问:
How to be sure about any parameterization?
This is the problem:

Find intersection of this two surfaces and then calculate its length.
$$x^2+4y^2=4$$
and

$$y\sqrt3 +z=1$$
Method 1 :

Common way is to take:

$$x=2\cos t  ,\:  y=\sin t ,\: z=1-\sqrt3 \sin t$$

So

$$r(t) =(2\cos t,\:\sin t,\:1-\sqrt3 \sin t)$$
and

$$r'(t) = (-2\sin t,\:\cos t,\:-\sqrt3\cos t) ,\: |r'(t)|=2$$

Finally the length is:

$$L = \int_{0}^{2\pi} 2\:dt=4\pi$$

But in another parameterization I find something strange

Method 2:

$$x=2\sqrt{1-t^2} ,\: y=t ,\: z=1-t\sqrt3$$
So

$$r(t) = (2\sqrt{1-t^2} ,\: t ,\: 1-t\sqrt3)$$
and

$$r'(t) = \left(\frac {-2t}{\sqrt{1-t^2}},1,-\sqrt3\right) ,\: |r'(t)|=\frac {2}{\sqrt{1-t^2}}$$ (and that's different from method 1 $|r'(t)|$)

Finally the length

$$L=\int_{0}^{2\pi} \frac {2}{\sqrt{1-t^2}}\:dt\\
=2\int_{0}^{\arcsin{2\pi}}\frac {1}{\cos{\theta}}\cos{\theta}\:d\theta\\
=2\int_{0}^{\arcsin{2\pi}}1\:d\theta$$

and because of $\arcsin(2\pi)$ its undefined it cant be solved.

Now my first question is that:

If we know from past that we can create many parameterizations for a single curve, so why the result of they length is different?

and the second one:

How to be sure that a parameterization is the correct one?

答:
The problem lies in the way you parametrize the curve in the second method.

The reason why you integrate from $0$ to $2\pi$ in the first method is because you have implicitly define you parametrization to go from $0$ to $2\pi$ because you are looking for a mapping $[0;2\pi) \longrightarrow \textrm{your intersection curve}$ such that that mapping is biyective with the entire curve.

If you take a look to what you limits of integrations are in the second method you see they are the same from the first one. That again implicitly define you parametrization to go from $0$ to $2\pi$ but this time (because of the equation $y=t$) you are saying that $y$ runs in the range $[0;2\pi)$ when in reallity in ranges $[-1;1)$ so you actually end up measuring the length of a totally different curve (one with that parametrization but with $y \in [0;2\pi)$ insted of $[-1;1)$).

Now given that $t$ should be in $[-1;1)$ the correct way to measure the lenght in the second method would be:
$$
\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt + \int\limits_{1}^{-1} \frac{2}{\sqrt{1-t²}} \ dt
$$

That is beacuse you can't map the entire curve with just the $[-1;1)$ range (in fact you only map one half of it). Thus the second integral.

So finally you have:

$$
2\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt = 4\pi
$$


问:
How is the formula for the volume of a body in $\mathbb{R}^n$ derived?
Here is a formula for the volume of a general body in $\mathbb{R}^n$ I came across, that uses spherical polar coordinates. I'm not sure how it came about:

Here, $K$ is a body with $0$ in its interior, and for each direction $\theta \in S^{n-1}$, let $r(\theta)$ be the radius of $K$ in that direction.
$$ \text{vol}(K) = nv_n \int_{S^{n-1}} \int_{0}^{r(\theta)} s^{n-1}ds d\sigma = v_n \int_{S^{n-1}} r(\theta)^n d\sigma(\theta)$$
$S^{n-1}$ is the sphere of radius $1$ in $n-1$ dimensions, and $v_n$ is the volume of the Euclidean ball of radius $1$ in $n$-dimensions.

I understand how the first integral goes to the second, but how did this formula come about in the first place?

I'm trying to start off with the regular Cartesian integral for volume which looks like $$\int \int ... \int dx_1 dx_2...dx_n$$ but I'm not able to proceed from here.

It has been a long while since I posted this. Could someone please add a detailed explanation? It would be really helpful.

答:
It is known from the Real Variables theory, see e.g. Theorem 2.49 in Folland's Real Analysis, that there exists a unique Borel measure $d\sigma$ on the unit sphere $\mathbb S^{d-1}=\{x\in\mathbb{R}^d:|x|=1\}$ such that
\begin{align}\tag{1}\label{eq:1}
\int_{\mathbb{R}^d} g(x) \,dx = \int_0^\infty \int_{\mathbb S^{d-1}} g(ry) \, d\sigma(y) \,  r^{d-1} dr.
\end{align}
Then the volume of the convex body $K$ can be computed as follows:
\begin{align*}
vol(K)&=\int_{\mathbb{R}^d} \chi_K(x) \,dx = \int_0^\infty \int_{\mathbb S^{d-1}} \chi_K (ry) \, d\sigma(y) \,  r^{d-1} dr
\\
=&\int_{\mathbb S^{d-1}}\int_0^\infty \chi_K (ry)\,  r^{d-1} dr \, d\sigma(y).
\end{align*}
Note that for a fixed unit vector $y$ we have $\chi_K (ry)=1$ iff $r\le r(y)$. Therefore, we can further write
\begin{align*}
=&\int_{\mathbb S^{d-1}}\int_0^{r(y)}  r^{d-1} dr \, d\sigma(y)=\frac{1}{d}\int_{\mathbb S^{d-1}} r(y)^{d} \, d\sigma(y),
\end{align*}
which gives the desired formula, up to normalization (in the above computation $\sigma$ is such that $\sigma(\mathbb S^{d-1})=\frac{2\pi^{d/2}}{\Gamma(d/2)}$).

ContentLicense: CC BY-SA 4.0


问:
Two cones intersecting in an ellipse (implicit form)
I have two cones with different opening angles, both open two the top and both with their tip on two different points on the xy-plane.

Two cones with those properties always intersect in a curve which, projected vertically onto the xy-plane, is a circle.

This property (the projection being a circle) is what I'm trying to prove.

My approach so far is, to get the implicit form for the intersection curve, projecting this onto the xy-plane and then showing that the result is circular.

The two cones are given in implicit form. Let the cones be called $C_1$ and $C_2$, the points of their tips $a_1, a_2$ and the parameter of their opening angle $b_1, b_2$. The implicit formula describing a cone is then:

$C_n: \{p \in \mathbb{R^3}: (x - a_{nx})^2 + (y - a_{ny})^2 - (b_1 \cdot z)^2 = 0, z > 0\}$

I'm now looking for the implicit formula of the curve describing their intersection to then project this curve back onto the xy-plane. How do I get there? I tried subtracting the formulas of the two cones:

$$(x-a_{1x})^2 + (y-a_{1y})^2 - (b_1z)^2) - [(x-a_{2x})^2 + (y - a_{2y})^2 - (b_2z)^2] = 0$$
$$\implies (2a_{1x}-2a_{2x})x + (2a_{2y}-2a_{1y})y + (b_2^2-b_1^2)z^2 + (a_{1x}^1 - a_{2x}^2 + a_{1y}^2 - a_{2y}^2) = 0$$

This result describes an extruded parabola - assumingly the bended plane that the curve is lying on.

(Please excuse my english, it's not my mother tongue)

答: You just have to eliminate $z$ from your two equations. So before you subtract them, multiply the first by $b_2^2$ and the second by $b_1^2$. Then $z$ disappears, and you get the equation of a circle in the $(x,y)$-plane.


问:
Why is the entire set $X$ open in every topology
So I've just picked up topology and i read that for any set $X$ the entire set is an element of every topology and so is an open set.
but why is this so?
I understand that the complement of $X$ is $\emptyset$ which is open by vacuous truth however how do we then show that $X$ is open?

Thanks in advance.

答:
It is possible to define topology by means of 'neighborhoods' in the first place.

Then a set $A$ is open if every point of $A$ has a neighborhood $B$ fully in $A$: $B\subseteq A$.

Since all these sets, including the neighborhoods (whatever they are) live inside $X$, the ambient set $X$ will satisfy this definition.

However, writing out the axioms for the neighborhood approach and working rigorously based on them would be more complicated than its cleaned up version that puts the open sets in the first place, and write up their basic properties as axioms.


问:
Minimum of $K(x,y) = 15x + 19y$ if $x^2+7xy+2y^2=3200$ and $x,y\geq 0$
I'm teaching calculus to someone and there is one problem I wasn't able to solve:

Determine the minimum of the function $K(x,y) = 15x+19y$ given that $x^2+7xy+2y^2=3200$ and $x,y\geq 0$

Usually, I'd rewrite the condition until it's of the form x = ... or y = ..., then substitute it into the function K, get a function in one variable and calculate its local extremum(s) by differentiating it and calculating the value for which it is zero. But I don't think it's possible to rewrite the condition in that manner, so any advice on how I would solve this would be appreciated!

The solution is supposed to be x = 18.74, y = 17.20 and I verified that it satisfies the condition.

答:
The corresponding Lagrange function is $L(x,y) = 15x+19y - \lambda \cdot (x^2+7xy+2y^2-3200)$

In order to find a local extremum, we look for values where the partial derivatives are equal to 0, so we get the equations
$15-2x-7y=0$ and $19-7x-4y=0$.
This set of equations has the solutions $x = \frac{73}{41\lambda}$ and $y = \frac{67}{41\lambda}$. Putting these values for x and y into the condition $x^2 + 7xy + 2y^2=3200$ and solving the equation for $\lambda$, we get $\lambda = \frac{\sqrt{1517}}{410}$.

As a result, $x = \frac{73}{41\cdot \frac{\sqrt{1517}}{410}} \approx 18.74$ and $x = \frac{67}{41\cdot \frac{\sqrt{1517}}{410}} \approx 17.2$

With these values, we get that $K(18.74,17.2) = 607.9$.

Now we look at the edge cases:

If $x = 0$, then $3200 = 2y^2 \Rightarrow y = 40 \Rightarrow K(0,40)=760$

If $y = 0$, then $3200 = x^2 \Rightarrow x = \sqrt{3200} \Rightarrow K(\sqrt{3200},0)\approx 848.53 $

In both cases, we ignored the negative solutions for x and y because they are both non-negative.

Since the values of $K$ at the edges are bigger than $K(18.74,17.2)=607.9$, the local minimum (18.74, 17.2) is the global minimum.


问:
What does a non-dimensional change of variable means in $\frac{dT}{dt} = 6.35397 \times 10^{-9} \frac{T_F^4 - T^4}{355.205 + 0.200822 T} $?
I have the following ODE:

$$\frac{dT}{dt} = 6.35397 \times 10^{-9} \frac{T_F^4 - T^4}{355.205 + 0.200822 T} $$

with the condition $T(0)=300$ and $T_F = 1200$.

A non-dimensional change of variable $\gamma = \frac{T}{T_F}$ is done resulting in:

$$\frac{d\gamma}{dt} = 10.980 \frac{1 - \gamma^4}{355.205 + 240.96\gamma}$$

I don't get the idea of what does a non-dimensional change of variable means. Neither how to do it. From where did they got to $10.980$? And, why would we want that kind of change in variables if we already have $T_F$? Thanks for the help.

答:
Heuristics for the step size in numerical solvers, if by human intuition or programmed controllers (which are codified human intuition), work best if the problem is transformed close to a standard situation where all variables have scales or ranges of magnitudes close to 1. In that sense, the scale factor for $T$ could also have been $100$ or $1000$.

Another more theoretical goal in non-dimensionalization is to reduce the number of parameters. This makes it easier to compile a catalogue of qualitative properties of solutions of the given class of equations, for instance in a bifurcation analysis. In that direction one coefficient can be cancelled from the denominator, and the leading factor can be removed by rescaling the time axis, so that
$$
\frac{dγ}{d\tau}=\frac{1-γ^4}{1+cγ}
$$
only has one remaining parameter.


问:
Uniqueness of an increasing sequence
I have an increasing sequence $(a_i)_{i=1}^{\infty}$$(a_i\in\mathbb{N})$ , satisfying the conditions

1) If  $a_n$ is a prime number, then $n$ is prime number.

2) For all $n$, $a_{2n}=a_n+n$.

Is $a_n=n$ the only possible sequence?

I haved proved that it is true for strictly increasing sequence. I suppose there is another solution if $(a_i)_i$ is only an increasing sequence, but I fail to construct one. If I assign $a_n$ with a value, I have to be sure that $a_n+n$ is not a prime.

答:
The answer is, perhaps surprisingly, negative if strictly increasing is replaced by only nondecreasing: first of all, let $k_0$ be the smallest known  Sierpiński number, that is, $k_0 = 78557.$ Let $A = \{2^l k_0, l \ge 0\}.$ Define then the sequence $\{a_i\}_{i \ge 1}$ by the following:

If $k \not \in A,$ then $a_k = k;$

If $k \in A,$ we let $a_k = k+1.$

Let us check that it satisfies all properties:

$\{a_i\}$ is nondecreasing: indeed, if $k,l \not \in A,k<l,$ the conclusion $a_k \le a_l$ follows trivially. If both $k,l \in A,$ it follows trivially as well from the way we defined the sequence. Moreover, if $k \in A, l \not \in A,$ then $a_k = k+1 = a_{k+1} \le a_l.$ Also, if $k \not \in A, l \in A,$ we have $a_k = k < l+1 = a_l.$

$a_i$ prime implies $i$ prime: indeed, if $i \not \in A,$ the conclusion is again obvious. If, on the other hand, $i \in A,$ then the definition of a Sierpiński number itself, then $a_i$ is never prime (as it is of the form $2^n k_0 + 1, n \ge 1.$)

$a_{2n} = a_n + n:$ notice that, by the way we defined $A,$ if $2n \in A,$ then $n \in A,$ and vice-versa. If $n \not \in A,$ the property is once more trivial. If $n \in A,$ then $a_n = n+1, a_{2n} = 2n+1 = (n+1) + n = a_n + n,$ as desired.

