问:
Problem understanding the definition of Quotient spaces
Quotient spaces are defined as $V/U := \{a + U | a \in V\} $ where $a + U := \{a + u | u \in U\}$ U is a subspace of V, so $u \in U \implies u \in V$ and $\{a + u | a \in V\}$ = V. How does this not mean that $V/U = V$ according to the definition?

答: Maybe it would help to think about the example where $V = \mathbb R^2$ and $U$ is the subspace $\{ (x,y) \mid y = x \}$. If $a \in \mathbb R^2$, then $a + U$ is a line parallel to the line $y = x$. So each element of $V/U$ is a line in $\mathbb R^2$ with slope $1$.


问:
Prove that this block matrix is positive definite
I'm trying to prove that this symmetric block matrix is positive definite
$$
    \begin{bmatrix}
    B & -I & 0 & \dots& 0 \\
    -I & B & -I & \ddots & \vdots\\
    0 & -I & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -I\\
    0 & \dots & 0& - I & B \\
    \end{bmatrix}
$$

with $I$ the identity matrix

$$B = \begin{bmatrix}
    4 & -1 & 0 & \dots& 0 \\
    -1 & 4 & -1 & \ddots & \vdots\\
    0 & -1 & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -1\\
    0 & \dots & 0& - 1 & 4 \\
    \end{bmatrix}$$

by proving that this term $x^T A x$ is positive, but given the fact that it is a block matrix I'm stuck. Could you help me please?

答:
As the comment on your post mentions, the Gershgorin circle theorem is enough to prove that the matrix is at least positive semidefinite.

However, it turns out that we can actually find an expression for the eigenvalues of this matrix without too much effort. Suppose that the block-matrix is $m \times m$, with each block an $n \times n$ matrix. Let $M_k$ denote the $k \times k$ matrix
$$
M_k = \begin{bmatrix}
    0 & -1 & 0 & \dots& 0 \\
    -1 & 0 & -1 & \ddots & \vdots\\
    0 & -1 & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -1\\
    0 & \dots & 0& - 1 & 0 \\
    \end{bmatrix}.
$$
Because $M_k$ is tridiagonal and Toeplitz, its eigenvalues can be easily found to be
$$
\lambda_j = 2 \cos\left(\frac{\pi j}{k + 1}\right), \quad j = 1,\dots,k.
$$
Using the Kronecker product and Kronecker sum, we can write this block matrix in the form
$$
A = 4 I_{mn} + M_m \otimes I_n + I_m \otimes M_n = 4I_{mn} + (M_m \oplus M_n).
$$
It is known that for square matrices $P,Q$, the eigenvalues of $P \oplus Q$ have the form $\lambda + \mu$ for eigenvalues $\lambda$ of $P$ and $\mu$ of $Q$. So, the eigenvalues of $A$ are given by
$$
\lambda_{p,q} = 4 + 2 \cos\left(\frac{\pi p}{m + 1}\right) + 2 \cos \left(\frac{\pi q}{n + 1}\right), \quad p = 1,\dots,m, \quad q = 1,\dots,n.
$$
We can see that these eigenvalues are non-negative because the minimum possible value for the above is $4 - 2\cdot 1 - 2 \cdot 1 = 0$. Thus, $A$ will always be positive semidefinite.

In fact, we see that $A$ is positive definite since neither of the cosines attain their minimum of $-1$.

ContentLicense: CC BY-SA 4.0


问:
Double integral properties used in theorem?
Ok this question is a about integral properties used to proof a theorem (a Expectation value for random variables theorem).

My question is only about calculation of what properties it used for double integrals.

The two photos show the complete theorem and its proof.

The photo in part2 with a arrow shows where the calculations are performed.

Expectation Theorem : part 1
 Expectation Theorem : part 2

In the photo n.r.v means continuous random variable

and n.r.d means discrete random variable

$f_{Y}$ is a density for the continuous random variable Y:

$ \int_{0}^{\infty} \  [\int_{y}^{\infty} f_Y(x)dx ] \ dy $

= $ \int_{0}^{\infty} \  [\int_{0}^{x} dy ] \ \ f_Y(x) dx $

 How are the calculations made for this part (What properties are used )?
 If you need any help about the text comment down.

答:
Assume for the moment that $f_Y(y)\equiv0$ when $y\notin[0,N]$. Then your question is about the chain of  equations
$$\int_0^N\left(\int_y^N f_Y(x)\>dx\right)\>dy=\int_0^N\left(\int_0^x f_Y(x)\>dy\right)\>dx=\int_0^Nx\,f_Y(x)\>dx\ .$$
Only the first equality sign needs an explanation. We are integrating here over the right triangle $T$ with vertices $(0,0)$, $(N,0)$, $(N,N)$ in the $(x,y)$-plane, first integrating along $x$ and then along $y$, then first integrating along $y$ and then along $x$. That these two ways lead to the same result is "Fubini's theorem".


问:
Show that the addition function, $f: \mathbb{R}^2 \longrightarrow R$, given by $f(x,y)=x+y$ is a continuous function.

Show that the addition function, $f: \mathbb{R}^2 \longrightarrow R$, given by $f(x,y)=x+y$ is a continuous function

If someone can give me a most clean proof this was very nice.
Thanks for read my proof

答:
Let $U=(a,b)$ where $a,b\in \mathbb{R},a<b$ a open in $R$ we should prove that $f^{-1}(U)$ is an open set in $\mathbb{R}^2$.\
Consider $f^{-1}(a,b)= \lbrace (x,y)\in \mathbb{R}^2:a<x+y<b\rbrace$ and notice that these region is a subset in the plane between $y=b-x$ and $y=a-x$,Namely $L$.
Let Fix $P=(x,y)\in L$ chose $\varepsilon>0$ such that $$a+\varepsilon<x+y<b-\varepsilon$$
and consider $B_{\epsilon/2}(P)$.Pick $Q\in B_{\varepsilon/2}(P)$we will prove that $Q \in L $ ie($a<q_1+q_2<b)$ by definition $$\sqrt{(q_1-x)^2+(q_2-y)^2}<\varepsilon/2$$
$$|q_1-x|<\varepsilon/2$$
$$|q_2-y|<\varepsilon/2$$
$$|q_1-x|+|q_2-y|<\varepsilon$$
$$|q_1+q_2-(x+y)|<|q_1-x|+|q_2-y|<\varepsilon$$
$$|q_1+q_2-(x+y)|<\varepsilon$$
$$-\varepsilon+(x+y)<q_1+q_2< \varepsilon+(x+y)$$
by our election of $\varepsilon$ we get
$$-\varepsilon+a+\varepsilon<q_1+q_2< \varepsilon+b-\varepsilon$$
$$a<q_1+q_2<b$$ therefore $Q\in L$ and $L=f^{-1}(U)$ is open in $\mathbb{R}^2$, Hence $f$ is a continuous function since $P$ was arbitrary in $f^{-1}(U)$.


问:
$f (x) ∈ Q [x]$ is irreducible if $f (x - 1)$ is irreducible.
I think the Eisenstein Irreducibility Criterion can work, but I don't know how to apply it.</p>

Body: If $f(x)=g(x)h(x)$, then $f(x-1)=g(x-1)h(x-1)$. If $f(x-1)=g(x)h(x)$, then $f(x)=g(x+1)h(x+1)$.


问:
What is the obvious mistake of this double integration?
There is a definite double integral in Cartesian coordinates as follows:

$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 2\sqrt{4-x^2}\ dydx\Rightarrow \int_{-2}^{2} 2y\sqrt{4-x^2}\rvert_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}  \ dx \Rightarrow 4\int_{-2}^{2} (4 - x^2)dx \Rightarrow 4(4x - \frac{x^3}{3})|_{-2}^{2} = \frac{128}{3} $$

and there is another double integral in polar coordinates which is supposed to be the exact same integral as the above, but the final answer is not 128/3. it took me several hours, but I cant figure out which step of this integration is wrong.

$$ \int_{0}^{2\pi}\int_{0}^{2} 2r\sqrt{4-r^2{\cos^2\theta}} drd\theta
 \xrightarrow[]{u = 4-r^2{\cos^2\theta}}  \int_{0}^{2\pi} - \frac{2}{3} \frac{(4-r^2{\cos^2\theta})^\frac{3}{2}}{\cos^2\theta}\Big|_{0}^{2}  \ d\theta \Rightarrow \int_{0}^{2\pi} (-\frac{2}{3}\frac{(4-4\cos^2\theta)^\frac{3}{2}}{\cos^2\theta} + \frac{2}{3}\frac{4^\frac{3}{2}}{\cos^2\theta} ) d\theta \Rightarrow \frac{16}{3}\int_{0}^{2\pi} \frac{1-(1-\cos^2\theta)^\frac{3}{2}}{\cos^2\theta} d\theta \Rightarrow  \frac{16}{3} \int_{0}^{2\pi} \frac{1-{\sin^3\theta}}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3}\int_{0}^{2\pi} \frac{1}{\cos^2\theta} - \frac{\sin^3\theta}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3}(\int_{0}^{2\pi} \frac{1}{\cos^2\theta} d\theta - \int_{0}^{2\pi} \frac{\sin^3\theta}{\cos^2\theta} d\theta) \Rightarrow \frac{16}{3}(\tan\theta - \int_{0}^{2\pi} \frac{\sin^3\theta}{\cos^2\theta} d\theta) \xrightarrow[]{u=\cos\theta} \frac{16}{3}(\tan\theta - \int \frac{1-u^2}{u^2} du) \Rightarrow \frac{16}{3}(\tan\theta - (-\sec\theta - \cos\theta))\Big|_{0}^{2\pi}\Rightarrow \frac{16}{3} ((0+1+1) - (0+1+1)) = 0 $$

I know there is a huge problem with the second one and my apologies if it's an elementary problem. But I would really appreciate if you could help me find the problem.

答:
The problem is that $(1-\cos^2)^\frac{3}{2}=|\sin^3\theta|$ and not $\sin^3\theta$. Then you have to consider intervals for $\theta$ to discard absolute value.

The second problem is at the second to the last line, the change of variable $u=\cos\theta \Rightarrow d\theta = -\frac{du}{\sin\theta}$ and you forgot a negative sign.

It is very nice that you try to verify your answer by considering other possible ways of solving and in this problem it is possible to solve it in both coordinates but sometimes it is not possible to find answer in other coordinates and the coordinate system that you choose is completely dependent on the geometry of problem, like finding the electrostatic potential function in spherical or cylindrical capacitors, just try to solve it with simplest way possible and be done with it dude.


问:
Let $f$ de multiplication function and complete the proof of the continuity of the function.

Let $f$ de multiplication function and complete the proof of the continuity of the function.
Any advice to do in my proof was very helpful

答:
Let $U=(a,b)$, $a<b$, $a,b\in \mathbb{R}$ a open in $\mathbb{R}^2$ we should prove $f^{-1}(U)$ is open in $\mathbb{R}^2$.By definition
$f^{-1}(U)=\lbrace a<xy<b \rbrace=L$ geometrically this is the region bounded by two hyperbolas.
Fix $P=(p,q)\in L $, by definition $a<pq<b$ and chose $m=Min\lbrace b-pq,pq-a \rbrace$ and pick $\delta>0$ such that $\delta|p|,\delta|q|,3\delta^2<\frac{1}{3}m^2$ without loss of generality suppose thet $m=Min\lbrace b-pq,pq-a\rbrace=b-pq$.
\
Consider the square $S=(p-\delta,p+\delta) \times (q-\delta,q+\delta)$  and $l=(l_1,l_2)\in S$ since $l$ is in $S$
$$p-\delta<l_1<p+\delta$$
$$q-\delta<l_2<q+\delta$$
$$(q-\delta)(p-\delta)<l_1 l_2<(p+\delta)(q+\delta)$$
$$pq-\delta p-\delta q+\delta^2<l_1l_2<pq+\delta p+\delta q+\delta^2$$
$$-\delta p-\delta<l_1l_2<\delta p+\delta q$$
since $\delta p< \delta |p|$ and $\delta q< \delta |q|$
$$-\frac{2}{3}m<l_1l_2< \frac{2}{3}m$$
$$-(b-pq)<l_1l_2<b-pq$$
notice that $b-pq<pq-a$
$$-(pq-a)<l_1l_2<b-pq$$
$$-pq+a<l_1l_2<b-pq$$
$$a<l_1 l_2<b$$
therefore $l\in L$ since $l$ was arbitrary point in $S$ then $l$ is open in $\mathbb{R}^2$ and the multiplication function is continuous function.\
Notice that we consider  the case where $0<a<b$ in the case where $a<b<0$ is analogue, since our square $S$ is the same, the case where $a<0<b$ not are possibly because the region $L=f^{-1}(U)$ only hcan be in the first and third quadrant


问:
Computing product of two cosets
Let in $S_3$, $H=\{(1),(12)\}$ show that $(13)H(23)H \neq (13)(23)H$

The solution claims $(13)H(23)H=\{(132),(23),(12)\}$ however I calculated $(13)H(23)H=\{(132),(23),(12),(1)\}$

The solution seems to omit the identity from the set. Am I correct here or is the solution correct?

答:
A faster way of solving this problem than explicitly calculating cosets:

Note since $() \in H$, $(13)(23) H \subseteq (13) H (23) H$. You can see this by replacing the first $H$ on the right-hand side by $()$. So it suffices to show that the $\subseteq$ is actually a $\subsetneq$.

Since $(13)(23) H$ is a coset of $H$, and $|H| = 2$, we have $|(13)(23) H| = 2$. Moreover, the two elements in $(13)(23) H$ must be $(13)(23) = (123)$ and $(13)(23)(12) = (23)$.

