问:
Number of phone locking patterns
LG android cell phones have locking screens with $9$ points to be traced in any pre-specified fashion (drawing pattern) so as to join $\geq 4$ points without including any points more than once.

Quoting from a Quora post...

You cannot go over an unlit dot without lighting it. For example, the
pattern [0 2 1 4] is illegal, because moving your finger between 0 and
2 will light 1.

Once a dot is lit, you can use it to reach another
unlit dot. For example, both [0 4 3 5] and [0 4 5 3] are legal.

The numbers are assigned as:

$$\begin{matrix}0&1&2\\3&4&5\\6&7&8\end{matrix}$$

Right off the bat, the maximum options available depend on the position within the grid:

The question is two-fold:

How many patterns are possible, given that not all points present the same number of choices for the next move, as well as a minimum ($4$) and maximum ($9$) number of points included in the pattern? And is there a way of obtaining a closed-form calculation as opposed to a computer simulation as in this Quora post?

Does the "pattern" inherent to the drawing on a rigid structure makes some patterns more likely than others (e.g. a letter "C")?

答:
Use concept of graph theory(Depth first search) where every dot on the lock screen is a node.Assume every dot in the lock screen is connected with all the the other dots. So there are total 9C2 =36 edges.

Now suppose you are at any dot on the lock screen.First we mark the dot as visited. From this dot you can only jump to those dots which have the following properties:

1.There are no unvisited dots on the straight line joining the two dots.

2.The dot we are gonna jump to must be unvisited

Proceeding same way; after searching through all the possible patterns from this dot when we leave the dot (i.e we move to parent dot or node), we mark the dot as unvisited.

Thus we visit through all the possible patterns.
then fianally we will get total number of pattern is 389112

Title: Number of phone locking patterns


问:
Solve equation $\cos (a)\cos(b)\cos(a+b) = -\frac18$ over $0 < a,b < \frac\pi2$
I have tried this question as follows:

$$4[2(\cos(a)\cos(b))\cos(a+b)] = -1$$

Then converted the $2\cos(a)cos(b)$ in $\cos(a+b) + \cos(a-b)$.

And so on but this got messy and big.

Another approach that I tried was that as $a$ and $b$ lie in the first quadrant then their cosine is positive this implies $\cos(a+b)$ is negative.

And without losing generality we can say each term be equal to 1/2 which implies

$a = b = π/3$.

I googled it but didn't find any solution.

Wolfram alpha after converting into the same thing as I have converted used complex numbers which was neat, but my question is:  can we solve this with a short method (using trigonometry)?

答:
Hint:

$$\cos a \cos b \cos (a+b)=\cos^2 a\cos^2 b\cdot(1-\tan a \tan b)=-\frac18$$

hence if $AB=\lambda<0 \iff A>0, B<0$ or $A<0, B>0$,

$$1-\tan a \tan b<0\iff \tan a \tan b>1$$ being $\cos^2 a\cos^2 b>0$ in $0 < a,b < \pi/2$ and $\cos a\cos b\neq 0$, $0 < a,b < \pi/2$.

ContentLicense: CC BY-SA 4.0


问:
Expected steps to solve a 3x3 jigsaw puzzle
Consider a 3x3 jigsaw puzzle, i.e., with 9 pieces. The first piece is randomly drawn from a bag and placed at its correct location. Subsequent pieces are drawn one at a time from the bag and they are placed only if they connect with any piece already placed. Pieces are returned to the bag if they do not connect with any piece already placed and another piece is drawn.  Including the first draw, what is the expected number of draws required to finish the puzzle?

Edit: Two pieces connect if they share an edge (as in a jigsaw puzzle).

答:
HINTS (updated for the Jigsaw puzzle case)

If you go brute force, you have 3 cases to consider:

First piece is at the center (weight $1/9$).

First piece is at the corner (weight $4/9$).

First piece is in the middle of a side (weight $4/9$).

In each case, the number of draws until you find a match is a geometric random variable (albeit the parameter will depend on the case), of which is easy to find the expected value. Then break down each of the cases based on what you get and continue on.

For example for case (1), probability of connecting the next piece is $p = 4/8=1/2$. Hence, we expect a connection in $2$ steps, which will always be the middle-of-the-side piece. Now probability of connecting is $5/7$, you expect $7/5 = 1.4$ steps and on connection you have 2 cases: (a) a corner piece (with weight $2/5$) or (b) a side piece (with weight $3/5$).

Can you finish?


问:
Conditional expectation property prove
Prove that if $E(Y|X) = X$ and $E(Y^2|X) = X^2$ than $Y$ and $X$ coincide everywhere
image

Title: Conditional expectation property prove

Body:
Consider
$$E((Y-X)^2 | X) = E(Y^2 - 2 YX + X^2|X) = E(Y^2|X) - 2 X E(Y|X) + X^2 = X^2 - 2X \cdot X + X^2 = 0,$$
thus $E(Y-X)^2  =E ( E((Y-X)^2|X)) =0 $. It follows that $Y=X$ a.s.


问:
If $x>0$ is a real number, show that $\lim_{n\rightarrow{\infty}}x^{1/n}=1$.
Suppose $x>0$ is a real number. I have to prove that $\lim_{n\rightarrow{\infty}}x^{1/n}=1$, and I'm working in the context of sequences, not continuous functions. Here's what I've done so far: there are two cases, $x\geq{1}$ and $0<x<1$. In the first case, it is possible to show that $(x^{1/n})_{n=1}^{\infty}$ is a decreasing sequence, and moreover is bounded from below by $1$, so we know it converges to a real number and furthermore, that: $$\lim_{n\rightarrow{\infty}}x^{1/n}=\inf(x^{1/n})_{n=1}^{\infty}$$
The problem then becomes to verify that the infimum of the sequence is indeed $1$. We already know that $1$ is a lower bound, so the task becomes to show that $1$ is the greatest lower bound. I have tried to employ a proof by contradiction of assuming we could find some lower bound $L$ such that $L>1$, but I'm not allowed to use logarithms so I'm stuck at this point (a similar argument to the above would establish for the case $0<x<1$ that $\lim_{n\rightarrow{\infty}}x^{1/n}=\sup(x^{1/n})_{n=1}^{\infty}$, etc, but I'm stuck for the same reason there). Any suggestions on how to move forward would be appreciated (in case you're curious, this is Exercise 6.5.3 in Tao's Analysis 1).
Note there are some duplicate questions, but all of the methods which were used to prove them are "too advanced" for my purposes. It needs be as elementary as possible, appealing to the most basic properties of suprema, infima, and possibly limit superior and limit inferior, etc.

答:
Suppose $x^{1/n}$ tends to some $L>1$. Then $L^2>L$, so there exists $N$ such that $x^{1/n}<L^2$ whenever $n\ge N$. But $x^\frac{1}{2n}=\sqrt{x^{1/n}}$, which is less than $L$ if $n\ge N$. This contradicts the assumption that $x^{1/n}$ tends monotonically to $L$ from above.

And for the case $0<x<1$, just apply the above to $1/x$.


问:
Orthogonal Matrix in Index notation
The following is an excerpt from Prof David Tong's notes on Classical Dynamics. This is in the section on Kinematics:

I am unable to understand how the equation circled in red, is the same as the condition for an orthogonal matrix given after 'in other words.' I may be missing something obvious, but I'm not sure what it is. What I have tried is the following:

$$R_{ac}R_{bc} = \delta_{ab} \implies R_{ac}R_{cb}^T = \delta_{ab} \implies RR^T = I$$

But how does this guarantee that $R^TR = I$ ?

Further, for two matrices $A_{m\times n}$ and $B_{n\times l}$ the product in index notation is given by:

$$(AB)_{ml} = \sum_n A_{mn}B_{nl}$$

For a general square matrix M, the condition on it being orthogonal is that:

$$M^TM = MM^T = I$$
If I try to write the first condition in index notation I seem to get:

$$(M^TM)_{jj} = \sum_iM_{ji}M_{ij}$$

But how do I get the non diagonal elements of this product at all? Do I change the index on one of the matrices?

Throughout the rest of this section wherever I expect a certain index ordering, I find that he has written down the opposite (for example here, I think it should be $R_{ac}R_{cb} = \delta_{ab}$) so I think there is something I don't understand with the notation, or something more fundamental that I am missing.

答:
You essentially have the right idea in the first and second equations you wrote down. I think you're just confused about how to apply index notation.

For what Tong wrote down (exclusively in index notation) we have

$$\delta_{ab} = R_{ac} R_{bc} $$

It is easier to understand this formula if we write it in expanded summation notation:

$$\delta_{ab} = \sum_{c} R_{ac} R_{bc}$$

Then we see that we have

$$\delta_{ab} = \sum_{c} R_{ac} R_{bc} = \sum_{c} R_{ac} R_{cb}^{T},$$

which by your third equation implies $\delta_{ab} = (RR^T)_{ab}$. Since $\delta_{ab}$ defines the components of the identity matrix, we have $I = RR^{T}$.

Title: Orthogonal Matrix in Index notation


问:
Let $A \in M_n(\mathbb{C})$, Find Canonical Jordan form of $A^2$ in correlation to Jordan form of $A$
No given Information since this question relies on typical Complex matrix.

I know that $J_n(x)= xI+J_n(0)$ but how does it help me? I have no idea what to do.

Thanks in advance.

答:
As you have already determined, it suffices to separately consider the Jordan blocks of $A$. For $x \neq 0$, $J_n(x)^2$ is indeed similar to $J_n(x^2)$.

The question remains: what happens when $x = 0$? We know that the only eigenvalue of $J_n(0)^2$ is zero, but it is not clear what the size of the blocks should be. I claim the following: for all integers $k \geq 1$, $J_{2k}(0)^2$ is similar to $J_{k}(0) \oplus J_k(0)$ and $J_{2k+1}(0)^2$ is similar to $J_{k}(0) \oplus J_{k+1}(0)$, where $\oplus$ denotes a direct sum.

One way to prove that this holds is by considering the Weyr characteristic (cf. also Matrix Analysis by Horn and Johnson). In other words, it suffices to note that $\operatorname{rank}((J_{n}(0)^2)^m) = n - 2m$ for all $m < n/2$ and $0$ for $m \geq n/2$. However, I find that a more satisfying approach is to construct a similarity, i.e. to find a matrix $P$ for which $P^{-1}(J_{2k+1}(0)^2)P = J_k(0) \oplus J_{k+1}(0)$.

I will only consider the case of $n = 2\cdot 2 + 1 = 5$, but the general construction is completely analogous. Note that
$$
J_{5}(0)^2 = \pmatrix{0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0}.
$$
Let $e_1,\dots,e_5$ denote the standard basis of $\Bbb R^5$. We note that the vectors $e_1,e_3,e_5$ for a Jordan chain.  That is, we have
$$
Je_5 = e_3, \quad Je_3 = e_1, \quad Je_1 = 0.
$$
Similarly, the vectors $e_2,e_4$ form a Jordan chain since we have
$$
Je_4 = e_2, \quad Je_2 = 0.
$$
With that, we can conclude if we take $P$ to be the (permutation) matrix with columns $e_2,e_4,e_1,e_3,e_5$, then we have
$$
P^{-1}J_5(0)^2P = \pmatrix{0&1&0&0&0\\0&0&0&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0}
 = J_2(0) \oplus J_3(0).
$$


问:
Quadratic variation of Ito integral : Relationship between Ito Integral and Brownian Motion
This is theorem 4.2.3 from Shreve's Stochastic Calculus for finance II.

Statement of the theorem:

The quadratic variation accumulated up to time $t$ by the Ito integral is:
$$ [ I,I](t) = \int_{0}^{t} \Delta^{2}(u) du.$$

The proof revolves around computing quadratic variation accumulated by the Ito integral on one of the subintervals $[t_j, t_{j+1}]$ on which $\Delta(u)$ is constant. They choose the partition points:

$$t_{j} = s_{0} < s_{1} < \ldots < s_{m} = t_{j+1} $$

and compute

$$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2 $$

I can't figure out how we got the aforementioned equality involving Ito integral and Brownian Motion. Any help appreciated.

Here we have the following definitions: if $t_{k} \leq t \leq t_{k+1}$ then
$$I(t) = \sum_{j=0}^{k-1}\Delta(t_{j})[W(t_{j+1} - W(t_j)] + \Delta(t_{k})[W(t) - W(t_{k})]$$
The process $I(t)$ is the Ito integral of the simple process $\Delta(t)$ that is written as
$I(t) = \int_{0}^{t}\Delta(u)dW(u).$

Suppose $\Delta(t)$ is an adapted stochastic process i.e. it is $\mathcal{F}(t)$ measurable for each $t \geq 0.$ Assume further that $\Delta(t)$ is constant in $t$ on each subinterval $[t_{j}, t_{j+1}).$ Then $\Delta(t)$ is a simple process.

Here u is just the dummy variable and W(t) is the Brownian Motion.

