问:
basic question about dual space in linear algebra
I've been reading 'Introduction to Lie Algebras and Representation Theory-Humphreys'.
The following is the text in page 19.

Let $s$=diag($a_1, a_2, ... , a_m$). Let E be the vector space spanned by $a_1, ... , a_m$. We want to show that $s=0$, equivalently, that $E=0$. Since $E$ has finite dimension, it will suffice to show that the dual space $E^*$ is $0$.

I have a question about the last sentence. If we show that $E^*=0$, then $(E^*)^*=0$. Since $E$ is finite dimensional, it is isomorphic to $(E^*)^*$. So we're done. Did I understand the text correctly?

Thanks in advance.

答:
I don't see why the hypothesis of finite dimension is needed.

If $E$ is non-zero, so is $E^\ast$, just pick $e \in E$ and consider the projection to $\langle e \rangle$ (I mean, write $E = \langle e\rangle \oplus S$ and consider $\lambda e +s \mapsto \lambda$). Conversely, if $E^\ast = 0$, then as you say $(E^\ast)^\ast  = 0$ and the canonical map $\jmath \colon x \in E \mapsto \mathbf{ev}_x \in (E^\ast)^\ast$ is always injective so $E = 0$.


问:
Let $A_n\subseteq\mathbb R$ for $n\ge1$ and $\mathcal X_n:\mathbb R\to\{0,1\}$ be defined in a certain manner, then which option/s is/are true?
Please verify my solution for the following problem and let me know if there be anything wrong in my solution and also provide if there be any other solution in other way of the same problem.

Problem : Let $~A_n\subseteq\mathbb R~$ for $~n\ge1~$ and $~\mathcal X_n:\mathbb R\to\{0,1\}~$ be the function $$\mathcal X_n(x)=\begin{cases}
      0 &\text{if}~~~~ x\not\in A_n \\
      1 &\text{if}~~~~ x\in A_n
         \end{cases}$$
Let $~g(x)=\lim_{n\to\infty}\sup\mathcal X_n(x)~$ and $~h(x)=\lim_{n\to\infty}\inf\mathcal X_n(x)~.$ Then which of the following option/s is/are true ?

$1)~~$If $~g(x)=h(x)=1~,$ then there exists $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$

$2)~~$If $~g(x)=1~$ and $~h(x)=0~,$ then there exists $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$

$3)~~$If $~g(x)=1~$ and $~h(x)=0~,$ then there exists a sequence $~n_1,~n_2,~\cdots~$ of distinct integers such that $~x\in A_{n_k}~$ for all $~k\ge1~.$

$4)~~$If $~g(x)=h(x)=0~,$ then there exists $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$

My Solution:

For option $\bf{(1)}$,  If $~g(x)=h(x)=1~,$ then by the definition of the $~\lim\sup~$ and $~\lim\inf~,$ after some stage $~\mathcal X_n(x)~$ becomes $~1~$ i.e., there exists an $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$
Hence option $(1)$ is correct.

For option $\bf{(4)}$,  If $~g(x)=h(x)=0~,$ then by the previous argument after some stage $~\mathcal X_n(x)~$ becomes $~0~$ i.e., there exists an $~m~$ such that for all $~n\ge m~$ we have $~x\not\in A_n~.$
Hence option $(4)$ is correct.

For option $\bf{(3)}$,  If $~g(x)=1~$ and $~h(x)=0~,$ then there exists two sub-sequences $~\mathcal X_{n_k}(x)~$ and $~\mathcal X_{n_l}(x)~$ such that $~\mathcal X_{n_k}(x)~$ converges to $~1~$ and $~\mathcal X_{n_l}(x)~$ converges to $~0~.$
i.e., there exists a sequence $~n_1,~n_2,~\cdots~$ of distinct integers such that for $~k\ge m~,$ we have $~x\in A_{n}~.$
 Hence option $(3)$ is correct.

And therefore only option $(2)$ is not true.

答: Your answers for 1), 3) and 4) are corect but you should also know why 2) is false. You need  a counter-example to show that 2) is false. Take two disjoint sets, say $A=(0,1)$ and $B=(1,2)$. Let $A_n=A$ for $n$ even and $A_n=B$ for $n$ odd.  Take $x \in A$. Then you can check that $g(x)=1$ and $h(x)=0$. But there is no $m$ such that $x \in A_n$ for all $n \geq m$.

ContentLicense: CC BY-SA 4.0


问:
Long equation problems
I am given this equation:

$$15x^6+31x^5-254x^4-352x^3+329x^2+9x-18=0$$

I know how to solve this question but the problem is it is too long to plug in numbers, what am I advised to do?

答:
An easy thing to check is if it has rational roots. You only need check for factors of the form $ax+b$ where $a$ is a factor of $15$ and $b$ is a factor (positive or negative) of $18$.

You should for example find that $5x-3$ is a factor i.e $\frac {3}{5}$ is a root.

There's also a much simpler factor for you to find.

Divide through by these factors to obtain a quartic $$3x^4+2x^3-50x^2+2x+3
$$ which has no more rational roots.

Title: Long equation problems


问:
Fixing a Calculation Mistake on p.51 in "Mathematical Analysis I" by Vladimir A. Zorich?
While reading "Mathematical Analysis I" by Vladimir A. Zorich. I came across this proof regarding the irrationality of $\sqrt{2}$ :

Let $X$ and $Y$ be the sets of positive real numbers such that $\forall x \in X\left(x^{2}<2\right)$ $\forall y \in Y\left(2<y^{2}\right) .$ Since $1 \in X$ and $2 \in Y,$ it follows that $X$ and $Y$ are nonempty sets. Further, since $(x<y) \Leftrightarrow\left(x^{2}<y^{2}\right)$ for positive numbers $x$ and $y,$ every element of $X$ is less than every element of $Y .$ By the completeness axiom there exists $s \in \mathbb{R}$ such that $x \leq s \leq y$ for all $x \in X$ and all $y \in Y$ We shall show that $s^{2}=2$ If $s^{2}<2,$ then, for example, the number $s+\frac{2-s^{2}}{3 s},$ which is larger than $s,$ would have a square less than $2 .$ Indeed, we know that $1 \in X,$ so that $1^{2} \leq s^{2}<2,$ and $0<\Delta:=2-s^{2} \leq 1 .$ It follows that :

$$
\boxed{\left(s+\frac{\Delta}{3 s}\right)^{2}=s^{2}+2 \cdot \frac{\Delta}{3 s}+\left(\frac{\Delta}{3 s}\right)^{2}<s^{2}+3 \cdot \frac{\Delta}{3 s}<s^{2}+3 \cdot \frac{\Delta}{3 s}=s^{2}+\Delta=2}
$$

The author made a simple mistake during the expansion as seen in the box above. Therefore, I attempt to fix the proof as follow :

Since $1 \leq s^2$ and $\frac{\Delta}{3}<1$ hold, then :
\begin{align*}
\left(s+\frac{\Delta}{3s}\right)^2&=s^2+2\cdot\frac{\Delta}{3}+\left(\frac{\Delta}{3s}\right)^2\\
&\leq s^2+2\cdot\frac{\Delta}{3}+\left(\frac{\Delta}{3}\right)^2\\
&<s^2+2\cdot\frac{\Delta}{3}+\frac{\Delta}{3}\\
&= s^2+3\cdot\frac{\Delta}{3}\\
&=s^2+\Delta \\
&= 2
\end{align*}

Hence, $(s+\frac{\Delta}{3s})^2<2$ is true as required.

Are my steps correct in my attempt to fix this part of the proof?

答: What year was your edition of Zorich published? A copy of the Russian edition from 2012 is here. It has the expansion you write about on the bottom of page number 77 and the calculations there match your correction.</p>


问:
Bounds on Hausdorff measure when extending a set
Let $(X, \rho)$ be a metric space that has Hausdorff dimension $d$, and let $S$ be an arbitrary Borel subset such that $H^d(S) = \alpha > 0$, where $H^d$ denotes the $d$-dimensional Hausdorff measure (assume this is nonzero). Is it possible to show that the quantity
$$H^d(\left\{x \in X \mid \rho(x, S) < \delta\right\})$$
must be nonzero and finite? In other words, can we show that some nonzero and finite amount of mass must be picked up in order to extend the set $S$ by a distance of $\delta$? Feel free to make any further assumptions on the space $(X, \rho)$ as needed.

答:
Denote $S_\delta = \{x \in X : \rho(x,S) < \delta \}$.  The claim that $H^d(S_\delta) < \infty$ is false -- a counterexample is given by $S = \mathbb{Z} \subseteq \mathbb{R}$.  In this case $d=1$, $H^d(S) = 0$, and $S_\delta$ is a countable union of intervals of length $2\delta$, so $H^d(S_\delta) = \infty$.

The claim that $H^d(S_\delta) > 0$ is also false.  Let $X = \{0\} \cup [1,2]$ and $S = \{0\}$.  Then $X$ is $1$-dimensional but $S_{1/2} = \{0\}$ is still measure zero according to $1$-dimensional Hausdorff measure.


问:
removing a point from a unit circle results in a connected space
This is an example our TA gave us.
"If we removed a finite numbers of points of the unit circle the resulting set is connected"
Can somebody explain why??</p>

答: Proof of the fact that the unit circle $S^{1}$ remains connected after removing one point from it: If $e^{i\theta}$ is removed from the circle then define $f: (0,1) \to S^{1}\setminus \{e^{i\theta}\}$ by $f(t)=e^{2\pi it +i\theta}$. Then $f$ is a homeomorhism . Since $(0,1)$ is connected so is the image $S^{1}\setminus \{e^{i\theta}\}$.


问:
Picard's Little Theorem Proofs
Picard's little theorem says that

If there exist two complex numbers $a,b$ such that $f: \Bbb{C} \to \Bbb{C}\setminus \{a,b\}$ is holomorphic then $f$ is constant.

I am interested in proofs for this theorem. Until now I found at least two, one in W. Rudin's Real and Complex Analysis and another one in S. Krantz, Geometric Function Theory. Both of them need some preparation before someone not very advanced in Complex Analysis could understand them, especially the one in S. Krantz's book.

My questions are

How many proofs are there for Picard's little theorem? (references if possible)

Is there a "simple" proof for Picard's little theorem? Simple means that it could be presented to an audience which had a one semester course in complex analysis.

Thank you.

答:
There is a clever proof using harmonic functions in Ransford's book Potential Theory in the Complex Plane.

The proof is originally by John Lewis. His paper is here. Note it also includes a short survey of other methods of proof.

ContentLicense: CC BY-SA 3.0

Title: Picard's Little Theorem Proofs


问:
convergence test for improper integrals
in an old book on calculus in the chapter about improper integrals there was this theorem that says when $\int_{a}^{\infty }f(x)dx $ can be written in the form
$\int_{a}^{\infty} \frac{\phi (x)}{x^k}dx$ you can test for convergence by verifying that $|\phi (x)|<M $
and $k>1$ what i don't understand is how do we write $f(x)$ in that form what is the choice of $\phi(x)$and$x^k$ depend on for example

$ \int_{0}^{\infty}e^{-x^2}dx$ can be written as $\frac{x^2 e^{-x^2}}{x^2}$ and therefor converges but why did we choose$x^2$not $x$ for example ?

答: Because one of the assumptions is that $k>1$. Besides, although you can indeed write $e^{-x^2}$ as $\frac{xe^{-x^2}}x$, since $\int_a^\infty\frac{\mathrm dx}x$ diverges, you deduce nothing from that.

Title: convergence test for improper integrals


问:
What is the probability that L does not intersect C1?
Let C1 and C2 be two concentric circles with radius r and 2r respectively. Given a fixed point Q on the circumference of C2, a line L is drawn at random going through Q. What is the probability that L does not intersect C1?

I tried using geometry to calculate probability whereby ratio of segment area to the overall area of C2 should be the answer. However I am not getting the right solution. It seems to be straightforward question unless I am missing something obvious.

答:
When looking all around, the angle scanned is $2 \pi$ radians. Area scanned is infinite, independent variable should be the azimuth scan angle $\theta$ that has to be partitioned.

$$ \theta= \sin^{-1}\frac12= \dfrac{\pi}{6}$$

What is seen is (white)$ 2 \theta$

What is not seen is (yellow) $ 2 \pi-2 \theta$

So the unseen yellow fraction is

$$1-\dfrac{1}{6}=\dfrac{5}{6}.$$


问:
Confusion in Argand plane
When I say a argand plane , in my textbook it says the plane having a complex number assigned to each of its point. I interpreted a meaning for this statement down below , let me know if it’s correct

Second is does it mean that when I talk about a plane. It is like x and z axis is one plane , x and y axis , y and z axis . So , it has to be 2D.  x,y and z axis all together can’t be an argand plane. Is this correct ?

Confusion in ordered pairs :

So , there is no squaring of numbers done here like R ^2 as written . That is my confusion.

答: The expression “Argand plane” simply means the set $\Bbb C$ of complex numbers, when we see it as a plane, that is, when we identify each complex number $a+bi$ ($a,b\in\Bbb R$) with the point $(a,b)\in\Bbb R^2$. So, yes, it is a $2$-dimensional thing.

