-a) (b-c) (x-b)}+\frac{1}{(c-a) (c-b) (x-c)}$$ So, the antiderivative is
$$\frac{\log(|x-a|)}{(a-b)
   (a-c) }+\frac{\log(|x-b|)}{(b-a) (b-c) }+\frac{\log(|x-c|)}{(c-a) (c-b) }$$ Now, computing the roots using the hyperbolic method for one real root
$$a=-\frac{2 }{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3
   \sqrt{3}}{2}\right)\right)$$ Now, the sum of the roots is $0$ and their product is $-1$; moreover $b$ and $c$ are complex conjugate; then, they are "simple" to compute. By the end, recombining the second and third terms, you will end with an arctangent and a logarithm.


问:
In what sense does Homotopy type theory "model types as spaces"
I've read at multiple points (e.g. here) that homotopy type theory "models" types as spaces. I can understand informally that we can "think of" types as spaces, in a vague sense. But what is formally meant by a type theory "modelling" something?

When I was taught a course on formal logic (using first order logic and a form of natural deduction) using the book "Mathematical logic by Ebbinghaus, Flum, Thomas", there was a formal semantics of logical propositions, in terms of structures and interpretations. An "interpretation of a first-order logical formula" was formally defined as a set, with mathematical functions corresponding to the symbols classified as "function symbols", similarly for mathematical relations and constants. i.e. "interpretation" has a formal definition (which can be found in that book).

Is there similarly a formal definition of "interpretation of a type theory" written down somewhere?

答:
A type theory models "something" when "something" gives an interpretation of the constructions (terms and types) of the type theory. More generally, a formal theory models "something" when the constructions of the theory can be interpreted by the "objects" of "something".

when the focus is on the (type) theory, it is said that "something" is a model of the (type) theory. Other models are possible: you can find different interpretations of the constructions of the theory.

when the focus is on "something", it is said that the (type) theory models "something". Other formalizations are possible: you can find different sets of rules and axioms to model "something".

At the end of the day, we could also say that a theory is a "formal model" of "something": but in practice we never say "A is a formal model of B" but "A models B", or "B is a model of A".

The interpretation is not just "something that you can think of, in a vague sense". A correct interpretation is justified by a systematic study of all possible constructions of the theory: each possible construction must have its own interpretation, functions and relations must have a consistent interpretation. When "something" is also formalized, and not just an intuition, then you should be able to build an interpretation function that converts any construction of the theory into a construction of the model.


问:
Why is the partial differential equation $u_{t} + a u_{x} = 0$ hyperbolic?
I don't quite understand why this advection equation, with some initial condition $u(0,x) = u_{0}(x)$, is considered hyperbolic (as for instance here). If I apply the test mentioned in Farlow, comparing it to
$$Au_{xx}+Bu_{xy} + Cu_{yy} + D u_{x} + Eu_{y} + Fu = G$$
then both $B^{2}$ and $AC$ are zero, and thus $B^{2} - 4AC = 0 \Rightarrow \text{parabolic}$. Or does a different criterion apply when there aren't any second-order terms?

答:
You have to look at the principal symbol
$$
a_1(x,t,\xi,\tau)=i\tau+ai\xi.
$$
For every fixed $\xi$ the equation $a_1(x,t,\xi,\tau)=0$ has exactly one real solution in $\tau$. Therefore the equation is hyperbolic with respect to $t$. Basically, this means that there is a real characteristic passing through each point. You can read more on the topic here:
https://people.sissa.it/~dabrow/IFM12small1.pdf


问:
Are propositions uniquely determined by their non-equivalent implications?
I was reading up on Pragmatism and the pragmatic maxim, which states: "Consider the practical effects of the objects of your conception. Then, your conception of those effects is the whole of your conception of the object".

I thought that it might be interesting to generalize this to mathematics, with some alterations.

So, here is my question:

"Let B be the set of all implications of a proposition $a$. Then there exists a function such that $f(a)=B-a.$  Is this function injective?"

答:
$\mathsf{Prop}$ with $\land$, $\lor$, $\lnot$ forms a boolean lattice, and the ordering is given by $x \leq y \iff$ $x \to y$.

Then asking if a propostion $a$ is determined by $\{ b \neq a ~|~ a \to b \}$ is asking if $a$ is determined by $\{ b > a \}$.

This is not true, though. For instance consider the lattice with four elements

Then $a$ and $\lnot a$ are both sent to $\{ T \}$ by your function, so it is not injective in general.

If we look at $\{b \geq a \}$ instead, which is your function but without removing $a$ from the set, we recover injectivity (for instance, by the Yoneda Lemma).

I hope this helps ^_^


问:
An analytic function on $\Bbb{C}$ has infinite radius of convergence about any point
I may be overlooking a small detail, but this question has me stumped. Suppose $f:\Bbb{C}\to \Bbb{C}$ is analytic. Then, by definition of analyticity, for every $z_0\in \Bbb{C}$, there exists a formal power series $S(X)=\sum\limits_{n\geq 0}a_nX^n$ with radius of convergence $\rho_{z_0}>0$ such that for $z\in \Bbb{C}$ with $|z-z_0|<\rho_{z_0}$, we have
\begin{align}
f(z)=\sum_{n\geq 0}a_n(z-z_0)^n.
\end{align}
I'm trying to show that since $f$ is analytic on $\Bbb{C}$, it is true that for every $z_0$, we have $\rho_{z_0}=\infty$.

I used to think this was a trivial consequence of the definitions, but now it doesn't seem to be so. I would appreciate it if someone could provide a proof (which hopefully doesn't use any complex integration theory, because I haven't gotten to that point yet).

Intuitively, this makes sense to me because if we fix $z_0$ as above and suppose that $\rho_{z_0}<\infty$, then we choose a point $z_1$ on the boundary of the disc $|z-z_0|\leq \rho_{z_0}$. Since $f$ is analytic on $\Bbb{C}$, we can find another formal series $\sum\limits_{n\geq 0}b_n X^n$ with radius of convergence $\rho_{z_1}>0$ such that for $|z-z_1|<\rho_{z_1}$, we have
\begin{align}
f(z)=\sum_{n\geq 0}b_n(z-z_1)^n
\end{align}
Now, there will be some points of overlap in the discs $\Delta_{\rho_{z_0}}(z_0)$ and $\Delta_{\rho_{z_1}}(z_1)$, so maybe this and the principle of uniqueness of analytic continuation somehow provide a contradiction to $\rho_{z_0}$ being finite? I'm not sure how to fill in the details or if this is even the right approach.

