问:
The 3-component hopf link covering the trefoil
Let $H$ be the complement of the 3-component Hopf link, which is homeomorphic to the complement of the 3-chain link if you prefer,

and let $T$ be the trefoil knot complement.

I recently encountered these 3-manifolds in the context of braid groups: the unordered configuration space of three points in the plane -- whose $\pi_1$ is the 3-strand braid group $B_3$ -- deformation retracts to a space homeomorphic to $T$. It seems that the ordered configuration space -- whose $\pi_1$ is the pure braid group $PB_3$ -- similarly deformation retracts to $H$.

So from covering space theory, I would expect that $H$ is the covering space of $T$ corresponding to the normal subgroup $PB_3 \triangleleft B_3$. The deck transformation group would be the symmetric group $B_3/PB_3 \cong S_3$ which permutes the points. (And that's isomorphic to the 6-element dihedral group $D_3$ if you prefer.) However, I'm having a tough time seeing this covering space. I see an action of $D_3$ on the Hopf link complement (there's a clear triangle there to play with), but this action isn't free on $H$ and so the quotient is an orbifold.

Can someone point out an action of $D_3 \cong S_3$ which is free on $H$ (though it could have fixed points on the link components)? Or does someone have a way to construct the 6-fold covering space of $T$ corresponding to $PB_3 \triangleleft B_3 = \pi_1(T)$?

答:
Let's start by fixing some notation.  The ordered configuration space  of three distinct points in $\mathbb{C}$ is
$$\operatorname{OConf}^3(\mathbb{C})=\mathbb{C}^3-\Delta^3\mathbb{C}$$
where $\Delta^3\mathbb{C}=\{(z_1,z_2,z_3)\in\mathbb{C}^3\mid \exists i\neq j, z_i=z_j\}$, the fat diagonal. Another way to write the fat diagonal is as the zero set of a polynomial.  Anticipating discriminants, we'll use $(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2$.

The symmetric group $S_3$ acts on $\mathbb{C}^3$ by permutation of coordinates, and furthermore it acts on $\Delta^3\mathbb{C}$. Hence, $S_3$ acts on $\operatorname{OConf}^3(\mathbb{C})$.  It does so in a properly discontinuous way, and thus the quotient yields a space $\operatorname{Conf}^3(\mathbb{C})$ covered by $\operatorname{OConf}^3(\mathbb{C})$.  This is a six-fold normal cover (in particular, a principal $S_3$-bundle).

The $S^3$ action on $\mathbb{C}^3$ is linear, and it splits as a direct sum $\mathbb{C}^3=V_1\oplus V_2$ where $V_1$ is spanned by $(1,1,1)$ and $V_2$ is the orthogonal complement, the set of $(z_1,z_2,z_3)$ such that $z_1+z_2+z_3=0$.  Define $\operatorname{OConf}^3_0(\mathbb{C})=V_2-\Delta^3\mathbb{C}$ to be the reduced ordered configuration space.  There is a deformation retract from $\operatorname{OConf}^3(\mathbb{C})$ to $\operatorname{OConf}^3_0(\mathbb{C})$ by collapsing $V_1$ to a point, so at least homotopically this reduced space is "the same."  This deformation retract respects the $S_3$ action hence yields a quotient space $\operatorname{Conf}^3_0(\mathbb{C})$.  Just to reiterate, these reduced configuration spaces are of triples of points whose sum is 0 -- what we did was remove a homotopically uninteresting translation symmetry ($V_1$).

There is another homotopically uninteresting symmetry to $\operatorname{OConf}^3_0(\mathbb{C})$, which is scaling a configuration by a positive real number.  That is, $(z_1,z_2,z_3)$ and $(rz_1,rz_2,rz_3)$ for $r\in\mathbb{R}_{>0}$ are essentially the same when $z_1+z_2+z_3=0$.  We can consider the intersection of $\operatorname{OConf}^3_0(\mathbb{C})$ with a sphere in $V_2$ centered at the origin, which has the effect of choosing some representative scaling for each triple.  Another description of this intersection is taking the complement of a link of the singularity at $0$ of $(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2$ restricted to $V_2$.

Consider the following orthonormal basis for $V_2$:
$$\left\{
\frac{1}{\sqrt{2}}\begin{bmatrix}
1\\-1\\0
\end{bmatrix},
\frac{1}{\sqrt{6}}\begin{bmatrix}
1\\1\\-2
\end{bmatrix}
\right\}$$
Using $(x,y)$ coordinates with respect to this basis, we have
$$(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2 = \tfrac{1}{2}x^2(x-\sqrt{3}y)^2(x+\sqrt{3}y)^2.$$
Let's check that the link of this singularity is a 3-component Hopf link.  The $S^3$ under consideration is the set of $(x,y)\in\mathbb{C}^3$ such that $\lVert x\rVert^2+\lVert y\rVert^2=1$, and in $\mathbb{C}^2$ the zero set is clearly the three complex lines respectively given by the equations $x=0$, $x=\sqrt{3}y$, and $x=-\sqrt{3}y$.  The intersection of a complex line with $S^3$ is a fiber of the Hopf fibration, so we've identified it.  But, for sake of making pictures, we can plot a stereographic projection just to be sure:

Mathematica:

```<code>stereo[a_, b_] := {Im[a], Re[b], Im[b]}/(1 - Re[a]);
components = {stereo[0, E^(I t)],
   stereo[Sqrt[3]/2 E^(I t), 1/2 E^(I t)],
   stereo[Sqrt[3]/2 E^(I t), -(1/2) E^(I t)]};
ParametricPlot3D[components, {t, 0, 2 Pi}, Axes -> False,
  Boxed -> False] /. Line[pts_] :> Tube[pts, 0.1]
</code>```

Now let's see how generators of $S_3$ act on $S^3$.  The first generator we'll consider is $\tau(z_1,z_2,z_3)=(z_2,z_1,z_3)$, given in the original $\mathbb{C}^3$ coordinates.  In $(x,y)$ coordinates, this is $\tau(x,y)=(-x,y)$.  Something we can deduce right now is that this transformation will leave the first component of the link fixed while swapping the other two components.

To visualize this, there is a family of transformations $\tau_t(x,y)=(e^{\pi i t}x,y)$ for $0\leq t\leq 2$ with $\tau_1=\tau$ and $\tau_0=\tau_2=\operatorname{id}$.  Here's a first animation involving the same stereographic projection from before:

It is somewhat misleading because there's an ambiguity with the axis of rotation: the axis of rotation is the first component!  This can be made clearer using a stereographic projection where one of the points of the first component goes to infinity:

From this point of view $\tau$ is a rotation by 180 degrees about the vertical axis.

Let's now look at the second generator, $\sigma(z_1,z_2,z_3)=(z_2,z_3,z_1)$.  In $(x,y)$ coordinates, this is $$\sigma(x,y)=(-\tfrac{1}{2}x+\tfrac{\sqrt{3}}{2}y,-\tfrac{\sqrt{3}}{2}x-\tfrac{1}{2}y),$$
which when restricted to $\mathbb{R}^2$ is a rotation by 120 degrees about the origin.  Evidently, this does a cyclic permutation of the three components.  We can visualize this transformation by a family of transformations given by rotations through various angles:
$$\sigma_\theta(x,y)=(\cos(\theta)x+\sin(\theta)y,-\sin(\theta)x+\cos(\theta)y)$$
where $\sigma_{2\pi/3}=\sigma$.  Using the first stereographic projection, we get this animation:

And using the second stereographic projection for comparison:

Recall that the isometry group of $S^3$ is $SO(4)$.  The matrix of $\sigma$ in the basis $\{(1,0),(0,1),(i,0),(0,i)\}$ is block diagonal with two $2\times 2$ rotations by 120 degrees clockwise.  Thinking of $\sigma$ as being the composition of these two commuting rotations (from setting one of the two blocks to the identity matrix), then we see that $\sigma$ is a composition of two rotations about geodesics.  One on the plane spanned by the first two vectors (parameterized by $(x,y)=(\cos t,\sin t)$) and the second on the plane by the last two (parameterized by $(x,y)=(i\cos t,i\sin t)$).  Plotting those axes should make it more clear:

After a small rotation, we can put those rotation axes into the "standard" 2-component Hopf link position:

There is another way to visualize this.  The 3-components of the link lie on a Clifford torus, and we can stereographically project in such a way that this torus is "standard."  The torus can be parameterized by $(x,y)=(\sin(\theta)e^{i t},\cos(\theta)e^{i t})$ where the three components of the link are given by $\theta=0,-2\pi/3,2\pi/3$.  Putting some red beads on the components to be able to keep track of the rotations, we see this:

(One application of $\sigma$ corresponds to a component rotating two components counter-clockwise.)