Title: Confusion in Argand plane


问:
Topology objective question based on homeomorphism .
Consider the following sets of $\Bbb R^2$ with usual topology

$$X=\{(x,y)\mid x^2+y^2<1\}\cup \{(x,y)\mid (x-2)^2+y^2<1\}$$

$$Y=\{(x,y)\mid x^2+y^2<1\}\cup \{(x,y)\mid (x-3)^2+y^2<1\}$$

$$Z=\{(x,y)\mid x^2+y^2<1,x\neq 0\}$$ then which of the following sets are homeomorphic ?

$1.$ all of $X,Y,$ and $Z$.

$2.$ none of $X,Y$ and $Z$.

$3.$ $X$ and $Z$ only .

$4.$ $ X$ and $Y$ only .

$ 5.$ $Y$ and $Z$ only .

For me it seems that all are homeomorphic to each other as all of $X$, $Y$ and $Z$ has two components and corresponding components are homeomorphic. Am I right ? Please suggest me for correct options. Thank you .

答: Indeed, all three spaces consist of two disjoint components, both of which are (homeomorphic to) an open unit disk in the plane. So they are all mutually homeomorphic.</p>


问:
Doubt in First Course in Probability by Sheldon Ross
Here is the example that I was solving from the book on probability.

A football team consists of $20$ offensive and $20$ defensive players. The players are to be paired in groups of $2$ for the purpose of determining roommates. If the pairing is done at random, what is the probability that there are no offensive-defensive roommate pairs? What is the probability that there are $2i$ offensive-defensive roommate pairs, $i=1,2,\ldots,10$?

Solution There are
$$\binom{40}{2,2,\ldots,2}=\frac{(40)!}{(2!)^{20}}$$
ways of dividing the $40$ players into $20$ ordered pairs of two each. (That is, there are $(40)!/2^{20}$ ways of dividing the players into a first pair, a second pair, and so on.) Hence, there are $(40)!/2^{20}(20)!$ ways of dividing the players into (unordered) pairs of $2$ each. Furthermore, since a division will result in no offensive-defensive pairs if the offensive (and defensive) players are paired among themselves, it follows that there are $[(20)!/2^{10}(10)!]^2$ such divisions. Hence, the probability of no offensive-defensive roommate pairs, call it $P_0$, is given by
$$P_0=\frac{\left(\frac{(20)!}{2^{10}(10)!}\right)^2}{\frac{(40)!}{2^{20}(20)!}}=\frac{[(20)!]^3}{[(10)!]^2(40)!}.$$

What I don't understand is why is the solution considering unordered pair for the calculation? Aren't we choosing sets without any permutation of elements within them? What are we over-counting here? Can someone explain what am I missing?

答:
Consider the $20$ rooms and assign players to these rooms as follows:-

Choose two players for the first room in   $\begin{pmatrix}40\\2\\\end{pmatrix}=\frac {40\times 39}{2\times 1}$  ways.

Now choose for the next room from the remaining 38 players etc.

This gives the number $$\frac {40!}{(2!)^{20}}.$$

The rooms themselves can be arranged in $20!$ ways and this arrangement is irrelevant to the pairings so the answer is divided by $20!$ to obtain the number of pairings.


问:
If 3 out of 5 randomly chosen reports used incorrect style, how many to expect out of 125

A teacher randomly reads 5 written reports from the 125 she has to grade. He finds that 3 reports did not follow the correct style. At this time same rate, how many of the 125 can she expect to not follow the correct style?

How do you do this question? You don't have to give the answer, just how you solve.

答: The problem tells you that 3 out of 5 of the papers the teacher looks at "did not follow the correct style".  Then the "rate" of papers not following the style is $3/5$.  To find the answer, you need to figure out how many out of 125 makes 3/5.  That is, find $X$ such that $X/125 = 3/5$.</p>


问:
Why does $\sqrt{n} (\bar X - \mu)/S$ have approximately a $t$-distribution?
Suppose that the random variables $X_1,\ldots, X_n$ are independent and identically distributed with mean $\mu$ and variance $\sigma^2$. Let $\bar X = (X_1 + \cdots + X_n)/n$ be the sample mean and $S = ((X_1 - \bar X)^2 + \cdots + (X_n - \bar X)^2)/(n-1)$ be the sample variance.

If the random variables $X_i$ are normally distributed, then $\sqrt{n}(\bar X - \mu)/\sigma$ is normally distributed and $\sqrt{n}(\bar X - \mu)/S$ has a $t$-distribution. Moreover, even if the random variables $X_i$ are not normally distributed, then $\sqrt{n}(\bar X - \mu)/\sigma$ is approximately normally distributed, according to the central limit theorem.

Question: In this case (where the random variables $X_i$ are not normally distributed), does the central limit theorem also imply that $\sqrt{n}(\bar X - \mu)/S$ has approximately a $t$-distribution?

This question arises in the context of constructing confidence intervals for the mean of a random variable, in the case where the variance is unknown.

The following passage appears in Ross's book Introduction to Probability and Statistics for Engineers and Scientists:

Our derivations of the $100(1-\alpha)$ percent confidence intervals
  for the population mean $\mu$ have assumed that the population
  distribution is normal. However, even when this is not the case, if
  the sample size is reasonably large then the intervals obtained will
  still be approximate $100(1-\alpha)$ percent confidence intervals for
  $\mu$. This is true because, by the central limit theorem,
  $\sqrt{n}(\bar X - \mu)/\sigma$ will have approximately a normal
  distribution, and $\sqrt{n}(\bar X - \mu)/S$ will have approximately a
  $t$-distribution.

I'm trying to understand the final statement about $\sqrt{n}(\bar X - \mu)/S$ having approximately a $t$-distribution.

答: In Ross's book, $\sqrt n (\overline{X} - \mu) / S$ has t distribution requires the nominator to be standard normal random variable (which under CLT satisfies). $\frac{(n-1)S^2}{\sigma^2}$ needs to have chi-square distribution (which satisfies according to this post using Taylor expansion with second order). The final condition will be the independence between $\overline{X}$ and $S^2$ (which I don't know...)


答:
It says Argand plane, not any volume, is n't it?

The Argand diagram  contains points  $z= x+iy$  and lines in the  the complex plane.

Title: Confusion in Argand plane

Body:
When I say a argand plane , in my textbook it says the plane having a complex number assigned to each of its point. I interpreted a meaning for this statement down below , let me know if it’s correct

Second is does it mean that when I talk about a plane. It is like x and z axis is one plane , x and y axis , y and z axis . So , it has to be 2D.  x,y and z axis all together can’t be an argand plane. Is this correct ?

Confusion in ordered pairs :

So , there is no squaring of numbers done here like R ^2 as written . That is my confusion.


问:
The using of Borsuk-Ulam theorem
Let a circle $k:(x-p)^2+(y-q)^2=r^2$, $p,q\in\mathbb{R},\thinspace r>0 $  and a point $(x_0,y_0)\in\mathbb{R}^2$ be given. Prove that there exists a square $[x_0-a,x_0+a]\times[y_0-a,y_0+a],\thinspace a>0$ whose boundary contains at least one pair of diametrically opposed points from the circle $k$.

I suppose that I can use Borsuk-Ulam theorem, but I don't know how to find the continuous function from $S^1$ to $\mathbb{R}^2$, and later from $\mathbb{R}^2$ to $\mathbb{R}$.

答:
As in the comments, without loss of generality the circle is $S^1$, write $p = (x_0,y_0)$.

Consider the taxicab norm $\|\cdot \|_1$ on the plane and $d$ the induced distance. Then $$[x_0-a,x_0+a] \times [y_0-a,y_0+a] = B_a^d(p),$$
a point $q$ is in that square precisely if $d(p,q) \leq a$ and it lies in the boundary iff $d(p,q) = a$.

Define $d(-,p) \colon \Bbb R^2 \to \Bbb R$ to be the taxicab distance to $p$. This is $\|\cdot\|_1$-continuous by its very definition, but by the equivalence of norms in $\Bbb R^2$ it is also continuous for the usual distance on the plane.

Now consider the restriction $\delta \colon S^1 \to \Bbb R$. By Borsuk-Ulam, there exists a point $x \in S^1$ such that $\delta(x) = \delta(-x) = a$. Note that $a \geq 0$ because $\delta \geq 0$, but moreover $a > 0$ or otherwise we would have $x = p = -x$ which is absurd, as $x \in S^1$.

By a previous remark, this means that $x,-x \in \partial B_a^d(p)$, as desired.

Title: The using of Borsuk-Ulam theorem


问:
Recursive inclusion exclusion principle
The inclusion exclusion principle is as follows:

$P\Bigl(\bigcup_{i=1}^n E_i\Bigr) = \sum_{i\le n} P(E_i) - \sum_{i_1<i_2}P(E_{i_1}\cap E_{i_2})+\sum_{i_1<i_2<i_3}P(E_{i_1}\cap E_{i_2}\cap E_{i_3}) \dots$

is there any way to write this equation recursively?

答:
Yes

based on Reduced Recursive Inclusion-exclusion Principle for the probability of union events

let

$π(Y_1, Y_2,...,Y_n) ≡ Pr(Y_1∪Y_2∪···∪ Y_n)$

be the function of probability of union events. According to the equivalent rearrangement, we have the following definition of recursive inclusion-exclusion principle:

$\pi(\emptyset) = 0$

$π(A_1) = Pr(A_1)$, for $n = 1$

$π(A_1, A_2,...,A_n) = \sum_{i=1}^n{Pr(A_i)−π(A_1 ∩ A_i, A_2 ∩ A_i,...,A_{i−1} ∩ A_i)}$, for $n > 1$

where $π(•)$ is a recursive function

Title: Recursive inclusion exclusion principle


问:
Metric assuming the value infinity
If we instead define a metric as $d:X\times X \rightarrow [0,\infty]$, do we lose any nice properties of metric spaces?

The reason I ask is that I saw this theorem: Given a finite measure space $(X,\mathcal{M},\mu )$, define equivalence relation $A \sim B$ if $\mu(A \mathbin{\Delta} B)=0$ for $A,B\in \mathcal{M}$. Then $d: \mathcal{M} /{\sim} \rightarrow [0,\infty)  $ defined by $d(A,B)=\mu(A \mathbin{\Delta}  B)$ is a metric.

But the same argument would go through if $X$ wasn't a finite measure space, the only place where it seems to fail is because metrics are defined not to take the value $\infty$.

答:
Nothing bad happens if you allow a metric to assume the value $\infty $. In fact, only good things happen.

Nothing bad happens since you can still talk about open balls, the induced topology, uniformly continuous functions, continuous functions, Lipschitz functions etc. and the theory looks pretty much the same as it would if the metric is not allowed to assume the value $\infty $. Moreover, any metric space where infinite distance is assumed naturally breaks down as a disjoint union of metric spaces where infinity is not assumed. The construction is very easy: define an equivalence relation on the points of the space by $x\sim y$ if $d(x,y)<\infty $. The equivalence classes are called galaxies, the space breaks down as the disjoint union of galaxies and each galaxy is metric space where all distances are finite.

That good things happen if you allow infinite distance are related to certain constructions that become nicer. For instance, if you have two metric spaces $M_1,M2$ and you wish to define their disjoint union as a metric space, then the most natural thing to do is require that the distance between a point in $M_1$ and a point in $M_2$ is infinite. If you disallow infinite distances, then you have to sweat unnecessarily to make this construction work. But with infinite distances allowed, the disjoint union of metric spaces becomes the trivial notion it should be. Also, reconciling different metric spaces on the same set becomes more transparent. For instance, suppose that $d_i$ is a family of distance functions on a set $X$. Even if it does not attain infinity, their point-wise supremum might become infinite, and thus fail to be a metric space. But if you allow infinity as a possible distance, then the supremum of any family of distance functions is again a distance function.

The reason that metric spaces with infinity allowed have better closure properties under familiar operations is that the set $[0,\infty ]$ has better closure properties than $[0,\infty )$ does.

Title: Metric assuming the value infinity


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答: Fractals everywhere by Michael Barnsley</p>

Body: Fractals everywhere by Michael Barnsley


问:
Linearly independent over $F_p$ to $Q$
Let $v_1, . . . , v_m$ be vectors with $n$ $(0,1)$-entries. Prove that if these vectors are linearly independent over $\mathbb{F}_p$ for some prime number $p$ then they are linearly independent over $\mathbb{Q}$.

My attempt:

If we have a $n$ by $m$ matrix $A$ whose column vectors are $v_1, . . . , v_m$, and they are linearly independent over the field $\mathbb{F_p}$ where p is a prime number. That implies $det(A)$ mod $p \neq 0$.

Then in the field $\mathbb{Q}$, there is also $det(A)$ mod $p \neq 0$, so the column vectors are linearly independent over the field $\mathbb{Q}$.

I am not sure it is ok.

答: Suppose that the vectors are $\Bbb Q$-linearly dependent. Clearing denominators, obtain a non-trivial $\Bbb Z$-linear combination of these vectors that equals zero. Now consider the residues modulo $p$.

Title: Linearly independent over $F_p$ to $Q$


答:
$\displaystyle \frac {\left(\begin{matrix}20\\2,2,...2\end{matrix}\right)\left(\begin{matrix}20\\2,2,...2\end{matrix}\right)/(10!)^2}{\left(\begin{matrix}40\\2,2,...2\end{matrix}\right)/20!}$

This is another way to think about it. Number of ways to pair up 20 people squared divided by the number of ways to pair up 40 people.

