问:
Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$
I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4}  }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot \sqrt[3]{4}  } |<ϵ.$$

I have this:

$\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac {1}{ 3\cdot  \sqrt[3]{4}} \cdot\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}} $

= $\dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}- \dfrac{\sqrt[3]{2}}{6} $

=$ \dfrac{6\cdot \sqrt[3]{2+h}-6\cdot \sqrt [3]{2}-\sqrt[3]{2}\cdot h}{6h} $

=$ \left |\dfrac{1}{(\sqrt[3]{2+h})^2 + \sqrt[3]{2+h}\sqrt[3]{2}+(\sqrt[3]{2})^2 } -  \dfrac {\sqrt[3]{2}} {6}\right | $

=$   $

I don't know how continue for use the fact $|h|<\delta$

The original exercise says: $f(x)=\sqrt[3]{x}$. Found $\displaystyle\lim_{h→0} \dfrac {f(2+h)-f(2)}{h}$

答:
The limit $3\sqrt[3]4$  is given by the problem and it is easy to calculate. What the problem asks is to use the definition of limit to show that this limit is indeed  $3\sqrt[3]4$. This presents some difficulty.

Let $\epsilon\gt0$ be given. We need to find out $\delta \gt0$ such that $|x|\le \delta\Rightarrow |3\sqrt[3]4-f(x)|\lt\epsilon$.

We’ll prove that $\boxed{\delta =f(-\epsilon)-f(\epsilon)}$ fits.

Let us first notice that $f$ is strictly decreasing in the neighborhood of $0$ (derivative negative).

Proposition.- Let us $f(x)=\dfrac{\sqrt[3]{2+x}-\sqrt[3]2}{x}$ then for $x\gt0$ and near zero we have $0\lt f(-x)-f(x)\lt x$.

Proof.- (This is graphically verified with any good calculator, for example Desmos). We have
$$0\lt f(-x)-f(x)=\dfrac{\sqrt[3]{2-x}-\sqrt[3]2}{-x}-\dfrac{\sqrt[3]{2+x}-\sqrt[3]2}{x}=\frac{2\sqrt[3]2-\sqrt[3]{2-x}-\sqrt[3]{2+x}}{x} \\0\lt f(-x)-f(x)=\frac{\sqrt[3]2}{x}\left(2-\sqrt[3]{1-\frac x2}-\sqrt[3]{1+\frac x2}\right)$$

Using Taylor series $$\sqrt[3]{1-t}=1-\frac t3-\frac{t^2}{9}-\frac{5t^3}{81}-\frac{10t^4}{243}+O(t^5)\\\sqrt[3]{1+t}=1+\frac t3-\frac{t^2}{9}+\frac{5t^3}{81}-\frac{10t^4}{243}+O(t^5)$$ It follows
$$ f(-x)-f(x)=\frac{\sqrt[3]2}{x}\left(\frac29(\frac x2)^2+\frac{10}{243}(\frac x2)^4+O((\frac x2)^6)\right)=\frac{\sqrt[3]2}{x}\left(\frac{x^2}{18}+\frac{5x^4}{1944}+O\left(\frac{x^6}{64}\right)\right)$$ $$f(-x)-f(x)= x\left(\sqrt[3]2\left(\frac{1}{18}+\frac{5x^2}{1944}+\frac{1}{x^2}O\left(\frac{x^6}{64}\right)\right)\right)=x\left(\frac{1}{9\sqrt[3]4}+\frac{5x^2}{972\sqrt[3]4}+\frac{\sqrt[3]2}{x^2}O\left(\frac{x^6}{64}\right)\right)\lt x$$
because near zero the factor of $x$ is clearly less than $1$.

Now the rest of the proof is immediate. By construction, being $y_0=3\sqrt[3]4$, if $|x|\lt\delta=f(-\epsilon) -f(\epsilon)$ we have $$|y_0-f(x)|\lt f(-\epsilon) -f(\epsilon)\lt\epsilon$$ See the attached figure for clarity.


问:
A problem on logic statements
I have series of statement that can logically be true($1$) or false($0$).
For example if I have just one statement i.e

Statement $1$ is false.

Statement $1$  says that itself is false which is contradictory, so I can conclude that the statement is, in conclusion, contradictory. But if I have $2$ statements such that

Statement $2$ is false

Statement $1$ is false

If statement $1$ is true, then statement $2$ must be false which implies that statement $1$ is true. If statement $2$ is true then statement $1$ must be false which in turn implies that statement $2$ is true. We see that there are no contradictions between these two statements.

But of course the two examples are simple cases. Assuming that I have the following $10$ statements (I can have more, but for example)

Statement $5$ is true.

Statement $7$ is false.

Statement $4$ is true.

Statement $5$ is false.

Statement $3$ is true.

Statement $7$ is false.

Statement $1$ is false.

Statement $8$ is true.

Statement $10$ is false.

Statement $1$ is false.

What procedures do I have to take to or simply how do I successfully determine the statements that correlate and those that contradicts?

答:
Let $a_1, a_2, \ldots, a_n$ be variables, each equal to $1$ (true) or $-1$ (false).  Then your statements become equations.

If Statement $i$ says Statement $j$ is false, that means $a_i=-a_j$.

If Statement $i$ says Statement $j$ is true, that means $a_i=a_j$.

In your example, $a_1=a_5, a_2=-a_7, a_3=a_4, a_4=-a_5, a_5=a_3, \ldots$.  Already you have a contradiction among $a_3, a_4, a_5$.

In general, if there is one solution, there will be at lesat two (everyone swaps true, false).  So pick $a_1$ to be true, and trace along the equations.  If $a_1$ is true, then $a_5$ is true, then $a_3$ is true, then $a_4$ is true, then $a_5$ is false (contradiction).  If you got to a cycle without a contradiction, look at the unused equations.  If any of them connect to what you've built so far, then continue; otherwise you have two disjoint components (e.g. $a_1=a_2, a_3=a_4=-a_5$).

ContentLicense: CC BY-SA 3.0

Title: A problem on logic statements


问:
Proof of First Quartile in Uniform Distribution
For an uniform distribution between [a,b], find the first quartile Q1, as P(X>Q1)=0.75.

How do I proof that in this uniform distribution f(x)=1/(b-a), the Q1 is 0.25?

答:
Let $U \sim \text{Unif}(0,1)$, then show that the quantile function $Q_U$ (the inverse of the CDF more or less) is the identity: $Q_U(p) := F_U^{-1}(p) = p$. That is, the $p$-th quantile is just $p$, hence the first quartile is 0.25.

Show that if $X = (b-a) U + a$ then $Q_X(p) = (b-a) Q_U(p) + a$. This should also be intuitively obvious: If a random variable undergoes an affine transformation, the quantile function undergoes the same. See also this post.

The second fact is more general. In your case, if $U \sim \text{Unif}(0,1)$, then $X \sim \text{Unif}(a,b)$.

ContentLicense: CC BY-SA 4.0


问:
Leibniz formula for the nth derivative of $f(x)=x^{n-1} \ln x$
Problem : Calculate the derivative of the function $f: ]0,+\infty\left[\longrightarrow \mathbb{R}\right.$ defined by $f(x)=x^{n-1} \ln x$.

Solution
Let $g_{1}(x)=x^{n-1}$ et $g_{2}(x)=\ln x .$ So we have$f=g_{1} g_{2}$
for all $k$ verifying $1 \leq k \leq n-1$ we have
$$
g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}
$$
et
$$
g_{1}^{(n)}(x)=0
$$
The Leibniz formula gives

$\begin{array}{c}
f^{(n)}(x)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) g_{1}^{(k)}(x) g_{2}^{(n-k)}(x) \\
=x^{n-1} \frac{(-1)^{n-1}(n-1) !}{x^{n}}+\sum_{k=1}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-1) \cdots(n-1-k+1) x^{n-1-k} \frac{(-1)^{n-k-1}(n-k-1) !}{x^{n-k}} \\
=\frac{(-1)^{n-1}(n-1) !}{x}+\sum_{k=1}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(-1)^{n-k-1}(n-1) !}{x}\\
=-\frac{(n-1) !}{x} \cdot \sum_{k=0}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}
\end{array}$

The sum above is the same as $\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}$ (prived of $k=n,$) which is equal to -1

We obtain
$$
f^{(n)}(x)=\frac{(n-1) !}{x}
$$

What I don't get:
What's the point of this step:

we have
$$
g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}
$$
et
$$
g_{1}^{(n)}(x)=0
$$

and which solution is wrong?

I spent a lot of time already on this, I think it's a matter of indices I can't point, both solutions seem correct to me.

How I calculated :

we have that

$\ln ^{(n)}(x)=\frac{(-1)^{n-1}(n-1) !}{x^{n}}$

and

$(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$

\begin{aligned}
f^{(n)} &=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \\
&=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! } \frac{(n-1-k) !(-1)^{n-1-k}}{x^{n-k}} \\
=& x^{-1} \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-1) !(-1)^{n-k-1}
\end{aligned}

$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}$

which should give me zero? since

$\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k} = (1+(-1))^{n}=0 $

答:
I know I'm not answering the question, but I can't resist the temptation to show you a trick to arrive at the solution faster (no Leibniz formula): Integrate once before differentiating.

Let $f_n(x)=x^{n-1}\ln x$. Compute one of its primitives (antiderivatives) by integrating by parts:
$$\int_0^xf_n(t)dt=\frac{x^n}n \ln x-\frac{x^{n-1}}{n^2}=\frac{f_{n+1}(x)}n-\frac{x^{n-1}}{n^2}$$
So to obtain $f_n^{(n)}(x)$, we just need to differentiate the above $n+1$ times (noting that this will cancel the second term as it's a polynomial of degree $n-1$):
$$f_n^{(n)}(x)=\frac{f_{n+1}^{(n+1)}(x)}n$$
Solving this recurrence formula gives you the result $$f_n^{(n)}(x)=(n-1)!f_1^\prime(x)=\frac{(n-1)!}{x}$$


问:
Every Banach space is connected?
Here is what I thought about this statement.

Let $V$ be a Banach space over real numbers. Consider $x\in V$ and $A_x=\{ax : a\in \mathbb{R}\}$.

Claim: $A_x$ is connected for each $x$ in $V$.

If we prove the above claim then we are done since $0 \in \cap_{x} A_x $ which is nonempty. So $\cup_{x} A_x = V$ will be connected

Suppose $A_x$ is not connected then there are open sets $G'$ and $H'$ such that $A_x = G' \cup H'$ where $G'$ and $H'$ are disjoint. Clearly $G'$ and $H'$ must contain scalar multiple of $x$. So there exist $G$ and $H$ such that $G' = \{ax : a \in G\}$ and $H' =\{ax : a \in H\}$

What I figured out is $G \cup H = \mathbb{R}$ and $G \cap H = \phi$. If I show $G$ and $H$ are open then we get a contradiction because $\mathbb{R}$ is connected.

I am not able to show $G$ and $H$ are open. Is my approach wrong? How do I use completeness of $V$? Or is there a Banach space which is not connected?

Any alternative approaches are also helpful.

答:
Consider the continuous surjective map $\Bbb R \to A_x, a\mapsto ax$. $\Bbb R$ is connected and the continuous image of a connected set is connected.

Also note that it's also easy to show that a Banach space (or indeed, any topological vector space) is path-connected directly: Let $v \in V$, then we have a path connecting $0$ and $v$ given by $[0,1] \to V, a \mapsto av$.

Title: Every Banach space is connected?


问:
What is the $n^{th}$ digit of this sequence?
This is from a closed question:
What is the $n$th digit of this sequence?: What's the $n$th digit of the sequence where all natural numbers are repeated the time equal to itself:

$$122333\cdots1010101010101010101011\cdots$$

My results so far:
First we compute how many digits are there from $10^{k-1}$ to $10^k-1, \forall k=1, 2, \ldots$

$$
a_k = k(10^{k-1}  + \cdots + (10^k-1))=k\frac{10^{k-1}+10^k-1}{2} \cdot (9\cdot 10^{k-1})
$$

Then we compute the partial sum (define $S_0=0$)
$$
S_m=\sum_{k=1}^m a_k = \sum_{k=0}^m \frac{9k}{2} \left(10^{2k-1}+10^{2k-2}-10^{k-1} \right)\\
=\frac{1}{198}(10^m-1)(99m \cdot 10^m - 10^m+10)$$
$$=5(10^m-1)\left(m\cdot 10^{m-1}-\frac{10^{m-1}-1}{10^2-1}\right)\tag 1
$$

I admit I cheated with WA.

For any $n \in \mathbb N$ we need to find $m$ such that $S_{m-1} < n \le S_m$, then we'd know the digit we are looking for is $$\text{the } \left(m- ((S_{m-1}-n) \pmod m ) \right)^{th} \text{ digit of the (} m \text{-digit number)
 }10^{m-1}-1 + \lceil \frac{n-S_{m-1}}{m}\rceil \tag 2$$

My questions:

a) Any single  formula to get the digit in terms of $n$? I doubt there is one because getting $m$ out of $S_m$ from $(1)$ looks difficult but who knows.

b) If a) is not possible, any better alternative to $(2)$ given $m$?

答:
Something tells me [viscerally] the digit D in the string 1223334444555556666667777777... 10101010101010101010 is given by itself modulo some index of the glyph in the pattern of the digit D.

