问:
Summing $\sum_{n=0}^{\infty}\dfrac{\cos(nx)}{\cos^{n}(x)}$
Summing a trigonometric series $$\displaystyle\sum_{n=0}^{+\infty}\dfrac{\cos(nx)}{\cos^{n}(x)}$$

This is just a curious tribute that I wish to dedicate to Student-A Level.

Let $C=1+\dfrac{\cos(x)}{\cos(x)}+\dfrac{\cos(2x)}{\cos^{2}(x)}+\dfrac{\cos(3x)}{\cos^{3}(x)}+...$

Let $S=\dfrac{\sin(x)}{\cos(x)}+\dfrac{\sin(2x)}{\cos^2(x)}+\dfrac{\sin(3x)}{\cos^3(x)}+...$

$C+iS=1+ \dfrac{\cos(x)+i\sin(x)}{\cos(x)}+\dfrac{\cos(2x)+i\sin(2x)}{\cos^2(x)}+\dfrac{\cos(2x)+i\sin(3x)}{\cos^3(x)}+...$

We have the geometric series:

$1+\left(\dfrac{e^{ix}}{\cos(x)}\right)+\left(\dfrac{e^{ix}}{\cos(x)}\right)^2+\left(\dfrac{e^{ix}}{\cos(x)}\right)^3+\left(\dfrac{e^{ix}}{\cos(x)}\right)^4...$

Using the geometric summing formula for infinite series

$\dfrac{\cos(x)}{\cos(x)+e^{ix}}=\dfrac{\cos(x)}{2\cos(x)+i\sin(x)}$

The real part is what I don't know.

Is this a valid derivation, or it is just rubbish?

Using this method, I cannot sum the infinte series, which is $C$, am I right?

I also have no clue whether it converges for diverge.

I hope to get a divergent series after doing this stuff to have one more example.

Wolfie won't sum this series.

答:
We can of course rationalize the denominator  and extract the real part

$$\dfrac{\cos x}{2\cos x+i\sin x}=\dfrac{\cos x(2\cos x-i\sin x)}{(2\cos x)^2+(\sin x)^2}$$

But unfortunately the infinite sum does not converge as  the common ratio $$=\left|\dfrac{e^{ix}}{\cos x}\right|=|\sec x|\ge1$$  for real $x$


问:
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

答:
$$(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$$

Let $x = \cos \theta$ and $y = \sin \theta$. Then we get

\begin{align}
   (y - x)(2+xy) &\leq 2 \\
   x^2 + y^2 &= 1
\end{align}

We can now argue
\begin{align}
      -(y-x)^2 &= -1 + 2xy \\
   5 - (y-x)^2 &=  2(2 + xy) \\
\hline
   (y-x)(5 - (y-x)^2) &\le 4
\end{align}

Let $z = y-x$ and we find

\begin{align}
   z(5 - z^2) &\le 4 \\
   z^3 - 5z + 4 \ge 0 \\
   (z-1)(z^2+z-4) \ge 0 \\
   \hline
   z \ge \dfrac{\sqrt{17}-1}{2} \\
   -\frac{\sqrt{17}+1}{2} \le z \le 1
\end{align}

The first answer is superfluous. The second needs to be modified to

$$-\sqrt{2} \le z \le 1$$

$$-\sqrt{2} \le \sin \theta - \cos \theta \le 1$$

$$-\sqrt{2} \le \sqrt 2 \sin\left(x - \frac{\pi}{4} \right) \le 1$$

$$-1 \le \sin\left(x - \frac{\pi}{4} \right) \le \frac{1}{\sqrt 2}$$

$$\theta \in \left[(2n-1)\pi, \frac{\pi}{2}(4n+1) \right] \qquad \forall n \in \Bbb Z$$

ContentLicense: CC BY-SA 4.0


问:
Winning strategy for the second player?
Suppose our two player game consists of them constructing a binary sequence (0’s and 1’s) by taking alternate turns choosing to write either zero or one at each turn, and thereby extending the sequence by one at each turn.

Before the game begins, we are given a number $n$. The game continues until some block of $n$ consecutive digits is encountered for the second time (the two occurrences  of the block may overlap). The player that completes this second/duplicate block loses.

Clearly this game is finite as a simplistic bound of $2^n+n-1$ on the number of digits in the sequence suffices as a proof by pigenholeing

If $n$ is odd, how does one prove that the second player (the player who writes the second number in the sequence) has a winning strategy irrespective of what the first player does? Is some kind of strategy stealing argument applicable here?

答:
Let $n$ be odd and let $s_t \in \{0,1\}$ be the $t^{th}$ binaray digit selected. The odd number index $t$ denotes player $1$'s turn and even index $t$ is player $2$'s turn. If player $2$ uses the strategy:

$$s_{q} = 1-s_{q-1} \qquad (\star)$$

for every even $q$ then player $2$ will win.
To show this, let $j$ be the smallest number where $s_{i}s_{i+1}\dots s_{i+n-1}$ is the first occurance of a block of $n$ digits with corresponding second occurrence $s_{j}s_{j+1}\dots s_{j+n-1}$.

First, notice that if $j$ is odd then player $1$ loses. Therefore, we will show if $j$ is even we will have a contradiction.

If  $j$ is even and $i$ is even then the sequence $s_{i-1}s_{i}\dots s_{i+n-2}$ and $s_{j-1}s_{j}\dots s_{j+n-2}$ are the same because $s_{j} = s_{i}$ and strategy ($\star$) implies that $s_{j-1}=s_{i-1}$. This contradicts the fact that $j$ is the smallest number where we have a block encountered for a second time.

If $j$ is even and $i$ is odd then,
\begin{align}
s_{j} &= 1-s_{j-1} &&= s_{i}  \\
s_{j+1} &=  &&= s_{i+1} = 1-s_{i} \\
s_{j+2} &= 1-s_{j+1} &&= s_{i+2}  \\
s_{j+3} &=  &&= s_{i+3} = 1-s_{i+2}  \\
s_{j+4} &= 1-s_{j+3} &&= s_{i+4}  \\
\vdots
\end{align}
Notice that $s_{j+4} = 1-s_{j+3} = 1-s_{i+3} = 1-(1-s_{i+2}) = s_{i+2} = s_{j+2}$ and similarly, $s_{i+5} = 1-s_{i+4} = 1-s_{j+4} = 1-(1-s_{j+3}) = s_{j+3} = s_{i+3}$. Following this pattern it can be shown that $s_{j}=s_{j+2}=s_{j+4}= \dots = s_{j+n}$ and $s_{i+1}=s_{i+3}=s_{i+5}= \dots = s_{i+n-1}$. Since $s_{i+1}=1-s_{i} = 1-s_{j}$ this implies that $s_{j}s_{j+1}\dots s_{j+n-1}$ is alternating $0$'s and $1$'s, i.e., $010101\dots$ or $101010\dots$.

Without loss of generality assume that $s_{i} = 1$. Since $n$ is odd we know $s_{i+n-1}=1$ and by $(\star)$ we know that $s_{i+n} = 0$. Additionally, we know that $s_{j} = 1$ and by $(\star)$ we know that $s_{j-1}=0$. Notice though that $s_{i+1}s_{i+2}\dots s_{i+n}$ is the same as $s_{j-1}s_{j}\dots s_{j+n-2}$. If $j-1 \neq i+1$ then this contradicts the fact that $j$ is the smallest number where we have a block encountered for a second time. If $i+1=j-1$ then $s_{i}\dots s_{i+n-1}$ is the same as $s_{i+2}\dots s_{i+n+1}$ and player $1$ loses.

Therefore, $s_{q} = 1-s_{q-1}$ for all even $q$ is the winning strategy when $n$ is odd.

Title: Winning strategy for the second player?


问:
Asking for a proof for the number of free variables in a consistent linear system
Given any system of linear equations, we can represent it by an augmented matrix and turn the matrix into (reduced) row echelon form.

Let $p$ denote the number of pivots in the (reduced) row echelon form and $n$ denote the number of unknowns in the system. If the system is consistent, then there are $n-p$ unknowns whose value can be assigned arbitrarily, and in particular, we can always be sure that if column $i$ of the (reduced) row echelon matrix does not contain a pivot, then we can choose to assign arbitrary value to the unknown $x_i$.

It's quite obvious to "see" that the above statement is true whenever we have an example of (reduced) row echelon matrix representing a consistent linear system. But may I ask how to prove the above statement in a rigorous manner?

Thanks in advance!

答:
Suppose you have the $n$ by $n$ matrix $A$, and that you were trying to find a vector $x$ such that $Ax=b,$ for some vector $b$ in $\mathbb{R}^{n}$.

Also suppose you were able to find a solution, call it $x_{p}$, by putting the augmented matrix $A|b$ in reduced row echelon form. This is only one solution. The complete solution $\hat x$ is given by $$\hat{x}=x_{p}+x_{n},\tag{$\star$}$$ where $x_{n}$ is ANY vector in the null space of $A$.

Remember that the null space is the set of all vectors $x$ such that $Ax=0$. Then $\hat{x}$ is indeed the complete solution since we have
$$
     \begin{alignat*}{2}
          A\hat{x}&=A(x_{p}+x_{n})\\
                  &=Ax_{p}+Ax_{n}\\
                  &=(b)+(0)\\
                  &=b.
     \end{alignat*}
$$

To answer your question:
Having $p$ column pivots means you have $p$ linearly independent column vectors, so the dimension of the column space of the matrix is $p$.

The dimension of the null space must then be $(n-p)$, so a list of $(n-p)$ independent vectors span the null space.

Suppose $x_{1},x_{2},...,x_{n-p}$ is such a list of $(n-p)$ independent vectors spanning the null space of A. Since the null space is a vector space, it is closed under addition and scalar multiplication. This means that any linear combination of $x_{1},x_{2},...,x_{n-p}$ will be in the null space of A. That is to say, the vector $$x_{n}=a_{1}x_{1}+a_{2}x_{2}+...+a_{n-p}x_{n-p}$$ is in the null space, for arbitrary coefficients $a_{i}$.

Then the complete solution from $(\star)$ becomes
$$
     \begin{alignat*}{2}
          \hat{x}&=x_{p}+x_{n}\\
                 &=x_{p}+(a_{1}x_{1}+a_{2}x_{2}+...+a_{n-p}x_{n-p}),
     \end{alignat*}
$$
where we can pick arbitrary values for $a_{1},...,a_{n-p}$. These are the $(n-p)$ arbitrary unknowns you mention in your question.


问:
Finding members of a ring
From Coding Theory: A first Course, by Ling and Xing

In example 3.2.8, it gives

$\mathbb Z_{2}[x]/(1+x^2)=\{0,1,x,1+x\}$, and $\mathbb Z_{2}[x]/(1+x+x^2)=\{0,1,x,1+x\}$

How did they find the elements of the set?  How would the method be extended beyond $\mathbb Z_{2}$?  The field $\mathbb Z_{2}$ seems to be making things appear simpler than they really are.  Based on this, I am supposed to find the set for $\mathbb Z_{16}[x]/(1+x+x^4)$, but I don't see how.

答:
These are residue classes in the ring of polynomials. So
$$f(x) \equiv g(x)$$ in $Z_k[x]/(m(x))$
if and only if their remainders upon division by $m(x)$ are the same.

Clearly then the equivalence classes of these polynomials can be represented by the polynomials in $Z_k[x]$ of degree up to $\deg(m(x))-1.$

Title: Finding members of a ring


问:
Hint/guidance on proving $a^x \leq 1 + (a-1)x$
I'm trying to prove that $a^x \leq 1 + (a-1)x$ for $a \geq 1$ and $x \in [0,1]$. Can someone please give me a hint or guidance on how to approach this in general?

答:
You can show it using the concavity of $\ln$.

First, note that your inequaliy is equivalent to

$$x\ln a \leq \ln(1+(a-1)x)$$

Now, noting that $x\in [0,1]$ you have

$$x\ln a = x\ln a + (1-x)\underbrace{\ln 1}_{=0}\stackrel{concavity}{\leq}\underbrace{\ln (x\cdot a + (1-x)\cdot 1)}_{ = \ln(1+(a-1)x)}$$

Done.


答:
Hint: Use the Division Algorithm. By definition, the elements of $\mathbb F_2[x] / (x^2 + 1)$ are the possible remainders $r(x)$ when $x^2 + 1$ is divided by another polynomial in $\mathbb F_2[x].$ What can those be?

Generally, if $R$ is a finite ring with $n$ elements and $f(x)$ is a monic polynomial with coefficients in $R,$ then the quotient ring $R[x] / (f(x))$ is finite. (Remember that the elements of $R[x] / (f(x))$ are the possible remainders when the Division Algorithm is run with $f(x).$ Each remainder is a polynomial of degree less than $f(x),$ and there are only finitely many choices for each coefficient.) Consequently, it is just a matter of writing down all of the possible remainders. For the field $R = \mathbb F_2,$ this is simple because there are only two choices for each coefficient. How many choices are there in $\mathbb Z / 16 \mathbb Z?$ What is the maximum degree of a polynomial in $(\mathbb Z / 16 \mathbb Z)[x] / (x^4 + x + 1)?$

Title: Finding members of a ring

Body:
From Coding Theory: A first Course, by Ling and Xing

In example 3.2.8, it gives

$\mathbb Z_{2}[x]/(1+x^2)=\{0,1,x,1+x\}$, and $\mathbb Z_{2}[x]/(1+x+x^2)=\{0,1,x,1+x\}$

How did they find the elements of the set?  How would the method be extended beyond $\mathbb Z_{2}$?  The field $\mathbb Z_{2}$ seems to be making things appear simpler than they really are.  Based on this, I am supposed to find the set for $\mathbb Z_{16}[x]/(1+x+x^4)$, but I don't see how.


