答:
By contradiction, assume that for any $p \in [0,1] \times [0,1]$ there exists $\epsilon_p >  0$ such that $E \cap B(p, \epsilon_p)$ is countable.

Consider the union $B = \bigcup B(p,\epsilon_p)$ over all rational points $p$ in $[0,1] \times [0,1]$ (that is, points where both components are rational). Then $E \subseteq B$ since $\mathbb{Q}$ is dense in $\mathbb{R}$. But $B$ is the countable union of countable sets, so $B$ is countable and thus $E$ is countable, a contradiction.

Title: For an uncountable subset $E$ in $[0,1]\times[0,1]$, show there is a condensation point of $E$.

Body:
I'm not sure how to approach this question:

Given $E\subseteq[0,1]\times[0,1]$ and $E$ is uncountable, show that there is some $p\in [0,1]\times[0,1]$ such that $\forall \epsilon>0, E\cap B(p,\epsilon)$ is uncountable.

I tried using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. I defined a set $A$ to be the collection of all open balls centered at rational points with some $\epsilon$ such that the unit square is completely covered. Surely at least one of the balls (centered at say $x$) contains uncountably many points in $E$. But, I don't know how to make this work (or even if it could work). I'm concerned about the overlap of the balls and how to find some $x$ that doesn't move around as $\epsilon$ changes. This is my introduction to analysis, so I don't have a good intuition for much of these concepts yet. Any help would be greatly appreciated!


问:
Elliptic integrals, general form
According to https://mathworld.wolfram.com/EllipticIntegral.html, Elliptic integrals are integrals of the form of equation (1).

I am trying to convert an integral of the form

$$\int_{|x_+|}^\infty \frac{dx}{\sqrt{(x^2-x_-^2)(x^2-x_+^2)}}$$

To express as a function of elliptic integrals, where $|x_+|>|x_-|$, we can take $x_+>0$ for simplicity. We can easily see that this integral converges, and that it is real.

I have trying following the method outline in the webpage I linked, with no success, because of the arising of imaginary roots in the process. However, plugging this expression into mathematica with $x_-=1$, $x_+=2$, yields me the expression :

$$\int_{2}^\infty \frac{dx}{\sqrt{(x^2-1)(x^2-4)}}=\frac{i}{2}K(\frac{3}{4})+K(4)$$

Where $K(m) = \int_0^{\pi/2}\frac{1}{\sqrt{1-m\sin^2(\phi)}}d\phi$. Now clearly the integral I'm looking to re-express is well-defined, and real, but it would appear I get an complex number as an answer. However, looking closer, $K(4)$ will be complex because the parameter $m>1$. Thus, if we numerically compute the expression the imaginary parts cancel and we are left with a real answer, which is correct.

So my question is as follows : when we say that Elliptic integrals (as defined in my link) are expressible as a function of the three "kinds" of elliptic integrals (+ elementary functions), do we allow "complex" elliptic integrals ? In other words, is there any way to express my example as a function of "well defined" elliptic integrals, in the sense that we only integrate real numbers ?

I am unsure if I expressed myself clearly, I hope it makes sense.

答:
The given integral is $K(1/2)/2$ (convention!), but i will try to follow also a geometric path, the canonical one, detailed in the given particular case $x_-=1$, and $x_+=2$, so that things are easier to type. There will be also numerical check (also for intermediate steps), so that i can calmly sleep tonight.

Let us recall the Complete elliptic integral of the first kind:
$$
K(k)
=
\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}
=
\int_0^1
\frac {dt}{\sqrt{(1-t^2)(1-k^2t^2)}}\ .
$$
See also DLMF on Elliptic integrals, where $K(k)=F(\pi/2,k)$, and in the definition of $F$ the above convention is used.

Note that the above convention differs from the convention used by Mathematica, as also remarked in loc. cit. .

Taking in the last integral $k=\frac 12$ we obtain an integral "similar" to the given one, at least under the integral, but the limits of integration differ. What can we do?

The lower integration limit $0$ is the point of symmetry for $(1-t^2)(1-k^2t^2)$, and $1$ is one of its roots. In our case, for $(x^2-1)(x^2-4)$ the value $2$ is also a root, so the substitution $t=2/x$ is natural. We get for the given integral $I$ (well, let us have this handy notation):
$$
\begin{aligned}
I&=
\int_{2}^\infty
\frac1{\sqrt{(x^2-1)(x^2-4)}}\; dx
\\
&=
\int_1^0
\frac1{\sqrt{\left(\frac 4{t^2}-1\right)\left(\frac 4{t^2}-4\right)}}\; \left(-\frac 2{t^2}\right)\; dt
\\
&=
\int_0^1
\frac1{\sqrt{(4-t^2)(4- 4t^2)}}\; 2\; dt
\\
&=
\frac12
\int_0^1
\frac1{\sqrt{(1-\frac 14t^2)(1- 1t^2)}}\; dt
\\
&=\frac 12K\left(\frac 12\right)\ .
\end{aligned}
$$

Numerical check using sage, which also suffers from the same convention as Mathematica.

```<code>sage: var('x,t');
sage: I = integral( 1/sqrt( (x^2-1)*(x^2-4) ), x, 2, oo, hold=1 )
sage: I.n()
0.8428751774069201
sage: def K(k):    return elliptic_kc(k^2)    # convention
sage: K(1/2).n() / 2
0.842875177406298
</code>```

pari/gp has the wiki convention, a numerical check using pari, taking advantage of the implementation of ellK, recently made public, October 2020,
is as follows:

```<code>? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%1 = 0.84287517740629802143560182889542910036

? \p 60
   realprecision = 77 significant digits (60 digits displayed)

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%2 = 0.842875177406298021435601828899538494750400447070248008383945

? ellK(1/2)/2
%3 = 0.842875177406298021435601828899538494750400447070544522059974
</code>```

It seems that Mathematica goes an other way when evaluating the given integral, thus delivering a value which reflects 19.7.3 in DLMF, Legendre's relations, explicitly:
$$K(1/k) = k(K(k)\mp iK(k')\ ,\qquad k^2 + {k'}^2=1\ .$$
In our case, with $k=1/2$, we get $K(2)=\frac 12(K(1/2)\mp iK(\sqrt{3/2}))$.
This is compatible with the result of Mathematica, written in its convention.

This is already an answer to the question.

$\square$

A possible digression: Please ignore, if it feels not related to the question.
This is inserted for my own purpose, so that i can find it next time as an explicit example. For me there is a natural relation, the most natural way to deal with elliptic integrals is to see them explicitly in relation with elliptic curves. We will obtain (after an expensive tour) an other realization of the given integral in terms of the complete elliptic integral, $K(\sqrt{8/9})/3$.

The given integral can be written as
$$
\color{blue}{
I = \int_{(2,0)}^{\infty}\frac{dx}y
} \ ,
$$
and it can be considered on the compact manifold $Q(\Bbb R)$ or $\Bbb C$, where $Q$ is the "quartic curve" (an algebraic variety, thus analytic, thus smooth manifold) with affine equation
$$
Q\ :\ y^2 =(x-1)(x+1)(x-2)(x+2)\ .
$$
(The infinity point is the point $(0:1:0)$ which verifies the projective form, the homogenized form
of this equation, $y^2z^2=(x-z)(x+z)(x-2z)(x+2z)$.)

The idea is to obtain a birational morphism to an elliptic curve.

We use the (birational) transformations:
$$
\left\{
\begin{aligned}
x &=-\frac 4Y(X-5)\ ,\\
y &=2-\frac 4{Y^2}X(X-5)^2\ ,
\end{aligned}
\right.
\qquad
\left\{
\begin{aligned}
X &=-\frac 4{x^2}(y-2)\ ,\\
Y &=\frac 1{x^3}(16(y-2)+20x^2)\ ,
\end{aligned}
\right.
$$
and then
$$
y^2- (x^2-1)(x^2-4)
=
\frac{16(X - 5)^3}{Y^4}\cdot(-Y^2+ X^3 - 5X^2- 16X + 80)\ ,
$$
so we have a passage to the elliptic curve with the equation from the last parentesis,
$$
Y^2 =(X-5)(X-4)(X+4)\ .
$$
We may (or may not) want to keep it simple (and unstructural), thus involve only the substitution
$$
x =
-\frac{4(X-5)}{\sqrt{(X-4)(X+4)(X-5)}}
\ .
$$
We have already made the "natural choice" for the square root.
Note that for $X=2$ we obtain $x=2$. And
$X\searrow -4$ implies $x\nearrow \infty$.

The structure or brute force computations shows now that we have:
$$
\color{blue}{
I=
\int_2^\infty\frac {dx}{\sqrt{(x^2-1)(x^2-4)}}
=
\int_{-4}^2
\frac {dX}{\sqrt{(X-4)(X+4)(X-5)}}\ .
}
$$
This is the passage from the integral on the quartic to an integral on the elliptic curve written in Weierstraß form.

Instead of a proof, here is a
numerical check using pari/gp:

```<code>? \p 60
   realprecision = 77 significant digits (60 digits displayed)
? intnum( x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ) )
%2 = 0.842875177406298021435601828899538494750400447070248008383945
? intnum( X=-4, 2, 1/sqrt( (X-4)*(X+4)*(X-5) ) )
%3 = 0.842875177406298021435601828899538494750400447070446045891423
</code>```

We are thus in the position to calculate the integral of the canonical differential on the elliptic curve $E$ with equation
$$
(E)\qquad
Y^2=(X-e_1)(X-e_2)(X-e_3)\ ,\qquad e_1,e_2,e_3=5,4,-4,\ e_1>e_2>e_3\ .
$$
The limit of integration $e_1=-4$ corresponds to one torsion point of order two, $(e_1,0)$. The other $2$-torsion points are $(e_2,0)$, and $(e_3,0)$. Can we give a structural sense to the limit of integration $2$ from the blue realtion?!

Yes, the points $(2,\pm 6)$ are ($\Bbb Q$-rational) points on $E$, and they are $4$-torsion points. For this reason we have the chain of relations:
$$
\begin{aligned}
I
&=
\int_2^\infty\frac {dx}{\sqrt{(x^2-1)(x^2-4)}}
\\
&=
\int_{-4}^2
\frac {dX}{\sqrt{(X+4)(X-4)(X-5)}}
=
\int_2^4
\frac {dX}{\sqrt{(X+4)(X-4)(X-5)}}
\\
&=\frac12\int_{e_3=-4}^{e_2=4}
\frac {dX}{\sqrt{(X-e_1)(X-e_2)(X-e_3)}}
\\
&=\frac 12\omega_1(E)\qquad\text{(real half-period of the elliptic curve $E$)}
\\
&=\frac 1{\sqrt{e_1-e_3}}K\left(\sqrt{\frac{e_2-e_3}{e_1-e_3}}\right)
\\
&=\frac 1{\sqrt{5-(-4)}}K\left(\sqrt{\frac{4-(-4)}{5-(-4)}}\right)
\\
&=\frac 13 K\left(\frac{2\sqrt 2}3\right)\ .
\end{aligned}
$$

Let us numerically check again, pari/gp.

```<code>? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%10 = 0.842875177406298021435601828899538494750400447070248008383945
? ellK(sqrt(8/9))/3
%11 = 0.842875177406298021435601828899538494750400447070544522059974
</code>```

Addendum. Computer algebra support for the elliptic curve computations.

Using sage:

```<code>sage: var('x,y,X,Y');
sage: def f(x,y): return y^2 - (x^2-1)*(x^2-4)
sage: factor( f(-4*(X-5)/Y, 2-4*X*(X-5)^2/Y^2) )
16*(X^3 - 5*X^2 - Y^2 - 16*X + 80)*(X - 5)^3/Y^4
sage: E = EllipticCurve(QQ, [0, -5, 0, -16, 80])
sage: E
Elliptic Curve defined by y^2 = x^3 - 5*x^2 - 16*x + 80 over Rational Field
sage: for T in E.torsion_points():
....:     if T != E(0):
....:         print(f"Order {T.order()} for the torsion point {T.xy()}")
....:
Order 2 for the torsion point (-4, 0)
Order 4 for the torsion point (2, -6)
Order 4 for the torsion point (2, 6)
Order 2 for the torsion point (4, 0)
Order 2 for the torsion point (5, 0)
Order 4 for the torsion point (8, -12)
Order 4 for the torsion point (8, 12)
</code>```

This structural aspect explains why, when we compute the integrals of the canonical differential form $\frac{dX}Y$ of $E$ on the intervals delimited (in $\Bbb R$) by the points
$$
-\infty\ ,\
-4\ ,\
2\ ,\
4\ ,\
5\ ,\
8\ ,\
+\infty
$$
we obtain numerically the values for the integral of $(\ (X-4)(X+4)(X-5)\ )^{-1/2}$ on the delimited intervals... (I am taking the absolute value to not confuse sage.)

