问:
Ito isometry proof
I have been reading Steven Shreve's Stochastic Calculus for Finance II. This question is from Chapter 4 Stochastic Calculus Page 129. This is theorem 4.2.2 (Proof of Ito isometry).

This theorem is about proving the following:

$$\mathbb E I^2(t)=\mathbb E\int_0^t \Delta^2(u)du$$

Here $\Delta(t)$ is an adapted stochastic process meaning that it is $\mathcal{F}(t)$- measurable for each $t \geq 0.$
I can't seem to understand how we got the following:

答: $\Delta$ is a simple process and is constant on $[t_j,t_{j+1}),$ i.e $\Delta(t)=\Delta(t_j)$ for all $t \in [t_j,t_{j+1}),$ Thus the integral $\int_{t_j}^{t_{j+1}} \Delta^2(u)du$ is just a Riemann integral and we have $\int_{t_j}^{t_{j+1}} \Delta^2(u)du=\Delta^2(t_j) \int_{t_j}^{t_{j+1}} du =\Delta^2(t_j)(t_{j+1}-t_j).$

Title: Ito isometry proof


问:
If $f: [0, 1] \to \mathbb R$ is defined by $() =  ^2$, check $f$ is bijective or not?how?
If $f:  [0, 1] \to \mathbb R$ is  defined by $() =  ^2$,   check $f$ is bijective or not?how?
I don't think it is onto because 6 doesn't have any pre image but 6 belongs to R

答:
No it's not surjective, to be bijective you  need it to be both injective and surjective

surjective $\forall$ y $\exists$ x such that $y=f(x)=x^2$

let $y=-3$ (or y=3) then $x=\sqrt(-3)$ but $\sqrt(-3) \notin [0,1]$

injective $f(x_1)=f(x_2)$ if and only if $x_1=x_2$

since the domain is in [0,1], yes it is

ContentLicense: CC BY-SA 4.0


问:
Find $\int_0^\infty \int_0^t e^{-t}\frac{\sin(\tau)}{\tau}d\tau dt$ using laplace transform.
Find $\int_0^\infty \int_0^t e^{-t}\frac{\sin(\tau)}{\tau}d\tau dt$ using laplace transform

My attempt:
I know that $\int_0^t f(\tau)d\tau=\mathscr{L}^{-1}(\frac{F(s)}{s}).$ Consider the integral $\int_0^\infty \int_0^t e^{-t}\frac{\sin(\tau)}{\tau}d\tau dt=\int_0^\infty  e^{-t} \int_0^t\frac{\sin(\tau)}{\tau}d\tau dt.$ Now let $I=\int_0^t\frac{\sin(\tau)}{\tau}d\tau=\mathscr{L}^{-1}(\frac{F(s)}{s}).$

Here $F(s)=\int_s^\infty \frac{1}{1+u^2}du= 1-tan^{-1}(s) (\because \mathscr{L}(\frac{f(t)}{t})=\int_s^\infty  \mathscr{L}(f(t))(u) du.$

Hence $I=\int_0^t\frac{\sin(\tau)}{\tau}d\tau=\mathscr{L}^{-1}(\frac{1-tan^{-1}(s)}{s})=\mathscr{L}^{-1}(\frac{1}{s}-\frac{tan^{-1}(s)}{s})=1-\mathscr{L}^{-1}(\frac{tan^{1}(s)}{s}).$

$\mathscr{L}^{-1}(\frac{tan^{1}(s)}{s})=\int_0^ t \mathscr{L}^{-1}(tan^{1}(s))(u)du.$ I don't know how to find  $\mathscr{L}^{-1}(tan^{1}(s)).$ Is there any shorter way to evaluate the above iterated integral using laplace transform?

答:
Well, we know that the Laplace transform is defined as:

$$\text{Y}\left(\text{s}\right)=\mathscr{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{y}\left(x\right)\exp\left(-\text{s}x\right)\space\text{d}x\tag1$$

In your case we are trying to find:

$$\mathcal{I}:=\left.\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}\right\vert_{\text{s}\space=\space1}\tag2$$

Using the time-domain integration property of the Laplace transform, we know that:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\mathscr{L}_x\left[\frac{\sin\left(x\right)}{x}\right]_{\left(\text{s}\right)}\tag3$$

Using the frequency-domain integration property of the Laplace transform, we know that:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\int_\text{s}^\infty\mathscr{L}_x\left[\sin\left(x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag4$$

Using the table of selected Laplace transforms, we know that:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\int_\text{s}^\infty\frac{1}{1+\sigma^2}\space\text{d}\sigma\tag5$$

We know that this integral represent the inverse tangent, so:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\lim_{\text{n}\to\infty}\left[\arctan\left(\sigma\right)\right]_\text{s}^\infty=$$
$$\frac{1}{\text{s}}\cdot\lim_{\text{n}\to\infty}\left(\arctan\left(\text{n}\right)-\arctan\left(\text{s}\right)\right)=\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\text{s}\right)\right)\tag6$$

So, for your problem we get:

$$\mathcal{I}=\left.\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\text{s}\right)\right)\right\vert_{\text{s}\space=\space1}=\frac{1}{1}\cdot\left(\frac{\pi}{2}-\arctan\left(1\right)\right)=$$
$$\frac{1}{1}\cdot\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{4}\tag7$$

Using the same method as above, we can show that:

$$\mathscr{S}_\omega\left(\text{s}\right):=\int_0^\infty\int_0^t\exp\left(-\text{s}t\right)\cdot\frac{\sin\left(\omega\tau\right)}{\tau}\space\text{d}\tau\space\text{d}t=\frac{\omega}{\text{s}}\cdot\int_\text{s}^\infty\frac{1}{\omega^2+\sigma^2}\space\text{d}\sigma\tag8$$

Subsitute $\text{u}:=\frac{\text{s}}{\omega}$, so we get (assuming that $\omega>0$):

$$\mathscr{S}_\omega\left(\text{s}\right)=\frac{1}{\text{s}}\cdot\int_\frac{\text{s}}{\omega}^\infty\frac{1}{1+\text{u}^2}\space\text{du}=\frac{1}{\text{s}}\cdot\lim_{\text{p}\to\infty}\left[\arctan\left(\text{u}\right)\right]_\frac{\text{s}}{\omega}^\text{p}=$$
$$\frac{1}{\text{s}}\cdot\lim_{\text{p}\to\infty}\left(\arctan\left(\text{p}\right)-\arctan\left(\frac{\text{s}}{\omega}\right)\right)=\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\frac{\text{s}}{\omega}\right)\right)\tag9$$

So:

$$\mathscr{S}_\omega\left(\text{s}\right):=\int_0^\infty\int_0^t\exp\left(-\text{s}t\right)\cdot\frac{\sin\left(\omega\tau\right)}{\tau}\space\text{d}\tau\space\text{d}t=\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\frac{\text{s}}{\omega}\right)\right)\tag{10}$$


问:
How to prove a series is not uniform convergent ? $\sum_{n=1}^\infty n^2xe^{-n^2x}$
We define the function $f_n(x), \, n=1,2,...$ by
\begin{align*}
    f_n(x) = n^2xe^{-n^2x} \qquad \qquad(x\in \mathbb{R})
\end{align*}

Show that the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent on $[a,\infty)$ if $a$ is a positive number.

Show that the series $\sum_{n=1}^\infty f_n(x)$ is not uniformly convergent on $[0,\infty)$

Idea: The convergence of series is defined in terms of the convergence of its sequence of partial sums. Here we use the Weierstrass M-test to check the uniform convergence.
Weierstrass M-test : Let $\{f_n\}$ be a sequence of function $f_n: A\xrightarrow[]{} \mathbb{R}$ and suppose that $\forall n \in \mathbb{Z}^+$ there exists a constant sequence $M_n$ such that,
\begin{align*}
    |f_n(x)| \leq M_n \qquad\qquad \forall x\in A
\end{align*}
If the series $\sum_{n=1}^\infty M_n$ is convergent then $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent.

Solution:

First find the $M_n$ using the Taylor series expansion for $e^{n^2x}$
\begin{align*}
    e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\\
    e^{n^2x} &= 1 + n^2x + \frac{(n^2x)^2}{2!} + \frac{(n^2x)^3}{3!} + ...\\
    e^{n^2x} &>  \frac{(n^2x)^2}{2!}\\
    \frac{1}{e^{n^2x}} &< \frac{2!}{(n^2x)^2}\\
    \bigg|\frac{n^2x}{e^{n^2x}}\bigg| &< \bigg|\frac{2!}{n^2x}\bigg| \leq \frac{2}{n^2a} \qquad\qquad \text{where} \,\,\, a > 0
\end{align*}

Take $M_n = \frac{2}{n^2a}$ and we need to check $\sum_{n=1}^\infty M_n$ is convergent on the given intervals
\begin{align*}
    \sum_{n=1}^\infty M_n =\sum_{n=1}^\infty \frac{2}{n^2a} = \frac{2}{a} \sum_{n=1}^\infty \frac{1}{n^2}
\end{align*}
We know by the p-series $\sum_{n=1}^\infty \frac{1}{n^p}$ where $p>1$ then the series is convergent. So in this case $p=2$ therefore the series $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent. Then the series $\sum_{n=1}^\infty M_n$ is convergent on $[a,\infty)$ therefore by the Weierstrass M-test the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent.

Is my answer for (1) correct? I tried to find the answer for (2) but failed. I could not find the partial sum of this series. Is there any other ways to prove that, can anyone give a little explanation I am really stuck here. Thanks

答:
Your answer for (1) looks fine to me.

Hint for (2):

If $(f_n)_n$ is uniformly convergent, then $f$ would be continuous and $f(0)=0.$

$f_n(x) \ge 0 $ for all $x \ge0$ and all $n.$

The sequence $a_n= (\frac{1}{n^2})_n$ satisfies
$$ \lim_{n\to \infty} a_n =0,$$ and
$$f_n(a_n) = e^{-1} > 0.$$


问:
Find missing angle in triangle

In this triangle, we are given that $AB=AD=CE$, $BE=CD$ and angle  $ECD=2*AED=2a$.

We are asked to find $AED=a$.

I have seen several problems similar to this but can't follow any of the solution methods.
The only thing which is obvious to me is that since $AD+CD = BE+EC$, the triangle is isosceles.

Therefore its base angles (the ones I have drawn in green) are equal.
Also triangle ABD is isosceles, therefore its base angles are equal.

I don't see any other relation to use.

(FYI this is not homework or anything - just practice and self-motivation).
Thank you very much.

EDIT: If this is of any help, through Geogebra I get that $a=10^{\circ}$, therefore $ECD=20^{\circ}$ and we have a 80-80-20 triangle.

答:
Using brute force.

Calling $u$ the segment with two marks and $v$ the segment with three marks we have

$$
\cases{
v^2=2(u+v)^2-2(u+v)^2\cos(2a)\Rightarrow a=\frac 12\arccos\left(\frac{2u^2+4u v+v^2}{2(u+v)^2}\right)\\
v^2=ED^2+AE^2-2ED\cdot AE\cos(a)\\
ED^2=u^2+v^2-2u v\cos(2a)\\
AE^2=u^2+v^2-2u v\cos(\phi)\\
\phi = \frac{\pi}{2}-a
}
$$

making substitutions we get at

$$
\sqrt{\frac{(2 u+v) (2 u+3 v) \left(u^3+u^2 v+v^3\right) \left(u^4-2 u^2 v^2+u v^3+v^4\right)}{(u+v)^5}}-\frac{v^3 (2 u+v)}{(u+v)^2}+2 u v-2 u^2=0
$$

making now $v = \lambda u$ we obtain

$$
\left(\sqrt{\frac{(\lambda +2) (3 \lambda +2) \left(\lambda ^3+\lambda +1\right) \left(\left(\lambda ^2+\lambda -2\right) \lambda ^2+1\right)}{(\lambda
   +1)^5}}-\frac{\lambda  \left(\lambda ^3-2 \lambda +2\right)+2}{(\lambda +1)^2}\right) u^2=0
$$

and the $\lambda$ factor reduces to

$$
\frac{\lambda ^2 \left(2 (\lambda +2) \lambda ^3+\lambda ^2+\lambda +1\right) \left(\lambda ^2 (\lambda +3)-1\right)}{(\lambda +1)^5}=0
$$

with the only feasible root

$$
\lambda = 0.532088886237956
$$

which corresponds to $a=10^{\circ}$

Title: Find missing angle in triangle


答:
On $\mathbb{R}^+$, $g(x)=x e^{-x}$ is a positive function attaining its absolute maximum, $\frac{1}{e}$, at $x=1$.

It follows that $f_n(x)=g(n^2 x)$ attains its absolute maximum, $\frac{1}{e}$, at $x=\frac{1}{n^2}$.

If $[a,+\infty)$ we have that $\sum_{n\geq 1}f_n(x)$, except for the first $\left\lceil \frac{1}{\sqrt{a}}\right\rceil$ terms, is a series of continuous and decreasing functions, with both $\lim_{x\to a}$ and $\lim_{x\to +\infty}\sum_{n\geq 1}f_n(x)$ being finite.

On the other hand we cannot have uniform convergence on $[0,+\infty)$ since $f_n(0)=0$ but the value of $\sum_{n\geq 1}f_n(x)$ is greater than $\frac{1}{e}$ for any $x$ of the form $\frac{1}{n^2}$.

As shown here through Gauss circle method we have
$$ \sum_{n\geq 1}e^{-n^2 x}\sim -\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{x}} $$
as $x\to 0^+$, hence
$$ \sum_{n\geq 1} n^2 x e^{-n^2 x}\sim \frac{\sqrt{\pi}}{4\sqrt{x}} $$
as $x\to 0^+$. The series is unbounded in a right neightbourhood of the origin.

Title: How to prove a series is not uniform convergent ? $\sum_{n=1}^\infty n^2xe^{-n^2x}$

Body:
We define the function $f_n(x), \, n=1,2,...$ by
\begin{align*}
    f_n(x) = n^2xe^{-n^2x} \qquad \qquad(x\in \mathbb{R})
\end{align*}

Show that the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent on $[a,\infty)$ if $a$ is a positive number.

Show that the series $\sum_{n=1}^\infty f_n(x)$ is not uniformly convergent on $[0,\infty)$

Idea: The convergence of series is defined in terms of the convergence of its sequence of partial sums. Here we use the Weierstrass M-test to check the uniform convergence.
Weierstrass M-test : Let $\{f_n\}$ be a sequence of function $f_n: A\xrightarrow[]{} \mathbb{R}$ and suppose that $\forall n \in \mathbb{Z}^+$ there exists a constant sequence $M_n$ such that,
\begin{align*}
    |f_n(x)| \leq M_n \qquad\qquad \forall x\in A
\end{align*}
If the series $\sum_{n=1}^\infty M_n$ is convergent then $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent.