答: Hint:  This is tailor made for the binomial theorem (Taylor series could probably also be used), because we have $a^x=(1+(a-1))^x$ as @J.W.Tanner notes. To use the binomial theorem note that it holds for fractional exponents, then take limits and use continuity (in case $x$ is irrational).

Title: Hint/guidance on proving $a^x \leq 1 + (a-1)x$

Body: I'm trying to prove that $a^x \leq 1 + (a-1)x$ for $a \geq 1$ and $x \in [0,1]$. Can someone please give me a hint or guidance on how to approach this in general?


问:
Informal Equivalents of Mathematica "Set" and "SetDelayed"
How would one distinguish between what is meant by Mathematica's "Set" and "SetDelayed" functions in informal mathematical notation? Is there a way to make this distinction any any reasonably standard formal logics? Also, how is informal mathematical notation and formal logic notation related to Mathematica's use of variables like 'x_'?  Here is the relevant part of the Mathematica documentation:

http://reference.wolfram.com/mathematica/howto/CreateDefinitionsForVariablesAndFunctions.html

答:
QUESTION 1: How would one distinguish between what is meant by Mathematica's "Set" and "SetDelayed" functions in informal mathematical notation?

Set vs. SetDelayed

f[x_] = Expand[x^2]

x^2

{f[3], f[3.5], f[x + 1]}

{9, 12.25, (x+1)^2}

f[x+1] is not expanded because that was evaluated when f was defined.

Also Set is evaluated once for certain definitions.

r1 = Random[];
{r1, r1, r1}

{0.937245,0.937245,0.937245}

SetDelayed is evaluated each time.

r2 := Random[];
{r2, r2, r2}

{0.687744,0.629629,0.732141}

Clear["`*"]

f[x_] := Expand[x^2]

Notice when you evaluate this expression, there is no output.  That's because the rhs is held until the function f  is evaluated.

{f[24],f[x + 2],f[(h - 4)^3]}

{576, x^2+4 x+4, h^6-24 h^5+240 h^4-1280 h^3+3840 h^2-6144 h+4096}

Now everything works because Expand is evaluated after f gets its input.

QUESTION 3:  Also, how is informal mathematical notation and formal logic notation related to Mathematica's use of variables like 'x_'?

The underscore is what Mathematica uses to show input variables.  It is used with Set and SetDelayed.  Even f[x_] is evaluated without :=.

f[x_]

x_^2

The human mind knows that the x in f(x) is the input.  Computers don't unless you tell them.

QUESTION 2: Is there a way to make this distinction any any reasonably standard formal logics?

I have never seen it.  The human mind would always evaluate like a SetDelayed definition, wouldn't it?

Here's a link to the notebook.  Hope this helps.

ContentLicense: CC BY-SA 3.0


问:
Volume of Revolution of area included between Sine and Exp curves
Question: Find the volume of the solid generated by rotating the region bounded by the curves below about the horizontal line $y=-2$.

$x= \frac{3\pi}{2}$,
$y=e^{x-3}+2$

$x = 0$,
$y=\sin x$

Here is what the graph look like.

I used washer.

\begin{align}
R_\mathrm{big} &= e^{x-3} + 2 + 2 \\
R_\mathrm{big} &= e^{x-3} + 4\\
R_\mathrm{small} &= \sin x+ 2 \\
V &= \pi \int_0^{\frac{3\pi}{2}} \left[ \left(e^{x-3} + 4\right)^2 - \left(\sin x+ 2 \right)^2 \right]\,\mathrm{d}x \stackrel{\text{calc}}{\implies} 343.9678\,\mathrm{units}^3.
\end{align}

I was wondering if my work is correct.

Thanks

答:
Trying to guess at your question:

Find volume by sweeping area between curves

$$y_1= e^{x-3}+2,\; y_2= \sin x $$

about a line of symmetry parallel to x-axis

$$y=-2$$

between x-limits $( 0, \frac{3 \pi}{2})$ by washer method.

Your integral setting up and result then seem to be alright.


问:
Apply Center Manifold Theorem
I am given the system

$x' = -x +y +x^2 +ax^3$

$y' = x -y +ax^2 +bxy +y^3$.

I have to determine the stability of the origin. After doing the linearisation I get the eigenvalues

$λ_1 = 0$ and $λ_2 =-2$.

According to my notes, in order to apply the Center Manifold Theorem, the system must be in the following form:

$x' = f(x,y)$

$y' = -y + g(x,y)$.

If the linear part of the system is not like the form shown before then I can use a transformation and bring it in the desired form. Which transformation should I use? Or am I missing something here?

Thanks for your effort !

答:
According to my notes, in order to apply the Center Manifold Theorem, the system must be in the following form:
$x' = f(x,y)
\quad$ $y' = -y + g(x,y)$.

Not exactly. The system must have the form
$$
y'= A_1 y+f(y,z),\qquad z'= A_2z +g(y,z),
$$
where all eigenvalues of $A_1$ have zero real parts and all  eigenvalues of $A_2$ have negative real parts. In the problem under consideration, $A_1$ and $A_2$ are real numbers, therefore, $A_1=0$, $A_2<0$.

The linearization matrix is
$$
A=\left(\begin{array}{rr}
-1 & 1\\
1 & -1\\
\end{array}\right).
$$
The matrix whose columns are eigenvectiors of $A$ is
$$
M=\left(\begin{array}{rr}
1 & 1\\
1 & -1\\
\end{array}\right),
$$
thus,
$$
\left(\begin{array}{c}x\\y\end{array}\right)=
\left(\begin{array}{rr}
1 & 1\\
1 & -1\\
\end{array}\right)
\left(\begin{array}{c}z_1\\z_1\end{array}\right),
$$
i.e.
$$
x=z_1+z_2,\qquad y=z_1-z_2
$$
or
$$
z_1=\frac12(x+y),\qquad z_2=\frac12(x-y).
$$
Indeed,
$$
z_1'= \frac12( x'+ y')=
\frac12( -x +y +x^2 +ax^3+x -y +ax^2 +bxy +y^3)
$$
$$
=\frac12( (z_1+z_2)^2 +a(z_1+z_2)^3 +a(z_1+z_2)^2 +b(z_1+z_2)(z_1-z_2) +(z_1-z_2)^3)
$$
$$
=f(z_1,z_2);
$$
$$
z_2'= \frac12( x'- y')=
\frac12(-x +y +x^2 +ax^3-x +y -ax^2 -bxy -y^3)
$$
$$
=-x-y+\frac12((z_1+z_2)^2 +a(z_1+z_2)^3-a(z_1+z_2)^2 -b(z_1+z_2)(z_1-z_2) -(z_1-z_2)^3)
$$
$$
=-2z_2+g(z_1,z_2).
$$

Title: Apply Center Manifold Theorem


问:
If a,b,c are positive real numbers such that a + b + c = 1 , then find the minimum value of $\frac{(1+a)(1+c)(1+c)}{(1-a)(1-b)(1-c)}$
I tried by solving inequality of numerator and denominator

Numerator:

$(1+a)(1+c)(1+c) \leq \frac{(3 + a + b + c)}{3} = \frac{(3 + 1)}{3} = \frac{4}{3}$

Denominator:

$(1-a)(1-c)(1-c) \leq \frac{3 - (a + b + c)}{3} = \frac{(3 - 1)}{3} = \frac{2}{3}$

By combining the 2 equations

$\frac{Numerator}{Denominator} \leq \frac{4/3}{2/3} = 2$

But that's not the answer (I know I found max value). Can anyone help me? Is the method I am following can be used in other types of question? Solving Numerator and denominator inequality and than dividing both inequatlity....

ANSWER: 8

答:
Method 1  $$\frac{(1+a)(1+c)(1+c)}{(1-a)(1-b)(1-c)}=\frac{(a+b+a+c)(b+c+b+a)(c+a+c+b)}{(a+b)(b+c)(c+a)}$$
Now apply AM-GM in numerator

 use $$(a+b+a+c)(b+c+b+a)(c+a+c+b)$$ $$ \ge 2\sqrt{(a+b)(a+c)}2\sqrt{(b+c)(b+a)}2\sqrt{(c+a)(c+b)}$$ $$=8(a+b)(b+c)(c+a)$$

Method 2

Apply Jensens inequality to $$f(x)=\log \left(\frac{1+x}{1-x}\right)$$


问:
Show that the set $\{x \in \mathbb{R}| f(x)=x\}$ is closed.
Show that the set $\{x \in \mathbb{R}| f(x)=x\}$ is closed, assuming that $f:\mathbb{R} \to \mathbb{R}$ is continuous. So here is my attempt:

Let $\{x_k\} \subseteq \mathbb{R}$ such that $x_k \in \{x \in \mathbb{R}| f(x)=x\}$ and $x_k \to x$ as $k \to \infty$.Then, as $f:\mathbb{R} \to \mathbb{R}$ is continuous by a theorem in my text, the sequence $\{y_k\}$ where $\{y_k\} = f(x_k)$ converges to $f(x)$. So we have that $f(x) = \lim_{x \to \infty} f(x_k) =x $.  So by definition $x$ is in the set, and we can conclude now that since the set contains all of its limit points, that the set is a closed subset of $\mathbb{R}$.

I have a feeling I made some flaws in logic somewhere around here, any help is appreciated. I only want some hints on what I did wrong so I can think about the problem longer instead of just having the solution.

答:
Here it is another way to approach it for the sake of curiosity.

Consider the function $g(x) = f(x) - x$. Since $f$ is continuous, $g$ is continuous as well.

Once $\{0\}$ is closed, we conclude that $g^{-1}(\{0\}) = \{x\in\mathbb{R}\mid f(x) = x\}$ is closed, and we are done.

Hopefully this helps!


问:
A question about quaternions and reference frames
I have 2 rotations represented by quaternions, quaternions A and B respectively. how would I go about finding the rotation from one refence frame to another, say finding the rotation of quaternion A in quaternion B's reference frame. Do I find the inverse of the quaternion? or something else?</p>

答:
If you apply your rotations like this:  $x\mapsto BxB^{-1}$, then I think you’re looking for $AB^{-1}$.

That’s the rotation you’d need to get to the result of $A$ beginning with the result of $B$.


问:
Equation $f(x,y) f(y,z) = f(x,z)$
How to solve the functional equation  $f(x,y) f(y,z) = f(x,z)$?</p>

答:
Suppose you know $f(x,z)= h_x(z)$ for some fixed $x$. Then you can work out

$$f(y,z) = f(x,z)/f(x,y)= h_x(z)/h_x(y)$$

That is, for all $y$ such that $h_x(y)\neq 0$ we find that $f$ is of the form $g(z)/g(y)$.

If $h_x(y)$ vanishes then $h_x(z)$ vanishes for all $z$. Then for any $a,b$ we have $$f(a,b)=f(a,x)f(x,b)=0$$

Hence the function is either identically zero, or nowhere zero and of the form $f(x,y)=g(y)/g(x)$ where $g$ is nowhere zero

Title: Equation $f(x,y) f(y,z) = f(x,z)$


问:
Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$

In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$

(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)

Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.

I was referred by a friend to the identities
$$\begin{align}
a+b+c &= 2s \tag{1} \\[4pt]
ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt]
abc &= 4Rrs \tag{3}
\end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.

Any other idea?

答:
Use the result $1+\frac rR=\cos A+\cos B+\cos C$ in the derivation below

\begin{align}
& \frac{s^2-(2R+r)^2}{4R^2}\\
= &\frac14\left(\frac{a+b+c}{2R}\right)^2-\frac14\left(2+\frac rR\right)^2\\
=& \frac14\left[ (\sin A +\sin B +\sin C)^2 -(1+ \cos A +\cos B +\cos C)^2 \right]\\
=& \frac14\left[(\sin^2A-\cos^2 A)+(\sin^2 B-\cos^2 B) + (\sin^2 C-\cos^2 C)\right.\\
& \left.-1 -2( \cos A +\cos B +\cos C)-2(\cos (A+B)+\cos (B+C) +\cos (C+A)) \right]\\
= &\frac14( -\cos2A - \cos2B -\cos2C-1)\\
= &\frac14( -2\cos(A+B)\cos(A-B) -2\cos^2C +1 -1)\\
= &\frac12\cos C(\cos(A-B)+\cos(A+B))\\
= &\>\cos A \cos B\cos C\\
\end{align}


Title: Volume of Revolution of area included between Sine and Exp curves

Body:
Question: Find the volume of the solid generated by rotating the region bounded by the curves below about the horizontal line $y=-2$.

$x= \frac{3\pi}{2}$,
$y=e^{x-3}+2$

$x = 0$,
$y=\sin x$

Here is what the graph look like.

I used washer.

\begin{align}
R_\mathrm{big} &= e^{x-3} + 2 + 2 \\
R_\mathrm{big} &= e^{x-3} + 4\\
R_\mathrm{small} &= \sin x+ 2 \\
V &= \pi \int_0^{\frac{3\pi}{2}} \left[ \left(e^{x-3} + 4\right)^2 - \left(\sin x+ 2 \right)^2 \right]\,\mathrm{d}x \stackrel{\text{calc}}{\implies} 343.9678\,\mathrm{units}^3.
\end{align}

I was wondering if my work is correct.

Thanks

Body:
Your integral is set up correctly. And

$$\begin{split}\pi\int_0^{3\pi/2}\left((e^{x-3}+4)^2-(\sin x +2)^2\right)dx&=\pi\int_0^{3\pi/2}\left(e^{2x-6}+8e^{x-3}+16-(\underbrace{\sin^2x}_{=\frac 1 2(1-\cos 2x)}+4\sin x+4)\right)dx\\
&=\pi\int_0^{3\pi/2}\left(e^{2x-6}+8e^{x-3}+\frac {23}{2}
+\frac 1 2\cos2x-4\sin x\right)dx\\
&=\pi\left[\frac 1 2e^{2x-6}+8e^{x-3}+\frac {23} 2x+\frac 1 4\sin 2x+4\cos x\right]_{0}^{3\pi/2}\\
&=\pi\left[\left(\frac 1 2e^{3\pi-6}+8e^{3\pi/2-3}+\frac{69}{4}\pi+0+0\right)-\left(\frac 1 2e^{-6}+8e^{-3}+0+0+4\right)\right]\\
&\approx343.97\end{split}$$

It looks to be correct.