As a final comment, notice that it is known that there are infinitely many Sierpiński numbers, and doing such a construction with each of them yields a different solution. Also quite related and, in some sense, analogous is the concept of a Riesel number, which also yields solutions of this problem.

Title: Uniqueness of an increasing sequence


问:
Observation and conjecture on Lychrel numbers
A palindromic number (also known as a numeral palindrome or a numeric palindrome) is a number that remains the same when its digits are reversed.

Consider a number $n>0$ in base $b\geq2$, where it is written in standard notation with $k+1$ digits $a_{i}$ as:

$${\displaystyle n=\sum_{i=0}^{k}a_{i}b^{i}}$$with, as usual, $0\leq a_{i}<b$ for all $i$ and $a_{k}\neq0$. Then $n$ is palindromic if and only if $a_{i}=a_{k-i}$ for all $i$. Zero is written $0$ in any base and is also palindromic by definition.

In this context, a Lychrel number is a natural number that cannot form a palindrome through the iterative process of repeatedly reversing its digits and adding the resulting numbers. It is conjectured that $196$ and other numbers that have not yet yielded a palindrome are Lychrel numbers, but no number in base 10 has yet been proven to be Lychrel.

I was curious about the sequence of "candidate Lychrel" numbers, which can be found at https://oeis.org/A023108. I tried to find some kind of pattern among them, and I found the following: every number of the form $99k-2$ is a candidate Lychrel for $1<k<9$. This “curious” fact led me to check if some similar property ocurred for higher candidate Lychrel numbers, and I have checked that every number of the form $999k-1$ is a candidate Lychrel for $1<k<10$ excepting $k=5$, which is a palindromic number; and every number of the form $9999k$ is a candidate Lychrel for $1<k<10$. I have not checked further, only that $99999*2+1$ is also a candidate Lychrel.

Clearly, it can be made the following rough conjecture:

Conjecture. Let it be some number in base $10$ of  the form $n=999...9k-m$. Let us denote with $j$ the number of "9" composing the form of $n$. Then, if $1<{k}<9$, $j\geq2$ and $m=4-j$, $n$ is either a palindromic number or a Lychrel number.

Any guess about the relationship between this clear pattern, and the dynamic of the add-and-reverse-digits process? Any counterexample to the conjecture?

Thanks in advance!

答:
There is a counterexample.

Take $(j,k)=(6,2)$ for which we have
$$n=(10^j-1)k-(4-j)=2000000$$

This is not a palindromic number.

Also, this is not a Lychrel number since under the "add reverse" operation, we have $$2000000+2=2000002$$ which is a palindromic number.

Added :

Let us call a number good if it is either a palindromic number or a Lychrel number.

According to A023108, we have the followings :

For $j=2$, $n=(10^2-1)k-2$ is good for $k=2,3,\cdots, 8$.

For $j=3$, $n=(10^3-1)k-1$ is good for $k=2,3,\cdots, 8,\color{red}{9}$.

For $j=4$, $n=(10^4-1)k-0$ is good for $k=2,3,\cdots, 8,\color{red}{9,10,11}$.

For $j=5$, $n=(10^5-1)k+1$ is good for $k=2$.

Also, we can see the followings :

For $j=6$ and $k=2$, $n=2000000$ is not good.

For $j=7$ and $k=2$, $n=20000001$ is not good.

For $j=7$ and $k=3$, $n=30000000$ is not good.

For $j=8$ and $k=3$, $n=300000001$ is not good.

For $j=8$ and $k=4$, $n=400000000$ is not good.

So, it seems that there are functions $F(j),G(j)$ such that $n=(10^j-1)k-(4-j)$ is good for $j\ge 2$ and $F(j)\le k\le G(j)$.

Something like
$$F(j)=2+(j-5)\left\lfloor\frac j6\right\rfloor,\qquad G(j)=j+6+\left\lfloor\frac j4\right\rfloor$$


答:
Eliminate $z$ from the equations of the cones.

$$x^2+y^2=p^2z^2,\\(x-a)^2+y^2=q^2z^2.$$

You immediately get

$$q^2x^2+q^2y^2=p^2(x-a)^2+p^2y^2$$ or $$(p^2-q^2)x^2+(p^2-q^2)y^2+2ap^2x-a^2p^2=0$$ which is the equation of a circle.

Title: Two cones intersecting in an ellipse (implicit form)

Body:
I have two cones with different opening angles, both open two the top and both with their tip on two different points on the xy-plane.

Two cones with those properties always intersect in a curve which, projected vertically onto the xy-plane, is a circle.

This property (the projection being a circle) is what I'm trying to prove.

My approach so far is, to get the implicit form for the intersection curve, projecting this onto the xy-plane and then showing that the result is circular.

The two cones are given in implicit form. Let the cones be called $C_1$ and $C_2$, the points of their tips $a_1, a_2$ and the parameter of their opening angle $b_1, b_2$. The implicit formula describing a cone is then:

$C_n: \{p \in \mathbb{R^3}: (x - a_{nx})^2 + (y - a_{ny})^2 - (b_1 \cdot z)^2 = 0, z > 0\}$

I'm now looking for the implicit formula of the curve describing their intersection to then project this curve back onto the xy-plane. How do I get there? I tried subtracting the formulas of the two cones:

$$(x-a_{1x})^2 + (y-a_{1y})^2 - (b_1z)^2) - [(x-a_{2x})^2 + (y - a_{2y})^2 - (b_2z)^2] = 0$$
$$\implies (2a_{1x}-2a_{2x})x + (2a_{2y}-2a_{1y})y + (b_2^2-b_1^2)z^2 + (a_{1x}^1 - a_{2x}^2 + a_{1y}^2 - a_{2y}^2) = 0$$

This result describes an extruded parabola - assumingly the bended plane that the curve is lying on.

(Please excuse my english, it's not my mother tongue)


问:
Scaling a parallelogram to fit a triangle
How do I, with straightedge and compass, scale the red parallelogram to the green one, given the blue triangle so that both top corners of the green parallelogram touches the triangle?

答: Start with the red parallelogram, draw on its upper side a yellow triangle similar to blue triangle, i.e., with parallel sides. Extend the sides until they reach the base of the parallelogram extended. Now compare the yellow triangle with the blue triangle. Keeping the same ratio, draw a smaller green parallelogram by first dividing the sides of the blue triangle in the same ratio, as in Euclid Book 6 Prop. 10.


Body:
Eliminating $z$ from

$$
\cases{
(x-x_1)^2+(y-y_1)^2-(b_1 z)^2=0\\
(x-x_2)^2+(y-y_2)^2-(b_2 z)^2=0
}
$$

we have

$$
b_2^2((x-x_1)^2+(y-y_1)^2)-b_1^2((x-x_2)^2+(y-y_2)^2)=0
$$

and after grouping

$$
(b_2^2-b_1^2)x^2+(b_2^2-b_1^2)y^2-2(b_2^2x_1-b_1^2x_2)x-2(b_2^2y_1-b_1^2y_2)y=b_1^2(x_2^2+y_2^2)-b_2^2(x_1^2+y_1^2)
$$

which is a circle equation as far as $\frac{b_1^2(x_2^2+y_2^2)-b_2^2(x_1^2+y_1^2)}{(b_2^2-b_1^2)} > 0$


问:
Continuity of the function $\begin{cases} \frac{x^3 y^3}{x^3+y^3}, \ \text{if} \ x+y \ne 0, \\ 0, \ \text{if} \ x+y=0 \end{cases}$
Study the continuity of
$$\begin{cases} \frac{x^3 y^3}{x^3+y^3}, \ \text{if} \ x+y \ne 0, \\ 0, \ \text{if} \ x+y=0 \end{cases}$$
$f$ is continuous for all $x+y \ne 0$ because it is a ratio of continuous function and the denominator is non zero if $x+y\ne0$, so the only points I must check with the definition are the ones such that $x+y=0 \iff y=-x$.

So let $a \in \mathbb{R}$, I must check if it is true or false that
$$\lim_{(x,y) \to (a,-a)} f(x,y)=0$$
My attempt: let $s:=x-a$ and $t:= y+a$, then $x=s+a$ and $y=t-a$; so I have to study $g(s,t)=f(s-a,t+a)$ for $(s,t) \to (0,0)$, meaning
$$\lim_{(s,t)\to (0,0)} \frac{(s+a)^3(t-a)^3}{(s+a)^3+(t-a)^3}$$
But this limit doesn't exist, since
$$\lim_{s \to 0} g(s,0)=\lim_{s \to 0} \frac{-a^3 (s^3+3s^3a+3sa^2+a^3)}{s^3+3s^3a+3sa^2+a^3-a^3}$$
$$=\lim_{s \to 0} \frac{-a^3 (s^3+3s^3a+3sa^2+a^3)}{s(s^2+3sa+3a^2)}=\begin{cases} -\infty, \ \text{if} \ s \to 0^+ \\ \infty, \ \text{if} \ s \to 0^-\end{cases}$$
So in particular the limit isn't $0$ and so $g$ is not continuous in the point $(0,0)$, so $f$ is not continuous on the point $(a,-a)$; since $a \in \mathbb{R}$ was arbitrary, $f$ is not continuous in $(a,-a)$ for all $a \in \mathbb{R}$.
However my lecturer doesn't use the substitution and he discuss the two separate cases when $a \ne 0$ and $a=0$, so I'm not sure if my approach is correct because maybe I'm missing something when $a=0$ or I should substitute the new variables in the set of definition of $f$ as well. Can someone tell me if my approach is correct and it is equivalent to discuss the two cases $a=0$ and $a \ne 0$? Thanks.

答: In terms of $u=\tfrac{x+y}{2}$ and $v=\tfrac{x-y}{2}$, $x=u+v$ and $y=u-v$, so for $u\ne0$ the function is $\tfrac{(u^2-v^2)^3}{2u(u^2+3v^2)}$, which is asymptotic to $\tfrac{-v^4}{6u}$ if $u\ll v$. So we do not have continuity at $u=0$.


问:
$B^n_2 \subset B^n_K$ implies that $\|x\|_K \le \|x\|$ for all $x\in\mathbb{R}^n$
I want to show that

$B^n_2$ is the Euclidean ball of the unit radius in $\mathbb{R}^n$, and $K$ is a centrally symmetric convex body. Prove that $B^n_2 \subset K$ implies that $\|x\|_K \le \|x\|$ for all $x\in\mathbb{R}^n$. Here, $\|x\|_K$ refers to the norm on which $K$ is the unit ball.

Note: In the title of this question, I have loosely used the notation $B^n_K$ to denote the unit ball under the norm $\|\cdot\|_K$ - this notation is by no means standard.

We know that every centrally symmetric convex body can be represented as the unit ball of some norm on  $\mathbb{R}^n$, and $\|x\|_K$ is precisely that norm for $K$. A proof can be found here. $\|x\|$ refers to the usual Euclidean norm.

How do I show this? $\|x\|_K$ is basically $$p_K(x) = \inf\{t>0, x\in tK\}$$
i.e. the Minkowski functional, so I believe I need to show
$$ \inf\{t>0, x\in tK\} \le \|x\| \text{ for all }x\in\mathbb{R}^n$$

I tried to start off a proof by contradiction, i.e. assume there is some $y\in\mathbb{R}^n$ such that $$ \inf\{t>0, y\in tK\} > \|y\|$$
but I'm not able to take it from here. I'd appreciate any help!

答: Let $||x||=a$. Then $x\in aB_2^{n}\subset aK$ (since containment is preserved under scaling), so we have $x\in aK$ and thus $||x||_k\le a=||x||$.


问:
Convergence of an increasing sequence given convergence of the Cesaro sum
I’ve been playing with Cesaro summation and I’m now stuck on a problem. Given an increasing sequence $(u_n)_{n\in\mathbb{N}}$ of real numbers, define it’s Cesaro sum as $C_n = \frac{1}{n} \sum\limits_{k=1}^n u_k$. I’ve been able to prove a few facts, ie, given any $(u_n)$, if $(u_n)$ converges to $l$ then $(C_n)$ does to, $(u_n)$ bounded implies $(C_n)$ is bounded as well and $(u_n)$ increasing implies $(C_n)$ increasing.

Here since we assume $(u_n)$ increasing we know $(C_n)$ is too.
I’d like to show that if we further assume that $(C_n)$ is convergent, then $(u_n)$ is too.

My guess was to try to show that if $(C_n)$ is convergent, then $(u_n)$ is bounded, and since it’s increasing we can conclude it is convergent. This works well enough for sequences of positive real numbers, but not sure how to make it work when the sequence is necessarily non-negative. I think it’s true, though it’s not obvious why. Any help is appreciated, thank you.

答:
Since $(u_n)$ is increasing, we have for $m < n$

$$C_n = \frac{1}{n}\sum_{k=1}^mu_k + \frac{1}{n}\sum_{k=m+1}^nu_k\geqslant\frac{1}{n}\sum_{k=1}^mu_k + u_m\left(1 - \frac{m}{n}\right)$$

If $C_n \to C$ as $n \to \infty$, then for any fixed $m$

$$u_m = \lim_{n \to \infty}u_m\left(1 - \frac{m}{n}\right) \leqslant \lim_{n \to \infty}C_n - \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^mu_k = C$$

Therefore, $(u_n)$ is bounded ( and convergent since it is an increasing sequence by hypothesis).


