问:
Product in $\mathbb{R}$ is not defined by addition
This is a follow up question on: Product not definable by addition in Th($\mathbb{R},+,\cdot,0,1,<)$.

I've shown that the only automorphism of $(\mathbb{R}, 0,1, +, - , \cdot)$ is the identity map.

But I'm having trouble to find an example non-trivial automorphism of $(\mathbb{R},0, 1, +)$ that fixes $\mathbb{R}$ pointwise  and show it's it's not definable.

Thanks in advance!

答:
You seem to have a bit of confusion about the relationship between definable subsets and automorphisms. The result we want to use here is that, if $A\subseteq M$ is a subset of an $\mathcal{L}$-structure $\mathfrak{M}$, then every subset of $M^n$ that is definable over $A$ (ie, that is definable by a formula in the expanded language $\mathcal{L}\cup A$) is fixed by any automorphism of $\mathfrak{M}$ that fixes $A$ pointwise. Can you see why this is?

In particular, if we're trying to show that multiplication is not definable in $(\mathbb{R}, 0, 1, +)$, (note that "definable" is just shorthand for "definable over $\emptyset$"), it suffices to find any automorphism of $(\mathbb{R}, 0, 1, +)$ that does not preserve multiplication. (Because any such map vacuously fixes $\emptyset$ pointwise.) In particular, we're not looking for automorphisms of $\mathbb{R}$ that "fix $\mathbb{R}$ pointwise", for as Alex points out any such map is the identity map by definition.

Now, if you know that $(\mathbb{R}, 0, 1, +, \cdot)$ has no non-trivial automorphisms, then we just need to find any non-trivial automorphism of $(\mathbb{R}, 0, 1, +)$, for any such map will not preserve multiplication. To do this, extend the element $x_0=1$ to a basis $(x_i)_{i\in I}$ of $\mathbb{R}$ considered as a $\mathbb{Q}$-vector space. Then any non-trivial permutation of the basis defines a non-trivial linear isomorphisms from $\mathbb{R}$ to itself; choose any such isomorphism that fixes $x_0=1$. By definition of linear maps, this map will fix $0$ and preserve addition, and since it also preserves $1$ by construction it is a non-trivial $\mathcal{L}$-automorphism of $\mathbb{R}$, as desired.


问:
Example of bijective function from $(-1,1)$ to $\Bbb R$
I'm looking for a bijective function $f : (-1,1) \to \mathbb{R}$. I'm having trouble finding an example.

答:
HINT: If the intent of this problem is for the codomain of $f$ to be all of $\mathbb{R}$,

Define the map $f : (-1,1) \to \mathbb{R}$ by $f(x) := \frac{x}{1-x^{2}}$.

ContentLicense: CC BY-SA 4.0


问:
A smooth (n-1)-surface $S \subset \mathbb {R}^n$ is orientable $ \implies S $ has a continuous field of normal vectors to it.
I would like to be able to show that: if a (n-1)-surface $S \subset \mathbb {R}^n$ of class $ C ^ 1 $ is orientable $ \implies S $ has a continuous field of normal vectors to it.
Below I show where I have arrived and where I stop.

$ S $ is a surface of class $ C ^ 1 $, so for each of its points $ x \in S $ there exists the tangent space to $S$ at that point, $ TS_x $. A basis for each of these spaces can be found by taking the columns of the Jacobian (evaluated in $ x $) of any chart that contains $ x $ in its domain of action. Given the basis, therefore, we can find the two normals to $ S $ in $ x $, that is the only two vectors with unit norm that are orthogonal to all vectors of the first basis found. Between these two normals, one can be chosen using the following criterion: the basis of $ \mathbb {R} ^ n $ formed by the (n-1) vectors of the basis of $ TS_x $ previously found plus the normal, must be oriented as the canonical basis of $ \mathbb {R} ^ n $. At this point it remains only to show that the normal thus selected, varies continuously on $S$. I have to show that:

$ \lim_ {S \ni x \to x_0 \in S} \mathbf {n} (x) = \mathbf {n} (x_0) $

Thanks to the fact that we are interested in the limit value, we can forget that $ S $ is described by several charts and consider any chart $ \phi \in \mathcal {A} $, fixed, which contains $ x_0 $ in its domain of action.
I feel that somehow I have to tie in the fact that $ J _ {\phi} (x) $ and $ J _ {\phi} (x_0) $ are $ \delta $-close (in any matrix norm) in order to conclude that the normal $ \mathbf {n} (x) $ and $ \mathbf {n} (x_0) $ are $\epsilon $-close, but I can't formalize this argument.

Can anyone help me, please?

EDIT: Proof attempt

$S$ is a surface of class $C^1$, therefore for each of its points $x\in S$ there exists the space tangent to it in that point, $TS_x$, of dimension $n-1$. Since the orthogonal complement of $TS_x$ in $\mathbb{R}^n$ has dimension 1, we can find the two normals to $TS_x$, ie the only two vectors with unit norm that are orthogonal to all vectors of a basis of $TS_x$. Between these two normals one can choose one using the following criterion: the basis of $\mathbb{R}^ n$ formed by the (n-1) vectors of the basis of $TS_x$ previously found and by the normal, must be oriented as the canonical basis of $\mathbb{R}^ n$. Such normal can be found practically as:

$$\mathbf{n}(x)=\frac{\det\begin{pmatrix}
\mathbf{e}_1 & |  & | & | \\
... & \mathbf{v}_1 & ... & \mathbf{v}_{n-1}  \\
\mathbf{e}_n & | & | & |
\end{pmatrix}}{\left | \det\begin{pmatrix}
\mathbf{e}_1 & |  & | & | \\
... & \mathbf{v}_1 & ... & \mathbf{v}_{n-1}  \\
\mathbf{e}_n & | & | & |
\end{pmatrix} \right |_{\mathbb{R}^n}}$$

where $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)\}$ are the columns of the Jacobian in $x$ of any chart that contains $x$ in its domain of action (they are a basis of $TS_x$).
We must therefore show that $\mathbf{n}(x)$ thus defined is such that:

$\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x),\mathbf{n}(x)\}$ is oriented like $\{\mathbf{e}_1 (x),…, \mathbf{e}_{n} (x)\}$;

The above formula for $\mathbf{n}(x)$ always gives the same vector if you use another map of the orienting atlas of $S$ (covering the point $x$), i.e. a different set $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)\}$;

$\mathbf{n}(x)$ is continuous $\forall x\in S$.

As for the first point, it follows from:

$$\det\begin{pmatrix}
|  & | & | & |\\
 \mathbf{v}_1 & ... & \mathbf{v}_{n-1} &\mathbf{n}  \\
 | & | & |&|
\end{pmatrix} = C_1^2+...+C_n^2>0$$

where the determinant has been developed along the last column (the normal) and the various $C_i$ are the cofactors of the Laplace development (which are also the components of $\mathbf{n}$, as the normal was first defined).
Since the possible normals for each $TS_x$ are only two, even changing the map used to define $\mathbf{n} (x)$, by point 1 one would always obtain the one capable of orienting $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x),\mathbf{n}(x)\}$ as $\{\mathbf{e}_1 (x),…, \mathbf{e}_{n} (x)\}$, so point 2 is proved.
Finally, the continuity $\forall x \in S$ can be inferred by taking any card covering x (thanks to point 2 and thanks to the fact that continuity is a local property) and noting that the single components of the various $\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)$  are continuous, being the chart of class $C^1$, and that $\det$ and $|\cdot |_(\mathbb{R}^n)$ are also continuous functions.

However, it seems that this 'proof' still does not go well, because if I am not mistaken it can still be applied to each step to the mobius strip. Still, what @neal told me to do I think I managed to prove it in point 2 above.

答:
Here I will use the definition (or characterization) that a manifold $M$ is orientable if (and only if) the line bunde $\Lambda^{\dim M} TM$ is trivial. It is a well-known fact and is not hard to show it is equivalent to having an orientation preserving atlas. I find it easier to use because there is no calculation, we just use the vocabulary of vector bundles.

All the following vector bundles will be $C^0$ because $S$ is $C^1$.

Let $i : S \to \mathbb{R}^n$ be the inclusion map. Let $i^*(T\mathbb{R}^n)$ be the pullback bundle of the tangent bundle of $\mathbb{R}^n$ on $S$ by $i$. Using the euclidean structure of $\mathbb{R}^n$, one can decompose $i^*\left(T\mathbb{R}^n\right)=TS \oplus N_p$ where $TS$ is the tangent bundle of $S$ and $NS$ is the normal bundle of $S$ in $\mathbb{R}^n$:
\begin{align}
NS = \left\{ (p,v) ~|~ v \in {T_pS}^{\perp}\right\}
\end{align}
The normal bundle of a hypersurface is a line bundle. We want to show that there exists a section of $NS$ that always have norm $1$. Notice that $i^*(T\mathbb{R}^n)$ is a trivial rank $n$ vector bundle, and $S$ is orientable, thus $\Lambda^n\left(i^*T\mathbb{R}^n\right)$ and $\Lambda^{n-1}\left(TS\right)$ are trivial line bundles. But it is easy to show, using the decomposition, that
\begin{align}
\Lambda^n\left(i^*T\mathbb{R}^n \right) = \Lambda^{n-1}\left(TS\right) \otimes \Lambda^1\left(NS\right)
\end{align}
and using the triviality of the two first line bundles show that $\Lambda^1\left(NS\right)$ is also a trivial line bundle over $S$: indeed, as the LHS is trivial, there exist a nowhere vanishing section $s\in \Lambda^n\left(i^*T\mathbb{R}^n\right)$. By the above decomposition, there exist $a_i,b_i$ sections of $\Lambda^{n-1}(TS)$ and $\Lambda^1(NS)$ respectively and $f_i$ $C^1$ functions with
\begin{align}
s = \sum_{i=1}^k f_i\cdot( a_i\otimes b_i)
\end{align}
As $\Lambda^{n-1}(TS)$ is trivial, there exist a nowhere vanishing section $a_0$ and smooth functions $g_i$ such that $a_i =g_ia_0$, and then
\begin{align}
s = \sum_{i=1}^k f_i\left(g_ia_0\right)\otimes b_i = a_0\otimes \left(\sum_{i=1}^kf_ig_ib_i \right)
\end{align}
The fact that $s$ and $a_0$ are nowhere zero shows that $b_0 = \sum_{i=1}^k (f_ig_i)b_i$ is a nowhere zero section of $\Lambda^1(NS)$. Consequently, $\Lambda^1(NS)$ is trivial as a line bundle.
But $\Lambda^1\left(NS\right)=NS^* \simeq NS$ (recall that $NS$ is of rank $1$), so $NS$ is a trivial line bundle over $S$: there exists a non-vanishing $C^1$ section $X \in \Gamma(NS)$.

To conclude, notice that $\nu = \frac{X}{\|X\|}$ is a $C^0$ normal unitary vector field over $S$.

Edit: another proof Let $S$ be an orientable $C^1$ hypersurface of $\mathbb{R}^n$, that is endowed with an orientable atlas $(U_{\alpha},\phi_{\alpha})$ - that is with $\det(\mathrm{d}(\phi_{\alpha}\circ{(\phi_{\beta}}^{-1}))>0$ if $U_{\alpha}\cap U_{\beta}\neq 0$. Let $\phi_{\alpha}: U_{\alpha} \to \phi_{\alpha}(U_{\alpha})\subset\mathbb{R}^{n-1}$ be such a chart. Let $V_1,\ldots,V_{n-1}$ be the pullback vector fields of the canonical basis of $\mathbb{R}^{n-1}$, that is
\begin{align}
V_i(p) = \mathrm{d}{\phi_{\alpha}}(p)^{-1} e_i,~ p \in U_{\alpha}
\end{align}
Then $(V_1(p),\ldots,V_{n-1}(p))$ is a basis of $T_pS$. By the Gram-Schmidtt process, one can construct on $U_{\alpha}$ an orthonormal frame $(E_1,\ldots,E_{n-1})$ of $TU_{\alpha}$, that is at each $p\in S$, $(E_1(p),\ldots,E_{n-1}(p))$ is an orthonormal basis of $T_pS$.

Notice that since the charts are $C^1$, the orthonormal frame above is $C^0$ ($\mathrm{d}\phi$ is $C^0$ and the Gram-Schmidtt algorithm is smooth).

Now, let $(E_1,\ldots,E_{n-1})$ be an orthonormal frame on $U_{\alpha}$. Fix $p\in U_{\alpha}$ and define $N_p$ to be the unique solution in $\mathbb{R}^n$ of the system:
\begin{align}
\forall i\in \{1,\ldots,n-1\},~ \langle E_i(p),N_p\rangle& = 0 & \det\left(E_1(p),\ldots,E_{n-1}(p),N_p\right) = 1
\end{align}
Then $N_p$ is the unit normal vector we are looking for. To conslude, we have to show two things:

the definition of $N_p$ does not depend on the chart $U_{\alpha}$, that is, if $p\in U_{\alpha}\cap U_{\beta}$, then the above construction does not depend on the orthonormal frame defined by $\phi_{\alpha}$ or $\phi_{\beta}$

$p \mapsto N_p$ is continuous

First, to show that $N_p$ is well defined, notice that if $V=(V_1,,\ldots,V_{n-1})$ and $W=(W_1,\ldots,W_{n-1})$ are the frames defined by $\phi_{\alpha}$ and $\phi_{\beta}$ respectively, then they define the same orientation on $U_{\alpha}\cap U_{\beta}$ because $\det_VW= \det\left(\mathrm{d}(\phi_{\alpha}\circ{\phi_{\beta}}^{-1})\right)>0$ (maybe it is the inverse of its determinant but the sign is the same, I did not make the full calculation). Thus, if $E$ and $F$ are the frame given by the ram-Schmidtt process from $V$ and $W$, they still define the same orientation, and thus,the definition of $N$ does not depend on $U_{\alpha}$ and $U_{\beta}$ of the chart $\phi_{\alpha}$ or $\phi_{\beta}$, but only on the orientation they define.