Note: this section is not complete and goes beyond the original question.  I want to eventually work out how to see the branched cover $S^3\to S^3$ with a trefoil branch locus dircetly.

A way to encode points of $\operatorname{Conf}^3_0(\mathbb{C})$ is via cubic polynomials by
$$[(z_1,z_2,z_3)]\mapsto (z-z_1)(z-z_2)(z-z_3).$$
This certainly respects the $S_3$ action so it is well-defined.  The roots of this polynomial recover the original triple (up to permutation by $S_3$).

Since $z_1+z_2+z_3=0$, then we have
$$(z-z_1)(z-z_2)(z-z_3) = z^3+(z_1z_2+z_1z_3+z_2z_3)z-z_1z_2z_3$$
and thus the map $(z_1,z_2,z_3)\mapsto (z_1z_2+z_1z_3+z_2z_3, z_1z_2z_3)$ identifies the space of unordered triples (allowing multiplicities) with $\mathbb{C}^2$.  In fact, we get an orbifold cover $V_2\to \mathbb{C}^2$ this way.  Our goal is to take the $S^3\subset V_2$ and show the image in $\mathbb{C}^2$ is another $S^3$, and then say that this gives a six-fold branched cover $S^3\to S^3$ with a trefoil branch locus, whose preimage in $S^3$ is the 3-component Hopf link.

Setting $e_2=z_1z_2+z_1z_3+z_2z_3$ and $e_3=z_1z_2z_3$, it is worth computing the image of $\Delta^3\mathbb{C}$ in $(e_2,e_3)$ coordinates.  It turns out that when $z_1+z_2+z_3=0$ then
$$(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2=-4e_2^3-27e_3^2.$$
This is the discriminant of the cubic polynomial $z^3+e_2z-e_3$.

In the $(x,y)$ basis, we can compute that
$$(e_2,e_3) = \left(-\tfrac{1}{2}(x^2+y^2), -\tfrac{1}{3\sqrt{6}}y(y-\sqrt{3}x)(y+\sqrt{3}x)\right)$$
(and one can verify this is invariant under $\tau$ and $\sigma$, as expected).

Unfortunately, $S^3$ doesn't map to anything that's obviously spherical, so I'll stop here for now.  At least, it's not so hard to work out that the link of the singularity $-4e_2^3-27e_3^2=0$ is a trefoil knot in $(e_1,e_2)$ coordinates.

By the way, this six-fold cover of the trefoil exterior is "Narnia" from the video Knots to Narnia with Bill Thurston.  The cover is associated to a homomorphism $\pi_1(S^3-3_1)\to S_3$, and one can check that by imposing the relation that a meridian squared is the identity (what I've heard called the "meridianal rank-2 knot group") that one gets a group isomorphic to $S_3$, hence there is a unique $S_3$ cover.  In particular, with the presentation $\langle a,b\mid aba=bab,a^2=1\rangle$ one can check that $b^2=1$ as well and so this is the $A_2$ Coxeter group, hence $S_3$.

For fun, here's a gif showing the motion of the roots of polynomials when taking a meridian path around the trefoil.  Each loop around swaps the two roots to the right, and so going around a meridian twice returns the roots to their original positions.

This gives some intuition for the branched cover -- the branch locus corresponds to when the polynomial has a root of multiplicity 2.

Body:
Let's start by fixing some notation.  The ordered configuration space  of three distinct points in $\mathbb{C}$ is
$$\operatorname{OConf}^3(\mathbb{C})=\mathbb{C}^3-\Delta^3\mathbb{C}$$
where $\Delta^3\mathbb{C}=\{(z_1,z_2,z_3)\in\mathbb{C}^3\mid \exists i\neq j, z_i=z_j\}$, the fat diagonal. Another way to write the fat diagonal is as the zero set of a polynomial.  Anticipating discriminants, we'll use $(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2$.

The symmetric group $S_3$ acts on $\mathbb{C}^3$ by permutation of coordinates, and furthermore it acts on $\Delta^3\mathbb{C}$. Hence, $S_3$ acts on $\operatorname{OConf}^3(\mathbb{C})$.  It does so in a properly discontinuous way, and thus the quotient yields a space $\operatorname{Conf}^3(\mathbb{C})$ covered by $\operatorname{OConf}^3(\mathbb{C})$.  This is a six-fold normal cover (in particular, a principal $S_3$-bundle).

The $S^3$ action on $\mathbb{C}^3$ is linear, and it splits as a direct sum $\mathbb{C}^3=V_1\oplus V_2$ where $V_1$ is spanned by $(1,1,1)$ and $V_2$ is the orthogonal complement, the set of $(z_1,z_2,z_3)$ such that $z_1+z_2+z_3=0$.  Define $\operatorname{OConf}^3_0(\mathbb{C})=V_2-\Delta^3\mathbb{C}$ to be the reduced ordered configuration space.  There is a deformation retract from $\operatorname{OConf}^3(\mathbb{C})$ to $\operatorname{OConf}^3_0(\mathbb{C})$ by collapsing $V_1$ to a point, so at least homotopically this reduced space is "the same."  This deformation retract respects the $S_3$ action hence yields a quotient space $\operatorname{Conf}^3_0(\mathbb{C})$.  Just to reiterate, these reduced configuration spaces are of triples of points whose sum is 0 -- what we did was remove a homotopically uninteresting translation symmetry ($V_1$).

There is another homotopically uninteresting symmetry to $\operatorname{OConf}^3_0(\mathbb{C})$, which is scaling a configuration by a positive real number.  That is, $(z_1,z_2,z_3)$ and $(rz_1,rz_2,rz_3)$ for $r\in\mathbb{R}_{>0}$ are essentially the same when $z_1+z_2+z_3=0$.  We can consider the intersection of $\operatorname{OConf}^3_0(\mathbb{C})$ with a sphere in $V_2$ centered at the origin, which has the effect of choosing some representative scaling for each triple.  Another description of this intersection is taking the complement of a link of the singularity at $0$ of $(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2$ restricted to $V_2$.

Consider the following orthonormal basis for $V_2$:
$$\left\{
\frac{1}{\sqrt{2}}\begin{bmatrix}
1\\-1\\0
\end{bmatrix},
\frac{1}{\sqrt{6}}\begin{bmatrix}
1\\1\\-2
\end{bmatrix}
\right\}$$
Using $(x,y)$ coordinates with respect to this basis, we have
$$(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2 = \tfrac{1}{2}x^2(x-\sqrt{3}y)^2(x+\sqrt{3}y)^2.$$
Let's check that the link of this singularity is a 3-component Hopf link.  The $S^3$ under consideration is the set of $(x,y)\in\mathbb{C}^3$ such that $\lVert x\rVert^2+\lVert y\rVert^2=1$, and in $\mathbb{C}^2$ the zero set is clearly the three complex lines respectively given by the equations $x=0$, $x=\sqrt{3}y$, and $x=-\sqrt{3}y$.  The intersection of a complex line with $S^3$ is a fiber of the Hopf fibration, so we've identified it.  But, for sake of making pictures, we can plot a stereographic projection just to be sure:

Mathematica:

stereo[a_, b_] := {Im[a], Re[b], Im[b]}/(1 - Re[a]);
components = {stereo[0, E^(I t)],
   stereo[Sqrt[3]/2 E^(I t), 1/2 E^(I t)],
   stereo[Sqrt[3]/2 E^(I t), -(1/2) E^(I t)]};
ParametricPlot3D[components, {t, 0, 2 Pi}, Axes -> False,
  Boxed -> False] /. Line[pts_] :> Tube[pts, 0.1]

Now let's see how generators of $S_3$ act on $S^3$.  The first generator we'll consider is $\tau(z_1,z_2,z_3)=(z_2,z_1,z_3)$, given in the original $\mathbb{C}^3$ coordinates.  In $(x,y)$ coordinates, this is $\tau(x,y)=(-x,y)$.  Something we can deduce right now is that this transformation will leave the first component of the link fixed while swapping the other two components.