答:
Indeed when $a=0$ you have that
$$
\lim_{s \rightarrow 0} \frac{-a^3(s^3+3s^2a+3sa^2+a^3)}{s(s^2+2sa+3a^2)}
$$

turns into:

$$
\lim_{s \rightarrow 0} \frac{-0³(s^3+3s^2(0)+3s\left(0^2\right)+0^3)}{s(s^2+2s(0)+3\left(0^2\right))} = \lim_{s \rightarrow 0} \frac{0}{s(s^2)} = 0
$$

And no contradiction arises from this calculation

Title: Continuity of the function $\begin{cases} \frac{x^3 y^3}{x^3+y^3}, \ \text{if} \ x+y \ne 0, \\ 0, \ \text{if} \ x+y=0 \end{cases}$

Body:
Study the continuity of
$$\begin{cases} \frac{x^3 y^3}{x^3+y^3}, \ \text{if} \ x+y \ne 0, \\ 0, \ \text{if} \ x+y=0 \end{cases}$$
$f$ is continuous for all $x+y \ne 0$ because it is a ratio of continuous function and the denominator is non zero if $x+y\ne0$, so the only points I must check with the definition are the ones such that $x+y=0 \iff y=-x$.

So let $a \in \mathbb{R}$, I must check if it is true or false that
$$\lim_{(x,y) \to (a,-a)} f(x,y)=0$$
My attempt: let $s:=x-a$ and $t:= y+a$, then $x=s+a$ and $y=t-a$; so I have to study $g(s,t)=f(s-a,t+a)$ for $(s,t) \to (0,0)$, meaning
$$\lim_{(s,t)\to (0,0)} \frac{(s+a)^3(t-a)^3}{(s+a)^3+(t-a)^3}$$
But this limit doesn't exist, since
$$\lim_{s \to 0} g(s,0)=\lim_{s \to 0} \frac{-a^3 (s^3+3s^3a+3sa^2+a^3)}{s^3+3s^3a+3sa^2+a^3-a^3}$$
$$=\lim_{s \to 0} \frac{-a^3 (s^3+3s^3a+3sa^2+a^3)}{s(s^2+3sa+3a^2)}=\begin{cases} -\infty, \ \text{if} \ s \to 0^+ \\ \infty, \ \text{if} \ s \to 0^-\end{cases}$$
So in particular the limit isn't $0$ and so $g$ is not continuous in the point $(0,0)$, so $f$ is not continuous on the point $(a,-a)$; since $a \in \mathbb{R}$ was arbitrary, $f$ is not continuous in $(a,-a)$ for all $a \in \mathbb{R}$.
However my lecturer doesn't use the substitution and he discuss the two separate cases when $a \ne 0$ and $a=0$, so I'm not sure if my approach is correct because maybe I'm missing something when $a=0$ or I should substitute the new variables in the set of definition of $f$ as well. Can someone tell me if my approach is correct and it is equivalent to discuss the two cases $a=0$ and $a \ne 0$? Thanks.


答:
Just observe that, for any given $x$
$$
\{t: x\in tB\}\subset\{t: x\in tK\}
$$
where $B$ is the Euclidean unit ball (because $tB\subset tK$). Then the inf of the right hand side is less than or equal to the one of the left one.

Title: $B^n_2 \subset B^n_K$ implies that $\|x\|_K \le \|x\|$ for all $x\in\mathbb{R}^n$

Body:
I want to show that

$B^n_2$ is the Euclidean ball of the unit radius in $\mathbb{R}^n$, and $K$ is a centrally symmetric convex body. Prove that $B^n_2 \subset K$ implies that $\|x\|_K \le \|x\|$ for all $x\in\mathbb{R}^n$. Here, $\|x\|_K$ refers to the norm on which $K$ is the unit ball.

Note: In the title of this question, I have loosely used the notation $B^n_K$ to denote the unit ball under the norm $\|\cdot\|_K$ - this notation is by no means standard.

We know that every centrally symmetric convex body can be represented as the unit ball of some norm on  $\mathbb{R}^n$, and $\|x\|_K$ is precisely that norm for $K$. A proof can be found here. $\|x\|$ refers to the usual Euclidean norm.

How do I show this? $\|x\|_K$ is basically $$p_K(x) = \inf\{t>0, x\in tK\}$$
i.e. the Minkowski functional, so I believe I need to show
$$ \inf\{t>0, x\in tK\} \le \|x\| \text{ for all }x\in\mathbb{R}^n$$

I tried to start off a proof by contradiction, i.e. assume there is some $y\in\mathbb{R}^n$ such that $$ \inf\{t>0, y\in tK\} > \|y\|$$
but I'm not able to take it from here. I'd appreciate any help!


问:
Improper Integral $\int_{1/e}^1 \frac{dx}{x\sqrt{\ln{(x)}}} $
I need some advice on how to evaluate it.
$$\int\limits_\frac{1}{e}^1 \frac{dx}{x\sqrt{\ln{(x)}}} $$
Thanks!</p>

答:
Here's a hint:
$$
\int_{1/e}^1 \frac{1}{\sqrt{\ln x}} {\huge(}\frac{dx}{x}{\huge)}.
$$
What that is hinting at is what you need to learn in order to understand substitutions.  It's all about the chain rule.  The part in the gigantic parentheses becomes $du$.</p>

ContentLicense: CC BY-SA 3.0


问:
Example of equivalence relations $S$ and $R$ such that $SR$ is an equivalence relation while $R \cup S$ isn't
I need to give an example of equivalence relations $S$ and $R$ such that $SR$ is an equivalence relation while $R \cup S$ isn't. I've already tried to find such relations on small sets of natural numbers but it looks like there aren't any.

答:
Consider the set $\{a,b,c,d\}$, with the equivalence relations
$$\alpha = \{a,b\}^2 \cup \{c,d\}^2
\quad\text{and}\quad
\beta = \{a,d\}^2 \cup \{b,c\}^2.$$
It's easy to see that $\alpha \cup \beta$ is not an equivalence relation, because it's not transitive: for example, $(a,b), (b,c) \in \alpha \cup \beta$, but $(a,c) \notin \alpha \cup \beta$.

Obviously, $\alpha \cup \beta \subseteq \alpha \circ \beta$, since the relations are reflexive.

From
$$
a \equiv_{\alpha} b \equiv_{\beta} c
\quad\text{and}\quad
c \equiv_{\alpha} d \equiv_{\beta} a,
$$
we conclude that $(a,c),(c,a) \in \alpha \circ \beta$.

Analogously we conclude that $(b,d),(d,b) \in \alpha \circ \beta$, and therefore, $\alpha \circ \beta = \{a,b,c,d\}^2$, an equivalence relation.


问:
Closure of difference of varieties is ideal quotient
For ideals $J_1,J_2$ of a commutative ring $R$, the ideal quotient is defined by $(J_1:J_2)=\{f\in R\mid fJ_2\subset J_1\}$. Suppose $X$ is an affine variety. I am trying to understand why
if $J_1$ and $J_2$ are radical ideal in the coordinate ring $A(X)=k[x_1,...,x_n]/I(X)$ (where $k$ is an algebraically closed field), then $\overline{V(J_1)-V(J_2)}=V(J_1:J_2)$.

The only tools I can think of to use are the fact that $\overline{S}=V(S)$ and the Nullstellensatz. But I still can't see why it's true. What am I missing?

答:
Suppose $f(x)=0$ for all $x\in W:=V(J_1)-V(J_2)$. Then for all $g\in J_2$ we have $fg(x)=0$ for all $x\in V(J_1)$, so $fg\in J_1$, and hence $f\in(J_1:J_2)$. Thus $I(W)\subseteq(J_1:J_2)$, and so $\overline W\supseteq V(J_1:J_2)$.

Conversely, if $f\in(J_1:J_2)$ and $x\in W$, then $fg(x)=0$ for any $g\in J_2$. In particular, since $x\not\in V(J_2)$, we know that there exists $g\in J_2$ with $g(x)\neq0$, and hence $f(x)=0$. Thus $W\subseteq V(J_1:J_2)$, so $\overline W\subseteq V(J_1:J_2)$.


问:
Limt and Convergence of Integral

If $\lim_{x \to \infty} x f(x)=0$, then $\int_a^\infty f(x) dx$ converges, where $a>0$. Is this statement true or is there any counterexample? I have seen the converse of this statement which is true when the function is monotonic.

If $\lim_{x \to \infty} f(x)=1$, then $\int_a^\infty f(x) dx$ may or may not converge. It is easy to find an example where the function diverges: $f(x)=1+\frac{1}{x}$. Is there any example where the function converges?

答:
Example 1

Take $f(x) = \frac{1}{x \ln x}$ as a counterexample. $f$ is monotonic by the way.

Example 2

The integral will always be divergent as $f(x) \gt 1/2$ for $x$ large enough.

Title: Limt and Convergence of Integral


问:
Linear equation problem (from text)
Tickets to attend a tennis tournament cost  $10,  20$ or $40$. The sales recipe of $10,000$ tickets is $180,000$.

From this information, write the system of equations that would allow you to determine the number of tickets sold in each category.

I have only found the following equation: $10x+20y+40z = 180000$

However, two other equations are missing and I am not able to find them, can you help me please?

答:
First of all, the problem can be simplified by noting that each ticket price is a multiple of 10.  Therefore, you can pretend that the ticket prices have all been divided by 10.

You now have the following three equations:

$x + y + z = 10,000.$

$x + 2y + 4z = 18,000.$

$x,y,z$ are all non-negative integers.

Subtracting (1.) from (2.) above, gives $y + 3z = 8000$.

Since $8000  \equiv 2\pmod{3}$, you know that $y \equiv 2\pmod{3}.$

Further, since $x,y,z$ are all non-negative integers, you know that $y$ must be an element in $S = \{2,5,8, \cdots, 7997, 8000\}.$

If I am not mistaken, each of the $2,667$ possible values of $y$ [re the set S has $2,667$ elements] will yield a (valid) distinct $(x,y,z)$ solution.

An alternative formulation is to recognize that $z \in \{0,1,2, \cdots, 2666\},$ and that once $z$ is chosen, you have that

$y = (8000 - 3z).$

$x = (10,000 - y - z).$

Title: Linear equation problem (from text)


问:
if $\lim _{x\rightarrow \infty}{f'(x)}=0$ then does $\lim_{x \rightarrow \infty}{f(x)}$ exist in the broad sense
Let $f$ be a differentiable a function in $\mathbb{R}$, and let $\lim _{x\rightarrow \infty}{f'(x)}=0$

Does $\lim_{x \rightarrow \infty}{f(x)}$ exist in the broad sense?

I'm really lost here. This exercise is from a section on MVT, and intuitively it seems to be correct, but I can't seem to find a lead. If someone could just give me a hint that would be great.

So far my best shot has been using Heine's definition of the limit, but no dice.

答:
Intuitively, what you want here is a function that iterates between (at least) two values and also gets straighter as time passes.

You can think of the usual sine function as a spring, and imagine stretching it to the right. Try to understand why this make the function get straight as $x$ increases.

Consequently, $\sin(\sqrt x)$ would work.

I suggest you draw it (using a graphing website) to see why.


问:
distribution of X*Y with X and Y both i.i.d. with a U(0,1) distribution
I have two random variables $X$ and $Y$ both being i.i.d. with a $U(0,1)$ distribution. I am trying to calculate the cumulative distribution function of $Z=X*Y$. Which would be the most efficient way to do this? Until now, I tried t put something together from here and try a similar approach: Finding distribution function of $Y/X$ and probability density function of $X+Y$

However, I am really unsure if I can work similarly in my case.
So this is my approach:

$$
F_{Z}(z) = P(X*Y \leq z) = \int_0^1 P \left( Y \leq \frac{z}{x} \right)*\frac{1}{f_X(x)}dx
$$

Is this approach still correct? Furthermore I am quite unsure how to continue from here, so any help is really appreciated.

答:
The law of total probability states that
$$P(XY \leq z) = \int_0^1 P(XY \leq z \: | \: X=x) f_X(x) \: dx,$$
where $f_X(x)=1$ for $x \in [0,1]$, since $X \sim \operatorname{Uniform}(0,1)$, and by independence we get that
$$P(XY \leq z \: | \: X=x) = P(xY \leq z \: | \: X=x) = P(Y \leq \frac{z}{x}).$$
Now by definition of the uniform CDF we get, that
$$P(Y \leq \frac{z}{x}) = \begin{cases} 1 & \text{ if $\frac{z}{x} \geq 1$} \\
                                        \frac{z}{x} & \text{ if $\frac{z}{x} \in (0,1)$} \\
                                        0           & \text{ if $\frac{z}{x} \leq 0$ }
   \end{cases}.$$
And since we only need to consider $z,x \in (0,1)$ we get, that $P(Y \leq \frac{z}{x})=1$ for $x \leq z$ and $P(Y \leq \frac{z}{x}) = \frac{z}{x}$ for $x> z$. Plugging this into the integral we get that
\begin{align*} P(XY \leq z) &= \int_0^z \: dx + \int_z^1 \frac{z}{x} \: dx \\
&= z - z \ln(z) \\
&= z \ln(\frac{e}{z})
\end{align*}


问:
What would synthetic linear algebra look like?
I'm aware of synthetic mathematical fields like synthetic differential geometry and synthetic topology where the area is developed axiomatically rather than deriving everything analytically from a foundation of sets.