答:
I think the easiest way to see this is Cauchy's Integral Formula. Write $\gamma_{a,r}$ for the circular contour of radius $r$ around $a$. Then, by Cauchy's Integral Formula, for $w \in B(a,r)$,

$$ f(w) =  \frac{1}{2 \pi i} \int_{\gamma_{a,r}} \frac{f(z)}{z-w} dz$$

Now note that

$$ \frac{1}{z-w} = \frac{1}{z-a} \frac{1}{1-\frac{z-w}{z-a}} = \frac{1}{z-a} \sum_{n=0}^\infty \left( \frac{z-w}{z-a} \right)^n$$

And we get absolute convergence since $|z-w|/|z-a| < 1$ for $z \in \partial B(a,r)$ (and you can get a uniform bound $ < 1$ using compactness). Since we have absolute convergence, you can change the order of integration (note that $f$ is bounded on a compact set), so we get

$$ f(w) = \frac{1}{2 \pi i} \sum_{n=0}^\infty (z-w)^n \int_{\gamma_{a,r}} \left( \frac{f(z)}{(z-a)^{n+1}} \right) dz$$

Since this is true for any $r > 0$, the radius of convergence must be infinite (by identity theorem you can show this expansion is unique).


问:
Proof regarding continuity and Dirichlet function.

I was given this assignment and I was wondering if my proof makes sense? Also I'd love some feedback and tips on how to improve!

Assume $f(x_{0}) = 0$, we prove $g$ is continuous on $x_{0}$.
Let $ϵ > 0$. We choose δ =1. Let $|x-x_{0}|<δ$, then $\left|g\left(x\right)-g\left(x_{0}\right)\right|<ϵ$.
$\left|g\left(x\right)-g\left(x_{0}\right)\right|=\left|f\left(x\right)\cdot D\left(x\right)-f\left(x_{0}\right)\cdot D\left(x_{0}\right)\right|=\left|f\left(x\right)\cdot D\left(x\right)-0\right|=\left|f\left(x\right)\cdot D\left(x\right)\right|<ϵ$

By the density of $\Bbb R\setminus\Bbb Q$ in $\Bbb R$, there exists $b∈\Bbb R$ such that $x_{0} - 1 < b < x_{0} + 1$ and $b∈\Bbb R\setminus\Bbb Q$. We choose $x = b$, then we have $\left|f\left(b\right)\cdot D\left(b\right)\right|=\left|f\left(b\right)\cdot0\right|=0<ϵ$

By the density of $\Bbb Q$ in $\Bbb R$, there exists $c∈\Bbb R$ such that $x_{0} - 1 < c < x_{0} + 1$ and $c∈\Bbb Q$. We choose x = c, then we have $\left|f\left(c\right)\cdot D\left(c\right)\right|=\left|f\left(c\right)\cdot1\right|=\left|f\left(c\right)\right|<ϵ$

Note that f is continuous at $x_{0}$, therefore $\left|f\left(c\right)-f\left(x_{0}\right)\right|<ϵ$. $\left|f\left(c\right)\right|=\left|f\left(c\right)-f\left(x_{0}\right)+f\left(x_{0}\right)\right|\le\left|f\left(c\right)-f\left(x_{0}\right)\right|+\left|f\left(a\right)\right|=\left|f\left(c\right)-f\left(x_{0}\right)\right|+\left|0\right|=\left|f\left(c\right)-f\left(x_{0}\right)\right|<ϵ$

Therefore $g$ is continuous at $x_{0}$.

Assume $g$ is continuous at $x_{0}$. We prove $f(x_{0})=0$. Note that f is continuous on $a$. Assume toward contradiction that $f(x_{0})≠0$. Then we have $g\left(x_{0}\right)=f\left(x_{0}\right)\cdot D\left(x_{0}\right)\ ➜\ \frac{g\left(x_{0}\right)}{f\left(x_{0}\right)}=D\left(x_{0}\right)$

By the algebra of continuous functions $\frac{g\left(x_{0}\right)}{f\left(x_{0}\right)}$ is continuous, but we proved in class that $D(x)$ is not continuous on any $x∈\Bbb R$. That means we have a contradiction, since we have a continuous function on one side of the equation, and a discontinuous function on the other which can't be.
Therefore $f(x_{0}) = 0$.

答:
Assume that $f(x_0) = 0$, and let $\varepsilon>0$. Then we need to find a $\delta>0$ such that for each $x \in (x_0-\delta,x_0+\delta)$ the following holds: $|f(x)D(x)| <\varepsilon$. Observe that the continuity of $f$ at $x_0$ gives us a $\delta_0 > 0$ such that $|f(x)| = |f(x)-f(x_0)|<\varepsilon$ whenever $x \in (x_0-\delta_0,x_0+\delta_0)$, so, we claim that $\delta := \delta_0$ works. Indeed, for each $x \in (x_0-\delta,x_0+\delta)$ we have two cases: either $x \in \mathbb Q$ or $x \notin \mathbb Q$.

If $x \in \mathbb Q$ then $|f(x)D(x)| = |f(x)| < \varepsilon$ because $x \in (x_0-\delta_0,x_0+\delta_0)$.

If $x \notin \mathbb Q$ then $|f(x)D(x)| = 0 < \varepsilon$.

Thus, we showed that for any $x \in (x_0-\delta,x_0+\delta)$ the following holds: $|g(x)-g(x_0)| = |f(x)D(x)| <\varepsilon$. Hence, $g$ is continuous at $x_0$.

Now suppose that $g$ is continuous at $x_0$, and for the sake of contradiction also suppose that $f(x_0) \neq 0$. Then $|f(x_0)|>0$ and the continuity of $f$ at $x_0$ gives us a $\delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $x \in (x_0-\delta,x_0+\delta)$; which implies that $|f(x)| > |f(x_0)|/2$ for all $x \in (x_0-\delta,x_0+\delta)$. Thus, $f(x) \neq 0$ for all $x \in (x_0-\delta,x_0+\delta)$, and then we can define $h : (x_0-\delta,x_0+\delta) \to \mathbb R$ by $$\forall x \in (x_0-\delta,x_0+\delta) : \quad h(x) = \frac{g(x)}{f(x)}.$$ Now, by the continuity of $f$ and $g$ at $x_0$, $h$ is also continuous at $x_0$, but $h$ is the same as the restriction of $D$ to $(x_0-\delta,x_0+\delta)$, a contradiction since $D$ is discontinuous in all $\mathbb R$.


问:
What is $2 - 1 + 1$?
$2-1+1$; a fairly straightforward question, but I (well, not me, but Henry Reich) found something strange.

Most people would evaluate it as $2+(-1)+1 = 2$; however, this goes against the famed, and fairly standard B.E.D.M.A.S./P.E.D.M.A.S., which states that addition goes first, and then subtraction.

If this is the case, then the answer is $2 - (1 + 1) = 2-2 =0$.

Which is the correct answer, and why is the conventional way (B.E.D.M.A.S./P.E.D.M.A.S.) so ambiguous?

答: Multiplication and division can be performed in any order, and the same holds for addition and subtraction. I learned it as PEMDAS, to drive the first point home. I suspect the only reason that addition always precedes subtraction in the mnemonic is that it makes the acronym simpler to say.</p>

Title: What is $2 - 1 + 1$?