Solution:

First find the $M_n$ using the Taylor series expansion for $e^{n^2x}$
\begin{align*}
    e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\\
    e^{n^2x} &= 1 + n^2x + \frac{(n^2x)^2}{2!} + \frac{(n^2x)^3}{3!} + ...\\
    e^{n^2x} &>  \frac{(n^2x)^2}{2!}\\
    \frac{1}{e^{n^2x}} &< \frac{2!}{(n^2x)^2}\\
    \bigg|\frac{n^2x}{e^{n^2x}}\bigg| &< \bigg|\frac{2!}{n^2x}\bigg| \leq \frac{2}{n^2a} \qquad\qquad \text{where} \,\,\, a > 0
\end{align*}

Take $M_n = \frac{2}{n^2a}$ and we need to check $\sum_{n=1}^\infty M_n$ is convergent on the given intervals
\begin{align*}
    \sum_{n=1}^\infty M_n =\sum_{n=1}^\infty \frac{2}{n^2a} = \frac{2}{a} \sum_{n=1}^\infty \frac{1}{n^2}
\end{align*}
We know by the p-series $\sum_{n=1}^\infty \frac{1}{n^p}$ where $p>1$ then the series is convergent. So in this case $p=2$ therefore the series $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent. Then the series $\sum_{n=1}^\infty M_n$ is convergent on $[a,\infty)$ therefore by the Weierstrass M-test the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent.

Is my answer for (1) correct? I tried to find the answer for (2) but failed. I could not find the partial sum of this series. Is there any other ways to prove that, can anyone give a little explanation I am really stuck here. Thanks


问:
If $a \not\equiv 0\ (\mod m\ )$ while m is prime, can I say that : $(a,m) = 1$?
It seems logical to me but I can't prove it.</p>

答:
Since $(a,m)$ divides the prime $\,m\,$ we infer $\,(a,m) = 1\,$ or $\,m\,$.

But $\,(a,m)=m\Rightarrow m\mid a,\,$ contra hypothesis. So $(a,m)=1$


问:
Asymptotic lines of a parametrized surface

Prove that the lines $v=k\pi$ are asymptotic lines for the surface parametrized by $$\varphi:(0,\pi )\times\mathbb{R}\rightarrow\mathbb{R}^{3}$$
$$\varphi (u,v)=(v,\cos(u),\sin(v))$$

Honestly, I don't know how to proceed because the professor talk with us only about the $v$ or $u-lines$ like $\varphi (u_{0}+t,v)$, $t\in\mathbb{R}$

Can someone axplain this concept? Thanks before!

答:
According to your description, we can plot the function using Mathematica:

I do not think it have any "asymptotic lines".


问:
Definition(s) of rational numbers
The definitions of rational numbers are somewhat confusing for me. The definition of rational numbers on wikipedia and most other sites is:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction $\frac pq$ of two integers, with the denominator $q$ not equal to zero.

All integers can be expressed as the quotient $\frac pq$. So if they are not written in the fraction form, will they still be called rational numbers?

Another definition which I've found is:

A rational number is a quotient $\frac mn$ where $m$ and $n$ are integers and $n \neq 0$.

These definitions might be the same. However, I'm not able to figure out how! Are rational numbers those numbers which can be expressed as the quotient $\frac mn$ of two integers or those which are directly the quotient of two integers? For example, if $n$ is an integer, so is $n$ a rational number? Or is $\frac n1$ a rational number?

答:
rational numbers is what you get when $\mathbb{Z}$ is closed to division.  Division is undefined when the divisor is $0$.

What this means is that if $p$ and $q$ are in the set, so is $p/q$, when allowed by division.  The set equates to the ratio of elements of $\mathbb{Z}$.

ContentLicense: CC BY-SA 3.0

Title: Definition(s) of rational numbers


问:
Riemann-Siegel Z and Theta Functions in PARI/GP
How can we use PARI/GP to calculate the Riemann-Siegel Theta function and/or the Riemann-Siegel Z function?

https://mathworld.wolfram.com/Riemann-SiegelFunctions.html

Body:
You can translate the formulas (2) and (1) from your source just literally:

(09:38) gp > rtheta(t)=imag(lngamma(1/4+I*t/2))-log(Pi)*t/2
%1 = (t)->imag(lngamma(1/4+I*t/2))-log(Pi)*t/2
(09:39) gp > z(t)=exp(I*rtheta(t))*zeta(1/2+I*t)
%2 = (t)->exp(I*rtheta(t))*zeta(1/2+I*t)
(09:40) gp > z(2.0)
%3 = -0.53963312564614487202970239667093128084 + 1.3224311446750734465 E-38*I


问:
Let $\mathscr F$ be a quasi-coherent sheaf on $X={\rm Spec}(A)$. How is $\mathscr F |_{D(g)} \cong \widetilde{M \otimes_B A_g}$?
From Hartshorne's Algebraic Geometry page 112:

Letting $f: B \to A_g$, I see that from Proposition 5.2 we can write $$f^*(\tilde M) \cong \widetilde{M \otimes_B A_g}$$ as sheaves of $\mathcal O_X |_{D(g)}={\rm Spec}(A_g)$-modules, where $$f^*(\tilde M) = \mathcal O_{{\rm Spec}(A_g)} \otimes_{f^{-1}\mathcal O_{{\rm Spec}(B)}} f^{-1}(\tilde M).$$

However, I don't see how $\mathscr F |_{D(g)} \cong \widetilde{M \otimes_B A_g}$.

答:
Note: Let $A$ be any commutative unital ring and let $E$ be any $A$-module and $R$ any $A$-algebra. Let $X:=Spec(A)$. Define for any open set $U\subseteq X$ the following:

D1. $\mathcal{E}(U):=lim_{D(f)\subseteq U}(A_f \otimes_A E).$

By definition $\mathcal{O}_X(U):=lim_{D(f)\subseteq U}A_f$.

Since the limit commutes with taking tensor product, there is an isomorphism

$\mathcal{E}(U) \cong (lim_{D(f)\subseteq U}A_f)\otimes_A E := \mathcal{O}_X(U)\otimes_A E$.

Definition D1 gives (this is Exercise 4.23 in Atiyah-Macdonald) a sheaf of $\mathcal{O}_X$-modules $\mathcal{E}$ on $X$. This is an equivalent definition to the definition given in Harsthorne.

Define

D2. $\mathcal{R}(U):=\mathcal{O}_X(U)\otimes_A R$.

It follows $\mathcal{R}$ is a sheaf of $\mathcal{O}_X$-algebras on $X$.

With this definition it follows $\mathcal{E}(D(f)):=A_f\otimes_A E \cong E_f$ and
$\mathcal{R}(D(f))\cong R_f$.

Doing the exercise in AM will improve your understanding of the construction of $\mathcal{E}$ from $E$. If $A$ is an integral domain it follows

$\mathcal{O}_X(U) \cong \cap_{D(f)\subseteq U}A_f$, since we may view all rings $A_f$ as sub rings of $K(A)$, and then we can take intersections inside this common ring ($K(A)$ is the quotient field of $A$) - "taking the limit" of the rings $A_f$ is a generalization of "taking intersections" $\cap A_f$.

Question: "However, I don't see how $\mathscr F |_{D(g)} \cong \widetilde{M \otimes_B A_g}$."

Answer: If you do the exercise in AM and prove that this is a definition equivalent to the definition in Hartshorne, you will get a better understanding of quasi coherent sheaves. With this new definition $\mathcal{E}(D(f))\cong E_f$
is "by definition".

This is also discussed here

Noetherian $R$-algebra corresponds to a coherent sheaf of rings on $\operatorname{Spec}(R)$


问:
Intermediate Value Theorem and polynomial of odd degree
As a consequence of Intermediate Value Theorem (IVT), it is illustrated in almost all calculus books and notes/courses that every polynomial of odd degree over real numbers have a real root.

I tried to see whether this result has been proved without the use of intermediate value property, and I did not find any link or question related to that on this and other sites.

Can one suggest me whether there is a proof of above mentioned consequence of IVT, which is done without using IVT?

I had seen that the fundamental theorem of algebra have been of interest to many people for proving just using algebraic techniques, but among all the known various proofs of FTA, epsilon amount of analysis is used.

So, my above question is of similar nature, but I did not see any comments on its proofs without IVT.

答:
One could use Descarte's Rule of Signs. So, below is the idea:

Assume that we have $$P(x) = \sum_{k=0}^{n} a_kx^k$$

If there is at least one sign change, then we are done. So we assume there is not.

Assume there are terms of both odd and even exponents, then we analyze $P(-x)$. There were no sign changes in $P(x)$, therefore, we must have in $P(-x)$ since $(-x)^{\text{odd}} = -x^{\text{odd}}$ and $(-x)^{\text{even}} = x^{\text{even}}$.

Assume there are only odd powers, then the polynomial is an odd function which means $x=0$ is a root.

We cannot assume all exponents are even since $P(x)$ is of odd degree.


问:
$A^3$ is similar to $B^3$ but $A^2$ is not similar to $B^2$
Find an example of $A,B$ such that $A^3$ is similar to $B^3$, and $A^2$ is not similar to $B^2$, and $A$ is not similar to $B$.

I only have some clues that $p_{A^2}(x) \neq p_{B^2}(x)$ and $p_{A}(x) \neq p_{B}(x)$

I first wanted to start from $A^3 = S^{-1} B^3 S$

Then let $S$ and $B$ to be some arbitrary matrices, where $S$ is invertiable, so that I could obtain $A^3$ which is similar to $B^3$. Thus, $A = \sqrt[\leftroot{-3}\uproot{3}3]{A^3}$
However, I realize I don't know how to find square root or cube root of matrices.

答:
I want to argue that, in some sense, the examples already given are the only things that can go wrong.

First, we consider the case when $A$ and $B$ are diagonalizable.  Diagonalizable matrices are determined (up to similarity) by their eigenvalues and multiplicities.  Let us group the eigenvalues together based on the value of $\lambda^3$.  For a given value of $\lambda$, the equation $x^3-\lambda^3=0$ has 3 solutions, $\lambda, \omega \lambda, \omega^2\lambda$, where $\omega$ is a primitive cube root of unity.  e.g., $\omega=e^{2\pi i/3}$.  For example, if $A=\operatorname{diag}(1,\omega, \omega, \omega^2, 2, 2\omega)$, we would break the eigenvalues into two multi-sets, $\{1,\omega, \omega, \omega^2\}$ and $\{2, 2\omega\}$

Each group of eigenvalues can contain up to 3 different eigenvalues in it, and we see that as long as the multiplicities are different for $A$ and $B$, $A$ and $A^2$ will not be similar to $B$ and $B^2$.

For a simple (rational!) example of this type, consider $A=I$, $B=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}.$  For a $2\times 2$ real (but not rational) example, again consider $A=I$ and $B$ a rotation of angle $2\pi/3$.

The thinking here actually gives a nice way to take square roots of diagonalizable matrices: diagonalize, then replace  every eigenvalue with one of its possible cube roots.  If $A$ is $n\times n$ and has no repeated eigenvalues, there are $3^n$ ways to do this.  However, repeated eigenvalues mean we can take multiple different eigenbases, which yield different cube roots.  For example, if $A=I_3$, pick any axis and and the rotation about that axis of angle $2\pi/3$ will be a cube root.  This method also does not allow for the possibility of non-diagonalizable square roots.

So one problem is that $A$ and $B$ can have different eigenvalues even though $A^3$ and $B^3$ have the same eigenvalues.  But what if they have identical characteristic polynomials? It isn't too hard to construct examples which are nilpotent, but that is essentially all that can go wrong, at least over $\mathbb C$.  This follows from the following result.

Lemma: If $A=J_n(\lambda)$ is an $n\times n$ Jordan block with eigenvalue $\lambda$, and $\lambda\neq 0$, then $A^{3}$ is similar to $J_n(\lambda^3)$.

Proof. Note that $(A-\lambda I)^n=0$ but $(A-\lambda I)^{n-1}\neq 0$ uniqutely characterizes $A$ up to similarity.  Since $A$ and $I$ commute, we can factor $(A^3-\lambda^3 I)=(A-\lambda I)(A-\omega\lambda I)(A-\omega^2 \lambda I)$, factors of which all commute with each other, and since $\lambda$ is the only eigenvalue of $A$, the latter two factors are invertible.  Then $(A^3-\lambda^3 I)^k=(A-\lambda I)^k(A-\omega\lambda I)^k(A-\omega^2 \lambda I)^k$.  Since the latter two terms are invertible, if $k<n$, the overall expression is not $0$, and if $k=n$, then the first factor is already $0$ so the entire expression is as well.

This shows that if $A$ and $B$ are invertible and $A^3$ and $B^3$ are similar, then we can pair off the Jordan blocks of $A$ and $B$ in such a way that corresponding blocks will have the same size and the ratio of corresponding eigenvalues is a cube root of unity.

This also says how to find all the cube roots of an invertible matrix $A$.  First, write the JNF of $A$ as a sum of Jordan blocks.  Second, replace the eigenvalue for each block with one of its 3 cube roots. Call this matrix $B$  Then $B^3$ is similar to $A$, and for every $P$ such that $PB^3P^{-1}=A$, we get $PBP^{-1}$ is a cube root of $A$.

For non-invertible matrices, we can canonically break them up as the sum of an invertible and nilpotent matrix, and it is a fun (but involved) combinatorial problem to understand what the square or cube roots of a nilpotent matrix are.


问:
An Olympiad geometry problem ; angle chasing / trig bashing / coordinate bashing
D is an inner point in equilateral triangle $ABC$ such that $\angle ADC=150^{\circ}$.Prove that the triangle formed by taking segments $AD$,$BD$ and $CD$ as its three sides is a right angled triangle. ...... [Source : Some European Olympiad in 2003]

This is the first time I am trying a problem like this (proving that these three lines make a right triangle).

I did manage to prove the special case when D $\in BM$ where M is midpoint of AC using simple sin rule, in which case I got $AD=CD=\frac{BD}{\sqrt2}=\frac{\sqrt3 - 1}{\sqrt 2}a$ where $a$ is the side length.

