Title: Finding the intervals of increase and decrease and concavity of $y=\frac{\sqrt{1+|x-2|}}{1+|x|}$

Body:
I have the function $f(x)=\frac{\sqrt{1+|x-2|}}{1+|x|}$, which I need to analyze. I have already proven that $\lim\limits_{x \to +-\infty}{f(x)}=0$, therefore the function has only a horizontal asymptote at $y=0$. Then I looked at three cases for the derivative

$x>2 \Longrightarrow f'(x)=\frac{1-3x}{2\sqrt{x-1}(1+x)^2}$

$x \in[0,2]\Longrightarrow f'(x)=\frac{x-7}{x(1+x)^2\sqrt{3-x}}$

$x<0 \Longrightarrow f'(x)=\frac{5-x}{2(1-x)^2\sqrt{3-x}}$

Now I need to look at the intervals of concavity of the function. I suppose I have to differentiate all derivatives for a second time and look at the results in each of the corresponding intervals. But this would require too much calculations. Isn't there a simpler way to find the intervals of increase/decrease and of concavity?

Body:
$$f(x)=\left\{
\begin{array}{ll}
 \frac{\sqrt{3-x}}{1-x} & x<0 \\
 \frac{\sqrt{3-x}}{x+1} & 0 \le x<2 \\
 \frac{\sqrt{x-1}}{x+1} & x\geq 2 \\
\end{array}\right.
$$
$$
f'(x)=
\left \{
\begin{array}{ll}
 \frac{5-x}{2 \sqrt{3-x} (x-1)^2} & x<0 \\
 \frac{x-7}{2 \sqrt{3-x} (x+1)^2} & 0<x<2 \\
 \frac{3-x}{2 \sqrt{x-1} (x+1)^2} & x\ge2 \\
\end{array}\right.
$$
$$
f''(x)=
 \left\{
\begin{array}{ll}
 \frac{-3 x^2+30 x-59}{4 (3-x)^{3/2} (x-1)^3} & x<0 \\
 \frac{3 x^2-42 x+83}{4 (3-x)^{3/2} (x+1)^3} & 0<x<2 \\
 \frac{3 x^2-18 x+11}{4 (x-1)^{3/2} (x+1)^3} & x>2 \\
\end{array}\right.
$$

Plot below, for reference

$$...$$


问:
A functional equation while attempting to integrate a function differently: $ f(x)-f(y) = z \bigl( f(xz) - f(yz) \bigr) $
Here is a functional equation :

For $f : \mathbb (0,\infty) \to \mathbb R$ continuous non-constant, $f$ has the property that for every $b,x,y>0$, $f(1) = -1$  and
$$
f(x) - f(y) = b \bigl( f(bx) - f(by) \bigr) \text .
$$

I want to show that the only possible solutions are $f(x) = a-\frac {a+1}x$ for some constant $a$, without using any form of calculus(with the exception of limits, if required).

For an approach, there are too many free variables, so I tried to set $b$ to a certain value, say $b=2$. Then I get:
$$
f(x)-f(y) =  2 \bigl( f(2x) - f(2y) \bigr) \text .
$$

but I don't know how to proceed : I can't use this to find a particular value of any single $x$, because the value of $x$ depends on the value of $2x$, and so on, so there's no way of proceeding I can think of. This also makes me feel that there is missing information, but that doesn't seem likely because in the presence of three constraining variables and domain, some kind of uniqueness must hold, right?

Setting $g(x) = xf(x)$ and trying to prove that it is constant has also proved futile, for the same reason as before.

Context : This post contains the question I am trying to answer.

Let $f : (0,\infty) \to \mathbb R$ be a function satisfying $f(x)-f(y) = \int_y^x \frac{\mathrm dt}{t^2}$ for all $y,x>0$. Under the fix $f(1) = -1$, we know that there is a unique function $f$ satisfying these conditions, given by $f(x) = -1 + \int_1^x \frac{1}{t^2}\,\mathrm dt = -\frac 1x$. I attempted to locate such an $f$ without using calculus, but rather using functional equations.

The logic was as follows : let $b>0$ be arbitrary, and in the expression $\int_x^y \frac{\mathrm dt}{t^2}$, make the change of variables $bt =u$, so that $b\,\mathrm dt = \mathrm du$. Accordingly, we get:
$$
f(x)-f(y) = \int_x^y \frac{\mathrm dt}{t^2} = \int_{bx}^{by} \frac{b^2 \,\mathrm du}{u^2} = b \int_{bx}^{by} \frac{\mathrm du}{u^2} = b \bigl( f(bx)-f(by) \bigr) \text .
$$

and thus the functional equation comes out. Now if I can use non-calculus techniques to prove that this functional equation is satisfied only by $-\frac 1x$ then I will have obtained the antiderivative of $\frac 1{x^2}$.

答:
In this case, more free variables is better: let's assume $$f(x)-f(y) = z(f(xz)-f(yz))$$ for $x,y,z>0$. Then,
$$f(x)-z\,f(xz)=f(y)-z\,f(yz),$$ meaning that expression depends only on $z$:
$$f(x)-z\,f(xz)=g(z).$$ Letting $x=1$, we see that $g(z)=-1-z\,f(z)$, i.e. we have
$$f(x)-z\,f(xz)=-1-z\,f(z).$$ Exchanging $x$ and $z$, we get
$$f(z)-x\,f(xz)=-1-x\,f(x),$$ multiplying the first equation by $x$ and the second one by $z$ and subtracting, we arrive at
$$x\,f(x)-z\,f(z)=-x+z+xz(f(x)-f(z)).$$ This equation has to hold for any $x,z>0$, substituting any constant value $z_0\neq1$ for $z$, we see that $x\,f(x)=ax+b$. Since $f(1)=-1$, this means $b=-a-1$, i.e. $x\,f(x)=ax-a-1$.

ContentLicense: CC BY-SA 4.0


问:
Prove $-a\cdot e\cdot \ln(x)\le x^{-a}$
Today during a Calc I exam we(me and my classmates) have been ask to prove $$-a\cdot e\cdot \ln(x)\le x^{-a}$$, $\forall x>0$ and $\forall a\in \mathbb{R}$. But, noneone in the room has known how to do it. Anyone knows how to?

Thanks in advance

答:
HINT

Try to find the maximum of $f(t):=\frac{\ln(t)}{t}$ and recall that your inequality is equivalent to
$$ \frac{\ln(x^{-a})}{x^{-a}}\le\frac{1}{e}.$$

Title: Prove $-a\cdot e\cdot \ln(x)\le x^{-a}$


问:
Comparing between regular polygons
In this experiment, I am adding the inradius (let's call it $A$) and circumradius (let's call it $B$) of different polygons with equal sides each equal $1$ (starting with a square and adding one side each time). The result is $A+B=C$ when side of polygon = 1.

When comparing the $C$ of one polygon with the $C$ of a polygon with one side more, the difference seems to go smaller, as if approaching a version of $\pi$ number with $0.$ before (possibly such as 0.314159265359...).

Can anyone confirm it or elaborate on it?

I can not go over a polygon with 1000 sides in my computation power, and would like to know what to expect while going towards a polygon with infinity sides.

Here are some examples:

4 sided polygon: $0.5 + 0.707106781 = 1.207106781$

5 sided polygon: $0.68819096 + 0.850650808 = 1.5388417680000002$ (Difference of 0.33173498700000015 from previous result)

6 sided polygon: $0.866025404 +1 = 1.866025404$$ (Difference of 0.3271836359999998 from previous result)

7 sided polygon: $1.0382607 + 1.15238244 = 2.1906431399999997$ (Difference of 0.32461773599999977 from previous result)

8 sided polygon: $1.20710678 + 1.30656296 = 2.51366974$ (Difference of 0.3230266000000004 from previous result)

9 sided polygon: $1.37373871 + 1.4619022 = 2.83564091$ (Difference of 0.32197116999999986 from previous result)

10 sided polygon: $1.53884177 + 1.61803399 = 3.15687576$ (Difference of 0.3212348500000002 from previous result)

11 sided polygon: $1.70284362+ 1.77473277 = 3.47757639$ (Difference of 0.3207006299999997 from previous result)

12 sided polygon: $1.8660254+ 1.93185165 = 3.79787705$ (Difference of 0.32030066 from previous result)

13 sided polygon: $2.02857974+ 2.08929073 = 4.11787047$ (Difference of 0.31999341999999986 from previous result)

14 sided polygon: $2.19064313 + 2.2469796 = 4.43762273$ (Difference of 0.31975226000000045 from previous result)

15 sided polygon: $2.35231505+ 2.40486717 = 4.757182220000001$ (Difference of 0.3195594899999996 from previous result)

...

999 sided polygon: $158.995264 + 158.99605 = 317.991314$

1000 sided polygon: $159.154419 + 159.155205 = 318.309624$ (Difference of 0.31830999999999676 from previous result)

答:
Consider that the apothem $a$, circumcircle radius $c$, and edge of the polygon form a right triangle.

That is, we have two legs, one of length $1/2$, the other of length $a$, and a hypotenuse of length $c$. Then we have:

$$a = \frac{\tan\left(\frac{(n-2)\pi}{2n}\right)}{2}$$

$$c = \frac{1}{2}\csc\left(\frac{\pi}{n}\right)$$

Since the angle $\angle ac$ is always $\pi/n$.

We may simplify $a + c = \frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$. Then what you seek to compute is:

$$\lim_{n \rightarrow \infty} \frac{1}{2}\left(\cot\left(\frac{\pi}{2n}\right) - \cot\left(\frac{\pi}{2n-2}\right)\right)$$

This in fact converges to $1/\pi$, which is $\approx 0.318$.

Title: Comparing between regular polygons


问:
If $G$ is simple and finite, and H subgroup then $|G|$ divides $[G:H]!/2$
I'm having trouble to prove that if $G$ is a finite simple group and $H$ is a subgroup of finite index, then $|G|$ divides $\frac{k!}{2}$, where $k$ is the index of H.

I've prove that in this conditions $G$ is isomorphic to a subgroup $F$ of the symmetric group of $k$, $S_k$. As we are trying to prove that $|G| = |F|$ divides $\frac{k!}{2}$ I think that I need to show that $F \leq A_k$ where $A_k$ is the alternating group, but I don't get to see clearly why this is true.

Also, I saw a similar question here but don't completely understand the answer. I would appreciate any help.

答: Hint:  Assume $|G|\gt2$ and $k\gt1$.  Let $G$ act by left multiplication on the set of cosets $G/H$.  Then we get a morphism $\varphi:G\to S_k$.  Furthermore, $\varphi(G)\cong G$, since the kernel has to be trivial.  Now $A_k\cap\varphi(G)$ is normal in $\varphi(G)$.  From there it takes a little more work to get that $\varphi(G)\leqslant A_k$.


问:
Find the integer solutions to $4^x - 9^y = 55$
I want to find the integer solutions of:

$$ 4^x - 9^y = 55$$

For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?

I tried to solve for $y$, but to no avail. WolframAlpha tells me that it is the only solution, but it doesn't provide an explanation.

答:
I like the explanations given in @MishaLavrov's answer, and I understand this only as a comment to his answer.

There, the second sentence introduces a bit obfuscation, I think: the bare idea of needing a crosscheck of $\{0,1,2,3,4,5\} \times
\{0,1,2,3\}$ insertions is somehow surprising (and misleading) after the introductory statement.

$$4^a - 9^b = 55 $$

This can be factored, and the factors of the lhs can be crosschecked with the two factorizings of the rhs, where the obvious sizes of the factors in the lhs are brought in agreement with the order of sizes of the factors of the rhs:
$$
  (2^a+3^b)(2^a-3^b)= \begin{cases}  (11) \cdot (5) & (a)\\  (55) \cdot (1) &(b) \end{cases} $$
Test $(a)$:

$$ (2^a+3^b)-(2^a-3^b)=2\cdot 3^b \overset?= 11-5 = 6 = 2\cdot 3^1 \implies b=1\\
   (2^a+3^b)+(2^a-3^b)=2\cdot 2^a\overset?= 11+5 = 16 = 2 \cdot 2^3 \implies a=3$$

$\qquad \qquad$ both ways of summing the corresponding factors give valide solutions.

Test $(b)$:
$$    (2^a+3^b)-(2^a-3^b)=2\cdot 3^b\overset?= 55-1 = 54 = 2\cdot 3^3 \implies b=3\\
   (2^a+3^b)+(2^a-3^b)=2\cdot 2^a\overset?= 55+1 = 56 = 2 \cdot 2^2 \cdot 7 \implies \text{error}
$$
$\qquad \qquad$ the second way of summing the corresponding factors does not agree with the problem.

Conclusion: We have only one solution: $4^3 - 9^1 = 55$


问:
What to do with an expression of the form: $\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$?
We are looking at the following expression:

$$\frac{d}{dx}\int_{u(x)}^{v(x)}f(x) dx$$

The solution is straightforward for this: $\frac{d}{dx}\int_{u(x)}^{v(x)}f(t) dt$. Do we evaluate the given expression in like manner? Do we treat the $f(x) dx$ as if it were $f(t) dt$?

答:
Be aware of the name of variables !

The $x$ in $v(x)$ is not the same as $dx$ lets call it $dt$

Consider your integral as a primitive

$$I(x)=F(v(x))-F(u(x))$$ with $F$ a primitive of   $f$. From here it is easy to use chain rule and conclude for an expression.


问:
Methods of function expansion
So far, I've learned about two kinds of orthogonal bases/kernels for expanding functions (on finite intervals):

(generalized) Fourier series, i. e. orthogonal sets of trigonometric functions.

Legendre polynomials.

