问:
Gradient proof for implicitly defined function
I am currently working on an assignment where I am seriously stuck. I need to prove that

$\nabla f(x) \cdot x = m \cdot f(x)$,

when $f: \mathbb{R^n} \rightarrow \mathbb{R}$ is differentiable and a $m \in \mathbb{N}$ with $f(t \cdot x) = t^m \cdot f(x)$ for all $t \neq 0$ and $x \in \mathbb{R^n}$ exists. So far I could deduce that $m=1$, since

$\nabla_x f(tx) = \nabla_x t^m f(x)$

$t \cdot \nabla_x f(x) = t^m \nabla_x f(x)$

$t = t^m$

$\Rightarrow m=1$

Furthermore I was able to show that f(0) = 0.
But now I am not sure how to proceed! Does anyone have an idea how I could go on?

Thanks for helping!

答:
Hint:

In general for a differentiable $f:\Bbb{R}^n\to\Bbb{R}$, $x,v\in\Bbb{R}^n$, we have by the chain rule that
\begin{align}
\langle\nabla f(x), v\rangle = \dfrac{d}{ds}\bigg|_{s=0}f(x+sv).
\end{align}
Now, take $v=x$, and use homogeneity of $f$.


问:
How to find the perimeter of a piece of paper which has been cut?
The problem is as follows:

From a rectangular piece of paper, $4$ straight cuts have been made.
These cuts are parallel to the diagonals of the rectangle. After
making the cuts, the 4 pieces are removed. The sum of the lengths of
the four cuts made is $\textrm{80 cm}$. Find the perimeter of the
piece of paper that is obtained.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{292 cm}\\
2.&\textrm{248 cm}\\
3.&\textrm{276 cm}\\
4.&\textrm{284 cm}\\
\end{array}$

For this particular situation. I'm stuck. Does it exist a way to get this perimeter?. The reason for why I'm stuck is that it isn't given the lengths of the corners of the paper neither full sides of the chunks which has been removed. Does it exist to get those, can this problem be solved?. Help please?

答:
Note that
$$\text{New perimeter} = \text{Old perimeter}-\text{Lengths of corners}+\text{Lengths of new edges}$$
Since the cuts are parallel to the diagonal of the rectangle, and one of the angle is $90^\circ$, by AA rule, every cut triangle is similar to the triangle formed by cutting the rectangle along one of its diagonals. So the sides of the cut triangles are in the ratio $3:4:5$ (as shown in the following image). So the new perimeter is
$$2(60+80)-3(x+y+z+w)-4(x+y+z+w)+5(x+y+z+w)$$
Using the constraint that $$5(x+y+z+w)=80$$ and putting in above expression, we get, the new perimeter to be $$280-2\times\dfrac{80}5=\boxed{248\ \mathrm{cm}}$$

ContentLicense: CC BY-SA 4.0


答:
Recognize that if the cuts are parallel to the diagonals, then the side lengths of the triangles that have been cut out are in proportion to a $3-4-5$ right triangle.

When a cut is made, a length equal to the sum of the two legs of the right triangle will have been taken out of the perimeter, but the length equal to the hypotenuse will be added (i.e. for every $5cm$ cut, $7cm$ gets taken out of the perimeter, but $5cm$ will be added to the perimeter, giving a net loss of $2cm$ per $5cm$ cut).

Since we started with a perimeter of $280cm$, and $80cm$ of cuts have been made, this gives us a total loss of $32cm$, resulting in a perimeter of $248cm$.

Title: How to find the perimeter of a piece of paper which has been cut?

Body:
The problem is as follows:

From a rectangular piece of paper, $4$ straight cuts have been made.
These cuts are parallel to the diagonals of the rectangle. After
making the cuts, the 4 pieces are removed. The sum of the lengths of
the four cuts made is $\textrm{80 cm}$. Find the perimeter of the
piece of paper that is obtained.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{292 cm}\\
2.&\textrm{248 cm}\\
3.&\textrm{276 cm}\\
4.&\textrm{284 cm}\\
\end{array}$

For this particular situation. I'm stuck. Does it exist a way to get this perimeter?. The reason for why I'm stuck is that it isn't given the lengths of the corners of the paper neither full sides of the chunks which has been removed. Does it exist to get those, can this problem be solved?. Help please?


问:
How to find the perimeter of a rectangle and an scalene triangle?
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{21 cm}\\
2.&\textrm{39 cm}\\
3.&\textrm{28 cm}\\
4.&\textrm{27 cm}\\
\end{array}$

I've attempted to follow a similar strategy of assigning values to the edges of the figures mentioned.

This was done as follows:

Since it was mentioned that the figure is a square then this means that the edges measure 6 cm.

Assuming that the inner sides of the scalene triangle are $a,\,b,\,c,\,d$ and $6$ (this starting from the left in clockwise direction).

This leads to:

$a+b+c+d+6=15$

The upper segment which has been cut by the triangle I'm assigning those as:
$e,\,f,\,g$. (These starting from the left to the right).

Thus $e+f+g=6$

Up here the perimeter of the orange shaded region would be.

$3\times 6 + a+b+c+d = P$

Thus $a+b+c+d=9$

Therefore:

$P=18+9=27\,cm$

Therefore this should be the perimeter. But does it exist a better way than doing this?. Can it be obtained in a faster way?.

答: Note that the orange region's border contains three sides of the square (three quarters of the square's perimeter), plus the entire triangle's border, except for the bottom side of the square. Thus the answer is $\frac3424+15-\frac1424=\frac1224+15=12+15=27$.


答:
You can simplify the calculations, at least, by thinking of it this way:

The perimeter of the orange shape consists of three sides of the square, along with two of the sides of the triangle. In particular, it consists of
$$(\text{perimeter of triangle})+(\text{perimeter of square})-2\times (\text{side of square}).$$
So, the answer is
$$15+24-2\cdot\frac{24}{4}=27\text{ cm}.$$

Title: How to find the perimeter of a rectangle and an scalene triangle?

Body:
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{21 cm}\\
2.&\textrm{39 cm}\\
3.&\textrm{28 cm}\\
4.&\textrm{27 cm}\\
\end{array}$

I've attempted to follow a similar strategy of assigning values to the edges of the figures mentioned.

This was done as follows:

Since it was mentioned that the figure is a square then this means that the edges measure 6 cm.

Assuming that the inner sides of the scalene triangle are $a,\,b,\,c,\,d$ and $6$ (this starting from the left in clockwise direction).

This leads to:

$a+b+c+d+6=15$

The upper segment which has been cut by the triangle I'm assigning those as:
$e,\,f,\,g$. (These starting from the left to the right).

Thus $e+f+g=6$

Up here the perimeter of the orange shaded region would be.

$3\times 6 + a+b+c+d = P$

Thus $a+b+c+d=9$

Therefore:

$P=18+9=27\,cm$

Therefore this should be the perimeter. But does it exist a better way than doing this?. Can it be obtained in a faster way?.


问:
Three players A,B,C throw dice in turns
Three players A,B, and C take turns to roll a dice; they do this in the order ABCABC....
Show that the probability that the first 6 to appear is thrown by A,the second 6 to appear is thrown by B, and the third 6 to appear is thrown by C is 46656/753571.

I'm new to this problem,the best clue I could find is a similar problem which calculating the probability A throws 6 first,B second and C third,but this problem is more complicate when A could throws 6 first and cann't throws multiple 6s until B did.I don't know how to tackle this case.Can you suggested me the way to do this problem.much appreciated

答:
In order for A to roll the first $6$, he must either roll a $6$ on his first turn (probability $\frac16$) or all $3$ players must miss on their first turn, and A roll $6$ on his second turn (probability $\frac16\left(\frac56\right)^3$), or all $3$ players must miss on their first two turns, and A roll $6$ on his third turn (probability $\frac16\left(\frac56\right)^6$) and so on.

The probability that A rolls the first $6$ is $$\frac16\sum_{k=0}^\infty\left(\frac56\right)^{3k}=\frac{1/6}{1-(5/6)^3}=\frac{36}{91}$$

Now once A rolls the first $6$, the players continue rolling in the order B,C,A and we want B to roll the first $6$.  Once again, the probability that this happens is $\frac{36}{91}$.  Once B rolls his $6$ the players are rolling in the order C,A,B and the probability that C rolls the first $6$ is $\frac{36}{91}$.

The probability of the event is $$\left(\frac{36}{91}\right)^3$$ which agrees with the answer given.

Title: Three players A,B,C throw dice in turns


问:
How to find the length of an unfolded piece of paper which has been folded by its diagonal?
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{68 cm}\\
2.&\textrm{130 cm}\\
3.&\textrm{63 cm}\\
4.&\textrm{112 cm}\\
\end{array}$

For this particular problem. I'm really stuck. It seems that there isn't any way to go further because the only thing which I'm getting is that as a consequence of the lenght of the diagonal the sum of the other sides must add up to 90 hence to be 50 cm.

But I cannot say for sure which side is greater or what?. Am I missing something?. Can someone help me here?.
For reference this problem was obtained from my Reason and logic book from 2000's. And it seems to be a reprinted copy of a Martin Gardner's 70's book Puzzle Carnival with some modifications.

Since I'm not good at this it could help me a lot if it would include some drawing to see what's happening?.

答:
Draw a vertical from the apex of the blue region to the diagonal fold. You should be able to see that the two right angled triangles you create are congruent and can be solved using what you have. Furthermore, those triangles are similar to the triangle which is half the page.

I get the fourth option as my answer, and Its at least possible ie it doesn't seem crazy when you look at all the other lengths you know.


问:
In the given figure find angle x.
I approached it these ways-

1)Constructing parallel lines and then using similarity

2)Trying exterior angle property

3)Constructing perpendiculars

I am not able to understand the suitable construction.

答:
This is almost similar to Math Lover's solution...

Draw $AH$ perpendicular to $BD$. In the right triangle AHD:
$$\widehat{HAD}=90^o-60^o=30^o \quad \Rightarrow \quad HD=\frac12AD=1 $$
Draw HC and observe that triangle $DHC$ is isosceles:
$$ HD=1=DC \quad \Rightarrow \quad \widehat{DHC} = \widehat{DCH} = \frac12(180^o-120^o)=30^o$$
Now observe that triangle $HAC$ is also isosceles:
$$\widehat{HAC}=\widehat{HCA}=30^0 \quad \Rightarrow \quad HA=HC \qquad(1)$$
In triangle $DBC$ note that $\widehat{DBC}=180^o-(120^o+45^o)=15^o$.

Now, observe that triangle $HBC$ is isosceles too:
$$\widehat{HCB} = \widehat{ACB}-\widehat{ACH}=45^o-30^o=15^o $$
$$\widehat{HCB}=\widehat{HBC}=15^o \quad \Rightarrow \quad HC=HB \qquad(2) $$
From (1) and (2) we conclude that $HA=HB$ and that the right triangle $AHB$ is isosceles, which means $\widehat{HAB}=45^o$. Now we can say that
$$\widehat{BAC}=\widehat{BAH}+\widehat{HAC}=45^o+30^o=75^o $$

Title: In the given figure find angle x.


问:
Can we relate the spectrum of a non-symmetric matrix to its (skew-) symmetric component?
Let $X$ be a non-symmetric square matrix and define $S=\frac{1}{2}(X+X^T)$ and $A=\frac{1}{2}(X-X^T)$.

Is there any result that somehow relates $\lambda(X)$, the spectrum of $X$, to $\lambda(S)$ and $\lambda(A)$?

答:
Here's a result from Bhatia's Matrix Analysis that you might find useful:

Theorem VI.2.2: Let $A$ be a Hermitian matrix and $B$ a skew Hermitian matrix. Let their eigenvalues $\alpha_1,\dots,\alpha_n$ and $\beta_1,\dots,\beta_n$ be arranged is such a way that
$$
|\alpha_1| \geq \cdots \geq |\alpha_n| \quad \text{ and } \quad |\beta_1| \geq \cdots \geq |\beta_n|.
$$
Then
$$
\max_{j} |\alpha_j - \beta_{n - j + 1}| \leq \|A - B\|.
$$

In the above, $\|A\|$ denotes the spectral norm (maximal singular value) of $A$. Now, if we set $A$ in the above to be $S$ and $B$ in the above to be $-A$, then we come to the conclusion that
$$
\|X\| = \|S - (-A))\| \geq \max_j|\alpha_j - \beta_{n-j+1}| = \max_j \sqrt{\alpha_j^2 + |\beta_{n-j+1}|^2},
$$
where the $\alpha_k,\beta_k$ denote the eigenvalues of $S$ and $A$ respectively indexed as in the theorem above. Note that $\|X\|$ is larger than the spectral radius of $X$ (maximal eigenvalue magnitude of $X$).


问:
How to evaluate the integral $\int e^{x^3}dx $
How to evaluate the integral
$$\int e^{x^3}dx \quad ?$$

I've tried to set $t=x^3$, but it seems to be a blind alley; I don't know what to do with $\int\frac{e^t}{3\sqrt[3]{t^2}}dt$.

ContentLicense: CC BY-SA 3.0

Body:
another try you can solve it with Gamma function

$$\int e^{x^3}dx=\frac{-1}{3}\int e^{-t}t^{\frac{1}{3}-1}dt$$

$$\frac{-1}{3}\int e^{-t}t^{\frac{1}{3}-1}dt=\frac{-1}{3}\int_{0}^{t}e^{-t}t^{\frac{1}{3}-1}dt+c$$

$$\frac{-1}{3}\int_{0}^{t}e^{-t}t^{\frac{1}{3}-1}dt+c=\frac{1}{3}(\int_{0}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt-\int_{t}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt)+c$$

$$\frac{-1}{3}(\int_{0}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt-\int_{t}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt)+c=\frac{1}{3}\Gamma (\frac{1}{3},t)+d$$

$$\frac{1}{3}\Gamma (\frac{1}{3},t)+d=\frac{1}{3}\Gamma (\frac{1}{3},-x^3)+d$$

so

$$\int e^{x^3}dx=\frac{1}{3}\Gamma (\frac{1}{3},-x^3)+d$$

where d and c are constant


问:
Sequence of rational numbers converging to an arbitrary irrational number $\beta$.
Here is the sequence of irrational number that converges to $\alpha\in \mathbb{Q}$.
Take $x_n=\alpha -\frac{\sqrt{2}}{n}$. Clearly $\{x_n\}\rightarrow\alpha .$

But I'm trying to find a sequence of rationals that converges to arbitrary irrational number $\beta$.
Can you give me such a sequence.