答:
Following the definitions, we have
$$I(s_{i+1})=I(t_j)+ \int_{t_j}^{s_{i+1}} \Delta(t_j)dW(u) = I(t_j)+\Delta(t_j)(W(s_{i+1})-W(t_j)).$$
Similarly
$$I(s_i)=I(t_j)+\Delta(t_j)(W(s_i)-W(t_j)).$$
Thus
$$I(s_{i+1})-I(s_i)=\Delta(t_j)(W(s_{i+1})-W(s_i))$$
and therefore
$$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2. $$

EDIT: In the first expression I am using the linearity property of the Ito integral. Remember that
$$t_j = s_0 < s_1 < ... < s_m = t_{j+1},$$
so
\begin{align}
I(s_{i+1})=\int_0^{s_{i+1}} \Delta(u)dW(u)&=\int_0^{t_j} \Delta(u)dW(u)+\int_{t_j}^{s_{i+1}} \Delta(u)dW(u) \\
&=I(t_j)+\int_{t_j}^{s_{i+1}} \Delta(u)dW(u)
\end{align}


问:
Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$
Let $N$ be a Poisson random variable with the mean parameter $m$.
We are interested in finding the following limit
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right].
\end{align}

Things that I tried:
First, I found an upper bound by using Jensen's inequality
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]\le  \lim_{m \to \infty} m  \log\left( \frac{ E \left[ N  \right]+\frac{1}{2}}{m+1} \right)= \lim_{m \to \infty} m  \log\left( \frac{ m+\frac{1}{2}}{m+1} \right)=-\frac{1}{2}.
\end{align}

For the lower bound I tried to use the CLT argument from a related question in here:
\begin{align}
m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] P[ N \ge \frac{m}{k} ]+ m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N < \frac{m}{k} \right] P[ N < \frac{m}{k} ]
\end{align}
for some $k>0$.  Via the CLT it can be shown that $P[ N < \frac{m}{k} ] \to 0$ for all $0<k <1$, and we have that
\begin{align}
\lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] .
\end{align}
Now if we use the lower bound at this point, we get
\begin{align*}
 \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right]  \ge   \lim_{m\to \infty}m  \log\left( \frac{ \frac{m}{k}+\frac{1}{2}}{m+1} \right) =-\infty,
\end{align*}
for all $k \in (0,1)$.

答:
This is an respectable answer, what a median in a Poisson distribution is. Your answer targets large, infinite such medians. This gives the formulae: Poisson distribution. This definition also relates the E symbol to the distribution. Expected value makes clear which operation is expected.

$$E[X]=\int_{0}^{\infty} x P(x)dx$$

$X$ is any function of the distributed parameter.

$$E[\log(\frac{N+\frac{1}{2}}{+1})]=\int_{0}^{\infty} \log(\frac{N+\frac{1}{2}}{+1}) P(N)dN$$

Now You named Poisson random variable so,

$$P(\lambda,N)=\frac{\lambda^N\exp(-\lambda)}{N!}$$

Median and $\lambda$ are : $m\approx\lambda-\frac{1}{3}-\frac{0.05}{\lambda}$. Since Your are interested in large medians only:

$$P(m,N)\approx\frac{(m+\frac{1}{3})^N\exp(-(m+\frac{1}{3}))}{N!}$$

For sufficiently high values the Poisson distribution is approximated by the standard distribution. This can only be evaluated numerically or literature.

A plot of the function over m looks like this:

$$\lim_{m->\infty} m E[\log(\frac{N+\frac{1}{2}}{+1})]=0$$

This might give some safety if the median, mean or parameter $\lambda$ has to be estimated. Since $N$ is natural a variation of $\frac{1}{2}$ is rather common as uncertainty. $\lambda$ is real and the variation in first order is estimated $1$. We approximate the entropies roughly and integration the difference between both over the "true" probability distribution with the mean as parameter. In first order the coefficient vanishes and show how good the approximation really is, if alone the Poisson distribution is selected.

This does not compare to other distribution such as the standard distribution. $\frac{1}{2}$ and $1$ might too named information uncertainty. The graph reflects teh general property of the Poisson distribution that goes over into the standard distribution. Mind this plot might have numerical errors. This is crude because it replaces the median for the parameter already $\lambda$. Use Poisson distribution for a good completion of formulae used.


问:
Reference for Poisson point process
I'm looking for a good reference book for PPP (Poisson point process) and the stochastic integration theory with regard to PPP.

I have tried "Stochastic Differential Equations and Diffusion Process" written by Shinzo Watanabe
&
the lecture notes for Stochastic Analysis written by Prof Andreas Eberle.

But I think they are both not strictly enough to introduce PPP in a mathematical way. (Indeed some conclusions seemed impossible under the former assumptions given in this two books!)

So I wonder if there is any notes or books which can provide a "strict" introduction of PPP and its stochastic integration theory.

答:
Poisson Processes by J.F.C. Kingman (Oxford University Press, 1993) is an excellent introduction.

A deeper dive can be found in  Lectures on the Poisson Process by G. Last and M. Penrose (Cambridge University Press, 2018).

Title: Reference for Poisson point process


问:
How should sequent calculus be extended if we add equality to the first order logic?
Are sequent calculus needed to be extended if we add equality to the first order logic? Is it not enough to say that we just have one more predicate that has all the properties of equality:

$\forall X . \forall Y . [E(X, Y) \implies E(Y, X)]$

$\forall X . \forall Y . \forall Z . [(E(X, Y) \land E(Y, Z)) \implies E(X, Z)]$

If it is not enough, how exactly sequent calculus has to be extended? Or, in other words, what inference rules have to be added?

By the "rules of inference" I mean rules like 12 rules given here in the green box.

答:
No, the rules you've written are not enough.

First, note that what you've written down - symmetry and transitivity - aren't enough to guarantee that $=$ ever holds. You also need reflexivity: that we're allowed to infer, from no hypotheses at all, $t_1=t_1$ for any term $t_1$.

More precisely, reflexivity is the rule which allows us to infer the sequent $\vdash t_1=t_1$ for any term $t_1$ from no hypotheses whatsoever. On that note, your own rules should be rephrased in sequent calculus language, e.g. from the sequents $\Gamma\vdash t_1=t_2$ and $\Gamma\vdash t_2=t_3$ we can infer the sequent $\Gamma\vdash t_2=t_3$.

More subtly, we need the substitution property, that from $t_1=t_2$ and $\varphi$ we can infer any formula $\psi$ gotten from $\varphi$ by replacing some instances of $t_1$ by $t_2$. After all, equality isn't just any old equivalence relation, it's the finest possible equivalence relation. And this needs to be reflected in the logical rules which govern it.

Again more precisely: whenever $t_1,t_2$ are terms, $\varphi,\psi$ are formulas, and $\psi$ is gotten from $\varphi$ by replacing some instances of $t_1$ with $t_2$, we have the rule which lets us infer the sequent $\Gamma\vdash\psi$ from the sequents $\Gamma\vdash\varphi$ and $\Gamma\vdash t_1=t_2$.

That this is in fact all we need is a consequence of the relevant completeness theorem (soundness being trivial), and working through the details is a good exercise. Note that the usual (Henkin) proof of (the contrapositive of) completeness should go through unchanged for all reasonable systems: given a consistent theory $T$, we whip up an extension $T'$ in a possibly larger language which is consistent, complete and has the witness property; we then form the term structure for $T'$ and show that it in fact is a model of $T'$ - at which point its reduct to the language of $T$ is a model of $T$, and so $T$ is satisfiable.

Finally, note as an aside that reflexivity and substitution on their own give symmetry and transitivity:

For symmetry, given $t_1=t_2$ we use reflexivity to introduce $t_1=t_1$. Using $t_1=t_2$ as the substituter and $t_1=t_1$ as the substitutee, we get $t_2=t_1$ by changing the first occurrence of $t_1$ to $t_2$ in the substitutee.

For transitivity, given $t_1=t_2$ and $t_2=t_3$ we use substitution with $t_2=t_3$ as the substituter and $t_1=t_2$ as the substitutee to get $t_1=t_3$.

So reflexivity and substitution are in fact the only equality rules we really need.


问:
Generate a frequency distribution function with given Mean, Median, Std Dev, 1st and 3rd Quartile
I want to generate plot of a function whose Mean, Median, Std Dev, 1st and 3rd Quartile are given.
How do I generate one?

I tried Gaussian Distribution but there Mean = Median and we don't have control over quartiles.

Edit 1:-
I'm helping my 10th grade cousin with statistics. In one of the question I need to compare the skewness of functions with two different sets of variables (Mean, Median, Std Dev, 1st and 3rd Quartile). I want to show him how the frequency distribution will look like.

I did some research and found some skewed function here https://www.desmos.com/calculator/ptnf1tbdgf

答:
Normal populations. The two parameters of a normal population are the mean $\mu$ and the standard deviation $\sigma$. As you say, the mean and median are the same. Also, once $\sigma$ is specified, the interquartile range (distance between quartiles) is determined.

For example, if a normal distribution has $\mu = 100, \sigma = 15.$ then its median is $100$ and its interquartile range is about 20.2347 (computation in R):

```<code>diff(qnorm(c(.25,.75), 100, 15))
[1] 20.23469
</code>```

Normal samples. Of course, samples do not exactly imitate the population from which they are drawn, but you need to keep the population characteristics in mind.

For example, if I have a large sample, say of size $n = 10\,000$ from the population $\mathsf{Norm}(100, 15),$ then we can expect its mean and median both to be near 100, its standard deviation to be near 15 and its IQR to be near 20.2.

```<code>set.seed(2020)
x = rnorm(10^4, 100, 16)
mean(x);  median(x)
[1] 99.83717         # sample mean
[1] 99.48441         # sample median
sd(x);  diff(quantile(x, c(.25,.75)))
[1] 15.9507          # sample SD
      75%
21.43506             # sample IQR
</code>```

Artificial samples to specification. If you want to contrive a normal sample of size $n=10$
to have $\bar X = 100$ and $S_X = 15$ (both almost exactly), then you can start with a sample of size ten from a standard normal distribution, and rescale it appropriately:

```<code>set.seed(1223)
z1 = rnorm(10)
</code>```

First, rescale so that the sample mean is $0$ and the sample SD is $1.$

```<code>sd(z1)
[1] 0.7214513
z2 = z1/sd(z1);  sd(z2)
[1] 1
mean(z2)
[1] -0.2631234
z = z2 - mean(z2)
mean(z); sd(z)
[1] 2.218549e-17  # essentially 0
[1] 1
</code>```

Then, rescale to make sample x with $\bar X = 100$
and $S_X = 15.$

```<code>x = 15*z + 100
mean(x);  sd(x)
[1] 100
[1] 15
</code>```

But now you're pretty much stuck with whatever the median and IQR
happen to be; in this particular example, about 99.38 and 23.34, respectively.

```<code>median(x);  diff(quantile(x, c(.25,.75)))
[1] 99.38192
     75%
23.33596
</code>```

[You might try to make adjustments to individual observations in the sample x to change the median and IQR, without changing $\bar X$ and $S_X,$ but that would be a tedious project, and strictly speaking, the resulting sample would no longer be normal.]

Body:
Normal populations. The two parameters of a normal population are the mean $\mu$ and the standard deviation $\sigma$. As you say, the mean and median are the same. Also, once $\sigma$ is specified, the interquartile range (distance between quartiles) is determined.

For example, if a normal distribution has $\mu = 100, \sigma = 15.$ then its median is $100$ and its interquartile range is about 20.2347 (computation in R):

diff(qnorm(c(.25,.75), 100, 15))
[1] 20.23469

Normal samples. Of course, samples do not exactly imitate the population from which they are drawn, but you need to keep the population characteristics in mind.

For example, if I have a large sample, say of size $n = 10\,000$ from the population $\mathsf{Norm}(100, 15),$ then we can expect its mean and median both to be near 100, its standard deviation to be near 15 and its IQR to be near 20.2.

set.seed(2020)
x = rnorm(10^4, 100, 16)
mean(x);  median(x)
[1] 99.83717         # sample mean
[1] 99.48441         # sample median
sd(x);  diff(quantile(x, c(.25,.75)))
[1] 15.9507          # sample SD
      75%
21.43506             # sample IQR

Artificial samples to specification. If you want to contrive a normal sample of size $n=10$
to have $\bar X = 100$ and $S_X = 15$ (both almost exactly), then you can start with a sample of size ten from a standard normal distribution, and rescale it appropriately:

set.seed(1223)
z1 = rnorm(10)

First, rescale so that the sample mean is $0$ and the sample SD is $1.$

sd(z1)
[1] 0.7214513
z2 = z1/sd(z1);  sd(z2)
[1] 1
mean(z2)
[1] -0.2631234
z = z2 - mean(z2)
mean(z); sd(z)
[1] 2.218549e-17  # essentially 0
[1] 1

Then, rescale to make sample x with $\bar X = 100$
and $S_X = 15.$

x = 15*z + 100
mean(x);  sd(x)
[1] 100
[1] 15

But now you're pretty much stuck with whatever the median and IQR
happen to be; in this particular example, about 99.38 and 23.34, respectively.

median(x);  diff(quantile(x, c(.25,.75)))
[1] 99.38192
     75%
23.33596

[You might try to make adjustments to individual observations in the sample x to change the median and IQR, without changing $\bar X$ and $S_X,$ but that would be a tedious project, and strictly speaking, the resulting sample would no longer be normal.]


问:
Arguing properties of trigonometric function with floor
$$\ \lim_{x\to 0}\lfloor\tan x\rfloor\cdot \cos x$$

Okay so when I first tried to solve it, I stated at the beginning of the proof

"Note that when

$-\frac{\pi}{4}<x<0, \lfloor\tan(x)\rfloor = -1$

and when

$0<x<\frac{\pi}{4} , \lfloor\tan(x)\rfloor = 0$ "

But I want to learn how to explain it more thoroughly , that will leave no doubt to the grader that I understood the situation.
The problem is, I don't quite understand it. I know when we do the one sided limits we can choose a delta such that delta = $\frac{\pi}{4}$, so am I correct in assuming that the expression of the floor can immediately be transformed into $-1/0$ as soon as I choose my delta, and only then I take the limit of the given expression?

