问:
Cohomology functors in a triangulated category with t-structure.
Given a triangulated category $\mathcal{D}$ with a t-structure $(\mathcal{D}_{\leq 0}, \mathcal{D}_{\geq 0})$ the cohomology functor can be defined as
\begin{equation}
H^k := \tau_{\geq 0}\circ \tau_{\leq 0} \circ [k] : \mathcal{D} \rightarrow \mathcal{A}
\end{equation}
where $\mathcal{A} := \mathcal{D}_{\leq 0} \cap \mathcal{D}_{\geq 0}$ is the heart of the t-structure. It is known that (e.g. https://en.wikipedia.org/wiki/T-structure#Cohomology_functors), given an exact triangle $A \rightarrow B \rightarrow C \rightarrow A[1]$ we get the long-exact-sequence
\begin{equation}
\cdots \rightarrow H^k(A) \rightarrow H^k(B) \rightarrow H^k(C) \rightarrow H^{k+1}(A) \rightarrow \cdots
\end{equation}
just like in the usual cohomological algebra. In the case of derived categories, I could see why this should be true (mainly because the cohomology functor simply computes the usual cohomology).

But is it easy to see that we get a long-exact-sequence from an exact triangle using only the definition of t-structures and the axioms of the general triangulated categories? Could someone suggest to me the easy way to do this?

答: This should probably be a comment, but I am not able to comment yet. This is proved as theorem 1.3.6 in Faisceaux Pervers by Beilinson, Bernstein and Deligne. It also appears in English in Topological invariants of stratified spaces by Banagl as Proposition 7.1.12.</p>


问:
Area Minimization of Circle
The radius of the circle having minimum area which touches the curve y=4-x² and the lines, y=|x| is?

I tried using the normal to the curve and satisfying it with the centre of the circle but it just gives me a complex 4th degree equation which I'm unable to solve...

答:
There is not enough constraints to choose a unique circle.  Here is a desmos graph of the set up:

To make this question work and get the intended answer as given by the JEE authorities, we need to say that the circle touches both $y=x$ and $y=-x $ of $|x|$ function.

ContentLicense: CC BY-SA 4.0

Title: Area Minimization of Circle


问:
Proof Explanation: Nonempty perfect sets of $(\Bbb R, |\cdot|)$ are uncountable
The proof I have some trouble with:

Suppose $\varnothing\ne A\subset\Bbb R$ is perfect. $A$ is perfect, so $A$ is closed and every point in $A$ is a limit point of $A$. Hence, $A$ cannot be finite. $A$ is infinite. We want to show that $A$ is uncountable. Suppose $A$ is countable. Then, $A = \{x_1,x_2,...\}$ is an enumeration of $A$. Consider $x_1$, and an interval $[a_1,b_1]$ s.t. $x_1\in (a_1,b_1)$. Since $x_1$ is a limit point of $A$, $(a_1,b_1)\setminus\{x_1\}\cap A\ne \varnothing$ (in fact, intersects at infinitely many points). Now, choose $x_{n_1}\in A$, s.t. $x_{n_1}\ne x_1,x_2$ and consider $(a_2,b_2) \ni x_{n_1}$ s.t. $(a_2,b_2)\subset (a_1,b_1)$ and $\displaystyle b_2-a_2 \le \frac{b_1-a_1}{2}$ and $x_1,x_2\notin [a_2,b_2]$.

Q1. Why can we find such a subset? i.e. why can we find $(a_2,b_2)$ with the above properties?

Again, $(a_2,b_2)\setminus \{x_{n_1}\}\cap A$ has infinitely many points. Choose $x_{n_2} \ne x_1,x_2,x_3$. Consider $(a_3,b_3)\ni x_{n_2}$ s.t. $(a_3,b_3)\subset (a_2,b_2)$ and $\displaystyle b_3 - a_3 \le \frac{b_2-a_2}{2} \le \frac{b_1-a_1}{2^2}$.

The same question again, I do not understand the construction of these intervals!

Continuing this way, one obtains a nested sequence of closed and bounded intervals $\{[a_k,b_k]\}_{k=1}^\infty$ s.t. $\displaystyle b_k - a_k \le \frac{b_1-a_1}{2^{k-1}}$ and $x_{n_{k-1}} \ne x_1,x_2,\ldots,x_k$ with $x_{n_{k-1}}\in (a_k,b_k)$ s.t. $x_1,\ldots,x_k \notin [a_k,b_k]$.

By the Nested Interval Theorem, we can find $x\in \bigcap_{k=1}^\infty [a_k,b_k]$ s.t. $x\in A$. Observe $x\ne x_n, n\ge 1$ by the construction above. This is not possible as $A = \{x_n\}_{n=1}^\infty$. Therefore, $A$ is uncountable.

Q2. How is the last step concluded? Where is the contradiction?

It'd be very helpful if someone could explain the proof to me in detail, I am not able to follow especially the construction of the intervals $[a_k,b_k]$ and what the contradiction is. Other than that, I was able to understand several intermediate steps. Thank you!

答:
Q1. The phrasing of the proof is quite poor. You need to start by choosing $x_{n_1} \in A$ *such that $x_{n_1} \in (a_1, b_1)$* in addition to the stated requirement that $x_{n_1} \neq x_1, x_2$. Then you can first pick any interval $(a_2, b_2)$ containing $x_{n_1}$ and contained inside $(a_1, b_1)$ where the length of the interval decreases by at least a factor of 2. If you shrink $(a_2, b_2)$ to be even closer to $x_{n_1}$, you can make sure $[a_2, b_2]$ doesn't contain $x_1$ or $x_2$, so do so.

Q2. The $\cup$ should be $\cap$. Anyway, the intersection of $[a_k, b_k]$ is a single point, say $x$. Since each $(a_k, b_k)$ contains a point of $A$ (namely $x_{n_{k-1}}$), we have that $x$ is a limit point of $A$. Since $A$ is perfect, $x \in A$. Since we assumed $A$ was countable, well $x = x_m$ for some $m$. But $x \in [a_m, b_m]$ while $x_1, \ldots, x_m \not\in [a_m, b_m]$, a contradiction.

The proof could definitely be better written. The indexing scheme isn't great, it's overly formal in places while using very confusing quantifiers in others. It could use another round of polish.

Edit: A rewrite of the argument was requested; here's mine.

Lemma: Let $\varnothing \neq A \subset \mathbb{R}$ be perfect. Then $A$ is uncountable.

Proof: Suppose to the contrary that $A$ is at most countable. If $A$ were finite, then $A' = \varnothing$, a contradiction since $A=A'$ is perfect, so $A = \{x_1, x_2, \ldots\}$ is countable.

We iteratively construct a nested sequence of closed intervals $I_1 \supset I_2 \supset \cdots$ with the following properties.

The midpoint of $I_k$ is in $A$.

The length of $I_k$ is positive and less than $1/k$.

$x_1, \ldots, x_k \not\in I_k$.

To begin the construction, pick any $x \in A$ different from $x_1$ and let $I_1$ be a closed interval centered on $x$ of positive length less than $1$ which is small enough to not contain $x_1$. Having constructed $I_1, \ldots, I_{k-1}$, pick $I_k$ as follows. Since the midpoint of $I_{k-1}$ is in $A$, which is perfect, it is a limit point of $A$. Hence there is some $y$ in the interior of $I_{k-1}$ different from $x_1, \ldots, x_k$. Let $I_k \subset I_{k-1}$ be an interval centered on $y$ of positive length less than $1/k$ which is small enough to ensure $x_1, \ldots, x_k \not\in I_k$.

By the Nested Interval Theorem and property (2), $\cap_{k=1}^\infty I_k = \{z\}$. Since the midpoint of each $I_k$ is in $A$, $z$ is a limit point of $A$, so since $A$ is perfect, $z \in A$. Hence $z=x_m$ for some $m$. But by (3), $x_1, \ldots, x_m \not\in I_m$, contradicting the fact that $x_m = z \in I_m$. $\Box$


问:
Define the bilinear form L on R2 by L(x,y) = det [x,y] and find its matrix
"Define the bilinear form L on R2 by L(x,y) = det[x,y],i.e. to compute L(x,y) we form a 2×2 matrix with columns x,y and compute its determinant. Find the matrix of L."

I'm still not clear on what a bilinear form is, but based on what I've read I need to find a matrix A where $x^{T}Ay$ is equal to the determinant of x,y. Since det[x,y] is $x_{1}y_{2}-y_{1}x_{2}$ the matrix should be

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

Is this correct?

答:
Bilinear forms are generalization of the scalar product of vectors: let $V$ be a real vector space, a function $b:V\times V\to\Bbb R$ (assigning a real number to each pair of vectors) is a bilinear form, if it is linear in both variables when the other variable is fixed on any value, that is, if it satisfies the following distribution-like properties:
$$b(u+\lambda v,\,w)=b(u,w)+\lambda\,b(v,w)\\
b(u,\,v+\lambda w)=b(u,v)+\lambda\,b(u,w)\,.$$
The determinant on $\Bbb R^2$ is a bilinear form. (In an $n$ dimensional space, it would be multilinear on $n$ arguments.)

Using bilinearity of $b$, one can see that it's uniquely determined by its values taken on pairs of basis elements for any fixed basis, and more specifically, as you write, letting $B:=(b(e_i,e_j))_{i,j}$ we obtain
$$b(v,w)=[v]^TB[w]$$
where $[v]$ is the column vector of coordinates of vector $v$ in basis $(e_i)_i$.


问:
How to calculate a infinite series divided by infinity?
I am confused in calculating this limit, is it infinite?
$$
\lim_{n\to\infty}\frac{1}{n}\sum^\infty_{k=1}\frac{1}{k}
$$

Here is where I find this question:

when
$$
f(x)=
\begin{cases}
    0,x<\pi\\
    \frac{(-1)^k}{k},x\in[k\pi,(k+1)\pi),k\in\mathbb{Z}^+
\end{cases}
$$

the limit

$$
\begin{aligned}
&\lim_{n\to\infty}\int^{+\infty}_{-\infty}f(x)\sin(nx)dx\\&=\lim_{n\to\infty}\sum_{k=1}^\infty\int^{(k+1)\pi}_{k\pi}\frac{(-1)^k}{k}\sin(nx)dx\\&=\lim_{n\to\infty}\sum^\infty_{k=1}\frac{(-1)^k}{k}\frac{1}{n}(\cos(nk\pi)-\cos(n(k+1)\pi))
\end{aligned}
$$

When $n$ is even, the limits is zero.

But when $n$ is odd, the limit equals to

$$
\lim_{n\to\infty}\frac{1}{n}\sum^\infty_{k=1}\frac{(-1)^k}{k}2(-1)^k
$$

which equals to

$$
2\lim_{n\to\infty}\frac{1}{n}\sum^\infty_{k=1}\frac{1}{k}
$$

答:
Correctly one should find
$$L=\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{k} =\lim_{n \to \infty} \frac{1}{n} (\ln n + \gamma).$$ Next, by L-Hospital $$\lim_{n \to \infty} \frac{1}{n}=0.$$
Here we have used $$\sum_{k=1}^{n} \frac{1}{k} \sim \ln n+\gamma$$


答: As the Harmonic series $\displaystyle\sum_{k=1}^\infty\dfrac 1k$ diverges, the limit is undefined. (You cannot divide infinity by anything.)

Title: How to calculate a infinite series divided by infinity?

Body:
I am confused in calculating this limit, is it infinite?
$$
\lim_{n\to\infty}\frac{1}{n}\sum^\infty_{k=1}\frac{1}{k}
$$

Here is where I find this question:

when
$$
f(x)=
\begin{cases}
    0,x<\pi\\
    \frac{(-1)^k}{k},x\in[k\pi,(k+1)\pi),k\in\mathbb{Z}^+
\end{cases}
$$

the limit

$$
\begin{aligned}
&\lim_{n\to\infty}\int^{+\infty}_{-\infty}f(x)\sin(nx)dx\\&=\lim_{n\to\infty}\sum_{k=1}^\infty\int^{(k+1)\pi}_{k\pi}\frac{(-1)^k}{k}\sin(nx)dx\\&=\lim_{n\to\infty}\sum^\infty_{k=1}\frac{(-1)^k}{k}\frac{1}{n}(\cos(nk\pi)-\cos(n(k+1)\pi))
\end{aligned}
$$

When $n$ is even, the limits is zero.

But when $n$ is odd, the limit equals to

$$
\lim_{n\to\infty}\frac{1}{n}\sum^\infty_{k=1}\frac{(-1)^k}{k}2(-1)^k
$$

which equals to

$$
2\lim_{n\to\infty}\frac{1}{n}\sum^\infty_{k=1}\frac{1}{k}
$$


问:
How do you find the maximum value of $r$ in a polar function?
I have $\, r=\cos\alpha +\sin2\alpha,\quad 0\le\alpha\le\frac{\pi}{2}.$

Do you then find $\dfrac{dr}{d\alpha}$ and let that $=0$ ?

I am after just a few set of instructions.

