答:
Use the fondamental theorm of aclculus to obtain
$$G'(x)=F(x).$$ Since $F$ is increasing, we have $G(x)=F(x)>0$ for all $x\geq0$. That means $G'>0$ and so $G$ is strictly increasing (meaning $a>b$ implies $G(a)>G(b)). A strictly increasing function is one to one.

Title: $F$ is an increasing function and $F(0) > 0$. For $x \geq 0$, define $G(x) = \int_{0}^{x} F$. Prove that $G$ is one-to-one

Body:
I am given a practice question where I am supposed to show that the function of G is one-to-one. Where $G(x) = \int_{0}^{x} F$, where $F$ is an increasing function and $F(0)>0$.

I thought about thinking about the second part of Fundamental Theorem of Calculus, but I don't know if that is the right method to show that the function is one to one. Also, I don't know if I can just say

Suppose that $F(0) = 1$ since $F(0) > 0$, then $G(x) = F(x) - 1$.

Any help will be great.

Edit: It was clarified that $x \geq 0$


问:
$F$ is an increasing function and $F(0) > 0$. For $x \geq 0$, define $G(x) = \int_{0}^{x} F$. Prove that $G$ is one-to-one
I am given a practice question where I am supposed to show that the function of G is one-to-one. Where $G(x) = \int_{0}^{x} F$, where $F$ is an increasing function and $F(0)>0$.

I thought about thinking about the second part of Fundamental Theorem of Calculus, but I don't know if that is the right method to show that the function is one to one. Also, I don't know if I can just say

Suppose that $F(0) = 1$ since $F(0) > 0$, then $G(x) = F(x) - 1$.

Any help will be great.

Edit: It was clarified that $x \geq 0$

答:
Differentiating under integral sign we have $G'(x)=f(x)>0$ as f(x) is an increasing function. Thus G(x) is an increasing function. We need to prove its montonically increasing for itto become one-to-one. Let's say for some interval $x_{0}-x_{1}$ f(x) has same value then $G(x_{0})=\int_0^{x_0} f(x)=H(x_{0})$. Now $G(x_{1})=H(x_{0})+\int_{x_{0}}^{x_{1}} f(x)$ but as we have assumed that here f(x) is constant and let it be equal to c.
we have $G(x_{1})=H(x_{0})+(x_{1}-x_{0})c$. Now I have assumed $x_1>x_0$ also c is >0 as f(x) is increasing and f(0)>0 . So $(x_1-x_0)c>0$. Hence $G(x)$ is one to one function.

ContentLicense: CC BY-SA 4.0


问:
Equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.
Let $f$ be differentiable function and $y=-2x+1$ be tangent line to graph of $y=f(x)$ at the point $x=2$.

How can I find the equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.

I created a function $f (x)$ myself, but it didn't work. Any ideas?

答:
Find $f(2)$ and $f'(2)$ as follows $$\begin{align} y \quad &= \quad f(2) + f'(2)(x-2)  \\&=\quad f'(2)x + f(2) - 2f'(2) \\&=\quad -2x + 1 \end{align}$$
Since the coefficients should be equal, we get
$$\begin{cases} f'(2) = -2 \\ f(2) = -3 \end{cases}$$
Now, take $g = f^2$. To find the tangent equation, we will need the derivative of $g$ at $2$. $$g'(2) = [f(x)^2]'\bigg|_2 = 2f(2)f'(2) = 2(-3)(-2) = 12$$
Hence, the required tangent is
$$\begin{align} y \quad &= \quad g(2) + g'(2)(x - 2)  \\&=\quad (-3)^2 + 12(x-2) \tag{$g(2) = f(2)^2$} \\&=\quad 12x - 15 \end{align}$$


答:
We know that the tangent line to a function $f$ at a point $(a, f(a))$ is given by the equation
$$(y - f(a)) = f'(a) (x - a) \Leftrightarrow y = f'(a) x + (f(a) - f'(a) a).$$
Since $y = -2x + 1$ and $a = 2$, you find $f'(2) = -2$ and $f(2) = -3$. Then, you compute the tangent line to $g(x) = (f(x))^2$ at $a = 2$,
$$(y - g(a)) = g'(a) (x - a).$$
The expression of $g'(2)$ is just an application of the chain rule.

Title: Equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.

Body:
Let $f$ be differentiable function and $y=-2x+1$ be tangent line to graph of $y=f(x)$ at the point $x=2$.

How can I find the equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.

I created a function $f (x)$ myself, but it didn't work. Any ideas?


问:
An integral problem?
How do you integrate $e^{e^x}$? I was able to get it down to du/(ln u) but I wasn't able to go further. Thanks!</p>

答:
$$\int e^{e^x}dx$$
Substitute $u = e^x$.  Then $du = e^x dx$

$$\int \frac{e^u}{u} du = \operatorname{Ei}(u) + C = \operatorname{Ei}(e^x) + C$$

Where $\operatorname{Ei}$ denotes the Exponential integral $\displaystyle \int_{-\infty}^{x} \frac{e^t}{t} dt$.

See http://en.wikipedia.org/wiki/Exponential_integral

ContentLicense: CC BY-SA 3.0

Title: An integral problem?


问:
Find the number of integral solutions $(x, y)$ to the equation $xy(x + y + 1) = 2019 ^{2020} + 1$.

Find the number of integral solutions $(x, y)$ to the following equation:$$xy(x + y + 1) = 2019 ^{2020} + 1$$

I've tried expanding to equate the corresponding terms, but that failed miserably and then I tried to take $\ln$ on both sides and I'm not sure how to continue after that. Can anyone give a little hint on how to start? Help would be appreciated!

答:
Thanks to the community for helping me. Here's the solution:

LHS:

$(2019^{2020} + 1)\equiv1\pmod3$

RHS:

(x and y cannot be zero)

if  $x\equiv1\pmod3$ and $y\equiv1\pmod3$,

$xy\equiv1*1\pmod3=1\pmod3$

$x+y\equiv2\pmod3$

$x+y+1\equiv3\pmod3=0\pmod3$

$\Rightarrow$ $xy(x+y+1)\equiv0\pmod3$

LHS $\ne$ RHS

Similarly, LHS and RHS won't be equal even if $x\equiv-1\pmod3$ and $y\equiv-1\pmod3$ and for any other values of x and y.

So there are no integral solutions for the given equation.


问:
A question related to Abel summation Formula
I am trying some questions from Apostol's Introduction to ANT and couldn't solve this particular problem on page 277.

I tried using Abel's summation Formula given in section 4.2 as follows:

$\frac{\chi(n) }{n^s}= \frac{S(M) }{M^s}+ s\int_{1}^{M}\frac{S(x) }{x^{s+1}} dx$  from which I got $L(s, \chi) = \sum_{n=M+1}^{\infty} \frac{\chi(n) }{n^s} + \frac{S(M) }{M^s}+s\int_{1}^{M}\frac{S(x) }{x^{s+1}} dx$

I am not able go to proceed from here as in integration limit is from 1 to $\infty$ and integrand is S(x) -S(N).

Can you please tell how should I proceed?

Thanks!!

答:
For $\sigma>0$ and $1\le N<M$, by Abel's summation formula
\begin{align}
\sum_{n=1}^{M}\frac{\chi(n)}{n^s}&=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}+\sum_{n=N+1}^{M}\frac{\chi(n)}{n^s}\\
&=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}-\frac{S(N)}{N^S}+\frac{S(M)}{M^S}+s\int_{N}^{M}\frac{S(x)}{x^{\sigma+1}}dx
\end{align}
Noting $\sigma>1$ and $S(x)=O(1)$, let $M\to\infty$. As a consequence $S(M)M^{-s}\to0$ and the integral converges. This means
\begin{align}
L(S,\chi)&=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}-\frac{S(N)}{N^s}+s\int_{N}^{\infty}\frac{S(x)}{x^{\sigma+1}}dx\\
&=\sum_{n=1}^{N}\frac{\chi(n)}{n^s}+s\int_{N}^{\infty}\frac{S(x)-S(N)}{x^{s+1}}dx
\end{align}
this completed the proof.


问:
What is the easiest way to sketch $\frac{1}{x} > \frac{1}{y}$?
When I do it, it never goes anywhere the first time because there are so many conditions and cases when x or y is positive or negative. The worked solutions suggested a "quadrant" based method but it is still confusing. Is there an even easier way or will a question like this naturally be difficult to solve?</p>

答: here are the worked solutions (as requested). I guess its relatively concise - is there an even better way?


答:
If $xy>0$ (the first and third quadrant excluding the vertical and horizontal axis), multiply $xy$ to both sides of the inequality, we have $y > x$ in the first and the third quadrant.

If $xy < 0$ (the second and fourth quadrant excluding the vertical and horizontal axis), multiply $xy$ to both sides of the inequality, we have $y < x$. This is just the entire fourth quadrant.

Title: What is the easiest way to sketch $\frac{1}{x} > \frac{1}{y}$?

Body: When I do it, it never goes anywhere the first time because there are so many conditions and cases when x or y is positive or negative. The worked solutions suggested a "quadrant" based method but it is still confusing. Is there an even easier way or will a question like this naturally be difficult to solve?</p>


答:
Multiply both sides of the inequality by $xy$.

If $xy > 0$, this means $x > 0$ and $y > 0$, or $x < 0$ and $y < 0$. We have $y > x$, so sketch the upper half of the 1st quadrant and the upper half of the 3rd quadrant.

If $xy < 0$, this means $x < 0$ and $y > 0$, or $x > 0$ and $y < 0$. This gives $y < x$, but this is violated when $x < 0$ and $y > 0$. Therefore sketch the entire 4th quadrant as both conditions are satisfied.


问:
Giving counterexample to statements about Quotient Groups
Suppose $H_i \trianglelefteq G_i$ for $i = 1, 2$.

Obs.: all statements below are false

(a) If $H$ and $G/H$ is abelian, then G is abelian.

(b) If every proper $H$ of $G$ is cyclic, $G$ is cyclic.

(c) If $G_1 \cong G_2 $ and $H_1 \cong H_2$ $\Rightarrow$ $G_1/H_1 \cong G_2/H_2$.

(d) If $G_1 \cong G_2 $ and $G_1/H_1 \cong G_2/H_2 \Rightarrow H_1 \cong H_2$.

(e) If $H_1 \cong H_2 $ and $G_1/H_1 \cong G_2/H_2 \Rightarrow G_1 \cong G_2$.

I think this question is very relevant to understand the structure of groups, and I have been trying to (a), and it just seems that I want to be clear if G/H has the identity element. And also I want a clear definition of what does it mean for a group to be isomorphic in the above cases, and examples would just be a way to comprehend better.

答:
Hints:

(a) Take a non-Abelian group $G$ with an Abelian subgroup $H$ of index $2$. The smallest example will work.

(b) Consider a group of order $4$.

(c) Consider $G_1 = G_2 = \mathbb Z/2 \times \mathbb Z/4$.

(d) See (c).

(e) Consider $G_1 = D_4$, the dihedral group of order $8$ and $H_1$ the center of $G_1$. Then consider $G_2 = \mathbb Z/2 \times \mathbb Z/4$. (There is also an example where $G_1, G_2$ have order $6$.)


问:
What are the constant solutions for separation of variables used for?
In my calc $2$ textbook there is an initial-value problem that requires the use of separation of variables.

The differential equation is $\frac{dy}{dx} = \left(2x+3\right)\left(y^2-4\right)$ with initial value $y(0)=1$.

In the list of steps to take to solve this equation using separation of variables, the very first step is "Check for any values of $y$ that make $g(y) = 0$. These correspond to constant solutions". When I solve $g(y)=0$ I get $g(y) = \pm 2$, but I don't really understand how these values are used to solve this equation?

答: In the separation of variables, the first step is division by $(y^2-4)$. As with any division, you want to prevent division by zero. So $y\ne \pm 2$. From the other direction, $y(t)=\pm2$ are constant solutions. The equation is locally Lipschitz, so all solutions are unique, do not cross or touch each other. So for any solution with an initial value not on these constant solutions, all other values will also stay away from these constants. Thus for the solution under consideration, the division is unproblematic.


问:
Question12.3 Introduction to analytic number theory
I am self studying number theory from Apostol's book and struck on this particular problem (Question 3 ).

$\Gamma(1-s) $is defined for every s and $\zeta(1-s) $ can also be defined for all s . So, F(a, s) is defined for all s.

Consider the equation F(a, s) = $\Gamma(1-s) $ ... in Question (2) .

Here$ F(a, s) =O(x^{-s}$ ). It is  not analytic in $\sigma \leq 1$.

$\Gamma(1-s) $ is not analytic at s=1, 2,...,n,...( n belongs to integers ) and $\zeta(1-s) $ is not analytic at s=2.

So, F(a, s) is entire as for all s as s=1, 2 ,..., n are entire .

how to prove that it is entire?

I don't have any ideas except this .

Kindly guide.

答:
For $0<a<1$,
$$\sum_{n=1}^{N}e^{2\pi ina}=\frac{e^{2i\pi a}(e^{2\pi iaN})}{e^{2i\pi a}-1}=O(1)$$
Using this when applying Abel’s summation formula shows
$$\sum_{n\le x}\frac{e^{2\pi ina}}{n^s}=O(x^{-s})$$
This means $F(s)$ is a convergent Dirichlet series for $\sigma > 0$, and hence is analytic there. Thus we only need to show the extended definition of $F(a, s)$ is analytic for $\sigma\le0$. Now in this region, $\Gamma(1 − s)$ is analytic everywhere and $\zeta(1 − s, ·)$ is analytic for $s\neq0$. This shows we only need to show $F(a, s)$ is analytic at $s = 0$ in order to show $F(a, s)$ is entire.

