问:
Extended Bertrand's Boxes Problem - 3 boxes (2 coins per box)
Apologies for the long text, but it is to give context to the exercise.

I am having some trouble understanding an extended version of Bertrand's paradox. It comes in a few questions I have answered most, but the last one, I couldn't determine if my answer is the correct one.

Basically, you have 3 boxes, each containing 2 coins, the first one contains 2 gold coins, the second contains one gold coin and one silver coin, and the third one contains 2 silver coins.

Question 1 : you pick one box at random, and take a coin (and DON'T LOOK) what is the probability that the remaining coin is a gold coin ?  Answer :  all events have the same probability so here it is 1/2

Question 2 : you open your palm, you see that you have a gold coin, what is the probability that the remaining coin is a gold coin ?  Answer : 2/3 (out of the 3 gold coins you could get in your palm only 2 are in a box where the remaining would be a gold coin).

Question 3 : you put the gold coin back, and IN THE SAME BOX, you pick a coin, and don't look again,  what is the probability that the remaining coin is a gold coin ?
Answer : Here you operate under the condition that you have picked a gold coin before, so 2/3 probability of having chosen the first box, and 1/3 probability of having chosen the second box, so the probability that the remaining coin is a gold coin is : $ 1* \frac{2}{3} +\frac{1}{2}*\frac{1}{3} = \frac{5}{6} $

Question 4: you open your palm and you see that you had a gold coin, what is the probability that the remaining coin is a gold coin ? Answer : Here you operate under the condition that you have picked a gold coin before, AND that you picked in the same box a gold coin, in the question 3, having a gold coin remaining, or picking a gold coin are equivalent, so here there can only be a gold coin remaining if you picked from the first box, so the answer is : $ \frac{\frac{2}{3}}{\frac{5}{6}} = \frac{4}{5} $

Question 5 : you get rid of the coin that you picked (there only remains 5 coins now), and pick a box (out of the 3) at random, pick a coin, what is the probability that the remaining coin is a gold coin ?

My Answer is : $  \frac{3}{5} $, basically, I compute conditional on the fact that we eliminated a gold coin from the first box, and conditional on the fact that we eliminated a gold coin from the second box.

First box case, the probability of eliminating that coin is $ \frac{4}{5} $, which as computed before, is the probability of having picked the "eliminated" gold coin from box #1. That said, let $B_{i} , i = 1,2,3$ be the event of picking a coin from box #$i$, we call $O$ the event of having a gold coin remaining in the chosen box, and $G_{i}, i = 1,2$ the event that the eliminated coin is coming from box #$i$ , we have , $P(G_{1}) = \frac{4}{5} $, and $P(G_{2}) = \frac{1}{5} $ and :

$ P(O) = P(O,G_{1}) + P(O,G_{2}) $, with :  $P(O,G_{1}) = P(O|G_{1}) P(G_{1}) = P(O,B1|G_{1}) P(G_{1}) + P(O,B2|G_{1}) P(G_{1})  + P(O,B3|G_{1}) P(G_{1})  =  0 * \frac{4}{5} + \frac{1}{2} * \frac{4}{5}  + 0* \frac{4}{5}  = \frac{2 }{5}$

And :  $P(O,G_{2}) = P(O|G_{2}) P(G_{2}) = P(O,B1|G_{2}) P(G_{2}) + P(O,B2|G_{2}) P(G_{2})  + P(O,B3|G_{2}) P(G_{2})  =  1 * \frac{1}{5} +0*\frac{1}{5}  + 0* \frac{1}{5}  = \frac{1 }{5}$

Total is  $\frac{3}{5}$, but the solution says $\frac{2}{5}$, because there are 2 gold coins out of 5, am I missing something ?

Thank you, and sorry again for the long text.

答:
Ah Indeed you are right, I see my mistake,  I forgot to compute conditional to $B_{i},G_{i}$ instead of what I did, and since $B_{i}$ and $G_{i}$ are independent, $P(B_{i}|G_{i}) = \frac{1}{3}$, so dividing by 3 in my calculation I get the same result as you

Thanks


问:
Proposition 2.2 in Eisenbud
Can someone explain to me the proof for "This is the case iff each element $u\in U$ is a non-zero-divisor mod $J$"?

I'm stuck on the forward implication. From what I understand, because we assume that $J = \varphi^{-1}(I)$ we obtain an injective homomorphism $R/J\to R[U^{-1}]/I$ along which the pullback of $U$ from $R[U^{-1}]/I$ into $R/J$ are going to be non-zero divisors? If this is correct, then it makes sense to me, but I'm having a hard time seeing what is happening. Could someone please explain to me what the right intuition is in this situation?

答: Recall that $\varphi:R\to R[U^{-1}]$ is defined by $x\mapsto x/1$. Let $u\in U$. We need to prove that $u$ is a nonzerodivisor on $R/J$. So assume that $u(x+J)=0$ for some $x\in R$. This means $ux\in J=\varphi^{-1}(I)$, that is, $\varphi(ux)\in I$. Now $\varphi(u)=u/1$ is a unit in $R[U^{-1}]$. Thus we obtain $$\varphi(u)^{-1}(\varphi(ux))=\varphi(u)^{-1}\varphi(u)\varphi(x)=\varphi(x)\in I.$$ In other words, $x\in \varphi^{-1}(I)=J$; hence $x+J=0$ as desired.

ContentLicense: CC BY-SA 4.0

Title: Proposition 2.2 in Eisenbud


问:
Are approximate minimisers the minimisers of a perturbed function?
Suppose we have a convex function $f(x)$ defined on some compact convex set $X \subset \mathbb R$ with minimiser $x^* \in X$. Without loss of generality $X$ contains the origin.

We run some algorithm to find a point $y^* \in X$ with $f(y^*)-f(x^*) < \epsilon$ where $\epsilon$ is a known small quantity.

Is it possible to  find a nearby function $g_\epsilon(x)$ such that $y^*$ minimises $g_\epsilon(x)$? By nearby I mean we can bound $\max \{|f(x)-g_\epsilon(x)|:x \in X\}$ in terms of $\epsilon$ and for $\epsilon=0$ we have $g_\epsilon \equiv 0$.

In the special case of a quadratic $f(x) = b\|x-a\|^2$ for $a \in X$ the answer is yes. If $f(y^*)-f(x^*) < \epsilon$ then $b \|y^*-a\| < \epsilon$ and since $\|\nabla^2 f(x)\|= bI$ we have $\|\nabla f(y^*)\| \le b \|y^*-x^*\| =b \|y^*-a\|\le \epsilon$. Hence we can take $g(x) = f(x) - \nabla f(y^*) \cdot x$ which has zero gradient at $y^*$. Also we have  $$\max \{|f(x)-g_\epsilon(x)|:x \in X\} =  \max \{|\nabla f(y^*) \cdot x|:x \in X\} $$ $$\le \max \{\|\nabla f(y^*)\| \|x\|:x \in X\}  \le \epsilon\max  \{\|x\|: x \in X\} \le \epsilon \,\text{diam}(X).$$

答: Yes. Given $f$ and $y^*$ define $g(x) = \max \{ f(x), f(y^*)\}$. This is convex as the max of two convex functions. It collapses the convex region $C = \{x \in X: f(x) \le f(y^*)\}$ into a flat section. The min and max of the original function over that region are $f(x^*)$ and $f(y^*)$. From this we see $\max |f(x)-g(x)| \le f(y^*)$ for all $x \in C$. For $x \in X-C$ we have $f(x)=g(x)$ and so $\max |f(x)-g(x)| \le f(y^*)$ for all $x \in X$.


问:
Long exact sequence for mapping torus from Mayer-Vietoris
Note: I have actually done the thing I am trying to do. My question is whether there is an easier way to do it.

Question:
Let $f:X\to X$ be a map of topological spaces, and define the mapping torus of $f$ to be the space
$$
T_f = X\times [0,1] /\sim
$$
where $\sim$ is the equivalence relation $(x,0)\sim (f(x),1)$ for all $x\in X$. Using the Mayer-Vietoris Theorem, show that there is a long exact sequence
$$
\ldots \to H_n(X) \overset{1 - f_*}{\to} H_n(X) \to H_n(T_f) \to H_{n-1}(X) \to \ldots.
$$

My attempt: Let $A$ and $B$ be the subspaces of $T_f$ defined by
$$
A = X \times [1/5, 4/5] / \sim, \quad B = (X\times [0, 2/5] \sqcup X\times [3/5,1])/\sim,
$$
as illustrated in the following picture:

Then the usual Mayer-Vietoris sequence gives us a long exact sequence
$$
\ldots \to H_n(X)\oplus H_n(X) \to H_n(X)\oplus H_n(X) \to H_n(T_f) \to H_{n-1}(X) \oplus H_{n-1}(X) \to\ldots,
$$
where the first map has matrix
$$
\begin{pmatrix}
1 & 1
\\
-1 & -f_*
\end{pmatrix}.
$$
I haven't really justified this, but it's not hard to do, and I'm happy it's true. Now from here, we can kind of check by hand that restricting to the second component of each $H_n(X)\oplus H_n(X)$ gives the desired long exact sequence, but this is a bit fiddly and involves some case-work. Is there an easier way to get from the Mayer-Vietoris Theorem to the desired result?

答:
I will try to answer the question in a way which you probably did not expect: Rather than giving an alternative quicker solutition I am trying to convince you that your solution is rather clear.

First of all, I would define $A=X\times ([0,1/3)\cup (1/3,1])$ and $A=X\times ([0,2/3)\cup (2/3,1])$ and then choose the orientation of the homology to be induced by the inclusion of $X\times \{1/2\}$. For me, this makes it even more obvious why the map $H_*(X)\oplus H_*(X)$ should be given by $F=\begin{pmatrix} 1&1\\1 & f_*\end{pmatrix}$. (Note: The sign in the last column differs from your solution, but this is up to personal preference of the orientation of the homology groups.)

By changing the basis and writing $(x_1,x_2)\in H_*(X)\oplus H_*(X)$ as $(x_1+x_2,x_2)$ and $(y_1,y_2)\in H_*(X)\oplus H_*(X)$ as $(y_1,y_1-y_2)$ one then gets $F$ written with respect to the new basis as $F'=\begin{pmatrix} 1&0\\0&1-f_*\end{pmatrix}$. Because the sequence is exact, we then can omit the first copy in each of the two terms of the l.e.s. and obtain the desired result.

In other words, I believe your solution is already the correct one. However, sometimes it is even harder to understand why a solution is quite simple than it is to write it down in the first place.


问:
Find all $x,y,z \in \mathbb R$ Such that $\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y=3 $
Find $x,y,z \in \mathbb R$ Such that :
$$\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y =3$$

My Attempt:
I’ve turned this into a system of equations :
$$\cases{\frac{1}{x}+y+z =3 \\ \frac{1}{y}+x+z =3 \\ \frac{1}{z}+x+y=3 } \iff \cases{1+xy+zx=3x \\1+xy+zy=3y \\1+xz+yz=3z}$$
Multiplying some equations by $-1$ and adding them together we get:
$$z(y-x)=3(y-x)\iff z=3$$
Notice that $y\ne x$.

You can play a little bit with $x,y$ ’s values, you will end up with :
$$(x,y,z) \in \{(3,\frac{-1}{3},3), (-3, \frac{1}{3},3)\}$$
Edit: user pointed out in comments that the first triple of my solution doesn’t work in all cases, but i don’t know why.

And the values of $x,y,z$ Can swap places because of the symmetry in the equations.

My question is what would happen if $x=y$.

And there is one more thing to notice is that one of the obvious solutions is $$x=y=z=1$$

Thank you.

答:
If $x = y$,
$$\dfrac{1}{z} + x + y = 2x + 1/z = 3 \implies z = \dfrac{1}{3-2x}$$
Hence $x + 1/y + z = x + 1/x + 1/(3-2x) = 3$. Multiply this term by $x(3-2x)$ for $x \notin \{0, 3/2\}$ and after rearranging you will get the cubic equation
$$2x^3 - 9x^2 + 10x - 3 =(x-1)(2x^2 - 7x + 3) =  0$$
$\implies x = 1$, $x = 3$ or $ x = 0.5$.


问:
How to maximize the product of scalar products?
Given $N$ unit vectors $v_1,\ldots,v_N$ of $\mathbb R^d$, I was curious about finding an explicit maximizer for the product mapping
$$
f(x) := \prod_{j=1}^N \langle x,v_j\rangle,\qquad x\in\mathbb R^d,
$$
assuming that, say, $\|x\|=1$.

If one replaces the product by a sum, then the Cauchy-Schwarz inequality would directly yield that the maximum is reached at the unit vector proportional to the sum $\sum_jv_j$, but for the product I was not able to find such a simple solution.

I've tried to use the tensor product formalism to write $f(x) = \langle x^{\otimes N},\otimes_{j=1}^N v_j\rangle$, so that we see that maximizing $f$ boils down to minimize
$$
g(x):=\|x^{\otimes N}-\otimes_{j=1}^N v_j\|^2,
$$
and thus the solution is the first tensor component of the "orthogonal projection" of $\otimes_{j=1}^N v_j$ onto the subset $\Delta := \{x^{\otimes N}: x\in\mathbb R^d\}\subset(\mathbb R^d)^{\otimes N}$. Unfortunately, $\Delta$ is not a vector space and is not even convex, hence the quotation marks. I don't know how to continue (and gradient derivations did not bring me anywhere). Any ideas?

I'm pretty sure that clever people have worked on how to approximate a tensor product by a tensor product of the same vector, but I guess I didn't give the good keywords to Google.

答:
Try Lagrange multipliers. As $f(x)=\prod_{j=1}^N\langle x,v_k\rangle$ making

$$
L = f(x)+\lambda (\|x\|^2-1)
$$

we have

$$
\nabla L = f(x)\left(\sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}\right)+2\lambda x=0
$$

now calling $\mu = \frac{2\lambda}{f(x)}$ we have the stationary points as the solutions for

$$
\cases{
\sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}+\mu x=0\\
\|x\|^2=1
}
$$

or $\mu = -N$

so we follow with

$$
\sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}-N x=0
$$

a set of $d$ equations with $d$ unknowns.