答:
To say the limit $\lim_{x\to a} f(x) = L$ is to say for any $\epsilon > 0$ there is a $\delta>0$ so that ....yadda, yadda, yadda.

Now we know that if $\delta_1$ leads to yadda, yadda, yadda  then for any $)<\delta < \delta_1$ we also get yadda, yadda, yadda.  So we can assume $\lim_{x\to a} f(x)$ that $x$ is as close to $a$ as we need it to be.

And in taking limits we always assume $x \ne a$.

So in calculating $\lim_{x\to 0}f(x)$ we may assume that $-\frac \pi 4 < x < \frac \pi 4$ and that $x\ne a$. (we can also assume $-0.0000000001 < x < 0.087$ and $x \ne 0$... we can assume $x$ is as close to zero as we want.  But as long as we assume it is within $\frac \pi 4$ of $0$ we will get something to simplify our work the way we want.)

And that's it

If $-\frac \pi 4 < x < 0$ then $\lfloor\tan x\rfloor\cos x = -\cos x$

And if $0< x < \frac \pi 4$ then $\lfloor\tan x\rfloor\cos x = 0$.

And because we can assume, for the purposes of taking limits that $x$ is within $\frac \pi 4$ of $0$ but not equal to $0$ we have

$\lim_{x\to 0}\lfloor\tan x\rfloor\cos x =$

$\lim_{x\to 0}\begin{cases} -\cos x& x< 0\\ 0&x > 0\end{cases}$

Now if we consider the upper and lower limits:

$\lim_{x\to 0^{-1}}\lfloor\tan x\rfloor\cos x$ we assume $x$ is within $\frac \pi 4$ of $0$; that $x \ne 0$; and we assume $x< 0$.

So

$\lim_{x\to 0^-}\lfloor\tan x\rfloor\cos x =$

$\lim_{x\to 0^-}\begin{cases} -\cos x& x< 0\\ 0&x > 0\end{cases}=$

$\lim_{x\to 0^-}-\cos x = -1$.

And for upper limits we assume $x > 0$ so

$\lim_{x\to 0^+}\lfloor\tan x\rfloor\cos x =$

$\lim_{x\to 0^+}\begin{cases} -\cos x& x< 0\\ 0&x > 0\end{cases}=$

$\lim_{x\to 0^+}0 = 0$.

And as $\lim_{x\to 0^-}\lfloor\tan x\rfloor\cos x \ne \lim_{x\to 0^+} \lfloor\tan x\rfloor\cos x$, $\lim_{x\to 0}[\tan x]\cos x$ doesn't exist.


问:
Eigenvalues of discrete second order derivative
I am trying to calculate the eigenvalues of the discrete second order derivative, with both Neumann and mixed Dirichlet-Neumann boundary conditions. I'm following the derivation here: https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#Neumann_case
But there are some things I don't understand for each derivation.

Neumann boundary conditions

For a grid of $n$ points, the problem corresponds to finding the eigenvalues of the $n\times n$ matrix

\begin{equation}
\frac{1}{h^2}\,
         \begin{pmatrix}
     -1 & 1 & & &  0\\
     1 & -2 & 1 & &  \\
      & 1 & \ddots & \ddots & \\
     & & \ddots & \ddots & 1 \\
     0 &&  & 1 & -1
     \end{pmatrix}
\end{equation}

where $h$ is the grid spacing, which I will set to be unity. Unfortunately, there are a few things I don't understand. They define $\vec{v}$ to be the eigenvector corresponding to an eigenvalue $\lambda$. They state that the eigenvalue problem corresponds to solving the problem

$$
v_{k+1} - 2v_k + v_{k-1} = \lambda v_k \quad \Rightarrow \quad v_{k+1} = \left(\lambda+2\right) v_k - v_{k-1}
$$

This is true, but not for all $k$ and it bugs me that this is not stated. Is this not important? Instead of working with $v_k$, they work with $w_k$, which is defined

$$
w_k = v_{k+1} - v_k \quad \Rightarrow \quad w_k- w_{k-1} = v_{k+1} - 2v_k + v_{k-1}
$$

This can be used to find a recursive expression for the $w$'s.

$$
w_{k+1} =(2+\lambda) w_k - w_{k-1} = 2\alpha w_k - w_{k-1}
$$

They then state that the boundary conditions are $w_0 = w_n = 0$. But why? $w_0 = v_1 - v_0 = v_1$ and $w_n = v_{n+1} - v_n = -v_n$. I don't see why either of these should be zero.

答:
The eigenvalue equation $v_{k+1} - 2v_k + v_{k-1} = \lambda v_k$ is actually true for all $k$. I think your confusion is in the introduction of the dummy variables $v_0$ and $v_{n+1}$ and in your mixing the Neumann and Dirichlet boundary conditions.

The variables $v_{0}$ and $v_{n+1}$ are introduced specifically to ensure that $v_{k+1} - 2v_k + v_{k-1} = \lambda v_k$ is true for all $k$, but they have different values for the Neumann and Dirichlet conditions:
$$ \text{Neumann condition:} \quad v_0 = v_1 \quad v_{n+1} = v_n.$$
$$ \text{Dirichlet condition:} \quad v_0 = 0 \quad v_{n+1} = 0.$$
What you wrote in your calculation of $w_0 = v_1 - v_0$ and $w_n = v_{n+1} - v_n$ was a combination of these two mutually exclusive conditions which is why you obtained the contradiction $w_0 = v_1 - v_0 = v_1 = 0$ and the analogous one for $w_n$.


问:
How can I prove the identity $\dfrac{1}{1+\sin(x)}\equiv \dfrac{\sec^2\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)+1\right)^2}$
I have to prove by working on the left hand side of the equation that

$$\dfrac{1}{1+\sin(x)}\equiv \dfrac{\sec^2\left(\frac{x}{2}\right)}{\left(\tan\left(\frac{x}{2}\right)+1\right)^2}$$
However I just cant seem to get it, I first started by rewriting the denominator using sin half angle formula

$$\frac{1}{1+2\sin(\frac{x}{2})\cos(\frac{x}{2})}=\frac{1}{\sin^2(\frac{x}{2})+\cos^2(\frac{x}{2})+2\sin(\frac{x}{2})\cos(\frac{x}{2})}$$

Then by simplifying further
$$=\dfrac{1}{(\sin(\frac{x}{2})+\cos(\frac{x}{2}))^2}$$
However I'm stumped here and do not know how to proceed, I even took an alternate path and tried rationalizing the fraction as such

$$\frac{1}{1+\sin(x)}*\frac{1-\sin(x)}{1-\sin(x)}=\frac{1-\sin(x)}{(1-\sin(x))^2}$$
And applying $\sin^2(x)+\cos^2(x)=1$
$$=\frac{1-\sin(x)}{\cos^2(x)}$$
$$=\sec^2(x)*(1-\sin(x))$$

But again I get stuck, is there some identity I am forgetting to implement or some other method to approach the left hand side with?

答:
We can express any value trigonometric ratio of a given angle $2\theta$ in terms of the tangent of half that angle. For example:

$$\sin2\theta\equiv\frac{2\tan\theta}{1+\tan^2\theta}$$
$$\cos2\theta\equiv\frac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$\tan2\theta\equiv\frac{2\tan\theta}{1-\tan^2\theta}$$
As an aside, this is useful in integration. So, you may want to use the first identity given to prove your identity.


问:
How do I prove that the union of two simply connected open sets whose intersection is path connected is simply connected?
I'm trying to understand Ronnie Brown's answer here: union of two simply connected open , with open and non empty intersection in $R^2$

Let $X$ be a topological space and $U,V$ be simply connected open subsets of $X$ such that $U\cap V $ is path-conneted.

Fix $x\in U\cap V$.

Let $r$ be a loop in $X$ based at $x$.

By Lebesgue number lemma, there is a partition $\{s_0,\cdots,s_n\}$ of $[0,1]$ such that $r([s_i,s_{i+1}])\subset U$ or $r([s_i,s_{i+1}])\subset V$.

Now, define $r_j(t)=r((1-t)s_j + t\cdot s_{j+1})$ for $1\leq i < n$. ($t\in [0,1]$)

Then, each $r_j$ is a path from $r(s_j)$ to $r(s_{j+1})$.

I completely understand till here.

However, he says that for each $j$, it is possible to choose a path $\alpha$ "in $U\cap V$" such that $\alpha(0)=x$ and $\alpha(1)=r(x_j)$.

How is that possible?

Since $r$ is an arbitrary path, it ranges over all $X$ not restricted in $U\cap V$. Since $U\cap V$ is path connected, if $r(x_j)$ is in $U\cap V$, then that makes sense, but it's possible that $r(s_j)\notin U\cap V$.

Would someone please complete the proof?

答:
We can show this property using the Seifert-van-Kampen's theorem.

Infact $U$,$V$ and $U\cap V$ are path connected for hypothesis.

Let's consider a point $x_0\in U\cap V$and the fundamental group of $X$ based at $x_0$ $$\Pi_1(X,x_0)=\langle\text{generators of }\Pi_1(U),\text{generators of }\Pi_1(V)|R_U\cup R_V\cup R_S\rangle=\langle S_1\cup S_2|R_1\cup R_2\cup R_S\rangle,$$
but $S_1,S_2=\emptyset\implies \Pi_1(X,x_0)=\langle\emptyset|\emptyset\rangle\implies X$ is simply connected, since its fundamental group is trivial.

ContentLicense: CC BY-SA 3.0


问:
Integration and dot product
Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ continuously differentiable.
We assume that there exists $L > 0$ such that
\begin{equation}
   \|∇f(x)-∇f(x')\| \leq  L\|x − x'\| \qquad \forall (x,x') \in \mathbb{R}^n \times \mathbb{R}^n.
\end{equation}

Show that
\begin{equation}
  |f(x + h) − f(x) − \langle ∇f(x), h\rangle | \leq \frac{L}{2} \|h\|^2 \qquad \forall (x, h) \in \mathbb{R}^n \times \mathbb{R}^n.
\end{equation}

I do not understand part of the demonstration of which here is:

As $f$ is continuously differentiable, we have from Taylor's formula at zero order with residual in integral form
\begin{equation}
  f (x + h) = f(x) + \int^1_0 \langle ∇f(x + th), h \rangle dt. \tag{1}
\end{equation}

Me, when I apply Taylor's formula at zero order with residual in integral form, we get
\begin{equation}
  f (x + h) = f(x) + \int_x^{x+h} ∇f(t) dt. \tag{2}
\end{equation}
What is the integration process that allows us to leave (2) to (1) ?

答:
By the fundamental theorem of calculus \begin{aligned}
f(\mathbf{x}+\mathbf{h}) - f(\mathbf{x}) &= \int_0^1\nabla f(\mathbf{x}+t\mathbf{h})^T\mathbf{h}\,dt \\
&=\nabla f(\mathbf{x})^T\mathbf{h}+ \int_0^1\big(\nabla f(\mathbf{x}+t\mathbf{h}) - \nabla f(\mathbf{x})\big)^T\mathbf{h}\,dt
\end{aligned}
and therefore
\begin{aligned}
|f(\mathbf{x}+\mathbf{h}) - f(\mathbf{x})-\nabla f(\mathbf{x})^T\mathbf{h}|
&=\Big|\int_0^1\big(\nabla f(\mathbf{x}+t\mathbf{h}) - \nabla f(\mathbf{x})\big)^T\mathbf{h}\,dt \Big|\\
&\leq \int_0^1\big|\big(\nabla f(\mathbf{x}+t\mathbf{h}) - \nabla f(\mathbf{x})\big)^T\mathbf{h}\big|\,dt \\
&\overset{\text{C.S.}}\leq \int_0^1\|\big(\nabla f(\mathbf{x}+t\mathbf{h}) - \nabla f(\mathbf{x})\big)\|\|\mathbf{h}\|\,dt  \\
&\leq\int_0^1tL\|\mathbf{h}\|^2\,dt \\
&\leq \|\mathbf{h}\|^2
\end{aligned}
where C.S. is Cauchy-Schwarz and the inequality following it is from the fact that $f$ is $L$-smooth.

For general knowledge, this inequality is known as the Descent Lemma, and it's very useful in convex analysis. Analysis of optimization algorithms strongly relies on the ability to quantify the change in function value when moving from some point $\mathbf{x}$ in a direction $\mathbf{h}$.

Title: Integration and dot product


问:
Function evaluation from area
Can we find a smooth continuous $  f(z)$ if

$$ \int_1^3 f(x) dx = \dfrac{26}{3},\; \int_2^4 f(y) dy=\dfrac{56}{3} ?$$

EDIT1:

What I actually wanted to ask was:

If $( f,g )$ are functions inverse of each other then find $f(x)$

$$ \int_1^3 f(x) dx + \int_2^4 g(y) dy=10 $$

Answer is known as any function as being "obvious" but how to go about showing the same?

Comments appreciated.

答: No constant $f$ will do, but the next simplest option is linear. Say $f(x)=ax+b$ so$$4a+2b=\frac{26}{3},\,6a+2b=\frac{56}{3}\iff a=5,\,b=-\frac{17}{3}.$$

Title: Function evaluation from area


问:
How to show that given inequality holds for some choices of a and n?
Show that there exists some $a\in \mathbb{R}$, $n\in \mathbb{N}\setminus\{0, 1, 2, 3\}$ such that $$\frac{(1+\sqrt[3]{|a|}\ \big)^{2n}}{n(n-1)(n-2)(2n-1)(2n-3)(2n-5)}\leq \frac{a^2}{90}.$$
I have tried by taking $a=8, 1000,  n=4, 5$ but not getting the result. Please help.

