问:
Show that $ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z) = ... $
Let $X, Y, Z$ be $2 \times 2$ matrices.  Show that these two matrix combinations are equal:

$ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z)   $

$ \text{Tr}(X) \, \text{Tr}(YZ) + \text{Tr}(YX)\,\text{Tr}(Z)+ \text{Tr}(Z)\,\text{Tr}(XY) $

There's lots of identities for matrix trace, here's the only other one that I know, that we can switch the order of the matrix:
$$ \text{Tr}(XYZ) = \text{Tr}(YZX) $$
I might specifically need to have $X,Y,Z \in \text{SL}_2(\mathbb{C})$.  The right side looks symmetric under the cycle permutation of $(XYZ)$ (basically a triangle) while the left side does not.

答:
(Edit: The result is true and can be derived from Cayley-Hamilton for $2\times 2$ matrices, and writing $det(A) = \frac{1}{2}[Tr(A)^2 - Tr(A^2)]$ for $A$ a $2\times 2$ matrix. See update 2 below.)

There is indeed a typo. We approach it brutally. Using what you have stated, SageMath (https://sagecell.sagemath.org/) gives the difference of your two expressions to be:

```
var('x1','x2','x3','x4','y1','y2','y3','y4','z1','z2','z3','z4')
X = matrix([[x1,x2],[x3,x4]])
Y = matrix([[y1,y2],[y3,y4]])
Z = matrix([[z1,z2],[z3,z4]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Z.trace()*(X*Y).trace())).expand()

---
output:
-x3*y2*z1 - x2*y3*z1 + x1*y4*z1 - x4*y4*z1 + x3*y1*z2 + x3*y4*z2 + x2*y1*z3 + x2*y4*z3 - x1*y1*z4 + x4*y1*z4 - x3*y2*z4 - x2*y3*z4
```

However, if we change the second expression to something more symmetric:

$$ \text{Tr}(X) \, \text{Tr}(YZ) + \text{Tr}(YX)\,\text{Tr}(Z)+ \color{blue}{ \text{Tr}(Y)\,\text{Tr}(XZ) }$$

Then the difference between this and $$ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z)   $$  is

```
var('x1','x2','x3','x4','y1','y2','y3','y4','z1','z2','z3','z4')
X = matrix([[x1,x2],[x3,x4]])
Y = matrix([[y1,y2],[y3,y4]])
Z = matrix([[z1,z2],[z3,z4]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Y.trace()*(X*Z).trace())).expand()

---
output:
0
```

(And thus a proof, albeit not enlightening.)

Remark. This identity is not true if $X,Y,Z$ are $3\times 3$ matrices:

```
var('x1','x2','x3','x4','x5','x6','x7','x8','x9','y1','y2','y3','y4','y5','y6','y7','y8','y9','z1','z2','z3','z4','z5','z6','z7','z8','z9')
X = matrix([[x1,x2,x3],[x4,x5,x6],[x7,x8,x9]])
Y = matrix([[y1,y2,y3],[y4,y5,y6],[y7,y8,y9]])
Z = matrix([[z1,z2,z3],[z4,z5,z6],[z7,z8,z9]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Y.trace()*(X*Z).trace())).expand()

---
output:
x9*y5*z1 - x8*y6*z1 - x6*y8*z1 + x5*y9*z1 - x9*y4*z2 + x7*y6*z2 + x6*y7*z2 - x4*y9*z2 + x8*y4*z3 - x7*y5*z3 - x5*y7*z3 + x4*y8*z3 - x9*y2*z4 + x8*y3*z4 + x3*y8*z4 - x2*y9*z4 + x9*y1*z5 - x7*y3*z5 - x3*y7*z5 + x1*y9*z5 - x8*y1*z6 + x7*y2*z6 + x2*y7*z6 - x1*y8*z6 + x6*y2*z7 - x5*y3*z7 - x3*y5*z7 + x2*y6*z7 - x6*y1*z8 + x4*y3*z8 + x3*y4*z8 - x1*y6*z8 + x5*y1*z9 - x4*y2*z9 - x2*y4*z9 + x1*y5*z9
```

UPDATE.(1/20/2021) Upon some reflection, we see that there is some permutation cycle structure in these expressions. Let us denote $(M) := \text{Tr}(M)$ to simplify the notation. Then we do indeed have for four $3\times 3$ matrices $X,Y,Z,W$, the following equality
$$
\sum_{\sigma\in \text{Perm}(X,Y,Z,W)}\text{sign}(\sigma) \sigma = 0
$$

where we interpret $\sigma$ first in cycle structure, and then interpret it as trace, (e.g., $\sigma = (X)(Y)(ZW) = \text{Tr}(X)\text{Tr}(Y)\text{Tr}(ZW)$). This is probably true for $n$ many matrices of size $k\times k$ where $k\le n-1$. The $n = 3$ case is this posted problem, and the $n=4$ case I verify it here:

```
var('x1','x2','x3','x4','x5','x6','x7','x8','x9','y1','y2','y3','y4','y5','y6','y7','y8','y9','z1','z2','z3','z4','z5','z6','z7','z8','z9','w1','w2','w3','w4','w5','w6','w7','w8','w9')
X = matrix([[x1,x2,x3],[x4,x5,x6],[x7,x8,x9]])
Y = matrix([[y1,y2,y3],[y4,y5,y6],[y7,y8,y9]])
Z = matrix([[z1,z2,z3],[z4,z5,z6],[z7,z8,z9]])
W = matrix([[w1,w2,w3],[w4,w5,w6],[w7,w8,w9]])

p4    = (X*Y*Z*W).trace() + (X*Y*W*Z).trace() + (X*Z*W*Y).trace()+ (X*Z*Y*W).trace()+ (X*W*Y*Z).trace() + (X*W*Z*Y).trace()
p1111 = X.trace()*Y.trace()*Z.trace()*W.trace()
p31   = (X.trace()*(Y*Z*W).trace() + X.trace()*(Y*W*Z).trace()
       + Y.trace()*(X*Z*W).trace() + Y.trace()*(X*W*Z).trace()
       + Z.trace()*(X*Y*W).trace() + Z.trace()*(X*W*Y).trace()
       + W.trace()*(X*Y*Z).trace() + W.trace()*(X*Z*Y).trace())
p22   = (X*Y).trace()*(Z*W).trace() + (X*Z).trace()*(Y*W).trace() + (X*W).trace()*(Y*Z).trace()
p211  = ((X*Y).trace()*Z.trace()*W.trace() + (X*Z).trace()*Y.trace()*W.trace() + (X*W).trace()*Z.trace()*Y.trace()
         +(Y*Z).trace()*X.trace()*W.trace() +(Y*W).trace()*X.trace()*Z.trace() + (W*Z).trace()*X.trace()*Y.trace())

(p1111 + p31 +p22 - p211-p4).expand()

---
output:
0
```

So a conjecture (at least to me):
Let $X_1,\ldots,X_n$ be $n$ many $k\times k$ matrices over a commutative ring where $k\le n-1$, then
$$
\sum_{\sigma\in \text{Perm}(X_1,\ldots,X_n)} \text{sign}(\sigma) \sigma = 0,
$$

where we interpret $\sigma$ in cyclic decomposition and then as a product of traces as before.
Spiritually this suggests there is a combinatorial sieve / inclusion-exclusion proof of this. It eludes me at this moment.

Update 2. Indeed this is true, and can be derived from Cayley-Hamilton. See Polynomial Identities and the Cayley-Hamilton Theorem by Formanek (he derives this particular one in Eqs (1)-(6)):

https://www.researchgate.net/publication/225640755_Polynomial_identities_and_the_Cayley-Hamilton_theorem

Body:
(Edit: The result is true and can be derived from Cayley-Hamilton for $2\times 2$ matrices, and writing $det(A) = \frac{1}{2}[Tr(A)^2 - Tr(A^2)]$ for $A$ a $2\times 2$ matrix. See update 2 below.)

There is indeed a typo. We approach it brutally. Using what you have stated, SageMath (https://sagecell.sagemath.org/) gives the difference of your two expressions to be:

var('x1','x2','x3','x4','y1','y2','y3','y4','z1','z2','z3','z4')
X = matrix([[x1,x2],[x3,x4]])
Y = matrix([[y1,y2],[y3,y4]])
Z = matrix([[z1,z2],[z3,z4]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Z.trace()*(X*Y).trace())).expand()

---
output:
-x3*y2*z1 - x2*y3*z1 + x1*y4*z1 - x4*y4*z1 + x3*y1*z2 + x3*y4*z2 + x2*y1*z3 + x2*y4*z3 - x1*y1*z4 + x4*y1*z4 - x3*y2*z4 - x2*y3*z4

However, if we change the second expression to something more symmetric:

$$ \text{Tr}(X) \, \text{Tr}(YZ) + \text{Tr}(YX)\,\text{Tr}(Z)+ \color{blue}{ \text{Tr}(Y)\,\text{Tr}(XZ) }$$

Then the difference between this and $$ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z)   $$  is

var('x1','x2','x3','x4','y1','y2','y3','y4','z1','z2','z3','z4')
X = matrix([[x1,x2],[x3,x4]])
Y = matrix([[y1,y2],[y3,y4]])
Z = matrix([[z1,z2],[z3,z4]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Y.trace()*(X*Z).trace())).expand()

---
output:
0

(And thus a proof, albeit not enlightening.)

Remark. This identity is not true if $X,Y,Z$ are $3\times 3$ matrices:

var('x1','x2','x3','x4','x5','x6','x7','x8','x9','y1','y2','y3','y4','y5','y6','y7','y8','y9','z1','z2','z3','z4','z5','z6','z7','z8','z9')
X = matrix([[x1,x2,x3],[x4,x5,x6],[x7,x8,x9]])
Y = matrix([[y1,y2,y3],[y4,y5,y6],[y7,y8,y9]])
Z = matrix([[z1,z2,z3],[z4,z5,z6],[z7,z8,z9]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Y.trace()*(X*Z).trace())).expand()

---
output:
x9*y5*z1 - x8*y6*z1 - x6*y8*z1 + x5*y9*z1 - x9*y4*z2 + x7*y6*z2 + x6*y7*z2 - x4*y9*z2 + x8*y4*z3 - x7*y5*z3 - x5*y7*z3 + x4*y8*z3 - x9*y2*z4 + x8*y3*z4 + x3*y8*z4 - x2*y9*z4 + x9*y1*z5 - x7*y3*z5 - x3*y7*z5 + x1*y9*z5 - x8*y1*z6 + x7*y2*z6 + x2*y7*z6 - x1*y8*z6 + x6*y2*z7 - x5*y3*z7 - x3*y5*z7 + x2*y6*z7 - x6*y1*z8 + x4*y3*z8 + x3*y4*z8 - x1*y6*z8 + x5*y1*z9 - x4*y2*z9 - x2*y4*z9 + x1*y5*z9

UPDATE.(1/20/2021) Upon some reflection, we see that there is some permutation cycle structure in these expressions. Let us denote $(M) := \text{Tr}(M)$ to simplify the notation. Then we do indeed have for four $3\times 3$ matrices $X,Y,Z,W$, the following equality
$$
\sum_{\sigma\in \text{Perm}(X,Y,Z,W)}\text{sign}(\sigma) \sigma = 0
$$

where we interpret $\sigma$ first in cycle structure, and then interpret it as trace, (e.g., $\sigma = (X)(Y)(ZW) = \text{Tr}(X)\text{Tr}(Y)\text{Tr}(ZW)$). This is probably true for $n$ many matrices of size $k\times k$ where $k\le n-1$. The $n = 3$ case is this posted problem, and the $n=4$ case I verify it here:

var('x1','x2','x3','x4','x5','x6','x7','x8','x9','y1','y2','y3','y4','y5','y6','y7','y8','y9','z1','z2','z3','z4','z5','z6','z7','z8','z9','w1','w2','w3','w4','w5','w6','w7','w8','w9')
X = matrix([[x1,x2,x3],[x4,x5,x6],[x7,x8,x9]])
Y = matrix([[y1,y2,y3],[y4,y5,y6],[y7,y8,y9]])
Z = matrix([[z1,z2,z3],[z4,z5,z6],[z7,z8,z9]])
W = matrix([[w1,w2,w3],[w4,w5,w6],[w7,w8,w9]])

p4    = (X*Y*Z*W).trace() + (X*Y*W*Z).trace() + (X*Z*W*Y).trace()+ (X*Z*Y*W).trace()+ (X*W*Y*Z).trace() + (X*W*Z*Y).trace()
p1111 = X.trace()*Y.trace()*Z.trace()*W.trace()
p31   = (X.trace()*(Y*Z*W).trace() + X.trace()*(Y*W*Z).trace()
       + Y.trace()*(X*Z*W).trace() + Y.trace()*(X*W*Z).trace()
       + Z.trace()*(X*Y*W).trace() + Z.trace()*(X*W*Y).trace()
       + W.trace()*(X*Y*Z).trace() + W.trace()*(X*Z*Y).trace())
p22   = (X*Y).trace()*(Z*W).trace() + (X*Z).trace()*(Y*W).trace() + (X*W).trace()*(Y*Z).trace()
p211  = ((X*Y).trace()*Z.trace()*W.trace() + (X*Z).trace()*Y.trace()*W.trace() + (X*W).trace()*Z.trace()*Y.trace()
         +(Y*Z).trace()*X.trace()*W.trace() +(Y*W).trace()*X.trace()*Z.trace() + (W*Z).trace()*X.trace()*Y.trace())

(p1111 + p31 +p22 - p211-p4).expand()

---
output:
0

So a conjecture (at least to me):
Let $X_1,\ldots,X_n$ be $n$ many $k\times k$ matrices over a commutative ring where $k\le n-1$, then
$$
\sum_{\sigma\in \text{Perm}(X_1,\ldots,X_n)} \text{sign}(\sigma) \sigma = 0,
$$

where we interpret $\sigma$ in cyclic decomposition and then as a product of traces as before.
Spiritually this suggests there is a combinatorial sieve / inclusion-exclusion proof of this. It eludes me at this moment.

Update 2. Indeed this is true, and can be derived from Cayley-Hamilton. See Polynomial Identities and the Cayley-Hamilton Theorem by Formanek (he derives this particular one in Eqs (1)-(6)):

https://www.researchgate.net/publication/225640755_Polynomial_identities_and_the_Cayley-Hamilton_theorem


问:
Determine whether or not the topological spaces are homeomorphic or not
• $(X, τ_X)$ where $X = \{(x, y) ∈ \mathbb{R}^2: x^2 +y^2 = 1\}$ with the topology $τ_X$ induced by
the Euclidean metric on $\mathbb{R}^2$
;

• $(Y, τ_Y )$ where $Y = \{(x, y) ∈ \mathbb{R}^2 : x^2 −y^2 = 1\}$ with the topology $τ_Y$ induced by
the Euclidean metric on $\mathbb{R}^2$
.

