问:
Closed subset of an affine variety... is it affine?

Preliminaries

So, first of all let me give you the definitions I'm dealing with. Let $k$ be an algebraically closed field, and $\mathbb{A}^n = k^n$.

An affine variety is a closed and irreducible subset of $\mathbb{A}^n$.

Here we endow $\mathbb{A}^n$ with the Zariski topology, so a closed subset is the zero set of an ideal in the polynomial ring $k[x_1,\dots,x_n]$. The irreducibility condition in the definition of affine variety requires that such an ideal is prime.

A k-space is a topological space $X$ together with a sheaf of $k$-algebras $\mathscr{O}_X$. We require that $\mathscr{O}_X$ is a subsheaf of the sheaf of $k$-valued functions on $X$.

An algebraic variety is an irreducible k-space $X$ such that $\exists$ an open cover $X=\bigcup_{i=1}^n U_i$ where each $U_i$ is an affine variety. Moreover we ask the diagonal $\Delta(X)$ to be closed in $X\times X$.

Now we use the term affine variety also for an algebraic variety which is isomorphic as a $k$-space to an affine variety as defined above.

The question

Given a closed subset of an affine variety, is this subset affine?

Remarks

My professor claimed this to be true, and I noticed some people take this fact as granted on this website as well, so I guess this should be true! Nevertheless I don't see why such a closed subset should be irreducible. But the point is: do we really care? My naive understanding of this affine/not affine business is that for an affine variety we can write down something like a global coordinate system (the coordinate ring), while for a general algebraic variety we have just local coordinates, ie we have many affines patched together in a way that cannot be described globally with the usual tools. Hence if a variety is reducible this is not a real obstruction to being affine: we can still find a global coordinate system. The important thing is that all the components lie in the same $\mathbb{A}^n$

As you can see I'm a bit confused... some clarification about these ideas would be really appreciated!

答: It is not true that any closed subset of an affine variety will be an affine variety. In fact, you may have seen that any affine variety $V\subset \mathbb{A}^n$ has a unique reduction into irreducible components; that is, we can write $V = \cup_{i=1}^n V_i$ such that each $V_i$ is irreducible and $V_i \subsetneq V_j $ for $i\neq j$ and this is unique up to reordering of the $V_i.$ If $V$ has 3 irreducible components, then the union of the first two is a closed subset that is not irreducible (it is already reduced into two pieces!). </p>


问:
Directional derivative of Cartesian coordinate functions
I'm going through the book "A Visual Introduction to Differential Forms and Calculus on Manifolds by Jon Pierre Fortney" and on page 61 of the book there is the following exercise:

Explain that the rate of change of the Cartesian coordinate function $x$ in the $x$-direction is indeed one, thereby showing that $\frac{\partial x}{\partial x} = 1$. It may be helpful to consider how $x(p)$ changes as the point $p$ moves in the $x$-direction.

I'm struggling to understand what the exercise actually wants me to do.

答:
I suppose the author is talking about the projection of a point to its first coordinate (named $x$).

In that case, let's go back to the definition of the directional derivative at a point $p$ of a scalar function in $\mathbb{R}^n \to \mathbb{R}$:

$$\lim_{h \to 0} \frac{f(p + hv) - f(p)}{h}$$

Now, $p$ is an $n$D vector, $h$ is a scalar, and $v$ is an $n$D vector. This might seem a bit complex to wrap your head around at first, if you're used to your original derivative, so I'll try to get into the essentials of the "why" behind this formula and its use.

When you're in $\mathbb{R} \to \mathbb{R}$, there is no possible ambiguity as to the (unsigned) direction of your derivative, as your tangent space to your curve is 1D. However, as soon as you get to a function $\mathbb{R}^n \to \mathbb{R}$ with $n > 1$, the usual derivative definition is ambiguous, as there are an infinity of directions to pick from, on a 2D plane or nD space, around a given input point $p$. Precisely, if you represent a function $\mathbb{R} \to \mathbb{R}$ as a 1D curve in 2D space; a (derivable) function $\mathbb{R}^n \to \mathbb{R}^m$ can be represented as an $n$-dimensional manifold embedded in an $(n+m)$-dimensional space. This means your derivatives exist in a tangent space to an $n$-dimensional manifold: these tangent spaces at each point $p$ are $n$-dimensional euclidean spaces. This is why there are infinite directions for derivatives in cases where $n > 1$.

Most of differential geometry and multivariable calculus's insight into this problem is to say "well, on one hand, we'll just recycle our 1D derivative by picking single direction; on the other hand, we'll do like we usually do for vector spaces: we'll make a basis of possible directions, and do linear combinations of this basis's elements to express all possible combinations, and thus all possible directions". The elements of this basis are your partial derivative operators ($\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$, etc). That's the "explain it like I'm 5" of how we generalize the derivative to multiple dimensions. From that, we can construct more "interesting" extensions of the derivative (the gradient for functions $\mathbb{R}^n \to \mathbb{R}$ and its extension, the jacobian, for functions $\mathbb{R}^n \to \mathbb{R}^m$).

Back to your question. We pick a single direction for our derivative, the $x$ axis. Here, your vector $v$ in the original formula is something like $v = (1, 0, 0, \dots, 0)$ (of dimension $n$), and you function $f$ is the projection of your vector $p$ to its $x$ coordinate $p_x$ (what the author calls $x(p)$). This is like ignoring everything but the 2D plane with the $x$ axis as input, and mapping the value for $p_x$ as output, and taking the normal derivative for this function $f_x$ which restricts $f$ to this simpler space. It is obvious that this function corresponds to $id_x : x \to x$, the identity. The derivative of the identity is $1$.

Now, say we were to keep everything identical, but study the derivative in the $y$ direction. Ie, we have $v = (0, 1, 0, \dots, 0)$, to represent the idea of $\frac{\partial}{\partial y}$. The function $f$ is now restricted to a function $f_y : p \to p_x$. However, this value $p_x$ is a constant in the $y$ direction. So its partial derivative respective to $y$ is null.

ContentLicense: CC BY-SA 4.0


问:
Evaluate the integral using Residue integration method
$$\int_0^{\pi} \frac{2 d\theta}{k - cos\theta}$$

What is k here?

I'm not sure if I understand the question.

I am told from my book the answer is $\frac{2 \pi}{\sqrt{k^2 - 1}}$ but not sure how we got here.

Body:
Let $I := \int_{0}^{\pi} \frac{2\, d\theta}{k-\cos(\theta)}$.

Then
\begin{align*}
I &= \frac{1}{2} \int_{0}^{2\pi} \frac{4}{2\left (k-\frac{1}{2}e^{i\theta}-\frac{1}{2}e^{-i\theta} \right )}\, d\theta \\
&= -2i \int_{0}^{2\pi} \frac{1}{2ke^{i\theta}-e^{2i\theta}-1}\, (ie^{i\theta}\, d\theta) \\
&= -2i \int_{\{|z|=1\}} \frac{1}{-z^{2}+2k-1}\, dz \\
&= 4\pi \sum_{\sigma \in \{|z|<1 \}} \mathrm{Res}(\sigma).
\end{align*}

Now $z^{2}-2k+1$ has roots $k \pm \sqrt{k^2-1}$. The sole residue inside $\{|z| < 1 \}$ is $k - \sqrt{k^2-1}$ with residue $\frac{-1}{(k-\sqrt{k^2-1})-k-\sqrt{k^2-1}} = \frac{1}{2\sqrt{k^2-1}}$.

Your integral should be
$$I = 4\pi \cdot \frac{1}{2\sqrt{k^2-1}} = \frac{2\pi}{\sqrt{k^{2}-1}}.$$


问:
Trivial homology groups and fundamental group
I was wondering if there are noncontractible spaces with trivial homology groups and trivial fundamental group. I can't think of any, but I suspect some must exist.</p>

答:
Actually if a simply connected space $X$ has trivial homology then it must be weakly contractible (all homotopy groups vanish) by the Hurewicz theorem. If $X$ is a CW complex then it follows from Whitehead's theorem that $X$ is contractible. So a counterexample can't be homotopy equivalent to a CW complex; in other words it will only fail to be contractible for point-set reasons, not "purely homotopy-theoretic" reasons.

A more general version of this argument can be used to prove Whitehead's theorem for homology, which says that if a map $f : X \to Y$ between simply connected CW complexes induces an isomorphism on homology then it must be a homotopy equivalence (the simply connected hypothesis is crucial here).


问:
How to show that $a_n=\left(1+\frac{1}{n}\right)^n$ monotonic growing
I have to show that

$a_n=\left(1+\frac{1}{n}\right)^n$

is monotonic increasing. I struggle alot and tryng to find a right form to epxress that,

$a_n\le a_{n+1}$

a little hint what might help would be great. Im in my first year in the introduction to real analysis, so tipps with derivative or more advanced arent usefull :/

I really only like to get a little hint :)

答:
Let
$$ f(x)=(1+\frac1x)^x $$
and show $f(x)$ is increasing.


问:
Trapezoid problem
A trapezoid in which the lengths of the two sides that are not parallel are equal. The length of the longer parallel side is $5$ times the length of the shorter parallel side, and the distance between the two parallel side is $3$ times the length of the shorter parallel side. The perimeter of the trapezoid is $135$ meters. Approximately what is the length, in meters, of the shorter parallel side?

I'm stuck on this question, not sure what I missed.

I can let the shorter side equal to $x$ so the longer side is $5x$ and two equal side is $2y$, sum up
$$6x + 2y = 135$$
not sure what I can do from here to go further, and the area is obviously
$$\frac{6x \times 3x}{2} = 9x^2 $$
The answer is $10$, any suggestions to help me to approach would be highly appreciated.

答:
Can you show that $$y = \sqrt{13}x$$ and solve for $x?$

Draw a perpendicular to the longer side from one end of the shorter side and consider the right angled triangle formed. It has two shorter sides of length $3x$ and $2x$. Hence the length of the hypotenuse $y$ must be $\sqrt{(3x)^2 + (2x^2)} = \sqrt{13}x$

Title: Trapezoid problem


问:
Error minimization for machine learning
Let consider $(X,Y)$  a couple of random variables with values in $\mathbb R^p \times \{0,1\}$ and a distribution $$\mathbb P(Y=k) = \pi_k>0 \text{ and } \mathbb P(X \in \text{d}x|Y=k)=g_k(x)\text{d}x, k\in\{0,1\}, x\in \mathbb R^p,$$ where $\pi_0 + \pi_1 = 1$ and $g_0,g_1$ are two probability densities in $\mathbb R^p.$

We define the cliassifier $h_*(x) = \mathbb{1}_{\{\pi_1g_1(x)>\pi_0g_0(x)\}}, x \in \mathbb R^p.$

I need to prove that the classifier $h_*$ fullfills $$\mathbb P(h_*(X)\ne Y)=\min_h\mathbb P(h(X)\ne Y).$$

I started by determining the distribution of $X$.

By the law of total probabilities, I've established that the distribution of $X$ is $$\forall A \in R^p, \mathbb P(X\in A) = \int_A(\pi_0g_0(x) + \pi_1g_1(x))\text{d}x.$$

I tried to calculate $\mathbb P(h(X)\ne Y \backslash h_*(X)\ne Y)$ but wasn't able to conclude anything.

Any help would be apreciated.

答:
You have
$$
P(h^*(X)\neq Y)=P(h^*(X)=1|Y=0)P(Y=0)+P(h^*(X)=0|Y=1)P(Y=1)=
$$
$$
=\pi_0\int\limits_{\pi_1g_1>\pi_0g_0} g_0(x)\,dx+\pi_1\int\limits_{\pi_1g_1\le \pi_0g_0} g_1(x)\,dx
$$
$$
=\pi_0\int\limits_{\pi_1g_1>\pi_0g_0} g_0(x)\,dx+\pi_1(1-\int\limits_{\pi_1g_1> \pi_0g_0} g_1(x)\,dx)=\pi_1+\int\limits_{\pi_1g_1> \pi_0g_0} (\pi_0g_0(x)-\pi_1g_1(x))\,dx
$$
Now take any other predictor $h$. Call $R$ the subset where $h=1$. A similar computation yields
$$
P(h(X)\neq Y)=\pi_1+\int\limits_{R} (\pi_0g_0(x)-\pi_1g_1(x))\,dx
$$
Now observe that the last integral is minimized if $R=\{\pi_1g_1> \pi_0g_0\}$. On any other subset part of your integral will be positive.

Title: Error minimization for machine learning


问:
How to solve/obtain this Differential Eq?
How do you solve/obtain the Differential Eq. of $\cot(p)(q′+1)+2q=0$, $q(\pi/4)=9$?
I tried answering since I think this is Linear but i cant seem to get the solution.

答:
I'll assume you are trying to solve $\cot(p)(q'(p)+1) + 2q(p) = 0$.

Rewrite as $q' + 2\tan(p)q = -1$.

Now find integrating factor $\mu(p) = e^{\int^{p} 2\tan(p)\, dp} = \sec^{2}(p)$ and solve for

$q(p) = \mu(p)^{-1} \int^{p} -\mu(p)\, dp = \cos^{2}(p)(C-\tan(p)) = -\sin(p)\cos(p)+C\cos^{2}(p)$.

Check the initial value to find $C$.


答:
The union of two points $\{1\} \cup \{2\}$ is a closed subset of $\Bbb{A}^1$ in the Zariski topology but it is obviously not irreducible.

To answer your querrey about the link in your question above, notice that Georges only claims that

$$X_1 \times \{a_1\} \subseteq X_1 \times X_2$$

is closed. However the fact that $X_1 \times\{ a_1\}$ is irreducible comes from it not being a closed subset, but rather because
it is homeomorphic to $X_1$ that is irreducible. Since irreducibility is preserved under homeomorphism thus $X_1$ irreducible implies that the slice $X_1 \times \{a_1\}$ is irreducible and thus affine.

