问:
Measure vs outer measure
Basically I feel like I'm having a chicken before egg problem.

An outer measure is a set function defined on some set X with the properties that empty returns 0, of sub additivity and of monotonicity.
From this outer measure we get a sigma algebra of measurable sets.
On this sigma algebra the outer measure is a measure with the property of additivity for unions of disjoint sets.

But we can also define a measure , a set function, on an algebra of sets with the properties that empty returns zero and of additivity of unions of disjoint sets. Then via caratheodory construction we can make an outer measure from the measure for the whole set.

So what comes first, or does neither come first?

Note: that this all began with a sample final question that I'm still stuck on that asked "if a measure is defined on the whole power set is it also an outer measure?" Which is why I marked this as homework.

Note 2: also is the answer to the question the following: Any measure can be transformed into an outer measure via caratheodory?

Thanks.

Edit: deleted the bit that is talked about below

答:
Here is standard way for constructing measures on a measurable space $(S,\mathcal{S})$ that can for example be used to construct Lebesgue measure; various closely related approaches exist. For simplicity, we look at finite measures only.

You start with an algebra $\mathcal{A}$ on $S$ such that $\mathcal{S}$ is the smallest $\sigma$-algebra that contains all elements of $\mathcal{A}$. You then define a finite, nonnegative and finitely additive set-function $\mu_0$ on $\mathcal{A}$ that is countably additive in the sense that if $(A_n)$ is a sequence of disjoint sets in $\mathcal{A}$ and $\bigcup_n A_n\in\mathcal{A}$, then $\sum_n \mu_0(A_n)=\mu_0\big(\bigcup_n A_n\big)$. This assumption is easier to check than for $\sigma$-algebras since it might well be the case that $\bigcup_n A_n\notin\mathcal{A}$.

Then you define an outer measure $\mu^*$ on all of the powerset $2^S$ by letting $$\mu^*(B)=\inf\bigg\{\sum_{A\in\mathcal{C}}\mu_o(A):\mathcal{C}\textrm{ is a countable subset of }\mathcal{A}\textrm{ such that }B\subseteq\bigcup_{A\in\mathcal{C}}A\bigg\}.$$
Let $$\mathcal{S}_\mu=\Big\{B\in 2^S:\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap S\backslash A)\textrm{ for all }A\in 2^S\Big\}.$$
It turns out that $\mathcal{S}_\mu$ is a $\sigma$-algebra and the restriction of $\mu^*$ to it, $\mu^*|\mathcal{S}_\mu$, is a measure. Moreover, $\mathcal{S}_\mu\supseteq \mathcal{S}\supseteq\mathcal{A}$, so $\mu=\mu^*|\mathcal{S}$ is a measure on $\mathcal{S}$ and moreover, $\mu^*|\mathcal{A}=\mu_0$.

Title: Measure vs outer measure


问:
What would be "tiefliegendes Ergebnis" in English?
In German textbooks one sometimes finds the term "tiefliegendes Ergebnis" for theorems whose proofs require strong methods. The literal translation would be "deep lying result" but I'm pretty sure that I have never read this phrase in any English textbook.

So does anyone know a correct translation?

答:
The word 'profound' works for both the figurative and literal meaning of 'tiefliegend':

1a : having intellectual depth and insight

  b : difficult to fathom or understand

2a : extending far below the surface

  b : coming from, reaching to, or situated at a depth

(source: Merriam Webster)

So 'profound result' may be what you're looking for.

ContentLicense: CC BY-SA 4.0


问:
How to define a non-abelian group over the set of integers $S=\{0, \cdots, 2^{128}-1\}$ with no commutativities nor self-inverse elements?
I am trying to solve a problem since a while, adding the remaining constraints along the process of learning the concepts. It's been a long journey.

The best approach (for the weak version of the problem) was nicely given in this answer here.

The remaining question seems to be how to ensure, if possible, no commutativities and no self-inverse elements - except the identity element, obviously.

The unitriangular approach is good to fit exactly the order needed for the group, but has many self-inverses due to symmetries along the secondary diagonal and probably due to other causes also.

The composition of permutations seems more resilient to such problems, but the group order does not fit exactly into $2^{128}$. It is either $34!$ or $35!$.

So, I am still in doubt if I am missing something obvious that a seasoned mathematician would spot right on. Is such problem trivial? Is it impossible?

答:
If $S$ is a group with $128$ elements, then it has an element of order $2$, by Cauchy theorem (as Berci already noticed), that is a non trivial self-inverse element.

Worse, it is known that the  center $Z(S)$ is non trivial. Hence $S$ has a non trivial element which commutes with every element of $S$. Applying Cauchy theorem to $Z(S)$ shows furthermore that you have a nontrivial self-inverse elements which commutes with all elements of $S$.


问:
How can I prove this inequality? $c\leq x\leq a+b$

Show that if ABC is a right triangle and curve $x$ is inside the ABC triangle then we have

$$c\leq x\leq a+b$$

Do not use integral.

答:
The inequality $c\leq x$ is just the statement "the shortest distance between any two points is a straight line." The upper bound $x\leq a+b$ is harder.

So far as I know, there isn't an elementary geometric argument. Suppose the graph curve is rectifiable (let's say with arclength $s$ rather $x$), and let $\gamma(x) = (x,f(x))$ parameterize the curve. Given $\epsilon>0$, the definition of rectifiability implies there are points $0=x_0<x_1<\dotsm< x_n=b$ such that
$$s-\epsilon<\sum_{i=0}^{n-1}d(\gamma(x_i),\gamma(x_{i+1}))$$
where $d$ is distance in $\mathbb{R}^2$. By triangle inequality we have
$$d(\gamma(x_i),\gamma(x_{i+1})) = \sqrt{(x_i-x_{i+1})^2+(f(x_i)-f(x_{i+1})^2}\leq |x_i-x_{i+1}|+|f(x_i)-f(x_{i+1})|.$$
Moreover, a continuous injective function is monotone (and hence decreasing in our case), so $|x_i-x_{i+1}|=x_{i+1}-x_i$ and $|f(x_i)-f(x_{i+1})| = f(x_i)-f(x_{i+1})$. Thus
$$s-\epsilon < \sum_{i=0}^{n-1}d(\gamma(x_i),\gamma(x_{i+1}))\leq\sum_{i=0}^{n-1}(x_{i+1}-x_i)+\sum_{i=0}^{n-1}\big(f(x_i)-f(x_{i+1})\big) = b+a.$$
Therefore $s<a+b+\epsilon$. Since $\epsilon>0$ was arbitrary in our argument, we conclude $s\leq a+b$, as desired.

As an aside, I believe the upper bound should be achievable with a construction similar to that of the Cantor function (whose graph in the unit square is nondecreasing and has length 2).


问:
Prove $E[\frac{1}{1+e^{t X}}] = 1/2$ for $X\sim N(0,1)$.
I'm trying to evaluate an expectation of the form
$$
E\left[\frac{\exp(t X_1)}{\sum_{i=1}^n \exp(t X_i)}\right],
$$
where each $X_i$ is independent standard normally distributed.

In the specific case $n=2$ this can be rewritten as
$$E\left[\frac{1}{1+e^{\sqrt{2} t X}}\right],$$
which numerically appears to be equal to $1/2$ for all real $t$.

I wonder if that might be possible to prove?, and if it could shed some light on the original integral?

If I series expand I get the sum
$$E\left[\frac{1}{1+e^{t X}}\right]
=\sum_{k\ge 0}(-1)^k e^{(kt)^2/2},
$$
which Mathematica evaluates as $\frac{1}{2} \left(\vartheta _4\left(0,e^{\frac{t^2}{2}}\right)+1\right)$.
I don't really know what this EllipticTheta[4, 0, E^(t^2/2)] function is, but if it is 0 it at least fits with my conjecture.

答: Let $M = E\left[\frac{\exp(t X_1)}{\sum_{i=1}^n \exp(t X_i)}\right]$.  Since $X_j$'s are IID, so $M = E\left[\frac{\exp(t X_j)}{\sum_{i=1}^n \exp(t X_i)}\right]$ for all $j = 1,\dots,n$.  Take summation over $j = 1,\dots,n$.  We get $nM = E[1] = 1$, so $M = 1/n$.


问:
How to find the differential of this implicit function to find the equation of the tangent?
$$x^2y^3 + xy^2 + y = 3$$

I am supposed to find the equation of the tangent to this curve. I have gotten my implicit differential answer but I dont think that I am correct. I know the method and concept of doing this but I just wish to confirm my answers here. I got:

$y^3 + 2x$ as the differential (derivative)....is that correct?

Body:
$x^2y^3 + xy^2 + y = 3\\
\frac {d}{dx}(x^2y^3 + xy^2 + y) = \frac {d}{dx}3\\
(2xy^3 + y^2) + (3x^2y^2 + 2xy + 1)\frac {dy}{dx} = 0\\
\frac {dy}{dx} = -\frac {2xy^3 + y^2}{3x^2y^2 + 2xy + 1}$


问:
Wrong way of using the squeeze rule

In the first part of the question, I was asked to prove $\lim_{x\to ∞} f(x)={1\over 2}$ using the squeeze rule.
But I did it in a much different way than "Student A" and I fail understand what exactly he did, and where he did wrong.. any thoughts?

答: The fallacy is that the number of terms is neither constant nor bounded, so a proof by induction on $n$ that $\lim_{x\to c}\sum_{i=1}^nf_i(x)=\sum_{i=1}^n\lim_{x\to c}f_i(x)$ doesn't extend in the way Student A hoped. The correct way uses $f(x)=\tfrac{\lfloor x\rfloor(\lfloor x\rfloor+1)}{2x^2}\in\left(\tfrac{x(x-1)}{2x^2},\,\tfrac{x(x+1)}{2x^2}\right]$.

Title: Wrong way of using the squeeze rule


问:
What's more robust than a structural homomorphisms?
This question isn't category theory; but, category theoreticians tend to be interested in mathematical structure, so I thought the answer might exist within that knowledge base.

Given two mathematical structures $X$ and $Y$ with the same pattern of airities, there is a natural notion of homomorphism $X \rightarrow Y$, described here, called a structural homomorphism. However, this notion lacks "robustness."

For instance, suppose that $X$ and $Y$ are partially ordered sets and that $f$ is a structural homomorphism $f : X \rightarrow Y$. Then $f$ is an order homomorphism. Suppose also that $X$ and $Y$ happen to be lattices. Then we can add meets and joins to the data of $X$ and $Y$, thereby obtaining new structures $X'$ and $Y'$. However, its feasible that $f$ might fail to be a homomorphism $X' \rightarrow Y'$, since not every order homomorphism is a lattice homomorphism.

So in general, if $f$ is a structural homomorphism $X \rightarrow Y$, and we extend $X$ and $Y$ by defining new relations/operations in terms of the old ones, thereby obtaining new structures $X'$ and $Y'$, well even if the new relations/operations are defined in terms of the old ones, using exactly the same definitions, nonetheless $f$ may fail to be a homomorphism $X' \rightarrow Y'$.

Now I originally thought that the notion of a "structural embedding" described here in that same article, might be robust with respect to the defining of new relations/operations. But that's completely wrong.

So my question is, what's more robust than a structural homomorphism?

答:
Note that the example given by Zev in the previous thread is a structural embedding. So the same example shows that embedding between posets need not preserve meet/join.

For a stronger notion you should probably talk about theories as well, and require the embedding to preserve truth values of sentences, in the sense of an elementary embedding. Then you can extend definable operations from one structure to the next.

ContentLicense: CC BY-SA 3.0


问:
Does probability change with time?
My question is will probability distribution change with time if some function is affected somehow by time. And in what way it will change.
e.g. consider i know a person who checks his messages at least after half an hour but there is also chance that he will check them at any moment in between half hour interval. If i send a message at random time without knowing when last time he checked it, how will probability of seeing message change at any instant or for seconds?</p>

答:
The short answer for your specific constraint is:  No

In your example, you make a single claim regarding time:  "person who checks his messages at least after half an hour".    So in this case, the probability that the message will be checked in the next 30 minutes is 1.

You give no basis for calculating any probability of the time dependent action variable "...if they have time...", so you have nothing to use.  The only time aspect you can calculate for your scenario is the probability of when within the 30 minute repeat cycle the message is received, which are all equally likely.

Certainly there are probability distributions that have time as a variable, but this occurs only when the time variable is well defined, just like any other domain quantity, something which your example fails to do.   You are basically asking "I pick a bead from a totally unknown quantity of beads, how does the probability of picking a white bead change when the total number of beads changes?"

Title: Does probability change with time?


问:
Analytic continuation and singularities
Suppose that $z\in\mathbb{C}$ and $f$ is a holomorphism on $U$ such that there is no holomorphism $g$ extending $f$ on  $V\supseteq U$ such that $z\in V$. Is it possible for there to be such a $g$ and a holomorphism $h$ on $W$ with $V\cap W\neq\emptyset$ such that $g|_{V\cap W}=h|_{V\cap W}$ with $z\in W$? That is, are singularities preserved by analytic continuation?

答: The answer to radius of convergence with Taylor series, functions with branch cuts is related, and indeed its negative answer can be easily seen to imply a negative answer to this question. The global analytic function $\frac{1}{\log(z)}$ is given as a counterexample. Since $\log(z)=0$ in only one branch, there is a singularity at $0$, but also a disappearing one.

Title: Analytic continuation and singularities


答:
The problem lies in the number of terms.
We have
$$\lim_{x\to\infty}f(x) = \frac{1}{x^2} + \cdots + \frac{\lfloor{x}\rfloor}{x^2} \leq \lim_{x\to\infty} ( \frac{1}{x} + \frac{1}{x} + \cdots + \frac{1}{x}),$$
where the number of terms in the parenthesis is $\lfloor{x}\rfloor$.

This is equal to
$$\lim_{x\to\infty}\sum_{i=1}^{i=\lfloor{x}\rfloor} \frac{1}{x},$$
and the lim cannot go through the sum, since the sum itself is not bounded and depends on $x$. You can still compute the limit and find that it is equal to
$$
\lim_{x\to\infty}\sum_{i=1}^{i=\lfloor{x}\rfloor} \frac{1}{x} = \lim_{x\to\infty}\frac{\lfloor{x}\rfloor}{x} = 1.
$$

Title: Wrong way of using the squeeze rule

Body:
In the first part of the question, I was asked to prove $\lim_{x\to ∞} f(x)={1\over 2}$ using the squeeze rule.
But I did it in a much different way than "Student A" and I fail understand what exactly he did, and where he did wrong.. any thoughts?