Secondly, to show that $N$ is continuous, it suffices to show that it is locally continuous, i.e that it is continuous on all charts $U_{\alpha}$ (recall that continuity in $\mathbb{R}^n$ is a local property). On $U_{\alpha}$, let $E_1,\ldots,E_{n-1}$ be the local orthonormal frame defined above. Let $f_{\alpha} : U_{\alpha}\times\mathbb{R}^n \to \mathbb{R}^n$ defined by
\begin{align}
f(p,X) = \left(\langle X,E_1(p)\rangle,\ldots,\langle X,E_{n-1}(p)\rangle, \det(E_1(p),\ldots,E_{n-1}(p),X)-1 \right)
\end{align}
One can show (easy but annoying calculations) that $f$ satisfies the hypothesis of the implicit function theorem around $f^{-1}(0)$ ($f$ is smooth in the second variable). Therefore, there exists a continuous function $p \in U_{\alpha} \mapsto X_p\in \mathbb{R}^n$ such that
\begin{align}
f(p,Y) = 0 \iff Y = X(p)
\end{align}
(maybe one has to restrict to a subset of $U_{\alpha}$ that is still open, this does not affect continuity as it is a local notion)
Consequently, $N = X$ and $N$ is continuous.


问:
How to you model a game when the rules you are given are wrong?
The title-question is the main focus of my confusion but I have several supporting questions at the end that try to clarify.

To illustrate my question, I will base it off of a game I played a while back during a team-building exercise at a university camp. I can't remember exactly how the rules were phrased. Basically all the students were blindfolded and led to a place where we were given a singular, long-taught rope to hold.

These were the constraints that were given to us or that could be inferred:

We were told that when the whistle blew we could begin moving.

We could move anywhere in space as long as we were touching the rope at any given moment (it was possible to pass other people with some talking and physical co-ordination).

If we thought we'd found the end of the rope we could raise our hand and some authority would tell us whether we had actually found it. There was no limit to how often we could do this.

If we found the end of the rope, we won.

We could ask to surrender; the game ended for us and we left with our blindfolds off.

It might not be the best formulation, but I have decided to model it as a one-player game with two actions, namely: surrender or continue searching for the end of the rope. Let $k$ be a quotient representing how much patience and/or resolve we have. And let $t$ be how much time has elapsed in minutes. Given the following decision matrix.

$$\begin{array}{c|c|}
\hline
Surrender & 0\\
Continue & 15-(t/k)\\
\hline
\end{array}$$

As time went on in the game, I heard as people asked if they had found the rope and being told that they hadn't. Soon, people with lower patience $k$ started surrendering. Everything fit the model of the rules explained so far.

The problem arose when I reached a knot in the rope. I recognised the knot from my starting point and realised that I had come full-circuit. The implication of this was that the theoretical win-condition stated at the beginning was un-achievable.

Much later, after the game, the facilitator told us that the solution was to surrender. The game was supposedly designed to show us how self-dependent we try to be. Before that, however, after I had realised that the actual rules were different to the way I had perceived them, I tried to formulate a new game with my new information. Since I wasn't satisfied to end with a solution that gave me the minimum utility.

Therefore, considering I was effectively playing against the patience of the facilitators, I adapted it as minimally as I could to a two-player game like the utility matrix bellow (where $j$ is the patience and/or resolve of the facilitator).

$$\begin{array}{c|c|c|}
& Surrender & Continue\\
\hline
Surrender & 0,0 & 0,5\\
Continue & 5,0 & 15-(t/k),15-(t/j)\\
\hline
\end{array}$$

The logic was that if I could motivate myself to have $k > j$, then I would have to win. Which, if this game was the measure, I eventually did, to the annoyance of the facilitator.

So what I want to know is, what was the actual game?

At what point did the rules change (if at all)?

If they did change, was my adaptation from the original one more reasonable or did the facilitator's "admission" about lying about the win condition not remove credence from their new win condition?

Is there a better way to model the two games (together maybe)?

This all borders on philosophy because it is unclear who determines the rules but I basically want to know what the "mathiest" way to talk about it is.

If there is any way this question can be clearer, feel free to edit it or suggest a change. I am just a hobbyist mathematician.

答:
Too long for a comment.

I think the situation which you describe is not related to mathematics or philosophy. Since the only scoring point provided was to win or to surrender, the numerical values in the modelling matrices are rather arbitrary. Also I doubt that so vague entities as patience or resolve can be numerically measured.

I think that the rules of the game was never changed. They were given to you by the declaration and it doesn’t matter which intentions had a facilitator about a game purpose.

Provided that the rope had no end, I agree with the facilitator that the solution was to surrender, to stop wasting time trying to find an end of the rope. In the game rules was not stated that the rope has an end (by the way, I don’t see much sense in Rule 3, because a player can clearly recognize whether he found an end of a rope).

On the other hand, I think given the rules it is natural to expect the rope has an end and it is not very fair from the facilitator to suggest to search a non-existing rope end. Even a quest for the Holy Grail looks optimistic compared to this task. So I see more game problem not in your tries for self-dependence, as was stated  by the facilitator, but in your initial credence to him. For instance, formulated as “This game is for fun, the facilitator is a fair guy and will not manipulate us”.


问:
Solving a PDE using a Fourier basis function
So I've got an equation

\begin{gather}
    \frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}\\
    \begin{cases}
    u(y,0) = 0\\
    u(0,t) = U\\
    u(h,t) = 0
    \end{cases}
\end{gather}
I need to change variable $u(y,t)$ to $v(y,t)$, where
\[
u(y,t) = U\Bigg(1-\frac{y}{h}\Bigg)+v(y,t)
\]
and use the Fourier basis function $sin(\frac{n\pi y}{h})$ in order to show that the solution is
\[
u = U\Bigg(1-\frac{y}{h}\Bigg)-\frac{2U}{\pi}\sum_{n=1}^{\infty}\frac{1}{n} e^{-n^2\pi^2\frac{\nu t}{h^2}}\sin\frac{n\pi y}{h}
\]

The new PDE we obtained is
\begin{gather}
   {\frac{\partial U}{\partial t}\Bigg(1-\frac{y}{h}\Bigg)}+\frac{\partial v}{\partial t} = \nu\frac{\partial}{\partial y}\Bigg(-\frac{U}{h}+\frac{\partial v}{\partial y}\Bigg)\\
    \frac{\partial v}{\partial t} = \nu\frac{\partial^2 v}{\partial y^2}
\end{gather}
New set of homogeneous boundary conditions
\begin{cases}
    v(0,t) = 0\\
    v(h,t) = 0
    \end{cases}

But somehow I do not know how to proceed exactly and use Fourier basis functions. I tried a bunch of series representation but do not seem to arrive at an answer. Could anyone outline the next steps for me please? In more detail if possible. I am looking at the theory but cannot put things together. Thanks!

答:
Basically, the Fourier expansion leads to a differential equation instead of PDE, because the second derivative of sin(k.y) is -k²sin(k.y).

Likewise, the time derivative will also simplify into algebraic operation, so that finally you get a linear algebraic equation.

Your first line does not look like an equation but it looks like a possible general solution of the diffusion equation below, according to the method of variable separation, but you need to add unknown coefficients $u_n(t)$ in front of sin(n.k.y).

U(1-y/h) is a particular solution (steady with uniform shear) so you rewrite the equation on the difference.

Your initial and boundary conditions seem inconsistent unless U=0.  There must be something wrong.

Actually, you are studying the linear stability of plane 2D Couette (shear) viscous flow.  Searching on the internet, I found the very source of your exercise, with the same notations.  I let you do the search.

The source is apparently Batchelor An introduction to fluid dynamics but I don't have it at home.

Assuming a parallel flow between two infinite plates with no slip conditions in the Navier-Stokes equations leads to your very equation.  The initial condition is indeed singular but the singularity is smoothed out by diffusion.

You may also be interested in Squire's theorem (see Wikipedia) and

Orszag, S., & Patera, A. (1983). Secondary instability of wall-bounded shear flows. Journal of Fluid Mechanics, 128, 347-385. doi:10.1017/S0022112083000518


问:
How to prove simple transitivity from faithful transitivity immediately from stabilizers?
The question comes from first few pages of Marcel Berger's textbook "Geometry I". Below is an excerpt in page 8:

1.5.3. One always has
$$G_{g(x)}=gG_xg^{-1}.$$
In other words, $G_{g(x)}$ and $G_x$ are conjugate subgroups of $G$,
and in particular are always isomorphic. Example 1.4.4.1 is an
$\underline{\textrm{immediate consequence}}$ of this remark.

where Example 1.4.4.1 is a proposition

1.4.4.1. Proposition. If $G$ is an abelian group, any faithful transitive action is simply transitive.

I have no difficulty in proving the claims in 1.5.3 and proposition 1.4.4.1 per se, but the last sentence of 1.5.3 (underlined in black) frustrated me. When $G$ is abelian, the two stabilizers merge into one, i.e., $G_x=G_{g(x)}$. But then I don't know how to proceed to establish simple transitivity. How to get a contradiction against faithfulness if there are another element in $G$ that sends $x$ to $g(x)$? I believe it will greatly deepen my understanding of concepts involved, so I greatly appreciate it if you can help me figure out how to prove the simple transitivity (immediately) using results of 1.5.3. Thank you.

答:
This is a general fact of group actions: if $g$ and $h$ do the same thing to $x$, that is, if $gx=hx$, then $h^{-1}g\in G_x$. This follows by applying $h^{-1}$ to both $gx$ and to $hx$, and using that in a group action $\sigma(\tau x) = (\sigma\tau)x$, and that $ex = e$.

This is a general fact about group actions: the intersection of the point stabilizers, $\cap_{x\in X} G_x$ is the kernel of the action; that is, the set (actually, normal subgroup) of all elements $g\in G$ such that $gx=x$ for all $x\in X$.

Now, from 1.5.3 you immediately get that if $x$ and $y$ are in the same orbit of the action of $G$, then $G_x$ and $G_y$ are conjugate.

In 1.4.4.1 you have the following hypotheses: the group $G$ is abelian; the action is transitive; the action is faithful.

Since the group is abelian, as you observe, if $x$ and $y$ are in the same orbit of $G$, then $G_x=G_y$. Okay: so, since the action is transitive there is only one orbit, so $G_x=G_y$ for all $x,y\in X$. Since the action is faithful, the kernel of the action is trivial. From the general fact in the second paragraph above, that means that the intersection of all point stabilizers is trivial; but that intersection is just equal to a single point stabilizer (because they are all the same subgroup!) So the one point stabilizers are all trivial.

But that means that if $gx=hx$ for any $x$, then $g=h$, by the “general fact” in the first paragraph.

This yields that the action is simply transitive (you already knew it was transitive).
So this follows pretty immediately from the assumptions that $G$ is abelian, and the action is faithful and transitive, since the two general facts I outline are pretty standard and basic about group actions.


问:
Proof Verification : Equivalent Condition for Locally Connected Space
A topological space $X$ is locally connected if

for every $x$ in $X$ and for every open set $V$ containing $x$, there is a connected open set $U$ with $x \in U \subset V$.

I think it is equivalent to the following statement:

For $X$, there is a basis all of whose elements are open and connected.

One way (Assume the latter, and prove that $X$ is locally connected) is trivial.

Let's prove the other way. (Assume that $X$ is locally connected.)

Let $\tau$ be the set $\{U : U ~\text{is open and connected}\}$.
We shall show that $\tau$ is a basis.

Becasue $X$ is locally connected, for every $x \in X$ (if we put $V = X$) there is a
connected open set $U_x$ with $x \in U_x \subset X$.

Hence, $\{U_x : x \in X\} \subset \tau$.
Because $\{U_x : x \in X\}$ covers $X$,
we know that $\tau$ covers $X$.

For any $V_1, V_2 \in \tau$, the set $V_1 \cap V_2$ is an open set.

Because $X$ is locally connected,
for every $x \in V_1 \cap V_2$,
there is $W_x$ which is open and connected
with $x \in W_x \subset V_1 \cap V_2$.

Note that $W_x$ belongs to $\tau$. Hence, $\tau$ satisfies all conditions for being a basis.

Please check my (above) proof is right.

And I think the same argument is available to the locally path-connected space. Please, give me an advice.

Thank you.

答:
What you prove is correct, but it is not that what is required.

In fact it is fairly obvious that the set $\mathcal U$
of all connected open $U \subset X$ forms a basis for the topolopy on $X$: For each  open $V \subset X$ and each $x \in V$ there exists $U \in \mathcal U$ such that $x \in U \subset V$.

What you prove is that $\mathcal U$ satisfies the conditions for an abstract basis on the set $X$, i.e. that $\mathcal U$ is the basis for some topology on $X$. See for example here.

For the locally path connected case everything works literally as above.


问:
How to interpret Newton's 6th Lemma?
In Newton's "Principia Mathematica" Book 1, Section 1 ("Of the Motion of Bodies") there is the following Lemma 6:

"LEMMA VI.

  If any arc ACB, given in position, is subtended by its chord AB, and in any point A, in the middle of the continued curvature, is touched
  by a right line AD, produced both ways; then if the points A and B
  approach one another and meet, I say, the angle BAD, contained between
  the chord and the tangent, will be diminished in infinitum, and
  ultimately will vanish.

