问:
Why does $A^TAx = A^Tb$ have infinitely many solution algebraically when $A$ has dependent columns?

This is a problem from least square approximation, where we solve the equation $A^TAx = A^Tb$ when $Ax = b$ is unsolvable.

The case I am dealing with is when A has dependent columns, i.e. A is an m by n matrix where the rank r is smaller than n. In this case $A^TA$ is a singular n by n matrix with dependent columns, the rank of which is also r (Rank($A^TA$)=Rank($A$)).

Now in the book Introduction to Linear Algebra of the legendary Gilbert Strang, he says and I quote, when A is singular, $A^TA$ is also singular, and the equation $A^TAx = A^Tb$ had infinitely many solutions, the pseudoinverse gives us a way to choose a "best solution" $x^+=A^+b$.

I understand why the equation has infinitely many solutions geometrically:

Because what the equation asserts geometrically is to find the projection of b, denoted by p, in the column space of A, then solve the new equation $A\hat x$ = p. Because we can always project b onto the column space of A, whether it's singular or not, we know there must be a solution to the equation $A\hat x$=p and if there is a solution, there are infinitely many because A is singular.

My question is how do we know that the equation have infinitely many solution algebraically, to make it clearer, I don't understand why the equation has at least one solution. I do understand that once it has at least one solution, it has infinitely many.

Algebraically, I understands that $A^Tb$ will take us to $C(A^T)$, and it will take away the part of b that lies in $N(A)$. But what does it has to do with   $C(A^TA)$ ? My hypothesis is there is some formula regarding $C(A^TA)$ and $C(A^T)$ that I am not aware of. For example, if $C(A^TA) = C(A^T)$, then my problem is solved.

Also, I found this How come least square can have many solutions?, I know what  $\hat x^TA\hat x$ in sums will looks like, but I don't know where it came from, but assuming that this is actually correct, I understand the arguments made in that thread.

Any instruction will be appreciated.

答:
Imagine you have two data points (2,4) and (2,6). Now, if we put the slope formula into matrix form, we will have
\begin{pmatrix}2 & 1 \\2 & 1 \\\end{pmatrix}
as our X matrix. First column is for the input X and second for the slope "b". We set "b" as one because we want to give it freedom to move (otherwise, 0 would kill the "b" out of the formula, mx = y). Then, we have the vector (M B) for our unknown coefficients. These two matrices multiplied will give the output vector (4 6).

Now, try to think of how you would graph through these two data points. The problem is that there are infinite solutions depending on what you set the bias "b" to be equal to. Why do we have infinite least square solutions? Well, The column space is not full rank and we have a dependency.

Now, b is a free variable then. The algebraic intuition would come from setting b with different values to obtain the equivalent solution.


问:
How to go around roots when trying to figure out if a sequence has a limit and if it converges
Consider this sequence

$$\lim_{n \to \infty} \sqrt[n]{n^5 + 3n^2}$$

Now what I did is i tried to get rid of the root by transforming it into an exponent meaning;

$$\lim_{n \to \infty} (n^5 + 3n^2)^n$$

Now I've multiplied n into the paranthesis and I have this
$$\lim_{n \to \infty} n^6 + 3n^3$$

And since n is moving to infinity,the numbers will be increasing infinitly so the limit is simply $$
 \infty$$

Now since I have no way to prove that this is the correct solution,and on a few similar problems,where I was able to check my work(they also had to deal with roots) I've made a mistake on most of them. Now I'd like if anyone has a tip or some "trick" on how to deal with roots when it comes to these type of problems?

Thank you!

答:
$$L=\lim_{n \to \infty} \sqrt[n]{n^5 + 3n^2}=\lim_{n \to \infty} (n^5+3n^2)^{1/n}$$
Now consider
$$\log L = \log \lim_{n \to \infty} (n^5+3n^2)^{1/n}=\lim_{n \to \infty} \log(n^5+3n^2)^{1/n}=$$
$$=\lim_{n \to \infty} \frac{1}{n}\log(n^5+3n^2)=\lim_{n \to \infty}\frac{\log(n^5+3n^2)}{n}=$$
That is an indetermination form $\frac{\infty}{\infty}$. Apply L'hopital rule
$$=\lim_{n \to \infty}\frac{5n^4+6n}{n^5+3n^2}=0$$
As $\log L =0$ we can conclude that $L=1$

ContentLicense: CC BY-SA 4.0


问:
Equivalence classes of the relation that the largest digit of integer a = largest digit of integer b.
Define the relation $\mathcal{R}$ on the set of all positive integers by: for all positive integers $a$ and $b$, $a\,\mathcal{R}\,b$ if and only if the largest digit of $a$ is equal to the largest digit of $b$. for example, $271\,\mathcal{R}\,770$ because the largest digit of $271$ is $7$ which is also the largest digit of $770$.

a) Find the number of equivalence classes of $\mathcal{R}$.

b) Find and simplify the number of positive integers between $100$ and $1000$ which are in the equivalence class $[271]$.

I have an idea of where to begin but I think I am wrong. For (a) would there be $9$ equivalence classes?
And for (b) I got my answer to be $56$.

答:
a) is fine as for each digit except $0$ we have an equivalence class

b) It is easy to count all numbers between $100\le n<1000$ which have only digits $\le 7$: There are $7\cdot 8\cdot 8$. However, some of them have all digits strictly less than $7$, namely those with all digits $\le 6$: There are $6\cdot 7\cdot 7$ of these. Take the difference.

ContentLicense: CC BY-SA 3.0


问:
Using trigonometric power formulas to derive an identity for $\cos^3(x)$
I am practicing with manipulating sigma notation and binomial coefficients right now. I am using the formula given here to derive the identity for $\cos^3(x)$

The identity for $\cos^3(x)$ is

$$\cos^3(x)=\dfrac{1}{4}[3\cos(x)+\cos(3x)]$$

Now, the formula for deriving the identity of $\cos^{2n+1}(x)$ is

$$\cos^{2n+1}(x)=\dfrac{1}{4^{n}}\sum_{k=0}^{n}\binom{2n+1}{k}\cos[(2n+1-2k)x]$$

Let $n=1$

I have

$$\cos^{3}(x)=\dfrac{1}{4}\sum_{k=0}^{1}\binom{3}{0}\cos(3x)$$

$$=\cos^{3}(x)=\dfrac{1}{4}\sum_{k=0}^{1}2\cos(3x)$$

I know that a summation notation denotes:

$$\sum_{k=0}^{n}x_k=k_0+k_1+k_2+...k_n$$

But $k$ varnishes because I set it as $k=0$, so I think I am wrong there.

I think my calculation is clearly incorrect. Could you show me full works of how to arriving at the final identity using the power reduction formula?

答:
This is way easier with Chebyshev polynomials.
From the property
$$T_n(\cos\theta)=\cos(n\theta)$$
We get
$$T_3(\cos\theta)=4\cos^3\theta-3\cos\theta=\cos(3\theta)$$
$$\cos^3\theta=\frac{\cos(3\theta)+3\cos\theta}{4}$$
Easy as pie if you understand the linked article.


问:
System of equations word problem
Andrea's Sequoia gets 12 miles per gallon while driving in the city, and 20 miles per gallon while on the highway. The last time she filled up the car it took 16 gallons of gas, and she had driven 280 miles since the previous fill-up. How many miles of that 280 were city driving?</p>

答:
This is a standard type of problem where you're given enough information to form two linear equations, with two unknowns. And the process outlined below is one that will generalize to many such situations. So this is a good time to learn and understand how you can approach such a problem:

Let $\,c =$ (the number of miles that Andrea drove in the city since her last fill-up).

Let $\,h = $ (the number of miles she drove on the highway since her last fill-up.)

We know Andrea drove a total of $280$ miles since her last fill-up, so $c$ and $h$ must add up to $280$.

So one equation relating $c$ and $h$ is

$$c + h = 280\tag{1}$$

We are also given Andrea gets $\,12\,$ miles per gallons while citi-driving, which means she used $\dfrac c{12}$ gallons driving in the city since her last fill up. And since she gets $20$ miles per gallon on the highway, she must have used $\dfrac h{20}$ gallons for highway-driving since her last fill-up. And we're given since her last fill-up, she used a total of $16$ gallons. So the gallons used while city-driving and the gallons used while highway-driving must add up to $16$. So our second equation relating $c$ and $h$ is given by:

$$\frac c{12} + \frac h{20}=16 \tag{2}$$

From equation $(1)$, we can express miles of city driving, $c$ as follows: $$c + h = 280 \implies c = 280 - h.\tag{c}$$

Now simply substitute $c = 280 - h$ into $c$ of equation $(2)$, and solve for $h$. Then you can easily determine what $c$ must be by using the equation for $(c)$.

Title: System of equations word problem


问:
Inverse of fourth root function
I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$

but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!

Title: Inverse of fourth root function

Body:
\begin{align}
y=a\sqrt[^4]{b^2-(x-c)^2} &\implies y^{4} = a^{4}(b^{2} - (x-c)^2)\\
&\implies \frac{y^{4}}{a^{4}} = b^{2} - x^{2}+2cx-c^{2}\\
&\implies x^{2} - 2cx + \left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right) = 0\\
&\implies x = \frac{2c \pm \sqrt{4c^2-4\left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right)}}{2}\\
&\implies\boxed{x = c\pm\sqrt{b^2-\frac{y^{4}}{a^{4}}}}
\end{align}


问:
How to plot / compute the normal at a given point for a differentiable parametric surface?
To assess my understanding of parametric surfaces and local coordinates differential geometry I am trying to compute and plot the normal of a parametric surface at a point.

Here is what I did :

To keep the example simple I choose a saddle surface parameterized by the map :
$$
S : \mathbb{R^2} \rightarrow \mathbb{R^3} \\
(u,v) \rightarrow (u,v,u^2 -v^2 )
$$
I computed the partial derivatives with respect to $ u$ and $v$:
$$
S_u = (1,v,2u -v^2) \\
S_v = (u,1,u^2 -2v)
$$
From my understanding of the definitions the normal $N$ at point $q$ should be
$$
N(q)= \frac{S_u \wedge S_v}{|S_u \wedge S_v|}
$$

So for a given point it should be easy to compute the normal we have :
$$
S_u \wedge S_v = (-3v^2-2u+u^2v,3u^2-uv^2 +2v, 1-vu)
$$

Lets take for example the point $q_1$ with coordinates $(1,1)$ and compute the normal vector :

$$
S_u(q_1) \wedge S_v(q_1) = (-4,4,0)\\
|S_u(q_1) \wedge S_v(q_1)| = 4\sqrt2\\
N(q_1) = (\frac{-1}{\sqrt2}, \frac{1}{\sqrt2},0) \approx (-0.7,0.7,0)
$$
Which is not the true since the normal a this point should point toward $z$ axis...

I have done the same in python with equivalently unconvincing results (red lines correspond to $S_u,S_v$and $N$ at point $q_1$ ). :