问:
Book on quadric surfaces with linear algebra
Most information that I can find about quadric surfaces is written from a calculus perspective - without using any matrices or vectors. However, I would like to have a reference that tells me the different representations for a quadric surface and also some background on why these representations are used. So, e.g., I want to see both the vector representation(s) and the scalar representation(s) of a cone. Are there any books that cover this?</p>

答: If you can find a copy, have a look at: "Solid Analytical Geometry and Determinants" by Arnold Dresden, c.1930, published by John Wiley and Sons. I have a copy of the Fourth Printing, 1948. (No ISBN in my copy.) </p>


问:
Sum of two closed sets in $\mathbb R$ is closed?
Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?

(By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)

Thanks!

答:
It's worth mentioning that :

if one is closed + bounded, another one is closed,then the addition is closed

Since closedness can be charaterized by sequence in $\Bbb{R}^n$,if $(x_n) \in A+B$  we need to show limit of the convergence sequence still lies in it.assume $A$ is compact $B $ is closed.

Since $x_n= a_n +b_n \to x$,compactness implies sequential compactness hence $a_{n_k} \to a\in A$  for some subsequence. now $x_{n_k} \to x$ which means subsequence $b_{n_k}\to x-a$ converge,since $B$ is closed,$x-a \in B$ ,hence $x = a+b \in A+B$,which means the sum is closed.


问:
Can a polynomial function limited by x axis values be convex?
Consider the given set:

$$ S = x ∈ E^2: x_2 − x_1^2 = 0, −1 \le x1 \le 1 $$

When I draw the set I find that it is a polynomial where the $$x_1$$ axis is cut at -1 and 1. The max value for $$x_2$$ is 1. It seems like a convex set to me but the book says it is not. Why is it not a convex set?

答: So, $a:=(\frac{1}{4},\frac{1}{2}) \in S$ and $b:=(\frac{-1}{4},\frac{1}{2}) \in S$ but $\frac{1}{2}(a+b) = (0, \frac{1}{2})$ which is not fulfilling $x_2-x_1^2=0$, hence not in $S$. So $\frac{1}{2}(a+b)$ is not in $S$, which renders $S$ is not konvex.


问:
Prove $ \sin ^{2} \frac{\pi}{x} $not uniform convergence
Negation of uniform convergence
$$
(\exists \varepsilon>0)(\forall N)(\exists x \in S)(\exists n>N)\left(\left|f_{n}(x)-f(x)\right| \geq \varepsilon\right)
$$
Let $f_{n}(x)=\left\{\begin{array}{l}0 \quad(x<1 /(n+1)) \\ \sin ^{2} \frac{\pi}{x} \quad(1 /(n+1) \leq x \leq 1 / n) \\ 0 \quad(1 / n<x)\end{array}\right.$

I need to prove $\left\{f_{n}\right\}$ converges to 0, but not uniformly. I've already proved it converges to 0.

My attempt:
This is not uniform convergence as $\exists \varepsilon=1, \forall n \in \mathbb{N}$ $\exists x=\frac{2}{2 n+1},\left|f_{n}(x)-0\right|=$ $\sin ^{2}\left(\frac{\pi}{\frac{2}{2 n+1}}\right)=\sin ^{2}\left(\frac{\pi(2 n+1)}{2}\right)=1$

However, Negation of uniform convergence should be $(\forall N\in\mathbb{N})$$(\exists n>N)$. I wonder if directly saying $\forall n \in \mathbb{N}$ is equivalent to $(\forall N\in\mathbb{N})(\exists n>N)$. Thanks

答:
To highlight the difference between $\forall n \in \mathbb{N}$ and $(\forall N\in\mathbb{N},\exists n>N)$, let $P(n)$ be the proposition "this problem solved at time $n$".

$\forall n \in \mathbb{N}, P(n)$ means that it's already solved at the very beginning $n = 1$.

$(\forall N\in\mathbb{N},\exists n>N: P(n))$ means that for every moment $N$, you think that "it can be  solved later OR it's already solved".  It might not be solved at the beginning $n = 1$.

Evening though (1) looks stronger than (2), in the context of proving uniform convergence, up to re-indexing of the sequence of functions $\{f_n\}_n$, it's possible to rephrase (2) as (1).  Note that in the definition of uniform convergence, $n$ is a function of $N$, so you may write $(n_N)_N$.  To skip "$\exists n > N$", you may

replace $\left|f_{n}(x)-f(x)\right|$ with $\left|f_{n_N}(x)-f(x)\right|$

replace $\{f_n\}_n$ with its subsequence $\{f_{n_N}\}_N$.  You may rename it to $\{g_N\}_N$ to make things clear.

Then the whole thing becomes
$$(\exists \varepsilon>0)(\forall N)(\exists x \in S)(\exists n>N)\left(\left|g_{N}(x)-f(x)\right| \geq \varepsilon\right).$$

As $n$ has no role in the last part of the above statement, you can safely cross out "$\exists n > N$", giving you a statement in (1) form.


问:
using Liouville-Ostrogradsky to solve differential equation
I have a differential equation:
$$y''+p_1\left(x\right)y'+p_0\left(x\right)y=0$$
for $x>0$ and it is given that $p_0,p_1$ are continuous for any $x\in \mathbb{R}$.

I am questioned if $y_1(x)=x$ and $y_2(x)=\ln(x)$ can be the solutions of this equation.

Answer:
They are linear independent, therefore using Liouville-Ostrogradsky:
$$W(y_1,y_2)(x)=ce^{-\int \:p_1\left(x\right)dx}$$
we get:
$$W(y_1,y_2)(x)=\left|\begin{pmatrix}x&\ln\left(x\right)\\ 1&\frac{1}{x}\end{pmatrix}\right|=\ln\left(x\right)-1$$
so, we can choose c=1 for comfort and get:
$$\ln\left(x\right)-1=e^{-\int p_1\left(x\right)dx}$$
therefore:
$$\int p_1\left(x\right)dx=\frac{1}{\ln\left(\ln\left(x\right)-1\right)}$$
Therefore, after deriving both sides:
$$p_1\left(x\right)=\frac{1}{x\cdot \left(\ln\left(x\right)-1\right)\cdot \ln^2\left(\ln\left(x\right)-1\right)}$$

Now the thing that bothers me, that is was said about $p_1$ that it is continuous for $x\in \mathbb{R}, x>0$, and here the denominator is defined only for any $x>e$, so I don't know if the authors of the question meant that it should be continuous for any $x>e$ and then we can continue and find $p_0$ or it contradicts the given statement and therefore those two functions cannot be a solution for the equation.

Am I right that those are the two options of answers for this diff. equation ?
If it is about $x>e$, then we can choose $$y=x+\ln(x)$$, calculate $y'$ and $y''$, put it all together in the first formula: $y''+p_1\left(x\right)y'+p_0\left(x\right)y=0$ and extract $p_0$.

Otherwise if the question means that $p_1(x)$ has to be continuous on $x>0$, then $p_1(x)$ doesn't satisfies that because it is defined only for $x>e$, is this correct ?

答:
If $t_1,y_2$ are two linearly independent solutions of a second order ODE. This ODE is given by: $$\begin{vmatrix} y'' & y'& y \\ y_1'' & y_1' & y_1 \\ y_2'' & y_2' & y_2 \end{vmatrix}=0$$ Here we have$$\begin{vmatrix} y'' & y'& y \\0 & 1 & x \\ -1/x^2 & 1/x & \ln x \end{vmatrix}=0$$
$$\implies (\ln x-1)y''-\frac{1}{x}y'+\frac{1}{x^2}y=0$$
So $$p_1(x)=(x(1-\ln x))^{-1}, p_0(x)=(x^2(\ln x-1))^{-1}$$


问:
Finding maximum value of perimeter of trapezium using tan θ
I study maths as a hobby. I am struggling with this problem.

In a trapezium ABCD the sides AB, DC are parallel and $\angle ABC$ is a right angle. If AD = DB = 10 cm and $\angle DAB = \theta$ find the value of $\tan \theta$ for which the perimeter p cm is a maximum. Hence find the range of values taken by p for $0\lt\theta\lt\frac{1}{2}\pi$

My working is as follows:

$\tan \theta = \frac{DE}{EA} = \frac{CB}{EA}$

$CB = EA.\tan \theta$

$EA = 10 \cos \theta \to CB = 10 \cos \theta \tan \theta$

$ABD$ is isosceles so $BE = EA$

$CD = EA = 10 \cos \theta$

$BA = 20 \cos \theta$

So perimeter P = $20 \cos \theta + 10 \cos \theta \tan \theta + 10 \cos \theta + 10$

$= 30 \cos \theta + 10 cos \theta \tan \theta + 10$

$\frac{dP}{d\theta} = - 30 \sin \theta + 10 \cos \theta \sec^2 \theta - 10 \sin \theta \tan \theta$

But now I am stuck.

答:
Once you have expressed the perimeter as
$$P(\theta) = 30\cos \theta + 10\sin \theta + 10,$$
you can even avoid calculus by just rewriting $P$ as
\begin{eqnarray}P&=&10 \left[3\cos\theta + \sin \theta + 1\right]=\\
&=&10 \left[\sqrt{10}\left(\frac{3}{\sqrt{10}} \cos\theta + \frac1{\sqrt{10}}\sin\theta\right)+1\right]=\\
&=&10\left[\sqrt{10}\sin\left(\theta +\arctan 3 \right)+1\right],
\end{eqnarray}
which reaches its maximum when the argument of $\sin$ is $\frac{\pi}2$, yielding
$$\theta_{max} = \frac{\pi}2 - \arctan 3= \arctan \frac13$$


问:
Relationship between representations over $\mathbb R$ and over $\mathbb C$
Let $\rho:G\rightarrow \text{GL}(V)$ be a representation of dimension $n$ over $\mathbb C$. Denote by $V^\mathbb R$ the space $V$ considered as a vector space over $\mathbb C$ of dimension $2n$. Denote by $\rho^\mathbb R$ the representation of $G$ in $V^\mathbb R$.

How do I show that $\chi_{\rho^\mathbb R}=\chi_\rho+\bar{\chi}_\rho$?

I mainly tried to prove $\rho^\mathbb R\cong\rho\oplus\bar{\rho}$ but with no luck.

I'm not quite sure how to approach it, but I sense it is not a difficult one..

答:
Asssume $\{v_1,...,v_n\}$ is a basis of $V$ over $\mathbb{C}$. Then $\{v_1, iv_1,...,v_n,iv_n\}$ is a basis of $V^{\mathbb{R}}$ over $\mathbb{R}$. For every $1\leq j\leq n$ let:

$\rho(g)(v_j)=c_{j1}v_1+c_{j2}v_2+...+c_{jn}v_n$

Then $\chi_{\rho}(g)=\sum_{j=1}^n c_{jj}$ by definition.

Now, write $c_{jk}=x_{jk}+iy_{jk}$. For every $j$ we have:

$\rho^{\mathbb{R}}(g)(v_j)=\rho(g)(v_j)=x_{j1}v_1+y_{j1}(iv_1)+...+x_{jn}(v_n)+y_{jn}(iv_n)$

$\rho^{\mathbb{R}}(g)(iv_j)=\rho(g)(iv_j)=i\rho(g)(v_j)=i(x_{j1}v_1+y_{j1}(iv_1)+...+x_{jn}v_n+y_{jn}(iv_n))=$

$=(-y_{j1})v_1+x_{j1}(iv_1)+...+(-y_{jn})v_n+x_{jn}(iv_n)$

In other words, the $v_j$-coefficient of the vector $\rho^{\mathbb{R}}(g)(v_j)$ is $x_{jj}$, and the $iv_j$-coefficient of the vector $\rho^{\mathbb{R}}(g)(iv_j)$ is also $x_{jj}$. So the trace of $\rho^{\mathbb{R}}(g)$ is:

$\sum_{j=1}^n 2x_{jj}=\sum_{j=1}^n 2Re(c_{jj})=\sum_{j=1}^n (c_{jj}+\overline{c_{jj}})=\chi_{\rho}(g)+\overline{{\chi_{\rho}}(g)}$


答: You could short-circuit the task from the start. Any solution, if not artificially restricted, has the whole of $\Bbb R$ as domain, as the coefficients are assumed as continuous everywhere. With the restriction to the positive half axis, this still gives a finite limit for $x\to0$, as in the extended solution this also is a point of continuity. Thus you can not get a solution $y=\ln x$ or $y=\ln|x|$ with its singularity at $x=0$.

Title: using Liouville-Ostrogradsky to solve differential equation

Body:
I have a differential equation:
$$y''+p_1\left(x\right)y'+p_0\left(x\right)y=0$$
for $x>0$ and it is given that $p_0,p_1$ are continuous for any $x\in \mathbb{R}$.

I am questioned if $y_1(x)=x$ and $y_2(x)=\ln(x)$ can be the solutions of this equation.

Answer:
They are linear independent, therefore using Liouville-Ostrogradsky:
$$W(y_1,y_2)(x)=ce^{-\int \:p_1\left(x\right)dx}$$
we get:
$$W(y_1,y_2)(x)=\left|\begin{pmatrix}x&\ln\left(x\right)\\ 1&\frac{1}{x}\end{pmatrix}\right|=\ln\left(x\right)-1$$
so, we can choose c=1 for comfort and get:
$$\ln\left(x\right)-1=e^{-\int p_1\left(x\right)dx}$$
therefore:
$$\int p_1\left(x\right)dx=\frac{1}{\ln\left(\ln\left(x\right)-1\right)}$$
Therefore, after deriving both sides:
$$p_1\left(x\right)=\frac{1}{x\cdot \left(\ln\left(x\right)-1\right)\cdot \ln^2\left(\ln\left(x\right)-1\right)}$$

Now the thing that bothers me, that is was said about $p_1$ that it is continuous for $x\in \mathbb{R}, x>0$, and here the denominator is defined only for any $x>e$, so I don't know if the authors of the question meant that it should be continuous for any $x>e$ and then we can continue and find $p_0$ or it contradicts the given statement and therefore those two functions cannot be a solution for the equation.

