问:
Relationship between two definitions of pro-representable functors
Edit:

I'm pretty sure that my conjecture
$$
\operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) = \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A),
$$
is true. To prove it, just use the construction of the colimit as a disjoint union modulo equivalence. It is then quite easy to set up a bijection explicitly. Therefore pro-representability in Mazur's sense implies pro-representability in the sense of the nLab. I suspect that the converse does not hold, and I intend to look for a counterexample when I have time. Until then, I will leave the question up, because I think it's interesting, and it doesn't get discussed much in the literature.

Original question:

Let $\mathcal{C}$ be the category of complete, local, Noetherian rings with residue field $k$ (where $k$ is fixed and finite). Let $\mathcal{C}^0$ be the full subcategory of rings that are also Artinian.

Mazur (and everyone else studying deformation rings, it seems) defines a functor $F:\mathcal{C}^0 \to \textbf{Set}$ to be pro-representable if there is some $R\in \mathcal{C}$ such that $F(A) = \operatorname{Hom}(R, A)$ as a functor $A\mapsto \textbf{Set}$.

The nLab defines a functor to be pro-representable if it is a filtered colimit of representables. I would like to show that these notions are equivalent in the case of functors $\mathcal{C}^0\to \textbf{Set}$.

For $R \in \mathcal{C}$, we have $R = \varprojlim_i R/\mathfrak{m}^i$, where $\mathfrak{m}$ is the maximal ideal of $R$. Thus, I would conjecture naively that
$$
\operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) = \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A),
$$
since the $R/\mathfrak{m}^i$ are in $\mathcal{C}^0$. However, I cannot see why this should be the case (and indeed I suspect that it is not), since the $\operatorname{Hom}$ functor does not play nicely with limits in the first argument. Using the universal properties, I am able to construct an injective map

$$
\operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A) \to \operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A)
$$
in the obvious way. However, I am not able to show that this map is surjective.

答:
This isn't a full answer because I don't know enough about noetherian complete local rings, but it's too long for a comment. Essentially I try to isolate the key-points you would need to show that they're the same.

It's a very nice question. The point is that in many cases, pro-objects in a given category $C$ have a nice description in terms of a variant of $C$.

For instance, profinite groups are the same thing as pro-(finite groups). That is, topological groups that are topologically inverse limits of finite (discrete) groups are a model for pro-objects in the category of finite groups. So if you have a functor $\mathbf{finGrp\to Set}$, it is pro-representable in the nLab sense if and only if there is a profinite group $G$ such that it is isomorphic to $\hom(G,-)$. But there is work behind this proof, namely the equivalence I mentioned earlier.

This equivalence works for a lot of neat algebraic gadgets (although I have never thought about the generality in which it was true... that would be a very nice question too)

You can indeed always define a map $\mathrm{colim}_i\hom(R/m^i,-)\to \hom(\lim_i R/m^i,-)$ , and you can ask when is it an isomorphism ? Essentially, you're asking whether the objects of $C^0$ are cocompact in $C$.

As you point out, it's pretty easy to see that this is always injective, so the question is about surjectivity: does a map $R\to A$ factor through $R/m^i$ for some $i$ ? I don't know a lot about these rings, but iirc the map $R\to A$ has to be continuous for the adic topologies, and so because the maximal ideal in $A$ is nilpotent, some power of $m$ has to be sent to $0$, thus proving the claim. Maybe I'm mistaken here, but something like that has to be true for the claim to hold anyways (otherwise you could find a morphism where no power of $m$ gets mapped to $0$ and get a nonsurjectivity claim)

How about the converse ? Well if I'm not mistaken about the previous paragraph, and if the proof can be adapted to prove the more general statement that any map $\lim_i R_i \to A$ factors through some $R_i$, where all the $R_i$'s are in $C^0$ and $A\in C^0$, then it would follow that the canonical functor $Pro(C^0)\to C$ is fully faithful. Since clearly any object of $C$ is an inverse limit of objects in $C^0$, it would follow that it's an equivalence, in other words, that the two definitions of pro-representable agree, just like in the case of groups !

Ah, maybe I should mention : for the canonical functor to even exist, one would need $C$ to be closed under inverse limits which doesn't seem completely obvious at first glance, but is perhaps still true.

As a conclusion (of this incomplete answer), here are the points you should try to work out:

1- Does any map $R\to A$ in $C$ factor through some $R/m^i\to A$ if $A\in C^0$ ? (probably yes, by some continuity things)

2- Does $C$ have inverse limits, and if so, is $R\cong \lim_i R/m^i$ in $C$ ? (be careful: this claim is true in the category of commutative rings by definition of complete, but it's not necessarily true that limits in $C$ are computed on the underlying rings ! )

If the answer to those 2 is "yes", then by the first part of 2- you get a canonical functor $Pro(C^0)\to C$, which is fully faithful by 1-, and is therefore an equivalence by the second part of 2-.

COnversely, if you can disprove any of 1- or 2-, it follows that $Pro(C^0)\not\simeq C$, and so the two definitions don't agree. In any case, you will get an answer to your question by solving 1 and 2.


问:
If two random variables are independent does that mean if one of them is squared then the independency still exists?
Suppose we have two random variables $X$ and $Y$ that are independent. Does that mean $X^2$ and Y are also independent?

答: @Hijaw, please see Here, Theorem 3 page 151 for the statement and proof</p>

ContentLicense: CC BY-SA 4.0


问:
Computing the expected value of a matrix?
This question is about finding a covariance matrix and I wasn't sure about the final step.

Given a standard $d$-dimensional normal RVec $X=(X_1,\ldots,X_d)$ has i.i.d components $X_j\sim N(0,1)$. Here we treat $X$ as a row vector. Now consider $Y:=\mu+XA$, where $\mu\in\mathbb{R}^m$, $A\in\mathbb{R}^{d\times m}$, so that $Y\in\mathbb{R}^m$. Then (we write the covariance matrix of Y as $C_Y^2$), $$C_Y^2=\mathbb{E}(Y-\mathbb{E}Y)^\intercal(Y-\mathbb{E}Y)=\mathbb{E}(XA)^\intercal XA$$ $$=\mathbb{E}A^\top X^\top XA=A^\top(\mathbb{E}X^\top X)A=A^\top IA=A^\top A.$$

I wasn't sure about how to compute $\mathbb{E}X^\top X$.  Since $X$ is a row vector then $X^\top X$ is a column vector multiplied by a row vector which resulted in a matrix, then how can we take the expected value of a matrix? Moreover how can we get the identity matrix as the product?

Also how to choose dimension d and m, it is said (without reason) that it is not good if m>d.

May anyone explain it to me? Thanks.

答: You do it term by term, which you can do because Expectation is a linear operator.  You get the identity matrix because the $X_i$ terms are independent and $N(0,1)$ so $E[X_i*X_i]=1$ but $E[X_i*X_j]=0.$</p>

ContentLicense: CC BY-SA 3.0


问:
Least Squares fit to closest convex function
I encountered a task where I have some discrete set of data points $(x_1, y_1), \ldots, (x_n, y_n)$, where
$x_1 < x_2 < \ldots < x_n$.

The discrete function graph they form should ideally be convex, but due to some noise and errors it deviates slightly from being convex everywhere.

I would therefore like to replace the $\{ y_i \}_{i=1, \ldots, n}$ with values  $\{ w_i \}_{i=1, \ldots, n}$ so that the new function graph given by the points $(x_1, w_1), \ldots, (x_n, w_n)$ is convex and the squared difference
$$
  \sum_{i=1}^n (y_i - w_i)^2
$$
is minimal.

I don't want to assign any parametric model for the $\{ w_i \}_{i=1, \ldots, n}$ unless I have to, since it should not be necessary to do so. This problem should be solvable anyway.

Maybe this is some well known stuff. I would then be thankful if anyone could point me to some good references.

One can of course write it as a general optimization problem in the $n$ unknown variables $\{ w_i \}_{i=1, \ldots, n}$ with a lot of inequalities as side conditions representing the convexity condition.

However, I would prefer some simpler solution, if it exists, or some algorithm that does not give the perfect optimal result, but still a convex graph that approximates the old graph "good enough".

答:
Using a cubic spline we can easily obtain good convex/concave fitting.

Follows a MATHEMATICA script which smooths the data imposing convexity/concavity as required. The quadratic spline has n equal spaced intervals, from 0 until tmax, with coefficients at each interval given by a[k], b[k], c[k] or as $s(k,t) = a_k + b_k(t-k\delta)+c_k(t-k\delta)^2$

```<code>Unprotect["`*"];
ClearAll["`*"];
fl[a_, b_, c_, delta_, t_, n_] := Sum[(UnitStep[t - k delta] - UnitStep[t - (k + 1) delta]) (a[k] + b[k] (t - k delta) + c[k] (t - k delta)^2), {k, 0, n}]
n = 10;
tmax = 2;
delta = tmax/n;
kappa = -0.3;
fl0 = fl[a, b, c, delta, t, n];
</code>```

After the spline formal definition the continuity conditions are introduced.

```<code>For[k = n - 1, k >= 0, k--,
  fl0 = fl0 /. {a[k + 1] -> a[k] + delta b[k] + delta^2 c[k] ,
                b[k + 1] -> b[k] + 2 delta c[k]}
]
</code>```

so the final independent parameters are given in

```<code>vars = Join[{a[0], b[0]}, Table[c[k], {k, 0, n}]];
</code>```

Now we define a function with concave epigraph, generating data added to noise

```<code>y[x_] := x (2 - x) + x/2
data = Table[{x, y[x] + RandomReal[{-0.1, 0.1}]}, {x, 0, tmax, 0.1}]
gr1 = ListPlot[data, PlotStyle -> {Thick, Red}];
</code>```

Follows a minimization procedure to determine de spline coefficients.

```<code>obj = Sum[Abs[(fl0 /. {t -> data[[k, 1]]}) - data[[k, 2]]], {k, 1,
Length[data]}];
</code>```

Due to epigraph concavity the condition is $c_k < \kappa$. Here $\kappa$ is the minimum acceptable to avoid the inclusion of line segments $(c_k = 0)$.

```<code>restr = Table[c[k] < kappa, {k, 0, n}];(* concave epigraph *)
</code>```

Now the minimization results: result1 with imposed concavity and result2 without conditions.

```<code>sol1 = NMinimize[Join[{obj, restr}], vars]
sol2 = NMinimize[obj, vars]
</code>```

Follows a graph shoving in red the data points, in dashed blue the unrestricted minimization and in blue continuous the conditioned minimization.

```<code>fl01 = fl0 /. sol1[[2]];
gr21 = Plot[fl01, {t, 0, tmax}];
fl02 = fl0 /. sol2[[2]];
gr22 = Plot[fl02, {t, 0, 2}, PlotStyle -> Dashed];
Show[gr1, gr21, gr22, PlotRange -> All]
</code>```

NOTE

The same data processed with a cubic spline. In this case the concavity/convexity is respectively $d_k > 0, \ \  d_k < 0$

Body:
Using a cubic spline we can easily obtain good convex/concave fitting.

Follows a MATHEMATICA script which smooths the data imposing convexity/concavity as required. The quadratic spline has n equal spaced intervals, from 0 until tmax, with coefficients at each interval given by a[k], b[k], c[k] or as $s(k,t) = a_k + b_k(t-k\delta)+c_k(t-k\delta)^2$

Unprotect["`*"];
ClearAll["`*"];
fl[a_, b_, c_, delta_, t_, n_] := Sum[(UnitStep[t - k delta] - UnitStep[t - (k + 1) delta]) (a[k] + b[k] (t - k delta) + c[k] (t - k delta)^2), {k, 0, n}]
n = 10;
tmax = 2;
delta = tmax/n;
kappa = -0.3;
fl0 = fl[a, b, c, delta, t, n];

After the spline formal definition the continuity conditions are introduced.

For[k = n - 1, k >= 0, k--,
  fl0 = fl0 /. {a[k + 1] -> a[k] + delta b[k] + delta^2 c[k] ,
                b[k + 1] -> b[k] + 2 delta c[k]}
]

so the final independent parameters are given in

vars = Join[{a[0], b[0]}, Table[c[k], {k, 0, n}]];

Now we define a function with concave epigraph, generating data added to noise

y[x_] := x (2 - x) + x/2
data = Table[{x, y[x] + RandomReal[{-0.1, 0.1}]}, {x, 0, tmax, 0.1}]
gr1 = ListPlot[data, PlotStyle -> {Thick, Red}];

Follows a minimization procedure to determine de spline coefficients.

obj = Sum[Abs[(fl0 /. {t -> data[[k, 1]]}) - data[[k, 2]]], {k, 1,
Length[data]}];

Due to epigraph concavity the condition is $c_k < \kappa$. Here $\kappa$ is the minimum acceptable to avoid the inclusion of line segments $(c_k = 0)$.

restr = Table[c[k] < kappa, {k, 0, n}];(* concave epigraph *)

Now the minimization results: result1 with imposed concavity and result2 without conditions.

sol1 = NMinimize[Join[{obj, restr}], vars]
sol2 = NMinimize[obj, vars]

Follows a graph shoving in red the data points, in dashed blue the unrestricted minimization and in blue continuous the conditioned minimization.

fl01 = fl0 /. sol1[[2]];
gr21 = Plot[fl01, {t, 0, tmax}];
fl02 = fl0 /. sol2[[2]];
gr22 = Plot[fl02, {t, 0, 2}, PlotStyle -> Dashed];
Show[gr1, gr21, gr22, PlotRange -> All]

NOTE

The same data processed with a cubic spline. In this case the concavity/convexity is respectively $d_k > 0, \ \  d_k < 0$


问:
The expression $i^{i^{i^2}}$ can be written $e^\alpha$ for some real number $\alpha \in [0, 2\pi)$. What is $α$?