ContentLicense: CC BY-SA 3.0

Title: Closed subset of an affine variety... is it affine?

Body:
Preliminaries

So, first of all let me give you the definitions I'm dealing with. Let $k$ be an algebraically closed field, and $\mathbb{A}^n = k^n$.

An affine variety is a closed and irreducible subset of $\mathbb{A}^n$.

Here we endow $\mathbb{A}^n$ with the Zariski topology, so a closed subset is the zero set of an ideal in the polynomial ring $k[x_1,\dots,x_n]$. The irreducibility condition in the definition of affine variety requires that such an ideal is prime.

A k-space is a topological space $X$ together with a sheaf of $k$-algebras $\mathscr{O}_X$. We require that $\mathscr{O}_X$ is a subsheaf of the sheaf of $k$-valued functions on $X$.

An algebraic variety is an irreducible k-space $X$ such that $\exists$ an open cover $X=\bigcup_{i=1}^n U_i$ where each $U_i$ is an affine variety. Moreover we ask the diagonal $\Delta(X)$ to be closed in $X\times X$.

Now we use the term affine variety also for an algebraic variety which is isomorphic as a $k$-space to an affine variety as defined above.

The question

Given a closed subset of an affine variety, is this subset affine?

Remarks

My professor claimed this to be true, and I noticed some people take this fact as granted on this website as well, so I guess this should be true! Nevertheless I don't see why such a closed subset should be irreducible. But the point is: do we really care? My naive understanding of this affine/not affine business is that for an affine variety we can write down something like a global coordinate system (the coordinate ring), while for a general algebraic variety we have just local coordinates, ie we have many affines patched together in a way that cannot be described globally with the usual tools. Hence if a variety is reducible this is not a real obstruction to being affine: we can still find a global coordinate system. The important thing is that all the components lie in the same $\mathbb{A}^n$

As you can see I'm a bit confused... some clarification about these ideas would be really appreciated!


问:
Show that the mapping $(Tf)(x) = \frac{1}{x}\int_0^{\infty}K(\frac{y}{x})f(y)dy$ is bounded
Let $K$ be a Lebesgue measurable function defined on $(0, \infty)$ and assume $\int_0^{\infty}\vert\,K(y)\,\vert y^{-\frac{1}{p}}dy < \infty$. Pick $f \in L_p(0, \infty) (p > 1)$ and define a linear mapping $T$ by:
$$(Tf)(x) = \int_0^{\infty}K(y)f(x y) dy = \frac{1}{x}\int_0^{\infty}K(\frac{y}{x})f(y)dy, x > 0$$

My questions are:

$\forall\,f \in L_p(0, \infty), Tf \in L_p(0, \infty)$ (solved)

$\|\,T\| \leq \int_0^{\infty}\vert\,K(y)\,\vert y^{-\frac{1}{p}}dy$ (solved) and the equality holds when $K$ is non-negative (not solved)

To prove the first two, by using Minkowski's integral inequality and replace $y$ by $t x$, I tried:
$$[\int_0^{\infty}\vert\,\int_0^{\infty}\frac{1}{x}K(t)f(t x)d t\,\vert^p dx]^{\frac{1}{p}} \leq \int_0^{\infty}[\int_0^{\infty}\vert\,\frac{1}{x}K(t)f(t x)\,\vert^p dx]^{\frac{1}{p}} dt = \\\int_0^{\infty}\vert\,K(t)\,\vert[\int_0^{\infty}\vert\,\frac{t}{x}f(x)\,\vert^p\frac{1}{t}dx]^{\frac{1}{p}} d t = \int_0^{\infty}\vert\,K(t)\,\vert t^{-\frac{1}{p}}\underline{[\int_0^{\infty}\vert\,\frac{t}{x}f(x)\,\vert^p dx]^{\frac{1}{p}}} d t$$

Then I can not compare the underlined part and $\|f\|_p$. If my attempt is on the right track, could you anyone provide hints on how to do the desired comparison? Also I was block by the second part of question 2 and the third question. I am not sure if I can apply Minkowski's inequality when $K$ is not absolutely integrable. Here is a similar question but I could not find it helpful ...

答:
It seems to me that the assumption on $K$ should be that $$\int^\infty_0 |K(y)|\,y^{-1/p}\,dy<\infty$$
As you pointed out, the generalized Minkowski inequality gives
\begin{align}
\left(\int^\infty_0\left|\int^\infty_0 K(y) f(xy)\,dy\right|^p\,dx\right)^{1/p} &\leq \left(\int^\infty_0\left(\int^\infty_0 |K(y)| |f(xy)|\,dy\right)^p\,dx\right)^{1/p}\\
&\leq\int^\infty_0\left(\int^\infty_0 |f(xy)|^p\,dx\right)^{1/p}|K(y)|\,dy\\
&=\|f\|_p\int^\infty_0y^{-1/p}|K(y)|\,dy
\end{align}

This is the version of the general Minkowski inequality that I am using:

Suppose $f:(X\times Y,\mathscr{F}\otimes\mathscr{G})\rightarrow(\mathbb{R},\mathscr{B}(\mathbb{R}))$ is measureable and that where $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{G},\nu)$ are $\sigma$--finite measures. Then,
\begin{align}
\Big(\int_X\Big|\int_Y f(x,y)\,
\nu(dy)\Big|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\leq
 \int_Y \Big(\int_X |f(x,y)|^p\,\mu(dx)\Big)^{\tfrac{1}{p}}\,\nu(dy)
\end{align}
for all $1\leq p<\infty$.

Comments:

Your problem is a particular case of the problem you quoted. Indeed, notice that
$$k(x,y):=\frac{1}{y}K(x/y)$$
is homogeneous of order $-1$, that
$$\int^\infty_0k(x,1) x^{-1/p}\,dx = \int^\infty_0K(x) x^{-1/p}\,dx,$$ and that
$$ Tf(y)=\int^\infty_0k(x,y)f(x)\,dx$$

Thus far we have proved that
$$\|T\|\leq\int^\infty_0|K(x)|x^{-1/p}\,dx$$
regardless of the sign of $K$.

The equality is not straight forward, even in the case $K\geq0$. Consider the particular case
$$ K(x)=\mathbb{1}_{(0,1]}(x)$$
which gives the Hardy operator
$$Hf(y)=\frac{1}{y}\int^y_0f(x)\,dx$$
Proving that $\|H\|=\frac{p}{p-1}=\int^1_0x^{-1/p}\,dx$ is not trivial. Here is a solution for this particular case


Title: How to show that $a_n=\left(1+\frac{1}{n}\right)^n$ monotonic growing

Body:
I have to show that

$a_n=\left(1+\frac{1}{n}\right)^n$

is monotonic increasing. I struggle alot and tryng to find a right form to epxress that,

$a_n\le a_{n+1}$

a little hint what might help would be great. Im in my first year in the introduction to real analysis, so tipps with derivative or more advanced arent usefull :/

I really only like to get a little hint :)

Body:
$$\frac{a_{n+1}}{a_n}=\frac{\left( 1+ \frac{1}{n+1} \right)^{n+1}}{\left( 1+ \frac{1}{n} \right)^{n}}=\frac{n+1}{n}\left( 1- \frac{1}{(n+1)^2} \right)^{n+1}>\\
>\frac{n+1}{n}\left( 1- \frac{1}{(n+1)} \right)=1$$


问:
Whether Pontryagin's maximum principle valid or not?
Let consider this system of the equation like $\dot{x}=p(1-e^{2x})$ where $x$ is position and $p$ is momentum. We want to minimize this payoff function $\int_{0}^{t}2a_1^2$ where $a_1\in[-\alpha,\alpha]$.

Now can we compute $H=f.w+r=p(1-e^{2x}).w+2a_1^2$ ? Note that $w$ is costate function and can we use Maximum principle to obtain time-optimal control?

答:
If I understood correctly, $a_1$ is inconsequential on the dynamics of the system, therefore it will bind to zero along the trajectory in order to minimize the integral.

It seems also that the system is uncontrollable, which means that the trajectory is fully defined given the initial state.

The only decision variable left is the final time which is also inconsequential because $a_1$ will bind to zero. In case $a_1$ is forced to be always positive, for instance, both $a_1$ and $t_f$ will bind to their lower boundaries in order to minimize the integral.

Hope this helps.


问:
What is the general equation of a cubic polynomial?
I had this question:

"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.

I took the general equation of a cubic equation, which was: $x^3 +bx^2/a + cx/a + d/a$.

Through simultaneous equations, I found what $b/a, c/a, d/a$ should equate to for my unknown cubic polynomial. Am I supposed to make $b/a, c/a, d/a$ all positive, then substitute it into the general formula?

Any help would be greatly appreciated, thanks.

Body:
Let $a,b,c$ be roots of $x^3 + 2x + 1 = 0$

Using Vieta's Formulas, $a+b+c=0,ab+bc+ca=2,abc=-1$

So, we need to find the cubic equation whose roots are $a^2,b^2,c^2$

i.e., $$(y-a^2)(y-b^2)(y-c^2)=0$$
$$\iff y^3-(a^2+b^2+c^2)y^2+(a^2b^2+b^2c^2+c^2a^2)y+a^2b^2c^2=0$$

Now, $a^2b^2c^2=(abc)^2=(-1)^2=1$

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0-2(2)=-4$

$a^2b^2+b^2c^2+c^2a^2=(ab)^2+(bc)^2+(ca)^2$
$=(ab+bc+ca)^2-2(ab\cdot bc+bc\cdot ca+ca\cdot ab)$
$=(ab+bc+ca)^2-2(a+b+c)abc=2^2-2\cdot0\cdot(-1)=4$


问:
How to solve this line integral?
I have this line integral $\int_{C}(x^2+y^2)ds$, where $C=\{(x,y)|x^2+y^2=ay\}$. I tried solving it by substituting the variables as $x=r\cos{t}$ and $y=r\sin{t}$. So when I differentiate, I get $dx=-r\sin{t}$ and $dy=r\cos{t}$.

So popping this into the integral we get:

$\int_{C}((\cos{t})^2+(\sin{t})^2)*\sqrt{(-r\sin{t})^2+(r\cos{t})^2}dt$

I could probably integrate this if I knew how to apply the given circle.

答:
Your value for $\mathrm{d}s$ looks off.  Here is a trigonometric approach.

$$
\begin{align}
\oint\left(x^2+y^2\right)\mathrm{d}s
&=\int_0^{2\pi}\left(\frac{a^2}2-\frac{a^2}2\cos(\theta)\right)\,\frac a2\,\mathrm{d}\theta\\
&=\frac{a^3}{4}\,2\pi\\[3pt]
&=\frac{\pi a^3}2
\end{align}
$$

Title: How to solve this line integral?


问:
$\begin{align}&(x\! -\! a,f_1(x),\ldots,f_r(x))\\ =\ &(x\! -\! a,f_1(a),\ldots,f_r(a))\end{align}\ $ [Ideal evaluation = mod reduction]
$\begin{align}&(x\! -\! a,f_1(x),\ldots,f_r(x))\\[.1em] =\ &(x\! -\! a,f_1(a),\ldots,f_r(a))\end{align}\ $  [Ideal evaluation = mod reduction]

I am solving Aluffi chapter 0. I am completely stuck on this question.

Let R be a commutative ring, $a \in R$, and $f_1(x),\ldots, f_r(x) \in R[x]$.

Prove the equality of ideals

$(f_1(x),\ldots,f_r(x),x - a) = (f_1(a),\ldots,f_r(a),x - a)$

Prove the useful substitution trick
$\frac{R[x]}{(f_1(x),\ldots,f_r(x), x - a)} \cong \frac{R}{(f_1(a),\ldots,f_r(a))}$

答:
Hint: as in Euclidean algorithm for gcds, ideals are preserved by: mod out a generator by another

$$x\!-\!a,\, f \in I \iff x\!-\!a,\!\!\!\!\! \overbrace{f\,-\, q\, (x-a)}^{\textstyle f(a) = f\bmod x\!-a}\!\!\!\!\!\in I $$

where we used Polynomial Remainder Theorem, i.e. $\,f(a) = f\bmod x\!-\!a\,$ .

Remark $ $ More generally ideals are prserved by any unimodular transformations of generators ("change of basis")


问:
2 jars with 50 balls each. Pick any ball with $>50\%$ probability.
You are blindfolded and placed in front a table with two jars. One jar has $50$ red balls and other has $50$ blue balls.

What should be your strategy so that you pick up the red ball with more than $50\%$ probability.

答:
take 49 red balls from one jar and put them in the other jar wit the blue balls,

now its 100 % he will pick red from the first jar, and 50% red from the mixed jar.


问:
Let $X\sim \text{Uniform}(0, \theta)$. What is the distribution of $Y= \theta - X$?
So cdf of $X$ is $F(X\leq x) = \frac{x}{\theta}$.

Let $X = \theta - Y$, subbing it into the cdf we now get:
$$F(\theta - Y \leq x) = \frac{x}{\theta}$$
$$F(-Y \leq x - \theta) = \frac{x}{\theta}$$
$$F(Y\geq \theta - x) = \frac{x}{\theta}$$
$$F(Y\leq \theta - x) = 1 - \frac{x}{\theta}$$
$$=1 - F(X\leq x)$$

This what I came up with but no idea if this is right.

