问:
Linear combination of random variables and independence
Given a set of real-valued normally distributed random variables $X_1, \dots, X_n$, how do we show the following are equivalent?

Any linear combination of $X_1, \dots, X_n$ is normally distributed

There is a linear mapping $A$ such that the transformed random variables $AX_1, \dots, AX_n$ are independent

Motivation: I want to reconcile two definitions of jointly Gaussian random variables. I believe a set of scalar Gaussian rvs $\{X_i\}$ can be shown jointly Gaussian under two characterizations: 1) $\{X_i\}$ are independent under some linear transformation, or 2) all linear combinations of $\{X_i\}$ are Gaussian-distributed. I don't know why these are equivalent, or how to prove this property for a given set of rvs $\{X_i\}$.

Edit: if the covariance matrix is known, how does this help? I think one case where 1 and 2 don't hold is if the covariance is singular.

答:
I have to admit that I have slightly misread your question (I thought you would assume Gaussian random variables for 2). In that case the characteristic function would be the way to go).

Re-readng the question, I think there is problem in the formulation of 2) as it is stated right now. To see why suppose that $X_1, X_2, \dots, X_n$ are identically but dependent univariate random variables and let $A$ be a linear transformation such that $AX_1, AX_2,\dots, AX_n$ are independent. In the univarite case, $A$ is just a scalar. Assume $A\neq 0$. If $AX_1, AX_2, \dots, AX_n$ are independent, so is $A^{-1}AX_1, A^{-1}AX_2,\dots, A^{-1}AX_n$ which yields a contradiction. If $A=0$, the distribution is degenerate. Thus, an $A$ as claimed stated in 2) does not exist.

I assume you meant there exists a linear transformation $A$ such that $A\mathcal X$ is independent, where $\mathcal X = (X_1, X_2, \dots, X_n)$. But that still does not suffice to characterize the normal distribution (see, for example, https://stats.stackexchange.com/questions/74410/for-which-distributions-does-uncorrelatedness-imply-independence).

Edit: The following holds:

Theorem. Suppose that $X_1, X_2, \dots, X_n$ are real-valued random variables. Then $\mathcal X = (X_1, X_2,\dots, X_n)$ is normally distributed iff $A\mathcal X$ is normally distributed for every full row rank linear transformation $A$.

Proof. Let $A$ be a full row rank linear transformation and assume that $\mathcal X$ is normally distributed with mean $\mu$ and vcov matrix $\Sigma$. Since $A$ has full row rank, for each $s$ ($\in\mathbb R^k$) there is a $t$ such that $s = A't$. The characteristic function of $A\mathcal X$ is then given by $$\begin{align*}\varphi(t) &= \mathbb E[\exp(it'A\mathcal X)] \\ &=\mathbb E[\exp(i(t'A)\mathcal X)]\\&= \mathbb E[\exp(is'\mathcal X)] \\ &=\exp\left(is'\mu - \frac 12s'\Sigma s\right) \\ &= \exp\left(it'A\mu - \frac 12 t'A\Sigma A't\right),\end{align*}$$ which is the characteristic function of a $\mathcal N(A\mu, A\Sigma A')$-distributed random variable (note that the full row rank ensures that $A\Sigma A'$ is positive definite). Thus, $A\mathcal X$ is normally distributed as the characteristic function uniquely defines the distribution.

Now assume that $A\mathcal X$ is normally distributed for every full row rank linear transformation $A$. Consider the canonical basis vectors for $A$ to obtain that each component of $\mathcal X$ is normally distributed with finite variance. Using the Cauchy Schwartz inequality, it follows that the covariances are also finite. Thus, the vcov matrix $\Sigma$ of $\mathcal X$ as well as the mean $\mu$ exist. Furthermore, the mean and the variance of $A\mathcal X$ are given by $A\mu$ and $A\Sigma A'$, respectively. The characteristic function of $\mathcal X$ is given by
$$
\begin{align*}
    \varphi(s) &= \mathbb E[\exp(is'\mathcal X)] \\
     &= \mathbb E[\exp(it'A\mathcal X)] \\
     &= \mathbb E[\exp(it'(A\mathcal X))] \\
     &= \exp\left(it'(A\mu) - \frac 12 t'(A\Sigma A')t\right) \\
     &= \exp\left(i(A't)'\mu - \frac 12(A't)'\Sigma(A't)\right) \\
     &= \exp\left(is'\mu - \frac 12s'\Sigma s\right),
\end{align*}$$ using again the fact that $A$ has full row rank. This is the characteristic function of a $\mathcal N(\mu,\Sigma)$-distributed random variable, and the characteristic function uniquely defines the distribution. $\square$

A special case is, for example, $A = a'$, for some $a\in\mathbb R^n$, $a\neq 0$. Another example is $A = \Sigma^{-\frac 12}$, which yields a linear combination of $\mathcal X$ that renders the components independent by de-correlation.


问:
Brownian motion dyadic construction convergence
Say $B$ is a Brownian motion and $B^n$ is the function that is equal to $B$ at all numbers of the form $2^{-n}\mathbb{N}$ and linear on the intervals $[k2^{-n},(k+1)2^{-n}]$. I am trying to show that $\lim_{n\to\infty} P(\sup_{[0,1]}|B-B^n|\leq\epsilon )=1$. I was thinking that $|B-B^n|$ can be bounded by $|B_{\frac{k}{2^n}}-B_{\frac{k+1}{2^n}}|$ which is $\mathcal{N}(0,\frac{1}{2^n})$ distributed. Is this correct? Any hint would be appreciated

答:
First, the process $W_t^{n,k}:=(B_t-B_t^n)$, $t\in[k2^{-n},(k+1)2^{-n}]$ is a Brownian bridge. Therefore, using for example this result,
\begin{align*}
&\mathsf{P}\!\left(\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}|W_t^{n,k}|>\epsilon\right) \\
&\qquad\le 2\mathsf{P}\!\left(\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}W_t^{n,k}>\epsilon\right)=2e^{-2^{n+1}\epsilon^2},
\end{align*}
and so,
\begin{align*}
\mathsf{P}\!\left(\sup_{t\in[0,1]}|B_t-B_t^n|> \epsilon\right)&= \mathsf{P}\!\left(\max_{0\le k\le 2^n-1}\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}|W_t^{n,k}|> \epsilon\right) \\
&\le \sum_{k=0}^{2^n-1}\mathsf{P}\!\left(\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}|W_t^{n,k}|> \epsilon\right) \\
&\le 2^{n+1}e^{-2^{n+1}\epsilon^2}\to 0 \quad\text{as}\quad n\to\infty.
\end{align*}

ContentLicense: CC BY-SA 4.0


问:
How to convert recursive to closed form
For example if $f(x+1)=af(x)+b$ is there a way to find the closed form?

Like what is $f(x+1)=\frac67f(x)+\frac67$ is closed form?

答:
You can first solve for $n$ integer with $u_n=f(n)$

if $a=1$ then $u_{n+1}-u_n=b$ is telescopic and $u_n=u_0+nb$

if $a\neq 1$ then $u_{n+1}=au_n+b$ has general solution of homogeneous equation $u_n=Ua^n$ and particular solution $c=ac+b\iff c=\frac b{1-a}$ therefore $u_n=Ua^n+\frac b{1-a}$ and $u_0=U+\frac b{1-a}$ therefore $u_n=a^n(u_0-\frac b{1-a})+\frac b{1-a}$

If you do not impose any more condition of regularity on $f$ then you'll have a piecewise function for all initial seeds $\{x\}\in[0,1)$, i.e $u_0=f(\{x\})$ and $f(x)$ the expression given for $u_n$ where $n$ is replaced by $x$.

Note: for $a<0$, the function $f$ is not well defined for $x<0$ (because of $a^x$)

Title: How to convert recursive to closed form


问:
Exterior derivative of a 2-form in Minkowski space
I'm working on a problem in Minkowski space that involves the exterior product of a two-form.
$$d(A\ dx\wedge dt)$$
Were this a problem in simple 3D space, the obvious answer would be to take the derivative of $A$ with respect to the third coordinate in the space. However, I am unsure of how to do this for a 4D space like the one I'm using.

答: The fact that you're living in Minkowski space is irrelevant, as the exterior derivative is not an operation defined from the metric tensor. If your coordinates are $(x,y,z,t)$ and $A$ depends on those coordinates, then $${\rm d}(A\,{\rm d}x \wedge {\rm d}t) = \partial_yA \,{\rm d}y\wedge{\rm d}x \wedge {\rm d}t + \partial_zA\,{\rm d}z \wedge \,{\rm d}x \wedge {\rm d}t,$$since ${\rm d}x \wedge {\rm d}x = {\rm d}t \wedge {\rm d}t = 0$.


问:
Markov Chain Equilibrium Distribution Question
Suppose I have the following equilibrium probability distribution:
$\vec π = ({2\over5} , {1\over5} , {3\over20},{1\over4})$, corresponding to states 0,1,2,3, respectively.From my possible states of {0,1,2,3}, my initial state is randomly determined by the equilibrium probability distribution $\vec π$. I am after the probability of being in state 2 after n processes.

My understanding is that $\vec π$ is telling me once I get to the distribution of the desired state e.g. either one of  $({2\over5} , {1\over5} , {3\over20},{1\over4})$   , it will stay in that distribution forever. So given that state 2 has been allocated ${3\over20}$ from the beginning it will stay there forever or it will be there after n processes. My final answer being ${3\over 20}$. $\\$

Might not be the easiest description to follow but a clarification would be great.

答:
For a Markov Chain, a discrete-time stochastic process, the following holds:
$$x^{(n+1)} = x^{(n)} P$$
where $x^{(n)}$ is the probability distribution at time $n$, $x^{(n+1)}$ the probability distribution at time $n+1$ and $P$ the transition matrix with elements $p_{ij}$ describing the probability of the next state being $j$ given the current state is $i$.

By definition, the stationary distribution $\pi$ is a vector such that
$$ \pi = \pi P$$
so once in the stationary distribution, the chain remains in this distribution. Therefore if the chain is initialised into the stationary distribution, the probability of being in state 2 at a later time is indeed $\frac{3}{20}$.

See also this worked example (with numerical values) in the Wikipedia article on Markov chains to help.

ContentLicense: CC BY-SA 3.0


问:
Proving $\sum_{n=1}^m\sin^2\left(n\frac\pi{m}\right)=\frac{m}2$ for $m\geq 2$ by induction

$$\sum_{n=1}^m \sin^2\left(n\frac{\pi}{m}\right) = \frac{m}{2}$$ for all $m\ge2$ positive integers

I'm doing a maths problem that involves proving by mathematical induction the sum of a sigma. I find it hard because the highest term is included in the sigma so it changes the actual sigma.

If anyone can help me, please respond it would be very useful!! Even just the smallest help will go a long way : )

Here is the Desmos link to show you this is true.

But I want to prove it using induction.

Thanks.

答:
Well I guess I found a proof without using induction. Perhaps it will be of any use to you so I will leave it here.
$$\sin\varphi=\frac{e^{i\varphi}-e^{-i\varphi}}{2i}$$
Let $\varphi_n=\frac{n\pi}{m}$
$$ \sum_{n=1}^{m}\sin^2{\varphi_n}=\sum_{n=1}^m\left(\frac{e^{i\varphi_n}-e^{-i\varphi_n}}{2i}\right)^2=\sum_{n=1}^m\frac{e^{2i\varphi_n}+e^{-2i\varphi_n}-2}{-4}=\frac{m}{2}-\left(\sum_{n=1}^me^{2i\varphi_n} +\sum_{n=1}^me^{-2i\varphi_n}\right)=\frac{m}{2}-\left(\sum_{n=1}^me^{2i\frac{n\pi}{m}} +\sum_{n=1}^me^{-2i\frac{n\pi}{m}}\right)=\frac{m}{2}-\left( \frac{e^{2\pi i}-1}{e^\frac{2\pi i}{n}-1}+\frac{e^{-2\pi i}-1}{e^{-\frac{2\pi i}{n}}-1}\right)=\frac{m}{2}-\left( \frac{1-1}{e^\frac{2\pi i}{n}-1}+\frac{1-1}{e^{-\frac{2\pi i}{n}}-1}\right)=\frac{m}{2}$$


问:
Argument of a complex number confusion - LaGrange's method and root finding
I am trying to use Lagrange's method to find the cubic roots of a polynomial with real coefficients. This involves taking the cubic root of a complex number and selecting the correct root (one of the several possible roots is appropriate, depending on certain conditions).

For instance, consider this case:

$arg\sqrt[3]x=\frac {2\pi}{3} + \frac 1 3 arg (x)$ if $arg(x)=\pi$

Ultimately, I need to calculate $\sqrt[3]x$ for some known value of $x$. For instance, $x=-1$. In that case, $arg(-1)=\pi$.

At the moment, I'm trying to replicate the results from this paper:

Ting Zhao, Dongming Wang, Hoon Hong.  Solution Formulas for Cubic Equations Without or With Constraints.  Journal of Symbolic Computation, Elsevier, 2011, 46 (8), pp.904-918.

The authors give the following example:
$\sqrt[3]{-1}=\sqrt[3]{e^{i\pi}}=e^{i\pi}=-1$

This example is very mysterious to me. From Wikipedia, I found the identity $x=\lvert x \rvert e^{i Arg(x)}$ but it seems that some steps are missing and I don't follow the logic. Clearly, the authors used this identity under the radical... what confuses me is that in the case I've given (and that they are using in this example) we had $arg\sqrt[3]{x}$. However, it seems like the argument has disappeared. Why is that? Is that correct?

I would like to see a step by step example of how to calculate $\sqrt[3]-1$ using the example case I've provided. Thanks!

答:
The notation for the real cube root and the complex cube root are, unfortunately, identical.  For the real cube root function, $\sqrt[3]{-1} = -1$.  For the complex cube root relation, $\sqrt[3]{-1} = \{-1, \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}, \frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}\}$, taking $-1$ to its three cube roots.  In order to turn the complex cube root relation into a function, we need a convention for choosing which of the three cube roots is returned by the function.

