答:
It always helps to draw a sample space diagram. In the below table, the entries show the difference between Ben and Betty's scores:
\begin{array}{cc|ccccc}
& & & & \mbox{Ben} \\
&  & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
& 0 & 0 & 1 & 2 & 3 & 4 & 5 \\
& 1 & 1 & 0 & 1 & 2 & 3 & 4 \\
\mbox{Betty} & 2 & 2 & 1 & 0 & 1 & 2 & 3 \\
& 3 & 3 & 2 & 1 & 0 & 1 & 2 \\
& 4 & 4 & 3 & 2 & 1 & 0 & 1 \\
& 5 & 5 & 4 & 3 & 2 & 1 & 0
\end{array}
Since the outcomes are equiprobable, we know that the average distance is
$$
\frac{\text{Sum of differences}}{\text{No. of outcomes}}=\frac{70}{36}=\frac{35}{18} \, .
$$
I found that it was possible to add up each row in my head, and also use the symmetries of the question to make the calculation easier. Notice how the rows sum to $15$, $11$, $9$, $9$, $11$, $15$; this happens because it is just as likely for Ben to get the number $x$ and Betty to get $y$ as it is for Betty to get $x$ and Ben to get $y$.

Title: Grade 8 Random Number Question - Which concept to use

Body:
I found a question in Math Challengers 2012 exam (Provencial, part2, Q8). It states

Betty and Ben each select independently and at random an integer between 0 and 5 (inclusive). What is the average non-negative difference between their numbers? Express the answer as a common fraction.

You get about 2 minutes to do this without a calculator. Which concept do you use? Does this belong to the Combinatorics/Probability Generating Functions?


问:
Why is there a point on the unit circle that is not represented by these parametric equations?
$$x=\frac{2t+1}{2t^{2}+2t+1}$$ $$y=\frac{2t^{2}+2t}{2t^{2}+2t+1}$$By squaring $x$ and $y$ and adding them up, I obtained $x^2 + y^2 = 1$ after some algebraic manipulation. But the question asks which point is not represented by these parametric equations. The first thing that came to my mind was to set the denominator does not equal to zero but obviously, it was the wrong this to do. What am I supposed to do in this situation?

Part 2 of the question asks whether the mapping from parameters to points is one to one or many to one. I have identifies that $x$ is odd, so it is definitely one-to-one. However, $y$ is represented by an even function so it cannot by one to one. The answer says overall it is indeed one-to-one - how can this be if $y$ is even? Unless the mapping is only defined by the $x$ function?

答:
For part $2$, $x$ is not odd: check your calculations again. It is easier to manipulate $y$ as follows:

$$y = 1 - \frac{1}{2x^2+2x+1} = 1 - \frac{1}{2(x^2+x+0.5)} = 1 - \frac{1}{2(x + 0.5)^2 + 0.5}$$

which is symmetric across $t = -0.5$. So for any given $t$, only $t$ and its reflection $t - 2(t + 0.5) = -t - 1$ have the same $y$-coordinate. If you can show that $x$-coordinates are different for these two values of $t$, this implies one-to-oneness.

ContentLicense: CC BY-SA 4.0


问:
In how many ways can the letters of the word ELEMENTARY be permutated such that no two E's are next to each other?
What I did was to separate out the 3 E's (EEE) from the word ELEMENTARY, which leaves L M N T A R Y. For no two E's to be next to each other, we can place the E's in 8 positions (between the alphabets
L M N T A R Y and also before L and after Y). Firstly, since all the letters L M N T A R Y are distinct the number of ways to permute them is 7! = 5040. Then to place the 3E's in the 8 locations; firstly the first E have 8 choices to choose from, then second E has 7 choices to choose from and 3rd E has 6 choices to choose from. Then using product rule, 5040 x 8 x 7 x 6 = 1693440.

However, my answer is wrong, the method above is correct till the "7! = 5040". In the solution that was given to me, there are indeed 8 positions to place the 3 E's. However, instead of using 8 x 7 x 6, the solution used C(8,3) = 56. Then, using product rule, 5040 x 56 = 282240 ways of permutating ELEMENTARY such that no two E's are next to each other.

Why is my answer wrong? Thank you! :")

答:
you have 7 letters {L,M,N,T,A,R,Y}, u have 7!=5040 ways to permute and create new string.

Now you have 8 places total( before, after and between them) to put 3 letters. C(8,3)=56

hence you have total ${5040  \cdot 56=282240}$ ways to create new string.

The number of subsets of k elements from a set with n elements, denoted by
${{N}\choose{k}}$

observe that we ask here for the number of subsets, and then the
order is irrelevant. Each collection of k elements can be obtained in k! different
orderings, so to obtain the number of subsets of size k, we need to take the number
of ordered subsets, and divide by k!, that is $${(n \times (n-1) \times... \times(n-k+1)) \mid k!  }$$

what you have taken is number of ordered subset.

The number of ordered subsets of k elements from a set with n elements is
equal to ${n \times (n-1) \times... \times(n-k+1)}$


问:
What is the working out to find "b" in the triangle below?
The image is attached to this link  https://i.stack.imgur.com/oVfOR.png 1. What is the working out to find "b"? I understand that I need to find an angle in the triangle but what steps can I take to first of all get an angle in the triangle? From this, I will be able to use trig to find b.

答: The circumference of the sector $s = r \theta$, where $\theta$ is in radians. Since you know both $s$ and $r$ you can find $\theta$.


问:
How to solve for theta in this equation
This is part of a large problem regarding double integrals with polar equations. I have successfully converted the needed equations to polar form, however I can't solve for theta to find where the two equations intersect.

The first equation is a circle centered at the origin, of radius $2$, so it's just $r = 2$.

The second equation is for a circle of radius $1$, centered at $(1, 1)$, which is

$$\cos(\theta) + \sin(\theta) + \sqrt {\sin(2\theta)}.$$

When I set them equal to each other, I have no idea what to do. I've tried a trig identity for $\sin(2 \theta)$, but it doesn't help much.

I've tried using basic online algebra calculators, and none can solve it. But I know there is a solution, since I can get the equation to be both above and below 2 (and the 2 equations intersect at 2 different points on a graph).

Any help would be greatly appreciated.

答:
Well, as user @ClaudeLeibovici suggested in his answer, the required values of x are the roots of the equation:

\begin{gather*}
\cos x+\sin x=2-\sqrt{\sin 2x}\\
\end{gather*}
Squaring both sides,

\begin{gather*}
(\cos x+\sin x)^{2} =\left( 2-\sqrt{\sin 2x}\right)^{2}\\
1+\sin 2x=4+\sin 2x-4\sqrt{\sin 2x}\\
4\sqrt{\sin 2x} =3\\
\sin 2x=\frac{9}{16}\\
\frac{2\tan x}{1+\tan^{2} x} =\frac{9}{16}\\
\end{gather*}
This is a quadratic in tan(x).
Simplifying,

\begin{gather*}
9\tan^{2} x-32\tan x+9=0\\
\tan x=\frac{16\pm 5\sqrt{7}}{9}\\
\end{gather*}
Can you take it from here?

Title: How to solve for theta in this equation


问:
Sequence of $C^2$ function with compact support generating borel-sigma-algebra.
I'm looking for a sequence of functions $(f_n)_{n\geq 0}$ all $\mathbb{R} \mapsto\mathbb{R}$, twice continuously differentiable and with compact support such that the sigma-algebra generated by them is the borel sigma-algebra on $\mathbb{R}$.
Does anyone know of a such?</p>

答:
First, show that given numbers $a<b$, we can find a smooth function with compact support $f$ such that $0\leqslant f\leqslant 1$ and $f^{-1}(\{1\})=[a,b]$.

Once this step is done, consider for rationals $r<r'$ a corresponding function $f_{r,r'}$. Then the $\sigma$-algebra generated by $\{f_{r,r'}\}$ is contained in the Borel $\sigma$-algebra and contains intervals $[r,r']$, hence intervals.

ContentLicense: CC BY-SA 3.0


问:
scaled $p$-norm, $p$-average, limit $p\to 0+$
Given a vector $v\in\mathbb{R^n}$, the $p$-average be given by
$$
\text{avg}_p := \left(\frac{1}{n} \sum |v_i|^p \right)^{1/p}.
$$
For $p=1$ this is the average of the $|v_i|$, for $p=2$ it's the root mean square.
Does this thing have an actual name? (Except scaled $p$-norm of course.)

Its limit $p\to\infty$ is $\max(|v|)$. What's the limit for $p\to 0+$? I played with uniformly randomly distributed entries and it seems the limit exists (as opposed to the $p$-norm):

答:
As answered by Kavi Rama Murthy:
$$
\lim_{p\to 0+}\left(\frac{1}{n} \sum |v_i|^p \right)^{1/p}
=
\left(\prod|v_i|\right)^{1/n}
$$
(The geometric mean.) See https://en.wikipedia.org/wiki/Generalized_mean.


答:
You are overcomplicating this. Recall the general equation for a circle with radius $r$ and centre $(h, k)$:

$$(x - h)^2 + (y - k)^2 = r^2$$

Therefore, you are solving:

$$(x - 0)^2 + (y - 0)^2 = 2^2 \Rightarrow x^2+y^2=4\tag{1}$$
$$(x-  1)^2 + (y - 1)^2 = 1^2 \tag{2}$$

Expand $(2)$ to get $x^2 - 2x + 1 + y^2 - 2y + 1 = 1$, and substitute $x^2+y^2 = 4$ from $(1)$ into this equation, which gives $4 -2x - 2y + 1 + 1 = 1 \Rightarrow 2x + 2y = 5, x + y = 2.5$. Now $y = 2.5 - x$, and substitute this into $(1)$ to get:

$$x^2 + (2.5 - x)^2 = 4$$

which is a quadratic in $x$. You should be able to solve this using the quadratic formula (remember $y = 2.5 - x$ at the end).

Title: How to solve for theta in this equation

Body:
This is part of a large problem regarding double integrals with polar equations. I have successfully converted the needed equations to polar form, however I can't solve for theta to find where the two equations intersect.

The first equation is a circle centered at the origin, of radius $2$, so it's just $r = 2$.

The second equation is for a circle of radius $1$, centered at $(1, 1)$, which is

$$\cos(\theta) + \sin(\theta) + \sqrt {\sin(2\theta)}.$$

When I set them equal to each other, I have no idea what to do. I've tried a trig identity for $\sin(2 \theta)$, but it doesn't help much.

I've tried using basic online algebra calculators, and none can solve it. But I know there is a solution, since I can get the equation to be both above and below 2 (and the 2 equations intersect at 2 different points on a graph).

Any help would be greatly appreciated.


问:
How to verify the funtion $\Bbb{P}_{a}$ , defined as $\Bbb{P}_{a}(B) = \int_{B} f d\lambda$ for any Borel set B, is a probability measure
I am trying some exercise on a measure theoretic probability text, and want to make sure if I am doing right.

The question is:

Let ($\Omega$, $\mathbf{F}$ , $\Bbb{P}$) be a probability space. Assume g : $\Omega \rightarrow \Bbb{R}$ is a nonnegative measurable function and let f be a nonnegative measurable function on $\Bbb{R}$ such that  $\int f d\lambda$= 1. Then, verify that, for any Borel set B, if $\Bbb{P}_{a}(B) = \int_{B} f d\lambda$, then the function $\Bbb{P}_{a}$ is a probability measure.

My argument is:

Show $\Bbb{P}_{a}(\Omega)$ is equal to 1 : very trivial by one of our assumption $\int_{\Omega} f d\lambda$ = $\int f d\lambda$= 1.

Show $\Bbb{P}_{a}(\emptyset)$ is equal to 0 : $\int_{\emptyset} f d\lambda $ = $\int (f I_{\emptyset})d\lambda $ = $\int (0)d\lambda$ = 0, because the indicator variable $I_{\emptyset}$ is equal to zero for all $\omega \in  \Omega$

Assuming $A_{1}, A_{2}, ... $ are (possibly countably infinite ) disjoint $\mathbf{F}$-measurable sets, show its countable additivity: This is the key part in this proof. As follows:
$\Bbb{P}_{a}(\cup_{i=1}^{\infty}{A_{i}})$ = $\int_{\cup_{i=1}^{\infty}{A_{i}}} f d\lambda $ = $\int (f I_{\cup_{i=1}^{\infty}{A_{i}}})d\lambda $ = $\int (f \sum_{i=1}^{\infty}I_{A_{i}})d\lambda $ = $\int (\sum_{i=1}^{\infty} f I_{A_{i}})$ =? = $\sum_{i=1}^{\infty} \Bbb{P}_{a}(A_{i})$

For the countable additivity part, the third equality holds because the indicator $I_{\cup_{i=1}^{\infty}{A_{i}}}$ = 1 only when one of  $A_{i}$'s contains $\omega$ and the fact that $A_{i}$ 's are disjoint implies $I_{\cup_{i=1}^{\infty}{A_{i}}}$ = $\sum_{i=1}^{\infty}I_{A_{i}}$. Also, the fourth equality holds, because $\sum_{i=1}^{\infty}I_{A_{i}}$ is either 0 or 1 (bounded), so that $f(\omega)\sum_{i=1}^{\infty}I_{A_{i}}(\omega) = \sum_{i=1}^{\infty} f(\omega)I_{A_{i} }$, by the property of convergent series.

The definition of abstract integration we learned in class is $\int_{A} f d\lambda $ = $\int (f I_{A})d\lambda $ using indicator variable. So, I used this definition to derive those.

Let me know if I derived so far is wrong or missing anything..I d appreciate.

What I do not know:

the question mark part in countable additivity proof. I want to interchange the integration and infinite series, but I am not quite sure this is legal here. I learned the property of linearity of integration symbol, but as far as I know, it only holds for finite case. Can I extend it to infinite sum case?

We are going to learn MonotoneConvergenceThm(MCT) and Fatou's Lemma, but when this exercise is assigned, MCT is not covered also.