问:
Direct product of simple groups case $G_1 \cong G_2$
I have the following exercise:

Find all normal non-trivial subgroups of $G_1 \times G_2$.  I managed to see that using projections, the normal subgroups are isomorphic to $G_1$ or $G_2$. But the professor say that we need to separate the case when $G_1\cong G_2$. But I don't understand why. Can you help me with this?

答:
You forgot the trivial group and the whole group, for starters.

Using projections, if $N\triangleleft G_1\times G_2$, then $\pi_1(N)=\{e\}$ or $G_1$, and $\pi_2(N)=\{e\}$ or $G_2$, because $G_1$ and $G_2$ are simple and $\pi_1(N)\triangleleft G_1$, $\pi_2(N)\triangleleft G_2$.

If both projections are trivial, then $N=\{e\}\times\{e\}$. If one projection is trivial and the other is nontrivial, then you get $N=G_1\times\{e\}$ and $N=\{e\}\times G_2$.

But if both projections are non-trivial, then what you have is a subdirect product. The subdirect products are described by Goursat’s Lemma. They correspond to isomorphisms between quotients of $G_1$ and quotients of $G_2$.

If $G_1$ is not isomorphic to $G_2$, then the only quotient of $G_1$ that is isomorphic to a quotient of $G_2$ is the trivial one. Goursat’s Lemma tells you that $N$ in that case is $G_1\times G_2$.

But if $G_1\cong G_2$, then you have other possibilities. Specifically, the missing subdirect products are of the following form: fix an isomorphism $f\colon G_1\to G_2$, and then you get the subgroup $N_f$, given by
$$N_f=\{(g_1,g_2)\in G_1\times G_2\mid f(g_1)=g_2\};$$
that is, the “graph” of $f$.

You should check whether these subgroups are normal in $G_1\times G_2$. As a hint, you may want to consider the case of $G_1\cong G_2$ abelian, and $G_1\cong G_2$ nonabelian, separately.

This is why you need to consider the case of $G_1\cong G_2$ separately.


问:
Volume of a parallelepiped with three adjacent vectors
Given a parallelepiped in $\mathbb{R}^3$ with the three adjacent vectors corresponding to three adjacent edges of the parallelepiped,

to find the volume, we just take any two vectors $\vec{u},\vec{v}$ from the three adjacent vectors and take the crossproduct $\vec{u}\times\vec{v}$ and then take the dot product with the other vector $\vec{w}$, that is, $(\vec{u}\times\vec{v})\cdot\vec{w}$.

And then taking the absolute value, we have the volume $|(\vec{u}\times\vec{v})\cdot\vec{w}|$.

The reason for the above argument is because

if we take any two adjacent vectors from the given three adjacent vectors, they form a base of a parallelepiped

and by taking the dot product with the other one, we have the volume or the negative the volume of the parallelepiped. So we take the absolute value.

Is this the correct argument?

答:
Enclosed volume is given by the scalar triple product and is unaffected when their order is unchanged.

$$|(\vec{u}\times\vec{v})\cdot\vec{w}|=|\vec{u}\cdot(\vec{v}\times\vec{w})|$$

When order is interchanged, the volume becomes negative.

Btw, triclinic mineral crystals form in nature with such a 3d edge geometry.


问:
Summing this series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$
I wish to sum the series $\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n}$.

I notice that by writing $\displaystyle\sum_{n=1}^{\infty}(2n-1)\dfrac{1}{2^n}$, $\dfrac{1}{2^n}$ is a geometric series and can be summed easily. But the next step I think I did wrong is

$$2n-1\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2^n}$$

I don't think I can pull out $2n-1$ like a constant because that will affect the sum. What should I have done. The series converges to $3$ according Wolfram Alpha.

答:
Here is yet another approach, again with no calculus involved

Firstly we split the sum, $\sum_{n=1}^\infty \frac{2n-1}{2^n}=2\sum_{n=1}^\infty \frac{n}{2^n}-\sum_{n=1}^\infty\frac{1}{2^n}$

as noticed, the rightmost term is GP and convergent to 1, what now remains is the fist term.

We expand the sum as--

$\frac{n}{2^n}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\frac{7}{128}+............\infty$

split the odd terms

$\frac{n}{2^n}=\frac{1}{2}+\frac{2}{4}+\frac{\left(2+1\right)}{8}+\frac{4}{16}+\frac{\left(4+1\right)}{32}+\frac{6}{64}+\frac{\left(6+1\right)}{128}+............\infty$

$=\frac{(1)}{2}+\frac{2}{4}+\frac{\left(2\right)}{8}+\frac{\left(1\right)}{8}+\frac{4}{16}+\frac{\left(4\right)}{32}+\frac{\left(1\right)}{32}+\frac{6}{64}+\frac{\left(6\right)}{128}+\frac{\left(1\right)}{128}+............\infty$

Take out the terms with 1

$=\left(\frac{\left(1\right)}{2}+\frac{\left(1\right)}{8}+\frac{\left(1\right)}{32}+\frac{\left(1\right)}{128}+.......\infty \right)+\left(\frac{2}{4}+\frac{\left(2\right)}{8}+\frac{4}{16}+\frac{\left(4\right)}{32}+\frac{6}{64}+\frac{\left(6\right)}{128}+............\infty \right)$

But we know that the sum $\left(\frac{\left(1\right)}{2}+\frac{\left(1\right)}{8}+\frac{\left(1\right)}{32}+\frac{\left(1\right)}{128}+.......\infty \right)$ is GP and converges to $\frac{2}{3}$

$\frac{\left(\frac{1}{2}\right)}{1-\frac{1}{4}}=\frac{2}{3}$

Hence, we get
$\frac{n}{2^n}=\frac{2}{3}+\left(\frac{2}{4}+\frac{\left(2\right)}{8}+\frac{4}{16}+\frac{\left(4\right)}{32}+\frac{6}{64}+\frac{\left(6\right)}{128}+............\infty \right)$

Simplifying the right terms, we get

$\frac{n}{2^n}=\frac{2}{3}+3\left(\frac{\left(2\right)}{8}+\frac{\left(4\right)}{32}+\frac{\left(6\right)}{128}+............\infty \right)$  -------(1)

Now we begin from the expansion, this time splitting the odd terms with minus one--

$\frac{n}{2^n}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+............\infty $

$\frac{n}{2^n}=\frac{\left(2-1\right)}{2}+\frac{2}{4}+\frac{\left(4-1\right)}{8}+\frac{4}{16}+\frac{\left(6-1\right)}{32}+\frac{6}{64}+............\infty $

$\frac{n}{2^n}=\frac{\left(2\right)}{2}-\frac{\left(1\right)}{2}+\frac{2}{4}+\frac{\left(4\right)}{8}-\frac{\left(1\right)}{8}+\frac{4}{16}+\frac{\left(6\right)}{32}-\frac{{\left(1\right)}}{32}+\frac{6}{64}+............\infty $

$\frac{n}{2^n}=\left(\frac{\left(2\right)}{2}+\frac{2}{4}+\frac{\left(4\right)}{8}+\frac{4}{16}+\frac{\left(6\right)}{32}+\frac{6}{64}+............\infty \right)-\left(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+........\infty \right)$

$\frac{n}{2^n}=\left(\frac{\left(2\right)}{2}+\frac{2}{4}+\frac{\left(4\right)}{8}+\frac{4}{16}+\frac{\left(6\right)}{32}+\frac{6}{64}+............\infty \right)-\frac{2}{3}$

$\frac{n}{2^n}=3\left(\frac{2}{4}+\frac{4}{16}+\frac{6}{64}+............\infty \right)-\frac{2}{3}$

Multiply and divide by 2

$\frac{n}{2^n}=6\left(\frac{2}{8}+\frac{4}{32}+\frac{6}{128}+............\infty \right)-\frac{2}{3}$  --------(2)

Solve 1 & 2 equations simultaneously, and we get $\frac{n}{2^n}=2$

Final answer

$\sum _{n=1}^{\infty }\frac{2n-1}{2^n}=2\sum _{n=1}^{\infty }\frac{n}{2^n}-\sum _{n=1}^{\infty }\frac{1}{2^n}$
=$2\cdot \left(2\right)-1=3$

P.S-- We can perform these operations ONLY as we know that the sum converges...Incredibly sorry if  I may have missed some finer higher-level details of why we can perform these operations for these infinite sums....


问:
Algebraic dimension of the (algebraic) complement of the Banach space Y in a Banach space X
Can we claim the dimension of the algebraic complement of a Banach space $Y$ in the Banach space $X$ never equals $\aleph_0?$ Here, $Y$ is a linear subspace of $X?$

答:
If you have a closed subspace $Y$ of Banach space $X$ ($Y$ will be closed if and only if it is a Banach space in its own right), we can form the quotient space
$$X / Y = \{x + Y : x \in X\}.$$ We can equip this space with a norm that will make it a Banach space as well.

Suppose $Z$ is a (not necessarily closed) complement to $Y$ in $X$ (i.e. $X = Y \oplus Z$). We can naturally identify $x + Y$ with $z$, where $y \in Y$ and $z \in Z$ are the unique vectors where $x = y + z$. This map is well-defined, and forms an isomorphism between $X/Y$ and $Z$ as pure (un-normed) vector spaces.

So, if $Z$ has alebraic dimension $\aleph_0$, then so must the Banach space $X / Y$, which is impossible.


问:
Ideal of height $>1$ containing no non-zero prime ideals.
Edit: Badam Baplan gives a remarkable example below, which answers the question completely. I would still appreciate any other examples that others may come up with; in particular, is an example still possible if we ask for $R$ to be Noetherian?

Let $R$ be a (commutative, unital) ring and $I<R$ an ideal. Recall that we define $$\operatorname{ht}I=\min\{\operatorname{ht}P:P\in\operatorname{Spec}R,P\geqslant I\},$$ where $\operatorname{ht}P$ is defined in the usual way for a prime ideal $P$. Is there a nice example of an integral domain $R$ and an ideal $I$ such that (i) $I$ contains no non-zero prime ideals, and (ii) $\operatorname{ht}I>1$? Note: by Krull's height theorem, if $R$ is Noetherian then $I$ cannot be principal.

If we drop the condition that $R$ is an integral domain, then we can get examples where $\operatorname{ht}I$ is arbitrarily large. For instance, let $F$ be your favorite field, and define $A$ to be the polynomial ring $F[x_1,\dots,x_n]$. Choose any non-zero ideal $M\leqslant A$, with some $o\in M\setminus\{0\}$ of minimal degree. (For later, note that $o\notin x_1M+\dots+x_nM$; call this fact $(\star)$.) Now, consider the ring structure on the product $R:=A\times M$ defined by

$(a,m)+(b,n)=(a+b,m+n)$

$(a,m)\cdot(b,n)=(ab,an+bm)$

for all $a,b\in A$ and $m,n\in M$. Note that every element $(0,m)$ is nilpotent, so any prime ideal of $R$ must contain $0\times M$. Conversely, if $P$ is a prime ideal of $A$, then $P\times M$ is a prime ideal of $R$. With this in mind, consider the ideal $I<R$ generated by the elements $(x_i,0)$ for $1\leqslant i\leqslant n$. First note that $I$ contains no prime ideals; indeed, one can explicitly compute $I$ to be the product $$(x_1A+\dots+x_nA)\times(x_1M+\dots +x_nM),$$ where both direct factors are considered as ideals of $A$. In particular, by fact $(\star)$, we have $(0,o)\notin I$, and so (by the remark above, since $(0,o)$ is nilpotent), $I$ cannot contain any prime ideal.

On the other hand, consider the prime ideal $P:=\langle x_1,\dots,x_n\rangle\times M>I$. This is the unique minimal prime lying above $I$, since $P=I+(0\times M)$ and every prime must contain $0\times M$. However, we have $\operatorname{ht}P=n$, since $$0\times M<\langle x_1\rangle\times M<\langle x_1,x_2\rangle\times M<\dots<P$$ is a strictly ascending chain of primes below $P$. Thus, since $P$ is the unique minimal prime above $I$, we have $\operatorname{ht}I=n$.

One can extend this example in the natural way to find an example where $\operatorname{ht}I=\infty$. So this resolves the question completely if we do not require $R$ to be an integral domain. However, I'm struggling to find an example where $R$ is an integral domain. Can anyone think of a nice example?

I tried attempting to consider the example above as a quotient of an appropriate polynomial ring, but didn't make any progress. My thinking was that we could perhaps find a desirable prime lying below the kernel of the projection map, and then quotient by that to find a domain with some similar behavior. However, this doesn't seem to work out very nicely.

As a smaller follow-up question, what is the geometric significance (if any) of such a ring?

