问:
"Trivial" part of the Portmanteau-Theorem
Let $(E,\mathcal{E})$ be a topological space equipped with the Borel-$\sigma$-algebra $\mathcal{E}$. Furthermore, let
$$
\mathcal{M}_{f}(E) := \{\text{finite measures on } (E,\mathcal{E})\}.$$
and
$$\mathcal{M}_{1}(E) := \{\mu \in \mathcal{M}_{f}(E): \mu(E) = 1\}.$$
Now we consider $\mu, \mu_{1}, \mu_{2}, \dots \in \mathcal{M}_{1}(E)$.

My question is: why is $\limsup_{n \to \infty} \mu_n(F) \leq \mu(F)$ for all measurable closed subsets $F$ equivalent to $\liminf_{n \to \infty} \mu_n(G) \geq \mu(G)$ for all measurable open subsets $G$?

答:
This statement is not true. The case $\mu_n = 0$ and $\mu$ any non-zero measure gives you a counterexample.

If you assume all $\mu_n, \mu$ have the same finite measure, then the statement is true.

Assume all measures have finite measure $M$ for example. Let $G$ be any open subset of $E$. Take $F = E-G$. It is a closed subset of $E$, and $E=F\sqcup G$. Then, since the measures are finite, $\mu_n(G) = \mu_n(E) - \mu_n(F) = M - \mu_n(F)$ and the same statement holds for $\mu$ instead of $\mu_n$.

Then $\lim \sup_{n\rightarrow +\infty}\mu_n(G) = M - \lim \inf_{n\rightarrow +\infty}\mu_n(F) \geq M - \mu(F) = M - (M - \mu(G)) = \mu(G)$.

The proof is similar for the other way of the equivalence.


问:
Deadly disease and probability to die
NHS records show that of patients suffering from a certain disease, 75% die of it.
In a specific hospital, there are 100 patients. What is the probability that 25 to 65 will die?

Clearly this is an example of use of binomial distribution.

I don't know, however, how to deal with the range 25 to 65. If the question was for exactly 25 patients to die, it would be

$P = \binom {100}{25}0.75^{25}.0.25^{75}$

but I don't know how to take into account the range.

Thank you.

答:
There are a couple of approaches we could take.

First, we can find the exact answer using the binomial distribution. Let $X\sim \mathscr{B}(100, 0.75)$. Then, the question is what is $P[25 \le X \le 65]$? The key to a problem like this is that the possible values of $X$ are mutually exclusive. If $X = 25$, then exactly 25 patients died, so we know that 26 did not die, 27 did not die, etc. Because of this mutual exclusivity, we can separate the probability of the union of each possible value of $X$ into into the sum of the probabilities of each possible value of $X$:

$$ P[25 \le X \le 65] $$
$$ = P[X = 25 \, \cup \, X = 26 \, \cup \, ... \, \cup \, X = 65] $$
$$ = P[X = 25] + P[X = 26] + ... + P[X = 65] $$
$$ = \sum_{x=25}^{65} \binom{100}{x}0.75^{x}\cdot 0.25^{100-x} . $$
$$ \approx 0.016 . $$

We could also use the normal approximation to the binomial distribution. Essentially, this says that under certain conditions on $n$ and $p$, $\mathscr{B}(n,p)$ can be approximated by $\mathscr{N}\left(np,np(1-p)\right)$.

My go-to rule of thumb is $np > 10$ and $n(1-p) > 10$, which both hold in this situation, so I'd feel comfortable using the approximation, but different practitioners have different preferences.

In this case, the mean of our normal approximation is $ \mu = np = 100\cdot 0.75 = 75 $, and the standard deviation is $\sqrt{100*0.75*0.25} = \sqrt{18.75}$.

We can then convert 25 and 65 into Z-scores and calculate the probability using the standard normal cdf from there:

$$ P[25 \le X \le 65] $$
$$ P\left[\frac{25-75}{\sqrt{18.75}} \le Z \le \frac{65-75}{\sqrt{18.75}}\right] $$
$$ \approx 0.010 . $$

ContentLicense: CC BY-SA 4.0

Title: Deadly disease and probability to die


问:
Finding $a$ such that $y_1=\frac{a}{2}x+\frac{1}{2}$ and $y_2=-\frac{3}{a}x+\frac{1}{a}$ are orthogonal. Impossible?

For which values of $a$ are the two lines orthogonal?
$$y_1=\frac{a}{2}x+\frac{1}{2} $$
$$y_2=\frac{-3}{a}x+\frac1a $$

The criterium for two lines to be orthogonal, is that the product of their slopes equals $-1$.

$$\frac{a}{2} \cdot \frac{-3}{a}=-1 $$
$$\frac{-3}{2}=-1$$
When I do the calculation I get an anomalous result. Does this mean that there really is no value for $a$ such that the lines are orthogonal to each other?

答:
$$y=\frac{a}{2}x+\frac{1}{2} $$
$$y=-\frac{3}{a}x+\frac1a $$
slopes are $$m_1=\frac{a}{2};\;m_2=-\frac{3}{a}$$

As $$m_1\cdot m_2=\left(\frac{a}{2}\right)\left(-\frac{3}{a}\right)=-\frac{3}{2}\ne -1$$ there is no value of $a$ such that the two values are perpendicular.

This is not an "anomalous" result. Your work was correct and your conclusion was right.


问:
Find rate of change of the area of a triangle at a particular moment
The side lengths of a right-angled triangle are $a, b$ and $c$ centimeters where $c$ is the length of the hypotenuse and $a, b$ and $c$ are changing as differentiable functions of time. It is known that the perimeter $a+b+c$ is always constant and at the moment $a=3$ and $b=4$ the rate of change of $a$ is $108$ cm/sec. What is the rate of change of the area of the triangle (in cm$^2$/sec units) at this moment?

This is a related rates question (the answer is $72$). What I tried and first of all I calculated the derivative of the area of a right triangle ($ab/2$) and everything is known except the derivative of side $b$. I need to find some relation between $b$ and $a$ but I am not sure how.

答:
$\frac{d a}{dt} + \frac{d b}{dt} + \frac{d c}{dt} = 0$ ...(i)

$c^2 = a^2 + b^2$ and that gives you

$c \frac{d c}{dt} = a \frac{d a}{dt} + b \frac{d b}{dt}$

Now when $a = 3, b = 4$, $c = 5$

$5 \frac{d c}{dt} = 3 \frac{d a}{dt} + 4 \frac{d b}{dt}$ ...(ii)

From (i) and (ii), you can find $\frac{d b}{dt} = -96$ as you know $\frac{d a}{dt} = 108$.

Now $\frac{d A}{dt} = \frac{1}{2} \big(a \frac{d b}{dt} + b \frac{d a}{dt} \big)$


问:
How to find Integral $\int_{0}^{1/2} (e^x - 1)/x \,dx$?
can you give me any advice how to caluclate $\int_{0}^{1/2} \frac{e^x - 1}{x} \,dx$ ? I need to use series. I tried to split it as
$$
\int_{0}^{1/2} \frac{e^x}{x} \,dx - \int_{0}^{1/2} \frac{1}{x} \,dx$$ and then I can use a series for the first equation but I get another $\int_{0}^{1/2} \frac{1}{x} \,dx$. Can I somehow evaluate these integrals or I need to use a completely different approach. Thanks!

Body: Hint: $\displaystyle x\ne0\implies\frac{e^x-1}x=\sum_{n=0}^\infty\frac{x^n}{(n+1)!}$.


问:
A solution for $\nabla^2\psi = k\lvert\nabla\phi\lvert^2$ with $\nabla^2\phi = 0$
So, I was resolving the problem 1.9.13 from "George B Arfken, Hans J Weber - Mathematical Methods For Physicists- Sixth edition", what a need to show is that: $\begin{equation}\psi = \frac{1}{2}k\phi^2\end{equation}$ is a solution for the equation $\nabla^2\psi = k\lvert\nabla\phi\lvert^2$ with $\nabla^2\phi = 0$, where $\phi$ and $\psi$ are scalar functions. I was able to show that by doing:

$\begin{equation}\nabla\cdot(\nabla\psi) = k\lvert\nabla\phi\lvert^2 = k\lvert\nabla\phi\lvert^2 + \underbrace{k\phi\nabla\cdot(\nabla\phi)}_{k\phi\nabla^2\phi = 0} = \nabla\cdot(k\phi\nabla\phi)\implies k\phi\nabla\phi = \nabla\psi\end{equation}$

Since $\nabla\phi\cdot d\vec{r} = d\phi$ and $\nabla\psi\cdot d\vec{r} = d\psi$:

$\begin{equation}k\phi d\phi = d\psi\end{equation}$ integrating both sides I have $\begin{equation}\int k\phi d\phi = \int d\psi \implies \psi = \frac{1}{2}k\phi^2\end{equation}$

But I have to say that I'm feeling a little guilty, I was only able to solve that because I already know the answer, so I first applied the gradient on both sides of $\begin{equation}\psi = \frac{1}{2}k\phi^2\end{equation}$ and I saw that I needed to put the null term $k\phi\nabla\cdot(\nabla\phi)$ on the equation in order to solve that. So I was wondering if is there another way to solve this equation supposing that I don't know already the answer to that. Could someone please help me with an alternative solution for that?

答:
Certainly, if $\psi =\frac12 k\phi^2$, then we have

$$\begin{align}
\nabla^2 \psi&=\frac12 k \nabla ^2 \phi^2\\\\
&=\frac12 k \nabla \cdot \nabla \phi^2\\\\
&=\frac12k \nabla \cdot (2\phi \nabla \phi)\\\\
&=k\nabla \cdot (\phi \nabla \phi)\\\\
&=k\phi \nabla^2\phi +k\nabla \phi \cdot \nabla \phi\\\\
&=k|\nabla \phi|^2
\end{align}$$

And we are done!

If we begin with the equality $\nabla^2 \psi=k|\nabla \phi|^2$, then using $\nabla^2\phi=0$, we find that

$$\begin{align}
0&=\nabla^2 \psi-k|\nabla \phi|^2\\\\
&=\nabla\cdot \left(\nabla \psi-k\phi \nabla \phi\right)\\\\
&=\nabla\cdot \nabla \left(\psi-\frac12 k\phi^2\right)\\\\
\end{align}$$

Note that both the curl and divergence of the vector $\nabla \left(\psi -\frac12 k\phi^2\right)$ are equal to $0$.  If $\lim_{r\to \infty}r\nabla(\psi-\frac12 k\phi^2)=0$, then appealing to Helmholtz's Theorem we assert that $\nabla \left(\psi -\frac12 k\phi^2\right)=0$.

This implies that $\psi=\frac12 k\phi^2 +C$, where $C$ is a constant.  If we demand that $\psi=\frac12k \phi^2$ at any point, then we conclude that $\psi =\frac12k\phi^2$.


问:
solving nonlinear equation with logarithmic terms
Let $0<a,b<1.$ How should one solve (for $x$)
$$
x+a \log x= \log y -b \log \log y.
$$
Please note the two opposite signs on $a,b.$ We can assume $y$ is very large and positive. Maybe the $W$ function is useful, but I am not an expert.

Even locating the solution in a manageable interval, whose two ends are in closed form would be good.

答:
Let $z$ be your right-hand side. Then some manipulations (divide by $a$, exponentiate, divide by $a$ again) yields
$$x+a \log x=z\implies (x/a)e^{x/a} = e^{z/a}/a\implies x/a = W(e^{z/a}/a)$$ where $W$ is the Lambert-W function. So $x=a W(e^{z/a}/a)$.


问:
Harmonic function defined through harmonic functions
Let $u(x,y)$ and $v(x,y)$ be a pair of conjugate harmonic functions, and let $x(u, v)$ be a harmonic function of the variables $u$ and $v$.
Prove that $\chi [u(x,y),v(x,y)]$ is a harmonic function of the variables $x$ and $y$.

I know that $u$ and $v$ satisfies the Cauchy Riemann ecuations, $\frac{\partial^2u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ (similar for $v$), and exist and are continuous $\frac{\partial^2 u}{\partial x^2}, \frac{\partial ^2 u}{\partial x \partial y}, \frac{\partial^2 u}{\partial y \partial x}, \frac{\partial^2u}{\partial y^2}$ (similar for $v$) and analogously for $x(u,v)$.

I guess I have to prove that the same occurs for $\chi$, but I can't see the relation between the partial derivatives of $ \chi $ and those of $ x (u, v) $.

I hope you can help me, thank you.

答: Let $f = u+iv$ so that $f$ is analytic. $\chi$ is harmonic so it is locally the real part of an analytic function $g$. Compositions of analytic functions are analytic and hence $\chi\circ f = \mathrm{Re}g\circ f$ is the real part of an analytic function, hence harmonic


问:
Likelihood function for data y under given distribution with probability p
I am struggling with the following question:

Consider a random variable (RV) Y that follows the distribution:
$$P(Y=y|p) = (1-p)^yp$$
where $y \in$ {0,1,2,..} is a non-negative integer, and p is the parameter of the distribution. Imagine we observe a sample of n non-negative integers y={$y_{1},....y_n$) and want to model them using the given distribution ( the data is independently and identically distributed).

Write down the likelihood function for the data y ( i.e, the joint
probability of the data under the given distribution with the
probability parameter p )

I have got as far as the following ( I think) but am lost on the next step.

$\prod_{i=0}^y p \prod_{i=0}^y(1-p)^y$

I have tagged maximum-likelihood because there is no likelihood tag.