答:
In the last equality, you're using the following theorem:

If $f_n:\Omega\to[0,\infty]$ is a sequence of measurable functions, then
\begin{align}
\int\sum_{n=1}^{\infty}f_n = \sum_{n=1}^{\infty}\int f_n
\end{align}

This is a consequence of the monotone convergence theorem, which says if you have a sequence of non-negative measurable functions $s_n$ which increase to a function $s$ (i.e $0\leq s_1\leq s_2\leq s_3\leq \dots \leq s:=\lim s_n$), then $\int s=\lim\int s_n$ (i.e you can interchange limits and integrals if the functions increase to a limit). Now, you just apply the monotone convergence theorem to the partial sums $s_n=\sum_{k=1}^nf_k$, and use the finite case.

Well, you mention that this is assigned before the monotone convergence theorem, which I find quite odd because this result is usually one of the first few applications of the monotone convergence theorem (after proving Fatou and DCT):)


问:
Deriving equation of parabola from general equation of conic
We can write general equation of conic as:

$$ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{a^2(1-e^2)} = 1$$

Where $a$ is some parameter and $e$ is eccentricity of conic

For e=0, it is a circle:

$$(x-h)^2 + (y-k)^2 = a^2$$

similarly,

$0 < e<1$, is it an ellipse

$ e>1$, it is a hyperbola

Now, I want to derive the equation of parabola from this , where $e=1$, however that leads me to blowing up the expression. So, I isolated the expression for $e$:

$$ \frac{(y-k)^2}{ a^2 - (x-h)^2} = 1-e^2$$

$$ e^2 = 1 - \frac{(y-k)^2}{a^2  - (x-h)^2}$$

If we send $e \to 1$, this equation becomes:

$$(y-k)^2 =0$$

Which is the equation of a straight line... not a parabola. Why is that the equation didn't reduce to parabola?

答:
What's wrong with your method: The ellipse/hyperbola has focus $(h+ c,k)$ and directrix $x=h+a^2/c$ where $c=\pm ae$.  In the limit $e\to 1$, the foci crashes into the directrix so you end up with the perpendicular to the directrix through the foci.

To do it properly: You need to scale $a$ as a function of $e$ to to keep them apart.

Method 1:  Fix the distance from vertex to directrix.

Fix the conic to have vertex at $(0,0)$ and the focus should be on the positive $x$-axis.  So set the directrix at $x=-a$ and the focus at $(ae,0)$.  The equation of the conic is
$$
(x-ae)^2+y^2=e^2(x+a)^2
$$
Now let $e\to 1$ and you get
$$
y^2=4ax
$$
the equation of the parabola.

Method 2: Fix both the directrix and the focus.
Let the focus be at $(a,0)$ and directrix $x=-a$.  Then the equation of our conic is
$$
(x-a)^2+y^2=e^2(x+a)^2
$$
Let $e\to 1$ gives
$$
y^2=4ax
$$
which isn't surprising.


问:
How does cosx = sinx equals to tan x = 1
So how does cos(x)=sin(x) can be rearranged to be tan(x)=1</p>

答:
Divide by cos(x) on both sides.

If cos(x) is not zero, or equivalently, if x in not an odd integral multiple of pi/2,
sin(x)/cos(x) equals tan(x), and hence from your equations, tan(x)=1.


问:
Solving for $x$ in a messy square root equation
I have these equations:
$$H(t)=h_A+2nh(t)\\
h(t)=\sqrt{\left(\frac{l}{2}\right)^2-\left(\frac A2-xt\right)^2}\\
(t_2-t_1)v_H=H(t_2)-H(t_1)$$
The problem is that squaring both sides in the last one will give me a product of square roots from which I cannot extract $x$.
Is there any other way of going about this? Maybe an approximation? My goal is $x=f(v_H, t_1, t_2)$.
$$\left(\frac{(t2-t_1)v_H}{2n}\right)^2=2\left(\frac{l}{2}\right)^2-x^2(t_2^2+t_1^2)+xA(t_2+t_1)-2\left(\frac A2\right)^2-2h(t_2)h(t_1)$$
Thank you for your time!

EDIT:

Defining $v_H$ as the derivative of $H(t)$ instead works, but the solutions are really, really ugly. https://www.wolframalpha.com/input/?i=solve+for+x+in+%284n%5E2t%5E2%29x%5E4+-+%284n%5E2At%29x%5E3+%2B+%5B%28%28A%2F2%29%5E2%294n%5E2%2Bt%5E2v%5E2%5Dx%5E2+-+%28Atv%5E2%29x+-+v%28l%2F2%29%5E2+%3D+0

答:
Squaring the equation

$$
\frac{v_H(t_2-t_1)}{2n}=\sqrt{l^2-(A-xt_2)^2}-\sqrt{l^2-(A-xt_1)^2}\tag1
$$
one obtains:
$$
\left[\frac{v_H(t_2-t_1)}{2n}\right]^2-2l^2+(A-xt_2)^2+(A-xt_1)^2=-2\sqrt{\Big[l^2-(A-xt_2)^2\Big]\Big[l^2-(A-xt_1)^2\Big]}
$$
which implies:

$$\left[\left[\frac{v_H(t_2-t_1)}{2n}\right]^2-2l^2+(A-xt_2)^2+(A-xt_1)^2\right]^2=4\Big[l^2-(A-xt_2)^2\Big]\Big[l^2-(A-xt_1)^2\Big].\tag2
$$

This is a polynomial equation of $4$-th order in $x$. It can be solved in radicals but the solution looks quite ugly, so I will not give it here.


问:
What is wrong in this choice of $\delta$?
Suppose I have a wrongly evaluated limit$-$
$$\lim_{x \to2} 3x+3=6$$Then we get $-$ $$3x+3-6<\varepsilon
$$
$$3x-3<\varepsilon
$$$$|x-1|<\frac\varepsilon {3}$$$$|x-2|<\frac\varepsilon 3 -1$$
$$|x-2|<\frac{\varepsilon-3} 3=\delta $$ What is wrong with this choice of $\delta$? (Also, please try to provide a diagram, which will help beginners like me!)

P.S I had posted another question, based on this question (which was closed recently). I wrote this post making the question explicit, so would expect a decent answer.Thanks!

答:
When you write $|x-2| < \varepsilon/3-1$ this means $$ 0\le|x-2| < \dfrac{\varepsilon-3}{3}\\
\varepsilon \ge 3 $$
For this inequality to hold true

But in the limit definition we require $ \forall \epsilon > 0$

EDIT

To prove that a sequence don't have a limit or in other terms it diverges we need to show that there doesn't exist any real number $l$ such that the limit definition holds which is the logical negation of the statement.

The above result is as general as the epsilon defition as it is just the negation of it.


问:
Can Two Compact, Non-Separating Subsets of the Plane be Separated by Jordan Curves?
Suppose $X, Y \subset \mathbb{R}^2$ are compact, mutually disjoint, and neither separates the plane.

How does one prove that there are disjoint Jordan curves $J_1, J_2$ such that $X$ is in the interior region of $J_1$, $Y$ is in the interior region of $J_2$, and neither of $J_1, J_2$ is contained in the interior region of the other?

The requirements on $X$ and $Y$ are necessary; for example, take two disjoint circles, one inside the other.  Or consider the graph of $|\frac{1}{x}|$.  The above theorem can be proven using the classification of non-compact surfaces, but that's too big a cannon for my purposes; I want something elementary.  Schoenflies, JCT, Zoretti etc. is fine, but nothing too beyond that!

If $X$ or $Y$ is connected then it's known due to either some algebraic topology results or Zoretti's Theorem, but in my case I'll be dealing with Cantor sets; I am concocting a proof of the Denjoy-Riesz Theorem that's actually readable and have put one together, but there's a somewhat tedious step that I'd like to simplify before I post it, if possible.

Thanks a lot!  The specific case that if $X, Y$ are Cantor Sets is also of interest; that's the tedious step in my Denjoy-Riesz proof.  I proved it using finite covers of open balls, some geometric arguments about these piecewise-nice neighborhoods, then further subdividing and filling in to avoid one neighborhood 'wrapping around' the neighborhood of the other Cantor set.  It's not an aesthetic proof.

答:
Here is a sketch how can do it. Drawing a picture will be helpful.

$X,Y$ have distance $d > 0$. Therefore we can cover $X,Y$ by disjoint compact subpolyhedra $P, Q$ of $\mathbb R^2$ (e.g. unions of sufficiently small squares). Next we take disjoint regular neigborhoods $M, N$ of $P, Q$. These are compact PL manifolds with boundary. Note that the boundaries of $M, N$ are disjoint from $X, Y$ since $X,Y$ are contained in the manifold interiors $\mathring{M}, \mathring{N}$.

Let us first consider $M$. It has only finitely many components; their interiors cover $X$. Let $C$ be one them; it is a compact PL manifold with boundary. Then $X_C = X \cap C$ is compact and does not intersect $\partial C \subset \partial M$.

We shall cut off pieces of $C$ to obtain a PL copy $D_C$ of the disk $D^2$ such that $X_C \subset D_C \subset C$ and $X \cap \partial D_C = \emptyset$.

It is known that $C$ has the form $C = D^* \setminus \bigcup_{i=1}^n \mathring D_i$ where
$D^*, D_1, \ldots, D_n$ are PL disks such that $D_i \subset \mathring D^*$ for $i=1,\ldots, n$ and $D_i \cap D_j = \emptyset$ for $i \ne j$. It is of course possible that $n = 0$, i.e. $C = D^*$, but this is the trivial case. So let $n > 0$. Pick $x^* \in \partial D^*$ and $x_1 \in \partial D_1$. Since $X$ does not disconnect $\mathbb R^2$, there is a path $u$ from $x^*$ to $x_1$ which avoids $X$. Thus the distance between $u(I)$ and $X$ is positive. There is a PL path $v$ from $x^*$ to $x_1$ which is so close to $u$ that it also avoids $X$. We can modify $v$ as follows: If $v$ enters some $D_j$ at time $t$ and leaves it at time $s > t$, we can replace $v$ on $[t,s]$ by a PL path in $\partial D_j$. Similarly, if $v$ enters $\mathbb R^2 \setminus \mathring D^*$ and leaves it at time $s > t$, we can replace $v$ on $[t,s]$ by a PL path in $\partial D^*$. Thus we may assume w.l.o.g. that $v$ is a path $C$ which avoids $X$. As the next step we choose a PL path $w$ in $C$ from $x^*$ to $X_1$ which so close to $v$ that it avoids $X$ and has the property $w((0,1)) \subset \mathring C$ ("pushing $v$ away from the boundary''). W.l.o.g. we may assume thet $w$ is a PL embedding (if it is not, we remove possible loops between points of self-intersection). Now choose a regular neigborhood $W$ ot $w(I)$ in $C$ which is so small that it avoids $X$ ("thickening of $w$ to a strip connecting $\partial D^*$ and $\partial D_1$''). There is a PL homeomorphism $h : I \times I \to W$ such that $h(\{0\} \times I) \subset \partial D^*$, $h(\{1\} \times I) \subset \partial D_1$ and $h((0,1) \times I) \subset \mathring C$. Then $C' =  C \setminus h(I \times (0,1))$ is again a PL manifold neigborhood of $X$ having the form $C' = D_1^* \setminus \bigcup_{i=2}^n \mathring D_i$ with a PL disk $D^*_1$. Proceeding inductively we end with a PL disk $D$  such that $X_C \subset D_C \subset C$ and $X \cap \partial D_C = \emptyset$.

Now let $C_1,\ldots, C_m$ be the components of $M$ and assume that they are reduced to disks as above. They cover $X$ and their boundaries do not intersect $X$. Similarly let $C'_1,\ldots, C'_r$ be the components of $N$ and assume that they are reduced to disks. By a similar approach as above we can find a a strip $W_1$ connecting $\partial C_1$ and $\partial C_2$ and avoiding all other $C_i$ and $C'_j$. Then $\bar C_2 = C_1 \cup W_1 \cup C_2$ is a PL disk whose boundary does not intersect $X$.  Proceeding inductively we end with a PL disk $\bar C_m$ containing $X$ in its interior and being disjoint from $N$. Doing the same with the $C'_j$ results in a PL disk $\bar C'_r$ containing $Y$ in its interior and being disjoint from $\bar C_m$. The boundaries of $\bar C_m$ and $\bar C'_r$ are the desired Jordan curves.


答:
I've changed my answer.

Your question as I see it, is like asking,

"Suppose we have a wrong equation:

$1 + 0 = 4$.

Then:

$1 + 0 + 1 = 4 + 1$

$2 = 5.$

What is wrong with the final equation?"

But perhaps I am not understanding the question?

Title: What is wrong in this choice of $\delta$?

Body:
Suppose I have a wrongly evaluated limit$-$
$$\lim_{x \to2} 3x+3=6$$Then we get $-$ $$3x+3-6<\varepsilon
$$
$$3x-3<\varepsilon
$$$$|x-1|<\frac\varepsilon {3}$$$$|x-2|<\frac\varepsilon 3 -1$$
$$|x-2|<\frac{\varepsilon-3} 3=\delta $$ What is wrong with this choice of $\delta$? (Also, please try to provide a diagram, which will help beginners like me!)

P.S I had posted another question, based on this question (which was closed recently). I wrote this post making the question explicit, so would expect a decent answer.Thanks!


问:
Alternating pattern of root 2 convergents
Is there an easy way to see why successive convergents to root 2 are alternately bigger and smaller than it? (I've looked at the continued fractions and the recursive formula and plotted them as gradients.)</p>

答:
If p/q is a convergent of root 2 [gcd(p,q) = 1], then (x,y) = (p,q) is a solution of the Brahmagupta-Pell's equation x^2 - 2y^2 = plusminus 1. For example, (x,y) = (3,2) is a solution of x^2 - 2y^2 = 1. This indicates that 3/2 will be larger than root 2. Similarly (x,y) = (7,5) is a solution of x^2 - 2y^2 = -1, indicating that 7/5 is lesser than root 2.

