答:
I have a quite tedious argument that this inequality holds:

Firstly, by taking $x \pmod 1$ we may assume $x \in [0,1)$. Let $k$ be the greatest integer, such that $\frac{k}{30} \leqslant x$ (so $k = \lfloor 30x \rfloor$). Then we would want to prove that

$$\lfloor 15x \rfloor = \bigg \lfloor \frac{k}{2} \bigg \rfloor.$$

Firstly, $\lfloor 15x \rfloor \geqslant \big\lfloor \frac{k}{2} \big\rfloor$, as $15x \geqslant \frac{k}{2}$. Suppose $\lfloor 15x \rfloor > \big\lfloor \frac{k}{2} \big\rfloor$, then we can choose an integer $l$ such that $15x \geqslant l > \frac{k}{2}$. But that would mean $\frac{2l}{30} \leqslant x$ and $2l > k$, which is a contradiction with a definition of $k$.

Using the same methods, we can transform our inequality to

$$ k+ \bigg\lfloor \frac{k}{30} \bigg\rfloor  \geqslant \bigg\lfloor \frac{k}{2} \bigg\rfloor + \bigg\lfloor \frac{k}{3} \bigg\rfloor + \bigg\lfloor \frac{k}{5} \bigg\rfloor $$

where $k$ is an integer between 0 and 29. That's $30$ cases to check, but they're all simple so it can be done quite quickly.

EDIT We can also transform RHS using $\lfloor a \rfloor + \lfloor b \rfloor \leqslant \lfloor a+b \rfloor$. Then we will obtain inequality $k + \big\lfloor \frac{k}{30} \big\rfloor  \geqslant \big\lfloor \frac{31k}{30} \big\rfloor$, which is not true in general case, but it holds for $k=0,\ 1,\ldots,29$.

Title: Sum of floors inequality

Body:
I'm trying to show the following inequality, for $x\geq0$:

$$
\lfloor30x\rfloor+\lfloor x\rfloor\geq\lfloor15x\rfloor+\lfloor10x\rfloor+\lfloor6x\rfloor.
$$

Intuitively this statement looks like it should be true because the LHS has two terms and the RHS has three terms - but I don't know how to formalize it. I am aware that $\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor$, but I don't know I can directly apply that here. Any help would be appreciated!


问:
show that the set $ A'=\left\{ x:x\in\partial(A\setminus\{x\}\right\} $is closed
I'm studying Calculus 3, and our teacher tends to give us extremely difficult questions.
one of those questions was this:
"let $A \subseteq \Bbb R^k $. show that the set $ A' = \{ x \in \Bbb R^k\mid x \in \partial(A \setminus \{ x \})\}$ is closed".
I've tried to show using sequence characterization or showing that $(A')^c$ is open, but I don't have a clue of how to start, or even if I'm in the right direction.
I could really use some help or a guiding hand with this.

答:
I'll use $d$ for the metric, and work in any metric space instead of $\Bbb R^n$.

Suppose that $x \notin A'$. This means that $x \notin \partial(A\setminus \{x\})$, so either there is an open ball $B(x,r_x), r_x>0$ such that $B(x,r_x) \subseteq A\setminus \{x\}$ or such that $B(x, r_x) \cap (A\setminus \{x\}) = \emptyset$; the former is impossible (contradicted by $x$) so the latter is the case.

My claim is that $B(x,r_x) \cap A' = \emptyset$, showing that $A'$ is closed.

Let $y \in B(x,r_x)$. If $y=x$, then we already know that $y \notin A'$.
So assume $y \neq x$. We can find $r_y>0$ so that $B(y, r_y) \subseteq B(x,r_x)$ and moreover $d(y,x) \ge r_y$ so that $x \notin B(y, r_y)$ as well. Now note that $B(y, r_y) \cap A\setminus \{y\} = \emptyset$ (being a subset of $B(x,r_x)$ avoids all points of $A$, except maybe $x$, but choosing $x$ outside of that ball takes care of that), and this means that $y \notin \partial(A\setminus \{y\})$ or $y \notin A'$, as required.

ContentLicense: CC BY-SA 4.0


问:
How is angular velocity of both these bodies equal here?
Here there is John who is looking at a bus at which behind there is Harry. My teacher says that Angular velocity I.e(Theta / time)of the bus and Harry is equal when they appear to be at same point by Swaraj.

Now , my question is how is it possible since time and theta for both (John and Harry)appear to be different. Also , the arc Length of both are different Is it some Euclidean postulate my teacher used? I am not able to get which one is it here. My teacher did not give a reason for this.

I believe theta 1 and theta 2 are different as they appear to be. So will their time be different to reach at those points

答:
The angular velocity can be given by $\frac{\Delta\theta}{\Delta t}$, i.e. rate of change of theta( the angle) with respect to time. Your confusion seems to arise from the fact that the theta's of the bus and K+G are different. I believe that the theta for both are the same and that is very clear from the diagram. Since both bus and K+G lie on the same line, they sweep the same angles. And since it stays at the same position w.r.t. the bus, it takes the same amount of time too.

Hence there angular velocity stays the same.


问:
How to evaluate limiting value of sums of a specific type
We know that if $f$ is integrable in (0,1) then

$$
\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k/n) = \int_{0}^{1}f(x)dx.
$$

Recently I found the following sum

$$
\lim_{n \to \infty} \sum_{k=1}^{n}\frac{k}{n^2 + k^2} = \frac{\ln 2}{2}.
$$

This sum cannot be expressed in the form $\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k/n)$. Rather it is of the form

$$
\lim_{n \to \infty} \frac{1}{n^2}\sum_{k=1}^{n} k f(k/n).
$$

Part 1

Is there any technique to evaluate limits of this form?

Part 2

Generalizing the question one step further, how can we asymptotics of summations of the form

$$
\sum_{k=1}^{n}  f(k)g\Big(\frac{k}{n}\Big).
$$

ContentLicense: CC BY-SA 3.0

Body:
$$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{k}{n^2+k^2}=\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{1}{n}\frac{nk}{n^2+k^2}$$
$$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{1}{n}\frac{nk}{n^2+k^2}=\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{1}{n}\frac{\frac{k}{n}}{1+(\frac{k}{n})^2}$$

so $$f(\frac{k}{n})=\frac{\frac{k}{n}}{1+(\frac{k}{n})^2}$$

so $$f(x)=\frac{x}{1+x^2}$$

and $$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{1}{n}\frac{nk}{n^2+k^2}=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k=1}^{n}\frac{nk}{n^2+k^2}=\int_{0}^{1}\frac{x}{1+x^2}dx=\frac{ln 2}{2}$$


问:
Does $Floor(1-\frac{1}{n})$ converge?
Does $x_n=Floor(1-\frac{1}{n})$ converge? I want to say no but I'm uncertain.

Suppose it converges. Then $\lim x_n=1 - \lim Floor(1/n))=1$
But there exists no N such that $|x_n-1|<\frac{1}{2} \ \forall n\ge N$, contradiction?

答:
Yes, it does converge, note that for every natural number $n$, we have $0 < 1-\frac{1}{n} < 1$. This means that $x_n = \lfloor{1-(1/n)}\rfloor = 0$ for every natural number.

Hence, $x_n \to 0$. That is, if you fix $\epsilon > 0$ and take $N = 1$, if $n > N$, then $|x_n| = |0| = 0 < \epsilon$

You made a mistake in assuming $\lfloor{1-(1/n)}\rfloor = 1 - \lfloor{1/n}\rfloor$

Title: Does $Floor(1-\frac{1}{n})$ converge?


问:
Find the probability that an employee is a drug user
A firm wants to know how many of its employees have drug problems. Realizing the sensitivity of this issue, the personnel director decides to use a randomized response survey.

Each employee is asked to flip a fair coin,

If head (H), answer the question “Do you carpool to work?”

If tail (T), answer the question “Have you used illegal drugs within the last month?”

Out of 8000 responses, 1420 answered “YES” (assuming honesty)

The company knows that 35% of its employees carpool to work. What is the probability that an employee (chosen at random) used illegal drugs within the last month?

I think the probability that I am trying to figure out is $\mathbb{P}(yes|T)$. From the problem, I was able to figure out that $\mathbb{P}(yes)=1420/8000$, $\mathbb{P}(T)=50%$ (because it's a fair coin) and that $\mathbb{P}(yes|H)=35%$. But for Bayes' theorem, I need to find $\mathbb{P}(T|yes)$, and that is where I am stuck.

I realized that I did not need Bayes' theorem as that would have made it more difficult.

答:
Heuristically:

Assume for convenience that we have $2000$ employees and $1000$ get heads and $1000$ get tails, equally distributed over (non-)carpoolers (of which we have $700$ by the given percentage). Suppose they all answer , we get $\frac{1420}{4} = 355$ yes answers (scaled from $8000$)

Out of the $1000$ heads, we will get $350$ yes answers from carpoolers, and out of the 500 tails we got an extra $5$ yes from drugs users. So we'd expect $\frac{5}{1000}$ as an estimate of the drug users fraction.


问:
$\sqrt a+\sqrt b$ is a root of polynomial, then $\sqrt a -\sqrt b$ is so over finite field
Let $a,b$ is not a square element in finite field.
If polynomial over the finite field has a root $\sqrt a+\sqrt b$, then the polynomial has $\sqrt a-\sqrt b$ as a root too?

Thank you for your help.

This question arises from concrete problem, $x^4-10x^2+1$（this is a minimal polynomial of $\sqrt{2}+\sqrt{3}$ over rational field） is reducible over arbitrary finite field.
$\sqrt{2}+\sqrt{3}$ is root over finite field, then $\sqrt{2}-\sqrt{3}$ is also a root（why?）.And $\sqrt2+\sqrt3$ and $\sqrt2-\sqrt3$ exists in $2$-degree extension of the finite field, so the minimal polynomial spilts. This leads minimal polynomial is at most $2$ degree, so $x^4-10x^2+1$ is reducible.

答: Consider $\Bbb F_5$ and consider $a=2$, $b=3$ and $\sqrt b=2\sqrt a$. Then, $x^2+2$ has roots $\pm(\sqrt a+\sqrt b)$, and neither $\sqrt a-\sqrt b$ nor $\sqrt b-\sqrt a$ are among them.


问:
Find the value of a function that satisfy $f(x+1)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$.
Let $f(x+1)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$. If $f(1)=8$ then find $f\left(\frac{2}{3}\right)$.

I'm trying to find it as below.

\begin{align}
f(1)=f(0+1)=f(0)\cdot f(y)=8\\
f\left(\frac{2}{3}\right) = f\left(-\frac{1}{3}+1\right)=f\left(-\frac{1}{3}\right)\cdot f(y).
\end{align}

Since I can't find $f(y), f(0), f\left(-\frac{1}{3}\right)$, now I confused how to find $f\left(\frac{2}{3}\right)$. Any hint to solve that?

答: $f(1)=f(0)f(y)$ for all $y$. So $f$ is a constant function but the only contsant functions satifying the hypothesis are $0$ and $1$ so we cannot have $f(1)=8$.


答:
With $y=1$, the functional equation becomes
$$f(x+1)=8f(x)$$ for all $x$.In particular, $f(2)=64$. Then with $x=1$, $y=2$, the functional equation becomes
$$ 64=8\cdot64$$
contradcition. No such $f$ exists.

Title: Find the value of a function that satisfy $f(x+1)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$.

Body:
Let $f(x+1)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$. If $f(1)=8$ then find $f\left(\frac{2}{3}\right)$.

I'm trying to find it as below.

\begin{align}
f(1)=f(0+1)=f(0)\cdot f(y)=8\\
f\left(\frac{2}{3}\right) = f\left(-\frac{1}{3}+1\right)=f\left(-\frac{1}{3}\right)\cdot f(y).
\end{align}

Since I can't find $f(y), f(0), f\left(-\frac{1}{3}\right)$, now I confused how to find $f\left(\frac{2}{3}\right)$. Any hint to solve that?


问:
Contruction of a triangle, knowing one angle and it perimeter. What is the locus of a point which splits the perimeter in two at a specific length?
Building a triangle (ABC) with a fixed perimeter (P) knowing just one angle (Summit A) is not that easy.
Having fixed one adjacent side (segment AB), we know that the sum of both the others sides is a constant (P-|AB|), and so the third summit (C) is on an ellipse whose focuses are A and B, and passing by the point I called “HalfPerimeter”.

Now, the problem is that I would like to know the locus of a point dividing the perimeter in two equal or not unequal parts.

By tracing it, It appears that if we cut it in two equal parts, the locus of this moving point is the arc of a circle, which we can draw by finding that both the degenerates cases of the triangle are tangent to this circle.

And, if we cut it in two unequal parts, the locus of the moving point is another arc, of a larger circle but with the same center.

However, I have no idea how to prove that this locus is a circle, nor how to prove that all those circles have the same center.

EDIT:

A consequence of this (Corollary of the first part), would be that in the following figure, the perimeter of ABC is independent of the position of M (the moving point), and both ABM and AMC have the same perimeter.

This become obvious, since AB is tangent to the circle in $T_B$, AC is tangent in $T_C$ and BC is tangent in M :

$|AB|+|BM| = |AB|+|BT_B| = (*) = |AC|+|CT_C| = |AC|+|CM|$

(*) those two are independents of M

(But It still doesn't solve the problem with the unequal parts)

答:
In this answer, we show how to construct a triangle without much ado using the given data, i.e. its perimeter $p$ and one of its angle $\alpha$, and, then, use it to prove the conjecture OP has stated in the problem statement.