So I wanted to ask if I'm missing any other common orthogonal base systems? (I'm only dealing with finite intervals)

I've also noticed that when using cylindrical coordinates $r, \theta$ for disk-like domains, the kernels of the Fourier series contain Bessel functions (c. f. https://en.wikipedia.org/wiki/Fourier%E2%80%93Bessel_series). Are these necessary? Or could I just use the same trigonometric/Legendre kernels as for Cartesian coordinates? By my understanding, a complete orthogonal basis should stay complete and orthogonal if the variable has a different name.

答: Sturm-Liouville theory provides many examples of complete orthogonal sets of functions. Many of the familiar orthogonal bases arise in this way, including Bessel functions and Legendre polynomials. There are many more well-known examples including the Hermite, Laguerre, Jacobi and Chebyshev polynomials.</p>

Title: Methods of function expansion


问:
Prove or disprove: If $v_1,v_2,...,v_n$ are vectors in $\mathbb{R}^n$...
and for every $v_i$ , $i=1,..,n$ the sum of coordinates of $v_i$ is $0$, then $v_1,...,v_n$ and linearly dependent.

What I tried to do: This questions was given after I learnt about matrix similarity and eigen vectors and eigen values, so I tried to think in that direction, I couldn't really get the question saying "the sum of coordinates of $v_i$ is $0$), because I always saw coordinates come with basis so I can use them? (would be appreciated if someone explains in what way should I understand that sentence).

But I proceed to think in that direction and reached something: I chose matrix $A$ to be all ones, ($a_{ij}=1\space \forall i,j) $, and let $T(v)=Av$ a linear transformation, and here I got stuck but my idea was to try to say by contradiction that they're linearly  independent and reach contradiction if I find that $A$, is similiar to the zero matrix (if I use the sum of coordinates), and then explain that they can't be similar with different ranks, and prove the statement. But  I have no idea how to continue and my weak understanding of the question scares me if I'm in the right direction so I came here to ask.

Appreciate all the help, Thanks in advance to everyone.

答:
When talking about the Euclidean space $\mathbb{R}^n$, the default basis is the standard basis.

Let $A$ be a matrix where the column vectors are $v_1,\cdots,v_n$. Then if you consider the row vector
$$
w=(\underbrace{1,\cdots,1}_{\textrm{$n$ terms}})
$$
it follows from the matrix multiplication that $wA=0$, which implies that $A$ is not of full rank and hence the column vectors are linearly dependent.

Notes.

If you write
$$
v_k=(a_{1k},\cdots,a_{nk})^T
$$
then the matrix $A$ is

\begin{pmatrix}
a_{11}&\cdots& a_{1n}\\
\vdots&\vdots&\vdots\\
a_{n1}&\cdots &a_{nn}
\end{pmatrix}

Note that the $k$-th column of $A$ is the vector $v_k$.


问:
Logical transforms of theorems
I'm trying to transform some theorems to different logical forms but having difficultly for some reason, it should be quite simple but my brain doesn't seem to be functioning lately.

I want to write an implication in terms of various forms to see the different versions of the theorem(s).

In fact I want to transform a definition: We can use the definition of a cts. function at a point as an example:

A function $f:X\rightarrow Y$ is called continuous at $a\in X$ if $\forall N \in N_{f(a)}$ then $f^{-1}(N) \in N_{a}$.

(Here $N_x$ means the set of neighborhoods of x, I can't get mathbb or others to change the font).

The idea is to change implication to the various other forms. The logic is simply $A\rightarrow B$. (So $A = \forall N \in N_{f(a)}$ and $B = f^{-1}(N) \in N_{a}$)

E.g., the contrapositive form $!B\rightarrow !A$ doesn't make any sense because N is defined afterwards:

A function $f:X\rightarrow Y$ is called continuous at $a\in X$ if $f^{-1}(N) \notin N_{a}$ then $\exists N \notin N_{f(a)}$.

of course we can translate/shift this so that we end up with something like(I moved the $\exists$ to try to make the implication work but...)

A function $f:X\rightarrow Y$ is called continuous at $a\in X$ if $\exists N \notin N_{a}$ then $f(N) \notin N_{f(a)}$.

But this too doesn't make sense and doesn't even logically follow from transformations.

I'm simply trying to turn these logical statements in to logical formula to manipulate but something is throwing me off. There are many ways to write an implication and I would like to figure out how to do so and retain the equivalent definition or theorem.

Any ideas?

答:
You haven't actually expresesd continuity correctly: "$\forall N\in N_{f(a)}$" is not a statement, so it can't be the hypothesis of an implication. The right version is $$\color{red}{\forall N(}N\in N_{f(a)}\rightarrow f^{-1}(N)\in N_a\color{red}{)},$$ that is, the entire implication occurs inside the scope of the universal quantifier. The hypothesis and conclusion of that implication are then "$N\in N_{f(a)}$" and "$f^{-1}(N)\in N_a$," respectively.

This is why you're running into trouble with the contrapositive. Applying the contrapositive to the correct version of the definition we get the equivalent property $$\color{red}{\forall N(}f^{-1}(N)\not\in N_a\rightarrow N\not\in N_{f(a)}\color{red}{)},$$ which is perfectly well-formed. Note that we're not touching the outermost "$\forall N$" here; the "contrapositive translation" only applies to the implication itself, not the quantifier structure it sits inside.

(That said, if you want there is a transformation we can apply to that quantifier, namely the duality of $\forall$ and $\exists$: a statement of the form "$\forall xP(x)$" is equivalent to "$\neg\exists x\neg P(x)$." But I don't see how that would be helpful in this context.)

Title: Logical transforms of theorems


问:
Calculate $\mathbb E(5 - 3X) - D(2 - X)$ with $Uniform$ ~ $[-3, 3]$
I have continuous uniform distribution $Uniform$ ~ $[-3, 3]$. And I want to calculate $\mathbb E(5 - 3X) - D(2 - X)$. But I am getting conflicting results:

First solution
$$\mathbb E(5 - 3X) - D(2 - X) = \mathbb E(5 - 3X) - \mathbb E((2-X)^2) + \mathbb E^2(2-X) = \mathbb E(5) - \mathbb E(3X) - [\mathbb E(4) - \mathbb E(4X) + \mathbb E(X^2)] + [\mathbb E(2) - \mathbb E(X)]^2 = 5 - 3\mathbb E(X) - 4 + 4\mathbb E(X) - \mathbb E(X^2) + 4 - 4\mathbb E(X) + \mathbb E^2(X) = 5 + \mathbb E(X) = 5$$

Second solution
$$\mathbb E(5 - 3X) - D(2 - X) = \mathbb E(5) - \mathbb E(3X) - D(X) = 5 - 3 \cdot0 - 3 = 2$$

I used the linearity of the expectation and the fact that $\mathbb E(X) = \frac{-3 + 3}{2} = 0$.

I can't understand what the error is.

答:
The second calculation is correct.  The first is incorrect.  It should be instead
$$\begin{align}
\operatorname{E}[5-3X] - \operatorname{Var}[2-X]
&= \operatorname{E}[5 - 3X] - \operatorname{E}[(2-X)^2] + \operatorname{E}^2[2-X] \\
&= \operatorname{E}[5] - \operatorname{E}[3X] - \operatorname{E}[4 - 4X + X^2] + (\operatorname{E}[2] - \operatorname{E}[X])^2 \\
&= 5 - 3 \operatorname{E}[X] - 4 + 4\operatorname{E}[X] - \operatorname{E}[X^2] + (2 - 0)^2 \\
&= 5 - 4 - \operatorname{E}[X^2] + 4 \\
&= 5 - \operatorname{E}[X^2].
\end{align}$$

But what is $\operatorname{E}[X^2]$?  It is certainly not $0$.  It is $$\operatorname{E}[X^2] = \int_{x=-3}^3 x^2 \cdot \frac{1}{6} \, dx = \left[\frac{x^3}{18}\right]_{x=-3}^3 = \frac{2(27)}{18} = 3.$$  Thus the result is $5 - 3 = 2$, same as with the second method.


答:
Let $C_n$ be the value of $C$ for a regular $n$-gon. I claim that $C_n/n=1/\pi+O(n^{-2})$ as $n\to\infty$.

Consider the triangle formed by two adjacent vertices and the centre of the polygon. This is an isosceles triangle with apex angle $2\pi/n$.

The inradius is $\frac{1}{2\tan(\pi/n)}$ and the circumradius is $\frac{1}{2\sin(\pi/n)}$. As $h\to0$, we have the results $\tan h=h+O(h^3)$ and $\sin h=h+O(h^3)$. So as $n\to\infty$,
$$n\tan(\pi/n)=\pi+O(n^{-2})\quad\text{and}\quad n\sin(\pi/n)=\pi+O(n^{-2}).$$
So we have
$$\frac{C_n}{n}=\frac{1}{2n\tan(\pi/n)}+\frac{1}{2n\sin(\pi/n)}=\frac{1}{2\pi}+\frac{1}{2\pi}+O(n^{-2})=\frac{1}{\pi}+O(n^{-2})$$
as claimed. Now
$$C_n-C_{n-1}=\frac{n}{\pi}-\frac{n-1}{\pi}+O(n^{-1})=\frac{1}{\pi}+O(n^{-1}).$$
So we get convergence to $\frac{1}{\pi}$.

Title: Comparing between regular polygons

Body:
In this experiment, I am adding the inradius (let's call it $A$) and circumradius (let's call it $B$) of different polygons with equal sides each equal $1$ (starting with a square and adding one side each time). The result is $A+B=C$ when side of polygon = 1.

When comparing the $C$ of one polygon with the $C$ of a polygon with one side more, the difference seems to go smaller, as if approaching a version of $\pi$ number with $0.$ before (possibly such as 0.314159265359...).

Can anyone confirm it or elaborate on it?

I can not go over a polygon with 1000 sides in my computation power, and would like to know what to expect while going towards a polygon with infinity sides.

Here are some examples:

4 sided polygon: $0.5 + 0.707106781 = 1.207106781$

5 sided polygon: $0.68819096 + 0.850650808 = 1.5388417680000002$ (Difference of 0.33173498700000015 from previous result)

6 sided polygon: $0.866025404 +1 = 1.866025404$$ (Difference of 0.3271836359999998 from previous result)

7 sided polygon: $1.0382607 + 1.15238244 = 2.1906431399999997$ (Difference of 0.32461773599999977 from previous result)

8 sided polygon: $1.20710678 + 1.30656296 = 2.51366974$ (Difference of 0.3230266000000004 from previous result)

9 sided polygon: $1.37373871 + 1.4619022 = 2.83564091$ (Difference of 0.32197116999999986 from previous result)

10 sided polygon: $1.53884177 + 1.61803399 = 3.15687576$ (Difference of 0.3212348500000002 from previous result)

11 sided polygon: $1.70284362+ 1.77473277 = 3.47757639$ (Difference of 0.3207006299999997 from previous result)

12 sided polygon: $1.8660254+ 1.93185165 = 3.79787705$ (Difference of 0.32030066 from previous result)

13 sided polygon: $2.02857974+ 2.08929073 = 4.11787047$ (Difference of 0.31999341999999986 from previous result)

14 sided polygon: $2.19064313 + 2.2469796 = 4.43762273$ (Difference of 0.31975226000000045 from previous result)

15 sided polygon: $2.35231505+ 2.40486717 = 4.757182220000001$ (Difference of 0.3195594899999996 from previous result)

...

999 sided polygon: $158.995264 + 158.99605 = 317.991314$

1000 sided polygon: $159.154419 + 159.155205 = 318.309624$ (Difference of 0.31830999999999676 from previous result)


问:
Define a unique linear map
The question in the textbook is the following: "Given the results of a linear map, is this linear map unique? If so, give the function rule".

I'm given $L_1 : \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Also the following results are given: $L_1(1, 2) = (0, -1)$ and $L_1(-1, 1) = (2, 1)$.

I have no idea how to even start the question. To find the correct function rule, I suppose I'll have to solve some kind of system of equations, but I'm not sure on how to start.

Title: Define a unique linear map

Body:
Let $$L_1=\left(
\begin{array}{cc}
 a & b \\
 c & d \\
\end{array}
\right)$$
$L_1(1,2)=(0,-1)$ means that $(a+2 b,c+2 d)=(0,-1)$

$L_1(-1,1)=(2,1)\to (b-a,d-c)=(2,1)$

So we have the linear system
$$
\begin{cases}
a+2b=0\\
c+2d=-1\\
b-a=2\\
d-c=1\\
\end{cases}
$$
which gives
$$a= -\frac{4}{3},b= \frac{2}{3},c= -1,d= 0$$
and finally
$$L_1=\left(
\begin{array}{cc}
 -\frac{4}{3} & \frac{2}{3} \\
 -1 & 0 \\
\end{array}
\right)$$


问:
Proving that $\mathbb{Q}(\omega)$ is the splitting field for $f(x)=x^4 - x^2 + 1$ over $\mathbb{Q}$
I have been given that $\omega = e^{\pi i\over{6}}$, a 12th root of unity. I have shown that $\omega$ is a root of the polynomial $f(t)=t^4 -t^2 +1$, as are $\omega^5,\omega^7,\omega^{11}$ and that $f$ is the minimal polynomial of $\omega$ over $\mathbb{Q}$. Now I must show that $$\Gamma(\mathbb{Q(\omega):Q})$$ has order 4 (and subsequently that it is isomorphic to the Klein 4-group). My immediate thought is to show that $\mathbb{Q(\omega)}$ is the splitting field of $f$ over $\mathbb{Q}$ and hence $\mathbb{Q(\omega):Q}$ is a normal extension then use the fundamental theorem of Galois Theory, but I am unsure how to prove this. Any advice is welcome.

答: As $\mathbb{Q}(\omega)$ is a field, i.e. closed under multiplication, then  $\omega^5,\omega^7,\omega^{11}\in\mathbb{Q}(\omega)$ and hence the splitting field of $f$: $$\mathbb{Q}(\omega,\omega^5,\omega^7,\omega^{11})=\mathbb{Q}(\omega)$$


答:
Hint.