For if that angle does not vanish, the arc ACB will contain with the
  tangent AD an angle equal to a rectilinear angle; and therefore the
  curvature at the point A will not be continued, which is against the
  supposition."

...this text is accompanied by the following diagram:

Noting, that the line rbd is parallel to the line RBD and the arc Acb appears to have a smaller curvature than the arc ACB on Newton's diagram, which of the following animated diagrams correctly depicts his Lemma ?

This diagram of mine?:

...or this diagram of mine?:

Note, that in the latter diagram, I have added the red colored depictions of the angles BAD and ABD.  These angle depictions do not appear on Newton's diagram (...but he writes about the angle BAD in the text of his 6th Lemma).

答:
The lemma can be found on archive.org.

I'd say that your second diagram captures the idea that a secant and tangent converge as $B$ approaches $A$.  But it's hard to say if that's what Newton was getting at, or something else.

Fortunately, there's lots of literature on this.  Let's start with an excerpt from Pourciau's The Preliminary Mathematical Lemmas of Newton's Principia.

In other words, you are not the first to wonder what Newton was talking about.  But you are in good company, including Leibniz.

Wright's Commentary On Newton's Principia, pg 14, referred to in the excerpt above, says this about Lemma VI:

In a Commentary on this Lemma if the demonstration be admitted, any
other definition than this is plainly inadmissible, and yet several of
the Annotators have stretched their ingenuity to substitute notions of
continued curvature, wholly inconsistent with the above. The fact is,
this Lemma is so exceedingly obscure, that it is difficult to make any
thing of it.

On the next page, in Article 24, Wright offers a substitute.

The trail doesn't necessarily go cold at that point.  Wright may be incorrect, and other annotators may have come up with reasonable interpretations. Pourciau, in the reference above, offers his interpretation later in the article, but it is hard for me to see how it explains Newton's version.

A good place to get another opinion is the History of Science and Mathematics site.

Title: How to interpret Newton's 6th Lemma?


问:
Relationship between matrix equations and determinants?
What is the relationship between a matrix equation and the determinants of the matrix?

Say if you have the matrix equation $A^3 - 4A = 0$, what can I learn about the matrix $A$'s determinants?

Or even if you have $A - 3B = 0$, does that imply $\det(A) = 3*\det(B)?$

答:
In general, an annihilating polynomial doesn't give you information about the determinant but rather the eigenvalues of the matrix. There is a well known theorem that says if $m$ is the minimal polynomial of $A$ and $p$ is any polynomial for which $p(A)=0$ then $m\mid p$. What this says is that the eigenvalues of the matrix necessarily forms a subset of the roots of any annihilating polynomial $p$.

For simpler cases, we can extract more information. For your first polynomial, we know that $A$ satisfies the polynomial $p(x)=x^3 - 4x$. This implies that the eigenvalues of $A$ are amongst the roots of $p$, being $\{0,\ 2,\ -2\}$. If the matrix has $0$ as an eigenvalue, that necessarily implies that the determinant is $0$. Otherwise, the determinant is of the form $\pm 2^n$ where $n$ is the size of the matrix. This is the most information that we can extract since all of these cases are possible.

For your second equation, we get $A = 3B$. Taking determinants gives $\det A = 3^n \det B$ where $n$ is the size of the matrix.

ContentLicense: CC BY-SA 3.0


问:
How do I make spirograh diagrams using software?
There are these "string math artworks" I believe they are called "spirographs" where people usually take a circle or other geometric shape and put in nails on the shape and then use yarn and weave the yarn in a particular way such as to make these artworks.

I am wanting to make similar diagrams using software instead of physically making them with nails and yarn. I found this one link where someone is using Tikz with Latex to generate a circle with nodes. This might work, but I don't know exactly how to use Latex and this Tikz (plug-in) software.

If someone can point me in the right direction that would be great. Thanks.

Here are some links:

https://tex.stackexchange.com/questions/415330/make-graph-where-nodes-and-connecting-edges-are-circles?newreg=03724c430452487a8f756a4c311bbdc3

https://www.youtube.com/watch?v=0K2YbYTyEqY

答:
You can draw a regular polygon in GeoGebra Classic, then connect the vertices with line segments.

You can compute the vertices of a regular polygon in a spreadsheet using
$$x=R \cos \frac {2 \pi k}n\\
y=R \sin \frac {2 \pi k}n$$
where $R$ is the distance from the origin to a vertex and $k$ runs from $1$ to $n$.  You can then calculate as many points as you like on some of the diagonals.  You can then plot the points as an $xy$ plot.


问:
Is $X(X^TX + \lambda I)^{-1}X^T$ invertible if $X$ is full rank?
Suppose $X$ is a $n \times p$ matrix with full rank. Is $X(X^TX + \lambda I)^{-1} X^T$ invertible? Note that $I$ is a $p \times p$ identity matrix.

答:
No, consider $X = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \lambda = 1$, then
\begin{align*}
X(X^TX + \lambda I)^{-1}X^T = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
\end{align*}
which is not invertible.

In fact, by the rank inequality $\textrm{rank}(AB) \leq \min(\textrm{rank}(A), \textrm{rank}(B))$, the rank of the matrix is at most $p$, which is the rank of $(X^TX + \lambda I)^{-1}$.


问:
FFT sampling: why is Nyquist frequency enough for inverse transform?
I am trying to understand an aspect of the Fast Fourier Transform (FFT) that is not making sense to me, but I'm struggling to find the flaw in my reasoning.

Suppose I model a continuous signal at a sampling rate $f$. According to Nyquist, the maximum frequency I will be able to "recover" from this signal via a FFT is $f/2$. How come when I apply the inverse transform, I can recover the original sample in real space with a sampling rate $f$, and not a "worsened" version of the original sample that only has frequencies up to $f/2$.

In other words: the intuition I am getting is that when applying an FFT, one looses half the frequency information. However, this cannot be right, otherwise one could not invert the transform.

Could someone help me realize where is the flaw here?

答:
You start with a signal having frequencies in the interval $[0,f]$.  You take the FFT and get a spectrum with frequencies in $[-f/2,f/2]$.  Then you apply the inverse FFT and recover the signal with frequencies in the interval $[0,f]$.  Why do you think half the information vanishes in the middle step?

You might ask, how can signal components varying as fast as $f$ be represented in frequencies from $[-f/2, f/2]$?  The answer is the wagon wheel effect and aliasing: all the spectrum originally in the interval $[f/2,f]$ is aliased into the $[-f/2,0]$ components.  Then the inverse transform recovers the aliases back to their original frequencies.

This is not, however, magic.  Time-domain frequencies in the interval $[f, 2f]$ also alias into the spectral interval $[-f/2,f/2]$ and your inverse transform will reconstruct them as if they were in the interval $[0,f]$.  But it is exceptionally important to realize -- the discrete-time samples for aliased frequencies are the same.  There is nothing in the time domain signal that distinguishes aliased frequencies.  So the infinite family of aliases are all sent to one spectral bin and then reconstructed as if they had the smaller (positive) frequency (although the reconstructed sample values are the same as if the frequency was correctly recovered).

To convince yourself of some of this, actually sample signals of sinusoids having frequencies $-f/2$, $3f/2$, and $5f/2$.  Then look at other sequences of aliases to firmly convince yourself that the sampled signals are identical.  So there is no way for the inverse transform to know which member of an alias family to reconstruct -- and it doesn't have to.  In the sample, there is no difference among the aliased frequencies.

So what is this all about?  Suppose the original source was actually a pair of sinusoids at aliased frequencies.  Apply the FFT and the inverse FFT and the result is at the lowest positive alias of both -- it is no longer a sum of two sinusoids, it's just one.


问:
Can't find lower bound for function.
I am looking to find the lower bound for a function $f(n) = \frac{n^5}{n^3 + 3n}$ where $g(n) = n^2$. I am looking to prove that $f(n)$ is $\theta(n^2)$ and have already proved that $f(n) = O(n^2)$ but cannot find the lower bound for this function and prove $f(n)=\Omega(n^2)$.

Title: Can't find lower bound for function.

Body:
For $n\ge 1$ we have

$$\frac{f(n)}{n^2}=\frac{n^3}{n^3+3n}\ge\frac{n^3}{(n+1)^3}=\left(\frac{n}{n+1}\right)^3\ge\frac18\,.$$


答:
$f:: (-1,1) \rightarrow \mathbb{R}$

$f: x \mapsto \tan(\frac{\pi}{2} x)$

Title: Example of bijective function from $(-1,1)$ to $\Bbb R$

Body: I'm looking for a bijective function $f : (-1,1) \to \mathbb{R}$. I'm having trouble finding an example.


问:
Block matrix notation
Given that $A$ is a real, rectangular matrix of dimension $m \times n$ and

$$\begin{align}
A = \left[\begin{array}{c} I \\ e^{\intercal} \\ -e^{\intercal}\end{array}\right]
\end{align}$$

is represented in block matrix notation. What does $A$ look like, given $e$ is a vector of $1$'s?

Title: Block matrix notation

Body:
It might look something like this:

$$\left[\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
1 & 1 & 1 & 1 & 1 & 1\\
-1 & -1 & -1 & -1 & -1 & -1
\end{array}\right]$$


问:
Let $a$ be a three digit integer number with digits $x; y; z $(in that order). Prove that $a$ is divisible by 9...
Let $a$ be a three digit integer number with digits $x; y; z $(in that order). Prove that $a$ is divisible by 9 if and only if $x + y + z$ is divisible by 9.

Following a proof of this:
Let $a; b; d; k$ be integers such that $a = dk + b$. Prove that $a$ is divisible by $d$ if and
only if $b$ is divisible by $d$.

答:
Notice that the number $xyz$ equals $$100x+10y+z = 9(11x+y)+(x+y+z),$$
which is clearly divisible by $9$ if and only if $x+y+z$ is.


问:
Torus action on compact manifolds
Let T = $S^1 \times ...\times S^1 $be a torus acting on a compact manifold M. Let $m\in M $ and $t_m = \lbrace X \in Lie(T), X.m = 0 \rbrace$.

Why is the set $ t=\lbrace t_m , m\in M \rbrace$ finite ? And does this result holds in a larger context, for example if M is not compact or if the group acting on M is an arbitrary lie group ?

答:
This is not totally obvious. The number of orbit type of the action of $T^n$ on a compact manifold is finite. A proof is in the book of Michelle Audin in the first chapter after the proof of the slice theorem of Koszul. Hamiltonian operation on symplectic manifolds (I translate the title in English, because I have this book in French) I believe it has been translated in English a long time ago.

The English tiltle: Torus action on symplectic manifolds. Michelle Audin.

Title: Torus action on compact manifolds


答:
Under the assumption that the bijection must be from

$(-1, 1) \to \mathbb{R}$:

$x \geq 0:~~$ Let

$$f(x) = - \log|x - 1|.$$

$x < 0:~~$ Let

$$f(x) = + \log| ~|x| - 1|.$$


答:
$f(n) = \underbrace{n^2}_{g(n)} \frac{n^3}{n^3} \frac{1}{1+3n^{-2}},$ so it is enough to bound $\frac{1}{1+3n^{-2}}$.
\begin{align*}
1 &\leq n < \infty  \\
1 = 1^2 &\leq n^2 < \infty  \\
1 &\geq n^{-2} > 0  \\
3 &\geq 3n^{-2} > 0  \\
4 &\geq 1+3n^{-2} > 1  \\
\frac{1}{4} &\leq \frac{1}{1+3n^{-2}} < \frac{1}{1} =  1  \text{.}
\end{align*}
Therefore,
$$  \frac{1}{4} g(n) \leq f(n) < g(n)  \text{.}  $$
(There's nothing special about "$1 \leq {}$" in the first line.  Pick another starting point or decide to have a strict inequality.  If you choose a larger lower bound or a strict inequality, you just raise the lower bound you end up with.)

Title: Can't find lower bound for function.

Body: I am looking to find the lower bound for a function $f(n) = \frac{n^5}{n^3 + 3n}$ where $g(n) = n^2$. I am looking to prove that $f(n)$ is $\theta(n^2)$ and have already proved that $f(n) = O(n^2)$ but cannot find the lower bound for this function and prove $f(n)=\Omega(n^2)$.


问:
why $\overline{ X-A}= \operatorname{int} X-A \cup \operatorname{Bd}(A)?$ How $\operatorname{Bd}(X-A)= \operatorname{Bd}(A)?$
Problem taken from Munkre topology book . Chapter $3$  -problem $6$ - page no : $150$

Let $A ⊂ X.$ Show that if $C$ is a connected subspace of $X$ that intersects both $A$
and $X − A$, then $C$ intersects $\operatorname{Bd} A$

Here  i got the answer  below  but i have some confusion in the given below answer my confusion marked in red box

My doubt : why $\overline{ X-A}= \operatorname{int} X-A \cup \operatorname{Bd}(A)?$ How $\operatorname{Bd}(X-A)= \operatorname{Bd}(A)?$

My thinking: It should $\overline{ X-A}= \operatorname{int} (X-A) \cup \operatorname{Bd}(X-A)$.

$\operatorname{Bd}(X-A)\neq \operatorname{Bd}(A)$ . Take $A= [0,1]$ and $X=\mathbb{R}$. Then $\operatorname{Bd}(\mathbb{R}\setminus [0,1])=\mathbb{R}$

$\operatorname{Bd}(A)=\{0,1\}$

答:
As Tito points out, $x \in \text{Bd}(A)$ if and only if every neighborhood $U$ of $x$ intersects both $A$ and $X - A$.