问:
A relationship between $a$ and $b$ related to algorithm Rock(1)-Scissor(0)-Paper(2)

This is related to algorithm Rock(1)-Scissor(0)-Paper(2)

I need to find a relationship between $a$ and $b$ so that
$$\begin{matrix}
a & b \\
1 & 0\\
0 & 2\\
2 & 1
\end{matrix}$$
is always true and
$$\begin{matrix}
a & b \\
1 & 2\\
0 & 1\\
2 & 0
\end{matrix}$$
is always false. I had read a book about this involving to binary and computer. I'm quite sure that we must do some analytics among
$$\begin{matrix}
01 & 00 \\
00 & 10
\end{matrix}$$
and
$$\begin{matrix}
00 & 10 \\
10 & 01
\end{matrix}$$
and
$$\begin{matrix}
10 & 01 \\
01 & 00
\end{matrix}$$
What should I do next ?? And I wanna know more about this knowledge, thanks a real lot !

Edit: another solution that is
$$a- b+ 2\equiv 0\mod 3$$

答:
I want some new observations for this problem like system of modulo-equations

This answer shows three systems.

Let us define $f(a,b):=pa+qa^2+rb+sb^2$ where $p,q,r,s$ are integers. (This is because $x^3\equiv x\pmod 3$ always holds.)

We want $f(a,b)$ to satisfy
$$f(1,0)\equiv f(0,2)\equiv f(2,1)\pmod 3,$$
i.e.
$$p+q\equiv -r+s\equiv -p+q+r+s\pmod 3,$$
i.e.
$$r\equiv -p+q\pmod 3,\qquad s\equiv -q\pmod 3$$

So, $f(a,b)$ can be written as
$$f(a,b)\equiv pa+qa^2+(-p+q)b-qb^2\pmod 3$$

For $(p,q)\equiv (0,1)\pmod 3$, the only solutions of $f(a,b)\equiv a^2+b-b^2\equiv 1\pmod 3$ are $(a,b)\equiv (1,0),(0,2),(2,1),(1,1),(2,0)\pmod 3$.

For $(p,q)\equiv (1,1)\pmod 3$, the only solutions of $f(a,b)\equiv a+a^2-b^2\equiv 2\pmod 3$ are $(a,b)\equiv (1,0),(0,2),(2,1),(0,1),(2,2)\pmod 3$.

For $(p,q)\equiv (1,2)\pmod 3$, the only solutions of $f(a,b)\equiv a-a^2+b+b^2\equiv 0\pmod 3$ are $(a,b)\equiv (1,0),(0,2),(2,1),(0,0),(1,2)\pmod 3$.

From these observations, we get the following three systems whose solutions are $(a,b)=(1,0),(0,2),(2,1)$.

(1)$$\begin{cases}a^2+b-b^2\equiv 1\pmod 3
\\\\a+a^2-b^2\equiv 2\pmod 3\end{cases}$$

(2)$$\begin{cases}a^2+b-b^2\equiv 1\pmod 3
\\\\a-a^2+b+b^2\equiv 0\pmod 3\end{cases}$$

(3)$$\begin{cases}a+a^2-b^2\equiv 2\pmod 3
\\\\a-a^2+b+b^2\equiv 0\pmod 3\end{cases}$$


问:
Five Cubes in Dodecahedron
I will demonstrate why the group of rotational symmetries of a Dodecahedron is $A_5$. For that, we have to find five nice objects, on which the group of symmetries acts.

One object is "Cubes" inscribed in Dodecahedron. But since my demonstration will be on "Black-Board" rather than "Slide-Show". Can anybody suggest me a simple way to describe five cubes in Dodecahedron, which are the principal objects of permutations in the group of rotations symmetries of Dodecahedron?

答:
Also, if you look at a main diagonal of the inscribed cube (there are 4 such diagonals), the ends of the diagonal are opposing vertices on the dodecahedron (see above graphic).

For each set of opposing vertices on the dodecahedron there are two inscribed cubes with a  main diagonal along the same axis.  Rotation along this axis by $2\pi/3$ fixes these two cubes and rotates the other three cyclically among themselves. In other words, this rotation of the dodecahedron is a three cycle among permutations the 5 cubes.  All 20 three cycles in $A_5$ are obtained this way, and they generate $A_5$.

Title: Five Cubes in Dodecahedron


问:
Measurable functions in a countable co-countable $\sigma$-algebra
I found a interesting problem that says: Let $(X,S)$ be a measurable space where $X=\mathbb R$ and $S$ is the countable co-countable $\sigma$-algebra in $\mathbb R$, i.e. $S=\{A\subset\mathbb R: A\ \lor\ \mathbb R-A\ \ is\ finite\ or\ countable\}$. So the point is to describe the S-measurable functions $\ f:X\to \Bbb R^*$, where $\Bbb R^*$ are the extended real numbers.</p>

答: Let $f$ be a measurable function. Divide the line into a countable number of half-closed intervals and look at their preimages under $f$. Exactly one of them is uncountable. Break that interval into two and look at their preimages. Exactly one of them is uncountable. Continue like this. By the nested interval theorem there is exactly one point whose preimage is uncountable and cocountable. So the function is constant except on a countable set.


问:
Can the null set be written like $(a,a)$?
Can the null set $\varnothing$ be written as $(a,a)$ for some real number $a$? It does not contain a. It does not contain anything else. So can I represent the null set as an interval like this?

答:
$(a,a)$ is usually referred to as a degenerate interval (since it actually contains no points), and is equivalent to the empty set.  However it would be clearer to write $\emptyset$ for the empty set as this is the more conventional notation.

The use of $(a,a)$ for the empty set normally arises when you are discussing intervals where the end points are chosen depending on other elements of the problem under discussion, and it is polite to note that the interval can degenerate to the empty set (and sometimes important to observe what difference this may make to the analysis).


问:
Avoid Loss of significance error
I want to compute $f(x)$to avoid loss of significance$$f(x)=\frac{1-\cos x}{x^2}$$

One way is writing Taylor series approximation for $\cos x$ about $x=0$. what about this approach:

$$f(x)=\frac{2\sin^2(\frac x2)}{x^2}$$

Is this ok ?

答:
Better than Taylor series would be the $[2n,2n]$ Padé approximants. Foe example, for $n=2$ you will have
$$f(x)=\frac{1-\cos (x)}{x^2}\sim\frac {65520-3780 x^2+59 x^4 } {131040+3360 x^2+34 x^4 } $$ Compared to a Taylor expansion, the difference is
$$\text{Padé- Taylor}=\frac{127 }{87178291200}x^{10}+O\left(x^{12}\right)$$

For example, for $x=10^{-6}$, the difference between the exact and approximated values is $1.46\times 10^{-69}$.