答:
Taking the first derivative of $r$ vs. $\alpha$ you find:
$$\frac{dr}{d\alpha}=-\sin(\alpha)+2\cos(2\alpha)$$
putting this to zero, you find:
$$\alpha=.6348668713, 2.506725783, -1.002966954, -2.138625700$$
Take the second derivative of $r$ which is:
$$\frac{d^2r}{d\alpha^2}-\cos(\alpha)-4\sin(2\alpha)$$
in these points. If it's negative, this give you a maximum</p>

ContentLicense: CC BY-SA 3.0


问:
Difficulty in plotting polar plot for trigonometric functions
I'm having some difficulty with the following question.

Sketch the graph of $r=\tan(\theta)$,where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ and $r=\frac{\sqrt{2}}{2}\csc(\theta)$ and find the area between the two curves.

I'm given the hint that the graph has two vertical asymptotes. This is a polar plot in case that wasn't clear.

I've put this into various software, but I am not really sure how to sketch the curves. I get that in cartesian coordinates, there are vertical asymptotes at $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ but I am not sure how that can help me in plotting the polar curve. Can someone help out here?

For the $\frac{\sqrt{2}}{2} \csc(\theta)$, it's just a horizontal line but I'm not really sure how that's being done either...

I feel like I could get the area if I knew how to plot this but I can't even plot this. Can someone please help out here?

答:
Plotting r=csc(θ) as a horizontal line

One way to plot $r(\theta) = \sqrt{2}/2 \cdot \csc{\theta}$ is by converting to Cartesian coordinates.

If you know the conversion formula $\langle x, y\rangle = \langle r\cos{\theta},\, r\sin{\theta}\rangle$, then you can rewrite:

$$r = A \cdot \csc{\theta} \quad \Longrightarrow\quad r=A\cdot\frac{1}{\sin{\theta}} \quad\Longrightarrow\quad r\sin{\theta} = A \quad \Longrightarrow y=A.$$

Now you can see that for any value of $A$, the polar plot of $r = A\cdot \csc{\theta}$ is a horizontal line of height $A$.

In this specific case, it is a horizontal line of height $y=\sqrt{2}/2$.

Plotting r=tan(θ) as a U shape

After thinking about it for a while, I suggest sketching $r=\tan{\theta}$ like this:

You know the value of $\tan{\theta}$ at special values of $\theta$. For example, you know that $\tan{0^\circ}=0$, $\tan{45^\circ}=1$, $\tan{-45^\circ} = -1$. You can start by plotting those special points. So:

When $\theta=0^\circ$, $r=0$. The graph passes through the origin $\langle 0,0\rangle$.

When $\theta=45^\circ$, $r=1$. You can show that  graph passes through the Cartesian point $\langle x=\sqrt{2}/2,y=\sqrt{2}/2\rangle$.

When $\theta=-45^\circ$, then $r=-1$. You can show that the graph passes through the Cartesian point $\langle x=-\sqrt{2}/2, y=\sqrt{2}/2\rangle$.

You can use the Cartesian-to-polar conversion formulas again to find the vertical asymptotes. If the equation is   $r = \tan{\theta}$, and you know that the cartesian conversion formula is $\langle x, y\rangle = \langle r \cos{\theta}, r\sin{\theta}\rangle$, then you can plug in the formula for $r(\theta)$ to get:

$$\langle x, y\rangle = \left\langle \frac{\sin{\theta}}{\cos{\theta}} \cdot \cos{\theta}, \frac{\sin{\theta}}{\cos{\theta}}\cdot \sin{\theta}\right\rangle.$$
This simplifies to:
$$x=\sin{\theta}, \quad y=\frac{\sin^2{\theta}}{\cos{\theta}}\\~\\\text{\{For this specific function }r(\theta)=\tan{\theta}\}$$

This formula tells us, for this specific function, what the $x$ and $y$ coordinates of the graph are for every value of $\theta$. And this formula actually tells us a lot about the shape of the graph! Notice:

The $x$ coordinate on this graph is equal to $\sin{\theta}$. This means that the whole graph lies between x=-1 and x=+1. The value of $x$ will approach but never go beyond those values because $\sin$ will never go beyond those values.
We've found the two asymptotes, at x=+1 and x=-1.

The $y$ coordinate on this graph is equal to $\sin^2{\theta}/\cos{\theta}$. You know that $\sin^2{\theta}$ is positive because it's a square, and $\cos{\theta}$ is positive as long as $\theta$ is between $-90^\circ$ and $90^\circ$. So the entire $y$ coordinate will always be positive when $\theta$ is in that range.
And as $\theta$ gets close to the limits $\pm 90^\circ$, the $y$ coordinate will increase toward infinity.

This is enough to sketch the graph: plot the three anchor points in the first step, then join them up with a symmetric U shape that increases toward infinity as it approaches the vertical asymptotes on either side at $x=\pm 1$.


问:
Point lies inside a triangle ABC with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ$
A triangle $ABC$ is given with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ.$ Point $M$ lies inside the triangle and $\measuredangle MAB=\measuredangle MBA=15^\circ.$ Find $\measuredangle BMC$.

I am not sure how to approach the problem. We can find some angles:$$\measuredangle ACB=180^\circ-45^\circ-30^\circ=105^\circ\\ \measuredangle CAM=45^\circ-15^\circ=30^\circ\\ \measuredangle MBC=30^\circ-15^\circ=15^\circ\\ \measuredangle AMB=180^\circ-2\cdot15^\circ=150^\circ.$$

I will be very grateful if we can see a solution without using trigonometry.

答:
I found a solution without trigonometry. The key is consider the midpoint of $BC$, which I have named $N$ in the picture below. Using the dashed lines and the observations that $|AD| = |CD| = |CN| = |DN| = |BN|$, it is easy to show that $\angle BAN = 15^{\circ}$. This means that $A$, $M$ and $N$ are colinear, and in addition that $\triangle NMB \sim \triangle NBA$. This gives the ratio $|NM| / |NB| = |NB| / |NA|$, and since $N$ was the midpoint of $BC$, we conclude that also $|NM| / |NC| = |NC| / |NA|$. This means that $\triangle NMC \sim \triangle NCA$, from which it follows that $\angle BMC = \angle BMN + \angle NMC =  30^{\circ} + 105^{\circ} = 135^{\circ}$, as desired.

A final remark: it is actually true that $|AM| = |AC|$, as follows easily from the above proof. However, I could not immediately find a solution that somehow exploits this fact, but it seems to me that it should be possible. Anyone up for it?


问:
Paradox: Roots of a polynomial require less information to express than coefficients?
A somewhat information theoretical paradox occurred to me, and I was wondering if anyone could resolve it.

Let $p(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0 = (x - r_0) \cdots (x - r_{n-1})$ be a degree $n$ polynomial with leading coefficient $1$. Clearly, the polynomial can be specified exactly by its $n$ coefficients $c=\{c_{n-1}, \ldots, c_0\}$ OR by its $n$ roots $r=\{r_{n-1}, \ldots, r_0\}$.

So the roots and the coefficients contain the same information. However, it takes less information to specify the roots, because their order doesn't matter. (i.e. the roots of the polynomial require $\lg(n!)$ bits less information to specify than the coefficients).

Isn't this a paradox? Or is my logic off somewhere?

Edit: To clarify, all values belong to any algebraically closed field (such as the complex numbers). And note that the leading coefficient is specified to be 1, meaning that there is absolutely a one-to-one correspondence between the $n$ remaining coefficients $c$ and the $n$ roots $r$.

答: What is happening here is just a consequence that an infinite set and a proper subset can be in bijective correspondence. That's an well known fact about infinite sets. And it is a paradox in the sense that it is anti-intuitive, but not in the sense that it leads to a contradiction.</p>


问:
Saddle-node behaviour of planar systems
I'm teaching a course on ODEs using Lawrence Perko's book. Having assigned Q3(c) of section 2.11, and both working out the solution using theorems in the sections and consulting the solution manual, I get an answer that seems to conflict with other calculations. The details are as follows:

The system is:
$$
\dot{x} = y, \qquad \dot{y} = x^4 + xy.
$$

According to a Thm 2.11.3 in Perko (see screenshot of Thms 2.11.2 and 2.11.3 below), taken from Andronov--Leontovich--Gordon--Maier, the critical point $(0,0)$ is a saddle-node, "because" the lowest order monomial of the form $x^k$ in the $\dot{y}$ equation is of even degree, the lowest order monomial of the form $x^ny$ has $n = 1$, with coefficient $b_1 = 1 \not= 0$, and $4/2 > 1$. The solution manual concurs.

Now from the equation, $\dot{x} > 0$ on $\{y > 0\}$, $\dot{x} < 0$ on $\{y < 0\}$. Moreover $\dot{y} = xy + O(|x,y|^4)$, which means $\dot{y} > 0$ in the 1st and 3rd quadrants, and $\dot{y} < 0$ in the 2nd and 4th quadrants close to $(0,0)$. And any way I draw this, it looks like a cusp near the critical point.

Finally, winplot confirms this: .

So my question is, is it known that Thm 2.11.3 is in fact not a theorem, or have I missed something?

答:
If we plot the phase portrait of our system of ODEs
$$
\dot x = y
,\qquad
\dot y = x (x^3+y)
,
$$
we see two quite obvious separatrices,
and it looks like they divide the plane into two hyperbolic sectors, so
that there's a cusp at the origin:

However, a closer look reveals that there is actually also a third separatrix, intersecting the $x$-axis at
$x_0 \approx -0.297$, such that the trajectories intersecting the $x$-axis
for $x_0 < x < 0$ all converge to the origin, forming a parabolic sector:

If this is correct, there is in fact a saddle-node at the origin (in agreement with the theorem).

Now the question is of course how we can be sure of this, and the trick is to change to new variables $(u,v)$
to “blow up” the phase portrait at the origin.
Exactly what new variables to take depends on the exponents appearing in the normal form of the system,
but with the exponents in this particular example, the suitable choice turns out to be
$$
x = u
,\qquad
y = u^2 v
,
$$
or equivalently
$$
u =x
,\qquad
v = y/x^2
,\qquad
\text{for $x \neq 0$}
.
$$
This change of variables of course behaves “badly” along the $y$-axis (the line $x=0$),
but that's the whole point.
It is already obvious what the trajectories of our original system $(\dot x, \dot y) = (y, x^4+xy)$
do along the $y$-axis (away from the origin):
they just cross it horisontally, since $\dot x \neq 0$ and $\dot y = 0$ there.
Any such trajectory, with $y$ approaching some $C \neq 0$ as $x \to 0$,
will be transformed into a curve (with two branches) in the $(u,v)$-plane that looks approximately like $v = C/u^2$ when $u$ is close to zero.
The interesting thing to investigate now is what becomes of the trajectories approaching the origin in the $(x,y)$-plane.

From the change of variables, we have $\dot x = \dot u$ and $\dot y = 2uv \dot u + u^2 \dot v$,
and hence
$$
\dot u = \dot x = y = u^2 v
$$
and (for $u \neq 0$)
$$
\dot v
= \frac{\dot y - 2 uv \dot u}{u^2}
= \frac{u(u^3 + u^2 v) - 2 uv \cdot u^2 v}{u^2}
= u(u+v-2v^2)
.
$$
So in the new variables, the system of ODEs becomes
$$
\begin{pmatrix} \dot u \\ \dot v \end{pmatrix}
= \begin{pmatrix} u^2 v \\ u (u + v - 2 v^2) \end{pmatrix}
= u \begin{pmatrix} u v \\ u + v - 2 v^2 \end{pmatrix}
.
$$
Since we're only really interested in the behaviour of the trajectories for $u \neq 0$,
we may as well remove the factor $u$ and consider the rescaled system
$$
\begin{pmatrix} \dot u \\ \dot v \end{pmatrix}
= \begin{pmatrix} u v \\ u + v - 2 v^2 \end{pmatrix}
$$
instead; this will not change the shape of the trajectories, since it only amounts to
a time reparametrization, plus a change of orientation for the trajectories in the left half-plane $u <0$.
The phase portrait for this system looks like this, with equilibria at $(0,0)$ and $(0,1/2)$:

Here the third separatrix (purple) is quite clearly visible. It, and the orange one too, emanates from the equilibrium at $(u,v)=(0,1/2)$, while the red one approaches $(u,v)=(0,0)$, but when you map things back to the $(x,y)$-plane, the whole $v$-axis gets squished into the point $(x,y)=(0,0)$ so that all three meet there.

A typical trajectory in the parabolic sector arises as follows (and I hope you'll excuse my ugly drawings here):

There's a pair of trajectories, one with $u>0$ and one with $u<0$, that are both asymptotic to $v = C/u^2$
with the same negative value of $C$ (more precisely, with $C_0 < C < 0$, where $C_0$ is the value
such that the lower part of the purple separatrix is asymptotic to $v = C_0/u^2$).
When you map back to the old variables, and change the orientation of the left
curve, they will join nicely at $(x,y)=(0,C)$, and form a single origin-bound trajectory of the original system:

Similarly, the curves in the left hyperbolic sector in the $(x,y)$-plane are formed by joining the images of trajectories in the $(u,v)$-plane asymptotic to $v = C/u^2$ with the same value of $C$, both for $C>0$ and for $C < C_0$.

Title: Saddle-node behaviour of planar systems


问:
Definable and $A$-invariant set is $A$-definable.
I know that this is a well known fact in model theory but I can not find a complete proof.

Let $A\subseteq \mathfrak{C}$ a small subset and $U\subseteq \mathfrak{C}$ a definable set. If $U$ is $A$-invariant (invariant under $Aut(\mathfrak{C}/A)$) then it is $A$-definable.