If p/q is a convergent of root 2, then p+2q/p+q is the next convergent.

We can prove that if (p,q) is a solution of x^2 - 2y^2 = 1 then (p+2q, p+q) is a solution of x^2 - 2y^2 = -1, and that if (p,q) is a solution of x^2 - 2y^2 = -1 then (p+2q, p+q) is a solution of x^2 - 2y^2 = 1. The proof of this is mere substitution.

This proves the pattern of the convergents of root 2 bouncing about root 2.


问:
How do I calculate a specific variation for a known value of the normal distribution function?
I am writing a Gaussian blur filter in graphics shader code.  I want to make the blur parameterized by radius from the users perspective.  The best method I can figure to do this is to pick a suitable stopping point for y, say .001, and solve for the variance to plug into the normal distribution function that will achieve that value of y.

Unfortunately I cannot for the life of me solve this equation for v...

$x = 10$ (blur radius)

$$.001 = \frac{1}{2 \pi v^2}e^{-\frac{x^2}{2v^2}}$$.

答:
To expand on the suggestion to use the Lambert function, I'll show how it arises in your equation of interest.

Starting with

$y=\frac1{2\pi v^2}\exp\left(-\frac{x^2}{2v^2}\right)$

we multiply both sides by $-\pi x^2$ to give

$-\pi x^2 y=-\frac{x^2}{2v^2}\exp\left(-\frac{x^2}{2v^2}\right)$

which can now be inverted to the Lambert function (recall that the Lambert function $W(z)$ is the inverse function of $z\exp(z)$, $W(z)\exp(W(z))=z$):

$-\frac{x^2}{2v^2}=W(-\pi x^2 y)$

which we can now solve for $v$

$v=\frac{x}{\sqrt{-2W(-\pi x^2 y)}}$

The choice of sign for the square root is motivated by the fact that variances are conventionally taken to be positive.

ContentLicense: CC BY-SA 2.5


问:
How to prove that $\lim\limits_{x \to 0 }\;x^{-a}e^{\frac{-1}{x^{2}}} =0$ for all a?
I'd like your help with this:

I tried using L'Hôpital's Rule and all kinds of arithmetic to prove that
$$\lim_{x \to 0 }\left(x^{-a}e^{\left(\frac{-1}{x^{2}}\right)}\right) = 0$$ for every $a$, and it didn't work.

($a=0$ is trivial)

Any hints?

Thank you.

答: Try setting $x = y^{-1}$. Then it should be easy. If you need more help, just ask.</p>


问:
Taylor expansion of $(1+x)^α$ to binomial series – why does the remainder term converge?
For $α ∈ ℝ$ the function $g_α \colon B_1(0) → ℝ, x ↦ (1+x)^α$ is $C^∞$ and $g_α^{(n)}(x) = n! \tbinom{α}{n}(1+x)^{α-n}$, where $\tbinom{α}{n} = \frac{α(α-1)\cdots(α-n+1)}{n!}$ is the generalized binomial coefficient.
I want to show that $g_α$ has a Taylor expansion $g_α (x) = \sum_{k=0}^∞ \tbinom{α}{k}x^k$. I can easily show that the series really converges for $|x| < 1$, but can I use this fact to show that the remainder term
$$R_n(x) = \int_0^x \frac{(x-t)^n}{n!} (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1} dt$$
or in Lagrange form
$$\quad R_n(x) = \frac{x^{n+1}}{(n+1)!} (n+1)! \binom{α}{n+1}(1+ξ)^{α-n-1} = x^{n+1} \binom{α}{n+1}(1+ξ)^{α-n-1}$$
converges to zero for $|x| < 1$?

How else can I show that the remainder term converges to zero for $|x| < 1$?

Progress: Using the Lagrange form, for $x ∈ (-1,1)$ and $n ∈ ℕ$ there is an $h=h(n,x) ∈ (0,1)$ such that:

$$|R_n(x)| = \Big| x^{n+1} \binom{α}{n+1}(1+hx)^{α-n-1} \Big| = \Big| \binom{α}{n+1}\big(\frac{x}{1+hx}\big)^{n+1}(1+ξ)^{α} \Big| $$
If $x > 0$, then $\big|\frac{x}{1+hx}\big| < {\big|\frac{x}{1+x}\big|}< 1$ and so for $q := \big|\frac{x}{1+x}\big|$, since $\sum_{k=0}^∞ \tbinom{α}{n} q^n$ converges:
$$|R_n(x)| ≤ \Big| \binom{α}{n+1} · q^{n+1} ·(1+ξ)^{α} \Big| \overset{n → ∞}{\longrightarrow} 0$$
The problem is still open for $x < 0$.
Mhenni suggested to use Stirling approximation, but I want to do this more elementary and I think it is possible.

答:
Let $f(z)=(1+z)^\alpha$ and $g(z)=\sum_{n\geq 0}\binom{\alpha}{n}z^n$.

If $\alpha$ is a nonnegative integer, it is clear that both functions coincide on $\mathbb{C}$ by the binomial theorem. So we assume that $\alpha\not\in\mathbb{N}$ from now on, so that $\binom{\alpha}{n}\neq 0$ for every $n\geq 0$.

Using the principal branch of the complex logarithm, we see that $g(z)=\exp(\alpha\log(1+z))$ is holomorphic on $\mathbb{C}\setminus (-\infty,-1]$, whence on the open unit disk $D$ in particular.

Now observe that
$$
\frac{\Big|\binom{\alpha}{n+1}\Big||z|^{n+1}}{\Big|\binom{\alpha}{n}\Big||z|^{n}}=\frac{|\alpha-n||z|}{n+1}\longrightarrow |z|.
$$
So the radius of convergence of $g$ is $1$ by the ratio test. Hence $g$ converges to a holomorphic function on $D$.

For every $x>0$, the remainder in the approximation of $f$ by its degree $n$ Taylor polynomial at $0$ is, by Taylor-Lagrange,
$$
R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\binom{\alpha}{n+1}(1+c)^{\alpha-n-1}x^{n+1}=(1+c)^\alpha\binom{\alpha}{n+1}\left(\frac{x}{1+c} \right)^{n+1}
$$
for some $0<c<x$. Since $0<\frac{x}{1+c}<1$, the convergence of $g$ at $\frac{x}{1+c}$ implies that the rhs above tends to $0$, since it is a constant times its general term. So the remainder of the Taylor approximation tends to $0$.

It follows that $f$ and $g$ coincide on $(0,1)$.

By the isolated zero principle, $f$ and $g$ coincide on the whole open unit disk $D$.


问:
Manifolds with boundaries: Why map boundary points onto boundary points?
The notion of a manifold with boundary has just been introduced in my script. If $\mathbb{R}^n_+$ is defined as $\mathbb{R}^n_+ := \{ (x^1, \ldots, x^n) \in \mathbb{R}^n \; | \; x^n \geq 0 \}$, then a manifold with boundary $M$ is an atlas whos charts $\phi : M \longrightarrow V \subseteq \mathbb{R}^n_+$.

Now, as far as I understand, it is required that the preimage of each point in $\mathbb{R}^n$ with $x^n = 0$ is a boundary point on $M$. In other words each boundary point of $M$ gets mapped onto the hyperplane defined by $x^n = 0$.

I don't really understand this requirement. I think my intuition of either this definition or of a manifold/coordinate chart is wrong. I always understood the coordinate charts as local coordinates of the manifold in Euclidean space. A ball in 3 dimensions for example can be described by $x,y,z$. But describing its boundary with $z = 0$ is pretty uncommon when parametrising a ball, isn't it?

How is that definition or the coordinate charts of a manifold to be interpreted with respect to this example?

答:
In addition to Hagen von Eitzen's great answer, let me add something basic.
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\B}{\mathbb{B}}$
$\newcommand{\pl}{\partial}$
$\renewcommand{\S}{\mathbb{S}}$

#1 The definition of a manifold (either with or without boundary) is based on comparison with some model manifold (with or without boundary, respectively). The convention is to take:

$\R^n$ as the model manifold without boundary,

$\R^n_+$ as the model manifold with boundary (with flat boundary $\{ x^n = 0 \} \cong \R^{n-1}$).

In principle, you could replace $\R^n$ and/or $\R^n_+$ with your favorite manifold (say, sphere $\S^n$ and closed ball $\B^n$), and the theory would look the same. Oversimplfying a bit, we choose our convention because the spaces $\R^n$ and $\R^n_+$ are easily parametrized and the boundary of $\R^n_+$ is also easily parametrized.

#2 The reason why the choice of model manifolds doesn't matter is the following: manifolds are locally the same. Of course, they are different globally, but the definition of atlas only considers small parts of the manifold.

And the same goes for manifolds with boundary: the $1$-dimensional manifold with boundary $M = [0,1]$ doesn't have the same boundary as $\R_+$, but near each of the boundary points $0,1$ our $M$ looks the same as $\R_+$. It's also similar for the closed ball $\B^2$, but now one needs to straighten out the boundary (note: this part only makes sense for submanifolds of $\R^n$ and not for general manifolds with boundary). To see that $\pl \B^2$ looks like $\pl \R^2_+$ near the point $(0,-1)$, consider the map
$$ \Phi \colon \B^2 \to \R^2_+, \qquad
\Phi(x,y) = (x,y+\sqrt{1-x^2}), $$
which maps the interior of $\B^2$ to the interior of $\R^2_+$, and the boundary of $\B^2$ (at least the lower part of it) to the boundary of $\R^2_+$.

#3 Let me also remark that it's necessary to treat the boundary separately. Consider the following two fake "definitions" that attempt to skip this point:

$M$ is a manifold with boundary if there's an atlas $\Phi_i \colon U_i \to \R^n$ on open subsets $U_i \subseteq \operatorname{int} M$ of the interior of $M$, covering all of $\operatorname{int} M$.

$M$ is a manifold with boundary if there's an atlas $\Phi_i \colon U_i \to \R^n$ on open subsets $U_i \subseteq M$ (possibly including the boundary), covering all of $M$, such that each $\Phi_i$ maps $U_i$ homeomorphically to an open subset of $\R^n$.

You can see that the first "definition" accepts more objects than we'd like (as it doesn't say anything about the boundary at all), while the second "definition" effectively accepts only manifolds without boundary (this is a good exercise).


问:
How do I find the value of k to make the equations of a line and a plane be parallel?
Given the equation of the line $r$:

$r:\frac{x-2}{k}=\frac{y-1}{2}=z$

And the equation of the plane $\alpha$:

$\alpha:3x-ky-z-2=0$

How do I determine $k$ such that the line $r$ and the plane $\alpha$ are parallel?

答:
Thanks guys, @Vectorizer and @Math Lover

So, by the equation of $\alpha$, I know that the normal vector is $(3, - k, - 1)$ and from the equation of $r$, I know that it's director vector is $(k, 2, 1)$

With that said, the dot product of both have to be $=0$, and so:

$3k-2k-1=0 \Leftrightarrow k=1$

I have to study harder!!


问:
Parameterizing both branches of a hyperbola
Recently I have been studying parametric equations of surfaces and curves, specifically hyperbolic functions. Given by the equations $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad\frac{(x-\alpha)^2}{a^2}-\frac{(x-\beta)^2}{b^2}=1$$

denoting hyperbolic functions centred at $(0,0)$ and $(\alpha,\beta)$ respectively. I am however, unable to derive a parametrisation that represents the entire function. Help to further understand this topic would be greatly appreciated.

答:
General equation of conic section :
$$ax^2+2bxy+cy^2+2dx+2fy+g=0$$
with constant coefficients $a,b,c,d,f,g$.

The curve is an hyperbola if :
$$b^2-ac>0\quad;\quad a\neq 0 \quad;\quad c\neq 0$$
For parametrisation see : https://mathworld.wolfram.com/QuadraticCurve.html


问:
Solve in the set of integers $2^x +5^x = 3^x + 4^x$

Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x +  4^x$$

With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$

It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$

So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.

If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).

Any easier solution?

Body:
For $x\geq2$ we have $3^{x-2},4^{x-2}\leq5^{x-2}$, and so from $5^2=3^2+4^2$ it follows that
$$2^x+5^x>5^x=5^{x-2}(3^2+4^2)=5^{x-2}3^2+5^{x-2}4^2\geq3^x+4^x.$$
For $x\leq-2$ we have $3^{x+2},4^{x+2}\leq2^{x+2}$, and so from $2^{-2}>3^{-2}+4^{-2}$ it follows that
$$2^x+5^x>2^x=2^{x+2}(3^{-2}+4^{-2})=2^{x+2}3^{-2}+2^{x+2}4^{-2}\geq3^x+4^x.$$
Then it suffices to check that the equality holds for $x=0$ and $x=1$, but not for $x=-1$.


问:
Relationship of the support of a test function with the support of distribution.

Let $\phi\in D(\Omega)$ and $f\in D'(\Omega)$. If $\phi$ is $0$ in a neighbourhood of $\operatorname{supp}(f)$, then how will we prove that $\langle f, \phi
\rangle$ is also $0$?
  Will it be sufficient if $\phi$ vanishes on $\operatorname{supp}(f)$?

Definition of $\operatorname{supp}(f)$:

Let $f$ be a distribution, and $U$ an open set in $\Bbb R^n$ such that, for all test functions $\phi$ with the support of $\phi$ contained in $U$, $f(\phi) = 0$. Then $f$ is said to vanish on $U$. Hence we can define the support of $f$ as the complement of the largest open set on which $f$ vanishes.

Alternatively, we can say $\operatorname{supp}(f)$ is the subset of $\Bbb R^n$ such that $x \in \operatorname{supp}(f)$, if and only if for all neighbourhoods $\mathcal U$ of $x$, there is $\phi \in D(\mathcal U)$ such that $\langle f, \phi\rangle \neq 0$.

