问:
Random variable and independence on unit interval
I have a simple probability question that has been confusing me.

Lets say I generate a random variable $X$ drawn from the uniform distribution on $[0,1]$. Now, lets define two real (nonrandom) numbers $p,q\in[0,1]$.
Define event $A$ to be the event that $X\leq p$.
Define event $B$ to be the event that $X \geq (1-q)$.

My question is, for what values of p and q are A and B independent?

My idea is that these two events will be dependent when there is a nonempty intersection between $X\leq p$ and $X \geq (1-q)$ and independent if this intersection is empty. So my answer is that $A$ and $B$ are independent when $p-(1-q)\leq 0$. I was hoping someone could help me know if I am correct and if this equality in my inequality answer relating $p$ and $q$ should be strictly less than? My confusion with this arises because of this boundary when $p=1-q$.

答: By definition, $A$ and $B$ are independent if and only if $\mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B)$.  In this case, $\mathbb P(A) = p$, $\mathbb P(B) = q$, $\mathbb P(A \cap B) = p - (1-q) = p+q-1$ if $p \ge 1-q$, $0$ otherwise.  So for independence you need $p q = p+q-1$ or $p q = 0$.  But $pq - (p+q-1) = (p-1)(q-1)$.  Thus the condition is that $p=0$ or $p=1$ or $q=0$ or $q=1$.


问:
Is every topological manifold completely metrizable?
Is every (second-countable) topological manifold completely metrizable?

It is known that every smooth manifold possess a complete Riemannian metric, hence in particular it is completely metrizable, however there are non smoothable manifolds.

答:
Express $M$ as a union of open sets $\{U_n\}$ with $clos(U_n)$ compact and contained within $U_{n+1}$ (always possible if $M$ is not itself compact).

Put a Riemannian metric $g_0$ on $M$.  Revise $g_0$ to $g_1$ as follows:  $g_1 = f_1 g_0$ for $f_1$ a positive scalar function obeying:

(a) $f_1 = 1$ on $U_1$ and

(b) on $U_3 - U_2$, $f_1 < 1/\sqrt{d_0(U_2, M - U_3)}$ (where $d_0$ = distance function generated by $g_0$).

Note that for $d_1$ the distance function generated by $g_1$, $d_1(U_2, M - U_3)$ is at least $1$; in particular, any curve starting in $U_1$ and escaping to infinity must have at least length $1$, because it crosses from the boundary of $U_2$ to the boundary of $U_3$, a distance of at least $1$.

Now continue inductively: treating $U_3$, $U_4$, and $U_5$ as we just did $U_1$, $U_2$, and $U_3$ and creating $g_2 = f_2 g_1$, with any curve from $U_1$ escaping to infinity having length at least $2$, as it crosses both from boundary of $U_2$ to $U_3$ (same metric as $g_1$) as well as from boundary of $U_4$ to boundary of $U_5$ (again with distance at least $1$); and so on.

We end with a well-defined metric $g_{\infty} = \lim g_n$.  Any curve escaping to infinity must cross an infinite number of bands of width at least $1$ in $g_{\infty}$, so it must have infinite length in $g_{\infty}$.  That means $(M, g_{\infty})$ is complete.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
2-Norm of the Gradient Mapping in Projected Gradient Descent
In Gradient Mapping, suppose $f(x):\mathbb{R}^d\rightarrow\mathbb{R}$ be a convex and L-smooth function, and $\mathcal{X}$ be a closed, convex, and nonempty set in $\mathbb{R}^d$. Denote the gradient mapping with parameter L as $G_L(x)=L(x-P_\mathcal{X}(x-\frac{1}{L}\nabla f(x)))$, where $P_{\mathcal{X}}$ is the Euclidean projection onto $\mathcal{X}$(in other words, $P_{\mathcal{X}}(x)$ is the vector that minimizes $\|y-x\|_2$ when $y\in\mathcal{X}$).

Denote $x_{k+1}=x_k-\frac{1}{L}G_L(x_k)$. How to prove that $\|G_L(x_{k+1})\|_2\leq\|G_L(x_{k})\|_2$?

I know that when proving $\|\nabla f(x_{k+1})\|_2\leq\|\nabla f(x_{k})\|_2$ in the standard gradient descent, I need to use that fact that $\langle\nabla f(x)-\nabla f(y),x-y\rangle\geq\frac{1}{L}\|\nabla f(x)-\nabla f(y)\|_2^2$ given f is a convex L-smooth function. Then in the project Gradient Descent case, is there an analogous result for $G_L(x)$?

答:
Let's notice that

$x_{k+1} = x_k-\frac{1}{L} G_L(x_k)=P_{\mathcal{X}}(x_k - \frac{1}{L}\nabla f(x_k))$.

The operator applied to $x_k$ on the righthand side is the projected gradient operator, also more generally known as the forward-backward operator. For notational simplicity, let's call it $F=P_{\mathcal{X}}\circ (\textrm{Id}-\frac{1}{L}\nabla f)$, where $\textrm{Id}$ denotes the identity. $F$ has some nice properties, but to show your desired inequality, it would suffice to show the following identity (I'll rewrite it in our notation using a few steps):

\begin{equation}
\begin{aligned}
\|G_Lx_{k+1}\|&=\|x_{k}-x_{k+1}\| & & \\
&=\|x_k-Fx_k\| & & \\
&=\|Fx_{k-1}-F\circ F(x_{k-1})\|&\leq \|x_{k-1}-Fx_{k-1}\|\\
& &=\|x_{k-1}-x_k\|\\
& &=\|G_L(x_k)\|
\end{aligned}
\end{equation}

In particular, it would suffice to show that $F$ is Lipschitz Continuous with a constant of $1$. Next, let us notice two things: firstly, since $P_{\mathcal{X}}$ is a projector onto a closed convex set, it has a Lipschitz constant of $1$ (this is an exercise in convex analysis). Furthermore, by the Baillon-Haddad theorem, $\frac{1}{L}\nabla f$ is firmly nonexpansive (which is the fact you've written), from which (using some tricks from Ch. 4 and Ch. 18 of Bauschke & Combettes' book on convex analysis) it follows that $\textrm{Id}-\frac{1}{L}\nabla f$ also has a Lipschitz constant of $1$.

Finally, since $F$ is the composition of two $1$-Lipschitz operators, it is also $1$-Lipschitz and therefore (*) holds.


问:
Local extremum points and inflection points
Find all values of $a, b$ for which the function $f(x)=a x^{3}+b x^{4}+$ $x^{6}$ has a local minimum at zero.

Can I conclude that $ f$ cannot have local minimum at $x=0$ since $f'(0)=f''(0)=0$ for any $a,b \in \mathbb R$.

答:
$f(x)=ax^3+bx^4+x^6$, then $f'(0)=0$ and  $f^{iii}(x)=6a$ (the first non-zero derivative of ODD order) so at $x=0$ is there is neither max nor min.

The Rule:

If $f'(a)=0$ and  the first non-zero derivative is of EVEN order at $x=a$ and is negative(positive), there is max (min) at $x=a$

If $f'(b)=0$ and the first non-zero derivative is of ODD order, there is neither min nor max at $x=b$.

If all the derivatives vanish at $x=c$, this method fails. Then check if the first derivative changes sign from positive (negative) to negative (positive) while passing through $x=c$, there is max (min) at $x=c$.
If the sign doesn't change there is neither max nor min at $x=c$


问:
Application of the transversality theorem
I am trying to do this question in Bredon's Topology and geometry about using the transversality theorem to show that the intersection of two manifolds is a manifold.

Now it goes as follows:

Let $f(x,y,z)=(2-(x^2+y^2)^{1/2})+z^2$ on $\mathbb{R}^3 - (0,0,z)$. Then one can show that $M=f^{-1}(1)$ is a manifold.

Now let $N = \{ (x,y,z) \in \mathbb{R}^3 | x^2+y^2=4 \}$. I need to show that $M \cap N$ is a manifolds by using the transversality theorem, but I'm not quite sure how to do this.

I thought about finding the tanget planes to each to these manifolds and then showing that there a points in one of them that aren't in the other, but I dont know if this will work. So I was wondering if I could get some hints on this.

Thank you

答:
I couldn't fit my response in the character limit of the comment box.

No, your first function needs to be the function $f$ you started with :) You're correct that I was guiding you just to use the regular value set-up. Of course, this is a special case of the transversality theorem. If you try to study the function $h(x,y,z)=x^2+y^2$ on $M$, you end up with a messy local-coordinates computation, so I always discourage my students from trying to show "directly" that the function $h\colon M\to\mathbb R$ has $4$ as a regular value. You could try to see that the inclusion map $\iota\colon M\to\mathbb R^3$ is transverse to the cylinder $N$: As you surmised, you just need to see that at any point $P\in M\cap N$, $T_PM+T_PN = \mathbb R^3$. The way I was leading you is doing this geometrically, by showing that $\nabla f(P)$ (which is the normal vector to $M$) and $\nabla h(P)$ (which is the normal vector to $N$) are linearly independent at every $P\in M\cap N$.


问:
Proving a set is open (with the condition that $z \neq c$)
Consider some set $A=\{(x,y,z) \in \mathbb R^3 : z \neq f(x,y)\}$, where $f:X \rightarrow \mathbb R$ is a continuous real-valued function and $X$  is a closed subset of $\mathbb R^2$.

Would I be correct in saying that if $z \neq f(x,y)$ then this means that $z > f(x,y)$ or $z < f(x,y)$ which are both open and since the union of open sets is open $\implies z \neq f(x,y)$ is open thus showing that $A$ is open?

答:
That is correct, but incomplete. If you use that approach, then the set$$\{(x,t,z)\in\Bbb R^3\mid z>f(x,y)\}\tag1$$is open because it is equal to $F^{-1}\bigl((0,\infty)\bigr)$ (with $F(x,y,z)=z-f(x,y)$), which is an open set, because $F$ is continuous (which follows from the continuity of $f$) and because $(0,\infty)$ is an open subset of $\Bbb R$. And the set $\{(x,y,z)\in\Bbb R^3\mid z<f(x,y)\}$ is open because it is equal to $F^{-1}\bigl((-\infty,0)\bigr)$, which is an open set, because $F$ is continuous and because $(-\infty,0)$ is an open subset of $(-\infty,0)$.

Of course, it would be simpler to say that $(1)$ is open because it is equal to  $F^{-1}(\Bbb R\setminus\{0\})$ and because $\Bbb R\setminus\{0\}$ is an open subset of $\Bbb R$.


问:
$\mathbb P[a<S\leq b]$ for $S=\sum_{i=1}^{N} X_i$ where $N \sim \mathrm{Poi}$ and $X_i$ iid $\sim \mathrm{Exp}$
Let

$S=\sum_{i=1}^{N} X_i$ where

$N \sim \mathrm{Poisson}(\mu)$

$X_i$ iid $\sim \mathrm{Exponential}(\lambda)$ and independent of $N$

I want to determine $\mathbb P[S=0]$ and $\mathbb P[a<S\leq b]$ for $0\leq a<b$.

I don't know how or where I should or could use the following hint, but the author states to use the series representation of the Bessel function $I_\nu$, i.e.

$$I_\nu(z)=(z/2)^\nu\sum_{k=0}^{\infty}\frac{(z/2)^{2k}}{k!\Gamma(v+k+1)}$$

I would appreciate any help to solve this problem.

答:
A completely different approach (which is why I though a new answer is necessary), is to use some properties of the conditional probability measure ($F_{\Gamma(k,1/\lambda)},f_{\text{Pois}(\mu)}(k)$ denote the CDF of the Gamma-distribution and the PMF of the Poisson-distribution respectively): \begin{align*}\textbf{P}(S\leq x)=\textbf{E}[\textbf{P}(S\leq x|N)]&=\textbf{E}[\textbf{P}(\sum^N_{i=1}X_i\leq x|N)]\\&=\textbf{E}[F_{\Gamma(N,1/\lambda)}(x)]=\sum^\infty_{k=0}F_{\Gamma(k,1/\lambda)}(x)f_{\text{Pois}(\mu)}(k)\end{align*}

In this case, we have used Adam's Law and the fact that the sum of independent identically Gamma-distributed random variables is again Gamma-distributed. This solution requires a bit more knowledge about conditional probabilities. In fact I am not sure if it works (would be great if someone could double check). It does look similar to your hint thought!


问:
Algorithm to get maximum number of n rectangles with given same width and height that fit into a rectangle with a given width and height
I am looking for an algorithm that can return the maximum number of rectangles in the same size with given width and height that fit in to a rectangle with given height and width.

For an example let's say I have a small rectangle with height of 50mm and width of 70mm and a bigger rectangle with height of 1200mm and width of 1000mm. I want to arrange the small rectangles inside the bigger rectangle in a fully utilize way. The small rectangles can be arrange in anyway (Small rectangle can rotated as well), I just want the number of small rectangles that I can put in the bigger rectangle in a maximum utilize way.

Thank you for your help!

答:
In your example, you can form a rectangle of 1,000 x 1,050 leaving 1,000 x 150, then a rectangle of size 150 x 980, leaving 150 x 20, which is less than a small rectangle, and therefore cannot be improved.

In general, this will be hard and involve systematically trying out various combinations.


问:
Using line integrals to find area
I am trying to find the area of a wall.

I have been given that the base is part of the circle of radius 1 centered at the origin, lying in the first quadrant.

I am also given that the height is equal to $$f(x, y) = y + 4x$$
How would I go about solving this?

I know that I have $$x^2 + y^2 = 1$$ and that I want dA using line integrals, but beyond this I'm a bit unsure as to what approach I should use.