Using the next approximant $(n=3)$,it would be
$$f(x)=\frac{1-\cos (x)}{x^2}\sim\frac {5491886400-346666320 x^2+7038360 x^4-45469 x^6} {24(457657200+9249240 x^2+86030 x^4+389 x^6 )} $$ Compared to a Taylor expansion, the difference is
$$\text{Padé- Taylor}=\frac{11321 }{438437062103040000}x^{14}+O\left(x^{16}\right)$$
For example, for $x=10^{-6}$, the difference between the exact and approximated values is $2.58\times 10^{-98}$.

Title: Avoid Loss of significance error


问:
Collatz Conjecture: Unclear on the last argument for Lemma 8 in the paper by Simon & Weger 2005
I am reading through the 2005 paper by Simons and Weger on the Collatz Conjecture.

Lemma 8 has the following conclusion:

$$\delta K < K +L < 1.000001\delta K$$

I am clear on most of the argument.  I am clear on this point:

$$0 < K+L - \delta K < \frac{m}{X_0\log 2} \le \frac{K}{X_0\log 2} < 10^{-17}K$$

I can see how "the inequalities readily follow" but when I do my calculations, I am reaching a conclusion significantly better.  I am wondering if I am making a mistake since it is not clear to me why the author would be reporting a weaker result.

When I add $\delta K$ to both sides, I get this:

$$\delta K < K + L < 10^{-17}K + \delta K = (10^{-17} + \delta)K < (6.31 \times 10^{-18} + 1)\delta K$$ which appears to be a better result than the conclusion.

Is my reasoning correct and the conclusion is provided to greatly simplify the result?  Did I make a mistake in my reasoning?

If my reasoning is correct, can this result then be restated for example as:

$$\delta K < K + L < 1.0000000001\delta K$$

答:
Your extra inequality of

$$(10^{-17} + \delta)K \lt (6.31 \times 10^{-18} + 1)\delta K \tag{1}\label{eq1A}$$

is, after dividing by $K$ and then subtracting $\delta$ from both sides, equivalent to

$$10^{-17} \lt 6.31 \times 10^{-18}\delta \iff \frac{10}{6.31} = 1.5847\ldots \lt \delta \tag{2}\label{eq2A}$$

Since the paper gives, near the top of page $53$ before its Lemma $1$, that

$$\delta = \frac{\log 3}{\log 2} = 1.5849\ldots \tag{3}\label{eq3A}$$

and assuming this definition is not changed anywhere later, which a scan of the paper seems to indicate this being the case, this means \eqref{eq2A} is true.

I don't know why the authors didn't use more decimal digits (e.g., up to $16$ zeroes before a $1$ instead of using just $5$ zeroes) to state a tighter upper bound, but possibly they felt there wasn't any need for the extra accuracy for their purposes.


问:
In the given figure find angle x.
I approached it these ways-

1)Constructing parallel lines and then using similarity

2)Trying exterior angle property

3)Constructing perpendiculars

I am not able to understand the suitable construction.

答:
I recently added two sections to the end of this answer: an Addendum which moderately refines the original answer, and an Addendum-1 which significantly refines the original answer.

Unclear if the following method is off-limits, given the tags of geometry and euclidean geometry.

The problem is routinely solved via double use of the law of sines.

Initially, since the small triangle at the bottom has angles of $45^\circ, 120^\circ, 15^\circ, ~$ you know that the base of the triangle is $$\frac{\sin(120^\circ)}{\sin(15^\circ)} \approx 3.346065215.$$

With the base of the triangle solved, you also know that the large triangle has angles $x^\circ, (135 - x)^\circ, 45^\circ.$

The second application of the law of sines yields

$$\frac{\sin(x^\circ)}{3.346065215} = \frac{\sin(135-x)^\circ}{3}.$$

You also know that

$$\sin(135-x)^\circ = \sin(135^\circ)\cos(x^\circ) -
\sin(x^\circ)\cos(135^\circ).$$

You also know that for any angle $\theta$,

$$\sin^2(\theta) + \cos^2(\theta) = 1.$$

Putting this all together, I computed that

$$x = 75^\circ.$$

A variation at the end would be to compute $\tan(x^\circ)$ and then feed in the computed number to a calculator's ArcTan function.

Note

A way of verifying that $x = 75^\circ$ is an exact answer is to note that

$$\sin(15^\circ) = \frac{\sqrt{2 - \sqrt{3}}}{2} ~\text{and}~ \sin(75^\circ) = \frac{\sqrt{2 + \sqrt{3}}}{2}.$$

Addendum

I was bothered by how inelegant my original answer was, so I thought that I'd take another crack at it.

There are three key identities:

$$\sin(15^\circ) \times \sin(75^\circ) = \frac{1}{4}. \tag1$$

$$\sin^2(75^\circ) = \frac{2 + \sqrt{3}}{4}.\tag2$$

$$3 + \sqrt{3} = \sqrt{6}\sqrt{2 + \sqrt{3}}. \tag3$$

As indicated in my original answer, the base of the triangle is

$$\frac{\sin(120^\circ)}{\sin(15^\circ)} = \frac{\sqrt{3}}{2} \times [4 \times \sin(75)^\circ] = 2\sqrt{3}\sin(75^\circ).$$

Denoting the left side of the large triangle by $c$, the law of cosines

[i.e. $c^2 = a^2 + b^2 - 2ab\cos(c)$]

may be used to compute

$$c^2 = [12\sin^2(75^\circ)] + 9 - \left[2 \times 3 \times 2\sqrt{3}\sin(75^\circ) \times \frac{1}{\sqrt{2}}\right]$$

$$= \left[12 \times \left(\frac{2 + \sqrt{3}}{4}\right)\right] ~+~ 9 ~-~
\left[12\left(\frac{\sqrt{3}}{\sqrt{2}}\right)
\times \frac{\sqrt{2 + \sqrt{3}}}{2}\right]$$

$$= \left[3 \times \left(2 + \sqrt{3}\right)\right] ~+~ 9 ~-~
\left[3\left(\sqrt{6}\right)
\times \left(\sqrt{2 + \sqrt{3}}\right)\right].$$

Using the identity of equation (3) above, this expression may be simplified to

$$= \left[3 \times \left(2 + \sqrt{3}\right)\right]
~+~ 9 ~-~
\left[3 \times \left(3 + \sqrt{3}\right)\right]$$

$$= 6.$$

In the large triangle, with the side length of $\sqrt{6}$ opposite the $45^\circ$ angle, and the side length of $2\sqrt{3}\sin(75^\circ)$ opposite the $x^\circ$ angle, the law of sines may again be used to compute

$$\sin(x^\circ) = 2\sqrt{3}\sin(75^\circ) \times
\frac{\sin(45^\circ)}{\sqrt{6}}$$

$$= \sin(75^\circ) \times \frac{2\sqrt{3}}{\sqrt{2} \times \sqrt{6}} = \sin(75^\circ) \implies $$

$$x = (75^\circ).$$

Addendum-1

Construct a (perpendicular) altitude from the lower left vertex of the triangle to the right side of the triangle.  Assume that the intersection of the altitude with the right side of the triangle creates a 45-45-90 right triangle, with each leg of the triangle equal to $(1 + y)$.

