问:
Moment generating function, change of variable
If $X=aY$ where $X, Y$ are random variables and $a$ is a constant. Do I have
$$M_X(t)=M_Y(t/a)$$ where $M$ is the moment generating function. How can I prove it?</p>

答:
Along the lines of what the commenter fgp said, we can go right to the definition of the moment generating function, $M_X(t) = E(e^{tX})$. Then we just use the fact that $X = aY$ as follows:

\begin{align*}
M_X(t) &= E(e^{tX})\\[5pt] &= E(e^{t(aY)})\\[5pt] &= E(e^{(ta)Y}) \\[5pt] &= M_Y(ta)
\end{align*}

Hope that helps.


问:
Pullback metric of inversion map
Let $I:\mathbb{R}^{n}\setminus\{0\} \longrightarrow \mathbb{R}^{n}\setminus\{0\}$ be the inversion map, i.e. $I(x)=\frac{x}{|x|^{2}}$ and equip $\mathbb{R}^{n}\setminus\{0\}$ with the standard euclidean metric $g_{eucl}=\langle\cdot,\cdot\rangle$. Show that for any $x\in \mathbb{R}^{n}\setminus\{0\}$ we have $I^{*}\langle\cdot,\cdot\rangle=\frac{1}{|x|^{4}}\langle\cdot,\cdot\rangle$ on the tangent space of $x$.

My work so far:

By definition, I have to show that for any $v,w\in T_{x}\left(\mathbb{R}^{n}\setminus\{0\}\right)$ we have $\langle d_{x}I(v),d_{x}I(w)\rangle = \frac{1}{|x|^{4}}\langle v,w\rangle$, where $d_{x}I$ is the differential of $I$ at $x$. Because we are in $\mathbb{R}^{n}\setminus\{0\}$, instead of choosing a curve $c$ that satisfies $c(0)=x$ and $c'(0)=v$ for $v\in T_{x}\left(\mathbb{R}^{n}\setminus\{0\}\right)$, I compute $d_{x}I$ by the product (or quotient) rule: $d_{x}I = \frac{1}{|x|^{4}}\left( |x|^{2}(\delta_{ij})_{i,j} -2(x_{i}x_{j})_{i,j}\right)\in \mathbb{R}^{n\times n}$. Now, it is no restriction if we set $v=e_{i}$ and $w=e_{j}$ to make the computation a little bit easier. Computing $\langle d_{x}I(e_{i}),d_{x}I(e_{j})\rangle$, I end up with an expression that containes $\frac{1}{|x|^{4}}\langle e_{i},e_{j} \rangle$ (which is good, I think) and something, that was not equal to zero (but it should). Can you give me some help with the last computation? I know it is probably an obvious mistake, but I don't see it.

答:
By definition of the pullback by $I$ of the euclidean metric, $(I^*\langle\cdot,\cdot\rangle)_x(v,w) = \langle\mathrm{d}I(x)v,\mathrm{d}I(x)w \rangle_{{I(x)}}$. The computation shows that $\mathrm{d}I(x): T_x\mathbb{R}^n \to T_{I(x)}\mathbb{R}^n$ is just, while canonically identifying these two tangent spaces to $\mathbb{R}^n$, equal to $$\mathrm{d}I(x)v = \frac{{v}}{\|x\|^2}-2\langle x,v\rangle \frac{x}{\|x\|^4}$$
Let $u$ and $v$ be two tangent vectors to $x$. Then
\begin{align}
(I^*g)_x(u,v) &= \left\langle \dfrac{u}{\|x\|^2}-2\frac{\langle u,x\rangle x}{\|x\|^4},\dfrac{v}{\|x\|^2}-2\frac{\langle v,x\rangle x}{\|x\|^4} \right\rangle \\
&=\frac{\langle u,v\rangle}{\|x\|^4} -2\dfrac{\langle u,x\rangle \langle x,v\rangle}{\|x\|^6} -2 \dfrac{\langle v, x \rangle\langle x ,u\rangle}{\|x\|^6} + 4 \dfrac{\langle u,x\rangle \langle v,x \rangle \|x\|^2}{\|x\|^8} \\
&= \dfrac{\langle u,v\rangle}{\|x\|^4} - 4 \dfrac{\langle x , u \rangle \langle x , v \rangle}{\|x\|^6} + 4 \dfrac{\langle x,u\rangle \langle x,v\rangle}{\|x\|^6}\\
&= \dfrac{\langle u,v\rangle}{\|x\|^4}
\end{align}
So you are done. Remark that you do not have to compute in any coordinate system or in any orthonormal basis.

ContentLicense: CC BY-SA 4.0

Title: Pullback metric of inversion map


问:
Calculus: Why can't a we define 1 as a critical point when we have $h(x) = \frac{4x^2}{x^2-1}$?
I'm trying to understand an explanation in Khan Academy on the topic "Finding the critical point".

I understand that we need to get the derivative down to either zero or undefined to find that critical point, but I don't get why we can't use use 1 in this equation's derivative. $h(x) = \frac{4x^2}{x^2-1}$

The derivative is $\frac{-8x}{(x^2-1)^2}$

Is it because the graph itself already has undefined at 1?

答:
If you have

$h(x) = \frac{4x^2}{x^2-1}$

What is $h(1)$ ? You have something divided by 0 which is undefined.

So 1 cannot be a critical point.

But if you have

$h(x) = |x|$

then $x=0$ is a critical point, because $h(0) = 0$ is well defined and $h'(0)$ is not well defined. So the point is that h is well defined at all critical points. $h'(0)$ can be undefined, but $h(0)$ should be well defined.

Another function:

$f(x) = \begin{cases}
      0, & \text{if}\ x=0 \\
      \frac{1}{x}, & \text{otherwise}
    \end{cases}$

This function is well defined everywhere. But $f'(0)$ is not well defined. So 0 is critical point here.

Another example:

$h(x) = \begin{cases}
      0, & \text{if}\ x=1 \\
      \frac{4x^2}{x^2-1}, & \text{otherwise}
    \end{cases}$

Here 1 is a critical point because now $h(1)$ is well-defined and $h'(1)$ is not well defined. So by making sure the function is defined at 1, we've made it a critical point now.


问:
Can I find length of bisector by knowing the position vectors?

$\vec A$ = $(\vec i - \vec j-3\vec k)$

$\vec B$ = $(2\vec i +\vec j-2\vec k)$

$\vec C$ = $(-5\vec i +2\vec j-6\vec k)$

To find the length of angle bisector of bac I marked the points $A(1,-1,-3)$ $B(2,1,-2)$ $C(-5,2,-6)$. How can I use the fact that the angle between the bisector and two adjacent sides is equal?

答:
$\vec{AB}=\vec b-\vec a=\vec r_1, \vec{AC}=\vec c-\vec a= \vec r_2$ then position vector of the angle bisector od BAC angle is $$\vec r=\frac{\hat r_1+ \hat r_2}{2}$$
The length of angle bisector is $|\vec r -\vec a|.$


问:
Real part of the square root of a complex number
I want to find the real part of the square root of a complex number $z$

$\operatorname{Re}(\sqrt(z))$

In general I am wondering if the following is true:

$\operatorname{Re}(\sqrt(z))$ = $\sqrt(\operatorname{Re}(z))$

Thanks

答:
Try with

$$z=2+i\to z^2=3+4i.$$

Do we have

$$2=\sqrt3\ ?$$


问:
Paths in a quiver form a $K$-algebra.
I am attending a course in foundations in representation theory and I am struggling to grasp the concept of path algebras, more explicitly why paths in a quiver form a $K$-algebra.

Let $Q$ be a quiver and $K$ be a field.
Why do paths in a quiver $Q$ form a $K-$algebra, denoted $KQ$?

I understand that the multiplication of paths is associative and that there exists an identity element.
Nonetheless no textbook I have studied explains why $KQ$ is a $K$-vector space and how the operation $+: KQ \times KQ \rightarrow KQ$, $(a,b) \mapsto a+b$ of paths is defined.
Furthermore they don't explain the scalar multiplication that is required for a $K$-algebra

答:
To construct the path algebra $KQ$, one first forms the $K$-vector space with basis the set of all paths in $Q$. This then tells you what the addition and scalar multiplication are. In particular, a general element of $KQ$ is a finite $K$-linear combination of paths.

We then need to define the multiplication in $KQ$. Since this must be $K$-linear and distributive, it is enough to declare what the multiplication of two basis elements, i.e. paths, is. One proves that this is associative by checking on paths, and that the trivial paths $e_i$ for vertices $i$ form a complete set of orthogonal idempotents.

In fact, the path algebra $KQ$ is naturally graded by path length, so the trivial paths $e_i$ have degree zero and each arrow has degree 1.

As an example, take the quiver $1\to2\to3$. Then the path algebra is isomorphic to the algebra of lower triangular $3\times3$ matrices over $K$
$$ \begin{pmatrix}\ast&0&0\\\ast&\ast&0\\\ast&\ast&\ast\end{pmatrix}$$
where $e_i$ corresponds to the elementary matrix $E_{ii}$, the arrow $a\colon1\to2$ corresponds to $E_{21}$, the arrow $b\colon2\to3$ to $E_{32}$, and the path $ba\colon1\to3$ to $E_{31}$.

Title: Paths in a quiver form a $K$-algebra.


问:
Let $X = [0,1]$ and $\sim$ have the following equivalence relation over $X$: is the quotient first countable?
Let $X = [0,1]$ and $\sim$ have the following equivalence relation over $X$:

$$x\sim y\iff x = y\text{ or }x, y \in {]}0, 1{[}$$

Show that $X/{\sim}$   is not first countable.

====================================================================

First countable means that for any distinct $x, y\in X$, there are $U, V \subset X$ open such that $x\in U, y\in V, x\notin V$ and $y\notin U$.

And

$[x] = \{y\in X : y\sim x\}$

$X/{\sim} = \{[x]: x\in X\}$

I know it should be a simple exercise, but I stuck, any help would be most welcome.
Thank you in advance for any assistance.

答:
$X/\sim$ has exactly 3 points and thus it is first countable. Just like any space with finitely many points or more generally countable number of open subsets.

First countable means that for any distinct $x, y\in X$, there are $U, V \subset X$ open such that $x\in U, y\in V, x\notin V$ and $y\notin U$.

No, this is not the definition of "first countable". What you've described is being $T_1$.

And indeed, $X/\sim$ is not $T_1$. First of all note that $X/\sim=\{[0],[1/2], [1]\}$ has $3$ points. Furthermore proper, nontrivial open subsets are precisely $\{[0],[1/2]\}$, $\{[1/2]\}$ and $\{[1/2],[1]\}$. So can you separate say $[0]$ from $[1/2]$?

More generally a finite space is $T_1$ if and only if it is discrete. And our $X/\sim$ is not discrete: neither $[0]$ nor $[1]$ are open.


问:
If $E(X|Y) = E(X)$ then $E(XY)=E(X)E(Y)$

If $E(X|Y) = E(X)$ then $E(XY)=E(X)E(Y)$.

I was attempting the discrete case. I was trying to show that $E(X|Y)E(Y)=E(XY)$. But I could not proceed much beyond the definition of expectations. Rather I feel that this equality may not always hold. I have failed to find any counter-example for $X$ and $Y$ either.

答:
What you need is the iterative law of conditional expectation and $E(XY|Y) = YE(X|Y)$:
\begin{align*}
E(XY) = E(E(XY|Y)) = E(YE(X|Y)) = E(YE(X)) = E(X)E(Y).
\end{align*}


问:
Is the following inequality correct and how to prove it?
We have $0 < m_n < 1$ and $m_{n+1}=m_{n}(1-\dfrac{m_{n}}{n})$ for all $n \in \mathbb{N}, n \ge 2$.

Does the inequality below hold true for all $n \in \mathbb{N}, n \ge 2$, and if it does, how to prove it ?
$$
\sum_{k=1}^{n} \frac{1}{k}<\frac{1}{m_{n}}<\sum_{k=1}^{n} \frac{1}{k}+1
$$

答:
Update: I wonder where OP got this problem. I found this one: Probabilistic Sieve of Eratosthenes and they are related because the harmonic series can be approximated by the logarithm function as shown in that problem.

This is not a complete answer but it's too long to be replied in the comment section.

Results so far: LHS inequality is true (with proof) as long as $m_2 < \frac{2}{3}$. RHS inequality seems to be true if $m_2 \geqslant 0.61 (0.6096)$ and false if $m_2 \leqslant 0.6095$ but is not proved. (We also need $m_1 > \frac{1}{2}$ but its value is mostly irrelevant.)

$\forall n>1$, we have
$$
\frac{1}{m_{n+1}}-\frac{1}{m_n}=\frac{1}{m_n\left(1-\frac{m_n}{n}\right)}-\frac{1}{m_n} = \frac{1}{m_n} \left( \frac{1}{1-\frac{m_n}{n}} - 1\right)=\frac{\frac 1n}{1-\frac{m_n}{n}}> \frac 1n,
$$

$$
\frac{1}{m_{n+1}}-\frac{1}{m_n}=\frac{\frac 1n}{1-\frac{m_n}{n}} < \frac{\frac 1n}{1-\frac 1n}= \frac{1}{n-1}
$$

Denote $S_n=\sum_{k=1}^n \frac 1k$, then $\forall N>2$ $$
S_{N-1}-1 < \frac{1}{m_N}-\frac{1}{m_2}  < \sum_{n=2}^{N-1} \frac{1}{n-1} = \sum_{n=1}^{N-2} \frac 1n = S_{N-2}. \tag 1
$$

The LHS inequality becomes
$$
\frac{1}{m_N} > \frac{1}{m_2} + S_N - 1 - \frac{1}{N} > \frac 12 -\frac{1}{N} + S_N \geqslant S_N, \text{ since } m_2<\frac 23.
$$

So long as $1+\frac 12 < \frac{1}{m_2}$, or $m_2<\frac{2}{3}$, then $S_N < \frac{1}{m_N}$ is always true.

The RHS inequality of (1) however, cannot be used to prove $\frac{1}{m_{n}}<S_n+1$.

