问:
The position of a ladder leaning against a wall and touching a box under it
I was reading a newspaper and there was a little math riddle, I thought "how funny, that's gonna be easy, let's do it" and here am I...

The problem goes as follow :
in a barn, there is a 1 meter cubic box against a wall and a 4 meter ladder is leaning against the wall, touching the box at its corner. Here is a picture :

So, the big triangle has a hypotenuse $FE$ of $4$, the square $ABDC$ has sides of length 1 and is basically "insquared" at the right angle, i.e. $D\in \overline{FE}$.
The question is "what is the length of the biggest cathetus", here $AF$.

So far, no problem.

Now here are my solutions:

By Thales' intercept theorem, $\frac{FB}{FA}=\frac{BD}{AE}$, by hypothesis, $FB=FA-1$ and $BD=1$. Now by Pythagoras, $FA^2+AE^2=FE^2$; by hypothesis, $FE=4$, so we end up with a system of equations, letting $h=FA, d=AE$: $$ \begin{align} &\frac{h-1}{h}=\frac{1}{d} \\ &h^2+d^2=4^2  \end{align} $$
Which solves (removing 3 non-relevant solutions) into $d \cong 1.3622$ and $h \cong 3.76091$.

Now, if I consider the "function" of the line : $f(x)=\frac{-h}{d}x+h$, I know that $f(1)=1$ and I end up with Pythagoras with the system :$$ \begin{align} &\frac{-h}{d}+h=1 \\ &h^2+d^2=4^2  \end{align} $$
it solves again into the same, again removing 3 non-relevant solutions

Okay, this means that using Pythagoras is no good since it ends up giving a quartic equation (4 answers of which 3 are "non-relevant").

Now if I consider the length of the arc $f(x)$ between $0$ and $d$ it has to be $4$ and again $f(1)=1$ I end up with the system:
$$ \begin{align} &\frac{-h}{d}+h=1 \\ &\int_0^d \sqrt{1+(f'(x))^2} dx =\int_0^d \sqrt{1+\left(\frac{-h}{d}\right)^2} dx = d \sqrt{1+\left(\frac{-h}{d}\right)^2}   \end{align} $$
Which solves again into the same answers, but this time removing only 2 non-relevant solutions (i.e. it gives a cubic equation instead of a quartic).

I tried also using the areas and the smaller trangles $FAD$ and $AED$ for example : $\frac{h \cdot d}{2} = \frac{h\cdot 1}{2}+\frac{d\cdot 1}{2}$

Yet I wasn't able to get to any "hand solvable" solution : if I were able to bring it down to some quadratic equation, that would be nice, since it is a common assumption, here, that everybody has seen the "general formula for solving quadratic equations" in school and so would be able to solve this, I may then see how it is seen as a funny riddle in the newspaper.

My best trial, with "just" a cubic equation, is way too complicated for the normal readers of this newspaper, so it's bugging me.

What am I missing? Some basic properties maybe? It's really bugging me, not being able to solve this without Wolfram.

答:
With $x=|FB|$ and $a=|CE|$ the total area of $AFE$ is

$A_{tot}=\frac {(1+x)(1+a)}{2}$

$A_{tot}$ is also given as the sum of $ABDC$, $FBD$ and $CDE$

$A_{tot}=1 + \frac {x}{2} + \frac{a}{2}$

Equating the two expressions for $A_{tot}$ results in $a=\frac{1}{x}$.

Substituting this in (Pythagoras for triangle $AFE$)

$(x+1)^2+(1+a)^2=16$

results in

$x^4+2x^3-14x^2+2x+1=0$

The largest positive root is $x \approx 2.7609$.


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
The imaginary number $i$ was discovered/invented/revealed as a solution the equation
$$
x^2+1=0\tag{$\ast$}
$$
That equation, and that equation alone, is the link between $i$ and the reals. However, there is another solution to $(\ast)$: $-i=-1\times i$ also satisfies $(\ast)$. Basically, this is because $(-1)^2=1$.

Since the link between $i$ and the reals is the same equation that is satisfied by $-i$, one can easily wonder, "which is which?" When we choose a solution to $(\ast)$, do we get $i$ or $-i$? In a sense, it doesn't matter; both satisfy $(\ast)$ and $i^2=-1$ is the property we use to do math in $\mathbb{C}$.

On the other hand, complex conjugation, swapping $i\leftrightarrow-i$, is an important isomorphism of $\mathbb{C}$. For instance, any polynomial with real coefficients that has $x+iy$ as a root, must also have $x-iy$ as a root. There are many ways to show this, but one of the most basic is by swapping $i\leftrightarrow-i$. This isomorphism does not affect the real coefficients of the polynomial, but it swaps $x+iy\leftrightarrow x-iy$.

In the end, we call one root of $(\ast)$, $i$, and the other, $-i$; it really doesn't matter which. They both satisfy $(\ast)$ and that is what is important.

ContentLicense: CC BY-SA 3.0

Title: What is $-i$ exactly?


问:
Prove that $z_1z_2z_3 = 1$ for complex numbers $z_1$, $z_2$, $z_3$.
Let $z_1$, $z_2$, $z_3$ be distinct complex numbers such that $|z_1| = |z_2| = |z_3| > 0$. If $z_1 + z_2z_3$, $z_2 + z_3z_1$, and $z_3 + z_1z_2$ are real numbers, prove that $z_1z_2z_3 = 1$.

I believe that this problem comes from 1979 Romanian Math Olympiad (State Competition, 10th grade).

I tried several approaches: The first was to substitute $a + ib$, $c + id$, and $f + ig$ for $z_1$, $z_2$, and $z_3$, respectively, and bash. I used $(1+z_1)(1+z_2)(1+z_3)$, which didn't really get me anywhere.

After a couple more attempts at substitutions and such, I hit a wall and didn't really know what to do.

答:
Let $r = |z_1| = |z_2| = |z_3| > 0$ be the common modulus and $P = z_1z_2z_3$. Then
$$
 z_1 + z_2 z_3 = \overline{z_1} + \overline{z_2} \cdot \overline {z_3} = \frac{r^2}{z_1} + \frac{r^4}{z_2 z_3} \, .
$$
Multiplication with $z_1$ gives
$$
z_1^2 + P = r^2 + \frac{r^4 z_1^2}{P}
\implies z_1^2 \left( 1 - \frac{r^4}{P}\right) = r^2 - P \, .
$$
The last identity holds for $z_2$ and $z_3$ as well.
It follows that $P = r^4$ because otherwise $z_1^2=z_2^2=z_3^2$, which is not possible for distinct $z_1, z_2, z_3$.

So we have $z_1z_2z_3 = r^4$. Taking the absolute value implies that $r=1$.

ContentLicense: CC BY-SA 4.0


问:
A way to find this shaded area without calculus?
This is a popular problem spreading around. Solve for the shaded reddish/orange area. (more precisely: the area in hex color #FF5600)

$ABCD$ is a square with a side of $10$, $APD$ and $CPD$ are semicircles, and $ADQB$ is a quarter circle. The problem is to find the shaded area $DPQ$.

I was able to solve it with coordinate geometry and calculus, and I verified the exact answer against a numerical calculation on Desmos.

Ultimately the result is 4 terms and not very complicated. So I was wondering: Is there was a way to solve this using trigonometry? Perhaps there is a way to decompose the shapes I am not seeing.

A couple of years ago there was a similar "Find the shaded area" problem for Chinese students. I was able to solve that without calculus, even though it was quite an involved calculation.

Disclosure: I run the YouTube channel MindYourDecisions. I plan to post a video on this topic. I'm okay posting only the calculus solution, but it would be nice to post one using only trigonometry as many have not taken calculus. I will give proper credit to anyone that helps, thanks!

Update: Thanks for everyone's help! I prepared a video for this and presented 3 methods of solving it (the short way like Achille Hui's answer, a slightly longer way like David K and Seyed's answer, and a third way using calculus). I thanked those people in the video on screen, see around 1:30 in this link: https://youtu.be/cPNdvdYn05c.

答:
The answer is $5+\frac34 (10^2\arctan(\frac12)-40) = 75\arctan(\frac12)-25 =\pm9.774$.

First I divided the large square in five equal areas of twenty (see image). A part of our shape occupies one fourth of the smaller central square with area 20. The remaining part of our shape consists of half a lense with radius 10 (A), minus a scaled copy of that lense with radius 5 (B). Hence the remaining part of our shape is $\frac34$ of area A.

Area A is obviously a circle segment with area $\frac{2\arctan(\frac12)}{2\pi}\times \pi 10^2 $ minus a triangle of area $40$. $\blacksquare$


问:
When not to treat dy/dx as a fraction in single-variable calculus?
While I do know that $\frac{dy}{dx}$ isn't a fraction and shouldn't be treated as such, in many situations, doing things like multiplying both sides by $dx$ and integrating, cancelling terms, doing things like $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ works out just fine.

So I wanted to know: Are there any particular cases (in single-variable calculus) we have to look out for, where treating $\frac{dy}{dx}$ as a fraction gives incorrect answers, in particular, at an introductory level?

Note: Please provide specific instances and examples where treating $\frac{dy}{dx}$ as a fraction fails

答:
Proposition 1 is an example of the situation when $\mathrm dy\over\mathrm dx$ can't be seen as a quotient.

(the definition of differential and derivative is at the end)

Proposition 1:

Forall function $f$ satisfied with $x=f(t)$, if its derivative at $t_0$ is equal to $\mathrm dx\over\mathrm dt$, $g$ satisfied with $y=g(x)$ is also derivable at $f(t_0)$, then the derivative of $g\circ f$ at $t_0$ is $${\mathrm dy\over\mathrm dt}={\mathrm dy\over \mathrm dx}\cdot{\mathrm dx\over\mathrm dt}$$
(End of proposition.)

The conclusion is that least one of ${\mathrm dy\over \mathrm dx}$ and ${\mathrm dx\over \mathrm dt}$ can not be understood as quotient. Below is the explanation. If both of them are quotient, then $\mathrm dx$ will be cancelled out; however, in this case, chain rule is not meaningful, it is not even a mathematics statement (only a sentence without maths meaning); nevertheless, chain rule is not a meaningless sentence. Let's look at the reason for if seeing both as a quotient, then chain rule is a meaningless sentence: not all function $f$ which has a derivative at $x_0$ is satisfied with the statement, for $\mathrm dx$ can be $0$ which can not be placed at denominator; furthermore, the quantity with the same notion (such as $\mathrm dx$ in that sentence) is seen as identical in maths. Therefor, there is at least one of them can not be seen as quotient.

Proposition 2 is an example of $\mathrm dy\over\mathrm dx$ can only be seen as a quotient.

Proposition 2:

Suppose function $f$ satisfied with $x=f(t)$ is derivable at $t_0$, $g$ satisfied with $y=g(u)$ is also derivable at $u_0$, then$${\mathrm dx\over\mathrm dt}\cdot{\mathrm dy\over\mathrm du}={\mathrm dy\over \mathrm dx}\cdot{\mathrm dx\over\mathrm du}$$
(End of proposition)

In this proposition, both of them can only be understood as quotients, otherwise, it is a fake statement or meaningless sentence. There is two situation if one understand it the other way round:

If $y$ is not with respects to $t$, both ${\mathrm dx\over\mathrm du}$ and ${\mathrm dy\over\mathrm dt}$ are not defined, the sentence is meaningful.

If $y$ is with respects to $t$, both ${\mathrm dx\over\mathrm du}={\mathrm dy\over\mathrm dt}=0$, the sentence is meaningful, but it is a fake statement.

It is concludeble that, both of ${\mathrm dx\over\mathrm du}$ and ${\mathrm dy\over\mathrm dt}$ in this proposition must be treated as quotient.

Definition of derivative:

Suppose $f$ of is a function defined on a subset of real number set, $D$; also, $x_0\in D$
$$\forall A(\forall\varepsilon\in\mathbf R_+\exists\delta\in\mathbf R_+\forall x\in\mathring U(x_0,\delta)(|{f(x)-f(x_0)\over x-x_0}-A|<\varepsilon)$$
$$\leftrightarrow  A\mathrm{\ equals\ the\ derivative\ of}\ f\mathrm{\ at}\ x_0)$$
$A$ is denoted as $\displaystyle{\mathrm df\over\mathrm dx}$

Definition of differential:

Suppose the derivative of function $f$ at the point $x_0$ is $A$,
$$\forall g(\forall h\in\mathbf R(g(x_0,h)=Ah)\leftrightarrow g\mathrm{\ equal\ the\ differential\ of\ }f\ \mathrm{at}\ x_0)$$
$Ah$ is denoted as $\mathrm df$, called the dependent variable of differential


问:
Quadrilateral $ABCD$ with chord of inscribed circle meeting $AC$ at $P$. Find ratio $AP$ : $PC$
In a quadrilateral $ABCD$, there is an inscribed circle centered at $O$. Let $F,N,E,M$ be the points on the circle that touch the quadrilateral, such that $F$ is on $AB$, $N$ is on $BC$, and so on. It is known that $AF=5$ and $EC=3$. Let $P$ be the intersection of $AC$ and $MN$. Find the ratio $AP:PC$.

I know that $AM=AF=5$ and $CN=CE=3.$ The answer is equivalent to the ratio of the areas $[ADP]:[DPC]$. I cannot continue on from this point. Would anyone please help?

