问:
Exercise 15.15 of Jech: forcing GCH
Jech uses the following forcing notion to force GCH. For each $\alpha$, let $P_\alpha$ be the notion of forcing that collapses $\beth_{\alpha+1}$ to $\beth_\alpha^+$. $P$ is then the Easton product of these notions. For simplicity, we'll assume there are no inaccessible cardinals. Let $P^{< \alpha}$ be the product below $\alpha$ and let $P^{\leq\alpha}$ be the product up to and including $\alpha$. So, $P^{<\alpha}$ is just the set of arbitrary functions from $\alpha$ to the corresponding partial orders $P_\beta$ for $\beta<\alpha$.

In part, my strategy is to prove that $P^{\leq \alpha}$ forces CH at $\beth_\beta$, for $\beta\leq\alpha$.

Let $\lambda$ be a limit ordinal and assume I've shown this for $\alpha<\lambda$ and I want to show it for $P^{\leq \lambda}$. I know how to show that $P^{<\lambda}$ forces CH at $\beth_\alpha$, for $\alpha<\lambda$. So the obvious thought is to force with $P_\lambda$ first, since that is $\beth_\lambda$-closed and will preserve all the relevant properties of $P^{<\lambda}$ whilst making $\beth_{\lambda+1} = \beth_\lambda^+$. So, all I need to show is that when I then force with $P^{<\lambda}$, $\beth_{\lambda+1} = \beth_\lambda^+$ is preserved. That's what I'm having difficultly seeing. In particular, I can't see why $\beth_{\lambda+1}$ might not change.

答:
The solution to this exercise is quite long. It can be broken down to the following steps.

Let $P_\alpha$ be $\text{Col}(\beth_\alpha^+, \beth_{\alpha+1})$. As in the hint, let $P$ be the Easton class product of the $P_\alpha$'s, i.e. bounded support below inaccessibles. And $P^{>\alpha}, P^{\geq\alpha}, P^{\leq\alpha}$ are defined in the familiar way. Also notice that Jech defines $\kappa$-closed with quantification over sequences of length $\leq\kappa$. Kunen's book would call this $\kappa^+$-closed. I am using Jech's convention for consistency with the lemmas he refers to. Another notational convention: in this answer the cardinal terms without superscript are evaluated in $V$.

Lemma 1. For each $\alpha$, $P_\alpha$ is $\beth_\alpha$-closed and has the $\beth_{\alpha+1}^+$-cc (the latter holds because by cardinal arithmetic calculation, $\beth_{\alpha+1}^{<\beth_\alpha^+}=\beth_{\alpha+1}$ and $|P_\alpha|=\beth_{\alpha+1}$).

Lemma 2. For each $\alpha$, $P^{\geq\alpha}$ is $\beth_\alpha$-closed. So in particular $P^{>\alpha}$ is $\beth_{\alpha+1}$-closed. Also, the whole product $P$, viewed as $P^{\geq 0}$ is then $\omega$-closed.

Lemma 3. For each $\alpha$, $P^{\leq\alpha}$ has the $\beth_{\alpha+1}^+$-cc.

1-3 are hints given in the exercise, the following are not.

Lemma 4. Each $(\beth_\alpha^+)^V$ is a cardinal in $V[G]$.

Proof. We prove the lemma by considering different cases. The case of $\alpha=0$ follows from $P$ being $\omega$-closed. For the case of $\alpha+1$, suppose for contradiction that $\beth_{\alpha+1}^+$ is not a cardinal in $V[G]$. Then in $V[G]$ we fix a bijection $f:\beth_{\alpha+1}\leftrightarrow \beth_{\alpha+1}^+$. By general facts of product forcing, we may view $V[G]$ as $V[G^{\leq\alpha}\times G^{>\alpha}]$, so Lemma 1 and 2 mean that we can apply Lemma 15.19 in Jech and deduce $f\in V[G^{\leq\alpha}]$. This means $\beth_{\alpha+1}^+$ is already not a cardinal in $V[G^{\leq\alpha}]$. This contradicts the fact that $P^{\leq\alpha}$ has the $\beth_{\alpha+1}^+$-cc. Almost the same argument, viewing $V[G]=V[G^{<\alpha}\times G^{\geq\alpha}]$ takes care of the case when $\alpha$ is inaccessible.

So it remains to show the case of $\alpha$ a singular limit. This means that $\beth_\alpha$ is singular in both $V$ and $V[G]$. Now suppose for contradiction that $\beth_\alpha^+$ is not a cardinal in $V[G]$. Consider what $V[G]$ thinks is the cardinality of $\beth_\alpha^+$. In $V[G]$, we have $|\beth_\alpha^+|^{V[G]}<\beth_\alpha^+$. Being a cardinal is downward absolute, so $|\beth_\alpha^+|^{V[G]}$ is a cardinal in $V$ as well. But since there's no cardinal between $\beth_\alpha$ and $\beth_\alpha^+$, we see that $|\beth_\alpha^+|^{V[G]}$ is at most $\beth_\alpha$. Notice that in $V[G]$, $|\beth_\alpha^+|^{V[G]}$ is at least $\beth_\alpha$, because $\beth_\alpha$ injects into $\beth_\alpha^+$, which in turn injects into $|\beth_\alpha^+|^{V[G]}$. This means $\text{cf}^{V[G]}(\beth_\alpha^+)\leq\text{cf}^{V[G]}(\beth_\alpha)<\beth_\alpha$. This tells us that $V[G]$ thinks $\beth_\alpha^+$ is singular. Splitting $V[G]$ into $V[G^{\leq\alpha}\times G^{>\alpha}]$ again, we can show using Lemma 15.19 that it must already be singular in $V[G^{\leq\alpha}]$, which contradicts the the cc property of $P^{\leq\alpha}$. $\square$

Lemma 5. In $V[G]$, each $(\beth_\alpha)^V$ has cardinality $\aleph_\alpha^{V[G]}$.

Proof. We show this by induction on $\alpha$. The case $\alpha=0$ is by absoluteness of $\omega$. The limit case is straightforward. So let's assume the lemma holds for $\alpha$, and consider $\alpha+1$.

First, notice that by the inductive assumption, we have that
$$
\aleph_\alpha^{V[G]}=|\beth_\alpha|^{V[G]}\leq \beth_\alpha<\beth_\alpha^+=|\beth_\alpha^+|^{V[G]}.
$$
where the last equality is by Lemma 4. Also, every $V[G]$-cardinal $\kappa<\beth_\alpha^+$ is a $V$-cardinal by absoluteness, so $\kappa\leq|\beth_\alpha|^{V[G]}=\aleph_\alpha^{V[G]}$. This shows $\beth_\alpha^+=\aleph_{\alpha+1}^{V[G]}$.

Now consider the projection $G_\alpha$ of $G$ onto $P_\alpha$. This gives us in $V[G]$ a function $f=\bigcup G_\alpha$ from $\beth_\alpha^+$ onto $\beth_{\alpha+1}$. THat is, $|\beth_\alpha^+|^{V[G]}=|\beth_{\alpha+1}|^{V[G]}$.  Putting things together, we've shown $|\beth_{\alpha+1}|^{V[G]}=\aleph_{\alpha+1}^V{[G]}$. $\square$

We are now ready to show the main claim.

Claim. $V[G]$ satisfies the GCH.

Proof. It suffices to show that $(2^{{\aleph_\alpha}^{V[G]}})^{V[G]}\leq \aleph_{\alpha+1}^{V[G]}$. We consider three different cases: successor, inaccessible, singular limit.

The successor case is a straightforward cardinality calculation: $(2^{{\aleph_{\alpha+1}^{V[G]}}})^{V[G]}=(2^{\beth_{\alpha+1}})^{V[G]}$ by Lemma 5. And $(2^{\beth_{\alpha+1}})^{V[G]}=(2^{\beth_{\alpha+1}})^{V[G^{\leq\alpha}]}$ by Lemma 15.19. Furthermore, $(2^{\beth_{\alpha+1}})^{V[G^{\leq\alpha}]}$ is at most $|B(P^{\leq\alpha})|^{\beth_{\alpha+1}}$ by counting names. Since each element of $B(P^{\leq\alpha})$ can be determined by an antichain from $P^{\leq\alpha}$, we have:
$$
|B(P^{\leq\alpha})|^{\beth_{\alpha+1}}\leq|P^{\leq\alpha}|^{\beth_{\alpha+1}}=\beth_{\alpha+1}^{\beth_{\alpha+1}}=2^{\beth_{\alpha+1}}=\beth_{\alpha+2}=\aleph_{\alpha+2}^{V[G]}
$$
This shows that $(2^{{\aleph_{\alpha+1}^{V[G]}}})^{V[G]}\leq \aleph_{\alpha+2}^{V[G]}$.

The inaccessible case is similar, by breaking $V[G]$ instead into $V[G^{<\alpha}\times G^{\geq\alpha}]$.

Now suppose $\alpha$ is a singular limit ordinal. Then $\beth_\alpha$ is singular. Take a confinal sequence $(\alpha_\beta\mid \beta<cf(\alpha)=\gamma)$ supping to $\alpha$ and consider $\beth_\alpha$ as $\bigcup_{\beta}\beth_{\alpha_\beta}$. We notice that $(2^{\beth_\alpha})^{V[G]}$ is at most $|\beth_\alpha|^\gamma$: if $X\in \mathcal{P}^{V[G]}(\beth_\alpha)$, then we map $X$ to the "stratifiying" sequence $(X\cap \beth_{\alpha_\beta}\mid \beta<\gamma)$. This map is injective. Taking stock, we've shown that $V[G]$ thinks $2^{|\beth_\alpha|}$ is at most $|\beth_\alpha|^\gamma$ (the latter being evaluated in $V[G]$). That is,

$V[G]\models 2^{{|\beth_{\alpha}|}}\leq\prod_{{\beta<\gamma}} 2^{{|\beth_{{\alpha_{\beta}}}|}}\leq\prod_{{\beta<\gamma}}\aleph_{{\alpha_{\beta}+1}}=\left(\sup_{{\beta<\gamma}}\aleph_{{\alpha_{\beta}}}\right) ^{\gamma}={|\beth_{{\alpha}}|}^{\gamma}$

This latter cardinal $|\beth_\alpha|^\gamma$ is at most $|\beth_\alpha^{\beth{\gamma+1}}|^{V[G]}$, which is in turn at most $|\beth_\alpha^{\beth{\gamma+1}}|^{V[G^{\leq\gamma}]}$. But now
$$
|\beth_\alpha^{\beth{\gamma+1}}|^{V[G^{\leq\gamma}]}\leq  |2^{\beth_\alpha\times\beth_{\gamma+1}}|^{V[G^{\leq\gamma}]} \leq |B(P^{\leq\gamma})^{\beth_\alpha\times\beth_{\gamma+1}}|= \beth_{\gamma+2}^{\beth_\alpha\times\beth_{\gamma+1}}=\beth_{\alpha+1}
$$

Using Lemma 5 again, $|\beth_{\alpha+1}|^{V[G]}=\aleph_{\alpha+1}^{V[G]}$, we conclude that $(2^{{\aleph_{\alpha}^{V[G]}}})^{V[G]}\leq{\aleph_{\alpha+1}}^{V[G]}$ for singular limit $\alpha$. This finishes the proof.

Title: Exercise 15.15 of Jech: forcing GCH


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

答:
\begin{align}
z & = e^{i\theta} \\[8pt]
dz & = ie^{i\theta}\,d\theta \\[8pt]
-i\frac{dz}{z} & = d\theta
\end{align}

$$
\int_0^{2\pi}\frac{d\theta}{3+\sin\theta+\cos\theta} = \int\limits_C \frac{-i\,dz}{z\left(3-\frac i2\left(z-\dfrac1z\right)+\frac12\left(z+\dfrac1z\right)\right)}
$$
$$
=\int\limits_C \frac{-i\,dz}{6z + (1-i)z^2 + (1+i)}.
$$
The quadratic polynomial on the bottom has two zeros: one outside the unit disk and one inside.  Specifially, they are at
$$
z = \frac{-3\pm2\sqrt{7}}{1-i} = \frac{-3\pm\sqrt{7}}{2}(1+i).
$$
You can write
$$
\frac{-i}{(1-i)z^2 + 6z +(1+i)} = \frac{A}{z+\left(\frac{3+\sqrt{7}}{2}(1+i)\right)} + \frac{B}{z+\left(\frac{3-\sqrt{7}}{2}(1+i)\right)}.
$$
You need to find $B$.  The integral of the second fraction around the circle will be $2i\pi B$.  Then integral of the first fraction around the circle will be $0$ since it doesn't vanish anywhere in the disk.

ContentLicense: CC BY-SA 3.0


问:
Prove limit converges in definition of $e.$
I've looked up several related questions, but they do not answer what I am looking for.  Please give link if this is a duplicate.

What I eventually want to know is why

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n$$

converges to a finite value.

I mean, we have to know that the limit converges before we can call it a constant, right?

I know that $f(n)=\left(1+\dfrac1n\right)^n$ is strictly increasing, but how do we know it is bounded above?

Also, I tried using the binomial theorem to write

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left[\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\right],\tag{1}$$

but I am unsure how to prove that the sum converges for $n\to\infty,$ because the general term depends on both $n$ and $k.$

Here is what I think will work.

Let $a_{n,k}=\dfrac{n!}{n^k k! (n-k)!}.$  Then $a_{n+1,k+1}=\dfrac{(n+1)!}{(n+1)^{k+1} (k+1)! (n-k)!}.$

Using the Ratio Test, we get

$$\lim_{n\to\infty\atop k\to\infty}\left|\frac{a_{n+1,k+1}}{a_{n,k}}\right|=\lim_{n\to\infty\atop k\to\infty}\left[\frac{1}{k+1}\left(\frac{n}{n+1}\right)^k\right].$$

where the limit on the RHS is less than 1.