So then all we need to do is think about what else could possibly be in $(13)H (23) H$. Note that $(13) H (23) H$ contains precisely the elements of the form $g := (13)h_1 (23) h_2$, where $h_1, h_2 \in H$. If $h_1 = ()$, then $g \in (13)(23) H$. So we look at elements $(13)(12)(23) h_2$. It suffices to pick one $h_2$ for which $g \notin (13)(23) H$, but you can calculate both to see what $(13)H (23) H$ actually looks like.

(And yeah, when $h_2 = (12)$, I also get $(13)(12)(23)(12) = ()$, so to answer your actual question, you are correct.)

I suppose the insight here is to try to save yourself grunt work whenever possible. Here the clever steps were noticing $|(13)(23) H| = |H|$ and $(13)(23) H \subseteq (13) H (23) H$.

Title: Computing product of two cosets


问:
Reordering vector product
If I have vectors $a, b, c \in \mathbb{R}^3$, and if we have e.g. $a = b\times c$, is there any way to express $b$ in terms of the other two? </p>

答: To elaborate on @Ralph Tandetzky's answer: Interpreting the cross product geometrically, assuming $a\ne 0$, $b$ must be in the plane orthogonal to $a$, and the signed area of the parallelogram spanned by $b$ and $c$ must be $\|a\|$. Thus, there is a line — parallel to $c$ — of solutions.</p>

ContentLicense: CC BY-SA 3.0

Title: Reordering vector product


问:
Least squares: How does the projection theorem lead to the normal equations?
I understand why $\vec{{P}}$ (the orthogonal projection of $\vec{{b}}$ on w)

is the closest vector to $\vec{{b}}$ (The vector that does not give us a solution to the
equation and that's why we need to do the process)

But I do not understand the equation:

$$A^TA\vec{{x}}=A^T \vec{{b}}$$
Why does multiplying each side of the original equation $A\vec{{x}}=\vec{{b}}$ (We know she has no answer)
by $A^T$ give us the closest answer?

Thank you

答: I'm not able to comment, but my answer to: How to formulate ordinary least squares regression in component formalism? may address your question. The reason multiplying each side by $A^T$ helps is that while $A$ may not have an inverse, $A^TA$ typically does, allowing you to solve for $\vec{x}$.


问:
Predicate Logic to Natural Language
Is my translation correct?

$P(x)$ - "$x$ is a program"

$F(x) $- "$x$ is a function"

$I(x,y)$ - "$x$ invokes y"

$ \forall x (P(x) \rightarrow \forall y (F(y) \rightarrow \neg I (x, y)))$

"For every $x$, if $x$ is a program then

For every $y$, if $y$ is a function then

$x$ does not invoke $y$"

$ \therefore $ "Every program does not invoke a function"

答:
Yes... but to make it a little more clear, I would say:

"Every program does not invoke any function"

Another option is:

"No program invokes a function"

or:

"Programs never invoke functions"

Title: Predicate Logic to Natural Language


问:
Benford's law and voting in Georgia
With Benford's law "the leading digits of the number found in real-world data sets" should make sense.  But this video

https://www.youtube.com/watch?v=DoF3WS42w3M&ab_channel=RobertA.Bonavito%2CCPA

claims that by applying Benford's law they can prove electoral fraud.

He claims that since Benford's law is "broken" in these graphs, it appears that there has been fraud committed, but I wonder if the artificiality of breaking things into counties and only having 156 of them violates the "naturally occuring" idea that is the foundation of Benford's Law.

I am curious why Benford's law does not seem to apply here.

答: Did you check Bedford's law against the number of registered voters in Georgia first? Because if that's off then the votes will be off also.</p>

Title: Benford's law and voting in Georgia


问:
How can $t \delta(t) = 0$ be proven using its defining propery?
How can $t\delta(t)=0$ be proven using it's defining principle, $\int f(t) \delta(t-a) dt = f(a)$?

I have a "proof" so far but I know its not even close to rigorous or proper.

$\int f(t)\delta(t-a)dt = f(a)$

Let $f(t) = t, a = 0$:

$\int t\delta(t)dt=f(0) = 0$

Differentiate both sides:

$\frac{d}{dt} \int t\delta(t)dt = \frac{d}{dt}0$

$\Longrightarrow t\delta(t)=0$

What would be a more formal proof?

答:
Since
$$
\int t\delta(t) \, f(t) \, dt
= \int \delta(t) \, tf(t) \, dt
= \left. tf(t) \right|_{t=0}
= 0 f(0)
= 0
= \int 0 \, f(t) \, dt
$$
for every test function $f$ we can conclude that $t\delta(t)=0.$

That's really all we need.

Above I wrote things using an integral sign. But in the formal theory, distributions are not functions that can be integrated. Instead they are linear functionals acting on a space of test functions. Usually this is taken to be $C_c^\infty,$ the space of infinitely differentiable functions with compact support. The action of a distribution $u$ on a test function $\varphi$ is often written $\langle u, \varphi \rangle$ or $\langle u(t), \varphi(t) \rangle,$ where the variable $t$ is purely formal; it shouldn't be interpreted as $u$ having a value at $t$. This formal notation corresponds to the abuse of notation $\int_{-\infty}^{\infty} u(t)\,\varphi(t)\,dt.$

The definitions we need for proving $t\,\delta(t)=0$ are the following:

If $f \in C^\infty$ and $u$ is a distribution, then $fu$ is the distribution defined by $\langle fu, \varphi \rangle = \langle u, f\varphi \rangle$ for every $\varphi\in C_c^\infty.$

$\langle \delta, \varphi \rangle = \varphi(0)$ for every $\varphi\in C_c^\infty.$

The zero distribution $0$ is defined by $\langle 0, \varphi \rangle = 0.$

Two distributions $u$ and $v$ are equal if and only if $\langle u, \varphi \rangle = \langle v, \varphi \rangle$ for every $\varphi\in C_c^\infty.$

For any $\varphi \in C_c^\infty$ we have
$$
\langle t\,\delta(t), \varphi(t) \rangle
= \{ \text{ def. 1 } \}
= \langle \delta(t), t\,\varphi(t) \rangle
= \{ \text{ def. 2 } \}
\\
= \left. t\,\varphi(t) \right|_{t=0}
= 0
= \{ \text{ def. 3 } \}
= \langle 0, \varphi \rangle.
$$
Thus, by def. 4, we have $t\,\delta(t) = 0.$


问:
Expression for power of a natural number in terms of binomial coefficients
Is there a general expression for the pth power of a natural number k in terms of binomial coefficients?

I found this identity in a high-school text, which was obtained by simply equating coefficients:

$k^3 = 6 \binom{k}{3} + 6 \binom{k}{2} + \binom{k}{1}$

I would like to know the general expression for $k^p$ in terms of binomial coefficients.

I would also appreciate it if someone could tell me the names of texts which might cover such expressions, and what they can be used for.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Also, from Knuth CM-book:
$$
\left\{\begin{array}{rcll}
\ds{n^{\underline{1}}} & \ds{=} & \ds{n,} &  \ds{n = n^{\underline{1}}}
\\
\ds{n^{\underline{2}}} & \ds{=} & \ds{n\pars{n - 1},} & \ds{n^{2} = n^{\underline{1}} + n^{\underline{2}}}
\\
\ds{n^{\underline{3}}} & \ds{=} & \ds{n\pars{n - 1}\pars{n - 2},} & \ds{n^{3} = n^{\underline{1}} + 3n^{\underline{2}} +
n^{\underline{3}}}
\\
\ds{\vdots} & \ds{\vdots} & \ds{\vdots} & \ds{\vdots}
\end{array}\right.
$$
For instance $\ds{\pars{~sum\ \mbox{looks like an}\ integration~}}$:
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{N}k^{2}} =
\sum_{k = 1}^{N}\pars{k^{\underline{1}} + k^{\underline{2}}} =
\pars{{1 \over 2}\,k^{\underline{2}} +
{1 \over 3}k^{\underline{3}}}_{1}^{N + 1}
\\[5mm] = &\
\bracks{{1 \over 2}\,\pars{N + 1}^{\underline{2}} +
{1 \over 3}\pars{N + 1}^{\underline{3}}}\ -\
\underbrace{\pars{{1 \over 2}\,1^{\underline{2}} +
{1 \over 3}1^{\underline{3}}}}_{\ds{= 0}}
\\ = &\
{1 \over 2}\pars{N + 1}N +
{1 \over 3}\pars{N + 1}N\pars{N - 1}
\\[5mm] = &\
\pars{N + 1}N\bracks{{1 \over 2} +
{1 \over 3}\pars{N - 1}}
\\[5mm] = &\
\bbx{N\pars{N + 1}\pars{2N + 1} \over 6} \\ &
\end{align}


问:
minus sign in the exponent of kernels of integral transformations
From PlanetMath and Wolfram, the Fourier-Stieltjes transform of a function $\alpha$ is defined as $\displaystyle \int_{\mathbb{R}} e^{itx} d(\alpha(t)).$ The kernel $\displaystyle e^{itx}$ is unlike the one $\displaystyle e^{-itx}$ used in Fourier transformation.

From Wikipedia, the Laplace-Stieltjes transform of a function $g$ is given as $ \displaystyle \int_{\mathbb{R}} e^{-sx}\,dg(x).$ The kernel ${e}^{-sx}$ is same as the one used in Laplace transformation.

I was wondering why the minus sign in the exponent of kernels for Fourier-Stieltjes transform and Fourier transform are not consistent, while consistent for Laplace-Stieltjes transform and Laplace transform?

Are there mistakes in the quoted sources, or other popular variant definitions?

答:
I guess this is mostly a matter of convention. I think the minus convention in Fourier Series comes from the fact that

$\int_0^{2 \pi} f(t) \, e^{-int} \, dt$

Is the hermitian scalar product of $f$ and $\epsilon^n : x \mapsto e^{inx}$, which makes Fourier inversion formula more natural :

$f = \sum_{n \in \mathbb{Z}} \langle f \,|\, \epsilon^n \rangle \, \epsilon^n$

If you look at planetmath and wolfram, their point of view is to take a function $f$ and write it as an integral of $\epsilon$ with regards to a certain $d \alpha$. On the other hand, Wikipedia starts from $\alpha$ and then defines $f$. I guess the (admittedly subtle) difference in point of view might explain their convention.


问:
Multidimensional infinitesimal generator of a jump-diffusion
Let $X=\{X_t\}_{t\geq0}$ be an $n$-dimensional Markov process, defined by the SDE

$$dX_t = \mu(t, X_t) \, dt + \sigma(t,X_t) \, dB_t+\beta(t-,X_{t-}) \, dN_t,$$

where $\mu, \sigma$ and $\beta$ are vector functions, $B_t$ is an $n$-dimensional Brownian motion and $N_t$ is an $n$-dimensional counting process, whose conditional intensity is given by $\lambda(t-, X_{t-})$, where $\lambda$ is a continuous function.

Does anyone know the explicit formula for the infinitesimal generator of $X_t$?

As always, help is greatly appreciated.

Edit: I should maybe mention that the result for the one-dimensional case is

$$ \mathcal{A}f = \frac{\partial f}{\partial t}(t,x) + \mu(t,x)\frac{\partial f}{\partial x}(t,x) + \frac{1}{2}\sigma^{2}(t,x)\frac{\partial^{2} f}{\partial x^{2}}(t,x) + f_{\beta}(t,x)\lambda(t,x)\text{,}$$

where

$$f_{\beta}(t,x) = f(t,x+\beta(t,x)) - f(t,x)\text{.}$$

The only thing that I really have no clue about is what the fourth term in the
previous equation looks like in the $n$-dimensional case.

答:
Perhaps the following will be of use. In "Levy Processes and Stochastic Calculus" by David Applebaum, Theorem 6.7.4, the following is argued: Consider a SDE of the form

$$
dY_t = b(Y_{t-})dt + \sigma(Y_{t-})dW_t
      + \int_{|x|<c} F(Y_{t-},x) \tilde{N}(dt,dx)
      + \int_{|x|\ge c} G(Y_{t-},x) N(dt,dx),$$

where $W$ is a $n$-dimensional Brownian motion, $N$ a $n$-dimensional Poisson random measure with intensity measure $\nu$ and compensator $\tilde{N}$, independent of $W$, and coefficients satisfying appropriate regularity conditions. Then this SDE has a unique solution, which is a Feller process with generator $A$ whose value on $C_c^2(\mathbb{R}^n)$ is

$$
Af(y) = \sum_{i=1}^n b_i(y)\frac{\partial}{\partial y_i}f(y)
       + \frac{1}{2}\sum_{i=1}^n \sum_{j=1}^n (\sigma(y)\sigma(y)^t)_{ij}\frac{\partial^2}{\partial y_i\partial y_j}f(y)\\
  +\int_{|x|\le c} f(y+F(y,x))-f(y) - F_i(y,x)\frac{\partial}{\partial x_i}f(y) d\nu(x)\\
  +\int_{|x|\ge c}f(y+G(y,x)) - f(y) d \nu(x).
$$

The only difference with your set-up appears to be that you have a nondeterministic intensity measure $\nu$.


问:
Proving a certain nonlinear ODE has a solution on $t \geq 0$
I'm trying to prove that the following ODE admits a solution on the interval $[0, \infty )$:

$\begin{align*}
\begin{cases}
x_1^\prime &= x_2 - x_1^3 \\
x_2^\prime &= \frac{1}{2}x_1 - x_2 + d \sin t
\end{cases}
\end{align*}$

With initial condition $x(0) = x_0$.

The usual approaches seem to fail: the derivative isn't bounded on $\mathbb{R}^2$ and the function does not seem Lipschitz on all of that domain either. Using energy functions doesn't seem to help either.

Any hints would be highly appreciated!

答:
We use the comment by Hans Engler to prove your result.