答: If $\phi$ vanishes only on $\operatorname{supp } f$, then this is not true anymore: Let $f=\delta'$ and and $\phi\in C_c(\mathbb{R})$ with $\phi(x)=x$ for $|x|\leq 1$. Then $\phi(0)=0$, but $f(\phi)= -\phi'(0) = -1$.</p>


问:
Exponentially weighted average, sum of weights
I'm just wondering how to actually prove that the weights of an exponentially weighted average sum to 1, like

$(1-\alpha)^n + \displaystyle\sum_{i=1}^n \alpha (1-\alpha)^{n-i} = 1$

where $\alpha \in (0, 1]$

This is mentioned in Reinforcement Learning, Barto & Sutton, first sentence of p33. Intuitively it holds true but I'd like to understand the proof of it.

答:
Just use the binomial theorem to calculate the coefficient of $\alpha^t$:

The coefficient of $\alpha^t$ in $\alpha(1-\alpha)^{n-i}$ is equal to (clearly for $n-i\geqslant t-1$ and $t>0$) the coefficient of $\alpha^{t-1}$ in $(1-\alpha)^{n-i}$ which is equal to:
$$(-1)^{t-1}\cdot\binom{n-i}{t-1}$$

So the coefficient of $\alpha^t$ in the whole thing is equal to:

$$(-1)^t\cdot\binom{n}{t}+(-1)^{t-1}\cdot\binom{n-1}{t-1}+(-1)^{t-1}\cdot\binom{n-2}{t-1}+\dotsb$$
$$=(-1)^t\Bigg(\binom{n}{t}-\binom{n-1}{t-1}-\binom{n-2}{t-1}+\dotsb\Bigg)$$
$$=(-1)^t\Bigg(\binom{n}{t}-\displaystyle\sum_{i=1}^{\small{n-t+1}}\binom{n-i}{t-1}\Bigg)$$
Now one can prove that $\displaystyle\sum_{i=1}^{\small{n-t+1}}\binom{n-i}{t-1}=\binom{n}{t}$ like the following:

Assume that we have $n$ people $p_1,p_2,\dotsb,p_n$ and their heights are $h_1<h_2<\dotsb<h_n$ and we put them in a row as the $i$-th person is $p_i$ and we want to choose $t$ people out of them. Casework on the tallest person we have chosen:

Since we're taking $t$ people, the tallest guy cannot be shorter that $h_t$. On the other hand if the tallest guy is to be $p_i$, the number of all ways to choose $t$ people so that the tallest one is $p_i$ is equal to $\binom{i-1}{t-1}$. So the number of all methods to choose $t$ people is $\displaystyle\sum_{\small{i=t-1}}^{n-1}\binom{i}{t-1}=\displaystyle\sum_{i=1}^{\small{n-t+1}}\binom{n-i}{t-1}$. On the other hand the number of ways to choose $t$ people out of $n$ people is clearly $\binom{n}{t}$ so finally $\displaystyle\sum_{i=1}^{\small{n-t+1}}\binom{n-i}{t-1}=\binom{n}{t}$.

Back to our problem, since we proved this identity, the coefficient of $\alpha^t$ in the whole expression ($\forall t\not=0$) is equal to zero. Hence it only has constant terms and we only have one constant term in the expression (which is in the $(1-\alpha)^n$ and it is equal to $1$. So the value of $(1-\alpha)^n + \displaystyle\sum_{i=1}^n \alpha (1-\alpha)^{n-i}$ is equal to $1$.


问:
Defining bounded operator on orthonormal basis.
Let $\{e_i : i \in I\}$ be an orthonormal basis for a Hilbert space $H$. Suppose we are given a collection of vectors $\{\xi_i\}_{i \in I}$ with $\sum_{i \in I}\|\xi_i\| < \infty$. I want to prove that there exists a unique bounded linear operator $T: H \to H$ such that $Te_i = \xi_i$.

This is a generalisation of an exercise in Conway's "A course in functional analysis".

Attempt: Uniqueness is clear because $T$ is uniquely determined on the linear span of $\{e_i: i \in I\}$, which is dense in $H$.

Let us focus on existence. Write $x= \sum_{i \in I} \langle x, e_i\rangle e_i$ where the series converges in norm. Then we want to define
$$Tx:= \sum_{i \in I} \langle x,e_i\rangle \xi_i$$
but it is not clear that this series converges. However,
$$\sum_{i \in I} \|\langle x, e_i\rangle \xi_i\| = \sum_i |\langle x, e_i\rangle|\|\xi_i\| \le \left(\sum_i |\langle x, e_i\rangle|^2\right)^{1/2}\left(\sum_i \|\xi_i\|^2 \right)^{1/2}< \infty \quad (*)$$
by Bessel's inequality and since $\sum_i\|\xi_i\| < \infty \implies \sum_i\|\xi_i\|^2 < \infty$.

Hence, the above series converges absolutely and thus $Tx$ makes sense. Since the inner product is linear in the first component, it follows that $T$ is linear. It remains to check that $T$ is bounded.

The equation $(*)$ also shows that $\|Tx \| \leq \|x\| \left(\sum_i \|\xi_i\|^2 \right)^{1/2}$ and thus $T$ is bounded. This establishes the result.

Questions: Is my proof correct? Am I correct that I can further weaken the hypothesis in the question by only requiring that $\sum_i\|\xi_i\|^2 < \infty$?

答: Your proof is fine and yes, you can change the hypothesis to the weaker condition $\sum_i \|\xi_i\|^{2} <\infty$.


问:
Calculating limit of integral using L'Hospital's rule

Calculate: $$\lim_{h\rightarrow  0} \frac{1}{h}\int_3^{3+h}e^{t^2}dt$$
I was thinking of using L'Hospital's rule, as this limit is of form $\frac{0}{0}$.

So the limit above equals(after derivatives): $$\lim_{h\rightarrow  0} \frac{e^{h^2}}{1}=1$$

Is it correct?

Body: The lmit is $\lim_{h \to 0}\frac {e^{(3+h)^{2}}} 1=e^{9}$.


答: You made a mistake in the limit it should be $e^{(3+h)^2}$

Title: Calculating limit of integral using L'Hospital's rule

Body:
Calculate: $$\lim_{h\rightarrow  0} \frac{1}{h}\int_3^{3+h}e^{t^2}dt$$
I was thinking of using L'Hospital's rule, as this limit is of form $\frac{0}{0}$.

So the limit above equals(after derivatives): $$\lim_{h\rightarrow  0} \frac{e^{h^2}}{1}=1$$

Is it correct?


答:
No this is false, let $f(h)=\int_0^h e^{t^2}dt$, then $\lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=f'(3)=e^9$. You can also use L'Hopital's rule :
$$ \lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h\rightarrow 0}\frac{f'(3+h)}{1}=f'(3)=e^9 $$


Body: No. If $G(h)=\int_3^{3+h}e^{t^2}dt$, then $G'(h)=e^{(3+h)^2}.$


答:
First of all you can put $\alpha$ and $(1-\alpha)^n$ in front of the sigma sign:

$$(1-\alpha)^n + (1-\alpha)^{n}\cdot \alpha\sum_{i=1}^n  (1-\alpha)^{-i} $$

Factoring out $(1+\alpha)^n$

$$(1-\alpha)^n \left(1+\alpha\sum_{i=1}^n  (1-\alpha)^{-i}\right) $$

That means, that the brackets must be $(1-\alpha)^{-n}$, since $(1-\alpha)^n\cdot (1-\alpha)^{-n}=1 \quad (\color{red}*)$

The sum is the partial sum of a $\color{grey}{\textrm{geometric series}}$, where $\sum\limits_{i=1}^n x^i=x\cdot \frac{x^n-1}{x-1}$. Thus

$$1+\alpha\sum_{i=1}^n  \left(\frac1{1-\alpha}\right)^{i}=1+\frac{\alpha}{1-\alpha}\frac{\left(\frac1{1-\alpha}\right)^{n}-1}{\frac1{1-\alpha}-1}$$

$(1-a)\cdot \left( \frac1{1-\alpha}-1\right)=1-1+a=a$. Therefore the denominator is $\alpha$.

$$1+\alpha\frac{\left(\frac1{1-\alpha}\right)^{n}-1}{\alpha}=1+\left(\frac1{1-\alpha}\right)^{n}-1=\left(\frac1{1-\alpha}\right)^{n}=(1-\alpha)^{-n}$$

This is was we wanted at $(\color{red}*)$

Title: Exponentially weighted average, sum of weights

Body:
I'm just wondering how to actually prove that the weights of an exponentially weighted average sum to 1, like

$(1-\alpha)^n + \displaystyle\sum_{i=1}^n \alpha (1-\alpha)^{n-i} = 1$

where $\alpha \in (0, 1]$

This is mentioned in Reinforcement Learning, Barto & Sutton, first sentence of p33. Intuitively it holds true but I'd like to understand the proof of it.


问:
Counting permutation of duplicate items
I am not really sure I fully understand the formula for finding the number of permutation of duplicate items.

The formula is:

$$\frac{n!}{p!q!r!\cdots}$$

Where do the factorials in the denominator exactly come from?

If we asssume the string "ANNA" and we want the count of the permutation of duplicate items.

We have $4$ characters so since we have $4$ options for the first character, $3$ for the second, $2$ for the third and $1$ for the last we have $4!$ different permutations.

But some of the characters are duplicates.

We have "A" and "N" repeated, meaning it adds more permutations to the outcome.

If $p=2$ is the number of occurences of "A" and if $q = 2$ is the number of occurences for "N"  then would the $p!$ and $q!$ essentially be selecting $1$ item at a time from $2$ (or $p$) duplicates? So essentially are the denominators the binomial $\binom{2}{1}$ (or generally $\binom{p}{1}$ where $p$ are the number of repeated characters) where order does not matter?

Is my understanding correct?

答:
$n!$ is, as you stated, the number of ways to permutate if no item is a duplicate.
However, let's take ANNA as an example.
Lets's number the letters by 1-4 according to their position ($1=4=A$, $2=3=N$)
So, within the $4!$ options, we have, for example, 1324 and 1234 which are identical (both spell ANNA).
The inner permutation between two identical items is what you see in the denominator; for this example, you have $2!$ since $1=4$, and $2!$ since $2=3$.
So, all in all you get $ \frac {4!} {2!2!} = 24/4 = 6$

Title: Counting permutation of duplicate items


问:
Derivatives of general absolute value functions
Let $f : (a, b) \to \mathbb{R}$ be differentiable at $c \in (a, b)$, and $f(c) \neq 0$.

How do I show that $|f|'(c) = f'(c)$ if $f(c)>0$?

I know that $|f|'(c) = $ $\frac{|f(x)| - |f(c)|}{x-c}$.

Thanks in advance!

答:
Case 1: $f(c)>0.$ then there is a neighborhood $U$ of $c$ with $U \subset (a,b)$ and $f(x)>0$ for all $x \in U.$

Then $|f|(x)=f(x)$ for all $x \in U.$ Thus $|f|'(c)=f'(c).$

Case 2: $f(c)<0.$ then there is a neighborhood $U$ of $c$ with $U \subset (a,b)$ and $f(x)<0$ for all $x \in U.$

Then $|f|(x)=-f(x)$ for all $x \in U.$ Thus $|f|'(c)=-f'(c).$


问:
The characteristic polynomial of a recurrence relation.
If I have a linear homogeneous recurrence relation

$$y_n=c_1y_{n-1}+\ldots+c_ky_{n-k},$$

I can get its characteristic equation, which is

$$r^k=c_1r^{k-1}+\ldots+c_k.$$

In particular for $y_n=cy_{n-1},$ I get $r=c.$ While I see that this obviously gives the right solution, something seems not right. For a differential equation $y'=cy,$ I get a similar characteristic equation: $r=c$, however, as I understand it, the the analogy between differential equations and recurrence relations is given by $y'\sim y_n-y_{n-1},$ not $y'\sim y_n.$ By this logic, I think the characteristic equation for the recurrence relation $y_n=cy_{n-1}$ should be $r=c-1,$ because the recurrence relation is equivalent to $y_n-y_{n-1}=(c-1)y_{n-1}.$ Could you help me understand why the analogy breaks down here (or why it doesn't)?

答:
The idea behind using the characteristic equation to solve the recurrence relation is to postulate exponential solutions, $y_n=r^n.$  The roots of the characteristic equation, assuming none are multiple roots, then give a set of basis functions, and the general solution is a linear combination of these basis functions,
$$y_n=\alpha_1r_1^n+\alpha_2r_2^n+\ldots+\alpha_kr_k^n.$$

The idea behind using the characteristic equation to solve the differential equation is similar.  We postulate an exponential solution $y=e^{rx}.$  The roots of the characteristic equation, again assuming no multiple roots, give a set of basis functions, and the general solution is a linear combination of these basis functions,
$$y=\alpha_1e^{r_1x}+\alpha_2e^{r_2x}+\ldots+\alpha_ke^{r_kx}.$$

Let's see how the two methods are related in your examples, $y_n=cy_{n-1}$ and $y'=cy.$

For the recurrence we get $r=c,$ and therefore the general solution $y_n=\alpha c^n$.