答:
You can use the standard argument saying that homeomorphisms preserve connectivity/path-connectivity. I want to give a related, but slightly different proof.

Note that $$\Bbb S^1 = \{(x,y) \in \Bbb R^2 \mid x^2 + y^2 =1\}$$ is the circle. It admits a continuous surjection given by
$$f:\Bbb R \rightarrow \Bbb S^1, t \mapsto (\cos t, \sin t).$$

On the other hand the hyperbola $$X = \{(x,y)\in \Bbb R^2 \mid x^2 -y^2 =1\}$$ can be decomposed into the two disjoint subspaces
$$Y_{-}=\{(x,y) \in \Bbb R^2 \mid x^2-y^2=1, x\leq -1\}$$ and $$Y_{+}=\{(x,y) \in \Bbb R^2 \mid x^2-y^2=1, x\geq 1\}.$$

Let $g:\Bbb R \rightarrow Y$ be any continuous map. Postcomposing it with the projection onto the $x$-coordinate
$$\pi_x:Y\rightarrow (-\infty, -1]\cup[1,\infty) \subseteq \Bbb R, (x,y) \mapsto x$$
we obtain a continuous function $g‘:\Bbb R \rightarrow \Bbb R$. Note that the image of $g‘$ indeed lies in $(-\infty,-1] \cup [1,\infty)$!

Now assume that $g‘$ admits some $t_{-} \in \Bbb R$ with $g‘(t_{-}) \in (-\infty,-1]$ and some $t_{+} \in \Bbb R$ with $g‘(t_{+})\in [1,\infty)$. By the intermediate value theorem then 0 is in the image of $g‘$ as well, a contradiction!

But this shows that the image of $g‘$ is contained either in $(-\infty,-1]$ or in $[1,\infty)$. Thus the image of $g$ is contained either in $Y_{-}$ or in $Y_{+}$. But this makes it impossible for $g$ to be a continuous surjection $\Bbb R \rightarrow Y$.

On a final note, if there were an homeomorphism $\Bbb S^1 \cong Y$ then precomposing it with the continuous surjection $f:\Bbb R \rightarrow \Bbb S^1$ would yield a continuous surjection $\Bbb R \rightarrow Y$, which we proved to be impossible.

ContentLicense: CC BY-SA 4.0


问:
Solving a system of equation:
Solve for $x,y$:
\begin{align}
x^3 + y^3&=2\\
x^2 +x + 9y - 3y^2&=8 \\
\end{align}

I can find $x=y=1$ by guessing. Please help me solve it without using computer.

Thanks

Edited, sorry, I have to change $x^2$ to $x^3$ to have a nice solution, but still can't done it by hand.

答: I think you can use Newton-Raphson or any other approximation methods for multivariate equations (this is an example)</p>

ContentLicense: CC BY-SA 3.0

Title: Solving a system of equation:


问:
Prove $\vert z\vert\leq \vert \mathfrak{Re}(z)\vert+ \vert \mathfrak{Im}(z)\vert$ where $z$ is a complex number
Prove $\vert z\vert\leq \vert \mathfrak{Re}(z)\vert+ \vert \mathfrak{Im}(z)\vert$ where $z$ is a complex number

Seems quite simple but I don't know how to.

答: This is the triangle inequality for the triangle of vertices $0, \, \Re z, \, z$.


问:
Implementation in MAGMA: Field extension over the p-adics with a polynomial which is neither inertial nor Eisenstein
Let $K = Q_3$ and $L = K(a)$ be the extension of $K$ defined by the polynomial $f = x^6+3x^5-2$ (i.e. this is the minimal polynomial of $a$ over $K$). Now I would like to obtain this field $L$ in MAGMA but I have issues with that.

More precisely, it seems that I can only define extensions over the p-adics by giving either an inertial or an Eisenstein polynomial, and $f$ is neither inertial nor Eisenstein. The code I tried is this one:

```<code>K := pAdicField(3, 100);
R<x> := PolynomialRing(K);
L := ext< K | x^6+3*x^5-2 >;
</code>```

which gives me the error message:

```<code>L := ext< K | x^6+3*x^5-2 >;
           ^
Runtime error in ext< ... >: Polynomial must be Eisenstein or inertial
</code>```

I also tried to work around that by defining the maximal unramified subextension F/K (which has degree 2, see https://www.lmfdb.org/LocalNumberField/3.6.6.1) like this

```<code>K := pAdicField(3, 100);
F := UnramifiedExtension(K,2);
R<x> := PolynomialRing(F);
f := x^6-x^4-5;
f_factor := Factorization(f)[1][1];
L := ext< F | f_factor >;
Degree(L,K);
RamificationIndex(L,K);
InertiaDegree(L,K);
</code>```

But in the end, Magma says that my extension $L/K$ is unramified of degree $6$ which cannot be true (cf. the LMFDB page where it says that the extension has inertial degree $2$ instead of $6$).

Could you please tell me if there is a way around this problem or what I did wrong?

答:
It looks like you have two different polynomials $f$. The original is $f=x^6+3x^5-2$. The one you typed in your second Magma session was $f=x^6-x^4-5$.

There are at least two ways to define your extension in Magma. One involves using the "LocalField" command:

```<code>K := pAdicField(3,30);
R<x> := PolynomialRing(K);
f := x^6+3*x^5-2;
L := LocalField(K,f);
RamificationIndex(L);
InertiaDegree(L);
</code>```