This is because if $\frac{1}{m_2} + S_{N-2} < S_{N}+1$ then $\frac{1}{m_2} < 1 + \frac{1}{N-1} + \frac 1N \Rightarrow \frac{1}{m_2} \leqslant 1$, as $N \to \infty$, a contradiction.

Mathematical induction utilizing both sides also didn't work.

Suppose $S_N < \frac{1}{m_N} < S_N + 1$, then
$$
\frac{1}{m_{N+1}}=\frac{1}{m_N (1-\frac{m_N}{N})} <  \frac{S_N+1}{1-\frac{1}{NS_N}}
$$

Unfortunately $\frac{S_N+1}{1-\frac{1}{NS_N}}<S_{N+1}+1=S_N+\frac{1}{N+1}+1$ is false if you expand all terms and simplify.

Numerical results I did run a script in R and compared $1/m_N$ with $S_N+1$ up to $N=10^8$ for various values of $m_2$ and I found that $1/m_n < S_n+1$ for $n=2,3,\ldots, 10^8$ if $m_2>0.61$.

Here's the script:

N=10^8;

L = cumsum(1/(1:N)); # S_N

R = L+1; # S_N+1

m=rep(NA,N);

m[1]=0.51;

m[2] = 0.61; # only need to change this line

for(i in 3:N) m[i]=m[i-1]*(1-m[i-1]/(i-1));

r=1/m;

all(L<r & r<R) # this should return true


问:
Which convex shape requires the longest slice to break in half?
Suppose you are given a convex unit area region of the plane, and you wish to minimize the length of a line segment that cuts the shape into two regions of equal area. Which shape results maximizes this length, i.e., is the hardest to slice into two equal regions. Is it the circle?

For generalizations, does anything change if we allow the slice to be any curve? What if we want to chop it into $n$ equal regions for $n>2$? What if we are minimizing the area of planes that slice higher dimensional regions in half? I'm curious in any generalizations along these lines.

答:
I'll only consider the particular case of an acute angled triangle with area $1$ and with angles $\alpha \le \beta \le \gamma$.

Let's first find the arc of circle inside an angle of angle $\phi$ and of area $\frac{1}{2}$.

If the radius of the arc is $r$ then we have
$$\frac{1}{2} r^2 \phi = \frac{1}{2}$$
so
$r= \frac{1}{\sqrt{\phi}}$,
and therefore the length of the arc is
$$r \phi = \sqrt{\phi}$$

Let us see when the arc corresponding to the angle $\alpha$ lies inside the triangle. This is true if and only if the distance from the vertex to the opposite side $h$ is $\ge \frac{1}{\sqrt{\alpha}}$.

Now, we calculate the area of the triangle
$$1 = \frac{1}{2} h^2 ( \cot \beta + \cot \gamma)$$
and therefore
$$h = \sqrt{\frac{2}{\cot \beta + \cot \gamma}}$$
and so the condition is
$$\frac{\cot \beta + \cot \gamma}{2}\le \alpha$$

Now, since the function $\cot $ is decreasing, we get
$$\frac{\cot \beta + \cot \gamma}{2}\le \cot \alpha$$
Therefore, if
$$\cot \alpha \le \alpha$$
(equivalently, $\alpha \ge \approx 49^{\circ}.\ldots$)
then we have the inequality.

Therefore, if the angle $\alpha$ is not too small, the shortest dividing line is of length $\sqrt{\alpha}$.

In the particular case of an equilateral triangle, the shortest line is $\sqrt{\frac{\pi}{3}} = 1.0233\ldots$

For the equilateral triangle, the radius halving the area equals
$$\frac{\sqrt{\pi}}{2 \sqrt[4]{3}}\cdot l= 0.6733\ldots \cdot l$$
where $l$ is the side of the triangle.


问:
Calculate the values of the "trigonometric" functions for the angles $a+b$ and $a-b$
Calculate the values of the "trigonometric" functions for the angles $a+b$ and $a-b$ if
$\sin a =\frac{3}{5} \, y\, \sin b= \frac{2\sqrt{13}}{13}$

I did for $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)= \frac{3}{5}\cos(b)+\frac{2\sqrt{13}}{13}\cos(a)$

but I don't know if am I the correct way and how to know how much is $\cos(b) $ and $\cos(a)$ how can I calculate it? just with calculator?

答:
There are 3 concepts (which I am assuming you already know):

1.) Definition of $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$

2.) Reciprocal-relations like $\sin{x}=\frac{1}{\operatorname{cosec}x}$

3.) Trigonometric identities which are based on Pythagoras theorem like $\sin^2 {x}+\cos^2{x}=1$ (note this identity was proved using Pythagoras theorem)

Now, we have 3 such reciprocal relations and 3 such identities based on the Pythagoras theorem (there are more trigonometric identities based on the Pythagoras theorem, but we can generally derive all of them from these 3 fundamental identities).

The best part of those above concepts is that they let us find any trigonometric ratio easily if we already know one. As:

If $\sin{x}=\frac{1}{2}$ then,
$\operatorname{cosec} x=\frac{1}{\sin x}=2$

and, $\sin^2 {x}+\cos^2{x}=1$
$\Rightarrow\cos^2{x}=1-\frac{1}{4}$
$\Rightarrow \cos^2 x= \frac{3}{4} \Rightarrow \cos x=\pm \frac {\sqrt3}{2}$

Now, if we have this angle $x$ only defined for a right-angled triangle, then we can say that $\cos x \neq -\frac {\sqrt 3}{2}$, since trigonometric ratios of angle $x$ where $0°\lt x \lt 90°$, are always positive. So, that gives us $\cos x= \frac {\sqrt3}{2}$

However, if $x$ is defined for any angle then both values of $\cos x$ is right. So, $\cos x= \pm \frac{\sqrt 3}{2}$

Now we can find $\sec x$ by reciprocal relation and $\tan x$ and $\cot x$ by their definition.

If we have taken $x$ as any angle, you can observe $x$ belongs from 1st or 2nd quadrant, since $\sin x$ which is $\frac {1}{2}$, is positive. So, we will get $\operatorname {cosec} x$ positive and $\cos x$, $\sec x$, $\tan x$ and $\cot x$ either positive or negative.

I also want to suggest that once you have found one trigonometric ratio of $a+b$, use that ratio to find other ratios of $a+b$ rather than writing compound angle formulae for each trigonometric ratio of $a+b$. (Similar with $a-b$)


问:
Solving a Difference Equation
Find the solution $Y \in \mathbb{R}^\mathbb{R}$  of difference equation $y(x+2) - y(x+1) + 2y(x) = x$ on  $\mathbb{R}$ such that $\forall x \in [0,2): Y(x) =x$

This is all the information of the given question that I am trying to solve.

I need some help/guidance to help me solve this question and understand the solving mechanism of difference equations.

Thank you all.

答: Hint: try getting rid of the $x$ by putting $g(x)+tx=y(x)$.</p>

ContentLicense: CC BY-SA 3.0

Title: Solving a Difference Equation


问:
What is the largest possible value of $E[X] - (E[X^{-1}])^{-1}$?

Suppose $X$ is a random variable on $[A; B]$ ($0 < A < B$). What is the largest possible value of $E[X] - (E[X^{-1}])^{-1}$?

On one side it is rather obvious, that $E[X] - (E[X^{-1}])^{-1} \leq E[X] \leq B$. But this bound is clearly not tight.

On the other side, the largest value for which I managed to build example for is $(\sqrt{B} - \sqrt{A})^2$. It is the random value $X$ with following distribution:

$$P(X = A) = \frac{\sqrt{A}}{\sqrt{A} + \sqrt{B}}$$

$$P(X = B) = \frac{\sqrt{B}}{\sqrt{A} + \sqrt{B}}$$

I suspect that this lower bound is tight, but do not know how to prove it.

答:
A loose application of Kantorovich's would give an upper bound of
$$  LHS \le \frac{ (A-B)^2}{4A}$$

Some manipulation can be used to have better estimation but I think that this bound is already ok.


答:
As pointed out in another answer, $\Re\sqrt{z} \ne \sqrt{\Re z}$ for general $z$.

In general, for $z = x + iy = re^{i\theta}$ with $r \ge 0$ and $\theta \in (-\pi,\pi]$, the principal square root of $z$ is defined to be the number $\sqrt{r}e^{i\frac{\theta}{2}}$. This means

$$\Re\sqrt{z} = \sqrt{r}\cos\frac{\theta}{2}$$
If you apply the half angle formula for cosine and use the fact  $$\frac{\theta}{2} \in\left (-\frac{\pi}{2},\frac{\pi}{2}\right)\implies \cos\frac{\theta}{2} \ge 0$$
You will find $\cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}}$ and hence

$$\Re\sqrt{z} = \sqrt{\frac{r + x}{2}} = \sqrt{\frac{\sqrt{x^2+y^2} +x }{2}}$$
By a similar argument, you will find
$$|\Im \sqrt{z}| = \sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}$$
The sign of $\Im\sqrt{z}$ will be the same as the sign of $\theta$ and hence $y$. As a result, the principal square root is given by following formula:

$$\sqrt{z} = \sqrt{\frac{\sqrt{x^2+y^2} + x}{2}} + \verb/sign/(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} i$$
where $\verb/sign/(y) = +1, 0, -1$ depends on whether $y$ is positive, zero or negative.

Title: Real part of the square root of a complex number

Body:
I want to find the real part of the square root of a complex number $z$

$\operatorname{Re}(\sqrt(z))$

In general I am wondering if the following is true:

$\operatorname{Re}(\sqrt(z))$ = $\sqrt(\operatorname{Re}(z))$

Thanks


问:
vicious circle of a circle area
I want to show that area of a circle $D$ with radius $r$ is $\pi r^2$. To do that I consider inscribed and circumscribed regular $n$-gons $I_n$ and $O_n$. Their areas equals to
$$
S(I_n) = \frac 12 n r^2 \sin\frac{2\pi}n
\quad\text{and}\quad
S(O_n) = nr^2 \tan \frac\pi n.
$$
As $I_n \subset D \subset O_n$, it follows that
$$
\frac 12  r^2 n\sin\frac{2\pi}n
\leq S(D) \leq
r^2 n\tan \frac\pi n.
$$
Then, by passage to limits,
$$
\frac 12 r^2 \cdot 2\pi \leq S(D) \leq r^2 \cdot\pi.
$$
However, I need the knowledge that $\lim_{n\to\infty} n\sin (1/n) = 1$.
Luckily, I know that for $x \in (0, \pi/2)$
$$
\sin x < x < \tan x.
$$
But the classic proof of a right inequality requires calculation of area of a circle segment!
What to do?

I am not sure that proofs by integration do not use the derivative of $\sin x$ (which is, again, follows from a limit $\lim_{x\to0} (\sin x)/x$). Same thing about tayloring to prove inequalities.

Shoul I somehow show that arc length is less than length of a tangent segment?

UPD
Now I see two ways, both using definition of length of a curve as supremum of inscribed polygonal chains.

One is to derive inequalty $x < \tan(x)$ from the fact that if $A \subset B$ are convex polygons, then the perimeter of $A$ is at most the perimiter of $B$. (Thanks to @Mark S.)

Second is to consider a sequence of polygons inscribed in circle whose perimeters converge to $2\pi r$. Area of each such polygon is
$$
\frac12 \sum a_i h_i,
$$
where $a_i$ are sides' lengths and $h_i$ are distances from sides to the center of circle. As perimeter of polygon tends to the perimeter of circle, $\max a_i$ tends to 0, so $\min h_i$ and $\max h_i$ tend to $r$.

答:
Comment:

$Lim_{n\rightarrow\infty}sin (\alpha)=lim_{n\rightarrow \infty}\frac{\alpha}n=d\alpha$

So: $dA=\frac 12 \cdot r\cdot r\cdot d\alpha=\frac 12 \cdot r^2 d\alpha$

$A=\frac12  r^2\int^{2\pi}_0 \cdot d\alpha=\frac 12 r^2\cdot \big[\alpha\big]^{2\pi}_0=\pi r^2$

Title: vicious circle of a circle area


问:
Need help with finding the limit of a function
I can't think of a way to solve this one. Any help or direction towards a solution is appreciated!

If $\lim_{x\to 1} \frac{f(x)-8}{x-1} = 10$, find $\lim_{x\to 1} f(x)$.

Thanks in advance!

P.S. Not any homework, just learning calculus on my own.

答:
Observe that
$$
f(x)-8=\frac{f(x)-8}{x-1}\cdot (x-1).
$$
And then find the limit of the left hand side as $x\to 1$.


答:
On direct substitution, we see that the denominator is $0$. For the limit to be $10$, the numerator should also be $0$.

Thus,
$$\lim_{x\rightarrow 1} \left[f(x)-8\right] =0$$
$$\implies \lim_{x\rightarrow 1} f(x) = 8$$

Hope this helps :)

Ask if you are confused at any step!

Title: Need help with finding the limit of a function

Body:
I can't think of a way to solve this one. Any help or direction towards a solution is appreciated!

If $\lim_{x\to 1} \frac{f(x)-8}{x-1} = 10$, find $\lim_{x\to 1} f(x)$.

Thanks in advance!

P.S. Not any homework, just learning calculus on my own.


问:
How many mappings are there from 0 to 1? from 0 to 0? from 1 to 0 and how many elements does 0 have?
This is my answer for the first three questions:
Since 1 is the terminal object there exists exactly one mapping from 0 to 1.
There is 1 mapping from 0 to 0 because o is the initial object in the category.
There is one mapping from 1 to 0 since 1 is the terminal set and then then there's another mappin from 1 to 0 because 0 is the initial object. Therefore, in total there are 2 mappings from 1 to 0.

I am struggling with the fourth part.Any tips would be much appreciated.

答:
You're right about $0\to 1$ and $0\to 0$, but not about $1\to 0$. The property of being a terminal object only talks about maps to $1$; it tells you absolutely nothing about what kinds of maps come from $1$. Similarly $0$'s property only tells you about maps from $0$. So neither of these properties can tell you anything about maps $1\to 0$.