Here is the python code :

```<code>import plotly.figure_factory as ff
import numpy as np
from scipy.spatial import Delaunay
import plotly.graph_objects as go

#Sampling the space of inputs (u,v)
samples =41
u = np.linspace(-10, 10, samples)
v = np.linspace(-10, 10, samples)
u,v = np.meshgrid(u,v)
u = u.flatten()
v = v.flatten()

#Create an array of 1 to fill the derivatives
onespace = np.ones(samples)
u_0,v_0 = np.meshgrid(onespace, onespace)
u_0 = u_0.flatten()
v_0 = v_0.flatten()

#Parametric equations of the sadle
x = u
y = v
z = u*u- v*v

#Partial derivatives
x_u = u_0
x_v= u
y_u= v
y_v=  v_0
z_u =2*u - v*v
z_v = u*u - 2*v

#Choosing a point p (u,v) on surface to compute normal
#ids (20,20) correspond to point (0.,0.)
u_test_id = 10
v_test_id = 10

#Display the parametric surface
points2D = np.vstack([u,v]).T
tri = Delaunay(points2D)
simplices = tri.simplices

fig = ff.create_trisurf(x=x, y=y, z=z,
                         simplices=simplices,
                         title="Sadle", aspectratio=dict(x=1, y=1, z=1))

#compute X at p and partial derivatives at p
Xp = [x[u_test_id+ v_test_id*samples] ,y[u_test_id+ v_test_id*samples],z[u_test_id+ v_test_id*samples]]
X_u  = np.array([x_u[u_test_id+v_test_id*samples], y_u[u_test_id+v_test_id*samples], z_u[u_test_id+v_test_id*samples]])
X_v  = np.array([x_v[u_test_id+v_test_id*samples], y_v[u_test_id+v_test_id*samples], z_v[u_test_id+v_test_id*samples]])

#computing non unit normals
N = np.cross(X_u,X_v)

#Display the frame using tubes
fig.add_trace(go.Streamtube(x=[Xp[0]],y=[Xp[1]],z=[Xp[2]],u=[X_v[0]],v=[X_v[1]],w=[X_v[2]],sizeref=0.1))
fig.add_trace(go.Streamtube(x=[Xp[0]],y=[Xp[1]],z=[Xp[2]],u=[X_u[0]],v=[X_u[1]],w=[X_u[2]],sizeref=0.1))
fig.add_trace(go.Streamtube(x=[Xp[0]],y=[Xp[1]],z=[Xp[2]],u=[N[0]],v=[N[1]],w=[N[2]],sizeref=0.1))

fig.show()
</code>```

答:
In order to work with surfaces, you can use gradient vectors, since they are by definition normal to the surface.

You can notice that your caracterization is equivalent to :

$$(x,y,z)\in S \Leftrightarrow z=x^2-y^2\\
\Leftrightarrow f(x,y,z)=x^2-y^2-z=0 $$

Then you know that $\nabla f(x,y,z) $ is the normal vector to your surface :

$$\nabla f(x,y,z)=(2x,-2y,-1) $$

By the way, your normal vector is only on the $z$-axis when $x=y=0$ (as you can see in your python plot)


答:
Taking the forth power, you get
$$
y^4 = a^4 (b^2-(x-c)^2) \Leftrightarrow \frac{y^4}{a^4}-b^2 = -(x-c)^2 \Leftrightarrow (x-c)^2 = b^2-\frac{y^4}{a^4}
$$

so, finally

$$
x = c \pm \sqrt{b^2-\frac{y^4}{a^4}}.
$$

We do know that $a \ne 0$ (if it was zero, then $y=0$ and there would be no inverse!), but now you must discuss how to pick the $\pm$ sign and what does that say about the existence  of the inverse in appropriate intervals.

If look into it, you'll see that you must have $x \in [c-b, c+b]$ and that $y(x)$ is invertible on $[c-b,c]$ and on $[c,c+b]$, but not on the full interval. Hence the choice of the sign.

Title: Inverse of fourth root function

Body:
I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$

but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!


问:
How to find a real sequence $x_n$ such that $n x_n \to 0$ as $n \to \infty$ and $\sum_{n=1}^{\infty} x_n$ diverges?
I'm stuck on trying to find a real sequence $x_n$ such that $n x_n \to 0$ as $n \to \infty$ and $\sum_{n=1}^{\infty} x_n$ diverges. Here's what I have tried so far:

I've started with trying to find $x_n$ as a real-valued function of $n$. In order to have $n x_n \to 0$, we require $$\frac{x_n}{1/n} \to 0$$ and since the denominator tends to $0$ as $n$ increases, we require that $x_n \to 0$. We also require that $x_n$ tends to $0$ "faster" than $1/n$ does, in order for the above quotient to go to $0$.

However if you try functions such as $x_n=1/n^2$ or $x_n=n/2^n$ (for example) which tend to $0$ faster than $1/n$ does, then you see that $\sum_{n=1}^{\infty} x_n$ converges and I haven't been able to find a function for $x_n$ that satisfies the above conditions and also has a divergent infinite sum. It's certainly possible that the form for $x_n$ might not be a simple elementary function of $n$, but I don't know how you'd go about finding any "unusual constructions" that work.

答: Take $x_n = \dfrac 1 {n\ln n }.$

Body: Take $x_n = \dfrac 1 {n\ln n }.$


问:
Double integral comes out different after changing $dx$ with $dy$.
$$\int_{0}^{1}\,dx \int_{0}^{1}\frac{x-y}{(x+y)^{3}}\,dy$$

My maths teacher gave this question to explain that if you change $dy$ with $dx$ the integral will have different value. My opinion is that the integral should come out as equal because this integral represents volume under a curve and the area over which we are integrating is same in both cases. Moreover this volume has to be $0$ because I plotted this curve in a 3d graph plotter and the surface is positive on one side of line $x=y$ and equally negative on the other side. BUT! after integrating the answer is coming as $1/2$ in one case and $-1/2$ in the other. Please explain.

答: If $\displaystyle \iint\limits_{[0,2]^2} \left| \frac{x-y}{(x+y)^3}\right| \, d(x,y) = +\infty$ then the two iterated integrals may have different (finite) values.


问:
Probability that $x$ elements belong to a group, $y$ to a second group and $z$ to a third one
I have been trying to solve a problem that goes like this:

After a software update, it is estimated that 10% of games will need complete modification of the code base, 60% won't need any modification and 30% will need only partial modification. What is the probability that out of 6 games, 2 will need complete modification, 1 will not need any modifications and 3 will need only partial modification?

I tried using a probabilistic model to approach this, yet it doesn't seem to fit any. A solution I thought of was to denote the games as $G_1,...,G_6$ and assign to each one a probability, such that 1 needs no modifications and 2 need complete modifications, take that probability, and then multiply with the number of permutations, so in the end I got to $6! \cdot (0.6 \cdot 0.1^{2} \cdot 0.3^{3})$. I think this is correct, but I'm not sure. Or, is there a better, more intuitive solution? Or one that uses a probabilistic model?

答:
After a software update, it is estimated that 10% of games will need complete modification of the code base, 60% won't need any modification and 30% will need only partial modification. What is the probability that out of 6 games, 2 will need complete modification and 1 will not need any modifications?

Chances that games 1 and 2 need complete modification, game 3 doesn't need any modification and games 4 through 6 need partial modification is

$$P = (0.1)^2 \times (0.6) \times (0.3)^3.$$

The number of ways that the above scenario can unfold is

$$T = \binom{6}{2} \times \binom{4}{1}.$$

Final Answer:

$$P \times T.$$


问:
(How) does lambda calculus encode/use associativity of function composition?
I'm just learning about Lambda Calculus, so apologies if this is an obvious question, but given how useful and fundamental the associativity property of function composition is, (how) is this associativity encoded/utilized in lambda calculus (which is apparently designed to abstract important patterns of functions)?

I'm not asking about the convention of left (or right) associativity in lambda or type notation, but the genuine and general associativity property of function composition, i.e. the general identity $f(gh)=(fg)h$.

The Hindley and Seldin bible on lambda calculus doesn't even mention associativity.

答:
In general it is not the case that given lambda terms $f,g,h$, then $(fg)h = f(gh)$. For example, let $I=\lambda x.x$, and take
$$f =\lambda x.\lambda y.x,\quad g=h = I$$
then $(fg)h = I$, however $f(gh) = \lambda y.I$.

However, you could define a composition term
$$c = \lambda f.\lambda g.\lambda x.f(gx)$$
and show that
$$((c((cf)g))h) = ((cf)((cg)h)).$$
Notice that this notation is hard to read, and for this reason it is common to associate application of lambda terms to the left, that is we can instead write
$$c(cfg)h = cf(cgh).$$

To show this we just expand $c$ as follows:
\begin{align*}
c(cfg)h &= \lambda x.(cfg)(hx)\\
&=\lambda x.(\lambda x.f(gx))(hx)\\
&=\lambda x. f(g(hx))\\
&=\lambda x.f((\lambda x.g(hx))x)\\
&=\lambda x.f((cgh)x)\\
&=cf(cgh).
\end{align*}


问:
Continuous functions and preservation of topological structure.
In some book I read "continuous fuctions between topological spaces preserve their topological structure" and they say that this is similar to the case when homomorphims perserve algebraic structures. But thinking about it, it seems to me that this structure preservation is made backwards.

For example, given $f:X \Rightarrow Y$, both $X$ and $Y$ topological spaces, it is true that the inverse image of finite intersection of open sets in $Y$ will be open in $X$. But this not happen with direct images. I might conclude that topological structure of $Y$ is preserved when the inverse images of $f$ send it to $X$.

It that true? Thanks in advance.

答:
In general, the inverse of a continuous function $f : X \rightarrow Y$ is not a function, so the claim that

I might conclude that topological structure of $Y$ is preserved when the inverse images of $f$ send it to $X$.

is not well-defined (or at least it's not obvious what well-defined statement you're trying to make).

However, the original statement is accurate. The definition of continuity, $O$ open in $Y \implies f^{-1}(O)$ open in $X$ turns out to be exactly what you need to prove that various properties are invariant under continuous functions. You can just take a look at examples like the proof for connectedness. There are many others.


问:
Marcus, Number Fields, problem 5(b), Chapter 4
Let $L/K$ be a Galois extension of number fields with Galois group G. Let $\mathfrak{p}$ be a prime of K (that is, $\mathfrak{p}$ is a prime ideal of the ring of integers $\mathcal{O}_L$ of L). We assume that $\mathfrak{p}$ is totally ramified in every intermediate field, but not totally ramified in $L$. We have to prove that no intermediate fields can exist. I think I have a proof. It would be helpful if someone could check my arguments and reassure me that everything is ok (assuming of course that my proof is correct).

So let $M$ be an intermediate field (we assume that $K \subsetneq M \subsetneq L$). It's true that there exists a prime $\mathfrak{q}'$ of $M$ such that $\mathfrak{q}'$ lies over $\mathfrak{p}$. It's also true that there exists a prime $\mathfrak{q}$ of $L$ such that $\mathfrak{q}$ lies over $\mathfrak{q}'$. Of course this implies that $\mathfrak{q}$ lies over $\mathfrak{p}$. Hence, it makes sense to talk about the ramification index of $\mathfrak{q}$ over $\mathfrak{p}$: $e(\mathfrak{q} \mid \mathfrak{p} )$. Using the multiplicativity of the ramification indices we have that $$ e(\mathfrak{q} \mid \mathfrak{p})= e(\mathfrak{q} \mid \mathfrak{q}') \hspace{1mm} e(\mathfrak{q}' \mid \mathfrak{p}). \hspace{10mm} (1) $$ By assumption, $\mathfrak{p}$ is totally ramified in $M$ and therefore $e(\mathfrak{q}' \mid \mathfrak{p})=[M:K]$. Thus relationship $(1)$ becomes $$e(\mathfrak{q} \mid \mathfrak{p})= e(\mathfrak{q} \mid \mathfrak{q}') \hspace{1mm} [M:K] .\hspace{10mm} (2) $$ IF we had that $e(\mathfrak{q} \mid \mathfrak{q}')=[L:M]$ then from $(2)$ we would have that $ e(\mathfrak{q} \mid \mathfrak{p}) = [L:K] $, which is wrong because we assumed that $\mathfrak{p}$ is not totally ramified in $L$. Therefore, in order to get our contradiction we have to prove that $e(\mathfrak{q} \mid \mathfrak{q}')=[L:M]$.

The extension $L/M$ is Galois and hence (look (for example) Theorem 21 by Marcus) $$ [L:M] = r' \hspace{1mm} e(\mathfrak{q} \mid \mathfrak{q}') \hspace{1mm} f(\mathfrak{q} \mid \mathfrak{q}') ,\hspace{10mm} (3) $$ where by $f(\mid)$ I denote the inertial degree and by $r'$ the number of primes of $\mathcal{O}_L$ that lie above $\mathfrak{q}'$. We shall prove that $r'$ and $f(\mathfrak{q} \mid \mathfrak{q}')$ are both equal to 1.

Consider the inertia field $L_E$ and it's prime $\mathfrak{q}_E$ (which lies above $\mathfrak{p}$ and under $\mathfrak{q}$). We know (look (for example) Theorem 28 by Marcus) that $f(\mathfrak{q} \mid \mathfrak{p})= f(\mathfrak{q}_E \mid \mathfrak{p} )$. But, by assumption, $\mathfrak{p}$ is totally ramified in $L_E$ and hence $f(\mathfrak{q}_E \mid \mathfrak{p})=1$. Therefore, $f(\mathfrak{q} \mid \mathfrak{p})=1$. Using now the multiplicativity of the inertial degrees we have: $f(\mathfrak{q} \mid \mathfrak{p})=f(\mathfrak{q} \mid \mathfrak{q}') \hspace{1mm} f(\mathfrak{q}' \mid \mathfrak{p}) $. That is, $1=f(\mathfrak{q} \mid \mathfrak{q}') \hspace{1mm} f(\mathfrak{q}' \mid \mathfrak{p}) $. Of course, this implies that $f(\mathfrak{q} \mid \mathfrak{q}')=1$, which is what we wanted.

Now we show that $r'=1$. Let's write $r$ for the number of primes of $\mathcal{O}_L$ tha lie above $\mathfrak{p}$. It's obvious that $r=1 \implies r'=1$. So we show that $r=1$. We have, $\mathfrak{p}$ totally ramified in $L_E \implies e(\mathfrak{q}_E \mid \mathfrak{p})=[L_E:K]$. But we know (look (for example) Theorem 28 by Marcus) that $[L_E:K]=r \hspace{1mm} f(\mathfrak{q} \mid \mathfrak{p})$. Above, we already showed that $f(\mathfrak{q} \mid \mathfrak{p})=1$. Therefore, $[L_E:K]=r$. Hence, $e(\mathfrak{q}_E \mid \mathfrak{p})=r$. But we know (look (for example) Theorem 28 by Marcus) that $e(\mathfrak{q}_E \mid \mathfrak{p})=1$. Hence, $r=1$. So, $r'=1$.

The above yield: $[L:M]=e(\mathfrak{q} \mid \mathfrak{q}')$. As we already explained, this leads us to a contradiction.

If something doesn't make sense please tell me. Thank a lot!

答:
I think that’s correct but there’s a way to present the proof in a shorter way?

Let $\mathfrak{q}$ be a prime in $L$ over $\mathfrak{p}$; let $I$ be the inertia group. If $I$ is the trivial subgroup, then $\mathcal{O}_\mathfrak{q}/\mathcal{O}_\mathfrak{p}$ is unramified so (as subextensions are ramified) there cannot be any subextension of $L/K$.

Assume $I$ is not the trivial subgroup (unramified extension, so it forces that $L/K$ without subextension). Let $M=L^I$. It’s an intermediate field smaller than $L$, so $\mathfrak{p}$ is totally ramified in $M$. Let $\mathfrak{q’}=\mathfrak{q} \cap M$, so $e(\mathfrak{q}’:\mathfrak{p})=[M:K]$.

But the Galois group of $L/M$ is the inertia group of $\mathfrak{q}$, so that $[L:M]=e(\mathfrak{q}:\mathfrak{q’})$. It follows $[L:K]=e(\mathfrak{q}:\mathfrak{p})$, which concludes.


问:
Why we count from 0 to 9?
In our daily lives, we generally use $10$ digits to define the numbers, i.e., $0$ to $9$. Based on these ten digits, we define all the natural, rational, and irrational numbers, etc. This representation is known as the decimal representation.

I know that one could also define the number system in other ways. For example:

Unary: 1 11 111 1111 ...

Binary: 0 1 10 11 100 ...

Hexadecimal: 0 1 2 3 4 5 6 7 8 9 A B C D E F ...

Similarly, I can define my own number system, for example say: $\alpha \,$ $\beta\,$ $\gamma\,$ $9\,$ $\alpha \alpha \,$ $\beta^{2} \alpha\,$...

But I am wondering, why do we use the decimal system to learn mathematics or say in our everyday lives. I know that a computer uses a binary number system since the most basic unit of storage is $0$ or $1$. However, I want to know, why humans use the decimal number system. Is there any logical advantage of using decimal representation over say Unary, Binary, or anything other? Thanks.

答:
We use the base $10$ because we have $10$ fingers, and we learn how to count with them.

Note that there were some people (https://en.wikipedia.org/wiki/Numeral_system) who used base $8$ to count, and it was due to the fact that they would count between knucles instead of counting on fingers

Title: Why we count from 0 to 9?


问:
Classifying the point $x=2$ for $f(x)=\cos(|x-2|)$
What is the point $x=2$ for this function?
$$f(x)=\cos(|x-2|)$$
I calculated
$$\lim_{x \to2^+} \frac{-\sin(|x-2|)|x-2|}{x-2}=0^-$$
$$\lim_{x \to2^-} \frac{-\sin(|x-2|)|x-2|}{x-2}=0^+$$
and it's a corner point.
If I calculate the limit of the incremental ratio
$$\lim_{h \to0^+} \frac{\cos(|h|)}{h}=+\infty$$
$$\lim_{h \to0^-} \frac{\cos(|h|)}{h}=-\infty$$
it's a cusp.
Can it just be a stationary point?

答:
Remember that
$\cos(x)=\cos(-x)$ so $\cos(|x|)=\cos(x)$.

Combine this with
$\cos(x-2)$ at $x=2$
being the same as
$\cos(x)$ at $x=0$
should give you your answer.


答:
When you say, "this integral represents volume under a curve and the area over which we are integrating is same in both cases," the integral you are speaking of is a double integral.  The two integrals you are computing are iterated integrals.  When the function is absolutely integrable, then the double integral is equal either one of the iterated integrals, and of course, both iterated integrals have the same value.  This is Fubini's theorem.

I haven't done the calculations here, but if what you say is true, it must be the case that the double integral is not absolutely integrable.  In fact, it must not exist.  What is the integral over the region $0\leq x \leq1, 0\leq y\leq x$?

Title: Double integral comes out different after changing $dx$ with $dy$.

Body:
$$\int_{0}^{1}\,dx \int_{0}^{1}\frac{x-y}{(x+y)^{3}}\,dy$$

My maths teacher gave this question to explain that if you change $dy$ with $dx$ the integral will have different value. My opinion is that the integral should come out as equal because this integral represents volume under a curve and the area over which we are integrating is same in both cases. Moreover this volume has to be $0$ because I plotted this curve in a 3d graph plotter and the surface is positive on one side of line $x=y$ and equally negative on the other side. BUT! after integrating the answer is coming as $1/2$ in one case and $-1/2$ in the other. Please explain.


问:
Dense implies strictly dense in TOP
A subspace $X$ of a convergence space $Y$ is strictly dense if for every $y \in Y$ and every filter $\mathcal F$ converging to $y$, there is a filter $\mathcal G$ containing $X$ that converges to $y$ such that $\bar{\mathcal G} \subseteq \mathcal F$. Here, $\bar{\mathcal G}$ is just the filter generated by the closure of all the elements of $\mathcal G$.

I want to prove that if $Y$ is a topological space and $X$ is dense is $Y$, then $X$ is strictly dense. This is my current argument: Suppose $\mathcal F$ converges to $y$. Then $\mathcal U_y \subseteq \mathcal F$, where $\mathcal U_y$ is the neighborhood filter of $y$. Since $X$ is dense in $Y$, it intersects every open set containing $y$, hence there is a filter $\mathcal G$ that contains $X$ and all the elements of $\mathcal U_y$. Since $\mathcal U_y \subseteq \mathcal G$, we have that $\mathcal G$ converges to $y$. Now somehow I need to show that $\bar{\mathcal G} \subseteq \mathcal F$, but I have no idea how to go about it. Any tips?

Addendum: The filter $\mathcal G$ in my argument above is the one generated by the sets of the form $U \cap X$, where $U$ is an open set containing $y$. The filter $\bar{\mathcal G}$ is generated by the closure of such sets.

答:
I agree with your choice of $\mathcal{G}$, as the filter generated by the sets $U \cap X$, where $U$ runs over the open neighbourhoods of $y$.

Now, for a dense subset $X$ of $Y$ and any open set $O$, $\overline{O \cap X} = \overline{O}$.

Proof: the left to right inclusion is obvious from $O \cap X \subset O$. To see the right to left inclusion, let $p$ be any point of $\overline{O}$ and let $U_p$ be any open neighbourhood of $p$. Then $O \cap U_p$ is non-empty (as $p \in \overline{O}$) and open, and so must intersect $X$, as $X$ is dense, and so $(O \cap U_p) \cap X = U_p \cap (O \cap X)$ is non-empty. Note that we have showed that every open neighbourhood $U_p$ intersects $O \cap X$, so $p \in \overline{O \cap X}$, which shows the reverse inclusion.

So if we have a generating set for $\overline{\mathcal{G}}$, which (you say) is of the form $\overline{U \cap X}$ for $U \in \mathcal{U}_y$, it is in fact also of the form $\overline{U}$, which is in $\mathcal{F}$ as an enlargement of $U \in \mathcal{F}$.

Title: Dense implies strictly dense in TOP


答:
An analogy might help.
$$
\sum_{n = 1}^\infty \frac{(-1)^n}{n} = -\log 2
$$
But by reordering the sum, we can get any real number we like. This is the Riemann series theorem.

A similar thing is happening with this integral. The moral is that order matters once infinity is involved. To restore our intuition about order, we have to make some additional hypotheses. In the case of series, we ask that they be absolutely convergent. In the case of integrals, we ask that they be absolutely integrable.


问:
Is every injective ring homomorphism an automorphism?
It seems a very interesting, innocuous and dubious result to me. Is the following true?

Consider the injective ring homomorphism $\phi :  R \rightarrow R$. Is it true that $\phi$ is an isomorphism? I am taking ring homomorphism to mean $f(ab)=f(a)f(b)$ and $f(a+b) = f(a)+f(b)$ with $f(1)=1$.

My attempt: $R \cong \phi (R) \subset R$. So, $\phi(R)=R$.

答: Your reasoning $R \cong \phi (R) \Rightarrow \phi(R)=R$ suggests me that you are misinterpreting cardinality with sets. For e.g. $tan^{}(\cdot) : (-\frac{\pi}{2}, \frac{\pi} {2} )  \rightarrow \mathbb{R} $ is a bijective function but the sets are not equal.


问:
Arithmetical Hierarchy
So i was reading the book "Turing Computability" of Soare.

I read about the Arithmetical Hierarchy.

There we it's defined that:

$$B \in \Sigma_n \iff  (\exists y)(\forall x_1)(\exists x_2) \dots (Qx_n)R(y,x_1,x_2 \dots x_n)$$ $$B \in \Pi_n \iff  (\forall y)(\exists x_1)(\forall x_2) \dots (Qx_n)R(y,x_1,x_2 \dots x_n) $$
where R is a computable prediacte.

So i found the folowing tasks: Show that:
$$Cof = \{x | W_x \text{ is cofinite}\} \in \Sigma_n$$
$$Cpl = \{x | W_x \equiv_T K \} \in \Pi_n$$
My attempt:
$$Cof = \{x | (\exists y)(\forall z)(\exists s){ (z \geq y \implies  z \in W_{x,s})}\}$$
$s$ is the number of steps. So this shows that cof is $\Sigma_3$. But I can't figure it out for CPL. Any tips?

答:
By definition of Turing equivalence, we have to study the complexity of saying $W_x\le_T K$ and $K\le_T W_x$. Since we know that $K$ is $\Sigma^0_1$-complete, it is enough to check whether $K\le_T W_x$ (the complexity would not change if you check both).

If we write $\varphi_e^A$ for the $e$-th computable function with oracle $A$ and identify a set with its characteristic function, then

$$K \le_T W_x \iff (\exists e)(\varphi_e^{W_x} = K)$$

The inner condition can be rewritten as

$$ (\forall n)((n\in K \rightarrow \varphi_e^{W_x}(n)=1) \land (n\notin K \rightarrow \varphi_e^{W_x}(n)=0))$$

The condition $\varphi_e^{W_x}(n)=k$ is $\Sigma^0_2$ (you need to ask that the machine converges to $k$, but you have a $\Sigma^0_1$ oracle), so the whole line is $\Pi^0_3$ and the condition $K \le_T W_x$ is $\Sigma^0_4$.

Of course every $\Sigma^0_n$ is a $\Pi^0_{n+1}$ set, so this shows that CPL is $\Pi^0_5$, but not sure why this is interesting. Maybe the idea was to write the condition as $(\forall y)(W_y \le_T W_x)$, but then again, this is more complicated than needed. The set is actually $\Sigma^0_4$-complete (check [1, Rem. 4.3.7]).

[1]: Soare, Robert I., Turing computability. Theory and applications, Theory and Applications of Computability. Berlin: Springer (ISBN 978-3-642-31932-7/hbk; 978-3-642-31933-4/ebook). xxxvi, 263 p. (2016). ZBL1350.03001.

Title: Arithmetical Hierarchy


问:
Find the area of the lateral surface defined by two sets
Let $S$ be the lateral surface of the set
$$
\left\{x^2 + y^2 +z^2 \leq 1, y \leq \frac{|x|}{\sqrt{3}}\right\} \bigcup
\left\{x^2 + y^2 +z^2 \leq 1, y \geq \frac{1}{2}\right\} = S_1 \cup S_2.\ \mbox{Evaluate} \left\vert\,{S}\,\right\vert.
$$

I think I need to define a curve and then evaluate the surface integral over this curve, but I don't know how to parametrize it.

In spherical coordinates, $x=r\sin\left(\phi\right)\cos\left(\theta\right), y = r\sin\left(\phi\right)\sin\theta, z = r\cos\left(\phi\right)$.
I came to the conclusion that

in $S_1$ whe have $0 \leq \phi \leq \pi,\ -7\pi/6 \leq \theta \leq \pi/6$

and in $S_2$ we have $0 \leq \phi \leq \pi/3,\ 0 \leq \theta \leq 2\pi$.

答:
First check if there is any common surface between $S_1$ and $S_2$ as $S = S_1 \cup S_2$.

Please note that if $y \geq \frac{1}{2}$, $|x| \geq y \sqrt3 \geq \frac{\sqrt3}{2}$ but $x^2+y^2+z^2 = 1$ and that shows $z = 0$ for $x = \frac{\sqrt3}{2}, y = \frac{1}{2}$. So we cannot simultaneously have $y \gt \frac{1}{2}, x \gt \frac{\sqrt2}{2}$.

Hence there is no common surface between $S_1$ and $S_2$. Once we find their surface area, we can simply add them.

You already found the right bounds. Saw your edit later.

To find $S_1$, please note that $y \leq \frac{|x|}{\sqrt3} \implies \sin\theta \leq \frac{|\cos\theta|}{\sqrt3}$. That leads to,

$0 \leq \theta \leq \frac{\pi}{6}, \frac{5\pi}{6} \leq \theta \leq 2\pi$. Bound of $\phi$ does not change: $0 \leq \phi \leq \pi$.

For $S_2$, my suggestion is to use $y-$axis for polar angle instead of $z$ given the plane is $y = \frac{1}{2}$. At $\rho = 1, \cos\phi = \frac{1}{2}, \phi = \frac{\pi}{3}$

So bounds are $0 \leq \phi \leq \frac{\pi}{3}, 0 \leq \theta \leq 2\pi$.

Edit:

As far as your question on integral, for a sphere it is simply $\iint \sin\phi \ d\phi \ d\theta$ as $\rho = 1$. You can find it as below,

for $S_1$,

$r(\theta, \phi) = (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi)$

$r'_{\theta} = (-\sin\theta \sin\phi, \cos\theta \sin\phi,0)$

$r'_{\phi} = (\cos\theta \cos\phi, \sin\theta \cos\phi,-\sin\phi)$

$|r'_{\theta} \times r'_{\phi}| = \sin\phi$

As you can see, it will be same for $S_2$.


问:
Euler Lagrange problem while keeping the integral of the function constant
I'm looking for a function $y(x)$ which minimizes following equation
$$
\int_0^L(y''(x)-f(x))^2dx
$$
under the constraints that
$$
\int_0^Ly(x)dx=A\\
y(x\in[0,L])>0 \\
y(0)=0\\
y(L)=0\\
$$
with $f(x)$ being known function that I will be plugging in later, however, I can't figure what equations I should be solving. If I understand correctly the Lagrange multipliers can't be used here, as my condition is for the value of the integral, and not for the value of the function for every given $x$.

Could someone point me in the right direction about what I should be reading about?

答:
You need to find the extremals of the functional
$$
\int\limits_0^L(y''(x)-f(x))^2+\lambda y(x))\,dx
$$
For a reference, you can check Gelfand-Fomin, section 12. (page 43 in my edition)

Observe that here $\lambda$ is a constant, not a function of $x$ as it is the case when you have finite (non-integral) subsidiary conditions.


问:
Birthday Paradox with Leap Year
I looked online, and found more than one and inconsistent answers to the Birthday Paradox when we throw the leap year into the mix. None of the answers I saw match with my own. I am posting my solution to see if it is correct, or if I am missing something.

Question:

Assume that the leap year occurs every four years. (i.e. ignore the 100 and 400 year rule). Also assume that the number of people born each day is the same. What is the probability that in a group of $n$ people (each one selected randomly), no two people share the same birthday?

My Solution:

Let $\mathcal{D}$ be the set of all possible dates in an year. (Thus $\mathcal{D}$ contains 366 elements. Note that these possibilities are not equally likely, since a person is four times as likely to be born on (say) Jan 1 than on Feb 29. This is true for any given day other than Feb 29, and it is encoded in the probability assignments given below.)

Now probability that a randomly selected person is born on Feb 29 is $\frac{1}{1 + 4*365} = \frac{0.25}{365.25}$.

Also, the probability that a randomly selected person is born on a given day other than Feb 29 is $\frac{1}{365.25}$.

Now, for a group of $n$ randomly selected people, the Sample Space of birthdays is $\mathcal{D}^n$. Let $\mathcal{A} \subset \mathcal{D}^n$ be the subset such that no two people share the same birthday.

Divide $\mathcal{A}$ into two disjoint sets $\mathcal{A}_1$ and $\mathcal{A}_2$ such that

$\mathcal{A}_1 = \{\xi: \xi \in \mathcal{D}^n \text{ and no two people have same birthday, and none is born on Feb 29} \}$, and

$\mathcal{A}_2 = \{\eta: \eta \in \mathcal{D}^n \text{ and no two people have same birthday, and exactly one is born on Feb 29} \}$

Now, $\mathbb{P}(\xi) = \frac{1}{(365.25)^n}$ for each $\xi \in \mathcal{A}_1$, and $\mathbb{P}(\eta) = \frac{0.25}{(365.25)^n}$ for each $\eta \in \mathcal{A}_2$.

Also, $|\mathcal{A}_1| = \; ^{365}P_{n}$, and $|\mathcal{A}_2| = \; n \; \cdot \; ^{365}P_{n-1}$

Finally, $\mathcal{A}_1$ and $\mathcal{A}_2$ being disjoint, it follows that

$$\mathbb{P}(\mathcal{A}) = \mathbb{P}(\mathcal{A}_1) + \mathbb{P}(\mathcal{A}_2) = \frac{^{365}P_{n}}{(365.25)^n} + \frac{0.25 \; \cdot \; n \; \cdot \; ^{365}P_{n-1}}{(365.25)^n}$$

PS:

As usual, if you want to find out the probability that at least two people share the same birthday then you would calculate $1 - \mathbb{P}(\mathcal{A})$. Interestingly, the number of people required so that this probability is more than 0.5 is still $n = 23$, same as the birthday paradox without leap year.

Please let me know if the solution above looks accurate.

答:
Answer below is mildly flawed, because I originally incorrectly computed $N_1$.  This caused me to doubt OP's answer.  With (my) flaw corrected, the OP's math looks good, and my only qualm of the OP's work is:

Having a sample space of 366 elements is not really a good approach, since the elements are not equally likely.

As indicated by the comment of J Moravitz, it is best to work with integers, rather than (for example) a denominator of $(365.25)^n.$

Another minor issue is that while the OP's math is correct, he approached it by examining the probability of various events occurring.  In problems like this, I think that it is better to approach it as

$$\frac{\text{Number of pertinent possibilities}}{\text{Number of total possibilities}}.$$

Anyway, given the original flaw in my thinking, I've left the flawed work intact, with the edit-patch pasted on as an example of my moderately going off the rails.

No, it's not correct.  In my opinion, the first comment of J Moravitz should be followed.  Therefore, your notion of a sample space of $\mathcal{D}^n$, where $\mathcal{D} = 366$ is not tenable.

Superficially examining your math, it looks like your math is (somehow) consistent with the idea of $\mathcal{D} = 365.25$  Therefore, although your math seems hard to follow, it might be correct; it's hard for me to tell.

I would have approached it as follows:

Probability of no two people out of $n$ sharing a birthday is

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$

where $D = (1461)^n.$

To calculate $N$ you must consider two possibilities : exactly one of the people is born on Feb 29, or none of the people is born on Feb 29.

Case-1

Without loss of generality, you can assume that Person-1 was born on Feb 29, and none of the other $(n-1)$ people were.

The number of ways that this can occur is therefore

$$N_1 = 1 \times (1460) \times (1456) \times \cdots \times (1468-4n).$$

Edit

Just realized that this is wrong.  Because the denominator is $(1461)^n$ the consistent enumeration of the numerator requires that order be regarded as important.  Therefore, I can not assume that Person-1 was the person born on Feb 29.  Therefore, the above computation of $N_1$ must be multiplied by the $\binom{n}{1}$ scalar.

The intuition is therefore: there are $\binom{n}{1}$ ways of choosing which person was born on Feb 29.  Having chosen this person, you can then line up the other $(n-1)$ people in a row, regarding them as Person-2, Person-3, ..., Person-n.

Then, there are $1460$ possible days to assign to Person-2.  Then, once this assignment is made, there are $1456$ possible days to assign to Person-3, and so forth.  This type of enumeration is consistent with the enumeration of $D = (1461)^n$, where (for example) Person-A on Jan 1 --- Person-B on Jan 2 is deemed distinct from Person-A on Jan 2 --- Person-B on Jan 1.  So the revised enumeration is :

$$N_1 = \binom{n}{1} \times (1460) \times (1456) \times \cdots \times (1468-4n).$$

Case-2

No one was born on Feb. 29.

The number of ways that this can occur is therefore

$$N_2 = (1460) \times (1456) \times (1452) \times \cdots \times (1464-4n).$$

Final Answer:

$$\frac{N_1 + N_2}{D}.$$

Title: Birthday Paradox with Leap Year


问:
Convergence of $\sum_{n=1}^{\infty} \left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n} \right)^{3n - \ln^2n}$
I need to examine convergence of that sum:
$$\sum_{n=1}^{\infty} \left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n} \right)^{3n - \ln^2n}$$

I can rewrite it as:
$$\sum_{n=1}^{\infty} \frac{\left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n}\right)^{3n}}{\left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n}\right)^{\ln^2n}}$$

But then, I can't even get:

$$\displaystyle \lim_{n \to \infty} \left(\frac{1+ 3\cos(n^2)}{7+2\sin^2n} \right)^{3n - \ln^2n} = 0$$

I know that is true but I don't know how to transform that expression to do something with trigonometric functions (how to merege them).

Additionally, I know that ratio test won't help me once I am done with limit. Same for root test. I think that it will be hard to use Abel's and Dirichlet's since trigonometric functions will equal to something periodic ( $\neq $ monotonic).

答: You could also notice that that $3n-\ln^2 n > n$ for large $n.$ For such $n$ the terms of the series are bounded above by $(4/7)^n.$ Since $\sum_{n=1}^\infty (4/7)^n<\infty,$ the original series converges by the comparison test.


问:
$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$ converges?
Does the series $$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$$ converges? I tried D' Alembert's quotient criterion but the $\displaystyle\lim_n \frac{a_{n+1}}{a_n}$ equals to $1$ and then by Cauchy's n-th root criterion $\displaystyle\lim_n \sqrt[n]{a_n}$ it gives an expession that is $\frac{\infty}{\infty}$ so I use the inequality $\sqrt[n]{n!}>2,\ \forall n\geq 4$ as a result $\displaystyle\lim_n \sqrt[n]{a_n}=+\infty$, and consequently the series diverges. However, I don't know if it is correct (if yes, then somebody may confirm my thoughts). Also, does anybody see another way to decide if the previous series converges or not? Thanks

答:
Set $a_n := \frac{(2n+1)!}{2^{2n}(n!)^2}$. I'm sure that you have done the following computation:

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(2n+3)! \cdot 2^{2n} \cdot (n!)^2}{(2n+1)! \cdot 2^{2n+2} \cdot ((n+1)!)^2}.$$

Now,

$$\frac{(2n+3)!}{(2n+1)!} = 2(n+1)(2n+3),$$
$$\frac{2^{2n}}{2^{2n+2}} = \frac14,$$
$$\frac{(n!)^2}{((n+1)!)^2} = \frac1{(n+1)^2},$$

so that the above expression simplifies to

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{2n+3}{2n+2} > 1.$$

This shows that $(a_n)$ is (strictly) increasing. Since $a_1 = \frac32$, this means that $\lim a_n \neq 0$. Thus, the series diverges.


答:
Hint:

The term $T_5$ is $$\frac{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11}{(2\cdot4\cdot6\cdot8\cdot10)^2}=\frac{3\cdot5\cdot7\cdot9\cdot11}{2\cdot4\cdot6\cdot8\cdot10}.$$ You see that every new factor is larger than $1$ and the general term does not decrease. (In fact, even the general term itself diverges.)

Title: $\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$ converges?

Body: Does the series $$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$$ converges? I tried D' Alembert's quotient criterion but the $\displaystyle\lim_n \frac{a_{n+1}}{a_n}$ equals to $1$ and then by Cauchy's n-th root criterion $\displaystyle\lim_n \sqrt[n]{a_n}$ it gives an expession that is $\frac{\infty}{\infty}$ so I use the inequality $\sqrt[n]{n!}>2,\ \forall n\geq 4$ as a result $\displaystyle\lim_n \sqrt[n]{a_n}=+\infty$, and consequently the series diverges. However, I don't know if it is correct (if yes, then somebody may confirm my thoughts). Also, does anybody see another way to decide if the previous series converges or not? Thanks


问:
Inverse Laplace of $p_1\cdot p_2/[(x+p_1)(x+p_2)]$
I can't find the inverse Laplace of :
$$\frac{p_1p_2}{(x+p_1)(x+p_2)}$$

I remove first $p_1,p_2$ and try to $$\mathscr{L}^{-1}\frac{1}{(x+p_1)(x+p_2)}$$
$$\frac{1}{(x+p_1)(x+p_2)}= \frac{A}{(x+p_1)} + \frac{B}{(x+p_2)}$$
$$1 = A(x+p_2) + B(x+p_1)$$

With $x = 0$

$$1 = Ap_2 + Bp_1$$

Then with $x = 1$ and $x = -1$, I get:
$$1 = A + B \\ A = 1-B \\ 1 = (1-B)p_2 + Bp_1\\
1 = p_2 -Bp_2 + Bp_1
\\ B = \frac{1-p_2}{p_1-p_2} \\
A =  1- \frac{1-p_2}{p_1-p_2} = \frac{p_1-1}{p_1-p_2}$$

Which would mean that $$\frac{p_1-1}{p_1-p_2} * \frac{p_1p_2}{x+p_1} + \frac{1-p_2}{p_1-p_2}\frac{p_1p_2}{x+p_2}$$
So I see that we should have $e^{-p_1x}$ and $e^{-p_2x}$ at some point but I don't know what to do with the factors  $\frac{p_1-1}{p_1-p_2}$ and  $\frac{1-p_2}{p_1-p_2}$. What is wrong and why?

Edit: I am adding a picture of the reply.

Body:
$$\frac{p_1 p_2}{(p_1+x) (p_2+x)}=\frac{A}{p_1+x}+\frac{B}{p_2+x}=\frac{Ap_2+Ax+Bp_1+Bx}{(p_1+x) (p_2+x)}$$

Therefore it must be
$$\frac{(Ap_2+Bp_1)+(A+B)x}{(p_1+x) (p_2+x)}=\frac{(p_1 p_2)+0\cdot x}{(p_1+x) (p_2+x)}$$
this means
$$
\begin{cases}
Ap_2+Bp_1=p_1p_2\\
A+B=0\\
\end{cases}
$$
$$
\begin{cases}
Ap_2-Ap_1=p_1p_2\to A(p_2-p_1)=p_1p_2\\
B=-A\\
\end{cases}
$$
$$
\begin{cases}
A=\frac{p_1p_2}{p_2-p_1}\\
B=-\frac{p_1p_2}{p_2-p_1}\\
\end{cases}
$$
therefore
$$\frac{p_1 p_2}{(p_1+x) (p_2+x)}=\frac{p_1p_2}{p_2-p_1}\left(\frac{1}{p_1+x}-\frac{1}{p_2+x}\right)$$
Inverse Laplace transform is
$$\frac{p_1p_2}{p_2-p_1} \left(e^{-p_1 t}-e^{-p_2 t}\right)$$


问:
Algebraic and definable closure of a vector space is a span of its subset
I am trying to show that for every infinite $K$-vector space $V$ (we use the first order language of vector spaces over $K$) and each $A \subseteq V$, $acl(A)$ is the $K$-linear subspace of $V$ spanned by $A$. The general notion I got is that it follows from strong minimality, yet I cannot see a connection between the two. How should I start thinking about it?

Is it also true for a definable closure?

EDIT [as suggested by @Berci in the comments]:

Definitions

The first-order language of vector spaces is: a constant $0$, a binary function "$+$", and a unary function symbol $F_a$ for each $a\in K$. Given a $K$-vector space $V$, $0$ is interpreted by the identity element of $V$, "$+$" is interpreted by the addition of $V$, and each $F_a$ is interpreted by the operation of scalar multiplication by $a$.

$acl(A)$ is the set of elements of $M$ that are algebraic over $A$ in an $L$-structure $\mathcal{M}$.

An element $a$ of $M$ is algebraic over $A$ in $\mathcal{M}$ if there is an $L$-formula $\phi(x,y_1,\ldots,y_n)$ and elements $e_1,\ldots , e_n$ of $A$ such that:

(i) $\mathcal{M}\models \phi [a,e_1,\ldots ,e_n]$

AND

(ii) $\{c\in M|\mathcal{M}\models[c, e_1, \ldots ,e_n]\}$ is finite.

答:
First, recall that the theory of vector spaces over an infinite field $K$ admits quantifier elimination.  That is, for all formulae $\psi$ there is a quantifier free formula $\psi'$ such that:
$$\psi \leftrightarrow \psi'.$$
Now, for any atomic formula $\varphi(x,y_1,y_2,...,y_n)$, the language can only express terms that are linear combinations, so either:

$\varphi(M) = \{\, m \in M \mid M \vDash \varphi(m,e_1,e_2,...,e_n)\,\}=M$ (occurs when $x$ appears with zero coefficient),

Or, $M \vDash \varphi(x,e_1,e_2,...,e_n)$ implies that $x$ is in the linear span of the $e_i$'s, and therefore unique.

If $v$ is a linear span of elements in $A$, say $v = \lambda_1 a_1 + ... \lambda_n a_n$, consider the atomic formula:
$$\varphi(x,y_1,...,y_n) \equiv x = F_{\lambda_1}y_1 + ... + F_{\lambda_n}y_n.$$
Clearly $M \vDash \varphi(v,a_1,...,a_n)$ and $\varphi(M) =\{\,v\,\}$.  So $v \in acl(A)$ and $v \in dcl(A)$.

Conversely, if $v \in acl(A)$, then there exists a formula $\psi$ such that:

$M \vDash \psi(v,a_1,...,a_n)$ ($a_i \in A$),

And $\psi(M)$ is finite.

By quantifier elimination, we can assume $\psi$ is quantifier free.  By the above remarks, we can thus deduce that $v$ also satisfies an atomic formula of type 2. and so $v$ is in the linear span of $A$.

The definable and algebraic closures coincide since $span(A) \subseteq dcl(A) \subseteq acl(A) = span(A)$.


问:
Inverse of fourth root function
I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$

but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!

答:
The equation is easier to work into the quadratic equation if we first get rid of radicals and avoid trying to "simplify" by using the $y^4/a^4$ rational. Expansion gives us the best view of a,b,c.

$$y=a\sqrt[^4]{b^2-(x-c)^2}\implies y^4=a^4\big(b^2-(c^2 - 2 c x + x^2)^2\big)\\
\implies 0=a^4 b^2 - a^4 c^2 + 2 a^4 c x - a^4 x^2-y^4\\
\implies      (a^4)x^2 - (2 a^4 c)x + (y^4 - a^4 b^2 + a^4 c^2) = 0\\
\implies x=  \frac{(2 a^4 c) \pm\sqrt{(2 a^4 c)^2-4(a^4)(y^4 - a^4 b^2 + a^4 c^2)}}{2a^4}\\