So I'm wondering if a synthetic linear algebra would make sense or if that's trivial. Is linear type theory essentially the same thing?

答: It would look like something out of principia. Instead of vectors being coordinated, one would have to define linear operators in terms of basic "geometric algebra".</p>


问:
Prove that matrix equation $AX-XA=I$ doesn't have a solution for any $A\in M_n(\mathbb{R})$
Prove that matrix equation $AX-XA=I$ doesn't have a solution for any $A\in M_n(\mathbb{R})$

If $n=2,$ $$A=
        \begin{bmatrix}
        x & y \\
        u & v  \\
        \end{bmatrix}
$$ $$X=
        \begin{bmatrix}
        \alpha \\
        \beta \\
        \end{bmatrix}
$$

$\Rightarrow AX-XA=O\neq I$

What is the more strict proof?

答:
Let $A\in M_n(F)$ where $\operatorname{char}(F)=0$ or $\operatorname{char}(F)>n$. If $[A,X]=I$, by mathematical induction we obtain $[A,X^k] = kX^{k-1}$ for each positive integer $k$. It follows that $[A,f(X)]=f'(X)$ for every polynomial $f$. In particular, if we take $f$ as the minimal polynomial of $X$ and $r=\deg(f)$ (so that $1\le r\le n$), then $f'(X)=[A,f(X)]=[A,0]=0$. But this contradicts the minimality of $m$, because the leading coefficient of $f'$ is $r$, which is nonzero over $F$ as we have either $\operatorname{char}(F)=0$ or $\operatorname{char}(F)>n\ge r$. Hence we conclude that $[A,X]=I$ is not solvable.

While this proof is (obviously) not as simple as the classical proof that relies on taking matrix traces, it illustrates the use of the mapping $X\mapsto[A,X]$ as a derivation, which is the key to the classical proof of Jacobson's lemma.


问:
Question about use of Banach fixed-point theorem
I am working through an exercise book and I have encountered a problem which I have very little idea of what to do, which involves Banach's fixed-point theorem.

Let $f\in C(\mathbb R^n \times \mathbb R^m, \mathbb R^n).$ Further, there exists an $\alpha \in [0,1)$ and to each $\lambda\in\mathbb R^m$ a $q(\lambda) \in [0,\alpha]$ with $$||f(x,\lambda) - f(y,\lambda)||\leq q(\lambda)||x-y||, \quad x,y\in\mathbb R^n .$$
I see here clearly that, through the fixed-point theorem, $f(\cdot, \lambda) $ gives us exactly one fixed point $x(\lambda)\in \mathbb R^n$ for each $\lambda \in \mathbb R^m$.

I am asked to show that $[\lambda \mapsto  x(\lambda)] \in C(\mathbb R^m, \mathbb R^n)$. What does this mean? How do I do this?  I suppose one could use that $\mathbb R^n \times \mathbb R^m$ is $\mathbb R^{m+n}$ but I'm not sure.

答:
The notation just means that the function $x:\mathbb{R}^m\to\mathbb{R}^n$ which maps each $\lambda\in\mathbb{R}^m$ to a point $x(\lambda)\in\mathbb{R}^n$ is a continuous function.

To get start on showing continuity, you might notice that for any two $\lambda\in\mathbb{R}^m$ that $f(x(\lambda),\lambda)=x(\lambda)$ because $x(\lambda_i)$ is defined to be the fixed point. You might also notice that the norm $||(y,\lambda)- (z,\gamma))||_{m}=\max(||y-z||,||\lambda-\gamma||)$ induces the standard topology on $\mathbb{R}^{n+m}$.


答:
To recap all that happened in the comments section:

Based on the initial problem of$$\int\limits_\frac{1}{e}^1 \frac{dx}{x\sqrt{\ln{(x)}}}$$

We perform a u-substitution with $u=\ln{x}$ and $du=\frac{dx}{x}$. Also, the bounds are converted to $\ln\frac{1}{e}=-1$ and $\ln{1}=0$. So we have:

$$\int\limits_{-1}^{0} \frac{du}{\sqrt{u}}=2\sqrt{u}\big|_{-1}^{0}=0-2i=-2i$$

And that's the solution!

Title: Improper Integral $\int_{1/e}^1 \frac{dx}{x\sqrt{\ln{(x)}}} $

Body:
I need some advice on how to evaluate it.
$$\int\limits_\frac{1}{e}^1 \frac{dx}{x\sqrt{\ln{(x)}}} $$
Thanks!</p>


问:
Orthogonal projection onto affine subspaces formula
Essentially what I want to understand is the derivation to this formula.

$π_L(x) = x_0 + π_U (x − x_0)$,

where $π_L$ is the projection mapping to affine space L, $π_U$ is the projection mapping to vector space U.
$L = x_0 + U$

In a 2D/3D space I can visualize how this formula works.

(This image is from the book Mathematics in Machine Learning)

Subtracting $x_0$ from $x$, I can find $π_U(x-x_0)$, the projection of $x-x_0$ onto U, and then use it to find the projection of $x$ onto L, by adding $x_0$ to it.

But I'm struggling to imagine, or derive this formula for vector spaces of more dimensions.

I would appreciate it if someone gave me a derivation (or an idea) for it, or maybe just an intuitive way to understand this concept.

答:
You can think of the term $x-x_0$ as moving the origin to $x_0$. Then the algorithm can be described in three steps.
Move the origin to $x_0$ so that the plane goes through the origin,
calculate the linear orthogonal projection onto the plane,
and finally move the origin back to $0$.

These steps are applied right to left in the formula. First, calculate $x_0-x$ to move the origin, then project onto the now linear subspace with $\pi_U(x-x_0)$ and finally move the origin back with $x_0 + \pi_U(x-x_0)$. Moving the origin back and forth without the projection would be the algorithm $x_0 + (x - x_0)=x$. In general in an affine space you can linearize it by choosing an origin $x_0$ and using $x-x_0$ as vectors since $x_0-x_0=0$.


问:
Symmetries of a graph with unconnected vertex
How do you determine the number of symmetries in the following graph:

https://i.stack.imgur.com/gviNB.png

I know how to determine the number of symmetries of this graph if the two middle vertices weren't there. How does an unconnected vertex change the number of symmetries?

Thanks

答:
You have two choices: leave those unconnected vertices alone, or swap them. This effectively doubles the total number of symmetries. In other words, if $\sigma$ is any symmetry of the main connected component, then a symmetry of the whole graph will be $\sigma e$ or $\sigma\pi$, where $e$ is the identity on the two vertices, and $\pi$ swaps them.

In general, if you had $n$ unconnected vertices, there would be $n!$ ways to permute them.


问:
Find $f$ such as its eigenvector associated to a given eigenvalue $-2$ is $v=(-1,1,1)$
Let $f:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ be an endomorphism such as

$f(1,0,1)=(1,2,1)$

$f(1,1,0)=(2,0,0)$

and $v=(-1,1,1)$ is an eigenvector respect to the eigenvalue $-2$

$1)$ Describe the $Im(f)$ and $Ker(f)$

$2)$ Describe explicitly $f$

$3)$ Determine if $f$ is diagonalizable

I've been give this exercise and I'm wondering if it lacks information in order to be solved or if it's me who lacks of information. Since I don't have a basis of the transformation defined and I can't either arbitrarily finish to define the function $f$ because must be a function that satisfies that the condition that $v=(-1,1,1)$ is an eigenvector. For the other side, I know that since $-2$ is eigenvalue satisfies that $(A+2Id)v=0$ but I don't know how to deduce the matrix of the transformation with this data! What am I missing here?

I'll be thankful if someone helps me!

Body: You know that$$f(1,0,1)=(1,2,1)\text{, that }f(1,1,0)=(2,0,0)\text{, and that }f(-1,1,1)=(2,-2,-2).$$Since $\{(1,0,1),(1,1,0),(-1,1,1)\}$ is a basis of $\Bbb R^3$, you know all that you need to know in order to work with $f$. For instance, since$$(1,0,0)=\frac13(1,0,1)+\frac13(1,1,0)-\frac13(-1,1,1),$$you know that$$f(1,0,0)=\frac13(1,2,1)+\frac13(2,0,0)-\frac13(2,-2,-2)=\left(\frac13,\frac43,1\right),$$an so on.


问:
Minimum of $K(x,y) = 15x + 19y$ if $x^2+7xy+2y^2=3200$ and $x,y\geq 0$
I'm teaching calculus to someone and there is one problem I wasn't able to solve:

Determine the minimum of the function $K(x,y) = 15x+19y$ given that $x^2+7xy+2y^2=3200$ and $x,y\geq 0$

Usually, I'd rewrite the condition until it's of the form x = ... or y = ..., then substitute it into the function K, get a function in one variable and calculate its local extremum(s) by differentiating it and calculating the value for which it is zero. But I don't think it's possible to rewrite the condition in that manner, so any advice on how I would solve this would be appreciated!

The solution is supposed to be x = 18.74, y = 17.20 and I verified that it satisfies the condition.

答:
Hint

Let $p=15x+19y\implies y=?$

Replace this value of $y$ in $$x^2+7xy+2y^2-3200=0$$ to form a quadratic equation in $x$

As $x$ is real, the discriminant must be $\ge0$


问:
Question about angles in a triangle having a median line

If PS = SR and x = y, is it true that angle QSR must be 90 degrees?

More generally, if PS = SR, is it true that x+y must equal 90?

I have tried drawing circumcircles, expanding the triangle into a parallelogram, and drawing an altitude from S onto QR.  I can't seem to prove QSR is 90.  Which makes me suspect it doesn't have to be.  On the other hand, I can't seem to visualize a triangle (or draw one) where it is clear that PS=SR and x=y, but angle QSR is not 90.  I also feel like using the laws of sines/cosines doesn't prove helpful.

If angle QSR is 90 and PS=SR, x need not equal y.  If angle PQR is 90 and PS=PR, x need not equal y.  If angle PQR is 90 and PS=SR and x=y, then, yes, angle QSR is 90.  There's a lot of combinations of givens and unknowns we can consider.  But for the particular 2 givens, PS=SR and x=y, and the unknown, angle QSR, I seem to be at a loss.  Any thoughts?

EDIT: I have solved this.  The answer to both questions is "no."  For a fixed value of y and a fixed length of QR, x can be written as a function of the length of PR, call it b.  Then taking the example of the fixed values y=30 and QR=2/sqrt(3), as b increases from 2, x strictly decreases from 60 (to 0), and angle QSR decreases from 90 degrees.  Then at some value of b>2 (approximately b=3.1), x will be equal to 30, while QSR will be less than 90.  Thus, there exists a triangle where PS=SR and x=y, but angle QSR is less than 90 degrees.  The same example shows that x and y can both be equal to 30, and thus x+y need not equal 90.

答:
No to everything that you have asked because there are too many free variables to force that in both cases. However, if you add the hypothesis PQR is a right angle then yes to everything you have asked.

It kind of looks like a right angle in the picture but you do not specify it.

You can use the sum of angle formula and law of sines formula to obtain the possible restrictions.

So:

$$QSR + QSP = 180$$

$$RQS + QSR + y = 180$$

$$QPS + QSP + x = 180$$

$$x + y + RQS + QPS =180$$

Gauss Elimination gives:

$$x + QSP + QPS = 180$$

$$y + QSR + RQS = 180$$

$$QSR + QSP = 180$$

The law of sines gives us:

$$sin(x)sin(y) = sin(QPS)sin(RQS)$$

This will leave two free variables. With the extra equation x=y we will have one free variable. And get the extra equation after elimination:

$$RQS - 2QSP - QPS = -180$$

To simplify we can use the product of sines trig identity and then linear substitution to transform the last formula:

$$2sin(x)sin(y) = cos(QPS-RQS) - cos(QPS+RQS) = cos( 180 -x -y - 2RQS ) - cos(180 - x - y)$$

Then in the first case we can go further:

$$2sin^2(x) = cos(180 - 2x -2RQS) - cos(180 -2x) = cos(2x) - cos(2*(x+RQS)) = $$
$$ = 1 - 2sin^2(x) -1 + 2sin^2(x+RQS)$$

Simplifying: $$2sin^2(x) = sin^2(x+RQS)$$

Which has solutions when

$$RQS = π - x - arcsin(sqrt(2) sin(x))$$ in radians.

Letting x be 30 degrees, We get that RQS is 105 degrees and y is forced to be 30 degrees making QSR 45 degrees. But there are infinitely many choices for x that work, it is a free variable in other words.

For the second question 30 degrees plus 30 degrees is 60 degrees not 90 so it is false.

If you enforce the extra condition that you did not mention that PQR is a right angle then the first statement is true which should be clear from the simplified form since

$$sin^2(x+RQS) = sin^2(RPQ) = 1.$$

And for the second statement continuing where we left off.

$$2sin(x)sin(y) = cos(QPS-RQS) - cos(QPS+RQS) = $$

$$= cos( 180 -x -y - 2RQS ) - cos(180 - x - y) = cos( x - y ) - cos(180 - x - y)$$

Simplifying: $$2sin(x)sin(y) = cos (x - y) + cos( x + y ) = 2cos(x)cos(y)$$

Which will happen when $$cos(x + y) = 0$$

Which means that either x+y = 90 degrees or x+y =270 degrees, but the second is too large for the interior of a triangle, and so under the additional constraint the more general question is true.


问:
Finding a parametrization of a curve from cartesian equations
I'm in need of some help to address this problem.

