问:
Hard power series problem

Consider the differential equation $$(1+t)y''+2y=0$$
with the variabel coefficient $(1+t)$, with $t\in \mathbb{R}$.

Set $y(t)=\sum_{n=0}^{\infty}a_nt^n$. What are the first 4 terms in the associated power series?

An Attempt

We write
$$(1+t)y''=\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}$$
$$2y=\sum_{n=0}^{\infty}2a_nt^{n} $$

Combining these three gives us

$$\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}+ \sum_{n=0}^{\infty}2a_nt^{n}$$

We now want all the powers to be $t^n$.
$$\sum_{n=0}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=0}^{\infty}2a_nt^{n} $$

We finally want the index of summation to all be the same.
$$2a_0+2a_2+ \sum_{n=1}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=1}^{\infty}2a_nt^{n}$$

Which simplifies to
$$2a_0+2a_2+ \sum_{n=1}^{\infty}\big[a_{n+2}(n+2)(n+1)+a_{n+1}n(n+1)+2a_n \big]t^n$$

Clearly, I'm not getting anywhere close to any of the answers given. I just can't find what I'm doing wrong. I have also tried with Maple, but that didn't give me any real result either.

I hope someone can help me out.

答:
If

$$y(t)=\sum_{n=0}^{\infty}c_n t^n$$

then

$$y''(t)=\sum_{n=2}^{\infty}n(n-1)c_n t^{n-2}$$

which is equivalent to

$$y''(t)=\sum_{n=0}^{\infty}(n+1)(n+2)c_{n+2}t^n$$

so

\begin{align*}
(1+t)y''(t) &= y''(t)+ty''(t)\\
&=\sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}t^n\right]+t\sum_{n=0}^{\infty}(n+1)(n+2)c_{n+2}t^n\\
&= \left(2c_2+2\cdot 3c_3 t+3\cdot 4c_4 t^2+\cdots\right)+t\left(2c_2+2\cdot 3c_3 t+3\cdot 4c_4 t^2+\cdots\right)\\
&= \left(2c_2+2\cdot 3c_3 t+3\cdot 4c_4 t^2+\cdots\right)+\left(2c_2 t+2\cdot 3c_3 t^2+3\cdot 4c_4 t^3+\cdots\right)\\
&= \left(1\cdot 2c_2+0\cdot 1c_1\right)+\left(2\cdot 3c_3+1\cdot 2c_2\right)t+\left(3\cdot 4c_4+2\cdot 3c_3\right)t^2+\cdots\\
&= \sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}\right]t^n
\end{align*}

It follows that

\begin{align*}
0 &= (1+t)y''(t)+2y(t)\\
&= \sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}\right]t^n+\sum_{n=0}^{\infty}2c_n t^n\\
&= \sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}+2c_n\right]t^n
\end{align*}

and upon equating coefficients,

$$(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}+2c_n=0$$

No initial conditions were given, so $c_0$ and $c_1$ can be chosen arbitrarily. The recurrence relation above then gives

\begin{align*}
c_2 &= -c_0\\
c_3 &= \frac{c_0-c_1}{3}\\
&\text{ }\vdots
\end{align*}

Thus,

$$y(t)=c_0+c_1 t-c_0t^2+\frac{c_0-c_1}{3}t^3+\cdots$$

so (d) is the correct answer.

Edit: I’m apparently blind! While writing this answer, I didn’t notice that (d) is the exact same series I derived. In light of this observation (thank you Carl!), I’ve edited out the last part of my original response, where I claimed that (b) was the correct answer.

Title: Hard power series problem


问:
Am I missing an equilibrium point?
I am trying to identify all the equilibrium points of this system, and am only concerned with the first quadrant:

$$x'=x(a-x-ay),
 \\ y'=y(b-bx-y)$$

Where a and b are parameters greater than 0. When I calculate the equilibrium points by setting x'=y'=0, I get (0,0), (a,0), (0,b). The problem is, when I plot the nullclines with varying values of a and b, I can get a fourth equilibrium point, as an x-nullcline and y-nullcline intersect. Why is this equilibrium point not showing up when I initially calculate them in terms of my parameters? Thank you for the help.

答:
Note that

$$x'=0 \implies x=0 \quad \text{or} \quad \frac{x}{a}+\frac{y}{1}=1$$

$$y'=0 \implies y=0 \quad \text{or} \quad \frac{x}{1}+\frac{y}{b}=1$$

There're $4$ combinations for choosing the conditions such that $x'=y'=0$.

The fourth equilibrium point is given by

$$
\left \{
  \begin{align}
    \frac{x}{a}+\frac{y}{1} &=1 \\
    \frac{x}{1}+\frac{y}{b} &=1
  \end{align}
\right.$$

which gives the solution $(x,y)=\left( \dfrac{ab-a}{ab-1},\dfrac{ab-b}{ab-1} \right)$

ContentLicense: CC BY-SA 4.0

Title: Am I missing an equilibrium point?


问:
For two parameterizations, normal unit vector coincides if the derivative of transition map has positive determinant
Let $S\subset R^3$ be a regular surface with $p\in S$. Let $(U,\phi)$ be a local chart such that $\phi(q) = p$ where $q\in U\subset R^2$ (Basically, $(U,\phi)$ is a local chart when $\phi$ is a local parameterization of $S$ that has $p$ in its image set).

We know that the tangent plane at $p$ $T_p(S)$ is defined as

$T_p(S) = \{\alpha'(0) \mid \alpha : (-\epsilon, \epsilon) \to S \text{ is smooth and } \alpha(0) = p\}$

We know that the tangent plane is spanned by $T_p(S) = \text{span} \{\phi_x(q), \phi_y(q)\}$

We define the normal unit vector to $T_p(S)$ with respect to local chart $(U, \phi)$ by

$$N_\phi(p) = \frac{\phi_x(q)\times \phi_y(q)}{\|\phi_x(q)\times \phi_y(q)\|} $$

Now, note that for another local chart $(V,\gamma)$ such that $\gamma(r) = p$,

$$N_\gamma(p) = \frac{\gamma_x(r)\times \gamma_y(r)}{\|\gamma_x(r)\times \gamma_y(r)\|}$$

is also unit normal to $T_p(S)$ at $p$.

Hence, $$N_\phi(p) = \pm N_\gamma(p)$$

The transition map between $\phi $ and $\gamma $ which is given by $\gamma^{-1}\circ \phi : \phi^{-1}(W)\to \gamma^{-1}(W)$ where $p\in W = \phi(U)\cap \gamma(V)$ is a diffeomorphism.

I want to prove that if $\text{det } [D(\gamma^{-1}\circ \phi )(q)] >0
 $ then we must have $N_\phi(p) = N_\gamma(p)$

How can I show this precisely?

Note that $D(\gamma^{-1}\circ \phi )(q)$ is a $2\times 2 $ matrix and doesn't have $0$ determinant because $\gamma^{-1}\circ \phi $ is diffeomorphism.

答:
Take $\gamma^{-1}\circ\phi$ as the change of variables
$$(x,y)^{\top}\mapsto(v,w)^{\top},$$
then $D(\gamma^{-1)}\circ\phi)$ is a $2\times 2$ matrix
which allows to express
$$\phi_x=P\gamma_v+Q\gamma_w,$$
$$\phi_y=S\gamma_v+T\gamma_w,$$
since both sets $\{\phi_x,\phi_y\}$ and $\{\gamma_v,\gamma_w\}$ are basis
for the tangent space.
Now you have
$$\phi_x\times\phi_y=(P\gamma_v+Q\gamma_w)\times (S\gamma_v+T\gamma_w)$$
$$\qquad=(PT-QS)\ \gamma_v\times\gamma_w,$$
which gives you control over the sign of vector $N$ in the two coordinates.


问:
For a diagonal matrix $M$, what is $e^M$?
For a diagonal matrix
$$
M=\left(\begin{array}{ccc}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right)
$$
show that
$$
e^M=\left(\begin{array}{ccc}
e^a & 0 & 0 \\
0 & e^b & 0 \\
0 & 0 & e^c
\end{array}\right).
$$</p>

答: Hint: $e^M$  is defined as the Taylor series. What does $M^2$ look like? Plug the powers of $M$ into the Taylor series.</p>

ContentLicense: CC BY-SA 3.0


问:
$k'(s_0)=0$ iff $k(s_0) = 1/r$ $\hspace{0.2cm}$ or $\hspace{0.2cm}$ $\tau(s_0)=0$
Let $\alpha : I \rightarrow \mathbb{R}^3$ be a curve parameterized by arc length (p.b.a.l.) with positive curvature and contained in a sphere of radius $r > 0$. If $s_0  \in  I$, show that the following
assertions are equivalent:

(i)$\hspace{0.1cm}$ $k'(s_0)=0$

(ii)$\hspace{0.1cm}$ $k(s_0) = 1/r$ $\hspace{0.2cm}$ or $\hspace{0.2cm}$ $\tau(s_0)=0$

My attempt

As usual, I want to prove that (i) implies (ii) and reciprocally that (ii) implies (1). But before this, I establish some general conditions:
If $ \alpha $ is contained in a sphere of radius $ r> 0 $ centered at the origin, then $ || \alpha || = r $, equivalently, $ || \alpha ||^2 = r ^ 2 $. Now, differentiating this expression three times, we have the following:

$$\frac{d}{ds}||\alpha||^2 = 2\alpha \cdot \alpha' = 0 \hspace{0.3cm}\text{so}\hspace{0.3cm} \alpha\cdot \alpha' = \alpha \cdot T = 0$$

$$\frac{d^2}{ds^2}||\alpha|^2 = \alpha \cdot T' + \alpha' \cdot T = \alpha \cdot T' + 1 = 0 \hspace{0.3cm} \text{(1)}$$

\begin{align*}
\frac{d^3}{ds^3}||\alpha||^2 &= k'(\alpha \cdot N) + k(\alpha \cdot N' + \alpha' \cdot N) \\
&= k'(\alpha \cdot N) + k(\alpha \cdot N')\\
&= k'(\alpha \cdot N) + k[\alpha \cdot(-kT - \tau B)]\\
&= k'(\alpha \cdot N) -k^2(\alpha \cdot T)-k\tau(\alpha \cdot B)\\
&= k'(\alpha \cdot N) - k\tau (\alpha \cdot B) = 0   \hspace{0.12cm} \text{(2)}
\end{align*}
Now if $k'(s_0)=0$, then the equation (2) implies that $\tau (s_0)(\alpha(s_0) \cdot B(s_0)) = 0$, but this does not give (ii).

Conversely, if $ k (s_0) = 1 / r $, clearly $ k '(s_0) = 0$. If $ \tau (s_0) = 0$ then equation (2) gives $ k '(s_0) (\alpha (s_0) \cdot N (s_0)) = 0 $, but I'm not sure if this implies that $ k' (s_0) = 0$.

I believe that at some point you must use equation (1). Where is my mistake in all this reasoning? I have not actually been able to conclude either of the two implications of the exercise.
Any help is appreciated

答:
Yes, you have to use (1). You want to analyze (at $s_0$) the components $\alpha\cdot N$ and $\alpha\cdot B$. How can $\alpha\cdot B=0$? What does this say if you use (1)? (Also, how do you justify your "clearly"?)

EDIT: OK, note firstly that (1) says $k(\alpha\cdot N) = -1$, so $\alpha\cdot N\ne 0$ and $\alpha\cdot N = -1/k$ precisely when $\alpha = -rN$ and $k=1/r$. On the other hand, $\alpha$ and $N$ are parallel if and only if $\alpha\cdot B = 0$. You already observed that
$$k' = 0 \implies \tau = 0 \quad\text{or}\quad \alpha\cdot B = 0,$$
and the latter occurs if and only if $\alpha$ and $N$ are parallel and so $k = 1/r$. This establishes (1) $\implies$ (2).

Can you now finish (2) $\implies$ (1)?


问:
"Continuity of determinant"
Im reading a proof of the Implicit function theorem in my class calculus book, and there's one claim that I cannot understand and it not explained in the book.

Denote :

$ F'_{y}\left(x,y\right)=\begin{pmatrix}\frac{\partial F^{1}}{\partial y^{1}} & \cdots & \frac{\partial F^{1}}{\partial y^{n}}\\
\vdots & \ddots & \vdots\\
\frac{\partial F^{n}}{\partial y^{1}} & \cdots & \frac{\partial F^{n}}{\partial y^{n}}
\end{pmatrix}\left(x,y\right) $

Where $ \left(x,y\right)=\left(x^{1},...x^{m},y^{1},...,y^{n}\right) $,

$ F $ is a function such that $ F:U\to\mathbb{R}^{n} $ where $ U\subseteq\mathbb{R}^{m+n} $

And $ F\in C^{(p)}\left(U;\mathbb{R}^{n}\right),p\geq1 $.

Assume that the determinant of $ F'_{y}\left(x,y\right) $ at a point $ (x_0,y_0) $ is not zero.

The claim in the book is that since the determinant at $ (x_0,y_0) $ is not zero, also in some neighborhood of the point $ (x_0,y_0) $ the determinant of the matrix $ F'_y(x,y) $ is not zero.

Why is this true? Im not familier with continuity of determinants

答:
I am elaborating on my comments, since you are confusing a few concepts with each other.

Suppose that $A \in \mathbb R^{n \times n}$ is an $(n \times n)$-matrix. If we identify $\mathbb R^{n \times n}$ with $\mathbb R^{n^2}$ (e.g. by reading the matrix row-wise and then column wise), then the determinant defines a function $\det \colon \mathbb R^{n \times n} = \mathbb R^{n^2} \to \mathbb R$. We show that this function is a polynomial function in the $n^2$ variables by induction on $n$. If we established this claim, then $\det$ is continuous, since it is a polynomial.

For $n = 1$, the matrix $A$ is of the form $(a_{11})$ and $\det(A) = a_{11}$, which is a polynomial in the only variable $a_{11}$. Hence the claim is true for $n = 1$.

Let now $n > 1$. Assume that the claim is already proved for $n-1$. By the Lapace formula, one has
$$\det(A) = \sum_{i=1}^n (-1)^{i+1}  a_{i1}\det(A_{i1}),$$

where $A = (a_{ij})$ and $A_{i1}$ is the matrix obtained from $A$ by deleting the $i$-th row and the $1$-st column. By induction hypothesis, $\det(A_{i1})$ is a polynomial in the entries of $A_{i1}$ for each $i$. Hence, using the above formula for $\det(A)$, one sees that $\det(A)$ is a polynomial, too.

As an example: if $n = 2$, then $\det(A) = a_{11}a_{22} - a_{12}a_{21}$ is a polynomial in the $a_{ij}$!

