问:
A property of radical ideals
Let $A$ be a commutative ring with $1 \neq 0$.

Theorem (Atiyah-MacDonald 1.13 (v)). Let $\mathfrak{a, b} \subseteq A$ be ideals. Then $\sqrt{\mathfrak{a + b}} = \mathfrak{\sqrt{\sqrt{a} + \sqrt{b}}}$.

Question. Is the following generalization true?

For any finite collection of ideals $\mathfrak{a}_{1}, \dots, \mathfrak{a}_{n} \subseteq A$, we have $$\sqrt{\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}} = \sqrt{\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}}.$$

I'm pretty sure that it is true since if $x_{j} \in \sqrt{\mathfrak{a}_{j}}$ so that $x_{j}^{n_{j}} \in \mathfrak{a}_{j}$ for chosen $n_{j} \geq 1$, each term of $(x_{1} + \cdots + x_{n})^{N}$ is $x_{1}^{r_{1}} \cdots x_{n}^{r_{n}}$ and not all $r_{j} < n_{j}$ if we take $N$ large. (This is a recycled argument of Proposition 1.7 in Atiyah-MacDonald.) I just want to make sure that this statement is correct as I don't have a reference for it.

I am interested in this statement because I'd like to show that given finitely many elements $f_{j} \in A$, we have $\sqrt{(f_{1}, \dots, f_{n})} = \sqrt{0}$ if and only if $f_{j} \in \sqrt{0}$ for all $j$.

It is true when $n = 1, 2$, as follows.

Lemma. Let $f \in A$. We have $\sqrt{(f)} = \sqrt{0}$ if and only if $f \in \sqrt{0}$.

Proof. Since $f \in \sqrt{(f)}$, we only need to show one implication.

Let $f \in \sqrt{0}$. Then $(f) \subseteq \sqrt{0}$. Since taking radical does not change the order of inclusion, we have $\sqrt{(f)} \subseteq \sqrt{\sqrt{0}} = \sqrt{0} \subseteq \sqrt{(f)}$. We showed that $f \in \sqrt{0}$ implies $\sqrt{(f)} = \sqrt{0}$. Q.E.D.

Corollary. Let $f, g \in A$. We have $\sqrt{(f, g)} = \sqrt{0}$ if and only if $f, g \in \sqrt{0}$.

Proof. Since $f, g \in \sqrt{(f, g)}$, we only need to show one implication.

Suppose that $f, g \in \sqrt{0}$. Then by Lemma, we have $\sqrt{(f)} = \sqrt{(g)} = \sqrt{0}$, so $\sqrt{(f, g)} = \sqrt{(f) + (g)} = \sqrt{\sqrt{(f)} + \sqrt{(g)}} = \sqrt{0}$. Q.E.D.

答:
$$\sqrt{\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}} = \sqrt{\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}}$$
Take into account that the radical of an ideal is the intersection of all prime ideals containing it. Now let $\mathfrak p$ be a prime ideal of $R$ containing $\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}$. This is equivalent to $\sqrt{\mathfrak{a}_{i}}\subseteq\mathfrak p$ for all $i$. But $\sqrt{\mathfrak{a}_{i}}\subseteq\mathfrak p$ iff $\mathfrak{a}_{i}\subseteq\mathfrak p$ and you are done. (You can also prove this by induction on $n\ge1$.)

$$\sqrt{(f_{1}, \dots, f_{n})} = \sqrt{0} \text{ if and only if } f_{j} \in \sqrt{0} \text{ for all } j=1,\dots,n$$
This is easy to prove without using the above equality for radicals. We have $(f_{1}, \dots, f_{n})\subseteq\sqrt{(f_{1}, \dots, f_{n})}=\sqrt 0$, so $f_j\in\sqrt 0$ for all $j$. Conversely, if $f_{j} \in \sqrt{0}$ for all $j$, then $(f_{1}, \dots, f_{n})\subseteq\sqrt0$ and therefore $\sqrt{(f_{1}, \dots, f_{n})}\subseteq\sqrt 0$. Since $(0)\subseteq (f_{1}, \dots, f_{n})$ we finally get an equality.

Title: A property of radical ideals


问:
Backward Differential Equation
There is a backward differential equation
$$\frac{\partial}{\partial t} F(x,t) + \frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2} F(x,t) = 0$$
for a function F(x,t). I see that the textbook rewrites this equation as an approximate difference on a two-step binomial x,t lattice with all transition probabilities = 1/2.

And it approximates that $$\frac{\partial}{\partial t} F(x,t) \approx \frac{F(x,t+2dt) - F(x,t)}{2dt}$$
I can understand this step. But it also shows that
$$\frac{\partial^2 F}{\partial x^2} = \frac{F(x+2dx,t+2dt) - 2F(x,t+2dt) + F(x-2dx.t+2dt)}{4\sigma^2dt}$$
I'm a little confused about how to achieve this step. And I'm curious why it represents a backward equation. The textbook says that if you write down the discrete difference equation that corresponds to the first equation, it will explain why. Could someone explain what this sentence means? Thank you very much.

答:
You can apply the same step to approximate in differences the second derivative from the first derivative, remembering that
$$\frac{\partial^2 F}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}\right).$$
I mean, you can use several finite differences approximations (forward, backward, or central differences -I recommend reviewing Wikipedia on this topic, https://en.wikipedia.org/wiki/Finite_difference -) to express your second derivative, like this
$$\frac{\partial }{\partial x}\left(\frac{\partial F}{\partial x}\right)\approx\frac{{\left(\frac{\partial F}{\partial x}\right)}_{forward} -{\left(\frac{\partial F}{\partial x}\right)}_{backward} }{2\mathrm{dx}}=\frac{\frac{F\left(x+2dx,t\right)-F\left(x,t\right)}{2\mathrm{dx}}-\frac{F\left(x,t\right)-F\left(x-2\mathrm{dx},t\right)}{2\mathrm{dx}}}{2\mathrm{dx}}$$
$$=\frac{F\left(x+2dx,t\right)-2F\left(x,t\right)+F\left(x-2\mathrm{dx},t\right)}{{\left(2\mathrm{dx}\right)}^2 }=\frac{F\left(x+2dx,t\right)-2F\left(x,t\right)+F\left(x-2\mathrm{dx},t\right)}{4\;\sigma^2 \;\mathrm{dt}}.$$
Hope it helps.

ContentLicense: CC BY-SA 4.0

Title: Backward Differential Equation


问:
Diffeomorphism from $T M$ to $M \times \mathbb{R}^2$.
Let $M$ be a two-dimensional manifold. Suppose there are two vector fields on $M$, $X(p) = (p,V (p))$ and $Y (p) = (p,W (p))$, such that
$$\text{span}\{V (p),W (p)\} = T_p M$$
for all $p \in  M$.
Construct a diffeomorphism from $T M$ to $M \times \mathbb{R}^2$.

Here is what I know so far:

Since $\text{span}\{V (p),W (p)\}$ span $T_p M$, then there is a diffeomorphism. Since

$T_pM=\cup_{p\in M} T_p M$

$M\times \mathbb{R}^2=\cup_{p\in M} \{p \} \times \mathbb{R}^2$

We need to find an isomorphism from $\mathbb{R}^2$ to $T_p M$ for all $p \in M$. But I still cannot find it. Any help would be appreciated. Thank you.

答:
For all $p\in M$ and $v\in T_pM$, since $\{X(p),Y(p)\}$ is a basis of $T_pM$, there exist unique real coefficients $a_p(v),b_p(v)$ such that $v=a_p(v)X(p)+b_p(v)Y(p)$. Now, it can be seen that $a,b\colon TM\to\mathbf{R}$ are smooth, since $X$ and $Y$ are, then $$TM\ni(p,v)\mapsto(p,a_p(v),b_p(v))\in M\times\mathbf{R}^2,$$
is the required diffeomorphism, whose inverse is given by $(p,x,y)\mapsto(p,xX(p)+yY(p))$.


问:
How to solve a nonhomogeneous Cauchy-Euler equation?
For this question, I'm not sure how to solve the non-homogeneous Cauchy-Euler equation. Here is what I have so far. I'm not sure how to continue. Can anyone please help me out?

$t^2 u''-2u = t^3$

$u''-\frac{2u}{t^2} = t$

Let $ u = t^m$

$u' = mt^{m-1}$

$u'' = m(m-1)t^{m-2}$

$ m(m-1)t^{m-2} - \frac{2t^m}{t^2} = t$

$ t^{m-2} (m(m-1)-2) = t$

Body:
Hint:

$$t^2 u''-2u = t^3$$
$$t^2 u''+2tu'-2tu'-2u = t^3$$
$$(t^2u')'-2(tu)'=t^3$$
$$(t^2u'-2tu)'=t^3$$
$$\left(t^4 \left(\dfrac u {t^2}\right)'\right)'=t^3$$
Integrate.


问:
$\lim_{x\to 0^{+}} \frac{\cos (\sinh x)- \cosh(\sin x)}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}$
I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\cos (\sinh x)- \cosh(\sin x)}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}$$

Applying Taylor's series.

$$\cos(\sinh x)=1- \frac{x^2}{2}+ o(x^2)$$

$$\cosh(\sin x)=1+ \frac{x^2}{2}+ o(x^2)$$

but I don't know how to treat the denominator because the derivatives are not defined in $0$.
Can someone please give me some hints?

答: Try to graph the function (you can use the trigonometric functions or the Taylor series as such) so that you can find the limit by approaching from the right (i.e. $0^+$).


问:
Understand rank-one operators in Hilbert space
Let $\{e_n \}$ be an orthonormal basis on a Hilbert space $H$. When we say that $T$ is a rank-one operator, it means that the range of $T$ has dimension $1$. Then for example, would $Te_n=e_n$ be of rank-one?

I think it's not of rank-one, since in this case we could have $Tx=x$ for $x\in H$. So the range is just $H$, so it doesn't see to be of rank-one.

So in that case would all rank-one operators be constants?

Could I get some examples of rank-one operators?

Thank you!

答:
It is easy to say in general what a rank-one operator on Hilbert space looks like. Suppose that $T$ is rank-one. So $\dim TH=1$. Let $y\in TH$ be nonzero. Then, for any $x\in H$,
$$
Tx=\lambda_xy
$$
for some $\lambda_x\in\mathbb C$. It is not hard to check that the map $x\longmapsto\lambda_x$ is linear and bounded; by the Riesz Representation Theorem, there exists $z\in H$ with $\lambda_x=\langle x,z\rangle$. So
$$
Tx=\langle x,z\rangle\,y.
$$
For some concrete examples, any rank-one matrix $A\in M_n(\mathbb C)$ gives you an example of a rank-one operator on $\mathbb C^n$.


答: You first solve the homogeneous equation using the characteristic polynomial $0=m^2-m-2=(m-2)(m+1)$. As the homogeneous basis is $t^{-1}, t^2$, the right side is not in resonance and the method of undetermined coefficients gives a particular solution in the form $y=At^3$.

Title: How to solve a nonhomogeneous Cauchy-Euler equation?

Body:
For this question, I'm not sure how to solve the non-homogeneous Cauchy-Euler equation. Here is what I have so far. I'm not sure how to continue. Can anyone please help me out?

$t^2 u''-2u = t^3$

$u''-\frac{2u}{t^2} = t$

Let $ u = t^m$

$u' = mt^{m-1}$

$u'' = m(m-1)t^{m-2}$

$ m(m-1)t^{m-2} - \frac{2t^m}{t^2} = t$

$ t^{m-2} (m(m-1)-2) = t$


答:
Hint:
Use equivalence of functions:

\begin{align}
\mathrm e^{\sqrt x}-5^{\sqrt x}&=\mathrm e^{\sqrt x}-\mathrm e^{\ln 5\sqrt x}=1+\sqrt x+o\bigl(\sqrt x\bigr)-1-\ln5\sqrt x+o\bigl(\sqrt x\bigr)\cr
&=(1-\ln 5)\sqrt x+o\bigl(\sqrt x\bigr)\sim_0(1-\ln 5)\sqrt x.
\end{align}
and therefore $\;\bigl(\mathrm e^{\sqrt x}-5^{\sqrt x}\bigr)^4\sim_0 (1-\ln 5)^4x^2$.

Body:
I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\cos (\sinh x)- \cosh(\sin x)}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}$$

Applying Taylor's series.

$$\cos(\sinh x)=1- \frac{x^2}{2}+ o(x^2)$$

$$\cosh(\sin x)=1+ \frac{x^2}{2}+ o(x^2)$$

but I don't know how to treat the denominator because the derivatives are not defined in $0$.
Can someone please give me some hints?


问:
Number of abelian groups Vs Number of non-abelian groups
I would like to see a table that shows the number of non-abelian group for every order n. It is a preferable if the table contains the number of abelian groups of order n (this is not necessary though). If anyone could provide me with such a table , I would be grateful.

Note: I already checked wikipedia. They have a table similar to the one I want but its small.

Here is a simliar table to the one I want:

http://oeis.org/wiki/Number_of_groups_of_order_n#Table_of_number_of_distinct_groups_of_order_n

Thank you

答:
Here are the GAP functions which will allow you to get these numbers:

NumberSmallGroups(n) : gives you the number of groups of order $n$

IdsOfAllSmallGroups : Gives you a list consisting of the ID of each of the groups satisfying the criteria given (type ?IdsOfAllSmallGroups to get a description of the syntax). Note that this will not work for certain orders, as there are just too many such groups. In those cases, you can compute the number of abelian groups by hand, since these just involve the number of partitions of some not too large numbers.

To get the number of the groups, use the function Size() which returns the number of elements in a list.

The above function can also be used to get groups with suitable other properties, such as solvable, nilpotent, M-groups and any other property you might want.

ContentLicense: CC BY-SA 3.0


问:
$ \lim_{\epsilon \to 0^{+}} \int_{a+\epsilon}^{b-\epsilon} f(x) dx = \int_{a}^{b} f(x) dx $
Prove that$$ \lim_{\epsilon \to 0^{+}} \int_{a+\epsilon}^{b-\epsilon} f(x) dx =   \int_{a}^{b} f(x) dx $$ Given that $f(x)$ is continuous in $(a+\epsilon,b-\epsilon) $ but is infinite at $a$ and $b$.