答:
About your example: preferring the notation $\mathfrak a$ to $M$, it is easier to think of your ring $A \times \mathfrak a$ as the Rees algebra of $\mathfrak a$ over $A$, namely $R = A[\mathfrak a t] = \bigoplus_{n \ge 0} \mathfrak a^n t^n$ (the variable $t$ is used to keep track of the degree of an element), modulo the ideal of all elements of $R$ of degree $\ge 2$. Under this notation, it is clear that if $\mathfrak p \in \mathrm{Spec}(A)$, then $(\mathfrak p, \mathfrak a t) \in \mathrm{Spec}(A)$ since $R/(\mathfrak p, \mathfrak a t) \simeq A/\mathfrak p$ is an integral domain. Now let $\mathfrak m = (x_1,\cdots,x_n)_A$ be the ideal of $A$ generated by the polynomial variables, and let us show that $\mathfrak b \overset{def}= (\mathfrak m)_R$ is a counter-example (as you did). Since one easily sees that $\mathfrak a t \subseteq \mathrm{Nil}(R)$, if $\mathfrak b$ contained a prime ideal $\mathfrak q$, then it would contain $\mathfrak a t \subseteq \mathrm{Nil}(R) \subseteq \mathfrak q$ as well, so we would expect
$$
\mathfrak a t = \mathfrak b \cap \mathfrak at = \mathfrak m (\mathfrak a t) = (\mathfrak m \mathfrak a)t
$$
because the only way to get an element of $t$-degree 1 by multiplying an element of $t$-degree $0$ in $\mathfrak m$ by something is if the element of $t$-degree $0$ in $\mathfrak m$ is multiplied by an element of $t$-degree $1$, and those all belong to $\mathfrak a t$. By Nakayama's Lemma (which we can use since $\mathfrak m$ is a maximal ideal of $A$), $\mathfrak a = \mathfrak m \mathfrak a$ implies $\mathfrak a = 0$, which is assumed false. So $\mathfrak b$ contains no prime ideal.

After that, as you mentioned, any minimal prime ideal above $\mathfrak b$ must contain $\mathfrak b \oplus \mathfrak a t$ since $\mathfrak a t \subseteq \mathrm{Nil}(R)$. If $\mathfrak q$ is a prime ideal containing $\mathfrak b$, then it contains $\mathfrak m$, and it contains $\mathfrak a t$ since those elements are nilpotent, so
$$
\mathfrak q = \mathfrak m \oplus \mathfrak a t \supsetneq \mathfrak m \oplus \mathfrak m \mathfrak a t = \mathfrak b
$$
because $\mathfrak q$ is maximal (and in fact $R/\mathfrak q \simeq A/\mathfrak m$). Therefore, $\mathfrak q$ is the unique prime ideal containing $\mathfrak b$, and thus $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak q)$. By construction, the ring extension $A \subseteq R$ is integral, and therefore $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak q) \ge \mathrm{ht}(\mathfrak m) > 1$.

I took care to not use anywhere the explicit expression of $\mathfrak m = (x_1,\cdots,x_n)_A$ that you picked in your example where $A = F[x_1,\cdots,x_n]$; nor did I exploit the fact that $\mathfrak a^2 = 0$, I only needed that $\mathfrak a t \subseteq \mathrm{Nil}(A)$. This means that for any non-zero ideal $\mathfrak a \trianglelefteq A$ that contains no non-zero prime ideals of $A$, the ideal $\mathfrak a t \subseteq A[\mathfrak a t]$ together with any maximal ideal $\mathfrak m$ of $A$ and the corresponding ideal $\mathfrak b = (\mathfrak m)_R$ provides an example of an ideal $\mathfrak b \trianglelefteq R$ containing no non-zero prime ideals and $\mathrm{ht}(\mathfrak b) \ge \mathrm{ht}(\mathfrak m)$, as long as $\mathfrak a t$ is nilpotent in $R$; furthermore, if $A$ is integrally closed, the integral extension $A \subseteq R$ satisfies the going-down theorem and $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak m)$.

In particular, since you didn't explicitly use the fact that $A$ was an integral domain but only the fact that $F$ was a field to have your maximal ideal of height $n$ (and help you through the proof), it suffices to pick $A$ a ring with an ideal $0 \subsetneq \mathfrak a \subseteq \mathrm{Nil}(A)$ (you can take equality iff $\mathrm{Nil}(A)$ is not prime), since such an ideal cannot contain any prime ideals by the definition of $\mathrm{Nil}(A)$ and $\mathfrak at$ automatically becomes nilpotent. If you pick $A$ integrally closed and a maximal ideal $\mathfrak m$ of height $n$ (or infinity), you can also find an example $(R,\mathfrak b)$ where $\mathfrak b \trianglelefteq R$ and $\mathrm{ht}(\mathfrak b) = n$.

As for the geometric significance of this construction, the Rees algebra $A[\mathfrak a t]$ corresponds to the blow-up of the affine scheme $\mathrm{Spec}(A)$ at the sheaf of ideals generated by $\mathfrak a$ (in algebraic geometry terms); if we restrict our attention to non-zero nilpotent ideals $\mathfrak a$ not equal to $\mathrm{Nil}(A)$, I can't really give you a geometric picture that's comfortable since the only comfortable blow-ups I know are those at a geometric point like a concrete point in $n$-dimensional affine space over a field or blowing up at some subvariety. Blowing up at a nilpotent ideal is a bit strange because it satisfies the two following properties:

Nilpotent elements $a \in \mathfrak a$ now satisfy $(at)^2 = 0$ in $A[\mathfrak at]$

The map $\mathfrak p \mapsto \mathfrak p \oplus \mathfrak at$ from $\mathrm{Spec}(A)$ to $\mathrm{Spec}(A[\mathfrak at])$ is a homeomorphism and comes from a closed immersion induced by the quotient map $A[\mathfrak at] \subseteq A[t] \to A$ sending $t$ to $0$. (In fact, seeing it like this, one notices that the property that $A$ is integrally closed is not really necessary in the above arguments; the homeomorphism property follows from the fact that $\mathfrak a$ is nilpotent.)

Let's look at a specific example to develop some intuition. Let $F$ be a field and consider $A = F[x_1,\cdots,x_n, x]/(x^n)$. We have $\mathrm{Nil}(A) = (x)/(x^n)$ which contains no non-zero prime ideals, so we can actually take $\mathfrak a = (x^k)/(x^n) \subseteq \mathrm{Nil}(A)$ for some $1 < k \le n$ (we can't take $k=1$ since in this case $(x^k)/(x^n)$ is prime). The Rees algebra $R$ of $\mathfrak a$ in $A$ is equal to $F[x_1,\cdots,x_n,x,x^kt]/(x^n) \subseteq F[x_1,\cdots,x_n, x,t]/(x^n)$. Clearly $x$ and $t$ are algebraically independent, so $R \simeq F[x_1,\cdots,x_n,x,u]/(x^n, u^{\lceil n/k \rceil})$ (just set $u = x^kt$). So the ring $F[x_1,\cdots,x_n,x]/(x^n)$ is like the point $0 \in \mathbb A^n_F$ that has been "fattened", and blowing-up at $(x^k)/(x^n)$ for $1 < k \le n$ is like making the point $0 \in \mathbb A^n_F$ "even fatter, but in a different direction" because you're blowing it up in this new direction $t$ while still preserving the fatness of that point you already had. But since we just fattened the already fat point, we keep the same points as before blowing up.

Our examples of ideals $\mathfrak b$ are sort of ideals representing "different degrees of fatness" of this point. They cannot contain any prime ideal $\mathfrak q$, because the inclusion $\mathfrak q \subseteq \mathfrak b$ is equivalent to the existence of a closed immersion $A/\mathfrak q \to A/\mathfrak b$, which would mean our "fat point" would contain a proper subvariety; $\mathfrak b$ is a fat point, so the only thing this ideal contains is "fatter points", and going fatter and fatter, you eventually obtain the zero ideal (the fattest point). But none of these fatter points are prime since they are all "fattened by nilpotents". However, the height of this fat point is the same as that of the corresponding point in the non-fattened variety.

Replace "geometric point in $\mathbb A^n_F$" by "some closed point of $\mathrm{Spec}(A)$" and you go back to the arbitrary setting of algebraic geometry and affine schemes. As a note, when $\mathfrak a \trianglelefteq A$ is an ideal, the associated graded ring $\mathrm{gr}_A(\mathfrak a) \overset{def}= (A/\mathfrak a)[(\mathfrak a/\mathfrak a^2)t]$ has also been extensively studied; in our context, $R/\mathfrak b = \mathrm{gr}_A(\mathfrak a) / \mathfrak m\mathrm{gr}_A(\mathfrak a)$.

Hope that helps,


问:
Problem with probability of hypothesis testing
Three coins are tossed. If all three show tails, $H_0$ is rejected,
otherwise, $H_0$ is not rejected. Using this procedure, what is the probability of committing a Type II error?

I have that $1 - [P(\text{get tail 1st toss}|H_1\text{ true})*P(\text{get tail 2nd toss}|H_1\text{ true})*P(\text{get tail 3rd toss}|H_1\text{ true})]$

But I am not sure of this $1 - ((1/8)(1/8)(1/8))$

答:
Something is missing in the text. Assuming that the null hypothesis is

$$H_0: \theta_0\leq\frac{1}{2}$$

Against

$$H_1: \theta_1 >\frac{1}{2}$$

Where $\theta$ is the probability to get tail, type II error is

$$\beta=1-\mathbb{P}[\text{TTT}|\theta_1]=1-\theta_1^3$$

Observe that $\beta$ decreases when $\theta_1$ increases in the Interval $\left(\frac{1}{2};1\right]$ as it must be.


问:
For $0 < p < 1$ will my integral converge?
Will this integral converge for $0 < p < 1$?

$$\int_{0}^{1}\frac{x^{p}}{\log(1+x)}{e^{-x}}\mathrm{d}x$$

I tried using limit comparison test and comparison test (the only 2 tests I have learnt so far). I tried taking

$$g(x) = \frac{x^{p}}{\log(1+x)}$$

but unfortunately I can't conclude anything. Please help.

答: $log (1+x) \sim x$ as $x \to 0$ and $\int_0^{1}x^{p-1}dx <\infty$. Hence the integral is finite.


问:
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

答:
Let $$A:=\sin\theta-\cos\theta,\qquad B:=\sin\theta\cos\theta$$

Then, we have
$$A^2=(\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=1-2B$$

To solve $$A(2+B)-2\le 0$$

let us see if $A(2+B)-2$ can be factorized.

To see if $A(2+B)-2$ can be factorized, let us add $k(-A^2-2B+1)$ (which equals $0$) to $A(2+B)-2$ where $k$ is a non-zero real number.

$$\begin{align}A(2+B) -2&=2A+AB-2
\\\\&=2A+AB-2+\underbrace{k(-A^2-2B+1)}_{=0}
\\\\&=k\bigg(-A^2+\frac 2kA+\frac{k-2}{k}+\frac Bk(A-2k)\bigg)\end{align}$$

Now we want

$$-A^2+\frac 2kA+\frac{k-2}{k}$$
to have $A-2k$ as a factor, so solving
$$-(2k)^2+\frac 2k(2k)+\frac{k-2}{k}=0$$
gives$$k=\frac 12,\frac{-1\pm\sqrt{17}}{4}$$
So, we see that $k=\frac 12$ works.

Therefore, we have

$$\begin{align}A(2+B)-2&=A(2+B)-2+\frac 12(-A^2-2B+1)
\\\\&=\frac 12\bigg(-A^2+4A-3+2B(A-1)\bigg)
\\\\&=\frac 12\bigg((A-1)(-A+3)+2B(A-1)\bigg)
\\\\&=\frac 12(-A+2B+3)(A-1)\end{align}$$

So, all we need is to solve
$$\frac 12(-A+2B+3)(A-1)\le 0$$


问:
Is it valid to use the Geometric Series Test for Power Series?
My textbook (Early Transcendentals 8th e., James Stewart) advises that in general to find the interval of convergence of a power series we should use the Ratio or Root Tests. However, I found that the interval of convergence could also be found by applying the Geometric Series Test: take the absolute value of the "common ratio", set it to less than $1$, and solve for $x$.

For example,

$$\sum_{n=1}^{\infty} \frac{x^n}{n^44^n}  =  \sum_{n=1}^{\infty} \frac{1}{n^4} (\frac{x}{4})^n$$

The "common ratio" is $r= \frac{x}{4}$ since it's the factor being raised to the power $n$.

A geometric series converges when $|r| < 1$

$$|\frac{x}{4}| < 1$$

$$-1 < \frac{x}{4} <1$$

$$-4 < x < 4$$

Which produces the same interval of convergence as when using the Ratio Test.

I found that this worked for the other power series presented in this section as well.

I'm aware that the power series in the example is NOT a geometric series because the coefficient of the series,  $c_n = \frac{1}{n^4}$ is not constant as $n\rightarrow\infty$ and thus it does not actually have a common ratio since $r$ changes depending on which terms in the series are used to calculate it. In fact, none of the power series in this section were geometric series because none had a constant coefficient nor a true common ratio, hence why I'm unsure why the Geometric Series Test seemed to work for the power series presented.

Is using the Geometric Series Test to find the interval of convergence for power series valid? If so, why
since not all power series are geometric series?

答:
Even though you call it the "Geometric Series Test," the actual argument your proof describes is clearly the Ratio Test:

For example,

$\sum_{n=1}^∞ \frac{x^n}{n^4 4^n} = \sum_{n=1}^∞ \frac{1}{n^4} \left( \frac{x}{4} \right)$

The "common ratio" is $r = \frac{x}{4}$
since it's the factor being raised to the power $n$.

Here, $a_n = \frac{x^n}{n^4 4^n}$, so applying Ratio Test gives $$r = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \left( \frac{x}{4} \right) \lim_{n \to \infty}  \frac{n^4}{(n+1)^4} = \frac{x}{4}.$$ The Ratio Test and the Root Test are both based on (and proven via) the condition for convergence of a geometric series. So it's not surprising that "pretending the series is geometric" works when the complicating factor is a rational function of $n$ like $\frac{1}{n^4}$, as that factor will multiply the limit by 1 in either the Ratio Test or the Root Test.


问:
Balls and Bins moving from Box A to box B
There are $100$ yellow balls and $300$ blue balls in urn I, and there are $300$ yellow balls and $100$ blue balls in urn II. You take half of the balls from I and transfer them to II. Then you chose one ball from II, what is the probability that this ball is yellow? Let $\mbox{Urn}_i$ (Yellow, blue) denote the number of Y and P balls in urn $i$.

So the first thing I did was use smaller numbers to get a better intuition.