Given that of string which represents the digit D, then let Dc represent that string i.e. 1, 9 or 10, then D is give by the index of the character in the string Dc at position Dc % D.

https://dotnetfiddle.net/rdtKBE - Proof

The justification is that there will always be D - 1 digits before any digital until D approaches 0.

I only need to solve for the index in the string D would otherwise represent which is given by the modulo operation.

I call
(n + (n - 1)) - n + 1) a Friedman Number ; its inverse or whatever the name were to be the term n.ToString().Length * n + indexOf(n - 1) - 1 which is going the other way and AFAIK multiplication is communitive, e.g. via addition but the subtraction is not hence the modulo.

This is like a sorta modulo for this problem set and its the Julian Friedman number or some Alpha in some Set.

Its term is similar to the case of for 10 but includes the extra term for the 0 case e.g. the identity which allows parametrization of... something else e.g. a Prisimetric en Volumeatrism.

See also:
What is the radix of a base zero and 1 number system. or Repunit.

If there is more succinct way to explain this I would really
appreciate that understanding and will happily remove my banter.
If there is anyway I can improve this question more importantly my
answer just lmk (formatting or otherwise) tyt tia


问:
prove using induction proving method
$$
\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{4\cdot5} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}
$$

I was wondering how to solve this equation using induction method, I am not quite sure where to begin as I am new to proving techniques.

答:
Firstly, write your equation in summation notation. It makes it easier to work with in my experience.

$$\sum_{k=1}^n\frac{1}{k(k+1)}=1-\frac{1}{n+1}\tag 1$$

Now, the induction proof is done in three steps.
Firstly, enumerate the base case $n=1$
$$\frac{1}{1\cdot 2}=\frac12=1-\frac12$$ so it is true.

Second, assume it true for a given $r\in \Bbb N$:
$$\sum_{k=1}^r\frac{1}{k(k+1)}=1-\frac{1}{r+1}\tag 2$$
and use this to prove it true for $n=r+1$, i.e. that $(2)$ implies: $$\sum_{k=1}^{r+1}\frac{1}{k(k+1)}=1-\frac{1}{r+2}$$

To do this, add the $(r+1)$th term to the sum of $(2)$.

Then conclude your proof by induction, explaining that having showed the statement true for the base case $n=1$, and the $n+1$th case when the $n$th case has been assumed, so it is true for all $n\in \Bbb N$

Title: prove using induction proving method


问:
Trigonometry Angle of Depression Word problem
I'm a $12^{\text{th}}$ Grader, in our trigonometry class, our prof gave this as an assignment. Please help me understand.

From the top of a building, the angle of depression of a point in the same horizontal plane with the base of the building is observed to be $47^{\circ} 18'$. What will be the angle of depression of the same point when viewed from a position halfway up the building?

I'm not that good at picturing out the figures or diagrams. Also, is it possible not to use the measurements of the sides (height of building etc.)?

答: Let $\triangle ABC$ be such that $A$ is the top of the building, and side $BC$ serves as the horizontal line on the ground. Thus $AB \perp BC$. Let $\angle ACB = \theta$ (angle of depression),$D$ be the midpoint of $AB$, and denote $\angle DCB = \beta$. We are to find $\beta$. Let $h = AB$ be the height of the building,and $a = BC$ be the distance from the point of interest on the ground to the building. We have: $\tan \beta = \dfrac{BD}{BC} = \dfrac{\frac{h}{2}}{a}, \tan \theta = \dfrac{AB}{BC} = \dfrac{h}{a}\implies \tan \beta = \dfrac{\tan \theta}{2}\implies \beta = \tan^{-1}\left(\dfrac{\tan \theta}{2}\right)$.


问:
Using the 6 sets of sets of 5 disjoint 2-cycles and the 3 sets of 5 pairwise disjoint triangles to exhibit an outer automorphism of $S_6$.
As a follow-up to the construction of a graph that $S_6$ acts on in Constructing a graph based on numbers of vertices, incident edges, and incident triangles, I am now specifically looking at the relationship between the 6 sets of sets of 5 disjoint 2-cycles (ie $\{(ab),(ac),(ad),(ae),(af)\}$ to the 3 sets of 5 pairwise disjoint 2-cycle triangles (ie $\{\{(ab),(cd),(ef)\},\{(ac),(be),(df)\},\{(ad),(bf),(ce)\},\{((ae),(bd),(cf)\},\{(af),(bc),(de)\}\}$. I think it's important that $\frac{6}{3} =2$ but I am still a bit lost on how to construct an outer automorphism from this info. I would appreciate it if someone could steer me in the right direction.

答:
Consider sets $A$ consisting of 5 sets of three disjoint transpositions, such that each transposition appears exactly once.

To find these sets, it makes sense to look at all transpositions having a common element. For example, look at $T = \{(12),(13),(14),(15),(16)\}$.
In a set $A$, all these transpositions must be present. Furthermore, since $A$ has $\binom{6}{2}/3 = 15/3 = 5$ elements, and no element of $A$ can contain two transpositions of $T$, each element of $T$ is contained in exactly one element of $A$. From this information, it is not too hard to construct all suitable sets $A$.

There are 6 such sets:
$$\{(12)(34)(56),(13)(25)(46),(14)(26)(35),(15)(24)(36),(16)(23)(45)\} \\
\{(12)(34)(56),(13)(26)(45),(14)(25)(36),(15)(23)(46),(16)(24)(35)\} \\
\{(12)(35)(46),(13)(24)(56),(14)(25)(36),(15)(26)(34),(16)(23)(45)\} \\
\{(12)(35)(46),(13)(26)(54),(14)(23)(56),(15)(24)(36),(16)(25)(34)\} \\
\{(12)(36)(45),(13)(24)(56),(14)(26)(35),(15)(23)(46),(16)(25)(34)\} \\
\{(12)(36)(45),(13)(25)(46),(14)(23)(56),(15)(26)(34),(16)(24)(35)\}
$$

Now any $\sigma\in S_6$ permutes this $6$-element set by applying it to all the numbers in the cycles simultaneously. Show that this gives rise to an outer automorphism $S_6\to S_6$.


问:
Convergence integral
Suppose sequence of the non-negative, measurable functions $\{f_n : \Omega \rightarrow \mathbb{R}\}_{n\in\mathbb{N}}$ with pointwise convergence $f_n \rightarrow f$. Moreover, suppose $\lim_{n \to \infty} \int_{\Omega} f_n d\mu = \int_{\Omega} fd\mu < \infty$. Prove that for any set $S$ in sigma algebra the following holds
$$\lim_{n \to \infty} \int_{S} f_n d\mu = \int_{S} fd\mu$$

My attempt: Note $\lim_{n \to \infty} \int_{\Omega} f_n d\mu = \int_{\Omega} fd\mu \iff \lim_{n \to \infty} (\int_{S} f_n d\mu + \int_{S^\complement} f_n d\mu)= \int_{S} fd\mu + \int_{S^{\complement}} fd\mu \iff\\ \lim_{n \to \infty} \int_{S} f_n d\mu + \lim_{n \to \infty} \int_{S^{\complement}} f_n d\mu =  \int_{S} fd\mu + \int_{S^{\complement}} fd\mu \ \ \ (1)\\$.

Moreover, by Fotou's Lemma we get the following
$$\int_{\Omega}\liminf_{n \to \infty} f_n \chi_S d\mu \leq \liminf_{n \to \infty} \int_{\Omega}  f_n \chi_S d\mu$$ which can be rewritten as
$$\int_{\Omega} f\chi_S d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n \chi_S d\mu$$
The same can be obtained for $S^{\complement}$, and if one of the integrals is strictly greater, then there will be a contrdiction with $(1)$. Therefore,
$$\int_{\Omega}\liminf_{n \to \infty} f_n \chi_S d\mu = \liminf_{n \to \infty} \int_{\Omega}  f_n \mathbf{1}_S d\mu$$
Is this proof correct? I am not particularly sure with the first part, where I split limit into two parts, as the sequence might not converge. Any corrections or suggestions are appreciated!

答:
The related result is Scheffé's lemma, and the proof goes as follows. First, note that

\begin{align*}
\int_{\Omega} 2f \, \mathrm{d}\mu
&= \int_{\Omega} \liminf_{n\to\infty} (f + f_n - \left| f - f_n\right|) \, \mathrm{d}\mu \tag{by assumption}\\
&\leq \liminf_{n\to\infty} \left( \int_{\Omega} f \, \mathrm{d}\mu + \int_{\Omega} f_n \, \mathrm{d}\mu - \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu \right) \tag{Fatou} \\
&= \int_{\Omega} f \, \mathrm{d}\mu + \int_{\Omega} f \, \mathrm{d}\mu - \limsup_{n\to\infty} \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu. \tag{by assumption}
\end{align*}

This shows that

$$ \limsup_{n\to\infty} \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu = 0. $$

Then for any measurable $S$,

\begin{align*}
\limsup_{n\to\infty} \left| \int_S f_n \, \mathrm{d}\mu - \int_S f \, \mathrm{d}\mu \right|
&\leq \limsup_{n\to\infty} \int_S \left| f_n - f \right| \, \mathrm{d}\mu \\
&\leq \limsup_{n\to\infty} \int_{\Omega} \left| f - f_n\right| \, \mathrm{d}\mu \\
&= 0,
\end{align*}

and so, the convergence follows.

Title: Convergence integral


答: Put $f : \mathbb R\to A_x$, $f(a) = ax$. Then $f$ is continuous and $G = f^{-1}(G')$ is open in $\mathbb R$.

Title: Every Banach space is connected?

Body:
Here is what I thought about this statement.

Let $V$ be a Banach space over real numbers. Consider $x\in V$ and $A_x=\{ax : a\in \mathbb{R}\}$.

Claim: $A_x$ is connected for each $x$ in $V$.

If we prove the above claim then we are done since $0 \in \cap_{x} A_x $ which is nonempty. So $\cup_{x} A_x = V$ will be connected

Suppose $A_x$ is not connected then there are open sets $G'$ and $H'$ such that $A_x = G' \cup H'$ where $G'$ and $H'$ are disjoint. Clearly $G'$ and $H'$ must contain scalar multiple of $x$. So there exist $G$ and $H$ such that $G' = \{ax : a \in G\}$ and $H' =\{ax : a \in H\}$

What I figured out is $G \cup H = \mathbb{R}$ and $G \cap H = \phi$. If I show $G$ and $H$ are open then we get a contradiction because $\mathbb{R}$ is connected.

I am not able to show $G$ and $H$ are open. Is my approach wrong? How do I use completeness of $V$? Or is there a Banach space which is not connected?

Any alternative approaches are also helpful.


答:
In general you cannot split up the limit like this unless you know that both of those limits exists. While this may be guaranteed by the hypotheses (it is guaranteed by the conclusion, so in a sense it is), I'm not sure.

You can prove this using (a modified version of) dominated convergence: if $h_n \to h$ pointwise, $g_n \to g$ pointwise, $|h_n| \leq g_n$, and if $\int g_n \to \int g$, then $\int h_n \to \int h$.

For this we let (having fixed $S$ in the $\sigma$-algebra) $h_n = f_n \chi_S$ and $g_n = f_n$. By assumption (everything is nonnegative) the $g_n$ satisfy the necessary convergence hypotheses, and $h_n \leq g_n$ by definition. Therefore $\int_\Omega h_n \to \int_\Omega h$, where $h = f \chi_S$, which is the desired result.

Title: Convergence integral

Body:
Suppose sequence of the non-negative, measurable functions $\{f_n : \Omega \rightarrow \mathbb{R}\}_{n\in\mathbb{N}}$ with pointwise convergence $f_n \rightarrow f$. Moreover, suppose $\lim_{n \to \infty} \int_{\Omega} f_n d\mu = \int_{\Omega} fd\mu < \infty$. Prove that for any set $S$ in sigma algebra the following holds
$$\lim_{n \to \infty} \int_{S} f_n d\mu = \int_{S} fd\mu$$

My attempt: Note $\lim_{n \to \infty} \int_{\Omega} f_n d\mu = \int_{\Omega} fd\mu \iff \lim_{n \to \infty} (\int_{S} f_n d\mu + \int_{S^\complement} f_n d\mu)= \int_{S} fd\mu + \int_{S^{\complement}} fd\mu \iff\\ \lim_{n \to \infty} \int_{S} f_n d\mu + \lim_{n \to \infty} \int_{S^{\complement}} f_n d\mu =  \int_{S} fd\mu + \int_{S^{\complement}} fd\mu \ \ \ (1)\\$.

Moreover, by Fotou's Lemma we get the following
$$\int_{\Omega}\liminf_{n \to \infty} f_n \chi_S d\mu \leq \liminf_{n \to \infty} \int_{\Omega}  f_n \chi_S d\mu$$ which can be rewritten as
$$\int_{\Omega} f\chi_S d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n \chi_S d\mu$$
The same can be obtained for $S^{\complement}$, and if one of the integrals is strictly greater, then there will be a contrdiction with $(1)$. Therefore,
$$\int_{\Omega}\liminf_{n \to \infty} f_n \chi_S d\mu = \liminf_{n \to \infty} \int_{\Omega}  f_n \mathbf{1}_S d\mu$$
Is this proof correct? I am not particularly sure with the first part, where I split limit into two parts, as the sequence might not converge. Any corrections or suggestions are appreciated!