答:
Construct line from C parallel to the chord MN, meeting the extension of AD at X. Then, XMNC is isosceles, implying XM = CN = CE. Note that the triangles AMP and AXC are similar, leading to

$$\frac {AP}{PC} =\frac {AM}{MX} = \frac {AF}{CE} = \frac 53
$$


问:
How to tell compactness in $\mathbb{R}^{n}$
Compactness is defined fundamentally using the Heine-Borel theorem. In terms of $\mathbb{R}$, we have a more straightforward theorem that tells compactness: a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded. How to extend this theorem to $\mathbb{R}^{n}$? The difficulty is, how to define "boundedness" in $\mathbb{R}^{n}$. I am not aware of a "less than" symbol for $\mathbb{R}^{n}$ that helps with the definition. Also, even with a proper definition, I am not sure if the theorem in $\mathbb{R}$ can be well extended to $\mathbb{R}^{n}$.

答:
Compactness is not defined using the Heine-Borel theorem; the usual definition using open covers makes sense in any topological space.

A set $A \subset \mathbb{R}^n$ is bounded if and only if there is an $R > 0$ so that $A \subset B(0, R) = \{ x \in \mathbb{R}^n : |x| < R\}$, where $|x|$ is the norm of $x$.


问:
Power set of $\{\varnothing, \{ \varnothing\}\}$
The power set of $\{\varnothing, \{ \varnothing\}\}$ is defined as follows:
$$\{\varnothing, \{ \varnothing\}, \{\{\varnothing\}\}, \{\varnothing, \{\varnothing\}\}\}$$

Why do we have the 3rd element in the answer $\{\{\varnothing\}\}$?

答:
If we have a set with two elements $X=\{a,b\}$, then the powerset of $X$ is: $\{∅, \{a\},\{b\},\{a,b\}\}$, just sub in your elements as $a$ and $b$.

I think the fact that the elements in the set are the empty set or the set containing the empty set is what is confusing you.

For more general practice consider $ Y = \emptyset $, then $\mathcal{P}(Y) = \{ \emptyset\}$

Further, let $Z = \{  \emptyset \}$, then  $\mathcal{P}(Z) = \{ \emptyset, \{ \emptyset \}\}$


答:
I don't know if this qualifies as an answer or not, but the problems come in the second derivative.  While $\frac{dy}{dx}$ can be used as a fraction, the standard notation for the second derivative, $\frac{d^2y}{dx^2}$, cannot, at least in that form.  An alternative notation for the second derivative, which can be used as a fraction, is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$, which can be derived simply from applying the quotient rule to the first derivative (which shows another place where $\frac{dy}{dx}$ can be treated as a quotient!).

Someone also mentioned partial derivatives.  You can treat those as a fraction, too, provided you use sufficient notation.  The reason why you can't split $\frac{\partial y}{\partial x}$ is because there is information about the numerator contained in the denominator.  In other words, the $\partial y$ in $\frac{\partial y}{\partial x}$ is a different term than the $\partial y$ in $\frac{\partial y}{\partial t}$.  If you used sufficient notation to distinguish these as different values (say, $\partial_ty$ and $\partial_xy$), then you can, in fact, split these fractions successfully.  If you do this, the problem specified by B. Goddard goes away.

Title: When not to treat dy/dx as a fraction in single-variable calculus?

Body:
While I do know that $\frac{dy}{dx}$ isn't a fraction and shouldn't be treated as such, in many situations, doing things like multiplying both sides by $dx$ and integrating, cancelling terms, doing things like $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ works out just fine.

So I wanted to know: Are there any particular cases (in single-variable calculus) we have to look out for, where treating $\frac{dy}{dx}$ as a fraction gives incorrect answers, in particular, at an introductory level?

Note: Please provide specific instances and examples where treating $\frac{dy}{dx}$ as a fraction fails


问:
How do I evaluate the following expression?
How to evaluate the following expression:  </p>

<p>$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}$</p>

答:
Observe that:

$(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1.$

Therefore,

$\dfrac1{\sqrt{2} + 1} = \sqrt{2} - 1.$

Similarly,

$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 2 - 1 = 1.$

And therefore,

$\dfrac1{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}.$

We keep doing this until we reach,

$\dfrac1{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8}.$

Now we add all the above equations to get:

$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}.$

$=\displaystyle (\sqrt{2} - 1) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4} - \sqrt{3}) +\cdots + (\sqrt{9}-\sqrt{8}).$

Notice that $+\sqrt{2}$ cancels with $-\sqrt{2}$. And this happens with all the numbers in the expression, except with $1$ and $\sqrt{9}$.

$= \sqrt{9} - 1 = 3 - 1 = 2.$


问:
Proving that for $x \ge 17$, $\prod\limits_{p < x\text{, prime, & }p \nmid (x^2+x)}p > x^2 + x$
Let:

$x$ be an integer

$p_n$ be the $n$th prime

$x\#$ be the primorial of $x$

$f(x) = x^2 + x$

$P(x) = \prod\limits_{p < x\text{, prime, & } p \nmid f(x)}p$

$m(p_n)$ be the minimum $P(i)$ where $p_n \le i < p_{n+1}$

Does it follow that $P(x) > f(x)$?

The challenge in the argument is to compare a strictly increasing function with a function that is not strictly increasing:

$P(x)$ is not strictly increasing.  In some cases $P(x) > P(x+1)$.

Example:  $P(4) = 3$ but $P(5) = 1$

$f(x)$ is strictly increasing since $x \ge 17$ and:

$$\frac{d}{dx}(x^2 + x) = 2x + 1$$

I handle this situation by leveraging Bertrand's Postulate and attempting to show:

$$m(p_n) > (2p_n - 1)^2 + 2p_n - 1 = (2p_n)^2 - 2p_n$$

Since $m(p_n) > \dfrac{p_n\#}{(p_n)^2}$ and $4(p_n)^2 > (2p_n - 1)^2 - 2p_n$, this will be demonstrated if:

$$p_n\# > 4(p_n)^4$$

(1)  Base Case: $p_n = 17$

$17\# = 510,510 > 4(17^4) = 334,084$

(2) Assume up to some $p_n \ge 17$, $p_n\# > 4(p_n)^4$

(3) Since $p_{n+1} \ge 19$, it follows that:

$$p_{n+1}\# \ge (19)(4(p_n)^4) > 4(2p_n)^4) > 4(p_{n+1})^4$$

I believe that the conclusion follows.  Please let me know if I made any mistakes in my reasoning or if there is a simpler way to make the same point.

答:
Your conjecture of $P(x) \gt f(x)$ for $x \ge 17$ is true, and your proof procedure is mostly correct, but there are several issues with it. Regarding your $P(x)$, I find it generally easier to deal with primes which are factors of a value rather than primes which are not factors. The product of all primes less than $x$ which don't divide $f(x)$ is the same as the product of all primes less than $x$ divided by the product of primes less than $x$ which do divide $f(x)$. Thus, for simpler algebra, by defining

$$g(x) = \prod_{p \lt x \text{, prime, & } p \mid f(x)}p \tag{1}\label{eq1A}$$

for $p_n + 1 \le x \le p_{n+1}$, we get

$$P(x) = \prod_{p \lt x \text{, prime, & } p \, \nmid f(x)}p = \frac{p_n\#}{g(x)} \tag{2}\label{eq2A}$$

Since $\gcd(x, x + 1) = 1$, the maximum value of $g(x)$ is $x(x + 1)$ when $x$ and $x + 1$ are both square-free integers. This gives, from \eqref{eq2A}, that

$$P(x) \ge \frac{p_n\#}{x(x + 1)} \tag{3}\label{eq3A}$$

With your $m(p_n)$ definition, it straddles $2$ cases in \eqref{eq2A}, with it using $n-1$ for $x = p_{n}$, and $n$ for the rest of the values of $x$. First, with $x = p_n$, since only factors less than $p_n$ are being considered, then $g(p_{n})$ doesn't include $p_{n}$, so we have

$$P(p_n) = \frac{p_{n-1}\#}{g(p_n)} \ge \frac{p_{n-1}\#}{p_n + 1} = \frac{p_{n}\#}{p_n(p_n  + 1)} \tag{4}\label{eq4A}$$

Thus, if $p_n + 1$ is square-free, then

$$m(p_n) \le P(p_n) = \frac{p_{n}\#}{p_n(p_n  + 1)} \lt \frac{p_{n}\#}{(p_n)^2} \tag{5}\label{eq5A}$$

This already provides a counter-example to your statement $m(p_n) \gt \dfrac{p_n\#}{(p_n)^2}$, plus other larger values where $x$ and $x + 1$ are both square-free will also be additional counter-examples. However, what is true is that

$$m(p_n) \gt \frac{p_n\#}{(p_{n+1})^2} \tag{6}\label{eq6A}$$

Perhaps this is what you had intended. The mechanics of the rest of your proof is correct, but it would need to be adjusted to use something like \eqref{eq6A} instead, which I'll leave to you to do.

I would approach solving this a bit differently. With $p_{n} + 1 \le x \lt p_{n+1}$, we have from \eqref{eq3A} that

$$P(x) \ge \frac{p_n\#}{x(x + 1)} \gt \frac{p_n\#}{(p_{n+1})^2} \tag{7}\label{eq7A}$$

For $x = p_{n+1}$, since $p_{n+1}$ is not included in $g(x)$ defined in \eqref{eq1A}, we get

$$P(p_{n+1}) \ge \frac{p_n\#}{p_{n+1} + 1} \gt \frac{p_n\#}{(p_{n+1})^2} \tag{8}\label{eq8A}$$

Thus, in both cases, we get

$$P(x) \gt \frac{p_n\#}{(p_{n+1})^2} \tag{9}\label{eq9A}$$

Since

$$p_{n+1}(p_{n+1} + 1) \ge f(x) \tag{10}\label{eq10A}$$

your conjecture will be true if we can prove

$$\frac{p_n\#}{(p_{n+1})^2} \gt p_{n+1}(p_{n+1} + 1) \iff p_n\# \gt (p_{n+1})^3(p_{n+1} + 1) = p_{n+1}^4 + p_{n+1}^3 \tag{11}\label{eq11A}$$

For $x = 17$, we have $n = 6$ and $p_n = 13$, so the left side of the second part of \eqref{eq11A} is $13\# = 30,030$ but the right side is $17^3(18) = 88,434$, so we need to check this value in particular. We have $P(17) = \frac{13\#}{2(3)} = 5,005$ and $f(17) = 17(18) = 306$, so your conjecture is true in this case.

With $18 \le x \le 23$, then $n = 7$ and $p_{n} = 17$, so the left side of \eqref{eq11A} is $17\# = 510,510$ and the right side is $19^3(20) = 137,180$, so the inequality holds.

As you basically did, using induction, this is the base step. Assume \eqref{eq11A} is true for all $n \le k$ for some $k \ge 7$. Then to check the $n = k + 1$ case, multiply the induction step inequality by $p_{k+1}$, use $p_{k+1} \ge 23$ and Betrand's postulate, to get

$$\begin{equation}\begin{aligned}
p_{k+1}\# & \gt p_{k+1}(p_{k+1}^4 + p_{k+1}^3) \\
& \gt 16(p_{k+1}^4) + 16(p_{k+1}^3) \\
& \gt (2p_{k+1})^4 + (2p_{k+1})^3 \\
& \gt p_{k+2}^4 + p_{k+2}^3
\end{aligned}\end{equation}\tag{12}\label{eq12A}$$

This shows \eqref{eq11A} is true also for $n = k + 1$, which proves by induction that it's always true for $n \ge 7$, and thus $x \ge 18$. Altogether, this proves your conjecture, i.e., $P(x) \gt f(x)$, is true for all $x \ge 17$.


问:
convolution and test function

Let $\varphi \in D(\mathbb{R}) $ and $f(x)=C \varphi(x)$ , C constant with $\int_{-\infty}^{+\infty}f(x)dx=1$.
Let $0 < \alpha < \frac{1}{2}$ and $\psi_{\alpha}(x)=\frac{1}{\alpha}f(\frac{x}{\alpha})$

I proved with a change of variables that $\int_{-\infty}^{+\infty}\psi(x)dx=1$.

Now let $\ell>0$ and
$$E(x)=\begin{cases}
1  \quad \text{if} \quad 0\le x\le \ell, \\
0 \quad \text{else.}\end{cases}
$$

My question is how prove that the convolution $h=\psi*E \in D(\mathbb{R}) $ and $h \equiv 1$ in $[\alpha; \ell-\alpha]$

答:
You are just willing to say that if $\phi \in L^1$ is supported on $[-r,r]$ then for $\ell > 2r$ $$\phi \ast 1_{[0,\ell]}$$ is constant $=\int_{-r}^r\phi(x)dx$ on $[r,\ell-r]$,

and $0$ on $(-\infty,-r]\cup [\ell+r,\infty)$.

If $\phi$ was $C^\infty$ then $\phi \ast 1_{[0,\ell]}$ is $C^\infty$ because its derivative is $\phi(x)-\phi(x+\ell)$.

Title: convolution and test function


答:
Power set is the set of all subsets. Each element is a subset of the initial set.

The initial set is $\{\color{red}\emptyset, \color{blue}{\{\emptyset\}}\}$

That is a set with two elements $\color{red}\emptyset$ and $\color{blue}{\{\emptyset\}}$

So what are the subsets of a set with two elements.  There are $2^2 =4$ such sets because each element can either be in as set or not.  So the four subsets are:

The set with neither $\color{red}\emptyset$ nor $\color{blue}{\{\emptyset\}}$.