Am I right that those are the two options of answers for this diff. equation ?
If it is about $x>e$, then we can choose $$y=x+\ln(x)$$, calculate $y'$ and $y''$, put it all together in the first formula: $y''+p_1\left(x\right)y'+p_0\left(x\right)y=0$ and extract $p_0$.

Otherwise if the question means that $p_1(x)$ has to be continuous on $x>0$, then $p_1(x)$ doesn't satisfies that because it is defined only for $x>e$, is this correct ?


问:
Basic maths question
If I have 25 apples and multiply them zero times, why is it that I end up with zero? I understand zero times means no times, right? I mean, multiply by zero is no multiplication. Can anyone explain, please?</p>

答:
Put it in a simple way. We know multiplication is just repeated addition. Then,

$$ 25\cdot0 = \underbrace{0+0+\dots + 0}_{25\text{ zeroes}} = 0 $$

In an apple context, this means trying to add up an amount of zero apples. You won't get any apples, because there aren't any apples to add up.

Title: Basic maths question


问:
How to evaluate $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?
How can I evaluate  $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?</p>

答:
I guess someone should mention the Taylor approximation approach:
$$\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right) = \sqrt{x}\left(\sqrt{1+ {1 \over \sqrt{x}}}-\sqrt{1- {1 \over \sqrt{x}}}\right) $$
$$= \sqrt{x}\bigg(\big(1 + {1 \over 2\sqrt{x}} + O({1 \over x})\big) - \big(1 - {1 \over 2\sqrt{x}} + O({1 \over x})\big)\bigg)$$
$$=\sqrt{x}\bigg({1 \over \sqrt{x}} + O({1 \over x})\bigg)$$
$$= 1 + O({1 \over \sqrt{x}})$$
So the limit is $1$.</p>


问:
How to find probability in when using an "or" statement
Can somebody please help me solve/understand the following problem?:

Consider two random variables $X$ and $Y$ with events $S_{X}=S_{Y}=\{1,2,3,4\}$ and probabilities:

$P(X=j)=\frac{1}{4} \quad$ for all $j \in \{1,2,3,4\}, \quad$ and $\quad P(Y=j)=\left\{\begin{array}{ll}\frac{1}{8}, & \text { if } j=1 \\ \frac{1}{4}, & \text { if } j=2 \\ \frac{3}{8}, & \text { if } j=3 \\ \frac{1}{4}, & \text { if } j=4\end{array}\right.$

Assume, that $X$ and $Y$ are independent.

Find the probablity $P(X \leq 2$ or $Y \leq 2)$.

First I tried to solve it like this:

$P(x \leq 2$ or $y \leq 2)=P(x \leq 2)+P(y \leq 2)=\frac{7}{8}$

But I realised that this method would give a probability higher than 1 if I considered $x\leq4$ or $y\leq 4$. So how do I solve this correctly, and why is this method wrong?

One thing, that I'm not sure of, is whether I should still count the possibilities when X and Y are $\leq2$ at the same time? Or if it can only be one of them at a time, for the statement to be true?

Secondly, I found this theorem in my book, that might be helpful (see below). Specifically, I think a) could be used? But I just can't seem to "translate" the theorem to this problem. The first problem is, that my theorem only considers one random variable, but my case has two. Secondly, I'm not certain how to use the union and intersection of sets in this problem. But if it's possible to use this theorem, I would really appreciate you helping me understanding how.

Thanks in advance

Theorem: Let $X$ be a random variable with the possible events $S_{X},$ and let $A$ and $B$ be subsets of $S_{X}$. Then the following rules can be used:

a. $P(X \in A \cup B)=P(X \in A)+P(X \in B)-P(X \in A \cap B)$

b. $P(X \in A \cup B) \leq P(X \in A)+P(X \in B)$.

c. If $A$ and $B$ are disjoint, then

$
P(X \in A \cup B)=P(X \in A)+P(X \in B)
$

d. $P(X \notin A)=P\left(X \in S_{X} \backslash A\right)=1-P(X \in A)$

e. If $A_{1}, A_{2} \ldots, A_{n}$ are disjoint subsets of $S_{X},$ then

$
P\left(X \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=P\left(X \in A_{1}\right)+P\left(X \in A_{2}\right)+\cdots+P\left(X \in A_{n}\right)
$

答:
You cannot use the theorem as it is, as for using the additivity you tried to use you have to know that $A$ and $B$ are disjoint, which you do not know.

You can use the fact that for independent variables $\mathbb{P}(A\cap B) = \mathbb{P}(A) \mathbb{P}(B)$. With $A^c$ denoting the complement of $A$ you can write $\mathbb{P}(A \cup B) = \mathbb{P}(((A\cup B)^c)^c)= \mathbb{P}((A^c\cap B^c)^c) = 1- \mathbb{P}(A^c\cap B^c)$.

If you just plug in $A= \{X\leq 2\}$ and $B= \{Y \leq 2\}$ you can calculate the answer.


问:
solution of GBM doubt
giving the dynamics of a GBM: $dX_t=\alpha X_tdt+\sigma X_tdW_t$, I know that the solution $X_t=X_0 e^{(\alpha-\frac{1}{2}\sigma^2)t+\sigma W_t}$ is given by ito applied to $\ln(X_t)$. My question is: Where is the mistake if I compute the solution in this other way:

$\frac{dX_t}{X_t}=\alpha dt +\sigma dW_t$

$\int_0^t\frac{dX_t}{X_t}=\int_0^t\alpha dt +\int_0^t\sigma dW_t$

$[\ln(X_t)]_0^t=\alpha t+\sigma W_t$

$\ln(\frac{X_t}{X_0})=\alpha t+\sigma W_t$

$X_t=X_0 e^{\alpha t+\sigma W_t}$

答:
For an Ito integral

$$\int_0^t \frac{dX_t}{X_t} \neq [\ln(X_t)]_0^t.$$

You need to apply Ito's lemma here.

Title: solution of GBM doubt


问:
CNF Rule hierarchy discovery
This is bothering me for some time. Consider that I have a set of CNF formulae:

$F_1 = \left( A \lor B \lor C \right) \land \left( C \lor D \lor E \right) \land \left( B \lor F \lor G \right)$

$F_2 = \left( B \lor F \lor G \right)$

$F_3 = \left( A \lor B \lor D \right)$

Now, given the values (T/F) of the literals ($A$, $B$, $\cdots$), I wish to evaluate these formulae.

However, the point is that, if we observe closely, we can see that formula $F_1$ subsumes formula $F_2$ (i.e., while evaluating $F_1$, I will be automatically evaluating $F_2$). If I evaluate $F_1$ first followed by $F_2$, I will be unnecessarily repeating the efforts (since I already evaluated the $3^{rd}$ clause in $F_1$, I could have used that result for $F_2$, if I had some way of knowing it). Again, in case of $F_1$ and $F_3$, they do share some parts of the $1^{st}$ clause.

So, the question is, whether I can re-use the work done while performing this evaluation, by discovering the relationships (or hierarchy) of these CNF rules. I would like some scheme which tells me to evaluate $\left(A \lor B \right)$, use that for $F_1$ and $F_3$, tells me to evaluate $F_2$ before $F_1$ and directly use that result while evaluating $F_1$ (and so on...)

Is anyone aware of such problems? I know concepts such as Junction Trees in machine learning, Memoization in DP, or data structures like Trie which loosely achieve the same, but I am not able to fit my problem to these formulations. Any help would be greatly appreciated.

Thanks,

Salil

(PS: hoping that I did not post this on the wrong forum on SE)

答:
This question has received an answer on CSTheory.SE by Vijay D, which I include below. Primary intent of this post is to remove this question from the Unanswered queue.

Efficient CNF Simplification based on Binary Implication Graphs, Marijn Heule, Matti Jarvisalo, and Armin Biere, 2011

"This paper develops techniques for efficiently detecting and removing redundancies from CNF (conjunctive normal form) formulas based on the underlying binary clause structure (i.e., the binary implication graph) of the formulas.

In addition to considering known simplification techniques (hidden tautology elim- ination (HTE), hyper binary resolution (HBR), failed literal elimination over binary clauses, equivalent literal substitution, and transitive reduction  of the binary implication graph), we introduce the novel technique of hidden literal elimination (HLE) that removes so-called hidden literals from clauses without affecting the set of satisfying assignments."

I recommend Section 2.1 of the paper which reviews known simplification techniques.

Title: CNF Rule hierarchy discovery


问:
Solving Prop 86 Euclid's Data
I was given the following problem:

Solve the equations of Prop-86 of the Data algebraically. Show that the two hyperbolas defined by the equations each have their axes as the asymptotes of each other.

All I can see about equation of Prop-86 is the system: $$xy=a\\\frac{y^2-b}{x^2}=\alpha$$

I don't understand what to do here. Am I supposed to solve this system of equations?

*Edit: This is the work I did.
I understand all of it, until the end where I stated: "Therefore, two hyperbolas as defined have their axes as the asymptotes of the other." I wrote that, because based on the question I know this will be the conclusion, but I have no idea how that result actually came to be.

答:
To me they appear to be two separate unconnected problems mentioned together, not in one system. First set/type is inclined to the axes and the second type/set is along the coordinate axis, is all the commonality that is there to it.

Plots of $$ \alpha x^2- y^2= \pm b, \;( \alpha= 3, b= \dfrac12) $$
show the desired hyperbolas pair.

Similarly in another statement of the same proposition  plots of

$$ xy= \pm b, \;(  b= \dfrac12) $$

show two different hyperbolas nested between (sharing) the same asymptote pair.

Title: Solving Prop 86 Euclid's Data


问:
Why isn't an adjacency matrix a graph invariant?
When working out if two graphs are isomorphic, why can't you use an adjacency matrix and match each pattern of row or column to the other graph's matrix?

The patterns seem to match up and everything works but obviously the answer is wrong; can you not do this?

答: You cannot make a conclusion by matching just rows or just columns of two graph's adjacency independently. What you need to do is to match both rows and columns after some permutation. So the conclusion is whether or not there is a permutation matrix $P$ which meets $A=PBP^{T}$ where $A$ and $B$ is the adjacency of two graphs, $P$ is the permutation matrix where it has just a single 1 on each row's and columns, rest of them is zero. So the complexity by brutal force is $n!$ which is ridiculously high. What all literature on graph isomorphism is to decrease this complexity to the linear if it is possible.


问:
For what values of $b\in\mathbb{R}$ this matrix is positive definite?
Let $b\in\mathbb{R}$ and consider the matrix

$\begin{vmatrix}
4 &-1 & b\\
-1& 2& -2\\
0& -2& 1
\end{vmatrix}$

The problem is: for what values of $b\in\mathbb{R}$ the matrix is positive definite?

I think the correct answer is that $b=0$, because for every $b\neq 0$, the martrix is even symmetric so it makes no sense to talk about positive definiteness.

Could anyone please help me to understand if I am doing it well or I am missing something?
Thank you in advance!

答:
$A$ can be decomposed in a sum $A=S+Q$ of a symmetric matrix $S=\frac 12(A+A^T)$ and a skew-symmetric matrix $Q=\frac 12(A-A^T)$.

When $A$ is symmetric, we just have $S=A$ and $Q=0$.

The general definition for definitive positiveness of a real matrix $M$ is:

$x^TMx>0$ for any vector $x\neq 0$.

Note: for the complex case change this to $x^*Mx$

We can notice that for skew-symmetric matrices, this product is always zero

$x^TQx\underbrace{=}_\text{this is a scalar}\left(x^TQx\right)^T=x^TQ^Tx=x^T(-Q)x=-(x^TQx)\implies x^TQx=0$

And since $\, x^TAx=x^T(S+Q)x=(x^TSx)+(x^TQx)=x^TSx\, $ then

$A$ is definite positive $\iff S$ is definite positive.

So we move on studying $\, S=\begin{bmatrix}4&-1&\frac b2\\-1&2&-2\\\frac b2&-2&1\end{bmatrix}$

We can now apply Sylvester minors criterion to the symmetric matrix $S$

$\Delta_1=\begin{vmatrix}4\end{vmatrix}=4>0$

$\Delta_2=\begin{vmatrix}4&-1\\-1&2\end{vmatrix}=8-1=7>0$

$\Delta_3=\begin{vmatrix}4&-1&\frac b2\\-1&2&-2\\\frac b2&-2&1\end{vmatrix}=-9+2b-\frac 12 b^2=-\frac 12(b^2-4b+18)<0$

Indeed the quadratic has discriminant $16-4\times 18=-56$ and leading coeff is negative.

Conclusion $A$ is never definite positive (nor semi-definite) for any value of $b$


问:
Absolute Value inequality Equivalence relation
For a,b ∈ R, check if $|a−b|\leq5$ is an equivalence relation.

Reflexive: a ~ a because $|a-a|=0\leq5$

Symmetric: Let $b~a$, $|b-a|≤5  \Rightarrow |a-b|≤5$, which means that a~b

Transitive: a~b and $b~c  \Rightarrow
|a-b|≤5  \Rightarrow -5≤ a-b ≤5$ (1)
$|b-c|≤5  \Rightarrow -5≤ b-c ≤5$ (2)

If I add (1),(2) i'll have: $-10≤ a-c ≤10  \Rightarrow |a-c|≤10$
Are the steps correct? Is the relation transitive or no?

答:
The first two steps are correct.

And, no, it is not transitive. In order to prove it, just provide an example. For instance: $0\sim4$ and $4\sim8$, but $0\not\sim8$.


问:
Lebesgue theorem on monotonic convergence
Can anyone give an example on (0,1) where the Lebesgue monotone convergence theorem is not true for decreasing sequences of non-negative functions? I can't find such an example, does anyone know? I would be grateful..

答:
The Lebegue Monotone Convergence theorem does not require the functions to have finite integral. But it requires the functions to be a non-decreasing sequence.

Here is the counter-example you are looking for:

Consider the function $f_n: (0,1) \to\mathbb{R}$, defined as $f_n(x)=\frac{1}{nx}$. Note that $\{f_n\}_n$ is a decreasing sequence of non-negative functions and  $f_n \to 0$ pointwisely, but, for all $n$, $\int_{(0,1)} f_n = +\infty $ does not converge to $0$.