To visualize this, there is a family of transformations $\tau_t(x,y)=(e^{\pi i t}x,y)$ for $0\leq t\leq 2$ with $\tau_1=\tau$ and $\tau_0=\tau_2=\operatorname{id}$.  Here's a first animation involving the same stereographic projection from before:

It is somewhat misleading because there's an ambiguity with the axis of rotation: the axis of rotation is the first component!  This can be made clearer using a stereographic projection where one of the points of the first component goes to infinity:

From this point of view $\tau$ is a rotation by 180 degrees about the vertical axis.

Let's now look at the second generator, $\sigma(z_1,z_2,z_3)=(z_2,z_3,z_1)$.  In $(x,y)$ coordinates, this is $$\sigma(x,y)=(-\tfrac{1}{2}x+\tfrac{\sqrt{3}}{2}y,-\tfrac{\sqrt{3}}{2}x-\tfrac{1}{2}y),$$
which when restricted to $\mathbb{R}^2$ is a rotation by 120 degrees about the origin.  Evidently, this does a cyclic permutation of the three components.  We can visualize this transformation by a family of transformations given by rotations through various angles:
$$\sigma_\theta(x,y)=(\cos(\theta)x+\sin(\theta)y,-\sin(\theta)x+\cos(\theta)y)$$
where $\sigma_{2\pi/3}=\sigma$.  Using the first stereographic projection, we get this animation:

And using the second stereographic projection for comparison:

Recall that the isometry group of $S^3$ is $SO(4)$.  The matrix of $\sigma$ in the basis $\{(1,0),(0,1),(i,0),(0,i)\}$ is block diagonal with two $2\times 2$ rotations by 120 degrees clockwise.  Thinking of $\sigma$ as being the composition of these two commuting rotations (from setting one of the two blocks to the identity matrix), then we see that $\sigma$ is a composition of two rotations about geodesics.  One on the plane spanned by the first two vectors (parameterized by $(x,y)=(\cos t,\sin t)$) and the second on the plane by the last two (parameterized by $(x,y)=(i\cos t,i\sin t)$).  Plotting those axes should make it more clear:

After a small rotation, we can put those rotation axes into the "standard" 2-component Hopf link position:

There is another way to visualize this.  The 3-components of the link lie on a Clifford torus, and we can stereographically project in such a way that this torus is "standard."  The torus can be parameterized by $(x,y)=(\sin(\theta)e^{i t},\cos(\theta)e^{i t})$ where the three components of the link are given by $\theta=0,-2\pi/3,2\pi/3$.  Putting some red beads on the components to be able to keep track of the rotations, we see this:

(One application of $\sigma$ corresponds to a component rotating two components counter-clockwise.)

Note: this section is not complete and goes beyond the original question.  I want to eventually work out how to see the branched cover $S^3\to S^3$ with a trefoil branch locus dircetly.

A way to encode points of $\operatorname{Conf}^3_0(\mathbb{C})$ is via cubic polynomials by
$$[(z_1,z_2,z_3)]\mapsto (z-z_1)(z-z_2)(z-z_3).$$
This certainly respects the $S_3$ action so it is well-defined.  The roots of this polynomial recover the original triple (up to permutation by $S_3$).

Since $z_1+z_2+z_3=0$, then we have
$$(z-z_1)(z-z_2)(z-z_3) = z^3+(z_1z_2+z_1z_3+z_2z_3)z-z_1z_2z_3$$
and thus the map $(z_1,z_2,z_3)\mapsto (z_1z_2+z_1z_3+z_2z_3, z_1z_2z_3)$ identifies the space of unordered triples (allowing multiplicities) with $\mathbb{C}^2$.  In fact, we get an orbifold cover $V_2\to \mathbb{C}^2$ this way.  Our goal is to take the $S^3\subset V_2$ and show the image in $\mathbb{C}^2$ is another $S^3$, and then say that this gives a six-fold branched cover $S^3\to S^3$ with a trefoil branch locus, whose preimage in $S^3$ is the 3-component Hopf link.

Setting $e_2=z_1z_2+z_1z_3+z_2z_3$ and $e_3=z_1z_2z_3$, it is worth computing the image of $\Delta^3\mathbb{C}$ in $(e_2,e_3)$ coordinates.  It turns out that when $z_1+z_2+z_3=0$ then
$$(z_1-z_2)^2(z_1-z_3)^2(z_2-z_3)^2=-4e_2^3-27e_3^2.$$
This is the discriminant of the cubic polynomial $z^3+e_2z-e_3$.

In the $(x,y)$ basis, we can compute that
$$(e_2,e_3) = \left(-\tfrac{1}{2}(x^2+y^2), -\tfrac{1}{3\sqrt{6}}y(y-\sqrt{3}x)(y+\sqrt{3}x)\right)$$
(and one can verify this is invariant under $\tau$ and $\sigma$, as expected).

Unfortunately, $S^3$ doesn't map to anything that's obviously spherical, so I'll stop here for now.  At least, it's not so hard to work out that the link of the singularity $-4e_2^3-27e_3^2=0$ is a trefoil knot in $(e_1,e_2)$ coordinates.

By the way, this six-fold cover of the trefoil exterior is "Narnia" from the video Knots to Narnia with Bill Thurston.  The cover is associated to a homomorphism $\pi_1(S^3-3_1)\to S_3$, and one can check that by imposing the relation that a meridian squared is the identity (what I've heard called the "meridianal rank-2 knot group") that one gets a group isomorphic to $S_3$, hence there is a unique $S_3$ cover.  In particular, with the presentation $\langle a,b\mid aba=bab,a^2=1\rangle$ one can check that $b^2=1$ as well and so this is the $A_2$ Coxeter group, hence $S_3$.

For fun, here's a gif showing the motion of the roots of polynomials when taking a meridian path around the trefoil.  Each loop around swaps the two roots to the right, and so going around a meridian twice returns the roots to their original positions.

This gives some intuition for the branched cover -- the branch locus corresponds to when the polynomial has a root of multiplicity 2.


问:
Improper Riemann Integral equal to Lebesgue Integral
Let $f: [a,b] \to [0,\infty)$ and $f$ is Riemann Integrable on every subinterval $[a + \epsilon,b]$ for $\epsilon > 0$. Suppose that the improper Riemann integral exists. That is
$$I = \lim_{\epsilon \to 0} \int_{a + \epsilon}^{b} f(x) dx < \infty$$
exists. Prove that $f$ is Lebesgue integrable on $[a,b]$ and that $\int_{[a,b]} f(x) dx = I$.

I found another posting of this problem written in the same way, but it used dominated convergence theorem, which I have not covered. The others I saw were written slightly differently and/or weren't making too much sense.

Also, maybe I'm not seeing something clearly, but here's something I thought:

The improper integral I is finite (it exists). If the improper integral exists, isn't it equal to the "proper" integral? That would mean the "normal" proof of showing that every Riemann Integrable function is Lebesgue integrable would apply. However, I have a feeling this is not the way to prove it or else this question wouldn't be asked.

Thanks for the help!

答:
Since $f$ is nonnegative, the Lebesgue integral must exist (but may be infinite).  With the existence of the improper Riemann integral we can show that the Lebesgue integral is finite and $f$ is "Lebesgue integrable" on $[a,b]$.