Title: Doubt in First Course in Probability by Sheldon Ross

Body:
Here is the example that I was solving from the book on probability.

A football team consists of $20$ offensive and $20$ defensive players. The players are to be paired in groups of $2$ for the purpose of determining roommates. If the pairing is done at random, what is the probability that there are no offensive-defensive roommate pairs? What is the probability that there are $2i$ offensive-defensive roommate pairs, $i=1,2,\ldots,10$?

Solution There are
$$\binom{40}{2,2,\ldots,2}=\frac{(40)!}{(2!)^{20}}$$
ways of dividing the $40$ players into $20$ ordered pairs of two each. (That is, there are $(40)!/2^{20}$ ways of dividing the players into a first pair, a second pair, and so on.) Hence, there are $(40)!/2^{20}(20)!$ ways of dividing the players into (unordered) pairs of $2$ each. Furthermore, since a division will result in no offensive-defensive pairs if the offensive (and defensive) players are paired among themselves, it follows that there are $[(20)!/2^{10}(10)!]^2$ such divisions. Hence, the probability of no offensive-defensive roommate pairs, call it $P_0$, is given by
$$P_0=\frac{\left(\frac{(20)!}{2^{10}(10)!}\right)^2}{\frac{(40)!}{2^{20}(20)!}}=\frac{[(20)!]^3}{[(10)!]^2(40)!}.$$

What I don't understand is why is the solution considering unordered pair for the calculation? Aren't we choosing sets without any permutation of elements within them? What are we over-counting here? Can someone explain what am I missing?


问:
about noetherian topological space
Definition. A topological space $X$ is called noetherian if it satisfies the descending chain condition for closed subsets: for any sequence $Y_{1} \supseteq Y_{2} \supseteq \ldots$ of closed subsets, there is an integer $r$ such that $Y_{r}=Y_{r+1}=\ldots$.

Show that the following conditions are equivalent for a topological space $X$ :

(i) $X$ is noetherian;

(ii) every nonempty family of closed subsets has a minimal element;

(iii) $X$ satisfies the ascending chain condition for open subsets;

(iv) every nonempty family of open subsets has a maximal element.

proof : i can show $(i) \iff (iii)$ (because complement of open set is close and if $A \subset B \implies B^c \subset A^c $ ) and $(ii) \implies (i)$ :

Suppose $(ii)$ holds and let $\{F_n\}_{n \in \mathbb N}$ be a sequence of decreasing closed subsets. By hypothesis, there is $k \in \mathbb N$ such that $F_k \subset F_n$ for all $n$. But since the sequence is decreasing, we have $F_n \subset F_k$ for all $n \geq k$. From here it follows directly $X$ is noetherian.

I think we can use  Zorn's Lemma for $(iii) \implies (iv)$.

答:
Take a pick on how to show the equivalence, they're all quite straightforward:

(i) implies (ii): Suppose $X$ is Noetherian and let $\mathcal{F}$ be a non-empty family of closed sets that has no minimal element (so we strive for a contradiction now). Let $F_0 \in \mathcal{F}$ (it's non-empty!). By assumption it's not minimal in $\mathcal{F}$, so there exists $F_1 \in \mathcal{F}$ such that $F_1 \subsetneq F_0$. If we have constructed (for some $m \ge 1$) $F_m \in \mathcal{F}$ such that $F_m \subsetneq F_{m-1}$, then by the non-minimality again we can pick $F_{m+1} \in \mathcal{F}$ such that $F_{m +1} \subsetneq F_m$, so we can construct by recursion a sequence $(F_n)_{n\ge 0}$ from $\mathcal{F}$ that is strictly decreasing by inclusion, and this directly contradicts $X$ being Noetherian. There is no need for full Zorn here (just DC, dependent choice, if we care about how much of AC we use).

(i) is equivalent to (iii) because we can take complements and turn an increasing sequence of open sets into a decreasing one of closed sets and vice versa.

(iii) implies (iv) by the same recursion argument but replacing non-minimal by non-maximal etc.

(iv) implies (iii) is easy: suppose that $(G_n)_n$ is an increasing by inclusion family of open sets. By (iv) $\{G_n\mid n \in \Bbb N\}$ has a maximal element $G_m$. But then if $n \ge m$, $G_m \subseteq G_n$ by increasingness and equality must hold by maximality of $G_m$. So the sequence stabilises at $m$ and (iii) is shown.

(ii) implies (i): exactly like (iv) implies (iii) with minimal instead of maximal.

(ii) and (iv) are equivalent by the same complements argument (complements reverse inclusions so maximal and minimal elements are interchanged).

Title: about noetherian topological space


问:
Why don't both 'branches' of a cyclic integral of $\sin(x)\cdot\cosh(x)$ give me an answer?
I am trying to calculate  $$\mathcal{C}=\int\sin(x)\cdot\cosh(x)\;dx$$
I realized that depending on which variables i treat as the 'to be differentatied/integrated' one i get different results.

case a)

$$\mathcal{C}=\sin(x)\cdot\sinh(x)-\int\cos(x)\cdot\sinh(x)\;dx=$$
$$=\sin(x)\cdot\sinh(x)-\left[\cos(x)\cdot\cosh(x)-\int-\sin(x)\cdot\cosh(x)\;dx\right]$$
$$=\sin(x)\cdot\sinh(x)-\cos(x)\cdot\cosh(x)-\underbrace{\int\sin(x)\cdot\cosh(x)\;dx}_{\mathcal{C}} $$
$$\mathcal{C}=\dfrac{\sin(x)\cdot\sinh(x)-\cos(x)\cdot\cosh(x)}{2}$$
This answer checks out.

case b)

$$\mathcal{C}=\int\sin(x)\cdot\cosh(x)\;dx=\sin(x)\cdot\sinh(x)-\int\cos(x)\cdot\sinh(x)\;dx=$$
$$=\sin(x)\cdot\sinh(x)-\left[\sin(x)\cdot\sinh(x)-\underbrace{\int\sin(x)\cdot\cosh(x)\;dx}_{\mathcal{C}}\right]$$
$$=\sin(x)\cdot\sinh(x)-\sin(x)\cdot\sinh(x)+\underbrace{\int\sin(x)\cdot\cosh(x)\;dx}_{\mathcal{C}}$$
$$\mathcal{C}-\mathcal{C}=0$$
$$0=0$$

Here i just get an identity, as $\mathcal{C}$ falls out.

Did I make a mistake?

Can i obtain a solution of a cyclic integral in only one way? Will the other way always result in an identity?

Edit: I left out the arbitrary constants, please ignore that.

答: You didn't make a mistake. It's perfectly possible to do integration by parts and get back exactly the same integral. That just means your particular application of integration by parts won't help you. The method gave you a true fact (that an integral is equal to itself); it's just that the fact was useless.</p>


问:
Family size process
The question I'm posing arises in the so-called "family-size process" studied by D.Kendall. It is an example of an open migration process, but describing carefully this model will take too much time, so let me go directly to my problem.

For $j \in \mathbb{N}$ define $n_j \sim \mbox{Poisson}(\alpha_j)$, where $\displaystyle \alpha_j:=\frac{\nu}{\lambda j} \left (\frac{\lambda}{\mu} \right )^j$.
All these processes are independent of each other and each random variable $n_j$ can be thought as the number of families of size $j$.

Define also the following random variables:

$\displaystyle N:=\sum_j n_j$ is the total number of families;

$\displaystyle M:=\sum_j j n_j$ is the total number of individuals.

Could you help me in proving that $\displaystyle E(N|M)=\sum_{i=1}^M \frac{\rho}{\rho+i-1}$ where $\displaystyle \rho:=\frac{\nu}{\lambda}$? Thanks a lot!

答:
Let $\beta = \lambda/\mu$. Let $P(x,y)$ be the generating function where the coefficient of $x^N y^M$ is the probability to get $N,M$. For an individual random variable, it is well-known that the generating function is $\exp(\alpha_j(xy^j-1))$. So
$$ P = \exp\left( \sum_{j=1}^\infty \frac{\rho}{j} \beta^j (xy^j - 1) \right). $$
Let's evaluate the constant term:
$$ \sum_{j=1}^\infty \frac{\rho}{j} \beta^j = -\rho\log(1-\beta). $$
Substituting this, and using the Taylor expansion of log once again, we get
$$ P = (1-\beta)^\rho \exp\left( x \sum_{j=1}^\infty \frac{\rho}{j} (\beta y)^j \right) = (1-\beta)^\rho \exp(-x\rho\log(1-\beta y)). $$
In order to get expectation with respect to $x$, we differentiate with respect to $x$ and substitute $x = 1$. First,
$$ \frac{\partial}{\partial x} P = (1-\beta)^\rho \exp(-x\rho\log(1-\beta y)) (-\rho \log(1-\beta y)).$$
Now substitute $x=1$ to get a univariate generating function $Q(y)$:
$$ Q = \rho (1-\beta)^\rho (1-\beta y)^{-\rho} \sum_{j=1}^\infty \frac{\beta^j y^j}{j}. $$
We can expand $(1-\beta y)^{-\rho}$ into a Taylor series as well:
$$ Q = \rho (1-\beta)^\rho \sum_{i=0}^\infty \binom{i+\rho-1}{\rho-1} \beta^i y^i \sum_{j=1}^\infty \frac{\beta^j y^j}{j}. $$
We can read off the coefficient of $y^M$:
$$ \rho (1-\beta)^\rho \sum_{j=1}^M \binom{M-j+\rho-1}{\rho-1} \beta^{M-j} \frac{\beta^j}{j} = \rho (1-\beta)^\rho \beta^M \sum_{j=1}^M \frac{\binom{M-j+\rho-1}{\rho-1}}{j}. $$
That's as far as I got...</p>

Title: Family size process


问:
Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$
This problem was already solved here (in different closed form).

But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $

Where $\displaystyle \operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $ is the the trilogarithm.

答:
The following solution is by Cornel Valean:

\begin{gather*}
 \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\textrm{d}x\overset{x= \frac1y}{=}\int_0^\infty\frac{\ln^2\left(\frac{y}{1+y}\right)}{1+y^2}\textrm{d}y\\
 \overset{\frac{y}{1+y}=x}{=}\int_0^1\frac{\ln^2(x)}{x^2+(1-x)^2}\textrm{d}x\\
\left\{\text{write $\frac{1}{x^2+(1-x)^2}=\Im \frac{1+i}{1-(1+i)x}$}\right\}\\
 =\Im \int_0^1\frac{(1+i)\ln^2(x)}{1-(1+i)x}\textrm{d}x\\
 =2\ \Im\{\operatorname{Li}_{3}(1+i)\}.
 \end{gather*}


问:
Long equation problems
I am given this equation:

$$15x^6+31x^5-254x^4-352x^3+329x^2+9x-18=0$$

I know how to solve this question but the problem is it is too long to plug in numbers, what am I advised to do?

Title: Long equation problems

Body:
$$f(x)=15x^6+31x^5-254x^4-352x^3+329x^2+9x-18$$
$$f(x)=(x+2) (5 x-3) \left(x^2-4 x+1\right) \left(3 x^2+14 x+3\right)$$ after factoring the quartic given by @S. Dolan


问:
Is the set $\{(x, y, a)\in\mathbb{R}^3\mid x^2+y^2\le 4 \}$ convex?
Is the set $\{(x, y, a)\in\mathbb{R}^3\mid x^2+y^2\le 4 \}$, where $a\in\mathbb{R}$ is a fixed parameter, a convex set?

I know that the set $\{(x, y)\in\mathbb{R}^2\mid x^2+y^2\le 4 \}$ is convex (in $\mathbb{R}^2$), but I don't know if it holds for $\mathbb{R}^3$, when the $z$ coordinate is fixed, and if it's true, why.

答:
Of course it remains convex if we add irrelevant coordinates:

Suppose a convex set of $\Bbb R^2$ is defined by $C=\{(x,y)\mid \phi(x,y)\}$, where we have some condition $\phi$ (whose exact form is irrelevant).

Then defining $C' = \{(x,y,z)\mid \phi(x,y)\}$ is also convex:

let $\vec{x}=(x,y,z),\vec{u}=(u,v,w) \in C'$ and $t \in [0,1]$. So $(x,y) \in C$ and $(u,v) \in C$ by definition so $(tx +(1-t)u, ty+(1-t)v) \in C$ as $C$ is convex.

So it follows that $(tx +(1-t)u, ty+(1-t)v, p) \in C'$ whatever $p$ is, so in particular, $t\vec{u} + (1-t)\vec{v} = (tx +(1-t)u, ty+(1-t)v, tz + (1-t)w) \in C'$ as well, and $C'$ is convex.

Or alternatively, appeal to a theorem that a product of convex sets is convex (we multiply $C$ by $\Bbb R$ here).

If the third coordinate is fixed, do the same, $\{a\}$ is also convex, and the first proof can be easily adapted too.


问:
Constructing an explicit map for the double cover of the Klein Bottle by a Torus
I know it's been asked before, but I wanted to know if my answer is correct (as I haven't been able to verify it).