= \frac{(2 a^4 c) \pm\sqrt{4 a^8 b^2 - 4 a^4 y^4}}{2a^4} \\
\implies x = c \pm \frac{\sqrt{(a^2 b - y^2) (a^2 b + y^2)}}{a^2 }
\quad a,b,c,y\in\mathbb{R}\land a\ne 0\\
 x\in\mathbb{R}\iff y\le \sqrt{a^2b}$$

Title: Inverse of fourth root function


答:
Let $$f(x) := (1-4x^2)^{-1/2} = \sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}.$$
The radius of convergence is $\tfrac{1}{2}$, so for $x<\tfrac{1}{2}$ we have:
$$\frac{d}{dx}(xf(x)) = \sum_{n=0}^{\infty}(2n+1)\binom{2n}{n}x^{2n}$$
which we can find analytically:
$$(xf(x))' = \frac{1+4x^2}{(1-4x^2)^2}$$
So as $x \to 1/2$ from below, the power series can be made as large as we like.


问:
Solving complex linear congruences
Find $x \in \mathbb{Z}[i]$ such that:

$(1+2i)x \equiv 1 \mod 3+3i$

How would you go about doing this? Best I can think of is keep guessing....

答:
Hint: $\ $ mod $\, 3(1\!+\!i)\!:\,\  0\equiv 3(1\!+\!i)(1\!-\!i) \equiv 6 \,\Rightarrow\, \color{#c00}{5\equiv -1}\,\ $ so, rationalizing denominators

 $$\, x\equiv \dfrac{1}{1\!+\!2i}\equiv \dfrac{1\!-\!2i}{\color{#c00}5} \equiv -1\!+\!2i$$

Written explicitly $\quad \begin{eqnarray} (2i\!+\!1)(2i\!-\!1) &=\,& 1 + \color{#0a0}{3(i\!+\!1)}(i\!-\!1)\equiv 1\!\!\pmod{\color{#0a0}{3(i\!+\!1)}} \\  {\rm via}\quad {-}4-1\ &=\,& 1+3(-2)\end{eqnarray}$

Remark: The point of rationalizing the denominator is that it transforms the problem of dividing by an $ $ irrational $ $ number $\,(1\!+\!2i)\,$ to the simpler problem of dividing by a $ $ rational $ $ number $(\color{#c00}5).\, $ To invert $\,5\,$ mod $\,w = 3\!+\!3i,\,$ it suffices to invert it mod any integer multiple $\, n = wv,\,$ since

$${\rm mod}\ n\!:\ 5j\equiv 1\ \Rightarrow\ 5j = 1\! +\! nk = 1\! +\! wvk\ \Rightarrow\ {\rm mod}\ w\!:\ 5j\equiv 1\ \Rightarrow\, j \equiv 1/5$$

We chose the obvious integer multiple $\, n = 3(1\!+\!i)(1\!-\!i) = 6.\,$ Any integer mutiple coprime to $\,5\,$ will do the trick, e.g. we could also use the norm $\,n = ww' = (3\!+\!3i)(3\!-\!3i) = 18.$

Title: Solving complex linear congruences


问:
How to contextualize an inverse function for interpolation purposes?
I have a function that describes the growth of a population in a logistic term:
$$N_t = rN_0(1-\frac{N_0}{\kappa})$$
I would like to interpolate the values so that, given a known $y$ I can get the related $x$. To make it easier, I linearized the equation by taking log($N_t$). I removed the point that were not in a straight line and the obtained a linear model:
$$log(N_t) = y = st + i = mx + b$$
Using R, I can interpolate x given the model y~x , as in this picture:

But what I need to do is the opposite:

Since in R is not possible to get y from the model, and it is not correct to invert the model as x~y, I found the inverse function as:
$$log(y) = sx+i$$ (where $s$ and $i$ are the slope and intercept of the first model
$$log(x) = sy+i$$
$$y = \frac{log(x) -i}{s}$$
But now I am lost on what to do next. Shall I use the inversed function on the original x or y values? How do I convert these values into the original space's one? In other words, I get something like this:

How do I use the inverted function?
Manually, I have found that if $y=1000000$, then $x=16.6$. The parameters are: $s=0.022$, $i=0.328$. If I solve, I get:
$$log(x) = s \cdot 10^6 -i$$
$$x = 10^{s \cdot 10^6 -i}= ∞$$ OR
$$x = 10^{s \cdot 6 -i}= 0.6$$
My apologies but I am not a mathematician.

Thank you

答:
I think there's some confusion around the meaning of the first equation. Let me write it this way

$$
N_1 = r N_0\left(1 - \frac{N_0}{\kappa} \right) \tag{1}
$$

this means if you know the population at time $t=0$ you can calculate the population at $t=1$ and that's it. So the question is, how do you calculate the value of the population at $t=2$? something very similar to Eq. (1)

$$
N_2 = r N_1\left(1 - \frac{N_1}{\kappa} \right) \tag{2}
$$

where the value of $N_1$ is already known. The pattern is: if I want to know the value of the population at instant $t$ I need to know its value at $t-1$

$$
N_{t} = r N_{t-1}\left(1 - \frac{N_{t-1}}{\kappa} \right) ~~~t=1,2,\cdots\tag{3}
$$

This is a recursive expression, to get the value $N_t$, you need to calculate all the preceding values, namely $N_{t - 1}, N_{t-2}, \cdots, N_1, N_0$, Here's an example of for $r=3$ and $\kappa=1$

$$\begin{array}{cc} \hline
t & N_t \\ \hline
0 & 0.4 \\
1 & 0.792 \\
2 & 0.543628 \\
3 & 0.818718 \\
4 & 0.489781 \\
5 & 0.824655 \\
\vdots & \\
995 & 0.479427 \\
996 & 0.823603 \\
997 & 0.479427 \\
998 & 0.823603 \\
999 & 0.479427 \\ \hline
\end{array}
$$

or in graphical form

From this, you immediately can see that your original question may not be solved, e.g. what is the value of $t$ for which $N_t = 0.479427$?

Additional comments

Here's another example for a different value of $r$

this is a textbook example of chaotic behavior, so in that regard, your question is similar to ask "which will be the times at which the temperature outside is 27 C?"


问:
Interpretation of a discrete channel and $(M,n)$ codes
I'm having trouble understanding what the right way to think about the presentation of discrete channels and codes in text by Cover and Thomas.

A discrete channel is a system consisting of an alphabet $\mathcal{X}$ and output alphabet $\mathcal{Y}$ and a probability transition matrix $p(y|x)$ expressing the probability of observing output symbol $y$ given that we send symbol $x.$

How do we typically think of the $\mathcal{X}$ and $\mathcal{Y}$ alphabets here? Are we mapping individual symbols from $\mathcal{X}$ to individual symbols of $\mathcal{Y}$? In this case, both alphabets should have the same cardinality? I don't see how we could take $\mathcal{X} = \{a,b,\ldots, z\}$ and $\mathcal{Y} = \{0,1\}$, for example.

Next, after introducing channel extensions, they define:

An $(M,n)$ code for the channel $(\mathcal{X}, p(y|x), \mathcal{Y})$ consists of

An index set $\{1,\ldots, M\}$

Encoding function $X^n:\{1,\ldots, M\} \rightarrow \mathcal{X}^n,$ yielding codewords $x^n(1), x^n(2), \ldots, x^n(M).$ The set of codewords is called the code-book.

Decoding function $g:\mathcal{Y}^n \rightarrow \{1,\ldots, M\},$ deterministic rule assigning guess to each possible received vector.

So we are taking some arbitrary "message" from the index set, encoding it using $n$ symbols from $\mathcal{X}$, sending those $n$ symbols over the channel, receiving a noisy version of $n$ symbols in $\mathcal{Y}$, and then trying to recover the original message based on those $n$ symbols?

Why do the codewords all have to have length $n$? What's the advantage of letting $\mathcal{X}$ be different from $\mathcal{Y}$?

I think I understand the formal stuff going on here, but my intuition as to why it's interesting is a bit lacking. Can someone shed some light on this?

答: It's not strictly necessary to have $\mathcal{X}$ equal to $\mathcal{Y}$. They can also be different sets. The purpose of a channel based on $(M,n)$-codes is to map a word of symbols from an alphabet to another word of symbols of another alphabet of the same length $n$ (i.e. so blocks of symbols). However, in a channel code you can also have communication errors, that are modeled by a conditional probability table $p(y\mid x)$. The advantage of letting $\mathcal{X}$ be different from $\mathcal{Y}$ is the fact that the communication channel can be defined in order to switch the alphabet from $\mathcal{X}$ to $\mathcal{Y}$, but not necessarily.


问:
closed linear span of $\{ e_n : n \in \mathbb{N} \}$ in $l^{\infty}$
If $A=\{ e_n : n \in \mathbb{N} \}$ where $e_n$ is a sequence whose $n$th term is $1$ and $0$ otherwise.

Then I need to find the smallest closed linear subspace in $l^{\infty}$ that contains $A$.

I think $c_0$ is the answer.

My attempt:

I  proved that $c_0$ is a closed linear subspace that containing $A$ and this proves one containment, but I couldn't prove that $c_0 \subset \bigcap_{A\subset E_i}{E_i}$.

答:
If $x=(a_1,a_2,a_3,\ldots)\in c_0$, then let

$x_1=(a_1,0,0,0,0,\ldots)$;

$x_2=(a_1,a_2,0,0,0,\ldots)$;

$x_3=(a_1,a_2,a_3,0,0,\ldots)$

and so on. Then $x=\lim_{n\to\infty}x_n$ (and, in order to prove this, the fact that $x\in c_0$ is essential) and each $x_n$ belongs to$$\operatorname{span}\bigl(\{e_n\mid n\in\Bbb N\}\bigr).\tag1$$Therefore every closed subspace of $\ell^\infty$ that contains $(1)$ also contains $c_0$. So, $c_0$ is indeed the smallest closed subspace of $\ell^\infty$ which contains $(1)$.


问:
Applying Eisenstein's criterion to $x^3 + x^2 − 2x − 1$?
Is it possible to apply a shift (to the variable $x$) and Eisenstein's criterion to show that the polynomial $f(x) = x^3 + x^2 − 2x − 1$ is irreducible over the rationals?

答: You can shift by 9 to get $x^3 +28x^2+259x+791$ which is irreducible by looking at divisibility by 7.


问:
Double integral comes out different after changing $dx$ with $dy$.
$$\int_{0}^{1}\,dx \int_{0}^{1}\frac{x-y}{(x+y)^{3}}\,dy$$

My maths teacher gave this question to explain that if you change $dy$ with $dx$ the integral will have different value. My opinion is that the integral should come out as equal because this integral represents volume under a curve and the area over which we are integrating is same in both cases. Moreover this volume has to be $0$ because I plotted this curve in a 3d graph plotter and the surface is positive on one side of line $x=y$ and equally negative on the other side. BUT! after integrating the answer is coming as $1/2$ in one case and $-1/2$ in the other. Please explain.

答:
I'm afraid your teacher is right. If we integrate over $x$ first,$$\int_0^1[(x+y)^{-2}-2y(x+y)^{-3}]dx=[y(x+y)^{-2}-(x+y)^{-1}]_0^1=-(1+y)^{-2}$$implies the result is $[(1+y)^{-1}]_0^1=-\frac12$. But if we integrate over $y$ first,$$\int_0^1[2x(x+y)^{-3}-(x+y)^{-2}]dy=[(x+y)^{-1}-x(x+y)^{-2}]_0^1=(x+1)^{-2}$$implies the result is $[-(1+x)^{-1}]_0^1=\frac12$.

Let's consider the geometry of this. The surface $z=\frac{x-y}{(x+y)^3}$ has parts with positive and negative $z$ for $(x,\,y)\in[0,\,1]^2$, and your hope was to define a net (above minus below) volume within this region for $x,\,y$. But the result is undefined, because the part-volumes above and below $z=0$ are each infinite, and $\infty-\infty$ is an indeterminate form. Simpler examples of this phenomenon (by which I mean they don't involve double integrals, but Wikipedia offers something similar that does) include $\int_{-a}^b\tfrac1xdx$ with $a,\,b>0$. As @CharlesHudgins notes, infinite series do this too.


答:
I would use equivalents * with Stirling's' formula:
\begin{align}
\frac{(2n+1)!}{2^{2n}(n!)^2}&=\frac{\sqrt{2\pi(2n+1)}\Bigl(\cfrac{2n+1}{\mathrm e}\Big)^{\!2n+1}}{2^{2n}2\pi\mkern1.5mu n\Bigl(\cfrac{n}{\mathrm e}\Bigr)^{\!2n}} \\
&=\frac{(2n+1)^{\tfrac 32}}{\sqrt{2\pi}\,n}\,\biggl(\frac{2n+1}{2n}\biggr)^{\mkern-5mu2n}\sim_\infty 2\mkern1.5mu  \mathrm e\sqrt{\cfrac n\pi },
\end{align}
which is tends to $\infty$, so the series diverges trivially.


问:
Question about the decomposability of the radical.
Let $A$ be an algebra over an algebraically closed field $K$ and $M$ an indecomposable $A$-module. Suppose that $M$ is indecomposable. Can we conclude that $\operatorname{rad}M$ is indecomposable? Thank you very much.</p>

答:
As Aaron noted it is quite easy to construct very small counterexamples like the path algebra of the Kronecker quiver $1\stackrel{2}{\to} 2$, where the $2$ indicates that there are two arrows from $1$ to $2$. Or the path algebra of the quiver $A_3$ with non-linear orientation: $2\leftarrow 1\to 3$ or the algebra $k[x,y]/(x^2,y^2,xy)$. In each case the projective module $P_1$ provides a counterexample to your question.

One may ask what algebras satisfy that the radical of every indecomposable (left and right) module $M$ is again indecomposable. These are precisely the Nakayama algebras. An algebra is called a Nakayama algebra if every left and every right indecomposable module has a unique composition series (there are several equivalent characterisations).

If $A$ is not a Nakayama algebra. Then there is a vertex in the quiver of $A$ in which two arrows start (or end, this is left and right case). Then take the projective module associated to the corresponding vertex. Its quite obvious that its radical splits into as many direct summands as arrows start in the quiver.

Suppose $A$ is a Nakayama algebra. Suppose there is an indecomposable module such that the radical splits. Then one reaches a contradiction as one can construct two composition series, one having the top of each of the summands as a composition factor.


问:
How to find a real sequence $x_n$ such that $n x_n \to 0$ as $n \to \infty$ and $\sum_{n=1}^{\infty} x_n$ diverges?
I'm stuck on trying to find a real sequence $x_n$ such that $n x_n \to 0$ as $n \to \infty$ and $\sum_{n=1}^{\infty} x_n$ diverges. Here's what I have tried so far:

I've started with trying to find $x_n$ as a real-valued function of $n$. In order to have $n x_n \to 0$, we require $$\frac{x_n}{1/n} \to 0$$ and since the denominator tends to $0$ as $n$ increases, we require that $x_n \to 0$. We also require that $x_n$ tends to $0$ "faster" than $1/n$ does, in order for the above quotient to go to $0$.

However if you try functions such as $x_n=1/n^2$ or $x_n=n/2^n$ (for example) which tend to $0$ faster than $1/n$ does, then you see that $\sum_{n=1}^{\infty} x_n$ converges and I haven't been able to find a function for $x_n$ that satisfies the above conditions and also has a divergent infinite sum. It's certainly possible that the form for $x_n$ might not be a simple elementary function of $n$, but I don't know how you'd go about finding any "unusual constructions" that work.

答:
Here's one way that you might plausibly come to an answer like $x_n=\frac1{n\log n}$: We know that $\sum_{i\lt n}\frac1i\approx\log n$. If we want a divergent series, then the partial sums of our series will still need to go to infinity, but they'll need to do it at an even slower rate than $\log n$. We could try a couple of natural targets for our partial sums, say $\sqrt{\log n}$ or $\log\log n$.

But how do we get a term $x_n$ that gives us partial sums that grow this way? This is where the Euler-Maclaurin formula comes in handy: we know that we can approximate $\sum_{i=1}^n f(i)$ by $\int_1^n f(x)\ dx$, and vice versa. So we want a function $f()$ with $\int_{1}^n f(x)\ dx\approx\sqrt{\log n}$. How do we find one?

Well, we turn to another tool: the Fundamental Theorem of Calculus! Since the integral is the antiderivative, we can start with our target function $g(n)=\sqrt{\log n}$, say, and take the derivative $f(x)=\frac{dg}{dx}$ of this; by the chain rule we get $f(x)=\frac1{2x\sqrt{\log(x)}}$. The constant factor of 2 doesn't matter for convergence here, so we can multiply it away; we get that $x_n=\frac1{n\sqrt{\log n}}$ will yield a divergent series because $\sum_{i=1}^nx_i\approx C\sqrt{\log n}$ for some constant $C$. At the same time, by the way we constructed our sequence as diverging 'more slowly' to infinity than $\log n$ we have $\lim_{n\to\infty}nx_n\to 0$ (as you can see for yourself). Doing the same with $g(n)=\log\log n$ will give you the suggested $x_n=\frac1{n\log n}$ term.

You sometimes need to be a little careful going back and forth between sums and integrals like this, but for general questions of convergence and approximation it's one of the most useful tools in the toolbox.


问:
$(x,y)\mapsto f(x)g(y)$ is continuous
Let $X$ and $Y$ be topological spaces; $X\times Y$ carries the product topology.
Can someone provide me with an easy argument that for $f\in C(X,\mathbb{R})$ and $g\in  C(X,\mathbb{R})$
the mapping
$$
h\colon X\times Y \to \mathbb{R},\ (x,y)\mapsto f(x)g(y)
$$

is continuous?
I somehow struggle with this more than i should...
My first idea would be using something like nets:

We have $(x_i,y_i)\to(x,y) \text{ [in $X\times Y$]}\Leftrightarrow x_i \to x \text{[in $X$] and } y_i \to y\text{ [in $Y$]}$ for a net $(x_i,y_i)_{i\in I}$ in the product topology.
Therefore it's just
$$
\lim_{(x_i,y_i)\to (x,y)} f(x_i)g(y_i) = \lim_{\substack{x_i\to x\\ y_i \to y}} f(x_i)g(y_i) = f(x)g(y)
$$
but this feels too sloppy and not very enlightening.
Any comments or different approaches would be great.

答: The map$$\begin{array}{ccc}X\times Y&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&\bigl(f(x),g(y)\bigr)\end{array}$$is continuous, and so is the map$$\begin{array}{ccc}\Bbb R^2&\longrightarrow&\Bbb R\\(x,y)&\mapsto&xy.\end{array}$$So, you can express your function as the composition of two continuous functions, and that proves that it is continuous.

Title: $(x,y)\mapsto f(x)g(y)$ is continuous


问:
Linear algebra find $k$
Given the linear system:

$$\begin{cases}
 x_1 + kx_2 − x_3 = 2\\
2x_1 − x_2 + kx_3 = 5 \\
x_1 +10x_2 −6x_3 =1
\end{cases}$$

for which values of $k$ has the system (2):
(a) No solutions
(b) A unique solution.
(c) Infinitely many solutions.

I've been trying echelon form where i switched $R_1$ with $R_3$ and then i switched $R_2$ with $R_3$

So I have $\left[\begin{array}{ccc|c}1&10&-6&1\\1&k&-1&2\\2&-1&k&5\end{array}\right]$

but then I'm stuck and don't know how to get any further.

答:
Matrix of coefficients is
$$A=\left(
\begin{array}{ccc}
 1 & k & -1 \\
 2 & -1 & k \\
 1 & 10 & -6 \\
\end{array}
\right)$$
$\det A=k^2+2 k-15$

If $\det A\ne 0\to k\ne 3;\;k\ne-5$ the system has one and only one solution.

If $k=3$ then $\det A=0$ and $\text{rank}(A)=\text{rank}(A|B)=2$ thus there are infinite solutions

$$A|B=\left(
\begin{array}{ccc|c}
 1 & 3 & -1 & 2 \\
 2 & -1 & 3 & 5 \\
 1 & 10 & -6 & 1 \\
\end{array}
\right)
$$
If $k=-5$
$$A|B=\left(
\begin{array}{cccc}
 1 & -5 & -1 & 2 \\
 2 & -1 & -5 & 5 \\
 1 & 10 & -6 & 1 \\
\end{array}
\right)$$
$\text{rank}(A|B)=3$ while $\text{rank}(A)=2$. As ranks are different, the system has no solutions.

see here for Rouchè-Capelli theorem

Title: Linear algebra find $k$


问:
How does this alternate construction of the Sierpinski triangle differ from the usual one?
Can you plot the Sierpinski triangle on the Cartesian plane and define the "removal of triangles" in terms of the points being removed? Imagine an equilateral triangle with vertices at
$(-1, 0), (1, 0), (0, \sqrt{3}).$

So for iteration 1, we remove a triangle with vertices at
$(-\frac{\sqrt{3}}{2}, \frac{1}{2}), (\frac{\sqrt{3}}{2}, \frac{1}{2}), (0, 0)$.

This makes sense to me geometrically, but I got to wondering, do these three vertices (and the points on the lines between them) also get removed, or can they stay behind? More casually, can you talk about recursively removing just "open" or just "closed" triangles? It seems each approach would leave behind two very different kinds of fractals.

答:
A Few Common Constructions

There are actually many different ways to construct the Sierpinski gasket, all of which produce the same object.  I will (briefly) go over three of them here.  I will note that in each of the constructions, I start with a triangle with side length $1$, in contrast to the original question, which starts with a larger triangle.  The distinction is mostly irrelevant.

Triangle removal

Let $T_0$ be an equilateral triangle with side length $1$, and let $T_n$ denote the $n$-th iterate in the construction of the gasket—$T_n$ consists of $3^n$ congruent equilateral triangles (each one of which has side length $2^{-n}$).  To obtain $T_{n+1}$ from $T_n$, remove the interior of an equilateral triangle with side length $2^{-(n+1)}$ from each of the triangles which forms $T_n$—this divides each of the triangles of side length $2^{-n}$ in $T_n$ into three triangles with side length $2^{-(n+1)}$.  The Sierpinski gasket is the limiting object of this construction—the precise way in which this limit is taken is actually kind of important, but I am going to elide that here.

Iterated function system

Define the three maps
$$ \varphi_1(x) = \frac{1}{2} x + \pmatrix{0 \\ 0}, \quad
\varphi_2(x) = \frac{1}{2} x + \pmatrix{1/2 \\ 0}, \quad\text{and}\quad
\varphi_3(x) = \frac{1}{2} x + \pmatrix{1/4 \\ \sqrt{3}/4 }, $$
where $x \in \mathbb{R}^2$.  This collection of maps is called an iterated function system.  Further define the map of sets
$$ \Phi(E) = \bigcup_{j=1}^{3} \varphi_j(E). $$
It follows from some abstract nonsense (essentially, the Banach fixed point theorem) that there is a unique, nonempty, compact set $A$ such that
$$ \Phi(A) = A. $$
This set, called the attractor of the iterated function system, is the Sierpinski gasket.

Graph construction

Let $G_0$ be the graph with vertices at the points
$$
(0,0), \quad
(1,0), \quad\text{and}\quad
(1/2, \sqrt{3}{2}),
$$
in the Cartesian plane, and undirected edges between each pair of vertices.  To construct $G_1$, add a vertex at the midpoint of each edge (dividing each edge into two edges), and join each pair of of new vertices with another edge.  To obtain $G_2$, again divide each edge into two edges at their midpoints, and add edges in the "obvious" manner.  Continue this process indefinitely ($G_4$ is shown below)—the Seirpinski gasket is the limiting object.

Note that I have been a little cagey about how to produce $G_{n+1}$ from $G_n$, and about the manner in which the limit is taken.  All of this can be made precise, but doing so would greatly increase the length of this answer.

Properties of the Sierpinski Gasket

Different properties of the Sierpinski gasket are more easily seen from the different constructions, but, in some sense, all of these constructions give the same object (with, perhaps, some additional different structures from one construction to another).  For example, the graph construction makes it reasonably straight-forward to show that the gasket is not just connected, but path connected.  Some of the important properties are

The Sierpinski gasket is compact in $\mathbb{R}^2$.  This can be seen either from the triangle removal process (each iterate is closed—we are always removing open sets—and the set is bounded), or from the iterated function system construction (via abstract nonsense).

The Sierpinski gasket is complete as a metric space (with the metric inherited from $\mathbb{R}^2$).  This follows from the iterated function system construction via more abstract nonsense.

The Sierpinski gasket is path connected.  This follows from the graph construction.

The Sierpinski gasket has Hausdorff dimension $\log(4)/\log(3)$, and has positive Hausdorff measure in this dimension.  Both claims can be proved in a straight-forward manner from the iterated function system construction.

There is a natural harmonic measure supported on the Sierpinski gasket.  This is important when thinking about performing analysis on the gasket, e.g. defining and solving differential equations on the gasket.  There is a lot of interesting theory here—both Jun Kigami and Robert Strichartz, along with numerous collaborators and advisees, have written extensively on this topic.

It is worth remarking, I think, that the Sierpinski gasket is maybe not all that interesting in and of itself.  It is commonly studied because it is a very simply example of a large class of objects which are of some general interest.  For example, the gasket is a very-not manifold object on which we can nevertheless describe some kind of smooth or differentiable structure.  By understanding how this structure works on the gasket, we hope to understand how analogous structures work in other similar spaces.

The "bvy Gasket"

In the question, it is suggested that the triangle removal procedure be modified to remove a closed, rather than open, triangle at each iteration.  I will refer to the object constructed in this manner as the "bvy gasket", after the author of the question.

For this construction, we must immediately throw out many alternative constructions, including the iterated function system and graph constructions—there just isn't a reasonable way to realize the bvy gasket as the attractor of an iterated function system, or as some kind of limit graph.

Considering the properties described above:

The bvy gasket is not compact in $\mathbb{R}^2$.  Indeed, it does not seem to have any nice topological properties—it is not closed, as numerous limit points have been removed at each iteration; and it is not open, as the limiting object contains no open balls (let alone an open ball around each point).

The bvy gasket is not complete as a metric space.  For example, we can construct a Cauchy sequence of points in the bvy gasket which converges to one of the removed vertices.  Such a vertex isn't in the bvy gasket, so Cauchy sequences don't necessarily converge in the space.  This makes any kind of analysis quite difficult, as we can't meaningfully take limits in many circumstances.

The bvy gasket is neither path connected nor connected.  Indeed, it appears to me that this gasket is totally disconnected.  Any two points in the bvy gasket must be contained in the interiors of triangles which are "left behind" at every stage of the construction.  These triangles are open in the subspace topology, and any two points will eventually be in distinct triangles, providing us with a disconnection.  Hence the only connected components are singleton points.

The bvy gasket has Hausdorff dimension $\log(4)/\log(3)$, and has positive Hausdorff measure in that dimension.  To show this, we just take advantage of what we know about the Sierpinski gasket.  At each stage of the process, we are removing a closed triangle, which differs from an open triangle by its boundary.  The boundary is the union of three segments, and segments have $0$ $s$-dimensional Hausdorff measure for any $s > 1$.  Hence the difference between an iterate of the Sierpinski gasket and an iterate of the bvy gasket is a null set, from which we can show that the $\log(4)/\log(3)$-dimensional Hausdorff measure of the bvy gasket is positive.

As the bvy gasket is totally disconnected, there aren't paths to speak of in the space.  This immediately rules out the possibility of building harmonic measures in "the usual manner" on this space.  It is not at all clear to me that there is any kind of "natural" notion of a harmonic measure on this space.  I am not sure that I see any kind of meaningful notion of analysis or differential structure here.


问:
How to draw samples of two independent iid random variables from real and psudo random number generators?
There are 2 independent and identically distributed random variables $X_1$ and $X_2$. I need 1000 drawings of each one of these variables (in a monte-carlo simulation setting). Given this problem, i am confused if below are equivalent -

(a) draw 1000 samples for $X_1$ and then draw 1000 samples for $X_2$

(b) draw 2000 samples and then assign half to each one of them

Is (a) same as (b) in an ideal scenario when random numbers are truly random?

What happens when, in practice, we use numbers that are psudo random? should I use a different seed in random number generator when following method (a) [i will call this method (c)]. for example, set a random seed -> generate 1000 samples for $X_1$ and then set another seed and then generate another 1000 samples for $X_2$.? How will that compare when using method (a) when i dont change random seed or with method (b) where i generate 2000 numbers and just assign them equally to $X_1$ and $X_2$?

I am adding here work some work that I did using python numpy.
First i generated, random numbers using np.random