Another way is to build the extension as a totally ramified cubic extension of the unramified quadratic extension, using the defining polynomials listed in the LMFDB link you gave:

```<code>K := pAdicField(3,30);
F<t> := UnramifiedExtension(K,2);
R<x> := PolynomialRing(F);
f := x^3+6*x^2+6*t*x+3*t;
L := ext<F|f>;
RamificationIndex(L,K);
InertiaDegree(L,K);
S<y> := PolynomialRing(L);
HasRoot(y^6+3*y^5-2);
</code>```

The last command shows your original polynomial $f$ has a root in the extension $L$, and thus is a defining polynomial for $L$.

Body:
It looks like you have two different polynomials $f$. The original is $f=x^6+3x^5-2$. The one you typed in your second Magma session was $f=x^6-x^4-5$.

There are at least two ways to define your extension in Magma. One involves using the "LocalField" command:

K := pAdicField(3,30);
R<x> := PolynomialRing(K);
f := x^6+3*x^5-2;
L := LocalField(K,f);
RamificationIndex(L);
InertiaDegree(L);

Another way is to build the extension as a totally ramified cubic extension of the unramified quadratic extension, using the defining polynomials listed in the LMFDB link you gave:

K := pAdicField(3,30);
F<t> := UnramifiedExtension(K,2);
R<x> := PolynomialRing(F);
f := x^3+6*x^2+6*t*x+3*t;
L := ext<F|f>;
RamificationIndex(L,K);
InertiaDegree(L,K);
S<y> := PolynomialRing(L);
HasRoot(y^6+3*y^5-2);

The last command shows your original polynomial $f$ has a root in the extension $L$, and thus is a defining polynomial for $L$.


问:
Subfields of an ultrapower $\mathbb{R}^\omega/U$ obtained by reducing $U$.
We promise that
$F(g_1, \ldots, g_k)$ denotes the composition for each $F \colon \mathbb{R}^k \to \mathbb{R}$ and $g_1,\ldots ,g_k \colon \omega \to \mathbb{R}$.
Let $M$ be a subset of $\mathbb{R}^\omega$ and  $M^\wedge$ denotes the set of all
$F(g_1, \ldots, g_k)$ where $k \in \omega$,   $g_1,\ldots, g_k \in M$, and $\ F \colon \mathbb{R} ^k \to \mathbb{R}$  where $F$ is definable in $\langle\mathbb{R},1,0, +,\cdot,<\rangle $ without parameters.
Notice that, considering the case $k=0$, $\emptyset^\wedge =$ the set of all definable elements in $\mathbb{R}$.
We say that a filter $\mathcal{F}$ on $\mathcal{P}(\omega)$ is an $M$-ultrafilter if either $\{i \in\omega \mid g(i) < g'(i) \}\in \mathcal{F}$, $\{i \in\omega \mid g(i) = g'(i) \}\in \mathcal{F}$, or $\{i \in\omega \mid g(i) > g'(i) \}\in \mathcal{F}$ for each $g,g' \in M$.
Let $f \in \mathbb{R}^\omega$ and $\mathcal{F}$ be an $M^\wedge \cup \{f\} $-ultrafilter.

My question is: Is $\mathcal{F}$ an $(M\cup \{ f\} )^\wedge$-ultrafilter?

I know that $M^\wedge/\mathcal{F}$ is a real closed field.
Thus, by the $o$-minimality of the theory of real closed fields, for each definable functions $F_1, F_2 \colon \mathbb{R}^{k+1} \to \mathbb{R}$ and $g,g_1,\ldots , g_k \in M$, there are $h_1^-,\ldots, h^-_m,h_1^+,\ldots, h_m^+, h_1,\ldots,h_n \in M^\wedge$ where maybe $h_1^-$ and $h_m^+$ are  $-\infty$ and $+\infty$, respectively,  such that
\begin{align*}
 F_1(g_1,\ldots, g_k,g) \leq_{\mathcal{F}} F_2(g_1,\ldots,g_k,g) \iff \bigvee_{1\leq l \leq m}h_l^- <_\mathcal{F} g <_\mathcal{F} h_l^+  \lor  \bigvee_{1\leq l \leq n} h_l =_{\mathcal{F}} g.
 \end{align*}
But I don't know whether this is valid if we replace $g$ by $f$.

答:
I have solved it.

Suppose that
\begin{align*}
 A := \{ i \in \omega \mid F(g_1(i),\ldots, g_k(i),f(i)) \leq 0 \} \notin \mathcal{F}.
 \end{align*}
Since $M^\wedge/ \mathcal{F} \models \mathrm{RCF}$ and $\mathrm{RCF}$ is an $o$-minimal theory, there are $h^-_1,\ldots, h^-_m,h_1^+,\ldots, h_m^+, h_1,\ldots,h_n \in M^\wedge$ where maybe $h_1^-$ and $h_m^+$ are  $-\infty$ and $+\infty$, respectively,  such that
\begin{align*}
 M^\wedge / \mathcal{F} \models
 \forall v \left( F_0([g_1]_\mathcal{F} ,\ldots, [g_k]_\mathcal{F} ,v) \leq 0
 \iff
 \left(
 \bigvee_{1\leq l \leq m}[h_l^-]_\mathcal{F} < v < [h_l^+]_\mathcal{F}  \lor  \bigvee_{1\leq l \leq n} [h_l]_\mathcal{F} = v.
 \right)
 \right)
 \end{align*}
Define
\begin{align*}
 X := \left\{ i \in \omega  \mid
 \mathbb{R} \models
 \forall v \left( F(g_1(i),\ldots, g_k(i) ,v) \leq 0
 \iff
 \left(
 \bigvee_{1\leq l \leq m}h_l^-(i) < v < h_l^+(i)  \lor  \bigvee_{1\leq l \leq n} h_l(i)= v
 \right)
 \right)
 \right\}
 \end{align*}
and
\begin{align*}
 Y := \left\{ i \in \omega \mid
 \mathbb{R} \models
 \left(
 \bigvee_{1\leq l \leq m}h_l^-(i) < f(i) < h_l^+(i)  \lor  \bigvee_{1\leq l \leq n} h_l(i)= f(i)
 \right)
 \right\}.
 \end{align*}
Since $\mathcal{F}$ is an $M^\wedge \cup \{f\}$ -ultrafilter, either $Y \in \mathcal{F}$ or $\omega \setminus Y \in \mathcal{F}$.
If $Y \in \mathcal{F}$, then  $X \cap Y \in \mathcal{F}$ and $\mathbb{R} \models F(g_1(i), \ldots , g_k(i) , f(i) )  \leq 0$ for each $i \in X\cap Y$, it is a contradiction.
Thus $\omega \setminus Y \in \mathcal{F}$.
For each $i \in X \setminus Y$, $\mathbb{R} \models F(g_1(i),\ldots , g_k(i), f(i) ) > 0 $ so that $[F(g_1,\ldots g_k , f)] _\mathcal{F} > _\mathcal{F} 0$.
Therefore, $(M \cup \{ f\} ) ^\wedge / \mathcal{F}$ is totally ordered.


问:
Equations of rotation of a vector
Let there be two completely coincident right-handed orthogonal frames  and then one of them is rotated CCW by $\theta$ degrees around the z axis. Let the unit vectors of the rotated frame be denoted as
$ x^{'} y^{'} z^{'} $ (those of stationary frame are $x y z$) The equations which express the rotated frame with respect to the stationary frame are

$x^{'}=x\cos\theta-y\sin\theta$

$y^{'}=x\sin\theta+y\cos\theta$

$z^{'}=z$

The rotation matrix is $$
    \begin{bmatrix}
    \cos \theta & -\sin\theta & 0 \\
    \sin\theta & \cos \theta & 0 \\
    0 & 0 & 1 \\
    \end{bmatrix}
$$
We know that this matrix is also a rotational operator and the operation above is equivalent to rotating a vector CCW around the z axis by the same amount.

I want to derive the rotation equations by geometrical means considering the picture above. The components of the vector before and after rotation are as depicted.

答:
Use the angle sum formulas for sine and cosine. If
$$
\begin{pmatrix}x\\y\\0\end{pmatrix}=\begin{pmatrix}r\cos\alpha\\r\sin\alpha\\0\end{pmatrix}
$$
Then
$$
\begin{pmatrix}x'\\y'\\0\end{pmatrix}=\begin{pmatrix}r\cos(\alpha+\theta)\\r\sin(\alpha+\theta)\\0\end{pmatrix}.
$$
Now,
$$
\cos(\alpha+\theta)=\cos\alpha\cos\theta-\sin\alpha\sin\theta
$$
and
$$
\sin(\alpha+\theta)=\sin\alpha\cos\theta+\sin\theta\cos\alpha
$$
so
$$
\begin{pmatrix}x'\\y'\\0\end{pmatrix}=r\cos\alpha\begin{pmatrix}\cos\theta\\\sin\theta\\0\end{pmatrix}+r\sin\alpha\begin{pmatrix}-\sin\theta\\\cos\theta\\0\end{pmatrix}.
$$
This formula is equivalent to multiplying $\begin{pmatrix}x\\y\\0\end{pmatrix}$ by $\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}$

Title: Equations of rotation of a vector


问:
What is the sum of all possible integer values of the leading cooefficient of $P(x)$?
Given

For every $x \in \mathbb{R},  P(x) \neq 0$;

$P(x)$ is a second degree polynomial;

$P(1) = 8$ and $P(2) = 2.$

答:
By $(2)$
$$ P(x)=ax^2+bx+c$$
with $a\ne 0$.
By $(1)$, certainly $a>0$, but more importantly, $P(x_0)\ge 0$ where $x_0=-\frac b{2a}$ is the point of minimum of $P$. This boils down to
$$ a\frac {b^2}{4a^2}+\frac{-b^2}{2a}+c\ge0$$
or:
$$ b^2-4ac\le 0.$$
Finally,  $(3)$ gives us two linear equations in three unknowns:
$$ a+b+c=8$$
$$ 4a+2b+c=2.$$
You can use these to express $b$ and $c$ in terms of $a$. Plug the resulting expression into the inequality and find out for which $a$ the inequality holds.


问:
For $T(z)=1/(1-z),$ find $M\in \operatorname{Aut}(\Sigma)$ $(\Sigma=\mathbb{C}\cup \{\infty\},$ the Riemann sphere) such that $MTM^{-1}=ze^{2i\pi/3}.$
I know that $ze^{2i\pi/3}$ is a rotation of $\Sigma$ by an angle $2\pi/3$ about the vertical axis through $0$ and $\infty$ and also that a Moebius function is determined by three points. This two facts are helpful?

答:
$$
\begin{array}{cc|cc}
z \mapsto ze^{2\pi i/3} & & & z \mapsto \dfrac 1 {1-z} \\[10pt]
\hline
\begin{array}{ccccc} 1 & & \longrightarrow & & e^{2\pi i/3} \\ & \nwarrow & & \swarrow \\ & & e^{4\pi i/3} \end{array} & \quad & \quad & \begin{array}{ccccc} 0 & & \longrightarrow & & 1 \\ & \nwarrow & & \swarrow \\ & & \infty\end{array}
 \\[4pt] {} \end{array}
$$
Say you start with $1,\,\,e^{2\pi i/3},\,\, e^{4\pi i/3}$ respectively. Then you feed them into $M^{-1}$ getting $0,\,\,1,\,\,\infty$ respectively. Then you apply $T,$ transforming them to $1,\,\,\infty,\,\,0$ respectively. Then push them through $M,$ getting $e^{2\pi i/3},\,\, e^{4\pi i/3},\,\, 1$ respectively.

So you want $M^{-1}(1)=0,$ $M^{-1}(e^{2\pi i/3}) = 1,$ $M^{-1}(e^{4\pi i/3})=\infty.$

To get $M^{-1}(1)=0,$ you need $M^{-1}(z) = \dfrac{\cdots(z-1)}{\cdots}.$

Then to get $M^{-1}(\infty)= e^{4\pi i/3}$ you need $M^{-1}(z) = \dfrac{e^{4\pi i/3}(z-1)}{z - \cdots}.$

Then to get $M^{-1}(e^{2\pi i/3}) = 1,$ put $e^{2\pi i/3}$ in place of $z$ in that last expression and set it equal to $1$ and see what you need to put in place of $\text{“} \cdots \text{”}$ in order to get that result.


问:
Prove that equation has exactly 2 solutions
Question:
Given $\lambda_1<\lambda_2<\lambda_3 \in \Bbb R, a_1,a_2,a_3>0 $

Show that this equation has exactly 2 solutions:

$\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}=0$

What I did:
Try 1: Since we are studying continuity, Weirstrass theorem and Mean value theorem at the moment, I think one of these theorems should somehow be involved in the proof...
Assuming that x is not equal to any of the lambdas I multiplied by the denumerators and got a quadratic equation- From this I figure that there must be at most 2 solutions. I thought about using the discriminant to prove that that there are exactly 2, but the calculation is very long, and I don't think this is what the lecturer intended for us to do, considering what we are studying now.
Try 2: Thought of using the Intermediate value theorem. So I defined a function $f(x)=\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}$ and I want to find when it equals 0.
but Intermediate value theorem only involves two lambdas, and also I don't know if the lambdas are negative or positive (some/all of them)- Should I divide to cases?

Hints are welcome at first :-)

答:
Yes, try to divide into cases.

More complete answer :

For  $x \in ]- \infty, \lambda_1[$, all your terms are negative -> No solutions

For  $x \in ]\lambda_1, \lambda_2[$ -> use the continuity of the function as you said on the try 2

For  $x \in ]\lambda_2, \lambda_3[$ -> use the continuity of the function as you said on the try 2

For  $x \in ]- \infty, \lambda_3[$, all your terms are positive -> No solutions


问:
$f(z)=\frac{1}{z}$ is not uniformly continuous in $|z|<1$, $z \ne 0$.
Show that $f(z)=\frac{1}{z}$ is not uniformly continuous in $|z|<1$, $z \ne 0$. Suppose by the sake of contradiction that it is, then for all $\epsilon>0$ exists $\delta(\epsilon)>0$ such that for all $z_1,z_2$ such that $|z_1|<1$ and $|z_2|<1$,  $|z_1-z_2|<\delta(\epsilon) \implies \left|\frac{1}{z_1}-\frac{1}{z_2}\right|<\epsilon$.

Suppose without loss of generality that it is $0<\delta(\epsilon)<1$, then since the definition of uniform continuity holds for all $z_1,z_2$ such that $|z_1|<1$ and $|z_2|<1$ I can choose $z_1=\delta(\epsilon)$ and (since $\frac{\delta(\epsilon)}{1+\epsilon} <\delta(\epsilon)<1$) I can choose $z_2=\frac{\delta(\epsilon)}{1+\epsilon}$ for it follows
$$\left|\frac{1}{\delta(\epsilon)}-\frac{1}{\frac{\delta(\epsilon)}{1+\epsilon}}\right|<\epsilon$$
But $\left|\frac{1}{\delta(\epsilon)}-\frac{1}{\frac{\delta(\epsilon)}{1+\epsilon}}\right|=\frac{\epsilon}{\delta(\epsilon)}$, so it is $\frac{\epsilon}{\delta(\epsilon)}<\epsilon \iff \delta(\epsilon)>1$ but this contradicts the fact that $\delta(\epsilon)<1$. So $f$ is not uniformly continuous in $|z|<1$. Is this correct? I am not sure because I'm not sure if all my supposition are correct, in particular: I'm not sure that I can suppose $\delta(\epsilon)<1$, because this is a proof by contradiction so I suppose that $f$ is uniformly continuous and I don't know if the $\delta(\epsilon)$ I get is actually $<1$ (but I think this is ok because if $\delta(\epsilon) \geq 1$ is the one that works for the uniform continuity then all the $\delta^*(\epsilon)<1$ work as well) and I'm not sure if the fact that when I get $\delta(\epsilon)>1$ I actually get a contradiction, since it was a supposition of mine that $\delta(\epsilon)<1$.

答: $f$ is not uniformly continuous as it doesn't send Cauchy sequences in the domain to Cauchy sequences, which is an easily verifiable property of uniformly continuous functions. Consider the Cauchy sequence $z_n = \frac{1}{n}$ in the domain, e.g.


问:
Conditional probability mixing Union and Intersect - $P(A\cap B\mid A\cup B)$
Could someone help with how the formula is applied to these sort of problems with conditional probability?

I have $P(A) = 0.5$, $P(B) = 0.3$, $P (A\cap B) = 0.1$
I was able to calculate $P(A\cup B)$ from set theory as $= 0.5+0.3-0.1 = 0.7$

But where do I go from here? Thank you!

答:
HINT

Since $A\cap B \subseteq A\cup B$, we can proceed as follows

\begin{align*}
\mathbb{P}(A\cap B|A\cup B) & = \frac{\mathbb{P}((A\cap B)\cap(A\cup B))}{\mathbb{P}(A\cup B)} = \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(A\cup B)}
\end{align*}

Can you take it from here?


问:
Showing that $U(\mathfrak{sl}_2)$ is a coalgebra
We know that there is a coalgebra structure on $U(\mathfrak{sl}_2)$ as follows for any $z\in \mathfrak{sl}_2$:
$$\Delta(z)=1\otimes z+z\otimes 1, \qquad \epsilon(z)=0.$$

Can someone be so kind to explain to me that why $\Delta$ is an algebra map?

Shouldn't we have $\Delta(zz´)=\Delta(z)\Delta(z´)$, for $z,z´\in \mathfrak{sl}_2$?