答:
The CDF of $X$ for different cases is really

$$\begin{align*}
\Pr(X \le x) = \begin{cases}
0 & x \le 0\\
\frac{x}{\theta}& 0 < x\le \theta\\
1& \theta < x
\end{cases}
\end{align*}$$

Let $X=\theta - Y$, then the CDF of $Y$ is

$$\begin{align*}
\Pr(Y \le y) &= \Pr(\theta - X \le y)\\
&= \Pr(X \ge \theta - y)\\
&= 1 - \Pr(X < \theta - y)\\
&= 1 -\begin{cases}
0 & \theta - y \le 0\\
\frac{\theta - y}{\theta}& 0 < \theta - y\le \theta\\
1& \theta < \theta - y
\end{cases}\\
&= \begin{cases}
0 & y < 0\\
\frac{y}{\theta}& 0 \le y < \theta\\
1& \theta \le y
\end{cases}\\
\end{align*}$$

Differentiate $\Pr(Y\le y)$ with respect to $y$ to obtain its PDF.


问:
When is $T$-Alg monoidal closed?
Given a category $\mathcal{V}$ and a monad $(T,\eta,\mu)$, what would be the sufficient conditions on $\mathcal{V}$ and $T$, for the category of $T$ algebras to be monoidal closed? </p>

<p>(I'm pretty sure that Kock proved that, if $T$ has strength and is commutative, then $T$-alg is closed; can we relax that condition, or change it in anyway?)</p>

答: You need that $\mathcal{V}$ is closed monoidal, $T$ is a monoidal monad and that certain coequalizers in $\mathcal{V}$ exist which commute with $T$. For details, see Tensors, monads and actions by Gavin Seal.</p>

Title: When is $T$-Alg monoidal closed?


问:
Hatcher 1.3.27: Clarification on left and right actions
This question has been asked here, and I am making a new question seeking clarification after a comment reply from Eric Wofsey.

I understand how both actions of $\pi_1(X,x_0)$ are defined on the fibre $p^{-1}(x_0)$, however I don't understand how the 'first action' (the action defined by lifting loops in $X$ at $x_0$ to points in the fibre) becomes a right action. As far as I can tell, it is a left action.

If I could get something more concrete for why this is true that would be an ideal answer.

答:
This is explained in Hatcher's book on p. 69. The elements of $\pi_1(X,x_0)$ are path homotopy classes of closed paths $\gamma : [0,1] \to X$ based at $x_0$ (i.e. $\gamma(0) = \gamma(1) = x_0$). Given $y \in  p^{-1}(x_0)$ and a path $\gamma$ based at $x_0$, we find a unique lift $\tilde \gamma$ of $\gamma$ such that $\tilde \gamma (0) = y$. Define $a(y, \gamma)  = \tilde \gamma (1)$. It is easy to verify that $\tilde \gamma (0)$ only depends on $[\gamma] \in \pi_1(X,x_0)$. That is, $a(y, [\gamma]) =  a(y, \gamma)$ is well-defined. The product of $[\gamma], [\delta] \in \pi_1(X,x_0)$ is given by $[\gamma] \cdot [\delta] = [\gamma  \cdot \delta]$, where the path $\gamma  \cdot \delta$ is defined by $(\gamma  \cdot \delta)(t) = \gamma (2t)$ for $t \le 1/2$ and $(\gamma  \cdot \delta)(t) = \delta (2t-1)$ for $t \ge 1/2$ (that is, we first travel with $\gamma$ and then with $\delta$). This shows that
$$a(y, [\gamma] \cdot [\delta]) = a(a(y, [\gamma]), [\delta]) .$$
In fact, we first lift $\gamma$ to $\tilde \gamma$ with $\tilde \gamma (0) = y$ to get $a(y, [\gamma]) = \tilde \gamma (1)$. Next we lift $\delta$ to $\tilde \delta$ with $\tilde \delta (0) = a(y, [\gamma]) = \tilde \gamma (1)$ to get $a(a(y, [\gamma]),[\delta]) = \tilde \delta (1)$. But the composed path $\tilde \gamma \cdot \tilde \delta$ is a lift of $\gamma \cdot \delta$ such that $(\tilde \gamma \cdot \tilde \delta)(0) = \tilde \gamma(0) = y$. Hence $a(y, [\gamma] \cdot [\delta]) = (\tilde \gamma \cdot \tilde \delta)(1) = \tilde \delta(1) = a(a(y, [\gamma]),[\delta])$. This shows that $a$ is a right action. See for example right group action. Writing suggestively $y  \cdot [\gamma] =  a(y, [\gamma])$ we get
$$y  \cdot ([\gamma] \cdot [\delta]) = (y  \cdot  [\gamma]) \cdot [\delta] .$$
This explains the name right action: The above "associative property" works via multiplication on the right side of $y$.

But isn't writing $y  \cdot [\gamma]$ quite arbitrary, coudn't we write $[\gamma] \cdot y$? Yes, we can, but then we get
$$([\gamma] \cdot [\delta]) \cdot y = [\delta] \cdot ([\gamma] \cdot y) $$
which doesn't look like a usual associative property.

This is the reason why Hatcher defines a left action by
$$[\gamma] * y = y \cdot [\gamma]^{-1} .$$
In fact
$$([\gamma] \cdot [\delta]) * y = y \cdot ([\gamma] \cdot [\delta])^{-1} = y \cdot ([\delta]^{-1} \cdot [\gamma]^{-1}) = (y \cdot ([\delta]^{-1}) \cdot [\gamma]^{-1} = ([\delta] * y ) \cdot [\gamma]^{-1} \\= [\gamma] * ([\delta] * y) .$$


问:
Show that the sequence $(a_n)_{n\ge 0}$ defined recurrence relation $a_0=25, a_{n+1}=2\sqrt{a_n}$ converges.
Show that the sequence $(a_n)_{n\ge 0}$ defined recurrence relation $a_0=25, a_{n+1}=2\sqrt{a_n}$ converges. I showed that it is bounded from below by $1$, but I'm struggling to show that the sequence is non-increasing. I tried induction, but the induction hypothesis is not helpful at all.

答:
Well, $a_1 = 10 < 25 = a_0$ so here we have our basic step.

For the inductive step, suppose $a_{n+1} < a_n$, we want $a_{n+2} < a_{n+1}$.

$a_{n+2} \stackrel{?}{<} a_{n+1}$ is equivalent to $2\sqrt{a_{n+1}} \stackrel{?}{<} 2\sqrt{a_n}$.

Can you finish now?


问:
Question about infinite integral convergence logic
For a question, I was given the continuous function f, who's lim as x goes to infinity = 2, and I have to determine if it's possible for the infinite integral of f(x) to be convergent. Would someone be able to help me with this? Not really sure how to prove it.</p>

答:
The improper integral will be divergent. Here's a proof of this fact:

Let's assume that $f$ is defined on $[a,\infty)$. If $\lim_{x\to\infty}f(x)=2$, then for any given $\varepsilon >0$, there will exist a $\delta$ such that $|f(x)-2|<\varepsilon$ for all $x>\delta$. As a special case, there will exist a $\delta_1$ such that $|f(x)-2|<1$ for all $x>\delta_1$.

The inequality $|f(x)-2|<1$ is equivalent to $-1<f(x)-2<1$, which in turn is equivalent to $1<f(x)<3$. It follows from basic integral properties and the fact that $f$ and $1$ are continuous and strictly positive for all real $t>\delta$ that

$$\int_{\delta}^{t}1\text{ }dx<\int_{\delta}^{t}f(x)\text{ }dx$$

It is clear that $\lim_{t\to\infty}\int_{\delta}^{t}1\text{ }dx=\infty$, so the inequality $\int_{\delta}^{t}1\text{ }dx<\int_{\delta}^{t}f(x)\text{ }dx$ implies that it must be true that $\lim_{t\to\infty}\int_{\delta}^{t}f(x)\text{ }dx=\infty$ (this argument can be generalized to prove part of the Comparison Test for Integrals). This means that $\int_{\delta}^{\infty}f(x)\text{ }dx$ is divergent. Since adding a finite, ordinary integral won't change this, we conclude that the whole improper integral $\int_{a}^{\infty}f(x)\text{ }dx$ will also be divergent.


问:
Sequence both converging and diverging?
I have the sequence sum 1 to infinity of 10^n/n!

If I plug it into symbollab, it says it converges by the ratio test.
If I plug it into wolfram alpha, it says it diverges by geometric series test.

Which is it ?!

答:
Based on the ratio test, it converges. Indeed,

\begin{align*}
\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| = \limsup_{n\to\infty}\frac{10^{n+1}}{(n+1)!}\times\frac{n!}{10^{n}} = \limsup_{n\to\infty}\frac{10}{n+1} = 0 < 1
\end{align*}

Which entry did you put at the WA website?

Title: Sequence both converging and diverging?


问:
Volumes of solids of revolution around x- and y-axis
What is the volume of the solid that results when the region under $y = 1 + x^3$ between x = 1 and x = 2 is revolved about the x- and y-axis?

For the x-axis:
$$ \int_1^2 \pi (1+x^3)^2 \,dx = \frac{373\pi}{14} $$

For the y-axis, I solved the equation in terms of y and I computed for the y-coordinates, so the volume is:
$$ \int_2^9 \pi [(y-1)^{\frac13}]^2 \,dx = \frac{93\pi}{5} $$

I only got 5/10 points in this problem, but I do not know which one is correct or wrong. Can you check these two answers?

答:
It's your second one that is wrong: you need to integrate the region under $y=1+x^3$ so you'd be best served using the shells formula in that case.  But as you mentioned in the comments you can't use that formula.  That's okay, but now you'd need two integrals to get the answer.  First do the lower portion which is just the region $1 \leq x \leq 2$ and $0 \leq y \leq 2$:
$$
\int_0^2 \pi(2^2-1^2)dy = 6\pi.
$$
Your second region is between $y=1+x^3, y=2$, and $x=2$ so use washer formula:
$$
\int_2^9 (2^2 - (y-1)^{2/3})dy = \frac{47\pi}{5}.
$$

That then adds up to the same answer you'd get from shells (much easier): $77\pi/5$.


问:
Determining the Jordan decomposition of a given matrix
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
\newcommand{\l}{\lambda}
$
I have the matrix given as follows:

$$A := \m{ -2 & -1 & 1 & 2 \\ 1 & -4 & 1 & 2 \\ 0 & 0 & -5 & 4 \\ 0 & 0 & -1 & -1 }$$

I wish to determine the Jordan decomposition for $A$. So far I have determined the following:

$A$ has characteristic polynomial $p(\l) = -(\l+3)^4$. (Wolfram agrees up to that leading coefficient which isn't relevant.) Thus, it has a single eigenvalue, $\l = -3$, with multiplicity $4$.

The matrix $A+3I$ has nullity $2$. Thus the algebraic and geometric multiplicities of $3$ differ, making it not diagonalizable in the traditional sense. (Wolfram agrees.)

I have found that there exist two ordinary eigenvectors for $A$, i.e. vectors $\vec p$ such that $(A+3I)\vec p = \vec 0$. Namely, the following:
$$v = (-4,0,2,1) \qquad w = (1,1,0,0)$$
(Wolfram agrees  with this being the basis of the null space.)

I still need two generalized eigenvectors to fill my roster. So let's start with squaring $A+3I$. In such a case, there is a single vector, by my eyes, which fits. Using the alternate characterization of generalizing a specific eigenvector, e.g. finding $v'$ such that $(A+3I)v' = v$ (or $w$), we can see that $v$ won't work in this way, but if we do it for $w$ we can obtain
$$w' = (1,0,0,0)$$
Wolfram does list this in the basis of the null space of $(A+3I)^2$. There isn't an issue picking $w'$ this way so far as far as I know: Wolfram agrees that the three are linearly independent. Thus, this is the only additional eigenvector we get from $(A+3I)^2$.

To find a fourth (generalized) eigenvector, $w''$, we will note that $(A+3I)^3$ is the zero matrix and thus has null space of $\Bbb R^4$. (Again, Wolfram agrees.) Thus, we can pick any vector we please that is linearly independent from the rest. As it happens, $w'' = (0,0,1,1)$ is such a vector. (Again, Wolfram agrees.)

From the structure of this problem, we see two Jordan chains:
$$v \qquad w \to w' \to w''$$
Thus, the Jordan form $J$ of $A$ should have one $1 \times 1$ block and one $3 \times 3$ block, each associated with $\l = -3$. Thus, we will have
$$J =
\m{ -3 & 0 & 0 & 0\\
0 & -3 & 1 & 0 \\
0 & 0 & -3 & 1 \\
0 & 0 & 0 & -3 }$$

Thus, we have our four eigenvectors to form the matrix $P$ such that $A = PJP^{-1}$.
$$P := \big[ \; v \; \big| \; w \; \big| \; w' \; \big| \; w'' \; \big] = \m{
-4 & 1 & 1 & 0 \\
0 & 1 & 0 & 0 \\
2 & 0 & 0 & 1 \\
1 & 0 & 0 & 1}$$

This is where I get confused. Performing the calculation of $PJP^{-1}$ does not give me quite $A$ back. Per Wolfram,

$$PJP^{-1} = \m{
-2 & -1 & 3 & -2 \\
1 & -4 & 4 & -4 \\
0 & 0 & -3 & 0 \\
0 & 0 & 0 & -3}$$

yet I'm not entirely sure where I've gone wrong. I've been trying to resolve this discrepancy for days with no luck, no matter what I look up. In fact, Wolfram gives a very different decomposition: namely, they claim

$$P = \m{1/4 & 8 & 2 & 0\\
5/4 & 8 & 2 & 0\\
1/2 & 0 & 4 & 0\\
1/4 & 0 & 2 & 1}$$

but at least we agree on $J$...

Can anyone enlighten me as to what I might have done wrong?

答:
There are two directions frequently taught. One way is to write down the genuine eigenvector for each block, then trickle down to the consistent generalized. That is reliable, but tends to introduce lots of fractions when the matrix elements and eigenvalues are integers.