Do I want to find $$\int \int y + 4x dx dy$$ with limits $$x = \pm \sqrt(1-y^2)$$ and $$y = \pm 1$$ or am I thinking about this in the wrong way?
Thanks in advance

答:
"The base is part of the circle of radius 1 centered at the origin, lying in the first quadrant" is parameterized as follow

$ \gamma :[0,\pi/2] \to \mathbb{R}^2 , \gamma(t) = (cos(t),sin(t))$

The height is equal to

$  f(x,y) = y+4x $

Then

$\int_{C} f \cdot \ d \gamma  = \int_{0}^{\pi/2} f(\gamma(t))||\gamma'(t)|| \,dt$

Title: Using line integrals to find area


问:
Volume of polygon-based pyramid
I read in a book a few months ago that the volume of a (solid) pyramid with a base that is ANY polygon (I'm not sure if it mentioned it being regular or not) is equal to
$$\frac{1}{3}\times A\times h$$
where $A$ is the area of the base (i.e. of the polygon) and $h$ is the height of the pyramid.

This seems to be true in many cases, such as when the base is a square, a triangle and when the base is a circle (ie the pyramid becomes a cone).

My question is, how can we prove this? I simply have no idea.

Thank you for your help.

答:
For the most general case in which the base doesn't even need to be a polygon, it is an application of Cavalieri's Principle.  Imagine a pyramid with a blob-shaped base and a square pyramid next to it with the same height and whose base has the same area as that of the first pyramid.  Since any level cross-section of the two pyramids would have to have the same area, the volumes must also be equal.

There is a marvelous image showing that the area of a square pyramid whose height is equal to its base size and whose apex is above one corner of the base is 1/3 the volume of a unit cube.  Long story short, you can construct three of those pyramids and see that they can be fit together to form a unit cube.  From there, the only remaining step is to convince yourself that the volume of a pyramid whose base is a unit square is directly proportional to the height of the pyramid.  At that point, you've proved it from a unit square pyramid with unit height to a unit square pyramid of arbitrary height to an arbitrary base and arbitrary height.

Title: Volume of polygon-based pyramid


问:
Find the length of the chord and the distance between parallel chords, given their angles
The radius of a circle is 21.4 meter. Find the length of the chord subtended by a central angle of 110 degrees 40 minutes and the distance between two parallel chords on the same side of the center subtended by central angles 118 degrees 40 minutes and 52 degrees 20 minutes.

Progress:  I know the radius is 21.4 meter, I set up a triangle but my final answer was wrong.

答: the radius of circle is 21.4m find (a) the length of the chord subtended by a central angle of 110°40' and (b) the distance between two parallel chords on the same side of the central subtended by central angles 118°40' and 52°20'</p>


问:
What is the largest volume box that can be shipped?
To send a box, a courier service requires that the length plus twice the width plus the
double the height does not exceed 280 centimeters. What is the largest volume box that can be shipped

I know that volume of a box is $(x)(y)(z)$ and I know that $2x+2y<=280cm$ and I think its using Lagrange multiplier but I don't know how to exactly use that is <=280 I used to use just "="

答:
$V = x y z = (280 - 2 y - 2 z) y z$. But clearly (by symmetry) the solution occurs when $y = z$.  To see this:  there is nothing in the problem that distinguishes the depth $z$ and the width $y$.  We could relabel $y$ with $z$ and the problem would be identical... as is clear from the functional form:  $(280 - 2 y - 2 z) y z$.  Given that there is a single optimum, that demands that $y = z$ at that optimum.

Thus $V = (280 - 4y) y^2$.  Set $\frac{dV}{dy}= 2 (280 - 4 y) y - 4 y^2 = 0$ to find $y = z = 140/3$, and resubstitute to find $x = 280/3$.  Then $V = 5488000/27$.


问:
I have tried to solve it by bulding a equation. But, my answer is differing from the answer given in the book.
A bottle $\frac34$ filled with a liquid weighs $\frac32 \text{ kg}$. The liquid alone weighs $\frac34 \text{ kg}$ more than the empty bottle. If the bottle is completely filled, how will the contents weigh?

答: Let the bottle have weight $x$ and liquid when full have weight $y$. We get $x+\frac{3}{4}y = \frac{3}{2}$ and $\frac{3}{4} y = x + \frac{3}{4}$. Solving gives us $x = \frac{3}{8}$ and $ y = \frac{3}{2}$. If the bottle is completely filled, we have the weight as $x+y = \frac{15}{8}$.


问:
The Functional Equation $ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) $, need solution have answer
Find all functions $ f: \mathbb R \to \mathbb R $ such that for all reals $ x $ and $ y $,
$$ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) \text . $$

I actually got the answer by guessing and checking, $ f ( x ) = a x ^ 2 + b x $, but I want to see the solution. My friend suggested surjectivity but I don't see how to continue.

$ f ( x ) = f \left( \frac x { f ( x ) ^ 2 } \right) $;
this is what I just got.

The answer is correct, just need a solution. Thanks!

答:
For differentiable functions, the only solution is as you have suggested (I can’t tell if differentiability can be dropped). To see this, first let $y=0$ and $x\ne 0$, and you obtain
$$x(f(x)-f(0))=xf(x)\implies f(0)=0\,.$$
Now, assume $xy\ne0$ and rewrite your identity as follows:
$$\frac{f(x+y)-f(y)}{x}+\frac{f(x)}{x}=\frac{f(x+y)-f(x)}{y}+ \frac{f(y)}{y}\,$$
which upon taking the limit as $y\to 0$ on the both sides gives us
$$\frac{f(x)}{x}+\frac{f(x)}{x}= f’(x) +f’(0)\implies f’(x)-\frac{2}{x}f(x)+f’(0)=0\,.$$
Using the integrating factor method, the solution to this becomes $$f(x)=ax^2+f’(0)x\,,$$ for some constant $a$.


问:
Difference of Divergent Series - Question Updated
If $\sum{a_n}$ and $\sum{b_n}$ are divergent infinite series, $\sum(a_n-b_n)$ may converge or diverge.

I am looking for interesting, nontrivial examples of both.

Context: I am teaching high school students with knowledge of differential and integral Calculus, who have recently been introduced to infinite series.

答:
Counterexample: $a_n = b_n$

The source contains useful theorems with regard to convergence and divergence of a sum of series, as well.


问:
How to find "n" by using weak law of large number?
Use the weak law of large numbers to find a number $n$ such that if one tosses a (fair) coin $n$ times the proportion of "heads" will lie in the range $[0.49, 0.51]$ with probability at least $0.95$.

I really have no idea on how to start working on this question as the value of variance is not provided. Can anyone help me for it?

答:
The weak law of large numbers only states that the expectation value $\hat{p}$ of your probability of observing head converges to the true value $p=0.5$ for $n\rightarrow \infty$. In order to calculate confidence intervals you need something more. The repeated toss of fair coins follows a binomial distribution, which is very well approximated by a normal distribution according to the central limit theorem. You then get the 95% confidence interval

$$
\hat{p}\pm 1.96 \sqrt{\hat{p}(1-\hat{p})/n}
$$

In your case $\hat{p} = 0.5$ (according to the weak law of large numbers) and we obtain the required number of tosses $n\approx 10^4$.


问:
Circle in a Square-Geometry
Let us say that there is a circle in a square. The circle has a radius $r$ while the square has a side length of $2r$.

Ok now we know that $\pi=4\frac{A_c}{A_s}$, where  $A_c$ and $A_s$ are the area of the square and the circle, respectively.

Just for fun, I thought I would see if there was a relation between the perimeter and the value of $\pi$. Turns out that $\pi=4\frac{P_c}{P_s}$ where P is perimeter or circumference.

Now we know that $\pi$ is constant so we could set these equal expressions equal to each other
$4\frac{A_c}{A_s}=4\frac{P_c}{P_s}$. Dividing out the $4$, we get that:

$\frac{A_c}{A_s}=\frac{P_c}{P_s}$.

But this doesn't make any sense? How can the perimeters and the areas equal?

Oh and yes, I did come across this after learning a bit about monte carlo methods.

答:
One way to think about it is to consider a small piece of perimeter (circumference), and join its two endpoints to the common centre of the square and the circle.

Notice how the perpendicular distance from the centre to that piece of perimeter is always $r$, for both the square and the circle.

For a piece of the square perimeter of length $s$, the area of the triangle it forms with the centre is

$$a = \frac 12 r\cdot s$$

For a piece of the circumference of arc length $s$, the area of the sector it forms with the centre is

$$a = \pi r^2 \cdot \frac{s}{2\pi r} = \frac 12 r\cdot s$$

So the area each small piece of perimeter (circumference) contributes to the total area is the same for both cases.

This is not saying that "the perimeters and the areas equal", but the ratio of perimeters and the ratio of the areas are equal $\left(\frac4\pi\right)$.

Title: Circle in a Square-Geometry


问:
Why are these angles the same?

I'm doing a problem for an hour now because I'm stuck. I decided to look at the solutions and saw this picture. I can't seem to figure out how the theta angles are the same? Can somebody please explain?

答:
Hint :

Construct the height from point $B$ , can you see which angle is equal to $\theta$?

To conclude, remember that the triangle is a right one.

Title: Why are these angles the same?


答:
If you want to take only positive integer sides into consideration, you can take a value that yields the highest possible quotient with a non-zero remainder. The best thing to do here (I am not that sure, but let it be so :D)is to take the remainder as $1$ , subtract it from the longer side of the larger rectangle and find the remaining number's smallest proper divisor. If the side minus one yields a length that has a prime number as its magnitude, subtract another $1$ from the value you have at hand and do the above steps.
But please take care not to end up in a place where the next dividend you take can be divided by the remainder at hand - that's where you can put a decision point whether to abort that iteration or not.

Another thing is that the smaller side must either be divisible or leave a small remainder when divided by the remainder.

I am not sure if this helps, but I felt this would work. Please do notify me if I did the wrong thing.

Title: Algorithm to get maximum number of n rectangles with given same width and height that fit into a rectangle with a given width and height

Body:
I am looking for an algorithm that can return the maximum number of rectangles in the same size with given width and height that fit in to a rectangle with given height and width.

For an example let's say I have a small rectangle with height of 50mm and width of 70mm and a bigger rectangle with height of 1200mm and width of 1000mm. I want to arrange the small rectangles inside the bigger rectangle in a fully utilize way. The small rectangles can be arrange in anyway (Small rectangle can rotated as well), I just want the number of small rectangles that I can put in the bigger rectangle in a maximum utilize way.

Thank you for your help!


答:
The perimeters and areas are not "equal".

You have two ratios, which are dimensionless numbers, and are equal. Nothing special.

Title: Circle in a Square-Geometry

Body:
Let us say that there is a circle in a square. The circle has a radius $r$ while the square has a side length of $2r$.

Ok now we know that $\pi=4\frac{A_c}{A_s}$, where  $A_c$ and $A_s$ are the area of the square and the circle, respectively.

Just for fun, I thought I would see if there was a relation between the perimeter and the value of $\pi$. Turns out that $\pi=4\frac{P_c}{P_s}$ where P is perimeter or circumference.

Now we know that $\pi$ is constant so we could set these equal expressions equal to each other
$4\frac{A_c}{A_s}=4\frac{P_c}{P_s}$. Dividing out the $4$, we get that:

$\frac{A_c}{A_s}=\frac{P_c}{P_s}$.

But this doesn't make any sense? How can the perimeters and the areas equal?

Oh and yes, I did come across this after learning a bit about monte carlo methods.


问:
Prove that $||x|-|y||\le |x-y|$
I've seen the full proof of the Triangle Inequality
\begin{equation*}
|x+y|\le|x|+|y|.
\end{equation*}
However, I haven't seen the proof of the reverse triangle inequality:
\begin{equation*}
||x|-|y||\le|x-y|.
\end{equation*}
Would you please prove this using only the Triangle Inequality above?

Thank you very much.

答:
$\left| |x|-|y| \right|^2 - |x-y|^2 = \left( |x| - |y| \right)^2 - (x-y)^2 = |x|^2 - 2|x| \cdot |y| +y^2 - x^2 + 2xy-y^2 = 2 (xy-|xy|) \le 0 \Rightarrow \left| |x|-|y| \right| \le |x-y|.$

$"=" \iff xy\ge 0.$ Q.E.D.

The Triangle Inequality can be proved similarly.

Title: Prove that $||x|-|y||\le |x-y|$


答:
Turning the entire figure counterclockwise by $\theta$ moves $OA$ to $OB$ ($O$ is the center of the earth), because the angle between the two lines is $\theta$.

Turning the entire figure counterclockwise by $\theta$ moves "to first sunset" to "to second sunset". Thus the angle between those two lines is also $\theta$.

Title: Why are these angles the same?

Body: I'm doing a problem for an hour now because I'm stuck. I decided to look at the solutions and saw this picture. I can't seem to figure out how the theta angles are the same? Can somebody please explain?


问:
Conceptually(not algebraically), why do I get a quadratic equation when I subtract a linear equation form a quadratic equation.
While simultaneously solving a quadratic equation and a linear equation, I noticed that their intercepts are actually equal to the x intercepts of another quadratic function. I find it hard to grasp that subtracting a linear equation from a quadratic equation results in a symmetric curve(parabola).

I understand that the result of the subtraction will give me a polynomial of degree 2 which forms a parabola, but why?

答:
A higher level understanding of this could come from the study
of affine geometry. In affine geometry all conics are either
ellipses, hyperbolas, or parabolas, and are not changed in
character by any affine transformation. Your notion of

subtracting a linear equation from a quadratic equation

is a special case of an affine transformation and and hence transforms a parabola to another parabola.