$\underline{\mathrm{1.\space Construction}}$

All we want to know is shown in $\mathrm{Fig.\space 1}$. After drawing the two legs emitting from the vertex $A$ and hemming the given angle $\alpha$, we mark the vertex $C$ on one of the legs a distance $b$ away from $A$. The value of $b$ is arbitrary, but need to satisfy the inequality
$$0 \le b\le \frac{1}{2}p \tag{1}.$$

On the other leg of the angle $\alpha$ we mark point $D$ so that $AD=p=a+b+c$. Then, the point $E$ is taken on $AD$, where $ED=AC=b$. Now, join $EC$ and draw its perpendicular bisector to meet $AD$ at $B$. Since $CBE$ is an isosceles triangle, we have
$$CA+AB+BC=CA+AB+BE=b+a+c=p.$$

Therefore, the perimeter of the $\triangle ABC$ is equal to $P$. We can construct infinite number of such triangles by varying the length $b$ within the limits specified by (1).

$\underline{\mathrm{2.\space A\space Lemma}}$

Before going on to find the locus mentioned in OP’s conjecture, we need to prove the following lemma

$\mathbf{\mathrm{Lemma\space :}}\quad$ All the triangles, which can be constructed using the above-described method, share one $A\mathrm{-excenter}$.

$\mathbf{\mathrm{Proof\space :}}\quad$ We illustrate the proof using $\mathrm{Fig.\space 2}$, where we extend the scenario depicted in $\mathrm{Fig.\space 1}$ by constructing another $\triangle APQ$. The perimeter of $\triangle APQ$ is equal to that of $\triangle ABC$. However, its vertex $Q$ is placed a distance of $d$ further away than the vertex $C$ from the common vertex $A$. As a result, point $T$ is located a distance of $d$ closer than $E$ to the common vertex $A$. The point of intersection of $AD$ and the perpendicular bisector of $QT$ is the vertex $P$.

We have also drawn the angle bisectors $AO$ and $CO$ of $\measuredangle FAD$ and $\measuredangle FCB$ respectively. They intersect each other at $O$, the $A$-excenter of $\triangle ABC$.  To facilitate the proof, it is necessary to draw the auxiliary line segments $OE$, $OQ$, and $OT$ as well.

It is obvious that the perpendicular bisector $MB$ of $CE$ serves also as the angle bisector of $\measuredangle CBE$. Therefore, $MB$ passes through $O$. Our aim is to show that the perpendicular bisector $NP$ of $QT$, which happens to be the angle bisector of $\measuredangle QPD$, also passes through $O$.

Let $\measuredangle BEC$ and $\measuredangle CEO$ equal $\phi$ and $\omega$.
$$\therefore\quad \measuredangle BEO = \phi + \omega. \tag{2}$$

$CBE$ is an isosceles triangle by construction. Therefore, we shall write $\measuredangle ECB = \phi$. $EOC$ is also an isosceles triangles, because $OM$ is the perpendicular bisector of its side $CE$.  So, we have $CO = EO$ and $\measuredangle BCO = \omega$.
We know that $CO$ is the angle bisector of $\measuredangle FCB$. Therefore,
$$ \measuredangle QCO = \measuredangle OCB = \measuredangle ECB + \measuredangle BCO = \phi + \omega. \tag{3}$$

Because of (2), (3), $QC = TE$, and $CO = EO$, the SAS rule can be applied to the pair of triangles $QCO$ and $OTE$ to show that they are congruent, whereby we have $OQ = OT$. This makes $OQT$ an isosceles triangle. Hence, $PN$, which is the perpendicular bisector of $QT$, passes through $O$. Since $PN$ is also the angle bisector of $\measuredangle QPT$, its meeting point $O$ with the angle bisector of $QAP$ is the $A$-excenter of $\triangle APQ$.

This proves that all the triangles, which can be constructed using the above-described method share one $A\mathrm{-excenter}$.

$\underline{\mathrm{3.\space A\space Conjecture}}$

$\mathbf{\mathrm{Conjecture\space :}}\quad$ If a point can be found on the side opposite of the vertex $A$ of certain individual triangles constructed using the above-described method to divide their perimeter in a given ratio, then, those points describe a circular arc having its center at the $A\mathrm{-excenter}$ shared by all those triangles.

$\mathbf{\mathrm{Proof\space :}}\quad$ As shown in $\mathrm{Fig.\space 3}$, an arbitrary point, i.e. $G$, is taken  on the line segment $AD$, which represents the perimeter $p$ of the triangle $ABC$. Let $G$ divide $AD$ so that $AG = \lambda AD$. In order to find the point $H$ that lies on the side $BC$ and divides the perimeter of $\triangle ABC$ such that $AB+BH= \lambda p$, we need to draw a line parallel to $CE$ through $G$ to intersect $BC$.

It is obvious that the perpendicular bisector $BO$ of $CE$ is also the perpendicular bisector of $GH$. Because of this, $OHG$ is an isosceles triangle, so we have
$$OH=OG. \tag{4}$$

The positions of the two points $O$ and $G$ depend only on the size of the perimeter of $\triangle ABC$, which is a constant. Therefore, length of $OG$ is also a constant. Now, according to (4), we can state syllogistically that the length of $OH$, which is equal to $OG$, is also a constant, even if the position of $H$ varies in space from triangle to triangle. Therefore, the locus describes by the perimeter-dividing point $H$ is nothing but a part of a circle with its center at $O$.

$\underline{\mathrm{4.\space Additional\space Information}}$

There is a very simple way to show that radius of the $A$-excircle of all triangles has the same length, if they also have the same perimeter $p$ and the same angle at the vertex $A$. Let this radius be $r_{A}$. Using trigonometry, it can be expressed as
$$r_A = \frac{p}{2}\tan\left(\frac{A}{2}\right).$$

Therefore, if the triangles have the same $p$ and $A$, they have the same $r_A$ as well.

Furthermore, it is possible to express $OH$ as a function of $p$, $A$, and $\lambda$.
$$OH=\frac{p}{2}\sqrt{4\lambda \left(\lambda - 1\right)+\sec^2 \left(\frac{A}{2}\right)}.$$

Since $OH$ does not depend on anything other than the mentioned constants, the locus of $H$ is a circular arc of radius $OH$ with its center at $O$.

You know that the locus coincides with the $A$-excircle when $H$ is the midpoint of $BC$, i.e. $\lambda = 0.5$. If you want to check the validity of (6), you can substitute $\lambda = 0.5$ into it and see whether it reduces to expression (5).


问:
Converse of the result about the composite of purely inseparable extensions
This post (Composite of two purely inseparable extensions is purely inseparable.) says that if $E < F$ is purely inseparable and $F < K$ is purely inseparable, then $E < K$ is separable.

However, does the converse hold? That is, if $E < K$ is purely inseparable and we have $E < F < K$, are $E < F$ and $F < K$ purely inseparable?

答: Yes, assume that $K/E$ is purely inseparable and let $F$ be an intermediate field. As $F\subseteq K$ the field extension $F/E$ is trivially purely inseparable. Let $x\in K$. Then the minimal polynomial $f$ of $x$ over $F$ is a divisor of the minimal polynomial  of $x$ over $K$. The latter one has by assumption only one root (in some extension field), hence also $f$ has only one root. It follows that $x$ and therefore also $K$ is purely inseparable over $E$.


问:
Is the permutation matrix P of PLU decomposition unique?
Let $A$ be a square matrix. Then there exists a permutation matrix $P$ such that $A=PLU$, where $L$ is a lower triangular matrix and $U$ is an upper triangular matrix. To further ensure the uniqueness, we assume that the main diagonal of $L$ (or $U$) to be 1. So, the question is, is the permutation matrix unique, i.e., can we find another $P'\ne P$ such that $A=PLU=P'L'U'$ where $L', U'$ are still triangular matrix? If yes, what is the condition for the uniqueness?

答:
No, the permutaion is not unqiue. Here is an example:
$$
\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix} =\underbrace{\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{L}\underbrace{\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{U}\\
\underbrace{\begin{pmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}}_{P}\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix} =\begin{pmatrix}
1 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix} =\underbrace{\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{L}\underbrace{\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{U}
$$
where $\displaystyle P$ is the permutation matrix that switches first and second row. It follows that we can find a different permutation matrix but has a different LU-decomposition.


答:
As the multiplicative group of a finite field $F$ is cyclic, it follows that $b=c^2a$ with come $c\in F^\times$ whenever $a,b$ are both non-squares. Then wlog. $\sqrt b=c\sqrt a$ and so $\sqrt a+\sqrt b$ is a root of one of $X^2-(1+ c)^2a$ whereas $\sqrt a-\sqrt b$ is a root of $X^2-(1- c)^2a$. These minimal$^1$ polynomials of $\sqrt a+\sqrt b$ and $\sqrt a-\sqrt b$ are different except in characteristic $2$.

$^1$ Okay, one of them is not minimal when $c=\pm1$, i.a., when $a=b$.

Title: $\sqrt a+\sqrt b$ is a root of polynomial, then $\sqrt a -\sqrt b$ is so over finite field

Body:
Let $a,b$ is not a square element in finite field.
If polynomial over the finite field has a root $\sqrt a+\sqrt b$, then the polynomial has $\sqrt a-\sqrt b$ as a root too?

Thank you for your help.

This question arises from concrete problem, $x^4-10x^2+1$（this is a minimal polynomial of $\sqrt{2}+\sqrt{3}$ over rational field） is reducible over arbitrary finite field.
$\sqrt{2}+\sqrt{3}$ is root over finite field, then $\sqrt{2}-\sqrt{3}$ is also a root（why?）.And $\sqrt2+\sqrt3$ and $\sqrt2-\sqrt3$ exists in $2$-degree extension of the finite field, so the minimal polynomial spilts. This leads minimal polynomial is at most $2$ degree, so $x^4-10x^2+1$ is reducible.


问:
Computing turning number of closed curve
I'm trying to compute the turning number of a closed plane curve, which should be a really straight forward computation. But my result is not a whole number, i.e. the total curvature is not a multiple of $2\pi$, which is impossible so I'm making a mistake somewhere. I know from plotting the curve that the turning number should be $-2$. I've been looking for hours but cannot seem to find it. Maybe someone can spot the mistake... Here is my work:

Let $\beta(t) = ((1+2\cos(t))\sin(t), (1+2\cos(t))\cos(t))$ be a regular closed plane curve. To compute the turning number $n_{\beta}$ I want to use the following:

$$n_{\beta} = \frac{1}{2\pi}\int_{0}^{L}K_{\beta}(t)dt\, ,$$

where $L$ is the period of $\beta$ and $K_{\beta}(t)$ is the curvature. I can compute the curvature like this:

$$K_{\beta} = \frac{1}{||\dot{\beta}(t)||^3}\cdot\det({\dot{\beta}(t), \ddot{\beta}(t)})$$

Computations:

$$\dot{\beta}(t) = (\cos t + 2\cos 2t, -\sin t -2\sin 2t)$$
$$\ddot{\beta}(t) = (-\sin t - 4 \sin 2t, -\cos t - 4 \cos 2t)$$
$$||\dot{\beta}(t))|| = (4\cos t + 5)^{1/2}$$
$$\det({\dot{\beta}(t),\ddot{\beta}(t)}) = -6 \cos t - 9$$

This is giving me the curvature

$$K_{\beta}(t) = \frac{-6 \cos t -9}{(4 \cos t + 5)^{3/2}}$$

At this point I integrated the curvature with the help of Wolframalpha, which did it numerically but I'm pretty sure it's $-2\pi-1$, when from the picture it should be $-4\pi$. Oddly enough, if I integrate the curvature without the power of $\frac{3}{2}$ in the denominator i actually get the correct result. Can anyone help me spot my mistake? Thanks a lot!

Edit: Here's a link to see the graph: https://www.desmos.com/calculator/vcxbpj2ios

答:
I found my mistake! Posting it as an answer, so future viewers can see what is happening. The issue was not the computation, but misuse of a formula. The thing is that the formula I wanted to use for the turning number, i.e.

$$n_{\beta} = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(t)dt$$

may only be used for curves that are parametrized by arclength. Since curvature is invariant under reparametrization I just assumed that it would work the same way. But the reparametrization-invariance actually gives us

$$K_{\beta \circ \varphi}(t) = K_{\beta}(\varphi(t)),$$

where $\beta \circ \varphi$ is a reparametrization of $\beta$ by arclength. So what I ignored was the $\varphi(t)$ as the argument of the curvature. If we account for this, by a change of variable we get

$$n_{\beta} = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(\varphi(t))\, dt = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(t)\cdot|\varphi'(t)|\,dt = \frac{1}{2\pi} \int_{0}^{L} K_{\beta}(t)\cdot||\dot{\beta}(t)||\,dt,$$

where for the last equality we used the fact that the reparametrization is given by $\varphi(s) = \int_{s_0}^{s} ||\dot{\beta}(\tau)||\,d\tau$, so $\varphi'(s) = ||\dot{\beta}(s)||$ by the fundamental theorem of calculus. Therefore, in my exercise one must integrate the function

$$K_{\beta}(t)\cdot||\dot{\beta}(t)|| = \frac{-6\cos t-9}{4\cos t + 5}$$

which evaluates to $-4\pi$, giving the turning number of -2, which is exactly what it should be.


问:
Question about Theorem 14.8 in Rudin's R&C analysis
In the proof of Theorem 14.8 in Rudin's real and complex analysis, which is

Every simply connected region $\Omega$ in the plane (other than the plane itself) is conformally equivalent to the open unit disc $U$,

while proving $\sum$, the class of all $\psi \in H(\Omega)$ which are one-to-one in $\Omega$ and which map $\Omega$ into $U$, is not empty,
he choose $\varphi \in H(\Omega)$ which is one-to-one and is an open mapping, so that $\varphi(\Omega)$ contains a disc $D(a;r)$ with $0 <r <|a|$. Then he said the disc $D(-a;r)$ fails to intersect $\varphi(\Omega)$.