答:
The likelihood function of the parameter $p$ given the data $(y_1,\cdots,y_n)$ is
$$
L(p\mid y)=P(Y_1=y_1,\cdots, Y_n=y_n\mid p)=\prod_{i=1}^n P(Y_i=y_i\mid p)
$$
where $Y_i$'s are iid copies of $Y$.

Now you simply do the substitution.


答:
Hint

$$c^2(t)=a^2(t)+b^2(t)\to c(t)c'(t)=a(t)a'(t)+b(t)b'(t)\quad (1)$$ and $$a(t)+b(t)+c(t)=k=cte\to a'(t)+b'(t)+c'(t)=0.\quad (2)$$

The area is given by

$$A(t)=\frac{a(t)b(t)}{2}\to A'(t)=\frac{a'(t)b(t)+a(t)b'(t)}{2}.\quad (3)$$

You have $a'(t)=108$ when $a(t)=3$ and $b(t)=4$ (it gives us that $c(t)=5$). Use that at $(1)$ and $(2)$ and find $b'(t)$. Then find your answer at $(3)$.

Title: Find rate of change of the area of a triangle at a particular moment

Body:
The side lengths of a right-angled triangle are $a, b$ and $c$ centimeters where $c$ is the length of the hypotenuse and $a, b$ and $c$ are changing as differentiable functions of time. It is known that the perimeter $a+b+c$ is always constant and at the moment $a=3$ and $b=4$ the rate of change of $a$ is $108$ cm/sec. What is the rate of change of the area of the triangle (in cm$^2$/sec units) at this moment?

This is a related rates question (the answer is $72$). What I tried and first of all I calculated the derivative of the area of a right triangle ($ab/2$) and everything is known except the derivative of side $b$. I need to find some relation between $b$ and $a$ but I am not sure how.


问:
Hartshorne Ex. II.3.21: computing height of an ideal.
Consider a DVR $R$ with the maximal ideal $(u)$ containing its residue field $k=R/(u)$. The exercise claims that the dimension of $\operatorname{Spec}(R[t])$ is not equal to the Krull dimension of all stalks $\mathcal{O}_p$  of the structure sheaf at closed points $p$.

As far as I understand, this is due to the fact that we have maximal ideals of different heights. In this case, the height of $(1-ut)$ is 1, while the height of $(u,t)$ is 2. But how can I show that the height of $(1-ut)$ is actually 1? And how does the requirement that $k\subset R$ enters this argument?

答:
Krull's height theorem is what you're looking for: any minimal prime ideal lying over an ideal with $n$ generators is of height at most $n$. Since $(1-ut)$ is generated by one element and is prime (the quotient is $\operatorname{Frac} R$, a domain), it must be of height at most one. It's not of height zero, since that would imply it's a minimal prime of $R[t]$ - as $R[t]$ is a domain, its unique minimal prime is $(0)\neq(1-ut)$.

$k\subset R$ is not actually important to the argument here. Letting $u$ be a uniformizer for $R$, we have $R[t]/(t,u)=k$, while $R[t]/(1-ut)=\operatorname{Frac} R$, regardless of whether $k\subset R$ or not. The reason Hartshorne points this out is to emphasize that the nice conclusions about dimension really do depend on finite type over a field - just being a scheme over a field is not enough.


问:
Finding the last two digits of $3^{400}$
I'm having some trouble with the following exercise:

Find the last two digits of the decimal representation of $3^{400}$.

This exercise Is under the Euler's theorem and Fermat's little theorem section of the worksheet so I think that we are supposed to use those to solve this. My problem is that this is the first time I'm seeing an exercise of this kind and I don't know how to proceed. Can someone give me a tip on how to do this or show me how it's done?

答:
Alternatively, you could use binomial expansion:

$3^{400}=(10-1)^{200}\equiv-200\times10+1\equiv1\bmod100$

(in fact, $\bmod 1000$).


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

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

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

For example, the first general solution presented reads:

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

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

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

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

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

答:
This sort of thing has a clear description in terms of Conway's topograph; I find it more convenient to use equivalent form $u^2 + uv - 5 v^2.$  The outcome for the original problem is sequences (note that the rule deals with every other element). For instance,
$5\cdot 9 - 2 = 43$  and $5 \cdot 14 -3 = 67$

$$   x_{n+4}= 5 x_{n+2} - x_n,$$
$$   y_{n+4}= 5 y_{n+2} - y_n,$$

There are two interleaved subsequences.

$$
\begin{array}{c}
1&1&2&3&9&14&43&67&206&321&987 &1538&4729& \ldots \\
3&2&1&1&2&3&9&14&43&67&206&321&987&\ldots \\
\end{array}
$$
Running the indices backwards leads to different solutions, but they are just transpositions of the ones above.

Let's see, given $u^2 + uv - 5 v^2 = -5$ and $x=u+3v, y=v$ gives $x^2 - 5 xy + y^2 = -5$ and the reverse holds as well

Apparently I drew one of these in 2016

Examples in previous answers:

Generate solutions of Quadratic Diophantine Equation
diagrams

Another quadratic Diophantine equation: How do I proceed?

How to find solutions of $x^2-3y^2=-2$?

Generate solutions of Quadratic Diophantine Equation

Why can't the Alpertron solve this Pell-like equation?

Finding all solutions of the Pell-type equation $x^2-5y^2 = -4$

If $(m,n)\in\mathbb Z_+^2$ satisfies $3m^2+m = 4n^2+n$ then $(m-n)$ is a perfect square.

how to solve binary form $ax^2+bxy+cy^2=m$, for integer and rational $ (x,y)$  ::::  69   55

Find all integer solutions for the equation $|5x^2 - y^2| = 4$

Positive integer $n$ such that $2n+1$ , $3n+1$ are both perfect squares

Maps of primitive vectors and Conway's river, has anyone built this in SAGE?

Infinitely many systems of $23$ consecutive integers

Solve the following equation for x and y:  <1,-1,-1>

Finding integers of the form $3x^2 + xy - 5y^2$ where $x$ and $y$ are integers, using diagram via arithmetic progression

Small integral representation as $x^2-2y^2$ in Pell's equation

Solving the equation $ x^2-7y^2=-3 $ over integers

Solutions to Diophantine Equations

How to prove that the roots of this equation are integers?

Does the Pell-like equation $X^2-dY^2=k$ have a simple recursion like $X^2-dY^2=1$?

http://math.stackexchange.com/questions/1737385/if-d1-is-a-squarefree-integer-show-that-x2-dy2-c-gives-some-bounds-i/1737824#1737824 "seeds"

Find all natural numbers $n$ such that $21n^2-20$ is a perfect square.

Is there a simple proof that if $(b-a)(b+a) = ab - 1$, then $a, b$ must be Fibonacci numbers? 1,1,-1; 1,11

To find all integral solutions of $3x^2 - 4y^2 = 11$

How do we solve pell-like equations?

Diophantine equation $x^2 + xy − 3y^2 = 17$ <1,1,-3>


问:
Solving the probability difference equation $p_{n} = p(1 - p_{n-1}) + (1 - p)p_{n-1}$ yields a questionable closed-form equation?
I am trying to solve the following difference equation by writing it in closed-form, which represents the probability that a person flips an even number of heads in $n$ tries, with $p$ being the probability that a person flips a head at any given time: $p_{n} = p(1 - p_{n-1}) + (1 - p)p_{n-1}$.

To derive this equation, I did a simple case analysis. There are 2 cases that $n$ flips yields an even number of heads:

odd # of heads $n - 1$ times, then a head: $(1 - p_{n - 1})p$

even # of heads $n - 1$ times, then a tail: $p_{n-1}(1 - p)$

Adding the cases up yields the difference equation I wrote above.

My strategy was to derive the characteristic equation, as this is a non-homogeneous linear equation:

$p_{n} = p - pp_{n-1} + (1 - p)p_{n-1}$

$p_{n} = (1 - 2p)p_{n-1} + p$

$x = (1 - 2p)$

$p_{n} = C(1-2p)^{n}$

We know that $p_{1} = p$:

$C(1 - 2p) = p$

$C = \frac{p}{1 - 2p}$

And plugging back into the original equation, we get:

$p_{n} = \frac{p}{1 - 2p}(1 - 2p)^n = p(1 - 2p)^{n - 1}$.

However, this does not seem to be correct, because if $p$ is $1/2$ (if it is a $1/2$ chance of getting a head on a flip), $p_{n}$ will always be 0.

答:
It is maybe worth pointing out that it's possible to compute this probability without using a recurrence, as follows. Let $E_n$ be the probability of flipping an even number of heads and let $O_n$ be the probability of flipping an odd number of heads. We have

$$E_n + O_n = 1$$
$$E_n - O_n = \sum_{k=0}^n (-1)^k {n \choose k} p^k (1 - p)^{n-k} = (1 - 2p)^n$$

which immediately gives

$$E_n = \frac{1 + (1 - 2p)^n}{2}, O_n = \frac{1 - (1 - 2p)^n}{2}.$$

This is a special case of a general trick for isolating the terms with exponent congruent to $a \bmod n$ in a generating function using the discrete Fourier transform, which is sometimes known in a contest-math setting as the "roots of unity filter." Here we are using the second roots of unity $\pm 1$ only, but if we used $k^{th}$ roots of unity then we could express e.g. the probability of flipping a number of heads divisible by $k$.


问:
What are the constructible angles and regular/uniform/semiregular/etc polytopes in n-dimensional integer lattices?
In a 2-dimensional regular square lattice, using only lattice points, lines connecting lattice points, and points where such lines intersect, I am aware it is impossible to make a 30-degree angle between two lines, and hence, impossible to make an equilateral triangle. Using a regular triangular lattice, however, you can make such angles and triangles, but you lose the ability to construct 45-degree angles, and hence, no squares.

In 3-dimensional space, however, using a cubical lattice with the same principles and extending to constructible planes given three non-colinear points, you can construct both 30-degree and 45-degree angles. For example, the points (0,0,0), (1,1,0), (1,0,1), and (0,1,1) together make a tetrahedron, which includes equilateral triangles.

One thing, iirc, you apparently cannot construct in 2-d or 3-d lattices of this type, despite being able to make 45-degree angles, are regular octagons. You can make equiangular ones, but not with equal side lengths as well.

So I’m wondering, does using regular lattices in four (or more) dimensions allow constructions of any new similar angles, polygons, or polyhedra, that can not be constructed in 3 dimensions? Also, just like how square and triangular lattices in 2-d allows different constructions, are there different regular lattices in 3-d that differ? It seems to me that everything that can be constructed in a 3d cubic lattice can be constructed by a 3d tetrahedral-octahedral one, and vice-versa, though tilted at a different angle from the axes (or am I incorrect in thinking that?). Finally, I know any lines with rational endpoints can only intersect at rational points in 2d Cartesian space, but does this hold in 3 dimensions? Anyone have any insights or resources on this topic?

答: Whilst traditional geometric construction relates to the plane with ruler and compass, iterative application of these simple methods reaches all constructible lengths and the 2d lattice can be generalised.  Viewed algebraically these finitely constructible lengths sit in a tower of field extensions as the roots of quadratics whose coefficients are the level below.  Pythagoras works in any number of dimensions because any three points in space are coplanar and form a triangle.</p>


问:
Solve equations $\sqrt{t +9} - \sqrt{t} = 1$
Solve equation:  $\sqrt{t +9} - \sqrt{t} = 1$

I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$

I squared both sides $(\sqrt{t+9})^2  = (1)^2 (\sqrt{t})^2$

Then I got $t + 9 = 1+ t$

Can't figure it out after that point.

The answer is $16$

答:
Hint. Let $x=\sqrt{t}$ and $y=\sqrt{t+9}$. Then
$$
y-x=1\implies y^2-x^2=y+x\implies 9=y+x
$$

ContentLicense: CC BY-SA 3.0


问:
Derive a recurrence relation with initial conditions for the sequence of hexagonal numbers
"hexagonal diagrams”
(diagrams for n = 1, 2, 3, 4 in picture)

Let H(n) be the n-th hexagonal number, which is given by the number of dots in the n-th hexagonal diagram. For example, H(1) = 1, H(2) = 6, and H(3) = 15.

Derive a recurrence relation with initial conditions for H(n).

I know that the hexagonal numbers can be derived using the formula: H(n) = n(2n-1) for the n-th hexagonal number however I am unsure of how to find a recurrence relation.

答:
Examine the diagrams to see how many dots are added to get from the diagram for $H_n$ to the diagram for $H_{n+1}$. They’re all added in the outermost hexagon. If we label the sides $1$ through $6$ counterclockwise, starting with the side rising up and to the left from the common vertex of the hexagons, we can see that side $1$ gets one new dot. Side $2$ gets that dot plus $n$ more. One of those $n$ dots is also on side $3$, and then there are $n$ more dots on side $3$. Keep going, and you’ll get an expression for $H_{n+1}-H_n$ in terms of $n$, which you can rearrange to get the desired recurrence. I’ve left the details in the spoiler-protected block below.

 To get the diagram for $H_{n+1}$ from the diagram for $H_n$, you add one dot at the upper right corner, then you add $n$ dots $4$ times, so you add altogether $4n+1$ dots. Thus, $H_{n+1}=H_n+4n+1$.


答:
You cannot split the integral into two unless at least one of them is convergent.

By the series definition of the exponential function, for $x\ne 0$
$$
\frac{e^x-1}{x}=\frac1x(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots )
=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots
$$
Now define the function $f:[0,1/2]\to\mathbb{R}$ with $f(x)=\frac{e^x-1}{x}$ if $x\ne 0$ and $f(0)=1$. Then the function $f$ is continuous on $[0,1]$.