That may take care of the case where $n\in\mathbb{N},$ so I suppose that (1) can be generalized using "Newton's generalised binomial theorem."

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

suppose $A=(1+\frac{1}{n})^n$ So:
$$ \lim_{n\to\infty} \ln(A) = \lim_{n\to\infty} \ln((1+\frac{1}{n})^n = \lim_{n\to\infty} n\ln(1+\frac{1}{n}) $$
Applying (1):
$$ \lim_{n\to\infty} n\ln(1+\frac{1}{n}) = \lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3}))$$
$$\lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3})) = \lim_{n\to\infty} (1-\frac{1}{2n}+nO(\frac{1}{n^3})) = 1 $$
Since $$ \lim_{n\to\infty} \ln(A) = 1 $$ Than $$ \lim_{n\to\infty} A = e $$
So e is the least upper bound.

ContentLicense: CC BY-SA 4.0


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

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

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

Thanks for suggestions

答:
Actually, the upper bound of this sequence is the standard definition of $e$. More formally,
$$
e=\lim_{n\to\infty} (1+{1\over n})^n\approx 2.71828182846
$$
To reject $14\over 5$ as an upper bound, note that $e< 14/5$.


问:
Computing the order, inverse, and parity of a permutation
How do you compute the order, inverse and parity of $\alpha=(12)(43)(13542)(15)(13)(23)$? Please explain all steps taken to get the answer.

I guess my thought process was to first put it into a permutation, but I didn't know how. So then I thought I should break down the middle cycle (idk if that's the correct terminology) 13542 into 2-cycles like the rest.  That's when I got (12)(43)(31)(51)(54)(52)(32)(42)(41)(43) but that seemed like I was doing way too much. So I guess I'm not really sure where or how to start.

答:
I would rather find out the cycle structure of $\alpha$ itself:

By convention (in most places) permutations are composed from right to left, just as functions ($\,f\circ g(x)=f(g(x))\,$). Input each given number to $\alpha$, they give

$\alpha(1)=5$ because $(23)$ doesn't touch $1$, then $1$ moves to $3$ then $3$ moves to $5$ in the middle cycle.

$\alpha(5)=4,\quad \alpha(4)= 1.$ That's the cycle $(154)$.

$\alpha(2)=3$, and $\alpha(3)=2$. Check these.

So, we get $\alpha=(154)(23)$.

It is odd ($\,\alpha=(15)(54)(23)\,$). Its inverse is $(145)(23)$. (Note that disjoint cycles commute with each other.)
Its order is neither $3$ nor $2$, but $6$.


问:
How is Maximum Independent Set maximum-degree-approximative?
a undirected graph $G=(V,E)$ is given and we want to find a maximum independent set $I$ with the following algorithm:

```<code>I = {}

for every vertex v in V:
    if v is not adjacent to any vertex from I:
        I.add(v)

return I
</code>```

I want to show that this algorithm is max-degree-aproximative. So how can we show, that $|I|\geq \frac{Opt}{max-degree}$ is valid. My first idea was to set $|V|$ as an upper bound for $Opt$ ($\Rightarrow |V|\geq Opt$).

Would be cool if someone could help me. :-)

答:
Welcome to MSE!

Your idea is a good one. Write $\Delta$ for the maximum degree. If we can show

$$\Delta |I| \geq |V| \geq \mathsf{OPT}$$
then we're done.

Say (towards a contradiction) that $\Delta |I| < |V|$. Then there must be some vertex $v$ which is not adjacent to any element of $I$ (do you see why?)

But then $v$ should have been added to $I$, and we contradict the definition of the algorithm.

I hope this helps ^_^


问:
Order of certain subgroups of $\text{GL}_3(\mathbb{F}_p)$.
Let $\mathbb{F}_p$ be the field with $p$ element. How many elements do the following subgroups of $\text{GL}_3(\mathbb{F}_p)$ have?

$G_1=\left\{ \begin{pmatrix} x & a & b \\ 0 & d & c \\ 0 & 0 & x \end{pmatrix} \right\}$,

$G_2=\left\{ \begin{pmatrix} x & a & b \\ 0 & x & 0 \\ 0 & d & c \end{pmatrix} \right\}$,

$G_3=\left\{ \begin{pmatrix} a & 0 & b \\ d & x & c \\ 0 & 0 & x \end{pmatrix} \right\}$.

Here GL is the general linear group, the group of invertible n×n matrices.

Could someone provide me with a complete solution for one of the three subgroups so that I know how to redo it with the others?

Thank you in advance.

答:
Let's do (3).

The only constraint on the variables is that the determinant is nonzero. The determinant in this case is $ax^2$, so we must require $a,x\ne 0$ but $b,c,d$ can then be arbitrary. That means $(p-1)^2$ choices for $a,x$ and $p^3$ choices for $b,c,d$ so there are $p^3(p-1)^2$ elements in this subset.


问:
The sum of quadratic residues mod $p$ is $0$
I'm trying to solve this problem:

If $p$ is congruent to $1$ mod $4$, then the sum of the quadratic residues mod $p$ is $0$.  (Where $p$ is an integer prime)

But I seem to have proven this holds for all prime $p$. Presumably my proof is wrong, but I can't see it; can anyone spot where?

Proof: As $p$ is a prime, there $\exists a $ primitive root. The quadratic residues are then $a^2, a^4, \dots, a^{p-1}$, as we know that there are $\phi(p)/2$ quadratic residues. Adding these gives $a^2(1+ (a^2)^1 + \dots + (a^2)^{\frac{p-3}{2}})$. As $(1-a^2, p) = 1$ (as $a$ is a primitive root it is not $\pm 1$ modulo $p$) we calculate $a^2(1+ (a^2)^1 + \dots + (a^2)^{\frac{p-3}{2}}) \times (1-a^2) = a^2\times (1-(a^2)^{\frac{p-1}{2}})$, which is divisible by $p$ due to Euler's theorem.

I haven't used that $p$ is congruent to $1$ mod $4$; have I gone wrong somewhere in my proof?

答:
According to https://oeis.org/A076409 (sum of quadratic residues), for primes $p\ge 5$, the sum of quadratic residues of $p$ is divisible by $p$. The comments points to Wolstenholme's theorem, but your proof is perfectly fine except for $p=2,3$, for which the primitive root $\equiv -1 \pmod p$, and $1$ is the only quadratic residue for these primes.

Remark: if you are to use $p \equiv 1 \pmod 4$, we must have $p\ge 5$, then your claim that $a \ne \pm 1\pmod p$ is true, and the proof can proceed.


问:
Finding asymptotes of a function using limits
I need to find asymptotes of a function. I didn't find any vertical asymptotes and I think there aren't any. But when trying to find a horizontal asymptote I evaluate the limit as $x$ approaches infinity but can't successfully find it.

$$f(x)=x\arctan(x)$$

答:
The function $f$ has an oblique asymptote $y=a x + b$ when $x\to\infty$ iff

$\lim_{x\to\infty} \frac{f(x)}{x} = a$

$\lim_{x\to\infty} (f(x)-a x) = b$

Similar conditions hold for the case $x\to -\infty$.

For $f(x)= x \arctan(x)$ we have when $x\to +\infty$
\begin{equation}
\frac{f(x)}{x} = \arctan x\to \frac{\pi}{2}
\end{equation}
\begin{equation}
f(x) - \frac{\pi}{2}x = x(\arctan x - \frac{\pi}{2}) = - x \arctan\left(\frac{1}{x}\right) \to -1
\end{equation}
Hence there is an oblique asymptote $y = \frac{\pi}{2} x - 1$ at $+\infty$.

What is the asymptote when $x\to -\infty$ ?


问:
What does area represent?
Since any two Euclidean shapes have an infinite number of points inside of them, shapes with different area have the same infinite number of points in them (and any object has the same number of points in it as are inside the plane it is inside). So area isn't a measure of the amount of points in an object, right? And if this is true, what does the area of a shape actually represent? In other words beyond the abstract idea of the amount of space in an object, what does an area of, for example, $4$, mean?</p>

答: It tells you that you could cover the region with 4 $1\times 1$ squares. If it's a weird shape, it would be more informative to say that if you cut it out of material and a $1\times 1$ square of that material weighs 1 ounce, then the shape you cut out would weigh 4 ounces.</p>

Title: What does area represent?


问:
Details of how many 3-element subsets of $\{1, 2, ..., 20\}$ exist such that no two consecutive integers are in the same subset?
I do realize that an answer exists already (https://math.stackexchange.com/a/1421851/797279), however I was wondering why we do not divide by $3!$, such that  the formula becomes $$\binom{20}{3} - \frac{19 \cdot 18}{3!} + 18$$

My reasoning is that when we choose a $3$-element subset with consecutive elements we go about it in such a process:

There are $19$ integers with consecutive numbers, thus we have $19$ choices for the first element

We then have $1$ choice for the second element as we want it to be consecutive number of the first element

We now have $18$ choices to work with for the third element

Order doesn't matter for the set thus we divide by the number of ways to permute the elements

答:
You don't divide by $3!$ because your way of choosing the triplets is in $1-1$ correspondance with the sets you are looking for.

Let me explain it a bit more in detail:

Frist, let us look at the case where in the set $\{a,b,c\}$ only $2$ of the elements are consecutive, that is (after relabeling if needed) $\{a,b,b+1\}$ with $a\neq b-1$, $a\neq b+2$. There is only one way you could have chosen that particular set.

For the sets of the form $\{a,a+1,a+2\}$ is a bit different because there are (exactly) $2$ ways you could have chosen this particular set. That is the reason of the last "$+18$" in the formula.


问:
Question about the Legendre's symbol: $\left(\frac{p-1}{5}\right) \overset?= \left(\frac{5}{p}\right)$
I was reading about Legendres symbol (number theory) and encountered the following equality:

I don't understand how they got
$$\left(\frac 3p \right) = (-1)^{\frac{p-1}{2}\frac{3-1}{2}}\left(\frac{p-1}{3}\right) \tag{underlined with red}$$
I thought $p-1$ in RHS should have been $p$ instead but saw the same thing in the other part of the section:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$

I tried to show that the last image illustrates the true identity as follows:
$$\left(\frac{p-1}{5}\right) = \left(\frac{2\cdot\frac{p-1}{2}}{5}\right) = \left(\frac{2}{5}\right)\left(\frac{\frac{p-1}{2}}{5}\right) =\left(\frac{\frac{p-1}{2}}{5}\right) $$

But I can't show that the last expression is equal to $(5/p)$.

Is there anything I missed while learning the theory behind this symbol?

答:
You are right, that should have been $p$ instead of $p-1$. It is certainly false in general that
$$\left( \frac 3p\right) = (-1)^{(p-1)(3-1)/4} \left( \frac {p-1}3\right) \,.$$
For example, when $p = 5$ the LHS is $-1$ and the RHS is $1 \cdot 1$.

Maybe the author got caught up in writing all the $p-1$'s in the exponents.


问:
Factor 43361 knowing $\phi(43361)$

It is given that the number $43361$ can be written as product of two distinct prime numbers $p_{1}$ and $p_{2}$. Further, assume that there are $42900$ numbers which are less than $43361$ and co-prime to it. Then find $p_{1}+p_{2}$.

A simple search on Google yielded $p_{1}$ and $p_{2}$ to be $131$ and $331$. But what would be the proper way to find it?

答:
Here is a method that I used, it helps you if you don't know the Euler's totient thing.
Since 42900 numbers before 43361 are divisible by neither p1 nor p2 we can tell that 462(inclusive of 43361) numbers are divisible by either p1 or p2 (with 43361 as an exception which is divisible by both).

let us say x numbers are divisible by p1 and y numbers by p2.

then, taking multiples of p1,

```<code>                        *p1, 2p1, 3p1, .....43361 is the series*    with a total of x
                                                                    numbers in it.
</code>```

we know from arithmetic progression that,
An= A+ (N-1)D

```<code>                                43361= p1+(x-1)p1
</code>```

solving, we get                     x= 43361/p1

similarly,