By Theorem 12.4, $\zeta(s, a)$ has a simple pole at $s = 1$ with residue $1$. Therefore there are entire functions $R_1(s, a)$ and $R_2(s, a)$ such that
$$\zeta(1-s,a)=-\frac{1}{s}+R_1(s,a)\text{  and  }\zeta(1-s,1-a)=-\frac{1}{s}+R_2(s,a)$$
Substituting shows there is an entire function $R_3(s, a)$ such that
\begin{align}
F(a,s)&=\frac{\Gamma(1-s)}{(2\pi)^{1-s}}\bigg{\{}-\frac{e^{\pi i(1-s)/2}+e^{\pi i(s-1)/2}}{s}+R_3(s,a)\bigg{\}}\\
&=\frac{\Gamma(1-s)}{(2\pi)^{1-s}}\bigg{\{}-\frac{2\sin(\pi s/2)}{s}+R_3(s,a)\bigg{\}}
\end{align}
We see $F(a, s)$ has a removable singularity at $s = 0$, hence by Riemann’s theorem on removable singularities $F(a, s)$ can be extended to an analytic function at $s = 0$.


问:
Stone compactification vs initial/final structure as in Bourbaki
Does the Stone-Cech compactification resemble final structure (such as quotient topology) as described in Bourbaki's Theory of Sets Ch.IV? I feel so since continuous $f:X \rightarrow K$ implies the existence of continuous $F:\beta X \rightarrow K$. Not sure I am on the right track though (probably not, but why?) I am not familiar with category theory so my feeling could be obviously wrong.

答:
This may not be what you're asking, but there is a sense in which the Stone-Cech compactification is "final" like the quotient topology... Unfortunately, the language is entirely muddled up because the initial structure category theoretically corresponds to having the most open sets you're allowed. Which means you're the biggest element in the lattice of possible topologies, and so you're called final.

Dually, of course, the topology a category theorist would call terminal or final is exactly the structure which is traditionally called initial. The language is entirely backwards. (If you'll continue indulging my rant -- this has permeated into the language of topologically concrete categories which is endlessly confusing).

To make the answer as reader-friendly as possible, I'll specify when I mean

categorical-initial (there is a map to every other object)

bourbaki-final (the same as above)

categorical-final (there is a map from every other object)

bourbaki-initial (same as above)

You need a mild technical assumption on $X$ (complete regularity), which I'll assume from here on out.

The reason you're seeing a similarity between $\beta X$ and the quotient topology is because they're both categorical-initial. Unfortunately, the quotient topology is called bourbaki-final, and $\beta X$ is typically called the "biggest" compactification... So at least they're both backwards.

To see why, we define a partial order on compactifications by setting

$$\alpha_2 \leq \alpha_1$$

whenever for some continuous $f$

That's not a typo: $\alpha_2 \leq \alpha_1$. The language is backwards.

If you like, this is the opposite poset structure that comes from the slice category under $X$. Since we're restricting attention to compactifications, and thus compact hausdorff spaces, we see $\alpha_1 \leq \alpha_2 \leq \alpha_1 \implies \alpha_1 X \cong \alpha_2 X$. So we really do get a poset.

Now one can show that $\beta$ is maximal in this poset. That is, it's categorical-initial in the category of compactifications under $X$ in a way that's entirely analogous to the quotient topology being categorical-initial in the category of "topologies identifying certain points" under $X$.

Rather interestingly, the one point compactification (when it exists), is smallest in this partial order. That is, the one point compactification is categorical-terminal, and is thus analogous to bourbaki-initial topologies in a similar sense.
You can find a proof of this here.

If you want an excellent blog post with proofs of the claims I glossed over, you should read this post from Dan Ma's topology blog. (You'll also notice I stole the above image from this site...). The whole series on the Stone-Cech compactification is good. In particular, though, you might find this post worthwhile too.

I hope this helps ^_^


问:
what does $x|1$ mean in a UFD?
so I'm working on a problem in a UFD and in some part of the problem I concluded that $x|1$? what does this tell us about $x$?
I was thinking well it means that for some $r$ we have $xr=1$ and that means $x$ is a unit now I have to get back to an equality: $d'=cdx$

I need to get rid of the x? how does $x|1$ help?.

答: Indeed, $x\mid 1$ means that $xr=1$ for some $r$. Then $x$ is a unit and has a unique inverse $r=x^{-1}$. Then you have $d'x^{-1} = cd$.

Title: what does $x|1$ mean in a UFD?


问:
Approaching Compact Sets of the Euclidean Space with the Product Topology
I was working on Croom's Principles of Topology with the product topology; and came across this problem:

Knowing that a subset $A$ of $\mathbb R$ is compact if and only if it is closed and bounded, use product space ideas to prove the same characterization of the compact subsets of $\mathbb R^n$.

Thus I'm supposed to show that a subset $A:=\prod_{i=1}^n A_i$ of $\mathbb R^n$ is compact iff it is closed and bounded. And sadly I am completely lost. I think I could "decompose" $A$ into each coordinate spaces, and work on $A_i$ to find out the properties for $\mathbb R^n$. However I can't see any way to do so.

Croom uses the box topology for the term "product topology" for finite product spaces, but it doesn't really matter since I'm working on $\mathbb R^n$.

答: If $A \subseteq \Bbb R^n$ is compact, it is closed and bounded: this is true in any metric space (the first even for just Hausdorff spaces, for the second consider the  cover by sets $B(a,n), n \in \Bbb N$ for some $a \in A$, which has a finite subcover etc.). If we start by assuming $A$ is closed and bounded we can find closed intervals $A_1,\ldots A_n \subseteq \Bbb R$ such that $A \subseteq \prod_{i=1}^n A_i$. The $A_i$ are compact by what we know about $\Bbb R$, and so Tychonoff tells us that $\prod_{i=1}^n A_i$ is compact and hence so is $A$, being a closed subspace of it. Hence the equivalence.


问:
Why is there no field with one element?
This has been asked here but marked as answered and I don’t feel like the question was ever answered, or at least was not clear to me.

I don’t understand why the set consisting only of the element $\{0\}$ along with the usual $+$ and $×$ does not satisfy the criteria, since $0$ acts as both the additive and multiplicative identity.

That is, letting $G = \{0\}$, then

$∀ g ∈ G, 0+g = g$ and

$∀ g ∈ G, 0·g = g$ (Since $0·0 = 0$ )

Similarly, it is both its own additive and multiplicative inverse. What is the problem at only the field level, without wishing it satisfy some additional properties for category theory or algebraic/arithmetic geometry?

答: Let $K := \{0\}$. Then $K \setminus \{0\}$ cannot be a multiplicative group, since there is no identity element contained in it.

Title: Why is there no field with one element?


答:
So, let's review: $(F,+,\cdot,0,1)$ is a field if

$(F,+,0)$ is an abelian group

$(F \setminus \{0\}, \cdot, 1)$ is an abelian group

What happens if $0 = 1$ and $F$ is the singleton containing that element? Then the latter characteristic is not satisfied, for $F \setminus \{ 0 \} = \varnothing$  yet all groups are nonempty by assumption. (Namely, the axioms of group imply the existence of an element in it, the identity element, so a group is always nonempty.)

Title: Why is there no field with one element?

Body:
This has been asked here but marked as answered and I don’t feel like the question was ever answered, or at least was not clear to me.

I don’t understand why the set consisting only of the element $\{0\}$ along with the usual $+$ and $×$ does not satisfy the criteria, since $0$ acts as both the additive and multiplicative identity.

That is, letting $G = \{0\}$, then

$∀ g ∈ G, 0+g = g$ and

$∀ g ∈ G, 0·g = g$ (Since $0·0 = 0$ )

Similarly, it is both its own additive and multiplicative inverse. What is the problem at only the field level, without wishing it satisfy some additional properties for category theory or algebraic/arithmetic geometry?


问:
Prove that for all positive integer n, the inequality $2n\choose n$ $<4^n$ holds
How do I prove that for all positive integer n, the inequality $2n\choose n$$<4^n$ holds?

Thank you!

答: Hint:  The LHS is the number of $n$-element subsets of $[2n]$, while the RHS is the number of all subsets of $[2n]$.</p>


问:
maximize $x^\top A x^\top$ subject to $\|x \| = 1$ and $ \langle x, v_1 \rangle$
Let $A$ be an $n \times n$ symmetric matrix.

Firstly, consider the optimization problem:
$$
maximize \ x^\top A x^\top \ subject \ to \ \| x\| = 1
$$

The optimization problem admits a solution denote by $v_1$. Here I know that $v_1$ is an eigenvector of $A$, since by Lagrange multipliers, $\nabla f = Ax+(x^\top A)^\top = Ax+A^\top x = 2Ax = \lambda \nabla g = \lambda 2x$, since $v_1$ is a solution to the optimization problem, $Av_1= \lambda v_1$.

Now consider this:
$$
maximize \ x^\top A x^\top \ subject \ to 、 \| x \| = 1 \ and \ \langle x, v_1 \rangle =0
$$
Let $v_2$ be the solution of this problem. How to show that $v_2$ is an eigenvector of $A$ that is orthogonal to $v_1$?

答:
Let $f(x)=x^TAx$, $g_1(x)=x^Tx-1$, $g_2(x)=x^Tv_1$. Lagrange multipliers says that to maximize $f(x)$ subject to $g_1(x)=g_2(x)=0$, we need to solve
$$\nabla f(x)=\lambda \nabla g_1(x)+\mu \nabla g_2(x)$$
together with $g_1(x)=g_2(x)=0$.

Explicitly,

$$2Ax=2\lambda x + \mu v_1, \langle x,x \rangle =1, \langle x,v_1 \rangle=0$$

If we take the first equation and take the inner product with $v_1$ and use the two constraints, we get $2\langle v_1,Ax \rangle = 2\lambda \langle v_1,x \rangle + \mu \langle v_1, v_1 \rangle$, which becomes $0=\mu$.  Going back to the original equation, we get

$$Ax=\lambda x.$$

So the $x$ which maximizes subject to the constraints is an eigenvector.


问:
How to use the constant rank theorem to show that there is no $C^1$ surjection $f : U \subset \mathbb R^n \to \mathbb R^m$ for $n < m$?
In a homework problem sheet, I had to prove that, if $f : U \to \mathbb R^m$ is a continuously differentiable surjection defined on an open subset $U \subset \mathbb R^n$, then $n \ge m$. The instructor intended us to use the constant rank theorem.

I did not see the instructor's intended solution. Instead, I used the fact that $f$ is locally Lipschitz to show that, if $n < m$, then $f(U)$ has measure zero, and therefore cannot be the whole of $\mathbb R^m$.

I am not going to change the solution I submitted, because it took me a lot of effort to come up with it already, and technically it does solve the problem. However, I would still like to know how this problem could have been solved using the constant rank theorem alone, without using any measure theory.

Suppose that $n < m$. Then $df_p$ is not surjective for any $p \in U$.

Since we intend to use the constant rank theorem, we initially restrict our attention to the points where the constant rank theorem can indeed be used. By definition, $p \in U$ is one such point if and only if it has a neighborhood $V_p \subset U$ such that the restriction $df \mid V_p$ has constant rank.

By the constant rank theorem, we may shrink each $V_p$ so that, after applying suitable changes of coordinates to both the domain and the codomain, $f \mid V_p$ admits a local representation of the form

$$(x_1, \dots, x_n) \longmapsto (x_1, \dots, x_k, 0, \dots, 0)$$

If $n < m$, then $k < m$, hence $f(V_p)$ is a meager subset of $\mathbb R^m$, in the sense of the Baire category theorem.

Now consider the set $V = \bigcup_p V_p$. By construction, $\{ V_p \}$ is an uncountable open cover of $V$. However, $V$ is Lindelöf, so we may replace $\{ V_p \}$ with a countable subcover. Therefore $f(V) = \bigcup_p f(V_p)$ is itself meager, because it is a countable union of meager subsets of $\mathbb R^m$.

This is fine and well, but I cannot for my life see why $f(U \setminus V)$ should be in some sense a “small” subset of $\mathbb R^m$ as well, other than using the measure-theoretic argument above. I did try some ideas, unsuccessfully:

Show that, if $p$ admits a neighborhood $V_p$ in which the rank of $df$ does not exceed $k$, then we may shrink $V_p$ so that $f(V_p)$ is contained in a $C^1$ submanifold of $\mathbb R^m$ of dimension $k$.

This does not work. Consider the function $\gamma : \mathbb R \to \mathbb R^3$ defined by

$$\gamma(t) = (t, e^{-1/t^2} u(-t), e^{-1/t^2} u(t))$$

where $u(t)$ is the characteristic function of the positive reals. Then $\gamma$ is of class $C^\infty$. Any putative $V_0$ may be shrunk to $I_\varepsilon = (-\varepsilon, \varepsilon)$ for some arbitrarily small $\varepsilon$. However, $\gamma(I_\varepsilon)$ is not contained in any $C^1$ submanifold of $\mathbb R^3$ of dimension $1$. It is too “pointy” at the origin.

Show that, if $p$ admits a neighborhood $V_p$ in which the rank of $df$ does not exceed $k$, then we may shrink $V_p$ so that, after applying suitable changes of coordinates, $df$ contains a fixed $k \times k$ minor such that every entry outside this minor is zero.

