问:
Proof verification: lim$(A)=$ lim$(A\setminus\{x\})$
Let $(X,\tau)$ be a $T_{1}$-topological space, $x\in X$ and $A\subset X$ be a subset. Show that lim$(A)= $ lim$\left(A\setminus\{x\}\right)$. (I know that the set of limit points is closed, but I want to prove this fact without using closedness)

Now, if $x\notin A$, then $\left(A\setminus\{x\}\right)=A$ and we are done. So, let $x\in A$. For the "$\supset$" part notice that for any $y\in A$ we have $\left(A\setminus\{x,y\}\right)\subset \left(A\setminus\{x\}\right)$, so clearly for every neighborhood $U_{y}$ we can conclude that

\begin{align*} \emptyset\neq U_{y}\cap \left(A\setminus\{x,y\}\right)\subset U_{y}\cap \left(A\setminus\{x\}\right). \end{align*} So $y$ fulfills the definition of being a limit point of $A$.

Now for the "$\subset$" part. Suppose that lim$\left(A\right)$ is not a subset of lim$(A\setminus\{x\})$. So there exists $y\in\text{lim}(A)$, but $y\notin\text{lim}(A\setminus\{x\})$, meaning that for all neighborhoods $U_{y}$ we have that $U_{y}\cap \left(A\setminus\{y\}\right)\neq\emptyset$, but there exists a neighborhood $W_{y}$ such that $W_{y}\cap\left(A\setminus\{x,y\}\right)=\emptyset$. It follows for the choice of $W_{y}$ that $W_{y}\cap \left(A\setminus\{y\}\right)\neq\emptyset$, but then we can conclude that $W_{y}\subset\{x\}$, which is absurd.

Is this a correct proof?

答:
If $X$ is Sierpiński space $\{0,1\}$ (with $0$ isolated) and $A=\{0\}$ and $x=0$ then $\lim (A) =\{1  \} \neq \emptyset = \lim (A\setminus \{x\})$.

If $X$ is $T_1$, then it does hold (Sierpiński space is merely $T_0$): assuem WLOG that $x \notin A$; if $y \in \lim(A\setminus\{x\})$ every neighbourhood $U$ of $y$ intersects $A\setminus \{x\})$ in a point distinct fom $y$, so in particular in a point of $A$ distinct from $y$, so $y \in \lim(A)$. If $y \in \lim(A)$, let $U$ be an open neighbourhood of $y$. Case 1: $y=x$ and $U$ intersects $A\setminus\{y\}=A\setminus\{x\} = A\setminus\{x,y\}$, which implies $y \in \lim(A\setminus\{x\})$. Case 2: $y \neq x$ and then $U\setminus\{x\}$ is also an open neighbourhood of $y$ by $T_1$-ness and hence intersects $A\setminus \{y\}$, so $U$ intersects $A\setminus\{x,y\}$ and again $y \in \lim(A\setminus\{x\})$. QED.


问:
I forgot how to solve two systems of equations, I am trying to solve these two equations?
$1 = \sqrt{r^{2}-(1-a)^{2}}$ and $2 = \sqrt{r^{2}-(4-a)^{2}}$.

ContentLicense: CC BY-SA 4.0

Body:
$$1=r^2-1+2a-a^2$$
$$4=r^2-16+8a-a^2$$
so $r^2=2-2a+a^2=20-8a+a^2 \iff 6a=18 \iff a=3 $
$$r^2=5 \iff r=\sqrt{5} \lor r=-\sqrt{5}$$


问:
Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant
Is the proof of these identities possible, only using elementary differential and integral calculus? If it is, can anyone direct me to the proofs? ( or give a hint for the solution )

1)$$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi} $$

2)$$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6} $$

3) $$\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx$$

and lastly,

4) $$\zeta(s) = \frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)} \prod_\rho \left(1 - \frac{s}{\rho} \right) e^{s/\rho}\!$$

I personally think the last is obtained from a simple use of the Weierstrass factorization theorem. I'm unsure as to what substitution is used.

$\gamma$ is the Euler-Mascheroni constant and $\zeta(s)$ is the Riemannian Zeta function.

Thanks in advance.

答:
The general recipe for integrals containing logarithms is to replace $\ln x$ by $\displaystyle\left[\frac{\partial}{\partial s}x^{s}\right]_{s=0}$, then to exchange the order of integration and differentiation, and evaluate the integral in terms of gamma functions. Then the derivatives with respect to $s$ will produce digamma functions and their derivatives (often related to particular values of $\zeta(x)$).

For example, the second integral can be written as
\begin{align}
\int_0^{\infty}e^{-x}\ln^2x\,dx&=\left[\frac{\partial^2}{\partial s^2}\int_0^{\infty}e^{-x}x^{s}dx\right]_{s=0}=\\
&=\left[\frac{\partial^2}{\partial s^2}\Gamma(s+1)\right]_{s=0}=\\&=\gamma^2+\frac{\pi^2}{6}.
\end{align}
A bit more sophisticated example of this technique can be found here.

ContentLicense: CC BY-SA 3.0


问:
For what $n$ is $\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ an integer?

For what $n$ is $\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ an integer?

I tried setting $a=\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}$ , $b=\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ and playing with their squares/fourth powers, but couldn't find anything.

答:
If you set the expression equal to $x$ and square the equation, you get $$x^2=25+2\sqrt{n}.$$ This is equivalent to $$n=\left(\frac{x^2-5^2}{2}\right)^2.$$ So just pick $x$ equal to an odd integer and then this above expression is $n.$

As Crostul points out, in order for the original expression to be well-defined, we need $n$ to be less than or equal to $\frac{625}{4}=156.25.$ And the fact that $n$ is a square means that it must be non-negative.


问:
My try of the Birthday Paradox
I have seen the solution of the paradox on Wikipedia, but before that I tried it out myself. But I got my answer wrong with my approach.

To calculate the probability of no two people in a group of 23 having the same birthday, I divided {the number of ways 23 different days in 365 days can be taken} by {the number of ways 22,21,20,...,2,1 different days can be taken multiplying with the number of ways 23 people can have 22,21,20,...,2,1  birthdays respectively}

I think that would make it a bit harder to understand what I wanted to convey.
And, I find using MathJax a bit tough as I'm new to this. So the link provides the image of what I tried on desmos:My expression of what I tried

答:
You have written $$1-\dfrac{{365\choose 23}}{\sum\limits_{i=1}^{23} i^{23-i}{365 \choose i}}$$ presumably as the probability that there are at least two people sharing a birthday, with $i$ iterating over the possible numbers of distinct birthdays.

Both the numerator and the denominator raise issues.  The numerator is a combination count but if you are counting equally likely events then you want permutations: if there are two people, it is twice as likely that one has a birthday on January 1 and another on January 2 compared with both having birthdays on January 1. So you should have $\frac{365!}{(365-23)!}$ as the numerator.

The other problem is the denominator.  It gives the correct number when $i=1$ of $365$ but not for other $i$.  For example when $i=2$ it gives $2^{22} {365 \choose 2}$ when the number of ways of allocating $23$ people between two days with both used  is $(2^{23}-2){365 \choose 2}$, almost twice as high.  But you do not need a sum over $i$ anyway: the number of equally likely ways of allocating $23$ birthdays (possibly duplicated) over $365$ days is $365^{23}$.

Putting these together means the probability of at least two people out of $23$ sharing a birthday  is $$1-\dfrac{\frac{365!}{(365-23)!}}{365^{23}} \approx 0.5073$$ i.e. just over a half.

Title: My try of the Birthday Paradox


问:
Difference between thin-plate and willmore energy
The thin-plate energy functional is defined by :
$$
E = \int \kappa_1^2 + \kappa_2^2
$$
while the Willmore energy functional is defined by :
$$
W = \int (\kappa_1 - \kappa_2)^2
$$

For a closed surface, if we develop $W$:
$$
W = \int \kappa_1^2 + \kappa_2^2 - 2\int K = E  -4\pi\chi
$$
where $\chi$ is the Euler characteristic of the surface that appears following the Gauss-Bonnet theorem.

In conclusion, for a closed surface, the thin-plate and the Willmore energies differ by $4\pi\chi$.

Is it correct?

If we look at the Euler-Lagrange equations related to these energies we have :
$$
\Delta^2 f = 0
$$
for the thin-plate energy, and :
$$
\Delta H + 2H(H^2-K) = 0
$$
for the Willmore energy.

But since these energies differ by a topological invariant of the surface, a critical point of one energy would be also critical for the other, which means the Euler-Lagrange equations above are equivalent.

Where am I wrong ?

答:
The biharmonic equation $\Delta^2 f = 0$ is NOT the Euler-Lagrange equation related to the energy defined by that first integral you posted.

Rather, it's what you get when you apply a thin plate approximation: $\nabla f \approx 0$. Under that, $\kappa_1^2 + \kappa_2^2 \approx f_{xx}^2 + 2 f_{xy}^2 + f_{yy}^2$. Applying Euler-Lagrange to that last expression leads to the (notably linear) biharmonic equation.


问:
Realizing a right group action as a contravariant functor
Let $G$ be a group regarded as a one-object category $\mathcal{G}$. Then a functor $F:\mathcal{G}\rightarrow \text{Set}$ consists of a set $S$ (the value of the unique object $G$), together with for each $g\in G$ a function $F(g) :S\rightarrow S$ satisfying the functoriality axioms. So, writing $F(g)(s)=g\cdot s$ it follows that $F$ amounts to a set $S$ together with a function $G\times S\rightarrow S$, $(g,s)\mapsto g\cdot s$ satisfying $(g'g)\cdot s = g'\cdot(g\cdot s)$ and $1\cdot s= s$ for all $g,g'\in G$ and $s\in S$. So we see that $F$ describes a set equipped with a left group action.

Now, I'm having trouble seeing why a functor $F:\mathcal{G}^{\rm op}\rightarrow \text{Set}$ describes a right action of $G$ on a set $S$. If $g,g'\in G$, then $F(gg')=F(g')g(g)$ since $F$ is now a contravariant functor. However, I'm having trouble seeing why this satisfies the axioms of a right group action. What am I missing?

答: Defining $S \times G \to S$ by $(s,g) \mapsto s \cdot g := F(g)(s)$, the axiom $(s \cdot g) \cdot h = s \cdot (gh)$ follows from $F(gh) = F(h) \circ F(g)$, since $$s \cdot (gh) = F(gh)(s) = \big( F(h) \circ F(g) \big)(s) = F(h) \big(F(g)(s) \big) = F(h)(s \cdot g) = (s \cdot g) \cdot h.$$


问:
About commutators and center o a certain group
Let $G$ be a group and $H, K\unlhd\ G$ (normal subgroups), with $K\leq H$ such that $\left[H, Aut(G)\right]\leq K$. Why does this imply that $H/K\leq Z(G/K)$? $Z(G)$ denotes the center of $G$.</p>

答:
Here is my comment as an answer with the details:

We know that for all $\varphi\in\rm{Aut}(G)$ and all $h\in H$ we have $h^{-1}\varphi(h)\in K$. In particular, this holds for the automorphisms $\varphi_g$ given by $\varphi_g(x) = g^{-1}xg$ (for $g\in G$).

Thus, we have that for all $h\in H$ and $g\in G$, $h^{-1}\varphi_g(h) = h^{-1}g^{-1}hg = [h,g]\in K$ which precisely means that $H/K\leq Z(G/K)$.


答:
If you wite

$$
x \cdot g = F(g)(x)
$$

then $x \cdot 1 = x$ and

$$
x \cdot gg' = F(gg')(x) = F(g')(F(g)(x)) = (x \cdot g) \cdot g'.
$$

A more conceptual way to think about this would be to realize that $B(G^{op}) \simeq (BG)^{op}$ and a right $G$-action is a left $G^{op}$-action.

Title: Realizing a right group action as a contravariant functor

Body:
Let $G$ be a group regarded as a one-object category $\mathcal{G}$. Then a functor $F:\mathcal{G}\rightarrow \text{Set}$ consists of a set $S$ (the value of the unique object $G$), together with for each $g\in G$ a function $F(g) :S\rightarrow S$ satisfying the functoriality axioms. So, writing $F(g)(s)=g\cdot s$ it follows that $F$ amounts to a set $S$ together with a function $G\times S\rightarrow S$, $(g,s)\mapsto g\cdot s$ satisfying $(g'g)\cdot s = g'\cdot(g\cdot s)$ and $1\cdot s= s$ for all $g,g'\in G$ and $s\in S$. So we see that $F$ describes a set equipped with a left group action.

Now, I'm having trouble seeing why a functor $F:\mathcal{G}^{\rm op}\rightarrow \text{Set}$ describes a right action of $G$ on a set $S$. If $g,g'\in G$, then $F(gg')=F(g')g(g)$ since $F$ is now a contravariant functor. However, I'm having trouble seeing why this satisfies the axioms of a right group action. What am I missing?