Let $C$ be a curve given by two equations:

$x^2+y^2-z^2-1=0$,

$x^2-y^2-z^2-1=0$

Express the curve by means of parametric equations.

Any ideas on how to work on it?

答:
If you take the first restriction and plug it into the second, you get

$$
x^2-y^2-(x^2+y^2-1) - 1 =0 \Leftrightarrow -2y^2 = 0 \Leftrightarrow y=0
$$

So, any point on the curve satisfies $y=0$ and $x^2-z^2 = 1$ and you can consider two separate parametrizations:

$$
(\sqrt{t^2+1}, 0, t), t\in \mathbb{R}
$$

$$
(- \sqrt{t^2+1}, 0, t), t\in \mathbb{R}
$$


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

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

答: The complex number $i$ is the representation of an angle of $\pi/2$ between the real axis and imaginary axis and $-i$ mean that the angle is $-\pi/2$. So there is a huge difference between $i$ and $-i$ as they have $\pi$ radians phase difference.</p>

Title: What is $-i$ exactly?


问:
Topological groups: if $A$ is an open subset of $G$, then $A^{-1}$ is open.
Let $G$ be a topological group with open subset $A$. I want to prove very thoroughly/precisely that   $A^{-1}:=\{a^{-1}:a\in A\}$ is also open.

Obviously the key to this proof is the fact that the inversion map $\phi:G\longrightarrow G$ is continuous.

$\phi:G\longrightarrow G$ is continuous, so since  $A\subseteq G$ is open in $G$, $\phi^{-1}(A)$ is open in $G$ (by definition of topological continuity).

$\phi^{-1}(A)$ consists of elements of the form $a^{-1}$, so $\phi^{-1}(A)=A^{-1}$. This looks like we are done.

I just want to check a small detail: we take $A$ to be in the codomain of $\phi$, which of course is $G$, but in particular, shouldn't the elements of $A$ be of the form "inverses of some elements of $G$", so $A=B^{-1}$ for some set $B\subseteq G$, say, and therefore $\phi^{-1}(A)=\phi^{-1}(B^{-1})=({B^{-1}})^{-1}=B$, So we do not necessarily obtain $A^{-1}$ back when we apply $\phi^{-1}$, unless $\phi^{-1}$ is actually an inverse of $\phi$, as opposed to a preimage?

I think I might be getting confused here. Could anyone help clarify on this?

Thanks

答:
The steps in your argument are correct, but I think you've confused yourself by not justifying the key step.

You say "$\phi^{-1}(A)$ consists of elements of the form $a^{-1}$" -- why is this? Let's prove it.

Let $x \in \phi^{-1}(A)$ be arbitrary. By definition, $x^{-1} = \phi(x) \in A$. Then $x = (x^{-1})^{-1}$ is the inverse of an element of $A$. Since $x$ was arbitrary, we have $\phi^{-1}(A) \subseteq A^{-1}$. Conversely, for any $a \in A$, $\phi(a^{-1}) = (a^{-1})^{-1} = a \in A$, so $a^{-1} \in \phi^{-1}(A)$. This shows $A^{-1} \subseteq \phi^{-1}(A)$, completing the proof.

Here's another way to do this: just note that $\phi$ is its own inverse! Thus $\phi^{-1}(A) = \phi(A) = A^{-1}$.


问:
Given a Biased Coin yielding a Flip Sequence, what is the expected number of even palindromes?
This is from a review of an exam.

You continue to play with one of your biased Vegas quarters by flipping it $2k > 0$ times in
a row. You write down these flips as a string of H and T. Recall from the earlier problem
that the probability any flip is H is $2/3$.

(a) Give an expression for the probability that the flip sequence is an even palindrome.

This should certainly be an expression written in terms of $k$.

(b) Even palindromes occur as a substring five times in the flip sequence TTHHTT: once
as the whole string TTHHTT, once each with the substrings THHT and HH. And then
there are two occurrences of TT, one starting at the first flip and one starting at the
fifth flip.
Give an expression for the expected number of even palindromes that occur as substrings within $2k > 0$ flips of the biased coin.

答:
For the part (a), it is useful to calculate how often two independent throws land with the same value. This is $(2/3)^2+(1/3)^2=5/9$. Now, if the coin was thrown $2k$ times, it will be an even permutation if and only if the pairs of throws $(k,k+1), (k-1,k+2),\ldots,(1,2k)$ all coincide, which happens with probability $(5/9)^k$.

For (b), I will not do the work all the way up to the end, but will show a hint what needs to be done. We will try to use linearity of expectation. For two indices $i,j$ such that $1\le i<j\le 2k$ and $j-i$ is odd, we will first calculate the expected number of palindromes starting precisely at $i$ and ending precisely at $j$. Because there can be only one or zero of those (depending precisely on whether this subsequence is a palindrome or not), the expectation is, as per (a): $E_{ij}=(5/9)^\frac{j-i+1}{2}$.

Thus, the total expectation of the number of even palindromes is:

$$E=\sum_{\begin{array}{c}i,j\\1\le i<j\le 2k\\j-i\text{ odd}\end{array}}E_{ij}=\sum_{\begin{array}{c}i,j\\1\le i<j\le 2k\\j-i\text{ odd}\end{array}}(5/9)^\frac{j-i+1}{2}$$

Now all that it takes is to do the summation, which can be done in various ways.


问:
Lipschitz-continuity of matrix-valued function
I would like to show the Lipschitz-continuity of a matrix-valued function $\sigma$.

Suppose $\sigma:\mathbb{R}^d\to\mathbb{R}^{d\times d}, \beta\mapsto \sqrt{\phi(\lVert \beta \rVert_2)}I_{d\times d} $ where the function
$$ \mathbb{R}\to\mathbb{R}, z\mapsto \sqrt{\phi(z)} $$
is Lipschitz-continuous.

I would like to show Lipschitz-continuity in the respective $\lVert \cdot \rVert_2$-norms. But for $x, y \in \mathbb{R}^d$ we have
$$
\lVert \sigma(x)-\sigma(y) \rVert_2 =  \max\{|\lambda| : \lambda \mathrm{~eigenvalue~of~} (\sigma(x)-\sigma(y)(\sigma(x)-\sigma(y)^T\} = \sqrt{ | \phi(\lVert x \rVert_2) - \phi(\lVert y \rVert_2)|}.
$$

I don't see, how the Lipschitz-continuity of $\sqrt{\phi(z)}$ is helping me, but the author seems to use it to conclude the L-cont. of $\sigma$. Could someone explain? Thanks a lot!

答:
By the Lipschitz-continuity of $\sqrt{\phi}$ there is some $L>0$ with $$|\sqrt{\phi(s)}-\sqrt{\phi(t)}|\leq L|s-t|$$ for all $s,t\in\Bbb R$. Hence we get for $x,y\in\Bbb R^d$ by the reverse triangle inequality: $$\|\sigma(x)-\sigma(y)\|=|\sqrt{\phi(\|x\|_2)}-\sqrt{\phi(\|y\|_2)}|\leq L\big|\|x\|_2-\|y\|_2\big|\leq L\|x-y\|$$
Hence $\sigma$ is Lipschitz-continuous. (Basically the composition of Lipschitz-continuous functions is again Lipschitz-continuous, here we have the functions $\Bbb R^d\to\Bbb R,x\mapsto\|x\|$, $\Bbb R\to\Bbb R,r\mapsto\sqrt{\phi(r)}$ and $\Bbb R\to\Bbb R^{d\times d},s\mapsto sI$.)


问:
How to find the derivative of $f(x)=\int_{-2020}^{x^{3}} \sin \left(\pi t^{2}\right) d t$?
By FTC, if the upper bound in the integral was simply $x$ then it would simply be $f'(x)=\sin \left(\pi t^{2}\right)$ right? But the upper bound is $x^3$, so how can I find the derivative?

答:
Hint :
$$f = g \circ h$$

where
$$g(x)=\int_{-2020}^{x} \sin \left(\pi t^{2}\right)dt \quad \text{and} \quad h(x)=x^3$$


问:
Equation for a circle that occurs when a plane goes through a sphere
So the problem I have is that I have a sphere $$x^2+y^2+z^2=1$$ and a plane $$x+y+z=0$$ that goes through the center of the sphere and creates a circle. Now, what I tried was plugging in $$z=-x-y$$ into the equation for the sphere and got $$x^2+y^2+yx= 1/2$$ But this isn't a circle, it's an ellipse so somewhere I must have gone wrong.

Also, the answer is supposed to be in spherical coordinates. Should I just convert the final answer to spherical coordinates then? That would mean I would get $$ r^2\sin(\phi) + r^2 \sin(\phi)^2\sin(\theta)\cos(\theta) = 1/2$$ after you plug in $$\sin(\theta)^2 + \cos(\theta)^2 = 1$$. Is that a correct transformation into spherical coordinates?

答:
The intersection of the sphere $x^2+y^2+z^2=1$ with the plane $x+y+z=0$ is simply described by those two equations. It certainly isn't $x^2+y^2+xy=\frac12$, since that defines a surface, not a line. You made no mistake in your computations, but you only have$$x^2+y^2+z^2=1\wedge x+y+z=0\implies x^2+y^2+xy=\frac12,$$and this implication is not an equivalence.

And the equation in spherical coordinates is$$\left\{\begin{array}{l}\rho=1\\\cos(\theta)\sin(\phi)+\sin(\theta)\sin(\phi)+\cos(\phi)=0.\end{array}\right.$$


问:
Implicit Differentiation of 3 variables
Find $\frac{dy}{dz}$ when

$$(-5x+z)^4-2x^3y^6+3yz^6+6y^4z=10.$$

I got an answer of

$$\frac{-24y^3(z-3)z^6+2x^36y^5}{4(-5x+z)^3+8yz^5+6y^4}.$$

Is this correct?

答:
You are considering the equation:
\begin{equation*} (-5x+z)^{4}-2x^{3}y^{6}+3yz^{6}+6y^{4}z=10\end{equation*}
and you wish to calculate $\frac{dy}{dz}$. It follows that
\begin{align*} 0&=\frac{d}{dz}{10}=\frac{d}{dz}\left[(-5x+z)^{4}-2x^{3}y^{6}+3yz^{6}+6y^{4}z\right] \\ &=4(z-5x)^{3}\left(1-5\frac{dx}{dz}\right)-2\left[6x^{3}y^{5}\frac{dy}{dz}+3x^{2}\frac{dx}{dz}y^{6}\right]+3\left[6yz^{5}+z^{6}\frac{dy}{dz}\right]+6\left[y^{4}+4zy^{3}\frac{dy}{dz}\right]\end{align*}
Can you take it from here to verify whether or not your answer is correct?

Title: Implicit Differentiation of 3 variables


问:
Closed form of $\cos(\frac{\pi}{7})$
I saw someone derive the closed form for $\cos(\frac{\pi}{5})=\frac{\varphi}{2}$, and got inspired to try to find a closed form expression for $\cos(\frac{\pi}{7})$ using the same method. In doing this, you get that $\cos(\frac{\pi}{7})$ is one of the solutions to the following polynomial.
$$8x^3-4x^2-4x+1$$
If there is an accessible closed form for the roots of this polynomial, I cant seem to find it. I have tried factoring, guessing and also looked at wolfram alpha, which only gives massive closed forms involving imaginary numbers.

So if anyone can find a closed form or show that none is accessible, that would be great!

答:
Maple says
$$
\frac{1}{12}\,\sqrt [3]{-28+84\,i\sqrt {3}}+\frac{7}{3}\,{\frac {1}{\sqrt [3]{-28+84\,
i\sqrt {3}}}}+\frac16
$$
despite that $i$ in there, the result is a real number.  Of course complex cube roots may be done trigonometrically, but this whole question is to avoid the trigonometry in $\cos\frac{\pi}{7}$.

Title: Closed form of $\cos(\frac{\pi}{7})$


问:
Find the volume of the solid by rotating the region bound by $y = 1 - x^2$, $y = 0$, and $x = 0$ about the line $y = 2$
I am not really sure if I am on the right track right now. But I am supposed to find the volume of the solid.

My Attempt (Volumes of Revolution):

So I am given the following equations:
$$\begin{align*}
y = 1-x^2 \quad (1)\\
y = 0 \quad (2)\\
x = 0 \quad (3)
\end{align*}$$

So first, I tried to find the points of intersection of these equations. For (1) and (2), I let them equal to each other and found that they intersect at $x = \pm 1$. And for (1) and (3), I sub (3) to (1) and I found that they intersect at $y = 1$.

I am not exactly sure where to proceed from here. Any help would be great!

答:
Firstly we must translate all points and curves $2$ units down, so we will end up rotating about the $x$ axis, which is manageable. So, the area we are concerned with is now bounded by the curves
$$\begin{align}
& x=0\\
& y=-2\\
& y=-1-x^2
\end{align}$$
Let's find the intersection points of the curves. Let $y=-2$. Then $x^2=1$ so $x=1$ (assuming you just mean the area bounded in the last quadrant, as there are $2$ possible quadrants you could be talking about, as noted in comments).

Hence the volume generated will be equal to

$$\pi\int_0^1 (-2)^2-(-1-x^2)^2dx$$
Can you finish? Please tell me if you need more help and I will be happy to comply.


答:
I've got sad news: we cannot express the solution to this equation using sums, products, and roots. The reasoning is strange to say the least: we can only find the solution to $\cos(\pi/p)$ where $p$ is on odd prime when $p$ is a Fermat prime. This means that we can assign $p$ to be something like $3, 5, 17, ...$, but $7$ is not an option. That's why we can find the solution to $\cos(\pi/5)$ but not for $\pi/7$. As an aside, this means we can find a solution for $\cos(\pi/17)$, which is really weird.