问:
Proof $(A-\{a\}) \sim (B - \{b\}) $
I must prove the following:

"let $A$ and $B$ two set, with $A \neq \emptyset$ and $B \neq \emptyset$ and $A \sim B$, and $a \in A$ and $b \in B$, then $(A-\{a\}) \sim (B - \{b\})$

The symbol $\sim$ is used for "equipotency"!!

Thanks in advance!

答:
Let $f:A\to B$ be a bijection. Intuitively you can think of $f$ as a bunch of strings tying each element of $A$ to an element of $B$ in such a way that each element of $B$ is tied to some member of $A$. If one of those strings ties $a$ to $b$ — i.e., if $f(a)=b$ — you can throw away $a$, $b$, and the string tying them together, and what’s left is a bijection from $A\setminus\{a\}$ to $B\setminus\{b\}$. More formally, $g=f\upharpoonright(A\setminus\{a\})$ is bijection from $A\setminus\{a\}$ to $B\setminus\{b\}$. Here’s a rough sketch of this nice situation.

$$\begin{array}{}
\begin{array}{ccc}
A&\overset{f}\longrightarrow&B\\ \hline
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
a&\longrightarrow&b\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet
\end{array}&&&
\begin{array}{ccc}
A\setminus\{a\}&\overset{g}\longrightarrow&B\setminus\{b\}\\ \hline
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\color{white}\bullet&\color{white}\longrightarrow&\color{white}\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\end{array}
\end{array}$$

But we can’t expect to be so lucky: most likely $a$ is tied to something other than $b$, and therefore $b$ is tied to something other than $a$. Suppose that $a$ is tied to $b'$, and $a'$ is tied to $b$; more formally, $f(a)=b'$, and $f(a')=b$. If I throw away $a$, that breaks the string tying $a$ to $b'$ and leaves $b'$ not attached to anything in $A$. And if I throw away $b$, that breaks the string tying $a'$ to $b$ and leaves $a'$ not attached to anything in $B$. All of the other strings remain intact. I now have one loose end coming from $a'\in A$ and another loose end attached to $b'\in B$. I can get rid of both loose ends by tying them together. In other words, I’ll define a function $g:A\setminus\{a\}\to B\setminus\{b\}$ that ties $a'$ to $b'$ and retains all of the unbroken threads from $f$:

$$g(x)=\begin{cases}
f(x),&\text{if }x\in A\setminus\{a,a'\}\\
b',&\text{if }x=a'\;.
\end{cases}$$

And here’s a sketch of what’s going on intuitively.

$$\begin{array}{}
\begin{array}{ccc}
A&\overset{f}\longrightarrow&B\\ \hline
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
a&\longrightarrow&b'\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
a'&\longrightarrow&b\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet
\end{array}&&&
\begin{array}{ccc}
A\setminus\{a\}&\longrightarrow&B\setminus\{b\}\\ \hline
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
&\color{red}\longrightarrow&\color{red}{b'}\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\color{red}{a'}&\color{red}\longrightarrow&\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\end{array}&&&
\begin{array}{ccc}
A\setminus\{a\}&\overset{g}\longrightarrow&B\setminus\{b\}\\ \hline
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\color{red}{a'}&\color{red}\longrightarrow&\color{red}{b'}\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\color{white}\bullet&\color{white}\longrightarrow&\color{white}\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\bullet&\longrightarrow&\bullet\\
\end{array}
\end{array}$$

Title: Proof $(A-\{a\}) \sim (B - \{b\}) $


答:
If the tangents at $A$ and $B$ intersects at point $X$ (just below where the $d$ is on the diagram), then consider the quadrilateral $OAXB$:

$\angle A = \angle B = 90^\circ$ and $\angle O = \theta$, and $\angle O + \angle A + \angle B + \angle X = 360^\circ$, so the external angle of $\angle X$ would also be $\theta$.

Title: Why are these angles the same?


问:
Symbolic definition for $i$?
Please help!

I am trying to find a definition for $i$ that doesn't work for $-i$.

let $j$ be either $i$ or $-i$.

saying $j^2=-1$ doesn't help since $(i)^2=-1$ and $(-i)^2 = -1$

saying $j=\sqrt{-1}$ doesn't halp since $(-1)^{\frac{1}{2}}$ is multivalued $i$ and $-i$

saying $j=e^{i\pi/2}$ doesnt help since $e^{i\pi/2}=i$ but $e^{(-i)\pi/2}=-i$ too (yeah I know I said j=$e^{i\pi/2}$ and not $j=e^{j\pi/2}$ I did that to hopefully make it more intuitive)

saying $j=ln(i)/(\pi/2)$ doesn't work since $i=ln(i)/(\pi/2)$ and $-i=ln(-i)/(pi/2)$ (check with a phase plotter, ln is multivalued but includes i and -i in their respective functions shown above http://davidbau.com/conformal/#log(z)%2F(pi%2F2)-z)

saying $Im(j) > 0$ doesn't work because $Im(z)=Re(z/i)$ and $Re(i/i)>0$ but $Re((-i)/(-i))>0$

Is this parity some sort of law? And yet $i\neq-i$ since $i=(-1)\cdot(-i)$. Unless...

Post Scriptum: Irionically, it is very easy to symbolically represent $1$ vs $-1$ even though $(1)^2=1$ and $(-1)^2=1$, we can just say $x$ such that $x=x^2$

EDIT: can quaternions help? Is $i$ a set of 2 numbers? Are they always equal?

答:
As @DanielFischer commented, $\mathbb C$ has the complex-conjugation automorphism that interchanges the two square roots of $-1$ (however we decide to label them). Thus, any algebraic relation (with real coefficients) that holds for one holds for the other. In different words, $\mathbb R(i)$ is constructed algebraically as $\mathbb R[x]/\langle x^2+1\rangle$, and the image of $x$ is a canonical square root of $-1$ in that model. (But/and, also, the image of $-x$ is another.) But/and I think this is not the type of distinction you want.

It gets worse in the Hamiltonian quaternions: there are infinitely-many square roots of $-1$, and they are all conjugate to each other in the quaternions.

Yes, if we choose to represent complex numbers as the real plane, we can label/name the square root of $-1$ that is in the upper half-plane "$i$". But, as in my first remark, flipping/interchanging upper and lower half-planes is an isomorphism of $\mathbb C$ to itself, and it is essentially impossible to distinguish... so this hasn't really accomplished anything.

For contrast, the case of $\sqrt{2}$ is somewhat different. It is the same, in the sense that the field $\mathbb Q[x]/\langle x^2-2\rangle$ is abstractly made by adjoining a square root of $2$ to $\mathbb Q$, because the image of $x$ in the quotient is a sort of canonical $\sqrt{2}$ in that model. BUT the distinction that matters in practice is that $\mathbb Q(\sqrt{2})$ admits two different imbeddings into $\mathbb R$, and in one the abstract square root of $2$ goes to $1.414...$ while in the other it is the negative of that. The "standard/canonical" square root of $2$ we usually refer to is actually the real number $1.414...$, rather than an abstract one.

Title: Symbolic definition for $i$?


问:
Prove that $f(x)$ have a maximum at $x=0.5$ for $0\leq x\leq 1$
It's an idea that I cannot substantiate. I want to show :

Let $a,b>0$ such that $a+b=1$ and $0\leq x\leq 1$ then we have :
$$f(x)=(ax+b(1-x))^{2(a(1-x)+bx)}+(bx+a(1-x))^{2(b(1-x)+ax)}=g(x)+h(x)
\leq 1.$$

So we examine :

$$f(x)=(ax+b(1-x))^{2(a(1-x)+bx)}+(bx+a(1-x))^{2(b(1-x)+ax)}=g(x)+h(x)$$

The problem is :

$$f(x)\leq f(\frac{1}{2})$$

So the idea is : prove that $x=0.5$ is a maxima .

Following a the natural way  we differentiate $f(x)$ and My claim is the following : the  derivative of $f(x)$ is positive for $0\leq x \leq 0.5$ and negative for $0.5\leq x \leq 1$ .

Inspecting the behavior of $f(x)$ it seems that for some value $a,b$ (for example $a=0.6$) the functions $g'(x)$ or $h'(x)$ have an maxima on $I=[0,1]$ . I mean in fact that on a part of this interval $I$, $g'(x)$ is increasing while $-h'(x)$ is decreasing wich is an advantage . Remains to examine the second derivative of $g'(x)$ or $h'(x)$ to find this local extrema  . Maybe we can use Lambert's function .

Also to prove the claim we can use a line such that :$f'(x)\leq ux+v$ and $f'(x)\geq ux+v$ if we are on $[0.5,1]$ or $[0,0.5]$.

Update 30/10/2020 :

For $0\leq x\leq 1$ and $0.6\geq a \geq b \geq 0.4$ such that $a+b=1$ we have the following approximation :

$$g(x)=(ax+b(1-x))^{2(a(1-x)+bx)}\simeq p(x)=x(g(1)-g(0))+g(0)$$

$$h(x)=(bx+a(1-x))^{2(b(1-x)+ax)}\simeq q(x)=x(h(1)-h(0))+h(0)$$

So we have :

$$f'(x)=g'(x)+h'(x)=g(x)u(x)+h(x)v(x)\simeq p(x)u(x)+q(x)v(x)$$

Update 31/10/2020

I start with a particulare case let $a=0.6$ and $b=0.4$

First fact $(1)$ :$0.33<0.4^{1.2}$

Second fact $(2)$ :

We have for $0\leq x \leq 0.5$

$$l(x)=2x(0.5-0.33)+0.33\leq (0.6 x+(1-x)0.4)^{2 (0.4 x+(1-x)0.6)}$$

Third fact $(3)$ :

$$0.67>0.6^{0.8}$$

Fourth fact $(4)$ :

We have for $0\leq x \leq 0.5$ :

$$t(x)=2x(0.5-0.67)+0.67\geq (0.4 x+(1-x)0.6)^{2 (0.6 x+(1-x)0.4)}$$

With facts $(2)$ and $(4)$ we have fact $(5)$ for $0\leq x \leq 0.5$:

$$f'(x)=g'(x)+h'(x)=g(x)u(x)+h(x)v(x)\geq g(x)l(x)+t(x)v(x)\geq 0$$

My question: How to prove the claim?

答: And why not substituting directly $b = 1-a$? If we do that, we want to find the global maximum for $f(a) = a^{2(1-a)} + (1-a)^{2a} = e^{2(1-a) \ln a}+e^{2a \ln (1-a)} $. If you plot this function, you'll see that the maximum is attained at $a = 0, \frac 12, 1$. ($f(0)$ and $f(1)$ are to be understood as a continuous extension).


问:
Simplifying sums over distinct indices
I'm currently working quite a bit with sums of the form

$$S = \sum_{a = 1}^N \sum_{b = 1}^N  \sum_{c = 1}^N \mathbf{I} \left[ a, b, c \text{ all distinct} \right] f (a, b, c) $$

where $f$ often has some symmetries with respect to its arguments.

For my purposes, it is often useful to rewrite such expressions in the form

$$S = \sum_{a = 1}^N \sum_{b = 1}^N  \sum_{c = 1}^N f (a, b, c) - 3 \sum_{a = 1}^N \sum_{b = 1}^N f(a, b, b) + 2\sum_{a = 1}^N  f (a, a, a)$$

(the above holds when $f$ is symmetric in its arguments) and similar.

My question is whether there is a good reference for simplifications of this form.

If it comes to it, I can derive these expressions carefully by hand by applying the inclusion-exclusion principle, and so on, but by the stage that I'm working with quadruple sums, I don't trust my own calculations as much. It would be useful to have a reference text against which to check these calculations.

答:
Firs, let us drop the symmetry assumption. Let $f$ have $n$ arguments. This can be solved using Möbius inversion over the poset of partitions of $n$, ordered by refinement (see the last example). It turns out there is a nice formula for this, which leads to a nice formula when the symmetry assumption is added back in.

Specifically, assuming symmetry, your sum ranges over all integer partitions $\lambda$ of $n$, where the size of each part indicates a number of arguments constrained to be equal. Given a partition, let $m_i$ be the number of occurrences of $i$ (the multiplicity of $i$). Then the coefficient of the summation corresponding to $\lambda$ is
$$
(-1)^{n-(m_1+\dots+m_n)}\frac{n!}{\prod_{i=1}^n i^{m_i}\cdot m_i!}
$$
Coincidentally (?), the absolute value of this coefficient is the number of permutations of $\{1,\dots,n\}$ whose cycle structure is $\lambda$.

This agrees with your example:
$$
\begin{array}{l|l|c}
\text{Summation} & \lambda & \text{Coefficient}\\\hline
\sum\sum\sum f(a,b,c) & (1,1,1)&(-1)^{3-(3)}\frac{3!}{1^3\cdot 3!}=1
\\
\sum\sum f(a,a,b) & (2,1)&(-1)^{3-(1+1)}\frac{3!}{1^1\cdot 1!\cdot 2^1\cdot 1!}=-3
\\
\sum f(a,a,a) & (3)&(-1)^{3-(1)}\frac{3!}{3^1\cdot 1!}=2
\end{array}
$$

I found a self-contained proof of why this works! Let's deal with a $d$-dimensional sum, your post was $d=3$. Furthermore, I will not assume $f$ is symmetric at first, then add that assumption in at the end.