```<code>sage: f = 1/sqrt( abs((X-4)*(X+4)*(X-5)) )
sage: integral(f, X, -oo, -4, hold=True).n()
1.07825782381822
sage: integral(f, X, -4, 2, hold=True).n()
0.8428751689072135
sage: integral(f, X, 2, 4, hold=True).n()
0.8428751680336899
sage: integral(f, X, 4, 5, hold=True).n()
1.0782578072972744
sage: integral(f, X, 5, 8, hold=True).n()
0.8428751637656617
sage: integral(f, X, 8, oo, hold=True).n()
0.8428751775376374
</code>```

And the above values are related to the periods of $(E)$:

```<code>sage: L = E.period_lattice()
sage: L
Period lattice associated to
    Elliptic Curve defined by y^2 = x^3 - 5*x^2 - 16*x + 80 over Rational Field
sage: L.gens()
(1.68575035481260, 1.07825782374982*I)
sage: L.gens()[0]/2
0.842875177406298
</code>```

ContentLicense: CC BY-SA 4.0

Title: Elliptic integrals, general form

Body:
The given integral is $K(1/2)/2$ (convention!), but i will try to follow also a geometric path, the canonical one, detailed in the given particular case $x_-=1$, and $x_+=2$, so that things are easier to type. There will be also numerical check (also for intermediate steps), so that i can calmly sleep tonight.

Let us recall the Complete elliptic integral of the first kind:
$$
K(k)
=
\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}
=
\int_0^1
\frac {dt}{\sqrt{(1-t^2)(1-k^2t^2)}}\ .
$$
See also DLMF on Elliptic integrals, where $K(k)=F(\pi/2,k)$, and in the definition of $F$ the above convention is used.

Note that the above convention differs from the convention used by Mathematica, as also remarked in loc. cit. .

Taking in the last integral $k=\frac 12$ we obtain an integral "similar" to the given one, at least under the integral, but the limits of integration differ. What can we do?

The lower integration limit $0$ is the point of symmetry for $(1-t^2)(1-k^2t^2)$, and $1$ is one of its roots. In our case, for $(x^2-1)(x^2-4)$ the value $2$ is also a root, so the substitution $t=2/x$ is natural. We get for the given integral $I$ (well, let us have this handy notation):
$$
\begin{aligned}
I&=
\int_{2}^\infty
\frac1{\sqrt{(x^2-1)(x^2-4)}}\; dx
\\
&=
\int_1^0
\frac1{\sqrt{\left(\frac 4{t^2}-1\right)\left(\frac 4{t^2}-4\right)}}\; \left(-\frac 2{t^2}\right)\; dt
\\
&=
\int_0^1
\frac1{\sqrt{(4-t^2)(4- 4t^2)}}\; 2\; dt
\\
&=
\frac12
\int_0^1
\frac1{\sqrt{(1-\frac 14t^2)(1- 1t^2)}}\; dt
\\
&=\frac 12K\left(\frac 12\right)\ .
\end{aligned}
$$

Numerical check using sage, which also suffers from the same convention as Mathematica.

sage: var('x,t');
sage: I = integral( 1/sqrt( (x^2-1)*(x^2-4) ), x, 2, oo, hold=1 )
sage: I.n()
0.8428751774069201
sage: def K(k):    return elliptic_kc(k^2)    # convention
sage: K(1/2).n() / 2
0.842875177406298

pari/gp has the wiki convention, a numerical check using pari, taking advantage of the implementation of ellK, recently made public, October 2020,
is as follows:

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%1 = 0.84287517740629802143560182889542910036

? \p 60
   realprecision = 77 significant digits (60 digits displayed)

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%2 = 0.842875177406298021435601828899538494750400447070248008383945

? ellK(1/2)/2
%3 = 0.842875177406298021435601828899538494750400447070544522059974

It seems that Mathematica goes an other way when evaluating the given integral, thus delivering a value which reflects 19.7.3 in DLMF, Legendre's relations, explicitly:
$$K(1/k) = k(K(k)\mp iK(k')\ ,\qquad k^2 + {k'}^2=1\ .$$
In our case, with $k=1/2$, we get $K(2)=\frac 12(K(1/2)\mp iK(\sqrt{3/2}))$.
This is compatible with the result of Mathematica, written in its convention.

This is already an answer to the question.

$\square$

A possible digression: Please ignore, if it feels not related to the question.
This is inserted for my own purpose, so that i can find it next time as an explicit example. For me there is a natural relation, the most natural way to deal with elliptic integrals is to see them explicitly in relation with elliptic curves. We will obtain (after an expensive tour) an other realization of the given integral in terms of the complete elliptic integral, $K(\sqrt{8/9})/3$.

The given integral can be written as
$$
\color{blue}{
I = \int_{(2,0)}^{\infty}\frac{dx}y
} \ ,
$$
and it can be considered on the compact manifold $Q(\Bbb R)$ or $\Bbb C$, where $Q$ is the "quartic curve" (an algebraic variety, thus analytic, thus smooth manifold) with affine equation
$$
Q\ :\ y^2 =(x-1)(x+1)(x-2)(x+2)\ .
$$
(The infinity point is the point $(0:1:0)$ which verifies the projective form, the homogenized form
of this equation, $y^2z^2=(x-z)(x+z)(x-2z)(x+2z)$.)

The idea is to obtain a birational morphism to an elliptic curve.

We use the (birational) transformations:
$$
\left\{
\begin{aligned}
x &=-\frac 4Y(X-5)\ ,\\
y &=2-\frac 4{Y^2}X(X-5)^2\ ,
\end{aligned}
\right.
\qquad
\left\{
\begin{aligned}
X &=-\frac 4{x^2}(y-2)\ ,\\
Y &=\frac 1{x^3}(16(y-2)+20x^2)\ ,
\end{aligned}
\right.
$$
and then
$$
y^2- (x^2-1)(x^2-4)
=
\frac{16(X - 5)^3}{Y^4}\cdot(-Y^2+ X^3 - 5X^2- 16X + 80)\ ,
$$
so we have a passage to the elliptic curve with the equation from the last parentesis,
$$
Y^2 =(X-5)(X-4)(X+4)\ .
$$
We may (or may not) want to keep it simple (and unstructural), thus involve only the substitution
$$
x =
-\frac{4(X-5)}{\sqrt{(X-4)(X+4)(X-5)}}
\ .
$$
We have already made the "natural choice" for the square root.
Note that for $X=2$ we obtain $x=2$. And
$X\searrow -4$ implies $x\nearrow \infty$.

The structure or brute force computations shows now that we have:
$$
\color{blue}{
I=
\int_2^\infty\frac {dx}{\sqrt{(x^2-1)(x^2-4)}}
=
\int_{-4}^2
\frac {dX}{\sqrt{(X-4)(X+4)(X-5)}}\ .
}
$$
This is the passage from the integral on the quartic to an integral on the elliptic curve written in Weierstraß form.

Instead of a proof, here is a
numerical check using pari/gp:

? \p 60
   realprecision = 77 significant digits (60 digits displayed)
? intnum( x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ) )
%2 = 0.842875177406298021435601828899538494750400447070248008383945
? intnum( X=-4, 2, 1/sqrt( (X-4)*(X+4)*(X-5) ) )
%3 = 0.842875177406298021435601828899538494750400447070446045891423

We are thus in the position to calculate the integral of the canonical differential on the elliptic curve $E$ with equation
$$
(E)\qquad
Y^2=(X-e_1)(X-e_2)(X-e_3)\ ,\qquad e_1,e_2,e_3=5,4,-4,\ e_1>e_2>e_3\ .
$$
The limit of integration $e_1=-4$ corresponds to one torsion point of order two, $(e_1,0)$. The other $2$-torsion points are $(e_2,0)$, and $(e_3,0)$. Can we give a structural sense to the limit of integration $2$ from the blue realtion?!

Yes, the points $(2,\pm 6)$ are ($\Bbb Q$-rational) points on $E$, and they are $4$-torsion points. For this reason we have the chain of relations:
$$
\begin{aligned}
I
&=
\int_2^\infty\frac {dx}{\sqrt{(x^2-1)(x^2-4)}}
\\
&=
\int_{-4}^2
\frac {dX}{\sqrt{(X+4)(X-4)(X-5)}}
=
\int_2^4
\frac {dX}{\sqrt{(X+4)(X-4)(X-5)}}
\\
&=\frac12\int_{e_3=-4}^{e_2=4}
\frac {dX}{\sqrt{(X-e_1)(X-e_2)(X-e_3)}}
\\
&=\frac 12\omega_1(E)\qquad\text{(real half-period of the elliptic curve $E$)}
\\
&=\frac 1{\sqrt{e_1-e_3}}K\left(\sqrt{\frac{e_2-e_3}{e_1-e_3}}\right)
\\
&=\frac 1{\sqrt{5-(-4)}}K\left(\sqrt{\frac{4-(-4)}{5-(-4)}}\right)
\\
&=\frac 13 K\left(\frac{2\sqrt 2}3\right)\ .
\end{aligned}
$$

Let us numerically check again, pari/gp.

? intnum(x=2, oo, 1/sqrt( (x^2-1)*(x^2-4) ))
%10 = 0.842875177406298021435601828899538494750400447070248008383945
? ellK(sqrt(8/9))/3
%11 = 0.842875177406298021435601828899538494750400447070544522059974

Addendum. Computer algebra support for the elliptic curve computations.

Using sage:

sage: var('x,y,X,Y');
sage: def f(x,y): return y^2 - (x^2-1)*(x^2-4)
sage: factor( f(-4*(X-5)/Y, 2-4*X*(X-5)^2/Y^2) )
16*(X^3 - 5*X^2 - Y^2 - 16*X + 80)*(X - 5)^3/Y^4
sage: E = EllipticCurve(QQ, [0, -5, 0, -16, 80])
sage: E
Elliptic Curve defined by y^2 = x^3 - 5*x^2 - 16*x + 80 over Rational Field
sage: for T in E.torsion_points():
....:     if T != E(0):
....:         print(f"Order {T.order()} for the torsion point {T.xy()}")
....:
Order 2 for the torsion point (-4, 0)
Order 4 for the torsion point (2, -6)
Order 4 for the torsion point (2, 6)
Order 2 for the torsion point (4, 0)
Order 2 for the torsion point (5, 0)
Order 4 for the torsion point (8, -12)
Order 4 for the torsion point (8, 12)

This structural aspect explains why, when we compute the integrals of the canonical differential form $\frac{dX}Y$ of $E$ on the intervals delimited (in $\Bbb R$) by the points
$$
-\infty\ ,\
-4\ ,\
2\ ,\
4\ ,\
5\ ,\
8\ ,\
+\infty
$$
we obtain numerically the values for the integral of $(\ (X-4)(X+4)(X-5)\ )^{-1/2}$ on the delimited intervals... (I am taking the absolute value to not confuse sage.)

sage: f = 1/sqrt( abs((X-4)*(X+4)*(X-5)) )
sage: integral(f, X, -oo, -4, hold=True).n()
1.07825782381822
sage: integral(f, X, -4, 2, hold=True).n()
0.8428751689072135
sage: integral(f, X, 2, 4, hold=True).n()
0.8428751680336899
sage: integral(f, X, 4, 5, hold=True).n()
1.0782578072972744
sage: integral(f, X, 5, 8, hold=True).n()
0.8428751637656617
sage: integral(f, X, 8, oo, hold=True).n()
0.8428751775376374

And the above values are related to the periods of $(E)$:

sage: L = E.period_lattice()
sage: L
Period lattice associated to
    Elliptic Curve defined by y^2 = x^3 - 5*x^2 - 16*x + 80 over Rational Field
sage: L.gens()
(1.68575035481260, 1.07825782374982*I)
sage: L.gens()[0]/2
0.842875177406298


问:
Am I doing Vieta's Formulas correctly?
I've been given this exercise: $x^2 - (m+3)x + m + 2$, I'm supposed to check for which real values of the parameter $m$ can be used for this to work:

$1/x_1 + 1/x_2 > 1/2$ (I multiply both sides by 2x1x2) and get the following:

$2(x_1 + x_2) -x_1x_2 > 0$

which results to $m > -4$ using Vieta's formulas

Afterwards I have another argument, $x_1^2 + x_2^2 < 5$

After solving I get that $m$ is in the interval $(-4,0)$

My book tells me the final result for possible $M$ solutions is in the interval $(-2,0)$.

What am I doing wrong?

答:
In inequalities, it's usually a good idea to combine the fractions if you are not sure about their signs. For examples see here and here.

Now
$$\frac{1}{x_1} + \frac{1}{x_2} > \frac 12 \iff \frac{2x_2+2x_1-x_1 x_2}{2x_1 x_2} >0 \\ \iff \frac{2(m+3)-m-2}{2(m+2)} = \frac{m+4}{2(m+2)}>0 \iff (m+2)(m+4) > 0 \\\iff m \in (-\infty, -4)\cup (-2, \infty)$$
And
$$x_1^2+x_2^2 < 5 \iff (x_1+x_2)^2-2x_1 x_2 < 5 \\\iff (m+3)^2-2(m+2)-5 = m(m+4) < 0\\ \iff -4<m<0
$$

Therefore $-2<m<0$.

Title: Am I doing Vieta's Formulas correctly?


问:
For an uncountable subset $E$ in $[0,1]\times[0,1]$, show there is a condensation point of $E$.
I'm not sure how to approach this question:

Given $E\subseteq[0,1]\times[0,1]$ and $E$ is uncountable, show that there is some $p\in [0,1]\times[0,1]$ such that $\forall \epsilon>0, E\cap B(p,\epsilon)$ is uncountable.

I tried using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. I defined a set $A$ to be the collection of all open balls centered at rational points with some $\epsilon$ such that the unit square is completely covered. Surely at least one of the balls (centered at say $x$) contains uncountably many points in $E$. But, I don't know how to make this work (or even if it could work). I'm concerned about the overlap of the balls and how to find some $x$ that doesn't move around as $\epsilon$ changes. This is my introduction to analysis, so I don't have a good intuition for much of these concepts yet. Any help would be greatly appreciated!

答:
$[0,1]\times [0,1]$ can be covered with $4$ closed squares (from here on out all squares will be closed) with side length $1/2$ so surely there is a square $S_1$ with side length $1/2$ that contains uncountably many points in $E$. By the same argument, their is a square $S_2 \subset S_1$ with side length $1/4$ that contains uncountably many points in $E$. Proceed inductively to get a nested sequence of squares $S_1 \supset S_2 \supset S_3\ldots $ such that each $S_n$ has side length $2^{-n}$ and contains uncountably many points in $E$.

Letting $s_n$ be the center of $S_n$ we see that the sequence $\{s_n\}_n$ is eventually contained in each $S_k$ and thus converges to an $s$. For the same reason $s$ lies in all $S_n$ squares. Any $B(s; \epsilon)$ ball centered will of course contain an $S_k$ square; in particular contain uncountably many points in $E$.


问:
Taylor Polynomial for $x^{1/3}$

a. Compute the Taylor polynomial $T_3(x)$ for the function $(x)^{1/3}$ around the point $x=1$.

  b. Compute an error bound for the above approximation at $x = 1.3$.

I'm having trouble figuring out what to do for this problem.

I have to find the first three derivatives and after that I'm not quite sure what to do.

答:
Hints:

A Taylor polynomial centered at $a$ has the form:
$$T_n(x) = \sum_{k=0}^n\frac{f^{(k)}(a)(x-a)^k}{k!}$$

So, when you compute the first three derivatives, you plug them into that formula:
$$T_3(x) = f(a)+ f'(a)(x-a) +\frac{f''(a)(x-a)^2}{2} + \frac{f'''(a)(x-a)^3}{6}$$

In your case, $a=1$.

For the error bound, we use the Taylor series remainder term:
$$R_n = \frac{f^{(n+1)}(a)(x-a)^{(n+1)}}{(n+1)!}$$