```<code>                     *p2, 2p2, 3p2, .....43361 is the series*    with a total of y
                                                                 numbers in it.
                                43361= p2+(y-1)p2
</code>```

solving, we get                      y= 43361/p2

since x+y= 462,

```<code>                             43361/p1 + 43361/p2 = 462

                              43361(p1+p2/p1.p2) = 462

                              43361(p1+p2/43361) = 462

</code>```

Solving, p1+p2= 462

Title: Factor 43361 knowing $\phi(43361)$

Body:
Here is a method that I used, it helps you if you don't know the Euler's totient thing.
Since 42900 numbers before 43361 are divisible by neither p1 nor p2 we can tell that 462(inclusive of 43361) numbers are divisible by either p1 or p2 (with 43361 as an exception which is divisible by both).

let us say x numbers are divisible by p1 and y numbers by p2.

then, taking multiples of p1,

                        *p1, 2p1, 3p1, .....43361 is the series*    with a total of x
                                                                    numbers in it.

we know from arithmetic progression that,
An= A+ (N-1)D

                                43361= p1+(x-1)p1

solving, we get                     x= 43361/p1

similarly,

                     *p2, 2p2, 3p2, .....43361 is the series*    with a total of y
                                                                 numbers in it.
                                43361= p2+(y-1)p2

solving, we get                      y= 43361/p2

since x+y= 462,

                             43361/p1 + 43361/p2 = 462

                              43361(p1+p2/p1.p2) = 462

                              43361(p1+p2/43361) = 462

Solving, p1+p2= 462


问:
Only at the vertex of the light-cone the vectors, that are tangent to the cone are all and only light-cone ones
Tangent vectors to a light-cone in the spacetime are not always vector timelike. The condition of tangency to a submanifolds is a linear equation in the tangent space, and this also applies to the light-cone at any point other than the vertex that is a singular point,while the equation of the light-cone in tangent space is quadratic.
Considering a light-cone with vertex in $x_{o}^\mu$a:
$$\eta_{\mu \nu}(x^\mu-x^\mu_{0})(x^\nu-x^\nu_{0})=0$$
Only at the vertex of the light-cone the vectors, that are tangent to the cone  are all and only light-cone ones The tangent vectors to the remainder of the cone may be  timelike or spacelike.
Someone cuold give me an explanation?</p>

答:
Consider the hypersurface $S=\{\eta_{\mu \nu} x^{\mu}x^{\nu}=0$} and a point $ P \in S$ different than the origin. A (tangent) vector $\mathbf{v}$ in $P$ is a lightlike vector if, by definition, $\eta(\mathbf{v},\mathbf{v})=0$. This is a quadratic equation. On the other hand the condition that defines the tangent space of $S$ in $P$ is linear. For simplicity, think of the lightcone as a surface of $M^3$, and suppose that $P$ lies in the future nappe of the cone. We can parametrize the cone $-x^2 +y^2 + z^2 =0$ as
$$\begin{cases}
x=r>0\\
y=r\cos \theta\\
z=r\sin \theta \text{.}
\end{cases} $$
A basis of the tangent space is given by $\mathbf{e}_1 = (1, \cos \theta, \sin \theta)$ and $\mathbf{e}_2=(0, -r\sin \theta, r \cos \theta)$. If we consider a linear combination $\lambda \mathbf{e}_1+ \mu \mathbf{e}_2$ we get (with a straightforward calculation):
$$\eta(\lambda \mathbf{e}_1+ \mu \mathbf{e}_2, \lambda \mathbf{e}_1+ \mu \mathbf{e}_2)= -\lambda^2 +\lambda^2 + r^2 \mu^2 \ge0 \text{.}$$
If, thus, $\mu \neq 0$ then the vector $\lambda \mathbf{e}_1+ \mu \mathbf{e}_2$ is spacelike. Note however that it can't be timelike. Intuitively, this is due to the fact that a cone is a convex surface. Also note that in the origin, which is a singularity, every tangent vector is lightlike (letting $r=0$).


问:
Rank of a matrix using colunm space
I'd like some verification of my solution method since I don't have any solution at hand. I'm given the following matrix and asked to determine the rank of it in function of  $x$:

$\begin{pmatrix}
2 & 2 & -6 & 8 \\
3 & 3 & -9 & 8 \\
1 & 1 & x & 4 \\
\end{pmatrix}$

I started with the fact that for any given matrix $A \in \mathbb{R}^{m \times n}$ the column space $C(A) = \mathbb{R}^m$. In this case, the column space is given by:

$$C(A) = vct \{\begin{pmatrix}
2 \\
3 \\
1 \\
\end{pmatrix}, \begin{pmatrix}
2 \\
3 \\
1 \\
\end{pmatrix}, \begin{pmatrix}
-6 \\
-9 \\
x \\
\end{pmatrix}, \begin{pmatrix}
8 \\
8 \\
4 \\
\end{pmatrix}\}$$ Since the column space $C(A)$ is $\mathbb{R}^3$ or $\mathbb{R}^4$ depending on $x$, I set out to find for which value of $x$ the vector $(-6, -9, x)^T$ would be a linear combination of the other vectors indicating that the column space would be equal to $\mathbb{R}^3$.

Solving the following system of linear equations:

$$ \begin{pmatrix}
2 & 2 & 8 & | & -6 \\
3 & 3 & 8 & | & -9 \\
1 & 1 & 4 & | & x \\
\end{pmatrix} \rightarrow
\begin{pmatrix}
1 & 1 & 4 & | & x \\
0 & 0 & -4 & | & -9 - 3x \\
0 & 0 & 0 & | & -6-2x \\
\end{pmatrix}$$

That the matrix is unsolvable when $-6-2x \neq 0$ or when $x \neq -3$. Thus I came to the conclusion that

$$ rank(A) = 3 \iff x = -3$$
$$ rank(A) = 4 \iff x \neq -3$$

I have thus no idea if this solution is correct. If not, could someone tell me how I should solve it?

答:
It can't be correct: the rank of $A$ can never be $4$ since the column space is a subspace of $\mathbb R^3$. The first and last column are linearly independent, so the rank is at least $2$.

Depending on $x$ it might be equal to $2$ or $3$ and your calculation shows that the rank is $3$ as long as $x\neq -3$ and $2$ when $x=-3$.

Title: Rank of a matrix using colunm space


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

答:
When a basis is orthonormal, then a vector is merely the sum of its orthogonal projections onto the various members of the basis.

That is not true of bases in general (see my comment below for a simple counterexample).


问:
Find all values of $c$ such that $\{(1,c,0),(1,0,1),(0,1,-c)\}$ are linearly independent.
Find all values of $c$ such that $\{(1,c,0),(1,0,1),(0,1,-c)\}$ are linearly independent.

So, my strategy is to consider these vectors  as column vectors, and then form a matrix with them as the columns. Once I do that, I will employ Gauss-Jordan elimination to find an equivalent matrix except in reduced row echelon form.

I typed the matrix into matlab and asked it to find the reduced row echelon form, and it gave me the identity matrix. Thus, I believe that no matter what value I pick for $c$, these vectors will be linearly independent.

However, upon trying to employ the Gauss-Jordan elimination by hand, I have a $1-c^2$ in the $(3,3)$ position. Then I got to thinking, if $c \in \{ \pm 1\}$ then this would be zero, and I could not use it to eliminate the entry above it, and thus I would not be able to reduce this to the identity matrix.

Then I thought some more, and wondered if I would have employed different row operations get to reduced row echelon form, then maybe a different situation like this would have arisen... So now I am very confused and don't know how to do the problem!!

Help is appreciated, thank you!

答:
Alternatively, should you wish to take a more..conservative approach for the vectors to be independent, they must satisfy the condition:

$$l_1(1,c,0)+l_2(1,0,1)+l_3(0,1,-c)=(0,0,0)\Rightarrow l_1=l_2=l_3=0$$

We notice that $$l_1(1,c,0)+l_2(1,0,1)+l_3(0,1,-c)\Rightarrow \\(l_1+l_2,cl_1+l_3,l_2-cl_3)=(0,0,0)\Rightarrow \\l_1=-l_2\\cl_1+l_3=0\\l_2-cl_3=0$$

or equivalently $$-cl_2+l_3=0\\l_2-cl_3=0$$

by multiplying the second relationship with $c$ and then adding them, we obtain $l_3-c^2l_3=0$ or $l_3(c^2-1)=0$. But we want $l_3=0$ to hold for any $c$, thus we must have $c^2\neq1$


问:
Proving that the induced connection on the dual bundle is indeed a connection
I have a hard time proving that the Leibniz property holds of the induced connection on the dual bundle. Let me first outline some definitions.

I will first give the definition of a connection on a vector bundle $E \to M$, after which I'll show the definition of the induced connection on the dual bundle $E^{*} \to M$. I will then show my work and at which part I am stuck.

Definitions

Definition 1. A connection on the vector bundle $E \to M$ is a bilinear map $\nabla$
$$
\mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E),
$$
satisfying
$$
\nabla_{fX}(s) = f \nabla_X(s), \qquad \nabla_X(fs) = f \nabla_{X}(s) + X(f)s.
$$
for all $f \in C^{\infty}(M), X \in \mathfrak{X}(M), s \in \Gamma(E)$.

Definition 2. Let $E^{*} \to M$ be the dual vector bundle of $E \to M$. The induced connection on the dual $E^*$ is given by
$$
\nabla^{*}_X(\xi)(s) = X(\xi(s)) - \xi(\nabla_X(s)),
$$
for all $s \in \Gamma(E), \xi \in \Gamma(E^*)$ and $X \in \mathfrak{X}(M)$.

A little note about the notation. Although I am not 100% sure, I suppose the notation $(\xi)(s)$ just means that for $x \in M$, $(\xi)(s)(x) := \xi(x)(s(x))$. In this sense $(\xi) \in \Gamma(E^*)$ must be linear when it comes to the sections $s \in \Gamma(E)$. This means that for $f \in C^{\infty}(M)$ and $s \in \Gamma(E)$, we have (where $x \in M$)
$$
(\xi)(fs)(x) = \xi(x)(f(x) s(x)) = f(x) \xi(x)(s(x))
$$
for all $x \in M$, which can be succinctly written as $\xi(fs) = f \cdot \xi(s)$.

Proving the induced connection is a connection

Bilinearity + First Condition (I managed to prove this)

Consider $f_1, f_2 \in C^{\infty}(M)$ and $X_1, X_2 \in \mathfrak{X}(M)$. We write
\begin{align}
\nabla_{f_1 X_1 + f_2 X_2}^*(\xi)(s) &= (f_1 X_1 + f_2 X_2)(\xi(s)) - \xi(\nabla_{f_1 X_1 + f_2 X_2}(s)) \\
&= f_1 X_1(\xi(s)) + f_2 X_2(\xi(s)) - \xi(f_1\nabla_{X_1}(s) + f_2\nabla_{X_2}(s)) \\
&=  f_1 X_1(\xi(s)) + f_2 X_2(\xi(s)) - f_1 \xi(\nabla_{X_1}(s)) + f_2\xi(\nabla_{X_2}(s)) \\
&= f_1 \nabla^*_{X_1}(\xi)(s) + f_2 \nabla^*_{X_2}(\xi)(s)
\end{align}

Take $a_1, a_2 \in \mathbb{R}$ and $\xi_1, \xi_2 \in \Gamma(E^*)$ (and $s \in \Gamma(E)$). We then write
\begin{align}
\nabla^*_X(a_1 \xi_1 + a_2 \xi_2)(s) &= X((a_1 \xi_1 + a_2 \xi_2)(s)) - (a_1 \xi_1 + a_2 \xi_2)(\nabla_X(s)) \\
&= X(a_1 \xi_1(s) + a_2 \xi_2(s)) - a_1 \xi_1(\nabla_X(s)) - a_2 \xi_2(\nabla_X(s)) \\
&= a_1 (X(\xi_1(s)) - \xi_1(\nabla_X(s))) + a_2 (X(\xi_2(s)) - \xi_2(\nabla_X(s))) \\
&= a_1 \nabla^*_X(\xi_1)(s) + a_2 \nabla^*_X(\xi_2)(s)
\end{align}

Leibniz Condition (Here I get stuck)

From here I am a bit confused about multiple things. Namely, we now want to prove that for $X \in \mathfrak{X}(M), \xi \in \Gamma(E^*)$ and $s \in \Gamma(E)$, we have
$$
\nabla^*_X(f \xi)(s) = f \nabla^*_X(\xi)(s) + X(f)\xi(s).
$$
I hope this is the case at least. From here if we write out from the LHS, we obtain
\begin{align}
\nabla^*_X(f \xi)(s) &= X(f \xi(s)) - (f \xi)(\nabla_X(s)) \\
&= f X(\xi(s)) - (f \xi)(\nabla_X(s))
\end{align}
From here I don't know how to continue. What mistakes am I making. Thanks in advance!

答:
While writing out this question, I realised the following: the equality $X(f \xi(s))  = f X(\xi(s))$ is not true. What is true though is, $X(f \xi(s))  = f X(\xi(s)) + \xi(s) X(f)$ due to the product rule. Writing out the Leibniz condition is thus:
\begin{align}
\nabla^*_X(f \xi)(s) &= X(f \xi(s)) - (f \xi)(\nabla_X(s)) \\
&= f X(\xi(s)) +  \xi(s) X(f) - f \xi(\nabla_X(s)) \\
&= f(X(\xi(s)) - \xi(\nabla_X(s))) + \xi(s) X(f) \\
&= f \nabla^*_X(\xi)(s)  + X(f)\xi(s),
\end{align}
which is exactly the property that I wanted to prove.


问:
$f(z)=\sum_{k=0}^\infty a_kz^{k}$ with $\sum |a_k|<\infty$ is of bounded variation on circle $|z|=1$
Let $f(z)=\sum_{k=0}^\infty a_kz^{k}$ be a power series. Show that if $\sum |a_k|<\infty$, then $f(z)$ is of bounded variation on every radius of the circle $|z|=1$. (If, e.g., the radius is $0\leq x\leq 1$ and the $a_k$ are real, then $f(x)=\sum a_k^+x^k-\sum a_k^-x^k$).  Problem copied from Wheeden & Zygmund Measure and Integral.

For example working in general with $a_k\in \mathbb{C}$, for all $k$, I was able to prove that $$S_\Gamma=\sum_{j=1}^m\left|f(z_j)-f(z_{j-1})\right|<M$$ for any partition $\Gamma=\{z_0,\cdots,z_m\}$ of the following radius: line joining $0$ and $1$, line joining $-1$ and $0$, line joining $0$ and $i$, line joining $-i$ and $0$. The interesting part, occurs when the radius is a line joining $0$ and other point in the circle $|z|=1$, let $z_m=x_m+iy_m$ be such point. Consider a partition $\Gamma=\{z_0,\cdots,z_m\}$ of this radius, where $z_j=x_j+iy_j$ and $z_0=0+i0$. If we can make $S_\Gamma<\infty$, we are done

$$S_\Gamma=\sum_{j=1}^m|f(z_j)-f(z_{j-1})| = \sum_{j=1}^m\left|\sum_{k=0}^\infty a_k\left((x_j+iy_j)^k-(x_{j-1}+iy_{j-1})^k\right)\right| = \sum_{j=1}^m\left|\sum_{k=0}^\infty a_k\sum_{l=0}^k \binom{k}{l}\left( (x_j)^{k-l}(iy_j)^k-(x_{j-1})^{k-l}(iy_{j-1})^k \right)\right|\leq $$
$$ \leq \sum_{j=1}^m \sum_{k=0}^\infty |a_k|\sum_{l=0}^k \binom{k}{l}\left| x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k \right|$$ since $|i^k|=1$.

Interchanging the order of summation we have $$S_\Gamma \leq  \sum_{k=0}^\infty |a_k|\sum_{l=0}^k \binom{k}{l}\sum_{j=1}^m\left| x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k \right| = \sum_{k=0}^\infty |a_k|\sum_{l=0}^k \binom{k}{l}x_m^{k-l}y_m^k =$$ $$= \sum_{k=0}^\infty |a_k|(x_m+y_m)^k  $$ in order to get convergence is enough that $|x_m+y_m|^k\leq 1$, since the initial condition is $|z_m|=1$ we have $x_m^2+y_m^2=1$. Replacing we get $|x_m+y_m|=|x_m+\sqrt{1-x_m^2}|$, but the function $|x+\sqrt{1-x^2}|>1$ on $0<x<1$. For this reason I cannot verify the convergence of the serie, so I can't conclude if it's of bounded variation.

Is my bounding approach too excessive? so that loses its property of convergence?
I think the problem could be the last two equalities, since I'm assuming $\left| x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k \right| = x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k $ which could be something that is not true all the time. Any comment will be appreciated.

答:
Your approach is too complicated. It suffices to consider the case of a ray joining $z=0$ and $z=1$:

Let $\Gamma=\{z_0,\cdots,z_m\}$ be a partition of the ray joining $z=0$ and $z=e^{i \alpha}$ on the unit circle. Then $z_j = x_j e^{i \alpha}$ with real numbers
$$
 0 = x_0 < x_2 < \cdots < x_m = 1 \, .
$$
Also
$$
\sum_{j=1}^m\left|f(z_j)-f(z_{j-1})\right| = \sum_{j=1}^m\left|F(x_j)-F(x_{j-1})\right|
$$
where $F$ is defined as $F(z) = \sum_{n=0}^\infty a_n e^{i n\alpha} z^n$.

The coefficients of $F$ satisfy the same condition: $ \sum_{n=0}^\infty |a_n e^{i n\alpha}| =  \sum_{n=0}^\infty |a_n| < \infty$. This shows that it suffices to show that $f$ is of bounded variation on the interval $[0, 1]$.

For a proof of that case, see for example Show that $\,f(x) = \sum_{n=0}^\infty a_nx^n$, for $x \in [0,1]$, is of bounded variation.


答:
Area (in $\mathbb R^2$) starts with agreeing that the area of a rectangle of sides $a$ and $b$ has area equal to $a\cdot b$. You may take that as an axiom. Now, the area of more complicated shapes is obtained by approximating the shape by rectangles. This can be done in different ways and there are plenty of things to prove here, as well as some surprises. For instance, one can approximate a given shape from the outside: cover it by countably many disjoint rectangles, and add their areas. Take the infimum over all such covers, and that will give you a measure of the area of the shape measured from the outside. Similarly, you can inscribe rectangles inside the shape, add their areas, and take the supremum over all such numbers. This will give you a measure of the area of the shape measured from the inside. </p>

<p>Your question is a prelude to measure theory, a very important part of modern analysis. One of the important surprises is that the concept of area (and even the concept of length) can't be extended sensibly to measure all subsets of the plane (line). A famous example is Vitali's Set. So, the resulting theory is quite subtle. </p>

Title: What does area represent?

Body: Since any two Euclidean shapes have an infinite number of points inside of them, shapes with different area have the same infinite number of points in them (and any object has the same number of points in it as are inside the plane it is inside). So area isn't a measure of the amount of points in an object, right? And if this is true, what does the area of a shape actually represent? In other words beyond the abstract idea of the amount of space in an object, what does an area of, for example, $4$, mean?</p>

Body:
Area (in $\mathbb R^2$) starts with agreeing that the area of a rectangle of sides $a$ and $b$ has area equal to $a\cdot b$. You may take that as an axiom. Now, the area of more complicated shapes is obtained by approximating the shape by rectangles. This can be done in different ways and there are plenty of things to prove here, as well as some surprises. For instance, one can approximate a given shape from the outside: cover it by countably many disjoint rectangles, and add their areas. Take the infimum over all such covers, and that will give you a measure of the area of the shape measured from the outside. Similarly, you can inscribe rectangles inside the shape, add their areas, and take the supremum over all such numbers. This will give you a measure of the area of the shape measured from the inside.

Your question is a prelude to measure theory, a very important part of modern analysis. One of the important surprises is that the concept of area (and even the concept of length) can't be extended sensibly to measure all subsets of the plane (line). A famous example is Vitali's Set. So, the resulting theory is quite subtle.


问:
Let $R$ be a ring and let $a \in R$. Let $n \in N$. Show that if $a^n$ has an inverse, a has an inverse.
I can say that $(a^n)^{-1}(a^n)=e$ but I do not know how to use that to show that a has an inverse. Not sure if I just have to manipulate the equation somehow to obtain the result.

Body: Hint: If $a^n b=1$, then $a a^{n-1} b = 1$. Also, $1=b a^n b=b a^{n-1}a$.


答:
Hint: if $n = 2$, we have $aau = a^2u = 1 = ua^2 = uaa$ for some $u \in R$. What's the inverse of $a$? Generalize this to any $n$.

 Concretely since $1 = a^{n}(a^{n})^{-1} = aa^{n-1}a^{-n}$ and likewise $1 = (a^n)^{-1}a^n = (a^n)^{-1}a^{n-1}a$ then $a$ has both a left and right inverse, and so it has an inverse (in general, if $ax = xb = 1$ then $a = axb = b$).

Title: Let $R$ be a ring and let $a \in R$. Let $n \in N$. Show that if $a^n$ has an inverse, a has an inverse.

Body: I can say that $(a^n)^{-1}(a^n)=e$ but I do not know how to use that to show that a has an inverse. Not sure if I just have to manipulate the equation somehow to obtain the result.


答:
We wish to show that $\exists$ $c\in R$ such that $ac=e$.

Let $b$ be the inverse of $a^n$.

Then we have $a^nb=e$.

But we can readily observe that $a(a^{n-1}b)=e$. So $c=a^{n-1}b$ is the inverse of $a$.


问:
Show that there's a unique minimum spanning tree if all edges have different costs

Show that there's a unique minimum spanning tree (MST) in case the edges' weights are pairwise different $(w(e)\neq w(f) \text{ for } e\neq f)$.

I thought that the proof can be done for example by contradiction, saying that we have $2$ different MST meaning that somewhere was possible to pick from more edges, so $w(e) = w(f)$ for $e\neq f$, contradiction. Apparently this is not correct.

How would you show that a graph has a unique MST if all edges have distinct weights?

答:
This is a (slightly) more detailed version of the currently accepted answer

(For the sake of contradiction) Let $T_1 = (V, E_1)$ and $T_2 = (V, E_2)$ be two distinct MSTs of the graph $G = (V, E)$

Note that both have the same vertex set $V$ since both are spanning trees of $G$

Consider the set $E_{\Delta} = E_1 \triangle E_2$

Let $e = (u, v)$ be the edge in $E_{\Delta}$ having the least cost (or weight)

Note that since all costs are unique, and $E_{\Delta}$ is non-empty, $e$ must be unique.

Without loss of generality, assume $e \in E_1$

Now, there must be a path $P$, with $e \notin P$, in $T_2$ connecting $u$ and $v$, since trees are by definition, connected.

Note that at least one edge (say $e'$) that occurs in $P$ must not be in $E_1$, thus, $e' \in E_{\Delta}$

This is because, if $P \subset E_1$,  $T_1$ will contain a cycle formed by the path $P$ and the edge $e$, this leads to a contradiction, since trees by definition are acyclic.

Note that by definition of $e$ and the fact that all costs are distinct, $\text{cost}(e) < \text{cost}(e')$

Now, consider $T_2' = (V, E_2')$, where $E_2' = (E_2 \backslash\{e'\})\cup\{e\}$, this has a strictly lesser total cost than $T_2$

As $T_2$ was a MST, this leads to a contradiction.


问:
Geometrical proof for property of chord of contact with a circle
Let their be a circle with center O, and radius r. Let their be a point P outside the circle at a distance d from O. Let a variable line passing through P cut the circle in 2 points, A and B. We define a point C between A and B such that PC is harmonic mean of PA and PB. Prove that the locus of C is chord of contact of P with respect to the circle.

I can do it through coordinate geometry, I just wanted to find a geometrical method. I tried looking at its inversion, and trying to find relations but I couldn't.
Any help is appreciated. Thanks.

答:
Let the circle have centre $O$.Let the points at which the tangents touch the circle be $D$ and $E$.A variable line through point $P$ cuts the circle at points $A$ and $B$. $DE$ cuts this line at $C$. So proving that $PC$ is the harmonic mean of $PA$ and $PB$ will prove the given statement.

Let $M$ be the midpoint of $AB$. Then $OM\perp AP$.

Now, observe that, pentagon $MODPE$ is cyclic. Hence, $PC\cdot CM=DC\cdot CE$.

Also, since $DC\cdot CE=AC\cdot CB$, $AC\cdot CB=PC\cdot CM$.

$AC=PA-PC$

$BC=PC-PB$

$CM=\frac {1}{2}(AC-BC)=\frac {1}{2}(PA+PB)-PC$

Plugging these values into the equation gives,

$PC\{\frac {1}{2}(PA+PB)-PC\}=(PA-PC)(PC-PB)$

$\Rightarrow   PC=\frac {2PA\cdot PB}{PA+PB}$

Hence,  $PC$ is the harmonic mean of $PA$ and $PB$.


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

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

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

Thanks for suggestions

Body:
If you are not allowed to use the definition of $e$, then by observing that the sequence terms are bound by (refer to this)
$$
\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\
&=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\
&\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3\end{align}
$$
Now to get a stricter bound, since
$$
\frac{1}{2^1} + \frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2} < \frac
{1}{2} \left( 1+\frac{1}{2^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right)
$$
we have
$$
\left(1+\frac 1n\right)^n < 1 + \frac{1}{2^0} + \frac
{1}{2} \left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right) < 2 + \frac{1}{2} \frac{1}{1-\frac{1}{3}} = \frac{11}{4} < \frac{14}{5}
$$
as required.


问:
polynomial polynomialfunction unique
polynomialfunction is unambigiously determined by polynomial but polynomial is not unambigiously determined by polynomialfunction.

for example $f:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,f(x)=x^2-x$ is the zerofunction. I.e. we also could have chosen the zeropolynomial. It was used as an example to show the second statement.

I find it difficult to understand the first statement: it says given a polynomial then the function is unambigiously determined.

Why can one polynomial not define two functions?

答:
"a polynomial function is unambiguously determined by polynomial" - This means that if you give me a polynomial, then it defines a unique function.

"a polynomial is not unambiguously determined by a polynomial function" - This means that if you give me a polynomial function, e.g. $f:\mathbb{Z}_2\to\mathbb{Z}_2$, then $f$ does not necessarily come from a single (unique) polynomial. The example you gave is correct: Let $p_1(X)$ and $p_2(X)$ be polynomials in $\mathbb{Z}_2[X]$, where

$$
p_1(X)=0\text{ (i.e. the zero polynomial) and }p_2(X)=X^2-X.
$$

Clearly, they define the same polynomial function $f:\mathbb{Z}_2\to\mathbb{Z}_2$, but $p_1(X)$ and $p_2(X)$ are not the same polynomials.

Title: polynomial polynomialfunction unique


答: Often orthogonality represents a form of independence (as in statistics, where it says that there is a lack of covariance, or linear algebra where direct sums of vector spaces have a canonical inner product for which the sum is orthogonal).  Orthogonal basis then means the ability to decompose an effect into separate, independent, non-interacting parts that simply add up to form the whole effect.  This kind of decomposition is hugely important in situations where it can be done.</p>

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


问:
Elliptic regularity estimate for Possion with bounded forcing term
Suppose I have $- \Delta u = f $ weakly on a bounded domain $\Omega$ with Dirichlet boundary conditions. I'm looking for an elliptic regularity result of the form $f \in L^\infty \implies u \in C^{1, \alpha}(\Omega)$. The usual Schauder/Newtonian potential results I know require $f \in C^{\alpha}(\Omega)$ but give $u \in C^{2, \alpha}(\Omega)$ Is there a result of the type I am looking for?

答: Since $\Omega$ is bounded, you have that $f\in L^p$ for $p>N$. Assuming the boundary is sufficiently regular and that the Dirichlet boundary condition is zero, you can apply Sobolev regularity, which tells you that $u$ is in $W^{2,p}$. Hence its gradient is in $W^{1,p}$. Now you can apply Morrey’s embedding theorem which says that a Sobolev function in $W^{1,p}$ for $p>N$ is Holder continuous (you need the boundary to be smooth). You can find the details in Gilbarg and Trudinger “ Second order elliptic equations”.


问:
Consequences of Taylor's theorem
Let $a$ be a complex number. Prove that there exists a power series
$$
f(z)=z+c_{2} z^{2}+c_{3} z^{3}+\ldots
$$
with a positive convergence radius $R,$ so that
$$
f(z)+a f(z)^{2}=z, \quad|z|<R
$$
How to prove such an existence? How the coefficients $c_2,~c_3,...$ can be calculated? How can Taylor's theorem be helpful?

答:
Since the question is tagged complex-analysis, here is such an argument:

If $a=0$ the statement is trivial, so assume that $a\ne0$. Now $f(z)+af(z)^2=z$ is equivalent to $$\left(f(z)+\frac{1}{2a}\right)^2=\frac{z}{a}+\frac{1}{(2a)^2}$$
Let $h(z)=\frac{z}{a}+\frac{1}{(2a)^2}$. Since $h(0)\ne0$ there is some neighborhood of $0$ in which $h$ has a square root $g$ where we may assume $g(0)=\frac{1}{2a}$.
Now set $f(z)=g(z)-\frac{1}{2a}$ and one can check that it satisfies the desired properties.

Title: Consequences of Taylor's theorem


问:
Desmos domain error?
Just curious if anyone has an idea why Desmos would claim that $f(0)=0$ for $f(x)=\frac{1}{\frac{1}{x}}?

As a note of interest, Geogebra commits the same error but both Symbolab and Wolfram get it right.

Any thoughts?

答:
They work on the Extended Number Line so
1/0 is ∞ then 1/∞ = 0 again

Note that 0/0 = undefined (NaN) is handled differently

https://en.wikipedia.org/wiki/Extended_real_number_line

Title: Desmos domain error?


问:
Trying to find a function for graph

Hello,
I am trying to write an equation for this graph but I cannot.
I have drawn the graph myself so it may not be a good one, sorry for that.

Until now I have tried to find a link between this graph and exponential graph but I could not.
Do you have any idea to help me?

答:
HINT

Try a function of the fom $a+\frac{b}{x}.$

Title: Trying to find a function for graph


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

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

答: If it is uniformly convergent then there exists $m$ such that $ |\sum\limits_{k=i}^{j}\frac {x^{k}} k| <1$ for all $x$ for all $j>i>m$. Let $x \to 1$ in this. Do you see a contradiction?


问:
Show that the function $y = e^{2x}(A \cos3x + B \sin3x)$ satisfies the differential equation $y'' - 4y' + 13y = 0$
Show that the function $y = e^{2x}(A \cos3x + B \sin3x)$ satisfies the differential equation $y'' - 4y' + 13y = 0$

My attempt:

I get

$y' = e^{2x}((2A + 3B) \cos3x + (2B - 3A) \sin3x)$.

$y'' = e^{2x} ((-5A + 12B) \cos3x + (-12A - 5B) \sin3x)$.

$y'' - 4y' + 13y = e^{2x} ((16A + 24B) \cos3x + (16B - 24A) \sin3x) = 0$.

$(16A + 24B) \cos3x + (16B - 24A) \sin3x = 0$.

$(2A + 3B) \cos3x + (2B - 3A) \sin3x = 0$

And I'm totally lost after this. It seems like this can't be true since I get

$\tan x = const$.

I'd appreciate if you could point me in the right direction. Thanks!

答:
Your derivatives $y'$ and $y''$ are correct, you've just made a mistake summing the whole thing.

$\begin{cases}
y'' & -5A+12B & -12A-5B\\
-4y' & -8A-12B & 12A-8B\\
13y & 13A+0B & 0A+13B\\\\
y''-4y'+13y & 0A+0B & 0A+0B
\end{cases}$


问:
Product of smooth varities is smooth
Let $X$ and $Y$ be smooth varieties, then $X\times Y$ is a smooth variety.
Here is my definition of smooth:

It is enough to show that if $X$ is smooth at $p$ and $Y$ is smooth at $q$ then the product is smooth at $(p,q)$. The smoothness of $X$ at $P$ gives us a $U,\phi,(f_1,..,f_a)$ where $a=n-d$ with the Jacobi having maximal rank. Simillary the smoothness of $Y$ gives a $V,\psi,(g_1,...g_b)$ with $b=m-d'$ such that the Jacobi has maximal rank.

By definition the the product $U\times V$ is open in $X\times Y$ and is also a variety.

I think $\phi\times\psi$ should give an isomorphism to $Z(f_1,..,f_a,g_1,..,g_b)$. But this is not obvious to me. It is also not clear why the jacobi of this should be of maximal rank.

答:
If the base field $k$ is algebraically closed and you are willing to use the criterion on the cotangent bundle (Hartshorne Prop III.10.4)