I tried assuming that we needed parameters outside of $A$ to get a contradiction with $A$-invariance but I am not able to finish the proof.

Also, is this still true if we replace definable set with type-definable set?

答:
There is a proof of this fact in A Course in Model Theory by Tent and Ziegler, Lemma 6.1.10. That proof is topological in nature, so I will translate it to logical terms here. Along the way we will also take a look at type-definable sets.

Claim 1. If $U$ is a type-definable $A$-invariant set then $U$ is type-definable over $A$.

Proof. Let $\pi(x)$ be a partial type over some $B$ defining $U$. We claim that the following partial type defines $U$:
$$
\Sigma(x) = \{ \psi(x, a) : a \in A \text{ and } \models \psi(c, a) \text{ for all } c \in U \}.
$$
Clearly, if $c \in U$ then $\models \Sigma(c)$. For the other direction, we assume that $c$ is such that $\models \Sigma(c)$ and we claim that $\operatorname{tp}(c/A) \cup \pi(x)$ is consistent. If not, then there is $\chi(x) \in \operatorname{tp}(c/A)$ such that $\pi(x)$ is inconsistent with $\chi(x)$. Then by construction $\neg \chi(x) \in \Sigma(x)$, but this contradicts $\Sigma(x) \subseteq \operatorname{tp}(c/A)$. So let $c'$ realise $\operatorname{tp}(c/A) \cup \pi(x)$. Then $c' \in U$ while $\operatorname{tp}(c/A) = \operatorname{tp}(c'/A)$, so by $A$-invariance $c \in U$, as required.

So the above claim already answers your question for type-definable sets. We can use the claim to show that if $U$ is actually definable, that then $\Sigma(x)$ can be assumed to be a formula.

Claim 2. If $U$ is a definable $A$-invariant set then $U$ is definable over $A$.

Proof. Let $\phi(x, b)$ define $U$. We apply claim 1 with $\pi(x) = \{\phi(x, b)\}$ to find a partial type $\Sigma(x)$ over $A$ that defines $U$. So clearly $\phi(x, b)$ implies $\Sigma(x)$. We have that $\Sigma(x)$ is inconsistent with $\neg \phi(x, b)$. So there is $\psi(x, a) \in \Sigma(x)$ that is inconsistent with $\neg \phi(x, b)$ (we may assume $\Sigma(x)$ to be closed under conjunctions). So $\psi(x, a)$ implies $\phi(x, b)$ and we see that $\psi(x, a)$ defines $U$.


问:
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

答:
As I said in a previous answer, finding out that such a simplification occurs is exactly as hard as finding out that $35+18\sqrt{-3}$ has a cube root in $\Bbb Q(\sqrt{-3})$ (well actually, $\Bbb Z[(1+\sqrt{-3})/2]$ because $35+18\sqrt{-3}$ is an algebraic integer).

Suppose $p,q,d$ are integers. Then $p+q\sqrt d$ is an algebraic integer, and so is its cube root, and we can look for cube roots of the form $(x+y\sqrt d)$ with $2x,2y \in \Bbb Z$

So here is what you can do :

compute analytically the $9$ possible values of $(p+q\sqrt{d})^{1/3}+(p-q\sqrt{d})^{1/3}$, using the polar form to take cube roots, so this involves the use of transcendental functions like $\arctan, \log, \exp$ and $\cos$. Does one of them look like an integer $x$ ? if so, let $y = 3qx/(x^3+p)$ and check algebraically if $(x+y\sqrt d)^3 = 8(p+q\sqrt d)$.

try to find a semi-integer root to the degree $9$ polynomial $64x^9-48px^6+(27dq^2-15p^2)x^3-p^3 = 0$. It will necessarily be of the form $\pm z$ or $\pm z/2$ where $z$ is a divisor of $p$, so you will need to decompose $p$ into its prime factors.

see if the norm of $p+q\sqrt d$ is a cube (in $\Bbb Z$). If it's not, you can stop. If you find that $p^2-dq^2 = r^3$ for some integer $r$, then try to find a semi-integer root to the degree $3$ polynomial $4x^3-3rx-p = 0$. Again you only need to check divisors of $p$ and their halves. The check that the norm was a cube allows you to make a simpler computation.

study the factorisation of $p+q\sqrt d$ in the ring of integers of $\Bbb Q(\sqrt d)$. This involves again checking that $p^2-dq^2$ is a cube $r^3$ (in $\Bbb Z$), then find the factorisation of the principal ideal $(p+q\sqrt d)$ in prime ideals. To do this you have to find square roots of $d$ modulo some prime factors of $r$ and find the relevant prime ideals. If $(p+q\sqrt d)$ is the cube of an ideal, you will need to check if that ideal is principal, say $(z)$ (so you may want to compute the ideal class group of $\Bbb Q(\sqrt d))$, and then finally compute the group of units to check if $(p+q\sqrt d)/z^3$ is the cube of a unit or not (especially when $d>0$, where that group is infinite and you need to do some continued fraction computation. otherwise it's finite and easy in comparison).

I would say that methods $2$ and $4$ are strictly more complicated than method $3$, that checking if a polynomial has some integer roots is not too much more evil than doing a prime factorisation to check if some integer is a cube or not. And if you can bear with it, go for method $1$, which is the best, and you don't need a high precision.

This is unavoidable. If you use Cardan's formula for the polynomial of the method $3$ you will end up precisely with the expression you started with, and that polynomial is the one you started with before considering simplifying that expression.

Finally, I think what you want to hear is There is no formula that allows you to solve a cubic real polynomial by only taking roots of real numbers.


问:
general definition of length
Let $\delta x$ be a line vector between two trajectories, could anyone tell me what is this transformation all about and why we would be interested in it?

$\delta z= \Phi(x,t) \delta x$

Reference: page 5, http://web.mit.edu/nsl/www/preprints/contraction.pdf

Thanks!

答:
The idea seems to be that $z(x,t)$ is a set of new coordinates, for instance with the intention to map solutions to parallel lines or similar. Such a transformation might not be possible to construct in general, but one might be able to construct local approximations of such a transform. As in a local context the first interest is in the linear parts, one finds the new relative coordinates for $x+δx$ as $δz=\Theta(x,t)δx$. As, again, the intent is to make the close solutions parallel and equally spaced, a less time-dependent distance of the solutions can be defined via the Euclidean norm of $δz$, which induces a variable metric tensor in the $δx$ coordinates.

The rest of that section is concerned with translating back and forth between properties in $δz$ and $δx$ and what properties are desired in the construction of useful coordinate transformation matrices $Θ(x,t)$.

Title: general definition of length


答:
The ordering of the roots doesn't give you any new information about the polynomial. You have a map
$$ {\mathbb C}^n \ni (r_1,r_2,\dots r_n) \mapsto (c_0,c_1\dots c_{n-1}) \in \mathbb{C}^n$$
This map is surjective, but not injective. That kind of things may happen because $\mathbb{C}^n$ is an infinte set. It is also true for other algebraically closed fields - no algebraically closed field is finite.

Title: Paradox: Roots of a polynomial require less information to express than coefficients?

Body:
A somewhat information theoretical paradox occurred to me, and I was wondering if anyone could resolve it.

Let $p(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0 = (x - r_0) \cdots (x - r_{n-1})$ be a degree $n$ polynomial with leading coefficient $1$. Clearly, the polynomial can be specified exactly by its $n$ coefficients $c=\{c_{n-1}, \ldots, c_0\}$ OR by its $n$ roots $r=\{r_{n-1}, \ldots, r_0\}$.

So the roots and the coefficients contain the same information. However, it takes less information to specify the roots, because their order doesn't matter. (i.e. the roots of the polynomial require $\lg(n!)$ bits less information to specify than the coefficients).

Isn't this a paradox? Or is my logic off somewhere?

Edit: To clarify, all values belong to any algebraically closed field (such as the complex numbers). And note that the leading coefficient is specified to be 1, meaning that there is absolutely a one-to-one correspondence between the $n$ remaining coefficients $c$ and the $n$ roots $r$.


问:
Principal value of $\int_0^\infty \frac{x^{-p}}{x-1}dx$ for $|p|<1$
If $|p|<1$, how to find the Cauchy Principal Value of $$\int_0^\infty \frac{x^{-p}}{x-1}dx$$

I tried spliting the integration from $0\to 1$ and $1 \to \infty$ and switching $x = 1/u$, but no luck getting desired result which the book says is $ \pi \cot p \pi$. Please help.

ADDED:: using above method I got $\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right) $, it turned out to be $p \cot p\pi$ from Mathematica, anyone any ideas?

ADDED:: It seems that $$\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right)  = \sum_{k=-\infty}^\infty \frac{1}{k +p} = - \text{Res}[ \pi \cot(z \pi)/(p+z), z=-p] = \pi \cot p\pi$$

Also I was having trouble doing this via contour. Is this the shape of contour?

答:
Note: we must restrict $p$ to $p \in (0,1)$ for the Cauchy PV of the integral to converge at infinity.

The idea is to consider an integral in the complex plane and use Cauchy's theorem.  That is, if we define a contour $C$ within which there are no poles, we have

$$\oint_C dz \frac{z^{-p}}{z-1} = 0$$

$C$ will take the form of a keyhole contour, except that we will indent with a semicircle about the singularity at $z=1$ each time we make a pass along the real axis.  Thus, the above contour integral is broken up into $8$ pieces:

$$\oint_C dz \frac{z^{-p}}{z-1} = \int_{\epsilon}^{1-\epsilon} dx \frac{x^{-p}}{x-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{(1+\epsilon e^{i \phi})^{-p}}{\epsilon e^{i \phi}} + \int_{1+\epsilon}^{R} dx \frac{x^{-p}}{x-1} +  \\i R \int_{0}^{2 \pi} d\phi \, e^{i \phi} \frac{R^{-p} e^{-i p \phi}}{R e^{i \phi}-1} + e^{-i 2 \pi p}\int_{R}^{1+\epsilon} dx \frac{x^{-p}}{x-1} + \\  i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(e^{i 2 \pi}+\epsilon e^{i \phi})^{-p}}{\epsilon e^{i \phi}} + e^{-i 2 \pi p} \int_{1-\epsilon}^{\epsilon} dx \frac{x^{-p}}{x-1} + \\ i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\epsilon^{-p} e^{-i p \phi}}{\epsilon e^{i \phi}-1} $$

Note that when we make the second pass along the real axis, we exploit the fact that we have traversed $2 \pi$ in argument, so we parametrize the real line using $z=x\, e^{i 2 \pi}$.  The multivaluedness of the integrand produces the nontrivial result we seek.

We take the limit as $\epsilon \to 0$ and $R \to \infty$.  The integrals over $x$ combine to form a factor times the Cauchy PV we seek.  The fourth and eighth integrals vanish in this limit because of the restriction over the value of $p$.  We are then left with the integrals over the bumps which do not vanish.  We thus have, after some simplification:

$$\left ( 1-e^{-i 2 \pi p}\right ) PV \int_0^{\infty} dx \frac{x^{-p}}{x-1} - i \pi \left ( 1+e^{-i 2 \pi p}\right ) = 0$$

Note that the second term comes from evaluating the second and sixth integrals and combining, while the first term comes from combining the first, third, fifth, and seventh integrals.

Solving for the Cauchy principal value, we get

$$PV \int_0^{\infty} dx \frac{x^{-p}}{x-1} = i \pi \frac{1+e^{-i 2 \pi p}}{1-e^{-i 2 \pi p}} = \pi \cot{\pi p}$$


问:
Prove that the radius of convergence of $ f(x)=\frac{1}{(x^4+2021)^2+16} $ is finite
Prove that the function $f:\Bbb R\to \Bbb R$,
$$
f(x)=\frac{1}{(x^4+2021)^2+16}
$$
can be represented as a real power series in a neighbourhood of any $x_0\in \Bbb R$, but for each $x_0$, the radius of convergence is finite.

Most of the times we convert equations by changing scale and origin and then using power series representation of $\frac 1z$ we derive the power series representation but this is a question of complex analysis and I think there is some trick that I am not getting. Can you please help?

答: That's an analytic function (it's the quotient of two analytic functions) and therefore, yes, it is the sum of its Taylor series centered at each $x_0\in\Bbb R$. But the radius of convergence cannot be greater than the distance from $x_0$ to the singularities of $f$ which are located at the fourth roots of the numbers of the form $-2021\pm4i$.


问:
How does integration find average?
I am slightly new to calculus. I haven't been able to get the concept that how plotting the area under the line gives us average value, not the instantaneous value. Then, how do you find the instantaneous value? I have seen few answers but they are not very beginner friendly.</p>

答:
In order to get that reasoning, it's key to understand integrals as infinite sums. I guess you see clear how a discrete sum of values, when divided by the total number of values turns out to be the (arithmetic) average of the set of numbers you summed. This is, if you have, say, a succession of $n$ numbers $\{a_n\}$, now:

$$\frac{1}{n}\sum_{i=1}^{n}a_i=\frac{1}{n}(a_1+a_2+...+a_n)=\bar{a}$$

Where $\bar{a}$ stands for the average of the ensemble. Just to make it clear, let $\{a_n\}={2,3,4}$, so that $n=3$. We would have precisely $\bar{a}=\frac{1}{3}(2+3+4)=3$, just as we knew it would turn out.