Now, the map you are considering is the composition of the map

$$U \to \mathbb R^{n \times n}, \ \ \ \ (x,y) \mapsto F'_y(x,y)$$

with the map
$$\mathbb R^{n \times n} \to \mathbb R, \ \ \ \ A \mapsto \det(A).$$

Both these maps are continuous, so their composition $(x,y) \mapsto \det(F'_y(x,y))$ is continuous.

Title: "Continuity of determinant"


问:
Calculate the volume bounded by surface
Calculate the volume bounded by the surface
$$(x^2+y^2+z^2)^2 = a^2(x^2+y^2-z^2)$$

Using the spherical coordinates
$$\begin{cases} x = rcos\varphi cos\theta & \\ y = rsin\varphi cos\theta \\z = rsin\theta \end{cases}$$
and substituting those into the original equation we get
$$r^2 = a^2(cos^2\theta - sin^2\theta) = a^2cos2\theta$$
and from that
$$0 \leq r \leq a\sqrt{cos2\theta}$$

Calculating Jacobian gives us
$$ J = r^2cos\theta$$

Given all that the target volume could be calculated as

$$V = \int_{0}^{2\pi}\,d\varphi \int_{0}^{\pi}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr$$

But this yields incorrect result, moreover the supposed answer should be calculated given the following integral
$$V = 8\int_{0}^{\frac{\pi}{2}}\,d\varphi \int_{0}^{\frac{\pi}{4}}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr $$

But I have trouble understanding where do the integrating boundaries for $\varphi$ and $\theta$ come from.

I undesrtand that given the fact that the surface and therefore target solid are symmetrical, we can integrate over a certain part of the solid and then multiply the result by a proper constant,
but if we use the following bounds for $\varphi$ and $\theta$

$$0 \leq \varphi \leq \pi/2 \\ 0 \leq \theta \leq \pi/4$$

how come we multiply by 8 and not by 16?

Any tips on what I'm doing wrong ?

答:
Note the plot of $r=a\sqrt{\cos2\theta}$ below

with range $\theta \in [-\frac\pi4, \frac\pi4]$. Thus, the volume integral is set up as
$$V = \int_{0}^{2\pi}\,d\varphi \int_{-\frac\pi4}^{\frac\pi4}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2\cos\theta dr\\
=8 \int_{0}^{\frac\pi2}\,d\varphi \int_{0}^{\frac\pi4}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2\cos\theta dr
$$

Title: Calculate the volume bounded by surface


答: If you multiply by $16$, you would get all possible values of $\theta$, from $0$ to $\pi$. What is the limit of integration for $r$ when $\theta=\pi/2$? You get $\cos(2\theta)=-1$, so the square root is not defined (in real numbers). In fact for $\theta\in(\pi/4,\pi/2)$ your upper limit is not defined.

Title: Calculate the volume bounded by surface

Body:
Calculate the volume bounded by the surface
$$(x^2+y^2+z^2)^2 = a^2(x^2+y^2-z^2)$$

Using the spherical coordinates
$$\begin{cases} x = rcos\varphi cos\theta & \\ y = rsin\varphi cos\theta \\z = rsin\theta \end{cases}$$
and substituting those into the original equation we get
$$r^2 = a^2(cos^2\theta - sin^2\theta) = a^2cos2\theta$$
and from that
$$0 \leq r \leq a\sqrt{cos2\theta}$$

Calculating Jacobian gives us
$$ J = r^2cos\theta$$

Given all that the target volume could be calculated as

$$V = \int_{0}^{2\pi}\,d\varphi \int_{0}^{\pi}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr$$

But this yields incorrect result, moreover the supposed answer should be calculated given the following integral
$$V = 8\int_{0}^{\frac{\pi}{2}}\,d\varphi \int_{0}^{\frac{\pi}{4}}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr $$

But I have trouble understanding where do the integrating boundaries for $\varphi$ and $\theta$ come from.

I undesrtand that given the fact that the surface and therefore target solid are symmetrical, we can integrate over a certain part of the solid and then multiply the result by a proper constant,
but if we use the following bounds for $\varphi$ and $\theta$

$$0 \leq \varphi \leq \pi/2 \\ 0 \leq \theta \leq \pi/4$$

how come we multiply by 8 and not by 16?

Any tips on what I'm doing wrong ?


答:
$\operatorname{diag}:\begin{array}{ll}\Bbb R^n \to M_n\left(\Bbb R\right)\\\left(x_1,\dots,x_n\right)\mapsto \begin{pmatrix}x_1& 0 & \dots&\dots & 0\\
0&x_2&\ddots&&\vdots\\
\vdots&\ddots&\ddots &\ddots&\vdots\\
\vdots&&\ddots&x_{n-1}&0\\
0&\dots&\dots&0&x_n\end{pmatrix}\end{array}$

I'll hide the proofs so that you can try to show the results by yourself, just move your mouse over the blank to show them.

 $\forall \left(x_1,\dots,x_n\right),\left(y_1,\dots,y_n\right)\in \Bbb R^n, \forall \lambda \in \Bbb R,\\\operatorname{diag}\left(x_1\lambda y_1,\dots,x_n+\lambda y_n\right)\\=\begin{pmatrix}x_1+\lambda y_1& 0 & \dots&\dots & 0\\0&x_2+\lambda y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}+\lambda y_{n-1}&0\\0&\dots&\dots&0&x_n+\lambda y_n\end{pmatrix}\\=\begin{pmatrix}x_1& 0 & \dots&\dots & 0\\0&x_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}&0\\0&\dots&\dots&0&x_n\end{pmatrix}+\lambda\begin{pmatrix}y_1& 0 & \dots&\dots & 0\\0&y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&y_{n-1}&0\\0&\dots&\dots&0&y_n\end{pmatrix}\\=\operatorname{diag}\left(x_1,\dots,x_n\right)+\lambda\operatorname{diag}\left(y_1,\dots,y_n\right)$

$\boxed{\operatorname{diag}\in \mathcal L\left(\Bbb R^n,M_n\left(\Bbb R\right)\right)}\tag{1}$

$\forall \left(x_1,\dots,x_n\right),\left(y_1,\dots,y_n\right)\in \Bbb R^n,\\\operatorname{diag}\left(x_1,\dots,x_n\right)\operatorname{diag}\left(y_1,\dots,y_n\right)\\=\begin{pmatrix}x_1& 0 & \dots&\dots & 0\\0&x_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}&0\\0&\dots&\dots&0&x_n\end{pmatrix}\begin{pmatrix}y_1& 0 & \dots&\dots & 0\\0&y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&y_{n-1}&0\\0&\dots&\dots&0&y_n\end{pmatrix}\\=\begin{pmatrix}x_1y_1& 0 & \dots&\dots & 0\\0&x_2y_2&\ddots&&\vdots\\\vdots&\ddots&\ddots &\ddots&\vdots\\\vdots&&\ddots&x_{n-1}y_{n-1}&0\\0&\dots&\dots&0&x_ny_n\end{pmatrix}\\=\operatorname{diag}\left(x_1y_1,\dots,x_ny_n\right)$

$\boxed{\forall \left(x_1,\dots,x_n\right),\left(y_1,\dots,y_n\right)\in \Bbb R^n,\operatorname{diag}\left(x_1y_1,\dots,x_ny_n\right)=\operatorname{diag}\left(x_1,\dots,x_n\right)\operatorname{diag}\left(y_1,\dots,y_n\right)}\tag{2}$

 By recurrence and $(2)$:

$\boxed{\forall \left(x_1,\dots,x_n\right)\in \Bbb R^n, \forall p \in \Bbb N,\operatorname{diag}\left(x_1,\dots,x_n\right)^p=\operatorname{diag}\left(x_1^p,\dots,x_n^p\right)}\tag{3}$

$\exp:\begin{array}{ll}M_n\left(\Bbb R\right)\to M_n\left(\Bbb R\right)\\M \mapsto \sum\limits_{k=0}^{+\infty}\cfrac{M^k}{k!}\end{array}$

 $\forall \left(x_1,\dots,x_n\right)\in \Bbb R^n,\\\exp\left(\operatorname{diag}\left(x_1,\dots,x_n\right)\right)\\=\sum\limits_{k=0}^{+\infty}\cfrac{\operatorname{diag}\left(x_1,\dots,x_n\right)^k}{k!}\\=\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{\operatorname{diag}\left(x_1,\dots,x_n\right)^k}{k!}\\=\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{\operatorname{diag}\left(x_1^k,\dots,x_n^k\right)}{k!}\\=\lim\limits_{m\to +\infty}\operatorname{diag}\left(\sum\limits_{k=0}^{m}\cfrac{x_1^k}{k!},\dots,\sum\limits_{k=0}^{m}\cfrac{x_n^k}{k!}\right)\\=\operatorname{diag}\left(\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{x_1^k}{k!},\dots,\lim\limits_{m\to +\infty}\sum\limits_{k=0}^{m}\cfrac{x_n^k}{k!}\right)\\=\operatorname{diag}\left(\sum\limits_{k=0}^{+\infty}\cfrac{x_1^k}{k!},\dots,\sum\limits_{k=0}^{+\infty}\cfrac{x_n^k}{k!}\right)\\=\operatorname{diag}\left(\exp\left(x_1\right),\dots,\exp\left(x_n\right)\right)$

Justification for each equality (try to find them yourself after reading the proof)

 The first inequality is the definition of $\exp$, the second is the definition of $\sum\limits_{k=0}^{+\infty}$, the third is $(3)$, the fourth is $(1)$, the fifth is the continuity of $\operatorname{diag}$ which comes from $(1)$ and the fact that both $\Bbb R^n$ and $M_n\left(\Bbb R\right)$ are finite dimensional, the sixth is the definition of $\sum\limits_{k=0}^{+\infty}$ and the seventh is the definition of $\exp$ over the real numbers.

$\boxed{\forall \left(x_1,\dots,x_n\right)\in \Bbb R^n,\exp\left(\operatorname{diag}\left(x_1,\dots,x_n\right)\right)=\operatorname{diag}\left(\exp\left(x_1\right),\dots,\exp\left(x_n\right)\right)}$

Title: For a diagonal matrix $M$, what is $e^M$?

Body:
For a diagonal matrix
$$
M=\left(\begin{array}{ccc}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right)
$$
show that
$$
e^M=\left(\begin{array}{ccc}
e^a & 0 & 0 \\
0 & e^b & 0 \\
0 & 0 & e^c
\end{array}\right).
$$</p>


问:
Polynomial Long Division:
My question essentially boils down to this: (it was part of a question about polynomial ring ideals)

Find an integer $b$ such that the rational function
\begin{equation}
\frac{x^5-bx}{x^2-2x}\in \mathbb{Z}[x]
\end{equation}
I've tried plugging in a bunch of values for $b$ using a calculator but none of them have come out to whole polynomials. Is there an easier way to do this?

答:
You may be guessing a long time at that rate! Write
$$
x^5 - bx - a = (x^2-2x)(-------),
$$
and fill in the divisor's bracket term by term. To multiply to $x^5$, the first term needs to be $x^3$:
$$
x^5 - bx - a = (x^2-2x)(x^3-----).
$$
Now $2x^4$ is being subtracted, so you better add $2x^2$ to the divisor to make that up.
$$
x^5 - bx - a = (x^2-2x)(x^3+2x^2----).
$$
Then
$$
x^5 - bx - a = (x^2-2x)(x^3+2x^2+4x--).
$$
So then
$$
x^5 - bx - a = (x^2-2x)(x^3+2x^2+4x+8),
$$
and you can read off $b = 16$, $a = 0$.

Title: Polynomial Long Division:


问:
Solving complex improper integrals

Evaluate the following integrals and show the details of your work.

$$\int_0^{2\pi} \frac{\cos^2\theta}{5 - 4\cos\theta}\, d\theta$$

So generally my approach is to find the roots because I need them to evaluate the residues which I can then use in the residue theorem.

So first thing's first is to find the roots...

The denominator factors to: $5(1-\frac{4\cos\theta}{5})$

But when does $\cos\theta = 5/4$ so that $1 - \frac{4 \cos \theta}{5} = 0$

My book says the answer is: $\frac{5\pi}{12}$ FWIW but I'm not sure how to get here.

Title: Solving complex improper integrals


问:
Approximating compactly supported $L^2$ functions with Schwartz functions "from within"?
Crossposted from MathOverflow.

It is well known that the class of Schwartz functions $\mathcal{S}$ in dense in all $L^p$ spaces therefore for each $f \in L^2$ there exists a sequence of Schwartz functions $(f_k)$ such that $\lVert f - f_k \rVert_{L^2} \to 0$ as $k \to \infty$.

If we suppose further that $f$ has compact support can we find a sequence of Schwartz functions $(f_k)$ such that $\lVert f - f_k \rVert_{L^2} \to 0$ as $k \to \infty$ (as above) and additionally $\operatorname{supp}(f_k) \subseteq \operatorname{supp}(f)$ for all $k$?

If so, what argument can I appeal to?

答:
Not if $$f = \lim_{N\to \infty} f_N, \qquad f_N=1_{[0,1]} \prod_{n=1}^N \prod_{k=0}^{2^n-1} (1-1_{\displaystyle [\frac{k}{2^n}-\frac1{2^{2n+2}},\frac{k}{2^n}+\frac1{2^{2n+2}}]})$$
$\lim_{N\to \infty} f_N$ converges in $L^p(\Bbb{R})$, the limit is non-zero since $\int_0^1 f_N(x)dx\ge 1-\sum_{n=1}^N 2^{-n-1}$

The support of $f$ doesn't contain any $k/2^n$ so it doesn't contain any open set.


问:
A minor point in Serre's Representation Theory
Let $R$ be a commutative ring of characteristic zero, $G$ a finite group. For each conjugacy class $[x]$ in $G$, define $e_{[x]} = \sum_{s\in[x]}s$. It is easily checked that these $\{e_{[x]}\}_{x\in G}$ form a basis for $Z(\mathbb{C}[G])$. Serre makes the claim that

"... each product $e_{[x]}e_{[y]}$ is a linear combination with integer coefficients of the $e_{[z]}$."

While I can see how these $e_{[x]}$'s form a basis for $Z(\mathbb{C}[G])$, for some reason I can't see this fact as easily. In fact, I can't even intuit why $e_{[x]}e_{[y]}$ should also be an element of $Z(\mathbb{C}[G])$ when written out as
$$
e_{[x]}e_{[y]} = \left(\sum_{g\in G}gxg^{-1}\right)\left(\sum_{h\in G}hyh^{-1}\right) = \sum_{g,h\in G}gxg^{-1}hyh^{-1}
$$
To me it's not clear at all that the above double sum is "a linear combination with integer coefficients of the $e_{[z]}$." Can someone provide some insight?