I did manage to find a solution online by AMAR_04 (Geo god on aops) (https://artofproblemsolving.com/community/c869463h1833925_build_up_your_understanding2_p21) but I am not very comfortable with rotation yet.

So, if anyone can please provide a different solution (trig bash/ coordinate bash or anything else except complex) OR explain the solution above, I would be really greatful!

Thanks!

答:
Here is the figure for the solution on AOPS.

Rotate the whole figure through $B$ by $60^\circ$ anticlockwise.

By construction, $CE = AE'$, $AE = DE'$, $BE = BE'$, $\angle EBE' = 60^\circ$.

The last two conditions assures that $\triangle EBE'$ is equilateral.

Hence $\triangle AEE'$ is a right-angled triangle, formed by the side lengths $AE, BE, CE$.


问:
How should I conceptualize differential forms as a generalization of scalar and vector fields?
I'm reading Munkres' "Analysis on Manifolds" at the moment, and I'm having a bit of trouble understanding what exactly differential forms are.

I understand that the point of integrating forms over manifolds is to generalize the concept of integrating functions over intervals/curves/surfaces, as you do in a standard multivariable calculus course.

Manifolds are much easier to generalize from surfaces to me. Taking out all the details, we're just considering subsets of $\mathbb{R}^n$ embedded in $\mathbb{R}^m$. Surfaces, but in more dimensions (to keep it brief).

Differential forms are MUCH harder to visualize for me. Perhaps it's because I don't tend to think of things from a linear algebra perspective, but the jump from a scalar field (a 0-form) or a vector field (a 1-form) to a $k$-form is WAY more difficult.

I just think of a scalar field as, for example, measuring mass or assigning certain parts of $\mathbb{R}^n$ as more "weighted" as other parts. I think of a vector field as, concretely, assigning every point in $\mathbb{R}^n$ to a vector, and a lot of times imagining it as "fluid flow" is a helpful way for students to visualize this stuff and what integrals may represent.

What is a $2$-form though? I understand that an example of one is the curl of a vector field. Is there something concrete that I can point to or think of to help me visualize what a $k$-form may represent? And, for the record, an integral? An integral of a scalar field just sums up over some subset of $\mathbb{R}^n$, everything where the $0$-form "weighs" it, in a sense. And an integral over a $1$-form, or a vector field, calculates this, where a greater magnitude signifies more "fluid flow", and takes direction into account. I think I'm missing the linear algebra portion of the definition of a $k$-form, but seeing the definitions with multi-indeces and sums just isn't useful to me.

Can someone shed some light on some physical interpretation of $k$-forms, and how I can think of generalizing scalar and vector fields in this fashion?

Thanks.

答:
If you're looking for a bit of geometric intuition, I've found it useful to think of 2-forms in terms of bivectors (or multivectors more generally): if one thinks of vectors in $TM$ as "infinitesimal oriented lengths", then bivectors, which are elements of $\Lambda^2TM$ can be thought of as "infinitesimal oriented areas" and so on. The wedge product fits nicely into this picture: for $u,v\in T_pM$ the element $u\wedge v\in\Lambda^2TM$ corresponds to the parallelogram spanned by $u$ and $v$ (with an induced orientation). Two different pairs of vectors have the same wedge product precisely if the paralellograms have the same area and are coplanar (in an oriented sense). A bit of algebra/geometry can show that this equivalence is related to the antisymmetry of the wedge product.

Differential forms, however, are not multivectors, they are instead dual to multivectors. That is, a $k$-form is a linear map from the space of $k$-vectors to the real numbers. They can be defined other ways (such as alternating linear maps on $TM$), but these definitions end up being isomorphic.

Integration fits nicely into this geometric picture: for an oriented $k$-dimensional (sub)manifold $N$ and a $k$-form $\omega$, one can think of $\int_N\omega$ as breaking up $N$ into a bunch of small $k$-dimensional volumes (expressed as multivectors), evaluating $\omega$ on these multivectors, and then summing. This is more or less analogous to Riemann integration in calculus.

Ultimately, this picture is a schematic, but it can be made quite precise at the cost of a lot more algebra and abstraction.


问:
Why is $\varphi$ called "the most irrational number"?
I have heard $\varphi$ called the most irrational number. Numbers are either irrational or not though, one cannot be more "irrational" in the sense of a number that can not be represented as a ratio of integers. What is meant by most irrational? Define what we mean by saying one number is more irrational than another, and then prove that there is no $x$ such that $x$ is more irrational than $\varphi$.

Note: I have heard about defining irrationality by how well the number can be approximated by rational ones, but that would need to formalized.

答:
The measure is indeed what you think. The simple continued fraction has "digits" which are all 1's. See, for example, Khinchin's little book on continued fractions.

Simple version, the convergent just before a large "digit" is a very good approximation, the relevant error being less than $$  \frac{1}{q_n q_{n+1}}  $$
where the $q$'s are the denominators. So, with a big digit, $q_n$ is of modest size but $q_{n+1}$ is quite large, so the error with denominator $q_n$ is small compared with $1/q_n^2.$


问:
Prove that multiplication is well defined

Let $M = \mathbb{N} \ \mathbb{x} \ \mathbb{N}$. We define the following relation on $M$. Let $(a,b)R(a',b')$ iff $a + b'=a'+b$
  We define the set of intergers $\mathbb{Z}$, to be the set of equivalence classes of $M$ under the equivalence relation above. We try to define the arithmetic operations of addition and multiplication in the following way $$[(a,b)] + [(a',b')] := [(a+a',b+b')]$$ and $$[(a,b)] \cdot [(a',b')] := [(aa'+ bb',a'b+ab')] $$

I should prove that multiplication is well defined. If I understand this correctly i should given that $$(a,b) \sim (c,d) \ i.e \ a+d =c+b$$ and $$(a',b') \sim (c',d') \ i.e \ a'+d'=c'+b'$$ I should prove that $$(ac+bd,cb+ad) \sim (a'c' +b'd',c'b'+a'd')$$ which means that $$ ac+bd+c'b'+b'd' = a'c' +b'd'+cb+ad$$
I dont know how to do this?

答:
I'll prove that multiplication is well-defined which is more difficult and leave addition as exercise.

You want to show that

$(a, b)\cdot (c, d) \sim (a', b')\cdot(c', d')$

supposing that

$(a,b) \sim (a', b') \Longleftrightarrow a + b'= b + a', \qquad   (c,d) \sim (c', d') \Longleftrightarrow c+d' = d+c' \quad (\mathbf{1.0})$

i.e. multiplication does not depend on the particular choice of representatives.

We start by showing that $(a,b)\cdot(c,d) \sim (a,b)\cdot(c',d')$ and then use commutativity of the product to show the previous follows.

By expanding the multiplication, it is equivalent to show that $(ac+bd,ad+bc) \sim (ac+bd, ad+bc)$, which by the definition of ~ is the same as showing that

$ac + bd + ad' + bc = ad + bc + ac' + bd' \quad (\mathbf{2.0})$

Multiplying $c+d' = d+c'$ by $a$ in 1.0 yields

$ac + ad' = ad + ac'$

Multiplying $d + c' = c + d'$ by $b$ in 1.0 yields

$bd + bc' = bc + bd'$

Summing up the two equations, you obtain 2.0.

By commutativity of the product (left as exercise to verify), we have also $(a, b)\cdot(c, d) \sim (a, b)\cdot(c', d') = (c', d')\cdot(a, b) \sim (c', d')\cdot(a', b') = (a', b')(c', d')$.


问:
Prove that $\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{b}{b-1}\right)}}$ for all $b>2$.
While solving a bigger problem, I've reduced it to an inequality $$\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{\color{red}b}{b-1}\right)}}$$for $b>2$, which looks plausible when looking at the plots. I've tried using Jensen inequality here with $x\mapsto x^{b-1}$, but not much luck.

I've also checked that the inequality works empirically with Wolfram Alpha.

Yes, this is written correctly, opposite to what some of the people in the comments below are trying to claim. I'm surprised this even needs claryfying.

Body:
My new proof (sketch)

Let $b > 2$. Prove that
$$1 + 2^{-1 + b^{\frac{b}{b-1}}} \ge \left(1 + 2^{-1 + b^{\frac{1}{b-1}}}\right)^b.$$

We first give the following auxiliary results (Facts 1-4).

Fact 1: $2^{b^{\frac{1}{b-1}}} \ge \frac{30b+52}{15b-2}$ for all $b \in [2, \infty)$.

Fact 2: Let $b_1 > 2$ be given. Let $p(x) = 1 + 2^{-1} x^{b_1} - (1 + 2^{-1} x)^{b_1}$.
Then $p(x)$ is strictly increasing on $(2, \infty)$.
(Hint: $p'(x) = \frac{1}{2} b_1 \left(x^{b_1-1} - \left(1+\frac{x}{2}\right)^{b_1-1}\right) > 0$. )

Fact 3: $\left(\frac{9}{16}\right)^{2/b} \ge \frac{363b - 168}{360b + 272}$ for all $2\le b \le 4$.

Fact 4: $9^{2/b} \ge \frac{(29b+50)^2}{36(5b-4)^2}$ for all $2\le b \le 4$.

Now, the desired inequality is written as
$$1 + 2^{-1}\left(2^{b^{\frac{1}{b-1}}}\right)^b \ge \left(1 + 2^{-1}2^{b^{\frac{1}{b-1}}}\right)^b.$$
By Facts 1-2, it suffices to prove that
$$1 + 2^{-1}\left(\frac{30b+52}{15b-2}\right)^b \ge \left(1 + 2^{-1}\frac{30b+52}{15b-2}\right)^b.$$

We split into two cases:

$b \ge 4$: It suffices to prove that
$$2^{-1}\left(\frac{30b+52}{15b-2}\right)^b \ge \left(1 + 2^{-1}\frac{30b+52}{15b-2}\right)^b$$
or
$$2^{-1/b} \ge \frac{15b-2}{30b+52} + 2^{-1}.$$
It is easy and thus omitted (Hint: Take logarithm and then take derivative).

$2 < b < 4$: Denote $Q = \frac{30b+52}{15b-2}$. By the power mean inequality, we have
\begin{align}
\left(1 + 2^{-1}Q^b\right)^{1/b} &= 9^{1/b}\left(\frac{1^b + (2^{-\frac{4}{b}}Q)^b\cdot 8}{9}\right)^{1/b}\\
&\ge 9^{1/b}\left(\frac{1^2 + (2^{-\frac{4}{b}}Q)^2\cdot 8}{9}\right)^{1/2}.
\end{align}
It suffices to prove that
$$9^{1/b}\left(\frac{1^2 + (2^{-\frac{4}{b}}Q)^2\cdot 8}{9}\right)^{1/2} \ge 1 + 2^{-1}Q$$
or
$$9^{2/b}\left(\frac{1^2 + (2^{-\frac{4}{b}}Q)^2\cdot 8}{9}\right) \ge (1 + 2^{-1}Q)^2$$
or
$$\frac{1}{9} 9^{2/b} + \frac{8}{9}\left(\frac{9}{16}\right)^{2/b}Q^2 -1 - Q - \frac{1}{4}Q^2 \ge 0.$$
By Facts 3-4, it suffices to prove that
$$\frac{1}{9} \cdot \frac{(29b+50)^2}{36(5b-4)^2} + \frac{8}{9}\cdot \frac{363b - 168}{360b + 272} \cdot Q^2 -1 - Q - \frac{1}{4}Q^2 \ge 0$$
or
$$\frac{(b-2)(-25504875b^4 + 147431700b^3 - 60455760b^2 - 298781264b + 181422064)}{324(5b-4)^2(45b+34)(15b-2)^2} \ge 0$$
which is true.

We are done.


问:
How many size 10 sequences can be formed with terms in $\{A,B,C\}$ and the number of terms A is odd?

How many size 10 sequences are there, all formed with terms belonging to $\{A,B,C\}$, in which the number of terms with A is odd.

My proposed solution:
$$\sum_{m\in \{1,3,5,7,9\}} {10 \choose m} 2^{10-m}=28804$$
That is for each $m\in \{1,3,5,7,9\}$ finding the possibilities for A and then filling the remaining spots with B and C (2 possibilities for each non-A spot).

But the given short answer is 2391484 (no solution provided).

Question: what am I missing? or is the given answer wrong?

答:
As has been remarked, the official solution can't be correct as it greatly exceeds the total number of ternary strings of length $10$.

To solve the problem recursively, let $A_n$ be the number of strings of length $n$ with an odd number of $A's$ and let $B_n$ be the number of strings of length $n$ with an even
number of $A's$.  Then $$A_n+B_n=3^n$$

And $$A_n=2A_{n-1}+B_{n-1}=A_{n-1}+\left(A_{n-1}+B_{n-1}\right)=A_{n-1}+3^{n-1}$$

Since $A_1=1$ we see that $$A_n=1+3+3^2+\cdots +3^{n-1}=\frac {3^n-1}2$$

Thus $$\boxed{A_{10}=\frac {3^{10}-1}2=29524}$$

Note that, as expected, $A_n$ is about half of $3^n$.

Your method makes sense logically, so I expect you just made an arithmetic error.  Wolfram Alpha uses your sum to match the recursive result.

Side note:  We remark that $A_{14}=2391484$ so perhaps the strings were intended to have length $14$.


问:
Any two perfect totally disconnected compact metric space are homeomorphic

Prove that any two perfect, totally disconnected , compact metric spaces are homeomorphic.

This was asked in my analyis quiz and althogh I tried to solve it at home also but I am unable to.
I took (X,d) and (Y,d') as metric spaces but I am not able to find homeomorphic map.

So, can you please help with that.

答:
Those are properties of the Cantor set, so in essence you are proving that any metric space with those properties is homeomorphic to the Cantor set. For the proof, for instance, see: https://arxiv.org/pdf/1210.1008.pdf . Apparently this is (in a slightly different form) a theorem of Brouwer. See also: https://en.wikipedia.org/wiki/Cantor_space#Characterization .

Out of curiosity, what sort of analysis quiz has this as a question, and what background knowledge are you supposed to have in order to answer it? Did your course cover the Cantor set and its characterisation above?


问:
First year french college. Is there a finite set of points of the plane A. Finite set of point A such as...
Is there a finite set of at least 3 points in the plane A, such as every 3 points of A are not aligned, and such as every circle going through 3 points of more of A, it’s center is in A ?
I believe that there can’t be such a set. So I tried showing that for every n point of the plane you can’t arrange them to be set A.

Being based on a number n I also tried doing it by doing a « mathematical induction » to show that there can’t be a set A for n points.