问:
Solve the equation $\tan(x-45 °)+\cot x =2$ for $0°\le x\le180°$
This is A levels trigonometry I am doing self studying so I have no teacher if someone could help me that would be great.</p>

答:
We have:

$$\tan(x - 45^{\circ}) + \cot(x) = 2$$

Using $\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}$:

$$\frac{\tan(x) + \tan(-45^{\circ})}{1 - \tan(x)\tan(-45^{\circ})} + \cot(x)=2$$

Using $\tan(-45^{\circ}) = -1$ and $\cot(x) = \frac{1}{\tan(x)}$:

$$\frac{\tan(x) - 1}{1+\tan(x)} + \frac{1}{\tan(x)} = 2$$

Cross-multiplying:

$$\tan(x)\big(\tan(x) - 1\big) + \big(1 + \tan(x)\big) = 2\tan(x)\big(1 + \tan(x)\big)$$

Simplifying:

$$\tan^{2}(x) + 2\tan(x) - 1 = 0$$

Using the Quadratic Formula:

$$\tan(x) = \frac{-2\pm\sqrt{2^{2} + 4(1)(1)}}{2}$$

$$\tan(x) = -1 + \sqrt{2}, -1 - \sqrt{2}$$

First, we will solve $\tan(x) = -1 + \sqrt{2}$. Since this doesn't correspond to any of the usual angles, we will see if it is a half-angle. We have:

$$\tan(2x) = \tan(x + x) = \frac{2\tan(x)}{1 - \tan^{2}(x)} = \frac{2\sqrt{2}-2}{1 - (\sqrt{2}-1)^{2}} = \frac{2\sqrt{2}-2}{2\sqrt{2}-2}=1$$

$\tan(y) = 1$ when $y = 225^{\circ}, 45^{\circ}$, so $\tan(2x) = 1$ means $x$ might equal $112.5^{\circ}, 22.5^{\circ}$. Checking these, we find that only $x=22.5^{\circ}$ works.

Now, we solve for when $\tan(x) = -1-\sqrt{2}$:

$$\tan(2x) = \frac{2\tan(x)}{1-\tan^{2}(x)} = \frac{-2-2\sqrt{2}}{1 - (-1-\sqrt{2})^{2}}=\frac{-2-2\sqrt{2}}{-2-2\sqrt{2}}=1$$

Once again, we find that $\tan(2x) = 1$, so $x$ could be $112.5^{\circ}$ or $22.5^{\circ}$. Checking, we find that only $112.5^{\circ}$ works.

Thus, $\boxed{x = 22.5^{\circ}, 112.5^{\circ}.}$


问:
Let $H, K \lhd G$, then $G = HK \iff G / (H \cap K) \cong G / H \times G / K$
This is a followup of this question

So, let $ (G,*) $ be a group.

Say $ H,K \lhd G $

If $ G $ is finite, then from the previously asked question we conclude: $ \ G = HK \iff G / (H \cap K) \cong G / H \times G / K$

Now, my question is for infinite groups $ G $, is the above conclusion still valid?

Can anyone please counter or prove it?

答:
There are groups $G$ with $G\cong G\times G$. For such a group take $H=K=\{e\}$. Then $HK=\{e\}$ and $G/H\cap K\cong G/H\times G/K$.

[For $G$ we can take (additively) the sum of countably many copies of $\mathbb{Z}$.]


问:
Lebesgue density of the random vector $(X,X+Y)$ where $X$ and $Y$ are exponentially distributed to different parameters
I want to find the density of the random vector $(X,X+Y)$ where $X\sim\text{Exp}_\alpha$ and $Y\sim\text{Exp}_\beta$ are exponentially distributed to different rate parameters $\alpha\ne \beta$.

I don't think I can use the product-density, as $X, X+Y$ are certainly not independent, but this is the only way I've ever seen a joint density done.

How can I proceed to calculate this density? Is there a rule I can use?

EDIT: More precisely, how can I get from $P(X\le a, X+Y\le b)$ to a place where I can actually use the known probability densities of $X,Y$? I can't split the probability because of dependence issues.

答:
You have to assume independence between $X$ and $Y$ then you can  use the fundamental transformation theorem (Jacobian).
For our knowledge, you have also to tell us if $\alpha$ and $\beta$ are "scale" or "rate" parameter because this modifies the density expression

Let's set

$$\begin{cases}
v=x+y\\
u=x
\end{cases}\rightarrow \begin{cases}
x=u\\
y=v-u
\end{cases}$$

The Jacobian is evidently $|J|=1$ thus

$$f_{UV}(u,v)=\alpha e^{-\alpha u}\beta e^{-\beta(v-u)}=\alpha \beta e^{-(\alpha-\beta)u}e^{-\beta v}$$

The only difficulty now is to understand which is the $(U,V)$ support. It is quite easy to understand that it must be $0<u<v<\infty$ thus the requested joint density is

$$f_{UV}(u,v)=\alpha \beta e^{-(\alpha-\beta)u}e^{-\beta v}\cdot\mathbb{1}_{[0;\infty)}(u)\cdot\mathbb{1}_{[u;\infty)}(v)$$

Useful observation

Now let's set $\alpha=\beta$ thus we have

$$f_{UV}(u,v)=\alpha^2 e^{-\alpha v}\cdot\mathbb{1}_{[0;\infty)}(u)\cdot\mathbb{1}_{[u;\infty)}(v)$$

Which integrated in $du$ gives us

$$f_V(v)=\int_{0}^v \alpha^2 e^{-\alpha v}du=\alpha^2 v e^{-\alpha v}\sim Gamma(2;\alpha)$$

as already known by Gamma distribution properties.


问:
Addition rule for mutually exclusive events
Why does the addition rule not work here?

$$P(\text{win this week}) = 0.5$$
$$P(\text{win next week}) = 0.5$$

They are mutually exclusive, and you can only win once (although it is possible that neither occurs). What is the probability that either occurs?

$P(A \cup B) = P(A) + P(B)$ clearly does not give the correct answer, unless you are guaranteed to win (but that is not the case here).

答:
The question makes sense in a scenario as under.

A school (say) gives a prize to some student every fortnight, but if someone has got the prize in the first fortnight, she is not considered for the prize in the second fortnight. Then, for any particular student,

P(doesn't get any prize that month) = $0.5\times0.5 = 0.25$

P(doesn't get a prize the $1^{st}$ fortnight, but does in the $2^{nd}) = 0.5\times 0.5 = 0.25$

P(gets a prize in the first fortnight) = $0.5$

As it must, the probabilities add up to $1$

Generalization added

Let probabilities of getting prize in first and second fortnight be $a$ and $b$ respectively, then

P(don't get any prize) = $(1-a)(1-b) = 1-a-b+ab$

P(get prize only first fortnight) = $a$

P(don't get $1^{st}$ fortnight, get in $2^{nd} = (1-a)b = b-ab$

Adding together, we get $1$, as we must


问:
Is this proof of Schröder-Bernstein Theorem correct?
If we know of the law of trichotomy, can we prove Schröder-Bernstein Theorem just by saying that if A $\preceq$B and B$\succeq$A, then it isn't be possible that A $\prec$B and it isn't possible that $A \succ B$, so we can conclude that $A \sim B$?

答: Well yes, but usually trichotomy is presented as a consequence of Schröder-Bernstein.So you just moved the problem to proving trichotomy.</p>


问:
Branch cut of the logarithm
I have a function $F$ holomorphic on some open set, and I have $ F(0) = 1 $ and $ F $ is non-vanishing. I want to show that there is a holomorphic branch of $ \log(F(z)) $.

Now, I'm getting confused. The principal branch of logarithm removes $ (-\infty, 0] $. But if the point 0 is missing from the plane, what happens when we take $ \log{F(0)} = \log{1} + 0 = 0 $? (I'm sure we can take the principal branch, because $ \exp(z) $ satisfies the conditions in question).

Any help would be appreciated. Thanks

答:
If $F(0)=1$, then $\log F(x)$ may be defined uniquely and is holomorphic in any open connected and simply connected set $S$ containing the point $x=0$ as long as this open set avoids all points $x$ for which $F(x)=0$. After all, in one such an open set - the disk around the origin - the logarithm may be calculated as the Taylor expansion
$$\log F(x) = (F(x)-1) -\frac{(F(x)-1)^2}2 + \frac{(F(x)-1)^3}{3} - + \dots $$
The radius of the convergence is the minimum of the values $|x_0^{(i)}|$ where $x_0^{(i)}$ are all points such that $F(x_0^{(i)})=0$.

It doesn't matter that $\log F(0) = 0$. The logarithn is just equal to zero at the origin but because we're not calculating another logarithm of the logarithm, it doesn't hurt. Near the origin, $\log F(x)$ may be approximated by $F(x) -1$. The full series is above.

Title: Branch cut of the logarithm


问:
Elementary theory of an algebraic structures
Could someone elaborate me what the sentence "The elementary theory of finite fields is decidable" means? I'm not sure that for example if I take $x\in \mathbb{F}_4$ and $y\in \mathbb{F}_5$ then can I form a first order predicate logic sentence that contains both $x$ and $y$? And what kind of inputs we can make to a Turing machine if we have elements from many fields? </p>

答:
This means that there is a Turing machine which decides whether or not a statement in the language of fields is provable from the theory of finite fields. If we consider the theory as closed under consequences, then it is the same as asking whether or not the statement is in the theory. The Turing machine takes a statement in the language, not a statement about models of the theory.

See also Wikipedia.

Note that theories live in the syntax, whereas $\Bbb F_4$ and $\Bbb F_5$ are semantics. You can't write a statement about different structures in first-order logic.


问:
Factors of ideal generated by odd integer in $\mathbb{Z}[\sqrt{-d}]$ are coprime
in a certain proof I'm working with the ring of integers of $\mathbb{Q}(\sqrt{-d})$ where $d=n^g-1$ is squarefree, $n>1$ is odd and $g>1$.

Then, as $d \equiv 2 \hspace{1mm} (\operatorname{mod} 4)$ I know that this ring is $\mathbb{Z}[\sqrt{-d}]$, and in this ring I have the following factorization of the ideal generated by $n$
$$(n)^g=(n^g)=(d+1)=(1+\sqrt{-d})(1-\sqrt{-d})$$
Know, I need to show that this two ideal factors are coprime and the text I'm using states that this fact derives from $n$ being odd but I don't see why.

I know that if a prime ideal $P$ divides both $(1+\sqrt{-d})$ and $(1-\sqrt{-d})$ then it divides $(n)$ and as the norm of $P$ is a power of a prime and the norm of $(n)$ is $n^2$ there is an odd prime dividing the norms of the two factors, which are both $1+d=n^g$, but I don't see any contradiction following this reasoning.

Any ideas?

答:
Hint: $\ (a,b)\supset (ab,a\!+\!b).\,$ You have $\,ab\,$ odd, $\,a+b\,$ even so $\ldots$

When $\,b = \bar a\,$ it says: $ $ if $\,a\,$ has coprime norm and trace then $\,a,\bar a\,$ are coprime.


问:
Show that the set of R-Algebra homomorphisms is Hausdorff
For an $\mathbb{R}$-Algebra $P$, let $M(P)$ be the set of $\mathbb{R}$-Algebra homomorphisms $\lambda: P \rightarrow \mathbb{R}$, w.r.t the initial topology generated by evaluation maps:
$$  \{ev_a : M(P) \rightarrow \mathbb{R};   \lambda
\rightarrow \lambda(a)  \}_{a \in P}$$

How do you show that $M(P)$ is a Hausdorff Space?

Trying to figure this out, but am stuck.

答:
By construction of the topology on $M(P)$, evaluation maps are continuous.

Suppose that $\lambda \neq \lambda' \colon P \to \mathbb{R}$ are distinct $\mathbb{R}$-algebra maps from $P$ to $\mathbb{R}$, and note that this occurs precisely if they are different functions. Hence we have some $x \in P$ for which $\lambda(x) \neq\lambda(x')$. Since $\mathbb{R}$ is Hausdorff, there are disjoint open sets $U_\lambda \ni \lambda(x), U_{\lambda'} \ni \lambda'(x)$ and so

$$
U = ev_x^{-1}(U_\lambda), \quad V = ev_x^{-1}(U_{\lambda'})
$$

are disjoint open sets separating $\lambda$ from $\lambda'$.

In general, we have shown that if $X$ is a Hausdorff space and $S \subset X^Y$ is a subset with some topology for which evaluation maps are continuous, then $S$ is also $T_2$. There is no need for the algebraic structure of $M(P)$ to make this argument work.


问:
Craps Math Question - Calculating the house advantage with multiple bets
I have a YouTube channel on throwing and playing craps. I like to play the hard way bets knowing there is a large house advantage when playing them. I had a comment on my channel that playing those bets is giving the house money. My strategy is treating it like a bonus when I have money on the 6 and/or 8 and money on the corresponding hard way. What I am trying to determine is what happens to the house advantage when you play both bets together.

Is it as simple as taking the expected outcome (house advantage) of throwing a 6 multiplied by the expected outcome of throwing a hard 6?

I am trying to calculate the house advantage with multiple bets, a 6 dollar place bet on the 6 and a 1 dollar bet on the hard 6 and understand the following to be true:

The True Odds of throwing a Hard 6 is 10:1 and pays 9:1 for a house advantage of 9.09%

```<code>(1/11)×9 + (10/11)×(-1) = -1/11 ≈ -9.091%
</code>```

The True Odds of throwing a Place Bet 6 is 6:5 and pays 7:6 for a house advantage of 1.52%

```<code>[(5/11)×7 + (6/11)×(-6)]/6 = (-1/11)/6 = -1/66 ≈ -1.515%
</code>```

I am trying to calculate the house advantage and I am coming up with the following formula:

```<code>((5/11)*7)+((1/11)*9) + ((6/11)*-6) + ((10/11)*-1)/6 = 0.575758
</code>```
```
Am I calculating this correctly?
The expected outcome of winning on any 6 is (5/11) x <span class="math-container">$7
The expected outcome of winning on a hard 6 is (1/11) x $</span>9
The expected outcome of losing to a 7 (6/11) * -<span class="math-container">$6
The expected outcome of losing the hard way to a 7 or easy 6 (10/11) * -$</span>1.