I visualize $P^2$ # $P^2$ as an upper hemisphere (with its boundary equator appropriately identified) connected through a small tube with a lower hemisphere (also with its boundary equator identified); basically, I visualized it explicitly as sort of a sandglass.

Now define $q$: $S^1\times S^1\mapsto $ $P^2\#P^2$ as follows:

Before defining the map partition, the upper half of the circle as follows: begins with a point at $(1,0)$, then let $U_1$ be an open arc in the first quadrant $($the one with $x,y > 0)$ starting at $(1,0)$ but not including it; then let $U_2$ be a closed arc that starts at the first quadrant (it starts and includes the point where $U_1$ ends) and continues to the second quadrant. And then let $U_3$ be an open arc, which starts where $U_2$ ends and ends at the point $(-1,0)$ (since it's an open arc, it doesn't contain the endpoints).

Now for the map:

$(1,0) \times S^1$ is sent to the boundary of the lower hemisphere (since the boundary has antipodal points identified, this will doubly cover it).

$U_1 \times S^1$ is sent to the lower hemisphere (minus the part that occupies the tube, of course).

$U_2 \times S^1$ is sent to the tube

$U_3 \times S^1$ is sent to the upper hemisphere.

$(-1,0) \times S^1$ is sent to the boundary of the upper hemisphere

Then repeating the same process to the low part of the circle covers the upper/low hemispheres and the tube again without covering the boundaries again.

Is this a correct map for the double cover of the Klein Bottle? This covers the Klein Bottle twice. Continuity follows via the gluing lemma. Am I missing something?

答: Consider the two-fold covering from Torus $\Bbb S^1\times \Bbb S^1$ to Klein-bottle$=\frac{[0,1]\times [0,1]}{(x,0)\sim(x,1)\text{ and }(0,y)\sim (1,1-y)}$ given by $$\big(e^{2\pi ix},e^{2\pi iy}\big)\longmapsto\begin{cases} [2x,y]_K&\text{ if }0\leq x\leq \frac{1}{2},\\ [2x-1,1-y]_K &\text{ if }\frac{1}{2}\leq  x\leq 1.\end{cases}$$


问:
Proof verification: if a set $K$ is compact, both $\sup K$ and $\inf K$ exist and are contained in $K$
Exercise 3.3.1 in Stephen Abbott's book poses the following question.

Show that if $K$ is compact and non-empty, then $\sup K$ and $\inf K$ both exist and are elements of $K$.

I attempted to write a simple proof, and would like someone to verify, if the proof is technically correct and rigorous.

My Attempt.

Let $K$ be compact and non-empty. By the theorem on the characterization of compactness, $K$ is closed and bounded. Since $K$ is a bounded subset of $\mathbf{R}$, by AoC, $K$ has a supremum and an infimum. Let $s = \sup K$.

There is a trichotomy. Either (i) $s \in int(K)$ (ii) $s \in \partial K$ (iii) $s \in ext(K)$.

(i) Suppose $s \in int(K)$. Then, there exists an $\epsilon$ such that the open interval $(s - \epsilon,s + \epsilon)$ is contained in $K$. Consider $s + \epsilon/2 \in (s, s+\epsilon) \cap K$. $s + \epsilon/2 > s$. But, this implies that $s$ is not an upper bound for $K$, which is a contradiction. Hence, $s \notin int(K)$.

(ii) Suppose $s \in ext(K)$. Then, there exists an $\epsilon > 0$ such that the open interval $(s - \epsilon,s + \epsilon)$ is contained in $K^C$. Consider $s - \epsilon/2 \in K \cap (s - \epsilon,s)$. $s - \epsilon/2 \ge x$ for all $x \in K$, and $s - \epsilon/2 < s$. So, $s$ is not an upper bound for $K$, which is a contradiction. Hence, $s \notin ext(K)$.

(iii) If $s \in \partial K$, because $K$ is a closed set, $\partial K \subseteq K$. So, $s \in K$.

A similar argument can be made for $\inf K$.

This closes the proof.

答:
Everything's fine except #2.  Just because it's not in the set doesn't mean it isn't an upper bound; indeed, if $s > k$ for every $k \in K$ then it's still an upper bound.  Also, it may be that $s$ is between components of $K$, or it could be less than every element of $K$ . . . being in the exterior of $K$ doesn't give you any info about if it's an upper/lower bound or neither.

The argument should look more like "$K$ is compact so closed, so ext($K$) is open, so $s$ is contained in an interval that doesn't intersect $K$."  Then if $s$ is supposed to be the sup, you should have elements (not necessarily distinct, as per your comment) of $K$ converging to $s$ from below.  But that's impossible, since the interval you constructed doesn't intersect $K$.

Make sense?


问:
Motivating the terminology of weaker vs stronger topology
Let $X$ be a set, $\tau_1,\tau_2$ two topologies on $X$, and consider the following statements

$\tau_1\subseteq \tau_2$ (i.e $\tau_1$ is coarser/weaker than $\tau_2$, or that $\tau_2$ is finer/stronger than $\tau_1$)

For every net $\langle x_i\rangle_{i\in I}$ in $X$ and $x\in X$, if $x_i\to x$ relative to the topology $\tau_2$ then $x_i\to x$ relative to the topology $\tau_1$.

It is clear to me that $(1)$ implies $(2)$. My question is whether the converse is also true, because then this would seem like a nice justification for the terminology "weak/strong" topology in the context of topological vector spaces.

答: Suppose (2) holds. Let $O \in \tau_1$. Then $X\setminus O$ is closed in $\tau_1$, and it's also closed in $\tau_2$: let $x$ be in the closure (w.r.t $\tau_2$) of $X\setminus O$, so there is a net $(x_i)_i$ of points from $X\setminus O$ such that $x_i \to x$ in $\tau_2$. By $(2)$ we also have that $x_i \to x$ in $\tau_1$, so that $x$ lies in $X\setminus O$ as that set is already closed in $\tau_1$. This shows that $X\setminus O$ is also closed under $\tau_2$ so $O \in \tau_2$ and we're done.


问:
Deriving the spherical law of sines using Clairaut's relation (without the law of cosines)
I'm studying geodesics in my differential geometry course. We derived and proved Clairaut's relation (see also Shifrin P73.), which states that for a surface of revolution, the geodesics satisfy the equation
$$r\cos\phi=\text{const}$$
where $r$ is the distance from the axis of revolution and $\phi$ is the angle between the geodesic and the parallel.

We have not done any spherical trig. so far, in particular, we haven't been introduced to the law of cosines.

However, following the section on Clairaut, an exercise asks for a proof of the spherical law of sines
$$\frac{\sin\alpha}{\sin a}=\frac{\sin\beta}{\sin b}=\frac{\sin\gamma}{\sin c}.$$
The hint is to put one corner of the triangle on a pole, then use Clairaut's relation. If we put one vertex on the pole, then two edges will be longitudinal, and thus parallels will intersect those edges at right angles. Right triangles are much easier to handle, but the parallels "cut off" the angle from its opposing side. Is something here perhaps?

As most sine-law proofs I've seen use the spherical law of cosines, I thought maybe I should prove that using the hint, then derive the sine-law from that. But then I might as well go through the standard derivation of the law of cosines, which is a.s. not the point of the exercise.

Is it possible to use Clairaut's relation to show the spherical law of sines more directly? If not, is it possible to use it to derive the cosine-law?

答:
Following the hint and @TedShifrin's comment, place $\alpha$ on the north pole, then we may apply Clairaut on the side $a$, which is part of a geodesic.
In particular, we may consider the intersections of $a$ with the parallels at the corners $\beta,\gamma$. Let $\theta_\beta, \theta_\gamma$ denote these angles and $r_\beta, r_\gamma$ denote the distance from the axis, we know
$$r_\beta\cos\theta_\beta=r_\gamma\cos\theta_\gamma$$

Wlog we may assume that $\beta\le\gamma$, so because parallels intersect longitudinal geodesics at right angles, we have
$$\frac\pi2=\theta_\beta+\beta=\theta_\gamma-\gamma$$
so $\cos\theta_\beta=\sin\beta, \cos\theta_\gamma=-\sin\gamma$.
Furthermore, using some elementary properties of circle-sections with the lengths $\lvert b\rvert, \lvert c\rvert$, we must have $r_\beta=\sin\lvert c\rvert, r_\gamma=\sin\lvert b\rvert$. The claim then follows when we release the orientation of angles and plug these results into the first equation.


问:
Lebesgue measure/Measurable sets
Question :

Let $f,g$ be measurable real valued functions on $\mathbb{R}$ such that  :

$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx$$

Let $E=\{x\in \mathbb{R} : f(x)\neq g(x)\}$ . Which of the followng statements are necessarily true?

$E$ is empty set

$E$ is measurable

$E$ has lebesgue measure $0$

For almost all $x\in \mathbb{R}$ we have $f(x)=0$ and $g(x)=0$

Explanation: What all I could see is that second bullet and third bullet are probably correct. Because :
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx$$
i.e.,
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx-2\int_{-\infty}^{\infty} f(x)g(x)dx=0$$
i.e.,
$$\int_{-\infty}^{\infty}(f(x)-g(x))^2dx=0$$

Though I have negative limits my function $(f(x)-g(x))^2$ is positive

So, I would see that $E=\{x\in \mathbb{R} : f(x)\neq g(x)\}$ is measurable and has measure $0$

Please tell me if what I have done is sufficient/clear.

答:
OP's explanation for second and third bullets are no-doubt correct but by giving a counter-example, we shall also have to explain why first and fourth bullets are false. Here I am going to provide it.

For the first bullet, let $~f(x)=5,~~g(x)=\begin{cases}
      5 &\text{if}~~~~ x\ne 0 \\
      1 &\text{if}~~~~ x= 0
         \end{cases}$

Clearly $~\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx~$ and $~0\in E\implies E=\phi~.$
hence first bullet is not correct.

For the fourth bullet, from the above example, as neither $~f(x)=0~$ nor $~g(x)=0~,$  we can conclude that the fourth bullet is also not correct.

Note: I know that I am attempting to answer an old question. But I think my answer will also useful for solving the problem to the future reader. So if you downvote my answer, please tell me, where I am wrong. Thank you.

Title: Lebesgue measure/Measurable sets


问:
Solve $\lim\limits_{x\to\infty}{\frac{1}{\sqrt{x^2-1}}\int_{0}^{x}{{\left(\arctan{t}\right)}^{2}\mathrm{d}t}}$
I have to solve the following limit: $\lim\limits_{x\to\infty}{\frac{1}{\sqrt{x^2-1}}\int_{0}^{x}{{\left(\arctan{t}\right)}^{2}\mathrm{d}t}}$. The problem for me is the definite integral. I tried $t=\tan{p}$ from where I get $\int_{0}^{x}{{p^{2}\mathrm{d}\tan{p}}}$ and then I integrated by parts but that didn't work. I also considered somehow inserting $\frac{1}{\sqrt{x^2-1}}$ inside the integral but I think that would complicate things even more. How can I solve the limit?

答:
You can apply l'Hôpital to any form $?/\infty$, where the question mark means any function that's differentiable where needed. On the other hand, it's not difficult to show that
$$
\lim_{x\to\infty}\int_0^x(\arctan t)^2\,dt=\infty
$$
so you can apply the standard form.

The fundamental theorem of calculus then allows you to compute
$$
\lim_{x\to\infty}\frac{(\arctan x)^2}{x/\sqrt{x^2-1}}
$$
and this is not indeterminate.


问:
Derivative of primitive function in high dimension
Suppose $x \in \mathbb{R}^n$, and $f : \mathbb{R} \rightarrow \mathbb{R}$, $g :\mathbb{R}^n \rightarrow \mathbb{R}$ are two functions. $F(u)$ is the primitive function of $f$, $F(u) = \int_0^u f(v) dv$.

I want to compute the gradient of $F(g(x))$. I know that I should use the chain rule, but I couldn't understand chain rule in higher dimensions.

答: The answer is $$(F'(g(x))\frac {\partial g} {\partial x_1},F'(g(x))\frac {\partial g} {\partial x_2},..., F'(g(x))\frac {\partial g} {\partial x_n})$$ $$=(f(g(x))\frac {\partial g} {\partial x_1},f(g(x))\frac {\partial g} {\partial x_2},..., f(g(x))\frac {\partial g} {\partial x_n})$$


答: Let $h(x)=F(g(x))$. The multivariate chain rule states $dh=\frac{\partial h}{\partial dx^i}dx^i$, with implicit summation over repeated indices. By the univariate chain rule, $\frac{\partial h}{\partial x^i}=f(g(x))\frac{\partial g}{\partial x^i}$, so $dh=f(g(x))\frac{\partial g}{\partial x^i}dx^i$. In particular, $\nabla h$ is the vector whose $i$th component is the $dx^i$ coefficient, so $\nabla h=f(g)\nabla g$, in agreement with @KaviRamaMurthy's answer.

Title: Derivative of primitive function in high dimension

Body:
Suppose $x \in \mathbb{R}^n$, and $f : \mathbb{R} \rightarrow \mathbb{R}$, $g :\mathbb{R}^n \rightarrow \mathbb{R}$ are two functions. $F(u)$ is the primitive function of $f$, $F(u) = \int_0^u f(v) dv$.