This means that the constructed altitude also creates a 60-30-90 triangle, with one of the legs equal to $(y)$.

Since the legs of the 45-45-90 triangle are equal, the length of the constructed altitude is also $(1+y).$

Since the tangent of $60^\circ = \sqrt{3},$ this implies that

$$\frac{1 + y}{y} = \sqrt{3} \implies y(\sqrt{3} - 1) = 1 \implies y = \frac{1}{\sqrt{3} - 1} $$

$$ = \frac{1}{\sqrt{3} - 1} \times
\frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{\sqrt{3} + 1}{2}.$$

This implies that the constructed altitude has also created a right triangle with one of the angles = $(x^\circ)$ such that

$$\tan(x^\circ) = \frac{1+y}{2-y} ~=~
\frac{1 + \frac{\sqrt{3} + 1}{2}}{2 - \frac{\sqrt{3} + 1}{2}}$$

$$= \frac{2 + \sqrt{3} + 1}{4 - \sqrt{3} - 1} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}} = \tan(75^\circ) \implies x = (75^\circ).$$

Title: In the given figure find angle x.


答:
Hint:

Draw $\angle ABE = 15^0$. Then what is $\angle BED$? Can you show $\triangle BDE, \triangle BEA \,$ and $\triangle BEC$ are isosceles? Also what is $\angle CED$? From there, it is straightforward to show $\angle C = 75^0$.

Just an observation - $E$ is the circumcenter of $\triangle ABC$.

Title: In the given figure find angle x.

Body:
I approached it these ways-

1)Constructing parallel lines and then using similarity

2)Trying exterior angle property

3)Constructing perpendiculars

I am not able to understand the suitable construction.


问:
Evaluating $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$
I am trying to evaluate $\int_0^{\infty}\frac{2}{x^2-8x+15}dx$. Factoring and using partial fraction decomposition, I have found that the indefinite integral is:
$$\ln{|x-5|}-\ln{|x-3|} + C$$

But when evaluating the improper integral, I don't know how to deal with the difference between two diverging limits, or even whether or not it even converges. My intuition tells me that since the denominator of the original integral is a second degree polynomial, it converges, but I'm not entirely sure.

$$\lim_{b\to\infty}{(\ln{(b-5)}-\ln(b-3))}$$

How do you evaluate this limit?

答:
Combine and you have
$$ \log(b -5) - \log(b - 3) = \log\left({b - 5 \over b - 3}\right).$$
Exploit the continuity of the log function at 1 to finish.
This is only useful if your start point for integrating is beyond the nasty places $x = 3 $ and $x = 5$.  Otherwise integrating across these points will render the integral nonexistent.</p>


问:
Prove $r_1 = r_2 \Leftarrow\Rightarrow Sr_1 = Sr_2$
With a construction of $\Bbb{R}$, to every real number $r\in\Bbb{R}$, $Sr:= (-\infty, r)\cap\Bbb{Q}$.

$Sr$ is the set of all rational numbers < r.

Prove, rigorously, $r_1 = r_2  \Leftarrow\Rightarrow Sr_1 = Sr_2$.

答:
$\bullet \quad r_1=r_2 \Longrightarrow Sr_1=Sr_2$ is true by definition

$\bullet  \quad Sr_1=Sr_2 \Longrightarrow r_1=r_2$ : you can suppose without loss of generality that $r_1 \geq r_2$. If by the absurd you had $r_1>r_2$ then you would have $Sr_1 \backslash Sr_2 = [r1,r2) \cap\mathbb{Q}$ but $[r1,r2) $ is a not empty interval of $\mathbb{R}$ and $\mathbb{Q}$ is dense in $\mathbb{R}$ therefore you could find an element in $Sr_1$ that is not in $Sr_2$ : contradiction. Hence $r_1=r_2$


问:
How to prove $\lim_{n\to \infty} \frac{n!}{n^n}$ is zero?
I am struggling in this limit: $$\lim_{n\to \infty} \frac{n!}{n^n}$$
I really worked on this for few days: clearly, it goes to zero but I couln't do it rigorously. I'd really appreciate your help. Thanks from now.

答: $\begin{aligned} y=& \lim _{n \rightarrow \infty} \frac{n !}{n^{n}} \\ \ln y &=\lim _{n \rightarrow \infty} \ln \frac{n !}{n^{n}} \\ &=\lim _{n \rightarrow \infty}(n \ln n-n-n \ln n) \\ &=\lim _{n \rightarrow \infty}-n \\ &\\ \ln y \rightarrow-\infty \\ \Rightarrow & y \rightarrow 0 \end{aligned}$


问:
Getting Standard Deviation of Population from s. d. of individual samples. Given mean and standard deviation of samples, with no access to data.
I know mean and standard deviation of individual samples (let say $10$ data points each, and no access to the underlying data); could I get the standard deviation of the population (let say $100$ samples) from the standard deviation of the individual samples? Which is the relationship?

EDIT

Thanks to BruceET's Answer. I've applied approach-$2$ to our data, and I'm just a little puzzled on results I get when calculating $S_{all}$. I would expect that $S_{all}$ is greater than the average of the individual $S_{n}$ (since there is a spread as well of the mean of subsets); on the contrary I get an $S_{all}$ that is smaller. How this should be understood?

Processed Data

答:
thanks a lot for your answer, I've applied approach 2 to our data, and I'm just a little puzzled on results I get when calculating Sall. I would expect that Sall is greater than the average of the individual Sn (since there is a spread as well of the mean of subsets); on the contrary I get an Sall that is smaller. How should that be understood?
In the annex you find an immage with some of the processed data. Thanks in advance and, hopefully ..., happy 2021.
Sergio

enter image description here


答:
You may use the quotient criterion for series:

$$a_n = \frac{n!}{n^n}\Rightarrow \frac{a_{n+1}}{a_n}=\frac 1{\left(1+\frac 1n\right)^n}\stackrel{n\to\infty}{\longrightarrow}\frac 1e<1$$

Hence, the series $\sum a_n$ is convergent and hence $\lim_{n\to\infty} a_n = 0$.

Title: How to prove $\lim_{n\to \infty} \frac{n!}{n^n}$ is zero?

Body:
I am struggling in this limit: $$\lim_{n\to \infty} \frac{n!}{n^n}$$
I really worked on this for few days: clearly, it goes to zero but I couln't do it rigorously. I'd really appreciate your help. Thanks from now.