But I can't see why this happens. Like what if $\varphi(\Omega)$ is a ball with center at $0$ and radius $|a| + 2r$?

What am I missing?

答: Before this he has proved that $\phi(z_1)=-\phi(z_2)$ cannot happen for distinct points  $z_1$ and $z_2$. Now suppose $D(-a,r)\cap \phi(\Omega)$ consatin a point $z$. Then there exists $c \in \Omega$ such that $z=\phi (c)$. Since $|z+a| <r$ it follows that $-z \in D(a,r)$. So $-z\in \phi(\Omega)$. If $-z=\phi \zeta)$ ($\zeta \in \Omega$) it follows that $\phi (\zeta) =-\phi (c)$.  But this is a contradiction. Note that $-z=z$ would imply $z=0$ and $|a|=|z+a|<r$ a contradiction to $0<r<|a|$.


答:
Set up a table like this, with $8000$ people in total:

$H$
$T$

Yes
$(1)$
$(2)$

No
$(3)$
$(4)$

As you have said, $(1) + (3) = 4000$ people, as the coin is fair. Now the $4000$ people in the 'heads' condition are a representative sample of the company, so $35 \%$ of those carpool as well. This means $(1) = 4000 \times 0.35 = 1400$, and since $1420$ answered 'Yes', $(2) = 1420 - 1400 = 20$:

$H$
$T$

Yes
$1400$
$20$

No
$4000-1400$
$4000-20$

and therefore the proportion of drug users is $\frac{20}{4000} = \frac{1}{200}$ or $0.005$.

Title: Find the probability that an employee is a drug user

Body:
A firm wants to know how many of its employees have drug problems. Realizing the sensitivity of this issue, the personnel director decides to use a randomized response survey.

Each employee is asked to flip a fair coin,

If head (H), answer the question “Do you carpool to work?”

If tail (T), answer the question “Have you used illegal drugs within the last month?”

Out of 8000 responses, 1420 answered “YES” (assuming honesty)

The company knows that 35% of its employees carpool to work. What is the probability that an employee (chosen at random) used illegal drugs within the last month?

I think the probability that I am trying to figure out is $\mathbb{P}(yes|T)$. From the problem, I was able to figure out that $\mathbb{P}(yes)=1420/8000$, $\mathbb{P}(T)=50%$ (because it's a fair coin) and that $\mathbb{P}(yes|H)=35%$. But for Bayes' theorem, I need to find $\mathbb{P}(T|yes)$, and that is where I am stuck.

I realized that I did not need Bayes' theorem as that would have made it more difficult.


问:
suggest a textbook on calculus

Possible Duplicate:

  What are the recommended textbooks for introductory calculus?

Hi, i am a software engg. and math was never my favourite. I somehow dragged thru the math i had to do in college to earn my degree. But my brother (7 years younger aged 15) does have an aptitude for math. i have seen him cruise thru his high school texts with little or no difficulty. He also says he wants to be a mechanical engineer (and backs up his claim by cleaning up and servicing my motorcycle incessantly) anyways he is 15 now and sadly his syllabus hasnt introduced to almost any calculus. I know he has to learn a lot of calculus if he is to excel in the Entrance exams for engg colleges in India.So what would you suggest is a good text to start with (in calculus) ? something that shall explain the basics but shall also take the reader to the more advanced and challenging places in calculus ?

TL; DR :

So what would you suggest is a good text to start with (in calculus) ? something that shall explain the basics but shall also take the reader to the more advanced and challenging places in calculus ?

答:
Good books on Calculus :

Calculus by Thomas Finney

Elementary Analysis: The theory of Calculus by Kenneth Ross.

Differential and Integral Calculus : Richard Courant.

You can see this link as well.

Title: suggest a textbook on calculus


问:
Why does a geodesic $\gamma$ have unit speed and $\ddot{\gamma}$ is normal to $S$ at each point
Hi i am reading about geodesics and there in the definition it is mentioned that

A curve $\gamma$ on a surface $S$ is called a geodesic if :

(a) it has unit speed

(b) $\ddot\gamma$ is normal to $S$ at each point.

My question is that is there any particular reason for choosing these conditions or are they arbitrary. Why unit speed and why is $\ddot\gamma$ is normal to $S$ at each point?

答:
First the unit speed is a choice of parametrization that is made mostly for convenience. One can define different curves $c_1: [0,1] \rightarrow M$, $c_2: [0,1] \rightarrow M$ that actually define the same set of points on the manifold $M$. Unit speed fixes a particular parametrization.

Second second derivative being normal to the curve is the property that makes a geodesic a geodesic (assuming it is unit speed). This is equivalent to saying that it locally minimizes distance. For any two points sufficiently close to each other a sufficiently short curve between them is a geodesic if and only if it minimizes the distance between the points. One has to prove that this is equivalent to the normal second derivative but this is the reason why we care about geodesics.


问:
$f$ and $g$ is homotopic $\iff$ $g-f$ is null-homotopic?
In the projective model structure on the category of chain complexes,
is the homotopicness between $f:A\to B$ and $g:A\to B$ is equivalent to null-homotopicness of $g-f:A\to B$ ?

I think it's intuitively true but can not find the proof.

答:
The claim is true for any model structure on any additive category.

Suppose we have a (weak) path object for $B$, i.e. an object $P$ and weak equivalences $j : B \to P$ and $q_0, q_1: P \to B$ such that $q_0 \circ j = q_1 \circ j = \textrm{id}_B$, and a homotopy from $f$ to $g$ w.r.t. this path object, i.e. a morphism $h : A \to P$ such that $q_0 \circ h = f$ and $q_1 \circ h = g$.
Then $h - j \circ f$ is a homotopy from $0$ to $g - f$: indeed, $q_0 \circ (h - j \circ f) = q_0 \circ h - q \circ j \circ f = f - f = 0$ and similarly $q_1 \circ (h - j \circ f) = q_0 \circ h - q \circ j \circ f = g - f$, as required.

Actually, more is true: even without a model structure, the localisation of an additive category is always additive and the localisation functor is additive.


问:
Simple calculation about derivative.
$M(s):=\{ e^s p + (1-p) \}^n$

My mathematics book says $M''(0)=np\{np+(1-p)\}.$
But my calculation is as follows.

\begin{align}
M'(s)&=n\{ e^s p + (1-p) \}^{n-1} e^s p. \\
M''(s)&=n(n-1)\{ e^s p + (1-p) \}^{n-2} e^s p +n\{ e^s p + (1-p) \}^{n-1} e^s p. \\
M''(0)
&=n(n-1)\{ e^0 p + (1-p) \}^{n-2} e^0 p +n\{ e^0 p + (1-p) \}^{n-1} e^0 p\\
&=n(n-1)p+np\\
&=n^2p.
\end{align}

Is my calculation wrong?

答:
Whithout doing any calculation it is self evident that the result of your book is correct!

In fact $M(s)$ is the moment generating function of a binomial distribution with second moment

$$\mathbb{E}[X^2]=np(1-p)+n^2p^2$$

that is exactly the second derivative in $s=0$

Anyway the first derivative is correct.

The second one is

$$M''(s)=np\{e^s[e^sp+(1-p)]^{n-1}+e^s(n-1)[e^sp+(1-p)]^{n-2}e^sp\}_{s=0}=$$

$$=np\{[p+(1-p)]^{n-1}+(n-1)p[p+(1-p)]^{n-2}\}=np[1+np-p]=np[np+(1-p)]$$

Title: Simple calculation about derivative.


问:
Like term reduction
In finding the derivative of $f(x) = 4x - x^2$ we first find the difference of the numerator $f(x + h) - f(x)$.  Therefore we have $f(x + h) = 4(x + h) - (x + h)^2 = 4x + 4h - x^2 - 2xh - h^2$ minus $f(x) = 4x - x^2$.  This leads to entire expression in the numerator being:

$$4x + 4h - x^2 -2xh -h^2 - (4x - x^2) $$

My book reduces this to $4h - 2xh - h^2$.  Now I understand how the $4x$ and the $-4x$ cancel.  But isn't $-x^2 - x^2 = 2x^2$?  How did they eliminate this term in the reduction?  If I plug in numbers (e.g. $x= 2 h = 3$) the reduction is not equal to the original equation.

In my experience I am wrong more than any math book so how am I going astray here?  Am I missing a sign somewhere?

答: You are missing that $- (-x^2) = +x^2$</p>

Title: Like term reduction

Body: You are missing that $- (-x^2) = +x^2$


问:
Generating Pythagorean Triples S.T. $b = a+1$
I am looking for a method to generate Pythagorean Triples $(a,b,c)$. There are many methods listed on Wikipedia but I have a unique constraint that I can't seem to integrate into any of the listed methods.

I need to generate Pythagorean Triples $(a,b,c)$ such that:

$$a^2 + b^2 = c^2$$
$$a\lt b\lt c \,; \quad a,b,c \in \Bbb Z^+$$
$$and $$
$$b=a+1$$

Is there a way to modify one of the listed methods to include this constraint?

答:
To produce Pythagorean triples where $a^2 + b^2 = c^2$ and $b = a + 1$

Start with $n = 1$

$\sqrt{n(1 + \sqrt{2})^4}$

Round this down to an integer, then square it.  This gets you $c^2$, which is $25$, so $c = 5$. Now take $c$ and divide it by $\sqrt{2}$.  $\dfrac{c}{\sqrt{2}}$, rounded to the nearest half, is $3.5$. $a$ and $b$ are the integers on either side of $3.5$.

Now repeat the process with $n = c^2$, to be precise, $n = 25$:

$\sqrt{25(1 + \sqrt{2})^4}$ rounded down, squared, makes $841$, so $c = 29$.  $\dfrac{c}{\sqrt{2}}$ rounded to the nearest half is $20.5$, so $a = 20, b = 21$

Again, repeat with $n = 841$ and keep repeating as long as you like, and you will find all of the infinite Pythagorean triples where $a^2 + b^2 = c^2$ and $b = a + 1$

Here's a really big one I found:

$a = 666554398276487455279750313289061631639$

$b = 666554398276487455279750313289061631640$

$c = 942650270102046130733437746596776286089$

I don't know if this is the kind of answer you were looking for, but I thought I'd share my method anyway.

Regards,

Josiah Hamilton

Edit: I had written "rounded to the nearest integer" but it was supposed to say "rounded DOWN ..."


问:
What Is This Measure Theory Rule Called? Is it Just the Substitution Rule in Disguise?
Let $(\Omega,\mathcal{F},\mu)$ be a measure space. For a $\mathcal{B}(\mathbb{R})$-measurable function $f \geq 0 : \Omega \rightarrow \mathbb{R}$ one can define a measure $\nu$ by:
$$
\nu(A) : = \int_A f d \mu = \int f \mathbb{1}_{A} d \mu
$$

$\nu$ is provably a measure according to this answer.

Now suppose we have some $\mathcal{B}(\mathbb{R})$-measurable function $g : \Omega \rightarrow \mathbb{R}$. And suppose $g$ is integrable w.r.t. $\nu$.

Question: Does this identity have a standard name in measure theory:

$$\int gf d \mu = \int g d \nu$$

Note: maybe the exact identity above isn't a named result in measure theory, but can be shown via the application of one or two standard measure theory results?

I think I can prove the identity using first principles of measure theory, but I'd rather use more standard results and my common sense is telling me the above is so simple it must have a standard name in measure theory that I'm just missing.

The proof of the identity I imagine is to use the approximation theorem for integrals, which asserts that a sequence of increasing simple functions approximates $\int g d \nu$. That is, we approximate with functions of the form:

$$\sum_{i=1}^n a_i \nu(A_i) = \sum_{i=1}^n a_i \int_{A_i} f d \mu$$

We can then invoke the approximation theorem again for the integrals inside the sum:

$$\sum_{i=1}^n a_i \sum_{j=1}^m b_{i,j} \mu(B_{i, j}) = \sum_{i=1}^n \sum_{j=1}^m a_ib_{i,j} \mu(B_{i, j})$$

Where $\bigcup_{j=1}^m B_{i, j} = A_i$. Since all the $B_{i,j}$ are disjoint, then the above is a simple function. Moreover, our invocations of the approximation theorem guarantee us that:

$\forall \omega \in B_{i,j} \subseteq A_i : a_i \leq g(\omega)$

$\forall \omega \in B_{i,j} : b_{i,j} \leq f(\omega)$

Therefore, we have:

$$\forall \omega \in B_{i,j} : a_ib_{i,j} \leq gf(\omega)$$

And so the term on the RHS is the integral of a simple function which is bounded by $gf$. And since the integral of $gf$ is the supremum over all such integrals of simple functions we have the bound:

$$\int gf d \mu \geq \int g d \nu$$

OK... so this is just an inequality rather than an identify. But I believe the proof can be fixed & completed by invoking the approximation theorem more fully: this will give us a sequence of simple functions converging to $f$ and another sequence converging to $g$. Then we can show that the resulting simple functions constructed above converges to $gf$ and so by the monotone convergence theorem the integrals will converge also.

But rather than complete this result in my own clunky way, I'd rather first see if there is some known result or results that can be applied here rather than go right back to first principles and simple functions.

Note: The induced measure $\nu$ is a measure on the original measurable space $(\Omega,\mathcal{F})$ - not a pushforward measure to $\mathcal{B}(\mathbb{R})$. So I can't see how something like change of variables / substitution can be applied. Unless I'm just not being creative enough...

Credit to this question for borrowed text.

答:
Thanks to the commenters for pointing me in the right direction. Since no formal answer has been posted I will do so on behalf of the commenters so that future readers may see the full argument. The crux of the argument is to use the Radon-Nikodym theorem, and crucially its uniqueness.