答:
First note that $a$ cannot be $0$.

Consider the even function$$f(a)=\frac{(1+|a|^{1/3})^{2n}}{a^2}$$For $a>0$,

$$\begin{align*}f'(a)=&\frac{2n(1+a^{1/3})^{2n-1}}{3a^{8/3}}-\frac{2(1+a^{1/3})^{2n}}{a^3}\\=&\frac{2(1+a^{1/3})^{2n-1}}{a^3}\left[\left(\frac n3-1\right)a^{1/3}-1\right]\end{align*}$$

The function attains global minimum at $a=\left(\frac n3-1\right)^{-3}$ with the minimum being$$f_\min(n)=\frac{n^{2n}}{3^6(n-3)^{2n-6}}$$

You can check that $\not\exists n\in\Bbb N_{\ge4}$ such that$$\begin{align*}f_\min(n)\le&\frac{n(n-1)(n-2)(2n-1)(2n-3)(2n-5)}{90}\\\iff0\le&\frac{(n-1)(n-2)(n-3)^{2n-6}(2n-1)(2n-3)(2n-5)}{n^{2n-1}}-\frac{10}{81}\end{align*}$$using a graphic calculator or program.


问:
A normed space $E$ is reflexive if and only if the weak- and weak*-topology coincide on $E'$
Let $E$ be a normed space. It is claimed in this answer that $E$ is reflexive (which is to say that the canonical embedding of $E$ into $E''$ is surjective) if and only if "the weak- and weak*-topology on $E'$ coincide".

I didn't found this result in the literature on functional analysis I'm usually consulting. So, could anyone recommend a good reference?

But more importantly: How do we need to understand this result at all? The "weak*-topology" is a topology on $E'$. So, that's fine. But the "weak-topology" is a topology on $E$. So, what does it mean that both topologies coincide on $E'$ when the latter is not even a topology on $E'$?

答:
I don't know a good reference to a proof of the equivalence you are looking but I think the proof is elementary.

Let $\mathcal{T} \subset 2^{E'}$ denote the weak topology on $E'$ and let $\mathcal{U} \subset 2^{E'}$ denote the weak$^{\ast}$ topology.

Assume first that $E$ is reflexive and aim to show that $\mathcal{U}= \mathcal{T}$. By definition, $\mathcal{T}$ is the weakest (smallest w.r.t inclusion) topology on $E'$ making all norm-continuous functionals on $E'$ continuous. Also by definition, $\mathcal{U}$ is the smallest topology making all point-evaluation functionals $\text{ev}_x:E' \rightarrow \mathbb{K}$, $x \in E$ on $E'$ continuous. Since $E$ is reflexive these sets of linear functionals on $E'$ are the same, so $\mathcal{T}= \mathcal{U}$.

Assume now that $\mathcal{T}= \mathcal{U}$ and aim to show that $E$ is reflexive. Let $\ell: E' \rightarrow \mathbb{K}$ be norm continuous. Then it is continuous w.r.t $\mathcal{T}$ and hence with respect to $\mathcal{U}$. The latter implies that the set $\{x' \in E'\,:\, |\ell(x')| < 1\}$ is weak$^{\ast}$ open.  It follows that there are $x_1, \dots, x_n \in E$ and $\varepsilon_1,\dots, \varepsilon_n > 0$ so that
$$
\bigcap_{i=1}^n{\{x' \in E' \,:\, |x'(x_i)| < \varepsilon_i\}} \subset \{x' \in E'\,:\, |\ell(x')| < 1\}.
$$
Hence
$$
\bigcap_{i=1}^n{\ker{(\text{ev}_{x_i})}} \subset \ker{\ell}.
$$
This implies that $\ell = \text{ev}_x$ for some $x$ in the linear span of the $x_i$'s . What I've just explained in the last paragraph is probably very standard: Every weak$^{\ast}$ continuous linear functional is a point-evaluation functional.

To answer your last paragraph: The weak topology can be defined on any normed vector space, in particular on dual spaces.


答:
Take any two smooth functions and form the linear combination $af(x)+bg(x)$.

Now solve the linear system

$$a\int_1^3f\,dx+b\int_1^3g\,dx=\frac{26}3,\\a\int_2^4f\,dx+b\int_2^4g\,dx=\frac{56}3.$$

I fail to see where the difficulty lies.

You can even start from antiderivatives and solve

$$a(F(3)-F(1))+b(G(3)-G(1))=\frac{26}3,\\a(F(4)-F(2))+b(G(4)-G(2))=\frac{56}3.$$

Title: Function evaluation from area

Body:
Can we find a smooth continuous $  f(z)$ if

$$ \int_1^3 f(x) dx = \dfrac{26}{3},\; \int_2^4 f(y) dy=\dfrac{56}{3} ?$$

EDIT1:

What I actually wanted to ask was:

If $( f,g )$ are functions inverse of each other then find $f(x)$

$$ \int_1^3 f(x) dx + \int_2^4 g(y) dy=10 $$

Answer is known as any function as being "obvious" but how to go about showing the same?

Comments appreciated.


问:
Darboux theorem for the symplectisation of a contact manifold
I'm wondering if there is some "nice" version of Darboux's theorem that can be applied in the case of a symplectisation of a contact manifold $(X,\alpha)$ with the canonical symplectic form. i.e. $(\Bbb R \times X, d(e^t\alpha))$, where $t$ is the $\Bbb R$-parameter.

More specifically, I'd like to know if we can assume that the first 2 coordinates in a Darboux neighbourhood are given by the Reeb vector field and the canonical Liouville vector field $\partial_t$.

I tried looking around but wasn't really able to find anything

答:
This is not possible. Let $\omega = d(e^t\alpha)$, and suppose $(x^1,\dots,x^n,y^1,\dots,y^n)$ are any Darboux coordinates for $\omega$. Then the vector fields $\partial _{x^i}$ and $\partial_{y^i}$ are all symplectic (or equivalently locally Hamiltonian), which means that the Lie derivative of $\omega$ with respect to each of them is zero. But $\partial_t$ is not symplectic, because
$$
\mathscr L_{\partial_t}\omega = d(i_{\partial_t}\omega) = d(e^t\alpha) = \omega.
$$


答:
We can find a lot of functions with this property.

Consider $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ or any other polynomial. It's easy to notice that we have
$$ \int_1^3 f(x) dx = \dfrac{26}{3},\; \int_2^4 f(y) dy=\dfrac{56}{3} $$
for many $f(x)$ of such type.

Title: Function evaluation from area


问:
If $X$, $Y$, and $\frac{X+Y}{2}$ are nonnegative and identically distributed, is $X=Y$ almost surely?

Suppose $X$ and $Y$ are nonnegative random variables such that the distributions of $X$, $Y$, and $\frac{X+Y}{2}$ are identical. Is it true that $X=Y$ almost surely?

$X$ and $Y$ are not necessarily assumed to be integrable. It is crucial that they are bounded on one side, since otherwise we may sample $X$ and $Y$ independently from the Cauchy distribution.

$X$ and $Y$ cannot be independent because one-sided $\alpha$-stable distributions do not exist for $\alpha \geq 1$. (Proved by Esseen and Ramachandran in the 1960s.) My gut feeling is that if $X$ and $Y$ cannot be independent then they cannot not be distinct at all, but this relies on how, when they are square-integrable, $Y$ can be written as a linear combination of $X$ and an independent copy of it. The proof itself is easy when we assume square-integrability:
$$ \mathbb{E}\left[ (X-Y)^2 \right] = \mathbb{E}\left[X^2\right] + \mathbb{E}\left[Y^2\right] - \mathbb{E}\left[(X+Y)^2\right] = 0. $$

I do not know how to proceed without this assumption. I don't think a truncation argument would work since $(X \wedge n) + (Y \wedge n) \neq (X+Y) \wedge (2n)$. Perhaps there is a simple counterexample I have overlooked.

For context, I have been reading a paper in which appears a continuous, strictly increasing process $(X_t)_{t\geq 0}$ with $X_0=0$ the following two properties: $X_s/s \stackrel{d}{=} X_t/t$ and $X_{s+t} - X_s \stackrel{d}{=} X_t$ for all $s,t>0$. The author claims that $\lim_{t\to 0} (X_t/t)$ exists almost surely because $X_t/t = \lim_{s\to \infty} (X_s/s)$ almost surely for all $t>0$. (The limit as $t\to\infty$ was claimed to exist almost surely by Birkhoff's pointwise ergodic theorem, although it was not assumed in the paper that $X_1$ is integrable.) I did not quite see how this is true, though I realized that the proof will hold even without the ergodic theorem if my question is answered to be true.

答:
If $g:[0,\infty)\to[0,\infty)$ is bounded and strictly concave then
$$
E[g(X)] = E\left[{g(X)+g(Y)\over 2}\right]\le E\left[g\left({X+Y\over 2}\right)\right]=E[g(X)].\qquad\qquad(\dagger)
$$
The inequality in ($\dagger$) is a consequence of Jensen's inequality,
$$
{g(X(\omega))+g(Y(\omega))\over 2}\le g\left({X(\omega)+Y(\omega)\over 2}\right),
$$
which also asserts that   strict inequality obtains unless $X(\omega) = Y(\omega)$.
In view of ($\dagger$), the two sides of this last display have the same expectation, so they are equal for a.e. $\omega$. It follows that $X=Y$ a.s.

A suitable choise for $g$ is $x\mapsto 1-e^{-x}$.


问:
Simplifying $\left|\left|\sqrt{-x^2}-1\right|-2\right|$
How do we simplify the expression $\left|\left|\sqrt{-x^2}-1\right|-2\right|$?

This is very confusing.  Do they cancel out and become just simply $\sqrt{-x^2}-1-2$?

答:
Assuming you are working with real numbers only, your expression is only defined if $x = 0$. Following on from that the answer would be $||0 - 1| - 2| = ||-1| - 2| = |1 - 2| = |-1| = 1$.

If we allow x to be complex or imaginary, it gets a bit more, well, complex. Let x be the complex number $a + bi$. Squaring this gives $a^2 + 2abi - b^2$. The negative sign changes this to $b^2 - 2abi - a^2$, and the expression has now become $||ai - b - 1| - 2|$.

If we suppose that the real part of this complex number is 0 (meaning it is a pure imaginary number), then we can get $||-b - 1| - 2|$. The answer would then depend on the value of b. Otherwise, there is no way to simplify this expression; although the absolute value of $a + bi$ is $\sqrt{a^2 + b^2}$ (can you figure out why?) we have flipped the real and imaginary parts, and also have a pesky 1 in the mix.

Note also that the absolute value of the absolute value is indeed the absolute value, but you cannot just "cancel out" the absolute value brackets. As you can see, you must take into account the operations in between; if you had evaluated the brackets as you say the answer would be -3, which is ridiculous, because absolute value gives only positive answers.


问:
How to solve $a+b+c=6; 2\le a\le 5, 2\le b\le 4, 1\le c\le 2$ with Generating Functions?
So here is the original problem

For the inter-hostel six-a-side football tournament, a team of 6 players is to be chosen from 11 players consisting of 5 forwards, 4 defenders and 2 goalkeepers. The team must include at least 2 forwards, at least 2 defenders and at least 1 goalkeeper. Find the number of different ways in which the team can be chosen.

Which I think can be written as $$ a+b+c=6; \quad 2\le a \le 5,\quad 2\le b \le 4\quad 1\le c \le 2 $$

I know how to solve the problem with combinations where we choose [(3,2,1), (2,3,1), (2,2,2)] which is of the form (#of forwards, #defenders, #goalkeepers), this gives answer as 260.

While solving the same problem with generating functions I arrived at following polynomial $$ x^2  (1+x+x^2+x^3) * x^2  (1+x+x^2) * x (1+x) $$ which is equivalent to $$ x^5 *\frac{(1-x^4)}{(1-x)} *\frac{(1-x^3)}{(1-x)} *\frac{(1-x^2)}{(1-x)}.$$

Now the problem is converted to finding $x^1$ in $ \frac{(1-x^4)}{(1-x)} *\frac{(1-x^3)}{(1-x)} *\frac{(1-x^2)}{(1-x)}$ and we can only get $x^1$ from $(1-x)^{-3}$ to which there are only 3 ways which is a wrong answer.

What is the correct method?

答:
Your $3$ solutions for $(a,b,c)$ are correct if players in the same position are indistinguishable.  Accounting for distinguishable players yields
$$\binom{5}{3}\binom{4}{2}\binom{2}{1}=10\cdot6\cdot2=120$$ solutions for $(a,b,c)=(3,2,1)$,
$$\binom{5}{2}\binom{4}{3}\binom{2}{1}=10\cdot4\cdot2=80$$ solutions for $(a,b,c)=(2,3,1)$, and
$$\binom{5}{2}\binom{4}{2}\binom{2}{2}=10\cdot6\cdot1=60$$ solutions for $(a,b,c)=(2,2,2)$, so $120+80+60=260$ solutions altogether.

Your generating function also yields the correct count of $3$ if players in the same position are indistinguishable.


问:
Abhyankar's Lemma with two wild ramifications
In the context of algebraic function fields, Abhyankar's Lemma states:

Let $F'/F$ be a finite separable extension of function fields. Suppose that $F'=F_1F_2$ is the compositum of two intermediate fields $F\subseteq F_1$, $F_2\subseteq F'$. Let $P'\in \mathbb{P}_{F'}$ be an extension of $P\in  \mathbb{P}_{F}$ and set $P_i:=P'\cap F_i$ for $i=1,2$. Assume that at least one of the extensions $P_1|P$ or $P_2|P$ is tame. Then $$e(P'|P)=lcm\{e(P_1|P), e(P_2|P)\}$$

To keep this question short, I can provide desired definitions in the comments.