答:
Consider $\{x_n\}$ where $x_n={[n\beta]+1 \over n}$  $\forall n \in \mathbb N$.

$[n\beta]<n\beta<[n\beta]+1$  $\forall n \in \mathbb N \Rightarrow \frac{[n\beta]}{n}<\beta<x_n$.

Again $\frac{[n\beta]}{n}<\beta \Rightarrow x_n=\frac{[n\beta]+1}{n}<\beta+{1 \over n}$. Thus $\beta<x_n<\beta+\frac  1n$  $\forall n \in \mathbb N$. By sandwich theorem $\lim_{n \to \infty} x_n =\beta$.


问:
How do I interpret $f=\{1+o(1)\}\phi$?

Prove: If $f$~$\phi$, then $f=\{1+o(1)\}\phi$

How do I interpret $f=\{1+o(1)\}\phi$?

I'm particularly confused because usually we write $f=o(1)$ when $f$ vanishes as $x\to\infty$ but in this case I'm not sure if I should say "$f$ is ANY FUNCTION that is the product of $\phi$ with the addition of $1$ and the CLASS of functions which vanish as $x$ tends to infinity." Are we allowed to divide both sides by $\phi$ and subtract $1$?

Since I don't understand what the equation exactly means, for proving the claim, I'm not sure if I should deal with when $\phi$ is zero, or $\phi$ tends to zero. I'd appreciate any help.

答:
Given equation means that the difference of $f$ and $\phi$ becomes negligible with respect to $\phi$ as $x\to\infty$. In mathematical terms, this means that:
\begin{equation}
\lim_{x\to\infty}\dfrac{f(x)-\phi(x)}{\phi(x)}=0
\end{equation}
As a concrete example, think of $f(x) = x$ and $\phi(x)=x+\dfrac1x$. So $f$ is asymptotically "like" $\phi(x)$.

Also note that the difference of $f$ and $\phi$ becomes negligible with respect to $f$ as well, since,

\begin{align}
\lim_{x\to\infty}\dfrac{f(x)-\phi(x)}{f(x)}&=\lim_{x\to\infty}\dfrac{f(x)-\phi(x)}{f(x)-\phi(x)+\phi(x)}\\
&=\lim_{x\to\infty}\dfrac{\ \dfrac{f(x)-\phi(x)}{\phi(x)}\ }{\ \dfrac{f(x)-\phi(x)}{\phi(x)}+1}\\
&=0
\end{align}

As another example, you can take $f(x)=x$ and $\phi(x)=x+\log x$. Here the difference of $f$ and $\phi$ is itself unbounded asymptotically, but still $f$ is "like" $\phi$ as $x$ grows large.

Title: How do I interpret $f=\{1+o(1)\}\phi$?


问:
Prove a set is open, closed, or neither
After proving $A=\left(\frac{2n+5}{3n+1}:n\in\Bbb N^+\right)= \left(\frac{7}{4},\frac{9}{7},\frac{11}{10},1,\frac{15}{16},\frac{17}{19},…\right)$ is convergent and a Cauchy sequence, I want to know if this set is an open set, closed set, or neither.

I know that the definition of an open set is "Definition 3.2.1 (Stephen Abbott “Understanding Analysis”) A set $O\subseteq \Bbb R$ is open if for all points $a \in O$ there exists an ϵ-neighbourhood $V_ϵ (a)\subseteq O$."

and for closed set is "Definition 3.2.7 (Stephen Abbott “Understanding Analysis”) A set $F\subseteq\Bbb R$ is closed if it contains its limit points"

$A$ isn't an open set because if we pick $a=1$ there's no $V_ϵ (1)$ in $A$. (Can someone please help me to word this more correctly?)

But for proving if $A$ is a closed set or not I had to find the limit points first based on the definition I used. I still don't understand how to find a limit point. By reading the definition in the book, I still couldn't understand how. Here's the definition : "Definition 3.2.4. A point $x$ is a limit point of a set $A$ if every ϵ-neighborhood
$V_ϵ(x)$ of $x$ intersects the set $A$ in some point other than $x$."

Then after proving that, I also need to find the isolated points too, so the limit point is still necessary for me to find.

答: Limit points precisely capture the intuition of a point "being the limit of a non-constant sequence". If you suspect this set is not closed (hint: you should suspect this), then what you should do is find a point outside of this set which is a limit of a sequence of points inside the set.</p>

Title: Prove a set is open, closed, or neither


问:
How can I find $P(A'\cap B')$ using De Morgan's laws?
$P(A'\cap B') = P(\overline{A\cup B})$ By De Morgan's laws.

I am given $P(A|B) = \frac{1}{5}$ and $P(B|A)=\frac{1}{4}$. Let $P(A\cap B)=p$

I want to find  $P(A'\cap B')$ in terms of $p$. Perhaps I don't need to use De Morgan's law in the first step ie convert to $P(\overline{A\cup B})$?

Anyway, I tried to sub it into the conditional probability formula so $P(\overline{A\cup B}) = P(\overline{A|B})\cdot P(\overline{B})$

Now I am stuck and I assume this method is not correct :(

Edit: In a previous section, I worked out that $P(A) = 4p$. That means I will need $P(A)$ to appear in my equation

答:
Nevermind I found the solution:

$(′∩′)=(\overline{∪})=P(A') + P(B') - P(\overline{A\cap B}) $

$=1-4p + 1-5p - (1-p)$

$= 1-8p$, as requested

However, I still welcome alternate solutions that don't initially use the De Morgan step


问:
Given functions $h,k:\Bbb R\to \Bbb R$, is it possible to determine whether $f,g:\Bbb R\to\Bbb R$ exist so that $g\circ f=h$ and $f\circ g=k$?
Let's say I have two functions $h,k:\Bbb R\to \Bbb R$.  I want to find $f,g:\Bbb R\to \Bbb R$ such that $g\circ f=h$ and $f\circ g=k$.  I know that $f,g$ might not exist (for example, Functional equation involving composition and exponents).  Do we know at least a condition for $h,k$ such that $f,g$ exist?

Which condition guarantees the uniqueness of $f,g$ (provided that they exist)?  Note that there are $h,k$ such that $f,g$ are not unique. For example, $h=k=0$, where $f=0$ works and $g$ is any function s.t. $g(0)=0$.  Or when $h=k$ is the identity function, and we take $f$ to be any bijection and $g=f^{-1}$.

At the very least, what do we know about this problem when $h,k$ are polynomial functions?  Is there a simple test that tells us there are polynomials $f,g$ that satisfy the conditions for a given pair of polynomials $h,k$?  Again, what about the uniqueness of polynomial solutions?

If the general problem is too hard, I am most interested in this specific problem. I want to find $f,g:\Bbb R\to\Bbb R$ such that
$$g\circ f(x)=x^3+1$$
and
$$f\circ g(x)=x^3+3x^2+3x+2.$$
Clearly $f,g$ are bijective functions if they exist.  So, can we determine the value of $g\circ f^{-1}(-7)$?

I found $f,g$ that almost work.  When $f(x)=x^3$ and $g(x)=x+1$, we have $g\circ f(x)=x^3+1$ but $f\circ g(x)=x^3+3x^2+3x+1$.  Unfortunately they don't quite work.  I know also that there are no polynomial functions $f,g$ that work.

Note that
$$f(x^3+1)=f(x)^3+3f(x)^2+3f(x)+2$$
and
$$g(x^3+3x^2+3x+2)=g(x)^3+1.$$
$\therefore$ if $a,b$ are the unique real numbers such that $a^3+1=a$ and $b^3+3b^2+3b+2=b$, we see that $f(a)=b$ and $g(b)=a$.  These are the only values of $f$ and $g$ that I know.  But I can also see that
$$ f^{-1}(-7)=g(-3)$$
if that helps.

Let $h(x)=x^3+1$ and $k(x)=x^3+3x^2+3x+2$.  Due to $f\circ g(x)$ and $g\circ f(x)$ are given; find $f$ and $g$, if $f=f_0$ and $g=g_0$ satisfy the conditions, then $f=f_0\circ \phi$ and $g=\phi^{-1}\circ g_0$ form a solution for any bijection $\phi:\Bbb R\to\Bbb R$ such that $h\circ \phi=\phi\circ h$.  Because any iteration of $h$ commutes with $h$, we can see that there are infinitely many $f$ and $g$, if $f_0,g_0$ exist.  How do I see whether $f_0,g_0$ exist?

答:
If $h= g\circ f$ and $k= f\circ g$, one of the $h,k$ is surjective, and the other injective, then $f$, $g$, $h$, $k$ are all bijective and
$$k = f\circ h \circ f^{-1}$$,
that is $h$, $k$ are conjugate. Conversely, if $h$, $k$ are conjugate, then you can find $f$, and then $g$. Now, the conjugation is an equivalence relation.

Now in our example $h(x) = x^3+1$, $k(x) = (x+1)^3 + 1$, so $k(x-1) + 1 = x^3+2$, a conjugate of $k$. So now we want to see whether $h_1(x) = x^3+1$ and $h_2(x) =x^3+2$ are conjugate. Note that both have a unique fixed point $\xi_1$, $\xi_2$, and for $x> \xi_i$ we have $h_i^{n}(x) \to \infty$ as $n\to \infty$, $h_i^{n}(x) \to \xi_i$, as $n\to -\infty$, while for $x< \xi_i$, we have $h_i^{n}(x) \to -\infty$ as $n\to \infty$, $h_i^{n}(x) \to \xi_i$, as $n\to -\infty$. Therefore, all the orbits of $h_i$ -except the one containing the fixed point- are infinite. So there exists a bijection $\phi\colon \mathbb{R}\to \mathbb{R}$ such that $h_2= \phi\circ h_1\circ \phi^{-1}$.  It is clearly not unique, so a nice $\phi$ would be desired. Note that $\phi$ takes the fixed point of $h_1$ to the fixed point of $h_2$.

It appears that both $h_1$, $h_2$ behave like the map $x\to 2 x$. Are they topologically conjugate to it?  Note that $l(x) = 2x$ is part of a $1$-parameter group of diffeomorphism of $\mathbb{R}$, $(t,x)\mapsto 2^{t}\cdot x$. If $h_1$, $h_2$ are conjugate to $l$, then they are also each part of a $1$-parameter group of homeomorphisms of $\mathbb{R}$. In particular, there exists $\psi$ a homeomorphism of $\mathbb{R}$ such that $\psi\circ \psi(x) = x^3+1$.  What would be such a homeomorphism?

$\bf{Added:}$ The case where both $k$, $k$ are bijections is simpler, it reduces to the question of when two maps are conjugate under a bijection.  They are if and only if the "graph" of the maps are isomorphic, where the graph consists of  vertices $x$, and edges $(x, h(x))$.  For bijections, their cycle structure has to be the same.

Consider for instance the maps $x\mapsto 2 x$, and $x\mapsto 4 x$. They are conjugate under the bijection $x\mapsto x^{2_+}\colon = x^2 \operatorname{sign} x$.  The maps $x\mapsto 2x$, and $x\mapsto 3x$ are conjugate under the map $x\mapsto x^{\log_2 3_+}$.


问:
Are boolean operations unique?
Are the well known operations of logic sum and logic product (defined by their two truth tables) the  unique couple of operations (defined on a two elements set) that realize the axioms of:
associativity, commutativity, absorption, identity, distributivity, complements?

(The axioms are reported here in section "definition")

答:
Welcome to Math.SE!

The usual truth value operations (disjunction, conjunction, negation, false, true) constitute the only structure $(\vee, \wedge, \neg, 0, 1)$ that realizes the properties of absorption, complements, commutativity and identity on a two-element set $\{0,1\}$. We show this by proving that the axioms determine the truth tables of $\neg, \vee, \wedge$.

First, observe that by complements $x \vee \neg x  = 1$, and by absorption $x \vee 0 = 0$. Thus, if we had $\neg 0 = 0$, we could set $x$ to $0$ and obtain $0 = 0 \vee 0 = 0 \vee \neg 0 = 1$, a contradiction. Similarly, we get $\neg 1 \neq 1$. Thus $\neg$ has to coincide with the usual negation.

By the identity $x \wedge 1 = x$ and commutativity we have $0 \wedge 1 = 1 \wedge 0 = 0$, and $1 \wedge 1 = 1$.

By the identity $x \vee 0 = x$ and commutativity, we have $0 \vee 0 = 0$ and $0 \vee 1 = 1 \vee 0 = 1$.

From here on, we get $0 \wedge 0 = 0 \wedge (0 \vee 0) = 0$ by absorption $x \wedge (x \vee y) = x$, and similarly $1 \vee 1 = 1 \vee (1 \wedge 1) = 1$ by absorption $x \vee (x \wedge y) = x$.

Having determined all values of the functions $\neg, \vee, \wedge$ on $\{0,1\}$, we see that they coincide with the usual truth value operations, as we claimed.

Title: Are boolean operations unique?


问:
Largest ellipsoid section
Suppose that we have an ellipsoid in $\mathbb{R}^3$ with centre at $0$. We consider all plane sections that pass through $0$ (which are all ellipses). Of all these ellipses which has the largest perimeter?

What happens in $\mathbb{R}^n$?

答:
For any geometric figure $X$ in a plane, let $\ell_X$ be its perimeter.
Recall

For any convex bodies $A, B$ in a plane, if $A \subseteq B$, then $\ell_A \le \ell_B$.

Let $\hat{z} = (0,0,1)$ and

$\mathcal{E}$ be the solid ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1$ with $a \ge b \ge c$.

$\mathcal{E}'$ be the solid ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{b^2} \le 1$.