The paper you cite specifies their convention on p. 906.  In that place, we see that they choose the argument of various complex quantities appearing in a cube root in the interval $(-\pi, \pi]$.  (This is the full range of "$\arg x$"s in the pieces of the piecewise function.  Any other $\arg x$ must be manipulated to be in this range.)  This means that $\sqrt[3]{-1} = \sqrt[3]{\mathrm{e}^{\mathrm{i}\pi}}$.  There are actually infinitely many choices for the argument, which I discuss below the horizontal line, below.

Since $\arg(-1) = \pi$ in their convention, we use the fifth choice in their piecewise definition of the argument of the cube root on that page:  $\arg \sqrt[3]{-1} = \frac{1}{3}\arg(-1) + \frac{2}{3}\pi = \frac{1}{3}\pi + \frac{2}{3}\pi$.  Of course, this, is just $\pi$.  We also know $|-1| = 1$ and $\sqrt[3]{1}$ has $\arg 0$ by their convention, so is using the real cube root.  Then $\sqrt[3]{1} = 1$ and we find
$$  |\sqrt[3]{-1}| = \sqrt[3]{|-1|} = \sqrt[3]{1}  = 1  \text{.}  $$
Therefore, the magnitude of $\sqrt[3]{-1}$ is $1$ and the argument is $\pi$.  I.e., by their argument convention, the cube root function gives
$$  \sqrt[3]{-1} = 1 \cdot \mathrm{e}^{\mathrm{i}\pi} = -1  \text{.}  $$

More generally, roots are a place where the standard, common notation hides a crucial fact.  Let's revisit the cube root that you recite,
$$  \sqrt[3]{-1} = \sqrt[3]{\mathrm{e}^{\mathrm{i}\pi}} = \mathrm{e}^{\mathrm{i}\pi} = -1  \text{,}  $$
but show the bit(s) that "everyone knows",
$$  \sqrt[3]{-1} = \sqrt[3]{\mathrm{e}^{\mathrm{i}\pi + 2\pi \mathrm{i}\Bbb{Z}}} = \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}\Bbb{Z}/3} \ni -1  \text{.}  $$

So what's this "$2\pi\mathrm{i}\Bbb{Z}$" stuff about?  Notice that $\mathrm{e}^{2\pi\mathrm{i} k} = 1$ for any choice of $k$ in the integers, typically denoted "$\Bbb{Z}$".  So the notation in the second expression in the display above is reminding us that there are infinitely many different powers of $\mathrm{e}$ that have the value $-1$.  Then we take the cube root by dividing the exponent by $3$.

This gives us an infinite list of powers of $\mathrm{e}$, but maybe it only gives one complex number, just written many different ways.  We're not that lucky.  In fact, it gives all three complex cube roots of $-1$, which, really, is what we should expect the cube root operation to do.  Let's make a table for our different choices of the integer in $\Bbb{Z}$:
\begin{align*}
k&  &&|&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}k/3}&  \\ \hline
 \vdots&  &&&  &\vdots  \\
 -2&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(-2)/3} &= \mathrm{e}^{-3\mathrm{i}\pi/3} = -1  \\
-1&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(-1)/3} &= \mathrm{e}^{-1\mathrm{i}\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}  \\
 0&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(0)/3} &= \mathrm{e}^{1\mathrm{i}\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}  \\
 1&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(1)/3} &= \mathrm{e}^{3\mathrm{i}\pi/3} = -1  \\
2&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(2)/3} &= \mathrm{e}^{5\mathrm{i}\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}  \\
\vdots&  &&&  &\vdots
\end{align*}
As we proceed through the integers, these values cycle through the three cube roots of $-1$.

The last bit of notation in the "show the bit(s)" display is the "$\ni$", which means $-1$ is one of the three values in the list of numbers produced by the cube root.

The paper you cite makes a choice of complex argument for each number you pass to a square or cube root.  Their choice has the nice property that it gives the roots of a cubic having real coefficients by following the usual equations, without having to try all the cube roots to see which ones actually do work and which ones don't.  So this convention on the argument is only useful for this one method of solving the cube root.


问:
Finding a palindromic number which is the difference of two palindromic numbers
Let $X$ and $Y$ be two $4$-digit palindromes and $Z$ be a $3$-digit palindrome. They are related in the way $X-Y=Z$. How can we figure out $Z$?