For the differential equation we also get $r=c.$  This gives the general solution $y=\alpha e^{cx}.$

Let's now solve a discrete approximation of the differential equation using the recurrence equation method.  Approximating the derivative, we have
$$\frac{y(x+\Delta x)-y(x)}{\Delta x}\approx cy(x).$$
If we discretize the $x$-axis, letting $x=n\Delta x,$ we get
$$\frac{y((n+1)\Delta x)-y(n\Delta x)}{\Delta x}\approx cy(n\Delta x).$$
If we write $y_n$ for $y(n\Delta x),$ we get
$$y_{n+1}-y_n\approx cy_n\Delta x$$
or
$$y_{n+1}\approx(1+c\Delta x)y_n.$$
The quantity in parentheses plays the role of $c$ in the recurrence relation solved above.  Adapting that solution gives $r=1+c\Delta x$ and therefore
$$y_n\approx\alpha(1+c\Delta x)^n.$$
Since $\Delta x=x/n,$ this is the same as
$$y_n\approx\alpha(1+cx/n)^n.$$
This is consistent with the exact solution to the differential equation since, for large $n,$ $(1+cx/n)^n$ is approximately $e^{cx}.$


问:
linearly isomorphic submodule
I'm currently learning about modules and am pretty new to the topic. I recently found this question:

Suppose that $R$ is an integral domain and not a field. Give, with proof, an eample of a proper submodule of $R$ that is linearly isomorphic to $R$.

I don't really know how to approach this question. Also I'm struggling with linearly isomorphic. From my notes linearly isomorphic means it only has the scalar multiplication property $(\phi(rm)=r\phi(m))$ but does not necessarily have the addition property $(\phi(m_1+m_2)=\phi(m_1)+\phi(m_2))$. But somehow i can't really find much on that topic either here or in the books that I'm using. Any help would be really appreciated!

答:
I don't think the definition of linearly isomorphic you're writing is standard. Addition is certainly linear! My best guess is that the problem is asking for a proper submodule of $R$ which is isomorphic to $R$ as an $R$-module.

Here's the example I have. As $R$ is not a field, take some nonzero nonunit element $a \in R$. You may make an objection about the zero ring, but this is not considered a domain. Anyways, consider the ideal $(a) \subseteq R$. As $a$ is not a unit this is a proper ideal. Furthermore, it is a submodule of $R$ (in fact, submodules of $R$ are precisely ideals of $R$). Now, consider the map $R \longrightarrow (a)$ via $r \mapsto ra$. This is certainly $R$-linear, and by definition of $(a)$ it is onto. Furthermore, as $a \neq 0$ and as $R$ is a domain, this map is injective. Hence, it is an isomorphism of $R$ with a proper submodule.

Title: linearly isomorphic submodule


答: Consider $R=\mathbb{Z}$ and consider the submodule $n\mathbb{Z}=\{nx,x\in\mathbb{Z}\}$.

Title: linearly isomorphic submodule

Body:
I'm currently learning about modules and am pretty new to the topic. I recently found this question:

Suppose that $R$ is an integral domain and not a field. Give, with proof, an eample of a proper submodule of $R$ that is linearly isomorphic to $R$.

I don't really know how to approach this question. Also I'm struggling with linearly isomorphic. From my notes linearly isomorphic means it only has the scalar multiplication property $(\phi(rm)=r\phi(m))$ but does not necessarily have the addition property $(\phi(m_1+m_2)=\phi(m_1)+\phi(m_2))$. But somehow i can't really find much on that topic either here or in the books that I'm using. Any help would be really appreciated!


答:
The denominators (p, q, r...) are permutations themselves. You mention C(p, 1) = p as the denominator which is not correct. The correct denominator is $p!$

A simpler way to understand this is: Let's take the string "ANNA". We have 4! (n! for n letters) permutations of this, ignoring repetitions. For each instance of these permutation, e.g., $A_1 A_2 NN$ and $A_2 A_1 N N$, where $A_1$ is the first A and so on. So the string $AANN$ is repeated twice (2!, in case of p repetitions of a letter it is p!), if we only take the duplicates due to the permutations of A. To discount the duplicates, you need to divide by p! If you follow the same line of logic, you get the formula you mentioned.

Title: Counting permutation of duplicate items

Body:
I am not really sure I fully understand the formula for finding the number of permutation of duplicate items.

The formula is:

$$\frac{n!}{p!q!r!\cdots}$$

Where do the factorials in the denominator exactly come from?

If we asssume the string "ANNA" and we want the count of the permutation of duplicate items.

We have $4$ characters so since we have $4$ options for the first character, $3$ for the second, $2$ for the third and $1$ for the last we have $4!$ different permutations.

But some of the characters are duplicates.

We have "A" and "N" repeated, meaning it adds more permutations to the outcome.

If $p=2$ is the number of occurences of "A" and if $q = 2$ is the number of occurences for "N"  then would the $p!$ and $q!$ essentially be selecting $1$ item at a time from $2$ (or $p$) duplicates? So essentially are the denominators the binomial $\binom{2}{1}$ (or generally $\binom{p}{1}$ where $p$ are the number of repeated characters) where order does not matter?

Is my understanding correct?


答:
From  L'Hospital's rule, we have

$\lim _{h\rightarrow 0}\left(\frac{\int _3^{3+h}\left(e^{t^2}\right)dt}{h}\right)=\frac{\frac{d}{dh}\left(\int _3^{3+h}\left(e^{t^2}\right)dt\right)}{\frac{d}{dh}\left(h\right)}$

From the Newton-Leibniz' Theorem--

$\frac{d}{dh}\left(\int _3^{3+h}\left(e^{t^2}\right)dt\right)=e^{\left(3+h\right)^2}\left(\frac{d}{dh}\left(3+h\right)\right)=e^{\left(3+h\right)^2}$

Hope it is clear now :~)
Still if any query feel free to ask


问:
Rigorous Derivation of Metropolis-Hastings Transition Density
The Metropolis-Hastings MCMC algorithm is as follows.  Set $X_0$ to some initial value in the support of the target density $f$ and choose a proposal density $q(y \mid x)$; a density in $y$ for each $x$ in the support of $f$.  For each time $t \in \mathbb{N}$, do the following:

Generate a candidate point $Y_t \sim q(\cdot \mid X_t)$

Set
$$
X_{t+1} =
\begin{cases}
Y_t, \text{ with probability } \rho(X_t, Y_t) \\
X_t, \text{ with probability } 1 - \rho(X_t, Y_t),
\end{cases}
$$
where
$$
\rho(x,y) = \min\left\{\frac{f(y)q(x \mid y)}{f(x)q(y \mid x)}, 1\right\}.
$$

These two steps define a Markov chain $\{X_t\}$ on an uncountable state space $\mathcal{S}$.  Our professor wrote the transition kernel $\pi$ for this Markov chain as
$$
\pi(x,y) = q(y \mid x)\rho(x,y) + (1 - r(x))\delta_x(y),
$$
where $r(x) = \int_{\mathcal{S}} q(y \mid x )\rho(x,y)\, dy$. I was unable to verify for myself this was indeed the transition density.  Is there a rigorous derivation/proof of this available?

I know I need to show the following.  Let $f_t$ be marginal density of the chain at time $t$.  Denote the conditional distribution $P( \cdot \mid X_t = x) : \mathcal{B}(\mathcal{S}) \to [0,1]$; a probability measure on the state space for each $x \in \mathcal{S}$.  For any sets $A,B \in \mathcal{B}(\mathcal{S})$ the conditional distribution is defined so as to satisfy
$$
P(X_{t+1} \in A, X_t \in B) = \int_B f_t(x)P(X_{t+1} \in A \mid X_t = x) \, dx.
$$
Thus I need to show
$$
P(X_{t+1} \in A, X_t \in B) = \int_B f_t(x) \left(\int_A \pi(x,y) \,dy \right)\, dx.
$$
For this I am at sea.

答:
To simplify things consider that your state space is $\mathbb{R}^n$. If you are at $x\in\mathbb{R}^n$ there are two ways of reaching a point $x'$ in the set $A\subseteq \mathbb{R}^n$:

You can propose a candidate $x'\in A$ by sampling from the proposal distribution $x'\sim q(x, x')$ and then you can accept it with probability $\alpha(x'\mid x)$.

Otherwise, you can reject whatever proposal but you are so lucky that the current state $x$ is actually already in $A$.

This means your kernel is given by

$$
K(x, A) = \underbrace{\int_Aq(x, x') \alpha(x' \mid x) dx'}_{\text{Probability of moving to a state in $A$}} + \underbrace{\left(1 - \int_{\mathbb{R}^d} q(x, x')\alpha(x' \mid x) dx'\right) \mathbb{I}_{A}(x)}_{\text{Probability of staying at $x$ when $x\in A$}}
$$

Let's inspect this:

In the first term we are summing over all candidates $x'\in A$ and the integrand is the probability that we move to $x'$ given that we were in $x$ and we proposed $x' \sim q(x, x')$.

In the second term we first compute the probability of leaving $x$ as
$$
a(x) = \int_{\mathbb{R}^d} q(x, x') \alpha(x'\mid x) dx'
$$
and then by taking $1-a(x)$ we compute the probability of rejection. We then multiply this by an indicator function that is $1$ only when $x\in A$.
$$
\mathbb{I}_A(x) = \begin{cases}
1 & x \in A \\
0 & x \notin A
\end{cases}
$$
These are the only two ways to move from a state $x\in\mathbb{R}^n$ to any state in $A\subseteq \mathbb{R}^n$.

Does it make sense?


问:
Integrally closed domain.
Suppose $A$ is a unique factorization domain, $a$ is an element of $A$. Is the ring $A[x,a/x]$ always integrally closed? ($x$ is a variable over $A$)

Thanks!

答:
Let $A$ be a normal domain, $a\in A$, then $B=A[x,a/x]$ is a normal domain.
In fact, let $y=a/x$, then $B=A[x,1/x]\cap A[y,1/y]$. The ring $A[x,1/x]$ is a localization of the normal domain $A[x]$, thus normal. So, $B$ is normal.

Title: Integrally closed domain.


问:
How to calculate module $a\cdot7 \equiv1\pmod8$?
How should i calculate following module ?

$a\cdot7 \equiv1\pmod8$

is value of $a = 1$ ?

Thanks.

答: For every natural number $n$ we have $(-1)*(n-1)  \equiv (-1)^2 = 1 \bmod n$. So in your example $a \equiv -1 \equiv 7 \bmod 8$.</p>


问:
Determine if the set $A= \{(x,y) \mid |y|\leqslant|x|, x \ne 0\}$ is closed?

Determine if the set $A= \{(x,y) \mid |y|\leqslant|x|, x \ne 0\}$ is closed?

I know that $A$ is closed if $A^c$ is open. Now $A^c$ is just $A^c = \{(x,y) \mid |y| >|x| \}$. To show that $A^c$ is open I would have to show that for every $x \in A^c$ there is a disk $B(x,r) \subset A^c$, but I'm not sure how to approach this?

答:
Actually, $A^{c}=\{(x,y): |y|>|x|\} \cup \{(0,y): y \in \mathbb R\}$.

$A$ is not closed: $(0,0) \in A^{c}$ and every neighborhood of $(0,0)$ contains points with $|y|\leq |x|$.

[Given a ball of radius $r$ around the origin consider the point $(\frac 1 4 r,\frac 1 2 r)$].


问:
Time of descent down a slope defined by a function
Let $\phi$ be a real, decreasing differentiable function defined on $[a, b]$. An object is dropped at the point $(a, \phi(a))$ and is accelerated only by gravity. How long does it take to reach $(b, \phi(b))$?

Using simple geometry, I believe I've correctly shown that the horizontal acceleration as a function of the horizontal position, x, is given by:

$$x''(t) = g^2\frac{\phi'(x)^2}{(1 + \phi'(x))^4}$$

Something like that anyway - I don't have my notes with me right now. The exact form of the expression doesn't matter, the point is it's a constant function (with respect to $t$) given by $\phi'(x)$.

Now I'd like to integrate this twice to get $x(t)$ (setting $x'(0) = 0$, $x(0) = a$), then set that equal to $b$ and solve for $t$. But I hit a snag. Conceptually, I can see that I need to do some sort of integration with respect to $x$ - thus "adding up" all of the values that the acceleration takes along the curve. But wait, maybe the integration should be with respect to $t$? After all, acceleration is a derivative with respect to $t$, not $x$. So how do I deal with the interactions between these two variables?

Could somebody explain the relevant methodology?

答:
I think the correct equation is
$$x''(t)=-g\frac{\phi'(x)}{ 1+(\phi'(x))^2 } \tag1$$

The exact form of the expression doesn't matter

I beg to differ. The exact form matters very much when we deal with nonlinear equations. (The right hand side is most likely a nonlinear function of the unknown $x$.) It would take a minor miracle to get an explicit solution of (1), and whether or not a miracle  occurs in your problem depends on what $\phi$ is.


问:
Is this subvariety affine?
Consider the following algebraic variety $X$:

Let $C\subset\mathbb{A}^3$ be the affine variety cut out by $xy-z^2=0$ (i.e. the affine cone over a projective conic). Let $B=\mathbb{A}^2$, with coordinate functions $x',z'$. Identify $C\setminus \{x=0\}$ with $B\setminus \{x'=0\}$ by the map $(x,y,z)\mapsto (x'^{-1},z')$.  Then let $X$ be $B\cup C$.

Now I believe that $y$ and $z$ (coordinate functions on $C$) extend to global sections on $X$. Therefore $y-z$ is a global section. My question is this:

Is the open subvariety $D(y-z)=\{p\in X\mid y(p)-z(p)\neq 0\}$ affine?

Some background (you can ignore this):

Why do I care about this question? My original motivation was that I was wondering if in general, for an arbitrary variety $X$, the set of $f\in \mathcal{O}_X(X)$ such that $D(f)$ is affine is an ideal of $\mathcal{O}_X(X)$, since it seems to me it is closed under multiplication by $\mathcal{O}_X(X)$. But I couldn't think of a reason for it to be closed under addition. I made up the variety $X$ in this question as a random test case. Then I realized I didn't have a way to check affineness. I now assume that the set of such $f$ is not an ideal in general, or somebody would already have answered this question. But I still hope to learn the answer, because I want to understand how I would determine affineness in this case.