For any unit vector $\hat{n} \in S^2$, let  $\Omega(\hat{n})$ and $\Omega'(\hat{n})$ be the intersection of $\mathcal{E}$ and $\mathcal{E}'$ with the plane $\hat{n}\cdot \vec{r} = 0$.
It is clear $\Omega(\hat{n})$ and $\Omega'(\hat{n})$ are solid ellipses, lying in same plane with $\Omega(\hat{n}) \subseteq \Omega'(\hat{n})$. Above result tell us $$\ell_{\Omega(\hat{n})} \le \ell_{\Omega'(\hat{n})}$$

Furthermore, $\Omega'$ is an ellipse with semi-minor axis $b$ and semi-major axis $a' \le a$. By suitable rigid motion, we can "move" and "fit" it inside $\Omega'(\hat{z})$. This means

$$\ell_{\Omega'(\hat{n})} \le \ell_{\Omega'(\hat{z})}$$

Combine these and notice $\Omega(\hat{z}) = \Omega'(\hat{z})$, we obtain

$$\ell_{\Omega(\hat{n})} \le \ell_{\Omega'(\hat{n})} \le \ell_{\Omega'(\hat{z})} = \ell_{\Omega(\hat{z})}$$

So the ellipses with largest parameter is the intersection of $\mathcal{E}$ with $xy$-plane (i.e. the plane containing the two longest axes).

The generalization to $\mathbb{R}^n$ for $n > 3$ is sort of obvious. One just replace

$\mathcal{E}$ by $\sum\limits_{k=1}^n \frac{x_k^2}{a_k^2} \le 1$ with $a_1 \ge a_2 \ge \cdots \ge a_n$.

$\mathcal{E}'$ by $\frac{1}{a_1^2}x_1^2 + \frac1{a_2^2}\sum\limits_{k=2}^n x_k^2 \le 1$.

Aside from one can no longer label the planes by unit vectors, above arguments continue to work.

Title: Largest ellipsoid section


问:
Proof regarding the limit of arctanx.

So we're very limited in what we can and cannot use regarding properties of trigonometric functions since we haven't proven much, so I was forced to proof it the long way which invites room for mistakes. I was wondering if you have any tips regarding shortening the proof or if something in the proof doesn't add up..

Consider the function $ f: R \to R \ $, $f(x) = \arctan(x).$

First we show the limits of $f$ exist.

Claim: $\lim _{x\to \infty }\left(\arctan\left(x\right)\right)=\frac{\pi }{2}$

Proof: Let $ϵ > 0$.

We choose $M = \max \left\{\tan\left(\frac{\pi}{2}-ϵ\right),1\right\}>0$. Then for every $x>M$ we have $|\arctan(x)-\frac{\pi}{2}|<ϵ$

Note that we've seen in class the function tan is strictly increasing for every $x ∈ (-\frac{\pi}{2},\frac{\pi}{2})$,

and since arctan is the inverse of tan, it's also strictly increasing.

Therefore ★$\arctan(x) > \arctan(y) \implies -\arctan(x) < -\arctan(y)$ for every $x > y$.

Also note we've seen in class that $-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$ for every $x ∈ R \ $, therefore $\arctan(x) - \frac{\pi}{2} < 0$.

$$\begin{align}|\arctan(x)-\frac{\pi}{2}|&=\frac{\pi}{2}-\arctan\left(x\right)\\
&\frac{★}{<} \ \frac{\pi}{2}-\arctan\left(M\right)\\
&=\frac{\pi}{2}-\arctan\left(\tan\left(\frac{\pi}{2}-ϵ\right)\right)\\
&=\frac{\pi}{2}-\frac{\pi}{2}+ϵ\\
&≤ϵ\end{align}$$

Therefore the limit holds.

Claim: $\lim _{x\to -\infty }\left(\arctan\left(x\right)\right)=-\frac{\pi }{2}$

Proof: Let $ϵ > 0$.

We choose $M = max \left\{\tan\left(-\frac{\pi}{2}+ϵ\right),-1\right\}>0$. Then for every $x<M$ we have $|\arctan(x)+{\pi}{2}|<ϵ$

Note we've seen in class that $-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$ for every $x ∈ R \ $, therefore $\arctan(x) + \frac{\pi}{2} > 0$.
$$\begin{align}|\arctan(x)-\frac{\pi}{2}|&=\frac{\pi}{2}+\arctan\left(x\right)\\
&\frac{★}{<} \ \frac{\pi}{2}+\arctan\left(M\right)\\
&=\frac{\pi}{2}+\arctan\left(\tan\left(-\frac{\pi}{2}+ϵ\right)\right)\\
&=\frac{\pi}{2}-\frac{\pi}{2}+ϵ\\
&≤ϵ\end{align}$$
Therefore the limit holds.

Now we show that $f$ has no extreme points.

Note that $f$ is strictly increasing, therefore

for every $x,y ∈ R$ such that $x < y$, we have $f(x) < f(y)$.

Assume toward contradiction that $f$ has a maximum, we will denote it $a$.

Since f is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,

in contradiction to the minimality of $a$.

Assume toward contradiction that $f$ has a minimum, we will denote it $c$.

Since f is strictly increasing, there exist a point $d ∈ R$ such that $d < c$ for which $f(d) < f(c)$,

in contradiction to the minimality of $c$.

答:
Consider the function $ f: R \to R \ $, $f(x) = \arctan(x).$

A much easier choice would have been something like
$$f(x) = \begin{cases}-\dfrac 1x,&\quad x\le -3\\
\dfrac 12 +\dfrac x{18},&\quad-3 < x < 3\\
1 - \dfrac 1x,&\quad 3\le x\end{cases}$$

Continuity is easy to establish (particularly if you already have the theorem about piecing together continuous functions that agree at the seam giving a continuous result). And it is also simple to show that it is strictly increasing. Showing the limits are $0$ and $1$ is also much simpler than yours.

Perhaps even easier would be $$f(x) = \dfrac x{|x| + 1}$$

Overall, your proof is good. Almost none of the things I bring up are logical errors, and the exceptions are trivial technical issues. But they do distract from your argument.

Claim: $\lim _{x\to \infty }\left(\arctan\left(x\right)\right)=\frac{\pi }{2}$

Proof: Let $ϵ > 0$.

You need to restrict $0 < \epsilon < \pi$ here. Otherwise some of your later claims become false. This is not an issue for the definition of a limit, because if something is $<$ this restricted $\epsilon$, it is also $<$ all larger values of $\epsilon$, which is all that is required for them.

We choose $M = \max \left\{\tan\left(\frac{\pi}{2}-ϵ\right),1\right\}>0$.

You make no use of the restriction that $M \ge 1$. In fact, later on, you ignore the possibility that $M$ could be $1$ instead of $\tan\left(\frac{\pi}{2}-ϵ\right)$. You should just set $M = \tan\left(\frac{\pi}{2}-ϵ\right)$ instead.

Then for every $x>M$ we have $|\arctan(x)-\frac{\pi}{2}|<ϵ$

Note that we've seen in class the function tan is strictly increasing for every $x ∈ (-\frac{\pi}{2},\frac{\pi}{2})$, and since arctan is the inverse of tan, it's also strictly increasing.

Note that you've already made use once of the increasing behavior of $\tan$ in saying that for $x > M, \left|\arctan x - \frac \pi2\right| < \epsilon$

Therefore ★$\arctan(x) > \arctan(y) \implies -\arctan(x) < -\arctan(y)$ for every $x > y$.

This is convoluted, and all you really need from it is that if $x > M$, then $\arctan x > \arctan M$, and therefore $\frac \pi2 - \arctan x < \frac \pi2 - \arctan M$. I suggest explaining that directly.

Also note we've seen in class that $-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$ for every $x ∈ R \ $, therefore $\arctan(x) - \frac{\pi}{2} < 0$.

You need to insert "Let $x > M$" here, as your last reference to $x$ is as an arbitrary element of $\Bbb R$. But you are about to do something that requires $x > M$.

$$\begin{align}|\arctan(x)-\frac{\pi}{2}|&=\frac{\pi}{2}-\arctan\left(x\right)\\
&\frac{★}{<} \ \frac{\pi}{2}-\arctan\left(M\right)\\
&=\frac{\pi}{2}-\arctan\left(\tan\left(\frac{\pi}{2}-ϵ\right)\right)\\
&=\frac{\pi}{2}-\frac{\pi}{2}+ϵ\\ &≤ϵ\end{align}$$

Therefore the limit holds.

Other than the weird symbol where you apparently mean $\le$, and forgetting that $M$ could be $1$, this is good.

Claim: $\lim _{x\to -\infty }\left(\arctan\left(x\right)\right)=-\frac{\pi }{2}$

The whole repeat proof is unnecessary. I don't know which of the many possible developments of trigonometric functions your course follows, but in all of them the fact that $\sin$ is odd and $\cos$ is even is almost trivial. From this it follows that $\tan$ is odd and therefore (with proper choice of domain) so is $\arctan$. Hence $$\lim_{x \to -\infty} \arctan x = \lim_{t \to \infty} \arctan(-t) = -\lim_{t\to\infty} \arctan t = -\frac\pi 2$$

Now we show that $f$ has no extreme points.

Note that $f$ is strictly increasing, therefore

for every $x,y ∈ R$ such that $x < y$, we have $f(x) < f(y)$.

Assume toward contradiction that $f$ has a maximum, we will denote it
$a$.

The way this is stated, $a$ would be the maximum value that $f$ takes on, but what you mean is "$f$ takes on a maximum at $a$", so it is $f(a)$ which is the maximum, not $a$.

Since f is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,

in contradiction to the <s>minimality</s> [maximality] of $a$.

Other than the wording errors, this argument and the minimum are correct, but overly verbose. Simply noting that for any $x$, if $y < x$ then $f(y) < f(x)$ and if $y > x$ then $f(y) > f(x)$ is sufficient to recognize that $x$ cannot be a maximum or minimum, even locally (because there are always $y$ above or below $x$ - if the domain were closed and bounded instead, then $f$ would have extrema at the endpoints).

Title: Proof regarding the limit of arctanx.


问:
Study of Set theory: Book recommendations?
Can you suggest a good book for set theory?

I have just started reading about Group theory and want to learn set theory on my own.

Thanks in advance

答: The best rigorous treatment that I’ve seen at the senior undergraduate/first-year graduate level is Karel Hrbacek & Thomas Jech, Introduction to Set Theory. The first edition was already very good, and the third has been expanded to include some important topics that were not originally covered. I agree with the very favorable Amazon review by Michael Greinecker (who also posts here at Math.SE).


答:
We can see that a $3-4-5$ right triangle is created here.

Knowing the hypotenuse of $3-4-5$, the lengths of the other two sides are $\frac{3+4}5=1.4$ times the length of the hypotenuse.

Since the triangle above and the triangle equal to half the piece of paper are similar, we can conclude that the length and width of the paper are $1.4$ times the length of the diagonal.  Thus, we can calculate:
$$40\times1.4\times2=112.$$

Title: How to find the length of an unfolded piece of paper which has been folded by its diagonal?

Body:
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{68 cm}\\
2.&\textrm{130 cm}\\
3.&\textrm{63 cm}\\
4.&\textrm{112 cm}\\
\end{array}$

For this particular problem. I'm really stuck. It seems that there isn't any way to go further because the only thing which I'm getting is that as a consequence of the lenght of the diagonal the sum of the other sides must add up to 90 hence to be 50 cm.

But I cannot say for sure which side is greater or what?. Am I missing something?. Can someone help me here?.
For reference this problem was obtained from my Reason and logic book from 2000's. And it seems to be a reprinted copy of a Martin Gardner's 70's book Puzzle Carnival with some modifications.

Since I'm not good at this it could help me a lot if it would include some drawing to see what's happening?.


问:
What kind of properties can be defined over the fundamental matrix of absorbing random walks?
A connected d-regular graph $G = (V,E,d)$ is given, when each node has exactly d outlinks reaching the other nodes. Also, a function $l$ maps nodes with positive and negative labels, where $V_+$ are the nodes having positive label (+1) and $V_-$ are the ones having negative label (-1). Assuming to have a random walk, where nodes with negative labels are considered as absorbing we can generate the fundamental matrix $\mathbf F$, where $\mathbf F= (\mathbf I - A_{++})^{−1}$, where $A_{++}$ is the matrix given by the transient (and positive) nodes.

I'm trying to understand which kind of properties can be defined over the matrix $\mathbf F$. In general, I would to understand if there is any kind of mapping with the adj matrix, like inequalities based on graph topology (e.g. neighbors or not neighbors may present higher or lower value). I would like to find out any nice property that can link edges distributions with distribution of values over $\mathbf F$.

答:
I think the most important property from a practical/computational point of view is that $A_{++}$ and $F$ commute and share eigenvectors.  The Perron-Frobenius eigenvectors carry a lot of information about the network.  You can write
$$ A_{++} = \sum_j \lambda_j v_j u^T_j $$
where $v_j$ and $u^T_j$ are the right and left eigenvectors associated with $\lambda_j$.  Then
$$ F = \sum_j \frac{1}{1 - \lambda_j} v_j u^T_j $$

For networks which do not have a lot of community structure, the Perron-Frobenius eigenvector is much bigger than the others, which means that in both of these expressions you can write
$$ A_{++} \approx \lambda_1 v_1 u^T_1 $$
and
$$ F \approx \frac{1}{1 - \lambda_1} v_1 u^T_1 $$
reducing everything to rank-1 matrices.  This can be incredibly helpful as an approximation.  See a couple of my other answers here and here for some more comments on the Perron-Frobenius eigenvector, random networks, and community structure.  Note that even if there is community structure, then you may end up with two or three eigenvalues dominating the others, in which case you just need to retain two or three terms in the approximation.

Unfortunately, because $A_{++}$ only represents a subset of the whole network, then we can't say anything nice about the eigenvectors.  (For the full transition matrix of any Markov process, then $1^T$ has to be a left eigenvector, but this is not true of $A_{++}$.)  So, in a computational setting, you would have to calculate them (using power iteration, the Lanczos method, or full diagonalization) in general.

However, there are additional simplifying assumptions that would allow you to approximate the Perron-Frobenius eigenvalue and eigenvectors using just degree information of the nodes.  This paper explores several such approximations.

Your question doesn't have enough context for me to know whether this is the kind of answer you are looking for, or whether you are interested in a more rigorous of graph-theoretical invaraints.  But I hope it is useful.


答:
By definition $o(f), x \to x_0$ is set of functions:
$$o(f) = \{g \colon \exists \delta>0, \exists \varepsilon(x), \lim\limits_{x \to x_0}\varepsilon(x)=0, \forall x \in U_\delta(x_0), g(x)= \varepsilon(x) \cdot f(x) \}$$

So, $f=\{1+o(1)\}\phi \Leftrightarrow f-\phi = o(\phi)$ means, that $f-\phi = \phi \varepsilon$, where $\varepsilon \to 0$.  Say strictly $f-\phi \in o(\phi)$.