it seems the result follows from the following argument. Let $X,Y \rightarrow Spec(k)$ be smooth varieties of dimension $d,e$. It follows from III.10.4 that
$\Omega^1_{X/k}, \Omega^1_{Y/k}$ are locally trivial of rank $d$ and $e$. There is the following formula (HH Ex II.8.3) for the cotangent bundle of the product:

$$C1.\text{    }p^*\Omega^1_{X\times_k Y/k}\cong p^*\Omega^1_{X/k}\oplus q^*\Omega^1_{Y/k}.$$

It follows $\Omega^1_{X\times_k Y/k}$ has rank $d+e$. Again by III.10.4 it follows $X\times_k Y$ is smooth since $dim(X\times_k Y)=d+e$. This result is mentioned in the relative setting in the same book HH III.10.1d: If $X,Y$ are smooth over $S$ of relative dimension $d,e$ it follow $X\times_S Y$ is smooth of relative dimension $d+e$.

If $X,Y$ are schemes of finite type over $S$, where $S$ a smooth scheme of finite type over $k$, the above argument applies since the formula $C1$ holds in general - hence the above argument holds in the relative setting if you work over an algebraically closed field $k$.

Example. It should be possible to write this out explicitly in terms ot the jacobian matrix as you try above. This may be a good exercise. If you choose open subschemes $U:=Spec(A), V:=Spec(B)$ with $p\in U, q\in V$ you may choose $A:=k[x_1,..,x_n]/(f_1,..,f_a)$ and $B:=k[y_1,..,y_m]/(g_1,..,g_b)$. It follows

$$ \Omega^1_{A/k}\cong A\{dx_1,..,x_n\}/A\{df_1,..,df_a\}.$$

Where

$$ df(x_1,..,x_n):=\frac{\partial f}{\partial x_1}dx_1+\cdots +\frac{\partial f}{\partial x_n}dx_n.$$

Formula $C1$ gives

$$ \Omega^1_{A\otimes_k B/k}\cong \Omega^1_{A/k}\otimes_k B \oplus A\otimes_k \Omega^1_{B/k}.$$

It may be this approach can  be used to prove smoothness at $(p,q)$. I believe this is done in the "Red book of varieties and schemes" - at least the book gives some hints.

Question: "By definition the the product $U\times V$ is open in $X\times Y$ and is also a variety. I think $\phi \times \psi$
should give an isomorphism to $Z(f_1,..,f_a,g_1,..,g_b)$. But this is not obvious to me. It is also not clear why the jacobi of this should be of maximal rank."

If $I\subseteq A, J\subseteq B$ are ideals in $k$-algebras $A,B$ it follows

$$ A/I\otimes_k B/J\cong A\otimes_k B/(I\otimes B + A\otimes J).$$

There is an isomorphism $U\times_k V \cong Spec(A\otimes_k B)$ by HH Theorem II.3.3. It seems this proves your open set $U\times_k V \cong Spec(A\otimes_k B)$

has (as you guessed) the following algebra of functions

$$A\otimes_k B\cong k[x_i,y_j](f_u, g_v).$$

Note: It is not always true that products of varieties are varieties. If $X:=Y:=Spec(\mathbb{C})$ and $k:=\mathbb{R}$,  it follows

$$X\times_k Y \cong Spec(\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C})$$

and

$$\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}\oplus \mathbb{C}$$

is not an integral domain. Hence the product of two integral schemes over a field need not be integral.

You find discussions and questions on varieties/schemes that are "geometrically integral" here:

https://math.stackexchange.com/search?q=geometrically+integral

Title: Product of smooth varities is smooth


Body:
Show that the function $y = e^{2x}(A \cos3x + B \sin3x)$ satisfies the differential equation $y'' - 4y' + 13y = 0$

My attempt:

I get

$y' = e^{2x}((2A + 3B) \cos3x + (2B - 3A) \sin3x)$.

$y'' = e^{2x} ((-5A + 12B) \cos3x + (-12A - 5B) \sin3x)$.

$y'' - 4y' + 13y = e^{2x} ((16A + 24B) \cos3x + (16B - 24A) \sin3x) = 0$.

$(16A + 24B) \cos3x + (16B - 24A) \sin3x = 0$.

$(2A + 3B) \cos3x + (2B - 3A) \sin3x = 0$

And I'm totally lost after this. It seems like this can't be true since I get

$\tan x = const$.

I'd appreciate if you could point me in the right direction. Thanks!

Body:
$\begin{align}y' &= e^{2x}(-3A\sin 3x+ 3B\cos 3x) + 2e^{2x}(A\cos 3x + B\sin 3x) \\&= 3e^{2x}(-A\sin 3x+B\cos 3x)+2y\end{align}$

$\begin{align}y''&=3e^{2x}(-3A\cos 3x - 3B\sin 3x) + 6e^{2x}(-A\sin 3x+B\cos 3x)+2y'\\& = -9y + 6e^{2x}(-A\sin 3x + B\cos3x) + 6e^{2x}(-A\sin 3x+B\cos 3x) + 4y \\& = -5y +12e^{2x}(-A\sin 3x + B\cos 3x) \\&= -13y +12e^{2x}(-A\sin 3x + B\cos 3x)  + 8y \\& =-13y+ 4y'\end{align}$

So, $y'' - 4y' +13y = 0$


问:
$u_t+(u(1-u))_x=a(1-2u)$, transients to steady solution
We consider the non conserving equation
$$u_t+(f(u))_x=af'(u)$$
where $a$ is a constant, 0$\leq$ x $\leq$ 1 and $f(u)=u(1-u)$.

The steady solution of this equation with boundary condition $u(0)=u_0$ and $u(1)=u_1$ can be obtained by putting $u_t=0$ and the solution obtained are $u=1/2, u=ax+u_0$ and $u=ax+u_1$

I am trying to solve this equation by method of characteristics to obtain the transient solution with the some given initial condition u(x,0).

By method of characteristics, I have $\displaystyle \frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{a(1-2u)}$, this means that the characteristics equation is $$\displaystyle \frac{dx}{dt}=1-2u$$ along with $\displaystyle \frac{du}{dx}=a, \displaystyle \frac{du}{dt}=a (1-2u).$

Solving these equations, I reached upto $u(x,t)=\displaystyle\frac{1-(1-2u_0)e^{-2at}}{2}$.

As I am trying to reach to the steady solution, I must apply $t \rightarrow \infty$ which gives me $u=1/2$.

But I also have another solution $u=ax+u_0$ which is obtained at a steady state.

Does changing the initial condition in some way can help me reach the other solution or I am not solving this equation in a correct way. I have also tried to solve this equation using the Reimann condition this post still, I reach the solution $u=1/2$ on applying $t\rightarrow \infty$.
Thanks in advance for any help.

答:
$$u_t+(1-2u)u_x=a(1-2u).$$
You correctly wrote the Charpit-Lagrange system of ODEs :
$$\frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{a(1-2u)}.$$
A first characteristic equation comes from solving $\quad \frac{dx}{1-2u}=\frac{du}{a(1-2u)}$
$$u-ax=c_1$$
A second characteristic equation comes from solving $\quad \frac{dt}{1}=\frac{du}{a(1-2u)}$
$$(2u-1)e^{2at}=c_2$$
The general solution expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$(2u-1)e^{2at}=F(u-ax)$$
Or alternatively
$$u-ax=G\big((2u-1)e^{2at}\big)$$
$F$ or $G$ are arbitrary functions, one inverse of the other. They are to be determined according to some condition.

IN ADDITION :

After the clarification of the boundary and initial conditions (In comments).

The specified conditions are
$\quad\begin{cases}
u(0,t)=u_0 \neq \frac12\\
u(1,t)=u_1 \neq \frac12\\
u(x,0)=u_0 \neq u_1
\end{cases}$

There is not much calculus to do to observe that there is no solution if $F$ or $G$ are continuous functions.

We have to consider piecewise functions. To some extend this is similar to the Burger's PDE where there is a transition frontier namely "shock". This might be not the right term in the present case but doesn't mater the denomination.

With piecewise functions they are a lot of different cases depending on the values of $u_0,u_1,a$. For exemple the schematic below gives a rough idea of the behaviour.

Along the border $x=0$ we have $U_t=0$. This is the case of constant above function $G$.

Along the border $x=1$ we have $U_t=0$. Again the function $G$ is constant, but different.

Along the border $t=0$ we have $U_x=0$. This is the case of constant above function $F$. In other words :  $u(x,0)=u_0\quad\implies\quad (2u_0-1)=F(u_0-ax)$. This is possible only if $F$ is a constant function : $F=2u_0-1$. Thus $\quad (2u-1)e^{2at}=2u_0-1$ and the solution is : $$u=\frac{(2u_0-1)e^{-2at}+1}{2}.$$

On the transition line which separates the fields green and blue on the figure, the function $u(x,t)$ is discontinuous. The values of $u$ are not equal on both sides of the line.

Below another presentation of the same example shows what happens for $t\to\infty$ in this case. The shapes can be different for other values of $a,u_0,u_1$.

Note that the above figures are not to intend to explain all cases. More work would be necessary for different cases depending on the values of $u_0,u_1,a$.


问:
Intuition behind stolz caesaro theorem
The stolz caesaro theorem seems to be a discrete analogue of L'hopitals rule. I can understand l'hopitals rule via the taylor series, that is:

$$ \lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{ P(a) + \frac{dP}{dx}|_a (x-a) + O ( (x-a)^2)  }{  Q(a) + \frac{dQ}{dx}|_a (x-a) + O ( (x-a)^2)}$$

Now if $P(a) = Q(a) = 0$, the limit becomes ratio of first order term/ the ratio of values nearby to the point of interest. So, Is there a similar 'series' intuition for stolz caesaro theorem?

答:
It seems trivial compared to Taylor series or L'Hopital, because it follows directly from a much more general theorem:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

Theorem. Take any $a,b,c∈ℂ$ and sequences $f,g$ from $ℂ$ such that both $\sum_{k=0}^n g(k) → ∞$ and $\lfrac{f(n)}{g(n)} → c$ as $n→∞$. Then $\lfrac{ a + \sum_{k=0}^n f(k) }{ b + \sum_{k=0}^n g(k) } → c$ as $n→∞$.

Proof. For convenience let $[ε]$ denote $\{ z : z∈ℂ ∧ |z|<ε \}$. Take any $ε>0$. Let $m∈ℕ$ such that $\lfrac{f(n)}{g(n)} ∈ c+[ε]$ for every $n∈ℕ_{≥m}$. As $n→∞$, $\lfrac{ a + \sum_{k=0}^n f(k) }{ b + \sum_{k=0}^n g(k) }$ $= \lfrac{ a + \sum_{k=0}^{m-1} f(k) + \sum_{k=m}^n f(k) }{ b + \sum_{k=0}^{m-1} g(k) + \sum_{k=m}^n g(k) }$ $\overset{*}∈ \lfrac{ (1+o(1)) · \sum_{k=m}^n f(k) }{ (1+o(1)) · \sum_{k=m}^n g(k) }$ $⊆ (1+o(1)) · \lfrac{ \sum_{k=m}^n \big( (c+[ε])·g(k) \big) }{ \sum_{k=m}^n g(k) }$ $⊆ (1+o(1)) · (c+[ε])$, where the $*$-marked relation holds because $\sum_{k=m}^n f(k) → ∞$ and $\sum_{k=m}^n g(k) → ∞$. Since this holds for every $ε>0$, therefore $\lfrac{ a + \sum_{k=0}^n f(k) }{ b + \sum_{k=0}^n g(k) } → c$ as $n→∞$.

−−−

Note that in the above theorem statement we allow division by zero but the assertion of a limit means that eventually the value is defined (no division by zero). If you wish to avoid that issue, simply impose that $g(n)$ and $b + \sum_{k=0}^n g(k)$ are never zero for every $n∈ℕ$.

Title: Intuition behind stolz caesaro theorem


问:
Existence of a closure of a subset
Let $M$ be a set and $\circ$ an operation defined on subsets of $M$ in the following way:

$A \circ B \subseteq M$ for any $A \subseteq M$ and $B \subseteq M$;

If $A' \subseteq A$ and $B' \subseteq B$, then $A' \circ B' \subseteq A \circ B$.

Let us say a subset $C$ of $M$ is closed under $\circ$ if $C \circ C \subseteq C$.

Clearly, $M$ is closed under $\circ$.

Let us say a subset $C$ is a closure of a subset $S$ of $M$ if:

$C$ is closed under $\circ$ and $S \subseteq C$;

If $S \subseteq C' \subseteq C$ for a subset $C'$ closed under $\circ$, then $C' = C$.

Clearly, $M$ is a closure of itself.

It is easy to show that a closure of a subset $S$ of $M$, if exists, is unique:

Assuming $C_1$ and $C_2$ are two closures of $S$;

Denoting $C$ = $C_1 \cap C_2$;

$C \subseteq C_1$, therefore $C \circ C \subseteq C_1 \circ C_1 \subseteq C_1$;

$C \subseteq C_2$, therefore $C \circ C \subseteq C_2 \circ C_2 \subseteq C_2$;

$C \circ C \subseteq C_1$ and $C \circ C \subseteq C_2$, therefore $C \circ C \subseteq C_1 \cap C_2$;

$C \circ C \subseteq C$, therefore $C$ is closed under $\circ$;

$C$ is closed under $\circ$, $S \subseteq C \subseteq C_1$, therefore $C = C_1$;

$C$ is closed under $\circ$, $S \subseteq C \subseteq C_2$, therefore $C = C_2$.

But how can we prove that the closure exists for any subset of $M$?

If it is not true, what could be an additional requirement on $\circ$ to guarantee existence of the closure?

答:
Let $S\subseteq M$ be arbitrary. Let $\mathcal{A}$ be the collection of all closed supersets of $S$. We know that $M\in\mathcal{A}$ so $\mathcal{A}\neq \emptyset$. Let us consider $C=\bigcap \mathcal{A}$.

First we will check that $C$ is a closed superset of $S$. Arguing as you did, for any $K\in \mathcal{A}$

$C\subseteq K$, therefore $C\circ C\subseteq K\circ K \subseteq K$.

As $K$ was arbitrary this means $C\circ C\subseteq C$ and hence $C$ is a closed superset of $S$. By construction $C$ is also the smallest such set and hence the closure of $S$.

Title: Existence of a closure of a subset


问:
Confusing moments about the Law of exponents for natural powers
Suppose that $a\in \mathbb{R}$ and $a>1$ and we define in the recursive way the following operation: $$a^1:=a,$$ $$a^{n+1}:=a^n\cdot a \quad\text{for each} \quad n\in \mathbb{N}.$$

I want to prove that $a^{n+m}=a^na^m$ for any $n,m\in \mathbb{N}$.

Proof: Let $E=\{m\in \mathbb{N}: a^{n+m}=a^na^m \ \text{for any} \ n\in \mathbb{N}\}$

For any $n\in \mathbb{N}$ we have $a^{n+1}=a^na=a^na^1$ which shows us that $1\in E$.

Suppose that $m\in E$ and we'll try to show that $m+1\in E$.

Indeed, for any $n\in \mathbb{N}$ we have $a^na^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{(n+m)+1}$.

As you see I am almost done but one probably stupid moment bothers me a lot. Probably I am too pedantic.

We know that $(n+m)+1=n+m+1$ by associativity of natural numbers. How it follows that $$a^{(n+m)+1}=a^{n+m+1} \ ?$$

I do not think the we can just substitute $(n+m)+1$ by $n+m+1$.

May be I cannot formulate my question quite explicitly but I'd be very thankful for your help!

答: Yes, this last step is fine. If $x=y$ then $a^x=a^y$ and in this case you know that $(n+m)+1=n+m+1$.


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

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

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

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

答:
How should you define $e^z$ for complex numbers so that it retains the good properties of the real function $e^x$?  You take a definition of $e^x$ for real $x$ and modify it to apply to complex values.  Then prove all the properties you want.

For each definition of the real case, you get a corresponding definition for the complex case.

One definition is the Taylor series.  So for complex $z$, define
$$
F(z) = \sum_{n=0}^\infty \frac{z^n}{n!},\quad z\in \mathbb C
$$

Another  definition is the differential equation.  So for complex $z$, define $F(z)$ to be the unique solution $F : \mathbb C \to \mathbb C$ for the differential equation
$$
F'(z)=F(z),\quad F(0)=1
$$

If you like another definition, try to use it...

Summary: for your original question, the answer will depend on which definition you used.

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


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

答: A polynomial that works is $x^6-2x^5+5x^2$. The second derivative is $10(3x^4-4x^3+10)$, and is always $\ge 0$. But the fourth derivative is $120(3x^2-2x)$, which is negative in the interval $\left(0,\frac{2}{3}\right)$. </p>


问:
X numbers that when subtracted will produce the same absolute value
Let's say I have x unique numbers and I choose one number y out of this set. Is it possible to create these x numbers such that the absolute difference between y and any other number in $x$ will always be the same? This should work for every value of y in x.

This is trivial for $x = 2$ ...