Now, imagine that, instead of having a discrete set of numbers (say {2, 3, 4}), you have a continous distribution of these. In order to make it easy, let's say you have the continuous straight line that joins $(x,y)=(2,2)$ to $(x,y)=(4,4)$, which can be mathematically expressed as $y(x)=x$ from $x=2$ to $x=4$.
If you were to sum all of these numbers one by one, you'd never see the end of the task! Since there's an infinity of them. Here's when integrals kick in.

Performing an integral of a function over some interval means summing all of the values of that function, which is continuous, over that interval. This clearly would go towards infinity, since you're summing infinite values... But there's a way around this. See how, in our previous sum, we took 2, 3 and 4 and divided that by 3? Now think about it this way: We took a sample of the interval we wanted to sum (from 2 to 4), calculated the value in 3 points, and then divided by the total value of points we calculated the value of. We could have taken, say, 15 points: {2, 2.2, 2.4, 2.6, 2.8, ..., 3.8, 4}, and we would have divided by 15 instead of 3. This is, now each number contributes with a smaller fraction to the total value of the sum. If you were to follow this argument towards infinity, you would now have an infinite sum of values that contribute $\frac{1}{\infty}$ to the total value of the sum. Those two effects combined converge to a finite value, and this is taken into account when you multiply the integrand by the inginitesimal displacement of $x$, $dx$.

So, back to our integral, we would now have: $\int_2^4xdx$, which would stand for the sum of all of the infinite values between two and four, pondered by their infinitesimally small contribution to the total sum. Performing this calculation gives $\frac{x^2}{2}\bigg\vert_2^4=\frac{4^2}{2}-\frac{2^2}{2}=6$. Dividing this by the interval we took into consideration, which is $\Delta(x)=4-2=2$, we get 3, which is precisely the mean we got when we first did our sum.

We can now do a generalization and see how the average of some continuous set of numbers x in the interval from point a to b, is given by $\bar x=\frac{1}{b-a}\int_a^bxdx$.

In our example, performing the mean by summing or by integrating yielded the same value. This is because all of the numbers we summed over were distributed over a straight line. Were them distributed according to a different function f(x), the values could have been different. It's straightforward to see though, that $\bar f(x)=\frac{1}{b-a}\int_a^bf(x)dx$ will hold for an arbitrary $f(x)$, and if you were to evaluate this function in $N$ different $x$ values and take the average of those, this discretely calculated average would tend to the value of the integral as N tends to infinity.

Hope this makes it clear enough. Good luck!

Title: How does integration find average?


答:
The complex function $f$ with
$$f(z) = \frac{1}{(z^4+2021)^2+16} $$
is meromorphic on $\mathbb C$ and has eight simple poles $z_i$ with nonvanishing imaginary part. Thus it is analytic everywhere else, in particular on the real line, and for all $x_0 \in \mathbb R$ the radius of convergence is the distance from $x_0$ to the nearest complex pole:
$$R_{x_0} = \inf_{z \in \{z_i\}} |x_0 - z|< \infty. $$

Title: Prove that the radius of convergence of $ f(x)=\frac{1}{(x^4+2021)^2+16} $ is finite

Body:
Prove that the function $f:\Bbb R\to \Bbb R$,
$$
f(x)=\frac{1}{(x^4+2021)^2+16}
$$
can be represented as a real power series in a neighbourhood of any $x_0\in \Bbb R$, but for each $x_0$, the radius of convergence is finite.

Most of the times we convert equations by changing scale and origin and then using power series representation of $\frac 1z$ we derive the power series representation but this is a question of complex analysis and I think there is some trick that I am not getting. Can you please help?


问:
Question about verification of the validity of the method of induction
Say whether the following verification of the method of induction is valid or not?

Proof: We have to prove that if

$A(1)$

$(∀n)[A(n) ⇒ A(n + 1)]$

then $(∀n)A(n)$.

We argue by contradiction. Suppose the conclusion is false. Then there will be a natural number n such that: not $A(n)$. Let m be the least such number. By the first condition, $m > 1$, so $m = n + 1$ for some n. Since $n < m, A(n)$. Then by the second condition, $A(n + 1)$, i.e., $A(m)$. This is a contradiction, and that proves the result.

The answer to this question is that the proof is valid.  But I do not understand the reasoning in the proof at:  Since $n < m, A(n)$.  Could someone please explain what we are considering here?  Why is it $A(n)$?

答: You are assuming that $m(=n+1)$ is the smallest natural for which $\neg A(m)$. Since $n<m$, $n$ is smaller that the smallest natural for which $\neg A(m)$, and therefore $A(n)$.


答:
$A(n)$ holds by the way you chose $m$. You write "Let $m$ be the least such number." this means $m$ is the least natural number such that $\neg A(m)$, or as a formula
$$ m = \min \{n : \neg A(n) \} $$
As $m$ is the least number for which $\neg A(m)$ we must have $A(n)$ for all smaller $n$, that is for all $n < m$ we have $A(n)$.

Title: Question about verification of the validity of the method of induction

Body:
Say whether the following verification of the method of induction is valid or not?

Proof: We have to prove that if

$A(1)$

$(∀n)[A(n) ⇒ A(n + 1)]$

then $(∀n)A(n)$.

We argue by contradiction. Suppose the conclusion is false. Then there will be a natural number n such that: not $A(n)$. Let m be the least such number. By the first condition, $m > 1$, so $m = n + 1$ for some n. Since $n < m, A(n)$. Then by the second condition, $A(n + 1)$, i.e., $A(m)$. This is a contradiction, and that proves the result.

The answer to this question is that the proof is valid.  But I do not understand the reasoning in the proof at:  Since $n < m, A(n)$.  Could someone please explain what we are considering here?  Why is it $A(n)$?


问:
Solve system of equations involving cos and sin
I have come up with the following system, I want to solve it for $a$ and $c$:

$
a \sin (x_0) - c \sin(x_0 - L) = 0\\
c \cos(x_0 - L) - a \cos(x_0) = 1
$

In this system $x_0$ and $L$ are arbitrary.

p.s. why can't $L$ be $n\pi$?

答:
Writing $P:= \sin x_0$, $Q := \sin(x_0-L)$, $R := \cos x_0$, $S := \cos(x_0-L)$, we solve this far-less-intimidating system ...
$$\begin{align}
\phantom{-}a P - c Q &= 0 \\
- a R + c S &= 1
\end{align}$$
... to get ...
$$a = \frac{Q}{PS-QR} \qquad c = \frac{P}{PS-QR}$$

Then, we invoke trig's angle-difference formula to simplify:
$$\begin{align}
PS-QR &= \sin x_0 \cos(x_0-L)-\sin(x_0-L)\cos x_0 \\[4pt]
&= \sin\left(x_0-(x_0-L)\right) \\[4pt]
&= \sin L
\end{align}$$
so that
$$a = \frac{\sin(x_0-L)}{\sin L} \qquad c = \frac{\sin x_0}{\sin L}$$
where $\sin L \neq 0$ (whence $L \neq n \pi$ for any $n$).


问:
Exercise 8.5.19 Tao's Analysis 1 (Well-Ordering Theorem)
I am having some trouble with the following problem from Terence Tao's Analysis 1 (Third Edition)

Let $X$ be a set, and let $\Omega$ be the space of all pairs $(Y,\leq)$, where $Y$ is a subset of $X$ and $\leq$ is a well-ordering of $Y$.
If $(Y,\leq)$ and $(Y',\leq ')$ are elements of $\Omega$, we say that $(Y,\leq)$ is an initial segment of $(Y',\leq ')$ if there exists an $x\in Y'$ such that $Y=\{y\in Y': y<'x\}$ (so in particular $Y\subset Y'$), and for any $y,y'\in Y$, $y\leq y'$ iff $y\leq ' y'$.
Define a relation $\preceq$ on $\Omega$ by defining $(Y,\leq)\preceq (Y',\leq ')$ if either $(Y,\leq)=(Y',\leq ')$, or if $(Y,\leq)$ is an initial segment of $(Y',\leq ')$.

I've shown that $\preceq$ is a partial ordering on $\Omega$. Now I am trying to use Zorn's lemma to show that there is a maximal element which will then show that the Well-Ordering Theorem is true. This is what I have so far

Let $Z$ be a totally ordered subset of $\Omega$. We define the set $Y=\bigcup_{(S,\leq_S)\in Z}S$ and a relation on Y $\leq_Y$ defined as follows: $y\leq_Yy'$ if and only if there exists a $(S,\leq_s) \in Z$ such that $y,y' \in S$ and $y \leq_s y'$. It is not hard to verify that this is a partial ordering on $Y$ and in fact it is a total ordering on $Y$.

I am having hard time figuring out how to show that this is a well-ordering on $Y$. I have tried to show that if $W$ is a non-empty set of $Y$ then there is a $(S,\leq_S)\in Z$ such that for all $w \in W$ we have $w \in S$ so in particular there is a $w \in W$ such that $w \leq_s w'$ for all $w' \in W$ and this would imply that $W$ has a minimal element. However I am struggling to make this idea work as well.

I will be very grateful for any help that is offered. In particular how could I go about showing that $\leq_Y$ is a well-ordering on $Y$ (if it is at all)?

答:
If $W \subseteq Y$ is non-empty, pick $w_0 \in W$ and $(S_0,\le_{S_0}) \in Z$ so that $w_0 \in S_0$. Then $W \cap S_0$ is a non-empty subset of the well-order $(S_0,\le_{S_0})$ so has a minimum $m \in S_0 \cap W$ for $\le_{S_0}$.

Now, if $w \in W$ is arbitrary, we know it's in some $(S,\le_S) \in Z$ by definition, and we have (as $Z$ is linearly ordered under $\le$), either $(S_0,\le_{S_0}) \le (S,\le_S)$ or the other way around. If $(S, \le_S) \le (S_0,\le_{S_0})$ we have that $w \in W \cap S \subseteq W \cap S_0$ so that $m \le_{S_0} w$ and thus $m \le_Y w$ follows immediately. If $(S_0,\le_{S_0}) \le (S,\le_S)$ holds, we have $m \in S$ and $m \le_S w$ because $w <_S m$ would have meant that $w \in S_0$ too (because we initial segments !) and that would have contradicted minimality in $S_0$.

So $(Y, \le_Y)$ is a well-order. Now show that $(Y, \le_Y)$ is not a proper initial segment of any $(S, \le_S) \in Z$, to see it is a maximal element of $Y$ in $\Omega$.


问:
How to find the center of a circle, given a point on the circumference of the circle, radius and arc length?
As shown in the figure below, I know:

a point $(x_1, y_1)$ on a circle

the length of an arc (of the same circle) starting at $(x_1, y_1)$

the angle $\theta$ subtended by this arc at the center of the circle

the radius $R$ of the circle

How can I find:

the coordinates of the center of the circle?

$(x_2, y_2)$?

答:
There are infinitely many possible answers.

The figure shows which direction $(x_1, y_1)$ and $(x_2, y_2)$ are, relevant to the centre. However, the text does not reference this information, so it is not clear whether we can rely on it.

Assuming we cannot rely on this information, Seyed’s comment applies (with a slight correction):

The center of the circle can be anywhere on the perimeter of a circle with its center at $(x_1, y_1)$ and radius $R$.


问:
minimizes the conditional expectation of a function
suppose $X\sim U(0,1), Y \sim U(0,x)$, find k that minimizes $E[(X-k)^2|Y=y]$.
My working below:
$E(f(x)|Y) = \int f(x)P(X|Y)dx$

$P(X|Y) = P(Y|X)P(X)/P(Y)$

$P(Y|X) = 1/x,P(X) = 1,P(X,Y)=1/x$

$P(Y)=\int_y^1 P(x,y) dx = -ln(y)$

$P(X|Y) = 1/(x\ln(y))$

$E((X-k)^2|y) = \int_y^1 (x-k)^2/(x\ln y)dx = k^2+k(2/\ln y-y/\ln y - 1/(2\ln y)+ y^2/\ln y)$

differentiate to obtain $g = (y-2)/(2 \ln y)$
clearly this is wrong as when I substitute $y =0.5, g > 1$

How do i fix it?

答:
First you missed a sign because

$$f_{X|Y}=-\frac{1}{x\log y}$$

Using the same procedure as yours I find

$$k=\frac{y-1}{\log y}$$


问:
Find sum of series $S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$, using $(1-x)S$
I study maths as a hobby. I am stuck on the following problem, which is an A level question from the Oxford board set probably in the '70s:

If $S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$, prove, by considering $(1 - x)S$ or otherwise, that if $x \ne 1$,

$$S = \frac{1 + x - (2n + 3)x^{n + 1} + (2n + 1)x^{n + 2}}{(1 - x^2)}$$

Firstly, I am puzzled that the nth term of S is given as $(2n + 1)x^n$.
For example, its 3rd term, $5x^2$ should surely be $7x^3$. Or does the numbering of n start at $n = 0$?

However,

$S =  1 + 3x + 5x^2 + ... + (2n + 1)x^n$

$Sx = x + 3x^2 + 5x^3 + ... + (2n + 1)x^{n + 1}$

$(1 - x)S = 1 + 2x + 2x^2 + ...+ (2n +1)x^{n+1}$

But I don't see how to get from this to the needed proof.

Body:
The numbering starts at $n = 0$.