Answering your question generally $\phi$ can have zeros in $U_\delta(x_0)$, which means we cannot divide both sides on $\phi$. For sure we can say only that if holds $f=\{1+o(1)\}\phi$, then  $f-\phi$ have same zeros as $\phi$.

Title: How do I interpret $f=\{1+o(1)\}\phi$?

Body:
Prove: If $f$~$\phi$, then $f=\{1+o(1)\}\phi$

How do I interpret $f=\{1+o(1)\}\phi$?

I'm particularly confused because usually we write $f=o(1)$ when $f$ vanishes as $x\to\infty$ but in this case I'm not sure if I should say "$f$ is ANY FUNCTION that is the product of $\phi$ with the addition of $1$ and the CLASS of functions which vanish as $x$ tends to infinity." Are we allowed to divide both sides by $\phi$ and subtract $1$?

Since I don't understand what the equation exactly means, for proving the claim, I'm not sure if I should deal with when $\phi$ is zero, or $\phi$ tends to zero. I'd appreciate any help.


问:
Understanding the answers of $x^2 \equiv 28 \pmod 6$

Solving  $x^2 \equiv 28 \pmod 6$

The answers are: $x \equiv 2 \pmod 6$ and $x \equiv 4 \pmod 6$.

When I plugged in to check the answers, it would be $2^2 \equiv 4 \pmod 6$ and $4^2 \equiv 16 \equiv 4 \pmod 6$. It is $4$ but not $28$, why is this, I don't understand? Anyone can help me understand this? Thanks a lot!

答:
Since $4\equiv28\pmod6$,

$x^2\equiv4\pmod 6\iff x^2\equiv28\pmod6$.


问:
Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$
Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$

If we let $X$ be the number of distinct outcomes in a given toss.

E.g. if we get the toss $1,2,2,4$ then $X=3$

Calculating some probabilities:

$P(X=1) = 6 \cdot 6^{-4}$ since we have six ways we can end up with all four dice having the same value face-up.

Now according to a solution I have for this:

$\displaystyle P(X = 2) = 6^{-4}\cdot 6 \cdot 5 \cdot \binom{4}{2}$

This is where I am not so sure; why $4$ choose $2$ as opposed to $6 $ choose $2$ ?

$P(X= 4) = 6^{-4}\cdot 6 \cdot 5 \cdot 4 \cdot 3$

and finally $P(X = 3) $ I haven't calculated as I am not clear on $P(X=2)$.

The issue for me seems to be the combinatorial part of the argument. i.e. The number of ways each event occurs.

答:
Your outcomes will be a quadruple ($X{=}1$), one triple and one single ($X{=}2$), two pairs (also $X{=}2$), one pair and two singles ($X{=}3$), or four singletons ($X{=}4$).

To evaluate the individual probabilities, select faces, then places.

$$\begin{align}\mathsf P(X{=}1)&=\frac 1{6^4}\binom 61\binom 44 &&\text{four of a kind}\\&=6/1296\\\mathsf P(X{=}2) &=\frac 1{6^4} \left(\binom 62\binom42\binom 22+\binom 61\binom43\,\binom 51\binom 11\right)&&\text{two pair or three of a kind}\\&=210/1296\\\mathsf P(X{=}3) &=\frac 1{6^4} \binom 61\binom 42\,\binom 52\binom 21\binom 11&&\text{one pair and two singletons}\\&=720/1296\\\mathsf P(X{=}4)&=\frac{1}{6^4}\binom 64\binom 41\binom 31\binom 21\binom 11&&\text{four singles}\\&=360/1296\end{align}$$

If you plot a histogram, there would be a strong peak at $X{=}3$ with a little skew to the right, so we should anticipate the expected value to lie just above $3$.   And indeed...

$$\mathsf E(X)=\dfrac{6+2\cdot210+3\cdot720+4\cdot360}{1296}=\dfrac{4026}{1296}=\dfrac{671}{216}$$

However, as stated by other answers, its easier to evaluate the expected count of faces shown.$$\begin{align}\mathsf E(X)&=\sum_{k=1}^6\mathsf P(X_k{>}0)\\&=6\cdot \left(1-\frac {5^4}{6^4}\right)\\&=\dfrac{6^4-5^4}{6^3}\\&\approx 3.1\end{align}$$


答:
First, note that because of symmetry, the triangle $AMB$, whose perimeter is 90cm, is isosceles. Therefore $AM=BM=\frac12(90-AB)=25cm$

Next, draw $MH$ perpendicular to $AB$ and note again by symmetry we have $AH=\frac12AB=20cm$. The sides of the right triangle $AHM$ are $25, 20\; \& MH=15$. You could calculate $MH$ using Pythagoras' theorem, or you could simply note that the lengths of the sides are 5 times those of the famous 3-4-5 right triangle.

Now, note the similarity of triangles $ADB$ and $AHM$:
$$\frac{AB}{AM} = \frac{AD}{AH} = \frac{DB}{HM}$$
$$AD=AH \frac{AB}{AM}=20 \times \frac{40}{25}=32 cm$$
$$DB=HM \frac{AB}{AM}=15 \times \frac{40}{25}=24 cm$$
Therefore the perimeter of the unfolded rectangle is $2 \times (32+24) = 112cm.$


问:
Calculating covariance of a jointly distributed probability function- solution verification

Let $X$ denote the size of a surgical claim and let $Y$ denote the size of the associated $2$
hospital claim. An actuary is using a model in which $E(X) = 5$, $E(X ) = 27.4, E(Y) = 7, 2
E(Y ) = 51.4$, and $Var(X+Y) = 8$. Let $C_1 = X+Y$ denote the size of the combined claims before the application of a $20\%$ surcharge on the hospital portion of the claim, and let $C_2$ denote the size of the combined claims after the application of that surcharge.
Calculate $Cov(C1, C2)$.

My attempt:

Note that $C_2 = 1.2 C_1$, so $$Cov(C_1, C_2) = Cov(C_1, 1.2C_1) = E(C_1\cdot 1.2C_1) - E[C_1] \cdot E[1.2C_1]$$
$$ = 1.2 E[C_1^2] - 1.2 \{E[C_1]^2\} = 1.2 \left\{E[C_1^2] - E[C_1]^2\right\} = 1.2 Var[C_1] = 1.2 Var(X+Y) =$$
$$ 1.2 \times 8 = 9.6$$

This is not correct. Can someone please point out why that is?

答:
$C_2\ne 1.2C_2$

The question says "20% surcharge on the hospital portion of the claim"

That is $20 $ % of $Y = 0.2Y$

Then it says
" C2 denote the size of the combined claims after the application of that surcharge"
Combined claim is addition of $C_1+0.2Y$

so

$C_2=X+Y+0.2Y=X+1.2Y$

$COV(C1, C2) = COV(X+Y, X+1.2Y ) = COV(X, X)+COV(X, 1.2Y )+COV(Y, X)+COV(Y, 1.2Y )$

$COV(X_1+X_2,X_1+X_2)=COV(X_1,X_1)+COV(X_1,X_2)+COV(X_2,X_1)+COV(X_2,X_2)$

$COV(aX_1,bX_2)=abCOV(X_1,X_2)$

$COV(X,X)=V(X)$

$Var X + 1.2 COV(X, Y ) + COV(X, Y ) + 1.2 V(Y) = Var X + 2.2 COV(X, Y ) + 1.2 V(Y)$

Can you do next?


问:
Given $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$, find the value of $f(1009, 2019)$

Let $f:(\mathbb{N}\ \cup \{0\})^2$ $\rightarrow\mathbb{N}\ \cup\{0\}$ be such that $f(0, x) = f(x, 0) = 1$ for all x $\in$ $\mathbb{N}\ \cup$ {$0$}.
Also for all $i, j \in\ \mathbb{N}$ we have $f(i, j) = f(i − 1, j − 1) + f(i, j − 1)$. Find $f(1009, 2019)$.

What I've tried:

I took two cases: $i>=j$ and $i<j$.

As $j$ is reduced by $1$ in both $f(i-1,j-1)$ and $f(i,j-1)$, if $j<=i$, then $j$ will reach $0$ first irrespective of $i$. So the value of $f(i,j)$ is $2^{j}$. Here's a branch diagram to illustrate what I meant

So $f(3,2)=2^2=4$

However I'm stuck at finding the pattern for the other case ($j>i$). Example for $f(i,j)$ where $i<j$:

$f(2,3)=7$

I do recognize that I can stop evaluating once I reach a point where $i=j$ but I can't find at which places and how many times they reach that point. It would be highly helpful If you would help me find the pattern. Also, am I going in the right direction? Is there any other way to solve this problem(a formula or something)? Help would be highly appreciable!

答:
A slightly simpler solution based on VIVID's work (it's essentially the same).

If we utilize the generalized binomial coefficients where $$\binom{n}{k} = 0, n, k \in \mathbb Z, k > n \text { or } k<0$$ then VIVID's formula becomes $$f(m,n) = \sum_{k=0}^m \binom{n}{k} \tag 1$$

We also know the following recursive relation is true for these generalized coefficients as well:
$$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \tag 2$$

Then, since $f$ is uniquely determined by $$f(0,x)=f(x,0)=1 \tag 3$$ and $$f(i, j) = f(i − 1, j − 1) + f(i, j − 1) \tag 4$$
we only need to show $(3)$ and $(4)$ are true when we plug in VIVID's formula $(1)$. Now, $(3)$ is obviously true. For $(4)$, indeed we have (via $(2)$)

$$f(i,j) = \sum_{k=0}^i \binom{j}{k} = \sum_{k=0}^i \binom{j-1}{k-1} + \sum_{k=0}^i \binom{j-1}{k}  \\
= \sum_{k=1}^i \binom{j-1}{k-1} + f(i,j-1)\\
= \sum_{k=0}^{i-1} \binom{j-1}{k} + f(i,j-1)\\
= f(i-1,j-1) + f(i,j-1).\blacksquare $$


问:
Derivatives with rational numbers
A definite integral written entirely in terms of rational numbers and functions of rational numbers does not always equal a rational number. For example, with the function $1/x$,
$$
\int_1^2 \frac{dx}{x}  = \ln 2$$

I am curious whether or not this is also true for derivatives. Given a function $f$ for which the image of $\mathbb Q$ is a subset of $\mathbb Q$, is the image of $\mathbb Q$ under $f'$ always a subset of $\mathbb Q$? I have verified that this is true for every function written as a combination of arithmetic operations, but I have not verified it for all rational functions.

答:
No; this is not true. One has a lot of freedom constructing differentiable functions $\mathbb Q\rightarrow \mathbb Q$ - this problem is susceptible to the usual tricks of enumerating the rational numbers and nudging them into place one by one.

For any desired real number $\alpha$, we will now construct a continuously differentiable $f:\mathbb R\rightarrow\mathbb R$ such that $f'(0)=\alpha$ and $f(\mathbb Q)\subseteq \mathbb Q$.

In broad strokes our proof will work as follows: we fix an enumeration $c_1,c_2,c_3,\ldots$ of the rational numbers, then prove the following:

Lemma: There is some sequence of continuously differentiable functions $f_0,f_1,f_2,f_3,\ldots$ such that:

$|f_i'(x)-f_{i+1}'(x)| < \frac{1}{2^i}$ for all $i$ and $x$.

$f(0)=0$.

$f'(0)=\alpha$.

$f_i(c_i)\in\mathbb Q$.

If $j\leq i$ then $f_{i}(c_j)=f_{i+1}(c_j)$.

From this lemma, it is straightforwards to see that $f(x)=\lim_{n\rightarrow\infty}f_n(x)$ is such that $f(\mathbb Q)\subseteq \mathbb Q$ and $f'(0)=\alpha$.

To construct such a function, we first need to choose some continuously differentiable function $t$ supported on $(-1,1)$ such that $t(0)=1$ - which are easy enough to come by, so we will not specify which. We also need to choose a sequence of radii $r_i$ such that the intervals $(c_i-r_i,c_i+r_i)$ don't contain any elements $c_j$ for $j<i$ and also don't contain $0$.

From this, we may define a function $t_i(x)=t\left(\frac{x-c_i}{r_i}\right)$ and propose that we define the desired sequence by, once we have constructed the first $i$ terms $f_0,f_1,\ldots,f_i$, extending the sequence by setting
$$f_{i+1}=f_i+\beta t_{i+1}$$
noting that this definition satisfies condition (2), (3), and (5) by choice of  radii $r_i$, satisfies condition (1) for all sufficiently small $\beta$ and satisfies (4) for every value in the coset $\mathbb Q - f_i(c_{i+1})$ - but, of course, we can choose a sufficiently small value in that coset, thus can always  extend such a sequence by one term. Note also that we can choose $f_0(x)=\alpha x$ as a starting point to this construction, as this sequence of one function satisfies the stated conditions. Therefore, we can inductively construct an infinite sequence satisfying the conditions and note that the limit of this sequence has the desired properties.

Title: Derivatives with rational numbers


问:
proving $n!>2^n\;\;\forall \;n≥4\;$ by mathematical induction
My teacher proved the following $n!>2^n\;\;\;\forall \;n≥4\;$ in the following way

Basis step: $\;\;4!=24>16$ ok

Induction hypothesis: $\;\;k!>2^k$

Induction step: $\qquad\qquad(k+1)!=k!(k+1)>(k+1)2^k>2^k\cdot 2=2^{k+1}$

I wonder how did he assume that $2^k(k+1)>2^{k}\cdot 2\quad\forall k≥4$?

Don't we have to show it by induction too?

答:
We need only take advantage of the proof's hypothesis, when we assume from the start that $k \geq 4$, so those are the only values of $k$ that need to be considered. Clearly,  $$\forall\;k\geq 4 \implies k + 1 \geq 4 + 1 =  5 > 2$$   </p>

<p>This is where we get that $$2^k \cdot \underbrace{(k + 1)}_{\large > 2} \;\gt \; 2^k \cdot 2 = 2^{k+1},\quad\text{as desired}.$$ </p>

Body:
We need only take advantage of the proof's hypothesis, when we assume from the start that $k \geq 4$, so those are the only values of $k$ that need to be considered. Clearly,  $$\forall\;k\geq 4 \implies k + 1 \geq 4 + 1 =  5 > 2$$

This is where we get that $$2^k \cdot \underbrace{(k + 1)}_{\large > 2} \;\gt \; 2^k \cdot 2 = 2^{k+1},\quad\text{as desired}.$$


问:
Out of 6 apples, 5 mangoes and 4 bananas how many selection of fruits can be made such that at least one fruit from each type is always included
According to me the answer should be $(6)(5)(4)-1= 119$ but given answer is $120$

I subtracted the 1 to eliminate the case where none of the fruits are picked.