Attempt:
$$\int_{a}^{b} f(x) dx=  \int_{a}^{a+\epsilon} f(x) dx  +  \int_{a+\epsilon}^{b-\epsilon} f(x) dx+  \int_{b-\epsilon}^{b} f(x) dx  $$
Now,$$\lim_{\epsilon \to 0^{+}} \int_{a}^{a+\epsilon} f(x) dx=0$$
$$\lim_{\epsilon \to 0^{+}} \int_{b-\epsilon}^{b} f(x) dx=0$$
So, $$ \lim_{\epsilon \to 0^{+}} \int_{a+\epsilon}^{b-\epsilon} f(x) dx =   \int_{a}^{b} f(x) dx $$

答:
Let $c\in(a,b)$. We have
$$ \int_{a+\epsilon}^{b-\epsilon} f(x) dx = \int_{a+\epsilon}^{c} f(x) dx + \int_{c}^{b-\epsilon} f(x) dx $$
By the definition of an improper integral, we have
$$ \lim_{\epsilon\to 0^+}\int_{a+\epsilon}^{c} f(x) dx =: \int_{a}^{c} f(x) dx $$
$$ \lim_{\epsilon\to 0^+}\int_{c}^{b-\epsilon} f(x) dx =: \int_{c}^{b} f(x) dx $$
assuming these limits exist. We have then
$$ \lim_{\epsilon\to 0^+}\int_{a+\epsilon}^{b-\epsilon} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx = \int_{a}^{b} f(x) dx  $$
Note that it is possible that $\lim_{\epsilon\to 0^+}\int_{a+\epsilon}^{b-\epsilon} f(x) dx$ exists even if $\lim_{\epsilon\to 0^+}\int_{a+\epsilon}^{c} f(x) dx$ and $\lim_{\epsilon\to 0^+}\int_{c}^{b-\epsilon} f(x) dx $ don't exist separately. This happens for example for $a=-1$, $b=1$, $f(x)=\frac{x}{1-x^2}$.


问:
Is there a non-Hausdorff space being a union of a countable increasing chain of "nice" spaces?
For a chain $X_0 \subseteq X_1 \subseteq X_2 ...$ of topological spaces (i.e the topology on $X_n$ coincides with the subspace topology induced by $X_m$ whenever $m > n$) we consider the space with the underlying set $X = \bigcup_{n\in \mathbb{N}} X_n$ and the topology such that $U\subseteq X$ is open if and only if for every $n\in \mathbb{N}$ the set $U \cap X_n$ is open in $X_n$.

The question is whether there exists a chain $X_0 \subseteq X_1 \subseteq ...$ of "nice" spaces (nice being $T_3$/$T_4$, or at least Hausdorff), such that the space $X$ defined as above is not even Hausdorff.

The genesis of my problem is such: in the short article "A Regular Space on which every Continuous Real-Valued Function is Constant" by T. E. Gantner the author seemingly concludes that a space $X$ as above is regular based solely on the fact that $X_n$s are regular. Perhaps he uses the fact, that each $X_n$ is closed in $X_{n+1}$ (in the proof in the article it holds), but it is not clear and this fact does not look helpful either way.

My thoughts:

If each $X_n$ is open in $X_{n+1}$, then the resulting space $X$ is always Hausdorff as long as all $X_n$s are Hausdorff.

Proof: It is easy to see that in fact $X_n$ is open in $X_{n+k}$ for every $n,k\in \mathbb{N}$ (say, by induction). Let $x,y \in X$ and let $m\in \mathbb{N}$ such that $x,y \in X_m$. Since $X_m$ is Hausdorff we can choose disjoint open $U_x, U_y\subseteq X_m$ such that $x\in U_x, y\in U_y.$ For every $n<m$ sets $U_x\cap X_n, U_y\cap X_n$ are open just by the definition of subspace topology, and for every $n>m$ sets $U_x\cap X_n = U_x, U_y\cap X_n = U_y$ are open by the assumption. So $U_x, U_y$ are in fact open in $X$ and they are disjoint open neighbourhoods of $x,y$, respectively. $\square$

That said, one can find a chain of "very nice" spaces $X_n$ such that $X$ is barely Hausdorff (i.e. is not $T_3$), namely let $X_n = \mathbb{R} \setminus \{\frac{1}{k} : k > n\}$, then $X$ is the very well known example of Hausdorff non-$T_3$ space (it is not possible to separate closed set $\{\frac{1}{k} : k > 0\}$ and point 0 with open sets). So the above cannot be strengthened, at least not without being very demanding about $X_n$s.

If we do not know anything about how $X_n$ lies in $X_{n+1}$ (or we know that they are closed subspaces - as I do not know how to reasonably use that assumption), then I can mainly say that $X$ is $T_1$ if only each of the $X_n$s is $T_1$ (and that for $T_2/T_3/T_4$ spaces $X_n$ space $X$ is $T_2/T_3/T_4$ with "open" in respective definition replaced with $\text{"$G_\delta$"}$, so it seems pretty weak).

The natural reasoning here for any $T_k$ is similiar, so I will show it for $T_1/T_2$ (simultaneously): for $x,y \in X$ take $m\in \mathbb{N}$ such that $x,y \in X_m$. Take $U^m_x \ni x$ (and $U^m_y \ni y$ for $T_2$) - open subset of $X_m$ as in the definition of $T_1(T_2)$ space. As before for $n< m$ set $U^m_x\cap X_n (U^m_y\cap X_n)$ is open in $X_n$, so we move to $n>m$, first $n=m+1$. We want open set $U^{m+1}_x (U^{m+1}_y) \subseteq X_{m+1}$ such that $U^{m+1}_x\cap X_m = U^m_x$ (and $U^{m+1}_y \cap X_m = U^m_y$ for $T_2$). For $T_1$ we do not want anything else, as $y\notin U^{m+1}_x$ as required, and we can just continue this way, declaring $U = \bigcup_{n=m}^\infty U^n_x\subseteq X$, which is open in $X$ and satisfies $T_1$ requirement for points $x,y$. For $T_2$ we would like something similiar but, regrettably, we have no guarantee that $U^{m+1}_x \cap U^{m+1}_y$ is empty. Since $X_{m+1}$ is $T_2$ we can just choose disjoint open $V^{m+1}_x \ni x$ and $V^{m+1}_y \ni y$, but they do not need to have open intersections with $X_m$. It looks like the best thing for us to do is to switch sets $U^m_x, U^m_y$ for $U^{m+1}_x \cap V^{m+1}_y$ and $U^{m+1}_x \cap V^{m+1}_y$, respectively, which are disjoint, open in $X_{m+1}$ and have open intersections with $X_m$. But then we have to follow with this kind of switching infinitely many times and it is (a priori) possible that we end up with non-open $G_\delta$s, which is not enough for us.

答:
Let $Y$ be any non-completely normal space and let $A,B\subseteq Y$ be separated sets that cannot be separated by open sets.  Let $X=(Y\times\mathbb{N})\cup\{a,b\}$ and define a topology on $X$ by saying a subset $U\subseteq X$ is open if it satisfies the following conditions:

$U\cap (Y\times\mathbb{N})$ is open in $Y\times\mathbb{N}$ (with respect to the product topology, where $\mathbb{N}$ is discrete).

If $a\in U$ then $A\times\{n\}\subseteq U$ for all but finitely many $n$.

If $b\in U$ then $B\times\{n\}\subseteq U$ for all but finitely many $n$.

This space is not Hausdorff: $a$ and $b$ cannot be separated by disjoint open sets, since that would give a separation of $A$ and $B$ by open subsets of $Y$.  However, note that if $U_n=(Y\setminus (A\cup B))\times (\mathbb{N}\setminus\{0,\dots n\})$, then $X$ is the colimit of the subspaces $X_n=X\setminus U_n$.  Moreover, each $X_n$ is "as nice as $Y$, $A$, and $B$": it's just a disjoint union of finitely many copies of $Y$, together with a copy of $A\times \mathbb{N}$ together with a "point at infinity" whose neighborhoods contain all but finitely many of the $A\times\{n\}$'s and similarly a copy of  $B\times\mathbb{N}$ together with a "point at infinity".  In particular, if $Y$ is Hausdorff, or regular, or completely regular, then so is each $X_n$, and if $Y$ and $A$ and $B$ are all normal, then so is each $X_n$.  Also, if $A$ and $B$ are closed in $Y$ (which is possible iff $Y$ is not normal), then each $X_n$ will additionally be closed in $X$.

(For an explicit example where $Y$ and $A$ and $B$ are all normal, you could take $Y$ to be the Tychonoff plank $(\omega+1)\times(\omega_1+1)$ with $A=\{\omega\}\times \omega_1$ and $B=\omega\times\{\omega_1\}$.)

These counterexamples are optimal in the following sense.  First, if $X$ is the colimit of an increasing sequence $(X_n)$ of closed normal subspaces, then $X$ is normal, as a consequence of the Tietze extension theorem (see Is the colimit of an expanding sequence of $T_4$ spaces $T_4$?).  Second, if $X$ is the colimit of an increasing sequence $(X_n)$ of completely normal subspaces, then $X$ is Hausdorff.  Indeed, let $a,b\in X$ be distinct points.  For any $n$ such that $X_n$ contains both $a$ and $b$, we can separate $a$ and $b$ by disjoint open subsets $U_n$ and $V_n$ of $X_n$.  Then $U_n$ and $V_n$ are separated, so we can separate them by disjoint open subsets $U_{n+1}$ and $V_{n+1}$ of $X_{n+1}$.  Continuing this recursively, we get increasing sequences $(U_m)_{m\geq n}$ and $(V_m)_{m\geq n}$ with each $U_m$ and $V_m$ disjoint open subsets of $X_m$.  The unions $U=\bigcup U_m$ and $V=\bigcup V_m$ will then be open subsets of $X$ that separate $a$ and $b$.


问:
What is the greatest integer that divides $p^4-1$ for every prime number $p$ greater than $5$?
What is the greatest integer that divides $p^4-1$ for every prime number $p > 5$?  This was on a practice math GRE so it's probably really easy.</p>

答:
$\rm\:p\nmid p^4-1\:$ so $\rm\:2,3,5\:$ are the only possible prime divisors. $240$ is the maximal product of such since $11^4-1 = 122\cdot 120 = 2^4\cdot3\cdot5\cdot 61\:.$ Now, as I mentioned on sci.math on Sep 1, 2003, for every n coprime to $\:2,3,5\:,\:$ hence certainly for every prime > $5$, we have that $\rm\:n^4 \equiv 1\ (mod\ 240)\:$ since $240$ has Carmichael lambda  function $\rm\:\lambda(240) = \lambda(2^4\cdot3\cdot 5) = lcm(2^{4-2},\phi(3),\phi(5)) = 4\:.$

See also my my post on Carmichael's generalizations of little Fermat and Euler's phi function, and many applications of such in prior posts. It's well-worth learning these generalizations since they are ubiquitous in number theory.


答:
As it's a reference-request and the amounts given there seem sufficient:

The number of goups of order $n$ is listed at OEIS as A00001. On its description page is a link to a table of $n$, $a_n$ for $n=1,\ldots, 2047$. For abelian groups, A000688 lsts the counts up to $n=10000$ as these are quite easily calculated.

Intriguingly, A060689 (non-abelian groups) gives only a list up to $n=2015$, sou you can earn merit by computing the differences from the tables and submit a bigger table.

Title: Number of abelian groups Vs Number of non-abelian groups

Body:
I would like to see a table that shows the number of non-abelian group for every order n. It is a preferable if the table contains the number of abelian groups of order n (this is not necessary though). If anyone could provide me with such a table , I would be grateful.

Note: I already checked wikipedia. They have a table similar to the one I want but its small.

Here is a simliar table to the one I want:

http://oeis.org/wiki/Number_of_groups_of_order_n#Table_of_number_of_distinct_groups_of_order_n

Thank you


问:
How to decompose a $3\times 3$ orthoginal matrix back into Reflection and Rotation components.
I am writing some CAD software that keeps track of the orientation of the modeled components.  The oriention is stored as a single $3\times 3$ orientation matrix that is a product of a reflection $R$ and a rotation $Q$, thus $T = RQ$.  The question is, given $T$, can I reconstruct the $R$ and $Q$ components?  And if so, is this deconstuction unique?

Notes:

$R$ is a Householder reflection matrix:

$R = I_3 - 2 n n^\text{T}.$ For some unit 3-vector $n$.

$\text{det}(R)$ = -1.

$Q$ is an Euler rotation matrix, $\text{det}(Q)$ = 1.

答:
Rewrite $$RT = Q$$

Now $$Q = {S_1}^{-1}\underset{C_1}{\underbrace{\left(\begin{bmatrix}a&b\\-b&a\end{bmatrix} \otimes 1\right)}}S_1$$
$$R={S_2}^{-1}\underset{C_2}{\underbrace{(1\otimes 1 \otimes -1)}}{S_2}$$

Here $\otimes$ denotes block diagonalization operator. In other words:

$$(a_1\otimes a_2 \otimes ... \otimes a_N) = \begin{bmatrix}a_1&&&\\&a_2&&\\&&\ddots&\\&&&a_N\end{bmatrix}$$

For two arbitrary transformations $S_1,S_2$ and $a=\cos(\phi),b=\sin(\phi)$

We can conclude that if $RT=Q$ holds, then $RT-Q =0$ and furthermore $\|RT-Q\|$ will be minimal.

Now we can create an alternating regularized linear least squares

$$\min\left\{\|RT-Q\|_2 + \|S_2R-C_2S_2\|_2 + \|S_1Q-(C_1)S_1\|_2\right\}$$


问:
GRE logic problem
So I am confused on the negation of the following:

Let $f,g$ be continuous on the reals.

Negate the statement:

For each $x \in \mathbb{R}$ there exists a $y \in \mathbb{R}$ such that if $f(x)>0$, then $g(y)>0$.

So the answer is

there exists an $x \in \mathbb{R}$ such that for each $y \in \mathbb{R}$  $f(x)>0$ and $g(y) \leq 0$.