问:
When two elements of a group generate the same subgroup
Let $G$ be a group and $x,y$ two elements from $G$. Suppose that $\langle xy \rangle = \langle yx \rangle$ , i.e. they generate the same subgroup of $G$. I want to find a counterexample in which $xy \neq yx$ and $xyyx \neq 1$. Maybe in a permutation group such a counterexample exists, but I haven't found one yet.

A counterexample for $xy \neq yx$ can be easily found in dihedral group $D_6$, for example $x=\rho^2\sigma$, $y=\rho\sigma$, where $\sigma^2=\rho^6=1$, but, in this case, $xyyx=1$.

Thank you in advance!

答:
Say $xy = z$. Then asking $\langle z \rangle = \langle xy \rangle = \langle yx \rangle$ is the same thing as asking $yx = z^k$, where $k$ had better be coprime to the order of $z$ (do you see why?).

Then you are asking for a group with elements $x$ and $z$ so that

$x^{-1} z x = z^k$

$z^n = 1$ for $n$ coprime to $k$.

Given such elements, we can set $y = x^{-1}z$, as then $xy = z$ is obvious, and $yx = x^{-1}zx = z^k$.

If $k \neq 1$ then we'll know $xy = z \neq z^k = yx$.

If $k \neq n-1$, then we'll know $xy = z \neq (z^k)^{-1} = yx$, and so $xyyx \neq 1$.

This leads us to a group $G$ with presentation

$$\langle x, z ~|~ x^{-1}zx = z^k, z^n = 1 \rangle$$

Ideally we'd get a finite group at the end of the day, so let's assert $x$ has finite order too.

$$\langle x, z ~|~ x^{-1}zx = z^k, z^n = 1, x^m = 1 \rangle$$

This group looks like two cyclic groups $\langle x ~|~ x^m = 1 \rangle$ and $\langle z ~|~ z^n = 1 \rangle$ that have been "glued together" in such a way that $x^{-1}zx = z^k$.... Luckily there's a construction called the semidirect product which exists precisely for gluing groups together in this way!

Then our group is $C_m \ltimes_k C_n$, for (almost) any $m$ and $n$ we like. For technical reasons we need $m \mid n-1$ if we want to get a real group out the other side. Here the action is given by $x^{-1}zx = z^k$ for any $k \neq n-1$ (this requirement ensures $(xy)(yx) = zz^k \neq 1$).

By $C_m \cong \mathbb{Z}/m$, I mean the cyclic group of order $m$ where we write our elements as $x^i$ (or $z^i$ for $C_n$) instead of integers.

This may not be a super satisfactory answer (since it's not the kind of group that one meets in an introductory group theory class), but hopefully this shows the kind of strategies which one might use to build these examples for yourself. As a (fun?) exercise, you should chase through the definition of a semidirect product and make sure you understand the multiplication table for some small choices of $m$, $n$, and $k$.

For instance, what are the group elements of $C_3 \ltimes_3 C_7$? It's a group of size $21$, so it's reasonable to get your hands on it. What is the group multiplication? Good luck!

I hope this helps ^_^


答:
A quick and dirty way to show it goes to 0 really fast:
$$
n! \leq n^{\lfloor n/2\rfloor}\cdot (n/2)^{\lceil n/2\rceil}
$$
So
$$
\frac{n!}{n^n} \leq \frac{1}{2^{\lceil n/2\rceil}}
$$


问:
What is the intuition behind the defintion of the degree of a field extension?
Longtime lurker and a first time poster.

This question has been bugging me A LOT, and I couldn't find many relevant posts about my question. To give more context I have been wanting to study the theory of field and extensions and ultimately Galois theory. Then I started using Dummite and Foote's treatment of it(From there abstract algebra book) and they didn't especially give much intuition behind the definition of the degree of a field extension. Adding examples to your answers is greatly appreciated!

答:
Whenever you have an inclusion $K\subset L$ of fields the larger field $L$ has a natural structure of vector space over the smaller field $K$.

Indeed $L$ is an abelian addtive group with a multiplication by elements in $K$ and the vector space axioms follow at once frm the fact that $L$ is a field and elements of $K% are also elements of $L$.

Then one defines the degree of the extension as the dimension of $L$ as $K$-vector space:
$$
[L:K]=\dim_K(L).
$$
This (possibly infinite) value is a raw index of complexity of the field $L$ relatively to $K$.

Its knowledge gives some information, for instance if $[L:K]=n$ we know that every element $z\in L$ satisfies a polynomial relation
$$
P(z)=0
$$
where $P(X)$ is a polynomial with coefficients in $K$ of degree $\leq n$.

Conversely, if $P(X)$ is an irreducible polynomial with coefficients in $K$ of degree $n$ and $z$ is a root, then the set
$$
L=\{a_0+a_1z+\cdots a_{n-1}z^{n-1}\,|\,a_i\in K\}
$$
is actually a field containing $K$ such that $[L:K]=n$ (this example is very relevant for arithmetic).


问:
Can different fractals have the same fractal dimensoins?
I first noticed that fractals can be defined by real numbers from this youtube video:
https://youtu.be/gB9n2gHsHN4

My question is that:

Are these fractal dimensions unique to the fractals, Can two different fractals have the same fractal dimensions?

$NOTICE:$

I am not using the word different based on translation, rotation and scaling of the fractal. I hope you understand that what I am trying to define when I use the word 'different' becuase I am not a mathematician and that's why I am struggling to define the word 'different'.

Since I am not a mathematician  I will request to not use very advance mathematics. Please use easy maths ('Easy' does not mean high school maths) and if it's possible then please make it a little bit intuitive.

答:
There are multiple notions of fractal dimension, but I'll focus on Hausdorff dimension.

The Hausdorff dimension is not unique to fractals. A Wikipedia article contains a list of fractals alongside their Hausdorff dimension, and it has "different" fractals with the same dimension.

For instance, the Hilbert curve and the Peano curve (which are both space filling curves) have dimension 2. They are clearly "different", but whether they are fractals depends on the definition (as they have integer Hausdorff dimension).

There are also fractals with non-integer dimensions. Both the Koch curve and the Cantor dust in 2 dimensions have the same dimension of $\log 4/\log 3$.


问:
How to calculate second derivative using definition?
I am currently reading Analysis on Manifolds by Munkres and this problem has me flustered.

$f(x, y) = \frac{xy(x^2 - y^2)}{x^2 + y^2}$ and $f(0) = 0 $. Show that $D_1D_2 f$ and $D_2 D_1f$ exist at $0$ but are not equal there?