答:
Well, since $e_{[x]}$ and $e_{[y]}$ are both in $Z(\mathbb{C}[G])$, so is their product.  Moreover, it is clear that $e_{[x]}e_{[y]}$ is an integer linear combination of elements of $G$ (since it is just some sum of elements of $G$).  But since it is in $Z(\mathbb{C}[G])$, the coefficients must be constant on each conjugacy class, and so it is in fact an integer linear combination of $e_{[z]}$'s.

To see this more explicitly from the expression $$e_{[x]}e_{[y]} = \sum_{g,h\in G}gxg^{-1}hyh^{-1},$$ note that $$k(gxg^{-1}hyh^{-1})k^{-1}=(kg)x(kg)^{-1}(kh)y(kh)^{-1}.$$ So, this sum is unchanged if you conjugate by $k$: the terms just get permuted, with the term indexed by $(g,h)$ becoming the term indexed by $(kg,kh)$.


问:
Suppose $A[x]$ is integrally closed then $A$ need not be UFD.
Let $A$ be a integral domain. Suppose $A[x]$ is integrally closed then $A$ need not be integrally closed?
(In particular $A$ is not UFD.)

I was trying to find a counter example. But after long time I couldn't find any. Please give me a hint.

(I was thinking of proving  $A[x]= \mathbb Z (\sqrt{-5})[x] $  is not integrally closed. But I could not do it.  I'm not sure if it's true either.
I end up with the following statement - if $A$ is integrally closed over $B$ then for ideal $I$ of B, $A/{A\cap I}$ is integrally closed over $B/I$ which I couldn't prove or disprove either -- sorry for not showing any work. I haven't made much progress to show)

答: As pointed out in the comments by user26857, if $A[x]$ is an integrally closed domain, then $A$ must also be an integrally closed domain. Suppose $b\in\operatorname{Frac} A$ is integral over $A$. Then $b\in \operatorname{Frac} A[x]$ is integral over $A[x]$, so it's in $A[x]$. But $A[x]\cap \operatorname{Frac} A=A$ inside $\operatorname{Frac} A[x]$, so $b\in A$.


问:
How to find a matrix $A$ given two vectors which are solutions to $Ax=0$ and two vectors which are not
How do I find a $3\times 4$ matrix $A$ given two $4\times 1$ vectors which are solutions to $Ax=0$ and two $4\times 1$ vectors which cannot be solutions?

I have this question and so far the only method I could think of is working out the value of each number in the matrix using simultaneous equations from the two linear systems which do actually give $Ax=0$. I found that I can't seem to work out how to do this simultaneously and I can't actually find the value of any of the numbers in $A$ without them being in terms of the other numbers in the matrix.

答:
Let $u, v \in \mathbb{R}^{4}$ be the given vectors for which should hold $Ax = 0$. We assume $u$ and $v$ are linearly independent.

Solve $[u \space v]^{T}x = 0$ and write your solution-space as a linear combination of 2 vectors, say $w, z \in \mathbb{R}^{4}$. $w$ and $z$ are linearly independent of course.

Now $A = [w \space z \space 0]^{T}$. It is straightforward to check $Au = 0$ and $Av = 0$.


问:
Proof that an embedding into $\ell^1$ is compact

Prove that any sequence $(x^{(n)})_{n\in\mathbb{N}}\subseteq\ell^1$ such that $\sum_{k=1}^\infty k\lvert x_k^{(n)}\lvert\leq1$ for all $n\in\mathbb{N}$ has a convergent subsequence.

My thoughts on this: Clearly $\lvert x^{(n)}_k\lvert\leq\frac{1}{k}$ uniform in $n$. Therefore the sequence $x_1^{(n)}$ has a convergent subsequence $x_1^{(\tilde{n}_k)}$. Further extracting subsequences for any fixed $N\in\mathbb{N}$ I find $({n_l})_{l\in\mathbb{N}}\subseteq\mathbb{N}$ such that $x_k^{(n_l)}$ converges to some $x_k$ for all $1\leq k\leq N$ as $n\rightarrow\infty$.

If for every $\epsilon>0$ I could find some $N\in\mathbb{N}$ such that $\sum_{k=N+1}^\infty \lvert x_k^{(n)}\lvert<\epsilon$ uniform in $n$ the proof would be complete. But I don't see why this should be true.

Can you give me some hint?

答:
Note: here is, I believe, the result you wanted to prove initially. The fact that the following operator is compact follows easily from the fact that it is the operator norm limit of finite rank operators.

Consider the bounded linear operator
$$
T:\ell^1\longrightarrow \ell^1\qquad (x_k)\longmapsto \left(\frac{x_k}{k}\right).
$$
Since $T$ is the operator norm limit of the finite rank operators,
$$T_n:(x_k)\longmapsto \left(x_1,\frac{x_2}{2},\ldots,\frac{x_n}{n},0,\ldots\right),$$
it is compact. Indeed, $\|T-T_n\|\leq \frac{1}{n+1}$. So if $B$ denotes the closed unit ball of $\ell^1$, then $T(B)$ is relatively compact in $\ell^1$.

Now just observe that
$$
T(B)=\{(y_k)\in\ell^1\;;\; \sum_{k=1}^{+\infty} k\,|y_k|\leq 1\}
$$
is the set you are considering. It is easily seen to be closed in $\ell^1$. Indeed, if $y^{(n)}$ is in $T(B)$ and converges to $y$ in $\ell^1$, then it converges pointwise to $y$ a fortiori. So for every $K$ and every $n$, we have
$$
\sum_{k=1}^K k\,|y^{(n)}_k|\leq 1\quad \Rightarrow \quad \sum_{k=1}^K k\,|y_k|\leq 1\quad \Rightarrow\quad \sum_{k=1}^{+\infty} k\,|y_k|\leq 1.
$$

Conclusion: your set $T(B)$ is a compact subset of $\ell^1$. What you want is just the sequential compactness of the latter: every sequence in $T(B)$ has a convergent subsequence in $T(B)$ (and not only in $\ell^1$).


答:
If $z=gxg^{-1}hyh^{-1}\in[x][y]$ and $\gamma\in G$ we have
$$
\gamma z\gamma^{-1}=(\gamma g)x(g^{-1}\gamma^{-1})(\gamma h)y(h^{-1}\gamma^{-1})\in[x][y]
$$
so that $[x][y]$ is a union of conjugacy classes.

The coefficient of $z\in G$ is the cardinality of the set
$$
P_z=\{(g,h)\in G\times G\,|\, z=gxg^{-1}hyh^{-1}\}.
$$
The same formula above says that multiplication by $\gamma$ embedded diagonally in $G\times G$ gives a bijection between $P_z$ ad $P_{\gamma z\gamma^{-1}}$ so that the coefficients are the same for all the elements in the same conjugacy class.

Title: A minor point in Serre's Representation Theory

Body:
Let $R$ be a commutative ring of characteristic zero, $G$ a finite group. For each conjugacy class $[x]$ in $G$, define $e_{[x]} = \sum_{s\in[x]}s$. It is easily checked that these $\{e_{[x]}\}_{x\in G}$ form a basis for $Z(\mathbb{C}[G])$. Serre makes the claim that

"... each product $e_{[x]}e_{[y]}$ is a linear combination with integer coefficients of the $e_{[z]}$."

While I can see how these $e_{[x]}$'s form a basis for $Z(\mathbb{C}[G])$, for some reason I can't see this fact as easily. In fact, I can't even intuit why $e_{[x]}e_{[y]}$ should also be an element of $Z(\mathbb{C}[G])$ when written out as
$$
e_{[x]}e_{[y]} = \left(\sum_{g\in G}gxg^{-1}\right)\left(\sum_{h\in G}hyh^{-1}\right) = \sum_{g,h\in G}gxg^{-1}hyh^{-1}
$$
To me it's not clear at all that the above double sum is "a linear combination with integer coefficients of the $e_{[z]}$." Can someone provide some insight?


问:
Is a closed set in a metric space necessarily an infinite intersection of open sets? If so, why?
This has been on my mind for a few days now. I know that an infinite intersection of open sets can be closed, but is a closed set necessarily an infinite intersection of open sets?</p>

答:
All of the answers so far have strived to write the given closed set as the intersection of countably many open sets, which is indeed a very useful thing to know, but the OP does not seem to mind the number of sets involved.

If we indeed ignore this point then every subset $C\subseteq X$, closed or not, is the intersection of open subsets, namely the sets  $$X\setminus \{x\},$$ where $x$ ranges in $X\setminus C$.


问:
Sufficient Statistic-Estimation
I have been given pdf $f$(x,$\theta$)=$\frac{\theta}{x^{2}}$, where $0<\theta \le x <\infty $ and I have to find the sufficient statistic. I have a ${({X}_{i})}_{i=1}^{n}$ i.i.d sample.

So joint pdf will be $\prod_{i=1}^{n}$$\frac{\theta}{{x}_{i}^{2}}$. As $\theta$ is independent of i, we can factorize the joint pdf into two functions, one being $\theta^{n}$ and $\prod_{i=1}^{n}$$\frac{1}{{x}_{i}^{2}}$. From here I am not sure how to apply factorization theorem and what is the sufficient statistic. Is it n? If anyone can guide, i'll be grateful.

答:
The joint pdf should be
$$\prod_{i=1}^{n}\frac{\theta}{x_i^2} I_{\{x_i \ge \theta\}} = \theta^n I_{\{\min\{x_i\} \ge \theta\}}\prod_{i=1}^n \frac{1}{{x}_{i}^{2}}$$
where $I$ is the indicator function.

Therefore your sufficient statistic is $\min\{ x_i \}$.

Title: Sufficient Statistic-Estimation


问:
For $\varphi$ strongly minimal, tuples independent over $\varphi(M)$ are independent over $M$
I'm having a bit of trouble with Lemma 6.3.2. of Tent and Ziegler's model theory textbook. Here is the lemma and their proof, transcribed verbatum.

Lemma. Let $T$ be $\omega$-stable, $M\prec N$ models of $T$, $\varphi(v)$ be strongly minimal and $b_i\in\varphi(N)$. If the $b_i$ are independent over $\varphi(M)$, they are independent over $M$.

Proof: Assume that $b_1,\ldots,b_n$ are algebraically independent over $\varphi(M)$ but dependent over $\bar{a}\in M$. Put $\bar{b}=(b_1,\dots,b_n)$. An argument as in the proof of Theorem 5.5.2 shows that we may assume that $M$ is $\omega$-saturated. Let $p$ be the type of $\bar{b}$ over $M$. We choose a sequence $\bar{b}^0,\bar{b}^1,\dots$ in $\varphi(M)$ such that $\bar{b}^{2i}$ is an $n$-tuple of elements algebraically independent over $\bar{b}^0,\dots,\bar{b}^{2i-1}$ and $\bar{b}^{2i+1}$ realizes $p\upharpoonright_{\bar{a},\bar{b}^0,\dots,\bar{b}^{2i}}$. Let $q$ be the type of $\bar{a}$ over the set $B$ of elements of $(\bar{b}^i)$. Since the sequence $(\bar{b}^i)$ is indiscernible, every permutation $\pi$ of $\omega$ defines a type $\pi(q)$ over $B$. If $\{i:\pi(2i)\text{ even}\}=\{i:\pi'(2i)\text{ even}\}$, we have $\pi(q)\neq\pi'(q)$. So there are uncountably many types over $B$ and $T$ is not $\omega$-stable.

Unfortunately, a few points of this proof are unclear to me. First, I think I have a sketch of a proof that we may assume $M$ is $\aleph_0$-saturated; we essentially enrich our language by including a single unary predicate symbol, call it $U$, such that $U(N)=M$. There are sentences expressing that $U$ is a proper elementary substructure, and we may further form a set of formulas over free variables $v_1,\dots,v_n$ expressing that $(v_1,\dots,v_n)$ is algebraically independent over $\varphi(U)$, by taking formulas of the form $$\forall\bar{w}(\bigwedge_j(\varphi(w_j)\wedge{U(w_j)})\rightarrow((\exists^{\leqslant k}\bar{u}\theta(\bar{u},\bar{w}))\rightarrow\neg\theta(\bar{v},\bar{w})))$$ for all $k\in\omega$ and $\theta\in\operatorname{Form}(\mathcal{L})$. Then $(b_1,\ldots,b_n)$ satisfies these formulas in $N$, and so if we take the union of an elementary chain $(N_i)_{i\in\omega}$ such that $N_0=N$ and all types over finite subsets of $U(N_i)$ are realized in $U(N_{i+1})$, we obtain a new structure $N'$ such that $M':=U(N')$ is $\aleph_0$-saturated and $(b_1,\dots,b_n)$ is not algebraic over $\varphi(M')$. Clearly $(b_1,\dots,b_n)$ will still be dependent over $\bar{a}\in M'$, witnessed by the same formula as in $N$, and so since $M'\prec N'$ we have that all the hypotheses of the theorem are satisfied. So, first question, is this sketch the right idea?

It's at this point that I start to get confused. Constructing the sequence $(\bar{b}^i)$ seems clear to me; the case where $i$ is odd is immediate, and for the even case we can do a similar trick as above, taking the partial type $$\Sigma(\bar{v})=\{\bigwedge_{i=0}^n\varphi(v_i)\}\cup\{(\exists^{\leqslant k}\bar{u}\theta(\bar{u}))\rightarrow\neg\theta(\bar{v}):\theta\in\operatorname{Form}(\mathcal{L}\cup\{\bar{b}^{0},\dots,\bar{b}^{2i-1}\})\}_{k\in\omega}.$$ If this were not finitely satisfiable, there would be algebraic formulas $\theta_1,\ldots,\theta_m\in\operatorname{Form}(\mathcal{L}\cup\{\bar{b}^{0},\dots,\bar{b}^{2i-1}\})$ such that  $$M\models\forall\bar{v}(\bigwedge_{i=0}^n\varphi(v_i)\rightarrow\bigvee_{j=1}^m\theta(\bar{v})),$$ which would mean $\varphi(M)^n$ is finite, an obvious contradiction. Thus $\Sigma(\bar{v})$ is consistent and so realized in $M$. Is this right?