I have absolutely no idea how to prove this, or whether it is even true. However, I suspect that any counterexamples to the previous item should also be counterexamples to this one.

Show that every $p$ admits neighborhood $V_p$ such that, after applying suitable changes of coordinates, $df$ contains a minor that is a diagonal matrix (or maybe in some sense “diagonally dominant”), and every entry outside this minor is zero.

I have no idea how to prove this, or whether it is even true.

Could someone give me a hint?

EDIT: Maybe this works. Prove that $V$ is dense in $U$.

All of this is leading me to the conclusion that the nice continuous surjections are those that are open maps. For example, if we let $X$ be the set of real numbers equipped with the discrete topology, then the identity function $f : X \to \mathbb R$ is a surjection, but sure as hell it locally does not look like one.

(Note that being a continuously differentiable open surjection is a strictly weaker condition than being a continuously differentiable submersion, e.g., consider the map $z \mapsto z^2$ in the complex plane.)

But then, $X$ is a rather weird object and not a manifold in the first place. If we consider spaces that can actually arise in practice, the closest thing to a non-open continuously differentiable surjection $f : M \to N$ is to take a foliation of $N$, and then identify $M$ with the disjoint union of countably many leaves. However, this still fails to be a surjection.

Since the original problem I posed is rather hard and I am starting to lose interest in it, I will post the original motivation for this question. The instructor claims that the following is a good enough proof.

By assumption, $f$ is surjective. If we take any open subset $B \subset \mathbb R^m$, and let $W = f^{-1}(B)$ be its preimage, then the restriction $f \mid W : W \to B$ is still surjective.

So far so good.

In particular, take a point $p \in U$ such that the rank of $df_p$ is maximal, and let $B$ be a neighborhood of $f(p)$. Then, after shrinking $B$, there exists a neighborhood $V_p$ of $p$ such that the restriction $f \mid V_p : V_p \to B$ admits a local representation of the form

$$(x_1, \dots, x_n) \mapsto (x_1, \dots, x_k, 0, \dots, 0)$$

So far so good. Then he makes what I think is a dubious step.

Since $f \mid V_p$ is surjective, then $m = k \le n$.

The problem is that $V_p$ need not be the whole of $W$. After all, we obtained $V_p$ by shrinking $W$! Or maybe he claims that we have redefined $B = f(V_p)$? Then $B$ need not be an open subset of $\mathbb R^m$ anymore! (This will turn out to be the case, but we cannot assume the conclusion.)

To summarize: We know that $f \mid W : W \to B$ is surjective. But we do not know that $f \mid V_p : V_p \to B$ is surjective.

Or maybe I am wrong and someone could tell me why?

答: If you invoke Sard's Theorem, then this problem is easy to solve: since $m > n$, the set of critical points must be $U$, but since $f$ is surjective, $f(U) = \Bbb R^m$. But this means that the set of critical values $f(U)$ does not have lebesgue measure zero which is a contradiction.


问:
How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables be a subring of $k[x,y]$?
I am trying to answer this question:

Let $k$ be a field and $k[x,y].$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.

But I have the following question:

How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables, in our case $A$ be a subring of $k[x,y]$ which is a ring of polynomials with coefficients in a field $k$, and in finitely many variables ? should not $A$ be at least a finite set? could anyone give me an example to clarify this point please?

答:
An element of $A$ can be written as a finite linear combination $\sum p_\alpha X^\alpha$ of monomials $X^\alpha$ where each monomial is of the form

$$X^\alpha = (x y^{i_1})^{j_1} \cdots (x y^{i_n})^{j_n}$$

At the end, you get an element of $k[x,y]$.

The variables of $A$, namely $x, xy, \dots$ are not independent.


问:
Response surface and variables importance
Disclaimer: I'm not expert in math; the solution to this problem may be trivial.

I have a process based on a deterministic computer simulation where two continuous input variables $x_1$ and $x_2$ (under my control) produce a continuous output variable $y$.

My objective is to determine which one of the two variables affects the variability of $y$ the most and I would like the answer to be generalizable to any number of independent variables.

Mathematically speaking, my model is $y=f(x_1,x_2)$, with $f:X \rightarrow R $ being a continous unknown function in $X$ which is a simply connected domain.

Lack of random variation in $y$ suggest me that a statistical approach is not the right way to address the problem .

If I knew $f$ ,$ \frac{\partial f }{\partial x_1}$ and $\frac{\partial f}{\partial x_2}$ answer my question but only if $ \frac{\partial f }{\partial x_1}$ depends solely on $x_1$ (and same for $ \frac{\partial f}{\partial x_2}$). If it's not the case I do not know how to properly answer the question even by knowing $f$.

My naive approach:

I was thinking about "meshing" the $X$ domain in regions where $f$ is approximated by a linear  o by a quadratic model.

Let us define:

$L(x_1,x_2)=\beta_0+\beta_1x_1 + \beta_2x_2$

$Q(x_1,x_2)=\beta_0+\beta_1x_1 + \beta_2x_2 + \beta_3x_1x_2 +\beta_4x_1^2 + \beta_5x_2^2 $

In domain regions where linear approximation holds, $\frac{\partial L}{\partial x_1}=\beta_1$ and $\frac{\partial L}{\partial x_2}=\beta_2$ and this should be nice because I can directly compare the coefficient in order to assess which of the two variable influences $y$ the most.

In those region where linear approximation does not hold, the presence of the interaction term disrupts my interpretation because $\frac{\partial Q }{\partial x_1} = \beta_1 +\beta_3x_2 + 2\beta_4x_1$ depends on $x_2$ (and similar for $\frac{\partial Q }{\partial x_2}$)

Some thougts: linear approximation may be incorrect where sub-regions contain minima, maxima or saddle points. Does avoiding them by partitioning the domain with a finite number of "holes" could solve the issue? (I may use an optimization algorithm such as gradient descent in order to find them)

答:
I believe your question is both utterly meaningless and highly relevant! The relevant part is clear to you because you wouldn't ask otherwise. For all others: One of the basic challenges to understand complicated computer models is to understand what is driving the results. A popular buzzword in this context is model "explainability".

So I agree, the problem itself and your thoughts on it are perfectly reasonable. That said, the question as stated is meaningless as a mathematical statement and hopeless in this generality from an applied perspective. You will need to sharpen your thinking around (at least) three issues:

The operational definition of your question

Assumptions on your function

Clarity about the inputs respectively your coordinates.

The operational definition of your question:

As the commentator already stated "which one of the two variables affects the variability of y the most" is too fuzzy to be workable. For the sake of argument I will sharpen this (somewhat) to: "Is $x$ or $y$ more important in causing $f$ to be large?" Mathematically the answer is now clear: Take a "large" threshold $\tau$, define a measure of "importance" and apply it to the set $f^{-1}([\tau,\infty[)\subset\mathbb{R}^2.$ The problem is, even if $f$ is really "nice", e.g. infinitely differentiable, $f^{-1}([\tau,\infty[)$ can have any crazy shape. This makes it impossible to come up with a natural notion of importance. So, unless you have a clear externally defined and operational notion of importance your problem will be infeasible.

Assumptions on your function:

As you noticed yourself, 1) above makes restricting your class of functions as much as possible a key concern. So what do you know about your function? Available to you may be properties like "smooth", "convex", "monotonic in x or y", "where are the critical points" and so on. Your idea of approximating $f$ with simple functions is a very good and valid one. Prior knowledge about $f$ then helps in finding good approximations.

Clarity about the inputs respectively your coordinates:

You seem to treat your $x$ and $y$ coordinates as fixed. This is most likely because they are externally given by the process you simulate. But is this really true? Take for example a physical process where stuff ($x$) is ground  at a certain temperature ($y$) to produce output $L(x,y)=\beta_0+\beta_1 x+\beta_2 y$. You would say $x$ is more important than $y$ iff $\beta_1 > \beta_2$. But the beta-factors depend on units of measurement. If you stop measuring the stuff you grind in tons but state $x$ in milligrams the new factor $\tilde{\beta}_1$ will be larger than $\beta_1$ by a factor of $10^{12}$. Often in applications another natural invariant is monotonic reparametrizations.

This can be driven further and actually might help a lot depending on the requirements of your problem. Take your linear $L$ example. Arguably it is not enough for either $x_1$ or $x_2$ to be large alone to achieve a large $L$. Both must be large at the same time. So the "really important" variable is $z=\beta_1 x+\beta_2 y.$ Now you can answer "What drives $L$?" easily because the result is driven clearly and only by "z". This idea works more generally. As long as the gradient of your function does not vanish, you can reparametrize it by its level surfaces. The coordinate orthogonal to those level surfaces is then your single variable $z$.

Further remarks:

Even though your problem is deterministic, probabilistic approaches may be very useful to you. This is because of two reasons. First, putting some probability distributions on your inputs is often a very intuitive model for variability. Furthermore, by averaging/integrating over inputs you will get rid of some of the complexity involved. In your case you could use normal distributions $X$ and $Y$ with variance $\sigma^2_i$ as inputs and calculate the variance of the random variable $f(X,Y)$ as a function of the variance of the inputs.

The second reason is that the people thinking about Uncertainty Quantification have been analysing these kind of problems for a long time.


问:
Evaluate $\lim\limits_{n\to\infty} \frac{\sin(1)+\sin^2(\frac{1}{2})+\ldots+\sin^n(\frac{1}{n})}{\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}}$
This was a recent problem on the Awesome Math Problem Column. The solution is given as follows:

We shall use Stolz-Cesaro Lemma. We have:

$$\lim_{n\to\infty} \frac{\sin(1)+\sin^2(\frac{1}{2})+\ldots+\sin^n(\frac{1}{n})}{\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}}$$
$$=\lim_{n\to\infty}\frac{\sin^n(\frac{1}{n})}{\frac{1}{n!}}=\lim_{n\to\infty}\frac{\sin^n(\frac{1}{n})}{\frac{1}{n^n}}\cdot\lim_{n\to\infty} \frac{n!}{n^n}.$$
Since
$$\frac{n!}{n^n}<\frac{1}{n}$$
we have
$$\lim_{n\to\infty}\frac{n!}{n^n}=0$$
So, the required limit is $0$.

I have two problems with the given proof that I can’t wrap my head around.

Firstly, from everything I have read so far the Stolz-Cesaro limit only applies to cases where the limit is of the form $\frac{0}{0}$ and the form $\frac{\cdot}{\infty}$. However, just from the wording of the theorem and the way the author presented this solution both seem to imply the following:
$$a_n=\sum_{k=1}^n \sin^k\Big(\frac{1}{k}\Big), \ \ \  \text{and} \ \ \ b_n=\sum_{k=1}^n \frac{1}{k!}$$
Then when “applying” the Stolz-Cesaro theorem you would have:
$$\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}=\lim_{n\to\infty} \frac{\sum\limits_{k=1}^{n} \sin^k(\frac{1}{k})-\sum\limits_{k=1}^{n-1} \sin^k(\frac{1}{k})}{\sum\limits_{k=1}^n \frac{1}{k!}-\sum\limits_{k=1}^{n-1} \frac{1}{k!}}=\lim_{n\to\infty}\frac{\sin^n(\frac{1}{n})}{\frac{1}{n!}}$$

The problem that I see is when $n\to\infty$ we know that $a_n$ converges by the root test, and $b_n$ converges to $e-1$. So by basic limit rules this would converge also. So it seems that $b_n$ does not satisfy either of the sequence requirements of the Stolz-Cesaro Theorem.

Secondly, according to wolfram $\displaystyle\sum_{n=1}^\infty \sin^n\Big(\frac{1}{n}\Big)\approx 1.11043$. So intuitively it seems to me that this limit would be approximately $\frac{1.11043}{e-1}\approx 0.646244$.

Does anyone know what is going on here?

答:
Note sure that this is a valide answer.

Let
$$a_k=\sin ^k\left(\frac{1}{k}\right)\implies \log(a_k)=k \log \left(\sin \left(\frac{1}{k}\right)\right)$$ Using Taylor
$$\log(a_k)=-k \log (k)-\frac{1}{6 k}-\frac{1}{180 k^3}+O\left(\frac{1}{k^5}\right)$$
$$a_k=k^{-k}\,\exp\left(-\frac{1}{6 k}-\frac{1}{180 k^3}+\cdots \right)$$ The summation of the $a_k$'s from $k=1$ to $n$ converges very fast to $1.110750$ which, divided by $e-1$ gives $0.646431$ while the result should be $0.646242$. Adding the next term in the exponential (namely $-\frac{1}{2835 k^6}$)  would give $0.646256$ and so forth since all terms in the exponential are negative.

Notice that
$$\sum_{k=1}^\infty k^{-k}=1.2912859970626635404\cdots$$ is a well know number (have a look here). Forgetting the exponential term, this would give an upper bound equal to $\frac 34$ for the limit.


问:
Prove there's no simple group of order $4p^a$
I'm trying to show/prove that there's no simple group of order $4p^a$ with $p \geq 5$ being an odd prime and $a \geq 1$.

Since $p$ is odd prime, $p \nmid 4$. By congruence and divisibility theorem, then $n_p \equiv 1$ (mod p) and $n_p | 4$. But from that $n_p$ can only be 1, which shows there's a Sylow p-subgroup, meaning that there exists a simple group with that order.

Am I doing anything wrong? I would appreciate any help on this.

答: You showed that $n_p = 1$. This means that there is a unique (hence normal!) $p$-Sylow subgroup. Hence any group of order $4p^a$ has a non-trivial normal subgroup, so it is not simple.