In general, for a map $L:V\to W$, in order to know $L$, you need to know $L(v)$ for every single element $v$ in $V$.

But if $L$ is a linear map where $V$ is a vector space, then it suffices to know $L(v)$ for some special elements, i.e., on a basis of $V$. Once $L$ is known on a basis, it is uniquely determined.

Once $L(v)$ for every element $v$ in a basis, then it is uniquely determined.

So for your question, it suffices to see whether $(1,2)$ and $(-1,1)$ form a basis of $\mathbb{R}^2$.

Title: Define a unique linear map

Body:
The question in the textbook is the following: "Given the results of a linear map, is this linear map unique? If so, give the function rule".

I'm given $L_1 : \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Also the following results are given: $L_1(1, 2) = (0, -1)$ and $L_1(-1, 1) = (2, 1)$.

I have no idea how to even start the question. To find the correct function rule, I suppose I'll have to solve some kind of system of equations, but I'm not sure on how to start.


答:
Let the sum of the inradius and circumradius of a regular polygon with $n$ sides be $C_n$. As $n\rightarrow \infty $, the regular polygon resembles a circle with circumference $\pi  C_n$. The actual circumference is $n$ for an $n-$sided polygon with side length 1. Thus $n \rightarrow \pi C_n$ as n gets bigger. This implies $$C_n \rightarrow \dfrac{1}\pi(n)$$
as $n$ gets bigger. Consequently, $C_{n+1} -C_{n} \rightarrow 1/\pi = 0.318309\dots$ as $n$ tend towards infinity.

Title: Comparing between regular polygons


问:
Understanding the second condition in Fritz John's Theorem

Fritz John's Theorem: Each convex body $K$ contains a unique ellipsoid of maximal volume. This ellipsoid is $B^n_2$ iff: $B^n_2 \subset K$ and (for some $m$), there are Euclidean unit vectors $(u_i)_1^m$ on the boundary of $K$ and positive numbers $(c_i)_1^m$ satisfying
\begin{equation}
    \sum_{i=1}^m c_i u_i = 0
\end{equation}
and
\begin{equation}
    \sum_{i=1}^m c_i \langle x,u_i\rangle^2 = \|x\|^2 \text{ for each }x\in\mathbb{R}^n
\end{equation}

This is an excerpt from Keith Ball's notes. He goes on to say that the second condition (involving $\|x\|^2$) is equivalent to $$x = \sum c_i\langle x,u_i\rangle u_i$$ for all $x\in\mathbb{R}^n$.
This is easily seen so I've skipped the proof.

"...The $(u_i)$ behave rather like an orthonormal basis in that we can resolve the Euclidean norm as a weighted sum of squares of inner products..."

I think this is because for two vectors $x,y$ we can write $$\langle x,y\rangle = \sum_i c_i\langle x,u_i\rangle \langle y,u_i\rangle$$

"This guarantees that the $(u_i)$ do not all lie close to a proper subspace of $\mathbb{R}^n$. If they did, we could shrink the ball a little in this subspace and expand it in an orthogonal direction, to obtain a larger ellipsoid inside $K$."

What does it mean to lie close to a proper subspace of $\mathbb{R}^n$?

Thanks for reading, I would appreciate any help! I hope to better understand this theorem and its consequences.

Edit:

Later in the notes, the author says to interpret the second condition as a rigidity condition:

One may get a bit more of a feel for the second condition in John’s Theorem by interpreting it as a rigidity condition. A sequence of unit vectors $(u_i)$ satisfying the condition (for some sequence $(c_i)$ has the property that if $T$ is a linear map of determinant $1$, not all the images $Tu_i$ can have Euclidean norm less than $1$.

How do we prove this? I am unable to come up with a proof by contradiction (as follows).
Assume $\|Tu_i\| < 1$ for all $1\le i\le m$. Now I'm trying to put $x = u_i$ in the second condition but haven't gotten anything fruitful yet. Thanks!

Update:
I tried putting $x = Tu_j$ in condition $2$ and got something redundant. If you're interested, you may read ahead:
$$\|Tu_j\|^2 = \sum_i c_i \langle Tu_j,u_i\rangle^2 \le \sum_i c_i \|Tu_j\|^2 \implies \|Tu_j\|^2 (1 - \sum_i c_i) \le 0$$ which is nothing new since we already have $\sum_i c_i = n$. Nothing special about $Tu_j$ here, we would have gotten this from any $x$.

Update 2:
I'm trying to use Hadamard's inequality but haven't gotten anywhere yet. The bounty is about to end and I hope someone gives it a shot?

答:
Write $\|x\|^2=\sum_i c_i\langle x,u_i\rangle^2$ in matrix form as
$$
x'x=\sum_i c_i x'u_iu'_ix,
$$
where the prime indicates transposition. Set $x'=t'_j$, the $j^\text{th}$ row of $T$. So
$$
t_j't_j=\sum_i c_i t'_ju_iu'_it_j.
$$
Summing $t'_ju_iu'_it_j$ over $j$ gives $\|Tu_i\|^2$. Therefore
$$
\sum_j\|t_j\|^2=\sum_ic_i\|Tu_i\|^2.
$$
Now $\sum_j\|t_j\|^2\ge n$ when $\det T=1$, while $\sum_i c_i=n$ and all $c_i$ are positive. This gives a contradiction if all $\|Tu_i\|$ are less than $1$. Note that the sum over $\|t_j\|^2$ is the square of the Hilbert–Schmidt or Frobenius norm of $T$.

We should explain where the properties just invoked come from. That $\sum_j\|t_j\|^2\ge n$ for $\det T=1$ should be intuitively clear: $\det T$ is the volume of an $n$-dimensional parallelepiped. For given side lengths of the parallelepiped, the volume is maximized when all sides are orthogonal, while the sum of the squares of the side lengths of an $n$-dimensional rectangular prism of fixed volume is minimized when all sides are equal (as can be shown, for example, by the method of Lagrange multipliers). So for unit volume, the parallelepiped that minimizes the sum of the squares of the side lengths is the unit cube, for which $\sum_j\|t_j\|^2=n$.

That the $c_i$ are positive is one of the given conditions. The condition $\sum_i c_i=n$ can be obtained by letting $x$ in the second condition equal each of the standard basis vectors of $\mathbf{R}^n$ in turn, and summing:
$$
n=\sum_j e'_j e_j=\sum_j\sum_ic_ie'_ju_iu'_ie_j=\sum_ic_i\|u_i\|^2=\sum_ic_i.
$$

I'm not sure, unfortunately, that I can say much about your original question (about the meaning of "close to a proper subspace"). I do think it's worth working out a variety of detailed examples, even in dimension $2$. For a rhombus, which is what Figure 14 in Ball's article shows, the condition fails, unless the rhombus is a square. The same is true, I believe, for rectangles, and parallelograms more generally. You could, for example, let the four points in Figure 14 where the inscribed circle touches the rhombus be $\left(\pm\frac{5}{13},\pm\frac{12}{13}\right)$. Then pick some random points for $x$—the points $(1,0)$ and $(0,1)$ will do— and try to solve for the $c_i$. You will gain an appreciation for why there is no solution.

So if parallelograms don't work (except for squares), try hexagons. You can cut off the two corners in Figure 14 on the horizontal axis by adding vertical line segments tangent to the circle at $(-1,0)$ and $(1,0)$ to form a hexagon. Now solve the modified system of equations. What allows this solution in this case is the addition of the two vectors $(-1,0)$ and $(1,0)$, which, unlike the other four vectors, are not close to the subspace along the vertical axis. Now if the two vertical sides were moved outward by a little bit, so that the circle was no longer tangent to those sides, we would be back in a situation similar to the rhombus.


问:
Question: are quantum computers a type of Turing machine or something else?
I read that in quantum computations you can not examine the internal state of the computation while it is happening. This is something I thought is always possible with Turing machines. Does this mean that quantum computers are not Turing machines? Also as an extra question: If not then what languages do they recognize in the sense of the hierarchy of language grammars?</p>

答:
Quantum computers are a distinct computation model from Turing machines. However, they have the same "ultimate" power: the languages accepted/recognized by the two coincide (respectively: computable and computably enumerable). This is basically Church's thesis, that no "actual" model of computation can be stronger than Turing machines. (Technically this isn't exactly what CT says, although in my opinion it's ultimately equivalent. The point stands, though.)

Note that this is a very coarse equivalence: for example, each model has a notion of "computation time," and if I look at what languages are accepted/recognized in some bounded amount of time the two models may behave very differently. In general the complexity theories (as opposed to computability theories) of the two models can be extremely different. But that's a separate issue.

Meanwhile, the claim of "unexaminability" of internal states of a quantum computation is an informal gloss on the nature of quantum computation and shouldn't be taken too literally. I'm not an expert so I don't want to get into the details here, but there's nothing magical about the definition of an (abstract) quantum computer.


问:
No differentiable function on $[-1,1]$ with derivative $\operatorname{sgn}(x)$
How can I prove for certain that there exists no differentiable function $g(x)\colon [-1,1]\rightarrow \mathbb{R}$ with

$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1,  & \text{for $x > 0$} \\
0, & \text{for $x=0$} \\ -1, & \text{for $x<0$} \end{cases} $$

I'm currently jumping to the absolute value function but that's not differentiable at $x = 0$ and that doesn't exactly disprove the existence of others.

Looking at the limits
\begin{align}
\lim_{h\to 0^{+}} g'(h) &=1\\
&= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\
&\neq  \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\
&= -1\\
&= \lim_{h\to 0^{-}}g'(h)\\
&\neq 0\\
&= g'(0)
\end{align}
which implies discontinuity at $h = 0$.

Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?

答:
Assumed a g(x) exists.

1)Solution for $x>0:$ $g(x)=x +a;$

2)Solution for $x <0:$ $g(x)=-x+b;$

Since differentiability implies continuity: $a=b=0;$

$3)\lim_{x \rightarrow \pm 0} \dfrac{g(x)-0}{x}$ does not exist,

hence not differentiable at $0$.


问:
Differentiable Operator Continuous?
Consider the space $C^{\infty}[a,b]$ with norm $||f|| = max_{[0,1]} |f(x)|$, with $f\in C^{\infty}[a,b]$. Is the differentiation operator $\frac{d}{dx}$ continuous on $C^{\infty}[a,b]$?

I'm very confused because it seems almost trivial-- there are plenty of examples of the derivative not being a continuous operator-- that's why it's so hard to study.. but in this space, isn't it defined to be only those functions whose derivatives are infinitely differentiable? How do I show this formally?

答:
Probably you don't want just the sup norm on smooth functions. Rather, topologize smooth functions by the family of seminorms given by sups of all derivatives. (This is a Frechet space, the projective limit topology on the Banach spaces $C^k[a,b]$ with sum of sups of derivatives up to order $k$.) Differentiation is continuous in that topology. (And differentiation is continuous from $C^k[a,b]$ to $C^{k-1}[a,b]$.)

A reason to understand that just the sup norm of values is not the right topology on smooth functions is that the space is not complete in that topology.

ContentLicense: CC BY-SA 3.0

Title: Differentiable Operator Continuous?


问:
Integrate $\int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx$
Need help integrating:
$$\int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx$$

I think I should multiply the $$xe^{-\frac{2x}{\theta }}$$ out and then use integration by parts but it is not really working for me?

答:
Hint.

Abstractly, all you need is to find integrals of the form:
$$
\int xe^{mx}dx
$$
which by a change of variables, you only need to find
$$
\int xe^{mx}dx
=\frac{1}{m^2}\int ue^udu=
\frac{1}{m^2}\int xe^{x}dx=\frac{1}{m^2}(xe^x-e^x)+C
$$

The integrand in your integral is
$$
xe^{\frac{-2}{\theta}\cdot x}-xe^{\frac{-3}{\theta}\cdot x}
$$
each term is of the form $xe^{mx}$ for some $m$.


问:
Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$?
Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$

Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?

I start with $AM-GM$

So $A \geq 8xyz$

Then I need to prove $xyz\geq 1$ right?

How to do that and if it's a wrong way, please help me to solve it !!!

答:
This answer uses a very similar approach to that of egreg, but it may be easier to follow for readers who are comfortable with single-variable differentiation but who aren't yet familiar with partial derivatives.

But first I'll show how we can use finite differences to verify that $(\frac12, \frac12, 2)$ gives a stationary value for $A=\frac{125}{16}$.

Recall that for small $\Delta x$,
$$f(x+\Delta x) \approx f(x) + f'(x)\Delta x$$
At a stationary point (a minimum, maximum, or point of inflexion), $f'(x)=0$.

We are given
$$A = (x^2+1)(y^2+1)(z^2+1)$$
with the constraint
$$x+y+z=3$$
So differentiating,
$$\Delta x +\Delta y +\Delta z = 0$$
For $f(x)=x^2+1, f'(x)=2x$, so
$$A+\Delta A = (x^2+1 + 2x\Delta x)(y^2+1 + 2y\Delta y)(z^2+1 + 2z\Delta z)$$
We want to show that $\Delta A=0$.

Note that when doing arithmetic with these deltas that terms like $\Delta x\Delta y$ and $(\Delta x)^2$ vanish.

Plugging in $(\frac12, \frac12, 2)$,
$$\begin{align}
A+\Delta A = & (\frac54 +\Delta x)(\frac54 +\Delta y)(5 +4\Delta z)\\
= & (5 +4\Delta x)(5 +4\Delta y)(5 + 4\Delta z)/16\\
= & (125 + 100(\Delta x+ \Delta y +\Delta z))/16\\
= & 125/16 = A\\
\end{align}$$
Thus $\Delta A$ is zero, and $A$ is indeed stationary at $(\frac12, \frac12, 2)$.