For all sufficiently large $n$ we have $[a+1/n,b] \subset [a,b]$. Since $f \chi_{[a+1/n,b]} \nearrow f$ as $n \to \infty$, it follows by the monotone convergence theorem that

$$\int_{[a,b]} f = \int_{[a,b]} \lim_{n \to \infty}f \chi_{[a+1/n,b]} =  \lim_{n \to \infty}\int_{[a,b]} f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{a+ 1/n}^b f(x) \, dx = \int_a^bf(x) \, dx < +\infty$$

Here we have applied the equivalence of the Lebesgue and Riemann integrals on the interval $[a+1/n,b]$

ContentLicense: CC BY-SA 4.0


问:
A second order differential equation, $y''=y^{-3}$
The exercise is about to solve $y''=1/y^3$ for $y=y(t)$ with initial values $y(0)=1$ and $y'(0)=2$.

My attempts:

I wrote up it as a first order diff.equation with $2$ variables by introducing $x:=y'$, so that we are looking for (the $y$-coordinate of) the trajectory of the vector field $(1/y^3, x)$, I could draw some vectors, and could 'guess' how the solution will look like.

Look for the solution in form $y=c\,t^\alpha$. I got that $\alpha=1/2$ and $c^4=-4$, that is $c=\pm1\pm i$. However, I guess, real function was asked.

I guess I'm missing something I should know about solving such a differential equation.

答:
Multiply $y'$,
$$
y'y''=y^{-3}y'$$
Integrate,
$$
\frac 1 2 (y')^2 = -\frac 1 2 y^{-2} + c$$
Using initial condition,
$$
2=-\frac 1 2 +c$$
Thus $c=\frac 5 2$. Thus, we have
$$(y')^2+y^{-2}=5.$$
Multiply $y^2$, and substitute $Y=y^2$, then
$$\frac {(Y')^2} 4 +1 = 5Y,$$
$$(Y')^2=20Y-4$$
I think it is easy to go from here. </p>

ContentLicense: CC BY-SA 3.0


问:
If $\lim_{n\to\infty}\sum_{i=1}^{n}w_i=1$, then $\lim_{n\to\infty}\sum_{i=1}^{n}w_i^{2}=0$?
If $\lim_{n\to\infty}\sum_{i=1}^{n}w_i=1$, then $\lim_{n\to\infty}\sum_{i=1}^{n}w_i^{2}=0$?

If it is true, then can you help prove it?

答:
Even more important of an example to note...

$$\sum\limits_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}} \approx -0.605$$

Letting the exact value of the above be $\alpha$ then

$$\sum\limits_{n=1}^\infty \frac{(-1)^n}{\alpha \sqrt{n}}=1$$

Yet if we were to square everything:

$$\sum\limits_{n=1}^\infty \frac{1}{\alpha^2 n}=\infty$$

Squaring termwise doesn't even necessarily retain convergence!


问:
Basic harmonic series (not sure)
The series I am presented with:
$$\sum_{n=1}^{\infty}\frac{n+17}{6n^3+4n^2+5}$$
How can I find the series $\sum_{n=1}^{\infty}b_n$ of the form $\sum_{n=1}^{\infty}b_n=c\sum_{n=1}^{\infty}\frac{1}{n^2}$ where $c\gt0$ such that $b_n\ge\frac{n+17}{6n^3+4n^2+5}$ for all $n\in\mathbb{N}$?
[Side note: I'm not entirely sure if this is in the Harmonic Series subject]

答:
Simplifying the following inequality
$$\begin{align}
&\qquad \ \  \ \ \frac{n+17}{6n^3+4n^2+5} \le \frac{c}{n^2} \\
&\iff n^3+17n^2 \le 6cn^3+4cn^2+5c \\
&\iff(6c-1)n^3 + (4c-17)n^2 + 5c \ge 0
\end{align}$$
Taking $4c > 17$ or $c > 5$ is enough.

You can see how it works with $c = 6$:

Title: Basic harmonic series (not sure)


问:
Different results obtained for $\int e^{-x}\sin x \mathrm dx$ using formula and using by parts.
So I have a very simple integral at hand which can be evaluated by parts directly, and the answer so obtained matches with the correct antiderivative.

$$\int e^{-x}\sin x\mathrm dx$$

My question arises when I try to use the formula for integrals of the type $\int e^{ax}\sin(bx+c)\mathrm dx$. Although the formula itself is derived from two applications of by parts technique, the result finally obtained for the said integral differs in sign. Here's how:

$$\begin{aligned}\int e^{ax}\sin (bx+c)\mathrm dx &= \frac{e^{ax}}{a^2+b^2}\left(a\sin(bx+c)-b\cos(bx+c)\right)\\ & = \frac{e^{ax}}{r}\sin(bx+c-\phi) \ , \ r=\sqrt{a^2+b^2} \ , \ \phi=\arctan\left(\frac{b}{a}\right)\end{aligned}$$

Plugging in $a=-1$, $b=1$ and $c=0$ for the integral at hand gives $1/\sqrt{2}\cdot \sin(x+\pi/4)$ which is negative of the correct antiderivative. What mistake am I making? Any pointers are appreciated. Thanks.

答:
Let's try it with complex numbers:

$$e^{ax} \sin(bx+c)=e^{ax} \frac{e^{i(bx+c)}-e^{-i(bx+c)}}{2i}=\frac{e^{(a+bi)x+ic}-e^{(a-bi)x-ic}}{2i}.$$

Now upon integration you get

$$\frac{\frac{1}{a+bi} e^{(a+bi)x+ic}-\frac{1}{a-bi} e^{(a-bi)x-ic}}{2i} + C \\
= \frac{e^{ax}}{r}  \frac{e^{i(bx+c)+i\phi_1} - e^{-i(bx+c)+i\phi_2} }{2i} + C.$$

Here $\phi_1$ is an argument of $a-bi$ and $\phi_2$ is an argument of $a+bi$. One can select these arguments to be each other's negative, because the numbers whose arguments are being taken are conjugates of each other. If you do that, then you can define $\phi:=\phi_1$ and then you can read off

$$\frac{e^{ax}}{r} \frac{e^{i(bx+c+\phi)}-e^{-i(bx+c+\phi)}}{2i}=\frac{e^{ax}}{r} \sin(bx+c+\phi).$$

Again in my notation $\phi$ was an argument of $a-bi$, so if you want to use an argument of $a+bi$ then you need a minus sign.

So the problem is that you should be using what mathematicians would call the argument of $a+bi$ and what programmers call "atan2(b,a)" to obtain the phase shift $\phi$. Indeed $-\pi/4$ is not a valid argument of $-1+i$, but $3\pi/4$ is.


问:
Find a weighted graph with $5$ vertices has exactly two minimal spanning trees

Draw a weighted graph with $5$ vertices has exactly two minimal spanning trees and justify that there are no other minimum spanning tree.

I could create a random graph which contain one minimal spanning tree but how to confirm it has another spanning tree?

Like the above graph has one minimal spanning tree. Another confusion is what does it mean when they say exactly two minimal spanning trees? Does those spanning tree should have same minimum cost? If not then isn't the latter spanning tree could be more or less cost?

Body:
Let $G = (V, E, w)$ be a weighted graph with $V = \{v_1, v_2, v_3, v_4, v_5\}$, $E = \{\{v_1, v_3\}, \{v_4, v_3\}, \{v_5, v_3\}, \{v_2, v_3\}, \{v_1, v_4\}\}$ and $$\begin{align}
w(\{v_1, v_3\}) &= 1 \\
w(\{v_4, v_3\}) &= 2 \\
w(\{v_5, v_3\}) &= 3 \\
w(\{v_2, v_3\}) &= 4 \\
w(\{v_1, v_4\}) &= 2
\end{align}$$

Then, it is easy to see $G$ has $3$ spanning trees by removing one of the $3$ edges $\{v_1, v_3\}, \{v_4, v_3\}, \{v_1, v_4\}$. Note that they have total weight $11, 10, 10$ respectively. Therefore, $G$ has exactly $2$ minimum spanning trees.