Here is another counter-example:

Consider the function $f_n: \mathbb{R} \to\mathbb{R}$, defined as $f_n=\chi_{[n,+\infty)}$. Note that $\{f_n\}_n$ is a decreasing sequence of non-negative functions and  $f_n \to 0$ pointwisely, but, for all $n$, $\int_{\mathbb{R}} f_n = +\infty $ does not converge to $0$.


问:
Calculating sum of consecutive powers of a number multiplied by a consecutive number
There is a serie that i want to calculate for a competitive programing question.
$$\sum_{i=1}^n \left(81i \cdot 10^{i-1} - 10^{i-1} + 1\right) $$
Most of it is easy but at the end i end up with:
$$\sum_{i=1}^n 10^{i-1} \cdot i $$

So it abstracts to: $$\sum_{i=1}^n P^i \cdot i : P\in ℤ_{>1}$$
I can implement an recursive function, but i get time limit for large numbers. How can i simplify this to a simple function, is it possible?

答:
You can use calculus;
$$\begin{align}
\sum_{i=1}^{n} P^i\cdot i &= P\sum_{i=1}^{n} P^{i-1}\cdot i\\
&= P\sum_{i=1}^{n} P^{i-1}\cdot i\\
&= P\frac{\mathbb d}{\mathbb dP}\left(\sum_{i=1}^{n} P^i\right)\\
&= P\frac{\mathbb d}{\mathbb dP}\left(P\cdot \frac{P^n-1}{P-1}\right)
\end{align}
$$
I think you can take it from here.


问:
Homology group of $(X,X)$
Let $H$ be a homology theory.

I want to show that $H(X,X) = 0 \; \forall p\in\Bbb Z$. Maybe someone can tell me whether my reasoning is correct.

From the exactness axiom I get a long exact sequence:
$$... \xrightarrow{\text{$\partial_*$}} H_p(X) \xrightarrow{\text{$i_*$}} H_p(X) \xrightarrow{\text{$j_*$}} H_p(X,X)  \xrightarrow{\text{$\partial_*$}}...$$

where $i_*$ is the map induced by the inclusion (identity in this case) $i:X\rightarrow X$. Therefore $i_* = id_*:H_p(X) \rightarrow H_p(X)$.

$j_*$ is induced by $id:X\rightarrow (X,X)$. Therefore $j_* = id_*$. Since this is a LES we have that $im(i_*)=ker(j_*)$, but $im(i_*)=H_p(X)$ and therefore $j_*$ sends everything to $0$. But $j_*$ is also the identity.

$\Rightarrow H_p(X,X) = 0$

答:
$j_*$ is induced by $id:X\to(X,X)$. Therefore $j_*=id_*$

This does not make sense.

The fact that you need to use is that, in the following sequence if the middle arrow is an isomorphism, then the other two arrows are $0$.

$A \to B \to C \to D$

In particular, in the LES,

$\cdots \xrightarrow{j_*} H_{n+1}(X, X) \xrightarrow{\partial_*} H_n(X) \xrightarrow{id_*} H_n(X) \xrightarrow{j_*} H_n(X, X) \xrightarrow{\partial_*} H_{n-1}(X) \xrightarrow{id_*} \cdots$

the maps $\partial_*$ and $j_*$ are zero, or that they factor through $0$. So, the following is exact:

$0 \to H_n(X, X) \to 0$

Title: Homology group of $(X,X)$


问:
If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$
If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$

I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$

From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$

Hence $1+9abc\ge 4(ab+bc+ac)$.

This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?

答:
This is an easy application of Lagrange as the first equality of $$3a^2+bcd=3b^2+acd=3c^2+abd=\lambda$$ gives $3(a+b)(a-b)=(a-b)cd$ so either $a=b$ or $3(a+b)=cd$. Thus there are two possibilities:

$a=b=c$ so that $f(a,b,c)=1/9+d/27$;

$a=b$ and $3(b+c)=ad$ so that $d=3/a-3$ since $c=1-2a$ giving $$f(a,b,c)=2a^3+(1-2a)^3+a^2(1-2a)\left(\frac3a-3\right)=3a^2-3a+1\ge\frac14.$$


答:
For your answer to be correct, you need to make sure you are adding probabilities of disjoint events. So you should look at it as only $\small X \leq 2$ occurs $+$ only $\small Y \leq 2$ occurs $+$ both occur.

So in your working, it should have been $\frac{1}{2} \times (1 - \frac{3}{8}) + \frac{3}{8} \times (1 - \frac{1}{2}) + \frac{3}{8} \times \frac{1}{2} = \frac{11}{16}$

If you want to apply the formula you showed later on,

$\small P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$

Please note $\small P(X \cap Y) = P(X) \, P(Y) \,$  for $X, Y \,$ are independent.

$ \small P(X \cup Y) = \frac{1}{2} + \frac{3}{8} - \frac{1}{2} \times \frac{3}{8}$

Or you could simply use the complement events to find the desired probability as,

$ \, \small P(X \cup Y) = 1 - P(X^c \cap Y^c) = 1 - (1 - \frac{1}{2}) \times (1 - \frac{3}{8})$

Title: How to find probability in when using an "or" statement

Body:
Can somebody please help me solve/understand the following problem?:

Consider two random variables $X$ and $Y$ with events $S_{X}=S_{Y}=\{1,2,3,4\}$ and probabilities:

$P(X=j)=\frac{1}{4} \quad$ for all $j \in \{1,2,3,4\}, \quad$ and $\quad P(Y=j)=\left\{\begin{array}{ll}\frac{1}{8}, & \text { if } j=1 \\ \frac{1}{4}, & \text { if } j=2 \\ \frac{3}{8}, & \text { if } j=3 \\ \frac{1}{4}, & \text { if } j=4\end{array}\right.$

Assume, that $X$ and $Y$ are independent.

Find the probablity $P(X \leq 2$ or $Y \leq 2)$.

First I tried to solve it like this:

$P(x \leq 2$ or $y \leq 2)=P(x \leq 2)+P(y \leq 2)=\frac{7}{8}$

But I realised that this method would give a probability higher than 1 if I considered $x\leq4$ or $y\leq 4$. So how do I solve this correctly, and why is this method wrong?

One thing, that I'm not sure of, is whether I should still count the possibilities when X and Y are $\leq2$ at the same time? Or if it can only be one of them at a time, for the statement to be true?

Secondly, I found this theorem in my book, that might be helpful (see below). Specifically, I think a) could be used? But I just can't seem to "translate" the theorem to this problem. The first problem is, that my theorem only considers one random variable, but my case has two. Secondly, I'm not certain how to use the union and intersection of sets in this problem. But if it's possible to use this theorem, I would really appreciate you helping me understanding how.

Thanks in advance

Theorem: Let $X$ be a random variable with the possible events $S_{X},$ and let $A$ and $B$ be subsets of $S_{X}$. Then the following rules can be used:

a. $P(X \in A \cup B)=P(X \in A)+P(X \in B)-P(X \in A \cap B)$

b. $P(X \in A \cup B) \leq P(X \in A)+P(X \in B)$.

c. If $A$ and $B$ are disjoint, then

$
P(X \in A \cup B)=P(X \in A)+P(X \in B)
$

d. $P(X \notin A)=P\left(X \in S_{X} \backslash A\right)=1-P(X \in A)$

e. If $A_{1}, A_{2} \ldots, A_{n}$ are disjoint subsets of $S_{X},$ then

$
P\left(X \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=P\left(X \in A_{1}\right)+P\left(X \in A_{2}\right)+\cdots+P\left(X \in A_{n}\right)
$


答:
Too much work!

By definition the chain complex $C_*(X,A)$ is, degreewise, the quotient $C_q(X)/C_q(A)$.

So $C_q(X,X) = C_q(X)/C_q(X) = 0$. Therefore the chain complex $C_*(X,X)$ is zero in every degree. Therefore its homology is zero.

Title: Homology group of $(X,X)$

Body:
Let $H$ be a homology theory.

I want to show that $H(X,X) = 0 \; \forall p\in\Bbb Z$. Maybe someone can tell me whether my reasoning is correct.

From the exactness axiom I get a long exact sequence:
$$... \xrightarrow{\text{$\partial_*$}} H_p(X) \xrightarrow{\text{$i_*$}} H_p(X) \xrightarrow{\text{$j_*$}} H_p(X,X)  \xrightarrow{\text{$\partial_*$}}...$$

where $i_*$ is the map induced by the inclusion (identity in this case) $i:X\rightarrow X$. Therefore $i_* = id_*:H_p(X) \rightarrow H_p(X)$.

$j_*$ is induced by $id:X\rightarrow (X,X)$. Therefore $j_* = id_*$. Since this is a LES we have that $im(i_*)=ker(j_*)$, but $im(i_*)=H_p(X)$ and therefore $j_*$ sends everything to $0$. But $j_*$ is also the identity.

$\Rightarrow H_p(X,X) = 0$


问:
Evaluate function if the result appears to be outside the domain?
I've been given a transformation $f,$ from $P_2$ to $P_2$ (the vector space of polynomials of maximum degree $2$) such that
$$f(p(x)) = p(x+1) +\int_{0}^{1} p(x)dx.$$
Here, $p(x)$ is just a polynomial in the vector space

Working with respect to the basis $\left\{1,x,x^2\right\}$, it appears as though when $p(x) =x^2$, that integral would be $\dfrac{1}{3}\cdot x^3$. Clearly, $x^3$ is of degree $3$ so is outside of the vector space.

Is my thinking correct? And how would I apply this to making a matrix with respect to the basis given?

答: The integral part is a definite integral, so results in a constant, but you have it as the indefinite integral.


问:
Computing $\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy$

If $\phi_\alpha(z)=(\alpha-z)/(1-\bar\alpha z)$ for $|\alpha|<1$,
prove that

$\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy=1$ and
$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1-|\alpha|^2}{|\alpha|^2}log\frac{1}{1-|\alpha|^2}$ where in the case $\alpha=0$ the expression on the right is understood
as the limit as $|\alpha|\to 0$.

[Hint: The first integral can be evaluated without a calculation. For
the second, use polar coordinates, and for each fixed $r$ use contour
integration to evaluate the integral in $\theta$]

I did the first question, but I have trouble doing the second question. I don't know how to use polar coordinates method here.

So I computed $\phi'_\alpha(z)=\frac{-1+\alpha\bar\alpha}{(1-\alpha z)^2}$ and by using polar coordinate method I know, I get $\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0|\frac{-1+\alpha\bar\alpha}{(1-\bar\alpha re^{i\theta)^2}}|rdrd\theta$, but it looks wrong. Can you give me any idea? Thank you.

答:
You can directly compute the integral without using the Poisson kernel. We assume that $\alpha$ is real. Using
$$ z=e^{i\theta}, \bar z=e^{-i\theta} $$
we have
\begin{eqnarray}
&&\frac{1}{\pi}\iint_\mathbb{D}|\phi'_\alpha|dxdy\\
&=&\frac{1-|\alpha|^2}{\pi}\int^1_0\int^{2\pi}_0\frac{1}{|1-\alpha re^{i\theta}|^2}r\,d\theta\,dr\tag1\\
&=&\frac{1-|\alpha|^2}{\pi}\int^1_0r\int_{|z|=1}\frac{1}{|1-\alpha rz|^2}\frac{dz}{iz}dr\tag2\\
&=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\int_{|z|=1}\frac{1}{(1-\alpha rz)(1-\alpha r\frac1z)}\frac{dz}{z}dr\tag3\\
&=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\int_{|z|=1}\frac{1}{(1-\alpha rz)(z-\alpha r)}dz\,dr\\
&=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\cdot2\pi i\frac{1}{1-\alpha^2r^2}dr\\
&=&2(1-|\alpha|^2)\pi\int^1_0\frac{r\,dr}{1-\alpha^2r^2}\\
&=&-\frac{1-|\alpha|^2}{\alpha^2}\ln(1-\alpha^2).
\end{eqnarray}
Update:

Since $z=e^{i\theta}$, we have $dz=ie^{i\theta}d\theta=izd\theta$ and hence $d\theta=\frac{dz}{iz}$. So we have (2).

Since $|z|^2=1$, we have $\bar z=\frac1z$ and hence
$$|1-\alpha r z|^2=(1-\alpha r z)(1-\alpha r\bar z)=(1-\alpha z)(1-\alpha r\frac1z). $$
So we have (3).


Title: Evaluate function if the result appears to be outside the domain?

Body:
I've been given a transformation $f,$ from $P_2$ to $P_2$ (the vector space of polynomials of maximum degree $2$) such that
$$f(p(x)) = p(x+1) +\int_{0}^{1} p(x)dx.$$
Here, $p(x)$ is just a polynomial in the vector space

Working with respect to the basis $\left\{1,x,x^2\right\}$, it appears as though when $p(x) =x^2$, that integral would be $\dfrac{1}{3}\cdot x^3$. Clearly, $x^3$ is of degree $3$ so is outside of the vector space.

Is my thinking correct? And how would I apply this to making a matrix with respect to the basis given?

Body: $f(x^2)=(x+1)^2+\int_0^1x^2dx=(x+1)^2+\left.\frac{x^3}{3}\right|_0^1=(x+1)^2+\frac{1}{3}$


答:
Hint:

From

$$1+2\cdot10+3\cdot 10^2+4\cdot 10^3+\cdots n\cdot 10^{n-1}$$ subtract $10$ times this number and simplify the equal powers of $10$. This leads you to the solution.

Title: Calculating sum of consecutive powers of a number multiplied by a consecutive number

Body:
There is a serie that i want to calculate for a competitive programing question.
$$\sum_{i=1}^n \left(81i \cdot 10^{i-1} - 10^{i-1} + 1\right) $$
Most of it is easy but at the end i end up with:
$$\sum_{i=1}^n 10^{i-1} \cdot i $$

So it abstracts to: $$\sum_{i=1}^n P^i \cdot i : P\in ℤ_{>1}$$
I can implement an recursive function, but i get time limit for large numbers. How can i simplify this to a simple function, is it possible?