First we show that $\nu \ll \mu$. This follows from that fact that if $\mu(A) = 0$ then by the definition of the integral we have:

$$\nu(A) = \int_A f d \mu = 0 $$

Since $\nu \ll \mu$, then the Radon-Nikodym theorem is in force. This means that there exists a function $f'$ such that for all measurable sets $A$ the following identity holds:

$$\nu(A) = \int_A f' d \mu$$

Furthermore, the Radon-Nikodym theorem gives us that $f'$ is uniquely defined up to a $\mu$-null set, meaning that we have:

$$f = f' \;\;\; \mathrm{a.e.}$$

And so for any function $g$ we have:

$$gf = gf' \;\;\; \mathrm{a.e.}$$

Now, another property of any Radon-Nikodym derivative $f'$, as stated on Wikipedia, is that:

$$\int g d\nu = \int gf'd\mu$$

Putting it all together then:

$$\int g d\nu = \int gf'd\mu = \int gfd\mu$$

Which completes the proof.


问:
If $x|y^2 ~~\text{ and } ~~ y|z^2$ then prove or disprove: $x$ divides $z^2$.
Let $x, ~y, ~ z$ be three integers such that $$x|y^2 ~~\text{ and } ~~ y|z^2.$$ Then prove or disprove: $x$ divides $z^2$.

My attempt: There exist integers $k$ and $l$ such that $$y^2=kx~~\text{ and } ~~ z^2= ly.$$ Then we have $~ x| z^4$ but not $z^2$ also I am unable to find the counterexample. Please help me to solve this.

答: A simple example to disprove $x$ divides $z^2$ is with $x = 16$, $y = 4$ and $z = 2$. Then $x \mid y^2$ becomes $16 \mid 4^2 = 16$, and $y \mid z^2$ becomes $4 \mid 2^2 = 4$. However, $x \mid z^2$ becomes $16 \mid 2^2 = 4$, which is not true.


问:
Examine convergence of $ \sum^{\infty}_{n=2}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}$
I need to examine convergence of the following sum:

$$\displaystyle \sum^{\infty}_{n=2}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}$$

I know that:
$$\displaystyle \lim_{n \to \infty}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n} = 0$$

And that all elements of sequence $a_n$ are positive. Therefore I can use Cauchy's root test:
$$\sqrt[n]{\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}}= \frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n}$$
$$\sqrt[n]{\frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n}}= \frac{\sqrt[n^2]{n!}(1+\frac{1}{2n})}{\sqrt[n]{n}}$$

But it gives me noting since I don't know what happens to $\sqrt[n^2]{n!}$ when $n \to \infty$. I tried also d'alembert's ratio test, but it gives me nothing. I think it may be LTC on limits.

答:
An alternative method. First note that
$$
\left( {1 + \frac{1}{{2n}}} \right)^{n^2 }  = \left( {\left( {1 + \frac{1}{{2n}}} \right)^{2n} } \right)^{n/2}  < e^{n/2}
$$
for all $n\geq 1$. Second,
\begin{align*}
\log n! & = \sum\limits_{k = 1}^n {\log k}  = \log n + \sum\limits_{k = 1}^{n - 1} {\log k}  < \log n + \sum\limits_{k = 1}^{n - 1} {\int_k^{k + 1} {\log tdt} } \\ & = \log n + \int_1^n {\log tdt}  = (n + 1)\log n - n + 1.
\end{align*}
Thus
$$
n! < en\frac{{n^n }}{{e^n }} \Rightarrow \frac{{n!}}{{n^n }}\left( {1 + \frac{1}{{2n}}} \right)^{n^2 }  < ene^{ - n/2} .
$$
But
$$
\sum\limits_{n = 1}^\infty  {ne^{ - n/2} }  = \frac{{\sqrt e }}{{(\sqrt e  - 1)^2 }} \approx 3.9
$$
converges, so does the original series.


问:
A geometric proof for the inequality $\frac{2x}{\pi} \le \sin(x)$
The inequality $\frac{2x}{\pi}\le \sin(x)\le x$ for $0 \le x\le \frac \pi 2$ is well known; it can be proved using calculus.

The second part can be proved for $x\in [0,\pi/2]$ by geometric arguments:

Take unit circle with center origin. Then compare areas of (sector with angle $x$) and (right angled triangle with height $\sin x$).

Q. Can we prove $\frac{2x}{\pi}\le \sin(x)$ for $x\in [0,\pi/2]$ by geometric arguments?

Note: There are proof of first inequality are available using calculus, but I want to know if there is a proof, not based on calculus (Rolle's theorem, or mean value theorem etc.), but with some basic geometric arguments as done in the proof of $\sin(x)\le x$.

答:
The following is taken from Wikipedia: Jordan's inequality where it is attributed to

Yuefeng, Feng. “Proof without Words: Jordan's Inequality 2x/π ≤ Sin x ≤ x, 0 ≤ x ≤π/2.” Mathematics Magazine, vol. 69, no. 2, 1996, pp. 126–126. JSTOR, https://www.jstor.org/stable/2690669.

(File: https://commons.wikimedia.org/wiki/File:Jordans_inequality.svg, Attribution: Kmhkmh, CC BY 3.0 https://creativecommons.org/licenses/by/3.0, via Wikimedia Commons.)

The arc from $C$ to $D$ on the unit circle has the length $x$, and the arc from $G$ to $D$ on the circle with radius $\sin(x)$ has the length $\frac \pi 2 \sin(x)$. It follows that
$$
x \le \frac \pi 2  \sin(x) \iff \frac 2 \pi x \le  \sin(x) \, .
$$


问:
Operations on two families of sets
When it comes to families of sets, I have a difficulty comprehending how the sum and the intersection works when there are two (indexed) families of sets. Here is an example:
$$\bigcap\limits^{}_{t}(F_t\cap G_t) \subset \bigcup\limits^{}_{t} F_t \cap \bigcup\limits^{}_{t} G_t
 $$
where $ (F_t)_{t \in T}, (G_t)_{t \in T}  $ are families of sets.

How can I conceptualize this property? Do I intersect each set of two families  i.e. $ F_1\cap G_1$, and then interset all the intersections? Even if, how does it even work?

答:
All we need are definitions:
$$\bigcap\limits^{}_{t \in A}F_t =\{x \colon \forall t \in A, x \in F_t  \}$$
$$\bigcup\limits^{}_{t \in A}F_t =\{x \colon \exists t \in A, x \in F_t  \}$$
Now using them and that $\forall$ is distributive against $\land$, we obtain,
$$x \in \bigcap\limits^{}_{t}(F_t\cap G_t) \Rightarrow (\forall t) (x \in F_t\cap G_t) \Rightarrow \\ \Rightarrow (\forall t) (x \in F_t \land x \in G_t) \Rightarrow (\forall t) (x \in F_t) \land (\forall t) (x \in G_t)\Rightarrow \\ \Rightarrow x \in\bigcup\limits^{}_{t} F_t \cap \bigcup\limits^{}_{t} G_t$$

Addition.

Let's take $F_n=\left(-1,\frac{1}{n}\right)$ and $G_n=\left(-\frac{1}{n}, 1\right)$ for $n \in \mathbb{N}$, then $E_n = F_n \cap G_n = \left(-\frac{1}{n},\frac{1}{n}\right)$. Considering left side gives example, that countable intersection of open sets can be not open.

Title: Operations on two families of sets


问:
Is there an easy way to see that all derivatives are bounded?

Show that all derivatives of $f:\mathbb{R}\to\mathbb{R}$ given by $$f(x):=\frac{1}{\sqrt{x^2+1}+1}$$ are bounded.

It's easy to see that all derivatives are continuous. So the only potential problem is that a derivative might blow up at $\infty$. Is there an easy way to see that this does not happen? I guess deriving an explicit form of the n-th derivative is not the way to do it (I'd doubt there is an easy closed form).

答:
It is clear by composition that $f$ is smooth. So we need only prove that the derivatives are bounded on, say $A=\{x\;;|x|\geq 2\}$. Since $f$ is even we can restrict to $I=[2,+\infty)$.

This form of $f$ is not tractable. Observe that for $x\neq 0$,
$$
f(x)=\frac{\sqrt{x^2+1}-1}{(\sqrt{x^2+1}+1)(\sqrt{x^2+1}-1)}=\frac{\sqrt{x^2+1}-1}{x^2}=\frac{\sqrt{x^2+1}}{x^2}-\frac{1}{x^2}.
$$
It is trivial to check that the derivatives of $\frac{1}{x^2}$ are all bounded on $I$, so we can focus on
$$
g(x)=\frac{\sqrt{x^2+1}}{x^2}=\frac{1}{x}\sqrt{1+\frac{1}{x^2}}.
$$
Now we can use the power series representation of $(1+u)^{1/2}$ at $0$, whose radius of convergence is $1$, to get
$$
g(x)=\frac{1}{x}\sum_{n\geq 0}\binom{1/2}{n}\left(\frac{1}{x^2}\right)^n=\sum_{n\geq 0}\binom{1/2}{n}x^{-2n-1} \quad \forall x\geq 2
$$
with normal convergence as $0\leq x^{-1}\leq \frac{1}{2}<1$. Term-by-term differentiation turns out well, as these yield after $k$ differentiations
$$
\frac{1}{x^{k+1}}\sum_{n\geq 0}\binom{1/2}{n}(-2n-1)(-2n-2)\cdots(-2n-k-1)(x^{-2})^n.
$$
Of course $\frac{1}{x^{k+1}}$ is bounded on $I$, and the remaining power series evaluated at $x^{-2}$ still has radius $1$. So the latter converges normally on $I$ for each $k$. Hence a simple induction justifies simultaneously term-by-term differentiation and boundedness on $I$.


问:
Showing that $B \rightarrow (C \rightarrow D) , (B \rightarrow C) \vdash_{L_1} (B \rightarrow D)$
First of all , I was able to prove that ,

$B \rightarrow (C \rightarrow D) , (B \rightarrow C) \vdash_{L_1} B \rightarrow (B \rightarrow D)$

But I wasn't able to prove from it that ,

$B \rightarrow (C \rightarrow D) , (B \rightarrow C) \vdash_{L_1}  (B \rightarrow D)$

Description of $L_1 :$

Primary connectives: $\lnot$ , $\lor$

$A\rightarrow B$ is $\lnot A \lor B$

Axioms:

(A1) $B \lor B \rightarrow B$

(A2) $B \rightarrow B \lor C$

(A3) $B \lor C \rightarrow C \lor B$

(A4) $(C \rightarrow D ) \rightarrow (B \lor C \rightarrow B \lor D)$

The only rule of inference is $MP$

One of my attempts:

1.$((B \rightarrow D) \rightarrow D) \rightarrow ((B \rightarrow (B \rightarrow D)) \rightarrow (B \rightarrow D))$

Here,if I was able to show that
 $B \rightarrow (C \rightarrow D) , (B \rightarrow C) \vdash_{L_1}((B \rightarrow D) \rightarrow D)$
  then I could have also shown that ,
 $B \rightarrow (C \rightarrow D) , (B \rightarrow C) \vdash_{L_1}  (B \rightarrow D)$

But I am not sure if it is even possible.

Can someone give me a clue about this?

答:
Hint

You have proved: $B→(C→D),(B→C)⊢_{L_1} B→(B→D)$.

The premises are the same; rewrite the conclusion as: $\lnot B \lor (\lnot B \lor D)$.

Apply Associativity (your previous post) to get: $(\lnot B \lor \lnot B) \lor D$ and the conclusion will follow using (A1), (A4) and Transitivity of $\to$ (result c. of your previous post).


答:
John Omielan answered the question by giving a counterexample. To complement his answer, note that the statement is true if $x$ is a prime number. Indeed, a prime number $x$ has the property that if $x|ab$, then $x|a$ or $x|b$.
Applied to the situation in the question, $x|y^2$ implies that $x|y$ if $x$ is prime. But $y|z^2$, so $x|z^2$.

Title: If $x|y^2 ~~\text{ and } ~~ y|z^2$ then prove or disprove: $x$ divides $z^2$.

Body:
Let $x, ~y, ~ z$ be three integers such that $$x|y^2 ~~\text{ and } ~~ y|z^2.$$ Then prove or disprove: $x$ divides $z^2$.

My attempt: There exist integers $k$ and $l$ such that $$y^2=kx~~\text{ and } ~~ z^2= ly.$$ Then we have $~ x| z^4$ but not $z^2$ also I am unable to find the counterexample. Please help me to solve this.


答:
Here is a generalized version of the previous counter example..

Let $x=t^4,y=t^2,z=t$, $\forall x \in Z^+$ then $x∣y^2$  becomes $t^4|(t^2)^2$ which is true,  $y∣z^2$ becomes  $t^2|(t)^2$ again true, and $x∣z^2$ becomes $t^4|(t)^2$ which is obviously false....


问:
Proving the Fibonacci identity $a^2_{n+1} - a_{n}a_{n+2} = (-1)^n$ using mathematical induction

Prove by Mathematical Induction that:
$$a^2_{n+1} - a_{n}a_{n+2} = (-1)^n$$

Here the terms are from the Fibonacci Sequence.

Body: A hint: let $a:=F_n,\,b:=F_{n+1}$ so$$\color{blue}{F_{n+2}^2-F_{n+1}F_{n+3}}+\color{red}{F_{n+1}^2-F_nF_{n+2}}=(a+b)^2-b(a+2b)+b^2-a(a+b).$$Simplify this.


问:
$\sqrt a+\sqrt b$ is a root of polynomial, then $\sqrt a -\sqrt b$ is so over finite field
Let $a,b$ is not a square element in finite field.
If polynomial over the finite field has a root $\sqrt a+\sqrt b$, then the polynomial has $\sqrt a-\sqrt b$ as a root too?