So you can write
$$
f(x)=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots
$$
and
$$
\int_0^{1/2} \frac{e^x-1}{x}\,dx=\int_0^{1/2}f(x)dx=
\int_0^{1/2}(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots)\,dx
$$

Now do integration term by term.

Title: How to find Integral $\int_{0}^{1/2} (e^x - 1)/x \,dx$?

Body:
can you give me any advice how to caluclate $\int_{0}^{1/2} \frac{e^x - 1}{x} \,dx$ ? I need to use series. I tried to split it as
$$
\int_{0}^{1/2} \frac{e^x}{x} \,dx - \int_{0}^{1/2} \frac{1}{x} \,dx$$ and then I can use a series for the first equation but I get another $\int_{0}^{1/2} \frac{1}{x} \,dx$. Can I somehow evaluate these integrals or I need to use a completely different approach. Thanks!


问:
Solutions to $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$

Find all positive integer solution pairs $(m,r)$ to $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$.

I thought about using Pell's equation, but that only tells us solutions such that $\sqrt{48r^2+1}$ is an integer. How do we deal with the $\dfrac{1}{18}$ and $-1$?

Body:
Whether or not it is a Pell-type equation, it is solveable.

$$m = \frac{\sqrt{48r^2+1}-1}{18}
\implies 18m = \sqrt{48r^2+1}-1$$

$$(18m+1)^2=48r^2+1\implies 48r^2=(18m+1)^2-1
$$
$$r= \sqrt{\frac{(18m+1)^2-1}{48}}$$

Notice in the $(m,r)$ samples below that $m$ occurs only half the time and  as a [step] funtion of $n$.

$$m=\frac{8n-1}{2}\mp 0.5\quad n\in\mathbb{N}$$

$$
( 3,\pm 63)\space
( 4,\pm 111)\qquad
( 7,\pm 336)\space
( 8,\pm 438)\qquad
( 11,\pm 825)\space
( 12,\pm 981)\\
( 15,\pm 1530)\space
( 16,\pm 1740)\qquad
( 19,\pm 2451)\space
( 20,\pm 2715)\qquad
( 23,\pm 3588)\space
( 24,\pm 3906)\\
( 27,\pm 4941)\space
( 28,\pm 5313)\qquad
( 31,\pm 6510)\space
( 32,\pm 6936)\qquad
( 35,\pm 8295)\space
( 36,\pm 8775)\\
( 39,\pm 10296)\space
( 40,\pm 10830)\qquad
( 43,\pm 12513)\space
( 44,\pm 13101)\qquad
( 47,\pm 14946)\space
( 48,\pm 15588)$$

The values $(3,4,\space 7,8,\space 11,12,\space 15,16,\space 19,20,\space 23,24,...)$ are part of series
A014601.


问:
error in integration by substitution
Can someone explain me what I get wrong here ?

I have this integral
\begin{equation}
\int \frac{1}{x(2 \ln x+1)} dx
\end{equation}

I set
$ u=\ln x $ and
$ du=\frac{1}{x} dx $
and so I obtain
\begin{equation}
\int \frac{1}{2u+1} du
\end{equation}
and then the solution seems to be
\begin{equation}
\ln|2u+1|=\ln|2 \ln x+1|+c
\end{equation}
However, the correct solution is
\begin{equation}
\frac{\ln|2 \ln x+1|}{2} +c
\end{equation}

答: That's because$$\int\frac1{2u+1}\,\mathrm du=\frac12\log|2u+1|.$$In general, if $F(x)$ is a primitive of $f(x)$, then $\frac1aF(ax)$ is a primitive of $f(ax)$ ($a\ne0$).

Title: error in integration by substitution


问:
Finding mean and variance through mean and variance of subsets
Suppose I have a big subset of smaller subsets where each small subset contains x amount of numbers.
How would I go about finding the mean and variance of the larger subset through the mean and variance of fragments of the smaller subsets?

For example:
Let subset A = [[1,3], [5, 7], [10, 19]]

Lets split subset A into the 2 smaller subsets -- B and C

B = [[1, 3]] and C = [[5, 7], [10, 19]]

If we know the mean and variance of B is 2 and 1 respectively,
and we know the mean and variance of C is 10.25 and 28.6875 respectively.

How would we find the mean and variance of A from the mean and variances of B and C?

Thanks!

答:
You seem to be using the $\frac1n$ version of variance

If $n_B$ and $n_C$ are the number of elements of $B$ and $C$, and $\mu_B, \mu_C, \sigma^2_B, \sigma^2_C,$ their means and variances then:

$n_A=n_B+n_C$

$\mu_A=\frac1{n_A}(n_B \mu_B+n_C\mu_C)$

$\sigma^2_A=\frac1{n_A}(n_B (\sigma^2_B +\mu_B^2)  +n_C (\sigma^2_C +\mu_C^2)) - \mu_A^2$


问:
Showing that $\mathbb{R}^n - D^n$ is path-connected

Show that $\mathbb{R}^n - D^n$ is path-connected, for $n>1$. Here $D^n$ is the closed ball centred at the origin $O$ with radius 1.

To be honest, I solved this problem. I spent around 30 minutes to solve this. I thought about many ways, but failed in all but one. I am curious to know other ways to solve this problem.

I won't post my solution intentionally. I am curious about other ways to solve this. This problem seems trivial for $\mathbb{R}^2$ and $\mathbb{R}^3$ but gets difficult for $n>4$.

Thank you.

答:
Solution 1:

Just for convencience sake, via a dilatation we can wlog assume that the space is rather $X = \mathbb{R}^n \setminus \frac{1}{2} D^n$.

Now,

The sphere $S^n$ is path connected: if $x,y \in S^n$ are not antipodal and $c(t)$ is the segment joining $x$ with $y$, then $c/\|c\|$ is a path from $x$ to $y$ contained in the sphere. If $x = -y$, pick a third point, use transitivity of path connectedness.

Once again; path connectedness is transitive so it is enough to note that any point can be connected to a point in $S^n$. If $x \in X$, so is $x/\|x\|$, and you can check that the segment joining $x$ and $x/\|x\|$ is contained in $X$.

Solution 2:

A techonological argument: path-connectedness is a homotopy invariant, hence it's sufficient to prove it for $Y = \mathbb{R}^n \setminus \{0\} \simeq X$. Now pick $x,y \in Y$. If $x \not \in \langle y\rangle$, the segment $\vec{xy}$ is contained in $Y$. Otherwise pick $z \not \in \langle y\rangle$ and by the exact same argument connect $x$ and $y$ to $z$ via their respective segments. Now use transitivity.


问:
What is wrong with my 'proof' of $i=1$?
So I recently learnt about $i$ and I can't wrap my head around the concept of $i^2=-1$ or that $\sqrt{-1}$ can even exist. Today I was thinking about $i$ again and thought of a "proof" that $i=1$.
$$i^4=1 \text{ and } 1^4=1 \text{ so } i^4=1^4$$
$$i^4=1^4 \to \sqrt[4]{i^4}=\sqrt[4]{1^4} \to i=1$$
But if $i=1$, then $i^2 \neq -1$. So I think I must have messed up something in the proof. Can someone point out where this went wrong? I know you can $\text{"prove" }1=2$ by accidentally dividing by $0$ and I suspect something similar is happening.

For anyone else having trouble with complex numbers @mrsamy commented this link and I found it quite helpful: https://www.math.toronto.edu/mathnet/answers/imaginary.html

答: This has nothing to do with complex numbers. By the same argument, $-1=1$, since$$(-1)^2=1^2\implies\sqrt{(-1)^2}=\sqrt{1^2}\implies-1=1.$$The error lies in assuming that $\sqrt{x^2}=x$. Actually, $\sqrt{x^2}=|x|$. In the case of complex numbers, it's even worst, since every complex number (other than $0$) has four fourth roots. So, the expression $\sqrt[4]z$ doesn't make sense unless and until you decide which fourth root of $z$ you have in mind. Even then, it will often be false that $\sqrt[4]{z^4}=z$.

Title: What is wrong with my 'proof' of $i=1$?


答:
Square roots don't work like that. With the same argument you used, you could have done
$$
(-1)^2=1^2\ \ \ \implies\ \ \ -1=1.
$$
It has nothing to do with complex numbers.

Title: What is wrong with my 'proof' of $i=1$?

Body:
So I recently learnt about $i$ and I can't wrap my head around the concept of $i^2=-1$ or that $\sqrt{-1}$ can even exist. Today I was thinking about $i$ again and thought of a "proof" that $i=1$.
$$i^4=1 \text{ and } 1^4=1 \text{ so } i^4=1^4$$
$$i^4=1^4 \to \sqrt[4]{i^4}=\sqrt[4]{1^4} \to i=1$$
But if $i=1$, then $i^2 \neq -1$. So I think I must have messed up something in the proof. Can someone point out where this went wrong? I know you can $\text{"prove" }1=2$ by accidentally dividing by $0$ and I suspect something similar is happening.

For anyone else having trouble with complex numbers @mrsamy commented this link and I found it quite helpful: https://www.math.toronto.edu/mathnet/answers/imaginary.html


答:
after squaring both sides you get:

$$t+9=1+2\sqrt{t}+t\implies4=\sqrt{t} \implies t=16$$

Title: Solve equations $\sqrt{t +9} - \sqrt{t} = 1$

Body:
Solve equation:  $\sqrt{t +9} - \sqrt{t} = 1$

I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$

I squared both sides $(\sqrt{t+9})^2  = (1)^2 (\sqrt{t})^2$

Then I got $t + 9 = 1+ t$

Can't figure it out after that point.

The answer is $16$


问:
Topology continuity
Let X={a,b,c,d} with topology

T={Φ,X,{a},{b},{a,b}, {b,c,d}} and let

F: (X,T) →(X,T) such that

f(a)=f(c)=b, f(b)=d, f(d)=c,

Choose the correct answer:

a) f is discontinuous everywhere.

b) f is not continuous at c.

c) f is not continuous at d.

d) f is continuous on X

Can someone help me solve this question.

If you need any clarification please ask in the comments.

答:
Recall the definition that f:X -> Y is continuous at x when

for all open V nhood f(x),

exists open U nhood x with

f(U) subset V.

Now check each point, a,b,c,d, and determine if f is continuous at that point.

Title: Topology continuity


问:
Solutions of $x'' + 2x' + x = 0$ such that $x(1) = 0$
I am trying to solve this question from an older exam that my teacher uploaded. The marks given for answering this question were very low, so I am thinking there is an easy way to solve it.
The question asks to find the solutions of $x'' + 2x' + x = 0$ such that $x(1) = 0$.
My approach is using the change $y=x'$, and therefore getting:
$$
x' = y
$$
$$
y' = -x -2y
$$
Which is a linear system. What I would normally do here is finding the eigenvalues (in this case there is only one eigenvalue, -1 with eigenvector (1,-1)), and solve the corresponding system using the Jordan matrix and the Jordan basis, but I was thinkin whether there was a faster approach, maybe playing with the fact that $x(1)=0$, or something else.

Thanks

答:
It is much too complicated, as this is a o.d.e. with constant coefficients. Its characteristic equation is $(\lambda+1)^2=0$.  As it has a double root, $\lambda=-1$, a basis of the space of solutions is made up of the functions $\mathrm e^{-t}$ and  $t\mathrm e^{-t}$. Therefore the general solution  is
$$x(t)=\mathrm e^{-t}(\alpha +\beta t),$$
and there remains to determine the relation between $\alpha$ and $\beta$ so that $x(0)=1$.


问:
finding the limit of the integral of an unknown function
I have a question from an exam written 18 years ago, and I can't solve it (I'm solving old questions to prepare for my own exam).

The Question

The question goes like so:

let  g be a function, which has a derivative defined $\forall x\in [-1,1]$, and $g(0)=0,g'(0)=1$.
Does the limit $$\lim_{x\rightarrow0}\frac{\sin\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{x^{8}}$$ converge?

What have I tried?

I thought about using L'Hôpital's rule, and F.T.C. so I wrote:

$$\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)'=2x\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x^{2}\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds$$

but that didn't help much. I can even show that the limit is of the form $"\frac{0}{0}"$ to use the rule.

I tested this on $g(x)=x$ and it seems like this limit does converge (to 0 if I'm not mistaken)

I also thought Taylor was somehow related, but I really doubt it. I am also unsure of where the information we have about $g(0),g'(0)$ comes in.

BTW

this is what I got after applying L'Hôpital once

$$\lim_{x\rightarrow0}\frac{\left(2\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds\right)\cos\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{8x^{6}}$$

答:
Let us write

\begin{equation}g (x) = x h (x)\end{equation}

so that $h (x) \rightarrow  1$ when $x \rightarrow  0$. Let us first substitute $s = t u$ in the integral

\begin{equation}I \left(x\right) \colon  = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}g \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}{s}^{2} h \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}{t}^{3} \left(\int_{0}^{1}u^2h \left({t}^{2} {u}^{2}\right) d u\right) d t\end{equation}

Now let us substitute $t = {x}^{2} v$, it follows that

\begin{equation}I \left(x\right) = {x}^{8} \int_{x}^{1}v^3\left(\int_{0}^{1} u^2h \left({x}^{4} {u}^{2} {v}^{2}\right) d u\right) d v \sim  {x}^{8} \int_{0}^{1}{v}^{3} \left(\int_{0}^{1}u^2 d u\right) d v = \frac{{x}^{8}}{12}\end{equation}

As $\sin  \left({\theta}\right) \sim  {\theta}$ when ${\theta} \rightarrow  0$ it follows that the limit is $1/12$.