Let $w(t) = x_1^2(t) + x_2^2(t) = || x ||_2^2$. Then,
$$\begin{align*}
w^\prime (t) &= 3x_1 x_2 - 2x_1^4 - 2x_2^2 + d x_2 \sin t \\
&\leq 3x_1 x_2 - 2x_2^2 + d |x_2|
\end{align*}$$

Using the elementary inequality $ab \leq \frac{1}{2} (a^2 + b^2)$, we get that:
\begin{align*}
w^\prime (t) &\leq \frac{3}{2} ( x_1^2 + x_2^2 ) - 2x_2^2 + \frac{1}{2} (x_2^2 + d^2) \\
&\leq \frac{3}{2} \left( w(t) + \frac{1}{3}d^2 \right)
\end{align*}

It follows by Grönwall's that $w(t)$ cannot blow up in finite time, which means that the norm of $x$, and hence $x$, cannot blow up in finite time either.


问:
Can $ y(t)=\left((t-1)^2+1\right)\cdot \cos(t) $ be a solution for this ODE
I've been asked if
$$
y(t)=\left((t-1)^2+1\right)\cdot \cos(t)
$$
can be a solution for the ODE:
$$
y''+y+(y-\cos(t))P(t,y,y') = 0
$$
when $P$ is smooth.

So I concluded that there exists a unique solution to the ODE, by Peano's theorem if:
$$
y_{1}=y,\quad y_{2}=y'.
$$
Then the ODE is of the form:
$$
y_{2}'=-y_{1}-P(t,y_{1},y_{2})(y_{1}-\cos(t)).
$$
$P$ is smooth and a function of $y_{2}$ and $y_{1}$, therefore the ODE has a unique solution, but I don't know how to continue any further.  Any help/hints would be appreciated!

答:
For a hint:

Compare the roots of $y'' + y$ and $y - \cos(t)$ for your given function $y$.


问:
$f$ is integrable $\iff$ for every sub-block $B$ we have that the function $f|_{B}$ is integrable, i.e. $\int_{A}f=\sum_{B}\int_{B}f|_{B}$
QUESTION: Let $f:A \rightarrow \mathbb{R}$ be a limited function and  let $P$ be a partition of the block $A$ ($A$ is a block in $\mathbb{R}^m$). Then $f$ is integrable $\iff$ for every sub-block $B$ we have that the function $f|_{B}$ is integrable and in this case, $$\int_{A}f=\sum_{B}\int_{B}f|_{B}$$.

REMARK: The professor allowed us to use the following concepts:

Proposition: Let $P_0$ be an arbitrary partition of the block $A$. In order to consider the upper and lower integrals of the limited function $f:A \rightarrow \mathbb{R}$, we just need to consider partition refinements of $P_0$. That is, we have $$\underline\int_{A} f(x) dx= \underset{P\supset P_0}{sup} s(f; P)$$ and $$\overline\int_{A} f(x) dx= \underset{P\supset P_0}{inf} S(f; P)$$

Theorem: The limited function $f: A \rightarrow \mathbb{R}$ is integrable $\iff$ for every $\epsilon>0$ it is possible to find a partition $P$ of the block $A$ such that $$\displaystyle\sum_{B\in P} \omega_{B}\cdot vol B<\epsilon$$
Where $\omega_{B}$ is the set of the oscillations, i. e., $$\omega_{B}:= sup\{|f(x)-f(y)|; x, y \in B\}$$

MY ATTEMPTY:

$(\Longrightarrow)$ Let $f: A \rightarrow \mathbb{R}$ be a limited function and let $P$ be a partition of the block $A$. Suppose that $f$ is integrable then $\forall \epsilon >0$ it is possible to obtain an partition  $P=P_1 \times \cdots \times P_n$ of $A$ such that $\displaystyle\sum_{B\in P} \omega_B \cdot \text{vol}B <\epsilon$, where $B$ are blocks in  $P$. Once $B$   are sub-blocks of $A$, let $P_0$ be an partition of  $B$. Therefore for every limited function $f|_{B}$ we just need to consider the refinement partitions of $P_0$. Indeed, let $B=\displaystyle\Pi_{i=1}^{n}[b_i, c_i] \subset A$ then for every $i=1, \cdots, n$ lets define $Q_i= P_i\cap[b_i, c_i]$ from this we have a new partition $Q = Q_1 \times \cdots \times Q_n$ of $A$ that is a refinement of $P$ and, furthermore, the blocks of $Q$ are contained in $B$ makes a partition $P_0$ of $B$. Thus $$\underbrace{\displaystyle\sum_{B'\in P_0}\omega_{B'}\cdot \text{vol} B'}_{(I)}\leq\displaystyle\underbrace{\sum_{B\in P}\omega_{B}\cdot \text{vol} B<\epsilon}_{(II)}$$  $(I) \subset (II)$ therefore $f|_{B}$ is intagrable.

$(\Longleftarrow)$ We just need to consider $P=P_1 \times \cdots \times P_n$ as a partition of the block $A$ and we also need to consider that this partition is a composition of the block $A$ in sub-blocks like $B=I_1 \times \cdots \times I_n$ where every $I_j$ is an interval of the partition $P_j$, where every sub-block $B$ is the block of partition $P$, i.e., $B\in P$. So, writting $A=\displaystyle\bigcup_{i=1}^{n}B_i$ and remembering that every $f|_{B}$ is integrable. Note that if  $P_i$ is a partition of $B_i$ we can consider $Q=\displaystyle\sum_{i=1}^{n}P_i$ as an refinement partition of $P$ thus $f:A \rightarrow\mathbb{R}$ is integrable.

Now we just need to show that: $$\int_{A} f \leq \displaystyle\sum_{B \in P} \int_{B} f|_{B}$$.

In $f:A \rightarrow \mathbb{R}$ considering the partition $P$ of the block $A$ we just need to consider refinement partitions of $P$, let $Q$ be an arbitrary partition of the block $A$ we can consider, for instance $P_0= P+Q$. It follows from upper integration definition that $$s(f, P)=\displaystyle\sum_{B \in P} m_B(f)\cdot \textbf{vol}B= \displaystyle\sum_{B \in P} m_{B}(f|_{B}) \cdot \textbf{vol} B$$. Then, for every $B$ we consider $B' \subset B$, the sub-blocks of $B$ resultants of the refinement of $P$, and $B=\bigcup B'$. Therefore,
\begin{align*}
    \int_{A} f = \displaystyle sup_{P_0\supset P} s(f, P_0)& = sup \left(\displaystyle\sum_{B\in P}m_{B}(f|_{B}) \cdot \textbf{vol} B\right)\\
    & = sup \left(\displaystyle\sum_{B\in P}m_{B}(f|_{B}) \displaystyle\sum_{B'\subset B} \textbf{vol} B'\right)\\
    & = sup \left(\displaystyle\sum_{B\in P}\displaystyle\sum_{B'\subset B}m_{B}(f|_{B})  \textbf{vol} B'\right)\\
    & \leq sup \left(\displaystyle\sum_{B\in P}\displaystyle\sum_{B'\subset B}m_{B'}(f|_{B})  \textbf{vol} B'\right)\\
    & = \displaystyle\sum_{B\in P} sup \left(\displaystyle\sum_{B'\subset B}m_{B'}(f|_{B})  \textbf{vol} B'\right)\\
    & = \displaystyle\sum_{B\in P} \underline{\int_{B}} f|_{B}\\
    & = \displaystyle\sum_{B\in P} \int_{B} f|_{B}
\end{align*}
Thus, $$\int_{A} f \leq \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$

Similarly, we can show for upper sum, and obtain $$\int_{A} f \geq \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$
And finally, conclude
$$\int_{A} f = \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$

MY DOUBT: Would you help me to improve my answer? Specialy in this $(\Longleftarrow)$ way.

答:
Suppose $B_1, \ldots, B_n$ are non-overlapping blocks (closed rectangles) such that $A = \cup_{k=1}^n B_k$ and $f|_{B_k}$ is Riemann integrable for all $k$.  For any $\epsilon > 0$ there are partitions $P_{B_1}, \ldots, P_{B_n}$ such that for $1 \leqslant k \leqslant n$,

$$U(f, P_{B_k}) - L(f, P_{B_k}) < \frac{\epsilon}{n}$$

The partitions of the individual blocks taken together form a  partition $P_A$ of $A$ where

$$U(f,P_A) = \sum_{k=1}^n U(f, P_{B_k}), \quad L(f,P_A) = \sum_{k=1}^n L(f, P_{B_k}),$$

and, thus,

$$U(f,P_A) - L(f,P_A) = \sum_{k=1}^n ( U(f, P_{B_k})-  L(f, P_{B_k})) < n \cdot \frac{\epsilon}{n} = \epsilon$$

Therefore, $f$ is Riemann integrable on $A$ (by the Riemann criterion).

Let $P_A$ be an arbitrary partition of $A$. Using vertices of the blocks $B_k$ we can construct a refinement $P'_A \supset P_A$ such that every sub-block of $P'_A$ is contained in some block $B_k$ and we have

$$L(f,P_A) \leqslant L(f,P'_A) = \sum_{k=1}^n L(f, P_{B_k})\leqslant  \sum_{k=1}^n U(f, P_{B_k})= U(f,P'_A) \leqslant U(f,P_A)$$

This implies that, for any partition $P_A$,

$$L(f,P_A) \leqslant \sum_{k=1}^n\int_{B_k} f|_{B_k}\leqslant U(f,P_A)$$

Since $f$ has been shown to be Riemann integrable, for any $\epsilon > 0$ there is a partition $P_A$ such  that $U(f,P_A) - L(f,P_A) < \epsilon$ and $L(f,P_A) \leqslant \int_Af \leqslant U(f,P_A)$.

Therefore, for every $\epsilon > 0$,

$$\left|\int_Af - \sum_{k=1}^n \int_{B_k}f|_{B_k} \right| < \epsilon \implies \int_Af = \sum_{k=1}^n \int_{B_k} f|_{B_k}$$


问:
Surface area ratio for cubic and spherical objects of equal volume
Given two objects of equal volume, one cubic and one spherical, which will have the greater outer surface area, and how much greater will it be versus the other object? I know this is a math forum, so a mathematical explanation is great but I'm also hoping for layman's terms and a simple ratio.

As a bonus, assuming the objects are constructed from identical material, if we know the mass of one can we know the mass of the other by using the same ratio? Why or why not?

答:
The volume of a sphere with radius $r$ is $V_{sphere} = \frac{4}{3}\pi r^3$.
The volume of a cube with side $s$ is $V_{cube} = s^3$.
If you set these equal, you see that:
$s^3 = \frac{4}{3}\pi r^3$
or
$s = (\frac{4}{3}\pi)^{1/3} r$.

The surface area of a sphere is $S_{sphere} = 4\pi r^2$, and the surface area of a cube is $S_{cube} = 6s^2$.
Using the expression for $s$ from above, $S_{cube} = 6(\frac{4}{3}\pi)^{2/3}r^2$.

Thus, $\frac{S_{cube}}{S_{sphere}}= \frac{6(\frac{4}{3}\pi)^{2/3}}{4\pi} \approx 1.2,$
so the surface area of the cube is greater.

This is not at all surprising since a sphere has the smallest surface area of all surfaces that enclose a given volume (see: https://en.wikipedia.org/wiki/Sphere). As an interesting aside, since animals give off heat in proportion to their surface area but generate heat in proportion to their volume, animals living in colder climates tend to be rounder than their relatives in warmer climates (see: https://www.amazon.com/Engineering-Animals-How-Life-Works/dp/0674048547).


问:
Direct Product of the $G_i $'s
I am a little confused in the interpretation of the product of groups. Here is what's written in my notes:

Given groups $G_1,G_2,...,G_n$, recall that $G_1\times G_2\times ...\times G_n=\{(g_1,g_2,...,g_n)\}\vert g_i \in G_i \}$ for all $i \in I$. More generally, for groups $G_i$ indexed $i \in I$, we have $\Pi_{i\in I}G_i=\{(g_i)_{i\in I}\vert g_i \in G_i$ for all $i\}$.

So if I let $G_1=(*,X)$ where $X=\{something\}$ and  $G_2=(*,Y)$ where $Y=\{something\}$, according to the definition above, what would $G_1 \times G_2 $ yield?

答:
If $G_{1}=(X,\ \ast)$ and $G_{2}=(Y,\square)$ are groups then $((G_{1}\times G_{2}, \cdot)$ is the group given by the set $X\times Y=\{(x,y):\ x\in X\ \land\ y\in Y\}$ and the operation $\cdot$  on it defined as
$$\forall (x,y),\ (w,z)\in X\times Y,\ (x,y)\cdot (w,z):=(x\ast w,\ y\square z).$$
If $G_{1}=G_{2}=(X, \ast)$ this just becomes the set $X\times X$ with the operation $(x,x_{1})\cdot (x',x_{1}')=(x\ast x', x_{1}\ast x_{1}')$ for all $(x,x_{1}),\ (x',x_{1}')\in X\times X$.</p>

Title: Direct Product of the $G_i $'s


问:
$ \cos x\geq 1-\frac{x^2}{2} $
Prove that for $x\in\mathbb{R}$
$$
\cos x\geq 1-\frac{x^2}{2}.
$$

My try:

Consider $g(x)=\cos(x)-1+\frac{x^2}{2}.$
If I differentiate $g(x)$ then we get $g'(0)>0$ so locally we get $g(x)>g(0)=0$ and then we can see that the function is increasing for any $x$ the function is increasing and hence we have $g(x)\geq 0$ for any $x \geq 0$. But I am getting that if $x<0$ then $g(x) \leq 0.$ So this inequality isn't true in general for all $x \in \Bbb R$.

But, if we use Taylor's theorem with Lagrange's remainder then also I am not sure what will be the point $\zeta\in [-x,0]$ where $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4}\cos(\zeta).$

答:
You can use the following corollary of the  Mean value theorem:

Let $f, g$ be differentiable functions on an interval $I$, $x_0\in I$, such that

$f(x_0)\ge g(x_0), $

$f'(x)\ge g'(x)$ $\:\forall x\ge x_0$.