Why do I think $C\setminus\{x=0\}$ and $B\setminus\{x'=0\}$ are isomorphic? Because the coordinate rings are, respectively, $\left(k[x,y,z]/(xy-z^2)\right)_x$ and $k[x',z']_{x'}=k[x',x'^{-1},z']$.  But $\left(k[x,y,z]/(xy-z^2)\right)_x$ is generated over $k$ by $x,x^{-1},y,z$ and $y$ satisfies $y=x^{-1}z^2$, so this is the ring $k[x,x^{-1},z]$. Then $x'\mapsto x^{-1},z'\mapsto z$ is an isomorphism of the rings.

Why do I think $y,z$ extend to global sections? The complement of $C$ in $X$ is the line $x'=0$ in $B$; this is where the functions need to be defined in order to become global sections.  But on $B\cap C$, $y=x^{-1}z^2 = x'z'^2$ which is a regular function on all of $B$, so $y$ extends. (Thus $y$ is actually identically zero on the line $x'=0$ in $B$.) Extending $z$ is even clearer because $z$ becomes $z'$ on $B$.

Aside: I described the variety $X$ in this previous question.

答:
Question: "Why do I care about this question? My original motivation was that I was wondering if in general, for an arbitrary variety $X$, the set of $f∈O_X(X)$ such that $D(f)$ is affine is an ideal of $O_X(X)$, since it seems to me it is closed under multiplication by $O_X(X)$."

Answer: For a projective variety $X \subseteq \mathbb{P}^n_k$ over a field $k$
it follows $dim_k(A):=dim_k(H^0(X, \mathcal{O}_X)) < \infty $ and hence $A$ is an Artinian ring with a finite number of maximal ideals. Im unsure if you get anything interesting when considering ideals in $A$.

If $X:=Spec(R)$ it follows $D(f) \cong Spec(B)$ where $B:=R[t]/(tf-1)$ which is affine. Hence the set of $f\in R \cong H^0(X, \mathcal{O}_X)$ with this property is all of $A$. Hence in the affine case you do not get anything interesting.

Title: Is this subvariety affine?


问:
Why is the integral between $\cos(x)$ and $\sin(x)$ between the intersections $\sqrt{2}$
I just recently noticed, that the integral
$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(x) - \sin(x) dx = \sqrt{2}.$$

I know who to calculate all of it but I was quite surprised that the result is so simple. Is there maybe a graphical intuition behind this? Maybe one can use the diagonal of a unit square? Or is it just a "coincidence" and only holds because of the properties of $\cos(x)$?

Thanks for any answers, it's just a question out of interest.

答:
Just for fun:
$$
 [-\pi/ 4,  \pi/ 4] \ni \phi \mapsto c(\phi) = \left( \frac 12 - \frac{\sin \phi}{\sqrt 2},\frac 12 + \frac{\sin \phi}{\sqrt 2} \right)
$$
is a parametrization of the diagonal from $(1, 0)$ to $(0, 1)$ in the unit square $[0, 1]^2$, and its length is
$$
 \sqrt 2 = \int_{-\pi/4}^{\pi/4} \Vert c'(\phi) \Vert \, d\phi = \int_{-\pi/4}^{\pi/4} \sqrt{\cos^2(\phi)} \, d\phi =  \int_{-\pi/4}^{\pi/4} \cos(\phi) \, d\phi \, .
$$
Also $\int_{-\pi/4}^{\pi/4}  \sin(\phi) d\phi = 0$ because the sine is an odd function.


问:
Natural transformation of groups
I am reading "Categories for the working mathematician" written by Mac Lane.
There is an exercise in natural transformation section: If $B$ and $C$ are groups (regarded as categories with one object each) and $S, T: B\to C$ are functors (homomorphisms of groups), show that there is a natural transformation $S\to T$ if and only if $S$ and $T$ are conjugate; i.e. iff there is an element $h\in C$ with $Tg=h(Sg)h^{-1}$ for all $g\in B.$

My attempt: Let $f$ be an endomorphisms from $Hom(B,B)$. If there is a natural transformation $\tau$, since there is only one component, the above diagram must be commutative.  $\Rightarrow Tf\circ \tau=\tau\circ Sf$.(I) Let $f=1_B$, therefore we have $(1_C\circ \tau)(g)=(\tau\circ 1_C)(g)$.

(I): If in the question $g$ was a morphism $\in Hom(B,B)$, from equation (I) we have $Tg=\tau(Sg)\tau^{-1}$

答:
Let's spell everything out. We view the group $B$ as a category with one object $*_B$, the only Hom-set is $\text{End}(*_B)=B$ and composition is given by the group law. Similar for $C$. If $S\colon B\to C$ is a functor, then $S$ consists of the following data: $S(*_B)=*_C$ and for each $b\in \text{End}(*_B)$, $S(b)\in \text{End}(*_C)$ such that $S(bb')=S(b)S(b')$.

A natural transformation $\eta\colon S\Rightarrow T$ consists of the following data:

$\eta_{*_B}\in \text{End}(*_C)=C$;

For all $c\in \text{End}(*_C)=C$, we have $T(c)\eta_{*_B}=\eta_{*_B}S(c)$.

The second condition is simply $$T(c)=\eta_{*_B}S(c)\eta_{*_B}^{-1}.$$ This is precisely saying that $S$ and $T$ are conjugate.

Title: Natural transformation of groups


问:
why do I invert the exponent (Economics)

This question refers to the Capital stock in macroeconomics, so K is the capital stock.

Hi everyone. I have this problem where I don't actually understand why you invert the exponent. You still have to remove the exponent in the 4th step so why invert it in the first place. Is this because of the negative exponent it will give because of alpha being so low?

答:
They want to go from the second line to fifth line (and then probably to $K = \dots$) and are just taking baby steps:

2 to 3: invert both sides;

3 to 4: simplify both sides;

4 to 5: raise both sides to the power $1/(1-\alpha)$.

They might as well have taken all those steps as one; the authors probably believe this is easier to understand for their readers.


问:
Real characters of odd degree
Suppose that $G$ is a finite group and $\chi$ is an irreducible real character, namely that $\chi(g) \in \mathbb{R}$ for every $g \in G$. Is it true that if $\chi(1)$ is an odd number, then $\chi$  is afforded by a representation over real numbers?

Motivation, let $m_{\mathbb{R}}(\chi)$ the Schur index of $\chi$ as defined in $10.1$ of Isaacs' "Character Theory of Finite Groups". If $\mathcal{X}$ is a complex representation that affords $\chi$, then there is an $\mathbb{R}$-representation $\mathcal{Y}$ of twice the dimention of $\mathcal{X}$ such that, viewed as complex, its character is $2\chi$. Hence the Schur $m_{\mathbb{R}}(\chi)$ should be $\le 2$. Nevertheless, by point $h)$ of Corollary $10.2$ of Isaacs' book, we have that $m_{\mathbb{R}}(\chi)$ divides $\chi(1)$ that is odd.

答: I think I found a quicker, more direct proof. By Brauer-Speiser Theorem (Isaacs' book page 171) we have that $m_{\mathbb{Q}}(\chi)\le 2$ and hence $m_{\mathbb{Q}}(\chi)=1$, since it divides $\chi(1)$ that is an odd number. By point $f)$ of 10.2 of the same book, $m_{\mathbb{R}}(\chi)$ divides $m_\mathbb{Q}(\chi)$.

Title: Real characters of odd degree


问:
Prove the limit equation: $\lim_{x\rightarrow\infty}e^{-x}\sum_{n=0}^{\infty}f_{n}\left(x\right)=-\gamma.$ FYI$,\quad\gamma$ is Euler's constant
Define a function sequence $\left \{ f_{n} \right \}_{n= 0}^{\infty}$ on $\left ( 0, \infty \right )$ as
$$f_{0}\left ( x \right )= \ln x,\quad f_{n+ 1}\left ( x \right )= \int_{0}^{x}f_{n}\left ( t \right ){\rm d}t\quad\left ( n= 0, 1, 2\cdots \right )$$
Prove the limiting equation:
$$\lim_{x\rightarrow\infty}e^{-x}\sum_{n= 0}^{\infty}f_{n}\left ( x \right )= -\gamma$$
For your information$,\quad\gamma$ is Euler's constant defined
$$\gamma:=\lim\left ( \sum_{k= 1}^{n}\frac{1}{k}- \ln n \right )$$
This problem belongs to my friend@twelve_sakuya, I haven't touched Euler's constant of course we have no approaches from me. But I had done a similar work. Maybe, there is a relationship between both problems, I need to the help. Here we go.

Given a function $f\left ( x \right )$ continuous on the real interval. Prove that
$$\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$
Source: Hungarian Math Contest Revision

Integral's limit, which I think that required us to divide and evaluate the integral function. By using the integral form of Cauchy-Schwarz inequality, we have
$$\left | e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq e^{-x}\left | \int_{-x}^{-\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{\frac{x}{2}}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq$$
$$\leq e^{-x}\left ( \int_{-x}^{-\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-x}^{-\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+ e^{-x}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+$$
$$+ e^{-x}\left ( \int_{\frac{x}{2}}^{x}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{\frac{x}{2}}^{x}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}\leq e^{-\frac{3}{2}x}\left ( \int_{-\infty}^{-\frac{x}{2}}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+ e^{-\frac{x}{2}}\left ( \int_{-\infty}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+$$
$$+ \left ( \int_{\frac{x}{2}}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}\underset{x\rightarrow\infty}{\rightarrow}0\therefore\left | \int_{-\infty}^{\infty}f^{2}\left ( x \right ){\rm d}x \right |< \infty\Rightarrow\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$

答:
This is not an answer, it's more like a hint, but it's too large for a comment. The $n$-th term of the sequence $\{f_n\}$ can be written as iterated integrals. For example
$$f_2 = \int_{0}^{x}\int_{0}^{t}\log u\,du \,dt.$$
The Cauchy formula for repeated integration states that if
$$I^n(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1,$$
this expression is equal to
$$I^n(x) = \frac{1}{(n-1)!} \int_a^x \left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.$$

Edit: I saw Yuris comment just now. Not a copy.


问:
Give an upper bound on the probability that you wait more than 30min for the bus.
Consider the time to wait for your bus is modeled by $X\text{~}Exp(\lambda)$.

What is the $90/100th$ quantile for $\lambda=1$?

Given $90/100th$ quantile is 20 min ($\lambda$ is unknown). Give an upper bound on the probability that you wait more than 30 min. (The upper bound must be less than $0.5$).

My attempt:

For the first part, I can just use the formula that $\mathbb{P}(X\geq x_p)=1-p$ where $p=90/100$, and the formula for Exponential distribution, which is $f(x)=\lambda e^{-\lambda x}$.
$$1e^{-1x}=1-90/100$$
$$x=-\ln(10/100)=2.3025$$

However, I don't understand what is meant by find the upper bound for the probability. Which formulas are involved and how should I approach this problem? Thank you.

Body:
$X \sim Exp(\lambda)$

The PDF $f(x) = \lambda.e^{-\lambda x}$

The CDF $F(x) = P(X \leq x) = 1 - e^{-\lambda x}$

The quantile function $F^{-1}(p) = \frac{1}{\lambda}\ln \frac{1}{1-p}$

For $\lambda=1$ the $\frac{90}{100}$ quantile is $\ln \frac{1}{1-\frac{9}{10}} = \ln 10 = 2.302585$ mins

Given,  $F^{-1}(90/100) = \frac{1}{\lambda}\ln \frac{1}{1-90/100} = \frac{\ln 10}{\lambda} = 20 \implies \lambda = \frac{\ln 10}{20}$

Note that it implies $P(X \leq 20) = F(20) = \frac{90}{100}$ and $P(X > 20) = \frac{10}{100}$

$\therefore P(X > 30)$

$= 1 - P(X \leq 30) = 1 - F(30) = e^{-\lambda. 30} = e^{-\frac{\ln 10}{20}. 30}$

$= (e^{-\ln 10})^\frac{3}{2} = \left(\frac{1}{10}\right)^\frac{3}{2} = 0.03162278$


问:
Why is it important that every (infinite) dimensional vector space has a (hamel) basis?
An argument often used in favor of the axiom of choice is that it is equivalent to every  infinite dimensional vector space having a hamel basis. However the article on wikipedia says that those basis are usually not very useful when they're uncountable, and that other concepts such as "orthogonal basis" are more important in these cases.

So why does it matter whether infinite dimensional vector spaces have hamel basis?

答: One interesting element is that if two vector spaces over the same field have the same dimension, they are isomorphic. So identifying the cardinality of a Hamel basis is a way to identify isomorphic vector spaces.</p>


答:
$x\ne 0$ screams at me that $(0,0)$ will be a limit point.

And indeed for every $r$ will be points $(x,y)$ so that $|y|\le |x|$ and $\sqrt{x^2+y^2} < r$.  For example $p= (\frac 1{\sqrt 2}r, \frac 1{\sqrt 2}r)\in A$ and $d(p,(0,0)) = \frac 12 r < r$.

So $(0,0)$ is a limit point of $A$ that is not in $A$.  SO $A$ is not closed as it doesn't contain all its limit points.

Body:
Determine if the set $A= \{(x,y) \mid |y|\leqslant|x|, x \ne 0\}$ is closed?

I know that $A$ is closed if $A^c$ is open. Now $A^c$ is just $A^c = \{(x,y) \mid |y| >|x| \}$. To show that $A^c$ is open I would have to show that for every $x \in A^c$ there is a disk $B(x,r) \subset A^c$, but I'm not sure how to approach this?


问:
Necessity for n-th roots of unity in (simple) Kummer extension
I'm currently studying cyclic Galois extensions (i.e. with a cyclic Galois group), specifically the special case where the characteristic is coprime to the extension degree. So consider some cyclic Galois extension $E/K$ with $n := [E:K]$ and $\mathrm{char}(K) \nmid n$.

Now, I know that if $K$ contains all $n$-th roots of unity (which makes $E/K$ into a special case of a so-called Kummer extension), then the extension is radical, i.e. there is some $\alpha \in E$ with $E = K(\alpha)$ and $\alpha^n \in K$.