And similarly, $x \in \text{Bd}(X - A)$ if and only if every neighborhood $U$ of $x$ intersects both $X - A$ and $X - (X - A) = A$.

Thus $x \in \text{Bd}(A)$ if and only if $x \in \text{Bd}(X - A)$. That is, $$\text{Bd}(A) =\text{Bd}(X - A)$$


答:
One of the properties of the determinant (more here: http://en.wikipedia.org/wiki/Determinant) is
$$
\det(AB) = \det(A)\det(B)
$$
and from here you can deduce
$$
\det(A^n) = (\det(A))^n
$$
Another property of determinant is
$$
\det(cA) = c^n\det(A)
$$
if $A$ is a $n \times n$ matrix.

If you apply the determinant on your first equation you get
$$
\det(A^3) = \det(4A)
$$
or
$$
(\det(A))^3 = 4^n \det(A)
$$
implying (if $\det(A)$ is not zero)
$$
(\det(A))^2 = 4^n
$$
and finally
$$
\det(A) = (\pm2)^n
$$
From the equation
$$
A-3B=0
$$
you can deduce
$$
\det(A) = 3^n \det(B)
$$
where $A$ and $B$ are square $n \times n$ matrices.

Title: Relationship between matrix equations and determinants?

Body:
What is the relationship between a matrix equation and the determinants of the matrix?

Say if you have the matrix equation $A^3 - 4A = 0$, what can I learn about the matrix $A$'s determinants?

Or even if you have $A - 3B = 0$, does that imply $\det(A) = 3*\det(B)?$


答:
The definition given in the book is $\operatorname{Bd}E = \overline E \cap \overline{X \setminus E}$, so
$$\operatorname{Bd}(X \setminus A) = \overline{X \setminus A} \cap \overline{X \setminus (X \setminus A)} = \overline{X \setminus A} \cap \overline A = \overline A \cap \overline{X \setminus A} =\operatorname{Bd} A.$$

Body:
Problem taken from Munkre topology book . Chapter $3$  -problem $6$ - page no : $150$

Let $A ⊂ X.$ Show that if $C$ is a connected subspace of $X$ that intersects both $A$
and $X − A$, then $C$ intersects $\operatorname{Bd} A$

Here  i got the answer  below  but i have some confusion in the given below answer my confusion marked in red box

My doubt : why $\overline{ X-A}= \operatorname{int} X-A \cup \operatorname{Bd}(A)?$ How $\operatorname{Bd}(X-A)= \operatorname{Bd}(A)?$

My thinking: It should $\overline{ X-A}= \operatorname{int} (X-A) \cup \operatorname{Bd}(X-A)$.

$\operatorname{Bd}(X-A)\neq \operatorname{Bd}(A)$ . Take $A= [0,1]$ and $X=\mathbb{R}$. Then $\operatorname{Bd}(\mathbb{R}\setminus [0,1])=\mathbb{R}$

$\operatorname{Bd}(A)=\{0,1\}$


问:
Using confidence interval versus using Hypothesis testing for inference
When is it more appropriate to use a confidence interval and more appropriate to use hypothesis testing when one wants to make inferences from a sample to a population?

I think that is what confuses me about the two ways of inference making.

Please, help me out.
Thanks..

答:
Suppose you have $n = 150$ normally distributed test scores from your school, summarized
(in R) as follows:

```<code>summary(x); length(x); sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  54.15   89.54  101.44  100.64  110.66  148.02
[1] 150        # sample size
[1] 17.32858   # sample SD
</code>```

The nationwide mean is 97.3. Is the average at our
school significantly larger? A t test rejects the null
hypothesis that our scores are the same, in favor of
the alternative that ours are higher, at significance level
1% because the P-value is smaller than $0.01.$

```<code>t.test(x, mu = 97.3, alt="g")

        One Sample t-test

data:  x
t = 2.3617, df = 149, p-value = 0.009743
alternative hypothesis:
  true mean is greater than 97.3
95 percent confidence interval:
 98.29965      Inf
sample estimates:
mean of x
 100.6415
</code>```

Then the issue arises whether the difference between nationwide 97.3 and our 100.6 is of practical importance.
This may be more a question for educators than for statisticians.

However, the one-sided 95% confidence
interval $(98.3, \infty),$ shows that, given the variability in our scores, our sample mean might be considered as consistent with an actual performance as low as 98.3. And this is only one point above the national
average.

So maybe it is best to interpret our higher
scores as a hint of progress, rather than a fantastic
breakthrough.