I like going from the ground up; the fractions wait until $P^{-1},$  where $J = P^{-1} A P.$  Here $(A + 3 I)^3 = 0$ but $(A+3I)^2 \neq 0.$  We need three column vectors $p,q,r$  such that $(A+3I)^3 r = 0$ but  $(A+3I)^2 r \neq 0.$  I took $r = (0,0,0,1)^T .$  Then $q = (A+3I) r = (2,2,4,2)^T $  and $p = (A+3I) q = (8,8,0,0)^T .$ Thus $p$ is a genuine eigenvector

The first column of $P$ is another genuine eigenvector


问:
Find rank for a given matrix
Let $v\in\mathbb{R}$ and let

$V=\begin{pmatrix}v&1&1&1\\1&v&1&1\\1&1&v&1\\1&1&1&v\end{pmatrix}$

now by using the row-echelon method, we switch row 1 for row 2

\begin{pmatrix}1&1&1&v\\1&v&1&1\\1&1&v&1\\v&1&1&1\end{pmatrix}

replace row $2$ by (row $2-$row $1$), replace row $3$ by (row $3-$row $1$), and replace row $4$ by (row $4-a$ row $1$)

\begin{pmatrix}1&1&1&v\\0&v-1&0&1-v\\0&0&v-1&1-v\\0&1-v&1-v&1-v^2\end{pmatrix}

replace row $4$ by (row $4 +$row $3+$row$2$)
\begin{pmatrix}1&1&1&v\\0&v-1&0&1-v\\0&0&v-1&1-v\\0&0&0&-v^2-2v+3\end{pmatrix}

Now, if rank $V\not=4$ then there should be at least one zero row, so we let

$-v^2-2v+3=0$

$(v+3)(v-1)=0$

$v=-3,\space v=1$

Thus, rank $V\not=4$ when $v\in\mathbb{R}$\ {1,-3}

Is my value for $v$ wrong or am I missing any values?

答:
You are ultimately correct here: there are indeed exactly two values of $a$ for which the rank of this matrix is less than $4$, and they are $1$ and $-3$. Your method is mostly sound, but you do leave one small hole here. When you get to
$$\begin{pmatrix}1&1&1&a\\0&a-1&0&1-a\\0&0&a-1&1-a\\0&0&0&-a^2-2a+3\end{pmatrix},$$
and conclude that there must be a $0$ row, you are implicitly assuming that the above matrix is already in row-echelon form. This is indeed the case, provided that the leading entries in the second and third rows are non-zero (i.e. $a \neq 1$). If they were $0$, then the second and third rows would be $0$, meaning that the rank of the matrix is certainly less than $4$.

Fortunately, $a = 1$ is also a root of $-a^2 - 2a + 3$, so no solutions were lost in the end. But, this is still a logical flaw in your argument. What you should do is consider separately the case where $a = 1$ and conclude that the matrix has rank less than $4$. Then, consider the case $a \neq 1$, say that the above matrix has row-echelon form, and conclude that the last row must be $0$, giving the other solution $a = -3$.

Title: Find rank for a given matrix


答:
A square matrix has full rank iff it is invertible. Thus we can approach such problem by calculating the determinant of $A$. More precisely, one has

\begin{align*}
\begin{vmatrix}
a & 1 & 1 & 1\\
1 & a & 1 & 1\\
1 & 1 & a & 1\\
1 & 1 & 1 & a
\end{vmatrix} =
\begin{vmatrix}
a + 3 & 1 & 1 & 1\\
a + 3 & a & 1 & 1\\
a + 3 & 1 & a & 1\\
a + 3 & 1 & 1 & a
\end{vmatrix}
& =
(a+3)
\begin{vmatrix}
1 & 1 & 1 & 1\\
1 & a & 1 & 1\\
1 & 1 & a & 1\\
1 & 1 & 1 & a
\end{vmatrix}\\\\
&
=
(a+3)
\begin{vmatrix}
1 & 0 & 0 & 0\\
1 & a-1 & 0 & 0\\
1 & 0 & a - 1 & 0\\
1 & 0 & 0 & a - 1
\end{vmatrix} = (a+3)(a-1)^{3}
\end{align*}

Since we are interested in the cases where $\text{rank}(A)\neq 4$, it suffices to set $\det(A) = (a+3)(a-1)^{3} = 0$.

Title: Find rank for a given matrix

Body:
Let $v\in\mathbb{R}$ and let

$V=\begin{pmatrix}v&1&1&1\\1&v&1&1\\1&1&v&1\\1&1&1&v\end{pmatrix}$

now by using the row-echelon method, we switch row 1 for row 2

\begin{pmatrix}1&1&1&v\\1&v&1&1\\1&1&v&1\\v&1&1&1\end{pmatrix}

replace row $2$ by (row $2-$row $1$), replace row $3$ by (row $3-$row $1$), and replace row $4$ by (row $4-a$ row $1$)

\begin{pmatrix}1&1&1&v\\0&v-1&0&1-v\\0&0&v-1&1-v\\0&1-v&1-v&1-v^2\end{pmatrix}

replace row $4$ by (row $4 +$row $3+$row$2$)
\begin{pmatrix}1&1&1&v\\0&v-1&0&1-v\\0&0&v-1&1-v\\0&0&0&-v^2-2v+3\end{pmatrix}

Now, if rank $V\not=4$ then there should be at least one zero row, so we let

$-v^2-2v+3=0$

$(v+3)(v-1)=0$

$v=-3,\space v=1$

Thus, rank $V\not=4$ when $v\in\mathbb{R}$\ {1,-3}

Is my value for $v$ wrong or am I missing any values?


问:
Limits and sequences of functions
Let the sequence of functions {$f_n$} define in [0,1] $\subset$ $\mathbb R$, where $f_{n}=nx^{n}(1-x^{n}), $ $\forall$x $\in$ [0,1]. Proof that:

$\lim\limits_{n}\int_{0}^{1}f_{n}(x)dx$ $\neq$ $\int_{0}^{1}\lim\limits_{n}f_{n}(x)dx$

答:
My attempt is the following:

$\lim\limits_{n}\int_{0}^{1}f_ndx$ = $\lim\limits_{n}\int_{0}^{1}nx^{n}(1-x^{n})$

$\lim\limits_{n}[n(\frac{1}{n+1}-\frac{1}{2n+1})]$

$\lim\limits_{n}\frac{n}{n+1}-\lim\limits_{n}\frac{1}{2n+1}$ = -1/2

$\lim\limits_{n}\int_{0}^{1}f_ndx$ = -1/2

The sencond part:

$\int_{0}^{1}\lim\limits_{n}f_ndx$ = $\int_{0}^{1}\lim\limits_{n}nx^{n}(1-x^{n})$ = not answer

I have trouble with the latest equation.

Title: Limits and sequences of functions


问:
Does Conway's Game of Life have attracting cycles (or equilibria)?
Is there a sequence $\:\langle S_0,S_1,S_2,...,S_{n-1},S_n\rangle\:$ of generations for Conway's Game of Life such that

[$\hspace{.01 in}S_0 = S_n$ $\:$ and $\:$ $S_0$ has at least one live cell $\:$ and $\:$ [for all members $i$ of $\:\{0,1,2,...,n-2,n-1\}\:$,

for all grid states $S$, if $S$ differs from $S_i$ on only one cell, then $S$ eventually evolves into $S_0$]]$\hspace{.04 in}$?

(I thought of the above question while I was reading

an article about self-organization and stability.)

答:
I would guess the answer is no, unless there happens to be a very simple (and therefore boring) solution. Something simple and unchanging like a $2\times2$ square, or with a very short period like a $3\times1$ "blinker"; however it turns out these are not examples. The reason for my guess is that any interesting repetitive pattern (with period${}>1$) will involve some region where interesting changes are going on, which means creation of "fertile" cells in some generations (a cell that vanishes improductively after creation does not contribute to reproduction of the population). Each cell creation will be undone if any one of the $8$ neighbour cells is toggled, which gives a large number of possible perturbances (and the number increases as the configuration gets larger), each of which must somehow magically be brought back into the right track. If small configurations do not provide examples, larger ones will have have an even harder time to do so.

There is of course nothing rigorous about this, but I feel that as one proceeds in exploring without finding examples, the probability that an example remains to be found will rapidly tend to $0$. It is like conjecturing that in the sequence of decimals of $\pi$, there is no initial subsequence that is immediately repeated a second time. An easy computation shows that the probability of such a repetition becomes vanishingly small after the first candidates have been checked (and ever smaller as more decimals are computed), even though I think it is also extremely unlikely that this conjecture can ever be proved (one could imagine some result about rational approximation of $\pi$ being applied, but for one thing I don't think any remarkable such properties are known, and on the other hands such results are usually asymptotic only, therefore independent of any particular isolated approximation).


问:
Trivial homology groups and fundamental group
I was wondering if there are noncontractible spaces with trivial homology groups and trivial fundamental group. I can't think of any, but I suspect some must exist.</p>

答:
The Warsaw circle $\mathbb{S}$ is such a space. To construct it start with the topologist's sine curve
$$X=\{(x,\sin(2\pi/x))\mid x\in(0,1]\}\cup(\{0\}\times[-1,1])$$
embedded in the euclidean plane. Recall that this space is often encountered as an example of a space which is connected, but not path-connected. Then the Warsaw circle is the union
$$\mathbb{S}=X\cup(\{0\}\times[-2,-1])\cup([0,1]\times\{-2\})\cup(\{1\}\times[-2,0]).$$
Thus $\mathbb{S}$ consists of the sine curve with its ends joined together by a simple arc in $\mathbb{R}^2$ which is disjoint from all other points of $X$. The result $\mathbb{S}$ is a compact metric space which is both connected and path-connected. However it is still highly pathological.

Basically, in terms of connectivity it thinks it is a circle, while in terms of path-connectivity is sees itself as a copy of the real line.

As a consequence all the homotopy groups and singular homology groups of $\mathbb{S}$ vanish. This is not difficult to see directly, but can be formally established as follows. There is an obvious bijective map $p:[0,\infty)\rightarrow\mathbb{S}$ that starts at $(0,1)$ and wraps itself anticlockwise around $\mathbb{S}$. For all intents and purposes this map acts like a covering projection with one-point fibres (though it is not locally-trivial - $\mathbb{S}$ is not locally connected). What this map is is the result of retopologising $\mathbb{S}$ by giving it the final topology with respect to all maps $M\rightarrow\mathbb{S}$ with locally path-connected domains. Thus in particular any map into $\mathbb{S}$ from a sphere $S^n$ lifts throught $p$ into $[0,\infty)$ and can be linearly null-homotoped. This gives us $\pi_*\mathbb{S}=0$, and working with simplices the same argument yields $\widetilde H_*\mathbb{S}=0$.

On the other hand, $\mathbb{S}$ is not contractible. In fact its first Cech cohomology group $\check H^1\mathbb{S}$ is nonzero. This can be computed as $\check H^1\mathbb{S}=[\mathbb{S},S^1]\cong\mathbb{Z}$, where $[\mathbb{S},S^1]$ is the group of homotopy classes of map $\mathbb{S}\rightarrow S^1$ (the group structure is induced by the multiplication on $S^1$). Thus you can very vividly see what is happening. For each $n\in\mathbb{Z}$ there is a map of $\mathbb{S}$ onto $S^1$ which wraps the 'pseudo-circle' $n$ times around the real circle. Because there are essential maps out of it, we see that $\mathbb{S}\not\simeq\ast$ .

For a formal calculation of the Cech group we can use Alexander duality, which states that if $A\subseteq\mathbb{R}^n$ is a compact subset, then there are isomorphisms
$$\check H^kA\cong \widetilde H_{n-k-1}(\mathbb{R}^n\setminus A),$$ where the group on the left is the Chech cohomology group, and the group on the right is ordinary (reduced) singular homology. Plugging in $\mathbb{S}\subseteq\mathbb{R}^2$ we get
$$\check H^1\mathbb{S}\cong \widetilde H_0(\mathbb{R}^2\setminus \mathbb{S}).$$
Now $\mathbb{S}$ cuts the plane into two connected components, and as such $H_0(\mathbb{R}^2\setminus \mathbb{S})\cong\mathbb{Z}^2$. In the reduced group we indeed retrieve
$$\check H^1\mathbb{S}\cong\mathbb{Z}.$$


问:
Convergece of Uniform process
Let $S_0=1$ and $S_n$ be uniformly distributed between $(0,S_{n-1})$.
How do we show that $m^nS_n \rightarrow0$ almost surely if $0 < m < e$
and $m^nS_n \rightarrow \infty$ almost surely if $m > e $

This is a bit confusing because I know X_n goes to zero a.s. For it to go to infinity does it mean it's growing faster than the uniform process? What's the best way to approach the problem? I was trying Borel Cantelli and the definition of almost sure convergence but both yielded no results.
Any hints or helpful resources will be appreciated.

答:
Let $U_n=\frac{S_n}{S_{n-1}}$.  This  is uniformly distributed on $(0,1]$ and each of the $U_n$s is independent of the others.

Let $V_n = \log_e\left(U_n\right)$.  Then both the CDF and density of $V_n$ is $e^t$ for $t \le 0$ and $\mathbb E[V_n]=-1$, and each of the $V_n$s is independent of the others.

Now let  $W_n = \log_e(m)+V_n$ so $\mathbb E[W_n]=\log_e(m)-1$, and each of the $W_n$s is independent of the others.  By the strong law of large numbers $\frac1n \sum\limits_{i=1}^n W_i \to \log_e(m)-1$ almost surely.  Then (in effect multiplying both sides by $n$)

if $\log_e(m) \lt 1$, i.e. $m \lt e$,  then  $\sum\limits_{i=1}^n W_i \to -\infty$ almost surely

if $\log_e(m) \gt 1$, i.e. $m \gt e$,  then  $\sum\limits_{i=1}^n W_i \to +\infty$ almost surely

But $\sum\limits_{i=1}^n W_i = \log_e(m^n S_n)$ so (in effect taking anti-logarithms)

if $m \lt e$,  then  $m^n S_n \to 0$ almost surely

if $m \gt e$,  then  $m^n S_n \to +\infty$ almost surely

Title: Convergece of Uniform process


问:
How to efficiently find out all clusters of a given size $k$ in an undirected graph?
For example, in an undirected graph as follows, I want to find out all clusters of size 4. How should I do it efficiently?