Time $0$:
$I(1,3)
\quad II(3,1)$

Transfer Case $1$ : $(1,1)$ from $I \to II \quad$ then
$I(0,2)
\quad II(4,2)$

Transfer Case $2$ : $(0,2)$ from $I \to II \quad$ then
$I(1,1)
\quad II(3,3)$

If $Y$ denotes the event that yellow is chosen from $II$ after the transfer, then

$Pr(Y)=Pr(Y|(1,1) \mathrm{\ transfer})+Pr(Y|(0,2) \mathrm{\ transfer})$

$Pr(Y|(1,1) \mbox{ transfer})=\frac 2{(4+2)}= \frac 13$ from the final numbers in $II$ after the transfer Case 1
$Pr(Y|(0,2) \mathrm{\ transfer})=\frac 3{3+3}= \frac 12$

So $Pr(Y)=\frac 56$ using the smaller scaled problem.
Now when I scale this to a larger multiples...I don't understand the pattern to get a better intuition.

I found this problem and I thought it was very similar, but I'm struggling with how to transfer the thought process from that Problem to a problem like this. However when I try and follow the process I am getting stuck.

Case 1: Event $Y$ s.t. $Pr(\mathrm{only \ Yellow\  transfered})= 0$
This is because you have $\frac{^{100}C_{200}}{^{400}C_{200}} = (0/\#)=0$. It also logically makes sense, if we have to move $200$ balls from I, and there are only $100$ Yellow balls in I, then there no way for only yellow balls to be transfered.

Case 2 Event $P$ s.t. $Pr(\mathrm{different \ are\  transfered})$
This is where I am having some trouble. There can be $\{(1 \mathrm{\ yellow \ ball, 199\  blue}), (2 Y, 198 B), ..., (99Y, 101B)\}$ and I am stuck on how to figure out the number of ways that different balls can be transfered. If I know this then the $Pr(P)=\frac{\#\{\mathrm{different\  ball\  combinations}\}}{^{400}C_{200}}$.

After this i should do probability of picking yellow $={\mathrm{number\  of \ yellow}}{600}$. I should divide by $600$ because that is the total number of balls in $II$ after the transfer.

Case 3 Event $B$ s.t. $Pr(\mathrm{only blue are transferred})$
this is $\frac{^{300}C_{200}}{^{400}C_{200}} = \frac{300!/(100!200!)}{400!/(200!200!)} =$ some number, but I don't understand how to simplify this. if the numbers were larger, i think i would have an overflow problem on my calculator.

However if $Pr($only  blue transfer$)=X$, then the probability of picking yellow after the transfer is $(300/600)=1/2$

Then the last step should be to add up the probabilities of each case.

答:
You have $\frac 23$ chance to draw a ball that was originally in urn II.  If you do, you have $\frac 34$ chance to get a yellow ball.  You have $\frac 13$ chance to get a ball that was originally in urn I.  If you do, you have $\frac 14$ chance to get a yellow ball.  The overall chance of a yellow ball is then
$$\frac 23 \cdot \frac 34 + \frac 13 \cdot \frac 14=\frac 7{12}$$


问:
Prove an open ball in $\mathbb{R}^n$ is not complete
I was looking at this proof that an open ball is not complete in $\mathbb{R}^n$. They showed that $\{u_k\}=\{(1-\frac{1}{n},0,0,0,...)\}$ is a sequence in $\mathbb{R}^n$ converging to $(1,0,0,...)$ by assuming without loss of generality, the open ball is the unit ball. Is this ok to do this or is there an easy way to generalize this to any ball in $\mathbb{R}^n$? They proved the sequence was cauchy by letting $\epsilon>0$ and choosing $N>\frac{2}{\epsilon}$ then for $k,l \geq N$ then $\sqrt{(1-\frac{1}{k}-(1-\frac{1}{l}))^2+(0-0)^2+\dots}=
|\frac{1}{k}-\frac{1}{l}|<\epsilon$. What would be a way to generalize this argument to any particular open ball?

答:
Given an open ball $B(a,r)$ centered at $a\in\mathbb{R}^{n}$ with radius $r > 0$, it suffices to take $b\in\partial B(a,r)$.

Indeed, since $b$ is an adherent point of $B(a,r)$, there exists a sequence $x_{n}\in B(a,r)$ which converges to $b$.

But $B(a,r)$ is open, which means that $B(a,r)\cap\partial B(a,r) = \varnothing$. In particular, $b\not\in B(a,r)$.

Since $x_{n}$ is convergent, it is Cauchy. Then we have just proven there exists a Cauchy sequence in $B(a,r)$ that does not converge within it, and we are done.

Hopefully this helps!


问:
how do I show that the set $\{x \in a : b \approx S_x \}$ is nonempty?
Here's a proof of the theorem from Charles Pinter's A book of set theory:

It seems intuitive but how do I show that the set $\{x \in a : b \approx S_x \}$ is nonempty? (I don't see how. Since $b< a$, there is an injection $b\to a$. How do I use this to establish that there is an $x\in a$ such that $b\approx S_x$?)

Also when the author says well order $a$, he means that there is a well ordering relation $\preceq$ on $A$ and shouldn't he mean $\varphi (d) \preceq \varphi (b)$ instead of the already used symbol $\leqslant$?

答:
You have an injection $f:b\to a$. Let $F=f[b]\subseteq a$, and recursively define a (possibly partial) function $h$ from a subset of $a$ onto $F$ as follows: if $h\upharpoonright S_x$ has been defined for some $x\in a$, and $h[S_x]\subsetneqq F$, let $h(a)=\min(F\setminus h[S_x])$. Informally $h$ sends the least element of $a$ to the least element of $F$, the next smallest element of $a$ to the next smallest element of $F$, and so on. In the end either $\operatorname{dom}h=a$, or there is an $x\in a$ such that $\operatorname{dom}h=S_x$. If $\operatorname{dom}h=a$, then $h^{-1}\circ f$ would be a bijection from $b$ to $a$, which is impossible, since $b<a$, so $\operatorname{dom}h=S_x$ for some $x\in a$, and $S_x\approx b$.

There’s no real harm here in using the same symbol for both orders, since it’s easy enough to keep straight which set one is working in.


问:
If a,b,c are positive real numbers such that a + b + c = 1 , then find the minimum value of $\frac{(1+a)(1+c)(1+c)}{(1-a)(1-b)(1-c)}$
I tried by solving inequality of numerator and denominator

Numerator:

$(1+a)(1+c)(1+c) \leq \frac{(3 + a + b + c)}{3} = \frac{(3 + 1)}{3} = \frac{4}{3}$

Denominator:

$(1-a)(1-c)(1-c) \leq \frac{3 - (a + b + c)}{3} = \frac{(3 - 1)}{3} = \frac{2}{3}$

By combining the 2 equations

$\frac{Numerator}{Denominator} \leq \frac{4/3}{2/3} = 2$

But that's not the answer (I know I found max value). Can anyone help me? Is the method I am following can be used in other types of question? Solving Numerator and denominator inequality and than dividing both inequatlity....

ANSWER: 8

答:
Although I am in favour of Albus' answer, here is another way to calculate the minimum directly using standard inequalities:

\begin{eqnarray*}\prod_{cyc}\frac{1+a}{1-a}
& \stackrel{GM-HM}{\geq} & \left(\frac{3}{\sum_{cyc}\frac{1-a}{1+a}}\right)^3\\
& = & \left(\frac{3}{3-2\sum_{cyc}\frac{a}{1+a}}\right)^3\\
& \stackrel{\frac 1{1+x}\, concave}{\geq} & \left(\frac{3}{3-2\frac{1}{1+\sum_{cyc}a^2}}\right)^3\\
& \stackrel{Cauchy-Schwarz}{\geq} & \left(\frac{3}{3-\frac{2}{1+\frac 13}}\right)^3\\
& = & 2^3 = 8
\end{eqnarray*}

All inequalities are "sharp" for $a=b=c=\frac 13$.


问:
What kind of distributions are the following
Let $Y_1, . . . , Y_5$ be a random sample of size 5 from a normal population with mean 0 and variance 1 and let $\bar{Y} = (1/5) \Sigma_{i=1}^5 Y_i$

Let $Y_6$ be another independent observation from
the same population. Finally, let $U = \Sigma_{i=1}^5 (Y_i - \bar{Y})^2$ and let $W = \Sigma_{i=1}^5 Y_i^2 $.

What is the distribution of

a. $\frac{\sqrt{5} * Y_6}{\sqrt{W}} $

b. $\frac{2 * Y_6}{\sqrt{U}} $

c. $\frac{2*(5\bar{Y}^2 + Y_6^2)}{U} $

Why?

I don't really get how to examine the problems. Maybe if I understood the answer to a. I could understand the other 2 but I don't have an answer to any of them right now.

Edit:
The distributions are probably either
T distribution
Chi distribution
normal
but it could be something else I guess

答:
You have to use the known distributions

a.

It's a Student's T with 5 d.o.f.

$$\frac{Y_6}{\sqrt{\frac{W}{5}}}\sim\mathcal{T}_5$$

This because $Y_6\sim\Phi$, $W\sim \chi_{(5)}^2$ and $Y_6\perp\!\!\!\perp W$

If you got a. the rest of the exercise follows as a consequence.

b. The distribution is

$$\frac{Y_6}{\sqrt{\frac{U}{4}}}\sim\mathcal{T}_4$$

I just did a copy & paste

c. Try yourself first...it's a Fischer distribution: $F(2;4)$


问:
Proving an equation relating to the Cauchy-Riemann Equations.
Here is what the question asks:

Let $\Omega \subset \mathbb{C}$ be a domain (meaning that it's a connected and open), and let $f: \Omega \rightarrow \mathbb{C}$ be holomorphic. Prove that if $f(z) = iv(x, y)$ for every $z \in \Omega$ (namely, if the real part of the function is $u \equiv 0$) then there is a constant $v_0 \in \mathbb{R}$ such that $f(z) ≡ iv_0 \in \Omega$

I have a vivid understanding of this problem, but I am not sure what the "$\equiv$" implies $u \equiv 0$ and $f(z) ≡ iv_0 $.

Although, here is how I went about solving it:

For the function to be Complex Differentiable, we know  that the Cauchy-Reimann equations must be satisfied.

Let $u(x,y)$ and $v(x,y)$ denote the real and imaginary parts of $f(z)$ respectively. Then since we know that $f(z)$ consists of only the imaginary part, $u(x,y) = 0$, hence $u_x=0, u_y=0$. The Cauchy-Riemann Equations imply: $$\begin{cases} u_x=v_y\\u_y=-v_x\end{cases}$$

Hence, the Cauchy-Riemann equations can be represented by the following matrix.
$$\begin {bmatrix} u_x &u_y\\-v_x & v_y \end {bmatrix}= \begin {bmatrix} 0 &0\\0 & 0\end {bmatrix}$$

So I have a few concerns at this point in the problem.

First, since the Jacobian matrix equals the $\vec{0}$, does that not imply that $(0,0)$ is the only place where this function is differentiable, hence contradicting the fact that $f$ is defined to be holomorphic (since there would be no other points in the neighborhood of $\Omega$)?

Second, if the Jacobian matrix is in fact the $\vec{0}$, then does that mean $v(x,y)=v_0=0$?

I would apprecaite some assistance on this problem. Thank you!

答: Any function $v$ with both partial derivative $0$ is a constant function. Since $v_x$ and $v_y$ are both $0$ it follows that $v$ is a constant. It appears that you are just over-thinking.


问:
How to show that $D$ is a Borel measurable set and $D_{f}$ is a Borel function.
How to show that $D_{f}$ is a Borel measurable function.

Well I have one Lipschitz function  $f:\Bbb{R}^{n}\to \Bbb{R}$ and I want to proof that

$D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ is Borel function, where $D=\{ x\in \Bbb{R}^{n}: f'(x)\quad \text{exist }\}$

I try with the definition to show that $\forall U$ Borel set in $ L(\Bbb{R}^{n},\Bbb{R})$ imply $D_{f}^{-1}(U)$ is Borel set.

Then let $U$ Borel set in $L(\Bbb{R}^{n},\Bbb{R})$ hence

$D_{f}^{-1}(U)=\{x\in \Bbb{R}^{n}: D_{f}(x)\in U\}$ but $D_{f}(x)$ is one linear transformation  hence

$D_{f}^{-1}(U)=\{x\in \Bbb{R}^{n}: T(x)\in U \}$ can to say : Like $T$ is continuous because is linear transformation then $D_{f}^{-1}(U)$ is measurable imply is Borel set?, can somebody help me please or give me one hint...thank you

答:
Let us prove the following result:

Let $f:\Bbb{R}^{n}\to \Bbb{R}$ be a Lipschitz function and $D=\{ x\in \Bbb{R}^{n}: f'(x)\: \text{exists }\}$. Let $D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ be the function that associates to each point $x\in D$  the value $f'(x)$. Then

$D$ is Borel subset of $\Bbb{R}^{n}$

$D_f$ is a Borel Measurable function.

(Note: $f'(x)$ is the derivative of $f$ at $x$ and $L(\Bbb{R}^{n},\Bbb{R})$ is the space of linear transformations from $\Bbb R^n$ to $\Bbb R$).

Since $f:\Bbb{R}^{n}\to \Bbb{R}$ is a Lipschitz function, there is $K>0$, such that for all $x, y \in  \Bbb{R}^n$, $|f(x)-f(y)| < K \|x - y\|$.  Note that $f$ is continuous and so it is a Borel measurable function.

Let $D=\{ x\in \Bbb{R}^{n}: f'(x)\: \text{exists }\}$

Let us prove that $D$ is a Borel measurable set.