Could someone explain this a bit more, that isn't the choice I selected and I am bothered I couldn't get the logic right here. Gotta be careful with quantifiers.

答:
My own attempt is as follows:

Let the given statement be

$$\forall x \in \mathbb{R}, \exists y \in \mathbb{R} \text{s.t.}$$

$$f(x) > 0 \Rightarrow g(y)>0.$$

Then negating becomes

$$\exists x \in \mathbb{R} \text{s.t.} \forall y \in \mathbb{R}$$

$$f(x) >0 \wedge g(y) \leq 0.$$

as needed.

Title: GRE logic problem


答: No, you cannot. You have $T = (I-2nn^T)Q$. Choose an arbitrary unit vector $e$ and set $U := (I-2ee^T)T$. Then $\det U = 1$, so $U$ is a rotation, and $T = (I-2ee^T)U$.

Title: How to decompose a $3\times 3$ orthoginal matrix back into Reflection and Rotation components.

Body:
I am writing some CAD software that keeps track of the orientation of the modeled components.  The oriention is stored as a single $3\times 3$ orientation matrix that is a product of a reflection $R$ and a rotation $Q$, thus $T = RQ$.  The question is, given $T$, can I reconstruct the $R$ and $Q$ components?  And if so, is this deconstuction unique?

Notes:

$R$ is a Householder reflection matrix:

$R = I_3 - 2 n n^\text{T}.$ For some unit 3-vector $n$.

$\text{det}(R)$ = -1.

$Q$ is an Euler rotation matrix, $\text{det}(Q)$ = 1.


问:
Principal ideal domain not euclidean
Can anyone give an example of a principal ideal domain that is not Euclidean and is not isomorphic to $\mathbb{Z}[\frac{1+\sqrt{-a}}{2}]$, $a = 19,43,67,163$?

I believe it is conjectured that no other integer rings of number fields have this property. What about other rings?

答:
One such simple example of a non-Euclidean PID is $ K[[x,y]][1/(x^2\!+\!y^3)]\,$ over any field $\,K,\,$ i.e. adjoin the inverse of $\,x^2\!+\!y^3$ to a bivariate power series ring over a field. For the proof, and a general construction method see

D.D. Anderson. An existence theorem for non-euclidean PID’s,

Communications in Algebra, 16:6, 1221-1229, 1988.

For number rings, by Weinberger (1973), assuming GRH, a UFD number ring R with infinitely many units is Euclidean, e.g. real quadratic number rings are Euclidean $\!\iff\!$ PID $\!\iff\!$ UFD.

Title: Principal ideal domain not euclidean


问:
Supposed counterexample to Liouville's theorem
I'm trying to understand Liouville's theorem, and I don't see why $f(z)=e^{-|z|^2}$ isn't a counterexample.  It's bounded ($0 < f(z) \leq 1$), so it must be somehow that it's not holomorphic.  Isn't it differentiable everywhere?</p>

答: Hint: $|z|^{2}=z\bar{z}$ but a complex function of complex variable $z\to\bar{z}$ is not differentiable (although its real and imaginary part are differentiable when considered as real functions).</p>


问:
There appears to be a contradiction while solving recurrence relation
Consider the equation :
$$
y[n]-\frac{1}{2}y[n-1]=x[n]
$$
where $x[n]:=\left(\frac{1}{3}\right)^{n}u[n]$. First, I was asked to find the homogenous part of the solution having a form $y_{h}:=A\frac{1}{2^{n}}$. Indeed the homogenous solution
$$
y_{h}[n]=\frac{1}{2^{n}}y[0]
$$
where I insist $A=y[0]$. Furthermore, The particular solution is given of the form $y_{p}[n]:=B\left(\frac{1}{3}\right)^{n}u[n]$ I shall determine $B$ as follow :
$$
B\left(\frac{1}{3}\right)^{n}u[n]-\frac{1}{2}B\left(\frac{1}{3}\right)^{n-1}u[n-1]=\left(\frac{1}{3}\right)^{n}u[n]
$$
\begin{align*}
\implies B&=\frac{\displaystyle\left(\frac{1}{3}\right)^{n}u[n]}{\displaystyle\left(\frac{1}{3}\right)^{n}u[n]-\frac{1}{2}\displaystyle\left(\frac{1}{3}\right)^{n-1}u[n-1]}\\ \\
&=\frac{\displaystyle\left(\frac{1}{3}\right)^{n}}{\displaystyle\left(\frac{1}{3}\right)^{n}-\displaystyle\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}}\\ \\
&=-2
\end{align*}

So now that we have $B=-2$, We are given the initial rest condition (a condition that states if $x[n]=0$, $\forall n<0$ then $y[n]=0$ $\forall n<0$). so what I did was sub $n=0$ in difference equation
$$
y[0]-\frac{1}{2}y[-1]=x[0]
$$
$$
\implies y[0]=1
$$
$$
**\text{So this means $A=y[0]=1$}**
$$
However, the major issue is that :
$$
y[0]-\frac{1}{2}y[-1]=x[0]
$$
$$
\implies y[0]=1
$$
$$
\implies y[n]:= y_{h}[n]+y_{p}[n] \implies y[0]=y_{h}[0]+y_{p}[0]=1
$$
$$
\implies A\left(\frac{1}{2}\right)^{0} -2\left(\frac{1}{3}\right)^{0}=1
$$
$$
\implies A=3
$$
So I have $A=1$ and $A=3$, an obvious contradiction and I hope someone can help me.

Remark : $u[n]$ is the unit-step function where :
$$
u[n]:=
\begin{cases}
1&\text{if $n\geq0$}\\
0&\text{if $n<0$}
\end{cases}
$$

答:
The mistake that you made is that you assumed $\displaystyle y_{h}=\left(\frac{1}{2}\right)^{n}y[0]$ while it should be $\displaystyle y_{h}=\left(\frac{1}{2}\right)^{n}y_{h}[0]$. Therefore :
$$
A=y_{h}[0]
$$
Moreover, in finding the value of $B$, you have two cases $n$ starting from $0$ and $n$ starting from $1$, you should specify where $n$ starts.


问:
Need help to prove that any non prime number that >= 4 can be written in the form: i^2 + i * k
Any non prime number that >= 4 can be written in the format:  i^2 + i * k.

Condition:  i>=2,  k >= 0 (i, k are integers)

4 = 2^2 + 2*0 -> Good

6 = 2^2 + 2*1 -> Good

10 = 2^2 + 2*3 -> Good
...

Thanks!

答:
I observe
$$  i^2 + ik = i(i+k)  \text{.}  $$

Let $n$ be a composite number, that is a non-prime number and let $i$ be the smallest prime dividing $n$.  We know $2 \leq i$.  Also, $n/i > 1$ and because $i$ is the smallest prime divisor of $n$, $n/i$ is an integer with $n/i > i$.  So we have
$$  2 \leq i \leq \frac{n}{i} < n  $$
Let $k = \frac{n}{i} - i$.  By the above inequality, this $k \geq 0$.  Also,
\begin{align*}
i^2 + ik &= i^2 + i\left(\frac{n}{i} - i \right)  \\
    &= i^2 + \left(n - i^2 \right)  \\
    &= n  \text{,}
\end{align*}
as desired.


问:
$E[X^2|\mathcal{G}]=2\int_0^{\infty}xP(X>x|\mathcal{G})dx$
Let $X$ a random variable on $(\Omega,\mathcal{F},P),\mathcal{G}$ a sub $\sigma$-algebra. Prove that, almost surely,

$$E[X^2|\mathcal{G}]=2\int_0^{\infty}xP(X>x|\mathcal{G})dx$$
The above formula was true for simple function $X,$ if we pick an increasing function $X_n$ to $X,$ there is a problem in applying the conditional monotone theorem in $\int_0^{\infty}xP(X_n>x)dx$ because $P(X_n>x|\mathcal{G}) \to P(X>x | \mathcal{G})$ on an event $E_x$ (which depends of $x>0$)such that $P(E_x)=1,$ so we can't change interchange limit-integral.

What do you suggest to solve this problem?

答:
Let $\nu$ denote the regular conditional distribution of $X$ given $\mathcal{G}$. Then for any suitable function $g$,
\begin{align}
\mathsf{E}[g(X)\mid \mathcal{G}](\omega)&=\int g(x)\nu(\omega,dx) \\
&=g(0)+\int_0^\infty g'(x)\nu(\omega,(x,\infty))\, dx-\int_{-\infty}^0 g'(x)\nu(\omega,(-\infty,x])\, dx \quad\text{a.s.}
\end{align}
(See, e.g., this question.)
Now, plug $g(x)=x^2$.


问:
Limits in functor categories
Let $C,C’,D$ be categories and $u:C\to C’$ be a functor. The functor $u^*:\mathbf{Hom}(C’^\circ,D)\to\mathbf{Hom}(C^\circ,D)$ that sends a functor $G$ to $G\circ u$ commutes with limits and colimits, according to SGA 4. Does this presume that $D$ is (co)complete? I know when $D$ is (co)complete, (co)limits are computed pointwise. If not, some (co)limits can exist ‘by accident.’

答:
Yes, it is best to assume that $\mathcal{D}$ is (co)complete for the purposes of this proposition.
Actually, there are two ways to make the proposition precise.
You know this one:

Proposition.
If $\mathcal{D}$ has limits (resp. colimits) of diagrams of shape $\mathcal{J}$ then $[\mathcal{C}'^\textrm{op}, \mathcal{D}]$ and $[\mathcal{C}^\textrm{op}, \mathcal{D}]$ also have limits (resp. colimits) of diagrams of shape $\mathcal{J}$ and $u^* : [\mathcal{C}'^\textrm{op}, \mathcal{D}] \to [\mathcal{C}^\textrm{op}, \mathcal{D}]$ preserves those limits (resp. colimits).

But here is another one:

Proposition.
If, for every object $c'$ in $\mathcal{C}'$, $\mathcal{D}$ has colimits (resp. limits) of diagrams of shape $(u \downarrow c')$ (resp. $(c' \downarrow u)$), then $u^* : [\mathcal{C}'^\textrm{op}, \mathcal{D}] \to [\mathcal{C}^\textrm{op}, \mathcal{D}]$ has a left (resp. right) adjoint.

Notice that in this proposition, existence of colimits (resp. limits) of particular shapes implies preservation of limits (resp. colimits) of arbitrary shapes, which may or may not be componentwise!

Title: Limits in functor categories


问:
How do I rotate a line around a central axis until it is parallel to x axis?
I have all the points shown in this image. I need to rotate the top line so that it is parallel to the x-axis but it has to rotate around the point (0, 0).

答:
Let $\vec{\ell} = (\ell_x, \ell_y)$ describe the line segment as a vector in the original coordinates.  If you construct unit vector $\hat{u} = (u_x, u_y)$,
$$\hat{u} = \left[\begin{matrix} u_x \\ u_y \end{matrix}\right] = \frac{\vec{\ell}}{\left\lVert\vec{\ell}\right\rVert} = \left[\begin{matrix}
\displaystyle \frac{\ell_x}{\sqrt{\ell_x^2 + \ell_y^2}} \\
\displaystyle \frac{\ell_y}{\sqrt{\ell_x^2 + \ell_y^2}} \\
\end{matrix}\right]$$
then rotation matrix $\mathbf{R}$ will rotate points so that $\vec{\ell}$ will end up on the positive $x$ axis (meaning the original line segment will be parallel to the $x$ axis):
$$\mathbf{R} = \left[\begin{matrix} u_x & u_y \\ -u_y & u_x \\ \end{matrix}\right] \tag{1a}\label{G1a}$$
In particular, if $\ell_x^2 + \ell_y^2 = 1$, then
$$\mathbf{R} \vec{\ell} = \left[\begin{matrix} \ell_x & \ell_y \\ -\ell_y & \ell_x \end{matrix} \right] \left[\begin{matrix} \ell_x \\ \ell_y \end{matrix} \right] = \left[\begin{matrix} \ell_x^2 + \ell_y^2 \\ - \ell_y \ell_x + \ell_x \ell_y \\
\end{matrix}\right] = \left[\begin{matrix} 1 \\ 0 \end{matrix}\right]$$
If, for any reason, you don't need the rotation matrix but the angle of the rotation $\theta$ (positive counterclockwise), then use
$$\theta = \operatorname{atan2}(-\ell_y ,\, \ell_x) \tag{1a}\label{G1b}$$

If you want to limit the rotation to -90° to +90°, i.e. a rotation $\mathbf{R}$ that brings $\vec{\ell}$ horizontal, use
$$\theta = \arctan\left(\frac{-\ell_y}{\ell_x}\right) \tag{2a}\label{G2b}$$
or whenever $u_x$ is negative, negate both $u_x$ and $u_y$, i.e.
$$\left[ \begin{matrix} u_x \\ u_y \end{matrix} \right] = \begin{cases}
\displaystyle \frac{1}{\sqrt{\ell_x^2 + \ell_y^2}}\left[ \begin{matrix} \ell_x \\ \ell_y \end{matrix} \right] , & \ell_x \ge 0 \\
\displaystyle \frac{-1}{\sqrt{\ell_x^2 + \ell_y^2}}\left[ \begin{matrix} \ell_x \\ \ell_y \end{matrix} \right] , & \ell_x \lt 0 \\
\end{cases} \tag{2a}\label{G2a}$$
(Apologies for the edits; had to iterate a couple of times to get the expressions right.)


答:
This isn't as deep as it looks.

$i^2 + ik = i(i + k)$.

And if you have any $m \le n$ then set $i = m$ and $k = n -m \ge 0$ and you are done.

If $J$ is not prime then $J = mn$ for some $1 < m\le n < J$.  So if $i = m \ge 2$ and $k =n-m \ge 0$ then we have $i(i+k) = mn  = K$.

That's all.

(although it might be strange to think that $J = mn = m^2 + m(n-m)$ but a little arithmetic shows that must be the case.)

======

Okay..... just for schlits and giggees.  Here's the hard way to do it:

Let $J = mn; m \le n$.

Then $m^2 \le mn = J \le n^2$

So $J = m^2 + W$ where $W = J - m^2$.  But $J= mn$ so $W = mn -m^2 = m(n-m)$ and so $m|W$.  Let $k = \frac Wm$.