I am getting a house advantage of 0.575758. Is this anywhere close?
```

UPDATE:
Is this as simple as multiplying the two expected outcomes together?

```<code>((1/11)×9 + (10/11)×(-1)) x ([(5/11)×7 + (6/11)×(-6)]/6 = (-1/11)/6) = 0.1377%
</code>```

Does that look correct?

I got the calculations from WizardOfOdds.com

答: If you bet $6$ on a bet that is losing $1.52\%$ and $1$ on a bet that is losing $9.09\%$ you are losing $6 \cdot 1.52\% + 1\cdot 9.09\%=0.0912+0.0909=0.1821$ on a bet of $7$.  The percentage loss is then $\frac {0.1821}7\approx 2.6\%$


问:
Thoughts about the use of $\epsilon$-$\delta$ language in proofs involving limits
The limit of a function $f: \mathbb{R} \to \mathbb{R}$ at $x_{0}$ is defined as follows:

$\lim_{x \to x_{0}} f\left(x\right) = L \iff \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon$.

In proofs of limits using the definition, we typically fix an arbitrary constant $\epsilon > 0$, and as long as we are able to prove
\begin{equation}
\exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon,
\end{equation}
we can conclude that $\lim_{x\to x_{0}}f\left(x\right) = L$. The key is to construct such a $\delta$ so that the above relation holds. However, as I saw in some textbooks, for an arbitrary $\epsilon > 0$, if we are able to find a $\delta > 0$ such that
\begin{equation}
\forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < g\left(\epsilon\right),
\end{equation}
where $g\left(\epsilon\right)$ is a term containing $\epsilon$ ($g\left(\epsilon\right) = 3\cdot\epsilon$, for example),
then we can still claim that $\lim_{x\to x_{0}}f\left(x\right) = L$.

I am wondering if there exists a simple proof for using the general $g\left(\epsilon\right)$ in proofs of limits?

答:
To formalize @DaniloGregorinAfonso's idea, you may set $\epsilon' = g(\epsilon)$, so as to cheer up yourself for the moment, giving a hope that you can "clean up the RHS".
$$\forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon'$$
Assume that $\epsilon = g^{-1}(\epsilon')$.  Note that $\delta$ is a function $\epsilon$, so you may write $\delta = \delta(\epsilon) = \delta(g^{-1}(\epsilon'))$.  You may find the notation $g^{-1}$ problematic, say, in case that $g(\epsilon) = \begin{cases}\epsilon (\sin(1/\epsilon) + 2021/2020) & \epsilon \ne 0 \\ 0 & \epsilon = 0. \end{cases}$  In such case, you may find a monotone increasing upper bound $h(\epsilon) \ge g(\epsilon)$ such that $h(\epsilon) \to 0$ as $\epsilon \to 0$, and do that same thing to get $\delta(\epsilon) = \delta(h^{-1}(\epsilon'))$.  Now you get a function $\delta$ which depends on the new $\epsilon'$.
$$\forall \epsilon > 0, \exists \delta(h^{-1}(\epsilon')) > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon'$$
Finally replace $\epsilon$ in "$\forall \epsilon > 0$" on the left-hand side by $\epsilon'$.

It remains to show that such monotone increasing upper bound $h$ for $g$ exists, but we may construct it like $h(\epsilon) = \sup\limits_{t \in (0,\epsilon)} |g(\epsilon)|$.  It's clear that $h$ is monotone increasing.  It's easy to verify that $h\searrow0$ as $\epsilon \searrow 0$.

 $$\forall \epsilon > 0, \exists \delta > 0, \forall t \in (0,\delta), |g(t)| < \epsilon \\ \implies \sup_{t \in (0,\delta)} |g(t)| < \epsilon \\ \implies \forall t \in (0,\delta), h(t) = \sup_{s \in (0,t)} |g(s)| \le \sup_{s \in (0,\delta)} |g(s)| < \epsilon$$

Add an extra term $\epsilon$ to $h$ guarantees that $h$ to be strictly increasing, i.e. the existence of $h^{-1}$, completing the proof.


问:
Find coordinates of trapezoid if 2 coordines and midpoint of opposite side is known
In the below picure A,B and O are known.I need to find C and D.
AB and CD are parallel and O is the midpoint of C and D
Any help is highly appreciated

The below is the actual scenario. Radius of 2 circles and center is known. Need to find C and D. Please help

答:
Welcome to MSE. Please try to use appropriate tools such as Geogebra or AutoCAD to produce your images. Your question becomes a lot easier to understand with this image:

We have points $A$ , $B$ and $P$ , which is the midpoint of $CD$. We also have the radii of the two circles. Suppose $CD$ is tangent to the larger circle at $P$. I am adding the assumption that the points $C$ and $D$ are on the extensions of $OB$ and $OA$ , respectively. The problem is to find the points $C$ and $D$.

Please note that without my added assumption, your problem would have an infinite number of solutions.

Please also be consistent in the way you name the elements of your figures. In the upper figure you marked the midpoint of $BC$ as $O$. In the lower figure you marked the midpoint of $BC$ as $P$ and you marked the center of the circle as $O$.

Please be mindful that if you do not care about your own question, then people in here may not be encouraged to help you with it.

You can often expect hints or comments on your own approach in here. You can often not expect people in here to solve your homework problems. I hope you enjoy your experience in MSE. Now, for the start, using the figure I just posted, tell us how you approach this problem, and we will help you further.


问:
Prove that if $(a + b + c)(ab + ac + bc) = abc$ then sum of some two numbers equals $0$
Prove that if $(a + b + c)(ab + ac + bc) = abc$ then sum of some two numbers equals $0$.

Without loss of generality let's suppose that $a=0$ then $(b + c)bc = 0 \Rightarrow b+c = 0$ or

$bc = 0 \Rightarrow b=0 \lor c = 0$ and $a+b=0 \lor a+c=0$ respectively. Q.E.D.

Now let's suppose $abc \not = 0$ and $a+b+c \not = 0$ then from initial equation we can get the following one that must hold $\dfrac{1}{a+b+c} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$

As $a,b,c$ are some arbitrary numbers (but not non-negative integers for example) I've got completely stuck here

Body:
Hint:

$$(a + b + c)(ab + ac + bc) - abc=(a + b)(b + c)(c + a)$$


问:
Solve $a^x + b^x + c^x = 1$ for $x$ when $a, b, c \in (0, 1)$
I'm looking to solve $a^x + b^x + c^x = 1$ for $x$ when $a, b, c \in [0, 1]$.

If $a+b+c>1$ the solution would be $x>1$, if $a+b+c<1$ then $x < 1$.

A closed formula would be best, but I actually want to do this in JavaScript, so algorithms are also helpful.

答:
Consider that you look for the positive zero of function
$$f(x)=a^x + b^x + c^x - 1$$ Its first derivative is always negative and the second derivative always positive.

The series expansion around $x=0$ is
$$a^x + b^x + c^x=3+\sum_{n=1}^\infty \frac {\log^n(a)+\log^n(b)+\log^n(c)}{n!} x^n$$ Using a few terms and series reversion, we have
$$x=t-\frac{ \log ^2(a)+\log ^2(b)+\log ^2(c)}{2 (\log (a)+\log (b)+\log(c))} t^2+O(t^3)$$ where $t=-\frac {4}{\log(abc)}$.

Let us try with $a=\frac 1 \pi$, $b=\frac 1 e$, $c=\gamma$. This would give, as an estimate, $x=1.25923$ while the "exact" solution given by Newton method is $1.28486$.

We could do better at the price of more terms and very lond expressions. But, starting from this guess, Newton method would work like a charm.
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.259229442 \\
 1 & 1.284569396 \\
 2 & 1.284861903 \\
 3 & 1.284861941
\end{array}
\right)$$


问:
An integral involving Fresnel integrals $\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$
I need to calculate the following integral:
$$\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$$
where
$$S(x)=\int_0^x\sin\frac{\pi z^2}{2}\mathrm dz,$$
$$C(x)=\int_0^x\cos\frac{\pi z^2}{2}\mathrm dz$$
are the Fresnel integrals.

Numerical integration gives an approximate result $0.31311841522422385...$ that is close to $\frac{16\log2-8}{\pi^2}$, so it might be the answer.

答:
This is not a solution yet, but I think it is a way forward.

Define

$$f(x) = \left (\frac12 - S(x)\right) \cos{\left (\frac{\pi}{2} x^2\right)} - \left (\frac12 - C(x)\right) \sin{\left (\frac{\pi}{2} x^2\right)} $$

$$g(x) = \left (\frac12 - C(x)\right) \cos{\left (\frac{\pi}{2} x^2\right)} + \left (\frac12 - S(x)\right) \sin{\left (\frac{\pi}{2} x^2\right)} $$

Then it is a straightforward exercise to show that

$$g(x) + i f(x) = e^{-i \pi x^2/2} \int_x^{\infty} dt \, e^{i \pi t^2/2}$$

and that

$$(2 S(x)-1)^2 + (2 C(x)-1)^2 = 4 [g(x)^2+f(x)^2] = 4 x^2 \int_1^{\infty} du \, \int_1^{\infty} dv \, e^{i \pi x^2 (u^2-v^2)/2}$$

I can convert the double integral to a single integral by changing coordinates to $p=u+v$, $q=u-v$, $p \in [2,\infty)$, $q \in [-(p-2),p-2]$.  The Jacobian is $1/2$ and we have

$$(2 S(x)-1)^2 + (2 C(x)-1)^2 = 2 x^2 \int_2^{\infty} dp \, \int_{-(p-2)}^{p-2} dq \, e^{i \pi x^2 p q/2}$$

which after evaluation of the inner integral and some rescaling, we get

$$(2 S(x)-1)^2 + (2 C(x)-1)^2 = \frac{8}{\pi} \int_1^{\infty} dp \, \frac{\sin{[2 \pi x^2 p (p-1)]}}{p}$$

I am not sure how to evaluate this integral analytically, nor am I sure that that would be the best move here.  The desired integral is therefore

$$\int_0^{\infty} dx \, x\, [(2 S(x)-1)^2 + (2 C(x)-1)^2]^2 =\\ \frac{64}{\pi^2} \int_0^{\infty} dx \, x\, \int_1^{\infty} dp \, \frac{\sin{[2 \pi x^2 p (p-1)]}}{p} \int_1^{\infty} dq \, \frac{\sin{[2 \pi x^2 q (q-1)]}}{q}$$

From here, I am not quite sure what to do.  Naturally, I would like to reverse the order of integration so that the integral over $x$ is interior, but I am not sure how to justify that, given that the integrals are not absolutely convergent.  Further, even if I can justify that, I think the resulting integral is some delta functions which do not look promising.  I am continuing to look at this, but I figure that maybe someone else can also contribute from here, or just tell me if I am off the mark.


问:
The proof of $\log(1+x) < x.$
I want to prove $\log(1+x) < x.$

Of course, I can prove this by letting $f(x):= x - \log(1+x),$ and calculating $f'(x)$, ... .

But I wonder if I can prove this using Maclaurin's expansion.

$\log(1+x)=x-\dfrac{1}{2}x^2+\dfrac{1}{3} x^3-\cdots.$

The right-hand side includes the term "$x$" so it seems that I can use this equality in order to prove $\log(1+x) < x$, but I cannot.

I would like you to give me some ideas.

答: If you accept (otherwise this can easily be proved) that $\log ()$ is a concave function, then it suffices to show (cf. Jensen) that $x$ is a tangent to $\log(1+x)$. But this is obvious: $x$ and $\log(1+x)$ touch at $x=0$ and have the same slope ($1$) there.

Title: The proof of $\log(1+x) < x.$


问:
MCMC method - burn-in chains
The Markov chain that consists of positions and the probabilities that a certain state is reached, which depends only on the previous state of the chain. Then the initial steps of the chain called burn-in chains are not taken into account because they are not relevant and not close to the converging phase. Why is it so? How would burn-in chains influence the result or process when the next state depends only on the previous one?

答: The previous step, in turn, depends on the step before, and so on. This way, although MC does not take into account the chain of all previous steps in the explicit calculation of the next step, the effect of the previous steps is implicit in the present step, which is used to build the next step.</p>

Title: MCMC method - burn-in chains


问:
How many different hues, saturations and values are possible when converting RGB to HSV?
If we specify a colour in the RGB space, there are $2^{24}$ different colours, as R, G and B are 8 bit numbers.

When a colour is converted to the HSV colour space (find the conversion formulas here for example), how many different values are possible for H, S and V?

Obviously, the total number of colours is equal in both spaces, as the conversion is a bijection. But how is the information distributed over the three dimensions of the HSV space?

答:
The answer ist 118920. The 2^24 RGB values result in 118920 different hues, 19821 saturations, and 256 values. The last one is kind of obvious, but the first two were interesting. I wouldn't have expected the difference to be that big.