Is my answer correct?

答:
The formula you are mentioning i.e $(p+1)(q+1)(r+1)2^s -1$ is when the total number of fruits is greater than $0$.
 Here fruit of each type is greater than $0$.

Number of ways of selecting at least 1 apple = 6

Number of ways of selecting at least 1 mange = 5

Number of ways of selecting at least 1 mange = 4

Total fruits = $6.5.4 = 120$


问:
Trouble finding the velocity of a projectile launched from a height
The following question has been giving me trouble for a while now, it states

A particle is thrown with speed $10 \space \text{ms}^{-1}$ from a point  $2\space \text{m}$ from the ground, at an angle of  $45°$ above the horizontal. What is the speed of the particle when it is at a height of $4 \space \text{m}$  for the second time in its motion?

I have tried the following,

$$v^2-u^2=2as \iff (v\sin(45))^2=(10\sin(45))^2+2(-10)(2)$$
$$v=\frac{\sqrt{10}}{\sin(45)} \implies v=4.47\space \text{ms}^{-1} $$

But I did not get the correct answer of $v=7.75\space \text{ms}^{-1}$, similarly I tried using the equation $s=\frac{1}{2}at^2+ut$  to first find the time taken and then plug it into $v=at + u$ to yield the velocity however this was to no avail.

Where am I going about wrong, I have a sneaking suspicion that it is to do with the value of $s$.

答:
In the horizontal direction, where you have no acceleration, $v_x=v_{x0}=v_0\cos45^\circ$. In the vertical direction $$v_y^2=v_{y0}^2-2gh=v_0^2\sin45^\circ-2gh$$
Then $$v^2=v_x^2+v_y^2=v_0^2-2gh$$
Notice that the angle does not matter. You could have gotten this from the conservation of energy.


问:
Triangle Inequality and linear algebra in geometry
Find the number of isosceles triangles with integral side lengths and having pe-rimeter 144 and only one side being largest.

My approach - If $a,a,b$ are sides such that $b>a$ then by triangle inequality $b<2a$. So $a<b<2a$, and by adding $2a$ on all sides, $3a<b+2a<4a$, so $3a<144<4a$.

So $36<a<48$ so $a$ has $11$ values from $37$ to $47$ and $11$ triangles are formed according to me but as per the book answer is $37$. Where am I wrong?

答:
You are correct and the book is wrong. We can list all partitions of $144$ into 3 positive integers, 2 of them equal:

$$(1,1,142)\\(2,2,140)\\:\\(36,36,72)\\(37,37,70)\\:\\(47,47,50)\\(48,48,48)\\(49,49,46)\\:\\(71,71,2)$$

There are a total of $71$ partitions, and as you have noted, only the ones from $(37,37,70)$ to $(47,47,50)$ are valid: the first $36$ violate the triangle inequality, and the last $24$ have two longest sides. The book's author probably forgot to account for the triangle inequality.


问:
Solving $\sin x\cos5x=\sin ax\cos3x$ without many addition formulas
I am trying to solve this trigonometric equation (with $a$ a constant)
$$\sin(x)\cos(5x)=\sin(ax)\cos(3x)$$
I tried expanding all terms using angle addition, but I got a huge Left and Right hand sides, and apart from the obvious $x=0$ solution,  I can't find the rest.

Is there an easy way to solve this? Or is it necessary to do the huge procedure?

答:
Consider that you look for the zero's of function
$$f(a,x)=\sin(x)\cos(5x)-\sin(ax)\cos(3x)$$ Since $f(a,x)+f(a,-x)=0$, we only care about the positive values of $x$. The linked coutour plot reveals the complexity of the problem (the same was noticed in @DatBoi's plot).

Excluding the trivial solution $x=0$, there are a few interesting observations

for any value of $a$, $f\left(a,(2 k+1)\frac{\pi}{2}\right)=0$

for any value of $a$, $f\left(a,k \pi\right)=(-1)^{k-1} \sin(a \pi)$ which would be $0$ every time $a$ will be an integer. The same would happen every time $\sin \left(\frac{\pi  k}{a}\right)=0$ or $\cos \left(\frac{5\pi  k}{a}\right)=0$.

As you may see, the situation is quite complex.

Taking into account the fact that $f(a,0)=0$ and $f\left(a,\frac{\pi}{2}\right)=0$, in that range there are one (if $a\geq 1$) or two (if $a\lt 1$) other solutions.

If $a<1$ we could perform a Taylor expansion around $x=0$ and have
$$\frac{f(a,x)} x=(1-a)+\frac{a^3+27 a-76}{6} x^2-\frac{a^5+90 a^3+405
   a-3376}{120} x^4+O\left(x^6\right)$$

Using the first term only, this would give as an estimate
$$x_{(1)}=\sqrt{\frac{6(1-a) }{76-27 a-a^3}}$$
Using the first and second terms, we need to solve a quadratic in $x^2$ (keep the $+\Delta$ solution) and this gives $x_{(2)}$ (it is too long to type but simple) (do not use it if $a<-2.724$ since it will make the estimate to be a complex number) .

Just a few results
$$\left(
\begin{array}{cccc}
a & x_{(1)} & x_{(2)} & \text{solution} \\
 +0.75 & 0.164654 & 0.171746 &  0.171508 \\
 +0.50 & 0.219308 & 0.236777 &  0.235572 \\
 +0.25 & 0.254944 & 0.283201 &  0.280237 \\
 +0.00 & 0.280976 & 0.319532 &  0.314159 \\
 -0.25 & 0.301027 & 0.348980 &  0.340785 \\
 -0.50 & 0.316889 & 0.373166 &  0.361993 \\
 -0.75 & 0.329568 & 0.393117 &  0.379004 \\
 -1.00 & 0.339683 & 0.409633 &  0.392699 \\
 -1.25 & 0.347644 & 0.423424 &  0.403741 \\
 -1.50 & 0.353738 & 0.435171 &  0.412640 \\
 -1.75 & 0.358184 & 0.445565 &  0.419791 \\
 -2.00 & 0.361158 & 0.455386 &  0.425501
\end{array}
\right)$$

Now, to polish the root, start with one of these estimates and use Newton method. For example, using $a=-2.345$, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.470197 \\
 1 & 0.432769 \\
 2 & 0.431443 \\
 3 & 0.431440 \\
\end{array}
\right)$$

I hope that you see the complexity of the problem. I suppose that it could become simpler only for integer values of $a$.

Edit

For the fun of it, I have been able to generate a still better estimate for the case where $a<1$. It writes
$$x_{(3)}=\sqrt{\frac{60 (a-1) \left(a^3+27 a-76\right)}{7 a^6+3 a^5+270 a^4-1250 a^3+6075
   a^2-29697 a+47632}   }$$


问:
How to evaluate $\int x\cdot\sin x\cdot\text{cosec} \,3x \,\mathrm dx$
Evaluvate integral $\int x\sin (x)\csc(3x) ~\mathrm{d}x$

My Try

Using a trigonometric identity I was able to simplify,

$$\int \frac{x\sin x}{3\sin x-4\sin^3x}\,\mathrm{d}x$$

$$=\int \frac{x}{3-4\sin^2x}~\mathrm{d}x$$

How can I proceed? Should I use a substitute for this? Any hint would be appreciated.  Thank you.

答:
\begin{gather*}
I=\int x\sin x\csc 3xdx\\
=\int \frac{xdx}{3-4\sin^{2} x}\\
Integrate\ by\ parts\ taking\ \\
F( x) =x\ as\ the\ first\ part\ and\ \\
G( x) =\frac{1}{3-4\sin^{2} x} as\ the\ second\ part\\
By\ integration\ of\ parts,\ \\
I=F( x)\int G( x) -\int F^{'}( x)\left(\int G( x) dx\right) dx\\
Now,\ \int G( x) dx=\int \frac{dx}{3-4\sin^{2} x} =\int \frac{\csc^{2} xdx}{3\csc^{2} x-4}\\
=\int \frac{\csc^{2} xdx}{3\left(\cot^{2} x+1\right) -4} =\int \frac{\csc^{2} xdx}{3\cot^{2} x-1}\\
Subsitute\ \cot x=t\\
\csc^{2} x=-dt\\
\ \int G( x) dx=\int \frac{-dt}{3t^{2} -1} =\frac{-1}{2\sqrt{3}}\ln\left(\frac{\sqrt{3} t-1}{\sqrt{3} t+1}\right) =\frac{-1}{2\sqrt{3}}\ln\left(\frac{\sqrt{3}\cot x-1}{\sqrt{3}\cot x+1}\right)\\
\\
=\frac{-1}{2\sqrt{3}}\ln\left(\frac{\sqrt{3}\cos x-\sin x}{\sqrt{3}\cos x+\sin x}\right) =\frac{-1}{2\sqrt{3}}(\ln(\cos\left( x+\frac{\pi }{6}\right) -\ln\left(\cos\left( x-\frac{\pi }{6}\right)\right)\\
=\frac{1}{2\sqrt{3}}\left(\ln\left(\cos\left( x-\frac{\pi }{6}\right)\right) -\ln\left(\cos\left( x+\frac{\pi }{6}\right)\right)\right)\\
So,\\
I=F( x)\int G( x) -\int F^{'}( x)\left(\int G( x) dx\right) dx\\
=\frac{x}{2\sqrt{3}}\left(\ln\left(\cos\left( x-\frac{\pi }{6}\right)\right) -\ln\left(\cos\left( x+\frac{\pi }{6}\right)\right)\right) -\int \frac{1}{2\sqrt{3}}\left(\ln\left(\cos\left( x-\frac{\pi }{6}\right)\right) -\ln\left(\cos\left( x+\frac{\pi }{6}\right)\right)\right) dx
\end{gather*}
All you need to do now is to compute the antiderivative of ln(cos(x-a)).


问:
How does the compound angle formula for sine apply to obtuse, reflex and negative angles?
My teacher showed us a proof for the compound angle formula by using a triangle and dropping a perpendicular line from an angle then getting the area of the triangle using sine rule (1) then getting it again by adding the area of the other two triangles (2) (created from the perpendicular line) then making (1) and (2) equal to each other.

I totally get the proof but not how it can be applied to obtuse, reflex and negative angles. I know about unit circle and how obtuse and reflex angles have sine just as their abtuse equivalent angles but I still don't get it. A visual proof will be tremendously appreciated!

答: A visual proof for obtuse/reflex angles in a triangle? I am not sure such thing exists(specifically for a triangle) but we can generalize the result by taking into account the fact that any angle(obtuse/reflex) can be reduced to a compound angle which I believe you are aware of.</p>


答:
We start with $x^2\equiv28\bmod6$.  As far as simply checking the two given answers, firstly replace $x$ by $2$ to get the true statement $2^2\equiv28\bmod6$ (since $2^2=4$ and $6|(28-4)=24$).

Similarly, replacing $x$ with $4$ gives us a true statement:  $4^2\equiv28\bmod6\iff0\equiv12\bmod6$.

Title: Understanding the answers of $x^2 \equiv 28 \pmod 6$

Body:
Solving  $x^2 \equiv 28 \pmod 6$

The answers are: $x \equiv 2 \pmod 6$ and $x \equiv 4 \pmod 6$.

When I plugged in to check the answers, it would be $2^2 \equiv 4 \pmod 6$ and $4^2 \equiv 16 \equiv 4 \pmod 6$. It is $4$ but not $28$, why is this, I don't understand? Anyone can help me understand this? Thanks a lot!


问:
convergence of the sum of reciprocals of "fake twin" primes
This question is inspired by the announcement of a proof that "fake twin" primes, i.e. pairs of consecutive primes differing by at most K, are -in infinite number- where K is a fixed integer which can be taken = 70,000,000.

It is not known OTOH whether the above would hold where K would be  = 2 instead (the twin prime conjecture).

Anyway, my question may be (hopefully is)  much simpler : it's been known after Viggo Brun that the sum of reciprocals of actual "twin primes" is finite (whether the number of terms in this sum is finite or infinite). Does the sum of reciprocals of the "fake twins" as defined above (for K=70 million) also converge ? Does it follow from an easy extension of Brun's method ?

I should make it clear I am no expert, just curious... I do not have accesss to a proof of Brun's original theorem, even a sketchy one.

答:
Your instinct is correct: the same proof that establishes the convergence of the sum of the reciprocals of the actual twin primes also establishes the convergence of the sum of the reciprocals of "$K$-twin primes" for any value of $K$.

To give a tiny amount of detail: sieve methods (which Brun pioneered) can show that the number of primes $p\le x$ for which $p+K$ is also prime is at most $c(K) x/\log^2 x$, for some constant $c(K)$ depending on $K$. This is enough to show that the sum of the reciprocals of such primes $p$ converges (consider the contribution of such primes between $2^n$ and $2^{n+1}$, then sum over $n$).


问:
$T^{eq}$ is a complete $\mathcal{L}^{eq}$-theory
Suppose $T$ is a complete $\mathcal{L}$-theory with infinite models. Enumerate the collection of all $\emptyset$-definable equivalence relations modulo $T$ as $(E_\alpha(\overline{v}_1,\overline{v}_2))_{\alpha\in\lambda}$ for some cardinal $\lambda$, where each $E_\alpha$ is an equivalence relation on $n_\alpha$-tuples; we may assume $E_0$ is equality of one-tuples. Then we define $\mathcal{L}^{eq}$ to be a $\lambda$-sorted language, with sorts $(S_\alpha)_{\alpha\in \lambda}$, and we refer to $S_0$ as the "home sort". $\mathcal{L}^{eq}$ contains a symbol for every symbol $\mathcal{L}$, interpreted as acting on $S_0$. (So, eg, if $\mathcal{L}$ has an $n$-ary relation symbol $R$, then $\mathcal{L}^{eq}$ has an $n$-ary relation symbol $R^{eq}$, defined on the sort $S_0$.) Also, for each $\alpha\in\lambda$, $\mathcal{L}^{eq}$ has a function symbol $\pi_\alpha:S_0^{n_\alpha}\to S_{\alpha}$, considered to be the projection map taking a tuple to its equivalence class modulo $E_\alpha$.