NOTE

This nonlinear system can be solved iteratively as follows

$$
x_{k+1}= x_k -\left(\nabla g(x_k)\right)^{-1}g(x_k)
$$

where

$$
\cases{
g(x) = \sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}-N x\\
\nabla g(x) = -\left(\sum_{k=1}^N\frac{v_k\otimes v_k}{\langle x,v_k\rangle^2}+N I_d\right)
}
$$

NOTE

Attached a MATHEMATICA script to perform the calculations. The iterative process is fired maxtries because the stationary points can be many.

```<code>Clear[dif, grad, iG, Y, Y1]
SeedRandom[1]
dim = 10;
n = 10;
v = RandomReal[{-1, 1}, {n, dim}];
V = Table[v[[k]]/Norm[v[[k]]], {k, 1, n}];
X = Array[x, dim];
obj = Product[(V.X)[[k]], {k, 1, n}];

dif[X_List] := Sum[V[[k]]/(V[[k]].X), {k, 1, n}] - n X;
grad[X_List] := -Sum[Outer[Times, V[[k]], V[[k]]]/(X.V[[k]])^2, {k, 1, n}] - n IdentityMatrix[dim];
iG[X_List] := Quiet@Inverse[grad[X]]

kmax = 20;
maxtries = 1000;
SOLS = {};

For[try = 1, try <= maxtries, try++,
  Y = RandomReal[{-1, 1}, dim];
  aerror = 10^-20;

  For[k = 1, k <= kmax, k++,
    Y1 = Y - iG[Y].dif[Y];
    If[Max[Abs[Y - Y1]] < aerror, AppendTo[SOLS, Y1]; Break[]];
    Y = Y1
  ]
]

precise = 10^15;
sols = N[Union[Round[precise SOLS] / precise]];
best = Sort[Table[{obj, sols[[k]]} /. Thread[X -> sols[[k]]], {k, 1, Length[sols]}]][[Length[sols]]]
</code>```

We can compare it with the result obtained from a nonlinear solver

```<code>solX = NMaximize[{obj, X.X == 1}, X, Method -> "DifferentialEvolution"]
obj /. Thread[X -> best[[2]]]
</code>```

Body:
Try Lagrange multipliers. As $f(x)=\prod_{j=1}^N\langle x,v_k\rangle$ making

$$
L = f(x)+\lambda (\|x\|^2-1)
$$

we have

$$
\nabla L = f(x)\left(\sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}\right)+2\lambda x=0
$$

now calling $\mu = \frac{2\lambda}{f(x)}$ we have the stationary points as the solutions for

$$
\cases{
\sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}+\mu x=0\\
\|x\|^2=1
}
$$

or $\mu = -N$

so we follow with

$$
\sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}-N x=0
$$

a set of $d$ equations with $d$ unknowns.

NOTE

This nonlinear system can be solved iteratively as follows

$$
x_{k+1}= x_k -\left(\nabla g(x_k)\right)^{-1}g(x_k)
$$

where

$$
\cases{
g(x) = \sum_{k=1}^N\frac{v_k}{\langle x, v_k\rangle}-N x\\
\nabla g(x) = -\left(\sum_{k=1}^N\frac{v_k\otimes v_k}{\langle x,v_k\rangle^2}+N I_d\right)
}
$$

NOTE

Attached a MATHEMATICA script to perform the calculations. The iterative process is fired maxtries because the stationary points can be many.

Clear[dif, grad, iG, Y, Y1]
SeedRandom[1]
dim = 10;
n = 10;
v = RandomReal[{-1, 1}, {n, dim}];
V = Table[v[[k]]/Norm[v[[k]]], {k, 1, n}];
X = Array[x, dim];
obj = Product[(V.X)[[k]], {k, 1, n}];

dif[X_List] := Sum[V[[k]]/(V[[k]].X), {k, 1, n}] - n X;
grad[X_List] := -Sum[Outer[Times, V[[k]], V[[k]]]/(X.V[[k]])^2, {k, 1, n}] - n IdentityMatrix[dim];
iG[X_List] := Quiet@Inverse[grad[X]]

kmax = 20;
maxtries = 1000;
SOLS = {};

For[try = 1, try <= maxtries, try++,
  Y = RandomReal[{-1, 1}, dim];
  aerror = 10^-20;

  For[k = 1, k <= kmax, k++,
    Y1 = Y - iG[Y].dif[Y];
    If[Max[Abs[Y - Y1]] < aerror, AppendTo[SOLS, Y1]; Break[]];
    Y = Y1
  ]
]

precise = 10^15;
sols = N[Union[Round[precise SOLS] / precise]];
best = Sort[Table[{obj, sols[[k]]} /. Thread[X -> sols[[k]]], {k, 1, Length[sols]}]][[Length[sols]]]

We can compare it with the result obtained from a nonlinear solver

solX = NMaximize[{obj, X.X == 1}, X, Method -> "DifferentialEvolution"]
obj /. Thread[X -> best[[2]]]


答:
Given such $x,y,z\in\Bbb{R}$, in particular you have
$$\frac1x+y+z=\frac1y+x+z,$$
and hence also $x-\tfrac1x=y-\tfrac1y$. By symmetry we see that
$$x-\frac1x=y-\frac1y=z-\frac1z=c,$$
for some constant $c$, and so $x$, $y$ and $z$ are all roots of
$$T^2-cT-1=0.\tag{1}$$
A quadratic polynomial has at most two real roots, so without loss of generality we have $x=y$. We distinguish two cases:

If $x=y=z$, then we get the identity
$$\frac1x+x+x=3,$$
and hence $2x^2-3x+1=0$. This quadratic factors as
$$2x^2-3x+1=(2x-1)(x-1),$$
yielding the two solutions $(x,y,z)=(1,1,1)$ and $(x,y,z)=(\tfrac12,\tfrac12,\tfrac12)$.

If $z\neq x$, then also $z\neq y$ and so $xz=yz=-1$ because $x$ and $z$ are distinct roots of the quadratic $(1)$. Then from
$$\frac1z+x+y=3,$$
we find that also
$$3z=1+xz+yz=1+(-1)+(-1)=-1,$$
which shows that $z=-\tfrac13$ and hence $x=y=3$. This yields the three solutions
$$(x,y,z)=(3,3,-\tfrac13),\qquad (x,y,z)=(3,-\tfrac13,3),\qquad(x,y,z)=(-\tfrac13,3,3).$$

Title: Find all $x,y,z \in \mathbb R$ Such that $\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y=3 $

Body:
Find $x,y,z \in \mathbb R$ Such that :
$$\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y =3$$

My Attempt:
I’ve turned this into a system of equations :
$$\cases{\frac{1}{x}+y+z =3 \\ \frac{1}{y}+x+z =3 \\ \frac{1}{z}+x+y=3 } \iff \cases{1+xy+zx=3x \\1+xy+zy=3y \\1+xz+yz=3z}$$
Multiplying some equations by $-1$ and adding them together we get:
$$z(y-x)=3(y-x)\iff z=3$$
Notice that $y\ne x$.

You can play a little bit with $x,y$ ’s values, you will end up with :
$$(x,y,z) \in \{(3,\frac{-1}{3},3), (-3, \frac{1}{3},3)\}$$
Edit: user pointed out in comments that the first triple of my solution doesn’t work in all cases, but i don’t know why.

And the values of $x,y,z$ Can swap places because of the symmetry in the equations.

My question is what would happen if $x=y$.

And there is one more thing to notice is that one of the obvious solutions is $$x=y=z=1$$

Thank you.


问:
Evaluate the given integral
I have got 2 questions I can't seem to get the same answer to as the book.

31) $\displaystyle \int \sin(\sqrt x)\,\mathrm{d}x = -2\sqrt x \cos(\sqrt x)+2\sin(\sqrt x )+c$.

I just get the integral to be $\frac{1}{2}\cos(\frac{1}{x})+c$

and

36) $\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} = -\frac{1}{4}x +\frac{1}{8}  \tan^{-1}\left(\frac{2}{x}\right)+c$

This one has me all over the place. The best I can come to is setting $x = 2\tan x$ but thats about it.

答:
Hints:

For the first one, take: $ t = \sqrt{x} $ and you get $$ 2\int{t \sin(t) dx} $$

For the second:

$$
\begin{align}
\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} &= \dfrac{1}{4} \displaystyle\int \frac{(x^2 + 4 - x^2) \ \mathrm{d}x}{x^2(4+x^2)} \\
&= \frac{1}{4} \left[ \displaystyle\int \frac{\mathrm{d}x}{x^2} - \displaystyle\int \frac{\mathrm{d}x}{x^2 + 4} \right]
\end{align}
$$

ContentLicense: CC BY-SA 3.0

Title: Evaluate the given integral


问:
Rotation of a wheel in the positive direction and clockwise?
The Question

I’m riding my bike at a constant speed of 20 miles per hour on flat
land when i ride over a dot of fresh yellow paint, thereby getting a
yellow spot on my front wheel, which is 2 feet in diameter. I continue
riding my bike in a straight line at 20 miles per hour. When t seconds
passed since running over the dot, how many feet is the yellow spot on
my tire from the yellow dot i ran over?

My Understanding

I faced some difficulties in understanding an exercise about parameterization (pre-calculus level). I understood the solution except this point; the spot starts at the bottom of the circle, corresponding to an angle of $\frac{3\pi}{2}$. We let the wheel move in the positive direction, so that the wheel rotates clockwise

Is it possible that we can move in the positive direction with a clockwise rotation?

答: If the bike is rolling ("moving") on the ground toward $+\infty$ (the positive $x$-direction, towards the right), then the wheel spins clockwise (negative rotation).


问:
Find the fundamental group of the torus with an open disc removed
I'm trying to find a fundamental group of $\mathbb{T} \setminus \mathbb{D}$, the $2$-torus $\mathbb{T}$ with an open disc $\mathbb{D}$ removed. Any help would be really appreciated. </p>

答: Big hint: View the torus as the square $I^2/\sim$ with the usual equivalence relation $\sim$ identifying opposite edges, and then removes a small disk from the interior of $I^2$. Can you see how this space is homotopy equivalent to a wedge of some number of circles? (I'll leave you to figure out how many)</p>


问:
Proving intersection of closed sets is closed (with a specific definition)
I need to prove that the intersection of any number of closed sets is a closed set itself. There are some proofs that come up upon doing some research but I want to do it with a very specific definition. Here it is:

A set A is closed if every point that is arbitrarily close to A is a member of A.

(Equivalently , if y --> x with y being an element of A, then x is an element of A)

I've been trying to figure out how to use this definition to prove that the intersection of any number of sets is closed as well.

答:
If $\{E_\alpha\}$ is a collection of closed set, then $\bigcap_{\alpha} E_\alpha $ is a closed set.

Proof:

$p_\alpha$ is a limit point of $E_\alpha$ implies that $p_\alpha \in E_\alpha$, since $E_\alpha$ is a closed set.

For every $\epsilon > 0$ there is a point $q_\alpha\in E_\alpha$ such that $\operatorname{d}(p_\alpha,q_\alpha) < \epsilon$, by the definition of limit points. Also if $p$ is a limit point of $\bigcap_\alpha E_\alpha$, there is a point $q\in  \bigcap_\alpha E_\alpha$ such that $\operatorname{d}(p,q) < \epsilon$. Hence
$$
\operatorname{d}(p_\alpha,q_\alpha) \leq \operatorname{d}(p,p_\alpha)+\operatorname{d}(p,q)+\operatorname{d}(q,q_\alpha)\leq\epsilon +\operatorname{d}(p,p_\alpha)+\operatorname{d}(q,q_\alpha)
$$
Also $\operatorname{d}(p_\alpha,q_\alpha) < \epsilon$, then $\operatorname{d}(p,p_\alpha) = \operatorname{d}(q,q_\alpha) = 0$. It concludes that $p\in E_\alpha$ for any $\alpha$, which completes the proof.


问:
How to calculate $\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du $
I am trying to calculate $\int_0^2\int_0^2 e^{-|t-u|} \,dt \,du $

I solved the first integral to obtain

$$\int_0^2 e^{-|t-u|} \, dt = \left|\frac{e^{-|t-u|}}{u}\right|_0^2 $$
$$ \Longrightarrow \int_0^2 e^{-|t-u|} \, dt = \frac{e^{-|2-u|}}{u} - \frac{e^u}{u} $$

Putting the answer above into the outer integral, I now have

$$ \int_0^2\int_0^2 e^{-|t-u|} = \int_0^2\left[\frac{e^{-|2-u|}}{u} - \frac{e^u}{u} \right]\, du $$
I tried separating the integrals and using integration by parts, but I could not get an answer.

Please can you help me with tips on how to solve this integral or show me where I'm wrong?

Body:
Thank you. I have seen my error. I made a mistake in the integration

Following @Quanto 's guide,

$\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du
= 2\int_0^2\int_0^t e^{-|t-u|} du \>dt = 2\int_0^2\int_0^t e^{-(t-u)} du \>dt =2 (1+e^{-2})$

$\int_0^t e^{(u-t)}du = \big|\frac{e^{(u-t)}}{1}\big|_0^t$

$\int_0^t e^{(u-t)}du = e^{t-t} - e^{-t} = 1 - e^{-t}$

Solving the outer integral gives

$2\int_0^2(1 - e^{-t})dt = 2\big[ t + e^{-t}\big]_0^2$

$2\int_0^2(1 - e^{-t})dt = 2\big[ (2 + e^{-2})-(0 + e^0)\big]_0^2$

$2\int_0^2(1 - e^{-t})dt = 2\big[ 2 + e^{-2} - 1\big]$

$2\int_0^2(1 - e^{-t})dt = 2(e^{-2} + 1)$

$=>\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du = 2(e^{-2} + 1)$

Also, following @DhanviSreenivasan 's solution,
$\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du = \int_0^2 \left(\int_0^t e^{(u-t)})du  + \int_t^2 e^{(t-u)}du\right)dt$

$\int_0^t e^{(u-t)}du  + \int_t^2 e^{(t-u)}du = \big[ \frac{e^{u-t}}{1}\big]_0^t - \big[\frac{e^{t-u}}{1}\big]_t^2$

$\int_0^t e^{(u-t)}du  + \int_t^2 e^{(t-u)}du = (e^0 - e^{-t}) - (e^{t-2} - e^0) = 2 - e^{t-2} - e^{-t}$

Applying the outer integral,

$\int_0^2 (2 - e^{t-2} - e^{-t})dt = \big[2t-e^{t-2}+e^{-t}\big]_0^2$

$\int_0^2 (2 - e^{t-2} - e^{-t})dt = 2(1+e^{-2})$

$=>\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du = 2(e^{-2} + 1)$

Thanks to those who helped me.