答:
Let $X=abba$, $Y=cddc$, and $Z=efe$. Then we can think of it as this:

```<code> abba
-cddc
=====
  efe
</code>```

Note that since $Z$ is positive, we must have $a\ge c$. However, we can't have $a = c$, because then the ones digits imply that $e=0$ so that $Z$ is actually a two digit number.

Thus, we must have $a>c$. But notice that in the leftmost column, there is a $0$, suggesting that the hundreds digit has to borrow $1$ from the thousands. Thus, we know that $b<d$ and $a=c+1$ so that $e=1$.

Trial and error gives us one possible solution: $\boxed{X=7007,Y=6886,Z=121}$.

It isn't unique though, as this also works: $\boxed{X=5005,Y=4884,Z=121}$

Body:
Let $X=abba$, $Y=cddc$, and $Z=efe$. Then we can think of it as this:

 abba
-cddc
=====
  efe

Note that since $Z$ is positive, we must have $a\ge c$. However, we can't have $a = c$, because then the ones digits imply that $e=0$ so that $Z$ is actually a two digit number.

Thus, we must have $a>c$. But notice that in the leftmost column, there is a $0$, suggesting that the hundreds digit has to borrow $1$ from the thousands. Thus, we know that $b<d$ and $a=c+1$ so that $e=1$.

Trial and error gives us one possible solution: $\boxed{X=7007,Y=6886,Z=121}$.

It isn't unique though, as this also works: $\boxed{X=5005,Y=4884,Z=121}$


问:
Finding exponential generating function of the family of specific trees using symbolic method
Consider the mapping-graphs given by $\mathcal{M}=Cycle(\mathcal{T})^{[∗]}$ (I'm not sure if this is read as $(Cycle(T))^{[∗]}$ or as $Cycle((T)^{[∗]})$? Which one does make more sense?), whereby $\mathcal{T}$ is the family of trees $\mathcal{T}={O}∗SET(\mathcal{T})={O}∗(\mathcal{T}^{[∗]})$.
I'd like to determine the formula for the exponential generating function M(z) with the symbolic method and calculate the number of mapping-graphs of size n.

I know that the corresponding equation for $T$
is $T(z)=z \exp(T(z))$, but I don't know how to combine this with the symbolic formula of $\mathcal{M}$, because it seems to me as if that expression would be quite complicated.
So I'd appreciate any help to get me on the right track!

答:
I know this is old, but...

We have
$$
\mathcal M = SET\left(CYC\left(\mathcal T\right)\right)
$$
so
$$
M(z) = \exp\left(\log\frac{1}{1-T(z)}\right) = \frac{1}{1-T(z)} = \frac{1}{1-ze^{T(z)}}
$$
The GF for $M$ involves the Lambert $W$ function, but we can find the coefficients using Lagrange inversion. Note that $T=ze^T$ implies $\frac{1}{1-ze^T} = zT'/T$. With $[z^n]$ meaning the coefficient at $z^n$, we get
$$
[z^n]M(z) = [z^n]z\frac{T'}{T} = [z^{n-1}]\frac{T'}{T} = [z^{n-1}](\log T)' = n[z^n]\log T
$$
Now Lagrange inversion / the Lagrange-Bürmann formula gives
$$
[z^n]M(z) = [u^{n-1}]\frac{1}{u}e^{nu} = [u^n]e^{nu} = \frac{n^n}{n!}
$$
Et voilà(!), the number of mappings from $n$ elements to $n$ elements is equal to $n^n$. (What a waste of energy to get that answer, huh?)


问:
The position of a car at time $t$ is given by the function $p (t)=t^2−t−7$. What is the velocity when $p(t)=5$?

The position of a car at time $t$ is given by the function $p (t)=t^2−t−7$. What is the velocity when $p(t)=5$? Assume $t\ge0$.

Setting $t^2-t-7=5$ yields $t=-3$ or $t=4$

Setting the derivative of $p(t)$ equal to $5$, which is $2t-1$, yields $t=3$. So the instantaneous velocity at $p(t)=5$ is $3$…

The test's correct answer is $7$. What am I doing wrong?

答:
When $p(t)=5$, you have $t=4$. And $p'(4)=2\times4-1=7$.

Solving $p'(t)=5$ tells you at which moment the speed is equal to $5$, but that is not what is being asked.


问:
Finding conditional distribution in matching ordering situation
Suppose we draw two values $x_1,x_2$ according to a CDF $F$. Independently, we draw another two values $y_1,y_2$ according to another CDF $G$. Both $F$ and $G$ has support $[0,1]$.

Among those four values, I first observe $x_1$ only. And then, I get to observe one of $y_1$ and $y_2$, depending on whether $x_1\leq x_2$ or $x_1>x_2$ following the rule below:

--- If $x_1\leq x_2$, then I observe $y_1$ or $y_2$, whichever is smaller (or equal to) than the other.

--- If $x_1>x_2$, I observe $y_1$ or $y_2$, whichever is greater than the other.

So, if my $x_1$ is smaller than the other $x$, my observation of $y$ equals the smaller value of $y_1$ and $y_2$. If my $x_1$ is larger than the other $x$, the $y$ I observe is the larger one between $y_1$ and $y_2$.

In this case, if I observe $x_1$ and some value $y$, what is the conditional distribution of the other value of $y$? (so, if $y=y_1$, then what is the distribution of $Y_2$?)

答:
[EDIT]

Let $Y$ be the unobserved $y_1$ or $y_2$. Let $y$ be the observed value among the two.
If you know only $x_1$ and $y$:

$$
\mathbb{P}[Y\leq t|x_1,y]
$$
$$
=\mathbb{P}[Y\leq t|x_1\leq x_2,x_1,y]\mathbb{P}[x_1\leq x_2|x_1,y]
+\mathbb{P}[Y\leq t| x_1>x_2,x_1,y]\mathbb{P}[x_1>x_2|x_1,y]
$$
Simplifying:
\begin{equation}
=\mathbb{P}[Y\leq t|x_1\leq x_2,y]\mathbb{P}[x_1\leq x_2|x_1]
+\mathbb{P}[Y\leq t| x_1>x_2,y]\mathbb{P}[x_1>x_2|x_1]
\end{equation}

If $x_1\leq x_2$, the missing $Y$ is $\max\{y_1,y_2\}$
Else, the missing $Y$ is $\min\{y_1,y_2\}$.

Part 1, calculation of $\mathbb{P}[Y\leq t|x_1\leq x_2,y]$.
We are interested in the CDF of $Y:=\max\{y_1,y_2\}$, conditional on $Y\geq \min\{y_1,y_2\}=:y$.
In symbols, we want to know for any $t\in\mathbb{R}$ the value of:
$$
\mathbb{P}[Y\leq t|x_1\leq x_2,y]=\mathbb{P}[Y\leq t|Y\geq y]
$$
This can be found as follows:
$$
=\frac{\mathbb{P}[Y\leq t\cap Y\geq y]}{\mathbb{P}[Y\geq y]}
=\frac{\mathbb{P}[y\leq Y\leq t]}{\mathbb{P}[Y\geq y]}
$$

$$
=\begin{cases}
\frac{\mathbb{P}[Y\leq t]-\mathbb{P}[Y<y]}{1-\mathbb{P}[Y< y]}, & \text{if }y\leq t,\\
0,&\text{else}.
\end{cases}
$$
To use your notation for the CDF, we assume $\mathbb{P}[Y\leq t]=\mathbb{P}[Y<t]$ (in other words, $G$ is absolutely continuous). Then:
$$
=\begin{cases}
\frac{G(t)-G(y)}{1-G(y)}, & \text{if }y\leq t,\\
0,&\text{else}.
\end{cases}
$$

If $F$ is also absolutely continuous, the first term in the initial equation is:
$$
\mathbb{P}[Y\leq t|x_1\leq x_2,y]\mathbb{P}[x_1\leq x_2|x_1]
=\begin{cases}
\frac{G(t)-G(y)}{1-G(y)}(1-F(x_1)), & \text{if }y\leq t,\\
0,&\text{else}.
\end{cases}
$$

The second term is calculated analogously.


问:
Linear Diophantine using substitution $15x+16y=17$
I'm not sure what part went wrong if someone could point out.
I did the question again with Euclidean algorithm and got answers $x=-17+16t$ and $y=17-15t$, so with that in mind, this is what I did

$16y \equiv 17\pmod{15}$

$16y \equiv 32\pmod{15}$

$y \equiv 2\pmod {15}$

Sub back into equation

$15x+16(2+15t)=17$

$x=1-16t$ and $y=2+15t$

since $t$ is an integer

$t=0\; x,y=(1,2)$ and $t=1\; x,y=(-15,17)$
by using $u=a/d$, $v=b/d$

my answer was
$(-15+16t,17-15t)$

答:
From $15x = (17-16\times 2)-(16\times 15t)$ you should have $x=-1-16t$

So you are missing a minus sign in your solution.

And since $t$ is an arbitrary integer you can replace by $u=1-t$ and get the same set of solutions

$$(x,y)=(-1-16t,2+15t)=(-17+16u,17-15u)$$


问:
Probability of an event occuring in a subinterval, given that some number of events ocurred on the interval
I've just started delving into probability, and I was presented this exercise:

Patients get to a hospital in a mean rate of $5$ patients per hour, following a Poisson process. Given that $13$ patients arrived between $12$PM and $3$PM, what is the probability that $10$ patients arrived between $12$PM and $2$PM.

How can I go about solving this?

答:
Write $N_{(a,b]}$ for the number of patients that arrive between time $a$ and $b$, where $a$ and $b$ are hours measured from noon. (Adjust the notation $N_{(a,b]}$ to fit your endpoint convention.) You are looking for
$$
P(N_{(0,2]}=10\mid N_{(0,3]} = 13)=\frac{P(N_{(0,2]}=10, N_{(0,3]} = 13)}{P( N_{(0,3]} = 13)}
$$
The event in the numerator can be written as the event $\{N_{(0,2]}=10, N_{(2,3]} = 3\}$, which is an intersection of independent events.


问:
The union of three subspaces of a vector space is a subspace iff one of the subspaces is contained in the other
I attempted to solve the exercise 1.c.13 of Axler's "Linear Algebra Done Right". The problem is to prove that the union of three subspaces of a vector space $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.

I know the question is asked and answered here. According to the comment by the author and the answers in the other page, the statement is false when the field only contains two elements. But my attempts did not explicitly used that assumption and to me it is a strong indication that my proof is flawed, but I could not find where. Can you help me to verify my proof?

In the proof, I used the following lemma which is also another exercise in the book and I could prove it.

Lemma 1 (exercise 12). Suppose $U_{1}$ and $U_{2}$ are subspaces of
$V$. Then $U_{1}\cup U_{2}$ is a subspace of $V$ if and only if
$U_{1}\subseteq U_{2}$ or $U_{2}\subseteq U_{1}$.

I also used a small lemma.

Lemma 2. Suppose $A$, $B$, and $C$ are sets. Then
$A\subseteq C$ and $B\subseteq C$ iff $A\cup B\subseteq C$.

Then the following is the theorem that I attempted to prove.

Theorem (exercise 13). Suppose $U_{1}$, $U_{2}$, and $U_{3}$ are subspaces of $V$. Then
$U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$ if and only if
one of $U_{1},U_{2},U_{3}$ contains the other two.

Proof.

($\leftarrow$) I could do this part.

($\to$) Suppose $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$. Since $U_{1}\cup U_{2}\cup U_{3}=\left(U_{1}\cup U_{2}\right)\cup U_{3}=\left(U_{1}\cup U_{3}\right)\cup U_{2}=\left(U_{2}\cup U_{3}\right)\cup U_{1}$, by Lemma 1, we have

$$\begin{align*}
\left(U_{1}\cup U_{2}\subseteq U_{3}\right)\lor\left(U_{3}\subseteq U_{1}\cup U_{2}\right), & \text{ and}\\
\left(U_{1}\cup U_{3}\subseteq U_{2}\right)\lor\left(U_{2}\subseteq U_{1}\cup U_{3}\right), & \text{ and}\\
\left(U_{2}\cup U_{3}\subseteq U_{1}\right)\lor\left(U_{1}\subseteq U_{2}\cup U_{3}\right).
\end{align*}$$

If $U_{1}\cup U_{2}\subseteq U_{3}$ or $U_{1}\cup U_{3}\subseteq U_{2}$
or $U_{2}\cup U_{3}\subseteq U_{1}$, by Lemma 2,
it means one of $U_{1},U_{2},U_{3}$ contains the other two. In the
following, we will consider the case when

$$\begin{align*}
U_{3} & \subseteq U_{1}\cup U_{2},\text{ and}\\
U_{2} & \subseteq U_{1}\cup U_{3},\text{ and}\\
U_{1} & \subseteq U_{2}\cup U_{3}.
\end{align*}$$

Base on these assumptions, we will prove

$$\begin{align*}
U_{1}\cup U_{2} & \subseteq U_{3},\text{ or}\\
U_{1}\cup U_{3} & \subseteq U_{2},\text{ or}\\
U_{2}\cup U_{3} & \subseteq U_{1},
\end{align*}$$

Suppose $U_{2}\cup U_{3}\nsubseteq U_{1}$ and $U_{1}\cup U_{3}\nsubseteq U_{2}$.
Then we have

$$\begin{align*}
U_{2}\nsubseteq U_{1} & \lor U_{3}\nsubseteq U_{1},\text{ and}\\
U_{1}\nsubseteq U_{2} & \lor U_{3}\nsubseteq U_{2}.
\end{align*}$$

Then we will prove $U_{1}\cup U_{2}\subseteq U_{3}$.
We need to prove the four cases but I will only prove the case when $U_{2}\nsubseteq U_{1}$ and $U_{1}\nsubseteq U_{2}$. The other cases can be proved similarly.

Since $U_{2}\nsubseteq U_{1}$ and $U_{1}\nsubseteq U_{2}$, we can choose $a\in U_{2}$ such that $a\notin U_{1}$ and $b\in U_{1}$
such that $b\notin U_{2}$. Since $U_{2}\subseteq U_{1}\cup U_{3}$, $a\in U_{1}\cup U_{3}$.
Since $a\notin U_{1}$, $a\in U_{3}$. Similarly, since $U_{1}\subseteq U_{2}\cup U_{3}$, $b\in U_{2}\cup U_{3}$.
Since $b\notin U_{2}$, $b\in U_{3}$.

Suppose $x\in U_{1}\cup U_{2}$. Then $x\in U_{1}$ or $x\in U_{2}$.

Case $x\in U_{1}$.
Since $U_{1}\subseteq U_{2}\cup U_{3}$,
$x\in U_{2}$ or $x\in U_{3}$. If $x\in U_{3}$, it is what we desired.