That is the set: $\emptyset$

The set that has  $\color{red}\emptyset$ but not $\color{blue}{\{\emptyset\}}$.

That is the set: $\{\color{red}\emptyset\}$.

The set that does not have $\color{red}\emptyset$ but has $\color{blue}{\{\emptyset\}}$

That is the set: $\{\color{blue}{\{\emptyset\}}\}$

The set that has both $\color{red}\emptyset$ and $\color{blue}{\{\emptyset\}}$.

That is the set $\{\color{red}\emptyset,\color{blue}{\{\emptyset\}}\}$

Title: Power set of $\{\varnothing, \{ \varnothing\}\}$

Body:
The power set of $\{\varnothing, \{ \varnothing\}\}$ is defined as follows:
$$\{\varnothing, \{ \varnothing\}, \{\{\varnothing\}\}, \{\varnothing, \{\varnothing\}\}\}$$

Why do we have the 3rd element in the answer $\{\{\varnothing\}\}$?


答:
Here's a ridiculously high-powered solution, because why not :)

By Brianchon's theorem on the "hexagon" $ABCEDM$, we get that $AE, BD, CM$ concur. So by Ceva's theorem,
$$\frac{AP}{PC}\cdot\frac{CE}{ED}\cdot\frac{DM}{MA}=1\implies\frac{AP}{PC}=\frac{MA}{CE}=\frac{5}{3}.$$

Title: Quadrilateral $ABCD$ with chord of inscribed circle meeting $AC$ at $P$. Find ratio $AP$ : $PC$

Body:
In a quadrilateral $ABCD$, there is an inscribed circle centered at $O$. Let $F,N,E,M$ be the points on the circle that touch the quadrilateral, such that $F$ is on $AB$, $N$ is on $BC$, and so on. It is known that $AF=5$ and $EC=3$. Let $P$ be the intersection of $AC$ and $MN$. Find the ratio $AP:PC$.

I know that $AM=AF=5$ and $CN=CE=3.$ The answer is equivalent to the ratio of the areas $[ADP]:[DPC]$. I cannot continue on from this point. Would anyone please help?


问:
Tranversal paths between two points
I have been studying differential topology from Hirsch, and sometimes in proofs he takes two points $x,y\in M$ a path between them and then just says that we can assume that this is transversal to a certain submanifold that we are interested in. Now I have tried to prove this myself but I don't see why this is true.

One attempt well let's take a path $\gamma $ between them such that $\gamma(0)=x$ and $\gamma(1)=y$, then using the fact  that we  have that transversal maps are dense we can approximate this map by transversal maps $h_n:[0,1]\rightarrow M$, but here I don't think I have any control over $h_n(0)$ and $h_n(1)$ and so this could be not $x$ and $y$.

Second attempt, we show that $\{\gamma \in C^r([0,1],M) : \gamma(0)=x, \gamma(1)=y\}$ is open in the strong topology, however this doesn't seem to be very true to me , since when we consider neighborhoods we lose control over the initial and final point again.

Another idea would be to consider the path an embedding which is possible, and then use the fact that $C_S^r([0,1],\{0,1\},\gamma([0,1]),\{x,y\})$ is dense, but here we would need the fact that the manifold that we want the map to be transverse too to be closed.

So I am out of ideas , and would appreciate some help with this. Thanks in advance.

答:
I like your first attempt and I bet it's possible to make it or some variant of it work, but I think Hirsch wants you to get your hands dirty and work in coordinates since the assertion follows by standard arguments there. He probably doesn't give details because they're pretty tedious. The real content can be found in the following special case.

Locally, a codimension $1$ submanifold of an $n$-manifold looks like a hyperplane in $\mathbf{R}^n$. More precisely, for each $p\in N$, there is a submanifold chart $(x,U)$ about $p$ for which $x(U\cap N)\subset \mathbf{R}^{n-1}\times\{0\}$. So, in the special case that $M=\mathbf{R}^n$ and $N\subset \mathbf{R}^{n-1}\times\{0\}$ where $x=(x^1,\ldots,x^n)$ and $y=(y^1,\ldots,y^n)$ with $x^n>0$ and $y^n<0$, this can be accomplished by joining the smooth linear path $\gamma_1\colon (x^1,\ldots,x^n)\to (x^1,\ldots,x^{n-1},y^{n})$ (which is certainly transverse to $N$) to the smooth linear path $\gamma_2\colon (x^1,\ldots,x^{n-1},y^{n})\to (y^1,\ldots,y^n)$ at their common juncture. You'll need to smooth the corner and if you've never done this sort of argument before it's definitely an instructive exercise.

This special case is the real content and the rest is just fiddly details. Here's one way the general case can proceed.

Pick any continuous path $\gamma$ from $x$ to $y$ and give $\mathop{\mathrm{Im}}\gamma$ an open cover by charts such that every chart in the cover that intersects $N$ is a submanifold chart. You can do this by noting that every point not in the closure of $N$ has a coordinate neighborhood that does not intersect $N$. Compactness of $\mathop{\mathrm{Im}}\gamma$ allows you to assume this covering is finite and the argument now proceeds by induction, the base case being essentially what was done in the special case above.

Title: Tranversal paths between two points


问:
Mean Number of Samples from Uniform Distribution to Achieve a Desired Total

Problem Statement

I'm interested in the mean number of samples from a given uniform distribution of integers I would need to take to reach a desired total or more.

For example, a six-sided dice is a uniform distribution of integers from $1$ through to $6$. The number of times you have to roll a six-sided dice ($r$) to achieve a desired total or greater ($t$) produces a distribution. For example, here is a graph produced by sampling dice rolls, aiming for a total of $80$ per run:

Graph of six-sided dice rolls with a target of $80$

The mean of this distribution is approximately $23.332448$. This is the number I am interested in deriving. Simply 23 would be too imprecise for what I need. Right now I am determining this by simply sampling large numbers of dice rolls and then calculating the average of that data set, however, I would like to instead understand the statistics involved and compute the mean, or a reasonable approximation of the mean.

One might try to derive the mean of the number of rolls needed $P(r=t)=0.5$ by simply using the mean of the uniform distribution, and dividing the total $t$ by that mean:

$$\frac{t}{(max-min)/2+min} = \frac{80}{(6-1)/2+1} \approx 22.857$$

I will call this the 'naive' approximation. As you can see, this result can be significantly off the desired result. Sometimes, it happens to be close to the right answer, however, I am interested in many cases where it is unacceptably imprecise.

Ultimately, I want to have a method or formula that takes an arbitrary continuous uniform distribution of integers (such as a die) and the desired total, which tells me the mean number of samples from that distribution I would need to take in order to reach that total.

Prior Research

I've looked at a few posts that talk about this subject, however, none of them were able to provide the complete answer to me. Here are two of the things I've found and read:

A Similar Question on This Forum

Number of dice rolls taken to reach a certain sum

This was useful in creating an approximation of the actual distribution I'm trying to find. I used this to turn the following recursive function into the following code (in rust, dependency on the cached crate for memoizing the recursive results to improve performance to something usable):

$$P(S_t=s) = \sum_{i=1}^6P(S_{t-1} = s-i)/6$$

With the following terminators:

$P(S_0=0) = 1$

$P(S_0<0) = P(S_{t>0}=0) = 0$

```<code>#[cached]
pub fn p(t: i32, s: i32) -> f64 {
    if s <= 0 && t <= 0 {
        return 1.0;
    }

    if s <= 0 || t <= 0 {
        return 0.0;
    }

    (1..(std::cmp::min(6, s) + 1)).map(|i| p(t - 1, s - i)).sum::<f64>() / 6.0
}
</code>```

Which, when run, can be used to create a graph comparing the frequency of the sampled results to the frequency predicted by that algorithm. In the graph, the blue line is the result of the algorithm, while the orange line is sampled data from a simple PRNG using a uniform distribution (the same data as in the first graph).

Note that to get this graph, you have to divide the results of this algorithm by the inverse of the mean of the uniform distribution (so $3.5^{-1}$ in this example) The author of the original answer suggests dividing by $P(S_t=36)$ (where $36$ was their target) however fails to realise that this is just an approximation of the inverse of the mean. I'm not actually sure why the mean of the uniform pops out here, so that's an interesting tidbit.

This looks pretty good, however, it does not provide the mean number of rolls as best I can tell. The author of the answer to the linked question states that the mean in their example is "about $10.5238577$", however, they don't state how they derived that number and I haven't been able to figure it out.

It's also not generalised to any uniform, however, I don't believe it would be difficult to generalise. I don't anticipate this to be an obstacle if someone finds out how to derive the mean from that answer.

A Similar Question on Reddit

On average, how many rolls of a fair die would it take to get a cumulative score above n?

This was useful in creating a different approach to solving this problem, where I iterated $r$ (the number of rolls) upwards from 1, each time producing an approximation of the distribution of the combined rolls using a normal distribution. The first normal distribution to have a cumulative distribution function for the given target of less than $0.5$ indicated when the integer mean number of rolls was being made. Again, however, like the previous solution, this only produced the integer number of rolls. I need to know the mean number of rolls more precisely than this.

There are a number of other answers than the top up-voted answer on this Reddit post with differing amounts of clear information. The answer from the user u/possiblywrong shows an interesting cumulative distribution function which I wonder if it would solve my problem. However, I am unable to figure out how to reproduce their ideas as code or fully understand them. For completeness sake, here is an image of that response in case it is deleted in the future.

Thanks

Thank you for any time you spend on this, this is doing my head in. It feels like it should be so easy to model, and yet I've lost an entire day researching it only to find vague answers that are either too complicated for me, or are simply missing information that I need to understand them. Hopefully, you can help.

答:
Let $f_{s}(t)$ be the expected number of rolls to get a sum greater than or equal to $t$ with an $s$-sided die. Then you have the recurrence $$f_s(t) = 1+\frac{1}{s}\sum_{k=1}^s f_s(t-k)$$ with the base cases of $f_s(t) = 0$ with $t \le 0$.

The solution to this (ignoring the base cases)
$$f_s(t) = \frac{2}{1+s}t+\sum_{n=1}^{s}c_nr_n^t$$

where $r_n$ goes over the distinct roots of $sr^{s+1}-(s+1)r^s+1$. Alternately, since $r=1$ is a double root, this becomes $$f_s(t) = \frac{2}{1+s}t+c_1+\sum_{n=2}^{s}c_nr_n^t$$

where $r_n$ instead goes over the roots of $$\sum_{m=0}^{s-1}(m+1)r^m = 1+2r+3r^2+...+sr^{s-1} \tag 1$$

To find the exact values of $c_1, ..., c_n$, you would need to solve the equations $f_s(t) = 0$ for $t = 1-s, 2-s, ..., 0$.

Alternately, you could just build up from $t = 0$ (using a computer) to find that $$f_6(80) = \frac{694905206512232068400374286878415315634626408538515111020013047}{29781651707669509088572079548239633038047628833600290523447296} \approx 23.333$$

If you don't need the exact value, since the magnitude of each root of $(1)$ is less than $1$, an approximation would be $f_s(t) \approx \frac{2}{1+s}t+c_1 = \frac{2}{1+s}t+O(1)$


问:
Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $
As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to  $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?

答: An alternative way of proving, is going into parametric equations. The cubic equation is a Follium of Descartes, and can be represented by $(\frac{\sqrt{2}t}{1+t^3},\frac{\sqrt{2}t^2}{1+t^3})$. With respect to the 1st quadrant we must now show that all points on the unit circle fall "outside" (that is on the "right" side) of the Follium. Note that the point$P(\sqrt{2}/2,\sqrt{2}/2)$ coincides with both curves, hence the need for the equality symbol. When we use the distance formula on the parametric representation (in squared form), we get $d^2=\frac{2t^2+2t^4}{(1+t^3)^2}$. We can optimize $d^2$ and show that for $t=1$, a local maximum is achieved for the distance. You need to apply the quotient rule, which is annoying algebra. The numerator will be $(1+t^3)(-4t^6-8t^4+8t^3+4t)$ and when you set this equal to zero, $t=-1$ doesn't yield anything productive (what's going on with $t=-1$?). But the sixth degree polynomial inside has a root $t=0$ (Origin) and $t=1$ (Rational Zero Theorem!) and thus is divisible by $(t-1)$, which (after factoring out $-4t$) gives $t^4+t^3+3t^2+t+1$ which has no real roots. It shouldn't be to hard to show that for $t=1$ a maximum distance (Point P) in the first quadrant is achieved. Here you have all the ingredients to set up the proof which is what I want you to do.


答:
After you substitute $(x,y)=(\cos{\theta},\sin{\theta})$, you just need to prove the followings:

$$
\begin{aligned}
\sin{\theta}+\cos{\theta}&\geq\sqrt{2}\\
\\
1-\sin{\theta}\cos{\theta}&\geq\sin{\theta}\cos{\theta}
\end{aligned}
$$

Which are easy. First one is well known, second one is just double angle sine identity

Title: Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $

Body:
As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to  $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?


问:
Decomposition of the matrix exponential (of the covariance matrix)
The sample covariance matrix $\Sigma$ (chosen because it is symmetric and, therefore, diagonalizable) can be decomposed as

$$\Sigma = \text{diag}(\sigma) C \text{diag}(\sigma)$$
where $\text{diag}(\sigma)$ is a matrix with volatilities along the diagonal, and $C$ is the correlation matrix.

Question

If we compute the matrix exponential of the covariance matrix, $$e^\Sigma = \sum_{k=0}^\infty \frac{1}{k!} \Sigma^k$$
does it too have a unique decomposition?

Attempt

Eigenvectors might indicate a possible answer regarding some link, since

$$\Sigma = PDP^{-1}$$
$$e^\Sigma = Pe^DP^{-1}$$

where $D = \text{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix whose entries are the eigenvalues of $\Sigma$, $e^D = \text{diag}(e^{\lambda_1}, ..., e^{\lambda_n})$, and $P$ contains eigenvectors corresponding to the eigenvalues.

This is unlike my case though because I am not interested in eigenvalues, only the actual matrix, or the matrix exponential of any symmetric matrix. Nevertheless, this is the only decomposition I know of so far, so someone could confirm that it is the only one out there

答:
Suppose $\Sigma = PDP^{-1}$. Look at your formula for $e^{\Sigma}$. Note that
$$\Sigma^k = \left(PDP^{-1}\right)^k=PDP^{-1}PDP^{-1}\dots PDP^{-1}=PD^kP^{-1}.$$
So we have $$e^{\Sigma}=\sum_{k=0}^\infty\frac{1}{k!}\Sigma^k=\sum_{k=0}^\infty\frac{1}{k!}PD^kP^{-1}=P\left(\sum_{k=0}^\infty\frac{1}{k!}D^k\right)P^{-1}=Pe^{D}P^{-1}.$$
And because $D$ is diagonal, $D^k$ and hence $e^D$ are really easy to compute: just take $D$, and raise the diagonal entries to the power of $k$ (for $D^k$) or exponentiate them (for $e^D$).


答:
HINT

Since $x > 0$ and $y > 0$, the proposed inequality is equivalent to

\begin{align*}
x^{3} + y^{3} \geq \sqrt{2}xy & \Longleftrightarrow (x^{3} + y^{3})^{2} \geq 2x^{2}y^{2}\\\\
& \Longleftrightarrow (x+y)^{2}(1 - xy)^{2} \geq 2x^{2}y^{2}\\\\
& \Longleftrightarrow (1 + 2xy)(1-xy)^{2} \geq 2x^{2}y^{2}\\\\
\end{align*}

If we make the change of variable $t = xy$, we obtain the following equivalent inequality:
\begin{align*}
(1 + 2t)(1 - t)^{2} \geq 2t^{2} & \Longleftrightarrow (1+2t)(1 - 2t + t^{2})\geq 2t^{2}\\\\
& \Longleftrightarrow 1 - 2t + t^{2} + 2t - 4t^{2} + 2t^{3} \geq 2t^{2}\\\\
& \Longleftrightarrow 2t^{3} - 5t^{2} + 1 \geq 0\\\\\
& \Longleftrightarrow (2t^{3} - t^{2}) - (4t^{2} - 1) \geq 0\\\\
& \Longleftrightarrow t^{2}(2t - 1) - (2t-1)(2t+1) \geq 0\\\\
& \Longleftrightarrow (2t-1)(t^{2} - 2t - 1) \geq 0
\end{align*}

Can you take it from here?


问:
Chinese remainder theorem issue
Let's say I have the following equations:

$$x \equiv 2 \mod 3$$
$$x \equiv 7 \mod 10$$
$$x \equiv 10 \mod 11$$
$$x \equiv 1 \mod 7$$

And I need to find the smallest x for which all these equations are correct. So:

$N = 3\times10\times11\times7 = 2310$

$N_1 = \frac{2310}{3} = 770$

$N_2 = \frac{2310}{10} = 231$

$N_3 = \frac{2310}{11} = 210$

$N_4 = \frac{2310}{7} = 330$

Then:

$\gcd (3, 770) = 1 = 257\times3 - 1\times770; \qquad x_{1} = -1 \equiv 2 \mod 3$
$\gcd(10, 231) = 1 = -23\times10 + 1\times231; \quad x_{1} = 1$

Etc., and finally:

$x = 2\times2\times N_1 + 7\times1\times N_2 + ...$

I understand the whole method except for one thing - when I solved $\gcd(3, 770)$ I got $x_{1} = -1$...why is there $-1 \equiv 2 \mod 3$, from which the final result is 2?

答:
What you have calculated is the LCM (least common multiple) of the moduli. However, this doesn't solve the problem, because they are all congruent to different values modulo m. For example, if the problem were as follows:

$$x \equiv b \mod 3$$
$$x \equiv b \mod 10$$
$$x \equiv b \mod 11$$
$$x \equiv b \mod 7$$

Your solution would be true if $b=0$. Similarly, if $b=1$, the solution would be $2371$.

Title: Chinese remainder theorem issue


问:
Proof $\overline X_n \rightarrow \mu$ a.s. given $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in distribution
Given $X_1...X_n...$ i.i.d. random variables with $E|X_1|<\infty$ and $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in law for a certain random variable Z and a certain real number $\mu$ (not necessarily equal to $EX_1$), I need to proof that $\overline X_n \rightarrow \mu$ a.s.

It is very similar to the central limit theorem but the problem is I don’t know if the variance is finite, so I cannot use it. Any hint on how can I approach this problem?

答:
I'm working under the impression that $\overline{X}_{n} = n^{-1} \sum_{j = 1}^{n} X_{j}$.

Claim: Assume $\{X_{n}\}_{n \in \mathbb{N}}$ are i.i.d. random variables, $\mathbb{E}(|X_{1}|) < \infty$, and $Z$ is a random variable with $\mathbb{P}\{|Z| < \infty\} = 1$.  Let $\mu \in \mathbb{R}$.  If $\sqrt{n}(\overline{X}_{n} - \mu) \to Z$ in law, then $\mu = \mathbb{E}(X_{1})$.

As a consequence of the claim, the strong law of large numbers implies $\overline{X}_{n} \to \mu$.

Proof of the claim: By the strong law of large numbers and Egoroff's Theorem, there is an $\omega : [0,\infty) \to [0,\infty)$ and an event $\mathcal{C}$ such that $\mathbb{P}(\mathcal{C}) \geq 1/2$, $\lim_{\delta \to 0^{+}} \omega(\delta) = 0$, and
\begin{equation*}
|\overline{X}_{n} - \mathbb{E}(X_{1})| \leq \omega(n^{-1}) \quad \text{on} \, \, \mathcal{C}.
\end{equation*}

At the same time, since $\mathbb{P}\{Z < \infty\} = 1$, Prokhorov's Theorem implies that there is an $M > 0$ such that
\begin{equation*}
\mathbb{P}\{|\sqrt{n}(\overline{X}_{n} - \mu)| \geq M \} \leq 1/4.
\end{equation*}

It follows that, for each $n \in \mathbb{N}$, $\mathbb{P}(\{|\sqrt{n}(\overline{X}_{n} - \mu)| < M\} \cap \mathcal{C}) > 0$.  On that event, we have
\begin{equation*}
\sqrt{n}(\mathbb{E}(X_{1}) - \mu) \leq \sqrt{n} \omega(n^{-1}) + \sqrt{n}(\overline{X}_{n} - \mu) \leq \sqrt{n} \omega(n^{-1}) + M.
\end{equation*}

Since the extreme sides of the inequality are constants, we deduce that, for each $n \in \mathbb{N}$,
\begin{equation*}
\mathbb{E}(X_{1}) - \mu \leq \omega(n^{-1}) + M n^{-\frac{1}{2}}.
\end{equation*}
Sending $n \to \infty$, we find $\mathbb{E}(X_{1}) \leq \mu$.

A similar argument shows that $\mathbb{E}(X_{1}) \geq \mu$.


问:
How to prove the inequality $f(z) = \varphi_1(-z) - \varphi_2(-z) = \frac{e^{-z} - 1}{-z} - \frac{e^{-z} + z - 1}{z^2} > 0, \forall z > 0$?
The exponential $\varphi(z)$ functions usually appear in the Butcher tables of exponential time difference Runge-Kutta methods for differential equations.

They are given by
\begin{align}
\varphi_j(z) = \frac{1}{z^j}\int_0^z\mathrm{e}^{(z - s)} \frac{s^{j-1}}{(j-1)!} \mathrm{d} s.
\end{align}

Let $\varphi_0(z) = \mathrm{e}^z$, it can be easily calculated that
\begin{align}
&\varphi_{k+1}(z) = \frac{\varphi_k(z) - \frac{1}{k!}}{z},  \quad \varphi_k(0) = \frac{1}{k!}, \\
&\varphi_1(z) = \frac{\mathrm{e}^z - 1}{z}, \\
& \varphi_2(z) = \frac{\mathrm{e}^z - z - 1}{z^2}.
\end{align}

Now consider the function $f(z) = \varphi_1(-z) - \varphi_2(-z), z > 0$,

It can be easily verified that
\begin{align}
\mathrm{lim}_{z\rightarrow 0} \varphi_1(z) = 1, \\
\mathrm{lim}_{z \rightarrow 0} \varphi_2(z) = \frac{1}{2}, \\
\mathrm{lim}_{z\rightarrow -\infty} \varphi_1(z) = 0, \\
\mathrm{lim}_{z \rightarrow -\infty} \varphi_2(z) = 0,
\end{align}
then we have
\begin{align}
\mathrm{lim}_{z \rightarrow 0} f(z) = \frac{1}{2},  \\
\mathrm{lim}_{z \rightarrow +\infty} f(z) = 0.
\end{align}