问:
matrix for this system of non-linear equation?
Let's consider the product of a square matrix and a vector: $
M{\times}v =
\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}
{\times}
\begin{bmatrix} a \\ b \\ c \end{bmatrix}=
\begin{bmatrix} 1 {\times} a + 2 {\times} b + 3 {\times} c \\
                4 {\times} a + 5 {\times} b + 6 {\times} c \\
                7 {\times} a + 8 {\times} b + 9 {\times} c \end{bmatrix}$

How can I write $M$ such that $M{\times}v=
\begin{bmatrix} 1 {\times} a + 2 {\times} b + a {\times} b \\
                4 {\times} a + 5 {\times} b + a {\times} c \\
                7 {\times} a + 8 {\times} b + b {\times} c \end{bmatrix}$

The difference is in the third column where it's not just some constant times $c$ but the product of two other inputs.

Update

Ben suggested using the input variables $a$ and $b$ directly in the matrix $M$, but then $M$ wouldn't be made of constants only (like in the first example)

答:
$\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}$Let
$$\eqalign{
P &= \m{1&2&3\\4&5&6\\7&8&9}\qquad a = \m{a\\b\\c} \\
}$$
Then what you have is a linear equation $(Mv = Pa)$ which you wish to solve for $M$. This can be accomplished using an arbitrary matrix $A,\,$ and $v^+$ the Moore-Penrose inverse of $v$.
$$\eqalign{
M &= (Pa)v^+ + A(I-vv^+) \\
  &= Pa\left(\frac{v^T}{v^Tv}\right) + A\left(I-\frac{vv^T}{v^Tv}\right) \\
  &= \left(\frac{Pav^T+(v^Tv)A-Avv^T}{v^Tv}\right) \\
}$$
Note that this isn't a unique solution, it's a whole family of solutions; one for each $A$ matrix.


问:
How to prove this with induction $1 + m + m^2 + m^3 +\dots+ m^{n−1} = \frac{m^{n }− 1}{ m − 1}$ , where $|m| \lt 1$
I am very confuse where to start and how to do it, I am very new to this kind of problems, would someone be able to guide me ?</p>

答: The inductive step would be:  $\color{blue}{1+m+\dots+m^{n-1}}+m^n=\color{blue}{\dfrac{m^n-1}{m-1}}+m^n=\dfrac{m^n-1}{m-1}+\dfrac{m^{n+1}-m^n}{m-1}=\dfrac{m^{n+1}-1}{m-1}$.


问:
How does one prove that $\text{Isom}(\mathbb{S}^2 \times \mathbb{R}) = \text{Isom}(\mathbb{S}^2) \times \text{Isom}(\mathbb{R})$?
Everywhere I'm looking a lot of authors claim (without any proof, but I'm sure they're right) that the group of isometries of $\mathbb{S}^2 \times \mathbb{R}$ and $\mathbb{H}^2 \times \mathbb{R}$ are $4$ dimensional and equal to $ \text{Isom}(\mathbb{S}^2) \times \text{Isom}(\mathbb{R})$ and $ \text{Isom}(\mathbb{H}^2) \times \text{Isom}(\mathbb{R})$, respectively. How can I prove those claims and what's the geometric intuition behind it?

答:
I am totally late to the party, but just to have it explained clearly on the Internet, here is a proof of a more general statement that covers your question for both $\mathbb{S}^2 \times \mathbb{R}$ and $\mathbb{R}H^2 \times \mathbb{R}$. Observe that whenever you have Riemannian manifolds $M_1, \ldots, M_k$ you have a natural embedding of groups $I(M_1) \times \ldots \times I(M_k) \hookrightarrow I(M_1 \times \ldots \times M_k)$. If all $M_i$'s have finitely many connected components, these are Lie groups and the embedding is smooth. In general, it may fail to be surjective (in the worst case, the dimension of the RHS may be greater than that of the LHS). However, we have

Proposition. Let $M$ be a connected Riemannian manifold with nonzero constant sectional curvature $R$ (for example, a round sphere, or a real hyperbolic/projective space, or a hyperbolic manifold; no assumption on the sign of $R$ here) and $N$ a connected flat manifold (e.g., a flat torus, or a Euclidean space, or a product of those). Then the natural embedding $I(M) \times I(N) \hookrightarrow I(M \times N)$ is an isomorphism.

Proof. Take an arbitrary isometry $\varphi$ of $M \times N$. Let's take a point $(m,n) \in M \times N$ and look at the linear isometry $d\varphi_{(m,n)} \colon T_{(m,n)}(M \times N) \cong T_mM \oplus T_nN \to T_{(m',n')}(M \times N) = T_{m'}M \oplus T_{n'}N$, where $(m',n') = \varphi(m,n)$. I claim that if we prove that it sends $T_mM$ onto $T_{m'}M$, then we're done. Indeed, if it does, then it sends $(T_mM)^\perp = T_nN$ onto $(T_{m'}M)^\perp = T_{n'}N$. But since an isometry maps geodesics to geodesics, $\varphi$ sends the totally geodesic submanifold $M \times \{n\}$ onto $M \times \{n'\}$, and similarly $\{m\} \times N$ onto $\{m'\} \times N$. (Completeness is not needed here, but it is essential that these slices are properly embedded submanifolds, so a geodesic in $M \times N$ that tangentially touches, say, $M \times \{n\}$ even once, will stay in it throughout the whole interval where it is defined.) Denote:

\begin{align}
\varphi_1 \colon M \to M \times \{n\} \xrightarrow{\varphi} M \times \{n'\} \to M, \\
\varphi_2 \colon N \to \{m\} \times N \xrightarrow{\varphi} \{m'\} \times N \to N.
\end{align}

Observe that $\varphi_1 \in I(M)$ and $\varphi_2 \in I(N)$, so we have a product isometry $(\varphi_1, \varphi_2)$ of $M \times N$. The isometries $\varphi$ and $(\varphi_1, \varphi_2)$ coincide on the "cross" $(M \times \{n\}) \cup (\{m\} \times N)$, which surely implies that they coincide at $(m,n)$ but also that their differentials at this point coincide as well:

\begin{multline}
d\varphi_{(m,n)} = d(\varphi_1)_m + d(\varphi_2)_n = d(\varphi_1, \varphi_2)_{(m,n)} \colon \\
T_{(m,n)}(M \times N) \cong T_mM \oplus T_nN \to T_{m'}M \oplus T_{n'}N \cong T_{(m',n')}(M \times N).
\end{multline}

But now it is a well-known fact that if two isometries of a connected Riemannian manifold coincide at some point together with their differentials, then they coincide globally (just apply these two isometries to geodesics emanating from that point and use the theorem on uniqueness of geodesics). So we are left to show that $d\varphi_{(m,n)}$ sends $T_mM$ onto $T_{m'}M$. Since isometries preserve sectional curvatures, it will immediately follow from the following claim:

Claim. If $L$ is a $2$-plane in $T_{(m,n)}(M \times N)$, its sectional curvature lies between $0$ and $R$ and only reaches $R$ if $L \subseteq T_mM$.

Proof of the claim. Clearly, if $L \subseteq T_mM$, then its sectional curvature is just $R$. Assume $L \nsubseteq T_mM$. Then there are two possibilities:

Case 1: $L$ intesects $T_nN$ nontrivially. Then take a basis for $L$ of the form $X = A, Y = W + B$, where $A, B \in T_nN$ and $W \in T_mM$. Then

$$
K(L) = \frac{Rm(X,Y,Y,X)}{\|X\|^2 \|Y\|^2 - \langle X \,|\, Y\rangle^2}.
$$

Since the product is Riemannian and $N$ is flat, we have

$$
Rm(X,Y,Y,X) = Rm(A,W+B,W+B,A) = Rm(A,B,B,A) + Rm(0,W,W,0) = 0,
$$

so $K(L) = 0$.

Case 2: $L$ intersects $T_nN$ trivially. Then we can take a basis $X = V+A, Y = W+B$ for $L$ such that $V,W$ is an orthonormal basis of the $2$-plane $\mathrm{pr}_{T_mM}(L) \subseteq T_mM$ and $A,B \in T_nN$ with at least one of them nonzero (since we assume $L \nsubseteq T_mM$). We compute:

$$
K(L) = \frac{Rm(X,Y,Y,X)}{\|X\|^2 \|Y\|^2 - \langle X \,|\, Y\rangle^2} = \frac{Rm(V,W,W,V) + Rm(A,B,B,A)}{(1 + \|A\|^2)(1 + \|B\|^2) - \langle A \,|\, B\rangle^2} = \frac{K(\mathrm{pr}_{T_mM}(L)) + 0}{1 + \|A\|^2 + \|B\|^2 + (\|A\|^2 \|B\|^2 - \langle A \,|\, B\rangle^2)} = \frac{R}{1 + \|A\|^2 + \|B\|^2 + (\|A\|^2 \|B\|^2 - \langle A \,|\, B\rangle^2)},
$$

which lies strictly between $0$ and $R$. $\square$ $\square$

Remark. There is also a nice fact directly pertinent to this discussion that if $M = M_0 \times M_1 \times \ldots \times M_k$ is the de Rham decomposition of a simply connected complete Riemannian manifold, then the natural morphism on the isometry groups above restricts to an isomorphism on the connected components of the idenitities: $I^0(M_0) \times I^0(M_1) \times \ldots \times I^0(M_k) \to I^0(M)$. Here $I(M)$ and each $I(M_i)$ are Lie groups in the compact-open topology, so we can take the connected component of the identity in each of them. For a proof, see Chapter VI, theorem 3.5 of

Kobayashi, Sh.; Nomizu, K., Foundations of differential geometry. I, New York-London: Interscience Publishers, a division of John Wiley & Sons. XI, 329 p. (1963). ZBL0119.37502.

EDIT. After giving it a thought, I managed to generalize the above proposition and remark by employing some ideas about holonomy from Kobayashi and Nomizu. In a nutshell, the following result says that if you have the product of a bunch of irreducible Riemannian manifolds and at most one flat manifold, then the full isometry group of the product is the product of the full isometry groups of the factors plus permutations of isometric factors. I am leaving the above proposition here because it already answers the MO's question, and its proof is rather elementary and doesn't require any sophisticated arguments involving holonomy.

Proposition. Let $M = M_0 \times M_1^{l_1} \times \ldots \times M_k^{l_k}$ be a Riemannian product, where all $M_i$'s are connected Riemannian manifolds, $M_0$ is flat, each $M_i, 1 \leqslant i \leqslant k,$ is irreducible, and $M_i^{l_i}$ simply means $\underbrace{M_i \times \ldots \times M_i}_{l_i \, \text{times}}$. (Here by irreducibility of $M_i$ we mean that it is not flat and its restricted holonomy representation is irreducible; if $M_i$ is complete, this is equivalent to asking that its universal Riemannian covering cannot be decomposed as a nontrivial Riemannian product.) We also assume that $M_i$ is not isometric to $M_j$ for $i \ne j$. Let $l = \sum_{i=1}^k l_i$, write $S_l$ for the symmetric group on $l$ elements, and let $S_l^{l_1, \ldots, l_k} \cong S_{l_1} \times \ldots \times S_{l_k}$ stand for the subgroup of permutations that permute the first $l_1$ elements with each other, the next $l_2$ elements with each other, and so on. Then the obvious map
$$
I(M_0) \times I(M_1)^{l_1} \times \ldots \times I(M_k)^{l_k} \rtimes S_l^{l_1, \ldots, l_k} \to I(M)
$$
is an isomorphism of Lie groups.

One obvious example of such decomposition is the de Rham decomposition of a complete simply connected Riemannian manifold. Note that we don't require the factors to be complete in the proposition.

Sketch of the proof. Let $p = (p_0, (p^1_i)_{i=1}^{l_1}, \ldots, (p^k_i)_{i=1}^{l_k}) \in M$, and consider the decomposition

$$
T_pM = T_{p_0}M_0 \oplus \bigoplus_{i=1}^{l_1} T_{p^1_i} M_1 \oplus \ldots \oplus \bigoplus_{i=1}^{l_k} T_{p^k_i} M_k.
$$

Let $\mathrm{Hol}^0(M,p)$ stand for the restricted holonomy group of $M$ at $p$, which is defined using parallel transport along contractible loops based at $p$ and is the identity component of the full holonomy group $\mathrm{Hol}(M,p)$. The decomposition of $T_pM$ above is obviously orthogonal, the first summand $T_{p_0}M_0$ is the subspace of invariants of $\mathrm{Hol}^0(M,p)$ in $T_pM$, and all the other summands are irreducible $\mathrm{Hol}^0(M,p)$-subrepresentations in $T_pM$ (see Chapter IV, Section 5 in Kobayashi&Nomizu for details on that). A decomposition of $T_pM$ satisfying these three properties is called a canonical decomposition in K&N. Now, part (4) of theorem 5.4 in Chapter IV, Section 5 of K&N says that a canonical decomposition of $T_pM$ is unique up to reordering of its factors. Caveat: K&N actually talk about decompositions with respect to the full holonomy group, and theorem 5.4(4) requires $M$ to be simply connected. But note that the restricted holonomy representation of $M$ is isomorphic to the full holonomy representation of its universal Riemannian covering $\widetilde{M}$ (just by lifting contractible loops). So the uniqueness of a canonical $\mathrm{Hol}(\widetilde{M}, \widetilde{p})$-decomposition for $\widetilde{M}$ ($\widetilde{p} \in \widetilde{M}$ any point over $p$) translates into the uniqueness of a canonical $\mathrm{Hol}^0(M, p)$-decomposition for $M$. It is also not hard to show that if $\varphi$ is an isometry of $M$, then $d\varphi$ sends a canonical decomposition of $T_pM$ to a canonical decomposition of $T_{\varphi(p)}M$ (basically because 'isometries commute with parallel transport'). Now, given such $\varphi$, a small adaptation of the first part of the proof of the proposition above (the one with geodesics) shows that $\varphi$ 'permutes the factors of the de Rham decomposition of $M$', or, formally, lies in the image of the morphism of isomerty groups above. $\square$


问:
Find the coefficient of $x^{100}$ in $\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$
Find the coefficient  of $x^{100}$ in $$\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$$
This problem came from finding  total number of 4-partition of an integer 100.
One uses mathematica and it maybe easy, but I want to find more Mathematical proof.

```<code>f[m_, n_] :=
  Coefficient[Series[x^m/Product[(1 - x^i), {i, 1, m}], {x, 0, n}], x,
    n];
f[4, 100]
</code>```

$7153$ is obtained. More general:

```<code>FullSimplify[
 SeriesCoefficient[x^4/Product[(1 - x^i), {i, 1, 4}], {x, 0, n}],
 Assumptions -> Element[n, PositiveIntegers]]

</code>```

$$
\frac{1}{288} \left(-32 U_n\left(-\frac{1}{2}\right)+(n+1) \left(2 n
   (n+2)+9 (-1)^n-13\right)+36 \cos \left(\frac{\pi  n}{2}\right)\right)$$

Body:
Partial fractions:
$$\begin{split}
\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)}
&=\frac{288x^4}{(1-x)^4(1+x)^2(1+x+x^2)(1+x^2)}\\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32}{1+x+x^2} + \frac{36}{1+x^2} \\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32(1-x)}{1-x^3} + \frac{36(1-x^2)}{1-x^4}
\end{split}$$
Taking coefficients:
$$\begin{split}
[x^n]\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)} &=
12\binom{n+3}{4}-12\binom{n+2}{3}+13\binom{n+1}{2} \\
&\quad + 9(-1)^n\binom{n+1}{2}  -32([3\mid n]-[3\mid(n-1)]) + 36([4\mid n]-[4\mid(n-2)])
\end{split}$$


问:
Understanding why Minkowski’s inequality doesn't hold true for $0 < p < 1$?
The triangle inequality given by

$\left(\sum_{i=1}^n |x_i+y_i|^p\right)^{1/p}\leq \left(\sum_{i=1}^n |x_i|^p\right)^{1/p} + \left(\sum_{i=1}^n |y_i|^p\right)^{1/p}$

is known as “Minkowski’s inequality" which holds true for $1\leq p <\infty$ while for $0 < p < 1$ it doesn't  hold.