问:
Existence of a measure space with a given condition
Motivated by this question and this question, among others, I wondered when can we say that there exists a measure space $(X, \mathcal{S}, \mu)$ such that
$$\operatorname{Im}(\mu) = A \cup\left\{0,+\infty\right\},$$
where $A \subset \mathbb{R}$ is a subset of the real numbers. This question may be too general, so in particular I am interested in the cases where $A = \mathbb{Q}$ and $A = \mathbb{R}\setminus \mathbb{Q}$. Thanks in advance!

答:
Thanks to Ramiro for pointing out a flaw in my original post.

I will stick with the case of measures, since this situation covers the all of the ideas. As it turns out, both cases in the question can be covered by a somewhat more general result. Using these ideas, one can probably come up with more classes of sets that aren't the ranges of measures, but I haven't yet come up with a good characterization.

Proposition: Let $A$ be dense in $\mathbb{R}$ with empty interior. Then there is no measure space $(X, \mathcal{S}, \mu)$ such that $\operatorname{Im}(\mu) = A \cup \{0, \infty\}$.

Proof: Suppose for contradiction that there is such a measure space. We will need the following claim:

Claim: For each $n \in \mathbb{N}$,
$$\sum_{\substack{E \text{ atom} \\ \mu(E) < \frac{1}{n}}} \mu(E) = \infty.$$
Assuming this claim, let $x \in \mathbb{R^{+}} \backslash A$. Since $A$ has empty interior and many sets of positive, finite measure, there are atoms with arbitrarily small measure. Using the claim with $n = 1$, we can find atoms $E_{1}, \dots, E_{m_{1}}$ such that
$$x - 1 < \sum_{k = 1}^{m_{1}} \mu(E_{k}) < x.$$
Applying the claim with $n = 2$, we can find atoms $E_{m_{1} + 1}, \dots, E_{m_{2}}$ such that
$$x - \frac{1}{2} < \sum_{k = 1}^{m_{2}} \mu(E_{k}) < x.$$
Some induction gives us atoms $E_{k}$ for $k \in \mathbb{N}$ with
$$\sum_{k = 1}^{\infty} \mu(E_{k}) = x.$$
Since the $E_{k}$'s are all atoms, the measure of their union is the sum of their measures, so
$$\mu(\cup_{k = 1}^{\infty} E_{k}) = x,$$
which is a contradiction, proving the proposition.

Proof of claim: Let $n \in \mathbb{N}$, and assume for contradiction that
$$\sum_{\substack{E \text{ atom} \\ \mu(E) < \frac{1}{n}}} \mu(E) < \infty.$$
Then there are countably many atoms $\{E_{k}\}$ with $\mu(E_{k}) < \frac{1}{n}$. Since their sum is finite, we can order the atoms such that $\mu(E_{k + 1}) \leq \mu(E_{k})$.

Let $E = \cup_{k} E_{k}$. We know that $\mu |_{X \backslash E}$ has no atoms of size smaller than $\frac{1}{n}$. Since the image of $\mu |_{X \backslash E}$ still has empty interior, we have that
$$\operatorname{Im}(\mu |_{X \backslash E}) \subseteq \{0\} \cup \bigg[\frac{1}{n}, \infty\bigg].$$
Hence, $\operatorname{Im}(\mu |_{E})$ is dense in $(0, \frac{1}{n})$. In particular, this means that for each $m > 2$,
$$\sum_{k = m}^{\infty} \mu(E_{k}) \geq \mu(E_{m - 1}).$$
Let $x \in (0, 1) \backslash \operatorname{Im}(\mu |_{E})$ be smaller than the measure of some $E_{k}$. Let $k_{1}$ be such that
$$\mu(E_{k_{1}}) < x < \mu(E_{k_{1} - 1}).$$
Find $k_{2}$ such that
$$\sum_{k = k_{1}}^{k_{2}} \mu(E_{k}) < x < \sum_{k = k_{1}}^{k_{2}} \mu(E_{k}) + \mu(E_{k_{2} + 1}).$$
This is possible since
$$\sum_{k = k_{1}}^{\infty} \mu(E_{k}) \geq \mu(E_{k_{1} - 1}) > x.$$
Find $n_{3}$ such that
$$\sum_{k = k_{1}}^{k_{2}} \mu(E_{k}) + \mu(E_{k_{3}}) < x < \sum_{k = k_{1}}^{k_{2}} \mu(E_{k}) + \mu(E_{k_{3} - 1}).$$
As for $k_{2}$, find $k_{4}$ such that
$$\sum_{k = k_{1}}^{k_{2}} \mu(E_{k}) + \sum_{k = k_{3}}^{k_{4}} \mu(E_{k}) < x < \sum_{k = k_{1}}^{k_{2}} \mu(E_{k}) + \sum_{k = k_{3}}^{k_{4}} \mu(E_{k}) + \mu(E_{k_{4} + 1}).$$
By induction, we'll end up with numbers $k_{m}$ such that
$$\sum_{m = 1}^{\infty} \sum_{k = k_{2 m - 1}}^{k_{2 m}} \mu(E_{k}) = x,$$
which is a contradiction since the union of these particular $E_{k}$'s is a set with measure $x$.


问:
Extending linear maps on modules.
Let $\{M_i\}_{i\in \mathbb{Z}}$ be an index family of $R$-modules. Then we know that the direct sum $ \bigoplus_{i\in \mathbb{Z}}M_i$ is a submodule of a direct product $ \prod_{i\in \mathbb{Z}}M_i$. Let $N$ be an $R$-module and if $$f:\bigoplus_{i\in \mathbb{Z}}M_i\to N$$ is any linear map. Can we always extend $f$ to a linear map $$\tilde{f}:\prod_{i\in \mathbb{Z}}M_i\to N .$$ That is, $\tilde{f}$ is a linear map such that $$\tilde{f}|_{\bigoplus_{i\in \mathbb{Z}}M_i} = f.$$ I know in vector spaces we can always construct $\tilde{f}$ with the help of a basis, but in general modules how do we proceed?

Secondly, if some $f$ has an extention to $\tilde{f}$, is $\tilde{f}$ unique?

答:
No, this is already false for $R = M_i = N = \mathbb{Z}$. Let me change the indexing set to $\mathbb{N}$ to avoid confusion. In this case linear maps

$$f : \bigoplus_{i \in \mathbb{N}} \mathbb{Z} \to \mathbb{Z}$$

can be arbitrary elements of the product $\prod_{i \in \mathbb{N}} \mathbb{Z}$, but it's known that linear maps

$$\tilde{f} : \prod_{i \in \mathbb{N}} \mathbb{Z} \to \mathbb{Z}$$

can conversely be identified with the infinite direct sum $\bigoplus_{i \in \mathbb{N}} \mathbb{Z}$ (this is due to Specker); in other words, a linear map $f$ has an extension $\tilde{f}$ iff it lies in the direct sum as a submodule of the direct product, and the extension is unique if it exists.

If $R$ is a field then extensions exist assuming the axiom of choice and they need not be unique. You can pick them by extending a basis of the direct sum to the direct product, and setting the values of the map arbitrarily on the new vectors in the basis.

Title: Extending linear maps on modules.


答:
Think of the complex numbers as a pair of real numbers $\mathbb{R}\times\mathbb{R}$, but with a really useful property: the product $(x, y)\cdot (0,1) = (-y, x)$. Which shows the rotation characteristic of the number i.

Thinking that way is clear that the number $(0,1)$ has the property of $(0,1)^2 = (-1,0)$.

Title: What is wrong with my 'proof' of $i=1$?


问:
breadth vs depth and time management in undergraduate math education
For the sake of time, is it bad to just accept some fast paced class's theorem's (such as MIT's algebra class) as true even if you don't completely understand the proof or can't remember the proof off top of your head (after a while has passed?). I often find myself wasting too much time trying to memorize proofs when that's not the point of the class (and I can actually wait to memorize the proof later). Sometimes, I get caught up in one detail of the proof for hours and end up not having time to learn how to actually use the theorem and do the homework. Also, is it bad to gain a complete and working understanding of something after you take the class. I am not mentally capable of fully absorbing both what the class want's us to get and thinking about it enough to have a complete and sufficiently deep understanding of the subject all in one semester. However, I feel bad if I wait to think deeply about the class material until after the course is done, but I simply don't have enough time to fully understand some things during the semester. I always hear this advice on making sure you understand everything when you are studying to practice/do mathematics, but that seems not practical if you are taking four or more classes and struggling to make sure you understand what you need to for your other classes have other obligations to attend to.(and also if English is not your first language). I feel like a lot of the advice I hear is for native English speakers (I came to the U.S. when I was four, so I'm practically a native English speaker, but not when it comes to understanding things well the first time through in math, or at least making the understand be thorough by a native English speaker's standards.)

The reason I ask this is because I can't say I fully understand something in English unless I can explain the proof verbally/descriptively to someone.

答:
At least for me, starting by trying to solve the homework questions, even if I hadn't fully grasped some proofs (or even full grasped the concepts involved) usually worked out better than trying to understand everything first and only then starting with the homework problems.

As long as I found a solution (where found means found it myself, though. I tried not to look it up before I hadn't spent considerable time pondering it), I wouldn't worry too much whether I understood all the theory - I just made sure I understood enough to see why this one solution worked.

At the end of the semester, before the final exam, I'd review all the theory. Only then did I strive to really get to the bottom of things. Quite often, concepts of proofs which had initially seemed inscrutable, suddenly seemed simple. It was often the insight I had gained later on during the course which helped me understand some of the concepts introduced earlier.

Also, I think it helps to try to understand the concepts first, and deal with the technicalities of the proofs later. Before working through a proof line-by-line, try to understand why the statement it proves is true. Once you have an idea of why something it's true, it's often much easier to understand the proof.

I must admit that for some courses, linear algebra in particular, I lot of things were a bit mysterious to me even after my final exam though. It was only when I learned functional analysis years later that I finally got to the bottom of them. Somehow, dealing with infinite-dimensional spaces and all the issues they can cause suddenly made the finite-dimensional case much easier to reason about.


答:
The problem boils down to parameterizing a path on the boundary of $D^n$ that connects two points $a, b$ on that boundary.  This is not difficult; let's assume that $\gamma_{a, b}(t)$ is such a parameterization, where $t$ increases from $0$ to $1$ and $\gamma_{a, b}(t)$ equals $a$ for $t=0$ and equals $b$ for $t=1$.

Now, if $p, q$, are two arbitrary points in $X = \mathbb{R}^n \setminus D^n$, two cases are possible:

The rectilinear segment from $p$ to $q$ is contained entirely in $X$, hence serves as a suitable path from $p$ to $q$.

The rectilinear segment from $p$ to $q$ intersects the boundary $D^n$ at two points, $a$ and $b$.  Here we would be tempted to concatenate: the rectilinear segment $p$ to $a$, then the path $\gamma_{a,b}$, then the segment $b$ to $q$.  The problem with this is that $\gamma_{a, b}$ does not lie in the open set $X$.  However, this is an easy fix: in the first paragraph of this answer, instead of $D^n$, use the closed ball centered at the origin and having radius $(1+\epsilon)$, where the positive $\epsilon$ is small enough to make sure that the latter ball does not contain $p$ or $q$.

Title: Showing that $\mathbb{R}^n - D^n$ is path-connected

Body:
Show that $\mathbb{R}^n - D^n$ is path-connected, for $n>1$. Here $D^n$ is the closed ball centred at the origin $O$ with radius 1.

To be honest, I solved this problem. I spent around 30 minutes to solve this. I thought about many ways, but failed in all but one. I am curious to know other ways to solve this problem.

I won't post my solution intentionally. I am curious about other ways to solve this. This problem seems trivial for $\mathbb{R}^2$ and $\mathbb{R}^3$ but gets difficult for $n>4$.

Thank you.


答:
In general,
$$\int a\frac{f'(u)}{f(u)}du=a\ln\lvert f(u)\rvert +C$$
So in your case, we have
$$\int\frac{1}{2}\frac{2}{2u+1}du=\frac{1}{2}\ln\lvert2u+1\rvert+C$$

By the way, without any substitutions you already had your integral in the above form, as
$$\int \frac{1}{x(2 \ln x+1)} dx=\frac{1}{2}\int\frac{2}{x(2 \ln x+1)} dx=\frac{1}{2}\int\frac{\left(\frac{2}{x}\right)}{2 \ln x+1}dx$$
$$=\frac{1}{2}\ln\lvert2\ln x+1\rvert+C$$

Title: error in integration by substitution

Body:
Can someone explain me what I get wrong here ?

I have this integral
\begin{equation}
\int \frac{1}{x(2 \ln x+1)} dx
\end{equation}

I set
$ u=\ln x $ and
$ du=\frac{1}{x} dx $
and so I obtain
\begin{equation}
\int \frac{1}{2u+1} du
\end{equation}
and then the solution seems to be
\begin{equation}
\ln|2u+1|=\ln|2 \ln x+1|+c
\end{equation}
However, the correct solution is
\begin{equation}
\frac{\ln|2 \ln x+1|}{2} +c
\end{equation}


问:
Finding power series for $f(x) = \frac{4x+53}{x^2-x-30}$

Given $f(x) = \dfrac{4x+53}{x^2-x-30}$, display it as a power series and find the radius of convergence. then calculate $f^{(20)}(0)$

So what I did was look at the Taylor Series Formula:
$$f(x) = \sum_{n=0}^{\infty} {\frac{f^{n}(0)}{n!} x^n } $$

$f(0) = \dfrac{53}{-30}$ and $f'(0) = \dfrac{-67}{(-30)^2}$.