It can be proved (see Remark below) that $x \in D$ if and only if

$$\forall\epsilon\in  \mathbb{Q}, \epsilon>0, \: \exists\delta \in  \mathbb{Q},\delta>0,\:\exists T\in\mathbb{Q}_{1\times n}, \Vert T\Vert \leqslant K ,\:\forall h \in \mathbb{Q}^{n},\Vert h\Vert<\delta, \:\\ | f(x+h)-f(x)-Th |\leq\epsilon\Vert h\Vert,$$
where $\mathbb{Q}_{1\times n}$ is the set of $1\times n$ rational matrices. It means that

$$ D= \bigcap_{\substack{\epsilon\in  \mathbb{Q} \\ \epsilon>0}}\bigcup_{\substack{\delta \in  \mathbb{Q} \\ \delta>0}} \bigcup_{\substack{T\in\mathbb{Q}_{1\times n} \\ \Vert T\Vert \leqslant K }}\bigcap_{\substack{ h \in \mathbb{Q}^{n} \\ \Vert h\Vert<\delta }} \{x \in \mathbb{R}^n: | f(x+h)-f(x)-Th |\leq\epsilon\Vert h\Vert\} \tag{1}$$

But it is easy to see that the sets $\{x \in \mathbb{R}^n: | f(x+h)-f(x)-Th |\leq\epsilon\Vert h\Vert\}$ are Borel measurable sets. So, since all intersections and unions in $(1)$ are countable intersections and countable unions, we have that $D$ is a Borel measurable set.

Now, let us prove that  $D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ is a Borel function.

First, note that Borel sets in $L(\Bbb{R}^{n},\Bbb{R})$ are the elements of the $\sigma$-algebra generate by the open sets of the norm topology of $L(\Bbb{R}^{n},\Bbb{R})$. Since $L(\Bbb{R}^{n},\Bbb{R})$ is isomorphic to $\Bbb{R}^{n}$, the Borel sets in $L(\Bbb{R}^{n},\Bbb{R})$ can be thought of as the Borel sets in $\Bbb R^n$.

Now, note that, by using the canonical basis in $\Bbb R^n$,  we have that
$$D_f= \left (\frac {\partial f}{\partial x_1}, \dots ,   \frac {\partial f}{\partial x_n} \right)$$

So it is enough to prove that any $i\in \{1,...,n\}$, the function

$\frac {\partial f}{\partial x_i}: D\to \Bbb R $ is a Borel function.

Let $(t_j)_{j \in \Bbb N}$ be a sequence of positive real numbers converging to $0$. Now,
for any $i\in \{1,...,n\}$ and any $j \in \Bbb N $, let us define $d_{i,j} : D\to \Bbb R $ by
$$d_{i,j}(x) = \frac{f(x+t_je_i) -f(x)}{t_j}$$

Since $f$ is a Borel measurable function, clearly $d_{i,j}$ are Borel measurable functions and, as $j\to \infty$, they converge pointwise to $\frac {\partial f}{\partial x_i}$ in $D$. So  $\frac {\partial f}{\partial x_i}$ are Borel measurable functions. So $D_f$ is a Borel measurable function.

Remark: let us prove that

$x \in D$ if and only if

$$\forall\epsilon\in  \mathbb{Q}, \epsilon>0, \: \exists\delta \in  \mathbb{Q},\delta>0,\:\exists T\in\mathbb{Q}_{1\times n}, \Vert T\Vert \leqslant K ,\:\forall h \in \mathbb{Q}^{n},\Vert h\Vert<\delta, \:\\ | f(x+h)-f(x)-Th |\leq\epsilon\Vert h\Vert,$$
where $\mathbb{Q}_{1\times n}$ is the set of $1\times n$ rational matrices.

Proof: Let $\mathbb{R}_{1\times n}$ be the set of $1\times n$ real matrices.

($\Rightarrow$) If $x \in D$ , then $D_f(x)$ is represented by a $1 \times n$ real matrice and $\Vert D_f \Vert \leqslant K$. Since $\mathbb{Q}_{1\times n}$ is a dense subset of $\mathbb{R}_{1\times n}$ and $D_f(x) \in \mathbb{R}_{1\times n}$, we can take $T \in \mathbb{Q}_{1\times n}$, such that $\Vert D_f(x) - T \Vert < \epsilon/2$. Now, since $\lim\frac{f(x+h)-f(x)-df(x)h}{\Vert h\Vert}=0$ there exists $\delta>0$ such that if $\Vert h\Vert<\delta$ then $\Vert f(x+h)-f(x)-df(x)h\Vert\leq\epsilon\Vert h\Vert/2$, so
\begin{align*}
\Vert f(x+h)-f(x)-Th\Vert &\leqslant \epsilon\Vert h\Vert/2+\Vert Th-df(x)h\Vert \\
& \leqslant \epsilon\Vert h\Vert/2+\Vert T-df(x)\Vert\Vert h \Vert \\
& \leqslant \epsilon\Vert h\Vert
\end{align*}

($\Leftarrow$) Let $\{\epsilon_n\}_n$ be a sequence of positive rational numbers converging to $0$. Then, there are $\delta_n \in  \mathbb{Q},\delta_n>0,\:\exists T_n\in\mathbb{Q}_{1\times n}, \Vert T_n\Vert \leqslant K$ such that
for all $ h \in \mathbb{Q}^{n},\Vert h\Vert<\delta_n$,

$$| f(x+h)-f(x)-T_n h |\leq\epsilon_n\Vert h\Vert \tag{2}$$
Since $f$ is continuous, $(2)$ holds for all  $ h \in \mathbb{R}^{n},\Vert h\Vert<\delta_n$.

Since $\{ A \in\mathbb{R}_{1\times n}: \Vert A \Vert \leqslant K\}$ is compact and, for all $n$, $T_n \in \{ A \in\mathbb{R}_{1\times n}: \Vert A \Vert \leqslant K\}$, we can, passing to a a subsequence if needed, assume that $T_n$ converges (in the norm topology) to some $T \in \mathbb{R}_{1\times n}$.

So, if $\Vert h\Vert<\delta_n$, then
\begin{align*}
\Vert f(x+h)-f(x)-T h\Vert&\leqslant \Vert f(x+h)-f(x)-T_n h\Vert+\Vert T_n h-T h\Vert\\
&\leq(\epsilon_n+\Vert T_n-T\Vert)\Vert h\Vert.
\end{align*}
Since $T_n$ converges to $T$ and $\epsilon_n$ converges to $0$, we have that $f$ is differentiable at $x$ with $df(x)=T$.


问:
Double summation proof
Prove that the sum of all the numbers in the triangular table
$$
a(1,1)\\
a(2,1) a(2,2)\\
a(3,1) a(3,2) a(3,3)\\
.\\
.\\
.\\
a(m,1) a(m,2) a(m,3)......a(m,m)\\
$$
can be expressed as $$\sum_{i = 1}^m \sum_{j = 1}^ i a(i,j) $$

答: When $m=3$ the outer sum has three terms.  The inner sum has $i$ terms.  When $i=3$ the inner sum gives $a(3,1)+a(3,2)+a(3,3)$ which is exactly row $3$ of your table.  Having seen this you should be able to convince yourself that the inner sum always gives row $i$ of the table, so the outer sum gives the sum of the whole triangle through row $m$.

Title: Double summation proof


问:
Why the $\nabla f(x)$ in the direction orthogonal to $f(x)$?
Why the $\nabla f$ in the direction orthogonal to $f$?

I can't understand it intuitively.

Is it because of the definition of gradient?

答:
Let's take an example to illustrate.
Let $f(\boldsymbol{x}) = f(x,y)$. You may think $z = f(x,y)$ and you can plot the graph of the function in $\mathbb{R}^3$.

Now take a contour of $f(x,y)$, i.e. $f(x,y) = c$, total derivative shows
$$\mathrm{d}f = \frac{\partial f}{\partial x}\mathrm{d}x+ \frac{\partial f}{\partial y}\mathrm{d}y = 0,$$
you can think $\mathrm{d}x, \mathrm{d}y$ as infinitesimal increments of $x,y$.
With notice the above equation can be written as
$$\nabla f\cdot\mathrm{d}\boldsymbol{x} =
\begin{bmatrix}
\frac{\partial f}{\partial x}\\
 \frac{\partial f}{\partial y}
\end{bmatrix}
\cdot
\begin{bmatrix}
\mathrm{d} x\\
\mathrm{d} y
\end{bmatrix}
=0,$$
we have shown that $\nabla f(x,y)$ is orthogonal to $f(x,y) = c $ for all $(x,y)$ and $c$.

Let $\boldsymbol{v}$ be a unit vector, $\nabla f\cdot \boldsymbol{v}$ shows how $f$ would change in that direction. Notice
$|\nabla f\cdot \boldsymbol{v}|\leq |\nabla f||\boldsymbol{v}| = |\nabla f| $ and equality only holds for $\boldsymbol{u} = \nabla f/|\nabla f|$. (You can argue what happen if $|\nabla f| = 0$, i.e. a plane $z = const.$)

So $\nabla f$ offers the information on how much and in which direction $f$ changes the most.


问:
Boundedness in a space implies boundedness in the subspace
Suppose that $X$ is a Banach space. And let $A$ be an unbounded operator with the domain $D(A)$ which is densely continuously embedded into $X$.

Recall that if $\|\cdot\|$ is a norm of $X$ then that of $D(A)$ is $\|x\|_{D(A)}=\|x\|+\|Ax\|$ for all $x\in D(A)$.

Now, let $U\subset D(A)$ be bounded in terms of $\|\cdot\|$. In other words, there exists $M>0$ such that $\|x\|\leq M$ for all $x\in U$.

Then, is it true that $U$ is bounded in terms of $\|\cdot\|_{D(A)}$?

答: If $A$ is unbounded then there is a sequence $(x_n)$ in $D(A)$ such that $\|x_n\| \to 0$ and $\|Ax_n\| \to \infty$. Take $U=\{x \in D(A):\|x\|\leq 1\}$. Clearly  $U$ is not bounded for $\|.\|_{D(A)}$.


问:
Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$
Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$

My work

Changing to cylindrical coordinates

For Paraboloid

$$x^2+y^2=az\;\;\;\implies r^2=az\;\;\;\implies z=\frac{r^2}{a}$$

For the cylinder

$$x^2+y^2=2ay\;\;\;\implies r=2a \sin \theta$$

Since this volume lies only in first two quadrants $\theta$ goes from $0$ to $\pi$

Volume=$\int_{\theta=0}^{\pi}\int_{r=0}^{2a\sin \theta}\int_{z=0}^{r^2/a}r\ dr\ d\theta \ dz$

答:
You may wish to re-order your integration.  You can also exploit the symmetry of the region of integration about the $ \ yz- $ plane by integrating over $ \theta $ from $ \ 0 \ $ to $ \ \frac{\pi}{2} \ $ .  Thus,

$$ V \ = \ 2 \ \int_{ 0}^{ \pi / 2}\int_{0}^{2a\sin \theta}\int_{0}^{r^2/a} \ \ \ dz \  \ r \ dr \ \  d\theta \ \ = \ \ 2 \ \int_{ 0}^{ \pi / 2}\int_{0}^{2a\sin \theta}  \ \  \frac{r^2}{a}  ·  r \ dr \ \  d\theta $$

$$ = \ \ \frac{2}{a} \ \int_{ 0}^{ \pi / 2}  \ \  \frac{1}{4} \ (2a \sin \theta)^4  \ \  d\theta \ \ = \ \ \frac{2·2^4·a^4}{4a} \ \int_{ 0}^{ \pi / 2}   \sin^4 \theta   \ \  d\theta \ \ = \ \ 8a^3 \ · \ \frac{3 \pi}{16} $$

$$  = \ \ \frac{3 \pi}{2}a^3 \ \ ,  $$

the last definite integral being of a kind $ \left( \ \int_{ 0}^{ \pi / 2}   \sin^n \theta   \ \  d\theta \  =  \ \int_{ 0}^{ \pi / 2}   \cos^n \theta   \ \  d\theta \ \right) $ for which a formula to evaluate it is available.


答:
let $t=\sin \theta -\cos \theta $ then we get $$\sin \theta \cos \theta =\frac{1-t^2}{2}$$ indeed we have tofind $\theta$ which satisfies the inequality $$t(2+\frac{1-t^2}{2})\le 2$$ $$\iff f(t)=t^3-5t+4\ge 0$$ As you say this came in an exam lets presume that the factorisation is not so complicated.Suppose it has an integer root then by rational root theorem we can say that any integer root $p$ (say) must divide $4$ .$p|4$ so if there is an integer root it has to be one of  $\pm1,\pm2\pm3\pm4$ We can skip $\pm 2\pm 3\pm4$ as $|t|\le \sqrt{2}$ .We are left with $\pm 1$ .By inspection we note that $f(1)=0$ hence we have found one factor i.e $t-1$ .The other factor can be easily found by long division ie $t^2+t-4$.

Now its easy to see $t^2+t-4<0$ for all $|t|\le ]\sqrt{2}$.

So $$t^3-5t+4\ge 0\implies (t-1)(t^2+t-4)\ge 0\implies t\le 1$$

Which  you can  be easily taken from here

Title: Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$

Body:
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too


问:
Question about definition of inverse function
So, I've been reading Precalculus books and I found a question in the definition of inverse function,
some of my books say:

Let $f\colon X\to Y$ be a real function, where $X,Y\subseteq\mathbb{R}$ such that: $g\colon Y\to X$ is the inverse function of $f$ if for each $x\in X$, such that $$g(f(x))=x,\ \forall x\in X\quad\text{and}\quad f(g(y))=y,\ \forall y\in Y.$$

And others of my books say:

Let $f\colon X\to Y$ be a real function, such that $g\colon Y\to X$ is the inverse functdion of $f$ if for each element $x\in X$, if always it’s true that
$$y=f(x)\quad\text{if and only if}\quad x=g(y),\qquad \forall x\in X\text{ and }\forall y\in Y.$$

So, my question is:
What of these is the true definition of inverse function? and
If posible to proof the other definition with the real definition of inverse function?
Further,  What if the proof the link a bijective function and inverse function?