In particular, every $\mathcal{L}$-sentence can be thought of as an $\mathcal{L}^{eq}$-sentence in a natural way, by considering it as a statement about the $S_0$ sort. Then we define $T^{eq}$ to be the union of all these sentences with the additional axioms
\begin{alignat*}{2}
 &\bullet\forall (w\in S_{\alpha}) &&\exists (v_1\dots v_{n_\alpha}\in S_0) [w=\pi_\alpha(v_1,\dots,v_{n_\alpha})] \\
 &\bullet\forall(v^{(1)}_1,\dots,&&v^{(1)}_{n_\alpha}\in S_0)\forall(v_1^{(2)},\dots,v_{n_\alpha}^{(2)}\in S_0) \\ & &&[\pi_\alpha(v^{(1)}_1,\dots,v^{(1)}_{n_\alpha})=\pi_\alpha(v_1^{(2)},\dots,v_{n_\alpha}^{(2)})]\leftrightarrow E_\alpha\left(\overline{v^{(1)}},\overline{v^{(2)}}\right)
\end{alignat*}

for every $\alpha\in\lambda$. It is easy to see that $T^{eq}$ is consistent; take any $M$ a model of $T$, and interpret each $S_\alpha^M$ as the equivalence classes of $M^{n_\alpha}$ under $E_\alpha^M$, with $\pi_\alpha$ the canonical projection map. Now, the claim is that $T^{eq}$ is complete. This seems obvious to me, but the resource I'm using (Tent and Ziegler) doesn't include a proof, and the only proof I can come up with uses some machinery that seems like it shouldn't be necessary:

For an arbitrary language $\mathcal{L}$, we say that an $\mathcal{L}$-structure $M$ of cardinality $\kappa\geqslant\aleph_0$ is "special" if $M$ is the union of an elementary chain $(M_\lambda)_{\lambda\in\kappa}$, where each $M_\lambda$ is $\lambda^{+}$-saturated.

Fact 1: If two special $\mathcal{L}$-structures of cardinality $\kappa$ are elementarily equivalent, then they are isomorphic. $\square$

Fact 2: Let $M$ be any infinite $\mathcal{L}$-structure. Then, for any cardinal $\lambda$ with $\beth_\lambda>|M|$, there is a special elementary extension of $M$ of size $\beth_\lambda$. In particular, any $\mathcal{L}$-theory with infinite models has arbitrarily large special models. $\square$

Lemma: The $\mathcal{L}^{eq}$-theory $T^{eq}$ is complete.

Proof: Let $T_1\supseteq T^{eq}$ and $T_2\supseteq T^{eq}$ be any complete $\mathcal{L}^{eq}$ theories. By fact 2, there are special $\mathcal{L}^{eq}$ structures $M_1\models T_1$ and $M_2\models T_2$ of the same cardinality.  By construction of $T^{eq}$, each $M_i$ can be considered as a model of $T$ in the natural way, by restricting attention to the $S_0$ sort. Reducts of $\mu$-saturated structures are $\mu$-saturated, so each $M_i$ is in fact a special $\mathcal{L}$-structure. Thus, since $T$ is complete and so $M_1$ and $M_2$ are elementarily equivalent $\mathcal{L}$-structures, by fact 1 there is an $\mathcal{L}$-isomorphism $f:M_1\to M_2$.

We claim we can extend $f$ to an $\mathcal{L}^{eq}$ isomorphism $\overline{f}:M_1\to M_2$ by taking $$\overline{f}(\pi^{M_1}_\alpha(m_1,\dots,m_{n_\alpha}))=\pi_\alpha^{M_2}(f(m_1),\dots,f(m_{n_\alpha})).$$ Note that, since the $\pi_\alpha$ are surjective, this defines $\overline{f}$ on all of $S^{M_1}_\alpha$. Also, for all $\overline{m}$ and $\overline{m}'$, we have
\begin{align}
\pi^{M_1}_\alpha(m_1,&\dots,m_{n_\alpha})=\pi^{M_1}_\alpha(m'_1,\dots,m'_{n_\alpha}) \iff E^{M_1}_\alpha\left(\overline{m},\overline{m}'\right) \iff E^{M_2}_\alpha\left(f(\overline{m}),f(\overline{m}')\right) \\
&\iff \pi^{M_2}_\alpha(f(m_1),\dots,f(m_{n_\alpha}))=\pi^{M_2}_\alpha(f(m'_1),\dots,f(m'_{n_\alpha})).
\end{align}
The forward direction of this shows that $\overline{f}$ is well-defined, and the backwards direction shows that $\overline{f}$ is injective. To show that $\overline{f}$ is surjective, let $\pi^{M_2}_\alpha(m_1,\dots,m_{n_\alpha})\in S_\alpha^{M_2}$ be arbitrary. By surjectivity of $f$, there are $m'_1,\dots,m'_{n_\alpha}$ such that $f(\overline{m}')=\overline{m}$, and then $$\pi^{M_2}_\alpha(m_1,\dots,m_{n_\alpha})=\overline{f}(\pi^{M_1}_\alpha(m'_1,\dots,m'_{n_\alpha})),$$ as desired. Finally, $\overline{f}$ commutes with the $\pi_\alpha$ by construction and takes each $S_\alpha^{M_1}$ to $S_\alpha^{M_2}$, so it is in fact an $\mathcal{L}^{eq}$ isomorphism, and – in particular – $M_1$ and $M_2$ are elementarily equivalent as $\mathcal{L}^{eq}$-structures. Thus $T_1=T_2$ and so $T^{eq}$ is itself complete, as desired. $\square$

Now, I have two questions. First, does this proof look right? Second, it seems to me that there should perhaps be a more immediate proof of the desired result. Special structures take a bit of effort to define (though no set-theoretic assumptions! in contrast with saturated structures); is there a more direct way of seeing the result?

答:
Your proof looks fine, but you're right that there's a more elementary approach, not using special or saturated models, or EF-games.

Show that for any model $M\models T^{eq}$, we have $M = (M\restriction \mathcal{L})^{eq}$.

Show that $\mathcal{L}^{eq}$-formulas can be "translated" back to $\mathcal{L}$-formulas in the following sense: Consider an $\mathcal{L}^{eq}$-formula $\varphi(x;y_1.\dots,y_k)$, where $x$ is a tuple of variables from the home sort and the $y_i$ are single variables from the new sorts, say $y_i$ comes from the sort $S_{\alpha_i}$. Then there is an $\mathcal{L}$-formula $\bar{\varphi}(x;z_1,\dots,z_k)$, where each $z_i$ is a tuple of length $n_{\alpha_i}$, such that for any $M\models T$ and any tuples $a,c_1,\dots,c_k$ from $M$, we have $$M\models \bar{\varphi}(a;c_1,\dots,c_k)\iff M^{eq}\models \varphi(a;\pi_{\alpha_1}(c_1),\dots,\pi_{\alpha_k}(c_k)).$$

Now any two models of $T^{eq}$ can be written as $M^{eq}$ and $N^{eq}$ for $M$ and $N$ models of $T$. For any $\mathcal{L}^{eq}$-sentence $\varphi$, we have $$M^{eq}\models \varphi \iff M\models \bar{\varphi}\iff N\models \bar{\varphi}\iff N^{eq}\models \varphi$$ by completeness of $T$. So $T^{eq}$ is complete.

Point 2 is a bit tedious to write out carefully, but it's worth doing: this is the fundamental fact about the eq construction that makes everything else work.


问:
How can I find $P(A'\cap B')$ using De Morgan's laws?
$P(A'\cap B') = P(\overline{A\cup B})$ By De Morgan's laws.

I am given $P(A|B) = \frac{1}{5}$ and $P(B|A)=\frac{1}{4}$. Let $P(A\cap B)=p$

I want to find  $P(A'\cap B')$ in terms of $p$. Perhaps I don't need to use De Morgan's law in the first step ie convert to $P(\overline{A\cup B})$?

Anyway, I tried to sub it into the conditional probability formula so $P(\overline{A\cup B}) = P(\overline{A|B})\cdot P(\overline{B})$

Now I am stuck and I assume this method is not correct :(

Edit: In a previous section, I worked out that $P(A) = 4p$. That means I will need $P(A)$ to appear in my equation

Body:
Solution without using DeMorgan's:

$$P(A' \cap B') = P(B')P(A'|B')\\
P(B') = 1 - P(B) = 1 - \dfrac{P(A \cap B)}{P(A|B)} = 1 - 5p\\
P(A'|B')=1-P(A|B') = 1-\dfrac{P(A \cap B')}{P(B')} \\ 1-\dfrac{P(B'|A)P(A)}{P(B')}=1-\dfrac{(1-P(B|A))P(A)}{P(B')}=1-\dfrac{(1-\frac{1}{4})(4p)}{1-5p}=\dfrac{1-8p}{1-5p}\\
P(A' \cap B')=1 - 5p\cdot\dfrac{1-8p}{1-5p}=\boxed{1-8p}$$

So yes, your answer is correct. Hope this helps!


问:
If the multiplication of two operators is hermitian, then will they commute?
It is proven that if two operators $\hat{X}$ and $\hat{Y}$ commute, then the multiplication of them will be hermitian, i.e. if $\hat{X}\hat{Y}=\hat{Y}\hat{X}$, then $\left(\hat{X}\hat{Y}\right)^\dagger=\hat{X}\hat{Y}$.

My question is that is the opposite true as well? In other words, if the multiplication of two operators is hermitian, then will they commute? If yes, I need a proof.

答:
The statement of your assertion is a little off, but essentially yes. The correct statement is that two Hermitian operators must commute if their product is also Hermitian. The proof is entirely straightforward as a Hermitian product implies $XY=(XY)^\dagger$ but
$$
(XY)^\dagger=Y^\dagger X^\dagger=YX
$$
using that $X$ and $Y$ are both Hermitian themselves. Hence, $XY=YX$.

So, in fact the full statement of the theorem would be given two Hermitian operators $X$ and $Y$, the operators commute if and only if their product is also Hermitian.

OwnerDisplayName: farhad mabrooki

OwnerDisplayName: Richard Myers


问:
A bag contains marbles of 6 different colors with 7 of each color. Four are chosen at random.
So given the premise in the title, part (a) asks "How many different ways can you choose 2 marbles of one color and 2 of another color?" and the answer is

${6\choose 2}{7\choose 2}{7\choose 2}$

If I am understanding this correctly, the first comes from the fact that we are "choosing" two colors first and then the ${7\choose 2}{7\choose 2}$ comes from choosing 2 marbles from out of the 7 (for both colors).

However, in the next part it asks "How many different ways can you choose 2 of one color, 1 of a different color, and 1 of yet another color?" and I thought it would be ${6\choose 3}{7\choose 2}{7\choose 1}{7\choose 1}$, but the video I was watching said that the answer was

${6\choose 1}{5\choose 2}{7\choose 2}{7\choose 1}{7\choose 1}$

The video explained this was the answer because the color where we get choose 2 marbles is "distinct," but honestly, I have no clue what they meant by that. Thank you in advanced for your help.

答:
Basically all they mean is that a situation where you pick $2$ red, $1$ green, and $1$ blue is different from where you pick $2$ green, $1$ red, and $1$ blue. If you just pick the three colors without this distinction, these two situations are counted as if they were the same. This isn't a necessary distinction in the first question because both colors got the same number of marbles, so we can treat them all the same.

The reasoning behind the video's solution is first you pick a color to be the color with $2$ marbles, so pick $1$ out of a possible $6$. Then, pick the other two colors from the remaining $5$. Then you choose the marbles out of each set, giving you the other three factors.

Hope this helps!


问:
Evaluating $\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$
I am trying to evaluate the following integral:

$$\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$$

I thought about using the substitution $\beta \doteq \tfrac{(t-a)}{C}$ so the integral would become:

$$\int_{0}^{\infty} \beta^2 \cdot \exp (- \beta^2 ) \cdot \exp(i w (\beta C - a)) \ C \ d\beta$$

However, I don't know how to evaluate this either.

Any idea on how to do this?

EDIT: As Caran-d'Ache pointed out, I did in fact lose a minus! I'm still just learning Mathematica, how did you enter the integral?

答:
Or there is another way.
One can look at $$\mathrm{I}=\alpha\int_{0}^{\infty} \beta^2 e^{- \beta^2}   e^{i \beta s } d\beta$$ as the Laplace transform $\mathcal{L}_{\beta }\left[\beta ^2 e^{-\beta ^2 } \right](p)$ with $p=-is$.
One can find (from the textbooks or using Mathematica) the answer.
And after some simplification get the same result:
$$\mathrm{I}=\frac{\tilde{\alpha}}{8} i \left(2 s-\sqrt{\pi } e^{-\frac{s^2}{4}} \left(s^2-2\right) \left(\text{erfi}\left(\frac{s}{2}\right)-i\right)\right)$$</p>


问:
find $\lim _{(x,y) \to (0,0)} \sqrt{(x^{2}+y^{2 }-1)}$
I have trouble finding a real solution.
What I get when I tried to change the form to polar form or just directly substitute the value is $\sqrt {-1}=i.$
Can anyone help me? Or the answer is just really $i$.

答: Hint. Check what happens for $x=0$ and $y\to 0.$ Then check what happens if $y=1$ and $x\to 0.$


问:
Covering space Hausdorff implies base space Hausdorff
There is an exercise problem in Hatcher's Algebraic Topology book asking to show that if $p:\tilde{X}\rightarrow X$ is a covering space with $p^{-1}(x)$ finite and nonempty for all $x\in X$, then $\tilde{X}$ is compact Hausdorff iff $X$ is compact Hausdorff.

I've managed to show each part of the statement except for $\tilde{X}$ Hausdorff $\Rightarrow X$ Hausdorff. The problem I encountered was that taking $x\neq y$ in $X$ and $\tilde{U},\tilde{V}$ disjoint open sets in $\tilde{X}$ with $p^{-1}(x)\subset\tilde{U}$ and $p^{-1}(y)\subset \tilde{V}$ (I'm using the assumption that $\tilde{X}$ is compact Hausdorff, thus normal), I could not use these to get disjoint open sets in $X$ separating $x,y$.