We will consider the case when $x\in U_{2}$. Since $x,a,b\in U_{1}\cup U_{2}\cup U_{3}$
and $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$, $x+a+b\in U_{1}\cup U_{2}\cup U_{3}$.

Case $x+a+b\in U_{1}$. Since $x,b\in U_{1}$ and $U_{1}$
is a subspace of $V$, $\left(x+a+b\right)-x-b\in U_{1}$ iff $a\in U_{1}$,
which contradicts $a\notin U_{1}$.

Case $x+a+b\in U_{2}$. Since $x,a\in U_{2}$ and $U_{2}$
is a subspace of $V$, $b\in U_{2}$, which contradicts $b\notin U_{2}$.

Case $x+a+b\in U_{3}$. Since $a,b\in U_{3}$ and $U_{3}$
is a subspace of $V$, $x\in U_{3}$.

Case $x\in U_{2}$. Since $U_{2}\subseteq U_{1}\cup U_{3}$,
$x\in U_{1}$ or $x\in U_{3}$. If $x\in U_{3}$, it is what we desired.
We can prove the case when $x\in U_{1}$ similarly as above.

Therefore, $U_{1}\cup U_{2}\subseteq U_{3}$, as required.

答: If you look at what happens in $V=\Bbb F_2^2$ with the subspaces $U_1=\{(0,0),(1,0)\}$, $U_2=\{(0,0),(1,1)\}$, $U_3=\{(0,0),(0,1)\}$, you'll see that $U_1\cup U_2\cup U_3=V$, but none of your six inclusions in the form $U_i\subseteq U_j\cup U_h$ or $U_i\cup U_j\subseteq U_h$ holds. Essentially, it is due to the fact that none of the $U_i\cup U_j$ (with $i\ne j$) is actually a subspace, and therefore lemma 1 doesn't work.


答:
Hint: Write $X = 1001a + 110b$, $Y = 1001c + 110d$ with $a,b,c,d \in \{0,\ldots, 9\}$, without loss of generality we may assume $a \ge c$. Then
$$ Z=  X-Y = 1001(a-c) + 110(b-d) $$
Now note: Can $a-c > 1$ hold? Can $a - c = 0$ hold? Now check for which value of $k := d-b$ we have that $1001 - 110k$ is a palindrome.

Title: Finding a palindromic number which is the difference of two palindromic numbers

Body: Let $X$ and $Y$ be two $4$-digit palindromes and $Z$ be a $3$-digit palindrome. They are related in the way $X-Y=Z$. How can we figure out $Z$?


问:
Counting the number of solution for the next inequality
How many series of $t$  integer numbers $(x_1,x_2,...,x_t)$, such that $\left|x_1\right|\le n$, and for each $1\le i \le t$ there exist $\left|x_{i+1}-x_{i}\right|\le n$, there are?

$\textbf{My attempt}$: by the given information we have - $-n\le x_1 \le n$. Also, we have that $-n\le x_2-x_1\le n$ which implies $-n+x_1\le x_2 \le n+x_1$ so we have found that $-2n\le x_2\le 2n$, and so on... Now we by summing all elements of the series, we have the next inequality: $-tn...-2n-n\le x_1+x_2+...+x_t\le n+2n+...+tn$, therefore: $$\left|x_1+x_2+...+x_t \right|\le \sum_{i=1}^{t}{i\cdot n}$$
Now, I don't know how to proceed.I will be glad to get some help from you. Thanks!

答:
Pretend you had a sorted list of all such sequences.  This list starts with a block of sequences starting $(-n, \dots)$ and ends with a  block of sequences starting $(n, \dots)$.  So there are $2n+1$ blocks, organized by value of $x_1$.  In the language of combinatorics, there are $2n+1$ choices for $x_1$.

In that first block, we start with
$$  (-n, -2n, \dots )  $$
and end with
$$  (-n, 0, \dots )  \text{.}  $$
In a generic block, we start with
$$  (x_1, x_1 -n, \dots)  $$
and end with
$$  (x_1, x_1 +n, \dots)  \text{.}  $$
That is, (for any choice of $x_1$) there are $2n+1$ choices for $x_2$.

It should be clear that these have sub-blocks exhibiting $2n+1$ choices for $x_3$ (for any choice of $x_1$ and $x_2$) and sub-sub-blocks exhibiting $2n+1$ choices for $x_4$ and so on for the subsequent $x_i$ for $4 < i \leq t$.

I agree that the particular range of choices of $x_2$ depends on $x_1$ and the particular range of choices of $x_3$ depends on $x_1$ and $x_2$ and, generally, the particular range of choices of $x_n$ depends on $x_i$ for all $1 \leq i < n$.  But, regardless of the particular range of choices, there are always $2n+1$ choices in that range.  So each $x_i$ is chosen from an interval containing $2n+1$ integers.

Therefore, the number of such sequences is
$$ (2n+1)^t  \text{.}  $$


问:
Counting/combinatorics problem: How many people do not eat either tacos, salmon, or a burger?
I'm working on this counting/combinatorics problem. Can figure if it's a trick question or what kind of technique I should use. Please help.

There are these people who eat either Tacos, Salmon or Burger:

11 eats tacos

11 eats salmon

11 eats burger

3 eats taco and salmon

4 eats taco and burger

7 eat salmon and burger

1 eats taco, salmon and burger

If there are 25 people in a set, how many do not eat either tacos, salmon or burger?

答:
We can solve the problem using a Venn diagram.

Let $B$ be the set of people who eat a burger, $S$ be the set of people who eat salmon, and $T$ be the set of people who eat a taco.

Start from the inside:  Since $|B \cap S \cap T| = 1$, we place a $1$ in $B \cap S \cap T$.

We are given that
\begin{align*}
|B \cap S| & = 7\\
|B \cap T| & = 4\\
|S \cap T| & = 3
\end{align*}
Since we have already accounted for the one person who eats all three dishes, we place a $6$ in $B \cap S$ outside $B \cap S \cap T$ since $B \cap S \cap T' = 6$.  Similarly, we place a $3$ in $B \cap T$ outside $B \cap S \cap T$ and a $2$ in $S \cap T$ outside $B \cap S \cap T$.

If we now look at set $B$, we see that we have accounted for $10$ of the $11$ people who eat a burger since $|B \cap S \cap T'| + |B \cap S' \cap T| + |B \cap S \cap T| = 6 + 3 + 1 = 10$.  Therefore, we must place a $1$ in $B \cap S' \cap T'$.

Since we have already placed $9$ of the $11$ people in set $S$, we must place a $2$ in $B' \cap S \cap T'$.  Since we have already placed $6$ of the $11$ people in set $T$, we must place a $5$ in $B' \cap S' \cap T$.

If we now add all the people we have placed in sets $B$, $S$, and $T$, we conclude that $1 + 6 + 3 + 2 + 1 + 2 + 5 = 20$ eat a burger, salmon, or a taco.  Since there are a total of $25$ people, $25 - 20 = 5$ eat none of those items.

If we simply add the people who eat a burger, the people who eat salmon, and the people who eat a taco, we will have counted each person who eats two of those dishes  twice, once for each dish they ate. However, we only want to count them once, so we must subtract them from the total.

$$|B| + |S| + |T| - |B \cap S| - |B \cap T| - |S \cap T|$$
However, doing so subtracts too much.  We counted the person who eats all three meals three times when we added the number of people who eat a burger, eat salmon, or eat a taco, once for each dish that person ate, then subtracted them three times, once for each of the three pairs of dishes they ate.  Thus, we have not counted that person at all.  Thus, we must add that person to the total.  Thus, the number of people who eat at least one of the dishes is
\begin{align*}
|B \cup S \cup T| & = |B| + |S| + |T| - |B \cap S| - |B \cap T| - |S \cap T| + |B \cap S \cap T|\\
& = 11 + 11 + 11 - 7 - 4 - 3 + 1\\
& = 20
\end{align*}
so the number of people who eat none of the dishes is
$$|U| - |B \cup S \cup T| = 25 - 20 = 5$$


问:
Isometry group of $\Bbb RP^3$
Is there a reference for a proof of the statement that the isometry group of the real projective space $\Bbb RP^3$ is isomorphic to $SO(3)\times SO(3)$? How does this group acts on $\Bbb RP^3$?

答:
First, you can think of $RP^3$ as the set of all unit tangent vectors to the unit sphere; I'll denote that $T_0 S^2$. That corresponds to a point $v \in S^2$ (which I'll regard as a unit vector) and a unit tangent vector at the "tip" of $v$, which is a unit vector $w$ that's orthogonal to $v$. So it really amounts to a pair of perpendicular unit vectors $(v, w)$ in 3-space.

Why is this the same as $\Bbb RP^3$? I'll say something at the end of my answer.

Suppose that $f$ is an isometry of $P = \Bbb RP^3$. It takes the pair
$(n, e) = (\pmatrix{0\\0\\1}, \pmatrix{1\\0\\0})$, i.e., an eastward-pointing vector at the north pole) to some other pair $(a, b)$. Extending this pair to a triple $(a, b, a \times b)$ using the cross product, we can read that triple as a matrix $M$ which is (you should check this) orthogonal, and has det = +1, so it's in $SO(3)$. If we look at $g(x) = M^{-1}f(x)$, we get a new isometry of $P$ that fixes the point $(n, e)$.

So: every isometry is a combination of a rotation (my $M$) and an isometry fixing the point $(n, e) \in P$. So now the question is "Why are isometries fixing a single point of $P$ isomorphic to $SO(3)$?"

Let's lift such an isometry $g: P \to P$ to $S^3$; I'm going to assume that $(n, e)$ lifts to $\pmatrix{0\\0\\0\\1}$.

That gets us an isometry $G: S^3 \to S^3$ that preserves antipodal pairs, and by composing with the antipodal map if necessary, we can assume that $G$ actually fixes the point $\pmatrix{0\\ 0 \\ 0\\ 1}$ rather than sending it to its antipode.

Well, if you fix that point (and of course its antipode) then your isometry must fix (setwise) the orthogonal great-sphere, which is just the set of points $(x,y,z,0)$, where $x^2 + y^2 + z^2 = 1$, i.e., it's the standard 2-sphere. And each rotation of that 2-sphere (i.e. each element of $SO(3)$) gives you an isometry of that 2-sphere. (And those are the only ones!)

So...I guess that's the correspondence: one element $SO(3)$ to take $f(n, e)$ back to $(n, e)$, and (after lifting) another element of $SO(3)$ to rotate the orthogonal 2-sphere in 3-space.

I don't have a reference for this, but I suppose that it's probably in Steenrod's Fiber Bundles book somewhere in disguise.

Wait! I promised an explanation of why $P = \Bbb RP^3$ is the same as $T_0 S^2$, didn't I? Well, the double cover of $P$ is certainly $S^3$, by the definition of the projective plane. And I've shown you the correspondence $(v, w) \mapsto (v, w, v \times w)$ between $T_0 S^2$ and $SO(3)$. And $SO(3)$ is also double-covered by $S^3$ (that actually is described in Steenrod), and some theorem involving universal covers or something shows that any two things double-covered by $S^3$ must be homeomorphic. But it's been almost 40 years since I had to know this stuff, so I cannot cite chapter and verse, alas.

Title: Isometry group of $\Bbb RP^3$


问:
Given a set of unitary matrices, can one find a vector whose images under these unitary matrices span the underlying Hilbert space?
Given a set of (linearly independent) $d\times d$ complex unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ with $n\geq d$, does there exist a vector $v\in \mathbb{C}^d$ such that $$\text{span} \{U_1v, U_2v, \ldots , U_nv\} = \mathbb{C}^d ?$$

The motivation for this question comes from the theory of mixed unitary quantum channels. A quantum channel $\Phi: M_d \rightarrow M_d$ is a completely positive and trace preserving linear map. Any such map admits a Kraus representation of the form $\Phi (X) = \sum_{i=1}^k A_i X A_i^*$, where $\{A_i \}_{i=1}^k \subseteq M_d$ and $\sum_{i=1}^k A_i^* A_i = \mathbb{I}_d$. We say that a quantum channel is mixed unitary if it can be expressed as a convex combination of unitary conjugations: $\Phi (X) = \sum_{i=1}^n p_i U_i X U_i^*$. Our aim then is to look for a rank one input projector $X = vv^*$ for some $v\in \mathbb{C}^d$ such that the output $\Phi (vv^*) = \sum_{i=1}^n p_i (U_i v) (U_i v)^*$ has full rank. This is possible only if $n\geq d$. To avoid trivial counterexamples, we can also assume that $\{ U_i\}_{i=1}^n \subseteq M_d$ is linearly independent.

Follow-up question: Since it has been shown below that the question can be answered in the negative for $d\geq 4$, the natural way of progression would be to ask if one can provide a classification of all the sets of (linearly independent) unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ which allow for the existence of $v\in \mathbb{C}^d$ such that $$\text{span}\{U_1v, U_2v, \ldots ,U_nv\}=\mathbb{C}^d.$$