But after that it only gets harder and harder to derive, thus making it harder to see a general form of the $n$th derivative.

Would be happy if someone gave me a direction.
Thank you

答:
Hint: Use partial fractions, and write it as the sum of geometric series.  Result:

$$\frac{7}{x-6}-\frac{3}{x+5}$$


答:
If this may reconcile you with complex numbers, consider this: a complex number is a pair of reals, and we define

addition as $(a,b)+(c,d):=(a+b,c+d)$,

multiplication as $(a,b)(c,d):=(ac-bd,ad+bc)$.

Then we accept the "shorthand" conventions $1=(1,0)$ and $i=(0,1)$. It is an easy matter to show that these rules form a consistent arithmetic with the four basic operation, where you can freely use these equivalences.

We also have the consequences

$$i^2=(0,1)(0,1)=(-1,0)=-1$$

and

$$i^4=(-1,0)(-1,0)=(1,0)=1.$$

But that means in no way that $i=1$. It just means that $i$ is one of the solutions of the equation

$$z^4=(1,0)=1.$$

All solutions are (see below)

$$(1,0)=1,\\(0,1)=i,\\(-1,0)=-1,\\(0,-1)=-i.$$

Unless you specify a convention, in the complex we don't know which solution $$\sqrt[4]1$$ denotes.

Final remark: in this discussion, we only used real numbers and the special symbol $i$.

$$z^4=(x,y)^4=(x^4-6x^2y^2+y^4,4yx^3-4xy^3)=(1,0)$$ requires $$x=0\lor y=0\lor x^2=y^2$$

(by cancelling the imaginary part).

By plugging these in the real part, we get
$$y^4=1\lor x^4=1\lor -4x^4=-4y^4=1.$$

As $x,y$ are real, the only options are

$$(\pm1,0),(0,\pm1).$$

Title: What is wrong with my 'proof' of $i=1$?


问:
Computation with a transformation of a random variable.
The random variable $X$ is uniformly distributed on the interval $[-4,4]$. Compute $P(X^2 ≤ 9)$.

I tried to define $Y = X^2$ and compute the density of this transformed variable. The PDF of the uniform distribution is $f_x(x) = \frac{1}{8}$. What I got as a transformed density is $f_y(y) = \frac{1}{16\sqrt(y)}$.

However, if I calculate the integral of that transformed density from $0$ to $9$, I get a wrong result. My guess is that it has something to do with the fact that the uniform can take negative values.

答:
Hint:
$$
P(X^2<9)=P(-3<X<3)=\int_{-3}^{3}f(x)\,dx
$$
where $f$ is the density function for $X$.

Now what you need is to find the correct density function for $X$ and then integrate it.

Notes.

It is fairly easy the see by intuitions of "uniform distribution" that
$$
P(-3<X<3)=\frac{\textrm{length of the interval }[-3,3]}{\textrm{length of the interval}[-4,4]}=\frac34
$$


问:
Fast method to convert Catalan path to lexicographic position.
I was looking through the questions: Fast way to get a position of combination (without repetitions) and (Fast way to) Get a combination given its position in (reverse-)lexicographic order for converting a combination to its lexicographic position and then going from the position to the combination. A logical extension of this is the generalized Catalan numbers which are basically combinations with an additional constraint.

Consider a gambler tossing a coin and winning 1\$ on heads and losing 1\$ on tails. We know he got $l$ tails and $l+k$ heads. The number of ways he could have done this is: ${k+2l \choose l}$ and if we wanted to generate the actual paths, we can do so using the approaches in the questions linked above (the binary representations of the combinations would be the tosses at which he got tails).

But now, we add the further constraint that he reaches $k\$$ for the first time after $k+2l$ tosses. The post: Probability that random walk will reach state $k$ for the first time on step $n$ shows that the number of ways to do this becomes: ${k+2l-1 \choose l} - {k+2l-1 \choose l-1}$. Now, I want to loop through these paths. So, just as the first two posts above show how to go from combinations to lexicographic position and back, is there an efficient way to do this for these paths as well (that doesn't involve storing all paths in memory). For this question, let's consider we're given a path in binary. Something like: [0,1,0,0,0,1,1] (1's are tosses where he got tails) and want to get an integer which is the lexicographic position if you list all the valid paths (the total number of which is the generalized Catalan number).

答:
Let $N(k,l)=\binom{k+2l-1}{l}-\binom{k+2l-1}{l-1}$ denote the number of paths. Note that
$$
N(k,l) = N(k-1,l) + N(k+1,l-1) \qquad (k>0)
$$
with the base cases $N(0,0)=1$, $N(0,l)=0$ for $l>0$. This comes from conditioning on whether the first step is a heads or tails. Furthermore, with the encoding Heads = 1, Tails = 0, all of the paths counted by $N(k+1,l-1)$ which begin with tails will occur lexicographically before those counted by $N(k-1,l)$. Therefore, to find the lexicographic position of a path, you can do as follows:

Maintain a variable accum, initially set to $0$.

Loop through the entries of the path from left to right.

Whenever the current entry of the path is heads, add the value $N(k' + 1, l' - 1)$ to accum, where $l'$ is the number of tails including and after the current entry, and $k'+l'$ is the number of heads including and after the current entry.

After the loop, accum will equal the desired lexicographic index.

Conversely, to convert from a lexicographic index to a path, you can use the following algorithm:

Maintain a variable accum, initially set to the given lexicographic index.

Maintain two variable $k'$ and $l'$, initially set to the given $k$ and $l$. As the path is specified, these will represent the values of $k$ and $l$ for the unassigned part of the list.

Maintain a list of length $k+2l$, whose entries are initially unassigned (this will be the output path).

For each entry of the path, do the following:

Check to see if $N(k' + 1, l' -1)$ is less than or equal to accum. If it is, then the current entry of the path should be set to heads. Furthermore, you should subtract $N(k' + 1,l' -1)$ from accum. Finally, subtract one from $k'$.

Otherwise, the current entry of the path should be set to tails, and accum is unchanged. Then, add one to $k'$, and subtract one from $l'$.

If your only purpose is to efficiently loop through all these paths in lexicographic order, then converting back and forth between paths and lexicographic indices is not necessary. All you need is a method that, given a path, finds the lexicographically next path. By calling this method repeatedly, you will loop through all paths.

Here is the idea of such a method. Given a path as a sequence of H and T, the next path is given by…

Finding an instance of TH (meaning tails immediately followed by heads), and replacing it with HT.

If there are multiple instances of TH, choose the rightmost one for which the switch TH -> HT still leaves a legal path. The switch is legal if and only if the accumulated wealth up to but not including TH is at most $k-2$.

Rearranging all of the flips to the right of that instance of TH so all tails occur before heads.