问:
$\int_0^t \frac{\sin x}{x}dx=\frac{1}{2i}\int_{-1}^1\frac{e^{iwt}}{w}dw$?
How the following identity is proved? $$\int_0^t \frac{\sin x}{x}dx=\frac{1}{2i}\int_{-1}^1\frac{e^{iwt}}{w}dw$$

答:
Start with the RHS: $$\frac{1}{2i} \int_{-1}^1 \frac{e^{iwt}}{w} dw \\ = \frac{1}{2i} \int_{-1}^1 \frac{\cos(wt)}{w} dw+\frac 12 \int_{-1}^1 \frac{\sin(wt)}{w} dw$$ The first integral evaluates to zero, as the integrand is odd. Substituting $wt=x \implies tdw=dx$ in the second gives $$\frac 12 \int_{-t}^t \frac{\sin x}{x}dx=\frac 12 \cdot 2\int_0^t \frac{\sin x}{x} dx$$

where the last step follows from the ‘even-ness’ of $\frac{\sin x}{x}$.


答:
$$I=\int_{0}^{1} \frac{\sin x}{x} dx= \int_{0}^{1}\frac{e^{ix}-e^{-ix}}{2ix}dx=\int_{0}^{1}\frac{e^{ix}}{2ix}dx-\int_{0}^{1}\frac{e^{-ix}}{2ix} dx.$$
In the second integral let $x=-y \implies dx=-dy$, them
$$I=\int_{0}^{1}\frac{e^{ix}}{2ix}dx+\int_{-1}^{0}\frac{e^{iy}}{2iy} dx =\int_{-1}^{1}\frac{e^{2iw}}{2iw} dw.$$
Lstly we have have put $x=w=y$.

Title: $\int_0^t \frac{\sin x}{x}dx=\frac{1}{2i}\int_{-1}^1\frac{e^{iwt}}{w}dw$?

Body: How the following identity is proved? $$\int_0^t \frac{\sin x}{x}dx=\frac{1}{2i}\int_{-1}^1\frac{e^{iwt}}{w}dw$$


问:
A doubt regarding nature of $F(a)$, where $F(a)$ is the intersection of all fields containing field $F$ and element $a$.
Proof in Herstein:
Let us consider elements like $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$. Here $f_{0},f_{1}\dots f_{s}\in F$. Now consider the quotient field $U$ generated by elements like $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$ dividing one another. Proving that this is a field is elementary.
Also, $U\subset F(a)$ as every field containing $F$ and $a$ has to contain elements like $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$, and $F(a)\subset U$ as $U$ is a field, and $F(a)$ is the smallest field containing $F$ and $a$. Hence $F(a)=U$.

My question is why does $U$ have to be a quotient group? Why can't it simply be the set of all possible $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$? Even in that case, $F(a)=U$!

答: $U$ has to be the quotient field because otherwise it wouldn't be a field. If it simply contained elements of the form $f_0+f_1a+f_2a^2+\cdots+f_sa^s$, then it wouldn't necessarily contain $a^{-1}$ (for example).</p>


问:
How can I prove $\lim_{(x,y)\rightarrow(3,5)}\sqrt[3]{x+y}=2$ by definition?
$$\lim_{(x,y)\rightarrow(3,5)}\sqrt[3]{x+y}=2$$

The cubic root over there makes this problem a little tricky for me, i don't really know how to proceed. Any ideas?

答:
HINT:
$f(x,y)$ is differentiable at $(a,b)$ if $(a,b)$ , $f_x(a,b)$,$f_y(a,b)$ exists and $\lim_{x,y \to a,b}\frac{f(x,y)-L(x,y)}{\sqrt{(x-a)^2 + (y-b)^2}}=0$  where $L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$

L is called linear approximation..


答:
Just another way to see it.
$$
\begin{aligned}
&\int_{0}^t \frac{\sin x}{x}dx\\
=&\frac{1}{2i}\int_{0}^t \frac{e^{i x}-e^{-ix}}{x} dx\qquad tw=x\\
=&\frac{1}{2i}\int_{0}^1 \frac{e^{itw}-e^{-itw}}{w} dw\\
=&\frac{1}{2i}(\int_{0}^1 \frac{e^{itw}}{w} dw-\int_{0}^1 \frac{e^{-itw}}{w} dw)\\
=&\frac{1}{2i}\int_{-1}^1 \frac{e^{itw}}{w} dw,
\end{aligned}
$$
where
$$\begin{aligned}
&-\int_{0}^1 \frac{e^{-itw}}{w} dw\quad -u=w\\
=&-\int_{0}^{-1} \frac{e^{itu}}{u} du\\
=&\int_{-1}^0 \frac{e^{itw}}{w} dw
\end{aligned}$$


问:
Finding the Euler Lagrange equation - differentiation
I'm teachin myself the basics of Calculus of variations. So far I know how to calculate the Euler Lagrange equation for simple functionals.
I'm now stuck on how to compute the total differentiation of the following problem:
$$I[y]=\int_0^1 (y\frac{dy}{dx})^2 -\lambda y^2 \ dx$$

To calculate the Euler Lagrange equation I have the following:
$$\frac{d}{dx}(2y^2y')-2y^2\frac{dy}{dx}+2\lambda y=0$$.

Is this correct? If so I'm unsure of how to evaluate the total differentiation part. That is taking the total derivative in this case.

答:
For a functional
  $$
I[y]=\int_a^bf(x,y,y')
$$
  the Euler-Lagrange equation is defined as
  $$
\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)=0
$$

For the $I$ you gave above, $f(x,y,y')=(yy')^2-\lambda y^2$, hence
the E-L equation for $I$ is
$$
2y\left(\frac{dy}{dx}\right)^2-2\lambda y-\frac{d}{dx}\left(2y^2\frac{dy}{dx}\right)=0
$$
Your only mistake is that $\frac{\partial f}{\partial y}=2\color{red}{y(y')^2}-2\lambda y$, not $2\color{red}{y^2y'}-2\lambda y$ as you have written. Does this help you go from here?

Edit. Since $f$ does not depend upon $x$, we can simplify the problem.

Lemma. For a functional
  $$
I[y]=\int_a^b f(y,y')
$$
  which is independent of $x$, we have that
  $$
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=0
$$

Proof. Just calculate the derivative
$$
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y''\frac{\partial f}{\partial y'}+y'\frac{d}{x}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial x}-y'\frac{\partial f}{\partial y}-y''\frac{\partial y}{\partial y'}
$$
Since $f$ is independent of $x$, $\frac{\partial f}{\partial x}=0$, and we can take out a factor of $y'$
$$
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y'\left(\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}\right)
$$
which is clearly equal to $0$ by E-L. QED

This means that
$$
y'\frac{\partial f}{\partial y'}-f=\text{constant}
$$
Applying this to your function $I$ gives
$$
\begin{align*}
y'(2y^2y')-(yy')^2+\lambda y^2&=C \\
y^2\left(\left(\frac{dy}{dx}\right)^2+\lambda\right)&=C \\
\frac{dy}{dx}&=\sqrt{\frac{C}{y^2}-\lambda}
\end{align*}
$$


问:
Weird geometry question ($AG^3=CE^2\times AB$)
$ABC$ is a triangle with $AB = 2AC$ and $E$ is the midpoint of $AB$. The point $F$ lies on the line $EC$ and the point $G$ lies on the line $BC$ such that $A, F, G$ are collinear and $FG = AC$. Show that $AG^3=CE^2\times AB$

答:
$AC=\frac{1}{2}BC=FG$ therefore $G$ coincides with $B$ and $F$ with $E$ (see the picture below).

From cosine law in triangle $AEC$ we have $CE^2=x^2+x^2-2x^2\cos 2t$

Applying the given relationship $AG^3=CE^2\cdot AB$ we get

$$(2 x)^3= \left(-2 x^2 \cos (2 t)+x^2+x^2\right)(2 x)$$
$$8x^3=8 x^3 \sin ^2 t\to \sin t=1\to t=90°\to 2t=180°$$
which is clearly impossible.
$$
...
$$


问:
path across a grid
Question : how many number of shortest path from the yellow bottom left corner to the yellow top right corner?

At first, I thought the answer will be $12!/(6!\cdot6!)$. But, I just realized that the combination of 6 consecutive to the north will not work here. Any idea to solve this problem?

答:
Clearly any shortest path can only go up or right.

So just work out the number of shortest paths to each point which may be on a correct route.  I have not checked the arithmetic, but something like

You could check this by considering those of the ${12 \choose 6}$ routes not through the blue circles (taking account of those going through more than one in the middle column).  This is ${12 \choose 6}-{4 \choose 0}{8 \choose 6} - {4 \choose 1}{8 \choose 5} -{8 \choose 6}{4 \choose 0} - \left({8 \choose 5}- {4 \choose 1}{4\choose 4}\right) {4 \choose 1}$

Title: path across a grid


问:
How to find the zeros of $z^{3} - iz^2 - z + i$
I know that the function  $z^{3} - iz^2 - z + i$ has zeros at $z = -1$ and $z = 1$. How do I calculate all its zeros in the form $a + bi$?

Body: Hint: $$= z^2(z-i)-(z-i) = ...$$


问:
For what $m,n$ is $D_{mn}$ isomorphic to $D_m \times Z_n$?
For what $m,n$ is $D_{mn}$ isomorphic to $D_m \times Z_n$?

Should I try to define an isomorphism between them? Where to start?

答: Recall that $D_n\cong \Bbb Z_n\rtimes\Bbb Z_2$.  So we want to know when $\Bbb Z_m\times(\Bbb Z_n\rtimes\Bbb Z_2)\cong \Bbb Z_{mn}\rtimes\Bbb Z_2$.  This will be true when $\Bbb Z_{mn}\cong\Bbb Z_m\times\Bbb Z_n$, or , when $(m,n)=1$.


Title: How to find the zeros of $z^{3} - iz^2 - z + i$

Body: I know that the function  $z^{3} - iz^2 - z + i$ has zeros at $z = -1$ and $z = 1$. How do I calculate all its zeros in the form $a + bi$?

Body:
The three zeros are:

$$z_1= -1 = -1 + 0i, \, z_2= 1 = 1 + 0i,\,  z_3 =-i = 0+ (-1) i$$

as the product $z_1 z_2 z_3$ is equal to $-i$.


答:
Proving this limit equation directly from the $\epsilon$,$\delta$ definition is possible, but tricky. It requires you to somehow deal with the inequality
$$\bigl| \sqrt[3]{x+y}-2 \bigr| < \epsilon
$$

Have you ever seen what to do when you have a square root inequality instead of a cube root? For example, let's consider the simpler problem of proving that $\lim_{(x,y) \to (1,3)} \sqrt{x+y}=2$ then you would have to deal with the inequality
$$\bigl| \sqrt{x+y}-2 \bigr| < \epsilon
$$
You can work with this inequality using the trick of "rationalizing the numerator", which is based on the formula $(a-b)(a+b)=a^2-b^2$ i.e.
$$|\sqrt{x+y}-2| = \left| \frac{(\sqrt{x+y}-2)(\sqrt{x+y}+2)}{\sqrt{x+y}+2} \right| = \left|  \frac{x + y - 4}{\sqrt{x+y}+2} \right|
$$
Then you can introduce appropriate lower bounds on the denominator: assuming, for example, that $|x+y-4| < 1$ it follows that $x+y > 3$ and that $\sqrt{x+y} + 2 > \sqrt{3}+1$ and so
$$\left|  \frac{x + y - 4}{\sqrt{x+y}+2} \right| \le \frac{1}{\sqrt{3}+2} |x+y-4|
$$
And, finally, you let $\delta = \min\{1,(\sqrt{3}+2)\epsilon\}$ and continue on with the proof.

So now, since what you actually have is a cube root inequality, you can use a similar "rationalizing the numerator" trick, based on the formula $(a-b)(a^2+ab+b^2)=a^3-b^3$. It's messy, but I'll write out the outcome with the middle step skipped:
$$|\sqrt[3]{x+y}-2| = \left| \frac{x+y-8}{(\sqrt[3]{x+y})^2 + 2 \cdot \sqrt[3]{x+y} + 4}   \right|
$$
And again, since it's only addition in the denominator, now you can introduce appropriate lower bounds on the denominator: if, say $|x+y-8|<1$ then $x+y > 7$ and that gives you a positive lower bound on the denominator which you can continue with, something like this (if I've done the arithmetic correctly):
$$(\sqrt[3]{x+y})^2 + 2 \cdot \sqrt[3]{x+y} + 4 > 7^{2/3} + 2 \cdot 7^{1/3} + 2
$$
and so you'll let $\delta = \min\{1, (7^{2/3} + 2 \cdot 7^{1/3} + 2) \epsilon\}$ and continue on with the proof.

Title: How can I prove $\lim_{(x,y)\rightarrow(3,5)}\sqrt[3]{x+y}=2$ by definition?

Body:
$$\lim_{(x,y)\rightarrow(3,5)}\sqrt[3]{x+y}=2$$

The cubic root over there makes this problem a little tricky for me, i don't really know how to proceed. Any ideas?


问:
Solving an equation with 2 square terms
Consider this equation;

$$ 3x^2 + 2dx - d^2 = 0 $$

Now I need to find the value of x.I've consider using the quadratic formula but I've got 2 squared terms. I've also tried forming the equation to get x squared on one side, but than I'm not sure what to next. Any help on this?

Thank you!

Title: Solving an equation with 2 square terms

Body: $3x^2 + 2dx - d^2=3(x+d)(x-d/3)$ so $x=-d$ or $x=d/3$


答:
If the question is to solve $3x^2+2dx-d^2=0$ for $x$ you can just apply the quadratic formula, getting
$$x=\frac {-2d \pm \sqrt{4d^2-4 \cdot 3 \cdot (-d^2)}}{2 \cdot 3}=\frac {-2d\pm 4d}6=\frac d3,-d$$

Title: Solving an equation with 2 square terms

Body:
Consider this equation;

$$ 3x^2 + 2dx - d^2 = 0 $$

Now I need to find the value of x.I've consider using the quadratic formula but I've got 2 squared terms. I've also tried forming the equation to get x squared on one side, but than I'm not sure what to next. Any help on this?

Thank you!


问:
$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$
Let $z=e^{i\frac{2\pi}{7}}$

Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$

How do I solve from here?

Body:
Since $z^7=1$ we have that it's equal to

$$z^2+z^5+z^6+z+z^3+z^4$$

$$=z^6+z^5+z^4+z^3+z^2+z$$ because $z^{7k+1}=z, z^{7k+2}=z^2,...$. This is equal to $-1$ since $z^7-1=(z-1)(z^6+..+1)=0$ and thus $z^6+..+1=0$ noting that $z -1\neq 0$.