It should be noted that Lawvere is not talking about an arbitrary category here, but the category of abstract sets, for which he has given some axioms. The most recent such axiom is that $0\not\simeq 1$, and this is what will answer your question.

Notice that if there is a map $x:1\to 0$ then the composite $1\overset{x}{\to}0\to 1$ is the identity on $1$, and $0\to 1\overset{x}{\to}0$ is the identity on $0$, meaning exactly that $0\simeq 1$. So there can be no such $x$. Since for Lawvere an element of $X$ is just a map $1\to X$, this also answers how many elements of $0$ there are.


问:
How can this sum sign be expanded?
How can this sum sign

$$\sum_{i=1}^{\log (n)} 2^i$$

be expanded to

$$2^{\log (n)+1}-2$$

Which rules do I need to look at to understand this?

答:
You can show $S_m=\sum\limits_{i=1}^m 2^i = 2^{m+1}-2$ by induction or as a geometric series.

For example $2S_m=\sum\limits_{i=2}^{m+1} 2^{i}$ and $S_m=2S_m-S_m = 2^{m+1}-2^1$

Then substitute $m= \log(n)$ to get $S_{\log(n)}$, at least when $\log(n)$ is an integer

Title: How can this sum sign be expanded?


答:
Just apply the formula for the geometric sum. The terms follow a geometric progression, first term $a = 2^1 = 2$, common ratio $ r = 2$ and number of terms $N = \log n$ (assuming $\log n$ is a positive integer, obviously).

$S_N = \frac{a(r^n - 1)}{r-1} = \frac{2(2^{\log n} - 1)}{2 - 1} = 2^{\log n + 1} - 2$

Title: How can this sum sign be expanded?

Body:
How can this sum sign

$$\sum_{i=1}^{\log (n)} 2^i$$

be expanded to

$$2^{\log (n)+1}-2$$

Which rules do I need to look at to understand this?


问:
Calculus: Isn't the velocity & tangent at a critical point = 0, and therefore a function is not increasing at that point?
I encountered a question from Khan Acad that asked, at what intervals of x does the function increase. My intuition is that all ranges except for 0 and 1, because tangents at those points are flat per green graph below.
Backed up by the red graph, showing h'(1)=0 & h'(0)=0, zero velocity.

I need help figuring out my knowledge gap:

Am I misinterpreting what "increasing" here?

Am I missing the point of what a critical point is..?

答:
Increasing functions and strictly increasing functions are both well-defined terms with different definitions.

From wikipedia:

In calculus, a function $f$ defined on a subset of the real numbers
with real values is called monotonic if and only if it is either
entirely non-increasing, or entirely non-decreasing. That is, a
function that increases monotonically does not exclusively have to
increase, it simply must not decrease.

A function is called monotonically increasing (also increasing or
non-decreasing), if for all $x$ and $y$ such that $x\leq y$ one has
$f(x)\leq f(y),$ so $f$ preserves the order. Likewise, a function is called
monotonically decreasing (also decreasing or non-increasing) if, whenever
$x\leq y$, then $f(x)\geq f(y)$, so it reverses the order.

If the order $\leq$ in the definition of monotonicity is replaced by
the strict order $<$, then one obtains a stronger requirement. A
function with this property is called strictly increasing. Again, by
inverting the order symbol, one finds a corresponding concept called
strictly decreasing. A function may be called strictly monotone if it
is either strictly increasing or strictly decreasing. Functions that
are strictly monotone are one-to-one (because for $x$ not equal to
$y$, either $x<y$ or $x>y$ and so, by monotonicity, either $f(x) <$
$f(y)$ or $f(x) > f(y)$ (thus $f(x) \neq f(y)$.)

As B.Martin pointed out in the comments, $f(x)=x^3$ is a strictly increasing function (!), with $f′(0)=0$. Also, the constant function $f(x)=1$ is an increasing (but not strictly increasing) function with a derivative that is $0$ everywhere. So one always has to be clear what one is talking/asking about: whether or not a function is increasing or strictly increasing? At a point in the domain, or a subset of the domain, or over the entire domain?

Your/Khan academy's question is: Is the function $f(x) = 6x^5 - 15x^4 + 10x^3$ increasing (but not necessarily strictly increasing) for all $x \in \mathbb{R}$?

The answer is yes. In fact, your function is strictly increasing for all $x \in \mathbb{R}$, exactly the same as how $f(x) = x^3$ is strictly increasing for all $x \in \mathbb{R}$.

We may also talk about a function being strictly increasing at a point. See: https://math.stackexchange.com/a/364619/29156 for details. (Although, Did's definition of increasing at a point is really a definition of "strictly increasing at a point", since we are going with wikipedia's definitions).
The domain $D$ of all points for which $f(x)$ is increasing is the set $D = $ { $x: f(x)$ is increasing at the point $x$ }.

The function $f(x) = 6x^5 - 15x^4 + 10x^3$ is in fact strictly increasing at the point $x=0$, and it's also true that $f'(0)=0$. However, we only need to show that this function is increasing, and we can prove this using the definition of "increasing", or we can use the fact that $f'(x) \geq 0$ $\forall x \in \mathbb{R}$, which can be seen from the graph, or by noting that $f'(x) = (\sqrt{30}x(x-1))^2 \geq 0 \ \forall x \in \mathbb{R}$.

Going back to your question, "... because tangents at those points are flat..."

What is your definition of "flat"?

Useful:

A (differentiable) function $f$ is increasing but not strictly increasing at a point $x=c$, i.e. $f'(c) = 0$, $\iff \exists \epsilon > 0 $ such that $f$ is constant on the interval $ (c - \epsilon, c + \epsilon)$. I think this fact also works for all functions, not just differentiable or continuous ones.


Title: How can this sum sign be expanded?

Body:
In binary,

$$1+10+100+1000+10000=11111=100000-1.$$

You can generalize.


问:
How to show this polynomial is irreducible over $\mathbb{Q}$?
QUESTION: Show that, $\sqrt{3}+\sqrt[3]{7}$ is algebraic over $\mathbb{Q}$ with degree $6$.

I'm allowed to use this definition:  We say that $a \in K$ is algebraic of degree $n$ over $F$ if the minimal polynomial of $a$ over $F$ has degree $n$, i.e., $\deg(Irr_{F}(a))(x)=n$.

MY ATTEMPT:
Defining $\alpha:=\sqrt{3}+\sqrt[3]{7}$  we are going to obtain a polinomial $p(x)$ such that $p(\alpha)=0$. Let's start:
\begin{align*}
\alpha = \sqrt{3}+\sqrt[3]{7} &\implies \alpha -\sqrt{3}=\sqrt[3]{7}\\
&\implies(\alpha -\sqrt{3})^3=7\\
&\implies\alpha^3-3\alpha^2 \sqrt{3}+9\alpha -3\sqrt{3}=7\\
&\implies (\alpha^3 +9\alpha -7)^2=3(3\alpha^2+3)^2\\
&\implies \alpha^6+9\alpha^4-14\alpha^3+27\alpha^2-126\alpha+22=0
\end{align*}

Therefore, $\alpha$ is a root of   $p(x)= x^6+9x^4-14x^3+27x^2-126x+22$, where $p(x)\in \mathbb{Q}[x]$ is monic polynomial.

MY DOUBT: Now, it is necessary to show that $p(x)$ is irreducible over $\mathbb{Q}$ in order to conclude this exercise. However here is my problem:

I can't use Eiseinstein criterion, because doesn't work, once there is not any p prime that is suitable to show irreducibility.

If I show all roots by use of De Moivre formula's is not enough. Once we have this result: If a polynomial is irreducible over $F$ then there is not any root of this polinomial over $F$. But, we do not have the opposite implication as a result! So, is not enough use De Moivre formula's.

Would someone help me with this part?

答:
Following Jyrki Lahtonen suggestion, I will use the approach described here.

First, I prove that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt{3})$:

Notice that: $\alpha = \sqrt{3} + \sqrt[3]{7} \implies \alpha - \sqrt{3} = \sqrt[3]{7} \implies (\alpha - \sqrt{3})^{3} = (\sqrt[3]{7})^{3} \implies \alpha^{3} - 3\alpha^{2}\sqrt{3} + 9\alpha - 3\sqrt[3]{3} = 7 \implies \alpha^3 + 9\alpha - 7 =  3\alpha^2 \sqrt{3} - 3\sqrt{3} \implies \alpha^3 + 9\alpha - 7 =   \sqrt{3} (3\alpha^2 - 3).$
Therefore $$\sqrt{3} = \frac{\alpha^3 + 9\alpha - 7}{(3\alpha^2 - 3)} (*)$$ and thus we conclude that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt{3})$.

Then, I prove that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt[3]{7})$:

Notice that $\sqrt[3]{7} = \alpha - \sqrt{3}$ and that $\mathbb{Q}(\alpha)$ contains both $\alpha$ and $\sqrt{3}$ (by the Equation ($*$) ). Thus, $\mathbb{Q}(\alpha)$ contains $\sqrt[3]{7}$ and therefore $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt[3]{7})$.

Since $[\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt[3]{7}): \mathbb{Q}] = 3$, we obtain that the $[\mathbb{Q}(\alpha): \mathbb{Q}] $ is a multiple of 6:

Notice that:

$[\mathbb{Q}(\alpha): \mathbb{Q}] = [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt{3})] [\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2 [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt{3})] $
$[\mathbb{Q}(\alpha): \mathbb{Q}] = [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt[3]{7})] [\mathbb{Q}(\sqrt[3]{7}) : \mathbb{Q}] = 3 [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt[3]{7})]$

and therefore, $[\mathbb{Q}(\alpha): \mathbb{Q}] $ is a multiple of both 2 and 3 (i.e. a multiple of 6).

As you found a monic polynomial $p$ with degree 6, we conclude.


答:
Assume that it is not true that $\lim_{x\to 1}f(x)=8$. Then there exists a positive number $\epsilon_0>0$ such that for every $\delta>0$ we can find a number $x$ such that $0<|x-1|<\delta$ but $|f(x)-8|\geq\epsilon_0$. Therefore, for every $n$ we can find $x_n$ such that  $0<|x_n-1|<1/n$ but $|f(x_n)-8|\geq\epsilon_0$. In other words, there is a sequence $x_n$ converging to $1$ such that
$$\frac{|f(x_n)-8|}{|x_n-1|}\geq\frac{\epsilon_0}{1/n}=n\epsilon_0\to\infty$$
and this contradicts the assumption. Therefore, our assumption that the limit of $f(x)$ as $x\to 1$ is not $8$ leads to a contradiction, so it must be the case that $\lim_{x\to 1}f(x)=8$.


问:
What is this limit called? Is it a different kind of derivative?
(first I should notice you this is not something I can look up in a textbook, because I'm learning partial derivatives, alike I do with most Maths, as a hobby. If something below is wrong, blame the internet and please let me know)

I know I can take a mixed partial derivative of a multi-argument function, for expl.:

$$
w = f(x; y; z) = x^2 y+ y^2z + xz^2\\
\frac{\partial^2 w}{\partial x\ \partial y} = \frac{\partial w}{\partial x}\frac{\partial w}{\partial y} = \frac{\partial}{\partial y}(2xy+z^2)= 2x = ^*\frac{\partial}{\partial x}(2yz+x^2)
$$
(* = double check)

But that's taking 2 partial derivatives. What If I want more than one variable to vary at once?
$$
\lim_{h \to 0} \left( \frac{f(x+h;y+h;z)-f(x;y;z)}{h}\right)=\\
=\lim_{h \to 0} \left( \frac{(x+h)^2 (y+h)+(y+h)^2 z+(x+h)^2z^2-(x^2 y+ y^2z + xz^2)}{h}\right)=\\
=\lim_{h \to 0} \left( \frac{(x^2+2xh+h^2)(y+h)+(y^2+2hy+h^2) z+(x+h)z^2-(x^2 y+ y^2z + xz^2)}{h}\right)=\\
=\lim_{h \to 0} \left( \frac{2hxy+h^2y+hx^2+h^2x+h^3+2hyz+h^2z+hz^2}{h}\right)=\\
=\lim_{h \to 0} \left( 2xy+hy+x^2+hx+h^2+2yz+hz+z^2\right)
=2xy+x^2+2yz+z^2
$$

Note: I took a while to write this, and only now I notice that the result of the above is the sum of $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$, while the mixed partial in the beginning is its multiplication

Nonetheless, the question remains: what is this limit called? Can it be seen as another type of derivative? What are its applications? Is there a notation in which I can write it such as the one for other derivatives?

Thank you readers in advance (and for reading, my questions are not usually short).

答:
For, what you're looking at is a special case of a directional derivative of a function from $\mathbb{R}^n$ to $\mathbb{R}$, i.e. a derivative along some direction (vector) $d$. In general, those are defined as $$
  f_d(x) = \lim_{t\to 0} \frac{f(x+td) - f(x)}{t} \text{,}
$$
where $x,d \in \mathbb{R}^n$ and $t \in \mathbb{R}$. You looked at the special case $d=(1,1,0)$, i.d. you let the first and the second argument vary simultaneously and the third not at all.

Sometimes (or often, depending on what functions you usually work with), the directional derivatives are a linear function of the direction. In that case, the function is simply called differentiable. In other words, you then have that $$
  f_{u+v}(x) = f_{u}(x) + f_{v}(x) \text{.}
$$
If that is so, then it's sufficient to know e.g. the derivatives in the direction of all the axes to compute any directional derivative. Since the derivatives in the direction of the axes are simply the ususal partial derivatives, this means that in this case $$
  f_{d} = \sum_{i=1}^n d_i \frac{\partial f}{\partial x_i}
  \quad\text{assuming that}\quad
  d = (d_1,\ldots,d_n).
$$
Which is exactly what you found - you computed $f_{(1,1,0)}$ and found it to be $1\frac{\partial f}{\partial x} + 1\frac{\partial f}{\partial y} + 0\frac{\partial f}{\partial z}$.