I want to compute the gradient of $F(g(x))$. I know that I should use the chain rule, but I couldn't understand chain rule in higher dimensions.


问:
Finding the ratios of triangle sides
I was asked to draw find the angles of a triangle with angles with the ratio of 10:7 and 7:3.

Multiplying all these numbers by nine gives angles of $90,63,$ and $27$. How can I now find the ratio of the sides?

This is my work: $$\frac{a}{\sin{\left(90\right)}}=\frac{b}{\sin{\left(63\right)}}=\frac{c}{\sin{\left(27\right)}}\\
 a=\frac{b}{.891}=\frac{c}{.45399}$$
Ratio of the sides: $$1:0.891 \text{ and } 0.891:0.45399$$

Does this make sense?

Title: Finding the ratios of triangle sides

Body:
You logic is sound because one Pythagorean triple that fits your picture is
$$(589,300,661)$$

where $\sin^{-1}\big(\frac{300}{661}\big)\approx 0.471089961 r\approx 26.99146656^\circ$

Side ratios:
$$A/C= 589/661\approx 0.89107413$$
$$B/C= 300/661\approx 0.453857791$$
$$A/B=589/300\approx 1.963333333$$


答:
You do not need to compute the integral. By applying de l'Hôpital's rule you find
\begin{align*}
\lim_{x \to \infty} \frac{1}{\sqrt{x^2-1}}\int_0^x (\arctan t)^2dt & = \lim_{x \to \infty} \frac{\arctan^2 x}{x(x^2-1)^{-1/2}} \\
& = \lim_{x \to \infty} \frac{|x|\left(1-\frac{1}{x^2}\right)^{1/2}}{x} \arctan^2 x \\
& = \pm \frac{\pi^2}{4}
\end{align*}
and the sign depends on wether you are taking the limit for $x \to \pm \infty$.

Title: Solve $\lim\limits_{x\to\infty}{\frac{1}{\sqrt{x^2-1}}\int_{0}^{x}{{\left(\arctan{t}\right)}^{2}\mathrm{d}t}}$

Body: I have to solve the following limit: $\lim\limits_{x\to\infty}{\frac{1}{\sqrt{x^2-1}}\int_{0}^{x}{{\left(\arctan{t}\right)}^{2}\mathrm{d}t}}$. The problem for me is the definite integral. I tried $t=\tan{p}$ from where I get $\int_{0}^{x}{{p^{2}\mathrm{d}\tan{p}}}$ and then I integrated by parts but that didn't work. I also considered somehow inserting $\frac{1}{\sqrt{x^2-1}}$ inside the integral but I think that would complicate things even more. How can I solve the limit?


问:
Show that $N=g_1 + g_2 + \cdots g_n$ is in the center of the group ring $RG$
There is a similar question here which I can understand but there is no mention of conjugacy class in my question so I'm somewhat confused.

Let $G=\{g_1, g_2, ..., g_n \}$ be a finite group, show that $N= g_1 + g_2 + \cdots +g_n$ is in the center of the group ring $RG$.

I tried multiplying out $(a_1g_1 + a_2g_2 + \cdots + a_ng_n)(g_1 + g_2 + \cdots + g_n)$ and $(g_1 + g_2 + \cdots + g_n)(a_1g_1 + a_2g_2 + \cdots + a_ng_n)$ and checking for equality.

Does it work out because $(a_1g_1 + a_2g_2 + \cdots + a_ng_n)(g_1 + g_2 + \cdots + g_n)= (a_1g_1)(g_1 + g_2 + \cdots + g_n)+(a_2g_2)(g_1 + g_2 + \cdots + g_n)+\cdots +(a_ng_n)(g_1 + g_2 + \cdots + g_n) = (g_1 + g_2 + \cdots + g_n)(a_1g_1)+(g_1 + g_2 + \cdots + g_n)(a_2g_2)+\cdots +(g_1 + g_2 + \cdots + g_n)(a_ng_n)= (g_1 + g_2 + \cdots + g_n)(a_1g_1 + a_2g_2 + \cdots + a_ng_n)$

where by definition $(ag_i)(bg_j)=(ab)(g_ig_j)$ and $g_i(g_1 + g_2 + \cdots + g_n)=(g_1 + g_2 + \cdots + g_n)g_i$, multiplying all the elements of $G$ on the left or right by the same element is a permutation of the elements of $G$ thus the sum is unchanged?

答:
The proof you refer to applies here in just the same way.

For any element $g$ of the group, consider $g^{-1}Ng=\sum g^{-1}g_ig$. The elements $g^{-1}g_ig$ are (in some order) precisely all the elements of the group and so $g^{-1}Ng=N$.

N.B. Your comment at the end of your proof is correct. This works because of the 'Latin Square' property of groups.


问:
Is second derivative of a convex function convex?
If $f$ is twice differentiable and convex, is it true that $f''$ is a convex function ?</p>

答:
Could you name a function that has continuous first and second derivative functions, has $ \ f'' > 0 \ $ everywhere, but for which $ \ f'' \ $ itself has zero curvature?  How about $ \ f(x) = x^2 \ $ ?

EDIT: It seems pretty hard to find (at least) an elementary function that is convex everywhere, for which the second derivative function is concave everywhere.  If you can be satisfied with just an interval, $ \ f(x) = -\arctan x \ $ for $ \ x > 0 \ $ is convex everywhere on the interval, but $ \ f''(x) \ $ is not.


问:
Fundamental group of a sphere with straight line joining the north and south poles
This problem comes from the book Topology and Geometry of Bredon :

Let $X$ be the union of the unit sphere in 3-space with the straight line segment from the north pole to the south pole. Find $\pi_1(X)$.

I am reading Atcher's book at the same time. He gives a CW-complex structure to this space, and since the line is contractible, he identifies this space to $S^2 \vee S^1$ from which we can compute the fundamental group using Seifert-van Kampen's theorem and we get
\begin{equation*}
\pi_1(X) \approx \mathbb{Z}.
\end{equation*}

Nevertheless, one should be able to use Seifert-van Kampen's theorem directly to this space without relying on some CW-complex structure as it is not yet introduced in the book of Bredon at this stage.
I have tried the following :

Consider the open cover $\{U, V\}$ where $U$ is a thickened neighborhood of the union of the north hemisphere and the middle-top part of the straight line, and where $V$ is a thickened neighborhood of the union of the south hemisphere and the middle-bottom part of the straight line. But, the way I understand it, they are both contractible (they contract to the north and south pole respectively), and thus the fundamental group vanishes by Seifert-van Kampen's theorem.

Where is the error in this argument?

答:
Consider $U$ and $V$ to be semispheres, i.e. $U$ goes from the north pole to a bit past the equator, and $V$ from a bit above the equator to the south pole. But don't imagine the line to be perpendicular to the equator plane but rather contained in it. Or if you will, construct $U$ and $V$ relative to Greenwich's meridian :)

If you stare at $U$ and $V$ for a while, they are (homotopic to) disks with a cord attached to two points of the boundary. Collapsing the disks, one gets moreover a homotopy equivalence with $S^1$ in both cases.

The intersection is now a band along the equation together with the line. Via a deformation retraction, this is homotopy equivalent to just the equator - i.e. $S^1$ - and the line, which is like a figure $8$ (collapse the line).

Hence by van Kampen we have $\pi_1(X) = \pi_1(U) \ast \pi_1(V) / \langle \langle  \iota_1(\omega)\iota_2(\omega)^{-1} : \omega \in \pi_1(U \cap V)\rangle\rangle$, and $\pi_1(U) = \pi_1(V) = \Bbb Z$.

The intersection $U \cap V$, as we have said, is homotopy equivalent to a figure $8$, whose fundamental group is the free group on two generators, one for each loop. Chasing the homotopies, we obtain that the generators for $\pi_1(U \cap V)$ correspond to going through the line and then to one half of each semicircle of the equator. If $a$ and $b$ are the generators of $\pi_1(U)$ and $\pi_1(V)$, then these loops correspond to $a$ and $b$ so the relation introduced is $a = b$.

Finally, this gives $\pi_1(X) = \langle a,b : a = b\rangle \simeq \Bbb Z$, as desired. Maybe a different choice of orientations changes the presentation a little bit (e.g. the relation could be $ab = 1$, which still yields the correct group).


问:
Is it true that for every connected matroid $M$ we have $\chi(M)=\chi(M^*)$?
If for any connected matroid $M$ on $E$ we let $\chi(M)$ be the minimum size of any partition of $E$ into independent sets of $M$, then is $\chi(M)=\chi(M^*)$? (where note $M^*$ is the dual of $M$ and $|E|>2$)

答:
This is not true. Consider a cycle graph $C_n$ and the corresponding matroid $M$. Then $\chi(M) = 2$. The dual is the matroid corresponding to the graph with two vertices and $n$ parallel edges between these vertices. It follows that $\chi(M^*) = n$.

More generally, if $M = U^r_n$ is a uniform matroid of rank $r$, then $\chi(M) = \lceil n / r\rceil$, while for $M^* = U_n^{n-r}$ we have $\chi(M^*) = \lceil n / (n-r) \rceil$, and these are not equal in general.


问:
Why $sgn(\sigma)=(-1)^{j-1} sgn(\tau)$?
Problem:

Suppose that $\sigma \in \mathcal{S}_k=\{f:\{1,...,k\} \to \{1,...,k\} \mbox{ is bijective} \}$. We know that $\sigma$ has $sgn(\sigma) \in \{-1,1\}$. Suppose that $\sigma(1)=j$.

Let us define $\tau \in \mathcal{S}_{k-1}$ in this way:

$$\tau(i)=
\begin{cases}
\sigma(i+1) \mbox{ if } \sigma(i+1)<j \\ \sigma(i+1)-1 \mbox{ if } \sigma(i+1)>j
\end{cases}  \mbox{      for all }i \in \{1,...,k-1\}$$

Why can we conclude $sgn(\sigma)=(-1)^{j-1}sgn(\tau)$?

答:
Let $\xi$ be permutation on $\{1,2,\ldots,k\}$ that leaves $\{1,2,\ldots,j-1\}$ unchanged and moves $\{j,j+1,\ldots,k\}$ to $\{k,j,j+1,\ldots,k-1\}$. Then $\xi \circ \sigma$ is basically $\tau$ with $k$ inserted to the beginning – so composing it with $k$-cycle we will get $\tau$ with additional $k$ at the end, which has the same sign as $\tau$. Then, since sign of $t$-cycle is $(-1)^{t-1}$,

$$(-1)^{k-1} \cdot (-1)^{k-j} \cdot sgn(\sigma) = sgn(\tau),$$

which is what we wanted to prove.

Title: Why $sgn(\sigma)=(-1)^{j-1} sgn(\tau)$?


答: For each value of $a \in \mathbb R$ you get a disc in the plane $\pi_a = \{(x,y,a): x,y \in \mathbb R\}$, and each disc is convex.

Title: Is the set $\{(x, y, a)\in\mathbb{R}^3\mid x^2+y^2\le 4 \}$ convex?

Body:
Is the set $\{(x, y, a)\in\mathbb{R}^3\mid x^2+y^2\le 4 \}$, where $a\in\mathbb{R}$ is a fixed parameter, a convex set?

I know that the set $\{(x, y)\in\mathbb{R}^2\mid x^2+y^2\le 4 \}$ is convex (in $\mathbb{R}^2$), but I don't know if it holds for $\mathbb{R}^3$, when the $z$ coordinate is fixed, and if it's true, why.


问:
Show that the derivative of an even function is an odd function and vice versa using the chain rule
Use the Chain Rule to prove the following.

(a) The derivative of an even function is an odd function.

(b) The derivative of an odd function is an even function.

My attempt:

I can easily prove these using the definition of a derivative, but I'm having trouble showing them using the chain rule.

(a) $f(x)$ is even $ \therefore f(-x) = f(x)$.

We need to show that $f'(-x) = -f'(x)$.

Let $u = -x$.

My reasoning for the next step is that if we want to find the derivative of $f$ with respect to $x$ from $f(u)$ we need to use the chain rule and first find the derivative of $f$ with respect to $u$ and multiply that with the derivative of $u$ with respect to $x$.

$f'(x) = f'(u) \cdot u' = - f'(u) = -f'(-x)$.

$f'(x) = -f'(-x)$.

$f'(-x) = -f'(x)$.

The problem with this is that I never used the fact that $f$ is even and I feel like there is a mistake in my approach, most likely where I explained my reasoning. I just can't see it. Can you please point out the mistake and point me to the right direction?

Thanks!

答:
$f$ being even means for every $x\in\Bbb{R}$, $f(x)=f(-x)$. Or, if we define the function $u:\Bbb{R}\to\Bbb{R}$ as $u(x)=-x$, then the condition for being an even function is that $f=f\circ u$. So, actually, you did use the evenness assumption when you started calculating derivatives, even if you didn't explain it very explicitly.

Now, the chain rule tells us that $f'=(f\circ u)' = (f'\circ u)\cdot u'$. Explicitly, this says for every $x\in\Bbb{R}$,
\begin{align}
f'(x)&= f'(u(x))\cdot u'(x)=f'(-x)\cdot (-1) = -f'(-x).
\end{align}
But this is precisely what it means for $f'$ to be odd. The other question is similar.