For example, suppose that $k=4$ and $l=3$, and the current path is

T H H T H H H T H H

There are three instances of TH. The accumulated wealth before each of them, going left to right, is $0$, $1$, and $3$. The ones where it would leave a legal path if you switch are $0$ and $1$; the last $3$ is too much, since $k-2=2$. Therefore, we switch the latest one. All of the heads after that are moved as far to the right as possible. Here is the next path, where the heads and tails which were switched are marked with ^.

```<code>T H H H T T H H H H
      ^ ^
</code>```

Body:
Let $N(k,l)=\binom{k+2l-1}{l}-\binom{k+2l-1}{l-1}$ denote the number of paths. Note that
$$
N(k,l) = N(k-1,l) + N(k+1,l-1) \qquad (k>0)
$$
with the base cases $N(0,0)=1$, $N(0,l)=0$ for $l>0$. This comes from conditioning on whether the first step is a heads or tails. Furthermore, with the encoding Heads = 1, Tails = 0, all of the paths counted by $N(k+1,l-1)$ which begin with tails will occur lexicographically before those counted by $N(k-1,l)$. Therefore, to find the lexicographic position of a path, you can do as follows:

Maintain a variable accum, initially set to $0$.

Loop through the entries of the path from left to right.

Whenever the current entry of the path is heads, add the value $N(k' + 1, l' - 1)$ to accum, where $l'$ is the number of tails including and after the current entry, and $k'+l'$ is the number of heads including and after the current entry.

After the loop, accum will equal the desired lexicographic index.

Conversely, to convert from a lexicographic index to a path, you can use the following algorithm:

Maintain a variable accum, initially set to the given lexicographic index.

Maintain two variable $k'$ and $l'$, initially set to the given $k$ and $l$. As the path is specified, these will represent the values of $k$ and $l$ for the unassigned part of the list.

Maintain a list of length $k+2l$, whose entries are initially unassigned (this will be the output path).

For each entry of the path, do the following:

Check to see if $N(k' + 1, l' -1)$ is less than or equal to accum. If it is, then the current entry of the path should be set to heads. Furthermore, you should subtract $N(k' + 1,l' -1)$ from accum. Finally, subtract one from $k'$.

Otherwise, the current entry of the path should be set to tails, and accum is unchanged. Then, add one to $k'$, and subtract one from $l'$.

If your only purpose is to efficiently loop through all these paths in lexicographic order, then converting back and forth between paths and lexicographic indices is not necessary. All you need is a method that, given a path, finds the lexicographically next path. By calling this method repeatedly, you will loop through all paths.

Here is the idea of such a method. Given a path as a sequence of H and T, the next path is given by…

Finding an instance of TH (meaning tails immediately followed by heads), and replacing it with HT.

If there are multiple instances of TH, choose the rightmost one for which the switch TH -> HT still leaves a legal path. The switch is legal if and only if the accumulated wealth up to but not including TH is at most $k-2$.

Rearranging all of the flips to the right of that instance of TH so all tails occur before heads.

For example, suppose that $k=4$ and $l=3$, and the current path is

T H H T H H H T H H

There are three instances of TH. The accumulated wealth before each of them, going left to right, is $0$, $1$, and $3$. The ones where it would leave a legal path if you switch are $0$ and $1$; the last $3$ is too much, since $k-2=2$. Therefore, we switch the latest one. All of the heads after that are moved as far to the right as possible. Here is the next path, where the heads and tails which were switched are marked with ^.

T H H H T T H H H H
      ^ ^


问:
Upper bound for $|\mathbb E[f(X)] - f(0)|$ when $X \sim N(0,\sigma^2)$ and $f:\mathbb R \to \mathbb R$ is piecewise-linear with $\|f'\|_\infty \le 1$.
Let $f:\mathbb R \to \mathbb R$ be a piecewise linear function such that $f(0)=0$ and $\|f'\|_\infty \le 1$ and let $X$ be a Guassian random variable with mean $0$ and variance $\sigma^2$. What is an upper-bound for $|\mathbb E[f(X)]|$ ?

Example

If $f(x) = \max(x,0)$, then $\mathbb E[f(X)]=\sigma/\sqrt{2\pi} = \mathcal O(\sigma)$.

答:
Turns out one can get a bound for general $f \in \mathcal C^1$ for which $\|f'\|_\infty \le 1$, via the Mean-Value Theorem and the Cauchy-Schwarz inequality for expectations.

Indeed, the MVT gives $f(X) = \pm Xf'(\theta(X))$ for some $\theta(X)$ between $\min(X,0)$ and $\max(X,0)$. Here $\theta(X)$ is a random variable which depends measurably on $X$. Thus
$$
|\mathbb E[f(X)]| \overset{\text{MVT}}{=} |\mathbb E[Xf'(\theta(X))]| \overset{\text{C.S}}{\le} \sqrt{\mathbb E[X^2]}\sqrt{\mathbb E[(f'(\theta(X)))^2]} \le \sqrt{\mathbb E[X^2]} = \sigma.
$$


答:
Your transformed density should be $f_Y(y) = \dfrac{1}{8\sqrt{y}}$ for $0 \lt y \le 4$.

You may have forgotten to account for two different values of $X$ corresponding to each value of $Y$

You then get $$\int_0^9 f_Y(y) \, dy = \frac34$$

Title: Computation with a transformation of a random variable.

Body:
The random variable $X$ is uniformly distributed on the interval $[-4,4]$. Compute $P(X^2 ≤ 9)$.

I tried to define $Y = X^2$ and compute the density of this transformed variable. The PDF of the uniform distribution is $f_x(x) = \frac{1}{8}$. What I got as a transformed density is $f_y(y) = \frac{1}{16\sqrt(y)}$.

However, if I calculate the integral of that transformed density from $0$ to $9$, I get a wrong result. My guess is that it has something to do with the fact that the uniform can take negative values.


问:
Why does the derivative of $\sin(kx) = k\cos(kx)$, practically
I was in a lecture in college and my lecturer mentioned the practical reason why the derivative of $ \sin(kx) = k\cos(kx) $. Although I didn't quite understand the reason totally.

Would anyone be able to explain why this derivative is so in practical terms and or logical terms, not with a mathematical proof (i.e. using the chain rule)? I was wondering if you could think of it in terms of the graphs of each of the functions?

I don't really know the reason, therefore any insight/help will be greatly appreciated.

Thank you :-)

答:
Informally, when we say that the 'derivative of sine is cosine', what we mean is that the following quotient is equal to $\cos x$:
$$
\frac{\sin(x+dx)-\sin(x)}{dx}
$$
where $dx$ is an infinitely small non-zero number. In other words, the line joining the two points $(x,\sin x)$ and $(x+dx,\sin(x+dx))$ has a slope of $\cos x$. Symbolically, if $f(x)=\sin x$, then $f'(x)=\cos x$. Now consider the graph of $g(x):=f(kx)$. The $y$-coordinates remain the same as before, but the $x$-coordinates are scaled by a factor of $1/k$. This means that the point $(x,\sin x)$ is mapped to $(x/k,\sin x)$, since $g(x/k)=f(x)=\sin(x)$. Similarly, the point $(x+dx,\sin(x+dx))$ is mapped to $(\frac{x+dx}{k},\sin(x+dx))$. Hence,
\begin{align}
g'(x/k) &= \frac{\sin(x+dx)-\sin(x)}{\frac{x+dx}{k}-\frac{x}{k}} \\[4pt]
&= k\cos x \, .
\end{align}
And if $g'(x/k)=k\cos x$, then $g'(x)=k\cos kx \; \blacksquare$


问:
Prove that a number is the product of two primes under certain conditions.
Suppose $p$ be the smallest prime dividing $n \in \mathbb{Z}^+$. Prove that if $n$ is not a perfect square and that $p<n<p^3$, then $n$ must be the product of two primes.

Clearly, the smallest $p$ can be is $2$ and $n$ must be an integer that is greater than $1$ in order to be divisible by a prime. I know that the Fundamental Theorem of Arithmetic (FTA) guarantees that every positive integer greater than $1$ is the product of two or more primes. So is it enough to argue that by the FTA, $n$ is the product of two primes?

答: The FTA doesn't say what you think it does, so let's be more formal about $n$'s prime factorisation. Since $n$ is neither a perfect power of $p$ nor large enough to be a product of the form $pqr$, $p^2q$ or $pq^2$ with primes $q,\,r$ distinct primes greater than $p$, it must instead be of the form $pq$.


问:
Expressing a probability as binomial coefficients
Here's the problem: A box contains 21 working calculators and 4 defective calculators. Two calculators are drawn from the box with replacement. What is the probability that one calculator is functional and the other one is defective?

I was able to solve it using three approaches ($x$ is the answer):

Intuitively: $\frac{21}{25} \cdot \frac{4}{25} \cdot 2 = x$

Solving for $x$ in the sample space: $(\frac{21}{25})^2 +(\frac{4}{25})^2+x = 1.$

Using the binomial distribution: $\binom{n}{k}a^kb^{n-k} = \binom{2}{1}(\frac{21}{25})^1(\frac{4}{25})^{2-1} = x$

Now, that is nice and all, but what I'm really wondering how you can express a solution to this problem in terms of binomial coefficients only (or if that's even possible). For example, if I want to find the probability of drawing exactly one ace from a 5-card hand in a standard 52-card deck, then I would do: $\frac{{4 \choose 1}{48 \choose 4}}{52 \choose 5}$

Which comes from the general rule of probability: $\text{number of ways an event can happen}\over\text{total number of outcomes}$

答:
Ways the event can happen $= 2\cdot 21\cdot 4 = 168$

Total number of outcomes $ = 25\cdot 25 = 625$

Probability of the event $ = 168/625 = 0.2688$


答:
To find the solutions of $x′′+2x′+x=0$ such that x(1)=0. My approach is using the change $y=x′$.

You can also choose $y=x'+x \implies y'+y=0$ or rewrite the DE as: $$(xe^t)''=0$$

Title: Solutions of $x'' + 2x' + x = 0$ such that $x(1) = 0$

Body:
I am trying to solve this question from an older exam that my teacher uploaded. The marks given for answering this question were very low, so I am thinking there is an easy way to solve it.
The question asks to find the solutions of $x'' + 2x' + x = 0$ such that $x(1) = 0$.
My approach is using the change $y=x'$, and therefore getting:
$$
x' = y
$$
$$
y' = -x -2y
$$
Which is a linear system. What I would normally do here is finding the eigenvalues (in this case there is only one eigenvalue, -1 with eigenvector (1,-1)), and solve the corresponding system using the Jordan matrix and the Jordan basis, but I was thinkin whether there was a faster approach, maybe playing with the fact that $x(1)=0$, or something else.

Thanks


问:
Covariant derivative do Carmo help clarification concept
Let $M$ be a Riemannian manifold and let $\nabla$ be a connection compatible with the metric. Then for any vector fields $V,W$ along the smooth curve $c:I\rightarrow M$ , we have

$\frac{d}{dt}\langle V, W\rangle =\langle \frac{DV}{dt},W\rangle+ \langle V, \frac{DW}{dt}\rangle$

However my question is: Why does $\frac{d}{dt} \langle V, W \rangle$ make sense?

$V$ and $W$ are vector fields along a curve, so I suppose $\langle V, W\rangle$ is viewed as a map $I\rightarrow \mathbb{R}$ , where $\langle V,W \rangle(t)=\langle V_t, W_t \rangle$ . But how do we show that it is smooth?  Also compatibility of the metric is defined on smooth vector fields.

答:
Indeed, $\langle V,W \rangle: I\rightarrow \mathbb{R}$  given by $\langle V,W \rangle (t)=\langle V_t,W_t \rangle_{c(t)}$ is indeed smooth.

Let $M$ be your manifold.
Let $t\in I$, So, $c(t)\in M$. Choose a frame $(U,e_1,.....,e_n)$ of $c(t)$ such that $\{$ $e_1,.....,e_n$ $\}$ forms an orthonormal local frame $U\rightarrow TM$ (Note $\langle e_i, e_j \rangle$  is smooth on $U$ for each $i,j$). Therefore $V_t=V^j(t)e_j$ and $W=W^k(t)e_k$ for smooth functions $V^j$ and $W^k$ on $I$. Hence, $\langle V_t , W_t\rangle= \sum_iV^i(t)W^i(t)$.


问:
Calculating amount of subgroups and normal groups of $G$ for $[G:Z(G)] = 6$ containing $Z(G)$
Let $G$ be a group for which $[G: Z(G)] = 6$. How many subgroups and normal groups does $G$ have that contain $Z(G)$.

I am trying to solve this problem, I had the following ideas:

Using the Class equation $|G| = |Z(G)| + \sum_{i=m+1}^n |Cl(a_i)|$: I first thought that I could use this equation, but the condition that $Z(G)$ has to be contained in the subgroups, leads me to think that I can't use the class equation.

$|G| = |C_G(x)|*|Cl(x)|$. Since $Z(G) = \cap_{x \in G} C_G(x)$ I thought  I could maybe use this to calculate the amount of subgroups or normal groups, but I don't really know how to continue this idea.

I think my second idea is maybe in the right direction but I don't know how to go further nor do I know if it's actually the right idea.

答:
Hint 1: There exists no group $\;G\;$ such that $\;G/Z(G)\;$ is non-trivial cyclic.

Hint 2: There exist only two groups of order $\;6\;$: an abelian (cyclic, in fact) one, and a non abelian one isomorphic with $\;S_3\;$ .


问:
Is there a nonrigid space X homeomorphic to X/Aut(X)?
Is there a topological space $X$ satisfying Conditions (a) and (b) below?

(a) $X$ has at least one nontrivial automorphism,

(b) $X$ is homeomorphic to the quotient $X/\!\operatorname{Aut}(X)$ of $X$ by the action of its automorphism group.

答:
Here is a general theorem that can be used to construct examples of such spaces.

Theorem: Let $X$ be a nonempty topological space.  Then there exists a topological space $Y$ such that $|Y|>|X|$ and a homeomorphism $Y/\operatorname{Aut}(Y)\to X$ which induces a bijection between the clopen subsets of $Y$ and the clopen subsets of $X$.  (In particular, if $X$ is connected, then so is $Y$.)