I’m a french student in my first year of college, I’m sorry by advance if my english is a bit wacky.

edit: forgot that it's a set of at least 3 points sorry

答: In the plane draw a circle, take  three points on the circle (I think you can take any  finite number of points but 3 will do). Call these points your set A. These 3 points are not aligned (not on the same line) and there is only one circle that goes through them and we know by construction that this circle is in the plane.</p>


问:
Binomial coefficient modulo a prime
I cannot find a reference for the following fact: "Lep $p$ be a prime. If the binomial coefficient of $d$ over $l$ is zero modulo $p$ for every $l \in \{1,\ldots,d-1\}$ then $d$ is a power of $p$." Do you have any suggestions? Thank you in advance.

Edit: I would like to thank both of you for your suggestions, but I would need a reference to put in the bibliography. Could you suggest any book or paper in which such a property is stated?

答:
If you prefer to ignore user10354138's comment, and prove it yourself, you will need:

Legendre's formula

prove that when $n$ has form $(p^r + k)$, where $k \in \{1,2,3, \cdots, [p^r - 1]\}$ that you can find at least one integer $l \in \{1,2,\cdots, [n-1]\}$ such that $p$ does not divide $\binom{n}{l}.$

One trap to avoid:

If $(ac) \equiv (bc) \pmod{p}$ and $c$ is not a multiple of $p$, then $a\equiv b \pmod{p}.$

The above implication does not necessarily hold when $c$ is a multiple of $p$.

Title: Binomial coefficient modulo a prime


问:
The Cauchy problem for a higher order differential equation
I need to solve the Cauchy problem for a differential equation:
$$yy''x^2+3yy'x+y^2=(y'x+y)^2(1-y'x^2-yx),\ y(1)=2,\ y'(1)=0$$

I solved this equation:
$$Let\ x=e^t,\ y=ze^{-t},\ y'=e^{-2t}(z'-z),\ y''=e^{-3t}(z''-3z'+2z)$$
$$=>zz''-(z')^2+(z')^3=0$$
$$Let\ z'=p, z''=pp'$$
$$=>zp'=p-p^2$$
$$\int \frac{dp}{p-p^2}=\int \frac{dz}{z}$$
$$\frac{p}{1-p}=zC$$
$$z'=p=\frac{zC}{zC+1}=\frac{z}{C+z}$$

What should I do next? How can I solve the Cauchy problem?

答:
So you set $y(e^t)=z(t)e^{-t}$ or $z(\ln x)=xy(x)$. Then indeed
\begin{align}
z'(t)&=e^{2t}y'(e^t)+e^ty(e^t)=x(xy(x))'\\
z''(t)&=e^{3t}y''(e^t)+3e^{2t}y'(e^t)+e^ty(e^t)\\
&=x(x^2y''(x)+3xy'(x)+y(x))\\
&=\frac{x}{y}\left((xy(x))'\right)^2(1-x(xy(x))')\\
&=\frac{z'(t)^2(1-z'(t))}{z(t)}
\end{align}
This can then be separated into integrable terms, as you did indirectly,
\begin{align}
\frac{z''}{z'(1-z')}=\frac{z'}{z}\\
\frac{z'}{1-z'}=Cz\implies z'=\frac{Cz}{1+Cz}
\end{align}
Now apply again separation-of-variables to get
$$\begin{align}
\frac{z'}{Cz}=1-z'\implies \frac{\ln|z|}{C}+z=t+D.
\end{align}$$
Apply the initial conditions to determine the constants, $z(0)=y(1)=2$, $z'(0)=y'(1)+y(1)=2$ which gives $C=-1$ and then $D=2-\ln(2)$. You can semi-formally solve this by applying the Lambert-W function,
$$
ze^{-z}=e^{-t-2+\ln 2}\iff -ze^{-z}=-2e^{-t-2}\implies z=-W_{-1}(-2e^{-t-2})
$$


问:
Asymptotics in recurrence relation
I have an recurrence relation $T(n)= n+ \sum_{k=1}^{n-1} (T(n-k)+T(k)) = O(3^n)$ and $T(1)=1$.

I self study on asymptotic and recurrence example. Lots of example I read but this one is very strange!

How we can solve this recurrence? i.e: reach to answer.

答:
I'll start from assumption $T(1)=1$ and $T(n)=\sum\limits_{i=1}^{n-1}(T(i)+T(n-i))$.

It's easy to see, that $T(n)=2\sum\limits_{i=1}^{n-1}T(i)$. Elaborating partial cases gives hypothesis $T(n)=2\cdot 3^{n-2}$, when $n\geqslant 2$:

$$\begin{array}{l}
T(2)=T(1)+ T(1) = 2 \cdot3^{0} \\
T(3)=2\big(T(1)+ T(2)\big) = 2 \cdot3^{1} \\
T(4)=2\big(T(1)+ T(2)+ T(3)\big) = 2 \cdot3^{2} \\
\cdots \\
T(n)=2\big(T(1)+ T(2)+ \cdots+ T(n-1) \big)=2 \cdot3^{n-2}\ (\text{hypothesis})\\
\cdots \\
\end{array}$$

General formula can be proved, for example, by induction:
$$T(n+1)=2T(1)+ 2\big( T(2)+ \cdots+ T(n) \big) = \\
=2T(1)+ 2\big(2\cdot3^{0} + 2 \cdot3^{1}+ \cdots+ 2 \cdot3^{n-2}\big)= \\
=2+2\frac{3^{n-1}-1}{3-1} = 2 \cdot 3^{n-1}$$
At last is clear, that $T(n) \in O(3^{n})$.

Title: Asymptotics in recurrence relation


问:
Given a triangle. Find the possible isosceles triangle.
Given a equilateral triangle ABC on a plane, and given a point P in the same plane.

How much points P is possible in such way that APC, ABP, PBC are isosceles and non degenerate?

By symmetry, i could find just seven. The answer say it is ten. but i can not find the another three.

The points i found was:
(The baricenter)

(6 points at the straight line that contains the median, these points are outside the triangle, i think it is i)

答:
There are two cases where an isosceles triangle can be formed:

It lies on some perpendicular bisector of the equilateral triangle;

The length of a new side is equal to the side length of the equilateral triangle.

so the 10 points are:


答:
Answer to be proved in details

Such a finite set of points $A$ can't exist.

Denote by $P_1, P_2, P_3$ three points for which the circumscribed circle $\mathcal C$ is achieving the minimum radius $R$ of all circumscribed circles. Let $O$ be its center. By hypothesis $O \in A$.

Assumption to be proven: one of the triangles $(O \ P_1 \ P_2)$, $(O \ P_2 \ P_3)$ or $(O \ P_1 \ P_3)$ has a circumscribed radius strictly less than $R$. A contradiction.

Title: First year french college. Is there a finite set of points of the plane A. Finite set of point A such as...

Body:
Is there a finite set of at least 3 points in the plane A, such as every 3 points of A are not aligned, and such as every circle going through 3 points of more of A, it’s center is in A ?
I believe that there can’t be such a set. So I tried showing that for every n point of the plane you can’t arrange them to be set A.

Being based on a number n I also tried doing it by doing a « mathematical induction » to show that there can’t be a set A for n points.

I’m a french student in my first year of college, I’m sorry by advance if my english is a bit wacky.

edit: forgot that it's a set of at least 3 points sorry


问:
Groups of transformations
I tried to find literature and articles about groups of transformations, but mostly what I found is either about groups or about transformations.

Can you suggest me literature where groups of transformations are described. Specifically I'm interested in groups of linear and orthogonal transformations.

P.S. I'd prefer books and articles in Russian.

答:
Mark A. Armstrong - Groups and Symmetry is definetely one of what you are looking for.

http://www.amazon.com/Groups-Symmetry-Undergraduate-Texts-Mathematics/dp/0387966757

Title: Groups of transformations


问:
Why is $a^{-1}$ used instead of $\frac{1}{a}$ in Calculus by Spivak?
In Calculus by Michael Spivak, why did he use the form $a^{-1}$ instead of $\frac{1}{a}$ to explain the multiplicative inverse (reciprocal) property? Is there a conceptual reason why the former is preferred rather than the latter? I know they are the same thing but I can't help but feel those who are not strong on proofs would have an easier time seeing the solution to problem 1(i) if it was used in that way rather than what was given. Any insight as to why his style is used or preferred in analysis is greatly appreciated.

答:
The question here is, how do we define exponentiation for negative integer powers? Given that our definition for $n\in\mathbb{N}$ is $$x^n=\underbrace{x\cdot ...\cdot x}_{n\text{ times}}$$
With the definition $x^0:=1$.

The property that we want exponentiation to satisfy is the equation $x^{a+b}=x^ax^b$. If we take this property as an axiom, then we can see
$$x^{1-1}=x^0=1$$
But on the other hand,
$$x^{1-1}=x^1x^{-1}=x\cdot x^{-1}$$
So we have
$$x\cdot x^{-1}=1$$
Hence we conclude
$$x^{-1}=\frac{1}{x}.$$
We can now extend this to any negative integer that we want. First write
$$x^{-n}=x^{\sum_{i=1}^n (-1)}$$
Then using the defining property of exponentiation,
$$=\prod_{i=1}^n x^{-1}={(x^{-1})}^n$$
We can go further! If we take as another axiom $x^{ab}={(x^a)}^b={(x^b)}^a$ (which we definitely want, since our previous work demands that $x^{mn}=x^m\cdot ...\cdot x^m$, $n$ times) then we can define exponentiation for reciprocals of natural numbers. The question is, what is $x^{1/n}$ ? Well,
$$x=x^1=x^{n/n}$$
Now using our other defining property of exponentiation,
$$x^{n/n}=(x^{1/n})^n=x$$
If we now define $\sqrt[n]{}$ as the inverse operation of exponentiation, we can say that
$$\sqrt[n]{(x^{1/n})^n}=\sqrt[n]{x}$$
But by the way we have defined this operation,
$$x^{1/n}=\sqrt[n]{x}$$
So in words, $x^{1/n}$ is the (I won't worry about uniqueness for now) number $a$ such that $a^n=x$. Now using the properties of exponentiation, we can extend this to any rational power:
$$x^{p/q}=\sqrt[q]{x^p}=\left(\sqrt[q]{x}\right)^p$$
Then of course we can extend this to any real number. Let $r\in\mathbb{R}$ be an irrational number. Let $\{p_n\},\{q_n\}$ be sequences of integers such that the sequence
$$r_n=\frac{p_n}{q_n}\to r\text{  as  }n\to\infty$$
We can then say
$$x^r:=\lim_{n\to\infty} x^{p_n/q_n}$$
While this definition is not particularly computationally efficient, it is the most intuitive.

Hopefully this serves as a good overview of why the notation is the way that it is and how, I think, exponentiation was defined in antiquity before fancy tools like $e^x$ and whatnot.


问:
A limit $\lim_{x \to 0} \sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 x^2} = \frac12$ and continuity of Kolmogorov distribution.
Problem: prove that

$$\sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 x^2} \to \frac12, x\to 0,$$
not using theta-function.

Motivation:

Kolmogorov–Smirnov statistic is well known. It's limit distribution has a continious distribution function $F(x)$ such that $F(x) = 0$ for $x \le 0$ and $$F(x) = 1 - 2 \sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 x^2}  = \frac{\sqrt{2\pi}}{x} \sum_{k=1}^{\infty} e^{-\frac{(2k-1)^2 \pi^2}{8x^2}}$$
otherwise.

The equivalence of two expressions $F_1(x) = 1 - 2 \sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 x^2}$ and  $F_2(x) = \frac{\sqrt{2\pi}}{x} \sum_{k=1}^{\infty} e^{-\frac{(2k-1)^2 \pi^2}{8x^2}}$ follows easily from transformation formula for theta-functions.

Using $F_2(x)$ it's easy to prove that $F(x)$ is continuous for all $x$. Using $F_1(x)$ it's easy to prove that $F(x)$ is continuous for all $x \ne 0$. My question was the next one: is there any simple way to prove that $F(x) $ is continuous at $x=0$, using the definition with $F_1(x)$?

As $F_1(x)$ is even then the statement "$F_1(x) \to 0, x \to 0+$" is equivalent to the statement "$F_1(x) \to 0, x \to 0$". So, my question is: is there any simple way (without theta-functions) to prove that
$$\sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 x^2} \to \frac12, x\to 0?$$

What do I know? I know that the statement  $\sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 x^2} \to \frac12, x\to 0$ follows from properties of theta-functions, but the question is about simple methods. I thought about the next idea: if $x=0$, then we have divergent series $S = 1 - 1 + 1 - 1 + 1 - \ldots = 1 - S$. It's a wrong way, but in some sense it's natural to write $S= 1 -S$ and hence $S = \frac{1}2$. Unfortunatelly, $S$ doesn't exist.

答:
Lots of questions about the Gauss circle problem today! Well, the number of lattice points in $x^2+y^2\leq N$ is $\pi N+O(\sqrt{N})$, and this implies that for $x\to 0^+$
$$ \sum_{n\geq 1} e^{-n^2 x} = -\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{x}}+O(\sqrt{x}) $$
as shown here, so
$$ \sum_{n\geq 1} e^{-2n^2 x} = -\frac{1}{2}+\frac{\sqrt{\pi}}{\sqrt{8x}}+O(\sqrt{x}) $$
$$ \sum_{n\geq 1} e^{-8n^2 x}=-\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{8x}}+O(\sqrt{x}) $$
as $x\to 0^+$. By absolute convergence
$$ \sum_{k\geq 1}(-1)^{k+1}e^{-2k^2 x}=-\sum_{k\geq 1}e^{-2k^2 x}+2\sum_{k\geq 1}e^{-8k^2 x}=\frac{1}{2}+O(\sqrt{x}) $$
as $x\to 0^+$. Actually, by the very same argument, all the right derivatives of the LHS at $x=0$ equal zero, so the LHS belongs to $(C^{\infty}\setminus C^{\omega})(\mathbb{R}^+)$.


答: Assuming $\text{Card} A\ge 3$, let $P_1, P_2, P_3$ be three points in $A$ such that the circle passing through the three points has maximal radius. Let C be the center of this circle. If any of the angles $P_i C P_j$ is $> 2\pi/3$, then the circle passing through $P_i, C, P_j$ has a larger radius, a contradiction. It follows that all the angles $P_i C Pj$ are equal to $2\pi/3$, but then the center of the circle passing through $P_1, C, P_2$ is aligned with $C$ and $P_3$, another contradiction. Hence there can be no such set with $\text{Card} A \ge 3$.