答:
You are correct, but it would have been better to say at the end that

…from that $n_p$ can only be 1, which shows that the Sylow $p$-subgroup is normal in the group of order $4p^a$, which means the outer group cannot be simple.

Title: Prove there's no simple group of order $4p^a$

Body:
I'm trying to show/prove that there's no simple group of order $4p^a$ with $p \geq 5$ being an odd prime and $a \geq 1$.

Since $p$ is odd prime, $p \nmid 4$. By congruence and divisibility theorem, then $n_p \equiv 1$ (mod p) and $n_p | 4$. But from that $n_p$ can only be 1, which shows there's a Sylow p-subgroup, meaning that there exists a simple group with that order.

Am I doing anything wrong? I would appreciate any help on this.


问:
Permutations of n pairs of elements accounting for relabelling symbols and a number of other symmetries
I am doing some work on gauss codes in higher genus surfaces (good resource on the planar case here if you are unfamiliar) and am trying to figure out how many distinct gauss codes there are for a given number of distinct letters. Gauss codes are a sequence of letters where each letter occurs exactly twice. Two codes are equivalent if you can get from one to the other using any combination of the following moves:

Cyclic permutation ie moving the last letter to the front. For instance, abcabc is the same as cabcab

Reversal ie reading the code backwards. For instance abcabc is the same as cbacba

Exchanging letters ie replacing all the instance of one symbol with another and vice versa. For instance abcdeabcde is the same as adcbeadcbe by switching b and d

These three moves seem to reduce it down to very few codes. There is obviously 1 option for 1 pair symbols (aa) and 2 for 2 pairs (abab and aabb). Less obviously, but still fairly clearly, there are 5 for 3 pairs (aabbcc, ababcc, abacbc, abbacc, abcabc). Beyond this I am not clear.

Some perhaps useful invariants I have noticed is that 'double letters (...aa...) cannot be changed by these moves except for putting 1 at the start and 1 at the end. Additionally, if you count the gaps between letters and their partners, that sequence doesn't change except for negating it mod 2n ie in 3 pairs, a gap of 4 is equivalent to a gap of 2 (in the other direction). There are a few others, but I am curious if there is a combinatorial way to reduce the permutations of these symbols (2n!/2^n) down to the set of equivalence classes under these rules.

Thank you!

答:
It looks like it is this sequence:

A054499:  Number of pairings on a bracelet; number of chord diagrams that can be turned over and having n chords.

Copied from the OEIS sequence page:

EXAMPLE

For n=3, there are 5 bracelets with 3 pairs of beads. They are represented by the words aabbcc, aabcbc, aabccb, abacbc, and abcabc. All of the 6!/(2x2x2) = 90 combinations can be derived from these by some combination of relabeling the pairs, rotation, and reflection. So a(3) = 5. - Michael B. Porter, Jul 27 2016

and the formula is derived from 2 other sequences:

FORMULA

a(n) = (2*A007769(n) + A047974(n) + A047974(n-1))/4 for n > 0.


问:
Simplify difference of two arctangents?
I have a problem, that I am trying to simplify, but there does not seem to be something obvious regarding it.

Very simply, I am trying to figure out if there is a way to 'open' the following:

$$
\tan^{-1} (\frac{a}{b}) - \tan^{-1} (\frac{c}{d})
$$

By 'open', I simply mean that I would like to be able to group $a$, $b$, $c$, and $d$ together somehow, so that I can continue with my work. I have scoured wiki for something that might allow me to do that, but could not find anything.

Disclaimer: I should also add that the $\tan^{-1}$ here refers to the 'Four Quadrant' inverse tangent, usually found in computer programs as 'atan2' instead of simply 'atan'.

Thanks for any insight.

答: note that: $$\tan(\alpha-\beta )=\frac{\tan\alpha- \tan\beta }{1+\tan\alpha \tan\beta } $$ $$\Longrightarrow \alpha -\beta =\tan^{-1}\left(\frac{\tan\alpha- \tan\beta }{1+\tan\alpha \tan\beta }\right).$$ Try letting $\alpha=\tan^{-1}\left(\frac{a}{b}\right)$ and $\beta =\tan^{-1}\left(\frac{c}{d}\right)$.  Then $\tan\alpha=\frac{a}{b}$ and $\tan\beta=\frac{c}{d}$.</p>

Title: Simplify difference of two arctangents?


问:
Translation to predicate logic
We have a statement to be translated into first order logic (not caring whether it is true or false in the real world):

There does not exist an even prime number which is the square of a prime number.

I have trouble translating this sentence to first order logic. These were my steps:

$\forall y \neg \exists x ((prime(2*x) \land prime(y)) \implies (2*x=y*y))$

So, for all prime numbers y, there does not exist an even prime number that is equal to the square of the prime number y. So far, it should be correct.

$\forall y \neg \exists x (\neg(prime(2*x) \land prime(y)) \lor (2*x=y*y))$

Using laws for implication transformation, I changed the expression.

$\forall y \neg \exists x (\neg prime(2*x) \lor \neg prime(y) \lor (2*x=y*y))$

Using De Morgan laws, I changed the expression again.

Now, I can change the negation of "there exist" to "for all" by negating the inside of the brackets:

$\forall y \forall x (prime(2*x) \land prime(y) \land \neg (2*x=y*y))$

This, however, is not correct, but I can't find the mistake in my reasoning. I would really appreciate if someone could point me the right way. Thank you.

答:
There does not exist an even prime number which is the square of a prime number.

First we should clarify what is meant by "even prime number". Below are two possible predicate definitions:

$\mbox{evenPrime}(x)\equiv x=2$

$\mbox{evenPrime}(x)\equiv (2\mid x\land\mbox{prime}(0.5 x))\lor x=2$

So now we can translate the above statement as the following
$$\forall x\lnot\exists y(\mbox{evenPrime}(y)\land \mbox{prime}(x)\land y=x^2)$$
If you apply the first definition of $\mbox{evenPrime}$, then the statement is true.

If you apply the second definition of $\mbox{evenPrime}$, then the statement is false.

Title: Translation to predicate logic


答:
For some $\,d\,$ we have $\,da=1,\,$ so $\,a\,$ is a unit. Now I need to get rid of $\,a\,$ in $\,ax = b$

Whenever $\,\color{#c00}{a\mid 1},\,$ scaling by $\,a^{-1}\,$ is a special case of cancelling $\,a,\,$ by   $\,c=1\,$ below.

Lemma $\ \ \,\color{#c00}{a\mid c},\ ax = b\!\:\color{#c00}c\ \,\Rightarrow\ x = b(c/a)\ $ if $\,a\,$ is cancellable.

Proof $\ \ \ \,\color{#c00}{da\! =\! c},\ ax = b\color{#c00}{da}\Rightarrow\, x = b\:\!d,\,$ by cancelling $\,a.\ \ \small\bf QED$

So the result is "obvious" once we understand how cancellation works. But inverses are so ubiquitous that it helpful to know this in both forms, i.e. scaling $\,ax = b\,$ by the inverse of $\,a,\,$ or cancelling $\,\color{#c00}a\,$ from its multiple $\color{#c00}1=a\:\!a^{-1}$ in $\,b = b\cdot\color{#c00}1$.

Title: what does $x|1$ mean in a UFD?

Body:
so I'm working on a problem in a UFD and in some part of the problem I concluded that $x|1$? what does this tell us about $x$?
I was thinking well it means that for some $r$ we have $xr=1$ and that means $x$ is a unit now I have to get back to an equality: $d'=cdx$

I need to get rid of the x? how does $x|1$ help?.


答:
I think it's a notational issue:

Let $S\le R$ be (commutative) rings, and $A=\{a_1,a_2,\dots\}\subseteq R$.

Then $S[a_1,a_2,\dots]$ simply denotes the subring generated by $S\cup A$, i.e. the smallest subring of $R$ that contains both $S$ and $A$.

Note that, however, there's a homomorphism $\varphi$ from the polynomial ring $S[x_1,x_2,\dots]$ to $R$ that maps $x_i\mapsto a_i$, and $S[a_1,a_2,\dots]$ is just the image of $\varphi$.

Title: How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables be a subring of $k[x,y]$?

Body:
I am trying to answer this question:

Let $k$ be a field and $k[x,y].$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.

But I have the following question:

How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables, in our case $A$ be a subring of $k[x,y]$ which is a ring of polynomials with coefficients in a field $k$, and in finitely many variables ? should not $A$ be at least a finite set? could anyone give me an example to clarify this point please?


问:
Adjunct for solution operator of Poisson problem
Consider the Poisson problem with Neumann boundary conditions

$\Delta u = f$ on $\Omega \subset \mathbb{R}^d$

$\partial_{\nu}u = 0$ on $\partial \Omega$

where $\Omega$ is Lipschitz and $f \in L^2(\Omega)$. Now we know that the problem has unique solution $u \in H_0^1(\Omega)$ and that the solution operator is linear and bounded.

We can then interpret the solution operator as a function

$S\colon L^2(\Omega) \to L^2(\Omega), \, f \mapsto u$

this is a well defined, bounded and linear map. My question now is: What is the Hilbert adjunct of $S$?

Per definition we have to find $S^\ast$ such that,

$\langle f, \, S^\ast v \rangle = \langle S f, \, v \rangle = \langle u, \, v \rangle$

for every $v \in L^2(\Omega)$. Remarkably I was unable to find anything about that in my literature.

Looking at the weak formulation of the problem, I might have guessed the adjunct is in direct relation to the differential operator, the Laplacian in this case, but I am not sure how, because $v$ is in $L^2(\Omega)$.

A related question would be if we could do something similar for the boundary data. That is having a problem

$\Delta u = 0$ on $\Omega \subset \mathbb{R}^d$

$\partial_{\nu}u = g$ on $\partial \Omega$

for $g \in L^2(\partial \Omega)$. Can now say something about the Hilbert adjunct $B^\ast$ of

$B\colon L^2(\partial \Omega) \to L^2(\Omega), \, g \mapsto u$?

答:
Let $f, g\in L^2(\Omega)$ and $u = Sf, v = Sg$ (to be understood in the weak sense). Then
\begin{align*}
\langle Sf, g\rangle & = \int_\Omega ug\, dx = \int_\Omega \nabla v\cdot\nabla u\, dx \\
\langle f, Sg\rangle & = \int_\Omega fv\, dx = \int_\Omega \nabla u\cdot\nabla v\, dx.
\end{align*}


问:
Understanding the meaning of $p_{01}$
In the following link for the historical problem (first part): in how many "regions" a finite set A of lines intersect the plane for the first he defines:

$p_{0}$ : is the number of intersection points of $A$ (without multiplicity)

$p_{1}$ : $|A|$ is the cardinal of $A$

$p_{01}$ : is the number of pairs consisting of a line of $A$ and one of its intersection points

If anyone can explain to me the meaning of $p_{01}$ because even with the following figure I'm having a hard time ^^.

Thanks in advance for your help.

答: $p_{01}$ is the number of "flags" in the configuration: an ordered tuple of a line and an intersection which the line partakes in. Three of the lines in the diagram have two intersections and the fourth has three intersections, so $p_{01}=9$ here.

Title: Understanding the meaning of $p_{01}$


问:
What is the minimum number of packs of each ingredient needed without left over?

Case 1

Ingredients A (3 kg per pack), B (4 kg per pack), and C (5 kg per
pack) must purchased in whole packs.

Given that 1 kg of A, 1 kg of B and 1 kg of C are needed to make a giant bread. We want to make some breads.

What is the minimum number of pack of each ingredient we have to purchase such that none is left over?

My attempt

First, find the least common multiple of 3 kg, 4 kg, and 5 kg. The LCM is 60 kg.

Second, find the number of packs of each ingredients.

$A=60/3=20$ packs

$B=60/4=15$ packs

$C=60/5=12$ packs

Third, additionally we can find the number of breads which is 60.

Case 2

Let's change to another recipe.

Ingredients A (3 kg per pack), B (4 kg per pack), and C (5 kg per
pack) must purchased in whole packs.

Given that 2 kg of A, 3 kg of B and 7 kg of C are needed to make a giant bread. We want to make some breads.

What is the minimum number of pack of each ingredient we have to purchase such that none is left over?

Here 60 cannot be divided by 7. What should I do next?

答:
Suppose the minimum number of packs of ingredients $A$, $B$, $C$ that we need to buy are $a$, $b$, $c$ respectively.  Then the total amounts of each ingredient are $3a$, $4b$, $5c$.  Let $n$ represent the number of breads that are made.  Then $2n$, $3n$, $7n$ are the total amounts of each ingredient that are used.  Therefore, we require
$$\begin{align}
3a &= 2n \\ 4b &= 3n \\ 5c &= 7n. \end{align}$$

Since $a, b, c, n$ are integers, we must have $n$ be multiples of $3$, $4$, and $5$ as you had observed in the first question; the least such common multiple is $60$, therefore let $n = 60k$ for some integer $k$.  Then we get
$$\begin{align}
a &= 40k \\ b &= 45k \\ c &= 84k. \end{align}$$
So long as $k$ is a positive integer, $a, b, c$ are also positive integers; thus $k = 1$ gives the minimum solution, $(a, b, c) = (40, 45, 84)$ packs.  This makes $60$ breads as in the first question.