If we plug in $(1, 1, 1)$, we get
$$8+\Delta A = 8 + 2(\Delta x+ \Delta y +\Delta z)$$
i.e., $\Delta A = 0$, so it's also a stationary point. In fact, it's a local maximum.

Now I'll show how to find these points. Let's temporarily treat one of the variables, eg $z$, as if it were a constant value. Geometrically, we want to know how the A function behaves on the $z=k$ plane as $x$ and $y$ vary. So we look at the function
$$B=(x^2+1)(y^2+1)$$
with $A=B(k^2+1)$.
If $A$ is stationary at $(x, y, z=k)$ then $B$ must also be stationary at $(x, y)$.

At $z=k$, we have
$$y=(3-k)-x$$
so $y$ and hence $B$ are simple functions of $x$, and we can use plain single-variable calculus on them. Firstly,
$$\frac{dy}{dx}=-1$$
Using the product and chain rules,
$$\frac{dB}{dx} = 2x(y^2+1) + 2y(x^2+1)\frac{dy}{dx}$$
$$\frac{dB}{dx} = 2x(y^2+1) - 2y(x^2+1)$$
At a stationary point, $$\frac{dB}{dx}=0$$
so
$$x(y^2+1) = y(x^2+1)$$
and for $x,y \ne 0$,
$$(y+1/y) = (x+1/x)$$
This is a quadratic, with solutions $x=y$ and $x=1/y$. We can see that they're the only solutions by factoring:

$$x(y^2+1) = y(x^2+1)$$
$$xy^2+x - yx^2-y = 0$$
$$xy(y-x) + x-y = 0$$
$$(xy-1)(y-x) = 0$$

So we've established that $B$ is stationary when $x=y$ or $xy=1$. That's true for any given $z=k$, regardless of whether A is also stationary at that point.

By symmetry, we can apply the same argument to a fixed $x$ or $y$ plane, and so we get a stationary point on $A$ when a point $(x,y,z)$ is stationary on all 3 planes simultaneously.

So now we just need to combine each of those solutions:

$x=y$ or $xy=1$,

$x=z$ or $xz=1$,

$y=z$ or $yz=1$

with $x+y+z=3$ to find the stationary points of A. We get either
$$x=y=z=1$$
or
$xy=1$ and $y=z$, and its symmetrical variations.
$$x+y+z=3$$
$$xy+2y^2=3y$$
$$2y^2-3y+1=0$$
$$(y-1)(2y-1)=0$$
So either
$$y=z=1,x=1$$
or
$$y=z=\frac12, x=2$$
and the symmetrical variations thereof.
Thus we end up with these four stationary points:

$(1,1,1),$

$(2,\frac12,\frac12),$

$(\frac12,2,\frac12),$

$(\frac12,\frac12,2)$

It would be nice to look at the graph of this function. Of course, that's a little difficult since we aren't four-dimensional. ;) But we can use our constraint $$x+y+z=3$$ to turn it into a viewable 3D function. We eliminate $z$, and use the 3rd axis for $A$.

Here's a graph of
$$A = (x^2+1)(y^2+1)((3-x-y)^2+1)$$
in the region $-0.5 \le x, y \le 3.5$. The function $A$ grows fairly quickly outside that region, and I've sliced off the boring bits above $A=10$ to focus on the interesting region containing the stationary points, which I've marked with coloured dots, red for the maximum, and blue for the minima.

Click on the image for an interactive version that you can zoom, rotate, etc.


问:
Does any sequence with values in a directed set have a monotone subsequence?

Definition: a directed set is a set $M$ together with a preorder $\geq$ (reflexive and transitive order) such that every pair of elements in $M$ has an upper bound ($\forall x,y \in M, \ \exists z \in M,\ z \geq x \wedge z \geq y$)

For sequences in the reals, there always exists a monotone subsequence. The proof of this fact uses the total order property of the real numbers. But I was wondering, would it work for directed sets? In other words, does a sequence with values in a directed set $(M, \geq)$ always have a monotone subsequence?

I tried to find a counterexample, but I have yet to work with directed sets to be able to visualize them at this level.

答: Let $M=\wp(\Bbb N)$, directed by inclusion, and let $a_n=\{n\}$ for $n\in\Bbb N$; then $\langle a_n:n\in\Bbb N\rangle$ is a sequence in $M$ with no monotone subsequence.


问:
How do I find the equation for a parabola when given 2 points and the maximum values?
I need an equation for a parabola that passes through $2$ points and has a specific maximum value. The points are $(0, 10)$ and $(7, 0)$, and the maximum is $y=45$.

答:
We start with "vertex form": $y = a(x - h)^2 + k$, where the point $(h,k)$ is the vertex.  This tells us that $k = 45$, because the maximum occurs at $y = 45$.  Now, we have $$y = a(x - h)^2 + 45$$ and two additional points that the graph has to pass through: $(0,10)$ and $(7,0)$.  We can use those two points to find the two remaining parameters, $a$ and $h$:
\begin{align}
10 &= ah^{2} + 45\tag{1}\\
0 &= a(7-h)^{2} + 45.\tag{2}
\end{align}
Now, solving for $a$ in $(1)$ gives us $$a = -\frac{35}{h^{2}}.\tag{3}$$  Substituting $(3)$ into $(2)$ gives us $$0 = -\frac{35}{h^{2}}(7 - h)^{2} + 45 = -\frac{35}{h^{2}}(49 - 14h + h^{2}) + 45.\tag{4}$$  Simplifying $(4)$ gives us \begin{alignat}{2}
\frac{9}{7}h^{2} = 49 - 14h + h^{2}&\implies\frac{2}{7}h^{2}+14h - 49 &&= 0\\
&\implies2h^{2} + 98h - 343 &&= 0.\tag{5}
\end{alignat}
We can solve $(5)$ using the quadratic formula:
$$h = \frac{-98 \pm \sqrt{(98)^2+(4)(2)(343)}}{4} = \frac{1}{2}\left(-49 \pm 21\sqrt{7}\right) \approx −52.28, \ 3.28.$$  Now, because $h$ is the $x$-coordinate of the vertex, we need to take $h>0$ based on the provided graph, so we have $$h = \frac{1}{2}\left(-49 + 21\sqrt{7}\right) = \frac{7}{2}\left(3\sqrt{7} - 7\right).$$  Lastly, then, we use this value of $h$ along with $(3)$ to get $$a = -\frac{35}{h^{2}} = -\frac{35}{\left(\frac{7}{2}\left(3\sqrt{7} - 7\right)\right)^{2}} = -\frac{10}{49}\left(3\sqrt{7}+8\right).$$  Altogether, then, your equation is $$\boxed{y = -\frac{10}{49}\left(3\sqrt{7}+8\right)\left(x - \frac{7}{2}\left(3\sqrt{7} - 7\right)\right)^{2} + 45}$$ Desmos confirms that this is correct:


问:
Is there a Mathematical system which is stochastic at its core?
Ok so I'm a programmer, I'm not a mathematician, I've got a minor in math but I didn't even do particularly well at it so please bear with some possibly really stupid thoughts.  Just please try to explain to me why what I'm asking is stupid so that I don't keep making a fool of myself.

I recently found myself thinking along the following lines:

If you take the stand that it's solipsism to talk about anything that cannot be put to use in  predicting an outcome then cannot you apply the same axiom to basic mathematical operations?  For example does addition really have any meaning outside of the context of placing two objects in a box and predicting how many are inside?  Yeah, I realize that's a dangerous question to ask a board full of mathematicians but I imagine you can ask this question about any axiom.

The problem is that in reality if you put one thing and then another into a box it does not mean that there will be two things in the box.  There will likely be two things of course but if the objects are point particles like electrons there's a chance there will be one or three, or a million electrons there.  Heck, the same is even possible with apples, just incredibly unlikely.

I don't have the mathematical wherewithal to think through what this would mean but intuitively it would seem that this might not have many implications for addition, but might for subtraction (unless you could somehow have negative amounts of particles which I won't rule out), and certainly for things like integration and derivation.

I would assume that someone's done work along these lines before?  Has anyone actually created a system around it?  Was it useful?  Are there any accessible books or articles about it?

Just interested

答:
Your question is undoubtedly philosophical in nature: it is predicated on the premise "it's solipsism to talk about anything that cannot be put to use in predicting an outcome".  But solipsism is a philosophical term (and more fundamentally, an inherently philosophical -- as opposed to mathematical or scientific -- idea)!

The philosophical view you advocate seems to be, roughly, that the truths of arithmetic are (only) empirical in nature.  As with many basic philosophical ideas, this has been propounded before and has gotten a lot of discussion.  Famously, it was the view of arithmetic espoused by John Stuart Mill.  For a pretty good quick summary of his positions, see

http://plato.stanford.edu/entries/mill/#GeoAri

Mill's ideas on arithmetic and geometry have generally been poorly received by later philosophers.  Notably, much of the work of Immanuel Kant is a reaction against Mill's ideas, especially his notion of the "synthetic a priori".

So far as I know, essentially no very few mathematicians or philosophers of mathematics have accepted Mill's ideas.  Rather, we have followed the path of Frege and viewed logic as being the foundation for mathematics.  (Later work by Russell, Godel and others have showed that there are some pitfalls and limitations of the logicist approach, but it has nevertheless been the one mathematicians have adopted for the last hundred years.)

[As Qiaochu says, if you really want to talk mathematics you should remove the overtly philosophical lead to your question.  Instead, you should say more about the mathematical aspects of your question: what mathematical problem are you trying to solve?]

ContentLicense: CC BY-SA 2.5


问:
Convergence of $\sum \frac{a_n}{(a_1+\ldots+a_n)^2}$
Assume that $0 < a_n \leq 1$ and that $\sum a_n=\infty$. Is it true that
$$ \sum_{n \geq 1} \frac{a_n}{(a_1+\ldots+a_n)^2} < \infty $$
?

I think it is but I can't prove it. Of course if $a_n \geq \varepsilon$ for some $\varepsilon > 0$ this is obvious.

Any idea? Thanks.

答:
There is mre general fact. Assume $a_k\geq 0$ for all $k\in\mathbb{N}$
and $\sum_{k=1}^\infty a_k$ diverges then for all $\delta>0$ the series
$$
\sum\limits_{n=1}^\infty\frac{a_n}{S_n^{1+\delta}}\tag{1}
$$
converges. Here we denoted $S_n=\sum_{k=1}^n a_k$. Fix $n\in\mathbb{N}$ Lets apply to the function $f(t)=t^{-\delta}$  Lagrange's mean value theorem on the interval $[S_{n-1},S_n]$. Then
$$
\frac{1}{S_n^\delta}-\frac{1}{S_{n-1}^\delta}=-\frac{\delta}{\xi_n^{1+\delta}}(S_n-S_{n-1})=-\frac{\delta}{\xi_n^{1+\delta}}a_n
$$
for some $\xi_n\in[S_{n-1}, S_n]$. In this case
$$
\sum\limits_{n=2}^\infty\frac{a_n}{S_n^{1+\delta}}\leq
\sum\limits_{n=2}^\infty\frac{a_n}{\xi_n^{1+\delta}}=
\sum\limits_{n=2}^\infty\frac{1}{\delta}\left(\frac{1}{S_{n-1}^\delta}-\frac{1}{S_n^\delta}\right)=\frac{1}{\delta}\left(\frac{1}{S_1^\delta}-\lim\limits_{n\to\infty}\frac{1}{S_n^\delta}\right)=\frac{1}{\delta}
$$
Hence the series $(1)$ converges.</p>


问:
How to calculate density functions from joint pdf with the following constraints?
Let $X,Y$ have joint density

$f(x, y) = 3xy/1250$ for $0 ≤ x, 0 ≤ y, x + y ≤ 10$
and $f(x,y) = 0 $ otherwise

Find the density $f(x)$ of $X$.

My attempt: $f(x) = \int_0^{10-x} \cfrac{3xy}{1250}dy = \cfrac{30}{1250}x^2-\cfrac{3}{1250}x^3$ for $0\leq x \leq 10-y$

But this does not match the answer in the book: $f(x) = \cfrac{3}{2500}x(10 − x)^2$
for $0 ≤ x ≤ 10 − y$

What am I misinterpreting about the bounds of the integral here? Where did I make a mistake?

Thank You!

答:
I think you have the bounds of integration correct, but there may be an issue in the execution of the integral. Consider this:

$f(x)=\int_{0}^{10-x}{\frac{3}{1250}xydy}$

$f(x)=\frac{3}{1250}\int_{0}^{10-x}{xydy}$

$f(x)=\frac{3}{1250}{\frac{xy^2}{2}}|_{0}^{10-x}=\frac{3}{1250}(x(10-x)^2-0)$

Which then simplifies to $f(x) =\frac{3}{2500}x(10-x)^2$ for $0 \leq x \leq 10-y $


问:
A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$
I saw this twitter post that reads:

Find all the pairs of positive integers $(x,y)$ satisfying $$  x^2 + y^2 - 5xy + 5 = 0 . $$

I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's interesting is that there are (at least?) 12 general forms that looks like they all resemble the solutions to Pell's equation.

For example, the first general solution presented reads:

$$ \begin{array} { r c l }
 x &=&
\dfrac1{42} \Big [ 21 \left(55 - 12\sqrt{21} \right)^n
- \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 21 \left(55 + 12\sqrt{21} \right)^n
+ \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big ] \\ \phantom0 \\
 y &=&
\dfrac1{42} \Big [ 63 \left(55 - 12\sqrt{21} \right)^n
- 13 \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 63 \left(55 + 12\sqrt{21} \right)^n
+ 13 \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big] \end{array} $$ for $n = 0,1,2,3\ldots $.

Plugging $n=0,1,2,3$ gives the first few solutions

$$(x,y) = (1,3), (67,321), (7369,35307), (810523,3883449)
$$

I tried converting these solutions to the form of $X^2 - DY^2 = A$ or something similar to Pell's equation but I got nothing.