答:
Actually, you are using the assumption that $f$ is even.

If $f$ is even, then as you say $f(-x) = f(x)$. By differentiating, the left hand side is $-f'(-x)$ by the chain rule, and the right hand side is simply $f'(x)$. Thus $-f'(-x) = f'(x)$, or $f'(-x) = -f'(x)$, meaning that $f'$ is odd. You can try with $f$ odd now.

Title: Show that the derivative of an even function is an odd function and vice versa using the chain rule

Body:
Use the Chain Rule to prove the following.

(a) The derivative of an even function is an odd function.

(b) The derivative of an odd function is an even function.

My attempt:

I can easily prove these using the definition of a derivative, but I'm having trouble showing them using the chain rule.

(a) $f(x)$ is even $ \therefore f(-x) = f(x)$.

We need to show that $f'(-x) = -f'(x)$.

Let $u = -x$.

My reasoning for the next step is that if we want to find the derivative of $f$ with respect to $x$ from $f(u)$ we need to use the chain rule and first find the derivative of $f$ with respect to $u$ and multiply that with the derivative of $u$ with respect to $x$.

$f'(x) = f'(u) \cdot u' = - f'(u) = -f'(-x)$.

$f'(x) = -f'(-x)$.

$f'(-x) = -f'(x)$.

The problem with this is that I never used the fact that $f$ is even and I feel like there is a mistake in my approach, most likely where I explained my reasoning. I just can't see it. Can you please point out the mistake and point me to the right direction?

Thanks!


问:
Evaluate $\int x \arcsin(x)dx$.

Evaluate:
$$\int x \arcsin(x)dx$$

My first guess was $u$ substitution but that didn't get me very far. I think using integration by parts is the correct way. Here's my attempt:

$$u = \arcsin(x), v' = x \Longrightarrow \int x \arcsin(x)dx = \arcsin(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2\sqrt{1-x^2}}dx$$

But I am stuck on how to continue from here.

答:
You are on the right track. As regards the remaining integral, note that
$$\int \frac{x^2}{2\sqrt{1-x^2}}\,dx=\frac{1}{2}\int \frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx
=\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}\,dx-\frac{1}{2}\int \sqrt{1-x^2}\,dx.$$
The first integral is easy, whereas for the second one you may use the substitution $x=\sin(t)$.

Can you take it from here?

Title: Evaluate $\int x \arcsin(x)dx$.


问:
If the limit of $f$ and $g$ is different from zero, can the limit of $fg$ be zero?
Suppose:
$$\lim_{x\to\infty}f(x)g(x)=0,$$
$$\lim_{x\to\infty}g(x)\ne0.$$

Is it has to be that
$$\lim_{x\to\infty}f(x)=0?$$

(Assume that the limits of $f$ and $g$ does not have to exist).

答:
You first write that $\lim g(x)\ne 0$. In that case,
$$ \lim f(x)=\lim\frac{f(x)g(x)}{g(x)}=\frac{\lim f(x)g(x)}{\lim g(x)}=0.$$
Later you write about a more general case, namely that the limit of $g(x)$ may not even exist. In that case, the limit of $f(x)$ need not exist either. As a simple example, consider
$$ f(x)=\begin{cases}1&x\in\Bbb Q\\0&x\notin\Bbb Q\end{cases}\qquad g(x)=1-f(x)$$


答: Since $(-x)^\prime=-1$, we can concisely prove both parts viz.$$f(-x)=\pm f(x)\implies f^\prime(-x)=-1\cdot\pm f^\prime(x)=\mp f^\prime(x).$$


问:
Prove or disprove: If $A\in \mathbb{R}^{m\times n}$ is a matrix and it's columns are basis for $\mathbb{R}^m$..
then it's rows are basis for $\mathbb{R}^n$.

I encountered this question and I answered wrong, and here's what I did:

I took a counter example: $A = \begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}_{2\times 3}$.

the columns $\{(1,0),(0,1)\}$, and they're a basis for $\mathbb{R}^2$, but the rows aren't a basis for $\mathbb{R}^3$.

Now I can't understand what's wrong with my counter example and why this is true.

I would appreciate it if someone can point out my mistakes and why my example doesn't work, and why the sentence is true.
 Thanks in advance.

答:
The linearly independent columns are those columns with leading $1$'s when the matrix is brought into $RREF$. Thus if the columns form a basis for $\mathbb{R}^m$, then necessarily there must be $n$ vectors, and so $m = n$.

For $m=n$ then the $RREF$ of $A$ is $I_{n\times n}$, the columns and rows are a basis for $\mathbb{R}^n$.


问:
Pullback of a function
I take the following example from the book of Introduction to Manifolds by Loring Tu page 198 example 17.16

Let $h : \mathbb{R} \rightarrow S^1 \subset \mathbb{R}^2$ be given by $h(t) = (x,y)=(\cos t,\sin t)$.

Why this is true: $h^*x = \cos t $ and $h^*y = \sin t $ ?

I'm a little confused, I tried to apply the definition of the pullback of a function but this doesn't give me the answer in the book!

答:
The following definition is from the book you mention but I couldn't find your example there.

If $\omega$ is a 1 -form on $M,$ we define its pullback $F^{*} \omega$ to be the 1 -form on $N$ given by
$$
\left(F^{*} \omega\right)_{p}\left(X_{p}\right)=\omega_{F(p)}\left(F_{*} X_{p}\right)
$$
for any $p \in N$ and $X_{p} \in T_{p} N$

You need to be certain of your definitions and objects you are working with (in this case your $x$ and $y$, pullback). An example calculation:

Consider the one-forms $\alpha_{(x_1,x_2)}=x_1dx_1+0dx_2$ and $\beta_{(x_1,x_2)}=0dx_1+x_2dx_2$. Let $h: \mathbb{R} \rightarrow \mathbb{R}^2$ be given by $h(t)=(\cos(t),\sin(t))$. Consider a general vector field $X$ on $\mathbb{R}$, $X=X^1 \frac{\partial}{\partial t}$. Jacobian matrix of $h_*$ is $(h_*)_{(t)}= \big(\begin{smallmatrix} -\sin(t) \\ \cos(t)  \end{smallmatrix}\big).$ Hence  $(h_*)_{(t)}X = -\sin(t)X^1\frac{\partial}{\partial x_1}+\cos(t)X^1\frac{\partial}{\partial x_2}.$

So $$\alpha_{h(t)}(h_*X)=\alpha_{(\cos(t),\sin(t))}(h_*X)=-\cos(t)\sin(t)X^1$$
$$\beta_{h(t)}(h_*X)=\beta_{(\cos(t),\sin(t))}(h_*X)=\sin(t)\cos(t)X^1$$
by comparing coefficients $$h^*\alpha = -\cos(t)\sin(t)dt \text{ and } h^*\beta = \sin(t)\cos(t)dt.$$

Title: Pullback of a function


问:
Intersection of all Annihilators of all nonzero elements of a simple R-module.
Need complete proof of the following result.

The intersection of all Annihilators of all nonzero elements of a simple $R$-module $M$ is a two sided ideal of $R$.

[Hint. $Ann(M)$ is a two sided ideal of $R$ and $\bigcap_{0_{M}\neq x\in M}Ann(x)=Ann(M)$.]

答:
We know that $a \in \bigcap Ann(x)$ iff $ax=0$ for any $x$. We just need to show, that

if $a,\ b \in \bigcap Ann(x)$, then $a+b \in \bigcap Ann(x)$ – then for any $x$ we have $(a+b)x=ax+bx=0+0=0$;

if $a \in \bigcap Ann(x)$ and $b \in R$, then for any $x$ we have $(ba)x=b(ax)=b \cdot 0 = 0$ and $(ab)x=a(bx)=0$ (as $bx$ is some element of $M$).

Therefore, $\bigcap Ann(x)$ is indeed two-sided ideal.


答:
You did use that $f$ is even. Let us look at the following more general situation that
$$ f = v \circ f \circ u .$$
For an even function we have $u(x) = -x$ and $v(x) = x$, for an odd function $u(x) = v (x) = -x$. Then
$$f'(x) = v'(f(u(x)))f'(u(x))u'(x) .$$
For $u(x) = -x$ and $v(x)  = (-1)^\epsilon x$ we get
$$f'(x) = (-1)^{\epsilon +1}f'(-x).$$


答:
$f(x) = -\sin x$ in $(0,\pi/2)$ is a counterexample: $f''(x) = \sin x > 0 $for all $x \in (0,\pi/2)$ while $f''''(x) = - \sin x < 0$ for all $x \in (0,\pi/2)$

this should probably be the easiest non trivial example :)

Title: Is second derivative of a convex function convex?

Body: If $f$ is twice differentiable and convex, is it true that $f''$ is a convex function ?</p>


问:
Is the derived set of the intersection of two sets contained in the intersection of the derived sets?
Given a topological set (X, $\tau$), and A and B subsets of X, is it true that

(A $\cap$ B)' $\subseteq$ A' $\cap$ B'

I know that (A $\cup$ B)' = A' $\cup$ B', but I don't know how to derived the result from here.

答:
Let $x\in (A\cap B)'$. Then every open neighbourhood of $x$ intersects $A\cap B$, hence intersects both $A$ and $B$. By the first, we conclude t$x\in A'$, by the second $x\in B'$, so $x\in A'\cap B'$. This shows
$$ (A\cap B)'\subseteq A'\cap B'.$$
In gerenal, we do not have equality, as witnessed in $X=\Bbb R$ by $A=\Bbb Q$, $B=\Bbb R\setminus \Bbb Q$ or by $A=[0,1)$, $B=[1,2]$.


问:
Contruction of a triangle, knowing one angle and it perimeter. What is the locus of a point which splits the perimeter in two at a specific length?
Building a triangle (ABC) with a fixed perimeter (P) knowing just one angle (Summit A) is not that easy.
Having fixed one adjacent side (segment AB), we know that the sum of both the others sides is a constant (P-|AB|), and so the third summit (C) is on an ellipse whose focuses are A and B, and passing by the point I called “HalfPerimeter”.

Now, the problem is that I would like to know the locus of a point dividing the perimeter in two equal or not unequal parts.

By tracing it, It appears that if we cut it in two equal parts, the locus of this moving point is the arc of a circle, which we can draw by finding that both the degenerates cases of the triangle are tangent to this circle.

And, if we cut it in two unequal parts, the locus of the moving point is another arc, of a larger circle but with the same center.

However, I have no idea how to prove that this locus is a circle, nor how to prove that all those circles have the same center.

EDIT:

A consequence of this (Corollary of the first part), would be that in the following figure, the perimeter of ABC is independent of the position of M (the moving point), and both ABM and AMC have the same perimeter.

This become obvious, since AB is tangent to the circle in $T_B$, AC is tangent in $T_C$ and BC is tangent in M :

$|AB|+|BM| = |AB|+|BT_B| = (*) = |AC|+|CT_C| = |AC|+|CM|$

(*) those two are independents of M

(But It still doesn't solve the problem with the unequal parts)

答:
So we have a triangle $ABC$, with a fixed angle $\angle BAC$ and fixed perimeter $p=2AH$. Its side $BC$ is then tangent to a circle whose center $O$ lies on the bisector of $\angle BAC$, with $OH\perp AH$.

We take then a point $K\in AH$ and let $B$ vary on $AK$. If we construct $N$ on $BC$ such that $BN=BK$, we also have $ON=OK$, because tangents $BH$ and $BC$, issued from the same point $B$, are symmetric about line $OB$. The locus of $N$ is therefore a circle of center $O$ and radius $OK$.

Note that this is not true if $B$ lies between $K$ and $H$: in that case the locus is a more complicated curve (green loop in figure below).


答:
It is true because your matrix $A$ has to in fact be square.

If the columns of $A$ are a basis for $\Bbb{R}^m$, then $A$ has $m$ linearly independent columns. But there are actually $n$ columns so it follows $m=n$.
Now recall that the column rank (which is $m$) is equal to row rank so there are $n$ linearly independent rows of $A$. They form a basis for $\Bbb{R}^n$.

Title: Prove or disprove: If $A\in \mathbb{R}^{m\times n}$ is a matrix and it's columns are basis for $\mathbb{R}^m$..

Body:
then it's rows are basis for $\mathbb{R}^n$.

I encountered this question and I answered wrong, and here's what I did:

I took a counter example: $A = \begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}_{2\times 3}$.

the columns $\{(1,0),(0,1)\}$, and they're a basis for $\mathbb{R}^2$, but the rows aren't a basis for $\mathbb{R}^3$.

Now I can't understand what's wrong with my counter example and why this is true.

I would appreciate it if someone can point out my mistakes and why my example doesn't work, and why the sentence is true.
 Thanks in advance.


问:
$f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective
I'm looking for a really simple example of $f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective. I thought about:
$$
f(x)=\begin{cases}
0 & x=0\\
\frac{x+1}{2} & x\text{ is odd}\\
-\frac{x}{2} & x\text{ is even}\wedge x\neq0
\end{cases}
$$
Is there a better and more simple  example I can use?

答: You can take, for instance, $f(n)=(-1)^n\left\lfloor\dfrac n2\right\rfloor$ (assuming, as you did, that $0\in\Bbb N$).