I can calculate $D_1D_2f$ and $D_2D_1f$ using partial differentiation. Thus
\begin{align}
D_1D_2 f &= \frac{\partial }{\partial x}\left( \frac{\partial f}{\partial y}\right) = \frac{(x^2 - y^2)(x^4 + 10x^2y^2 + y^4)}{(x^2 + y^2)^3}\\
D_2D_1f &= \frac{\partial }{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{(x^2 - y^2)(x^4 + 10x^2y^2 + y^4)}{(x^2 + y^2)^3}
\end{align}

So I know for a fact that this is not the way to go. However, how do I actually calculate $D_1 D_2 f(\bf 0)$ using definition of derivative? I can calculate $D_2 f(\bf 0)$ as follows:
\begin{align}
D_2 f({\bf 0}) &= \lim\limits_{\lambda \to 0}\frac{f({\bf 0} + \lambda(0 , 1)) - f({\bf 0})}{\lambda} = \lim\limits_{\lambda \to 0} \frac{f(0, \lambda) - 0}{\lambda} \\
&= 0
\end{align}

How do I get a closed form function for $D_1 f$ 0r $D_2 f$ so I can take a second derivative?

答:
You have

$$\begin{cases}
\frac{\partial f}{\partial x}(0,y)=-y\\
\frac{\partial f}{\partial y}(x,0)=x
\end{cases}$$

From which it follows that:

$$\frac{\partial^2 f}{\partial x \partial y}(0,0)=1\neq −1= \frac{\partial^2 f}{\partial y \partial x} (0,0)$$


问:
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$?
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.

答:
The positive root of the equation $1-\cos x = \frac{x^2}{3}$ is approximately $2.16$
so the inequality is valid on a larger interval, say $[-2, 2]$. To show this consider
$$\frac{1-\cos x}{\frac{x^2}{3}} = \frac{2 \sin^2 \frac{x}{2}}{\frac{x^2}{3}}= \frac{3}{2} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2$$
Now, the function $t\mapsto \sin t$ is concave on $[0, \pi]$, so $\frac{\sin t}{t}$ is decreasing on $[0, \pi]$. Therefore, on the interval $[-2, 2]$ we have $\frac{\sin \frac{x}{2}}{\frac{x}{2}} \ge \sin 1$ and so
$$\frac{1-\cos x}{x^2/3} \ge \frac{3}{2} \sin^2 1= 1.06.. > 1$$

On the interval $[-1,1]$ the inequality is
$$1- \cos x \ge 2 \sin^2\frac{1}{2}\cdot  x^2$$
where $2 \sin^2 \frac{1}{2} = 0.45969\ldots$
and this is the best estimate on $[-1,1]$.


问:
Can anyone explain or/and provide a proof of the following statement?
Can anyone explain or/and provide a proof of the following statement?

I am reading a book on Combinatorics. In the book the author provides the following statement along with an explanation:

The cardinality of the set of all the functions from an n-set to an m-set is $m^n$.

The explanation given is:

Let $D = \{1, 2 .... , n\}$ be the n-set and let $R = \{r_1, r_2, ....., r_m\}$. Given any function $f$ from $D$ to $R$, we write down the sequence $(f(1), f(2), ... , f(n))$ of length $n$ with entries from $R$.

I am not able understand the explanation so I am looking for a more elaborate or a more visual proof/explanation of the statement.

答:
It may help to consider some small examples. For example, $m=2, n=3$: let $A=\{a,b,c\}$ and $B=\{x,y\}$. Here are all the maps $A\to B$ (all $2^3=8$ of them):

$$\begin{pmatrix}a&b&c\\x&x&x\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\x&x&y\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\x&y&x\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\x&y&y\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\y&x&x\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\y&x&y\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\y&y&x\end{pmatrix}$$
$$\begin{pmatrix}a&b&c\\y&y&y\end{pmatrix}$$

To generalise: for the first element in $A$ you have $m$ choices (in this case, you could've chosen $a\mapsto x$ or $a\mapsto y$). For each of those choices, you have $m$ choices to map the second element of $A$. This gives you $m^2$ different maps of the first two elements. For each of those, you have $m$ ways to map the third element - in total $m^3$ maps for the first three elements. And so on - eventually you conclude that the number of maps of all $n$ elements of $A$ is $m^n$.


答:
Every such sequence $(f(1),...,f(n))$ determines a function $f: D \to R,$
and all such functions can be obtained in this way.
For each value of $f$, or in other words, for each sequence element, there are $m$ choices. Since we need to define $f$ on $n$ elements, there are in total $m \times m \times ... \times m = m^n$ possible functions $f$.

Another way to phrase it: to define $f$ is the same as to specify its values on $D,$ and there are $m^n$ distinct ways to do this.

Body:
Can anyone explain or/and provide a proof of the following statement?

I am reading a book on Combinatorics. In the book the author provides the following statement along with an explanation:

The cardinality of the set of all the functions from an n-set to an m-set is $m^n$.

The explanation given is:

Let $D = \{1, 2 .... , n\}$ be the n-set and let $R = \{r_1, r_2, ....., r_m\}$. Given any function $f$ from $D$ to $R$, we write down the sequence $(f(1), f(2), ... , f(n))$ of length $n$ with entries from $R$.

I am not able understand the explanation so I am looking for a more elaborate or a more visual proof/explanation of the statement.


问:
Can someone check my calculation, if it's wrong can you explain where I went wrong. Thank you
$\text{Percentage decrease }= \dfrac{\text{actual decrease}}{\text{original value}} \times 100$

Decrease between $2003$ and $2018 = 5.5 \cdot 10^6 \text{ km}^2 – 4.6 \cdot 10^6 \text{ km}^2 = 0.9 \cdot 10^6 \text{ km}^2$

Percentage decrease $2003$ of $2018$ value $=\dfrac{ 0.9 \cdot 10^6 \text{ km}^2}{4.6 \cdot 10^6 \text{ km}^2} = 19.6\%$

答:
Well I assume you are looking for the following:

The decrease of land between $2003$ and $2018$ is $5.5 \times 10^6 \text{ km}^2-4.6\times 10^6 \text{ km}^2=0.9\times10^6 \text{ km}^2$.

Therefore the percentage decrease is $(\frac{0.9\times 10^6}{5.5 \times 10^6}) =16.4\%$.

The percentage decrease is $\frac{\text{amount of decrease}}{\text{original value}} \times 100$ by definition.

Here your original value is the area of land in $2003$ that is $5.5\times 10^6 \text{ km}^2$


问:
Is probability influenced by order of actions?
I'm wondering if probability is influenced by order of actions. For instance, lets have a problem:

Suppose there are two holes close together in a golf course. The golfer can hit the ball into either hole with equal probability. He is allowed to hit 4 times. Assuming he can always get the ball into one of the two holes. What is the probability that he could get 2 balls in each hole?

My solution is 1/5. This is result of (accepted results)/(possible outcomes). This seems a little simple to me, is it possible that the probability is influenced by order in which the player hits the shots?. Could anyone tell me if the result is correct please? Thanks in advance.