As we have $\Delta(z)\Delta(z´)=1\otimes zz´+z´\otimes z+z\otimes z´+zz´\otimes 1$ which as I understand is not equal to $\Delta(zz´)$!

Thanks!

答: This is equal to $\Delta(z z')$ by definition. Remember that $z z'$ is not an element of $\mathfrak{sl}_2$. This definition of the comultiplication is hiding an extra step: it is defining the comultiplication on $\mathfrak{sl}_2$ only, and from here the comultiplication extends uniquely to $U(\mathfrak{sl}_2)$ precisely because it must be an algebra homomorphism, so must satisfy $\Delta(z z') = \Delta(z) \Delta(z')$.


问:
Showing that a graph is planar
G is a graph with 5 vertices with degrees 3,4,5,6,7 (one each), n vertices with degree 2, and m vertices with degree 1.

Prove that G is planar.

I know that to prove G is planar, according to Kuratowski's and Wagner's theorems, I need to show it does not contain any minor of $K_5$ or $K_{3,3}$. But am unsure how to go about it.
For $K_5$, If I'm not wrong, I could use the fact there are only 4 vertices for which $deg_G(v)\geq4$, So it cannot contain a minor of $K_5$.
About $K_{3,3}$, maybe I could show that G contains an odd circle, therefore is not bipartite and does not contain any minor of a bipartite graph. But I do not know how to show it contains an odd circle.

I would appreciate any clarifications about this question! Thank you!

EDIT: Is it true that for G to include a minor of $K_{3,3}$ it needs to have at least 6 vertices with degree at least 3 each? If so, then it sorts that out for me. Thank you.

答:
The traditional $K_{3,3}$-or-$K_5$ argument also works.

It is easier to use Kuratowski's theorem than Wagner's, because we can immediately say that

a subdivision of $K_5$ must contain $5$ vertices of degree $4$, and

a subdivision of $K_{3,3}$ must contain $6$ vertices of degree $3$.

However, our graph only has $5$ vertices of degree $\ge 3$ and only $4$ vertices of degree $\ge 4$, so it cannot contain either subdivision.

For minors, the reasoning would be trickier: it is not true that if a graph $H$ has a vertex of degree $d$, then that vertex must be represented by a degree-$d$ vertex in any $H$-minor. In particular, for any graph $H$ (including $K_5$) there is an $H$-minor with maximum degree $3$. The Petersen graph is a $K_5$-minor. It might still be possible to apply Wagner's theorem, by reasoning about how many vertices of each degree such graphs would need, but it's not worth it.

Title: Showing that a graph is planar


问:
Range of the distribution of $(1-X)$ when $X$ follows Beta distribution as $X\sim beta(p,q)$
if $X$ follows beta distribution with parameter $p$ and $q$  where $p>0\quad ,  q>0$

then $1-X$ follows beta distribution with parameters $q$ and $p$,

that is if $X\sim beta(p,q)$ then $(1-X)\sim beta(q,p)$

i have come up with the result but the range of the distribution of $1-X$ is coming 1 to 0,that is,

$1<(1-x)<0$

i computed the limit thus:

if $x=0$ ,then , $(1-x)=(1-0)=1$

and if $x=1$ ,then , $(1-x)=(1-1)=0$

But the range should be $0<(1-x)<1$  otherwise i can't conclude  with the result that

if $X\sim beta(p,q)$ then $(1-X)\sim beta(q,p)$
.

答:
Decomposing the double inequality:

$x\lt1\iff 1-x\gt0$

$x\gt0\iff 1-x\lt1$

$0\lt x\lt1\iff$ $___$ $\lt1-x\lt$ $___$

Or: plot the graph of the function $u:[0,1]\to\mathbb R$, $x\mapsto u(x)=1-x$.


问:
Can a subgroup be a subfield but not a subring?
I was doing my homework and came across a question with several true or false parts. One of the parts asked the question, "Every subring of $\mathbb{C}$ is a subfield of $\mathbb{C}$", which I assumed was false, and the next part asked, "Every subfield of $\mathbb{C}$ is a subring of $\mathbb{C}$", which I assumed was true. The solution I found was the opposite, though; the first suggestion was labeled true and the second false.

I assumed the first statement was false because the working definition I have of a field is, 'a subring that has inverses', which would imply that if something is a field then it is also a ring. It also tells me that a group being a subring is not sufficient to prove the group a subfield. What am I missing, and are there any good examples that prove the truth or falsity

答:
Every field is a ring, so a subfield of $\mathbb{C}$ is in particular a subring.

$\mathbb{Z}$ is a subring of $\mathbb{C}$ which is not a field.


问:
Show that $(\mathbf{a}·\mathbf{a})(\mathbf{b}·\mathbf{b})-(\mathbf{a}·\mathbf{b})^2=(\mathbf{a}\times\mathbf{b})·(\mathbf{a}\times\mathbf{b})$?
I need to prove that, for two linearly independent vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$,

$$(\mathbf{a} · \mathbf{a}) (\mathbf{b} · \mathbf{b}) - (\mathbf{a} · \mathbf{b})(\mathbf{a} · \mathbf{b}) = (\mathbf{a} \times \mathbf{b})·(\mathbf{a} \times \mathbf{b})$$

Could someone give me a demonstration of this identity? Or a hint to prove it?

答:
For a unit vector $u$, $|u\cdot v|$ is the length of the projection of $v$ onto $u$ (which is also the distance of $v$ from the plane perpendicular to $u$) and $|u\times v|$ is the length of the projection of $v$ onto the plane perpendicular to $u$ ($u\times v$ is the projection of $v$ onto the plane rotated by $\frac\pi2=90^{\large\circ}$ counter-clockwise).

The Pythagorean Theorem says
$$
|u\cdot v|^2+|u\times v|^2=|v|^2\tag1
$$
Due to the linearity of "$\cdot$" and "$\times$", we can remove the restriction on $u$ by making the equation homogeneous; here, substitute $u\mapsto\frac{u}{|u|}$ (which changes nothing if $|u|=1$) and multiply by $|u|^2$:
$$
|u\cdot v|^2+|u\times v|^2=|u|^2|v|^2\tag2
$$


问:
Find the radius of convergence of the power series representation of $\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$
We are given the following function:
$$\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$$
How do we find the radius of convergence for the power series, representing that function around $z = 0$?

The power series representation for $tan(\frac{z}{2})$ is:
$$\frac{z}{2} + \frac{z^3}{24} + \frac{z^5}{240} + O(z^7)$$
The power series representation for $\frac{1}{(z+2)(z+3)}$ is:
$$\frac{1}{6} - \frac{5z}{36} + \frac{19z^2}{216} + O(z^3)$$

Then I multiply both to obtain:
$$\frac{z}{12} - \frac{5z^2}{72} + \frac{11z^3}{216} - \frac{5z^4}{162} + \frac{697z^5}{38800} + O(z^6)$$

I don't know how to find the radius of convergence from that. Is there a way to convert it to a representation of summation and the nth term in the power series, so that for example, we can use the ratio test to find the radius or is there another way to find it?

答: The closest singularities of the function are located at $z=-2,-3,\pm\pi$. The closest one to $0$ is $-2$, thus the radius of convergence of the series centered at $0$ would be $2$.


问:
Prove that if $\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty$ for $\forall \delta > 0$, then $X_n \to 0$ almost surely.
Let $\{X_n\}$ be a sequence of independent random variables. Assume that $\forall \delta > 0$ the following sum converges:
$$\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty.$$

How can I prove that $X_n \to 0$ almost surely? In other words, I want to prove that

$$\mathbb{P}(\lim\limits_{n\to +\infty} X_n = 0) = 1.$$

To prove that, I think I need to use Borel–Cantelli lemma. It states that for a sequence of events $\{A_n\}_{n=1}^{\infty}$ such that $A = \bigcap\limits_{N=1}^{\infty} \bigcup\limits_{n > N} A_n $ and $\sum\limits_{n=1}^{\infty}\mathbb{P}\left(A_n\right) < \infty$, it is true that $P(A) = 0$.

From problem statement, I have $A_n = \{|X_n| \geq \delta\}$ and in order to solve the problem I need to put $A = \{\lim\limits_{n\to +\infty} X_n \neq 0\}$. So I need to make sure that this equation holds:

$$\{\lim\limits_{n\to +\infty} X_n \neq 0\} = \bigcap\limits_{N=1}^{\infty} \bigcup\limits_{n > N} \{|X_n| \geq \delta\} $$

I would say the equation above is true if I put $\delta := \frac{1}{N}$, for example, but I'm not allowed to make $A_n$ dependent on $N$, right? I don't know how to use the fact that $\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty$ for every $\delta > 0$.

Any clue how to finish the proof? I would appreciate any help to proceed from here.

答:
Let $E_\delta$ be the set of $\omega$ for which  $\sum_n 1_{[|X_n| \ge \delta]}(\omega)$ is unbounded. You know that $P E_\delta = 0$ for all $\delta >0$.
Let $E = \cup_m E_{1 \over m}$ and note that $P E = 0$.

Suppose $\epsilon>0$ and $\omega \notin E$ then choose ${1 \over m} < \epsilon$ and
so $\omega \notin E_{1 \over m}$. Then $\sum_n 1_{[|X_n| \ge {1 \over m}]}(\omega) < \infty$ and so there is some $M$ such that if $m \ge M$ we have $|X_n(\omega)| < { 1\over m} < \epsilon$.


问:
Coxeter: An opposite isometry with no invariant point is a glide reflection. Me: What about a reflection followed by a rotation?
In Coxeter's "Introduction to Geometry", I was reading the section about Glide Reflections (page 44, section 3.3), which states

Every opposite isometry with no invariant point is a glide reflection.

I see no issues with the proof, however I seem to have a "counter example" of a reflection followed by a rotation not centred on the line of reflection, specifically:

Consider a line l and a point O not on l, then let S be the isometry of reflection through l and then a rotation of any non-zero angle about O (call it 30 degrees to be concrete)

As this is a product of an opposite and direct isometry, it is an opposite isometry.

There can't be a fixed point:

Any point on l is fixed by the reflection and as the center of rotation O is not on l, it unfixes all of these points, so none of the points on l are fixed by S

For any point not on l, a rotation that mapped its reflection through l back to itself must have be centred at l, but O is not centred at l. Hence none of the points not on l are fixed by S either.

So S is an opposite isometry with no fixed points, but it is evidently not a glide reflection as a rotation is not a translation, so we have a "contradiction".

I would like to know where exactly my reasoning fails and why.

答: Actually S will be a glide reflection, although its axis of reflection will not be parallel to l. For an example see https://www.geogebra.org/geometry/gqx3a7kx . The quadrilateral $CDEF$ is reflected in the line $AB$ to give $C'D'E'F'$. This is then rotated 30 deg. clockwise about point $O$ to give $C''D''E''F''$. I have then reflected this last one in the second straight line giving a translate of our original polygon.


问:
Derivative of linear map?
Given $f: \mathbb{R}^{n\times n}\to \mathbb{R}^{n\times n}$, $f(X)=X^TX$, how do I know that $df(A)H=A^TH+H^TA$ for $A\in O(n,\mathbb{R})$ and $H\in \mathbb{R}^{n\times n}$ where $O(n,\mathbb{R})$ is the orthogonal group.

答:
$$
f(X+\epsilon H) = (X+\epsilon H)^T(X+\epsilon H) = \underbrace{X^T X}_{f(X)} + \underbrace{(X^TH+H^T X)}_{f'(X)\cdot H}\epsilon + \mathcal O(\epsilon^2)
$$

Title: Derivative of linear map?


问:
Quartic function sharing three common roots with another function
So the question is

"The quartic function f(x) = (x^2+x-20)(x^2+x-2) has three roots in common with the function g(x) = f(x-k), where k is a constant. Find the two possible values of k."

So my understanding so far is that function of g is translated k units. Function of g would look something like

g(x) = ((x-k)^2+(x-k)-20)*((x-k)^2+(x-k)-2)

Since these two functions share the same coordinates, therefore to find x:

f(x) = g(x)

(x^2+x-20)(x^2+x-2) = ((x-k)^2+(x-k)-20)((x-k)^2+(x-k)-2)

When expanded, this produces 45 terms and when like terms are added/subtracted, I get:

-4kx^3 + 6k^2x^2 - 6kx^2 - 4k^3x + k^4 + 7k^2x - 2k^3 - 21k^2 + 22k = 0

Maybe I think I have not taken the right direction so any help will be much appreciated.
And my apologies in advance for any confusion caused.

Thank You

答: The roots of $f(x)=(x^2+x-20)(x^2+x-2)$ are $-5$, $-2$, $1$, and $4$. So, the roots of $f(x-k)$ are $k-5$, $k-2$, $k+1$, and $k+4$. So, those two possible values of $k$ are $k=3$ (in which case the roots of $f(x-k)$ are $-2$, $1$, $4$, and $7$) and $k=-3$ (in which case the roots of $f(x-k)$ are $-8$, $-5$, $-2$, and $1$).


问:
Understanding the concept of hereditary sets
I have just started set theory and I am having one big issue with understanding what a pure/hereditary set is.

The definition I had is that a set is pure/hereditary if the elements of such set is also a pure/hereditary set.

The way I understand this definition is that a pure set is a set such that its element is also a set.

Then I was thinking, is $\{1\}$ a pure set? Clearly the only element is just $1$ and I do not think $1$ on its own is a set and thus $\{1\}$ is not a pure set?

Could someone please give me several examples of pure sets and not pure sets so that I could better my understanding!

Many thanks!

答:
No, a hereditary (or pure) set is a set whose elements are all hereditary sets. Thus, a set that is not hereditary has some element that is not hereditary. This means that the empty set is vacuously hereditary: it has no elements, so it certainly has no elements that are not hereditary sets. From this we can construct the set $\{\varnothing\}$, the set whose only element is the empty set. Since the only element of $\{\varnothing\}$ is hereditary, $\{\varnothing\}$ is itself hereditary.

In the usual von Neumann construction of the ordinal numbers, $1$ is defined to be $\{\varnothing\}$, so $\{1\}$ is indeed hereditary: its only element is $1=\{\varnothing\}$, which, as we just saw, is hereditary.

If one works in $\mathsf{ZF(C)}$ set theory, the only objects are sets, so all sets are hereditary. There are, however, axiomatizations of set theory that allow objects, called atom or urelements, that are not sets but can be elements of sets. It is in those versions of set theory that the distinction between hereditary sets and non-hereditary sets comes into play.