Thank you for your help.

This question arises from concrete problem, $x^4-10x^2+1$（this is a minimal polynomial of $\sqrt{2}+\sqrt{3}$ over rational field） is reducible over arbitrary finite field.
$\sqrt{2}+\sqrt{3}$ is root over finite field, then $\sqrt{2}-\sqrt{3}$ is also a root（why?）.And $\sqrt2+\sqrt3$ and $\sqrt2-\sqrt3$ exists in $2$-degree extension of the finite field, so the minimal polynomial spilts. This leads minimal polynomial is at most $2$ degree, so $x^4-10x^2+1$ is reducible.

答:
Let's look at the concrete example, to see what is going on, and assume we are not in characteristic $2$.

Then, using the normal quadratic formula if $$x^4-10x^2+1=0$$ then $$x^2=\frac {10\pm \sqrt{100-4}}{2}=5\pm 2\sqrt 6$$ and if $6$ has a square root in your finite field, then the original equation is reducible.

The roots in a splitting field are $\pm \sqrt 2\pm \sqrt 3$ and the factorisation already obtained corresponds to one pairing of the roots: $$(x^2-5-2\sqrt 6)(x^2-5+2\sqrt 6)=$$$$=(x-\sqrt 2-\sqrt 3)(x+\sqrt 2+\sqrt 3)\cdot(x+\sqrt 2-\sqrt 3)(x-\sqrt 2+\sqrt 3)$$

If we were instead to choose other pairings we'd get $$(x-\sqrt 2+\sqrt 3)(x+\sqrt 2+\sqrt 3)\cdot(x-\sqrt 2-\sqrt 3)(x+\sqrt 2-\sqrt 3)=$$$$=(x^2+2\sqrt 3 x+1)(x^2-2\sqrt 3x+1)$$ or $$(x-\sqrt 2+\sqrt 3)(x-\sqrt 2-\sqrt 3)\cdot(x+\sqrt 2-\sqrt 3)(x+\sqrt 2+\sqrt 3)=$$$$=(x^2-2\sqrt 2 x-1)(x^2+2\sqrt 2x-1)$$ and these give factorisations in the case that $\sqrt 3$ or $\sqrt 2$ exist in your finite field - ie that $3$ or $2$ are squares.

Now, you may also know that the product of two non quadratic residues modulo $p$ is a quadratic residue. From this it is easy to deduce that at least one of $2,3, 6$ has a square root in a field of characteristic $p$. So the polynomial is reducible. If two of the three have roots then the third does and the polynomial splits into linear factors.

Modulo $5$ we have that $6$ is a square while $2$ and $3$ are not. Modulo $7$ we have that $2$ is a square while $3$ and $6$ are not. Modulo $11$ we have that $3$ is a square and $2$ and $6$ are not. Modulo $23$ we have that $2, 3, 6$ are all squares. So all four possibilities occur.

Finally it is easy to show that in characteristic $2$ we have $x^4-10x+1=x^4+1=(x+1)^4$.

So I've worked this in some detail in case it helps you to see what is going on. You can generalise to $a$ and $b$. Obviously there are more efficient expositions, but sometimes longhand helps.


问:
Can we show that if $T'$ is injective (surjective), then $T$ is surjective (injective)?
If $T$ is a linear operator between vector spaces $X,Y$ and $T^\ast$ denotes the algebraic adjoint, i.e. $T^\ast:Y^\ast\to X^\ast,y^\ast\mapsto y^\ast\circ T$, it is easy to see that if $T$ is injective (surjective), then $T^\ast$ is surjective (injective). In particular, if $T$ is bijective, then $T^\ast$ is bijective.

Now I've read that if $X,Y$ are Banach spaces and $T':Y'\to X'$ denotes the topological adjoint, then $T$ is bijective if and only if $T'$ bijective. How can we show this and why is it important that we are dealing with Banach spaces and that we've replaced the algebraic with the topological duals?

And are we only able to show the equivalence or does it even hold that if $T'$ is injective (surjective), then $T$ is surjective (injective)?

答:
The implication: ``$T$ surjective  $\Rightarrow$ $T'$ or $T^*$ is injective'' is quite easy to prove:

Assume $T'f=T'g$. Then $(T'f)(x)=(T'g)(x)$ and $f(Tx)=g(Tx)$ for all $x\in X$. Since $T$ is surjective, $f=g$ follows. Same for $T^*$. Note that when dealing with continuous operators here, surjectivity of $T$ can be replaced by: image of $T$ is dense in $Y$.

The implication: ``$T$ injective  $\Rightarrow$ $T'$ surjective'' is false in general: take $T$ to be an injective compact operator between infinite-dimensional Banach spaces. Then $T'$ is compact as well, and the range of $T'$ is not closed. I have no idea/intuition about the algebraic dual $T^*$ for this implication.

The implication: ``$T'$ injective  $\Rightarrow$ $T$ surjective'' is false as well. Take $T$ compact with dense range, then $T'$ is injective, but $T$ has not closed range.

The implication ``$T$ bijective  $\Rightarrow$ $T'$ or $T^*$ bijective'' on the other hand is true, and can be proven by showing that $(T')^{-1}=(T^{-1})'$ and $(T^*)^{-1}=(T^{-1})^*$.

The implication ``$T'$ bijective  $\Rightarrow$ $T$ bijective'' is a consequence of the closed range theorem: By the closed range theorem, the range of $T$ is closed and the annihilator of the nullspace of $T'$, so $T$ is surjective. Injectivity of $T$ is clear. Here, $X,Y$ have to be Banach spaces.


答: In your calculation of the second derivative $M^{''}(s)$, you forgot a factor of $e^sp$ in your first term. So you should have $$M''(s)=n(n-1)\{ e^s p + (1-p) \}^{n-2} e^{2s} p^2 +n\{ e^s p + (1-p) \}^{n-1} e^s p$$

Title: Simple calculation about derivative.

Body:
$M(s):=\{ e^s p + (1-p) \}^n$

My mathematics book says $M''(0)=np\{np+(1-p)\}.$
But my calculation is as follows.

\begin{align}
M'(s)&=n\{ e^s p + (1-p) \}^{n-1} e^s p. \\
M''(s)&=n(n-1)\{ e^s p + (1-p) \}^{n-2} e^s p +n\{ e^s p + (1-p) \}^{n-1} e^s p. \\
M''(0)
&=n(n-1)\{ e^0 p + (1-p) \}^{n-2} e^0 p +n\{ e^0 p + (1-p) \}^{n-1} e^0 p\\
&=n(n-1)p+np\\
&=n^2p.
\end{align}

Is my calculation wrong?


问:
Find the value of $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$
Can this limit be solved without using L'Hopital's rule :

$$\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$$

Answer of this limit is : $1$

答:
Here is another approach

$$ \lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}=\lim_{x \rightarrow \infty}e^{\frac{\ln\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}} =e^{\lim_{n\to \infty}\frac{\ln\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}} $$

$$ = e^{\lim_{n\to \infty}\frac{ -\frac{1}{1+x^2} }{1}}=e^0=1.$$

Notice that, We used the L'hopital's rule.


问:
How is angular velocity of both these bodies equal here?
Here there is John who is looking at a bus at which behind there is Harry. My teacher says that Angular velocity I.e(Theta / time)of the bus and Harry is equal when they appear to be at same point by Swaraj.

Now , my question is how is it possible since time and theta for both (John and Harry)appear to be different. Also , the arc Length of both are different Is it some Euclidean postulate my teacher used? I am not able to get which one is it here. My teacher did not give a reason for this.

I believe theta 1 and theta 2 are different as they appear to be. So will their time be different to reach at those points

答:
Consider these four pairs of arcs:

In each picture, the angles match: $\theta_{red} = \theta_{blue}$.

The arc length $L_{red}$ and $L_{blue}$ does vary, because $L = r \theta$, and the radiuses $r_{red}$ and $r_{blue}$ differ.

Angular velocity is defined as the rate of change in the angle $\theta$ with respect to time.  So, if we assume picture (1) with $\theta = 5°$ is taken at time $t = 1\,s$, and that picture (4) with $\theta = 20°$ is taken at time $t = 4\,s$, the average angular velocity is the same for objects represented by the blue arc as they are for objects represented by the red arcs:
$$\omega = \frac{\Delta \theta}{\Delta t} = \frac{20° ~-~ 5°}{4\,s ~ - ~ 1\,s} = \frac{15°}{3\,s} = 5°/s$$

The actual velocities (or speeds) do differ, average velocity being the distance travelled over the time taken.  Since the object represented by the blue arc travels a much longer distance than the one represented by the red arc in the same time, the velocity of the object represented by the blue arc is much higher than the velocity of the object represented by the red arc.  However, their angular velocities are the same.


问:
n unique sums of squares for 2p^(2n) for prime p (1 mod 4) and n natural number >= 1?
After looking at many sums of squares, this is something I am hypothesizing. I'd love to know if there is any work related to this, if it is easy to solve or even if there exist counterexamples.

Conjecture: Let $p$ be some prime number where $p \equiv 1 \bmod 4$. Then, for $n \in \mathbb{N}$ and $n \geq 1$, $2p^{2n}$ will give $n$ unique pairs $a,b \in \mathbb{N}$ such that:

\begin{align*}
    a^2 + b^2 &= 2p^{2n}
\end{align*}

In the image below I have provided 3 different examples along with the $a,b$ sums of squares for more details.

Ex 1: $p=617$ and $n=1$.

Ex 2: $p = 13$ and $n=3$.

Ex 3: $p = 5$ and $n=7$.

link to image with examples described above

答: There’s an exact formula for the number of ways to express given integer as a sum of two squares(by the theory of modular forms). You can use this to prove your statement.</p>


问:
On maximizing $\operatorname{Trace}(AX)$ subject to $\|X\|=1$
Let $\Bbb S_n (\Bbb R)$ denote the set of $n \times n$ real symmetric matrices. Given $A \in \Bbb S_n (\Bbb R)$, let

$$ X^* := \arg\max_{X \in\Bbb S_n (\Bbb R)} \operatorname{Trace}(AX) \quad\mbox{s.t.}\quad \|X\| = 1 $$

For what set of matrix norms (e.g., Frobenius, spectral, $p$-Schatten, etc) in the problem above, would the maximizer have the following form?

$$ X^* = \frac{1}{\|A\|} A $$

答:
Let $A=\|a_{ij}\|\ne 0$ and $X=\|x_{ij}\|$. Then $\operatorname{Trace} AX=\sum_{i,j=1}^n a_{ij}x_{ji}$. If the Frobenius norm $\|X\|_F=\sqrt{\sum_{i,j=1}^n x_{ij}^2}$ of the matrix $X$ equals $1$ then by the Cauchy-Schwarz inequality
a set $\mathcal X$ of $\arg\max_X\operatorname{Trace}(AX)$ consists of a unique element  $X^*=\|A\|_F^{-1}A^t$.

For other matrix norms this conclusion can fail.

For instance, for $L_{1,1}$ norm of $X$, we have if $\|X\|=1$ then $X\in\mathcal X$ iff for each $1\le i,j\le n$ we have $a_{ij}x_{ji}\ge 0$, if $(i,j)$ is $\arg\max_{i’,j’} |a_{i’,j’}|$ and $x_{ji}=0$, otherwise. Since for each $p\ge 1$, $L_{1,1}$ norm coincides with $L_{p,1}$ norm for diagonal matrices, $X^*$ can fail to belong to $\mathcal X$ for these norms either.

For max norm of $X$, we have $X\in\mathcal X$ iff for each $1\le i,j\le n$ such that $a_{ij}\ne 0$ we have $x_{ji}=\operatorname{sgn} a_{ij}$. Since max norm coincides with $\|\cdot \|_1$ and $\|\cdot \|_\infty$ norms for diagonal matrices, $X^*$ can fail to belong to $\mathcal X$ for these norms either.


问:
Quick Check on Bayes Rule in Terms of Sets
Recently, I took a quiz with a question asking me to find the conditional probability of X given Y in terms of sets.

I solved the question as:

$$Pr(X|Y) = \frac{Pr(Y|X)Pr(X)}{Pr(Y)}$$
$$Pr(X|Y) = \frac{Pr(X \cap Y)}{Pr(Y)}$$

The answer was revealed to be "the probability of the intersection of X and Y, divided by the probability of X." Have I made a mistake somewhere? Any way I think about it, $Pr(Y)$ should be the denominator.

答:
Let's take definition of conditional probability  of X given Y, when $P(Y) \gt 0$
$$P(X|Y)=\frac{P(X \cap Y)}{P(Y)}\quad (1)$$
Obviously, when $P(X) \gt 0$ we have
$$P(Y|X)=\frac{P(Y \cap X)}{P(X)}\quad (2)$$
Now solving $P(Y \cap X)$ from second $P(Y \cap X) =P(Y|X) \cdot P(X)$ and putting it in (1), using $Y \cap X = X \cap Y$, gives
$$P(X|Y)=\frac{P(X \cap Y)}{P(Y)}=\frac{P(Y|X) \cdot P(X)}{P(Y)}\quad (3)$$

As to the your last question in comments, then generally, when $P(X) \ne P(Y)$ and $P(X \cap Y) \ne 0$, then of course $P(X|Y) \ne P(Y|X)$.


问:
Why is $\pi $ equal to $3.14159...$?
Wait before you dismiss this as a crank question :)

A friend of mine teaches school kids, and the book she uses states something to the following effect:

If you divide the circumference of any circle by its diameter, you get the same number, and this number is an irrational number which starts off as $3.14159... .$

One of the smarter kids in class now has the following doubt:

Why is this number equal to $3.14159....$? Why is it not some other irrational number?