Note: Fake data for this illustration simulated in R as shown
below. If you use the same seed for the sampling, you will get exactly the
same data.

```<code>set.seed(2020)
x = rnorm(150, 100, 15)
</code>```

Body:
Suppose you have $n = 150$ normally distributed test scores from your school, summarized
(in R) as follows:

summary(x); length(x); sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  54.15   89.54  101.44  100.64  110.66  148.02
[1] 150        # sample size
[1] 17.32858   # sample SD

The nationwide mean is 97.3. Is the average at our
school significantly larger? A t test rejects the null
hypothesis that our scores are the same, in favor of
the alternative that ours are higher, at significance level
1% because the P-value is smaller than $0.01.$

t.test(x, mu = 97.3, alt="g")

        One Sample t-test

data:  x
t = 2.3617, df = 149, p-value = 0.009743
alternative hypothesis:
  true mean is greater than 97.3
95 percent confidence interval:
 98.29965      Inf
sample estimates:
mean of x
 100.6415

Then the issue arises whether the difference between nationwide 97.3 and our 100.6 is of practical importance.
This may be more a question for educators than for statisticians.

However, the one-sided 95% confidence
interval $(98.3, \infty),$ shows that, given the variability in our scores, our sample mean might be considered as consistent with an actual performance as low as 98.3. And this is only one point above the national
average.

So maybe it is best to interpret our higher
scores as a hint of progress, rather than a fantastic
breakthrough.

Note: Fake data for this illustration simulated in R as shown
below. If you use the same seed for the sampling, you will get exactly the
same data.

set.seed(2020)
x = rnorm(150, 100, 15)


问:
Total no. of ordered pairs $(x,y)$ in $x^2-y^2=2013$
Total no. of ordered pairs $(x,y)$ which satisfy $x^2-y^2=2013$

My try:: $(x-y).(x+y) = 3 \times 11 \times 61$

If we Calculate for positive integers Then $(x-y).(x+y)=1.2013 = 3 .671=11.183=61.33$

my question is there is any better method for solving the given question.

thanks

答:
You can solve this pretty quickly, since you essentially need to solve a bunch of linear systems. One of them is e.g.

$$x-y = 3\times 11$$
$$x+y = 61$$

Just compute the inverse matrix of

$$\left[\begin{array}{cc}
1 & -1\\
1 & 1
\end{array}\right]$$

and multiply that with the vectors corresponding to the different combinations of factors.

(ps. How do I write matrices???)


问:
Suppose $A = U\Sigma U^T$. If $A$ is not invertible, is it correct to write $A^{-1} = (U\Sigma U^T)^{-1}$?
Suppose $A$ is symmetric and thus, orthogonally diagonalizable: $A = U\Sigma U^T$, where $UU^T = I$ and $U^TU = I$. $D$ is a diagonal matrix.

Suppose $A$ is not invertible, then would it be correct to write the following?
$$A^{-1} = (U\Sigma U^T)^{-1} = U\Sigma^{-1}U^T$$

A follow-up question is about how one might express the pseudoinverse of $A$ in terms of $U$ and $\Sigma$. Please see the edit below.

Edit: Suppose $A = U\Sigma V^T$ by singular value decomposition. Now $A$ may or may not be symmetric. How can I express the Moore-Penrose pseudoinverse $A^{+}$ in terms of $U, \Sigma, V$?

答:
$A$ will fail to be invertible when one of its eigenvalues is $0$. Then that will be one of the diagonal entries of $\Sigma$, which is thus itself not invertible.

Since the transformation is just changing the underlying coordinate system, rewriting the matrix this way can't change any geometric property the linear transformation.


答:
Your matrix is symmetric, but not necessarily definite positive. If it isn't definite (has a null eigenvalue) it is not invertible.

In the case it IS definite, then it is invertible, and by the law $(AB)^{-1} = B^{-1}A^{-1}$; you can deduce that:

$$A^{-1} = (U\Sigma U^T)^{-1} = (U^{T})^{-1}\Sigma^{-1}U^{-1} = U\Sigma^{-1}U^T$$

If it isn't invertible/definite; then this equation does not hold, since $\Sigma^{-1}$ is undefined.

Title: Suppose $A = U\Sigma U^T$. If $A$ is not invertible, is it correct to write $A^{-1} = (U\Sigma U^T)^{-1}$?

Body:
Suppose $A$ is symmetric and thus, orthogonally diagonalizable: $A = U\Sigma U^T$, where $UU^T = I$ and $U^TU = I$. $D$ is a diagonal matrix.

Suppose $A$ is not invertible, then would it be correct to write the following?
$$A^{-1} = (U\Sigma U^T)^{-1} = U\Sigma^{-1}U^T$$

A follow-up question is about how one might express the pseudoinverse of $A$ in terms of $U$ and $\Sigma$. Please see the edit below.

Edit: Suppose $A = U\Sigma V^T$ by singular value decomposition. Now $A$ may or may not be symmetric. How can I express the Moore-Penrose pseudoinverse $A^{+}$ in terms of $U, \Sigma, V$?


问:
Example of bijective function from $(-1,1)$ to $\Bbb R$
I'm looking for a bijective function $f : (-1,1) \to \mathbb{R}$. I'm having trouble finding an example.

答:
For a different example, $$f(x) = \frac{1}{x - 1} + \frac{1}{x + 1}$$ has this property:

It is injective: if $x,y\in(-1,1)$ and $f(x) = f(y)$, then we have $$\frac{1}{x-1}+\frac{1}{x+1} = \frac{1}{y-1} + \frac{1}{y+1}.$$ Multiplying by $(x-1)(x+1)(y-1)(y+1)$ gives
$$(x+1)(y-1)(y+1) + (x-1)(y-1)(y+1) = (x-1)(x+1)(y+1) + (x-1)(x+1)(y-1).$$ Now, we can simplify that down to $$x(y - 1)(y + 1) = y(x - 1)(x + 1).$$
At this point, we look to be stuck with expanding the brackets, so we'll do that, obtaining
$$xy^2 - x = yx^2 - y,$$
or
$$xy(y - x) = x - y.$$
If $x = y$, we're done, so suppose not (so that we can divide by $x - y$). Then we have $xy = -1,$ so $|xy| = 1$. But we know that $|x| < 1$ and $|y| < 1$, so $|xy| = |x||y| < 1$, a contradiction. Thus, $f$ is injective.

It is surjective: for this, we shall simply show that it is unbounded above and below, and appeal to a version of the intermediate value theorem. First, note that, for $x = 1 - \varepsilon$ and $\varepsilon < 1$, we have $$f(x) = \frac{1}{-\varepsilon} + \frac{1}{2 + \varepsilon} = \frac{2}{-\varepsilon(2 + \varepsilon)}\leq \frac{-2}{3\varepsilon} \to -\infty$$ as $\varepsilon \to 0$ from above. Similarly, for $x = -1 + \varepsilon$, we have $f(x) \geq \frac{2}{3\varepsilon} \to +\infty$ as $\varepsilon \to 0$ from above.

Thus, for any $y \in \mathbb{R}$, there are some $x_0, x_1$ such that $f(x_0) < y < f(x_1)$ (and, indeed, we see that $x_0 > x_1$). Applying the intermediate value theorem (after noting that $f$ is continuous on $(-1, 1)$ as a sum of rational functions whose denominators have no zeroes in that interval) to $[x_1, x_0]$ then gives us some $x_2 \in [x_1, x_0] \subset (-1, 1)$ such that $f(x_2) = y$, and so $f$ is surjective.


问:
Finite Quotient Groups $\mathbb{Z}/\!\!\sim$
Given an equivalence relation $\sim$ on $\mathbb{Z}$, we can talk about the set $\mathbb{Z}/\!\!\sim\;= \{[x]\ \vert\ x\in\mathbb{Z}\}$ where $[\cdot]$ are the equivalency classes of $\sim$, that is, $[x] = \{y\in\mathbb{Z}\ \vert\ x\sim y\}$. Letting the binary operation $+\colon\ \mathbb{Z}/\!\!\sim\times\  \mathbb{Z}/\!\!\sim\ \to\ \mathbb{Z}/\!\!\sim$ be defined by $[x] + [y] = [x+y]$, we can treat $\mathbb{Z}/\!\!\sim$ as a group. My question is whether or not there exists some $\sim$ such that $\mathbb{Z}/\!\!\sim\;\not\cong \mathbb{Z}/n\mathbb{Z}$ but that $\mathbb{Z}/\!\!\sim$ is finite, cyclic and abelian. Also, can anyone provide examples of $\sim$ such that $\mathbb{Z}/\!\!\sim\;\neq \mathbb{Z}/n\mathbb{Z}$ but that $\mathbb{Z}/\!\!\sim$ is finite, cyclic and abelian if we allow $\mathbb{Z}/\!\!\sim\;\cong \mathbb{Z}/n\mathbb{Z}$?

答: Any cyclic group is abelian: $g^ng^m = g^{n+m} = g^{m+n} = g^mg^n$. Also, any finite cyclic group $G$ is isomorphic to $\mathbb Z/n\mathbb Z$, where $n=|G|$. To see this, note that $G$ can be written as $$G = \{1,g,\dots,g^{n-1}\}$$ and then the map $G \to \mathbb Z/n\mathbb Z$ given by $g^k \mapsto k+n\mathbb Z$ is a group isomorphism (check this!).


问:
Prove that iterated $x_{i + 1} = x_i - \log x_i$ is sublinear.
Define a recurrence as: $$x_0 = n, x_{i + 1} = x_i - \log x_i, \forall i \geq 0.$$ Let $N_n$ be such that $x_{N_n} < 1$. Is it possible to show that $N_n \in o(n)$?

Some thoughts: if $x_{i + 1} = x_i - f(x_i)$ and $f$ is constant then $N_n = \frac{n}{f(1)}$ and is thus not sublinear. If $f$ is $\Omega(n)$ then $N_n$ is constant and thus trivially sublinear.  Is $f(x) = \log x$ sufficient to make $N_n$ sublinear?

Edit: As pointed out in the comments, the criteria $x_{N_n} < 1$ can never be met for this question. The correct question should be

Let $N_n$ be the smallest integer such that $x_{N_n} < 10$. Is it possible to show that $N_n \in o(n)$?

答:
This is one approach using power series. Define sequence
$$ x_i := 1+y_i, \tag{1}$$ and function
$$ g(x) := x - \log(1+x). \tag{2}$$ Using the recursion we have
$$ y_{i+1} = g(y_i). \tag{3}$$
From previous experience with these iterations, such as
Newton's method, we expect that
$$ f(x^2) = g(f(x)) \tag{4}$$ for some function expressed as
a power series
$$ f(x) := a_1\,x + a_2\,x^2 + a_3\,x^3 + \dots \tag{5}$$
where we solve for the "undetermined coefficients" $\,a_i\,$
one at a time using equation $(4).$
The result of the computations is that we must have
$$ f(x) := 2x + 4/3x^2 + 8/9x^3 + 112/135x^4 + 76/135x^5 + \dots. \tag{6}$$
Thus,
$$ y_0 = n-1 = f(t_0), \tag{7}$$ for some $\,t_0\,$ and
$$ y_n = f(t_0^{\,2^n}) \approx 2t_0^{\,2^n} \tag{8}$$ which implies
$$ x_n \approx 1+2t_0^{\,2^n}. \tag{9}$$
The convergence of $\,x_n\,$ to $1$ is called "quadratic"
or of "second order".


问:
Denote a ``limiting path'' by $ p := \bigcap_{n \in \mathbb{N}} \overline{\bigcup_{\ell \geq n} p_\ell}.$ Show that $p \in P$.
Fix $x \neq y \in \Bbb R^n$. Denote by the class of "paths" connecting $x$ and $y$ the collection of sets

$$ P\equiv P(x,y) := \{ p \subset \Bbb R^n: \text{$p$ connected and compact},\ x,y \in p\}.$$
Assume that $(p_n)_{n \in \mathbb{N}} \subset P(x,y)$ is "uniformly bounded" in the sense that there exist $R > 0$ such that $p_n \subset B(0,R)$ for all $n \in \mathbb{N}$.

Denote a "limiting path" by

$$ p := \bigcap_{n \in \mathbb{N}} \overline{\bigcup_{\ell \geq n} p_\ell}.$$
Show that $p \in P$.

My try:

Each $\overline{\bigcup_{\ell \geq n} p_\ell}$ is closed and bounded and connected. [As they are subset of $\overline{B(0,R)}$ and $p_l$ has two points $x,y$ in common.] Then
$$ p := \bigcap_{n \in \mathbb{N}} \overline{\bigcup_{\ell \geq n} p_\ell}.$$ is compact.

I am having trouble to understand why $p$ is connected. In generally for nested connected set, the intersection is not connected. So I am guessing that I have to use compact somehow. Please help.

答: HINT: If $p$ is not connected, we can write it as $H\cup K$, where $H$ and $K$ are disjoint, non-empty closed sets. There are disjoint open sets $U$ and $V$ such that $H\subseteq U$ and $K\subseteq V$. Let $n\in\Bbb N$; the connectedness of $p_n$ implies that $p_n\nsubseteq U\cup V$ (why?), so there is an $x_n\in p_n\setminus(U\cup V)$. The sequence $\langle x_n:n\in\Bbb N\rangle$ is bounded, so it has a convergent subsequence. Consider the limit of that subsequence.


问:
Is there any official, specific convention that defines whether an expression is considered "Simplified"?
I see all the time in high school math textbooks problems saying to "simplify" an expression. Their explanations of what it means for an expression to be "simplified" is a bit vague and does not reflect the rigor mathematics usually provides:

"To simplify an expression means to do all the math possible." (OpenStax Intermediate Algebra)

I have encountered numerous situations in which there is ambiguity as to what looks "simpler" to people. For example:
$$
\text{Simplify: }\frac{a^2b^{-3}c}{c^2b}
$$

Some people would say the most "simple" way to write this is $a^2b^{-4}c^{-1}$. However, most textbooks would consider $\frac{a^2}{cb^4}$ to be the "correct" answer.

From my perspective, problems beginning with "Simplify" don't really make sense (and are frankly unfair to the students) unless there is a clear definition of what it means for the expression to be "simplified" (especially with teachers who are picky about that kind of thing).

Is there some sort of convention for simplification that I don't know about that makes this work?

答:
This is right:

From my perspective, problems beginning with "Simplify" don't really
make sense (and are frankly unfair to the students) unless there is a
clear definition of what it means for the expression to be
"simplified" (especially with teachers who are picky about that kind
of thing).

In K-12 "simplify" usually means "manipulate so that you end up with an expression of the form we've seen as the answer to similar problems".

That is indeed ambiguous. It does however give students practice with the kinds of manipulations they are supposed to learn how to do. It makes exams easier to grade. How much it helps students learn interesting and important mathematics is doubtful. Good students understand the ambiguity, but still do what the teacher expects. If they have good teachers they can discuss the ambiguity in a way that does not disturb their classmates who just want to get the right answer.


问:
Let $p$ be any prime greater than or equal to $5$. Prove that $a^p ≡ a \pmod{3p}$
I am unsure how to prove the following statement, although I am fairly sure it has something to do with Fermat's Little Theorem, which states that $a^p \equiv a \pmod{p}$ for a prime, $p$. Any help is greatly appreciated.

答:
The second part can also be done like this:

Fermat's little theorem gives $a^3\equiv a \pmod{3}$, so
$$a^p = a^{p-3} a^3 \equiv a^{p-3} a \equiv a^{p-2} \pmod{3} \\