If this works, it's custom to arrange the partial derivatives into a vector (or matrix, if the function also produces multiple values). The derivative of a function $f \,:\, \mathbb{R}^n \to \mathbb{R}$ is thus the vector $$
  \frac{df}{dx} = f' = \left(\frac{\partial f}{\partial x_1},\ldots, \frac{\partial f}{\partial x_n}\right).
$$
Similarly to the one-dimensional case, you can then approximate $f$ around $x$ by (note that $\cdot$ is the scalar product!) $$
  f(x+d) \approx f(x) + f'(x)\cdot d \text{.}
$$

Note that this doesn't always work - one example is $$
  f(x,y) = \begin{cases}
    \frac{xy}{\sqrt{x^2 + y^2}} &\text{if $(x,y)\neq(0,0)$} \\
    0                           &\text{otherwise}
  \end{cases}
$$
at $(0,0)$. Both partial derivatives are $0$ at $(0,0)$, yet the derivative in direction $(1,1)$ isn't. This function therefore is said to be non-differentiable at $(0,0)$, even though the partial derivatives do exists. But they don't convey all the information there is to know about the function's behaviour around $(0,0)$.


问:
Prove that $\sigma \mapsto A_\sigma $ is a group homomorphism $S_n\to GL_n(\mathbb R)$.
For $\sigma \in S_n$, let $A_\sigma $ the matrix associated de the linear map $\lambda_\sigma  :\mathbb R^n\to \mathbb R^n$ defined by $\lambda (x_1,...,x_n)=(x_{\sigma (1)},...,x_{\sigma (n)}).$ Prove that $\varphi :S_n\to GL_n(\mathbb R)$ defined by $$\varphi (\sigma )=A_\sigma .$$
is a group homomorphism. Describe the action on $GL_n(\mathbb R)$ of $S_n$ induced by $\varphi $. I recall that $GL_n(\mathbb R)$ are the $n\times n$ invertible matrix and $S_n$ is the group of permutation of $\{1,...,n\}$.

Q1) For the group homomorphism, let $\sigma ,\tau\in S_n$. So indeed $$\varphi ( \sigma )\varphi (\tau)(x_1,...,x_{n})=\varphi (\sigma )(x_{\tau(1)},...,x_{\tau(n)})=(x_{\sigma \circ\tau(1)},...,x_{\sigma \circ \tau(n)})=\varphi (\sigma \circ \tau)(x_1,...,x_n).$$

But I don't understand why as follow it's wrong : set $(y_1,...,y_n)=(x_{\tau(1)},...,x_{\tau(n)})$. Then $$\varphi (\sigma )\varphi (\tau)(x_1,...,x_n)=\varphi (\sigma )(x_{\tau(1)},...,x_{\tau(n)})=\varphi (\sigma )(y_1,...,y_n)=(y_{\sigma (1)},...,y_{\sigma (n)}).$$

And since $y_{i}=x_{\tau(i)}$ for all $i$ we have that $y_{\sigma (i)}=x_{\tau(\sigma (i))}$, and thus
$$\varphi (\sigma )\varphi (\tau)(x_1,...,x_n)=(x_{\tau\circ\sigma (1)},...,x_{\tau\circ\sigma (n)})=\varphi (\tau\circ \sigma )(x_1,...,x_n).$$
I can see that the last argument is not correct, but I don't understand what's wrong. Any idea ?

Q2) What means Describe the action on $GL_n(\mathbb R)$ of $S_n$ induced by $\varphi $. ? Would it be : it permute row and column of $A$ ?

答:
I think I disagree with the question as stated. I claim that your first argument is incorrect, and that your second argument proves that the given map is not a group homomorphism. A slight modification gives what I think is the correct map.

First of all, I'd like to say in words what I think the given map $\varphi$ does:

$\varphi(\sigma)$ takes the tuple $(x_1,\ldots ,x_n)$, and for each $i$, moves the element in position $\sigma(i)$ to position $i$.

Now, to address argument 1. I disagree with the statement

$$\varphi (\sigma )(x_{\tau(1)},...,x_{\tau(n)})=(x_{\sigma \circ\tau(1)},...,x_{\sigma \circ \tau(n)}).$$

The definition of the map in words makes it clear that $\varphi(\sigma)$ should put in position $i$ the element in position $\sigma(i)$ from its argument. This is clearly the element $x_{\tau(\sigma(i))}$.

You may counter by saying "Well we defined $\varphi(\sigma)$ to map $x_{i} \mapsto x_{\sigma(i)}$. So we just apply $\sigma$ to the index. How is that wrong?". The reason is that that sentence does not properly define a function. To take an example, consider the case $n=3$ and the permutation $(1 2 3)$. Clearly, the definitions agree:

$$\varphi((123))(x_1,x_2, x_3) = (x_2,x_3,x_1)$$
So, $$A_{(123)} = \pmatrix{0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0}.$$
But what if we labelled the variables differently? What is $\varphi((123))(x_2,x_1,x_3)$? Well, we simply swapped two variables, so it should be $(x_1, x_3, x_2)$, which is what the wordy definition predicts. If we follow our more recent definition, however,

$$\varphi((123))(x_2,x_1, x_3) = (x_{\sigma(2)}, x_{\sigma(1)}, x_{\sigma(3)}) = (x_3, x_2, x_1)$$

I hope it is therefore clear that the definition leading to argument 1 cannot be correct, and that is where the argument fails. On the other hand I think your second argument is good: we therefore have the property

$$\varphi(\sigma \circ \tau) = \varphi(\tau) \circ \varphi(\sigma)$$

which is not a group homomorphism. However, as is pointed out in the comments, the related map $\psi(\sigma) = \varphi(\sigma^{-1})$ is a group homomorphism:

$$\psi(\sigma \circ \tau) = \varphi(\tau^{-1} \circ \sigma^{-1}) = \varphi(\sigma^{-1}) \circ \varphi(\tau^{-1}) = \psi(\sigma) \circ \psi(\tau).$$

As for describing the action, I suggest you work through an example. It permutes the rows but not the columns.


问:
Finding the number of 7 letter words from A, B, C, D, E, F, G in which A, B, C each occurs at least once.
I've attempted doing this question using the inclusion-exclusion principle. But I have an answer which apparently isn't right. So I think my understanding might be a bit off. If someone could clarify where I went wrong or show me an alternate solution I'd appreciate it.

Finding the number of 7 letter words from A, B, C, D, E, F, G in which A, B, C each occurs at least once.

Total words = $7^7$

Words which don't include A, B, C = $3(6^7) - 3(5^7) - 4^7$

答:
Let $X$ be the set of all seven letter words using these letters with no further restrictions.  Let $N_A$ be the set of such words which contain no $A$'s.  Let $N_B$ be the set of words which contain no $B$'s and $N_C$ no $C$'s.

You are trying to calculate $|N_A^c\cap N_B^c\cap N_C^c|$, the number of words containing at least one $A$ and at least one $B$ and at least one $C$.

We recognize that this is $|X\setminus (N_A\cup N_B\cup N_C)|=|X|-|N_A\cup N_B\cup N_C|$ which expands further by inclusion-exclusion principle as:

$$|X|-|N_A|-|N_B|-|N_C|+|N_A\cap N_B|+|N_A\cap N_C|+|N_B\cap N_C|-|N_A\cap N_B\cap N_C|$$

Each term of this should be familiar how to calculate at this point.

 $7^7 - 3\cdot 6^7 + 3\cdot 5^7 - 4^7$

Note in particular how the signs change when expanding via inclusion-exclusion.


问:
A general form of a circle throw common points of two circles
We have two non intersecting in $\mathbb R^2$ circles:
$
C_1(x,y)\equiv x^2+y^2-2mx-2ny+p=0,
$
$
C_2\equiv x^2+y^2-2m'x-2n'y+p'=0,
$
where circles are with real coefficients and positive radiuses.
Then they intersects in some $(a,b), (a',b')\in \mathbb C^2$.
Let $C$ be an arbitrary circle, with or without real points, with real coefficients, passing throw $(a,b), (a',b')$.
I wish to know, is $C$ is the form  $\alpha C_1(x,y)+\beta C_2(x,y)$, for some real $\alpha, \beta$ ?

Thanks

答:
The answer is YES

Given
$$C_1:x^2+y^2-2mx-2ny+p=0;\;C_2:x^2+y^2-2m'x-2n'y+p'=0$$
The linear combination

$$\mathcal{C}:\lambda C_1+\mu C_2=0$$
$\mathcal{C}$ gives the family of all circles, real or complex, passing through the intersection points of the two given circles.

For the special values $\lambda=1;\;\mu=-1$ we get the equation of the line joining the intersection points $(a,b), (a',b')$ wich is perpendicular to the line containing all the centers of $\mathcal{C}$

$(2m'+2m)x+(2n'-2n)y+p-p'=0$


问:
Elementry probability question - counter probability or addition.
Say there are 5 red balls, 7 blue balls and 8 green balls, the probability of taking 2 balls (each time one ball) of different colour could be calculated

chance of taking red and blue  $ \Biggl(\frac{5}{5+7+8}*\frac{7}{5+7+8-1}\Biggr)*2$

chance of taking red and green $ \Biggl(\frac{5}{5+7+8}*\frac{8}{5+7+8-1}\Biggr)*2$

chance of taking blue and green $ \Biggl(\frac{7}{5+7+8}*\frac{8}{5+7+8-1}\Biggr)*2$

Adding these all up gives a probability of roughly 0.54

counter-probability (1-the probability of the second ball being the same as the first): $$ P= 1-\left(\frac{5}{5+7+8}\right)\left(\frac{4}{5+7+8-1}\right)-\left(\frac{7}{5+7+8}\right)\left(\frac{6}{5+7+8-1}\right)-\left(\frac{8}{5+7+8}\right)\left(\frac{7}{5+7+8-1}\right) = 0.69 $$

Now, I'm wondering how these two different methods yield different answers. They should be the same right?  Can someone explain why this happens?

Thank you very much

答:
Both methods should indeed give the same result.  You must have an arithmetic error.

"Adding these all up gives a probability of roughly $0.54$"

No they don't.  Perhaps you forgot to multiply the last term by two?


问:
Show that $\exp: \mathfrak{sl}(n,\mathbb R)\to \operatorname{SL}(n,\mathbb R)$ is not surjective
It is well known that for $n=2$, this holds. The polar decomposition provides the topology of $\operatorname{SL}(n,\mathbb R)$ as the product of symmetric matrices and orthogonal matrices, which can be written as the product of exponentials of skew symmetric and symmetric traceless matrices. However I could not find  out the proof that  $\exp: \mathfrak{sl}(n,\mathbb R)\to\operatorname{SL}(n,\mathbb R)$ is not surjective for $n\geq 3$.</p>

答: I came up with an example. Consider diag(-2,-1/2, 1, ..., 1) in SL(n,R). Clearly the lattice of logarithm of this does not intersect sl(n,R).</p>


问:
Issue finding the models of a formula using semantic tableux method.
Using semantic tableux method, decide what kind of formula U is. If it is consistent, find its models.

U = (p V q → r) → (p V r → q).

This is the semantic tableux I got, is it correct?

DNF(U) = ( ¬ r ^ p) V (¬ r ^ q) V ( ¬ p ^ q) V ( ¬ r ^ q) by absorption law

≡ ( ¬ r ^ p) V (¬ r ^ q) V ( ¬ p ^ q)

Is it correct so far? Then how do I find its models?

答:
Your tableau is correct, though I don't understand why afterwards you compute the DNF (is this asked in the question?).

To construct a model from an open branch, define the valuation such that it assigns all unnegated propositional variables on the branch $\text{True}$ and all negated propositoinal variables on the branch $\text{False}$. For example, for the formula $p \to (q \lor r) \land (q \to \neg r)$:

Since not all propositional variables always occur on an open branch, the resulting models may be partial models, meaning that some variables remain unspecified. In this case their truth value does not matter, so you can extend the partial model to a full model by adding arbitrary truth value assignments for the undefined variables. For instance, from $\mathcal{S}_{10}$ above you can obtain two total models by defining $\mathcal{V}_{10.1} : p \mapsto 1, q \mapsto 0, r \mapsto 1$ and $\mathcal{V}_{10.2} : p \mapsto 0, q \mapsto 0, r \mapsto 1$ respectively.


问:
Is the boundary of a subset a subset of the supersets boundary
Suppose there is a set $X\subset \mathbb{R}^3$. Define $(A,B)$ to be the separation of $X$ iff:

$A, B\subset X$

$A \cup B=X$

$A\cap B=\partial A\cap \partial B$

Define the "cut" of separation $(A,B)$ to be $C=A\cap B$.

Does this imply that $\partial C\subset\partial X$?

Intuitively it seems true, but I can't seem to prove it. I'm new to topological proofs, so I'm probably missing something quite obvious. Any help would be appreciated. Thanks!

答: No, here's a counterexample. Take $X$ to be the closed unit ball. Take $A$ and $B$ both to be a closed half of $X$. Then $C$ is a disc that divides $X$ in two halves, but this is not part of the boundary of $X$.


答: No. For example, if $X=[0,1]^3$, $A=[0,\frac 12]\times [0,1]^2$, $B=[\frac 12, 1]\times[0,1]^2$, then $C=\partial C=\{\frac 12\}\times[0,1]^2\not\subseteq \partial X$.

Title: Is the boundary of a subset a subset of the supersets boundary

Body:
Suppose there is a set $X\subset \mathbb{R}^3$. Define $(A,B)$ to be the separation of $X$ iff:

$A, B\subset X$

$A \cup B=X$

$A\cap B=\partial A\cap \partial B$

Define the "cut" of separation $(A,B)$ to be $C=A\cap B$.

Does this imply that $\partial C\subset\partial X$?

Intuitively it seems true, but I can't seem to prove it. I'm new to topological proofs, so I'm probably missing something quite obvious. Any help would be appreciated. Thanks!


问:
Fields and Irreducible Polynomials
I'm trying to make sense of this problem and solution.

Let $K$ be a field, and let $P(X)$ be an irreducible polynomial of degree $n$ with coefficients in $K .$ Let $Q(X)$ be a polynomial with coefficients in $K$. Show that every irreducible factor of $P(Q(X))$ has degree divisible by $n$.