```<code>import numpy as np

sample_size = 100000

# method a - drawing samples for x1 and x2
np.random.seed(0)
x1_a = np.random.rand(sample_size)
x2_a = np.random.rand(sample_size)

# method b - drawing 2 * samples and then allocating to x1 and x2
np.random.seed(0)
x_b = np.random.rand(sample_size * 2)
x1_b = x_b[0:sample_size]
x2_b = x_b[sample_size:2 * sample_size]

# method c - drawing samples for x1 and x2 but with different random seeds
np.random.seed(0)
x1_c = np.random.rand(sample_size)
np.random.seed(120)
x2_c = np.random.rand(sample_size)

# now calculating correlations between x_1 and x_2
corr_a = np.corrcoef(x1_a, x2_a)
corr_b = np.corrcoef(x1_b, x2_b)
corr_c = np.corrcoef(x1_c, x2_c)
print(f" correlations are: {corr_a[0][1]}, {corr_b[0][1]}, {corr_c[0][1]}")
</code>```

this returns following

```<code> correlations are: 0.004584252381322772, 0.004584252381322772, -0.0012275661688751346
</code>```

Results make me wonder if :

is coefficient of .005 as witnessed in method a and b good
enough to assume that $X_1$ and $X_2$ are independent with zero
correlation

method c seems to be giving lower correlation. Is method c better in
that case?

答:
Comment:

For distribution $\mathsf{Binom}(10, .4).$ Computation in R. By using the same seed for the pseudorandom generator in R, you will get exactly the same results I did.  In R, rbinom generates random variables of the indicated number and with the indicated parameters.

Method 1:

```<code>set.seed(129)
x1 = rbinom(1000, 10, .4)
summary(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   3.000   4.000   3.941   5.000  10.000
x2 = rbinom(1000, 10, .4)
summary(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   0.00    3.00    4.00    4.04    5.00    9.00
</code>```