问:
How would one show the evaluation map to $K(\alpha)$ is surjective directly?
Let $K$ be a field and $\alpha$ be an algebraic element. Consider the following ring homomorphism by which $\phi:K[X]\to K(\alpha)$ by which $\phi(X)=\alpha.$ I was wondering how would one show such $\phi$ must be a surjection directly? ($K(\alpha)$ denotes the smallest field to contain both $K$ and $\alpha$.)

I wanted to show this is a surjection thus the minimal polynomial of $\alpha$ must be irreducible. (I know other ways of proving this statement but I thought it was interesting trying out this way)

I know that we can use FIT and thus $\frac{K[X]}{\langle m_{\alpha}(X)\rangle}\cong K(\alpha)$; however, via this route we need to use $m_\alpha$, the minimal polynomial, is irreducible and thus I do not want to end up in a circular argument. Hence I was wondering if there exists other clever way of showing such map must be surjective.

Many thanks in advance!

答: The image of $\phi$ is by definition $K[\alpha]$, a integral domain that is an integral extension of the field $K$. Hence $K[\alpha]$ is a field, so $K(\alpha)\subseteq K[\alpha]\subseteq K(\alpha)$.


问:
What is the general structure of an argument?
On Wikipedia  it is given that

In logic and philosophy, an argument is a series of statements (in a natural language), called the premises or premisses (both spellings are acceptable), intended to determine the degree of truth of another statement, the conclusion.

Here nowhere is implied that it also contains the conclusion, so, Does an argument contain conclusion? If yes, then how many conclusive statements?
What is the general structure of an argument?

答:
The terminology is not used entirely uniformly.

If one considers "argument" as synonymous to "inference", i.e. as a claim that may be valid or invalid, then:

An argument is a tuple $\langle \Gamma, \phi \rangle$ where

$\Gamma$ is a set of sentences, the premises,

$\phi$ is a single sentence, the conclusion.

So an argument has exactly one conclusion and an arbitrary number of premises.

In principle a set can be empty, and all definitions will work in the same way for the borderline case where there are no premises, but in this case one may not want to call it "argument", but rather simply a (valid or invalid) sentence.

If one considers "argument" as a series of reasoning steps, i.e. as a proof, rather than just a pair of premises and conclusion, then:

An argument is a sequence of sentences $(\phi_1, \ldots, \phi_n)$, where

each reasoning step $\phi_i$ is either a premise or the result of applying an inference rule on an appropriate number of previous steps $\phi_j, j < i$,

$\phi_n$ is the conclusion.

Note that axioms/lemmata/world knowledge can be viewed as special kinds of inference rules that require no previous steps.

In any event, an argument contains a conclusion, and premsies if applicable.


问:
How to label a cube
I can't come up with a clever way to label a cube's faces, such that if I turned it in one of four directions, "Up", "Down", "Left", "Right", I'd know the resulting face by applying a function to the current face number, e.g., right(f) = f + 1, left(f) = f - 1. Any ideas?

Please excuse my terrible ASCII cube. I needn't label them 1 - 6.

Cube:

```<code>     ____
    / 5 /|   3 (back)
   /___/_|
4  | 1 |2|
   |___|/

     6 (underside)
</code>```

答:
A general function is simply a mapping, and the suggestion in your drawing can do that almost verbatim in any language:

$$
\begin{eqnarray}
R(1) &=& 2;\\
R(2) &=& 3;\\
R(3) &=& 4;\\
R(4) &=& 1;\\
R(5) &=& 2;\\
R(6) &=& 4;\\
\end{eqnarray}
$$

That is, (in C style)

```<code>R(x) = (x==1) ? 2 : (x==2) ? 3 : (x==3) ? 4 : (x==4) ? 1 : (x==5) ? 2 : 4;
</code>```

You could create a function for L, U, D, too (and presumably O for opposite), and this would work for any labeling. Note though that you should have a preferred orientation for each face/label (so that you can't rotate the cube keeping the same face forward). This you already knew.

What you are seeking is a labeling such that you can have a simpler function and corresponding labeling that fits, one that does not involve laboriously listing out the assigned values for each function and face. For some reason a case-by-case version is just not as elegant as a pre-calculus function. For example, to mimic the above function for $R$ and your labeling, you might use:

$$
\begin{eqnarray}
R(x) &=& (x \bmod 4) + 1\\
L(x) &=& (x-2 \bmod 4) + 1\\
U(x) &=& 5\\
D(x) &=& 6\\
\end{eqnarray}
$$
(note that the range of mod is from 0 to 3).

These functions work very well for $x=1,2,3,4$ but not at all when $x=5,6$.

The difficulty as you've probably already noticed is how to assign labels and some other functions so that $U$ and $D$ are likewise easy to compute so that it will work for $x=5$ and $x=6$ too.

Again, one could use an arbitrary labeling and for each function come up with a polynomial that fits all 6 values. Suppose $R(5) = 2$ and $R(6) = 4$.

Lagrange Interpolation is a one possible technique that already does this. For $n$ $x,y$ points you fit an $n-1$ degree polynomial to get

$$
\begin{eqnarray}
R(x) &=& 2\cdot\frac{(x-2)(x-3)(x-4)(x-5)(x-6)}{(1-2)(1-3)(1-4)(1-5)(1-6)}\\
     &+& 3\cdot\frac{(x-1)(x-3)(x-4)(x-5)(x-6)}{(2-1)(2-3)(2-4)(2-5)(2-6)}\\
     &+& 4\cdot\frac{(x-1)(x-2)(x-4)(x-5)(x-6)}{(3-1)(3-2)(3-4)(3-5)(3-6)}\\
     &+& 1\cdot\frac{(x-1)(x-2)(x-3)(x-5)(x-6)}{(4-1)(4-2)(4-3)(4-5)(4-6)}\\
     &+& 2\cdot\frac{(x-1)(x-2)(x-3)(x-4)(x-6)}{(5-1)(5-2)(5-3)(5-4)(5-6)}\\
     &+& 4\cdot\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{(6-1)(6-2)(6-3)(6-4)(6-5)}\\
\end{eqnarray}
$$

(notice the cancellation of all terms to either 0 or 1) and after a lot of tedious calculation and simplification (or rather after Mathematica does it), you get:

$$R(x) = \frac{1}{120} (4800 - 10062 x - 7755 x^2 - 2635 x^3 - 405 x^4 - 23 x^5)$$

Using Zhen Lin's suggestion in the comment, one can take this mod 7 to get more manageable coefficients:

$$R(x) = (5 + 4 x + 6 x^2 + 4 x^3 + 6 x^4 + 5 x^5) \bmod 7$$

So two things to note...not only was that calculation tedious (just checking it is intractable by hand (more than it's worth)), but the resulting function is not particularly nice, and looks even more complex than just the plain old list.

At this point, maybe we could find a simpler function for a particular idiosyncratic labeling like Qiaochu's, using mod or some other simple functions, and it might even be simpler to do the finding, but I have no flash of insight right yet that it would allow it to work for all faces.

So I think the moral of the story is that, really, the easiest thing to code (and also quickest to execute, and easiest to understand later) will be the 'list-of-values' function.

The desire for a non-case based function is a natural one, like capturing a recurrence relation with a generating function, or fitting a function to a set of points. But in this case the result is as or more complex than the simple list.

Title: How to label a cube

Body:
A general function is simply a mapping, and the suggestion in your drawing can do that almost verbatim in any language:

$$
\begin{eqnarray}
R(1) &=& 2;\\
R(2) &=& 3;\\
R(3) &=& 4;\\
R(4) &=& 1;\\
R(5) &=& 2;\\
R(6) &=& 4;\\
\end{eqnarray}
$$

That is, (in C style)

R(x) = (x==1) ? 2 : (x==2) ? 3 : (x==3) ? 4 : (x==4) ? 1 : (x==5) ? 2 : 4;

You could create a function for L, U, D, too (and presumably O for opposite), and this would work for any labeling. Note though that you should have a preferred orientation for each face/label (so that you can't rotate the cube keeping the same face forward). This you already knew.

What you are seeking is a labeling such that you can have a simpler function and corresponding labeling that fits, one that does not involve laboriously listing out the assigned values for each function and face. For some reason a case-by-case version is just not as elegant as a pre-calculus function. For example, to mimic the above function for $R$ and your labeling, you might use:

$$
\begin{eqnarray}
R(x) &=& (x \bmod 4) + 1\\
L(x) &=& (x-2 \bmod 4) + 1\\
U(x) &=& 5\\
D(x) &=& 6\\
\end{eqnarray}
$$
(note that the range of mod is from 0 to 3).

These functions work very well for $x=1,2,3,4$ but not at all when $x=5,6$.

The difficulty as you've probably already noticed is how to assign labels and some other functions so that $U$ and $D$ are likewise easy to compute so that it will work for $x=5$ and $x=6$ too.

Again, one could use an arbitrary labeling and for each function come up with a polynomial that fits all 6 values. Suppose $R(5) = 2$ and $R(6) = 4$.

Lagrange Interpolation is a one possible technique that already does this. For $n$ $x,y$ points you fit an $n-1$ degree polynomial to get

$$
\begin{eqnarray}
R(x) &=& 2\cdot\frac{(x-2)(x-3)(x-4)(x-5)(x-6)}{(1-2)(1-3)(1-4)(1-5)(1-6)}\\
     &+& 3\cdot\frac{(x-1)(x-3)(x-4)(x-5)(x-6)}{(2-1)(2-3)(2-4)(2-5)(2-6)}\\
     &+& 4\cdot\frac{(x-1)(x-2)(x-4)(x-5)(x-6)}{(3-1)(3-2)(3-4)(3-5)(3-6)}\\
     &+& 1\cdot\frac{(x-1)(x-2)(x-3)(x-5)(x-6)}{(4-1)(4-2)(4-3)(4-5)(4-6)}\\
     &+& 2\cdot\frac{(x-1)(x-2)(x-3)(x-4)(x-6)}{(5-1)(5-2)(5-3)(5-4)(5-6)}\\
     &+& 4\cdot\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{(6-1)(6-2)(6-3)(6-4)(6-5)}\\
\end{eqnarray}
$$

(notice the cancellation of all terms to either 0 or 1) and after a lot of tedious calculation and simplification (or rather after Mathematica does it), you get:

$$R(x) = \frac{1}{120} (4800 - 10062 x - 7755 x^2 - 2635 x^3 - 405 x^4 - 23 x^5)$$

Using Zhen Lin's suggestion in the comment, one can take this mod 7 to get more manageable coefficients:

$$R(x) = (5 + 4 x + 6 x^2 + 4 x^3 + 6 x^4 + 5 x^5) \bmod 7$$

So two things to note...not only was that calculation tedious (just checking it is intractable by hand (more than it's worth)), but the resulting function is not particularly nice, and looks even more complex than just the plain old list.

At this point, maybe we could find a simpler function for a particular idiosyncratic labeling like Qiaochu's, using mod or some other simple functions, and it might even be simpler to do the finding, but I have no flash of insight right yet that it would allow it to work for all faces.

So I think the moral of the story is that, really, the easiest thing to code (and also quickest to execute, and easiest to understand later) will be the 'list-of-values' function.

The desire for a non-case based function is a natural one, like capturing a recurrence relation with a generating function, or fitting a function to a set of points. But in this case the result is as or more complex than the simple list.


问:
What are the two main ways to prove that a matrix $\Bbb R^{n\times n}$ is definite positive.
What are the two main ways to prove that a real $n\times n$ matrix is definite positive?

Is the first way: If a matrix is $n \times n$ symmetric matrix, then the associated quadratic form is positive definite iff the eigenvalues of the matrix are all positive?

答:
Quoting from the wikipedia entry in the link jkn gave, your matrix is positive definite if and only if

"Its leading principal minors are all positive. The $k$th leading principal minor of a matrix $M$ is the determinant of its upper-left $k\times k$ sub-matrix. It turns out that a matrix is positive definite if and only if all these determinants are positive. This condition is known as Sylvester's criterion, and provides an efficient test of positive-definiteness of a symmetric real matrix."


问:
Integrating a Uniform Joint PDF with $P[|X-Y|<1/2]$
This question came up, and I guess it's both a calculus review question and a continuous probability one, because I don't really get why in part b, the limits of integration for $x$ is $0$ to $1/2-y$, because if y then goes from $1/2$ to $1$, $1/2-1/2 = 0$. Could someone walk through why we chose those bounds? Also if we chose to not do the complement, what would the integral be?

答:
We want to integrate the density over the region satisfying the simultaneous conditions

$$0 \le X \le 1 \\ 0 \le Y \le 1 \\ Y \ge X \\ |X - Y| \ge 1/2.$$  The first three conditions represent a triangle with vertices $(0,0), (0,1), (1,1)$ in the Cartesian coordinate plane.  Within the unit square, the last condition represents two triangles:  one with vertices are $(0, 1/2), (0, 1), (1/2, 1)$, and another one that is the first's reflection about $y = x$, with vertices at $(1/2, 0), (1,0), (1, 1/2)$.  So the intersection of both of these regions is the single triangle with vertices $$(0, 1/2), (0, 1), (1/2, 1).$$  This gives us two ways to write the double integral:  either we may write it as $$\int_{x=0}^{1/2} \int_{y=x+1/2}^1 f(x,y) \, dy \, dx,$$ or as $$\int_{y=1/2}^1 \int_{x=0}^{1/2-y} f(x,y) \, dx \, dy.$$  The provided solution corresponds to the second choice.