Let $V=\{1,\dots,N\}^d$ be the total set of index vectors, and let $W$ be the set of index vectors where the indices are distinct. Furthermore, given $v\in V$ and $\pi \in S_d$, let $\pi(v)$ denote the vector obtained by permuting the indices of $v$ according to $\pi$. I claim that
$$
\sum_{w\in W}f(w)=\sum_{\substack{v\in V,\pi \in S_d\\\pi(v)=v}}(-1)^{\text{sign}(\pi)} f(v)
$$
To be clear, the second summation is over all ordered pairs $(v,\pi)$ which satisfy $\pi(v)=v$. The idea is that for any $v$ with repeated indices, the permutations fixing $v$ come in pairs with opposite sign, so the net contribution of $v$ is zero, while permutations with distinct indices appear exactly once as $(v,\text{id})$.

Therefore, letting $P$ range over partitions of $V$ and $V_P$ be the set of index vectors such that $v_i=v_j$ when $i$ and $j$ are in the same part of $P$, then
\begin{align}
\sum_{w\in W}f(w)
  &=\sum_{\pi}(-1)^{\text{sign}(\pi)}\sum_{\pi(v)=v}f(v)
\\&=\sum_{P} (-1)^{n-|P|}(\text{# $\pi$ w/ cycle structure $P$})\sum_{v\in V_P}f(v)
\\&=\sum_{P} (-1)^{n-|P|}\left(\prod_i (P_i-1)!\right)\sum_{v\in V_P}f(v)
\end{align}
Finally, if we add in the assumption that $f$ is symmetric, then letting $V_{\lambda}$ be the set of vectors where the first $\lambda_1$ are equal, and the next $\lambda_2$ are equal, etc, and letting $m_1,m_2,\dots,$ denote the multiplicities of this $\lambda$, then
$$
\begin{align}
\sum_{w\in W}f(w)
&= \sum_{\lambda}(-1)^{n-\text{len}(\lambda)}\left(\prod_i (\lambda_i-1)!\right)(\text{# $P$ w/ shape $\lambda$})\sum_{v\in V_\lambda}f(v)
\\&= \sum_{\lambda}(-1)^{n-\text{len}(\lambda)}\left(\prod_i (\lambda_i-1)!\right)\binom{d}{\lambda_1,\lambda_2,\dots}\frac1{m_1!m_2!\cdots}\sum_{v\in V_\lambda}f(v)
\\&= \sum_{\lambda}(-1)^{n-\text{len}(\lambda)}\frac{d!}{\prod_i i^{m_i}\cdot m_i!}\sum_{v\in V_\lambda}f(v)
\end{align}
$$

Title: Simplifying sums over distinct indices


问:
Baby Rudin Exercise 11.15
I am working through the last chapter of baby Rudin I have a question about exercise 11.15. The question is

Let $\mathfrak{R}$ be the ring of all elementary subsets of $(0,1]$. If $0 < a \leq b \leq 1$ define
$$\phi([a,b]) = \phi([a,b)) = \phi((a,b]) = \phi((a,b)) = b-a $$
but define
$$\phi((0,b)) = \phi((0,b]) = 1+b $$
Show that this gives an additive set function $\phi$ on $\mathfrak{R}$, which is $\textbf{not regular}$ and which cannot be extended to a countably additive set function on a $\sigma$-ring.

I have been looking at solutions, such as here, but I am having issues with the way it is proved that $\phi$ is not regular and I cannot think of an alternative proof. They say that $\phi$ is not regular because if we take a set like $(0, 0.5]$ then this cannot be approximated from below by a closed set because a closed set cant contain $0$ as the end point. But, isn't $(0, 0.5]$ closed in $(0,1]$? So it seems that this argument doesn't work. But, I also cannot think of an alternative argument to show that $\phi$ isn't regular.

答:
Here is a screen shot of the page where Rudin proves that every regular set function  can be
extended to a countably additive set function.

I really don't understand where did the "compactness" assumption came from, but it is surely being used!

Moral:  I think that Definition (11.5) of regularity must be amended by requiring that the set $F$ mentioned there be compact, rather than just closed.  Otherwise Theorem (11.8) does not hold, and a counter example is precisely the one described by the OP!

Title: Baby Rudin Exercise 11.15


问:
Bhaskara approximation of cos(x)
In my knowledge the best aproximation of $\sin$ and $\cos$ is Bhaskara approximation which is :
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
and
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$

what is the best approximation of $\cos(x)$ for $ -\pi\leq x\leq 0$ (by a regular function ) ?

答:
There is no such thing as a "best" approximation for a function, there are always minor tweaks you can make to produce better approximations.

However, to your question, you can massage the approximation you gave by noticing that $$-\pi\leq x\leq 0 \iff -\frac \pi2\leq x+\frac \pi2\leq \frac \pi2$$

Title: Bhaskara approximation of cos(x)


问:
Convergence of oscillating sequence
I was trying to do a problem to prepare for my Real Analysis test but i got stuck. The problem states the following:

Suppose that $x_{n} \geq 0$ for all $n \in \mathbb{N}$ and that $\lim \left((-1)^{n} x_{n}\right)$ exists (Let's suppose $\lim \left((-1)^{n} x_{n}\right) = \alpha$). Show that $\left(x_{n}\right)$ converges.

What have i tried so far? My most promissing attempt went like this:

Define $z_{n}:=(-1)^nx_{n}$ We already know that $z_n \xrightarrow{\
} \alpha$. If we can prove that both subsequences $z_{2n}$ and $z_{2n - 1}$ converge to $\alpha$ we'll be done. For that, consider the following subsequence: $$z_{2n} = (-1)^{2n}x_{2n} = x_{2n}$$ We can say that $\lim x_{2n} = \alpha$ because $z_{2n} \xrightarrow{\
} \alpha$ as every subsequence of $z_{n}$ should converge to the same limit $\alpha$.

On the other hand, consider the subsequence: $$z_{2n-1} = (-1)^{2n - 1}x_{2n-1} = -x_{2n - 1}$$ We know that $\lim z_{2n-1} = \lim -x_{2n - 1} = -\lim x_{2n-1} = \alpha$. Therefore, $\lim x_{2n-1} = -\alpha$.

However, i wanted $$\lim x_{2n-1} = \alpha$$ for $\alpha$ bigger than zero in order to finish the proof, because then both odd and even subsequences would be converging to the same limit. Can anyone help me finish this exercise? If this idea cannot be fixed, can anyone suggest a solution?

Thanks in advance!

答:
Suppose $\{(-1)^n x_n\}$ is convergent to $x>0$.

$\implies \forall\  \epsilon \ \exists\  N_0\  \forall n>N_0,\ |(-1)^n x_n-x|<\epsilon $
$\implies \forall n>N_0,\ x-\epsilon< (-1)^n x_n<x+\epsilon$.
Choosing $\epsilon$ in such a way that $x-\epsilon>0$ we get that

$\forall n>N_0,\ 0<x-\epsilon< (-1)^n x_n$.

But this is impossible as for odd $m>N_0$, $(-1)^m x_m<0$. $\therefore x\not>0$
Similarly we can prove that $x\not<0$.

Therefore $x = 0$.Therefore $(-1)^nx_n \to 0$

$\implies \forall\  \epsilon \ \exists\  N_1\  \forall m>N_1,\ |(-1)^m x_m|<\epsilon/2,\exists\  N_2\  \forall n>N_2\ |(-1)^n x_n|<\epsilon/2 $.

So for $\forall m,n>\max(N_1,N_2), |(-1)^m x_m|<\epsilon/2, |(-1)^n x_n|<\epsilon/2$

Thus $\forall\  \epsilon \ \exists\  N\  \forall m,n>N,\ |x_m-x_n|<|x_m|+|x_n|<\epsilon $.

Therefore $x_n$ is cauchy. And since Cauchy sequences in $\mathbb R$ are convergent, so $x_n$ is convergent.

Now if $x_n \leq 0, Let z_n=-x_n\geq 0$, by above argument $z_n$ converges which in turn implies that $x_n$ converges.

Title: Convergence of oscillating sequence


问:
Decomposition of a stochastic matrix
Let $T$ be a square row (not doubly) stochastic matrix. Assume that $T$ is diagonalizable. Then write $T$ as $P^{-1}DP$. Consider now the matrices $D_1=P^{-1}D^{\frac{1}{2}}P$ and $D_2=P^{-1}D^{\frac{1}{2}}P$ whose product is the initial stochastic matrix $T$. Can you prove that $D_i$ is stochastic for $i=1,2$? Note that the converse is true, that is if $D_i$ were stochastic for $i=1,2$ then also $T$ would be stochastic.

答:
The statement here is False.

Let $T$ be any $2\times 2$ stochastic matrix such that $\text{trace}\big(T\big) \lt 1$.  Then $T$ has $\lambda_1 =1$ and $\lambda_2 \lt 0$, both real and distinct so $T$ is diagonalizable.  $T=P^{-1}DP$

It is true that $\mathbf 1$ is still an eigenvector of

$T^\frac{1}{2}= D_1 =P^{-1}D^\frac{1}{2}P$

(and I assume $\sqrt{\lambda_1}=1$)

that is $T^\frac{1}{2}\mathbf 1 = \mathbf 1$

What is also true is that a stochastic matrix is defined as having real non-negative components.  But
$T^\frac{1}{2}$ has a natural eigenvalue of $\sqrt{\lambda_1} = 1$  and an unnatural eigenvalue given by $\sqrt{\lambda_2}$, which is non-zero and purely imaginary, irrespective of branch chosen.  Thus for some $b\in \mathbb R-\{0\}$ we have

$1 +b\cdot i\notin\mathbb R=\text{trace}\big(T^\frac{1}{2}\big)\implies T^\frac{1}{2}\text{has non-real diagonal}\implies T^\frac{1}{2}\text{not stochastic}$

Title: Decomposition of a stochastic matrix


问:
X-sided dice probability
I have an $x$-sided dice and I roll it $y$ times.

What is the probability that I will see at least $z$ of the $x$ sides?

e.g. If I rolled a $6$-sided dice $8$ times, what is the probability I see at least $4$ of the sides?

But I want to do with much bigger numbers

Specifically, I want to know  what is probability I see at least $z$ of the $160,000$ sides if I rolled a $160,000$-sided dice $25,000$ times.

Thanks in advance!

答:
The chance of a side missing out is $$\left({x-1\over x} \right)^y$$

The chance of two sides missing out is $$\left({x-2\over x}\right)^y$$
The usual approximation by an exponential causes a large error.

I will use the binomial distribution.

On average, you will see $$\mu\approx x(1-e^{-y/x})$$ different sides.  The expected value of the square is
$$x(1-\left({x-1\over x}\right)^y)+ \\
(x^2-x)(1-2\left({x-1\over x}\right)^y+\left({x-2\over x}\right)^y) $$
So the variance is
$$\sigma^2= x(\left({x-1\over x}\right)^y-\left({x-2\over x}\right)^y \\
+x^2(\left({x-2\over x}\right)^y-\left({x-1\over x}\right)^{2y})$$
This is about 1500 for your numbers, compared with 19000 when I neglected the x^2 term.  As I mention in comments, this is close to $$\sigma^2\approx xe^{-y/z}-(x+y)e^{-2y/x}$$
For large $x$ and $y$, it will be very close to a normal distribution with that mean and variance.  Sixty-eight percent of the time $z$ will be between $\mu-\sigma$ and $\mu+\sigma$ (note $\sigma$ is the square-root of $\sigma^2$ defined above), and ninety-five percent of the time it will be between $\mu-2\sigma$ and $\mu+2\sigma$.

Title: X-sided dice probability


问:
Proof that $\sin x=4^{-x/\pi}\sum_{n=0}^\infty (-1)^n\binom{4x/\pi}{2n+1}$ for $x\ge 0$
Is there any proof of the identity
$$\sin x=4^{-x/\pi}\sum_{n=0}^\infty (-1)^n\binom{4x/\pi}{2n+1}$$
where $x\ge 0$?

If we replace $4x/\pi$ by $x$ and write the binomial in terms of the falling factorial $(x)_n$, we get
$$\sum_{n=0}^\infty (-1)^n \frac{(x)_{2n+1}}{(2n+1)!},$$
which is remarkably similar to the power series for $\sin x$ centered at $0$:
$$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}.$$

答:
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\left.\sin\pars{x}
\vphantom{A^{A}}\,\right\vert_{\ x\ \geq\ 0}\,\,\, =\,\,\,
4^{-x/\pi}\sum_{n = 0}^{\infty}\pars{-1}^{n}
{4x/\pi \choose 2n + 1}}:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{4^{-x/\pi}
\sum_{n = 0}^{\infty}\pars{-1}^{n}{4x/\pi \choose 2n + 1}} =
-\,4^{-x/\pi}\,\,\,\ic\sum_{n = 0}^{\infty}\ic^{2n + 1}
{4x/\pi \choose 2n + 1}
\\[5mm] = &\
-\,4^{-x/\pi}\,\,\,\ic\sum_{n = 1}^{\infty}\ic^{n}
{4x/\pi \choose n}{1 - \pars{-1}^{n} \over 2} =
4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}\ic^{n}
{4x/\pi \choose n}
\\[5mm] = &\
4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}\ic^{n}
\,{\Gamma\pars{4x/\pi + 1} \over
\Gamma\pars{n + 1}\Gamma\pars{4x/\pi - n + 1}}
\\[5mm] = &\
4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}\ic^{n}\
{\Gamma\pars{4x/\pi + 1} \over \Gamma\pars{n + 1}}
\ \times
\\[1mm] &\ \phantom{4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}}
{\Gamma\pars{n - 4x/\pi}\sin\pars{\pi\bracks{n - 4x/\pi}} \over \pi}
\\[5mm] = &\
-\,{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\,\Im\sum_{n = 1}^{\infty}\pars{-\ic}^{n}\
{\Gamma\pars{n - 4x/\pi}\Gamma\pars{4x/\pi + 1} \over \Gamma\pars{n + 1}}
\\[5mm] = &\
-\,{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\,\Im
\sum_{n = 1}^{\infty}\pars{-\ic}^{n}
\int_{0}^{1}t^{n - 4x/\pi - 1}\,\,\pars{1 - t}^{4x/\pi}
\,\,\,\dd t
\\[5mm] = &\
-\,{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\,\Im
\int_{0}^{1}t^{-4x/\pi - 1}\,\,\pars{1 - t}^{4x/\pi}
\sum_{n = 1}^{\infty}
\pars{-\ic t}^{n}\,\,\,\dd t
\\[5mm] = &\
{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\
\underbrace{\int_{0}^{1}
{t^{-4x/\pi}\,\,\pars{1 - t}^{4x/\pi} \over
1 + t^{2}}\dd t}
_{\ds{4^{x/\pi}\,\pi\,\csc\pars{4x}\color{red}{\sin\pars{x}}}}
\\[5mm] = &\
\bbx{\sin\pars{x}} \\ &
\end{align}

The last integral is straightforward evaluated with the change $\ds{{1 - t \over t} \mapsto t}$.