As for a cluster, points should be connected. For example {1,2,3,4},{1,2,3,13},{1,2,3,14},{1,2,14,17},{1,14,16,17} are clusters; while {1,2,4,13}, {1,2,4,14} are not clusters because the four points in a cluster are not connected (i.e. there are isolated tree(s)/node(s)).

答:
Your term "clusters" is improper.  You're seeking either complete subgraphs (i.e., subgraphs of size $4$ where each vertex is connected to all the others) or connected subgraphs (i.e., subgraphs of size $4$ that are connected).  None of my dozen or so books on graph theory mention this novel use of "cluster" (nor this novel use of "pivot").

Anyway....

In Mathematica:

```<code>g = RandomGraph[{30, 100}, VertexLabels->"Name"]
</code>```

The potential subsets of four vertices:

```<code>subs = Subsets[Range[30], {4}];
</code>```

Check all these subgraphs to see which are complete subgraphs (of size $4$):

```<code>Select[
      Table[
      Subgraph[g, subs[[i]],
      VertexLabels -> "Name"],
      {i, Length[subs]}],
   CompleteGraphQ]
</code>```

If instead you are looking for merely connected subgraphs, then the last function should be:

```<code>Select[
          Table[
          Subgraph[g, subs[[i]],
          VertexLabels -> "Name"],
          {i, Length[subs]}],
       ConnectedGraphQ]
</code>```

The above code is certainly efficient enough for your task.  For the graph above, all connected subgraphs of size $4$ were found in 1.54531 seconds on a Mac laptop.  For a random graph with $40$ vertices and $150$ edges, all $7068$ connected 4-subgraphs were found in less than a minute.

It is very unclear whether the OP allows (non-tree) subgraphs of size four connected in a loop, such as this:

Body:
Your term "clusters" is improper.  You're seeking either complete subgraphs (i.e., subgraphs of size $4$ where each vertex is connected to all the others) or connected subgraphs (i.e., subgraphs of size $4$ that are connected).  None of my dozen or so books on graph theory mention this novel use of "cluster" (nor this novel use of "pivot").

Anyway....

In Mathematica:

g = RandomGraph[{30, 100}, VertexLabels->"Name"]

The potential subsets of four vertices:

subs = Subsets[Range[30], {4}];

Check all these subgraphs to see which are complete subgraphs (of size $4$):

Select[
      Table[
      Subgraph[g, subs[[i]],
      VertexLabels -> "Name"],
      {i, Length[subs]}],
   CompleteGraphQ]

If instead you are looking for merely connected subgraphs, then the last function should be:

Select[
          Table[
          Subgraph[g, subs[[i]],
          VertexLabels -> "Name"],
          {i, Length[subs]}],
       ConnectedGraphQ]

The above code is certainly efficient enough for your task.  For the graph above, all connected subgraphs of size $4$ were found in 1.54531 seconds on a Mac laptop.  For a random graph with $40$ vertices and $150$ edges, all $7068$ connected 4-subgraphs were found in less than a minute.

It is very unclear whether the OP allows (non-tree) subgraphs of size four connected in a loop, such as this:


问:
Changing Basis question: how is this the right answer?
First part is fine, I get 2/5 but how is the instructor getting 11/5 and not 55/7?

Formula being used:

Changing Basis Question:

Thanks

答:
The answer indicated is correct, because $\dfrac25\pmatrix{3\\4}+\dfrac{11}5\pmatrix{4\\-3}=\pmatrix{10\\-5}$.

You should divide by the square of the magnitude of the basis vector,

which is $(4)^2+(-3)^2=25$, not $(4)^2-(3)^2=7$.


问:
Is the intersection a line, plane or an empty set?
I'm wondering about a geometric interpretation of the equation system shown below:
$$x_1+x_2+x_3+x_4=6\\
x_1+x_3+x_4=4\\
x_1+x_3 = 2$$
I've calculated the echelon form of those equations and got these results:
$$x_1 = 2- x_3\\x_2 = 2\\ x_3 = x_3 \\ x_4 = 2$$
I know that the system has infinitely many solutions. Can you explain me how to intuitively find out if the intersection is a line, plane or an empty set?

答:
First of all, since the system has solutions (and even many of them!), you already know that its solution set is not empty.

To understand this solution set geometrically, note that it has one free variable in it, which effectively tells us that the solution set is one-dimensional. (I'm being a bit nor rigorous here, since this solution set is not a vector subspace; but it is an affine subspace, which is indeed one-dimensional.) So it's a line.

To actually see that this is a line, and more specifically what this line is, let's represent the solutions in vector form:
$$\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix} 2-x_3 \\ 2 \\ x_3 \\ 2 \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 0 \\ 2 \end{bmatrix}+\begin{bmatrix} -x_3 \\ 0 \\ x_3 \\ 0 \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 0 \\ 2 \end{bmatrix}+x_3\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}.$$

Now we can see that it's a line thru the point $(2,2,0,2)^T$ and parallel to the direction vector $(-1,0,1,0)^T$.


问:
If the acceleration is constant, does it lie on a plane?
The Calculus 3 book Apostle Volume 1 has a group of questions I don't quite know how to approach. It says that for the next two statements about a curve traced out by a particle moving in 3-space to give a proof or exhibit a counter example.

(a) If the acceleration is constant, the curve lies in a plane.

(b) If the velocity is perpendicular to the acceleration, the curve lies in a plane.

How do I approach these problems? What does it mean to lie in a plane? A solution to either would be appreciated as it would help me understand better.

答:
If acceleration is constant, then the object takes a parabolic path through space (see constant gravitational acceleration), which thus lies on a plane.

If velocity is perpendicular to acceleration, then the object travels in a circle (see this video for a full explanation: https://www.youtube.com/watch?v=x3u53wVeWfA)
For this second case, you can imagine swinging a ball on a string in a circle over your head. Acceleration is always directed inwards, perpendicular to the velocity of the ball, which is tangent to the curve of the circle.


答: $x_3=x_3$ means $x_3$ can be anything we like. $x_2$ and $x_4$ are constant, so we cant "move" along those axis. Now imagine sliding $x_3$ around. Think of it as an independent variable, if you like. $x_1$ is a line in terms of $x_3$, so the relationship between them will form a line in the $x_1 - x_3$ plane. Thus the solution set forms a line.

Title: Is the intersection a line, plane or an empty set?

Body:
I'm wondering about a geometric interpretation of the equation system shown below:
$$x_1+x_2+x_3+x_4=6\\
x_1+x_3+x_4=4\\
x_1+x_3 = 2$$
I've calculated the echelon form of those equations and got these results:
$$x_1 = 2- x_3\\x_2 = 2\\ x_3 = x_3 \\ x_4 = 2$$
I know that the system has infinitely many solutions. Can you explain me how to intuitively find out if the intersection is a line, plane or an empty set?


问:
Prove $U(|f|, P)-L(|f|, P) \le U(f, P) - L(f, P)$
This was one of the problems I had on an exam and it's been bothering me because I couldn't figure it out. I was only able to show $-L(|f|, P)\le-L(f, P)$ and $U(f, P)\le U(|f|, P)$, but adding the inequalities doesn't give the desired result.

"Let $f$ be a bounded integrable function on $[a,b]$. Let $P=\{x_0, ..., x_n\}$ be a partition of $[a,b]$. Prove that $U(|f|, P)-L(|f|, P) \le U(f, P) - L(f, P)$."

Here are the relevant definitions:

$\Delta x_i =x_i -x_{i-1}$

$M_i(f, P)=\text{sup}\{f(x): x\in [x_{i-1}, x_i]\}$

$m_i(f, P)=\text{inf}\{f(x): x\in [x_{i-1}, x_i]\}$

$U(f, P)= \sum \limits_{i=1}^{n}M_i(f, P)\Delta x_i$

$L(f, P)= \sum \limits_{i=1}^{n}m_i(f, P)\Delta x_i$

答:
By the reverse triangle inequality $|\,|f(u) - |f(v)|\,| \leqslant |f(u) - f(v)|$.

Hence, for any partition interval $I_j = [x_{j-1},x_j]$ we have

$$M_j(|f|,P)- m_j(|f|,P) = \sup_{u,v \in I_j}|\,|f(u)| - |f(v)|\,| \leqslant \sup_{u,v \in I_j}|f(u) - f(v)| = M_j(f,P)- m_j(f,P),$$

and

$$U(|f|,P) - L(|f|,P) = \sum_{j=1}^n(M_j(|f|,P)- m_j(|f|,P)) \Delta x_j \leqslant \sum_{j=1}^n(M_j(f,P)- m_j(f,P)) \Delta x_j\\ = U(f,P) - L(f,P)$$


Body: Show that the sequence $(a_n)_{n\ge 0}$ defined recurrence relation $a_0=25, a_{n+1}=2\sqrt{a_n}$ converges. I showed that it is bounded from below by $1$, but I'm struggling to show that the sequence is non-increasing. I tried induction, but the induction hypothesis is not helpful at all.

Body: Note that $\frac{a_{n+1}}{4}=\left(\frac{a_n}{4}\right)^{\frac 12}$, therefore $\frac{a_n}{4} = \left(\frac{a_0}{4}\right)^{\frac{1}{2^n}} = \left(\frac{25}{4}\right)^{\frac{1}{2^n}} \to 1$, as $n \to +\infty$.


问:
$x^3 + y^3 = p^2$ over the integers
$x^3 + y^3 = p^2$ has a solution over the integers for some three digit prime p. Find all p that satisfy.

The first thing I did was factorize the left hand side, getting $(x+y)(x^2 - xy + y^2) = p^2$

I then considered the case $x^2 - xy + y^2 = p^2$, which gave me $p^2 = 3y^2 -3y +1$, which can be naturally factorized into $(p+1)(p-1) = 3y(y-1)$, but how do I do anything with this? I don't see a continuation from this point onwards.

Any hints or solutions would be greatly appreciated. Thanks!

答:
Take $u = 6y-3.$  The first few values of $p$ in $u^2 - 12 p^2 = -3$  are
$$ 1, 13, 181, 2521, 35113, ...  $$
subject to
$$ p_{j+2} = 14 p_{j+1}  - p_j   $$

and $p=181$ is a three digit prime

Here is a portion of the Conway Topograph for $u^2 - 12 p^2.$  A continuation a little bit in the direction of increasing $(u,p)$ would show the next occurrence of $u^2 - 12 p^2 = -3$  at $(45, 13).$ That recipe is
$$ (u,p) \mapsto (7u+24p, 2u + 7p) $$
This is multiplication of a column vector by the square matrix
$$
\left(
\begin{array}{cc}
7&24 \\
2&7
\end{array}
\right)
$$
which has determinant $1$ and trace $14.$  It follows from Cayley-Hamilton for this matrix that

$$ u_{j+2} = 14 u_{j+1}  - u_j   $$

$$ p_{j+2} = 14 p_{j+1}  - p_j   $$
You can also work it out by hand from

$$
\left(
\begin{array}{c}
u_{j+1} \\
p_{j+1}
\end{array}
\right) =
\left(
\begin{array}{cc}
7&24 \\
2&7
\end{array}
\right)
\left(
\begin{array}{c}
u_j \\
p_j
\end{array}
\right)
$$
write out the values at $j+2$ and substitute some things