Indeed, if given a primitive $n$-th root of unity $\zeta_n \in K$, the above statement can be proven by using Hilbert's Satz 90 on $\zeta_n^{-1}$. Now, I'm guessing that the statement does not hold if we drop the assumption of $K$ containing all $n$-th roots of unity. I have tried to come up with a counterexample, however I have so far not been able to find one, so I would be thankful for any hints on how to get to one.

Specifically, I am looking for a cyclic Galois extension $E/K$ of degree $n = [E:K]$ and with $\mathrm{char}(K) \nmid n$ such that $K$ does not contain all $n$-th roots of unity and $E/K$ is not a radical extension, i.e. there is no $\alpha \in E$ with $E = K(\alpha)$ and $\alpha^n \in K$.

答:
$[E:K]=n$ and $E=K(a)$ with $a^n\in K$ means that $E/K$ is purely radical: $x^n-a^n$ is the minimal polynomial of $a$ and $E/K$ is normal iff $x^n-a^n$ splits completely iff $x^n-1$ splits completely in $E$.

To me $E/K$ is radical means that $E=K(a)$ with $a^m\in K$ for some $m$.

$a$ is a root of $x^m-a^m$ so its minimal polynomial is $\prod_{j=1}^n (x-\zeta_m^{c_j}a)$ for some $c_j$ and $E/K$ is Galois means that the $\zeta_m^{c_j}$ are in $E$.

With $K=\Bbb{Q},E=K(\cos(2\pi / 7))$ this implies that $\zeta_m^{c_j}=\pm 1$ thus in fact $n=2$, a contradiction since $[E:K]=3$. Whence $E/K$ is not radical.

On the other hand $E$ is contained in the tower of purely radical extensions $\Bbb{Q}(\zeta_3,\cos(2\pi / 7))/\Bbb{Q}(\sqrt{-3})/\Bbb{Q}$.


答:
Another possibility:

Let the sequence $((x_n,y_n))$ be defined by $(x_n,y_n)=(1/n,1/n).$ Then $((x_n,y_n))$ is a convergent sequence in $A$ with limit $(0,0)$. But $(0,0) \notin A.$ Hence $A$ is not closed.

Title: Determine if the set $A= \{(x,y) \mid |y|\leqslant|x|, x \ne 0\}$ is closed?


问:
Variance of the sum of the even numbers rolled
I tried to solve the following problem for my probability theory class :

A dice is rolled $100$ times. If $X$ is the sum of the even numbers
rolled, find the expected value and the variance of $X$.

My attempt:

Let $X_i$ be the number of times we rolled $i$ $\left(\text{for  } i = \overline{1,6}\right)$. For example, if after 100 tries, we rolled $5$ seven times, then $X_5 = 7$.

Then $X=2X_2 + 4X_4 + 6X_6$, so by the linearity of expectation, we have:

$$ \mathbb{E}(X) = \mathbb{E}(2X_2 + 4X_4 + 6X_6) = 2\mathbb{E}(X_2)+4\mathbb{E}(X_4) + 6\mathbb{E}(X_6) $$

Also,
$$
\mathbb{P}(X_i = k) = {100 \choose k} \left( \dfrac{1}{6} \right)^k \left( \dfrac{5}{6} \right)^{100-k}
$$
Therefore, $X_i \sim \text{Bin}\left(100, \dfrac{1}{6} \right) $, so we have $\mathbb{E}(X_i) = \dfrac{100}{6}$
$$
\Rightarrow \mathbb{E}(X) = \dfrac{100}{6}(2+4+6) = 200.
$$

To find the variance, I know that for two independent random variables $A$ and $B$, we have $$\text{Var}(A+B) = \text{Var}(A)+\text{Var}(B),$$
which would eventually solve my problem, but somehow, I can't figure out whether $X_2, X_4, X_6$ are independent or not... Or is this a wrong approach?

Thanks in advance for any help!

答:
$X_i$ are NOT independent, because $\sum\limits_{i=1}^6 X_i = 100$ for example.

Consider $Y_i\sim U\{1,2,3,4,5,6\}$, represent each dice, and $Z_i=1\{Y_i \text{ even}\}$.

Now, you want to calculate expectation and variance of $\sum\limits_{i=1}^{100}Y_iZ_i$

Try to prove:

$Y_iZ_i$ and $Y_jZ_j$ are independent if $i\neq j$

$E(Y_iZ_i)=2P(Y_i=2)+4P(Y_i=4)+6P(Y_i=6)$

For variance, use that $Var(X)=E(X^2)-E(X)^2$ and try to analyze the variable $(Y_iZ_i)^2$. What are its possible values?


问:
First moment inequality implies tail distribution inequality?
Let $U,V$ be two continuous random variables, both with continuous CDF. Suppose that $\mathbb E V \geq \mathbb E U$. Can one conclude that $\mathbb P(V> x) \geq \mathbb P(U>x)$ for all $x\geq 0$? If not, what additional conditions are needed?</p>

答: Here is an argument with may make think it's not true: if $Y$ is a positive random variable, then $E(X)=\int_0^{+\infty}P(X>t)dt$. The fact that $P(U>x)\geqslant P(V>x)$ seems much stronger than $E(U)\geqslant E(V)$.</p>


答:
$$f_n(x)=\frac{1}{(n-1)!}\int_0^x (x-t)^{n-1} \ln t ~dt$$

$$\sum_{n= 1}^{\infty}f_{n}\left ( x \right )=\int_0^x \ln t ~dt~\sum_{n= 1}^{\infty} \frac{(x-t)^{n-1}}{(n-1)!}=\int_0^x \ln t ~dt~e^{x-t}$$

$$\lim_{x \to \infty} e^{-x}\sum_{n= 1}^{\infty}f_{n}=\int_0^\infty e^{-t} \ln t ~dt=-\gamma$$

The summation could be started with $1$ and not with $0$, as $e^{-x} f_0(x)$ simply gives zero in the limit.

Title: Prove the limit equation: $\lim_{x\rightarrow\infty}e^{-x}\sum_{n=0}^{\infty}f_{n}\left(x\right)=-\gamma.$ FYI$,\quad\gamma$ is Euler's constant

Body:
Define a function sequence $\left \{ f_{n} \right \}_{n= 0}^{\infty}$ on $\left ( 0, \infty \right )$ as
$$f_{0}\left ( x \right )= \ln x,\quad f_{n+ 1}\left ( x \right )= \int_{0}^{x}f_{n}\left ( t \right ){\rm d}t\quad\left ( n= 0, 1, 2\cdots \right )$$
Prove the limiting equation:
$$\lim_{x\rightarrow\infty}e^{-x}\sum_{n= 0}^{\infty}f_{n}\left ( x \right )= -\gamma$$
For your information$,\quad\gamma$ is Euler's constant defined
$$\gamma:=\lim\left ( \sum_{k= 1}^{n}\frac{1}{k}- \ln n \right )$$
This problem belongs to my friend@twelve_sakuya, I haven't touched Euler's constant of course we have no approaches from me. But I had done a similar work. Maybe, there is a relationship between both problems, I need to the help. Here we go.

Given a function $f\left ( x \right )$ continuous on the real interval. Prove that
$$\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$
Source: Hungarian Math Contest Revision

Integral's limit, which I think that required us to divide and evaluate the integral function. By using the integral form of Cauchy-Schwarz inequality, we have
$$\left | e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq e^{-x}\left | \int_{-x}^{-\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{\frac{x}{2}}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq$$
$$\leq e^{-x}\left ( \int_{-x}^{-\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-x}^{-\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+ e^{-x}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+$$
$$+ e^{-x}\left ( \int_{\frac{x}{2}}^{x}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{\frac{x}{2}}^{x}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}\leq e^{-\frac{3}{2}x}\left ( \int_{-\infty}^{-\frac{x}{2}}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+ e^{-\frac{x}{2}}\left ( \int_{-\infty}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+$$
$$+ \left ( \int_{\frac{x}{2}}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}\underset{x\rightarrow\infty}{\rightarrow}0\therefore\left | \int_{-\infty}^{\infty}f^{2}\left ( x \right ){\rm d}x \right |< \infty\Rightarrow\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$


问:
Solve combinatorics tasks

The Question

The student forgot the four-digit identification code of his credit card. What is the probability that a student will receive his money by
typing a code at random if he remembers that:

$\quad$ a) all digits of the code are different;

$\quad$ b) the code does not contain the digits $0$ and $1$?

$\quad$ P.S. a and b are two separated cases

$15$ parcels came to the orphanage. In four of them - winter things, in one - a leather jacket, in the rest - books. Workers open three
parcels randomly. What is the probability that:

$\quad$ a) in two of them - winter clothes and in one - a leather jacket;

$\quad$ b) all three parcels - with books?

$\quad$ P.S. a and b are two separated cases

My Understanding

I tried solving them using formulas and logic

1(a) the total number of possible combinations is $\frac{10!}{ 6!} = 5040$, and the probability $\frac{1}{5040}$?

1(b) there are only $8$ possible digits, repetitions are possible, so the probability is $\frac{1}{4096}$.

2(a)$\frac{2}{455}$

2(b)$\frac{24}{91}$

I've done all calculations using fractions and multiplication because I don't know how to do these using formulas and which

答:
Your answers are correct except for $2$(a). You do not need to apply much of formula anyway. Some of them did not have the working behind the answer so not sure how you arrived at them.

1(a). As you said each of the four digits are different so there are $10$ choices for the first digit, $9$ for the second, $8$ for the third and $7$ for the fourth digit. That is $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ codes in total and hence the probability that they will enter the right code in one attemtp $ = \frac{1}{5040}$.

1(b). There are $8$ choices for each of the four digits and that is $8^4 = 4096$ codes. So probability $ = \frac{1}{4096}$.

2(a). You seem to have made a mistake in this one. We select $3$ random boxes out of $15$. We need to find probability that one of the boxes has leather jacket and other two have winter clothes. As there is just one box with leather jacket, there is just one way to choose the box with leather jacket. But there are $4 \choose 2$ ways to choose two boxes with winter clothes.

So the probability should be $ \displaystyle \frac{4 \choose 2}{15 \choose 3} = \frac{6}{455}$.

2(b) As we need to find probability that all $3$ boxes have books,

The answer should be $ \displaystyle \frac{10 \choose 3}{15 \choose 3} = \frac{24}{91}$.

Title: Solve combinatorics tasks


问:
Prove whether the M.o.M. estimator is consistent or not
Let $X_{1}, X_{2}, \cdots, X_{k}$ to be random samples from $Binomial(n, p)$.

And denote k'th moment of distribution as $\mu_{k}$.

Then the first and second moments are:

\begin{align}
\mu_{1} = np, \quad \mu_{2} = np(1-p) + n^{2}p^{2}
\end{align}

So we can get,

\begin{align}
p = 1 - \frac{\mu_{2}-\mu_{1}^{2}}{\mu_{1}}, \quad n = \frac{\mu_{1}}{p} = \frac{\mu_{1}^{2}}{\mu_{1}-\mu_{2}+\mu_{1}^{2}}
\end{align}

And now I use the method of moments to estimate $n$ and $p$.

\begin{align}
\hat{p} = \frac{\bar{X} - \frac{1}{k}\sum{(X_{i}-\bar{X})^{2}}}{\bar{X}}, \quad \hat{n} = \frac{\bar{X}^{2}}{\bar{X} - \frac{1}{k}\sum{(X_{i}-\bar{X})^{2}}}
\end{align}

What I wonder is whether $\hat{p}$ and $\hat{n}$ are consistent estimators.

My poor assumption is - I refered to Slutsky's theorem(https://en.wikipedia.org/wiki/Slutsky%27s_theorem) -

$\bar{X}$ is the consistent estimator for $\mu$ and $\frac{1}{k}\sum{(X_{i}-\bar{X})^{2}}$ is the consistent estimator for $\sigma^{2}$

so by using Slutsky's theorem, $\bar{X} - \frac{1}{k}\sum{(X_{i}-\bar{X})^{2}}$ is going to converge to $\mu-\sigma^{2}$ as $k$ goes to infinity.

Since both numerator and denominator converges, $\hat{p}$ is going to converge at $\frac{\mu-\sigma^{2}}{\mu}$ and $\hat{n}$ is going to converge at $\frac{\mu^{2}}{\mu-\sigma^{2}}$ as $k$ goes infinity.

And I concluded that they are consistent estimator as those values are equivalent to $p$ and $n$.

Is it correct or not? I'm quite new to mathematical statistics so I might wrong while developing my logic, especially at the procedure using Slutsky's theorem.

答:
In other words, assuming your calculations as good, you get

$$\hat{p}=1-\frac{\frac{1}{n}\Sigma_i X_i^2}{\overline{X}}+\overline{X}$$

Thus using the Strong Law of Large Numbers (SLLN) and the continuous mapping theorem you get

$$\hat{p}\xrightarrow{\text{a.s.}}1-\frac{np(1-p)+n^2p^2}{np}+np=p$$

and similarly for the other estimator

$$\hat{n}=\frac{(\overline{X})^2}{\overline{X}-\frac{1}{n}\Sigma_i X_i^2+(\overline{X})^2}\xrightarrow{\text{a.s.}}\frac{n^2p^2}{np-np(1-p)-n^2p^2+n^2p^2}=n$$


问:
How to create an inverted $S$ cruve
I'm looking for a way to create an inverted $S$ curve function. Below is an image of what I want. Left is a regular S curve, on the right is the kind of curve I want. Here are the allowed set of operations that I can use:

答:
I found a function which does what I wanted to do:

$$
f(x)=-\frac{(1-2x)^3}{2}+0.5
$$

Image of the curve

Title: How to create an inverted $S$ cruve


问:
Composite function gradient
Suppose I have smooth maps $g:R^m \rightarrow R^n$ and $f:R^n \rightarrow R$. Then I think $$\nabla (f \circ g) = (\nabla g)^T (\nabla f) \circ g,$$ where $\nabla g$ is the $n \times m$ Jacobian matrix. This is the only plausible answer that makes the dimension work out, and if I use row vectors instead of columns, it nicely resembles the univariate chain rule: $$\nabla (f \circ g) = ((\nabla f) \circ g) (\nabla g).$$ But can anyone prove it? And is there some way this can work with column vectors that doesn't involve transposing the Jacobian? Thanks!