问:
Number of characteristic polynomial?
By theorem (it takes too much space to write their proof so I just show it's conclusion),

characteristic polynomial of a matirx $A \in M_{2*2}(\mathbb{Z}_{2})$ is

$$f(t)=t^2-\operatorname{tr}(A)t+\det(A)$$

Then determine the number of distinct characteristic polynomial.

The answer is 4 and the reason is that the coefficient could be 0 or 1.

I can't figure out what the coefficient is. It means $\operatorname{tr}(A)$? Then why it could be 0 or 1?

Title: Number of characteristic polynomial?

Body:
$$f_1(t)=t^2$$
$$f_2(t)=t(t+1)$$
$$f_3(t)=t^2+1$$
$$f_4(t)=t^2+t+1$$


问:
Number of paths for random walk with n steps and an absorbing barrier at -1
I have a random walker that starts at $x = 0$. They can step either left or right one unit. However, if they reach $x = -1$ they fall off a cliff (say), so those paths are invalid -- I've heard this called an absorbing barrier. How do I count the number of possible valid paths if the walker has taken $n$ steps?

As a bonus question, how can I count the number of valid paths which end at $x  =0$?

答:
$$
\begin{array}{|r|c|c|}
\hline
&\text{Without absorbing barrier}&\text{With absorbing barrier at $-1$}
\\\hline
\text{Number of paths of length $n$}&2^n&\displaystyle\binom{n}{\lfloor n/2\rfloor}
\\\hline
\text{Number of paths of length $n$ ending at $0$*}&\displaystyle\binom{n}{n/2}&\displaystyle\frac1{n/2+1}\binom{n}{n/2}
\\\hline
\end{array}$$
${}^*$Assuming $n$ is even.

I present the data this way to highlight the strange coincidence that the lower left entry and the upper right entry are the same (when $n$ is even).


问:
Single transformation equivalent with the following composition
I always have trouble visualizing these as I am trying to figure out the single transformation that is equivalent to the below compositions.

A 90 degree rotation about the origin, followed by a 90 degree rotation about the point (0,6).

    R
    (0,6),90°

  ∘

    R
    (0,0),90°

My idea is to break the two rotations into two reflections each (so 4 reflections). I am assuming two of the reflections would cancel and be left with two reflections, which would be a single rotation, but I am stuck on what reflections they break down to.

I think I am on the right track, but who knows. Any help would be very much appreciated.

答:
I claim that the first composition is the $180^\circ$ rotation around the point $(3,3)$.

After the first rotation (the rotation around the origin), we have $(x,y) \mapsto (-y,x)$.

Now, we want to rotate the point $(-y,x)$ $90^\circ$ around (0,6). The vector pointing from $(0,6)$ to $(-y,x)$ is
$$ \vec{v} = (-y,x-6) $$
To rotate this vector $90^\circ$, we do the same transformation as before (I mean $(a,b) \to (-b,a)$), so the rotated vector is
$$ \vec{w} = (6-x,-y) $$
We should be thinking of this vector based at the point $(0,6)$, so its endpoint is at
$$ (0,6) + \vec{w} = (6-x,6-y) $$

So in conclusion, the composition of these two rotations sends any point $(x,y)$ to the point $(6-x,6-y)$. Notice that the only point which is fixed is $(3,3)$. You can see this just by solving the equations $6-x=x$ and $6-y=y$. Since there is only one fixed point, this can't be a reflection (reflections fix all the points on a line), so it must be some rotation around $(3,3)$. You can do a little more work to see that it's a 180-degree rotation.


问:
If $S_{1},S_{2}$ be subsets of complex numbers then $\inf\lbrace |z_{1} - z_{2}| \rbrace > 0\; (z_{1} \in S_{1}\text{ and }z_{2} \in S_{2})$.
Problem: "Suppose that $S_{1}$ be a compact subset of $\Bbb C$ (complex numbers) and  $S_{2}$ be a closed subset of $\Bbb C$(complex numbers). If $S_{1} \cap S_{2} = \varnothing$, then show that $\inf\lbrace |z_{1} - z_{2}| \rbrace  > 0\; (z_{1} \in S_{1}\text{ and }z_{2} \in S_{2})$".

I supposed that $A=\lbrace|z_{1} - z_{2}| : z_{1} \in S_{1} , z_{2} \in S_{2} \rbrace $ and $a = \inf A$.

If $a \in A$ then $a > 0$.
For "If $a \notin A$" I can't show that $a > 0$. Can you help me? Or have you another way for this problem? Can I solve this problem using Heine-Borel theorem?

答: Suppose that $\inf\{|z-y|:\textrm{ }z\in S_1, y\in S_2\}=0$, so we can find a sequence $(z_n)$ on $S_1$ and a sequence $(y_n)$ on $S_2$ such that $|z_n-y_n|\rightarrow 0$, now use the compactness of $S_1$ and that $S_2$ is closed to show that there is a point on $S_1\cap S_2$ this show the result.


问:
Matrix exponential when $(Av.v) \leq -(v.v)$
I have a question from a previous homework that I never ended up figuring out:

If $(Av.v) \leq -(v.v)$ for all $v \in V$ (where $V$ is finite dimensional vector space with an inner product), show that $e^{tA}(v)$ converges to $0$ as $t \to \infty$ for all $v\in V.$

What I think will be useful is:

$(Av.v) \leq (v.v) \implies ((A+I)v.v) \leq 0.$ Also, since $A$ and $I$ commute, then $e^{t(A+I)} = e^{tA}\circ e^{tI}.$ So, showing $e^{t(A+I)}(v)$ converges to zero might be the way to go.

To do this, I can show $||e^{t(A+I)}(v)|| \to 0$ which means $(e^{t(A+I)}v.e^{t(A+I)}v) = 0.$

This is where I get stuck because I have tried manipulating the sum, but I don't know how to deal with the $(A+I)^k(v)$ term. Do I need to use the matrix exponential sum anyways? Or can I used properties of the matrix exponential/inner products/adjoints/etc to prove this?

答:
One approach is to show that $A$ must have eigenvalues with negative real part and proceed. See this post for details.

Alternatively, we have
$$
\frac d{dt} \left\|e^{At}v \right\|^2 = 2 \left( e^{At}v,\frac d{dt} e^{At}v \right)
= 2 \left( [e^{At}v],A [e^{At}v] \right) \leq -2 (e^{At}v,e^{At} v) = -2 \left\|e^{At}v \right\|^2.
$$
Thus, $\|e^{At}v\|$ is either $0$ or strictly decreasing for all $t$. The conclusion follows.


问:
How to find the change of basis matrix using inconvenient vectors
I have the linear transformations T:
Vector transformations

I know how to find the matrix that induces the transformation using the standard basis vectors (i.e. RREF the vectors = standard basis vector, and apply the transformation to them)

What I can't figure out is how to find the matrix using the change of basis matrix (i.e. find B in A = SBS^-1). When I was taught, I always used convenient vectors (on the plane or perpendicular to it), but never with other vectors

I know that for A = SBS^-1, S is
$$
    \begin{bmatrix}
    1 & -2\\
    -1 & -1\\
    \end{bmatrix}
$$
And S^-1 is
$$
    \begin{bmatrix}
    1/3 & -2/3\\
    -1/3 & -1/3\\
    \end{bmatrix}
$$

But I don't understand how to find B.

Would appreciate any help I can get.

答:
If you know $A$, $S$, $S^{-1}$ and are given $A = S B S^{-1}$, you can find $B$ by transforming both sides of the equation: left-multiply by $S^{-1}$ and right-multiply by $S$. You will obtain $ S^{-1} A S =  \underbrace{S^{-1} S}_{=I} B \underbrace{S^{-1} S}_{=I} = B$, which gives you $B$.

P.S. I hope this is what you were having a problem with (if I understood correctly).


问:
Surjective when A is smaller than B
Wikipedia uses this picture for injective and surjective functions.

The top right is non-surjective. I have added a mapping on, is this now surjective? Why/why not?

答: Some authors would call it a surjective relation. However, very few (but never too few) would call it a surjective function (or mapping) because the common definition of function would be that each element in the domain is related to exactly one element of the codomain.

Title: Surjective when A is smaller than B


答: No, because it's not a function. A function can only map each number in the domain to exactly one element in the codomain. Since in the top right, you've drawn 3 to map to both C and A, it ceases to be a function.</p>

Title: Surjective when A is smaller than B

Body:
Wikipedia uses this picture for injective and surjective functions.

The top right is non-surjective. I have added a mapping on, is this now surjective? Why/why not?


问:
Club sets of $\omega_1$ have the same order type as $\omega_1$
This is a short lemma that I've come up with that I need to prove in a solution to a homework problem. I'm sure that it's true, but I'm not sure how to prove it.

What I need to do is show that if $C \subset \omega_1$ is club, then there is an increasing bijection $f:\omega_1 \to C$.

So I should try to inductively construct this $f$. So for $0$, define $f(0) = \inf(C)$. Then, for each $\alpha + 1\in \omega_1$, define $f(\alpha +1) = \inf\{ C \setminus \{f(\beta) : \beta < \alpha\}\}$. Then if $\lambda$ is a limit ordinal, define $f(\lambda) = \sup\{ f(\beta): \beta < \lambda\}$. Since $C$ is closed, $f(\lambda) \in C$.

Does this construction work?

答:
That works but it's unnecessarily complicated. In fact every unbounded subset $S$ of $\omega_1$ has the same ordertype as $\omega_1$.

To see this we use a simpler version of your bijection: recursively set $f(\alpha)=\min(S\setminus \{f(\beta): \beta<\alpha\})$ for all $\alpha<\omega_1$, successor or limit.

By induction together with the fact that $S$ is unbounded you can prove that for all $\alpha<\omega_1$ the ordinal $f(\alpha)$ exists (that is, $S\setminus \{f(\beta): \beta<\alpha\}$ is nonempty and so has a minimal element) and is $<\omega_1$; this uses as a key step the fact that every countable subset of $\omega_1$ is bounded below $\omega_1$, which is a good exercise.


问:
Alternate proof of integral equality using MVT
Here is the original problem:

Let $f:[0,1]\to \mathbb{R}$ be a positive continuous function. Show that for every positive integer $n$ there is a unique $a_n\in (0,1]$ with
$$
\int _0^{a_n} f(x)\,dx = \frac{1}{n} \int _0^1 f(x)\,dx.
$$
Further, compute $\lim_{n\to\infty} n a_n.$

My work:

Let $F(x) = \int _0^x f(t)\,dt.$ We have $F(0)=0$ and by the FTC $F$ is continuously differentiable, and strictly increasing since $f$ is positive. Note also that $F(1) = \int _0^1 f(x)\,dx = M<\infty.$ Since $F$ is differentiable (hence continuous), by the Intermediate Value Theorem there is some point $a_2\in (0,1)$ such that $F(a_2) = M/2;$ further, since $F$ is increasing, $a_2$ is unique. The generalization to $a_n$ is immediate.

Since $F$ is strictly increasing and differentiable, it has a well-defined inverse on $[0,1].$ Then the limit is
$$
\lim_{n\to \infty} n a_n = \lim _{n\to \infty} n\cdot F^{-1}\left(\frac{M}{n}\right)
$$Make the substitution $u=M/n$:
$$
= M\lim _{u\to0^+} \frac{F^{-1}(u)}{u}.
$$Since $F(0)=0,$ $F^{-1}(0)=0$ and since $F$ is differentiable, we may use L'Hopital's Rule:
$$
= M\lim_{u\to 0^+} \frac{d}{du}{F^{-1}(u)} = \lim _{u\to 0^+} \frac{M}{F'(F^{-1}(u))}= \frac{M}{f(F(0))}=\frac{1}{f(0)}\int_0^1 f(x)\,dx.
$$

The problem provided the following hint:

First show the existence of $\{a_n\}.$ Then show that $a_n\to 0 $ and $\int _0^{a_n} f(x)\,dx=a_n f(\xi_n)$ for some $0<\xi_n<a_n.$

I think the idea is to use the hint and the First MVT for Integrals to show that $\lim_{n\to\infty}na_n =\frac{1}{f(0)}\int_0^1 f(x)\,dx$ but I'd be curious to see how and if there are other methods of proof as well.

答:
For the first part, you don't need MVT. The IVT already suffices: let $I=\int_0^1f(x)dx$ and $g(a)=\int_0^af(x)dx$. Since $f$ is positive and continuous, $g$ is strictly increasing and continuous. Also, $g(0)=0<\frac1nI\le g(1)$. Therefore, by IVT, there exists some $a_n\in[0,1]$ such that $g(a_n)=\frac1nI$. The value of $a_n$ cannot be zero because $g(0)<I$. It is also unique because $g$ is strictly increasing.