Now, my question is: what if we take two wild ramifications? Obviously, Abhyankar's Lemma does not hold. I found an example where it doesn't hold but I want to prove why it doesn't work. Unfortunately, I have no clue about the proof.

I would be grateful for any help.

答:
I would build such an extension as a compositum of two Artin-Schreier extensions.

Let $F=\Bbb{F}_2(x)$ be the rational function field over $\Bbb{F}_2$. Consider the two quadratic extensions
$$
F_1=F(y), y^2+y=x,
$$
and
$$
F_2=F(z), z^2+z=x^3.
$$
Both $F_1/F$ and $F_2/F$ are Galois, the respective Galois groups generated by the order two automorphisms $y\mapsto y+1$ and $z\mapsto z+1$. Therefore the compositum $K=F_1F_2$
is also a Galois extension of $F$ with the Galois group isomorphic to the Klein four. The third intermediate field is $F_3=F(y+z)$, where $(y+z)^2+(y+z)=x^3+x$.

The standard tools (I use results from Stichtenoth's book, mine is the earlier edition from 1993) then imply easily that the infinite place $F_\infty$ of $F$ is totally (hence wildly) ramified in $F_i/F$ for all $i=1,2,3$. Therefore all the automorphisms of $Gal(K/F)$ belong to the inertia group. It follows that above $F_\infty$ there is a single place $K_\infty$ of $K$, and $e(K_\infty|F_\infty)=4$. Clearly $e(F_{i,\infty}|F_\infty)=2$
$F_{i,\infty}=K_\infty\cap F_i$, for all $i=1,2,3$.


问:
Proof that $\left \langle 0\mid \hat{S}\mid 0 \right \rangle = e^{\sum (\text{all connected vacuum diagrams})}$
In my book on QFT  (Lancaster & Blundell) they state the following equation, without proof: $$\left \langle 0\mid \hat{S}\mid 0 \right \rangle  = e^{\sum (\text{all connected vacuum  diagrams})}$$ where $\hat{S}$ is the S-matrix. This is what I tried:
If we expand the expression above in terms using dyson's series, we get:$$\left \langle 0\mid \hat{S}\mid 0 \right \rangle = \left \langle 0\mid Te^{-i\int d^{4}z\hat{H}_{I}(z)}\mid0  \right \rangle$$ and then maclaurin expanding the exponential gives us the following:$$\left \langle 0\left | T \left [ 1-i\int d^{4}z\hat{H}_I(z)+\frac{(-i)^{2}}{2}\int d^{4}yd^{4}w\hat{H}_I(y)\hat{H}_I(w)+... \right ]\right | 0\right \rangle$$    I am not sure where to go from here, however. Can anyone help me complete this proof?

答:
Note that each term in your MacLaurin$^\dagger$ series represents one or more diagrams. Since they are sandwiched between the vacuum $|0>$, the only nonzero diagrams are those with no external lines. In particular, there will be 'disconnected' diagrams where not all vertices are connected. If you agree that the series generates all the diagrams, then the linked cluster theorem tells you that the exponential of those are the connected ones (where connected means that all vertices in the diagram are connected).

$\dagger$ Most people would call it a Dyson series.

If this is more appropriate as a comment, feel free to move/remove it. I haven't posted enough to comment yet.


问:
Calculate initial value given different percentages and the total
Let's imagine an invoice:

Sub total: 100 (what we don't know)

Tax1 21%: 21 (this increases the total)

Tax2 10%: 10 (this de-increase the total)

TOTAL: 111

For a single tax we could do:

```<code>100 (total) * 100 / (21 + 100) = 82.64 (sub total)
</code>```

But how can we calculate the sub total when having different percentages?

Note, every tax affects the original price only (without other taxes)

答:
```<code>subTotal = total * 100 / (totalTaxes - totalWithholdings + 100)
</code>```