What is very interesting is the scatter plots of hue, saturation and values of the colours.

scatter plots of hue, saturation a value of RGB colours

It turns out that there are very few distinct hue values for low saturations and values, which makes sense, because hue is not "important" for greys and dark colours.

What I found most astonishing were the patterns in the hue-saturation plot.

Anyway, that was an interesting and fun exercise.


问:
Is $\begin{bmatrix} a &b \\0 &1 \end{bmatrix}$ a cyclic group?
I appreciate all the help I can get with this task.
$$
G=\left \{ \begin{bmatrix}
a &b \\0
 &1
\end{bmatrix},a,b\in \mathbb{Z}_3, a\neq 0 \right \}
$$

Is G a cyclic group?

Does a subgroup with 3 elements exists?

All elements in G:
$$\begin{bmatrix}
1 &0 \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &1  \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &2 \\0
 &1
\end{bmatrix},\begin{bmatrix}
2 &0 \\0
 &1
\end{bmatrix},\begin{bmatrix}
2 &1 \\0
 &1
\end{bmatrix},\begin{bmatrix}
2 &2 \\0
 &1
\end{bmatrix}$$

What do I do now?

EDIT: Thanks for your comments.

Is this one subgroup of order 3? $a=1$ generates 3 matrices.
$$\begin{bmatrix}
1 &0 \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &1  \\0
 &1
\end{bmatrix},\begin{bmatrix}
1 &2 \\0
 &1
\end{bmatrix}$$

From my book

A group is said to be cyclic if it contains an element x such that every member of G is a power of x.

Can $x$ be one of the six elements above?

EDIT: Thanks for all the help! I appreciate it very much.

答:
A quicker way to see that $G$ is not cyclic: note that
$$
\pmatrix{2 & 0\\0 & 1} \pmatrix{1 & 1\\0&1} = \pmatrix{2 & 2\\0 & 1}, \\
\pmatrix{1 & 1\\0 & 1} \pmatrix{2 & 0\\0 & 1} = \pmatrix{2 & 1\\0 & 1}.
$$
That is, we have found elements $g,h \in G$ with $hg \neq gh$. Because $G$ is not abelian, it cannot be cyclic.


问:
calculate $\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$
How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get

$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
and this doesn't get me further. Any ideas?

答:
This is not quite a solution, more of a comment about the assertion that the L'Hospital's Rule calculation
$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$
does not get us any further.

Let $L$ be the original limit, assumed to exist and be non-zero.  Now look at the right-hand side of the expression you reached. The $\pi$'s cancel. The term $\cos(\pi t)$ sedately approaches $-1$, cancelling the minus sign. And the rest has limit $\dfrac{1}{L}$! What you saw as a flaw becomes a virtue.

We conclude that $L=2\cdot \dfrac{1}{L}$. Thus $L=\pm \sqrt{2}$, and a quick examination of signs shows that we  need the negative one, since a little past $\pi$, the sine is negative.

I do not advocate this approach, since there are details of existence to fill in, and one could easily reach an incorrect conclusion. Anyway, there is a simple non-L'Hospital calculation that quickly yields the answer.


答: Imagine this in the context of a weighted random walk on the plane, where the probability that we move closer to the origin is dependent on how far we are from it; for instance, from $\langle x,y\rangle$ with $x, y\geq 0$ we might move to $\langle x+1, y\rangle$ with probability $\frac12\cdot\frac1{x^2+2}$, to $\langle x-1, y\rangle$ with probability $\frac12\cdot\left(1-\frac1{x^2+2}\right)$, and similarly for the two moves to $\langle x, y\pm 1\rangle$ (and then extended symmetrically to negative $x$ and $y$). Here the probability that we move from $\langle x,y\rangle$ to any of the four neighbors only depends on the point where we're at, and not how we got there — but the probability that we're at that point is a sum of all the possible paths we might have taken to get to that point. The purpose of ignoring the burn-in is to move past the 'transient' portion that corresponds to, e.g., the fact that it's impossible to ever be at $\langle 5,3\rangle$ within the first six steps (starting from the origin) and get to the steady-state distribution (or a good approximation to it).

Title: MCMC method - burn-in chains

Body: The Markov chain that consists of positions and the probabilities that a certain state is reached, which depends only on the previous state of the chain. Then the initial steps of the chain called burn-in chains are not taken into account because they are not relevant and not close to the converging phase. Why is it so? How would burn-in chains influence the result or process when the next state depends only on the previous one?


问:
Extreme value theorem for functions defined on weakly compact sets.
So I start this by saying that I know very littler about topology and the related topics in this question, so  this question may be trivial/ill-possed, etc.

I've read in many places that it is a well known result in functional analysis that a Banach space $X$ is reflexive if and only if the unit ball is weakly compact (compact in the weak topology).

So assume we have a real separable Hilbert space $H$ (this space is a reflexive Banach space), and consider the unit sphere $S=\{f\in H:\|f\|=1\}$, then $S$ is weakly compact.

I would like to use (some generalization of...) Weierstrass extreme value theorem to prove that a continuous function $F:S\to \mathbb R$ attains a maximum.

A little bit of background, here $H$ is a quantum Hilbert space, $S$ is the family of "quantum states". Then I wanted to prove that a (continuous) function of the quantum states will attain a maximum.

Is it possible to do something like this?

Thanks in advance!

答:
If $F: S \to \Bbb R$ is continuous (where $S$ has the weak topology, and $\Bbb R$ the standard topology), a standard fact in topology is that $F[S]$ is also compact in $\Bbb R$ and by standard facts about the reals a compact set like $F[S]$ has a maximum and a mimumum. QED.

So a combination of a standard "compactness preserving" fact and properties of the reals. Both are easily shown from definitions.


问:
Prove that function is non-monotonic and is invertible

We have  $$f(x)=\begin{cases}\frac x2,&x\in\Bbb Q\\-\frac x2,&x\in\Bbb R\setminus\Bbb Q\end{cases}$$
Prove that function is non monotonic and is invertible.

I tried to take points from $\Bbb R\setminus\Bbb Q$ and $\Bbb Q$ and prove that is decreasing but couldn't. Can you help?

答:
Hint:

To prove non-monotonicity, pick $0<x_1<x_2<x_3$ such that $x_1,x_3\in\mathbb{Q}$ and $x_2\in\mathbb{R}\setminus\mathbb{Q}$.

To prove that it is invertible, show that this function is injective and surjective.

for "injective", try $x,y$ both irrational, both rational, and exactly one rational.

for "subjective", follow the definition.


问:
Finite Fields of orders which are four consecutive integers
As I was preparing properties of Finite Fields I realised that there are Finite Fields of orders 2,3,4 and 5 which happen to be four consecutive integers. I am just curious to know if there are four other consecutive integers which are orders of Finite Fields! Any thoughts on this please.</p>

答: This is not possible. The order of finite fields are prime powers and there are no four consecutive prime powers greater that $2$ since two of them would have to be even.


答:
You found the only solution.

Two of those four consecutive integers are even. If they are cardinalities of a finite field, they must both be powers of two. But they differ from each by two, and $2$ and $4$ are the only powers of two satisfying that condition.

Actually it is difficult to find even three consecutive integers that would be prime powers. By the above argument, the one in the middle has to be even, and hence a power of two. This implies that one of the odd neighbors is divisible by three, hence a power of three. By Catalan's conjecture (now Mihailescu's theorem) the only adjacent powers of $2$ and $3$ are $8=2^3$ and $9=3^2$.

Therefore, in addition to the four you found, the only other triple is $7,8,9$.

Title: Finite Fields of orders which are four consecutive integers

Body: As I was preparing properties of Finite Fields I realised that there are Finite Fields of orders 2,3,4 and 5 which happen to be four consecutive integers. I am just curious to know if there are four other consecutive integers which are orders of Finite Fields! Any thoughts on this please.</p>


问:
Question involving polynomials and fields
I'm studying Galois Theory and I'm stuck in a problem, let's see:

Let $P(x) = X^5 - p \in \mathbb{Q}[X]$, where $p$ is prime. Find a Splitting field $L$ of $P$, five subfields of $L$ and the factoration of $P$ in that fields.

Now, let $\epsilon = e^{2i\pi/5}$, of course $L=\mathbb{Q}(\sqrt[5]{p},\epsilon)$ and the subfields may be $\mathbb{Q}_i(\epsilon^i \cdot \sqrt[5]{p})$, for $i=0,1,2,3,4$.

I'm stuck on the factorization, for instance, if $i=0$ then in $\mathbb{Q}_{0}(\sqrt[5]{p})$ we can write $$P(x) = (x - \sqrt[5]{p})(x^4 + x^3(\sqrt[5]{p}) + x^2 (\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4)$$

My question is:

Is $x^4 + x^3(\sqrt[5]{p}) + x^2 (\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4$ irreductible over $\mathbb{Q}_0(\sqrt[5]{p})$??

I've tried to solve a system or thinking in Binomial Theorem, but well... I'm here. Can you give me a hint? Also, I need to find that in another four fields...

答: As we have $x^4 + x^3(\sqrt[5]{p}) + x^2(\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4 = (x-\sqrt[5]{p}\epsilon)(x-\sqrt[5]{p}\epsilon^2)(x-\sqrt[5]{p}\epsilon^3)(x-\sqrt[5]{p}\epsilon^4)$ implies if one of the linear reducible factor lies in $\mathbb{Q}(\sqrt[5]{p})$ then $5th$ primitive root of unity also lies in $\mathbb{Q}(\sqrt[5]{p})$ which is not true. Then we have only possibility left is degree 2 irreducible factors. As none of the $5th$ primitive roots of unity lies in $\mathbb{Q}(\sqrt[5]{p})$, so we can club the roots $\sqrt[5]{p}\epsilon$ and $\sqrt[5]{p}\epsilon^4$ together and see whether $x^2 - \sqrt[5]{p}(\epsilon + \epsilon^4)x + (\sqrt[5]{p})^2$  but $(\epsilon + \epsilon^4) \notin \mathbb{Q}(\sqrt[5]{p}) $ as if you adjoin $\epsilon + \epsilon^4$ in $\mathbb{Q}$ the resulting field have even dimension over $\mathbb{Q}$. So the polynomial $x^4 + x^3(\sqrt[5]{p}) + x^2(\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4$ is irreducible over $\mathbb{Q}[\sqrt[5]{p}]$.


问:
Local Truncation Error on Improved Euler's Method
I need to reach the conclusion that
$$\tau(x,h) = \frac{\phi(x + h) - \phi(x)}{h} - \frac{1}{2}[f(x,\phi(x)) + f(x+h,\phi(x) + h f(x,\phi(x)))]  = O(h^2)$$

So I tried using Taylor expansions and reached the following

$$\phi(x+h) = \phi(x) + h\phi'(x) + \frac{h^2}{2}\phi''(x) + O(h^3)$$

$$\frac{\phi(x+h) - \phi(x)}{h} = \phi'(x) + \frac{h}{2}\phi''(x)$$

$$f(x+h,\phi(x) + h f(x,\phi(x))) = f(x,\phi(x)) + hf_x(x,\phi(x)) + hf(x,\phi(x))f_y(x,\phi(x)) $$

$$\frac{1}{2}[f(x,\phi(x)) + f(x+h,\phi(x) + h f(x,\phi(x)))] = f(x,\phi(x)) + \frac{h}{2}[f_x(x,\phi(x)) + f(x,\phi(x))f_y(x,\phi(x))]$$

$$\tau(x,h) = \phi'(x) + \frac{h}{2}\phi''(x) -  f(x,\phi(x)) - \frac{h}{2}[f_x(x,\phi(x)) + f(x,\phi(x))f_y(x,\phi(x))]$$

In the end I reach this which is still $O(h)$

$$\tau(x,h) = \frac{h}{2}[\phi''(x) - f_x(x,\phi(x)) - f(x,\phi(x))f_y(x,\phi(x)) ] $$

What am I doing wrong?

答:
If you compute the derivative of $ϕ'(x)=f(x,ϕ(x))$ you get
$$
ϕ''(x)=f_x(x,ϕ(x))+f_y(x,ϕ(x))ϕ'(x)=f_x(x,ϕ(x))+f_y(x,ϕ(x))f(x,ϕ(x)).
$$
The term in question cancels to zero, so the next terms are quadratic in $h$.

This is also called Heun's (2nd order) method or explicit trapezoidal method. For a similar question with a slightly different approach see Heun's method: Why is it of second order?


问:
Proof of $(a+b)^{n+1}$
I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.

I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.

I however have trouble with the second one, I don't know what to start exactly. Any help/hints?