It is worth mentioning that we need to be careful:  if, for instance, the ingredient ratio had been $2 : 4 : 7$ instead of $2 : 3 : 7$, then we only need $n$ to be a common multiple of $3$ and $5$, since the second equation in the system reduces:
$$\begin{align}
3a &= 2n \\ 4b &= 4n \color{red}{\iff b = n} \\ 5c &= 7n \end{align}$$
This gives us $n = 15k$ and for $k = 1$ we get $$(a, b, c) = (10, 15, 21).$$


问:
Finding integrating factor for non exact differential equation
I am trying to solve non exact differential equation but I am unable to find the integration factor.

I solve the following question
$$-3y\frac{dy}{dx}+2x=0 $$

and I get the integrationg factor $x^{-5/2}$ and I use the proper way to find the integrating factor, but in our text book the integrating facot for above differential equation is finded and that is $y/x^4$.

Now I am completely confused that why please help me out with this.

答:
$$-3y\:dy+2x\:dx=0$$
is an EXACT differential equation.

In other words the integrating factor is $1$.
$$d\left(-\frac32 y^2+x^2\right)=0$$
$$-\frac32 y^2+x^2=\text{constant}$$


问:
find (′) at point (0,1/e).
find $ln(y')$ at point (0,1/e) where $xy^6-ln(x^8+y)=1$

I could not solve the question by taking normal derivatives. Is there a theorem here? how can i solve it.

Title: find (′) at point (0,1/e).

Body:
Check that $y(0)=1/e$, D. w.r.t. $x$

$$xy^6-\ln(x^8+y)=1, \implies y^6+6xy^5y'+\frac{8x^7+y'}{x^8+y}=0$$
$$\implies y'=\frac{-8x^7+x^8y^6+y^7}{6xy^5(x^8+y)+1} \implies y'(0)=e^{-7}$$
So $\ln y'=-7$.


问:
In how many ways can you choose $k$ out of $n$ people standing in line, So there's a space of at least 3 people between them
In how many ways can you choose $k$ out of $n$ people standing in line, So there's a space of at least $3$ people between them.

Actually, I don't even know how to start on this one.

答:
Let $x_i$ denote the distance between the men with numbers i and i+1. Then $x_i\ge 3$ and

$$
  x_1+x_2+\ldots x_{k-1} \le n
$$

Put $y_i = x_i-2$. Then $y_i\ge 1$ and

$$
  y_1+y_2+\ldots y_{k-1} \le n - 2(k-1)
$$

The last inequality is the problem of choosing $k$ out of $n-2(k-1)$ people whiout any restrictions. So the answer is $C_{n-2(k-1)}^k = \binom{n-2(k-1)}{k}$


问:
Proof of ADMM special case
I was reading the textbook about ADMM and came across some special cases without proof, which is really interesting so I was trying to prove them. I did a couple but had some trouble getting this one so please help. Thank you in advance

Considering x-update when A=I

$x^+=argmin_x(f(x)+(ρ/2)∣∣x−v∣∣_2^2)=prox_{f,ρ}(v)$.

The special case to be proved is
$f(x)=(1/2)x^TPx+q^Tx+r, x^+:= (P+{\rho}A^TA)^{-1}({\rho}A^Tv-q)$

The attached photo shows what I have tried so far, I think I am close but there is something not right.

答:
Hint (almost full solution)

Assuming you want to solve the following update

\begin{align}
x^{+} = \arg\min_{x} \frac{1}{2} x^TPx + q^Tx + r + \frac{\rho}{2}\| Ax - b - v \|_2^2.
\end{align}

Compute the gradient and set it to zero to obtain the $x$-update solution.

Gradient of $\frac{1}{2} x^TPx + q^Tx+ r$ is $\frac{1}{2}(P+P^T)x + q$ (see example here How to take the gradient of the quadratic form?).  Assuming $P$ is symmetric, the gradient simplifies to $Px + q$.

Gradient of the last piece, i.e., $\frac{\rho}{2}\|Ax-b-v\|^2$, is $\rho A^T(Ax-b-v)$.

To this end, add the gradients and set them to zero such that
\begin{align}
Px + q + \rho A^T(Ax-b-v) = 0
\end{align}

Can you take it from here?

Title: Proof of ADMM special case


问:
Sum of piecewise odd/even complex power series
I'm trying to solve the following problem:
Given

$$g(z)=\sum_{n=0}^{\infty}a_nz^n$$

Where
$$a_n= f(x)=
\begin{cases}
 (n+1)(n+2)&\text{if}\,\ n\ \text{odd}\\
 \ 1&\text{if}\,\ n\ \text{even}\\
\end{cases}$$

Show that the radius of convergence ($\rho$) is 1 and that
$$ g(z)=\frac{1}{1-z^2}+(\frac{z^3}{1-z^2})^{''}$$
for $|z|<\rho$

I think I've managed the first part by splitting the series up into its even and uneven parts like this:
$$g(z)_{even}=\sum_{{n=0}}^{\infty}2n^2+6n+2\cdot z^{2n}$$
$$g(z)_{odd}=\sum_{n=0}^{\infty}z^{2n+1}$$

Both converge with radius of convergence 1, allowing me to conclude $g(z)$ converges with radius of convergence 1

But I am not sure about how to get started with the second part. I know that I 'just' need to differentiate $g(z)$ term by term, but I'm unsure about how to present $g(z)$ using the work I've already done.

Any tips are appreciated.

答:
You have the right idea of splitting it up into the even and odd exponents.

We have:
\begin{alignat*}{2}
g(z) &= \sum_{n=0}^\infty a_n z^n \\
&= \sum_{\substack{n=0\\n \ \text{even}}}^\infty z^n+\sum_{\substack{n=0\\n \ \text{odd}}}^\infty (n+1)(n+2) z^n&&\qquad \text{by splitting $g$ up}\\
&=\sum_{n=0}^\infty z^{2n}+\sum_{n=0}^\infty ((2n+1)+1)((2n+1)+2) z^{2n+1} &&\qquad \text{by reindexing}\\
&=\sum_{n=0}^\infty z^{2n}+\sum_{n=0}^\infty (2n+2)(2n+3) z^{2n+1} \\
&=\sum_{n=0}^\infty z^{2n}+\left(\sum_{n=0}^\infty  z^{2n+3}\right)''\\
\end{alignat*}

Now, recall that the geometric series is $$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$$
and so we mold our results into something similar to this series:
\begin{alignat*}{2}
g(z) &= \sum_{n=0}^\infty z^{2n}+\left(\sum_{n=0}^\infty  z^{2n+3}\right)''\\
&=\sum_{n=0}^\infty (z^2)^n+\left(z^3 \sum_{n=0}^\infty (z^2)^n\right)''\\
\end{alignat*}
Which gives you your answer after plugging it into the geometric series above.


答:
In short, focus on minimum number of breads we need to make. Everything else follows from there.

To make a number of giant breads with given ingredients and quantities and no leftover ingredient, you need to ensure two things -

i) Considering each ingredient individually - you are purchasing quantity of each ingredient in multiple of what one bread requires

For example, if I can purchase ingredient $C$ only in multiple of $5$ kg packs and a bread requires $7$ kg, I need to purchase $35$ kg of the ingredient (making $5$ breads) or in multiple of that otherwise I would have leftover.

$5$ kg packs and $7$ kg required for a bread - so you need to purchase $7$ packs and you need to make $5$ breads.

ii) Now combining all ingredients - you are also purchasing quantity of each ingredient such that they can all make same number of breads.

Taking the above example, ingredient $C$ will have no leftover if I make $5$ breads.
But now if I take $B$, ingredient $B$ is available in $4$ kg pack and each bread requires $3$ kg. So ingredient $B$ will have no leftover if make $4$ breads and so purchase $3$ packs of $4$ kg each.

Ingredient $C$ requires me to make breads in multiple of $5$. Ingredient $B$ requires me to make breads in multiple of $4$. What does that mean? I of course need to make $5 \times 4 = 20$ (LCM) breads so I have no leftover of either the ingredient $B$ or of $C$.

Now if I combine $A$ into the mix, $A$ requires me to make breads in multiple of $3$ ($3$ kg packs and each bread requires $2$ kg).

That means I need to make $5 \times 4 \times 3 = 60 \,$ breads (LCM of $5, 4, 3$).

Now that we know the number of breads, we can find number of packs I need to purchase for each ingredient.

So to end it, for each ingredient, I find the minimum number of whole breads I need to make given how much is required for each bread and the package size it is available in. Then LCM of minimum number of whole breads I need to make for each ingredient gives me the number of breads I need to make. Rest follows from there.

Title: What is the minimum number of packs of each ingredient needed without left over?

Body:
Case 1

Ingredients A (3 kg per pack), B (4 kg per pack), and C (5 kg per
pack) must purchased in whole packs.

Given that 1 kg of A, 1 kg of B and 1 kg of C are needed to make a giant bread. We want to make some breads.

What is the minimum number of pack of each ingredient we have to purchase such that none is left over?

My attempt

First, find the least common multiple of 3 kg, 4 kg, and 5 kg. The LCM is 60 kg.

Second, find the number of packs of each ingredients.

$A=60/3=20$ packs

$B=60/4=15$ packs

$C=60/5=12$ packs

Third, additionally we can find the number of breads which is 60.

Case 2

Let's change to another recipe.

Ingredients A (3 kg per pack), B (4 kg per pack), and C (5 kg per
pack) must purchased in whole packs.

Given that 2 kg of A, 3 kg of B and 7 kg of C are needed to make a giant bread. We want to make some breads.

What is the minimum number of pack of each ingredient we have to purchase such that none is left over?

Here 60 cannot be divided by 7. What should I do next?


问:
Counterexample to $\operatorname{Ln}\frac{1}{z}=-\operatorname{Ln}z$ for $z\ne 0$
Is there a counterexample to
$$\operatorname{Ln}\frac{1}{z}=-\operatorname{Ln}z$$
where $z\in\mathbb{C}\setminus \{0\}$ and
$$\operatorname{Ln}z=\operatorname{Ln}|z|+i\operatorname{Arg}z$$
where $\operatorname{Ln}|z|$ is the unique real solution $x$ of $e^{x}=|z|$ and $-\pi\lt\operatorname{Arg}z\le\pi$?

It was easy to find a counterexample to
$$\operatorname{Ln}wz=\operatorname{Ln}w+\operatorname{Ln}z$$
which is $w=-1$ and $z=i$. But I have trouble finding a counterexample to $\operatorname{Ln}\frac{1}{z}=-\operatorname{Ln}z$.

答:
For $z= -1$ you have
$$
\operatorname{Ln}(z) = \operatorname{Ln} \frac 1z = i \pi \, .
$$

More precisely,
$$
\operatorname{Ln} \frac 1z = \ln \left|\frac 1z\right| + i\operatorname{Arg} \frac 1z
= - \ln |z| + i\operatorname{Arg} \frac 1z
$$
and
$$
\operatorname{Arg} \frac 1z = - \operatorname{Arg} z
$$
unless $\operatorname{Arg} z = \pi$.

So $\operatorname{Ln}\frac{1}{z}=-\operatorname{Ln}z$
holds exactly for all $z \in \Bbb C \setminus (-\infty, 0]$, that is for all non-zero complex numbers which are not on the negative real axis.


问:
Continuous Probability and Central Limit Theorem -- Approximating sum of coin tosses distribution
I ran into a problem I know I have to use central limit theorem and continuous probability for, but I'm not sure how to really apply those. Would appreciate some guidance :))

Assume there are 1 through k people with coins. Person i's coin has
sides i and -i. They toss their coins n times and sum the outputs into
a number $X_i$. For a large n, what is the distribution of $(X_1 + ... +
> X_k)/\sqrt{n}$ approximately equal?

答:
Let's call the result of person $i$'s $j$th toss $Y_{i,j}$ which has mean $0$ and variance $i^2$.

Now let $Z_j = Y_{1,j}+\cdots + Y_{k,j}$, which will have mean $0$ and variance $\frac16 k(k+1)(2k+1)$.

Applying the central limit theorem to $Z_j$, we get that $\frac1{\sqrt{n}}(Z_1+ \cdots + Z_{n})$ converges in distribution to $\mathcal N\left(0, \frac16 k(k+1)(2k+1)\right)$.

Since  $Z_1+ \cdots + Z_{n} = X_1+\cdots +X_k$, you get that $\frac{X_1+\cdots +X_k}{\sqrt{n}}$ also converges in distribution to $\mathcal N\left(0, \frac16 k(k+1)(2k+1)\right)$.


问:
Finding the determinant of a generalised matrix
For this $n \times n$ matrix I am trying to find the determinant:

$\left[\begin{array}{cccc}
1 & 2 & ... & n \\
1 & 2^3 & ... & n^3 \\
. & . & ... & . \\
1 & 2^{2n-1} & ...& n^{2n-1}
\end{array}\right]$

I am having abit of trouble with row reduction hence not getting anywhere. I think I need to transform it somehow as it looks similar to a vandermonde matrix but I am not sure how can someone help me out?

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain why you are stuck.)

Hint: Show that the matrix whose entries $a_{ij} = b_i ^j$ has determinant $ \prod_{i > j } ( b_i - b_j)$.

Hint: Hence, conclude that the answer is $n! \prod_{i > j} ( i^2 - j^2 ) $.

This matches up with your calculated values when $ n = 1, 2, 3, 4$.

Note: The way the numerical answer is written up is the hint of what matrix you should be considering.


问:
Orthogonal complement of functions where Fourier coefficients are zero

Let W be a subspace of $L^2[0, 2\pi]$ of all functions that $\hat{f}(n)=0$ for all $n \leq 0 $ (where $\hat{f}(n)=\frac{1}{2 \pi }\int_\limits{0}^{2\pi}f(t)e^{-int}dt$ denotes n-th Fourier coefficient of $f$). What is orthogonal complement of W and what form does orthogonal projection onto W take?