The expression $i^{i^{i^2}}$ can be written $e^\alpha$ for some real number $\alpha \in [0, 2\pi)$. What is $α$?

I am not sure what this is asking, could I get a hint on where to start?

答: The calculation $i^{i^{i^2}}=i^{-i}=e^{-i(\pi i/2+2n\pi i)}=e^{2n\pi+\pi/2},\,n\in\Bbb Z$ is multi-valued, but for your purposes we take the $n=0$ branch.


问:
Multivariable calculus - Conservative vector field
We have vector field $\vec{F}=(-\frac{y}{x^2+y^2}),(\frac{x}{x^2+y^2})$

And the domains

$D_1=\{(x,y):4\leq x^2+y^2\leq 9, y\geq 0\}$

$D_2=\{(x,y):4\leq x^2+y^2\leq 9, y\leq 0\}$

$D_3= D_1\cup  D_2$

Now the vector field is conservative in $D_1$ and $D_2$ but not $D_3$

Why is that?

答: Basically, because $\;D_1\cup D_2\;$ isn't simply connected: it "wraps" around $\;(0,0)\;$ and the field isn't even defined at this point. In each of $\;D_1,\,D_2\;$ it is just fine...


问:
Where is the mistake in a point closest to cross section of two planes and a line problem
The problem:

Find the point that lies on the line that is the intersection of two planes:

plane one: $$3x-2y+z = 2$$

plane two: $$x+y+2z = 4$$

The point also  needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$

My try:
First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.

What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$

I got: $z = 7y +2$ and $x = \frac{-5y}{3}$

Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$

So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$

Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$

Then I wanted to position the plane($O$...origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$

So the equation of this plane is: $4z = 4*\frac{1}{2}=2$

I got $z_0$ form the last coordinate from the point.

Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.

The correct solution is: $T(1,1,1)$

Where did I go wrong?

答:
One error is in "I got: $z = 7y +2$ and $x = \frac{-5y}{3}$".

Hint.

You are doing the problem in a much more complicated way than necessary.

The second condition implies that the point lies in the plane that is perpendicular to the vector $(0,0,4)$, and contains the point $(1,-1,1)$. This plane is given by
$
z=1
$.

Now you have two equations in $x$ and $y$:
$$
3x-2y=1,\quad x+y=2
$$


答: It's because the midpoint between $A$ and $B$ is $(1,-1,1)$. So, the plane which consists of all the points that have the same distance from $A$ and $B$ is the plane $z=1$, not the plane $z=\frac12$.

Title: Where is the mistake in a point closest to cross section of two planes and a line problem

Body:
The problem:

Find the point that lies on the line that is the intersection of two planes:

plane one: $$3x-2y+z = 2$$

plane two: $$x+y+2z = 4$$

The point also  needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$

My try:
First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.

What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$

I got: $z = 7y +2$ and $x = \frac{-5y}{3}$

Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$

So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$

Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$

Then I wanted to position the plane($O$...origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$

So the equation of this plane is: $4z = 4*\frac{1}{2}=2$

I got $z_0$ form the last coordinate from the point.

Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.

The correct solution is: $T(1,1,1)$

Where did I go wrong?


问:
Do the matrix ring endofunctors have left adjoints?
Given any positive integer $n$, does the endofunctor $R \mapsto M_n(R)$ of the category of (unital) rings have a left adjoint?

One possible idea for constructing the left adjoint $L(R)$ at a ring $R$ is to consider the quotient of the free ring on $R \times \{1,2,...,n\} \times \{1,2,...,n\}$ by the two-sided ideal generated by the following elements (to ensure that the map $\eta_{R}:R \to M_n(L(R))$, $r \mapsto [(r,i,j)]_{1 \le i,j \le n}$, is a ring homomorphism):

$(r,i,j)+(s,i,j)-(r+s,i,j)$ for $r,s \in R$ and $1 \le i,j \le n$ (necessary for $\eta_{R}$ to be an additive group homomorphism)

$\big(\sum_{k=1}^{n} (r,i,k)(s,k,j)\big)-(rs,i,j)$ for $r,s \in R$ and $1 \le i,j \le n$ (necessary for $\eta_{R}$ to preserve multiplication)

$(1,i,i)-1$ for $1 \le i \le n$ (one of the necessary conditions for $\eta_{R}(1)$ to be the identity matrix)

$(1,i,j)$ for $1 \le i,j \le n$ with $i \neq j$ (the other necessary condition for $\eta_{R}(1)$ to be the identity matrix)

But does the above quotient ring idea really give a left adjoint?

答:
A silly answer: it preserves limits and is accessible, so because the category of rings is presentable, it has a left adjoint by the adjoint functor theorem.

Slightly less silly: It can be described as $\hom(\mathbb Z[X_{ij}, 1\leq i,j\leq n], -)$ with a certain ring structure, which is given by universal formulas, in other words, by a co-ring object structure on $U=\mathbb Z[X_{ij}, 1\leq i,j\leq n]$ ( a ring structure on that object in the category $\mathrm{Ring}^{op}$).

In particular, $\hom(S,M_n(R)) \cong \hom(S, \hom(U, R))$ and this is definitely representable by some quotient of the free ring on $S\times \{1,...,n\}^2$. One can then work out the relations to impose on this free ring, the ones you outlined are definitely among them, and I think it should be enough.

The conditions you need to impose are exactly those such that if you start from a ring morphism from this quotient to $R$, then $s\mapsto (f(s,i,j))_{i,j}$ is a ring morphism $S\to M_n(R)$.

Then you can work these out and see that exactly what you wrote comes out.


答:
You have incorrectly solved the system of equations of planes since $(-5/3,1,9)$ doesn't lie on plane $2$. On solving you get $z=2-y,x=y$ so the correct choice of point lying on both planes for $y=1$ is $(1,1,1)$.

Your second error is in finding the equation of plane containing points equidistant from $A,B$: you wrote $OA+\frac12AB$ but found $OA+\frac12\color{red}{OB}$. The correct point lying on the plane is $$OA+\frac12AB=\frac{OA+OB}2=(1,-1,1)$$giving the required plane as $z=1$. Can you continue?


问:
Calculate the limit of two interrelated sequences?
I'm given two sequences:

$$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$

as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.

Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:

$$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$

Seeing no pattern, I did the same once more:

$$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$

While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.

How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!

Edit:

While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.

答:
The answer is $a_n \to 5$ , $b_n \to \infty$.

I'm trying to prove that and I will edit this post if I figure out something.

EDIT:
I would write all this in comment instead in answer, but I cannot find how to do it.. maybe I need to have more reputation to do this (low reputation = low privileges:P)

Anyway, I still didn't solved it, but maybe something of that will help you. I will edit it when I think something out.

EDIT:
After many transformations and playing with numbers, I think that the limit, for $a<b$, is $$ \frac{ab + 2a +1}{b-a}$$

But still cannot prove it.

(In statement above: $a = a_1 $ , $ b = b_1 $)


问:
'Algebraic' computation of the differential of a Lie algebra homomorphism (using vector fields)
The following is an exercise from section 1.5.4 of Goodman and Wallach's textbook.
Let $$ \phi(A) = \begin{bmatrix} \det(A)^{-1} & 0 \\ 0 & A \end{bmatrix}$$ for $A \in GL(n,\mathbb{C})$.
Show that $$d\phi(X) =  \begin{bmatrix} -tr(X) & 0 \\ 0 & X \end{bmatrix}$$ for $X \in \mathfrak{g}$.

Using the definition of the differential given by the exponential map, $$ d \phi(X) = \frac{d}{dt} \phi ( \exp(tX)) \vert_{t=0},$$ then a straightforward application of $\det(\exp(X)) = \exp(\text{tr}(X))$ gives the desired identity.

On the other hand, the 'algebraic' definition of $d\phi$ using vector fields is the following.
For $A \in \mathfrak{gl}(V)$, define the derivation $X_A$ by $$X_A f (u) = \frac{d}{dt} f(u(I+tA))\vert_{t=0}$$ for $u \in G$ and $f \in \mathcal{O}[G]$.
The map $A \mapsto X_A$ gives an isomorphism between the Lie algebra of $G$ and the Lie algebra of left-invariant vector fields on $G$.
Then for any $A \in \mathfrak{g}$ and any regular function $f$ on $G$, we have $$ X_A (f \circ \phi) = (X_{d\phi(A)}f) \circ \phi.$$

I'm not sure though how to use this definition to compute $d\phi(X)$ as above.

答:
Let $f \in \mathcal{O}[\operatorname{SL}(n+1, \mathbb{C})], u \in \operatorname{GL}(n, \mathbb{C}), A \in \mathfrak{g}$ and $X_A$ the corresponding derivation. For convenience, let us denote $d\phi(X_A)$ by  $Y_A$, $\phi(u) = w$, and $h = f\circ \phi$. By definition, we have that
\begin{align}
Y_Af(w) :=&\ X_Ah(u)= \frac{d}{dt}h(u\cdot(I+tA))\bigg|_{t=0} \\
=&\ \frac{d}{dt}f\left( w\cdot \phi(I+tA)\right)\bigg|_{t=0}
\end{align}
since $\phi:\operatorname{GL}(n, \mathbb{C})\rightarrow \operatorname{SL}(n+1, \mathbb{C})$ is a group homomorphism. Recall that $f$ is a function of $(n+1)^2+1$ variables, i.e.
\begin{align}
f(\det(x)^{-1},x_{01},\ldots,x_{n0} ,x_{11}, \ldots, x_{n\ n}).
\end{align}
Then, by the chain rule of differential calculus, we have that
\begin{align}
\frac{d}{dt}f\left( \phi(u)\phi(I+tA)\right)\bigg|_{t=0} =&\ \frac{d}{dt}f\left( \begin{bmatrix}
\det(u)^{-1} & 0\\
0 & u
\end{bmatrix}
\begin{bmatrix}
\det(I+tA)^{-1} & 0\\
0 & I+tA
\end{bmatrix}
\right)\bigg|_{t=0}\\
=&\ \left(-\frac{\operatorname{tr}A}{\det(u)}\frac{\partial }{\partial \det(x)^{-1}}
+\sum_{i=1}^{n}\sum_{j=1}^n (u\cdot A)_{ij}\frac{\partial }{\partial x_{ij}}\right)f(w).
\end{align}
Furthermore, we can identify $Y_A = d\phi_u(X_A)$ with a $(n+1)\times (n+1)$ matrix just like how we identify $X_A$ with $A$, that is,
\begin{align}
d\phi_u(X_A) =
\begin{bmatrix}
-\det(u)^{-1}\operatorname{tr}  A & 0\\
0 & u\cdot A
\end{bmatrix}.
\end{align}
Since we only care about $u=e$, then we have the desired result.