答:
Consider,

$$ e^x e^y = e^{x+y}$$

Expand out both sides as power series:

$$ \sum_i \frac{x^i}{i!} \sum_j \frac{y^j}{j!} = \sum_{k=0}^{\infty} \frac{(x+y)^k}{k!} \tag{1}$$

Consider sum on LHS,

$$ \sum_{ i, j \geq 0}^{\infty}\frac{x^i y^j}{i! j!} $$

Now, consider summing the series as sets of homogenous polynomials that is set the condition, $i+j=k$ this makes our sum:

$$ \sum_{k=0}^{\infty} \left(\sum_{i=0}^k \frac{ x^i y^{k-i} }{i! (k-i)!} \right) \tag{2}$$

Equate RHS of (1) to (2):

$$  \sum_{k=0}^{\infty} \left(\sum_{i=0}^k \frac{ x^i y^{k-i} }{i! (k-i)!} \right) = \sum_{k=0}^{\infty} \frac{(x+y)^k}{k!}$$

Rewrite this sum as:

$$ \sum_{k=0}^{\infty} Q_k = \sum_{k=0}^{\infty} P_k$$

Now notice that $Q_k$ is a homogenous polynomial of degree $k$, and $P_k$ is also a homogenous polynomial of degree $k$, hence these two quantities must be equal  Refer

$$ \sum_{i=0}^k \frac{ x^i y^{k-i}}{i!(k-i)!} = \frac{(x+y)^k}{k!}$$

Or rearranging,

$$ \sum_{i=0}^k \binom{k}{i} x^i y^{k-i} = (x+y)^k$$

Title: Proof of $(a+b)^{n+1}$


答:
Note that monotonic entails both increasing and decreasing. Non-motonic means neither increasing nor decreasing.

$0<\sqrt2,f(0)=0>f(\sqrt2)=-\frac1{\sqrt2}$ so the function is not monotonically increasing.

$0<1,f(0)=0<f(1)=\frac12$ so the function is not monotonically decreasing.

The function is surjective over its range $\Bbb R$.

For injectivity: let $f(x_1)=f(x_2)$. Note that the function maps rational numbers to rational numbers and irrational numbers to irrational images. Thus both $f(x_1),f(x_2)$ are either rational (in which case $f(x_i)=x_i/2\implies x_1/2=x_2/2\iff x_1=x_2$) or irrational (in which case $f(x_i)=-x_i/2\implies-x_1/2=-x_2/2\iff x_1=x_2$).

Title: Prove that function is non-monotonic and is invertible

Body:
We have  $$f(x)=\begin{cases}\frac x2,&x\in\Bbb Q\\-\frac x2,&x\in\Bbb R\setminus\Bbb Q\end{cases}$$
Prove that function is non monotonic and is invertible.

I tried to take points from $\Bbb R\setminus\Bbb Q$ and $\Bbb Q$ and prove that is decreasing but couldn't. Can you help?


问:
$ G/H \cong K $ and $ G/K \cong H\implies G \cong H \times K $
Let $ (G, *) $ be a group.

Say, $ H, K \lhd G $ such that $ G/H \cong K $ and $ G/K \cong H $

Example:$\ G = Z_{4}, \ H = \{ 0, 2\}, \ K = \{ 0, 2\}  \ $ satisfies the above condition, but $ G \not\cong \{(0,0),(0,2),(2,0)(2,2) \} = H \times K$,

Here we observe that $H \cap K \not= \{(0,0)\}$

Example: $ \ G = GL(n,R), H = SL(n,R), K = Z(G)$ also satisfies the above condition, and $G = H \times K$,

Here we observe that $H \cap K = \{I_{n \times n}\}$

So, intutively I can feel that
$ \ \ G \cong H \times K \iff H \cap K = \{ e \} $

Can anyone prove or counter it?

(In case the statement is wrong, please provide an iff condition for $ G \cong H \times K$ where $ H, K \lhd G $ such that $ G/H \cong K $ and $ G/K \cong H $)

答:
I think I have a counterexample. Let $D = \langle a,b \mid b^2=(ab)^2 = 1 \rangle$ be the infinite dihedral group, and let $X = \langle a_1,b_1 \rangle \times \langle a_1,b_2 \rangle$ be the direct product of two copies of $D$.

Now let $G = \langle a_1^2,b_1,a_2^2,b_2,a_1a_2 \rangle \le X$, and note that $|X:G| = 2$.

Let $H = \langle a_1^2,b_1 \rangle$ and $K = \langle a_2^2,b_2 \rangle$. Then $H \cong K \cong D$, and $H \times K \le G$ with $|G:(H \times K)| = 2$.

Now $G/H$ is generated by the images in $H$ of $a_1a_2$, $a_2^2$ and $b_2$, but $(a_1a_2)^2H = a_2^2H$ and $(a_1a_2b_2)^2H = H$, so $G/H \cong D \cong K$ and similarly $G/K \cong H$.

Now, since $[b_1,a_1a_2] = [b_1,a_1] = a_1^2$ and $[b_2,a_1a_2] = a_2^2$, we see that $G/[G,G]$ is a $3$-generator group (of order 8) generated by the images of $b_2$, $b_2$ and $a_1a_2$.

But  the  abelanization of $H \times K$ is a $4$-generator group of order $16$, so $G \not\cong H \times K$.


问:
Simple ordinal number question
I got an excercise to add some ordinals, and one of given ordinals was: $$\omega\cdot4\omega^2+1.$$
I know that $4\omega = \omega$ and that multiplication is associative, so I think this simply equals to $\omega^3+1$? </p>

<p>I don't feel confident about this and so I would like to hear some confirmation. </p>

答:
To get this off the unanswered list,

Yes. Your argument is correct.

Title: Simple ordinal number question


问:
Is it possible to extend a homomorphism from an etale algebra into the separable closure?
I am currently learning etale algebras and they seem to be generalizations of separable algebraic field extensions, so I wanted to try to explore which properties hold and which don't. My question is as follows:

Let $L$ be an etale algebra over a field $K$,$L'$ a subalgebra of $L$ and $f:L' \longrightarrow K_{sep}$ a morphism of $K$ algebras from $L'$ into the separable closure of of $K$. Is it possible to extend $f$ into a morphism from $L$ to $K_{sep}$?

答:
I think I found an answer.

Consider $L=L'[\theta ]$. If we manage to prove this case, we can then proceed by induction.
First of all an etale algebra is by definition a product of separable extensions of $K$: say $L=F_1 \times \cdots \times F_m$. Now consider $\pi_i$ to be the natural projection of $L$ onto $F_i$.

$ker(f)$ is a maximal ideal, since $\frac{L'}{ker(f)}$ is isomorphic to $im(f)$, which has to be a field.
For all $i$ $ker(\pi_i) \cap L'$ is a maximal ideal of $L'$ for the same reason. Now let us show that there exists $i_0$, such that $ker(\pi_{i_0})\cap L' \subseteq ker(f)$. If it wasn't the case, we could for every $i$ find an $a_i \in ker(\pi_i)$,such that $f(a_i) \neq 0$. Then $f(\prod_i a_i )$ is non zero, since $f$ is a morphism. However $\prod_i a_i$ has to be zero, since all of its components are null, contradicting that $f$ is a morphism.

Now consider the ideal of polynomials $I=\{P\in L'[X]| P(\theta )=0 \}$. Then write $f(I)=\{f(P) \in f(L')[X]|P \in I\}$. Given that $f$ is surjective onto the field $f(L')$, $f(I)$ is an ideal in the polynomial ring $f(L')[X]$. Now let us prove, that $f(I)$ is not the whole $f(L')[X]$.

By contradiction assume it is. Then there exist $a_0, \cdots , a_n \in L'$, such that $f(a_0)+f(a_1)X+\cdots +f(a_n)X^n=1$ and $a_0+a_1 \theta+ \cdots a_n \theta^n=0$. From the first equation we deduce that $f(a_0)=1$ and $f(a_i)=0$ for $i>1$. Since $ker(f)=L'\cap ker(\pi_{i_0})$, we deduce that $pi_{i_0}(a_i)=0$ for $i>1$. Applying that to the second equation, we get that $\pi_{i_0}(a_0)=0$. However since $f$ and $\pi_{i_0}$ have the same kernel, we get that $f(a_0)=0$, which is a contradiction.

Now that we know that the ideal $f(I)$ is not the whole $f(L')$, we know that it is generated by a non constant polynomial $f(Q)$ with $Q\in I$. Since it is non constant, we can take $u$ a root of $Q$ in $K_{sep}$. Then finally we extend $f$ by the formula $f(P(\theta ))=f(P)(u )$.


问:
Let a, b and c be positive real numbers such that abc=1. Let d = {a^(b+c)} * {b^(c+a)} * {c^(a+b)}. Find the maximum possible value of a.
Let a, b and c be positive real numbers such that abc=1. Let d = {a^(b+c)} * {b^(c+a)} * {c^(a+b)}.
Find the maximum possible value of d.

I found that the answer is 1 by using some hacks/fakesolve but I am unable to prove it.

答:
The answer from S. Dolan is totally correct, but from your comment I guess you would like to see more intermediate steps.

If you want to find the maximum of $f$ it is the same as if you search for the maximum of $\ln (f)$. Therefore you can search for the maximum of $\ln (\frac{1}{a^a}\frac{1}{b^b}\frac{1}{c^c})$.

And this is just $-a\ln a - b\ln b - c\ln c$. Rather than maximising this you can also minimise this multiplied with minus 1.

And now you need to look up how you minimise a function with constraints by applying the method of Lagrange multipliers. You have to add to your function the constraint multiplied by $\lambda$:

$$ f =  a\ln a + b\ln b + c\ln c + \lambda (1 - abc) $$

This function you have to derive wrt $a, b, c$ and $\lambda$. This gives for $a$:

$$\frac{\delta f}{\delta a} = \ln a + 1 - \lambda bc = 0$$ or $$a \ln a + a = \lambda \hspace{1cm} (bc = \frac{1}{a})$$ and you get the same for $b$ and for $c$.

And since $a \ln a + a$ is a continuously increasing function, which means f(x) = f(y) => x = y, we can conclude that $a = b = c$. And from $abc = 1$ it follows that $a = b = c = 1$.

Is this now clear?


问:
Continuous function on a closed set
Let $f: F \to \mathbb R$ be defined in a closed set $F \subset \mathbb R$. Show that $f$ is continuous if and only if for all $c \in \mathbb R$, the sets $E[f \le c]=\{x \in F; f(x) \le c\}$ and $E[f \ge c]=\{x \in F; f(x) \ge c\}$ are closed.</p>

答:
HINT: One direction is very easy. For the other, assume that the sets $E[f\le c]$ and $E[f\ge c]$ for $c\in\Bbb R$ are closed. Then the sets $\{x\in F:f(x)>c\}$ and $\{x\in F:f(x)<c\}$ are relatively open in the subspace topology on $F$ for each $c\in\Bbb R$. This implies that if $a,b\in\Bbb R$ with $a<b$, then

$$\{x\in F:a<f(x)a\}\cap\{x\in F:x<b\}$$

is open. Use this to show that $f$ must be continuous.

Title: Continuous function on a closed set

Body:
HINT: One direction is very easy. For the other, assume that the sets $E[f\le c]$ and $E[f\ge c]$ for $c\in\Bbb R$ are closed. Then the sets $\{x\in F:f(x)>c\}$ and $\{x\in F:f(x)<c\}$ are relatively open in the subspace topology on $F$ for each $c\in\Bbb R$. This implies that if $a,b\in\Bbb R$ with $a<b$, then

$$\{x\in F:a<f(x)<b\}=\{x\in F:x>a\}\cap\{x\in F:x<b\}$$

is open. Use this to show that $f$ must be continuous.


问:
Split group of courses into two with constraints
The school I work for is trying to split the district of students into an A group and a B group while taking into consideration students from the same family and balancing class sections to be as close to 50/50 as possible.

I have a spreadsheet of students that is similar to the following:

Could anyone point me into a direction that would allow me optimize this list into two groups and maintaining the family to be in the same group? Meaning - All of Family ID 286 need to be in the same group A or B? All while balancing CourseID to be as close to a 50/50 split as possible? The end goal is to have each course ID split as equally as possible and maintain families within the same group.

Unfortunately I don't even know what kind of math this would be - so not sure what tag to assign it to?
Thanks!

答:
Number the students from $1$ to $n$.  For each student $k$ we'll have a variable $x_k$ that takes the value $0$ or $1$, where $0$ indicates that the student is assigned to group A, and $1$ that he is assigned to group B.  For each class, we need an equation, or an inequality, indicating that about half the class is in group B.

For example, suppose one class is made up of students $1,2,4,8,9,11,14$.  There are $7$ students in the class, so ideally we want $$3\leq x_1+x_2+x_4+x_8+x_9+x_{11}+x_{14}\leq4$$ When students are in the same family, their variables must take the same value.  If students $4,8,17$ are in the same family, then we have the equations $$x_4=x_8\\x_4=x_{17}$$

This then is a problem in $0$-$1$ integer linear programming, in which only the constraints must be satisfied, and unfortunately, this is known to be NP-complete.

In the form I have stated the problem, it isn't necessarily true that a solution exists, so we should perhaps loosen the requirements.  For each class, define the "assignment error" to be the difference between the number of group B students in the class and half the class size.  For the example class above, the error would be $$\left\lvert x_1+x_2+x_4+x_8+x_9+x_{11}+x_{14}-3.5\right\rvert$$  Then we would seek to minimize the sum of the assignment errors over all classes, subject to the family constraints.

Still we have $0$-$1$ variables, so I believe, though I'm not certain, that it's still NP-complete.  I am not at all expert in this area, as I imagine has been apparent from the discussion.  I've edited the tags in hopes of attracting the attention of an expert.

How many students and classes do you have?  You should add as much detail as possible to your question, so that if an expert does look at it, he'll be able to give you good advice.

You'll almost surely need software to do the problem, unless it's very small, and I'm sure there would be a learning curve.  Here's a list of such software to get you started.