Also we can solve using sum of geometric progression.

$\frac{a(r^n-1)}{r-1}=\frac{z(z^6-1)}{z-1}=\frac{z^7-z}{z-1}=\frac{1-z}{z-1}=-1$


Title: $z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$

Body:
Let $z=e^{i\frac{2\pi}{7}}$

Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$

How do I solve from here?

Body:
Since $$100\equiv _7 2$$ and $$300\equiv _7 -1$$ and  $$500\equiv _7 3$$ We have \begin{align} &=z^{2}+z^{-2} + z^{-1}+z^{1} + z^{3}+z^{-3} \\
&= {z^5+z+z^2+z^4+z^6+1\over z^3}\\
& ={z^6+z^5+z^4+\color{red}{z^3}+z^2+z+1-\color{red}{z^3}\over z^3}\\ &= {{z^7-1\over z-1} -z^3\over z^3} \\
 &= {0-z^3\over z^3} =-1\end{align}


答:
I suppose that we move along lines (not cells), starting at S and ending at E:

I take for granted that we know that all possible shortest paths for a full $n \times m$ grid is ${n+m \choose m} = {n+m \choose n}$ (see for example here).

Let $N(A,B)$ the number of paths from point $A$ to point $B$.

We have then:

$$N(S,E) = [N(S,1)-N(S,2)]N(1,4)N(4,E)+[N(S,2)-N(S,3)][N(2,4)-N(2,5)]N(4,E)+[N(S,2)-N(S,3)]N(2,5)N(5,E)+N(S,3)[N(3,4)-N(3,5)]N(4,E)+N(S,3)[N(3,5)-N(3,6)]N(5,E)+N(S,3)N(3,6)N(6,E) = \left[{6 \choose 2}-{5 \choose 2}\right]{2 \choose 2}{4 \choose 2}+\left[{5 \choose 2}-{4 \choose 2}\right]\left[{3 \choose 2}-{2 \choose 2}\right]{4 \choose 2}+\left[{5 \choose 2}-{4 \choose 2}\right]{2 \choose 2}{5 \choose 2} + {4 \choose 2}\left[{4 \choose 2}-{3 \choose 2}\right]{4 \choose 2} + {4 \choose 2}\left[{3 \choose 2}-{2 \choose 2}\right]{5 \choose 2}+{4 \choose 2}{2 \choose 2}{6 \choose 2} = 436$$

Where, for example, $N(S,1)-N(S,2)$ count all paths from $S$ to $1$ not passing by $2$, and similarly for other subtractions.

The result agrees with the answer by @Henry.

Title: path across a grid

Body:
Question : how many number of shortest path from the yellow bottom left corner to the yellow top right corner?

At first, I thought the answer will be $12!/(6!\cdot6!)$. But, I just realized that the combination of 6 consecutive to the north will not work here. Any idea to solve this problem?


问:
Can this counter example disprove the statement?
Consider $X$ is a path connected space. Choose an element $[a]$ from it's fundamental group $\pi_{1}(X)$, and then choose  a loop $a\in[a]$. This $a$ can be taken as a map from $S^{1}$ to $X$. Now we attach a 2-cell to $X$ via $a$ to obtain space $Y$. The statement is $Y$ is simply connected.

My counter example: Consider a disk removing two point in the interior, let $a$ be a loop enclosed one point. Then the space $Y$ is homotopy equivalent to a disk removing one point, which is not simply connected.

Does the counter example works? If not, what's wrong?

答:
What you are doing here is known as the method of killing homotopy groups. See here. Let us see what is going on in dimension $1$ if we take an element $g \in \pi_1(X,x_0)$, represent it by a map $\gamma : S^1 \to X$ and attach a $2$-cell to $X$ via $\gamma$. If $i : X \to X' = X \cup_\gamma D^2$ denotes inclusion, then
$$i_* : \pi_1(X,x_0) \to \pi_1(X', x_0) $$
is an epimorphism whose kernel is the normal subgroup $N(g)$ of $\pi_1(X,x_0)$ generated by $g$. This shows that in general $X'$ is not simply connected. For example, if $X = S^1$ and $g$ is the equivalence class of a loop of degree $n$, then $\pi_1(X') \approx \mathbb Z_{\lvert n \rvert}$.

The above result can be easily proved by the Seifert - van Kampen theorem.

We have $X' = A \cup B$ with $A  = X' \setminus \{\text{point in the interior of the attached } 2\text{-cell}\}$, $B =$ interior of the attached $2$-cell. But $B$ is contractible, $X$ is a strong deformation retract of $A$ and $A \cap B \approx S^1 \times (0,1) \simeq S^1$. We get
$$\pi_1(X') \approx (\pi_1(A) * \pi_1(B)) / N$$
where $N$ is the normal subgroup generated by all words of the form $i_A(g)i_B(g)^{-1}$ with $g \in \pi_1(A \cap B)$ and $i_A: \pi_1(A \cap B) \to \pi_1(A)$ and $i_B: \pi_1(A \cap B) \to \pi_1(B)$ induced by the inclusions $\iota_A: A \cap B \to A, \iota_B: A \cap B \to B$. But $\pi_1(B) = 0$ and thus all  $i_B(g)^{-1} = 0$. We conclude that
$$\pi_1(X') \approx \pi_1(A)/H ,$$
where $H$ is the (normal) subgroup generated by all $i_A(h)$ with $h \in \pi_1(A \cap B)$. But now we have a commutative diagram
$\require{AMScd}$
\begin{CD}
\pi_1(A) @>{r_*}>> \pi_1(X)   \\
@A{i_A}AA @A{\gamma_*}AA  \\
\pi_1(A \cap B) @>{\rho_*}>> \pi_1(S^1) \end{CD}
with retraction $r : A \to X$ and the obvious $\rho : A \cap B \to S^1$. Both $r_*$ and $\rho_*$ are isomorphisms, thus we see that $\pi_1(A)/H \approx \pi_1(X)/ N(g)$.


问:
Concluding the proof of the Baire category theorem
Let $(M,d)$ be a complete metric space, $A_n\subseteq M$ be nowhere dense for $n\in\mathbb N_0$, $x_0\in M$ and $r_0>0$.

We can show that there is a $(x_n)_{n\in\mathbb N}\subseteq M$ and $(r_n)_{n\in\mathbb N}\subseteq(0,\infty)$ such that $$\forall n\in\mathbb N:r_n<\frac{r_0}{2^n}\text{ and }\overline B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\setminus\overline{A_n}\tag1.$$

In fact, since $A_1$ is nowhere dense, there is a $x_1\in B_{r_0}(x_0)\setminus\overline{A_1}\ne\emptyset$. And since $B_{r_0}(x_0)\setminus\overline{A_1}$ is open, there is a $r_1\in\left(0,\frac{r_0}2\right)$ with $$\overline B_{r_1}(x_1)\subseteq B_{r_0}(x_0)\setminus\overline{A_1}\tag2.$$ Continuing this procedure, we obtain $(x_n)_{n\in\mathbb N}$ and $(r_n)_{n\in\mathbb N}$ as claimed.

Now, by construciton, $$\forall n\in\mathbb N:\overline B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\subseteq\overline B_{r_{n-1}}(x_{n-1})\tag3$$ and hence $$\forall n\ge m\ge0:x_n\in\overline B_{r_n}(x_n)\subseteq\overline B_{r_m}(x_m)\tag4.$$ Moreover, $$\forall n\in\mathbb N:\overline B_{r_n}(x_n)\subseteq B_{r_0}(x_0)\setminus\bigcup_{i=1}^n\overline{A_i}=B_{r_0}(x_0)\cap\bigcap_{i=1}^n\overline{A_i}^c\tag5$$ and hence $$\bigcap_{n\in\mathbb N}\overline B_{r_n}(x_n)\subseteq B_{r_0}(x_0)\cap\bigcap_{n\in\mathbb N}\overline{A_n}^c=B_{r_0}(x_0)\setminus\bigcup_{n\in\mathbb N}\overline{A_n}\tag6.$$ By $(4)$, $$d(x_m,x_n)\le r_{\min(m,\:n)}\xrightarrow{m,\:n\:\to\:\infty}0\tag7$$ and hence $(x_n)_{n\in\mathbb N_0}$ is Cauchy. Since $M$ is complete, there is a $x_\infty\in M$ with $$d(x_n,x_\infty)\xrightarrow{n\to\infty}0\tag8.$$ By $(4)$ and $(6)$, $$x_\infty\in\bigcap_{n\in\mathbb N_0}\overline B_{r_n}(x_n)\subseteq B_{r_0}(x_0)\setminus\bigcup_{n\in\mathbb N}\overline{A_n}\tag9.$$

The desired conclusion is that $M\ne\bigcup_{n\in\mathbb N}A_n$. Or, the even stronger conclusion, that $M\setminus\bigcup_{n\in\mathbb N}A_n$ is dense.

However, we should even have shown that $M\setminus\bigcup_{n\in\mathbb N}\overline{A_n}$ is dense, since we have shown that for arbitrary $x_0\in M$ and arbitrary small $r_0>0$, there is a $x_\infty\in M$ with $d(x_0,x_\infty)<r_0$.

From this it clearly follows that $M\setminus\bigcup_{n\in\mathbb N}A_n\supseteq M\setminus\bigcup_{n\in\mathbb N}\overline{A_n}$ is dense as well; which in turn clearly implies that $M\ne\bigcup_{n\in\mathbb N}A_n$.

However, my question is: If our goal is only to show the original claim (i.e. the denseness of $M\setminus\bigcup_{n\in\mathbb N}A_n$ wouldn't it be sufficient to choose the radii such that $B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\setminus\overline{A_n}$ (instead of $\overline B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\setminus\overline{A_n}$) or would the given argument fail at some point?

答: For each natural $n\le i$, $x_i\in\overline{B_n}$, so $x=\lim_{i\to \infty}  x_i\in\overline{B_n}$. Therefore $x\in\bigcap_{n=1}^\infty \overline{B_n}\subset M\setminus\bigcup_{n=1}^\infty A_n$.


问:
What's the $\lim \limits_{x \to 0} \frac{\sin x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$
This exercise is kind of hard for me to understand. Could somebody clarify it step by step please?  $$\lim \limits_{x \to 0} \frac{\sin x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$$

答:
Method 1:

It can be solved using L'Hospital rule:

$$
\begin{aligned}
\lim \limits_{x \to 0} \frac{\sin x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}} &= \lim \limits_{x \to 0}\frac{2 x \cos \left(x^2\right)}{\frac{\sin (x)}{2 \sqrt{\cos (x)}}+\frac{\sin (x)+x \cos (x)}{2 \sqrt{x \sin (x)+1}}} \\
&=\lim \limits_{x \to 0} \frac{4 x \cos \left(x^2\right)}{\frac{\sin (x)}{\sqrt{\cos (x)}}+\frac{\sin (x)+x \cos (x)}{\sqrt{x \sin (x)+1}}} \\
&=\lim \limits_{x \to 0} \frac{4 \cos \left(x^2\right)-8 x^2 \sin \left(x^2\right)}{\sqrt{\cos (x)}+\frac{\sin ^2(x)}{2 \cos ^{\frac{3}{2}}(x)}-\frac{(\sin (x)+x \cos (x))^2}{2 (x \sin (x)+1)^{3/2}}+\frac{2 \cos (x)-x \sin (x)}{\sqrt{x \sin (x)+1}}}\\
&=\dfrac{4}{3}
\end{aligned}
$$

Method 2: We can use Taylor Series:

$$
\begin{aligned}
\lim \limits_{x \to 0} \frac{\sin x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}} &= \lim \limits_{x \to 0}\dfrac{x^2+o\left(x^3\right)}{\frac{3 x^2}{4}+o\left(x^3\right)} \\
&=\lim \limits_{x \to 0}\dfrac{x^2}{\frac{3 x^2}{4}}\\
&=\dfrac{4}{3}
\end{aligned}
$$

See here for a tutorial in how to evaluate limits with Taylor Series.


问:
Asymptotic behaviour of a sequence.
Let $\{p_n\}_n\subset \mathbb{R}$ and $\{N_n\}_n\subset \mathbb{N}$ be sequences with $0\leq p_n\leq 1$ and $\lim_{n\to \infty} N_n=\infty$ increasing. More over, let $M\in \mathbb{N}$, $0\leq M\leq N_n$ and $M = N_n p_n + \mathcal{O}(\sqrt{N_n p_n (1-p_n)})$.

How can I prove that

$$\limsup {N_n\choose M} p_n^M(1-p_n)^{N_n-M}<1 $$

Any help will be appreciated.

Title: Asymptotic behaviour of a sequence.

Body:
Just:
$$\binom{N_n}{M}p_n^M(1-p_n)^{N_n-M} \leq \sum\limits_{i=0}^{N_n} \binom{N_n}{i}p_n^i(1-p_n)^{N_n-i} = (p_n + (1-p_n))^{N_n} = 1$$


答:
I think it's $\sin^2 x$, here's an attempt :
Let :
$$L=\lim \limits_{x \to 0} \frac{\sin^2 x}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$$
\begin{align}
L&=\lim_{x\to 0}\frac{(\sin^2 x)(\sqrt{1+x\sin x} +\sqrt{\cos x})}{1+x\sin x - \cos x}
\\
&=2\times\lim_{x\to 0} \frac{\sin^2 x}{1+x\sin x - \cos x}\\
&=2\times \lim_{x\to 0} \frac{1}{\frac{1+x\sin x - \cos x}{\sin^2 x}}\\
&=2\times\lim_{x\to 0} \frac{1}{\frac{x\sin x}{\sin^2 x}+\frac{1-\cos x}{\sin^2 x}}\\
&=2\times\lim_{x\to 0} \frac{1}{\frac{x}{\sin x}+\frac{x^2}{\sin^2 x}\frac{1-\cos x}{x^2}}\\
&=2\times \frac{1}{1+1\times \frac{1}{2}}\\
&=\frac{4}{3}
\end{align}
Hence (if it's in the expression $\sin^2 x$) :
$$L=\frac{4}3$$
Otherwise you can see @FFjet's answer, but I'm not quiet sure that you'll understand it if you're a high school student !

Good luck.