For example, for:

```<code>sub total: X

+tax1 21%

+tax2 10%

-tax1 15%

-tax2 7%

TOTAL: 150
</code>```

subTotal = 150 * 100 / (31 - 22 + 100)

subtTotal = 137.61

Body:
subTotal = total * 100 / (totalTaxes - totalWithholdings + 100)

For example, for:

sub total: X

+tax1 21%

+tax2 10%

-tax1 15%

-tax2 7%

TOTAL: 150

subTotal = 150 * 100 / (31 - 22 + 100)

subtTotal = 137.61


问:
Solve equation $\cos (a)\cos(b)\cos(a+b) = -\frac18$ over $0 < a,b < \frac\pi2$
I have tried this question as follows:

$$4[2(\cos(a)\cos(b))\cos(a+b)] = -1$$

Then converted the $2\cos(a)cos(b)$ in $\cos(a+b) + \cos(a-b)$.

And so on but this got messy and big.

Another approach that I tried was that as $a$ and $b$ lie in the first quadrant then their cosine is positive this implies $\cos(a+b)$ is negative.

And without losing generality we can say each term be equal to 1/2 which implies

$a = b = π/3$.

I googled it but didn't find any solution.

Wolfram alpha after converting into the same thing as I have converted used complex numbers which was neat, but my question is:  can we solve this with a short method (using trigonometry)?

答:
Hint:

$$-\dfrac14=(\cos(A+B)+\cos(A-B))\cos(A+B)$$

$$\iff\cos^2(A+B)+\cos(A+B)\cos(A-B)+\dfrac14=0$$

which is a Quadratic Equation in $\cos(A+B)$

As $\cos(A+B)$ is real, the discriminant must be $\ge0$

i.e., $\cos^2(A-B)-4\cdot\dfrac14\ge0\iff\sin^2(A-B)\le0$

$\implies\sin(A-B)=0\implies A-B=n\pi$

But for the given ranges of $A,B;$ $$0<A-B<\pi\implies n=0$$


问:
Isomorphy of $2$-dimensional $L$-modules

Let $L$ be the $2$-dimensional Lie algebra over $\mathbb C$ with basis $x, y$ and $[xy] = x$. Let $V$ be a vector space over $\mathbb C$ with basis $v_1, v_2$. Then $V$ is an $L$-module if we define multiplication by
     $$xv_1 = 0,\quad xv_2 = v_1,\quad yv_1 = -v_1,\quad yv_2 = v_1$$
  Show that $V$ is isomorphic to the adjoint module for $L$.

I have been trying to work with explicit $L$-module isomorphisms such as
   $$f:V\to L,\ \ \alpha v_1 + \beta v_2\mapsto \alpha x + \beta y$$

but when I try to validate the homomorphism condition $f(lv)=lf(v)$ for $l\in L,\ v\in V$ I always run into a dead end, e.g.
 $$f(y(\alpha v_1+\beta v_2))=(\beta-\alpha)x\ \ \ \ne\ \ -\alpha x=[y,f(\alpha v_1+\beta v_2)]$$

Any hints as to how I could show isomorphy?

答:
The eigenvectors for $y$ acting on $V$ are $v_1$ with eigenvalue $-1$ and $v_1+v_2$ with eigenvalue $0$. So these have to go (up to scalars) to $x$ (eigenvalue $-1$) and $y$ in the adjoint representation. Now you just have to check what (if any) scalar multiples work to match up the $x$ action.

That is, define a linear map $\phi: V \rightarrow L$ by $\phi(v_1)=ax$ and $\phi(v_1+v_2)=by$ for scalars $a$ and $b$. By design $\phi$ commutes with the action of $y$. Write out the equations $\phi(x v_1)=x \phi(v_1)$ and $\phi(x(v_1+v_2))=x \phi(v_1+v_2)$ to find a choice of $a,b$ that will work.


问:
Positive property of the mixture of two convex functions
Let $f_1, f_2$ be two convex functions on $[0, 1]$ and let $\epsilon>0$. Assume that for every $x\in[0, 1]$, we have
$$
\max\{f_1(x), \, f_2(x)\}>\epsilon.
$$
Then does there exist a $\lambda\in[0, 1]$ such that
$$
\lambda f_1(x)+(1-\lambda)f_2(x)>0,\,\,\,\forall x\in[0, 1]?
$$
I'm not sure if one can prove or disprove this using only calculus, or one can apply some functional analysis argument to obtain it.

答:
Let
\begin{equation}
m = \max(\min_{x\in[0,1]} f_1(x), \min_{x\in[0,1]} f_2(x))
\end{equation}

If $m \ge \epsilon$, we have $\lambda f_1 + (1 - \lambda) f_2\ge \epsilon$ for $\lambda = 0$ or $\lambda = 1$.

Let us now suppose $m<\epsilon$. Let $[x_0, x_1]$ and $[x_2, x_3]$ be the non-empty subintervals of $[0, 1]$ where $f_1(x)\le m$ and $f_2(x)\le m$ respectively. Their intersection is empty because of the condition $\max\{f_1, f_2\}>\epsilon$ everywhere. Upon perhaps swapping the two functions, we can suppose that $x_1 < x_2$. As $f_1 - f_2$ is negative at $x_1$ and positive at $x_2$, there is a point $a\in (x_1, x_2)$ such that $f_1(a) = f_2(a) = u \ge \epsilon$. Let $\alpha > 0$ and $\beta >0$ be such that
\begin{equation}
\forall x\in [0, 1],\quad\left\{
\begin{array}\cr
f_1(x)\ge u + \alpha (x -a)\cr
f_2(x)\ge u - \beta (x-a)\cr
\end{array}\right.
\end{equation}
that is to say $\alpha$ and $-\beta$ are subderivatives of $f_1$ and $f_2$ at $x=a$. Now let $\lambda = \frac{\beta}{\alpha+\beta}$ and $1-\lambda= \frac{\alpha}{\alpha+\beta}$. One has
\begin{equation}
\lambda f_1(x) + (1-\lambda)f_2(x) \ge u \ge \epsilon
\end{equation}
Hence the result is true, we have shown that the minimum of $\lambda f_1 + (1-\lambda)f_2$ can be chosen larger than $\epsilon$.

Note that we don't even need $\epsilon > 0$ and we have shown that
\begin{equation}
\max_{\lambda\in[0,1]}\min_{x\in[0,1]}(\lambda f_1(x) + (1-\lambda)f_2(x))\ge \min_{x\in[0,1]}\max\{f_1(x), f_2(x)\}
\end{equation}

Edit
Formulated this way, this results appears to be only a special case of Von Neumann's minimax theorem because if we define
\begin{equation}
f(\lambda, x) = \lambda f_1(x) + (1-\lambda) f_2(x)
\end{equation}
then $f(\cdot, x)$ is concave for all $x$ and $f(\lambda, \cdot)$ is convex for all $\lambda\in[0,1]$. The theorem says that in this case,
\begin{equation}
\max_{\lambda\in[0,1]}\min_{x\in[0,1]}f(\lambda, x)= \min_{x\in[0,1]}\max_{\lambda\in[0,1]} f(\lambda, x)
\end{equation}
which is stronger than what we proved above.


问:
Is Zero ring a Boolean ring?
I have a simple doubt. Is zero ring a Boolean ring? ( We do have $0 ^2 =0$ ). Or do we assume a Boolean ring to be non-zero?. The doubt came in my mind while showing that in a Boolean ring every prime ideal is maximal.

答:
There is no requirement that a Boolean ring have more than one element. (Contrast this with fields, where we explicitly require that the additive and multiplicative identities be distinct). So yes, the trivial ring is a Boolean ring.

Re: your edit, note that "prime = maximal" is (appropriately enough) trivially true for the trivial Boolean ring: both prime and maximal ideals are required to be nonempty proper subsets of the ring itself, and the trivial Boolean ring has no nonempty proper subsets at all.

(Actually, it looks like there's a subtlety here: being a proper subset of the ring is not included in the wikipedia definition of prime ideals in the context of noncommutative rings. I suspect that this is an omission on wikipedia's part, but if not then "prime = maximal" using this notion of "prime ideal" does indeed fail for the trivial ring.)

Title: Is Zero ring a Boolean ring?


问:
Is it fair to say that set theory is meta–mathematics?
Is it fair to say that set theory is meta–mathematics?

(Understanding Set Theory (definition))

Instead of distinguishing the properties of one 'thing' ... a number or numerical proxy such as "x" ... we're talking about properties of ALL things and in this way, a more 'pure' or 'core' symbolic representation of reality?

答:
I wouldn't say so. It is precisely the essence of mathematics to make statements of about all objects satisfying some property. For example, a typical theorem is:

Let $V$ be a vector space. Then $V$ has a basis.

or in other words:

For all objects $V$ that are vector spaces, it is true that $V$ has a basis.

However, it is reasonable to say that mathematics is the study of (more or less) formal theories - Zermelo-Fraenkel set theory is one of those. With this in mind, there are ways of considering the properties of theories themselves and make statements about classes of theories. Such are called meta-theories and are probably closer to what you want.


问:
Help - construct Hermite interpolation polynomial
Construct Hermite interpolation polynomial if:
$ f(1)=0, f'(1)=3, f''(1)=4, f(3)=1, f''(3)=4 $.
Can you give me some tips how to do this task?

Body:
$a+bx+cx^2+dx^3+ex^4$

a+b+c+d+e=0,

b+2c+3d+4e=3,

2c+6d+12e=4,

a+3b+9c+27d+81e=1,

2c+18d+108e=4

$49/16-14x+(133/8)x^2-(13/2)x^3+(13/16)x^4$


答:
Not the way we normally intend it.

Metamathematics comprises metatheories, studying mathematics with mathematical methods. To me, that means model theory or something like it. Set theory is just another of the branches of mathematics that might be thereby studied. But it's an understandable question, because you're likely to run into metamathematics alongside a discussion of set theory.

For example, if you're reading about $\mathsf{ZFC}$, the author might talk about proper classes, and in doing so they're really telling you what we learn in a metatheory, and talk about in a metalanguage of $\mathsf{ZFC}$. The reason this happens is simply that it's very convenient at times to talk about "the class of all sets meeting this condition, whether or not there is a set with exactly those sets as elements, but if there is we'll say said class 'is' that set, by which we really mean they have the same elements".

Or at the easier end of the scale, let's say we were just talking about numbers, when all of a sudden sets of numbers get mentioned. These "bunch together" numbers in the same way classes bunch together sets, in that you can have a first-order theory talking about one kind of object, only for a metatheory to let us aggregate them. You might say that's an exercise in a metatheory, but it's a theory that practice has made part of "mainstream" enough mathematics to make the prefix meta- feel inappropriate here.

Sets are not a "special" enough way of "going up a level" to capture what meta- usually means. Ultimately, "metamathematics" is a term we use when we want to say things like, "this statement is undecidable in this theory because the latter meets certain criteria we can only talk about once you've learned enough metatheoretical jargon to follow a discussion of theories, rather than of more ordinary mathematical objects in a theory."

Body:
Is it fair to say that set theory is meta–mathematics?

(Understanding Set Theory (definition))

Instead of distinguishing the properties of one 'thing' ... a number or numerical proxy such as "x" ... we're talking about properties of ALL things and in this way, a more 'pure' or 'core' symbolic representation of reality?


问:
How to compute next/previous representable rational number?
An (approximate) non-negative rational number representation is a pair of natural numbers each not greater than some fixed limit M (and of course denominator being non-zero).

With this condition there is finite number of representable rational numbers. This means that for each such number we can name previous and next number in the set (of course except for smallest 0 and largest 1). How to compute them?

In other words lets have a set of numbers

$$
R(M)=\{\frac{p}{q} : p,q\in\mathbb{N},q\neq0,p\leq q\leq M\}
$$

where $M\in\mathbb{N}_+$

and given number $\frac{p_1}{q_1}\in R(M)$.

I want to find numbers $\frac{p_S}{q_S}\in R(M)$ and $\frac{p_L}{q_L}\in R(M)$ (if they exist) such that

$$
\frac{p_S}{q_S}<\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_S}{q_S}<\frac{p}{q}<\frac{p_1}{q_1}
$$

and smilarly

$$
\frac{p_L}{q_L}>\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_1}{q_1}<\frac{p}{q}<\frac{p_L}{q_L}
$$

答:
All this needs is a single modular inversion, with complexity $O(\log M \log \log M)$ (or something like that).

We are given $p/q$ in lowest terms, with $p, q \in \mathbb{N}$ and $q > 0$. We want the nearest neighbours of $p/q$, $a/b < p/q < c/d$, subject to $a,b,c,d \in \mathbb{N}$ and $c,d > 0$. Assume for the moment that $p < q$.

Consider $c/d$. It is the element after $p/q$ in the Farey sequence $F_n$, which means that $cq - dp = 1$. In particular, $dp+1 = 0 \pmod q$, i.e. $d = -1/p \pmod q$. So let

$r = 1/p \pmod q$

Now make $d$ as big as possible, subject to (i) $d = -r \pmod q$ and (ii) $d \leq n$. Then just put $c = (dp+1)/q$.

Similarly, to calculate $a/b$, make $b$ as big as possible subject to (i) $b = +r \pmod q$ and (ii) $b \leq n$. Then put $a = (bp-1)/q$.

An example  Suppose $p/q = 28/61$ and $n=100$. Then the inverse of $28 \pmod{61}$ is $r=24$. So $d = 37 \pmod{61}$, and we can take $d=98$, with $c=(98.28+1)/61 = 45$. And $b = 24 \pmod{61}$, so we can take $b = 85$, with $a = (85.28-1)/61=39$. So we get:

$39/85 < 28/61 < 45/98$

We only have to modify this slightly to handle the case $p > q$: instead of computing the largest $d \leq n$ that satisfies the modulo requirement, we find the corresponding modulo requirement for $c$, and then look for the largest such $c \leq n$. You can fill in the details for yourself.


问:
Check existence of cycle in graph with only number of vertices and their degree
I know that we can check existence of cycle in graph by simply traversing the graph and check whether the vertex has been visited.

However, if there are only number of vertices and their corresponding degree known, can we check if the graph contains at least a cycle?
Provided that there is no self-loop and no cycle involving exactly only 2 vertices with 2 degrees each. So it doesn't matter how they are connected to each other.

Is that possible without first randomly connecting each vertex according to the their degrees and then traversing the graph?

Edit

Actually the problem is to find a maximum value from a set of vertices if the vertex connected to other vertex. So let say I'm provided with a set of number $M$ and $|M| = |V| - 1$ with $V$ being a set of vertices, $M_n$ being the value that a vertex can have if the vertex is connected to $n$ other vertices (degree).

Eg.

$$M = (4, 1, 1, 3, 2)$$
In that case the $|V|=6$ and let say $V=(V_1,....,V_6)$.
That means if any of those vertices is connected to a vertex the value of both vertices would be $4$ ($M_1=4$, degree = 1) but if connected to other 3 vertices the value of that particular vertex would be $1$ ($M_3=1$, degree = 3).

From that, I'd like to find which combination yields the most value without introducing a cycle in the graph.

At first I was thinking of trying each combination one-by-one and constructing a graph from each combination then look for a cycle. However that approach wouldn't be feasible if there are a lot of vertices. So now I'm thinking of sorting the degree ($n$) first by the value ($M_n$) then start from there and thinking if it'd be possible to detect a cycle without constructing a graph.

答:
If you have an $n$-vertex graph with $n$ or more edges, there must be a cycle.  You can calculate the number of edges from the degree sequence using the Handshaking Lemma.

If the $n$-vertex graph has $n-1$ or fewer edges, provided there are two vertices of degree $2$ or more, it may contain a cycle (or it may not).  For example, these two graphs have the same degree sequences but only the first one has a cycle:

If you also knew the number of components $c(G)$ of the graph $G=(V,E)$ you could determine whether or not there is a cycle.  Specifically, a graph is acyclic if and only if $c(G)=|V|-|E|$.


问:
Expected steps to solve a 3x3 jigsaw puzzle
Consider a 3x3 jigsaw puzzle, i.e., with 9 pieces. The first piece is randomly drawn from a bag and placed at its correct location. Subsequent pieces are drawn one at a time from the bag and they are placed only if they connect with any piece already placed. Pieces are returned to the bag if they do not connect with any piece already placed and another piece is drawn.  Including the first draw, what is the expected number of draws required to finish the puzzle?

Edit: Two pieces connect if they share an edge (as in a jigsaw puzzle).