Also, if you don't have any luck here, you might try https://scicomp.stackexchange.com/ which seems to have related questions.

Sorry I can't be of more help.  Good luck.


问:
Using residue theorem to evaluate integral and calculating residues.
Evaluate the integral $\int_{0}^{2\pi} \frac {\cos^2(x)}{13+12\cos(x)} \,dx$ using the residue theorem.

I have managed to make a start on this problem by putting this problem in a complex analysis setting i.e. by changing the integral to $\int_{\gamma} -\frac {\frac {i}{4} (z+\frac {1}{z})^2}{\left(13+6z+\frac {6}{z}\right)z} \,dz$ where $\gamma$ is the path $ t \mapsto e^{it}$. I then found that the singularities of the function $f(z) =  -\frac {\frac {i}{4} \left(z+\frac {1}{z}\right)^2}{\left(13+6z+\frac {6}{z}\right)z}$ are at $z=0,z=-\frac {2}{3}, z = -\frac {3}{2}$.

I know that I need to now calculate the residues for $z=0, z= -\frac {2}{3}$ (and not $z=-\frac {3}{2}$ since this singularity lies 'outside' the path) however I am unsure of how to do this.

答: You want to compute$$\int_0^{2\pi}R(\cos\theta)\,\mathrm d\theta,\tag1$$with $\displaystyle R(x)=\frac{x^2}{13+12x}$. So, consider\begin{align}\frac1zR\left(\frac{z+1/z}2\right)&=\frac{(z^2+1)^2}{4z^2(6z^2+13z+6)}\\&=\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)},\end{align}and then $(1)$ is equal to$$2\pi\left(\operatorname{res}_{z=0}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right)+\operatorname{res}_{z=-2/3}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right)\right).$$In order to compute$$\operatorname{res}_{z=0}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right),\tag2$$define$$g(z)=\frac{(z^2+1)^2}{24\left(z+\frac23\right)\left(z+\frac32\right)},$$and then $(2)=g'(0)=-\frac{13}{144}$. And, in order to compute$$\operatorname{res}_{z=-2/3}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right),\tag3$$define$$h(z)=\frac{(z^2+1)^2}{24z^2\left(z+\frac32\right)}=\frac{(z^2+1)^2}{12z^2(2z+3)},$$and then $(3)=h\left(-\frac23\right)=-\frac{169}{720}$. So, $(1)$ is equal to $2\pi\left(-\frac{13}{144}+\frac{169}{720}\right)=2\pi\frac{13}{90}=\pi\frac{13}{45}$.


问:
How to calculate trend line
I have the following dataset

```<code>5 338.45
6
7
8
9
10 350.75
11
12
13
14
15 380.70
</code>```

I need to calculate the missing values by plotting a linear trend line between the values I know (5, 10 and 15)

I did that by doing (350.75-338.45)/6 = 2.05 then adding that between 5 and 10 so:

```<code>5 338.45
6 340.5
7 342.55
8 344.6
9 346.65
10 350.75
</code>```

Then same between 10 and 15 by adding 4.99

```<code>10 350.75
11 355.74
12 360.73
13 365.72
14 370.71
15 380.70
</code>```

My question is once all that is calculated, if I had new 5 and 15 values, how do I apply those same trendlines to it

I only have new 5 and 15, not the 10.

```<code>5 295.99
6
7
8
9
10
11
12
13
14
15 309.99
</code>```

Any help appreciated, thanks in advance!

答:
What you did seems fine.

If you have new values for 5 and 15 and they replace the old ones, you just need to repeat the process which you already did while using the new values.

But if you mean... you need to add the new values for 5 and 15 to the old dataset, then this would make no sense.

You are building a model here. You need to specify if you expect 5 (or 10) to have a single value or can have more than 1 values associated with it.

It really depends on what you're trying to model here and what your initial assumptions are.

Title: How to calculate trend line


问:
Describe using polar coordinates the contour line of $f(x,y)=\frac{2xy}{x^2+y^2}$
Describe using polar coordinates the contour line of $f(x,y)=\frac{2xy}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $0$ if $(x,y)=(0,0)$

I did $\frac{2xy}{x^2+y^2}=k$ the contour line

then $\frac{2(r \cos\theta)(r \sin\theta)}{r^2  \left(cos^2(\theta)+\sin^2(\theta)\right)}=2 \cos\theta \cdot \sin \theta=k$ is the countourr line in polar coordinates is right?

答: Correct. Equivalent forms such as $\sin2\theta=k$ may also be useful.


问:
Clarification on Proof that "for $p$ a prime, the elementary abelian group of order $p^2$ has exactly $p+1$ subgroups or order $p$"
I'm reading in Dummit's Abstract Algebra the example (3) p. 156, where they prove the claim "for $p$ a prime, the elementary abelian group of order $p^2$, $E$ (i.e. $Z_{p} \times Z_{p}$, where $Z_{p}$ is the cyclic group of order $p$) has exactly $p+1$ subgroups or order $p$".

They start proving this by saying that, since obviously each nonidentity element of $E$ has order $p$ (since it happens for every element of $Z_{p}$), then they generate a cyclic subgroup of order p. Noting that each of this subgroups has exactly $p-1$ generators (since $p$ is prime) and that by Lagrange's Theorem distinct subgroups of order $p$ intersect trivially, they deduce there are $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$.

I agree with the fact we have found $p+1$ subgroups of order $p$, but I don't know why they suppose they are all the subgroups of order $p$.

答: Each of the $(p+1)$ subgroups has order $p$, and they intersect trivially.  Thus we count $(p+1)(p-1)+1=p^2$ elements in the group.


答: As @saulspatz described, you can solve the problem exactly via integer linear programming.  An alternative heuristic approach to this partition problem is to apply a greedy algorithm that is guaranteed to be no more than 17% above optimal and is typically closer than 17%.  First merge all students from the same family, yielding one positive integer count per family.  Then sort the families in decreasing order of this count.  Finally, loop through this list, adding each family to the group that has the smaller number of people.</p>

Title: Split group of courses into two with constraints

Body:
The school I work for is trying to split the district of students into an A group and a B group while taking into consideration students from the same family and balancing class sections to be as close to 50/50 as possible.

I have a spreadsheet of students that is similar to the following:

Could anyone point me into a direction that would allow me optimize this list into two groups and maintaining the family to be in the same group? Meaning - All of Family ID 286 need to be in the same group A or B? All while balancing CourseID to be as close to a 50/50 split as possible? The end goal is to have each course ID split as equally as possible and maintain families within the same group.

Unfortunately I don't even know what kind of math this would be - so not sure what tag to assign it to?
Thanks!


问:
How will the plot of CDF and PDF look like for $A\ =\ \sqrt{X_1^2\ +\ X_2^2}$ where $X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.
I know Chi-Square distribution. Here the expression is similar to Chi-square distribution with the degree of freedom being 2, except for the square root.

Given:
$X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.

$A\ =\ \sqrt{X_1^2\ +\ X_2^2}$

How to find the CDF $F_A(x)$ and PDF $f_A(x)$?

The PDF of $\chi^2$ distribution for $\nu$ degrees of freedom is given
by -

$f(\nu^2)\ =\ \frac{1}{2^{\nu / 2}.\Gamma(\nu /
 2)}.e^{-\chi^2/2}.(\chi^2)^{\nu/2-1}\ \ ;\ \chi^2\ \geq\ 0$

Can I substitute 2 in place of $\nu$ to get the expression for PDF $f_A(x)$?

答: If $X_1$ and $X_2$ are independent, $A$ is Rayleigh. If not, the dependency must be determined. For instance, if $X_1$ and $X_2$ are jointly normal, their correlation suffices to calculate the distribution of $A$. If not, the joint distribution is required. If the dependency is unknown, the distribution of $A$ is also unknown.


问:
Showing how many times 3 shows up in compositions of n greater than or equal to 4
Suppose $n \geq  4$. Show that in a list of all $2^{n−1}$
compositions of $n$, the integer 3 occurs exactly $n\cdot 2^{n-5}$
times.
(Hint: As in lectures, look at ways of drawing lines between $n$ dots.)

I understand that for 4, say, you draw a line between the 1st and 2nd dot and then you have the partition (1,3) and also drawing a line between the 3rd and 4th dot gives you the partition (3,1). There's the two examples for 4.

It's when you get to the examples for $n \geq 4$ that I struggle. I know that there will be a sum ...+3+... and that the numbers on either side of the 3 are important but how this then gets to $n \cdot 2^{n-5}$ is where I lose track.

I don't have an example like this in my lecture notes. Only the formulas to say that the number of compositions of n is $2^{n-1}$ and a diagram explaining how the dots and bars works to find total compositions of n but nothing on how these could be used to find the number of compositions containing a specific integer.

答:
Imagine if you will making this "dots and bars" representation for every possible composition and laying them out in a grid.

$$\begin{array}{cccccccc}\bullet &\bullet &\color{red}{\mid \bullet} &\color{red}{\bullet} &\color{red}{\bullet} &\color{red}{\mid} \bullet &\mid\bullet&\cdots\\\bullet &\bullet &\mid \bullet &\mid\bullet &\bullet &\mid \bullet& \bullet&\cdots\\\bullet &\mid\bullet &\mid \bullet &\mid\bullet &\bullet &\mid \bullet &\bullet&\cdots\\\bullet&\color{red}{\mid\bullet}&\color{red}{\bullet}&\color{red}{\bullet}&\color{red}{\mid}\bullet&\bullet&\mid\bullet&\cdots\\&&\vdots&&\vdots\end{array}$$

The above example corresponding to $2+\color{red}{3}+1+\dots,~2+1+2+\dots,1+1+1+2+\dots,1+\color{red}{3}+2+\dots$ etc...

Now... let us count the number of $3$'s in a partition that occur not by organizing our thoughts into rows of the grid... but rather organize our thoughts in terms of columns.

We count how many partitions have a pattern $\color{red}{\mid \bullet\bullet\bullet\mid}$ which started in column $k$ for each $k$ from $2$ to $n-3$.  (We focus our attention at the moment on those which occur central in the grid, not on the edge.  We consider the literal edge cases later)

Well, to count these, we note that the bar on the left of the pattern must be present, the two places for bars in the middle must be empty and the bar on the right must be present.  All other positions of bars can either be present or not be present with no restriction.  There are $n-1$ possible positions for bars in general, $4$ of which have their state pre-determined since we are considering where those positions gave rise to a $3$ in the composition, leaving $n-1-4=n-5$ positions remaining to decide whether they have or do not have a bar.  This gives rise to $2^{n-5}$ compositions who have a $3$ in the desired columns within our grid.

Note, we do not care that we have double-counted some of these compositions which had multiple $3$'s because we wanted such compositions to have contributed to our total multiple times... once for each of the $3$'s within the composition which is exactly what we have done.

Since there are $n-4$ different choices for which "inner columns" we could have considered, this gives us $(n-4)\cdot 2^{n-5}$ occurrences of $3$ within compositions we have counted so far.

Now, we still need to count the edge cases where our composition began with a $3$ or ended with a $3$.  In these scenarios, we only have $n-4$ remaining positions for bars to worry about whether they were included or not, not $n-5$, giving $2^{n-4}=2\cdot 2^{n-5}$ for those which began with a $3$ and the same for those ending with a $3$.

This gives us a final total of:

$$(n-4)\cdot 2^{n-5}+2\cdot 2^{n-5}+2\cdot 2^{n-5}=(n-4+2+2)\cdot 2^{n-5}=n\cdot 2^{n-5}$$

occurrences of $3$'s within the collection of all compositions of $n$.


问:
Simple numerical methods for calculating the digits of $\pi$
Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?</p>

答:
Check out the
questions tagged 'pi' at Stack Overflow.
Many of them discuss simple numerical methods for calculating any number of digits of pi.</p>

ContentLicense: CC BY-SA 2.5


问:
Given $2$ randomly chosen integers $x,y$ what is $P(k=gcd(x,y))$?
Given $2$ randomly chosen integers $x,y$ what is the probability that a integer $k$ is the greatest common divisor of $x$ and $y$?

I know that the probability of $x,y$ being coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$ this could be a starting point. I think one could try to analyse $P(x/k $  and  $ y/k$ coprime $| k$ divides $x$ and $y)$ But then the distribution of $x$ and $y$ is no longer uniform. Maybe there is some helping hand here?

答:
Just a copy...

Another approach which is easier and is more probabilistic from the beginning...

One may use the honest probability $P(X=n)=n^{-s}/\zeta(s)$ for $s>1$.
From this we may prove that:

1 Events $E_{p}=\{X$ is divisible by $p\}$ are independent.

2 $P(gcd(X,Y)=n)=n^{-2s}/\zeta(2s)$ for independent $X$ and $Y$. If $s\to 1$ for $n=1$ we obtain the claim.

See ex. 4.2. in Probability with martingales by Williams.