Searching through net I have found that  this inequality holds for $1\leq p <\infty$  is related to the observation that for such $p$ the function $x \to x^p$ for $x \geq 0$ is convex.

The failure of the triangle inequality is related to the observation that for $0 <
p < 1$,the function $x \to x^p$ for $x \geq 0$ is not convex.

I am not able to understand these points. Could anybody explain me?

Thanks

答: I have a good(well, I think it's good) interpretation. The Minkowski's inequality simply suggests that $L^p$ is a normed space. But you can show that $L^p$ isn't even locally convex space if $0<p<1$.</p>


问:
Does a Lie algebra generate the entire identity component?
Suppose we have a Lie group $G$ with Lie algebra $\mathfrak{g}$. If $G$ is connected, is it true that $\exp\mathfrak{g} = G$? If $G$ is not connected, does $\exp\mathfrak{g} = G_0$ where $G_0$ is the identity component of $G$?

If we take
$
A = \begin{pmatrix}
-1 & 1 \\ 0 & -1
\end{pmatrix} \in \mathsf{SL}(2; \mathbb{C})
$, I think there does not exist any $X\in M_2(\mathbb{C})$ such that $\mathrm{e}^X = A$ which hints that the Lie algebra does not "generate" the entire (connected) Lie group. That said, is this true, and are there are any other statements we can make about how much of a Lie group is generated by it's Lie algebra?

答:
The image of the exponential mapping only "generates" the identity component of the group in the sense of group theory, i.e. the smallest subgroup of $G$ that contains all elements of the form $\exp(X)$ with $X\in\mathfrak g$ is the connected component $G_0$ of the identity. Otherwise put, any element of $G_0$ can be written as a products of finitely many exponentials.

In general, this does not imply that $\exp:\mathfrak g\to G_0$ is surjective and you have a nice counterexample in your question. There is a general result that $\exp:\mathfrak g\to G_0$ is surjective if $G$ is compact, but this is much more difficult.


答:
Base step: $n=1$ then $\frac{m^1-1}{m-1}=1$

Induction hypothesis: suppose it is valid for $n = k$.

Inductive step: Let
\begin{align}
(1+m+m^2+\ldots+m^{k-1})+m^{(k-1)+1}&=\frac{m^k-1}{m-1}+m^k\\
&=\frac{m^k-1+m^{k+1}-m^k}{m-1}\\
&=\frac{m^{k+1}-1}{m-1}\\
\end{align}
with this you can already argue

Title: How to prove this with induction $1 + m + m^2 + m^3 +\dots+ m^{n−1} = \frac{m^{n }− 1}{ m − 1}$ , where $|m| \lt 1$

Body: I am very confuse where to start and how to do it, I am very new to this kind of problems, would someone be able to guide me ?</p>


问:
Inequalities regarding odd degree polynomial's coefficients
Define an odd degree polynomial as a function $f: \mathbb{R} \to \mathbb{R}$ such that

$f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$

where n is odd.

We all know odd degree polynomials have special characteristics which they don't share with even degree polynomials, here I propose four of them, which I think are true, but I'm unable to prove. My final goal is to prove something I already know, which is that all odd degree polynomials have at least one real root.

i) $\mid x \mid \gt 1 \implies \mid \frac{a_{n-1}}{x} + ... +\frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \mid \lt \frac{1}{\mid x \mid}(\mid a_{n-1} \mid + \mid a_{n-2} \mid) + ... + \mid a_1 \mid + \mid a_0 \mid$

ii) $\mid x \mid \gt \max\{ 1,2\sum_{i=0}^{n-1}\mid a_i \mid \}\implies \mid \frac{a_{n-1}}{x} + ... +\frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \mid \lt \frac{1}{2}$

iii) $x_1 \gt \max\{ 1,2\sum_{i=0}^{n-1}\mid a_i \mid \} \implies f(x_1) \gt 0$

iv) $x_2 \lt -\max\{ 1,2\sum_{i=0}^{n-1}\mid a_i \mid \} \implies f(x_2) \lt 0$

Any help or proof is appreacieted.

Thanks

答:
Why do you need the inequalities to prove the statement that an odd degree polynomial with real coefficients has at least one real root ?

This statement follows from the intermediate value theorem and from having/knowing these two limits:

$ \lim_{x \to +\infty} f(x) = +\infty $

$ \lim_{x \to -\infty} f(x) = -\infty $

So with some basic real analysis knowledge the statement can be proved easily.

Right?


问:
Invertibility of inference rules with quantifiers
In the $LK$ system for sequent calculus, the inference rules with quantifiers are four: $\forall R$, $\exists L$, $\forall L$ and $\exists R$. In the course we specified that the first two are invertible, and the last two are not (because, clearly, when you apply those rules from bottom to top, you "lose some informations"). Now, in order to maintain the invertibility, the professor said that one should use contraction, and then apply one of the non invertible rules. My question is: if one applies $\forall L$ or $\exists R$ substituting a variable not used before (so basically if one apply the non invertible rules with the same restrictions that are used for the invertible ones), do those rules become invertible too?

答:
No; for example, let $a,b$ be two distinct variables and $P$ a unary predicate. Then the $(\forall L)$ inference
$$
{
P(a)\vdash P(b)
\over
\forall x P(x) \vdash P(b)
}
$$
satisfies the eigenvariable condition ($a$ does not occur freely in the lower sequent) and its conclusion is valid, while its premise clearly is not.


问:
The order of a matrix and its eigenvalues.
Suppose we have a family of intervitble  matrices $\{M_\epsilon\}_{\epsilon>0} \subset \mathbb{R}^{d\times d}$ , for which $M_\epsilon $ is $O(\epsilon)$, which is to say there exists a constant $C$ such that $(M_\epsilon)_{i,j} \leq C \epsilon,$ for all $i,j=1,\ldots, d$. What can we say about the eigenvalues of $M_\epsilon^{-1}$ ? will they be $O(\frac{1}{\epsilon})$.

Answer attempt :

If we know $(M_\epsilon^{-1})_{i,j}=O(\frac{1}{\epsilon})$ for all $i,j=1,\ldots,d$ then for any eigenvalue $\lambda$ of $M_\epsilon^{-1}$

$$
\lambda \leq \| M^{-1}_\epsilon\|_{op}\leq \sup_{x~:~\|x\|=1}\|A\|_2\|x\|=\|A\|_2\leq d \max_{i,j}(M_\epsilon^{-1})_{i,j}=O(\frac{1}{\epsilon})
$$

where $\|\cdot\|_{op}$ is the operator norm and $\|\cdot\|_2$ the 2 norm. $\textbf{But}$ is it obvious that $(M_\epsilon^{-1})_{i,j}=O(\frac{1}{\epsilon})$ just because $(M_\epsilon )_{i,j}=O(\epsilon)$?

答:
If $Av=\lambda v$ then $\frac{1}{\lambda}v=A^{-1}v$. So eigenvalues of $A^{-1}$ are inverse of those of $A$.

If $|A_{ij}|<\epsilon$ then $||Av||<C\epsilon||v||$ (for some $C$ depending on the order or the matrix) so eigenvalues of such an $A$ are bounded above by $C\epsilon$. Therefore eigenvalues of $A^{-1}$ are bounded below by $1/C\epsilon$.


问:
How do you explain the appearance of a sine in the integral for calculating the surface area of a sphere?
Let's say I want to calculate the surface area of a sphere. For simplicity, let's just use the unit sphere. A naïve argument might go like this. Let's say I mark the north and south "poles" and draw half of a great circle, which has length $\pi$. I could say that since I need to go all the way around the sphere, I need to multiply this by $2\pi$ (the circumference of the equator). Therefore, the surface area of the unit sphere is $2\pi^2$.

Now, as we all know it should be $4\pi$. Let's say we do an integral, using the following parametrization:

$$
T(\theta, \phi) = \begin{pmatrix} \sin \phi \cos \theta \\
\sin \phi \sin \theta \\
\cos \phi
\end{pmatrix},
$$

with $0 \le \phi \le \pi$ and $0 \le \theta \le 2\pi$. If we work out all the formulas, we get that
$$Area(S^2) = \int_0^{2\pi} \int_0^\pi \sin \phi\ \mathrm{d}\phi \ \mathrm{d}\theta = 2\pi \int_0^\pi \sin \phi\ \mathrm{d}\phi.$$

The $2\pi$ is there all right, but it multiples not $\pi$ but $\int_0^\pi \sin \phi\ \mathrm{d}\phi$, which equals $2$. Where does this come from? In other words, why is it wrong to just multiply $2\pi$ by half the length of a great circle? It would be great if there was a geometric explanation, with as little calculus as possible involved.

答: There is a geometrical argument for it, all you need to do is construct a coordinate system in spherical coordinates in the Cartesian system, in order to move along a line of constant radius we must change the angular dependence in $\phi$ and $\theta$ (assuming the usual definition that $\theta$ is the azimuthal angle and $\phi$ is the polar angle), to move in the  $\phi$ direction requires that surface traversed is simply $ dl = R d\phi$, but to move in the theta direction we have to use a bit of trig, a circular segment in the $\theta$ direction has a radius $ r = R \sin \phi $ ( we can construct a triangle with the z axis as one leg, the hypotenuse is the radius R, and the horizontal remaining leg is given by the preceding formula), in order to get the length of a line in the $\theta$ direction its the same relationship as before ($s = r \times \text{ angle}$) so that $dl = r \sin \phi d \theta$. thus multiplying these to get the surface area element gives $dA = R^2 \sin \phi d \theta d \phi$ When $\theta$ encompasses a complete unit circle ( $R= 1$) this yields $ 2 \pi \int \sin \phi d \phi $. If you want a more non-calculus representation replace every "d" with a $\Delta$</p>


问:
How do I interpret the definition of a distribution in this book by Joshi & Friedlander?
I was wondering about the following definition in the book Introduction to the Theory of Distributions, by Joshi and Friedlander:

In the Appendix the bracket operation is defined as

To me this definition of a distribution doesn't make sense if we view the bracket operation as an inner product, e.g., an integral, since the domain of $u$ is $C_c^\infty(X)$ whereas the domain of $\phi$ is $X$. So then how do we make sense of $\int u(x)\phi(x) dx$?

I would understand if $u$ were, say, a continuous function from $X$ to $\mathbb{R}$. Then $u$ would determine a linear form through $\langle u, \phi \rangle$. Is that what the definition means? We take some object $u$, then $u$ determines a linear form through the pairing $\langle u, \phi \rangle$? And then we refer to $u$ as a linear form because somehow the linear form determines $u$ uniquely? Perhaps this bit from the book is relevant:

答:
The notation $\langle u,\varphi\rangle$ is just a fancy way of writing $u(\varphi)$ for a linear from $u\colon C^\infty_c(X)\to\mathbb{C}$. Examples of linear forms which are not defined by functions are $\delta_0$ or $\partial^\alpha\delta_0$ which are defined as
$$
\langle \delta_0, \varphi\rangle = \varphi(0) \qquad \text{and}\qquad \langle \partial^\alpha\delta_0, \varphi\rangle = (-1)^{|\alpha|}(\partial^\alpha\varphi)(0).
$$
For these two it is easy to check that they satisfy the condition stated above. A slightly more complicated example might be
$$
\langle \mathrm{vp}\frac{1}{x}, \varphi\rangle = \lim_{\varepsilon\to 0} \int_{\mathbb{R}\setminus(-\varepsilon,\varepsilon)} \frac{\varphi(x)}{x}\,\mathrm{d}x.
$$
Note that none of these distributions can be represented by a continuous function.


问:
Are complete metric spaces of finite Hausdorff dimension locally compact
Is every complete metric $(M,d)$ of Hausdorff dimension $n<\infty$ a locally compact space?

More generally, given a complete metric space of finite Hausdorff dimension, can we always find a locally identical metric $d'$ on $M$ so that $(M,d')$ has the Heine-Borel property, i.e. so that every bounded and closed subset is compact?

答: Consider $M=\left\{v\in\Bbb R^2\,:\, v_x=0\lor \exists q\in\Bbb Q, v_y=qv_x\right\}$ with the metric $$d(a,b)=\begin{cases}\lVert a\rVert+\lVert b\rVert&\text{if }a,b\text{ linearly independent}\\ \lVert a-b\rVert&\text{if }a,b\text{ linearly dependent}\end{cases}$$ It's a connected complete metric space, and the origin has no compact neighbourhood. However, it is countable union of subspaces which are isometric to $\Bbb R$, and therefore its Hausdorff dimension is $1$.