答:
Claim. Let $X,Y$ be sets amd $f\colon X\to Y$, $g\colon Y\to X$ be functions. Then the following are equivalent

$$\tag1\forall x\in X\colon g(f(x))=x\;\land\;\forall y\in Y\colon f(g(y))=y$$
$$\tag2 \forall x\in X,\forall y\in Y\colon y=f(x)\leftrightarrow x=g(y)$$

Proof.
Assume $(1)$. To show $(2)$, let $x\in X, y\in Y$ be given. We wan to show $y=f(x)\iff x=g(y)$. So first assume $y=f(x)$. Then by the first part of $(1)$, $g(y)=g(f(x))=x$. In summary, $y=f(x)\to x=g(y)$. Conversely assume $x=g(y)$. Ten by the seoncd part of $(1)$, $f(x)=f(g(y))=y$. In summary, $x=g(y)\to y=f(x)$. IN overall summary, $y=f(x)\leftrightarrow x=g(y)$. As $x, y$ were arbitrary elements of $X$ and $Y$ respectively, we have shown $(2)$ from $(1)$.

Next, Assume $(2)$. To show the first part of $(1)$, let $x\in X$ be given. As $f$ is a function $X\to Y$, we can let $y=f(x)\in Y$. Specialize $(2)$ to $x$ and $y$ to find $y=f(x)\leftrightarrow x=g(y)$. By choice of $y$, the left side holds, hence $x=g(y)=g(f(x))$, or: $g(f(x))=x$. As $x$ was an arbitrary element of $X$, this shows $\forall x\in X\colon g(f(x))=x$.
To show the first part of $(1)$, let $y\in Y$ be given. As $g$ is a function $Y\to X$, we can let $x=f(y)\in X$. Specialize $(2)$ to $x$ and $y$ to find $y=f(x)\leftrightarrow x=g(y)$. By choice of $x$, the right side holds, hence $y=f(x)=f(g(y))$, or: $f(g(y))=y$. As $y$ was an arbitrary element of $Y$, this shows $\forall y\in Y\colon f(g(y))=y$.
Now that we have shown both $\forall x\in X\colon g(f(x))=x$ and $\forall y\in Y\colon f(g(y))=y$, we have in fact shown $\forall x\in X\colon g(f(x))=x\;\land\;\forall y\in Y\colon f(g(y))=y$, i.e., we have shown $(1)$ from $(2)$.

In summary, $(1)$ and $(2)$ are equivalent. $\square$


问:
If a,b are real numbers. Find the least value of $a^2 + ab + b^2 - a - 2b$
I tried simple $AM \geq GM$ inequality But no luck

$a^2 + ab + b^2 - a - 2b \geq \sqrt{a^2 * ab * b^2 * (-a) * (-2b)} = \sqrt{2a^4b^4} = a^2b^2\sqrt{2}$

Anwer is $-1$

答:
Just do some square completion:

\begin{eqnarray*}a^2+ab+b^2-a-2b
& = & \left(a+\frac b2\right)^2 + 3\left(\frac b2\right)^2 - a - 2b \\
& = & \left(a+\frac b2\right)^2 + 3\left(\frac b2\right)^2 - \left(a+\frac b2\right) - 3\frac b2 \\
& \stackrel{u=a+\frac b2, v= \frac b2}{=} & \left(u-\frac 12\right)^2 -\frac 14+3\left(v-\frac 12\right)^2-\frac 34 \\
& \geq & -1
\end{eqnarray*}


问:
How is "n+n/2+n/4....1" equal to "2n-1" using the formula for geometric series?
I never knew not having good knowledge of basic maths will be so crippling!! So please help me out this time. I'll be working on my maths from today on.

I was discussing about complexity of an algorithm on StackOverflow and I was told that the series $n+n/2+n/4 + \dots + 1$ evaluates to $2n-1$ and I was linked to the following formula on Wikipedia:

Even after trying hard, I regret to say I still don't get it how using this formula I can conclude that my series evaluates to $2n-1$. Please help me out as I am sure it will take only a few seconds for you.

https://stackoverflow.com/questions/16748454/complexity-for-nested-loops-diving-by-2?noredirect=1#comment24124682_16748454

答:
Assuming $n$ is a power of $2$, say $n = 2^p$. Then the sum is
$$
S = n + \frac n2 + \frac n4 + \ldots + 1 =
2^p + 2^{p-1} + 2^{p-2} + \ldots + 1.
$$
Multiply by $2$ on both sides to get
$$
2S_p = 2^{p+1} + 2^p + 2^{p-1} + \ldots + 2.
$$
Compute the difference $2S_p - S_p$ using the two equations above:
\begin{align*}
2S_p - S_p & = 2^{p+1} - 1 \\
\therefore S_p & = 2^{p+1} - 1 = 2n - 1.
\end{align*}

For a discussion of the general series, suppose $a \ne 1$ and $p \in \mathbb N$ are given, and suppose we want to compute
$$
S = 1 + a + a^2 + \ldots + a^p = \sum_{i=0}^p a^i.
$$
We can use the same idea as above: multiply by $a$ to get
$$
aS = a + a^2 + a^3 + \ldots + a^{p+1} = \sum_{i=1}^{p+1} a^i.
$$
Compute the difference:
\begin{align*}
aS - S & = a^{p+1} - 1 \\
(a - 1)S & = a^{p+1} - 1\\
S & = \frac{a^{p+1} - 1}{a - 1}.
\end{align*}


问:
How to show that the sequence $ \left\lbrace 4+3(-1)^n \right\rbrace $ diverges by $ \varepsilon -N $ definition?
Preface

We will use the following facts

i) The sequence $ \left\lbrace a_n \right\rbrace $ is convergent to $a$ if for each $ \varepsilon >0$ there exists $ N>0 $ such that $ \vert a_n -a \vert < \varepsilon $ whenever $ n>N$.

(ii) $ |A+B| \leqslant |A|+|B| $; and $ \vert -A+B \vert = \vert A-B \vert $

(iii) $ (-1)^{2n+1} =-1$ and $ (-1)^{2n}=1 $

(iv) The statement $ 8<8 $ is a contradiction. Contradiction=false in every way.

Proof

Suppose the sequence is convergent to $ \omega $. Then for each $ \varepsilon >0 $, there exists $ N>0 $, such that $ \vert 4+3(-1)^n - \omega \vert < \varepsilon $, whenever $ n>N $. When $ n $ is even, we have $ \vert 7-\omega \vert < \varepsilon $, whenever $ n>N $. When $ n $ is odd, we have $ \vert 1-\omega \vert < \varepsilon $, whenever $ n> N $. We choose $ \varepsilon=1 $ and  consider \begin{align*}
 8 &=\vert 7+1 \vert=\vert 7 -\omega + \omega +1\vert\\
  &=\vert (7-\omega)+4 -3+\omega \vert \\
  & \leqslant \vert 7 -\omega \vert + \vert 4 \vert+ \vert -3+\omega \vert \\
  &= \vert 7 -\omega \vert + \vert 4 \vert+ \vert -2 -1+\omega \vert \\
  & \leqslant \vert 7 -\omega \vert + \vert 4 \vert+\vert -2 \vert + \vert -1+\omega \vert \\
  &= \vert 7 -\omega \vert + 6 + \vert 1-\omega \vert \\
  & < 2 \varepsilon +6 \\
  &= 8
 \end{align*}
This is a contradiction.

Question: Do you like this proof? Is it correct?

答:
Everything looks good through choosing $\varepsilon=1$. Once you have done this, you have that $$|7-\omega|<1\text{ and }|1-\omega|<1.$$
The first inequality has a solution set of $6<\omega<8$ and the second has a solution set of $0<w<2$. No real need for the extended calculation when you can simply solve for $\omega$ and show no solution exists.

Edit: Technically, after you choose $\varepsilon=1$, you should say something like "Now, let $N$ be arbitrary." Then any even $n>N$ gives us $|7-\omega|<1$ and any odd $n>N$ gives us $|1-\omega|<1$. This shows super clearly what your logic is showing.


问:
How to show that $D_{f}$ is a Borel function
How to show that $D_{f}$ is a Borel function.

Well I have one Lipschitz function  $f:\Bbb{R}^{n}\to \Bbb{R}$ and I want to proof that

$D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ is Borel function, where $D=\{ x\in \Bbb{R}^{n}: f'(x) \text{ exists in the Fréchet sense }\}$

I try with the definition to show that $\forall U$ Borel set in $ L(\Bbb{R}^{n},\Bbb{R})$ imply $D_{f}^{-1}(U)$ is Borel set.

Then let $U$ Borel set in $L(\Bbb{R}^{n},\Bbb{R})$ hence

$D_{f}^{-1}(U)=\{x\in \Bbb{R}^{n}: D_{f}(x)\in U\}$ but $D_{f}(x)$ is one linear transformation (using or not using Fréchet sense) so

$D_{f}^{-1}(U)=\{x\in \Bbb{R}^{n}: T(x)\in U \}$ can to say : Like $T$ is continuous because is linear transformation then $D_{f}^{-1}(U)$ is measurable imply is Borel set?, can somebody help me please or give me one hint...thank you

答:
Let us prove the following result:

Let $f:\Bbb{R}^{n}\to \Bbb{R}$ be a Lipschitz function and $D=\{ x\in \Bbb{R}^{n}: f'(x)\: \text{exists }\}$. Let $D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ be the function that associates to each point $x\in D$  the value $f'(x)$. Then
$D_f$ is a Borel measurable function.

(Note: $f'(x)$ is the derivative of $f$ at $x$ and $L(\Bbb{R}^{n},\Bbb{R})$ is the space of linear transformations from $\Bbb R^n$ to $\Bbb R$).

Since $f:\Bbb{R}^{n}\to \Bbb{R}$ is a Lipschitz function, there is $K>0$, such that for all $x, y \in  \Bbb{R}^n$, $|f(x)-f(y)| < K \|x - y\|$.  Note that $f$ is continuous and so it is a Borel measurable function.

First, note that Borel sets in $L(\Bbb{R}^{n},\Bbb{R})$ are the elements of the $\sigma$-algebra generate by the open sets of the norm topology of $L(\Bbb{R}^{n},\Bbb{R})$. Since $L(\Bbb{R}^{n},\Bbb{R})$ is isomorphic to $\Bbb{R}^{n}$, the borel sets in $L(\Bbb{R}^{n},\Bbb{R})$ can be thought of as the Borel sets in $\Bbb R^n$.

Now, to prove that  $D_{f}:D\to L(\Bbb{R}^{n},\Bbb{R})$ is Borel function, note that, by using the canonical basis in $\Bbb R^n$,  we have that
$$D_f= \left (\frac {\partial f}{\partial x_1}, \dots ,   \frac {\partial f}{\partial x_n} \right)$$

So it is enough to prove that any $i\in \{1,...,n\}$, the function

$\frac {\partial f}{\partial x_i}: D\to \Bbb R $ is a Borel function.

Let $(t_j)_{j \in \Bbb N}$ be a sequence of positive real numbers converging to $0$. Now,
for any $i\in \{1,...,n\}$ and any $j \in \Bbb N $, let us define $d_{i,j} : D\to \Bbb R $ by
$$d_{i,j}(x) = \frac{f(x+t_je_i) -f(x)}{t_j}$$

Since $f$ is a Borel measurable function, clearly $d_{i,j}$ are Borel measurable functions and, as $j\to \infty$, they converge pointwise to $\frac {\partial f}{\partial x_i}$ in $D$. So  $\frac {\partial f}{\partial x_i}$ are Borel measurable functions. So $D_f$ is a Borel measurable function.

Remark: It is also possible to show that $D$ is a Borel measurable set.


问:
Proving that there is a unique real number x such that for every real number $y$, $xy+x-4=4y$
The question is from problem 2 of chapter 3.6 of Velleman's How To Prove It (2nd edition).
The question has been asked by another user before, but I couldn't find any thread that addresses the specific problem I have with my proof/problem.

My proof:

Existence
Let $x = 4$, then $xy+x-4 = 4y+4-4 = 4y$

Uniqueness
Let a and y be arbitrary and suppose $∀(+−4=4)$ and $∀(b+b−4=4)$.

From $∀(+−4=4)$, Suppose $y = b$, then
$ab + a -4 = 4b$.

Then from $∀(b+b−4=4)$, suppose y = a, thus $ba+b-4 = 4a$.

Subtracting $ba+b-4 = 4a$ from $ab + a -4 = 4b$, we get $4b-4a = 0$, thus, $a = b$.

Since $a$ and $b$ are arbitrary, we can conclude that $a$ = $b$ = $x$ = $4$. Thus, $x$ is unique.

Q.E.D.

My issue with the proof:

I think this is correct, but something about the existence part of the proof is bothering me. Specifically the fact that
we infer $x = 4$ by solving $xy+x-4=4y$ for $x$ like so:

$xy+x-4=4y$ iff

$x(y+1)=4y+4$ iff

$x = 4(y+1)/(y+1)$.

The $(y+1)$ factors cancel out and we are left with $4$. However, isn't $x$ undefined for $y=-1$?

So how is the statement "There exists a real number $x$ for all $y$ such that $xy+x-4=4y$" actually true? (As $x$ is undefined for $y = -1$?)
Plugging in $-1$ for $y$ and $4$ for $x$ still makes the equation work, so I think the existence part of my proof is possibly correct, but how?

答:
Actually, when you try to find $x$ from the given property, you are more in the uniqueness part than the existence part ("If the property holds, then $x$ must be ..."); and indeed your second "iff" in that derivation is not correct as it works only in one direction unless we poystulate $y\ne -1$. Ultimately, you show existence just as you did: you plug in the value $x=4$ and win. How you arrived at that guess, doesn't matter for the existence proof. You could have derived it "legally", or with a dirty trick of ignoring a sometimes undefined expression, or played around with a few values until things work out, our have dreamt about it, or whatnot.