Edit: Notice that
\begin{align}
 f(\phi(u)\cdot \phi(I+tA))= f(x_{00},x_{01},\ldots,x_{n0} ,x_{11}, \ldots, x_{n\ n})
\end{align}
where
\begin{align}
x_{00} =&\ \det(u\cdot (I+tA))^{-1}, \ \ x_{0i} = x_{i0} =0 \text{ for }  i=1, \ldots, n,\\
x_{ij} =&\ (u\cdot (I+tA))_{ij}  \ \text{ for }\ i, j=1, \ldots, n.
\end{align}
Then, by chain rule, we have that
\begin{align}
\frac{d}{dt}f\left( \phi(u)\phi(I+tA)\right) =&\ \sum^n_{i=0}\sum^n_{j=0}\frac{\partial f}{\partial x_{ij}}\frac{dx_{ij}}{dt}
= \frac{\partial f}{\partial x_{00}}\frac{dx_{00}}{dt}+\sum^n_{i=1}\sum^n_{j=1}\frac{\partial f}{\partial x_{ij}}\frac{dx_{ij}}{dt}.
\end{align}

Hence, it suffices to compute $d x_{ij}/dt$ and $d x_{00}/dt$. By Jacobi's formula, we see that
\begin{align}
\frac{d}{dt} \det(u\cdot (I+tA))^{-1} =&\ -\det(u\cdot (I+tA))^{-2}\frac{d}{dt} \det(u\cdot (I+tA))\\
=&\ -\det(u\cdot (I+tA))^{-2}\det(u) \frac{d}{dt}\det(I+tA)\\
=&\ -\det(u\cdot (I+tA))^{-2}\det(u)\operatorname{tr}\left(\operatorname{adj}(I+tA)A \right).
\end{align}
Finally, set $t=0$ yields
\begin{align}
\frac{d}{dt} \det(u\cdot (I+tA))^{-1}\bigg|_{t=0} = -\det(u)^{-1}\operatorname{tr}\left(\operatorname{adj}(I)A \right) =-\det(u)^{-1}\operatorname{tr}\left(A \right).
\end{align}

Next, notice that
\begin{align}
\frac{dx_{ij}}{dt} =&\ \frac{d}{dt} (u\cdot (I+tA))_{ij} = \frac{d}{dt} \sum^n_{k=1} u_{ik}(I+tA)_{kj}\\
=&\ \frac{d}{dt} \sum^n_{k=1} u_{ik}(\delta_{kj}+tA_{kj}) = \sum^n_{k=1} u_{ik}A_{kj} = (u\cdot A)_{ij}.
\end{align}


答:
Step 1. Find a closed path in $D_3$ whose line integral is non-zero.

Step 2. If then you remove $D_2$ from the domain, what happens to the path you found in Step 1?

Title: Multivariable calculus - Conservative vector field

Body:
We have vector field $\vec{F}=(-\frac{y}{x^2+y^2}),(\frac{x}{x^2+y^2})$

And the domains

$D_1=\{(x,y):4\leq x^2+y^2\leq 9, y\geq 0\}$

$D_2=\{(x,y):4\leq x^2+y^2\leq 9, y\leq 0\}$

$D_3= D_1\cup  D_2$

Now the vector field is conservative in $D_1$ and $D_2$ but not $D_3$

Why is that?


问:
Difference between simultaneous and ordered N.E. in the problem.
Consider the following game. There are two players and one of them could be in two states:
$A$ with probability $p$ and $B$ with probability $1 - p$. Both players actions are $1, 2,$ and to not participate.

If the state is $A$ then players get $\$1$ if 1 is majority action, 1/2 if they tie, and 0 if 2 is the majority. Likewise, we determine the game in the state $B$.

Now we have a question to find Bayes-Nash equilibrium (pure) in two situations: (1) simultaneous game and (2) first go first, and then second take part.

I've decided that in first case we have: N.E. not participate for first player and $(1 , 2)$ (say $1$ in situation $A$ and say $2$ in situation $B$). However I've got the same for simultaneous game. There is a problem?

答:
Your description of the game is a bit vague, so I have to make a few assumptions. First, you say "Likewise, we determine the game in the state $B$." I am assuming here the payoffs are flipped (payoff of 1 iff both choose "2"), otherwise payoffs would be state independent and there is no point in using Bayes Nash equilibrium.

Second, in the sequential game, I assume the actions of the first moving player is observable for the second (otherwise it would be equivalent to a simultaneous move game).

Third, you say "one of them could be in two states". It can make a difference which player observes those states and which player moves first in the sequential game. Suppose player 2 is the one who privately observes the state, but player 1 doesn't, and player 2 moves second. Then this situation is identical to the simultaneous move game. However, if player 1 were to move second, then it changes things, because the action of player 2 could then "reveal" the state to the other player.

Regarding equilibrium: You are right that the informed player choosing $(1,2)$ and the uninformed choosing "no participation" is a BNE. Any deviation by the informed player would strictly reduce his payoff, given the uninformed player abstains. And a deviation by the uninformed player cannot improve things but can decrease his payoff (if the incorrect action is chosen which does not match the state). You are also right that in the sequential game this is a BNE as well - after all, both players get the maximum payoff in every state with probability 1, so there is no way to improve on this.

There are additional equilibria in some cases. For example, in the sequential game, if player 1 moves first and is informed, then the following strategy is also a BNE: $(1,2)$ for player 1 and for player 2: 1 if first player chose 1 and 2 if first player chose 2. The outcomes in this equilibrium are the same, however, the uninformed player does not actually abstain. This equilibrium does not exist if the order of moves is reversed; this is why I pointed out above it might matter.


问:
What is the background to inequality $|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$?
$\alpha=(x_1,x_2,\dotsc,x_n)$, $\beta=(y_1,y_2,\dotsc,y_n)\in\Bbb R^n$, $\alpha \cdot\beta=x_1y_1+x_2y_2+\dotsb+x_ny_n$.

$|\alpha|=\sqrt{x_1^2+x_2^2+\dotsb+x_n^2} $, $|\beta|=\sqrt{y_1^2+y_2^2+\dotsb+y_n^2}, $

Show that : there exists a constant $L\gt 0$ such that
$$|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$$
for all $\alpha,\beta$

I don't know where to start

The question produces background . What is the background?

答:
$\textbf{Hint}$: First prove that

$$|e^{ix}-e^{iy}|\leq L|x-y|$$

for $x,y\in\Bbb{R}$


答:
Well for $t\in\mathbb{R}$
$$|e^{it}-1|^2=(cos(t)-1)^2+\sin(t)^2=2-2\cos(t)<2|t|^2$$
so $|e^{it}-1|\leq\sqrt{2}|t|$ for all $t$.
Now for $t=\alpha\cdot\beta$ we have that
$$|e^{i\alpha\beta}-1|\leq\sqrt{2}|\alpha\cdot\beta|$$
and using the famous Cauhcy-Schwarz inequality $|\alpha\cdot\beta|\leq|\alpha|\cdot|\beta|$ we get the desired estimate.

Title: What is the background to inequality $|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$?

Body:
$\alpha=(x_1,x_2,\dotsc,x_n)$, $\beta=(y_1,y_2,\dotsc,y_n)\in\Bbb R^n$, $\alpha \cdot\beta=x_1y_1+x_2y_2+\dotsb+x_ny_n$.

$|\alpha|=\sqrt{x_1^2+x_2^2+\dotsb+x_n^2} $, $|\beta|=\sqrt{y_1^2+y_2^2+\dotsb+y_n^2}, $

Show that : there exists a constant $L\gt 0$ such that
$$|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$$
for all $\alpha,\beta$

I don't know where to start

The question produces background . What is the background?


问:
Distinction between series and sum of infinite terms
Look, we all know that $$\sum_{k=1}^\infty a_k$$ represents an infinity series. Also it represents a sum.

So my question is:

How can one distinguish between the two ? Is there any difference between the ordinary addition of the terms with the value of the series ?

By ordinary addition, I mean the normal  addition what we done in number system, e.g., $2+3=5$ type.

My thoughts: After getting useful information from @JWP_HTX and also searching from various books, I think, the sum (which is used here in a very special sense) of a convergent series is the limit of the sequence of partial sums that is not obtained by ordinary addition. Again the sum gives only a number and it is not capable of being convergent or divergent.

Note: I had already made an edit of my question (on the same date of posting) and there is no other additional information with me. So please help and vote to reopen the question so that I can get the correct explanation of my doubts.

答:
I learned that the notation $\sum_{n=1}^{\infty}a_n$ is shorthand for an ordered pair of sequences, the first being the sequence of "terms" $a_1,a_2,a_3,\dots$ and second, the sequence of "partial sums" $s_1,s_2,\dots,$ where $s_n= a_1+\cdots+a_n.$

Given that, writing something like $\sum_{n=1}^{\infty}1/2^n = 1$ doesn't make too much sense. How can a pair of sequences equal the number $1?$ They seem to live in different universes.

Of course, $\sum_{n=1}^{\infty}1/2^n = 1$  is defined to mean $\lim_{n\to \infty}\sum_{k=1}^{n} 1/2^k=1.$ Kind of an abuse of the equal sign if you ask me. But we do this all the time in mathematics. Writing $\sum_{n=1}^{\infty}a_n = S$ is just too tempting and perfect to avoid. There's no problem with it if we just keep in mind what's happening behind the scenes.


问:
Algebra Equation with equals on left hand side
$6 = 4 - 2x =$

Show the answer with the mechanics of working out.

答:
It does not matter left or right.

We can add $2x$ on both sides to get $$6+2x=4-2x+2x$$ $$\implies 6+2x=4$$ then subtract 6 on both sides we get $$6-6+2x=4-6$$ $$\implies 2x=-2$$ then divide 2 on both sides we get $$\frac{2}{2}x=\frac{-2}{2}$$ $$\implies x=-1.$$

If we have an equation, it is "legal" to perform addition, subtraction, multiplication and division (except by 0) by equal number on both sides. Later you will also learn that it is also "legal" to take the log, power, derivative and integral and so on, on both sides.


问:
Prove that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$ if and only if $a$ does not divide $b$ in $R$.
Let $R$ be a commutative ring with $1$. Suppose that $a, b \in R$, $b \notin R^\times$ and $\langle a \rangle$ is a maximal ideal of $R$. Prove that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$ if and only if $a$ does not divide $b$ in $R$. [Notice that $R^\times = \{ r \in R ~|~ r \text{ is a unit.} \}$].

$\textbf{My Attempt:}$

Prove of "$\implies$":

Which want to prove that if $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$, then $a$ does not divide $b$ in $R$.

By contra-positive, we can prove that if $a$ divide $b$ in $R$, then $a + \langle b \rangle$ is not a unit in the quotient ring $R/\langle b \rangle$.

Assume $a$ divide $b$ in $R$.

Then, there exist $k \in R$ such that $b = ak$.

Since, $b \in R$ and $b \notin R^\times$.

Then, $b = b \cdot 1 \in \langle b \rangle$.

Since, $a \in R$. So, $ak = b \in \langle b \rangle$.

Then, $(a + \langle b \rangle)(k + \langle b \rangle) = ak + \langle b \rangle = \langle b \rangle = 0 + \langle b \rangle \neq 1 + \langle b \rangle$.

So, $a + \langle b \rangle$ is not a unit in the quotient ring $R/\langle b \rangle$.

So, this proves that if $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$, then $a$ does not divide $b$ in $R$.

Prove of "$\Longleftarrow$":

Which want to prove that if $a$ does not divide $b$ in $R$, then $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$.

Assume $a$ does not divide $b$ in $R$.

Then, there exist $k \in R$ such that $b = ak + r$, where $r \in R$ and $r \neq 0$.

Which I am stuck on how to prove that $r = -1$. Which then I can prove that $ak - 1 \in \langle b \rangle$ and conclude that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$.

Also, is the prove of "$\implies$" correct ?

And, I don't know where to apply the given condition that $\langle a \rangle$ is a maximal ideal of $R$.

答:
Hint: it's straightforward using $\rm\color{#c00}{QR}$ = Quotient Reciprocity (Third Isomorphism Theorem)

$$\begin{align}
&(a)\!=\!(1)\,\ {\rm in}\,\ R/b\\[.2em]
\iff\ & (R/b)/a = (0)\\[.2em]
\iff\ & (R/a)/b = (0),\ {\rm by\ \color{#c00}{QR}}\,\\[.2em]
\iff\ & (b)\!=\!(1)\ \,{\rm in}\,\ R/a\\[.2em]
\iff\ &\ \,b\not \in (a),\ {\rm by}\ R/a\,\ {\rm a\ field,\ by}\ (a)\ {\rm max}
\end{align}\qquad$$


问:
Finding $k$ such that $\binom{-2}{k}$ is the direction vector of the line $y=\frac17(4x+1)$
I'm new here, I asked my friend about this question and he told me to go on this math forum. If someone can get me started on these questions it would make my day. Thank You

Find the value of $k$ such that $\dbinom{-2}{k}$ is a direction vector of the line with equation $y=\dfrac{4x+1}{7}$.

Find an equation for the line with vector form $\dbinom{5}{-2}+t\dbinom{2}{6}$ in the form $y=mx+b$.

答:
On a plane, any line can be presented in two ways:

A point on the line and the line direction vector, pointing along the line. Like $\mathbf{x}=\mathbf{a}+\mathbf{v}t$.

A linear equation about the coordinates of a point, belonging to the line -- the equation is true for a point on the line and false everywhere else. These equations looks like $Ax+By+C=0$ or $\mathbf{n}.(\mathbf{x}-\mathbf{a})=0$, where $\mathbf{n}$ is the line normal vector (and perpendicular to the line), $\mathbf{a}$ is a point on the line and $\mathbf{x}$ is any point we're checking whether it's on the line or not.

So here we just have to convert one equation form to another. To convert from 1 to 2 we rewrite $\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x_0\\y_0\end{pmatrix}+\begin{pmatrix}v_x \\v_y\end{pmatrix}t$ as $\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}(=t)$ and convert it to $Ax+By+C=0$ by just clearing the denominator.