问:
In a transitive action there is a bijection between the fixed points of a stabilizer of a and the lateral clases of the stabilizer in his normalizer
the question is the given above, specially in the case infinite:

If the action of $G$ is transitive, then there is a bijection between the fixed points of the stabilizer of a element $a$ and the lateral classes of the stabilizer of $a$ in the normalizer of the stabilizer of $a$?.

My bijection is $G_{ah}$(this denotes a lateral class of the stabilizer of a) to $ha$ where $a$ is the point of the stabilizer, this function is well-defined and is one to one. Since $G$ acts transitively then all point $y$ is equal to a $ga$ with $g \in G$, and if this is in the fixed points then the $g$ is such that the conjugate of $G_a$ is contained in $G_a$. In orden to prove that the function is a surjection, i need that the $g$ given above normalizes $G_a$, and this is precisely the problem. I can assure something about the $g$?

Thank you for the further answers

答:
Addendum OP made me realize that in the finite case there is a fine point here. So I believe that in the general case the bijection asked for does not involve the normalizer
$$
N_{G}(G_{\alpha}) = \{ g \in G : g G_{\alpha} g^{-1} =  G_{\alpha} \},
$$
but the related object
$$
\tilde{N}_{G}(G_{\alpha}) = \{ g \in G : g G_{\alpha} g^{-1} \subseteq  G_{\alpha} \}.
$$
Note that $\tilde{N}_{G}(G_{\alpha})$ is a submonoid of $G$, but need not be a subgroup of $G$. (I will provide an example later.) It coincides with the normalizer, and thus it is a subgroup, if $G$ is finite, or a torsion group.

Let $G$ act transitively on a set $A$, let $a \in A$, and let $B$ the set of points fixed by $G_{a}$.

Let $b \in B$. Then there exists $g$ such that $a g = b$. Let $h \in G_{a}$. Then $a g h = b h = b = a g$, so that $g h g^{-1} \in G_{a}$. In other words, $g \in \tilde{N}_{G}(G_{a})$.

Conversely, if $g \in \tilde{N}_{G}(G_{a})$, then $b = a g$ is fixed by $G_{a}$.

Now we are ready to set the correspondence. Start with the map
$$
\tilde{N}_{G}(G_{a}) \to B
$$
given by $g \mapsto a g$. We have shown that this is surjective.

Now $a g = a g'$ if and only if $g' g^{-1} \in G_{a}$, so this yields a bijection as requested between the cosets of $G_{a}$ in $\tilde{N}_{G}(G_{a})$ and $B$.

Addendum: an example Consider the additive abelian group $$A = \Bbb{Z}\left[ \dfrac{1}{2}\right] = \left\{ \dfrac{z}{2^{n}} : z \in \Bbb{Z}, n \ge 0 \right\}.$$ Let $c$ be the automorphism of $A$ given by $a \mapsto 2 a$, and let $G = A \langle c \rangle$ be the natural semidirect product.

Let $H = \Bbb{Z} \le A$, and consider the action of $G$ on the cosets of $H$. Let $\alpha = H$. Then $G_{\alpha} = H$. Now $N_{G}(G_{\alpha}) = A$, whereas $\tilde{N}_{G}(G_{\alpha})$ is the submonoid generated by $A$ and $c$, that is
$$\tilde{N}_{G}(G_{\alpha}) = \{ a c^{n} : a \in A, n \ge 0 \}.$$


答:
The first thing you could have noticed is f(X) is being given as product of two quadratic polynomials. And more astonishing is that those two quadratic polynomials are themselves factorised. So you easily get the roots of f(X). Now comes the play of g(X). Put X =x-k everywhere in f(X) such that you get the function g(X). And now you can easily get roots of g(X) in terms of k. After that it's just manually checking values of k. And yes, don't directly jump into expansion because it is feasible upto cubic but after that it's much more complicated.
Thank You!</p>

Title: Quartic function sharing three common roots with another function

Body:
So the question is

"The quartic function f(x) = (x^2+x-20)(x^2+x-2) has three roots in common with the function g(x) = f(x-k), where k is a constant. Find the two possible values of k."

So my understanding so far is that function of g is translated k units. Function of g would look something like

g(x) = ((x-k)^2+(x-k)-20)*((x-k)^2+(x-k)-2)

Since these two functions share the same coordinates, therefore to find x:

f(x) = g(x)

(x^2+x-20)(x^2+x-2) = ((x-k)^2+(x-k)-20)((x-k)^2+(x-k)-2)

When expanded, this produces 45 terms and when like terms are added/subtracted, I get:

-4kx^3 + 6k^2x^2 - 6kx^2 - 4k^3x + k^4 + 7k^2x - 2k^3 - 21k^2 + 22k = 0

Maybe I think I have not taken the right direction so any help will be much appreciated.
And my apologies in advance for any confusion caused.

Thank You


答:
Given $\epsilon =1$, let $1>\delta >0$.

If $x= \delta /2$ and $y=\delta$ then $|x-y|<\delta$ but $|f(x)-f(y)|=1/\delta>\epsilon.$

Therefore $f$ is, by definition, not uniformly continuous.

Body:
Show that $f(z)=\frac{1}{z}$ is not uniformly continuous in $|z|<1$, $z \ne 0$. Suppose by the sake of contradiction that it is, then for all $\epsilon>0$ exists $\delta(\epsilon)>0$ such that for all $z_1,z_2$ such that $|z_1|<1$ and $|z_2|<1$,  $|z_1-z_2|<\delta(\epsilon) \implies \left|\frac{1}{z_1}-\frac{1}{z_2}\right|<\epsilon$.

Suppose without loss of generality that it is $0<\delta(\epsilon)<1$, then since the definition of uniform continuity holds for all $z_1,z_2$ such that $|z_1|<1$ and $|z_2|<1$ I can choose $z_1=\delta(\epsilon)$ and (since $\frac{\delta(\epsilon)}{1+\epsilon} <\delta(\epsilon)<1$) I can choose $z_2=\frac{\delta(\epsilon)}{1+\epsilon}$ for it follows
$$\left|\frac{1}{\delta(\epsilon)}-\frac{1}{\frac{\delta(\epsilon)}{1+\epsilon}}\right|<\epsilon$$
But $\left|\frac{1}{\delta(\epsilon)}-\frac{1}{\frac{\delta(\epsilon)}{1+\epsilon}}\right|=\frac{\epsilon}{\delta(\epsilon)}$, so it is $\frac{\epsilon}{\delta(\epsilon)}<\epsilon \iff \delta(\epsilon)>1$ but this contradicts the fact that $\delta(\epsilon)<1$. So $f$ is not uniformly continuous in $|z|<1$. Is this correct? I am not sure because I'm not sure if all my supposition are correct, in particular: I'm not sure that I can suppose $\delta(\epsilon)<1$, because this is a proof by contradiction so I suppose that $f$ is uniformly continuous and I don't know if the $\delta(\epsilon)$ I get is actually $<1$ (but I think this is ok because if $\delta(\epsilon) \geq 1$ is the one that works for the uniform continuity then all the $\delta^*(\epsilon)<1$ work as well) and I'm not sure if the fact that when I get $\delta(\epsilon)>1$ I actually get a contradiction, since it was a supposition of mine that $\delta(\epsilon)<1$.


答:
Just to present an alternative with the notation you used, note that by negating the limit definition, $X_n \not\to 0$ if and only if there exists a $m \in \mathbb{Z}_+$ such that for all $N \in \mathbb{Z}_+$ there exists an integer $n \geq N$ such that $|X_n| > \frac{1}{m}$.

Let's define the event $A_n(\epsilon) = \{ |X_n| > \epsilon\}$. By assumption $\sum_{n \geq 1} \mathbb{P}(A_n(\epsilon)) < \infty$ for all $\epsilon > 0$, and therefore by the Borel-Cantelli lemma $\mathbb{P}(\limsup A_n(\epsilon)) = 0$. (We can also say $A_n(\epsilon)$ occurs infinitely often or i.o., which intuitively says that almost surely $|X_n|$ exceeds $\epsilon$ only finitely many times.)

Therefore,
$$
A = \{ \lim_{n \to \infty} X_n \ne 0 \} = \bigcup_{m \geq 1} \bigcap_{N \geq 1} \bigcup_{n \geq N} \{ |X_n| > \frac{1}{m} \} = \bigcup_{m \geq 1} \limsup A_n(1/m) .
$$
Thus $\mathbb{P}(A) \leq \sum_{m \geq 1} \mathbb{P}(\limsup A_n(1/m))) = 0$, and we conclude $\mathbb{P}(A^c) = 1$, or equivalently $X_n \to 0$ almost surely.

Title: Prove that if $\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty$ for $\forall \delta > 0$, then $X_n \to 0$ almost surely.

Body:
Let $\{X_n\}$ be a sequence of independent random variables. Assume that $\forall \delta > 0$ the following sum converges:
$$\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty.$$

How can I prove that $X_n \to 0$ almost surely? In other words, I want to prove that

$$\mathbb{P}(\lim\limits_{n\to +\infty} X_n = 0) = 1.$$

To prove that, I think I need to use Borel–Cantelli lemma. It states that for a sequence of events $\{A_n\}_{n=1}^{\infty}$ such that $A = \bigcap\limits_{N=1}^{\infty} \bigcup\limits_{n > N} A_n $ and $\sum\limits_{n=1}^{\infty}\mathbb{P}\left(A_n\right) < \infty$, it is true that $P(A) = 0$.

From problem statement, I have $A_n = \{|X_n| \geq \delta\}$ and in order to solve the problem I need to put $A = \{\lim\limits_{n\to +\infty} X_n \neq 0\}$. So I need to make sure that this equation holds:

$$\{\lim\limits_{n\to +\infty} X_n \neq 0\} = \bigcap\limits_{N=1}^{\infty} \bigcup\limits_{n > N} \{|X_n| \geq \delta\} $$

I would say the equation above is true if I put $\delta := \frac{1}{N}$, for example, but I'm not allowed to make $A_n$ dependent on $N$, right? I don't know how to use the fact that $\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty$ for every $\delta > 0$.

Any clue how to finish the proof? I would appreciate any help to proceed from here.


问:
Integration using summation
How do you integrate $\sqrt{x}$ from an arbitrary constant $a$ to another $b$ by summation ?</p>

答:
Let us divide the region $a\le x\le b$ into $n$ sub-region as $a,ar,ar^2,\cdots ar^n=n$ such that $\delta_i=ar^i-ar^{i-1}$

If real $m\ne-1,$

$$\int_a^b x^mdx=\lim_{n\to\infty}\left(\sum_{1\le i\le n}(ar^i)^m\delta_m\right)$$

$$=\lim_{n\to\infty}\left(\sum_{1\le i\le n} (ar^i)^m(ar^i-ar^{i-1})\right)$$

$$=a^{m+1}\lim_{r\to1}(r-1)\sum_{1\le i\le n} (r^{i-1})^{m+1} $$

$$=a^{m+1}\lim_{r\to1}(r-1)\sum_{1\le i\le n} (r^{m+1})^{i-1} $$

$$=a^{m+1}\lim_{r\to1}(r-1)\cdot1\cdot\frac{(r^{m+1})^n-1}{r^{m+1}-1} $$

$$=a^{m+1}\lim_{r\to1}\frac{\left(\frac ba\right)^{m+1}-1}{\frac{r^{m+1}-1}{r-1}} $$ as $\frac ba= r^n\implies (r^{m+1})^n=(r^n)^{m+1}=\left(\frac ba\right)^{m+1}$

$$=(b^{m+1}-a^{m+1})\cdot\frac1{\lim_{r\to1}\left(\frac{r^{m+1}-1}{r-1}\right)}$$

$$=(b^{m+1}-a^{m+1})\cdot \frac1{(m+1)\cdot1^m}$$

$$\text{  as }\lim_{x\to a}\frac{x^n-a^n}{x-a}=nx^{n-1} \text{ using L'Hospital Rule }$$

Title: Integration using summation


问:
Finding least upper bound and greatest lower bound of the set
I am given the following set:

$A = \left\{ \frac{1}{n} + (-1)^{n}: n \in \mathbb{N} \right\}$

I am asked to find the greatest lower bound and least upper bound, but am having difficulty. I tried to split the set up into when $n$ is even and when $n$ is odd. So, when $n$ is even, $A = {(1/n) + 1}$ and when $n$ is odd, $A = (1/n) - 1$.

Where does one go from here?

答: Each element of $A$ is smaller than or equal to $\frac32\left(=\frac12+(-1)^2\right)$, which is an element of $A$. So, $\sup A=\frac32$. On the other hand, each element of $A$ is greater than $-1$. But, if $-1<a<0$, there is some odd number $n$ such that $\frac1n<a+1$, and then $(-1)^n+\frac1n<a$. Therefore, $a$ is not a lower bound of $A$ and so $-1$ is the greatest lower bound. So, $\inf A=-1$.


答: Hint: It seems clear: if you denote $a_n=\frac1n+(-1)^n$, all terms with even index are positive, $>1$ and make up a decreasing sequence, whereas all terms with odd index are  nonpositive and $>-1$ and also make up a decreasing sequence. Can you conclude from these observations?

Title: Finding least upper bound and greatest lower bound of the set

Body:
I am given the following set:

$A = \left\{ \frac{1}{n} + (-1)^{n}: n \in \mathbb{N} \right\}$

I am asked to find the greatest lower bound and least upper bound, but am having difficulty. I tried to split the set up into when $n$ is even and when $n$ is odd. So, when $n$ is even, $A = {(1/n) + 1}$ and when $n$ is odd, $A = (1/n) - 1$.

Where does one go from here?


问:
Big-O analysis of ratio of log
Based on various trial-and-error in desmos, it seems like the following statement is true but kind of unsure about a formal proof.

Any helps/hints are appreciated.

Claim:

$$\forall \epsilon\in(0,1); \frac{\ln(\frac{n^2\epsilon^2}{n-1}+1)}{\ln(\frac{n\epsilon^2}{n-1}+1)}\in O(\frac{\ln(n)}{\epsilon})$$

答:
Hint:

You can more easily say $O(\ln(n))$ here, because Big-O ignores multiplicative constants (and $\epsilon$ is a constant here).

The denominator clearly converges to a constant. So, you only need to worry about the numerator.

The inside of the logarithm is visibly asymptotic to $\epsilon^2n$. Try factoring $\epsilon^2n$ out (and using $\ln(ab)=\ln(a)+\ln(b)$ and easy limits) to rewrite this expression in terms of $\ln(n)$.

Title: Big-O analysis of ratio of log


问:
Why is this integral diverging? $\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$
$$\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$$

I can easily prove that this integral is diverging by taking the limit over the following proper integral,

$$\lim_{R_1,R_2\to \infty} \int\limits^{R_2}_{-R_1} \,\frac{x}{x^2+1} dx$$

Mathematically, this makes sense to me, but intuitively I am not able to absorb this.

If we observe the function $y=\frac{x}{x^2+1}$, it is clearly an odd function. And since, integrals return signed areas, an integral to an odd function having limits that are the negatives of each other, should evaluate to zero.

As clearly seen by the following plot.

Can someone please clarify this?

Thank you.

答:
so:
$$\frac{x}{x^2+1}=\frac{1}{x+1/x}$$
which means that for large $x$,
$$\frac{x}{x^2+1}\approx\frac1x$$
which the integral diverges for. what you have is the integral of the odd function so in other words,
$$\int_{-a}^af(x)dx=0\,a\to\infty$$
Normally the condition for convergence is considered to be that the interval can be split into parts and still converge, this cannot but we can take the cauchy principle value which does go to zero


答:
Do you mean something like this:

$f(x)= \sqrt{x}$,

$$\int_a^b \sqrt{x}=\lim_{n\to \infty} \dfrac{b-a}{n} \sum_1^n\sqrt{x_i}$$