Then $f(x)\ge g(x)$ $\:\forall x\in I, x\ge x_0$.

Now, as both sides of the inequality are even functions, it is enough to prove it for $x\ge 0$.

Let $f(x)=\cos x$, $\:g(x)=1-\dfrac{x^2}2$, $x_0=0$.  We indeed have  $f(0)=g(0)$, and $f'(x)=-\sin x$, $g'(x)=-x$, and it is well-known that $\sin x \le x$ on the interval $\bigl[0,\frac\pi 2\bigr]$, because $\sin x$ is concave on this interval. As $\frac\pi 2>1$, a fortiori, we have $\sin x \le x $ on $\mathbf R^+$. Therefore
$$f'(x)=-\sin x\ge g'(x)=-x\qquad \forall x\ge 0$$

Title: $ \cos x\geq 1-\frac{x^2}{2} $


问:
Limit doesn't turn out as expected
I'm trying to find the limit of the following function at $x=\frac{\pi}{4}$ and as $n$ approaches infinity:
$$\frac{\ln(\tan x)-\tan^{n}x\lfloor\sin(\tan\frac{x}{2})\rfloor}{1+\tan^{n}x}.$$

The left-hand limit is $\ln(\tan x)$ as expected, but apparently the right-hand limit is $-\lfloor\sin(\tan\frac{x}{2})\rfloor$.

Now, since the $\tan^{n}x$ in the numerator, which approaches infinity, is multiplied with $0$ (since $\sin(\tan\frac{x}{2}$) lies between $0$ and $1$, and the floor function turns it to $0$), shouldn't it end up as $0$? And while the log value is a small positive value, the denominator approaches infinity, so shouldn't the final right hand limit come out as zero?

What's the error I'm making here?

答:
On the interval $\left(\frac{\pi}{8},\frac{3\pi}{8}\right)$, we find that $\lfloor \sin(\tan(x/2))\rfloor=0$, so on that interval we have that

$$ \frac{\ln(\tan x)-\tan^{n}x\lfloor\sin(\tan\frac{x}{2})\rfloor}{1+\tan^{n}x}=\frac{\ln(\tan x)}{1+\tan^nx} $$

But this function is continuous on that interval, for any integer $n$ and equals $0$ when $x=\frac{\pi}{4}$.

So

$$ \lim_{n\to\infty}\lim_{x\to\pi/4}\frac{\ln(\tan x)-\tan^{n}x\lfloor\sin(\tan\frac{x}{2})\rfloor}{1+\tan^{n}x}=0 $$

Title: Limit doesn't turn out as expected


答: The canonical construction of the product of two groups yields the group $ (G_{1} \times G_{2},*,1_{G_{1} \times G_{2}},^{-1})$ where the underlying set is the cartesian product $G_{1} \times G_{2}$ and the binary operations is defined by $(g_{1},g_{2})*(h_{1},h_{2}):=(h_{1}*g_{1},h_{2}*g_{2})$ (I use the same symbol for those operations but their meaning should be clear from the context). Moreover $(g_{1},g_{2})^{-1}=(g_{1}^{-1},g_{2}^{-1})$.</p>

Title: Direct Product of the $G_i $'s

Body:
I am a little confused in the interpretation of the product of groups. Here is what's written in my notes:

Given groups $G_1,G_2,...,G_n$, recall that $G_1\times G_2\times ...\times G_n=\{(g_1,g_2,...,g_n)\}\vert g_i \in G_i \}$ for all $i \in I$. More generally, for groups $G_i$ indexed $i \in I$, we have $\Pi_{i\in I}G_i=\{(g_i)_{i\in I}\vert g_i \in G_i$ for all $i\}$.

So if I let $G_1=(*,X)$ where $X=\{something\}$ and  $G_2=(*,Y)$ where $Y=\{something\}$, according to the definition above, what would $G_1 \times G_2 $ yield?


问:
Simplify the following sum $\sum_{i=1}^n\frac1{n-(i-1)}$
Extremely simple question, but one I am struggling with (I haven't taken a math class for a couple years so this may be very easy).
I just need to simplify the following sum, but can't seem to figure out how:

$\sum_{i=1}^n\frac1{n-(i-1)}$

I am really struggling to find where to start, since it seems you would need to break the sum into parts to simplify, but I don't think you can. My best guess would be doing something like below, but I don't think you can break the fraction into parts like this (but can't think of any other way to solve)

$\sum_{i=1}^n\frac1{n-(i-1)} =\sum_{i=1}^n\frac1{n} - \sum_{i=1}^n\frac1{i} - \sum_{i=1}^n1 = 1 - H_n - n$

I have also heard from a semi-reliable force that this equals $H_n$ but not sure if this is right/how it was reached.

答: Set $j=n-(i-1)$, then substituting, the first sum is over $j=n\to1$ And it is given by $\sum_{j=1}^n\frac 1 j$


问:
Finite sum of $1/(n!)$
I know that an infinite series $1/(n!)$ converges to $e$, but how do I find finite sums? For example, the first 100 members, or the first 1000, and so on?

答:
I am not sure I understand your question, but finite sums are often referred to as partial sums.

One can obtain a partial sum by simply setting a finite number as the upper limit of the sum (compared to infinity).

This way, one merely iterates through the appropriate values of $x$ (in this case) to obtain the partial sum (or finite series)

For example:

$\sum_{x=1}^3 \frac{1}{x!} = \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=1+\frac{1}{2}+\frac{1}{6}=1+\frac{2}{3}$

This can be generalized as:

$\sum_{x=n}^m f(x) = f(n)+f(n+1)+f(n+2)...+f(m)$

where $f$ is an function from the integers to the reals.

Title: Finite sum of $1/(n!)$


答:
As far as I know there is no analytical expression, but using the remainder formula for the Taylor expansion of the exponential,

$$e-\sum_{k=0}^n\frac1{k!}=\frac c{(n+1)!}$$ where $1<c<e$. For $n=100$ or $n=1000$, this is microscopic.

Update:

From the link given by Raymond, we have

$$c=(n+1){\{e\,n!\}}=1+\frac1{n+2}+\frac1{(n+2)(n+3)}+\cdots<1+\frac1{n+1},$$ which obviously tends to $1$.

Title: Finite sum of $1/(n!)$

Body: I know that an infinite series $1/(n!)$ converges to $e$, but how do I find finite sums? For example, the first 100 members, or the first 1000, and so on?


问:
Let X and Y be topological spaces and f: X $\to$ Y a continuous function. The graph of f is the set $\Gamma_f $ = {(x, f (x)): $x \in X$}.
Let X and Y be topological spaces and f: X $\to$ Y a continuous function. The graph of f is the set  $\Gamma_f $ = {(x, f (x)): $x \in X$}.

a) Show that if Y is hausdorff then $ \Gamma_f $ is closed at X $ \times $ Y.

b) Show that X is homeomorphic to $ \Gamma_f $ where $ \Gamma_f $ is seen as a subspace of X $ \times $ Y.

====================================================================

a) We know that f is continuous and Y is hausdorff , so there is a point (x,y) outside of the graph and  exist an open neighborhood that does not intersect the graph. So the complement of the graph is open, and, therefore the graph is closed.

Let f(x) $\in$ V and y $\neq$ f(x) $\in V'$ where v, v'are two disjoint open neighborhoods, because Y is Hausdorff.But f is continuous, so, U= $f^{-1}$(V) is open and x $\in$ U  and U $\times$ V' is an open neighborhood of (x,y) such that it does not intersect the graph, so (z,f(z))$\in U \times V'$, and that implies f(z) $ \in V \cap V'$.

b) We say that f is a homeomorphism if f is bijective, continuous and its inverse is also continuous.

That's what I've done so far.

答:
I would rewrite your proof for part a as follows.

Part a:
We proceed by showing that a point $(x,y)$ that is not on the graph has a neighborhood that does not intersect the graph. We then conclude that the complement of the graph is open, which means that the graph is closed.

Let $(x,y)$ be such a point. Because $y \neq f(x)$ and $Y$ is Hausdorff, there exist disjoint open sets $V,V'$ with $f(x) \in V$ and $y \in V'$. $f$ is continuous, so $U = f^{-1}(V)$ is open with $x \in U$.

We see that $U \times V'$ is an open neighborhood of $x,y$ that does not intersect the graph. Indeed, if we suppose (for contradiction) that $(z,f(z)) \in U \times V'$, then it would follow that $f(z) \in V'$. However, $z \in U = f^{-1}(V)$, which means that $f(z) \in V$. Thus, $f(z) \in V \cap V'$, contradicting the fact that $V,V'$ are disjoint.

For part $b$, the homeomorphism between $X$ and the graph needs to be a map $g:X \to \Gamma_f \subset X \times Y$. However, you never define such a map.

Define $g$ by $g(x) = (x,f(x))$. Why is $g$ invertible, even if $f$ is not?


问:
Help with proof concerning ordered fields and smallest elements
I have been trying to prove that every ordered field has no smallest positive element for a while now, and I think I have it worked out. However, I feel like something is missing in my proof. Here is what I have so far:

Let $F$ be an ordered field

Let $x$ exist in $F$ such that $0$ is less than $x$

Then $x/2$ is less than $x$

Therefore, $F$ can have no smallest element

This does seem to prove that the field has no smallest element, but it seems like there are some holes. Am I correct in thinking this way? I saw a similar proof involving the positive real numbers, but $x/2$ exists in the positive reals.  How do I know that $x/2$ exists in $F$?  I am just concerned that this is not enough.  Any helpful pushes would be most appreciated. Thank you!

答:
A lot of this proof is spent taking care of preliminaries so that we can smoothly apply an argument of the form you gave. You asked whether $x/2$ exists, so first we give a proof of that; then, we prove it is positive and less than $x$. Finally, we argue that there is no smallest positive element.

Every ordered field $F$ has characteristic zero. In fact, a proof by induction shows that $n>0$ for each $n=\sum_{i=1}^n1$. In particular, $2$ is greater than $0$ and it is invertible. Thus, for any $x\in F$, $x/2$ exists.

Now, suppose $x>0$ but $x/2<0$, then $-x/2>0$. Hence, $-x=2(-x/2)>0$. However, this is a contradiction, since $0=x+(-x)>0$. Therefore, $x/2>0$. On the other hand, $0<x/2$ implies $x/2<x/2+x/2=x$. Thus, for any positive $x$, we are guaranteed a positive $y<x$.

We proceed to prove that there is no smallest positive element. Suppose for the sake of contradiction that $x$ is the smallest positive element of $F$. Since $x$ is positive there is some $y\in F$ such that $0<y<x$. This is a contradiction.


问:
Do all generalised eigenvalue problems with Hermitian $N \times N$ matrices $\mathbf{K}$, $\mathbf{M}$ have $N$ linearly independent eigenvectors?
Consider the generalised eigenvalue problem

\begin{equation}
\mathbf{K}\mathbf{u} = \lambda \mathbf{M}\mathbf{u}
\end{equation}

where both $N \times N$ matrices are Hermitian.

I am trying to prove that one can form an $\mathbf{M}$-orthonormal basis using $N$ eigenvectors, regardless of whether there exists any repeated eigenvalues. ($\mathbf{M}$-orthogonality refers to $\mathbf{a}^*\mathbf{M}\mathbf{b} = 0$, and normalisation of eigenvectors is according to $\mathbf{a}^*\mathbf{M}\mathbf{a} = 1$)

However, in the process of doing so, I realised that I had been assuming that solving the generalised eigenvalue problem will always spits out $N$ linearly independent eigenvectors. Indeed, this is not always true in the general case, as seen for solving the ordinary eigenvalue problem for the non-Hermitian matrix,

\begin{equation}
\begin{bmatrix}
1 & 1 \\ 0 & 1
\end{bmatrix}
\end{equation}

which has eigenvectors that span only one dimension.

So, for generalised eigenvalue problems for Hermitian matrices $\mathbf{K}, \mathbf{M}$, is it possible to prove that it's always possible to obtain $N$ linearly independent eigenvectors? If not, under what additional conditions to Hermiticity are required?

答:
With a bit extra poking around, I've found on the Wikipedia page Definite system matrix that, for the case where

$\mathbf{K}$ is an $N \times N$ Hermitian matrix,

$\mathbf{M}$ is an $N\times N$ Hermitian matrix and positive definite,

both matrices can be diagonalised simultaneously. From this, it can then be shown that $N$ linearly independent eigenvectors can be obtained. (Furthermore, it is possible to select the eigenvectors such that they are all $\mathbf{M}$-orthogonal.