For the second part, as Daniel Fischer commented, the hint was likely intended to mean that $na_n=\frac{1}{\xi_n}\int_0^1f(x)dx$.

Alternatively, note that $(a_n)$ is a decreasing sequence that is bounded below. Therefore it converges and $g\left(\lim_{n\to\infty}a_n\right)=\lim_{n\to\infty}g(a_n)=\lim_{n\to\infty}\frac1nI=0$. Since $g>0$ on $(0,1]$, $\lim_{n\to\infty}a_n$ must be zero. Therefore, by the first FTC,
\begin{aligned}
I=\lim_{n\to\infty}I
&=\lim_{n\to\infty}na_n\frac{g(a_n)}{a_n}\\
&=\lim_{n\to\infty}na_n\cdot\lim_{n\to\infty}\frac{g(a_n)-g(0)}{a_n-0}\\
&=\lim_{n\to\infty}na_n\cdot\lim_{a\to0}\frac{g(a)-g(0)}{a-0}\\
&=\lim_{n\to\infty}na_ng'(0)\\
&=\lim_{n\to\infty}na_nf(0)
\end{aligned}
and hence $\lim_{n\to\infty}na_n=\frac{1}{f(0)}I$.


问:
Example of uniform but not normal convergence
Let's consider a function series defined as follows.
For every $n\in \mathbb N, n\gt 1$, let $ f_n(x):= \frac{1} {n} $ if $x=\frac{1} {n}$ and $f_n(x)=0$ $x\neq\frac{1} {n}$.
Prove, using the definition of uniform convergence and the Weierstrass M-test, that $$\sum_{n=1}^{+\infty }f_n(x) $$ converges uniformly but not normally in $\mathbb R$.
(I assume a series is normally convergent if $\sum_{n=1}^{+\infty } sup \{ \vert f_n(x) \vert \} $ converges)

My attempt.

For every $n$, $sup \{ \vert f_n(x) \vert; x \in \mathbb R \}= \frac{1} {n} $ and the harmonic series diverges. So we don't have normal convergence.
Though, I'm not sure how to do with the uniform convergence. It may seem a trivial question, but I got lost with the epsilons and the supremums. I suspect that it converges to the 0 function, but $ sup \{ \vert \sum_{k=1}^{n} f_k(x) \vert; x \in \mathbb R\} \ge 1=f_1(1)=\sum_{k=1}^{n} f_k(1)$, so it does not tend to 0.

Any help, using the definition or the epsilon?

答:
Given $\epsilon > 0$, note that with $m > n > N$ and $\frac{1}{N} < \epsilon$

$$\left|\sum_{j=n+1}^m f_j(x) \right| = \begin{cases}0, &x \neq \frac{1}{p} \, \text{ where }\,\, p \in \mathbb{N},\\ 0, &x = \frac{1}{p}\, \text{ where }\,\, p \in \mathbb{N},  p \not\in(n,m]\\ \frac{1}p, & x= \frac{1}{p} \, \text{ where }\,\, p \in \mathbb{N}, n < p \leqslant m \end{cases} $$

Since $\frac{1}{p} < \frac{1}{n} < \frac{1}{N} < \epsilon$, it follows that for all $m > n > N$ and all $x \in \mathbb{R}$,

$$\left|\sum_{j=n+1}^m f_j(x) \right| < \epsilon$$

and we have uniform convergence of the series by the Cauchy criterion.


问:
Stationary sets in $\kappa$ of countable cofinality
This is exercise 47 of chapter 7 of "Introduction to Modern Set Theory" by Judith Roitman.
The problem is:

Show that if $\kappa$ has countable cofinality then $S \subseteq \kappa$ is a stationary subset of $\kappa$ iff $\kappa \setminus S$ is bounded below $\kappa$.

I am trying to prove the forward direction (starting with a stationary set and showing there is a bound on $\kappa \setminus S$).

$S \subseteq \kappa$ is said to be stationary iff $S \cap C \neq \emptyset$ for all $C$ club in $\kappa$. A club set is one that is closed and unbounded in $\kappa$. This text defines closed as: $C \subseteq \kappa$ is closed in $\kappa$ iff for all $A \subseteq C$ if A is not cofinal in $\kappa$ then $\sup(A) \in C$.

If $\operatorname{cf}(\kappa) \lt \omega$ then $\kappa$ has a greatest element $\alpha$, and $\{ \alpha \}$ is club in $\kappa$ so $\alpha \in S$ and $\kappa \setminus S$ is bounded below $\kappa$.

But I run into trouble if $\operatorname{cf}(\kappa) = \omega$. Here is what I can figure out:

Suppose for the sake of a contradiction that $\kappa \setminus S$ is unbounded. Then $\kappa \setminus S$ would be club if for every $A \subseteq \kappa \setminus S$ not cofinal in $\kappa$, $\sup(A) \in \kappa \setminus S$. But $\kappa \setminus S$ cannot be club since then $S$ would intersect it. Thus there exists $A \subseteq \kappa \setminus S$ not cofinal in $\kappa$ such that $\sup(A) \notin \kappa \setminus S$, hence $\sup(A) \in S$. If $|A| \lt \omega$, then $A$ has a greatest element $\alpha$, and $\sup(A)=\alpha \in A \subseteq \kappa \setminus S$. Thus $|A| \geq \omega$. The problem is solved if $\kappa = \omega$ since every set not cofinal in $\omega$ is finite, meaning $\kappa \setminus S$ would be club and we would have a contradiction. But for $\kappa \gt \omega$ I'm stumped.

Any help would be greatly appreciated. Thanks!

答: HINT: Suppose that $\kappa\setminus S$ is unbounded. Since $\operatorname{cf}\kappa=\omega$, there is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $\kappa$ that is cofinal in $\kappa$. Use this to construct a strictly increasing sequence $\langle\beta_n:n\in\omega\rangle$ in $\kappa\setminus S$ that is cofinal in $\kappa$; then $\{\beta_n:n\in\omega\}$ will be a club set disjoint from $S$.


问:
Show that the continuous ranked probability score is strictly proper
I am having trouble verifying a claim from this paper (section 4.2).

Let $\mathcal P$ be the set of all probability measure on the Borel algebra of $[0,1]$. The continuous ranked probability score for $P \in \mathcal P$ at $x \in [0,1]$ is defined as
$$S(P,x) = \int_0^1 \big(P([0,y]) - \mathbf 1\{y \geq x\}\big)^2dy,$$
where $\mathbf 1\{y \geq x\} = 1$ if $y \geq x$ and $=0$ otherwise.

Claim. For all $P, Q \in \mathcal P$ such that $P \neq Q$
$$\int S(P,x) P(dx) < \int S(Q,x)P(dx).$$

Can someone prove this or provide a reference?

答:
Adding and subtracting $Q([0,y])$, one gets
\begin{align}
S(P,x)&=\int_0^1(P([0,y])-1_{[0,y]}(x))^2\, dy=S(Q,x) \\
&\qquad- 2\int_0^1(P([0,y])-Q([0,y]))(1_{[0,y]}(x)-Q([0,y]))\, dy \\
&\qquad+\int_0^1(P([0,y])-Q([0,y]))^2\, dy
\end{align}
Then, taking expectations and interchanging the order of integration,
\begin{align}
\int_0^1 S(P,x)P(dx)&=\int_0^1 S(Q,x)P(dx)-\int_0^1(P([0,y])-Q([0,y]))^2\, dy.
\end{align}


问:
Is it possible to characterize artinian modules by their support?
Rings are commutative Noetherian (and local if needed). modules are not necessarily finitely generated

There are several characterization for a modules to be artinian.
but I want to know whether there is a characterization for artinian modules $M$ based on $\operatorname{Supp}M$?

I can prove that if $M$ is artinian, then all elements of $\operatorname{Supp} M$ are maximal (and $|\operatorname{Supp} M|< \infty$). Is the converse true?

Thank you.

答: No. Let $k$ be any field, and let $V$ be an infinite-dimensional $k$-vector space. Then $V$ is neither Artinian nor Noetherian, but $\operatorname{Supp}(V) = \{0\}$ is a finite set of maximal ideals of $k$.


问:
Value of a double indexed product
Let $n$ be a positive integer.
While trying to calculate the product $\prod_{1\leq i\leq j \leq n}^{} ij $, my approach was to write :
$$\prod_{1\leq i\leq j \leq n}^{} ij=\prod_{j=1}^{n} \prod_{i=1}^{j} ij$$
But I am not sure if this is correct or not. Can someone please enlighten me?

答:
Yes, this is valid. You are just traversing the same set of $(i,j)$ that constitute your index set in a different order. But you should note that

This would be true for any $a_{i,j}$, not just $a_{i,j} = i \cdot j$.

You couldn't automatically do this if your index set was infinite.

As @CameronWilliams' comment points out, while it is valid it may or may not be useful.

Title: Value of a double indexed product


问:
Show that $\langle 2,x \rangle$ is not a principal ideal in $\mathbb Z [x]$
Hi
I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please.

The ring that I am talking about is $\mathbb{Z}[x]$ so $\langle 2,x \rangle$ refers to $2g(x) + xf(x)$ where $g(x)$, $f(x)$ belongs to $\mathbb{Z}[x]$.

答:
I want to record a somewhat less elementary, but perhaps more conceptual answer.

Note first that $\langle 2 \rangle$, $\langle x \rangle$ and $\langle 2, x \rangle$ are all prime ideals of $\mathbb{Z}[x]$.  Indeed, the quotients by these ideals are isomorphic, respectively, to $(\mathbb{Z}/2\mathbb{Z})[x]$, $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, which are all integral domains.

So in particular we have a proper inclusion of nonzero prime ideals

$0 \subsetneq \langle x \rangle \subsetneq \langle x, 2 \rangle$

in which the smaller ideal is principal.  Now let $R$ be any integral domain and let
$I \subset J$ be a proper inclusion of nonzero prime ideals, with $I$ a principal ideal.  Then $J$ cannot be principal.  Indeed, suppose $I = \langle x \rangle$ with $x$ a prime element.  Suppose also $J = \langle y \rangle$.  Then $x \in J$, so that
there exists $a \in R$ with $x = ay$.  Since $ay = x \in I$ and $I$ is prime, we
have either $a \in I$ or $y \in I$.  If $a \in I$, then $a = bx$, so $x = byx$ or
$x(1-by) = 0$ in the domain $R$; since $x \neq 0$ we conclude $by = 1$, i.e., $y$
is a unit and therefore $J = R$, contradiction.  Similarly if $y \in I$,
then $y = bx$, so $x = abx$ and we conclude that $a$ is a unit and thus $I = J$, contradiction.

Added: A variant on the above argument is: if $0 \subsetneq I = \langle a \rangle \subsetneq \langle b \rangle = J$ with both $I$ and $J$ prime, then $a$ and $b$ are both irreducible elements and $b$ properly divides $a$, contradiction.  This is technically a stronger fact because in an arbitrary domain a generator of a principal prime ideal is necessarily an irreducible element but the converse generally does not hold.  However, the easiest way to show that an element $a \in R$ is irreducible is to show that $\langle a \rangle$ is prime, or equivalently that $R/\langle a \rangle$ is a domain.  To show that a nonprime element $x$ is irreducible is more delicate.

Remark: If $R$ is a commutative Noetherian ring, then if $J$ is any nonzero principal prime ideal, there cannot be any nonzero prime ideal $I$ -- principal or otherwise -- with $0 \subsetneq I \subsetneq J$.  This is a special case of Krull's Principal Ideal Theorem.


问:
Closed-form for $ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $ where $p\in\mathbb{N}$
Is there a closed form to the sum $$ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $$
where $p$ is a positive integer? I know that for $p = 1$ the series sums to $xe^x$, but not sure if there's a simple expression for $p > 1$. If there isn't a closed form, is this series related to any special functions? Thanks!

答:
BTW, the proof of this identity goes like this:

$$\sum_{n\ge 0} n^p \frac{x^n}{n!}
= p! [z^p] \sum_{n\ge 0} \frac{x^n}{n!} \exp(nz)
= p! [z^p] \exp(x\exp(z))
\\ = \exp(x) p! [z^p] \exp(x(\exp(z)-1)).$$

Observe that $\exp(z)-1 = z + \cdots$ so only the initial segment of
the outer exponential contributes:

$$\exp(x) p! [z^p] \sum_{k=0}^p \frac{x^k (\exp(z)-1)^k}{k!}.$$

Thus when $p=0$ only $k=0$ contributes and we get $\exp(x)$ as
expected. Otherwise with $p\ge 1$ we find

$$\exp(x) \sum_{k=1}^p x^k {p\brace k}$$

as claimed. Here we have used the fact that set partitions have
combinatorial class

$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z}))$$

which gives the EGF

$$\exp(u(\exp(z)-1))$$

so that

$$\sum_{n\ge k} {n\brace k} \frac{z^n}{n!} =
[u^k] \exp(u(\exp(z)-1)) = \frac{(\exp(z)-1)^k}{k!}.$$


问:
A question on probability theory with 3 urns with return
Each of $3$ urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is the probability that the second ball drawn from the same urn is white?

I think, this is $\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$ by Bayes'theorem and Law of total probability, but can't be sure.

Thanks for any help.

答:
I am unsure of where your math came from, and I am unfamiliar with the "Law of total probability".  The following is how I would compute the probability.