To find the error at $x=1.3$, plug in that value.

ContentLicense: CC BY-SA 3.0

Title: Taylor Polynomial for $x^{1/3}$


问:
Hungarian Algorithm: Sequence of Choices
I'm thinking about utilizing the Hungarian algorithm for a research problem. I need to better understand what's happening behinds the scenes. I'm utilizing the matrix version.

The algorithm proceeds first with a "Step 0" in which row/column minimums are subtracted out. It then proceeds with Steps 1 and 2, which repeat until a solution is found. See details here.

Here are some observations about the Hungarian algorithm that I'm trying to confirm:

It seems that Step 0 (the row/column subtractions) pins down at least one match that ends up in the final set of assignments. Once that's in the set, the algorithm never reconsiders the assignments made in Step 0.

I think the same is true for each iteration of Steps 1 and 2: Specifically, each interaction of Step 1+2 pins down at least one additional match that eventually ends up in the final assignment. Also: Once those are made, they are never un-done in later stages.

Are these intuitions correct? It would greatly help understand what's going on behind the scenes.

答:
Observation 1 is true

Given a cost matrix $(a_{ij})$, the Hungarian algorithm computes potential vectors $(x_i), (y_j)$ such that $(b_{ij}) = (a_{ij} - x_i - y_j)$ is a nonnegative matrix containing a matching made of $0$s.  Consider what step 0 does to the row $i$ containing the element $a_{ij}$ of the final matching with the smallest $y_j$: for every element $a_{ij'}$ in the row, $a_{ij'} \ge x_i + y_{j'} \ge x_i + y_j = a_{ij}$, so $a_{ij}$ will immediately become $0$.

Observation 2 is false

Here’s a counterexample.  I’ve drawn boxes around elements of the final matching.  Observe that the number of boxed $0$s sometimes decreases: $3, 3, 3, 2, 3, 4, 5, 6$.

$$\begin{bmatrix}
\boxed{4} & 8 & 5 & \color{red}{0} & 8 & 8 \\
3 & 5 & 2 & \color{red}{0} & \boxed{4} & 5 \\
5 & 8 & 5 & \boxed{\color{red}{0}} & 8 & 8 \\
3 & \boxed{4} & 1 & \color{red}{0} & 5 & 5 \\
\color{red}{0} & \color{red}{5} & \boxed{\color{red}{0}} & \color{red}{0} & \color{red}{4} & \color{red}{4} \\
\color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{3} & 7 & 4 & \color{red}{0} & 7 & 7 \\
2 & 4 & 1 & \color{red}{0} & \boxed{3} & 4 \\
4 & 7 & 4 & \boxed{\color{red}{0}} & 7 & 7 \\
\color{red}{2} & \boxed{\color{red}{3}} & \color{red}{0} & \color{red}{0} & \color{red}{4} & \color{red}{4} \\
\color{red}{0} & \color{red}{5} & \boxed{\color{red}{0}} & \color{red}{1} & \color{red}{4} & \color{red}{4} \\
\color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{1} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{2} & 6 & \color{red}{3} & \color{red}{0} & 6 & 6 \\
1 & 3 & \color{red}{0} & \color{red}{0} & \boxed{2} & 3 \\
3 & 6 & \color{red}{3} & \boxed{\color{red}{0}} & 6 & 6 \\
2 & \boxed{3} & \color{red}{0} & \color{red}{1} & 4 & 4 \\
\color{red}{0} & \color{red}{5} & \boxed{\color{red}{0}} & \color{red}{2} & \color{red}{4} & \color{red}{4} \\
\color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{2} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{\color{red}{1}} & 5 & \color{red}{3} & \color{red}{0} & 5 & 5 \\
\color{red}{0} & 2 & \color{red}{0} & \color{red}{0} & \boxed{1} & 2 \\
\color{red}{2} & 5 & \color{red}{3} & \boxed{\color{red}{0}} & 5 & 5 \\
\color{red}{1} & \boxed{2} & \color{red}{0} & \color{red}{1} & 3 & 3 \\
\color{red}{0} & 5 & \boxed{\color{red}{1}} & \color{red}{3} & 4 & 4 \\
\color{red}{1} & \color{red}{0} & \color{red}{2} & \color{red}{3} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{1} & 4 & 3 & \color{red}{0} & 4 & 4 \\
\color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{0} & \boxed{\color{red}{0}} & \color{red}{1} \\
2 & 4 & 3 & \boxed{\color{red}{0}} & 4 & 4 \\
\color{red}{1} & \boxed{\color{red}{1}} & \color{red}{0} & \color{red}{1} & \color{red}{2} & \color{red}{2} \\
\color{red}{0} & \color{red}{4} & \boxed{\color{red}{1}} & \color{red}{3} & \color{red}{3} & \color{red}{3} \\
\color{red}{2} & \color{red}{0} & \color{red}{3} & \color{red}{4} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{\color{red}{0}} & 3 & 2 & \color{red}{0} & 3 & 3 \\
\color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{1} & \boxed{\color{red}{0}} & \color{red}{1} \\
\color{red}{1} & 3 & 2 & \boxed{\color{red}{0}} & 3 & 3 \\
\color{red}{1} & \boxed{\color{red}{1}} & \color{red}{0} & \color{red}{2} & \color{red}{2} & \color{red}{2} \\
\color{red}{0} & 4 & \boxed{1} & \color{red}{4} & 3 & 3 \\
\color{red}{2} & \color{red}{0} & \color{red}{3} & \color{red}{5} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{\color{red}{0}} & 2 & \color{red}{1} & \color{red}{0} & 2 & 2 \\
\color{red}{1} & \color{red}{1} & \color{red}{0} & \color{red}{2} & \boxed{\color{red}{0}} & \color{red}{1} \\
\color{red}{1} & 2 & \color{red}{1} & \boxed{\color{red}{0}} & 2 & 2 \\
\color{red}{2} & \boxed{1} & \color{red}{0} & \color{red}{3} & 2 & 2 \\
\color{red}{0} & 3 & \boxed{\color{red}{0}} & \color{red}{4} & 2 & 2 \\
\color{red}{3} & \color{red}{0} & \color{red}{3} & \color{red}{6} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$
$$\begin{bmatrix}
\boxed{\color{red}{0}} & \color{red}{1} & \color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{1} \\
\color{red}{2} & \color{red}{1} & \color{red}{1} & \color{red}{3} & \boxed{\color{red}{0}} & \color{red}{1} \\
\color{red}{1} & \color{red}{1} & \color{red}{1} & \boxed{\color{red}{0}} & \color{red}{1} & \color{red}{1} \\
\color{red}{2} & \boxed{\color{red}{0}} & \color{red}{0} & \color{red}{3} & \color{red}{1} & \color{red}{1} \\
\color{red}{0} & \color{red}{2} & \boxed{\color{red}{0}} & \color{red}{4} & \color{red}{1} & \color{red}{1} \\
\color{red}{4} & \color{red}{0} & \color{red}{4} & \color{red}{7} & \color{red}{0} & \boxed{\color{red}{0}} \\
\end{bmatrix}$$


问:
What's the specific definition of a direct sum of vector spaces?
I understand a couple people have posted about this before, but different sources are giving me different answers on what exactly a direct sum is with regards to vector spaces. I've seen both of these definitions for $V \oplus W$:

$$V \oplus W = \{ (v,w) \mid v \in V, w \in W \}$$

$$V \oplus W = \{ v + w \mid v\in V, w\in W, v+w \space \text{distinct} \}$$

These definitions are all given that addition and scalar multiplication done component-wise. In the second definition, different sources note that $v + w$ being distinct is only for the direct sum. Is $V \oplus W$ a $2$-tuplet, or a set of vectors like $V$ and $W$ are?

In addition, if $V$ has dimension $x$, and $W$ has dimension $y$, what is the dimension of $V \oplus W$, and why?

I'm actually pretty certain the definitions I gave above are imprecise and somewhat incorrect, but if someone could help me with this, that'd be great. Thanks.

答:
Your first definition is the correct definition (along with a suitable definition of vector addition and scaling). It works for any two vector spaces $V$ and $W$.

On the other hand, if $V$ and $W$ are both subspaces of the same, larger vector space $U$, then $v+w$ for $v\in V, w\in W$ makes sense to talk about. With this, we can construct $V+W$ as the subspace
$$
V+W=\{v+w\mid v\in V, w\in W\}\subseteq U
$$
Then one can ask: When is $V\oplus W$ isomorphic to (i.e. "essentially the same as") $V+W$? And the answer is when $V\cap W=\{0\}$. When this is true, $V+W$ is also called "the direct sum", because it's a bit pedantic to insist on maintaining the distinction (at least once they have been shown to be isomorphic), and it's often more hassle than its worth to keep using the ordered pairs when we already have the elements of $U$ to work with.

Direct sums add together dimensions. it's not difficult to see that if $\{v_1, v_2, \ldots, v_x\}$ is a basis for $V$ and $\{w_1,\ldots,w_y\}$ is a basis for $W$, then
$$
\{(v_1,0),(v_2,0),\ldots,(v_x,0),(0,w_1),(0,w_2),\ldots,(0,w_y)\}
$$
is a basis for $V\oplus W$.


答:
The first definition is of an external direct sum, whereas the second definition is of an internal direct sum. See here, for example. These definitions are isomorphic.

The external direct sum does result in tuples. The dimension in this case sum since the tuples are the result of the Cartesian product of the basis vectors.

External direct sums builds up new vector spaces. For example, the vector space of polynomials of the form $a_0+a_1x+a_2x^2$ has basis $V=\left\{1,x,x^2\right\}$ can be direct summed to the vector space of polynomials of the form $b_3x^3+b_4x^4+b_5x^5$  with basis $W=\left\{x^3,x^4,x^5 \right\}$ to yield the vector space $V \oplus W$ of polynomials of the form $c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5$ of dimension $\small \text{dim}(V) + \text{dim}(W)=6.$

The vector spaces added in the external direct sum $V$ and $W$ do not belong to the sum. For instance the associated vector space to $\mathbb R^2$ can be direct summed to $\mathbb R^3$ to yield $\mathbb R^5.$ The basis vectors of this external direct sum are represented in a block matrix (assuming the standard basis for each one) - the basis vectors of $\mathbb R^2$ are "padded" with three added zeros at the bottom, while the basis vectors of $\mathbb R^3$ need two added zeros at the top:

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

The dimension stays unchanged in the interior direct sum if $V$ and $W$ are subspaces without intersection other than the zero vector.

Example: In this case $V$ may be the subspace of $\mathbb R^3$ given by vectors of the form $\small\begin{bmatrix}a\\b\\0 \end{bmatrix}$ and $W$ be the subgroup of vectors of the form  $\small\begin{bmatrix}0\\0\\c \end{bmatrix}.$ The internal direct sum will span the ambient space, and be formed by vectors of $\small\begin{bmatrix}a\\b\\0 \end{bmatrix}  + \small\begin{bmatrix}0\\0\\c \end{bmatrix}=\small\begin{bmatrix}a\\b\\c \end{bmatrix}$ where $a,b,c\in \mathbb R.$

Title: What's the specific definition of a direct sum of vector spaces?

Body:
I understand a couple people have posted about this before, but different sources are giving me different answers on what exactly a direct sum is with regards to vector spaces. I've seen both of these definitions for $V \oplus W$:

$$V \oplus W = \{ (v,w) \mid v \in V, w \in W \}$$

$$V \oplus W = \{ v + w \mid v\in V, w\in W, v+w \space \text{distinct} \}$$

These definitions are all given that addition and scalar multiplication done component-wise. In the second definition, different sources note that $v + w$ being distinct is only for the direct sum. Is $V \oplus W$ a $2$-tuplet, or a set of vectors like $V$ and $W$ are?

In addition, if $V$ has dimension $x$, and $W$ has dimension $y$, what is the dimension of $V \oplus W$, and why?

I'm actually pretty certain the definitions I gave above are imprecise and somewhat incorrect, but if someone could help me with this, that'd be great. Thanks.


问:
Monoton function proof of discontinuity
Let  $f:[a,b] \to \mathbb {R}$ be nondecreasing.

(a) Recall from Thm. 2.1.9 (c) that  $f(x-)$ and $f(x+)$ exist at every $x\in [a,b]$,
and conclude that $f$ is discontinuous at $x\in S $ iff $f(x-) < f(x+)$.
Im assuming this true because the if the $f(x-)$ < the x+ then we would have a constant function not an increasing funtion. the second part is more obvious since f is nondecreasing then $f(x+) \geq f(x-)$ or it would violate the original statement. how would i try and lay this out with $ e$ and $ \delta$'s?

(b) Give an upper bound for the number of points $x$ at which the `jump' $(f(x+)- f(x-))$ is greater than
a given $r>0$.

one point?

(c) Prove that the set $S = \{ x\in [a,b]: f \text{ is discontinuous at }x \}$ is at most countable.

Think i got this one figured out, suppose that S is uncountable  then S must contain a Limit Point, but f is discontinuous at x so there exists a B(r,x) r>0 s.t x is isolated for all x in S if this wasn't the case then f(x) would be continuous thus S is countable.

答:
For (a) let $x_n$ be a non decreasing arbitrary sequence converging to $x$. Then $f(x_n)$ is a non decreasing bounded sequence, with Bolzano Weierstrass we also know that there is a convergent subsequence. Now we look at $a_n=x-\frac{1}{n}$, wlog we say that $f(a_n)$ converges (we could take a subsequence to do so). We gonna call this limit $f(x-)$ and obviously $f(x-)\leq f(x)$. We show now that for every non decreasing to $x$ converging sequence $x_n$  $f(x_n)$ converges to $f(x-)$.
As $x_n$ converges to $x$ we know that for all $m$ there a $N$ such that for all $n>N$ $|x_n-x| < |x-a_m|$. Hence
$$ f(a_m) \leq f(x_n) \leq f(x-).$$
With sandwhiching we get $\lim f(x_n)= f(x-)$.

For (b) we gonna denote $n$ as the number of possible jumps. As $f$ is not monotone decreasing we have that $n\cdot r \leq f(b)-f(a)$. Hence an upper bound will be
\[ n \leq \frac{f(b)-f(a)}{r}\]

As $f$  is non decreasing we surely now that $f(x)\leq f(b)$ for all $x$. A non decreasing function can only have jump discontinuities. Let $y_i=f(x_i+)-f(x_i-)$ for all $i \in I$. With $I$ I mean the indexing set of our discontinuites. The $y_i$ measures the jumps of the function.