Then $J= m^2 + W = m^2 + mk$.

Title: Need help to prove that any non prime number that >= 4 can be written in the form: i^2 + i * k

Body:
Any non prime number that >= 4 can be written in the format:  i^2 + i * k.

Condition:  i>=2,  k >= 0 (i, k are integers)

4 = 2^2 + 2*0 -> Good

6 = 2^2 + 2*1 -> Good

10 = 2^2 + 2*3 -> Good
...

Thanks!


问:
Improper Integral Calculation with Lots of Constants
I have an improper integral:

$$T=4\sqrt{m}\int_0^\sqrt{2E/k}\frac{1}{\sqrt{2E-kx^2}}dx$$

I said it was improper because if you plug in the upper bound, I believe the denominator = 0.
I am supposed to evaluate to show that:

$$T=2\pi\sqrt{m/k}$$

I am just not even sure how to start, any help is appreciated!!

答:
The idea is to rewrite it this way : Let's keep $4\sqrt{m}$ away for a moment :

\begin{align}
\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{2E-kx^2}}}dx
&=\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{2E}}\frac{1}{\sqrt{1-kx^2/2E}}}dx\\
&=\frac{1}{\sqrt{2E}}\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{1-[(\sqrt{k/2E})x]^2}}dx}
\end{align}
If we dentoe $\alpha=\sqrt{k/2E}$ then we'll have
$$\frac{1}{\sqrt{2E}}\int_{0}^{1/\alpha}{\frac{1}{\sqrt{1-(\alpha x)^2}}dx}$$

Which is the $Arcsin$ primitive :
$$\frac{1}{\sqrt{2E}}\int_{0}^{1/\alpha}{\frac{1}{\sqrt{1-(\alpha x)^2}}dx}=\frac{1}{\sqrt{2E}}\Bigg[\frac{1}{\alpha}arcsin(\alpha x)\Bigg]_0^\frac{1}{\alpha}$$
Which yields :
$$\frac{1}{\sqrt{2E}}\Bigg[\frac{1}{\alpha}arcsin(\alpha x)\Bigg]_0^\frac{1}{\alpha}=\frac{1}{\sqrt{2E}}\Bigg[\frac{1}{\alpha}arcsin(1)-\frac{1}{\alpha}arcsin(0)\Bigg]$$

Since $arcsin(1)=\frac{\pi}{2}$ and $arcsin(0)=0$, and by replacing $\alpha$ with $\sqrt{k/2E}$, we get
$$\int_{0}^{\sqrt{2E/k}}{\frac{1}{\sqrt{2E-kx^2}}}dx =\frac{1}{\sqrt{2E}}\frac{1}{\sqrt{k/2E}}\frac{\pi}{2}=\frac{1}{\sqrt{2E}}\sqrt{\frac{2E}{k}}\frac{\pi}{2}$$
We are left with
$$\frac{\pi}{2\sqrt{k}}$$
By multiplying by the $4\sqrt{m}$ left at the begining, we finally get :
$$T=4\sqrt{m}\frac{\pi}{2\sqrt{k}}=2\pi \sqrt{\frac{m}{k}}$$


问:
Complex function f is differentiable implies |f| is harmonic.
Let's say $u(x,y)+iv(x,y)=f: D\rightarrow \mathbb{C}$ is differentiable. Is function $|f|$ harmonic on $D$?

Function $f$ is said to be harmonic if $\Delta f \equiv 0.$
If $f$ is differentiable then Cauchy-Riemann equations hold.

I tried to directly compute $|f|$, but I don't think it is possible in general case. Any tips?

答:
No. For example, the function $\;f(z)=z\;$ is analytic and thus (infinitely) differentiable, yet (with $\;z=x+iy\,$)

$$g(x,y)=|f(z)|=\sqrt{x^2+y^2}\;\;\text{ isn't harmonic since}$$

$$\begin{cases}g_x=\frac x{\sqrt{x^2+y^2}},&g_{xx}=\frac{x^2}{(x^2+y^2)^{3/2}}\\{}\\g_y=\frac y{\sqrt{x^2y^2}},&g_{yy}=\frac{y^2}{(x^2+y^2)^{3/2}}\end{cases}$$

and thus

$$\nabla^2g=g_{xx}+g_{yy}=\frac1{\sqrt{x^2+y^2}}\neq0$$


问:
Calculating the integral with an undefined function f(x)
I am having a problem with an excersice of calculating the integral

I= $\int_{0}^{a}\frac{1}{1+f(x)}\,dx$

and we know that $f(x)f(a-x)=1$ , $a>0$  and $f(x)$ is continous and positive on the interval $[0,a]$

i've tried manipulating the expression with the fact that $f(x)= \ \frac{1}{f(a-x)} $ in order to find something to cancel or something that makes me solve the integral "nicely"

However after 20 minutes i have no idea what to do, or what i am missing

so if anyone could give me a tip or point me in the right direction i would appreciate it

答:
$$
\int_0^a\frac{1}{1+f(x)}\,dx = \int_0^a\frac{1}{1+f(a-x)}\,dx = \int_0^a\frac{1}{1+\frac 1{f(x)}}\,dx = \int_0^a\frac{f(x)}{1+f(x)}\,dx,
$$
where in the first transformation we changed variables $x = a-y$. Hence,
$$
2I = I + I = \int_0^a\frac{1}{1+f(x)}\,dx + \int_0^a\frac{f(x)}{1+f(x)}\,dx = \int_0^a1\,dx = a.
$$
Thus, $I = \frac a2$.


答:
No, it's nowhere differentiable (except at the origin). First off, the modulus function $f(z)=|z|$ is nowhere differentiable in $\mathbb{C}$, and since any point (other than the origin) is contained in some open set with a holomorphic branch of logarithm and square root defined, if $g(z) = e^{-|z|^2}$ were differentiable then so too would $(-\log(g(z)))^{1/2}=|z|$, which is a contradiction.

At the origin, $g(0)=1$ so:
$$\frac{g(re^{i\theta})-g(0)}{re^i\theta}=\frac{e^{-r^2}-1}{re^{i\theta}} $$

So taking the limit as $r\rightarrow 0$ we get:
$$e^{-i\theta}\frac{d}{dx}\bigg(e^{-x^2}\bigg)\biggr\rvert_0$$
Where we take the real function $e^{-x^2}$, which is (real) differentiable at the origin. As pointed out in the comments below, this derivative is equal to zero so the function is differentiable at the origin with derivative $0$.

Title: Supposed counterexample to Liouville's theorem

Body: I'm trying to understand Liouville's theorem, and I don't see why $f(z)=e^{-|z|^2}$ isn't a counterexample.  It's bounded ($0 < f(z) \leq 1$), so it must be somehow that it's not holomorphic.  Isn't it differentiable everywhere?</p>


问:
Finding all monic complex polynomials $P(x)$ such that $P(x)|P(x^2)$

Find all monic complex polynomials $P(x)$ such that $P(x)|P(x^2)$.

My progress so far is that I have find that for degree 1, $P(x)=x, x^2$ are the only ones.

For degree 2, they are $P(x)=x^2+x+1, x^2, x^2-1, x^2-x, x^2-2x+1$.

I also prove that these are only solutions for degree 1 and 2. However I do not see how this generalizes. Any help please?

答: Hint: If $P(c)=0$ then $0=P(c^{2})=P(c^{4})=P(c^{8})=...$ and  a polynomial can only have finite number of roots. Can you take over from here?


问:
Proving two set are homeomorphic
I want to deduce that the following sets: $\Sigma_1=\{(x,y,z,w)\in\Bbb{R^4}|x^2+y^2=1,z^2+w^2=1\}$ and $S^1\times S^1$ (where $S^1$ is simply the unit circle) are homeomorphic. The question hints that first I need to prove that:

"If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$ then the product topology on $A\times B$ is the same as the topology $A\times B
$ inherits as a subspace of $X\times Y$".

My thought was simply to define $f(x,y,z,w)=((x,y),(z,w))$ which obviously is $1-1$ and onto. Now I can't see how what I proved helps me to prove continuity of the function I defined. Any help would be apperciated.

答: Your map $f: \Sigma_1 \to S^1 \times S^1$ works fine. It's clearly a bijection and continuity follows from the universal property for products (see my post here where I also show the hint you were given): $p_1 \circ f = \pi_{1,2}\restriction_{\Sigma_1}$, where $p_1,p_2: S^1 \times S^1 \to S^1$ are the two projections on the codomain, and $\pi_{1,2}: \Bbb R^4 \to \Bbb R^2$ is just the "projection" $(x_1,x_2,x_3,x_4) \to (x_1,x_2)$ which is also continuous by standard facts (restrictions of continuous maps are continuous etc), and $p_2 \circ f = \pi_{3,4}\restriction_{\Sigma_1}$ (similar considerations). So $f$ is continuous. I won't bother with doing the same for the inverse, because that is already "automatically" continuous as $f$ is closed, going from the compact domain $\Sigma_1$ to a Hausdorff (even metric) $S^1 \times S^1$. That's all there is to it. I don't see how your hint would actually be needed here. It's just index juggling on products, continuous by the universal property.

Title: Proving two set are homeomorphic


问:
Finding the distribution $(_=)$ of $k$ heads when throwing a die n times and only when 4 or 1 is rolled tossing a coin.
Player $A$ throws the die $n$ times. Whenever 1 or 4 is thrown, he tosses an unbiased coin. Let $X_n$ be the random variable which counts the number of heads from the Bernoulli experiment of tossing a coin. I am trying to find the distribution $P(X_n=k)$ for every $k$, its mean and variance (giving me a hint towards the distribution would be good enough though).

I noticed that the number of $4$'s and $1$'s could be counted using a binomial approach, so let $Y$ be a random variable which counts the number of throwing $4$ or $1$.

Since $X$ counts the number $k$ heads I have to throw at least $k$ times $4$ or $1$.

Giving us $P(Y \geq k) = \sum\limits_{i = k}^n {n \choose i} {2 \over 6}^i {4 \over 6}^{n-i}$

My next steps was thinking about how to count the number of heads.

The naive approach was just to multiplying $P(Y \geq k)$ with $1 \over 2$ but I can see no possibility to count heads now since I dont know whether a head or tails is flipped.

The next thing I came up with was combining this with binomial distribution again for counting k heads.

Something along this way: $P(X = k) =  \sum\limits_{i = k}^n {n \choose i} {2 \over 6}^i {4 \over 6}^{n-i} {n \choose k} {1 \over 2}^k {1 \over 2}^{i-k}$

I am not sure whether this is going into the right direction or not and it doesn't seem handy at all. As the next step I would try to use the binomial theorem, but I can't find a way to get this simpler along this way because I have to shift k indices.

答:
You have $Y \sim B(n,\frac{1}{3})$. Continuing the calculation in your way (which is pretty straight forward),

$P(X_n = k)$

$=  \sum\limits_{i = k}^n P(X_n = k|Y=i).P(Y=i)$

$ = \sum\limits_{i = k}^n {i \choose k} \left({1 \over 2}\right)^i \left({1 \over 2}\right)^{i-k}. {n \choose i} \left({1 \over 3}\right)^i \left({2 \over 3}\right)^{n-i}$

$ = \left({1 \over 3}\right)^n \sum\limits_{i = k}^n {i \choose k} \left({1 \over 2}\right)^i . {n \choose i} 2 ^{n-i}$

$ = \left({1 \over 3}\right)^n.{n \choose k} \sum\limits_{i = k}^n {n-k \choose i-k} 2 ^{n-2i}$

$ = \left({1 \over 3}\right)^n.{n \choose k} \sum\limits_{j = 0}^{n-k} {n-k \choose j} 2 ^{n-2j-2k}$   (let $j=i-k$)

$ = \left({1 \over 3}\right)^n. 2^{n-2k}.{n \choose k} \sum\limits_{j = 0}^{n-k} {n-k \choose j} \left({1 \over 4}\right) ^j$

$ = \left({1 \over 3}\right)^n. 2^{n-2k}.{n \choose k} (\frac{5}{4})^{n-k}$

$= {n \choose k} \left(\frac{1}{6}\right)^n.5^{n-k}$

$= {n \choose k} \left(\frac{1}{6}\right)^k.\left(\frac{5}{6}\right)^{n-k}$ (p.m.f. of a Binomial distribution)

$\Rightarrow X_n \sim B(n, \frac{1}{6})$

$\Rightarrow E[X_n] = n.\frac{1}{6}=\frac{n}{6}$ and
$ V[X_n] = n.\frac{1}{6}\frac{5}{6}=\frac{5n}{36}$


问:
False logic argument
I have to proof the following theorem $$(w ⇒ p ∨ s) ∧ (t ∧ p ⇒ r) ⇒ (t ∧ ¬ r ⇒ ¬ q)$$

Using the assumptions, I would like to proof that $(t∧p⇒r)⇒(t∧¬r⇒¬q)$, but I got only $(p⇒r)⇒(t∧¬r⇒¬q)$. I thought I could use $(t∧p⇒r) ⇒ (p⇒r)$, but it seems to be false. How can I modify/finish this proof?

The answer of @DonAntonio is great, but I can't work with this. I would need a complete, linear proof.

答:
Shorter and probably  simpler proof by reductio ad absurdum: suppose that isn't true that $\;(t\wedge\neg r)\rightarrow\neg q\;$ . Then it must be that $\;t\wedge\neg r=T\;$  and $\;\neg q=F\iff q= T \;$ , and both $\;t=T\,,\,\neg r= T\iff r=F\;$

But then, since $\;(t\wedge p)\rightarrow r=T\;$ (assumption given), it must be that $\;t\wedge p=F\;$ . Since $\;t=T\;$ , this means that it must be $\;p=F\;$.

Also, we have the assumption $\;s\rightarrow\neg t=T\;$ . But $\;\neg t=F\;$ , so it must be that also $\;s=F\;$.

And we then get our contradiction with the first assumption, since $\;q=T\;$ yet $\;s\vee p=F\;$ as both $\;s= p= F\;$ .

Title: False logic argument


问:
Need help simplifying complicated rational expression.
Studying for my final and I can't figure this out.