Here are the first few values of $p$ after $1.$  Some of them are prime.

```<code>13  =   13
181  =   181
2521  =   2521
35113  =   13 37 73
489061  =   489061
6811741  =   6811741
94875313  =   13 61 181 661
1321442641  =   1321442641
18405321661  =   18405321661
256353060613  =   13 757 2521 10333
3570537526921  =   277 3037 4244329
</code>```

Title: $x^3 + y^3 = p^2$ over the integers

Body:
Take $u = 6y-3.$  The first few values of $p$ in $u^2 - 12 p^2 = -3$  are
$$ 1, 13, 181, 2521, 35113, ...  $$
subject to
$$ p_{j+2} = 14 p_{j+1}  - p_j   $$

and $p=181$ is a three digit prime

Here is a portion of the Conway Topograph for $u^2 - 12 p^2.$  A continuation a little bit in the direction of increasing $(u,p)$ would show the next occurrence of $u^2 - 12 p^2 = -3$  at $(45, 13).$ That recipe is
$$ (u,p) \mapsto (7u+24p, 2u + 7p) $$
This is multiplication of a column vector by the square matrix
$$
\left(
\begin{array}{cc}
7&24 \\
2&7
\end{array}
\right)
$$
which has determinant $1$ and trace $14.$  It follows from Cayley-Hamilton for this matrix that

$$ u_{j+2} = 14 u_{j+1}  - u_j   $$

$$ p_{j+2} = 14 p_{j+1}  - p_j   $$
You can also work it out by hand from

$$
\left(
\begin{array}{c}
u_{j+1} \\
p_{j+1}
\end{array}
\right) =
\left(
\begin{array}{cc}
7&24 \\
2&7
\end{array}
\right)
\left(
\begin{array}{c}
u_j \\
p_j
\end{array}
\right)
$$
write out the values at $j+2$ and substitute some things

Here are the first few values of $p$ after $1.$  Some of them are prime.

13  =   13
181  =   181
2521  =   2521
35113  =   13 37 73
489061  =   489061
6811741  =   6811741
94875313  =   13 61 181 661
1321442641  =   1321442641
18405321661  =   18405321661
256353060613  =   13 757 2521 10333
3570537526921  =   277 3037 4244329


问:
What is the joint distribution of $(X,Y)$ under the $\mathbb Q$ equivalent probability measure?

Let $X$ and $Y$ be two independent standard random variables. Let $\theta$ be a
constant with $\theta^2 \in (0, 1)$ and define the $\mathbb P$-equivalent probability measure, $\mathbb Q$, by
$\frac{d\mathbb Q}{d\mathbb P}=\frac{e^{\theta XY}}{\mathbb E[e^{\theta XY}]}>0$. What is the joint distribution of $(X,Y)$ under $\mathbb Q$? Are $(X,Y)$ also independent under $\mathbb Q$?

My try:

Compute that
\begin{align}
\mathbb P(X<x_0, Y<y_0)&=\frac {1}{\mathbb E(e^{\theta XY})}\int_{-\infty}^{x_0}dx\int_{-\infty}^{y_0} e^{\theta xy}\frac 1{2\pi}e^{-\frac{x^2+y^2}{2}}dx\\
&=\frac{1}{2\pi\mathbb E(e^{\theta XY})}\int_{-\infty}^{x_0}e^{-x^2/2}dx\int_{-\infty}^{y_0}e^{-\frac 12(y-\theta x)^2-\frac 12\theta^2 x^2}dy
\end{align}
Let $t=y-\theta x$, then
$$ \mathbb P(X<x_0, Y<y_0)=\frac{1}{\sqrt{2\pi}\mathbb E(e^{\theta XY})}\int_{-\infty}^{x_0}e^{-\frac 12x^2-\frac 12\theta^2 x^2}dx\int_{-\infty}^{y_0-\theta x}\frac{1}{\sqrt{2\pi}}e^{-\frac 12 t^2}dy. $$

But I have not idea how to proceed. Also, the hint says maybe we can consider the moment generating function but we have the same type of issue that we cannot seperate $e^{\theta xy}$ and use the independence of $X$ and $Y$ under $\mathbb P$.

答:
If you want to consider the moment generating function approach, compute the MGF by conditioning on  (say) $X$. This permits you to "separate" the $XY$. If $E$ denotes expectation under $P$ and $E_Q$ expectation under $Q$, then
$$
\begin{align}
E_Q\exp(sX+tY)&=k(\theta)E\exp(sX+tY+\theta XY)\\
& = k(\theta)E\left[ E\left(\exp(sX+tY+\theta XY)\mid X\right)\right]\\
& =k(\theta)E\exp\left(sX + \frac12(t+\theta X)^2\right),
\end{align}
$$
the last equality using the univariate MGF $E\exp (uY)=\exp \frac12( u^2 )$ when $Y$ is standard normal:
$$\begin{aligned}
E(\exp(sX+tY+\theta XY\mid X=x)&=E(\exp(sx+tY+\theta xY\mid X=x)]\\
&=E(\exp(sx+tY+\theta xY))\\
&=e^{sx}E(\exp((t+\theta x)Y)\\
&=e^{sx}e^{(t+\theta x)^2/2}
\end{aligned}
$$ Now you are left with a straightforward expectation involving $X\sim N(0,1)$, which, when the smoke clears, should represent the MGF of a  familiar bivariate distribution (also hinted at by the comment by @sudeep5221 and the answer by @ChristopherK). To perform the expectation you can prove and use the fact
$$
E\exp\left(\frac a2X^2+bX+c\right) = \exp\left(c +\frac{b^2}{2(1-a)}\right)
$$
when $X$ is standard normal.


问:
Need help with evaluating an integral
Given that $f$ is continuous and positive, and

$$I=\int_0^\frac{\pi}{2} \frac{f(\sin (x))}{f(\sin (x)) + f(\cos(x))}\mathrm dx $$

I Know that $I=\dfrac{\pi}{4}$.

答:
HINT

If you make the change of variable $y = \pi/2 - x$, you get

\begin{align*}
I = \int_{0}^{\pi/2}\frac{f(\cos(x))}{f(\cos(x)) + f(\sin(x))}\mathrm{d}x
\end{align*}

Then add both expressions of $I$ and see what happens.

Title: Need help with evaluating an integral


问:
Expected value of position of the first $i$ where $X_i$ is greater than $X_0$
Say there's an infinite list of numbers drawn uniformly from $[0,100]$. Let's call the first number in this list $X_0$, and the $i^\text{th}$ number $X_i$. What's the expected value of $i$, where $i$ is the first number where $X_i > X_0$?

Here's how I've approached this problem; where am I wrong?

All numbers after $X_0$ have a $\frac{1}{2}$ chance of being larger or smaller than $X_0$. Therefore, we can model this as an infinite series, where $X_1$ happens to be the first number larger than $X_0$ with probability $\frac{1}{2}$, $X_2$ happens to be the first number larger than $X_0$ with probability $\left(\frac{1}{2}\right)^2$, $X_3$ with probability $\left(\frac{1}{2}\right)^3$, etc...

This results in the following series:

$$ \frac{1}{2}\cdot 1 + \frac{1}{4}\cdot 2 + \frac{1}{8}\cdot 3 \dotsb = 2$$

答:
Let $N = \min\{ n \geq 1 : X_n > X_0\}$ (it's a random variable).

We want $\mathbb{E}[N]$.

Let's start with something easier, $\mathbb{E}[N|X_0]$. In this case, since $X_0$ is "fixed", we have $N_{|X_0} \sim Geo(1-\frac{X_0}{100})$. Therefore $\mathbb{E}[N|X_0] = \frac{100}{100-X_0}$.

Now $\mathbb{E}[N] = \mathbb{E}[\mathbb{E}[N|X_0]] = \mathbb{E}[\frac{100}{100-X_0}] = +\infty$


问:
Problem about almost everywhere convergence in measure theory
I am having trouble with the following problem

Let $(X, \mathcal{F}, \mu)$ a measure space where $\mu (X)<\infty.$  Let $f,f_n:X \to \mathbb{C}$ be measurable. Set $A_n=\{ |f_n-f|\geq a_n\}$ where $a_n>0$ and $a_n \to 0$. Show that if $\sum_n \mu (A_n)<\infty,$ then  $f_n\xrightarrow{a.e.} f.$

I've been trying a lot this problem. For example, I tried to show that $\mu (\{f_n \nrightarrow f\})<\varepsilon$ for all $\varepsilon>0$ using facts as $\mu(A_n) \to 0$ (because the series is convergent) and even asumming that $(a_n)$ could be taken strictly decrasing.
In my "closer" attempt I showed that every $x \in \{f_n \nrightarrow f\}$ is contained in infinitely many of the sets $A_n$. But at the end, it doesn't worked.

At every attempt I made, I though "I'm very very close to the solution"... but something failed.

Could you please help me solving this problem?

答:
First observe that the set where $f_n$ does not converge to $f$ is measurable and it can be written as $A:=\{f_n\not\to f\}=\bigcap_{k=1}^{\infty}\bigcup_{n\ge k}A_n.$

Now note that showing $f_n\to f$ almost everywhere is equivalent to showing that $A$ has measure $0.$  To do this, we first observe that
$$\mu(A)\le \mu(\bigcup_{n\ge k}A_n)\le \sum_{n\ge k}A_n.$$

Since $\sum_{n=1}^{\infty}\mu(A_n)$ is finite, by choosing $k$ large, we can make $\sum_{n=k}^{\infty}\mu(A_n)$ arbitrarily small. It follows that $\mu(A)=0.$


问:
Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

Find the equation of a surface obtained by rotating of a circle $^2 + ^2 − 4 + 3 = 0$ around $$-axis.

Find the intersection of surfaces $^2 + ^2 = 2$ and $^2 + ^2 + ^2 = 8$. What kind of curve is it?

What I supposed to do? In 1) I need to find some transformation, which I don't know. In second I just tried to substitute equation variables one into another? but still not sure.

答:
The point $(a,b)$ traces a circle of radius $a$ in the plane $y=b$ after rotating around the $y$-axis, i.e.
$$x^2+z^2=a^2,\>\>\>\>\>y=b$$
Then, substitute $a$ and $b$ into $(x-2)^2+y^2=1$ to obtain the 3D surface
$$(\sqrt{x^2+z^2}-2)^2+y^2 =1$$
which is a ring torus.

Substitute $z=\frac12(x^2+y^2) $ into $x^2+y^2+z^2=8$ to get

$$(x^2+y^2)^2 +4(x^2+y^2)-32=0$$
which leads to $x^2 +y^2 = 4 $ and $z=2$, i.e. a circle of radius $2$ in the plane $z=2$.


问:
Find eigenvectors of $A$
Let $A=\begin{pmatrix} 0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2 \end{pmatrix}$

I find eignevalues $-1,0,1,2$ and $0$ has multiplicity of $2$.

But I only find one eigenvector of $0$ and all the other eigenvalues have eigenvectors, so anyway $A$ has four eigenvectors and diagonalizable.  But I think $0$ must have $2$ eigenvectors and one of eigenvalue can't have eigenvector. What's wrong with me?

I corrected matrix!!!!!

Then the C.P is $x(x^3-2x^2-x+2)=x(x+1)(x-1)(x-2)$ and the eigenvalues that I got is right.

Title: Find eigenvectors of $A$

Body:
$$\begin{align} \det(A-xI) &= \det(\begin{pmatrix}-x&0&0&0\\1&-x&0&-2\\0&1&-x&1\\0&0&1&2-x\end{pmatrix})\\ &=-x(-x(-x(2-x)-1)-2) \\&=-x(2x^2-x^3+x-2)\\&=-x(-x^2(x-2)+x-2) \\&=-x(x-2)(-x^2+1)\end{align}$$

so $x=0,x=2,x=1,x=-1$ are eigenvalues

then we must find X such that :

$(\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2\end{pmatrix}-\lambda_iI)X=0$

$\to$

$\lambda$=0$ :\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2\end{pmatrix}X=0 \to x_1=2x_4$ and $x_2=-x_4$and $x_3=-2x_4$ our vector can be $(-2x_4,x_4,2x_4,x_4)$as(2,-1,-2,1)

$x=2:$

$\begin{pmatrix}-2&0&0&0\\1&-2&0&-2\\0&1&-2&1\\0&0&1&2-2\end{pmatrix}X=0$$\to $$x_1=0,x_2=-x_4,x_3=0\,\,$as $\,\,(0,-1,0,1)$

$x=1:$

$\begin{pmatrix}-1&0&0&0\\1&-1&0&-2\\0&1&-1&1\\0&0&1&2-1\end{pmatrix})X=0$ $\to $
$x_1=0,x_2=-2x_4,x_3=-x_4$
as $\,\,(0,-2,-1,1)$

$x=-1:$

$\begin{pmatrix}1&0&0&0\\1&1&0&-2\\0&1&1&1\\0&0&1&2+1\end{pmatrix}X=0$$\to$$x_1=0,x_2=2x_4,x_3=-3x_4\,\,$as $\,\,(0,2,-3,1)$


问:
Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing

Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.

I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$

I am trying to go ahead of this step.

答:
My two cents (as later similar question closed):

$$\frac{a_{n+1}}{a_n}=\frac{\left( 1+ \frac{1}{n+1} \right)^{n+1}}{\left( 1+ \frac{1}{n} \right)^{n}}=\frac{n+1}{n}\left( 1- \frac{1}{(n+1)^2} \right)^{n+1}>\\
>\frac{n+1}{n}\left( 1- \frac{1}{(n+1)} \right)=1$$


问:
Proof that equicontinous and suriective dynamical system is distal
Let us have dynamical system $(X,f)$ (what means that $X$ is compact metric space with metric $d$ and $f: X \to X$ is continuous function). Moreover $f$ is surjective and equicontinous so when we take $x,y \in X$ we have $\forall \epsilon > 0 \ \exists \delta > 0 \ \forall \ n \in \mathbb{N}: d(x,y) < \delta \Rightarrow d(f^n(x),f^n(y)) < \epsilon$. I want to show that such dynamical system is distal, so  it does not contain proximal pair $\{x,y\}$ such that $x \ne y$ and $\lim \inf_{n \to \infty} d(f^n(x),f^n(y)) = 0$. My idea of proof is to take increasing sequence $n_k$ such that $\lim_{k \to \infty}d(f^{n_k}(x),f^{n_k}(y)) = 0$. This is of course equivalent to definition of proximal pair. Then using assumptions I want to show that $x = y$. Of course $\forall \epsilon > 0 \exists n_k \ d(f^{n_k}(x),f^{n_k}(y)) < \epsilon $. I have also suriectivity of $f$. I have problem though with formal proof. I wolud be grateful for any hint.

答:
In addition to the above, suppose the family of functions $\{f^{-n}: n \in \mathbb{N}\}$ is equicontinuous on $X$.

Assume we have $x,y \in X$ such that $x\neq y$ and $\underset{n\to \infty}\liminf d(f^n(x),f^n(y)) = 0$. So there is a sequence of natural numbers $\{n_k\}$ such that $d\left(f^{n_k}(x),f^{n_k}(y)\right)<\delta$ for all $\delta>0$.

Let $x_k:=f^{n_k}(x)$ and $y_k:=f^{n_k}(y)$   $\, (k=1,2,\ldots)$.

Let $\delta>0$ be given. So we have $d\left(x_k,y_k\right)<\delta$ and $d\left(f^{-n_k}(x_k),f^{-n_k}(y_k)\right)=d(x,y)>0$.


问:
About maximal subfields of a division ring
Given the division ring $D$ and a maximal subfield $K\subset D$, I'm trying to prove that $K=C_D(K)$, where $C_D(K)=\{d\in D:\forall k\in K, dk=kd\}$. In particular, I'm stuck in "$C_D(K)\subset K$". I found some notes where the argumentation is as follows:

Let $a\in C_D(K)$ and let $T$ be the division ring generated by $K$ and $a$. Since $a$ commutes with all elements of $K$, $T$ is a field, then $K\subset T\subset D$ implying that $T=K$ with $a\in T=K$. Hence, $C_D(K)\subset K$.

This argument is confusing for me because I don't see why $T$ results to be a field. I only know the following: Since $a$ commutes with all elements of $K$ the set $B=\{\sum_n ka^n:k\in K\}$ is an integral domain and it's possible to construct the field of fractions $\hat{B}$. I'm trying to see $\hat{B}$ as a subset of $D$ but I fail because the natural inclusion of $(a,b)\in\hat{B}$ via $ab^{-1}$ doesn't seen well-defined.

Is there a theorem that asure the existence of a subfield $T$ with $K\cup\{a\}\subset T\subset D$?

答: Hint: As you have noted, $K[a]$ is a commutative ring. It follows that for any nonzero $p,q\in K[a]$ you have $q/p=p\backslash q=q\cdot (1/p)=(1/p)\cdot q$. This can be used to show that the ring generated by $K[a]$ and the inverses of all its nonzero elements is still commutative. It follows that it is isomorphic to the field of fractions of $K[a]$ (so that in particular, it is a field).


问:
Prove $\sum b_n$ converges if $\sum a_n$ converges and $\lim \frac{a_n}{b_n}=6$
I recently received this problem on a real analysis exam and I'm still having a hard time understanding it.

"Let $(a_n)$ and $(b_n)$ be sequences of $\mathbb{R}^+$. Suppose $\lim \frac{a_n}{b_n}=6.$ Prove that if the infinite series $\sum a_n$ converges, then $\sum b_n$ converges."

Here's what I understand so far: we can use the Algebraic Limit Theorem to simplify and get $\lim (a_n) = 6*\lim(b_n)$. Also, since we know $\lim \frac{a_n}{b_n}=6$, we know that division by $\lim(b_n)$ is valid, so $\lim(b_n) \ne 0$.

What's confusing me is that there is a theorem that says if an infinite series $\sum s_n$ converges, then $\lim(s_n) =0$. If we attempt the problem and suppose that $\sum a_n$ converges, wouldn't that mean that $\lim(a_n) =0$, so $\lim \frac{a_n}{b_n}=6$ can't occur? What am I missing?

Edit: I understand now that the Algebraic Limit Thm may not come into play with this problem. However, I'm still confused how to use the given information to prove that $\sum b_n$ converges.

答: Hint: Let $\epsilon>0$. Since $\frac{a_n}{b_n} \to 6,$ there exists $N$ such that $\frac{a_n}{b_n}>6-\epsilon$ for all $n\geq N$. Show that $\sum b_n$ is bounded above. It is obviously monotone.


答:
To prove it, I would do the following:

$\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = 6 \implies \exists N \in \mathbb{N}$ such that $\frac{a_n}{7} < b_n < \frac{a_n}{5} \quad \forall n \geq N.$

Can you finish this now?

$$$$

Formal proof of the above statement, as requested by OP:

By definition of the limit of a sequence, given $\varepsilon>0, \ \exists N \in \mathbb{N}$ such that $\left|\frac{a_n}{b_n} - 6\right| < \varepsilon \quad \forall \ n \geq N.$

Let $\varepsilon = \frac12.$

Then $ \exists N$ such that:

$\left|\frac{a_n \ - \ 6b_n}{b_n}\right| < \frac12 \quad \forall n \geq N,\ $ which implies that

$ \left|a_n-6b_n\right| < \frac12 b_n \quad \forall \ n \geq N \quad (*),$

where $|b_n| = b_n$ is justified because $b_n \in \mathbb{R}^+$ was assumed in the question.

If $a_n > 6b_n,\ $ then $\left|a_n-6b_n\right| = a_n - 6b_n$, so $(*) \implies a_n < 6 \frac12 b_n\ < 7b_n \quad \forall \ n \geq N.$

Else if $a_n \leq 6b_n\ ,$ then $\left|a_n-6b_n\right| = -a_n + 6b_n$, so $(*) \implies a_n > 5 \frac12b_n\ > 5b_n \quad \forall \ n \geq N.$

So $5b_n < a_n < 7b_n \quad \forall \ n \geq N$, i.e.:

$\frac{a_n}{7} < b_n < \frac{a_n}{5} \quad \forall n \geq N.$

Title: Prove $\sum b_n$ converges if $\sum a_n$ converges and $\lim \frac{a_n}{b_n}=6$

Body:
I recently received this problem on a real analysis exam and I'm still having a hard time understanding it.

"Let $(a_n)$ and $(b_n)$ be sequences of $\mathbb{R}^+$. Suppose $\lim \frac{a_n}{b_n}=6.$ Prove that if the infinite series $\sum a_n$ converges, then $\sum b_n$ converges."

Here's what I understand so far: we can use the Algebraic Limit Theorem to simplify and get $\lim (a_n) = 6*\lim(b_n)$. Also, since we know $\lim \frac{a_n}{b_n}=6$, we know that division by $\lim(b_n)$ is valid, so $\lim(b_n) \ne 0$.

What's confusing me is that there is a theorem that says if an infinite series $\sum s_n$ converges, then $\lim(s_n) =0$. If we attempt the problem and suppose that $\sum a_n$ converges, wouldn't that mean that $\lim(a_n) =0$, so $\lim \frac{a_n}{b_n}=6$ can't occur? What am I missing?

Edit: I understand now that the Algebraic Limit Thm may not come into play with this problem. However, I'm still confused how to use the given information to prove that $\sum b_n$ converges.


问:
$\mathbb{N}$ is not definable in a non-standard model

Let $\mathcal{M}$ be a non-standard model of arithmetic and consider the set of natural numbers $\mathbb{N}$. Show that $\mathbb{N}$ is NOT a definable subset of $\mathcal{M}$.

My attempt so far is to note that since $\mathcal{M}$ is a model in $Th(\mathbb{N})$ then every definable subset must have a least element. Moreover, since definable sets are closed under complement if $\mathbb{N}$ is definable then $M\setminus\mathbb{N}$ should be too, but $M\setminus \mathbb{N}$ is a set of infinite numbers and so, it doesn't have a least element.

What I want to understand is why every definable subset in $Th(\mathbb{N})$ must have a least element?

答:
Let $X \subseteq \mathcal{M}$ be defined by some formula $\phi_X$. Then, since every subset of $\mathbb{N}$ has a least element (it's well ordered) we have

$$\mathbb{N} \models \exists a . \phi_X(a) \land \forall x . (\phi_X(x) \to a \leq x)$$

This is a first order sentence saying that $a$ is a minimal element of $X^{(\mathbb{N})}$.

But now, we find $\exists a . \phi_X(a) \land \forall x . (\phi_X(x) \to a \leq x)$ is in $\text{Th}(\mathbb{N})$, and so any model of $\text{Th}(\mathbb{N})$ must also satisfy it. In particuler, $\mathcal{M}$ models it, and so $X$ has a least element in $\mathcal{M}$.

This is an application of transfer in first order logic. If we know something about some model of $\text{Th}(\mathbb{N})$, and everything in sight is first order, then we can transfer that result to other models of $\text{Th}(\mathbb{N})$ as well. So by proving the theorem for some special choice of model (in this case we chose $\mathbb{N}$, because it's well ordered), we actually get to claim the result for all models.

Indeed, we had a definable set $X$. Then we transferred our question about it $\mathbb{N}$, where the problem was easy to solve, then we transfered the result back to $\mathcal{M}$.

More generally, you can run this argument with any complete theory. You can transfer first order properties from one model to another. This is one of the big ways that model theory is used in "mainstream" mathematics.

I hope this helps ^_^


问:
Probability that leftmost lamp is not chosen and rightmost lamp is chosen
Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?

Here's my thought process. There are $\frac{6!}{3!3!}$ ways to arrange the lamps. Of those $20$ ways, there are $\binom{6}{3}$ ways to turn the lamps on. This gives us $400$ different possibilities. From here, I'm not sure what to do. Help is greatly appreciated.

答: Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, then turns 3 random lamps on. What is the probability that the leftmost lamp on the shelf is red, and the leftmost lamp which is turned on is also red?</p>


问:
Norm and matrix exercise

Let $T \in \mathcal{L}(\mathbb{R^{n}},\mathbb{R^{m}})$ and $A=[a_{ij}] \in \mathcal{M}_{m\times n}$ the matrix asocciated to $T$ (relative to the canonical base). Let $$\mu =: \max_{i \in [m], \, j \in [n]} |a_{ij}|$$ Prove that $$\mu \leq \| T \| \leq \mu \sqrt{mn}$$

答:
For any matrix $A$ the following inequality holds

$$ \|A \|_{\text{max}} \leq \| A\|_{2} \leq \sqrt{mn}\|A\|_{\text{max}} $$

where $\|A\|_{\text{max}} = \displaystyle \max_{ij} |a_{ij}| $. When you set $\mu = \|A \|_{\text{max}}$ we get

$$ \mu \leq \| A\|_{2} \leq \mu  \sqrt{mn}$$

It comes from Hölder's inequality when $p=q=\infty$

Title: Norm and matrix exercise


问:
Closed subset of an affine variety... is it affine?

Preliminaries

So, first of all let me give you the definitions I'm dealing with. Let $k$ be an algebraically closed field, and $\mathbb{A}^n = k^n$.

An affine variety is a closed and irreducible subset of $\mathbb{A}^n$.

Here we endow $\mathbb{A}^n$ with the Zariski topology, so a closed subset is the zero set of an ideal in the polynomial ring $k[x_1,\dots,x_n]$. The irreducibility condition in the definition of affine variety requires that such an ideal is prime.

A k-space is a topological space $X$ together with a sheaf of $k$-algebras $\mathscr{O}_X$. We require that $\mathscr{O}_X$ is a subsheaf of the sheaf of $k$-valued functions on $X$.

An algebraic variety is an irreducible k-space $X$ such that $\exists$ an open cover $X=\bigcup_{i=1}^n U_i$ where each $U_i$ is an affine variety. Moreover we ask the diagonal $\Delta(X)$ to be closed in $X\times X$.

Now we use the term affine variety also for an algebraic variety which is isomorphic as a $k$-space to an affine variety as defined above.

The question

Given a closed subset of an affine variety, is this subset affine?

Remarks

My professor claimed this to be true, and I noticed some people take this fact as granted on this website as well, so I guess this should be true! Nevertheless I don't see why such a closed subset should be irreducible. But the point is: do we really care? My naive understanding of this affine/not affine business is that for an affine variety we can write down something like a global coordinate system (the coordinate ring), while for a general algebraic variety we have just local coordinates, ie we have many affines patched together in a way that cannot be described globally with the usual tools. Hence if a variety is reducible this is not a real obstruction to being affine: we can still find a global coordinate system. The important thing is that all the components lie in the same $\mathbb{A}^n$

As you can see I'm a bit confused... some clarification about these ideas would be really appreciated!

答:
This is just a matter of terminology.  In both books I have to hand (Hartshorne, and Eisenbud's "Commutative algebra..."), the authors define an 'affine algebraic set' to be any subset of $\mathbb{A}^n$ given by the vanishing of polynomials, and an 'affine algebraic variety' to be an irreducible such set.

What is perfectly clear (and is possibly what 'The question' really asks, given the absence of the word 'variety' at the end) is that any closed subset of an affine algebraic set is again an affine algebraic set.

In practice though, the word 'variety' often seems to be used more generally, without the irreducibility requirement.


问:
How is the îto solution calculated?
Currently I'm learning about SDE's. In my document there is a SDE. It is a stochastic BOD model:

$\begin{aligned} \frac{d B_{t}}{d t} &=-K_{1} B_{t}+s_{t}-B_{t} \sigma N_{t} \\ B_{t_{0}} &=B_{0} \end{aligned}$

This is written in terms of a Wiener process as:

$d B_{t}=\left(-K_{1} B_{t}+s_{t}\right) d t-B_{t} \sigma d W_{t}$

$B_{t_{0}}=B_{0}$

From this the Îto formulation is given.

The explaination is:

"since the white noise process in this stochastic model is a mathematical approximation of a noise
process with a relatively short correlation scale, this SDE has to be interpreted in the
Stratonovitch sense. since the Euler scheme (14) can only be used for İto equations, the model
above is first rewritten as an Ito SDE:"

(14):$X_{t}+f\left(X_{t}, t\right) \Delta t+g\left(X_{t}, t\right)\left(W_{t+\Delta t}-W_{t}\right)$

Ito formulation:

\begin{array}{l}
d B_{t}=\left(-K_{1} B_{t}+s_{t}+\frac{1}{2} B_{t} \sigma^{2}\right) d t-B_{t} \sigma d W_{t} \\
B_{t_{0}}=B_{0}
\end{array}

My question is:

How should I've known to interpretate the SDE as Stratonovitch?

答:
Short answer

If your SDE is derived directly from a physic equation involving the derivative term  $\frac{d}{dt}$( or not as often, $\frac{d}{dx}$ ), it is kinda sure that your SDE is in Stratonovitch sense.

Reference

In fact, the choice is casewise, depends largely on the pragmatic aspect of the choice and also the interpretation of the phenomenon through that SDE. More discussion here https://physics.stackexchange.com/questions/245692/it%C3%B4-or-stratonovich-calculus-which-one-is-more-relevant-from-the-point-of-view

Title: How is the îto solution calculated?


问:
Find $\mathbb{E}[X(X-1)]$ for the geometric distribution without using derivation

I'm trying to find the $\mathbb{E}[X(X+1)]$ for the geometric distribution.

Everywhere I've looked explains how to do it using the derivative but I have not been taught that method. Is there another way to do this? I know that $\mathbb{E}[X(X+1)] = E[X^2]+\mathbb{E}[X]$, but I am unable to calculate $E[X^2]$. Any tips would be greatly appreciated.
Thank you in advance!

答:
Note that $$\mathbb{E}[X(X+1)]=\mathbb{E}[X^{2}+X]=\mathbb{E}[X^{2}]+\mathbb{E}[X]$$
Now, we know that $$\mathbb{V}ar[X]=\mathbb{E}[X^{2}]-[\mathbb{E}[X]]^{2}$$
Using that, we can see that $$\mathbb{E}[X(X+1)]=\mathbb{V}ar[X]+[\mathbb{E}[X]]^{2}+\mathbb{E}[X]$$
So, since that  $X\sim \textbf{Geometric}(p)$ so, $$\mathbb{E}[X]=\frac{1}{p} \quad \text{and} \quad \mathbb{V}ar[X]=\frac{1-p}{p^{2}}$$

Other approach is by definition, we can see that
$$\mathbb{E}[X(X+1)]=\mathbb{E}[X^{2}+X]=\mathbb{E}[X^{2}]+\mathbb{E}[X]$$
Now, we know that $\mathbb{E}X=\frac{1}{p}$ and we can see that $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]$$
with $x=1,2,\ldots$ and $\mathbb{P}[X=x]=p(1-p)^{x-1}$.
So, you need to find $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]=\sum_{x=1}^{\infty}x^{2}p(1-p)^{x-1}$$
Using that $0<p\leq 1$, so we can find that
$$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]=\sum_{x=1}^{\infty}x^{2}p(1-p)^{x-1}=\frac{2-p}{p^{2}}$$


Title: Norm and matrix exercise

Body: Let $T \in \mathcal{L}(\mathbb{R^{n}},\mathbb{R^{m}})$ and $A=[a_{ij}] \in \mathcal{M}_{m\times n}$ the matrix asocciated to $T$ (relative to the canonical base). Let $$\mu =: \max_{i \in [m], \, j \in [n]} |a_{ij}|$$ Prove that $$\mu \leq \| T \| \leq \mu \sqrt{mn}$$

Body:
Hints:

$\ \|T\|=\sup_\limits{\|u\|=1,\|v\|=1\\}|u^\top Tv|\ $.

If $\ |a_{st}|=\max \big\{|a_{ij}|: i=1,...,m; j=1,...,n \big\}\ $, $\ u^*_i=\delta_{is}\ $, and $\ v^*_i=\delta_{it}\ $, then $\ |\,u^{*\top} Tv^*\,|=\mu\ $.

By the Cauchy-Schwartz inequalty,
\begin{align}    \|Tv\|^2&=\sum_{i=1}^m\left(\sum_{j=1}^n a_{ij}v_j\right)^2\\
&\le\sum_{i=1}^m\|v\|^2\sum_{j=1}^n a_{ij}^2\ .
    \end{align}


问:
Understanding Example 1 [p. 436] in Munkres Topology (fundamental group of a wedge of circles)

Hi. I'm trying to understand this example from Munkres's book and there is a part that I can't understand what the definition of the $g$ loop is. More precisely, I don't understand "For each $n$, define $g$ on the interval $[1/(n+1),1/n]$ to be positive linear map of this interval onto $[0,1]$ followed by $f_n$"

What would the definition of $g$ look like? what value does the function $g$ take in that interval?
I can't understand this part.
Thanks.

答:
$$g(x) = \begin{cases}
p & x = 0
\\
f_n\left(\frac{x-\frac{1}{n+1}}{\frac{1}{n} - \frac{1}{n+1}}\right)
& x \in \left[\frac{1}{n+1}, \frac{1}{n}\right], n \ge 1
\end{cases}$$

The map $x \mapsto \frac{x-\frac{1}{n+1}}{\frac{1}{n} - \frac{1}{n+1}}$ is the "positive linear map" from $[1/(n+1), 1/n]$ that is surjective onto $[0,1]$.


答:
We define $g_n$ to be the function $g_n: [1/(n+1), 1/n] \to [0,1]$ defined by
$$
g_n(x) = n(n+1)x-n
$$
This is a line of positive slope connecting the points $(1/(n+1)),0)$ and $(1/n,1)$. By design, $g_n$ is bijective.  We can then form the composition $f_n \circ g_n: [1/(n+1), 1/n] \to C_n$ (this is what he means by "followed by $f_n$").  Doing this for all $n$ and patching them together defines the loop $g$ everywhere but $0$, so he defines $g(0)=p$ to complete the loop continuously.

Title: Understanding Example 1 [p. 436] in Munkres Topology (fundamental group of a wedge of circles)

Body:
Hi. I'm trying to understand this example from Munkres's book and there is a part that I can't understand what the definition of the $g$ loop is. More precisely, I don't understand "For each $n$, define $g$ on the interval $[1/(n+1),1/n]$ to be positive linear map of this interval onto $[0,1]$ followed by $f_n$"

What would the definition of $g$ look like? what value does the function $g$ take in that interval?
I can't understand this part.
Thanks.


问:
Recursive function that outputs its own code
This problem is probably a rather trivial one, since I have the impression, that it is a textbook-style one, but nonetheless somehow it won't give in. Here it is:

I have to show that there exists a (unary) recursive function, that has code $c$ and also takes the constant value $c$ (i.e., it outputs its own code).

I am pretty sure, I have to use Kleene's (second) recursion theorem, that says that for a given (total) recursive function $f:\mathbb{N} \rightarrow \mathbb{N}$ there is a number/code $c$ such that $\phi_c=\phi_{f(c)}$ (where $\phi_a$ is the partial recursive function that has code $a$), but I can't figure out how...

答:
You want $\phi_c$ to be "the function which outputs $c$". Hence, you want $\phi_{f(c)}$ also to be a function that outputs $c$. The difference is that you can control $\phi_{f(c)}$ since you control $f(x)$.

So you want to define $f(x)$ to be the code of the constant function $x$ (i.e. returns $x$ for every input). Now Kleene's theorem gives you the following: There exists $c$ such that the function coded by $c$ is exactly the constant function $c$.


问:
Evaluate $\int \frac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$
I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue.
$$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$

$$\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\,\operatorname d\!x &= \int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{1-\cos^2(x)} \\
&=\int \sin^{-2}(x)\,\operatorname d\!x + \int \cos(x)\,\operatorname d\!x - \int \frac{\operatorname d\!x}{\cos(x)}\end{align}$$
Any suggestions how to continue?

Thanks!

答:
$$\int \frac{1+\cos(x)}{\sin^2(x)}dx=\int \frac{dx}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}dx$$

$$=\int \csc^2xdx+\int\csc x\cot xdx=-\cot x-\csc x+C$$

Alternatively, $$\int \frac{1+\cos(x)}{\sin^2(x)}dx=\int \frac{1+\cos(x)}{1-\cos^2(x)}dx=\int \frac{dx}{1-\cos x}$$

$$\text{Use }\cos x=\frac {1-\tan^2\frac x2}{1+\tan^2\frac x2}$$ and put $\tan\frac x2=u$ (Weierstrass substitution formulas)


问:
Determine the linear function $f: R \rightarrow R$
There's a linear function $f$ whose correspondence rule is $f(x)=|ax^2-3ax+a-2|+ax^2-ax+3$. State the values of the parameter $a$ that fully define the function $f$.

According to the problem condition the function $ f (x) $ is linear by
so the coefficient that accompanies $ x ^ 2 $ must be equal to 0, this is
meets when:
\begin{align*}
  ax^2-3ax+a-2 & < 0\\
 a(x^2-3x) &<2-a \\
  \
    a\underbrace{\left(x^2-3x+\frac{9}{4}\right)}_{
  \left(x-\frac{3}{2}\right)^2≥ 0} &<2-a +\frac{9}{4}a\\
  a \left(x-\frac{3}{2}\right)^2 &<2+\frac{5}{4}a\\
  0 & <2+\frac{5}{4}a \\
  -\frac{8}{5} &
\end{align*}

What else is missing to analyze?

答:
For any $a$, in order for the $x^2$ term to vanish, the inequality $ax^2-3ax+a-2\le0$ must hold for all values of $x$. Using your analysis, this is equivalent to

$$a\left(x-\frac32\right)^2\le2+\frac54a$$

If $a > 0$, there is some $x$ large/small enough such that LHS > RHS. Hence $a \le 0$.

Moreover, if $2 + \dfrac54 a < 0$, the above inequality fails for $x = \dfrac32$. Hence $a \ge -\dfrac 85$.

This gives the range $-\dfrac 8 5 \le a \le 0$; indeed in this range, LHS is nonpositive for all $x$ while RHS is nonnegative, so our initial inequality holds, and the function is linear.


问:
At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?
I can't figure out how to properly solve this. I attempted but got confused.

The radius of a sphere is increasing at a constant rate of 0.04 centimeter per second. At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

答:
Well, you have

$$ 4\pi r^2 (0.04\text{cm/sec})=(0.04\text{cm}^3/\text{sec}) $$

so

$$ r^2=\frac{1}{4\pi}\text{cm}^2 $$

Therefore

$$r=\frac{1}{2\sqrt{\pi}}\text{cm}$$


问:
Probability of two independent events...
Suppose $A$ and $B$ are independent with $P(A \cap B') = 0.2$ and $P(A' \cap B) = 0.3$. Find $P(A \cup B)$.

This problem came with this hint: ''there are two correct answers.''

My attempt: $P(A \cup B) = 0.2+0.3+P(A \cap B)$.

Also,  $P(A \cap B) = P(A) \cdot P(B)$.

How can I proceed? I tried drawing the Venn diagram but that didn't get me anywhere.

答:
HINT

We know that $P(A) = P(A\cap B) + P(A\cap B^{c})$ and $P(B) = P(B\cap A) + P(B\cap A^{c})$.

Consequently, one has that
\begin{align*}
\begin{cases}
P(A\cap B) = P(A) - P(A\cap B^{c})\\\\
P(A\cap B) = P(B) - P(B\cap A^{c})
\end{cases}
\end{align*}

Due to the independence between $A$ and $B$, we can rewrite the above-mentioned system as follows
\begin{align*}
\begin{cases}
P(A)P(B) = P(A) - P(A\cap B^{c})\\\\
P(A)P(B) = P(B) - P(B\cap A^{c})
\end{cases} \Longleftrightarrow
\begin{cases}
P(A) - P(A)P(B) = P(A\cap B^{c}) = 0.2\\\\
P(B) - P(A)P(B) = P(B\cap A^{c}) = 0.3
\end{cases}
\end{align*}

After solving such system of equations, you will be able to find the desired probability.


答:
So you have that
$$
V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.
$$
You want to solve for when
$$
\frac{dV}{dt} = \frac{dr}{dt} \Rightarrow \frac{dV}{dt} / \frac{dr}{dt} = 1,
$$
so just solve $4\pi r^2 = 1$.  The constant rate is irrelevent.

Body:
I can't figure out how to properly solve this. I attempted but got confused.

The radius of a sphere is increasing at a constant rate of 0.04 centimeter per second. At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?


答:
If $A$ and $B$ are independent, so are the pairs $A, B'$ and $A', B$.

Suppose $P(A) = p$ and $P(B) = q$. We have

$$p(1-q)=0.2, \quad (1-p)q = 0.3$$

From these equations we have $q-p=0.1$, and thus $p(0.9-p)=0.2$, which is a quadratic in $p$ (hence the two solutions). With that we can find the corresponding $q$ and finish the rest.

Title: Probability of two independent events...

Body:
Suppose $A$ and $B$ are independent with $P(A \cap B') = 0.2$ and $P(A' \cap B) = 0.3$. Find $P(A \cup B)$.

This problem came with this hint: ''there are two correct answers.''

My attempt: $P(A \cup B) = 0.2+0.3+P(A \cap B)$.

Also,  $P(A \cap B) = P(A) \cdot P(B)$.

How can I proceed? I tried drawing the Venn diagram but that didn't get me anywhere.


问:
Show that the localization $\mathbb{R}[x]_{(x-1)}$ is not an integral extension of $\mathbb{R}[x^2 -1]_{(x-1)\cap \mathbb{R}[x^2-1]}$
Let $R' = \mathbb{R}[x]$, $R= \mathbb{R}[x^2 -1] \subset R'$, $P' =  (x-1) \subset R'$, and $P = P' \cap R$.

The given hint suggested to consider the element $\frac{1}{x+1} \in R'_{P'}$.

So the following is what I've tried:

Assume that $\frac{1}{x+1}$ were integral over $R_P$, i.e., there exist $a_{n-1}, ..., a_0 \in R_P$ such that

$(\frac{1}{x+1})^n + a_{n-1} (\frac{1}{x+1})^{n-1}+ ... + a_0 = 0$

Then multiplying both sides with $(x^2 - 1)^n$ gives us

$(x-1)^n + a_{n-1} (x^2-1) (x-1)^{n-1} + ... + a_0 (x^2 - 1)^n=0$

Then plugging in $x=-1$, it follows that $(-2)^n = 0 $,

which is a contradiction.

Here is the point:

I'm not sure if it is okay to plug in $x=-1$, since $R'_{P'}$ is well-defined near $x=1$, not $x=-1$.

Edit: This approach seems to be wrong, since the $x+1$ is at the denominator , plugging in $x=-1$ is not allowed.

Could you give me a hint? Thank you.

答: $b=(x-1)^n + a_{n-1} (x^2-1) (x-1)^{n-1} + ... + a_0 (x^2 - 1)^n$ is an element of $\Bbb{R}[x^2-1]_{(x^2-1)}[x-1]$ which clearly has a well-defined $\Bbb{R}$-valued homomorphism $x\to -1$. It shows that $b\ne 0$ in $\Bbb{R}[x^2-1]_{(x^2-1)}[x-1]$ thus in $\Bbb{R}(x)$ which is a contradiction.