Let $p(k)$ represent the probability that the 1st ball came from urn $k : k \in \{1,2,3\}.$

Then, the chance that the new drawing will also be a white ball, since the sampling is done with replacement is

$$\left[p(1) \times \frac{10}{20}\right]
~+~ \left[p(2) \times \frac{6}{20}\right]
~+~ \left[p(3) \times \frac{2}{20}\right].$$

Let $D$ (i.e. denominator) =
$$\left[\frac{1}{3} \times \frac{10}{20}\right]
~+~ \left[\frac{1}{3} \times \frac{6}{20}\right]
~+~ \left[\frac{1}{3} \times \frac{2}{20}\right].$$

Then:

$$p(1) = \frac{\frac{1}{3} \times \frac{10}{20}}{D}.$$

$$p(2) = \frac{\frac{1}{3} \times \frac{6}{20}}{D}.$$

$$p(3) = \frac{\frac{1}{3} \times \frac{2}{20}}{D}.$$


问:
Show a polynomial is irreducible in $\Bbb C[x,y]$
If I have a polynomial with more than one variable, for example $\Bbb C[x,y]$ (with $\Bbb C$ the complex numbers), how can Ii see whether it is irreducible or not?
For example, if I have $x^2-y^2-1$, what can I do to know whether it is irreducible or not? In this example I tried to used Eisenstein's theorem in $\Bbb C[x][z]$, but I don't know how to do it.
Thanks!

答:
Yes, Eisenstein's criterion is a possibility: rewrite the polynomial as
$$X^2-(Y^2+1)\in \mathbf  C[Y][X]. $$
$Y+\imath$ is an irreducible factor of all coefficients, but the leading coefficient, and $(Y+ \imath)^2=Y^2-2\imath Y-1$ does no divide the constant term.


问:
Computing (Simplicial) Homology of 2 holed torus
Here's the $\Delta$ complex. I would like to compute the simplicial homology for this complex ($\mathbb{Z}$-coefficients). There's several steps I'm not sure about. To start off with, I'm confused with orientations in $\Delta$-complexes; are the orientations marked in my picture (for the triangles) correct? Beyond that, here's what I have so far:

$C_0 = \langle v \rangle$ ($v$ is the vertex)

$C_1 = \langle a, b, c, d, e, f, g, h, i \rangle$

$C_2 = \langle A, B , C, D, E, F \rangle$

Our chain complex is:

$$...0 \xrightarrow{\partial_3} C_2 \xrightarrow{\partial_2} C_1 \xrightarrow{\partial_1} C_0 \xrightarrow{\partial_0} 0$$

$im(\partial_0) = 0, ker(\partial_0) = C_0$

$im(\partial_1) = 0$ since there's only one vertex (all generators begin and end with $v$) and thus $ker(\partial_1) = C_1$

$\partial_2$ is something I'm wary of. $\partial_2(A) = e-c-d$, etc, so I thought of expressing $\partial_2$ in terms of a $9\times6$ matrix:

$\begin{bmatrix} 0&0&0&0&-1&1 \\ 0&0&0&-1&0&1\\-1&0&1&0&0&0 \\-1&1&0&0&0&0\\1&-1&0&0&0&0\\0&1&-1&0&0&0\\0&0&1&-1&0&0\\0&0&0&1&-1&0\\0&0&0&0&1&-1 \end{bmatrix}$

and then simply putting it through an online calculator to get the null space and column space. Is this a valid technique (I know $\mathbb{Z}$ technically isn't a field/vector space but the methods should translate)? Is there a simpler way to do this? It feels rather strange to be using a calculator in algebraic topology.

Anyway, this yielded:

$ker(\partial_2) = \langle A+B+C+D+E+F \rangle$ (which I suppose was obvious)

and $im(\partial_2) = \langle e-d-c, d-e+f,c-f+g, h-g-b, i-h-a \rangle$ (basically the last column is the only one linearly dependent on the rest)

Finally $im(\partial_3) = 0$

So, finally, our homology groups are:

$H_0(X, \mathbb{Z}) = C_0 \cong \mathbb{Z}$

$H_1(X, \mathbb{Z}) = \langle a, b, c, d, e, f, g, h, i \rangle / <e-d-c, d-e+f, c-f+g, h-g-b, i-h-a \rangle$

$H_2(X, \mathbb{Z}) = \langle A+B+C+D+E+F \rangle \cong \mathbb{Z}$ again

This leaves me with two final questions: Does this seem right? And how do I simplify $H_1(X, \mathbb{Z})$!?

TLDR: My main concerns are about the following: Are the orientations of the triangles marked out in the picture correct/consistent? How does one simplify the $H_1$ I obtained (assuming it's correct)? And I would appreciate if any errors could be pointed out.

答:
Sure. Those orientations are fine.

Using a single vertex is a bit shaky...more on this later.

Your matrix looks OK at first glance. rows 3-8 and columns 1-5 give a submatrix whose determinant is obviously nonzero, so its rank is at least 5, and as you observe, the last column is a linear combination of earlier ones, so the rank is exactly 5.

Your computation of $H_1$ is OK, but it's not really a great thing to look at, is it? I mean, is there a $Z/2Z$ factor in there? It's hard to tell.

It turns out that $H_1$ is actually $\Bbb Z \oplus \Bbb Z \oplus \Bbb Z \oplus \Bbb Z$, so let's see how to get there.

From the last item in the quotient (the generator $i - h - a$) we can say that in our group, $i$ is the same as $h+a$, so let's just get rid of it:

\begin{align}
H_1(X, \mathbb{Z})
&= \langle a, b, c, d, e, f, g, h, i \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b, i-h-a \rangle \\
&= \langle a, b, c, d, e, f, g, h \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b \rangle \\
\end{align}

and after that, you can do the same with h, and then $g$, then $f$, then $e$:

\begin{align}
H_1(X, \mathbb{Z})
&= \langle a, b, c, d, e, f, g, h \rangle / \langle e-d-c, d-e+f, c-f+g, h-g-b \rangle \\
&= \langle a, b, c, d, e, f, g \rangle / \langle e-d-c, d-e+f, c-f+g \rangle \\
&= \langle a, b, c, d, e, f \rangle / \langle e-d-c, d-e+f \rangle \\
&= \langle a, b, c, d, e \rangle / \langle e-d-c\rangle \\
&= \langle a, b, c, d \rangle  \\
\end{align}
at which point the group is evidently the free abelian group on four generators. You can probably, at this point, see how to do all those operations by messing with integer row operations on matrices, but I figured I'd do it out without that.

Back to item 2: what you've got here is not actually a simplicial complex, because each 1-simplex should have as boundary a pair of 0-simplices, but your 1-simplexes all have $v - v$ as their boundaries, and that's not allowed in the definitions.

On the other hand, it all worked out OK, right? How can that be? Well, you've kind of computed the cellular homology of the 2-hold torus, and there's a great theorem that says that this gives the same result as the simplicial homology. But do to it right, you really should turn your octagon into a 16-gon, then put a concentric octagon inside, and a vertex at the very center, and then confirm that every triangle, for instance, has three distinct vertices. Your matrix will be much larger...but the operations on it will go nice and fast and very soon you'll get rid of most of the rows and have something no more complicated than the one you have above.


问:
Continuous function on $[a, b]$ with bounded upper and lower derivatives on $(a, b)$ is Lipschitz.
This question was posted previously, but without the critical assumption that $f$ be continuous, and so this question was left unanswered.

I am using the following definitions:

Upper derivative of $f$ at $x$:

$\overset{-}{D} \, f(x) = \displaystyle \lim_{h \to 0} \left [\displaystyle\sup_{0 < |t|\leq h} \frac{f(x + t) - f(x)}{t} \right ]$

Lower derivative of $f$ at $x$:

$\underset{-}{D} \, f(x) = \displaystyle \lim_{h \to 0} \left [\displaystyle\inf_{0 < |t|\leq h} \frac{f(x + t) - f(x)}{t} \right ]$.

My idea is that if, say, $\overset{-}{D}$ is bounded then the $\sup$ in the definition is finite call it $C$, and we can use this as the constant in the Lipschitz condition. Similarly for $\underset{-}{D}$; call its bound $C'$. However, if $\underset{-}{D} \neq \overset{-}{D}$ then $f$ is not differentiable, then these bounds are different. In this case, can I just take $\textrm{max} (|C|, |C'|)$ as the Lipschitz constant?

Or am I way off?

PS: as a side question, why do we need $t$ in absolute value in the definitions? Is this because we want to approach from the left as well?

答:
Since lower and upper derivatives are bounded, then there exists $M > 0$ such that
\begin{align*}
  -M \leq \underline{D}f(x)\leq \overline{D}f(x)\leq M,\quad \forall x\in [a,b].
 \end{align*}
Recall
\begin{align*}
 \overline{D}f(x) = \lim_{h\rightarrow 0^+}\left(\sup_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}\right)
 \quad\text{and,}\quad
 \underline{D}f(x) = \lim_{h\rightarrow 0^+}\left(\inf_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}\right)
 \end{align*}
Claim 1 For every $x\in (a,b)$, there exists an open interval $I\subseteq[a,b]$ such that $x\in I$ and $|f(y)-f(x)|\leq M|y-x|\}$ for all $y\in I$.

Proof of Claim 1. Let $x\in [a,b]$ and $\varepsilon > 0$. Then there exist $h_0$ and $h_1$ such that $h\leq \min\{h_0,h_1\}$ implies
\begin{align*}
  \sup_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}< M  +\varepsilon,
  \quad\quad
  \inf_{0 < |t|\leq h} \frac{f(x+t)-f(x)}{t}> -M-\varepsilon.
 \end{align*}
It further implies that
\begin{align*}
  \left|{\frac{f(x+t)-f(x)}{t}}\right| <M + \varepsilon,\quad\forall |{t}| \leq h.
 \end{align*}
We can also write it as $|f(y)-f(x)| \leq M|{y-x}|$ for all $|{y-x}| \leq h$ and $y\in [a,b]$.
Let $I = (x-h, x+h)\cap (a,b)$, then $I$ is the desired open interval.

Claim 2 For every $[c,d]\subseteq[a,b]$, we have $|{f(c)-f(d)}|\leq M |{c-d}|\}$.

Proof of Claim 2. Define
\begin{align*}
  S = \{x\in[c,d]:|{f(c)-f(x)}|\leq M |{c-x}|\}.
 \end{align*}
Since $c\in S$, then $S$ is nonempty. Let $u = \sup S$. Since $f$ is a continuous function, then $S$ is closed. Hence, $u\in S$ and $u\leq d$. Suppose $u\neq d$, that is, $ u < d$. From Claim 1, there exists an open interval $I$ containing $u$. Let $y\in I$ with $u < y < d$. Then
\begin{align*}
  |{f(y) - f(c)}| \leq |f(y)-f(u)| + |f(u)-f(c)|\leq M(y-u) + M(u-c) = M(y-c).
 \end{align*}
Therefore, $y\in S$ and $y > u$, which contradicts $u = \sup S$. Thus, $u = d$.

Suppose there exists $x,y\in [a,b]$ such that $|{f(x)-f(y)}| > M|{x-y}|$, which contradicts Claim 2, since $[x,y]\subseteq [a,b]$.


问:
Double Summation - am I interpreting this assignment correctly?
I'm working on an assignment that says

$\sum_{i=2}^{10}\sum_{j=1}^5(i$+j)

From what I've gathered I should now expand the inner summation, and if I'm doing this correctly it should be like this:

(i+1) + (i+2) + (i+3) + (i+4) + (i+5)

And this leaves me with

$\sum_{i=2}^{10}(15 + 5i$)

Now I just want to make sure that I've been following the assignment correctly, and if so my next steps should be to do the following:

(15+10) + (15+15) + (15+20) + (15+25) + (15+30) + (15+35) + (15+40) + (15+45) + (15+50)

Which leaves me with 450 as the correct answer. But I'm not sure if I have been following the steps correctly here.

I did try this website to see if I came up with the same answer but that online calculator actually gave me 405 as the result which made me insecure about my own results and I wanted to see if anyone here could give me a nod in the right direction if I have made some mistakes along the way.

** EDIT ** For anyone looking into this question, I made a simple calculation error so the procedure that shows how the calculations are performed are correct, I just made a stupid error when putting the numbers correctly so the correct answer is 405, not 450 as I posted in this thread initially ****

Body:
Your calculation is right until the summation:

$(15+10) + (15+15) + (15+20) + (15+25) + (15+30) + (15+35) + (15+40) + (15+45) + (15+50)=405$

But you can also use that $\sum\limits_{i=1}^n  k=n\cdot k$ and $\sum\limits_{i=1}^n c\cdot i=c\cdot \frac{n\cdot (n+1)}{2}$. Thus

$$\sum_{i=2}^{10}(15 + 5i)=\sum_{i=2}^{10} 15 + 5\cdot \sum_{i=2}^{10} i=(10-1)\cdot 15+5\cdot \left(\frac{10\cdot 11}{2}-1\right)$$ $$=9\cdot 15+5\cdot 54=135+270=405$$


问:
To prove the interior of a circle is a path connected region
I'm trying to prove that the interior of a circle is a path connected region.

To prove that, we take two random points $A \neq B$ in the interior of a circle we don't have any point of the circumference in line segment $AB$ and I come to the following result which I can't prove:

Given $\triangle ABC$ and a point $X \in \overline{BC}$ then $AX < \max(b,c)$.

I'm taking a point $P \in \overline{AX}, AP = \max (b,c)$ and trying to use triangular inequality on some triangle, but it is not working so far

答:
$AP$ is a minimum at the foot $D$ of the perpendicular from $A$ to $BC$. It's a monotonic function of $P$ on each of the intervals $BD$, $CD$.