Title: Integration using summation

Body: How do you integrate $\sqrt{x}$ from an arbitrary constant $a$ to another $b$ by summation ?</p>


答:
De Facto way- even though writing $\epsilon$ in $\frac{\ln(n)}{\epsilon}$ is redundant.

For simplicity: $ \epsilon = t$

$$ f(n) = \frac{\ln( \frac{n^2 t^2}{n-1} +1)}{\ln( \frac{n t^2}{n-1} +1)}$$

We need to prove that $$ f(n) = O( \frac{\ln(n)}{t}) \\ \text{A.K.A} \\ \lim_{n \to \infty} \frac{f(n)}{ \ln(n) /t } \ne \infty$$

$$\lim_{n \to \infty}  \frac{f(n)}{\ln (n) / t} = \lim_{n \to \infty} \frac{t\ln( \frac{n^2 t^2}{n-1} +1)}{\ln( \frac{n t^2}{n-1} +1) \ln (n)} = t\lim_{n \to \infty} \frac{1}{\ln( \frac{t^2 n + n- 1}{n-1})} \cdot \lim_{n \to \infty} \frac{\ln ( \frac{t^2 n^2 + n - 1}{n-1})}{\ln(n)} $$

First limit is not that hard to calculate: $$\lim_{n \to \infty}\frac{1}{\ln( \frac{t^2 n + n- 1}{n-1})} =  \frac{1}{\ln(t^2 + 1)}$$

Second limit can be broken up into again, $2$ limits:

$$\lim_{n \to \infty} \frac{\ln ( \frac{t^2 n^2 + n - 1}{n-1})}{\ln(n)} = \lim_{n \to \infty} \frac{\ln(t^2 n^2 +  n-1) - \ln(n-1)}{\ln(n)} \\= \lim_{n \to \infty} \frac{\ln(t^2 n^2 +  n-1)}{\ln(n)} - \frac{\ln(n-1)}{\ln(n)} \\ = \lim_{n \to \infty} \frac{\ln(n^2(t^2 + t^{-1} - n^{-2}))}{\ln(n)} - \lim_{n \to \infty} \frac{\ln(n-1)}{\ln(n)}  \\ = \lim_{n \to \infty} \frac{2\ln(n) + \ln(t^2 + t^{-1} - n^{-2}))}{\ln(n)} - \lim_{n \to \infty} \frac{\ln(n-1)}{\ln(n)} = 2 - 0 -1 = 1$$

Thus this whole limit is equal to:

$$ \frac{1}{\ln(t^2 + 1)} $$

Which is not infinity as $t \ne 0$ as it is not in the interval $(0,1)$

Title: Big-O analysis of ratio of log

Body:
Based on various trial-and-error in desmos, it seems like the following statement is true but kind of unsure about a formal proof.

Any helps/hints are appreciated.

Claim:

$$\forall \epsilon\in(0,1); \frac{\ln(\frac{n^2\epsilon^2}{n-1}+1)}{\ln(\frac{n\epsilon^2}{n-1}+1)}\in O(\frac{\ln(n)}{\epsilon})$$


问:
Dealing with numbers bigger than Realmax
I have a code in Matlab, and in one step it calculates Gamma function for different values. But when I try to find Gamma(172.5)  I get inf as it is bigger than realmax. My question:  is there a away that I can get and calculate  big numbera like Gamma (172.5) in Matlab without getting inf?

Thanks

答:
In Matlab? No. You just tried it and surpassed the software maximum, right? If you simply want the value of the function applied to some high value you could try calculating it in other software (for instance: https://www.wolframalpha.com/input/?i=gamma%28172.5%29)

Now if you are asking it so that you can operate with such things, such as dividing such number by gamma applied to a second value and obtain the result, then it helps to remember the properties of the gamma function and see if you can come up with a way to rewrite the desired expression in matlab so that it doesn't need to evaluate such a high value.

Another way of dealing with high numbers is scaling. Imagine you have an expression full of huge numbers. Divide the expression by the biggest of them, now you only have small numbers to deal with. Solve the expression. Then multiply it back again by the maximum of the expression to obtain the desired result.


问:
What is the sum of all possible integer values of the leading cooefficient of $P(x)$?
Given

For every $x \in \mathbb{R},  P(x) \neq 0$;

$P(x)$ is a second degree polynomial;

$P(1) = 8$ and $P(2) = 2.$

Body:
There are an infinite number of solutions
of the form
$P(x)
=(ax+b)^2+c^2
$
where
$2 \le a^2 \le 18$,
$b
=-3(\dfrac{1}{a}+\dfrac{a}{2})
$,
$c^2
=-\dfrac{1}{4a^2}(a^2-18)(a^2-2)
$.

Theleading coefficient is
$a^2$ where
$2 \le a^2 \le 18$
so the sum is
$2+3+...+18
=18(19)/2-1
=170
$.

If we want
$P(x) = 0$ for some $x$ then
$a^2 = 2$ or $a^2 = 18$.

Derivation.

From 1 and 2,
$P(x)
=(ax+b)^2+c^2
$.

From 3,
$(a+b)^2+c^2 = 8,
(2a+b)^2+c^2 = 2
$.

$6
=(a+b)^2-(2a+b)^2
=(-a)(3a+2b)
$
so
$2b
=-\dfrac{6}{a}-3a
$,
$b
=-\dfrac{3}{a}-\dfrac{3a}{2}
=-3(\dfrac{1}{a}+\dfrac{a}{2})
$,
$a+b
=-\dfrac{3}{a}-\dfrac{a}{2}
$.

$\begin{array}\\
c^2
&=8-(a+b)^2\\
&=8-(-\dfrac{3}{a}-\dfrac{a}{2})^2\\
&=8-(\dfrac{9}{a^2}+3+\dfrac{a^2}{4})\\
&=-(\dfrac{9}{a^2}-5+\dfrac{a^2}{4})\\
&=-\dfrac{1}{4a^2}(36-20a^2+a^4)\\
&=-\dfrac{1}{4a^2}(a^2-18)(a^2-2)\\
\end{array}
$

Therefore
$2 \le a^2 \le 18
$.

Then get $b$ and $c$.

For example,
try $a = 3, a^2 = 9$.

$c^2
=-\dfrac1{36}(-9)(7)
=\dfrac74
$.

$b = -3(\dfrac13+\dfrac32)
=-3(\dfrac{11}{6})
=-\dfrac{11}{2}
$.

$P(x)
=(3x-\dfrac{11}{2})^2+\dfrac74
$.

$P(1)
=(3-\dfrac{11}{2})^2+\dfrac74
=\dfrac{25}{4}+\dfrac74
=\dfrac{32}{4}
=8
$.

$P(2)
=(6-\dfrac{11}{2})^2+\dfrac74
=\dfrac1{4}+\dfrac74
=2
$.


问:
Function satisfying both these limits
What is an example of a function $f(x)$ that satisfies both

$\lim_{x\rightarrow0} f(x)/x^2 = C$, any finite non-zero real number

and

$\lim_{x\rightarrow0} f(x)/x = 1$?

I'm struggling to find anything that works after trying lots of different functions.

答:
Suppose $\ \lim_{x\to0}\left(\frac{f(x)}{x}\right) = 1$.

Then given $\varepsilon=\frac{1}{10},\ \exists\ \delta\ $ with $\ 0<\delta<\frac{1}{10}$ such that $\left|1-\frac{f(x)}{x}\right| < \frac{1}{10} \quad \forall\ |x|<\delta.\ $ So we have,

$\frac{9}{10} < \frac{f(x)}{x} < \frac{11}{10}$ and $0<|x|<\frac{1}{10},\ $ which implies $\ \large{\left| \frac{\frac{f(x)}{x}}{x} \right| } >\ $$9\quad \forall\ |x|<\delta.$

Now we may rewrite all of the above by replacing $10$ with any real number $r >1$.

Given $\varepsilon_r=\frac{1}{r}\ (r>1),\ \exists\ \delta_r\ $ with $\ 0<\delta_r<\frac{1}{r}$ such that $\left|1-\frac{f(x)}{x}\right| < \frac{1}{r} \quad \forall\ |x|<\delta_r.\ $ Since $r>1,$ we have:

$0<\frac{r-1}{r} < \frac{f(x)}{x} < \frac{r+1}{r}$ and $0<|x|<\frac{1}{r},\ $ which implies $\ \large{\left| \frac{\frac{f(x)}{x}}{x} \right| } >\ $$r-1\quad \forall\ |x|<\delta_r.$

This is almost the definition of $\lim_{x \to 0}\large{\left| \frac{\frac{f(x)}{x}}{x} \right| } = \infty,$ which tells us, among other things, that $\large{\left| \frac{\frac{f(x)}{x}}{x} \right| }$ does not converge as $x \to 0$.

Title: Function satisfying both these limits


问:
Borel sets and clopen decompositions
Let $X$ be a locally compact (Hausdorff) space, and suppose I can write $X$ as the disjoint union of a family of open subsets $(X_i)_{i\in I}$ (so each $X_i$ is actually clopen).  A typical example I have in mind is an uncountable disjoint union of copies of $\mathbb R$ (so I am certainly not wishing so assume $X$ is $\sigma$-finite, etc.)

Let's be careful: let $B$ be the Borel $\sigma$-algebra on $X$, so this is the intersection of all $\sigma$-algebras containing the open sets.  For each $i$, treat $X_i$ as a locally compact space in its own right, and let $B_i$ be the Borel $\sigma$-algebra.  Let $E\subseteq X$ be such that $E\cap X_i\in B_i$ for each $i$.  Is $E\in B$?

The converse is true: if $\Omega=\{ E\subseteq X : \forall i, E\cap X_i\in B_i\}$ then $\Omega$ is a $\sigma$-algebra, and $\Omega$ contains all the open subsets of $E$, so $B\subseteq\Omega$.  I want to show that $\Omega=B$, which seems much harder??

答: There are plenty of spaces for which (using your notation) $\Omega$ is strictly larger than $B$, and one such space is $\omega_1 \times \mathbb{R}$ (with the topology you described).  To build a set in $\Omega \backslash B$, just paste a set of Borel rank $\alpha$ in the $\alpha^{\mathrm{th}}$ copy of $\mathbb{R}$. </p>

Title: Borel sets and clopen decompositions


问:
Convergence of the Power Method
Given a vector $x^{(0)}$ with $\left\|x^{(0)}\right\|_{2}=1$, I wish to show that the power method :
$$
x^{(k)}=\frac{A x^{(k-1)}}{\left\|A x^{(k-1)}\right\|_{2}}, \quad \mu^{(k)}=\left(x^{(k)}\right)^{T} A x^{(k)}
$$
generates the pair $\left(x^{(k)}, \mu^{(k)}\right)$ such that $\displaystyle x^{(k)}=\frac{A^{k} x^{(0)}}{\left\|A^{k} x^{(0)}\right\|_{2}}$ for any matrix $A$.

My attempt by induction:

Base Case: For $k=1, \displaystyle x^{(1)}=\frac{A x^{(0)}}{\left\|A x^{(0)}\right\|_{2}}$ by definition.

For the inductive step, it seems trivial but I can't seem to figure it out, any help is much appreciated!

答:
[Note] : The inductive step is indeed trivial, now assume that $\displaystyle x^{(k)}=\frac{A^{k} x^{(0)}}{\left\|A^{k} x^{(0)}\right\|_{2}}$ all you have to do is work things out simply by substituting $x^{(k)}$ in $x^{(k+1)}$ due to its recursive nature as you can see :
$$
x^{(k+1)}=\frac{A x^{(k)}}{\left\|A x^{(k)}\right\|_{2}}=\frac{\displaystyle\frac{A^{k+1} x^{(0)}}{\left\|A^{k} x^{(0)}\right\|_{2}}}{\displaystyle\frac{\left\|A^{k+1} x^{(0)}\right\|_{2}}{\left\|A^{k} x^{(0)}\right\|_{2}}}=\frac{A^{k+1} x^{(0)}}{\left\|A^{k+1} x^{(0)}\right\|_{2}}
$$
Since $\displaystyle A x^{(k)}=A \frac{A^{k} x^{(0)}}{\left\|A^{k} x^{(0)}\right\|_{2}}=\frac{A^{k+1} x^{(0)}}{\left\|A^{k} x^{(0)}\right\|_{2}}$ and $\displaystyle\left\|A x^{(k)}\right\|_{2}=\frac{\left\|A^{k+1} x^{(0)}\right\|_{2}}{\left\|A^{k} x^{(0)}\right\|_{2}} .$

Title: Convergence of the Power Method


问:
How to show that $A$ is increasing?

Suppose that $f$ is a twice differentiable real function such that
$f''(x)>0$ for all $x\in[a,b]$. Find all numbers $c\in[a,b]$ at which
the area between the graph $y=f(x)$, the tangent to the graph at
$(c,f(c))$, and the lines $x=a$, $x=b$, attains its minimum value.

I understand that the area mentioned above is

$$A(c)=\int\limits_a^b (f(x)-f(c)-f'(c)\cdot(x-c))dx,$$

and using this formula I can get that $c= \frac{a+b}{2}$ is a critical point of $A$, but I'm falling to see why this has to be where $A$ attains, specifically, a minimum. If anyone could enlighten me, that would be wonderful.

答:
You have
$$
A'(c)=-f''(c)\int\limits_a^b(x-c)\,dx.
$$
Clearly, $A'(c)<0$ if $c<(a+b)/2$ and $A'(c)>0$ if $c>(a+b)/2$. Therefore $A$ is decreasing before the midpoint and increasing after, attaining a minimum at $c=(a+b)/2$.

Title: How to show that $A$ is increasing?


问:
Deciding if these two improper integrals diverge or converge. (comparison test)
I have two integrals to decide and prove if they converge or diverge, I solved them (or tried), but I'm unsure of my solution and got some questions:

$1)$ $$\int_1^3\frac {\sin(1-x)}{(x-1)^2}\,dx $$
What I tried to do: I said that this integral is less than $$\int_1^3\frac{1}{(x-1)^2}\,dx,$$ and if I substitute $t= x-1$, I will get $$\int_0^2\frac {1}{t^2}\,dt,$$ which diverges. Now I know that I can't say that my integral diverges using this fact, sadly I couldn't finish it.

$2)$ $$\int_1^\infty\frac{\ln(1+e^x)-x}{x^2}\,dx$$

What I did: I found the limit
$$\lim_{x\to\infty}\frac {\ln(1+e^x)}{x} = 1,$$ so I decided to take the function $\frac {1}{x}$ (I checked the limit with my integrand and got $1$), and $\frac {1}{x}$ diverges so my integral has to diverge.

I feel like I went a little far, I tried a couple functions, and most of them failed or weren't good for comparison test, I would really appreciate any explanation and if someone can approve my answers or point out my mistakes.
 Thanks in advance.

答:
For the first integral notice that

$$\frac{\sin(1-x)}{(x-1)^2} = \frac{\sin(-(x-1))}{(x-1)^2} = -\frac{\sin(x-1)}{(x-1)^2} \sim -\frac{1}{x-1} ~~~~~ \text{as} ~~~~~ x\to 1$$

Hence the first integral does diverge.

About the second one we have

$$\frac{\ln(1 + e^x) - x}{x^2} \to 0$$

as $x\to +\infty$ hence the second integral does converge.


问:
I need help understanding factoring better for this problem
Here is the equation:
$$\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}$$

I am getting the wrong answer every time.
\begin{align*}
f & = \frac{-3(x^2+3xh+h^2)-6x+6h+4+3x^2+6x-4}{h}\\
f & = \frac{-3x^2-6xh-3h^2-6x+6h+4+3x^2+6x-4}{h}\\
f & = \frac{-6xh-3h^2}{h}\\
f & = 3(-2x-h^2)
\end{align*}

Two questions, what am I doing wrong when factoring down this problem and where is the best place to really learn factoring super well?

答:
It is easy to make errors, expanding, collecting  terms and cancelling.
To reduce typos, I often use WolframAlpha for intermediate results such as here, here, and here.

$$f=\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}\\
=\frac{\big(-3 h^2 - 6 h x - 3 x^2\big)-\big(6 h + 6 x\big)+\big(4+3x^2+6x-4\big)}{h}\\
=\frac{-3 h (h + 2 x + 2)}{h}=-3(h + 2 x + 2)$$


问:
Finding the variance of estimators?
Given a  normally distributed random variable X ∼ N(µ, 1) with an unknown mean, µ
but known variance = 1. Consider an IID  random sample (n = 2) from this distribution, {X1, X2}. Now, consider the following two estimators of µ based on this sample:

µ1 = (X1 + X2)/2

µ2 =(X1 + 2X2)/3

1. Calculate each estimators bias

2. Calculate each estimator's variance

3.  Which estimator should you select and why?

Hi everyone! Although I think I've managed to solve question 1 (I believe both estimators are unbias, but can't seem to figure out how I might solve the variance for these estimators, I don't believe I can integration is possible here. I also believe that 3 is reliant on MSE, but can't solve anything until I've figured out 2! Any help would be greatly appreciated =)

答:
Use the fact that the $X_i$ are independent and identically distributed; therefore, the variance of a linear combination of them is equal to the linear combination of the individual variances; i.e., $$\operatorname{Var}[c_1 X_1 + c_2 X_2] \overset{\text{ind}}{=} \operatorname{Var}[c_1 X_1] + \operatorname{Var}[c_2 X_2] = c_1^2 \operatorname{Var}[X_1] + c_2^2 \operatorname{Var}[X_2],$$ where $c_1, c_2$ are scalar constants.  The "ind" over the equals sign means the equality holds under the assumption that $X_1$ and $X_2$ are independent random variables.  If they were not independent, then there would be an additional covariance term required.