To convert from $Ax+By+C=0$ to 1 we can take any vector, perpendicular to $\mathbf{n}=\begin{pmatrix}A \\B\end{pmatrix}$ (e.g. $\begin{pmatrix}-B \\A\end{pmatrix}$) as direction vector and also we have to find a point, belonging to the line, e.g. by intersection with $x=0$ or $y=0$ or any other convinient line.

$\begin{pmatrix}-2 \\k\end{pmatrix}$ is orthogonal to $\begin{pmatrix}4 \\-7\end{pmatrix}$ when $k=-\frac{8}{7}$ and $\frac{x-5}{2}=\frac{y-2}{6}$ is $y=3x-17$.


问:
Given matrices $A,B,C$ such that $BC=A$ find $\det(CB)$ and $\operatorname{Tr}(CB)$

Let $A = \begin{pmatrix} 2&-2 & 3\\ 0&0&3 \\ 0&0&3 \\ \end{pmatrix}$. $B$ is a $3\times 2$ matrix and $C$ is a $2 \times 3$ matrix such that $BC=A$. Find $\det(CB)$ and $\operatorname{Tr}(CB)$.

I read this thread but i have not learned about characteristic polynomials yet and the answer that does not use those is tedious.

I also read this but in my case it is the other way around, i know the determinant which equals $0$, but i do not know how to find the "other one".

Also, I am not that familiar with spaces and dimensions in linear algebra so i would prefer a solution that does not involve deep knowledge in this area.

Any tips ?

答:
Since $A$ has rank $2$, $B$ has full column rank. It follows that if $Av=BCv=0$ for some vector $v$, we must have $Cv=0$.

Now let $v=(1,1,0)^T$. Then $Av=0$. Hence $Cv=0$. Let $u=(0,1,0)^T$. Then $u^Tv=1$. Therefore
\begin{aligned}
\det(CB)
&=\det\pmatrix{CB&0\\ \ast&1}\\
&=\det\left[\pmatrix{C\\ u^T}\pmatrix{B&v}\right]\\
&=\det\left[\pmatrix{B&v}\pmatrix{C\\ u^T}\right]\\
&=\det(A+vu^T)\\
&=\det\pmatrix{2&-1&3\\ 0&1&3\\ 0&0&3}=6.
\end{aligned}
Finally, as the trace of a product of matrices is invariant under cyclic permutation of the multiplicands, we have $\operatorname{tr}(CB)=\operatorname{tr}(BC)=\operatorname{tr}(A)=5$.


问:
How to solve such a trigonometric equation?
If we have a right triangle, which we know the Hypotenuse only, and let's assume H=3, A (adjacent) and O (opposite) are both unknown, as well as both $\theta_{1}$ and $\theta_{2}$ angles are unknown. But we know the angle $\theta_{3} = 90°$ as always, so we get the following two equations to solve for Opposite side and Adjacent side:

$\sin{(\frac{O}{3})} = 90°$

$\cos{(\frac{A}{3})} = 90°$,

which; we then need to solve for A and O...

By using arcus sinus, we get the following:

$\arcsin{(\sin{(\frac{O}{3})})} = \arcsin{(90°)} \implies O = 3 \cdot \arcsin{(90°)}$,

But arc sine can-not work with degrees, only with radians where $-1 > x < 1$, so if we try to convert $90°$ into radians, we must multiply by $π$ and then divide by $180$, which we get the following:

$\frac{90π}{180} \approx 1.57$,

which the following dominates:

$-1 > x > 1$,

while we need $-1 > x < 1$...

So... is there any solution for that???

Thanks for the answers.

答: Your question as stated cannot be answered, because there are many right triangles with hypotenuse 3, with different values of the lengths of the sides.  Also, you have your input and output reversed in that sin(90º) = 1, sin(30º) = 1/2 and so forth.  Trigonometric functions are defined for degrees and radians.  It is when you do calculus that the difference becomes important.</p>


问:
If $ x \in \Bbb{R} \setminus \Bbb{Z}, n \in \Bbb{N} $ for $ y = \sqrt[n]{x} $ then $ y \in \Bbb{R} \setminus \Bbb{Z} $?
Is this statement true ?

If:

$$ y = \sqrt[n]{x} $$
$$ x \in \Bbb{R} \setminus \Bbb{Z}, n \in \Bbb{N} $$
Then:
$$ y \in \Bbb{R} \setminus \Bbb{Z} $$

答:
By contraposition:
$$
y\in\Bbb Z \Rightarrow x=y^n \in \Bbb Z.
$$


问:
Is the completion of regular ring regular?
This is probably an elementary problem but I wasn't able to figure it out. Given a regular ring $A$ and an ideal $I$ which you can assume $A/I$ is also regular. Is the completion ring $\hat{A}$ necessarily regular? This is true if $A$ is local but unfortunately I am not sure whether this can be turned into the local case. Because I'm not sure whether it is possible to figure out what all maximal ideal of $\hat{A}$ look like. If not true is it true if I assume $I$ is a minimal prime ideal generated by a single element?

Maybe the problem can be turned into the local case. I was thinking given any maximal ideal of $\hat{A}$ like $m$. Its image under projection to $A/I$ is going to be an ideal which will be inside another maximal ideal $m'$ of $A/I$. Now $m'$ induces a maximal ideal on $\hat{A}$ which contains $m$. So if $m$ was maximal they have to be the same thing. This means every maximal ideal of $\hat{A}$ is induced from a maximal ideal in $A$. If this is true now we can turn the problem to the local case.

答: See EGA 0 (in $\text{IV}_{1}$), Lemme (17.3.8.1) and also Lemma 5.6 of Greco, "Henselization of a ring with respect to an ideal" (link).


问:
Coupon Collecting Problem
There are $c$ different types of coupon, and each coupon obtained
is equally likely to be any one of the $c$ types. Find the probability that the first $n$ coupons which you collect do not form a complete set.

It is clear that the probability of getting a new coupon on the first draw is 1, and that the probability of getting a new coupon on the second is $(c-1)/c$, or $(c-i)/c$ for the $i$-th new coupon after the initial draw (or $(c-(i-1))=c$ if you include the first draw as a new $i$-th coupon) I don't know where to go from here.

答: Here is another answer: The probability that none of the n coupons are of type 1 is ${(\frac{c-1}{c})}^n$.  Likewise for all the other types.  The sum of these is an overcount of the total probability. We can use the inclusion exclusion principle. Subtract the probability that two types are excluded, then add the probability that three types are, and so forth until all possibilities are covered.

Title: Coupon Collecting Problem


问:
Evaluate $\int \int_S z \space dS$ where $S$ is the surface of a hemisphere.
I am trying to evaluate a surface integral

$$\int \int_S z \space dS$$

where

$S$ is the surface

$$x^2+y^2+z^2=a^2, \space z \in [0,a]$$

I am a little confused on what to do here, because although my first instinct was to use spherical coordinates I only have two integrands.  If I try to use $x=a\cos(\theta), \space y=a\sin(\theta), z=z$ then the equation does not work out.

It seems as though I am missing something very basic but I cannot get my mind around it.

May I have some assistance, please?

答:
If you are parametrizing in spherical coordinates,

For a sphere of radius $a$, $|r'_{\theta} \times r'_{\phi}|$ is simply $a^2 \sin \phi$. If you do not remember it, you can parametrize as below and take derivative wrt $\theta$, $\phi$ and do the cross product.

$r(\theta, \phi) = (a \cos\theta \sin\phi, a\sin\theta \sin\phi, a \cos\phi)$

$|r'_{\theta} \times r'_{\phi}| = a^2 \sin \phi$

So your integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi/2} a \cos\phi \cdot a^2 \sin \phi \ d\phi \ d\theta$

Similarly if you parametrize in cylindrical coordinates, it will be

$r(\rho,\theta) = (\rho \cos\theta, \rho \sin \theta, \sqrt{a^2-\rho^2})$

Now can you find $|r'_{\rho} \times r'_{\theta}|$ and integrate in cylindrical coordinates? Let me know if you have questions.


问:
Expand $\frac{\Gamma\left(\frac{x}{2}\right)}{\Gamma\left(\frac{x-1}{2}\right)}$ at $x=\infty$
I used Wolfram Alpha for this problem and it gives me a "Puiseux expansion": $$\sqrt{x}-\frac{3 \sqrt{\frac{1}{x}}}{8}-\frac{7}{128}\left(\frac{1}{x}\right)^{3 / 2}-\frac{9\left(\frac{1}{x}\right)^{5 / 2}}{1024}+O\left(\left(\frac{1}{x}\right)^{3}\right)$$
which is exactly what I need.

My only question is how one would be able to obtain this expansion manually.

答:
Note that the expansion you gave is for $\Gamma(x)/\Gamma(x-1/2)$. Using the beta function and an appropriate change of integration variables, we obtain
\begin{align*}
&\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi  }}\frac{{\Gamma \left( {\frac{x}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {\frac{{x + 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi  }}B\left( {\frac{x}{2},\frac{1}{2}} \right) = \frac{{x - 1}}{{2\sqrt \pi  }}\int_0^1 {\frac{{t^{x/2 - 1} }}{{\sqrt {1 - t} }}dt}
\\ &
 = \frac{{x - 1}}{{\sqrt \pi  }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - 2s} } }}ds}  = \frac{{x - 1}}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - xs} s^{ - 1/2} \sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}} ds} .
\end{align*}
By the definition of the Nörlund numbers (cf. https://mathworld.wolfram.com/NorlundPolynomial.html),
$$
\sqrt {\frac{{2s}}{{1 - e^{ - 2s} }}}  = \sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}s^n }
$$
for $|s|<\pi$. Thus, by Watson's lemma (cf. http://dlmf.nist.gov/2.3.ii),
\begin{align*}
\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} & \sim \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\int_0^{ + \infty } {e^{ - xs} s^{n - 1/2} ds} }
\\ &
 = \frac{{x - 1}}{{\sqrt {2\pi } }}\sum\limits_{n = 0}^\infty  {( - 1)^n \frac{{2^n B_n^{(1/2)} }}{{n!}}\Gamma \left( {n + \frac{1}{2}} \right)\frac{1}{{x^{n + 1/2} }}}
\\ &
 = \sqrt {\frac{x}{2}}  + \sum\limits_{n = 0}^\infty  {\frac{{( - 1)^{n + 1} }}{{2\sqrt \pi  n!}}\Gamma \left( {n + \frac{1}{2}} \right)\left( {\frac{{2n + 1}}{{n + 1}}B_{n + 1}^{(1/2)}  + B_n^{(1/2)} } \right)\left(\frac{2}{{x}}\right)^{n+1/2}}
\end{align*}
as $x\to +\infty$. It is possible to simplify this result to
$$
\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} \sim \sqrt {\frac{x}{2}}  + \sum\limits_{n = 0}^\infty  {
   \binom{ 1/2}{n + 1}B_{n + 1}^{(3/2)} \left(\frac{2}{{x}}\right)^{n+1/2}} .
$$
Note that this is a divergent asymptotic expansion. See http://dlmf.nist.gov/5.11.iii for more general results and references.


问:
Harmonic Function Transformation Help
Consider the harmonic function $$u(x,y)=1-y+\frac{x}{x^2+y^2}$$ on the upper half plane $y>0$.

What is the corresponding harmonic function on the first quadrant $x>0$, $y>0$, under the transformation $z\mapsto z^2$?

Do I have to use Cauchy-Reimann equations to find the corresponding harmonic function? Any help would be appreciated. Thanks.

答:
If I understand correctly, all that you should do is plug in $z^2$ instead of $z$. No need for the Cauchy-Riemann equations.

First, write $u$ in terms of $z$:
$$
u(z)=1- \operatorname{Im} z + \operatorname{Re} \frac{1}{z}
$$
Then plug in $z^2$:
$$
u(z^2)=1- \operatorname{Im} z^2 + \operatorname{Re} \frac{1}{z^2}
$$
Optionally,  return to $z=x+iy$ notation:
$$
u(z^2)=1- \operatorname{Im} z^2 +  \frac{\operatorname{Re} z^2}{|z|^4}
=1-2xy+\frac{x^2-y^2}{(x^2+y^2)^2}
$$

Title: Harmonic Function Transformation Help


问:
A Vandermonde-like identity, new or existing?
From my effort of finding Vandermonde-like identities, I found out that if $n \le m-2$, then  $$\sum_{r=1}^{n+1} \frac{\binom{2r}{r}\binom{m+n-2r}{n+1-r}}{r+1}=\binom{m+n}{n}.$$
I am not sure if this is new or not. Any idea?

Note: $\binom{n}{r}=0$ if $n<r$ or $n<0$.