Is it a coincidence that this innocuous-looking quadratic Diophantine equation is actually a caboodle of Pell's equation in disguise? If so, how can we derive them all?

答:
The question as given is perfect for a technique from contest mathematics called Vieta Jumping. This is a special case of automorphism of quadratic forms. It has the virtue that it can be justified using nothing worse that the quadratic formula, and it does not require the use of  square roots either. If we have a solution in positive integers to $x^2 - 5xy + y^2 = -5$ we get new ones using either
$$  (x,y) \mapsto (5y-x,y) $$  or

$$  (x,y) \mapsto (x,5x -y) $$

Note that repeating one of the "jumps" twice in a row goes back to the original solution.

A "fundamental" solution is one that minimizes $x+y$ as much as possible. That is, fundamental when both
$$  x+y \leq 5y - x + y  $$ and
$$  x+y \leq x +5x - y . $$
The first one becomes
$$  2x \leq 5 y, $$ the second
$$  2y \leq 5x .$$

Altogether, fundamental solutions are on the arc $x^2 - 5xy + y^2 $ with
$$\frac{2x}{5}  \leq  y \leq \frac{5x}{2} $$

As you can see, there are the two integer points between the slanted lines, those being $(2,1)$  and $(1,2).$  Every solution in positive integers  reduces to one of these; in turn, jumping up from one of the fundamental solutions generates all solutions.

The linear recurrence $w_{n+2} = 5 w_{n+1} - w_n$  that applies to every other number in a sequence  comes from the trace and determinant  of
$$
\left(
\begin{array}{r}
5 & -1 \\
1 & 0 \\
\end{array}
\right)
$$
being $5$ and $1.$


问:
Given that $\sum_{n=1}^\infty x_n$ converges, show that $\sum_{n=1}^\infty \frac{\sqrt{x_n}}{n}$ converges
Given that $\sum_{n=1}^\infty x_n$ converges and $x_n \ge 0$, show that $\sum_{n=1}^\infty \frac{\sqrt{x_n}}{n}$ converges.

I tried showing it as follows:

Since $x_n$ is positive it must be the case that $(x_n)$ is a monotonically decreasing sequence, then $x_n > x_{n+1}$. We'll show that $\left(\frac{\sqrt{x_n}}{n}\right)$ is a monotonically decrasing sequence with a lower bound of $0$, and therefore, converges.

Since $x_n > 0 \to \sqrt{x_n} > 0 \to \frac{\sqrt{x_n}}{n} > 0$, therefore it is bounded bellow by $0$.

Now all we have to show is that $\frac{\sqrt{x_{n+1}}}{n+1} \le \frac{\sqrt{x_n}}{n}$:

$\frac{\sqrt{x_{n+1}}}{n+1} \le \frac{\sqrt{x_n}}{n}$

$\frac{\sqrt{x_{n+1}}}{n+1} - \frac{\sqrt{x_n}}{n} \le 0$

$-n\sqrt{x_{n+1}} + n\sqrt{x_n} + \sqrt{x_n} \ge 0$

$n(\sqrt{x_n}- \sqrt{x_{n+1}} ) + \sqrt{x_n} \ge 0$

Since $n>0 \land \sqrt{x_n}>0\land\sqrt{x_{n+1}}>0$

if $\sqrt{x_n}- \sqrt{x_{n+1}} \ge 0 \to n(\sqrt{x_n}- \sqrt{x_{n+1}} ) + \sqrt{x_n} \ge 0$

but

$\sqrt{x_n}- \sqrt{x_{n+1}} \ge 0 \iff \sqrt{x_n} \ge \sqrt{x_{n+1}} \iff x_n \ge x_{n+1}$

Since $x_n \ge x_{n+1}$ we have shown that $\left(\frac{\sqrt{x_n}}{n}\right)$ converges.

But as is the case with the harmonic series, this is not enough to show that $\sum_{n=1}^\infty \frac{\sqrt{x_n}}{n}$ converges. I tried solving it using the comparison test but I'm having a hard time showing that $\frac{\sqrt{x_n}}{n} \le cx_n$ (I don't even know how to tackle showing this inequality).

Note:

Also in my attempt above I've assumed that it is the case that $(x_n)$ must be monotonically decreasing, but I'm not sure if this is true.

答:
Hint: Cauchy's inequality

$\sum_n\frac{\sqrt{x_n}}{n}\leq\Big(\sum_n x_n\Big)^{1/2}\Big(\sum_n\frac{1}{n^2}\Big)^{1/2}
$

I leave the details for the OP


问:
proving that if $z$ is a root of $A+B$, it is not a root of $B$ & finding common roots of $A$ and $B$
for $n \in \mathbb{N}$, we have $A$ and $B$ two polynomials as such:

$A\left(X\right)=nX^{n+1}-\left(n+1\right)X^n+1$

$B\left(X\right)=X^n-nX+\left(n-1\right)$

I'm struggling with the two following questions:

Show that if $z \in \mathbb{C}, z \not= 1$ and z is a root of $(A+B)(X)$, then $B(z) \not= 0$

Find the common roots of $A$ and $B$

Any hints?

答:
For part 1, we know that $(A+B)(X) = n(X^{n+1}-X^n-X+1) = n(X-1)(X^n-1)$. So for any root of $A+B$ we have $X^n=1$.

For part 2, observe that if $X$ is a root of both $A$ and $B$, it is also a root of $A+B$.


问:
Lebesgue integrable implies measurable.
Suppose we define Lebesgue integral like this, but without assuming that $f$ is measurable. Suppose there exists a finite integral $\int_Xfd\mu$. Does it follow that $f$ is measurable?

UPD. $f$ is integrable, if $\Xi(f) = \xi(f)$.

答:
So if you can construct a non-measurable set $A \subset (0,1)$ and a function

$f(x)= 1$ for $x \in A$

$f(x)=-1$ for $-x \in A$

$f(x)=0$ $otherwise$

over the interval [-1,1]. f is an odd function so if you can define the integral without measurable f this has a good chance of having a defined integral of 0 over [-1,1].

If your definitions force the preimage of 1 to be measurable it is likely those same restrictions will force f to be measurable.

Title: Lebesgue integrable implies measurable.


问:
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$
I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?

Thanks.

答:
Consider the derivative of $f(x)=1+a^2+ax-x^3$, that is
$$
f'(x)=a-3x^2
$$
This vanishes at $x=\pm\sqrt{a/3}$ and we can compute the values at the relative minimum $-\sqrt{a/3}$; it's simpler if we set $\sqrt{a/3}=b$, so $a=3b^2$:
$$
f(-b)=1+9b^4-3b^3+b^3=1-2b^3+9b^4
$$
The function decreases from $\infty$ to the relative minimum then, goes up to the relative maximum and then crosses once the $x$-axis decreasing to $-\infty$. So, if we prove that $f(-b)>0$, we're done.

Consider $g(x)=1-2x^3+9x^4$; its derivative is $g'(x)=-6x^2+36x^3=6x^2(6x-1)$, so the function has an absolute minimum at $x=1/6$. Since
$$
g(1/6)=1-\dfrac{1}{108}+\dfrac{1}{144}=\dfrac{431}{432}>0
$$
we're done.

By the way, for $a\le0$ the function is globally decreasing, so the equation has a single real solution also in this case.


问:
Are all disjoint events dependent?
This is a basic question about probability theory.

My reasoning goes as follows:

If $A$ and $B$ are independent events, the probability they both happen is their multiplication:
$$\Pr(A \text{ and } B) = \Pr(A) \times \Pr(B)$$

If their marginal probability is not impossible, also their product is non-zero:
$$\Pr(A) > 0,\, \Pr(B) > 0 \implies \Pr(A \text{ and } B) > 0$$

Hence, independent events cannot be disjoint

Hence, only dependent events can be disjoint

Hence, all disjoint events are dependent.

Can you help me point out the error in my argument?

答:
Yes, as long as A and B are mutually exclusive and sum up to 1 or 100%,  P(A) + P(B) = 1.
Example : You are either dead or alive. The chance of being alive or dead always sums up tp 1 or 100%, no matter what condition you live in.</p>

Title: Are all disjoint events dependent?


问:
Taylor series expansion or polynomial regression which is the correct term here for finding an analytical estimate for an unknown function?
I have a question regarding the Taylor expansion of an unknown function and I would be tanksful to have your comments on that.

Suppose we want to find an analytical estimate for an unknown function. The available information for this function is; its exact value at $x_0$ $(f_0)$ and first derivative at $x_0$ $(f_0')$, and its exact value at $x=x_1$ $(f_1)$ which is far from $x_0$. This means that we only need to find the second derivative through solving $$f_1 = f_0 + f_0'(x_1-x_0) + \frac12 {f_0}''(x_1-x_0)^2$$ to form the truncated Taylor series for $f$.

My question is that whether the obtained quadratic function is the result of Taylor series expansion or polynomial regression? Is this a proper way of approximating a function considering that f"0 is found by matching the function value at $x_1$ which is far from $x_0$?

答:
Thanks for your answer Claude and sorry for the late response.

I thinks that for this case the quadratic polynomial and second-order Taylor expansion are identical as you can see for the both functions in the below figs.

Moreover, as it is clear the result is not accurate even locally when the Taylor series coefficients are calculated based on two different points, i.e. by knowing f(x0), f'(x0) and f(x1), as opposed to f(x0), f'(x0) and f"(x0). Obviously, this is much more pronounced for an exponential function.

To sum up, when estimating a function by Taylor expansion, the coefficients must be calculated from a single point.

In the following figs, x0=2 and x1=6. the Taylor expansions are either calculated based on f(2), f'(2) and f"(2) or f(2), f'(2) and f(6).


问:
Solving $V_t=V_{xx}$ on $\{(x,t):x\in\mathbb{R}\}, t>0$ and $V(x,t)=e^tf(x)$ for some function $f$.

Find solutions to $V_t=V_{xx}$ on $\{(x,t):x\in\mathbb{R}\}, t>0$ and $V(x,t)=e^tf(x)$ for some function $f$.

So I have that $V_{xx}(x,t)=e^tf''(x)$ and $V_t(x,t)=e^tf(x)$

But from the equation I get $e^tf''(x)=e^tf(x)$?

So I should have $f''(x)=f(x)$?

I'm not really sure what to be doing from here, or if what I've done even makes sense.

But solving $y''-y=0$
I get $r^2-r=r(r-1)=0$
So I have $r=0,r=1$
So $f(x)=c_1e^{x}+c_2$ is a solution?

But this doesn't seem to work with me given equations. Since if $V(x,t)=c_1e^te^x+c_2e^t$

then $V_t=c_1e^te^x+c_2e^t\neq V_{xx}$

答:
So l take it that we are asked for solutions of

$V_t = V_{xx} \tag 1$

on the set

$\Omega = \{ (x, t): x \in \Bbb R, 0 < t \in \Bbb R \} \tag 2$

which take the form

$V(x, t) = e^tf(x), \tag 3$

where

$f: \Bbb R \to \Bbb R \tag 4$

is sufficiently differentiable.  How differentiable?  Well, it is implicitly indicated that

$V_{xx} = (e^t f(x))_{xx} = e^t f''(x) \tag 5$

exists, hence

$f(x) \in C^2(\Bbb R, \Bbb R). \tag 6$

Now it follows from (3) that

$V_t = e^tf(x), \tag 7$

and this in concert with (1) and (5) yields

$e^t f''(x) = e^tf(x), \tag 8$

whence, since

$e^t \ne 0, \tag 9$

we have

$f''(x) = f(x), \tag{10}$

which of course may be solved in the usual fashion: setting

$f(x) = e^{rx}, \tag{11}$

it is easy to see that

$f'(x) = re^{rx}, \tag{12}$

$f''(x) = r^2e^{rx}, \tag{13}$

and thus via (10), (11), and (13),

$r^2e^{rx} = e^{rx}, \tag{14}$

from which, dividing by $e^{rx} \ne 0$,

$r^2 = 1, \tag{15}$

that is,

$r = \pm 1. \tag{16}$

We conclude that $e^{\pm x}$ are two linearly independent solutions of (10), and since (10) is a second order equation, these two functions span its entire space of solutions.  Thus we may write

$f(x) = c_+ e^x + c_-e^{-x}; \tag{17}$

now combining (3) and (17) we find that

$V(x, t) = e^t(c_+ e^x + c_-e^{-x}), \tag{18}$

the requisite result.  Note that we have also shown that

$f(x) \in C^\infty(\Bbb R, \Bbb R) \tag{19}$

in the course of the derivation of (18).


Title: Integrate $\int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx$

Body:
Need help integrating:
$$\int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx$$

I think I should multiply the $$xe^{-\frac{2x}{\theta }}$$ out and then use integration by parts but it is not really working for me?

Body:
$$I=\int_0^{\infty } \frac{6 x}{t} e^{-\frac{2 x}{t}} \left(1-e^{-\frac{x}{t}}\right) \, dx$$
Set $$e^{-\frac{x}{t}}=u\to x=-t\log u;\;dx=-\frac{t}{u}$$
$$I=\int_1^0-6 (1-u) u^2 \log u\left(-\frac{t}{u}\right)\,du=$$
$$=6t\int_0^1 (u-1) u \log u\,du=6t[-\frac{u^3}{9}+\frac{1}{3} u^3 \log u+\frac{u^2}{4}-\frac{1}{2} u^2 \log u]_0^1=\frac{5 }{6}t$$
As $$\underset{u\to 0}{\text{lim}}\left(-\frac{u^3}{9}+\frac{1}{3} u^3 \log u+\frac{u^2}{4}-\frac{1}{2} u^2 \log u\right)=0$$

TL;DR $I=\frac{5 }{6}t$


问:
Get $\pi$ decimals manually
I know that there was a guy that could get 100 decimal digits of $\pi$ before computers were able to get thousands.