\Rightarrow a^p \equiv a^{p-2} \equiv a^{p-4} \equiv \cdots \equiv a \pmod{3}$$.


问:
Recurring Sequence with Exponent
Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have  no clue on how to approach this problem.

I've solved the Fibonacci sequence by subtracting the two pervious terms and shifting the sequence, but it does not seem to work here.

I'm particularly working with $a_n = 2a_{n-1} + 5^{n-1}$, but the sequence expands extremely fast. The base case, $a_{0} = 1$.

Any help would be appreciated!

Title: Recurring Sequence with Exponent

Body:
Because the one with which you’re now working is first-order, you can simply ‘unwind’ it:

$$\begin{align*}
a_n&=2a_{n-1}+5^{n-1}\\
&=2\left(2a_{n-2}+5^{n-2}\right)+5^{n-1}\\
&=2^2a_{n-2}+2\cdot5^{n-2}+5^{n-1}\\
&=2^2\left(2a_{n-3}+5^{n-3}\right)+2\cdot5^{n-2}+5^{n-1}\\
&=2^3a_{n-3}+2^2\cdot5^{n-3}+2\cdot5^{n-2}+5^{n-1}\\
&\;\;\vdots\\
&=2^ka_{n-k}+\sum_{i=0}^{k-1}2^i5^{n-1-i}\\
&\;\;\vdots\\
&=2^na_0+\sum_{i=0}^{n-1}2^i5^{n-1-i}\\
&=2^na_0+5^{n-1}\sum_{i=0}^{n-1}\left(\frac25\right)^i\\
&=2^na_0+5^{n-1}\cdot\frac{1-\left(\frac25\right)^n}{1-\frac25}\\
&=2^na_0+\frac{5^n-2^n}3
\end{align*}$$


Title: Recurring Sequence with Exponent

Body:
Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have  no clue on how to approach this problem.

I've solved the Fibonacci sequence by subtracting the two pervious terms and shifting the sequence, but it does not seem to work here.

I'm particularly working with $a_n = 2a_{n-1} + 5^{n-1}$, but the sequence expands extremely fast. The base case, $a_{0} = 1$.

Any help would be appreciated!

Body:
We use ordinary generating functions. Let $A(x) = \sum_{i=0}^n a_n x^n$. Then we have (summing from $n=1$)

\begin{align}
a_n &= 2a_{n-1} + 5^{n-1},\\
\sum_{n=1}^\infty a_nx^n &= \sum_{n=1}^\infty 2a_{n-1} x^n + \sum_{n=1}^\infty 5^{n-1} x^n,\\
A(x) - a_0 &= 2x\sum_{n=1}^\infty a_{n-1} x^{n-1} + x\sum_{n=1}^\infty 5^{n-1} x^{n-1},\\
A(x) - 1 &= 2x\sum_{n=0}^\infty a_{n} x^{n} + x\sum_{n=0}^\infty 5^{n} x^{n},\\
A(x) - 1 &= 2xA(x) + \frac{x}{1-5x},\\
A(x) - 2xA(x) &= \frac{x}{1-5x} + 1,\\
A(x) &= \frac{x}{(1-2x)(1-5x)} + \frac{1}{1-2x}.\\
\end{align}
Now we use partial fraction decomposition and a bit of algebra to obtain

\begin{align}
A(x) &= \frac{1}{3(1-5x)} - \frac{1}{3(1-2x)} + \frac{1}{1-2x}\\
&= \frac{1}{3} \left(\frac{1}{(1-5x)} + \frac{2}{(1-2x)}\right)\\
&= \frac{1}{3} \left(\sum_{n=0}^{\infty} 5^nx^n + 2\sum_{n=0}^{\infty}2^nx^n \right).
\end{align}

From here we see $$a_n = \frac{5^n + 2^{n+1}}{3}.$$


问:
Matrix representation of multiplication of elements in Galois Field
I would appreciate it if you could please explain how in the following example the mod by f(x) results in the suggested answer (orange frame). Example at the top of page 3, https://eprint.iacr.org/2019/1090.pdf.

答:
how the mod by f(x) results in the suggested answer (orange frame)

If $x^3+x+1=0$ in characteristic $2$, then $x^3=x+1$ and $x^4=x^2+x$.

Can you take it from there?


问:
If $\mathbb{E}(X_n)$ is finite for $n=1,2,...$, are $\mathbb{E}|X_n|$ and $\mathbb{E}\left(\sum_{i=1}^n|X_i|\right)$ also finite?
I'm not sure that I can think of a counter example for the first one. It seems intuitive to me that this is the case, but perhaps there could exist a sequence of random variables whereby mean of the sum of them is 0, but the mean of the sum of their absolute values is infinite. Or am I totally off the ball here?</p>

答:
$\mathbb{E}\left[\sum\limits_{i=1}^n|X_i|\right] =\sum\limits_{i=1}^n\mathbb{E}\left[|X_i|\right]$ so is finite iff each of the $\mathbb{E}\left[|X_i|\right]$ is finite

If $\mathbb{E}\left[|X_i|\right]$ is infinite then $\mathbb{E}\left[X_i\right]$ is either infinite or undefined, while if $\mathbb{E}\left[|X_i|\right]$ is finite then $\mathbb{E}\left[X_i\right]$ is finite

So each of the $\mathbb{E}\left[X_i\right]$ being finite implies that each of the $\mathbb{E}\left[|X_i|\right]$ is finite and that $\mathbb{E}\left[\sum\limits_{i=1}^n|X_i|\right]$ is finite


答:
If we require that $+ \colon \mathbb{Z}/{\sim} \times \mathbb{Z}/{\sim} \rightarrow \mathbb{Z}/{\sim} $ is well-defined (that is, it takes the same values no matter which class representatives you pick) you can show that $[0]$ has to be a (normal) subgroup of $\mathbb{Z}$:

$0 \in [0]$ by definition.

If $x, y \in [0]$ then $[x+y] = [x] + [y] = [0] + [0] = [0]$ so $x + y \in [0]$.

If $x \in [0]$ then $[-x] = [0] + [-x] = [x] + [-x] = [0]$ so $-x \in [0]$.

Similarly, you can show that the equivalence classes $[n]$ must be cosets of $[0]$. (Give it a try!)

Therefore, for $\mathbb{Z}/{\sim} $ to be a group with that operation it must be the quotient of $\mathbb{Z}$ by a subgroup.

Title: Finite Quotient Groups $\mathbb{Z}/\!\!\sim$

Body: Given an equivalence relation $\sim$ on $\mathbb{Z}$, we can talk about the set $\mathbb{Z}/\!\!\sim\;= \{[x]\ \vert\ x\in\mathbb{Z}\}$ where $[\cdot]$ are the equivalency classes of $\sim$, that is, $[x] = \{y\in\mathbb{Z}\ \vert\ x\sim y\}$. Letting the binary operation $+\colon\ \mathbb{Z}/\!\!\sim\times\  \mathbb{Z}/\!\!\sim\ \to\ \mathbb{Z}/\!\!\sim$ be defined by $[x] + [y] = [x+y]$, we can treat $\mathbb{Z}/\!\!\sim$ as a group. My question is whether or not there exists some $\sim$ such that $\mathbb{Z}/\!\!\sim\;\not\cong \mathbb{Z}/n\mathbb{Z}$ but that $\mathbb{Z}/\!\!\sim$ is finite, cyclic and abelian. Also, can anyone provide examples of $\sim$ such that $\mathbb{Z}/\!\!\sim\;\neq \mathbb{Z}/n\mathbb{Z}$ but that $\mathbb{Z}/\!\!\sim$ is finite, cyclic and abelian if we allow $\mathbb{Z}/\!\!\sim\;\cong \mathbb{Z}/n\mathbb{Z}$?


问:
Number Theory: $x^2+y^2=a^2$
Is there a coprime triple $(x,y,z)$ such that $x^2+y^2=a^2, x^2+z^2=b^2, y^2+z^2=c^2$, where $a,b,c$ are integers

P.S. such solution doesn't exist for $a,b,c<1000$, as the computer says

P.P.S. the statement above is WRONG

答:
It's called Euler brick,the smallest one is $x=44,y=117,z=240$

http://en.wikipedia.org/wiki/Euler_brick

Title: Number Theory: $x^2+y^2=a^2$


答:
You're missing a condition on $\sim$ for $\mathbb{Z}/{\sim}$ to support the group structure you'd like: $\sim$ needs to be compatible with $+$ and $-$, in the sense that for all $n,m,p,q$, if $n\sim m$ and $p\sim q$, then $n+p \sim m+q$ and $-n \sim -m$. In the language of universal algebra, we require that $\sim$ be a congruence.

Let $H$ be the $\sim$-equivalence class containing $0$. We make two claims: $n \sim m$ if and only if $n-m \in H$, and $H$ is a subgroup of $\mathbb{Z}$. For the first claim, if $n\sim m$, then $-m \sim -m$ and compatibility of $\sim$ with $+$ implies $n-m \sim m-m = 0$. Conversely, if $n-m \sim 0$, then $m \sim m$ and compatibility of $\sim$ with $+$ implies $n = (n-m)+m \sim 0+m = m$. To prove the second claim, given $n, m \in H$, we have $n \sim 0$ and $m \sim 0$. Compatibility with $+$ implies $n+m \sim 0+0 = 0$, i.e., $n+m \in H$. Compatibility with $-$ implies $-n \sim -0 = 0$, i.e., $-n \in H$.

What this shows is that $\mathbb{Z}/{\sim}$ as a group is the same as the quotient group $\mathbb{Z}/H$. The only nontrivial subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ for $n>0$, so it follows that $\mathbb{Z}/{\sim}$ is always the group of integers mod $n$ for some $n$.


问:
Munkres section 41, expansion lemma and paracompact spaces

[Ex5] Let $X$ be paracompact. We proved a "shrinking lemma" for arbitrary indexed open coverings of $X$. Here is
an "expansion lemma" for arbitrary locally finite indexed families in
$X$.

Lemma. Let $\{ B_ \alpha \} _ { \alpha \in J}$ be a locally finite
indexed family of subsets of the paracompact Hausdorff space $X$. Then
there is a locally finite indexed family $\{ U_ \alpha \} _ { \alpha
 \in J}$ of open sets in $X$ such that $ B_ \alpha \subset U_ \alpha$
for each $\alpha$.

I read Brian M. Scott's answer in Expansion lemma for paracompact hausdorff space. and Showing a uniformity is complete. But I couldn't understand it because I haven't studied about barycentric refinement, star refinement, uniform spaces, completeness, and filters(also they haven't appeared in Munkres text yet). Are there any other ways to prove this?

答:
Here is an answer which uses only the basic definition of paracompactness and does not require a Hausdorff assumption. There are some gymnastic involved but it's not too messy.

Let $\mathcal{A}$ be a locally-finite family of closed subsets of a paracompact space $X$. Since the closures of the members of a locally-finite family themselves form a locally-finite family, there is no loss of generality in posing the additional assumption.

Notation: Write $Fin(\mathcal{A})$ for the set of finite nonempty subsets $\alpha=\{A_1,\dots,A_n\}\subset\mathcal{A}$. For a subset $U\subseteq X$ write $\mathcal{A}_U=\{A\in\mathcal{A}\mid A\cap U\neq\emptyset\}$. For $x\in X$ we write $\mathcal{A}_x=\mathcal{A}_{\{x\}}$. Since a locally-finite family is point finite, we have $\mathcal{A}_x\in Fin(\mathcal{A})$ for each $x\in X$.

Claim: There is a locally-finite family $\mathcal{U}=\{U_A\}_{A\in\mathcal{A}}$ of open subsets of $X$ such that $A\subset U_A$ for each $A\in\mathcal{A}$.

Proof: For $\alpha\in Fin(\mathcal{A})$ write $V(\alpha)=X\setminus\bigcup(\mathcal{A}\setminus\alpha)$. By the local-finiteness of $\mathcal{A}$, each $V(\alpha)$ is open. Moreover, if $x\in X$, then $x\in V(\mathcal{A}_x)$, and thus the family $\{V(\alpha)\}_{\alpha\in Fin(\mathcal{A})}$ is an open covering of $X$. Let $\mathcal{V}$ be a locally-finite open refinement of it.

Now for $A\in\mathcal{A}$ put $U_A=St(A;\mathcal{V})=\bigcup\{V\in\mathcal{V}\mid V\cap A\neq\emptyset\}$. Then $A\subseteq U_A$ and $U_A$ is open since the sets in $\mathcal{V}$ are. We claim that $\mathcal{U}=\{U_A\}_{A\in\mathcal{A}}$ is locally-finite. In showing this we will complete the proof.

So fix a point $x\in X$. Let $W\subseteq X$ be an open neighbouhood of $x$ which meets only finitely many sets in $\mathcal{V}$, say $V_1,\dots,V_n$. For each $i=1,\dots,n$ choose $\alpha_i\in Fin(\mathcal{A})$ such that $V_i\subseteq V(\alpha_i)$ and write $\alpha=\{A\in\mathcal{A}\mid W\cap U_A\neq\emptyset\}$. Then $\alpha\subset\bigcup_{i\leq n}\alpha_i$. In particular $\alpha\in Fin(\mathcal{A})$, which shows that $W$ meets only finitely many sets in $\mathcal{U}$. $\square$


问:
Connection between irreducibly diagonally dominant matrices and positive definiteness?
I am attempting to prove a proposition that I found in Cottle's "The Linear Complementarity Problem" book in which the proof has been omitted.

I will start by introducing some definitions.

${\bf Def.}$ A matrix $A\in\mathbb{R}^{n\times n}$ is said to be irreducible if and only if for any two distinct indices $1\le i,j\le n$, there is a sequence of nonzero elements of $A$ of the form
$$
\{a_{ii_1},a_{i_1i_2},\ldots,a_{i_mj}\}.
$$

${\bf Def.}$ A matrix $A\in\mathbb{R}^{n\times n}$ is irreducibly diagonally dominant if it is irreducible, weakly diagonally dominant, $|A_{ii}| \ge \sum_{j\neq i}|A_{ij}|$ for all $i$, and there is at least one row or column where strict diagonal dominance holds, that is $\exists$ $i$ such that $|A_{ii}| > \sum_{j\neq i}|A_{ij}|$.

The statement is as follows.

${\bf Prop.}$ Let $A\in\mathbb{R}^{n\times n}$ be symmetric, strictly or irreducibly diagonally dominant, and $A_{ii}>0$ for all $i$, then $A$ is positive definite.

${\bf Pf.}$ (attempt, sketch) My attempt used the Gershgorin Circle Theorem which allowed me to show that the eigenvalues of the matrix were all nonnegative, with one strictly positive. However, I do not know how to use irreducibility or the fact that $A$ has strictly positive diagonal entries to conclude that it is positive definite.

答:
In your mentioned proposition, since $A$ is real symmetric and diagonally dominant and it has a positive diagonal, it is positive semidefinite (Gershgorin discs in play here). The role of irreducibility (coupled with diagonal dominance) is to ensure that $A$ is nonsingular. Every nonsingular positive semidefinite matrix is, of course, positive definite.

To prove that $A$ is nonsingular, suppose $Ax=0$. Since $A$ is irreducible, it has no zero rows. Therefore the diagonal entries of $A$ are nonzero, because $A$ is diagonally dominant. (This fact, of course, also follows from the given assumption that $A$ has a positive diagonal, but here we wish to infer the invertibility of $A$ using only irreducibility and diagonal dominance. So, we deduce the nonzeroness of the diagonal of $A$ from these two assumptions too.)

Now, suppose $x_i$ has the largest modulus among all entries of $x$. By assumption, strict diagonal dominance of $A$ occurs on some row $j$ and there is a path $i_0=i\to i_1\to \ldots\to i_k=j$ such that $a_{i_mi_{m+1}}$ is nonzero for each $m$. Since $|x_{i_0}|$ has a maximum modulus, by (weak) diagonal dominance of row $i_0=i$, for every $\ell$ such that $a_{i_0\ell}\ne0$, we must have $|x_\ell|=|x_{i0}|$. In particular, $|x_{i1}|=|x_{i0}|$. Apply the same argument recursively, we get $|x_j|=|x_{i_k}|=|x_{i_{k-1}}|=\cdots=|x_{i_0}|=|x_i|$. Therefore $x_j$ also has a maximum modulus. But then the strict diagonal dominance of $A$ on row $j$ implies that the $j$-th entry of $Ax$ is nonzero, which is a contradiction. Therefore $A$ must be nonsingular.


问:
Prove that if $\gcd(a,480)=1$, then $a^{16} ≡ 1 \pmod{480}$.
I would like to prove it using Euler's Theorem, which states that if $\gcd(a,n)=1$,then $a^{\phi(n)}=1 \pmod{n}$. Then, $480$ should be factored, but I am unsure how to proceed with the proof. Thanks in advance.

答:
Hint: $480 = 2^5 \cdot 3 \cdot 5$. By the Chinese remainder theorem, we see that $a^k \equiv 1 \pmod {480}$ if and only if we have