Furthermore we know that
$$f(b) \geq f(a) + \sum_{i\in I} y_i. $$
As $f(b)$ is a real number it is finite and hence
$$\sum_{i \in I} y_i$$
must converge. This implies that $I$ must be countable, cause an uncountable sum of positive numbers must diverge (why?)

Title: Monoton function proof of discontinuity


问:
Intuitive explanation of Homotopy Equivalence vs Homeomorphism geometrically
I am learning these two notions and I was convinced that homotopy equivalence is a weaker condition(thus more spaces can be homotopy equivalent to one another) than homeomorphism in describing relationship between two topological spaces. Thus, homotopy equivalence is used to determine whether two spaces are NOT homeomorphism by finding two spaces are NOT homotopy equivalent.

Intuitively, on the other hand, when determining whether two spaces are homeomorphism or not geometrically, I noticed that people check to see if they can "transform" one space to the other. For example, the author of the book I'm reading claims that hemisphere(boundary not included) in $S^n$ is homeomorphic to $B_1(0)$ by projection. Can we make a similar claim for homotopy equivalence? Or, is this approach accurate for determining whether two spaces are homeomorphic or not?

答:
To verify both homeomorphism and homotopy equivalence you have a similar process: you must verify the existence of a function that satisfies certain properties. One common way of doing this is to construct the appropriate function (i.e. to write down a formula for that function), and then to verify the appropriate properties. So yes, verifying homotopy equivalence is similar to verifying homeomorphism. However, as you can see when you examine the definitions, it is similar but more complicated.

So to prove $X$ and $Y$ are homeomorphic, you need to construct a bijective function $f : X \to Y$ and verify that $f$ is continuous and $f^{-1} : Y \to X$ is continuous.

But to prove $X$ and $Y$ are homotopy equivalent is more complicated. You need to construct continuous function $f : X \to Y$, and $g : Y \to X$, and $F : X \times [0,1] \to X$, and $G : Y \times [0,1] \to Y$, and you must verify that $F(x,0)=g \circ f(x)$, $F(x,1)=x$, $G(y,0) = f \circ g(y)$, and $F(y,1)=y$.

The functions $F$ and $G$ are "homotopies", and they make rigorous the intuition of "deforming".


问:
How many solutions does the equation have? (Inclusion-exclusion principle)
How many solutions does the equation $x_1+x_2+x_3+x_4=1$ have? $x_i$ are integers between $-3$ and $3$.

Hints please!

答:
This answer will be easier to understand if you first read this.

First, you biject the problem to:

$x_1 + x_2 + x_3 + x_4 = 13$, where each $x_i \in \{0,1,2,3,4,5,6\}.$

Then, you use inclusion exclusion.

$T_k$ will represent the following computation:

each of the $\binom{4}{k}$ possible combinations of $k$ of the variables will be examined.

for each such combination, it will be assumed that the minimum value of the pertinent variables is $7$ rather than 0.  That is, the $k$ variables will all be presumed to be out of bounds.

the $4 - k$ remaining variables will be left unconstrained (i.e. minimum value = 0, they may or may not be out of bounds).

the number of solutions possible for each such combination will be computed by bijecting the existing equation to one that adjusts for the pertinent variables being forced to be out of bounds.

$T_k$ will represent the sum of the number of solutions possible for each of the $\binom{4}{k}$ combinations.

Consistent with Inclusion-Exclusion the answer will be

$T_0 - T_1 + T_2 - T_3 + \cdots - \cdots$

$T_0 = \binom{16}{3}.$

By symmetry

$T_1 = 4 \times$ the number of solutions where $x_1 \geq 7.$

Using the analysis in the cited article,

$T_1 = 4 \times \binom{9}{3}.$

At this point, the analysis stops, because it is impossible to have 2 or more constraints out of bounds and still have the sum be $\leq 13.$

Final answer:

$$\binom{16}{3} - [4 \times \binom{9}{3}].$$


问:
Probability of picking exactly 5 'E' Scrabble tiles when drawing 7 from a standard bag of 100?
Given a bag of 100 Scrabble tiles, 12 of which are E tiles, what is the probability of drawing 7 tiles where exactly 5 of them are Es and 2 of them are not Es?

Based on these three questions, I took a stab at the answer:

There are 12 Es to start, so for the event of drawing 5 Es, the number of outcomes for that would be: $12 \times 11 \times 10 \times 9 \times 8$

For the other 2 tiles, there are 88 non-E tiles to start, so the number of outcomes would be: $88 \times 87$

Together, the numerator becomes: $12 \times 11 \times 10 \times 9 \times 8 \times 88 \times 87$

For the denominator, the total number of outcomes is based on how many tiles there are for each draw. So: $100 \times 99 \times 98 \times 97 \times 96 \times 95 \times 94$

Which gives us this fraction:

$$\frac{12 \times 11 \times 10 \times 9 \times 8 \times 88 \times 87}{100 \times 99 \times 98 \times 97 \times 96 \times 95 \times 94} = \frac{727,626,240}{80,678,106,432,000} = 0.000009018880985 = 0.0009018880985\%$$

As X out of Y:

$\approx$ 9 out of 1,000,000 OR

$\approx$ 1 out of 110,879

Is that right?  And whether it is or not, how could I better articulate the answer, because I don't really know what I'm doing in this space, and it strikes me as sloppy and bad.

答:
So you need to choose 5 E tiles out of 12.

The number of ways of doing this is $\binom{12}{5}$.

You need to choose 2 non E tiles out of 88.

The number of ways of doing this is $\binom{88}{2}$.

So the number of ways to get the 7 tiles you need = $\binom{12}{5}  $ $\binom{88}{2}$

So the probability = $\frac{\binom{12}{5}\binom{88}{2}}{\binom{100}{7}} = \frac{2871}{15,158,675}$ which is approx $0.000189$

EDIT:

If you take the answer you posted and multiply by $\binom{7}{5}$ you get the answer above. Reason being... to get all the permutations, you need to first make a choice of which 5 positions out of the 7 get E's. For each of those choices, you get $12 \times 11 \times 10 \times 9 \times 8 \times 88 \times 87$ permutations.


问:
Functional Equation $ f ^ { f ( x ) } ( y ) = f ( x ) f ( y ) $

Find all functions $ f $ taking real numbers to positive integers, such that
$$ f ^ { f ( x ) } ( y ) = f ( x ) f ( y ) $$
holds true for all real numbers $ x $ and $ y $.

Here $ f ^ n ( x ) $ is the $ n $-th iteration of $ f $ applied to $ x $; i.e. $ f ^ 0 ( x ) = x $ and $ f ^ { n + 1 } ( x ) = f \big( f ^ n ( x ) \big) $.

I think I can use $ f ( 0 ) = x $ but that just gets me stuck, as you can plug in $ x = 0 $ and $ y = 0 $ to get $ x ^ 2 = f ^ x ( 0 ) $, which I don't know how to solve. Any ways someone recommends to start this off?

答:
It is clear that the constant function $f = 1$ satisfies the equation.

We show that there is no other possibility, by showing that the function $f$ cannot take any value greater than $1$.

Suppose that there exists $a$ such that $f(a) = k > 1$.

We put $x = a$ in the original equation: $$f^k(y) = kf(y)\tag{1}$$ for all $y$.

Put $y = a$ in (1): $f^{k - 1}(k) = k^2$.

Put $y = f^{k - 2}(k)$ in (1): $f^{2k - 2}(k) = kf^{k - 1}(k) = k^3$.

Put $y = f^{2k - 3}(k)$ in (1): $f^{3k - 3}(k) = kf^{2k - 2}(k) = k^4$.

Etc. This procedure continues, and we may show by induction that $$f^{m(k - 1)}(k) = k^{m + 1} \tag{2}$$ for all integer $m \geq 1$.

Now put $x = f^{k - 2}(k)$ in the original equation: $$f^{k^2}(y) = k^2 f(y)\tag{3}$$ for all $y$.

Put $y = a$ in (3): $f^{k^2 - 1}(k) = k^3$.

However, putting $m = k + 1$ in (2) gives $f^{k^2 - 1}(k) = k^{k + 2} > k^3$. This is a contradiction.


问:
Subsets $A$ of $\Bbb{Z}$ such that $xy \in A \implies x \in A, $ or $y \in A$. Primality of arb. subsets.
If

$$
A \subset \Bbb{Z}
$$

is such that $xy \in A \implies x \in A, $ or $y \in A$.  Then $A$ is either a prime ideal or ?

Can we describe all "prime subsets" of $\Bbb{Z}$ that aren't prime ideals in one fell swoop?

It's not just a subset closed under taking divisors, though those are counted, and so it's also not just a subset of negative prime numbers together with $\{0, -1\}$ either.

答:
Notice that the implication
$$
xy \in A \implies \big( x \in A \text{ or } y \in A \big)
$$
is equivalent to its contrapositive
$$
\big( x \notin A \text{ and } y \notin A \big) \implies xy \notin A.
$$
In other words, $A$ has the given property if and only if $\Bbb Z\setminus A$ is closed under multiplication.

In particular, there is a wide variety of examples of such sets $A$, such as:

$A = \{-m,\dots,m\}$ for any $m\in\Bbb N$;

$A = \{n\in\Bbb Z\colon n$ cannot be written in the form $n = 13^a2020^b\}$;

$A = \{n\in\Bbb Z\colon n$ is not a perfect square$\}$;

$A = \{n\in\Bbb Z\colon \text{there exists } p\mid n \text{ such that } p^p\nmid n\}$....


问:
Infinite shortest paths in graphs
From Wikipedia:

"If there is no path connecting two
  vertices, i.e., if they belong to
  different connected components, then
  conventionally the distance is defined
  as infinite."

This seems to negate the possibility that there are graphs with vertices connected by an infinite shortest path (as opposed to being not connected).

Why is it that for every (even infinite) path between two vertices there is a finite one?

Note that infinite paths between vertices do exist - e.g. in the infinite complete graph -, but they are not the shortest.

答:
Traditional graph theory focuses on finite graphs.  Two vertices are considered connected iff there is a finite walk between them (basically a sequence of vertices, each one adjacent to the last).

If we start considering infinite graphs, we have to consider some interesting consequences.  Suppose we want to consider vertices x,y as connected by an infinite walk.  Then we can order this walk linearly, and get a sequence $x=x_1,x_2,x_3,\ldots,x_\infty=y$.  If this is an infinite sequence, it must contain some limit point -- some point $x_k$ must have either no immediate predecessor or no immediate successor.  This idea doesn't match the old definition of a walk between vertices.

Although I don't know of any references, I do think this idea is interesting as a way to extend connected concepts to infinite graphs.

For example, consider the tree of surreal numbers.  Think of the surreals as sign sequences; each is a map from an ordinal to $\{-,+\}$.  Say that $s$ and $t$ are adjacent if the domain of $s$ is one more or one less than the domain of $t$.  So 0 is adjacent to -1 and 1, 1 is adjacent to 0, 2, and 1/2, etc.

This looks like a standard tree until we get surreals with infinite domains.  Consider some $s$ with domain $\omega$, and the sequence $s_i$ where $s_i$ has domain $i = \{j:j<i\}$, and $s_i(j)=s(j)$.  It feels like the $s_i$ walk should connect 0 with $s$.  A new definition to capture this idea would be interesting.

Title: Infinite shortest paths in graphs


Title: Taylor Polynomial for $x^{1/3}$

Body:
a. Compute the Taylor polynomial $T_3(x)$ for the function $(x)^{1/3}$ around the point $x=1$.

  b. Compute an error bound for the above approximation at $x = 1.3$.

I'm having trouble figuring out what to do for this problem.

I have to find the first three derivatives and after that I'm not quite sure what to do.

Body:
$$(x^{\frac13})' = \frac13 x^{-\frac23}$$
$$(x^{\frac13})'' = -\frac29 x^{-\frac53}$$
$$(x^{\frac13})''' = \frac{10}{27} x^{-\frac83} $$

Taylor expansion:
$$
x^{\frac13} = (1)^{\frac13} + \frac13 1^{-\frac23}\frac{x-1}{1!} -\frac29 1^{-\frac53} \frac{(x-1)^2}{2!} + \frac{10}{27} 1^{-\frac83}  \frac{(x-1)^3}{3!} + \dots
$$
You have
$$
T_3(x) = 1 + \frac13 (x-1) - \frac29 \frac{(x-1)^2}{2!} + \frac{10}{27} \frac{(x-1)^3}{3!}
$$
Plugging in $x=1.3$ you get
$$
T_3(1.3) = 1.1117
$$
Your error is
$$
T_3(1.3) - (1.3)^{\frac13} = 1.1117 - 1.0914 = 0.0203
$$


问:
How to show $\frac{\sigma(m!)}{m!}>ln(m)$?
I am trying to show that $\frac{\sigma(m!)}{m!}>ln(m)$, where $m$ is a positive integer, and I can't seem to get anywhere in my proof. Where $\sigma(n)$ is the sum of the divisors of $n$.

I've tried to write $\sigma(m!)$ after taking an arbitrary prime power decomposition for m! and can't seem to get anywhere.

I'm personally thinking this has something to do with Merten's function since: $log(x)=\sum_{n\leq x} \frac{\Lambda(n)}{n}-O(1)$ with $\Lambda(n)=log(p)$ if $n$ is a prime power and 0 otherwise. I've tried expanding this function and playing around with it but still can't make headway relating this back to $\sigma$.

Any suggestions or hints would be greatly appreciated!

答: Hint: Note that the numbers $m!, \dfrac{m!}{2}, \dfrac{m!}{3},\ldots,\dfrac{m!}{m-1}, \dfrac{m!}{m}$ are all distinct divisiors of $m!$. So $$\sigma(m!) \ge m!+\dfrac{m!}{2}+ \dfrac{m!}{3}+\ldots+\dfrac{m!}{m-1}+\dfrac{m!}{m}.$$ See if you can use that along with elementary bounds on the harmonic numbers to solve the problem.


问:
Cross simplification property: does ax=ya $\Rightarrow$ x=y implies conmutativity?
Let $R$ be a (not necessarily unitary), we know that if R doesn't have zero divisors then it satisfies the cancellation laws: $\forall a,x,y \in R$ with $a \neq 0$ $ax=ay$ implies $x=y$ and similarly,  $xa=ya$ implies $x=y$ but $xa=ay$ does not imply $x=y$