Simplify:

$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$

答:
Hint: Is there something that you could multiply both the numerator and denominator by in order to make this expression easier to tackle? Then it might be easier to simplify and reduce the expression.

(Bigger hint: multiply by $\frac{x^{a}y^{b}}{x^{a}y^{b}}$ for some appropriate choice of $a, b$ and then simplify!)


问:
Truth value of a statement
I have a problem:

Let P be the statement " x $\in$ A and x $\in$ $\mathbb{Z}$ "

statement: ($\forall$x)P $\Longrightarrow$ ($\exists$x)P

Is there a set A for which the truth value of the above statement is false? Explain.

My approach is:

The statement is only false when the antecedent is true and the consequent is false.

This is not possible because ...

Can someone help, please?

NOTE: Please forgive me : our assumption is that x is in the universal set R

答:
In first-order logic as presented in standard textbooks and standard introductory courses, the universe of discourse (i.e., the set of things that we talk about in our logical language) is non-empty. $\forall x\,P(x)$ means that $P(x)$ holds for every element of the universe and, as that universe is non-empty, $\exists x\,P(x)$ must hold if $\forall x\,P(x)$ does. So $(\forall x\,P(x)) \Rightarrow (\exists x\,P(x))$ is valid, for any predicate $P(x)$. In particular, $(\forall x\,(x \in A \land x \in \Bbb{Z})) \Rightarrow (\exists x\,(x \in A \land x \in \Bbb{Z}))$ is valid for any set $A$. So you correct in concluding that there is no set $A$ that makes this proposition false.

To address Prime Mover's concerns: if we are working in a first-order language that can talk about sets and set membership, then it is convenient to use relative quantifiers: i.e., to write $\forall x \in A\,Q(x)$ as an abbreviation for $\forall x \, (x \in A \Rightarrow Q(x))$ and to write $\exists x \in A, Q(x)$ as an abbreviation  for $\exists x(x \in A \land Q(x))$. Using this short-hand $\forall x \in \emptyset\,Q(x))$ is true and $\exists x \in \emptyset\,Q(x)$ is false, but this does not contradict the statement that $(\forall x\,P(x)) \Rightarrow (\exists x\, P(x))$, since when we expand the abbreviations, the matrix (body) of the two quantified subformulas are not the same.

Finally, I should say that there are logics that do not assume that quantification ranges over non-empty sets. Tennant's Natural Logic is one such system. Martin-Löf type theory is another. Such logics are, in some sense, more precise than traditional first-order logic, but the extra precision has to be paid for by more onerous proof obligations.

Title: Truth value of a statement


答:
Same answer, different approach.  By symmetry,
$$
\begin{align}
\int_0^a\frac1{1+f(x)}\,\mathrm{d}x
&=\frac12\left(\int_0^a\frac1{1+f(x)}\,\mathrm{d}x+\int_0^a\frac1{1+f(a-x)}\,\mathrm{d}x\right)\tag1\\
&=\frac12\int_0^a\left(\frac1{1+f(x)}+\frac1{1+f(a-x)}\right)\,\mathrm{d}x\tag2\\
&=\frac12\int_0^a\frac{\color{#C00}{1}+f(x)+\color{#C00}{1}+f(a-x)}{\color{#090}{1}+f(x)+f(a-x)+\color{#090}{f(x)f(a-x)}}\,\mathrm{d}x\tag3\\
&=\frac12\int_0^a\frac{\color{#C00}{2}+f(x)+f(a-x)}{\color{#090}{2}+f(x)+f(a-x)}\,\mathrm{d}x\tag4\\[3pt]
&=\frac a2\tag5
\end{align}
$$
Explanation:

$(1)$: by substituting $x\mapsto a-x$, we get $\int_0^a\frac1{1+f(x)}\,\mathrm{d}x=\int_0^a\frac1{1+f(a-x)}\,\mathrm{d}x$

$\phantom{\text{(1):}}$ average the two

$(2)$: sum of integrals is the integral of the sum

$(3)$: $\frac1u+\frac1v=\frac{u+v}{uv}$

$(4)$: the sum of the integrands is $1$ since $f(x)f(a-x)=1$

$(5)$: integrate

Title: Calculating the integral with an undefined function f(x)

Body:
I am having a problem with an excersice of calculating the integral

I= $\int_{0}^{a}\frac{1}{1+f(x)}\,dx$

and we know that $f(x)f(a-x)=1$ , $a>0$  and $f(x)$ is continous and positive on the interval $[0,a]$

i've tried manipulating the expression with the fact that $f(x)= \ \frac{1}{f(a-x)} $ in order to find something to cancel or something that makes me solve the integral "nicely"

However after 20 minutes i have no idea what to do, or what i am missing

so if anyone could give me a tip or point me in the right direction i would appreciate it


问:
Absolute value in solution of first order linear differential equation
Let’s solve the first order linear equation $ y’ + {1\over{x}}y = 0$.

$ y’/y = -1/x $

$ ln|y| = -ln|x| + C $

$ |y| = {C\over{|x|}} $

So it seems to me, I have to choose every y that equals to $C/|x|$ if I get the absolute value of y.

Both $C/|x|$ and $C/x$ seems to satisfy this requirement. However, my textbook only has $C/x$ as the solution.

Did I do something wrong?

Or is it a convention to just ignore the absolute value? If so, why?

答:
I appreciate that you thought deeply about this question and raised it again after our comments of a couple of months ago.

As you noted, things can become complicated with the absolute value. The functions $y=\frac{C}{x}$ and $y=\frac{C}{|x|}$ are not the same. Note that in general the value of $C$ can be positive or negative, which produces different functions. Which function should we choose as the solution to the equation? The answer depends on the constraints of the problem.

Let's look more carefully at the following combinations of functions and $C$ signs. For simplicity, I chose $|C|=1$ but that doesn't reduce the generality of the arguments that follow. Clearly, $x=0$ is excluded from the domains.
$$y = \frac 1x $$

Notice that $x$ and $y$ always have the same sign and that the function is always descending ($y' < 0$) .
$$ $$
$$y = \frac {-1}{x} $$

Notice the contrast with the previous function. Here, $x$ and $y$ always have opposite signs and the function is always ascending ($y' > 0$) .
$$ $$
$$y = \frac {1}{|x|} $$

$y$ is always positive. When $x < 0$ , $x$ and $y$ have opposite signs. When $x > 0$ , $x$ and $y$ have the same sign. The function is ascending for $x < 0$ and descending for $x > 0$ .
$$ $$
$$y = \frac {-1}{|x|} $$

$y$ is always negative. The function is descending for $x < 0$ and ascending for $x > 0$ .

Notice how fundamentally different the absolute value functions are from the non-absolute value functions. Now, which one(s) of these should we choose as the solutions to the original equation? Let's look at the equation more closely. We have:
$$y' = - \frac yx $$
The sign of $y'$ is the opposite of the sign of $\frac yx$ .

If $\frac yx > 0$ , that is, if $x$ and $y$ have the same sign, then $y' < 0$ and the curve of the function is descending.

If $\frac yx < 0$ , that is, if $x$ and $y$ have opposite signs, then $y' > 0$ and the curve of the function is ascending.

Now, if all over the domain of $x$ only one of the above two cases applies, that is, if $\frac yx > 0$ for all $x$ in the domain, or if $\frac yx < 0$ for all $x$ in the domain, then the sign of $y'$ does not change in the domain. You can see from the figures above that in these cases, $y = \frac Cx$ is the solution to the equation. You have also verified that this solution satisfies the equation.

Alternatively, it is possible that $x$ and $y$ have the same sign in some part of the domain, opposite signs in another part of the domain. There is nothing mentioned in the original problem that prevents this case. In such a case, as the last two graphs show, the solution of the equation is $y = \frac{C}{|x|}$ , and you have verified that this, too, satisfies the equation.

Your textbook says $y = \frac Cx$ is the answer. User @yves-daoust gives an answer that seems to be always positive, so suggesting that $y = \frac {C}{|x|}$ is the answer. I would say they are all right, as long as they are not exclusive answers. To give you a specific example, if we know that the curve of the function passes through point $(1,1)$ then the solution function may be $y = \frac 1x$ , and it may as well be $y = \frac {1}{|x|}$ . We saw in the figures that these two functions are not the same, and yet they both satisfy the original equation. I would say, in order to choose one of these two functions, we need one more piece of information.

Such additional information can be a second point, whose $x$ has the opposite sign of the first point. For example, if we are told that $(-1,-1)$ is the second point on the curve of the function, then we are certain that $y = \frac 1x$ is the only solution to the equation. Conversely, if we were told that $(-1,1)$ is the second point on the curve of the function, then $y = \frac{1}{|x|}$ would be the solution. However, all of this may sound weird because we expect to need only one point of the curve (and not two) in order to find the solution to a 1st order ODE.

Alternatively, the additional piece of information to help us specify the solution can be the range of $y$ . If we are told that we should find a solution that ranges in $R^+$ only (or in $R^-$ only), then we could say that the answer involves the absolute value function. Conversely, if we are told that the solution ranges in $R^+$ and $R^-$ , then we know that the answer does not include the absolute value function.


答: Find the LCM of all four denominators.  It is just the product of the highest power of $x$ and the highest power of $y$ that appear.  Now multiply numerator and denominator by that LCM.  It will clear all the fractions except the main one.</p>

Title: Need help simplifying complicated rational expression.

Body:
Studying for my final and I can't figure this out.

Simplify:

$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$


问:
Metric space and closed sets
Let $\mathbb{R}^n$ have a Euclidian metric $d$ and $\emptyset \neq A \subseteq \mathbb{R}^n$ be a subset such that $(A, d_{|A \times A})$ is a metric space.

(a) Let $x \in A$ and consider the set $D := \{y \in A: d(x, y) \geq r\}$. Is $D$ closed in $A$?

(b) Let $x \in A$ and consider the set $E := \{y \in A: d(x, y) > r\}$. Is $D$ contained in the closure of $E$ in $A$?

Justify your answers.

I am reviewing solutions to various problems for an upcoming examination and this was a problem in a practice set but there are no solutions available. Any assistance with a solution is much appreciated.

答:
Let $x$ be in $A$. We have

$D^c=\{y\in A, d(x,y)<r\}=B(x,r)$

$D^c$ is just the open ball having $x$ as a center and $r$ as a radius. It's an open set so its complementary is closed. Thus $D$ is closed.

Suppose now that $y\in E$, then by definition :
$$y\in E \implies d(x,y)>r$$

Logically speaking, if $d(x,y)>r$ then $d(x,y)\geq r$

So $y\in E \implies y\in D$

And since $D$ is a closed set, we have the following property :
$$ D \text{  is a closed set} \iff \bar{D}=D$$

So D is a closed set, and it's equal to its closure, and since $y\in D$ then

$$y\in \bar{D}$$

Title: Metric space and closed sets


问:
Definition of Boolean truth value and its expression in meta theory.
In Bell's set theory book, the statement 'the axiom scheme of separation is true in $V^{B}$(Boolean valued model) for any complete Boolean algebra $B$' is proved in $ZFC$
by fixing arbitrary formula $\phi$
and showing
$\forall u\exists v\forall x[x\in v \iff x\in u \land\phi(x)]$ is true in $V^{B}$.

And he says that $ZFC$ cannot completely formalize the construction of the Boolean value $\Vert{\sigma}\Vert^{B}$ for arbitrary sentence $\sigma$.

My question is, how can I express the above statement in first order arithmetic($PA$)?

I wish to express in $PA$ that if $\phi(x)$ is a formula in which $v$ is not free, then $ZFC\vdash\Vert\forall u\exists v\forall x[x\in v \iff x\in u \land\phi(x)]\Vert=1$

But I don't know whether $\Vert\forall u\exists v\forall x[x\in v \iff x\in u \land\phi(x)]\Vert$ can be expressed for arbitrary $\phi$ in the meta theory $PA$.

答:
There is no difficulty expressing $\Vert\forall u\exists v\forall x[x\in v \iff x\in u \land\phi(x)]\Vert$ in the metatheory.  Specifically, you can write down an algorithm that takes a formula $\sigma$ in the language of ZFC and outputs a formula $\varphi$ in the language of ZFC that defines (with $B$ as a parameter) $\Vert{\sigma}\Vert^{B}$ (that is, $\Vert{\sigma}\Vert^{B}$ is the unique $x$ such that $\varphi(x,B)$).  This is just the usual definition of $\Vert{\sigma}\Vert^{B}$ by recursion on formulas.

What ZFC cannot formalize is that ZFC cannot define a function that takes a formula $\sigma$ as an input and outputs the value of $\Vert{\sigma}\Vert^{B}$.  It can take $\sigma$ and output a formula that defines this value, but it can't actually evaluate this defining formula to get a value of $B$.  This is an instance of the non-definability of truth: ZFC can't define a function that takes a formula $\varphi$ and outputs the unique set that $\varphi$ defines (if $\varphi$ defines a unique set), because ZFC can't define what it means for a set to satisfy $\varphi$.  This is not a problem for expressing a purely syntactic statement like $ZFC\vdash\Vert\forall u\exists v\forall x[x\in v \iff x\in u \land\phi(x)]\Vert=1$, though, since this only requires a formula that defines $\Vert\forall u\exists v\forall x[x\in v \iff x\in u \land\phi(x)]\Vert$ in the language of set theory.


问:
Prove the functor is not representable
Let $F$ be the functor from rings to sets that takes a ring R to the set of all pairs $(a,b)\in R \times R$ such that $aR + bR = R$.  Show that $F$ is not representable.

This is a 2019 UCLA qualifying exam question.  I have tried using that if $(a,b)$ is a universal element in a ring $R$, then so is $(b,a)$.

答:
For convenience, let $k$ be a field.

Lemma As subrings of $k(x,y)$, we have $k[x,y] = k[x,y,x^{-1}] \cap k[x,y,y^{-1}]$.