Your evaluation is not quite complete.  Instead,

$$\begin{align}
S &= 1 + 3x + 5x^2 + 7x^3 + \cdots + \hphantom{(\vphantom{} - 1)}(2n + 1)x^n \\
xS &= \hphantom{1 + 3}x + 3x^2 + 5x^3 + \cdots + (2(n-1)+1) x^n + (2n + 1)x^{n+1} \\
(1-x)S &= 1 + 2x + 2x^2 + 2x^3 + \cdots + \hphantom{((n-1)+1)}2x^n \color{red}{- (2n+1)x^{n+1}}.
\end{align}$$

Therefore, $$(1-x)S + 1 + (2n+1)x^{n+1} = 2 + 2x + 2x^3 + \cdots + 2x^n = 2(1 + x + x^2 + \cdots + x^n).$$  Since we know that the sum of a finite geometric series with common ratio $x$ is $$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1} - 1}{x - 1},$$ it follows that $$(1-x)S = \frac{2(x^{n+1} - 1)}{x-1} - 1 - (2n+1)x^{n+1},$$ and dividing by $1-x$ and simplifying gives the desired value of $S$.

For the OP's benefit:

$$\begin{align}
S &= \frac{1}{1-x} \left( \frac{2(x^{n+1} - 1)}{x-1} - 1 - (2n+1)x^{n+1} \right) \\
&= -\frac{2(x^{n+1} - 1) - (x-1) - (2n+1)x^{n+1}(x-1)}{(x-1)^2} \\
&= -\frac{2x^{n+1} - 2 - x + 1 - (2n+1) x^{n+2} + (2n+1) x^{n+1}}{(x-1)^2} \\
&= \frac{(2n+1) x^{n+2} - (2n+3)x^{n+1} + x + 1}{(x-1)^2}.\end{align}$$


答:
Here's an intuitive way to look at it:

Suppose the function $f(t)$ represents something physical, like the height of a water level over time. If the graph of $f(t)$ is piecewise constant (made of horizontal lines) like this:

then you can compute its average value over any time period using algebra (no calculus needed). It's a weighted sum:
For each height level, you measure the height and multiply it by whatever percent of the time the function is at that height (in other words, the length of time the function is at that height, divided by the total time). Then you add all those weighted values together to get the average height overall.

For example, if the function is at height 6m for 3 out of 10 total seconds, and at height 21m for 7 out of 10 total seconds, then the average height is 6*0.3 + 21 * 0.7 = 16.5m.

Your answer has the right physical units: height. And it has the right behavior in extreme cases, like if the function is at height 6 for 99% of the time, the average value will be close to 6.

Another way of writing this weighted sum is like this: for each height, you multiply it by the length of time the function is at that height. You add up all of those values, then divide by the total length of time. This gives your average height just like the earlier formula.
To reuse the earlier example, we could also have computed the average as (6*3 + 21*7)/10.

This second formula is convenient because it corresponds to something in the picture. It says that to compute the average height of this function, just compute the area of each of the rectangles. Add up those areas, then divide by the total width of the time interval (!).

We've arrived at a formula: To compute the average height of a function over a certain interval of time, compute the area under that function during that interval, then divide by the length of that interval.

You can think of integration, in part, as a tool for calculating the area under curves that are not made of straight lines. If a curve has the formula $f(t)$, then its integral $\int_{a}^b f(t)\,dt$ tells you the area under that curve.

Intuitively, the formula for average height should still be the same: compute the area under the curve, then divide by the length of the interval.  Turns out, that's exactly right: if $f(t)$ is a function, then its average height over the interval between $t=a$ and $t=b$ is its area divided by the length of the interval, or in symbols:

$$f_{ave} = \frac{1}{b-a} \int_{a}^b f(t)\,dt$$

(Nitpick: technically the areas are *signed*, so areas under the curve count as negative, but I didn't want to clutter the discussion with that detail.)

Title: How does integration find average?

Body: I am slightly new to calculus. I haven't been able to get the concept that how plotting the area under the line gives us average value, not the instantaneous value. Then, how do you find the instantaneous value? I have seen few answers but they are not very beginner friendly.</p>


问:
How to prove the set of group axioms is independent?
Hey I am having a hard time solving a logic problem. It one of the exercises in Ebbinghaus, Mathematical Logic (exercise 4.14, p. 39):

A set $\phi$ of S-formulas is independent if there is no $\varphi \in \phi$ s.t. $\phi  $/$  \{\varphi\} \vDash \varphi$. Let $\mathcal{S}=\{\circ, e\}$ and $$\phi := \{ \forall x \forall y \forall z(x \circ (y \circ z) \equiv (x \circ y) \circ z), \forall x \ x \circ e \equiv x, \forall x \exists y \ x \circ y \equiv e \}\text{.}$$ Show that  $\phi$ is independent.

My thoughts: I need to find three models $M1$, $M2$, $M3$ s.t. $M1$  satisfies the second and third formulas, but cannot satisfy the first formula $\forall x \forall y \forall z(x \circ (y \circ z) \equiv (x \circ y) \circ z)$, so the first formula is independent.

$M2$, $M3$ are similar cases. i,e., $M2$ can satisfy the first formula and the third formula, but cannot satisfy the second formula, so the second one is independent. $M3$ can satisfy the first formula and the second formula, but cannot satisfy the third formula, so the third one is independent.

Like, consider M3, let the domain of M3 is $\mathbb{Z}$, $\varphi(\circ)$ $= +$ on $\mathbb{Z}$，and $\varphi(e) = 0$, so $M3$ satisfies the first formula (associative law) and the second $\mathbb{Z}+0=0$. But $M3$ cannot satisfy the third one, so the third formula is independent.

But I'm wondering :

(1) Could this kind of specific model works? I mean I just find a "specific model" to satisfy two formulas in the set. I just feel I need to prove independence in a more general way, right? But I have no idea how to do.

(2) Could anyone think examples of $M1$ and $M2$?

Any ideas please? I appreciate your help.

答:
I'll just write in a very non-rigorous way but you can get some ideas.

For $M_3$, take any monoid that is not a group, such as $(\mathbb N,+)$.

For $M_1$, consider $(\mathbb Z,-)$. More generally, for most groups $(G,\cdot)$, define a new operation $x\circ y=x\cdot y^{-1}$, and use it on $G$. (Won't work so easily if your axioms are two-sided though. Maybe the octonions?)

For $M_2$, take any group but interpret $e$ differently (not identity). For example take $e=1$ in $(\mathbb Z,+)$.


问:
How can I prove that the Core of $H$ in $G$ is a normal subgroup?
Given $G$ a group and $H$ a subgroup of $G$, how can I prove that ${\rm Core}_H G:=\bigcap_{g\in G} gHg^{-1}$ is a normal subgroup of $G$?

I've proved that is a subgroup, but how can I prove that it is a normal subgroup?

If $x \in{\rm Core}_H G,\ y \in G$, how can I prove that $yxy^{-1}$ is in ${\rm Core}_H G$?

As $x \in{\rm Core}_H G,\ \forall g\in G,\ \exists a\in H:x=gag^{-1} $
Moreover, as ${\rm Core}_H G \subseteq H$, (since it is subgroup), $x\in H$

How can I continue and show that $yxy^{-1}\in {\rm Core}_H G$

答: Yet another way to get this result: $\operatorname{Core}_G(H)$ is the kernel of the $G$-action by left multiplication on left quotient set $G/H$ (see here for a proof). As such, it is normal in $G$.


问:
Weak convergence of a dominated sequence in $L^1$
We are given a sequence of functions $\{f_n\}$ on $L^1[0,1]$ so that for each $g \in L^\infty[0,1]$, we have $\int f_ng$ converges. We are also given that there exists $u \in L^1$ such that $\lambda(|f_n|>M)\leq \lambda(|u| >M)$ for all $M>0$. How to prove that $f_n$ converges weakly to some $f \in L^1$?

My approach goes like this: first consider $|f_n|$ being bounded: $|f_n|\leq M$ for some $M$, and consider $T: L^1 \to (L^\infty)^*$ by $T(f_n)(g) := \int f_ng$ for $g \in L^\infty$, so that $T(f_n)$ converges in $(L^\infty)^*$. By boundedness of the sequence and Alaoglu theorem, we prove that $f_n \rightharpoonup f$. Does it makes sense?

Now for unbounded sequence, consider $\tilde{f}_n := f_n \chi_{|f_n|\leq M}$, so that $\tilde{f}_n$ is bounded and weak limit exists. How should I proceed?

答: Let $\mu (A) =\lim \int_A f_n =\lim \int f_n g$ where $g=\chi_A$. Then $\mu$ is a real measure and $\mu << m$ where $m$ is the Lebesgue measure. Let $f=\frac {d\mu} {dm}$. Now $\int |f_n| =\int_0^{\infty} m\{x:|f_n(x)| >M\} dM \le \int_0^{\infty} m\{x:u(x) >M\} dM=\int u $. Hence, $\sup_n \int |f_n| <\infty$. We now have $\int f_n g =\int fg$ for any simple function $g$. Any $g \in L^{\infty}$ is a uniform limit of simple functions. Now you can use triangle inequality to show that $\int f_n g =\int fg$ for any $g \in L^{\infty}$.


问:
Verify integral using Residue Theorem
Verify the following equation:

$$
 \int_{0}^{2\pi}{\log{\left(\sin^{2}{2\theta}\right)}\,\mathrm{d}\theta}  =  4\int_{0}^{\pi}{\log{\left(\sin{\theta}\right)}\,\mathrm{d}\theta} = -4\pi \log{2}$$

I am unable to think of a closed rectifiable curve so that I can apply the residue theorem to prove the above equation. At first I thought we can put $z=e^{i\theta}$ and take $0\leq \theta \leq 2\pi $ but then I could not understand the singularities and analyticity of the log in that.

答:
I did not find a way to perform the integration by evaluating a residue at a singularity of a function. However I will present below a hint how to calculate the integral using contour integration of an analytic function.

$$\begin{align}
\int_0^{\pi} \log(\sin x)\,dx
&=\int_0^{\pi} \log\left(\frac{e^{ix}-e^{-ix}}{2i}\right)\,dx\\
&=\int_0^{\pi} \left[\log\left(1-e^{-2ix}\right)+ix-\log(2i)\right]dx\\
&=\int_0^{\pi}\log\left(1-e^{-2ix}\right)\,dx+\frac{i\pi^2}2-\pi\log2-\frac{i\pi^2}2\\
&=\int_0^{\pi}\log\left(1-e^{-2ix}\right)\,dx-\pi\log2,
\end{align}$$
where we used $\log(i)=\frac{i\pi}2$.

So we have to prove $\int_0^{\pi}\log\left(1-e^{-2ix}\right)\,dx=0$. This can be done by the integration over the following path:
$$\begin{array}{lllll}
(1):&\quad \epsilon&\to&\pi-\epsilon\\
(2):&\quad \pi-\epsilon&\to& \pi+i \epsilon\\
(3):&\quad \pi+i\epsilon&\to& \pi+iR\\
(4):&\quad \pi+iR &\to& iR\\
(5):&\quad iR&\to& i\epsilon\\
(6):&\quad i\epsilon&\to&\epsilon\\
\end{array}$$
Observe that the integrals along $(3)$ and $(5)$ cancel each other, hence it remains to show that the integrals along $(2),(4),(6)$ tend to $0$ as $\epsilon\to0,\; R\to\infty$, which is left as an exercise.

Title: Verify integral using Residue Theorem


答:
As $x$ varies over $[a,b],\;$ $f(x)$ continuously attains different
values.

The average $A_{[a,b]}$ of all these values $\{f(x):a\leq x\leq
   b\}$ is their sum divided by the amount/length $(b-a).$

But this (infinite) sum is just the signed area
between the curve $y=f(x),$ the $x$-axis, and the vertical lines
$x=a$ and $x=b,$ isn't it? This equals $\int_a^bf(x)\,\mathrm{d}x.$

Hence, $$A_{[a,b]}=\frac1{b-a}\int_a^bf(x)\,\mathrm{d}x.$$

The instantaneous value of $f(x)$ at $\alpha$ is just
$f(\alpha).$

So, if $f(t)$ represents instantaneous velocity at time $t$,

then the average velocity over $[t_1,t_2]$ is $$\frac1{t_2-t_1}\int_{t_1}^{t_2}f(t)\,\mathrm{d}t.$$

Title: How does integration find average?


答:
among radius, arc length and subtended angle, one parameter can be deduced from the others;

as you don't know the coordinates of the point $p_2$, the information about this point cannot be exploited !

a circle of radius $r$ and passing through $p_1$ has its center somewhere on the circle of radius $r$ centered at $p_1$.

This is all you can say.

Title: How to find the center of a circle, given a point on the circumference of the circle, radius and arc length?

Body:
As shown in the figure below, I know:

a point $(x_1, y_1)$ on a circle

the length of an arc (of the same circle) starting at $(x_1, y_1)$

the angle $\theta$ subtended by this arc at the center of the circle

the radius $R$ of the circle

How can I find:

the coordinates of the center of the circle?

$(x_2, y_2)$?