How did the guy do that?

答:
Check this, you can find n-th PI digit without computing previous ones :)
It was a big surprise when this formula was discovered.

http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula

Title: Get $\pi$ decimals manually


问:
Prove/disprove $(\int_0^{2 \pi} \!\!\cos f(x) \, d x)^2+(\int_0^{2 \pi}\!\!\! \sqrt{(f'(x))^2+\sin ^2 f(x)} \, dx)^2\ge 4\pi^2$

Let $f(x)$ be a differentiable function on $[0,2\pi]$ s.t. $0\leq f(x)\leq 2\pi$ and $f(0)=f(2\pi)$. Prove or disprove that
$$
\left(\int_0^{2 \pi} \cos f(x) \,d x\right)^2+\left(\int_0^{2 \pi} \sqrt{(f'(x))^2+\sin^2 f(x)} \, d x\right)^2 \geq(2 \pi)^2
$$

It seems that when $f$ is an arbitrary constant, the left side equals $(2\pi)^2$ and seems to be the minimum. But how can I show that there's no other $f$ that makes the left side equal (or be less than) $(2\pi)^2$?

A geometric interpretation of the inequality has been found: Consider a closed curve on a sphere: $C=\{(\cos x\cdot\sin f(x),\,\sin x\cdot\sin f(x),\,\cos f(x))\mid x\in[0,2\pi)\}$, we have its perimeter $\displaystyle L=\int_0^{2\pi}\sqrt{(f'(x))^2+\sin^2 f(x)}\,dx$ and its area $\displaystyle S=2\pi-\int_0^{2\pi}\cos f(x)\,dx$. From spherical isoperimetric inequality $L^2\ge S(4\pi-S)$, we have $(2\pi-S)^2+L^2\ge(2\pi)^2$, and the equality holds iff $C$ is any circle on the sphere. In this way we get the original inequality in the sense of geometry.

Now the question is, how to prove the inequality with only pure analysis methods?

答:
Also not an answer, but my train of thoughts. We can take advantage of the fact that for a complex number $z$ and its conjugate $\bar{z}$, it holds that $z\bar{z}=|z|^2$.

Thus, if we associate $$z(x)=\int_0^{2\pi}\left(\cos f(x)+i\sqrt{(f'(x))^{2}+\sin ^{2} f(x)}\right)dx=\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx,$$

the LHS becomes $z(x)\overline{z(x)}$ and thus we want to show that $$\left|\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx\right|^2 \geq (2\pi)^2$$

Now, somewhat fishy, but I believe since the function $f(x)$ is positive on the interval $(0, 2\pi)$, and also $\sqrt{1+(f'(x))^2}$ is positive, the following equality should hold:
$$\color{gray}{\left|\int_0^{2\pi}\sqrt{1+(f'(x))^2}e^{i\phi(x)}dx\right|=\int_0^{2\pi}\left|\sqrt{1+(f'(x))^2}e^{i\phi(x)}\right|dx=\int_0^{2\pi}\sqrt{1+(f'(x))^2}dx=L}$$

Then, by the same argument as in the answer by @Joshua Woo, this is a length of the curve $f(x)$, which must be greater than the length of interval ($0$, $2\pi$), i.e., $L=2\pi$ (due to the condition $f(0)=f(2\pi)$), therefore we get the required inequality.

But I will be happy to learn where my argument is incorrect.

EDIT: Thanks for pointing out the error. I will keep the post struck-through just for the transparency.


问:
Derivation on $C(X)$
If we define derivation on $C(X)$, where $X$ is a compact Hausdorff space, will the dimension of $Der(C(X))$(if it is well-defined) equal to the (topological) dimension of $X$? My guess is this is true since derivation is just differentiation of function, but I cannot prove it.

答:
Suppose $D$ is a derivation and $f\in C(X)$. For $x\in X$ look at $(Df)\,(x)$, since $D$ applied to any constant function is necessarily $0$ you have
$$(Df)\,(x) = D(f-c )\, (x)$$
with $c$ a constant function. Now for $c=f(x)$ you have that $f-c$ is $0$ at $x$. You may check that there exist $g,h$ continuous with $f-c= g\cdot h$ and $g(x)=h(x)=0$ (this is a feature of continuous functions that fundamentally fails for any notion of differentiable functions). Then

$$(Df)\, (x) = D(f-c)\,(x) =D(g\cdot h)\,(x)= g(x) (Dh)\,(x)+(Dg)\,(x)\,h(x) =0$$

hence $Df$ is zero at $x$ for any $x\in X$. This implies $D=0$ so there are no non-zero derivations.

Title: Derivation on $C(X)$


问:
For $0 \leq a_n, b_n$: Is $\varliminf a_n \varlimsup b_n \leq \varlimsup a_n b_n $ true?
The sequences are assumed to be bounded!

I suspect it is, since $\varliminf a_n + \varlimsup b_n \leq \varlimsup (a_n+b_n)$ is. But I have not managed to prove it.

I thought it possible to mirror the following trick when proving the inequality for sums:

$\varlimsup b_n = \varlimsup (a_n + b_n - a_n) \leq \varlimsup (a_n + b_n ) + \varlimsup(-a_n) = \varlimsup (a_n + b_n )  -\varliminf(a_n)$

But the incipient idea fell short rather fast:

$0 = \varlimsup (a_n b_n - a_n b_n) \leq  \varlimsup (a_n b_n) + \varlimsup (-a_n b_n)$

I was hoping to manipulate $\varlimsup (-a_n b_n)$ in some ingenious way, but have not succeded.

Would love to read any ideas or hints.

答:
Under the assumption that the sequences $(a_n)$ and $(b_n)$ are bounded:

Let $\alpha = \liminf a_n$ and $\beta = \limsup b_n$ (the boundedness assumption telling us that $-\infty < \alpha,\beta < \infty)$.  Let $\epsilon > 0$.  By definitions, we know that $a_n > \alpha - \epsilon$ for all $n$ sufficiently large and that $b_n > \beta - \epsilon$ for infinitely many $n$.  Therefore also $a_n b_n > (\alpha - \epsilon)(\beta - \epsilon) = \alpha\beta - \epsilon(\alpha+\beta) + \epsilon^2$ for infinitely many $n$.  This implies that $\limsup (a_n b_n) > \alpha\beta - \epsilon(\alpha+\beta) + \epsilon^2$.  This holds for arbitrary $\epsilon$, so we conclude $\limsup(a_n b_n) \geq \alpha \beta$ as desired.

If either of the sequences is not bounded, there may be some other cases to check, but I am a bit too lazy to check them all :)


答:
A cubic has only one real root iff its discriminant is negative. (See Wikipedia.)

The discriminant of $1 + a^2 + ax - x^3$ is $-27 a^4 + 4 a^3 - 54 a^2 - 27$.

Now, $-27 a^4 + 4 a^3 - 54 a^2 = a^2 (-27 a^2 + 4 a - 54)$ and $-27 a^2 + 4 a - 54=-27 \left(a - \frac{2}{27}\right)^2 - \frac{1454}{27} < 0$ for all $a$.

Therefore, $-27 a^4 + 4 a^3 - 54 a^2 - 27 < 0$ for all $a$.

Title: Show that there is only one real root of the equation $1 + a^2 + ax - x^3$

Body:
I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?

Thanks.


答:
Both solutions are incorrect: the marginal density must not contain the variable $y$. By definition:
$$
f_X(x)=\int_{-\infty}^\infty f(x,y)dy
$$

The density of $X$ is given as follows.

For $x\not\in (0,10)$, $f_X(x)=0$.

For $0\le x\le 10$,
$$
f_X(x)=\int_{-\infty}^\infty f(x,y)dy
=\int_0^{10-x}\frac{3xy}{1250}dy
=\frac{3x}{1250}\cdot \frac12 (10-x)^2
$$

It is useful to draw a picture for the region where $f$ is nonzero.

Title: How to calculate density functions from joint pdf with the following constraints?

Body:
Let $X,Y$ have joint density

$f(x, y) = 3xy/1250$ for $0 ≤ x, 0 ≤ y, x + y ≤ 10$
and $f(x,y) = 0 $ otherwise

Find the density $f(x)$ of $X$.

My attempt: $f(x) = \int_0^{10-x} \cfrac{3xy}{1250}dy = \cfrac{30}{1250}x^2-\cfrac{3}{1250}x^3$ for $0\leq x \leq 10-y$

But this does not match the answer in the book: $f(x) = \cfrac{3}{2500}x(10 − x)^2$
for $0 ≤ x ≤ 10 − y$

What am I misinterpreting about the bounds of the integral here? Where did I make a mistake?

Thank You!


问:
Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.

Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.

We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?

答:
Continuing your idea, let $a = 1+ \dfrac 1t, t > 0.$ Then, you can write your inequality as $xy-x-y > e^2-2e$, which is in terms of the variable $t:$

$$f(t) = t\ln\left(1+\frac 1t\right) + \ln\left(\left(1+\frac 1t\right)^{t+1}-2-\frac 1t\right) > 1+\ln(e-2).$$

Wolfram says that this is indeed a decreasing function in $t$, so the infimum will be:
$$\lim\limits_{t\to\infty} f(t) = 1+ \ln(e-2).$$

Then, the only thing to show now is the fact that $f(t)$ is decreasing, but I suspect this will take an ugly computation.

EDIT: Actually, the derivative did not turn out to be too bad:

\begin{align}f'(t) &= \\
&\dfrac{\dfrac{1}{t^2} + \left(1+\dfrac 1t\right)^{t+1}{\left(2\ln\left(1+\frac 1t\right)-\dfrac 1t-\dfrac{1}{t+1}\right)} + \left(2+\dfrac{1}{t}\right)
\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\left(1+\dfrac 1t\right)^{t+1} - \dfrac 1t-2} &
\\<0\end{align}

Finally, the above is equivalent to:

$$\left(1+\dfrac 1t\right)^{t+1} > \dfrac{\dfrac{1}{t^2} + \left(2+\dfrac 1t\right)\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\dfrac{1}{t}+\dfrac{1}{1+t} - 2\ln\left(1+\dfrac{1}{t}\right)},$$

which I think should be doable by brute-force expanding with a Taylor's series etc.


问:
Inverse image of the canonical projection
Let $R$ a ring, $I$ one of its ideal and π the canonical projection. Let $A$ a subring of $R$ not  containing $I$. I need to prove that

$\pi^{-1}(\pi(A)) = A + I$

where π is a ring homomorphism between from $R \to R/I$,

I already proved that when $A$ is a subring (ideal) of $R$ then $\pi(A)$ is a subring (ideal) of $R/I$ and that

$\pi(A) = (A+I)/I$

by replacing the last into the original equation we must have

$\pi^{-1}((A+I)/I) = A + I$

but if A is mapped to $(A+I)/I$ by $\pi$, I don't understand how the preimage can possibly be the set  $A + I$.

答:
More generally, if $\phi : R \to S$ is a ring homomorphism, then $$\phi^{-1}(\phi(A)+B) = A+\phi^{-1}(B)$$ for any subring $A$ of $R$ and any subring $B$ of $S$. Indeed, since $A \subseteq \phi^{-1}(\phi(A)) \subseteq \phi^{-1}(\phi(A)+B)$ and $\phi^{-1}(B) \subseteq \phi^{-1}(\phi(A)+B)$, it follows that $A+\phi^{-1}(B) \subseteq \phi^{-1}(\phi(A)+B)$. On the other hand, if $r \in \phi^{-1}(\phi(A)+B)$ then $\phi(r) = \phi(a)+b$ for some $a \in A$ and $b \in B$, and so $r-a \in \phi^{-1}(B)$, meaning that $r = a+(r-a) \in A+\phi^{-1}(B)$.

In particular, if you take $B = \{0\}$, then $\phi^{-1}(\phi(A)) = A+\ker \phi$.


问:
Integrate $\int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx$
Need help integrating:
$$\int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx$$

I think I should multiply the $$xe^{-\frac{2x}{\theta }}$$ out and then use integration by parts but it is not really working for me?

答:
Integrate by parts as follows

\begin{align}
& \int _0^{\infty }\:\:\frac{6}{\theta}xe^{-\frac{2x}{\theta }}\left(1-e^{-\frac{x}{\theta }}\right)dx\\
=& \frac6{\theta} \int _0^{\infty }
x\> d \left( -\frac {\theta}2 e^{-\frac{2x}{\theta } }
 + \frac {\theta}3e^{-\frac{3x}{\theta } } \right)
= \int _0^{\infty }
\left( 3e^{-\frac{2x}{\theta } }
 -2 e^{-\frac{3x}{\theta } } \right)dx
= \frac 56{\theta}
\end{align}


问:
Is that true that for every linear transformation $\phi : V^* \to W^*$ there is a linear transformation $\psi: W \to V$ such that $\psi^* = \phi$?
$V$ and W are finitely dimensional linear spaces over the field $K$. Is that true that for every linear transformation $\phi : V^* \to W^*$ there is a linear transformation $\psi: W \to V$ such that $\psi^* = \phi$?

W* means dual space of W, that is a space of all functionals $W \to K$

V* means dual space of V, that is a space of all functionals $V \to K$

It is like an invert definition of dual mapping. It must be something simple, I just can't think of it anymore.

答:
Yes, it is true. You can consider the transpose map $\phi^*\colon W^{**}\to V^{**}$ and the canonical isomorphisms $\omega_W\colon W\to W^{**}$, $\omega_V\colon V\to V^{**}$.

Set $\psi=\omega_V^{-1}\circ\phi^*\circ\omega_W$ and finish up.


问:
what is $\lim_{x\to6}\frac{5}{(x-6)^2}$
$$\lim_{x\to6}\frac{5}{(x-6)^2}$$

Is it undefined or infinity and why ?