My friend is in a fix as to how to answer this in a sensible manner. Could you help us with this?

I have the following idea about how to answer this: Show that the ratio must be greater than $3$. Now show that it must be less than $3.5$. Then show that it must be greater than $3.1$. And so on ... .

The trouble with this is that I don't know of any easy way of doing this, which would also be accessible to a school student.

Could you point me to some approximation argument of this form?

答:
I wrote a blog article about this a few years ago in response to a friend who asked why $\pi$ came out to be approximately 3, rather than some completely different number.  I didn't know, but tried to make a list of potentially relevant lines of investigation. I came up with 8 items, some obviously more pertinent than others.  In approximately descending order of quality, they were:

Inscribe a regular hexagon in a circle. The hexagon obviously has a perimeter of 6. The circle passes through the same six points but instead of going straight from one point to the next it takes a circuitous route. So it must be a bit longer.  Therefore $\pi$ must be a bit more than 3.

The shortest curve that can enclose a unit area has length $2\sqrt\pi$.

A penny on the table can be touched by at most six other pennies at the same time; this is closely related to the fact that 6 is the largest integer less than $2\pi$, the circumference of the penny.

$\pi$ is the smallest positive root of the equation $$x-\frac{x^3}6 +\frac{x^5}{120} - \frac{x^7}{5040} +\cdots = 0$$

I mentioned Buffon's needle, but I should have left it out, since it's closely related to $\pi$'s appearance as the perimeter of a unit circle and so is a repeat of earlier items.

The probability that two randomly-selected integers are relatively prime is $\frac6{\pi^2}$, but I think on closer examination this turns out to be related to the circles again.

$$1+\frac14+\frac19 + \cdots = \frac{\pi^2}6$$ but this seems closely related to the previous item.

$\pi$ appears not just in circles but also in spheres, but this is obviously scraping the bottom of the barrel.

People have since written to me pointing out that $\pi$ appears in certain statements of physical law, but these all seem clearly and simply related to its appearance as a circle and sphere constant.  For example in Coulomb's law: the  electric field E at distance $r$ from a charge $q$ is $$\frac1{4\pi\epsilon_0}\frac{q}{r^2}.$$  But the appearance of $4\pi$ in the denominator here is solely because it is the surface area of a unit sphere, and so we divide  by it to get the electric flux per unit area.

I had a bit more to say in the original article.  I hope this is some help.

Title: Why is $\pi $ equal to $3.14159...$?


问:
Show that $(\mathbb{Q}(\theta)/\mathbb{Q})$ is a divisor of 32.
Let $\theta:=\frac{\sqrt{2}+\sqrt{3}+i\sqrt{3}}{\sqrt{5}+\sqrt{7}+i\sqrt{7}}\in \mathbb{C}$. Show that $(\mathbb{Q}(\theta)/\mathbb{Q})$ is a divisor of 32.

Studying for my upcoming algebra exam, I stumbled upon this exercise from an old algebra exam.

My idea:

$\mathbb{Q}(\theta)$ from $\mathbb{Q}$ can be constructed out of a series of quadratic field extensions. But I'm not sure how to do that properly so maybe someone could give me a hint? Thanks in advance!

Edit:

SoI think the the quadratic field extensions should look something like this

$\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{5}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7},i)$

Is this better, since the $i$ only appears once?

答:
Multiply numerator and denominator by $(\sqrt{5}+\sqrt{7}-i\sqrt{7})$ to get a real denominator. Then multiply numerator and denominator by a suitable number (which one ?) to get a rational denominator.

Then, simplifying numerator if necessary, you should get that you only need to add five square roots: $\sqrt{2},\sqrt{14},\sqrt{15},\sqrt{21}$ and $i$. Since $2^5=32\cdots$


问:
Proof that Expression Involving Moment Generating Function is Expectation
Suppose we have a random variable (RV) $X$ defined on a measurable space $\mathcal{M} = (\Omega, \Sigma)$. Suppose we equip the measurable space with a probability measure $P$ and associated expectation operator $E$ such that for all $\theta \in \mathbb{R}$ we have $E[e^{\theta X}] < \infty$. Then, for all bounded, continuous $f : \mathbb{R} \rightarrow \mathbb{R}$ define $F_{\theta}$ as follows:

$$F_{\theta}(f) = \dfrac{E[f(X)e^{\theta X}]}{E[e^{\theta X}]}$$

Question: Show that there exists a probability measure $P_{\theta}$ defined on $\mathcal{M}$ such that for all bounded continuous functions $f$ the following identity holds:

$$F_{\theta}(f) = E_{\theta}[f(X)]$$

where $E_{\theta}[Y]$ is the expectation of any RV $Y$ w.r.t. $P_{\theta}$.

Background: Around 45 minutes into this video, the lecturer defines an operator $E_{\theta}$ w.r.t to a moment generating function and makes a brief argument as to why this operator is in fact the expectation of an implicitly defined probability measure $P_{\theta}$. I can't see how this conclusion can be made.

Note: If the above result is, or relies on, some known theorem I'd be delighted if someone could point me in the direction of that.

Note: in expressions such as $f(X)$ and $e^{\theta X}$ we are composing the RV $X$ with continuous functions to build new RVs. Also in the expression $f(X)e^{\theta X}$ we use both composition with $X$ and mutliplication of two RVs to get a new RV.

Note: I'm not sure if maybe we need to restrict $\theta > 0$. I'm thinking about it...

答:
This answer and the subsequent determination of the answerer in the comments helped guide me towards the right solution. However, I still don't fully follow the use of LOTUS which is unfamiliar to me so I will rewrite the basic ideas of that answer in a way that I understand.

Given an arbitrary $\theta$ we are looking for an existential witness $P_{\theta}$ satisfying the condition in the question. With that motivation in mind, let's proceed. We start by defining $h_{\theta}$ as follows:

$$h_{\theta} = \dfrac{e^{\theta X}}{E[e^{\theta X}]}$$

Now, since $h_{\theta} \geq 0$ then by this answer we have that $P_{\theta}$ as defined below is a measure:

$P_{\theta}(A) = \int_A h_{\theta} dP$

However, in order for $P_{\theta}$ to qualify as an existential witness to our problem, we must show that it is a probability measure i.e. we want to show that $P_{\theta}(\Omega) = 1$. Well using the definitions of $P_{\theta}$ and expectation we can reason as follows:

$$P_{\theta}(\Omega) = \int_{\Omega} h_{\theta} dP = \int \dfrac{e^{\theta X}}{E[e^{\theta X}]}dP = \dfrac{\int e^{\theta X}dP}{E[e^{\theta X}]} = \dfrac{E[ e^{\theta X}]}{E[e^{\theta X}]} = 1$$

OK, so now we have a probability measure $P_{\theta}$. Therefore it has an associated expectation $E_{\theta}$. Applying $E_{\theta}$ to the RV $f(X)$ and expanding out the definition of expectation we get:

$$E_{\theta}[f(X)] = \int f(X) dP_{\theta}$$

Now by the identity in this answer and expanding out the definition of $h_{\theta}$ we get:

$$\int f(X) dP_{\theta} = \int f(X) h_{\theta} dP = \int f(X) \dfrac{e^{\theta X}}{E[e^{\theta X}]} dP = \dfrac{\int f(X) e^{\theta X}dP}{E[e^{\theta X}]} = \dfrac{E[f(X) e^{\theta X}]}{E[e^{\theta X}]} = F_{\theta}(f)$$

Which completes the proof.


问:
How do you visualise a subspace of 2 by 2 matrices(and higher up, 3 by 3 etc)
If a subspace is a linear combination of some vectors, how can a subspace of matrices(which act to stretch a space) exist?</p>

答:
$m\times n$-matrices with coefficients in a field $K$ form a $K$-vectorspace. You can add and subtract them componentwise and you have a multiplication with scalars in $K$ (again componentwise). These two operations can be easily checked to satisfy the axioms of a $K$-vectorspace. Moreover the matrices with a 1 in exactly one entry can be seen to form a basis, so we immediately get that $\operatorname{Mat}_{m,n}(K)$ is $mn$-dimensional.

Any vector space has subspaces. The easiest one being the zero subspace. If your vector space has dimension $\geq 2$, then any single nonzero vector generates a nontrivial subspace of dimension 1. So in particular for $mn\geq 2$ we can choose a nontrivial matrix (ie a vector in the vector space of matrices) and get a subspace of matrices out of it.

As $mn\geq 4$ rather easily we find that visualizing subspaces of matrices is rather difficult. I don’t know if anyone can really visualize anything of dimension $>4$. It is common to think of it like it would be 3-dimensional.


问:
Can we deduce that $x$ is a common divisor of $y$, $z$ and $u$
Let us consider the following relations between integers:

$x$ divides $yw$

$x$ divides $zt$

$x$ divides $uq$

and

$x$ is not a common divisor of $w$, $t$ and $q$. Here $w$, $t$ and $q$ are composite.

Can we deduce that $x$ is a common divisor of $y$, $z$ and $u$? If not, can we add some conditions to guarantee this property?

PS: We can consider $x$ as the product of three primes. We assume that those primes are factors of $w$, $t$, and $q$ respectively.

答:
No. For example, take $x$ to be any product of three primes $abc$ and set $y=a,w=bc$ and $z=b,t=ac$, etc.)

I don't see a natural extra condition that forces this (other than $x$ being coprime to each of $w,t,q$ in which case it is not a very interesting statement). For example, you can certainly have $w,t,q$ being pairwise coprime without it working, by taking $x=abc,y=ab,z=bc,u=ac,w=cd,t=ae,q=bf$ where $a$ to $f$ are distinct primes.


答:
This is why the abstraction  provided by algebraic structures is useful.

The "vectors" of a vector space $E$ need not be actual, geometric vectors (as in $\vec{AB}$ is a vector of $\mathbb R^2$). Saying that $E$ is a vector space (say with underlying field $\mathbb R$) simply means that you have some operation $+$ in $E$, some neutral element $0 \in E$, that you can multiply elements of $E$ by any real number $\lambda$ ... and a bunch of other properties.

We do not care what the elements of $E$ actually are: number, matrices, functions, matrices of functions acting on polynomial on some finite field ... who cares ? This is the power that algebra provides, as long as your space satisfies this set of axioms, then all the theorems apply without ever having to look at what you are actually applying them to.

This allows you to think about any (finite dimensional) vector space as $\mathbb R^2$ or $\mathbb R^3$ (in order to get good intuitions - obviously, this is not a rigorous visualisation ...) and make conjectures, observations ... that might be true just because you are in a vector space. This is because what matters is the structure, the relations between the points, not the actual points of the space.

Obviously, a lot of questions are specific to some vector spaces, but some of them are really about the structure of the space ... in which case you don't even need to know what you are working with, the fact that you have a vector space means that you can think of it as $\mathbb R^2$, get an easy intuition on how things work, and immediately transpose it to the general setting.

Title: How do you visualise a subspace of 2 by 2 matrices(and higher up, 3 by 3 etc)

Body: If a subspace is a linear combination of some vectors, how can a subspace of matrices(which act to stretch a space) exist?</p>


问:
Computing time of flight in a medium with a varying index of refraction
The index of refraction, $n$, for some medium, is the speed of light in a vacuum divided by the speed of light in the medium ($n = \frac{c}{v}$).

Consider some light source, $p_s$ (at coordinates $[x_p,y_p,z_p]$), and some light receiver, $p_r$ (at coordinates $[x_r,y_r,z_r]$). Neglecting the effects of refraction, we might calculate the time of flight for a light signal, moving in a straight line from $p_s$ to $p_r$, as:

$T = \frac{1}{c} \cdot \sqrt{(x_p - x_r)^2 + (y_p - y_r)^2 + (z_p - z_r)^2 }$

For a constant index of refraction, we simply replace $c$ by $\frac{c}{n}$. How, however, might we calculate the time of flight for a light signal traveling in a "straight line" when the index of refraction varies as a function of some coordinate (say, depth ($z$))? In this case, of course, light doesn't actually travel in a straight line, in the common sense.

What I have so far...

Note: This could be very wrong. It's just where I've gotten so far.

Using the principle of least time (see), we have that:

$T = \int dt$

We have that $v = \frac{ds}{dt}$. Thus, $T = \int \frac{ds}{v}$

Suppose $n$ is a function of $z$. Then, $n = n(z)$. Using $v = \frac{c}{n(z)}$, we have that:

$T = \int \frac{n(z)}{c} ds$

Now, by Pythagoras $ds^s = dx^2 + dy^2 + dz^2$. So, $ds = \sqrt{dx + dy +dz}$. Then,

$T = \int \frac{n(z)}{c} \sqrt{dx + dy + dz}$

What I want, in the end, is to have a formula such that I can plug in $[x_p,y_p,z_p]$, $[x_r,y_r,z_r]$, $n(z)$, $c$, and get $T$.

EDIT

I've found a reference, here. Unfortunately, I don't quite follow everything. I recreate the argument here. I use the above example, where $n$ is a function of depth, $z$. We simplify this to a problem in 2-dimensions.

We know that the index of refraction, $n$, at a point $z,x$ is equal to $c/v$, where $v$ is the velocity of light at that point. We parameterize the path by equations $x = x(u)$ and $z = z(u)$. The "optical path length" from a point $A$ to a point $B$ is then given by:

$$L = \int_A^B n\;ds = \int_A^B n\;\sqrt{z^2 + x^2}\;du$$

I don't quite follow what is being done at this point. I know that we want to minimize $L$... I'm not sure, however, how this step achieves this.