问:
How would I go about finding $\theta$ and suggesting an unbiased estimator?

a) Suppose $N$ is a Poisson random variable with mean $\lambda$. Show that $E[(1+a)]^N]=e^{a\lambda}$ for any constant $a$.

b) Telephone calls enter a switchboard in a Poisson process of rate $\lambda$ every $10$ minutes. In a 10 minute period, $N$ calls enter the switchboard. In order to decide whether he can chance an unofficial tea-break, the operator wants to estimate the probability $\theta$ that there will be no incoming calls in the next $20$ minutes, using this single observation
$N$.

Write down an expression for $\theta$ in terms of $\lambda$. Use this, together with the result in part (a), to suggest an unbiased estimator $T(N)$ of $\theta$. Is this estimator sensible?

I have done the first part of the question but I am unsure on how to find $\theta$ in the next part and how to find an estimator. I am thinking that I should set $\alpha$ to be something but I am unclear on how these estimators work.

答:
The probability that there are no calls in the next $20$ minutes is simply $\Pr[N(2) = 0] = e^{-2\lambda}$, where $N(t) \sim \operatorname{Poisson}(\lambda t)$ is the random number of calls in $t$ time increments of $10$ minutes.  So $$\theta = e^{-2\lambda}$$ is the desired relationship.  This would seem to suggest letting $a = -2$ in part (a), which would give you $$\operatorname{E}[(-1)^N] = e^{-2\lambda} = \theta,$$ and so the estimator $$T(N) = (-1)^N$$ is unbiased for $\theta$.  But when it comes to asking whether it's a good estimator, is it?  It would mean that if $N$ is odd, you would choose $\theta = -1$, whereas if $N$ is even, you would choose $\theta = 1$, and since $\theta$ is a probability--remember, it is the probability of no calls in the next $20$ minutes--this is an awful choice for an estimator.

Is there a better estimator we can choose, even if it is possibly biased?


问:
Out of the box & WOW effect Probability
It is a common knowledge, that problems from graph theory/combinatorics can be often solved via short and
astonishing probabilistic reasoning. In contrast to this, it is much harder to find such "out of the box"/"WOW effect" solutions for analysis, topology or algebra problems. So, what are your favorite problems like this, how do you prove them?

This question was inspired by a very nice analysis problem:

evaluate $ \ \lim_{n\to \infty} e^{-n} \cdot \sum_{k=0}^{n}\frac{n^k}{k!}$.

with the following, extremely beautiful solution:

This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$.

By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$.

Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED.

Both, question and solution come from Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$.

I think  that such a list might be helpful for educational purposes. Such problems act on the imagination of students/listeners and much more likely raise their interest.

答:
I have lately found a nice problem that matches the description:

Consider $m,n \in \mathbb{Z}_+$ and  $p,q\ge 0$ with $p+q=1$. Show
that $$(1-p^n)^m+(1-q^m)^n \ \ge 1.$$

It is quite elementary, which makes it actually even more fun.

Title: Out of the box & WOW effect Probability


问:
Caculus of $j^j$, $i^j$ and $j^i$ (where $i^2=-1$, $j^2=1$, $j \neq -1$ and $j\neq 1$), and others like $Ln(i+j)$
First of all, I am do math only for fun, and I am only amateur, so excuse me because of my lack on knowledge in this field.
I was curious about some answers and I want to know if my judgment was ok or wrong.
First I wanted to find out the answer of:
$$i^j$$ (where $i^2=-1, j^2=1, j≠-1, j≠1$), j is split complex number
$$j^j$$
$$j^i$$

Maybe I was wrong to combine complex number with split numbers. I don’t know.
Please let me know If my calculus is ok or there are some mistakes in it.

Thank you.

First:
$$\ln(j)$$
I use the formula for logarithm of split complex numbers:
$$\ln (a + bj) = ½ (\ln ((a + b)(a - b)) + ½ j( \ln (a + b)/(a-b))$$

I know that $(a + b)(a - b) < 0$, we will get a complex number in our answer and that that the split complex logarithm is not closed.
Of course for all formulas I take into consideration only the principal branch.

So:
$$Ln(j)= ½((\ln (0+1)(0-1))+1/2 j (\ln (0+1)/(0-1))$$
$$=1/2 \ln (-1) + ½ j \ln (-1)$$
$$= ½ (i \pi ) +1/2 j ( i \pi)$$

BUT, when I wanted to verify I am correct I discover that, the answer $ ½ (i \pi ) +1/2 j ( i \pi)$ is actually $\ln(-j)$.

I also discover that are at least two correct answer for $\ln(j)$ and for $\ln(-j)$.

Ex. for $\ln(j)$ :
$$Ln(j)=i\pi/2 – j i\pi/2$$ and $$Ln(j)=-i\pi/2 + j i\pi/2$$

Proof:
$$e^{i\pi/2–ji\pi/2}$$

$$=(e^{i\pi/2})*(\cosh(-i\pi/2)+j\sinh(-i\pi/2))$$

$$=(e^{i\pi/2})(0 -ij)$$

$$=(-ij)e^{i\pi/2})$$

$$=(-ij)(\cos(\pi/2)+i\sin(\pi/2)$$

$$=-iji=j$$

I hope is nothing wrong in my calculus.

So $$j^j=e^{j\ln(j)}$$

$$=$$

1) $$e^{(i\pi/2–ji\pi/2)j}=e^{ji\pi/2–i\pi/2}=-j$$

$$=e^{(-i\pi/2)}(cosh(i\pi/2)+jsinh(I\pi/2)) =e^{(-i\pi/2)}(0-ij)=-ij(cos(-\pi/2)+isin(-\pi/2))=-ij(0-1i)=-j$$

2) $$e^{(-i\pi/2+ji\pi/2)j}= e^{-ji\pi/2+i\pi/2}=-j$$
because $$Ln(j)=i\pi/2 – j i\pi/2$$ and $$Ln(j)=-i\pi/2 + j i\pi/2$$

$$=e^{(i\pi/2)}(cosh(-I\pi/2)+jsinh(-i\pi/2))= e^{(i\pi/2)}(0+ij) =ij(cos(\pi/2)+isin(\pi/2))=ij(0+1i)=-j$$

So... $$j^j=-j$$....
Am I wright?

$$Ln(i)= i\pi/2$$

$$j^i=e^{(ilnj)}$$

1) $$e^{(i\pi/2 – j i\pi/2)i}=e^{-\pi/2 + j pi/2}= (e^{-\pi/2})(\cosh (\pi/2)+ j \sinh (\pi/2))$$

$$=0.521606...+j0.478393...$$

2) $$e^{(-i\pi/2 + j i\pi/2)i}= e^{i\pi/2 - i\pi/2} = (e^{\pi/2})(\cosh (-\pi/2)+ j \sinh (-\pi/2))$$

$$=12.070346...-j11.070346$$

Am I wright?

$$i^j$$
$$=e^{ji\pi/2}$$

$=\cosh(i\pi/2)+j(\sinh(i\pi/2)$

$=0+j*i$

So ... $i^j=ij$....
Am I right?

Proof:

$i^j=ij$

$$jlni=ln(ij)$$

$$ln(ij)=1/2(ln (0+i)(0-i)) +1/2j ((ln (0+i)/(0-i))$$

$$ln(ij)=1/2ln1 +1/2j ln (-1)$$

$$ln(ij)= (ipi/2)j$$

$$jlni=ln(ij)$$

$$j(ipi/2)=(ipi/2)j$$

Is it possible to find $ln(i+j)$? Cand I find $\ln(i+j+\varepsilon)$, where $\varepsilon^2=0$, with  $\varepsilon \neq 0$?

Thank you.

答:
Split-complex numbers can be represented as 2x2 matrices: $a+bj=\left(
\begin{array}{cc}
 a & b \\
 b & a \\
\end{array}
\right)$.

So, $j^j=\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)^ \left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)=\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)=j$

Verify it with Wolfram Alpha.

It seems you possibly made a mistake with sign.

The combination of complex and split-complex numbers is called tessarines, they can be represented as 2x2 matrices $a+bj=\left(
\begin{array}{cc}
 a & b \\
 b & a \\
\end{array}
\right)$ with complex elements $a$ and $b$.

So, in tessarines, you are correct that $i^j=ij$.
The $j^i$ is $\left(\frac{1}{2}-\frac{e^{-\pi }}{2}\right) j+\frac{e^{-\pi }}{2}+\frac{1}{2}$


问:
generalized derivative of Wiener process
Defined a standard Wiener process $W = (W_t , \mathcal F_t)_{t≥0}$ and a deterministic, continuously differentiable function $f : [0, ∞) → \mathbb R$. Prove that
$$f(t)W_t=\int_0^tW_sf'(s)ds+\int_0^tf(s)dW_s$$</p>

答:
If you know Ito Lemma, then you just consider $F(t,x) = f_t\cdot x$ and the differential of $F$ is
$$
  \mathrm dF(t,W_t) = f'_tW_t\mathrm dt + f_t\mathrm dW_t + \mathrm d[f,W]_t
$$
and since $f$ is deterministic and continuous, the latter term is zero. Hence
$$
  \mathrm dF(t,W_t) = f'_tW_t\mathrm dt + f_t\mathrm dW_t \quad\Leftrightarrow \quad F(T,W_T) = F(0,W_0) + \int_0^T f'_tW_t\mathrm dt + \int_0^Tf_t\mathrm dW_t.
$$</p>


问:
Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.
Let $R$ be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.

My attempt: Let $I$ be an ideal of $R$. Then we have $I$ is maximal $\Leftrightarrow$ $R/I$ is a finite field $\Leftrightarrow$ $R/I$ is a finite integral domain $\Leftrightarrow$ $I$ is a prime ideal.

Is my proof valid ?

答:
If a finite commutative ring $R$ is without unity, then only one direction of your statement is true, that is

$I$ prime ideal $\iff$ $R/I$ finite integral ring $\iff$ $R/I$ finite field $\Longrightarrow$ $I$ maximal ideal

Why can't be the last implication reverted? Example: $R=2\mathbb Z / 4 \mathbb Z = \{0,2\}$ and $I = \{0\}$. Then $I$ is maximal but not prime because $2\cdot 2 = 0 \in I$, but $2 \notin I$. In other words: $R/I = R$ is not a field.


问:
How to do this proof of sets?
The proof example preceding the question seems a bit random to me, if someone could explain the proof I would understand all questions of the type

Let $A,B,C,D$ be arbitrary sets. Prove that: $$(A\times B)\cap (C\times D)=(A\times C)\cap (B\times D).$$

Thanks!

答:
Take $A=B=\{1\}$ and $C=D=\{2\}$.

Then

$A\times B = \{(1,1)\}$

$C\times D = \{(2,2)\}$

so $(A\times B)\cap (C\times D) = \emptyset$

$A\times C = \{(1,2)\}$

$B\times D = \{(1,2)\}$

so $(A\times C)\cap (B\times D) = \{(1,2)\}$

The statement is false.

Title: How to do this proof of sets?


问:
compute $\log_e(j)$ of split complex number $j$
I am trying to calculate the value of $\ln j$ where $j^2=1, j\ne\pm1$($j$ is split complex).

This is how I did it:

given $e^{j\theta}=\cosh\theta+j\sinh\theta$ I can set $\cosh\theta=0\implies \theta = i\pi n - \frac{i \pi}2, n \in \Bbb Z,i$ is the imaginary number, for convenience sake i'll take $n=1$, using this i can calculate $\sinh\theta$ and get $e^{ji\frac\pi2}=0+ji=ji$(which also implies that $\cos\left(j\frac\pi2\right)=0$ and $\sin\left(j\frac\pi2\right)=j)$

now divide it by $i$(or multiply by $-i$) and i get $\frac{e^{ji\frac\pi2}}i=j$

so: $$\ln j=\ln\left(\frac{e^{ji\frac\pi2}}i\right)=\ln\left(e^{ji\frac\pi2}\right)-\ln i=ji\frac\pi2-i\frac\pi2$$

Am I right by doing this?

答:
Split-complex numbers are not closed under logarithm, so one has to assume tessarines (a combination of complex and split-complex numbers). They can be represented as 2x2 matrices of the form $a+bj=\left(
\begin{array}{cc}
 a & b \\
 b & a \\
\end{array}
\right)$ with complex elements $a$ and $b$.

So, we can verify your answer with Wolfram Alpha.
The result is $\ln j=i\frac\pi2-ji\frac\pi2$, it differs from your result in sign, but it depends on the branch choice, as $1/j=j$.


问:
Problem using Bayes' Theorem
I know I'm going wrong somewhere just not sure where

I haven't used Bayes' theorem much before so any help would be greatly appreciated.

Suppose you are given following data.

$5\%$ of the population have heart disease.

If you have heart disease, the probability that you have high blood pressure is $90\%$

If you do not have heart disease, the probability that you have high blood pressure is $15\%$

What is the probability that a person chosen at random from the population
 has high blood pressure?

$$P(B) = P(B|H)P(H) + P(B|H')P(H')$$

$$P(B) = (.9)(.05) + (.15)(.95) = .1875$$

Using Bayes Theorem calculate the probability that the person has heart disease, if they have high blood pressure.

$$P(H|B) = \frac{P(B|H)P(H)}{P(B)}$$

$$(.9)(.05)/.1875 = .24$$

Using Bayes Theorem calculate the probability that the person has heart disease, if they do not have high blood pressure.

$$P(H|B') = \frac{P(B'|H)P(B')}{P(H)}$$

When I sub in for this part I'm getting an invalid answer

答:
The reason why you got an incorrect answer for the 3rd sub-part is because of the incorrect implementation of the Bayes rule.