However, I have absolutely no idea why the sequence $(\bar{b}^i)$ is indiscernible. We wish to show that, for any $\mathcal{L}$-formula $\psi(\bar{v}_1,\dots,\bar{v}_m)$, and any $i_1<\dots i_m\in\omega$ and $j_1<\dots j_m\in\omega$, we have $$M\models\psi(\bar{b}^{i_1},\dots,\bar{b}^{i_m})\leftrightarrow \psi(\bar{b}^{j_1},\dots,\bar{b}^{j_m}).$$ Certainly it suffices to show the $\rightarrow$ implication, and we clearly need to divide into cases depending on whether $i_m$ and $j_m$ are odd or even. My attempt at this has been with induction on $m$; for instance, the case $m=0$ is vacuous. For the case $m=1$, we first consider the case where $i_1$ and $j_1$ are both odd. Then $$M\models \psi(\bar{b}^{i_1})\iff\psi(\bar{v})\in p\upharpoonright_{\emptyset}\iff M\models \psi(\bar{b}^{j_1}),$$ as needed.
Next we might consider the case where $i_1$ and $j_1$ are both even, but even here I seem to get stuck. We know that $M\models \psi(\bar{b}^{i_1})$ forces $\psi(\bar{v})$ to be non-algebraic, but this alone is certainly not enough to conclude that $M\models\psi(\bar{b}^{j_1})$. I imagine that we want to use strong minimality of $\varphi$ here somehow, to show that if $M\models\neg\psi(\bar{b}^{j_1})$ then $\bar{b}^{j_1}$ would be algebraic, but how can we exploit this when $\psi$ has more than one free variable? Indeed, $\bigwedge_{i=1}^n\varphi(v_i)$ will not be a strongly minimal formula when $n>1$ (for instance, consider its intersection with the formula $v_1=c$ for any $c\in \varphi(M)$, which is infinite and coinfinite), and I don't see an obvious way to "extract" a strongly minimal formula in $n$ free variables from a strongly minimal formula in $1$ free variable. So I'm not sure how to proceed even in the case $m=1$, let alone cases where $m>1$. I get even further stuck when $i_m$ and $j_m$ have different parities, as I can't see a way to link the construction of odd entries in the sequence with even entries in the sequence.

Furthermore, even if the $(\bar{b}^i)$ were indiscernible, why would this mean that $\pi(q)$ is a type? The only thing I can imagine they might mean by $\pi(q)$ is $$\{\psi(\bar{v},\bar{b}^{\pi(i_0)},\dots,\bar{b}^{\pi(i_n)}):\psi(\bar{v},\bar{b}^{i_0},\dots,\bar{b}^{i_n})\in q\},$$ but for general sequences of indiscernibles I can't see why this construction would possibly yield types. For instance, consider $(\mathbb{Q},<)$ and any infinite sequence of distinct elements $B\subseteq\mathbb{Q}$. If $b_1<b_2\in B$, then any type over $B$ will contain the sentence $b_1<b_2$, so the "type" induced by any permutation $\pi:B\rightarrow B$ that swaps $b_1$ and $b_2$ will contain the sentence $b_2<b_1$ and hence not be consistent. I'm clearly missing something here; what do they actually mean by $\pi(q)$?

答:
Q1: Yes, your sketch of why we may assume $M$ is $\aleph_0$-saturated is correct.

Q2: Yes, your construction of $\bar{b}^i$ when $i$ is even is correct. But there is a typo here in Tent and Ziegler: $\bar{b}^i$ should be chosen to be algebraically independent over $\bar{a}, \bar{b}^1,\dots,\bar{b}^{i-1}$, not just $\bar{b}^1,\dots,\bar{b}^{i-1}$ (so that $\text{tp}(\bar{a}/\bar{b}^i)$ contains a formula witnessing algebraic dependence of $\bar{b}^i$ over $\bar{a}$ when $i$ is odd, but no such formula when $i$ is even).

Q3: Why is the sequence $(\bar{b}^i)_{i\in \omega}$ indiscernible? Well, it comes down to the fact that in a strongly minimal set, for each $m$, there is a unique type satisfied by any algebraically independent $m$-tuple (this is Corollary 5.7.4 in Tent and Ziegler). Now for any $i_1<\dots<i_m$, we have that $\bar{b}^{i_1}\dots\bar{b}^{i_m}$ is an algebraically independent $(mn)$-tuple, so its type does not depend on the choice of $i_1<\dots<i_m$.

Note that the distinction between the odd and even cases has disappeared, since we're just asking for indiscernibility (over $\varnothing$), not indiscernibility over $\bar{a}$. If we forget about $\bar{a}$, each new $\bar{b}^i$ is just a tuple which is algebraically independent over $\bar{b}^1,\dots,\bar{b}^{i-1}$. In the even case, we explicitly chose it that way, and in the odd case, that's part of the type $p\restriction \bar{b}^1,\dots,\bar{b}^{i-1}$, since $\bar{b}^1,\dots,\bar{b}^{i-1}\in \varphi(M)$ and $\bar{b}$ is algebraically independent over $\varphi(M)$.

Some superfluous information: This kind of indiscernible sequence, where each element of the sequence is chosen to be independent over the previous elements, is called a Morley sequence. The concept of Morley sequence, and especially its generalizations outside the strongly minimal context (where we need to use more general notions of independence than algebraic independence - most commonly non-forking independence) are hugely important in model theory.

Q4: Your understanding of what is meant by $\pi(q)$ is correct. So why is $\pi(q)$ a type? The crucial thing that Tent and Ziegler have neglected to tell you here is that $(\bar{b}^i)_{i\in \omega}$ is not just an indiscernible sequence, it's an indiscernible set. This means that for any tuple $i_1,\dots,i_m$ of pairwise distinct elements of $\omega$, and any other tuple $j_1,\dots,j_m$ of pairwise distinct elements of $\omega$, $\text{tp}(\bar{b}^{i_1},\dots,\bar{b}^{i_m}) = \text{tp}(\bar{b}^{j_1},\dots,\bar{b}^{j_m})$. This is just the same as the definition of indiscernible sequence, but we no longer require $i_1,\dots,i_m$ and $j_1,\dots,j_m$ to be increasing tuples, i.e., we forget about the ordering of the sequence. An indiscernible set is also called a totally indiscernible sequence - see the definition at the beginning of Section 9.1 of Tent and Ziegler.

It follows that for any permutation $\pi$ of $\omega$, $\text{tp}((\bar{b}^i)_{i\in \omega})) = \text{tp}((\bar{b}^{\pi(i)})_{i\in \omega})$, and thus if $p$ is a consistent type over $(\bar{b}^i)_{i\in \omega}$, then $\pi(p)$ (which I would denote by $\pi_*p$) is also consistent (by the lemma on pushforward types, which we've already discussed here).

Ok, so why is $(\bar{b}^i)_{i\in \omega}$ an indiscernible set? The key point here is that algebraic independence is a symmetric notion: if $a_1,\dots,a_k$ is a tuple of algebraically independent elements, then so is $a_{\sigma(1)},\dots,a_{\sigma(k)}$ for any permutation $\sigma$ of $\{1,\dots,k\}$. So for any tuple $i_1,\dots,i_m$ of pairwise distinct elements of $\omega$, we have that $\bar{b}^{i_1}\dots\bar{b}^{i_m}$ is an algebraically independent $(mn)$-tuple, and this does not depend on the choice of tuple $i_1,\dots,i_m$ (and in particular not on its ordering).

Some more superfluous information:

There's a higher-level reason why $(\bar{b}^i)_{i\in \omega}$ is an indiscernible set: We're working in the context of an $\omega$-stable theory. And in any stable theory (a theory $T$ is stable if there is some cardinal $\kappa$ such that $T$ is $\kappa$-stable), every indiscernible sequence is an indiscernible set. In fact, this condition characterizes stability! See Lemma 9.1.1 and Exercise 9.1.5 in Tent and Ziegler.

Let $\varphi(\bar{a},\bar{y})$ be the formula witnessing algebraic dependence of $\bar{b}$ over $\bar{a}$. Note that because $(\bar{b}^i)_{i\in \omega}$ is an indiscernible set, it doesn't actually matter that the truth value of $\varphi(\bar{a},\bar{y})$ alternates on the sequence (true on odds, false on evens) - what matters is just that there are infinitely many $i$ such that $\varphi(\bar{a},\bar{b}^i)$ is true and infinitely many such that it is false. In fact, this is another characterization of stability: A theory $T$ is stable if and only if for every indiscernible sequence $(\bar{b}^i)_{i\in \omega}$ and every formula $\varphi(\bar{a},\bar{y})$, the set $\{i\in \omega\mid \varphi(\bar{a},\bar{b}^i)\}$ is finite or cofinite. So stability is characterized by a kind of minimality condition on indiscernible sequences.

Let's say we didn't know that $(\bar{b}^i)_{i\in \omega}$ was an indiscernible set. We could still get continuum-many types, we would just have to restrict ourselves to using order embeddings instead of permutations. Let $X\subseteq \omega$ be an arbitrary subset. Then there is an order embedding $\sigma_X\colon \omega\to\omega$ such that $\sigma^{-1}_X(\text{Odds}) = X$. We build $\sigma_X$ by induction: define  $\sigma_X(n)$ to be the least odd number greater than $\text{ran}(\sigma_X\restriction \{0,\dots,n-1\})$ if $n\in X$ and the least even number greater than $\text{ran}(\sigma_X\restriction \{0,\dots,n-1\})$ if $n\notin X$. Now we have an inverse order-isomorphism $\sigma_X^{-1}\colon \text{ran}(\sigma_X)\to \omega$, which induces a partial elementary map $(\bar{b}^i)_{i\in \text{ran}(\sigma_X)}\to  (\bar{b}^i)_{i\in \omega}$. Letting $p_X = \text{tp}(\bar{a}/(\bar{b}^i)_{i\in \text{ran}(\sigma_X)}$ and $q_X = (\sigma_X^{-1})_* p_X$, we have $\{i\in  \omega\mid \varphi(\bar{x},\bar{b}^i)\in q_X\} = X$, where again $\varphi(\bar{a},\bar{y})$ is the formula witnessing algebraic dependence of $\bar{b}$ over $\bar{a}$. So if $X\neq X'$, then $q_X \neq q_{X'}$, and we get continuum-many types. Note that this time (in the construction of $\sigma_X$) it was actually important that the truth value of $\varphi(\bar{a},\bar{y})$ alternated on the sequence. It turns out that for a complete theory $T$, the existence of an indiscernible sequence $(\bar{b}^i)_{i\in \omega}$ and a formula $\varphi(\bar{a},\bar{y})$ such that $\varphi(\bar{a},\bar{b}^i)$ is true if and only if $i$ is odd characterizes $T$ having the independence property. Theories without the independence property (called NIP theories for "Not the Independence Property") are a class of theories which contain the stable theories but also theories with definable orders, like the theory DLO of dense linear orders without endpoints or the theory RCF (real closed fields) of $\mathbb{R}$ as a field.


问:
Number of all possible non-redundant comparisons that can be formed in a set of 12 elements
We have a set of 12 elements (A, B, C, D, E....).
I would like to know the number of ways to compare them in a non-redundant way.
For example, some comparisons could be:
A vs others;
B vs others;
A and B vs others;
A , B, E vs others;
A vs B vs E vs others...
So on until all possibilities are exhausted
Logically, redundant comparisons must be removed (A vs others equals others vs A).
Is there any formula for this?</p>

答:
The answer is $2^{12} - 1$. For a given set with $n$ elements, there are $2^n$ subsets (plus the empty set).

correction: $2^{12} - 12 - 1$ since you're not comparing elements with themselves.


问:
Asymptotic behavior of $\sum\limits_{n \leq x}\frac{\phi(n)}{n}$
I'm working with the asymptotic behavior of square-free integers and have already proven that if $f$ is an arithmetic function with $F(n) := \sum\limits_{d \mid n}f(d)$, then
\begin{align*}
\sum\limits_{n \leq x}F(n)=x\sum\limits_{d \leq x}\frac{f(d)}{d}+O\left(\sum\limits_{d \leq x}|f(d)|\right).
\end{align*}

I'm trying to use this fact to prove that $\sum\limits_{n \leq x}\frac{\phi(n)}{n}=\frac{6}{\pi^2}x+O(\log{x})$.

I know that $\phi=\mu \ast E$ where $\ast$ is the usual Dirichlet convolution $f \ast g(n)=\sum\limits_{d \mid n}f(d)g\left(\frac{n}{d}\right)$, so I can write
\begin{align*}
    \phi(n)=\sum\limits_{d \mid n}\mu(d)E\left(\frac{n}{d}\right)=\sum\limits_{d \mid n}\mu(d)\frac{n}{d}=n\sum\limits_{d \mid n}\frac{\mu(d)}{d},
\end{align*}
leading me to conclude that $\frac{\phi(n)}{n}=\sum\limits_{d \mid n}\frac{\mu(d)}{d}$.

Using my identity from before, I can write
\begin{align*}
    \sum\limits_{n \leq x}\frac{\phi(n)}{n}&=x\sum\limits_{d \leq x}\frac{\frac{\mu(d)}{d}}{d}+O\left(\sum\limits_{d \leq x}\left|\frac{\mu(d)}{d}\right|\right) \\
    &=x\sum\limits_{d \leq x}\frac{\mu(d)}{d^2}+O\left(\sum\limits_{d \leq x}\left|\frac{\mu(d)}{d}\right|\right) \\
&=x\sum\limits_{d \geq 1}\frac{\mu(d)}{d^2}-x\sum\limits_{d \geq x}\frac{\mu(d)}{d^2}+O\left(\sum\limits_{d \leq x}\left|\frac{\mu(d)}{d}\right|\right) \\
&=x\cdot\frac{1}{\zeta(2)}-x\sum\limits_{d \geq x}\frac{\mu(d)}{d^2}+O\left(\sum\limits_{d \leq x}\left|\frac{\mu(d)}{d}\right|\right) \\
&=\frac{6}{\pi^2}x-x\sum\limits_{d \geq x}\frac{\mu(d)}{d^2}+O\left(\sum\limits_{d \leq x}\left|\frac{\mu(d)}{d}\right|\right).
\end{align*}

My problem is that I don't understand how to get the $O(\log{x})$. Can someone walk me through this?

答:
First $\displaystyle\sum_{d\leqslant x}\left|\frac{\mu(d)}{d}\right|\leqslant\sum_{d\leqslant x}\frac{1}{d}=\mathcal{O}(\log x)$. On the other hand,
$$ \left|\sum_{d>x}\frac{\mu(d)}{d^2}\right|\leqslant\sum_{d>x}\frac{1}{d^2}=\mathcal{O}\left(\int_x^{+\infty}\frac{dt}{t^2}\right)=\mathcal{O}\left(\frac{1}{x}\right) $$
which finally leads to $\sum_{n\leqslant x}\frac{\varphi(n)}{n}=\frac{6}{\pi^2}x+\mathcal{O}(\log x)$.