问:
Solve for area of A1 when volume and A2 are known for a frustum pyramid
I'm wondering how to solve out the equation so that A1 is on the left side of the equation when the volume and A2 are known. Any thoughts?

答:
HINT:  Here are steps to remove the square root  sign

$V=\frac{h}{3}(A_1+A_2+\sqrt{A_1A_2})$

$\frac{3V}{h}=(A_1+A_2+\sqrt{A_1A_2})$

$\frac{3V}{h}-A_1-A_2=\sqrt{A_1A_2}$

$(\frac{3V}{h}-A_1-A_2)^2=A_1A_2$

$(\frac{3V}{h})^2+A_1^2+A_2^2+2(A_1A_2+\frac{3V}{h}A_1+\frac{3V}{h}A_2)=A_1A_2$

The above is a quadratic equation about $A_1$ and hope you can continue now.


问:
Adding powers of $i$
I've been struggling with figuring out how to add powers of $i$.

For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...

It has to do something with division by 4, since the "power cycle" of $i$ repeats every fourth power.

Thank you for any clues.

Title: Adding powers of $i$

LastEditorDisplayName: anonymous

Body:
HINT $\rm\quad\quad i^3 + \: i^4 \; + \:\;\cdots\;\: + \; i^k = 0\ \:\iff\: k\:\equiv\: 2 \:\pmod 4$

Generally, suppose that $\rm\: z \:$ has order $\rm m>1\:.$ Therefore $\rm\; z^n = 1 \iff\ m\:|n\;\;\:$ hence:

LEMMA $\quad\rm z^j + z^{j+1} + \:\cdots + z^k = 0\;\; \iff \rm\: k \:\equiv\;\; j \:-\: 1 \:\pmod m $

Proof: $\;\;\;\;\rm \displaystyle z^j \  (1+z+\cdots + z^{k-j}) \;=\; z^j \: \frac{1-z^{k-j+1}}{1-z} = 0 \;\iff\; \rm m\:|\:k-j+1\quad\;$

ContentLicense: CC BY-SA 2.5


问:
Evaluate $\int_{0}^{\pi} \frac{x\coth x-1}{x^2}dx$
I've been trying to evaluate certain series recently, and I found that
$$\sum_{r=1}^{\infty}\frac{1}{r}\arctan\frac{1}{r}=\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
Therefore, I would like to know how to evaluate $$\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
The integral is approximately equal to $0.895004...$ but even Wolfram Alpha gives no closed form solution. Any ideas? Part of the problem is that the indefinite integral has no elementary antiderivative. I've tried using 'Feynman's trick' on this integral too but without success.

Thank you for your help.

答:
An inelegant method (for pure mathematicians) is to use infinite series forms for both sides of the claimed equality to meet at some identical intermediate result:

Starting with the infinte series expansion for $\arctan x$
$$\arctan x = \sum _{k=1}^{\infty } \frac{(-1)^{k-1} }{2 k-1}x^{2 k-1}$$

immediately gives

$$\sum _{r=1}^{\infty } \frac{1 }{r}\arctan\left(\frac{1}{r}\right)=\sum _{r=1}^\infty \frac{1}{r} \sum _{k=1}^{\infty } \frac{(-1)^{k-1} }{2 k-1} \frac{1}{r^{2 k-1}}\tag{1}$$

swapping the order of the double sum gives

$$\sum _{r=1}^{\infty } \frac{1 }{r}\arctan\left(\frac{1}{r}\right)=\sum _{k=1}^{\infty } \frac{(-1)^{k-1} \zeta (2 k)}{2 k-1}\tag{2}$$

Then using the infinite series expansion for $\coth x$
$$\coth(x)=\frac{1}{x}+2 \sum _{k=1}^{\infty } \frac{  (-1)^{k-1} \zeta (2 k)}{\pi ^{2 k}}x^{2 k-1}$$

We can integrate
$$\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
term by term to give
$$\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx=\sum _{k=1}^{\infty } \frac{(-1)^{k-1} \zeta (2 k)}{2 k-1}\tag{3}$$


问:
Compute the maximal proper submodule of $Z\left(\frac{1}{2}\alpha\right)$
Let $\alpha$ be the positive root of the Lie algebra $sl(2,F)$, where $F$ is algebraically closed, of characteristic $0$.

I think the verma module $Z\left(\frac{1}{2}\alpha\right)$ should have this weight space decomposition:
$$Z\left(\frac{1}{2}\alpha\right)=\bigoplus_{n=0}^\infty V_{\frac{1}{2}\alpha-n\alpha} \quad \text{ for } \ \ \  V_{\frac{1}{2}\alpha-n\alpha}=\text{Span}(y^nv^+)$$

Since $\frac{1}{2}\alpha$ is a dominant integral weight, the quotient of the maximal proper submodule should be finite dimensional. But I can't figure out what the max submodule is, since $Z\left(\frac{1}{2}\alpha\right)$ seems irreducible to me. Could someone show me how to determine the max submodule?

答:
Hint: The irreducible finite-dimensional representation you want to find as quotient of this Verma module is the $2$-dimensional one with weights $\pm\frac{1}{2}\alpha$.

To get there, use $h v =(xy-yx)v$ to show inductively that for $ w \in V_{\frac12\alpha-n\alpha}$, we have $x w = n(2-n) w'$ for some $w' \in V_{\frac12\alpha-(n-1)\alpha}$. So what is $x v$ for $v \in V_{-\frac32 \alpha}$? Can you now see an invariant submodule of codimension $2$? (Compare also this section of the WP article on Verma modules, just adapt the notation.)


问:
How to prove the non-existence of solution of a contour problem with ordinary differential equation methods?
Let
$$\begin{cases}
-y^{''}(x)-y(x)= \sin(x), \forall x \in [0, \pi] \\
y(0)=y(\pi)=0
\end{cases}\label{eq1}\tag{1}$$

Prove that it has no solution.

It is a non-homogeneous linear ordinary differential equation of the second order, so the general solution can be written as: $ y (x) = y_ {h} (x) + y_ {p} (x) $ where

$y_{h} (x) $ is the solution of the homogeneous equation $ -y^{''} - y = 0 $, which is $ y_{h} (x) = c_{1} \cos (x) + c_{2} \sin (x) $.

$ y_{p} (x) $ is the particular solution that satisfies the non-homogeneous equation $ -y^{''}(x) -y(x) = \sin (x) $, then we assume it as $y_{p} (x) = A \cos (x) + B \sin (x) $, deriving and substituting we get that: $ -y_{p}^{''} (x) -y_{p}(x) = \sin(x) \longrightarrow A \cos(x) + B \sin(x) - A \cos(x) - B \sin (x) = \sin(x) $,

i.e. $ 0 = \sin (x) $
but this is not true for all $ x \in [0, \pi] $.

Would this prove that $ (1) $ has no solution?

And if not, can you tell me how to test it?

答: $ y_{p} (x) = Ax \cos x + B \sin x \rightarrow A= 1/2 $ and $B=0$


问:
Hypothesis testing for reaction time activity
I need help with completing a HT with the following data:

Table 1: Experiment 1 - Reaction Time Activity
\begin{array}{|l|c|c|c|}
\hline
& \text{Completion time} & \text{Completion time}  & \text{Difference}  \\
&  \text{(Dominant Hand)} & \text{(Non-Dominant Hand)}  & \text{(Dominant - Non)}  \\ \hline
\text{Number of} &  &  & \\
\text{observations} & 1491 & 1491 & 1491\\ \hline
\text{Sample mean}  & 0.4259 & 0.482 & -0.056\\ \hline
\text{Sample standard}  &  & & \\
\text{deviation}  & 0.153 & 0.191& 0.140\\ \hline
\end{array}

"Perform a hypothesis test to assess whether there is a difference in completion time, on average, when undergraduate students use their dominant vs. non-dominant hand, using a $5\%$ significance level. Be sure to include:

Definition of the parameter of interest in context

Your hypotheses and results of your test.  You can include either a screenshot from R or a table of important values (test statistic, distribution, and $p$-value) calculated by hand.