One can also try to answer this question for (linearly independent) sets of arbitrary complex matrices: $\{ A_i \}_{i=1}^k \subseteq M_d$.

答:
There is a simple necessary and sufficient condition for the absence of such $v$.

Let $A_1,\ldots,A_n$ be linearly independent $d\times d$ matrices. Define the completely positive map $T(X)=\sum_{i=1}^nA_iXA_i^*$.

Now, let $u=\sum_{i=1}^de_i\otimes e_i$, where $e_1,\ldots,e_d$ stands for the canonical basis of $\mathbb{C}^d$.

It is not hard to prove that for every $A,B\in M_{d\times d}$, $$tr(A\otimes B\  uu^*)=tr(AB^t).$$

Thus, we have the following equivalent conditions:

$tr(T(vv^*)ww^*)=0$

$tr((T(vv^*)\otimes \overline{w}w^t) uu^*)=0$

$tr((vv^*\otimes \overline{w}w^t)B)=0$, where $B$ is the non-normalized state $B=T^*(\cdot)\otimes id (uu^*) $.

Since $A_1,\ldots,A_n$ are linearly independent, $\operatorname{rank}{B}=n$.

So the absence of such $v$ is equivalent to the fact that for every $v\in\mathbb{C}^d$ there is a vector $w\in \mathbb{C}^d\setminus\{\vec{0}\}$ such that $v\otimes w\in \ker(B)$.

In particular, it implies that $\operatorname{rank}(B)\leq d^2-d$, since there are at least $d$ linearly independent product vectors in its kernel.

Therefore, if the number of matrices above, $n$, is greater than $d^2-d$ then there exists such $v$.


问:
Prove that $\frac{\sqrt{N}}{(\frac{N(N - 1)}{2} + 1)^2} \le 1$ for $N \ge 2$.
I am trying to prove that for natural numbers $N \ge 2$, the following expression would always be true:

$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2} \le 1.
$$

I can clearly see that when I calculate. For example: when $N = 2$,
$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2}  = \frac{\sqrt{2}}{\Big(\frac{2(2 - 1)}{2} + 1\Big)^2}  = .3536.
$$
Or when $N = 10$,
$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2}  = \frac{\sqrt{10}}{\Big(\frac{10(10 - 1)}{2} + 1\Big)^2}  =.0015.
$$
But how do I prove it for the general case? TIA.

答:
There are many ways to show this. For example, for all $\ N\geq2,\ $

$$\frac{N(N - 1)}{2}\geq \frac{N}{2}>0$$

Therefore, $$ \frac{N(N - 1)}{2}+1\geq \frac{N}{2} + 1>0$$

which implies

$$ \left(\frac{N(N - 1)}{2}+1\right)^2\geq \left(\frac{N}{2} + 1\right)^2 = \frac{N^2}{4} + N + 1 > N.$$

Finally we have

$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2} < \frac{\sqrt{N}}{N}< 1.
$$


问:
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?
Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?

I thought it would be 12 this as per pemdas rule:

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$

Wanted to confirm the right answer from you guys. Thanks for your help.

答:
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$

since priority of multiplication $\times$ or $\cdot$ is greater than addition $+$ so expression will be:

$1+1+1+1+1+1+1+1+1+1+1+0+1\implies 12$

This is the order of operation:

$1$ B:- Brackets first

$2$ O:- Orders (i.e. Powers and Square Roots, etc.)

$3$ DM:- Division and Multiplication (left-to-right)

$4$ AS:- Addition and Subtraction (left-to-right)

I think this will helpful :

http://www.magicalmaths.org/bodmas-starter-5-5-5-5-5-5-5-5-x-0-i-bet-more-people-will-answer-it-wrongp-give-a-try/


问:
Solve the following system in $\mathbb{N}$
I want to find $(x,y,z)\in\mathbb{N}^3$ such that $xy-z^2=161$ and $xz-yz=7$. I know that the solution is $x=15,y=14,z=7$ by using elementary aritmethics. But, is there a geometric way to solve this problem? I mean, a not traditional way.

答:
$\mathbb{N}^3$ will have 'gaps' in between the numbers unlike any Euclidean space. Thus, you can visualize this as:

First plot the graphs of the equations in the standard $\mathbb{R}^3$ setting.

In $\mathbb{R}^3$ the intersection of the two curves would be your solution.

Now, 'cut' down the unwanted elements, $\mathbb{N}^c$ to be precise where $\mathbb{N}\subset \mathbb{R}$. This would give discontinuous curves which intersect only at the points which are in $\mathbb{N}^3$, if any.

That's the idea behind solving it 'geometrically' from what I have understood from your question. Hope this helps.


答:
You say:

Suppose $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$. Since $U_{1}\cup U_{2}\cup U_{3}=\left(U_{1}\cup U_{2}\right)\cup U_{3}=\left(U_{1}\cup U_{3}\right)\cup U_{2}=\left(U_{2}\cup U_{3}\right)\cup U_{1}$, by Lemma 1, we have
$$\begin{align*}
\left(U_{1}\cup U_{2}\subseteq U_{3}\right)\lor\left(U_{3}\subseteq U_{1}\cup U_{2}\right), & \text{ and}\\
\left(U_{1}\cup U_{3}\subseteq U_{2}\right)\lor\left(U_{2}\subseteq U_{1}\cup U_{3}\right), & \text{ and}\\
\left(U_{2}\cup U_{3}\subseteq U_{1}\right)\lor\left(U_{1}\subseteq U_{2}\cup U_{3}\right).
\end{align*}$$

Now, Lemma 1 says that if you have two subspaces $V_1$ and $V_2$, and $V_1\cup V_2$ is a subspace, then either $V_1\subseteq V_2$ or $V_2\subseteq V_1$. You seem to be trying to apply Lemma 1 to $U_1\cup U_2\cup U_3$ by, for example, thinking of the union as $(U_1\cup U_2)\cup U_3$, and then thinking of $U_3$ as $V_2$, and thinking of $U_1\cup U_2$ as $V_1$. There's a problem with that:

You don't know that $U_1\cup U_2$ is a subspace.

That means that you cannot yet apply Lemma 1 to the expression $(U_1\cup U_2)\cup U_3$ by setting $V_1=U_1\cup U_2$ and $V_2=U_3$; you would first need to prove that $U_1\cup U_2$ is a subspace. Likewise, you cannot apply it to $(U_1\cup U_3)\cup U_2$, nor to $(U_2\cup U_3)\cup U_1$. So this assertion is unjustified.

Note that, by Lemma 1, the only way that $U_1\cup U_2$ could possibly be a subspace would be if either $U_1\subseteq U_2$ or $U_2\subseteq U_1$. So you would be assuming part of what you are trying to show.

As the rest of your attempt flows from this set of conjunctions/disjunctions, the rest of the argument is doomed and I will not review it. This is probably why your argument does not use anything about the field.

Title: The union of three subspaces of a vector space is a subspace iff one of the subspaces is contained in the other

Body:
I attempted to solve the exercise 1.c.13 of Axler's "Linear Algebra Done Right". The problem is to prove that the union of three subspaces of a vector space $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.

I know the question is asked and answered here. According to the comment by the author and the answers in the other page, the statement is false when the field only contains two elements. But my attempts did not explicitly used that assumption and to me it is a strong indication that my proof is flawed, but I could not find where. Can you help me to verify my proof?

In the proof, I used the following lemma which is also another exercise in the book and I could prove it.

Lemma 1 (exercise 12). Suppose $U_{1}$ and $U_{2}$ are subspaces of
$V$. Then $U_{1}\cup U_{2}$ is a subspace of $V$ if and only if
$U_{1}\subseteq U_{2}$ or $U_{2}\subseteq U_{1}$.

I also used a small lemma.

Lemma 2. Suppose $A$, $B$, and $C$ are sets. Then
$A\subseteq C$ and $B\subseteq C$ iff $A\cup B\subseteq C$.

Then the following is the theorem that I attempted to prove.

Theorem (exercise 13). Suppose $U_{1}$, $U_{2}$, and $U_{3}$ are subspaces of $V$. Then
$U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$ if and only if
one of $U_{1},U_{2},U_{3}$ contains the other two.

Proof.

($\leftarrow$) I could do this part.

($\to$) Suppose $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$. Since $U_{1}\cup U_{2}\cup U_{3}=\left(U_{1}\cup U_{2}\right)\cup U_{3}=\left(U_{1}\cup U_{3}\right)\cup U_{2}=\left(U_{2}\cup U_{3}\right)\cup U_{1}$, by Lemma 1, we have

$$\begin{align*}
\left(U_{1}\cup U_{2}\subseteq U_{3}\right)\lor\left(U_{3}\subseteq U_{1}\cup U_{2}\right), & \text{ and}\\
\left(U_{1}\cup U_{3}\subseteq U_{2}\right)\lor\left(U_{2}\subseteq U_{1}\cup U_{3}\right), & \text{ and}\\
\left(U_{2}\cup U_{3}\subseteq U_{1}\right)\lor\left(U_{1}\subseteq U_{2}\cup U_{3}\right).
\end{align*}$$

If $U_{1}\cup U_{2}\subseteq U_{3}$ or $U_{1}\cup U_{3}\subseteq U_{2}$
or $U_{2}\cup U_{3}\subseteq U_{1}$, by Lemma 2,
it means one of $U_{1},U_{2},U_{3}$ contains the other two. In the
following, we will consider the case when

$$\begin{align*}
U_{3} & \subseteq U_{1}\cup U_{2},\text{ and}\\
U_{2} & \subseteq U_{1}\cup U_{3},\text{ and}\\
U_{1} & \subseteq U_{2}\cup U_{3}.
\end{align*}$$

Base on these assumptions, we will prove

$$\begin{align*}
U_{1}\cup U_{2} & \subseteq U_{3},\text{ or}\\
U_{1}\cup U_{3} & \subseteq U_{2},\text{ or}\\
U_{2}\cup U_{3} & \subseteq U_{1},
\end{align*}$$

Suppose $U_{2}\cup U_{3}\nsubseteq U_{1}$ and $U_{1}\cup U_{3}\nsubseteq U_{2}$.
Then we have

$$\begin{align*}
U_{2}\nsubseteq U_{1} & \lor U_{3}\nsubseteq U_{1},\text{ and}\\
U_{1}\nsubseteq U_{2} & \lor U_{3}\nsubseteq U_{2}.
\end{align*}$$

Then we will prove $U_{1}\cup U_{2}\subseteq U_{3}$.
We need to prove the four cases but I will only prove the case when $U_{2}\nsubseteq U_{1}$ and $U_{1}\nsubseteq U_{2}$. The other cases can be proved similarly.

Since $U_{2}\nsubseteq U_{1}$ and $U_{1}\nsubseteq U_{2}$, we can choose $a\in U_{2}$ such that $a\notin U_{1}$ and $b\in U_{1}$
such that $b\notin U_{2}$. Since $U_{2}\subseteq U_{1}\cup U_{3}$, $a\in U_{1}\cup U_{3}$.
Since $a\notin U_{1}$, $a\in U_{3}$. Similarly, since $U_{1}\subseteq U_{2}\cup U_{3}$, $b\in U_{2}\cup U_{3}$.
Since $b\notin U_{2}$, $b\in U_{3}$.

Suppose $x\in U_{1}\cup U_{2}$. Then $x\in U_{1}$ or $x\in U_{2}$.

Case $x\in U_{1}$.
Since $U_{1}\subseteq U_{2}\cup U_{3}$,
$x\in U_{2}$ or $x\in U_{3}$. If $x\in U_{3}$, it is what we desired.
We will consider the case when $x\in U_{2}$. Since $x,a,b\in U_{1}\cup U_{2}\cup U_{3}$
and $U_{1}\cup U_{2}\cup U_{3}$ is a subspace of $V$, $x+a+b\in U_{1}\cup U_{2}\cup U_{3}$.

Case $x+a+b\in U_{1}$. Since $x,b\in U_{1}$ and $U_{1}$
is a subspace of $V$, $\left(x+a+b\right)-x-b\in U_{1}$ iff $a\in U_{1}$,
which contradicts $a\notin U_{1}$.

Case $x+a+b\in U_{2}$. Since $x,a\in U_{2}$ and $U_{2}$
is a subspace of $V$, $b\in U_{2}$, which contradicts $b\notin U_{2}$.

Case $x+a+b\in U_{3}$. Since $a,b\in U_{3}$ and $U_{3}$
is a subspace of $V$, $x\in U_{3}$.

Case $x\in U_{2}$. Since $U_{2}\subseteq U_{1}\cup U_{3}$,
$x\in U_{1}$ or $x\in U_{3}$. If $x\in U_{3}$, it is what we desired.
We can prove the case when $x\in U_{1}$ similarly as above.

Therefore, $U_{1}\cup U_{2}\subseteq U_{3}$, as required.


问:
Finding the argument of a complex number
So I have the following complex number: $z = -2 + 4i$ and I'm trying to find the argument where $\theta$ is between $-\pi < \theta \leq \pi$, which I've been told would be $\arctan(\frac{b}{a})$ yet when I do just that, I get  $\arctan(\frac{4}{-2})$. The answer to that question is $-\arctan(\pi + 2)$ which I don't quite understand. I would appreciate some help, thanks in advance!

答: The formula $\arg(a+bi)=\arctan(b/a)$ only works if $b>0$. You can avoid using formulae entirely by constructing a a right-angled triangle that has vertices at $A(0,0)$, $B(-2,0)$, and $C(-2,4)$. Then, find the acute angle at $A(0,0)$, and subtract this angle from $\pi$ to find the obtuse angle made with the positive real axis. This will give you the principal argument of $-2+4i$.


问:
How to calculate co-ordinates after deformation
I am working in Objective-C which allows me to deform an image by pushing/pulling a single corner.  This allows me to take this image:

And deform it so that line AB is horizontal (points A & B have same y-value) thusly:

Notice how the deformation causes BOTH points A & B to be moved.  Now that they have moved, I want to know their new y-values.

My attempt has been:

```<code>1) Find angle E in triangle DEF
2) Use that to find angle D in triangle HDG
3) Use DG / sin(E) = HG / sin(D)
HG = (DF / sin(E)) * sin(D)
</code>```

to solve for HG, which would give me the distance from the original y-coordinate for point A.

But these results are not correct.  What is the correct method?

答:
General approach

Your deformation looks like a projective transformation. Since projective transformations preserve neither angles nor lengths nor even length ratios, your attempts using elementary trigonometry are probably bound to fail. A projective transformation in the plane is uniquely defined by four points and their images, as long as no three points lie on a common line. This answer gives details on how you can compute such a transformation.

Note that if you already know the coordinates of the “pulled” rectangle corner ($E$ in your question), then you can simply use your four corners as the four defining points. Since finding the correct position for $E$ is non-trivial, I didn't make that assumption. So the main task here is finding four suitable pairs of points. You can use the resulting machinery not only without prior knowledge of $E$, but in fact even as a tool to compute $E$ in the first place.

Coordinates for four defining points

Suppose your original rectangle has width $w$ and height $h$. Then you can assume the following homogenous coordinates. I'll place the origin of the coordinate system in the lower left corner of your rectangle, and the positive $y$ axis pointing upwards. If you use a different convention, these coordinates should be easy to adapt.

\begin{align*}
P_1 &= \begin{pmatrix}0\\0\\1\end{pmatrix} &
P_2 &= \begin{pmatrix}w\\0\\1\end{pmatrix} &
P_3 &= \begin{pmatrix}w\\h\\1\end{pmatrix} &
P_4 &= \begin{pmatrix}0\\h\\1\end{pmatrix} \\
A &= \begin{pmatrix}a\\h\\1\end{pmatrix} &
B &= \begin{pmatrix}w\\b\\1\end{pmatrix} &
H &= \begin{pmatrix}1\\0\\0\end{pmatrix} &
V &= \begin{pmatrix}0\\1\\0\end{pmatrix} &
\end{align*}

The points $P_i$ are the corners of your rectangle, $A$ and $B$ are as in your images, while $H$ and $V$ are the points at infinity through which all horizontal resp. vertical lines must pass. Now we need some additional points. Using homogenous coordinates, you can use the cross product to express both the line joining two points and the point at the intersection oftwo lines. So the line $AB$ will intersect the bottom line of your rectangle in a point

$$Q = (A\times B)\times(P_1\times P_2)$$

Now you can define your transformation using the following:

\begin{align*}
P_1 &\mapsto P_1 \\
P_2 &\mapsto P_2 \\
P_4 &\mapsto P_4 \\
V &\mapsto V \\
Q &\mapsto H
\end{align*}

So three corners remain where they are, vertical lines remain vertical, and the two lines intersecting at $Q$ will both become horizontal. These are five points, not four, but some of them are collinear. So I'd keep the first three conditions, and use some point constructed from several of these defining points as the fourth condition:

$$ (P_2\times V)\times(P_4\times Q) \mapsto
(P_2\times V)\times(P_4\times H) $$

This gives you a fourth defining pair of points which (barring degenerate situations) does not lie on the same line as two of the other defining points. So now you have four pairs of preimage and image points, and can compute the transformation matrix from these.

Using the transformation

Applying that transformation matrix to $A$ or $B$ (and perhaps $P_3$) will give you their new coordinates, which you'll probably want to dehomogenize. But these steps are also described in my other post.

Explicit result

All of the above computations could be done symbolically using some computer algebra system. the resulting expressions aren't that bad. The $y$ coordinate of $E=P_3'$ is

$$E_y = \frac{a\,b\,h-w\,h^2}{a\,b-b\,w}$$

and the $y$ coordinate of the transformed line $AB$ is

$$H_y = \frac{a\,b-h\,w}{a-w}$$

So if you only need these, but need them in a number of situations, then you might simply want to implement this final formula. Understanding the whole process used to obtain these will however be useful in dealing with other, similar situations.

Sage code

I obtained the above formulas using the following piece of sage code:

```<code>def joinMeet(a, b, c, d):
    return a.cross_product(b).cross_product(c.cross_product(d))
def pm1(a, b, c, d):
    M = Matrix([a , b, c]).transpose()
    f = M.solve_right(d)
    return M*diagonal_matrix(f)
def pm(a1, a2, b1, b2, c1, c2, d1, d2):
    return pm1(a2, b2, c2, d2)*(pm1(a1, b1, c1, d1).adjoint())
def dehom(p):
    return vector([p[0]/p[2], p[1]/p[2]])
var('w h a b')
P1 = vector([0, 0, 1])
P2 = vector([w, 0, 1])
P3 = vector([w, h, 1])
P4 = vector([0, h, 1])
A = vector([a, h, 1])
B = vector([w, b, 1])
H = vector([1, 0, 0])
V = vector([0, 1, 0])
Q = joinMeet(A, B, P1, P2)
R1 = joinMeet(P2, V, P4, Q)
R2 = joinMeet(P2, V, P4, H)
M = pm(P1, P1, P2, P2, P4, P4, R1, R2)
print(dehom(M*P3)[1].simplify_full())
print(dehom(M*A)[1].simplify_full())
</code>```

Body:
General approach

Your deformation looks like a projective transformation. Since projective transformations preserve neither angles nor lengths nor even length ratios, your attempts using elementary trigonometry are probably bound to fail. A projective transformation in the plane is uniquely defined by four points and their images, as long as no three points lie on a common line. This answer gives details on how you can compute such a transformation.

Note that if you already know the coordinates of the “pulled” rectangle corner ($E$ in your question), then you can simply use your four corners as the four defining points. Since finding the correct position for $E$ is non-trivial, I didn't make that assumption. So the main task here is finding four suitable pairs of points. You can use the resulting machinery not only without prior knowledge of $E$, but in fact even as a tool to compute $E$ in the first place.

Coordinates for four defining points

Suppose your original rectangle has width $w$ and height $h$. Then you can assume the following homogenous coordinates. I'll place the origin of the coordinate system in the lower left corner of your rectangle, and the positive $y$ axis pointing upwards. If you use a different convention, these coordinates should be easy to adapt.

\begin{align*}
P_1 &= \begin{pmatrix}0\\0\\1\end{pmatrix} &
P_2 &= \begin{pmatrix}w\\0\\1\end{pmatrix} &
P_3 &= \begin{pmatrix}w\\h\\1\end{pmatrix} &
P_4 &= \begin{pmatrix}0\\h\\1\end{pmatrix} \\
A &= \begin{pmatrix}a\\h\\1\end{pmatrix} &
B &= \begin{pmatrix}w\\b\\1\end{pmatrix} &
H &= \begin{pmatrix}1\\0\\0\end{pmatrix} &
V &= \begin{pmatrix}0\\1\\0\end{pmatrix} &
\end{align*}

The points $P_i$ are the corners of your rectangle, $A$ and $B$ are as in your images, while $H$ and $V$ are the points at infinity through which all horizontal resp. vertical lines must pass. Now we need some additional points. Using homogenous coordinates, you can use the cross product to express both the line joining two points and the point at the intersection oftwo lines. So the line $AB$ will intersect the bottom line of your rectangle in a point

$$Q = (A\times B)\times(P_1\times P_2)$$

Now you can define your transformation using the following:

\begin{align*}
P_1 &\mapsto P_1 \\
P_2 &\mapsto P_2 \\
P_4 &\mapsto P_4 \\
V &\mapsto V \\
Q &\mapsto H
\end{align*}

So three corners remain where they are, vertical lines remain vertical, and the two lines intersecting at $Q$ will both become horizontal. These are five points, not four, but some of them are collinear. So I'd keep the first three conditions, and use some point constructed from several of these defining points as the fourth condition:

$$ (P_2\times V)\times(P_4\times Q) \mapsto
(P_2\times V)\times(P_4\times H) $$

This gives you a fourth defining pair of points which (barring degenerate situations) does not lie on the same line as two of the other defining points. So now you have four pairs of preimage and image points, and can compute the transformation matrix from these.

Using the transformation

Applying that transformation matrix to $A$ or $B$ (and perhaps $P_3$) will give you their new coordinates, which you'll probably want to dehomogenize. But these steps are also described in my other post.

Explicit result

All of the above computations could be done symbolically using some computer algebra system. the resulting expressions aren't that bad. The $y$ coordinate of $E=P_3'$ is

$$E_y = \frac{a\,b\,h-w\,h^2}{a\,b-b\,w}$$

and the $y$ coordinate of the transformed line $AB$ is

$$H_y = \frac{a\,b-h\,w}{a-w}$$

So if you only need these, but need them in a number of situations, then you might simply want to implement this final formula. Understanding the whole process used to obtain these will however be useful in dealing with other, similar situations.

Sage code

I obtained the above formulas using the following piece of sage code:

def joinMeet(a, b, c, d):
    return a.cross_product(b).cross_product(c.cross_product(d))
def pm1(a, b, c, d):
    M = Matrix([a , b, c]).transpose()
    f = M.solve_right(d)
    return M*diagonal_matrix(f)
def pm(a1, a2, b1, b2, c1, c2, d1, d2):
    return pm1(a2, b2, c2, d2)*(pm1(a1, b1, c1, d1).adjoint())
def dehom(p):
    return vector([p[0]/p[2], p[1]/p[2]])
var('w h a b')
P1 = vector([0, 0, 1])
P2 = vector([w, 0, 1])
P3 = vector([w, h, 1])
P4 = vector([0, h, 1])
A = vector([a, h, 1])
B = vector([w, b, 1])
H = vector([1, 0, 0])
V = vector([0, 1, 0])
Q = joinMeet(A, B, P1, P2)
R1 = joinMeet(P2, V, P4, Q)
R2 = joinMeet(P2, V, P4, H)
M = pm(P1, P1, P2, P2, P4, P4, R1, R2)
print(dehom(M*P3)[1].simplify_full())
print(dehom(M*A)[1].simplify_full())


问:
Properties of Miquel point
Let $ABCD$ be a quadrilateral with $P=AD \cap BC$ and $Q=AB\cap CD$. Let $M$ be the miquel point of the quadrilateral. Prove the following-
$\text{1)}$ If $O_1$ and $O_2$ be the centres of $\triangle PAB$ and $\triangle PDC$ then $MO_1O_2 \sim MAD$.
 $\text{2)}$ $O_1,O_2$ and the circumcircles of $\triangle QBC$ and the circumcircle of $\triangle QAD$ are concyclic.

答:
I think in the second question you're referring to the circumCENTER of $\triangle QBC$ and $\triangle QAD$,which we denote by $O_3$ and $O_4$ respectively.

For the first part, we use the following lemma:

$M$ is the center of the spiral similarity sending $AB$ to $DC$.

Proof:

since $A,B,P,M$ and $D,C,P,M$ are concyclic, we get that
$$\measuredangle AMB=\measuredangle APB=\measuredangle DPC=\measuredangle DMC$$
Similiary $$\measuredangle ABM=\measuredangle DCM$$
and we've proved the lemma.

Now since this spiral similarity sends $AB$ to $DC$,it also sends the circumcenter of $\triangle MAB$ to $\triangle MDC$, namely it sends $O_1$ to $O_2$.

Hence $MO_1O_2$ is sent to $MAD$ by a spiral similarity and this leads to the fact that they're similar.

For the second part we use angle chasing.

Known that $\measuredangle MO_2O_1=\measuredangle MDA$ and $M,Q,D,A$ are concyclic,we can compute $\measuredangle MO_2O_1=\measuredangle AQM=\frac{1}{2}\measuredangle AO_4M$

The last thing we need to notice is that $O_1O_4$ is the perpendicular bisector of $MA$ beacause the latter is the radical axes of two circles with center $O_1$ and $O_4$.

With this fact in mind, we get $\measuredangle O_1O_4M=\frac{1}{2}\measuredangle AO_4M$.

Therefore $\measuredangle MO_2O_1=\measuredangle MO_4O_1$ and these four points are concyclic.Similiarly we can prove that $O_3$ also lies on this circle.

Title: Properties of Miquel point


问:
How can I prove that the Core of $H$ in $G$ is a normal subgroup?
Given $G$ a group and $H$ a subgroup of $G$, how can I prove that ${\rm Core}_H G:=\bigcap_{g\in G} gHg^{-1}$ is a normal subgroup of $G$?

I've proved that is a subgroup, but how can I prove that it is a normal subgroup?

If $x \in{\rm Core}_H G,\ y \in G$, how can I prove that $yxy^{-1}$ is in ${\rm Core}_H G$?

As $x \in{\rm Core}_H G,\ \forall g\in G,\ \exists a\in H:x=gag^{-1} $
Moreover, as ${\rm Core}_H G \subseteq H$, (since it is subgroup), $x\in H$

How can I continue and show that $yxy^{-1}\in {\rm Core}_H G$

答:
For all $k \in G$,  \begin{align*}
k \mathrm{Core}_G(H)k^{-1} &= k \left(\bigcap_{g \in G} g H g^{-1} \right) k^{-1}  \\
&= \bigcap_{g \in G} kg H (kg)^{-1}
\end{align*}
As $g$ ranges through $G$, $kg$ ranges through $G$.  (That is, left-multiplication by $k$, $\varphi_k: G \rightarrow G: g \mapsto kg$, is a bijection.)  So the expression on the right is the same intersection as in the definition of the core, with the sets permuted.  Intersection is invariant under permutation of its arguments.  Therefore,
$$  \forall k \in G, k \mathrm{Core}_G(H) k^{-1} = \mathrm{Core}_G(H)  \text{,}  $$
so $\mathrm{Core}_G(H)$ is normal in $G$.


问:
Evaluating $\lim_{x\to-\infty} xe^{-x}$
Not sure how to think about this one.

$$\lim_{x\to-\infty}  xe^{-x}$$

If I place "$-\infty$" in place for $x$ I get
$$(-\infty)e^{\infty}$$
the limit of which would seem to be $0$ as $x$ approaches $-\infty$.

But in fact the limit is $-\infty$ (if looking at the drawn graph) as $x$ approaches $-\infty$.

答:
Intuitively we can see the limit is $-\infty$:

To make it rigorous we want to find  $x_0(\epsilon)<0$ (a function in terms of $\epsilon<0$) such that

$$x<x_0(\epsilon)\implies xe^{-x}<\epsilon$$
We know that $$e^x \ge1+x \implies e^{-x}\ge 1-x$$So for $x\le 0 $ we have $$xe^{-x}\le x(1-x)$$Let $$x(1-x)\lt \epsilon \implies -x^2 +x - \epsilon \lt 0 \implies x^2 - x+ \epsilon\gt 0$$By choosing $x_0(\epsilon) = \frac{1 - \sqrt{1 - 4\epsilon}}{2} - 1$, the proof is complete. We can check it easily $$x_0(1-x_0) =\epsilon - \sqrt{1 - 4\epsilon} - 1 \lt \epsilon \iff 1 + \sqrt{1 - 4\epsilon} \gt 0$$which is true since $\sqrt{1 - 4\epsilon} \gt 0$.

Title: Evaluating $\lim_{x\to-\infty} xe^{-x}$


问:
How to summarise pattern in binomial-type expansion into a single expression
I have a series of polynomials as follows:

\begin{eqnarray}
&1\\
&1+4x^3+x^6\\
&1+20x^3+48x^6+20x^9+x^{12}\\
&1+54x^3+405x^6+760x^9+405x^{12}+54x^{15}+ x.^{18}\\
& \vdots\tag{1}
\end{eqnarray}

and I want to find a general expression that will generate these.

I have noted that the coefficients follow a recognised (and documented) pattern:

```<code>1;
1, 4, 1;
1, 20, 48, 20, 1;
1, 54, 405, 760, 405, 54, 1;
1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
</code>```

My initial researches have identified that the coefficients of my polynomials are those listed in the first set of numbers (above),which is the known integer sequence A181544 as listed at https://oeis.org/A181544.

The listing goes on to give the generating function for the polynomial coefficients:

```<code>Row generating functions begin:
(1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
(1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
(1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
(1 + 54*x + 405*x^2 + 760*x^3 + 405*x^4 + 54*x^5 + x^6) = (1-x)^10*(1 + 4^3*x + 10^3*x^2 + 20^3*x^3 + 35^3*x^4 +...); ...
</code>```

and provides a combinatorial formulae for doing so.  For row $n$ the $k^{th}$ term where $0<k<2n$ is:

\begin{equation}
\left\{\sum_{k=0}^{2n} \binom{n+k-1}{k}^3 x^{k}\right\} \left(1-x\right)^{3n+1}
\end{equation}

I have two problems:  (1) I am unable to generate the given example polynomials from this expression - how is that done ? and (2) my polynomials are in powers of $x^3$, (see (1) above). How do I modify this algorithm to generate these?

答:
The expression given in the OEIS header seems to have a mismatch between the $n$ in the binomial coefficient and the $n$ in the exponent on $1-x$. If you change the displayed formula to

$$\left(\sum_{k\ge 0}\binom{n+k-1}k^3x^k\right)(1-x)^{3n-2}\,,$$

it will give you the correct results for row $n$. For instance, for row $2$ you get

$$\left(\sum_{k\ge 0}\binom{k+1}k^3x^k\right)(1-x)^4=(1-x)^4\sum_{k\ge 0}(k+1)^3x^k\,,$$

exactly as in the second line in the table of row generating functions. If you number the rows starting with $0$ instead of $1$, it becomes

$$\left(\sum_{k\ge 0}\binom{n+k}k^3x^k\right)(1-x)^{3n+1}\,;$$

it appears that they inadvertently mixed the expressions for the two origins.

It seems to me that the simplest way to generate your polynomials is to generate these and then multiply every exponent on $x$ by $3$.


问:
Proving $\mathbb{Q\times Z}$ is countable
I want to prove that  $\mathbb{Q\times Z}$ is countable. I found bijection from $\Bbb N\times \Bbb N$ to $\Bbb N$. Can I use it?

答:
What you can do is find explicit bijections $\psi_\mathbb{Q}$ & $\psi_\mathbb{Z}$ between $\mathbb{N}$, $\mathbb{Q}$ and $\mathbb{N}$, $\mathbb{Z}$, respectively.

Then you can define a map $\psi : \mathbb{N} \times \mathbb{N}\rightarrow \mathbb{Q} \times \mathbb{Z}$ s.t.
$(n,m)$ $\mapsto$ $(\psi_\mathbb{Q}(n), \psi_\mathbb{Z}(m))$, $\forall (n,m) \in \mathbb{N}\times\mathbb{N}$.

I think this would work. You can verify that $\psi$ is indeed a bijection. Hope this helps.


问:
Graphical solution of LOP with three variables and a parameter
For what values of $\lambda$ does the following linear optimization problem has no solution?
$$
 x_2 - \lambda x_3 \to \operatorname{min}
$$
subject to
\begin{align*}
 -x_1 + 3x_2 + x_3 & ~=~ 3 \\
 -x_1 +  x_2       & ~\le~ 4 \\
               x_i & ~\ge~ 0, \quad i = 1,2,3
\end{align*}
Solve this graphically! (in $\mathbb{R}^2$).

I drawed an area of admissable solutions for different, fixed values of $x_1$ and came to the conclusion that it is always solvable, regardless of the values of $\lambda$ because the two restrictions always have a nonempty region of admissable solutions. Is this correct that it's solvable for all $\lambda$? But that makes this exercise quite trivial...

EDIT: It's $x_2$!

答:
Having a nonempty region of admissible $x$ is necessary for existence of minimum. But it's not sufficient. The solution will also fail to exist if the function $x_2-\lambda x_3$ is not bounded from below on the admissible region.

To make the problem two-dimensions, you can eliminate $x_1$ from it, replacing it by $3x_2+x_3+3$. The constraints become
$$
\begin{align*}
 -x_1 + 3x_2 + x_3 & ~=~ 3 \\
 -2x_2-x_3    & ~\le~ 7 \\
3x_2+x_3& ~\ge -3  \\
               x_i & ~\ge~ 0, \quad i = 2,3
\end{align*}
$$
After drawing the region, you should determine whether or not it is

empty (if so, there is no solution)

nonempty and bounded (if so, there is a solution)

unbounded (then there is a solution if and only if the region does not contain any half-line with the direction vector $(-1,\lambda)$)


答:
Yes, but you should make sure you don't skip any important steps.

If you can prove (or already have results that show that) there is a bijection from $\mathbb{Q} \times \mathbb{Z}$ to $\mathbb{N} \times \mathbb{N}$ (let's call it $f$), and a second bijection from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$ (let's call it $g$), then the function $g \circ f$ will be a bijection from $\mathbb{Q} \times \mathbb{Z}$ to $\mathbb{N}$ and hence will demonstrate that it is countable. (It's also sufficient if $f$ and $g$ are injections, as long as you can prove that there is also a separate injection from $\mathbb{N}$ to $\mathbb{Q} \times \mathbb{Z}$, which should be pretty easy to do.)

I'm guessing that you've already proven that $\mathbb{Q}$ and $\mathbb{Z}$ are countable, so you should be able to create $f$ from that fact.

It's also helpful to see that you can apply this to any countable sets $A$ and $B$, as pointed out in one of the comments.

Title: Proving $\mathbb{Q\times Z}$ is countable

Body: I want to prove that  $\mathbb{Q\times Z}$ is countable. I found bijection from $\Bbb N\times \Bbb N$ to $\Bbb N$. Can I use it?


问:
Subsets partitioning $\mathbb R^d$ into $2^k$ disjoint sets
I have been going through Terrance Tao's Introduction to Measure Theory book and I have been having trouble developing intuition behind a certain statement he makes after defining a (complex-valued) Simple function as a linear combination of indicator functions of Lebesgue measurable sets and complex numbers.

Within the definition, he makes the comment that it is not required that the Lebesgue measurable sets be disjoint but that it can be easy enough to arrange this. He says that any $k$ subsets $E_1, \ldots, E_k$ of $\mathbb R^d$ partition $\mathbb R^d$ into $2^k$ disjoint sets, each of which is an intersection of $E_i$ or the complement $\mathbb R^d - E_i$. I do not understand how the $k$ subsets partition $\mathbb R^d$ into $2^k$ disjoint sets and I am not able to create a simple case which can help me understand the general one. Any help would be much appreciated. Thank you in advance.

答:
For each $I\subseteq[k]=\{1,2,\ldots,k\}$ let

$$E_S=\bigcap_{i\in S}E_i\setminus\bigcup_{i\in[k]\setminus S}E_i\,;$$

$E_S$ consists of those point of $\Bbb R^d$ that are in $E_i$ if and only if $i\in S$. For instance, $S_\varnothing=\Bbb R^d\setminus\bigcup_{i=1}^kE_i$, $S_{[k]}=\bigcap_{k=1}^kE_i$, and $E_{\{1\}}$ is the set of points of $\Bbb R^d$ that are in $E_1$ but not in any of $E_2,\ldots,E_k$. Since $[k]$ has $2^k$ subsets, there are $2^k$ of these sets $E_S$.

Show that if $S,T\subseteq[k]$, and $S\ne T$, then $E_S\cap E_T=\varnothing$.

(It might be helpful to think about $3$ sets in $\Bbb R^2$ and draw a typical Venn diagram: it divides the plane into $2^3=8$ regions.)


问:
Line tangent to an ellipse in 3d space at a given point
I have the ellipse defined by the equation $\frac{x^2}{2} + \frac{z^2}{8} = 1$, and I need to find the parametric equation for a line tangent to the ellipse at $(1,2,2)$

I know that the parametric equation for the ellipse is $x = 2cos(t), y=2, z=8sin(t)$, but I'm not sure what to do next.

The original question was

Find parametrization of curve of intersection of following two surfaces
Surface
$4x^2+2y^2+z^2=16$ intersects plane
$y = 2$
and generates an ellipse

Find the parametric equation of the tangent line to this ellipse at
( 1 , 2 , 2 )

答:
Intersection of the given surface and the plane is,

$\displaystyle \frac{x^2}{2} + \frac{z^2}{8} = 1, y = 2$ which can be parametrized as

$r(t) = \langle\sqrt2 \ cos (t), 2, \sqrt8 \ sin (t) \ \rangle $

So $r'(t) = \langle - \sqrt2 \ sin (t), 0, \sqrt8 \ cos (t) \ \rangle$

At $(1, 2, 2), x = \sqrt2 \cos (t) = 1, z = \sqrt 8 \sin (t) = 2 \implies t = \frac{\pi}{4}$.

$r'(t = \frac{\pi}{4}) = (-1, 0, 2)$

So the equation of tangent line is $(1,2, 2) + (-1, 0, 2) t$

or $ \ x = 1 - t, y = 2, z = 2 + 2t$


问:
Norm of $p$-adic number well defined.
After the construction of the $p$-adic numbers as equivalent Cauchy sequences of rational numbers with respect to the $p$-adic absolute value, we define the norm of $\lambda\in\mathbb{Q}_p$ to be $|\lambda|_p=\operatorname{lim}_{n\rightarrow\infty}|x_n|_p$, where $(x_n)$ is any Cauchy sequence representing $\lambda$. But how do we know the limit exists? Can we say that from knowing $(x_n)$ is Cauchy that $|x_n|$ is Cauchy (and so converges since $\mathbb{R}$ is complete?) If so, why is this true?

答:
Use reverse triangle law.

$$ | |x_n|_p - |x_m|_p | \leq |x_n - x_m|_p$$

So $|x_n|_p$ is Cauchy.

Title: Norm of $p$-adic number well defined.


问:
Unique representation of simple function
Claim: Consider a simple function $f : X \to \mathbb{R}$ where $(X,M,\mu)$ is a measure space. If we represent $f$ as $\sum_{i=1}^{n} a_{i} \chi_{A_i}$, with $A_i \cap A_j = \varnothing$ and $a_i \ne a_j$ for $i \ne j$, then is this representation of the simple function is unique.

I intuitively feel it should be true, but I'm not able to prove it.

My attempt at proof: Suppose $f = \sum_{i=1}^{n} a_{i} \chi_{A_i} = \sum_{j=1}^{m} b_{j} \chi_{B_j}$. Then we need to prove the following: $n=m$, $a_i = b_{p_i}$ and $A_i = B_{p_i}$, where $p_1,p_2,\ldots,p_n \in \{1,2,\ldots,n\}$ for all $i=1,2,\ldots,n$ and $p_i \ne p_j$ if $i \ne j$. However, I'm stuck at this point and do not know how to proceed.

I have two questions:

Could you suggest how do I complete the proof or some other method to prove the claim, or a counterexample to the claim?

If the claim is true, does it hold without can $a_i \ne a_j$ be removed from the hypothesis, and then established as a part of the proof?

答:
Of course it's easy to construct examples showing that you need to assume $$a_j\ne0$$ and $$b_k\ne0.$$ IF you assume that:

HINT Show that $f(X)\setminus\{0\}=\{a_1,\dots,a_n\}$.

Hence $$\{b_1,\dots,b_m\}=\{a_1,\dots,a_n\},$$ and now since the $a_j$ are distinct and the $b_k$ are distinct you're on your way...

Note another way to fix it is to assume that $\bigcup A_j=X$. (If you assume that you had better not assume $a_j\ne0$.) The stuff at the start maybe comes out more elegant with the second version, but once we've moved on the second seems silly, requiring $\chi_A+0\chi_{X\setminus A}$ in place of a simple $\chi_A$.


问:
Prove or disprove a proposition on Taylor series
Let $f(x) = \sum_{k\geq 0} a_kx^k$ be a real analytic function, converging on $-R < x < R$, whose $n^{th}$ Taylor polynomial is $f_n(x) = \sum_{k=0}^n a_kx^k$.  If there exist $N\in\mathbb{N}$ and a sequence $\epsilon_n$ of values in $(0, R)$ such that $f_n > 0$ on $x\in(0, \epsilon_n)$ for $n\geq N$, then there exists $\epsilon\in(0,R)$ such that $f > 0$ on $x\in(0,\epsilon)$.

Is this statement true or false?  If true, can it be strengthened?  If false, is a weaker version true?

Edit: the statement is true provided that no subsequence of $\epsilon_n$ has limit 0.  Certainly $f(x)\geq 0$ on $x\in\left(0,\liminf_{n\rightarrow\infty}\epsilon_n\right)$ by definition of convergence.  Moreover, since $f$ is analytic, by the coincidence theorem for analytic functions there can be no sequence $x_n\rightarrow 0$ such that $f(x_n) = 0$; otherwise $f\equiv 0$, contradicting the hypothesis.  So a suitable $\epsilon > 0$ must exist.

I still need to figure out what happens if a subsequence of $\epsilon_n$ approaches $0$.

答:
The statement is true.

If $f(0) > 0$ then this follows by continuity.  Otherwise $f(x)=x^kg(x)$ for some $k\in\mathbb{N}$ and real analytic $g(x)$ such that $g(0)\neq 0$.  Since the lowest-order nonzero derivative of $f$ at $x = 0$ must be strictly positive, it follows that $g(0) > 0$, so that by continuity $f(x) > 0$ for small $x > 0$.


问:
combinatorics equality proof
if $k , n ,q $ are  positive integers and $n \geq k $. I want to prove that:

$ {\begin{bmatrix} n+1 \\ k \end{bmatrix}}_q =   {\begin{bmatrix} n \\ k \end{bmatrix}}_q + q^{n+1-k} {\begin{bmatrix} n \\ k-1 \end{bmatrix}}_q$ .

I think it may done buy using schur theorem.

Title: combinatorics equality proof

Body:
Alternatively, for a strictly algebraic proof:
$$
\begin{align}
\require{cancel}
\def\qbinom(#1,#2){\left[#1\atop #2\right]_q}
\qbinom(n+1,k)
  &=\frac{(q^{n+1}-1)(q^{n}-1)\cdots(q-1)}{(q^k-1)\cdots(q-1)(q^{n-k+1}-1)\cdots (q-1)}
\\\\&=\frac{(q^{n+1}\color{red}{-q^{n-k+1}+q^{n-k+1}}-1)(q^{n}-1)\cdots(q-1)}{(q^k-1)\cdots(q-1)(q^{n-k+1}-1)\cdots (q-1)}
\\\\&=\frac{(q^{n+1}\color{red}{-q^{n-k+1}})(q^{n}-1)\cdots(q-1)}{(q^k-1)(q^{k-1}-1)\cdots(q-1)(q^{n-k+1}-1)\cdots (q-1)}
\\\\&\quad+\frac{\cancel{(\color{red}{q^{n-k+1}}-1)}(q^{n}-1)\cdots(q-1)}{(q^k-1)\cdots(q-1)\cancel{(q^{n-k+1}-1)}(q^{n-k}-1)\cdots (q-1)}
\\\\&=\frac{q^{n+1}-q^{n-k+1}}{q^k-1}\cdot \qbinom(n,k-1)+\qbinom(n,k)
\\&=q^{n+1-k}\cdot \qbinom(n,k-1)+\qbinom(n,k)
\end{align}
$$


问:
what values of b_2 result in solving this system?
Solve the system of equations:
$$ \begin{pmatrix}
0  & 1 & 0 & 3  \\
0  & 2 & 0 & 6  \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\    x_3\\  x_4 \end{pmatrix}
 = \begin{pmatrix} b_1 \\ b_2\end{pmatrix}$$

What value of $b_2$ results in this system having a solution?

I did my row reduction and came up with:

$$\begin{pmatrix}
0 & 1 & 0 & 3 & b_1\\
0 & 0 & 0 & 0 & 2b_1 - b_2\\
\end{pmatrix} $$

therefore, $2b_1=b_2$

Is it really that simple?

Any help is greatly appreciated!

答:
Your answer is correct and your reasoning is good (although I think most people would come up with $b_2-2b_1$ rather than $2b_1-b_2$ in the reduced row echelon form). However, in this particular case, there is a conceptually simpler solution (I emphasize "particular" because in general, your way of solving the problem is preferred). Note that
$$
\underbrace{\begin{pmatrix}0&1&0&3\\ 0&2&0&6\end{pmatrix}}_{A}
\begin{pmatrix}x_1 \\ x_2 \\ x_3\\ x_4\end{pmatrix}
=x_1\begin{pmatrix}0\\ 0\end{pmatrix}
+x_2\begin{pmatrix}1\\ 2\end{pmatrix}
+x_3\begin{pmatrix}0\\ 0\end{pmatrix}
+x_4\begin{pmatrix}3\\ 6\end{pmatrix}
=(x_2+3x_4)\begin{pmatrix}1\\ 2\end{pmatrix}.
$$
Therefore the system of linear equations is solvable if and only if $(b_1,b_2)^T$ is a scalar multiple of $(1,2)^T$, i.e. iff $b_2=2b_1$.

Using the language of linear algebra, $A\mathbf{x}=\mathbf{b}$ is solvable iff $\mathbf{b}$ lies in the column space of $A$. Now the column space of $A$ is the line spanned by $(1,2)^T$. Hence the result.


问:
Domain and set notation
I looked through many threads and I found some different answers to this, but I want to know which one is the most 'proper'. Suppose I want to mark the domain of a function as the set of all real numbers ($\mathbb{R}$) except for all real numbers that satisfy $\frac{\pi}{2} + n\cdot\frac{5}{6}$ where $n$ is an integer. What is the best way to write this in math notation?

答: "Let $D$ be the set of all real number that are not of the form $\frac{\pi}{2} + n\cdot\frac{5}{6}$ for some integer $n$" would be usual. If you mean "how do I write it in some format with curly brackets at either end", then $$\left\{x \in \mathbb{R} : \not\exists n \in \mathbb{Z} : x = \frac{\pi}{2} + n \cdot \frac{5}{6}\right\}$$ or $$\mathbb{R} \setminus\left\{\frac{\pi}{2} + n\cdot\frac{5}{6} : n \in \mathbb{Z}\right\}$$ works, though neither is as clear as the version with more words, to me.

Title: Domain and set notation


问:
For the binomial distribution, why does no unbiased estimator exist for $1/p$?
Suppose that $X \sim \mathrm{Binomial}(n,p)$ for $0 < p < 1$

Why does no unbiased estimator exist for $1/p$?

My approach:

We try to find the structure of $E_p(U(x))$, where $U(x)$ is any estimator of $1/p$.

Now, we will have:

$$\sum{U(x)\binom{n}{x}\theta^x(1-\theta)^{n-x}}<\sum{U(x)\binom{n}{x}}<M(n)<\infty$$

so that the expectation is bounded above. So this is supposed to mean that if $\theta < 1/M(n)$, then the expectation cannot attain $1/\theta$ but I am not sure why the above argument even makes sense and what being bounded means for the expectation.