问:
Basic result on convex sets
I am trying to understand the concept of the support function of convex sets. I want to verify if the following result is correct.

Let $w(\cdot)$ be the support function of a convex set $C\in
\mathbb{R}^{n}$. Suppose $x\not\in C$, let $s\in C$ be the closest
point to x contained in C. Let $\vec{d}=\vec{Ox}-\vec{Os}$. Then, it would be correct to say $w(\vec{d})=\vec{Os}\cdot \vec{d}$?, if it is correct, why?

答:
Your intuition has not lead you astray here; yes this condition holds.

The classic result behind it is sometimes known as Kolmogorov's inequality. Specifically, suppose $C$ is a closed, non-empty, convex subset of $\Bbb{R}^n$ (though the result holds in any real Hilbert space). Suppose $x \in \Bbb{R}^n$ and $z \in C$. Then $z$ is the closest point in $C$ to $x$ if and only if
$$(x - z) \cdot (c - z) \le 0$$
for all $c \in C$.

Why does this hold? I'm not going into a full proof here, but it means that the vector from $z$ to $x$ and the vector from $z$ to any other $c \in C$ must be an obtuse (or possibly right) angle. If you think about it, the angle can't ever be an acute angle, because if this were the case, then moving along the line segment between $z$ and $c$ (which is contained in $C$), you will get points that are strictly closer to $x$ than $z$. This would contradict $z$ being the nearest point to $x$ in $C$.

So, if $w$ is the support function of $C$, then we have
$$w(x - z) = \sup_{c \in C} (x - z) \cdot c.$$
Kolmogorov's inequality implies
$$(x - z) \cdot c \le (x - z) \cdot z$$
for all $c \in C$. Since $z \in C$ too, this means $(x - z) \cdot c$ achieves its maximum over $c \in C$ at $c = z$. Thus,
$$w(x - z) = (x - z) \cdot z,$$
as expected.

Title: Basic result on convex sets


答:
If $x^2-2x=x(x-2)$ divides $x^5-bx=x(x^4-b)$, it means $2$ is a root of $x^4-b$, whence $b=2^4$.

Indeed, you easily check that
$$x^5-16x=(x^2-2x)(x^3+2x^2+4x+8).$$

Title: Polynomial Long Division:

Body:
My question essentially boils down to this: (it was part of a question about polynomial ring ideals)

Find an integer $b$ such that the rational function
\begin{equation}
\frac{x^5-bx}{x^2-2x}\in \mathbb{Z}[x]
\end{equation}
I've tried plugging in a bunch of values for $b$ using a calculator but none of them have come out to whole polynomials. Is there an easier way to do this?


问:
$\prod (1+a_k)$ converges $\Leftrightarrow \sum a_k$ converges if $a_k\geq 0$ or $a_k\leq 0$ for each k.
Basically the title. I'mo working on a proof right now and I'm not quite sure where to begin. I thought about writing out the $n$th partial product and then taking logarithms but that doesn't get rid of the $1$ in each of the terms. Just a nudge in the right direction would be very helpful. Basically I have to prove that if $a_k\leq 0$ or $a_k\geq 0$ then:

$$\prod_k^\infty c_k=\prod_k^\infty(1+a_k)=C\Leftrightarrow \sum_k^\infty a_k=S$$

Where $C$ is the limit of the $n$th partial product and $S$ is the limit of the $n$th partial sum.

Body: Hint: $\displaystyle\log\left(\prod_{k=1}^N(1+a_k)\right)=\sum_{k=1}^N\log(1+a_k)$. Besides, $\displaystyle\lim_{k\to\infty}a_k=0\implies\lim_{k\to\infty}\frac{\log(1+a_k)}{a_k}=1$.


问:
2 identities of Shanks from Apostol's Book
I am trying exercises of Ch-14 partitions from Tom Apostol Introduction to ANT and unable to Solve (a) part of Question 5.

5. If $x\ne 1$ let $Q_0(x)=1$ and for $n\ge 1$ define
$$ Q_n(x) = \prod_{r=1}^n \frac{1-x^{2r}}{1-x^{2r-1}}. $$
(a) Derive the following finite identities of Shanks:
$$ \sum_{m=1}^{2n} x^{m(m-1)/2} = \sum_{s=0}^{n-1} \frac
   {Q_n(x)}{Q_s(x)} x^{s(2n+1)}, $$
$$ \sum_{m=1}^{2n+1} x^{m(m-1)/2} = \sum_{s=0}^{n} \frac
   {Q_n(x)}{Q_s(x)} x^{s(2n+1)}. $$
(b) Use Shank's identities to deduce Gauss' triangular-number
theorem:
$$ \sum_{m=1}^\infty x^{m(m-1)/2} = \prod_{n=1}^\infty
   \frac{1-x^{2n}}{1-x^{2n-1}} \text{ for } |x|<1. $$

I have question in proof of (a) as I am unable to derive any of the identity. I tried by evaluating $Q_n(x) / Q_s(x) $ and if I multiply it by $ x^{s(2n+1) }$ I don't see anything to get LHS in (I).

So, kindly shed some light on how to prove it by proving any of the identity.

I shall be really thankful for your help.

Original image: https://i.stack.imgur.com/kUJbT.jpg

答:
Main theme: We set  for $n\geq 1$
\begin{align*}
A_n(x)=\sum_{s=0}^{n-1}\frac{Q_n(x)}{Q_s(x)}x^{s(2n+1)}\qquad\qquad
S_n(x)=\sum_{m=1}^{2n}x^{\frac{m(m-1)}{2}}\\
\end{align*}
and find a representation of $A_n(x)$ via $A_{n-1}(x)$ by also separating terms which are summands from $S_n(x)$.
\begin{align*}
\color{blue}{A_n(x)}&=A_{n-1}(x)+x^{(n-1)(2n-1)}+x^{n(2n-1)}\\
&\,\,\color{blue}{=A_{n-1}(x)+S_{n}(x)-S_{n-1}(x)}\tag{1}
\end{align*}
We so derive by induction
\begin{align*}
A_n(x)-S_n(x)&=A_{n-1}(x)-S_{n-1}(x)=\cdots\\
&=A_{1}(x)-S_{1}(x)\\
&=\frac{1-x^2}{1-x}-(1+x)=0
\end{align*}
and the claim follows.

Proof: The derivation of (1) is elementary but technically tricky.

We start with an algebraic identity. The following holds:
\begin{align*}
\left(1-x^{2n}\right)x^{s(2n+1)}&=\left(1-x^{2n-1}\right)x^{s(2n-1)}+\left(1-x^{2s+1}\right)x^{(s+1)(2n-1)}\tag{2}\\
&\qquad-\left(1-x^{2s}\right)x^{s(2n-1)}
\end{align*}

The identity (2) can be shown easily my multiplying out and comparing corresponding terms. The idea is to represent the left-hand side by an expression where $n$ is consequently replaced with $n-1$. This will be used as building block to obtain the sums in (1). Division of (2) by $1-x^{2n-1}$ gives
\begin{align*}
\frac{1-x^{2n}}{1-x^{2n-1}}x^{s(2n+1)}&=x^{s(2n-1)}+\frac{1-x^{2s+1}}{1-x^{2n-1}}x^{(s+1)(2n-1)}\tag{3}\\
&\qquad-\frac{1-x^{2s}}{1-x^{2n-1}}x^{s(2n-1)}
\end{align*}

Recalling $Q_n(x) = \prod_{r=1}^n \frac{1-x^{2r}}{1-x^{2r-1}}$ we have $$Q_n(x)=Q_{n-1}(x)\frac{1-x^{2n}}{1-x^{2n-1}}$$ We obtain from (3) by expanding with $\frac{Q_{n-1}(x)}{Q_s(x)}$:
\begin{align*}
\frac{Q_n{x)}}{Q_{s}(x)}x^{s(2n+1)}&=\frac{Q_{n-1}(x)}{Q_{s}(x)}x^{s(2n-1)}
+\underbrace{\frac{1-x^{2s+1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{(s+1)(2n-1)}}_{\alpha_{s,n}(x)}\\
&\qquad-\underbrace{\frac{1-x^{2s}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}}_{\beta_{s,n}(x)}\tag{4}
\end{align*}

Intermezzo:

We obtain for $s=0,\ldots,n-2$:
\begin{align*}
\color{blue}{\beta_{s+1,n}(x)}&=\frac{1-x^{2s+2}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_{s+1}(x)}x^{(s+1)(2n-1)}\\
&=\frac{1-x^{2s+2}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}\,\frac{1-x^{2s+1}}{1-x^{2s+2}}x^{(s+1)(2n-1)}\\
&=\frac{1-x^{2s+1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{(s+1)(2n-1)}\\
&\,\,\color{blue}{=\alpha_{s,n}(x)}\tag{5}
\end{align*}

We also have $\beta_{0,n}(x)=0$ and
\begin{align*}
\alpha_{n-1,n}(x)=\frac{1-x^{2n-1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_{n-1}(x)}x^{n(2n-1)}=x^{n(2n-1)}\tag{6}
\end{align*}

Summing (4) from $s=0$ to $n-1$ we obtain with (5) and (6)
\begin{align*}
\color{blue}{A_n(x)}&=\sum_{s=0}^{n-1}\frac{Q_{n}(x)}{Q_s(x)}x^{s(2n+1)}\\
&=\sum_{s=0}^{n-1}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+\sum_{s=0}^{n-1}\alpha_{s,n}(x)-\sum_{s=1}^{n-1}\beta_{s,n}(x)\tag{7}\\
&=\sum_{s=0}^{n-1}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+\sum_{s=0}^{n-1}\alpha_{s,n}(x)-\sum_{s=0}^{n-2}\beta_{s+1,n}(x)\tag{8}\\
&=\sum_{s=0}^{n-1}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+\alpha_{n-1,n}(x)\tag{9}\\
&=\sum_{s=0}^{n-2}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+x^{(n-1)(2n-1)}+x^{n(2n-1)}\tag{10}\\
&\,\,\color{blue}{=A_{n-1}(x)+S_{n}(x)-S_{n-1}(x)}
\end{align*}
which proves the claim. The other identity in (a) follows by adding the term with $s=n$ to both sides of the first identity.

Comment:

In (7) we use the identity (4) and sum from $s=0$ to $s=n-1$. We set the lower index of the right-most sum to $s=1$ since $\beta_{0,n}(x)=0$.

In (8) we shift the index of the right-most sum by one to start with $s=1$.

In (9) we use the telescoping property $\alpha_{s,n}(x)-\beta_{s+1,n}(x)=0$ for $s=0,\ldots, n-2$.

In (10) we separate the term with $s=n-1$ from the sum and use the identity (6).

Note: This proof was given by D. Shanks in Two Theorems of Gauss in 1958.


问:
Show $|f(x + \delta_n) - f(x)| < \epsilon_n$ for $\epsilon_n$ independent of $x$
For a sequence $\delta_n \rightarrow 0$ I need to show that there exists another sequence $\epsilon_n \rightarrow 0$ such that:

$$\sup_x|f(x + \delta_n) - f(x)| < \epsilon_n$$

What is required for this property to hold? Does it have a name?

If this did not have the supremum over $x$, I can find such an $\epsilon_n$ if the left and right limits are equal at all $x$. Is it sufficient that $f$ is bounded on it's domain? Is that too strong?

答:
Here is why uniform continuity is enough:

Let $\epsilon_n = 2 \sup | f (x+\delta_n) - f x |$.

This obviously satisfies the first condition you want. So we just need to show that $\epsilon_n \to 0$.

Let $\epsilon^* > 0$. By uniform continuity, we can find a $\delta^*$ so that whenever $\delta_n < \delta^*$, $\sup | f(x+\delta_n) - f x | < \epsilon^*$.

Then since $\delta_n \to 0$, we know $\delta_n < \delta^*$ for $n$ large enough.

But this means, for $n$ large enough, $\epsilon_n = 2 \sup | f(x+\delta_n) - f x | < 2 \epsilon^*$.

So $\epsilon_n \to 0$.

I hope this helps ^_^


问:
solution without using the L'Hopital
I want to complete the solution without using the L'Hopital. I got here

$\lim _{x \rightarrow \infty} \frac{\log \left(x^{4}+1\right)+1}{\log \left(x^{3}+2\right)-3}$

$\lim _{x \rightarrow \infty}\left(\frac{\log \left(x^{4}\left(1+\frac{1}{x^{4}}\right)\right)+1}{\log \left(x^{3}\left(1+\frac{2}{x^{3}}\right)\right)-3}\right)$

$\lim _{x \rightarrow \infty}\left(\frac{\log \left(x^{4}\right)+\log \left(1+\frac{1}{x^{4}}\right)+1}{\log \left(x^{3}\right)+\log \left(1+\frac{1}{x^{3}}\right)-3}\right)$

答:
Your are almost done :

$\lim _{x \rightarrow \infty}\left(\frac{\log \left(x^{4}\right)+\log \left(1+\frac{1}{x^{4}}\right)+1}{\log \left(x^{3}\right)+\log \left(1+\frac{1}{x^{3}}\right)-3}\right)$

$= \lim _{x \rightarrow \infty} \frac{4log(x)}{3log(x)}$

$=\frac{4}{3}$

Title: solution without using the L'Hopital


问:
Is there a Divisibility Metric for Numbers?
Both prime numbers and highly divisible numbers have a common characteristic: divisibility. The former are divisible by as few lower numbers as possible, and the latter by as many as possible, like two poles on a scale.
I'm interested into fitting all the other non-prime and non-highly-divisible whole numbers into such a scale too.

Any suggestions for creating a formula that translates whole numbers to the range of [0,1], where prime numbers result in 0 and highly divisible numbers in 1, and all the other numbers in between?

Have there already been attempts to do this?

答:
Well, you can look at the Prime Omega function : $\Omega(n)$ maps to the number of prime factors of n. This is not a function that goes from 0 to 1, but such a function is not possible I think.

It is impossible for one major reason : Take, for exemple the number $n!$, it has as many prime factors as you want (you just have to increase $n$). So, if you create a number $\psi \in \mathbb{N}$ such that $f(\psi)=1$ where $f$ is the function of your dreams. Then what is $f(\psi!)$ ? Knowing that $\psi!$ has much, much more prime factors than $\psi$, we will have $f(\psi!)>f(\psi)=1$ so your function isn't bounded.

Now, if what you want is a function that say "how much" a prime is divisible, you can take $k(n) = (\Omega(n)-1)/n$ which is very interesting too ! This function measure how much a number is divisible compared to the value of this number.

If $p$ is a prime number, then $k(p)=0$ and if $k(n)=1$ for some $n$, then it means $n$ is extremely divisible. In fact, this number is so divisible, that it has more factors than its own value, which I think is totally impossible.