答:
Rewriting the LHS as:
$$
\sum_{r=0}^n \frac{1}{2+r}{2+2r \choose 1+r}{m+n-2-2r \choose n-r}
=\sum_{r=0}^n\frac{1}{1+r}{2+2r \choose r}{m+n-2-2r \choose n-r}
$$
one recognizes that this is a particular case ($a=2,b=2,c=m+n-2$) of the Rothe–Hagen identity:
$$
\sum_{k=0}^n\frac a{a+bk}\binom{a+bk}k\binom{c-bk}{n-k}=\binom{a+c}n,
$$
valid for any complex numbers $a,b,c$.


问:
Question about proper subgames in game theory.
Consider the following game presented as a tree.
As we can see there are $3$ players and the question is to find all N.E and all S.P.N.E. But I have a doubt about one thing. As we can see, if the first player selects $L$ then there are no actions for the second player. Am I right that in this game there is no $(L,L,L)$ strategy? Or if there is such strategy, then it's utility is (3,3,2)? If the last variant is valid, then how can we produce any proper subgame for this tree?

答:
Strategies in seqential games of perfect information are phrased a bit differently than in a simultaneous move game - after all, actions of the players moving later can condition on what happened before.

Hence, the stategies of the third player in such a sequential game are more appropriately described as follows: $L$ if 1 chose $L$, $R$ if 1 chose $R$ and 2 chose $L$.

Similarly, the strategy of the second player would then be described as: $L$ if 1 chose $R$. This makes clear that player 2 does not get to move if 1 chose $L$.

So a strategy profile in this sequential game would not be $(L,L,L)$, nor even $(L,L)$, but indeed $(L, L\textrm{ if ... and } R \textrm{ if ...}, L \textrm{ if ... and } R \textrm{ if ...})$ (for example).

This conditional phrasing of strategies is important when describing Nash equilibrium. It needs to be clear what would happen in another branch of the tree if that were chosen, hence strategies in the off-equilibrium paths need to be specified as well, even if the game never ends up there in equilibrium. But it is important to establish that going there is not a profitable deviation, i.e., is important to establish the equilibrium.

For example, $(L, R \textrm{ if 1 chose }R, L \textrm{ if 1 chose }L \textrm{ and anything otherwise})$ is a Nash equilibrium. That it is a NE depends very much on the strategy of player 2 who does not even get to move on the equilibrium path in this equilibrium, because the outcome is $L$ for 1 and $L$ for 3. So if 2's strategy was $L$ instead, then 1 might profitably deviate from $L$ to $R$, which shows that even players who do not get to move on the equilibrium path are still important for the equilibrium, because all players react to each other.


问:
how many monotonically increasing functions are in $\Omega = \{ \space f \mid f: A \rightarrow B\space \}$
Given $A=\{1,2,3..,n\}$,  $B=\{1,2,3,..,m\},$$\space \space $while $5<m,n$ $\space$ and $\space \Omega = \{ \space f \mid f: A \rightarrow B\space \}$

How many functions are monotonically increasing in $\Omega$ while $m>n$?

My answer is $m-n+1$.

Basic example is $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5,6\}$.
hence:
$$f_1(1) = 1, \space \space f_1(2) = 2, \space \space f_1(3) = 3, \space \space f_1(4) = 4, \space \space f_1(5) = 5$$
$$f_2(1) = 2, \space \space f_2(2) = 3, \space \space f_2(3) = 4, \space \space f_2(4) = 5, \space \space f_2(5) = 6$$

So we get 2 functions, and while the diffrence between $m$ and $n$ getting bigger, the result's getting bigger.
Am I missing somthing?

答:
In question, number of monotonically increasing functions from $A$ to $B$ is asked, NOT that of "strictly increasing".

The question can be thought as number of ways of slicing $A$ to $m$ pieces while keeping order of its elements. So we have to make $m-1$ cuts to cut $A$ to $m$ pieces such that if $a$ is in $kth$ slice, $f(a)= k$. Imagine slices as follows:

```<code>_ /_ /_ /_ ... / _
</code>```

Here there are $m$ slots (_) and $m-1$ slicers (/). We are after the number of ways to order this slicers and the elements of A. For example suppose $m = 4$ and  $A = \{1,2,3\}$ One possible slicing would be:

```<code>1,2/ _ /_ / 3
</code>```

which means: $f(1)=f(2)=1, f(3)=4$

So we have $n$ elements from $A$ and $m-1$ slicers to order. To place all these, now imagine we have $n+m-1$ slots. If we choose $m-1$ slots among these to place the slicers, there will be one way to place elements from $A$, since we must keep the order among them.

So the number of ways to place all n+m-1 slicers and elements of A is  ${n+m-1\choose n-1}$

Body:
In question, number of monotonically increasing functions from $A$ to $B$ is asked, NOT that of "strictly increasing".

The question can be thought as number of ways of slicing $A$ to $m$ pieces while keeping order of its elements. So we have to make $m-1$ cuts to cut $A$ to $m$ pieces such that if $a$ is in $kth$ slice, $f(a)= k$. Imagine slices as follows:

_ /_ /_ /_ ... / _

Here there are $m$ slots (_) and $m-1$ slicers (/). We are after the number of ways to order this slicers and the elements of A. For example suppose $m = 4$ and  $A = \{1,2,3\}$ One possible slicing would be:

1,2/ _ /_ / 3

which means: $f(1)=f(2)=1, f(3)=4$

So we have $n$ elements from $A$ and $m-1$ slicers to order. To place all these, now imagine we have $n+m-1$ slots. If we choose $m-1$ slots among these to place the slicers, there will be one way to place elements from $A$, since we must keep the order among them.

So the number of ways to place all n+m-1 slicers and elements of A is  ${n+m-1\choose n-1}$


问:
Tangent integration $\int \tan (2x)\tan (3x)\mathrm{d}x$
I know how to calculate $\displaystyle\int \tan (x) \tan (2x)\tan (3x)\mathrm{d}x$ with the tangent sum formula
$\tan 3x\tan 2x\tan x =\tan3x -\tan 2x -\tan x$. But I don't know how I can solve

$$\displaystyle\int \tan (2x)\tan (3x)\mathrm{d}x$$

I tried u sub, trig sub, integration by parts. But I couldn't make it.

Thanks.

答: Bicohe's rules tell us to evaluate $\int\tan x\tan 2x\tan 3x\mathrm dx,\,\int\tan 2x\tan 3x\mathrm dx$ with the respective substitutions $c=\cos 2x,\,t=\tan x$. Since you can do one, I'll do the other:$$\begin{align}\int\tfrac{4t^2(3-t^2)}{(1-t^2)(1-3t^2)(1+t^2)}dt&=\int\left(\tfrac{4}{1-3t^2}-\tfrac{2}{1-t^2}-\tfrac{2}{1+t^2}\right)dt\\&=\tfrac{4}{\sqrt{3}}\operatorname{artanh}(t\sqrt{3})-2\operatorname{artanh}t-2\arctan t+C.\end{align}$$


问:
How to prove that a set of polynomials spans a vector space?
Let $V_n$ denote the vector space of polynomials of degree at most $n$ over a field $F$, and suppose $\{f_0, f_1, \cdots, f_{n-1}, f_n\}$ is a set of polynomials such that for all $j$ between $0$ and $n$, we have that $f_j$ has degree $j$. Show that the $\{f_j\}$ form a basis for $V_n$.

Let $B = \{f_0, f_1, \cdots, f_{n-1}, f_n\}$. We want to show that $B$ is linearly independent and spans $V_n$.
For linear independence, suppose $$\sum_{i=0}^{n}a_if_i = \sum_{i=0}^{n}a_i(\sum_{j=0}^{i}a_{ij}x^j) = \sum_{i=0}^{n}(\sum_{j=0}^{i}a_ia_{ij}x^j) = \sum_{j=0}^{n}(\sum_{i=j}^{n}a_ia_{ij})x^j = 0$$ Then, $\sum_{i=j}^{n}a_ia_{ij} = 0$ for all $j$. Specifically, for $j=n$, $a_na_{nn} = 0$. $a_{nn}$ cannot be 0 since $x^j$ for $j=n$ must be of degree $n$, and so $a_n = 0$. Thus, we can rewrite as $$\sum_{i=0}^{n}a_if_i = \sum_{i=0}^{n-1}a_if_i +a_{n}f_n = \sum_{i=0}^{n-1}a_if_i + 0 =   \sum_{j=0}^{n-1}(\sum_{i=j}^{n-1}a_ia_{ij})x^j = 0$$ We do the same process for $j = n-1$ to conclude that $a_{n-1}=0$ until $j=0$. Thus, $a_i=0$ for all $i$, and so $B$ is linear independent.

For spanning ...

EDIT: Attached below is answer key's proof, though I'm a bit confused by it.

答:
Since $\dim V_n=n+1$ and since $B$ has $n+1$ linearly independent vectors, $B$ spans $V_n$.

This has nothing to do with your specific situation. If $V$ is a $n$-dimensional vector space over some field $F$ and if $S\subset V$ is a set with $n$ elements, then the following assertions are equivalent:

$\operatorname{span}(S)=V$;

$S$ is linearly independent;

$S$ is a basis of $V$.


问:
Why are there problems when interpolating $f(x)=\arctan(10x)$?
Given $f(x)=\arctan(10x)$, there would be a problem when we interpolate it by using Lagrange's method. This would have something to do with the derivatives of $f(x)$. I plotted some derivatives of $f$ but I did not come up with an answer. Can anybody tell me what goes wrong if we interpolate $arctan(10x)$ and why this comes from the derivatives of $f(x)$. thanks :-)

Oh.. second question: I also would like to know why this problem does not occur when using cubic Hermite interpolation... In addition, I want to know why this problem does not occur when using cubic spline interpolation ? I can not figure this out

答:
If you do the interpolation intelligently (i.e. not at equally-spaced points) then you can avoid the Runge phenomenon, and you can approximate any reasonably smooth function very well with a polynomial on a bounded interval.

An excellent tool for constructing polynomial approximations is the Chebfun system. The Chebfun folks love to rant about how Lagrange interpolation has been criticized unfairly by people who don't know what they're talking about (including the authors of some textbooks on numerical analysis). The rants are fun to read.

If you want a polynomial approximation to be good on the entire real line, rather than a bounded interval, then you have a problem that's much worse than the Runge phenomenon. The value and the derivatives of any polynomial $p(x)$ will tend to $\pm\infty$ as $x \to \infty$ (unless it's linear). And for your function $\arctan(10x)$, the function value tends to $\pi/2$ and the derivative approaches zero as $x \to \infty$, so there is a dramatic mismatch that's difficult to handle. This is the problem you referred to in your question, I suppose.

The way forward is to use splines or rational approximations.


答:
I think are you on the right track to evaluating this directly, so I will instead show you a fast way to do the problem. The unit normal vector for a sphere is given by $$\frac{1}{a}(x,y,z)$$ You can either recognize this formula straight away or derive it by taking the gradient of the sphere equation and normalizing it. With this we can rewrite this scalar surface integral as a vector one:

$$\iint_SzdS=\iint_S(0,0,a)\cdot ndS$$

Because the divergence of this vector field is $0$, this integral is equivalent to the intgeral of the same vector field over any other surface that shares the same boundary and orientation as this one by Stokes' theorem. So let's change the integral to be over a disk in $xy$ plane oriented up

$$\iint_{z=0\cap x^2+y^2\leq a^2} (0,0,a)\cdot (0,0,1)dS = \pi a^3$$

Title: Evaluate $\int \int_S z \space dS$ where $S$ is the surface of a hemisphere.

Body:
I am trying to evaluate a surface integral

$$\int \int_S z \space dS$$

where

$S$ is the surface

$$x^2+y^2+z^2=a^2, \space z \in [0,a]$$

I am a little confused on what to do here, because although my first instinct was to use spherical coordinates I only have two integrands.  If I try to use $x=a\cos(\theta), \space y=a\sin(\theta), z=z$ then the equation does not work out.

It seems as though I am missing something very basic but I cannot get my mind around it.

May I have some assistance, please?


问:
Does $\sin(x)$ and $\cos(x)$ have representations as hypergeometric series?
Through wolfram and wiki, I've learnt that these elementary functions have a representation as hypergeometric series:

$$_2F_1(\color{red}{1,1;2;-x})=\ln(x+1)$$

$$_2F_1(\color{red}{\color{red}{\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2})}=\arcsin(x)$$

$$_2F_1(\color{red}{1,1;1;x})= \dfrac{1}{1-x}$$

I have never seen the hypergeometric series representation for $\sin(x)$ and $\cos(x)$. I wonder if trigonometric and hyperbolic functions have hypergeometric series' representation as those three functions? Do you know any table from mathematical handbook or references that list these special values of the Gauss hypergeometric function?

答:
Any sufficiently well-behaved function has a unique power series representation at a given point (such as the origin). The theory about that is known as Complex Variable theory and is one of the spectacular jewels of mathematics. I strongly recommend taking a basic course in it.