Now, I know that $f \in L^2[0, 2\pi]$ can take form of $f(x) = \sum_\limits{n=-\infty}^{\infty} \hat{f}(n)e^{inx}$ and ${e^{izx}}_{z \in \mathbb{Z}}$ is complete orthogonal set in $L^2[0,2\pi]$.

If W contains those functions with all Fourier coefficients $n \leq 0$ equal to 0, then its orthogonal complement would be functions where $\hat{f}(n) = 0$ for all $n>0$. Is that correct?

Now, the projection. We know that
$$
L^2[0, 2\pi]\ni f(x) = \sum_\limits{n=-\infty}^{\infty} \hat{f}(n)e^{inx}  = \sum_\limits{n=-\infty}^{0} \hat{f}(n)e^{inx} + \sum_\limits{n=1}^{\infty} \hat{f}(n)e^{inx}
$$
Does it mean that projection onto W will be just:
$$proj(f) = \sum_\limits{n=-\infty}^{0} \hat{f}(n)e^{inx}$$

If so, why? I have seen similar example from $L^2[\mathbb{R}]$ with set V being all f such that $f = 0$ on $(0, \infty)$, its orthogonal complement was all functions $f = 0$ on $(-\infty, 0)$ and the projection onto V was defined as $proj(f) = f \mathbb{1}_{(0, \infty)}$. So I kinda guessed it will be the same idea applied here.

I don't see clearly why a projection of any function $g$ from $L^2[0, 2 \pi]$ (not necesarilly with all coefficients $n \leq 0$ of $g$ equal to 0) will be projected onto $\sum_\limits{n=-\infty}^{0} \hat{g}(n)e^{inx}$. Any help and explanation appreciated.

答: In general $(M^{\perp})^{\perp} =M$ for any closed subspace of a Hilbert space. If $(e_n)$ is an orthonormal basis for  a Hilbert space $H$ then  the orthogonal complement of $\{e_i: i \in A\}$ is given by $\{\sum_{i \notin A} a_i e_i: \sum |a_i|^{2} <\infty\}$. And the orthogonal complement of this  space is exactly the closed subspace spanned by $\{e_i: i \in A\}$.


问:
convolution of sums
I have to find an infinite power series $B(x)$ in accordance to an already given infinite power series $A(x)$ so that $B(x)A(x) = 1 $

$A(x)$ is defined as follows:
$$1 + \sum_{n \ge 1 } a_n Z^n$$
when one considers the $0$. term as 1 then
$$a_0Z^0 + \sum_{n \ge 1 } a_n Z^n = \sum_{n \ge 0 } a_n Z^n$$

so far I tried to define $B(x)$ as
$$ B(x) :=  \sum_{n \ge 0 } b_n Z^n$$
with these limitations: $b_0 = 1 ; b_1= -1; b_i = 0$ for every $i > 1$

under the use of the convolution of sums
$$\sum_{n \ge 0}^{} (\sum_{k=0}^{n} a_k b_{n-k})$$
I could cancel many terms, but not the problem is that I also cancel the first and the second $1$ with $-1$

$$ a_0b_0 + (a_0b_1 + a_1b_0) + (a_0b_2 + a_1b_1 + a_2b_0) + ... = 1 + (-1 +a_1) + (0 + -a_1+a_2) + ...$$

as you can see every last term is being canceled with the term before the last one in the next brackets but the two first cancel each other as well.
How can one do it in this method so that it will fit nevertheless?

Thank you for your help

答:
Trying to guess the $b_n$ coefficients in that manner will not work. There is a useful trick in problems like these though. Put
$$C(z): = \sum_{n=1}^\infty a_nz^n,$$
so $A(z) = 1+C(z)$. In some sufficiently small neighborhood of $0$ we may assume $|C(z)|<1$, so by uniform convergence of geometric series
$$\frac{1}{A(z)} = \frac{1}{1+C(z)} = \frac{1}{1-(-C(z))} = \sum_{j=0}^\infty (-1)^j(C(z))^j.$$
In other words,
$$B(z) = 1-\bigg(\sum_{n=1}^\infty a_nz^n\bigg)+\bigg(\sum_{n=1}^\infty a_nz^n\bigg)^2-\bigg(\sum_{n=1}^\infty a_nz^n\bigg)^3+\dotsm.$$
Writing out an explicit formula for the coefficients would be quite tedious, but do note that $z^k$ only appears in the firt $k+1$ summands by the way the powers increase. We have $b_0=1$, and for $k\geq 1$ we have
$$b_k = -a_k+(C^2)_k-(C^3)_k+\dotsm+(-1)^k(C^k)_k$$
where $(C^j)_k$ denotes the sum of all terms $a_{i_1}\dotsm a_{i_j}$ where $i_r\geq 1$ and $i_1+\dotsm+i_j = k$.

Title: convolution of sums


问:
Expected value of area of rectangle within a unit square given arbitrary point $E$

Question: Suppose we have a unit square ABCD. Arbitrarily, pick a point within the interior of the unit square and call that point E. Let the line through E parallel to AB intersect AD at F and let the line through E parallel to BC intersect DC at G. What is the expected value of the area of the rectangle EFDG?

To start, I plotted everything on the coordinate plane such that $D = (0,0), C = (1,0), B = (1,1), A = (0,1),$ and $E = (x,y)~\text{where}~ 0 < x,y < 1.$ We can observe how the area is just $xy$ and the probability of this occurring would be $\frac{xy}{1} = xy$. Therefore we have to solve for $$E = \sum_{0 < x,y < 1} x^2y^2.$$ This is where I am stuck. How are we supposed to evaluate this summation? Any help would be appreciated.

答:
You want a double integral not a sum, as Greg Martin commented

and the area is $xy$ not $x^2y^2$

so you want $$\int_0^1\int_0^1 x\,y\, f(x,y)\,dx \,dy$$ where $f(x,y)=1$ since you have a uniform distribution in a unit square


问:
Computing a probability of 5 or more same values when rolling 21 fair dice 12 sides each
The way I'd like to approach the problem is to calculate separate probabilities (P(1), P(2), P(3), P(4)) and then calculate P(>=5) = 1 - (P(1) + P(2) + P(3) + P(4)) where:

P(1) is the probability of having exactly one side of the dice represented and 20 others different. For example: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 (1 is repeated exactly once)

P(2) is the probability of having one side of the dice represented exactly 2 times and 19 others different. For example: 5 1 2 3 4 2 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 (5 is repeated 2 times)

P(3) is the probability of having one side of the dice represented exactly 3 times and 18 others different. For example: 6 6 1 6 2 3 4 5 7 8 9 10 11 12 1 2 3 4 5 7 8 (6 is repeated 3 times)

P(4) is the probability of having one side of the dice represented exactly 4 times and 17 others different. For example: 7 1 2 7 3 4 7 5 6 7 8 9 10 11 12 1 2 3 4 5 6 (7 is repeated 4 times)

My issue is that I having trouble calculating the number of combinations for each example yet calculating the overall number of combinations is trivial (${12^{21}}$).
What I came up with so far is as follows:

P(1) = $\binom{21}{1}⋅12⋅{11^{20}}$

P(2) = $\binom{21}{2}⋅{12^2}⋅{11^{19}}$

P(3) = $\binom{21}{3}⋅{12^3}⋅{11^{18}}$

P(4) = $\binom{21}{4}⋅{12^4}⋅{11^{17}}$

I know duplicates aren't ruled out in these formulas but I am not sure I can come up with calculating the number of these duplicates.
To illustrate the point, suppose we have the following combination: 1 5 6 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9.
We can see that it matches the case of P(1) for 1, 5, 6, 7 and 9 which means that combination could have been repeated at least 4 times, however, the number of repetitions varies depending on the combination.
The question is similar to this one, however, I can't properly eliminate duplicates.
Hints and points would be greatly appreciated.

答:
Well, I am describing  how to compute the number of ways you can have $\le4$ repetitions w/o using inclusion-exclusion, but have to say it'll be laborious.

I arrange the configurations to be excluded in descending order, eg

$4-4-4-4-4-1-0-0-0-0-0-0$

$4-4-4-4-3-2-0-0-0-0-0-0$

$4-4-4-4-2-1-1-0-0-0-0-0\;\;$ etc down to

$2-2-2-2-2-2-2-2-2-1-1-1$

for each such configuration, find the product of two multinomial coefficients, eg for the third one in the line,

$\dbinom{21}{4,4,4,4,2,1,1,0,0,0,0,0}\dbinom{12}{4,1,2,5}$

Add up the figures to obtain ways to be excluded = say, $N$,

then $Pr = 1 - \frac{N}{12^{21}}$


问:
Does this Series Converge? and how does it converge/diverge, what method should i use
Does the series of $$\frac{n!2^n}{n^n}$$ as $n \to \infty$ converge?

I found this question while studying series and sequences and I can't find an answer for it. I tried several criteria but no luck yet. I also have just started learning this material. Maybe that's why I can't quickly detect what criteria fits these type of series. Any help would be appreciated.

答:
Note that $a_n=\dfrac{n!2^n}{n^n}>0$ $\forall n \in \mathbb N$.

If we study the series with the ratio test we have, for $n \to \infty$
$$\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!2^{n+1}}{(n+1)^{n+1}}\dfrac{n^n}{n!2^n}=\dfrac{2{ {(n+1)} }n!2^n}{(n+1)(n+1)^n}\dfrac{n^n}{n!2^n}=2\dfrac{n^n}{(n+1)^n}=$$ $$=2\bigg(\dfrac{n}{n+1}\bigg)^{n}=2\dfrac{1}{\bigg(1+\dfrac{1}{n}\bigg)^n}\longrightarrow_{n\to \infty} \dfrac{2}{e}<1$$
We can conclude that $\sum a_n$ is absolutely convergent .


答:
Do you know that $\det (AB)=\det A\cdot \det B$ for any two square matrices $A,B$?

If so, notice that $-A=(-I)A$ where $-I$ is the matrix with $-1$ on the diagonal and $0$ elsewhere. Clearly, $\det(-I)=(-1)^n$.

More generaly, $\alpha A=(\alpha I)A$ and hence $\det(\alpha A)=\alpha^n\det A$ for any scalar $\alpha$.

Title: Why is $\det(-A)=(-1)^n\det(A)$?

Body: Why is $\det(-A)=(-1)^n\det(A)$?</p>


问:
Determine the multiplicity of a root in determinant
Consider the determinant
$$\det(A(x,z))=\begin{vmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{vmatrix}.$$
It is easy to see that if we write the determinant as a bivariate polynomial $p(x,z)$, then $x=0$ and $z=0$ are its roots, i.e. $p(0,z)=p(x,0)=0$. But how can we show that these roots have multiplicity 2?

Specifically, I'm interested if it is possible to determine the multiplicity of a root in a determinant without computing the determinant explicitely. One possibility would be to use the Jacobi formula to compute, e.g., $\frac{\partial}{\partial x}\det(A(x,z))$, but this task seems to be even more complicated. Probably there are some tricks that allow us to make a conclusion about the multiplicity of a root with less effort?

EDIT: The solution is supposed to follow this scheme:

Show that $x=0$ ($z=0$) is a root of the polynomial.

Show that $x=0$ ($z=0$) is a double root of the polynomial.

Deduce that $p(x,z)=x^2z^2$.

It is the second item, where I have a problem.

答:
Adding or subtracting one row to another does not change the value of the determinant. So $$\det\begin{bmatrix}1-x&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}=\det\begin{bmatrix}1&1&1&1+z\\0&x&0&-z\\0&0&z&z\\0&0&x&x+xz\end{bmatrix}=x^2z^2$$

NB It is $x=0$ and $z=0$ that are the roots of the polynomial.

EDIT: Let's expand $\det(A(x))$ as a series in $x$:

$\det(A(0))=0$ since it contains a repeated column of $1$s. This shows that $x=0$ is a root of $p(x,z)$.

To get the first order terms in $x$, recall that determinants are linear in each column separately, hence expanding the first and second columns gives \begin{align}\det(A(x))&=\det\begin{bmatrix}1&1&1&1\\1&1+x&1&1\\1&1&1-z&1\\1&1&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&1&1&1\\0&1+x&1&1\\0&1&1-z&1\\0&1&1&1+z\end{bmatrix}\\
&=\det\begin{bmatrix}1&0&1&1\\1&x&1&1\\1&0&1-z&1\\1&0&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&1&1&1\\0&1&1&1\\0&1&1-z&1\\0&1&1&1+z\end{bmatrix}-\det\begin{bmatrix}x&0&1&1\\0&x&1&1\\0&0&1-z&1\\0&0&1&1+z\end{bmatrix}\\
&=-x^2\det\begin{bmatrix}1-z&1\\1&1+z\end{bmatrix}\end{align} In the second equation, the first two determinants cancel out by noticing that the second can be obtained from the first by swapping the first two columns and rows. It is this cancellation that annihilates the $x$ term and produces the double root $x^2$.


问:
if $C$ is a vector space over a finite field $\mathbb{F_q}$ where q is prime, and $|C|=q^n, n\in \mathbb{N}$, then $dim(C)=n$?
If $C$ is a vector space over a finite field $\mathbb{F_q}$ where q is prime, and the number of different vectors in the vector space is $q^n, n\in \mathbb{N}$, then $dim(C)=n$ ?

It seems to me very simple, since the if $dim(C)>n$ then $|C|>q^n$ which contradicts that $|C|=q^n$, and $dim(C)<n$ then $|C|<q^n$ which again contradicts that $|C|=q^n$. The question is whether it's the right way, and if I need to prove something in addition to what I said?