问:
proving the divergence or convergence of an improper integral
$\int_0^1$${\frac {e^{x^2}}{{\sqrt[3]{x}} - \tan x}}$$dx$

I have to determine whether the above integral diverges or converges. Well, over the interval (0, 1) the denominator has another root apart from $0$ , say $\alpha$. First of all since $e^{x^2}$ is always positive an never zero by the $\lim$ $\frac fg$ formula I proved that the divergence or convergence of the above integral is equivalent to $\int_0^1$${\frac {1}{{\sqrt[3]{x}} - \tan x}}$$dx$  . other than that i don't know how to find an easier function for the denominator nothing comes to mind. Any help?

答:
The first ptoblem would be around $x=0$.

Use taylor expansion of the integrand
$$\frac {e^{x^2}}{{\sqrt[3]{x} - \tan (x)}}=\frac{1}{\sqrt[3]{x}}+\sqrt[3]{x}+x+2 x^{5/3}+O\left(x^{7/3}\right)$$ So, no problem (integrate termwise)

Now, the problem is at $x=\frac \pi 4 < 1$


问:
Recurrence formula (wolfram alpha error?!)
I tried to identify the simplified reccurence formula for the following function:g(0) = x ; g(n) = g(n - 1) + 1/2 (1 - sqrt(8 g(n - 1) + 1))

But the answer given by wolfram alpha is
g(n) = 1/2 n (n + sqrt(8 x + 1)) + x which is clearly wrong because for g(0)=10 we have g(10)=0
Thank you in advance,

https://www.wolframalpha.com/input/?i=g%280%29%3Dx%2C+g%28n%29%3Dg%28n-1%29-%28%28%288g%28n-1%29%2B1%29%5E%281%2F2%29%29-1%29%2F2

答:
I presume the problem with the program must be with the taking of the square root. This is perhaps illustrated by a neat 'by hand' solution is as follows.

Let $8g_n+1=y_n^2$. Then $y_{n+1}=|y_n-2|.$

Your series $10,6,3,1,0,0,0...$ then corresponds to $9,7,5,3,1,1,1,...$.

However, we could have taken $y_{n+1}=-|y_n-2|$ and obtained a different series.


问:
When would a graphing calculator make an error while plotting a function?
It is possible for a graphing utility/calculator to plot a function incorrectly...

What are some possible errors that may occur?

What will cause those errors?

答:
One example is the following:
$$ \sqrt{x-y} = 0 $$
Any point $(\alpha, \alpha)$ should be lying on this curve $\forall ~\alpha \in \mathbb{R}$  , however, when you try to plot it on Desmos, GeoGebra, or Wolfram|Alpha, the graph shows no real points for some reason.


问:
Why is it important that a basis be orthonormal?
Why is it important that a basis be orthonormal? The requirement for Orthonormal basis is so often repeated in linear algebra that it seems linear algebra depends on it as requirement. What is gained (or lost) if the basis are not orthonormal? Is it just for convention and convenience or is it just to restrict to the type of problems that assumption of orthonormality is beneficial? </p>

答:
The question is somewhat equivalent to asking "why is it important to use the standard basis in $\mathbb R^n$". The answer is that sometimes it is important and sometimes it is not, but clearly it is convenient to use the standard basis. So, what do you do if you have a vector space that does not have a standrad basis? or, more fundamentally, what is so standard about the standard basis?

Well, the standard basis is an orthonormal basis with respect to a very familiar inner product space. And any orthonormal basis has the same kind of nice properties as the standard basis has.

As with everything, the choice of the basis should be made with consideration to the problem one is trying to solve. In some cases, orthonormal bases will help. If there is no particular reason to prefer one base over another, then choosing an orthonormal one is likely to be easier to compute with.


问:
Show that $a\gamma_B(v)=v$ for all $v\in \mathbb{R}^n$
Let $1\leq n\in \mathbb{N}$, $E=\left (e_1, \ldots , e_n\right )$ the standardbasis of $\mathbb{R}^n$, $B=\left (b_1, \ldots , b_n\right )$ a basis of $\mathbb{R}^n$ and $a:=\left (b_1\mid b_2\mid \ldots \mid b_n\right )\in M_n(\mathbb{R}^n)$.

Show the following:

$a\gamma_B(v)=v$ for all $v\in \mathbb{R}^n$.

$a$ is invertible and it holds that $a^{-1}=\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \ldots \mid \gamma_B(e_n)\right )$.

I have done an example and I have seen that it holds that $\left (b_1\mid b_2\mid b_3\right )\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \gamma_B(e_3)\right )=I_{3\times 3}$.

But how can we show that it holds?

We have that $$\gamma_B(v)=v \Rightarrow c_1b_1+c_2b_2+\ldots c_nb_n=v$$ then multiplying with the matrix $a$ and knowing that $B$ is a basis we get the result, but how exactly can we show that?

答:
Let $v=c_1 b_1+\dots+c_n b_n$ so that $\gamma_B(v) = (c_1,\dots,c_n)^T$. We then have
$$
a\,\gamma_B(v) = (b_1\mid \dots\mid b_n)(c_1,\dots,c_n)^T = c_1 b_1 + \dots + c_n b_n = v.
$$
Since the columns of $a$ form a basis of $\mathbb R^n$, the matrix has full rank and hence is invertible.

Now note that $a\,\gamma_B(v) = v$ is equivalent to $\gamma_B(v) = a^{-1} v$. Consider $v=e_i$ to obtain $\gamma_B(e_i) = a^{-1} e_i$. Since in general $Me_i$ is the $i$-th column of the matrix $M$, we conclude that
$$
a^{-1} = (\gamma_B(e_1)\mid\dots\mid\gamma_B(e_n)).
$$


问:
How to find a upper bound of the set $A= \{ (1+\frac{1}{n})^n : n \in \mathbb{N}^{*} \} $
i have the set $A= \{(1+\frac{1}{n})^{n} : n \in \mathbb{N}^* \}$ and the exercise ask me to find at least $2$ upper bound least equal to $\frac{14}{5}$.

My first question is, how i know $\frac{14}{5}$ is a upper bound of $A$?

My attempt is show doesn't exist $m \in \mathbb{N}^*$ such that $(1+\frac{1}{m})^{m} = \frac{14}{5}$, but i'm stuck,  for the other upper bound numbers, i could reason the same way?

Thanks for suggestions

答:
First you should know that:
$$ (1) \ln(1+x)=x-\frac{x^2}{2}+O(x^3)$$
Also consider that ln is an injective function meaning if $\ln(A)=1$ then $A=e$

So:
$$ \lim_{n\to\infty} \ln(A) = \lim_{n\to\infty} \ln((1+\frac{1}{n})^n = \lim_{n\to\infty} n\ln(1+\frac{1}{n}) $$
Applying (1):
$$ \lim_{n\to\infty} n\ln(1+\frac{1}{n}) = \lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3}))$$
$$\lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3})) = \lim_{n\to\infty} (1-\frac{1}{2n}+nO(\frac{1}{n^3})) = 1 $$
Since $$ \lim_{n\to\infty} \ln(A) = 1 $$ Than $$ \lim_{n\to\infty} A = e $$
So e is the least upper bound and any number greater than e is an upper bound (Note: $\frac{14}{5} > e $)


问:
How do I find the values of $\theta_0$ and $\theta_1$ in linear regression?
Suppose I am provided with a dataset of $X$ and $Y$ values, and if I would like to use linear regression $\left( hθ(x)=\theta_0+\theta_1 x \right)$, to estimate $Y$ as a function of $X$, then how can I find values for both $\theta_0$ and $\theta_1$ mathematically?</p>

答:
The deviation of the approximated function from the actual is given by the difference $y_i-\theta_0-\theta_1x_i$ . Squaring it to get a positive quantity so that after summing it for all data points $i=1,3,...n$, you have

$S=\sum_{i=1}^n (y_i-\theta_0-\theta_1x_i)^2$

You need to minimize $S$ for which

$\partial S/\partial \theta_0=0$ and $\partial S/\partial \theta_1=0$

$ \implies \sum_{i=1}^ny_i=\theta_0 n+     \theta_1\sum_{i=1}^nx_i$......(1)

and

$\implies \sum_{i=1}^n x_iy_i=\theta_0\sum_{i=1}^nx_i+\theta_1\sum_{n=1}^nx_i^2$................(2)

Can you proceed from here?


答:
Ok, we can temporarily suspend disbelief and assume that $a$ is a vector and not a matrix. If $v = c_1b_1+\cdots + c_nb_n$, we have
$$a\gamma_B(v) = \begin{bmatrix} b_1 & b_2 & \cdots & b_n\end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}  = c_1b_1 + c_2b_2 +\cdots c_nb_n = v.$$

Now, using this we get that $$a\gamma_B(e_i) = e_i, \quad 1 \le i \le n$$
so
$$a\begin{bmatrix}\gamma_B(e_1)  & \cdots & \gamma_B(e_n)\end{bmatrix} = \begin{bmatrix}a\gamma_B(e_1)  & \cdots & a\gamma_B(e_n)\end{bmatrix} = \begin{bmatrix} e_1 & \cdots & e_n\end{bmatrix} = I$$
so $a$ is invertible with inverse $a^{-1} = \begin{bmatrix}\gamma_B(e_1)  & \cdots & \gamma_B(e_n)\end{bmatrix}$.

Title: Show that $a\gamma_B(v)=v$ for all $v\in \mathbb{R}^n$

Body:
Let $1\leq n\in \mathbb{N}$, $E=\left (e_1, \ldots , e_n\right )$ the standardbasis of $\mathbb{R}^n$, $B=\left (b_1, \ldots , b_n\right )$ a basis of $\mathbb{R}^n$ and $a:=\left (b_1\mid b_2\mid \ldots \mid b_n\right )\in M_n(\mathbb{R}^n)$.

Show the following:

$a\gamma_B(v)=v$ for all $v\in \mathbb{R}^n$.

$a$ is invertible and it holds that $a^{-1}=\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \ldots \mid \gamma_B(e_n)\right )$.

I have done an example and I have seen that it holds that $\left (b_1\mid b_2\mid b_3\right )\left (\gamma_B(e_1)\mid \gamma_B(e_2)\mid \gamma_B(e_3)\right )=I_{3\times 3}$.

But how can we show that it holds?

We have that $$\gamma_B(v)=v \Rightarrow c_1b_1+c_2b_2+\ldots c_nb_n=v$$ then multiplying with the matrix $a$ and knowing that $B$ is a basis we get the result, but how exactly can we show that?


问:
A separable, regular space which has cardinality of the continuum but is not first countable?
Actually the title says it all. Is there such a topological space which is separable, regular, has cardinality of the continuum but is not first countable?

If so, is there also an example of a topological vector space having this property?
Thank you!

答:
Example: Let $X=\mathbb R^{s}$, where $\mathbb R$ with the usual topology and $\omega<  s < \mathfrak c$. There is a result from Pondiczery, Hewitt and Marczewski

If there are not more than $\mathfrak{c}$ ( which is the continuum), separable topological spaces, then their product is still separable.

It is easily to see that $X$ is regular, separable, however it is not first countable.


答:
Assuming $(i)$ we get $(ii)$ by direct computation:

$$
a(\gamma_B(e_q)\mid\cdots\mid\gamma_B(e_n)) = (a\gamma_B(e_q)\mid \cdots \mid a\gamma_B(e_n)) = (e_1 \mid \cdots \mid e_n) = I
$$

so $a^{-1} = (\gamma_B(e_q)\mid\cdots\mid\gamma_B(e_n))$. Here we use that for square matrices $ab = 1$ implies $ba = 1$.

Now, for $(i)$, if $\gamma_B(v) = (\lambda_1,\ldots,\lambda_n)$ then this precisely means that $v = \sum_i \lambda_i b_i$, which is the same as

$$
v_j = \sum_i \lambda_i (b_i)_j = \sum_i \lambda_i a_{ij} = \sum_i a_{ij}\lambda_i = [a\cdot (\lambda_1, \ldots, \lambda_n)]_j = [a \gamma_B(v)]_j
$$

for all $j$, and so $v = a\gamma_B(v)$.


问:
Homotopy equivalence in Propositional Logic
Let $\scr S, \scr S'$ be two signatures with $S$ and $S'$ being two models in the language $\scr S, \scr S'$

Let $f$ and $f'$ be two reconstrual extentions as follows:

$f: S \to S'$ and $f': S' \to S$

The functions $f$ and $f'$ are just transformations preserving deductive entailments:

Let's say, $\phi \vdash_{\scr S} \psi$ for $f(\phi) \vdash_{\scr S' } f(\psi)$

Or, in general, $S \vdash \phi \implies S' \vdash f(\phi)$

Let's say there is an Identity function $I_{s,s'}$ mapping $S \to S$ and $S' \to S'$

If $f(f')=I_s$ and $f'(f)=I_{s'}$ then models $S$ and $S'$ are said to be homotopy equivalent.

What does homotopy equivalence mean in this context (Or, in general, when speaking of Models)?  If my understanding is correct, there is a notion of continuity at play when we speak of homotopies. However, in this case how can we even speak of continuity since for propositional logic our universe of discourse, in most cases, is countable at most.