The more $n$ is divisible compared to its own value, the more $k(n)$ will be high.

Edit : notice that $k(1) = -1$ because $1$ is not a prime number but it has many properties of it, for exemple, it is divisible by $1$ and itself, but no other number. $1$ is a very special case.


问:
Extension of Mean Value Property of holomorphic function from single interval on a boundary circle to the double integral on the whole disc.
The following is the mean value property of holomorphic function:

Let $f(z)$ be a holomorphic function on the disc $D(z_{0},R)$, then $$f(z_{0})=\dfrac{1}{2\pi}\int_{0}^{2\pi}f(z_{0}+re^{i\theta})d\theta,\ \ \text{for all}\ \ 0<r<R.$$

Basically it tells us the value of the holomorphic function at the center of the disc is the average mean of it on the boundary circle of any smaller disc.

I am wondering if we can extend this to a double integral, something like $$f(z_{0})=C\int_{D(z_{0}, R)}f(x+iy)dxdy=C\int_{0}^{2\pi}\int_{0}^{R}f(z_{0}+re^{i\theta})rdrd\theta,$$ for some constant $C$.

The reason I want something like this because I want to prove that if $f(z)$ is holomorphic function on $D(z_{0}, R)$, then $$|f(z_{0})|\leq \dfrac{C}{R}\Big(\int_{D(z_{0}, R)}|f(x+iy)|^{2}dxdy\Bigg)^{\frac{1}{2}}.$$ I have tried several method like using the power series expansion of $f\overline{f}$, which kinds of work, but somehow in the end you need to argue that you can interchange $\int_{0}^{R}$ with the infinite series, and I am not sure after so many manipulations if I can still argue the interchange by uniform and absolute convergence.

After several try, I thought perhaps working on series is not really a good approach, so I tried to use the Mean-Value property, but it only has a single integral.

If we can have a double integral like above, then I can directly replace $f:=f^{2}(z)$, and take square root.

Thank you!

答:
Thanks to Conrad's inspiring suggestion (I was being really dumb), here is the solution.

I want to prove

If $g$ is holomorphic on $D(b,\epsilon)$, then $$|g(b)|=\dfrac{C}{\epsilon}\Bigg(\int_{D(b,\epsilon)}|g(x+iy)|^{2}dxdy\Bigg)^{\frac{1}{2}}.$$

As $g$ is holomorphic on $D(b,\epsilon)$, then so is $g^{2}(z)$. Hence, by the Mean-Value Property, we have $$g^{2}(b)=\dfrac{1}{2\pi}\int_{0}^{2\pi}g^{2}(b+re^{i\theta})d\theta.$$ Hence, $$\int_{D(b,\epsilon)}g^{2}(x+iy)dxdy=\int_{0}^{\epsilon}r\int_{0}^{2\pi}g^{2}(b+re^{i\theta})d\theta dr=\int_{0}^{\epsilon}r\cdot 2\pi g^{2}(b)dr=\pi\epsilon^{2}g^{2}(b).$$

Hence, $$g^{2}(b)=\dfrac{1}{\pi\epsilon^{2}}\int_{D(b,\epsilon)}|g(x+iy)|^{2}dxdy$$ and thus $$|g(b)|=\dfrac{C}{\epsilon}\Bigg(\int_{D(b,\epsilon)}|g(x+iy)|^{2}dxdy\Bigg)^{\frac{1}{2}},$$ where $C=\frac{1}{\pi}$.

In particular, the Mean-value property can be extended to a double integral:  $$\int_{D(z_{0},R)}f(x+iy)dxdy=\int_{0}^{R}r\int_{0}^{2\pi}f(z_{0}+re^{i\theta})d\theta dr=2\pi f(z_{0})\int_{0}^{R}rdr=\pi R^{2}f(z_{0}).$$


问:
What is the sign of the integral $\int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx$?
Let $I_n = \int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx$ for a natural $n$.
I would like to prove that for $n$ odd this integral vanishes, while for $n = 4k$ for some natural $k$, the integral is always positive and for $n = 4k+2$ it is always negative.

I was able to do a few integrations by parts, and provided I didn't make mistakes, the result was the following rather ugly recurrence relation, valid for $n > 1$: $$I_{n+2} = -\frac1{n-1}\left[(4n^3-2n)I_n+(n+1)I_{n-2}\right].$$ Also we have $I_1 = I_3 = 0$ (the first is obvious from the integral, the second follows by integration by parts). And from here, clearly it is immediate to see that it vanishes for every $n$ odd, just by induction. However I'm not able to see why $I_{4k}$ is always positive or $I_{4k+2}$ is always negative from this relation. What I know is that $|I_n| \le I_0 \approxeq 7.95$.

I also tried to rewrite the integral using $e^x = 1 + x + \ldots$, moving the integral inside the summation, but still I couldn't come up with a solution. What do you suggest?

Note: this integral comes from the calculation of the Fourier coefficients of $e^{\sin(x)}$, and analogous properties hold for the coefficients of the sine terms. Also by a well-known theorem (Riemann-Lebesgue), $I_n$ goes to zero as $n\to\infty$, and hence it's not possible to find a fixed constant lower bound for $|I_n|$, making any estimation in the recurrence relation more difficult.

答:
If you just want to find the sign of the integral, you can do it directly as follows:

the substitution $x \to -x$ shows that

$\int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx=\int_{0}^{2\pi}e^{-\sin(x)}\cos(nx)\,dx$ but now for odd $n$ the substitution $x \to x+\pi$ sends $\int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx$ to $-\int_{0}^{2\pi}e^{-\sin(x)}\cos(nx)\,dx$ hence the integral is zero.

Using the well known (and easy to prove fact) that $\cos^q x=\sum c_{k,q}\cos kx, c_{kq} \ge 0, c_{q,q}>0$ it immediately follows that $\int_{0}^{2\pi}e^{\cos(x)}\cos(nx)\,dx >0$

(edited later - this follows by an easy induction since $2\cos a \cos b=\cos(a+b)+\cos(a-b)$ hence the coefficients stay non-negative if they start so; we have $\cos^2x=(\cos 2x+1)/2$ and then $\cos^3x=\cos x (\cos 2x+1)/2=...$ and all the terms remain non-negative when expanded, while the highest order term in $\cos^q x$ is clearly $\cos qx$ with coefficient $1/2^{q-1}$ since that appears only once in the inductive computation with half the previous highest term coefficient; then using that $e^{\cos x}=\sum \frac{\cos^qx}{q!}$ with uniform convergence in $[0, 2\pi]$ we can integrate term by term and all the coefficients of $\cos nx$ that appear in the expansions are non-negative while at least the one from $\cos^n x$ is positive, so the integral is positive by orthognality of the cosines)

But the substitution $x \to \pi/2-x$ sends $\int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx$ into $(-1)^{n/2}\int_{0}^{2\pi}e^{\cos(x)}\cos(nx)\,dx$ for even $n$ hence the required sign result follows.


问:
Expected Value of (1/X) for Poisson X for x>0
I know how to calculate the E(1/(x+1) for Poisson X, but it gets more complicated when it is just E(1/X). Does the answer reduce to a closed form, or is it an infinite sum?</p>

答: $\Pr[X = 0] = e^{-\lambda} > 0$, therefore $$\operatorname{E}[1/X] = \sum_{x=0}^\infty \frac{1}{x} \Pr[X = x] = \frac{1}{0} e^{-\lambda} + \sum_{x=1}^\infty \frac{1}{x} \Pr[X = x] = \infty.$$


问:
Domain of composite of function and its inverse
I'm looking for the domain of definition of the  composite function of one function and its inverse.

Let $$H(x)= \frac{e^x -1}{e^x +1} $$ and its inverse $$G(x)=\ln{ \frac{x+1}{1-x}}$$

Is $-1 <x <1 $ the domain of definition of $H(G(x))$ ?

What is weird is that by going through the computer algebra tool Mathematica, I get $\mathbb{R}$ as the domain.
I think the tool does a simplification and gets the identity function, and that explains why I get $\mathbb{R}$ as the domain.
Am I right?

The graph of $$ \frac{x+1}{1-x}: $$

The graph of $H(x)$ in blue and $G(x)$ in orange:

答:
This is one of those places in mathematics where the language gets a little muddy, and where authors often use words differently.  From reading your question I can infer some things, but it would not be unreasonable to consult the source from which you are working.

Usually, we require three data when defining a function:  (1) a domain, (2) a codomain, and (3) a (unique) pairing of each element in a the domain to some element of the codomain.  However, as a practical matter, we often elide the an explicit statement of the domain and/or codomain, and take the domain to be the largest set in our "universe" for which the formula makes sense.  This gives us a domain of definition.

In this case, it appears that you are working over $\mathbb{R}$.  Thus both $G$ and $H$ are functions which act on subsets of the real numbers.  The function $H$ is defined by the formula
$$ H(x) = \frac{\mathrm{e}^x-1}{\mathrm{e}^x + 1}. $$
This formula gives a well-defined real number for any real number $x$, and so we say that the domain of definition for $H$ is $\mathbb{R}$.

The function $G$ is defined by the formula
$$ G(x) = \ln\left( \frac{x+1}{1-x} \right). $$
This function does not give a well-defined real number for all real numbers $x$.  Indeed, if $(x+1)/(x-1) \le 0$, then the logarithm is undefined.  Thus the domain of definition for $G$ cannot contain any values of $x$ for which this fraction is nonpositive.  Some quick algebra shows that
$$ \frac{x+1}{x-1} > 0 \iff x \in (-1, 1). $$
For any such $x$, the formula defining $G$ gives a well-defined real number, and so the domain of definition for $G$ is $(-1,1)$.

Finally, given two function $f$ and $g$ defined by formulae, the domain of definition of the composition $f\circ g$ consists of all of the points in the domain of definition of $g$, minus those points that are not mapped into the domain of definition of $f$.  That is,
$$ D(f\circ g) = D(g) \setminus \{ x\in D(f) \mid g(x) \not\in D(f) \}. $$
For the given functions $G$ and $H$,
$$ D(H\circ G) = D(G) \setminus \underbrace{\{ x\in D(G) \mid G(x) \not\in D(H) \}}_{\text{$=\varnothing$, see below}} = (-1,1)\setminus \varnothing = (-1,1).$$
The indicated term is the empty set, as there are no real values of $x$ (in the domain of $G$ or otherwise) which are not in the domain of $H$.

In other words, $H\circ G$ is only defined on the interval $(-1,1)$, and so this interval is the domain of definition of $H\circ G$.  On this interval, it is the identity function, i.e. $(H\circ G)(x) = x$.

Observe that if we define a function $F$ by $F(x) = x$, then this function has $\mathbb{R}$ as its domain of definition, because the defining formula is well-defined for all real inputs.  However, $F$ and $H\circ G$ are not the same function:  $F$ is defined by a formula with its domain understood implicitly, while $H\circ G$ is defined by the composition of two functions with previously understood domains of definition.  If you are using a computer algebra system (CAS) to compute the domain of the composition, it is very likely that the CAS is first simplifying the resulting formula for the composition, and then determining the domain.  This highlights how careful you need to be when working with computers.


问:
Lebesgue integral iteration with the limit
For every n, let $f_{n}:[0;1]\rightarrow{\mathbb R}$ a measurable Lebesgue function:  $\int_{[0;1]} {\left| f_{n}\right\| ^{2}}\mathrm{d}m\leq{5}
$

Is true that $\lim\limits_{n\to\infty} \int_{[0;1]} {\left| f_{n}\right\| ^{2}}\mathrm{d}m=0$? Is true that $\lim\limits_{n\to\infty} \int_{[0;1]} {\left| f_{n}\right\|}\mathrm{d}m=0$?

答:
No.

Consider $f_n(x) = \sqrt{5}$ for all $x$ for all $n$.

Then $\int |f_n|^2 \mathrm{d}m \leq 5$, but both
$\lim \int |f_n|^2 \mathrm{d}m$ and $\lim \int |f_n| \mathrm{d}m$
are nonzero.

I hope this helps ^_^


问:
What is the probability that an event will happen when the probability decreases exponentially?
So I am just a middle student who started studying probability in my stats class and I see the questions on this site are little advanced so I don't hope this one is too basic.

Anyways lets say that the probability of an event happening was some number(like 1%), and each time a trial is run the of the event happening decreases by half(so 1% probability for the first trial, .5% probability for the second, .25% probability, and so on).
What is the probability that the event will happen after 1 trial, 10 trials, 100 trials, and an infinite amount of trials?

答:
An interesting problem, that has many practical applications.

Moreover, considering a physical model will help to fix clearly the universe of events which might be bewildering
when dealing with probability of infinite sets.

We can think in fact of having a quantity of objects (radioactive nuclei, reacting molecules, mutating viruses, etc.) of type  $A$, which
decay (transform, mutate,..) into objects of type $B$ at a constant rate, so that
$$
\eqalign{
  & A(t) = A(0)\,e^{\, - \,t/T} \quad A(0) + B(0) = A(t) + B(t)  \cr
  & B(t) = B(0) + A(0) - A(t) = B(0) + A(0)\left( {1 - \,e^{\, - \,t/T} } \right) \cr}
$$

Starting with a given concentration at time $t=0$, then the concentration will vary in time as
$$
c_{\,0}  = {{A(0)} \over {A(0) + B(0)}}\quad  \Rightarrow \quad c(t) = c_{\,0} \,e^{\, - \,t/T}
$$
Now we "scan" each single object every $\Delta t$ and ask which is the probability of
revealing an $A$ object at scan $n$, letting $n$ to start from $0$.

Let's denote with $\overline P \left( n \right)$ the cumulative probability of revealing only type $B$ till  scan $n$ incluse, then

$$
\eqalign{
  & \overline P \left( n \right)
 = \prod\limits_{k = 0}^n {\left( {1 - c_{\,0} e^{\, - \,k\,\Delta t\,/\,T} } \right)}
  = \prod\limits_{k = 0}^n {\left( {1 - c_{\,0} e^{\, - \,\,k/K} } \right)}  =   \cr
  &  = \prod\limits_{k = 0}^n {\left( {1 - e^{\, - \,\,\left( {k + a} \right)/K} } \right)}
  = \prod\limits_{k = a}^{n + a} {\left( {1 - e^{\, - \,k/K} } \right)}
  = {{\prod\limits_{k = a}^{n + a} {\left( {e^{\,k/K}  - 1} \right)} }
 \over {\prod\limits_{k = a}^{n + a} {e^{\,k/K} } }} \cr}
$$
and taking the logarithm
$$
\ln \overline P \left( n \right)
 = \sum\limits_{k = a}^{n + a} {\ln \left( {1 - e^{\, - \,k/K} } \right)}
 \quad  \Rightarrow \quad \left\{ \matrix{
  f(x) =  - \ln \left( {1 - e^{\, - \,x/K} } \right) \hfill \cr
   - \ln \overline P \left( n \right) = \sum\limits_{x = a}^{n + a}
 {f(x)}  \hfill \cr}  \right.
$$
where we changed the sign of the log so that $f(x) \; |\,0<x$ is positive, monotonically decreasing and convex.