Unfortunately, maths is such a vast field that the hypergeometric and other "classical" functions have got squeezed out of most undergraduate math courses. There are relatively few modern textbooks on them. The old classic is Whittaker and Watson (A course in Modern Analysis - but it is far from modern in its flavour).

If you really want to look up details the NIST Handbook of Mathematical Functions is a great (but large and fairly expensive) reference book. Chapter 15, p386 gives plenty of examples of elementary functions that can be written as special cases of the hypergeometric function.

But not all "elementary" functions can be written as hypergeometric functions (and anyway "elementary" is not a preciselyy defined term). To go into this properly you have to look at the associated differential equations. But I strongly recommend against racing ahead until you have done a little more of the undergraduate syllabus.


问:
The quotient group of the Heisenberg group and its center is isomorphic to $(\mathbb{R}^2, +)$.

Let $$ H = \left\{ \;\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 &
 1 \end{bmatrix} : x,y,z \in \mathbb{R}\right\}, \qquad Z(H) = \left\{
 \begin{bmatrix} 1 & 0 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} :
y \in \mathbb{R}\right\}, $$ then show that $H/Z(H) \cong (\mathbb{R}^2, +)$.

Proof: First I show that $Z(H) \cong (\mathbb{R}, +)$ using the isomorphic map $\phi(h) = y$ for $h \in Z(H)_y$. Then I show that $H \cong (\mathbb{R}^3, +)$ using the isomorphic map $\Phi(h) = (x, y, z)$ for $h \in H_{(x,y,z)}$. So we have
$$
H/Z(H) \cong (\mathbb{R}^3, +)/(\mathbb{R}, +).
$$
We can construct a homomorphic map $\varphi\left((x,y,z)\right) = x + y + z$ with $$\text{ker} \varphi = \{ x,y \in \mathbb{R}: z = -x -y\}$$ which is clearly $(\mathbb{R}^2, +)$ so by the first isomorphism theorem we have
$$
H/Z(H) \cong (\mathbb{R}^3, +)/(\mathbb{R}, +) \cong (\mathbb{R}^2, +)
$$
finishing the proof. Does this sort of iterative isomorphic map proof work? These concepts are new to me so I'm unsure.

答:
Here is the correct proof:

Proof: For $h \in H$ with
$$
h = \begin{bmatrix}
1 & x & y \\
0 & 1 & z\\
0 & 0 & 1
\end{bmatrix}
$$
we have $\phi : H \to (\mathbb{R}^2, +)$ with $\phi(h) = (x,z)$. Clearly $\phi$ is homomorphic as $\phi(h_1 h_2) = \phi(h_1) + \phi(h_2)$. Furthermore,
$$
\text{ker} \phi = \{ (x,z) : x = z = 0 \} = Z(H).
$$
Lastly we see that $\text{im} \phi = (\mathbb{R}^2, +)$ so by the first isomorphism theorem
$$
H/Z(H) \cong (\mathbb{R}^2, +).
$$


问:
Computing problems and generating functions

Q. Find the generating function for the sequence $\{a_n\}$, where $a_n$ is the number of solutions to the equation: $a+b+c=k$ when $a, b, c$ are non-negative integers such that $a\ge2, 0\le b\le3$ and $2\le c\le5.$

As far as I have understood, here, $a_n$ must be the coefficient of $x^k$ in the expansion of $(x^2+x^3+x^4+\cdots)(1+x+x^2+x^3)(x^2+x^3+x^4+x^5).$

This follows because we obtain the term $x^k$ in the product by picking a term $x^a$ in the first sum, $x^b$ in the second and $x^c$ in the third, where the exponents $a, b, c$ satisfy the equation $a+b+c=k$ along with the given constraints.

But since the positive constant $k$ is not specified here so I'm not aware of how to obtain the desired coefficient in the above product. Does there exist any closed form for the same? Please suggest.. Give me some insights. It would be so helpful. Thanks in advance...

答:
Let's do a few simplifications:

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

$1 + x + x^{2} + x^{3} = \dfrac{1-x^{4}}{1-x}$ (by the sum of a geometric series)

$x^{2} + x^{3} + x^{4} + x^{5} = x^{2}(1 + x + x^{2} + x^{3}) = x^{2} \cdot \dfrac{1-x^{4}}{1-x}$

So really, you have the expression:

$$x^{4} \cdot (1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}}$$

It suffices to look for the coefficient of $x^{k-4}$ in the expansion of:

$$(1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}}$$

Now using the Binomial and (Derivative of a) Geometric Series Expansions, we have that:
\begin{align*}
(1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}} &= \left( 1 -2x^{4} + x^{8} \right) \cdot \left( \sum_{i=0}^{\infty} \binom{i+2-1}{i} x^{i} \right)
\end{align*}

By expanding things out, we have that the coefficient of $x^{k-4}$ is:
\begin{align*}
\binom{(k-4)+2-1}{k-4} - 2\binom{(k-8)+2-1}{k-8} + \binom{(k-12) + 2 - 1}{k-8}
\end{align*}


问:
Express $169$ as the sum of $1,2,3,4,5$ non-zero squares
I'm trying to solve the following exercise.

Show that $169$ can be expressed as a sum of $1,2,3,4,5$ non-zero squares, and deduce that any $n \ge 169$ is the sum of five non-zero squares.

The latter part of the exercise is clear from Lagrange's four squares theorem, namely that every integer can be expressed as the sum of four integers. (There's also an answer Integers which are the sum of non-zero squares, detailing the exact steps.) I'm having trouble with the first part of the exercise.

For expressing $169$ as a sum of one square, we have $169=13^2$. Now, $13 = 2^2 + 3^2$, so we can use the identity $(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2$ to get $13^2 = (6+6)^2 + (2^2-3^2)^2$. But what about the rest? Is there any way of obtaining $169$ as a sum of $3,4,5$ non-zero squares, except by brute force, perhaps? (Brute force answers are also welcome, especially if they employ intuition or clever tricks! I will upvote them too)

There is one numerical answer to this in the link above, but I'm looking for intuition or a systematic way of doing things, not just a number solution expressing $169$ as a sum of non-zero squares.

Body:
I posted this in comments, so I might as well document the results. If one has access to Mathematica, then Select[DeleteDuplicates[Sort/@Tuples[Range[13],n]],#.#==169&] lists all ways to write $169$ as a sum of n squares. (This can probably be improved but it works fine for the present purpose.) The results for $n=1,2,3$ are just those found above:
$$169=13^2=5^2+12^2=3^2+4^2+12^2$$
For $n=4$ and $n=5$ we respectively obtain {{1,2,8,10},{2,4,7,10},{4,4,4,11},{4,5,8,8},{4,6,6,9}} and {{1,2,2,4,12},{1,2,6,8,8},{1,4,4,6,10},{2,2,2,6,11},{2,2,4,8,9},{2,2,5,6,10},{2,4,6,7,8},{3,4,4,8,8},{5,6,6,6,6}}. These give the representations
\begin{align}
169
&=1^2+2^2+8^2+10^2\\
&=2^2+4^2+7^2+10^2\\
&=4^2+4^2+4^2+11^2\\
&=4^2+6^2+6^2+9^2\\
&=1^2+2^2+2^2+4^2+12^2\\
&=1^2+2^2+6^2+8^2+8^2\\
&=1^2+4^2+4^2+6^2+10^2\\
&=2^2+2^2+2^2+6^2+11^2\\
&=2^2+2^2+4^2+8^2+9^2\\
&=2^2+2^2+5^2+6^2+10^2\\
&=2^2+4^2+6^2+7^2+8^2\\
&=3^2+4^2+4^2+8^2+8^2\\
&=5^2+6^2+6^2+6^2+6^2.\\
\end{align}
The reader may note the prevalence of solutions with repeated squares. If we require distinct squares, then the code can be simplified to Select[Subsets[Range[13]],#.#==1692&] and produces all such decompositions (regardless of number of terms). In this manner we obtain
{{13},{5,12},{3,4,12},{1,2,8,10},{2,4,7,10},{2,4,6,7,8},{1,2,3,5,7,9}}, or

\begin{align}
169
&=13^2\\
&=5^2+12^2\\
&=3^2+4^2+12^2\\
&=1^2+2^2+8^2+10^2\\
&=2^2+4^2+7^2+10^2\\
&=2^2+4^2+6^2+7^2+8^2\\
&=1^2+2^2+3^2+5^2+7^2+9^2.
\end{align}
Note that there are two decompositions of $169$ into four distinct squares and we've also found a decomposition of $169$ into six distinct squares. (If we allow repeated squares, then the code above produces a total of 20 such decompositions into 6 squares.)


答:
If you decide to go with spherical coordinates, parametrize $S$ by $F: U \to \mathbb{R}^{3}$, where $$
F(\phi, \theta) = \left(\begin{array}{c}
a\cos\phi \cos\theta\\
a\cos\phi \sin\theta\\
a\sin\phi
\end{array}
\right)
$$
for an appropriate $U \subset \mathbb{R}^{2}$ (which I leave to you to determine). Compute $F_\phi := \frac{\partial F}{\partial \phi}$ and $F_\theta := \frac{\partial F}{\partial \theta}$ as well as
$$
d\sigma = \left \|F_\phi \times F_{\theta} \right\| d\phi \, d\theta,
$$
which is known as surface element. Then,
$$
\int_{S} z \, d\sigma  = \int_{U}a \sin \phi \left \|F_\phi \times F_{\theta} \right\| d\phi \, d\theta
$$
by definition of the surface integral.


问:
Parametric equation of a cone
I usually use the following parametric equation to find the surface area of a regular cone $z=\sqrt{x^2+y^2}$:
$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$z=r$$

And make $0\leq r \leq 2\pi$, $0 \leq \theta \leq 2\pi$.

I've now have a cone $z=\sqrt{2x^2+2y^2}$ and I think the parametric equation I normally use won't work anymore. Which would be a more suitable one in this case? Is there any generic parametric equation for cones, because one of the form $z=\sqrt{4x^2+y^2}$ would also have a different one.