An evaluation of the $p$-value and conclusion in context

Define alpha and power in context

Explain two ways in which you could increase the power of this test

My work so far:

We need to conduct a two-sided HT to assess whether there is a difference in completion time, on average, when undergraduate students use their dominant vs non-dominant hand.

$H_0$: There is no difference in average completion of the reaction time assessment based on a student using their dominant or nondominant hand in the population of undergrads. $\mu_1 - \mu_2 = 0$

$H_a$: There is a difference in average completion of the reaction time assessment based on a student using their dominant or nondominant hand in the population of undergrads. $\mu_1 - \mu_2 \neq 0$

I am unsure of what to do next. Do I conduct a T-test (I would be using R Studio, so using the t.test() function)?

To increase the power of this test, would I just decrease the significance level?

Thank you!

答:
The choice of test statistic depends on the sampling methodology.  In other words, how is the data collected?  What does the data mean?

To illustrate, we could sample the data like this:  we recruit a random sample of $n_1$ students and each student does one reaction time assessment with their dominant hand.  Then we recruit a random sample of $n_2$ students and each of these  does the reaction time assessment with their non-dominant hand.  This would be two-sample independent $t$-test.

Or we could recruit $n$ students, and each student does the reaction time assessment twice, once with the non-dominant hand, and once with the dominant hand.  This would be a paired $t$-test.

Or we could ask each student to perform the test $2m$ times:  $m$ times each with the dominant hand and $m$ with the non-dominant hand.  Now we have a repeated measures paired $t$-test.  You could even design it as a crossover study.

Without specifying how the data is collected, it is not clear how the test should be performed.

Since you specified that the data is collected in a paired manner, then the appropriate R syntax is

t.test(x, y, paired = TRUE, alternative = "two.sided")

Note that the alpha level is the maximum probability of a Type I error you would like to allow--in other words, the probability that you would conclude to reject the null hypothesis, given that there is no difference in reaction times, is at most $\alpha$.

The power is the probability that you do correctly reject the null, given that there is a difference in reaction times.

To increase the power of the test, you would increase the sample size.  Decreasing $\alpha$ actually reduces the power of the test, because by decreasing your tolerance for incorrectly rejecting the null, you are raising the standard by which you must observe a difference in the data in order to reject.  This also has the effect of making it harder to reject when there does in fact exist a difference, which means the power decreases.

Instead, what you can do is keep the $\alpha$ the same, and increase the sample size.  This will increase the power because with a larger sample size of students, the variance of your estimate of mean reaction times will decrease, thus giving you more precision and ability to discriminate whether the mean reaction times are in fact truly different.  Or, you could relax your criterion for rejecting the null hypothesis by increasing $\alpha$, but this also has the unfortunate effect of increasing the probability of concluding that the reaction times are different when they might not be.

Second edit.  Now that I can see the table in which the data is presented, you cannot use t.test at all because that requires a list of individual observations.  What you are given is the relevant summary statistics.  So what you must do is perform the test manually, using the formulas for a paired test.

Since the sample size is very large--in the thousands--I would advise performing the paired test as a $z$-test, not $t$.  The difference is negligible.  The test statistic is simply $$z = \frac{\Delta}{s/\sqrt{n}},$$ where $\Delta$ is the sample mean of the paired differences, and $s$ is the sample standard deviation of the paired differences.  This ratio is approximately standard normal.  You would compute this value, compare it against the critical value for a test of level $\alpha$ (e.g., if $\alpha = 0.05$, then the critical value for a two-sided test is $z_{\alpha/2}^* \approx 1.96$.  If $|z| > 1.96$ then you would reject $H_0$.  If not, you fail to reject.  The $p$-value of the test is $$2 \Pr[Z > |z|],$$ twice the probability that a standard normal random variable is larger than the absolute value of your test statistic.


问:
how to prove the existence of solution of system of equations using the degree of Brouwer?
Help me!

Using the theorem (Existence and uniqueness of the Brouwer degree). There is a unique map $\deg_{B}: \Sigma \rightarrow \mathbb{Z}$ (Brouwer degree), where $\mathbb{Z}$ represents the set of integers, such that the homotopy invariance holds.

I must prove the existence of the solution of the following system, using the topological degree (Brouwer degree)

$$\begin{cases}
8x+6y+ \dfrac{\log (x^{2}+2)}{x^{2}+y^{2}+1}-\sin (y-7)+3=0 \\
3x-y+ e^{(-y^{2}-4)}+ \cos(xy)+2=0
\end{cases}\label{eq1}\tag{1}$$

I've done this:

Step (a)

Take the linear part of the system and propose a function
$f: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$
$$(x,y) \longrightarrow (8x+6y,3x-y)= \begin{bmatrix}
8 & 6 \\
3 & -1
\end{bmatrix} \begin{bmatrix}
x\\
y
\end{bmatrix} $$

i.e the linear system can be written in a matrix form where $A= \begin{bmatrix}
8 & 6 \\
3 & -1
\end{bmatrix}$

We know that if $\deg_{B} (f, \Omega, 0) \neq 0$ the equation $f(x)=0$ has a solution, and $\deg_{B} (f, \Omega, 0) = \mathop{\rm sgn}(\det (A))$, but $\det (A)=-26$, then:

$\deg_{B} (f, B_{\mathbb{R}^{2}}(0,r), 0) = \mathop{\rm sgn}(\det (A)) = -1  \neq 0$, in the case $\Omega = B_{\mathbb{R}^{2}}(0,r)$

Step (b)

Take the nonlinear part of the system and propose a function
$g: \mathbb{R}^{2} \longrightarrow \mathbb{R}^{2}$

$$(x,y) \longrightarrow  \left( \dfrac{\log (x^{2}+2)}{x^{2}+y^{2}+1}-\sin (y-7)+3,   e^{(-y^{2}-4)}+ \cos(xy)+2 \right)$$

Then $(\ref{eq1}) \Leftrightarrow (f+g)(x,y)=(0,0)$.

So I need to prove that $\deg_{B} (f+g, B_{\mathbb{R}^{2}}(0,r) , 0) \neq 0 $, so I propose a homotopy from $f$ to $f + g$ of the form $f + tg$, $\forall t \in [0,1]$:

$$\begin{aligned}
H:[0,1]\times B_{\mathbb{R}^{2}}(0,r) &\longrightarrow \mathbb{R}^{2} \\
(t,x,y) &\longrightarrow  f(x,y) + tg(x,y) \\
(t, \chi) &\longrightarrow  f(\chi) + tg(\chi)
\end{aligned}$$

For this homotopy to be valid, it must be true that: $\forall \chi=(x,y): \Vert \chi \Vert =r, \forall t \in [0,1] \Rightarrow H(t, \chi) \neq 0 $

I test this using "Methods of a priori dimensions", that is, if $H(t, \chi) = 0 $
$ \forall t \in [0,1]$, $\chi \in \mathbb{R}^{2} \Rightarrow \Vert \chi \Vert < M$

Since we take $ \Omega = B_{\mathbb{R}^{2}} (0, r) $, it must be true that $ M \leq r $, then $\Vert \chi \Vert < M$, and it is true, since the system can be written as:

$$ \begin{aligned}
A &\begin{bmatrix}
x\\
y
\end{bmatrix} + t \begin{bmatrix}
\dfrac{ \log (x^{2}+2)}{x^{2}+y^{2}+1}-\sin (y-7)+3\\
e^{(-y^{2}-4)}+ \cos(xy)+2
\end{bmatrix}\\ \Rightarrow &\begin{bmatrix}
x\\
y
\end{bmatrix} = A^{-1}\begin{bmatrix}
-t \dfrac{ \log (x^{2}+2)}{x^{2}+y^{2}+1}+t\sin (y-7)-3t \\
-te^{(-y^{2}-4)} -t\cos(xy)-2t
\end{bmatrix}\\ \Rightarrow &\begin{bmatrix}
x\\
y
\end{bmatrix} \leq M
\end{aligned} $$

Is my solution correct? or can you give me an idea how to do it.