Given this theorem, start with a nonempty rigid connected space $X_0$ (say, a singleton).  Recursively use the theorem to obtain a sequence of connected spaces $X_n$ such that $X_n\cong X_{n+1}/\operatorname{Aut}(X_{n+1})$ and $|X_{n+1}|>|X_n|$.  Now let $X$ be the disjoint union of all these spaces $X_n$, together with an additional copy of $X_0$.  No two $X_n$'s are homeomorphic (since they all have different cardinalities), so any automorphism of $X$ must map each $X_n$ to itself (except that it can swap the two $X_0$'s).  So the quotient $X/\operatorname{Aut}(X)$ will turn $X_n$ into $X_{n-1}$ for each $n>0$ and will identify the two copies of $X_0$ together into one.  Thus, $X/\operatorname{Aut}(X)$ is a disjoint union one copy of each $X_n$ together with an extra copy of $X_0$, so it is homeomorphic to $X$.

Proof of Theorem: Consider the following general construction.  Suppose $(S_x)_{x\in X}$ is a family of nonempty totally ordered sets indexed by points of our space $X$.  We define a topological space $Y$ as follows.  The underlying set is the disjoint union $\bigsqcup_{x\in X} S_x$.  Let $p:Y\to X$ be the map sending each $S_x$ to $x$.  We topologize $Y$ by saying a set $U\subseteq Y$ is open if it satisfies the following two conditions:

$p(U)$ is open in $X$.

For all $x\in X$ and $a,b\in S_x$, if $a\leq b$ and $b\in U$, then $a\in U$.

In particular, for $A\subseteq X$, this means $A$ is open iff $p^{-1}(A)$ is open, so $p$ is a quotient map.  Also, each $S_x$ is connected as a subset of $Y$, so $p$ induces a bijection between the clopen subsets of $Y$ and the clopen subsets of $X$.

Now assume additionally that the totally ordered sets $S_x$ are all homogeneous, have more than one point, and are pairwise nonisomorphic.  I claim that then the orbits of the action of $\operatorname{Aut}(Y)$ on $Y$ are exactly the sets $S_x$, so that $p$ realizes $X$ as the quotient $Y/\operatorname{Aut}(Y)$.

First, if $f_x:S_x\to S_x$ is an order automorphism of $S_x$ for each $x$, then these maps combine to give an automorphism of $Y$.  So, each orbit of $\operatorname{Aut}(Y)$ is a union of sets $S_x$, since each $S_x$ is homogeneous.  Now note that for $a,b\in Y$, every open set containing $a$ contains $b$ iff $a$ and $b$ are in the same $S_x$ and $a\geq b$. (Here we use the assumption that every $S_x$ has more than one point to construct an open set containing $a$ but not $b$ if they are in different $S_x$'s.)  It follows that any automorphism of $Y$ must map each $S_x$ to some other $S_y$ by an order isomorphism.  But since the $S_x$'s are pairwise nonisomorphic, this means any automorphism must map each $S_x$ to itself.  Thus each $S_x$ is an orbit, as desired.

All that remains to be shown is that such a family $(S_x)$ of homogeneous non-isomorphic totally ordered sets actually exists (and moreover that such a family can be chosen such that $|Y|>|X|$).  This follows from the following lemma, which enables us to choose the $S_x$'s to all have different cardinalities which are as large as we like.

Lemma: For any cardinal $\kappa$, there exists a homogeneous totally ordered set of cardinality greater than $\kappa$.

Proof: Take $\mathbb{Z}^\kappa$ with the lexicographic order.  It is homogeneous since for any $a\in\mathbb{Z}^\kappa$, the map $b\mapsto b+a$ is an order-automorphism.


问:
Confusion about calculating cdf and pdf of random variable
Let $X$ be a random variable with pdf:
$$p_X(x)=\begin{cases}
1-|x| & x\in[-1,1] \\
0 & \text{Otherwise}
\end{cases}$$

Let $Y$ be
$$Y=g(X)=\begin{cases}
-X & x<0 \\
2X & x\geq 0
\end{cases}$$
Find cdf and pdf of $Y$.
This is my attempt:
If $0\leq y\leq 1$, then $\mathbb{P}(Y\leq y)=\mathbb{P}(-X\leq y)=\mathbb{P}(X\geq -y)=1-\mathbb{P}(X\leq -y)$. Furthermore, $\mathbb{P}(X\leq -y)=\int_{-1}^{-y}(1+x)dx=\dfrac{1}{2}-y+\dfrac{y^2}{2}$, hence $\mathbb{P}(Y\leq y)=\dfrac{1}{2}+y-\dfrac{y^2}{2}$. On the other hand, $\mathbb{P}(Y\leq y)=\mathbb{P}(2X\leq y)=\mathbb{P}(X\leq \dfrac{y}{2})=\int_0^{y/2}(1-x)dx=\dfrac{y}{2}-\dfrac{y^2}{8}$. So, that would yield
$$P_y(Y)=\begin{cases}
\dfrac{1}{2}+\dfrac{3y}{2}-\dfrac{5y^2}{8} & y\in[0,1] \\
\dfrac{y}{2}-\dfrac{y^2}{8} & y\in(1,2] \\
0 & \text{Otherwise}
\end{cases}$$

Then, I can take the derivative and thus find the pdf. However, I have several problems, because there is a jump at $y=0$, namely $P_y(0^-)=0$, but $P_y(0^+)=\dfrac{1}{2}$; also $P_y(1)=1.375$, which doesn't make sense. I know I am doing something wrong, but I am not sure where or what. Could someone give me some hint? Thanks!

Edit:
Since I know the pdf of $Y$ and I know that the cdf is just the integral and its value final value should be $1$, I calculated:
$$P_y(Y)=\begin{cases}
\dfrac{3y}{2}-\dfrac{5y^2}{8} & y\in[0,1] \\
\dfrac{1}{2}+\dfrac{y}{2}-\dfrac{y^2}{8} & y\in(1,2] \\
0 & \text{Otherwise}
\end{cases}$$

Everything seems to work, but I still can't "formally" get the cdf.

答:
I didn't check your arithmetic, but I got different results, which are consistent.

$P(Y\le y)=\frac{3y}{2}-\frac{5y^2}{8}$ for $0\le y\le 1$.

$P(Y\le y)=\frac{y}{2}-\frac{y^2}{8}+\frac{1}{2}$ for $1\le y\le 2)$.

Note that it is continuous at $y=1$ where $P=\frac{7}{8}$ so getting density is no problem.

Derivation: $Y\le 1$,  $\int_{-y}^0(1-|x|)dx+\int_0^\frac{y}{2}(1-|x|)dx$

$1\le Y\le 2$, $\int_\frac{1}{2}^\frac{y}{2}(1-|x|)dx+\frac{7}{8}$


问:
What is wrong with my 'proof' of $i=1$?
So I recently learnt about $i$ and I can't wrap my head around the concept of $i^2=-1$ or that $\sqrt{-1}$ can even exist. Today I was thinking about $i$ again and thought of a "proof" that $i=1$.
$$i^4=1 \text{ and } 1^4=1 \text{ so } i^4=1^4$$
$$i^4=1^4 \to \sqrt[4]{i^4}=\sqrt[4]{1^4} \to i=1$$
But if $i=1$, then $i^2 \neq -1$. So I think I must have messed up something in the proof. Can someone point out where this went wrong? I know you can $\text{"prove" }1=2$ by accidentally dividing by $0$ and I suspect something similar is happening.

For anyone else having trouble with complex numbers @mrsamy commented this link and I found it quite helpful: https://www.math.toronto.edu/mathnet/answers/imaginary.html

答:
The thing is that if $m= n$ then we can conclude $f(m)  = f(n)$.

But if $f(m) = f(n)$ we can not conclude $m = n$ unless $f$ is one to one.

If $f$ is not one to one is perfectly possible to have $m \ne n$ but $f(m) = f(n)$.

A simple example is $f(x) = x^2 - 3x +6 $  If $x=2$ we get $f(2) = 4 - 6 + 6 = 4$.  And if $x=1$ we get $f(1)=1-3 + 6 = 4$.  So $f(1) = f(2) =4$ but $1\ne 2$.

An even easier example is $f(x) = x^4$ and $f(-1) = f(1) = f(i) = f(-i)$ will give us $1^4 = 1$ and $(-1)^4 = 1$ and $i^4 = (i^2)^2 =(-1)^2 =1$ and $(-i)^4 = ((-i)^2)^2=((-1)^2i^2)^2 = (1*(-1))^2 = 1$.

the problem is when sloppy teachers give to impressionable students this INCORRECT definition:

THIS IS WRONG:  $\sqrt[k]{m}$ is equal to the $x$ so that $x^k = m$.

The problem is that there are $k$ different $x$ that give $x^k = m$ and so that isn't actually a definition of a single value.

Ex.   $\sqrt{16}$ iss the $x$ so that $x^2= 16$.  Well $4^2 = 16$ and $(-4)^2 = 16$.  So which is it?  is $\sqrt{16} = 4$ of is $\sqrt{16} = -4$.

Well, the answer is we define that $\sqrt{m}$ is the positive value $x$ so that $x^2 = m$.

But two things come about from this.

One $\sqrt{x^2} \ne x$.  That is just wrong.  $\sqrt{x^2} =|x|$ because we don't know that $x$ is positive.

And 2)  $i^2 = 1$ is a property of $i$ but that does not mean $\sqrt{-1} = i$. Because $i$ is neither positive nor negative.

Anyway.....

tl;dr

There are four values, all different of $x$ so that $x^4 = 1$.  They are $1^4 =1; (-1)^4 = 1; (-i)^4=1$

$\sqrt[4]{x^4} \ne x$.  IF $x^4$ is a positive real number then $\sqrt[4]{x^4} = |x|$ and indeed $|1| = |-1| = |i| = |-i|$ but $\sqrt[4]{x^4} \ne x$.

And if $x^4$ is not a positive real number we don't actually have a definition for $\sqrt[4]{x^4}$.

Title: What is wrong with my 'proof' of $i=1$?


问:
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degree $3$.

Body:
$K$ has $2^4$ elements:
$$ax^3+bx^2+cx+d, \ (a,b,c,d) \in {\mathbb{Z}_2}^4$$
We have $$g(0)^2=g(1)^2=1$$
So the only possible candidates for $g$ are
$$1\\
x^3+x^2+1\\
x^3+x+1\\
x^2+x+1
$$
First we calculate
$$x^4\equiv x^2+x$$
$$x^6\equiv (x^4+x^2+x)x^2 +x^4+x^3\equiv x^3+x^2+x$$
We can skip the constant polynomial $1$ immediately.  We use $$(a+b)^2=a^2+b^2 \bmod(2)$$
to check the remaining 3 candidates.

$$(x^3+x^2+1)^2=x^6+x^4+1=(x^3+x^2+x)+(x^2+x)+1=x^3+1$$
$$(x^3+x+1)^2=x^6+x^2+1=(x^3+x^2+x)+x^2+1=x^3+x+1$$
$$(x^2+x+1)^2=x^4+x^2+1=(x^2+x)+x^2+1=x+1$$


问:
The smallest set of well formed formulas in Tourlakis' Mathematical Logic
In page 13 of Mathematical Logic, Tourlakis defines the set of all well-formed-formulae as:

... the smallest set of strings, WFF, that satisfies: 1) All Boolean variables are in WFF, and so are the symbols ⊤ and ⊥. 2) If A and B are strings in WFF, then so are the strings (¬A), (A∧B), (A∨B), (A→B), (A≡B).

Capture from the book

But as I understand it, the set of Boolean variables is infinite.

So, how can a set be a smallest one when the cardinality is infinite?

In this question somebody asked about it and the answer was

The "smallest set" condition there is crucial: if we omit it, there are lots of sets satisfying the definition

and I would go even further as to say, indeed there are infinite sets that satisfy the condition of being sets of Well Formed formulas, it's not unique because I can always find an element (Boolean variable) that belongs to another set of WFF.

答:
The natural numbers $\mathbb{N} = \{0,1,2,3,\ldots\}$ is the smallest set $I$ with the property that $0 \in I$ and whenever $n \in I$ then $n+1 \in I$ ($I$ is an inductive set). What do we mean when we say 'smallest' though? In this context we aren't talking about cardinality, but about set containment. So saying that $\mathbb{N}$ is the smallest set with the aforementioned property means that if $I$ were another such set then $\mathbb{N} \subseteq I$.

This is the same sort of thing with $\mathsf{WFF}$ - we specify some initial set of things that must belong to it (the Boolean variables and $\bot,\top$ in this case) and then require that it be closed under a certain set of operations (in this case, the Boolean connectives followed by surrounding with parentheses). $\mathsf{WFF}$ is the smallest (with respect to containment, not cardinality) set with the aforementioned properties.

How do we know there is such a set? One can show that the intersection of all such sets (either all inductive sets in the case of $\mathbb{N}$, or all sets containing the Boolean variables and so on in the case of $\mathsf{WFF}$) also has the corresponding properties. Since we took the intersection of every set with those properties, it must be the smallest in terms of containment.


问:
finding the limit of the integral of an unknown function
I have a question from an exam written 18 years ago, and I can't solve it (I'm solving old questions to prepare for my own exam).

The Question

The question goes like so:

let  g be a function, which has a derivative defined $\forall x\in [-1,1]$, and $g(0)=0,g'(0)=1$.
Does the limit $$\lim_{x\rightarrow0}\frac{\sin\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{x^{8}}$$ converge?

What have I tried?

I thought about using L'Hôpital's rule, and F.T.C. so I wrote:

$$\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)'=2x\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x^{2}\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds$$

but that didn't help much. I can even show that the limit is of the form $"\frac{0}{0}"$ to use the rule.

I tested this on $g(x)=x$ and it seems like this limit does converge (to 0 if I'm not mistaken)

I also thought Taylor was somehow related, but I really doubt it. I am also unsure of where the information we have about $g(0),g'(0)$ comes in.

BTW

this is what I got after applying L'Hôpital once

$$\lim_{x\rightarrow0}\frac{\left(2\int\limits _{0}^{x^{2}}g\left(s^{2}\right)ds-3x\int\limits _{0}^{x^{3}}g\left(s^{2}\right)ds\right)\cos\left(\int\limits _{x^{3}}^{x^{2}}\left(\int\limits _{0}^{t}g\left(s^{2}\right)ds\right)dt\right)}{8x^{6}}$$

答:
This is an answer based on my comments.

Let $$G(t) =\int_{0}^{t}g(s^2)\,ds,F(u)=\int_{0}^{u}G(t)\,dt$$ Then we have $$F'(u) =G(u), G'(t) =g(t^2)$$ and hence by two applications of LHospital Rule we get $$\lim_{u\to 0}\frac{F(u)}{u^4}=\lim_{u\to 0}\frac{G(u)}{4u^3}=\lim_{u\to 0}\frac{g(u^2)}{12u^2}=\frac {1}{12}$$ The limit in question is $$\lim_{x\to 0}\frac{\sin(F(x^2)-F(x^3))}{x^8}$$ Since $F(x^2)\to 0,F(x^3)\to 0$ we can write the expression under limit as $$\frac {\sin(F(x^2)-F(x^3))}{F(x^2)-F(x^3)}\cdot \frac {F(x^2)-F(x^3)}{x^8}$$ and the first factor tends to $1$ so the limit equals the limit of second factor.

Now we can write $$\frac{F(x^2)-F(x^3)}{x^8}=\frac {F(x^2)}{x^8}-x^4\cdot\frac{F(x^3)}{x^{12}}$$ which tends to $$\frac {1}{12}-0\cdot\frac {1}{12}=\frac {1}{12}$$

LHospital Rule is a very powerful tool for evaluation of limits, but it has become infamous due to the crappy ways in which it is used frequently by beginners.

It is possible to avoid LHospital and use $\epsilon, \delta$ and integrating inequalities.


问:
What is $\mathfrak m_P\mathcal L_P$?
I stumbled upon the term  $\mathfrak m_P\mathcal L_P$ in Hartshorne's book. The statement was (in short)

"s separates points, i.e. for distinct points $P$ and $Q$, $s\in \mathfrak m_P\mathcal L_P$ but $s\not \in\mathfrak m_Q\mathcal L_Q$ ?"

So what is the $\mathfrak m_P\mathcal L_P$  here? (I know that $\mathfrak m_P=\{s\mid s(P)=0\}$)