Relevant definitions:

Definition 1.4.3:

Let $T$ be a theory in $\Sigma$, let $T'$ be a theory in $\Sigma'$ and let $f: \Sigma \to \Sigma'$ be a reconstrual. We say that $f$ is a translation or interpretation of $T$ to $T'$, written $f:T \to T'$, just in case:

$$T \vdash \phi \implies T' \vdash f(\phi)$$

Definition 1.4.6:

"For each theory $T$ the identity relation $1_T: T \to T$ is given by the identity reconstrual on $\Sigma$. If $f: T\to T'$ and $g: T'\to T$ are translations (basically reconstrual extentions as given earlier) we let $gf$ denote the translation from $T \to T$ given by $(gf)(p)=g(f(p))$ for each atomic sentence $p$ of $\Sigma$. Theories $T$ and $T'$ are said to be homotopy equivalent, or simply equivalent, just in case there are translations $f: T \to T'$ or $g: T' \to T$ such that $f(g) \simeq T'$ and $g(f)\simeq T$

Source screen-shots:

Book: The Logic in Philosophy of Science, by Halvorson.

答:
Theorem 4.6.17, attributed to Barrett by Halvorson says:

"Let T1 and T2 be theories with a common  definitional extension. Then there are translations F:T1→T2 and G:T2→T1 that form a homotopy equivalence."

The scholia on the proof says: "We prove the converse of this theorem in 6.6.21."

Thus, if there exist two translations that form a homotopy equivalence, the two theories will have a common definitional extension.

The text on p. 126 reads: "The previous results show, first, that a definitional extension is conservative: it adds no new results in the old vocabulary. In fact, Proposition 4.6.13 shows that a definitional extension is, in one important sense, equivalent to the original theory. You may want to keep that fact in mind as we turn to a proposal that some logicians made in the 1950s and 1960s, and that was applied to philosophy of science by Glymour (1971). According to Glymour, two scientific theories should be considered equivalent only if they have a common definitional extension."

So, if two scientific theories should be considered equivalent, then they will have a common definitional extension.

Thus, Halvorson would appear to interpret Glymour's position that if two theories should be considered equivalent, then they will have a homotopy equivalence.  Conversely, if no homotopy equivalence can get formulated (in other words, no homotopy equivalence exists), then the two theories should get considered distinct.

Consequently, it appears that Halvorson would argue that Glymour's position on scientific theories (and perhaps Halvorson's also?) comes as the separation/distinctness of scientific theories becomes clear from the impossibility of a homotopy equivalences, and that if two scientific theories are the same, then there must exist a homotopy equivalence.  Apparently, homotopy equivalence provides a way to test whether two scientific are distinct or equivalent.


问:
Is power series $\sum_{n=1}^\infty\frac{x^n}{n}$ uniformly convergent in $(-1,1)$
I know that if $\sum a_nx^n$ has radius of convergence R then this power series converges uniformly for every compact subset of $(-R,R)$ and now no doubt converges pointwise in $(-R,R)$ but little bit confused what can we say about uniform convergence in $(-R,R)$.

As for example what can we say about uniform convergence of series $\sum_{n=1}^\infty\frac{x^n}{n}$ in (-1,1)?

答:
Here's a mostly painless way to show unboundedness, in order to apply Arthur's hint.

As you note $s = \sum_j x^j/j$ converges over compact sets, and moreover so does the series of term-wise derivatives $\sum_{j \geq 1}x^{j-1} = \sum_{j \geq 0}x^j = 1/(1-x)$. A standard theorem of series gives $s' = 1/(1-x)$ and so integrating and using that $s(0) = 0$ one gets

$$
s(x) = -\mathbf{ln}(1-x)
$$

and then $s(x) \xrightarrow{x \to _+1} +\infty$.


问:
parametrisation of sphere
The sphere given by $x^2+y^2+z^2=r^2,$ its parametric repression is given by
$$ \Phi(u,v)= \left(r \sin u \cos v, r \sin u \sin v, r \cos u \right), \quad  (u,v)\in [0, \pi] \times [0, 2\pi].$$ (This seems to be standard). (I'm curious to know why the range of $u$ and $v$ differs)

My question is: in the above parametrisation can we tale the range of $(u, v)\in [0, 2\pi] \times [0, 2\pi] $? if so, what goes wrong?

答:
Start with cylindrical coordinates

$$\begin{cases}x = r\cos\theta \\ y = r\sin\theta \\ z = z \end{cases}$$

which is a polar coordinates substitution in the $xy$ plane. The way one retrieves spherical coordinates from cylindrical is by the further "polar polar" coordinates substitution

$$\begin{cases}r = \rho\sin\phi \\ z = \rho\cos\phi \end{cases}$$

in the $rz$ plane. However, the $rz$ plane differs from the $xy$ plane because it is only a half plane. By definition, the radial coordinate of a polar substitution must be strictly positive. Thus the angular restriction is to accommodate this restriction to the right half plane where $\sin\phi \geq 0$

Title: parametrisation of sphere


答: Hint: Any partial sum is bounded, but the series converges pointwise to something unbounded.

Title: Is power series $\sum_{n=1}^\infty\frac{x^n}{n}$ uniformly convergent in $(-1,1)$

Body:
I know that if $\sum a_nx^n$ has radius of convergence R then this power series converges uniformly for every compact subset of $(-R,R)$ and now no doubt converges pointwise in $(-R,R)$ but little bit confused what can we say about uniform convergence in $(-R,R)$.

As for example what can we say about uniform convergence of series $\sum_{n=1}^\infty\frac{x^n}{n}$ in (-1,1)?


问:
A naive inquiry of Godel's incompleteness--or why does mathematics need proofs of unprovability?
My question stems from reading Swetz, 1994 (mostly excerpts from the journal Mathematics Teacher) and Berlinski, 2005 (a popular book on 10 most important mathematical breakthroughs in history).

1) I'm having difficulty understanding why Godel's theorem (if I understand what I have read) requires both formulas provable and unprovable. My naive concept of mathematics is a system containing only "positive" theorems and laws and I imagine the same for science. That the scientific method culls anything proven wrong, so as not to carry failures as baggage together with its laws. Where failure, mistakes, and problems are separated as historical study or challenge. Why do unprovable statements need be admitted to a proper system?

In anticipation of my enlightenment, I have a few follow-up Qs,

2) Its my understanding the Hilbert project meant to take the point of view that mathematics and formalism needed to take a meta-point of view for the purpose of separating formulas (as symbols) from the discussion (in natural language) about mathematics. If these unprovable formulas are necessary to prove other provable proofs, then why not segregate them to yet another meta-level?

3) What I gather about Godel's argument from these sources includes a point about mathematics' use of symbols (representing numbers, variables for formulas, variables for sets and sets of sets, etc.). Then is it his argument to suggest that where a proof can be admitted to the system which can be neither proved or disproved, the substitution of variables will propagate this error?

Answers need not be technical, only sufficient to sort my "Godel baggage".

Chris

PS. I found this January post very useful for its references and look forward to finding a volume for my naive appetite for the history of formalism:Understanding Gödel's Incompleteness Theorem---

答:
As your quotes from Berlinski suggest, the heart of Godel's statements is arguably not in the unprovability itself, but in the fact that the system is able to 'talk about itself' - that is, that statements about provability can be cast as strictly arithmetic questions.  It might help to familiarize yourself with another undecidability result which goes through roughly the same route: Matiyasevich's Theorem about the solvability of diophantine equations (i.e., the question 'does this polynomial in $x$, $y$, $z$, $w$, etc. ever evaluate to zero at some integer values of $x$, $y$, $\ldots$?').  The root of the proof lies in showing that diophantine equations are expressive enough to represent all recursively enumerable sets; what Godel does is essentially the same, showing that Peano Arithmetic (or any similar system) is expressive enough that questions about 'provability' can be cast as arithmetic questions.

As for why unprovable statements need admission, the point isn't simply that they're unprovable but that they're unprovable and true - that is, that the set of 'things we can prove' will always be a proper subset of 'things that are true'.  Obviously this isn't a crippling discovery; mathematics didn't stop as of Godel!  But it is arguably a profound one, and one that has substantial practical implications: for instance, as noted above there can't possibly be an algorithm for solving diophantine equations; there can't be algorithms for finding a 'minimal forbidden minor' categorization of many sets of graphs; there can't be algorithms for solving the Word Problem for groups; etc, etc.


问:
Cauchy's Theorem- Trigonometric application
any help would be very much appreciated. The question asks to evaluate the given integral using Cauchy's formula. I plugged in the formulas for $\sin$ and $\cos$ ($\sin= \frac{1}{2i}(z-1/z)$ and $\cos= \frac12(z+1/z)$) but did not know how to proceed from there.<br>
$$\int_0^{2\pi} \frac{dθ}{3+\sinθ+\cosθ}$$
Thanks. </p>

答:
$$\begin{align} \int_0^{2 \pi} \frac{d\theta}{3+\sin{\theta}+\cos{\theta}} &= -i \oint_{|z|=1} \frac{dz}{z} \frac{1}{3+(z-z^{-1})/(2 i) + (z+z^{-1})/2}  \\&=  -2 i \oint_{|z|=1}  \frac{dz}{(1-i) z^2+6 z+(1+i)}  \end{align}$$

The poles of  the above integrand  are at

$$z_{\pm}=\frac{-3 \pm \sqrt{7}}{2}(1+i)$$

Of these, only $z_+$  is inside the unit circle.  The residue at this pole is

$$\frac{-2 i}{ (1-i)((\sqrt{7}-3)(1+i)+6} = \frac{-i}{\sqrt{7}}$$

Therefore, by the residue theorem

$$\int_0^{2 \pi} \frac{d\theta}{3+\sin{\theta}+\cos{\theta}}  = \frac{2 \pi}{\sqrt{7}}$$


问:
Prove that you can choose three of spotlights.
The plane is illuminated by several spotlights, each of which illuminates a half-plane. Prove that you can choose three of these spotlights, which also illuminate the entire plane?</p>

答:
One of the half planes must cover the final uncovered space (which is convex).

The convex shape it covers must have a closest point to the covering line (or a parallel in which case we can solve with two spotlights).

The covering line and the two lines radiating from the closest point cover the plane.


问:
Prove that $\overline{e^{ix}}=e^{-ix}$
I am trying to construct properly the imaginary exponential as an extension of the real exponential. For this, I need to show that  $\overline{e^{ix}}=e^{-ix}$.

I know that there is a very simple proof using an infinite sum of powers. But I am really looking for something purely algebraic using mainly if not only the properties of the real exponential extended to complex numbers.

One can easily show that $1=e^{ix}\cdot e^{-ix}$ and therefore that if $z=e^{ix}$, then $e^{-ix}=\lambda\cdot \overline z$ with $\lambda\in\mathbb R$, but how can we show that $\lambda=1$ ?

Once again, I am really looking for a proof that doesn't use any infinite sum.

答:
Just use Euler's formula:
$$\overline{e^{ix}}=\overline{\cos x+i\sin x}=\cos x-i\sin x=\cos(-x)+i\sin(-x)=e^{-ix}.$$

Title: Prove that $\overline{e^{ix}}=e^{-ix}$


问:
if $f\in C(\mathbb{\overline{D}})$ and $\forall z\in\partial\mathbb{D}.f(z)=z$ then $\exists z\in\mathbb{D}.f(z)=0$
if $f\in C(\mathbb{\overline{D}})$ and $\forall z\in\partial\mathbb{D}.f(z)=z$ then $\exists z\in\mathbb{D}.f(z)=0$

f is not analytical only Continuous .

solve with complex analysis tools

I thought looking at Winding  numbers , but I reached a deadend.

答:
Suppose $f$ never vanishes. Then $\gamma(s,t):=f(se^{it})$ is a homotopy equivalence between the constant curve $f(0)$
and the circle in $\mathbb{C}-\{0\}$. This, however, is impossible: the winding number around zero of the two curves is different and we know it is homotopically invariant.

(The fact that the winding number is homotopically invariant follows from Cauchy's integral theorem)


问:
Section : Cyclotomic extension Hungerford Algebra
This question is from Algebra by Thomas Hungerford page 301 (question 5) and I was unable to solve it.

Let $g_n (x) $ be cyclotomic polynomial over $\mathbb{Q}$. If p is a prime and $k \geq 1$ then show that  $g_{p } (x^p) = g_{p^2}(x)$.

So, I used $g_{p^2}(x)= \frac{x^{p^2}-1 }{ g_1(x) g_p(x)} $ = $\frac{ x^{p^2 -1} + x^{p^2 -2} + ...+1} {x^{p-1} +...+ x+1}$ and if I put $u =x^p $ then I will get $g_p{x^p} = \frac{ u^p -1} {u-1}  = u^{p-1} +...+1 = x^{p(p-1} + x^{p(p-2)} +...+1 $.

But I don't think why both of them are equal. I don't think I am making mistake in the concepts as I have used fair amount of time in solving.

Can u please tell How should I proceed now? If there is another short way that is  also wecome.

Body: $$\Phi_{p^{k+1}}(x)=\frac{x^{p^{k+1}}-1}{x^{p^k}-1}= \frac{(x^{p^k})^p-1}{x^{p^k}-1}$$