Thanks!

答:
Suppose $X_1,...,X_n$ iid Bernoulli($\theta$). If we want to estimate $\tau(\theta)=\frac{1}{\theta}$, we then find that no unbiased estimator exists.

We can use MLE of $\tau(\theta)=\frac{1}{\theta}$, which is $\tau(\hat\theta)=\frac{1}{\bar x}$, by invariance principle. But this is biased.

Now I provide the proof to show that no unbiased estimator exists. We can use contradiction.

Suppose W(x) is an unbiased estimator of $\frac{1}{\theta}$. That means whatever estimator we choose (not only MLE), it cannot be unbiased. Let M=supW(x). Recognize that $M<\infty$, since M is from finite sample. Note that our sample x($X_1,...,X_n$) is either 0 or 1. We have $2^n$ possible data sets. It is finite, thus $M<\infty$. (In analysis, we can also say sup is just your maximum due to finiteness.)

Then $E_\theta$ W(x) $\leq M$ for all $\theta$. In other words, M is an upper bound.

By unbiased definition, $E_\theta$ W(x) = $\frac{1}{\theta}. $But it is$ \rightarrow \infty$ as $\theta \rightarrow 0$. Note that it is allowed to let $\theta \rightarrow 0$ because $\theta$ is in (0,1) and we can let $\theta$ be very small. And when $\theta$ is very small, then $ \frac{1}{\theta}\rightarrow \infty$.

But previously, we show that $E_\theta$ W(x) $\leq M$ for all $\theta$, including the case when $\theta$ is very small. Hence, this is contradiction.


答: Define $A=${$x_n|x_n=\frac{pi}{2}+\frac{5n}{6}$} and let $B=R-A$, $B$ is the set you want.

Title: Domain and set notation

Body: I looked through many threads and I found some different answers to this, but I want to know which one is the most 'proper'. Suppose I want to mark the domain of a function as the set of all real numbers ($\mathbb{R}$) except for all real numbers that satisfy $\frac{\pi}{2} + n\cdot\frac{5}{6}$ where $n$ is an integer. What is the best way to write this in math notation?


问:
Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$
As the title insinuates, in my readings I can across this isomorphism $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ and i cannot see why this is the case. Can someone help me see why this is an isomorphism. Here $\mathbb{H}$ is referring to the quaternions.

答: You should clarify certain what field you are tensoring over and how you view the quaternions as a complex vector space (left or right: they are not in the center of the quaternions). See Example 2.6 and Exercise 5 of Section 2 here for a starting point.</p>


答:
$\Bbb{H=C+Cj}$ is a 2-dimensional complex vector space so $End_\Bbb{C}(\Bbb{H})\cong M_2(\Bbb{C})$ as complex algebras.

Then we are willing to check that $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ as complex algebras, where $\Bbb{H\otimes_R C }$ comes with its own complex algebra structure.

For this, send $1\otimes a+i\otimes b+j\otimes c+ij\otimes d$ to $a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$

Title: Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$

Body: As the title insinuates, in my readings I can across this isomorphism $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ and i cannot see why this is the case. Can someone help me see why this is an isomorphism. Here $\mathbb{H}$ is referring to the quaternions.


答:
If $\phi:\mathbb N\times \mathbb N\to N$ is the bijection you have found.

And $\psi:\mathbb Q \to \mathbb N$ is a bijection not found yet.

And $\chi:\mathbb Z \to \mathbb N$ is a bijection not found yet then

$F:\mathbb Q \times \mathbb Z \to N$ via $F(q,n) = \phi(\psi (q), \chi(n))$ will be a bijection by the lemma that compositions of bijections are bijections.

So now all you need to do is find $\psi$ and $\chi$.  Prove $Q$ and $\mathbb Z$ are countable is a very important result so it shouldn't be hard to to google an example.

.....

$\mathbb Z \to N$ is easy.  Map the positive numbers to the evens, and $0$ and the negatives to the odds.

$\chi(n) = \begin{cases}2n& n>0\\-(2n-1)&n\le 0\end{cases}$.

The "picture" demonstration of $\psi$ is a little handwavey but if we consider $\gamma: \mathbb N \to \mathbb Q^{+}$ as.

$\gamma: 1\to \frac 11; \require{cancel} 2\to \frac 21; 3\to \frac 12; 4\to \frac 31;\cancel{\frac 22};5\to \frac 13; 6\to \frac 41; 7\to \frac 32; 8\to \frac 23; 9\frac 14; 10\to \frac 51,\cancel{\frac 42},\cancel{\frac 33},\cancel{\frac 24},11\to \frac 15; 12\to \frac 61; 13\to \frac 52; 14\to \frac 43; 15\to \frac 34;16\to \frac 25; 17\to \frac 16 etc.$

If it not clear what going on:  If $\gamma (n) = \frac ab; \gcd(a,b) = 1$ and $a > 1$ then $\gamma(n+1) = \frac {a-1}{b+1}$ assume $\gcd(a-1,b+1) = 1$.  If not... just skip it and move to the next one $\frac {a-2}{b+2}$ (assuming $a-2> 0$).

If $\gamma (n) = \frac 1b$ then $\gamma(n+1) =\frac {b+1}1$ and keep going. By induction we can do this for even $n$ and $\gamma$ is a bijection.

Then let $\delta: \mathbb Q \to \mathbb Z$ be a bijection by $\delta(q) =\begin{cases}\gamma^{-1}(q)& q > 0\\0& q= 0\\-\gamma^{-1}(|q|)& q < 0\end{cases}$

And finally $\psi: \mathbb Q \to \mathbb Z \to N$ via $\psi(q) = \chi(\delta(q))$.

There are more clever was to find a function to take the place of $\psi$.


问:
Limit distribution of gaussian norm and Paley-Zygmund inequality (solution verification)
Let $X_i \sim \mathcal N(0, I_{d(n)})$ be iid normal random vectors taking values in $\mathbb R^{d(n)}$ where ${d(n)}$ is a sequence (dependent on a positive integer $n$) that grows such that ${d(n)}/n \rightarrow \gamma >0$ as $n\rightarrow \infty$ for some positive constant $\gamma \in (0,1)$. Consider the following quantity:

$$Y_n = \left\|\frac{1}{\sqrt n} \sum_i^n X_i\right \|_2^2$$

Does $Y_n$ have a limiting distribution? If so, what is it?

What I've tried

Rewriting, we have:

$$Y_n = \left\|\sqrt n\,\frac{1}{n} \sum_i^n X_i\right \|_2^2 = \left\|\sqrt n\, \bar X\right \|_2^2$$

where $\bar X \sim \mathcal N\left(0, I_{d(n)}/n\right)$ so that $Y_n$ is clearly chi squared distributed with $d(n)$ degrees of freedom. The chi squared distribution does not have a limiting distribution as the number of degrees of freedom increases to $\infty$. This shows that $Y_n$ does not converge in distribution.

The homework problem suggests to apply the Paley-Zygmund inequality, but I don't understand where that might be relevant. Any hints would be appreciated.

Paley-Zygmund:

$$\mathbb P[X\ge\theta \mathbb EX]\ge (1-\theta)^2\frac{\mathbb E[X]^2}{\mathbb E[X^2]}$$

答:
As pointed out in the comments, the lower bound from the Paley-Zygmund inequality allows us to show that the limit distribution is not tight, and so is not actually a probability measure.

Evaluating the terms in the expression:

$$\mathbb P[Y_n\ge\theta d]\ge (1-\theta)^2\frac{d^2}{2d + d^2} \rightarrow 1/2$$

and so the limit distribution of $Y_n$ is not tight and so does not exist.


问:
Span of the union of two sets - Halmos
If $\mathbb{H}$ and $\mathbb{K}$ are any two subspaces and if $\mathbb{M}$ is the subspace spanned by $\mathbb{H}$ and $\mathbb{K}$ together, them $\mathbb{M}$ is the same as the set of all vectors of the form $x+y$, with $x \in \mathbb{H}$ and $y \in \mathbb{K}$.

I understand how the span is defined; I just don't how a vector of $\mathbb{M}$ can be simply writen as $x+y$.

答:
If $u \in \mathbb{M}$, then $u = \sum_i \alpha_i x_i$; since $x_i \in \mathbb{H} \cup \mathbb{K}$, we can separate the vectors in two groups, respectively.
Separating, we have $u = \sum_i \alpha_i x_i = \sum_{x_i \in \mathbb{H}} \alpha_i x_i + \sum_{x_i \in \mathbb{K}} \alpha_i x_i$; but, since $\mathbb{H}$ and $\mathbb{K}$ are subspaces, these two sums are simply two vector from those same subspaces. In other words, $u = \sum_i \alpha_i x_i = \sum_{x_i \in \mathbb{H}} \alpha_i x_i + \sum_{x_i \in \mathbb{K}} \alpha_i x_i = x + y$, where $x \in \mathbb{H}$ and $y \in \mathbb{K}$.

I think it's right now...

Title: Span of the union of two sets - Halmos


问:
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?
Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?

I thought it would be 12 this as per pemdas rule:

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$

Wanted to confirm the right answer from you guys. Thanks for your help.

答:
If you type $1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$ into most pocket calculators you will get the answer $1$. This is because the calculator applies the next operation to the result of the previous one, effectively computing$$(1+1+1+1+1+1+1+1+1+1+1+1)⋅0+1=0+1=1$$

When using a calculator you have to respect the logic of the machine. When writing mathematics you need to be aware of the standard conventions for brackets, which give the answer $12$ in this case. I suspect that the confusion arises because the two are not the same. Understanding why will help you to make better use of a calculator and to write correct mathematical equations.


问:
Mahalanobis distance for vectors from different distributions
Given that I want to calculate the distance of a vector x (say from the blue distribution) from the centroid of a different distributions than x's - say centroid of the red vectors:

I want to use the Mahalanobis distance:

$M^2 = (x-\mu)^T \Sigma^{-1}(x-\mu)$

Is the covariance matrix $\Sigma^{-1}$ than calculated only from the distribution of the here red vectors? If so, how is the distribution of the blue datapoints considered?

答:
It depends on what your model for this data is. Judging from the image you provided, a Gaussian Mixture Model is applied, that is we have

$$x\mid A \sim \mathcal N(\mu_A, \Sigma_A) \qquad x\mid B\sim\mathcal N(\mu_b, \Sigma_B)$$

and in total $p(x) = p(x\mid A)p(A) + p(x\mid B)p(B)$. One commonly writes $\pi_A$ and $\pi_B$ for $p(A)$ and $p(B)$. You can now either

Consider the case when both populations have the same variance $\Sigma =\Sigma_A=\Sigma_B$. This is known as Linear Discriminant Analysis (LDA). In you can either

estimate $\Sigma$ by averaging the population covariances, weighted by frequency: $\hat\Sigma=\hat\pi_A\hat\Sigma_A + \hat\pi_B\hat\Sigma_B$

estimate $\Sigma$ using the pooled variance technique

In this case, the Mahalabonis distances from a point to distribution $B$ is $(x-\mu_B)^T\Sigma^{-1}(x-\mu_B)$. Here, the blue distribution would have an effect on the distance, since $\Sigma$ is estimated using data from both populations.

Assume both populations have distinct covariances $\Sigma_A\neq\Sigma_B$. This is known as Quadratic Discriminant Analysis (QDA)

In this case, the Mahalabonis distances from a point to distribution $B$ is $(x-\mu_B)^T\Sigma_B^{-1}(x-\mu_B)$. Here, the blue distribution would have no effect on the distance, since $\Sigma_B$ is estimated using only data from the red population.


问:
Are the Hermite polynomials independent?
I believe they are not independent as obviously,
$$X := He_1(x) = x \; \text{and} \; Y:=He_2(x) = x^2-1$$ then
$$P(Y\in A | X \in A) \neq P(Y \in A) \; \; \text{for some set } A \subset \mathbb{R}.$$

This is a counter-example (assuming the x's are distributed with Gaussian measure.)

But this seems too trivial, and so I suspect I may be wrong. Could anyone please verify my proof?

答: If $X$ is a random variable with a continuous distribution and $f$ is any non-constant polynomial, $X$ and $f(X)$ are never independent.  Indeed, the conditional distribution of $f(X)$ given $X=x$ is the unit mass at $f(x)$, and this depends on $x$.


问:
Suppose $f(x) = x^5 + 2x^4 − x^3 + 3x^2 + 5x + 1$. Find a polynomial $g$ of degree $2$ or less such that $f(x) \equiv g(x) \pmod 3$.
Suppose $f(x) = x^5 + 2x^4 − x^3 + 3x^2 + 5x + 1$. Find a polynomial $g$ of degree $2$ or less such that $f(x) \equiv g(x) \pmod 3$.

答: Just apply Fermat's little theorem, $x^3 \equiv x \pmod 3$. Also any term with a coefficient that's a multiple of $3$ will vanish modulo $3$.


问:
Divergence in terms of Levi-Civita connection
The divergence of a vector field $X$ on a manifold $M$ is defined usually as the function $\text{Div}(.)$ such that $(\text{Div} X) \;\mu =L_X \mu$ for $\mu$ a volume form.

I know that there is also an alternative expression for $\text{Div} \; X$ in terms of the covariant derivative:

$\text{div} \; X= \text{trace}(\nabla X)= \nabla^aX_a.$

Can anyone explain to me how to prove this equivalence?

答:
Though I probably would not have found this without Travis proof above, there is actually a way to show this without having to recourse to coordinate calculations. As far as I can tell, this is the most elegant way to show this:

It is correct that the covariant derivative has to be torsion-free for the calculation to work out like this:
$$[X,Y] = \nabla_XY - \nabla_Y X$$

Using an argument I found in O'Neill's book "Semi-Riemannian Geometry", the first step in the proof is to realize that $\mu$ is a top degree form and so is $\mathcal{L}_X \mu$. Therefore, there has to exist a smooth (possibly zero) function $f$ such that
$$\mathcal{L}_X \mu = f \, \mu$$
This is true regardless of the torsion of $\nabla$.
Introducing a local frame field $(Y_1, \dots, Y_n)$ with
$$\mu (Y_1, \dots, Y_n) =1 \, ,$$
we locally find
\begin{align}
f &= \left( \mathcal{L}_X \mu \right) (Y_1, \dots, Y_n) \\
&=  \mathcal{L}_X \left( \mu (Y_1, \dots, Y_n) \right) -
  \mu (\mathcal{L}_X Y_1, Y_2, \dots, Y_n) - \dots -
\mu (Y_1, \dots, Y_{n-1}, \mathcal{L}_X Y_n)
\\
&= \nabla_X \left( \mu (Y_1, \dots, Y_n) \right) -
  \mu ([X, Y_1], Y_2, \dots, Y_n) - \dots -
\mu (Y_1, \dots, Y_{n-1}, [X,Y_n])
\\
&= \left(\nabla_X  \mu \right) (Y_1, \dots, Y_n) +
  \mu (\nabla_{Y_1} X, Y_2, \dots, Y_n) + \dots +
\mu (Y_1, \dots, Y_{n-1}, \nabla_{Y_n}X)
\end{align}
For the last equality, one uses the 'product rule' for $\nabla_X$ and its torsion-freeness. Since $\mu$ is parallel, the first summand vanishes. For the other summands, only the component with respect to the $i$th 'missing' $Y_i$ matters. Thus indeed
$$f = \sum_{i=1}^n \left( \nabla_{Y_i} X \right)^i = \operatorname{tr} \nabla X \, ,$$
which clearly has to hold everywhere.


问:
If $X$ is any convex set in $\mathbb{R}^n$ ,then show that $\pi_1(X,x_0)=0$
If $X$ is any convex set in $\mathbb{R}^n$ ,then show that  $\pi_1(X,x_0)=0$

My attempt : By using the theorem any convex subspace  $X$ of $\mathbb{R}^n$ is contractible

There is  a  continuous map $F : X \times [0,1] \to X$ such that $F(\cdot,0) = f$ and $F(\cdot,1) = c$.

This implies $[f] =[c] $

since $[f] \in \pi_1(X,x_0) \implies [c] \in  \pi_1(X,x_0) $

Therefore  $ \pi_1(X,x_0) =[c]=0$

Is my proof is correct or not ?

答:
Let $X$ be a convex subset of $\mathbb{R}^n$. Fix $x_0\in X$ and let $I=[0,1]$.  Let
$$
F:I\times X\rightarrow X
$$
be defined by $F(s,x)=(1-s)x+sx_0$ (we can do this because $X$ is convex).  Then $F(0,\cdot )=id_X$ and $F(1,\cdot)=x_0$.  What we've shown so far is that a convex subset of $\mathbb{R}^n$ is a contractible space.

Now let $c: I\rightarrow X$ be a (continuous) curve such that $c(0)=c(1)=x_0$.  Let $G:I\times I\rightarrow X$ be defined by
$$
G(t,s):=F(s,c(t)).
$$
Then $G(t,0)=c(t)$ and $G(t,1)=x_0$.  Hence, $c(t)$ is homotopic to the constant curve.  Since $c(t)$ was arbitrary, we have $\pi_1(X,x_0)=\{[x_0]\}$, i.e., its trivial.

In proving that $\pi_1(X,x_0)$ is trivial, we only made use of the fact that $X$ is contractible.  Hence, we actually proved something more general: if $Y$ is any contractible space, then $\pi_1(Y,y_0)$ is trivial.


问:
Looking for a hypergeometric 2F1 identity
I'm trying to express
$$_2F_1\left(\frac{1}{2} , 1;\,\frac{3}{2}+m;\,z\right), \quad m\in\mathbb{Z}\quad\text{and}\quad m\geq0$$
in terms of $\tanh^{-1}(\sqrt{z})$ and $\sqrt{z}$, basically generalizing these results.

Any ideas?

Body:
Using Maple, I am getting
$$
{}_2F_1\left(\frac12,1;\frac32+m;z\right) = A_m(z)\operatorname{atanh}\sqrt{z}+B_m(z)
$$
where
$$
A_m(z) = \frac{2(z-1)^m\Gamma(m+\frac32)}{z^{m+1/2}\sqrt{\pi}\;\Gamma(m+1)}
$$
and
$$
B_m(z) = \frac{\Gamma(m+\frac32)}{\Gamma(m+1)}\;
\sum_{k=1}^m\frac{(z-1)^{m-k}\Gamma(k)}{z^{m+1-k}\Gamma(k+\frac12)}
$$

For example,  $$
A_{10}(z) = {\frac {969969\, \left( z-1 \right) ^{10}}{262144\,{z}^{21/2}}},
\\
B_{10}(z) = \frac{1}{{3932160\,{z}^{10}}}\; \big(68025825\,{z}^{9}-382331775\,{z}^{8}+1168982220{z}^{7}-2255541300\,{z}^{6}+2918514950{z}^{5}-2585198330{z}^{4}+1554721740\,{z}^{3}-609140532{z}^{2}+140645505z-14549535\big)
$$