```<code>Numbers: -1, 1
|(-1) - 1| = 2
|1 - (-1)| = 2
</code>```

But with $x > 2$ this seems more complicated (or impossible?). What are your thoughts?

Is it possible to accomplish this with $x > 2$ so the difference is at least almost the same?

答:
The answer is yes.

Take three numbers $a, b, c.$ Say that $a+b=c$. From that information, we can make a set of fact families:

$$a+b=c$$
$$b+a=c$$
$$c-b=a$$
$$c-a=b$$

These are the traditional equations in a fact family. But we can add one more.

$$b-c=-a$$

This is the most important one for our problem. We know it is true, because we can subtract $c$ and subtract $a$ from both sides of the first equation.

Now we can look at two important equations:

$$c-b=a$$
$$b-c=-a$$

These have a lot of relation to our first problem. We can put absolute value in:

$$|c-b|=a$$
$$|b-c|=a$$

And there you have it. No matter what $b$ and $c$ are, the two equations above will always be true.


答:
I think you answered your own question. Because of associativity, $(n+m)+1=n+(m+1)$, so using your calculation, $a^na^{m+1}=\cdots=a^{(n+m)+1}=a^{n+(m+1)}$, so the inductive step is done.

Recall that associativity means $(a+b)+c=a+(b+c)$.

Title: Confusing moments about the Law of exponents for natural powers

Body:
Suppose that $a\in \mathbb{R}$ and $a>1$ and we define in the recursive way the following operation: $$a^1:=a,$$ $$a^{n+1}:=a^n\cdot a \quad\text{for each} \quad n\in \mathbb{N}.$$

I want to prove that $a^{n+m}=a^na^m$ for any $n,m\in \mathbb{N}$.

Proof: Let $E=\{m\in \mathbb{N}: a^{n+m}=a^na^m \ \text{for any} \ n\in \mathbb{N}\}$

For any $n\in \mathbb{N}$ we have $a^{n+1}=a^na=a^na^1$ which shows us that $1\in E$.

Suppose that $m\in E$ and we'll try to show that $m+1\in E$.

Indeed, for any $n\in \mathbb{N}$ we have $a^na^{m+1}=a^n(a^ma)=(a^na^m)a=a^{n+m}a=a^{(n+m)+1}$.

As you see I am almost done but one probably stupid moment bothers me a lot. Probably I am too pedantic.

We know that $(n+m)+1=n+m+1$ by associativity of natural numbers. How it follows that $$a^{(n+m)+1}=a^{n+m+1} \ ?$$

I do not think the we can just substitute $(n+m)+1$ by $n+m+1$.

May be I cannot formulate my question quite explicitly but I'd be very thankful for your help!


问:
Confusion in least argument of z
$a$ and $b$ are ordered pairs of a point on the Argand plane. For tan $\theta = b/a$. It’s $z = a+ib$

My sir taught me that unique value of theta = $\tan ^{-1}|b/a|$ where theta belongs to $0,\pi/2$. Since then only it is present in the first quadrant. And tan inverse function has ranged from $-\pi/2$ to $\pi/2$.

Then , for the least + argument of $z$:

unique value of theta = $\tan ^{-1}|b/a|$ such that  $0<\theta<2\pi$.

Why is the theta different for least + argument of z and unique value of theta?

Q Q Find least + argument for z = -1-i$\sqrt{3}$ .

Here , mod of z = 2. So ,angle theta = 60 degree

But least + argument is in -pi/3 which we write as -pi+3 + 2pi = 5pi/3.

Here theta is less than 60 but isn’t it a coincidence.

答:
On the Argand plane a complex number is given by $z=a+ib$, and also by $z=r\ cis\theta$. The two expressions are connected by $a=r\ cos\theta$ and $b=r\ sin\theta$. We also have $r=\sqrt{a^2+b^2}$ and $tan\theta=b/a$, but this last expression does not give us enough information to find $\theta$.

One way to find $\theta$ is to find $\alpha=tan^{-1}|b/a|$ and then use the quadrant to decide which of the possible values for $\theta$ is most appropriate. The value for $\theta$ in $0\le\theta\lt2\pi $ (the least positive value) will be one of $\alpha,\ $ $\pi-\alpha,\ $ $\pi+\alpha,\ $ $2\pi-\alpha.\ $ For your example we have $a=-1$ and $b=-\sqrt{3}$, so we find $\alpha=tan^{-1}\sqrt{3}=\pi/3$ ($60$ degrees). The point $z$ is in the 3rd quadrant, so $\theta=\pi+\alpha=\pi+\sqrt{3}=4\pi/3$.

You have found $\theta=5\pi/3$. This is a 4th quadrant angle, and would be correct for $z=1-i\sqrt{3}$.

Title: Confusion in least argument of z


问:
Calculating $ \int_{E}(y-3) \, dx \, dy \, dz$
So I have the set $E$ as the smallest polyhedron that contains the points $(-1,0,0), (0,2,0), (0,-2,0), (2,0,0)$ and $(0,0,3)$. And I have to calculate $$ \int_{E}(y-3) \, dx \, dy \, dz$$ In all of these types of exercises is for me an extremely hard thing to find the borders of $x$, $y$ and $z$.

The set $E$ according to what I get would be a pyramid, and I got the borders directly from the given points and the way how this pyramid looks when I draw it and I calculated the Integral:

$$ \int_{0}^{3}\int_{-2}^{2}\int_{-1}^{2}(y-3) \, dx \, dy \, dz =-108$$

The right answer according to my book would be $-18$. What am I doing wrong? What is the problem with my borders and what would be some tips to be able to calculate them in these types of exercises?

Thanks in advance

答:
Given set $E$: If we take the base of the pyramid in $XY$ plane, it is formed by lines joining the points $A(-1, 0, 0), B(0,2,0), C(2,0,0)$ and $D(0,-2,0)$. The vertex of the pyramid is $P(0,0,3)$.

In $XY$ plane this translates to,

$AB: y-2x = 2, BC: x+y = 2, CD: x-y=2, DA: 2x+y = -2$

Given the vertex $P$ of the pyramid, the equation of the plane that face $ABP$ lies on is $y-2x+\frac{2}{3}z = 2$

Equation of the plane that face $BCP$ lies on is $x+y+\frac{2}{3}z = 2$

We will not need the other two planes below $x-$axis.