答:
If $x$ is near but not equal to $6$, then $(x-6)^2$ is a small positive number, and $\frac5{(x-6)^2}$ is therefore a large positive number. The closer $x$ gets to $6$, the larger this number gets, and there is no upper bound on how large it can be. Thus, we write

$$\lim_{x\to 6}\frac5{(x-6)^2}=+\infty\;.\tag{1}$$

Since $+\infty$ is not a real number, the limit in question does not exist as a real number; in the set $\Bbb R$ of real numbers the limit does not exist. However, it fails to exist in a very particular way: the function $f(x)=\frac5{(x-6)^2}$ ‘explodes’ upwards as $x$ gets closer and closer to $6$. The notation $(1)$ is an abbreviated statement of this fact, and in words we often say simply that the limit is $+\infty$, understanding that this is not a real number.

At a more advanced level one deals with the extended real numbers, which do include $-\infty$ and $+\infty$, and in that context the limit in $(1)$ actually does exist. That, however, is not the usual first-year calculus context.

Title: what is $\lim_{x\to6}\frac{5}{(x-6)^2}$


问:
The Automorphism Group $\Gamma(\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q})$ is trivial if $n$ is odd.
The full question is: Given a group $F_n=\mathbb{Q}(\sqrt[n]{2})$ then prove that:

(A) If $n$ is odd then $\Gamma(F_n:\mathbb{Q})$={$\rm{id}$} and

(B) If $n$ is even then $\Gamma(F_n:\mathbb{Q})\cong\mathbb{Z_2}$

I am confused as to how to find either, I know that $F_n:\mathbb{Q}$ is a non-normal extension. We were given the hint:

"For any $\tau\in\rm{Aut}_\mathbb{Q}(F_n)$ we must have that $\tau(\sqrt[n]{2})\in F_n\subseteq\mathbb{R}$ is again a real $n$-th root of 2"

答: The $\mathbb Q$-automorphisms of $F_n$ are completely determined by where they send $\sqrt[n]2$, and they must send it to another $n$-th root of $2$. But if $n$ is odd, then no other $n$-th root of $2$ is in $F_n$, and if $n$ is even, then only one other $n$-th root of $2$ is in $F_n$ (you might need to prove these facts beforehand). So there are only one/two possible $\mathbb Q$-automorphisms.


问:
Deriving $-nE\left[\frac{\partial ^2\ln\left(f\left(X\right)\right)}{\partial \theta ^2}\right]$
The information about θ in a random sample of size n is also given by
$$-nE\left[\frac{\partial ^2\ln\left(f_\theta\left(X\right)\right)}{\partial \theta ^2}\right]$$

where $f_\theta(x)$ is the value of the population density at x, provided that the extremes of the region for which $f_\theta(x)$ is not equal to 0 do not depend on θ. The derivation of this formula takes the following steps:

Differentiating the expressions on both sides of
$$\int f_\theta\left(x\right)dx=1$$
with respect to $\theta$, show that
$$\int \frac{\partial \ln\left(f_\theta\left(x\right)\right)}{\partial \theta }\left(f_\theta\left(x\right)\right)dx=0$$
by interchanging the order of integration and differentiation.

Differentiating again with respect to θ, show that
$$E\left[\left(\frac{\partial \ln\left(f_\theta\left(X\right)\right)}{\partial \theta }\right)^2\right]=-E\left[\frac{\partial ^2\ln\left(f_\theta\left(X\right)\right)}{\partial \theta ^2}\right]$$

I'm a little confused about what the question is asking. I am not sure how the $\ln$ comes in, in the first part and then how I relate that to the second part of the question?

答:
Notice that
$$\frac{\partial}{\partial\theta}\ln(f_\theta(x))=\frac{1}{f_\theta(x)}\frac{\partial}{\partial\theta}f_\theta(x)$$
and so,
$$\Big(\frac{\partial}{\partial\theta}\ln(f_\theta(x))\Big) f_\theta(x)=\frac{1}{f_\theta(x)}\Big(\frac{\partial}{\partial\theta}f_\theta(x)\Big) f_\theta (x)=\frac{\partial}{\partial\theta}f_\theta(x)$$

So, assuming that he order of differentiation and integration can be change one has, for the first step,

$$0=\frac{\partial}{\partial\theta}\int_R f_\theta(x)\,dx=\int_R\frac{\partial}{\partial\theta} f_\theta(x)\,dx=\int_R \Big(\frac{\partial}{\partial\theta}\ln(f_\theta(x))\Big) f_\theta(x)\,dx=\mathbb{E}_\theta\Big[\frac{\partial}{\partial\theta}\ln(f_\theta(X))\Big]
$$

where the last identity follows  from the definition of expectation of functions with respect to probability measures with densities, namely
$$\mathbb{E}_\theta[g(X)]=\int_R g(x) f_\theta(x)\,dx$$


答:
By Eisenstein's criterion, we know that $X^n-2$ is always irreducible over $\Bbb Q$. Hence $m(\sqrt[n]{2},\Bbb Q) = X^n-2$ and the roots of this polynomial are of the form $\xi_n^j\sqrt[n]{2}$ with $\xi_n$ an $n$-th primitive root and $1 \leq j \leq n$.

As you note, a morphism $\sigma \colon F_n/\Bbb Q \to \overline{\Bbb Q}$ will be determined by the image of $\sqrt[n]{2}$ in $\Bbb Q$ and, by the previous remark, this ought to be an element of the form $\xi_n^j \sqrt[n]{2}$. But if we want $\mathsf{im} \ \sigma$ to be in $\Bbb R$, can't we say something about $n$ and $j$?

 We need to have $\xi_n^j \in \Bbb R$. But a real root of unity is either $-1$ of $1$. Thus the options are $j = n, \xi_j^n = 1$ and $n \in 2\Bbb Z, j = n/2, \xi_n^j = -1$. When $n$ is even, both options can happen, hence the Galois group is cyclic generated by $\sigma$ such that $\sqrt[n]{2} \mapsto - \sqrt[n]{2}$. When $n$ is odd, the first case is the only one possible, thus every automorphism fixes $\sqrt[n]{2}$... but this means they were the identity to begin with.

Title: The Automorphism Group $\Gamma(\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q})$ is trivial if $n$ is odd.

Body:
The full question is: Given a group $F_n=\mathbb{Q}(\sqrt[n]{2})$ then prove that:

(A) If $n$ is odd then $\Gamma(F_n:\mathbb{Q})$={$\rm{id}$} and

(B) If $n$ is even then $\Gamma(F_n:\mathbb{Q})\cong\mathbb{Z_2}$

I am confused as to how to find either, I know that $F_n:\mathbb{Q}$ is a non-normal extension. We were given the hint:

"For any $\tau\in\rm{Aut}_\mathbb{Q}(F_n)$ we must have that $\tau(\sqrt[n]{2})\in F_n\subseteq\mathbb{R}$ is again a real $n$-th root of 2"


问:
Shell method to compute volume
I'm currently learning about the shell method to compute the volume of a solid of revolution. I am working on the following problem:

Find the volume of the solid obtained by rotating the region by $y = x - x^2$ and $y = 0$ about the line $x = -2$.

I've plotted the three lines, and I'm still really unsure about how to figure this out. I saw some examples on this website: https://brilliant.org/wiki/shell-method/. However, they do not seem very similar to the problem that I have.

I would appreciate it if someone could please help me with this problem.

I think a cylindrical shell formed by rotating about the line $x = -2$ has radius $2 + x$ and circumference $2\pi(2 - x)$ and height $(x - x^2)$, which gives

$$V = \int_{0}^{1} 2\pi (2 + x)(x - x^2)\mathop{dx} $$

Is this correct?

答:
Yes, your integral is correct as the shell integration:
$$
V=2\pi\int_a^b (x-h)f(x)dx,\quad \textrm{for } h\le a<b
$$
where $h=-2$, $a=0$ and $b=1$ and $f(x)=x-x^2$ in your example.

It is useful to draw a picture.

Title: Shell method to compute volume


问:
"Math Lotto" Tickets - finding the minimum winning set

"Math lotto" is played as follows:  a player marks six squares on a 6x6
square. Then six  "losing squares" are drawn. A player wins if none of the losing squares
are marked on his lottery ticket.

1)Prove that one can complete nine lottery tickets in such a way that at least one of
them wins.

2)Prove that this is not possible with only eight tickets.

My attempt is as follows;

First I divided the square into 6 rectangles (figure 1).

If one rectangle doesn't contain a cross then some ticket (ticket 1 to ticket 6) would win the game.

Now we consider the case where each rectangle has one cross each.

Now take the two rectangles on the top left of the square (figure 2). These have a total of two crosses.

The first two columns together contains one cross and the third and fourth columns together contains one cross.

There are four cases and we need at least four tickets (ticket 7 to ticket 10) to ensure win.

I am only getting a minimum of ten tickets. How do I prove only nine tickets is required and for eight tickets it is not possible?

Reference: Combinatorics by Stephan Wagner, Page 42, Problem 49
.
https://math.sun.ac.za/swagner/Combinatorics.pdf

答:
Here is a proof that, given a set of 8 tickets, we can always find 6 losing squares such that each ticket has marked at least one losing square.

If there is a square that is marked on 3 tickets, we choose that square along with one square from each of the other 5 tickets to be losing. So suppose instead that each square is marked on at most 2 tickets. By the pigeonhole principle, there must be a square marked on exactly two tickets; call it Square A. The remaining 6 tickets avoid square A, so applying the pigeonhole principle again, there must be a different square marked on two of the remaining 6 tickets; call it Square B. As losing squares, we then choose Square A, Square B, and one square from each of the remaining 4 tickets.

A very similar argument can prove that, if we have a collection of 9 tickets with the property that at least one will be a winner, then no square can be marked on 3 or more of those tickets.


问:
Proof of the duality of the dominance order on partitions
Could anyone provide me with a nice proof that the dominance order $\leq$ on partitions of an integer $n$ satisfies the following: if $\lambda, \tau$ are 2 partitions of $n$, then $\lambda \leq \tau \Longleftrightarrow \tau ' \leq \lambda '$, where $\lambda'$ is the conjugate partition of $\lambda$ (i.e. the transpose of the set of 'dots' which $\lambda$ represents).

I feel like there must be a nice clever and concise/intuitive proof of this, but all I can come up with is an ugly brute force approach based on the definition of sums of the components $\lambda_i$. Could anyone suggest a particularly nice way to obtain this result? Many thanks.

答:
This question is pretty old, but I would like to propose "a nice clever and concise/intuitive proof" as asked, although it may be considered not formal enough. It is a variant of the answer by Hjalmar Rosengren.

In his 1973 paper, Brylawski shows that all partitions of integer $n$ and the dominance order are obtained from the initial partition $(n)$ by iterating the following two rules on any partition (assuming that the last part is followed by a null part):

if its $i$-th part is strictly larger than its $i+1$-th one, then decrease the $i$-th part by $1$ and increase the $i+1$-th one by $1$;

if its $i$-th part is equal to its $i+1$-th part plus $1$, and its $i+1$-th part is equal to its $j$-th part plus $1$, then decrease the $i$-th part by $1$ and increase the $j$-th one by $1$.

This may be represented graphically on Ferrers diagrams as follows:

In the case $n=7$, for instance, we obtain:

Then, just notice that applying the first Brylawski rule to a partition is equivalent to applying the second rule to its conjugate; and conversely, applying the second Brylawski rule to a partition is equivalent to applying the first rule to its conjugate.

From this, the equivalence in the asked question is obvious: a partition $\lambda$ is reachable from a partition $\tau$ by iterating the Brylawski rules if and only if the conjugate $\tau'$ is reachable from the conjugate $\lambda'$ by iterating these rules.

Several studies envision the lattice of integer partitions from this iterated rules point-of-view, see for instance this one.


问:
Approximating measures by open sets and compact sets.
I'm having trouble starting two similar proofs:

Let $\epsilon > 0$. And let $E$ be a measurable set of finite measure.

Prove that there is an open set $U$ containing $E$ such that $m(U \setminus E) < \epsilon$.

Similarly, prove there is a compact set $K$ contained in $E$ such that $m(E \setminus K) < \epsilon$.

Any hints are much appreciated.

NOTE: $m$ is the Lebesgue outer measure. And $E \subseteq \mathbb{R}$ is measurable if $m(A) \geq m(A \cap E) + m(A \setminus E)$.

答:
Let me know if this works.
If $m_*(E)=\infty$, then taking $O=\mathbb{R}$ give the desired result. Suppose then that $m_*(E)<+\infty$ and let $\epsilon>0$ be given. By definition of the exterior measure we are guaranteed the existence of a countable collection of closed cubes $\{I_j\}_j$ with
$$E \subseteq \bigcup_j I_j$$
such that
$$\sum_j \vert I_j \vert \leq \vert E \vert + \frac{\epsilon}{2}.$$
For each $j$, let $I^*_j$ be a closed cube such that
$$I_j \subseteq \mathrm{int}(I^*_j).$$
and
$$\vert I^*_j \vert \leq \vert I_j \vert + \frac{\epsilon}{2^{j+1}}.$$
Then we can take our open set to be
$$O = \bigcup_j \mathrm{int}(I^*_j)$$
which is open as arbitrary unions of open are open and it is clear to see $E \subset O.$ We can thus compute
\begin{eqnarray*}
m_*(O) &=& \inf \sum_j \vert \mathrm{int}(I^*_j)\vert\\
&\le&   \sum_j \vert I^*_j \vert \\ &\le&  \sum_j (\vert I_j \vert + \frac{\epsilon}{2^{j+1}}) \\
&\le& \vert E \vert + \epsilon,
\end{eqnarray*}
as needed.


答:
Your set up is correct. Due to the confusion, I am adding the diagram for your method.

You can use alternate method (washer) to check your working if you already know.