The derivative of $f(z)$ is
\begin{equation}
\begin{aligned}
f'(z) &= \frac{ z^2 e^{-z} + 2z e^{-z} + 2e^{-z} -2}{z^3}  \\
 & = \frac{2e^{-z}(z^2 / 2 + z + 1 - e^z)}{z^3} \\
& < 0, \quad \forall z> 0
\end{aligned}
\end{equation}

Q: How to prove that $f(z) > 0, \forall z > 0$?

Thank you very much!

答:
You basically have it down?
If $f'(z)<0$ for all $z>0$, then $f$ is strictly decreasing on $[0,+\infty)$. Let's proceed arguing by contradiction.

Suppose there were some $z_0>0$ with $f(z_0) \leqslant 0$.
We would then have $f(z) < f(z_0)$ for all $z > z_0$.

You could then take any $z_1 > z_0$, so that $f(z_1) < 0$ (strictly).
Finally, this would mean $f(z) < f(z_1) < 0$ for all $z>z_1$.

This would contradict $\lim_{z\to+\infty}f(z) = 0$ and imply that our original assumption (that there is some $z_0>0$ with $f(z_0)\leqslant 0$) is wrong, proving hence that $f(z)>0$ for all $z>0$.


问:
Multiset identity.
How does one get to the following combinatorial identity: $$\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{i}=\binom{n+k-1}{k}$$

I'm well aware of the definition of a multiset, as well as of the derivation of cardinalities of such. The closest one I came across is a special case of the Chu-Vandermonde identity, namely for any integers $i,k$ and $n$ satisfying $0\leq i\leq k \leq n$ the following holds true $$\sum_{m=0}^n\binom{m}{i}\binom{n-m}{k-i}=\binom{n+1}{k+1}$$

Any help is appreciated. Thanks in advance.

答:
$n$ boys and $k-1$ girls. How many ways to pick a team of $k$?

Ignoring the gender distinction, the answer is just $\binom{n+k-1}{k}$.

The number of boys picked is between $1$ and $k$. If I pick $i+1$ boys, then I need to pick $k-i-1$ girls. So $$\binom{n+k-1}{k}=\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{k-i-1}=\sum_{i=0}^{k-1}\binom{n}{i+1}\binom{k-1}{i}.$$

Title: Multiset identity.


Body:
$(x+y)^2
=x^2+2xy+y^2
=2xy+1
$
so
$x+y
=\sqrt{2xy+1}
$.

$\begin{array}\\
x^3+y^3
&=(x+y)(x^2-xy+y^2)\\
&=\sqrt{2xy+1}\dfrac{2x^2-2xy+2y^2}{2}\\
&=\sqrt{2xy+1}\dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\
&=\sqrt{2xy+1}\dfrac{1+(x-y)^2}{2}\\
&\ge\dfrac12\sqrt{2xy+1}\\
\end{array}
$

so we want

$\dfrac12\sqrt{2xy+1}
\ge xy\sqrt{2}
$
or
$2xy+1
\ge 8(xy)^2
$
or,
with $z = xy$,
$8z^2-2z-1
\le 0
$.

$8z^2-2z-1
=8(z+\frac14)(z-\frac12)
$.

$0
\le (x-y)^2
=x^2-2xy+y^2
$
so
$2xy
\le x^2+y^2
=1
$
or
$0 \le z \le \frac12
$.

Therefore
$8z^2-2z-1
=8(z+\frac14)(z-\frac12)
\le 0
$

which is what we want.


问:
Does the function is coercive?
How to prove the following function is coercive?

\begin{align*}
f(x,y)= x^{2n}+y^{2n}-nx^2+2nxy-ny^2
\end{align*}
I found the definition of a coercive function and tried some problems and figure out how to prove a function is coercive but in this case, there is the term $n$ is a bit confusing to me.

Definition: A continuous function $f(x)$ that is defined on $\mathbb{R}^n$ is coercive if,
\begin{align*}
\lim_{||x||\xrightarrow[]{}\infty} f(x)= +\infty
\end{align*}
that is for any $M>0$ there exists a constant $R_{M}>0$ such that $||f(x)||>M$ whenever $||x||>R_{M}$

Solution: \begin{align*}
f(x,y)= x^{2n}+y^{2n}-nx^2+2nxy-ny^2 &= x^{2n}+y^{2n}-n(x-y)^2\\
x^{2n}+y^{2n}-n(x-y)^2 &\leq x^{2n}+y^{2n}
\end{align*}

then how do we prove $x^{2n}+y^{2n} \xrightarrow[]{} \infty$ as $||x||\xrightarrow[]{} \infty$

Is my approach is correct if it is not could anyone please show me the right path? Thank you

答:
hint 1): Since $f$ is a function on $\mathbb{R}^2$, if we translate the definition of coercivity in terms of both its arguments $x$ and $y$, the limit is actually $\|(x,y)\|_{\mathbb{R}^2}\to\infty$, not just $\|x\|_{\mathbb{R}}\to\infty$.

hint 2) The inequality you have is correct. However, we actually want to show directly that the limit on $f$ goes to $\infty$, or that a lower bound on $f$ goes to $\infty$. The upper bound $x^{2n}+y^{2n}$ that is currently written in your question will not prove what we need -- this is because, even if it is coercive, it will only imply
$$\lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}f(x,y)\leq \lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}x^{2n}+y^{2n}=\infty,$$
which is not sufficient to show the limit on the left is equal to $\infty$. If, instead, we can show some lower bound $L(x,y)\leq f(x,y)$ is coercive, then we will have

$$\infty=\lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}L(x,y)\leq \lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}f(x,y)\leq\infty,$$
from which the desired equality follows.

Title: Does the function is coercive?


问:
Proving a statement about $k$-colouring of a graph

Prove that a graph is $k$-colourable iff its edges can be oriented in such a way that the resulting directed graph does not contain a path of length $k$.

It seems to me that the '$\Leftarrow$' implication might be slightly easier to show if we were to consider directed acyclic graphs (then we could assign subsequent colors to vertices as we traverse along a path and do some rearranging when we visit an already coloured vertex)... But I am stuck at finding a valid argument that justifies the implication in either direction, not to mention I am actually interested in a stronger statement.

答: Perhaps you could use the Gallai-Hasse-Roy-Vitaver Theorem.</p>


答:
(I agree that most of the algebraic steps here are not obvious nor intuitive. It was a lot of wishful thinking that this method could work, in part because of the background of the problem.)

For the sextic inequality, from the graph, you know that $\sqrt{2} - 1 $ is a double root, so $ ( x - (\sqrt{2} - 1) )^2$ is a factor of the polynomial.

(Alternatively, we know from the work you did that that's the equality case.)

So, we have

$$\frac{ x^6 - 2\sqrt{2} x^5 - 3 x^4 - 8 x^3 + 3 x^2 + 2 \sqrt{2} x - 1} { ( x - (\sqrt{2} - 1) )^2} \\ = x^4 - 2x^3 - 2 ( 1 + \sqrt{2} ) x^2 - 2 ( 3 + 2 \sqrt{2} ) x - 2\sqrt{2} - 3.$$

We make the observation (by staring really hard) that the quartic factors as

$$ (x^2 - ( 2 + \sqrt{2} ) x - \sqrt{2} - 1 ) ( x^2 + \sqrt{2} x + \sqrt{2} + 1 ) $$

We can now easily show that on $  0 < x < 1$, this value is negative. (First term is negative, second term is positive.)

Hence, the original expression is $ \leq 0$, with equality at $ \sqrt{2} - 1 $.


问:
Square of the Error Function
The author defines the probability integral as follows
$$\Phi(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}\mathrm{d}t$$.
(S)he instructs the reader to derive the following integral representation of the square of the probability integral by transforming it to polar coordinates
$$\Phi(z)^2=1-\frac{4}{\pi}\int_0^1\frac{\exp(-z^2(1+t^2))}{1+t^2}\mathrm{d}t$$.
I am uncertain how transforming the integral from cartesian coordinates to polar coordinates would yield the integral representation. Can someone explain this to me?
$$I^2=\frac{4}{\pi}\int_{0}^{z}\int_{0}^{z}e^{-(x^2+y^2)}dydx=-\frac{2}{\pi}\int_{0}^\frac{\pi}{2}\int_{0}^{z}-2re^{-r^2}drd\theta$$?

答:
$\operatorname{erf}(z)^2$ can be expressed as an integral over the region $[0,z]\times[0,z]$. By symmetry, you can calculate half of its value by looking at the lower half triangle (i.e. with the additional bound $y\leq x$). I believe the correct coordinate transformation is $R=x^2+y^2, t=\frac yx$ (you could think of the former as $r^2$ and the latter as $\tan\theta$ in polar coordinates).  Express the region in the new coordinates, calculate the Jacobian, and the result should follow.

Full solution: $\def\pdv#1#2{\frac{\partial #1}{\partial #2}}$

 First, $erf^2(z)=\frac4\pi\int_{[0,z]^2}e^{-(x^2+y^2)}dxdy=\frac8\pi\int_{D}e^{-(x^2+y^2)}dxdy$, where $D=\{(x,y)\in[0,z]^2 \mid y\leq x\}$. Write $I=\int_{D}e^{-(x^2+y^2)}dxdy$.

 The inverse of the Jacobian (since this is easier to calculate) is $\begin{pmatrix}\pdv Rx&\pdv Ry\\ \pdv tx&\pdv ty\end{pmatrix}$ with determinant $2(1+t^2)$, so its inverse has determinant $\frac1{2(1+t^2)}$. The triangular region $D$ becomes the region $\{(R,t)|0\leq t\leq1, 0\leq R\leq z^2(t^2+1)\}$. (You need to notice $\frac1{\cos^2\theta}=\tan^2\theta+1$.)

 At last the calculation. $$I=\int_0^1\int_0^{z^2(t^2+1)}e^{-R}\cdot\frac1{2(1+t^2)}dRdt=\frac12\int_0^1(1-e^{-z^2(1+t^2)})\frac1{1+t^2}dt=\frac12(\frac\pi4-\int_0^1\frac{e^{-z^2(1+t^2)}}{1+t^2}dt),$$ so $$erf^2(z)=\frac8\pi I=1-\frac4\pi\int_0^1\frac{e^{-z^2(1+t^2)}}{1+t^2}dt$$ as desired.

Title: Square of the Error Function


问:
rules of inference help
Let p, q, r, s and t be statements variables. Use the valid argument forms to deduce the conclusion, ¬q, from the premises, giving a reason for each step.

(a) ¬p ∨ q → r.

(b) s ∨ ¬q

(c) ¬t

(d) p → t

(e) ¬p∧r → ¬s

——————————————————

(f) ∴ ¬q

I have tried modus pollen on (c) and (d) but I can't find anyway to link to the other premise.

I have also tried double negation law and de morgan's law  on (e),  followed by hypothesis syllogism resulting in  p∨¬r → ¬q but it also got me to nowhere.

答: You are on the right track.  You get $\neg p$ from modus tollens on (d).  Now (a) gives you $r$. Now that you have $\neg p$ and $r$ statement (e) gives $\neg s$.  (b) now forces $\neg q$.

Title: rules of inference help


问:
The number of non-isomorphic spanning trees in K4
K4 has 16 spanning trees. I believe there are two non-isomorphic spanning trees in K4. Is this because half of the spanning trees have the sequence (1,2,2,1) as the degrees of their vertices, while the other half have (1,3,3,1)? Or is there some other reason why just two of the spanning tree graphs of K4 are non-isomorphic? </p>

答:
You're essentially asking for the number of non-isomorphic trees on $4$ vertices.  Here they are:

We can verify that we have not omitted any non-isomorphic trees as follows.

The total number of labelled trees on $n$ vertices is $n^{n-2}$, called Cayley's Formula.  When $n=4$, there are $4^2=16$ labelled trees.

The number of labelled graphs isomorphic to a graph $G$ is given by $$\frac{n!}{|\mathrm{Aut}(G)|}$$ where $n$ is the number of vertices of the graph, and $\mathrm{Aut}(G)$ is the automorphism group of the graph.  This is a consequence of the Orbit-Stabiliser Theorem.  In the above two cases, the automorphism groups have size $6$ (and is isomorphic to $S_3$) and $2$, respectively.

So the $16$ labelled $4$-vertex trees partition into $24/6=4$ trees isomorphic to the tree on the left and $24/2=12$ trees isomorphic to the tree on the right.  Since $16=4+12$, we have accounted for all isomorphism classes of trees.

(Note: more than half of the labelled $4$-vertex trees are isomorphic to the graph on the right.)


问:
Basic question on Open Sets
I'm trying to prove the following:

Let $X$ be a topological space, and $U \subset X$ be an open set, and $C \subset X$. Then, every open in $C \cap U$ is open in $C$.

My proof: $U \cap C$ is open in $C$, directly by definition of the subspace topology. Thus for every open $A \subset (U \cap C)$, there $\exists B \subset C$ such that $B \cap (U \cap C) = A$. So $A$ is open in $C$ as an intersection of two open sets.

Firstly, is this proof correct?

Also, apparently there's another proof that goes: Every open in $U$ is open in $X$, as $U$ is open. It follows that every open in $C \cap U$ is open in $C$. I don't understand how it follows; could someone demistify this (in a more step-by-step manner)? Thanks.

答:
If $A$ is an open set of the subspace $U \cap C$, then $A = (U\cap C)\cap B$, for some open set $B$ in $X$. $B$ need not be a subset of $C$.

But from there the conclusion holds, because $B \cap U$ os open (they are both open in $X$), hence $A$ is open in the subspace topology of $C$.

The second argument is really very similar: By saying that every open set in the subspace topology on $U$ is open in $X$, we are really saying that $U \cap O$ is open in $X$, for every open $O$ in $X$. Then consider an open set, $H$, of $C \cap U$: $H = (C \cap U) \cap O$, for some open $O$ in $X$. But $U \cap O$ is an open set of $U$ and hence an open set of $X$. Hence $H$ must be open in $C$, by the definition of the subspace topology on $C$.

Unless I'm missing something, this second argument is essentially redundant. It's just asserting the fact that $U \cap O$ is open in $X$. But this is immediate from the openness of both $U$ and $O$ anyway; you don't really need to invoke the subspace of $U$ to achieve this.

Title: Basic question on Open Sets


问:
$\frac{1}{n}\sum_{k=1}^n(X_k-\mathbb E[X_k])$ converges a.s. to $0$
Let $X_1,X_2,\dots$ be a sequence of independent random variables, such that the series $$\sum_{n=1}^\infty\frac{\operatorname{Var}(X_n)}{n^2}$$ converges. Show that as $n\to\infty$, $$\frac{1}{n}\sum_{k=1}^n(X_k-\mathbb E[X_k])$$ converges almost surely to $0$.

There is a quick solution via the martingale convergence theorem: we have that $$Y_n=\sum_{k=1}^n\frac{X_k-\mathbb E[X_k]}{k}$$ is martingale, and $\sup_n\mathbb E[|Y_n|]$ is finite, so $Y_n$ converges almost surely to some random variable $Y$, and we can finish with Kronecker's lemma.

I'm interested though in any approaches avoiding the heavy machinery of the martingale convergence theorem. I feel like defining the $Y_n$'s as I did above could be fruitful. For example, the Kolmogorov inequality gives the bound
$$\mathbb P\left(\max_{1\leq i\leq n}|Y_i|>\varepsilon\right)\leq\frac{1}{\varepsilon^2}\mathbb E\left[Y_n^2\right]=\frac{1}{\varepsilon^2}\cdot\sum_{k=1}^n\frac{\operatorname{Var}(X_k)}{k^2}.$$
But I'm unsure what to make from all this.

答:
A proof, which you kind of started, can be derived from Kolmogorov inequality,

For $n < m$, Kolmogorov inequality applied to the martingale $Y_k-Y_n$, $n \leq k \leq m$, gives that, for all $\epsilon > 0$