Your RHS gives $P(B'|H)$ which can be calculated as $1-P(B|H)=1-0.9=0.1$

Here $P(H|B')=\frac{P(H)P(B'|H)}{P(B')} = \frac{P(H).[1-P(B|H)]}{1-P(B)} = \frac{0.05*0.1}{0.8125} = \frac{50}{8125}=0.0061$

Title: Problem using Bayes' Theorem


问:
Finding out the argument a complex number
So this is something I'm still having issues with. Lets say I have the following:
$$z^6 = -8$$
Now if I take $\frac{1}{6}$ power on both sides, I get $z = \sqrt{2}i$. Now, if I look at the point that this is in, I mean the value for the real number is $0$ so it must be at angle $\frac{\pi}{2}$ right? For some reason though, it should be the $arg(z) = \frac{\pi}{6} + \frac{\pi}{3}k$ which I don't quite understand. Would appreciate some help, thanks in advance!

Body:
$$z^6=-8=2^3e^{i\pi}=2^3e^{(2n+1)\pi i}$$ where $n$ is any integer

$$\implies z=2^{3/6}e^{(2n+1)\pi i/6}$$  where $n=0,1,2,3,4,5$


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答: If $f(x)$ is the function then $E = \{ x \colon f(x) = 0 \} $ is a closed set. Do you know of a nontrivial, by your standards, closed set with uncountable many points? The complement of $E$ is open. Is there a way to describe the complement of $E$ so that it would be easy to construct the remainder of the function in such a way that the function was never $0$ on the complement of $E$.</p>


问:
Principal value of $\int_0^\infty \frac{x^{-p}}{x-1}dx$ for $|p|<1$
If $|p|<1$, how to find the Cauchy Principal Value of $$\int_0^\infty \frac{x^{-p}}{x-1}dx$$

I tried spliting the integration from $0\to 1$ and $1 \to \infty$ and switching $x = 1/u$, but no luck getting desired result which the book says is $ \pi \cot p \pi$. Please help.

ADDED:: using above method I got $\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right) $, it turned out to be $p \cot p\pi$ from Mathematica, anyone any ideas?

ADDED:: It seems that $$\displaystyle \sum_{k=0}^\infty \left( \frac{1}{k + p} + \frac{1}{p - (k+1)} \right)  = \sum_{k=-\infty}^\infty \frac{1}{k +p} = - \text{Res}[ \pi \cot(z \pi)/(p+z), z=-p] = \pi \cot p\pi$$

Also I was having trouble doing this via contour. Is this the shape of contour?

答:
Though Ron Gordon was here first, let me describe another (hopefully simpler) approach.

Complete the contour of integration for principal value (i.e. $(0,1-\epsilon)\cup(1+\epsilon,\infty)$) by a half-circle of radius $\epsilon$ in the upper half-plane (centered at $1$). This adds  $-i\pi$ to the initial integral.

Next rotate the integration contour by $\pi$ counterclockwise so that it becomes the negative real axis. The integral then transforms into
$$e^{-i\pi p}\int_{0}^{-\infty}\frac{(-x)^{-p}d(-x)}{(-x)+1}=e^{-i\pi p}\int_{0}^{\infty}\frac{y^{-p}dy}{y+1}=e^{-i\pi p}\frac{\pi}{\sin\pi p}.$$
Therefore the initial principal value integral is equal to
$$e^{-i\pi p}\frac{\pi}{\sin\pi p}+i\pi=\pi \cot\pi p.$$


问:
Intuition explanation of the gauge function in optimization
The gauge function is defined as $\gamma(x|C) = \inf\left\{\lambda  \ge 0 | x \in \lambda C \right\}$ where $C$ is a convex set. I read that it is a generalization of norm concept but I can't understand that. Is there an intuitive  explanation of what this function does in optimization?

答:
First, usually there is a condition that $0 \in \mathring C$ or something similar ; else there are some points who will have infinite gauge value.

The relation to the norm is this : take $C = B_N(0,1)$ the ball for the norm $N$, then $N(x) = \gamma(x|C)$. Indeed, it's the smallest value $\lambda$ such that $x \in B_N(0,\lambda)$.

Actually, if the convex set $C$ is bounded and symmetric ($x \in C \Leftrightarrow -x \in C)$, then $\gamma(\cdot,C)$ is a norm (without the bounded attribute, it's only a seminorm). So the concepts are very similar. You can simply see a gauge function as an asymmetric seminorm.


问:
Is there a compact flat isotropic manifold?
A Riemannian manifold $M$ is flat if it has constant curvature zero. It is called isotropic if all directions are equal under symmetry, that is for $p\in M$ and unit vectors $v,w\in T_p M$ there is a $\phi\in\mathrm{Aut}(M)$ with $\phi(p)=p$ and $\phi_*(v)=w$.

Question: is there a compact flat isotropic manifold of dimension $d\ge 2$?

I suspect that it is safe to say that this is not possible in dimension $d=2$, so I care most about $d=3$.

Compactness is neccessary as otherwise we can choose $\Bbb R^d$.

答:
First, your definition is not quite right, isotropy requires $v, w$ to be unit tangent vectors.

There are no such manifolds in any dimension $n\ge 2$. If such a manifold $M$ were to exist, it would be the quotient of the flat ${\mathbb R}^n$ by a discrete (and torsion-free) group $\Gamma$ of isometries. Lift the isometry group of $M$ to the universal covering; the result is a subgroup $G$ of Euclidean isometries normalizing $\Gamma$. By the isotropy assumption, for each (enough, some) $q\in {\mathbb R}^n$ its stabilizer $G_q$ would act transitively on the unit tangent sphere. It is a nice exercise to conclude from this that the group $\Gamma$ then would be non-discrete (this is easier if you already know Bieberbach's theorems, which imply that $\Gamma$ contains a nontrivial translation). A contradiction. qed


问:
proving 2 limits are equal
How do I prove that: $$\lim_{h\to0}f(x_0+h) = \lim_{x\to x_0}f(x)$$ using $\epsilon$-$\delta$ definition of limit.

I have seen a few proofs online for similar questions. But they assume a case such as $x_0+h=x$, or that $\delta$s for both functions are equal. Which I don't understand.

Intuitively they both approach the same point (Assuming the limits exist), so why are such conditions necessary?

One of the proofs I found online:
Epsilon-Delta Proof of Limits Being Equal

答:
The method you described is essentially the way. I suspect that your own issues are with using the limit definition itself. As you'll no doubt be aware, to say $\lim_{x \to x_0} f(x) = L$ is to say:

For all $\varepsilon > 0$, there exists some $\delta > 0$ such that
$$0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon.$$

So, no matter what positive number $\varepsilon > 0$ I choose, you should be able to provide me with a $\delta > 0$ so that, whenever $x \neq x_0$ is between $x_0 - \delta$ and $x_0 + \delta$, we have $f(x)$ is between $L - \varepsilon$ and $L + \varepsilon$.

When you're proving a limit, the onus is on you to find such a $\delta$. For most limits, this $\delta$ will change depending on the $\varepsilon$ provided. So, it is natural to have it as a function of $\varepsilon$.

When you're using an established limit, you're allowed to pick whatever positive $\varepsilon$ you want, and can assume that there is a $\delta$ to match (that's what assuming the limit's existence gives you). Then, for any given $x$ satisfying $0 < |x - x_0| < \delta$, you can then assume $|f(x) - L| < \varepsilon$.

So, let's look at the given problem. To start with, we should assume one of the limits exists, and give it a name. We'll stick with $L$, for simplicity. Let's assume that $\lim_{h \to 0} f(x_0 + h) = L$, and try to prove that $\lim_{x \to x_0} f(x) = L$ as well.

To prove the latter limit, we have suppose we have some arbitrary $\varepsilon > 0$, and find a way to construct a $\delta > 0$ such that, whenever $0 < |x - x_0| < \delta$, we also have $|f(x) - L| < \varepsilon$. So, the onus is on us to provide the $\delta$.

We are assuming that $\lim_{h \to 0} f(x_0 + h) = L$. That is, if we set $\varepsilon'$ to any strictly positive number we wish, we can automatically assume that there is some $\delta' > 0$ such that, whenever $0 < |h - 0| < \delta'$, we also have $|f(x_0 + h) - L| < \varepsilon'$.

How do we do it? We must, as usual, start by supposing $\varepsilon$ is some arbitrary strictly positive number. We have no control over its value, but we know it's positive.

Next, we use the definition of $\lim_{h \to 0} f(x_0 + h) = L$, taking $\varepsilon' = \varepsilon$ (we can set $\varepsilon'$ to any positive number, so why not $\varepsilon$?). We then get a $\delta'$ such that
$$0 < |h - 0| < \delta' \implies |f(x_0 + h) - L| < \varepsilon' = \varepsilon. \tag{$\star$}$$
Remember, we need to find a $\delta$ such that
$$0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon.$$
I claim (as do all the other answers you've seen) that $\delta = \delta'$ (the value with the certain properties that we are allowed to assume exists) will do the trick.

How do we prove this claim? Start by assuming $0 < |x - x_0| < \delta = \delta'$. Recall that $(\star)$ tells us that, whenever any real number $h$ satisfies $0 < |h| < \delta'$, then we automatically have $|f(x_0 + h) - L| < \varepsilon'$. So, our assumption says that the particular value of $h$, $x - x_0$, satisfies this inequality. Thus, using $(\star)$ in the particular case $h = x - x_0$, we get
$$0 < |x - x_0| < \delta = \delta' \implies |f(x_0 + (x - x_0)) - L| < \varepsilon' = \varepsilon.$$
Once we simplify the right hand side, this is precisely the implication we wanted to prove. Thus, choosing $\delta = \delta'$ works; we have a guaranteed $\delta > 0$ such that
$$0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon.$$
Technically speaking, we aren't done yet, as we should also assume that $\lim_{x \to x_0} f(x)$ exists, and use that to prove $\lim_{h \to 0} f(x_0 + h)$ exists and has the same value, but I'll leave that to you.

Title: proving 2 limits are equal


问:
Is there a closed-form solution of this ODE (computing a flow on an ellipse)?
Can we find an explicit closed-form solution of the following ODE?

$\beta:[0,\epsilon] \to (0,2\pi]$, $\beta(0)=\beta_0$, and
$$\dot \beta(t)=\frac{c}{R(\beta(t))+c_0},$$ where $c,c_0$ are constants, and
$$R(\theta):=\frac{1}{\sqrt{1-\big(e \cos(\theta)\big)^2}}, 0<e<1.$$

The motivation for this ODE comes from trying to compute a flow of a given vector field on an ellipse: $(R\theta),\theta))$ are essentially the polar coordinates of the ellipse, $e$ is its
eccentricity.

答:
$$\frac{d \beta}{dt}=\frac{c}{R(\beta(t))+c_0}\quad;\quad
R(\beta)=\frac{1}{\sqrt{1-\big(e \cos(\beta)\big)^2}}$$

$$\left(\frac{1}{\sqrt{1-\big(e \cos(\beta)\big)^2}}+c_0\right)\frac{d \beta}{dt}=c$$

$$\int \frac{d\beta}{\sqrt{1-\big(e \cos(\beta)\big)^2}}+c_0\beta=c(t-t_0)$$

$$t=t_0+\frac{1}{c}\left(\frac{1}{\sqrt{1-e^2}}F\left(\beta\:\Bigg|\:{\frac{e^2}{e^2-1}} \right)+c_0\beta \right)$$
$F$ is the elliptic integral of the first kind.
With initial condition $\beta(0)=\beta_0$ :

$$\boxed{t(\beta)=\frac{1}{c\:\sqrt{1-e^2}}\left( F\left(\beta\:\Bigg|\:{\frac{e^2}{e^2-1}}\right)-F\left(\beta_0\:\Bigg|\:{\frac{e^2}{e^2-1}}\right) \right)+\frac{c_0}{c}(\beta-\beta_0)}$$
This is the solution $t$ as a function of $\beta$.

The inverse function $\beta(t)$ cannot be written with a finite number of available standard functions. Numerical calculus is recommended.


问:
how to calculate $E(X+Y|X<x,Y<y)$
consider $X\sim U(0,1), Y\sim U(0,1-x))$
for $E(X+Y|X<x,Y<y)$ I'd consider $E(X|X<x,Y<y)+E(Y|X<x,Y<y)$
$f_{x,y} = f_{y|x}f_x = 1/(1-x)$

$E(X|X<x,Y<y) = E(X|X<x) = \int_0^x xdx = x^2/2$

$E(Y|X<x,Y<y) = \int_0^{y}\int_0^{x}y/(1-x)dxdy = \int_0^{x}1/(1-x)dx\int_0^{y}ydy = \frac{-\ln(1-x)y^2}{2}$

Why is this logic wrong?

答:
Don't use $x,y$ as the bound variables, when you have them in the conditions.  Integrate with bound variables of $s,t$ to avoid confusion.