答:
You could reason as follows: any finite dimensional vector space $V$ of dimension $r$ over a field $K$ is necessarily isomorphic to $K^r$, and thus when $K$ is finite, we will have $|V|=|K^r|=|K|^r$.

Therefore if $|C|=q^n$, and $|K|=|\mathbb F_q|=q$, then the dimension of $C$ must be $n$.

Of course, your argument also works (and is probably equivalent), but you need to explain why $\operatorname{dim}C>n$ implies $|C|>n$ and similarly for the other case. It's not so clear how the logic flows, as it currently stands.


答:
The point $E (x,y)$ is randomly selected with uniform probability within  the unit square with $A(0,0), B(1,0), C(1,1), D(0,1)$.

Expectation $ \,E(X(1-Y)) \, = x(1-y)$ (area of rectangle $\, DGEF$).

By the way please note, $ \,E(X(1-Y))$ and $E(XY)$ are same given a square.

Joint PDF $f(x,y) = f(x) \,f(y) = 1 \, (0 \leq xy \leq 1) \,$ as,

$f(x) = 1, (0 \leq x \leq 1)$ and same for $f(y)$.

$\displaystyle E(X-XY) = \int_{y = 0}^{1} \int_{x = 0}^{1} f(x,y) \, (x-xy) \, dx \,dy$

Title: Expected value of area of rectangle within a unit square given arbitrary point $E$

Body:
Question: Suppose we have a unit square ABCD. Arbitrarily, pick a point within the interior of the unit square and call that point E. Let the line through E parallel to AB intersect AD at F and let the line through E parallel to BC intersect DC at G. What is the expected value of the area of the rectangle EFDG?

To start, I plotted everything on the coordinate plane such that $D = (0,0), C = (1,0), B = (1,1), A = (0,1),$ and $E = (x,y)~\text{where}~ 0 < x,y < 1.$ We can observe how the area is just $xy$ and the probability of this occurring would be $\frac{xy}{1} = xy$. Therefore we have to solve for $$E = \sum_{0 < x,y < 1} x^2y^2.$$ This is where I am stuck. How are we supposed to evaluate this summation? Any help would be appreciated.


问:
Evaluate this limit of integration in Complex analysis
I am trying assignments of Complex analysis and I couldn't evaluate this particular integral.

Find the limit $\int_{|z|=R} |\sin(z)/z | |dz| $ as $R\to\infty$.

I am unable to see which result I should use and also there is confusion due to dz being in modulus and also $\sin(z) /z$.

Can you please tell significance of modulus here and how to compute the limit.

Thanks!!

答:
First note that
$$
 \int_{|z|=R} \frac{|\sin(z)|}{|z|} |dz| = \int_0^{2 \pi} \frac{|\sin(R e^{it})|}{|R e^{it}|} R \, dt = \int_0^{2 \pi} |\sin(R e^{it})| \, dt \, .
$$

Generally, for $z= x+iy$,
$$
 |\sin(z)|^2 = \sin^2(x) + \sinh^2(y) \implies |\sin(z)| \ge |\sinh(y)| \, .
$$
If we restrict the integration to the arc $z = R e^{it}$ with $\pi/4 \le t \le 3\pi /4$ then
$$
|\sin(R e^{it})| \ge \sinh(R \sin(t)) \ge \sinh(\frac{R}{\sqrt 2}) \, .
$$
(This is what @reuns meant with the comment that “$|\sin(z)|$ is large near $iR$”).

It follows that
$$
\int_0^{2 \pi} |\sin(R e^{it})| \, dt \ge
\int_{\pi/4}^{3 \pi/4} |\sin(R e^{it})| \, dt \\
\ge \int_{\pi/4}^{3 \pi/4} \sinh(\frac{R}{\sqrt 2}) \, dt = \frac \pi 2 \sinh(\frac{R}{\sqrt 2}) \, ,
$$
and that grows to infinity for $R \to \infty$.


问:
What is this question in laymen terms, and how do I interpret it? "What is x in terms of a if the product $x*(a-x)$ is a maximum?"
Not sure if this is an appropriate math question because it's to do more with comprehension(my lack thereof).

From the answer, it seems like the question is asking "for the maximum value" x in terms of a. Although it doesn't state this outright.

How do we interpret this question without having first seen the answer..?

Where does the first sentence, $a=x+(a-x)$ come into play solving this?

答: You can interpret that you have a stick of fixed lenght, $a$. Then the question is how can you break it in two in such a way that you can construct the rectangle with the bigger area.


问:
Prove that $a^{(p-1)/2} $$\equiv$-1 (modp). Deduce that if $a, b$ are primitive roots modp, then $a\times b$ is NOT a primitive root mod p.

Let $a$ be a primitive root mod the odd prime p.  Prove that $a^{(p-1)/2} $$\equiv$-1 (modp).  Deduce that if $a, b$ are primitive roots modp, then $a\times b$ is NOT a primitive root mod p.

Here we define that, $a$ is a primitive root of $n$ if $(a,n)=$1 and $a$ has order exactly $\phi(n)$  (where the order refers to the least such $k$ where $a^k$ $\equiv$ 1 (modn), $k \geq 1$) (and $\phi$ denotes Euler's totient function.)

I have started by noting that $a^{(p-1)/2}$ $\equiv \pm 1 $ mod $p$ by Fermats Little Theorem and equating $y=a^{(p-1)/2}$

Also Eulers criterion tells us that $(\frac{a}{p})$ is congruent to  $a^{(p-1)/2}$ mod p

Also, Legendre tells us that $(\frac{a}{p})$ =-1 if $a$ is a quadratic non-residue of $p$

So putting this together, am I to show that $a$ being a primitive root mod p implies that it is a quadratic non-residue mod p?

Can't quite get my head round this problem.  Help appreciated,  thanks in advance.

答:
Using Fermat's Little Theorem, $$a^{p-1}\equiv1\pmod p$$

$\implies p$ divides $(a^{p-1}-1)\implies p$ divides $(a^{\frac{p-1}2}-1)(a^{\frac{p-1}2}+1)$

If $a$ is a primitive root, $ord_pa=p-1\implies a^r\not\equiv 1\pmod p$ for $r<p-1$

$\implies p$ does not divide $a^{\frac{p-1}2}-1$

$\implies p$  divides $a^{\frac{p-1}2}+1$

If $a,b$ are primitive roots $a^{\frac{p-1}2}\equiv-1\pmod p$ and $b^{\frac{p-1}2}\equiv-1\pmod p$

$\implies (ab)^{\frac{p-1}2}\equiv(-1)^2\pmod p\equiv1$

So, $ord_p(ab)$ divides $\frac{p-1}2<p-1\implies ord_p(ab)<p-1$


问:
Making single product into double product
I have this product: $\prod_{i > j} ( i^2 - j^2 ) $ and I am trying to turn it into either a double product with one for $j$ and the other for $i$ or keep as a single product and rearrange $j$ in terms of $i$. (Note the $i,j$ are for matrix indexes).

I first thought $\prod_{i = 1} ( i^2 - (i-1)^2 ) $ however testing this a few times shows that it does not work... can anyone help me? (The products are up to $n$)

Note I am trying to remove $i>j$ from my product by splitting into two or some other way

Body: You can write it as $\prod_{i=1}^n\prod_{j=1}^{i-1}(i^2-j^2)$ or $\prod_{j=1}^n\prod_{i=j+1}^n(i^2-j^2)$.


问:
Question about a set of distinct odd integers and their least common multiple
Let $S=\{x_1,x_2,...,x_n\}$ be some set of distinct odd integers greater than 1, such that $\forall  x_i\in S$ it holds that $x_i<\sqrt{\text{lcm} (x_1,x_2,...x_n)}$ and $x_i\mid \text{lcm} (x_1,x_2,...x_n)$ (where $\mid$ means "divides").

I was wondering if:

is it possible that $1+\sum_{k=1}^{n}x_k = x_ix_j$, where $x_i\in S$ and $x_j\in S$, $x_ix_j >\sqrt{\text{lcm}\{x_1,x_2,...,x_n\}}$ and $x_ix_j\mid \text{lcm} (x_1,x_2,...x_n)$ ?

I am trying to find an example of this equality, but have not been able yet. I am wondering if this kind of equality is even possible, but I am stuck when trying to prove if such a set is possible or not. Any help on this would be really welcomed!

答:
(It's late and it might be possible that this doesn't work. If so, let me know and I'd delete it.)

Show that this works:

$ S = \{3, 5, 9, 21, 25, 27, 35, 63\}$

$LCM = 4725, \sqrt{LCM} \approx 68.7$

$x_i = 9, x_j = 21$.


答: The Legendre symbol is unimportant here.  $a^{(p-1)/2}\equiv \pm 1$, and it's not $1$ since $a$ is a primitive root, so it's $-1$.  </p>

Body:
Let $a$ be a primitive root mod the odd prime p.  Prove that $a^{(p-1)/2} $$\equiv$-1 (modp).  Deduce that if $a, b$ are primitive roots modp, then $a\times b$ is NOT a primitive root mod p.

Here we define that, $a$ is a primitive root of $n$ if $(a,n)=$1 and $a$ has order exactly $\phi(n)$  (where the order refers to the least such $k$ where $a^k$ $\equiv$ 1 (modn), $k \geq 1$) (and $\phi$ denotes Euler's totient function.)

I have started by noting that $a^{(p-1)/2}$ $\equiv \pm 1 $ mod $p$ by Fermats Little Theorem and equating $y=a^{(p-1)/2}$

Also Eulers criterion tells us that $(\frac{a}{p})$ is congruent to  $a^{(p-1)/2}$ mod p

Also, Legendre tells us that $(\frac{a}{p})$ =-1 if $a$ is a quadratic non-residue of $p$

So putting this together, am I to show that $a$ being a primitive root mod p implies that it is a quadratic non-residue mod p?

Can't quite get my head round this problem.  Help appreciated,  thanks in advance.


答:
Usually in the context of calculus, if someone is asking for the maximum, they want to know the largest possible value of a given quantity. In this case, that quantity is $x \cdot (a-x)$. Note that in this quantity, we think of $a$ as being a fixed value. This means the quantity varies only when $x$ varies too.

So the question is asking you: what is the value of $x$ that gives you the maximum possible value of $x \cdot (a-x)$ (the quantity)?

I'm not sure what relevance the sum has; it's trivial and not needed to complete the problem. (trivial in the sense that even if they hadn't said it, I'm sure you can immediately see it's true!)

Title: What is this question in laymen terms, and how do I interpret it? "What is x in terms of a if the product $x*(a-x)$ is a maximum?"

Body:
Not sure if this is an appropriate math question because it's to do more with comprehension(my lack thereof).

From the answer, it seems like the question is asking "for the maximum value" x in terms of a. Although it doesn't state this outright.

How do we interpret this question without having first seen the answer..?

Where does the first sentence, $a=x+(a-x)$ come into play solving this?


问:
Definable sets properties that uniquely determine an $\mathscr{L}$-structure
$\newcommand{\Def}{\mathsf{Def}}$
It's known that the set $\Def(\mathfrak{M}, S)$ of $S$-definable sets over an $\mathscr{L}$-structure $\mathfrak{M}$ is the smallest subset of $\bigcup_n M^n$ which

is closed under Boolean set operations ($\bigcap$, $\bigcup$, $\cdot^C$), cartesian product, permutation of the coordinates, and linear projections $\pi: \begin{array}{ccc}M^{n+m} &\to& M^n\\(\bar{x}, \bar{y}) &\mapsto& \bar{x}\end{array}$ ;

contains $\varnothing$, $M^n$ for $n \in \mathbb{N}$, the diagonals $\Delta^n_{i, j} =
\{\bar{a} \in M^n : a_i = a_j\}$, the singletons $\{s\}$ for all $s \in
S$, and the sets $\{c^\mathfrak{M}\}$ for all constant symbols $c$, the graph of $f^\mathfrak{M}$ for all function symbols $f$ and $R^\mathfrak{M}$ for all relation symbols $R$.

Now I came across a remark saying that given a language $\mathscr{L}$ and a set $M$, a subset $\mathscr{D} = \bigcup_n \mathscr{D}_n$ of $\bigcup_n M^n$ which satisfying both properties above, determines uniquely an $\mathscr{L}$-structure on $M$. Namely, the interpretation of symbols in $\mathscr{L}$ are specified by $\mathscr{D}$.

Question: How one could "recover" the interpretation of constants, functions and relations symbols from $\mathscr{D}$ ?

答:
I'm not sure where you got this remark from, but it is fairly easy to see that it is not true as written.

For example, if you have a language with two unary predicates $P_1, P_2$ and $X\supseteq Y_1, Y_2$ are any sets with $Y_1\neq Y_2$, then $M_1=(X,Y_1,Y_2)$ and $M_2=(X,Y_2,Y_1)$ give rise to exactly the same $\mathscr D$.

There are two similar observations which are actually true.

One, if you have a $\mathscr D$ and you know which sets correspond to which function or relation symbols, then you can recover the underlying $\mathscr L$-structure. But this is quite trivial, since this is basically the same as handing you the interpretations of symbols, plus some extra data which you don't care about.

The other, slightly deeper one, is that given a family $\mathscr D_0$ of subsets of $\bigcup_n M^n$ satisfying the obvious axioms (being closed under projections, products and Boolean operations, containing each $M^n$ and each diagonal --- although it is enough to include the diagonal in $M^2$, as the others follow), you can endow $M$ with a first order structure such that the resulting $\mathscr D$ is equal to $\mathscr D_0$, and moreover, this structure is unique up to interdefinability.