Now, unfortunately, there is not a closed form for the product or sum.

The series development of
$$
P(x,n) = \prod\limits_{k = 1}^n {\left( {1 - x^{\,k} } \right)}
$$
is quite unmanageable : see OEIS sequence A231599.

However we can squeeze the logarithm into two inequalities
$$
{1 \over 2}\left( {f(a) + f(a + n)} \right) + \int_{x = a}^{n + a} {f(x)dx}
  < \sum\limits_{x = a}^{n + a} {\,\,f(x)}
  < \int_{x\, = \,a - 1/2}^{n + a + 1/2} {f(x)dx}
$$
by comparing the continuous integral with the discrete sum, represented with trapezia
and rectangles as shown in the sketch.

The indefinite integral can be expressed through the Polylogarithm as
$$
\int {f(x)dx}  =  - K\int {\ln \left( {1 - e^{\, - \,x/K} } \right)d\left( {x/K} \right)}
  =  - K\,{\rm Li}_2 \left( {e^{\, - \,x/K} } \right)
$$

Therefore the inequality becomes
$$
\eqalign{
  & {1 \over {2K}}\left( { - \ln \left( {1 - e^{\, - \,a/K} } \right)
 - \ln \left( {1 - e^{\, - \,\left( {a + n} \right)/K} } \right)} \right)
 - {\rm Li}_2 \left( {e^{\, - \,\left( {a + n} \right)/K} } \right)
 + \,{\rm Li}_2 \left( {e^{\, - \,a/K} } \right)  \cr
  &  <  - {{\ln \overline P \left( n \right)} \over K} <   \cr
  &  - {\rm Li}_2 \left( {e^{\, - \,\left( {n + a + 1/2} \right)/K} } \right)
 + {\rm Li}_2 \left( {e^{\, - \,\left( {a-1/2} \right)/K} } \right) \cr}
$$
an example of which is sketched below.

In the limit for $n \to \infty$ we get
$$
K\,{\rm Li}_2 \left( {e^{\, - \,a/K} } \right) - {1 \over 2}\ln \left( {1 - e^{\, - \,a/K} } \right)
 <  - \mathop {\lim }\limits_{n\, \to \,\infty } \ln \overline P \left( n \right)
 < K\,{\rm Li}_2 \left( {e^{\, - \,\left( {a - 1/2} \right)/K} } \right)
$$
which is graphed below


问:
Taylor Polynomial for $x^{1/3}$

a. Compute the Taylor polynomial $T_3(x)$ for the function $(x)^{1/3}$ around the point $x=1$.

  b. Compute an error bound for the above approximation at $x = 1.3$.

I'm having trouble figuring out what to do for this problem.

I have to find the first three derivatives and after that I'm not quite sure what to do.

答: Hint: Expand $x^{\frac{1}{3}}$ as a Taylor series till the appropriate term and ignore the remaining terms. Ignoring the remaining terms results in an approximate value of the function for $x=1.3$. Then compare the approximation with the true value.</p>

Title: Taylor Polynomial for $x^{1/3}$


问:
In $\triangle PQR$ with $QR^2+PR^2=5PQ^2$, medians $PM$ and $QN$ meet at $G$. Find $\angle QGM$.
This problem becomes much simpler if we solve it using coordinate geometry but i am seeking a solution which involves pure geometry.

Consider a $\triangle PQR$ in which the relation
$$QR^2+PR^2=5PQ^2$$
holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then find the value of $\angle QGM$.

答:
From the triangle median theorem
$$NQ^2= \frac12( RQ^2+ PQ^2)-\frac14RP^2, \>\>\>
MP^2= \frac12( RP^2+ PQ^2)-\frac14 RQ^2
$$
Combine the two equalities
$$NQ^2 +MP^2 = PQ^2 +\frac14(RQ^2+RP^2)= \frac94 PQ^2$$

Then, substitute it into the cosine rule below
\begin{align}
\cos\angle MGN= \frac{NG^2+MG^2- NM^2}{2NG\cdot MG}= \frac{\frac19 (NQ^2+MP^2)- \frac14 PQ^2}{2NG\cdot MG}=0
\end{align}Thus, $\angle QGM = \angle MGN =90^\circ$.


问:
Infinite product formula for $\sum_{n \geq 0} p_e(n)\cdot x^n$
If $n$ is an integer and $p_e(n)$ is the number of partitions for $n$ such that all parts are even, what would be an infinite product formula for $\sum_{n \geq 0} p_e(n)\cdot x^n$

答: Replace $x$ with $x^2$ in the product formula for all partitions.


问:
Is there a solution to the functional equation $f(x)=2\log(x)^2f\left(x^\frac{3}{8}\right)^2f\left(x^\frac{1}{4}\right)^2$?
I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:

Solve for $f$: $$f(x)=2\log(x)^2f\left(x^\frac{3}{8}\right)^2f\left(x^\frac{1}{4}\right)^2$$

No other information or context is given, but I'm assuming that $f$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $x$ would require $f(x)$ to be complex), and that $\log$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).

For curiosity's sake, I present it as a challenge to either solve for $f$ or prove that <s>a solution does not exist</s> there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.

Edit:

In their answer, Chrystomath provides a solution:

$$f(x)=a\log(x)^{-\frac{2}{3}}\quad:\quad a\in\left\{z\in\mathbb{C}\Bigg| z^9=\frac{6^4}{8^9}\right\}$$

Which is quite possibly the most multivalued multivalued function I've ever seen.

I don't know whether or not the solution is unique, and I would still be interested in any other solutions.

答:
Here, we will only seek for real-valued solutions on an appropriate domain. We transform the equation by substituting

$$ f(x) = a \left| \log x \right|^k \exp( g(\log x) ),$$

where $k$ and $a$ are to be chosen later. Writing $u = \log x$ and plugging this to the functional equation,

$$ a \left| u \right|^k \exp( g(u) )
= 2 a^4 \bigl(\tfrac{3}{32}\bigr)^{2k} \left| u \right|^{4k+2} \exp\bigl( 2g\bigl(\tfrac{3}{8}u\bigr) + 2g\bigl(\tfrac{1}{4}u\bigr) \bigr). $$

As in @Chrystomath's solution, we can choose $k$ and $a$ as

$$ k = -\frac{2}{3} \qquad \text{and} \qquad a = 3^{4/9}2^{-23/9}. $$

Then the above equation reduces to

$$ g(u) = 2g(\tfrac{3}{8}u) + 2g(\tfrac{1}{4}u), \tag{*} $$

Now we analyze this equation.

@Chrystomath's solution corresponds to the trivial solution $g(u) \equiv 0$.

Setting $g(u) = c \left| u\right|^{\alpha}$ and plugging to $\text{(*)}$, the equation boils down to
$$2\left( (3/8)^{\alpha} + (1/4)^{\alpha} \right) = 1$$
This equation has a unique real root $\alpha \approx 1.195882190\cdots$. So, with this $\alpha$ and for any $c$,
$$ f(x) = a \left| \log x \right|^{-2/3} e^{c \left| \log x \right|^{\alpha}}$$
also solves the functional equation.

$\text{(*)}$ can relate the values of $g(u)$ and $g(u')$ only when $u$ and $u'$ satisfy $u' = \left(\frac{3}{8}\right)^m\left(\frac{1}{4}\right)^n u$ for some $m, n \in \mathbb{Z}$. So by using the Axiom of Choice, we can find infinitely many solutions of $\text{(*)}$.


问:
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$
Number Theory : Solve the system of congruence

(1)    $28x+17y\equiv 18 \pmod{41}$

(2)   $31x+11y\equiv 35\pmod{41}$

Attempt :

we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic

lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.

(1)    $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$

(1) $868x+527y\equiv 558\pmod{41}$

(2)   $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$

(2) $868x+308y\equiv 980\pmod{41}$

So finally we can subtract equation (1) from (2) we get:

$219y\equiv -422\pmod{41}$

lets check the $\gcd(219,41)$ by Euclidian algorithm :

$219 = 41\cdot 5 + 14$

$41= 14\cdot 2 + 13$

$14= 13\cdot 1 + 1$

$\gcd(219,41)=1$

Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.

$219y\equiv 1\pmod{41}$

$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.

I don't know how to continue from here .

答:
You used the Euclidean algorithm to find $\gcd(219,41)=1$;

you could then use the extended Euclidean algorithm to find that

$1=14\cdot3-1=14-(41-14\cdot2)=3\cdot14-41=3(219-41\cdot5)-41=3\cdot219-16\cdot41$,

from which we know that $3\cdot219\equiv1\bmod41$,

so $219y\equiv-422\equiv29$ means $y\equiv3\cdot219y\equiv3\cdot29=87\equiv5\bmod41$.

Can you take it from here?


问:
Similar cones - volumes and lateral areas
Two similar cones have volumes 9$\pi$ and 72$\pi$. If the lateral area of the larger cone is 32$\pi$, what is the lateral area of the smaller cone?

I did the following...

$\frac {(9\pi)^3} {(32\pi)^2} = \frac {x}{(32\pi)^2}$

resulting in a lateral area of $4\pi$. Is this right?

答:
Since the two cones are similar the ratios of the radius, height and slant height of the larger cone is some multiple of the radius, height and slant height of the smaller cone. Let's call it $k$. This gives $r^{\prime} = kr$, $h^{\prime}=kh$ and $l^{\prime}=kl$ where the primed variables are the dimensions of the larger cone.

We know that $\frac{1}{3}\pi r^2h = 9\pi\Rightarrow r^2h=27$. Likewise, $r'^2h'=216$. Substituting $r' = kr$ and $h' = kh$ gives $(kr)^2kh=216\Rightarrow k^3r^2h=216\Rightarrow k^3(27)=216\Rightarrow k^3=8\Rightarrow k=2$.

Therefore, the larger cone is twice as big as the smaller cone in linear dimensions. We know that the lateral surface of the larger cone is $\pi r'l'=32\pi$ so $\pi (kr)(kl)=32\pi$ from which we get $k^2\pi rl=4\pi rl = 32\pi\Rightarrow \pi rl= 8\pi$.

Therefore, the lateral surface area of the smaller cone is $8\pi$.


问:
Am I doing Vieta's Formulas correctly?
I've been given this exercise: $x^2 - (m+3)x + m + 2$, I'm supposed to check for which real values of the parameter $m$ can be used for this to work:

$1/x_1 + 1/x_2 > 1/2$ (I multiply both sides by 2x1x2) and get the following:

$2(x_1 + x_2) -x_1x_2 > 0$

which results to $m > -4$ using Vieta's formulas

Afterwards I have another argument, $x_1^2 + x_2^2 < 5$

After solving I get that $m$ is in the interval $(-4,0)$

My book tells me the final result for possible $M$ solutions is in the interval $(-2,0)$.

What am I doing wrong?

答:
You can't multiply by $x_1x_2$ since you don't know if it's a positive or negative quantity (remember the sign of the inequality would have to swap if it were negative, and stay the same otherwise).

Remember what Viete's formulae tell you, that $x_1+x_2 = m+3$ and that $x_1x_2 = m+2$. You can use these if you simplify the left-hand side:
$$\frac 1{x_1} + \frac1{x_2} = \frac{x_1+x_2}{x_1x_2} = \frac{m+3}{m+2},$$
so you want to ensure that $m$ is such that
$$\frac{m+3}{m+2}>\frac12.$$
We can't multiply throughout by $m+2$ since we don't know its sign. We can multiply by $(m+2)^2$, this is surely non-negative. This gives us
$$(m+3)(m+2)>\frac12(m+2)^2$$
which simplifies to
$$(m+2)(m+4)>0.$$
A product of two numbers is $>0$ either if they are both $>0$, or if they are both $<0$.

In the first case (when $m+2$ and $m+4$ are both positive), we have $m>-2$ and $m>-4$, which is simply equivalent to saying $m>-2$.

In the second case (when they are both negative), we have $m<-2$ and $m<-4$, which is the same as saying that $m<-4$.

So in summary, your condition is equivalent to saying that $$\boxed{\text{$m<-4$ or  $m>-2$}}.$$

Title: Am I doing Vieta's Formulas correctly?


问:
Copy and paste keeps positive semi-definiteness
Let $S \in \mathbb{R}^{n \times n}$ be a positive semi-definite matrix. We can show that
$$\bar{S} \triangleq\begin{pmatrix}
S & S \\
S & S
\end{pmatrix}$$
is positive semi-definite as well. To show this algebraically, considering the decomposition of $S = V^2$ with $V \in \mathbb{R}^{n \times n}$, we have
$$\begin{pmatrix}
S & S \\
S & S
\end{pmatrix}
= \frac{1}{2} \cdot
\begin{pmatrix}
V & V \\
V & V
\end{pmatrix}^2.
$$
Although I derived the algebraic proof, I still haven't fully understood it yet.

I wonder if there is any simple and direct interpretation that just demonstrates that $\bar S$ is positive semi-definite?

答:
Assume $S$ is positive definite. Consider $\bar Sz\cdot z$ for any $z\in\Bbb R^{2n}$. Write $z=(x,y)\in\Bbb R^n\times\Bbb R^n$. Since $S$ is positive definite, $Sv\cdot v\ge 0$ for all $v$, with equality holding only for $v=0$. Thus, we have
$$\bar Sz\cdot z = S(x+y)\cdot (x+y) \ge 0,$$
with equality holding iff $x=-y$. Thus, $\bar S$ is positive semidefinite. (If $S$ is only semidefinite, we still have the inequality here, but equality may hold for $x+y\ne 0$.)


问:
Help with an algebraic geometry result
I am studying this part of algebraic geometry and I have come to this proposition. I understand the basic idea well but there are two details that escape me.