Method 2:

```<code>set.seed(129)
x = rbinom(2000, 10, .4)
x1 = x[1:1000]
summary(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   3.000   4.000   3.941   5.000  10.000
x2 = x[1001:2000]
summary(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   0.00    3.00    4.00    4.04    5.00    9.00
</code>```

Same results either way.

Body:
Comment:

For distribution $\mathsf{Binom}(10, .4).$ Computation in R. By using the same seed for the pseudorandom generator in R, you will get exactly the same results I did.  In R, rbinom generates random variables of the indicated number and with the indicated parameters.

Method 1:

set.seed(129)
x1 = rbinom(1000, 10, .4)
summary(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   3.000   4.000   3.941   5.000  10.000
x2 = rbinom(1000, 10, .4)
summary(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   0.00    3.00    4.00    4.04    5.00    9.00

Method 2:

set.seed(129)
x = rbinom(2000, 10, .4)
x1 = x[1:1000]
summary(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   3.000   4.000   3.941   5.000  10.000
x2 = x[1001:2000]
summary(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   0.00    3.00    4.00    4.04    5.00    9.00

Same results either way.


问:
Analytic expression for $\mathbb E_X[(X^\top u)^p (X^\top v)^p]$, where $u$ and $v$ are fixed vectors in $\mathbb R^d$ and $X$ is uniform on sphere
Let $u$ and $v$ be fixed vectors in $\mathbb R^d$. Let $X$ be a random vector uniformly distributed on the unit-sphere in $\mathbb R^d$, and let $f_p$ be a real polynomial of degree $p \ge 1$ (for simplicity, we might simply consider $f_p(t) \equiv t^p$).

Question. What is an analytic expression for the correlatetion $c(u,v) := \mathbb E_X[f(X^\top u)f(X^\top v)]$ ?

Special case $p=1$

In this case, a simple computation gives

$$
c(u,v) = \mathbb E[X^\top uv^\top X] = \mbox{trace}(\mbox{cov}(X)uv^\top) = \mbox{trace}((1/d) I_d uv^\top) = \frac{u^\top v}{d}
$$

Is it too crazy to conjecture that in general, $c(u,v) = K_{d,p} \cdot (u^\top v)^p$, for some constant $K_{d,p}$ which only depends on $d$ and $p$ ?

答:
Here's a strategy for computing $c_{p,q}(u,v)=\mathbb{E}[(X^T u)^p(X^T v)^q]$. First compute the generating function $G(\lambda)=\mathbb{E}(e^{i\lambda^TX})$, then set $\lambda=au+bv$. The result is that we can express the quantities above as derivatives of the generating function:

$$c_{p,q}(u,v)=(-i)^{p+q}\frac{\partial^{p+q}}{\partial a^p \partial b^q}G(au+bv)\Bigg|_{a=b=0}$$

Calculating the generating function is fairly simple for $X$ uniformly distributed on the sphere by performing a rotation so that the vector $\lambda$ is in the $x_1$ direction:

$$G(\lambda)=\int e^{i\lambda\cdot x}\frac{d\Omega_{d-1}}{\mu(S_{d-1})}=\frac{1}{\mu(S_{d-1})}\int_{R^{d}}d^dx \delta(|x|-1)e^{i\lambda\cdot x}=\frac{1}{\mu(S_{d-1})}\int_{R^{d}}d^dx \delta(|x|-1)e^{i|\lambda| x_1}$$

Reverting back to spherical coordinates
$$G(\lambda)=\int e^{i|\lambda|\cos\theta}\frac{d\Omega_{d-1}}{\mu(S_{d-1})}=\frac{\mu(S_{d-2})}{\mu(S_{d-1})}\int_{0}^{\pi}\sin^{d-2}\theta~ e^{i|\lambda|\cos\theta}d\theta$$
We recognize the last integral as the Poisson integral for Bessel functions, and hence $G(\lambda)$ can be succinctly written in closed form

$$G(\lambda)=2^{d-2/2}\Gamma\left(\frac{d}{2}\right)\frac{J_{\frac{d-2}{2}}(|\lambda|)}{|\lambda|^{\frac{d-2}{2}}}$$

Using the series representation of the Bessel function one can show that

$$G(au+bv)=\sum_{n=0}^{\infty}\frac{(-1)^n \Gamma(d/2)}{n!\Gamma(n+d/2)2^{2n}}(a^2|u|^2+b^2|v|^2+2ab (u\cdot v))^n$$
We see immediately that the result is non-zero only when $p+q$ is even. It is also obvious that the conjecture cannot hold since

$$c_{2,2}(u,v)\propto 4(u\cdot v)^2+2 |u|^2 |v|^2$$

I have not found a satisfactory algebraic formula for arbitrary $p,q$, to be continued!

EDIT: I came up with a good way to expand this sum.

Obviously when one choose particular values $p+q=2m$, only the term with $n=m$ in the sum will contribute. All we need to do is expand the multivariable polynomial in parentheses above for any value of $n$. The trinomial theorem is not very illuminating so we pick a different route. First rewrite with $z=b/a$

$$(a^2u^2+b^2v^2+2ab(u\cdot v))^m=a^{2m}|v|^{2m}(z+\lambda e^{i\theta})^m(z+\lambda e^{-i\theta})^m$$

where $\lambda=|u|/|v|$, and, $|u||v|\cos\theta=u\cdot v$, $|u||v|\sin\theta=\sqrt{u^2 v^2-(u\cdot v)^2}$

Now we can use the binomial expansion on each of the individual terms

$$(a^2u^2+b^2v^2+2ab(u\cdot v))^m=a^{2m}|u|^{2m}\sum_{kl}{m\choose k}{m\choose l}\left(\frac{z}{\lambda}\right)^{k+l}(e^{i\theta})^{l-k}$$

All we need to do now is reindex the sum so that everything is written as a coefficient of $z^s$:

$$(a^2u^2+b^2v^2+2ab(u\cdot v))^m=a^{2m}|u|^{2m}\sum_{s=0}^{2m}z^s(\lambda e^{i\theta})^{-s}\sum_{l=\max(s-m,0)}^{\min(s,m)}{m\choose s-l}{m\choose l}e^{i2l\theta}$$

One can explicitly show that this expression is real. We take the real part of the right hand side and since $\cos(n\theta)=T_n(\cos\theta)$ are the Chebyshev polynomials

$$(a^2u^2+b^2v^2+2ab(u\cdot v))^m=a^{2m}|u|^{2m}\sum_{s=0}^{2m}z^s\lambda^{-s}\sum_{l=\max(s-m,0)}^{\min(s,m)}{m\choose s-l}{m\choose l}T_{2l-s}\left(\frac{u\cdot v }{|u||v|}\right)$$

with the understanding that $T_{-n}(x)=T_n(x), n>0$. Finally, we conclude that

$$c_{2m-p,p}(u,v)=\frac{\Gamma(d/2)\Gamma(p+1)\Gamma(2m-p+1)}{2^{2m}\Gamma(m+d/2)\Gamma(m+1)}|u|^{2m-p}|v|^p\sum_{l=\max(p-m,0)}^{\min(p,m)}{m\choose p-l}{m\choose l}T_{2l-p}\left(\frac{u\cdot v }{|u||v|}\right)$$

I haven't attempted to try and simplify the Chebyshev polynomials much further. Further simplification occurs when $m=p$ but not to the point of a compact expression better than

$$c_{p,p}=\frac{\Gamma(d/2)\Gamma(p+1)}{2^{2p}\Gamma(p+d/2)}|u|^p |v|^p\sum_{l=0}^{p}{p \choose l}^2T_{2l-p}\left(\frac{u\cdot v }{|u||v|}\right)$$


问:
Proof that the set of all bijective functions over the natural numbers is uncountable.

Problem: Prove that the set of all bijective functions over the natural numbers is uncountable.

I have already seen a lot of answers to this problem and there seems to be always a part where the natural numbers are partitioned into two countable infinite subsets of $\mathbb{N}$.
Am I missing something? I have come up with the following proof and I am suspicious that there might be a mistake somewhere.

I use the fact that there exists a bijection between the set of all bijective functions on $\mathbb{N}$ and the set of all permutations of $\mathbb{N}$. That part should not be the problem as far as I understand.

I use Cantor's diagonalization argument in the following way:
Let $\pi_{i}$ be the i-th permutation in a list of all permutations on $\mathbb{N}$. The ability to list all permutations would give us a enumeration making the set countable, we therefore assume that such list exist and make a proof by contradiction:

Let the list of all permutations be $(\pi_{1},\pi_{2},\pi_{3},\pi_{4},...)$, we now define a permutation $\pi$ in the following way:

$\pi(0) \in $ $\mathbb{N} \backslash \{\pi_{1}(0)\} $

$\pi(1) \in $ $\mathbb{N} \backslash \{\pi(0), \pi_{2}(1)\} $

$\pi(2) \in $ $\mathbb{N} \backslash \{\pi(0), \pi(1), \pi_{3}(2)\} $

.
.
.

$\pi(k) \in $ $\mathbb{N} \backslash \{\pi(0), \pi(1), ...,\pi(k - 1), \pi_{k+1}(k)\} $

I would then conclude the proof by stating that above permutation is not in the list, which is a contradiction to our assumption and therefore the set of all permutations cannot be countable, therefore being uncountable. Is this correct? I appreciate the help.

答:
You have to assure that you can define $\pi$ in such a way that it is a permutation. Here is one way to do that. Let a list of all permutations be $(\pi_{1},\pi_{2},\pi_{3},\pi_{4},\dots)$. $\pi$ will also be defined recursively.

For every $n$, define $\pi(2n)$ to be the least natural number in $\mathbb{N} \backslash \{\pi(0), \pi(1), ...,\pi(2n - 1), \pi_n(2n)\}$. And define $\pi(2n+1)$ to be the least natural number in $\mathbb{N} \backslash \{\pi(0), \pi(1), ...,\pi(2n - 1), \pi(2n)\}$. $\pi$ will be injective, and the definition in the odd numbers forces it to be surjective. It is also different from all the $\pi_n$.


问:
Unary predicates theory with no principal types
Definition

Take a satisfiable set of $L$-sentences $\Sigma$ and variables $x = x_1, \ldots, x_n$. Denote by $S_x(\Sigma)$ the set of all $\Sigma$-realizable $x$-types in $L$.

A type  $p(x) \in S_x(\Sigma)$ is principal if it contains a $\Sigma$-complete formula. Equivalently, the singleton $\{p(x)\}$
is an open set in $S_x(\Sigma)$, or principal $x$-types are exactly the isolated
points of $S_x(\Sigma)$.

Problem

I want to find a theory with no principal types realizable in $T$. I was suggested theory of infinitely many independent unary predicates is such a theory.

Attempt

I realize I'd have to prove that no formula in $\mathcal{L}$ is $T$-complete. To that end, I think I would need to show that if and only if $\phi(x)$ is a single-variable formula not containing a unary predicate symbol $U_n$ then both $\phi(x) \wedge U_n(x)$ and $\phi(x) ∧ \neg U_n(x)$ are realized in $T$. It seems intuitive, yet I do not know how to prove it.

How could I go about proving it?

答: You have the right idea, so I will just work out your argument. Basically $T$ just says nothing. So any $\mathcal{L}$-structure is a model. So suppose for a contradiction that a type $p(x)$ is isolated by $\phi(x)$ and let $a \in M$ realise $p(x)$. Let $U_n$ be a predicate that does not occur in $\phi(x)$. Now let $\mathcal{L}'$ be the language $\mathcal{L}$ without the predicate $U_n$. Let $M'$ be the $\mathcal{L}'$-reduct of $M$. Note that still $M' \models \phi(a)$. We define an $\mathcal{L}$-structure $M_1$ by taking $M'$ and interpreting $U_n$ as just $\{a\}$. We define another $\mathcal{L}$-structure $M_2$ by taking $M'$ and interpreting $U_n$ as the empty set. Now $M_1 \models \phi(a) \wedge U_n(a)$ and $M_2 \models \phi(a) \wedge \neg U_n(a)$, contradicting that $\phi(a)$ isolates $p(x)$.


问:
Unexplainable Divergent area?
Somebody came up to me recently with the following problem. Consider $y = \dfrac{1}{x^2}$ with $x>0$. Now there is this square $S$ sitting under the curve, connecting $(0,0) , (1,0) , (1,1)$ and $(0,1)$. The area under the curve from $x = 1$ to infinity is equal to $1$ through an improper integral. This area sits on the right of $S$. Due to symmetry, the same area can be found on top of $S$. This area is then between the $y$-axis and the curve. The area of $S$ is finite ($1$ of course) and so the area of $S$ plus the area on top of it should be $2$. But evaluating the integral from $0$ to $1$ is divergent. How can this be explained?</p>

答: There is no such symmetry.  The inverse function would be in this range $x=1/\sqrt{y}$, which when integrated over $[1,\infty)$, gives a divergent result.</p>

Title: Unexplainable Divergent area?


问:
proof verification $\left \| f-p \right \|_{\infty }<\varepsilon $ and $\left \| {f}'-{p}' \right \|_{\infty }<\varepsilon $
$f:[0,1]\rightarrow \mathbb{R}$ continuous and differentiable, show that for every $\varepsilon >0 $ it exists a polynomial $p$ such that  $\left \| f-p \right \|_{\infty }<\varepsilon $ and  $\left \| {f}'-{p}' \right \|_{\infty }<\varepsilon $

proof:

since $f:[0,1]\rightarrow \mathbb{R}$ continuous from the Stone-weierstrass theorem we know it exist a polynomial sequence  $p_n$ such that $p_n \xrightarrow[]{uni}f$ so it exist a $n_0 \in\mathbb{N}$  $\left \| f-p_{n_0} \right \|_{\infty }<\varepsilon $ consider  $p_{n_0}=p \Rightarrow $ we have  $\left \| f-p \right \|_{\infty }<\varepsilon/2 (1)$ so if $x_0 \in[0,1]$ such that $\left \| p(x_0)-f(x_0) \right \|_{\infty }<\varepsilon/2 (2)$

$(1)+(2)$   $\left \| f-p \right \|_{\infty }+ \left \| p(x_0)-f(x_0) \right \|_{\infty }<\varepsilon   $

$\left \| (f(x)-f(x_0))-(p(x)-p(x_0))  \right \|\leq $  $\left \| f-p \right \|_{\infty }+ \left \| p(x_0)-f(x_0) \right \|_{\infty }<\varepsilon   $

then if  $\frac{1}{x-x_0} \leq1 \Rightarrow $ $\left \| \frac{(f(x)-f(x_0))}{x-x_0}-\frac{(p(x)-p(x_0))}{x-x_0}  \right \|=\left \| {f}'-{p}' \right \|_{\infty }< \varepsilon$

答:
Since ${f}'$ is continuous (it is i checked again the original problem) frome S-W theorem we have $q_n\xrightarrow[]{uni} {f}'\Rightarrow  \left \|q_n- {f}'  \right \|< \varepsilon
 $ lets call $q_n={p_n}'$ from before we saw that

$\left | {f}'-{p}' \right |< \varepsilon \Rightarrow  \int_{0}^{1}\left | {f}'-{p}' \right |dx<\varepsilon \Rightarrow \left | \int_{0}^{1}{f}'-{p}'dx \right |\leq \int_{0}^{1}\left | {f}'-{p}' \right |dx< \varepsilon$

$\left | \int_{0}^{1}{f}'-{p}'dx \right |=\left | f-p \right |$


问:
Is there a chain of finitely generated modules $M_i$ not satisfying ascending chain condition s.t. $\cup_i M_i$ is finitely generated?
It is easy to find such a chain s.t. $\cup_i M_i$ is not finitely generated. So I am curious about the other situation. Can $\cup_i M_i$ be finitely generated?

Thanks in advance.

答:
For the record: you are putting more information in the subject line than in the body. It should be the other way around! The body should be self-contained and include all relevant information, while the subject line gives a rough idea of what the question is about.

First, say $\{M_i\}_{i\in\omega}$  is an ascending chain of modules, and let $M=\cup_{i\in\omega}$ be their union, which is also a module. The chain "stabilizes" if and only if there exists $N\in\omega$ such that for all $m\geq N$, $M_m=M_N$.

First, let us put no conditions on any of the $M_i$. We have the following three examples:

The chain stabilizes and $M$ is finitely generated. Just take your favorite finitely generated module (say, a cyclic one), and let $M_i=M$ for all $i\in\omega$.

The chain stabilizes and $M$ is not finitely generated. Take your favorite non-finitely generated module (e.g., the free module on the set $\{x_i\}_{i\in\omega}$), and let $M_i=M$ for all $i\in\omega$.

The chain does not stabilize and $M$ is not finitely generated. Take your favorite non-finitely generated but countably generated module $M$ (e.g., the one mentioned above), let $X=\{m_i\}_{i\in\omega}$ be a generating set for $M$, and let $M_i=\langle x_0,\ldots,x_i\rangle$ for each $i\in\omega$. The chain never stabilizes, and the union is $M$, which is not finitely generated.

However,

Proposition 1. If $M$ is finitely generated, then the chain stabilizes.

Proof. Let $m_1,\ldots,m_k$ be a generating set for $M$. For each $j$, $1\leq j\leq k$, there exists $i_j\in\omega$ such that $m_j\in M_{i_j}$. Let $N=\max\{i_1,\ldots,i_k\}$. Then $m_j\in M_{i_j}\leq M_N$ for all $j$, hence for all $n\geq N$,
$$M = \langle m_1,\ldots,m_k\rangle \leq M_N\leq M_n \leq M,$$
so we have equality throughout and in particular the chain stabilizes. $\Box$

If we also require each $M_i$ to be finitely generated, then example 2 no longer applies. In that case, we have:

Proposition 2. Let $\{M_i\}_{i\in\omega}$ be a chain of finitely generated modules. Then the chain stabilizes if and only if $M=\cup_{i\in\omega}M_i$ is finitely generated.

Proof. If the chain stabilizes, then there exists $N\in\omega$ such that $M=M_N$, and since $M_N$ is finitely generated, so is $M$. The converse is Proposition 1. $\Box$


问:
Rotation in $\mathbb{R}^n$
Let $E_1:=\{(x_1,\ldots,x_n)\in \mathbb{R}^n : x_n = 0\}$ be a hyperplane and let $E_2$ be any hyperplane that goes through the origin. I want to find a rigid motion $A: \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $A(E_2)=E_1.$ Is the following procedure correct?

Since $E_2$ goes through the origin, it is a $(n-1)$-dimensional subspace. By Gram-Schmidt we can find a orthonormal system $(v_1,\ldots,v_{n-1})$ such that $E_2=\langle v_1,\ldots,v_{n-1}\rangle$. Let $w:= v_1 \times \ldots \times v_{n-1}$ (cross product) be the vector that is orthogonal to $v_1,\ldots,v_{n-1}.$
Then $A$ is given by $$x \mapsto\begin{pmatrix}\mid &  \mid&\cdots  & \mid\\
w &  v_1 & \ldots & v_{n-1}\\
\mid & \mid&\ldots&  \mid  \end{pmatrix}^{-1}x.$$

Is $A$ the correct rotation?

答:
Welcome to MSE. Your argument is almost correct, but there's a problem in your definition for $w$: cross product is only defined for vectors in $\mathbb R^3$!

So, how do we find this vector $w$ that is orthogonal to $(v_1, \cdots, v_n)$, the orthonormal basis of the plane? One solution is via orthogonal projection. Namely, choose any vector $w'$ not in the hyperplane, and subtract off its parallel components in the directions of $v_1, \dots, v_n$ to make it perpendicular to all of them. Then, your matrix $A$ will work.

Title: Rotation in $\mathbb{R}^n$