Proof. Clearly, we have $k[x,y] \subseteq k[x,y,x^{-1}] \cap k[x,y,y^{-1}]$. Next, let $p \in k[x,y,x^{-1}] \cap k[x,y,y^{-1}]$ be arbitrary. Then we have $p = f/x^n$ and $p = g/y^m$ for some $f,g \in k[x,y]$ and $n,m \geq 0$. This yields $y^m f = x^n g$, so $y^m$ divides $x^n g$, whence $y^m$ divides $g$, so $p = g/y^m \in k[x,y]$. $\square$

Corollary The following is a pullback square in the category of (commutative) rings:
$\require{AMScd}$
\begin{CD}
k[x,y] @>>> k[x,y,y^{-1}]\\
@VVV @VVV \\
k[x,y,x^{-1}] @>>> k(x,y)
\end{CD}

Now, suppose for contradiction that $F$ is representable. Then $F$ preserves limits, so in particular the following would be a pullback square in the category  of sets:

\begin{CD}
F(k[x,y]) @>>> F(k[x,y,y^{-1}]) \\
@VVV @VVV \\
F(k[x,y,x^{-1}]) @>>> F(k(x,y))
\end{CD}

Moreover, we see directly that $F$ preserves inclusions (if $f : R \to S$ is an inclusion, then $F(f) = (a,b) \mapsto (f(a),f(b)) : F(R) \to F(S)$ is an inclusion). Thus, we have
$$F(k[x,y]) = F(k[x,y,x^{-1}]) \cap F(k[x,y,y^{-1}])$$
However, this is not true, since $(x,y) \in F(k[x,y,y^{-1}]) \cap F(k[x,y,x^{-1}])$ but $(x,y) \notin F(k[x,y])$.

Title: Prove the functor is not representable


问:
Meaning of P(dw)
I'm reading Shreve's Stochastic Calculus for Finance.

On page 35, he gives this example:

Example 1.6.4.

Recall Example 1.2.4 in which $\Omega =  [0, 1]$, $\mathbb{P}$ is the
uniform (i.e., Lebesgue) measure, and

$$\widetilde{\mathbb{P}}[a, b] = \int_a^b 2\omega d\omega = b^2 - a^2,
> 0 \leq a \leq b \leq 1.$$

We may use the fact that $\mathbb{P}(d\omega)=d\omega$ to rewrite the
above as

$$\widetilde{\mathbb{P}}[a, b] = \int_{[a, b]} 2\omega d\mathbb{P}(\omega)$$

I'm confused by "$\mathbb{P}(d\omega)=d\omega$". In my mind, $\mathbb{P}$ is a measure, $d\omega$ outside of an integral is meaningless, and $\mathbb{P}(d\omega)$ symbolically doesn't have any meaning.

How do I interpret $\mathbb{P}(d\omega)$? How do I make sense of Shreve's substitution?

Edit: What set is the equivalence relation $=$ in "$\mathbb{P}(d\omega)=d\omega$" defined over?

答: $\mathbb{P}(d\omega)$ means $d\mathbb{P}(\omega)$. It's just another notation. Since $\mathbb{P}$ is the Lebesgue measure here, we have $d\mathbb{P}(\omega) = d\omega$.

Title: Meaning of P(dw)


问:
Characteristic function of the variable X - Y
Let $\phi(X)$ be a characteristic function of $X$. Let $Y$ be independent of $X$ and has the same distribution as $X$. How to find a characteristic function of a variable $X-Y$? I know for a variable $X+Y$ we have: $\phi_{X+Y} (t) =\phi_{X} (t) \phi_{Y} (t)$ but in this case I don't know how to use it..

答: $X$ and $-Y$ are also independent. So $Ee^{it(X-Y)}=Ee^{it(X+(-Y))}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is  nothing but the complex conjugate of $\phi_Y(t)$. Hence, $\phi_{X-Y}(t)=\phi_X (t) \overline {\phi_Y(t)}=|\phi_X(t)|^{2}$.


问:
$L^p$-space isomorphisms
Let $G$ be a compact hausdorff abelian group.

There is a Haar integral $\int : C(G) \rightarrow \mathbb{R}$ on $C(G)$, the $C^*$-algebra of continuous functions from $G$ to $\mathbb{R}$. We can complete this $C^*$-algebra in various ways to get $L^p(G)$, the completion of $C(G)$ under the norm $||f||_p = \sqrt[p]{\int_G |f|^p}$.

We have a map
$$C(G) \rightarrow C(\hat{G})$$
sending $f : G \rightarrow \mathbb{R}$ to the map $\hat{f} : G \rightarrow \mathbb{R}$ sending $\chi$ to $\int_G f \chi$.

My question is about completing on one side or another by various norms. For which $p$ and $q$ do we have a map
$$L^p(G) \rightarrow L^q(\hat{G})$$
and for which $p$ and $q$ is this an isomorphism?

I suspect that for $p$ and $q$ such that $1/p + 1/q = 1$, we have such a map, and that it is an isomorphism. But a reference would be nice.

答: Hint: If $1 < p < \infty$, $L^p$ is reflexive. Otherwise not.

Title: $L^p$-space isomorphisms


问:
Proof of Lagrange theorem - Order of a subgroup divides order of the group
The Lagrange theorem states:

If $G$ is a finite group, and $H$ a subgroup of $G$, then the order of $H$ will divide the order of $G$. More precisely, $|G| = |H| \cdot (\text{number of left cosets of }H)$:

$$|G| = |H| \cdot (G:H)$$

The proof I have in my notes says:

$G$ consists of $\{G:H\}$ cosets, each of them consists of $|H|$ elements, the cosets are disjoint.

That's it. How does this prove the theorem?

答:
One simple approach is to show that the (left) cosets are distinct. Once we have that, the rest follows fairly obviously.

Step 1: We are given that $H$ is a group (a subgroup of $G$). In particular for every $h\in H$ we also have $h^{-1}\in H$. So for some element $j\in G, j  \not \in H$ we have that $jh \not \in H$, since otherwise $jhh^{-1}=j\in H$.

Step 2: Form the coset $jH$ from $\bigcup_{h\in H} jh$, and note $|jH| = |H|$. Of course since $e\in H$, $j\in jH$. Now consider the coset $kH$ with $k\in jH$. We must have $k=jh_0$ so also $j=kh_0^{-1}$. Then any $jh_i = kh_0^{-1}h_i$ and of course $h_0^{-1}h_i \in H$ so $kH = jH$.

Step 3: Thus cosets are distinct (since any common member implies that the cosets are identical), all are of size $|H|$, and every member of $G$ is in a coset, so $|H|$ divides $|G|$


问:
Upper bound for a Fresnel Integral
I was wondering if there is an upper bound strictly less than 1 to the function $x \mapsto \int_{0}^{1} \sin (x^2 + t^2) dt$ for $x \in [0,1]$. And if it is how to derive it.

Thank you for your answer,

答:
Use a trig identity to write

$$
\sin(x^2+t^2)=\sin(x^2)\cos(t^2)+\sin(t^2) \cos(x^2)
$$

Now you have two Fresnel integrals; S and C, leading to

$$
I(x)=\cos(x^2)\operatorname{S}(1) +\sin(x^2)\operatorname{C}(1)
$$

Over $0 \leq x \leq 1$, this is maximum at $x=1$, so

$$
I_{\text{max}}=I(1)=\cos(1)\operatorname{S}(1) +\sin(1)\operatorname{C}(1) \approx 0.93
$$

Title: Upper bound for a Fresnel Integral


问:
Expectation of minimum of uniformly random subset of $\{2^1,2^2,⋯,2^{10}\}$?
Choose a random subset of  $\{2^1,2^2,⋯,2^{10}\}$  by selecting each of the 10 elements independently with probability  $\frac{1}{2}$. Find the expected value of the smallest element in the subset (e.g. the subset can be  $\{2^1,2^3,2^4,2^7\}$. The smallest element is $2^1$).

I am not sure how to approach this question.  I did however try to do $\{(2^1*1)+(2^2*2)+\cdots+(2^{10}*10)\}$ and did not get the right answer.

Any assistance would help.

答:
The probability that $2$ is chosen is $\frac12$.

The probability that $2^2$ is the smallest element chosen is $\left(\frac12\right)^2$. We do not pick $2$ and then pick $2^2$.

In general, the probability of $2^k$ is the smallest element  is $\left( \frac12 \right)^k$.

Hence it should be evaluated to be $$\sum_{k=1}^{10}2^k\cdot  \left( \frac12\right)^k=10$$


答:
The expected value is $\sum_{2^k\in \{2^1,2^2,....,2^{10}\}} P(2^k$ is the smallest item in the set$)\times 2^k$.

The way I see it, the way that the least element is $2^k$ is if all the $2^{j_{j< k}}$ are not chosen and $2^k$ is.  The probability of that is $\frac 1{2^{k-1}}\times \frac 12 = \frac 1{2^k}$. So $P(2^k$ is the smallest item in the set$)= \frac 1{2^k}$.  So the expected value is $\sum_{k=1}^{10} \frac 1{2^k}\times 2^k = 10$.

This assumes if the set has no elements we consider the least of zero elements as $0$.  If we simple exclude this for the sample space, then our probabilities are all bumped up by a facto for $\frac {2^{10}}{2^{10} - 1}$ and our expected value is $10 \frac {2^{10}}{2^{10} - 1} = 10.0097...$

Title: Expectation of minimum of uniformly random subset of $\{2^1,2^2,⋯,2^{10}\}$?

Body:
Choose a random subset of  $\{2^1,2^2,⋯,2^{10}\}$  by selecting each of the 10 elements independently with probability  $\frac{1}{2}$. Find the expected value of the smallest element in the subset (e.g. the subset can be  $\{2^1,2^3,2^4,2^7\}$. The smallest element is $2^1$).

I am not sure how to approach this question.  I did however try to do $\{(2^1*1)+(2^2*2)+\cdots+(2^{10}*10)\}$ and did not get the right answer.

Any assistance would help.


答:
The shortest route is to note that the least common multiple of all four denominators of the ‘small’ fractions is $x^3y^4$ and to multiply the expression by $1=\frac{x^3y^4}{x^3y^4}$:

$$\large{\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}\cdot\frac{x^3y^4}{x^3y^4}}\;.$$

I’ll let you check that this does the trick. It’s analogous to seeing that the least common multiple of $2,3,4$, and $6$ is $12$, so that multiplying

$$\large\frac{\frac12+\frac23}{\frac14+\frac56}$$

by $\frac{12}{12}$ will clean it up nicely.


问:
Cylinder geometry
I want to measure the surface contour of a cylinder. I am using a laser line scanner to measure it. I am currently facing a problem with x-tilt - a misalignment caused when the measurement laser line is not parallel to the plane of rotation.

For instance, say the two points of laser line on the cylinder surface are $(x_{1},\theta_1)$ and $(x_2,\theta_2)$. If $\theta_1 = \theta_2$ then the alignment is correct. If I measure a perfectly smooth cylinder with a properly aligned sensor, z (the distance between the sensor and the cylindrical surface) as a function of x (axial length of the cylinder) would be flat line. However, if   $\theta_1 \neq \theta_2$, then z vs x would be a concave curve with peak value equal to the z_correct_aligned.

The question: How can we extract the angle by which the sensor should be rotated to ensure proper alignment when we have the curve measurement and know the radius of the cylinder?

Attached image illustrates the problem.

答:
The laser beam when moved defines a plane. If the laser is properly aligned, the plane contains the axis of the cylinder. If it is misaligned in a way that it creates an angle with the axis, the plane it defines interescts this axis at some angle $\alpha$. Note that it is also possible to have the laser misaligned in such a way, that it defines a plane parall to the axis, but not containing it; this isn't the case you're asking about, so I won't be talking about it.

If a plane intersects the axis of a cylinder, then the crossection of the plane with the cylinder is an ellipse. It's projection on $(x,z)$ plane (which is measured) will also be an ellipse. It can be easily checked that the short axis (iun the direction $z$) of this ellipse is $R$ and the long axis (in the direction $x$) is $\frac{R}{\tan\alpha}$.

The profile you measure will be an arc of this ellipse. In a perfect situation, assuming that the maximum of the profile is at the point $(x_0,z_0)$, the profile should be given by the formula
$$ \frac{(x-x_0)^2}{R^2/\tan^2\alpha} + \frac{\big(z-(z_0-R)\big)^2}{R^2} = 1$$
that is
$$ z = z_0 + \sqrt{R^2-(x-x_0)^2\tan^2\alpha} - R$$
In a realistic situations with all possible measurement errors in the way, you don't expect full agreement with the formula, and you just need to find the value of $\alpha$ that gives the best fit. Or you can just look at $z$ around the maximum; we expect
$$ z \approx z_0 -\frac{\tan^2\alpha}{2R} (x-x_0)^2 $$
Which means that
$$ -\frac{\tan^2\alpha}{R} = z''(x_0)$$
$$ \tan^2\alpha = -R\, z''(x_0)$$
$$ \alpha = \arctan \sqrt{R\, |z''(x_0)|}$$
Fitting an ellipse to a measured profile will probably give you a better approximation for $\alpha$ than just looking at the neighbourhood of the maximum.