问:
Completing a proof on integrability: $f$ is Riemann integrable under the mesh definition iff $\sup L_\Gamma = \inf U_\Gamma$
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ve}{\varepsilon}
\newcommand{\d}{\delta}
\newcommand{\g}{\Gamma}
\newcommand{\norm}[1]{\left \Vert  #1   \right \Vert}
\newcommand{\volume}{\mathrm{volume}}
\newcommand{\gs}{{\g_\ast}}
$Definitions:

Considering we're working in $\R^n$, and there are a lot of competing/equivalent definitions for Riemann integrability (as I learned the hard way in trying to find some help here and elsewhere), I want to get all of these out of the way first, so we're on the same page. These are the definitions I'm working with, in their essence taken from Measure & Integral: An Introduction to Real Analysis by Richard Wheeden and Antoni Zygmund.

Riemann Integrable: Let $f : \mathcal I \subseteq \R^n \to \R$ be bounded. We say $f$ is Riemann integrable on $\II$ if
$$
\lim_{\norm \g \to 0} R_\g
$$
exists. In particular, we call this value $L$, and say $L = \int_\II f(x) \, \mathrm{d}x$.

Formalization: Formally, the above is equivalent to

$$(\forall \ve >0)(\exists \d > 0)(\forall \g : \norm \g < \d)(|R_\g - L| < \ve)$$

Boxes: $\II$ denotes a box. You can write it a couple of different ways: for given $a_i,b_i$,
$$
\II := \Big\{ (x_1,\cdots,x_n) \in \R^n \; \Big| \; x_i \in [a_i,b_i] \Big\} = \prod_{i=1}^n [a_i,b_i]
$$

Volume of Boxes: The volume of a box is given in the usual, expected way:
$$
\volume(\II) = \prod_{i=1}^n (b_i - a_i)
$$

Partitions: Let $\g$ denotes a partition of $\II$ into finitely many nonoverlapping boxes $\II_k$. (That is, $\mathrm{int}(\II_i) \cap \mathrm{int}(\II_j) = \varnothing$ for all $i \ne j$, and $\bigcup \II_k = \II$.)

Norm/Mesh of Partitions: The mesh or norm of a partition is defined to be
$$
\norm \g := \max_i \mathrm{diam}(\II_i)
$$

Diameter of Set: The diameter of a subset of a metric space is the supremum of possible distances between points. In $\R^n$ under the $2$-norm,

$$\mathrm{diam}(\II) := \sup_{x,y \in \II} \norm{x - y}_2$$

Tags: We take representative points (or tags) $\xi_i \in \II_i$ for each $\II$.

Riemann Sum: We define the Riemann sum to be (implicitly dependent on the partitioning and tags)
$$
R_\g := \sum_i f(\xi_i) \cdot \volume(\II_i)
$$

Upper & Lower Sums: We define the upper and lower sums as so:
$$
U_\g := \sum_i \sup_{x \in \II_i} f(x) \cdot \volume(\II_i) \qquad L_\g := \sum_i \inf_{x \in \II_i} f(x) \cdot \volume(\II_i)
$$
Trivially, $L_\g \le R_\g \le U_\g$ for any choice of tags $\xi_i$.

My Goal: I would like to show that
$$
\text{$f$ is Riemann integrable} \iff \inf_\g U_\g = \sup_\g L_\g
$$
More specifically, I would like to show the backwards direction; I've already proven the forwards I believe. I'd also in particular like as direct a proof as possible (rather than circumnavigating through the other definitions of integrability).

My Attempt So Far:

So far, I feel like I've hit on the right idea, with a minor snag in the way.

So, let $\ve > 0$, and let
$$
\sup_\g L_\g = \inf_\g U_\g =: L
$$
From the definition of supremum/infimum, we may find two partitions $\g_1,\g_2$ such that
$$
0 \le \sup_\g L_\g - L_{\g_1} < \frac \ve 2 \qquad 0 \le U_{\g_2} - \inf_\g U_\g < \frac \ve 2
$$
Form their common refinement $\gs$ of $\g_1$ and $\g_2$. (This would just be their union in the $\R^1$ case where we can equivalently use points as our partition, but not the case in general.) Then we easily see that
$$
0 \le \sup_\g L_\g - L_{\gs} < \frac \ve 2 \qquad 0 \le U_{\gs} - \inf_\g U_\g < \frac \ve 2
$$
Recalling the assignment of $L$, we have
$$
0 \le L - L_{\gs} < \frac \ve 2 \qquad 0 \le U_{\gs} - L < \frac \ve 2
$$

Where I Get Lost:

This is where I get lost. My end goal is to ultimately show $|R_\g - L| < \ve$ (independent of the choice of $\xi_i$ namely). At this point, I feel like I'm meant to use the inequality $L_\g \le R_\g \le U_\g$, but I'm not certain that it would be valid. If it were viable, then I would intend to have
$$
0 \le L - R_{\gs} < \frac \ve 2 \qquad 0 \le R_{\gs} - L < \frac \ve 2
$$
but would it not be possible for $R_\gs$ to fall on one side of $L$ or the other, and thus outside of the bound of $0$? (Though on the other hand, would that even matter? Is the bound of $0$ important for this?)  From a situation like this, I would then like to claim the conclusion.

Another worrying part is that I have no particular $\d$ brought up. Maybe I could just arbitrarily set a partition $\g_0$ such that $\norm{\g_1},\norm{\g_2} < \norm{\g_0} = \d$, but that doesn't really feel right.

Can anyone give me an idea as to how to complete this proof, even if just a nudge in the right direction? Or am I completely off-base? Thanks for any insights you can give!

答:
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\R}{\Bbb{R}}
\newcommand{\g}{\Gamma}
\newcommand{\vol}{\operatorname{volume}}
\newcommand{\gs}{{\Gamma_\ast}}
$

If $U \subseteq V$, then $$\inf_V f \le \inf_U f \le \sup_U f \le \sup_V f$$

Since $\gs$ is a refinement of $\g_1$, every box $\II_1 \in \g_1$ is the union of a collection $\mathscr B(\II_1)$ of boxes in $\gs$, and $$\vol(\II_1) = \sum_{\II_*\in \mathscr B(\II_1)} \vol(\II_*)$$

So $$\begin{align}L_{\g_1} &= \sum_{\II_1\in \g_1} \inf_{\II_1}f\cdot\vol(\II_1)\\
&= \sum_{\II_1\in \g_1} \sum_{\II_*\in \mathscr B(\II_1)} \inf_{\II_1}f\cdot\vol(\II_*)\\
&\le \sum_{\II_1\in \g_1}\sum_{\II_*\in \mathscr B(\II_1)} \inf_{\II_*}f\cdot\vol(\II_*)\\
&= \sum_{\II_*\in \gs}\inf_{\II_*}f\cdot\vol(\II_*)\\
&=L_{\gs}\end{align}$$

Similar remarks hold for the upper sums and for $\g_2$


问:
How to calculate the angle of 2 curves on a surface
Calculate the angle of the curves $(C1) : u = v-1$, $(C2): u=3-v$ on the surface $(S):$
$$\textbf{r}(u,v) = u\cos v\ \textbf{i} + u\sin v\ \textbf{j} +u^2\ \textbf{k}$$

I know the angle of two coordinate curves is the angle of the tangent lines at the intersection point, but I'm not sure how to apply this.

答:
The first thing to do is to find the intersection point of the two curves.

The tangent vectors at a point $\textbf{r}(u_0,v_0)$ have the nice formula of
$$
\frac{\partial\textbf{r}}{\partial u}\bigg\vert_{(u_0,v_0)}\ \ \ \text{ and  }\ \ \ \frac{\partial\textbf{r}}{\partial v}\bigg\vert_{(u_0,v_0)}\ .
$$
Think you can take it from there?


问:
PDE with Initial Values and no solutions
I need some help understanding PDE with initial values. The exercise ask us to find all solutions to the PDE $u_x - 2u = 0$ that satisfies the following conditions:

$u(0,y) = y^2$

$u(1,y) = y^2$

$u(x,1) = x^2$

My attempt is the following: I start by finding $u(x,y)$.
$$
u_x - 2u = 0 \implies \frac{u_x}{u} = 2
$$

$$
\implies \int \frac{u_x}{u}dx = \int 2dx
$$

$$
\implies u(x,y) = k(y) e^{2x}
$$

Now, for the initial conditions I substitute the $x$ and $y$ values given and would get the following :

$u(x,y) = k(y) e^{2x}$ and $u(0,y) = y^2 \implies k(y) = y^2$. Therefore the solution must be $u(x,y) = y^2 e^{2x}$.

$u(x,y) = k(y) e^{2x}$ and $u(1,y) = y^2 \implies k(y) = \frac{y^2}{e^{2}}$. Therefore the solution must be $u(x,y) = \frac{y^2}{e^{2}} e^{2x}$.

Has no solution.

I'm not sure I'm understanding correctly, is this what I'm supposed to do? I'm not sure what the actual solution should look like.

答:
Its right.

Regarding the 3rd one: the equation can be thought of as a transport equation whose characteristics (c.f.: the method of characteristics, wikipedia link) are lines parallel to the $x$ axis.

What is happening is that they asked you to put initial data right on a characteristic curve. Therefore, unless the initial data is prepared correctly, there will be no solutions.

For example, if they instead asked you for solutions solving $u(x,1)=e^{2x}$, you should be able to write down at least two different solutions (!) with the same values on the line $y=1$.


问:
Breaking down numbers to $2$'s and $3$'s
How can I prove that all numbers greater than 3 could be written as a sum of $2$'s and $3$'s only?

答:
Let $n$ be a number, $n>3$. There are only two possibilities, either $n$ is even or odd. If $n$ is even, then $\exists\;k$, s.t. $2k=n$. Then it is clear that, $n$ can be written as sum of $k$ twos. On the other hand, if $n$ is odd, then $\exists\;l$, s.t. $2l+1=n\Rightarrow 2(l-1)+3=n$. Then it is clear that, $n$ can be written as sum of $l-1$ twos, and one $3$.

Note that numbers $1$ and $3$ can't be written as sum of $2$ and $3$. That is why the condition greater than $3$  is needed.


问:
Integrality property of the root system of a semisimple Lie algebra
For $\mathfrak{g}$ a semisimple Lie algebra, $\mathfrak{h}$ a choice of Cartan subalgebra, $\Phi$ the set of roots, there is the integrality property that if $\alpha$ is a root then $c \alpha \in \Phi$ if and only if $c = \pm 1$.
I am trying to understand the proof of the 'only if' direction as it appears in my textbook Goodman-Wallach (although Humphreys has a similar proof) (I have also seen the proof here, but I want to do it the way it is presented in my book).

To begin with, $s(\alpha) = \text{span}\{e_\alpha, f_\alpha, h_\alpha\} \cong \mathfrak{sl}_2$ is an isomorphic copy inside $\mathfrak{g}$.
Construct $$ M_\alpha = \mathbb{C}h_\alpha + \sum_{c \neq 0} \mathfrak{g}_{c\alpha}.$$
Then the eigenvalues of $h_\alpha$ on $M_\alpha$ (under the adjoint representation) are 0 and $2c$.
Since $M_\alpha$ is completely reducible and the weights of $\mathfrak{sl}_2$-irreps are known, we conclude that every $2c$ must be an integer, so $c$ is an integral multiple of 1/2.
By the classification of $\mathfrak{sl}_2$-irreps (eigenvalues are integers which differ by 2), this then tells us that the set $\{ 2c : \mathfrak{g}_{c\alpha} \in M_\alpha\} $ either consists of all even or all odd integers.
In particular, there are two cases: either every $c$ is an integer, or else every $c$ is of the form $p+1/2$ for some positive integer $p$ (they are very similar, so I am posting them as one question).

In the first case, my textbook says that since $s(\alpha)$ contains the zero eigenspace in $M_\alpha$ (I agree with this), then it follows that $c\alpha$ is not a root for any integer $c$ for which $\vert c \vert > 1$.
I don't see how this conclusion follows at all.
I understand that $s(\alpha)$ is one such irreducible component of $M_\alpha$, and that its eigenvalues are $0, \pm 2$.
I also recognise that this exhausts the occurrences of the zero weight, since 0 has multiplicity one in $M_\alpha$.
However I don't understand how either of these statements imply the conclusion.

In the second case, if $(p+1/2)\alpha \in \Phi$, then the eigenvalues of $h_\alpha$ are $2p+1, 2p-1, \dots, 3, 1$ on $M_\alpha$, and then somehow this implies $(1/2)\alpha$ is also a root.
I also don't see how this follows, although maybe if my first question is answered then this will also follow.

答:
Your agrument goes wrong where you say "... by the classification of $\mathfrak{sl}_2$-irreps ...". What you can conclude there is that $M_\alpha$ is a direct sum of finitely many non-trivial irreducible representations. Now consider irreducible components of odd dimension. On these, $h_\alpha$ has only even eigenvalues and always has a one-dimensional zero-eigenspace. Since the zero-eigenspace in $M_\alpha$ is one-dimensional and $s(\alpha)\subset M_\alpha$ is one irreducible component, you conclude that there may not be any other odd dimensional irreducible components. Hence in all other irreducible components $h_\alpha$ has only odd eigenvalues. In particular, the eigenspace for the eigenvalue $4$ in $M_\alpha$ has to be trivial, so $\mathfrak g_{2\alpha}=\{0\}$ and hence $2\alpha$ cannot be a root.

But now if there were any even dimensional irreducbible components, each of them would have a non-trivial eigenspace for the eigenvalue $1$. But this would imply that $\mathfrak g_{\alpha/2}$ would be non-zero and hence $\alpha/2$ would be a root. But this is a contradiction, since from above, this would imply that $2(\alpha/2)=\alpha$ is not a root.


问:
Last three digits of a large exponent, without using modular arithmetic
If i have a really large number like \begin{gather} 3^{762259784987} \end{gather} or \begin{gather} 7^{836793257}\end{gather}
Is there a way to find the last three digits without using modular arithmetic or is that the only method. If their is could someone please outline a method because i would be very curious. I don't want to use modular arithmetic, despite it probably being the most obvious method, because we haven't covered the topic yet but I still received a similar problem which leads me to believe that their could be another method that relies more on problem solving skills. But I could be very wrong and maybe the only method involves modular arithmetic. I worked out how to find the last digit because of the cyclical nature but I haven't had any luck with figuring out all of the last three digits.