问:
Permutations and number of lists
How many permutations $a_1, a_2, a_3, a_4$ of $1, 2, 3, 4$ satisfy the condition that, for $k = 1, 2, 3$, the list $\{a_1, \dots, a_k\}$ has a number greater than $k$?

I am not sure if I understand what the question is asking. Probably not, since my answer is wrong.

What I found is:

For $k = 1$,

$\{2\},\{3\},\{4\}$

for $k = 2$,

$\{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}$

for $k = 3$,

$\{1,2,4\}, \{1,3,4\}, \{2,3,4\}$

That is $11$, but the answer is $13$.

答:
There are only $24$ permutations of four elements, so we might as well use brute force.

Suppose $a_1=4$, then all three conditions are automatically satisfied – for all of $k=1,2,3$, it is true that the first $k$ elements of the permutation have a number larger than $k$. This gives $6$ permutations.

Suppose $a_2=4$. Then the $k=2,3$ conditions are satisfied and the only remaining restriction is $a_1\ne1$. This gives $4$ permutations.

The $k=3$ condition is not satisfied if $a_4=4$, so let $a_3=4$. If $a_1=3$, all conditions are satisfied ($2$ more ways). If $a_4=3$, the $k=2$ condition cannot be satisfied. If $a_2=3$, the only way to satisfy all conditions is $a_1=2$ and $a_4=1$.

Summing it all up gives $13$ admissible permutations.

Title: Permutations and number of lists


答:
Assume such a set $A$ exists.

Let $X,Y$ be two points in $A$ that are the closest of all pairs of points in $A$.

Look at the bisector $l$ of the segment $XY$. For each point $Z\in A\setminus\{X,Y\}$ we can find the circumcentre of $\triangle XYZ$ and it will be on $l$. Thus, some points in $A$ will be on $l$, and so let's choose the point $O\in l$ so that it is the closest to the line $XY$.

Case 1: $O$ is on the line $XY$, and so $A$ contains three colinear points - a contradiction.

Case 2: $O$ is not on the line $XY$. Note that $OX=OY\ge XY$ so the angle $\angle XOY$ (opposite to the shortest side $XY$) is acute, and so all three angles of the isosceles triangle $\triangle OXY$ are acute. This means that the circumcentre $O'$ of $\triangle XOY$ lies inside $\triangle XOY$, and therefore $O'\in l$ and is closer to $XY$ than $O$ - a contradiction.


问:
Help me prove this inequality :
How would I go about proving this?

$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$

Thank you! I've tried so many things. I've tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn't work out, please could someone give me a little hint? Thank you!

Title: Help me prove this inequality :

Body:
Here is the proof that  $(1+1/x)^x$ is concave for $x\ge 1$.

The second derivative of $(1+1/x)^x$ is $(1+1/x)^x$ times $$p(x)=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}$$
Now for $x\ge 1$, we have $$\ln(1+1/x)-\frac{2}{1+x}\le \frac{1}{x}-\frac{2}{1+x}=\frac{1-x}{x(1+x)}\le 0$$ and $$\ln(1+1/x)\ge \frac{1}{x}-\frac{1}{2x^2}\ge 0,$$ so \begin{align*}p(x)&= \ln^2(1+1/x)-\frac{2\ln(1+1/x)}{1+x}+\frac{1}{(1+x)^2}-\frac{1}{x(1+x)^2}\\
&=\ln(1+1/x)(\ln(1+1/x)-2/(1+x))+\frac{x-1}{x(1+x)^2}\\
&\le \left(\frac{1}{x}-\frac{1}{2x^2}\right)\left(\frac{1}{x}-\frac{2}{1+x}\right)+\frac{x-1}{x(1+x)^2}\\
&=-\frac{(x-1)^2}{2x^3(1+x)^2}\le 0
\end{align*}
proving that $(1+1/x)^x$ is concave for $x\ge 1$.


答:
Since by definition $$
T(n)= n+ \sum_{k=1}^{n-1} (T(n-k)+T(k)) = n + 2\sum_{k=1}^{n-1} T(k)
$$
you can rewrite, for any $n$, that
$$
T(n+1)= n + 1 + 2\sum_{k=1}^{n} T(k) = 1+ 2T(n)+ \underbrace{n+ 2\sum_{k=1}^{n-1} T(k)}_{T(n)}
= 1+3T(n)
$$
This is easily solved: $T(1)=1$ and $T(n+1)=3T(n)+1$ have the solution
$$
\boxed{\forall n\geq 1,\qquad T(n) = \frac{1}{2}\cdot 3^{n}-\frac{1}{2}} \tag{$\ast$}
$$
which is indeed $O(3^n)$.

You can derive $(\ast)$ yourself by setting
$S(n) = T(n)+\frac{1}{2}$, so that $S(1) = \frac{3}{2}$ and observing that then
$$S(n+1) = T(n+1)+\frac{1}{2}=3T(n)+1+\frac{1}{2} = 3\left(T(n)+\frac{1}{2}\right) = 3S(n)$$ this is a geometric progression, so $S(n) = 3^{n-1}S(1)$. This gives you $T(n)$.

Title: Asymptotics in recurrence relation

Body:
I have an recurrence relation $T(n)= n+ \sum_{k=1}^{n-1} (T(n-k)+T(k)) = O(3^n)$ and $T(1)=1$.

I self study on asymptotic and recurrence example. Lots of example I read but this one is very strange!

How we can solve this recurrence? i.e: reach to answer.


问:
Convergence proof on Newton's method from Boyd & Vandenberghe's book on convex optimization
From Boyd & Vandenberghe's book, page 491:

Applying the Lipschitz condition,we have
$$\begin{aligned}
\left\|\nabla f\left(x^{+}\right)\right\|_{2} &=\left\|\nabla f\left(x+\Delta x_{\mathrm{nt}}\right)-\nabla f(x)-\nabla^{2} f(x) \Delta x_{\mathrm{nt}}\right\|_{2} \\
&=\left\|\int_{0}^{1}\left(\nabla^{2} f\left(x+t \Delta x_{\mathrm{nt}}\right)-\nabla^{2} f(x)\right) \Delta x_{\mathrm{nt}} d t\right\|_{2} \\
& \leq \frac{L}{2}\left\|\Delta x_{\mathrm{nt}}\right\|_{2}^{2} \\
&=\frac{L}{2}\left\|\nabla^{2} f(x)^{-1} \nabla f(x)\right\|_{2}^{2} \\
& \leq \frac{L}{2 m^{2}}\|\nabla f(x)\|_{2}^{2}
\end{aligned}$$
to get the inequality assumption (9.33), which is:

$$\frac{L}{2 m^{2}}\left\|\nabla f\left(x^{(k+1)}\right)\right\|_{2} \leq\left(\frac{L}{2 m^{2}}\left\|\nabla f\left(x^{(k)}\right)\right\|_{2}\right)^{2}$$

I dont understand how he get the first two lines of the derivation?

My understanding is that $\nabla f\left(x^{+}\right)$ equals $\nabla f\left(x+\Delta x_{\mathrm{nt}}\right)$ already when t=1, so where do the rest of two terms come from?
please anyone help me, really appreciate.

答: $-\nabla f(x)-\nabla^{2} f(x) \Delta x_{\mathrm{nt}}=0$ since $\Delta x _{nt}$ is the Newton search direction.


问:
Proving that $f(x)−x^p = g(x)$ has an infinite number of solutions for any continuous $f(x)$
Taken from an exam paper of analysis:

Let $g : (0, 1] \to [−1, 1]$ be the function defined by $g(x) := \sin (\frac {\pi}{2x})$.

(i) Prove that for every continuous function $f : (0, 1] \to [0, 1]$ and for every $p\in\mathbb N$, the equation $f (x)−x^p = g(x)$ has an infinite number of solutions.

[Hint: Apply the Intermediate Value Theorem to the function $h(x) :=
f (x) − x^p − g(x)$ on suitable intervals.]

Got no idea how to proceed, even with the IVT.  Should I seek a proof by contradiction?

答:
The hint you have referenced goes a long way:  first observe that as $n$ varies the intervals $\left[\frac{1}{4n+1}, \frac{1}{4n-1}\right]$ are mutually disjoint.  This is easy to verify as the right hand end of one interval $n$ is less than left hand end of the interval $n-1$, and similarly the left hand end of interval $n$ is greater than the right hand end of interval $n+1$, as follows from
$$ 4n-1 > 4(n-1)+1, \text{ when }n > 1 \quad{and}\quad
4n+1 < 4(n+1)-1, \text{ when } n\geqslant 1.$$

Also,
$$ g\left(\frac{1}{4n+1}\right) = \sin\left(2n\pi+\frac{1}{2}\pi\right) = 1$$ and
$$g\left(\frac{1}{4n-1}\right) = \sin\left(2n\pi - \frac{1}{2}\pi\right)=-1.$$
Then, when $n \geqslant 1$, and using the given fact $0 \leqslant f(x) \leqslant 1$,
$$f\left(\frac{1}{4n+1}\right) - \frac{1}{(4n+1)^p}-g\left(\frac{1}{4n+1}\right) \leqslant 1-\frac{1}{(4n+1)^p}-1<0,$$
while
$$f\left(\frac{1}{4n-1}\right) - \frac{1}{(4n-1)^p}-g\left(\frac{1}{4n-1}\right) \geqslant 0-\frac{1}{(4n+1)^p}+1>0.$$
By the intermediate value theorem $f(x) - x^p -g(x)$ will have a zero inside the interval.  Each interval is disjoint from the others, so you obtain the result you needed.


问:
Multidimensional signal synchronization
I am trying to find ways of synchronizing multidimensional waves gotten from the brain. For one dimensional signals, I used Cross correlation and was able to synchronise, but for multi dimensional signal, After searching online, I couldn't find somewhere to start from, can someone help with a list of different methods I can use.
Materials that can help will also be appreciated.</p>

答:
Abdulmueez—that's an interesting problem statement. Here is one idea (maybe you already tried it)—Could you just generalize the one-dimensional case to $p$-dimensional?

Assuming stationary sequences, the cross-correlation can be defined as $R_{xy}(m){\triangleq}E[x(m)y^\ast(0)]$ (where $E$ is the expectation operator). If $x(m)$ and/or $y(m)$ are random sequences, then $R_{xy}(m)$ is also a random sequence$\ldots$ but of course it can be estimated. One of the most common estimators $\hat{R}_{xy}(m)$ of the random sequence $R_{xy}(m)$ is the humble average such that
$$\hat{R}_{xy}(m)\triangleq\frac{1}{N}\sum_{n=1}^N x(n+m)y^{\ast}(n)$$
If the dimension is $p=2$, then at this point you could try just letting $x(m)$ and $y(m)$ be complex valued (sequences of ordered pairs) and using the real part for one dimension and the imaginary part as the other.

If $p\ge3$, then for the $p$-dimensional $x(m_1,m_2,\ldots,m_p)$ and $y(m_1,m_2,\ldots,m_p)$ you could define a $p$-dimensional cross-correlation function as $$R_{xy}(m_1,m_2,\ldots,m_p){\triangleq}E[x(m_1,m_2,\ldots,m_p)y^\ast(0,0,\ldots,0)]$$ and compute a $p$-dimensional estimate $\hat{R}_{xy}(m_1,m_2,\ldots,m_p)$ of this $p$-dimensional random sequence $R_{xy}(m_1,m_2,\ldots,m_p)$ by again averaging:
$$\hat{R}_{xy}(m_1,m_2,\ldots,m_p)\triangleq\frac{1}{N^{p}}\sum_{n_1=1}^N \sum_{n_2=1}^N\cdots\sum_{n_p=1}^N x(n_1+m_1,n_2+m_2,\ldots,n_p+m_p)y^{\ast}(n_1,n_2,\ldots n_p)$$

Title: Multidimensional signal synchronization


答:
This question seems to work best by looking at it from sub-lists that are excluded.

For $k=1$, we cannot start a permutation with $1$, so we lose $6$ from the starting $24$.

Then for $k=2$, we cannot have $[1,2], [2,1]$, but we have already eliminated $[1,2]$, and so we can safely remove $2$ more permutations ($[2,1,3,4], [2,1,4,3]$).

And finally for $k=3$, from the six permutations of the invalid $[1,2,3]$, we can remove $[1,.,.]$ and $[2,1,.]$, which means that we need to remove $[2,3,1,4],[3,1,2,4]$ and $[3,2,1,4]$, another $3$.

So the final answer is $24-6-2-3=13$.

Title: Permutations and number of lists

Body:
How many permutations $a_1, a_2, a_3, a_4$ of $1, 2, 3, 4$ satisfy the condition that, for $k = 1, 2, 3$, the list $\{a_1, \dots, a_k\}$ has a number greater than $k$?

I am not sure if I understand what the question is asking. Probably not, since my answer is wrong.

What I found is:

For $k = 1$,

$\{2\},\{3\},\{4\}$

for $k = 2$,

$\{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}$

for $k = 3$,

$\{1,2,4\}, \{1,3,4\}, \{2,3,4\}$

That is $11$, but the answer is $13$.


问:
Two proper maps inducing same map at the level of fundamental groups are properly homotopic
I am dealing with the following problem:

Let $X$ and $Y$ be two connected non-compact surfaces, possibly with
boundary. Let $f_0,f_1:(X,x_0)\to (Y,y_0)$ be two proper maps with
$f_{0*}=f_{1*}:\pi_1(X,x_0)\to \pi_1(Y,y_0)$, then $f_0$ is properly homotopic
to $f_1$.

Thoughts: Since $Y$ is a non-compact surface, its universal cover is a convex subset of $\Bbb R^2$, i.e., $Y=K(G,1)$ for some group $G$. If one goes through the proof the Hatcher's proposition 1B.9. on page 90, then the construction of the homotopy is dependent on $f_0$ and $f_1$, i.e., homotopy itself is proper if we are given $f_0$ and $f_1$ as proper maps.

Any help will be appreciated. Thanks in advance.

答:
This is (a) false, and (b) obviously false.

I will explain (b) by showing that your question fails in the first non-trivial example. (You should therefore try to think of examples...)

Take $X = Y = S^1 \times \Bbb R$, with $x_0 = y_0 = (1, 0)$. Set $f_0$ to be the identity map, and $f_1(x,t) = (x,-t)$. Then these clearly induce the same map on the fundamental group --- they are the identity on the central circle, and $X$ deformation retracts to its central  circle --- but they are also obviously not properly homotopic, as $\text{End}(X)$ is a two-point set, and proper maps induce maps on $\text{End}(X)$ which are equal if the original maps are properly homotopic. But $f_0$ is the identity on $\text{End}(X)$ and $f_1$ swaps the two ends.