In your case, for the first estimator of $\mu$, your choice of constants is $c_1 = c_2 = 1/2$, and for the second estimator, $c_1 = 1/3$, $c_2 = 2/3$.

Alternatively, you may compute $$\operatorname{E}[(c_1 X_1 + c_2 X_2)^2],$$ the second moment of the linear combination of $X_1$ and $X_2$, by expanding the binomial and using linearity of expectation to obtain $$c_1^2 \operatorname{E}[X_1^2] + 2c_1 c_2 \operatorname{E}[X_1 X_2] + c_2^2 \operatorname{E}[X_2^2].$$  Then use the fact that $X_1$ and $X_2$ are independent to write $\operatorname{E}[X_1 X_2] = \operatorname{E}[X_1]\operatorname{E}[X_2]$, and use $\operatorname{E}[X_1^2] = \operatorname{Var}[X_1] + \operatorname{E}[X_1]^2$.  Then finally use $$\operatorname{Var}[c_1 X_1 + c_2 X_2] = \operatorname{E}[(c_1 X_1 + c_2 X_2)^2] - \operatorname{E}[c_1 X_1 + c_2 X_2]^2.$$  This of course is much more tedious than directly using the first equation, because in essence this second approach is a proof of the validity of the first equation.

Title: Finding the variance of estimators?


问:
Calculating number of possible password combinations given specific assumptions
I am trying to calculate the number of possible password combinations, but with specific assumptions.

These are my assumptions:

The password contains one word that is 8 characters in length

The first character of this word is uppercase. It is always the first character, nowhere else.

There is a single number and a single special character at the end of the password.

Numbers: only the numbers 1, 2, 3 can be used

Special Characters: only the special characters !, ?, $, #, @, *, &, ^, % can be used

The number and special character must be at the end of the password, but can be in any order

For example, the number can come before the character, or the character before the number

So, an example password may be: Bookworm2$ or Icecream!1

My current attempt is as follows:

26 possible uppercase characters

26^7 lowercase characters (in any order)

3 possible numbers

9 possible special characters

Becoming: 26 + (26^7) + (3 * 9 * 2), which is 8,031,810,256 different combinations.

I am looking to seek confirmation that this is correct, and if not what I need to do to correct it.

答:
First letter is always uppercase thus you have $26$ options for it.

Then, we know it is the prefix of a word of length $8$ thus we need to complete it by choosing $7$ lowercased letters:

$$ \overbrace{26}^{\text{uppercase}} \cdot \overbrace{26^7}^{\text{lowercase}} = 26^8$$

Then we need to end the password by choosing a place for the special character or the number, choosing one position immediately reveals the position of the other - thus the total number of options for them is:

$$ \binom{2}{1} \binom{9}{1}  \binom{3}{1} = 54$$

We need to then multiply everything together and get:

$$ 26^8 \cdot 54 $$

As for every single choice we have a new password to "complete", for example, if we choose $B$ over $D$ to be the first letter, we have all the sub-passwords that begin with $D$ (we then complete the password for each choice) In other words every choice is independent.


问:
Approximating $\int_2^4 e^{\frac{1}{x}} dx$

Estimate the integral precisely up to the error of $0.001$:$$\int_2^4 e^{\frac{1}{x}} dx$$

I started by dividing the segment $[2,4]$ into some pieces $(n=5)$ as in the following figure:

$\qquad\qquad\quad$

Now, I have $$\Delta x = \frac{4-2}{5} = 0.4$$ and therefore, $$S \approx \sum_{i=1}^5 f(x_i)(x_i - x_{i-1}) = \Delta x \sum_{i=1}^5 e^{\frac{1}{x_i}}$$ where $x_i = 2 + 0.4i$.

Then, approximating the area boils down to approximating the exponential function at $5$ points. What is the best way to proceed? And how do I know that the final result will give the error at most $0.001$?

答:
$$\exp(1/x)=\sum_{n=0}^\infty\frac{x^{-n}}{n!}$$
$$\int\exp(1/x)dx=\sum_{n=0}^\infty\frac{x^{1-n}}{(1-n)n!}+C$$
then sub in your limits and take a few terms, so thats one option

$$\int_2^4e^{1/x}dx=\int_{1/4}^{1/2}\frac{e^u}{u^2}du=\int_{1/4}^{1/2}\sum_{n=0}^\infty\frac{u^{n-2}}{n!}du$$
so you have two choices here


问:
Is there a method for decomposing a complex function into the sum of its real and imaginary part?
I know that for every function $f:X \to \mathbb{C}$ there exists a pair of functions $u, v:X \to \mathbb{R}$ such that $f(x) = u(x) + iv(x)$. This is of course trivial because you can just say $u(x) = \Re (f(x))$ and $v(x) = \Im(f(x))$.

However my question would be if there exists an algorithm or method to compute more meaningful functions for $u$ and $v$. By more meaningful I mean that the result will be more efficient than just calculating $f$ twice and then discarding half of the results.

An example of this would be decomposing $$f(z) = z^2$$ into $$u(z) = x^2 - y^2 \\v(z) = 2xy$$, where $x = \Re(z)$ and $y = \Im(z)$; source.

I am especially interested in functions $\mathbb{C} \to \mathbb{C}$. Any help would be much appreciated.

答: $\Re{f}= \frac{f+\overline{f}}{2}$ and you can express the imaginary part in a similar fashion.


问:
Does There Exist a Term for the Unique Nonpositive Square Root of a Nonnegative Real Number?
The term "principal square root" describes the unique nonnegative square root of a nonnegative real number.

Does there exist a term to describe the unique nonpositive square root of a nonnegative real number?

答: I have heard "negative square root."  It's about the simplest and most aptly descriptive thing I have heard.</p>


问:
How to overcome the wastefulness of uniform distributions in a hash function
Recently on the programming StackOverflow I posted a query about my hash function - I was looking for mistake when there was none. I was alarmed by an empty bucket count of around 36%, as I thought, for an uniform distribution, that that number was erroneously high. Later today I figured it out using the binomial distribution and I realised that the counter-intuitive (similar to birthday problem) 36% “inefficiency” makes perfect sense. Since something cannot surely be more fair than a uniform distribution, how can a real world example of a hashing system reliably and evenly distribute data across ALL its buckets - is it even possible? The programmers at StackOverflow tell me that using a modulus of a large prime would help, but mathematically that wouldn’t change the binomial result of 0.3679... - or would it? Please enlighten me or tell me that I’m worrying over nothing and that the 0.367... is unavoidable! Thanks

Note: I am aware that a real world example would have more data points than buckets which would make the number of empty buckets decrease to zero (or would it? I’ve lost confidence in my probability intuitions) - I’m curious about the one-to-one case, where we have near identical numbers in search of a uniform spread

The equations I used to calculate probabilities were: (excuse the mobile phone formatting)

Let N = the number of buckets in the table

Let C = the number of repeat entries or collisions.

The probability of any bucket having C collisions is:

nCr(N, C) * (1/N^C) * (N-1/N)^(N-C) ≈ 0.368,  when C = 0 [empty, wasted bucket], as N tends to infinity.

答:
If we have $N$ buckets and place $M$ entries in them, with equal probability, the number of elements in each bucket, for large $N,M$, approximately follows ("Poissonization") a Poisson distribution $P_k= e^{-\lambda} \frac{\lambda^k}{k!}$ with $\lambda=M/N$.

If we denote $X_i=1$ if the bucket $i$ is empty, $0$ otherwise, we have $E[X_i]\approx   e^{-M/N}$ , and if $Z=\sum X_i$ denotes the number of empty buckets we have

$$E[Z] \approx N e^{-M/N}$$

Hence the proportion of empty buckets is approximately $e^{-M/N}$. If $M=N$ we get $0.368$, the $36\%$ you are getting. Notice, however, that this goes to $13.5\%$ for $M=2N$ and so on,

You cannot expect having a better occupation rate for a "universal" hashing function.

Only in special cases (where your key domain has some restrictions) you can do better, see eg https://en.wikipedia.org/wiki/Perfect_hash_function

Edit: To prove that the uniform distribution minimizes the expected number of empty buckets you might write

$$E[Z]=\sum_{i=1}^N E[X_i]= \sum_{i=1}^N (1-h_i)^M $$

where $h_i$ is the hashing distribution function. You need only to prove that $h_i=1/N$ attains the minimum, subject to the restrictions $ \sum_{i=1}h_i=1$, $h_i \ge 0$.


问:
How to solve $y'' -2\,y'+2\,y = \cos(x)\,e^x$
So to come to a quick end:

For the particular solution I've chosen the ansatz

$y_p  = a\,x\,\cos(x)\,e^x$

in orientation to the formula.
However, if I plug this in the Differential equation I get:

$-2\,a\,e^x\,\sin(x)$

I see no way how this could be equal to

$\cos(x)\,e^x$

so did I pick a wrong approach - try again?

答:
$$y'' -2\,y'+2\,y = \cos(x)\,e^x$$
$$(ye^{-x})'' +ye^{-x} = \cos(x)$$
$$w'' +w = \cos(x)$$
Where $w=e^{-x}y$.

Try for the particular solution:
$$w_p= Ax \sin x $$
You will find that $A=\dfrac 12$ then:
$$w_p=\dfrac 12 x \sin x$$
$$\implies y_p=\dfrac 12e^x x \sin x$$


问:
Unique Bounded Operator $A \in B(H)$ as a Unilateral Weighted Shift Operator
In Arveson's book "A Short Course on Spectral Theory", the concept of a unilateral weighted shift operator is introduced in one of the excercises in the Section 1.6.

Given a bounded sequence $(a_n)$ of complex numbers and a Hilbert space $\mathcal{H}$ with orthonormal basis $\{e_i\}$, we want to show that there is a ($\textbf{necessarily unique}$) bounded operator $A \in B(\mathcal{H})$ satisfying $Ae_n = a_n e_{n+1}$ for $n = 1,2,...$. Such an operator is called a unilateral weighted shift.

I can see that this weighted shift is the product of a shift ($Se_n = e_{n+1}$) and a compatible diagonal operator ($Pe_n = a_ne_n$). The derivation of the spectrum $\sigma(A)$ is also fairly straightforward. However, I do not see why it is necessarily unique? Why there is a bounded operator satisfying the above? I feel like I am missing something rather obvious. Any guiding in the darkness would be appreciated.

答:
$\|A(\sum c_ne_n)\|=\|\sum a_nc_n e_n\|=\sqrt {\sum |a_nc_n|^{2}}\leq M \|\sum c_ne_n\|$ where $M=\sup_n |a_n|$. Hence, $A$ is a bounded opeartor with norm at most $M$. Uniqueness is because $A(\sum c_ne_n)=\sum a_nc_n e_n$.

[Recall that $\sum c_ne_n$ converges if and only if $\sum |c_n|^{2} <\infty$].


答:
When you want to use the method of undetermined coefficients to find particular solution for an ODE of this form :
$$
a_{n}(x)\frac{d^{n}y}{dx^{n}}+a_{n-1}\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_{1}(x)\frac{dy}{dx}+a_{0}(x)=g(x)
$$
For a non-zero $a_{n}(x)$, it depends on how $g(x)$ is defined. You claim $y_{p}:=ax\cos(x)e^{x}$ when you should have tried :
$$
y_{p}:=(c_{1}x+c_{0})e^{x}(\cos(x)+\sin(x))
$$

Title: How to solve $y'' -2\,y'+2\,y = \cos(x)\,e^x$

Body:
So to come to a quick end:

For the particular solution I've chosen the ansatz

$y_p  = a\,x\,\cos(x)\,e^x$

in orientation to the formula.
However, if I plug this in the Differential equation I get:

$-2\,a\,e^x\,\sin(x)$

I see no way how this could be equal to

$\cos(x)\,e^x$

so did I pick a wrong approach - try again?


问:
Minimum possible distance between $n$ grid points
We are given a grid and a set $S$ of $n$ points on it (i.e points in the plane with integer coordinates).
 We define the diamatar $diam(S)$ of $S$ to be the maximum possible distance between two points in $S$, that is, $diam(S) = \max\{distance(a, b):a,b \in S \}$.
 What is the minimal possible value of $diam(S)$ for general $n$? Is there a closed formula, in terms of $n$?
 For example, for $n=1$ the answer is $0$. For $n=2$ the answer is $1$ and for $n=5$ the answer is $2$.

I would also like to know closed formulas for special cases, for example, for the case where $n$ is a perfect sqaure.

答:
A circle is clearly a sensible shape to consider for a 'close-packing' of integer points. For a circle of given radius $r$, (diameter $2r$), the numbers of points can easily be calculated. The solutions for small integer values of $r$ are given in A000328 - OEIS.

For example the value of $5$, as you've already determined, is given for $r=1$  and $13$ is the value for $r=2$.

If you've not already found it, https://en.wikipedia.org/wiki/Gauss_circle_problem is also a site you should visit. In particular it gives an approximate relation between the diameter, $d$,
and $n$.

$$\pi d^2\approx 4n.$$

As you will see the general problem is difficult, without precise formulae.


问:
Defining a real number
No answers, please, hints only.
I want to express every nonzero $x\in \mathbb R$ as a product of two numbers that are not rational.

My attempt is $x=(a_1+b_1i)(a_2+b_2i)$ where $a_1, b_2\in \mathbb R$

Am I correct or do I need a hint?

答:
A bit of cardinality might help.

For a given $x$ consider all then $w \in \mathbb R$ so that $w \ne 0$.

Only a countable number of $w$ will give us $\frac xw$ is rational, and an uncountable number of $w$ will give us $\frac xw$ is irrational.  And only a countable number of $w$ are rational and an uncountable number of $w$.  So there are an uncountable number of $w$ that are irrational and and have $\frac xw$ are irrational.

So just pick one of those.  Then $w$ is irrational and $\frac xw$ is irrational and $x = \frac xw\cdot w$.

So we know it can always be done.  But how to do it?

Well, if $x$ is rational, then we can pick $w$ be any irrational, say $w=\pi$.  Then $\pi$ is irrational, and $\frac x{\pi}$ is irrational, and $\pi\cdot \frac x\pi = x$.

ANd if $x$ is irrational.... well either $\frac x\pi$ is rational or irrational.  If $\frac x\pi = q\in \mathbb Q$ then $x = \pi*q$.  So we need to pick an irrational $w$ so that $\frac xw = \frac {\pi *q}{w}$ is irrational.