How do we determine these regions bounded by the aforementioned inequalities?  This is something you should know how to do from high school algebra.  You plot the corresponding equalities, which are lines in the plane, and then determine using test points the region(s) that satisfy the respective inequalities.


问:
In a Rubik's cube what is the probability of randomly getting a face blue?
When we solve a Rubik's cube we make planned moves and easily get the desired results but here I am talking about getting a colour  by randomly moving a Rubik's cube without making an effort to solve it.
I am getting 1/269,438,400 as the probability but I am not sure that it's correct.

I solved the problem this way. I started with the middle block and considered the probability of it being blue as 1/6. Then I considered the blocks located b/w 2 corner blocks. For the first such block the probability is 4/24 that is 1/6 as 4 such blocks are available and a total of 24 blocks of that position are available but for the next such block I considered the probability as 3/22 as in the previous step 2 of the total such blocks because unavailable and one desired (blue) block of that position became unavailable. This way I got the probabilities as 1/10 and 1/18.
I did the same for corner blocks. I then multiplied all the values I got as they are all interdependent.

答:
This is a rather involved problem which I rather doubt you would have correctly taken into account everything.  It is not clear where your calculation comes from.

As for an approach... you can imagine the center pieces never move and use those as a central focal point.  Now... for the side with the blue center square to have all of the blue stickers on that face, two things must happen.  The four corners must all be the corners who have a blue side.  Then, they must all be oriented such that they have the blue side on the correct side.  If we were to assume each collection of four corners is equally likely to be chosen and each orientation of each is equally likely to be used this occurs with probability $\frac{1}{\binom{8}{4}}\times\frac{1}{3^4}$.

Next, the four edges must all be the edges who have a blue edge and they must all be oriented the correct way.  If we were to again assume each collection of four edges are equally likely to be chosen and each orientation of each is equally likely this would be occurring with probability $\frac{1}{\binom{12}{4}}\times \frac{1}{2^4}$.

I get as an estimation then:

$$\frac{1}{\binom{8}{4}\binom{12}{4}6^4}\approx \frac{1}{44906400}\approx 2.23\times 10^{-8}$$

rather far from your $\approx 3.7\times 10^{-9}$

Now, to be fair, we know that there must be some dependency between the events if we were to have extended it to all of the positions elsewhere on the cube including on the sides not touching the intended blue side since we know that there are arrangements which cannot be achieved via standard moves, however I believe if looking solely at the blue face they may yet be independent.  Ensuring that we remained within the orbit can be done by enforcing the orientation of one of the opposite corners and opposite edges.


问:
List of graph data points, need accurate calculation/estimate formula for excel
So I'm working on a small project that will use a calculation in excel, I'm nearly there and the only piece missing is being able to accurately estimate a data point from the following set of data points.
I have tried using online calculators to generate formulas, but they are accurate to 3d.p. at best, which unfortunately isn't any good.
I have tried the FORECAST tool in Excel which again, estimates some values below 0.75, when that's clearly where they converge too.

Any ideas? Or even if this is possible. Data points are:

0 0.7894735

0.1 0.7857143

0.3 0.78

0.5 0.77586205

0.75    0.7720588

1   0.7692306

1.5 0.7653061

2   0.762712

3   0.759494

4   0.757575757

5   0.7563025

6   0.7553956

7   0.754717

8   0.75419

9   0.753769

11  0.753138

13  0.752688

15  0.752351

19  0.7518795

29  0.7512519

39  0.7509388

49  0.7507508

99  0.75037517

249 0.75015001

999 0.7500375

Many Thanks

答: On the off chance that this is still useful, I've found a formula that describes that data set very accurately, to at least six decimal places: $$f(x)=\frac{0.0375003}{x+0.950011}+0.75$$ I used Desmos to find that formula, and you can check out what I did here. In cell 8, you can see that the greatest amount by which this formula's prediction differs from the data is about $3\times10^{-7}$.


问:
Proving the Law of the Unconscious Statistician
The proof seems a little too easy.  I am wondering if I misunderstood something.

Let $Y = g(X)$. Prove the $$\mathbb E(Y) = \sum_x g(x)f_x(x)$$
  provided that the sum converges absolutely.

By definition: $\mathbb E(Y) = \sum_y yf_y(y)$. Since $g^{-1}(y) = \{x_1, x_2,\dots\}$ $$f_y(y) = \mathbb P(Y=y) = \sum_i \mathbb P(X=x_i)$$ where $g(x_i) = y$. Hence $$\mathbb E(Y) = \sum_y y \sum_i \mathbb P(x_i) = \sum_y \sum_i g(x_i) \mathbb P(x_i) = \sum_i g(x_i) \mathbb P(x_i) = \sum_x g(x)f_x(x)$$
My understanding is that you need to capture every single $x_i \in g^{-1}(y)$. Repeat the process for every $y$ and then add up the terms.

答:
I found this pretty confusing myself until I spelled it out for myself in slow motion (using Iverson brackets to simplify the bounds on the summations):

\begin{align}
\mathbb EY &= \mathbb Eg(X)\\
&= \sum_z z \cdot \mathbb P(g(X) = z) &&\text{Definition of expected value}\\
&=\sum_z z\sum_w \mathbb P(X=w)[g(w) = z] &&\text{Sum probability of basic events s.t. $g(w) = z$}\\
&=\sum_z\sum_w z\cdot \mathbb P(X=w)[g(w)=z] &&\text{Distribute the $z$}\\
&=\sum_w\sum_z z\cdot \mathbb P(X=w)[g(w)=z] &&\text{Associativity: swap order of summation}\\
&=\sum_w\sum_{z = g(w)}g(w)\cdot \mathbb P(X=w) &&\text{Non-zero inner sum occurs when $z = g(w)$}\\
&=\sum_w g(w)\cdot \mathbb P(X=w)\sum_{z = g(w)}1 &&\text{Inner product depends only on $w$; factor it}\\
&=\sum_w g(w)\cdot \mathbb P(X=w) &&\text{$z$ takes on one value so the inner sum is trivially 1}
\end{align}


问:
General topology Bourbaki Exercise 3 sec. 1 ch. 1 (How to use hint?)
The exercise reads as follows.

Let X be a topological space. For any subset A of X, let $\alpha$(A)=$\mathring{\overline{A}}$ and $\beta$(A)=$\overline{\mathring{A}}$.

b) Show that for any subset $A$ of X, $\alpha (\alpha (A))$=$\alpha (A)$ and $\beta (\beta (A))$=$\beta(A)$.

d) Show that if A, B are two open sets such that $A \cap B=\emptyset$, then  $\alpha(A) \cap \alpha (B) = \emptyset$.  [use b)]

I solved this problem and I'm interested in the another solution of d).

This is the sketch of my solution of d)

[Let A, B are two open sets such that $A \cap B=\emptyset$. Then, for any open set $G$,
($G \subset \overline A \cap \overline B \rightarrow G \subset A \cap B=\emptyset$).

So, there's no interior points in ($\overline A \cap \overline B$) and $\alpha(A) \cap \alpha (B)= \mathring {\overline A} \cap \mathring {\overline B} = \mathring {(\overline A \cap \overline B)}=\emptyset$ ]

But I didn't use b) and I tried to think how to use b) to solve d) such as $\mathring A =A$ and  $\alpha(A) \cap \alpha (B)= \alpha (\beta (A) \cap \beta (B))$,

but I couldn't find a solution using b). How can I use b) to solve d)?

答:
This is a fairly indirect and trivial use of (b), but it’s the only one that occurs to me at the moment.

$A$ is open, so $\overline{\alpha(A)}=\beta(\overline{A})=\beta\big(\beta(A)\big)=\beta(A)=\overline{A}$. Thus, $A$ is dense in $\overline{\alpha(A)}$, and since $A\subseteq\alpha(A)\subseteq\overline{\alpha(A)}$, $A$ is dense in $\alpha(A)$. Similarly, $B$ is dense in $\alpha(B)$. Let $U=\alpha(A)\cap\alpha(B)$. $U$ is open, so $U\cap A$ is dense in $U\cap\alpha(A)=U$. Similarly, $U\cap B$ is dense in $U\cap\alpha(B)=U$. But then

$$(U\cap A)\cap(U\cap B)=U\cap(A\cap B)=\varnothing$$

is dense in $U$, so $U=\varnothing$.


问:
Why is the graph of 4 nodes and 2 edges not self-complementary?
I am having some trouble seeing why a graph of 4 nodes and 2 edges is not self-complementary such that $G$ is isomorphic to $\overline{G}$ (G complement) (please see the attachment below). I know that the number of edges in a self-complementary graph must be $\frac{1}{4}n(n-1)$, (half of that of a $K_n$ graph) and so for a graph of 4 nodes, the number of edges must be 3. However, why is the graph attached not self-complementary?

The number of edges, nodes, the degree sequence, length of cycles, etc., are the same, and could I not map $G$ and $G$ complement like:
$f(d) = b$; $f(b) = d$; $f(a) = a$; $f(c) = c$?

I must be missing something about isomorphism here - if someone could point it out, that'd be great - thanks so much!

答: As David pointed out in a comment, the second graph is simply not the complement of the first.</p>


问:
Prove that (x+1)! is not O(x!)
Discrete math question which is as follows:

Prove that $(x+1)!$ is not $O(x!)$ using only the definition of big-O notation. (Hint!: $\log(ab) = (\log a + \log b)$)

I used a proof by contradiction saying that if it was $O(x!)$, by definition: $(x+1)! \leq (c)x!$ for $x > k$ for some $k$. This simplifies to $(x+1)(x!) \leq c(x!)$ where the factorials cancel and we get $x+1 \leq c$ for $x \gt k$ for some $k$. We can always pick an $x$ that breaks this condition no matter what. Therefore $(x+1)!$ is not $O(x!)$.

Is this correct? I did not need to use the $\log(ab)$ property at all, and therefore am skeptical.

答:
Your proof is correct. Note that factorial is particularly amenable to solution using the "limit" definition of big-O:

$$
f(n) = O(g(n))\iff \limsup_{n\to\infty} \frac{|f(n)|}{g(n)} \lt +\infty.
$$

Substituting, we get

$$
\limsup_{n\to\infty} \frac{(n+1)!}{n!} = \limsup_{n\to\infty} \,(n+1) = +\infty
$$

so $(n+1)!$ is not $O(n!)$.

Title: Prove that (x+1)! is not O(x!)


问:
Representing distances in high-dimensional space
I have a set of n points $P_i$ $(i=1..n)$ in an d-dimensional space. I can calculate the distances of every point to each other point.

What is the best way to represent the distances of these points?

What is the best way to see whether some of the points group together?

(Primarily i'm searching for the idea how this can be done in general, not specifically for a software implementation.)

Example:

Let n=5, d=4:
$$P_1=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}, P_2=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}, P_3=\begin{pmatrix}0\\5\\0\\0\end{pmatrix}, P_4=\begin{pmatrix}0\\6\\0\\0\end{pmatrix}, P_5=\begin{pmatrix}0\\5.5\\1\\0\end{pmatrix}$$

I can simply calculate the distance $\Delta_{i,j}=||P_i - P_j||_2$, and get
$$\Delta^2_{1,2}=1, \Delta^2_{1,3}=25, \Delta^2_{1,4}=36, \Delta^2_{1,5}=31.25,\\ \Delta^2_{2,3}=26, \Delta^2_{2,4}=37, \Delta^2_{2,5}=32.25\\ \Delta^2_{3,4}=1, \Delta^2_{3,5}=1.25,\\ \Delta^2_{4,5}=1.25$$

Here, $P_1$ and $P_2$ as well as $P_3$, $P_4$, $P_5$ are grouped together (their mutual distance is small).

Is there a way to represent this behaviour graphically?

答:
Converting Roy's comment into an answer so this question can be removed from the "Unanswered" queue

This question may be NP-Complete. There is an NP-Complete problem called CLUSTERING which is similar to this.


问:
Fermat Last Theorem for non Integer Exponents
We now that Fermat's last theorem is true so there are not positive integer solutions to
$$x^n+y^n=z^n$$
for $n\in\mathbb{N}$ and $n>2$.

But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?

Body: $1782^n + 1841^n = 1922^n$ with $n \approx 11.999999995097161$


问:
How to get the right hand side
I am confused with how to get from $$\frac{x}{e^{x}-1}=\sum_{n=0}^{\infty}B_{n}\frac{x^{n}}{n!}$$

In the book im studying from it says its a definition but idk how to generate terms from the right hand side or anything. I guess i should expand the left side in a series but idk what to do really. Thank you for your help

答:
Indeed, the definition of $(B_n)_{n\geqslant 0}$ is the relation $\displaystyle\frac{x}{e^x-1}=\sum_{n=0}^{+\infty}\frac{B_n}{n!}x^n$. Therefore
$$ x=\frac{x}{e^x-1}(e^x-1)=\left(\sum_{n=0}^{+\infty}\frac{B_n}{n!}x^n\right)\left(\sum_{n=1}^{+\infty}\frac{x^n}{n!}\right)=\sum_{n=1}^{+\infty}\left(\sum_{k=0}^{n-1} \frac{B_k}{k!(n-k)!}\right)x^n $$
Thus $B_0=1$ and for all $n\geqslant 1$,
$$ B_n=-n!\sum_{k=0}^{n-1}\frac{B_k}{k!(n-k+1)!}=-\frac{1}{n+1}\sum_{k=0}^{n-1}\binom{n+1}{k}B_k $$
This recursive relation makes it possible to compute the Bernoulli numbers.