$$
\begin{cases}
a^k \equiv 1 \pmod{2^5},\\
a^k \equiv 1 \pmod 3,\\
a^k \equiv 1 \pmod 5.
\end{cases}
$$


问:
Why does $f^{(n)}(x)=\sin(x+\frac{n\pi}{2})$ for $f(x)=\sin(x)$?
I'm not quite sure how does $f^{(n)}(x)=\sin(x+\frac{n\pi}{2})$ for $f(x)=\sin(x)$. Taking the initial derivatives I get,
\begin{align}
f'(x)&=\cos(x)\\
f''(x)&=-\sin(x)\\
f'''(x)&=-\cos(x)\\
f^{4}(x)&=\sin(x)
\end{align}
I'm not quite understanding this relation.

答:
Just see how sin line change its value when rotating 90degrees counter clock wise and compare with the fact that every 4 derivatives you back to the same place (ups!, you got the same function).

Using maths:

The usage of $\sin(x+n\frac{\pi}{2})=\sin(x)\cos(n\frac{\pi}{2})+\cos(x)\sin(\frac{\pi}{2})$ is a must, because the value of functions with argument $\frac{\pi}{2}$ takes $1,0$ or $-1$, so when one takes value $0$, the other can take values $1$ or $-1$ which are the neccesary for having the right derivative. Of course, this is not casuality, this is because the derivate for sin is a idempotent operator of order $4$ (i.e $T^4=T$).

Also, since $\sin$ is periodic, its derivative will be periodic too:

$$f'(x+c) = \displaystyle\lim_{h\to0}\frac{(f((x+h)+c)-f(x+c))}{h}$$
$$= \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h} = f'(x)$$


答:
Easy alternative (I hate math).

Known that $(3k+r)^n \equiv r^n\pmod{3}.$

$1^p \equiv 1\pmod{3} \implies (1^p - 1) \equiv 0\pmod{p}.$

Since $p$ is odd, $(p-1)$ is even.

$2^{(\text{any even #)}} \equiv 1 \pmod{3} \implies 2^{(p-1)} \equiv 1\pmod{3}.$

Therefore, $2^{p} \equiv 2^1\pmod{3}\implies (2^p - 2) \equiv 0\pmod{p}$

Therefore, since $a$ must be congruent (mod 3) to either $0,1,$ or $2$, $(a^p - a) \equiv 0\pmod{3}.$

Title: Let $p$ be any prime greater than or equal to $5$. Prove that $a^p ≡ a \pmod{3p}$

Body: I am unsure how to prove the following statement, although I am fairly sure it has something to do with Fermat's Little Theorem, which states that $a^p \equiv a \pmod{p}$ for a prime, $p$. Any help is greatly appreciated.


问:
Connection between left null space and matrix
This question comes from Strang's "Introduction to Linear Algebra, 5th edition" (3.5, exercise 20)

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 4 & 2 & 0 & 1\\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} = ER$

How many independent solutions to $A^T \mathbf{y} = \mathbf{0}$? Why does $\mathbf{y}^T = \text{row } 3 \text{ of } E^{-1}$?

Based on the two matrices $E$ and $R$ I can see $\text{rank}(A) = 2$ so the left null-space of $A$ (i.e. the nullspace of $A^T$) has dimension $3 - 2 = 1$, so it has one "independent solution" to $A^T \mathbf{y} = \mathbf{0}$.

However, I am stuck in showing how $\mathbf{y}^T$ is equal to row 3 of $E^{-1}$. The instrutor manual solution for the exercise states: "$E^{-1}A = R$ has a zero row, which is just the transpose of $A^{T}\mathbf{y} = \mathbf{0}$". I am unsure what he means here.

Thus, I wanted to ask if anyone could clarify what is stated in the manual or any tips on approaching the problem.

Thank you in advance.

答:
Write $\mathbf{y}^TER=\mathbf{0}$. We can define $\mathbf{u}^T=\mathbf{y}^TE$ so that $\mathbf{u}^TR=\mathbf{0}$, and it is clear that the solution set is generated by $\mathbf{u}^T=\begin{bmatrix}0 & 0 & 1\end{bmatrix}$. Because $E$ is invertible, we then have $\mathbf{y}^T=\mathbf{u}^TE^{-1}$, from which the result follows.

Added: I've been trying to make sense of the solution manual's statement. It is indeed the case that $E^{-1}A=R$ has a zero row, namely its third row, a row of four zeroes. It is also true that $\mathbf{y}^{T}A$, which is the transpose of $A^T\mathbf{y}$, is a row of four zeroes. So the third row of $E^{-1}A$ is the same as $\mathbf{y}^{T}A$. Now the third row of $R$ is the result of multiplying the third row of $E^{-1}$ by $A$. So we can take $\mathbf{y}^T$ to be the third row of $E^{-1}$ (or any scalar multiple thereof).


Title: Recurring Sequence with Exponent

Body:
I love to telescope.

If
$a_n = ua_{n-1} + vc^{n}
$,
then
$\dfrac{a_n}{u^n}
= \dfrac{a_{n-1}}{u^{n-1}} + v(c/u)^{n}
$.

Let
$b_n = \dfrac{a_n}{u^n}$.
Then
$b_n
=b_{n-1}+vd^n
$
where
$d = c/u$.

Then
$b_n-b_{n-1}
=vd^n
$.

Summing,

$\begin{array}\\
b_m-b_0
&=\sum_{n=1}^m (b_n-b_{n-1})\\
&=\sum_{n=1}^m vd^n\\
&=v\dfrac{d-d^{m+1}}{1-d}\\
&=vd\dfrac{1-d^{m}}{1-d}\\
\end{array}
$

so

$\begin{array}\\
\dfrac{a_m}{u^m}
&=a_0+vd\dfrac{1-d^m}{1-d}\\
\text{or}\\
a_m
&=a_0u^m+\dfrac{vc}{u}u^m\dfrac{1-(c/u)^m}{1-c/u}\\
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
\end{array}
$

In  this case,
$u=2, c=5, v = \frac15, a_0 = 1
$
so
$a_m
= 2^m + \dfrac{2^m-5^m}{2-5}
= 2^m + \dfrac{5^m-2^m}{3}
$.

This can be rewritten as

$\begin{array}\\
a_m
&=a_0u^m+vc\dfrac{u^m-c^m}{u-c}\\
&=\dfrac{(u-c)a_0u^m+vc(u^m-c^m)}{u-c}\\
&=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}\\
\end{array}
$

Again, we get
$=\dfrac{(a_0(u-c)+vc)u^m-vc^{m+1}}{u-c}
=\dfrac{(-3+1)2^m-5^{m}}{-3}
=\dfrac{2\cdot 2^m+5^{m}}{3}
$.


答:
The question is incorrect as written: there is a line of solutions to $A^T \mathbf y = 0$, one of these is such that $\mathbf y^T$ is row $3$ of $E^{-1}$. Also, the solution from the manual is incomprehensible (at least to me).

Here is what I think Strang is getting at. Note that $A^T\mathbf y = 0 \iff \mathbf y^TA = 0$, which we can write as
$$
(\mathbf y^T E)R = \mathbf 0.
$$
Now, a convenient way to select $\mathbf y$ is to select it so that the above product becomes
$$
[0 \ \ 0 \ \ 1] \ R,
$$
since for any matrix $M$ of $3$ rows, $[0 \ \ 0 \ \ 1] M$ is simply the third row of $M$. In other words, an easy solution is to choose $\mathbf y$ such that $\mathbf y^T E = [0 \ \ 0 \ \ 1]$.  Now, note that if we let $\mathbf r_1, \mathbf r_2, \mathbf r_3$ denote the rows of $E^{-1}$, then we have
$$
E^{-1}E = I \implies \pmatrix{\mathbf r_1\\ \mathbf r_2\\ \mathbf r_3} E = \pmatrix{1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1} \implies
\pmatrix{\mathbf r_1 E\\ \mathbf r_2 E\\ \mathbf r_3 E} = \pmatrix{1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1}.
$$
Now, if we consider the third row in this last equation, we see that $\mathbf r_3 E = [0 \ \ 0\ \ 1]$. That is, taking $\mathbf y^T$ to be row $3$ of $E^{-1}$ gives us exactly what we're looking for.

Title: Connection between left null space and matrix

Body:
This question comes from Strang's "Introduction to Linear Algebra, 5th edition" (3.5, exercise 20)

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix} \begin{bmatrix} 4 & 2 & 0 & 1\\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} = ER$

How many independent solutions to $A^T \mathbf{y} = \mathbf{0}$? Why does $\mathbf{y}^T = \text{row } 3 \text{ of } E^{-1}$?

Based on the two matrices $E$ and $R$ I can see $\text{rank}(A) = 2$ so the left null-space of $A$ (i.e. the nullspace of $A^T$) has dimension $3 - 2 = 1$, so it has one "independent solution" to $A^T \mathbf{y} = \mathbf{0}$.

However, I am stuck in showing how $\mathbf{y}^T$ is equal to row 3 of $E^{-1}$. The instrutor manual solution for the exercise states: "$E^{-1}A = R$ has a zero row, which is just the transpose of $A^{T}\mathbf{y} = \mathbf{0}$". I am unsure what he means here.

Thus, I wanted to ask if anyone could clarify what is stated in the manual or any tips on approaching the problem.

Thank you in advance.


问:
Finding the x-intercept of a cubic function
How would one go about solving for x in this cubic equation?

$f(x) = x^3 + x - 4$

I don't think factoring works in this situation:

$f(x) = x^2(x + 1) - 4$

and I don't know how to find the value of $x$ where f(x) = 0.

答:
Alright, define real numbers $u,v$  with, say, $u \geq v,$ then
$$ u^3 + v^3  = 4, $$
$$  3uv = -1$$
so that $$  u^3 v^3 = \frac{-1}{27}$$
This is enough information to solve for $u^3, v^3$ separately. Then $u,v$ separate. Meanwhile, the quantity $u+v$ is the real root of your original cubic.


问:
Suppose $A = U\Sigma U^T$. If $A$ is not invertible, is it correct to write $A^{-1} = (U\Sigma U^T)^{-1}$?
Suppose $A$ is symmetric and thus, orthogonally diagonalizable: $A = U\Sigma U^T$, where $UU^T = I$ and $U^TU = I$. $D$ is a diagonal matrix.

Suppose $A$ is not invertible, then would it be correct to write the following?
$$A^{-1} = (U\Sigma U^T)^{-1} = U\Sigma^{-1}U^T$$

A follow-up question is about how one might express the pseudoinverse of $A$ in terms of $U$ and $\Sigma$. Please see the edit below.

Edit: Suppose $A = U\Sigma V^T$ by singular value decomposition. Now $A$ may or may not be symmetric. How can I express the Moore-Penrose pseudoinverse $A^{+}$ in terms of $U, \Sigma, V$?

答:
$\newcommand{\diag}{\mathrm{diag}}$

To its most generality, assume $A \in \mathbb{C}^{m \times n}$, and let $A = U\Sigma V$ be the SVD of $A$, where $U$ and $V$ are order $m$ and $n$ unitary matrices respectively, and
$$\Sigma = \diag(\diag(\mu_1, \mu_2, \ldots, \mu_r), 0, \ldots, 0), $$
and $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_r > 0$ are all (positive) singular values of $A$.

It is then straightforward to verify that
$$A^+ = V^*\diag(\diag(\mu_1^{-1}, \mu_2^{-1}, \ldots, \mu_r^{-1}), 0, \ldots, 0)U^*$$
is the Moore-Penrose inverse of $A$. Here $V^*$ means the conjugate transpose of $V$.


问:
Trees with vertex set
I am having hard time understanding and solving the following question:

There are exactly three trees with vertex set {1,2,3}. Note that all these trees are paths; the only difference is which vertex has degree 2. How many trees have a vertex set {1,2,3,4}?

So how I understand it is we have $K_3$ trees for the vertex set {1,2,3}. So since a tree is a path, we have a walk that repeats no vertex twice, or a path. When I drew it out, every time, only one vertex at a time has 2 degrees. (So far correct?) So what I don't understand is how we can determine the number of trees we can find for a vertex set $K_4$.

答:
There is a general statement for the number of trees on $n$ vertices given by Cayley's formula: $n^{n-2}$.  But if we didn't know about this, we could go about it in the following way:

Remember that on $4$ vertices, every tree must have $3$ edges.  There are two options: if the tree is not a path, then there's a vertex of degree $3$ (can't be more) that accounts for all edges.  So the graph must be $P_4$ or $K_{1,3}$.

For $K_{1,3}$, we have $4$ choices for the central vertex: then all other choices are forced.  For $P_4$, we have $\left( \begin{array}{c} 4 \\ 2 \end{array} \right)$ choices for the central vertices (we simply choose any two out of $4$ labeled vertices), and then two ways to connect the two remaining leaves, so $12$ choices altogether.

This gives us $16$ distinct trees ($2$ up to isomorphism).

Title: Trees with vertex set


问:
Prove by induction that $D^{\alpha}\hat{f}(x)=(-2\pi i)^{|\alpha|}[\widehat{(\cdot)^{\alpha}f(\cdot)}](x)$
Let $S(\mathbb{R}^{n})$ be the Schwartz space on $\mathbb{R}^{n}$. For $f\in S(\mathbb{R}^{n})$ and $\alpha=(\alpha_1,\cdots,\alpha_n )\in\mathbb{N}$ and $|\alpha|=\alpha_1+\cdots+\alpha_n$.

Prove by induction that $$D^{\alpha}\hat{f}(x)=(-2\pi i)^{|\alpha|}[\widehat{(\cdot)^{\alpha}f(\cdot)}](x)$$ where $D$ is the partial differential operator given by $$D^{\alpha}f=\frac{\partial^{\alpha}}{\partial x^{\alpha}}f=\left(\frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}\cdots\frac{\partial^{\alpha_2}}{\partial x_2^{\alpha_2}} \frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\right)f$$

Now since $|\hat{f}(x)|=|\int_{\mathbb{R}^n}f(\xi)e^{-2\pi ix\xi}d\xi|\leq \int_{\mathbb{R}^n}|f(\xi)|d\xi$ which is the $L^1$ norm, all derivatives of $\hat{f}$ will be $L^{\infty}$, but I'm not too sure how I should proceed to prove

答:
Hints:  You want to induct on the order of the multi-index. For the base case $(|\alpha|=1)$, $D^\alpha=\partial_j$ for some $1\leq j\leq n.$ Check that you get the right result by differentiating under the integral sign.

Now, suppose that it holds for all $|\alpha|=m,$ and consider $D^\beta$, where $|\beta|=m+1.$ Observe that it must be the case that $D^\beta=D^\alpha D^\gamma$, where $|\alpha|=m$ and $|\gamma|=1.$ Apply the base case and the inductive step to conclude.

By the way, it's customary to use $x$ to denote the variable of the original function and $\xi$ to denote the variable of the Fourier-transformed function.


问:
$\mathcal{B}(\mathcal{X},\mathcal{Y})$ is Banach space iff $\mathcal{Y}$ is Banach space.
Taken from Conway's A course in Functional Analysis Chapter 3 Section 2 Problem 1

Problem Statement: Show that for $\mathcal{B}(\mathcal{X}, \mathbb{F})\neq (0)$, $\mathcal{B}(\mathcal{X},\mathcal{Y})$ is a Banach space if and only if $\mathcal{Y}$ is a Banach space.

In the forward direction, we suppose that $\mathcal{B}(\mathcal{X}, \mathbb{F})$ is a Banach space. Then $\{T_n\} \in \mathcal{B}(\mathcal{X}, \mathbb{F})$ is a Cauchy sequence and therefore it converges to $T\in \mathcal{B}(\mathcal{X}, \mathbb{F})$. This means that $\sup\{||(T_n - T)(x)||: ||x|| = 1\} \rightarrow 0$ as $n \rightarrow \infty$.

Now consider a sequence $y_n \in \mathcal{Y}$. We hope to show that this sequence converges to $y\in \mathcal{Y}$. If I can write $y_n = T_n(x)$ then since $T_n$ is Cauchy, it must follow that $T_n(x)$ converges to some $y \in \mathcal{Y}$. But I'm not sure how to make this clear.

Any tips is greatly appreciated!

答:
Sketch (fill in the details):