It is clear that this implication is true if $R$ is conmutative, so then I thought if the converse is true, but I could't either prove it nor disprove it.

So the exact statement is:
is a ring $R$ without zero divisors and for which $ax=ya$ hapens only when $x=y$ or $a=0$ necessarily commutative?
Can you prove this or give a couterexample ring?

答: Suppose $x,a \not= 0$ then $a(xa) = (ax)a \implies xa = ax$ then is a commutative ring


问:
Derived series of Lie algebra in reverse
Consider the first element in the derived series of a Lie algebra $L$,
defined by $L^{(1)}:=[L,L]$. For a given Lie algebra $\tilde{L}$, is
there always a Lie algebra $L$ such that $L^{(1)}=\tilde{L}$? If yes, how do I
construct $L$?

答:
Suppose that $\mathbb{k}$ is a field of characteristic zero. It follows from Lie’s theorem and Engel’s theorem that a finite-dimensional $\mathbb{k}$-Lie algebra $\mathfrak{g}$ is solvable if and only if its derived Lie algebra $[\mathfrak{g}, \mathfrak{g}]$ is nilpotent.

Every finite-dimensional, solvable, non-nilpotent $\mathbb{k}$-Lie algebra is therefore not the derived Lie algebra of another $\mathbb{k}$-Lie algebra. Examples for this are $\mathfrak{t}(n, \mathbb{k})$ with $n \geq 2$.


问:
Given that $g$ is only continuous at $0$, not on $[0,1]$, show $\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.
Important Notice: Observe that although this question is very similar to many other questions such as the one in proving $\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$ when f is continous on [0,1], this problem only assumes continuity at $0$, but not on $[0,1]$.

Here is the question:
Assume $g$ is (Riemann) integrable on $[0, 1]$ and continuous at $0$. Show
$\lim_{n \to \infty} \int_0^1 g(x^n) dx = g(0)$.

My attempt at this question: I split up the integral into two parts from $[0,1-\alpha]$ and $[1-\alpha,1]$. For $[0,1-\alpha]$, there is uniform continuity, so the limit $n \to \infty$ can be shifted inside the integral to get $g(0)$.

But how do I keep the integral over $[1-\alpha,1]$ to be small?

Note: This is Exercise $7.4.10$ in Abbott, Understanding Analysis, 2nd edition.

答:
The proof is straightforward if we assume Lebesgue integration
theory: Since $g$ is Riemann integrable, it is bounded. Choose $M>0$
such that $|g(x)|\leq M$ for all $x\in[0,1]$. For each $x\in[0,1)$,
$g(x^{n})\rightarrow g(0)$ as $n\rightarrow\infty$ because $g$
is continuous at $0$. Moreover, $|g(x^{n})|\leq M$. By Dominated
Convergence Theorem, we have
\begin{eqnarray*}
 &  & \lim_{n\rightarrow \infty}\int_{0}^{1}g(x^{n})dx\\
 & = & \int_{0}^{1}\lim_{n\rightarrow\infty}g(x^{n})dx\\
 & = & \int_{0}^{1}g(0)dx\\
 & = & g(0).
\end{eqnarray*}

Alternative approach that does not invoke Lebesgue integration theory:
Since $g$ is Riemann integrable, it is bounded. Choose $M>0$ such
that $|g(x)|\leq M$ for all $x\in[0,1].$ Let $\varepsilon>0$ be
given. Choose $\delta>0$ such that $|g(x)-g(0)|<\varepsilon$ whenever
$x\in[0,\delta]$. Choose $a\in(0,1)$ such that $a>1-\varepsilon$
. Observe that $a^{n}\rightarrow0$, so there exists $N$ such that $a^{n}\in[0,\delta]$
whenever $n\geq N$. Note that for any $x\in[0,a]$, we have that
$0\leq x^{n}\leq a^{n}$, so $x^{n}\in[0,\delta]$ whenever $n\geq N$
and $x\in[0,a]$. Consider
\begin{eqnarray*}
 &  & \left|\int_{0}^{1}g(x^{n})dx-g(0)\right|\\
 & = & \left|\int_{0}^{1}g(x^{n})dx-\int_{0}^{1}g(0)dx\right|\\
 & \leq & \int_{0}^{1}\left|g(x^{n})-g(0)\right|dx\\
 & = & \int_{0}^{a}\left|g(x^{n})-g(0)\right|dx+\int_{a}^{1}\left|g(x^{n})-g(0)\right|dx.
\end{eqnarray*}
If $n\geq N$, and $x\in[0,a]$, we have $x^{n}\in[0,\delta]$, so
$|g(x^{n})-g(0)|<\varepsilon.$ It follows that $\int_{0}^{a}\left|g(x^{n})-g(0)\right|dx\leq\int_{0}^{a}\varepsilon dx\leq\varepsilon$.
On the other hand,
\begin{eqnarray*}
 &  & \int_{a}^{1}\left|g(x^{n})-g(0)\right|dx\\
 & \leq & \int_{a}^{1}2Mdx\\
 & = & 2M(1-a)\\
 & \leq & 2\varepsilon M.
\end{eqnarray*}
That is, $\left|\int_{0}^{1}g(x^{n})dx-g(0)\right|\leq\varepsilon(2M+1)$
whenever $n\geq N$. This shows that $\int_{0}^{1}g(x^{n})dx\rightarrow g(0)$.


问:
Oscillatory integral with absolute value
Suppose $f:[0,1]\to[0,+\infty)$ is a continuous function. Are there general conditions under which the following limit exists?

$$\lim_{n\rightarrow \infty} \int_0^1 |\cos(nf(x))|dx$$

Obviously if $f$ is a constant that is not an integer multiple of $\pi$ the limit does not exist. Does the limit exist if $f'(x)\neq 0$ for all $x \in [0,1]$? If so, what if we drop differentiability and only require $f$  not be locally constant?

When $f$ is not differentiable, can one still show
$$\liminf_{n \rightarrow \infty} \int_0^1 |\cos(nf(x))|dx >0$$
under suitable conditions on $f$?

答:
We can generalize and perhaps gain some insight. I'll start with a generalization of the Riemann Lebesgue lemma (GRLL):

GRLL: Suppose $g$ is continuous and periodic on $\mathbb R,$ with period $p>0.$ Let $f$ be continuous on $[a,b].$ Then

$$\lim_{n\to \infty} \int_a^b g(nx)f(x)\,dx = \frac{\int_0^p g(x)\,dx }{p}\cdot\int_a^b f(x)\,dx.$$

This is not too hard to prove; I'll omit the proof for now.

Corollary: Assume in addition that $f\in C^1[a,b]$ and $f'$ is nonzero in $[a,b].$ Then

$$\tag 1 \lim_{n\to \infty}\int_a^b g(nf(x))\,dx = \frac{(b-a)}{p}\int_0^p g.$$

Proof: On the left side of $(1),$ let $x=f^{-1}(y).$ We get

$$\tag 2 \int_a^b g(nf(x))\,dx = \int_{f^{-1}(a)}^{f^{-1}(b)} g(ny)\frac{1}{f'(f^{-1}(y))}\,dy.$$

Apply GRLL to the right side of $(2)$ to get

$$\tag 3\frac{\int_0^p g(x)\,dx }{p}\int_{f^{-1}(a)}^{f^{-1}(b)}\frac{1}{f'(f^{-1}(y))}\,dy.$$

Now let $y=f^{-1}(z)$ to see $(3)$ equals the right side of $(1).$ The proof of the Corollary is complete.

Let's compare this result with Alex Ravsky's, where we have $g(y)=|\cos y|$ and $a=0,b=1.$ Here $g$ has period $\pi.$ The Corollary gives

$$\frac{1-0}{\pi}\int_0^\pi |\cos x|\,dx =  \frac{1}{\pi}\cdot 2 = \frac{2}{\pi}$$

for the answer, which agrees with Ravsky.

I believe we can extend these results further: Again suppose $f\in C^1$ but this time we assume only $m(\{x\in [a,b]: f'(x)=0\})=0.$ Then we get the same answer:

$$ \lim_{n\to \infty}\int_a^b g(nf(x))\,dx = \frac{(b-a)}{p}\int_0^p g.$$

I'll leave this unproven at the moment as time is slipping by.


问:
A zero sum matrix game is given: $A=\begin{bmatrix}13&29&8\\18&22&31\\23&22&19\\ \end{bmatrix}$ Reduce that game to a 2 x 2 game using dominance.
I know that a 2 x 2 matrix is as follows:

$A=\begin{bmatrix}18&31\\23&19\\ \end{bmatrix}$

But it is not possible to reduce the original matrix by row or by columns, how is this result achieved?

Thanks in advance for any help.

答:
Try this:

Suppose that in a mixed strategy of the column player he chooses Col 1 with a probability of $p$ and Col 3 with a probability of $1-p$. You can find that there exists a $p$ such that this mixed strategy dominates the pure strategy of Col 2.


问:
A special subset of uniformly distributed numbers is still uniformly distributed?
Assume that I have a value range [1,1000].

I uniformly choose 10 numbers from [1,1000].

Assume that the chosen numbers are a1, a2, ..., a10.

Besides, assume that they are ordered so that a1< a2< ...< a10.

Here comes my question.

If I always choose 10 numbers, order these 10 numbers, and always pick the first three numbers (e.g., a1, a2, and a3) from the ordered result, then can I still claim that these three numbers are uniformly chosen from [1,1000]?

On the other hand, if I always choose 10 numbers, order these 10 numbers, and always pick those at even-positions (e.g., a2, a4, a6, a8, and a10) from the ordered result, then can I still claim that these five numbers are uniformly chosen from [1,1000]?

答:
Unfortunately you can't.

If you have $k$ uniformly distributed random variables in $[1,N]$ then the expected value of $a_1 + a_k$  will always be $N$.

To see this, for every $i$ set
$$b_i = N+1 - a_{k+1-i}.$$
So $b_1 = N+1 - a_k$, $b_2 = N +1 - a_{k-1}$ up to $b-k = N+1-a_1$.

Now if I choose a uniformly distributed random variable $a\in[1,N]$ then $b=N+1-a$ is also uniformly distributed in $[1,N]$ so my $b_i$ are distributed exactly the same as by $a_i$.

So $E(b_i) = E(a_i)$ for every $i$ and in particular $E(a_1 + a_k) = E(b_1 + a_k) = N+1$.

In both your guesses you remove the lowest number, but not the highest so the expected value of the lowest remaining number plus the highest remaining number must be strictly greater than $1000$ and the remaining numbers cannot be independent uniforms.

The only way of removing some of your $a_i$ such that the remaining values are iid uniform is to remove each of them independently with a fixed probability $p$.


问:
Algorithmic simplification of expressions with $\arctan$, e.g. $2\arctan\frac{5-\sqrt2}{5+\sqrt2}=3\arctan\frac{1}{2\sqrt2}$.
This is a problem I encountered in another (unrelated) question.

I got an expression which is a bit ugly, $2\arctan\frac{5-\sqrt2}{5+\sqrt2}$, but when changing the method I got another prettier expression, $3\arctan\frac{1}{2\sqrt2}$.

Since they are both equal to the same integral, they must be equal as well (and we can easily prove that by doing some simple calculation). But what makes it interesting is, a lot of common mathematics software cannot do this kind of simplification (I tried SageMath and WolframAlpha, they both did nothing).

So let's do it by hand to see what happened.

\begin{align}
2\arctan\frac{5-\sqrt2}{5+\sqrt2} &= \arctan\frac{23}{10\sqrt2}\tag{automatic}\\
&= \arg\left(10\sqrt2+23i\right)\tag{automatic}\\
&= \arg\left(2\sqrt2+i\right)^3\tag{miracle}\\
&= 3\arctan\frac{1}{2\sqrt2}.\tag{automatic}
\end{align}

(In the general case, we should also count how many $\pi$ should be added at the end.)

As you can see, the key step here is the factorization $10\sqrt2+23i=\left(2\sqrt2+i\right)^3$. This is a factorization in $\mathbb Z[\sqrt2,i]$, which might be difficult even for computers (I'm not sure).

What's more, normally there's not just $\sqrt2$ that appears, e.g.
$$\arctan\frac{\sqrt5-77\sqrt2}{7+11\sqrt{10}}=3\arctan\sqrt5+5\arctan\sqrt2-3\pi.$$

So here is my question:

Is there an algorithmic way to do this kind of simplification?

Let me explain a little where this problem comes from.

I got an integral, for $a>0$:

$$\int_0^\infty e^{-at}\left(\operatorname{erf}\sqrt t\right)^3\,dt=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac{1-b}{1+b},$$

where $b=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}}$.

I wanted to simplify this expression but failed. The example mentioned before is just the case $a=1$.

I'd be glad if anyone can simplify this expression for any $a>0$, and I'll take it as an acceptable answer since I think this expression is already complicated enough.

I added my answer to the integral problem only.

$$\int_0^\infty e^{-at}\left(\operatorname{erf}\sqrt t\right)^3\,dt=\frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).$$

And it didn't give us a general method to deal with all cases.

答:
As you noted, this question is equivalent to factoring numbers in the ring $\mathbb Z[\sqrt2,i]$. I don't know whether or not unique factorization holds here, even though it does in $\mathbb Z[\sqrt 2]$ and $\mathbb Z[i]$. If it does  not we may have many ways of writing a number as a product of irreducible elements, and some might be more interesting than others. Another complication is that this ring has infinitely many units which are related to the solutions of the Pell equation $x^2-2y^2=\pm 1$. It is possible to do this, but it's violent and terrible. As an appetizer have the identity:

$$2\sqrt 2+i=-i(1+i\sqrt 2)^2$$

$$\arctan\left(\frac 1{2\sqrt 2}\right)=2\arctan(\sqrt 2)-\frac\pi 2$$

Consider a general number in our ring

$$x_1=a+b\sqrt 2+ci+di\sqrt 2$$

Its algebraic conjugates are:

$$x_2=a-b\sqrt 2+ci-di\sqrt 2$$

$$x_3=a+b\sqrt 2-ci-di\sqrt 2$$

$$x_4=a-b\sqrt 2-ci+di\sqrt 2$$

And define the norm

$$N(x_1)=x_1x_2x_3x_4=a^4 - 4a^2b^2 + 2a^2c^2 + 4a^2d^2 - abcd + 4b^4 + 4b^2c^2 + 8b^2d^2 + c^4 - 4c^2d^2 + 4d^4$$

By grouping the quartet of conjugates two by two and applying the definition we also get

\begin{align*}
N(x_1)=(x_1x_3)(x_2x_4)&=(a^2+2b^2+c^2+2d^2+(2ab+2cd)\sqrt 2)(a^2+2b^2+c^2+2d^2-(2ab+2cd)\sqrt 2)\\
&=(a^2+2b^2+c^2+2d^2)^2-2(2ab+2cd)^2
\end{align*}

And also

\begin{align*}
N(x_1)=(x_1x_2)(x_3x_4)&=(a^2-2b^2-c^2+2d^2+(2ac-4bd)i)(a^2-2b^2-c^2+2d^2-(2ac-4bd)i) \\
&=(a^2-2b^2-c^2+2d^2)^2+(2ac-4bd)^2
\end{align*}

So we immediately get that the norm of any element is the sum of two squares (at least one even) and also a square minus twice a square. We start with

$$N(5+\sqrt 2+5i-i\sqrt 2)=2916=2^2 3^6$$

By putting the leading $2$ into the arctangent you effectively square this number, which makes things harder even though there's some cancelation

$$N(10\sqrt 2+23i)=531441=3^{12}$$

Since any number must have a norm which is the sum of two squares at least one of which must be even, it's easy to see that none has norm 2 or 3. Here's a search for elements with norm 9 with c++