Proof (adapted from Wikipedia article)

If $\mathbf{M}$ is positive definite, it is invertible, and its inverse $\mathbf{M}^{-1}$ is also positive definite. Furthermore, since $\mathbf{M}$ is Hermitian, so is its inverse. Every positive definite Hermitian matrix has a Cholesky decomposition, so we can express a Cholesky decomposition for $\mathbf{M}^{-1}$:

\begin{equation}
\mathbf{M}^{-1} = \mathbf{L}\mathbf{L}^*
\end{equation}

where $\mathbf{L}$ is a lower triangular matrix with real positive entries on the diagonal. Therefore, $\mathbf{L}$ is also positive definite, and can be inverted. Therefore, we can see that

$$\mathbf{M} = \left(\mathbf{L^{-1}}\right)^* \mathbf{L}^{-1}$$

or, equivalently

$$\mathbf{L}^* \mathbf{M}\mathbf{L} = \mathbf{I} \tag{1}\label{1}$$

This looks reminiscent of a similarity transformation from $\mathbf{M}$ to the identity $\mathbf{I}$. However, it is not quite, as $\mathbf{L}$ is generally not a unitary matrix. We can define a new matrix $\mathbf{A}$ by doing the same operation to $\mathbf{K}$:

$$\mathbf{L}^* \mathbf{K}\mathbf{L} = \mathbf{A} \tag{2}\label{2}$$

$\mathbf{A}$ is generally not diagonal. However, it can be readily shown that since $\mathbf{K}$ is Hermitian, then $\mathbf{A}$ is Hermitian. (In fact, if we let $\mathbf{u} = \mathbf{L} \mathbf{v}$, we can use Equations \eqref{1} & \eqref{2} to convert our generalised eigenvalue problem $\mathbf{K}\mathbf{u} = \lambda \mathbf{M}\mathbf{u}$ to the standard eigenvalue problem $\mathbf{A}\mathbf{v} = \lambda \mathbf{v}$)

Any complex matrix has a Schur decomposition $\mathbf{A} = \mathbf{Q}\mathbf{T}\mathbf{Q}^*$, where $\mathbf{T}$ is a triangular matrix, and $\mathbf{Q}$ is a unitary matrix. However, since in our case $\mathbf{A}$ is Hermitian, the triangular matrix is also Hermitian. A Hermitian triangular matrix must be a diagonal matrix! We will represent this matrix with $\mathbf{D}$:

$$\mathbf{A} = \mathbf{Q} \mathbf{D} \mathbf{Q}^*$$

or, equivalently,

$$\mathbf{Q}^* \mathbf{A} \mathbf{Q} = \mathbf{D}$$

Trivally, we may also express $\mathbf{I}$ as

$$\mathbf{Q}^* \mathbf{I} \mathbf{Q} = \mathbf{I}$$

Substituting Equations \eqref{1} & \eqref{2} into the above, and letting $\mathbf{U} = \mathbf{L}\mathbf{Q}$, we can now see that the operations that simulataneously diagonalise $\mathbf{M}$ and $\mathbf{K}$ are

$$\mathbf{U}^* \mathbf{M}\mathbf{U} = \mathbf{I} \tag{3}\label{3}$$

and

$$\mathbf{U}^* \mathbf{K}\mathbf{U} = \mathbf{D} \tag{4}\label{4}$$

By letting $\mathbf{u} = \mathbf{U}\mathbf{q}$, and substituting this and Equations \eqref{3} & \eqref{4}, we obtain the straightforward eigenvalue problem

$$\mathbf{D}\mathbf{q} = \lambda \mathbf{q}$$

whose $n^\text{th}$ solution is readily obtained: $\lambda_n = D_{nn}$ and $\mathbf{q}_n = \mathbf{e}_n$, where $\mathbf{e}_n$ is the unit column vector whose $n^\text{th}$ element is 1, with all other elements equal to 0.

Then, since $\mathbf{u} = \mathbf{U}\mathbf{q}$, we can see that the $n^\text{th}$ eigenvector of our original problem has the form

$$\mathbf{u}_n = \mathbf{U}\mathbf{e}_n$$

In other words, the columns of $\mathbf{U}$ are the eigenvectors of $\mathbf{K}\mathbf{u} = \lambda \mathbf{M}\mathbf{u}$.

Since $\mathbf{U} = \mathbf{L}\mathbf{Q}$, and $\mathbf{L}$ and $\mathbf{Q}$ are both invertible, it follows that $\mathbf{U}$ must also be invertible. As a result, this means that $\mathbf{U}$ must be full rank, which implies that the columns comprising the matrix must all be linearly independent.

Therefore, it is indeed possible to express $N$ linearly independent eigenvectors.

(In fact, since $\mathbf{U}^* \mathbf{M}\mathbf{U} = \mathbf{I}$, it is also clear that the linearly independent eigenvectors are also $\mathbf{M}$-orthogonal. Note that this $\mathbf{M}$-orthogonality still applies even between eigenvectors corresponding to repeating eigenvalues.)


Title: Surface area ratio for cubic and spherical objects of equal volume

Body:
Given two objects of equal volume, one cubic and one spherical, which will have the greater outer surface area, and how much greater will it be versus the other object? I know this is a math forum, so a mathematical explanation is great but I'm also hoping for layman's terms and a simple ratio.

As a bonus, assuming the objects are constructed from identical material, if we know the mass of one can we know the mass of the other by using the same ratio? Why or why not?

Body:
$$V_{cube}=a^3=\frac{4{\pi}r^3}{3}=V{sphere}$$

$$\frac{a^3}{r^3}=\frac{4{\pi}}{3}$$

$$\frac{a}{r}=\sqrt[3]{\frac{4{\pi}}{3}}$$

$$\frac{{Area_{cube}}}{{Area_{sphere}}}=\frac{6a^2}{4{\pi}r^2}=\sqrt[3]{\frac{216\times 16{\pi}^2}{9 \times64{\pi}^3}}=\sqrt[3]{\frac{6}{\pi}}>1$$

$$Area_{cube} > Area_{sphere}$$


问:
Kernel functions

Shown is the kernel function $k(x,x')$ for $x'=0.$
Their point is to show the localization of kernel func. But if $x'=0$, how is it varying over space. Shouldn't the dot product be zero?

Can someone please explain the graphs in the 2nd row in the picture.This is taken from CM Bishop.

The polynomial kernel  function k(x,x') is defined as (x(trpose)*x' + c)^2

答: There is an error in the figure, which was fixed by the authors in the book's errata. You can find the correct figure here: https://www.microsoft.com/en-us/research/wp-content/uploads/2016/05/prml-errata-1st-20110921.pdf</p>

Title: Kernel functions


问:
Solving the equation $y'^2-2xy'+2y-x^2=0$
Solve the following differential equation:
$$y'^2-2xy'+2y-x^2=0$$

First, I tried to solve it w.r.t $y'$:
$$y' = \frac{2x \pm \sqrt{4x^2 - 4(2y-x^2)}}{2} \\[5mm]y' = x\pm\sqrt{2x^2-2y}$$
Now, I don't know how to solve this one. Still I have attempted as follows
$$y'-x = \pm \sqrt{2x^2-2y} \\[5mm] z = y-x^2/2 \implies z' = y'-x \\[5mm] z' = \pm\sqrt{-2z+x^2}$$
Then, I got stuck.

答:
$$y'^2-2xy'+2y-x^2=0$$
Change the variable to $v=x^2$:
$$4v(y'^2-y')+2y=v$$
$$y(v)=v  \left( 2 \left(y'-y'^2\right)+\dfrac 12 \right)$$
Is on the form:
$$y=vf(y')$$
This is D'Alembert differential equation
$$y=vf(y')+g(y')$$


问:
Can all possible angles on a rational triangle be represented as a rational multiplied by Pi?
The reason I ask: I was wondering if it was possible to find the angle of a rational triangle by only using the lengths of its sides and knowledge of $\pi$ (that is, no inverse trig functions).

So, that probably isn't possible unless every angle $\theta$ such that $\sin (\theta)$ is a rational number can be represented as $\theta = a \pi$ with $a$ is some rational number.

Clarification edit: A rational triangle is a triangle where every ratio of sides is rational.

答:
Some observations before I try to answer your question. First, a “rational triangle” will always be similar to a triangle with all sides integers, so I’ll speak rather of “integral triangles”. Second, an angle that’s a rational multiple of $\pi$ is one whose degree measure is a rational number of degrees, and I’ll speak of these as “rational angles”. The issue for angles whose radian measure is a rational number involves mathematics much more advanced than I’ll bring up here, as @Gerry and @Erick have suggested.

Third, and most significantly, it looks like you didn’t bring into the question the Law of Cosines, according to which if the angles of a triangle are $\alpha_i$ for $i=1,2,3$, opposite the sides $s_i$ in the usual way, then
$$
\cos\alpha_1=\frac{s_2^2+s_3^2-s_1^2}{2s_2s_3}\,.
$$
This tells you that in an integral triangle, all the cosines of the angles are rational numbers. It’s not hard to show, conversely, by an analytic-geometry argument, that every angle $\theta$ with $0^\circ<\theta<180^\circ$ and $\cos\theta$ rational appears as vertex angle of an integral triangle. For instance, you see that the $(3,3,2)$ triangle and the $(9,10,11)$ triangle both have an angle with cosine equal to $1/3$.

Therefore, the question you asked in the title boils down to asking which angles $\theta$ with rational cosine are rational (in degrees, remember). The answer is that there are only three of them, $60^\circ$, $90^\circ$, and $120^\circ$. Now let’s see why.

First step is to note that an angle $\theta$ is rational if and only if there is a positive natural number $n$ with $n\theta$ a multiple of $\pi$ (radian measure now), in other words with $\cos(n\theta)=\pm1$.

Second step is to bring in the Čebyšev (Chebyshev) polynomials (of the first kind), $T_n(X)$, which tell you, among many other things, the relation between $\cos(n\theta)$ and $\cos\theta$. When you use the addition formulas for cosine, you see that not only is $\cos(2\theta)=2\cos^2\theta-1$, but that every $\cos(n\theta$) can be expressed as a polynomial in $\cos\theta$, and this polynomial is $T_n$. You can read about these polynomials in Wikipedia, but the important fact for us is that there’s a recursive definition of them, $T_0=1$, $T_1(X)=X$, and $T_{n+1}(X)=2XT_n(X)-T_{n-1}(X)$. You should check that they do what you want for the cosines, at least in the case of $n=2$ and $n=3$. From the recurrence relation, you see that, first, the Čebyšev polynomials have their coefficients in the integers $\mathbb Z$, and, second, that the leading (highest) coefficient of $T_n(X)$ is $2^{n-1}$, for $n>0$.

Now let’s look at an angle whose cosine $\lambda$ is rational, different from $0$ and $\pm1/2$. There are two cases, either the denominator of $\lambda$ is divisible by an odd prime $p$, or it’s divisible by $4$. Let’s look at the first case, that the “$p$-order” of $\lambda$, often written $v_p(\lambda)$, is negative. You get this “order” of $\lambda$ by writing it as a product of primes with exponents, in fact you can write
$$
\lambda=\pm\prod_i p_i^{v_i}\,,
$$
where for convenience here only I’ve written $v_i$ instead of $v_{p_i}$ As an example, $-105/52=-2^{-2}3^15^17^113^{-1}$. So $v_2(-105/52)=-2$, $v_{11}(-105/54)=0$. The important thing to remember is that when $\lambda$ and $\mu$ are rational numbers with different $p$-orders, then the $p$-order of $\lambda+\mu$ is the smaller of the two $p$-orders.

Now we’re ready to roll. You see that if $v_p(\lambda)=m<0$, then $v_p(T_n(\lambda))=nm<0$. In other words, if there’s $p$ in the denominator of the cosine of your angle $\theta$, there’s at least $p^n$ in the denominator of $\cos(n\theta)$. In particular, it can’t be $\pm1$, so that $\theta$ is not a rational number of degrees.

The other case is that $4$ divides the denominator of $\lambda=\cos\theta$, i.e. $v_2(\lambda)\le-2$. Then you see that the leading term of $T_n(X)$ has the most important contribution to the $2$-order of $T_n(\lambda)$, because $v_2(2^{n-1}\lambda^n)=n-1+nv_2(\lambda)$, which is still negative, because of the condition on $v_2(\lambda)$.

The upshot? If $\theta$ is an angle of an integer triangle, then $n\theta$ is not a multiple of $180^\circ$ no matter what positive $n$ you may take. This with the exception of the angles $60^\circ$, $90^\circ$, and $120^\circ$ only.


问:
maximum of the function $S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi\sqrt{2+2\cos\phi}$
In a problem I have the function:
$$S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi \sqrt{2+2\cos\phi}$$
in which $\phi \in [0,\pi]$.

I have to find the point of maximum and calculate the maximum value of the function.

$\frac{dS}{d \phi}= 2 \pi*sin(\phi) cos(\phi)+\frac{\pi*((3cos(\phi))^2+2cos(\phi)-1))}{\sqrt{2+2cos(\phi)}}$

Putting this derivative equal to $0$ we have the equation:
$-8 (cos\phi)^5-17 (cos\phi)^4-4(cos\phi)^3+(cos\phi)^2+4cos\phi-1=0\Rightarrow  (cos\phi+1)^3*(-8(cos\phi)^2+7cos\phi-1)=0$
The maximum of this function is for $\cos \varphi = \frac{7+ \sqrt{17}}{16} $

Then $$\sin\varphi=\frac{\sqrt{190-14 \sqrt{17}}}{16}$$

Now I'm trying to find the maximum value of the function but I'm lost in the calculation and I don't know what is the mistake.
The result should be  $$S_{\max}= \frac{\pi(107+51 \sqrt{17})}{128}$$

答:
Note

\begin{align}
S(\phi) &= \pi \sin^2\phi+\pi\sin\phi \sqrt{2+2\cos \phi}\\
&= \pi \sin^2\phi+2\pi\sin\phi \cos \frac\phi2\\
&= 4\pi\left(\sin\frac\phi2 + \sin^2\frac\phi2-\sin^3\frac\phi2-\sin^4\frac\phi2\right)\tag1
\end{align}
Set $S’(\phi)= 0$ to get
\begin{align}
&2\pi \cos\frac\phi2\left(
4\sin^3\frac\phi2 +3 \sin^2\frac\phi2 - 2\sin\frac\phi2-1\right)\\
= &2\pi \cos\frac\phi2 \left(\sin\frac\phi2+1\right)\left( 4\sin^2\frac\phi2- \sin\frac\phi2-1\right)=0
\end{align}
and the valid solution $\sin\frac\phi2=\frac{1+\sqrt{17}}8$. Then, plug it into (1) to obtain the maximum value
$$S_{max}=\frac{107+51\sqrt{17}}{128}\pi$$


问:
super martingale which dominates
Could anyone tell me what it means by ''it dominates $\{E (Z|G_k), G_k\}, E(|Z|)<\infty$'' ? in the following statement? Thanks.