To make this integral an extremum, let $f$ denote the integrand function:

$f(z,x,\dot{z},\dot{x}) = n(z,x)\sqrt{z^2+x^2}$

Then the Euler equations are:

$$\frac{\partial n}{\partial z} = \frac{d}{du}\Bigg(\frac{\partial f}{\partial \dot{z}}\Bigg)$$

$$\frac{\partial n}{\partial x} = \frac{d}{du}\Bigg(\frac{\partial f}{\partial \dot{x}}\Bigg)$$

Which gives:

$\frac{\partial n}{\partial z} \sqrt{\dot{z}^2 + \dot{x}^2} = \frac{d}{du} \Bigg[\ \frac{n\dot{z}}{\sqrt{\dot{z}^2 + \dot{x}^2}}\Bigg]$

$\frac{\partial n}{\partial x} \sqrt{\dot{z}^2 + \dot{x}^2} = \frac{d}{du} \Bigg[\ \frac{n\dot{x}}{\sqrt{\dot{z}^2 + \dot{x}^2}}\Bigg]$

If we define parameter $u$ as spatial path length $s$, then $z^2 + x^2 = 1$, and the above equations reduce to:

$$\frac{\partial n}{\partial z} = \frac{d}{ds}\Bigg(n \frac{dz}{ds}\Bigg)$$
$$\frac{\partial n}{\partial x} = \frac{d}{ds}\Bigg(n \frac{dx}{ds}\Bigg)$$

I could be looking at this wrong, however this is my interperetation...

So, since (in my example) $n$ is a function of $z$, it occurs to me that we can drop the partial derivatives, and have:

$$\frac{dn}{dz} = \frac{d}{ds}\Bigg(n \frac{dz}{ds}\Bigg)$$

Then:

$$L = \frac{d}{ds} = \frac{\frac{dn}{dz}}{\Bigg(n \frac{dz}{ds}\Bigg)}$$

Suppose $n(z) = e^z$, for example...

$$L = \frac{e^z}{\Bigg(e^z \frac{dz}{ds}\Bigg)}$$

$\frac{dz}{ds}$ should be a differential equation, yes? Perhaps I'm being slow about this, but it isn't immediately clear to me how I'd then use starting and ending values for $x$,$y$, and $z$ to obtain an $L$. As I recall, the derivative can't simply be "flipped" (i.e. this isn't $= \frac{ds}{dz}$)... but maybe I'm wrong.

答:
Short answer:

There is no formula for arbitrary $n(z)$.  However, for many specific functional forms of $n(z)$ (e.g. $n(z) = e^z$) you can derive a formula for $T$ as a function of $[x_p,y_p,z_p]$ and $[x_r,y_r,z_r]$ (and $c$ of course).  To derive such a formula you need to solve a differential equation involving $n(z)$.

Background:

Finding paths of least time is one of the classic problems of a subject called calculus of variations.  Basically what we try to do in the calculus of variations is find a function which minimizes some integral.  This is similar to what we do in ordinary calculus when we try to find the maximum or minimum value of some function, but slightly more complicated.  (The idea is either case is to look for where the derivative is zero.)  You've already written down the integral that we want to minimize in this case, namely $$T=\int_{\text{path}} \frac{n(z)}{c} ds$$ Once we know which path minimizes this integral, we can just plug it in and evaluate the integral to find the minimum time.

How to solve it:

There is a set of differential equations for the path of least time, which you've also written down: $$\begin{align}\frac{\partial n}{\partial z} = \frac{d}{ds}\left(n \frac{dz}{ds}\right) \\ \frac{\partial n}{\partial x} = \frac{d}{ds}\left(n \frac{dx}{ds}\right) \end{align}$$
Here's how to interpret these formulas: $x=x(s)$ and $z=z(s)$ are coordinates of the path we want to find, parametrized by the path length $s$.  Then $n=n(z)=n(z(s))$ is also implicitly a function of $s$.  If we know the form of $n(z)$ then we can try to solve these differential equations.  E.g. let's take $n(z) = e^z$.  Then the two equations above become
$$\begin{align}
e^z = \frac{d}{ds}\left(e^z \frac{dz}{ds}\right) \\
0 = \frac{d}{ds}\left(e^z \frac{dx}{ds}\right)
\end{align}$$
The first of these equations can be written $e^z = \frac{d^2}{ds^2} e^z$, which implies $e^z = c_1 e^s + c_2 e^{-s}$, or $z(s) = \log\left(c_1 e^s + c_2 e^{-s}\right)$ (for some integration constants $c_1, c_2$).  Then the second equation above can be integrated once to give $c_3 = e^z \frac{dx}{ds}$, which means $\frac{dx}{ds} = c_3 e^{-z} = c_3/\left(c_1 e^s + c_2 e^{-s}\right)$.  This can be integrated to give a formula for $x(s)$.  In the end we find
$$\begin{align}
z(s) = \log\left(c_1 e^s + c_2 e^{-s}\right) \\
x(s) = \tan^{-1}\left(\sqrt{\frac{c_1}{c_2}} e^s\right)\frac{c_3}{\sqrt{c_1 c_2}} + c_4
\end{align}$$
You can see how this is a bit messy already, but doable.  It will get messier as we proceed to the next steps, and in the interest of clarity I will only say what you need to do, rather than writing out all the math.

The next step is to invert these formulas for $x(s)$ and $z(s)$ to find the constants of integration $c_i$ in terms of the starting and ending points $[x_p,z_p]$ and $[x_r,z_r]$.  Two of your equations for this step will be $x(0)=x_p$, $z(0)=z_p$.  There is also a consistency condition $x'(0)^2 + z'(0)^2 =1$ which comes from the fact that we parametrized by arc length $s$.  Then there will be the equations for the final point $x(s_f) = x_r$ and $z(s_f) = z_r$ which must be solved for both the remaining unknown integration constants and the final path length $s_f$.

Once you have found all of the $c_i$'s and the final path distance $s_f$, you can plug all of this into the formula for $T$, $T=\int \frac{n(z)}{c} ds$, to get your final formula.

Summary:

As you can see, the hard part of this comes from solving the differential equation for the path of least time.  It can get algebraically heavy, and for some forms of $n(z)$ it will have no analytical solution.  That's why there will be no general formula for time $T$.  However, for many forms of $n(z)$ it is solvable, and in general you can solve the differential equations numerically if you really need that information.

If you want to learn more about this, I would recommend looking into any book on classical mechanics.  Lagrangian mechanics, in particular, proceeds very much along the same lines.


问:
Normal modes in physics - solution of set of 2nd order linear differential equations
In physics we study normal modes for vibrating system. Mathematically speaking we have a set of linear differential equations. For example we have something like this:

$$
\begin{align*}
m \frac{dx_1}{dt^2} &= k(x_2-x_1) \\
M \frac{dx_2}{dt^2} &= k(x_3-x_2) - k(x_2-x_1) \\
m \frac{dx_3}{dt^2} &= -k(x_3-x_2)
\end{align*}
$$

We seek solutions for $x_i(t)$. A general solution is to assume to look for these functions to be sines, cosines or exponents for all $x_i(t)$ having the same frequency $\omega$. For above example our solution would be something like:

$$
\begin{align*}
x_1 (t) &= A e^{\omega t + \varphi} \\
x_2 (t) &= B e^{\omega t + \varphi} \\
x_3 (t) &= C e^{\omega t + \varphi}
\end{align*}
$$

I could not find a clear explanation why we assume that we seek solution with the same $\omega$ (normal modes), and later why the general solution is the superposition of all such solutions (we obviously find . Many thanks for comments and explanation.

答:
The system modeled as

$$
\begin{align*}
m \frac{d^2x_1}{dt^2} &= k(x_2-x_1) \\
M \frac{d^2x_2}{dt^2} &= k(x_3-x_2) - k(x_2-x_1) \\
m \frac{d^2x_3}{dt^2} &= -k(x_3-x_2)
\end{align*}
$$

can be represented shortly as

$$
\ddot x = A x
$$

Normally in this kind of problems $A$ can be represented as $T^{-1}\Lambda T$ with $T^{-1}T = I$ where $\Lambda$ is a diagonal matrix with the eigenvalues from $A$. So we can follow with

$$
\ddot y = \Lambda y,\ \ \ y = T x
$$

This system doesn't contain attenuation terms so it represents an oscillatory system. This system will remain forever oscillating along their natural frequencies  associated to the eigenvalues $\lambda_k$. The most plausible system with the properties you cited is an actuated dissipative system with the structure

$$
\ddot x + 2B \dot x + A x = u
$$

with $u = f(\omega t)$ in which $f$ represents a sinusoidal adequate function. The solution for this system can be written as

$$
y_k(t) = e^{-b_k t}\left(c_k e^{-\sqrt{b_k^2-\lambda_k}t}+\bar c_k e^{\sqrt{b_x^2-\lambda_k}t}\right)+ y_p(\omega t)
$$

For passive systems, the homogeneous component goes asymptotically to zero remaining the particular $y_p(\omega t)$


问:
Properties of non-singular matrix
Let's say I have matrix A and its n x n and non-singular. Which of the following always true?

The linear system Ax = 0 has only one non-trivial solution.

A row echelon form of A has no non-pivot column

det(A) = n

rank(A) can be any non-zero integer value not more than n.

From what we've done in class, I think it should be 2.

It's not 1 as Ax=0 only has the trivial solution as its non-singular. Its not 3 as det(A) does not always equal n and since A is non-singular, it should be full rank.

答:
You're right.

False, because if the matrix is non-singular, then the system $Ax=0$ has only the trivial solution (and consequently no non-trivial solutions). This is because the matrix being non-singular implies that every system $Ax=b$ has unique solution, and $x=0$ is always a solution to $Ax=0$, so it's unique in the case of $A$ being non-singular.

True (consecuence of the matrix having determinant different from $0$, and also with the fact said in point 4, because if it had a non-pivot column, then it would not have full rank and it would be a singular matrix).

False, the determinant can be anything different from $0$, but in general it's not equal to $n$ (take for example $I_2$, the $2\times 2$ identity matrix, then $|I_2|=1\neq 2$).

False. If the determinant is different from $0$, then the column vectors of $A$ are linearly independent, and then you conclude that $\text{rank}(A)=n$ (full rank).

Title: Properties of non-singular matrix


问:
Is it possible to rationalize a denominator containing two cube roots?
The fraction in question is

$$-\frac{12}{\sqrt[3]{12\sqrt{849} + 108} - \sqrt[3]{12\sqrt{849} - 108}}$$

And was reached in calculating the solution to $x^4 - x - 1 = 0$. I've tried all the standard methods, including $(a+b)(a-b) = a^2 - b^2$, but that doesn't work for cube roots, because once you have the square of one the two middle terms will not cancel each other out.

答:
As imranfat suggests in his comment, you should use the identity
$$
\frac1{\sqrt[3]a-\sqrt[3]b} = \frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}{a-b}
$$
which can be verified via cross-multiplication. In your case, take $a=12\sqrt{849} + 108$ and $b=12\sqrt{849} - 108$ and then work through simplifying the resulting expression. When I do so, I obtain
$$
-\frac{48+\left(12 \sqrt{849}-108\right)^{2/3}+\left(108+12 \sqrt{849}\right)^{2/3}}{18}.
$$</p>


问:
Calculating the tangent plane on a point (x, y, z) when the function given has no z?
So I have this function that looks something like $2x^2-x-5y=17$ and I need to find out the tangent plane for this surface on the point $P=(x_1,y_1,z_1)$. I understand that the gradient for a function on a point is the normal vector to the level surface on that point. Given that normal-vector I can define a plane that will be tangent to the level surface.

Now what I do not understand is that this function does not contain a z-component. And to find out the level surface for a point of a surface I basically need to take the funtion $f(x,y)$ such that $f(x,y)=C$. Am I supposed to bring the whole expression to one side, as in my example such that $2x^2-x-5y-17=0$ and then define $f(x,y)$ such that $f(x,y)=2x^2-x-5y-17$ and furthermore $f(x,y)=z$? Is this $C$ that is supposed to be $C=z_1$?

Of course the functiongraph is not equal to the function but when I plot both $2x^2-x-5y=17$ and $z=2x^2-x-5y-17$ I (as I expect) get different results in geogebra.

So my question is, how do I calculate the tangentplane on the point $(x_1,y_1,z_1)$ of a surface, but the function is unbound by one component?

答: Consider the curve $2x^2-x-5y=17$ on the plane $z=0$. Now compute the tangent line to that curve passing through $(x_1,y_1,0)$. You will get a line $ax+by=c$ on the plane $z=0$. Then your tangent plane will be the plane $ax+by=c$, which is the vertical plane containing the line that you got.


问:
How to find the conditional variance?
Suppose I have the following time series process with $a_t$ being an independent white noise with mean $0$ and variance $\sigma^2$:

$$
z_t = \frac{1}{3}a_t + \frac{1}{3}a_{t-1} + \frac{1}{3}a_{t-2}
$$

How do I compute the variance conditional on $a_{t-2} = 5$

Also, the unconditional variance in this problem is $\frac{\sigma^2}{3}$ correct? Thank you!

答:
$\frac{1}{3}a_t$ has variance $\frac19\sigma^2$

$a_t,a_{t-1},a_{t-2}$ are mutually independent so

$\mathrm{Var}(z_t)=\frac39\sigma^2=\frac13\sigma^2$

$\mathrm{Var}(z_t\mid a_{t-2}=5)=\frac29\sigma^2$

Title: How to find the conditional variance?


问:
How to find the second derivative of an expression in tensor form.
I would like to calculate $\Box\phi$

whereby $\phi = exp(ip_{\mu}x^{\mu})$ and $\Box = \partial_{\mu}\partial^{\mu}$  and whereby $\partial_{\mu} =   \frac{\partial}{\partial x^{\mu}} $ and $\partial^{\mu} = \frac{\partial}{\partial x_{\mu}}$.