Title: What's the $\lim \limits_{x \to 0} \frac{\sin x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$

Body: This exercise is kind of hard for me to understand. Could somebody clarify it step by step please?  $$\lim \limits_{x \to 0} \frac{\sin x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$$


问:
Prove theorem of Bolzano-Weierstrass by least upper bound property
Let be $(a_n)_{n\in\mathbb{N}}$ a bounded sequence of real numbers and we want to show that it contains a convergent subsequence.

My approach:

We know, by the least upper bound property, that the set $M:=\{a_n\mid n\in\mathbb{N}\}$ has a supremum $S$. As $S$ is the least upper bound of $M$, for each $\epsilon>0$ there exists a $a_n$ such that $S-\epsilon<a_n<S+\epsilon$.

Let be $\epsilon>0$ then there exists a $k_0\in\mathbb{N}$ such that $\frac{1}{k_0}<\epsilon$. We define a subsequence $\left( a_{n_k}\right)_{k\in\mathbb{N}}$ as follows:

There exists a $a_n$ which we set $a_{n_{k_0}}:=a_n$ such that

$$
S-\frac{1}{k_0}<a_{n_{k_0}}<S+\frac{1}{k_0}.
$$
For all $k>k_0$ it follows that
$$
S-\frac{1}{k_0}<S-\frac{1}{k}<a_{n_k}<S+\frac{1}{k}<S+\frac{1}{k_0}\\
\implies |a_{n_k}-S|<\frac{1}{k}<\epsilon.
$$

Hence, $(a_n)_{n\in\mathbb{N}}$ is convergent.

My tutor said that it is flawed!? I have spent hours revising it and couldn't find any mistakes?

答:
The flaw is that the selected term may always be the same one. In particular if the difference of one term with the other ones is greater than a fix number.

Example. Take $\{a_n\}$ the always vanishing sequence except that $a_1 =1$. Then $S = 1$ but you won't be able to select a converging subsequence to $1$. Whatever $k \in \mathbb N$ you take, the only $a_n$ satisfying $
\vert a_n - 1 \vert \lt 1/k$ is $a_1$. Remember than for a subsequence $\{a_{n_k}\}$, $n_k$ has to be strictly increasing with $k$.

You can however build a subsequence converging to $\limsup\limits_{n} a_n$.


问:
Solve $\sin z = 2$

How to solve $\sin(z) = 2$,where $z$ is a complex number.

Another approach I tried:

$$\sin z=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right)=2$$

$$e^{iz}-e^{-iz}=4i$$

Let $x=e^{iz}$ so that $x^2-4ix-1=0$. Then, I would use the quadratic formula to solve would this be correct?

答: Continuing your answer, we have now $\sin x\cosh y=a$ and $\cos x\sinh y=0.$ From the second equation $\cos x=0$ or $\sinh y=0.$ If $\cos x=0$ then $x=\pi/2(2n+1)$. But the first equation then gives that $\cosh y=a$ which is not possible because $\cosh y\geq1.$ Hence $\sinh y=0$ and therefore $y=0$.

Title: Solve $\sin z = 2$


问:
Prove that there are at least two points a, b ∈ [0, 2] whose distance is equal to 1 and for which h (a) = h (b) is verified
I have been thinking about this problem but I am not able to find a solution.

Ler h be a continuous function in the range [0, 2] and such that h (0) = h (2). Prove that there are at least two points a, b ∈ [0, 2] whose distance is equal to 1 and for which h (a) = h (b) is verified.

Best regardes.

答: Hint: Consider $g\colon[0,1]\to\Bbb R$ given by $g(x)=h(x)-h(x+1)$.


问:
Show that this process is a local martingale
Let $B$ be a Brownian Motion, $a>0$ and $b>0$.
We set $\tau_{a,b}:=inf\{t\geq0:B_t\in\{-a;b\}\}$.

How can I show that:
$$
M_t=sh(\theta(B_t+a))exp(-\frac{\theta^2}{2}t)
$$
is a local martingale $\forall \theta \in \mathbb{R}$.

thanks!

答:
Just use Ito's formula.

Let $f(t,x) = sh(\theta(x+a))e^{-\frac{\theta^2}{2}t}$.

Then $f_x(t,x) = \theta ch(\theta(x+a))e^{-\frac{\theta^2}{2}t}$, $f_{xx}(t,x) = \theta^2 sh(\theta(x+a))e^{-\frac{\theta^2}{2}t}$ and $f_t(t,x) = -\frac{\theta^2}{2} sh(\theta(x+a))e^{-\frac{\theta^2}{2}t}$.

Then
$$df(t,B_t) = f_xdB_t + f_tdt + \frac{1}{2}f_{xx}dt = \theta ch(\theta(B_t+a))e^{-\frac{\theta^2}{2}t}dB_t$$ Thus $f(t,B_t)$ is a local martingale.


答: An alternative: write $w:=\exp(iz)$ so $w^2-2iaw-1=0$, with roots $w=ia\pm\sqrt{1-a^2}$. Since the surd is real, $|w|^2=a^2+1-a^2=1$, so $z\in\Bbb R$.

Title: Solve $\sin z = 2$

Body:
How to solve $\sin(z) = 2$,where $z$ is a complex number.

Another approach I tried:

$$\sin z=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right)=2$$

$$e^{iz}-e^{-iz}=4i$$

Let $x=e^{iz}$ so that $x^2-4ix-1=0$. Then, I would use the quadratic formula to solve would this be correct?


问:
How to evaluate $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?
How can I evaluate  $\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)$?</p>

答:
Almost always when you have a limit of $\sqrt{A}-\sqrt{B}$ type. It is good idea to multiply it by $\frac{\sqrt{A}+\sqrt{B}}{\sqrt{A}+\sqrt{B}}$. So you have to do limit of $\frac{A-B}{\sqrt{A}+\sqrt{B}}$

Thus for your limit you get:
$$\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} = \frac{ (x+\sqrt{x} ) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}}=\frac{ 2\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} \rightarrow 1$$ as $x \rightarrow \infty$.


问:
Least square method to fit data according to $g(x) = a - x + bx^2$, where $a, b \in \mathbb{R}$
If I have a set of points $(x_i, f(x_i)) \in \mathbb{R^2} $ and I want to use the method of the the least square fitting to calculate the free parameters $a$ and $b$ of the function
$$
g(x) = a - x + bx^2
$$

By looking for the function $g$, one possible basis of function would be:
$$
\phi_1(x) = 1,  \phi_2(x)  = x^2
$$
And the missing function of $g$ would cause a translation of the linear space of the solutions. Is this correct?

I don't know to proceed with this exercise.
Could anyone give me a hint?
Thanks in advance.

答:
$$h(x)=g(x)+x$$
First fit the function $h(x)$ for $a,b$ :
$$h(x)=a+bx^2$$
The LINEAR Least Mean Square Regression leads to approximate $a$ and $b$ :
$$\left(\begin{matrix}
  a  \\
  b
 \end{matrix}\right) \simeq
\left(\begin{matrix}
  n & \sum_{i=1} ^{i=n} x_i^2 \\
  \sum_{i=1} ^{i=n} x_i^2 & \sum_{i=1} ^{i=n} x_i^4
 \end{matrix}\right)^{-1}
\left(\begin{matrix}
  \sum_{i=1} ^{i=n} h_i \\
  \sum_{i=1} ^{i=n} h_ix_i^2
 \end{matrix}\right)\quad \text{with}\quad h_i=f_i(x_i)+x_i$$


问:
I found two definitions for an atom in a boolean algebra, but cannot find a proof of their equivalence.
I want to show that the following definitions for an atom $a$ are equivalent for a nonzero element $a$ in a Boolean algebra $\mathcal{B}$:

for all $x\in\mathcal{B},a\leq x$ or $x\land a=0$

for all $x\in\mathcal{B},x\leq a\Rightarrow x=0$ or $x=a$

I can proof that the first statement indicates the second, but I could not figure out how to proof the first statement starting from the second.

What I have so far is (proof that first implies second):

$a\leq x\Rightarrow x\land a=a$, so we know that either $x\land a=a$ or $x\land a=0$

First possibility: $x\land a=a$

Since $x\leq a$ implies $x\land a=x$, we find $x=a$

Second possibility: $x\land a=0$

Since $x\leq a$ implies $x\land a=x$, we find $x=0$

Can someone find the second part of the proof?

答:
Suppose that $a \in B$ is such that for all $x \in B$, if $x \leq a$ then $x=0$ or $x=a$.

Take $x \in B$, whatever.

Given that $x \wedge a \leq a$, we conclude that $x \wedge a = 0$ or $x \wedge a = a$;
in the second case, $a \leq x$.

Update.
I found your proof that (1) implies (2) rather confusing (starting in the first sentence), so I'm adding my own version too.

Suppose that $a \in B$ is such that for all $x \in B$, we have that $a \leq x$ or $a \wedge x = 0$.

In order to prove that for all $x \in B$, if $x \leq a$ then $x=0$ or $x=a$, take $x \leq a$, in $B$.

By hypothesis, either $a \leq x$ or $a \wedge x = 0$.

In the first case, $x=a$;
in the second, since $x = a \wedge x$ (because $x \leq a$), we obtain $x=0 $.


问:
Is the linear transformation $T: P_{n} \to \mathbb{R}^{n+1}$ an isomorphism?
So, I need to prove that the linear transformation $T: P_{n} \to \mathbb{R}^{n+1}$, defined as $T(p(x)) = \{p(0),p(1),...,p(n)\}$ is an isomorphism, with $P_{n}$ being the polynomials with a degree equal or lesser than $n$, but I have a problem.

I tried to check if $T$ is injective, and it is since any polynomial can only have at max $n$ roots, so any polynomial that gets sent to $\underbrace{(0,0,0,...,0)}_\text{n+1 times}$ must be the zero polynomial. Then, $T$ is injective. And, since $P_{n}$ and $\mathbb{R}^{n+1}$ have the same dimension, then $T$ is an isomorphism.

Then, I tried to see what $T$ does to the base of $P_{n}$. I got a bunch of vectors and tried to see if they're linearly independent, so I got a matrix, and since the first row is all $0$, then the determinant is $0$, and since the determinant is $0$, then these vectors are not a base for $\mathbb{R}^{n+1}$. Then, since an isomorphism has to send a base to a base, $T$ is not an isomorphism.

What went wrong?

Edit: I got the vectors $(1,1,1,...,1), (0,1,2,...,n),(0,1,4,...,n^{2}),...,(0,1,2^{n},...,n^{n})$, so I did the equation $$a_{0}(1,1,1,...,1) + a_{1}(0,1,2,...,n) + ... + a_{n}(0,1,2^{n},...,n^{n}) = 0$$ and then got this:
\begin{align}
a_{0} + a_{1}(0) + a_{2}(0) +...+ a_{n}(0) &= 0\\
a_{0} + a_{1}(1) + a_{2}(1) +...+ a_{n}(1) &= 0\\
\vdots &\\
a_{0} + a_{1}(n) + a_{2}(n^{2}) +...+ a_{n}(n^{n}) &= 0\\
\end{align}
and got the coefficient matrix:
\begin{matrix}
0 & 0 & 0 &...& 0\\
0 & a_{1} & a_{2} &...& a_{n}\\
\vdots & \vdots & \vdots & \vdots & \vdots \\
0 & a_{1}(n) & a_{2}(n^{2}) &...& a_{n}(n^{n})
\end{matrix}

答:
The first colunm is all of 1's (if you take as the first element of the basis of $P_n$ the polynomial $p_0(x)=1$, then $p_0(0)=p_0(1)=\ldots=p_0(n+1)=1$), so there no problem, or I mean, you have to prove that the determinat of the matrix

$$
\begin{pmatrix}
1 & 0 & 0 &...& 0\\
1 & 1 & 1 &...& 1\\
1 & 2 & 4 &...& 2^n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & n & n^{2} &...& n^{n}
\end{pmatrix}
$$
is not $0$ (which is true, and not so difficult to prove).

But maybe, what you are trying is not the easier way to prove that the function is surjective, you can just use the definition.

Indeed, let $(y_0,\ldots,y_{n+1})\in \mathbb{R}^{n+1}$, and consider the polynomial

$$
p(x):=\frac{y_0}{(n+1)!}\cdot \prod_{i=0,\ i\neq 0}^{n+1} (i-x)+\frac{y_1}{-(n)!}\cdot \prod_{i=0,\ i\neq 1}^{n+1} (i-x)+\ldots +\frac{y_{n+1}}{(-1)^n(n)!}\cdot \prod_{i=0,\ i\neq n+1}^{n+1} (i-x)=
\sum_{m=0}^{n+1} \left( \frac{y_m}{(-1)^m(n+1-m)!m!} \cdot \prod_{i=0,\ i\neq m}^{n+1} (i-x) \right)
$$

then it is easy to see that $p$ has degree $n$ (so it is in $P_n$) and $T(p)=(y_0,\ldots,y_{n+1})$.

Indeed, if $0\leq j\leq n$, then evaluating $p(j)$ you have that all the addends in the sum are $0$ except for the $j$-th, which is

$$
\frac{y_j}{(-1)^j(n+1-j)!j!} \cdot \prod_{i=0,\ i\neq j}^{n+1} (i-j)=\frac{y_j}{(-1)^j(n+1-j)!j!}\cdot (-j\cdot (-j+1)\cdot \ldots\cdot  -1\cdot 1\cdot \ldots \cdot (n+1-j))=y_j.
$$

Maybe you are also interested in Vandermonde matrices.


问:
system of intermediate fields of extension is infinite
I need help to prove that system of intermediate fields of extension $\mathbb{k}(x,y)$, of field $\mathbb{k}(x^p,y^p) \ char(\mathbb{k}) = p > 0$ is infinite

答:
For $m\ne n$ if
$$k(x^p,y^p,x^{mp+1} +y)= k(x^p,y^p,x^{np+1} +y)$$
then
$$k(x^p,y^p,x^{mp+1} +y,x^{np+1} +y)=k(x,y)$$

has degree $p$ over $k(x^p,y^p)$ which is a contradiction.