Thanks!

答: The solution I mention is correct and complete.</p>


问:
How one vector space can have more than one basis?
What I know is that a  basis contains the maximum number of linearly independent vectors in the vector space that spans the vector space and the # of basis vectors is the dimension of the vector space, if a vector space have two basis then wouldn't  the dimension of vector space change?</p>

答:
If you are looking for some intuition, then try to think like this : Think of 3-dimensional (euclidean) space. This is a vector space with respect to $\mathbb{R}$. There are infinitely many ways to define a coordinate axis with the origin unchanged, for this. Especially any rotation of the original coordinate system gives us a new coordinate system. The unit vectors along the original coordinate axes is a basis whereas the unit vectors along the new axes is another basis. Since there are many (infinitely many) ways (angles) to rotate the axes, there are infinitely many bases possible even though number of dimensions remain 3.

PS: I am not claiming that the axes obtained by rotation are the only bases for $\mathbb{R}^3$. In fact it is not. It is much more richer than that. Any collection of vectors obtained from the original axes by a bijective transformation- such as rotation, reflection, or 'squeezing' are all other bases.


问:
Is there a definition for the LTE in implicit numerical ODE solvers?
While writing down some basic definitions for my thesis I noticed that the Local Truncation Error (LTE), and, therefore, consistency, is defined only for explicit numerical integration methods. Take wikipedia as an example but on my reference book (Quarteroni Sacco Saleri, Numerical Mathematics) and over the web I couldn't find an analogus for implicit methods.

My reference book says that for an explicit method of the form
$$
u_{n+1} = u_n + h\,\Phi(t_n, u_n, f_n, h),
$$
having $f_n=f(t_n, u_n)$, and $u_n$ the $n$-th numerical approximation. Assuming we got the exact solution $y(t)$, with $y_n=y(t_n)$ we can write:
$$
y_{n+1} = y_n + h\,\Phi(t_n, y_n, f(t_n, y_n), h) + \varepsilon_{n+1},
$$
where $\varepsilon_{n+1}$ is the residual error generating in $t_{n+1}$. Thus we can write $\varepsilon_{n+1} = h\,\tau_{n+1}(h)$ and we can define $\tau_{n+1}(h)$ as the LTE, and iff $\tau_{n+1}(h)\to0$ for $h\to0$ the method is consistent.

Could there be an analogus for a generical implicit method? My guess is that every implicit method has got its own definition for the LTE.

答:
I'm digging up an old question here, but since I was working on this problem a while ago, I might help you (or someone else) by contributing to this question.

As you mentioned,

we can define $\tau_{n+1}(h)$ as the LTE, and iff $\tau_{n+1}(h)$
$\rightarrow 0$ for $h \rightarrow 0$ the method is consistent. Could there be an analogus for a generical implicit method?

Regarding this question, if I understood you correctly and you want to "measure" the LTE to see, if an implicit method is converging, then yes, we can do something similar for implicit methods too. For this we will rely on the theorem which says: Consistency + zero-stability implies convergence.

We know that for the forward Euler method, the LTE is $\mathcal{O}(h^{2})$. In general, higher-order schemes have a lower LTE for the same step size. Normally, we do not know the exact solution of our problem, so the global error is not easy to calculate. Nevertheless, we say that a convergent numerical method is the one which approaches the true solution as $h \rightarrow 0$. However, we can also show convergence if the coefficient of our scheme satisfy a specific property called absolute stability.

Let us consider the IVP
\begin{align*}
\begin{cases}
y^\prime &= \lambda y \\
y(0) &= 1.
\end{cases}
\end{align*}

We know that the solution to this IVP is $\mathrm{e}^{\lambda t}$ for $\lambda \in \mathbb{C}$. Thus, $\lambda = \mathrm{Re}{(\lambda)} + \mathrm{i}\cdot\mathrm{Im}(\lambda t)$. The solution now can be rewritten as
\begin{align*}
y(t) = \mathrm{e}^{\lambda t} &= \mathrm{e}^{(\mathrm{Re}\lambda)t}\mathrm{e}^{\mathrm{i}(\mathrm{Im}\lambda)t} \\
&= \mathrm{e}^{(\mathrm{Re})t} \left[\cos(\mathrm{Im}\lambda t) + \mathrm{i} \cdot \sin(\mathrm{Im}\lambda t)\right].
\end{align*}

The key observation here is that if $\mathrm{Re}\lambda < 0 \Rightarrow y(t) \rightarrow 0$ for $t \rightarrow +\infty$. In this case we say that the ODE is stable which means in other words that the equilibrium $y=0$ is asymptotically stable. (Think of a pendulum that stops swinging with time progress). So far so good, for analytical solutions. The question is, do we retain this property of the ODE also in discrete settings? In other words, after discretising the ODE with our favourite scheme, do we observe the same property, that if $\mathrm{Re}\lambda < 0$ we actually get to $0$? In general, no we do not! But if this property holds, then we can say that the scheme is absolutely stable, which means that $u_n \rightarrow 0$ for $n \rightarrow +\infty$.

I will give you an example for the implicit Euler scheme $u_{n+1} = u_n + hf_{n-1}$ applying to the IVP problem above.

\begin{align}
u_{n+1} &= u_n + h\lambda u_{n+1} \\
(1-h\lambda)u_{n+1} &= u_n \\
u_n &= (1-h\lambda)^{-n} u_0,
\end{align}
so the condition is $|1-h\lambda|^{-1} < 1$ or $|1-h\lambda| > 1$. Now, let's say $z = h\lambda$, we can draw a unit circle in the complex plane to visualise the region of absolute stability. If you do so, see here, you will see that the region of abs. stability for the implicit Euler is the entire area around the unit circle centred at 1, so the complement of that circle. Hence, for implicit Euler, if $\mathrm{Re}\lambda < 0$, then $\forall h$, $u_n \rightarrow 0$, which is the entire left half of the complex plane. This is a very important fact, because, no matter how small/large your $h$ is, we will always end up in the absolute stability region. Therefore we say that the implicit Euler is unconditionally stable. This makes them especially suitable for the stiff problems.

And that, in my opinion, is the beauty of implicit schemes, because they very often come with absolute stability ''for free''. However, the problem with implicit schemes is that we have to solve an implicit system, which is way more expensive computationally wise. BUT, because implicit schemes are unconditionally stable, we can choose a larger $h$ to reduce the computation. So, even though we did not calculate the LTE per se, or discussed consistency, we are able to observe important properties regarding the convergence and stability of ODE systems.


问:
Checking if a number is a Fibonacci or not?
The standard way (other than generating up to $N$) is to check if $(5N^2 + 4)$ or $(5N^2 - 4)$ is a perfect square.  What is the mathematical logic behind this? Also, is there any other way for checking the same?  </p>

答: If $X^2 \pm nX -(n^2-1)=0 $ has positive integer solutions for $X$ then $n$ is in the sequence and the solutions are the numbers before and after it in the sequence. This is because if you take any number in the sequence to the $4$th power and subtract the product of the two numbers before it and the two numbers after it the answer will be $1$.


问:
Why is the functor that associates to a scheme $S$ the set of $S$-isomorphism classes of elliptic curves over $S$ not representable?
A few days ago I heard an online presentation about elliptic curves and the presenter claimed that the functor which assigns to a scheme $S$ the isomorphism classes of elliptic curves over $S$ is not representable, and it seemed this was taken for granted by everybody.

Can anybody explain to me why this is the case or give a good reference?

答:
I'll elaborate a little on my comment, this answer is not really formal, but it gives the idea which can be made so.

Suppose that your functor were represented by a scheme $X$ over $\mathbb{Q}$, say. Then take two points $p, q \in X(\mathbb{Q})$ corresponding to a pair of elliptic curves $E_p$ and $E_q$ which are nontrivial quadratic twists. Then $p, q$ are distinct, since the twist is nontrivial.