If you want a proper map of a surfaces to be properly homotopic to the identity, you should (a) assume it is the identity on $\pi_1$, (b) assume it preserves ends, (c) assume it induces the identity on the fundamental group at infinity of each end. (For instance, this last condition distinguishes between the identity on $\Bbb R^2$ and the reflection map on $\Bbb R^2$.)


问:
Category $\mathcal{O}$ is closed under tensor product
I'm reading the proof of lemma 19.1 in Carter's Lie algebras of finite and affine type, which states that the category $\mathcal{O}$ is closed under tensor products. I'm having a bit of trouble understanding a couple of steps in the proof.

We first suppose $V_1,V_2\in\mathcal{O}$ (the category $\mathcal{O}$). Then $V_1=\bigoplus_{\lambda\in H^*}(V_1)_\lambda$ and $V_2=\bigoplus_{\lambda\in H^*}(V_2)_\lambda$ with $(V_1)_\lambda, (V_2)_\lambda$ finite-dimensional and it's weights bounded by a finite subset of $H^*$. Thus we have $V_1\otimes V_2=\bigoplus_{\lambda_1,\lambda_2\in H^*}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$. Now one have $((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})\subset (V_1\otimes V_2)_{\lambda_1+\lambda_2}$ which is also clear to me, but why does it imply that $V_1\otimes V_2=\bigoplus_{\lambda\in H^*}(V_1\otimes V_2)_\lambda$ with $(V_1\otimes V_2)_\lambda=\sum_{\lambda_1+\lambda_2=\lambda}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$. How does the inclusion implies that $(V_1\otimes V_2)_\lambda=\sum_{\lambda_1+\lambda_2=\lambda}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$? Wouldn't one need to show that $(V_1\otimes V_2)_\lambda\subset\sum_{\lambda_1+\lambda_2=\lambda}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$?

Now if there seems to be an easier proof of why the category $\mathcal{O}$ is closed under the tensor product, I'm very happy to see it (or maybe a reference to it).

答:
It's worthwhile to work slightly more generally, since this comes up whenever you have modules graded by an abelian group.

First, we have that tensor products commute with direct sums, so $$\left(\bigoplus_{i\in I} M_i \right) \otimes \left( \bigoplus_{j\in J} N_j \right)\cong \bigoplus_{(i,j)\in I\times J} M_i \otimes N_j.$$  For a quick proof, tensor products are left adjoint to hom, so they commute with arbitary colimits, and in particular coproducts (which is what direct sums are).

If the indexing sets are both some abelian group $A$, then we can further decompose
$$\left(\bigoplus_{a\in A} M_a \right) \otimes \left( \bigoplus_{b\in A} N_b \right)\cong \bigoplus_{(a,b)\in A\times A} M_i \otimes N_j\cong \bigoplus_{c\in A}\bigoplus_{a+b=c} M_a\otimes N_b.$$

At a more concrete level, given an element of the left hand side, we can write it as a sum of pure tensor products (things of the form $m\otimes n$), and we can decompose each $m$ and $n$ into homogenous components and then split things apart into a sum of pure tensors of homogenous elements.

Once we have this abstract decomposition, let's show that it's actually the decomposition of $V_1\otimes V_2$ into weight spaces.  It's a straight forward calculation that $(V_1)_{\lambda_1}\otimes (V_2)_{\lambda_2}$ is homogenous of weight $\lambda_1+\lambda_2$, and so we have an inclusion $$\bigoplus_{\lambda_1+\lambda_2=\lambda} (V_1)_{\lambda_1}\otimes (V_2)_{\lambda_2}\subset (V_1\otimes V_2)_{\lambda}$$, but summing over all $\lambda$ and using the above isomorphism, we have
$$V_1 \otimes V_2 \cong \bigoplus_{\lambda}\bigoplus_{\lambda_1+\lambda_2=\lambda} (V_1)_{\lambda_1}\otimes (V_2)_{\lambda_2}\subset \bigoplus_{\lambda}(V_1\otimes V_2)_{\lambda}\cong V_1\otimes V_2.$$

Moreover, since the isomorphisms come from the natural inclusions, this forces the inclusion in the middle to be equality.

If you wish to abstract out the idea here, if you have $A_i\subset B_i$ for all $i\in I$, and if $\bigoplus_i A_i = \bigoplus B_i$, then $A_i=B_i$ for all $i$.

Again, the reason this abstraction is what is happening is because the isomorphisms above are coming from the natural inclusions.

As far as addressing local finiteness, given a sum $v=\sum v_{\alpha}$, we have an inclusion $\mathfrak nv \subset \sum \mathfrak n v_{\alpha}$, and the above isomorphism tells us we can write any $v\in V_1\otimes V_2$ as a sum of elements of the form $v_{\lambda_1}\otimes v_{\lambda_2}$ where $v_1$ and $v_2$ are homogenous of weight $\lambda_1$ and $\lambda_2$ respectively, but $\mathfrak n(v_1\otimes v_2)\subset (\mathfrak n v_1)\otimes (\mathfrak n v_2)$, which is a tensor product of finite dimensional vector spaces, and therefore finite dimensional.  Since a finite sum of finite dimensional vector spaces is finite dimensional, the result follows.


问:
How to describe weak topologies T on the $\mathbb{R}$
I was unable to solve this question  of Exercise 2.4 (Weak Topologies)

Question : Let U be the usual topology on $\mathbb{R}$ . Describe the weak topology T on $\mathbb{R}$ induced by each of the following families of functions.

(a)The family of constant functions that map $\mathbb{R}$ into $(\mathbb{R} ,U)$

(b) The family that consists only of the function $i: \mathbb{R} \to (\mathbb{R}, U)$ defined by i(x)=x.

Now  the probelm is that i am not able to use the definition of weak topology in here.

Definition given in the book :Let {$X_{\alpha} , T_{\alpha}$}: $\alpha \in \Lambda$ be a indexed family of topological spaces, and for each $\alpha \in \Lambda$  let $f_{\alpha} : X\to X_{\alpha}$ , and for each $\alpha \in \Lambda$ let $ f_{\alpha}: X \to X_{\alpha} $be a function. The weak topology on X induced by {${ f_{\alpha} : \alpha \in \Lambda}$} is smallest topology for which each $f_{\alpha}$ is continuous.

Consider answering (a) part
I am unable to understand what should I take $f_{\alpha}$, should I take $\alpha \in \mathbb{R}$  ? and  fuamily of functions by $f_{\alpha} $ by $f_{\alpha}(x)=\alpha$ for all x in real numbers.

then i will get the weak  topology =$\mathbb{R}$.
Am I right?

For (b), I think it should also be $\mathbb{R}$ by the sam elogic as used for answering (a).

Am I right?

答:
For (a): As I think you were trying to say, the weak topology is indeed $\{\mathbb{R}, \emptyset\}$ (you just said $\mathbb{R}$, but every topology must contain the empty set as well, and moreover the empty set is needed here for continuity: for instance, if $f_0$ is the constant function mapping to $0$, then $f^{-1}(1, 2) = \emptyset$ needs to be open)

For (b): The weak topology is again $U$ itself. For any $U$-open set (meaning an open set in the standard topology $U$) $A \subseteq R$, $i^{-1}(A)$ must be open in the weak topology. But $i^{-1}(A) = A$ itself. So the weak topology must contain all the open sets of $U$. Since clearly $i: (\mathbb{R}, U) \rightarrow (\mathbb{R}, U)$ is continuous, and the weak topology is the smallest topology making $i$ continuous, the weak topology is $U$ itself.


问:
Why is $\varphi$ called "the most irrational number"?
I have heard $\varphi$ called the most irrational number. Numbers are either irrational or not though, one cannot be more "irrational" in the sense of a number that can not be represented as a ratio of integers. What is meant by most irrational? Define what we mean by saying one number is more irrational than another, and then prove that there is no $x$ such that $x$ is more irrational than $\varphi$.

Note: I have heard about defining irrationality by how well the number can be approximated by rational ones, but that would need to formalized.

答:
How well can a number $\alpha$ be approximated by rationals?
Trivially, we can find infinitely many $\frac pq$ with $|\alpha -\frac pq|<\frac 1q$, so something better is needed to talk about a good approximation.
For example, if $d>1$, $c>0$ and there are infinitely many $\frac pq$ with $|\alpha-\frac pq|<\frac c{q^d}$, then we can say that $\alpha$ can be approximated better than another number if it allows a higher $d$ than that other number. Or for equal values of $d$, if it allows a smaller $c$.

Intriguingly, numbers that can be approximated exceptionally well by rationals are transcendental (and at the other end of the spectrum, rationals can be approximated exceptionally poorly  - if one ignores the exact approximation by the number itself). On the other hand, for every irrational $\alpha$, there exists $c>0$ so that for infinitely many rationals $\frac pq$ we have $|\alpha-\frac pq|<\frac c{q^2}$. The infimum of allowed $c$ may differ among irrationals and it turns out that it depends on the continued fraction expansion of $\alpha$.
Especially, terms $\ge 2$ in the continued fraction correspond to better approximations than those for terms $=1$. Therefore, any number with infinitely many terms $\ge 2$ allows a smaller $c$ than a number with only finitely many terms $\ge2$ in the continued fraction. But if all but finitely many of the terms are $1$, then $\alpha$ is simply a rational transform of $\phi$, i.e. $\alpha=a+b\phi$ with $a\in\mathbb Q, b\in\mathbb Q^\times$.


问:
Find missing angle in triangle

In this triangle, we are given that $AB=AD=CE$, $BE=CD$ and angle  $ECD=2*AED=2a$.

We are asked to find $AED=a$.

I have seen several problems similar to this but can't follow any of the solution methods.
The only thing which is obvious to me is that since $AD+CD = BE+EC$, the triangle is isosceles.

Therefore its base angles (the ones I have drawn in green) are equal.
Also triangle ABD is isosceles, therefore its base angles are equal.

I don't see any other relation to use.

(FYI this is not homework or anything - just practice and self-motivation).
Thank you very much.

EDIT: If this is of any help, through Geogebra I get that $a=10^{\circ}$, therefore $ECD=20^{\circ}$ and we have a 80-80-20 triangle.

答:
Using Sine law,

Say, $\angle DAE = u, AB = x, DE = y, AE = z$

In $\displaystyle \triangle ADE, \frac{\sin u}{y} = \frac{\sin a}{x} \implies y = \frac{x \sin u}{\sin a}$

In $\displaystyle \triangle CDE, \frac{\sin 2a}{y} = \frac{\sin (a+u)}{x} \implies y = \frac{x \sin 2a}{\sin (a+u)}$

This leads to, $\, \sin u \sin(a+u) = \sin a \sin 2a$

Using $\cos(A - B) - \cos(A+B) = 2 \sin A \sin B,$ we get $u = a$.

Now in $\triangle ADE,$

$\displaystyle \frac{\sin 2a}{z} = \frac{\sin a}{x} \implies z = 2x \cos a$

In $\triangle ABE,$

$\angle ABE = 90^0 - a, \angle BAE = (90^0 - a) - a = 90^0 - 2a, \angle BEA = 3a$

$\displaystyle \frac{\sin 3a}{x} = \frac{\sin (90^0 - a)}{z}$

$\displaystyle \sin 3a = \frac{1}{2} \implies a = 10^0$

Title: Find missing angle in triangle


问:
Binomial coefficient modulo a prime
I cannot find a reference for the following fact: "Lep $p$ be a prime. If the binomial coefficient of $d$ over $l$ is zero modulo $p$ for every $l \in \{1,\ldots,d-1\}$ then $d$ is a power of $p$." Do you have any suggestions? Thank you in advance.

Edit: I would like to thank both of you for your suggestions, but I would need a reference to put in the bibliography. Could you suggest any book or paper in which such a property is stated?

答:
$$(1+x)^d -1-x^d = \sum_{l=1}^{d-1}{d\choose l} x^l$$
$p$ divides all the ${d\choose l}$ iff $$(1+x)^d -1-x^d=0\in \Bbb{F}_p[x] \tag{1}$$

Write $d= np^k$ where $p\nmid n$. From the obvious $p | {p\choose l}$ for $1\le l\le p-1$ we get by induction on $k$ that $$(1+x)^{p^k}= 1+x^{p^k}\in \Bbb{F}_p[x]$$

If $d$ is not a power of $p$ then $$(1+x)^d=(1+x^{p^k})^n = 1+n x^{p^k}+ \sum_{m=2}^n{n\choose m} (x^{p^k})^m\ne 1+x^d \in \Bbb{F}_p[x]$$
Thus $(1)$ holds iff $d$ is a power of $p$.

Title: Binomial coefficient modulo a prime


问:
$Y=UX$ with $U \sim \mathcal{U}[0,1]$ and $X \sim \gamma(n+1)$
Let :

$U \sim \mathcal{U}[0,1]$

$X \sim \gamma(n+1)$ that is to say :

$X= \sum_{k=0}^n T_k$ with $(T_k)_k$ iid and follow an exponential distribution with parameter $1$

$Y=UX$

$U$ and $X$ independent.

We want to prove that : $f_Y(y)= \frac{1}{n} \sum_{k=0}^{n-1} \frac{ y^k e^{-y} }{ k!}$ for $y >0$

My attempt :

$C= \ln(UX) = \ln U + \ln X$

$f_C(t)= \int_{ - \infty}^{\infty} e^{t-u} e^{u} f(e^u)  \mathbb{1}_ {u <0} du = e^t \int_{t}^{+ \infty} f(e^u) du$

$ f_Y(t) = \frac{1}{t} f_C( \ln t) = \int_{ \ln t}^ { \infty} f(e^u) du = \int_{t}^{\infty}f(u) du $

\begin{align*}
f_Y(t)
& = \int_{t t} ^{ \infty} f_X(x) dx   \\
&=   \int_{ t} ^{ \infty} \frac{1}{n !} e^{-x}x^n dx   \\
&= \frac{1}{n!}  \frac{1}{t} ( t ) ^n +  \int_{ t} ^{ \infty} \frac{1}{(n-1) !} e^{-x}x^{n-1} dx  \\
\end{align*}

答:
Setting

$$\begin{cases}
y=ux\\
z=x
\end{cases}\rightarrow\begin{cases}
x=z\\
u=\frac{y}{z}
\end{cases}$$

The Jacobian is $|J|=\frac{1}{z}$ and thus

$$f_{YZ}(y,z)=\frac{1}{n!}z^{n-1}e^{-z}\cdot\mathbb{1}_{[0;\infty)}(y)\mathbb{1}_{[y;\infty)}(z)$$

Thus

$$f_Y(y)=\frac{1}{n!}\int_y^{\infty}z^{n-1}e^{-z}dz\cdot\mathbb{1}_{[0;\infty)}(y)$$

You can leave the density in this way or observing that

$$f_Y(y)=\frac{\Gamma(n,y)}{n!}\cdot\mathbb{1}_{[0;\infty)}(y)$$

probably this expression can be expressed in what you are looking for...