Title: How to get the right hand side


问:
applying law of large numbers to matrices?
Let us consider a random matrix X such that 1) entries are independently drawn, 2) its diagonal terms come from mean $1$ distribution 3)non-diagonal terms from mean $0$ distribution.

$X_n$ is iid copy of $X$.

In this case, I am wondering for a given matrices $A,B$ whether it holds or not.
\begin{align}
\frac{1}{n} \sum_{i=1}^n A X_n B \to AB
\end{align}

I think it holds because everything is linear and entries are independently drawn, but want to confirm it. Thanks!

答:
Yes. It has to hold. Observe that it's enough to show that $$\dfrac{1}{n}\sum_{i=1}^n X_i \longrightarrow \text{Id}_n \quad~ \text{a.s.}$$

Remember that $X_i$'s are matrix-valued random variables, i.e. each of its elements are random variables too.

Take a sequence of diagonal elements of the $(i,i)^\text{th}$ position, i.e. $\{X_{ii}^{(k)}\}_{k\in\mathbb{N}}$ . By SLLN, $$\dfrac{1}{n}\sum_{k=1}^n {X_{ii}^{(k)}} \longrightarrow \mathbb{E}[X_{ii}] = 1\quad \text{a.s.} $$

Take a sequence of non-diagonal elements of the $(i,j)^\text{th}$ position, i.e. $\{X_{ij}^{(k)}\}_{k\in\mathbb{N}}$ . By SLLN, $$\dfrac{1}{n}\sum_{k=1}^n {X_{ij}^{(k)}} \longrightarrow \mathbb{E}[X_{ij}] = 0\quad \text{a.s.} $$

Since all the entries of all the matrices are drawn independently, thus we have $$\dfrac{1}{n}\sum_{i=1}^n X_i \longrightarrow \text{Id}_n\quad~\text{a.s.}$$ which is what we were needed to prove.


问:
Confusion concerning a proof of Schroeder-Bernstein Theorem
Set up:
Let $A,B$ be sets, and $f: A \longrightarrow B$, $g: B \longrightarrow A$ be injective functions.

We call $b \in B$ lonely, if $b \neq f(a)$ for any $a \in A$.

We call $b \in B$ a densecndent of $b_0$, if $b = (f \circ g)^n(b_0)$, where $n$ is some natural number.

Define $h: A \longrightarrow B$; it equals $g^{-1}(a)$ if $f(a)$ is a descendent of a lonely point, and it equals $f(a)$ otherwise.

I'm trying to understand this proof: https://artofproblemsolving.com/wiki/index.php/Schroeder-Bernstein_Theorem, but I'm confused by this part:

"We first note that for any $a \in A$, the point $h(a)$ is a descendent of a lonely point if and only if $f(a)$ is a descendent of a lonely point.".

Here's my attempt to verify this:

Let $h(a)$ be a descendent of a lonely point; I want to show that this implies that $f(a)$ is a descendent of a lonely point. Assume that $f(a)$ is not a descendent of a lonely point; then, by definition, $h(a) = f(a)$, and so we've a contradiction.

The other way:

Assume that $f(a)$ is a descendent of a lonely point, say $b_0$; then, by definition, $h(a) = g^{-1}(a)$. Since $f(a)$ is a descendent of $b_0$, we can write $f(a) = (f \circ g)(b_0)$, and since $f$ is injective, it follows that $a = g(b_0)$, but then $$h(a) = g^{-1}(a) = g^{-1}(g(b_0)) = b_0.$$

Hence $h(a)$ is lonely, but that means that $h(a)$ is not a descendent of any lonely point, since that would mean that $$\forall \alpha \in A: h(a) \neq f(\alpha) \\ \text{and} \\ h(a) = f(g(\beta)),$$ where $\beta$ is some lonely point in $B$; but this is a contradiction, since $g(\beta) \in A$.

I must've made a mistake; I'd appreciate pointing it out.

答:
You turn too much to proof by contradiction here. I think the more natural way to prove this is

$f(a)$ is not a lonely descendent $\implies h(a)$ is not a lonely descendent.

$f(a)$ is a lonely descendent $\implies h(a)$ ia a lonely descendent.

The first case is trivial. If $f(a)$ is the not a lonely descendent, then by definition, $h(a) = f(a)$ which is not a lonely descendent.

For the second case, we know that $f(a) = (f\circ g)^n(b_0)$ for some lonely $b_0$ and some $n$. We know that $n$ cannot be $0$, since that would make $b_0 = f(a)$, so $b_0$ would not be lonely. But then $a = g((f\circ g)^{n-1}(b_0))$, and $$h(a) = g^{-1}(a) = (f\circ g)^{n-1}(b_0)$$
which is a descendent of a lonely number (note that we must include $b_0$ as a descendent of itself for this to be always true: $n-1$ could be $0$).


问:
Evaluating the integral: $\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx$
I am interested in evaluating the following integral:
$$
\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx
$$
Using Matlab, Numerically it seems that the integral is convergent,
but I'm not sure about it. How can we prove that the integral is
convergent or not?

Many Thanks in advance.

答:
The integral converges. There is potential trouble at $0$ and "at" infinity. So we split the integral into (i) the part from $0$ to $1$, and (ii) the part from $1$ to $\infty$.</p>

<p>(i) Look at the power series expansion of $2-2\cos x-x\sin x$. The first few terms are $2-2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right) -x\left(x-\frac{x^3}{31}\right)$.  There is cancellation, and the first non-zero term is the $x^4$ term. So our function behaves well as $x$ approaches $0$ from the right: It approaches $\frac{1}{12}$. If we define your integrand to be $\frac{1}{12}$, the resulting function is continuous on $[0,1]$, and hence integrable. </p>

<p>(ii) For large $x$, the integrand is $\lt \frac{K}{x^3}$ for some constant $K$, and we know that $\int_1^\infty \frac{dx}{x^3}$ converges.  </p>

Body:
The integral converges. There is potential trouble at $0$ and "at" infinity. So we split the integral into (i) the part from $0$ to $1$, and (ii) the part from $1$ to $\infty$.

(i) Look at the power series expansion of $2-2\cos x-x\sin x$. The first few terms are $2-2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right) -x\left(x-\frac{x^3}{31}\right)$.  There is cancellation, and the first non-zero term is the $x^4$ term. So our function behaves well as $x$ approaches $0$ from the right: It approaches $\frac{1}{12}$. If we define your integrand to be $\frac{1}{12}$, the resulting function is continuous on $[0,1]$, and hence integrable.

(ii) For large $x$, the integrand is $\lt \frac{K}{x^3}$ for some constant $K$, and we know that $\int_1^\infty \frac{dx}{x^3}$ converges.


问:
The norm of the difference of normal linear functionals
Let $M$ be a von Neumann algebra. Suppose that $\varphi_1, \varphi_2\in M_{*}^{+}$, then we have

$||\varphi_1–\varphi_2||=\text{inf}\{2\rho(1)–\varphi_1(1)–\varphi_2(1)：\rho\geq \varphi_1,\rho\geq \varphi_2 \}$.

The above conclusion is from Haagerup and Stormer's paper.
I wonder whether the infimum in the above relation can be attained.

In Haagerup and Stormer's paper, they mention that the infimumu can be attained, I cannot prove the conclusion.

答: In the lemma where they prove the equality, they explicitly say that the infimum is attained when $\rho=\varphi_1\vee \varphi_2=\varphi_2+(\varphi_1-\varphi_2)^+$.


问:
Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$
I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.

However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.

Does non-uniform convergence of individual terms imply non-uniform convergence of series?

答:
Say $\sum f_n(x) $ is uniformly convergent. Then there exists an $e$ such that for all x there exists integer N such that
$$|f_{n+1}(x)+ f_n(x)|<e \forall n\ge N$$

Taking $e=1/3$ and $n=m>N$ and $x=1/m$ we get a contradiction. Hence series isn't uniformly convergent.


问:
The quantifier "most"
It is well known that the quantifier "most", understood as "more than half the ps" is not first-order definable. This is one of the results Barwise and Cooper (1981: p122-3), in "Formal Semantics: the essential readings" by Portner and Partee) discuss. Furthermore, many so-called proportional quanitifers have been proven not to be first order-definable (Keenan 1996: "The semantics of determiners"). My question is whether "most" can be defined in plural (first-order) logic. Is there a proof that "most" and other proportional quantifiers can be defined in plural (first-order) logic? Does a definition of these quantifiers require full 2nd order logic?</p>

答:
That depends on whether you have the predicate is larger than $>$ defined for plural individuals. The two-place most quantifier (most ps are qs) is definable if and only if $>$ is definable.

First, a word on notation. Let an unadorned letter $a, b, c$ refer to a plural individual in either a free or bound variable. Let a letter with a subscript zero $x_0$ refer to a singular individual. $a_0$ and $a$ are not related in any way.

We write two-place most in the following way:

$$ \mathop{\text{M}} x_0 : x_0 \in p \mathop. x_0 \in q $$

Or in prose For most values of the variable x such that x is in p, x is in q.

Let $a > b$ mean that a plural individual $a$ is strictly larger than a plural individual $b$.

There are more things in $a$ than there are in $b$ if most of the things in $a$ or $b$ but not both are in $a$.

Intuitively, the common elements of $a$ and $b$ don't count for or against our claim of mostness since they count for the $a$ side and the $b$ side equally. We need to examine the elements in the symmetric difference.

$$ a > b \;\;\Longleftrightarrow\;\; \mathop{\text{M}} x_0 : x_0 \in a \oplus x_0 \in b \mathop. x_0 \in a $$

We can define mostness in terms of $>$ if we allow ourselves to use the relative complement $\setminus$.

$$ \mathop{\text{M}} x_0 : x_0 \in a \mathop. x_0 \in b \;\;\Longleftrightarrow\;\; b > a \setminus b $$

The relative complement is itself definable.

$$ x_0 \in a \setminus b  \;\;\Longleftrightarrow\;\; x_0 \in a \land x_0 \notin b $$

We can rewrite the RHS by introducing two new variables

$$ \forall c \mathop. b > c \lor (\exists x_0 \mathop. x_0 \in c \land (x_0 \notin a \setminus b)) $$

$$ \forall c \mathop. b > c \lor (\exists x_0 \mathop. x_0 \in c \land (x_0 \notin a \lor x_0 \in b)) $$

So, putting this all together in prose:

Most $a$s are $b$s if and only if for all groups $c$ that are subsets of the non-$b$ $a$s, $c$ is strictly smaller than $b$.

If we have either $>$ or the mostness quantifier, we can rewrite our expression and get the other.

The most natural way to get $>$ to is to define in terms of equality and the existence of injections (which is second order).

If you have the ability to talk about two plural individuals having the same size $a \simeq b$, then you can construct $>$ by claiming that $b$ is the same size as a proper subset of $a$ and $a$ is not the same size as a proper subset of $b$.

But ... we also need to quantify over functions to talk about two plural individuals being the same size.

Title: The quantifier "most"


问:
Integral of $\cot^2 x$?
How do you find $\int \cot^2 x \, dx$? Please keep this at a calc AB level. Thanks! </p>

答:
Since the integrand $$\cot^2 x=\frac{\cos^2 x}{\sin^2 x}$$ is a rational fraction of $\sin x,\, \cos x$, you could use a universal standard substitution  called the Weirstrasse substitution
$$
\begin{equation*}
\tan \frac{x }{2}=t,\qquad x =2\arctan t,\qquad dx =\frac{2}{1+t^{2}}dt
\end{equation*},
$$

which converts the integrand into a rational function of $t$ whose evaluation is by partial fractions expansion. From the double-angle formulas and the identity $\cos ^{2}\frac{x}{2}+\sin ^{2}\frac{x}{2}=1$ we get:

$$
\begin{eqnarray*}
\cos x &=&\cos \left(2\cdot\frac{x}{2}\right)=\cos ^{2}\frac{x}{2}-\sin ^{2}\frac{x}{2}=\frac{\frac{\cos ^{2}
\frac{x}{2}-\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}
\frac{x}{2}+\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{1-\tan ^{2}
\frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}=\frac{1-t^2}{1+t^2}, \\
&& \\
\sin x &=&\sin \left(2\cdot\frac{x}{2}\right)=2\sin \frac{x}{2}\cos \frac{x}{2}=\frac{\frac{2\sin \frac{x}{2}
\cos \frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}\frac{x}{2}+\sin ^{2}
\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{2\tan \frac{x}{2}}{1+\tan ^{2}
\frac{x}{2}}=\frac{2t}{1+t^2}.
\end{eqnarray*}
$$

Then
$$
\begin{eqnarray*}
\int \cot ^{2}xdx &=&\int \frac{\left( t^{2}-1\right) ^{2}}{2t^{2}\left(
1+t^{2}\right) }\,dt=\int \frac{1}{2}+\frac{1}{2t^{2}}-\frac{2}{1+t^{2}}dt \\
&=&\frac{1}{2}t-\frac{1}{2t}-2\arctan t+C \\
&=&\frac{1}{2}\tan \frac{x}{2}-\frac{1}{2\tan \frac{x}{2}}+C \\
&=&-\cot x+C,
\end{eqnarray*}
$$

because
$$
\begin{equation*}
\cot x=\frac{\cot ^{2}\frac{x}{2}-1}{2\cot \frac{x}{2}}=\frac{1}{2}\cot
\frac{x}{2}-\frac{1}{2\cot \frac{x}{2}}=\frac{1}{2\tan \frac{x}{2}}-\frac{1}{2}\tan \frac{x}{2}.
\end{equation*}$$

Title: Integral of $\cot^2 x$?