Part of why this is not doable is that this cubic equation has three roots, and when a cubic has three roots we might not have an approach to finding them. Cardano's method finds a root if the cubic has only one root, but we can employ his method with complex numbers in an attempt to find the remaining roots. Once you employ complex numbers you will need to find the cosine of some strange angles, and after performing some simplifications, you will need the value of...you guessed it, $\cos(\pi/7)$. Thus solving this problem is circular and there is nothing we can do about it.

We can find the solution in terms of complex numbers if interested, but the solution is not particularly helpful.

Hope this helps!

Title: Closed form of $\cos(\frac{\pi}{7})$

Body:
I saw someone derive the closed form for $\cos(\frac{\pi}{5})=\frac{\varphi}{2}$, and got inspired to try to find a closed form expression for $\cos(\frac{\pi}{7})$ using the same method. In doing this, you get that $\cos(\frac{\pi}{7})$ is one of the solutions to the following polynomial.
$$8x^3-4x^2-4x+1$$
If there is an accessible closed form for the roots of this polynomial, I cant seem to find it. I have tried factoring, guessing and also looked at wolfram alpha, which only gives massive closed forms involving imaginary numbers.

So if anyone can find a closed form or show that none is accessible, that would be great!


问:
Number of balls touching one ball at the center.
Let $B$ be a ball of radius one in $\mathbb{R}^d$. I'm looking for references about the following problem :

How many disjoint balls $B_i$ of radius one is it possible to put in contact (tangential contact) to the ball $B$ ?

For example, if $d=2$, we can put a maximum of $6$ balls touching $B$ (each $B_i$ will then be in contact with $B$ + two other neighbor balls, see picture below).

I found out that this problem is linked to the question of "compact arrangement of balls" arising in physics and chemistry (especially in crystallography), but I don't know the classic references for a mathematical approach. Does this problem have a mathematical name ? Do you have references to share ?

Any help is welcomed, please feel free to re-tag as appropriate.

答: It is called the kissing number.</p>

Body: It is called the kissing number.


问:
Comparing Risk of Estimators
I would like to compare the risk of 2 estimators.
Let $X_1,..,X_n$ are $\mathcal{N}(\theta,1)$. I have to show that the average $\frac{1}{n}\sum X_i$ is not admissible if we have prior knowledge that $\theta \in [-1,1]$. Therefore I need to show that it exits an estimator with smaller Risk and we assume quadratic risk: $R(\theta, T) = E[(\theta-T(X_1, .., X_n)^2]$.

I tried it with $n=1$ and with the estimator $T(x) = \begin{cases} X & if & |X| < 1
\\ 0  & else \end{cases}$

and arrive at:
$R(\theta, \bar{X}) = R(\theta, T) -2E[X\mathbb{1}_{\{|X| > 1\}}] + E[X^2\mathbb{1}_{\{  |X| > 1\}}]$. But I am not able to show that the last 2 terms are larger than 0. Or rather not able to show that $-2E[X\mathbb{1}_{\{|X| > 1\}}] < E[X^2\mathbb{1}_{\{  |X| > 1\}}]$

Any help is appreciated.

答:
Your choice of estimator basically suggests choosing the sample mean if it is between $-1$ and $1$; otherwise, choose $0$.  But why $0$?  If for instance $\theta = 1$, and we happen to observe many "large" values of $X_i$ in the sample, then your estimator will claim $\hat \theta = 0$ when in fact $\hat \theta = 1$ is an intuitively better estimate.  A similar situation holds when $\theta = -1$, yet you would again choose to say the parameter is zero.  That is the reason why you are having difficulty demonstrating that your estimator has smaller quadratic risk--because it might not when compared to the raw sample mean.

In light of this, what estimator do you think would perform better than the sample mean in this case?  What does it look like?

Title: Comparing Risk of Estimators


答:
Subtract the two equations. We get $y=0$

Plug in the first
$$x^2-z^2=1$$
A parametrization is
$$(x=\cosh t,y=0,z=\sinh t); (x=-\cosh t , y=0, z=-\sinh t)$$
In the image below the two surfaces and their intersection.

$$...$$

Title: Finding a parametrization of a curve from cartesian equations

Body:
I'm in need of some help to address this problem.

Let $C$ be a curve given by two equations:

$x^2+y^2-z^2-1=0$,

$x^2-y^2-z^2-1=0$

Express the curve by means of parametric equations.

Any ideas on how to work on it?


问:
Number of ways to arrange knights in a round table such that the right of each knight is different than it was last time
$n$ knights sat for lunch at a round table with $n$ indistinguishable chairs.

In the evening, before dinner, the king asked his assistant to make sure that when the knights sit back at the table for dinner, every knight must not sit on the right of a knight he sat on the right to at lunch.

How many ways can the king's assistant select places for the knights at the table for dinner?

I tried using inclusion-exclusion here but can't really get it right, stuck at the beginning.

答:
I think your idea of inclusion-exclusion works.

Since the chairs are indistinguishable we can fix the position of one particular knight throughout.

Suppose a particular set of $k$ of the knights are on the right of the same knight as before. Note that there are   $\begin{pmatrix}n\\k\\\end{pmatrix}$ ways of choosing these knights.

We now have $n-k$ 'blocks' of knights to seat. The block with the 'special' knight is fixed and the number of arrangements is then $(n-k-1)!$ ways. A special case is when $n=k$ and then the number of arrangements is $1$ and not $(-1)!$

So, by inclusion-exclusion, the number of arrangements is
$$ (n-1)!-(n-2)!n+(n-3)!\begin{pmatrix}n\\2\\\end{pmatrix}-(n-4)!\begin{pmatrix}n\\3\\\end{pmatrix}+...\pm 1$$


问:
Simplifying Rational Expression (Excluded Values)
I would like some opinions on a problem that I am working. The problem is below.

Simplify the expression.

$$\frac{3n+6}{n+6}\cdot \frac{2n-10}{3n+6}$$
So, I simplified and got
$$\frac{2n-10}{n+6}$$

I understand why the two expressions above are not equal when n = -2. I would like to know if the two expressions are equal when n = -6. Is $$\frac{264}{0}=\frac{-22}{0}$$ a true statement? I know that the two fractions written above are undefined. Is it okay to say the the two rational expressions above that are equivalent when n = -6 since they both produce similar results (undefined = undefined)?

答:
Not unless you define what you mean by 'equivalent'. Division by $0$ traditionally has no meaning, so I guess they could be 'equivalent' in their 'meaninglessness'.

Notice that regardless of how you define 'equivalent', do not mistake 'equivalent' for 'equal'. The equality symbol has well defined meaning and properties, and division by $0$ does not interact properly with it.


问:
Three logical rewrittings that are giving me a headache

~q ⋀ r ⋀ (~s -> t) using only  ∨ , ~
Ive got

~(~(q ∨ ~ r) ∨  (s ∨ t)

but this is wrong according to the truth tables

p ∨ (~q ⋀ (r ⩖[exclusive or] s)) using only -> and ~

This one completely stumped me

If there is justice in this life, then there is no need for a future life. If, on the other hand, there is no justice in our earthly life, then we have no reason to believe that God is righteous. But if we have no reason to believe that God is righteous, then we have no reason to believe that He will provide us with a future life. So either we don't need a future life, or we have no reason to believe that God will provide us with such a life.