Proof: Let $T(X)$ be an irreducible factor of $P(Q(X)),$ and consider the field $K(\alpha)$ where $\alpha$ is a root of $T(X) .$ Then $Q(\alpha)$ is a root of $P(X),$ so $K(Q(\alpha))$ has degree $n$ over $K .$ Since $K(Q(\alpha))$ is clearly a subfield of $K(\alpha),$ the degree of $K(\alpha)$ over $K$ (which is equal to the degree of $T(X)$) is divisible by $n$.

Question 1: I was taught that $K(\alpha)$ was an "abuse of notation", in that $\alpha := x + K[X]/\langle T(X) \rangle$ and $K(\alpha):= K[X]/\langle T(X) \rangle$. In the content above, $K(\alpha)$ almost seems to be treated as the field of fractions of the ring $K[\alpha]$, where $\alpha$ is to be treated not as $\alpha := x + K[X]/\langle T(X) \rangle$ but as an element in a larger field $L$, where $K \subset L$. Is my proof written down below correct?

Question 2: I do not understand how $\alpha$ being a root of $T(X)$ implies $Q(\alpha)$ being a root of $P(X)$. (I do not know how to justify this rigorously, in the language of $\alpha := x + K[X]/\langle T(X) \rangle$ and $K(\alpha):= K[X]/\langle T(X) \rangle$. If there is a simpler way to view things, please teach me!)

Please consider writing the answers out in detail, I get confused when they are too compact...

For question 1, I think this approach is relevant: As $K[X]/\langle T(X) \rangle$ is a field due to the irreducibility of $T(X)$ in the PID $K[X]$. $\alpha := x + K[X]/\langle T(X) \rangle$ is an element of that field. Clearly $K, \alpha \subset K[X]/\langle T(X) \rangle$, thus $K'\subset K[X]/\langle T(X) \rangle$ where $K'$ is the field of fractions of $K[\alpha]$. Elements of $K[X]/\langle T(X) \rangle$ are a $K$-linear dependence of $\alpha^n$, thus $K'\supset K[X]/\langle T(X) \rangle$.

Partial answers, e.g. only answering question 1, are very welcome!

答:
If $T(\alpha)=0$ then $P(Q(\alpha))=0$ thus $Q(\alpha)$ is a root of $P(X)$.  Here $\alpha$ is an element of a splitting field for $f(X)=P(Q(X))\in K[X]$, those splitting fields exist and are all isomorphic.

That a splitting field exists is because we can factorize $f=\prod_j f_j$ in $K_0=K$, pick a non-linear factor, let $K_1=K_0[x_1]/(f_i)$, and factorize $f=\prod_j g_j$ in $K_1$, pick a non-linear factor, let $K_2=K_1[x_2]/(g_j)$, and so on until $f$ splits completely in $K_m$.

That the splitting fields are all isomorphic is because for $b$ in a field containing $K$ and algebraic with $K$-minimal polynomial $f(X)$ then $K(b)\cong K[t]/(f(t))$, then apply again an inductive argument.

That $K(Q(\alpha))$ is a subfield of $K(\alpha)$ is obvious, why do you think $|K|\le |L|$ appears somewhere ? Yes it is the fraction field of $K[Q(\alpha)]$, but since $Q(\alpha)$ is algebraic then $K[Q(\alpha)]=K(Q(\alpha))$.

Title: Fields and Irreducible Polynomials


问:
Proof the following trig series
Prove that
$$\frac{ \sin x}{ \cos x}+\frac{\sin2x}{\cos^{2}x}+\frac{\sin3x}{\cos^{3}x}+\cdots+\frac{\sin nx}{\cos^{n}x}=\cot x-\frac{\cos(n+1)x}{\sin x \cos^{n}x}$$

I am not necessarily looking for a solution to this question but a way to approach it, i.e. when I get a question like this in the test what should I do, what should I look for?

答:
$\textbf{HINT}:$

$\text{Use the fact that}$
$$\cos((n+1)x) = \cos(nx) \cos(x) - \sin(nx) \sin(x)$$
$\text{and obtain}$
$$\sin(nx) \sin(x) = \cos(nx) \cos(x) - \cos((n+1)x)$$
$\text{to conclude that}$
$$\dfrac{\sin(nx)}{\cos^n(x)} = \dfrac{\cos(nx)}{\sin(x) \cos^{n-1}(x)} - \dfrac{\cos((n+1)x)}{\sin(x) \cos^{n}(x)}$$

 $\text{Do a telescopic summation to get the result.}$

Title: Proof the following trig series


答:
The equation of a circle is

quadratic,

with equal coefficients of the squares, $x^2$ and $y^2$,

without mixed term $xy$.

Now the combination $\alpha C_1+\beta C_2$ is

quadratic,

with equal coefficients of the squares, $x^2$ and $y^2$,

without mixed term $xy$

hence it is the equation of a circle.

Plus, any point that belongs to both circles satisfies $\alpha C_1+\beta C_2=0$ and also belongs to that circle. So if the two given circles make two intersections, the whole family shares these intersections.

Notice that the coefficients of the squares can vanish, leaving the equation of a straight line as a degenerate case. For the same reason as below, this is the line through the two intersection points.

Title: A general form of a circle throw common points of two circles

Body:
We have two non intersecting in $\mathbb R^2$ circles:
$
C_1(x,y)\equiv x^2+y^2-2mx-2ny+p=0,
$
$
C_2\equiv x^2+y^2-2m'x-2n'y+p'=0,
$
where circles are with real coefficients and positive radiuses.
Then they intersects in some $(a,b), (a',b')\in \mathbb C^2$.
Let $C$ be an arbitrary circle, with or without real points, with real coefficients, passing throw $(a,b), (a',b')$.
I wish to know, is $C$ is the form  $\alpha C_1(x,y)+\beta C_2(x,y)$, for some real $\alpha, \beta$ ?

Thanks


问:
The Birthday Paradox with combinatorics
Assume a year has $365$ days, how many are required to have a $50%$ chance of $2$ people having the same birthday?

According to Scientific American, there are $23$ people needed to achieve the goal.

$${\begin{pmatrix} 23 \\ 2\end{pmatrix}}=253$$
$$1-(1-\frac{1}{365})^{253}\approx 0.50048$$

However, I have a different approach but I'm not sure if this is correct.

One could be any day in a year. And $23$ people would be $365^{23}$ possibilities.

Suppose no one in $23$ people has the same birthday.
That would be :- ${\begin{pmatrix} 365 \\ 23 \end{pmatrix}} \cdot23!$

The possibility of having at least 2 people having the same birthday is :-

$$1-\frac{{\begin{pmatrix} 365 \\ 23 \end{pmatrix}} \cdot23!}{365^{23}}\approx0.5072972$$

Although $0.50048\approx 0.5072972$, these two numbers are not equal to each other.

Can anyone explain this difference and the reason behind this phenomenon?

答:
The Scientific American's calculation is not quite accurate: it supposes that every non-match between two people is independent from any other non-match between two people, but that is not the case.

To see this, consider just 3 people: A, B, and C.

What is the chance that they don't share a birthday?

Scientific American's approach is to say: well, we need the following three events:

E1. A and B don't share a birthday

E2. A and C don't share a birthday

E3. B and C don't share a birthday

Now, individually, the probability of each of those events is $\frac{364}{365}$

So, they say: $P(E1 \cap E2 \cap E3) = P(E1) \cdot P(E2) \cdot P(E3) =  (\frac{364}{365})^3$

But again, that is assuming these three events are independent ... and they are not! For example, once we know that $A$ does not share a birthday with either $B$ or $C$ (i.e. once we have events $E1$ and $E2$) the probability of $B$ not sharing a birthday with $C$ (i.e. event $E3$) is no longer a simple $\frac{364}{365}$, because it is no longer possible for $B$ or $C$ to have their birthday on the day that $A$ has their birthday.

Indeed, the more people we know do not share a birthday with $B$ and $C$, the fewer options become available for $B$ and $C$ to have their birthdays, and this will in fact increase the chances of $B$ and $C$ to share a birthday, and thus decrease the chance of them not sharing a birthday.

To illustrate how the probability of $B$ and $C$ not sharing their birthday is affected once we know they don't share the same birthday with $A$ (how, indeed, that probability has gone down), consider what happens when we consider only three possible days that each of $A$, $B$, and $C$ can have their birthday.

First of all, the probability of $B$ and $C$ sharing their birthday not knowing anything about $A$ is $\frac{1}{3}$, and hence there is a $\frac{2}{3}$ probability that they don't share their birthday.

OK, but now let's add the information that $B$ nor $C$ share their birthday with $A$. That means that there are now only two days left for $B$ and $C$ to have their birthday, meaning that now there is a $\frac{1}{2}$ probability that they share their birthday, and hence a $\frac{1}{2}$ probability that they don't share their birthday. So, indeed, the probability that $B$ and $C$ don't share their birthday has gone down once we know they don't share their birthday with $A$.

And indeed, note what happens when we have even fewer possible birthdays: if we get to a point where we have more people than possible birthdays, then the probability of two people not matching their birthdays given that no one else shares their birthday with any of these two people simplybecomes $0$.

In sum: the probability that Scientific American assumes for each non-match is too high.

Now, with the actual birthday problem we are not considering $3$ people and $3$ possible birthdays, but $23$ people with $365$ possible birthdays. And with so many more birthdays than people, it turns out that Scientific American's calculations are not going to be that far off. But still: the $\frac{364}{365}$ is a little too high for the remaining non-matches once you already know other non-matches, meaning that the $(\frac{364}{365})^{253}$ is a little too big, and thus when subtracted from $1$, the Scientific American's calculation ends up being a little too low.

But of course, it would be hard to tell this beforehand (note that the $\frac{364}{365}$ does get raised to the $253$-th power!) ... It happened to get 'close enough' for this problem ... but I really don't like to think that it was an a priori justified simplification. You certainly can't use this kind of approximation to establish that $23$ people was the magic 'breaking point' as opposed to $22$ people, or maybe even fewer. Indeed, they we're lucky that with their calculation for $23$ people it just stayed above $0.5$.

Your calculation, however, is completely correct. Good job!!

Title: The Birthday Paradox with combinatorics


问:
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$?
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ?
i know that its "1", but why?
How can i calculate this?

Thank you very very much =)

答:
Let $ \log x = u $.

Then $ u \to 0^+ $ and you have the limit:

$\lim_{u\to 0^+} u^{u}$

which can be calculated if we consider the fact that

$ u^u = e^{ulnu}$ for $u>0$

and then apply L'Hopital's rule to $\lim_{u\to 0^+}ulnu$ :)


答:
The SciAm article seems to be wrong. The reason is, I believe, this: let's say with three (rather than $23$) birthdays $b_1,b_2,b_3$ they treat $b_1\ne b_2$, $b_2\ne b_3$ and $b_1\ne b_3$ as independent events, which they are not:

$$P(b_1\ne b_3\mid \lnot(b_1\ne b_2), b_2\ne b_3)=1\ne P(b_1\ne b_3)$$

for example. (Knowing that $b_1=b_2$ and $b_2\ne b_3$ makes it a certainty that $b_1\ne b_3$.)

Title: The Birthday Paradox with combinatorics

Body:
Assume a year has $365$ days, how many are required to have a $50%$ chance of $2$ people having the same birthday?

According to Scientific American, there are $23$ people needed to achieve the goal.

$${\begin{pmatrix} 23 \\ 2\end{pmatrix}}=253$$
$$1-(1-\frac{1}{365})^{253}\approx 0.50048$$

However, I have a different approach but I'm not sure if this is correct.

One could be any day in a year. And $23$ people would be $365^{23}$ possibilities.

Suppose no one in $23$ people has the same birthday.
That would be :- ${\begin{pmatrix} 365 \\ 23 \end{pmatrix}} \cdot23!$

The possibility of having at least 2 people having the same birthday is :-

$$1-\frac{{\begin{pmatrix} 365 \\ 23 \end{pmatrix}} \cdot23!}{365^{23}}\approx0.5072972$$

Although $0.50048\approx 0.5072972$, these two numbers are not equal to each other.

Can anyone explain this difference and the reason behind this phenomenon?


问:
Cambridge Admissions Exam 1987 Paper
The problem I am attempting is the following

I got pretty far into it, however, I was not able to do the last step of the solution.  The given solution is here

What I don't understand: I don't understand where  the last equation comes from. Can someone clarify where it comes from?

答:
It's just the moment of force balance. In the statement of the problem, it's said that the distance from the axis to the center of mass is $\cfrac{3\pi a\sin \theta}{16\theta}$.

Basically, in the following formula $$F_1 l_1 = F_2 l_2$$ we have $F_1 = \cfrac{Mg\theta}{\pi}$,$~ l_1 =\cfrac{3\pi a\sin \theta}{16\theta}$ and $F_2 = 2T\sin \theta$, $~l_2 = a$.

Title: Cambridge Admissions Exam 1987 Paper


问:
Concentration Results for the Range of Gaussian
Let $X_1,..., X_n$ be an i.i.d. standard normal.

Let $X_{(1)}= \min \{ X_1,X_2,\ldots, X_n \}$ and $X_{(n)}= \max \{ X_1,X_2,\ldots, X_n \}$  and define the range as $R=|X_{(n)}-X_{(1)}|$.

Are there any concentration results for the range?  That is can we find bound on
\begin{align}
\mathbb{P}[R  \ge t ] \le ????
\end{align}

I know that there are concentration results for the maximum like this is on
\begin{align}
\mathbb{P}[ X_{(n)} \ge \sqrt{  2 \log (2n) }+t ] \le 2e^{- \frac{t^2}{2}}.
\end{align}

I had a feeling that because of symmetry the range can be estimated by $R$ approximate by $2 | X_{(n)}|$. Not sure how to do this though.

答:
Maybe you want a tighter bound than the straightforward one in the following?
$$ \mathbb{P}[ R_n \ge 2\sqrt{2 \log(2n)}+2t ] \le 4e^{-t^2/2} $$
Which I doubt to have any kind of better estimation.