```<code>const constexpr double sqrt2 = std::sqrt(2);
const constexpr std::complex<double> half = (0.5,0.0);
for(int a = 0; a != 201; ++a) {
  for(int b = 0; b != 201; ++b) {
    for(int c = 0; c != 201; ++c) {
      for(int d = 0; d != 201; ++d) {
        std::complex<double> x1 (a+b*sqrt(2), c+d*sqrt(2));
        std::complex<double> x2 (a-b*sqrt(2), c-d*sqrt(2));
        std::complex<double> x3 (a+b*sqrt(2),-c-d*sqrt(2));
        std::complex<double> x4 (a-b*sqrt(2),-c+d*sqrt(2));
        auto prod = x1 * x2 * x3 * x4 + half;
        int n = prod.real();
        if(n == 9) std::cout << a << " " << b << " " << c << " " << d << "\n";
      }
    }
  }
}
</code>```

I ran this both 4 and 9 to find the following. This is the output for 4. I chose to try $1+i$ first because it's the simplest looking one, and it was a success

$$5+\sqrt 2+5i-i\sqrt 2=(1+i)(5-i\sqrt 2)$$

When trying the next step the first try fails because $5-i\sqrt 2$ is not a multiple of $\sqrt 2+i$, but the next one succeeds

$$5+\sqrt 2+5i-i\sqrt 2=(1+i)(1+i\sqrt 2)(1-2i\sqrt 2)$$

It works thrice in fact

$$5+\sqrt 2+5i-i\sqrt 2=-(1+i)(1+i\sqrt 2)^3$$

Which corresponds to

$$\arctan\left(\frac{5-\sqrt 2}{5+\sqrt 2}\right)=3\arctan(\sqrt 2)-\frac {3\pi}4$$

Body:
As you noted, this question is equivalent to factoring numbers in the ring $\mathbb Z[\sqrt2,i]$. I don't know whether or not unique factorization holds here, even though it does in $\mathbb Z[\sqrt 2]$ and $\mathbb Z[i]$. If it does  not we may have many ways of writing a number as a product of irreducible elements, and some might be more interesting than others. Another complication is that this ring has infinitely many units which are related to the solutions of the Pell equation $x^2-2y^2=\pm 1$. It is possible to do this, but it's violent and terrible. As an appetizer have the identity:

$$2\sqrt 2+i=-i(1+i\sqrt 2)^2$$

$$\arctan\left(\frac 1{2\sqrt 2}\right)=2\arctan(\sqrt 2)-\frac\pi 2$$

Consider a general number in our ring

$$x_1=a+b\sqrt 2+ci+di\sqrt 2$$

Its algebraic conjugates are:

$$x_2=a-b\sqrt 2+ci-di\sqrt 2$$

$$x_3=a+b\sqrt 2-ci-di\sqrt 2$$

$$x_4=a-b\sqrt 2-ci+di\sqrt 2$$

And define the norm

$$N(x_1)=x_1x_2x_3x_4=a^4 - 4a^2b^2 + 2a^2c^2 + 4a^2d^2 - abcd + 4b^4 + 4b^2c^2 + 8b^2d^2 + c^4 - 4c^2d^2 + 4d^4$$

By grouping the quartet of conjugates two by two and applying the definition we also get

\begin{align*}
N(x_1)=(x_1x_3)(x_2x_4)&=(a^2+2b^2+c^2+2d^2+(2ab+2cd)\sqrt 2)(a^2+2b^2+c^2+2d^2-(2ab+2cd)\sqrt 2)\\
&=(a^2+2b^2+c^2+2d^2)^2-2(2ab+2cd)^2
\end{align*}

And also

\begin{align*}
N(x_1)=(x_1x_2)(x_3x_4)&=(a^2-2b^2-c^2+2d^2+(2ac-4bd)i)(a^2-2b^2-c^2+2d^2-(2ac-4bd)i) \\
&=(a^2-2b^2-c^2+2d^2)^2+(2ac-4bd)^2
\end{align*}

So we immediately get that the norm of any element is the sum of two squares (at least one even) and also a square minus twice a square. We start with

$$N(5+\sqrt 2+5i-i\sqrt 2)=2916=2^2 3^6$$

By putting the leading $2$ into the arctangent you effectively square this number, which makes things harder even though there's some cancelation

$$N(10\sqrt 2+23i)=531441=3^{12}$$

Since any number must have a norm which is the sum of two squares at least one of which must be even, it's easy to see that none has norm 2 or 3. Here's a search for elements with norm 9 with c++

const constexpr double sqrt2 = std::sqrt(2);
const constexpr std::complex<double> half = (0.5,0.0);
for(int a = 0; a != 201; ++a) {
  for(int b = 0; b != 201; ++b) {
    for(int c = 0; c != 201; ++c) {
      for(int d = 0; d != 201; ++d) {
        std::complex<double> x1 (a+b*sqrt(2), c+d*sqrt(2));
        std::complex<double> x2 (a-b*sqrt(2), c-d*sqrt(2));
        std::complex<double> x3 (a+b*sqrt(2),-c-d*sqrt(2));
        std::complex<double> x4 (a-b*sqrt(2),-c+d*sqrt(2));
        auto prod = x1 * x2 * x3 * x4 + half;
        int n = prod.real();
        if(n == 9) std::cout << a << " " << b << " " << c << " " << d << "\n";
      }
    }
  }
}

I ran this both 4 and 9 to find the following. This is the output for 4. I chose to try $1+i$ first because it's the simplest looking one, and it was a success

$$5+\sqrt 2+5i-i\sqrt 2=(1+i)(5-i\sqrt 2)$$

When trying the next step the first try fails because $5-i\sqrt 2$ is not a multiple of $\sqrt 2+i$, but the next one succeeds

$$5+\sqrt 2+5i-i\sqrt 2=(1+i)(1+i\sqrt 2)(1-2i\sqrt 2)$$

It works thrice in fact

$$5+\sqrt 2+5i-i\sqrt 2=-(1+i)(1+i\sqrt 2)^3$$

Which corresponds to

$$\arctan\left(\frac{5-\sqrt 2}{5+\sqrt 2}\right)=3\arctan(\sqrt 2)-\frac {3\pi}4$$


问:
Understanding the position operator in Reed and Simon Mathematical physics vol 1.
I am studying the subject of unbounded operators and I cannot understand some statements that appear in the book by Reed, Simon

Question 1. The statement "support near plus or minus infinity" means that $\text{supp}(\varphi)=\left\{x\in\mathbb{R}:\varphi(x)\neq 0\text{ and } x\geq R \text{ some } R>0 \text{ sufficiently large } \right\}$?

Question 2. Why we can make $\left\|T\varphi\right\|$ as large as we like while keeping $\left\|\varphi\right\|=1$?

答: I think "support near $\pm \infty$" is a poor choice is words. I would instead just take $\phi$ to be the characteristic function of the interval $[n,n+1]$ for $n$ large enough, and a simple computation would give $\|T\phi\|>n$, hence pretty large.


问:
$\int_{[1,- i]} \frac{1}{z}$ where $[1,- i]$ is the straight line joining the first point to the second point

$\int_{[1, -i]} \frac{1}{z}$ where $[1, -i]$ is the straight line
joining the first point to the second point

I am trying to evaluate this using the fundamental theorem of calculus for path integrals

If I use the branch of log $(-\pi, \pi)$ I get the answer as $(\frac{-\pi}2)i$

If I use the branch of log $(\frac{\pi}2, \frac{5\pi}{2})$, I get the answer as $\frac{3\pi}2$

For the first case, I have $\log(z) = \ln|z| + i\theta$ where $\theta = \text{arg} z \in (-\pi, \pi)$

$$\int_{[1,-i]} \frac{dz}{z} = \log(-i) - \log (1) \\= \ln|-i| + i\theta_{-i} - \ln|1| - i\theta_1 = 0 + (\frac{-\pi}2)i - 0 - 0i \\= (\frac{-\pi}2)i$$

For the second case, I have $\log(z) = \ln|z| + i\theta$ where $\theta = \text{arg} z \in (\pi/2, 5\pi/2)$

$$\int_{[1,-i]} \frac{dz}{z} = \log(-i) - \log (1) \\= \ln|-i| + i\theta_{-i} - \ln|1| - i\theta_1 \\= 0 + \frac{3\pi}2i - 0 - 0i = \frac{3\pi}2i$$
Are both these answers correct?

答: The point $i = \exp(\frac{\pi}{2} i) = \cos(\pi/2) + i \sin(\pi/2)$ is on the branch cut for your second computation, so that computation is not legal.


问:
Show that $p(x)=T_{n,x_0}(x)$
Let $f\in C^n(I)$ and $x_0\in I$. Suppose that, for some polynomial $p(x)$ of degree $n$, the inequality $|f(x)-p(x)|\leq c|x-x_0|^{n+1}$ holds for all $x\in\ I$ and some constant $c$. Show that $p(x)=T_{n,x_0}(x)$, i.e. the Taylor polynomial of $f$ of degree $n$ centered at $x_0$.

I'm having a really hard time trying to solve the previous  exercise. The hypothesis reminds me of Lipschitz continuity, however; I don't know how to use the hypothesis. How should I proceed?

答:
The trick here is to use characterizations of the remainder in Taylor's theorem. The most basic version of Taylor's theorem applied to $f$ on the interval $I$ with expansion centered at $x_0$ gives $f(x) = T_{n,x_0}(x)+h_n(x)(x-x_0)^n$ where $\lim_{x\to x_0} h_n(x)=0$. Thus
$$|f(x) - p(x)| = |(T_{n,x_0}(x)-p(x))+h_n(x)(x-x_0)^n|\leq c|x-x_0|^{n+1}.$$
Of course, this implies
\begin{align*}|T_{n,x_0}(x)-p(x)|&\leq |(T_{n,x_0}(x)-p(x))+h_n(x)(x-x_0)^n|+|h_n(x)(x-x_0)^n|\\
&\leq c|x-x_0|^{n+1}+|h_n(x)||x-x_0|^n,\end{align*}
so $g(x):=T_{n,x_0}(x)-p(x)$ is a polynomial of degree $n$ that is $o(|x-x_0|^n)$ as $x\to x_0$. Such a polynomial must be identically $0$. Indeed, let $k\leq n$ be the smallest nonnegative integer such that $a_k\neq 0$ in the expansion $g(x) = \sum_{j=0}^n a_j(x-x_0)^j$. Then $g(x)/(x-x_0)^k$ is a polynomial with constant term $a_k$, and we should have
$$|a_k| = \lim_{x\to x_0}|g(x)|/|x-x_0|^k.$$
On the other hand, the limit on the right must be zero as $g(x)/(x-x_0)^k$ is $o(|x-x_0|^{n-k})$ as $x\to x_0$, a contradiction. This completes the proof.

Title: Show that $p(x)=T_{n,x_0}(x)$


问:
System of Differential Equations Question Assistance
The following question has just left me confused with no real decent avenue of attack so any assistance on this would be appreciated.

For the system of equations

$t {\frac{d \vec x}{dt}} = A\vec x $

￼where A is an n×n matrix and $\vec x = [x_1,x_2,...,x_n]^T$. Assuming that $ \vec x = \vec wt^r$, where $ \vec w $ is a constant vector and r is a real number, show that $ \vec w $ and r must satisfy: $ (A − rI)\vec w $ = $\vec 0$.

Thanks!

答: Substitute $\vec{x} = \vec{w}t^r$ into the equation and subtract the LHS.</p>


问:
Whether a group $G$ is cyclic or not.
Let $G$ be a group. If for any two subgroups $A,B$ in $G$, either $A\subset B $ or $B\subset A$ , then $G$ is cyclic.

This asked in a exam, but I give the answer as no, as because in cyclic group , there is only unique subgroup of a particular order.

But, what is the actual conclusion and am I right?

答: If $G$ is finite the result is true. Let $x$ with the maximal order, for every $y\in G$, the group generated by $y$ $gr(y)\subset gr(x)$ implies that $G=gr(x)$.

Title: Whether a group $G$ is cyclic or not.


问:
Linear Algebra Done Right 8.5 $V$ is the direct sum of $nullT^{dimV}$ and $rangeT^{dimV}$ Proof

8.5 $V$ is the direct sum of $nullT^{\dim V}$ and $range T^{\dim V}$

Suppose $T \in \mathcal{L}(V)$. Let $n=\dim V$. Then

$V=nullT^n \oplus range T^n$

The book first proves $(null T^n)\cap(range T^n)={0}$. It's easy to understand.