Let $(Y_k, G_k)$ be a super-martingale and it dominates $\{E (Z|G_k), G_k\}, E(|Z|)<\infty$,

答:
In general, we say that a random variable $Y$ dominates a random variable $X$ if
\begin{align}
|X(\omega)|\leq Y(\omega) \quad a.s.
\end{align}
This language is used for example in the Dominated Convergence Theorem.

In your case, you have a supermartingale $\{Y_k\}$ and a stochastic process $\{E[Z|G_k]\}$. Thus, that $\{Y_k\}$ dominates $\{E[Z|G_k]\}$ simply means that for all $k$ we have
\begin{align}
\big|E[Z|G_k]\big|\leq Y_k \quad a.s.
\end{align}

Title: super martingale which dominates


问:
Finding a contradiction to $G$ being abelian when $G\cong M\times L$
I am trying to disprove/prove the following:

"Let $M, L \subseteq G$, $G \cong M \times L$. Then if all the elements of $G$ commute with all the elements of $L$, we have that $G$ is abelian."

For this to be true, it would be necessary that $(ab, cd) = (ba, dc)$,$\ \ \forall a,b \in M, \forall c,d \in L$. As we don't necessarily have $ab=ba$ (since we don't know if $M$ is abelian, then I believe this statement would be false). However I cannot find a counterexample to show this so I'm not sure if I'm perhaps missing out on something.

So I'm just wondering if there's a neat way to disprove/prove the statement, since I can't really seem to find a counter example.

答: Let $L=\{e\}$ and $M=D_3$, the dihedral group of order $6$.


问:
Independent and identically distributed random variable questions.

So I am currently studying Probability and Statistics and was wondering if anyone could help me to understand this questions.

This is what I have gathered from my notes so far to aid me in answering this but I don't know if this is correct as my notes don't have any examples or text specifically relating to these sort of questions:

Let $Mn=max(X1,X2,…,Xn)$. where $X1,…,Xn$ are independent and uniformly distributed.

The distribution function of the maximum is the joint probability that:

$Xk≤x$ for all $k
FM(x)=P(Mn≤x)=P(X1≤x,…,Xn≤x)=xn
for 0≤x≤1$. Also $FM(x)=0 for x<0

and FM(x)=1 for x>1$.

The density function is

$fM(x)=F′M(x)=nxn−1$

So I'm not sure whether I should be using this and saying that:

$Y_3=max(X_1,X_2,X_3)$

and that I need to find the distribution function in terms of $Y$ rather than $X$

and then let $F'_Y(x)=nxn-1$

And this is where my ideas/knowledge of this question end unfortunately. Can anyone advise how to further tackle this please.

答:
$P(Y>1) = 1- P(Y\le1)$ so it suffices to compute $P(Y \le 1)$

$$P(Y \le 1) = P(X_1 \le 1, X_2 \le 1, X_3 \le 1) = P(X_1 \le 1) P(X_2 \le 1) P(X_3 \le 1)$$ (make sure you know how to justify each equality above)

$P(X_i \le 1) = F(1)$ for each $i$


问:
exponential inequality: $[r/(r-2)]^{1-2/r}\leq \exp\{2/r\}$ holds?
Does the following hold:

$$\bigg(\frac{r}{r-2}\bigg)^{1-2/r}\leq\exp\bigg\{\frac{2}{r}\bigg\}, \ r>2?$$

How to show it?

I see that the maximum of $x^{1/x}$ occurs at $x=e$.

答:
let $\frac 2r = t > 1$, then your inequality is equivalent to:

$$\left(1 + \frac{t}{1-t}\right)^{1-t} = \left(\left(1+\frac{t}{1-t}\right)^{\frac{1-t}{t}}\right)^t\leq e^t$$
because $\left(1+\frac 1x\right)^x \leq  e$ for for any positive $x.$


问:
Meyer's Theorem in Williams & Rogers
In Diffusions, Markov Processes and Martingales Volume 2 by Rogers and Williams they state the following theorem due to Meyer:

$\mathbf{Theorem }$ Le $M\in\mathcal{M}^2_0$. Then there exists a unique increasing process $[M]$ null at $0$ such that

$M^2-[M]$ is a uniformly integrable martingale

$\Delta [M]=(\Delta M)^2$ on $(0,\infty)$

By definition $\mathcal{M}^2_0$ is the space of martingale null at $0$, which are bounded in $L^2$. I'm confused about the following: Clearly $[M]$ is adapted, by $1.$, but I also read (in a comment in my lecture notes) that it is even predictable. Why do Rogers and Williams not mention this? First I was also confused that they do not mention the adaptness explicitly. Usually they write something like this: "...exists a unique adapted increasing process...". I just want to be sure, that I do not miss something important here. Thanks for clarification.

答:
The result you cite regards the existence "ordinary" quadratic variation process, $[M]$, of a square-integrable martingale. It is adapted, but not in general predictable. To "compensate" for its lack of predictability and still ensure uniqueness, the requirement on the jumps is made. There is another process, $\langle M\rangle$, the predictable quadratic variation, which is characterized by having $M^2 - \langle M\rangle$ be a local martingale and having $\langle M\rangle$ be predictable (but no requirement on the jumps, in contrast to the theorem you cite). The process $\langle M\rangle$ is given as the dual predictable projection of $[M]$.

See for example "Semimartingale theory and stochastic calculus" by He, Wang and Yan for more on all processes mentioned above.

Title: Meyer's Theorem in Williams & Rogers


问:
Prove that this block matrix is positive definite
I'm trying to prove that this symmetric block matrix is positive definite
$$
    \begin{bmatrix}
    B & -I & 0 & \dots& 0 \\
    -I & B & -I & \ddots & \vdots\\
    0 & -I & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -I\\
    0 & \dots & 0& - I & B \\
    \end{bmatrix}
$$

with $I$ the identity matrix

$$B = \begin{bmatrix}
    4 & -1 & 0 & \dots& 0 \\
    -1 & 4 & -1 & \ddots & \vdots\\
    0 & -1 & \ddots & \ddots & 0\\
    \vdots & \ddots & \ddots & \ddots & -1\\
    0 & \dots & 0& - 1 & 4 \\
    \end{bmatrix}$$

by proving that this term $x^T A x$ is positive, but given the fact that it is a block matrix I'm stuck. Could you help me please?

答:
Call your matrix $A$. By Gerschgorin disc theorem, all eigenvalues of $A$ lie inside the union of the Gerschgorin discs $X=\overline{B}(4,4),\,Y=\overline{B}(4,3)$ and $Z=\overline{B}(4,2)$, where $\overline{B}(z,r)$ denotes the closed disc centred at $z\in\mathbb C$ with radius $r$. Since these eigenvalues are also real (because $A$ is real symmetric), they must be nonnegative.

Now, by Taussky's extension to Gerschgorin disc theorem, since $A$ is irreducibly diagonally dominant (i.e. $A$ is irreducible, diagonally dominant and it is strictly diagonally dominant on at least one row), it is nonsingular. Hence all eigenvalues of $A$ are positive and $A$ is positive definite.


问:
How to find the direction of maximum increase?
For a room of side length $k$ the vertices of which are

$(0,0,0),(k, 0, 0),
(0, k, 0), (0, 0,k), (k, k, 0), (k, 0, k), (0, k, k), (k, k, k)$

The temperature within this rooms is given by the scalr field $T(x,y,z) =T_{0}xyz$ where $T_{0}$ is a constant.

I have three question.

where is the hottest point in the room

In the centre of the room what is the direction of maximum increase of temperature?

Find the function $T_{x=0}(yz)$ and $T_{x=a}(yz)$

My working so far:

The hottest point is at (k,k,k)

$\nabla T= T_0yzi +T_0xzj+T_0xyk $

So the direction of maximum increase of temperature is in positive x,y,z direction?

3)
$T_{x=0}(yz) =0$

$T_{x=a}(yz)=T_0 ayz$

答:
You've got the gradient correct; it's $$yz\mathbf i +xz\mathbf j + xy \mathbf k$$  At the center of the cube, we have $x=y=z=\frac k2$, so you need to substitute these values in the formula for the gradient.  Actually, since $x=y=z$, we can see that the direction is $\mathbf i+\mathbf j+\mathbf k$ without doing any substitution.

Sometimes, "direction" is taken to mean a unit vector.  If your instructor follows this convention, you'll have to normalize the vector.


问:
Another evaluating limit question: $\lim\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$
How do I begin to evaluate this limit: $$\lim_{n\to \infty}\ \frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}\;?$$

Thanks a lot.

答:
Well known equality :
$$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\leqslant \frac{1}{\sqrt{2n}}$$
Proof of right side. Having
$$\frac{1}{2}<\frac{2}{3},\frac{3}{4}<\frac{4}{5},\frac{5}{6}<\frac{6}{7},\cdots,\frac{2n-3}{2n-2}<\frac{2n-2}{2n-1},\frac{2n-1}{2n}<1$$
and multiplying we obtain
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}<\frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7} \cdots \frac{2n-2}{2n-1}$$
Now multiply both sides on $ \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}$ gives
$$ \left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\right)^2<\frac{1}{2n}$$
and we obtain
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} <\frac{1}{\sqrt{2n}}$$
Analogical way works for left side.


问:
What is this notation (series)?
I'm asked to show the convergence or divergence of the following series using the ratio test:

$$\sum_{k=1}^{\infty} \frac{1\cdot 4\cdot 7\cdot\cdot\cdot (3k-2)}{2\cdot 4\cdot 6\cdot\cdot\cdot (2k)}$$

What does the "$\cdot\cdot\cdot$" represent, and how would I go about applying the ratio test to such a series (which is, of course, 'trivial' for something less.. obscure(?)) ?

Title: What is this notation (series)?

Body:
hint

If
$$a_k=\frac{1.4.7.10....(3k-2)}{2.4.6.8...(2k)}$$

then

$$a_{k+1}=\frac{1.4...(3k-2)(3(k+1)-2)}{2.4....(2k)(2(k+1))}$$


答:
$...$ is used to imply some sort of pattern.

On the numerator, the meaning of $ 1\cdot 4\cdot 7\dots(3k-2)$ means the product of all numbers $1$ more than a multiple of $3$, ending at $3k-2$. For example when $k=5$ it is the product $1\cdot4\cdot7\cdot10\cdot13.$

Similarly $ 2\cdot 4\cdot 6\dots(2k)$ would be the product of even numbers up to a certain point. For $k=5$ it would be $2\cdot4\cdot6\cdot8\cdot10$.

Applying the ratio test should be pretty simple given that the top and bottom both have $k$ terms in the product.

Title: What is this notation (series)?

Body:
I'm asked to show the convergence or divergence of the following series using the ratio test:

$$\sum_{k=1}^{\infty} \frac{1\cdot 4\cdot 7\cdot\cdot\cdot (3k-2)}{2\cdot 4\cdot 6\cdot\cdot\cdot (2k)}$$

What does the "$\cdot\cdot\cdot$" represent, and how would I go about applying the ratio test to such a series (which is, of course, 'trivial' for something less.. obscure(?)) ?


问:
Bijection between $P(E)^F$ and $P(E*F)$
Here * means the cartesian product, and P(X) is the set of all subsets of $X$. $E^F$ is the set of all applictions from $F$ to $E$.
The question is contained in the title : i wish to find a natural bijection between those sets.
Here is my try :

Let $f:F \rightarrow P(E)$ be a random function of $P(E)^F$.
We call $F_0=\{x\in F, f(x)\neq \emptyset\}$.
Then using the axiom of choice, we can define a function that associates, to every $x\in F$,
some $y_x\in f(x)$ and get a pair $(x,y_x)$.

Let's call $X_f=\bigcup_{x\in F_0}\{(x,y_x)\}$ for every function $f$. It is clearly a subset of $E*F$.

My idea would be to consider after that the application $g:P(E)^F\rightarrow P(E*F)$ such that :
$\forall f\in P(E)^F$, $g(f)=X_f$.

Now i get stucked because i cannot prove that $g$ is bijective. To do so, i would have to make further suppositions about the choices of the $y_x$ that were made for every function $f$, in order to get both surjectivity and injectivity. My idea to get surjectivity was to consider the size of $F_0$ (in order to control the size of $X_f$ by controlling the elements of $F$ that i could put into a pair that is in $X_f$). However, i feel like this is not the right method. Could anyone give me a hint about this ? I would be delighted if you could ! :)

答:
Currying is normally written like this(this is known to be bijection):
$$ hom(A,hom(B,C)) == hom(A \times B,C) $$

This can be written more conviniently like this:
$$ A\rightarrow B \rightarrow C = (A \times B) \rightarrow C$$

Now if we use $C=2$, $A=E$, $B=F$ we arrive at:
$$ E \rightarrow F \rightarrow 2 = (E \times F) \rightarrow 2$$

Now using $E\rightarrow X = X^E$
We can get:
$$(F\rightarrow 2)^E = (E \times F) \rightarrow 2$$

Now the last thing is to notice that $P(X) = X\rightarrow 2$, and we'll have the desired property:
$$P(F)^E = P(E\times F)$$

NOTE: The question asked for $P(E)^F$, but we arrived at $P(F)^E$, so either the question is wrong or we don't have all the necessary steps available.

Title: Bijection between $P(E)^F$ and $P(E*F)$


问:
Probability of stopping experiment with throw of unbalanced coin?
We have two bags  such that the first bag contains  $ 1 $  black balls and  $ 2 $  red balls and the second bag contains  $ 5 $  black balls and  $ 1 $  red balls.
We throw an unbalanced coin such that  the probability of getting a "PILLE" is equal to one third of the probability of getting a "FACE". If we get the back " PILLE" we pull a ball from the second bag and if we get "FACE" we pull a ball from the first bag. The process stops when we pull a red ball. What is the probability of stopping the experiment ?