Kindly,

reinbot

答: If $\mathcal L$ is a line bundle, then in particular it is a sheaf of $\mathcal O_X$-modules, and localising, $\mathcal L_P$ is an $\mathcal O_{X,P}$-module. Thus it makes sense to multiply elements of $\mathcal L_P$ by elements of $\mathfrak m _{X,P}\subseteq\mathcal O_{X,P}.$ In particular, what is "meant" in this passage is that the section $s$ (of the line bundle $\mathcal L$) vanishes at $P$ but not at $Q.$</p>

Title: What is $\mathfrak m_P\mathcal L_P$?


问:
$p_{_{F}}\circ p_{_{G}}$ has at least one strictly positive eigenvalue.

let $(E,\langle\cdot,\cdot\rangle)$ an euclidean vector space with $\dim(E)=2n$. Let F and G two non-orthogonal linear subspaces of E with $\dim(F)=\dim(G)=n$. Let $p_{_{F}}$ be the orthogonal projection on F and $p_{_{G}}$ be the orthogonal projection on G. Prove that $p_{_{F}}\circ p_{_{G}}$ has at least one strictly positive eigenvalue.

The characteristic polynomial of $p_{_{F}}\circ p_{_{G}}$ and $p_{_{G}} \circ p_{_{F}}\circ p_{_{G}}$ are the same since $p_{_{G}}^2 = p_{_{G}}$.

Since $p_{_{G}}$ and $p_{_{F}}$ are orthogonal projections, they are self-ajoint operators.                     Let $(x,y)\in E^2$
$$
\langle (p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,y\rangle = \langle ( p_{_{F}}\circ p_{_{G}})(x) ,p_{_{G}}(y)\rangle = \langle p_{_{G}}(x) ,(p_{_{F}} \circ p_{_{G}})(y)\rangle = \langle x ,(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})(y)\rangle$$
So $p_{_{G}} \circ p_{_{F}} \circ p_{_{G}}$ is a self-ajoint operator

Ad absurdum, let's suppose that $p_{_{G}} \circ p_{_{F}} \circ p_{_{G}}$ has only negative eigenvalues, ie $-(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})$ is a positive self-adjoint operator ie $\forall x \in E, \langle -(p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,x\rangle \geq 0$

if $ F \cap G \neq \lbrace{0}\rbrace$  , let  $x\in  (F \cap G)\setminus \lbrace{0}\rbrace$
$$\langle -(p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(x) ,x\rangle = -\langle x,x \rangle < 0$$
hence the contradiction because $-(p_{_{G}} \circ p_{_{F}} \circ p_{_{G}})$ is a  positive self-adjoint operator

I could not find the case where $ F \cap G = \lbrace{0}\rbrace$

答:
Since F and G are non-orthogonal, there is $g_0 \in G \setminus F^\perp$
$$\langle (p_{_{G}} \circ p_{_{F}}\circ p_{_{G}})(g_0) ,g_0\rangle = \langle (p_{_{G}}( p_{_{F}}(p_{_{G}}(g_0) ),g_0\rangle = \langle (p_{_{F}}(p_{_{G}}(g_0) ),p_{_{G}}(g_0)\rangle \:\:\text{($p_{_{G}}$ is self-adjoint)}$$
$$=\langle (p_{_{F}}(p_{_{G}}(g_0) ),p_{_{F}}(p_{_{G}}(g_0))\rangle \:\:\text{($p_{_{F}}$ is an orthogonal projection)}$$
$$ = \langle (p_{_{F}}(g_0),p_{_{F}}(g_0)\rangle > 0 \: because \: g_0 \notin F^\perp $$
hence the contradiction


问:
Quadrature for integrating a univariate function twice
I'm looking to investigate quadrature-like rules that can evaluate
$$
\int_a^b \int_a^s f(t) \, dt \, ds
$$
for sufficiently smooth real univariate functions $f$ by only using point evaluations of $f$. I assume that this is a classical problem, but I cannot seem to find the right keywords to properly search it. What are some known methods, and what is the correct keyword to search for in the literature?

Background: Let $f$ be a smooth function on $\mathbb{R}$. A quadrature rule is a method to approximate a definite integral
$$
\int_a^b f(x) \, dx \approx \sum_{i=1}^n w_i f(x_i)
$$
for appropriate values of $w_i, x_i$, accurate given that $f$ is sufficiently smooth. Thus a quadrature approximates an integral using point evaluations of $f$. If $f$ is a velocity function, then its definite integral can be viewed as the total position displacement from time $a$ to $b$.

I'm interested in the equivalent situation where $f$ is an expression for acceleration. If $f$ is acceleration, then the total displacement from time $a$ to $b$ is given by
$$
v_a(b-a) + \int_a^b \int_a^s f(t) \, dt \, ds,
$$
where $v_a$ is the initial velocity at time $a$.

答:
Just note that,
$$
\int_a^b \underbrace{\int_a^s f(t) dt}_{g(s)} ds = \int_a^b g(s) ds \approx  \sum_{i=1}^n \omega_i g(s_i)
$$

On the other hand,

$$
g(s_i) = \int_a^{s_i} f(t) dt \approx \sum_{j=1}^{n_i} \omega_{ij} f(t_j)
$$

So, finally

$$
\int_a^b \int_a^s f(t) dt\, ds\approx \sum_{i=1}^n \omega_i \sum_{j=1}^{n_i} \omega_{ij} f(t_j).
$$

Unless I'm missing something, there is really nothing nonstandard here.

Here's another idea... In general,

$$
\int_a^b \int_a^s f(t,s) dt ds = \int_a^b \int_t^b f(t,s) ds dt
$$

but, if $f=f(t)$ does not depend on $s$, this becomes

$$
\int_a^b f(t)(b-t) dt \approx \sum_{i=1}^n \omega_i f(t_i) (b-t_i).
$$


问:
Getting the kinematic equation $v^2=v_0^2+2a(x−x_0)$
As I've searched through the internet, looking for explanation. I found an answer with transformation of this formula:

$$
x = x_0 + v_0t + \frac{1}{2}at^2 ⟹ x = x_0 + \frac{1}{2}a(t + \frac{v_0}{a})^2 - \frac{v^2_0}{2a}
$$
I don't understand how to get $\frac{1}{2}a(t + \frac{v_0}{a})^2$
Can anyone help please?

Body:
hint

$$At^2+Bt=A(t^2+\frac{Bt}{A})$$

with

$$t^2+\frac{Bt}{A}=(t+\frac{B}{2A})^2-(\frac{B}{2A})^2$$

Replace $ A $ by $ \frac 12a $ and

$B $ by $ v_0$.


答:
Hint

You just have to complete the square. See that

$$\frac{1}{2}a(t + \frac{v_0}{a})^2=\frac{1}{2}a(t^2 + 2\frac{tv_0}{a}+\frac{v_0^2}{a^2})=\frac{at^2}{2}+tv_0+\frac{v_0^2}{2a}$$
and then

$$\frac{1}{2}a(t + \frac{v_0}{a})^2-\frac{v_0^2}{2a}=\frac{at^2}{2}+tv_0.$$

Just move backward.

Title: Getting the kinematic equation $v^2=v_0^2+2a(x−x_0)$

Body:
As I've searched through the internet, looking for explanation. I found an answer with transformation of this formula:

$$
x = x_0 + v_0t + \frac{1}{2}at^2 ⟹ x = x_0 + \frac{1}{2}a(t + \frac{v_0}{a})^2 - \frac{v^2_0}{2a}
$$
I don't understand how to get $\frac{1}{2}a(t + \frac{v_0}{a})^2$
Can anyone help please?


问:
Show that the largest connected exists (with respect to the inclusion of sets) $C_{x} \subseteq X$ containing $x$.
If $X \subseteq \mathbb{R^n}$ and $x \in X$, show that the largest connected exists (with respect to the inclusion of sets) $C_{x} \subseteq X$ containing $x$.

The connected set $C_{x}$ is called the connected component of $x$ with respect to $X$.

I already know that the largest connected subset $C_{x}$ of $X$ exists as just the union of all the connected subsets of $X$ that contain $x$, but now I don't know how to use that to prove the above. Can someone help me?

答:
It's trivial if you know how to construct $C_x$.

So you define $$C_x = \bigcup\{C\subseteq X\mid x \in C, C \text{ connected }\}$$

This is non-empty, as we at least have $C=\{x\}$ as one of the sets in the union, contains $x$ for that same reason and is connected as a union of connected sets that all intersect at $x$ (this is a standard theorem on connectedness).

Now if $C$ is any connected set that contains $x$, it's by definition one of the sets we are taking the union of, so trivially $C \subseteq C_x$, whch is exactly the maximality under inclusion that you were looking for. It's no more than that. The fact that we're in $\Bbb R^n$ has no special relevance; this works in any topological space.

Also note that as $C_x$ is connected and the closure of a connected set is connected (another standard theorem) $\overline{C_x}$ is in particular such a connected set containing $x$ and so by that maximality $\overline{C_x} \subseteq C_x$ which implies $C_x$ is closed. So components are closed in $X$.


问:
similar triangles, trigonometry and pythagorean theorem

This is a Rectangle (i didn't name it because it's not important), $ABD$ and $ABC$ are both right triangles, such that: $AB=5, BD=7, AC=9$
The question is to find the length of $MH$?

My Attempt:
First, calculate $AD$ and $BC$:
$$AD^{2} = BD^{2}-AB^{2} \Leftrightarrow AD = \sqrt{49-25}=\sqrt{24}$$
$$BC^{2}= AC^{2}-AB^{2} \Leftrightarrow BC = \sqrt{81-25}=\sqrt{56}$$

Now let's look at angles :
$$\tan{\alpha} = \frac{AD}{AB}=\frac{MH}{BH}=\frac{\sqrt{24}}{5}\Leftrightarrow MH = \frac{BH\sqrt{24}}{5}$$
$$\tan{\theta} = \frac{BC}{AB}=\frac{MH}{AH}=\frac{\sqrt{56}}{5}\Leftrightarrow MH = \frac{AH\sqrt{56}}{5}$$

This leaves us with the following system of equations:
$$\begin{cases}  MH = \frac{DH\sqrt{56}}{5}\\MH = \frac{AH\sqrt{24}}{5} \end{cases}$$
$$\implies \begin{cases}  MH = \frac{AD\sqrt{56}-\sqrt{56}AH}{5}\\MH = \frac{AH\sqrt{24}}{5} \end{cases}$$
$$\implies 5\sqrt{56}-AH\sqrt{56} = AH\sqrt{24}$$
and by some basic arithmetic we got:

$$AH = \frac{5\sqrt{56}}{\sqrt{24}+\sqrt{56}}$$
and now we will plug this result to one of the equations:
$$MH = \frac{5\sqrt{56}}{\sqrt{24}+\sqrt{56}} \cdot \sqrt{24} = \frac{7\sqrt{6}-3\sqrt{14}}{2}$$
But i still wondering if this answer is right because it appears to be irrational?

Any Help will be appreciate

答:
I stumbled upon a nice observation, while trying to see for a more straight forward solution.

From triangle similarity, we get

$\frac{HB}{AB} = \frac{MH}{AD}$ and $\frac{AH}{AB} = \frac{MH}{BC}$

adding the two

$\frac{HB}{AB} + \frac{AH}{AB} = \frac{MH}{AD} + \frac{MH}{BC}$
, then

$\frac{1}{MH} = \frac{1}{AD} + \frac{1}{BC} $,

Which is interesting because it says the length of $MH$ is independent $AB$,

Now you can solve for AD and BC with pythagorian theorem and find $MH$

...

$AD = \sqrt{24}$ and $BC=\sqrt{56}$

$\frac{1}{MH} = \frac{1}{\sqrt{24}} + \frac{1}{\sqrt{56}}$

$MH = \frac{\sqrt{24}\sqrt{56}}{\sqrt{56}+\sqrt{24}}$


问:
How to differentiate functions in matrices
I have always been wondering about this. How does one actually differentiate a polynomial/function in matrices? I can see for elementary functions such as $f(X)=AX+B$ where $X$ is the matrix as a variable, you get $f'(X)=A$, but what about functions such as
$$
f(X)=AXA^{-1},
$$
how can I calculate the derivative?

答:
Your function $f$ is still linear, so it's still its own derivative. In general you can do it by plugging in an infinitesimal matrix $X \mapsto X + \epsilon$ such that the coefficients of $\epsilon$ all square to zero. This is, to my mind, by far the fastest and cleanest way, and I don't know why it isn't taught like this. It is more or less equivalent to using big-O notation and expanding everything out to $O(\| \epsilon \|^2)$ but saves some notation.

Example #0: The derivative of $X^2$ is given by writing

$$(X + \epsilon)^2 = X^2 + \epsilon X + X \epsilon$$

so the derivative at $X$ is the linear map $\epsilon \mapsto \epsilon X + X \epsilon$ (you might have guessed that it's $2X \epsilon$ but this only works if $X$ and $\epsilon$ commute).

Example #1: The derivative of $X^{-1}$ is given by writing

$$(X + \epsilon)^{-1} = \left( X(1 + X^{-1} \epsilon) \right)^{-1} = (1 + X^{-1} \epsilon)^{-1} X^{-1} = (1 - X^{-1} \epsilon) X^{-1}$$

so the derivative at $X$ is the linear map $\epsilon \mapsto -X^{-1} \epsilon X^{-1}$ (you might have guessed that it's $-X^{-2} \epsilon$ but again this only works if $X$ and $\epsilon$ commute).

Example #2: The derivative of $\det(X)$ is given by writing

$$\det(X + \epsilon) = \det(X) \det(1 + X^{-1} \epsilon) = \det(X) \left( 1 + \text{tr}(X^{-1} \epsilon) \right)$$

so the derivative at $X$ is the linear map $\epsilon \mapsto \det(X) \text{tr}(X^{-1} \epsilon)$, and in particular when $X = 1$ it's the linear map $\epsilon \mapsto \text{tr}(\epsilon)$.

Example #3: The derivative of $\exp(X)$ is given by writing

$$\exp(X + \epsilon) = \sum_{n=0}^{\infty} \frac{(X + \epsilon)^n}{n!} = \sum_{n=0}^{\infty} \frac{X^n + \sum_{k=1}^n X^{k-1} \epsilon X^{n-k}}{n!}$$

so the derivative at $X$ is the linear map

$$\epsilon \mapsto \sum_{n \ge 0} \frac{\sum_{k=1}^n X^{k-1} \epsilon X^{n-k}}{n!}$$

which I don't believe can be simplified in general, although it ought to have some nice expression in terms of commutators, maybe after multiplying by $\exp(-X)$. If $X$ and $\epsilon$ commute this simplifies to $\exp(X) \epsilon$, and in particular if $X = 0$ this simplifies to $\epsilon$. Formalizing this example would require proving various things about convergence and exchange of limits but it can be done.


问:
Action on orbits defined by another action
Let $(G,\Omega)$ be a transitive action of $G$ on $\Omega$. Let now $H\le G$ be a subgroup of $G$ and define $\Omega/H=\{\omega\cdot H|\omega\in\Omega\}$ as the set of the orbits generated by the action of $H$ on $\Omega$. I've already showed that
$$(\omega\cdot H)\cdot g=(\omega\cdot g)\cdot H,\text{ }\forall\omega\cdot H\in \Omega/H\text{ and }g\in G. $$is a group action of $G$ on $\Omega/H$.

I've problems in defining the stabilizer of a generic $\omega\cdot H\in\Omega/H$ in the action described.

Could you give me some advice? Thank you.

答:
If $G$ acts on $\Omega $ transitively then, choosing some $\omega _0$ in $\Omega $, and letting $K$ be the stabilizer of $\omega _0$,  the map
$$
  g\in  G\mapsto  \omega _0g\in \Omega
  $$
factors through the quotient
$
  K\backslash G = \{Kg: g\in G\},
  $
providing a equivariant bijection
$$
  K\backslash G \to  \Omega ,
  $$
so we may assume WLOG that $\Omega =K\backslash G$.

Now, the orbit space of the right action of the given subgroup $H$ on  $K\backslash G$ consists of the set of all "double
cosets"
$$
  \Omega /H = K\backslash G/H = \{KgH: g\in G\}.
  $$

Thus,  unless $H$ or $K$ are  normal,  as proposed by @Berci,  there is no sensible action of $G$ on that double coset space.  The
fundamental parameter "$g$" in $KgH$, protected on the left and right by $K$ and $H$, is too inaccessible to be acted upon in any meaningful way.