Likewise, you could simplify your uniqueness part by testing $a$ vs. $b$ only at $y=0$, which means $a-4=0$ and $b-4=0$, so $a=b$ ($=4$, but we don't care here). This ignores whether $a$ and $b$ are also good for $y=42$, but that gap doesn't matter for the uniqueness part.


答:
You wrote:

$$x(y + 1) = 4y + 4 \iff x = 4(y + 1) / (y + 1).$$

I disagree. While $x = 4(y + 1) / (y + 1) \implies x(y + 1) = 4y + y$ by multiplying both sides by $y + 1$, the reverse implication is false in the particular case where $y = -1$. This is why you're getting this small uncertainty regarding $y = -1$.

Now, obviously when $y = -1$, there is no way to determine a specific value of $x$. Indeed, if $y = -1$, then any value of $x$ will satisfy the equation. But, the question specifically asks you to show there is a unique $x$ so that any $y$ satisfies the equation. Certainly $y = -1$ is the easiest case, but there will be only one $x$ that satisfies the equation for, say, $y = 3$ (or indeed any other value of $y$ except $y = -1$).

Title: Proving that there is a unique real number x such that for every real number $y$, $xy+x-4=4y$

Body:
The question is from problem 2 of chapter 3.6 of Velleman's How To Prove It (2nd edition).
The question has been asked by another user before, but I couldn't find any thread that addresses the specific problem I have with my proof/problem.

My proof:

Existence
Let $x = 4$, then $xy+x-4 = 4y+4-4 = 4y$

Uniqueness
Let a and y be arbitrary and suppose $∀(+−4=4)$ and $∀(b+b−4=4)$.

From $∀(+−4=4)$, Suppose $y = b$, then
$ab + a -4 = 4b$.

Then from $∀(b+b−4=4)$, suppose y = a, thus $ba+b-4 = 4a$.

Subtracting $ba+b-4 = 4a$ from $ab + a -4 = 4b$, we get $4b-4a = 0$, thus, $a = b$.

Since $a$ and $b$ are arbitrary, we can conclude that $a$ = $b$ = $x$ = $4$. Thus, $x$ is unique.

Q.E.D.

My issue with the proof:

I think this is correct, but something about the existence part of the proof is bothering me. Specifically the fact that
we infer $x = 4$ by solving $xy+x-4=4y$ for $x$ like so:

$xy+x-4=4y$ iff

$x(y+1)=4y+4$ iff

$x = 4(y+1)/(y+1)$.

The $(y+1)$ factors cancel out and we are left with $4$. However, isn't $x$ undefined for $y=-1$?

So how is the statement "There exists a real number $x$ for all $y$ such that $xy+x-4=4y$" actually true? (As $x$ is undefined for $y = -1$?)
Plugging in $-1$ for $y$ and $4$ for $x$ still makes the equation work, so I think the existence part of my proof is possibly correct, but how?


答:
You have an algebraic error in your uniqueness proof: subtracting $ba+b-4=4a$ from $ab+a-4=4b$ yields $a-b=4b-4a$, not $4b-4a=0$. Of course then you get $5b-5a=0$, which also leads to the desired conclusion that $a=b$.

Solving for $x$ as you did is harmless provided that you then check the case $y=-1$ to be sure that the desired identity still holds, though your last iff is not actually correct, precisely because the fraction is undefined for $y=-1$. However, you can avoid the problem altogether and get the uniqueness at the same time: $xy+x-4=4y$ if and only if $(x-4)y+(x-4)=0$, i.e., if and only if $(x-4)(y+1)=0$. This in turn is the case if and only if $x=4$ or $y=-1$. In particular, if $x=4$ it is true for all $y$. If, on the other hand, $x\ne 4$, then it is true if and only if $y=-1$, so it is not true for all $y$.


答:
Using your idea a cleaner way to show the contradiction is noticing that

$$
2>|7-w|+|1-w|\geqslant |7-w+w-1|=6
$$

Title: How to show that the sequence $ \left\lbrace 4+3(-1)^n \right\rbrace $ diverges by $ \varepsilon -N $ definition?

Body:
Preface

We will use the following facts

i) The sequence $ \left\lbrace a_n \right\rbrace $ is convergent to $a$ if for each $ \varepsilon >0$ there exists $ N>0 $ such that $ \vert a_n -a \vert < \varepsilon $ whenever $ n>N$.

(ii) $ |A+B| \leqslant |A|+|B| $; and $ \vert -A+B \vert = \vert A-B \vert $

(iii) $ (-1)^{2n+1} =-1$ and $ (-1)^{2n}=1 $

(iv) The statement $ 8<8 $ is a contradiction. Contradiction=false in every way.

Proof

Suppose the sequence is convergent to $ \omega $. Then for each $ \varepsilon >0 $, there exists $ N>0 $, such that $ \vert 4+3(-1)^n - \omega \vert < \varepsilon $, whenever $ n>N $. When $ n $ is even, we have $ \vert 7-\omega \vert < \varepsilon $, whenever $ n>N $. When $ n $ is odd, we have $ \vert 1-\omega \vert < \varepsilon $, whenever $ n> N $. We choose $ \varepsilon=1 $ and  consider \begin{align*}
 8 &=\vert 7+1 \vert=\vert 7 -\omega + \omega +1\vert\\
  &=\vert (7-\omega)+4 -3+\omega \vert \\
  & \leqslant \vert 7 -\omega \vert + \vert 4 \vert+ \vert -3+\omega \vert \\
  &= \vert 7 -\omega \vert + \vert 4 \vert+ \vert -2 -1+\omega \vert \\
  & \leqslant \vert 7 -\omega \vert + \vert 4 \vert+\vert -2 \vert + \vert -1+\omega \vert \\
  &= \vert 7 -\omega \vert + 6 + \vert 1-\omega \vert \\
  & < 2 \varepsilon +6 \\
  &= 8
 \end{align*}
This is a contradiction.

Question: Do you like this proof? Is it correct?


问:
machine learning model vs time series model?
A general question about machine learning models vs classical time series model in prediction tasks? When to use machine learning models, when to use time series models? What is the advantage and disadvantage of them?

My first thought is time series models are more widely used when there is 1 predictor. However, there are multivariate time series models so this argument doesn't really stand.

Another thought is about the assumption of the model. For example, many machine learning model assumes independence of data while time series model does not. But what else? when should we use machine learning models for prediction task instead of time series models?

答:
Check out this paper on arXiv. It expounds on the following:

We demonstrate how by applying deep learning techniques to forecasting, one can overcome many of the challenges faced by widely-used classical approaches to the problem.

I think you're right in terms of the one or several time series being predicted: at least Amazon Web Services claims the following for the above arXiv-presented model:

When your dataset contains hundreds of feature time series, the DeepAR+ algorithm outperforms the standard ARIMA and ETS methods

I hope that this gives you some (at least close to) cutting edge work done on machine learning for time series. After all, if "classical" time series models e.g. ARIMA didn't have room for improvement, there would be no comparison to make!


问:
Existence and uniqueness of $SL(n,\mathbb R)$ left-invariant Borel probability measure on $SL(n,\mathbb R)/SL(n,\mathbb Z)$.
Consider the action of $SL(n,\mathbb R)$ on the homogeneous space $SL(n,\mathbb R)/SL(n,\mathbb Z)$ by left translation. Are there any good refences where I can find the proof of the existence and uniqueness of the left $SL(n,\mathbb R)$-invariant Borel probability measure on $SL(n,\mathbb R)/SL(n,\mathbb Z)$.

If there is some powerful theorem addressing this for general $G$ and $G/\Gamma$, please let me know how $SL(n,\mathbb R)$ and $SL(n,\mathbb R)/SL(n,\mathbb Z)$ satisfy the hypothesis of that theorem.

答:
For the $SL(n,\mathbb{R})$ case, this is the content of the fact that $SL(n,\mathbb{Z})$ is a lattice. There are two well-known approaches; one is by way of building good fundamental domains ("Siegel sets") and the other is Margulis' unipotent orbit argument.

As a starting point Witte Morris' Introduction to Arithmetic Groups is probably appropriate (available at https://arxiv.org/abs/math/0106063). See Chapter 7 there. The references there would point toward generalizations too I presume.


问:
why is an annulus close to it's boundary when it's boundary curves are close?
This is the motivating question for the rather vague question here: when is the region bounded by a Jordan curve "skinny"?

Suppose we are given two Jordan curves in the plane, one inside the other and each contained in an epsilon neighborhood of the other. How can we conclude that the annular region between the two curves is itself contained in the two epsilon neighborhoods of the original curves?

Note: by epsilon neighborhood of a curve I mean the set of all points which lie within a distance epsilon from some point in the image of the curve.

Follow up: Since the statement above is false (see Hagen's answer) we might ask whether it is true if we assume point-wise closeness of chosen parametrizations of the curves.  This is answered in the affirmative by user7...

答: We can't. Consider an arbitrarily large closed disk. At two points of its circumference let two thin strips emerge that first wind araound back and forth very close to the disk boundary and then meet. The contour arount this figure consists of two Jordan curves making a counterexample.


问:
Prove that $P(A\cup B) = P (B) + P (A)\cdot P (B^c)$
Suppose A and B are independent events. Show that $P(A\cup B) = P (B) + P (A)\cdot P (B^c)$

Could someone tell me how i can prove this?

Remembering that two events are independent if $P (A \cap B) = P (A)\cdot  P (B)$

答:
Make use of the formula $P(A\cup B) = P(A)+P(B)-P(A\cap B)$.

You have also provided $P(A\cap B) = P(A)\cdot P(B)$ since $A$ and $B$ are independent.

You can use them to express the terms in the problem.


问:
(Wrong site) Can't find mistake in resolution of second order non-homogenous linear differential equation
I'm trying to solve: $y'' + 10y' + 25y = (3x+2)e^{-5x}$

I start by rewriting as $(D^2 + 10D + 25)y = (3x + 2)e^{-5x}$

We have a non-distinct real root of multiplicity 2 which is -5.

The homogeneous solution is then $y_{h} = (C_{1} + C_{2})e^{-5x}$

We now find the right candidate for the undermined coefficients method.

$F(x) = (3x+2)e^{-5x}$

$(3x+2)$ is in the form of a first-degree polynomial and $e^{-5x}$ is an exponential function.

So our particular solution is $y_{p} = (Ax + B)e^{-5x}$

My problem is that when I replace $y_{p}$ into the differential equation of line 1, I get 0, so I can't continue the problem...

Thank you.

答:
Homogeneous part:

If we propose $y=e^{\lambda x}$ as a solution, then the characteristic equation is given by $\left(\lambda + 5 \right)^{2}=0$, whose solution is $\lambda=-5$ with multiplicity two. So, the solution to the homogeneous part has the following form
$$ y_{h}\left(x\right)= \left(c_{1} + c_{2} x\right)e^{-5x}. $$

Inhomogeneous part:

The inhomogeneous term should be considered to be a polynomial of degree one by $e^{-5x}$, also $e^{-5x}$ and $xe^{-5x}$ are solutions of the homogeneous part. We propose as a particular solution the next function
$$y_{p}=x^{2}\left( A x + B\right)e^{-5x}, $$
with $A$ and $B$ to be determined. After some calculations, we obtain the following linear system for $A$ and $B$
$$2 B + 6 A x = 2 + 3 x, $$
whose solution is: $A=\displaystyle \frac{1}{2}$ and $B=1$. Therefore, the particular solution is
$$ y_{p}\left(x\right)=\frac{1}{2} e^{-5 x} x^2 (x+2),$$
and the general solution is given by
$$y_{g}\left( x\right)=\frac{1}{2} e^{-5 x} \left(2 (c_{1}+c_{2}x)+x^2(x+2)\right).$$


问:
My paradox in determining the probability of distributing similar balls into different boxes
Inspired by the discussion between @Matthew Pilling and me in this post What is the probability that there more rabbits than chickens in each of these three cages?, I am trying to find and understand the answer to a simple problem I am thinking for a while. (Feel free to read the comments under my answer in that post if you are interested and want to know more.)

Let's say I have two identical balls and I want to distribute them into two different boxes (box1 and box2). What is the probability of having one ball in each box?

Approach 1:
We can talk about the total number of solutions for $n_{1} + n_{2} = 2$, where $n_{1},n_{2} \in W $. And, then, the number of solutions to $n_{1} + n_{2} = 2$, where $n_{1},n_{2} \ge 1$. The probability will be calculated as: $\frac{1}{3}$.

Approach 2:
Or, we can think of this problem as follows: we have two boxes. The probability of choosing the first box is $\frac{1}{2}$. So, there is $\frac{1}{4}$ probability of choosing the first box for both balls. Same reasoning is true for choosing the second box for both balls. So, in total, there is $\frac{1}{2}$ probability to have both balls in only one box. And, thus, the probability of having only one ball in each box is $\frac{1}{2}$.

Could someone please help me understand what's happening here? Are we dealing with two different problems (in these two approaches) or one of the approaches is wrong because ... ?

Thanks for any input you may provide.

Nima

答: The probabilities of each solution in Approach 1 are not equal. That's like saying "When I roll two six-sided die, the probability of rolling a sum of $2$ is $1/11$ because there are $11$ options: $2$ through $12$."


问:
Coin flipping in a row
If we are flipping a coin and the probability of:

Heads: $\frac{1}{3}$

Tails: $\frac{2}{3}$

$W$: 2 results in a row: $T-T$ or $H-H$

If we are flipping until we have two in a row, what is $E[W]$?

答:
Write $E_H(W)$ for the waiting time remaining  after the first  toss was H.
Then $E_H(W)=1+1/3+(2/3)E_T(W)$ and  $E_T(W)=1+2/3+(1/3)E_H(W)$, so $E_H(W)=22/7$  and $E_T(W)=19/7$. Finally $E(W)=1+E_H(W)/3+2E_T(W)/3=27/7.$

Title: Coin flipping in a row