答:
It's overkill in this case since the series
$$
\sum_{n=1}^\infty \frac{nx}{1+n^2x^2}
$$
is not even pointwise convergent for $x\neq 0$, but in general:

If you have uniform convergence over $E$ of a series with partial sum $S_n=\sum_{k=1}^n f_n$ to a limit function $S$, then
$$\begin{align*}
\sup_{x\in E}|f_n(x)| &= \sup_{x\in E} |S_{n}(x)-S_{n-1}(x)| \leq \sup_{x\in E} |S_{n}(x)-S(x)|+ \sup_{x\in E} |S_{n-1}(x)-S(x)| \\&\xrightarrow[n\to\infty]{} 0+0=0
\end{align*}$$
so $(f_n)_n$ must converge uniformly (on $E$) to the zero function.

Title: Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$

Body:
I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.

However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.

Does non-uniform convergence of individual terms imply non-uniform convergence of series?


问:
Why the induced orientation need the manifold is connected
Picture below is from 18th page of  Do Carmo's Riemannian Geometry. In my view, the induced orientation does not connected. Assume $(U_\alpha, x_\alpha)$ is orientation of $M_1$, then $(\varphi(U_\alpha), \varphi\circ x_\alpha)$ is a orientation of $M_2$. Seemly, the connected is not used in here.

答:
The connectedness assumption isn't required to define the induced orientation, but is instead required to make sense of of the last sentence: in the case where $M,N$ are not connected, a diffeomorphism $\varphi:M\to N$ may be neither orientation reversing nor orientation preserving.

As a general note, in many topics it's common to assume connectedness as a matter of convenience, even when it is not strictly necessary. Extending to the nonconnected case (by treating connected components individually) is usually trivial and/or not particularly useful.


问:
integral representation of matrix logarithm
Consider the following integral representation of $ln(X)$:
$$\ln(X) = \int_{0}^{\infty}[\frac{1}{1+t}I - (X+tI)^{-1}]dt.$$
I do not understand how they derived this formula.

The only formula I found is $\ln(A) = (A-I)\int_{0}^{1}[s(A-I)+I]^{-1}ds$ from page 13 in the link http://scipp.ucsc.edu/~haber/ph251/exp19.pdf

I appreciate if someone could send a reference of this formula, or its derivation. Thank you

答:
Intuitively, this is nothing but the matrix-version of the following identity

$$ \int_{0}^{\infty}  \left( \frac{1}{1+t} - \frac{1}{\lambda + t} \right) \, \mathrm{d}t = \log \lambda $$

for $\lambda$ in $ \Omega := \mathbb{C}\setminus(-\infty, 0]$. In light of functional calculus, this identity will remain true if we replace $\lambda$ by a square matrix $X$ with all of its eigenvalues lying in $\Omega$.

Let $\mathcal{X}_n$ denote the set of all $n\times n$ complex matrices for which the eigenvalues lie in $\Omega$.

1st Solution. Note that the set of all diagonalizable $X \in \mathcal{X}_n$ is dense in $\mathcal{X}_n$ and the integral in question is continuous on $\mathcal{X}_n$, it suffices to prove the claim when $X$ is diagonalizable. Then the general case follows by the continuity argument.

If $X \in \mathcal{X}_n$ is diagonalizable, we may write $X = P\operatorname{diag}(\lambda_k:1\leq k\leq n)P^{-1}$ for some invertible $P$ and for $\lambda_1, \dots, \lambda_n \in \mathbb{C}\setminus(-\infty, 0]$. Then

\begin{align*}
&\int_{0}^{\infty} \left( (1 + t)^{-1}I - (X + tI)^{-1} \right) \, \mathrm{d}t \\
&= P \operatorname{diag}\left( \int_{0}^{\infty}  \left( \frac{1}{1+t} - \frac{1}{\lambda_k + t} \right) \, \mathrm{d}t : 1 \leq k \leq n \right) P^{-1} \\
&= P \operatorname{diag}\left( \log \lambda_k : 1 \leq k \leq n \right) P^{-1} \\
&= \log X.
\end{align*}

Therefore the desired identity follows.

2nd Solution. By the holomorphic functional calculus, for each $X \in \mathcal{X}_n$,

$$ \log X = \frac{1}{2\pi i} \oint_{\Gamma} (z I - T)^{-1} \log(z) \, \mathrm{d}z $$

for any piecewise-smooth simple closed curve $\Gamma \subset \Omega$ enclosing all the eigenvalues of $X$. Since $\log(I) = 0$, we may as well write

$$ \log X = \frac{1}{2\pi i} \oint_{\Gamma} f(z) \, \mathrm{d}z, \qquad f(z) := \left[ (z I - T)^{-1} - (zI - I)^{-1} \right] \log(z). $$

This makes the integrand decay fast enough as $z \to \infty$, namely

$$ \left\| f(z) \right\| = \mathcal{O}\biggl( \frac{\log\left|z\right|}{\left|z\right|^2} \biggr) \quad\text{as } \left|z\right|\to\infty. $$

So by deforming the contour $\Gamma$ in the shape of keyhole contour surrounding the branch cut $(-\infty, 0]$ and letting the radius to infinity, only the infinite contour surrounding the branch cut remains, yielding

\begin{align*}
\log X
&= \frac{1}{2\pi i} \biggl( \int_{-\infty + 0^+i}^{0^+i} f(z) \, \mathrm{d}z + \int_{- 0^+i}^{-\infty-0^+i} f(z) \, \mathrm{d}z \biggr) \\
&= \frac{1}{2\pi i} \int_{0}^{\infty} \bigl( f(-t + 0^+i) - f(-t - 0^+i) \bigr) \, \mathrm{d}t \\
&= \int_{0}^{\infty} \left( -(t I + T)^{-1} + (tI + I)^{-1} \right)  \, \mathrm{d}t.
\end{align*}

In the third line, we used the fact that $\log(-t + 0^+i) - \log(-t - 0^+i) = 2\pi i$ for $t > 0$.


问:
How do you integrate $\int_0^\infty \exp(it^k)\,\mathrm dt$ for $k \in \Bbb N$?
My problem is with the integral $$\int^\infty_0 e^{it^k}\,\mathrm dt$$ with $k\in\mathbb{N}$.

Somehow it can be evaluated by use of Cauchy's theorem. But I don't see how. The best thing I can imagine is drawing a fourth part of a pizza with radius $r$ in for instance the north east quadrant of the complex plane and then do a contour integral around the boundary of the pizza slice and finally see what happens when $r$ is send to infinity.

What seems to be especially nasty is the complex line integral around the circular curved part of the contour of the pizza slice since it is going to contain some difficult "$\leq k$"-powers of angle functions. But to be honest I am also not sure about the rest of the contour integral.

Kindly appreciated,

Aris

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{\ic t^{k}}
\,\dd t}
\,\,\,\stackrel{t^{k}\ \mapsto\ t}{=}\,\,\,
{1 \over k}\int_{0}^{\infty}t^{\color{red}{1/k} - 1}\,\,\expo{\ic t}\,\dd t
\end{align}
The integral will be evaluated with Ramanujan Master Theorem. Note that
$\ds{\expo{\ic t} =
\sum_{n = 0}^{\infty}{\pars{\ic t}^{n} \over n!} =
\sum_{n = 0}^{\infty}\color{red}{\expo{-\ic n\pi/2}}
\,\,{\pars{-t}^{n} \over n!}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{\ic t^{k}}
\,\dd t} =
{1 \over k}\,\Gamma\pars{\color{red}{1 \over k}}
\expo{-\ic\pars{-\color{red}{1/k}}\pi/2}
\\[5mm] = &\
\bbx{{1 \over k}\,\Gamma\pars{1 \over k}
\exp\pars{{\pi \over 2k}\,\ic}} \\ &
\end{align}


问:
Map from $n$-cube to smash product of spheres
I have a question regarding the map:
$$\Phi: I^n \to S^1 \wedge ... \wedge S^1.$$
It is induced by the map $\pi: I \to I / \partial I \cong S^1$.
The wegde is taken $n$ times.

I know that this smash product is homeomorphic to $S^n$. So I could also write down the map $$\Psi: I^n \to S^n.$$ I obtain this map by composing $\Phi$ with the homeomorphism mentioned above.

I'm having difficulties understanding what this map looks like. I know that if $n=1$, then this $\Psi$ is just the quotient map $I \to I/\partial I$. In the case of dimension $n$, I imagine this map to be the one that lies the cube over the sphere and identifies the boundary of the cube with the south pole. Is this the correct way to think about it? Is there an explicit formula for $\Psi$?

答: Restricting this map to $I^n\setminus\partial I^n$ gives a homeomorphism to $S^n\setminus *$, as this restriction is the product of the homeomorphisms $I\setminus\partial I\to S^1\setminus *$. All the other points are identified and sent to $*$. It is not hard to see that this is a quotient map: the map $I^n\to (S^1)^n$ is a product of quotient maps between locally compact spaces, and the map $(S^1)^n\to S^n$ is a quotient by definition. So this is just the map collapsing the boundary of $I^n$.


问:
(Integration) Problem 7.3 from The Cauchy-Schwarz Master Class
The below picture comes from "The Cauchy-Schwarz Master Class". I am interested in the argument in the red box. Note that the two circles indicate a typo by the author. They should be replaced by $x+h$ and $h$ respectively.

My question is that how does one arrive at the last inequality in the red box? I've tried to integrate it out and apply the inequalty $h^{1/2}\leq f(x)/D(x)$ but this doesn't work.

答:
The author uses 7.13 together with the conditions on $h$ above the box:

$$ |f(x+t)| \geq |f(x)| -\sqrt t D(x)$$

and

$$0\leq h \leq \frac{f(x)}{D(x)} \
\Rightarrow f(x) \geq hD(x) \geq 0 \Rightarrow |f(x)| = f(x)$$

Hence,

$$\int_0^{h}(x+t)^2|f(x+t)|^2dt \geq \int_0^{h}(x+t)^2||f(x)| -\sqrt t D(x)|^2dt $$
$$\stackrel{f(x)\geq 0}{=}\int_0^{h}(x+t)^2|f(x) -\sqrt t D(x)|^2dt$$
$$\stackrel{f(x) \geq hD(x) \geq 0}{\geq}hx^2\left(f(x)-\sqrt h D(x)\right)^2$$

The last inequality hold because for $t\in [0,h]$ you have

$$(x+t)^2|f(x) -\sqrt t D(x)|^2 \stackrel{f(x) \geq hD(x) \geq 0}{\geq} x^2\left(f(x)-\sqrt h D(x)\right)^2$$

This is independent of $t$ - the variable of integration. Hence, the factor $h$ in the integral inequality.


问:
definite integral by parts with discontinuous point
Let $f(x)=\int_1^{x^2} \frac{\sin t}{t} dt$, can we find $\int_0^1 xf(x) dx$? via integral by parts since the function $\frac{\sin t}{t}$ is discontinuous at the point $x=0$.

A possible answer was given as
$$\int_0^1xf(x)dx=\frac{1}{2}\int_0^1 f(x) d (x^2)=\frac{1}{2}[x^2 f(x) ]_0^1-\frac{1}{2}\int_0^1 x^2 d f(x)=\frac{1}{2}f(1)-\frac{1}{2}\int^1_0 x^2f'(x)dx=-\frac{1}{2}\int^1_0 x^2f'(x)dx.$$
and
$$
\int^1_0 x^2f'(x)dx=\int_0^1 x^2 \frac{2\sin  x^2}{x}=\int_0^1\sin x^2 d(x^2)=[-\cos x^2]_0^1=1-\cos 1.
$$

答: Note that $f$ has a finite limit at $0$. For this define $g(t)$ as $\frac {\sin t} t$ for $ t \neq 0$ and $1$ for $t=0$. $g$ is a continuous function. Above analysis shows that $\int_0^{1} xg(x)dx=1-\cos 1$. But $ xf(x)$ becomes Riemann integarble if you define its value at $0$ as $1$ and then we get $\int_0^{1}xf(x)dx= \int_0^{1}xg(x)dx=1-\cos 1$.


问:
Covering $\mathbb{C}$ with open sets
I'm trying to cover $\mathbb{C} \setminus \{0\}$ with countably many open sets $\{U_n\}$ such that the map $\phi(z) = z^2$ is injective on each of the open sets. But this seems trickier than I had imagined. Would anyone have a nice way to do this?

Here is my closest attempt: for $z = r e^{i\theta}$, the map $\phi$ essentially rotates $z$ by $\theta$ and scales by $r = |z|$. So I tried cutting the complex plane into Quadrants (like cutting a pie/pizza), and then cut each quadrant in half such that each "pie slice" had a $45$ degree angle at the origin. I tried to claim that these were the open sets, but this doesn't work; (some tinkering with them i.e. taking unions may give a covering, but $\phi$ is not injective at the 4th Quadrant slices.)