Title: Cylinder geometry


问:
Find $x$ for $\overline{x}\oplus(\overline{7}\odot\overline{11}) = \overline{2} \odot (\overline{x}\oplus \overline{9})$

Find $x \in \mathbb{Z}$ such that the equality $\overline{x}\oplus(\overline{7}\odot\overline{11}) = \overline{2} \odot (\overline{x}\oplus \overline{9})$ is true in $\mathbb{Z_{12}}$

My reasoning:

Initially, defining an equivalence relation in $\mathbb{Z}$ with, such that $a\thicksim b \Leftrightarrow a - b$ is multiple of $n$ and $n>1$. So we have $a\equiv b \text{ (mod }n)$. For $x\in\mathbb{Z}$ and its equivalent class denoted as $\overline{x}$, we have
$$\overline{x} = \{a\in\mathbb{Z}: a\equiv z \text{ (mod }n)\}$$

So, the equivalent classes are

$\overline{0} = \{..., -2n, -n, 0, n, 2n, ...\}\\
\overline{1} = \{..., -2n+1, -n+1, 1, n+1, 2n+1, ...\}\\
\vdots\\
\overline{n-1} = \{..., -n-1, n-1, n-1, ...\}
$

And the congruent modulo $n$ in $\mathbb{Z}$ gives

$\mathbb{Z}_n = \mathbb{Z}/\sim = \{\overline{0}, \overline{1}, \overline{2}, ... , \overline{n-1}\}$

Once the equality given is true in $\mathbb{Z_{12}}$, such that $\mathbb{Z_{12}} = \{\overline{0}, \overline{1}, \overline{2}, ... , \overline{11}\}$, we know $x\in [0, 11]$. Also, once $n > 1$ the following properties are true:

$\overline{a}\oplus(\overline{b}\oplus \overline{c}) = (\overline{a}\oplus\overline{b})\oplus \overline{c} \quad (\text{Associativity of } \oplus)\\
\overline{a}\oplus\overline{b}= \overline{b}\oplus\overline{a} \quad (\text{Commutativity of } \oplus)\\
\overline{a}\odot(\overline{b}\odot\overline{c}) = (\overline{a}\odot\overline{b})\odot \overline{c} \quad (\text{Associativity of } \odot)\\
\overline{a}\odot\overline{b}= \overline{b}\odot\overline{a} \quad (\text{Commutativity of } \odot)\\
\overline{a}\odot(\overline{b}\oplus\overline{c}) = (\overline{a}\odot\overline{b})\oplus(\overline{a}\odot\overline{c}) \quad (\text{Distributivity of } \odot)\\
\text{For } \overline{a}, \overline{b}, \overline{c} \in\mathbb{Z}_n
$

Also,

$\overline{a}\oplus\overline{b} = \overline{a+b}$

and

$\overline{a}\odot\overline{b} = \overline{ab}$

Then, for the given question I tried the following

$\overline{x}\oplus(\overline{7}\odot\overline{11}) = \overline{2} \odot (\overline{x}\oplus \overline{9})$

As $\overline{7}\odot\overline{11} = \overline{77}$, we have $\overline{72} = \overline{0}$ for $\mathbb{Z}_{12}$, then $\overline{77} = \overline{5}$.

$\overline{x}\oplus\overline{5} = \overline{2} \odot (\overline{x}\oplus \overline{9}) \implies \overline{x}\oplus\overline{5} = (\overline{2}\odot\overline{x})\oplus(\overline{2}\odot\overline{9}) $

As $\overline{2}\odot\overline{9} = \overline{18}$, we have $\overline{12} = \overline{0}$ for $\mathbb{Z}_{12}$, then $\overline{18} = \overline{6}$.

$\overline{x}\oplus\overline{5} = (\overline{2}\odot\overline{x})\oplus(\overline{2}\odot\overline{9}) \implies \boxed{\overline{x}\oplus\overline{5} = (\overline{2}\odot\overline{x})\oplus\overline{6}}$

From the last step I couldn't finish it. Taking all the solutions for $x$ I could find that $x=11$, but I was not able to arrange the equality to get the solution.

For $\color{red}{\overline{x}\oplus\overline{5}} = \color{blue}{(\overline{2}\odot\overline{x})\oplus\overline{6}}$

$x = 1: \color{red}{\overline{6}} = \color{blue}{\overline{8}}\\
x = 2: \color{red}{\overline{7}} = \color{blue}{\overline{10}}\\
x = 3: \color{red}{\overline{8}} = \color{blue}{\overline{12} = \overline{2}}\\
x = 4: \color{red}{\overline{9}} = \color{blue}{\overline{14} = \overline{2}}\\
\vdots\\
x = 11: \color{red}{\overline{16} = \overline{4}} = \color{blue}{\overline{28} = \overline{4}}\\
$

How can I solve the question?

Sorry for the simple question, but I was not sure about how to proceed in the boxed step. In this context, what I could have done?

答:
Guide:

Modulo operations is almost like regular algebraic manipulation with caution when it comes to division.

If the question is $x+5 = 2x + 6$, we would add $-x-6$ on both sides.

Now looking at

$$\overline{x} \oplus \overline{5} =(\overline2 \odot \overline{x}) \oplus \overline6$$

Try to perform $\oplus (\overline{11} \odot \overline{x}) \oplus \overline{6}$ on both sides. I will leave the remaining steps to you.

Another comment is personally I like to write $\pmod{12}$ as a reminder to what we are working on.


问:
For the Lagrange Multiplier Theorem, why must the number of constraints be less than the number of variables?
In Fitzpatrick's "Advanced Calculus", the beginning of the General Lagrange Multiplier Theorem is given by:

Let $U$ be an open subset of $\mathbb{R}^n$ and suppose that $f: U \rightarrow \mathbb{R}$ is continuously differentiable. Let $k$ be a positive integer less than $n$ and suppose that the mapping $G: U \rightarrow \mathbb{R}^k$ is also continuously differentiable."

The function $G$ is a vector valued function, where each component function is a constraint. Why must $k < n$, that is, why must the number of constraints be less than the number of variables?

答: Each constraint defines a hypersurface, which is by definition of dimension $n-1$. Hence, $k$ constraints define an intersection of $k$ such hypersurfaces, which is a manifold of dimension $n-k$.


问:
Find subset of integers that sum to a specific value
I have a column in an Excel spreadsheet that is composed of positive and negative integers (credits and debits from an accounting ledger). I also have a specific value that is the sum of some subset of the integers from the aforementioned column. I want to find out which subset of these integers sum to produce this specific value in question.

In linear algebra terms, I interpret this problem as the following:

$$ \begin{bmatrix} a_1 & a_2 & a_3 & \ldots & a_n \end{bmatrix}
\begin {bmatrix} x_1 \\ x_2 \\ \ldots \\ x_n \end{bmatrix} = y $$

where:

$a_1,a_2,\ldots,a_n$ are integers (positive or negative)

$x_1,x_2,\ldots,x_n$ are restricted to either $0$ or $1$

$y$ is an integer

Is there a way to solve this programmatically? I am willing to try Wolfram Alpha, Excel tools, some Python script, or any other method. The best I can brainstorm at the moment is a Python script that loops through trying $2^n$ number of possibilities (0 or 1 for each $x_i$). There must be a better way, no?

答: This is the subset sum problem, and you can solve it via integer linear programming, with binary variables $x_i$ and linear constraint $\sum_{i=1}^n a_i x_i = y$.


问:
Is there a bijection from $[0,1]$ to the Cantor set $C$?
Does there exist a bijective function $\phi$ from the unit interval $[0,1]$ to the Cantor set $C$? If so, how can it be constructed? I could then proceed to build a measure space ($C$, $\mathcal{M}_\phi$, $m_\phi$) where $m_\phi(E)$ = $m(\phi^{-1}(E))$ for $E \subset C$. How would $m_\phi(C) = 1$? </p>

答:
There happens to be a very simple near-bijection between the two. In short, consider the binary decimal representation of all the numbers in $[0,1]$ and consider the ternary representation of all the elements of the Cantor Set. Since the Cantor Set can be attained by removing the middle third repeatedly, the elements of the Cantor Set are exactly those whose ternary representations have only 0s and 2s.

Then a number such as $0.202202...$ in the Cantor set could be associated with the number $0.101101...$ in $[0,1]$. So the function is simply to send $c \to (\frac{c}{2})$, taking account of the obvious change of base of course.

There is a problem with double-representation, as pointed out by Q. Yuan. Namely, both $0.0222...$ and $0.2$ (in ternary) get sent to $0.1$ (in binary). But perhaps there is a quick work-around here.

However, this does not complete the second half of your question. And, in addition, this is not in the least continuous.


问:
Comparing $2^n$ to $n!$
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?

答:
Consider for $n \ge 4$:
$$
2^n = 2 \cdot 2 \cdot 2^{n - 2} < 2 \cdot 3 \cdot 4 \cdot \ldots \cdot n = n!
$$</p>

Title: Comparing $2^n$ to $n!$


问:
Calculating the integral with an undefined function f(x)
I am having a problem with an excersice of calculating the integral

I= $\int_{0}^{a}\frac{1}{1+f(x)}\,dx$

and we know that $f(x)f(a-x)=1$ , $a>0$  and $f(x)$ is continous and positive on the interval $[0,a]$

i've tried manipulating the expression with the fact that $f(x)= \ \frac{1}{f(a-x)} $ in order to find something to cancel or something that makes me solve the integral "nicely"

However after 20 minutes i have no idea what to do, or what i am missing

so if anyone could give me a tip or point me in the right direction i would appreciate it

答:
If the answer is independent of $f(x)$, then you can choose one consistent with the stated conditions, e.g., $f(x)=1$, in which case we find:

$$\int\limits_{x=0}^a \frac{1}{1 + 1} dx = \frac{a}{2}.$$


问:
Show $\frac{(n-1)S^2}{\theta}$ is pivotal quantity of random sample $Y_1,...,Y_n$ from $N(\theta,\theta)$ then derive confidence interval

Show $\frac{(n-1)S^2}{\theta}$ is pivotal quantity of random sample $Y_1,...,Y_n$ from $N(\theta,\theta)$ then derive confidence interval $(1-\alpha)100%$ confidence interval for $\theta$

So I can show that $\frac{(n-1)S^2}{\theta}\sim \chi^2_{n-1}$ since $S^2=\sum_{k=1}^n \frac{(Y_i-\bar{Y})}{n-1}$

But I'm not sure how to find the confidence interval?

I believe it is supposed to look like $P(\chi^2_{n-1}\leq \frac{(n-1)S^2}{\theta}\leq \chi^2_{n-1})=1-\alpha$

And then solving for $\theta$ in the middle. But I know $\chi^2_{n-1}$ is wrong, I don't understand what the bounds are supposed to be.

答:
It is correct. Find the 2 quantiles of the chi-squared assuming equiprobable tails and solve w.r.t. $\theta$

Your pivotal qty is not the only one possible.

Check the definition of $S^2$ because the one you posted is $=0$


问:
Mapping reflections in 3D
I am working on a program in python that takes a point $\begin{pmatrix} x_i \\y_i\\z_i \end{pmatrix} $, generates a displacement vector $\begin{pmatrix} x_d \\y_d \\z_d\end{pmatrix} $ and adds it to the initial pont to determine the next point$\begin{pmatrix} x_f \\y_f \\z_f\end{pmatrix} = \begin{pmatrix} x_i \\y_i\\z_i \end{pmatrix}+\begin{pmatrix} x_d \\y_d\\z_d \end{pmatrix} $

My trajectory is confined to a rectangular box $\begin{bmatrix} x_0, x_1 \end{bmatrix} \times \begin{bmatrix} y_0, y_1 \end{bmatrix} \times \begin{bmatrix} z_0, z_1 \end{bmatrix} $
So when the trajectory hits a wall it bounces off. My question is how can I construct a function that takes the initial point and the displacement vector (and the boundaries of the box) and returns the point where my trajectory will end up after bouncing off said wall?

答:
Although I posted a description of the algorithm, the mathematical formulae used are easily lost in the wall of text.  So, I decided to write a separate answer to list the basic vector algebra operations one needs here.

Define the walls as planes (properly, halfspaces), using an unit normal $\hat{n}$ ($\lVert\hat{n}\rVert = 1$) and signed distance from origin $d$, with the unit normal pointing away from the wall.

For an axis aligned box enclosure $(x_\min, y_\min, z_\min) - (x_\max, y_\max, z_\max)$ the six walls with unit normals pointing inside the box are
$$\begin{array}{r|r}
\hat{n} ~ ~ ~ ~ & d ~ ~ ~ ~ \\
\hline
(1, 0, 0) & x_\min \\
(-1, 0, 0) & -x_\max \\
(0, 1, 0) & y_\min \\
(0, -1, 0) & -y_\max \\
(0, 0, 1) & z_\min \\
(0, 0, -1) & -z_\max \\
\end{array} \tag{1}\label{G1}$$

Distance from sphere of radius $r$ ($r = 0$ for a point) centered at $\vec{p}$, to halfspace $\hat{n}$, $d$, is
$$L = \vec{p} \cdot \hat{n} - d - r\tag{2}\label{G2}$$
This will be positive outside the halfspace, and zero when the sphere touches the surface of the halfspace.

A sphere of radius $r$ (for a point, $r = 0$) located outside the halfspace, at $\vec{p}$ at time $t=0$, having a velocity vector $\vec{v}$, and therefore its location as a function of time being
$$\vec{p}(t) = \vec{p} + t \vec{v}$$
will touch halfspace $\hat{n}$, $d$, at time $t \gt 0$ if and only if
$$\hat{n} \cdot \vec{v} \lt 0$$
where
$$t = \frac{d + r - \hat{n} \cdot \vec{p}}{\hat{n} \cdot \vec{v}} \tag{3}\label{G3}$$
When dealing with multiple spheres and/or walls, solve all positive $t$, and handle them in order of increasing $t$.  In a simulation, this lets you skip ahead in the simulation to the next bounce event, without having to calculate any intermediate steps.  If you store $\vec{p}_0$, $\vec{v}$, $t_0$ (so $\vec{p}(t_0) = \vec{p}_0$), and the time $t_1$ for its next bounce (and the wall it bounces from) for each sphere, you only need to update that one sphere for each bounce. The position of the sphere at time $t_0 \le t \le t_1$ is then
$$\vec{p}(t) = \vec{p}_0 + (t - t_0) \vec{v}, \quad t_0 \le t \le t_1$$

To reflect vector $\vec{v}$ (outside the halfspace) with respect to halfspace $\hat{n}$, $d$,
$$\begin{aligned}
\vec{v}_r &= \vec{v} - 2 \left( \vec{v} \cdot \hat{n} \right) \hat{n} \\
~ &= \vec{v} - \hat{n} \bigl( 2 \hat{n} \cdot \vec{v} \bigr) \\
\end{aligned} \tag{4}\label{G4}$$
Note that the part in parentheses evaluates to a scalar (a real number, not a vector).

A sphere of radius $r$ ($r = 0$ for a point) traveling between $\vec{p}_o$ outside the halfspace and $\vec{p}_i$ inside the halfspace $\hat{n}$, $d$, will touch the halfspace at $\vec{p}_b$,
$$\begin{aligned}
L_o &= \vec{p}_o \cdot \hat{n} - d - r \\
L_i &= \vec{p}_i \cdot \hat{n} - d - r \\
\vec{p}_b &= \displaystyle \vec{p}_o + \frac{L_o}{L_o - L_i} \left( \vec{p}_i - \vec{p}_o \right)
 = \displaystyle \frac{L_o \vec{p}_i - L_i \vec{p}_o}{L_o - L_i} \\
\end{aligned} \tag{5a} \label{G5a}$$
noting that $L_o \gt 0$ because $\vec{p}_o$ is outside the halfspace, and $L_i \lt 0$ because $\vec{p}_i$ is inside the halfspace.