答:
When the problem says "either hole with equal probability", they mean that every time he hits a ball, the probability it goes into a specific hole is 0.5. In other words, no, the problem does not allow for the probability to be influenced by what order shots are taken (this is completely irrelevant to whether or not it is the case in real life that the arrangement of grass or whatnot could influence the probability between shots).

Besides, each shot itself should be indistinguishable from the rest; otherwise, it's not "4 occurrences of one shot", but "1 occurrence each of 4 slightly different shots", making the problem impossible to analyze without knowing the exact probabilities related to each shot. Because the shots are indistinguishable, there is no real meaning to what "order" they are taken in.

By the way, this means that the correct answer is $\frac{\binom{4}{2}}{2^4}=\frac{3}{8}$.


问:
What is $2 - 1 + 1$?
$2-1+1$; a fairly straightforward question, but I (well, not me, but Henry Reich) found something strange.

Most people would evaluate it as $2+(-1)+1 = 2$; however, this goes against the famed, and fairly standard B.E.D.M.A.S./P.E.D.M.A.S., which states that addition goes first, and then subtraction.

If this is the case, then the answer is $2 - (1 + 1) = 2-2 =0$.

Which is the correct answer, and why is the conventional way (B.E.D.M.A.S./P.E.D.M.A.S.) so ambiguous?

Title: What is $2 - 1 + 1$?

Body:
Take it like a succesion,
$$2+(-1)+1=2$$
first
$$2+(-1)=1$$ then
$$1+1=2$$
like this
$$2+(-1)+1=2$$
$$2-1+1=2$$
$$2=2$$


答:
A function from a set $A$ to a set $B$ assigns to every element in $A$ precisely one element in $B$. If $A=\{a_1,\ldots,a_n\}$ and $B=\{b_1,\ldots,b_m\}$, this means that for every $a_i$ we can choose one of the $b_1,\ldots, b_m$ as its image.

Lets give an easy example. Suppose you have bought two plain white t-shirts $t_1$ and $t_2$ from the store, and you have bought three different colours of paint; blue, yellow and red. Then the a function from $T=\{t_1,t_2\}$ to $C=\{$blue, red, yellow$\}$ is a choice of a color for each shirt. For the first shirt $t_1$ there are $3$ different choices. For the second shirt $t_2$ there are again $3$ choices. These choices are mutually independent so we get $3\times 3=3^2$ choices of functions.

We can generalise this to $n$ t-shirts $t_1,\ldots, t_n$ and $m$ colours $c_1,\ldots, c_m$. For the first shirt we can choose between all the $m$ colours, for the second shirt we can again choose between all these colours, and so we end up with $m\times m\times \ldots \times m$ different combinations, which is precisely $m^n$.

Title: Can anyone explain or/and provide a proof of the following statement?


答: Let $D = \{1, 2 .... , n\}$ be the n-set and let $R = \{r_1, r_2, ....., r_m\}$, and let $f: D \to R$ be a function from $D$ to $R$. The first element $1$ can be sent by $f$ in any element of $R$, meaning that you have $m$ possible images (for $1$). For $2$ the same concept applies, and so for every possible image of $1$, there are $m$ choices for the image of $2$, and in total $m\times m$ possible combinations for both the images of $1$ and $2$. You can now prove the statement by induction.


问:
Verify Divergence Theorem for $F(x,y,z)=(xz,0,0)$ on a solid bounded by the paraboloid $z=x^2+y^2$ and the plane $z=1$
Here is the problem: "Verify Gauss Theorem for $F:\mathbf{R}^3 \rightarrow\mathbf{R}^3$ with $F(x,y,z)=(xz,0,0)$ in the positively oriented solid $S$, bounded by the paraboloid $\{x^2+y^2\}$ and the plane $\{z=1\}$"

Also, here it is the definition of the theorem i have:

Gauss Theorem: Let $F=(P,Q,R):U \subset \mathbf{R}^3\rightarrow{\mathbf{R}^3}$ which is $C^1$ in $U$, and $\Omega \subset U$ a Jordan measurable set such that  $S=Fr(\Omega)=\partial \Omega$ it is a surface by pieces and $\Omega \cup S \subset U$ then we have that:
$$\int_{\Omega}^{} \nabla \cdot \, F=\int_{S=\partial \Omega}^{}F \cdot d\sigma $$
where $\sigma$ is a parametrization.

The thing is that i think it verifies Gauss Theorem but i don't get that when i do the integrals, here is my attempt:

Let's define:
$$\Omega=\{(x,y,z): 0 \leq z \leq x^2+y^2 \}, z \in [0,1] $$
We want to calculate:
$$\int_{\partial \Omega }^{}F \cdot d\sigma $$
We define the following parametrization:
$\sigma_{1,2}:B\subset\mathbf{R}^2\xrightarrow{}\mathbf{R}^3$
$$\sigma_1 (u,v)=(u,v,u^2+v^2) $$
$$\sigma_2(u,v)=(u,v,0) $$
We then proceed to find the normal vectors $\frac{\partial \sigma_i}{\partial u} \times \frac{\partial \sigma_i}{\partial v} $

$$\frac{\partial \sigma_1}{\partial u} \times \frac{\partial \sigma_1}{\partial v}=(-2u,-2v,1) $$

Then i multiply this one for a $-1$ because i need it to point the other direction, so what i get is $(2u,2v,-1)$.
Also
$$\frac{\partial \sigma_2}{\partial u} \times \frac{\partial \sigma_2}{\partial v}=(0,0,1) $$
We do calculate $F\cdot d\sigma_1=2u^2(u^2+v^2)$ and $F\cdot d\sigma_2=0$.
We proceed to calculate the integral (which i will do it in polar coordinates):
$$ \int_{B}^{} F \cdot d\sigma_1 \cdot (\frac{\partial \sigma_1}{\partial u} \times \frac{\partial \sigma_1}{\partial v})=\int_{0}^{2 \pi}\int_{0}^{1} 2 r^5 cos^2 \theta  dr d\theta= \frac{\pi}{3}$$
(i also did use symbolab to compute it just  to make sure i did not make an error myself).
By the other hand, we need to calculate the following integral:
$$\int_{\Omega}^{} \nabla \cdot F $$
$\Omega$ is the same as before, we then proceed to calculate the integral in cylindrical coordinates, we also have that $\nabla \cdot F= z$ so
$$\int_{\Omega}^{}  \nabla \cdot F=\int_{0}^{2 \pi}\int_{0}^{1} \int_{0}^{r^2} hr \, dh dr d \theta=\frac{\pi}{6}$$

And as i said at the beginning, i do not know what i am doing wrong, did i messed up at the parametrization? or it does not satisfies the conditions of the theorem?... any hint will be useful, thank you.

答:
Please note your second integral is set up incorrectly. Otherwise you working is fine.

So it should be $\, \displaystyle \int_{0}^{2 \pi}\int_{0}^{1} \int_{r^2}^{1} \, h \, r \, dh \, dr \, d\theta$

Note the limits of $h$.