This is only slightly less trivial: namely, it is enough to consider $\mathscr L=\{R_D\mid D\in \mathscr D_0 \}$, and then you can interpret each $R_D$ as $D$. It is easy to see that this yields $\mathscr D=\mathscr D_0$.

The uniqueness part is actually a separate observation: if $\mathfrak M_1,\mathfrak M_2$ share the same universe and give rise to the same $\mathscr D$, then they are interdefinable, pretty much by definition.


问:
How to solve the following geometry problem?
The diameters of the circumcircle of triangle $ABC$ drawn from $A$,$B$ and $C$ meet $BC$, $CA$ and $AB$, respectively, at $L$, $M$ and $N$. Prove that: $$\frac{1}{ AL} + \frac{1}{BM}+\frac{1}{CN}=\frac{2}{R}$$ I tried with formulas based on properties of triangle and some with geometry properties with circumcircle but at the end not able to reach the result, I earnestly urge for your help.

答:
We have $$\dfrac{OA}{AL}=\dfrac{[AOC]}{[ALC]}=\dfrac{[AOB]}{[ALB]}$$$$=\dfrac{[AOC]+[AOB]}{[ALC]+[ALB]}=\dfrac{[AOC]+[AOB]}{[ABC]}$$

Similarly $$\dfrac{OB}{BM}=\dfrac{[AOB]+[BOC]}{[ABC]}$$

$$\dfrac{OC}{CN}=\dfrac{[BOC]+[AOC]}{[ABC]}$$

Summing up we get $$\dfrac{R}{AL}+\dfrac{R}{BM}+\dfrac{R}{CN}=\dfrac{2[ABC]}{[ABC]}=2$$


问:
Why I'm not getting contradiction for $B$?
Prove that $\mathbb{R}$ is connected.

My attempt: Suppose $\mathbb{R}$ is not connected. Then we can write $\mathbb{R}= A\bigcup B$ where $A$ and $B$ are non empty  open disjoint set .
Fixed $a \in A$ and $b \in B$

Take $G=\{ x\in \mathbb{R} | [ a, x] \subseteq A\}$.Here $G$ is bounded by $b$  .Now used the completeness axiom  of $\mathbb{R}$ , then $G$ has  upper  bound  call it $c$  .Now we have   $\sup G = c$

Since $ A$ and $B$ cover all of $\mathbb{R}$ , so $c$ must be lie  either in $A$ or $B$

Since  $A$ is open , so there exist  open ball $B_{\epsilon}(c) \subseteq A$.Now for any $y \in B_{\epsilon}(c)$ some $ y$ such that  $c < y <c + \epsilon$  we have  $[a,y] \subseteq A$.
This put $y$ also in $G$ since $B_{\epsilon}(c) \cap G\neq \emptyset$
But here $c+ \epsilon $ is the upper bound so we are getting contradiction that $c$ is suppose to be upper bound .
Therefore $ c\notin A$

Now for $B$

Since  $B$ is open , so there exist  open ball $B_{\epsilon}(c) \subseteq B$.Now for any $y \in B_{\epsilon}(c)$ some $ y$ such that  $c-\epsilon < y <c $  we have  $[a,y] \subseteq A$.
This put $y$ also in $G$ since $B_{\epsilon}(c) \cap G\neq \emptyset$.Here $c$ is the upper bound as same as upper bound of $G$

Therefore $c$ must be lie on $B$

But here I have to contradicts both $A$ and $B$ .

I'm not able to find out my mistake. Where am I doing mistake?

My doubt: why I'm not getting contradiction for $B$?

答: Since $c-\epsilon<\sup G$, there is some $y\in G$ such that $c-\epsilon<y<c$. Since $y\in G$ you know $[a,y]\subseteq A$. In particular, $y\in A$. But $y\in B_\epsilon(c)\subseteq B$. So $y\in A\cap B$.


问:
A point $p$ of a variety $X$ is smooth if and only if it is smooth in $U\subset X$ an open sub variety.
I want to show the following:

Let $X$ be an algebraic variety, let $U$ be an open subvariety of $X$ containing $p$. Then $X$ is smooth at $p$ if and only if $U$ is smooth at $p$.

Here is my definition of being smooth at a point

Definition: Let $X$ be an algebraic variety. We say $X$ is smooth at $p$ if there is an open subvarity $U$ of $X$ containing $p$ and an isomorphism $\varphi: U \rightarrow Z(f_1,..,f_{n-d})\subset \mathbb A^n$ for some $d\leq n$ for some $f_1,..,f_{n-d}$ such that the rank of $$\left(\frac{\partial f_i}{\partial x_j}(\varphi(p)\right)$$
is maximal.

Attempt:

($\Leftarrow$)
Let $U$ be smooth at $p$, then there is some $\overline{U}$ an open subvariety of $U$ and $\varphi:U\rightarrow Z(f_1,...,f_{n-d})$ satisfying the smoothness criterion. Since $U$ is a subvariety of $X$ so to is $\overline{U}$. Again this $\varphi:U\rightarrow Z(f_1,...,f_{n-d})$ still satisfies the smoothness crieterion so we are done.

($\Rightarrow$)
Let $X$ be smooth at $p$, then there is some $\overline{U}$ an open subvariety of $X$ and an isomorphism $\varphi:\overline{U}\rightarrow Z(f_1,..f_{n-d}$ satisfying the smoothness criterion.

We want an open subvariety of $U$, define $V=U\cap\overline{U}$ this certainly is an open subvariety of $U$. And $\varphi\mid_V$ is an isomorphism onto its image. But how do we know its image is of the form $Z(f_1,...,f_{n-d})$ and how do we know that the Jacobian of the restriction still has full rank?

答:
If $X:=V(I)\subseteq \mathbb{A}^n_k$ is an affine algebraic variety over a field $k$ of characteristic zero (the ideal $I$ is assumed to be a prime ideal) it follows from Hartshorne's book (HH) and Thm I.3.2 that $\Gamma(X,\mathcal{O}_X)=A$ where $A:=k[x_1,..,x_n]/I$. Again by   HH Thm I.3.2 it follows $\mathcal{O}_{X,p} \cong A_{\mathfrak{p}}$ where $\mathfrak{p} \subseteq A$   is the prime ideal corresponding to the point $p\in X$. By HH.II.1 (the section on stalks) it follows the stalk $\mathcal{O}_{X,p}$ is the limit of $\mathcal{O}_X(V)$ over all open subsets $V$ containing $p$. We may restrict the sheaf $\mathcal{O}_X$ to the open set $U$ to get the structure sheaf $\mathcal{O}_U$ on the open sub-variety $U$. It follows from the defnition there is an isomorphism $\mathcal{O}_{X,p} \cong \mathcal{O}_{U,p}$.
Hence the sheaves $\mathcal{O}_X$ and $\mathcal{O}_U$ have the same stalk at the point $p$. Again by HH.I.5 we define the point $p$ to be non-singular using the Jacobian matrix of a set of generators $f_i$ of the ideal $I$ defining $X$. This definition is independent of choice og generators $f_i$ of the ideal $I$. By HH.I.5.1 it follows $p$ is non-singular iff the local ring $\mathcal{O}_{X,p}$ is a regular local ring. Since $\mathcal{O}_{X,p}\cong \mathcal{O}_{U,p}$ it follows we may define non-singularity using any open set $U$ containing $p$. Hence we may use the local ring at $p$ to define non-singularity for any variety of finite type over $k$, where $k$ is a field of characteristic zero.

Example 0. If $X$ is a (projective) variety of finite type over $k$ and $p \in X$ is a closed point, we may choose an open affine set $p\in U \subseteq X$. By definition $U:=Spec(A)$ with $A=k[x_1,..,x_n]/I$, where $I:=\{f_1,..,f_l\}$ is a set of polynomials. Then we may use the Jacobian criterion to check if $p$ is a non-singular point of $X$. This does not depend on the choice of open affine subset $U$ containing $p$ and the choice of generators $f_i$ of the ideal $I$.

Let us prove your claim using the Jacobian criterion. Let $U:=Spec(A)$ and $V:=Spec(B)$ be two open affine subvarieties containing $p$ with $A:=k[x_1,..,x_n]/(f_1,..,f_l)$ and $B:=k[y_1,..,y_n]/(g_1,..,g_t)$. Let $U$ be smooth/non-singular at $p$ by the Jacobian criterion. It follows from HH that $\mathcal{O}_{U,p}$ is a regular local ring. There is an isomorphism of local rings $\mathcal{O}_{U,p}\cong \mathcal{O}_{V,p}$ (this is a formal property of limits, and the proof may be found in Atiyah-Macdonalds (AM) book on commutative algebra), hence it follows $\mathcal{O}_{V,p}$ is a regular local ring. Again by HH it follows the Jacobian criterion holds for the generators $g_i$ at the point $p \in V$. The converse is similar and the claim is proved.

Note: This proof uses Chapter I in HH which is "elementary", a formal result on limits in AM - this is also "elementary" - and the definition of the stalk of a sheaf in Chapter II in HH. This is also an "elementary" definition.  Hence this proof is in a sense "elementary". To students: Do you think this proof is "elementary enough" or is there a more "elementary" proof "somewhere out there"?

Example 1. There are problems arising in characteristic $p>0$. Let $k$ be a field of characteristic $p$ and let $f(x,y) \in k[x^p,y^p]$ be any polynomial. It follows $\partial f/\partial_x=\partial f/\partial_y=0$, hence the Jacobian matrix of $f$ is identically zero.

If $k$ is any algebraically closed field and $f:X \rightarrow Y$ a morphism of schemes of finite type over $k$, we say "f is smooth of relative dimension $n$" if the following holds:

$f$ is flat.
For any irreducible components $X' \subseteq X$ and $Y' \subseteq Y$ with $f(X') \subseteq Y'$ it follows $dim(X')=dim(Y')+n$.
For any point $x\in X$ it follows $dim_{\kappa(x)}(\Omega_{X/Y}\otimes \kappa(x))=n$.

Theorem III.10.2 in Hartshorne says that $f$ is smooth of relative dimension $n$ iff $f$ is flat and for any $y \in Y$ it follows the fiber $X_y:=f^{-1}(y)$ is geometrically regular, which means that the fiber product $X_y \times Spec(\overline{\kappa(y)})$ is regular (meaning all local rings are regular local rings). Here $\overline{\kappa(y)}$ is the algebraic closure of $\kappa(y)$. Hence if $Y=Spec(k)$ with $k$ an algebraically closed field it follows the morphism $f:X \rightarrow Spec(k)$ is smooth iff the local ring $\mathcal{O}_{X,p}$ of $p \in X$ is a regular local ring for any point $p$. Hence if the base field $k$ is algebraically closed we may use the module of relative differentials $\Omega^1_{X/Y}$ to define "regularity/non-singularity": We define $X$ (a scheme of finite type over an algebraically closed field $k$) to be "non-singular" iff the canonical map $f:X\rightarrow  Spec(k)$ is smooth of relative dimension $dim(X)$.

Example 2: Let $k$ be an algebraically closed field of characteristic $p>0$ and let $f(x_1,..,x_n)\in k[x_1^p,...,x_n^p]$ be a polynomial. It follows there is a polynomial $g(x_1,..,x_n)$ with $g^p=f$, hence $f$ is never irreducible.

The module of Kahler differentials is "functorial" in the following sense: If $f: X \rightarrow Y$ is a morphism of schemes and $i:U \rightarrow X, j:V \rightarrow Y$ are open subschemes with $f(U) \subseteq V$, it follows there is an isomorphism of sheaves $i^*\Omega^1_{X/Y} \cong \Omega^1_{U/V}$. Hence for any point $p \in U$ it follows there are isomorphisms at the stalks $\Omega^1_{U/V,p}\cong \Omega^1_{X/Y,p}$ and at the fiber $\Omega^1_{U/V}\otimes \kappa(p) \cong \Omega^1_{X/Y}\otimes \kappa(p)$.  Hence the stalk and fiber of the module of Kahler differentials at $p$ does not depend on the choice of open subscheme $U \subseteq X$ containing $p$. We may choose $U:=Spec(B)$ and $V:=Spec(A)$ and then $i^*\Omega^1_{X/Y}$ is the sheafification of the module $\Omega^1_{B/A}$ of relative differentials.

Example 3: If $f(x_1,..,x_n)\in k[x_1^p,..,x_n^p]$ is any polynomial and $char(k)=p>0$ it follows $\partial f/\partial_{x_i}=0$ for all $i$. It follows the differential $df:=\sum_i \partial f/\partial_{x_i} dx_i=0$. If $X:=Spec(k[x_1,..,x_n]/(f))$ it follows the cotangent bundle $\Omega^1_{X/k}$ is a trivial $\mathcal{O}_X$-module of rank $n$.


答:
Hint 1: For odd primes $p\geq 5$, $\frac{p-1}{2}$ is an integer $\geq 2$. If $a^{\frac{p-1}{2} } = 1 $, then $a$ is not a primitive root by definition.

Check $p=3$ and see if the statement holds.

Hint 2: $ab^{\frac{p-1}{2} } = a^{\frac{p-1}{2} } \times b^{\frac{p-1}{2} } $.

Title: Prove that $a^{(p-1)/2} $$\equiv$-1 (modp). Deduce that if $a, b$ are primitive roots modp, then $a\times b$ is NOT a primitive root mod p.