Does anyone have a nice answer to this (covering $\mathbb{C} \setminus \{0\}$ with $U_i$, and not necessarily my failed attempt)? Thanks.

答:
The four open sets
$\{ z | \operatorname{re} z > 0 \}$,
$\{ z | \operatorname{re} z < 0 \}$,
$\{ z | \operatorname{im} z > 0 \}$,
$\{ z | \operatorname{im} z < 0 \}$
form an open cover of $\{0\}^c$, and $f$ is injective on each one.

Title: Covering $\mathbb{C}$ with open sets


答:
Your statement is correct, but really not very relevant to the author's purpose of giving a meaningful definition of orientation-preserving vs orientation-reversing diffeomorphisms.

If $\varphi:M_1\to M_2$ is a diffeomorphism and say $M_1$ is not connected, notice that for any connected component $C$ of $M_1$ that the restriction $\varphi\vert_C : C\to \varphi(C)$ is a diffeomorphism between connected manifolds. The restriction can preserve or reverse orientation by the author's definition.

So for each connected component, $\varphi$ may preserve or reverse orientation. This is the usual way to discuss disconnected manifolds: Break them down into their connected components and talk about those.

Title: Why the induced orientation need the manifold is connected

Body: Picture below is from 18th page of  Do Carmo's Riemannian Geometry. In my view, the induced orientation does not connected. Assume $(U_\alpha, x_\alpha)$ is orientation of $M_1$, then $(\varphi(U_\alpha), \varphi\circ x_\alpha)$ is a orientation of $M_2$. Seemly, the connected is not used in here.


问:
Examples of when Newton's Method will fail?
I'm currently working on Newton's Method, and my instructor gave four instances where Newton's Method will fail.

(A) Newton's method converges to another solutions x=b such that f(b)=0 instead of converging to the desired solution x=a.

(B) Newton's method eventually gets into the never ending cycle, bouncing between the same two approximations $x_i$ and $x_{i+1}$.

(C) Eventually, each next approximation $x_{i+1}$  falls further from desired solution $x_a$ than the previous approximation $x_i$  determined by the Newton's method.

(D) Newton's method is not able to find the next approximation $x_{i+1}$  because f'($x_i$)=0 or f'($x_i$) Does Not Exist.

However, there aren't any examples of when this happens. Would anyone be willing to provide examples of these instances?

答:
For $A$, suppose you have a function with multiple zeros like $f(x) = \sin x$. Depending on your starting point, you will see that you could converge to any of the roots ($x = n\pi$), even if you want it to converge to a particular $x = \pi$ for example - say you start with $x_0 = 7\pi / 4$

For $B$, imagine a function whose gradient is very sharp - almost vertical. Then there is a chance that the tangent will cut at a point beyond the root, and then when you repeat at the new point, it will come back to the original point (Something like $f(x) = \arctan x$ at $x = 0$)

For $C$, if your starting point is too close to a stationary point, the next point is very far away, and will keep pushing the point further and further from the root


问:
How to obtain the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$
I need to give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$ and show that $P_n(X)=(2n+1)\prod_{k=1}^{n}(X^2-\frac{1}{t_{n,k}^{2}})$ with $t_{n,k}^{2}=\tan(\frac{k\pi}{2n+1})$.

Here is what I have done so far:
$$P_n(X)=\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{2k}X^{2k}=(2n+1)\left(X^{2n}+(-1)\frac{(1-2n)}{3}X^{2n-2}+...+(-1)^{n-1}nX^2+\frac{(-1)^n}{2n+1}\right)$$

But now I don't know how to progress further to get the asked result. I know that i can factorize a polynomial using the dominant coefficient $(2n+1)$ followed by the products of $(X-t_1)(X-t_2)...$ where $t_i$ are the roots, but how can i get the factorization with $X^2$ ?

答:
In this answer to your question we showed that
$x= 1/\tan \theta_k$ where $\theta_k=\frac{k\pi}{2n+1}$ are the solutions to $P_n(x)=0.$  There we showed $k\ne0$ and suggested $k\in\{1,2,\cdots,2n\}$.  Another set of $k$'s that work are $\{ \mp1,\mp2,\cdots,\mp n\}.$  We also showed that $P_n(x) = (2n+1) (x^{2n} + \text{lower order terms}).$
Thus

$$\begin{aligned}
P_n(x) &= (2n+1) \prod_{k=1}^n \left(x- \frac{1}{\tan \theta_k}\right)\left(x+ \frac{1}{\tan \theta_k}\right)\\
&=(2n+1) \prod_{k=1}^n \left( x^2-\frac{1}{\tan^2 \theta_k}\right)\\
&=(2n+1) \prod_{k=1}^n \left( x^2-\frac{1}{\tan^2 \frac{k\pi}{2n+1}}\right)\\
&=(2n+1) \prod_{k=1}^n \left( x^2-{\cot^2 \frac{k\pi}{2n+1}} \right).
\end{aligned}$$


问:
Probability for packet length and bit error rate
Question: A wireless transmission system has bit error p bits per sec (0 < p < 1) and packet length n bits. What is the probability that a packet has an error?

Answer: 1 - (1-p)^n

I'm not sure how to get this answer for this practice problem. Wouldn't the probability just be p^n?

答:
$p^n$ is the probability that all of the $n$ bits are in error.

Likewise, $(1-p)^n$ is the probability that all of them are error free.

You seek the probability that at least one among them are in error.

That is the probability that not all of them are error free.

So we use the rule of complements.

$$1-(1-p)^n$$


问:
$k$-th moment of product of gaussian and sinc
I would like to calculate the following integrals:

$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx$$

$$\int_{-\infty}^{+\infty} \quad x^k\quad \left(\frac{\sin(\pi a x\pm\pi)}{\pi ax\pm\pi}\right)^2\quad \exp(-bx^2) \,dx$$

Thanks!

答:
First one:

The only non-zero moments correspond to even $k=2m$. In this case we have
$$I_k=\int_{-\infty}^{\infty}x^k\left(\frac{\sin(\pi a x)}{\pi ax}\right)^2\quad \exp(-bx^2)\,dx=\frac{(-1)^{m-1}}{\pi^2a^2}\frac{\partial^{m-1}}{\partial b^{m-1}}\int_{-\infty}^{\infty}\sin^2\pi a x\, e^{-bx^2}dx$$

But the last integral can be written as
\begin{align}
\int_{-\infty}^{\infty}\sin^2\pi a x\, e^{-bx^2}dx=\frac14\int_{-\infty}^{\infty}\left(2-e^{2\pi i a x}-e^{-2\pi i a x}\right)e^{-bx^2}dx=\\
=\frac12\sqrt{\frac{\pi}{b}}\left(1-e^{-\pi^2a^2/b}\right),
\end{align}
where at the last step we have used the gaussian integral $\displaystyle \int_{-\infty}^{\infty}e^{-\beta x^2+2\alpha x}dx=\sqrt{\frac{\pi}{\beta}}\,e^{\alpha^2/\beta}$.

Therefore one finds
$$I_{2m}=\frac{(-1)^{m-1}}{2\pi^2a^2}\frac{\partial^{m-1}}{\partial b^{m-1}}\left[\sqrt{\frac{\pi}{b}}\left(1-e^{-\pi^2a^2/b}\right)\right].$$

Concerning the integrals of the 2nd type, consider the change of variables $y=x\pm a^{-1}$ and try adapt the above, it's not difficult.


问:
Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$
Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable

Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$

答:
Another possibility

Switch variables to make
$$-\frac {x''}{[x']^3}+y=\frac 1 {y^3}$$ Reduction of order $p=x'$ gives a separable equation
$$-\frac {p'}{p^3}=\frac 1 {y^3}-y\implies p=\pm\frac{y}{\sqrt{c_1  y^2-y^4-1}}$$ Now
$$x+c_2=\pm \int \frac{y}{\sqrt{c_1  y^2-y^4-1}}\,dy=\pm \frac 12 \int \frac{dt}{\sqrt{c_1  t-t^2-1}}$$ and you will arrive to some arctangent.


问:
Silverman *Arithmetic of Elliptic Curves* Problem 1.12 (a)
I am trying to understand Problem 1.12 (a) in Silverman's Arithmetic of Elliptic Curves. Here is the problem

Let $K$ be a perfect field, $V/K$ be an affine variety, and let $G_K = \operatorname{Gal}(\overline{K}/K)$. Prove that $K[V] = \{f\in\overline{K}[V]:f^\sigma=f\;\;\forall\sigma\in G_K\}$.

Here is the hint: the $\subset$ direction is clear. Conversely, if $F\in\overline{K}[X]$ is a representative of $f$, the map $G_K\to\mathcal{I}(V)$ via $\sigma\mapsto F^\sigma-F$ is a 1-cocycle (indeed, if $f^\sigma=f$ in $\overline{K}[V]$, then $F^\sigma-F\in\mathcal{I}(V)$ and $F^{\sigma\tau}-F = (F^{\sigma}-F)^\tau+ (F^\tau-F)$). It's the next point that I do not understand. Using $H^1(G_K,\overline{K}^+) = 0$, I'm supposed to deduce that there is some $G\in\mathcal{I}(V)$ such that $F+G\in K[X]$. I see that this would solve the problem, since in that case $f = (F+G)+\mathcal{I}(V)\in K[V]$.

Here's what I don't understand. Since $H^1(G_K,\overline{K}^+) = \frac{Z^1(G_K,\overline{K}^+)}{B^1(G_K,\overline{K}^+)}$, so if $H^1 = 0$, this means $Z^1(G_K,\overline{K}^+) = B^1(G_K,\overline{K}^+)$, so the function $\sigma\mapsto F^\sigma-F$ out to come from a boundary.

Question 1: It seems that $\sigma\mapsto F^\sigma-F$ is already in $B^1(G_K,\overline{K}^+)$ by definition, so how am I supposed to use the condition that $H^1 = 0$?

Question 2: In addition, it seems to me that $H^1(G_K,\overline{K}^+)$ is also the wrong group to be looking at. We're treating $\overline{K}[X]$ as an additive $G_K$-module, not $\overline{K}^+$, so shouldn't I be looking at $H^1(G_K,\overline{K}[X])$ or $H^1(G_K,\overline{K}[V])$?

Any elucidation would be much appreciated.

答:
$f\in \overline{K}[X]$

$ c(\sigma)=f^\sigma-f$ is in $Z^1(G_K,\overline{K}[X])$

We are told that $f^\sigma \equiv f\in \overline{K}[V]$ ie. $c\in Z^1(G_K,I(V)\overline{K}[X])$.

We are told that $H^1(G_K,\overline{K})=0$.

This implies that $H^1(G_K,I(V)\overline{K}[X])=0$. This is because a $K$-vector space basis of $I(V)$ gives a $G_K$-invariant $\overline{K}$-vector space basis of $I(V)\overline{K}[X]$.

This gives that $c(\sigma)=g^\sigma-g$ with $g\in I(V)\overline{K}[X]$

so that $(f-g)^\sigma-(f-g)=0\in \overline{K}[X]$ ie. $f-g\in K[X] $ and $f\equiv f-g\in K[V]$


问:
How to show that there exists a scalar $\lambda$ so that $A = \lambda I$?
Let $V$ be a finite dimensional vector space, $V \ne \{ 0 \}$ and $A\in L(V)$ an operator which commutes with every operator from $L(V)$. Show that there is a scalar $\lambda$ so that the following applies: $A = \lambda I$.

$L(V)$ = The set of all linear mappings (linear operators) from V to V

If $B\in L(V)$ is arbitrary and if I understood the assignment correctly, this would mean $AB = BA \implies AB - BA = 0 $

I believe, a matrix representation of the linear operators would apply here but I don't have any idea what to do here.

EDIT (Hint by Jared)

I tried to examine it on 5x5 matrices:  example 1, example 2.

For example, for:

$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}
\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}
\\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35}
\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55}\end{bmatrix}
\begin{bmatrix} 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 1 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0\end{bmatrix} =
\begin{bmatrix} 0 & 0 & 0 & a_{12} & 0
\\ 0 & 0 & 0 & a_{22} & 0
\\ 0 & 0 & 0 & a_{32} & 0
\\ 0 & 0 & 0 & a_{42} & 0
\\ 0 & 0 & 0 & a_{52} & 0\end{bmatrix}$

$\begin{bmatrix} 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 1 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}
\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}
\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}
\\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35}
\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55}\end{bmatrix}
 =
\begin{bmatrix} 0 & 0 & 0 & 0 & 0
\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}$

which means $a_{22}=a_{44}$.
For a 1 on the $ij$th and $ji$th position, I get 2 equations (actually 4 but 2 of them are equal to the other 2).

But I still don't understand what I can conclude from that. I reviewed older homework assignments and found the same problem but with a hint:
"Show that A has at least one eigenvalue and observe the corresponding subspace of the eigenvalue. Before that, show that every subspace of an eigenvalue of the operator $T \in L(V)$ is invariant for every operator $S \in L(V)$ which commutes with $T$."