In general, it's easy to prove that any convex set is path connected. That's a stronger property than connectedness: https://en.wikipedia.org/wiki/Connected_space#Path_connectedness .


问:
Prove that there is no rational $x$ such that $x-1 = 1/x$
As said in title,

I need to prove that there doesn't exist a rational number $x$ that satisfies $x-1 = 1/x$.

I remember doing something like this a while back at school but I can't recall how to do it. Am I right in starting out by simplifying it to:

$x^2-x=1$?

答:
By Ruffini's Theorem (or Rational root Theorem), all possible rational solutions of $$x^2-x-1=0$$ are those whose numerator divides 1 (the independent term) and whose denominator divides 1 (the leading coefficient).

So the only possible rational solutions are $\pm1$ and, trivially, they aren't.


答: Write $x=a/b$ in its lowest terms so$$x=\frac{1}{x-1}=\frac{b}{a-b},$$forcing a contradiction if $|x|>1$. See if you can do the same for $|x|<1$.

Title: Prove that there is no rational $x$ such that $x-1 = 1/x$

Body:
As said in title,

I need to prove that there doesn't exist a rational number $x$ that satisfies $x-1 = 1/x$.

I remember doing something like this a while back at school but I can't recall how to do it. Am I right in starting out by simplifying it to:

$x^2-x=1$?


问:
Spectrum (eigenvalues) and their relation to Projection Valued Measures(PVM)
I'm going to ask 2 questions but they're (I believe) related.

1) Let $z_0$ be an eigenvalue and $\psi$ a corresponding normalized eigenvector. Compute $μ_{\psi}$.

This is the whole question but we could assume operator $A$ which we know has eigenvalue $z_0$ is self-adjoint and $\mu_{\psi}(\Omega)$ is given by $\langle\psi,P_A(\Omega)\psi\rangle$ where $P_A$ is associated PVM to $A$.

2) Show that $z_0$ is an eigenvalue if and only if P({$z_0$})$\neq0$. Show that $Ran(P(\{z_0\}))$is the corresponding eigenspace in this case. Again we could assume $A$ is self-adjoint.

I've already (kind of) answered both of the questions but reason for why i'm asking these questions is both questions seems very intuitive like the moment i saw the first question i immediately said if $\Omega$ contains $z_0$ it is $||\psi||^2$ if not it is $0$. And i would like to answer them more formally \

3) Under the shadow of these 2 questions i would like to ask another question.We know spectrum $$\sigma(A)=\{z\in\mathbb{R} : P_A(z-\epsilon,z+\epsilon)\neq0 \enspace \forall\epsilon>0\}$$ but this does not differentiate between different types of spectrums pure point,point embedded in continuum and purely continuous.Clearly eigenvalues are in point part of the spectrum and above results shows us $P(\{z_0\})$ is an eigenspace but what happens if $z_0\in$ purely cont.?Could someone give me an example(maybe example from physics like quantum harmonic oscillator etc.)?

Related links Show that eigen-vectors belong in range of projection valued measure. Spectral measure associated to eigenvector of self-adjoint operator

Any hint and solution is appreciated THANKS!

答:
Claim: $A\psi=\lambda\psi$ for some $\psi\ne 0$ iff
$$
           P(\{\lambda\})\psi = \psi.
$$
Proof: First assume that $A\psi=\lambda\psi$ for some $\psi\ne 0$. Then
$$
       0=\|(A-\lambda I)\psi\|^2=\int_{-\infty}^{\infty}|\mu-\lambda|^2d\rho(\mu),
$$
where $\rho(S)=\langle P(S)\psi,\psi\rangle= \|P(S)\psi\|^2$ is the measure associated with $\psi$. It follows that $\rho$ is concentrated only at $\lambda$; indeed, if $\rho$ had positive mass on $\mathbb{R}\setminus\{\lambda\}$, then the above equality could not hold. Therefore,
$$
        \psi=P(\{\lambda\})\psi.
$$
Conversely, if $\psi=P(\{\lambda\})\psi$ for some $\psi\ne 0$, then $P(\mathbb{R}\setminus\{\lambda\})=0$, which gives
$$
                 A\psi=\int_{\mathbb{R}}\mu dP(\mu)\psi=\int_{\{\lambda\}}\mu dP(\mu)\psi=\lambda P(\{\lambda\})\psi=\lambda \psi.
$$
Q.E.D.


问:
Defining an ideal in the tensor algebra
In the wikipedia article about exterior algebra:

The exterior algebra $Λ(V)$ over a vector space $V$ over a field  $K$ is defined as the Quotient algebra of the tensor algebra by the two-sided Ideal $I$ generated by all elements of the form $x \otimes x$ such that $x \in V$.

I want to show that $I$ is an ideal of $T(V)$ but i was confused about the elements of $I$, i mean what is exactly $I$ as a set and more particularly as a subset of $T(V)$.

答:
egreg says:

I'm afraid there's not much more that can be said, even in the finite dimensional case

Actually a fairly precise characterization of the ideal generated by the tensors of the form $x \otimes x$ is possible. The ideal is spanned, as a subspace, by tensors of the form

$$x_1 \otimes \dots \otimes x_n \in V^{\otimes n}$$

where the vectors $\{ x_1, \dots x_n \}$ are linearly dependent! This is a nice exercise. It implies, among other things, that a wedge product $x_1 \wedge \dots \wedge x_n \in \wedge^n(V)$ in the exterior algebra vanishes iff the vectors $\{ x_1, \dots x_n \}$ are linearly dependent, which is a standard and useful feature of the exterior algebra.

Title: Defining an ideal in the tensor algebra


答:
The only solutions for x are GoldenRatio and 1-GoldenRatio, which are both irrational.

GoldenRatio =( 1+Sqrt(5) ) /2


答:
Using the quadratic formula on $x^2-x=1$ we have

$$x = \frac {1\pm\sqrt5}2$$

If $x$ can be rational, so can $2x-1 = \pm \sqrt 5$. But the irrationality of $\pm\sqrt 5$ can be proved by infinite descent, for instance.


问:
Standard Basis of Function Space $\mathbb{R}^\mathbb{R}$
I have seen $\mathbb{R}^\mathbb{R}$, the set of functions from $\mathbb{R}$ to $\mathbb{R}$, described as a vector space (with the usual operations $(f+g)(x)≔f(x)+g(x)$ and $(a\cdot f)(x)≔a\cdot f(x)$). However, I have been unable to find what the standard basis for this space is. Intuitively, I would think that the set of all functions that can be defined as $f_r(x)≔\begin{cases}1&\text{if }x=r\\0&\text{if }x\ne r\end{cases}$ for some $r\in\mathbb{R}$ would be the standard basis, as it is orthonormal and looks similar to the standard bases of the $\mathbb{R}^n$ spaces; however, I could not find this definition anywhere and was wondering if this might not be the case for some reason.

答:
You have a couple of misconcenptions. Firstly, there is no the basis, a vector space can have many different bases (plural of basis). Secondly, what do you mean by orthogonal? I don't think the set of all functions from $\mathbb{R}$ has a particularly natural inner product.

Also, the set of functions you describe are certainly linearly independent, they don't span all the functions. Fx. $f(x)=x$. I'm pretty sure you won't be able to find any nice basis of this space, as the very existance of a basis uses the axiom of choice, which means it might not even to explicitly describe a basis.


问:
Showing $\left\lbrace x: \lim f_n(x) \text{ exists} \right\rbrace$ is measurable
Assuming $\left\lbrace f_n\right\rbrace$ is measurable for all n, I was able to show using the fact that sequences of real numbers converge iff they are Cauchy that:

$\left\lbrace x:  \lim f_n(x) \text{ exists} \right\rbrace =\bigcap_{k=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{m=N}^{\infty} \bigcap_{n=N}^{\infty} \left\lbrace x:  f_n(x)-f_m(x) < \frac{1}{k} \right\rbrace \cap \left\lbrace x:  f_m(x)-f_n(x) < \frac{1}{k} \right\rbrace$

Since $f_n$ is measurble for all n, $\left\lbrace x:  f_n(x)-f_m(x) < \frac{1}{k} \right\rbrace $ should be a measurable set for all $n,m,$ and $k$, and then since there's a bunch of countable intersections and unions of all these sets, I should have that $\left\lbrace x:  \lim f_n(x) exists \right\rbrace$ is measurable as well. What I don't understand is why $f_n$ being measurable immediately implies that $\left\lbrace x:  f_n(x)-f_m(x) < \frac{1}{k}\right\rbrace $ is measurable. We just started learning about measurable functions and I know that $f:X \rightarrow Y$ is $M,N-$measurable if $f^{-1}(E) \in M$ for all $E \in N$ where $M$ is the sigma-algebra on $X$ and $N$ the sigma algebra on $Y$, but I don't see how to apply that definition here.

答:
An easier way to do this so that you won't have to take unions/intersections over both $m$ and $n$:

Define $f(x) = \limsup_{n \to \infty} f_{n}(x)$. Now, if $f_{n}(x)$ converges at $x$, it must converge to $f(x)$ (recall that convergence implies that your limit is equal to your limsup and liminf). Define
$$
A_{m,n} = \{x: |f_{n}(x) - f(x)| < \frac{1}{m}\}
$$
And notice that
$$
\{x: f_{n}(x) \text{ converges}\} = \bigcap_{m \in \mathbb{N}} \bigcup_{N \in \mathbb{N}}\bigcap_{n \geq N} A_{m,n}
$$
Showing that $f$ and consequetly $|f_{n} - f|$ are measurable isn't too hard, and if you're doing this problem, you've probably already proved this in your class/textbook.


问:
how to solve Lazy janitor problem

Let, $T_n=\sum_{k=1}^{n}(X_k+Y_k)$. Then, $T_n/n\to EX_1+EY_1$ a.s.

By definition, $R_t=\sup\{n:T_{n-1}+X_n\leq t\}$

Then, $T(R_t-1)+X(R_t)\leq t< T(R_t)$ which implies, $\dfrac{T(R_t-1)+X(R_t)}{R_t}\leq \dfrac{t}{R_t}< \dfrac{T(R_t)}{R_t}$

After that, I don't understand how to get the answer. Thanks for any help.

答:
I've been reading Rick Durret's book and i went through this problem.

Since no one have posted an answer i'll give it for completion in case if
anyone is intrested in the future and search for this.

Let $Z_n=X_n+Y_n$. Then $Z_n$ corresponds to the total time that the system operated
with $n$ light bulbs until the $n+1$ light bulb started working. Then $N_t=\sup\{n : Z_1+...+Z_n\leq t\}$ is the largest number of cycles that the system undergone from a working light bulb until the next light bulb started operating. Equivalently, $N_t$ also denote the number of light bulbs that we will have to change up to time $t$.

Now let $S_n=X_1+...+X_n$ denote the total time that $n$ light bulbs operate. Then, $R_t=S_{N_t}$ corresponds to the total amount of time that the light was working in the interval $[0,t]$. So, we need to find $\lim_{t\to \infty}R_t/t$.

Since, all $X_i$'s,$\ Y_i$'s are independent with each other and they have $F,G$ as distributions respectively, it follows that $Z_i$'s are independent and all have the same distribution (i.e. $F*G$ the convolution of $F,G$).

Now using the Renewal Theorem ( Example 2.4.1 in Durret ) we deduce that
$$\tag{1}\lim_{t\to \infty}\frac{N_t}{t}=\frac{1}{EZ_1}=\frac{1}{EX_1+EY_1}$$
with probability 1, and $N_t\nearrow \infty$ as $t\to \infty$. Now, by the strong law of large numbers we also have
$$\tag{2}\lim_{n\to \infty}\frac{S_n}{n}\to EX_1$$
again with probability 1. Now, using $(1),(2)$ and $N_t\to \infty$ as $t\to \infty$ with probability 1 we have
$$\lim_{t\to \infty}\frac{R_t}{t}=\lim_{t\to \infty}\frac{S_{N_t}}{N_t}\cdot \frac{N_t}{t}=\frac{EX_1}{EX_1+EY_1}$$

Meaning that if we have $EY_1\ggΕΧ_1$ we will probably have to walk in the dark!

Title: how to solve Lazy janitor problem


问:
Can you have a matrix with no zero rows that has an eigenvalue equal to zero
Can you have a matrix with no zero rows that has an eigenvalue equal to zero?

So for $A \in \mathbb{R}^{n \times n}$ with $0$ as an eigenvalue, is the $\text{rank}(A)$ always smaller than $n$?

答:
$$\pmatrix{1&1\\1&1}\binom1{-1}=\;?$$

The answer to your second question, however, is yes, because the nullity of $A$ is positive; use the rank-nullity theorem.


问:
How do I come up with a formula to calculate the possible combinations of this set?
Let's say I have the letters a, a, a, b, c, d. I can only arrange them in pairs of 2. If I use up a letter, I cannot use it again. Order doesn't matter.

So the possible sets of pairs I can get are:

(a,a)(a,b)(c,d)

(a,b)(a,c)(a,d)

(a,c)(a,a)(b,d)

(a,d)(a,a)(b,c)

How do I come up with a way to calculate that there would be 4 possiblities? I tried the combination and permutation calculators on the internet but they didn't work because they would either give me 15 probabilities or 30 probababilities.

答: You could consider the cases for whether there is a pair with two $a$'s or not. Then clearly there is only $1$ case where no pair has two $a$'s If a pair does have two $a$'s then another pair must have exactly one and then there are only $3$ ways to arrange the remaining letters. Sometimes there are too many conditions and you must use a combination of techniques (or a more general result than what you currently know).