问:
Plotting parametric equations in Maple
I have the following problem. Let $\alpha:=\alpha(v)$ and $\beta:=\beta(v)$, where $v\in\mathbb{R}$. Using MAPLE, I would like to plot $\alpha$ and $\beta$ in the $(\alpha,\beta)$ space as $v$ varies continuously. I suspect that this will involve creating lists in Maple. I need some help! Here is what I am thinking at this time:

```<code>restart:

alpha:=v-> some function of v:
beta:=v-> some function of v:

for v from 0.01 to 10 by 0.001 do
  alpha:=alpha(v):
  beta:=beta(v):
  ? put the alpha and beta into a list?
od:

?plot the sequence of points in the list?
</code>```

I need some direction/help from Maple users out there, please.

答:
You can plot a list of points (where each 2D point is represented as a list),

```<code>restart:
alpha:=v->sin(v):
beta:=v->cos(v):

plot([seq([alpha(v),beta(v)], v=0.01..10.0, 0.001)]);
</code>```

Or you could make a parametric plot, optionally forcing the number of points if you wish,

```<code>restart:
alpha:=v->sin(v):
beta:=v->cos(v):

plot([alpha(t),beta(t),t=0.01..10.0]);

plot([alpha(t),beta(t),t=0.01..10.0],adaptive=false,numpoints=9991);
</code>```

See also the plot,parametric help-page.

Title: Plotting parametric equations in Maple

Body:
You can plot a list of points (where each 2D point is represented as a list),

restart:
alpha:=v->sin(v):
beta:=v->cos(v):

plot([seq([alpha(v),beta(v)], v=0.01..10.0, 0.001)]);

Or you could make a parametric plot, optionally forcing the number of points if you wish,

restart:
alpha:=v->sin(v):
beta:=v->cos(v):

plot([alpha(t),beta(t),t=0.01..10.0]);

plot([alpha(t),beta(t),t=0.01..10.0],adaptive=false,numpoints=9991);

See also the plot,parametric help-page.


问:
Non complete connected Riemannian manifold
In John M. Lee "Riemannian manfidolds: an introduction to curvature", page 108, as a previous motivation for Hopf-Rinow theorem, there is an example of non geodesically complete manifold:

"$\mathbb{R}^n$ with the metric $(\sigma^{-1})^{*}\mathring{g}$ obtained from the sphere by stereographic projection"

In the book says that there are geodesics that escape to infinity in finite time.

I am trying to prove it for the case $n=2$. I was intending to compute explicity the pull back via the inverse of stereographic projection (https://en.wikipedia.org/wiki/Stereographic_projection#First_formulation) of the round metric in the sphere, but I did not get anywhere. Do you know how to prove it? (Without using Hopf-Rinow theorem, I want to see this as a motivation for that theorem).

答:
Let $\mathbb{S}^n$ be the $n$ dimensional sphere, $N = (0,\ldots,0,1)$ its north pole and $S$ its south pole. Let $g$ be the usual round metric on $\mathbb{S}^n$. Then $\left(\mathbb{S}^n \setminus\{N\},g\right)$ is a Riemannian manifold.

Suppose $v$ is a unit tangent vector at $S$. Define
$$
\forall t \in \left(-\pi,\pi\right),~\gamma(t) = \cos t \cdot S + \sin t \cdot v
$$
Then $\gamma$ is a geodesic. It goes "out" of $\mathbb{S}^n\setminus\{N\}$ in finite time (at $t = \pi$).

Now, let $f : \mathbb{S}^n\setminus\{N\} \overset{\sim}{\to} \mathbb{R}^{n+1}$ be the stereographic projection. It is a diffeomorphism, which allow us to push forward the metric $g$ on $\mathbb{R}^{n+1}$, say $h = f_*g$ (or $h = (f^{-1})^*g$), so that it becomes an isometry.

Being an isometry, $f$ maps geodesics to geodesics, and thus, $f\circ \gamma : \left(-\pi,\pi\right) \to \mathbb{R}^{n+1}$ is a geodesic. It goes to infinity in finite time ($t = \pi$).

Remark that if $(M,g)$ is a connected Riemannian manifold and if $x \in M$, then $(M\setminus\{x\},g)$ cannot be complete. Any geodesic of $M$ passing through $x$ can be viewed, by a sort of local stereographic projection, as a geodesic in $M\setminus\{x\}$ going out of $M\setminus\{x\}$ in finite time. Here, the stereographic projection identify $x$ with "infinity".


问:
Computing $\int_B e^{(x^2+y^2+z^2)^{\frac32}} \ dx \ dy \ dz$, where $B$ denotes is the unit sphere.

Compute the integral $$\int_B e^{(x^2+y^2+z^2)^{\frac32}} \ dx \ dy \ dz$$ where $B= \{(x,y,z) \mid x^2+y^2+z^2 \leqslant 1 \}.$

My approach was to do a spherical coordinate transform and proceed from there, but had trouble finding the limits of integration...

After the transformation I'm left with $$\int_B e^{{r^2}^{\frac32}} r^2\sin(\varphi) \ dr \ d\varphi \ d\theta$$

Since I'm dealing with the unit sphere and I have that $r \in [0,1]$ I considered $\theta\in[0, 2\pi]$ and $\varphi \in [0, \pi]$ which would cover the whole sphere and thus

$$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} e^{{r^2}^{\frac32}} r^2\sin(\varphi) \ dr \ d\varphi \ d\theta$$

which will evaluate to $4\pi \cdot \frac{e-1}{3}$. I'm not sure that I got the limits right though...

Shouldn't I only consider the positive quadrant, since the integrand isn't never negative?

答: You're almost there: $\int_0^1 r^2e^{r^3}dr=[\tfrac13e^{r^3}]_0^1=\tfrac13(e-1)$ is the hard part.


问:
Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.

This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?

答:
Partial answer :

Remarking that the function $f(x)=\frac{e^x}{e^x(n-1)+1}$ is concave on $(-\ln(n-1),\infty)$ and rewriting the inequality like :

$$S=\sum_{i=1}^{n}\frac{e^{x_i}}{e^{x_i}(n-1)+1}\leq 1$$

We apply Jensen's inequality to get the value $1$ since $e^{\sum_{i=1}^{n}x_i}=1$ and $x_i\in(-\ln(n-1),\infty)$


问:
How many Borel conjectures are there
The following may be referred to as Borel conjecture:

Every strong measure zero set of reals is countable.

On the other hand Wikipedia refers to the following as the Borel conjecture:

Let  $M$ and $N$ be closed and aspherical topological manifolds, and let
$f:M\to N$
be a homotopy equivalence. The Borel conjecture states that the map $f$  is homotopic to a homeomorphism.

The two seem to be unrelated. Is the naming confusing coincidence? Or are they related? Moreover are they really still conjectures or were they subsequently solved? The article I link to does not mention the status of the "conjecture(s)".

答:
These Borels are different people.

The first one is Émile Borel, whereas the second one is Armand Borel. From the latter's Wikipedia page,

He used to answer the question of whether he was related to Émile Borel alternately by saying he was a nephew, and no relation.

Title: How many Borel conjectures are there


问:
Confusion about the space of functions on $D(x) \cup D(y)$ in $\mathbb A^2_k$
The following questions concern notes from Vakil's notes on Algebraic Geometry (page 140):

How is it clear that the rational functions with only powers of $x$ in the denominator, and only powers of $y$ in the denominator, are the polynomials? The function $1/y$ is not even in $D(x)$.

Also, if this is the case, then how is $\Gamma(U, \mathcal O_{\mathbb A^2_k})=k[x,y]$?

答:
Suppose $s$ is a rational function with only powers of $x$ in the denominator. So $s = g/x^m$ and we can assume $x \not\mid g$. Now suppose $s$ also has only powers of $y$ in the denominator. So $s = h/y^n$ and again we may as well assume $y \not\mid h$.

So $gy^n = hx^m$ and since $x \not\mid g$ we know $x^m \mid y^n$. Thus $m = 0$. Likewise $n = 0$. Therefore $s = g = h$ is a polynomial.


问:
Derivative of $f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$
$f:{\mathbb{R}}^{+}\cup \left\{0\right\}\to \mathbb{R}$

$f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$ then what is the derivative of function $f?$

A)$\frac{-1}{\sqrt{x}}$

B)$\frac{1}{\sqrt{x}}$

c)$\frac{1}{2\sqrt{x}}$

Here is my solution: I found derivative of function is
$$\frac{-1}{2\sqrt{x}}\left(\frac{1}{\sin\sqrt{x}}+\frac{1}{\cos\sqrt{x}}\right).$$

But I can not reach the answer that given in question. Any help will be appreciated.

答:
Hint:

$$f(x)=\arcsin(\cos\sqrt x)+\dfrac\pi2-\arcsin(\sin\sqrt x)$$

$$=\dfrac\pi2+\arcsin(\cos\sqrt x)+\arcsin(-\sin\sqrt x)$$

Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,

$$f(x)=\dfrac\pi2+\arcsin\left(\cos\sqrt x|\sin\sqrt x|-\sin\sqrt x|\cos\sqrt x|\right) $$

as here $x=\cos\sqrt x,y=-\sin\sqrt x\implies x^2+y^2=1$

Now  use $|u|=\begin{cases}u&\mbox{if } u>0 \\-u &u\le0\end{cases}$

For example if $\cos\sqrt x,\sin\sqrt x$ have the same sign $\iff n\pi\le\sqrt x\le n\pi+\dfrac\pi2$ where $n$ is any integer

$$f(x)=\dfrac\pi2+\arcsin0=?$$

Can you take it from here?


答:
Applying the chain rule we get:
$\\ f'(x)=\frac{1}{\sqrt{1-(\cos\sqrt{x})^2}}\cdot (-\sin\sqrt{x})\cdot\frac{1}{2\cdot\sqrt{x}}-\frac{1}{\sqrt{1-(\sin\sqrt{x})^2}}\cdot (\cos\sqrt{x})\cdot\frac{1}{2\cdot\sqrt{x}}=\\=-\frac{1}{2\cdot\sqrt{x}}\cdot(\frac{\sin\sqrt{x}}{|\sin\sqrt{x}|}+\frac{\cos\sqrt{x}}{|\cos\sqrt{x}|})$

So the correct answer, as @Bernard Massé was saying, is the first one, but only in the interval $(0,(\frac{\pi}{2})^2)$, where both $\sin\sqrt{x}$ and
$\cos\sqrt{x}$ are positive..

Title: Derivative of $f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$

Body:
$f:{\mathbb{R}}^{+}\cup \left\{0\right\}\to \mathbb{R}$

$f(x)=\arcsin\left(\cos\sqrt{x}\right)+\arccos\left(\sin\sqrt{x}\right)$ then what is the derivative of function $f?$

A)$\frac{-1}{\sqrt{x}}$

B)$\frac{1}{\sqrt{x}}$

c)$\frac{1}{2\sqrt{x}}$

Here is my solution: I found derivative of function is
$$\frac{-1}{2\sqrt{x}}\left(\frac{1}{\sin\sqrt{x}}+\frac{1}{\cos\sqrt{x}}\right).$$

But I can not reach the answer that given in question. Any help will be appreciated.


问:
Well ordered Isomorphic sets
(ZF) Let A be a infinite set. Prove that if there is a well order
≤1 in A, then there is a well order ≤2 in A, for which  and
are not isomorphic.

My attempt:
I start with the use of the definition of well ordered set, hence for every non empty B in A there exists a minimal element to ≤:
(∀ B)[B != ∅ & B ⊆ A → (∃ y)[y ∈ A & (∀ x )[x ∈ A → y ≤ x ]]].\

Also we say that  is isomorphic to  if there is a bijection f:A1-> A2 such that if for all x and all y, <x,y> is in R1, then <f(x1),f(x2)> is in R2.

If I asume that A1 and A2 are well ordered, then they have a minimal element x0,y0, respectively.Then if A1,A2 are isomorphic there is bijection f such that f(x0)=z1, z1 in A2. Since f preserves the order=> z1 != y0 is a minimal element-contradiction. Hence A1 and A2 are not isomorphic.

I beg someone to check my attemp. I will be thankful to anyone who share their ideas.

答:
There is no actual argument here. You start by assuming that there is an isomorphism $f$ from $\langle A,\le_1\rangle$ to $\langle A,\le_2\rangle$, and then you simply assert that this isomorphism is in fact not an isomorphism, because $f(\min_{\le_1}A)\ne\min_{\le_2}A$. Assuming something and then asserting its negation is not a proof by contradiction: it is simply asserting two contrary things independently.

Moreover, the result holds only for infinite sets, and nowhere have you used the fact that $A$ is infinite.

What you must do is assume that $\le_1$ well-orders $A$ and use $\le_1$ to produce a new well-order $\le_2$ that is not isomorphic to $\le_1$. I will tell you one way to do this, but I will leave it to you to prove that it works.

Let $a=\min_{\le_1}A$, the least element of $A$ with respect to the order $\le_1$. To form the new order $\le_2$, move $a$ to the other end. That is, define $x\le_2y$ if and only if $x\le_1y$ and $x\ne a$, or $y=a$. Pictorially, if $\langle A,\le_1\rangle$ looks like this:

$$a\;\bullet\;\bullet\;\bullet\;\cdots$$

then $\langle A,\le_2\rangle$ looks like this:

$$\bullet\;\bullet\;\bullet\;\cdots\;a$$

Title: Well ordered Isomorphic sets


问:
Trouble with partial derivatives and changing variables
I am struggling with how to bring a factor of $\frac{1}{r}$ inside of a derivative in the following question:

Spherical polar coordinates $(r,\theta,\phi)$ are defined in terms of cartesian coordinates $(x,y,z)$ by

$$x=r\ \mathrm{sin}\ \theta\ \mathrm{cos}\ \phi, \ y=r\ \mathrm{sin}\ \theta\ \mathrm{sin}\ \phi,\ z = r\ \mathrm{cos}\ \theta$$
Given that $f$ is a function of $r$ only, independent of $\theta$ and $\phi$, show that
$$\frac{\partial^2f}{\partial x^2} = \frac{1}{r}\frac{df}{dr} + \frac{x^2}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{df}{dr}\right)$$