Well there are an uncountable number of irrational $w$ so that $\frac \pi w$ is irrational.  Can we come up with one?  Well $\sqrt \pi$ is irrational and $\frac {\pi}{\sqrt \pi} = \sqrt \pi$ is irrational.

So

If $\frac x{\pi}\not \in \mathbb Q$ let $w = \frac x{\pi}$ and $\pi \cdot \frac x{\pi} = x$.

If $\frac x{\pi} \in \mathbb Q$ then $\frac x{\sqrt{\pi}}\not \in \mathbb Q$.  Then $\sqrt \pi\cdot \frac x{\sqrt \pi} = x$.

That'll do it.

Title: Defining a real number


问:
How to get the least amount of transfers to get $4L$?
The problem is as follows:

There is a bucket, fully filled, with 8 liters of water and two 5 and
3 liter empty jugs. All three containers have no markings that allow
measurements and they are not of a regular shape. Using the bucket and
two jugs and without spilling the water at any time, how many
transfers must be done, at a minimum, to make the bucket and one of
the jars contain 4 liters of water, each?

The given alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{7 transfers}\\
2.&\textrm{6 transfers}\\
3.&\textrm{9 transfers}\\
4.&\textrm{8 transfers}\\
\end{array}$

I'm confused exactly how to approach this problem?. So far what I have been able to get was only 2L in one of the vessels while the others have 3L and 3L

Assuming I begin as follows:

The first row represents the labels of each container:

8L 5L 3L

8L 0L 0L

3L 5L 0L

3L 2L 3L

That's where I end, with just 2L in one of the containers.

But that's it. I'm stuck. Can this puzzle be solved from here?. The problem only mentions three of these containers. But upon trying on different ways I can't get the right number of transfers. Can someone guide me on here on a method or way to get the requested 4L?.

It would help me a lot an answer which could include a step by step approach on how to get this right.

答:
You are close to solving it! I solved it in 5 additional steps continuing from where you left off.

As a hint, if you can get either 7/8L in the first jug or 2/3L in the third jug, then you will have a solution by filling the second jug to 5/5L and then transferring 1L to either the first or third jug.

Discussion of the generalized version of this problem found here: Measuring water puzzle


问:
Simple $R$-module equivalent statement
I am trying to show that the two statements are logically equivalent.

a) $X$ is a simple $R$-module.

b) For any $x , y \in X$, $x,y \neq 0$ , there exists $r \in R$ such that $rx = y$.

In other words,  I am trying to show that $X$ is a simple $R$-module $\iff$ For any $x , y \in X$, $x,y \neq 0$ , there exists $r \in R$ such that $rx = y$.

Here $R$ is a commutative ring. Furthermore, by a simple $R$-module I mean a non zero module that has non proper submodules.

I am very new to the study of modules so any push in the right direction would be helpful.

答:
$(a)\to (b)$: Let $0\ne x,y\in X$. Then $Rx=\{rx: r\in R\}$ is a submodule of $X$. This submodule is nonzero, as it contains the element $x$. Since $X$ is simple it follows that $Rx=X$, and in particular $y\in Rx$.

$(b)\to (a)$: Let $Y$ be a nonzero submodule of $X$. So take any $0\ne y\in Y$. We will show that $Y=X$. Let $x\in X$. By assumption there is some $r\in R$ such that $x=ry$ (if $x=0$ you can take $r=0$), so in particular $x\in Y$.

Title: Simple $R$-module equivalent statement


问:
Mysterious morphism
Until I get a clearer understanding of what follows, and it is the aim of this post, I am sorry I cannot make the question title more explicit.

In the setting of algebraic groups I found the following notations written many times, and I don't figure out what they are, both intuitively and formally. $G$ is a reductive group. $X(G)$ is the additive group of homomorphisms from $G$ to $GL(1)$, it is a free abelian group. Then we consider
$$\mathfrak{a}_G = Hom_{\mathbb Z} (X(G), \mathbb R)$$
which notation makes me think of a Lie algebra. Is there a natural interpretation as such? As e.g. the Lie algebra of the connected component of the central subgroup (usually denoted $A_G$)?

Most importantly, we define an homomorphism
$$H_G : G(\mathbb A) \to \mathfrak{a}_G$$
by
$$\langle H_G(x), \chi \rangle = |\log (x^\chi)|$$

How to make this definition more explicit, e.g. in the simple cases of $GL_2$ or $GL_n$? Is it a kind of norm?

答:
Even though your tags imply this is in the realm of algebraic groups proper, it's really an object which appears more naturally in the context of automorphic representations.

The point is that in the theory of automorphic representations of $G$ (a reductive group over $\mathbb{Q}$) one wants to do harmonic analysis on the locally compact Hausdorff group $G(\mathbb{A})$ or, more precisely, to study the harmonic analysis of the measure space $G(\mathbb{A})/G(\mathbb{Q})$ (the measure coming from the quotient measure of the Haar measure on $G(\mathbb{A})$). In particular, as is usually the case when doing harmonic analysis, one wants to study something like the space of square-integrable complex valued functions on $G(\mathbb{A})/G(\mathbb{Q})$. This is slightly complicated though by the fact that $G(\mathbb{A})/G(\mathbb{Q})$ doesn't have finite volume (with the Haar measure). That said, there is a way of harmlessly modifying $G(\mathbb{A})/G(\mathbb{Q})$ to have finite volume.

To do this we need, in some sense, the set of points of 'bounded norm' in $G(\mathbb{A})$. Of course, we need to explain what we mean by the 'norm' here. Note that for any character $\chi$ in $X^\ast(G)$ one gets an induced map

$$|\cdot|_\chi:G(\mathbb{A})\to \mathbb{R}:g\mapsto |\chi(g)|$$

where, as per usual, for an element $(a_v)\in \mathbb{A}^\times$ we set

$$|(a_v)|:=\prod_v |a_v|$$

which makes sense since $a_v\in\mathcal{O}_v^\times$ for all but finitely many $v$ and so, consequently, $|a_v|=1$ for all but finitely many $v$.

One can then form the subgroup $G(\mathbb{A})^1$ as follows:

$$G(\mathbb{A})^1:=\left\{g\in G(\mathbb{A}):|g|_\chi =1\text{ for all }\chi\in X^\ast(G)\right\}$$

Note that $G(F)\subseteq G(\mathbb{A})^1$ which is precisely the product theorem from basic valuation theory.

One then has the following spectacular theorem:

Theorem(Borel, Conrad, Oesterle, Harder, [1, §5.3]): The quotient $G(\mathbb{A})^1/G(\mathbb{Q})$ has finite volume.

Now to explain why this modification is 'harmless' it's useful to understand $G(\mathbb{A})^1$ in terms of the objects you described. So, let us preliminarily take your definitions

$$\mathfrak{a}_G:=\mathrm{Hom}_\mathbb{Q}(X^\ast(G),\mathbb{R})$$

and (I think you have a typo here)

$$H_G:G(\mathbb{A})\to \mathfrak{a}_G$$

defined by letting $H_G(g)$ be the $\mathbb{Q}$-linear map $X^\ast(G)\to\mathbb{R}$ which associates to $\chi$ the element $\log|\chi(g)|\in\mathbb{R}$.

We then note that

$$\begin{aligned}\ker(H_G)&=\left\{g\in G(\mathbb{A}): H_G(g)=0\right\}\\ &= \left\{g\in G(\mathbb{A}):H_G(\chi)=0\text{ for all }\chi\in X^\ast(G)\right\}\\ &= \left\{g\in G(\mathbb{A}):\log|\chi(g)|=0\text{ for all }\chi\in X^\ast(G)\right\}\\ &= \left\{g\in G(\mathbb{Q}):\chi(g)=1\text{ for all }\chi\in X^\ast(G)\right\}\\ &= G(\mathbb{A})^1\end{aligned}$$

To continue it is useful to address your other question. Namely, you astutely noted that the notation $\mathfrak{a}_G$ looked like the notation for a Lie algebra. The reason for that is as follows. Let $A_G$ denote the largest split subtorus of $Z(G)$. Then, we have that

$$\mathrm{Lie}(A_G)\otimes_\mathbb{Q}\mathbb{R}\cong \mathrm{Hom}_\mathbb{Q}(X^\ast(G),\mathbb{R})$$

Indeed, note that

$$X^\ast(G)=X^\ast(G^\mathrm{ab})\cong X^\ast(Z(G))\cong X^\ast(A_G)$$

where the first isomorphism is obvious, the second is from the fact that the natural map $Z(G)\to G^\mathrm{ab}$ is an isogeny, and the final is since $Z(G)$ is isogenous to $A_G\times T'$ where $T'$ is an anisotropic subtorus of $Z(G)$. But, we then see that

$$\mathrm{Hom}_\mathbb{Q}(X^\ast(G),\mathbb{R})=\mathrm{Hom}_\mathbb{Q}(X^\ast(A_G),\mathbb{R})=X_\ast(A_G)\otimes_\mathbb{Q}\mathbb{R}\cong \mathrm{Lie}(A_G)\otimes_\mathbb{Q}\mathbb{R}$$

where the last isomorphism associates to a cocharacter $\mu:\mathbb{G}_m\to A_G$ the element $(d\mu)(1)\in \mathrm{Lie}(A_G)$.

In fact, using this one can show the following:

Proposition: The map $$G(\mathbb{A})^1\times A_G(\mathbb{R})\to G(\mathbb{A}):(g,a)\mapsto ga$$ is an isomorphism.

In particular, the reason that $G(\mathbb{A})^1/G(\mathbb{Q})$ is functionally not much different than $G(\mathbb{A})/G(\mathbb{Q})$ is that they differ by the term $A_G(\mathbb{R})$ which, since this is central and a direct factor of $G(\mathbb{A})$, can be twisted away in practice.

References:

[1] Platonov, V., Rapinchuk, A. and Rowen, R., 1993. Algebraic groups and number theory. Academic press.

Title: Mysterious morphism


问:
Does $\sum_{k = 1}^{n-1} \frac{\binom{2k}{k}\binom{2n-2k}{n-k}}{k(n-k)}$ have a simpler expression?

The following constants have appeared in my recent research and I was wondering if they have a simpler expression $?$.

I have calculated several values but I don't see any obvious combinatorial answer.

$$
\sum_{k = 1}^{n-1}
\frac{\displaystyle\binom{2k}{k}\binom{2n - 2k}
{n - k}}{k\left(n-k\right)} =\ ?
$$
and more general
$$
\sum_{\substack{k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{r}\ =\ n
\\[1mm] k_{1}\,,\ k_{2}\,,\ \ldots\ k_{r}\ \ge\ 1}}
\frac{\displaystyle\binom{2k_{1}}{k_{1}}
\binom{2k_{2}}{k_{2}}\cdots\binom{2k_{r}}{k_{r}}}{k_{1}\,k_{2}\ldots k_{r}} =\ ?
$$

答:
This sum has the form of a convolution so can be done by taking a product of generating functions. We can get the generating function we need by starting from

$$\sum_{k \ge 0} {2k \choose k} x^k = \frac{1}{\sqrt{1 - 4x}}$$

(which can be proven using e.g. the generalized binomial theorem) and subtracting $1$, dividing by $x$, and then integrating to give

$$F(x) = \sum_{k \ge 1} {2k \choose k} \frac{x^k}{k} = \int_0^x \frac{1 - \sqrt{1 - 4t}}{t \sqrt{1 - 4t}} \, dt.$$

The substitution $u = \sqrt{1 - 4t} \Rightarrow u^2 = 1 - 4t, 2u \, du = -4 \, dt$ gives

$$\int_1^{\sqrt{1 - 4x}} \frac{1 - u}{ \left( \frac{1 - u^2}{4} \right) u} \left( - \frac{u}{2} \right) \, du = -2 \int_1^{\sqrt{1-4x}} \frac{1}{u+1} \, du = 2 \ln 2 -2 \ln (1 + \sqrt{1 - 4x}).$$

This can be rewritten

$$2 \ln \frac{2}{1 + \sqrt{1 - 4x}} = 2 \ln \frac{1 - \sqrt{1 - 4x}}{2x}$$

where the expression inside the logarithm is, curiously, the generating function of the Catalan numbers. Writing $f_r(n)$ for the sum you're interested in, we then have

$$\boxed{ \sum_{n \ge 0} f_r(n) x^n = F(x)^r = \left( 2 \ln \frac{1 - \sqrt{1 - 4x}}{2x} \right)^r }.$$

If we consider a two-variable generating function summing over all $r$ this gives

$$\boxed{ \sum_{r, n \ge 0} f_r(n) x^n \frac{t^r}{r!} = \exp(F(x) t) = \left( \frac{1 - \sqrt{1 - 4x}}{2x} \right)^{2t} }.$$

These are both funny generating functions and I can't say that I've seen them before. I don't think there will be nice closed forms but there should be asymptotics at least. This second generating function might have a nice combinatorial interpretation but I'd have to think for a bit about what it is.


问:
Understanding Perron's formula by calculation of residues
I'm trying to get an intuitive understanding of Perron's formula, which assets that, for the Dirichlet series $\alpha(x)=\sum_{n=1}^{\infty}a_n n^{-s}$ and the coefficient sum $A(x)=\sum_{n\leq x} a_n$ we have:

$A(x)= \frac{1}{2\pi i} \int_{\sigma_0-i\infty}^{\sigma _0+i \infty} \alpha(s) \frac{x^s}{s} ds $

for $\sigma _0 >0$.

My text book (H. L. Montgomery and R. C. Vaughan. Multiplicative number theory. I. Classical theory, chapter 5.1 page 137) says that we should expect the above formula to be true, as we can calculate the integral using calculus of residues. It further says that:

$ \frac{1}{2\pi i} \int_{\sigma_0-i\infty}^{\sigma _0+i \infty} y^s \frac{1}{s} ds
= $ $\{ 1 \text{ if } y>1, \quad 0 \text{ if } 0<y<1,\quad \frac{1}{2} \text{ if } y=1 \} $

I have two problems with this:

When I learned about residues I learned that

$ \frac{1}{2\pi i} \int_{\gamma}f(z) dz = \sum_{j=1}^n \operatorname{Res}\left(f,a_j\right)$

only when $\gamma$ is a simple closed path in a simply connected domain. But the path from $\sigma_0-i\infty$ to $\sigma_0+i\infty$ is not simply connected?

If we say that the equation for the residue holds, we'll now need to calculate the residue. The pole, as far as I can see, is simple so we need to find the limit:

$\lim _{s \longrightarrow 0} \frac{y^s}{s}(s-0)=\lim _{s \longrightarrow 0} y^s $

But I can't see how the limit of $y^s$ can be anything but $1$. How does this work out?

答:
Indeed, there is a possible method for us to derive Perron's formula from scratch. If $\alpha(s)$ converges at $\Re s=\sigma_0$, then using OP's notation, we can obtain the following relation if we consider Stieltjes integration for $\Re s\ge\sigma_0$.

$$
\begin{aligned}
\alpha(s)
&=\int_{1^-}^\infty{\mathrm dA(x)\over x^s} \\
&=\left.{A(x)\over x^s}\right|_{1^-}^\infty+s\int_1^\infty A(x)x^{-s-1}\mathrm dx \\
&=s\int_0^\infty A(e^t)e^{-st}\mathrm dt
\end{aligned}
$$

This indicates that a Laplace transform relationship exists between the summatory function and the Dirichlet series:

$$
{\alpha(s)\over s}=\mathcal L\{A(e^t)\}(s)
$$

Consequently, by applying Mellin's inversion formula, we obtain

$$
A(e^t)={1\over2\pi i}\int_{\sigma_0-i\infty}^{\sigma_0+i\infty}\alpha(s){e^{st}\over s}\mathrm ds
$$

Now, plugging in $x=e^t$ gives the Perron's formula

$$
A(x)={1\over2\pi i}\int_{\sigma_0-i\infty}^{\sigma_0+i\infty}\alpha(s){x^s\over s}\mathrm ds
$$

In particular, if $x$ is an integer then the integral is interpreted as the Cauchy principal value:

$$
{A(x-0)+A(x+0)\over2}
$$