(Note: Someone rightfully pointed out that the title and the question asked for two different things, my fault and sorry for any confusion but I was originally thinking about the last three)

答:
Update: Last three digits of $3$ requires lots of work but not impossible as has already been in the comments. With the last three digits of $7$, we are a little bit luckier, though:

```<code>1 7 <-
2 49
3 343
4 401
5 807
6 649
7 543
8 801
9 607
10 249
11 743
12 201
13 407
14 849
15 943
16 601
17 207
18 449
19 143
20 1
21 7 <-
</code>```

So, the last three digits are repeated per $20$ steps.

Looking at the two last digits of powers of $3$ until the repetition, we get the following list:

```<code>1 3 <-
2 9
3 27
4 81
5 43
6 29
7 87
8 61
9 83
10 49
11 47
12 41
13 23
14 69
15 7
16 21
17 63
18 89
19 67
20 1
21 3 <-
</code>```

So, the last two digits are repeated per $20$ steps.

Looking at the two last digits of powers of $7$ until the repetition, we get the following list:

```<code>1 7 <-
2 49
3 43
4 1
5 7 <-
</code>```

So, the last two digits are repeated per $4$ steps.

Body:
Update: Last three digits of $3$ requires lots of work but not impossible as has already been in the comments. With the last three digits of $7$, we are a little bit luckier, though:

1 7 <-
2 49
3 343
4 401
5 807
6 649
7 543
8 801
9 607
10 249
11 743
12 201
13 407
14 849
15 943
16 601
17 207
18 449
19 143
20 1
21 7 <-

So, the last three digits are repeated per $20$ steps.

Looking at the two last digits of powers of $3$ until the repetition, we get the following list:

1 3 <-
2 9
3 27
4 81
5 43
6 29
7 87
8 61
9 83
10 49
11 47
12 41
13 23
14 69
15 7
16 21
17 63
18 89
19 67
20 1
21 3 <-

So, the last two digits are repeated per $20$ steps.

Looking at the two last digits of powers of $7$ until the repetition, we get the following list:

1 7 <-
2 49
3 43
4 1
5 7 <-

So, the last two digits are repeated per $4$ steps.


问:
Is every hereditarily unicoherent plane continuum non-separating?
Let $X$ be a connected compact subset of the plane.  Suppose that $A\cap B$ is connected for every two closed connected subsets $A,B\subseteq X$.  Then is $X$  non-separating?  That is, is $\mathbb R^2\setminus X$ connected?

It seems like the answer should be yes, and I have tried proving it but do not know how.

答:
The answer is no.  Take the two endpoint Knaster continuum and identify the two endpoints as shown in the image below:

[2-endpoint Knaster Continuum with Endpoints Identified][1]

The image is from my paper "Atriodic acyclic continua and class(W)" which appeared in Proceedings of the AMS, Vol. 90, No. 3, 1984,  starting on p. 477.

Jim Davis
[1]: https://i.stack.imgur.com/N16g3.png


问:
When does pointwise convergence imply uniform convergence?
On an exam question (Question 21H), it is claimed that if $K$ is compact and $f_n : K \to \mathbb{R}$ are continuous functions increasing pointwise to a continuous function $f : K \to \mathbb{R}$, then $f_n$ converges to $f$ uniformly. I have tried proving this claim for the better part of an hour but I keep coming short. I suspect a hypothesis on equicontinuity has been omitted — partly because the first half of the question is about the Arzelà–Ascoli theorem — but I don't have access to the errata for the exam so I can't be sure.

Here is my attempted proof: let $g_n = f - f_n$, so that $(g_n)$ is a sequence of continuous functions decreasing pointwise to $0$. Clearly, $0 \le \cdots \le \| g_n \| \le \| g_{n-1} \| \le \cdots \le \| g_1 \|$, so we must have $\| g_n \| \longrightarrow L$ for some constant $L$. $K$ is compact, so for each $g_n$, there is an $x_n \in K$ such that $g_n(x_n) = \| g_n \|$, and there is a convergent subsequence with $x_{n_k} \longrightarrow x$ for some $x \in K$. By hypothesis, $g_n(x) \longrightarrow 0$, and by construction $g_{n_k} (x_{n_k}) \longrightarrow L$. I'd like to conclude that $L = 0$, but to do this I would need to know that the two sequences have the same limit. This is true if, say, $\{ g_n \}$ is an equicontinuous family, but this isn't one of the hypotheses, so I'm stuck.

答:
Your argument is very close.

Fix $\epsilon > 0$.  Since $g_n(x) \downarrow 0$, there exists $m$ such that $g_m(x) \le \epsilon$.  Now, for all $n_k \ge m$ we have $g_{n_k}(x_{n_k}) \le g_m(x_{n_k})$.  As $k \to \infty$, the left side converges to $L$ by your construction.  And by the continuity of $g_m$, the right side converges to $g_m(x) \le \epsilon$.  So $L \le \epsilon$.  Since $\epsilon$ was arbitrary, the proof is complete.

Edit: Incidentally, a uniformly convergent sequence is always equicontinuous, so equicontinuity does in fact hold.  One could also prove it directly from the assumptions given (try it!).

Edit: As noted below by Monstrous Moonshine, your proof assumes that $K$ is sequentially compact, so it works for, say, metric spaces.  But there is a gap if $K$ should be compact but not sequentially compact.


问:
$y'=\frac{x-y+2}{x+y-1}$
First, I substitute $x=\xi + h$, $y=\eta+k$. The equation becomes $$y'=\frac{\xi+h-(\eta+k)+2}{\xi+h+\eta+k-1}$$

and this becomes homogenous if we let $h=-\frac{1}{2}$ and $k=\frac{3}{2}$ ( We solve the constant terms such that they become zero ). I think it's correct this far, but what next? At this point we'd have $$\eta'=\frac{\xi-\eta}{\xi+\eta}$$

which is homogenous, and I tried substituting $\eta = u\xi$. This would lead to the equation $$u'\xi+u=\frac{1-u}{1+u}$$

I reduce $u$ from both sides. This will give $$u'\xi=\frac{1-u}{1+u}-u=\frac{1-2u-u^2}{1+u}$$

or $$\frac{du}{d\xi}\xi=\frac{1-2u-u^2}{1+u}$$

inverting this gives the relation $$\frac{1}{\xi}d\xi= \frac{1+u}{1-2u-u^2}du$$

So there has been some mistake made. The correct answer does not involve logarithms, and $\ln(\xi)$ will inevitably be a part of the solution. Can somebody spot my mistake?

答:
Just take expinential on both sides. Here is another way:
$$y'=\frac{x-y+2}{x+y-1}$$
$$y'\color {red} {+1}=\frac{x-y+2}{x+y-1} \color {red} {+1}$$
$$(y+x-1)'=\frac{2x+1}{x+y-1}$$
$$(y+x-1)d(y+x-1)=( {2x+1})dx$$
Integrate both sides:
$$(y+x-1)^2=2x(x+1)+C$$

Title: $y'=\frac{x-y+2}{x+y-1}$


问:
drawing numbers game - probability of winning
Someone asked me this interview question he got recently, but I am not sure if I get it right.

A and B play a number drawing game. A can draw 3 numbers from 1-9 without replacement, but B can only draw 3 numbers from 1-8 without replacement. After numbers are drawn, they are sorted such that the 3-digit number formed by those 3 numbers are maximized, for example, if 3 numbers drawn are 1,3,8 then the number A/B will get is 831. what's the probability that A will have a greater number than B?

I tried to solve with conditional probability:

P(A wins) = P(A wins|9 is not selected)P(9 is not selected) + P(A wins|9 is selected)P(9 is selected)

P(A wins|9 is not selected)=1/2

P(9 is not selected)=C(8 choose 3)/C(9 choose 3) = 2/3

P(A wins|9 is selected)=1

P(9 is selected)=C(8 choose 2)/C(9 choose 3)=1/3

so P(A wins)=2/3. Am I correct?

答:
With probability $1/3$, A draws a $9$ and wins.

With probability $2/3$, his digits are drawn from $1$ to $8$, just as B's.  But as noted in a comment, A only wins if his score is strictly greater than B's; in effect, B wins ties.  The probability of a tie is $1/\binom{8}{3} = 1/56$. If there is no tie, A's and B's chances are equal, by symmetry. So if $A_W$ is the event that A wins in this case,
$$P(A_W) + 1/56 + P(A_W) = 1$$
and solving, $P(A_W) = 55/112$.

So A's overall probability of winning is
$$\frac{1}{3} + \frac{2}{3} \cdot \frac{55}{112} = \frac{222}{336} \approx 0.660714$$


问:
optimization with constraint
The topic we are giving is optimization with restrictions. So, the exercise they gave me is the following:

minimize $f(x,y,z)=x^{2}+y^{2}+z^{2}$

subject to the following restrictions: $x^{2}+y^{2}+z^{2}-5=0$ and $x^{2}+y^{2}+z^{2}-2x-3=0$.

I proceed to use the lagrange multiplier, so I have the following system of equations:

$(Ec.1)$  $2x-\lambda(2x)-\mu(2x-2)=0$

$(Ec.2)$  $2y-\lambda(2y)-\mu(2y)=0$

$(Ec.3)$  $2z-\lambda(2z)-\mu(2z)=0$

$(Ec.4)$  $x^{2}+y^{2}+z^{2}-5=0$

$(Ec.5)$  $x^{2}+y^{2}+z^{2}-2x-3=0$

solving this system in matlab get the following results $x=\lambda=1$, $y=z=1.4142$, $\mu={-7.2176e^{-11}}$

Now I must see the optimality condition but I don't know how to proceed, someone please help me on how I can finish the exercise

答:
As it was mentioned in the comments, since one of the constraints sets the value of the objective function to $5$, you only have two possibilities:

The feasible set is empty and there is no solution, i.e. the two restrictions are incompatible (This is not the case).

The objective function is constant over the feasible set and the minimum an maximum are both 5.

If you want to ignore this and just look at the system you wrote:

The last two equations yield directly $x = 1$, $y^2+ z^2 = 4$.

Substituting $x=1$ in the first equation you get $\lambda = 1$.

Equations (2),(3) then give you $\mu y = 0$ and $\mu z = 0$

So, taking $\mu = 0$, you get as a solution every point such that $x = 1$ and $y^2+z^2=4$. If $\mu \ne 0$, then $z = y = 0$, which is impossible.

Finally, each point in the feasible set is a critical point of the Lagrangian... Since the feasible set is compact, and the objective function in continuous, there will be a minimum and maximum over the set. Additionally, considerations on the Jacobian matrix of the restricitons guarantee that the min/max are critical points of the Lagrangian. But the value of $f$ over all those critical points is the same, and we are done.

You may also use criteria on the "bordered Hessian" to get to the same conclusion.

Title: optimization with constraint


问:
What is the distribution of the first zeros of the Bessel function?
Let $j_{m,n}$ be the $n^{th}$ zero of the $m^{th}$ order Bessel function $J_{m}(x)$. I would like to know the asymptotic behaviour of the first zeros of the Bessel function $j_{m,1}$, as the order $m$ of the Bessel function tends to infinity.

Numerically it seems to go very roughly as $j_{m,1}\sim m$, but I am interested in the specific behaviour and this is something I'm struggling to find, or work out, myself.

I am also interested in the same behaviour for the zeros $j_{m,n}'$ of the derivative of the $m^{th}$ order Bessel function, $J_{m}'(x)$.

Thanks in advance for any help.

答:
Hoping that my memory is still correct, I think that they are given by something like
$$j_{m,1}=m+\sum_{k=0}^p a_k \,m^{\frac{1-k}3}$$ but I do not think that the coefficients are known. Tricomi proposed in $1949$
$$j_{m,k}=m+a_k\, m^{\frac 13}+O\left(m^{-\frac 13}\right)$$

I generated a table for $m$ from $10$ to $1000$ by steps of $10$  and made a curve fit. Its results are
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a_0 & +1.85576 & 3.26\times 10^{-8} &   \{+1.85576,+1.85576\} \\
 a_1 & +1.03359 & 8.54\times 10^{-6} &   \{+1.03358,+1.03361\} \\
 a_2 & -0.00601 & 7.72\times 10^{-5} &   \{-0.00616,-0.00586\} \\
 a_3 & +0.02915 & 2.40\times 10^{-4} &   \{+0.02867,+0.02963\} \\
 a_4 & -0.08568 & 2.44\times 10^{-4} &   \{-0.08616,-0.08519\} \\
\end{array}$$ Over the considered range the error is less than $10^{-6}$.


问:
An introductory measure theory problem
Let $(X,\mathcal{M},\mu)$ be a measure space. Define $\mu_0: \mathcal{M} \rightarrow[0,\infty]$ by
$\mu_0(E):=\sup\{ \mu(F): F \in \mathcal{F}(E) \}$, where $\mathcal{F}(E):= \{ F \in \mathcal{M}: F \subseteq E \text{ and } \mu(F) < \infty \}$.

Prove that if $\mu(E) < \infty$, then $\mu_0(E) = \mu(E)$

It is obvious to me that $\mu_0(E) \geq \mu(E)$ as $\mu_0(E)$ is defined through supremum, but I have no clue how to prove the reverse inequality.

Thank you.

答:
Let $E \in \mathcal{M}$.  By definition of supremum, for any $\epsilon>0$ given there exists $F \in \mathcal{F}(E)$ such that
$$ \mu_{0}(E) - \epsilon < \mu(F).$$
(Observe that if $\mu(E) = \infty$ the above claim is not possible since $\mu_0(E)\geq \mu(E)$ and $\mu(F)< \infty$ by definition)
Since $F \subseteq E$, we get that $\mu(F) \leq \mu(E)$ and then
$$ \mu_0(E) - \epsilon < \mu(E).$$
Since this is valid for any $ \epsilon >0$, we can conclude that $\mu_{0}(E) \leq \mu(E).$

Title: An introductory measure theory problem


问:
Probability that 4 socks selected at random from 4 pairs will form 2 pairs