However, the book then gives $\dim nullT^n + \dim rangeT^n = dimV$ directly from the Fundamental Theorem of Linear Maps(3.22). Why?

答:
$\newcommand{\null}{\operatorname{null}} \newcommand{\range}{\operatorname{range}}$First, $\null T^n \cap \range T^n = \{0\}$ implies that$^\color{blue}1$ $\null T^n + \range T^n = \null T^n \oplus \range T^n$. Now, $\dim(\null T^n \oplus \range T^n) = \dim(\null T^n) + \dim(\range T^n)$$^\color{blue}2$, but the Fundamental Theorem of Linear Maps applied to $T^n$ also gives us $\dim V = \dim(\null T^n) + \dim(\range T^n)$. So, $\null T^n \oplus \range T^n$ is a subspace of $V$ with the same dimension, which means that they are the same.

$^\color{blue}1$If $W_1$ and $W_2$ are two subspaces of $V$, $W_1+W_2 = W_1 \oplus W_2$ if and only if $W_1 \cap W_2 = \{0\}$. Recall: we say that $W_1+W_2 = W_1 \oplus W_2$ if every $v \in W_1+W_2$ has a unique expression of the form $v=w_1+w_2$, with $w_i \in W_i$.

$^\color{blue}2$In general, $\dim(W_1+W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$, so, when $W_1+W_2 = W_1 \oplus W_2$ we have $\dim(W_1+W_2) = \dim(W_1)+\dim(W_2)$.


问:
Differentiating polar functions using complex numbers
I was wondering, given some polar function $r(\theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?

For example take the polar function $r=\cos(a\theta)$, also known as a rose curve for $a\in\mathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?

答:
Let $f:\mathbb{R}\to\mathbb{R}$. Furthermore, let $r(f) = \cos(f) \:\mathrm{or} \:\sin(f)$.
Form Eulers formula we know
$$e^{if} = \cos f+i\sin f.$$
We know that $\operatorname{Re}e^{if} = \cos f$ and $\operatorname{Im}e^{if} = \sin f$.

Since derivative and real/imaginary part commute it is allowed to do what you have stated.


问:
Approximating compactly supported $L^2$ functions with Schwartz functions "from within"?
Crossposted from MathOverflow.

It is well known that the class of Schwartz functions $\mathcal{S}$ in dense in all $L^p$ spaces therefore for each $f \in L^2$ there exists a sequence of Schwartz functions $(f_k)$ such that $\lVert f - f_k \rVert_{L^2} \to 0$ as $k \to \infty$.

If we suppose further that $f$ has compact support can we find a sequence of Schwartz functions $(f_k)$ such that $\lVert f - f_k \rVert_{L^2} \to 0$ as $k \to \infty$ (as above) and additionally $\operatorname{supp}(f_k) \subseteq \operatorname{supp}(f)$ for all $k$?

If so, what argument can I appeal to?

答:
If $E$ is a compact set with positive measure and empty interior, and if $f$ is the characteristic function of $E$, then $f$ is nonzero in $L^2$ and there is no nonzero continuous function whose support is contained in the support of $f$.

To construct such a set $E$, find an enumeration of $[0,1]\cap\mathbb Q$, say $\{q_n\}_{n=1}^\infty$, and for each $n$ choose an open interval $I_n$ around $q_n$ with length $\varepsilon /2^n$.

Then $U:= \bigcup I _n$ is an open set whose measure is at most $\varepsilon$, and
$$
E:= [0,1]\setminus U
$$
is a compact set with empty interior and measure at least $1-\varepsilon$.


问:
Polynomial Long Division:
My question essentially boils down to this: (it was part of a question about polynomial ring ideals)

Find an integer $b$ such that the rational function
\begin{equation}
\frac{x^5-bx}{x^2-2x}\in \mathbb{Z}[x]
\end{equation}
I've tried plugging in a bunch of values for $b$ using a calculator but none of them have come out to whole polynomials. Is there an easier way to do this?

答:
Proof $1\!:\ $ $x(x\!-\!2)\mid xf \overset{{\rm cancel}\ x}\iff x\!-\!2\mid f \!\!\!\overset{\rm\large\color{#c00}{RT}\!\!\!}\iff f(2)=0 \overset{f = x^4-b\!}\iff b=2^4\,$ by $\color{#c00}{\small \rm RT}$ = Rem Theorem.

Proof $2\!:\  0 = xf \bmod x(x\!-\!2) \overset{\color{#90f}{\rm DL}}=\, \color{#0a0}x(f \bmod x\!-\!2) \overset{\rm\large\color{#c00}{RT^{\phantom{|}}}_{\phantom |}\!\!\!}= x f(2)\! \iff\! f(2)=0\!\iff \ldots$

upon applying $\color{#90f}{\rm DL}\!: \   gf\bmod gh^{\phantom{|^{|^|}}}\!\!\! =\, \color{#0a0}g(f\bmod h)\,$ to factor  out $\,\color{#0a0}{g\!=\!x},\,$ $\color{#90f}{\rm DL}$ = mod Distrib. Law.

Title: Polynomial Long Division:


问:
Almost sure uniqueness of conditional expectation
Let $Z(t)$ and $E_N^0(f(Z(0)))$ be given as in the snippet below.
I would like to know how follows this: $$E[\lambda(0)\ | \ Z(0)]=E[\lambda(0)]$$
from the a.s. uniqueness of the conditional expectation and from this:
$$E_N^0[f(Z(0))] = E[f(Z(0))]$$

for all non-negative measurable functions $f:E\to \mathbb{R}$

答:
For random variables $X:\Omega\rightarrow\mathbb{R}$ and $Y:\Omega\rightarrow\mathbb{R}$, we can form a new random variable $E[X|Y]:\Omega \rightarrow\mathbb{R}$, called the conditional expectation of $X$ given $Y$ (also called the conditional expectation of $X$ given the sigma algebra $\sigma(Y)$).  For details, you can google "conditional expectation given a sigma algebra."

The conditional expectation $E[X|Y]$ is always a deterministic function of $Y$, so that
$$ E[X|Y] = g(Y)$$
for some deterministic and measurable function $g:\mathbb{R}\rightarrow\mathbb{R}$.  Since all random variables $X, Y$ are really functions of the outcome $\omega \in \Omega$, we can emphasize such dependency by writing $X(\omega)$ and $Y(\omega)$ and
$$ E[X|Y](\omega) = g(Y(\omega))$$

It turns out that there can be infinitely many versions of the conditional expectation $E[X|Y]$.  So we can have three different versions $g_1(Y)$, $g_2(Y)$, $g_3(Y)$, all valid versions of $E[X|Y]$.  So then what do we mean when we write $E[X|Y]$?  Which version do we use: $g_1(Y)$, or $g_2(Y)$, or $g_3(Y)$, or some other version?  It turns out that it does not matter since it can be proven that any two versions are the same, except possibly on a set of measure zero.  So
$$ g_1(Y(\omega)) = g_2(Y(\omega))=g_3(Y(\omega)) \mbox{ for almost all $\omega \in \Omega$} $$
where "almost all" means "except possibly on a set of measure zero."

This is similar in spirit to the fact that the exponential random variable with parameter $\lambda>0$ has an infinite number of PDF functions (nonnegative functions $f_X:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy $P[X\leq x] = \int_{-\infty}^x f_X(t)dt$ for all $x \in \mathbb{R}$), but each one of them differs from the standard one
$$ f_X(x)=\left\{\begin{array}{ll} \lambda e^{-\lambda x} & \mbox{ if $x \geq 0$} \\ 0 & \mbox{ else} \end{array}\right.$$
only on a set of measure zero. None of the integration properties are changed if we change the function value at only one point, or two points, or a set of points that has measure zero.  So the following function $\tilde{f}_X$ is also a valid PDF:
$$ \tilde{f}_X(x) = \left\{\begin{array}{ll} \lambda e^{-\lambda x} & \mbox{ if $x \geq 0, x \neq 12.6$} \\ 42 & \mbox{ if $x=12.6$}\\ 0 & \mbox{ else} \end{array}\right.$$

So if we say "assume $E[X] = E[X|Y]$ almost surely" it means that $E[X|Y]$ is almost surely equal to the constant $E[X]$.  That is, all versions of $E[X|Y]$ differ from the constant $E[X]$ only on a set of measure zero.  That is, if $g(Y)$ is a version of $E[X|Y]$ then
$$ P[\omega \in \Omega : g(Y(\omega)) \neq E[X]] = 0$$

Practice exercise:
Define
$$\Omega = \{pink, yellow, green, blue\}$$ Assume all outcomes are equally likely, so \begin{align}
&P[\{pink\}]=P[\{yellow\}]= P[\{green\}] \\
&= P[\{blue\}]=1/4
\end{align}
Define $X:\Omega\rightarrow\mathbb{R}$ and $Y:\Omega\rightarrow\mathbb{R}$ by
\begin{align*}
Y(pink) = 0 , \quad & X(pink) = -1\\
Y(yellow) = 0, \quad & X(yellow) = 1\\
Y(green) = 1, \quad & X(green) = -2\\
Y(blue) = 1, \quad & X(blue) = 2
\end{align*}

a) Compute the real number $E[X]$.

b) Compute the real numbers $E[X|Y=0]$ and $E[X|Y=1]$. Call these real numbers $m_0$ and $m_1$, respectively.

c) Are $X$ and $Y$ independent?  Are they uncorrelated?

d) Note that $E[X|Y]$ can be defined as follows:  Define the function
$$g(y) = \left\{\begin{array}{cc}
m_0 & \mbox{ if $y=0$} \\
m_1 & \mbox{ if $y=1$}
\end{array}\right.$$
Then $E[X|Y]$ can be defined as $g(Y)$.  Is it true that $E[X|Y](\omega)=E[X]$ for all $\omega \in \Omega$?

e) Change the problem by making $\Omega = \{pink, yellow, green, blue, tan\}$, and assume
\begin{align}
P[\{pink\}]&=1/4\\
P[\{yellow\}]&= 1/4\\
P[\{green\}]&=1/4\\
P[\{blue\}]&=1/4\\
P[\{tan\}]&= 0
\end{align}
Define $X$ and $Y$ the same, except we make $X(tan) = Y(tan)=42$.  You will find that none of the expectations are different, and now $E[X|Y]$ is "almost surely" $E[X]$, and you can define infinitely many versions of $E[X|Y]$ as $g(Y)$ via the function:
$$g(y) = \left\{\begin{array}{cc}
m_0 & \mbox{ if $y=0$} \\
m_1 & \mbox{ if $y=1$} \\
AnythingYouWant & \mbox{ if $y=42$}
\end{array}\right.$$
You then ask "Why do we care about part (e), since probability zero outcomes don't matter"?  Exactly.  You should not be overly concerned about part (e). The practical meat of a problem will not change if you intuitively interpret "almost surely" as "surely." You can always change a probability space by throwing away a set of outcomes that has measure zero, which reduces the sample space but has no impact on any probability or expectation.


问:
How do I prove that there are no repeating integers and all numbers from 0 to b-1 are present?
While messing with modulus in python,

lomod = [num % 27 for num in [i * 8 for i in range(27)]]

print('Unsorted:', lomod)

lomod.sort()

print('Sorted:', lomod)

I found something interesting.

For $a,b\in\mathbb{Z}$, and $a,b$ are relatively prime, and $0 \le x < b$, $A =ax\mod b$, then $A = \{0,1...,b-1\}$ with no repeating integers.

For example in the case of $a = 8, b=27$, for values of $x$ such that $0\le x < 27$

$A =8x\mod27$ ,  then

$A =$ [0, 8, 16, 24, 5, 13, 21, 2, 10, 18, 26, 7, 15, 23, 4, 12, 20, 1, 9, 17, 25, 6, 14, 22, 3, 11, 19]

sorted,

$A =$ [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]

Which contains all values from $0$ to $26$ once.

However, I don't know how to prove this generally. Is there anyone that can help me?

答:
Let $x_1, x_2$ be integers, and assume they have the same remainder $r = ax_1 \bmod b = ax_2\bmod b$.

Then for some integer quotients $q_1, q_2$,

$$\begin{align*}
ax_1 &= bq_1+r\\
ax_2 &= bq_2 + r\\
ax_1 - ax_2 &= bq_1 - bq_2\\
a(x_1 - x_2) &= b(q_1 - q_2)\\
\end{align*}$$

Since $a$ and $b$ are relatively prime, they share no factors. But the right hand side is a multiple of $b$, so $x_1-x_2$ is also a multiple of $b$.

If the range of $x_1$ and $x_2$ is between $0$ and $b-1$ inclusive, then

$$\begin{align*}
-(b-1) \le x_1-x_2 &\le b-1\\
x_1 - x_2 &= 0b\\
x_1 &= x_2.
\end{align*}$$

Or restating in contraposition, different $x_1 \ne x_2$ (in that range) lead to different remainders $ax_1\bmod b \ne ax_2\bmod b$.


问:
Sampling Random Rotations in $n$ Dimensions Using Bivectors
I understand how bivectors can be used to represent rotations in 2 and 3 dimensions, where we start with a normalized bivector which defines the plane of rotation and an angle through which to rotate points. (I'm assuming rotations about the origin because it's simpler to understand).

I also understand that in 4+ dimensions, bivectors are not necessarily "simple" (they cannot be represented by a single wedge product of vectors). Unfortunately, this shift in complexity happens after I'm able to visualize the rotations, so I'm having trouble extending the notion to higher dimensions. I know there are single and double rotations in 4 dimensions (Wikipedia). I think, but am not sure, that these cover the cases where the bivector of the rotation is simple or not?