However, I am stuck on how to do this since I was always told that we can never have more than two (a covariant and contravariant component) of the same indices in the same expression but the expressions we were given will mean I will get multiple of the same index in one term.

How would I go about solving this problem and avoiding that. I tried doing it so that I define $\Box = \partial_{\nu}\partial^{\nu}$ but it jut does not give me the correct answer, if someone could walk me through this I would greatly appreciate it.

答:
As far as I understand you operate in four-dimensional flat Minkowski space. In this case all indices lifting is done by means of the metric tensor $g^{ik} =diag (1 -1 -1 -1)$. For example, $\partial^i=g^{ik}\partial_k$. Descent of indices is done via $g_{ik} =diag (1 -1 -1 -1)$; $g_{ik}g^{kl}=\delta_i^l =diag (1 1 1 1)$.
Everywhere the summation over repeated indices is supposed.

Now you can easily take the derivatives:

$\partial_{k}x^i=\delta_k^i$

$\partial_{k}\exp(ip_{m}x^{m})=\exp(ip_{m}x^{m})ip_l\delta_k^l=$$\exp(ip_{m}x^{m})ip_k$

$\Box=g^{ik}\partial_{i}\partial_{k}\exp(ip_{m}x^{m})=g^{ik}\partial_{i}(\exp(ip_{m}x^{m})ip_k)=(ip_k)(ip_i)g^{ik}\exp(ip_{m}x^{m})$$=-p_kp^k\exp(ip_{m}x^{m})$, etc.

$p_kp^k =p_0p^0+p_1p^1+p_2p^2+p_3p^3=g^{ik}p_ip_k=p_0p_0-p_1p_1-p_2p_2-p_3p_3$

$g^{ik}\partial_{i}\partial_{k}=\partial_0\partial_0-\partial_1\partial_1-\partial_2\partial_2-\partial_3\partial_3$


问:
Infinite product is entire and has order 1/2
Prove that the following function is entire and has order $\frac{1}{2}$.

\begin{equation}
        f(z) = \prod_{k = 1}^\infty \left(1 + \frac{z}{k^2}\right)
    \end{equation}

I think this product is supposed to equal some familiar function in complex analysis but I'm not sure what that is. Some hints would be greatly appreciated!

答:
We can find the order without any prior knowledge

$$\frac{f'(z)}{f(z)}=\sum_{k\ge 1}\frac1{z+k^2}$$
$\sup_{|z|=r}|f(z)|=f(r)$, for $r\in (0,\infty)$ we have

$$\frac{f'(r)}{f(r)}\sim \int_1^\infty \frac1{r+x^2}dx=
 \frac1{2i\sqrt{r}}\int_1^\infty (\frac1{x+i\sqrt{r}}-\frac1{x-i\sqrt{r}})dx
$$ $$=\frac{\log(1-i\sqrt{r})-\log(1+i\sqrt{r})}{2i\sqrt{r}}\sim \frac{\log(i)-\log(-i)}{2i\sqrt{r}}=\frac{\pi}{2\sqrt{r}}$$

Whence $$\log f(r)\sim \pi \sqrt{r}$$
therefore $f$ has order $1/2$.


问:
Sequence of operators in $L^2(\mathbb{R})$
I am trying to solve the following exercise:

In the space $L^2(\mathbb{R})$ define a sequence of operators $(A_n)$ that are such that $||A_nf||=||f||$ and $\langle A_nf,g\rangle\to0$ for every $f,g\in L_2(\mathbb{R})$.

I have thought of defining $(A_n)f(x)=f(x+n)$ but I don't know how to proceed, can someone give me a hint? Thanks a lot.

答:
It seems like this question has had many iterations, but I'll react to what I see now in the hopes that I am right. My answer would be to essentially look at $A_n$, where it is  the $n$-shift operator on $\ell^2_{\mathbb{R}}(\mathbb{N})$.

Take a countable orthonormal basis of $L^2(\mathbb{R})$, and denote it by $\{ \phi_n\}_{n=1}^\infty \subseteq L^2(\mathbb{R}) $. For every $f\in L^2(\mathbb{R})$ there is a unique sequence $\{ a_n(f) \}_{n=1}^\infty$ such that

$$ f=\sum_{k=1}^\infty a_k(f)\cdot \phi_k.$$

$A_n(f)$ will simply be

$$ A_n(f)=\sum_{k=1}^\infty a_{k}(f)\cdot \phi_{n+k}. $$

Notice that

$$ \langle A_n(f),\phi_{k} \rangle= \begin{cases} a_{k-n}(f) & ,k> n\\ 0 & ,k\leq n \end{cases}  $$

Hence, $\langle  A_n(f), \phi_k\rangle=0$ for $n$ large enough.

By Paresval's identity, we know that

$$\Vert A_n(f) \Vert^2= \sum_{k=1}^\infty \vert a_k(f)\vert^2. $$
Any finite linear combination

$$ g' = \sum_{j=1}^N c_j \cdot \phi_{k_j}, $$
satisfies that

$$  \langle A_n(f) ,\sum_{j=1}^N c_j \cdot \phi_{k_j}\rangle=0  $$

for $n$ large enough. Since the closure of the linear span of $\{ \phi_n \}$ is equal to $L^2$, you can deduce that

$$ \langle A_nf,g\rangle\to0 $$

for all $g\in L^2(\mathbb{R})$. Notice that $f$ was arbitrarily chosen here.


问:
Lifespan estimate for system of ODEs
My lectures notes on ODE mention the following theorem:

Proposition (Lifespan): Let $I \subseteq \mathbb{R}$ be an interval, $D \subseteq I \times \mathbb{R}$ open, $(t_0,y_0) \in D$ and $f: D \to \mathbb{R}$ continuous and locally Lipschitz with respect to $y$.
Let $U(t_0,y_0)$ be the neighbourhood of $(t_0,y_0)$ from the local
Lipschitz condition and $h,c>0$ and define

$$Q:=[t_0,t_0+h] \times [y_0-c,y_0+c] \subseteq U(t_0,y_0),$$

then the maximal interval of existence of the solution to the IVP

$\left\{\begin{array}{} y'(t)=f(t,y(t)), t \in I\\ y(t_{0})=y_{0}\\ \end{array} \right.$

contains $[t_0,t_0+a]$ where $a=min\{h,c/M\}$ with $M=\max \limits_{(t,y) \in Q}  \{\lvert f(t,y) \rvert\}$.

Proof: We define the function

$\tilde{f}(t,y) = \left\{\begin{array}{} f(t,y_0-c), t \in [t_0,t_0+h], y<y_0-c\\
f(t,y), t \in [t_0,t_0+h], y_0-c<y<y_0+c\\
f(t,y_0-c), t \in [t_0,t_0+h], y_0+c<y\\ \end{array} \right.$

then for all $(t,y) \in [t_0,t_0+h] \times \mathbb{R}$ the argument of
$\tilde{f}$ lies in $Q$, and thus in $U(t_0,y_0)$. So $\tilde{f}$
satisfies a global Lipschitz condition on $[t_0,t_0+h]$, so the IVP

$\left\{\begin{array}{} y'(t)=\tilde{f}(t,y(t)), t \in [t_0,t_0+h]\\
y(t_{0})=y_{0}\\ \end{array} \right.$

has a unique solution $v$ on $[t_0,t_0+h]$. So for $t \in [t_0,t_0+a]$
(note that $a \leq h$)

$$\lvert v(t)-y_0 \rvert=\lvert y_0 + \int \limits_{t_0}^{t} v'(s) ds
> - y_0\vert=\lvert \tilde{f}(s,v(s)) ds \rvert.$$

In addition, $\tilde{f}$ is bounded by $M$ by construction, so we have

$$\lvert v(t) - y_0 \rvert \leq M \lvert t-t_0\rvert \leq Ma \leq M
> \frac{c}{M}=c$$

It follows that $v(t) \in [y_0-c,y_0+c]$ for all $t \in [t_0,t_0+a]$.
But this means that for all $t \in [t_0,t_0+a]$, we have
$\tilde{f}(t,v(t))=f(t,v(t))$, and so

$$v'(t)=\tilde{f}(t,v(t))=f(t,v(t))$$

and $v(t_0)=t_0$. But then $v$ is a solution to the IVP given in the
theorem on $[t_0,t_0+a]$. So the maximal interval of existence must
contain at least $[t_0,t_0+a]$.

The author also states that this can be generalized to systems of ODEs, but I could not find anything on google. The idea of the proof for a single ODE is to extend the function to a globally Lipschitz function by changing its values outside of the cube $Q$. My guess is that the same line of argument works for systems, i.e. we need to choose suitable values $h,c_1,...,c_d$ such that the solution does not leave the cube $Q$ in any direction. I think we can restrict ourselves to directions parallel to the axes since these are the shortest paths to get outside of $Q$. However, I cannot see how to define a globally Lipschitz function $g$ that agrees with $f$ on $Q$ in the case of a $d$-dimensional system.

Could anyone please point me to a source that proves this theorem for systems or give a proof for this case?

Thanks a lot!

答:
Define
$$
\psi(y)=y_0+\frac{c}{\max(c,\|y-y_0\|)}(y-y_0).
$$
This is a continuous function with $\psi(y)=y$ for $\|y-y_0\|\le c$ and constant on rays through $y_0$ outside this "ball".

Now define $\tilde f(t,y)=f(t,\psi(y))$ and continue in the same fashion as before.

One might have to use
$$
\left\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\right\|
\le\left\|\frac{a}{\|a\|}-\frac{a}{\|b\|}\right\|+\left\|\frac{a}{\|b\|}-\frac{b}{\|b\|}\right\|
=\|a\|\,\frac{|\|b\|-\|a\||}{\|b\|\,\|a\|}+\frac{\|a-b\|}{\|b\|}
\le\frac{2}{\|b\|}\|a-b\|
$$
or better to prove the global Lipschitz property.

Title: Lifespan estimate for system of ODEs


问:
SKI Calculus prefix notation of odd/even number
I'm working on a homework with SKI calculus. I saw the hints in this very useful post.

We basically defined SKI functions as:

```<code>def tt  = K;
def ff  = S K;

def inc = S (S (K S) K);
def _0  = S K;
def _1  = inc _0;
def _2  = inc _1;
def _3  = inc _2;

def not  = S (S (K (S (K S) K)) S) (K K);
(not tt) x y;  # y
(not ff) x y;  # x
</code>```

By the hint provided by the above post, I found out that _3 f x = f(f(f(x))).

If I do _3 not ff, I can basically get the answer I'm looking for: (_3 is_odd) x y and it will output x, therefore is_odd = not ff.

However, the hw requirement is to write the function as (is_odd _3) x y -- a prefix notation.

I tried to use the reverse order operator def reverse = S (K (S I)) K; where reverse x y -> y x, but when you apply this to three arguments: reverse (not ff) _3 -> _3 (not ff), which is not the right order (_3 not) ff.

Am I going down the wrong path here? How to deal with the prefix notation problem in SKI calculus?

答:
Your on the right track here with the _3 not ff and the reverse operator. But you seem to be have a small amount of confusion with the bracketing. _3 not ff is (_3 not) ff instead of the of the _3 (not ff) you seem to be implying by setting is_odd to not ff.

Since reverse x y is equivalent to y x 1 we can transform (_3 not) ff into reverse ff (_3 not). We can apply the same transformation to the subterm _3 not to get reverse ff (reverse not _3) This is almost but not quite what we what: here _3 all the way on the right, but buried inside some parentheses. We can fix this by noticing we have a composition of the functions reverse ff and reverse not applied to _3. The composition of a function f with a function g can be made as S (K f) g. Applying this to reverse ff (reverse not (_3)) gives (S (K (reverse ff)) (reverse not)) _3. Note we didn't use anything special about _3 so we could do the same transformation for any numeral.

Overall we have is_odd can be defined as (S (K (reverse ff)) (reverse not)).

1 I'm assuming that claim from your question is true, I haven't checked myself


问:
Polya's Urn for Three Colors Instead of Two
Is it possible to extend Polya's Urn problem to balls with $3$ different colors instead of just $2$?

ie. An urn contains $1$ red, $1$ blue, and $1$ green ball. At each turn you draw one ball and put it back along with another ball of the same color. Let $X_R$ be the number of red balls in the urn after turn $n$. What is the distribution for $X_R$?

From this post, in a regular Polya's urn with two colors, the number of red balls in the urn after $n$ draws is uniform over $\{1, ..., n+1\}$ ie:
$$P(X_R = k) = \frac{1}{n+2} \text{ for } 1 \le k \le n+1$$

How would I need to change this for my scenario in which there are three colors? Any help would be great!

答:
After $n$ turns the probability distribution for $1,2,3,...,n+1$ red balls is
$$\frac{2(n+1)}{(n+1)(n+2)},...,\frac{6}{(n+1)(n+2)},\frac{4}{(n+1)(n+2)},\frac{2}{(n+1)(n+2)}.$$
In other words, the probability of there being $r$ red balls is proportional to $n+2-r$.

This is straightforward to prove by induction. For the inductive step, consider the probability of having $r+1$ balls after $n+1$ turns.

This is the probability of having $r$ balls after $n$ turns multiplied by the probability a red ball is then chosen, $\frac{r}{n+3}$, plus the probability of having $r+1$ balls after $n$ turns  multiplied by the probability a red ball is not then chosen, $\frac{n+2-r}{n+3}$.

The probability is therefore proportional to $$r(n+2-r)+(n+2-r)(n+1-r)=(n+2-r)(n+1)$$ and is thus proportional to $$n+2-r=(n+1)+2-(r+1).$$