I have come across this problem in a book I am studying out of interest.

A drawer contains 4 different pairs of socks. Find the probability that (a) if 2 socks are selected at random they will form a pair, (b) if 4 socks are selected at random they will form 2 pairs.

I solved (a) by saying the number of different selections of socks is $\binom{8}{2} = 28$

Number of different matching combinations = 4

$\implies probability = \frac{4}{28} = \frac{1}{7}$

To solve (b) I have looked at similar questions on this forum but can find none which I have been successfully able to apply.

I know that when choosing the second sock there is a $\frac {1}{7}$ probability of it matching the first. But then I get confused.

The answer in the book is $\frac{3}{35}$

答:
So for (a) you want the second sock to match the first with probability $\frac17$.

Do something similar for (b):

Second sock matches first and fourth sock matches third with probability $\frac17\times \frac15$

Second sock does not match first, third sock matches first and fourth sock matches second with probability $\frac67\times \frac16\times \frac15$

Second sock does not match first, third sock matches second and fourth sock matches first with probability $\frac67\times \frac16\times \frac15$

and add these up to give $\frac3{35}$


问:
How to build linear order on $\mathcal P({\bf N})$
How to build linear order on $\mathcal P({\bf N})$?

Had idea about inclusion relation, but it does not satisfy linearity.

答:
Hint:

Given two sets of natural numbers, $A,B$ we can say that $A\prec B$ if the minimal element of $A\triangle B$ is a member of $A$.

Another alternative is to find an injective function from $\mathcal P(\Bbb N)$ into $[0,1]$ and use that to define the order.

A third alternative is to think about it as infinite binary sequences, and order them lexicographically.


问:
Distance of an observer above the surface of the earth to his Horizon
The earth's diameter $D$ is approx. $12742$ km.

a) Under the assumption of an exact spherical shape of the earth, show that the distance $f(h)$ of an observer who is at height $h> 0$ above the surface of the earth to his Horizon is given by
$f (h) = \sqrt{Dh}\sqrt{1+\frac{h}{D}}$.

b) Show that $f(h)=\sqrt{Dh}(1+r(h))$, where $0 <r (h) <\frac{h}{2D}$, and evaluate the approximation $f (h) \approx \sqrt{Dh}$ for $h = 10\ m$.

Show that in this specific case the error $f(h)-\sqrt{Dh}$ is smaller than $1\ cm$.

$$$$

I haven't really understood the description. What does it mean that the distance to his horizon?

Body:
$$f(h)=\sqrt{\left(\frac{D}{2}+h\right)^2-\frac{D^2}{4}}=\sqrt{h (D+h)}=\sqrt{D h} \sqrt{1+\frac{h}{D}}$$

$$\sqrt{1+x}\sim 1+x/2;\;\text{ as }x\to 0$$
Therefore, as $h\ll D$ we have $\frac{h}{D}\sim 0$
$$\sqrt{1+\frac{h}{D}}\sim 1+\frac{h}{2D}$$
And finally $$f(h)\approx \sqrt{Dh}\left(1+\frac{h}{2D}\right)$$

If $h=10$m$=0.010$km then

$$f(0.01)=\sqrt{12742 \times0.01} \sqrt{1+\frac{0.01}{12742}}=11.2880556\text{km}$$
the approximation gives
$$f(0.01)=\sqrt{12742\times 0.01}\approx 11.2880512$$
The difference is about $8.86\times 10^{-6}$km that is $0.886$cm

$$...$$


问:
Help with inverse trigonometric substitutions $ \int x^2\sqrt{a^2+x^2}\,dx $.
how would I go about integrating this? It is a lecture exercise.

$$
\int x^2\sqrt{a^2+x^2}\,dx
$$
I used a substitution of $x=a\tan\theta$ and ended up with $$\int a^4\tan^2\theta \sec^3\theta \,d\theta.$$ I was thinking by parts but then I would have to integrate a $\ln|\sec\theta+\tan\theta|$. Can't seem to think of the appropriate identity to collapse it. I am given the answer to verify but I don't know any mental algorithms to even begin doing inverse trigonometric substitution. The answer given is:
$$
\frac{x}{8}(a^2+2x^2)\sqrt{a^2+x^2}-\frac{a^2}{8}\ln\left(x+\sqrt{a^2+x^2}\right)
$$
Also, what are some things I should be thinking about first when looking at such questions when I need to substitute trigonometric functions into the x-variable? Should I attempt to make odd powers into even? How do I do so?
Thank you very much for your help.

答:
Integrate by parts directly without substitutions

\begin{align}
\int x^2\sqrt{a^2+x^2}dx
= &\ \frac14 \int \frac x{\sqrt{a^2+x^2}}d[(a^2+x^2)^2]\\
\overset{ibp}= &\ \frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}4\int \frac{\sqrt{a^2+x^2}}{2x}d(x^2)\\
\overset{ibp} =& \ \frac{2x^3+a^2x}8{\sqrt{a^2+x^2}}-\frac{a^4}8\int \frac{dx}{\sqrt{a^2+x^2}}\\
\end{align}
where $\int \frac{dx}{\sqrt{a^2+x^2}}= \sinh^{-1}\frac xa$.


问:
Representation of the sum of simple functions
In the book Elementary Introduction to the Lebesgue Integral  by Steven Krantz, the author considers two simple functions  $\varphi$ and $\psi$ and says that the sum of these two functions has the representation given in equation $(3.3.4)$. I think that in the second term on the RHS of $(3.3.4)$ (respectively third) there is no the sum $\sum_{l=1}^{m}$ (respectively $\sum_{j=1}^{k}$).

$\def\phi{\varphi}$ Suppose now that $$\phi = \sum_{j=1}^k a_j\cdot \chi E_j \qquad \text{and} \qquad \psi = \sum_{\ell=1}^mb_\ell\cdot \chi F_\ell$$
with the $E_j$ pairwise disjoint and the $F_j$ pairwise disjoint. Then $\phi+\psi$ has the representation:
$$\scriptsize{\phi+\psi = \sum_{j=1}^k\sum_{\ell=1}^m(a_j+b_\ell)\cdot\chi_{ E_j\cap F_\ell} + \sum_{j=1}^k a_j\cdot \chi_{ E_j\setminus(F_1\cup\dots\cup F_m)}+ \sum_{\ell=1}^m} b_\ell\cdot\chi_{ F_\ell\setminus(E_1\cup\dots\cup E_k)} \tag{3.3.4}$$
This last representation for $\phi+\psi$ is a bit confusing since different occurences of $a_j+b_\ell$ may be equal (so that (3.3.4) is not necessarily the standard representation of simple function given the original definition). Let $c_p$, $p=1,2,\dots,r$ be the distinct numbers in the collection $\{a_j+b_\ell : j=1,\dots,k ; \ell=1,\dots,m\}$. Let $H_p$ be the union of all those seys $E_j\cap F_\ell\neq \emptyset$ so that $a_j+b_\ell=c_p$. Then
$$p(H_p) = \sum_{(p)} \mu(E_j\cap F_\ell).$$

see reference

My question: I would like to know how to prove that this representation is valid.

I started to separate in cases, but I couldn't find an elegant way to write all the cases.

答:
The reason you can write the sum in the form of $(3.3.4)$ is basic set theory in disguise. The function $\varphi + \psi$ is supported on $(\cup_j E_j) \cup (\cup_l F_l)$ since the individual pieces are supported on $\cup_j E_j$ and $\cup_l F_l$, respectively. We do not know anything about these sets, so to capture the full behavior, there are cases to consider. We know in general that for any two sets $A$ and $B$, $A = (A\cap B) \cup (A\setminus B)$ (check this via element memberships if you're not sure). Thus

$$\varphi = \sum_{j=1}^k a_j \chi_{E_j} = \sum_{j=1}^k a_j \chi_{(E_j \cap F_l) \cup (E_j \setminus F_l)}.$$

The indicator function of the union of two distinct sets (which $A\cap B$ and $A\setminus B$ are for any sets $A$ and $B$) is the sum of the indicator functions on the component pieces, i.e. $\chi_{(A\cap B)\cup(A\setminus B)} = \chi_{A \cap B} + \chi_{A\setminus B}$. Thus

$$\varphi = \sum_{j=1}^k a_j \chi_{E_j \cap F_l} + \sum_{j=1}^k a_j \chi_{E_j\setminus F_l}.$$

Repeat this process for each $F_l$. Then do the same for $\psi$ and you'll recover each piece in $(3.3.4)$ after summing them all up and combining like terms.


Title: Solving recurrence with substitution

Body: I am trying to solve this recurrence $T(n) = 4T(n − 2) + 2^{2n}$ by using substitution and knowing $T(1)=1,T(2)=2$ and here is my attempt:Expanding $T(n)=4T(n-2)+2^{2n}$ using $T(n-2)=4T(n-4)+2^{2(n-2)}$, we get $T(n)=4(4T(n-4)+2^{2(n-2)})+2^{2n}=4^2T(n-4)+2^{2n-2}+2^{2n}=4^2(4T(n-6)+2^{2(n-4)})+2^{2n-2}+2^{2n}=4^3T(n-6)+2^{2n-4}+2^{2n-2}+2^{2n}=...=4^k(T(n-2k)+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$.Substituting $T(n-2k)=2, i.e. T(2)=2$, $n-2k=2$ then $k=\frac{n-2}{2}$, so our equation looks like $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$ with $k=\frac{n-2}{2}$, $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}=2^{n-2}+2^{n+4}+...+2^{2n}=2^{n-2}+\frac{2^{n+4}(1-2^{\frac{n-4}{2}})}{1-2}=2^{n-2}+2^{\frac{3n+4}{2}}-2^{n+4}$. However I think this answer doesn't satisfy $T(2)=2$ and I really don't know where went wrong. Is there another way to do this problem?

Body:
Hint

$$T(n)=4T(n-2)+4^{n}$$
$$T(n)=4^2T(n-4)+4^{n}+4^{n-1}$$

by induction, find that

$$T(n)=4^kT(n-2k)+4^{n}+4^{n-1}+...+4^{n-(k-1)}.$$

Now, take $n=2k$ and $n=2k+1$ and find your general formula.


问:
Ideals of quotient algebras.
Suppose $I$ and $J$ are ideals of a Lie Algebra L.
I know that we have the fact that:

$\frac{I+J}{J} \cong \frac{I}{I\cap J}$

Prove that the ideals of $\frac{L}{I}$ - the quotient algebra of L defined by $x + I$ $x \in
L$ are of the form $\frac{K}{I}$ where K is an ideal of L containing I.

答:
If $f\colon L\to L'$ is a surjective homomorphism of Lie algebras, then, for any ideal $I$ of $L$,

$$
f^\to(K)=\{f(x):x\in K\}
$$

is an ideal of $L'$.

For any ideal $K'$ of $L'$,
$$
f^\gets(K')=\{x\in L:f(x)\in K'\}
$$
is an ideal of $L$ (here surjectivity is not needed), containing $\ker f$.

If an ideal $K$ of $L$ contains $\ker f$, then $f^\gets(f^\to(K))=K$. Moreover, for any ideal $K'$ of $L'$, $f^\to(f^\gets(K'))=K'$. The proofs of these two statements are easy.

Therefore there is a bijection, given by $f^\to$ and $f^\gets$ between the ideals of $L$ containing $\ker f$ and the ideals of $L'$.

Apply this to the canonical homomorphism $L\to L/I$, when $I$ is an ideal of $L$.

Title: Ideals of quotient algebras.


问:
Verify that the set of polynomials over $\mathbb{Q}$ is not a field
First let $\mathbb{F} = (\mathbb{Q}[x], +_\mathbb{F},\times_\mathbb{F})$ be the structure such $\mathbb{Q}[x]$ is the set of all polynomials over $\mathbb{Q}$ and $+_\mathbb{F}$ and $\times_\mathbb{F}$ are the ordinary addition and multiplication operations on polynomials.

I already have proved by other exercises that $\mathbb{F}$ is a integral domain, thus i will skip the axioms for integral domains and start by checking the existence of a multiplicative inverse for every element of $\mathbb{Q}[x]$.

Let $f$ be a polynomial in $\mathbb{Q}[x]$, if there is some other polynomial $g$ which is the multiplicative inverse of $f$, then we can satisfy $fg = 1_\mathbb{F}$, that is $[fg(0) = 1_\mathbb{Q} \land fg(n)=0, \forall n\gt0]$, but if $f(0) = 0_\mathbb{Q}$, we have $fg(0) = 0_\mathbb{Q}\times g(0) = 0_\mathbb{Q} \neq 1_\mathbb{Q}$

Therefore any polynomial over $\mathbb{Q}$ which have the first coefficient equal $0_\mathbb{Q}$, dont have a multiplicative inverse, I think this suffice to show that $\mathbb{F}$ is not a field.

答:
Your argument works, but you can make it even simpler - all you need is one non-zero element without an inverse. We will show that $x$ has no inverse.

Suppose that $x f(x) = 1$ for some polynomial $f(x)$. Then $f(x) = \sum_{i=0}^n a_ix^i$ for some elements $a_i\in \mathbb{Q}$. Then
$$
xf(x) = \sum_{i=0}^na_ix^{i+1} = 1.
$$
In particular (by definition of polynomials), comparing constant coefficients gives $0=1$, a contradiction, so we are done.

Note: We could have made things even easier by evaluating at zero (i.e. if $xf(x) = 1$ then $0f(0) = 1$ so $0 = 1$), but I think it's better to compare coefficients, because then we only make use of the definition of polynomials. Evaluating the polynomial automatically makes use of (slightly) more sophisticated machinery, which in this case is not needed.


问:
nullity of matrices
If A ,is an n x m matrix and B is an m x n matrix, then nullity(AB) =
nullity(BA).

Determine if the statement is true.

For this question, how do I go about finding the solution? What's the thought process? I never get the type of questions where the required answer is to prove a statement right or wrong, it feels like plain guess and check.

答:
Let $n=m=2$ and

$$A:=\begin{pmatrix}0& 0\\
   0 & 1\\ \end{pmatrix} \ \ \text{and} \ \
B:=\begin{pmatrix}0&1\\
   0 & 0\end{pmatrix}.\tag{2}$$

Then we have $AB=0$ but $BA \ne 0$.This shows that

$$nullity(AB)  \ne  nullity(BA).$$

Title: nullity of matrices


问:
Maximization of a Function over a Sequence of Nested Compact Sets
Given a continuous function $f:\mathbb{R}\to\mathbb{R}$ and a sequence of nested compact subsets of $\mathbb{R}$, $A_{n}\supseteq A_{n+1}$ for all $n\ge 1$. Let $A_{\infty}=\cap_{n\ge 1}A_n$. Want to show $$\lim_{n\to\infty} \max_{x\in A_n}f(x)=\max_{x\in A_{\infty}}f(x).$$