Update

You see , by symmetry, $-X_{(1)}$ also has the same distribution as $X_{(n)}$ .

Then we have this straight forword probability relation for all $T>0$
$$ \mathbb{P}( R_n > 2T ) = \underbrace{\mathbb{P}( R_n > 2T, X_{(n)}>T )}_{ \le \mathbb{P}( X_{(n)}>T )}+\underbrace{\mathbb{P}( R_n > 2T, X_{(n)}\le T, -X_{(1)} > T )}_{\le \mathbb{P}(-X_{(1)} > T) } + \underbrace{\mathbb{P}( R_n > 2T, X_{(n)}\le T, -X_{(1)} \le T )}_{=0}$$
Thus $$ \mathbb{P}( R_n > 2T )\le 2\mathbb{P}( X_{(n)}>T )$$

By choosing $T:= \sqrt{2 \log(2n)}+t$, we'll the above result.


问:
Identify the isolated singularities of the $f(z) = \frac{\cos(1/z)}{(z-1)^3}$. If the isolated singularity is a pole, find its order.
I am trying to solve this problem.

Identify the isolated singularities of the $f(z) = \frac{\cos(1/z)}{(z-1)^3}$. If the isolated singularity is a pole, find its order.

First what I did is I solve the zero for $(z- 1)^3$ and found out that the zero is $z = 1$.  I don't have to be bother with the $\cos \frac{1}{z}$ because it is in the numerator so whatever is, it does not affect the analyticity of the function $f$. Right?

Now what I did instead on the numerator is the $\frac{1}{z}$ of the $\cos \frac{1}{z}$ because such function won't be analytic when $z = 0$.

Does it mean that the points of singularities of $f$ are z = 0, 1? IF YES.  Is that a pole? meaing the order of $z = 0$ is 1 and the order of $z = 1$ is 3?

Please help me. Thanks

答: The singularities of $f$ are located at $z=0$ and at $z=1$, yes. And $1$ is a pole of order $3$, since $\cos(1)\ne0$. On the other hand, $f$ has an essential singularity at $0$, not a pole. In fact, if $f$ had a pole or a removable singularity at $0$, then $(z-1)^3f(z)$ would also have a pole or a removable singularity there. But$$\cos\left(\frac1z\right)=1-\frac1{2!}z^{-2}+\frac1{4!}z^{-4}-\frac1{6!}z^{-6}+\cdots$$


答:
If you note that
$$
\cos \frac 1z = \sum_{n=0}^{\infty} \frac{(1/z)^{2n}}{(2n)!}
$$

you'll see that the Laurent series has an infinite number of negative order terms. Therefore, $z=0$ is not a pole, it is an essential singularity.

Body:
I am trying to solve this problem.

Identify the isolated singularities of the $f(z) = \frac{\cos(1/z)}{(z-1)^3}$. If the isolated singularity is a pole, find its order.

First what I did is I solve the zero for $(z- 1)^3$ and found out that the zero is $z = 1$.  I don't have to be bother with the $\cos \frac{1}{z}$ because it is in the numerator so whatever is, it does not affect the analyticity of the function $f$. Right?

Now what I did instead on the numerator is the $\frac{1}{z}$ of the $\cos \frac{1}{z}$ because such function won't be analytic when $z = 0$.

Does it mean that the points of singularities of $f$ are z = 0, 1? IF YES.  Is that a pole? meaing the order of $z = 0$ is 1 and the order of $z = 1$ is 3?

Please help me. Thanks


问:
Image of domain under map $f(z) = z^6$?
I am given the domain $\Omega := \{z \in \mathbb{C}: \operatorname{Im}(z) > \sqrt{3}|\operatorname{Re}(z)|\}$. I want to know what the image $f(\Omega)$ of this domain is under the mapping $f(z)=z^6$. I'm not sure if my method is correct, here is my attempt:

Having sketched the domain, I think I can rewrite $\Omega$ as:
$$\Omega = \{z=re^{i\theta} \in \mathbb{C}: r>0, \frac{\pi}{3} < \theta \leq \frac{\pi}{2}\} \cup \{z=re^{i\theta} \in \mathbb{C}: r>0, \frac{\pi}{2} < \theta < \frac{2\pi}{3}\}$$
Then, using the reasoning that $f$ sends $z=re^{i\theta}$ to $z^6=r^6e^{6\theta i}$, I have said that the domain is mapped to:
$$f(\Omega) = \{z=re^{i\theta} \in \mathbb{C}: r>0, 2\pi < \theta \leq 3\pi\} \cup \{z=re^{i\theta} \in \mathbb{C}: r>0, 3\pi < \theta < 4\pi\}$$
which I am interpreting as: if $f(\Omega)$ was a sketch on the complex plane, it would be the whole plane minus the positive real axis.

Is this correct? If it is, is there a quicker or more intuitive/interesting way to arrive here?

答:
You have the right idea, but I'm a bit confused by your description of the domain. Surely the union of those two sets is simply
$$\Omega=\{re^{i\theta}:\ r>0,\ \tfrac13\pi<\theta<\tfrac23\pi\}.$$
Either way, your reasoning from here is correct, and then the image $f(\Omega)$ can similarly be described simply as
$$f(\Omega)=\{re^{i\theta}:\ r>0,\ 2\pi<\theta<4\pi\},$$
which is indeed the entire complex plane minus the positive real axis. Also, don't forget that you're missing $0\in\Bbb{C}$ is well; perhaps it would be less ambiguous to say that it's the entire complex plane minus the nonnegative real axis.

Another way to see this, though less rigorously, is to note that $\Omega$ is an open region bounded by two half-lines. These two half-lines are both mapped to the nonnegative real axis, and so $\Omega$ is mapped to an open region bounded by the nonnegative real axis. This must then be precisely $\Bbb{C}$ minus the nonnegative real axis. This can of course be made rigorous as well.

Here's an interactive piece of geometry that shows what $f(\Omega)$ looks like for $f(z)=z^a$, as $a$ varies. The red region is the part of $f(\Omega)$ that is inside the unit disk. It breaks at $a=6$, unfortunately.

Title: Image of domain under map $f(z) = z^6$?


问:
Number of ways to distribute $n$ items to $k$ groups given number of items required for each group.
The problem goes as follows:

we have 20 workers and we need to distribute them to 4 departments,
department 1 needs 6 workers, department 2 needs 4 workers, department
3 needs 5 workers, department 4 needs 5 workers, by how many ways can
distribute these workers?

My attempt:

since workers are not identical so the order is important then the
answer should be   = $$4^6 \times 3^4 \times 2^5 \times1^5$$

but I do not believe that to be the correct answer as one department can have the same workers but different order and still be considered another way to arrange the workers,

I tried fixing this by dividing by possible arrangments of workers in
each department

= $$(4^6 \times 3^4 \times 2^5 \times1^5) \over 6! \times  4! \times5!  \times 5!$$ $$$$

I ended up having a decimal number so my answer is of course incorrect

what has gone wrong? is my intuition incorrect?

thanks in advance.

NOTE: I know the problem is not the same as what the title says but I
wanted to generalize

答:
Notice that when you do $4^6$ you are essentially taking $6$ workers out of nowhere to place them in one of the $4$ departments. This never guaranties that you have $6$ workers in the first department.

Edit: When you do $4^6$ means that you are going to list tuples of length $6$ where each entry can have one of $4$ values. So you are counting $(d_1,d_2,d_3,d_4,d_5,d_6)$ where $d_i$ is the department in which the $i-$th person is working. Notice that these $6$ people are not specified, so you have to specify them by picking them. Furthermore, when you pick them, nothing guarantees that you will choose the tuple $(4,4,4,4,4,4)$ so in these arrangement you have put all your six workers in the department 4 (but this is not possible because department $4$ has just $5$ people).

The correct approach is, you have your workers, then lets say you want to build department $1,$ then you choose $6$ workers out of the total $20$ in $\binom{20}{6}$ ways. How many workers are left? $20-6,$ now choose the workers for department $2.$ Use the multiplication principle.

You should get something like $$\binom{20}{6}\binom{14}{4}\binom{10}{5}=\frac{20!}{6!\cdot 4!\cdot 5!\cdot 5!}.$$


问:
Show that $\log({n\choose n/2})=O(n)$
Show that $\log({n\choose n/2})=O(n)$,

I've tried to use stirling's formula but didn't get any close

答:
Hint

$$\binom{n}{\frac{n}{2}}=\frac{n!}{\left(\frac{n}{2}!\right)^2}$$
$$\log \left(\binom{n}{\frac{n}{2}}\right)=\log(n!)-2\log \left(\frac{n}{2}!\right)$$

Use Stirling approximation twice and continue with Taylor expansion.

Title: Show that $\log({n\choose n/2})=O(n)$


问:
About Galois theory
I have difficulty with the proof of this fact:

Prop: Let $L/K$ be an infinite Galois extension. Let $H$ be a subgroup of $Gal(L/K)$. Then $Gal(L/L^H)=\overline{H}$.

Proof of [$\overline{H} \subseteq {Gal(L/L^H)}$]:

Clearly we have $H \subseteq Gal(L/L^H)$.
Also, note that $Gal(L/L^H)$ is closed, so we have $\overline{Gal(L/L^H)}={Gal(L/L^H)}$.
Take closure, then we have $\overline{H} \subseteq \overline{Gal(L/L^H)}={Gal(L/L^H)}$.
$\checkmark \checkmark \checkmark$

Proof of [${Gal(L/L^H)} \subseteq \overline{H} $]:
Equivalently we will show that if $\sigma \not\in \overline H$, then $\sigma \not\in {Gal(L/L^H)}$:

Let $\sigma \not\in \overline H$.
We know that $\overline{H}$ is closed, so its complement is open.
Also, we know that the topology is Hausdorff.
So there is an open neighborhood of $\sigma$ which is disjoint with $\overline{H}$,
i.e. there is a finite Galois extension $F$, such that $\sigma Gal(L/F)\cap\overline{H}=\emptyset$.
Also notice that
$\sigma Gal(L/F)= \{\tau\mid \tau\restriction_F = \sigma\restriction_F \}$.
Also, the fact that $\sigma Gal(L/F)$ and $\overline{H}$ are disjoint,
implies that
$\sigma Gal(L/F)\cap{H}=\emptyset$.
So we can conclude that $\sigma\restriction_F\neq h\restriction_F$ for any $h \in H$.

My Question (The point I am stuck in):
Why there exists an $\alpha \in F$ such that $\sigma(\alpha)\neq \alpha$
and $h(\alpha)=\alpha$ for any $h \in H$.
Please explain it very clear, because I am totally confused.

I can not see the reason for the existence of such an element.
If we can find such an $\alpha$, then $\alpha \in L^H$.
So $\sigma \not\in Gal(L/L^H)$.
(Otherwise, $\sigma$ should fix any element of $L^H$,
in particular it should fix $\alpha \in L^H$,
for which we know that $\sigma(\alpha)\neq \alpha$).
$\checkmark \checkmark \checkmark$

If I can see the reason for that special question, then everything is solved.

答:
If $h\mid_F$ were to equal $\sigma\mid_F$, then $h$, by what you wrote previously, would appear in $\sigma\operatorname{Gal}(L/F)$ in which case it would be an element in the intersection $H\cap \sigma\operatorname{Gal}(L/F)\subseteq \bar{H}\cap\operatorname{Gal}(L/F)=\emptyset$.

Maybe your uncertainty arises in the imprecise notation with which $\sigma\operatorname{Gal}(L/F)$ is written, note that the $\tau$ here are considered to be inside of $\operatorname{Gal}(L/K)$

Title: About Galois theory


问:
proof by contrapositive if $x^2(y^2-2y)$ is odd, then $x$ and $y$ are both odd
proof by contrapositive,

prove if $x^2(y^2-2y)$ is odd, then $x$ and $y$ are both odd

Attempt

The contrapositive of the given statement is if $x$ or $y$ even then $x^2(y^2-2y)$ is even

so then I considered three cases

Case 1, Assumed $x$ even and $y$ is odd then proved $x^2(y^2-2y)$ is even

Case 2, Assumed $y$ even and $x$ is odd then proved $x^2(y^2-2y)$ is even

Case3, Assumed $x$ even and $y$ is even then proved $x^2(y^2-2y)$ is even

Then I said contrapositive statement is true and so given statement is true

Is my proof correct?

答:
You don't need three cases, just two: in your first case, you don't use at all the assumption that $y$ is odd and similarly in the second case.

If $x$ is even, then $x^2(y^2-2y)=x\bigl(x(y^2-2y)\bigr)$ is even.

If $y$ is even, then $x^2(y^2-2y)=y\bigl(x^2(y-2)\bigr)$ is even.

Your logic is fine, of course, but the aim of exercises of this kind is to test your skill at applying hypotheses when making deductions. While you certainly may at first consider the three cases, when attacking the problem, but then you should realize that two cases are enough.


问:
After many years, my Grandad told me his 'magic trick' isn't magic... but why does it work so often?
Take a standard pack of playing card, no jokers and no magic shop gimmicks. The 'Magician' invites you to name 2 cards, but without their suit; eg "two" and "queen" (ie but not "two of hearts" and "queen of spades") and says that he can make them 'stick together'.

From the pack face down, he turns over the first card off the top and then second... and third... and keeps going.  Almost every time (and that's the point of the question!) - almost every time, long before he's got to over the last card, he'll have dealt a "two" immediately followed by a "queen" or vice-versa. There they are, 'stuck together'.

The audience is impressed and thinks there is sleight of hand but no, he hands them the pack. They shuffle it and the 'magician' names 2 cards which they turn over themselves. It works! Someone says that the magician must have already known of 2 cards that would come up; so the magician asks the volunteer to choose anyone else to name 2 cards and they deal the cards out themselves a second time. It works again and the magician isn't even involved!

Sometimes, if it doesn't work,  the magician blames the 'novice' magician and berates them for not concentrating. Worst case, the 2 cards are normally only separated from being "stuck together" by one stray card. "Try again!" says the magician - and it works!  This delights children and confounds adults!