The "untraveled" part that is reflected off the wall is
$$\begin{aligned}
\vec{p}_r &= \frac{-L_i}{L_o - L_i} \left( \vec{p}_i - \vec{p}_o \right) = \vec{p}_i - \vec{p}_b \\
\end{aligned}$$
where $\vec{p}_b$ is the above calculated position where the sphere touches the halfspace.

In other words, if the sphere does not bounce off any other walls, it will end up at $\vec{p}_e$,
$$\begin{aligned}
\vec{p}_e &= \vec{p}_b - \hat{n} \bigl( 2 \hat{n} \cdot ( \vec{p}_i - \vec{p}_b ) \bigr) \\
\end{aligned} \tag{5b}\label{G5b}$$
where the part in larger parentheses evaluates to a scalar (a real number, not a vector).

Title: Mapping reflections in 3D


问:
Need help simplifying complicated rational expression.
Studying for my final and I can't figure this out.

Simplify:

$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$

Body:
$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$

$$\large\frac{\frac{3y^3+5x^2}{x^3y^4}}{\frac{5-4x^2}{x^3y}}$$
$$\dfrac{3y^3+5x^2}{x^3y^4}\cdot{\dfrac{x^3y}{5-4x^2}}$$
$$\dfrac{3y^3+5x^2}{{y^3}\cdot{(5-4x^2)}}$$


问:
Open, closed or connected subsets in $\mathbb{R}$
Given that $(\mathbb{R}, |\cdot|)$ is a metric space, consider the following subsets of $\mathbb{R}$:

(a) $\{(-1)^n(\frac{n + 1}{n})\}$

(b) $\cup_{n \in \mathbb{N}} [n, \frac{n^2 + 1}{n}]$

(c) $\cup_{n \in \mathbb{N}} [\frac{1}{2^{n+1}}, \frac{1}{2^{n}}]$

(d) $\mathbb{Q}$

(e) $\mathbb{Q} \cap [0, 1]$

For each subset above, determine if they are open, closed, or not open and not closed, connected or not connected. Then, give the set of accumulation points for each of them. Don't forget to justify your answers.

My attempt:

(a) Closed. Not connected.

(b) Unsure.

(c) Unsure.

(d) Neither open nor closed. Not connected.

(e) Neither open nor closed. Not connected.

I am preparing for an exam and this problem was part of a practice set but there are no solutions available. Any assistance with a solution, preferably with an explanation, is much appreciated.

Edit: $\mathbb{N}$ in this case, does not include $0$.

答:
(a) $n$ usually denotes an integer.  If so, you have a totally disconnected set.  EDIT: Wrong here: "You are right that it is closed.  It has one accumulation point, can you see it?"  There are actually two accumulation points, and they are not in the set, therefore it is not closed.  Thanks to those who pointed that out.

To help see this, list the first several terms: $-2/1, 3/2, -4/3, 5/4, -6/5, 7/6, -8/7, 9/8, ...$. The odd terms are all negative: $-2, -4/3, -6/5, -8/7, -10/9...$ and approach $-1$.  Meanwhile the even terms are all positive: $3/2, 5/4, 7/6, 9/8...$ and approach $1$.

(b) actually makes no sense, as all the sets are empty except the first, which is $[1,2]$.  If that is what it actually says, then work with the interval $[1,2]$.  EDIT: Now that the intervals are defined more clearly, $[n, n+ 1/n]$ is a set of disjoint intervals, closed but not connected.

For a closed interval, every point in it is an accumulation point, and the same is true for this union of closed intervals.

(c) This one is $[1/4, 1/2] \cup [1/8, 1/4] \cup ... $. Sketch this out if you need to to see whether it is connected.  The union forms the interval (0, 1/2]. So every point in the set, and 0 as well, are accumulation points.

(d) and (e) you have correct.  Remember the definition of the reals when looking for accumulation points.


问:
Why is a compact codomained function with compact fibers closed?
Let $Y$ be a compact topological space and $f:X \rightarrow Y$ a function such that
\begin{equation}
f^{-1}(y) \text{ is compact in } X \quad \forall y \in Y,
\end{equation}

i.e. fibers are compact. Then $f$ is a closed function.

Why?

Source of the statement is theorem 4.44 from Marco Manetti's "Topologia" coursebook.

EDIT: the theorem reported there is false, and it has been fixed in a later version
of the book (ISBN 8847007569,9788847007567).

答:
(This proof is already known to be wrong, please don't downvote.)

Requirements

Directed set

A set $A$ is said to be a directed set if and only if for every $\alpha_1$, $\alpha_2 \in A$ there exists $\alpha \in A$ such that $\alpha > \alpha_1$ and $\alpha > \alpha_2$.

Net

Let $X$ be a set. A function from $A$ to $X$ is called net and is denoted with the symbol $\{x_\alpha\}_{\alpha \in A}$.

Convergent net

Let $X$ be a topological space and $A$ a directed set.

A net $\{x_\alpha\}_{\alpha \in A}$ is said to be convergent to $x \in X$ (and $x$ is said to be its limit) if for every neighborhood $V$ of $x$ there exists an $\alpha_V \in A$ such that $x_\alpha \in V$ for every $\alpha > \alpha_V$.

Closed sets characterization

A set $F$ in a topological space $X$ is closed if and only if and only if every $F$-valued convergent net has its limit in $F$.

Proof

We can choose $f$ surjective because even if $Y \setminus f(X) \neq \emptyset$ the hypothesis stays
untouched, because

\begin{equation}
f^{-1}(X \setminus f(X)) = \emptyset
\end{equation}

is indeed a compact set.

Suppose $C \neq \emptyset$ is a closed subset of $X$ such that $f(C)$ is not closed.

Let $\overline{y}$ be a point in $\overline{f(C)} \setminus f(C) \neq \emptyset$ and $\{y_\alpha\}_{\alpha \in A}$ be an $\overline{y}$-convergent net in $f(C)$.

Because every $y_\alpha$ is in $f(C)$ for each one of them we can choose $x_\alpha \in C$ such that $y_\alpha = f(x_\alpha)$ and $\{x_\alpha\}_{\alpha \in A}$ is a net in $C$ convergent to a point in $f^{-1}(\overline{y})$.

Since $C$ is closed, $f^{-1}(\overline{y})$ is compact and they are disjoined, there exists an open set $D \supseteq f^{-1}(\overline{y})$ which is still disjoined from $C$.

We get an absurd because there is no $\alpha \in A$ such that $x_\alpha \in D \supseteq f^{-1}(\overline{y})$ and so $C$ would not be a closed set.

Don't trust this answer until you get a review.


答:
$$2^n \;=\;\; \underbrace{\;2\cdot 2 \cdot  2\cdot 2\cdot\, \cdots \,\cdot 2 \; \cdot \;2\;}_{\text{n $ $factors of $2$}}$$

$$n! = \underbrace{1\cdot 2\cdot 3 \cdot 4\cdot\, \cdots\,\cdot (n - 1)\cdot n}_{\text{n factors}}$$

When we compare the products $\,$ factor-by-factor,$\,$ we can easily see that when $n \geq 4$, $\;n!\;>\;2^n$.

This can, of course, be formalized using induction on $n$.

Title: Comparing $2^n$ to $n!$

Body:
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?


问:
Irreducible unitary representation of compact group is contained in the left-regular representation
Given a compact metric topological group $G$ and an irreducible unitary representation $\pi$ of $G$, I would like to show that $\pi$ is contained in the left-regular representation $\lambda$ of $G$.

One of my reference books is 'A Course in Abstract Harmonic Analysis' by Folland. In chapter 5. the Peter-Weyl Theorem is exposed. However, I content myself with the weaker statement above. I was hoping that you could explain how to prove the result above (maybe by using matrix coefficients or otherwise).
Thank you very much in advance.

答:
Let $u$ be any  representation of the compact group $G$  on the Hilbert space $H$.

Fixing a unit vector $\xi $ in $H$,  define the linear map
$$
  F:H\to L^2(G),
  $$
by
$$
  F(\eta )|_g = \langle \eta , u_g(\xi )\rangle,  \quad\forall \eta\in H, \quad\forall g\in G.
  $$
For  $h$ in $G$, notice that
$$
  \lambda _h(F(\eta ))|_g =
  F(\eta )|_{h^{-1}g} =
  \langle \eta , u_{h^{-1}g}(\xi )\rangle  = $$$$ =
  \langle u_h(\eta ), u_g(\xi )\rangle  =
  F(u_h(\eta ))|_g,
  $$
so we see that
$$
  \lambda _h\circ F=  F\circ u_h,
  \tag 1
  $$
meaning that $F$ is a covariant map.

Assuming from now on that $u$ is irreducible, notice that (1) implies that $F^*F$ commutes with $u$, so the Schur Lemma
implies that $F=cI$, for some constant $c$.  Defining $V=c^{-1/2}F$, we have that $V$ is an isometric operator  and clearly
$\lambda _h\circ V=  V\circ u_h$.

The conclusion is that $V$ establishes a unitary  equivalence between $u$ and the restriction of $\lambda $ to the range of $V$.


答:
This answer doesn't seem to belong to Math.SO, but here it goes in R (consider it an extended comment):</p>
```<code>install.packages('utils')
require('utils')

set.seed(2)
(vec = unique(round(runif(10,-30, 30)))) # "positive and negative integers"
(x = sample(vec, 5, replace=F))
(S = sum(x)) # "a specific value that is the sum"

for(i in 1:length(vec)){
total.com = combn(vec, i, simplify = T)
if(sum(total.com[,which(colSums(total.com)==S)])!=0) print(total.com[,which(colSums(total.com)==S)])
}
</code>```

Title: Find subset of integers that sum to a specific value

Body:
I have a column in an Excel spreadsheet that is composed of positive and negative integers (credits and debits from an accounting ledger). I also have a specific value that is the sum of some subset of the integers from the aforementioned column. I want to find out which subset of these integers sum to produce this specific value in question.

In linear algebra terms, I interpret this problem as the following:

$$ \begin{bmatrix} a_1 & a_2 & a_3 & \ldots & a_n \end{bmatrix}
\begin {bmatrix} x_1 \\ x_2 \\ \ldots \\ x_n \end{bmatrix} = y $$

where:

$a_1,a_2,\ldots,a_n$ are integers (positive or negative)

$x_1,x_2,\ldots,x_n$ are restricted to either $0$ or $1$

$y$ is an integer

Is there a way to solve this programmatically? I am willing to try Wolfram Alpha, Excel tools, some Python script, or any other method. The best I can brainstorm at the moment is a Python script that loops through trying $2^n$ number of possibilities (0 or 1 for each $x_i$). There must be a better way, no?

Body:
This answer doesn't seem to belong to Math.SO, but here it goes in R (consider it an extended comment):

install.packages('utils')
require('utils')

set.seed(2)
(vec = unique(round(runif(10,-30, 30)))) # "positive and negative integers"
(x = sample(vec, 5, replace=F))
(S = sum(x)) # "a specific value that is the sum"

for(i in 1:length(vec)){
total.com = combn(vec, i, simplify = T)
if(sum(total.com[,which(colSums(total.com)==S)])!=0) print(total.com[,which(colSums(total.com)==S)])
}


问:
Ovaloid is orientable
Given a compact, connected regular surface $S$ in Euclidean space which has everywhere positive gaussian curvature = an ovaloid. It is a theorem of Hadamard that ovaloids are diffeomorphic to the sphere (discussed in Klingenberg), but the proof of this follows the proof of their orientability = that there exists a continuous, unit normal field defined globally on $S$. Unfortunately, there is no proof of this in doCarmo (as far as I can tell) and the proof in Klingenberg is said to be obvious, but the details are not explained. (He says the second order surface approximating $S$ near any point is an elliptic paraboloid, but it is not explained how to define the unit normal field globally)

Does anyone know a full proof of this theorem? Ovaloids were extensively studied by Blaschke in his book (kreis und kugel), unfortunately I do not speak German. I would be extremely grateful for any references on ovaloids in English (but only classical differential geometry)

答:
Here is a possible proof: you can use the Jordan-Brouwer Separation Theorem (found on page 89 of Guillemin and Pollack's book of Differential Topology) which says

The complement of a compact and connected hypersurface $X$ in $\Bbb{R}^n$ consists of two open sets, the "inside" $D_1$ and the "outside" $D_0$. Moreover, $\overline{D}_1$ is a compact manifold with boundary and $\partial \overline{D}_1=X$.

In particular, being the boundary of a compact manifold, $X$ inherits a boundary orientation. I.e. we define a tangent vector $V_p\in T_pX$ to be outward facing if it is the trajectory of a curve in $D_1$. In particular, this gives a well-defined notion of global outward normal vector field for the surface $X$ (take $n=3$).

Anyway, this implies that $X$ is orientable.

Addendum: It depends on what your notion of "classical" differential geometry is, but I think that the Jordan-Brouwer theorem is a fairly classical theorem of differential topology/geometry.

Title: Ovaloid is orientable


问:
Gradient and hessian of a function
For function
$$J: \Bbb{R}^d\to\Bbb{R}$$
defined in the following way:
$$
J(x) = c + b^Tx + x^TAx
$$
where:

c is a scalar

b is d-dimensional real vector

A is a positive definite matrix

How to find gradient and hessian of function J? Any tips or a simple example with d=2 is much appreciated

答: If $J(x) = c + b^Tx + x^TAx$, then $$DJ(x)(v) = b^Tv + v^TAx+x^TAv = (b+2Ax)^Tv,$$since $(v^TAx)^T=v^TAx$ (it is a real number). This means that $$\nabla J(x) = b+2Ax.$$The Hessian is the total derivative of the above, so $${\rm Hess} J(x)(v,w) = v^T (2A)w = 2v^TAw.$$

Title: Gradient and hessian of a function