$$
P(\max_{n \leq  k \leq m} |Y_k - Y_n| > \epsilon) \leq \frac{1}{\epsilon^2} \sum_{k = n}^m \frac{Var(X_k)}{k^2}.
$$

So, for all $\epsilon > 0$

$$
P(\inf_l \max_{n,m \geq l} |Y_m - Y_n| > \epsilon) \leq \lim_{l \rightarrow \infty} P(\max_{n,m \geq l} |Y_m - Y_n| > \epsilon) = 0.
$$

Now take $\epsilon_p \rightarrow 0$. Define
$$
\Omega_p = \{ \inf_l \max_{n,m \geq l} |Y_m - Y_n| \leq \epsilon_p \},
$$ and $\Omega' = \bigcap_p \Omega_p$.
Then $P(\Omega') = 1$ and, for all $\omega \in \Omega'$, $Y_n(\omega)$ is a Cauchy sequence, therefore converges.
Kronecker's Lemma then finishes the proof, as before.

Comment

This does not really "avoid the martingale machinery", though. This argument substitutes martingale convergence by Kolmogorov inequality, which is a maximal inequality for martingales.

It does not seem easy to get away from using the martingale property in one way or another.

The alternative argument proposed by previous answer uses the fact that a series of independent summands that converges in probability must also converge almost surely. The standard proof of this fact is a stopping time argument, that also uses the stronger independence property.


问:
On re-writing polynomials of the form, $P(x) = x^n - b^n$
This is a follow up question to this answer, in which it says:

Let $b$ be fixed, and consider the polynomial $P(x) = x^n - b^n$. Then there is a polynomial $Q(x)$, and a constant $r$, such that

$P(x) = (x - b)Q(x) + r$

Could someone please tell me why this is true (without using the factorization formula for $a^n - b^n$ or the fact that $b$ is a root, since that's what the answer is trying to prove)? (Does it involve any theorems or facts from congruence-class arithmetic or am I just thinking too hard about this?)

答:
Given polynomials $P(x)$ and $x-b$,

by Euclidean division of polynomials there exist polynomials $Q(x)$ and $r(x)$

such that $P(x)=Q(x)(x-b)+r(x)$, where the degree of $r(x)$ is less than the degree of $x-b$.

Since $x-b$ is linear (first degree), $r(x)$ must be a constant.


答:
Polynomial remainder theorem:

If a polynomial $f(x)$ is divided by $x-a$, the remainder is $f(a)$.

Proof.  Suppose $f(x)$ is divided by $x-a$.  Then we can write $$f(x)=(x-a)q(x)+r(x)$$ where $q(x)$ is the quotient and $r(x)$ is the remainder.  Since the remainder is at most one degree less than the divisor, we can conclude that $r(x)$ is a constant and rewrite the above as $$f(x)=(x-a)q(x)+R.$$
By substitution $x=a$, we have $$f(a)=(a-a)q(a)+R\implies f(a)=R.$$

See here.

Title: On re-writing polynomials of the form, $P(x) = x^n - b^n$

Body:
This is a follow up question to this answer, in which it says:

Let $b$ be fixed, and consider the polynomial $P(x) = x^n - b^n$. Then there is a polynomial $Q(x)$, and a constant $r$, such that

$P(x) = (x - b)Q(x) + r$

Could someone please tell me why this is true (without using the factorization formula for $a^n - b^n$ or the fact that $b$ is a root, since that's what the answer is trying to prove)? (Does it involve any theorems or facts from congruence-class arithmetic or am I just thinking too hard about this?)


问:
Why is $\frac{987654321}{123456789} = 8.0000000729?!$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729\ldots$.

I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!

My life has gone downhill since then:)

My questions are:

Why is this so?

What happens beyond the "$729$"?

What happens in bases other than $10$?

Body:
Just to add to the excellent answers above, some examples:

${987654321\,/\,123456789}\approx 8.00000007290000066339$

${{87654321}_9\,/\,{12345678}_9}\approx {7.000000628000056238}_9$

${{7654321}_8\,/\,{1234567}_8}\approx {6.0000052700046137}_8$

${{654321}_7\,/\,{123456}_7}\approx {5.00004260036036}_7$

${{\mathrm{fedcba987654321}}_{16}\,/\,{\mathrm{123456789abcdef}}_{16}}\approx {\mathrm{e.0000000000000d2f00000000000c693f}}_{16}$


问:
The weather on any given day can either be sunny, cloudy, or partially cloudy.
Each day is also classified as dry or rainy.The
probability of a sunny day is $0.48$, and the probability of a cloudy day is $0.39$.
The probability of having a sunny and dry day is $0.48$. The probability of a
cloudy and dry day is $0.14$. The probability of a partially cloudy and dry day
is $0.09$.

Find the probability of a cloudy day given that it's a rainy day.

My work:
Let $D$ be the event that the day is dry, $S$ be the event that the day is sunny, $R$ be the event that the day is rainy, $C$ be the event that the day is cloudy, $PC$ be the event that the day is partly cloudy.

So $P(C) = 0.39, P(S) = 0.48, P(PC) = 0.13 $

First I found Probability that it's a dry day $P(D) = P(S\cap D) + P(C \cap D) + P(PC \cap D) = .48+ .14 + .09 = .71$

Then Using the complement I found the probability that it's a rainy day $P(R) = 1 - P(D) = 1 - .71 = .29$

But How do I use this info to find $P(C | R)$?

I tried Bayes' Theorem $P(C | R) = \cfrac{P(R | C)P(C)}{P(R)} = \cfrac{P(R|C)0.39}{.29}$ But I can't find $P(R|C)$?

Can someone please tell me what I need to do to find $P(R|C)$?

Thank You!

答:
It may be better to organize the data into a joint probability table.  This is what we are given in the question.
$$\begin{array}{c|c|c|c|c}
& S = \text{Sunny} & P = \text{Partly Cloudy} & C = \text{Cloudy} & \text{Total}\\
\hline
D = \text{Dry} & 0.48 & 0.09 & 0.14 & \Pr[D] \\
\hline
R = \text{Rainy} & \Pr[S \cap R]  & \Pr[P \cap R] & \Pr[C \cap R]& \Pr[R]\\
\hline
\text{Total} & 0.48 & \Pr[P]& 0.39 & 1 \\
\end{array}$$

Since we must have the overall total equal to $1$, we require $$0.48 + \Pr[P] + 0.39 = 1,$$ or $\Pr[P] = 0.13$.  We also require $$0.48 + 0.09 + 0.14 = \Pr[D],$$ so $\Pr[D] = 0.71$.  This takes care of the row sums.  Now for the column sums, we must have $$\Pr[D] + \Pr[R] = 1,$$ so $\Pr[R] = 1 - 0.71 = 0.29$, and similarly, $\Pr[S \cap R] = 0$, $\Pr[P \cap R] = 0.13 - 0.09 = 0.04$, and $\Pr[C \cap R] = 0.39 - 0.14 = 0.25$.  So now the complete table is

$$\begin{array}{c|c|c|c|c}
& S = \text{Sunny} & P = \text{Partly Cloudy} & C = \text{Cloudy} & \text{Total}\\
\hline
D = \text{Dry} & 0.48 & 0.09 & 0.14 & 0.71 \\
\hline
R = \text{Rainy} & 0  & 0.04 & 0.25 & 0.29\\
\hline
\text{Total} & 0.48 & 0.13 & 0.39 & 1 \\
\end{array}$$
Now all that remains is to use the table to compute the desired probability:  $$\Pr[C \mid R] = \frac{\Pr[C \cap R]}{\Pr[R]} = \frac{0.25}{0.29} \approx 0.862069.$$


问:
Prove that $z_1z_2z_3 = 1$ for complex numbers $z_1$, $z_2$, $z_3$.
Let $z_1$, $z_2$, $z_3$ be distinct complex numbers such that $|z_1| = |z_2| = |z_3| > 0$. If $z_1 + z_2z_3$, $z_2 + z_3z_1$, and $z_3 + z_1z_2$ are real numbers, prove that $z_1z_2z_3 = 1$.

I believe that this problem comes from 1979 Romanian Math Olympiad (State Competition, 10th grade).

I tried several approaches: The first was to substitute $a + ib$, $c + id$, and $f + ig$ for $z_1$, $z_2$, and $z_3$, respectively, and bash. I used $(1+z_1)(1+z_2)(1+z_3)$, which didn't really get me anywhere.

After a couple more attempts at substitutions and such, I hit a wall and didn't really know what to do.

答:
Alternative approach:

$x$ and $y$ say hello from kludgiville.

Given that $$(z_1\overline{z_1}) = (z_2\overline{z_2}) = (z_3\overline{z_3}). \tag1$$

Let $(x_k + iy_k)$ denote $z_k : k \in \{1,2,3\}.$

From the constraints:

$(x_1 + iy_1) + (x_2 + iy_2)(x_3 + iy_3) = a \in \mathbb{R}.$

$(x_2 + iy_2) + (x_1 + iy_1)(x_3 + iy_3) = b \in \mathbb{R}.$

$(x_3 + iy_3) + (x_1 + iy_1)(x_2 + iy_2) = c \in \mathbb{R}.$

Let $s_1 = (a-z_1), s_2 = (b-z_2), s_3 = (c-z_3).$

Then $$(z_1 z_2 z_3) = (s_1 z_1) = (s_2 z_2) = (s_3 z_3).\tag2$$

Therefore:

$s_1 = (a - z_1) = a - (x_1 + iy_1) = a + (x_1 - iy_1) - 2x_1 = [a + \overline{z} - 2x_1] \implies $

$$(s_1 z_1) = (az_1) + (z_1\overline{z_1}) - 2x_1z_1 = (z_1\overline{z_1}) + z_1(a - 2x_1).\tag3$$

Similarly:

$$(s_2 z_2) = (z_2\overline{z_2}) + z_2(b - 2x_2).$$
$$(s_3 z_3) = (z_3\overline{z_3}) + z_3(c - 2x_3).$$

Using equations (1) and (2) above, this implies that

$$z_1(a - 2x_1) = z_2(b - 2x_2) = z_3(c - 2x_3).$$

Suppose that $(a - 2x_1) \neq 0.$

$$\text{Then} ~\frac{b - 2x_2}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k$$

$$\text{and} ~z_1 = kz_2 ~\text{and} |z_1| = |z_2|.$$

The only way that this is possible is if $z_1 = z_2$.

This is a contradiction, since it violates the constraint that $z_1, z_2, z_3$ are all distinct.

Edit

The above analysis is definitely the right approach, but contains a flaw.  See the comment/analysis of JyrkiLahtonen immediately following this answer.  Applying his comment/analysis, the explicit remedy is as follows:

Exploring the hypothesis that $(a - 2x_1) \neq 0$, and keeping in mind that none of $z_1,z_2,z_3$ can $= 0$ [else, by equation (1) above, they all $= 0$ and are therefore not distinct], you have that

$$~\frac{b - 2x_2}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k_1$$

$$\text{and} ~\frac{c - 2x_3}{a - 2x_1} ~= ~\text{a real non-zero scalar} ~~k_2.$$

This implies that $~z_1 = k_1z_2 = k_2z_3,~~$ with
$~~|z_1| = |z_2| = |z_3|.$

This implies that $~|k_1| = 1 = |k_2|.$

Since, in this hypothetical $~~k_1, k_2 \in \mathbb{R},$

this implies that $~k_1,k_2~$ are both $~\in \{+1, -1\}.$

In order for $z_2$ and $z_3$ to both be distinct from $z_1$, it is necessary for $k_1 = -1 = k_2.$  However, in that case, you have that $z_2 = z_3$.

Therefore, the assumption that $(a-2x_1) \neq 0$ implies that $z_1, z_2, z_3$ can not all be distinct.  Therefore, one can conclude that $(a-2x_1) = 0.$

Therefore, from equation (3) above,
$$(z_1 z_2 z_3) = (s_1 z_1) = z_1 \overline{z_1} \implies
(z_2 z_3) = \overline{z_1}. \tag4 $$

Using equation (1) above, this implies that
$$|z_1|^2 = |z_2 z_3| = |\overline{z_1}| = |z_1| \implies |z_1| = 1.$$

Therefore, by equation (4) above, $$(z_1 z_2 z_3) = 1.$$


问:
Find the transform
I have the paper with 3 points on it. I have also a photo of this paper. How can I determine where is the paper on the photo, if I know just the positions of these points? And are 3 points enough?
It doesn't look like a linear transform: the paper turns into trapeze on the photo. But it should be able to be written mathematically. I was unsuccessfully looking for some function $\mathbb{R}^2\rightarrow\mathbb{R}^2$, which turns the coordinates of the points on the default paper to their coordinates on photo, so I could just apply this function to the paper corners, but I have no idea, what kind of transform it should be.

Any help is appreciated :)

答:
I believe a projective transformation is exactly right. But with a projective transformation you can take any four points in general position (no three collinear) to any other four. So you'd better add a fourth point to reconstruct the transformation.

In terms of the linear algebra, you want to construct a $3\times 3$ matrix $A$, up to scalar multiples. You think of a point in $\mathbb R^2$ as a vector $X=\begin{bmatrix}1\\x\\y\end{bmatrix}\in\mathbb R^3$ and its image will be the vector $AX$ rescaled so that its first coordinate is likewise $1$. So see that you can determine $A$ from knowing where four points map, think of it this way. A linear map from $\mathbb R^3$ to $\mathbb R^3$ is determined by where it sends a basis. Given our three points in general position, they correspond to vectors $P$, $Q$, $R\in\mathbb R^3$ up to scaling.
But now, given our fourth point in general position, we can rescale those vectors so that the fourth point is given by $P+Q+R$. Doing the corresponding construction with the four image points will now allow us to specify precisely where $A$ sends our basis vectors, up to one common scalar multiplication.

Title: Find the transform


问:
How to calculate the derivative of $\frac{d}{dx}\left[\frac{7}{(3x)^{-2}}\right]$?
Can anyone explain how to calculate the derivative of $\frac{d}{dx}\left[\frac{7}{(3x)^{-2}}\right]$? The correct answer given was $126x$, however, I have gotten $42x$.

$\frac{d}{dx}\left[\frac{7}{(3x)^{-2}}\right]=7\cdot(3x)^2=7\cdot2\cdot3x=42x$

答:
The derivative of $7\times(3x)^2=63x^2$ is $2\times63x=126x$.

As imranfat commented,

$\dfrac d {dx} \left[7\times u^2\right]=7\times2\times u\times\dfrac {du}{dx}=7\times2\times3x\times3$,

where $u=3x$.


问:
Solving System of equations and L-U decomposition of matrices
I just read about the L-U decomposition method to decompose matrices to upper and lower triangular ones. And I saw some examples of solving systems of equations using this method.

My question: with what conditions can we safely use this method to solve the system of equations?

For example should the coefficient matrix be necessarily a square matrix? In other words when can one write the matrix $A$ as $LU$ ?

答:
The purpose of LU decomposition is to solve a system of equations without going through the route of matrix inversion. If the coefficient matrix is not square, then the system of equations is either under-constrained or over-constrained depending on the number of equations available (i.e, number of rows) and number of unknowns (i.e, number of columns).
Also, for a non-square coefficient matrix, you can't decompose it into two triangular matrices.</p>


问:
Show that $f'(x) \simeq \frac{1}{12h} [-f(x+2h) + 8f(x+h) -8f(x-h) + f(x-2h)]$?
How would you show that $f'(x) \simeq \frac{1}{12h} [-f(x+2h) + 8f(x+h) -8f(x-h) + f(x-2h)]$?

Here is what I have tried:

It would make sense to me that Richardson extrapolation should/could be used here because the question came up in a chapter about this topic - although it does not have to be. Anyway, so I thought pretty hard but the closest approximation I could come up with was $f'(x) \simeq \frac{1}{12h} [-2f(x+2h) + 8f(x+h) -8f(x-h) + 2f(x-2h)]$. This was derived as follows:

$$
D(n,0) = \frac{1}{2h} [f(x+h/2^n)-f(x-h/2^n)]\\
D(n,k) = \frac{4^k}{4^k-1} D(n,k-1) \frac{1}{4^k-1} D(n-1,k-1)\\
D(0,1) = \frac{4^1}{4^1 - 1} D(0,0) - \frac{1}{4^1-1} D(-1, 0)\\
\Rightarrow D(0,1) = \frac{4}{3} \frac{1}{2h} [f(x+h)-f(x-h)] - \frac{1}{3} \frac{1}{2h} [f(x+2h)-f(x-2h)]\\
= \frac{4}{6h} [f(x+h)-f(x-h)] - \frac{1}{6h} [f(x+2h)-f(x-2h)]\\
= \frac{1}{12h} [-2f(x+2h) + 8f(x+h)-8f(x-h) + 2f(x-2h)]
$$

I was about to get happy but realized that even though it was close, it was not the requested result.

How can I derive $f'(x) \simeq \frac{1}{12h} [-f(x+2h) + 8f(x+h) -8f(x-h) + f(x-2h)]$?

答:
This is a 4 point derivative approximation. WLOG assume $x = 0$. Then the approximation takes the form
$$D_h(f) = \frac{w_1f(-2h) + w_2f(-h) + w_3f(h) + w_4f(2h)}{h},$$
Where $w_1, w_2, w_3, w_4$ are to be determined. To get the coefficients, set $D_h(f) = 0$ for $f = 1, x, x^2, x^3$ to get the linear system
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\
-2 & -1 & 1 & 2 \\
4 & 1 & 1 & 4 \\
-8 & -1 & 1 & 8 \\
\end{pmatrix}
\begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
1 \\
0 \\
0 \\
\end{pmatrix}
$$
which has unique solution
$$w_1 = \frac{1}{12}, w_2 = -\frac{2}{3}, w_3 = \frac{2}{3}, w_4 = -\frac{1}{12}.$$
Let $E_h(f) = f'(0) - D_h(f)$. Note that $E_h$ is a linear map. By symmetry, $E_h(x^4) = 0$. But $E_h(x^5) = 4h^4$. This implies that the method has error
$$E_h(f) = \frac{f^{(5)}(0)}{5!} \cdot 4h^4 + O(h^5),$$
This can easily be derived as follows. Assume $f$ has $6$ continuous derivatives on $[-2h, 2h]$. Then by Taylor's theorem,
\begin{align}
E_h(f) &= E_h\left(\sum_{n = 0}^{5}\frac{f^{(n)}(0)}{n!}x^n + \frac{f^{(6)}(\xi(0, x))}{6!}x^6\right) \\
&= \sum_{n = 0}^{5}\frac{f^{(n)}(0)}{n!}E_h(x^n) + E_h\left(\frac{f^{(6)}(\xi(0, x))}{6!}x^6\right) \\
&= \frac{f^{(5)}(0)}{5!}E_h(x^5) + O(h^5) \\
&= \frac{f^{(5)}(0)}{30}h^4 + O(h^5).
\end{align}
This procedure can be used for other approximation methods, such as integration.


答:
You're applying the rule $$\frac{d}{dx}(\alpha x^\beta)=\alpha\beta x^{\beta -1}$$

and using this to conclude that $$\frac{d}{dx} (\alpha(ax)^\beta)=\alpha\beta(ax)^{\beta-1}$$

This is not true, indeed by the chain rule$$\frac{d([f(x)]^\beta)}{dx}=\frac{d[f(x)]}{dx}\cdot\frac{d([f(x)]^\beta)}{d[f(x)]}=f'(x)\cdot\beta(f(x))^{\beta-1}$$

(This might seem confusing, but set $u=f(x)$ and it looks more manageable.

Here $f(x)=ax\implies f'(x)=a$, so $$\frac{d}{dx} (\alpha(ax)^\beta)=a\alpha\beta(ax)^{\beta-1}$$ is correct.

Title: How to calculate the derivative of $\frac{d}{dx}\left[\frac{7}{(3x)^{-2}}\right]$?

Body:
Can anyone explain how to calculate the derivative of $\frac{d}{dx}\left[\frac{7}{(3x)^{-2}}\right]$? The correct answer given was $126x$, however, I have gotten $42x$.

$\frac{d}{dx}\left[\frac{7}{(3x)^{-2}}\right]=7\cdot(3x)^2=7\cdot2\cdot3x=42x$


问:
How are Jacobians of genus $3$ curves different from one another?
There are two types of smooth projective (complex) curves of genus $3$: plane quartics, and hyperelliptic curves. The Torelli morphism $M_3\to A_3$, assigning a curve to its (principally polarized) Jacobian, is bijective on points. Thus $A_3$ consists of Jacobians of genus $3$ curves. Because of the "partition" of $M_3$ as {plane quartics} $\amalg$ {hyperelliptics}, I would expect this to be reflected somehow in $A_3$. Whence my questions:

Question 1. What are the main differences between Jacobians of plane quartics and of hyperelliptic curves?

(I am intentionally vague here, as I am curious about any sort of difference).

It would be also interesting to know if the deformation theory of the curve inside its Jacobian is sensibly different according to the curve itself; this leads to a second question:

Question 2. (Deformations inside the Jacobian) What do the normal bundles $\mathcal N_{C/Jac(C)}$ look like, when $C$ is a plane quartic or a hyperelliptic curve, respectively?

Thanks for any clue on this.

答:
For Question 2, a quick description of the normal bundle $N_{C|J(C)}$ is (See Lemma 3.8 of [1]):

$$h^0(C,N_{C|J(C)})=h^1(C,N_{C|J(C)})=
\begin{cases}
3 ~\text{for} ~C~\text{non-hyperelliptic}\\
4 ~\text{for} ~C~\text{hyperelliptic}.\tag{1}\label{1}
\end{cases}$$

This tells us some information of the deformation theory of the pair $(J(C),C)$. Let $\mathcal{D}$ be the deformation space of such pairs, for $C\in M_3$ and $J(C)$ the Jacobian of $C$, then the tangent space of $\mathcal{D}$ at $(J(C),C)$ is the hypercohomology $\mathbb{H}^1(J(C),C)$, which fits into the exact sequence

$$0\to H^0(J(C),T_{J(C)})\to H^0(C,N_{C|J(C)})\to \mathbb{H}^1(J(C),C)\xrightarrow{d} H^1(J(C),T_{J(C)})\to H^1(C,N_{C|J(C)})\to 0.$$

Using $T_{J(C)}\cong \mathcal{O}^{\oplus 3}_{J(C)}$, one has $h^0(J(C),T_{J(C)})=3$ and $h^1(J(C),T_{J(C)})=9$, together with $(\ref{1})$, one has $\dim \mathbb{H}^1(J(C),C)$ is always equal to $6$, so $\mathcal{D}$ is smooth, no matter $C$ is hyperelliptic or not.

However, the distinction comes from the rank of $d$, which is the answer to your Question 2:

If $C$ is non-hyperelliptic, $d$ is of maximal rank, so when $J(C)$ deforms in the 6-dimensional p.p.a.v. locus, the curve $C$ deforms with $J(C)$; If $C$ is hyperelliptic, $d$ has a one-dimensional kernel, so $C$ will be obstructed when $J(C)$ deforms normal to the 5-dimensional hyperelliptic locus.

The proof of $(\ref{1})$ is essentially a classical fact known as Noether's theorem, which states that $$\text{Sym}^2H^0(C,K_C)\to H^0(C,2K_C)$$ is an isomorphism when $C$ is non-hyperellptic curve and has a one-dimensional kernel when $C$ is hyperellptic.

Question 1 is essentially answered in rfauffar's comment, namely  $\tau(C)$ coincides with $C$ up to a translation in torus if and only if $C$ is hyperelliptic, where $\tau:J(C)\to J(C)$ is the involution given by $(-1)$-map and we identify $C$ with its Abel-Jacobi image in $J(C)$.

I'd like to mention a theorem related to this fact. Note that $\tau$ always induces identity map on $H_2(J(C),\mathbb Z)$, so $\tau(C)$ is homologous to $C$ in $J(C)$, however, they are in general not algebraically equivalent:

Theorem: (G. Ceresa [2], 1983) When $C$ is general in $M_3$, $C$ is not algebraically equivalent to $\tau(C)$.

Based on this result, F. Bardelli and M. Nori showed

Theorem: (F. Bardelli [1], 1989, M. Nori [3], 1989) Let $A$ be a general abelian threefold, the Griffiths group $\mathcal{G}_1(A)$ (the group of algebraically trivial 1-cycles modulo homological equivalence) is not finitely generated.

[1] Bardelli, Fabio.
Curves of genus three on a general abelian threefold and the nonfinite generation of the Griffiths group. Arithmetic of complex manifolds (Erlangen, 1988), 10–26,
Lecture Notes in Math., 1399, Springer, Berlin, 1989.

[2] Ceresa, G.
$C$ is not algebraically equivalent to $C^−$ in its Jacobian.
Ann. of Math. (2) 117 (1983), no. 2, 285–291.

[3] Nori, Madhav V.
Cycles on the generic abelian threefold.
Proc. Indian Acad. Sci. Math. Sci. 99 (1989), no. 3, 191–196.


问:
Calculating a Point that lies on an Ellipse given an Angle
I need to find a point (A on this diagram) given the center point of the ellipse as well as an angle. I've been melting my brain all day (as well as searching through questions here) testing out different equations. What's the best way to do this?

I intend to grab point A at $225^o$ as well as another point at approximately $250^o$ using the same math. These need to be fetched regardless of elliptic width and height.

答:
You can also use parametric equations:

$$x=a\cos(\theta)$$
$$y=b\sin(\theta)$$

Where $a$ is the radius on the horizontal axis, and $b$ is the radius on the vertical axis.


问:
Does the mathematical meaning of "up to" relate to the ordinary English meanings?
Does "up to" defined by Pete L. Clark or
Tobias Kildetoft relate to the 2020 English meanings from OED?

If not, why did mathematicians utilize "up to"?

答:
In my experience, the most common meaning of "up to" in mathematical literature is "when you do not distinguish the objects that share some property". For example, the following statement

There are 2 groups of order 4, up to isomorphism.

means that when we do not distinguish isomorphic groups, there are only 2 groups of order 4 ($\Bbb Z_4$ and $\Bbb Z_2\times \Bbb Z_2$).


问:
Parametric plot of a cycle - transition between two lines - dealing with a jump
I have the following problem. I am modelling a cycle that consist of "loading" and "unloading" parts.

Firstly, the material is being loaded from the origin on the "blue line", that is the simple linear function ( $y = 2x$ ), and then I would like to perform transition to the bottom orange line that represents "unloading" of the material (now it is the linear function $y = x$ ):

I created the following solution based just on the ratios of the linear coefficients,

but it is not accurate enough. I would like to have steeper transition between the blue and orange line. I tried to use $tanh$ function as suggested here but in my situation there is a jump between linear functions and I do not know how to overcome it.

Please, do you have any advices?

Thank you very much in advance!

EDIT:

I would like to get a curve, that comes from the origin to the point $[1;2]$, then I need some transition from the upper line to the bottom line (that I need to be steep enough, using hyperbolic tangens?) and from the point that is as close as possible to the point $[1;1]$ on the bottom line go back on the curve to the origin.

Actually, I have a solution (it is my black curve), but it is not precise enought. I would like to kindly ask if anybody knows some improvement of the transition...

答:
I will show you how to think about this type of problems. Let's start from $\tanh$ function:$$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$
Notice that this function varies between $-1$ and $1$. I want to change it to vary from $0$ on the negative side. To do that, we add $1$. So $$\tanh x+1=\frac{2e^x}{e^x+e^{-x}}$$
The transition between $0$ and $2$ happens around $x=0$. So if we want the transition to happen at different $x$, you just need to change the variable. In this case replace $x$ with $x-1$. If you want to change the slope to be more steep, multiply $x-1$ with a large number $c$ (say $c=10$).

Now you have two choices to connect the curves. We start smoothly from the lower curve in both cases. You can approach smoothly with the upper curve, or you can just truncate the $\tanh$ function. It seems like in this case you go with the second option. So we choose to truncate our connecting function at $x=1$, where $$\tanh c(x-1)+1=1$$
Then you add to the $y=x$ this function multiplied by the gap between the curves at $x=1$. In this case the gap is $1$.

So the function you have is $$y=x+1\cdot(\tanh 10(x-1)+1)$$


问:
How has the author mathematically modeled a double pendulum in this study?
I'm looking for project ideas for my final year dissertation for a computing degree. Here is the study (A double pendulum model of tennis strokes. Rod Cross. Uni of Sydney, 2006) - http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/49.%20TennisDPend.pdf. My matlab files and spreadsheet files are linked at the bottom of the post.

My goal was to investigate a link between arm length/mass and the angular velocity/swingspeed of a forehand, mapped by a pendulum .

I've looked into this before, but I wasn't able to solve the calculus. I'm a fairly strong coder but I'm terrible with calculus. I've coded the pendulum in an executable matlab code, but the angular velocity peaks too early in time, making the pendulum swing too fast, suggesting there is an issue with the formulas. Is there any other pendulums that I could use, that are solvable.

I've put all ODE's into a spreadsheet. The author doest state which numerical method was used. So I implemented the formulas using the RK4 method, to try and investigate the formulas, and why the velocity is rising too early. It's not made clear in the paper how L2, M2, h2 and I2 are calculated. It's also not mentioned whether the hand was included in the calculations for the lower pendulum, which would really complicate it, as h2 and the moment of inertia could differ, depending on the position of the hand on the grip, the mass distribution of the hand and the size of the hand.

I also tried to replicate the results from this study , which is based on the same equations from Rod Cross' study to see if that helped - https://arxiv.org/pdf/1505.01916.pdf . I had the exact same problem of the chart looking similar but rising too high, too early. .https://arxiv.org/pdf/1505.01916.pdf. I emailed the authors of both studies to check what method/timestep they used, but never heard back. I also asked whether the hand was included.

Matlab files are here - https://ufile.io/f/s3jmi

Spreadsheet is here. - https://drive.google.com/file/d/1uzjtXV2bweHj7ADU47lX3NaNEX7mZPEK/view?usp=sharing

答:
Seems like the author in the first paper is using standard Newtonian analysis.

(EDIT: If you get sick of pendula, too, you can always look at masses on springs.)

Not sure if I'll get downvoted for this move, but I will paste some code below that solves and plots (in real time) the double pendulum using the RK4 method. Problem is, it's QB64 code, which you're welcome to attain and run yourself - but I'm sure you want the equations (just for comparison to your own) more than the program. Here it all is in context:

```<code>DEFSNG A-Z
pi = 3.141593
r = 100
a1 = -pi / 2
a2 = -pi / 2
h = 0.001
m = 1000
g = 1000
sw = 640
sh = 480

DIM p1 AS LONG
p1 = _NEWIMAGE(sw, sh, 12)
SCREEN _NEWIMAGE(sw, sh, 12)

DO
    a1p = (6 / (m * r ^ 2)) * (2 * m * v1 - 3 * COS(a1 - a2) * m * v2) / (16 - 9 * (COS(a1 - a2)) ^ 2)
    a1k1 = a1p
    a1k2 = a1p + h * a1k1 / 2
    a1k3 = a1p + h * a1k2 / 2
    a1k4 = a1p + h * a1k3
    a2p = (6 / (m * r ^ 2)) * (8 * m * v2 - 3 * COS(a1 - a2) * m * v1) / (16 - 9 * (COS(a1 - a2)) ^ 2)
    a2k1 = a2p
    a2k2 = a2p + h * a2k1 / 2
    a2k3 = a2p + h * a2k2 / 2
    a2k4 = a2p + h * a2k3
    na1 = a1 + h * (a1k1 + 2 * a1k2 + 2 * a1k3 + a1k4) / 6
    na2 = a2 + h * (a2k1 + 2 * a2k2 + 2 * a2k3 + a2k4) / 6
    v1p = -0.5 * r ^ 2 * (a1p * a2p * SIN(a1 - a2) + 3 * g * SIN(a1) / r)
    v1k1 = v1p
    v1k2 = v1p + h * v1k1 / 2
    v1k3 = v1p + h * v1k2 / 2
    v1k4 = v1p + h * v1k3
    v2p = -0.5 * r ^ 2 * (-a1p * a2p * SIN(a1 - a2) + g * SIN(a2) / r)
    v2k1 = v2p
    v2k2 = v2p + h * v2k1 / 2
    v2k3 = v2p + h * v2k2 / 2
    v2k4 = v2p + h * v2k3
    nv1 = v1 + h * (v1k1 + 2 * v1k2 + 2 * v1k3 + v1k4) / 6
    nv2 = v2 + h * (v2k1 + 2 * v2k2 + 2 * v2k3 + v2k4) / 6
    a1 = na1
    a2 = na2
    v1 = nv1
    v2 = nv2

    _DEST p1
    PSET (sw / 2 + r * COS(a1 - pi / 2) + r * COS(a2 - pi / 2), sh / 2 - r * SIN(a1 - pi / 2) - r * SIN(a2 - pi / 2)), 12
    _DEST 0
    _PUTIMAGE , p1

    LINE (sw / 2, sh / 2)-STEP(r * COS(a1 - pi / 2), -r * SIN(a1 - pi / 2))
    CIRCLE STEP(0, 0), 5
    LINE -STEP(r * COS(a2 - pi / 2), -r * SIN(a2 - pi / 2))
    CIRCLE STEP(0, 0), 5

    _LIMIT 200
    _DISPLAY
LOOP UNTIL _KEYHIT = 27
SYSTEM

</code>```

Body:
Seems like the author in the first paper is using standard Newtonian analysis.

(EDIT: If you get sick of pendula, too, you can always look at masses on springs.)

Not sure if I'll get downvoted for this move, but I will paste some code below that solves and plots (in real time) the double pendulum using the RK4 method. Problem is, it's QB64 code, which you're welcome to attain and run yourself - but I'm sure you want the equations (just for comparison to your own) more than the program. Here it all is in context:

DEFSNG A-Z
pi = 3.141593
r = 100
a1 = -pi / 2
a2 = -pi / 2
h = 0.001
m = 1000
g = 1000
sw = 640
sh = 480

DIM p1 AS LONG
p1 = _NEWIMAGE(sw, sh, 12)
SCREEN _NEWIMAGE(sw, sh, 12)

DO
    a1p = (6 / (m * r ^ 2)) * (2 * m * v1 - 3 * COS(a1 - a2) * m * v2) / (16 - 9 * (COS(a1 - a2)) ^ 2)
    a1k1 = a1p
    a1k2 = a1p + h * a1k1 / 2
    a1k3 = a1p + h * a1k2 / 2
    a1k4 = a1p + h * a1k3
    a2p = (6 / (m * r ^ 2)) * (8 * m * v2 - 3 * COS(a1 - a2) * m * v1) / (16 - 9 * (COS(a1 - a2)) ^ 2)
    a2k1 = a2p
    a2k2 = a2p + h * a2k1 / 2
    a2k3 = a2p + h * a2k2 / 2
    a2k4 = a2p + h * a2k3
    na1 = a1 + h * (a1k1 + 2 * a1k2 + 2 * a1k3 + a1k4) / 6
    na2 = a2 + h * (a2k1 + 2 * a2k2 + 2 * a2k3 + a2k4) / 6
    v1p = -0.5 * r ^ 2 * (a1p * a2p * SIN(a1 - a2) + 3 * g * SIN(a1) / r)
    v1k1 = v1p
    v1k2 = v1p + h * v1k1 / 2
    v1k3 = v1p + h * v1k2 / 2
    v1k4 = v1p + h * v1k3
    v2p = -0.5 * r ^ 2 * (-a1p * a2p * SIN(a1 - a2) + g * SIN(a2) / r)
    v2k1 = v2p
    v2k2 = v2p + h * v2k1 / 2
    v2k3 = v2p + h * v2k2 / 2
    v2k4 = v2p + h * v2k3
    nv1 = v1 + h * (v1k1 + 2 * v1k2 + 2 * v1k3 + v1k4) / 6
    nv2 = v2 + h * (v2k1 + 2 * v2k2 + 2 * v2k3 + v2k4) / 6
    a1 = na1
    a2 = na2
    v1 = nv1
    v2 = nv2

    _DEST p1
    PSET (sw / 2 + r * COS(a1 - pi / 2) + r * COS(a2 - pi / 2), sh / 2 - r * SIN(a1 - pi / 2) - r * SIN(a2 - pi / 2)), 12
    _DEST 0
    _PUTIMAGE , p1

    LINE (sw / 2, sh / 2)-STEP(r * COS(a1 - pi / 2), -r * SIN(a1 - pi / 2))
    CIRCLE STEP(0, 0), 5
    LINE -STEP(r * COS(a2 - pi / 2), -r * SIN(a2 - pi / 2))
    CIRCLE STEP(0, 0), 5

    _LIMIT 200
    _DISPLAY
LOOP UNTIL _KEYHIT = 27
SYSTEM


问:
Derangement question but you stop once you've found a matching pair
It's rather common knowledge to know that if we want to rearrange $N$ distinct items' positions such that none of them appears in its original position, then it is a derangement problem. So if we rearrange 7 distinct items' position, then the probability that none of them appears in their original position is $ \frac{!7}{7!} = \frac{1854}{5040}$.

Here's a follow-up question:

Suppose this time we have $N$ distinct items arranged from left to right. We start from the left to right and count how many items are not in their original position, and we stop counting once we have found an item in the original position. $$$$ To illustrate this, consider a permutation of the 5 items $(A,B,C,D,E)$ to be $(B,A,C,E,D)$. This tells us that we've only found 2 wrong positions, instead of 4 (because we stopped searching after discovering that $C$ is in its original position). $$$$ Let $\mathsf B_N$ denote the discrete probability distribution for the number of these "partial derangements" after permutating $N$ items. It's easy to construct $\mathsf B_N$ for small $N$'s: $$ \mathsf  B_2 = \begin{cases} 0, \quad \mathsf{Pr} = 1/2 \\ 2, \quad \mathsf{Pr} = 1/2 \end{cases} \quad\quad, \quad\quad  \mathsf  B_3 = \begin{cases}
0, \quad \mathsf{Pr} = 2/6 \\
1, \quad \mathsf{Pr} = 1/6 \\
2, \quad \mathsf{Pr} = 1/6 \\
3, \quad \mathsf{Pr} = 2/6 \\
\end{cases}   $$ Here's the directive: For all integers $n\geqslant 2$, prove the probability mass function $$ \mathsf{Pr}[\mathsf B_n = k] =
\left \{
\begin{array} {l l }
\left [ \sum \limits_{j=0}^k (-1)^j \cdot \binom kj \cdot (n-j-1)! \right ]/n! \,,& k = 0,1,2,\ldots , n-2 \\
!(n-1) /n! \, , & k=n-1 \\
!n /n! \,, & k=n
\end{array}  \right .
$$

Here are a few obvious observations:

$\mathsf{Pr}[\mathsf B_n = 0 ] $ simplifies to $ \frac1n$, which is true because this means that the first item (after shuffled) is already in the original position.

$\mathsf{Pr}[\mathsf B_n = n ] $ is simply a derangement problem.

$\mathsf{Pr}[\mathsf B_n = n-1 ] $ is simply a derangement problem as well but only the rightmost item is in the correct position.

How did I come up with this? I was playing around with slightly larger $N$'s, and with some Monte Carlo, I'm able to construct the empirical formulas for $\mathsf B_N$. Then I noticed that the numerators of the probability fractions for each of these specific $N$'s follow a nice sequence. E.g.:

$$ \mathsf   B_8 = \begin{cases}
0, \quad \mathsf{Pr} = 7!/8!\\
1, \quad \mathsf{Pr} = (7! - 6!)/8! \\
2, \quad \mathsf{Pr} =(7! - 2\cdot 6! + 5!)/8! \\
3, \quad \mathsf{Pr} = (7! - 3\cdot 6! + 3\cdot 5! - 4!)/8!\\
4, \quad \mathsf{Pr} =  (7! - 4\cdot 6! + 6\cdot 5! - 4\cdot 4! + 3!)/8! \\
5, \quad \mathsf{Pr} = (7! - 5\cdot 6! + 10\cdot 5! - 10\cdot 4! + 5 \cdot 3! - 2!)/8!  \\
6, \quad \mathsf{Pr} =  (7! - 6\cdot 6! + 15\cdot 5! - 20\cdot 4! + 15 \cdot 3! - 6\cdot 2! + 1!)/8!  \\
7, \quad \mathsf{Pr} = (!7)/8!  \\
8, \quad \mathsf{Pr} = \, !8/8! \\
\end{cases}
$$

答:
Essentially the same inclusion-exclusion argument that yields the yields the result

$$d_n=n!\sum_{i=0}^n\frac{(-1)^i}{i!}$$

also yields the result

$$\mathsf{Pr}[\mathsf{B}_n=k]=\frac1{n!}\sum_{i=0}^k(-1)^i\binom{k}i(n-i-1)!$$

for $k=0,\ldots,n-2$. For $i\in[k]$ let $S_i$ be the set of permutations of $[n]$ that fix $i$ and $k+1$. Then

$$\begin{align*}
\left|\bigcup_{i=1}^kS_i\right|&=\sum_{\varnothing\ne I\subseteq[k]}(-1)^{|I|+1}(n-1-|I|)\\
&=\sum_{i=1}^k(-1)^{i+1}\binom{k}i(n-i-1)!\,,
\end{align*}$$

So the number of permutations of $[n]$ for which $k+1$ is the smallest fixed point is

$$\begin{align*}
&(n-1)!-\sum_{i=1}^k(-1)^{i+1}\binom{k}i(n-i-1)!\\
&\qquad=\sum_{i=0}^k(-1)^i\binom{k}i(n-1-i)!\,,
\end{align*}$$

as desired.