问:
Question about the definition of representability of a quadratic form
Say I have an integral quadratic form $q$ on $\mathbb{Z}^r$ and another integral binary quadratic form $\tilde{q}$. What does it mean for $q$ to primitively represent $\tilde{q}$? I can't seem to find the definition anywhere. Of course $q$ primitively represents an integer $m$ if there exists a primitive element $\alpha\in\mathbb{Z}^r$ such that $q(\alpha)=m$, but I'm not sure what it means to represent another quadratic form (especially when the other quadratic form is not defined on the same amount of variables!).</p>

答:
You take the Gram matrix, which is generally taken to be the Hessian matrix of second partial derivatives. Let $A$ be the Gram matrix for your $q,$ and $B$ be the Gram matrix for the other one. Representation is a rectangular matrix $P$ such that $$ P^T A P = B.   $$ Primitive representation is if the GCD of all the entries of $P$ is $1.$ Where are you reading about this?

The case with which you are familiar is when $P$ is a column vector and $B$ is a 1 by 1 matrix, that is, a number.

On page 18, remark 35, you need Estes and Pall (1973), which I put at TERNARY. The principal forms are the squares in the class group. The spinor kernel is the fourth powers in the group. Your $\bar{s}(D)$ is the number of spinor genera per genus, which is the same as the number of squares divided by the number of fourth powers.


问:
For what $m,n$ is $D_{mn}$ isomorphic to $D_m \times Z_n$?
For what $m,n$ is $D_{mn}$ isomorphic to $D_m \times Z_n$?

Should I try to define an isomorphism between them? Where to start?

答:
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Let $D_{k} = \Span{ a, b : a^{k} = 1, b^{2} = 1, b^{-1} a b = a^{-1} }$. When is it that $D_{k}$ has a non-trivial (cyclic) direct factor?

If $k = 1$, then $D_{1} \cong Z_{2}$. If $k = 2$, then $D_{ 2} \cong Z_{2} \times Z_{2} \cong D_{1} \times Z_{2}$.

If $k \ge 3$, it is easy to see that no element $c$ outside $\Span{a}$ is in the centre, as $c^{-1} a c = a^{-1} \ne a$. Now if an element $a^{t} \ne 1$ is in the centre, we must have $a^{-t} = b^{-1} a^{t} b = a^{t}$, hence $a^{2 t} = 1$, that is, $a^{t}$ is an involution, which can only happen when $k$ is even, and then $a^{t} = a^{k/2}$. And in this case $a^{k/2}$ is a direct factor in $\Span{a}$, and thus in $D_{k}$, if and only if $k/2$ is odd,

The moral of the story is that this happens iff $n = 2$ and $m$ is odd.


问:
Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$.

Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$, but $\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}$ is not a compact set.

(Can we use definition of compact set for this problem? )

答: We have to use the definition of a compact set here. The latter set is not compact since you can come up with its open cover $B(\frac1n,2^{-n})$ which does not have a finite subcover (why?). The former set is compact since it is closed and bounded.


问:
Simplify $(AB^{-1}A^T)^{-1}$ where $A$ is a non-square matrix
Suppose matrices $A\in\mathbb{R}^{2\times3}$ and $B\in\mathbb{R}^{3\times3}$ are available for computing $(AB^{-1}A^T)^{-1}$.

If $A$ was square we could simplify it to $A^{-T}BA^{-1}$. Is a simplification possible if $A$ is non-square?

Background

The application here is to calculate the precision matrix (inverse covariance matrix) of a Gaussian random variable $y=Ax$ where $y\in\mathbb{R}^2$ and $x\in\mathbb{R}^3$~$N(\mu,B^{-1})$, where $B$ is known and could be close to singular (hence the desire to work with precision matrices rather than covariance matrices).

答:
The pseudo-inverse of a non-square matrix can be (amongst some other methods) computed using the singular value decomposition:
Suppose $A = U\Sigma V^T$, where $U \in \mathbb{R}^{2\times2}, V\in \mathbb{R}^{3\times3}$ are unitary matrices and $\Sigma \in \mathbb{R}^{2\times3}$ is a matrix with on the diagonal the singular values of $A$. the inverse can then be estimated as:
$$A^\dagger = V \Sigma_{-1} U^T\in \mathbb{R}^{3\times2}$$
Where $\Sigma_{-1}$ means that you take the numerical inverse of every non-zero number in $\Sigma$ and transpose the result (if anybody got a better notation for this, let me know). As $A$ is not square, only postmultiplying this pseudo-inverse with $A$ will yield an identity matrix of size $\mathbb{R}^{2\times2}$. So the inverse of $AB^{-1}A^T$ is a bit of a conundrum: no pre-multiplication or post-multiplication with $A^\dagger$ will isolate $B^{-1}$ in this equation. Therefore, I think that one cannot simplify this inverse any further.


问:
Is the limit definition of the derivative symmetric?
I want to show that $f'(x+h)=f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$. When I consider the secant line formed by the two points in the limit, I see no reason to dismiss the idea that the slope of the secant line also approaches the slope of the tangent line at $(x,f(x+h))$ as well. If we let $c=x+h$, substituting into the conventional definition gives $f'(x)=\lim_{c\rightarrow x}\frac{f(c)-f(x)}{c-x}=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=f'(c)=f'(x+h)$. My question may boil down to whether the input to the limit is reversible (is ${c\rightarrow x}$ equivalent to ${x\rightarrow c}?)$. The idea that one point is stationary and the other one is "moving" seems like an unnecessary construct. What really matters is that the distance between the points approaches zero, right?

答:
You're wrong. When you write down the following formula $$f'(x)=\lim_{c \to x}\frac{f(c)-f(x)}{c-x},$$
$x$ is always a fixed number, and $c$ is a variable. Hence you can't obatain $$\lim_{c \to x}\frac{f(c)-f(x)}{c-x}=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}.$$


问:
Is this multivariable calculus?
The following is a question from Calculus made easy by Silvanus P. Thompson

The volume of a right circular cylinder of radius $r$ and height $h$ is given by the formula $V=πr^2h$. Find the rate of variation of volume with the radius when $r = 5.5 \text{in}$. and $h = 20 \text{in}$. If $r = h$, find the dimensions of the cylinder so that a change of 1 in. in radius causes a change of 400 cubic inches in the volume.

I understand the solution to the problem apart from the second part when $r=h$. I got the first solution for that when I took h as a constant but there is second answer for the question in which $h$ varies with $r$ shown below:

$$\frac {dV}{dr}=3πr^2$$
 $$=400$$

and then we solve for $r$ which is equal to $h$. My question is isn't this multivariable calculus and is this method correct?

答:
Short answer: No.

Long(ish) answer: This is not multivariable calculus, because multivariable calculus deals with multiple independent variables, whereas the problem in question deals with dependent variables.

To help illustrate my point, here is an example.

It would be multivariable calculus to find the maximum values of the function $$f(x,y)=\frac{\sin(x+y)}{x^2+y^2+1}$$
for all real numbers $x$ and $y$.

By contrast, it would be a single-variable calculus problem to find the maximum values of $$\frac{\sin(x+y)}{x^2+y^2+1}=c,$$
where $c$ is a constant. This is because, by setting the function $f$ equal to a constant, you are implicitly defining $y$ in terms of $x$, and thus there is only one independent variable to work with, namely $x$.

Let me know if this helps. :)

Title: Is this multivariable calculus?


问:
A module that has a basis shorter than its rank?
Rank has been defined to me as the maximum number of elements of M which are linearly independent, allowing the rank to be $0$ if the module is of torsion, or $\infty$ if there is no maximum. Now, the course is based pretty much on the book Abstract Algebra by Dummit and Foote, and there the rank is defined to be as the length of a basis of $M$ (assume $M$ has a finite basis and that $R$ is commutative with $1$ here to have a well defined rank).

We now are seeing the following theorem:

Let $R$ be a PID and $M$ a free module of rank $n$. Then any submodule $N$ of $M$ satisfies:

(1) $N$ is free of rank less than $n$.

(2) There is a basis $y_1,...,y_n$ of $M$ such that $r_1y_1,...,r_ty_t$ is a basis for $N$, for some $t \le n$.

I think that, according to my definition, the length of the basis' of M doesn't have to be n. It certainly can't be more than n, cause then the elements would be l.d., but it could be less than n. In consequence, the assertion (2) makes no sense a priori.

My question is whether the fact that $R$ is a PID and $M$ is a free module over $R$ implies that both definitions of rank coincide (i.e., that the maximum number of l.i. elements is also the length of a basis's of $M$), or whether this was just a mistake due to my professor giving a different definition of rank.

答:
Just to add on Richard, the result is still true even if $R$ is not an integral domain, i.e.

Given a free module of rank $n$ over $R$, any $(n+1)$ elements in $M$ are algebraically dependent.

Assume by contradiction that there are $(n+1)$ elements, then the span is isomorphic to $R^{n+1}$. We then have an injective linear morphism from $i: A^{r+1}\rightarrow A^{r}$. This is not possible:

We acknowledge the lemma that an $A$-module morphism $ A^r\rightarrow A^r$ is injective $\iff$ the determinant is not a zero divisor. If $i$ exists, by adding a roll of zero to the determinant, the induced map $A^{r+1}\rightarrow A^{r+1}$ is still injective, but the determinant is zero. Thus we have a contradiction. Therefore there is no $A^{r+1}$ in $A^r$, as one would hope.


问:
Given an LP problem how to transform the problem?
Given the following problem:

$$ \max. -3x_1 -2x_2 - \frac{9}{2}$$
$$ \text{s.t.} -x_1 + x_2 \le 0$$
$$      5x_1 + 3x_2 \le 8$$
$$ x_1,x_2 \ge 0 $$

When we convert the LP into standard form and then write down the simplex tableau we see that the starting solution is already optimal:$$  x_3 ,x_4  \rightarrow \text{basic variables} $$
$$ x_1, x_2 \rightarrow \text{nonbasic variables}$$ Is it possible to transform the original problem so that the starting solution will be another extreme point?

答:
Notice that $(1,1,0,0)$ is also a basic feasible solution.

That is we pick the basis using the first tow columns rather than the last two columns, that is $B=\begin{bmatrix} -1 & 1 \\ 5 & 3\end{bmatrix}$.

Rather than $Ax=b$, we have

$$(B^{-1}A )x = (B^{-1}b)$$


问:
Matrices and power series
For any square matrix $A$, we can define $\sin A$ using the formal power series as follows:
\begin{equation}
\sin A=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}A^{2n+1}
\end{equation}
Prove or disprove: there exists a $2\times2$ real matrix $A$ such that
\begin{equation}
\sin A=\begin{bmatrix}1 & 2020 \\0& 1\end{bmatrix}
\end{equation}

答:
Notice that

\begin{equation}
B=\begin{bmatrix}1 & 2020 \\0& 1\end{bmatrix}
\end{equation}

is not diagonalizable. If it was $1$ would be the only eigenvalue and $B$ would be equal to the identity matrix.

Therefore if $\sin A = B$, $A$ isn't diagonalizable either. The (complex) Jordan normal form of $A$ would be

\begin{equation}
\overline A=\begin{bmatrix}\lambda & 1 \\0& \lambda\end{bmatrix}
\end{equation}

And

\begin{equation}
\sin \overline A=\begin{bmatrix}\sin \lambda & \cos \lambda \\0& \sin \lambda\end{bmatrix}
\end{equation}

Hence $\lambda$ belongs to $\pi/2 + 2\pi \mathbb Z$  and

\begin{equation}
\sin \overline A=\begin{bmatrix}\sin \lambda & 0 \\0& \sin \lambda\end{bmatrix} = \begin{bmatrix}1 & 0 \\0& 1\end{bmatrix}
\end{equation}

is diagonalizable. A contradiction.

Title: Matrices and power series


问:
finite difference equations
i havent had a response to this question in a while, could someone please help me. Im struggling to understand the concepts of forward/backward/central differences on finite difference equations.

i have formulas for:

first order with respect to x (spatial)- forward (explicit), backward(implicit) and central

second order with respect to x- forward (explicit), backward(implicit) and central

first order with respect to t (time)- forward (explicit), backward(implicit) and central

what are the formulas for second order with respect to t- forward (explicit), backward(implicit) and central?

i have a one second order equation for time but not sure what it specifys- is it the forward (explicit)?:

$$ \frac{d^2u}{dt^2} = \frac{u_{i}(m+1)-2u_{i}(m)+u_{i}(m-1)}{(\delta t)^2} $$

i know how to solve questions just need the second order equations with respect to time. thanks in advance for any help- been looking all over the web but not really getting a grasp. thanks.

答:
From what I understood of your question, that is quite confusing.

A simple reference for formulas for finite differences, or their coeficientes, is found here in Wiki , for backward, forward and central.

Regarding being explicit, implicit:

You didn't say where you want to apply those finite difference discretizations. What is the equation, depending on what you choose backward/forward/central for each term of your equation you get a implicit or explicit schema.

If you specify the equation will be easier, for example for the heat equation:

$$ U_t = \alpha U_{xx} $$

Using a forward difference at time and a second-order central difference for the space derivative at position $u_j$ (FTCS) we get explicit schema:

$$ \frac{u_j^{t+1}-u_j^t}{\Delta t} = \alpha \frac{u_{j+1}^t - 2u_j^t + u_{j-1}^t}{h^2} $$

where $ \Delta t $ is the discretization in time and $ h $ in space.

Title: finite difference equations


问:
A question about functional in normed space and Hahn Banach theory
Let $X\neq {0}$ a normed space, $x_n\in X$ a sequence in $X,x\in X$.

Assume for every $x^*\in X^*$:
$x^*(x_n)\to x^*x$.

Show that:
$\|x\|\leq \liminf_{n\to \infty} \|x_n\|$.

I tried to use Hahn Banach's corollary that says:
If $X$ is a normed space and $0\neq x\in X$, there is a functional $x^*\in X$ such that $\|x^*\|=1$ and $x^*x=\|x\|$.

We can notice that $x$ that satisfies $x^*(x_n)\to x^*x$ is unique.

Because $X\neq 0$ then the $x$ satisfying this is $\neq 0$.
Then, by the corollary mentioned above and the given information we get:
$$\|x\|=x^*x=\lim_{n\to \infty} x^*x_n.$$
Now, how can I use the fact that $\|x^*\|=1$?

答:
As you said, there is some $f\in X^*$ such that $||f||=1$ and $|f(x)|=||x||$. Note that since $f$ has norm $1$ it follows that:

$|f(x_n)|\leq ||x_n|| \ \ \forall n\in\mathbb{N}$

And so $||x||=|f(x)|=\liminf_{n\to\infty} |f(x_n)|\leq \liminf_{n\to\infty} ||x_n||$.