Now coming to the integral we need to work upon is $\displaystyle \iiint_E (y-3) \ dv$

But if you look at the given set $E$, it is completely symmetric to $y-$plane ($XZ$ plane) and given $y$ is an odd function, integral of $y$ over the volume of set $E$ will be zero. So our integral reduces to,

$I = -3 \displaystyle \iiint_E  dv$, in other words if we just find the volume of the pyramid, we are done which is,

$V =  2 \displaystyle \int_0^3 \int_{0}^{(6-2z)/3} \frac{6 - 3y - 2z}{2} \ dy \ dz = 6$

(we integrate over the region above $y-$ plane and multiply the volume by $2$ as the set is symmetric to $y-$plane as I mentioned earlier.

So, $I = -3 V = -18$


问:
Algorithm to find the exact roots of solvable high-order polynomials?
It is not generally possible to determine the roots of a polynomial whose grade is bigger than 4 in terms of roots and basic operations. But I heard that it is possible to give a criteria whether a polynomial has such a solution.

For instance Wikipedia tells that the roots of a polynomial of degree $5$ are expressible in terms of roots and basic operations, if it is representable in the form

$$x^5 + \frac{5\mu^4(4\nu + 3)}{\nu^2 + 1}x + \frac{4\mu^5(2\nu + 1)(4\nu + 3)}{\nu^2 + 1} = 0,$$

where $\mu$ and $\nu$ are rational numbers.

Given the case that the polynomials roots are expressible in such a form, is it possible to give an algorithm that computes the solution in this form?

I am not an expert on that topic, just a student who is interested in maths.

答:
While one cannot express the terms of polynomials of degree 5 and higher with basic operations and roots (Abel-Ruffini), it is possible to explicitly calculate the roots of polynomials of degree five and higher if you use other tools.

See this MO question.


问:
Proof of that the graph of subdifferential is closed when $f$ is closed proper convex function on $\mathbb{R}^n$.
In the Theorem 24.4 of Convex Analysis written by R. Tyrrell Rockafellar, if $f$ is a closed proper convex function on $\mathbb{R}^n$, then the graph of $\partial f$ is a closed subset of $\mathbb{R}^n \times \mathbb{R}^n$.

In the proof of the theorem, the author takes "lim inf" and uses the fact that $f$ and $f^*$ are closed.

I don't fully understand the proof because I'm not familiar with detailed convex analysis. Can anybody explain it in detail for me?

For example, why do we have to take "lim inf" instead of "lim"? Further, how is the closedness of $f$ and $f^*$ used there?

答:
I kept considering this problem, and I think that I found the detailed proof.
Here's the ingradients:

For any sequence $(a_i)$ and $(b_i)$, if $a_i \geq b_i, \forall i \Rightarrow \liminf_{i \rightarrow \infty} a_i \geq \liminf_{i \rightarrow \infty} b_i$.

Let sequences $(x_i)$ and $(x_i^*)$ such that $x_i \rightarrow x$ and $x_i^* \rightarrow x^*$. Then, $\liminf_{i \rightarrow \infty} \left< x_i, x_{i}^{*} \right> = \left <x, x^{*} \right>$ cuz $<\cdot, \cdot>$ is continuous function.

"$f$ is closed" means that all sublevel sets are closed. Note that "all sublevel sets are closed" $\Rightarrow$ "$f$ is lower semi-continuous (l.s.c.)" (according to wikipedia. I didn't actually know this).

A function $f$ is l.s.c. if $\liminf_{i \rightarrow} f(x_i) \geq f(x)$ where $x_i \rightarrow x$.

Then, the proof of the theorem that I asked is very natural :)


问:
Generate all possible combinations of 3 digits without repetition
It's possible to generate all possible combinations of 3 digits by counting up from 000 to 999, but this produces some combinations of digits that contain duplicates of the same digit (for example, 099). I could obtain all combinations of 3 digits without repetition by counting up from 000 to 999, and then removing all numbers with duplicate digits, but this seems inefficient. Is there a more efficient way to accomplish the same task?</p>

答:
You have 10 options for the first, then 9 (10 - the first) for the second, and 8 for the third. That makes $10 \cdot 9 \cdot 8$. But they can be shuffled in $3!$ ways, so the result is:
$$
\frac{10 \cdot 9 \cdot 8}{3!} = 120
$$</p>


问:
Grothendieck point of view of algebraic geometry
Given a ring $R$ and $I\subseteq R[x_1,\dots ,x_n]$ an ideal, define the functor $V_R(I):\operatorname{Alg}_R\to \operatorname{Sets}$, that sends a $R$-algebra $A$ in the subset of points $\mathbf a \in A^n$ such that $f(\mathbf a)=0\ $for all $f\in I$. Recall that in the case $R=k$ an algebraically closed field, $V_k(I)(k)$ is in bijective correspondence with the maximal ideals of $k[x_1,\dots ,x_n]$.

Now, in the course, we talked about the geometric points of a ring $A$, defining them as the equivalence classes of ring homomorphisms $A\to K$, where $K$ is a field (not fixed). The motivation that we were given is that this is a generalization of the construction above, since in this way we can obtain a correspondence between homomorphisms and prime ideals, not only maximal ideals; moreover, one can consider any ring, not necessarily a $k$-algebra.

However, I don't understand in what sense this is a generalization: it seems to me that they are just different constructions. For example, even if I take $k[x_1,\dots ,x_n]$, $k$ an algebraically closed field, in order to capture all its spectrum I need to consider not only the homomorphisms in $k$, but also in some trascendental extension $k\subset K$; so we are not generalizing, we are just adding something in my opinion, because these two constructions (i.e. homomorphisms in $k$ fixed and homomorphisms in any field) don't reduce to the same when considering the case of finitely generated $k$-algebras. Am I right? Moreover, I have not clear the sense of arriving to the prime ideals, but not to any ideal (i.e. why geometric points of a ring $A$ are not defined as the classes of homomorphisms from $A$ to any ring). Thanks for any clarify, I know that these question may sound stupid but I'm quite new to this things.

答:
Question: "I mean, it would be a generalization if one could capture every prime ideal of $k[x_1,…,x_n]$ just by considering the homomorphisms with values in $k$, but still we need to consider homomorphisms in any field in order to capture every prime ideal of a generic ring; however as I said this is not the case."

In general the ring $A$ is not a $k$-algebra for $k$ a field.

Lemma: If $A$ is a commutative unital ring and $\mathfrak{p}\subseteq A$ is a prime ideal,
there is a field $K$ and a map of unital rings $\phi: A \rightarrow K$ with $\mathfrak{p}=ker(\phi)$.

Proof. Let $u:A\rightarrow A_{\mathfrak{p}}$ be the canonical map into the localization of
$A$. It follows prime ideals $\mathfrak{q}* \subseteq A_{\mathfrak{p}}$ corresponds 1-1 to a prime ideal $\mathfrak{q}\subseteq \mathfrak{p}$ in $A$. Hence the zero ideal $(o) \subseteq \kappa(\mathfrak{p})$ in the residue field $\kappa(\mathfrak{p})$ has inverse image $\mathfrak{p}$ under the canonical map $\phi_{\mathfrak{p}}: A \rightarrow \kappa(\mathfrak{p})$. Hence you may choose $\phi:=\phi_{\mathfrak{p}}$.

Hence any prime ideal in $A$ is the kernel of some map into a field $K$. The problem is that the "set of all fields $K$" is a "very large set" - it is not even a set! The set of prime ideals in $A$ is "smaller" and this is one reason to work with $Spec(A)$: Since $A$ is a set it follows $Spec(A)$ is a set  - it is a subset of the set of all subsets of $A$.

Example. If $A$ is a $k$-algebra of finite type over a field $k$, a non-maximal prime ideal $\mathfrak{p}$ will in general have a residue field $\kappa(\mathfrak{p})$ that is no longer a finite extension of $k$. If $\mathfrak{p}$ is maximal it follows $k \subseteq \kappa(\mathfrak{p})$ is a finite extension.

Here you find a discussion on the problem of defining algebraic "varieties" avoiding prime ideals and instead using maps to fields or the space of maximal ideals:

https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961

Note that if $A,B$ are commutative unital rings that are finitely generated over a field $k$ or the integers $\mathbb{Z}$ the following holds:

Lemma.Let $\phi:A \rightarrow B$ be a map of rings and let $\mathfrak{m}\subseteq B$ be a maximal ideal. it follows $\phi^{-1}(\mathfrak{m})\subseteq A$ is a maximal ideal.

Proof. Assume $k$ is a field and let $\mathfrak{n}:=\phi^{-1}(\mathfrak{m})$. We get an inclusion of rings

$$ k \subseteq \kappa(\mathfrak{n}) \subseteq \kappa(\mathfrak{m})$$

and $\kappa(\mathfrak{n})$ is an integral domain. Since $\kappa(\mathfrak{m})$ is a finite extension of $k$ the following holds: Let $0\neq x\in \kappa(\mathfrak{n})$ be an element. There is a canonical surjective map

$$ \psi: k[t]\rightarrow k(x)$$

defined by $\psi(t):=x$, with $ker(\psi):=(p(t))$ for a non-zero ideal $p(t)\in k[t]$. Since $k(x)$ is an integral domain it follows $(p(t))$ is a maximal ideal, hence $x$ is a unit in $\kappa(\mathfrak{n})$, and it follows $\mathfrak{n}$ is a maximal ideal. A similar proof holds when $k$ is the ring of integers. QED.

If we let $Specm(A)$ denote the set of maximal ideals in $A$ we get by the Lemma an induced map

$$ \phi_m: Specm(B) \rightarrow Specm(A)$$

defined by $\phi_m(\mathfrak{m}):=\phi^{-1}(\mathfrak{m})$, and the map $\phi_m$ is continuous in the topology induced by the Zariski topology. Hence the classical map defined for the affine spectra

$$\phi: Spec(B) \rightarrow Spec(A)$$

induce a map at the level of maximal ideals.

There is a class of rings  - Hilbert-Jacobson rings - where the following result holds:

Theorem. Let $A$ be a Hilbert-Jacobson ring and let $\phi: A\rightarrow B$ be a map of commutative unital rings where $B$ is a finitely generated $A$-algebra. It follows $B$ is a Hilbert-Jacobson ring, and for any maximal ideal $I\subseteq B$ it follows $J:=\phi^{-1}(I)\subseteq A$ is a maximal ideal and the canonical map $A/J \rightarrow B/I$ is a finite extension of fields.

A ring $A$ is a Hilbert-Jacobson ring iff every prime ideal in $A$ is the intersection of the maximal ideals containing it.

For this class of rings you get a well defined continuous map at the level of max-spectra and you can from such a ring $A$ construct a locally ringed space where the underlying topological space $X^m$ is the set of maximal ideals in $A$ with the Zariski topology, and where the structure sheaf $\mathcal{O}_{X^m}$ has nilpotent elements. Hence the ringed space $(X^m, \mathcal{O}_{X^m})$ is in a sense a "compromise" between a "classical algebraic variety" and a "scheme".  If the base-ring is finitely generated over a field $k$ it follows the points in the topological space corresponds to solutions of systems of polynomial equations with values in a finite extension of $k$, hence the definition is in a sense "more intuitive" and "pedagogical" and it may be interesting for people teaching algebraic geometry.


问:
Area bounded by $2 \leq|x+3 y|+|x-y| \leq 4$

Find the area of the region bounded by $$2 \leq|x+3 y|+|x-y| \leq 4$$

I tried taking four cases which are:
$$x+3y \geq 0, x-y \geq 0$$
$$x+3y \geq 0, x-y \leq 0$$
$$x+3y \leq 0, x-y \geq 0$$
$$x+3y \leq 0, x-y \leq 0$$
But it becomes so confusing to draw the region. Any better way ?

答:
Well, if you're looking for a plot:

Now, if we want to find the area of the part where $y\ge0$ and $x\ge0$, we can take a look at the following picture:

We can find the shaded area by finding:

$$3\cdot\left(\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}\right)+\frac{1}{2}\cdot\frac{1}{2}+\int_1^2\left(2-x\right)\space\text{d}x=\frac{9}{8}\tag1$$

And the place where $y\ge0$ and $x\le0$, we can take a look at the following picture:

We can find the shaded area by finding:

$$3\cdot\frac{1}{2}\cdot1+\frac{1}{2}\cdot\frac{1}{2}+\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}=\frac{15}{8}\tag 2$$

So, the total area is given by:

$$\mathscr{A}=2\cdot\frac{9}{8}+2\cdot\frac{15}{8}=6\tag3$$


问:
Integrate $\int_0^1e^{3y}(-1+e^3)dy$ which step is wrong?
I'm not sure what I'm doing wrong here when trying to integrate $\int_0^1e^{3y}(-1+e^3)dy$

My steps:

Factor out the constant so Integrate $(-1+e^3)\int_0^1e^{3y}dy$

$u = 3y$ and $du = 3 dy => (-1+e^3)\int_0^3e^{u}du$

$(-1+e^3)(e^3) - (-1+e^3)(e^0)$

$e^6 - e^3 + 1 - e^3$

$e^6 - 2e^3 + 1$

So my answer is wrong and I'm not sure what exactly is wrong here in my computation.
Would appreciate some help, thanks in advance!

答:
If $du = 3dy $ then $dy = \frac{du}3$. You missed this step.

So the value is $\frac{e^6-2e^3+1}{3} = \frac{(e^3-1)^2}{3}$


问:
Eigenvectors basis and orthonormal basis for of linear transformation $T$
Is the following true:

Let $V$ be an inner product space, and $T:V\to V $ is a linear transformation, suppose that we have eigenvectors basis of $T$. Thus, there exist orthonormal eigenvectors basis of $T$.

I suspect that this statement true iff $T$ is self adjoint.