$x-x^2 = y \implies x = \frac{1}{2} \pm \sqrt{\frac{1}{4}-y}$

So, $\frac{5}{2} - \sqrt{\frac{1}{4}-y} \leq r \leq \frac{5}{2} - \sqrt{\frac{1}{4}-y}$

$0 \leq y \leq \frac{1}{4}$

So, $\displaystyle \pi \int_0^{0.25} (2.5+ \sqrt{0.25-y})^2 - (2.5 - \sqrt{0.25-y})^2 dy = \frac{5 \pi}{6}$. This is same that you would get from your integral.

Title: Shell method to compute volume

Body:
I'm currently learning about the shell method to compute the volume of a solid of revolution. I am working on the following problem:

Find the volume of the solid obtained by rotating the region by $y = x - x^2$ and $y = 0$ about the line $x = -2$.

I've plotted the three lines, and I'm still really unsure about how to figure this out. I saw some examples on this website: https://brilliant.org/wiki/shell-method/. However, they do not seem very similar to the problem that I have.

I would appreciate it if someone could please help me with this problem.

I think a cylindrical shell formed by rotating about the line $x = -2$ has radius $2 + x$ and circumference $2\pi(2 - x)$ and height $(x - x^2)$, which gives

$$V = \int_{0}^{1} 2\pi (2 + x)(x - x^2)\mathop{dx} $$

Is this correct?


问:
Which "known" ring is $K = \frac{\mathbb{Z}_5[X,Y]}{(Y-X^2, XY + Y + 2)}$ isomorphic to?
Which "known" ring is $K = \frac{\mathbb{Z}_5[X,Y]}{(Y-X^2, XY + Y + 2)}$ isomorphic to?

I am not 100% sure how to solve this. So far this is what I have:

I know that $(Y-X^2, XY + Y + 2)$ is the ideal, which means that $Y-X^2$ gets 'projected' to 0, in other words $Y=X^2$.

This then gives me that $XY + Y + 2 = X*X^2 + X^2 +2 = X^3 + X^2 + 2 = 0$.

But now I don't know what I have to do.
I have checked and 1,2,3,4 are not roots of the equation in $\mathbb{Z}_5$. So I'm a bit stuck on how to solve this question. Is what I have done so far correct, and how do I continue?

答:
Very good.

You can prove that it's isomorphic to $\Bbb Z_5[X]/(X^3+X^2+2)$ and you already showed that polynomial has no roots in $\Bbb Z_5$.

Hence, it is irreducible (being of degree 3), so the quotient is the field of $5^3$ elements.


答:
Note that by your remark $(Y-X^2,XY+Y+2) = (Y-X^2,X^3+X^2+2)$.

The assignment $X \to X, Y\to X^2$ defines an arrow $\tau \colon \Bbb Z_5[X,Y]/(Y-X^2) \to \Bbb Z_5[X]$. Surjectivity is clear, as for injectivity: pick $p(X,Y)$ in the kernel of $\tau$ and
via the identification $\Bbb Z_5[X,Y] = \Bbb Z_5[X][Y]$ divide $p$ by $Y-X^2$. Note that even though polynomial rings in several variables need not have euclidean algorithms, we can always divide by monomials (as in this case). Hence we have

$$
p(X,Y) = (Y-X^2)q(X,Y) + r(X).
$$

But then applying $\tau$ we get $r(X) = 0$ as desired.

Finally, by the third isomorphism theorem we have

$$
\Bbb Z_5[X,Y]/(Y-X^2, X^3+X+2) \simeq \Bbb Z_5[X]/(X^3+X^2+2).
$$

Note that the polynomial $X^3+X^2+2$ has no roots modulo $5$ and is thus irreducible in $\Bbb Z_5$. Therefore, the quotient is a field. To compute the order, you can note for example that it is a $\Bbb Z_5$ vector space of dimension $3$ over $\Bbb Z_5$.

Hence the quotient can be characterized as the field with $5^3$ elements.

Body:
Which "known" ring is $K = \frac{\mathbb{Z}_5[X,Y]}{(Y-X^2, XY + Y + 2)}$ isomorphic to?

I am not 100% sure how to solve this. So far this is what I have:

I know that $(Y-X^2, XY + Y + 2)$ is the ideal, which means that $Y-X^2$ gets 'projected' to 0, in other words $Y=X^2$.

This then gives me that $XY + Y + 2 = X*X^2 + X^2 +2 = X^3 + X^2 + 2 = 0$.

But now I don't know what I have to do.
I have checked and 1,2,3,4 are not roots of the equation in $\mathbb{Z}_5$. So I'm a bit stuck on how to solve this question. Is what I have done so far correct, and how do I continue?


问:
Error in the proof of $420 > 422$
I came across a puzzle where we need to determine the error in the following proof.

False Theorem. $420 > 422$

Proof.
We will demonstrate this fact geometrically. We begin with a $20 \times 21$ rectangle which has area $420$. Now we cut along the diagonal and slide the upper piece parallel to the cut until it has moved exactly $2$ units leftward. This leaves a couple stray corners which are $2$ units wide and just over $2$ units high.
Finally, we snip off the two corners and place them together to form an additional small rectangle.
Now we have two rectangles, a large one with area just over $(20 + 2) \times 19 = 418$ and a small one with area just over $2 \times 2 = 4$. Thus, the total area of the resulting ﬁgure is a bit over $418 + 4 = 422$. By conservation of area, $420$ is equal to just a little bit more than $422$.

Can someone please explain what the error is in the above argument ?

Thank You!

答:
Consider the statement

"if the numbers I give you are correct, area is not conserved."

We'll show that this statement is true via contradiction. Observe that this statement and a similar statement claiming that the area is conserved cannot both be true.

Let $L_1, W_1$ be the height and width of the original rectangle and let $L_2$ be the height of the triangles formed from sliding the upper half of the original rectangle $W_2$ units horizontally along its diagonal.

The negation of our consequent is that we "conserve area," that is

$L_1W_1 = (L_1 + L_2) * (W_1 - W_2) + L_2W_2$.

Expanding the RHS of the equation gives us

$L_1W_1 = L_1W_1 - L_1W_2 + L_2W_1$.

Then

$L_1W_2 = L_2W_1$, or $\frac{L_1}{W_1} = \frac{L_2}{W_2}$, i.e. the triangles formed by the diagonal of the original rectangle and the "stray corners" must be similar triangles.

If we plug in the numbers from our antecedent, however, we get

$\frac{20}{21} = \frac{L_2}{2}$, which is absurd since $L_2 > 2$.

Title: Error in the proof of $420 > 422$


问:
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$
I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?

Thanks.

答:
Oh... for shucks and giggles.

All cubic have at least one real solution.  So let $x=r$ be one real root.

Then $f(x)=(1+a^2) + ax -x^3 = (x-r)(Ax^2 + Bx + C) = Ax^3 + (B-Ar)x^2 + (C-Br)x -Cr$ for some

$A=-1; B+r=0; C-Br= a; -Cr=1+a^2$

So $B = -r; C = -\frac {1+a^2}r$  (assuming $r \ne 0$ but as $f(0) = 1+a^2 \ne 0$ we know $r \ne 0)$.

And $-\frac {1+a^2}r + r^2 = a$.

So $f(x) = (x-r)(-x^2 -rx -\frac {1+a^2}r)$

Any other root of $f(x)$ must be $x = \frac {-r \pm \sqrt{r^2 -4\frac {1+a^2}r}}2=\frac {-r\pm \sqrt{r^2 - 4(r^2 -a)}}2$

But $r^2 \ge 0$ and $a > 0$ so  $r^2 - 4(r^2 -a)=-3r^2 -a=-(r^2 + a)  < 0$. so those solutions can never exist.


答:
Just to provide a concrete choice of 9 tickets that will have at least one winning ticket:

Suppose to the contrary that all of these are losing tickets. The yellow ones forces 3 losing squares on the top half. Then 3 remaining losing squares will fill the bottom half. By pigeonhole, at least 2 of these losing squares will be in the bottom left 3x3 or the bottom right 3x3.

Say there are at least two losing squares in the bottom left 3x3, which means at most 1 losing square is in the bottom right 3x3. In this case one of the red tickets will not have a losing square, as the losing square is in one of the 3 columns.

And vice versa, if at least two losing squares are in the bottom right 3x3, then one of the purple tickets will not have a losing square.

Edit. A reference for this problem can also be found here https://www.cut-the-knot.org/pigeonhole/MathLotto.shtml

Title: "Math Lotto" Tickets - finding the minimum winning set

Body:
"Math lotto" is played as follows:  a player marks six squares on a 6x6
square. Then six  "losing squares" are drawn. A player wins if none of the losing squares
are marked on his lottery ticket.

1)Prove that one can complete nine lottery tickets in such a way that at least one of
them wins.

2)Prove that this is not possible with only eight tickets.

My attempt is as follows;

First I divided the square into 6 rectangles (figure 1).

If one rectangle doesn't contain a cross then some ticket (ticket 1 to ticket 6) would win the game.

Now we consider the case where each rectangle has one cross each.

Now take the two rectangles on the top left of the square (figure 2). These have a total of two crosses.

The first two columns together contains one cross and the third and fourth columns together contains one cross.

There are four cases and we need at least four tickets (ticket 7 to ticket 10) to ensure win.

I am only getting a minimum of ten tickets. How do I prove only nine tickets is required and for eight tickets it is not possible?

Reference: Combinatorics by Stephan Wagner, Page 42, Problem 49
.
https://math.sun.ac.za/swagner/Combinatorics.pdf


问:
Distribution for Response Times
I have samples from a response time population for a web transaction. I want to be able to use them to describe a distribution for the population but don't know a proper one to use. I have shied away from a Normal since it would result in some probability of getting a negative time. Could someone suggest a reasonable approach?

Please go easy with the responses; I'm an engineer, not a math guy.

Thank you.

答: You might want to try fitting a log-normal distribution. If you have more interesting data, it's possible that some sort of power-law distribution would fit better. See, for instance, the Lévy distribution.</p>

Title: Distribution for Response Times


问:
Is the number of Lebesgue covering dimensions a topological property?
Is it correct to characterize the number of dimensions that a topological space has as a 'topological property'? By 'dimensions,' I have in mind Lebesgue covering dimensions.

If the answer is yes, then why? If the answer is no, then what kind of a property is the number of dimensions?

答: Isn’t it obvious from the definition that the covering dimension is a topological invariant? it’s defined purely in terms of open sets and set-theory concepts.</p>


问:
$L^p$ space: measure theory
Suppose that $(X,m,\mu)$ is a measure space.

Prove that
$$L^{1}(\mu)=L^{\infty}(\mu)  \iff \dim L^{1}(\mu)<\infty$$

It's obvious that if $\mu(X)<\infty$, then $L^{\infty}(\mu)\subseteq L^{1}(\mu)$ which does not hold in this case. Also don't know how to use the finiteness of $\dim L^{1}(\mu)$.

Any help would be appreciated. Thanks in advance!

答:
This is not quite true. Let $X = \{0\}$, and let $\mu(\{0\}) = \infty$. Then $L^{1}(\mu)$ contains only the zero function, hence has finite dimension, but $L^{\infty}(\mu)$ contains every function $f \colon \{0\} \rightarrow \mathbb{R}$.

However, like many results in measure theory, the statement becomes true if we restrict our attention to $\sigma$-finite measures. We can get away with less, if we want, but I'm going to stick with $\sigma$-finiteness.

A useful result here is that $L^{1} \nsubseteq L^{\infty}$ if and only if $X$ contains sets of arbitrarily small positive measure, and $L^{\infty} \nsubseteq L^{1}$ if and only if $\mu(X) = \infty$ (as you noted). This is essentially the content of Exercise 6.1.5 of Folland's Real Analysis.

With this in hand, proving the following three claims will give us our solution:

If $X$ contains sets of arbitrarily small positive measure, then $\dim L^{1} = \infty$.

If $\mu(X) = \infty$, then $\dim L^{1} = \infty$.

If $X$ does not contain sets of arbitrarily small postive measure and $\mu(X) < \infty$, then $\dim L^{1} < \infty$.

Proof of 1:

Let $E_{1}$ be a set such that $0 < \mu(E_{1}) < 1$. Inductively, define $E_{n}$ such that $0 < \mu(E_{n}) < \frac{\mu(E_{n - 1})}{2}$. This is possible since $X$ contains sets of arbitrarily small positive measure. Set $F_{n} = E_{n} \backslash \cup_{k = n + 1}^{\infty} E_{k}$. Then the $F_{n}$'s are all disjoint and have positive finite measure, so the indicator functions $1_{F_{n}}$ are independent functions in $L^{1}$. Thus, $\dim L^{1} = \infty$.

Proof of 2:

Since $\mu$ is $\sigma$-finite, we can find $(E_{n})_{n \in \mathbb{N}}$ such that $\mu(E_{n}) < \infty$ for all $n$ and $\cup_{n = 1}^{\infty} E_{n} = X$. Let $F_{n} = E_{n} \backslash \cup_{k = n + 1}^{\infty} E_{k}$. Then the $F_{n}$'s are disjoint, have finite measure, and $\cup_{n = 1}^{\infty} F_{n} = X$. Since $\mu(X) = \infty$, there are infinitely many $F_{n}$'s with positive measure. Thus, the indicator functions $1_{F_{n}}$ contain an infinite family of independent functions in $L^{1}$.

Proof of 3:

Any measure which does not contain sets of arbitrarily small positive measure is purely atomic. Thus, there is a countable partition of $X$ into atoms (see wikipedia for slightly more information). Since $\mu(X) < \infty$ and there is a lower bound on the measure of any subset of $X$, that countable partition of $X$ into atoms is actually a finite partition. The indicator functions of these atoms is a finite family of functions that spans $L^{1}$, so $\dim L^{1} < \infty$.

Title: $L^p$ space: measure theory