答:
Correct formula (from: https://mathworld.wolfram.com/Cone.html) in Mathematica Language is:

```<code>ParametricPlot3D[{r*(h - u)/h Cos[t], r*(h - u)/h Sin[t], u}, {u, 0,
  h}, {t, 0, 2 Pi}]
</code>```

where:
$h$ is height of cone, $r$ is radius of cone (constants);
$t, u$ are parameters.

Title: Parametric equation of a cone

Body:
Correct formula (from: https://mathworld.wolfram.com/Cone.html) in Mathematica Language is:

ParametricPlot3D[{r*(h - u)/h Cos[t], r*(h - u)/h Sin[t], u}, {u, 0,
  h}, {t, 0, 2 Pi}]

where:
$h$ is height of cone, $r$ is radius of cone (constants);
$t, u$ are parameters.


问:
Inverse of the tangent map for tangent bundles
(Physicist here trying to understand some more involved differential geometry).
In the context of differential geometry we define the tangent bundle as

$ TM = \bigsqcup_{p \in M} T_p(M) $ where a general element is given by the cartesian product $(p,v_P)$ for $p \in M$ and $v_p \in T_p(M)$.

We have the natural map $\pi: TM \rightarrow M$, explicitly $\pi(p,v_p)=p$, called the tangent map.

However, we then define the inverse tangent map as $\pi^{-1}: M \rightarrow TM$ but I dont understand how we can do this. The tangent map is surjective and so its inverse does not uniquely map onto an element in the tangent bundle.

答:
As Thorgott said we cannot have a true inverse function from $M$ to $TM$ because $\pi : TM \to M$ is not injective.

However, vector fields $X$ (i.e. sections of $TM$) are a "half inverse" to $\pi$ in the sense that they give you a map from $M$ back up to $TM$. Specifically, for each $p \in M$ the vector field $X$ gives a vector $X_p \in T_pM \subset TM$. Formally, vector fields satisfy the equation $\pi \circ X = \text{id}_M$, so they are right inverses to $\pi$.


问:
Where is the mistake in a point closest to cross section of two planes and a line problem
The problem:

Find the point that lies on the line that is the intersection of two planes:

plane one: $$3x-2y+z = 2$$

plane two: $$x+y+2z = 4$$

The point also  needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$

My try:
First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.

What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$

I got: $z = 7y +2$ and $x = \frac{-5y}{3}$

Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$

So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$

Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$

Then I wanted to position the plane($O$...origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$

So the equation of this plane is: $4z = 4*\frac{1}{2}=2$

I got $z_0$ form the last coordinate from the point.

Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.

The correct solution is: $T(1,1,1)$

Where did I go wrong?

答:
I would have done this a slightly different way.  The two planes are
3x−2y+z=2 and x+y+2z=4.  From the equation of the second plane, x= 4- y- 2z.  Replace x in the first equation by that:  3(4- y- 2z)- 2y+ z= 12- 3y- 6z- 2y+ z= 12- 5y- 5z= 2 so 5y+ 5z= 10 and y+ z= 2 so z= 2- y and x= 4- y- 2(2- y)= 4- y- 4+ 2y= y.

Taking y= t, parametric equations for the line of intersection are x= t, y= t, z= 2- at.  Any point on that line of intersection is of the form (x, y, z)= (t, t, 2- t).

The distance, squared, from A(1, -1, -1) to (t, t, 2- t) is (t- 1)^2+ (t+ 1)^2+ (3- t)^2.

The distance, squared, from B(1, -1, 3) to (t, t, (2- t)) is (t- 1)^2+ (t+ 1)^2+ (1+ t)^2.

For a point on the line equally distant from A an(td B, we must have (t- 1)^2+ (t+ 1)^2+ (3- t)^2= (t- 1)^2+ (t+ 1)^2+ (1+ t)^2.

Those first two squares on the left cancel the first two on the right leaving (3- t)^2= (1+ t)^2.  Taking the square root of both sides 3- t= 1+ t so 2t= 2 and t= 1.  Taking 3- t= -(1+ t)= -1- t does not give a solution.

So the point on the line equidistant from A and B is (t, t, 2- t)= (1, 1, 2- 1)= (1, 1, 1).

Check:  Is (1, 1, 1) on the plane 3x- 2y+ z= 2?  Yes,3- 2+ 1= 2.
Is (1, 1, 1) on the plane x+ 2y+ z= 4?  Yes 1+ 2+ 1= 4.

The distance, squared, from (1, 1, 1) to A(1,-1,- 1) is (1- 1)^2+ (1-(-1))^2+ (1- (-1))^2= 0+ 4+ 4= 8.
The distance, squared, from (1, 1, 1) to B(1, -1, 3) is (1- 1)^2+ (1- (-1))^2+ (1- 3)^2= 0+ 4+ 4= 8.

Yes, (1, 1, 1) is equally distant from A and B.


问:
What is the background to inequality $|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$?
$\alpha=(x_1,x_2,\dotsc,x_n)$, $\beta=(y_1,y_2,\dotsc,y_n)\in\Bbb R^n$, $\alpha \cdot\beta=x_1y_1+x_2y_2+\dotsb+x_ny_n$.

$|\alpha|=\sqrt{x_1^2+x_2^2+\dotsb+x_n^2} $, $|\beta|=\sqrt{y_1^2+y_2^2+\dotsb+y_n^2}, $

Show that : there exists a constant $L\gt 0$ such that
$$|e^{i \alpha\cdot\beta}-1|\leq L |\alpha||\beta|$$
for all $\alpha,\beta$

I don't know where to start

The question produces background . What is the background?

答:
$e^{\,i\, \alpha \cdot \beta}$ is always less than $2$, whatever is $\alpha \cdot \beta=|\alpha| \, |\beta|\, \cos \theta$.

So we have to consider the case in which $\alpha \cdot \beta$ is small.

Since the arc $\alpha \cdot \beta$ is comprised between $\pm |\alpha|\,|\beta|$, then the chord $|e^{\,i\, \alpha \cdot \beta}|$ is not greater that the chord $|e^{|\alpha| \, |\beta|}|$, which in turn is less then the arc $|\alpha| \, |\beta|$, which remains true also when the arc equals $\pi$.

Thus $L=1$


问:
Is the conjecture true? $3^n - 2^m = 1$ has infinitely many solutions, where n, m are natural numbers.
Is the conjecture true?

$3^n - 2^m = 1 $ has infinitely many solutions, where n, m are natural numbers.

More generally,
$P^n-Q^m=1$ has infinitely many solution for n, m ϵ {1,2,3,…} where P is odd, Q is even, P and Q have no common factors.

$P^n-Q^m=2$ has infinitely many solution for n, m ϵ {1,2,3,…} where P, Q  are both odd, P and Q have no common factors.

This is my first post - I joined this community to look into this.

答:
Mihăilescu's theorem, before 2002 known as Catalan's conjecture, states that:

The only natural numbers $m,n>1$ and $P,Q>0$ satisfying $$P^n-Q^m=1,$$are $(m,n,P,Q)=(2,3,3,2)$.


问:
Estimating the length of a curve with the Cauchy-Crofton formula
This is a comment from Do Carmo's Differential Geometry of Curves and Surfaces:

Equation $\iint n dp d\theta=2l$ (where $n$ is thenumber of intersection points of a curve $C$ with all straight lines in the plane that meet $C$) can also be used to obtain an efficient way of estimating
lengths of curves. Indeed, a good approximation for the integral $\iint n dp d\theta$
is given as follows. Consider a family of parallel straight lines such that
two consecutive lines are at a distance $r$. Rotate this family by angles of
$\pi/4, 2\pi/4, 3\pi/4$ in order to obtain four families of straight lines. Let $n$ be the
number of intersection points of a curve $C$ with all these lines. Then $\frac{1}{2}{nr} \frac{\pi}{4}$ is an approximation to the integral $ \iint n dp d\theta=\hbox{length of }C$.

My question: How can I prove it and if it is true that $\lim_{r \to 0}\frac{1}{2}n(r)r\frac{\pi}{4}=\hbox{length of }C$?

I couldn't understand how the author arrived at the formula $\frac{1}{2}{nr} \frac{\pi}{4}$.

答:
In the ultimate reference on integral geometry, Santaló's Integral Geometry and Geometric Probability, he gives this formula, originally due to Steinhaus (one reference is "Length, shape and area," Colloq. Math. 3 (1954), 1-13), on pp. 31-32. If we turning the grid through multiples of $\pi/m$, we obtain
$$\ell\approx \frac{N\pi r}{2m}.$$
If we let $r\to 0$ and $m\to\infty$, then we can suspect with some hypotheses that this is calculating  the integral $\dfrac12\iint n\,dp\,d\theta$.

Here's a reference for the error estimates: P.A.P. Moran, "Measuring the length of a curve," Biometrika 53 (1966), 359-364.


问:
Sums of roots of unity
If the integral linear combination of some $n$th roots of unity has magnitude 1, does this necessarily imply that this linear combination is some root of unity as well? More precisely,

Let $\zeta_1, \ldots \zeta_k$ be $n$th roots of unity. If
  $$|\sum_{i=1}^k n_i \zeta_i| = 1,$$
  where $n_i \in \mathbb{Z}$, does this imply that $\sum_{i=1}^k n_i \zeta_i$ is an $n$th root of unity? What about if the $n_i$ are Gaussian integers?

答:
Here's an alternative proof. It seems more elementary to me, but that might just be because I don't see the connections to the results used in David's proof and am reinventing the wheel. If there are such connections, I'd appreciate if someone could point them out in the comments. My proof will use David's result that $|\sum a_k \zeta^{k}|=1$ implies $|\sum a_k \zeta^{kr}|=1$ for $r$ relatively prime to $n$; the only "sophisticated" result he used for that was the irreducibility of the cyclotomic polynomials (and implicitly unique factorization).

First, note that the question has changed and the answer to the new question is "no". As I wrote in the comments, two cube roots of unity sum to a square root of unity, so the restriction to $n$-th roots of unity makes the statement false.

The true statement in the original question was "If the sum of some $n$th roots of unity has magnitude $1$, this implies that this sum is some root of unity as well" (where the same $n$-th root of unity may occur multiple times in the sum, to make this equivalent to the integer linear combination formulation). This is equivalent to the same statement with "$n$-th" deleted, since for every set of roots of unity there is a common denominator $n$ for which they are all $n$-th roots.

Now for the proof, consider the free vector space on the set of the $n$-th roots of unity. Any linear combination of $n$-th roots of unity acts on this vector space by multiplication. The corresponding matrix with respect to the canonical basis is a circulant matrix with the first column given by the coefficients $a_i$ of the linear combination. The eigenvectors of this circulant matrix are the Fourier modes $v_r=(1,\omega^r,\omega^{2r},\dotsc$), with $\omega=\mathrm e^{2\pi\mathrm i/n}$, and the corresponding eigenvalues are the Fourier transforms of the coefficients, $\sum a_k \omega^{-kr}$.

Now consider the natural map $f$ from this vector space to $\mathbb C$ sending each vector to the corresponding linear combination of the roots of unity. All but one of the Fourier modes are in the kernel of this map. Only the $r=-1$ mode isn't, and the corresponding eigenvalue is our number $\alpha=\sum a_k \omega^k$.

Now let's start with the vector $(1,0,\dotsc,0)$, corresponding to unity, and successively multiply by the circulant matrix. This initial vector has equal components $1/n$ of each Fourier mode. Consider what happens if we add the components corresponding to the primitive roots for some divisor $d\mid n$ together. If $d\neq n$, all of these are in the kernel of $f$. The entries in their sum are Ramanujan's sums, which are integers. Thus, by dropping these components, we don't change the complex number that the vector maps to, and we don't change the fact that all entries in the vector are integer multiples of $1/n$.

Now all that's left is the sum of the components corresponding to the primitive $n$-th roots. We know that the eigenvalue for $r=-1$ has magnitude $1$, and by David's result that implies that all the others also have magnitude $1$. So we have a vector whose entries are integer multiples of $1/n$ and we keep multiplying it by an integer matrix whose eigenvalues (as far as this vector is concerned) have magnitude $1$. It follows that there are only finitely many values this vector can take, so it has to eventually get back to where it started, which implies that $\alpha$ is a root of unity.

Title: Sums of roots of unity


问:
Intuition behind picking group actions and Sylow
A common strategy in group theory for proving results/solving problems is to find a clever group action. You take the group you are interested in (or perhaps a subgroup), and find some special set that your group can act on, usually by left multiplication or conjugation. Somehow studying this action makes the result you want clearer to see.

The most obvious example of this is with proving the Sylow theorems.

To prove the First Sylow theorem (that a $p$-subgroup exists) you can let $G$ act by left multiplication on all subsets of $G$ that have size $p^n$.

To prove the Second Sylow theorem (that Sylow $p$-subgroups are conjugate) you let $Q$ be any $p$-subgroup, $P$ a Sylow $p$-subgroup, and let $Q$ act on $G/P$ by left multiplication.

To prove the Third Sylow theorem (that the number of Sylow $p$-subgroups is $1\pmod p$ and divides $|G|$) you can consider both $G$ and some Sylow $p$-subgroup $P$ acting on $\operatorname{Syl}_p(G)$ by conjugation.

My question: is there any intuitive or natural way to come up with these group actions? To me, all three proofs seem magical -- if someone had told me which group action to consider, I could probably have completed the proofs myself, but I would never have come up with the appropriate action myself.

More generally, are there any ways to "see" which group action might help for solving a specific problem?

答:
The Sylow Theorems definitely have a lot of non-obvious ideas, and I think expecting someone to come up with these ideas on their own is not a reasonable ask. With that said, here are some pieces of intuition I have.