EDIT 2

For $B=E_{ij}$ I conclude that $ AB=BA \implies a_{ii}=a_{jj}, a_{ij} = 0$ if $i \ne j$

(In my example, i is 2 and j is 4 so it turned out $a_{ii}=a_{22}=a_{44}=a_{jj}$)

For $B = E_{ij} + E_{ji}, AB = BA \implies a_{ii} = a_{jj},  a_{ij} = a_{ji},  a_{kl} = 0$ for $ k=l \ne i,$ and $ k=l \ne j $ and $ k\ne i \space \& \space l\ne j$ and $ k\ne j \space \& \space l\ne i$

But we can write B in a different way:

$B = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij}$

$AB = A(\sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij}) =$ associativity of matrix and scalar multiplication & distributivity over matrix addition $=\sum_{i=1}^n\sum_{j=1}^n(\beta_{ij}A)E_{ij} = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}(AE_{ij})$

$BA = (\sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij})A = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij}A = \sum_{i=1}^n\sum_{j=1}^nE_{ij}(A\beta_{ij}) = \sum_{i=1}^n\sum_{j=1}^n(E_{ij}A)\beta_{ij}$

$AB = BA \implies \beta_{ii}a_{ii} = \beta_{jj}a_{jj},  \forall i,j\in \{1, ... , n\} \space \& \space \beta_{ij}a_{ij} = 0 \space , i \ne j$

Which would give us a system of $n^2$ linear equations:

$\beta_{11}a_{11} = \beta_{11}a_{11}$

$\beta_{11}a_{11} = \beta_{22}a_{22}$

...

$\beta_{11}a_{11} = \beta_{nn}a_{nn}$

$\beta_{22}a_{22} = \beta_{11}a_{11}$

$\beta_{22}a_{22} = \beta_{22}a_{22}$

...

$\beta_{22}a_{22} = \beta_{nn}a_{nn}$

...

$\beta_{nn}a_{nn} = \beta_{11}a_{11}$

$\beta_{nn}a_{nn} = \beta_{22}a_{22}$

...

$\beta_{nn}a_{nn} = \beta_{nn}a_{nn}$

$\implies \beta_{11}a_{11} = \beta_{22}a_{22} = ... = \beta_{nn}a_{nn} = \beta a$
$\implies A = \lambda I$ for $\lambda = \beta a$

答:
Note that $Tx=ax\implies T(Sx) = S(Tx) = a (Sx)$ and hence $Sx$ is in the $a$ eigenspace of $T$ along with $x$.

Hence if $T$ has one eigenvalue, by choosing an $S$ that maps a corresponding eigenvector to an arbitrary $y$, we see that $y$ has the same eigenvalue. Hence $T\propto I$.

Now you just need $T$ to have one eigenvalue. Lots of ways to do this. For example, choose $S$ to have the maximal $n$ distinct eigenvalues. Then $T$ must map each eigenvector to a multiple of itself, and hence has not just one but $n$ eigenvectors.

This result is essentially Schur's Lemma.


答:
Example for Case (A):  $$f(x) = \frac{1}{1+x^2} - \frac{1}{2},$$ which has roots at $x \in \{-1,1\}$.  The initial choice $x_0 = 2$ converges to the negative root.

Example for Case (B):  $$f(x) = \begin{cases}\sqrt{x}, & x \ge 0 \\ -\sqrt{-x}, & x < 0 \end{cases}$$ has the peculiar property that for any initial guess $x_0 \ne 0$, the orbit is trapped in a cycle of period $2$, with $x_k = -x_{k-1}$.  This is quite easy to prove and is left as an exercise for the reader.

Example for Case (C):  $$f(x) = x^{1/3}.$$  The Newton's method recursion has no fixed point except for the initial guess $x_0 = 0$.

Title: Examples of when Newton's Method will fail?

Body:
I'm currently working on Newton's Method, and my instructor gave four instances where Newton's Method will fail.

(A) Newton's method converges to another solutions x=b such that f(b)=0 instead of converging to the desired solution x=a.

(B) Newton's method eventually gets into the never ending cycle, bouncing between the same two approximations $x_i$ and $x_{i+1}$.

(C) Eventually, each next approximation $x_{i+1}$  falls further from desired solution $x_a$ than the previous approximation $x_i$  determined by the Newton's method.

(D) Newton's method is not able to find the next approximation $x_{i+1}$  because f'($x_i$)=0 or f'($x_i$) Does Not Exist.

However, there aren't any examples of when this happens. Would anyone be willing to provide examples of these instances?


问:
Is $\{v^1+w^1,\cdots,v^k+w^k\}$ linearly independent if $\{v^1,\cdots,v^k\}$ and $\{w^1,\cdots,w^k\}$ are?

If $\{v^1,\cdots,v^k\}$ and ${w^1,\cdots,w^k\}$ are linearly independent sets in $V$, then is $\{v^1+w^1,\cdots,v^k+w^k\}$ a linearly independent set in $V$?

This may be a basic question but I don't know if this statement is correct or not. Can I get some help?
As far as I can see this statement is not true because it is possible

$$a_1v^1+\cdots+a_kv^k+a_1w^1+\cdots+a_kw^k=0$$

to have another solution rather than all scalars to be $0$, say all $v$ terms are kinda multiple of $w$ terms.
Can I get some thoughts on this?
Thanks

答: If $\{v_1,v_2,...,v_k\}$ is linearly independent then so is $\{-v_1,v_2,...,v_k\}$. But $\{0,2v_2,...,2v_k\}$ is not  linearly independent.


问:
Showing $\zeta(s)-{1\over s-1}$ is analytic
It is well known that Euler-Mascheroni constant has an alternative definition in terms of zeta function:

$$
\gamma=\lim_{s\to1^+}f(s)\equiv\lim_{s\to1^+}\left[\zeta(s)-{1\over s-1}\right]
$$

Using this definition, I would like to derive the Laurent series expansion for Riemann zeta function, but I am currently stuck because I am in a hard time trying to show the convergence of the following limit:

$$
\lim_{s\to1^+}{\mathrm{d}^n\over\mathrm{d}s^n}f(s)
$$

My attempt was to show that the function $f(s)$ is analytic in $\mathbb{C}\setminus\{1\}$ and continuous in $\mathbb{C}$ (by setting $f(1)=\gamma$), and prove that it is entire using Morera's theorem. Using analytic continuation, I am able to express zeta functon in terms of a contour integral:

$$
\zeta(s)=\Gamma(1-s)g(s)\equiv{\Gamma(1-s)\over2\pi i}\oint_C{z^{s-1}\over e^{-z}-1}\mathrm{d}z
$$

where the contour $C$ is shown below:

Therefore, $f(s)=\Gamma(1-s)g(s)-{1\over s-1}$, which is already analytic every where except for $z=1$, so how can I show that it is continuous at $z=1$ so that I can prove $f(s)$ to be entire using Morera's theorem?

答:
After learning more knowledge in this area, I found that Riemann-Stieltjes integration does the job. For $\Re(s)>1$, we have

$$
\begin{aligned}
\zeta(s)
&=1+\int_1^\infty{\mathrm d\lfloor x\rfloor\over x^s} \\
&=1+\int_1^\infty{\mathrm dx\over x^s}-\int_1^\infty{\mathrm d\{x\}\over x^{s+1}} \\
&={s\over s-1}-s\int_1^\infty{\{x\}\over x^{s+1}}\mathrm dx
\end{aligned}
$$

Since the last integral converges for all $\Re(s)>0$:

$$
\begin{aligned}
\left|\int_1^\infty{\{x\}\over x^{s+1}}\mathrm dx\right|
&\le\int_1^\infty{\mathrm dx\over x^{\Re(s)+1}}={1\over\Re(s)}<\infty
\end{aligned}
$$

we conclude that

$$
\zeta(s)-{1\over s-1}=1-s\int_1^\infty{\{x\}\over x^{s+1}}\mathrm dx
$$

converges and is continuous at $s=1$.


问:
Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$
I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.

However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.

Does non-uniform convergence of individual terms imply non-uniform convergence of series?

答:
A simple comparison test shows the divergence of the series for $x\neq 0$:

For $x\neq 0$ you have $\frac 1{xn}\stackrel{n\to\infty}{\longrightarrow}0$. Hence, for $N$ large enough you can estimate

$$\sum_{n=N}^{\infty}\frac{nx}{1+n^2x}=\sum_{n=N}^{\infty}\frac{1}{\frac 1{nx}+n}\geq \sum_{n=N}^{\infty}\frac{1}{1+n}=\infty$$


问:
Why do we use the the Chain Rule for the Fundamental Theorem of Calculus Part 1?
So part 1 of the Fundamental Theorem of Calculus states: If $f$ is continuous on $[a,b]$, then the function $g$ defined by
\begin{equation}
g(x) = \int_{a}^{x} f(t) \;\mathrm{d}x \qquad a \leq x \leq b
\end{equation}
is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x) = f(x)$.

In my opinion the derivative should be more general as:
\begin{equation}
g'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\left[\int_{a}^{u} f(t) \;\mathrm{d}x\right]\frac{\mathrm{d}u}{\mathrm{d}x} \qquad a \leq x \leq b
\end{equation}
Where $u$ is some function of $x$.

The reason I write the above use of the chain rule is because one can encounter such a scenario often, for example:

Find $h'(x)$ where
\begin{equation}
h(x) = \int_{1}^{e^{x}} \mathrm{ln}(t) \;\mathrm{d}t
\end{equation}
\begin{align}
h'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{1}^{e^x} \mathrm{ln}t \;\mathrm{d}t \right]
\end{align}
By the chain rule $\textit{Let}$ $e^x = u$
\begin{align}
h'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{1}^{u} \mathrm{ln}t \;\mathrm{d}t \right]\frac{\mathrm{d}u}{\mathrm{d}x}
\end{align}
Where
\begin{align}
\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x} \left[e^{x}\right]\\
                                &= e^{x}
\end{align}
$h'(x)$ becomes
\begin{align}
h'(x) &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{1}^{u} \mathrm{ln}t \;\mathrm{d}t \right]\left(e^{x}\right)\\
      &= \mathrm{ln}(u)e^{x}
\end{align}
Substituting $e^{x}$ for $u$
\begin{align}
h'(x) &= \mathrm{ln}(u)e^{x}\\
      &= \mathrm{ln}\left(e^{x}\right)e^{x}\\
      &= xe^{x}
\end{align}

What confuses me is the reason that we are able to apply the chain rule in this way. My understanding was that the chain rule is applied to take the derivative of a composite function e.g. $\frac{\mathrm{d}}{\mathrm{d}x}f(g(x))$. So since we are using the chain rule in this way, does that mean that the integral is a function...? My understanding was that the integral was just an operator. Even still this appears to raise more questions for me, like why is the $u$-substitution being appied to the interval of the integral? What does the interval have to do with the chain rule...? I suspect that I perhaps simply don't have a solid enough understanding of the chain rule.

答:
The integral itself is not a function, but it does define a function. When I first started learning calculus, I made this concrete in my head by writing

$$h(x)=F(e^x)$$

instead of

$$h(x)=\int_{1}^{e^x}\ln(t)\text{ }dt$$

where

$$F(x)=\int_{1}^{x}\ln(t)\text{ }dt$$

It then follows from the chain rule that

$$h'(x)=F'(e^x)\cdot\frac{d}{dx}e^x=F'(e^x)e^x$$

But $\text{FTC}1$ implies that $F'(x)=\ln(x)$, so we can write

$$h'(x)=\ln(e^x)e^x=xe^x$$

I hope this makes applying $\text{FTC}1$ with the chain rule more intuitive!


问:
Give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$
I need to give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$ and show that $P_n(X)=(2n+1)\prod_{k=1}^{n}(X^2-\frac{1}{t_{n,k}^{2}})$ with $t_{n,k}^{2}=\tan(\frac{k\pi}{2n+1})$.

Here is what I have done so far:
$$P_n(X)=\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{2k}X^{2k}=(2n+1)\left(X^{2n}+(-1)\frac{(1-2n)}{3}X^{2n-2}+...+(-1)^{n-1}nX^2+\frac{(-1)^n}{2n+1}\right)$$

but now i don't know how to progress further to get the asked result.

答:
Please see here for the solution.

Also, please see the discussion here.


问:
Linear Algebra determinant and rank relation
True or False?

If the determinant of a $4 \times 4$ matrix $A$ is $4$
   then its rank must be $4$.

Is it false or true?

My guess is true, because the matrix $A$ is invertible.

But there is any counter-example?

Please help me.

答: If it is invertible, it is full-rank. The rank is the dimension of $\operatorname{Im} A$.</p>


答: Please see the discussion here where it is mentioned that $z=x+i$, $z^\ast=x-i$ and $z^{2n+1}={z^{\ast}}^{2n+1}$.  Another way to write this is $$\left(\frac{x+i}{x-i}\right)^{2n+1} = 1.$$

Title: Showing that if $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$

Body:
I need to show that if $P_n(X)=\frac{(X+i)^{2n+1} -  (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$.

Here is what i have done:
$$P_n(X)=\frac{(X+i)^{2n+1} -  (X-i)^{2n+1}}{2i} = 0 \\ (X+i)^{2n+1}-(X-i)^{2n+1} = 0*2i = 0 \\ (X+i)^{2n+1} = (X-i)^{2n+1} \\ \frac{(X+i)^{2n+1}}{(X-i)^{2n+1}} = 1 \\ \left(\frac{X+i}{X-i}\right)^{2n+1} = 1$$

Is my reasoning correct ?