So here's my thought for randomly sampling rotations in higher dimensions (without having to generate massive matrices and factor them):

Generate a bivector $B$ by sampling the standard basis coefficients (which I think in $n$ dimensions there should be $n \choose 2$ of) from the surface of a hypersphere (so that the coefficients are normalized)

Generate an angle $\theta$ in $[0, 2\pi)$

Build the rotor $R = e^{B/2}$ associated with the two (interior part $\cos(\theta/2)$ and exterior parts $\sin(\theta/2) B$)

Rotating a vector $v$ is accomplished by doing the geometric product $R v R^{-1}$, where $R^{-1}$ simply has the coefficients of the the exterior part negated

Is this a good approach? Am I misunderstanding something fundamental? My main sources for learning were this great interactive post and the Wikipedia page on bivectors. My main goal is for doing random rotations in Euclidean spaces, but it would be very cool if this approach could be extended to more general manifolds.

答:
The exponential of an arbitrary bivector $B$ in a $n$-D geometric algebra is a multivector called a versor. A versor is an operator that transforms a vector to a vector, or the (sub-)space as a whole. In a nutshell, versors represents orthogonal transformations.

As a multivector, a versor is represented as a sum of k-vectors. We call it an odd versor if k is odd and even versor if k is even. If we consider the algebra of sub-spaces (i.e., lines and planes passing through the origin) then odd versors represents reflections and even versors represents rotations.

You want to construct even versors. That means, to construct a rotor the exponential must be a 2-blade (2-blade is not the same as 2-vector). Since 2-blades coincide with bivectors in 2-D and 3-D, you don't need to care about those distinctions in those dimensions. But in 4 and more dimensions that is not the case. However, Riesz shows that in the Euclidean space $R^{n,0,0}$ (the space of your interest) a rotor can be expressed as the exponential of an arbitrary bivector. The reason is that in that n-dimensional Euclidean space, an arbitrary bivector can be written as the sum of $commuting$ 2-blades.

Also, in a Euclidean $R^{n,0,0}$ space, a rotor is the geometric product of an even number of unit vectors. So you can also do this to generate random rotors.


问:
How do I prove Poisson appraches Normal distribution
I want to prove why the mean and variance of a $\operatorname{Poisson}(\lambda)$, is different when the time index approaches infinite (it's approximated by the mean and variance of a Normal).

For example:
$$
N_k = N_1 + (N_2 - N_1) + (N_3 - N_2) + ... + (N_k - N_{k-1})
$$

Using CLT:
$\frac{N_k - k\lambda}{\sqrt{k\lambda}}$ is normally distributed (in the limit).

I want to answer why is that a Poisson R.V. characterized by $\mathbb{E}[X] = \lambda$ and $\operatorname{Var}[X] = \lambda$.
Because when it approaches a normal distribution, $\mathbb{E}[Z] = \mu$ and $\operatorname{Var}[Z] = \sigma^2$.

答: The Poisson distribution does not approach the normal distribution, the centered Poisson distribution does. More precisely, if $X_\lambda$ is Poisson with parameter $\lambda$, then $Y_\lambda$ converges in distribution to a standard normal random variable $Z$, where $Y_\lambda=(X_\lambda-\lambda)/\sqrt{\lambda}$. In particular, for every $\lambda$, $E[Y_\lambda]=E[Z]=0$ and $\mathrm{var}(Y_\lambda)=\mathrm{var}(Z)=1$ (in your language, $\mu=0$ and $\sigma^2=1$).


问:
Bounded Random Variables are Subgaussian
I am looking at a proof that states bounded random variables are subgaussian. I have been able to follow most of it but am missing the following line.

We are letting $X$ be a bounded random variable and $X^*$ is an independent exact copy.

Then

$$E_X[\text{exp}(tX)]=E_X[\text{exp}(t(X-E[X^*])]$$

This I can follow since X is mean zero and I am just adding a constant so does not require joint. The next step is where I am confused

$$E_X[\text{exp}(t(X-E[X^*])]\le E_{X,X^*}[\text{exp}(t(X-X^{*})]$$ by  jensens inequality and the convexity of exp.

Body:
Jensen's inequality implies $e^{-t E[X^*]} \le E_{X^*}[e^{-t X^*}]$

$E_X[e^{t(X-E[X^*])}] = E_X[e^{tX}] e^{-tE[X^*]} \le E_X[e^{tX}]E_{X^*}[e^{-tX^*}]$


问:
Theorems you wish you knew in complex analysis
I recently stumbled upon a theorem of Landau, which states

Let $f(z)=\sum a_nz^n$ such that $f'(0)\neq 0$ and such that $f^{-1}(\{0,1\})=\emptyset$. The radius of
convergence of $f$ is bounded by a constant $C(f(0),f'(0))$

The theorem is not hard to prove once one knows the Ahlfors-Schwarz-Pick lemma, but it is an extremely surprising statement, and a quite elementary one. In particular, one can easily state and explain it to a student taking a first course in complex analysis. A similar situation occurs with Picard's theorems and Schottky's theorem.

My question is: What other "relatively unknown" theorems do you think fit this description?

Before this question gets closed as "opinion based", let me state some criteria for what I am looking for:

The statement is elementary (as explained before)

The theorem is not usually taught (or stated) in a first course in complex analysis

The proof requires nontrivial machinery from more advanced math(e.g. A-S-P for Landau's theorem)

答:
One of mine that you might enjoy:

There is a rational function $f$ such that for every holomorphic function $g$ on the open unit disk $\mathbb D$, $g$ or $g−f$ has a zero in $\mathbb D$.

This is American Mathematical Monthly problem 6520, solution here


问:
Gre algebra question, percentage
During a certain month, $20$ percent of all the electricity used by a household was used by the water heater. The cost per kilowatt-hour of the electricity used by the water heater was half the cost per kilowatt-hour of the rest of the electricity used. For that month, the cost of the electricity used by the water heater was what fraction of the cost of the electricity used by the household?

I don't think I understand the question but I will share my approach here.

If we have $20\%$ of electricity used was water heater, then $80\%$ is what the rest of the electricity, and water heater costs is half of the rest costs. Let $E$ denotes electricity usage and $C$ is the total costs.

Maybe we have,

$$ 20\%E = \frac{1}{2}C$$
$$ 80\%E = C$$

But it doesn't seem make sense to me even I wrote it, haven't solve this kind of algebra for long time, any suggestions?

答:
You can pretend the household used 100 kilowatt-hours, and that the water heater cost is \$1/kwh and that the other cost is \$2/kwh.

Water heater used 20 kilowatt-hours; each kilowatt-hour costs \$1; total cost from water heater is \$20.

Rest of the household used 80 kilowatt-hours; each of these kilowatt-hours costs \$2; total cost from rest of household is \$160.

Title: Gre algebra question, percentage


问:
Simplifying $\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$
My teacher gave me the following expression to change into a single expression:

$$\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$$

This is what I got after working it out:

$$\log_x 25 - \log_5 \left(\frac{x}{4x^2}\right)$$

Body:
Hint:

$$\log_{x}25=\dfrac1{\log_{25}x}$$


问:
A 10x10x10 cube is comprised of smaller 2x2x2 white cubes - Math Question
A 10x10x10 cube is comprised of smaller 2x2x2 white cubes. The outer 6 faces of the large cube are painted red. How many faces of the smaller cubes are in red?

I get that there are 1000/8 = 125 small cubes, but then how I calculate how many red cubes are red? Anyone could explain it step by step? thank you!

答: $\frac{10\cdot10\cdot10}{2\cdot2\cdot2}=125$ small cubes, that's true. However, that's not what they are asking. They are asking how many faces of smaller cubes are red. Therefore, just count how many faces are red. There are 6 faces of the large cube, and each face is made up of $5\cdot 5$ faces of smaller cubes. Therefore, there are $6\cdot5\cdot5=150$ red faces of smaller cubes.


问:
filtration on the (co)homology of a space from the filtration of a space
Fix $n\!\in\!\mathbb{N}$. Let $X$ be a topological space and $X_0\subseteq X_1\subseteq X_2\subseteq \ldots$ subspaces of $X$. Let $\iota_k:X_k\rightarrow X$ be the inclusion. Let
$F_k:=\mathrm{Im}\,(\iota_{k\ast}: H_nX_k\rightarrow H_nX)$ and
$F^k:=\mathrm{Ker}\,(\iota_k^\ast: H^nX\rightarrow H^nX_k)$.

Question: Is $(F_k)_{k\in\mathbb{N}}$ a filtration on $H_nX$ and $(F^k)_{k\in\mathbb{N}}$ a filtration on $H^nX$, i.e. does there hold $$F_0\subseteq F_1\subseteq F_2\subseteq\ldots \text{ and } F^0\supseteq F^1\supseteq F^2\supseteq\ldots\text{, and why?}$$

If not, then for a fibration $F\rightarrow E\rightarrow B$, how is the convergence of $H_p(B; H_q(F))$ to $H_{p+q}E$ in the Leray-Serre spectral sequence defined when homology is over a ring $R$? I am using McCleary's A User's Guide to Spectral Sequences, 2nd edition. Could you find a specific reference in that book? I've searhed through chapter 5, but to no avail.

答:
Of course!

For homology: We have that $X^k \subset X^{k+1} \subset X$, so by functoriality we get $H_n(X^k)\rightarrow H_n(X^{k+1}) \rightarrow H_n(X)$. So if I'm in the image of the composite, certainly I'm in the image of the second map. Indeed, denote the first map by $i$, then we have $i_{k} = i_{k+1} \circ i$, so if $i_k(x) = y$ then $i_{k+1}(i(x)) = y$, whence $\text{im}(i_{k+1})$ contains $y$. So $F_k \subset F_{k+1}$.

For cohomology: We have $H^n(X) \rightarrow H^n(X^{k+1}) \rightarrow H^n(X^k)$ and, by a similar argument, if I'm in the kernel of the first map then certainly I'm in the kernel of the composite - if I'm zero after the first application then I better stay zero under the second! So $F^k \supset F^{k+1}$.

This is how you get the filtration on $H^*(X)$, in fact on $C^*(X)$, by the same argument... and now you can apply the general formalism of a spectral sequence associated to a filtered complex to get a spectral sequence converging to $H^*(X)$ with the given filtration and with $E_1$ term $H^*(X^k, X^{k-1}).$

In the special case when the filtration comes from lifting the skeletal filtration of the base to the total space of a fibration, one has to check that the $E_2$-page is what you want for the Serre spectral sequence, and this takes a little work but isn't too bad (it's in McCleary, for example.)


答:
I don't understand what you mean by "but I give the answer as no, as because in cyclic group , there is only unique subgroup of a particular order."

The claim is true, and I would prove it by contradiction. Suppose $G$ is not cyclic. So a minimal collection of generators has at least two generators; pick two. What can you say about the two subgroups generated by these? Is either subgroup contained within the other?

Title: Whether a group $G$ is cyclic or not.

Body:
Let $G$ be a group. If for any two subgroups $A,B$ in $G$, either $A\subset B $ or $B\subset A$ , then $G$ is cyclic.

This asked in a exam, but I give the answer as no, as because in cyclic group , there is only unique subgroup of a particular order.

But, what is the actual conclusion and am I right?


问:
Prove equivalence between [-4,0) and (0,12)
I need to prove that $[-4,0)\sim(0,12)$ using the Cantor-Bernstein theorem and also by constructing a bijection.

From what I understood, in order to prove it with the Cantor theorem, I need to construct an injection between the two. From looking at one example I found, I saw that they did it by using a function.

Example was, from $(0,1)$ to $(0,1]$, $f(x) = x$, $x \in (0,1)$, because $(0,1) \subset (0,1]$

I tried to apply the same logic to my question, but got stuck..

I'm very new to this topic and I don't have a clue about where to start.

答: Let's firstly consider $y=f(x)=3x+12$ which maps $(-4,0)$ to $(0,12)$. We can take some countable set from $(-4,0)$, for example $\left\{-\frac{1}{n}\right\}=\left\{x_n\right\}$ for $n\in \mathbb{N}$. By function $f$ these set goes to set $\left\{-\frac{3}{n}+12\right\}=\left\{y_n\right\}$ as $x_n \to y_n$. If now we consider set $\left\{4, x_n\right\}$, then is possible to define bijection $4 \to y_1, x_n \to y_{n+1}$.


问:
What happens when you take the limit with a big-oh error?
What happens if you have some error in your expression and you wish to take the limit? Does the big-oh go away or are you unable to take a limit at all?

This may seem very obvious as it is likely I am misunderstanding the definition of big-oh?

答:
Check your definitions. Big-Oh and limits should commute:

$\begin{align*}
  \lim_{n \to \infty} (f(n) + O(g(n)))
     = \lim_{n \to \infty} f(n) + O(\lim_{n \to \infty} g(n))
\end{align*}$


答:
Logarithms to different bases are interchangeable.  More precisely they are proportional, and one can "change base" as follows:
\begin{align*}
x&=b^{\log_bx}&\text{(definition)}\\
\log_ax&=\log_bx\log_ab&\text{(take logs of both sides)}\\
\log_bx&=\frac{\log_ax}{\log_ab}&\text{(rearrange)}.
\end{align*}
In the problem you pose, we have
$$
\log_x25=\frac{\log_{5}25}{\log_{5}x}=\frac{2}{\log_{5}x},
$$
changing between bases $x$ and $5$.

Title: Simplifying $\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$

Body:
My teacher gave me the following expression to change into a single expression:

$$\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$$

This is what I got after working it out:

$$\log_x 25 - \log_5 \left(\frac{x}{4x^2}\right)$$


问:
How to solve this integral for a hyperbolic bowl?
$$\iint_{s} z  dS $$ where S is the surface given by $$z^2=1+x^2+y^2$$ and $1 \leq(z)\leq\sqrt5$ (hyperbolic bowl)</p>

答:
The shape of your bowl $S$ is given by
$$z(r)=\sqrt{1+r^2}\quad (0\leq r\leq2)\ ,\tag{1}$$
where $r:=\sqrt{x^2+y^2}$. The bowl can be considered as a union of "infinitesimal lampshades" of area
$$dS=2\pi r\>ds\ ,$$
where $ds$ denotes arclength along the shape curve $(1)$. Therefore
$$ds=\sqrt{1+z'^2(r)}={1\over z(r)}\>\sqrt{1+2r^2}\ dr\ .$$
In this way your integral ($=: J)$ becomes
$$J=2\pi\int_0^2 z(r)\> r\ ds=2\pi\int_0^2 r\>\sqrt{1+2r^2}\ dr={\pi\over3}\bigl(1+2r^2\bigr)^{3/2}\biggr|_0^2={26\pi\over3}\ .$$</p>