Suppose $(T_n)$ is Cauchy in $B(X,Y)$. Then $(T(x)_n)$ is Cauchy in $Y$ for every $x\in X\Rightarrow \lim_{n \to \infty} T_n(x) = y_x.$ This defines a map $x\mapsto y_x$, which is evidently linear. Furthermore, $\|T_m(x)-T_n(x)\|\le \|T_m-T_n\|\|x\|\le \epsilon \|x\|$ if $m,n$ are large enough. Let $m\to \infty$. Then, $\|T-T_n\|\le \epsilon $, which shows that $T_n\to T$ and that since $T-T_n\in B(X,Y)$ so $T=T_n+T-T_n$ is, too.

Now, suppose $(y_n)$ Cauchy. First choose a $\phi\in X^*$ and an $x_0\in X$  such that $\phi(x_0)=1.$ For each $y\in Y$, define a map $g_y:X\to Y:x\mapsto \phi(x)y$ and show that $(g_{y_n})\subseteq B(X,Y)$ is Cauchy and therefore $g_{y_n}\to g\in B(X,Y)$. Finally, show that $y_n\to g(x_0).$


问:
Find the residue of $1/(z^2+1)^{n+1}$ at $z=i$.
I'm trying to find the residue of $\frac{1}{(z^2+1)^{n+1}}$ at $z=i$.  I believe this should be approached combinatorially but I am getting stuck in the combinatorics.  Here's what I have so far.

$$ \frac{1}{(z^2+1)^{n+1}} = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(z-i+2i)^{n+1}} $$

$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}(1-[-z+i]/(2i))^{n+1}} $$

$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}}\cdot \left[\sum_{k=1}^\infty \left((-z+i)/(2i)\right)^k\right]^{n+1} $$

$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}}\cdot \left[\sum_{k=1}^\infty \left(\frac{-1}{2i}\right)^k(z-i)^k\right]^{n+1} $$

To get the residues we want the $(z-i)^{-1}$ term, which we can determine once we know the $(z-i)^n$ term of the factor which is expressed above as a series.  This will come from making $n+1$ choices of a term from each of the $n+1$ copies of the series, such that the exponents sum to $n$.  However, this now gets into a pretty famous combinatorial problem of finding the ways to partition an integer.  That sounds like a lot of tricky combinatorial stuff when you're learning about residues.  Am I doing this the hard way--is there an easier one?

Body:
If we write $$g(z) = g(z_0)+g^\prime(z_0)(z-z_0) + g^{\prime \prime}(z_0)(z-z_0)^2 + \cdots$$ then $$\frac{g(z)}{(z-z_0)^{n+1}}= \frac{g(z_0)}{(z-z_0)^{n+1}}+\cdots+\frac{g^{(n)}(z_0)}{n!\,(z-z_0)}+ \frac{g^{(n+1)}(z_0)}{(n+1)!}+\frac{g^{(n+2)}(z_0)}{(n+2)!}(z-z_0)+\cdots,$$ so if $g(z)$ does not have poles or zeros at $z=z_0$, and $$\bbox[5px, border: 1pt solid green]{f(z)=\frac{g(z)}{(z-z_0)^{n+1}},}$$ then the residue of $f(z)$ at $z=z_0$ is $$\bbox[5px, border: 1pt solid blue]{\text{Res}_{z=z_0}f(z) = \frac{g^{(n)}(z_0)}{n!}.}$$

We have
$$\bbox[5px, border: 1.5pt solid green]{ f(z) = \frac{1}{(z+i)^{n+1}} \frac{1}{(z-i)^{n+1}}=\frac{g(z)}{(z-i)^{n+1}}}$$The residue of $f(z)$ at $z=i$ is thus

$$\bbox[5px, border: 1.5pt solid blue]{\text{Res}_{z=i} f(z) = \frac{ g^{(n)}(i)}{n!} = \frac{1}{i}\frac{(2n)!}{n!n!}\frac{1}{2^{2n+1}}.}$$

Here is one way of computing the derivatives $\cdots$

$$g(z) = (z+i)^{-n-1}$$
$$g^\prime(z) = -(n+1) (z+i)^{-n-2}$$
$$g^{\prime\prime}(z) = (-1)^2 (n+1) (n+2) (z+i)^{-n-3}$$
$$\cdots$$
$$g^{(n)}(z) = (-1)^n \frac{(2n)!}{n!}(z+i)^{-2n-1}$$
$$g^{(n)}(i) = \frac{(2n)!}{n!} \frac{2^{-2n-1}}{i}$$


Title: Number Theory: $x^2+y^2=a^2$

Body:
Is there a coprime triple $(x,y,z)$ such that $x^2+y^2=a^2, x^2+z^2=b^2, y^2+z^2=c^2$, where $a,b,c$ are integers

P.S. such solution doesn't exist for $a,b,c<1000$, as the computer says

P.P.S. the statement above is WRONG

Body:
x = 720
y = 132
z = 85
a = 732
b = 725
c = 157


问:
Group of real numbers under addition
I saw this statement in a paper (not a mathematical paper):

"Although the operations of addition and multiplication are defined in the field of real numbers, multiplication is undefined in the group of real numbers under addition."

I would be grateful if someone could explain this in very simple terms. I am absolutely ignorant about this area of mathematics, though I have read the definitions of group and field.

答:
The answer to this question is rather simple. I will present briefly what is going on, without giving the definition of a group. You can go search for that easily, but I think you can follow the answer below without knowing the details, even tho a group is defined quickly, and you do not need much mathematics to understand it. So I recommand you to look it up.

A group is by definition a set $G$ with a binary operation $\circ: G\times G\to G$, which has to fulfill some properties. Here $\circ$ is just a sign. The group operation vary. Sometimes we have a group with $+$. Sometimes we have groups with $\cdot$ as operation. If we look of sets of matrices, we have a group with matrix addition $\oplus$, and so on.

So a group just has one operation, and not two (like the two operations $+$ and $\cdot$ in $\mathbb{R}$).

If we want a group structure on $\mathbb{R}$, we can take the operation $+$.
This fulfills every property there is.

But $\mathbb{R}$ with multiplication $\cdot$ would not give a group, as this does not have every needed property.

It would fail, because one property is that every element needs an inverse, so for every $a\in\mathbb{R}$ there has to be a $b\in\mathbb{R}$ with $a\cdot b=b\cdot a=1$ (1 would here be the so called 'neutral element').

But as you know $0$ has no such inverse. As $a\cdot 0=0$ for every $a\in\mathbb{R}$, and not $1$.

For a more elaborate look on groups you can check the wikipedia page about groups.

However, we could define a group structure on $\mathbb{R}\setminus\{0\}$ denoted of $\mathbb{R}^\ast$, with muliplication as operation.

Title: Group of real numbers under addition


Title: Recurring Sequence with Exponent

Body:
First make it homogeneous.

$$a_n-2a_{n-1} = 5^{n-1} $$
$$a_{n+1}-2a_n = 5^n $$
$$\Rightarrow a_{n+1}-2a_n=5(a_n-2a_{n-1}) \tag 1$$
$$\Rightarrow a_{n+1}-5a_n=2(a_n-5a_{n-1}) \tag 2$$

Both (1) and (2) are geometric sequences, so

$$
a_{n+1}-2a_n = 5^n (a_1-2a_0) = 5^n (3-2)= 5^n \tag 3
$$
$$
a_{n+1}-5a_n = 2^n (a_1-5a_0) = 2^n (3-5)= -2^{n+1} \tag 4
$$
(3)-(4)
$$
a_n = \frac{1}{3} (5^n + 2^{n+1}). \blacksquare
$$

(Please see my post Show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$)


问:
Cyclic Inequality involving four variables
Here is an inequality that I think is true for all nonnegative $x,y,z,t$:
$$(x^3+y^3+z^3+t^3)^2 \geq (x+y+z+t)(x^4y+y^4z+z^4t+t^4x).$$
If any two of the variables are equal, I can show the inequality holds (by letting for example $x=y=1$, $z=1+p$, and $t=1+p+q$ for $p,q \geq 0$, and expanding). I also tried using the method of Lagrange multipliers by using the constraint $x+y+z+t=1$, but the algebra gets out of hand. See also a similar inequality I posted earlier and its solution. I could not replicate the ideas of that solution. So my question is that is this inequality true?

答: Your inequality does not true, please check $x = \frac{7}{10}, \,y = \frac 23, \,z = 1,\,t = \frac{10}{11}.$


问:
Limit definition of a derivative
If I have a continuously differentiable function $u(\theta)$ on $\mathbb{R}$, is the following true? $$\lim_{\delta\rightarrow0}\frac{u(\theta+\delta)-u(\theta)}{\delta}=\lim_{\delta\rightarrow0}\frac{u(\theta)-u(\theta-\delta)}{\delta}=u’(\theta)$$

I understand that the left-most expression is the limit definition of the derivative. I am not sure about the second. What exactly does the second define? Thank you.

答: The second part is the first with the substitution $\theta_1=\theta+\delta$. Since $\delta\to 0$, we have that $\theta_1\to\theta$ so the two are equivalent.

Title: Limit definition of a derivative


答:
Instead of finding the slope between the points:

$(,u())$ and $(+,u(+))$

as  approaches 0 (getting the two points as infinitely close as possible),

We are finding the slope between the points:

$(-,u(-))$ and $(,u())$

As long as u() is differentiable, then this will calculate the same slope as both $(+,u(+))$ and $(-,u(-))$ as  approaches 0, are both equally infinitely close to the point $(,u())$.

Title: Limit definition of a derivative

Body:
If I have a continuously differentiable function $u(\theta)$ on $\mathbb{R}$, is the following true? $$\lim_{\delta\rightarrow0}\frac{u(\theta+\delta)-u(\theta)}{\delta}=\lim_{\delta\rightarrow0}\frac{u(\theta)-u(\theta-\delta)}{\delta}=u’(\theta)$$

I understand that the left-most expression is the limit definition of the derivative. I am not sure about the second. What exactly does the second define? Thank you.


问:
Evaluating integral with difference argument inside sumation
I am working with an expression that looks like this ($n,m$ are integers, whereas $a,b$ are constants, real numbers):

\begin{eqnarray}
\sum_{n,m}\int_{-\infty}^{+\infty}dx\int_{-\infty}^{+\infty}dy f_{n}(x-na)g_{m}(y+mb)
\end{eqnarray}
My question is if the following change of variables $w=x-na$ and $z=y+mb$ is allowed in order to write the expression as:
\begin{eqnarray}
\sum_{n,m}\int_{-\infty}^{+\infty} dw\int_{-\infty}^{+\infty}dz f_{n}(w)g_{m}(z)
\end{eqnarray}
Since the limits of integration are not changed, this should in principle not change the value of the sum; however, the change of variables contains explicitly the integers $n,m$ and this is my doubt, I would appreciate some help, thanks.

答:
Yes, that is allowable.

For each fixed $n, m$, the change does not affect the value of the integral. And since the values being summed are the same before and after the change, the sum itself will not change.


问:
Constructing a Quadrilateral
I am given three sides of length 4.5 and one more side of length 4.2 (dimensionless entities). Also the area is given as 19.575. Now I have been given the task to construct a quadrilateral with these figures. I don't know whether it is trivial or complex. Since the type of quadrilateral is not specified, I'm concerned whether such a polygon can be constructed (I tried considering trapezium), and if possible, whether it is unique! And when unique, how to construct. I need help with this. Any online source for construction is also welcome. The main goal is to know angles between the sides. Thanks in advance.

答:
I am going to assume that the quadrilateral is  constructible.

The first thing to do is check out Heron's formula from here.

Draw a main diagonal and assume its length is $x$.  Then you have two triangles.  You can compute the area of both triangles using Heron's formula, as a function of $x$.  Since the quadrilateral's area is given, you know the area of the two triangles formed by the main diagonal.

Therefore, you can solve for the length of the main diagonal.  That is, you can solve for $x$.  Once this is done, you can solve for two of the angles via the Law of Cosines.

Then, rinse and repeat, to get the length of the other main diagonal, and consequently the other two angles.

Addendum

Never having done this, it is unclear to me whether the computation of each diagonal will yield a unique value.  From my perspective, that is actually irrelevant because when you identify each of the 4 angles that correspond to a pair of (possible diagonal lengths), you have the constraint that the 4 angles must sum to $360^{\circ}.$

Title: Constructing a Quadrilateral


问:
Relation between exponential random variables and their min
If
$$U=exponential(\lambda), \quad V = exponential(\mu)$$
and if they are independent, then $T = \min (U,V)$ is exponential $(\lambda + \mu)$.

How do we show that $P(T = U) = \lambda / (\lambda  + \mu)$? Any hint would be much appreciated.

答:
\begin{align}
& \Pr(T=U) \\[10pt]
= {} & \Pr(U<V) = \iint\limits_{u,v\,:\, u\,<\,v} e^{-\lambda u} e^{-\mu v} (\lambda\mu\,d(u,v)) \\[10pt]
= {} & \int_0^\infty \left( e^{-\lambda u} \int_u^\infty e^{-\mu v} (\mu\,dv) \right) (\lambda \, du) \\[10pt]
= {} & \frac\lambda{\lambda+\mu}.
\end{align}


答:
The word "group" in "group ... under addition" means (by the definition of a group) that you are going to use only one operation, and in this case it will be addition. The word "field" means (again by definition) that you are going to be considering two operations, addition and multiplication.

I like analogies, so here's one for this case: When you listen to an MP3 of a song, you have no idea what the color of the lead guitar was, but if you watch a video of that song's performance you probably will. If someone starts telling you how much they liked the cherry red guitar in the MP3 of "The Cradle Will Rock", you're going to say "Wait, you mean in the video, right? You can't see the color of the guitar in the MP3.". When we restrict ourselves to the group $\langle \mathbb R, +\rangle$ we can't see multiplication. If you object that "but multiplication still exists" then perhaps it's better to say that we don't look at it, or don't consider it in the group setting.

BTW, this statement reflects the modern approach to abstract algebra. We try to cut mathematical objects down to as few operations as possible and make as few assumptions about them as possible, and see what we can deduce from them. You can always add more operations, or more assumptions, but we are excruciatingly clear about what we are including, and often use completely different names. So if you said "In the group of the reals, when we multiply ..", my mathematically trained eyebrow would go up, and I would rudely interrupt you with "Multiply? So you mean the field of the reals, not the group?". Like I said, excruciatingly clear.

Title: Group of real numbers under addition

Body:
I saw this statement in a paper (not a mathematical paper):

"Although the operations of addition and multiplication are defined in the field of real numbers, multiplication is undefined in the group of real numbers under addition."

I would be grateful if someone could explain this in very simple terms. I am absolutely ignorant about this area of mathematics, though I have read the definitions of group and field.