答:
Take distinct $y_1 \neq y_2$ in $X$. Then $F_1 = f^{-1}[\{y_1\}]$ and $F_2 = f^{-1}[\{y_2\}]$ are disjoint, finite and non-empty subsets of $\tilde{X}$. By Hausdorffness we can find disjoint open sets $U_1$ and $U_2$ around $F_1$ and $F_2$. Let $O_1$ be an evenly covered neighbourhood of $y_1$, and $O_2$ one of $y_2$.

Now, $f^{-1}[O_1] = \cup_{i=1}^k V_i$, where the $V_i$ are disjoint and open and $f|V_i$ is a homeomorphism between $V_i$ and $O_1$ for all $i$; define $V'_i = V_i \cap U_1$, and define $O'_1 = \cap_{i=1}^k f[V'_i]$, which is open, as a finite intersection of open sets.

Similarly we define $W'_i$ from the inverse image of $O_2$ and $U_2$, and the intersection of their images is called $O'_2$.

These $O'_1$ and $O'_2$ are the required disjoint open neighbourhoods as can be easily checked.


问:
Compute $P(9) - Q(10)$ given some equalities of values of $P(x)$ and $Q(x)$.
Let $P(X)$ and $Q(x)$ be cubic polynomials such that

i.) $P(1) = Q(2)$

ii.) $P(3) = Q(4)$

iii.) $P(5) = Q(6)$

iv.) $P(7) = Q(8) + 13$

Compute $P(9) - Q(10)$.

I wanted to set $R(x) = P(x) - Q(x+1) = (x-1)(x-3)(x-5)$, but  I don't think that works.

答:
You are almpost there

set  $$R(x)=P(x)-Q(x+1)=a(x-1)(x-3)(x-5)$$ where $a$ is a constant

now use $P(7)-P(8)=13$ to find $a$ and then $R(9)$ should be easy to find


问:
How far can you go with the Martingale Roulette strategy
I am currently arguing with my friends that just learned the Martingale roulette technique for the casino that consist of doubling your initial bet as much time that you lose (with a budget of course).
So if you have 5000\$ you could initially bet 1\$ on a color (with 48% aprox. chances of winning) and if you lose you bet 2\$, then 4\$, then 8\$... up to the point where you don't have the money to double on and you lose it all.

I am certain that in the end you are losing using this technique. Yes your probability of winning a percentage of your budget are great up until a limit of wins using this technique. I'm almost certain that you hit that limit before 5000 times ( winning more than what you were willing to lose ). So in general even with millions of people who would use this the casino is still winning in total and with any maximum budget (except infinity of course). My friend base a lot of their thoughts behind the fact that this is banned from a lot of casinos along with all sort of techniques or strategies but i think it's more of a collateral damage and in general you still lose with that.

So I'm asking your help to explain why I am or they are wrong about this.
Thank you !

答:
You have the right idea.  Your expected loss on any bet is $\frac {1}{19}$ of your wager in an American casino and $\frac {1}{37}$ in Monte Carlo.  Your betting strategy does not change this.

If you had an infinite endowment (and the casino had no maximum bet size) even if you are down a very large sum of money at some point, eventually your double-or-nothing bet will hit and almost surely you will eventually hit a win and be up one dollar.  But you don't have an infinite endowment and the casino will not allow you to make bets of unlimited size.

Supposing you have a budget of $\$5,000$ for example, if you hit a streak of 12 consecutive losses, you can not afford to double your bet.  We can calculate your expected loss with a strategy of play to the first win, or walk away after losing 12 in a row.

$(1 - \frac {20}{38}^{12}) - \frac {20}{38}^{12}(2^{12}-1)\approx - 0.85$ in Vegas.

$(1 - \frac {19}{37}^{12}) - \frac {19}{37}^{12}(2^{12}-1)\approx - 0.38$ in Europe.

Spending enough time employing this strategy, almost surely you will bust.


问:
Solving $x^2 \equiv 3 \pmod{2003}$

Solving $x^2 \equiv 3 \pmod{2003}$

I think it is like solving $x^2 \equiv 2006 \pmod{2003}$ because $2006 \equiv 3 \pmod{2003}$. The possible value for $x$ are from $1$ to $2002$, so I can just insert $x$ to check until I get the right answers, but that only works for small modulo, this modulo is very big, which is impossible to check from $1$ to $2002$. Any hints to solve this equation? Thanks!

答: Lucky for you, $2003$ is a prime $\equiv3\bmod4$. If a solution $x$ exists, $x\equiv\pm3^{(2003+1)/4}\bmod2003$. We compute the modular exponent as $385\bmod2003$, which indeed squares to $3\bmod2003$. So $x\equiv\pm385\bmod2003$.

Title: Solving $x^2 \equiv 3 \pmod{2003}$


问:
Two questions on finite abelian groups
Which of the following are true?

1.Every group of order $6$ abelian.

2.Two abelian groups of the same order are isomorphic

答:
none of them is true.
for 1. consider $S_3$.

for 2. consider $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$

Title: Two questions on finite abelian groups


答:
Parcly has correctly deduced that $3^{501}$ will be a solution, and now we show how this can be reduced by hand. Note that $3^{512} = 6561^{64} \equiv 552^{64} = 204704^{32} \equiv 248^{32} = \dots = 555 \mod 2003.$

Now $3|2004,$ so we obtain $t = 3^{-1} = 2004/3 = 668 \mod 2003.$ Lastly, another round of repeated squaring using $11_{10} = 1011_2$ gives $t^{11} \equiv -524 \mod 2003.$

Thus, at last $3^{501} \equiv 3^{512} \cdot t^{11} \equiv 555 \cdot -524 = -290820 \equiv -385 \mod 2003.$

Title: Solving $x^2 \equiv 3 \pmod{2003}$

Body:
Solving $x^2 \equiv 3 \pmod{2003}$

I think it is like solving $x^2 \equiv 2006 \pmod{2003}$ because $2006 \equiv 3 \pmod{2003}$. The possible value for $x$ are from $1$ to $2002$, so I can just insert $x$ to check until I get the right answers, but that only works for small modulo, this modulo is very big, which is impossible to check from $1$ to $2002$. Any hints to solve this equation? Thanks!


问:
Getting $3$ (but not $4$) of the same in a row at least twice when flipping $100$ coins

$100$ coins are flipped. What is the probability of getting $3$ of the
same in a row at least twice, but not $4$ of the same in a row?

I tried using recursion, but I am having trouble finding the relation to the previous one. I think there's a better solution.

答:
Recurrence is the only way I've been able to do this.  It's a little tricky, because we can't confine our attention solely to strings of heads and tails that meet the criteria.

Call a string of H and T a $0$-string if it has no occurrences of either HHH or TTT.  Let

$a_n$ be the number of $n$-character $0$-strings that end in HT,

$b_n$ be the number of $n$-character $0$-strings that end in TH,

$c_n$ be the number of $n$-character $0$-strings that end in TT, and

$d_n$ be the number of $n$-character $0$-strings that end in HH.

By symmetry, $a_n=b_n$ and $c_n=d_n$

We have $$a_{n+1}=b_n+d_n= a_n+c_n,\tag1$$

since we can add a T to any $0$-string that ends in H to get a $0$ string that  ends in HT, and all $0$-strings that end in HT arise in this way.  We have also $$c_{n+1}=a_n\tag2$$ because if we add a $T$ to a string that ends in TT, we will no longer have a $0$-string.

From $(1)$ and $(2)$ we get $$a_{n+1}=a_n+a_{n-1},\ n\geq3$$  We have $a_2=1,\ a_3=2$ and we see $$c_{n+1}=a_n= F_n\tag3$$ where $F_n$ is the $n$th Fibonacci number.

Call a string of H and T a $1$-string if it has exactly one occurrence of either HHH or TTT, and no occurrence of either HHHH or TTTT.  Let

$e_n$ be the number of $n$-character $1$-strings that end in HT,

$f_n$ be the number of $n$-character $1$-strings that end in TH,

$g_n$ be the number of $n$-character $1$-strings that end in HTT,

$h_n$ be the number of $n$-character $1$-strings that end in THH,

$i_n$ be the number of $n$-character $1$-strings that end in TTT and

$k_n$ be the number of $n$-character $1$-strings that end in HHH.

We have $e_n=f_n$, $g_n=h_n$, $i_n=k_n$ by symmetry.

As above, we get $$\begin{align}
e_{n+1}&=f_n+h_n+k_n\\
e_{n+1}&=e_n+g_n+i_n\\
g_{n+1}&=e_n\\
i_{n+1}&=c_n=F_n
\end{align}$$ The last equation expresses the fact that to get a $1$-string ending in TTT, we must add a T to a $0$-string ending in TT.  This gives
$$e_{n+1}=e_n+e_{n-1}+F_{n-2}\tag4$$ for $n\geq 5$.

Call a string of H and T a $2$-string if it has at least two occurrences of either HHH or TTT, and no occurrence of either HHHH or TTTT.  Let

$p_n$ be the number of $n$-character $2$-strings that end in HT,

$q_n$ be the number of $n$-character $2$-strings that end in TH,

$r_n$ be the number of $n$-character $2$-strings that end in HTT,

$s_n$ be the number of $n$-character $2$-strings that end in THH,

$t_n$ be the number of $n$-character $2$-strings that end in TTT, and

$u_n$ be the number of $n$-character $2$-strings that end in HHH.

We have $p_n=q_n$, $r_n=s_n$, $t_n=u_n$ by symmetry.

We get $$\begin{align}
p_{n+1}&=q_n+s_n+u_n\\
p_{n+1}&=p_n+r_n+t_n\\
r_{n+1}&=p_n\\
t_{n+1}&=r_n+g_n
\end{align}$$
because we can get a $2$-string ending in TTT either by adding a T at the end of a $2$-string ending in HTT, or at the end of a $1$-string ending in HTT.

Observe that the total number of $2$-strings of length $n$ is $$T_n:=2p_n+2r_n+2t_n=2p_{n+1}$$

We have $$\begin{align}
p_{n+1}&=p_n+r_n+t_n\\
&=p_n+p_{n-1}+r_{n-1}+g_{n-1}\\
&=p_n+p_{n-1}+p_{n-2}+e_{n-2}\tag5
\end{align}$$

or $$T_n=T_{n-1}+T_{n-2}+T_{n-3}+2e_{n-2}\tag6$$

With initial data, $$
F_0=0,\ F_1=1\\
e_0=e_1=e_2=e_3=0,\ e_4=1,\ e_5=2 \\
T_0=T_1=T_2=T_3=T_4=T_5=0,\ T_6=2$$
and formulas $(4)$ and $(6)$, I have checked the values produced by the recurrence against a brute-force count of $2$-strings, up through $n=20$.

Carrying through the recurrence gave me $$T_{100}=360772317646371359457234176$$ which results in a probability of success of $$\boxed{0.0002845991770771989}$$

It should be possible, at least in theory, to find a closed-form expression for $T_n$.  Equation $(4)$ is a second-order linear recurrence, with constant coefficients, and $F_n$ has a well-known formula as a linear combination of two exponentials, so we should be able to solve $(4)$ completely, and we expect a solution in the form of a linear combination of exponentials.  Then equation $(6)$ is of the same type as equation $(4)$, expect that it is third-order rather than second-order.  WolframAlpha will solve the characteristic equation exactly, but the roots look unpleasant.  I wouldn't want to try it without a CAS.

I did notice however, that near $n=100$, we have $$\frac{T_{n+1}}{T_n}\approx1.8393$$ which is very close to the real root found by WolframAlpha.  It shouldn't be hard to get a good empirical formula for $T_n$.

EDIT

It occurred to me that there is a better way of approaching the problem of finding a closed-form solution.

From $(4)$ we have $$\begin{align}
e_{n+3}&=e_{n+2}+e_{n+1}+F_{n}\\
-e_{n+2}&=-e_{n+1}-e_{n}-F_{n-1}\\
-e_{n+1}&=-e_{n}-e_{n-1}-F_{n-3}\\
\end{align}$$

Adding these equations gives $$e_{n+3}-2e_{n+2}-e_{n+1}+2e_n+e_{n-1}=F_n-F_{n-1}-F_{n-2}=0\tag7$$ so that $e_n$ satisfies a homogeneous linear recurrence.

Now we can try a similar procedure with equation $(6)$ .  We have $$\begin{bmatrix}
.5 &-.5 &-.5 &-.5&0&0&0&0\\
0&-1&1&1&1&0&0&0\\
0&0&-.5&+.5&+.5&+.5&0&0\\
0&0&0&1&-1&-1&-1&0\\
0&0&0&0&.5&-.5&-.5&-.5
\end{bmatrix}
\begin{bmatrix}
T_{n+5}\\
T_{n+4}\\
T_{n+3}\\
T_{n+2}\\
T_{n+1}\\
T_{n}\\
T_{n-1}\\
T_{n-2}\\
\end{bmatrix}=
\begin{bmatrix}
e_{n+3}\\
-2e_{n+2}\\
-e_{n+1}\\
2e_n\\
e_{n-1}\\
\end{bmatrix}$$
Adding these $(5)$ equations, and clearing fractions gives $$T_{n+5}-3T_{n+4}+4T_{n+2}+2T_{n+1}-2T_n-3T_{n-1}-T_{n-2}=0\tag8$$
This recurrence has characteristic polynomial
$$x^7-3x^6+4x^4+2x^3-2x^2-3x-1=(x^2-x-1)^2(x^3-x^2-x-1),$$ so that WolframAlpha is able to solve it exactly.

This still seems to require a CAS, but at least we have a homogeneous recurrence, so there's no need to guess a particular solution.  I may try to do this with sympy tomorrow.


问:
Die and coin variance of random variable question.
Die and coin. Roll a die and flip a coin. Let $Y$ be the value
of the die. Let $Z = 1$ if the coin shows a head, and $Z = 0$ otherwise. Let
$X = Y + Z$. Find the variance of $X$.