答:
It is convenient to approach this kind of problems with a more general and convenient rule.

Using the chain rule for differentials $d(f\circ g)(x, dx)=df(g(x),dg(x,dx))$:
$$d(f\circ g)(x, dx)=\nabla (f\circ g)\cdot dx=df(g(x),dg(x,dx))=(\nabla f)\circ g\cdot dg(x,dx)=$$
$$=(\nabla f)\circ g\cdot\nabla gdx=\operatorname{tr}([(\nabla f)\circ g]^T\nabla gdx)=(\nabla g)^T (\nabla f)\circ g\cdot dx$$
$\cdot$ is the dot product, also i have used the property of trace $\operatorname{tr}(A^TB)=A\cdot B$.

So we have from the derivation above that
$$\nabla (f\circ g)=(\nabla g)^T (\nabla f)\circ g$$

Proof sketch for the chain rule above:

Using the definitions of differentials
$$f\circ g(x+h)=f\circ g(x)+d(f\circ g)(x, h)+o(||h||)$$
$$g(x+h)=g(x)+dg(x,h)+o(||h||)$$
Putting them in the composition
$$f\circ g(x+h)=f(g(x)+dg(x,h)+o(||h||))=$$
$$f\circ g(x)+df(g(x),dg(x,h)+o(||h||))+o(||dg(x,h)+o(||h||)||)=$$
$$=f\circ g(x)+df(g(x),dg(x,h))+df(g(x),o(||h||))+o(||dg(x,h)+o(||h||)||)=$$
$$=f\circ g(x)+df(g(x),dg(x,h))+o(||h||)$$
because $df(g(x),o(||h||))+o(||dg(x,h)+o(||h||)||) \in o(||h||)$, i omit here tedious epsilon-delta manipulations to prove this fact, but it can be shown using continuity and linearity of $df, dg$ in $h$.

So from the first formula and the last and from the uniqueness of differential we have
$$d(f\circ g)(x, h)=df(g(x),dg(x,h))$$

Title: Composite function gradient


问:
Does all proofs by contrapositive have a direct proof that is similar in length?
I know this may be a question not suitable to ask here or is a duplicate, but I'm just wondering if all proofs by contrapositives can be rephrased into a direct proof, without using the fact that they are logically equivalent?

Are there instances when a proof by contrapositive is not too hard, but direct proof is way more involved or literally impossible to do?

答:
I assume below that by "proof by contrapositive" you mean the principle

You can prove $A\rightarrow B$ by showing that $\lnot B$ implies $\lnot A$.

or in other words

From $\lnot B\rightarrow \lnot A$ you may infer $A\rightarrow B$.

[EDIT: The above lines were added following Rob Arthan's comments]

Proof by contrapositive is a reasoning principle which is valid in classical logic, but not in intuitionistic logic. So if you have a proof by contrapositive of a theorem which is not intuitionistically valid (e.g., $\lnot\lnot A\rightarrow A$), you cannot really get rid of the contrapositive reasoning: At best (depending on the precise formulation of your proof system) you can replace it by another equivalent "non-constructive" principle such as double negation elimination or proof by excluded middle.

So how about classical theorems which are also intuitionistically valid? Here you can remove all non-constructive reasoning steps in proofs, but this might lead to much larger proofs. A nice and simple example of this is blowup is mentioned in the Introduction of:

M. Baaz, A. Leitsch and G. Reis, "A Note on the Complexity of Classical and Intuitionistic Proofs," 2015 30th Annual ACM/IEEE Symposium on Logic in Computer Science, Kyoto, Japan, 2015, pp. 657-666, doi: 10.1109/LICS.2015.66.

For an intuitionistically valid theorem $A$, simply consider the formula $B:=A\lor\lnot A$. An intuitionistic proof of $B$ will essentially go through the proof of $A$ and the conclude $A\lor \lnot A$ in the last step, and so it can be as large as intuitionistic proofs can get (there is no computable upper bound on the size of intuitionistic proof of a (first-order) theorem $A$ relative to the size of $A$)! On the other hand, $B$ has a trivial classical proof as it is an instance of the law of excluded middle.


问:
Proof that the largest eigenvalue of a stochastic matrix is $1$
The largest eigenvalue of a stochastic matrix (i.e. a matrix whose entries are positive and whose rows add up to $1$) is $1$.

Wikipedia marks this as a special case of the Perron-Frobenius theorem, but I wonder if there is a simpler (more direct) way to demonstrate this result.

答:
Say $A$ is a $n \times n$ row stochastic matrix. Now:
$$A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} =
\begin{pmatrix}
\sum_{i=1}^n a_{1i} \\ \sum_{i=1}^n a_{2i} \\ \vdots \\ \sum_{i=1}^n a_{ni}
\end{pmatrix}
=
\begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}
$$
Thus the eigenvalue $1$ is attained.

To show that the this is the largest eigenvalue you can use the Gershgorin circle theorem. Take row $k$ in $A$. The diagonal element will be $a_{kk}$ and the radius will be $\sum_{i\neq k} |a_{ki}| = \sum_{i \neq k} a_{ki}$ since all $a_{ki} \geq 0$. This will be a circle with its center in $a_{kk} \in [0,1]$, and a radius of $\sum_{i \neq k} a_{ki} = 1-a_{kk}$. So this circle will have $1$ on its perimeter. This is true for all Gershgorin circles for this matrix (since $k$ was taken arbitrarily). Thus, since all eigenvalues lie in the union of the Gershgorin circles, all eigenvalues $\lambda_i$ satisfy $|\lambda_i| \leq 1$.


问:
$\vdash (\lnot \chi \to \lnot \theta) \to (\theta \to \chi) $
For $\theta, \chi$ $\mathcal{L}$-formulas in predicate logic, I'm trying to prove: $\vdash (\lnot \chi \to \lnot \theta) \to (\theta \to \chi) $.

I've managed to prove $\vdash \lnot\lnot \theta \to \theta$, which was the hint for this problem. I have already proved $\vdash (\theta \to \chi) \to (\lnot \chi \to \lnot \theta) $. My attempt was to substitute $\lnot \chi$, $\lnot \theta$ into $\theta, \chi$ in $\vdash (\theta \to \chi) \to (\lnot \chi \to \lnot \theta) $. This gives $\vdash (\lnot \chi \to \lnot \theta) \to (\lnot \lnot \theta \to \lnot \lnot \chi) $. Finally, I was trying to show $\vdash (\lnot \lnot \theta \to \lnot \lnot \chi) \to (\theta \to \chi)$, but was not having much luck. Is there a nice way of doing this?

Furthermore, how does one go about doing things like this? I have seen natural deduction proofs that are very complicated, or are written in a way such that I suspect that there is some motivation that I am surely missing. But I can't figure out how to get the intuition/motivation for natural deduction proofs in general. My attempts purely seem to be brute-forcing, which is not so good.

答:
I’d do this the following way similarly to what @Mauro has been hinting at in comments but I’m not sure if this is correct in the system for natural deduction you are using.

I’ll use numbers to indicated the levels of „nested proofs”, this is the method of natural deduction I am used (so there are rules for introducing and eliminating different connectives which one can apply)

We want $\vdash (\lnot p \to \lnot q) \to (q \to p)$.

Let’s assume $(\lnot p \to \lnot q)$. Goal: $q\to p$. So we want $\lnot p \to \lnot q \vdash q \to p$

1.1. Lets assuume $q$. Goal: $p$, ie. $\lnot p \to \lnot q, q \vdash p$

edit: it seems I have messed up the final part. Perhaps this would be better:

1.1.1. proof by contradiction since $\vdash \lnot\lnot p \to p$. So as we recall our goal now was $p$. Lets assume $\lnot p$. Then we have $\lnot p \to \lnot q$ hence we can eliminate the implication and get $\lnot q$.

1.1.2. So now we have both $q$ and $\lnot q$. That's a contradiction -- we achieved the goal ie $\lnot p \to \lnot q, q, \lnot p \vdash \bot$

1.2. So assuming $\lnot p$ we got $\bot$, so we can intro an implication $\lnot p \to \bot$ so $\lnot p \to \lnot q, q\vdash \lnot p \to \bot$ which is equivalent to $\lnot p \to \lnot q, q\vdash \lnot\lnot p$ (intro of negation). Thus we have proven $\lnot\lnot p\leftrightarrow p$ and got our goal

we managed to prove $p$ assuming $q$ so we intro $q \to p$ -- we have achieved our current goal. $\lnot p \to \lnot q \vdash q \to p$

So from assuming $\lnot p \to \lnot q$ we have proven $q\to p$. Intro: $\vdash (\lnot p \to \lnot q) \to (q\to p)$

Hope that is correct. My intuition to natural deduction is to realise it s natural so I can just think: when is an implication true? When the antecedent is false (the triv case) or when both parts are true. Let's assume it is true and try proving the rest with those assumptions using those rules for introducing or eliminating different things. Then I simply do it recursively with goals and subgoals etc. Jaśkowski's boxes are good for that as they resemble a computer program with their nested structure.


问:
Why is the set of all smooth sections of a smooth vector bundle closed under addition?
Let $\pi:E\to M$ be a smooth vector bundle, where $E$ and $M$ can have nonempty boundaries. Let $\sigma_1$ and $\sigma_2$ be smooth global sections of $E$. I want to show that $\sigma=\sigma_1+\sigma_2$ defined by $\sigma(p)=\sigma_1(p)+\sigma_2(p)$ is also smooth.

It suffices to show that $\sigma$ is smooth on an open neighborhood of each point $p$ in $M$. I can do that when $p$ is an interior point of $M$. Because then there is a smooth local trivialization $(U,\phi)$, where $p\in U\subseteq\operatorname{Int}M$ and $\phi:\pi^{-1}(U)\to U\times\mathbb R^k$ is a diffeomorphism. Then $\sigma|_U$ is smooth because it is a composition of the smooth maps
\begin{equation}
U\xrightarrow{(\sigma_1,\sigma_2)}\pi^{-1}(U)\times\pi^{-1}(U)\xrightarrow{\phi\times\phi}(U\times\mathbb R^k)\times(U\times\mathbb R^k)\xrightarrow{+}U\times\mathbb R^k\xrightarrow{\phi^{-1}}\pi^{-1}(U)\text{,}
\end{equation}
where $(\sigma_1,\sigma_2)(p)=(\sigma_1(p),\sigma_2(p))$ and $+((q_1,v_1),(q_2,v_2))=(q_1,v_1+v_2)$. But if $p$ is a boundary point of $M$, then I cannot argue as above, since $\pi^{-1}(U)$ could have a nonempty boundary, so $\pi^{-1}(U)\times\pi^{-1}(U)$ does not make sense.

答:
I found the answer myself. Let $\pi:E\to M$ be a smooth vector bundle. If $(U,\phi)$ is a smooth local trivialization and $\tau:U\to\pi^{-1}(U)$ is a rough local section of $\pi$, then
\begin{equation}
\phi\circ\tau(p)=(p,\tau^1(p),\ldots,\tau^k(p))
\end{equation}
for some functions $\tau^i:U\to\mathbb R$. $\tau$ is smooth if and only if each $\tau^i$ is smooth.

In my case, $\sigma^i=\sigma_1^i+\sigma_2^i$. Since $\sigma_1^i$ and $\sigma_2^i$ are smooth, $\sigma^i$ are
also smooth. Thus $\sigma$ is smooth.

I would be grateful for any feedback.


答:
First of all, the series $\sum_{n\geqslant 0}f_n(x)$ converges for all $x>0$ because one can show by induction that $|f_n(x)|\leqslant\frac{(x+1)^{n+1}}{(n+1)!}$. The case $n=0$ being straightforward, let $n$ such that $|f_n(x)|\leqslant\frac{(x+1)^{n+1}}{(n+1)!}$ for all $x>0$, then
$$ |f_{n+1}(x)|\leqslant\frac{1}{(n+1)!}\int_0^x (t+1)^{n+1}dt=\frac{(x+1)^{n+2}}{(n+2)!} $$
Moreover, it follows that $\sum_{n\geqslant 1} f_n'=\sum_{n\geqslant 0}f_n$ is normally convergent on all compact of $(0,+\infty)$, therefore, if $\varphi:=\sum_{n\geqslant}f_n$, then $\varphi$ is differentiable and
$$ \varphi'(x)=\sum_{n\geqslant 0}f_n'(x)=\sum_{n\geqslant 0}f_n(x)+f_0'(x)=\varphi(x)+\frac{1}{x} $$
Thus there exists $\lambda\in\mathbb{R}$ such that $\displaystyle\varphi(x)=\lambda e^x+\int_1^x\frac{e^{x-t}}{t}dt$ so that
$$ \lim\limits_{x\rightarrow +\infty}\varphi(x)e^{-x}=\lambda+\int_1^{+\infty}\frac{e^{-t}}{t}dt $$
But since $f_n(0)=0$ for all $n\geqslant 1$, then $\lim\limits_{x\rightarrow 0}(\varphi(x)-\log x)=0$. Using an integration by parts, we have
$$ \int_1^x\frac{e^{x-t}}{t}dt=\log x+\int_1^x e^{x-t}\log tdt $$
Therefore $\lambda=\int_0^1e^{-t}\log tdt$. Finally,
$$ \lim\limits_{x\rightarrow +\infty}\varphi(x)e^{-x}=\int_0^{+\infty}e^{-t}\log tdt=\Gamma'(1)=-\gamma $$