Let $f^*_n=\max_{x\in A_n}f(x)$. Since $A_n$'s are nested, $f^*_n$ is a non-increasing sequence. By monotone convergence theorem, $\lim_{n\to\infty}f^*_n$ exists, which we denote it by $L$. The next step is to show $L=\max_{x\in A_{\infty}}f(x)$.

Since $L=\lim_{n\to\infty}f^*_n$, we know that for any $\epsilon>0$, there exists an $N$ such that $|\max_{x\in A_n}f(x)-L|<\epsilon$ for all $n\ge N$. As the inequality is true for all $n\ge N$, we may say that $|\max_{x\in \cap_{n\ge N} A_n}f(x)-L|<\epsilon$ and hence $|\max_{x\in A_\infty}f(x)-L|<\epsilon$.

I feel the last step of my approach is not rigorous. I wonder if anyone can point out the problem or give me other suggestion to prove this result. Thanks in advance.

答:
Well, the problem is that you say "As the inequality is true for all $n\geq N$, we may say that $|\max_{x\in \cap_{n≥N}A_n}f(x)−L|<\varepsilon$" without providing a justification. The entire point of the exericse is establishing a link between the quantities $\max_{x\in A_n} f(x)$ and $\max_{x\in \cap_{n\geq N} A_n}f(x)$. Indeed, just consider the non-compact case where
$$
f(x)=\begin{cases}1 & x\in  [0,\infty) \\ -1 & else\end{cases}
$$
and $A_n=(-\infty,0)\cup [n,\infty)$. Then, the $A_n$ are nested, $\max_{x\in A_n}f(x)$ exists for every $n$ and is non-decreasing (it's constantly equal to 1). However, its limit is not equal to $\max_{x\in \cap_{n\geq 1} A_n} f(x)=-1$, even though the latter quantity exists.

So clearly, we need compactness for more than just the the abstract existence of a limit of the $f_n^*$.

Let $\varepsilon$ be given and consider the Hausdorff distance
$$
d(A_N,\cap_{n=1}^{\infty}A_n)=\max\{\sup_{x\in A_N}\inf_{y\in \cap_{n\geq 1}A_n} |x-y|,\sup_{x\in \cap_{n\geq 1} A_n} \inf_{y\in A_N} |x-y| \}
$$
We claim that $d(A_N,\cap_{n=1}^{\infty}A_n)\to 0$ as $N\to\infty$. Indeed, if that were not the case, there would exist $\varepsilon>0$ and sequences $x_N\in A_N$ and $y_N\in \cap_{n\geq 1}A_n$ such that either $|x_N-y|\geq \varepsilon$ for all $y\in \cap_{n\geq 1}A_n$ or $|y_N-x|\geq \varepsilon$ for all $x\in A_N$. Since $\cap_{n\geq 1} A_n\subseteq A_N,$ the latter case is absurd. Hence, we can focus on the $x_N$. Since $x_N$ is a sequence in $A_1$, we can extract a convergenct subsequence $x_{N_k}$ with limit $x_{\infty}$. For given $K,$ $x_{N_k}\in A_K$ for $k$ large and thus, we get that $x_{\infty} \in A_K$, since $A_K$ is closed. Accordingly $x_{\infty}\in \cap_{n=1}^{\infty}, A_n$. However, that is absurd, since then, $|x_N-x_{\infty}|\geq \varepsilon$ for all $N$. Thus, we must have $d(A_N,\cap_{n=1}^{\infty} A_n)\to 0$.

Hence, given $\varepsilon>0$, fix $\delta>0$ such that $|x-y|\leq \delta$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y\in A_1$. Then, fix $N$ large enough that $d(A_N,\cap_{n=1}^{\infty} A_n)<\delta$. Then, for any $x\in A_N$, there exists $y\in \cap_{n\geq 1} A_n$ such that $|x-y|<\delta$. However, then $|f(x)-f(y)|<\varepsilon$. In particular, $f(x)\leq \max_{y\in \max \cap_{n\geq 1} A_n} f(y)+\varepsilon$. Since $x$ was arbitrary, we get that
$$
\max_{y\in \max \cap_{n\geq 1} A_n} f(y)\leq \max_{x\in A_N} f(x)\leq \max_{y\in \max \cap_{n\geq 1} A_n} f(y)+\varepsilon
$$
for $N$ sufficiently large (note that the first inequality is obvious). Since $\varepsilon$ was arbitrary, this finishes the proof.


问:
How parabolic subgroups are determined by roots?
I am interested in understanding the general construction of important subgroups in reductive groups, and how they are parametrized (Borel, Levi, parabolic, etc.). But for simplicity I take the example of $SL(3)$ over a field $k$.

There are six roots, three positive roots and thus three "root" hyperplanes (do they have a name?). I saw many times the statement that parabolic subgroups correspond to subsets of positive roots, and I would like to understand how this correspondence work. Give a parabolic or a subset of roots, how can I construct the corresponding element on the other side?

More precisely, let $M$ be a standard minimal Levi subgroup. I would like to understand why

the set of parabolic having Levi component $M$ is in bijection with the (6) open Weyl chambers (i.e. the connected sectors between hyperplanes)

the set of Levi subgroups containing M is in bijection with the (5) "subspaces" in the diagram: the plane, the three line, and the origin

the set of parabolics containing $M$ is in bijection with the (13) "half-subspaces": the half-lines, the chambers and the origin.

I am really troubled by these statements but I would like to develop a geometric version of this reductive structure. What is a good reference for it (with lots of examples)?

答:
I should say right away that I am only familiar with the case that $k=\mathbb C$, I still hope that the answer is helpful.

Parabolics do not correspond to subsets of positive roots but to subsets of simple roots. There is a description in terms of saturated subsets of positive roots (i.e. subsets $\Phi\subset\Delta^+$ such that if $\alpha,\beta\in\Phi$ and $\alpha+\beta$ is a root, then $\alpha+\beta\in\Phi$), but it turns out that such a subset is uniquely determined by the simple roots it contains. This leads to a rather simple description of parabolic subalgebras in the Lie algebra, but it sounds like this is not quite what you are after.

On a more group theoretic level, the best description that comes to my mind is to start from an irreducible representation $V$ of your group and consider the associated action on the projective space $\mathcal P(V)$. Then the highest weight line of $V$ determines a point in $\mathcal P(V)$ and the stabilizer of that point is a parabolic subgroup. Indeed, the orbit of that point $\mathcal P(V)$ turns out to be the unique closed orbit in there. In appropriate conventions this is the group corresponding to the subset of all those simple roots which are not orthgonal to the highest weight of $V$. So maximal parabolic subgroups can be obtained from fundamental representations.

In the case of $SL(3,\mathbb C)$ you get 3 parabolics, the stabilizer of a line, the stabilizer of a plane and the stabilizer of a full flag in the standard representation.


问:
Understanding why Minkowski’s inequality doesn't hold true for $0 < p < 1$?
The triangle inequality given by

$\left(\sum_{i=1}^n |x_i+y_i|^p\right)^{1/p}\leq \left(\sum_{i=1}^n |x_i|^p\right)^{1/p} + \left(\sum_{i=1}^n |y_i|^p\right)^{1/p}$

is known as “Minkowski’s inequality" which holds true for $1\leq p <\infty$ while for $0 < p < 1$ it doesn't  hold.

Searching through net I have found that  this inequality holds for $1\leq p <\infty$  is related to the observation that for such $p$ the function $x \to x^p$ for $x \geq 0$ is convex.

The failure of the triangle inequality is related to the observation that for $0 <
p < 1$,the function $x \to x^p$ for $x \geq 0$ is not convex.

I am not able to understand these points. Could anybody explain me?

Thanks

答:
A function $h(x)$ is convex if it satisfies

$$h\left ( \alpha x_1 + (1-\alpha)x_2 \right) \le \alpha h(x_1) + (1-\alpha)h(x_2),
\quad\alpha\in [0,1]\tag{1}$$

Geometrically this means that for $x_1\le x\le x_2$ the graph of $h(x)$ must lie under the line connecting the points $h(x_1)$ and $h(x_2)$. Inspecting the graph of $h(x)=x^p$ it is easy to see that the function $x^p$ is convex for $p\ge 1$, but not for $0<p<1$.

In the proof of Minkowski's inequality they make use of exactly this fact. The proof starts by proving that the sum of some functions $f$ and $g$ has finite $p$-norm if both $f$ and $g$ do. We get

$$\left | \frac{f}{2}+\frac{g}{2}\right |^p\le
\left | \frac{|f|}{2}+\frac{|g|}{2}\right |^p$$

This inequality has nothing to do with convexity but only follows from taking the magnitudes of $f$ and $g$. In the following step, however, we make use of convexity:

$$\left | \frac{|f|}{2}+\frac{|g|}{2}\right |^p\le
\frac{|f|^p}{2}+\frac{|g|^p}{2}$$

Here we made use of (1) with $\alpha=1/2$ and $h(x)=x^p$, $p\ge 1$.
The rest of the proof builds on this result, that's why convexity (and therefore $p\ge 1$) is necessary for Minkowsi's inequality to hold.


问:
Normal form of a non linear system
Let the system:
$\begin{cases}
\dot{x_1}=a\sin{x_2}\\
\dot{x_2}=-x_1^2+u\\
y=x_1
\end{cases}$

that is:

$\begin{cases}
\dot{x}=
\begin{bmatrix} a\sin{x_2}\\
-x_1^2
\end{bmatrix}+\begin{bmatrix} 0\\
1
\end{bmatrix}u=f(x)+g(x)u\\
y=x_1
\end{cases}$

Since the relative degree is equal to $2=n$, I think, looking at the system, that since the input appears only in $\dot{x_2}$, the system is in normal form. But to be sure I have tried to write how it should be the system in normal form, so:
$\begin{cases}
\dot{\xi_1}=\xi_2\\
\dot{\xi_2}=\mathcal{L}_f^2h(x)+\mathcal{L}_g\mathcal{L}_fh(x)u\\
\end{cases}$

(where I have used the lie derivative for the notation and $y=\xi$)

The $\mathcal{L}_fh(x)$=$a\sin{x_2}$ and so $\mathcal{L}_f^2h(x)$=$a\cos{x_2}(-x_1^2)$.

Then $\mathcal{L}_g\mathcal{L}_fh(x)u$=$a\cos{x_2}u$, so $\dot{\xi_2}=a\cos{x_2}(-x_1^2+u)$.

So my question is: why there is the factor $a\cos{x_2}$ in $\dot{\xi_2}$ that there is no in $\dot{x_2}$, even if the original system seems to be normal? The system written in $\xi$ should be exactly equal, right?

答:
They are equal. What you basically do is a nonlinear state (and input) transformation to obtain a linear system basically. So you define a new input $v$ as
$$v = L^2_fh(x)+L_gL_fh(x)u$$
such that
$$\begin{align}
\dot\xi_1 & = \xi_2; \qquad (1a) \\
\dot\xi_2 & = v \qquad (1b)
\end{align}
$$
So if you draw this in a block diagram (I'll do it in text for now), you obtain:

--> [ NL transformation ] --> [original system] --> [NL transformation]$^{-1}$ -->

which is equal to

--> (1) -->

edit:
What you do to obtain a normal form is you first differentiate the output, such that you can obtain something like $y^{(n)}(t)+\cdots+\dot y(t) + y(t) = v(t)$. So:
\begin{align}
y &= x_1 = \xi_1 \quad \text{(first 'normal' state)} \qquad (2a)\\
\dot y & = \dot x_1 = \xi_2 \quad \text{(second 'normal' state)}\quad(2b)
\end{align}
We only have two state, so two 'normal' state should suffice. Furthermore, the jacobian of $[y, \dot{y}]^\top$ w.r.t. the state is full rank around the origin, so this is fine.

Next you calculate your dynamics of the normal system, which requires to determine the time-derivative of (2). This implies taking the derivative w.r.t. $x_2$ of $a\sin x_2$ (which is obviously $a \cos x_2$)

I hope this explains your question

Title: Normal form of a non linear system


问:
Calculate length of module
Let $A$ be the ring $ A := \mathbb{R}[X] / (X^3 - X^2 + X -1)$. How would I go about calculating $l_{\mathbb{R[X]}}(A)$ the length of $A$ as $\mathbb{R[X]}$ module?</p>

答:
Since ideal $I=(x^3-x^2+x-1)$ is contained in two maximal ideals $m_1=(x-1)$ and $m_2=(x^2+1)$ we can construct (at least) two composition series. Notice that $I=m_1m_2=m_1\cap m_2$ since $m_1$ and $m_2$ are coprime $m_1+m_2=R:=\mathbb{R}[x]$. First composition series is
$$
0=M_0 \subset M_1=m_2/I \subset M_2=R/I.
$$
In this case $M_2/M_1 \cong R/m_2 \cong \mathbb{C}$ and $M_1 = m_2/m_1 \cap m_2\cong (m_1+m_2)/m_1 = R/m_1 \cong \mathbb{R}$. Second composition series is
$$
0=M_0 \subset M_1=m_1/I \subset M_2=R/I.
$$
So, $M_2/M_1 \cong R/m_1 \cong \mathbb{R}$ and $M_1 = m_1/m_1 \cap m_2\cong (m_1+m_2)/m_2 = R/m_2 \cong \mathbb{C}$.

We can see that quotients are isomorphic but permuted as predicted by Jordan-Holder theorem. But in any case, the length is two $\ell_R(R/I)=2$.

Title: Calculate length of module


问:
Example of a weak inverse that is not an inverse

In the theory of semigroups, a weak inverse of an element $x$ in a semigroup $(S, \cdot)$ is an element $y$ such that $y \cdot x \cdot y = y$.

What is an example of this that is not also an inverse?

答:
Let $S$ be a semigroup and let $x \in S$. A weak inverse of $x$ is an element $y$ such that $yxy = y$. It is an inverse of $x$ if it also satisfies the equality $xyx = x$. In particular, every idempotent is its own inverse.

Let $S = \{a, 0\}$ be the commutative semigroup defined by $a^2 = 0$ and $0^2 = a0 = 0$. Then $0$ is a weak inverse of $a$, since $0a0 = 0$, but is not an inverse of $a$ since $a0a = 0 ≠ a$.

Finally observe that if $y$ is a weak inverse of $x$, then $xyx$ is an inverse of $y$ since $(xyx)y(xyx) = x(yxy)xyx = xyxyx = xyx$ and $y(xyx)y = yxyxy = yxy$.