Proposition: Let $G$ be a finite group acting linearly over $X$,
affine closed. Then, exsist an affine closed $Y$ and a surjective
morphism $\psi:X \rightarrow Y$ such that $K[Y]=K[X]^G$ and $\psi^*$
is the inclusion map $K[X]^G \subset K[X]$.  Furthermore, for all
$p,q$ in $X$  $$\psi(q)=\psi(p) \iff \exists g \in G :\;\;\;q=gp$$

Proof of the second part (within my notes)

$\Leftarrow$  Let $p,q$ points in $\mathbb{A}^n$ such that $\;\exists g \in G :\;q=gp. \;\;$ For all $f \in K[X]^G$ we have $f(q)=f(gp)=f^g(p)=f(p)$, so
$$f(\psi(q))=\psi^*(f(q))=f(q)=f(p)=\psi^*(f(p))=f(\psi(p))$$ Then
$\psi(p)=\psi(q)$.

$\Rightarrow$ Let $p,q$ points in $\mathbb{A}^n$ such that
$\psi(p)=\psi(q)$. Supposte that $\not\exists g \in G: \;\; q=gp$.

Choice $f \in I_X(p)$  [i.e.$f\in K[X], f(p)=0$]   but $f(gq)\neq0
> \;\; \forall g \in G$

So $h=\prod_{g\in G} f^g$ is such that $h \in K[X]^G$, $h(p)=0$
because $f$ is a factor, $h(q)=\prod_{g\in G}    f^g(q)=\prod_{g\in G}
> f(gq) \neq 0$

This implies $h(\psi(q))=\psi^*(h(q))=(h(q))\neq
> h(p)=\psi^*(h(p))=h(\psi(p))$. Absurd because $\psi(p)=\psi(q)$.

I didn't understand these two steps:

Why $f(\psi(p))=f(\psi(q))$ implies $\psi(p)=\psi(q)$?
I don't understand how this works, since f is non-injective

What guarantees me that such a function exists?
I tried to prove this with a proof by contradiction but I couldn't.

For 2.  had thought of proceeding like this:

Being $O(p)$ and $O(q)$, i.e. orbits of $p$ and $q$, G-invariant and disjoint closed of X, their respective ideals are G-invariants:
$$f(p)=0 \Rightarrow f \in I(O(p))\Rightarrow Z(f) \subset Z(I(O(p)))=O(p)$$
$$f(gq)=f^g(q)=0 \Rightarrow f^g \in I(O(q))\Rightarrow f \in I(O(q)) \Rightarrow q \in Z(f) \subset O(p)$$
absurd since $q \in O(q)$ and $O(q)\cap O(p)=\emptyset$

All this only works if and oly if $O(p)$ and $O(q)$ are closed

BUT CAN I SAY THAT $O(p)$ AND $O(q)$ ARE CLOSED?

答:
I'm guessing that this is about varieties over an algebraically closed field. If I'm right, then the answers are:

This proves that for every $f\in K[X]^G$, $f(\psi(p))=f(\psi(q))$. One of these functions is not necessarily injective, but the first part of the proposition says that $K[X]^G$ is an affine closed variety, i.e. a closed subset in some affine space. Therefore, there are functions $f_1,\ldots ,f_N$ such that $(f_1,\ldots ,f_N):Y\to \mathbb A^N$ is a closed embedding (in particular injective), and since $(f_1,\ldots ,f_N)(\psi(p)) = (f_1,\ldots ,f_N)(\psi(q))$, this means that $\psi(p)=\psi(q)$.

We are proving the contrapositive. If $p\neq gq$ for all $q$, then $\psi(p)\neq \psi(q)$. The first claim we want to prove is that if $p$ is not in the finite set $G\cdot q = \{gq\colon g\in G\}$, then there is a function $f$ that vanishes at $p$ but nowhere in $G\cdot q$. In fact, we can take a linear function: say we have coordinates of $\mathbb A^n$, $(x_1,\ldots ,x_n)$, such that $p$ is the origin. Then we are looking for a function of the form $f = \sum a_ix_i$. The condition that $f(gq)=0$ is a nonzero linear equation in the $a_i$'s. If we say that $\overline{a} = (a_1,\ldots ,a_n)\in \mathbb A^n$, to $gq$ corresponds a hyperplane $H_g$ in $\mathbb A^n$, and $f(gq)\neq 0\Leftrightarrow \overline a\notin H_g$. So, there exists $\overline{a}$ such that $f(gq)\neq 0$ for all $g$ if and only if $\mathbb A^n\setminus (\bigcup_{g\in G} H_g)$ is not empty. But affine space is not a finite union of hyperplanes*, so we are done.

*(It's crucial that $K$ is algebraically closed, and therefore infinite: if $K$ is finite, then affine space is a finite union of hyperplanes)

Why affine space is not a finite union of hyperplanes: We prove this for $\mathbb A^n$ by induction on $n$. For $n=1$, hyperplanes are points and if $K$ is infinite, then we are done. Now, for general $n$, say our hyperplanes are $H_1,\ldots , H_m$. There are infinitely many planes parallel to $H_1$ (because the field is infinite), so one of them, say $H'$, must be different from all of $H_1,\ldots ,H_m$. This means that $H'\cap H_j\subsetneq H'$ for all $j$. $H'$ is an affine space of dimension $n-1$, and each $H'\cap H_j$ is either a hyperplane in $H_j$ or it is empty (as it can't equal $H'$). By induction, there is a point $p\in H'$ not contained in any of the hyperplanes.

Title: Help with an algebraic geometry result


问:
Why does $\sum_p \sum_{k \geq 2} \frac{1}{kp^{ks}} \leq \frac{1}{2} \sum_p \sum_{k \geq 2} \frac{1}{p^{k}}$?
In his short note on Euler and the prime harmonic series, Professor of Mathematics Paul Pollack, uses the following step without explanation other than $s>1$.

$$
\sum_p \sum_{k \geq 2} \frac{1}{kp^{ks}} \leq \frac{1}{2} \sum_p \sum_{k \geq 2} \frac{1}{p^{k}}
$$

I understand that $\frac{1}{kx^s} < \frac{1}{x}$ if $s>1$ and $k>1$, but ..

Q1. I don't understand where the $\frac{1}{2}$ comes from.

Q2. I don't understand why the inequality is $\leq$ and not $<$.

I would appreciate any help. I am not a trained mathematician so avoiding technical terms would be a bonus.

答: For $k\geqslant 2$, we have $\frac{1}{kp^{ks}}\leqslant\frac{1}{2p^{ks}}$ and for $s>1$ we have $\frac{1}{p^{ks}}<\frac{1}{p^k}$ so we have $\frac{1}{kp^{ks}}<\frac{1}{2p^s}$. The inequality is $<$ so it is also $\leqslant$, it seems that the weaker inequality $\leqslant$ is sufficient to prove what the note wants to prove.


问:
Find a property of groups that is inherited by subgroups but not by quotients.
This is Exercise 4.6b of Roman's "Fundamentals of Group Theory: An Advanced Approach". Part a is here. According to this search and Approach0, it is new to MSE.

The Details:

From p.116 of Roman's book,

Definition: Let $\mathcal{P}$ be a property of groups. [ . . . ] We write $G\in \mathcal{P}$ if $G$ has property $\mathcal{P}$. [ . . . ] A property $\mathcal{P}$ of groups is inherited by subgroups if $$G\in \mathcal{P}\text{ and }H\le G\implies H\in \mathcal{P},$$ and $\mathcal{P}$ is inherited by quotients if $$G\in \mathcal{P}\text{ and }H\unlhd G\implies G/H\in \mathcal{P}.$$

The Question:

Find a property of groups that is inherited by subgroups but not by quotients.

Thoughts:

The property cannot be "virtually $\mathcal{Q}$" for some property $\mathcal{Q}$ by this answer by Arturo Magidin to some other question, where "$G$ is virtually $\mathcal{X}$" means that there exists $H\le G$ of property $\mathcal{X}$ such that $[G:H]$ is finite. So, in particular, it cannot be finite.

Please help :)

答:
One of the answers to your other question uses the following general pattern: the condition that a certain type of equation is always solvable in a group (e.g. the condition that every element have a square root) is inherited by quotients but usually not by subgroups.

There's a sort of dual to this: the condition that a certain type of equation is not solvable in a group is inherited by subgroups but usually not quotients. For example, the condition that there exist no nontrivial solutions to $g^n = 1$ (which is exactly the condition of being torsion-free) works.


问:
Subspace dimension, polynomials
When finding the dimension of subspace A = {a + bx + cx2 + dx3| a + b + c + d = 0} with space of polynomials of degree at most 3.

How do I go about this? I do know how to proceed when given vectors or matrices, but I am not sure how to exactly proceed with this subspace.

答:
Based on the proposed constraint, one has that $p(x)\in A$ iff
\begin{align*}
p(x) & = a + bx + cx^{2} - (a+b+c)x^{3}= a(1 - x^{3}) + b(x - x^{3}) + c(x^{2} - x^{3})
\end{align*}

Consequently, the set $\{1-x^{3},x-x^{3},x^{2}-x^{3}\}$ spans $A$ and is LI. Thus $\dim(A) = 3$.

Title: Subspace dimension, polynomials


问:
What makes an approximation to a constant good , bad or moderately good?
I have always had this question , and yet I do not have a good solid answer.

Is it the number of accurate decimal places?

Here : https://mathworld.wolfram.com/AperysConstantApproximations.html

$$\zeta(3) \approx \gamma^{\frac{-1}{3}}+\pi^{-4} \left(1+2\gamma-\frac{2}{130+\pi^{2}}\right)^{-3} \tag{1}$$

is given but is only accurate to 4 decimal places.

while I may provide for example :

$$\Gamma \left(\frac{-1}{4}\right) \approx -2\,(2\pi)^{\frac{3}{4}}\sqrt{\frac{4+\ln(2)}{2\sqrt{2}\,\pi+\frac{4(\pi-1)(6\pi-5)}{35\pi^2+8\pi-22}+\sqrt{2}\,\pi\ln(2)}} \tag{2} $$

which is accurate up to 10 decimal places .. and yet I consider the approximation given in $(1)$ to be more beautiful due to its simplicity.

Q = So what is the criteria that must be taken into account when deciding if an approximation to a constant is good/bad or worth mentioning or not?

答:
In both cases, it's preferable a formula which shows - or allows to deduce - a reasonable bound for the accuracy that can be attained, which is lacking in both formulas you presented.

Keeping firm the above, even more preferable is a formula that allows you to extend the computation till the accuracy you need, e.g. a series, a continued fraction, a recursion ...


问:
For an uncountable subset $E$ in $[0,1]\times[0,1]$, show there is a condensation point of $E$.
I'm not sure how to approach this question:

Given $E\subseteq[0,1]\times[0,1]$ and $E$ is uncountable, show that there is some $p\in [0,1]\times[0,1]$ such that $\forall \epsilon>0, E\cap B(p,\epsilon)$ is uncountable.

I tried using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. I defined a set $A$ to be the collection of all open balls centered at rational points with some $\epsilon$ such that the unit square is completely covered. Surely at least one of the balls (centered at say $x$) contains uncountably many points in $E$. But, I don't know how to make this work (or even if it could work). I'm concerned about the overlap of the balls and how to find some $x$ that doesn't move around as $\epsilon$ changes. This is my introduction to analysis, so I don't have a good intuition for much of these concepts yet. Any help would be greatly appreciated!

答:
Hint:I would divide the square into four smaller $1/2\times 1/2$ (closed) squares: one of them must contain uncountably many elements of $E$. Then, I would divide that square further into four (closed) $1/4\times 1/4$ squares, etc.

We get a chain of ever smaller squares, which (due to compactness of the original square) must have nonempty intersection. Let $p$ be a point in all those squares: it should now be an easy check that $p$ satisfied the requirement from your question.


问:
Two finite positive measures are singular iff a condition is satisfied
Q) Let $\mu,\nu$ be two finite positive measures on $X$. Prove that $\mu\perp \nu \iff \mu=(\mu-\nu)^+$

To prove that $\mu\perp \nu\implies \mu=(\mu-\nu)^+$, I can consider the signed measure $\lambda=\mu-\nu$, by uniqueness in Jordan's decomposition theorem, $(\mu-\nu)^+=\lambda^+=\mu$ but here I am not using the fact that $\mu\perp \nu$?

To prove that $\mu=(\mu-\nu)^+\implies \mu\perp \nu$, I know that I need to find disjoint $E_1,E_2\subset X$ such that $E_1\cup E_2=X$ and $\mu(A\cap E_2)=\nu(A\cap E_1)=0$ for any measurable set $A$ and am not sure how to find these $E_1,E_2$?

Any suggestions?

答:
It's helpful to keep in mind the uniqueness condition for the Jordan decomposition theorem: If $\lambda$ is a signed measure and there are positive measures $\lambda_1$, $\lambda_2$ with $\lambda = \lambda_1-\lambda_2$ and $\lambda_1\perp\lambda_2$, then $\lambda_1=\lambda^+$ and $\lambda_2 = \lambda^-$. This is where the assumption $\mu\perp\nu$ comes into play; setting $\lambda = \mu-\nu$ then implies $\mu = \lambda^+ = (\mu-\nu)^+$.

For the other direction one can use the Hahn Decomposition Theorem (which is a precursor of Jordan decomposition). Since $\lambda = \mu-\nu$ is a signed measure, this theorem provides a $\lambda$-positive set $P$ and $\lambda$-negative $N$ such that $X = P\cup N$ and $P\cap N=\emptyset$. These sets further have the property $\lambda^+(A)=\lambda(A\cap P)$ and similarly for $\lambda^-$ in the Jordan decomposition $\lambda = \lambda^+-\lambda^-$.

Taking $P$ and $N$ as above, I claim $E_1=P$ and $E_2=N$ work in showing $\mu\perp\nu$. Indeed, $\mu = \lambda^+$ implies $\mu(A\cap N) = \lambda^+(A\cap N) = \lambda((A\cap N)\cap P)= 0$ for any measurable set $A$. Since $\mu = \lambda^+$ implies $\nu = \lambda^-$, we also have $\nu(A\cap P) = \lambda^-(A\cap P) = \lambda((A\cap P)\cap N) = 0$. This completes the proof.


问:
$f(x+y) +y$ does not exceed $f(f(f(x)))$
Find all real to real functions for which $f(x+y) +y$ does not exceed $f(f(f(x)))$ for all real $x,y$

I don't feel like substituting values is that useful, except probably substituting $x=0$, which gives a really nice bound on $f(y)$, but other than that, I can't quite find anything else useful, at least not in my opinion.

答:
Choosing $y:=f(f(x))-x$,we imply :
$$f(f(x))\le x$$
Thus $$f(x+y)+y \le f(x)$$
Thus $$f(x+y)+y \le f( (x+y)-y) \le f(x+y)+y$$
Hence, $$f(x+y)+y=f(x)$$
Hence forth the conclusion.

Comment

It seems to me that if we impose that $y$ must be nonnegative in the very first hypothesis, we will have non linear solutions.