If there is justice in this life-p

then there is no need for a future life.-q

If, on the other hand, there is no justice in our earthly life,-  ~p

we have no reason to believe that God is righteous -r

we have no reason to believe that He will provide us with a future life - s

we don't need a future life -q

or we have no reason to believe that God will provide us with such a life. -s

I dont know how to connect the sentences here

ive got p->q

```<code>    ~p->r

     r->s

     q⩖(exclusive or) s
</code>```

答:
Question 2. There might be a better way, or shorter ways, but this is what I got:

Using the definition of exclusive or:

$$\begin{align} r \overbrace{\oplus}^{\text{xor}} s \equiv \\( \sim r \land s) \lor (\sim s \land r) \equiv\\ \sim( \sim r \land s) \rightarrow (\sim s \land r) \equiv\\ (r \lor \sim s)\rightarrow ( \sim ( \sim (r \land \sim s))) \equiv \\ ( \sim r \rightarrow \sim s) \rightarrow ( \sim (\sim r \lor s)) \equiv\\ (\sim r \rightarrow \sim s) \rightarrow (\sim ( r \rightarrow s))  \tag{0} \end{align}$$

The known facts:
$$\begin{align} A \rightarrow B \equiv (\sim A \lor B) \tag{1} \\ A \lor B \equiv \sim A \rightarrow B \tag{2} \\ \sim ( \sim A) \equiv A \tag{3} \end{align}$$

And De-Morgan's law:

$$ \begin{align}\sim (A \lor B) \equiv (\sim A) \land (\sim B) \tag{4} \\ \sim (A \land B) \equiv (\sim A) \lor (\sim B) \tag{5} \end{align} $$

Thus we have:

$$\begin{align} p \lor ( \sim q \land ( r \oplus s)) \equiv \\\sim p \rightarrow ( \sim q \land ( r \oplus s)) \equiv  \tag{2} \\\sim p \rightarrow ( \sim ( \sim (\sim q \land (r \oplus s))))  \equiv \tag{3} \\\sim p \rightarrow ( \sim (q \lor \sim (r \oplus s))) \equiv \tag{4} \\\sim p \rightarrow ( \sim ( \sim q \rightarrow  \sim (r \oplus s))) \end{align}$$

Substituting what we got for the xor above, we get:

$$ \equiv  \tag{0}\\ \sim p \rightarrow ( \sim ( \sim q \rightarrow  \sim ((\sim r \rightarrow \sim s) \rightarrow (\sim ( r \rightarrow s))))) $$


问:
How do I divide a set of data samples which follow a logarithmic distribution?
I'm working for the first time with Logarithmic distribution. I have a set of samples which follow logarithmic distribution. I extracted the maximum and the minimum values from the set and defined the interval as [min,max]. Now I need to partition this range into 5 parts. Say [min,a], [a,b], [b,c], [c,d] and [d,max]. </p>

<p>If I calculate the mean, then I get two splices. I need 5 such. How do I go about it? Any pointers on articles, web pages and sources that describe and explain related concepts are also well appreciated.  </p>

答: Two natural ways are equal intervals, so $a=\min+(\max-\min)/5$ and so on, and equal ratios, so define $r=\left(\frac {\max}{\min}\right)^{\frac 15}$, then $a=r\min, b=r^2 \min$,etc.  These work for any distribution.  Ratios doesn't work if $\min \lt 0 \lt \max$</p>


问:
Coordinates of a functional in dual space with one qiven coordinate.
$f \in (\mathbb{R}^{3})^*$ is a functional of the form $f(x_1, x_2, x_3) = x_1 - 3x_2 + x_3$.

$\mathcal{A} = \{α_1, α_2, α_3\}$ is a basis of the linear space $\mathbb{R}^{3}$ and $\mathcal{A}^* = \{g_1, g_2, g_3\} ⊂ (\mathbb{R}^{3})^*$ is a dual basis.

$g_1(x_1, x_2, x_3) = x_1 - x_3$

$g_2(x_1, x_2, x_3) = x_1 - 2x_2$

Assuming that the third coordinate of the functional $f$ in the basis $\mathcal{A}^∗$ is equal to $1$, find the vector $α_3$.

So I don't know $g_3$ but I know that "the third coordinate of the function $f$ in the basis $\mathcal{A}^∗$ is equal to $1$". I think that I need to use that information to get full $\mathcal{A}^∗$, but I'm not sure how. My first idea was to express $g_3(x_1, x_2, x_3) = ax_1 + bx_2 + cx_3$ in matrix:

\begin{bmatrix}
1 & 1 & a & 1\\
0 & -2 & b & -3\\
-1 & 0 & c & 1
\end{bmatrix}

with an aim to get:

\begin{bmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & ?\\
0 & 0 & 1 & ?
\end{bmatrix}

And then somehow find those $3$ variables. But that does not work for me and I don't know what to do next.

答:
Are you sure the dual basis isn't just an arbitrary basis but the one induced by the basis of $\mathbb{R}^3$? Your language of "a basis" is unusual; it should probably be "the corresponding basis." Once you pick a basis of $V$ you get a naturally induced basis of $V^*$.

If you do assume the $g_i$ are induced by the $\alpha_i$, then $g_i(\alpha_j)=\delta_{ij}$, the Kronecker delta (which is $1$ when $i=j$ and $0$ otherwise).

Thus $g_1(\alpha_3)=0$, $g_2(\alpha_3)=0$, and $g_3(\alpha_3)=1$.

Assume $\alpha_3=(a,b,c)$.

The first equation above tells you $a=c$, so $\alpha_3=(a,b,a)$.

The second equation tells you $a=2b$, so $\alpha_3=(2b, b, 2b)$.

Now, we also know that the third coordinate of $f$ with respect to the $g_i$ is 1, so

$$f=c_1g_1+c_2g_2+g_3$$

Thus

$$2b-3b+2b=f(\alpha_3)=c_1\underbrace{g_1(\alpha_3)}_{\text{zero}}+c_2\underbrace{g_2(\alpha_3)}_{{\text{zero}}}+\underbrace{g_3(\alpha_3)}_{\text{one}}=1$$

So $b=1$.

Thus $\alpha_3=(2,1,2)$.


问:
Uniqueness of pre-abelian structures on categories
For a pre-additive cateory we need the structure of an abelian group on the Hom sets of a category, with respect to which composition is bilinear. Is this structure, when it exists, unique? More generally, does every category come endowed with at most one abelian category structure?</p>

答:
As Zhen Lin says, for a pre-additive category with one object you're asking whether the addition on a ring is uniquely determined (not up to isomorphism but as a function) by its multiplication. This is false; to prove this it suffices to exhibit a bijection from a ring to itself which preserves multiplication but not addition, and for example we can consider multiplicative bijections $\mathbb{Z} \to \mathbb{Z}$ which permute the primes.

(A harder and also interesting question is to ask if you can find two rings with isomorphic multiplicative monoids but nonisomorphic additive groups. We can do this with $\mathbb{Q}$ and $\mathbb{Q}(x)$; their additive groups can be distinguished by dimension over $\mathbb{Q}$, but their multiplicative monoids consist of their multiplicative groups together with a zero element, and their multiplicative groups are both $C_2 = \{ \pm 1 \}$ times the free abelian group on countably many generators.)

However, the following structures on a category are all unique:

semiadditive

additive

abelian

These all follow more or less from the first one; the key point is that biproducts are on the one hand a limit and a colimit, and on the other hand uniquely determine the addition of morphisms. There is a slight subtlety here involving zero objects and morphisms which is needed to make this sketch really work out; we need to consider the nullary case first and separately and as far as I know this is unavoidable. This is all done in detail in my blog post A meditation on semiadditive categories.


问:
Creating a summation equation
I was asked to solve a problem involving finding the number of days it will take to journey 2955 miles when: On the first day 1.5 miles are traveled and every day following he increases his journey by $\frac16$ of a mile.

I have not yet figured this out, and am looking for some help. But, I think I can set it up like this: $$\sum_{n=0}^x1.5+\frac16n$$.

This seems like it can work, but I am afraid I am missing something here.

In addition - I can solve this using a CAS calculator, but how can I do it without?

答:
Yes if $n$ is the number of days, you can write it as

$\sum \limits_{i=1}^n 1.5 + \frac16(i-1) = 2955$

or, $\ 1.5 n - \frac{n}{6} + \frac{1}{6}\sum \limits_{i=1}^n  i = 2955$

and $\sum \limits_{i=1}^n  i = \frac{n (n+1)}{2}$

Title: Creating a summation equation


问:
Equations for Mandelbrot bifurcation diagram
The Mandelbrot set is the set of all complex numbers $c$ that cause the function $z_{n+1} = z_n^2 + c$ to remain bounded within a circle of radius 2 when iterated from $z_0 = 0$. Looking at only the real numbers, all values between positive $\frac{1}{4}$ and negative 2 are part of the set. If we plot the attractive fixed point value on the vertical axis against the input $c$ value on the horizontal axis we get a classic bifurcation diagram. Going from right to left, the diagram starts with the function $\frac{1 - \sqrt{-4c + 1}}{2}$ where $\frac{1}{4} ≥ c ≥ -\frac{3}{4}$. It then splits into two branches as $\frac{-1 ± \sqrt{-4c + 3}}{2}$ where $-\frac{3}{4} ≥ c ≥ -\frac{5}{4}$. Both branches divide again and again, over increasingly narrow intervals, until the plot dissolves into chaos around $-\frac{7}{5}$.

Are there closed-form equations for any of the divisions beyond the first two I've listed? For example, the period-4 region or the period-8 region? They don't seem to line up with quadratic expressions or any other expressions I've tried. I also can't find closed-form expressions for the end-points of the intervals for each division, that is, the $c$ value for each bifurcation point.

The closest I've been able to get so far for the topmost period-four section is $\frac{(-4c - 5)^{\frac{1}{2.05}}} {2.5} - \frac{1 - \sqrt{2}} {2}$, which is pretty darn close, but it's not exact.

I do know that the sweeping curves visible in the chaotic portion of the bifurcation diagram, the critical curves, are of the form $(c² + c)² + c$, $((c² + c)² + c)² + c$, $(((c² + c)² + c)² + c)² + c$, and so on, but these curves don't line up with the bifurcating parts either.

答:
If we let
$$f_c(z)=z^2+c$$
then the lines on that diagram (in the column corresponding to $c$) are exactly at attractive cycles of $f_c$ - which is to say that they are certain solutions of
$$f_c^n(z)=z$$
where $f_c^n$ is the $n$-fold composition of $f_c$ with itself - and where $n$ will be the number of lines seen on the plot. More specifically, they are the solutions for which the derivative $|(f_c^n)'(z)| \leq 1$.

So, the curves are described as some of the roots of the following degree $2^n$ polynomial in $z$ (but note that cycles of every length appear in that plot - so certain pieces will require arbitrarily large $n$ to describe):
$$f_c^n(z)-z.$$
You can slightly reduce this by noting that if $m$ divides $n$ then $f_c^m(z)-z$ divides $f_c^n(z)-z$, which at least lets us factor the polynomial a little bit. As far as I know, the only $n$ for which the solutions of this have a form with radicals is $n=1$ (the fixed point) and $n=2$ (the two-cycle).

You can certainly, however, solve these equations using numerical methods for solving polynomials - and even something simple such as Newton's method is going to converge a lot faster than finding the cycles just by iterating a single point (as is how bifurcations diagrams are usually made) under $f_c$ itself.


问:
Compactness of semisimple Lie algebra
I want to prove that on a semisimple Lie algebra $\mathfrak{g}$ over ${\bf R}$:

$\mathfrak{g}$ is compact if and only if the Killing form is strictly
  negative definite.

Here the Lie algebra is compact if Int$(\mathfrak{g})$, which is a Lie group of $\{ ad_X | X\in \mathfrak{g}\}$, is compact.

I am reading a Helgason's book : The proof is short and I cannot catch the strategy.

The proof : If Killing form $B$ is strictly negative definitie then $O(B)$ is a set of all linear transformations which leave $B$ invariant. Then $O(B)$ is compact.
I cannot understand.

Thank you in advance.

答:
The key is to look at just the vector space structure of $\mathfrak g$ and then use some nice topological properties.

So for now, think of $\mathfrak g$ as just a real vector space, say $\mathbb R^n$, so that the Killing form is essentially an inner product. Since the Killing form is negative definite, there is a an element $A \in GL_n(\mathbb R)$ with all negative eigenvalues such that $B(x,y) = x^T A y$. The collection of all linear transformations which preserve B will be this more generalized orthogonal group $O(B)$, which is closed and bounded by precisely the same arguments as those showing that $O(n)$ is compact (just no longer using the Euclidean inner product).

Now $B$ is adjoint invariant and all Lie algebra automorphisms of $\mathfrak g$ must preserve the commutator (and hence the adjoints) we get that $\operatorname{Aut}(\mathfrak g) \subseteq O(B)$. Furthermore, $\operatorname{Int}(\mathfrak g)$ is a closed subgroup of $\operatorname{Aut}(\mathfrak g)$ and so is also a closed subspace of $O(B)$ and hence compact. Thus the Lie algebra is also compact as required.

Title: Compactness of semisimple Lie algebra


问:
Proving continuity of $f(x)=\sqrt{1−x^3}$ on $[0,1]$
I have to prove the continuity of the function with $\epsilon, \delta$. For this I have to prove that the function is continuous at every $x \in [0, 1]$ by proving:

$$\forall 0 < \epsilon, \exists 0 < \delta, \forall a \in [0, 1]: |x - a| < \delta \implies \left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right| < \epsilon.$$

This is where I am right now:

\begin{align}
\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|&= \frac{\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{x^3 - a^3}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{(x - a)(x^2 + ax + a^2)}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3 + 1 - a^3}\right|}
\end{align}

Any hint is welcome.

答:
$$\frac{\mid x^3-a^3\mid}{\mid\sqrt{1-x^3}+\sqrt{1-a^3}\mid}$$

I'm taking it up from here...

Fix a $\delta'>0$. For simplicity, take $\delta'=1$. Then, $\forall\mid x-a\mid<1$,
$$\frac{\mid x-a\mid\mid x^2+ax+a^2\mid}{\mid\sqrt{1-x^3}+\sqrt{1-a^3}\mid}\leq\frac{\mid x-a\mid\mid x^2+ax+a^2\mid}{\mid\sqrt{1-a^3}\mid}<\frac{\mid x-a\mid\mid \left(a+1\right)^2+a\left(a+1\right)+a^2\mid}{\mid\sqrt{1-a^3}\mid}$$
Thus, for $\mid f(x)-f(a)\mid<\epsilon$, $\mid x-a\mid<\frac{\epsilon\mid\sqrt{1-a^3}\mid}{\mid \left(a+1\right)^2+a\left(a+1\right)+a^2\mid}$.

Now just take $\delta:=\min\left\{1,\frac{\epsilon\mid\sqrt{1-a^3}\mid}{\mid \left(a+1\right)^2+a\left(a+1\right)+a^2\mid}\right\}$, and you've found a $\delta>0$, for which the conditions of continuity hold for any arbitrary $\epsilon>0$.


问:
How to be sure about any parameterization?
This is the problem:

Find intersection of this two surfaces and then calculate its length.
$$x^2+4y^2=4$$
and

$$y\sqrt3 +z=1$$
Method 1 :

Common way is to take:

$$x=2\cos t  ,\:  y=\sin t ,\: z=1-\sqrt3 \sin t$$

So

$$r(t) =(2\cos t,\:\sin t,\:1-\sqrt3 \sin t)$$
and

$$r'(t) = (-2\sin t,\:\cos t,\:-\sqrt3\cos t) ,\: |r'(t)|=2$$

Finally the length is:

$$L = \int_{0}^{2\pi} 2\:dt=4\pi$$

But in another parameterization I find something strange

Method 2:

$$x=2\sqrt{1-t^2} ,\: y=t ,\: z=1-t\sqrt3$$
So

$$r(t) = (2\sqrt{1-t^2} ,\: t ,\: 1-t\sqrt3)$$
and

$$r'(t) = \left(\frac {-2t}{\sqrt{1-t^2}},1,-\sqrt3\right) ,\: |r'(t)|=\frac {2}{\sqrt{1-t^2}}$$ (and that's different from method 1 $|r'(t)|$)

Finally the length

$$L=\int_{0}^{2\pi} \frac {2}{\sqrt{1-t^2}}\:dt\\
=2\int_{0}^{\arcsin{2\pi}}\frac {1}{\cos{\theta}}\cos{\theta}\:d\theta\\
=2\int_{0}^{\arcsin{2\pi}}1\:d\theta$$

and because of $\arcsin(2\pi)$ its undefined it cant be solved.

Now my first question is that:

If we know from past that we can create many parameterizations for a single curve, so why the result of they length is different?

and the second one:

How to be sure that a parameterization is the correct one?

答:
As a complement to @Axel's answer