$\begin{align}\mathsf E(X+Y\mid X<x,Y<y) &= \mathsf E((X+Y)\mathbf 1_{X<x,Y<y})/\mathsf P(X<x,Y<y)\\  &=\dfrac{\int_{0}^{x}\int_0^{\min\{y,1-s\}}(s+t)/(1-s)\,\mathrm d t\,\mathrm d s}{\int_{0}^{x}\int_0^{\min\{y,1-s\}}1/(1-s)\,\mathrm d t\,\mathrm d s}\end{align}$

Title: how to calculate $E(X+Y|X<x,Y<y)$


问:
Which sets are by definiton countable/uncountable?
Are there some general rules on whether a set is countable or not? Thus far I only know about $\mathbb{R}$ and $\mathbb{C}$ being uncountable, so I wonder if there are possibly any theorems that ascertain the countability/uncountability for sets, i.e., the cartesian product of a set, the power set of a set... etc. Is there maybe an intuitive way to see that? Other than going by the definition of countability itself.

Thank you very much!

答:
If a set $S$ contains an uncountable set, then $S$ is uncountable too. So, every set that contains, say, $\Bbb R$, is uncountable. Also, if $S$ is an infinite set, the set $\mathcal P(S)$ of all subsets of $S$ is uncountable. In particular, $\mathcal P(\Bbb N)$ is uncountable.

On the other hand, any subset of a countable set is coubtable. And uny countable union of countable sets is also countable.


问:
Cauchy Principal Value for Poles stronger than 1
While calculating the Cauchy Principal Value of $\int_{-\infty}^{+\infty}\frac{\sin x}{x}dx$ we take an indented semicircle (in the upper half plane) of radius $\delta$ centered at the simple pole of $\frac{e^{iz}}{z}$ at $z=0$. Then we Laurent expand and by integrating and taking $\delta\rightarrow0$ we obtain over the small semicircle (in $\pi\rightarrow0$ clockwise sense): $$\int\frac{e^{iz}}{z}dz=-i\pi\qquad\text{Residue at}\,z=0=-i\pi$$ (As done in Arfken $7^{th}$ edition,  $11.75$).

But this method fails when we have poles stronger than of order $1$, e.g., $\frac{e^{iz}}{z^{2}}$. How do I approach such cases?

答:
$\oint z^n dz =0$ except for $n=-1$. So expand $e^{iz}$ as a series
$$\oint \frac{e^{iz}}{z^2} dz =\oint \frac{1+iz-z^2+\dots}{z^2} dz =i\oint \frac{1}{z} dz=-2\pi$$
More generally, if $f(z)$ is regular at $z=0$
$$\oint \frac{f(z)}{z^2} dz =2\pi i f'(0)$$

As pointed out in the comments, the question is actually about a semicircle. Let me show this is still a (partially) valid way to calculate it in that case for the function $f(z)=e^{iz}$

$$\oint \frac{e^{iz}}{z^2} dz =\int_+ \frac{e^{iz}}{z^2} dz + \int_- \frac{e^{iz}}{z^2} dz$$
The plus and minus refer to upper and lower semicircles. We can bring the lower semicircle integration to an upper semicircle integration by changing variables from $z\rightarrow -z$
$$\oint \frac{e^{iz}}{z^2} dz =\int_+ \frac{e^{iz}-e^{-iz}}{z^2} dz =\int_+ \frac{2i \sin z}{z^2} dz$$
So it is true that
$$\oint \frac{e^{iz}}{z^2} dz =\oint \frac{i\sin z}{z^2} dz=2\int_+ \frac{i \sin z}{z^2} dz$$
In principle though there is a part coming from $\int_+ \frac{ \cos z}{z^2} dz$ which cancels on the full closed integral (since it comes with opposite sign). This part is logarithmically divergent as the radius of semicircle integration goes to zero, so without specifying the regularization of the integral further you can't proceed.

OwnerDisplayName: octonion


问:
$\mathbb{Q}$ is in the Borel $\sigma$-algebra

Prove that the set of rational numbers $\mathbb{Q}$ belongs to the Borel σ-algebra of $\mathbb{R}$

I was just thinking about the method I would go about proving this

I was thinking if we can write

$$\mathbb{Q} = \bigcup_{x \in \mathbb{Q}} \left \{ x \right \}$$

And then if we prove the {x} is in the Borel σ-algebra then the union is too?

If there's a method to prove this, It would really help thank you

答:
Yes, your argument is right. Here are the details.

It is enough to prove that for all $x\in \Bbb Q$ (in fact, for all $x\in \Bbb R$), $\{x\}$ is in the Borel $\sigma$-algebra.  Then, since
$$\mathbb{Q} = \bigcup_{x \in \mathbb{Q}} \left \{ x \right \}$$
and since such union is countable (because $\Bbb Q$ is countable), we have that $\Bbb Q$ is in the  the Borel $\sigma$-algebra.


问:
Implied prior with relationship $y=\text{arccot}(x)$
I'm trying to solve an exercise, which I think I have almost managed to solve but not quite. Any help would be appreciated!

So, what we have is a vector which we obtain by norming the vector $\lambda=(x, 1)$, where $x \in (-\infty, \infty)$, into "half unit circle". I assume that this means that the vector we get is:

$$\lambda^*=\left(\frac{x}{\sqrt{x^2+1}}, \frac{1}{\sqrt{x^2+1}}\right)$$

Expressed in terms of the polar coordinates, the vector is

$$
\lambda^*=(\cos y, \sin y)
$$
where $y\in[0,\pi)$. If $y$ is assigned the uniform prior, what is the implied prior for $x$? A hint is that $y=\text{arccot}x$.

My way of reasoning is this. First, the cdf of $y$ is
$$
F_Y(y)=\frac{y}{\pi}, \quad 0\leq y <\pi.
$$

Furthermore, we have the following relationship:
$$
\sin y=\frac{1}{\sqrt{x^2+1}}
$$
which means

$$
\sqrt{x^2+1}=\frac{1}{\sin y}\\
x=\left(\frac{1}{\sin y}\right)^2-1
$$

Using the transformation theorem, we can obtain

$$
F_X(x)=Pr(X\leq x)=Pr\left(\frac{1}{(\sin Y)^2}-1\leq x\right)=Pr\left(\frac{1}{x+1}\leq(\sin Y)^2\right)
$$
This is where I think I go wrong. To continue and get $Y$ by itself, I need to take the square root and then the inverse of sin. But then I'll have $\sqrt{x+1}$, which will yield non-real numbers for $x<-1$ (remember that $x\in(-\infty, \infty)$.

Any ideas how to attack this?

答:
One considers $Y$ uniform on the interval $(0,\pi)$ and $X=\cot Y$. There are several ways to compute the distribution of $X$, one is to start from the fact that $[X\leqslant x]=[Y\geqslant\cot^{-1} x]$ for every real number $x$, hence
$$
F_X(x)=1-\frac1\pi\cot^{-1}x,
$$
and, by differentiation,
$$
f_X(x)=\frac1\pi\frac1{1+x^2}.
$$</p>


问:
Existence of a limit and a counterexample
For homework, my group was given this false statement which we were supposed to find an example for:

If a function $g$ is bounded from top and bottom on an open interval $(a,b)$, then there exists a real number (non infinite) limit:
$$\lim_{x\to b^-} f(x) $$

We were trying to come up with a solution but we think that there is no counterexample, as when a function is bounded from both sides, it must have a real number lmit at the ends of the function. Are we wrong?

答:
(As requested by the OP)

It seems to me that $f(x) = \sin(1/x)$ considered on the $(-1,0)$ interval provides a counter-example, since $f(x)$ is bounded on the interval, and $\lim_{x \rightarrow 0^-} f(x)$ is undefined.


答:
$$
\begin{aligned}
\dot{\beta}{(\theta)} &= \frac{c}{R(\beta{(\theta)}) + c_0} \Leftrightarrow \\
&\Leftrightarrow \left(R(\beta{(\theta)}) + c_0\right)\dot{\beta}{(\theta)} = c \Leftrightarrow \\
&\Leftrightarrow \int\limits_0^\theta\left(R(\beta{(\tau)}) + c_0\right)\dot{\beta}{(\tau)}d\tau = c\int\limits_0^\theta d\tau = c\theta
\end{aligned}
$$

Now, we have to evaluate the integral

$$
\begin{aligned}
\int\limits_0^\theta\left(R(\beta{(\tau)}) + c_0\right)\dot{\beta}{(\tau)}d\tau &= \left|u = \beta{(\tau)} \Rightarrow du = \dot{\beta}(\tau)d\tau\right| = \\
&= \int\limits_{\beta{(0)}}^{\beta{(\theta)}}\left(R(u) + c_0\right)du = \\
&= \int\limits_{\beta{(0)}}^{\beta{(\theta)}}\frac{du}{\sqrt{1-\left(e\cos{u}\right)^2}} + c_0(\beta{(\theta)} - \beta{(0)}) = \left|\right|
\end{aligned}
$$

Title: Is there a closed-form solution of this ODE (computing a flow on an ellipse)?

Body:
Can we find an explicit closed-form solution of the following ODE?

$\beta:[0,\epsilon] \to (0,2\pi]$, $\beta(0)=\beta_0$, and
$$\dot \beta(t)=\frac{c}{R(\beta(t))+c_0},$$ where $c,c_0$ are constants, and
$$R(\theta):=\frac{1}{\sqrt{1-\big(e \cos(\theta)\big)^2}}, 0<e<1.$$

The motivation for this ODE comes from trying to compute a flow of a given vector field on an ellipse: $(R\theta),\theta))$ are essentially the polar coordinates of the ellipse, $e$ is its
eccentricity.


问:
Show that $e^x\ge x^e$ for all $x\ge 0$
Show that $e^x\ge x^e$ for all $x\ge0.$

I was playing around with splitting a number into parts (that add to the original number) and then multiplying the parts to get the result, and I found that the product is highest when the parts tended towards $e$. This inequality is a consequence of that. Is there a proof or intuitive explanation for this?

答:
Having that $\dfrac{\log x}{x}$ is a increasing function with $x\leq e$
$$\therefore\frac{\log e}{e}\geq\frac{\log x}{x}$$
Excluding the logarithm signs in both side, we have what we need to show.

Title: Show that $e^x\ge x^e$ for all $x\ge 0$


答: Basically a set $\mathcal{S}$ can be called countable, if there exists a bijective function $f$ between $\mathcal{S}$ and the set of natural numbers: $\mathbb{N}$ (i.e. $\mathcal{S} \overset{f}{\sim} \mathbb{N}$). In this case we can say that sets $\mathcal{S}$ and $\mathbb{N}$ are equinumerous, and that is any set, which is equinumerous to $\mathbb{N}$ is called countable.

Title: Which sets are by definiton countable/uncountable?

Body:
Are there some general rules on whether a set is countable or not? Thus far I only know about $\mathbb{R}$ and $\mathbb{C}$ being uncountable, so I wonder if there are possibly any theorems that ascertain the countability/uncountability for sets, i.e., the cartesian product of a set, the power set of a set... etc. Is there maybe an intuitive way to see that? Other than going by the definition of countability itself.

Thank you very much!


问:
Metric space of binary strings
Hi I am trying to prove that the set of binary strings is a metric space. I define $S$ as the binary string (0/1). For two different strings I look at the metric space for the number of entries in which the two strings differ.

So $x=10000$ and $y=11000$ would be $1$ clearly. I am trying to prove the triangle inequality for a metric space on this but I am having trouble doing it vigorously enough. I want to show:

$$d(a,b)\le d(a,c)+d(b,c)$$

I said that $A,C$ differ by $K$ entries and $B,C$ differ by $J$ entries. I think that $A,B$ cannot differ by more than $K+J$ but I can't prove it can someone help me out?

答:
When you add any two members of same length, say $n$, some of the non-zero entries may become zero (due to the fact that string is binary). Therefore, given any two vectors, their sum has definitely fewer non-zero entries than the sum of their individual non-zero entries. In other words,

Given any $x$ an $y$ in the said domain,

$d(0,x)+d(0,y)\geq d(0,x+y)$.

Obviously,

$a+b=a+c+b+c$, since the string is binary and $c+c=\theta$.

So,

$d(0,a+b)\leq d(0,a+c)+d(0,b+c)$. Note that $d(0,a+b)=d(a,b)$. Therefore,

$d(a,b)\leq d(a,c)+d(b,c)$.

Title: Metric space of binary strings


问:
Definition of inertia group in infinite extensions
In algebraic number theory we learn the definition of inertia group of a finite Galois extension. What is the definition in the case of an infinite extension? (say in the algebraic closure of Q).</p>

答:
as the global case was only parenthetically asked for, I will answer the (1-dimensional) local case:

local case: say $K$ is a discrete valuation field. then one can attempt the same definition as in the above answer but $K$ must be assumed to be henselian (e.g. in the case of a so-called 'local field', $K$ is then assumed to be complete (hence henselian) [and with perfect (or simply finite) residue field]). Then, assuming $K$ henselian, if $K^s$ denotes a fixed separable closure of $K$ and $G_K$ the Galois group $Gal(K^s/K)$, one can define the inertia subgroup $I_K\leq G_K$ as:

$$I_K = Gal(K^s/K^{ur}),$$ where $K^{ur}$ is 'the' maximal unramified extension of $K$ in $K^s$. I'll leave following as a fun and instructive exercise:

Suppose $K$ is a discrete valuation field and consider the henselization $K^h$ and completion $\hat{K}$ of $K$ with respect to the valuation on $K$. Is it true that $I_{K^h} = I_{\hat{K}}$?