Attempts: Probability of stopping the experiment is the probility of pulling red ball , I have tried to use conditional probability such that $P(A|B)=\dfrac{p(A\cap B)}{p(B)}$, "B" is the probability of throwing the coin " , "$A|B $ is the probability  when the experiment stop" , Now we have $p(B)=1/6$ because we have  equiprobaility if we pull any ball from any above bags which is $\frac12$, now my problem is what is  $p(A\cap B)$? Events here are dependents according to this exercices this means that we can't have $p(A|B)=p(B) $.? Any help ?

答:
You want to use the law of total probability.  Let R be the event of drawing a red ball, P, be the event of throwing Pille, and F the event of throwing face.  Then

$$\Pr(R)= \Pr(P)\Pr(R|P)+\Pr(F)\Pr(R|F)$$

Can you take it from here?


问:
Proving a certain nonlinear ODE has a solution on $t \geq 0$
I'm trying to prove that the following ODE admits a solution on the interval $[0, \infty )$:

$\begin{align*}
\begin{cases}
x_1^\prime &= x_2 - x_1^3 \\
x_2^\prime &= \frac{1}{2}x_1 - x_2 + d \sin t
\end{cases}
\end{align*}$

With initial condition $x(0) = x_0$.

The usual approaches seem to fail: the derivative isn't bounded on $\mathbb{R}^2$ and the function does not seem Lipschitz on all of that domain either. Using energy functions doesn't seem to help either.

Any hints would be highly appreciated!

答:
Hint.

We have

$$
x_2'=x_1'' + 3x_1^2x_1'
$$

substituting into the second DE we have

$$
x_1''+\left(3x_1^2+1\right)x_1'+\left(x_1^2-\frac 12\right)x_1=d\sin t
$$

In this equation, $|x_1|$ can't grow without limit, being stable for $x_1^2>\frac 12 $ and consequently $x_2$ is also stable.


问:
Do we really need polynomials (In contrast to polynomial functions)?
In the following I'm going to call

a polynomial expression an element of a suitable algebraic structure (for example a ring, since it has an addition and a multiplication)
that has the form $a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $x$ is
some fixed element of said structure,

a polynomial function a function of the form $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ (where the same algebraic considerations as above apply)

a polynomial an element of the form $a_{n}X^{n}+\cdots+a_{1}X+a_{0}$ with indeterminates $X$ (these
can be formalized, if we are in a ring, with strings of $0$s and
a $1$).

Note that when we are very rigorous/formal, polynomial expressions
and polynomials are something different (although in daily life we
often use them synonymously). Polynomial functions and expressions
are also different from each other although in this case the relationship
is a closer one, since every polynomials expression can be interpreted
as a polynomial function evaluated at a certain point (thus "polynomial
functions" are something more general than "polynomial expressions").

My question is: Why do we use polynomials ? It seems to me that every
piece of mathematics I have encountered so far, one could replace
every occurrence of polynomials with polynomial expressions/functions
without any major change in the rest of the proof/theorem/definition
etc.

My thoughts:

The only reasons that I can see to use polynomials are the following
two:

After one makes the idea precise that one can plug "suitable"
elements into polynomials (which may lie a ring containing the ring
in which the coefficients live in), one can save time in certain setting,
by handling the "plugging of suitable elements into polynomials"
more elegantly: For example in case of the theorem of Cayley-Hamilton,
which in its "polynomial function" version would look like:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial (function) is $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$.
Then
$$
a_{n}A^{n}+\cdots+a_{1}A+a_{0}I=0.
$$

whereas the "polynomial" version looks more elegant:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial is $p_{A}\in K\left[X\right]$. Then
$$
p_{A}\left(A\right)=0.
$$

The only thing that polynomials can "do", but algebraic expressions/functions can't, is to be different, when the algebraic expressions/functions are the same (i.e. there's a theorem that tells us that the mapping of polynomials to polynomials expressions/functions
isn't injective, if the field is finite). Maybe this small difference makes a big enough difference
to consider polynomials after all, but as I said, I haven't encountered
any situation in which this difference could manifest itself.

(I'm guessing that maybe cryptography or higher number theory really needs
polynomials and not just polynomial expressions/functions. Since I don't know
anything about these subjects, I would be very happy with an example
of a theorem (whose content isn't merely technical as it is the case with the theorem above) involving polynomials, where these absolutely cannot
be replaced by polynomial expressions/functions. Conversely I would also be happy with an authoritative statement from someone knowledgeable that indeed we could dispense of polynomials, if we wanted to.)

答:
If the mapping from polynomials to polynomial functions is not injective, then one cannot talk about the coefficients of a polynomial function. Without coefficients, one cannot do Euclidean division. Without Euclidean division, not much of what one does readily with polynomials over a field can be done (defining minimal polynomials of linear operators is one such thing; there are many others). In short, in algebra one works with polynomials, not with polynomial functions.

Note that the ring of polynomial functions over a finite field $F$ (which is in fact the ring of all functions $F\to F$) is not an integral domain; it is in fact a rather ugly object from the algebraic point of view (a product ring of $\#F$ copies of $F$).

However, if you want a statement from someone knowledgeable that indeed we could dispense of polynomials, note that Serge Lang in Introduction to Linear Algebra (2) Chapter IX defines polynomials over an arbitrary field $K$ as polynomial functions, and goes on stating a Theorem$~1$ that the mapping from coefficient sequences to polynomials (i.e., functions) is injective (which as observed in the question fails for finite fields). The proof of the theorem is conveniently only given for the case where $K$ is the real or complex numbers. I certainly would not want to exclude Serge Lang from the set of knowledgeable people, but it seems that in this particular instance he made a regrettable error where he should have known better.

Added: Leafing though the cited book, I recently found out a detail that might explain what Lang did. I thought I knew without having to look it up what was meant by "a field $K$"; however, in section II.2 of the book it says "Let $K$ be a subset of the complex numbers $\mathbf C$. We shall say that $K$ is a field if it satisfies the following conditions:" (requiring closure under addition, multiplication, negation and inversion of nonzero$~x$, and containment of $0,1$). The proof later given of uniqueness of polynomial coefficients is by considering the asymptotic behaviour of the polynomial function for large arguments. The proof is immediately followed by the cryptic sentences "We have used the notion of limit. In Chapter$~$XII, $\S1$ we shall prove the same statement for more general fields, by a method which is completely algebraic." This leaves me wondering what "more general" fields are intended; given the earlier definition one might assume "general subfields of $\mathbf C$", but the proof using limits would appear to do fine for any such field. Anyway I cannot see which proof is being referred to. There is corollary 3 saying that a polynomial in $K[t]$ of degree $n$ has at most $n$ roots in$~K$, but that is not the same statement. More importantly, it assumes that degree has been defined, and it uses Euclidean division; as these notions use the existence of (unique) polynomial coefficients, this would appear to be circular.


答:
There is no need to apply that formula here as we have not been given an event that has occurred with certainty.

Second, the question does not state if we continue to draw and if so with replacement or without. If we have to find it for the first draw, here is how we can do -

Probability that we will pick a ball from bag $1$ is $3$ times of that we would pick from bag $2$ depending on the biased coin toss.

So probability that we would pick from bag $1$ is $\frac{3}{4}$. Probability that we would pick from bag $2$ is $\frac{1}{4}$. You know the number of balls in each bag to calculate the probability.

The probability of stopping the experiment after first draw = $\frac{3}{4} \times P(R|1) + \frac{1}{4} \times P(R|2)$.

$P(R|1)$ is probability of picking a red ball from a draw from bag $1$ and $P(R|2)$ is probability of picking a red ball from a draw from bag $2$.

Title: Probability of stopping experiment with throw of unbalanced coin?

Body:
We have two bags  such that the first bag contains  $ 1 $  black balls and  $ 2 $  red balls and the second bag contains  $ 5 $  black balls and  $ 1 $  red balls.
We throw an unbalanced coin such that  the probability of getting a "PILLE" is equal to one third of the probability of getting a "FACE". If we get the back " PILLE" we pull a ball from the second bag and if we get "FACE" we pull a ball from the first bag. The process stops when we pull a red ball. What is the probability of stopping the experiment ?

Attempts: Probability of stopping the experiment is the probility of pulling red ball , I have tried to use conditional probability such that $P(A|B)=\dfrac{p(A\cap B)}{p(B)}$, "B" is the probability of throwing the coin " , "$A|B $ is the probability  when the experiment stop" , Now we have $p(B)=1/6$ because we have  equiprobaility if we pull any ball from any above bags which is $\frac12$, now my problem is what is  $p(A\cap B)$? Events here are dependents according to this exercices this means that we can't have $p(A|B)=p(B) $.? Any help ?


问:
Diffeomorphism from $n$-cone to $\mathbb{R}^{n+1}$ -- parametric equation of geodesic?
Let $\mathbb{S}^n=\{\mathbf{x}\in\mathbb{R}^{n+1}\colon\lVert\mathbf{x}\rVert^2=1\}$ be the $n$-sphere. We know that a bijection from $\mathbb{S}^n\backslash \{N\}$ to $\mathbb{R}^n\times\{0\}\subset\mathbb{R}^{n+1}$, where $N$ is, for instance, the north pole of $\mathbb{S}^n$, is given by stereographic projection. Moreover, we know that a great circle on $\mathbb{S}^n$ can be given analytically, parametrized for instance by an angle $\theta$ and with respect to some basis where this geodesic belongs.

I am trying to figure out if any of the following can be defined in a similar manner:

Can an $n$-dimensional cone be defined similarly to the $n$-sphere, i.e., as a subset of $n+1$-dimensional Euclidean space, given a formula?

If so, is there any injection (more specifically a diffeomorphism) from the $n$-cone manifold to the Euclidean $\mathbb{R}^{n+1}$?

Can a geodesic on $n$-cone be parametrized in a similar way? For instance, if we define a $2$-plane using a set of $n+1$-dimensional vectors, can we parametrized the conic section in the general case?

答:
Let $f : \mathbb{R}^{n+1} \to \mathbb{R}$ be defined by $f(x_1,\ldots,x_{n+1}) = \sum_{k=1}^n {x_k}^2 - {x_{n+1}}^2$. Writing $(x_1,\ldots,x_n,x_{n+1})=(X,x_{n+1})$, $f(X,x_{n+1}) = \|X\|^2 - {x_{n+1}}^2$. Then $C = \left\{ v \in \mathbb{R}^{n+1} ~|~ f(v) = 0\right\}$ is a cone of revolution. It is not a submanifold of $\mathbb{R}^{n+1}$ because of the origin, but $C\setminus \{0\}$ is a submanifold. Define $C^+$ to be the set of all $(X,t)\in C$ with $t >0$ (the upper-half cone).

Then $C^+$ is a diffeomorphic to $\mathbb{R}^n\setminus\{0\}$ by the diffeomorphism $(X,t)\in C^+ \mapsto X \in \mathbb{R}^n$. With the induced metric of $\mathbb{R}^{n+1}$, the above diffeomorphism is obviously not an isometry.

One can distinguish two kinds of geodesics on $C^+$:

the first kind of geodesics are those that are obtained by intersecting $C^+$ with a vertical 2-plane of $\mathbb{R}^{n+1}$

the others geodesics, which are not contained in such a vertical 2-plane

To visualize this, consider a point $p\in C^+$ and a tangent vector $v \in T_pC^+$. If $v$ is in the vertical plane $\mathrm{span}(e_{n+1},p)$, then the euclidean geodesic of $\mathbb{R}^{n+1}$defined by $\gamma(t) = p + tv$ is included in $C^+$ on an open interval. In fact, if $v$ points "upward", it is defined on $(t_0,+\infty)$ and if $v$ points "downward", on $(-\infty,t_0)$ for a certain number $t_0$ ($t_0$ is the parameter for which $\gamma(t_0) = 0$).

The second kind of geodesics are much more complicated. If $v$ is not contained in the vertical $2$-plane $\mathrm{span}(p,v)$, the geodesic in $C^+$ with $\gamma(0) =p$  and $\gamma'(0)=v$  turns around the cone, going upward or downward depending on $v$, and looks like a spiral expanding upward or retracting downward.There is no really simple way to compute them. Look at this picture to visualize them.

We could have choosen another cone of revolution (with another angle).

For the sake of completeness, I have to add that cones of revolution are flat hypersurfaces. You can cut the cone along a vecrtical geodesic (the first kind of geodesics) and developp the surface to flatten it. Then the cone is just an angular sector of $\mathbb{R}^n\setminus\{0\}$. More than that, this cut along a vertical geodesic provides an isometry of the cone minus the vertical geodesics to the angular sector. That means that geodesics are just straight lines on the flatten angular sector. You can visualize it here (the reference is in french, but the pictures just talk for themselves).

I encourage you to look for some references about hypersurfaces of revolution in euclidean spaces. The $3$ dimensional case is really instructive.


问:
Given X+Y+Z = 0, find $\frac{(X^3+Y^3+Z^3)}{XYZ}$
The result can be found using the equation :

$(X^3+Y^3+Z^3) - 3XYZ = (X+Y+Z)(X^2 - XY +Y^2 - YZ +Z^2 - XZ)$

Since X+Y+Z = 0, the right side of the equation is equal to 0. Therefore $X^3+Y^3+Z^3 = 3XYZ$ and the answer to the problem is 3.

However what if we calculate $X^3+Y^3+Z^3$ as $(X+Y+Z)^3 - 3X^2Y - 3Y^2X - 3X^2Z - 3z^2X - 3Y^2Z - 3Z^2Y$. $(X+Y+Z)^3 = 0$, so $X^3+Y^3+Z^3$ can be replaced with $- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y$.

$\frac{- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y}{XYZ}$ = $\frac{-3X - 3Y}{Z} - \frac{-3X-3Z}{Y} - \frac{-3Y-3Z}{X}$

If X+Y+Z = 0, consequenty 3X+3Y+3Z = 0. Our expression can be rewritten as $\frac{-3Z}{Z} + \frac{-3Y}{Y} +\frac{-3X}{X}$, so the answer is -9.