Just as an example, with $n=3$ you get

$$f_Y(y)=\frac{1}{6}\int_y^{\infty}z^2e^{-z}dz=\frac{e^{-y}}{6}[y^2+2y+2]\cdot\mathbb{1}_{[0;\infty)}(y)$$

which is a nice density


问:
What kind of geometric object does $3x^2+4y^2-12x+8y+17z=0$ represent?
I am given the following equation $$3x^2+4y^2-12x+8y+17z=0$$ and I have to find what kind of geometric object this is.

I don't think this is a sphere because I can't find the radius or the center from the given information. And I also don't think that this is a cylindrical surface because it contains all the variables $x$, $y$, and $z$. But I may be wrong.

Could you please explain what this is? Thank you very much

答:
It is elliptical paraboloid:

For each fixed $z\leq {16\over 17}$ you get an ellipse:

$$3(x-2)^2 + 4(y+1)^2 = 16-17z$$

For each fixed $x$ or $y$ you get a parabola.


问:
cantor set prove verfication
Question:

My proof:

Can anyone help me verify my proof?

答:
This proof is correct (apart from, as noted in comments, the first line on the $2$nd page should still say $x\in C$ as opposed to $x\in C^c$).

I would probably simplify the first part of the proof by always choosing the "left" end of the subinterval, rather than caring a lot whether to take the left or the right end.

Commentary: In essence, if you interpret Cantor set as the real numbers in $[0,1]$ written in the ternary system without using the digit $1$, then, for $x\in C$ you have the ternary expansion such as:

$$x=0.a_1a_2\ldots a_{n-1}a_na_{n+1}\ldots$$

where $a_i\in\{0,2\}$, and in the first case you can use:

$$x_n=0.a_1a_2\ldots a_{n-1}a_n$$

(a finite expansion - that is what I am suggesting above), or:

$$x_n=0.a_1a_2\ldots a_{n-1}a_na_na_n\ldots$$

(which is what you are doing in your original proof). In the second case, you are using

$$x_n=0.a_1a_2\ldots a_{n-1}111\ldots$$

which is perfectly ok.

Title: cantor set prove verfication


答:
It is an elliptic paraboloid.

If you know the canonical equation of an elliptic paraboloid, you can show this by completing the square.

Title: What kind of geometric object does $3x^2+4y^2-12x+8y+17z=0$ represent?

Body:
I am given the following equation $$3x^2+4y^2-12x+8y+17z=0$$ and I have to find what kind of geometric object this is.

I don't think this is a sphere because I can't find the radius or the center from the given information. And I also don't think that this is a cylindrical surface because it contains all the variables $x$, $y$, and $z$. But I may be wrong.

Could you please explain what this is? Thank you very much


问:
$\iint f(x,y)\,dxdy$ and $\iint f(x,y)\,dydx$ exist but $f$ not integrable on $[0,1]\times[0,1]$
I want to look for a function $f(x,y)$, whose support is inside $[0,1]\times[0,1]$, such that $\int_0^1\!\int_0^1\!f(x,y)\,dxdy$ and $\int_0^1\!\int_0^1\!f(x,y)\,dydx$ both exist, but $f(x,y)$ is not Riemann-integrable (or Darboux-integrable) on $[0,1]\times[0,1]$.

By Riemann-Lebesgue Theorem, I know that the set of discontinuities of $f$ in $[0,1]\times[0,1]$ cannot be contained in a set of measure $0$ in $\mathbb{R}^2$, but with each fixed $x$ (or fixed $y$), the set of continuities of $f_x(y)$ or $f_y(x)$ can be contained in a set of measure $0$ in $\mathbb{R}^1$.

I'm unable find a set of continuities in $[0,1]\times[0,1]$ that satisfy this, thus unable to find a function like this. Please help. Thank you very much.

Further question: is it possible that $\int_0^1\!\int_0^1\!f(x,y)\,dxdy=\int_0^1\!\int_0^1 \!f(x,y)\,dydx$ but $f(x,y)$ is not Riemann-integrable on $[0,1]\times[0,1]$?

答:
There are less complicated examples (at least this won't require a page! :)). Try
$$f(x,y) = \begin{cases} 1\,, & x = k/q \text{ and } y = \ell/q \text{ for some integers } k,\ell \text{ with $q$ prime} \\ 0\,, & \text{otherwise}\end{cases}\,.$$
Then each horizontal/vertical line segment contains (at most) finitely many discontinuities of $f$. But you need to do the exercise of showing the set $\{(k/q,\ell/q): k,\ell\in\mathbb N,\ q \text{ prime}\}$ is dense in $[0,1]\times [0,1]$.

Oh, and this function answers your second question. Both iterated integrals are $0$.


答:
I believe it's an elliptic parabloid.

I think if you complete the square you will get something more decipherable like this:

$$3(x-2)^2 +4(y+1)^2 +17z -16 = 0$$


问:
Intermediate Value Theorem and polynomial of odd degree
As a consequence of Intermediate Value Theorem (IVT), it is illustrated in almost all calculus books and notes/courses that every polynomial of odd degree over real numbers have a real root.

I tried to see whether this result has been proved without the use of intermediate value property, and I did not find any link or question related to that on this and other sites.

Can one suggest me whether there is a proof of above mentioned consequence of IVT, which is done without using IVT?

I had seen that the fundamental theorem of algebra have been of interest to many people for proving just using algebraic techniques, but among all the known various proofs of FTA, epsilon amount of analysis is used.

So, my above question is of similar nature, but I did not see any comments on its proofs without IVT.

答:
Every polynomial $P$ can be split (not sure of the word, the one in French is "scindé") in $\mathbb{R}$ : $P = \prod (X^2-a_k X + b_k) \prod (X- \lambda_k)$. As the degree of $P$ is odd, there is at least one factor $(X-\lambda)$, so there is at least one real root.

As for Vivid's proof, the proof of Descarte's Rule of Signs is based on IVT, so it is not really a different proof.

(Sorry if my words are incorrect, I'm French and I'm not used to writing maths in English.)


问:
Is there any analytical approximation to this integral?
I have an integral:$$ I(x) = x\int_{x}^{\infty}K_{5/3}(\xi)\,\rm{d}\xi~,$$
where $K$ is modified Bessel function.
I need to evaluate this integral for tens of thousand different $x$ ultimately to use it for some numerical calculation. In the extreme limits of $x\gg 1$ or $x\sim 0$, the approximations are:
$$\lim_{x\rightarrow 0}I(x) = \frac{4\pi}{\sqrt{3}\Gamma(1/3)}\left(\frac{x}{2}\right)^{1/3}~,$$
and
$$\lim_{x\rightarrow \infty}I(x) = \sqrt{\frac{\pi x}{2}}e^{-x}~.$$

However, I would like to know if there are any approximate expressions $I_\text{approx}(x)$ valid for all $x$ such that $$\frac{|I_\text{approx}(x)-I(x)|}{I(x)}\lesssim 0.1~,$$
for all $x$.

Body:
Using hypergeometric functions, the antiderivative exists. So,

$$I(x)=-\frac{\pi  x}{\sqrt{3}}-\frac{2^{2/3} \sqrt{3} \pi  x^{1/3} \,
   _1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};\frac{x^2}{4}\right)}{\Gamma
   \left(-\frac{2}{3}\right)}-$$ $$\frac{81 x^{11/3} \Gamma \left(\frac{4}{3}\right) \,
   _1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};\frac{x^2}{4}\right)}{320\
   2^{2/3}}$$

As series,
$$_1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};u\right)=\sum_{n=0}^\infty \frac{\Gamma \left(-\frac{2}{3}\right)}{(1-3 n) n! \Gamma \left(n-\frac{2}{3}\right)}u^n$$
$$_1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};u\right)=\sum_{n=0}^\infty \frac{4 \Gamma \left(\frac{8}{3}\right)}{(3 n+4) n! \Gamma \left(n+\frac{8}{3}\right)} u^n$$


答:
Mathematica image

p = 7 ;  ContourPlot3D[
3 x^2 + 4 y^2 - 12 x + 8 y + 17 z = = 0, {x, -p, p}, {y, -p,
p}, {z, -p, p}]

Could be reducible to form

$$\dfrac{z}{c} =  1 -\dfrac{( x-p)^2}{a^2}-\dfrac{ (y-q)^2}{b^2} ...$$

Elliptic paraboloid as the sections parallel to axes are ellipses, parabolas..


问:
How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$?
in a theoretical physics book, the author makes the following claim:

$$\frac{1}{e^z + 1} = \frac{1}{2} + \sum_{n=-\infty}^\infty \frac{1}{(2n+1) i\pi - z}$$
and justifies this as

These series can be derived from a theorem which states that any meromorphic function may be expanded as a summation over its poles and residues at those poles

What's the name of that theorem? It's not really a Laurent series, since the Laurent series is for an expansion around one particular point only. I can see that the poles occur whenever $z = (2n+1)i\pi$ for $n \in \mathbb{N}$, but then where does that constant $1/2$ come from?

EDIT: Well, it appears that the general claim isn't valid, so now I'd be interested in a justification for the expansion in my particular example...

答:
Mittag-Leffler's theorem guarantees the existence of a meromorphic function $g(z)$ whose poles and principal parts are given by any values specified. Then, if $f(z)$ is a meromorphic function, then $f(z) - g(z)$ is holomorphic, and it remains to compute this difference. In practice this is probably nontrivial, because $g(z)$ is not uniquely determined, but for functions with nice poles and principal parts, this is possible.

Such a possibility applies in your case with $f(z) = 1/(e^z + 1)$. We can justify the formula you gave in your question by using an approach based on a discussion between me and one of my friends, so I do not claim the credit for these ideas.

In order to properly handle the convergence of the infinite sum, we should first symmetrize the infinite sum you gave, so instead let
$$ g(z) = \sum_{k > 0 \text{, odd}} \left( \frac{1}{k i \pi - z} - \frac{1}{k i \pi + z} \right) = -\sum_{k > 0 \text{, odd}} \frac{2z}{z^2 + k^2 \pi^2}$$

We can check that $g(z)$ is a meromorphic function whose poles and principal parts match that of $f(z) = 1/(e^z + 1)$, so it follows that $f(z)- g(z)$ is entire. It remains to compute this difference. First, notice that both $f(z)$ and $g(z)$ are $2 \pi i$ periodic.  So to check the growth of $f(z), g(z)$, we need only check the behavior as $\mathrm{Re}(z) \rightarrow \pm \infty$. Notice that $f(z) \rightarrow 1,0$ as $\mathrm{Re}(z) \rightarrow -\infty, +\infty$, respectively. Thus it follows that $f(z)$ is in fact uniformly bounded away from its poles. To check $g(z)$, split the sum as
$$ g(z) = \sum_{0<k<2|z|/\pi, \text{ odd}} + \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{-2z}{z^2 + k^2 \pi^2} = S_1(z) + S_2(z)$$
Now, notice that for $\mathrm{Re}(z)$ sufficiently large,
\begin{align}
|S_1(z)| & =  \left|\frac{-2}{z} \sum_{0<k<2|z|/\pi, \text{ odd}} \frac{1}{1 + k^2 \pi^2/z^2} \right| \le \frac{2}{|z|} \frac{2|z|}{\pi} = 4,\\
|S_2(z)| & = \left| \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{-2z}{z^2 + k^2 \pi^2} \right| \\
& \le \frac{8}{3} |z| \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{1}{ \pi^2 k^2} \\
& \le \frac{8}{3} |z| \int_{-1 + 2|z|/\pi}^{\infty} \frac{1}{\pi^2 s^2} \, ds \\
& \le \frac{8}{3} \frac{|z|}{-1 + 2|z|/\pi} \le C
\end{align}
for $C > 0$ a constant. Thus $g(z)$ is also uniformly bounded away from its poles. Then, it follows that the difference $f(z) - g(z)$ is uniformly bounded, and being entire, then by Louiville's theorem it must be constant. Now, we compute that
$$ f(0) - g(0) = \frac{1}{2} - 0 = \frac{1}{2} $$
and hence $f(z) - g(z) \equiv 1/2$.


问:
What is the probability of $X <Y <Z$?
I have been trying to derive the probability $Pr[X<Y<Z]$, where $X$, $Y$, and $Z$ are independent and follow exponential distribution with parameters $\lambda_x$, $\lambda_y$, and $\lambda_z$, respectively.

What I did is as follows:

$\Pr \left[ {X < Y < Z} \right] = \Pr \left[ {X < Y,Y < Z} \right]\\
 = \int\limits_0^\infty  {\Pr \left[ {X < y,Z > y} \right]{f_Y}(y)dy} \\
 = \int\limits_0^\infty  {\Pr \left[ {X < y} \right]\Pr \left[ {Z > y} \right]{f_Y}\left( y \right)dy}$

Is it right or wrong?
If it is wrong, how can we solve this problem?
Thank you very much.

答:
You can integrate the joint pdf over the region of integration in 3!=6 different ways:

$\begin{gathered}
  \int\limits_{z = 0}^{z = \infty } {\int\limits_{y = 0}^{y = z} {\int\limits_{x = 0}^{x = y} {{f_{X,Y,Z}}\left( {x,y,z} \right)dxdydz} } }  \hfill \\
  \int\limits_{z = 0}^{z = \infty } {\int\limits_{x = 0}^{x = z} {\int\limits_{y = x}^{y = z} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydxdz} } }  \hfill \\
  \int\limits_{y = 0}^{y = \infty } {\int\limits_{x = 0}^{x = y} {\int\limits_{z = y}^{z = \infty } {{f_{X,Y,Z}}\left( {x,y,z} \right)dzdxdy} } }  \hfill \\
  \int\limits_{y = 0}^{y = \infty } {\int\limits_{z = y}^{z = \infty } {\int\limits_{x = 0}^{x = y} {{f_{X,Y,Z}}\left( {x,y,z} \right)dxdzdy} } }  \hfill \\
  \int\limits_{x = 0}^{x = \infty } {\int\limits_{z = x}^{z = \infty } {\int\limits_{y = x}^{y = z} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydzdx} } }  \hfill \\
  \int\limits_{x = 0}^{x = \infty } {\int\limits_{y = x}^{y = \infty } {\int\limits_{z = y}^{z = \infty } {{f_{X,Y,Z}}\left( {x,y,z} \right)dzdydx} } }  \hfill \\
\end{gathered} $

You may be able to find a close form expression among these. I haven't try it myself yet !