My work:

$E(Y) = 1 \cdot \cfrac{1}{6} + 2 \cdot \cfrac{1}{6} +  3 \cdot \cfrac{1}{6} +  4 \cdot \cfrac{1}{6} +  5 \cdot \cfrac{1}{6} + 6 \cdot \cfrac{1}{6} = \cfrac{7}{2}$

$E(Z) = 0 \cdot \cfrac{1}{2} + 1 \cdot \cfrac{1}{2} = \cfrac{1}{2}$

So $E(X) = E(Y + Z) = E(Y) + E(Z) = \cfrac{7}{2} + \cfrac{1}{2} = 4$

$E(Y^2) = 1^2 \cdot \cfrac{1}{6} + 2^2 \cdot \cfrac{1}{6} +  3^2 \cdot \cfrac{1}{6} +  4^2 \cdot \cfrac{1}{6} +  5^2 \cdot \cfrac{1}{6} + 6^2 \cdot \cfrac{1}{6} = \cfrac{91}{6}$

$E(Z^2) = 0^2 \cdot \cfrac{1}{2} + 1^2 \cdot \cfrac{1}{2} = \cfrac{1}{2}$

so $E(X^2) = E(Y^2 + Z^2) = E(Y^2) + E(Z^2) = \cfrac{91}{6} + \cfrac{1}{2} = \cfrac{47}{3}$

$Var(X) = E(X^2) - (E(X))^2 = \cfrac{47}{3} - 4^2 = -\cfrac{1}{3}???$

Where did I go wrong, variance can't be negative so clearly my work is wrong, but I have no idea where I went wrong? Can someone point me in the right direction of how to do this question?

Body:
$Y\perp\!\!\!\!\!\!\perp Z$ thus $V(X)=V(Y)+V(Z)$

Your error is this

$$E(X^2)=E[(Y+Z)^2] \ne E(Y^2)+E(Z^2)$$


Title: Die and coin variance of random variable question.

Body:
Die and coin. Roll a die and flip a coin. Let $Y$ be the value
of the die. Let $Z = 1$ if the coin shows a head, and $Z = 0$ otherwise. Let
$X = Y + Z$. Find the variance of $X$.

My work:

$E(Y) = 1 \cdot \cfrac{1}{6} + 2 \cdot \cfrac{1}{6} +  3 \cdot \cfrac{1}{6} +  4 \cdot \cfrac{1}{6} +  5 \cdot \cfrac{1}{6} + 6 \cdot \cfrac{1}{6} = \cfrac{7}{2}$

$E(Z) = 0 \cdot \cfrac{1}{2} + 1 \cdot \cfrac{1}{2} = \cfrac{1}{2}$

So $E(X) = E(Y + Z) = E(Y) + E(Z) = \cfrac{7}{2} + \cfrac{1}{2} = 4$

$E(Y^2) = 1^2 \cdot \cfrac{1}{6} + 2^2 \cdot \cfrac{1}{6} +  3^2 \cdot \cfrac{1}{6} +  4^2 \cdot \cfrac{1}{6} +  5^2 \cdot \cfrac{1}{6} + 6^2 \cdot \cfrac{1}{6} = \cfrac{91}{6}$

$E(Z^2) = 0^2 \cdot \cfrac{1}{2} + 1^2 \cdot \cfrac{1}{2} = \cfrac{1}{2}$

so $E(X^2) = E(Y^2 + Z^2) = E(Y^2) + E(Z^2) = \cfrac{91}{6} + \cfrac{1}{2} = \cfrac{47}{3}$

$Var(X) = E(X^2) - (E(X))^2 = \cfrac{47}{3} - 4^2 = -\cfrac{1}{3}???$

Where did I go wrong, variance can't be negative so clearly my work is wrong, but I have no idea where I went wrong? Can someone point me in the right direction of how to do this question?

Body:
$E(X^2)\ne E(Y^2+Z^2)$

$E(X^2)=E(Y^2+Z^2+2YZ)=E(Y^2)+E(Z^2)+2E(YZ)=\dfrac{91}{6}+\dfrac{1}{2}+2\cdot\dfrac{7}{2}\cdot \dfrac{1}{2}=\dfrac{47}{3}+\dfrac{1}{2}=\dfrac{97}{6}$

$V(X)=E(X^2)-E(X)^2=\dfrac{97}{6}-16=\dfrac{97-96}{6}=\dfrac{1}{6}$

An alternate way:

Since X and Y are independent events covariance term vanishes therefor
$V(X)=V(Y)+V(Z)=E(X^2)-E(X)^2+E(Z^2)-E(Z)^2$


问:
Calculating population size at any future time
If current population size is $P$ and average life-time of any member is $L$. Any pair of member of the population is allowed to have $C$ children on an average. They can have children between the age of $A$ to $B$.Then how can I tell population size at any  future time?</p>

答: Though this is still not a well-specified problem, here are some thoughts.  You start with $\frac{P}{L}$ people of each age, which will be strange as the distribution will not stay uniform.  If the childbearing is random over the span, you will have $\frac{3PL}{2(B-A)}$ children born in the first year and $\frac{P}{L}$ die.  These numbers will hold for $A$ years so you will have $P+\left(\frac{CPL}{2(B-A)}-\frac{P}{L}\right)N$ in year $N$</p>


答: HINT: At this point you should have seen two groups of order $6$, $\Bbb Z/6\Bbb Z$ (sometimes called $\Bbb Z_6$) and $S_3$, and two groups of order $4$, $\Bbb Z/4\Bbb Z$ (sometimes called $\Bbb Z_4$) and the Klein $4$-group. Are both of those groups of order $6$ Abelian? Are those two groups of order $4$ isomorphic? How do you know?</p>

Title: Two questions on finite abelian groups

Body:
Which of the following are true?

1.Every group of order $6$ abelian.

2.Two abelian groups of the same order are isomorphic


问:
$f: \mathbb{N} \rightarrow [1,2,3, ..., n] $
Is there a function that is of the form $f: \mathbb{N} \rightarrow [1,2,3, ..., n] $ with $ n \in \mathbb{N} $?

I am looking for a way to assign a natural number to a natural number, but this second it falls in an interval $ [1, n] $
Is this possible?

答:
If I understand your question correctly, there are many such functions.

For example,

$$f(k) = min(k,n)$$

or

$$f(k) = modulo(k,n)$$


问:
Limit Supremum of Sequence Created by Interleaving 2 Bounded Sequences
I'm studying for an upcoming departmental exam, and this was one of the questions given on a previous year's exam.

Let $(a_{n})$ and $(b_{n})$ be real, bounded sequences, and let $(c_{n})$ be the sequence created by interleaving the terms of $(a_{n})$ and $(b_{n})$,
$$(c_{n}) = (a_{1}, b_{1}, a_{2}, b_{2},\cdots)$$
Prove that
$$\limsup(c_{n})\leq \max\{\limsup(a_{n}), \limsup(b_{n})\}.$$
Further, if both $(a_{n})$ and $(b_{n})$ both converge to $\gamma$, then $(c_{n})$ converges to $\gamma$ also.

The second part is pretty straightforward, no problems there. I've tried a couple of different things to solve the first part, mostly playing around with the various definition of $\limsup$ (supremum of subsequential limits, limit of the supremum of the $m$-tail, the "eventually less than $t$" definition, etc), as I am not sure one would work best here. What really confused me was at one point, I thought it should be the opposite, that the inequality should be reversed:

If you let $A, B,$ and $C$ be the set of subsequential limits of $(a_{n}), (b_{n})$, and $(c_{n})$, respectively $(b_{n})$, then clearly, both $A\subseteq C$ and $B\subseteq C$, as both $(a_{n})$ and $(b_{n})$ are already subsequences of $(c_{n})$. This would then imply that $\sup A\leq\sup C$ and $\sup B\leq\sup C$, and hence, that $\max\{\sup A, \sup B\}\leq\sup C$, which is the opposite of what we are to prove.

I'm probably completely wrong with this and missing something painfully obvious, but if anyone can come up with a proof of the original problem and tell me why my "proof" is wrong, I would greatly appreciate it. Bonus points if you can use the first part to imply the second part, which is what I would imagine the test wants.

答:
You have not done anything wrong. You just proved the easy inequality but you are asked to prove that the reverse inequality is also true.

Consider any subsequence of $(c_n)$. Then it includes a subsequence of $(a_n)$ or  it includes a subsequence of $(b_n)$. Can you use this to finish the proof?


答:
If $n = 1$, this is trivial: $f(x) = 1$.

If $n > 1$, then I suggest we use the modulo function, let's call it $mod(x,n)$ where $x$ and $n$ are integers. Essentially, the modulo function gives us the remainder if we were to divide $x$ by $n$. More formally speaking, $mod(x,n)$ is the smallest nonnegative value $k$ such that we can write $x = mn + k$ for some integer $m$. For example, $mod(67,10) = 7$ because $67 = 6\cdot10+7$, and this is the smallest such value for which we can write an expression of this form.

The codomain for this function is natural numbers in the interval $[0, n-1]$: by definition the value must be nonnegative, so the minimum is zero, and any value greater than or equal to $n$ cannot be correct because we could subtract $n$ to get a smaller viable nonnegative solution:

$$k\geq n, x = mn+k\to k-n \geq 0, x = (m+1)n + (k-n)$$

It is also clear that this function can only output integers because for $mod(x,n) = k$ we have that there is some integer $m$ such that $k = x - mn$, and because for our discussion $x$ and $n$ are also always integers, $k$ is an integer by the closure properties of integers on addition and multiplication.

Finally, the surjectivity of $mod$ is trivial because for any $0 \geq x \geq n-1$, $mod(x,n) = x$.

So for any natural $n > 1$, $f(x) = mod(x,n) + 1$ satisfies your desired properties.

Body:
Is there a function that is of the form $f: \mathbb{N} \rightarrow [1,2,3, ..., n] $ with $ n \in \mathbb{N} $?

I am looking for a way to assign a natural number to a natural number, but this second it falls in an interval $ [1, n] $
Is this possible?


问:
Explanation of the statement of the fundamental theorem of arithmetic.
It's from the book Mathematics Made Difficult.

If $L^+(P,N_0)$ is the set of functions $f:P\rightarrow N_0$ with a property such that
$$\exists\; n_0 \in N_0 \; \forall \; p \in P \;$$

$$ p\ge n_0 \implies f(p) = 0 $$
then there exists a bijection $N_1 \rightarrow L^+(P,N_0) $ such that if $n \rightarrow f$ then
$$n = \prod_{p\in P} p^{f(p)}$$
Here $P$ is the prime numbers and $N_1 = N_0 - \{0\} $

I know this is <s>related to</s> the fundamental theorem of arithmetic and the book mentioned above made this statement a bit complicated, intentionally. Out of curiosity, I am interested in what this statement is actually saying?

答: The statement essentially says that each natural number is a unique product of prime numbers.</p>


问:
$\gamma$ is line of curvature $\iff$ $\frac{d\theta}{ds} + \tau = 0$
I have the following problem in differential geometry.

Suppose $S\in\mathbb R^3$ regular surface, and $\gamma:[0,L]\rightarrow S$ regular curve parametrized by arc-length.

Suppose that $\theta(s)$ is the angle between normal of the curve $n(s)$ and normal of the surface $N(\gamma(s))$.

I need to show that $\gamma$ is line of curvature $\iff$ $\frac{d\theta}{ds} + \tau = 0$ when $\tau$ is curve torsion.

my problem is that I don't know how to caculate $\frac{d\theta}{ds}$.

P.s.

I found here that this equality is called geodesic torsion

My attempt:

(Updated solution thanks to @TedShifrin)

I know that $cos\theta=<n,N>$ and by differentiating this equation I get:

$-sin\theta*\frac{d\theta}{ds} = <n\prime, N> + <n, N\prime>$.

Now using Fernet-serret I get:

$-sin\theta*\frac{d\theta}{ds} = <-\kappa\gamma\prime + \tau b(s), N> + <N\prime, n> = <\kappa\gamma\prime, N> + \tau(s)*sin\theta + <N\prime,n>$ (and this is because I know that the bi-normal is perpendicular to the curve normal).

Also I know that $Sp(\gamma\prime) = k_i \gamma\prime$ because it is line of curvature, and that $<\gamma\prime,N(\gamma)> = 0$ because $\gamma\prime \in TpS$.

Putting all together I get that:

$-sin\theta*\frac{d\theta}{ds} = \tau(s)*sin\theta$ and in case locally we have that $\theta \neq \pi*n$ we can divide by $sin\theta$ and get what we desired.

答: HINT: Differentiate $N\cdot n = \cos\theta$.


问:
Dirichlet function
I thought maybe there is any connection to that "one" in the numerator :

$$ \text{let }  f(x):[-2,2]→(0,∞)\quad \text{be a continuous function for every } x  $$
$$\text{Prove that}\quad f(x)=D(x)\quad\text{ has  a solution in the  section }  [-2,2]  $$

$$f(x+\sqrt{2})=\frac{1}{f(x)}$$
$$D(x)= \text{Dirichlet  function}$$

On the one hand, I can say that $f(x)$ is continuous,  the "image" can get any value and in particular for non-$\Bbb Q$ numbers → Dirichlet function gets a  value of 1 and I finish. I have a feeling this is not the way to solve it and there is a meaning to the limits they gave in the question.
I am incompetent. If anyone can help me, I will appreciate it a lot!

答: Since $1$ is between $t$ and $1/t$ for any positive number $t$, by the intermediate value theorem, $f(x)=1$ for some $x\in[0,\sqrt 2]$. If $x$ is irrational, then we're done. Otherwise, $x-\sqrt 2$ is in $[-2,2]$ and is irrational.

Title: Dirichlet function


问:
Denoting tolerable range for common values in two sequences
Given set X and set Y, common values can be denoted in a new set via set intersection, so: Z = X ∩ Y. However, let's say set X = {1,2,3,4} and set Y = {1.3, 2.3, 2.7 4.3}. All values in set X are within .3 of some value in set Y, yet Z = X ∩ Y would be empty since the values aren't exactly the same. How can you specify that set Z includes the intersection of values in X and Y, within the tolerable range of plus or minus .3?

答:
You can define the so-called $0.3$ neighborhood  $A^{(0.3)}$ of $A$ as the set of all numbers $x$ such that $|x-a| \leq 0.3$ for some $a \in A$. Then the property you have is $ B \subseteq A^{(0.3)}$.

General definition:  $\epsilon$ neighborhood  $A^{(\epsilon)}$ of a set $A$ is the set of all numbers $x$ such that $d(x,a) \leq \epsilon$ for some $a \in A$. [ $(d(x,a)$ is the distance between $x$ and $a$]. This definition makes sense in any metric space.