Let $t_j$ denote the $j$th method's $t$ so $t_2=\sin t_1$ and $\tfrac{dt_2}{\sqrt{1-t_2^2}}=d\arcsin t_2=\pm dt_1$. If for example $t_1$ increases from $0$ to $2\pi$, $t_2$ is monotonic in three parts, cut at $t=\pi/2$ and $t=3\pi/2$. The monotonic pieces have alternating $\pm$, in this case with the pattern $+-+$, so the length is$$\begin{align}\int_0^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}+\int_{-1}^0\frac{2dt_2}{\sqrt{1-t_2^2}}&\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\\&\color{red}{=4\pi}\\&\color{blue}{=\left(\int_0^{\pi/2}+\int_{\pi/2}^{3\pi/2}+\int_{3\pi/2}^{2\pi}\right)2dt_1}\\&=\color{blue}{4\pi},\end{align}$$where the $t_2$-space calculation can continue from the black expression with either the red or the blue treatment. If $t_1$ varies over another length-$2\pi$ interval the monotinicity division details are different, but any choice requires at least one such cut, succumbs to the above technique, and gets the same answer. Let's check another example from $t_1=-\pi/2$ to $t_1=3\pi/2$, giving two pieces (meeting at $t=\pi/2$) where the $\pm$ are $+-$:$$\int_{-1}^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\color{blue}{=\left(\int_{-\pi/2}^{\pi/2}+\int_{\pi/2}^{3\pi/2}\right)2dt_1}.$$


答:
You don't talk of two expressions being equal, by talking about if they are equal  at the points they are both undefined.  You talk of two expressions being equal if the are equal at all the points where they are defined and if they are both defined and/or undefined at the same points.

$\frac{3n+6}{n+6}\cdot \frac{2n-10}{3n+6}$ and $\frac{2n-10}{n+6}$ are not both equal at $n= -6$.  They are both undefined at $n=-6$.

To be equal they must i) both be equal at all points they are both defined.  And if $n \ne -6$ and $n\ne -2$ they are equal.  ii) be (un)defined at the same points.  And they are both undefined at $n= -6$ and they are both defined for all $n\ne -6; n\ne -2$. BUT $\frac{2n-10}{n+6}$ is defined at $n=-2$ (it is equal to $-\frac 72$) whereas $\frac{3n+6}{n+6}\cdot \frac{2n-10}{3n+6}$ is not definied at $n=-2$.  So the expressions are not equal.

....

Perhaps a clear example of what you are trying to express.

$\frac {x+1}{x^2 + 2x + 1} = \frac 1{x+1}$.

These expression are equal because if $x \ne -1$ then we have the terms are equal.

But  if $x$ does equal $-1$ we don't have the two expressions equal.  And saying $\frac 00 = \frac 10$ is FALSE.

They are not equal at $x=-1$.

But they are both undefined at $x= -1$.  And so, at all points we have either the two expressions both equal or they are both undefined.  So they are equal.

....

Anyway.... we don't have $\frac{3n+6}{n+6}\cdot \frac{2n-10}{3n+6}=\frac{2n-10}{n+6}$ but we do have $\frac{3n+6}{n+6}\cdot \frac{2n-10}{3n+6}=\frac{2n-10}{n+6}_{n\ne -2}$ where that notation, $_{n\ne -2}$ means we are restricting $\frac {2n-10}{n+6}$ only to values where $n\ne -1$.

These expressions are equal as they are both undefined and $n=-6$ and at $n=-2$ (the second is not defined ant $n=-2$ because we said it wasn't) and they are equal everywhere else.

....

P.S.  You wrote $2n+10$ rather than $2n-10$.  I'm assuming that was a typo.

Title: Simplifying Rational Expression (Excluded Values)

Body:
I would like some opinions on a problem that I am working. The problem is below.

Simplify the expression.

$$\frac{3n+6}{n+6}\cdot \frac{2n-10}{3n+6}$$
So, I simplified and got
$$\frac{2n-10}{n+6}$$

I understand why the two expressions above are not equal when n = -2. I would like to know if the two expressions are equal when n = -6. Is $$\frac{264}{0}=\frac{-22}{0}$$ a true statement? I know that the two fractions written above are undefined. Is it okay to say the the two rational expressions above that are equivalent when n = -6 since they both produce similar results (undefined = undefined)?


问:
If $4$ distinct integers $a, b, c, d$ are randomly selected from $n$ distinct integers, what is the probability that $a \lt b \lt c \lt d$?

a. If $i$ distinct integers $r_1, r_2, \ldots,r_i$, i $\leq n$, are randomly selected (one after other) from n distinct integers, what is the probability that $r_1 < r_2 < \cdots < r_i$?

b. Does the probability you got in part a of the question change if < is replaced with $\leq$?

I know that a question similar to part 'a' of this  question can be solved using the concept of Decision tree that orders the elements of the list, an application of which is to analyse the complexity of comparison-based sorting algorithm.

I came across a solution to part 'a' of the question that says the required probability is $C(n, i)/P(n,i)$, which is nothing but equal to the solution of the Decision tree based approach, which is $1/i!$.

What I don't understand is this - how $C(n,i)$ is equal to favourable number of outcomes and $P(n,i)$ is equal to total number of outcomes?

Figure: A decision tree for sorting three distinct elements.

Image taken from Discreet Mathematics and its Applications by Kenneth Rosen

答:
$C(n,i)$ chooses $i$ integers from $n$ integers without giving importance to their order. Given such a set of $i$ integers, we can first order them in ascending order, and then assign the corresponding values to $r_1,r_2\dots r_i$ so that $r_1 \lt r_2 \lt \dots \lt r_i $.

$P(n,i)$ counts the total ways. Why? The number of choices for $r_1$ is $n$, for $r_2$ is $n-1$ and so on. That gives $$n(n-1)(n-2)\dots (n-i+1)=\frac{n!}{(n-i)!} = P(n,i) $$


问:
Difference between geometrical and graphical representations.
My books has a sub-topic called geometric meaning of differential equations. It mentions that  slope of tangent at every point in 2d space is from given differential equation. According to it, without solving the equation we can find the nature of solutions as they are family of curves passing through these points in 2d space following the direction of tangents at those points.
According to my understanding what we are doing here is the graphical analysis of a differential equation. I watched 3blue1brown's video on geometric representation of derivatives where he used geometric figures and their sizes to explain the derivatives.
So, please help me understand geometrical meaning of differential equations. How the slope of tangent at every point in 2d space and curves passing through them give geometric meaning of differential equaitons?</p>

答:
In 3blue1brown's video, he is looking at the change in area as you give a little increase to the sidelength of a square, thinking of the derivative the amount of area bump per sidelength bump: $\frac{dA}{dx} = \frac{d(x^2)}{dx}$.  He also looked at the amount of volume bump per sidelength bump: $\frac{dV}{dx} = \frac{d(x^3)}{dx}$.  He is using geometrical figures as standins for functions, and showing the derivative relationship between $x^2$ and $2x$ through their shapes.

A derivative is an operation you can do to functions.  You put a function into the derivative, you get a function out.  But a differential equation is asking, "What function behaves like this?"  For example, "What function is twice its derivative?" would be $y = 2 * \frac{dy}{dx}$.  To extend 3blue1brown's video, we might ask, "What function is $\frac{2}{x}$ times its derivative?  What shape does that?"  The $x^2$ square does that: $x^2 * \frac{2}{x} = 2x$

The problem here, though, is the analogy will only extend so far.  Geometrical analogies for $x^2$ and $x^3$ work for degrees of 2 and 3, but don't help much at higher dimensions.  And these analogies only really help in visualizing the change in $x^n$.  We don't usually ask ourselves what shape might double its constant while it decreases a dimension.  And that geometrical meaning of decreasing a dimension only applies to $x^n$, whereas our question might be also answered by another non-$x^n$ function.  There are many other functions we want to take derivatives of and ask questions about.  But if you're wanting to just get across the idea of the derivative, looking at how it works for these special geometrical $x^n$ cases can help.

Your book is taking a different approach to visualizing a differential equation.  We may know how to describe a derivative as a function of x and y.  We can visualize that by plotting slopes (the derivatives) on a graph and asking, "What function would act like that?  What function has slopes like this at these points?"  This more naturally extends into some of the physical questions that use differential equations, looking at forces in the instant to build up a more complete behavior on the whole.


问:
When do $L^p$ and $L^q$ contain the same functions?
From Rudin:

Given that X is a finite measure space, when do $L^p$ and $L^q$ contain the same functions?

What I know is that $L^p$  contains $L^q$ if p<q.

I was thinking

How to make a function which is in $L^p$ but not in $L^q$, assuming it is possible to do so? And what fails when it is not possible.

I appreciate any help.

答:
I am not sure if this completely determines when $L^p=L^q$, but it is a case. Throughout all of this we can assume without loss of generality that $\mu(X)=1$ and that $p < q$. Using Holders we can show that $||f ||_p \leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}||f||_q=||f||_q$. This shows that $L^q\subset L^p$.

Assume that $$L^p\subset L^\infty $$. Then for any $f\in L^p$ we have that
$$||f||_{p+1}^{p+1}=\int_\Omega |f|^{p+1}d\mu\leq \int_{\Omega}|f|^p ||f||_\infty d\mu = ||f||_p^p ||f||_\infty$$
Since $f\in L^p\subset L^\infty$, then $f\in L^{p+1}$. We can actually show by induction using this argument that $f\in L^{p+n}$ for any $n\in\mathbb{N}$. Now we can find an $n\in\mathbb{N}$ so that $q < p+n$. Then since $||f||_q\leq ||f||_{p+n}$ and $f\in L^{p+n}$, it follows that $||f||_q<\infty$ and so $f\in L^q$. Hence $L^q=L^p$.

So the condition I found is that if we are on a finite measure space, $p<q$ and $L^p\subset L^\infty$, then $L^p=L^q$.


答:
For any given choice of $i$ integers from among the $n$ possible integers, only one permutation of these $i$ integers produces a valid result, because the $n$ integers are distinct.

There are $C(n,i)$ such choices of $i$ integers from among $n$ integers, and $P(n,i)$ permutations of $i$ integers from among $n$ integers.

Title: If $4$ distinct integers $a, b, c, d$ are randomly selected from $n$ distinct integers, what is the probability that $a \lt b \lt c \lt d$?

Body:
a. If $i$ distinct integers $r_1, r_2, \ldots,r_i$, i $\leq n$, are randomly selected (one after other) from n distinct integers, what is the probability that $r_1 < r_2 < \cdots < r_i$?

b. Does the probability you got in part a of the question change if < is replaced with $\leq$?

I know that a question similar to part 'a' of this  question can be solved using the concept of Decision tree that orders the elements of the list, an application of which is to analyse the complexity of comparison-based sorting algorithm.

I came across a solution to part 'a' of the question that says the required probability is $C(n, i)/P(n,i)$, which is nothing but equal to the solution of the Decision tree based approach, which is $1/i!$.

What I don't understand is this - how $C(n,i)$ is equal to favourable number of outcomes and $P(n,i)$ is equal to total number of outcomes?

Figure: A decision tree for sorting three distinct elements.

Image taken from Discreet Mathematics and its Applications by Kenneth Rosen


问:
How many coefficients of the polynomial $P_n(x_1,...,x_n) = \prod_{1\leq i < j \leq n}(x_i+x_j)$ are odd?
How many coefficients of the polynomial $P_n(x_1,...,x_n) = \prod_{1\leq i < j \leq n}(x_i+x_j)$ are odd?

I computed a couple small cases first. for $n = 1$ the problem doens't make sense. For $n=2$ there are 2 odd coefficients, and for $n = 3$ there are 6. I'm pretty sure I need to plug something in for $x_i$ and do some modular arithmetic but I'm not sure what I need to do.

Some hints would be greatly appreciated.

答:
First, we convert this to a combinatorial problem.  Suppose we have $n-1$ balls of each of $n$ colors, and $\binom{n}2$ buckets, each with two balls in it, such that every color combination occurs exactly once.  We draw a ball from each bucket, and record the result as an ordered $n$-tuple $(c_1,c_2,\dots,c_n)$ where $c_k$ is the number of times we drew color $k$ for $k=1,\dots n$.

There is an obvious correspondence between the tuple $(c_1,c_2,\dots,c_n)$ and the term $x_1^{c_1}\cdots x_n^{c_n}$, and we are asked how many tuples can be obtained in an odd number of ways.

I claim that only the $n!$ tuples comprising the numbers $0,1,,2,\dots n-1$ in some order can be obtained in an odd number of ways, and that in fact, they can only be obtained in one way.

It is easy to see that each of these tuples can be obtained in exactly one way.  If $c_k=n-1$, we must draw all $n-1$ balls of color $k$.  Then there remain $n-2$ balls of each color other than $k$, so if $c_j=n-2$, then we must draw all the balls of color $j$ and so on.

It remains to show that every other combination can be obtained in an even number of ways.  If the numbers in the tuple $(c_1,c_2,\dots,c_n)$ are not all different, then there must be two nonzero numbers that are the same, because the sum has to be $\binom n2$.  Suppose that $0<c_i=c_j$ where $i\neq j$.  We must draw a ball from the bucket containing $i$ and $j$, and I claim that there are just as many ways to achieve $(c_1,c_2,\dots,c_n)$ by drawing $i$ as there are by drawing $j$.  Indeed, this is clear, because the problem is symmetric in $i$ and $j$.


问:
Integral basis for a number field
I need some help in solving the following problem:

Suppose $K$ is a number field and $K=\mathbb{Q}(\theta)$ where $\theta\in\mathfrak{O}_K$, the ring of integers of $K$. Now among the elements in $\mathfrak{O}_K$ of the form $$\frac{1}{d}(a_0+\cdots+a_i\theta^i)$$($0\ne a_i$; $a_0,\ldots a_i\in\mathbb{Z})$, where $d$ is the discriminant of $K$, pick one with the minimum value of $|a_i|$ and call it $x_i$. Do this for $i=1,\ldots ,n=\dim_{\mathbb{Q}}K$. I need to show that $\lbrace x_1,\ldots ,x_n\rbrace$ is an integral basis for $K$.

Here are my thoughts: If we can show that the discriminant of $\lbrace x_1,\ldots ,x_n\rbrace$, which can be shown to be a $\mathbb{Q}$-basis, is less than or equal to $d$ then we are done; I tried to show this but have not succeed. I have to somehow use the fact that $|a_i|$ is the minimum which I am unable to see how. Any help will be appreciated.

答:
Can you show that $|\Delta(x_1,\dots,x_n)|$ is minimal amongst all integers? If so, suppose that this isn't an integral basis, so that there is some $\alpha\in\mathfrak{O}_K$ such that $\alpha=b_1x_1+\cdots+b_nx_n$, for $b_1,\dots,b_n\in\mathbb{Q}$, but at least one is not in $\mathbb{Z}$, say $b_1\notin\mathbb{Z}$. Then $b_1=b+r$, where $b\in\mathbb{Z}$, and $0<r<1$, and we can construct a new set of integers, $y_1=\alpha-bx_1$, and $y_i=x_i$, for $i=2,\dots,n$. Can you then show that $|\Delta(y_1,\dots,y_n)|<|\Delta(x_1,\dots,x_n)|$?

Edit. Sorry, scratch the method I was thinking of. Although I still think that you can do it that way, it would be considerably more involved than I think is necessary.

Let $G=\mathbb{Z}x_1+\cdots+\mathbb{Z}x_n\subseteq\mathfrak{O}$. Clearly $x_1,\dots,x_n$ span $K/\mathbb{Q}$, so we can write $\alpha=b_1x_1+\cdots+b_nx_n$, for some $b_1,\dots,b_n\in\mathbb{Q}$. Write each $b_i=c_i+r_i$, where $c_i\in\mathbb{Z}$, and $0\le r_i<1$. Then
$$
\alpha-\underbrace{c_1x_1+\cdots+c_nx_n}_{\in~G}=r_1x_1+\cdots+r_nx_n
$$
Now look at the coefficients of the powers of $\theta$, and don't forget that $[\mathfrak{O}_K/\mathbb{Z}[\theta]]$ divides $d$.

Title: Integral basis for a number field