The Sylow Theorems are inherently combinatorial. It is natural to ask for an underlying combinatorial object (or collection of such objects) upon which the group should act. Let $n := p^{\alpha}m$, where $p$ does not divide $m$. The Lucas Congruence tells us that $p$ does not divide $\binom{p^{\alpha}m}{p^{\alpha}}$. The binomial coefficient tells us that it might be worth looking at the $p^{\alpha}$ subsets of the group. In terms of the action, left multiplication and conjugation are the natural things to try. The left action is a bit easier to analyze; so if I hadn't seen the proof before, I'd start with that.

When analyzing a group action, the orbits and stabilizers tell us a lot. Once we have found that the stabilizer of the above action is a Sylow $p$-subgroup, it is very natural to ask whether we can get all of the Sylow $p$-subgroups from the stabilizer. This would actually be very nice if we could do so; as otherwise, we have a lot more structural and combinatorial analysis to do, which can be messy. Note that conjugation gives us automorphisms. So the conjugates of the stabilizer are also Sylow $p$-subgroups, but we don't know if we have enumerated all the Sylow $p$-subgroups.

To get that the Sylow $p$-subgroups are conjugate, we have a few observations. First, the $p$-group fixed point theorem that if a $p$-group $P$ acts on a finite set $X$, where $p$ does not divide $|X|$, then the action has a fixed point. We use this fact, together with the intuition that every $p$-subgroup of $G$ should be a subgroup of a Sylow $p$-subgroup to get the conjugation action. Precisely, if $P$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $|G/P|$. Let $Q$ be a $p$-subgroup of $G$. We let $Q$ act on $G/P$ by left multiplication. The conjugacy condition comes from analyzing the fixed points. There are technicalities to this in the proof, but these are the main ideas (at least, to me).

The condition that $n_{p} \equiv 1 \pmod{p}$ and $n_{p}$ divides $m$ falls out from the Orbit-Stabilizer Theorem. Once we have that $G$ acts transitively on $\text{Syl}_{p}(G)$ by conjugation, we have that for $P \in \text{Syl}_{p}(G)$, $\text{Stab}(P) = N_{G}(P)$.


问:
Eliminating $x$, $y$, $z$ from $\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$ and $ax+by+cz=0$
Here is a Math Tripos problem I cannot solve.

Eliminate $x$, $y$, $z$ from the equations
$$\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$ and
$$ax+by+cz=0$$

I dont know if there is a general (practical) method for such problem. I think you just manipulate the equations to eliminate $x,y,z$ but have been unable to find the path in this question.

The answer is $a^3+b^3+c^3=a^2(b+c)+b^2(a+c)+c^2(a+b)$ if that yields any insight.

答:
Let's denote $t$ be equal to

$$t = \frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$

We have:
\begin{align}
t &=\frac{(y^2-yx-yz)+(z^2-zx-zy)-(x^2-xy-xz)}{b+c-a} \\
  &=\frac{(y^2-2yz+z^2)-x^2}{b+c-a} \\
 &=\frac{(y-z)^2-x^2}{b+c-a} \\
  &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \\
\end{align}
Hence,
\begin{align}
t^2 &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \times \frac{x^2-xy-xz}{a} \\
    &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \times \frac{-x(y+z-x)}{a} \\
&=(x+y-z)(y+z-x)(z+x-y) \times \frac{x}{a(b+c-a)} \\
\end{align}

We have then
$$\frac{a(b+c-a)}{x} = \frac{b(c+a-b)}{y} = \frac{c(a+b-c)}{z}  = u $$
$$\left(\text{both equal to  } u =\frac{(x+y-z)(y+z-x)(z+x-y)}{t^2} \right)$$

We can deduce that
$$u = \frac{a^2(b+c-a)}{ax} = \frac{b^2(c+a-b)}{by} = \frac{c^2(a+b-c)}{cz} $$
Hence:
$$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c) = u(ax+by+cz) = 0$$
(because $ax +by +cz = 0$)

Finally, we can conclude that
$$a^2(b+c)+b^2(c+a)+c^2(a+b) = a^3 + b^3+c^3$$


问:
Unique symmetric covariant $k$-tensor satisfying $(\operatorname{Sym} T)(A,...,A)=T(A,...,A)$ for all $A \in V$
Let $T$ be a covariant $k$-tensor on a finite dimensional vector space $V$. I want to prove that the symmetrization of $T$ is the unique symmetric $k$-tensor satisfying the following condition:

$(\operatorname{Sym} T)(A,\ldots,A)=T(A,\ldots,A)$ for all $A \in V$.

Definition. Symmetrization of $T$ is defined as $$(\operatorname{Sym}T)(A_1,\ldots,A_k)=\frac{1}{k!} \sum_{\sigma \in S_k} T(A_{\sigma(1)},\ldots,A_{\sigma(k)})$$ where $S_k$ is the symmetric group on $k$ letters.

I assumed that there exists another symmetric $k$-tensor $\tilde{T}$ which satisfies the condition. Since $\tilde{T}$ is symmetric, it is equal to its symmetrization $\operatorname{Sym} \tilde{T}$. Then I tried to show that $(\operatorname{Sym}T)(A_1,\ldots,A_k)=\tilde{T}(A_1,\ldots,A_k)$, or equivalently, $(\operatorname{Sym}T)(A_1,\ldots,A_k)=(\operatorname{Sym}\tilde{T})(A_1,\ldots,A_k)$ but I couldn't.

Thanks in advance.

答:
Suppose there is another symmetric tensor $(\text{Sym}~\alpha)'$ such that $$(\text{Sym}~\alpha)'(v,\dots,v) = \alpha (v,\dots,v),~ \forall v\in V$$
Let's define a tensor $\beta\in T^k(V^*)$ by: $\beta = (\text{Sym}\alpha) - (\text{Sym}\alpha)' $. We observe that $$\beta(v,\dots,v)=0, ~\forall v \in V$$
Moreover, we clearly see that $\beta$ is actually symmetric, since $\Sigma^k(V^*)$ is a vector space. Define $\gamma_v:(-\epsilon,\epsilon)\ni t\mapsto v+t w_1 \in V$, where $v,w_1\in V, ~\epsilon >0$. Then, we have:
$$ \beta(\gamma_v(t),\dots,\gamma_v(t)) = 0, \quad \forall t\in (-\epsilon,\epsilon) \implies \frac{d}{dt}(\beta(\gamma_v(t),\dots,\gamma_v(t))) = 0, \forall t\in (-\epsilon,\epsilon)$$
Now, we compute:
$$\frac{d}{dt}(\beta(\gamma_v(t),\dots,\gamma_v(t)))\big|_{t=0} = \beta(w_1,v,\dots,v) + \beta(v,w_1,v,\dots,v) + \cdots + \beta(v,\dots,v,w_1) = 0 $$
Since $\beta$ is symmetric, we conclude that $\beta(w_1,v,\dots,v)=0, \forall w_1,v\in V$. We repeat this argument $k-1$ times, each time replacing $v$ at some slot with an arbitrary $w_i\in V$ until we conclude that $\beta(w_1,\dots,w_k)=0, ~\forall w_i\in V$. This means that $\beta\equiv 0$, i.e. $(\text{Sym}\alpha) = (\text{Sym}\alpha)'$


问:
Is there a formal definition of "continuous at a point for all practical purposes" or "continuous at a point within a given tolerance"?
Consider a function $f\colon D \to \mathbf{R}$ such that at some inner point $a$ of its domain $D\subseteq \mathbf{R}$
$$\lim_{x \to a^-}f(x) = L, \qquad \lim_{x \to a^+}f(x) =R, \qquad L \ne R \ ,$$
and the value $f(a)$ may be equal to either limit or have a yet different value.

Obviously the function isn't continuous at $a$.

But suppose that we are studying it with some numerical method, and the difference between $L$, $R$, and $f(a)$ is less than the precision used in the method. Say, for example, that $R=2+ \mathrm{e}^{-1000}$ and $L=2$. From the point of view of the numerical method the function can, so to speak, be considered continuous at $a$. A similar situation can arise with functions meant to fit empirical data.

Is there some notion of "approximate continuity at $a$" that applies in this case? An intuitive definition would be something like $\lvert \lim_{x \to a^-}f(x) - \lim_{x \to a^+}f(x)  \rvert <\delta$ and $\lvert \lim_{x \to a^-}f(x) - f(a) \rvert <\delta$ and so on, for some tolerance $\delta>0$. But I have no idea whether some definition of this kind is used in computational mathematics and how it is formally called.

I'd be thankful for any references you know of about some notion similar to this.

Note: The expression "approximate continuity" I use here has nothing to do with the definition of approximate continuity connected with measure theory, of course; but I couldn't find a better expression. Also, I'm aware that in computable analysis all computable functions are continuous.

答:
So in engineering and applied mathematics, when two functions $f,g$ are equal in all but a finite number of points, it makes some sense to consider them the same function. In other words, almost everywhere (a.e.) equality is good enough. A more technical way of saying this is that in engineering fields, engineers and researchers like to work with functions modulo a.e. equality.

This is due to a lot of good reasons. First, the derivatives of $f,g$ are also a.e. equal. Most importantly, $\|f-g\|^2=\int_a^b |f(x)-g(x)|^2 dx=0$. this nicely implies that $f,g$ share the same Fourier series. It follows from these properties many real analysis theorems for functions hold when you use a.e. equality instead of equality. I mean think about it, they are the same thing except at this 1 infinitesimally tiny point. Other than that one point, the information of the function is fixed. Your line of thinking is correct.

There are definitely generalizations of continuity for a.e. equal functions. One example is piecewise continuity.


问:
What is the best method to complete a set to a basis [Linear Algebra]
What is the best method to complete a set to a basis?

For example, given two vectors in $\mathbb{R}^4$: $(1,1,1,1),(1,0,1,0)$, and I want to complete it to a basis of $\mathbb{R}^4$, this is what I do:

Building a matrix with the 2 vectors, and adding the standard basis to it:

\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
1 & 0 & 0 & 1 & 0 & 0\\
1 & 1 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}

Than by reducing the matrix we get:

\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 1 & 1 & -1 & 0 & 0\\
0 & 0 & 1 & 0 & -1 & 0\\
0 & 0 & 0 & 1 & 0 & -1
\end{bmatrix}

So we have leading entries in the first 4 columns, so we can go back to the original matrix and take $c_3,c_4$ to complete the set to a basis, so we get:

$B=\left \{ (1,1,1,1),(1,0,1,0),(1,0,0,0),(0,1,0,0) \right \}$

But I find it with a lot of work that needed to be done. Any better way that you can suggest me?

Thanks a lot!

答:
One efficient way is to use the determinant. Adding two row vectors gives the matrix
$$
A=\begin{pmatrix}
1 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 \\
a_1 & a_2 & a_3 & a_4 \\
b_1 & b_2 & b_3 & b_4
\end{pmatrix}.
$$
The four row vectors are a basis if and only if the determinant is nonzero, where
$$
\det(A)= - a_1b_2 + a_1b_4 + a_2b_1 - a_2b_3 - a_4b_1 + a_4b_3 + 3b_2 - 3b_4.
$$


答:
What you're doing is the standard, possibly easiest way to do it.

Consider that if you are adding a new vector to the existing set, you will want to know if it can be expressed in terms of those vectors, which is asking if a system of equations has a solution. The most straightforward way to check that is by using row reduction - the exact row reduction, in fact, that you did.

As a special case, though, if you have two vectors in $\mathbb R^3$ and need a third, you can take the cross product. The determinant definition of cross product generalizes to telling you the last vector, if you have $n-1$ vectors in $\mathbb R^n$. But this is only questionably an improvement: for large $n$, finding that $n \times n$ determinant is best done by, you guessed it, row reduction.

Title: What is the best method to complete a set to a basis [Linear Algebra]

Body:
What is the best method to complete a set to a basis?

For example, given two vectors in $\mathbb{R}^4$: $(1,1,1,1),(1,0,1,0)$, and I want to complete it to a basis of $\mathbb{R}^4$, this is what I do:

Building a matrix with the 2 vectors, and adding the standard basis to it:

\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
1 & 0 & 0 & 1 & 0 & 0\\
1 & 1 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}

Than by reducing the matrix we get:

\begin{bmatrix}
1 & 1 & 1 & 0 & 0 & 0\\
0 & 1 & 1 & -1 & 0 & 0\\
0 & 0 & 1 & 0 & -1 & 0\\
0 & 0 & 0 & 1 & 0 & -1
\end{bmatrix}

So we have leading entries in the first 4 columns, so we can go back to the original matrix and take $c_3,c_4$ to complete the set to a basis, so we get:

$B=\left \{ (1,1,1,1),(1,0,1,0),(1,0,0,0),(0,1,0,0) \right \}$

But I find it with a lot of work that needed to be done. Any better way that you can suggest me?

Thanks a lot!


问:
Why an affine morphism is separated?
I'm working on the basic property of affine morphism that they are all separated.

The proof on stacks project uses this lemma but I want to find a direct method.
Here is my attempting:

for an affine morphism $f:X\rightarrow Y$ of schemes, cover $Y$ with open affines $V_i$, and put $U_i=f^{-1}(V_i)$ which is an open affine covering of $X$ . Now we can naturally factor the diagonal morphism as $X\rightarrow\coprod U_i \rightarrow \coprod U_i \times_{V_i}U_i \rightarrow X\times _Y X$. But I can't figure out why the last morphism is closed immersion(is it?).

And does anyone have a brief proof in other ways?

答:
Cover $Y$ with affine opens $\def\Spec{\operatorname{Spec}}\Spec B$: then $$f^{-1}(\Spec B)\times_Y f^{-1}(\Spec B) = f^{-1}(\Spec B)\times_{\Spec B} f^{-1}(\Spec B)=\Spec A\times_{\Spec B}\Spec A$$ cover $X\times_Y X$.
Since a morphism of affine schemes is separated, we have that $\Delta_{X/Y}: X\to X\times_Y X$ is locally a closed immersion on each of these sets. As the condition of being a closed immersion is affine-local on the target, we see that $\Delta_{X/Y}:X\to X\times_Y X$ is a closed immersion and thus $X\to Y$ is separated.

Title: Why an affine morphism is separated?


问:
Divisibility by 7 Proof by Induction
Prove by Induction that

$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$

Base case:

$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.

Now, having $n=k$, we assume that:

$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$

We have to prove that for $n=k+1$ that,

$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$

We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.

Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$

I get stuck here, please help me.

Title: Divisibility by 7 Proof by Induction

Body:
$7|8\cdot 2^{2^{k}}+2\cdot 2^{2^{k}}-2\cdot 2^{2^{k}}-2^{2^{k}}$

$7| 2^{2^{k}}\cdot (8+2-2-1)$

$7| 2^{2^{k}}\cdot (7)$

I think I proved it.