So, clearly down to probability and not magic but it almost always works.  Question is... what is the probability of it working on any given 'deal'?

Thanks
Mike

答:
Let us assume without loss of generality that the cards selected are Kings and Queens.

Consider the event $K_{\spadesuit}$ that the king of spades is adjacent somewhere in the deck to a some queen.  Similarly define $K_{\heartsuit},K_{\clubsuit}$ and $K_{\diamondsuit}$.

For the purposes of estimation, let us assume these are independent events and also let us assume that the deck "wraps around" so the top card is said to be adjacent also to the bottom card.  I know they are not, but it will help simplify calculations considerably and cause the numbers to be off by a relatively insignificant amount.

We try to find $\Pr(K_\spadesuit \cup \dots \cup K_{\diamondsuit})$ which expands by incusion-exclusion as $\Pr(K_\spadesuit)+\dots+\Pr(K_\diamondsuit) - \Pr(K_\spadesuit\cap K_{\heartsuit}) - \dots - \Pr(K_{\clubsuit}\cap K_{\diamondsuit}) + \Pr(K_\spadesuit \cap K_\heartsuit \cap K_{\clubsuit}) +\dots - \Pr(K_{\spadesuit}\cap \dots \cap K_{\heartsuit})$

Now, $\Pr(K_{\spadesuit}) = \frac{8}{51}-\frac{12}{51\cdot 50}\approx 0.152$

$\Pr(K_{\spadesuit}\cap K_{\heartsuit})$ then by our simplifications/assumptions is about $0.152^2$ and so on...

This gives our estimated probability as being about $4\cdot 0.152 - 6\cdot 0.152^2 + 4\cdot 0.152^3 - 0.152^4\approx 0.4828$, an astounding (to some) almost half of the time.  If we were to include allowing being separated by a card, that raises $\Pr(K_{\spadesuit})$ up to $\approx 0.286$ bringing the total to closer to $0.7401\dots$

Getting an exact value for this seems like a lot of work.  Approximations like the one used here should get close enough (I expect within $0.05$ of the true value).


问:
Show that if $u_n + u_{n+1} \sim \dfrac{1}{n}$ then $u_n \to 0$
Saw this question in my textbook : Let $(u_n)$ be a real decreasing sequence such that :

$$u_n + u_{n+1} \sim \dfrac{1}{n}$$
Show that $(u_n)$ converges to $0^{+}$ and give a simple sequence equivalent to $(u_n)$

My attempt : for every $\epsilon > 0$, we have for every $n\ge N $ :$$2u_n < u_n + u_{n+1} \le |u_n +u_{n+1}|\le \epsilon$$
It seems like I need to show that the terms of this sequence are all positive for this to work, I just don't know how to procede.

答:
$(u_n) $ is decreasing, so to prove it converges, you just need to show that

$$(\exists m\in\Bbb R)\;: (\forall n\in\Bbb N)\; u_n\ge m$$

Assume that

$$(\forall m\in \Bbb R)\;(\exists N\in\Bbb N) \;:\; u_N<m$$

then
$$(\forall n\ge N)\; u_n\le u_N<m$$
and with $m=-1$,
$$(\forall n\ge N)\; u_n+u_{n+1}<-2$$

thus $$\lim_{n\to+\infty}(u_n+u_{n+1})\ne 0$$
but
$$u_n+u_{n+1}\sim \frac 1n$$

Finally, if we put $$\lim u_n=L$$
we will have
$$L+L=0\; and\; L=0^+$$


问:
Koebe distortion theorem for general domains
I am currently studying the theory of elementary univalent functions, and am stuck on how to generalise the famous Koebe 1/4 theorem. I have seen the Koebe 1/4 theorem provided in many books, in the form that if $f:\Bbb{D}\rightarrow\Omega$ be a function of the elementary univalent class $S$ (defined as usual). Then $dist(0,\delta\Omega)\in [\frac{1}{4},1]$.

I now want to extend this, as many books do without proof, to the general case where $f:\Omega\rightarrow\Omega'$ is a univalent map, and $z$ is some point in $\Omega$. Then the inequality that follows reads that
$$\frac{1}{4}dist(f(z),\delta\Omega')\leq|f'(z)|dist(z,\delta\Omega)\leq4dist(f(z),\delta\Omega')$$

I am not sure how to relate these two inequalities, following the proof for the original Koebe inequality my instinct is to find some function, using our univalent $f$ so that our new function, call it $\phi$, also belongs to $S$, to that we can use Bieberbach's theorem to create boundsin the same way? I just am really unsure on how to go about this proof, and can't find it explicitly in the literature. Any help, or pointing in the right direction, would be really appreciated. Thank you.

答:
Fix $w \in \Omega$; then $g(z)=\frac{f(dist(w,\delta\Omega)z-w)-f(w)}{|f'(w)|dist(w,\delta\Omega)}$ maps the unit disc in an univalent fashion to some domain $W$, while $g \in S$ by our normalizations, hence by the usual Koebe theorem we have $dist(0,\delta W) \ge 1/4$

Noting that $dist(f(w),\delta\Omega') \ge dist(0,\delta W)|f'(w)|dist(w,\delta\Omega)$ we get the RHS inequality in the OP for $w$ arbitrary in $\Omega$:

$|f'(w)|dist(w,\delta\Omega)\leq 4dist(f(w),\delta\Omega')$

But now apply the result above with $f^{-1}$ at the point $f(w)$ (here obviously we must assume $f$ is onto as otherwise, the result is not true since we can take $\Omega$ the unit disc, $f=z$, $\Omega'$ a huge disc containing zero) and the result follows by a simple computation, again under the required assumptions that $f(\Omega)=\Omega'$ since ${f^{-1}}'(f(w))f'(w)=1$


问:
Disc with two points identified
Is a disc $D^2$ with two points on the boundary identified, same as $D^2 \vee D^2$ ? They both have boundary $S^1 \vee S^1$. I am confused because an exercise in Hatcher seems to ask the same question about the two spaces (i.e. to prove that there doesn't exist a retraction from each of this space to $S^1 \vee S^1$).</p>

答:
The two spaces are not even homotopy equivalent. I'll describe the intuition for the homotopy type of each space.

For $D^2$ with two (distinct) points on the boundary identified, visualize it as the cylinder $S^1 \times [0,1]$ with one part of the wall "pinched" to a point. It's easy to see this is homotopy equivalent to $S^1$.

$D^2 \vee D^2$ is contractible since we can just contract each copy of $D^2$ to the basepoint.

Title: Disc with two points identified


问:
If $X$ is a random variable and $a$ is a positive number then does $Y = \max (a,X)$ is a legitimate random variable?
If $X$ is random variable with a given PDF and $a$ is a positive number then could $Y = \max (a,X)$ be considered as a legitimate random variable ?

Mathematica result look strange !

For example, when $Y = \max (3,X)$ and $X$ follow the exponential distribution with parameter $\lambda$, could $Y$ be considered as a random variable ?

If so then how can I interpret the result attach in the picture ?

答:
A) Yes, $Y$ is a legitimate random variable as obtained through a legitimate operation (max) between two random variables: $X$ (with its exponential distribution) and a constant random variable equal to $a$ (with its pdf $\delta(x-a)$).

B) Let us work with the cdf of $Y$:

$$F_Y(x)=P(Y<x)=\begin{cases}0&\text{if }x<a\\P(X<x)=F_X(x)=1-e^{\lambda x}&\text{otherwise}\end{cases}\tag{1}$$

You can write (1) under the form:

$$F_Y(x)=e^{-\lambda x}U(x-a)\tag{2}$$

where $U$ is the unit step function at the origin ($0$ for negative arguments, $1$ for positive arguments, with derivative the $\delta(x)$ density function).

If we differentiate the cdf expression (2), we get the pdf into two parts :

$$f_Y(x)=\underbrace{P(X<a)\delta(x-a)}_{\text{accounting for the jump in } x=a (*)}+e^{-\lambda x}$$

(*) : with $P(X<a)=F_Y(a)=(1-e^{-\lambda a})$. Intuitively, by taking the "max", all the "mass" $1-e^{-\lambda a}$ before $x=a$ has been moved  into $x=a$.

It remains a little issue: Mathematica gives for the first term:

$$(1-e^{-\lambda  x})\delta(x-a) \ \text{instead of  } \ \ (1-e^{-\lambda a})\delta(x-a)$$

This is in fact the same thing because, for any function $\varphi$, $\varphi(x)\delta(x-a)$ is the same as $\varphi(a)\delta(x-a)$

(intuition: $\delta(x-a)$ ignores what is outside $x=a$).

Mathematica's result is right!


答:
Suppose that there exists $n_0$ such that $u_{n_0}<0$, this implies that for every $n>n_0, u_n\leq u_{n_0}$, we deduce that if  $n>n_0, |u_n+u_{n+1}|>2|u_{n_0}|$. Take $N$ such that $N>n_0$ and ${1\over N}\leq {{u_{n_0}}\over 4}$. For $n>N$,

$|u_n+u_{n+1}-{1\over n}|\geq 2|u_{n_0}|-{1\over n}\geq 2|u_{n_0}|-{{u_{n_0}}\over 4}={3\over 4}u_{n_0}$. Contradiction.

Since the sequence it positive and decreasing, it as a limit $l$, and $lim(u_n+u_{n+1})=2l=lim {1\over n}=0$ implies that $l=0$.

Title: Show that if $u_n + u_{n+1} \sim \dfrac{1}{n}$ then $u_n \to 0$

Body:
Saw this question in my textbook : Let $(u_n)$ be a real decreasing sequence such that :

$$u_n + u_{n+1} \sim \dfrac{1}{n}$$
Show that $(u_n)$ converges to $0^{+}$ and give a simple sequence equivalent to $(u_n)$

My attempt : for every $\epsilon > 0$, we have for every $n\ge N $ :$$2u_n < u_n + u_{n+1} \le |u_n +u_{n+1}|\le \epsilon$$
It seems like I need to show that the terms of this sequence are all positive for this to work, I just don't know how to procede.


问:
Using Ito calculus to show $d(\frac{1}{3} W_t^3) = W_t^2 dW_t + W_t dt$
I am trying to self study some stochastic calculus and have come upon a result I do not understand. Can anyone show me how
\begin{align}
d(\frac{1}{3} W_t^3) = W_t^2 dW_t + W_t dt
\end{align}

where $W_t \in \mathbb{R}$ is Brownian motion. My notes (from my supervisor) say that the result is found using Ito's rule. I am not sure how this is applied in this case though?

答:
Apply the Ito’s rule
$$df(W)= f’(W_t)d W_t + \frac12f’’(W_t)dt$$
to $f(W_t)=\frac13 W_t^3$. Then, $f’(W_t)= W_t^2$, $f’’(W_t)=2 W_t $ and
$$d(\frac{1}{3} W_t^3) = W_t^2 dW_t + W_t dt$$


问:
Real part of the square root of a complex number
I want to find the real part of the square root of a complex number $z$

$\operatorname{Re}(\sqrt(z))$

In general I am wondering if the following is true:

$\operatorname{Re}(\sqrt(z))$ = $\sqrt(\operatorname{Re}(z))$

Thanks

答:
Let $\sqrt{z}=x+iy$, and suppose that $\Re(\sqrt z)=\sqrt{\Re(z)}$, then $$x=\sqrt{\Re(x^2-y^2+i2xy)}=\sqrt{x^2-y^2}$$

Thus it holds if and only if $\sqrt{z}\in\Bbb R_{\geq 0}$.


问:
Product topology for $\mathbb{R}^{\mathbb{N}}$ and generating subbasis
The product topology on $\mathbb{R}^{\mathbb{N}}$ is defined to be the topology generated by the subbasis:

$$S_n := \{ \pi_n ^{-1} (U) :  U \subseteq \mathbb{R} \text{ open}\} $$

where $\pi_n$  are the projection maps from the product space to the component space. Now I learned the product topology as the topology with least number of open sets such that the projection maps are continuous (although I never understood the "why" for this definition). However, I don't quite get why the subbasis above would generate the product topology (I actually don't even know how can a subbasis "directly" generate a topology since I always though that a topology is generated by the union of basis elements, so first we would need to construct the basis from the subbasis and then get the topology).

答:
The product topology on $\mathbb{R}^{\mathbb{N}}$ is defined to be the topology generated by the subbasis:
$$S_n := \{ \pi_n ^{-1} (U) :  U \subseteq \mathbb{R} \text{ open}\} $$

Although you're saying the correct thing, this notation is not correct. You want to take into the subbasis the sets $\pi_n ^{-1}(U)$ for all $n$'s. So the right notation would be
$$\mathcal S := \{ \pi_n ^{-1} (U) :  U \subseteq \mathbb{R} \text{ open}, n\in\mathbb N\}.$$
(The main thing I object to in the notation you wrote down is that it looks as if the subbasis depends on $n$.)

the topology with least number of open sets such that the projection maps are continuous (although I never understood the "why" for this definition)

Well, if you want $\pi_n$ to be continuous, then $\pi_n^{-1}(U)$ has to be open for every open set $U$. This is basically the definition of continuity.

So you want the topology which contains all such sets, i.e., a topology which contains $\mathcal S$. Saying that $\mathcal S$ is a subbasis is simply saying that you are taking the smallest topology which contains $\mathcal S$.

I actually don't even know how can a subbasis "directly" generate a topology since I always though that a topology is generated by the union of basis elements

If you have a subbasis $\mathcal S$, then the corresponding basis is obtained as set of all finite intersections of elements from $\mathcal S$.
$$\mathcal B=\{S_1\cap \dots \cap S_k; k\in\mathbb N, S_i\in\mathcal S \text{ for }i=1,2,\dots,k\}.$$
Here, if you work with projections, the basis for the product topology on $\prod\limits_{i\in I} X_i$ can be written simply in this way
$$\mathcal B=\{\pi_{i_1}^{-1}(U_1) \cap \pi_{i_2}^{-1}(U_2) \cap \dots \cap \pi_{i_k}^{-1}(U_k); k\in\mathbb N, U_j\text{ is open in }X_{i_j}\text{ for }i=1,2,\dots,k\}.$$

In you case the index set is $\mathbb N$ and each $X_n$ is equal to $\mathbb R$. So you're dealing with sequences of real numbers. A reasonable ways to visualize the basic open sets is that you have chosen finitely many coordinates $i_1,\dots,i_k$ and for each of those coordinates you've chosen some open sets. And you're taking only the sequences where the $i_k$-th term belongs to the chosen open set $U_k$. (On coordinates outside of the set $\{i_1,\dots,i_k\}$, the sequences belonging to this particular basic open set can behave arbitrarily.)