答: Writing it down as a Markov process with all its states and calculating from the leaves of the tree should give you the answer. I have done it for the centrepiece for some intuition. Notice that many states are essentially the same due to symmetry of rotation. Hope it helps.


问:
a Poset Over a Set of Partial Orders
I've been scratching my head at this for quite a while, the question is as follows:

(pardon my weird wording, I'm not studying math in English so I might get some proof wording conventions wrong)

let $X$ be a set, and $\mathcal{R}$ be the set of all non-strict partial orders over set $X$.

let $\trianglelefteq,\leqslant \in \mathcal{R} $

We'll define Partial order $\sqsubseteq$ over $\mathcal{R} $ as follows:

$\leqslant \sqsubseteq \trianglelefteq \ \Leftrightarrow \forall x,y\in X\ \left( x\leqslant y\ \rightarrow x\trianglelefteq y\right)$

is $(\mathcal{R},\sqsubseteq)$ a poset?

The first thing that baffled me is that if I want to disprove this, I have to give a counterexample, but to do that, I need to give a partial order over $X$, only I don't know what $X$ is.

On the other hand, to prove that this is a poset, I've tried showing that it's antisymmetric to no avail, and got stuck on that front.
for example:

$let\,\, R_1,R_2\in\mathcal{R}\,    such\  that\,     R_{1} \sqsubseteq R_{2}$
By the assumption, for all $x,y$ the condition is satisfied, in particular for $\hat{x} ,\hat{y} \in X\ \left(\hat{x} R_{1}\hat{y} \ \rightarrow \hat{x} R_{2}\hat{y}\right)$

then in order for $R_{2} \sqsubseteq R_{1}\,$ to also occur, the condition must also apply for all $x,y\,$, in particular $\hat{x} R_{2}\hat{y} \ \rightarrow \hat{x} R_{1}\hat{y}$

But this is where I got stuck, I can't seem to get to a conclusion from this point.

Much appreciated

答:
There's another way of looking at this that makes it very easy.  If we think of a relation as a set of ordered pairs, then the relation $\sqsubseteq$ is the same as $\subseteq$.  $$xR_1y\rightarrow xR_2y$$ means the same thing as
$$(x,y)\in R_1\rightarrow (x,y)\in R_2$$ which, of course, is equivalent to $$R_1\subseteq R_2$$

So, the answer is "yes".

Title: a Poset Over a Set of Partial Orders


问:
Do you know this function?
I'm looking for a real-valued function with the following properties:

$\lim \limits_{x \to 0} f(x) = 1$

$\lim \limits_{x \to 0^+} f'(x) = -\infty$

$\lim \limits_{x \to \infty} f(x) = 0$ and $0<f(x)\leq 1$ always

$1-\sqrt{x}$ fits the first two criteria and may just need the right factor. I would also appreciate ideas on how to approach problems like this generally. Thanks in advance.

答:
Any $$f(x) = \frac{c + \sqrt{\vert x \vert}}{1+ a x^{2p}} \text{ or } \left(c + \sqrt{\vert x \vert}\right) e^{-a x^2}$$ with $a >0$ and $p \ge 1$ integer will do the job.

Approach:

Find a function satisfying the first two criteria.

Multiply it by a function that converges at $\infty$ fast enough to zero, that is positive and has a zero derivative at zero.

Side element: notice that such a function will never be differentiable at zero.

Title: Do you know this function?


问:
Trying to unify techniques used to compute derived functors
I have noticed a pattern when dealing with different types of derived functors.

Let $\mathcal A$ be an abelian category with enough injectives. If $F: \mathcal A \rightarrow \mathcal A$ is a left exact functor we can define its derived functor $R^*F(A)$ by finding an injective resolution of the object $A$, applying $F$, taking cohomology and shifting the complex on step to the left assuming the boundary arrows go from left to right.

We can describe this situation in terms of model categories since $Ch^*(\mathcal A)$ is a model category in the sense of the nlab, theorem $2.2$. I.e, weak equivalences are quasi-isomorphisms and cofibrations are the monomorphisms.

Now $R^*F$ is simply the right derived functor of the induced functor $F':Ch^*(\mathcal A) \rightarrow Ch^*(\mathcal A)$ restricted along $\mathcal A \rightarrow Ch^*(\mathcal A)[\mathcal W^{-1}]$ which sends an object $c$ to the cochain complex consisting of $c$ in degree $0$ and $0$ everywhere else.

Switching gears now. Let $C$ be a right proper model category, we can construct the homotopy limit functor as the right derived functor of the limit functor $\text{lim}: C^I \rightarrow C$ for an appropriate model category structure on $C^I$.

For both $F$ and $\text{lim}$ we essentially constructed the right derived functor valued at $X$ by finding a fibrant replacement $X \rightarrow RX$ and applying the functor to that fibrant replacement. However in both of these cases we see that in practice it suffices to take a weaker form of replacement.

For $F$ we often use the fact that an $F-$acyclic resolution can take the place of an injective resolution.

And for $\text{lim}$ we let $I$ be the category $c \xrightarrow{f} a \xleftarrow{g} b$, then a diagram $\eta: I \rightarrow C$ is fibrant if both $\eta f$ and $\eta g$ are fibrations. But if $C$ is right proper we see that the pullback diagram

is in fact a homotopy pullback if at least one of the morphisms $\eta f$ or $\eta g$ is a fibration. Both need not be fibrations. This is theorem 4.4 on this nlab page.

We see that in both cases, to get the derived functor of a functor $G$ valued at an object $X$ we can take a "weak" fibrant replacement $X \rightarrow (WQ)X$ and apply $G$ to $(WQ)X$ and we will see that $G((WQ)X) \approx G(QX)$ in the homotopy category.

Sorry for the long build up but what is actually going on here category theoretically? Is there some way to unify these two notions of "weak" fibrant replacements in a nice way? I can't for the life of me figure it out.

Thank you in advance!

答:
As mentioned this is examples of computing derived functors using deformations.

A left deformation on category $\mathscr{C}$ with weak equivalences is a functor $Q: \mathscr{C}\to \mathscr{C}$ with a natural weak equivalence $q:Q\to \mathrm{id}_{\mathscr{C}}$. Note this implies that $Q$ preserves weak equivalences.

Now suppose we a given a functor $F:\mathscr{C}\to \mathscr{D}$ between categories with weak equivalences, such that $F$ preserves weak equivalences when restricted to the full subcategory spanned by the essential image of $Q$. In this case we say that $Q$ is a left deformation for $F$.

Now one can show that in this case the left derived functor $\mathbb{L}F$ of $F$ can be computed by the formula

$$
\mathbb{L}F(X)=FQ(X).
$$

Now I have not defined by left derived functors, but i assume this is what you where interested in. I really recommend reading through the 2nd chapter of Emily Riehl's "Categorical homotopy theory" in which she explains this in great detail. She also proves the existence of homotopy colimits (that is a left derived functor for the colimit functor $\mathscr{C}^{I}\to \mathscr{C}$) using the ideas presented in chapter 2.


问:
about homeomorphism on subspace topology of $ \mathbb{R}$
Consider $X=\mathbb{R} \setminus \{0\} \cup \{p,q\}$, that is $X$ is the union of the reals minus $0$, and two points. Consider sets of the type

$U_a = (-a,0) \cup \{p\} \cup (0,a)$

$V_a=(-a,0) \cup \{q\} \cup (0,a)$

where $a >0$. And let

$\mathcal{B}=\{U_a\}_{a>0} \cup \{V_a\}_{a >0} \cup \{ \text{all open intervals of} \hspace{2mm} \mathbb{R} \hspace{2mm} \text{not containing the origin} \}$

Then $\tau=\{\bigcup_{\alpha} B_{\alpha} \big | B_{\alpha} \in \mathcal{B} \}$.

1-Show that $(X,\tau)$ is locally Euclidean and second countable, but not Hausdorff.

For any two positive real numbers $c,d$ define $$I_p(-c,d)=]-c,0[\cup\{p\}\cup]0,d[.$$

2-Prove that the interval $I_{p}(-c, d)$ is homeomorphic to the interval $(-c, d)$

for 1) $(X,\tau)$ is locally Euclidean and second countable because $ \mathbb{R}$ is locally Euclidean and second countable ($X$ is subspace topology of $ \mathbb{R}$ with Euclidean meter). also $(X,\tau)$ is not Hausdorff  because any neighbourhood $U$ of $p$ and any neighbourhood $V$ of $q$ then $ U \cap V \neq \emptyset$.

for 2) let $h\colon I_p(-c,d)\to (-c,d)$ defined by
\begin{split}
h(x)&=x\quad\text{for}\;x\in]-c,0[\;\cup\;]0,d[\\
h(p)&=0
\end{split}

then Clearly $h$ is bijective and continuous (with  pasting lemma).

Now, we must prove that $h^{-1}$ is continuous or equivalently, that $h$ is an open map.then $h$ is a homeomorphism

答:
Any open set is a union of open intervals, and open intervals of $I_p(-c,d)$ are mapped to (basically the same) open intervals of $(-c,d)$.

By the way, $h$ can be extended to the whole space as $X\to\Bbb R$ (also set $h(q)=0$, of course), and that's continuous and open as well.


问:
Finite morphism that is not projective
There are two definitions of a projective morphism.

Hartshorne: A morphism $f: X\to Y$ is projective if it factors as $f=gi$, where $i: X\to P_Y^n$ is closed imbedding and $g: P_Y^n\to Y$ is canonical.

Ravi's notes: Morphisms of the form $Proj(L)\to Y$ where $L$ is a quasicoherent sheaf of graded $O_Y$-algebras, finitely generated in degree 1.

In Ravi's notes, it is shown that finite morphisms are projective. The corresponding algebraic fact is that $Proj(A\oplus B\oplus B\oplus B\cdots)=Spec(B)$ for a finite morphism $A\to B$ of rings. In Ravi's notes, it is also mentioned that a finite morphism need not be projective in Hartshorne's sense. Is there an easy counterexample showing this?

I apologize if this turns out to be a simple question.

best,
minimax

答:
The following seems to hold: Let $\mathcal{L}:=\oplus_{i\geq 0}\mathcal{L}_i$ with $\mathcal{L}_0:=\mathcal{O}_Y$ and $\mathcal{L}$ generated by $\mathcal{L}_1$ as sheaf of graded $\mathcal{O}_Y$-algebras. Let $U:=Spec(A)\subseteq Y$ be an open subscheme with $\mathcal{L}_1(U):=L$ and let $z_1,..,z_n\in L$ be elements generating $\mathcal{L}(U)$ as graded $A$-algebra. It follows $\mathcal{L}(U)\cong B:=A[x_1,..,x_n]/I$ where $I$ is a homogeneous ideal in $A[x_1,..,x_n]$. Let $\pi: \mathbb{P}(\mathcal{L})\rightarrow Y$ be the projection morphism. It follows there is an isomorphism $\pi^{-1}(U)\cong \mathbb{P}(\mathcal{L}_U) \cong \operatorname{Proj}(B)$. Here $\mathcal{L}_U$ is the restriciton of $\mathcal{L}$ to the open set $U$. There is over $U$ a closed embedding

$i: \pi^{-1}(U)\cong \operatorname{Proj}(B) \rightarrow \mathbb{P}^n_U.$

Hence locally you can embed the scheme $\mathbb{P}(\mathcal{L})$ into a projective space over $U$, but not globally.

Hartshorne starts with a trivial sheaf $\mathcal{T}:=\mathcal{O}_Y^{n+1}$ of rank $n+1$, and constructs $\mathbb{P}^n_Y:=\operatorname{Proj}(Sym_{\mathcal{O}_Y}^*(\mathcal{T}^*))$. Then defines a morphism $f:X\rightarrow Y$ to be projective if $X$ globally has a closed embedding $i: X\rightarrow \mathbb{P}^n_Y$ with $p\circ i=f$. Hence it seems Vakils definition generalize Hartshornes definition to the case when $\mathcal{L}$ is no longer on the form $Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*)$ for a trivial rank $n+1$ $\mathcal{O}_Y$-module $\mathcal{E}$.

Your question: "In Ravi's notes, it is also mentioned that a finite morphism need not be projective in Hartshorne's sense. Is there an easy counterexample showing this?"

Answer to your question: It seems Vakils definition is more general: the sheaf of graded algebras $\mathcal{L}$ need not be on the form $Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*)$ for a trivial $\mathcal{O}_Y$-module $\mathcal{E}$ of rank $n+1$.

Here I define $Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*)$ as follows:

$$Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*):=\oplus_{i \geq 0} Sym_{\mathcal{O}_Y}^i(\mathcal{E}^*).$$

Title: Finite morphism that is not projective


问:
Finding the area of a circle with changing radius?
Hi guys I need some help with a math question I have. It is about a tethered dog and calculating the accessible area of the dog on a given leash. Let us say the pole has dimensions radius=0.1 meters and lets say the leash is legnth=2 meters. The leash of the dog is tied at the very top of the pole (facing north) and it is free to walk around on its leash. You would see that the rope would interact with the rope as the dog walks around the pole in such a way that the leash gets shorter which changes the radius of accessible circle of the dog. How would I go about solving this accessible area for the dog? I know its a circle but the radius (leash) of the circle gets shorter as he walks around the pole as the leash gets tied around it. If you have further questions, always ask me! I would really appreciate some help!!! Thanks everyone!!!  - Skillzz9</p>

答:
After the dog has turned by angle $\theta$, the length of the available leash is $(2 - \frac{\theta}{10}) \,$ where $\theta$ is the angle in radians ($n$ complete turns will be $2n \pi$) and $\frac{1}{10}$ is the radius of the pole.

So the accessible area $A = \pi (2 - \frac{\theta}{10})^2$

At $ \theta = 20 \,$ radians, the leash would be completely wrapped around the pole and the accessible area will be zero.

Also, the rate of change in accessible area wrt change in angle at a given angle $\theta$,

$\displaystyle \frac{dA}{d\theta} = - \frac{\pi}{5}(2 - \frac{\theta}{10})$


问:
Find minimal polynomial of given matrix
I want to find the minimal polynomial of this matrix
$$\left[\begin{array}{ccccc}
2 & 0 & 0 & 0 & 0\\
4 & 2 & 0 & 0 & 0\\
1 & 4 & 2 & 0 & 0\\
0 & 0 & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 3
\end{array}\right]$$

I found it charastic polynomial $\chi_{A}=\left(2-x\right)^{3}\left(3-x\right)^{2}$

I also know that the minimal polynomial irreducible factors have the same irreducible factors as charastic polynomial so
$$m_{A}\left(x\right)\in\left\{ \left(2-x\right)^{3}\left(3-x\right)^{2},\left(2-x\right)^{3}\left(3-x\right),\left(2-x\right)^{2}\left(3-x\right),\left(2-x\right)\left(3-x\right),\left(2-x\right)\left(3-x\right)^{2}\right\}  $$
but I don't know how to continue from here

答: You can see in a glance that the generalised eigenspace for $\lambda=2$ is spanned by the first three vectors of the canonical basis, and that for $\lambda=3$ by the last two; the minimal polynomial is the product of the minimal polynomials of the restrictions to these generalised eigenspaces. For the former the minimal polynomial is $(X-2)^3$ because $(A-2)^2$ does not kill the first standard basis vector; for the latter the restriction is $3$ times the identity, so the minimal polynomial is $X-3$. Their product is $(X-2)^3(X-3)$.

Title: Find minimal polynomial of given matrix


问:
Prove that $d(f,Y) \lt 1,$ for any $f \in X$ with $\|f\|_{\infty} = 1.$

Let $X = \left \{f \in C[0,1]\ \big |\ f(0) = 0 \right \}$ and $Y = \left \{f \in X\ \bigg |\ \displaystyle {\int_{0}^{1} f(t)\ dt = 0} \right \}.$ Show that there does not exist any $f \in X$ with $\|f\|_{\infty} = 1$ such that $d \left (f,Y \right ) \geq 1.$

If for any $f \in X$ with $\|f\|_{\infty} = 1$ we can able to find $g_f \in Y$ such that $\left \|f - g_f \right \|_{\infty} \lt 1$ then we are through. But for a given $f \in X$ how to find such a $g_f\ $? Any help in this regard will be appreciated.

Thanks for your time.

答:
Let $f \in X$ with $\|f\|_{\infty} = 1$. Consider $C = \int_{0}^1 f(t)dt$.  Why is $|C|<1$? (The sign is strict)

Now find a $f' \in X$ such that $\int_0^t f' > C$ and $\|f'\|_{\infty} = 1$ (use a function that goes from $0$ to $1$ very fast, then stays at $1$, so the integral goes above $C$) and let $h = f - \frac{cf'}{\int_0^1 f'(x)dx}$. Then see that $\|f-h\|_{\infty}< 1$.


问:
Show that the Boolean lattice $B_{n}$ is isomorphic to the $n$-fold product of the chain $[2]$.