Title: What is the probability of $X <Y <Z$?


问:
Prove that dual space of $\ell^1$ is $\ell^{\infty}$
Prove that dual space of $\ell^1$ is $\ell^{\infty}$

My attempt : I got the answer Here but im not able to understand the answer

we  know that the norm of  $ x\in \ell^1$ is given by  $||x||_1=\sum_{k=1}^{\infty}|a_k|$

norm of  $ x\in \ell^{\infty}$ is given by  $||x||_{\infty}=\sup_{k\in \mathbb{N}}|a_k|$

Now here my proof start :

Since  $\ell^1$  is infinite dimensional because it contains the infinite sequence in the form $(0,0,\dots,1,0,\dots)$

So there exists a basis $\{e_1,e_2,\dots,e_k\dots\}$ of $\ell^1$  where $e_k=M_{jk}=\begin{cases} 1 &\text{ if } j=k \\ 0 & \text{ if } j \neq k. \end{cases}$

This implies that every $x \in \ell^1$ can be written as $x=a_1e_1+a_2e_2+\dots$

Now take  a bounded linear functional  $f$ of $\ell^1$

$f: \ell^1 \to \mathbb{R}$ defined by $f(x)= f(a_1e_1+a_2e_2+\dots)= a_1f(e_1)+a_2 f(e_2)+\dots=\sum_{k=1}^{\infty}a_kf(e_k)$

After that I am not able to proceed further..

答:
Clearly, every element of $v\in\ell^\infty$ defines an element of the dual of $\ell^1$, since if $v=(v_j)$ and $x=(x_j)\in\ell^1$, then
$$
v(x)=\sum_j v_jx_j\quad\text{and}\quad |v(x)|\le \sum_j |v_j||x_j|\le \big(\sup_j |v_j|\big)\sum_j|x_j|=\|v\|_\infty\|x\|_1
$$
Let $\varphi\in(\ell^1)^*$ and set $v_j=\varphi(e_j)$ and $v=(v_j)$. Clearly
$$
|v_j|=|\varphi(e_j)|\le \|\varphi\|_*\|e_j\|_1=\|\varphi\|_*
$$
and hence $v\in\ell^\infty$ and $\|v\|_\infty\le \|\varphi\|_\infty$. It remains to show that $\varphi(x)=v(x)$, for all $x\in\ell^1$ and $\|v\|_\infty= \|\varphi\|_*$.

Clearly, $\varphi(x)=v(x)$, for $x=e_j$ and for all $x$'s which are finite linear combinations of the $e_j$'s. They are also both bounded linear functionals, and they agree on a dense subset of $\ell^1$, and hence the agree everywhere, i.e., $v\equiv \varphi$.

For the final part, it remains to show that $\|v\|_\infty\ge\|\varphi\|_*$.
Now, for every $\epsilon>0$, there exists a unit vector $w=(w_j)\in\ell^1$, such that
$$
|\varphi(w)|>\|\varphi\|_*-\epsilon
$$
and also there exists $n\in\mathbb N$, such that
$\|w-w^n\|_1<\varepsilon$, where $w^n=(w_1,w_2,\ldots,w_n,0,0,\ldots)$ and $v(w^n)=\varphi(w^n)$, while $\|w^n\|_1\le 1$. So
$$
\|v\|_\infty\ge |v(w^n)|=|\varphi(w^n)| \ge
|\varphi(w)|- |\varphi(w-w^n)| \ge
\|\varphi\|_*-\varepsilon-\|\varphi\|_*|w-w^n|_1 \\
\ge \|\varphi\|_*-\varepsilon-\varepsilon\|\varphi\|_*=
\|\varphi\|_*-\epsilon(1+\|\varphi\|_*)
$$
and this is true for all $\varepsilon>0$, which implies that $\|v\|_\infty\ge\|\varphi\|_*$.


问:
Measure of the set of real numbers that can be approximated in this way
Let $$A = \{x \in \mathbb{R}\mid \exists\,\text{infinitely many pairs of integers $p,q$ such that $|x-p/q| \leq 1/q^3$}\}.$$  Is the measure of $A$ equal to $0$?  Any ideas?</p>

答:
Let's restrict to $(0,1]$ and let $A_{q}=\bigcup_{p=1}^q B(\frac pq,\frac 1{q^3})$. Then $\mu(A_{p,q})\le \frac 1{q^2}$.
We are looking for
$$ \mu(A\cap(0,1])\le \mu\left(\bigcap_{n=1}^\infty\bigcup_{q=n}^\infty A_q\right)\le \mu\left(\bigcup_{q=n}^\infty A_q\right)\le \sum_{q=n}^\infty\mu(A_q)\le\sum_{q=n}^\infty\frac1{q^2}.$$
Since the last expression is the tail of the convergent series $\sum_{q=1}^\infty\frac1{q^2}(=\frac{\pi^2}{6})$, it tends $\to0$ as $n\to\infty$. We conclude $\mu(A\cap(0,1])=0$ and as $A$ is the countable union of translated copies of $A\cap(0,1]$, we also have $\mu(A)=0$.</p>


问:
Finding the equation of a line in a plane, concurrent with two lines that are themselves determined by pairs of planes
I need to find a vectorial line equation that is contained in the plane $\pi : x - y + z = 0$ and the line t is concurrent with the lines
$$ r : \left\{ \begin{array}{lcr} x+y+2z-2 & = & 0 \\ x-y & = & 0 \end{array} \right. \qquad s : \left\{ \begin{array}{lcr} x-z+2 & = & 0 \\ y & = & 0 \end{array} \right.$$

I found the vectorial line equation of r and s, but I don't know what to do after this.

$$\begin{align}
r : (\phantom{-}2,2,-1) + \alpha(\phantom{-}2,2,-2) \\
s : (-1,0,\phantom{-}1) + \beta(-1,0,-1)
\end{align}$$

答:
Hint.

Given the lines $r,s,t$ and the plane $\pi$

$$
\cases{
r\to p=p_1 + \lambda_1\vec v_1\\
s\to p=p_2 + \lambda_2\vec v_2\\
\pi\to (p-p_0)\cdot\vec n_0=0 }
$$

we have

$$
\cases{r\cap \pi\to \lambda_1^* = \frac{(p_0-p_1)\cdot\vec n_0}{\vec v_1\cdot\vec n_0}\\
s\cap \pi\to \lambda_2^* = \frac{(p_0-p_2)\cdot\vec n_0}{\vec v_2\cdot\vec n_0}\\
t\in \pi\to p = p_1 + \lambda_1^*\vec v_1+\mu\left(p_2 + \lambda_2^*\vec v_2-p_1 - \lambda_1^*\vec v_1\right)\\
}
$$

Here

$$
\cases{
p_1 = (1,1,0),\ \
\vec v_1 = (1,1,-1)\\
p_2 = (-2,0,0),\ \
\vec v_2 = (1,0,1)\\
p_0 = (1,1,0),\ \
\vec n_0 = (1,-1,1)
}
$$

NOTE

In red $t$, in blue $s, r$, in black the intersection points and in yellow $\pi$. Attached a MATHEMATICA script to build the graphics.

```<code>r1 = {1, 1, 2};
r2 = {1, -1, 0};
s1 = {1, 0, -1};
s2 = {0, 1, 0};
p0 = {1, 1, 0};
n0 = {1, -1, 1};
v1 = Cross[r1, r2]
v2 = Cross[s1, s2]
p1xy = Solve[{x + y + 2 z - 2 == 0, x - y == 0}, {x, y}][[1]];
p1 = {x, y, z} /. p1xy /. {z -> 0}
p2xy = Solve[{x - z + 2 == 0, y == 0}, {x, y}][[1]];
p2 = {x, y, z} /. p2xy /. {z -> 0}
l1 = (p0 - p1).n0/v1.n0
l2 = (p0 - p2).n0/v2.n0
p10 = p1 + l1 v1
p20 = p2 + l2 v2
t = p10 mu + (1 - mu) p20
gr1 = ParametricPlot3D[p1 + lambda v1, {lambda, -2, 2}];
gr2 = ParametricPlot3D[p2 + lambda v2, {lambda, -2, 2}];
gr0 = ContourPlot3D[x - y + z == 0, {x, -4, 4}, {y, -4, 4}, {z, -4, 4}, Mesh -> None, ContourStyle -> Directive[Yellow, Opacity[0.6],Specularity[White, 30]]];
grt = ParametricPlot3D[t, {mu, 0, 1}, PlotStyle -> Red];
grp1 = Graphics3D[{Black, Sphere[p10, 0.1]}];
grp2 = Graphics3D[{Black, Sphere[p20, 0.1]}];
Show[gr1, gr2, gr0, grt, grp1, grp2]
</code>```

Body:
Hint.

Given the lines $r,s,t$ and the plane $\pi$

$$
\cases{
r\to p=p_1 + \lambda_1\vec v_1\\
s\to p=p_2 + \lambda_2\vec v_2\\
\pi\to (p-p_0)\cdot\vec n_0=0 }
$$

we have

$$
\cases{r\cap \pi\to \lambda_1^* = \frac{(p_0-p_1)\cdot\vec n_0}{\vec v_1\cdot\vec n_0}\\
s\cap \pi\to \lambda_2^* = \frac{(p_0-p_2)\cdot\vec n_0}{\vec v_2\cdot\vec n_0}\\
t\in \pi\to p = p_1 + \lambda_1^*\vec v_1+\mu\left(p_2 + \lambda_2^*\vec v_2-p_1 - \lambda_1^*\vec v_1\right)\\
}
$$

Here

$$
\cases{
p_1 = (1,1,0),\ \
\vec v_1 = (1,1,-1)\\
p_2 = (-2,0,0),\ \
\vec v_2 = (1,0,1)\\
p_0 = (1,1,0),\ \
\vec n_0 = (1,-1,1)
}
$$

NOTE

In red $t$, in blue $s, r$, in black the intersection points and in yellow $\pi$. Attached a MATHEMATICA script to build the graphics.

r1 = {1, 1, 2};
r2 = {1, -1, 0};
s1 = {1, 0, -1};
s2 = {0, 1, 0};
p0 = {1, 1, 0};
n0 = {1, -1, 1};
v1 = Cross[r1, r2]
v2 = Cross[s1, s2]
p1xy = Solve[{x + y + 2 z - 2 == 0, x - y == 0}, {x, y}][[1]];
p1 = {x, y, z} /. p1xy /. {z -> 0}
p2xy = Solve[{x - z + 2 == 0, y == 0}, {x, y}][[1]];
p2 = {x, y, z} /. p2xy /. {z -> 0}
l1 = (p0 - p1).n0/v1.n0
l2 = (p0 - p2).n0/v2.n0
p10 = p1 + l1 v1
p20 = p2 + l2 v2
t = p10 mu + (1 - mu) p20
gr1 = ParametricPlot3D[p1 + lambda v1, {lambda, -2, 2}];
gr2 = ParametricPlot3D[p2 + lambda v2, {lambda, -2, 2}];
gr0 = ContourPlot3D[x - y + z == 0, {x, -4, 4}, {y, -4, 4}, {z, -4, 4}, Mesh -> None, ContourStyle -> Directive[Yellow, Opacity[0.6],Specularity[White, 30]]];
grt = ParametricPlot3D[t, {mu, 0, 1}, PlotStyle -> Red];
grp1 = Graphics3D[{Black, Sphere[p10, 0.1]}];
grp2 = Graphics3D[{Black, Sphere[p20, 0.1]}];
Show[gr1, gr2, gr0, grt, grp1, grp2]


问:
Unable to find weak topology T in $\mathbb{R}$
I am self studying topology and I was unable to solve this question.

Wayne Patty's Question 2.4.5.

Let $U$ be usual topology on $\mathbb{R}$ and let $T$ be the weak Topology on $\mathbb{R}$ induced by the family of all bounded real valued continuous functions that map $(\mathbb{R},U)  $ to $(\mathbb{R}, U)$. Describe $T$.

The problem is, how do I know which topology will be smallest? Also $f_{\alpha}$ 's can be constant functions and more functions also like $\sin x$.

Also, what approach should one follow to find smallest topologies which make the all $f_{\alpha}$'s continuous?

Can you please help me with this?

答:
If I understand your question right, it seems to me that any interval $(a,b)$ is in $T$.

This is because you can have a (bounded) function such as:

$$f(x)=\begin{cases}-1&xb\end{cases}$$

and $f^{-1}((-1,1))=(a,b)$. So, as the interval $(-1,1)$ is open in $U$, the interval $(a,b)$ must be open in $T$ if $f$ is to stay continuous as a map $(\mathbb R, T)\to(\mathbb R, U)$.

However, then it follows that every open set in $U$ is also in $T$, i.e. $T=U$.

Body:
If I understand your question right, it seems to me that any interval $(a,b)$ is in $T$.

This is because you can have a (bounded) function such as:

$$f(x)=\begin{cases}-1&x<a\\\text{linear: }\frac{2x-(a+b)}{b-a}&a\le x\le b\\1&x>b\end{cases}$$

and $f^{-1}((-1,1))=(a,b)$. So, as the interval $(-1,1)$ is open in $U$, the interval $(a,b)$ must be open in $T$ if $f$ is to stay continuous as a map $(\mathbb R, T)\to(\mathbb R, U)$.

However, then it follows that every open set in $U$ is also in $T$, i.e. $T=U$.


问:
The easiest way possible to evaluate $\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$
I want to solve this integral the easiest way possible, how should I do it?
$$\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$$

I tried with substitution, but it didn't work out. And I think that it isn't possible to use integration-by-parts method - at least I couldn't find a way to do it.

答:
Hint. Since $D\left(\frac{1}{\tan(x/2)}\right)=-\frac{1}{1-\cos(x)}$, by integrating by parts we have that
$$\int\frac{\ln(\cos(x))}{1-\cos(x)}\,dx=-\frac{\ln(\cos(x))}{\tan(x/2)}
-\int \frac{\tan(x)}{\tan(x/2)}\, dx.$$
Then use the Weierstrass substitution.
Can you take it from here?