答:
If $T$ is not self-adjoint, there exist $v,w\ne 0$ such that
$$\langle Tv,w\rangle\ne \langle v,Tw\rangle\,.$$
Then there must be a pair $b_i,b_j$ of the given eigen-basis which also satisfies this inequality (since if all pairs would give equality, then it would hold for all vectors as well, by linearity).

This means $\lambda_i\,\langle b_i,b_j\rangle \ne \bar\lambda_j\,\langle b_i,b_j\rangle$. In particular, $b_i\not\perp b_j$ follows.

On the other hand, if $T$ is self-adjoint, the same argument, $\lambda_i\,\langle b_i,b_j\rangle = \bar\lambda_j\,\langle b_i,b_j\rangle$ shows that its eigenvalues must be real, and that the eigenvectors corresponding to different eigenvalues are orthogonal.


问:
finite transcendence degree but not finitely generated field extension
Sorry for my bad English.

Let $K\subset L$ be field extension such that transcendence degree tr.deg $L/K<\infty$.

Now, can we say $L/K$ is  finitely generated field extension?

Please tell me proof or contradiction, thanks.

答: No, for example the algebraic closure $L$ of $\Bbb Q$ is algebraic over $\Bbb Q$, hence $L/\Bbb Q$ has transcendence degree $0$ but it cannot be finitely generated as it is an infinite algebraic extension.


问:
Prove this polynomial to be either irreducible or competely splitting in K[x].
The following question was asked in my quiz of field theory(over now) and I was unable to solve it and so I am posting here. So, I tried it again today but couldn't make any progress.

Let p be a prime and assume either (i) char $K = p$ or (ii) char $K \neq  p$ and $K$
contains a primitive $p$th root of unity. Then $x^p - a \in K[x]$ is either irreducible or
splits in $K[x]$.

So, 1st I let that char $K= p$ and that $x^p-a$  is  reducible in $K[x]$ so , let a be it's root Now I have to prove that $x^p -a$ actually splits in $K[x]$, $x^p -a = f(x) (x-a)$ and degree of $f(x)$ is $n-1$ but I am unable to think using what result I should proceed now.

And alas similar problem is occuring when I tried to solve (ii) of the question when I assumed that polynomial is not irreducible then how I should do it.

I think I will not be able to do this without any help and so do you mind giving your time on this.

答:
We write $b=\sqrt[p]{a}$, $\omega$ a $p$-th root of unity and $h(x)=x^p-a$.

If char $K=p\ (\neq 0)$, then $p$ is prime as $K$ is a field. So $$h(x)=(x-b)^p$$ (by freshman's dream). Thus, if there is a $b\in K$ our polynomial splits. If there is no $b\in K$, then $h(x)$ must be irreducible as it can only be factorised into powers of $(x-b)$, contradicting $b\not\in K$.

If char $K\neq p$, then we don't have the freshman's dream factorisation $h(x)=(x-b)^p$, so the approach must be different. We're given $\omega\in K$.

If there is a $b\in K$, then $h(x)$ splits: $$h(x)=(x-b)(x-\omega b)(x-\omega^2 b)\cdots (x-\omega^{p-1}b).$$

If there is no $b\in K$, we need to show $h(x)$ is irreducible. We have $[K(b):K]=p$ because we have a basis $\{1,b,b^2,\cdots,b^{p-1}\}$ (using $\omega\in K$). Suppose $h(x)$ does factorise, say, $h(x)=g(x)k(x)$. Then take any root $\alpha$ of $g(x)$, by above we know that $\alpha=\omega^ib$, for some $0\leq i<p$, then we have $$p=[K(b):K]=[K(b):K(\alpha)][K(\alpha):K]$$ by the tower law. Since $p$ is prime, $[K(\alpha):K]$ can only be $1$ or $p$. It cannot be $1$ as that would imply $K(\alpha)=K$, contradicting the assumption that $b\not\in K$. If it is $p$, then in fact $K(b)=K(\alpha)$, so all roots of $h(x)$ are in $K(\alpha)$, so $h(x)$ splits. So to summarise, we've shown that if $h(x)$ factorises over $K$, then it must split.


答:
Fix a root $\omega$ of $f = X^p-a$ in some algebraic closure.

First suppose $\mathbf{char}(K) = p$. Then $f = X^p-\omega^p = (X-\omega)^p$, as the comments indicate. Hence splitting and having a root in $K$ are equivalent conditions.

Likewise, if $\mathbf{char}(K) \neq p$ and $\xi_p \in K$ is a $p$-th primitive root of unity, then $f = X^p-a$ is separable its roots are $\xi_p^j \omega$ for all $0\leq j \leq p-1$. Thus if some root $\xi_p^j\omega$ lies in $K$, dividing by $\xi_p^j \in K$ we get $\omega \in K$ and so $p$ has all its roots in $K$, hence it splits.

What we have to do now, in both cases, is showing that splitting and reducibility are equivalent. By our previous discussion, this is the same as showing that if $p$ is reducible then $\omega \in K$ in each case.

The case $\mathbf{char}(X) = p$ is  the easiest: if $f = gh$ in $K[X]$ for $g,h \in K[X]$ of positive degree, then - in some algebraic closure -  we have $g = (X-\omega)^s$ with $s < p$. In particular, we know that $\omega^s \in K$. But then $s$ and $p$ are coprime so choosing $k,l$ such that $ks+pl = 1$ we obtain

$$
\omega = \omega^1 = \omega^{sk}+\omega{pl} = (\omega^s)^k + a^l \in K.
$$

On the other hand, in the characteristic $\neq p$ setting having $f = gh$ for positive degree $g$ and $h$ shows that a proper product of $\{\xi_p^j\omega\}_j$ lies in $K$, but since we are assuming $\xi_p \in K$, dividing by its powers we still have $\omega^s \in K$ with $s <p$.

Title: Prove this polynomial to be either irreducible or competely splitting in K[x].

Body:
The following question was asked in my quiz of field theory(over now) and I was unable to solve it and so I am posting here. So, I tried it again today but couldn't make any progress.

Let p be a prime and assume either (i) char $K = p$ or (ii) char $K \neq  p$ and $K$
contains a primitive $p$th root of unity. Then $x^p - a \in K[x]$ is either irreducible or
splits in $K[x]$.

So, 1st I let that char $K= p$ and that $x^p-a$  is  reducible in $K[x]$ so , let a be it's root Now I have to prove that $x^p -a$ actually splits in $K[x]$, $x^p -a = f(x) (x-a)$ and degree of $f(x)$ is $n-1$ but I am unable to think using what result I should proceed now.

And alas similar problem is occuring when I tried to solve (ii) of the question when I assumed that polynomial is not irreducible then how I should do it.

I think I will not be able to do this without any help and so do you mind giving your time on this.


问:
Laplace transform to determine system output
I have the following question:
Below, differential equations that describe dynamic systems are given together with system inputs and initial conditions. Use the Laplace transform to determine the system outputs.

$$
\begin{aligned}
\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} y}{\mathrm{~d} t}+2 y &=\sigma(t) \\
\sigma(t) &=\left\{\begin{array}{ll}
0, & t<0 \\
1, & t \geq 0
\end{array}\right.\\
\frac{\mathrm{d} y}{\mathrm{~d} t}(0)=y(0) &=0
\end{aligned}
$$

After using the  $\mathcal{L}$-transform and partial fraction decomposition, I come to :

$$
Y(s) = \frac{1}{s+1} - \frac{1}{s+2}$$
which I tranform back to $\mathcal{L}^{-1} = e^{-t} - e^{-2t}$.
The correct answer is : $\frac{1}{2} -e^{-t} + \frac{1}{2} e^{-2t}$.

I am self-taught so I probably missed something very basic here, but after looking up many examples, I don't understand why this is the right answer.
Thanks in advance for help!

答:
The Laplace transform of $1$ is $\frac1s$. Probably you'd have taken it as $1$ and so obtained the wrong result.

It is actually, $$Y(s) = \frac{1}{\color{blue}{s}(s^2+3s+2)} = \frac{1}{s(s+1)(s+2)} = \frac1{2s} - \frac1{s+1} + \frac1{2(s+2)}$$

So, $$\boxed{y(t) = \frac12 - e^{-t} + \frac12e^{-2t}}$$


问:
Projection on the eigenspace
Let A be a normal operator on a Hilbert space $\mathcal H$ and dim($\mathcal H$) = n . Let $\alpha$ be a eigen value of A and again $E_α$ be the orthonormal projection onto the α-eigenspace . Let {$u_1$ , $u_2$ , ... ,$u_m$ } be an orthonormal basis of the eigen space . Then $E_\alpha$ = $\sum_{i=1}^m u_iu_i^*$ .
I know that $E_\alpha$($\beta$) =  $\sum_{i=1}^m <\beta ,u_i>u_i/||u_i||^2$ . But how can I show this .

答:
Hint: complete $\{u_i\}$ to an orthonormal basis, then compute both expressions on a given basis vector. This is a mere formality, though, as it suffices to evaluate in each $u_i$: we already know that $E_\alpha$ and each $u_i^\ast$ are zero in the orthogonal complement of the $\alpha$-eigenspace.

 Indeed, since $u_j$ are in the eigenspace, their are fixed by $E_\alpha$, and the same holds for $\sum_i u_iu_i^\ast(u_j) = \sum_i \delta_{ij}u_i = u_j$.

Title: Projection on the eigenspace


答:
The transformation
$$
\begin{align}
u &= x + y \\
v &= 2y
\end{align}
$$
transforms the region
$$
2 \leq|x+3 y|+|x-y| \leq 4
$$
to
$$
 2 \leq|u+v|+|u-v| \leq 4
$$
and increases the area by a factor of two (why?). The new region is limited by two squares with side length $2$ and $4$, respectively (compare How to solve $|a+b|+|a-b|=c$?),  so that its area can be computed easily.

Title: Area bounded by $2 \leq|x+3 y|+|x-y| \leq 4$

Body:
Find the area of the region bounded by $$2 \leq|x+3 y|+|x-y| \leq 4$$

I tried taking four cases which are:
$$x+3y \geq 0, x-y \geq 0$$
$$x+3y \geq 0, x-y \leq 0$$
$$x+3y \leq 0, x-y \geq 0$$
$$x+3y \leq 0, x-y \leq 0$$
But it becomes so confusing to draw the region. Any better way ?


答:
Here is how I would resolve the absolute equations.

$2 \leq|x+3 y|+|x-y| \leq 4$

If $x \geq y,$

$2 \leq|x+3 y|+ x-y \leq 4$

For $x+3y \geq 0, x+3y \leq 0,$ we get

$1 \leq x+y \leq 2, - 1 \leq y \leq -\frac{1}{2}$

Similarly for $x \leq y$, we have

$-2 \leq x+y \leq -1, \frac{1}{2} \leq y \leq 1$

That gives you the below region to find area of,

Finding area may be easier with a transformation and the transformation choice is obvious but given the focus of the question, I am not getting there.


问:
Properties of $\mathbb{C}+(h) \subseteq \mathbb{C}[x]$
Let $h=h(x) \in \mathbb{C}[x]-\mathbb{C}$.
Write: $h=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$, where
$c_j \in \mathbb{C}$, $1 \leq j \leq n$.

Denote by $R$ the following $\mathbb{C}$-subalgebra of $\mathbb{C}[x]$:
$R= \mathbb{C}+\langle h \rangle$, where $\langle h \rangle$ is the ideal of $\mathbb{C}[x]$ generated by $h$.

Question 1: Is it possible to characterize, in terms of the $c_j$'s or in another way, all $h$'s such that :

(1) $R$ is a UFD or at least $R$ is integrally closed in its field of fractions (=normal)?

(2) $R \subseteq \mathbb{C}[x]$ is flat?

Remarks:

(1) if $\deg(h)=1$, then $h=ax+b$, for some $a \in \mathbb{C}^{\times}, b \in \mathbb{C}$, so $R=\mathbb{C}[x]$ which is of course a UFD and $R=\mathbb{C}[x] \subseteq \mathbb{C}[x]$ is flat.

(2) Actually, the field of fractions of $R$ is $\mathbb{C}(x)$, since
$x=\frac{xh}{h}$. Therefore, $R$ is normal if and only if $R=\mathbb{C}[x]$...
since $x$ is obviously integral over $R$.
So the answer to my first question is trivial:
$R$ is normal iff $R=\mathbb{C}[x]$ iff $h=ax+b$, $a,b \in \mathbb{C}$.

(3) It is not difficult to see that $h=x^2$ yields: $R=\mathbb{C}+\langle x^2 \rangle=\mathbb{C}[x^2,x^3]$ which is not a UFD, since $x^2x^2x^2=x^3x^3$ are two different factorizations of $x^6$. It is also not integrally closed in its field of fractions $\mathbb{C}(x)$, since $x$ belongs to $\mathbb{C}(x)$ but $x$ does not belong to $R$.
It is well-known that $\mathbb{C}[x^2,x^3] \subseteq \mathbb{C}[x]$ is not flat.

What about $h=x^2+x+1$? $h=x^2+1$? According to the above remark (2), $R$ of those $h$'s is not normal. Still what about flatness?

The following questions are perhaps relevant: i and ii.

Question 2: Assuming we have a complete answer to Question (1) or (2); is it true that the same answer holds for $\mathbb{C}+ \langle h,y_1\ldots,y_m \rangle \subseteq \mathbb{C}[x,y_1,\ldots,y_m]$? I think the answer is positive.

Any hints and comments are welcome!

答:
Perhaps (please correct me if I am wrong! Truly, I am not sure) according to the anwer to this question (first comment):
$R \subseteq \mathbb{C}[x]$ is flat iff $R=\mathbb{C}[x]$.
(Without assuming separability of $R \subseteq \mathbb{C}[x]$?).