Could you please tell me which way of solving this problem is right and why

答:
In your second set of calculations, where you're subtracting terms from $(X + Y + Z)^3$, you're missing a $-6XYZ$ term. Also, going from $3X + 3Y + 3Z = 0$ gives $-3X - 3Y = 3Z$, not $-3Z$, with this sign error occurring for the other $2$ terms as well. When you include the $-6XYZ$ term, you'll get an additional $\frac{-6XYZ}{XYZ} = -6$ value, and correcting for the second mistake with the term sign values, it becomes $9$ instead of $-9$.

Thus, the end result becomes $9 - 6 = 3$, as you got originally. As such, both methods give the same end result, as expected.


问:
Particular additive / multiplicative structures of quotient rings?
I've come across these questions with solutions asking about the +/x tables of quotient rings, and I can't figure out where the get the solutions from.

The first example: The quotient ring $\mathbb{Z}_3 [x]/(x^2+1)$ has additive structure $(a+bx+I)+(c+dx+I)=a+b+(c+d)x+I$, and multiplicative structure given by setting $x^2=2$. I cannot figure out where either of those structures come from.

The second example: The field $\mathbb{Z}[\alpha]:= \mathbb{Z}_2[x]/(x^3+x+1)$ has multiplicative structure defined by $(a+b\alpha+c\alpha^2)(a'+b'\alpha+c'\alpha^2)=aa'+bc'+b'c+x(b(a'+c')+b'(a+c)+cc')+x^2(ca'+ac'+bb'+cc')$.

Once again I have no idea where that structure comes from. I asked my prof and the only thing she said was "Since $\alpha$ is a root of $x^3+x+1$, then $\alpha^3=\alpha+1$" I also have no idea where that comes from.

While I definitely want to know where these problems come from, I really just don't have an understanding in general of how to get to those. Really no explanation is given for these problems. Thank you!

答:
The additive and multiplicative structure are defined as follows in any quotient ring $R/I$:
$$(r+I)+(r'+I) = (r+r')+I$$
$$(r+I)(r'+I) = (rr') + I$$

If we look at the first example, any equivalence class in $\Bbb{Z}_3[x]/(x^2+1)$ has a unique representative of the form $(a+bx)+I$. By the generic definition above,

$$(a+bx+I)+(c+dx+I) = (a+bx+c+dx)+I = (a+c)+(b+d)x + I.$$

For multiplication,

$$(a+bx+I)(c+dx+I) = (a+bx)(c+dx)+I = ac + (ad+bc)x + bdx^2 + I,$$

and since $$x^2 +1 \in I \implies -bdx^2-bd \in I \implies I = -bdx^2-bd + I, $$

the product above may be written as $$(a+bx+I)(c+dx+I) = ac+(ad+bc)x - bd + I = (ac - bd) + (ad+bc)x + I.$$

Finally in $\Bbb{Z_3}$, we have $-1 = 2$, so we may also write $(ac + 2bd) + (ad+bc)x + I$.


答:
The first attempt is correct, assuming that $X, Y, Z \neq 0$. You can verify this with a simple example, say $X = Y = 1, Z = -2$. The mistake in the second attempt is due to the fact that you have missed a term in your expression of $X^3 + Y^3 + Z^3$. Specifically,
$$(X + Y + Z)^3 = X^3 + Y^3 + Z^3 + 3(X^2 Y + Y^2 X + X^2 Z + Z^2 X + Y^2 Z + Z^2 Y) + \color{blue}{6XYZ}$$
To avoid errors like the one you made, always check that the number of terms in your expansion is correct! We expect $3^3 = 27$ terms in the expansion of $(X + Y + Z)^3$, and your expansion only gave $21$. You can then plug this into your second method and you will see that you get the correct answer of $3$.

Title: Given X+Y+Z = 0, find $\frac{(X^3+Y^3+Z^3)}{XYZ}$

Body:
The result can be found using the equation :

$(X^3+Y^3+Z^3) - 3XYZ = (X+Y+Z)(X^2 - XY +Y^2 - YZ +Z^2 - XZ)$

Since X+Y+Z = 0, the right side of the equation is equal to 0. Therefore $X^3+Y^3+Z^3 = 3XYZ$ and the answer to the problem is 3.

However what if we calculate $X^3+Y^3+Z^3$ as $(X+Y+Z)^3 - 3X^2Y - 3Y^2X - 3X^2Z - 3z^2X - 3Y^2Z - 3Z^2Y$. $(X+Y+Z)^3 = 0$, so $X^3+Y^3+Z^3$ can be replaced with $- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y$.

$\frac{- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y}{XYZ}$ = $\frac{-3X - 3Y}{Z} - \frac{-3X-3Z}{Y} - \frac{-3Y-3Z}{X}$

If X+Y+Z = 0, consequenty 3X+3Y+3Z = 0. Our expression can be rewritten as $\frac{-3Z}{Z} + \frac{-3Y}{Y} +\frac{-3X}{X}$, so the answer is -9.

Could you please tell me which way of solving this problem is right and why


问:
Filling Riemannian manifolds, Gromov, Proposition 5.1.B
In Gromovs paper Filling Riemannian manifolds, https://projecteuclid.org/download/pdf_1/euclid.jdg/1214509283 it states on page 46:

5.1.B Proposition

Let $v$ be a surface with complete Finsler metric, fix a point $v\in V$ and let $R\in[\frac{1}{2}h(v),\frac{1}{2}Sys(V,v)]$. Then
\begin{equation}
Area \ B(v,R) \ \geq \ \frac{1}{2}(2R-h(v)).
\end{equation}
Here $Sys(V,v)$ denotes the length of a shortest noncontractible loop based at $v$ and
\begin{equation}
h(v)=\inf\{tension(\gamma)|\gamma  \ \text{is either a noncontractible loop or an infinite path through  }v\},
\end{equation}
where
\begin{equation}
tension(\gamma)=\sup\{\delta>0|\text{there is an homotopy from }\gamma \ \text{to } \overline{\gamma}, \text{s.t. } length(\overline{\gamma})=length(\gamma)-\delta\}.
\end{equation}
Now the proof works as follows:

Suppose $R<\frac{1}{2}Sys(V,v)$, then $B(v,R)$ is contractible. Thus for every connected component of its boundary $S_i\subset \partial B(v,R)$, there is a surface $D_i$ homeomorphic to a disk, s.t. $\partial D_i=S_i$. Denote by $B^+(v,R)$ the union of $B(v,r)$ with all the $D_i's$.

Let $\gamma$ be either a noncontractible loop or an infinite path through $v$. Thus $\gamma$ intersects $\partial B^+(v,R)$ at least two times. Choose the first $x_1$ and the last $x_2$ of this interesections when walking along $\gamma$. This divides $\partial B^+(v,R)$ in two parts $L_1,L_2$ with $l_1=length(L_1)\leq length(L_2)=l_2$. Now $\gamma$ is homotopic to the loop $\gamma'$ obtained by the part of $\gamma$ between $x_1$ and $x_2$ union $L_1$. Thus in particular
\begin{equation}
tension(\gamma)\geq 2R-l_1,
\end{equation}
Since $\gamma'$ is at least $2R-l_1$ shorter then $\gamma$. Thus we get
\begin{equation}
tension(\gamma)\geq 2R-\frac{1}{2}length(\partial B^+(v,R)),
\end{equation}
which is equivalent to
\begin{equation}
length(\partial B^+(v,R))\geq 4R-2tension(\gamma).
\end{equation}
In the next line it says that we can thus say
\begin{equation}
length(\partial B^+(v,R))\geq 4R-2h(v).
\end{equation}
I don't see how this follows. Isn't by definition $h(v)\leq tension(\gamma)$?

Thank you for any hint!

答:
Ok, so in most cases the inequality

$$\mathrm{tens}(\gamma) ≥ 2 R - l_1$$
is going to be rather worthless, since the term $2R-l_1$ might as well be negative. You have correctly made this inequality even weaker:
$$\mathrm{len}(\partial B^+(v,R)) ≥ 4 R - 2\mathrm{tens}(\gamma)$$
The next step is not to bound $\mathrm{tens}(\gamma)≤h(v)$, this is certainly true but won't help you. Rather you want to make this weak inequality as strong as possible by looking only at those $\gamma$ where it is helpful. So you take the $\sup$ over all admissible $\gamma$:
$$\mathrm{len(\partial B^+(v,R)) = \sup_{\gamma}}\ \mathrm{len}(\partial B^+(v,R))≥ \sup_{\gamma} [4R - 2\mathrm{tens}(\gamma)] = 4R - 2\inf_\gamma\ \mathrm{tens}(v) = 4R-2h(v)$$


问:
Sum of binomial probabilities
One of my friends is building a game where the player will get questions from 6 different categories. Each category has a total of 50 questions. A single game consists of answering one question from each category. Questions are selected randomly and independent of previous games. In each game the drawings from each category are also independent.

From observations we have concluded that
1) either there's a high probability for a player to get the same questions multiple times when e.g. 20 games have been played.
2) or our random function is not random at all as people do get the same questions regularly.

I have a basic understanding of probability theory and have concluded that we are dealing with multiple binomial distributions (the categories) where the probability for a particular event to occur is p=0.02

If we were dealing with only one category there are many online calculators to put in relevant values for N, k and p http://www.vassarstats.net/textbook/ch5apx.html.

I'm just not sure how to deal with the multiple categories we have.

Does the rule for sums of binomials apply?
http://en.wikipedia.org/wiki/Binomial_distribution#Sums_of_binomials

答:
This is a variant of the Birthday problem, played six ways simultaneously.  Playing the game $n$ times, the probability of no repetition is $$\left(\frac{50}{50}\frac{49}{50}\frac{48}{50}\cdots\frac{51-n}{50}\right)^6$$

While the probability of a repetition is $1$  minus that value.

With even $n=4$, the probability of no repetition is about 47%.  With $n=3$, the probability of no repetition is about 69%.  If you want to avoid repetition, you'll need to make these choices non-randomly (i.e. remembering and avoiding previous selections).

Title: Sum of binomial probabilities


问:
Formula for Combinations With Replacement
I understand how combinations and permutations work (without replacement). I also see why a permutation of $n$ elements ordered $k$ at a time (with replacement) is equal to $n^{k}$. Through some browsing I've found that the number of combinations with replacement of $n$ items taken $k$ at a time can be expressed as $(\binom{n}{k})$ [this "double" set of parentheses is the notation developed by Richard Stanley to convey the idea of combinations with replacement].

Alternatively, $(\binom{n}{k})$ = $\binom{n+k-1}{k}$. This is more familiar notation. Unfortunately, I have not found a clear explanation as to why the above formula applies to the combinations with replacement. Could anyone be so kind to explain how this formula was developed?

答:
A useful way to think about it:

Imagine you have $n$ different cells form left to right. A combination without replacement of $k$ objects from $n$ objects would be equivalent to the number of ways in which these $k$ objects can be distributed among the cells with at most one object per cell. The key here is that due to the fact that there is no replacement there is only one or zero objects in each cell. It is easy to see that this corresponds to a combination without replacement because if we represent the occupied cells with a black circle and the empty cells with a white one there would be $k$ black circles in the row and $(n-k)$ white ones in the row, so the permutations of this row is precisley:

$$\frac{n!}{(n-k)!k!}=\binom{n}{k}$$

$$[][x][x][][x](...)[x][]= \bigcirc \bullet \bullet \bigcirc \bullet (...)\bullet \bigcirc $$

Note here that necessarily $k\leq n$.

Using the same analogy for combinations with replacement we have $k$ objects that we want to distribute into this $n$ cells but now we can put more than one object per cell (hence with replacement) also note that there is no bound on $k$ because if $k>n$ then we can just put more than one object in each cell. Now here comes the tricky part, we can count the permutations of this set by cleverly assigning circles. Let the division between the cells be a white circles and the objects black circles, then there would be $(n-1)$ white circles and $k$ black ones. It's not so hard to see that each permutation of these circles corresponds to a different way of putting each these $k$ objects into the $n$ cells. We have a total of $(n-1)+k$ circles, $(n-1)$ white and $k$ black, so the number of permutations of this row of circles is percisley:

$$\frac{(n-1+k)!}{(n-1)!k!}=\binom{n+k-1}{k}$$

$$[][xxxx][xx][][x][(...)][x][]=$$

$$\bigcirc \bullet \bullet \bullet \bullet \bigcirc \bullet \bullet \bigcirc \bigcirc \bullet \bigcirc(...)\bigcirc \bullet \bigcirc  $$

This is not really a rigorous way of proof but I think it illustrates the concept well. If you want a slightly more detailed explanation and exercises I recommend the book Introduction to Combinatorics published by the United Kingdom Mathematics Trust (UKMT) available at their webpage. It covers many interesting topics with a problem solving approach to them.


问:
how to find the number of solutions for $\ln(x) =x^2/8$ using simple math principles?
how to find the number of solutions for  $\ln(x) = x^2/8$ with high school level math (basic logarithm rules or something of that nature)( and without graphing $\ln(x)$ and $x^2/8$). (sorry for the poor formatting, feel free to edit)

I tried to change the equation to someting familiar to use a logarithm rule:

$$\displaylines{
x^2/8 = \ln(x) \cr
x/8^{0.5} = lan(x)^{0.5} \cr
x = 8^{0.5}*lan(x)^{0.5}
}$$

Changing the representation of the equation using the basic logarithm rule
$$\displaylines{
x^2/8 = \ln(x) \cr
x = e^{x^2/8}
}$$
But my tries got me nowhere. Any suggestions?

答:
Let $y:=x^2$ (as $x>0$, no solution is introduced). The equation can be written

$$y=e^{y/4}.$$

The exponential is a convex function, so it has at most two intersections with a straight line.

Now as $0<e^{0/4},4>e^{4/4}$ and $12<e^{12/4}$, we have these two intersections.