For two posets $\left(\Pi_{1}, \preceq_{1}\right)$ and $\left(\Pi_{2}, \preceq_{2}\right)$ we define their (direct) product with underlying set $\Pi_{1} \times \Pi_{2}$ and partial order
$$
\left(x_{1}, x_{2}\right) \preceq\left(y_{1}, y_{2}\right) \quad: \Leftrightarrow \quad x_{1} \preceq_{1} y_{1} \quad \text { and } \quad x_{2} \preceq_{2} y_{2}
$$
Show that the Boolean lattice $B_{n}$ is isomorphic to the $n$-fold product of the chain $[2]$.

I need to show this statement based on first principles. I have a rough idea, but there are some points that I can’t turn into rigorous mathematical statements, so I’d like to ask for some help.

By the definition above, an element of the $n$-fold product of the chain $[2]$, for example in the case $n = 3$, would look like $((a_1,a_2),a_3)$, where $a_1,a_2,a_3 \in \{0,1\}$. In my argument, I want to identify these elements with vertices of a $n$-cube, i.e. the $n$-tuples $\{0,1\}^n$, by just “removing the extra parentheses”. Is this a clear enough bijection?

Assuming that we have the transformation above, my idea for the isomorphism between $B_n = ([n], \subseteq)$ and the vertices of the $n$-cube is as follows: a set $S \subseteq [n]$ would be mapped to a vertex containing $|S| \ 1$’s, with the positions of the $1$’s indexed by the elements in $S$. For $n = 3$, for example, we would have $\{1\} \mapsto (1,0,0)$, $\{2\} \mapsto (0,1,0)$, $\{1,3\} \mapsto (1,0,1)$, $\{1,2,3\} \mapsto (1,1,1)$, etc. How do I make this into precise function definition?

答:
You are right in both points.

A possible way to define your desired map is to take
$$S \mapsto (u_1, \ldots, u_n),$$
where $u_i \in \{0,1\}$ and $u_i = 1$ iff $i \in S$.


问:
Why can ellipses be expressed as $y=Ax$?
I am reading linear algebra textbook. One question is it said for vectors in the unit circle $||x||=1$. I can understand this. However, it said the vectors $y=Ax$ in the ellipse will have $||A^{-1}y||=1$. Why it is an ellipse?

I only know the ellipse can be expressed in this way:
$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 $

答:
A partial answer: your formula for an ellipse holds if the (semi-minor and semi-major) axes of the ellipse are horizontal and vertical. More generally, the ellipses can point in any two perpendicular directions.

With that in mind, suppose that $\vec u,\vec v$ are orthogonal unit vectors. The component of $(x,y)$ along $\vec u = (u_1,u_2)$ is the dot-product $(u_1,u_2)\cdot (x,y) = \vec u \cdot \vec x$. We can produce the equation of an ellipse with axes in the direction of $u$ and $v$ by using the component of the vector $\vec x$ along $\vec u$ instead of the $x$-value and the component along $\vec v$ instead of the $y$-value. Doing this gives us the equation
$$
\frac{(\vec u \cdot \vec x)^2}{a^2} + \frac{(\vec v \cdot \vec x)^2}{b^2} = 1
$$
Now, let $M$ denote the matrix
$$
M = \pmatrix{u_1/a & u_2/a\\ v_1/b & v_2/b}.
$$
Verify that $M \vec x$ = $((\vec u\cdot \vec x)/a, (\vec v \cdot \vec x)/b)$. It follows that the equation of our ellipse is given by
$$
\|M \vec x\|^2 = 1 \implies \|M\vec x\| = 1.
$$
In other words, we see that every ellipse can be written in the form $\|M \vec x\| = 1$ for some matrix $M$.

The original question was why is it that $\|A^{-1}x\| = 1$ defines an ellipse for every (invertible) matrix $A$. So far, my question only answers this in the case that the rows of $A^{-1}$ happen to be orthogonal to each other.

One approach to answering this question is to combine our analysis above with the existence of the singular value decomposition.


问:
Simplifying $\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$
How do I simplify:

$$\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$$

Should I use modulos or should I factor them?  Or any I suppose to use combinatorics?  Any one have a clue?

答:
Nice question there!

Let $x=\underbrace{11\cdots1}_{n\ 1's}$, then

\begin{align}
\therefore\underbrace{11\cdots1}_{2n\ 1's}-\underbrace{22\cdots2}_{n\ 2's}&=\underbrace{11\cdots1}_{n\ 1's}\times10^n+\underbrace{11\cdots1}_{n\ 1's}-2\times\underbrace{11\cdots1}_{n\ 1's}
\\
&=x\times10^n+x-2x
\\ \\
&=x\times10^n-x
\\ \\
&=x(10^n-1)
\\ \\
&=x\times\underbrace{99\cdots9}_{n\ 9's}
\\
&=x\times(9\times\underbrace{11\cdots1}_{n\ 1's})
\\
&=9x^2
\end{align}

\begin{align}
\therefore\text{The original equation is equal to: }\sqrt{9x^2}=3x=\boxed{\underbrace{33\cdots3}_{n\ 3's}}
\end{align}


问:
$X, Y, Z$ square-integrable rvs. Let that $Z$ is $\sigma(Y)$-measurable and $X$ is not $\sigma(Y)$-measurable. Prove the equivalence of statements
Let $X, Y, Z$ are square-integrable random variables. Let also that $Z$ is $\sigma(Y)$-measurable and $X$ is not $\sigma(Y)$-measurable.

Are the following statements equivalent?

$X-Z$ not correlate with any square-integrable $\sigma(Y)$-measurable random variable $W$

$X-Z$ is independent with any square-integrable $\sigma(Y)$-measurable random variable $W$

I tried to solve the problem through a conditional expectation, and used the fact that if $\mathbb E(X \mid Y) = const$ then $X$ and $Y$ are independent. It is also true in the opposite direction.
But it seems that this does not work in general, and I am at a dead end.

答:
This example is another form of idea from Peter Morfe's answer.

Put $Y = \pm 1$ with probability $\frac12$, $\xi$ is independent of $Y$ and has the same distribution, $X = \xi (Y+1)$, $Z=0$, $W_0 = Y$.

It's easy to see that $X-Z = X$ and $W_0 = Y$ are dependent: if $Y=-1$ we have $X = 0$ a.s. and if $Y=1$ then $X \ne 0$ a.s. Hence second statement is false.

But $$E(X-Z)W = EXW = E \xi Y W = E E(\xi Y W| Y) = E YW (E (\xi | Y)) = E YW (E \xi) = 0$$
and hence the first statement is true.

It follows that conditions 1 and 2 are not equivalent.


问:
If a sequence converges then the sequence is bounded?
I am taking an online real analysis class. In the class on sequences and series, the professor said the above statement is true since when a series converges to a point it is bounded in its neighborhood.

But 1/(n-2) is clearly a series that converges to 0 but is not bounded in the neighborhood of 2. (considering n to be a rational number). Since many pointed out that the series is not defined at 2. I am defining it to be 0 for n=2. So it seems like I am missing something here. Or is the above statement false?

Here is the link to the video https://www.youtube.com/watch?v=VEx3Ys6JAJo&list=PL0E754696F72137EC&index=15

答: Suppose $s_{n}$ converges to $s$. Let $\epsilon > 0$ be given. Then there exists $N$ such that $|s_{n} - s| < \epsilon$ for all $n \geq N$. This means $s_{n} \in (s - \epsilon, s + \epsilon)$ for all $n \geq N$.


问:
Prove convergence of a sequence and find it's limit if $x_{0} = 2$
I'm studying for my end of semester exam. And I'm having trouble understanding how to prove convergence of a sequence and finding it's limit.
$$x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) $$
I need to prove that this sequence converges and find it's limit if $x_{0} = 2$. $$$$ I had an idea to prove it by Cauchy's criteria, but I'm not sure how to do it here.

答:
Hint

Prove by induction that the sequence $\{x_n\}$ is decreasing and bounded below by one. Therefore it converges.

To one which is the only positive root of the equation

$$x=f(x) = \frac{1}{2}\left(x + \frac{1}{x}\right)$$ where $f$ is continuous.


问:
Is there a mathematical notation for restricting the depth of a factorial?
Say I only want to go $3$ "levels" deep for $5!$ and end up with just $5\times 4\times 3$. Is there a notation for this?

答:
Yes, this is called a "falling factorial":

$$
n^{\underline{3}} = n(n-1)(n-2)
$$
for example.

I pronounce $n^{\underline{3}}$ as "$n$ fall $3$".


答:
This is an iteration formula. (More specifically, it is Heron's Formula for approximating square roots). Iteration works by shrinking the distance between each successive $x_{n-1}, x_n$ to as close to $0$ as possible, so you can find it's limit by setting $x_{n}-x_{n-1}=0\implies x_n=x_{n-1}$. In this case, the limit is $\lambda=1$

To prove this is convergent, show that it is both decreasing (i.e. $x_{n}<x_{n-1}$, or simply that $\frac12(x+\frac 1x)<x$ when $x\in(1,2]$) and that $x_t>1$ for all $t$. With a decreasing sequence that is bounded below, we have convergence.

Title: Prove convergence of a sequence and find it's limit if $x_{0} = 2$

Body:
I'm studying for my end of semester exam. And I'm having trouble understanding how to prove convergence of a sequence and finding it's limit.
$$x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) $$
I need to prove that this sequence converges and find it's limit if $x_{0} = 2$. $$$$ I had an idea to prove it by Cauchy's criteria, but I'm not sure how to do it here.


答:
You seem to be confusing the definition of a sequence. A sequence is a countable list of real numbers (possibly finite or infinite). Thats it. It has a $1$ term, a $2$ term, a $3$ term, and so on.

When you say: what about the sequence $\frac{1}{n-2}$ for $n\in\mathbb{N}$, at $n=2$? The answer is that this is not a sequence. In fact, it is a sequence for $n\geq 3$, but you cannot call an undefined value as part of a sequence.

But you say, what about the sequence $\frac{1}{n-2}$ for all $n\in\mathbb{R}^+$ except for $n=2$? You are correct, this function is unbounded around $n=2$. However, a sequence takes as inputs natural numbers, not real numbers. Thus, what you have described is again not a sequence.

I think a main point you are misunderstanding is that generally, $n$ is taken to be a natural number. That is, $n\in\mathbb{N}$. It is sloppy notation to define a sequence as $a_n=\frac{1}{n-2}$ without also saying what happens at $n=2$. However, mathematicians will generally just ignore this undefined term (or let it be $0$).

But you say, what if you let $n$ run over all rational numbers except for $2$? Well, then what we are really doing is defining

$$a_n=\frac{1}{b_n-2}$$

where $b_n$ is any enumeration of the rationals except for $2$. But then this sequence does not converge. Let $M>0$ be given. Then there is some $k$ such that

$$b_k=\frac{4\lceil M\rceil+1}{2\lceil M\rceil}$$

Then

$$a_k=\frac{1}{b_k-2}=\frac{1}{\frac{4\lceil M\rceil+1}{2\lceil M\rceil}-2}=2\lceil M\rceil>M$$

Thus, the sequence does not converge. That is, the sequence definately converges (ignoring the issue at $n=2$) if $n$ goes over the naturals. However, if $n$ runs over some other countable set then the sequence may not converge.

Title: If a sequence converges then the sequence is bounded?

Body:
I am taking an online real analysis class. In the class on sequences and series, the professor said the above statement is true since when a series converges to a point it is bounded in its neighborhood.

But 1/(n-2) is clearly a series that converges to 0 but is not bounded in the neighborhood of 2. (considering n to be a rational number). Since many pointed out that the series is not defined at 2. I am defining it to be 0 for n=2. So it seems like I am missing something here. Or is the above statement false?

Here is the link to the video https://www.youtube.com/watch?v=VEx3Ys6JAJo&list=PL0E754696F72137EC&index=15


问:
Conversion of daily averages to an hourly average
I'm having some trouble with a basic question regarding Poisson distributions.

Question: the average rate of admission to the ICU is 48 patients per day, what is the probability of less than two patients being admitted in an hour?

With the daily $\lambda = 48$ and hourly $X \leq 1$ what would be the chosen distribution?

I don't think it would be right to assume an equal rate for hours and choose $\lambda = 2, X \leq 1$, then again even $\lambda = 48, X \leq 24$ seems too optimistic since it does not take into account hours where $X = 0$ (assumes an $X = 1$ for each hour).

答: With Poisson processes, you can scale up or down the time intervals as needed to get to the interval asked about in the problem. So you can convert your average 48 events per day to an average of 2 events per hour and use $ \lambda = 2 $.


问:
Conditional expectation- biased coin, fair coin and die
Suppose you have a fair six-sided die, one fair coin, and one biased coin. The biased coin has probability 0.25 of showing Heads. First you roll the die. If the number on the die is less than three, you then toss the fair coin until you see a Head; otherwise, you toss the biased coin until you see a Head. Compute the expectation of the number of tosses, giving your answer correct to two decimal places.

I have no clue how to approach this.any hints would be helpful

答:
If the probability of showing a Head (H) in a flip of a coin is $p$ then expected number of flips to see the first H is $\displaystyle \frac{1}{p}$.

Using that, find expected number of flips $\small E(F)$ for the fair coin and expected number of flips for the biased coin $\small E(B)$.

Now probability of die showing less than $3, \small P(X \lt 3)= \displaystyle \frac{2}{6}$ (die shows either $1$ or $2$)

So expected number of flips to see first H $ \small = P(X \lt 3) \times E(F) + (1-P(X \lt3)) \times E(B)$


答:
Alternatively, you could write
$$
\frac{5!}{2!} \, ,
$$
with the $2!$ cancelling out the unwanted terms. This is actually quite common. For example, it can be used to interpret the '$n$ choose $r$' formula:
$$
{n \choose r} = \frac{n!}{r!(n-r)!} \, .
$$
Say you have $n$ objects and you want to pick $r$ of them. Then, you have $n$ choices at first, then $(n-1), (n-2), \ldots ,$ and finally $(n-r+1)$. This gives us the
$$
\frac{n!}{(n-r)!}
$$
part of the formula. Then, the $r!$ removes duplicates.

Title: Is there a mathematical notation for restricting the depth of a factorial?

Body: Say I only want to go $3$ "levels" deep for $5!$ and end up with just $5\times 4\times 3$. Is there a notation for this?


答:
Here is a more fundamental way to prove:

First, we use the AM-GM to find the lower bound:
$$x_{n} = \dfrac{1}{2}\cdot\left(x_{n-1} + \dfrac{1}{x_{n-1}}\right) \ge \frac{1}{2}\cdot2\sqrt{x_{n-1}\cdot\frac{1}{x_{n-1}}} = 1$$
Now, we will show that it is decreasing:
$$
x_n - x_{n-1} = \frac{1}{2}\left(-x_{n-1} + \frac{1}{x_{n-1}}\right) = \frac{1-x_{n-1}^2}{2x_{n-1}} \le \frac{1-1}{2x_{n-1}^2} = 0 \\ x_n \le x_{n-1}$$
Hence, it is a decreasing sequence and lower bounded with $1$ which proves the convergence.

Now, let the limit of $x_n$ be $L$. Then, limiting the given recurrence we get
$$L = \frac{1}{2}\left(L + \frac{1}{L}\right) \implies 2L^2 = L^2 + 1 \implies L^2 = 1$$ Since $L$ must be positive, we have $L = 1$.


问:
How do I solve the following recursive equation?
$$\begin{cases}
    p(1) = \frac{1}{100} \\
    p(t) = p(t-1)\left(1+\frac{1}{101-t}\right)
\end{cases}$$

I feel like I could use the Telescoping Technique, but I do not quite see how.

答: Hint: Write it as a product and use $1+\frac{1}{101-n}=\frac{102-n}{101-n}$, then you can apply the telescoping technique.


答:
Here it is the standard proof of such result.

Proposition

Let $x_{n}\in\mathbb{R}$ be a sequence which converges to $x_{0}\in\mathbb{R}$, then it is bounded.

Proof

Let $\varepsilon = 1$. Then, according to the definition of limit, there exists $n_{1}\in\mathbb{N}$ such that
\begin{align*}
n\geq n_{1} \Rightarrow |x_{n} - x_{0}| < 1 \Rightarrow |x_{n}| - |x_{0}| < 1 \Rightarrow |x_{n}| < |x_{0}| + 1
\end{align*}

Thus, if we take $M = \max\{|x_{1}|,|x_{2}|,\ldots,|x_{n_{1}-1}|,|x_{0}| + 1\}$, we can conclude that $|x_{n}| \leq M$.

That is to say $x_{n}$ is bounded, and we are done.

Hopefully this helps!