Still, even if you prefer to use basis to generating topologies, it might be useful to work (at least sometimes) with subbasis. For example, here in the case of product topology, you can see that the description using subbasis is much simpler.


问:
Obtain $x$ from exponential equation
I'm looking for $x$ from the following equation:

$$\frac1{(1+x)^{\frac1{p}}}+\frac1{(1+x)^{\frac2{p}}} + \cdots \frac1{(1+x)^{\frac{n}{p}}} = 0$$

Or, what's the same:

$$\sum_{i=1}^n\frac1{(1+x)^{\frac{i}{p}}} = 0 $$

where $n$ and $p$ are given (say, $n=16$ and $p=52$, for example)

Is there a way to get $x$ from this?

答: You have an expression, which can be rewritten using the geometric series, but $x$ is not determined by it. You may actually choose any $x$ but $-1$, since the sum is finite.</p>

Title: Obtain $x$ from exponential equation


问:
Proving a statement in the form $A \implies (B \implies C )$
Say I have a statement in the form

$$A \implies (B \implies C )$$

how can I prove it? Should I assume $A$ and $B$ and then try to prove $C$ from them ?

It seems that it could be the right thing because if $A = \bot $ then the statement is always true, so I assume $A$, now if $B = \bot$ then again I have an always true statement because $A$ is true and $B \implies C$ is true so I assume also $B$.

Is my reasoning right? Is there an intuitive explanation of this 'iterative' process?

答:
This may not be the most intuitive, but you can observe that this statement is equivalent to $(A \land B) \implies C$, either by truth table or natural deduction:

\begin{align}&\quad \ \ A \implies (B \implies C)\\
&\equiv\neg A \lor (\neg B \lor C) &\text{(Material Implication)}\\
&\equiv(\neg A \lor \neg B) \lor C &\text{(Associativity of $\lor$)}\\
&\equiv\neg(A \land B) \lor C &\text{(De Morgan's Law)}\\
&\equiv (A \land B) \implies C &\text{(Material Implication)}\\
\end{align}

so one can prove the statement is true by proving $C$ holds if both $A, B$ hold.


问:
Chess board corners probability problem.
We have a chess board (8*8) and we distribute 8 different chess pieces (4 different white pieces and 4 different black pieces) what is the probability of having only black on the corners of the board.

So I thought the probability space would be ${64 \choose 8}$ since we have 64 squares and we need to distribute 8 pieces to it.
and for the number of combinations to chose 4 black for the 4 corners I seem to be lost , I tried $4!$${4 \choose 1}$$4$ since we have 4! ways to sort the black ones and we need to pick one out of 4 for each corner ${4 \choose 1}$ and multiply by 4 for the white ones

The final answer should be $1.574*10^{-6}$ according to the book but I am not even close to this..

Appreciate any help and tips! thank you

答: Just place the black pieces in order.  What is the chance the first goes into a corner?  Now that a corner is occupied, what is the chance that the second goes into a corner?  Keep going.</p>


问:
Exclusive or with three and more terms
How to define in general an n-ary operator $⊕^n$ such that $⊕^n(A_1,…,A_n)$ is true when exactly one of its disjuncts is true, because using binary xor as in $A⊕^2B⊕^2C$, does not have the desired truth conditions.

答:
You can just straightforwardly define

(syntax) If $A_1, \ldots, A_n$ are formulas, then $\oplus^n(A_1, \ldots, A_n)$ is a formula.

(semantics) $[\![\oplus^n(A_1, \ldots, A_n)]\!]^V := 1 \text{ iff for exactly one } i, [\![A_i]\!]^V = 1$.

You can also invent symbol like $\oplus^n$ for it if you want.

If your question is how to define $\oplus$ via other connectives, we have

$\oplus^n(A_1, \ldots, A_n) \equiv \underbrace{(\underbrace{A_1}_{j=1} \land \neg \underbrace{A_2}_{j=2} \land \ldots \land \neg \underbrace{A_n}_{j=n})}_{i=1} \lor \underbrace{(\neg \underbrace{A_1}_{j=1} \land A_2 \land \neg A_3 \land \ldots )}_{i=2} \lor \ldots \lor \underbrace{(\ldots \land \neg A_{n-1} \land A_n)}_{i=n}$.

i.e. a disjunction of cases where in each disjunct, exactly one formula is true and the others are all false, expressed by a conjunction of negated and unnegated formulas.

If you want a more concise notation, you can write the above using generalized conjunction and disjunction:

$\oplus^n(A_1, \ldots, A_n) \equiv \bigvee_{i=1}^{n} \bigwedge_{j=1}^{n} \begin{cases} A_j & \text{iff } i = j\\ \neg A_j & \text{iff } i \neq j \end{cases} $

meaning that you have a disjunction of conjunctions in which each formula occurs unnegated if it is that literal which under the current disjunction ($i$) iteration is supposed to be the only true formula, and negated otherwise.

Title: Exclusive or with three and more terms


问:
Complex analysis for real valued integrals.
I'm asked to compute

$$\int_{-\infty}^{\infty}\frac{\sin(ax)}{x(1+x^4)}dx$$

To do this, I consider the function

$$f(z)=\frac{e^{aiz}}{z(1+z^4)}$$

and the contour composed by the segment on the real positive line going from $\epsilon$ to $R$, the half-upper circumference, the segment from  $-R$ to $-\epsilon$ and the half-lower circumference around zero.
When the radius of the small circumference tends to zero, the border of the contour contains a pole of the function, $0$ in particular. What should I do in this case?

Can you please show me how to compute this integral?
Thanks in advance.

答:
Define
$$f(z)=\frac{e^{aiz}}{z(1+z^4)}$$
Let $C_R, C_\epsilon$ to be the half-upper circumferences with radius $R,\epsilon$, respectively and $C_1, C_2$ to the segments from $-R$ to $-\epsilon$, $\epsilon$ to $R$, respectively. For big $R$ and small $\epsilon$, inside $C_R$-$C_1$-$C_\epsilon^-$-$C_2$, $f(z)$ has two poles
$$ z_{1,2}=\pm\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, z_{1,2}^4=-1. $$
So
$$ \int_{C_R}f(z)dz+\int_{C_1}f(z)dz -\int_{C_\epsilon}f(z)dz+\int_{C_2}f(z)dz=2\pi i\bigg[\text{Re}(f,z_1)+\text{Re}(f,z_2)\bigg]. $$
Note
\begin{eqnarray}
\text{Re}(f,z_1)+\text{Re}(f,z_2)=\frac{e^{aiz_1}}{4z_1^4}+\frac{e^{aiz_2}}{4z_2^4}=-\frac{1}{2}e^{-\frac{\sqrt2}{2}a}\cos(\frac{\sqrt2}{2}a) \\
\int_{C_1}f(z)dz=\int_{-R}^{-\epsilon}f(x)dx, \int_{C_2}f(z)dz=\int^{R}_{\epsilon}f(x)dx.
\end{eqnarray}
Since
$$ \bigg|\int_{C_R}f(z)dz\bigg|\le\int_{C_R}|f(z)||dz|\le\frac{e^{-aR\sin\theta}}{R(R^4-1)}\pi R=\frac{\pi e^{-aR\sin\theta}}{R^4-1} $$
and
$$ \int_{C_\epsilon}f(z)dz=\pi i\text{Re}(f,0)=\pi i$$
letting $R\to\infty,\epsilon\to0$, one has
$$ \int_{-\infty}^\infty f(x)dx-\pi i= -\pi ie^{-\frac{\sqrt2}{2}a}\cos(\frac{\sqrt2}{2}a) $$
Taking imaginary parts in both sides gives
$$ \int_{-\infty}^{\infty}\frac{\sin(ax)}{x(x^4+1)}dx=\pi- \pi e^{-\frac{\sqrt2}{2}a}\cos(\frac{\sqrt2}{2}a) $$


问:
How to derive Lobachevsky's formula for the angle of parallelism
I'm lightly studying some non-Euclidean Geometry and in the book I am reading there is no proof or derivation from where the Lobachevsky formula for angle of parallelism comes from:

$$\Pi(x)=2\tan^{-1}\left(e^{-x}\right)$$

Any help? Thanks

P.S. I couldn't find anything by google search either.

答:
Lemma 1: Small scale Lobachevsky is Euclidean.

Lemma 2: The area of a polygon is $$\rho^2\times defect$$

First, let us derive a helpful small scale formula. [Picture 1][1]

This is small scale so $d\approx\tan\alpha \Delta x$.
The trapezoid's area is $$-\rho^2\Delta\alpha\approx xd\approx x\tan\alpha \Delta x $$
Let's solve this differential equation:$$\int\cot\alpha\ d\alpha=-\int\frac{x}{\rho^2}dx+C\Rightarrow\ln\sin\alpha=-\frac{x^2}{2\rho^2}\Rightarrow\sin\alpha=e^{-\frac{x^2}{2\rho^2}}\Rightarrow\underline{\alpha\approx\frac{\pi}{2}-\frac{x}{\rho}} $$
Now we can find the general formula. [Picture 2][2]

The smallest triangle's area is $\rho^2(\phi-\frac\pi2-\Delta\alpha)$. It is negligible compared to $\Delta x$ and $\Delta\alpha$ so we can write that it is zero. Using our first, small-scale formula we get the following differential equation:$$0=\phi-\frac\pi2-\Delta\alpha\approx-\frac{\Delta x\sin\alpha}{\rho}-\Delta\alpha$$ It only remains to solve it:$$\int{d\alpha\over\sin\alpha}=-\int{dx\over\rho}+C\Rightarrow\ln\tan\frac{\alpha}{2}=-{x\over\rho}+C\Rightarrow\underline{\alpha=2\arctan(\cot{\beta\over 2}e^{-{x\over\rho}})}$$
This can be used to derive everything in Lobachevskian Geometry. It's the first and the most important formula and it can be written in many ways:$$\cot{\alpha\over 2}\cot{\beta\over 2}=e^\frac{x}{\rho}$$ $$\tanh{x\over\rho}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}$$ $$\tanh{x\over 2\rho}=\frac{\cos\alpha+\cos\beta}{1+\cos(\alpha-\beta)}=\frac{1+\cos(\alpha+\beta)}{\cos\alpha+\cos\beta}$$ $$\sinh{x\over\rho}=\frac{\cos\alpha+\cos\beta}{\sin\alpha\sin\beta}$$ $$\cosh{x\over\rho}=\frac{1+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}$$
[1]: https://i.stack.imgur.com/SkWSa.jpg
[2]: https://i.stack.imgur.com/szDHH.jpg


问:
Proving that if a sequence of continuous real functions is uniformly convergent on a compact real subset then it is uniformly equicontinuous
Let $\{f_n\}$ be a sequence of real-valued continuous functions on $[a,b]$. I am trying to show that if $\{f_n\}$ is uniformly convergent, it is also uniformly equicontinuous.

My attempt at this seems like it must be too simple. I will outline it below. So, in class, we proved this theorem, followed by another proposition we proved:

Theorem: Let $\{f_n\}$ be a sequence of continuous real functions on a compact set $K\subset X$. Then $\{f_n\}$ is equicontinuous and pointwise convergent if and only if it is uniformly convergent.

Proposition: Let $K\subset X$ be compact. If $\mathcal{F}\subset C(K)$ is equicontinuous at every point in $K$, then $\mathcal{F}$ is uniformly equicontinuous in $K$.

So, my idea is that since $[a,b]$ is a compact set, the proof follows from simply combining these two statements. This feels too simple to be correct to me, so I was hoping if someone could tell me if this will suffice or if there is any other simple way to prove this without the need of these theorems? Thanks.

答:
This is essentially answered here.

Let $\epsilon>0$ be given. Since $f_n$ converges uniformly, it is uniformly Cauchy: there is $n_0 \in \mathbb N$ so that
$$ |f_n(x) - f_{n_0}(x)|<\epsilon/3, \ \ \ \forall n\ge n_0, x\in [a, b].$$
Since $f_1, \cdots, f_{n_0}$ are continuous on $[a, b]$, they are all uniform continuous. Thus there are $\delta_1, \cdots, \delta_{n_0}$ so that
$$|f_k(x) - f_k(y)|<\epsilon/3$$
whenever $x, y \in [a, b]$ and $|x-y|<\delta_k$. Let
$$\delta=\min_{k=1, \cdots, n_0} \delta_k.$$
Then for all $n\ge n_0$ and $|x-y|<\delta$,

$$|f_n(x)-f_n(y)|\le |f_n(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(y)|+|f_{n_0}(y)-f_n(y)|<\epsilon.$$

if $n \le n_0$ and $|x-y|<\delta$, then $|f_n(x) - f_n(y)|<\epsilon/3 < \epsilon$ by the choice of $\delta$.

Thus
$$|f_n(x) - f_n(y)|<\epsilon$$
for all $n$ and for all $x, y\in [a, b]$ such that $|x-y|<\delta$. Thus $\{f_n\}$ is uniform equicontinuous.


问:
The last 2 digits of $7^{7^{7^7}}$
What is the calculation way to find out the last $2$ digits of $7^{7^{7^7}}$? WolframAlpha shows  $...43$.</p>

Title: The last 2 digits of $7^{7^{7^7}}$

Body:
Since

$$7^{2k+1} \equiv3\pmod {4}$$
$$ 7^{7^{7^7}}\equiv3\pmod {4}$$
and $7^{3} \equiv43\pmod {100}$ &  the cycle has period $4$, then $7^{7^{7^7}}\equiv43\pmod {100}_\blacksquare$