given that
$$\frac{\partial f}{\partial x} = \frac{x}{r}\frac{df}{dr}$$.

I can take a partial derivative of $\frac{\partial f}{\partial x}$ with respect to $x$ using the product rule:

$$\frac{\partial^2f}{\partial x^2} = \frac{df}{dr}\frac{\partial}{\partial x} \frac{x}{r} + \frac{x}{r}\frac{\partial}{\partial x}\frac{df}{dr}$$
$$= \frac{1}{r}\frac{df}{dr} +\frac{x}{r}\frac{\partial r}{\partial x}\frac{\partial}{\partial r}\left(\frac{df}{dr}\right) $$
$$\frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2+z^2} = \frac{x}{r}$$

$$ \rightarrow \frac{1}{r}\frac{df}{dr} + \frac{x^2}{r^2}\frac{d}{dr}\left(\frac{df}{dr}\right)$$
and I don't see why this is equivalent to the "show that" expression.
Can anyone help me see why? Is it because my naive attempt at a chain rule has messed things up?

答:
We start with the property

$$\frac{\partial f}{\partial x}=\frac{x}{r}\frac{df}{dr}$$

Taking the partial derivative with respect to $x$,

$$\frac{\partial^2f}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{x}{r}\frac{df}{dr}\right)=\frac{1}{r}\frac{df}{dr}+x\frac{\partial}{\partial x}\left(\frac{1}{r}\frac{df}{dr}\right)$$

Here, all we did was employ the product rule. The function $\frac{1}{r}\frac{\partial f}{\partial r}$ is a function that only depends on $r$, so we can employ our starting equation again, where the partial derivative with respect to $x$ is equivalent to the total derivative with respect to $r$ then multiplying by $x/r$. Thus,

$$\frac{\partial^2f}{\partial x^2}=\frac{1}{r}\frac{df}{dr}+x\frac{\partial}{\partial x}\left(\frac{1}{r}\frac{df}{dr}\right)=\frac{1}{r}\frac{df}{dr}+x\frac{x}{r}\frac{d}{dr}\left(\frac{1}{r}\frac{df}{dr}\right)$$

and we are done. The error you have made is in factoring the $1/r$ out when taking the derivative with respect to $x$, which is not correct, since $r$ is also a function of $x$. Hope this helps!

Edit: As an explanation as to why your derivation was incorrect, we need to dive a bit into the notation that we use to represent derivative operators and how these operations change. It all has to do with coordinates, which are different from points. A point is just that; a point in a greater space. When working with PDEs, a point is just a location inside a domain, or along the boundary of the domain. A coordinate is a collection of numbers used to identify the point. You and I can have completely different coordinates but describe the same set of points, and so long as our equations are consistent then there is no issue.

The derivative operators we know and love so much cause an issue because of this. If we wanted a language we could both agree on, we would describe the change in a function as we move between points, but points are abstract and don't have native $\mathbb{R}^n$ representations, so there is no way to describe derivatives concretely without using coordinates (this not entirely true, since we can describe derivatives in terms of other properties, but the notion of direction then gets complicated). This means that if I want to talk about derivatives, I should stick with one set of coordinates and try not to mix them. If I do, I have to try to think of the quantities in question as functions of just one set of coordinates, not both.

We start with the function $f$. I can think of it as a function of $(x,y,z)$ and compute the quantity $\frac{\partial f}{\partial x}$. To find out how to compute this value in your coordinates, we instead assume that $f$ is a function of $(r,\theta,\phi)$ and do

$$\frac{\partial }{\partial x}f(r,\theta,\phi)=\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\frac{\partial\theta}{\partial x}+\frac{\partial f}{\partial\phi}\frac{\partial\phi}{\partial x}$$

We are told that $f$ only has radial dependency, so we again abuse notation and wriite this as

$$\frac{\partial f}{\partial x}=\frac{df}{dr}\frac{\partial r}{\partial x}$$

We have

$$r^2=x^2+y^2+z^2\implies2r\frac{\partial r}{\partial x}=2x\implies\frac{\partial r}{\partial x}=\frac{x}{r}$$

This last part mixes coordinates, so it's less than ideal. When we take future derivatives, we either have to think of the whole thing as a function of $(x,y,z)$, or as a function of $(r,\theta,\phi)$. For example,  if we wanted the derivative with respect to $x$ (like we do), we would do

$$\frac{\partial}{\partial x}\left(\frac{x}{r(x,y,z)}\right)$$

If instead we wanted a derivative with respect to $r$, we would have to write

$$\frac{\partial}{\partial r}\left(\frac{x(r,\theta,\phi)}{r}\right)$$

Ultimately, we have to make sure that when taking a derivative in one coordinate system, we treat the other coordinate system as functions. This all comes from the fact that we are abusing the notation and treating the coordinates as the points, which is not correct. To do this all rigorously, we would have to define functions that map points to coordinates, then take derivatives of composite functions, and the notation would get cumbersome. Instead, we settle for slightly confusing writing to simplify the nasty true writing.

I can dive even deeper if you'd like, but that's your quick and dirty explanation.


问:
Is the solution that Wolfram Alpha gives really the only solution to this problem?

My formula

$$(z-y)(z^2+zy+y^2)=(y-x)(y^2+yx+x^2)$$
$$z>=y+2$$
$$y>=x+2$$
$$x>=2$$

I am trying to understand a geometric problem better by plugging my formula into Wolfram Alpha. Normally I get a result which helps me understand the issue more graphically, but this time it is only spitting out a single solution.

This formula on Wolfram Alpha

The solution from Wolfram Alpha written out below:

$$x>=2$$
$$y>=\frac{\sqrt[3]{x^3+\sqrt{x^6+96x^3+256}+48}}{\sqrt[3]{2}}+\frac{8\sqrt[3]{2}}{\sqrt[3]{x^3+\sqrt{x^6+96x^3+256}+48}}+2$$
$$z=(-1)^{2/3}\sqrt[3]{2y^3-x^3}$$

My questions

Is this the only solution?

How would I know or check this?

Since z has a complex component (-1)^{2/3} in this solution it can only ever be a complex number, correct?

Anything that improves my understanding is greatly appreciated!

答:
Your equation says $x^3 + z^3 = 2 y^3$.  Note that you want $x,y,z$ to be all $\ge 2$.
With $y = ((x^3 + z^3)/2)^{1/3}$, your inequalities are equivalent to
$$ \eqalign{\frac{x^3}{2} + \frac{z^3}{2} &\le (z-2)^3\cr
             (x+2)^3 &\le \frac{x^3}{2} + \frac{z^3}{2}\cr
              x \ge 2\cr}$$
Now the resultant of the polynomials $(x^3 + z^3)/2 - (z-2)^3$ and $(x+2)^3 - (x^3+z^3)/2$ with respect to $x$ is $-108 (3 z^2 - 6 z + 4) (z-2)^3$ whose only real root is $2$.  Thus in the region $z
> 2$, the boundaries of the regions represented by the first and second inequalities don't intersect.  We can then check that the second inequality is implied by the first in this region.  So the answer can be written as

$$ \eqalign{y &= ((x^3 + z^3)/2)^{1/3}\cr
            2 \le x  &\le (2(z-2)^3 - z^3)^{1/3} \cr}$$

Note finally that $2 \le (2(z-2)^3 - z^3)^{1/3}$ if and only if $z$ is greater than the real root of $4 + z^2/2 = (z-2)^3$, which is
$$ (28 + 4 \sqrt{17})^{1/3} + \frac{8}{(28+4 \sqrt{17})^{1/3}} + 4$$


答:
Shortcut solution:

$$\arcsin\left(\cos(\sqrt{x})\right)+\arccos\left(\sin(\sqrt{x})\right)
\\=\arcsin\left(\sin\left(\frac\pi2-\sqrt{x}\right)\right)+\arccos\left(\cos\left(\frac\pi2-\sqrt{x}\right)\right).$$

Using this transformation, the rest is easy.


问:
$5\times 5$ Matrix $A$ such that $A^5 = I_5$ (in $R^5$)
I'm stuck on an exercise I got in my linear algebra class where you have to find a $5\times5$ matrix $A$ such that $A$ multiplied by itself five times is the identity matrix $I_5.$ $A$ must not be the identity matrix itself.

My problem is that I don't know how to go about this in a clever way. I know that it makes sense to put only ones on the main diagonal of A and only zeroes below the ones but I don't know how to construct the rest of the matrix. It seems tedious to just try out random examples and I want to do it systematically.
Any ideas would be appreciated!

答:
A straightforward example would be a matrix that rotates on two of the dimensions by some multiple of $72^\circ$ - as noted in comments here by @BenGrossmann this is a Givens rotation matrix.

... and I now noticed Anthony Saint-Criq's permutation suggestion & subsequent discussion in main comments, that's even easier (as well as being numerically stable), e.g.
$\tiny\begin{pmatrix}
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
\end{pmatrix}$ - ensuring you get a 5-cycle as described (not eg 3+2 length cycles, although that would be a neat way to have $A^{\color{violet}{6}}=I_5$ ).


问:
RSA cryptosystem: If $k$ is a multiple of $\phi(N)$, then $k=2^t r$ with $r$ odd and $t\geq1$
I'm reading Twenty Years of Attacks on the RSA Cryptosystem by Dan Boneh and trying to understand the proof of the Fact 1 on page 3.

Fact 1: Let $(N,e)$ be an RSA public key. Given the private key $d$, one can efficiently factor the modulus $N = pq$.

Dan says that since $\phi(N)=(p-1)(q-1)$ is even, $k=de-1=2^t r$ with $r$ odd and $t\geq1$. $k$ is a multiple of $\phi(N)$. Well, I tried to derive that formula for $k$, but didn't succeed. Any ideas?

答:
The condition $de \equiv 1 \mod \phi(N)$ can also be written as
$$ de - 1 = k = m\phi(N)
$$
with some integer $m$. Now $\phi(N)$ is divisible by 4 since $p-1$ and $q-1$ are both even, hence $k$ is divisible by 4. Keep dividing $k$ by 2 until you get an odd quotient. Then $k = 2^tr$ with $t \ge 2$ and odd $r$.  </p>


问:
Clarification on Proof that "for $p$ a prime, the elementary abelian group of order $p^2$ has exactly $p+1$ subgroups or order $p$"
I'm reading in Dummit's Abstract Algebra the example (3) p. 156, where they prove the claim "for $p$ a prime, the elementary abelian group of order $p^2$, $E$ (i.e. $Z_{p} \times Z_{p}$, where $Z_{p}$ is the cyclic group of order $p$) has exactly $p+1$ subgroups or order $p$".

They start proving this by saying that, since obviously each nonidentity element of $E$ has order $p$ (since it happens for every element of $Z_{p}$), then they generate a cyclic subgroup of order p. Noting that each of this subgroups has exactly $p-1$ generators (since $p$ is prime) and that by Lagrange's Theorem distinct subgroups of order $p$ intersect trivially, they deduce there are $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$.

I agree with the fact we have found $p+1$ subgroups of order $p$, but I don't know why they suppose they are all the subgroups of order $p$.

答:
An alternative is to think of this in terms of linear algebra. The elementary abelian group of order $p^n$ is an $n$-dimensional vector space over $\mathbb{F}_p$; the subgroups of order $p$ are exactly the $1$-dimensional subspaces. Each one dimensional subspace contains $p$ points, and each nonzero point can function as a basis for the subspace. Moreover, the intersection of two distinct $1$-dimensional subspaces must be trivial (dimension $0$) by dimensional considerations.

Thus, each nonzero element generates a one-dimensional subspace, and every one-dimensional subspace is thus obtained,  and we are counting each of them $p-1$ times. Thus, the number of distinct $1$-dimensional subspaces of the elementary abelian group of order $p^n$ is $\frac{p^n-1}{p-1}$.

One can easily generalize this for $k$-dimensional subspaces, i.e. subgroups of order $p^k$: you need to select a linearly  independent subset with $k$ elements. The first one can be picked arbitrarily, $p^n-1$ choices; the next one must be outside the one-dimensional subspace that first generates, $p^n-p$ choices; the next $p^n-p^2$ choices, and so on until the $k$th one has $p^n-p^{k-1}$  choices.  But then we have to account for the overcount. How many distinct ordered bases does a $k$-dimensional space over $\mathbb{F}_p$ has? Well,  using the same counting technique we see that we can pick  an ordered basis in
$$(p^k-1)(p^k-p)\cdots(p^k-p^{k-1})\quad\text{ways.}$$
Thus, the number of $k$-dimensional subspaces is
$$\frac{(p^n-1)(p^n-p)\cdots(p^n-p^{k-1})}{(p^k-1)(p^k-p)\cdots(p^k-p^{k-1})}.$$
In particular, for $n=2$ and $k=1$ we get
$$\frac{p^2-1}{p-1} = p+1.$$