But since $X$ represents our moduli problem, $p$ and $q$ must be the same point in $X(\bar{\mathbb{Q}})$ because they become isomorphic over $\bar{\mathbb{Q}}$. This is absurd, because $X$ is a scheme, and schemes are nice.


问:
Why does the discriminant of a ODE in the canonical form $a_{11}G^2_x+2a_{12}G_xG_y+\:a_{22}G^2_y=0$ tells us if it is an elliptic/etc equation?
Having

$$a_{11}G^2_x+2a_{12}G_xG_y+\:a_{22}G^2_y=0$$

We divide by $G^2_y$, setting $t=\frac{G_x}{G_y}$ gives us

$$a_{11}t^2+2a_{12}t+\:a_{22}=0$$

The discriminant of this quadratic equation tells us that:

if it's greater than zero, it is a hiperbolic equation

if it's equal to zero, it is parabolic

if it's less than zero, it is elliptic

My question is: Why? Our teacher just gave this piece of information to us, without mathematically proving this

答:
The naming scheme simply refers to the shape of the quadratic curves
$$
a_{11}x^2+2a_{12}xy+a_{22}y^2=c
$$
Changing coordinates to $\tilde x=x+\frac{a_{12}}{a_{11}}y$ gives
$$
\iff a_{11}\tilde x^2-\frac{a_{12}^2-a_{22}a_{11}}{a_{11}}y^2=c
$$
which should make the names clear in the elliptic and hyperbolic case. In-between is always the parabolic case, even if this is somewhat degenerate here. It would probably work better if the right side was set to a generic linear expression $b_1x+b_2y+c$.


问:
Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?
We know that $f(x)=x^4+1$ is a polynomial irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$.

My question is:

Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ has a linear factor over $\mathbb{Q}$?

Edit: Thanks for @Jyrki Lahtonen's answer, I want to do some modifications:

Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?

Thanks in advance!

答:
The idea used by user14972 in their answer can be used to find other examples:

$f(x)=(x^2+3)(x^3+2)$. How many zeroes the cubic factor has in $\Bbb F_p$ depends on which quadratic form of discriminant $D=-108=-3\cdot6^2$, if any, represents $p$; if 3 zeroes, $a^2+27b^2$; if no zeroes, $4a^2+2ab+7b^2$; if one zero, neither quadratic form. If that cubic factor has no zeroes, then $-108$ is a square $c^2$; $p>3$ so 6 has an inverse, and $(6^{-1}c)^2+3=0$.

Similarly, $f(x)=(x^2+3)(x^3+3)$, $D=-243=-3\cdot9^2$.

user14972's example, and the two above, use the characteristic cubic of quadratic forms whose discriminants have some value $D$ where $h(D)=3$, i.e. where quadratic forms of discriminant $D$ are in three classes. But it is not necessary to use such concepts of quadratic form theory.

If $p=3$ or $p=1\mod 6$, then $p$ has the form $a^2+3b^2$, so $0=(a^2+3b^2)b^{-2}=(ab^{-1})^2+3$. So, if $x^2+3$ has no zeroes in $\Bbb F_p$, then $p=2$ or $p=5\mod6$, so everything is a cube.[1] So we may take $f(x)=(x^2+3)(x^3+a)$ where $a$ is any integer except a cube (so that $f$ has no linear factor in $\Bbb Q$).

[Brandl][2] gives the example $f(x)=(x^2+x+1)(x^3-2)$. The following proof that $f$ has a zero in each $\Bbb F_p$ is adapted from [Brandl]. If $p\ne 1\mod 3$, then the order of the multiplicative group of $\Bbb F_p$ is not divisible by 3. Hence 2 is a cube modulo $p$.[1] If $p=1\mod 3$, then there exists a primitive cube root of unity $x$ modulo $p$. [This follows from Sylow's first theorem.]
This is obviously a root of the polynomial $x^2+x+1$. [This is because $0=x^3-1=(x-1)(x^2+x+1)$ but $x\ne 1$.]

[1] Proof. If $p=2$ or 3, $a^3=a$. If $p=6k+5$, then, by Fermat's little theorem, $a=a^p=a^{6k+5}$, so $a=r^3$ where $r=a^{4k+3}$. Alternatively, if any element of $\Bbb F_p$ were not a cube, then, by the pigeonhole principle, at least two unequal elements $a, b$ must have equal cubes. Then $0=a^3-b^3=(a-b)(a^2+ab+b^2)$; $a-b\ne 0$ so $a^2+ab+b^2=0$. The only element whose cube is 0 is 0, so $b\ne 0$, so $b$ has an inverse.
So $(a^2+ab+b^2)b^{-2}=0$ so $c^2+c+1=0$ where $c=ab^{-1}$. So $0=(c^2+c+1)(c-1)=c^3-1$ so $c^3=1$. $a\ne b$ so $c\ne 1$, so $c$'s order is 3, so $3\mid p-1$, contradiction.

[2] Rolf Brandl, Daniela Bubboloni and Ingrid Hupp. Polynomials with roots mod $p$ for all primes $p$. J. Group Theory 4 (2001), p.233--239.


问:
Drawbacks of gradient descent
In the method of gradient descent, which in each iteration we reduce the problem to a single variable optimization problem in order to find $\alpha$ (called the learning rate?) using the update rule:

$$\mathbf{x}_{t+1} = \mathbf{x}_t-\alpha\nabla f(\mathbf{x}_t)$$

Drawbacks:

computational: we must evaluate the gradient, of at least estimate it using finite differences.

order of convergence, due to the fact that $$\nabla f (\mathbf{x}_{t+1})^T\nabla f(\mathbf{x}_t) = 0$$ we will obtain a "zig-zag" pattern which will increase the number of iterations.

When will the gradient descent do a $90$ degrees turn? On the one hand we see that

$$\nabla f (\mathbf{x}_{t+1})^T\nabla f(\mathbf{x}_t) = 0$$

On the other hand we know that the gradient is perpendicular to the level set?

I see that in some case we start with a random starting point and set a learning rate $\alpha$ and in other cases, we just start with staring point, and find $\alpha$ using line search methods, is this the same optimization method?

答: I recommend a reading of Chapter 3 of the review paper "Optimization Methods for Large-Scale Machine Learning" (published on SIAM Review), which contains a through overview of classical gradient descent method (although the emphasis is on the stochastic gradient descent and its various variants/refinements). A more basic introduction to the analysis of (stochastic) gradient descent can be found in Chapter 14 of the book "Understanding Machine Learning - From Theory to Algorithms" written by Shai Shalev-Shwartz and Shai Ben-David.

Title: Drawbacks of gradient descent


问:
Conditions of polynomial irreducibility
I am in abstract algebra and am currently reading on a theorem which describes equivalent conditions relating to polynomial irreducibly. Here is the theorem:

Let $\mathbb{F}$ be a field and $p(x)$ a non-constant polynomial in $\mathbb{F}[x]$. Then the following conditions are equivalent:

$p(x)$ is irreducible

if $b(x)$ and $c(x)$ are any polynomials such that $p(x)|b(x)c(x)$, then $p(x)|b(x)$ or $p(x)|c(x)$

if $r(x)$ and $s(x)$ are any polynomials such that $p(x) = r(x)s(x)$, then $r(x)$ or $s(x)$ is a nonzero constant polynomial.

When the textbook proves condition $3$ by using condition $2$, they say the following:

"If p(x) = r(x)s(x), then p(x)|r(x) or p(x)|s(x), by 2." It goes on to show that this will lead to either r(x) or s(x) being a unit, and thus a constant polynomial. What I'm not getting, though, is where it says that if p(x) = r(x)s(x), then p(x)|r(x) or p(x)|s(x). Condition 2 hinges on p(x)|r(x)s(x), meaning r(x)s(x) = t(x)p(x) for some t(x) in F[x]. Wouldn't p(x)=r(x)s(x) mean that r(x)|p(x) and s(x)|p(x), rather than the other way around?

I had the idea that perhaps p(x)|r(x) or p(x)|s(x) when t(x)=1 in r(x)s(x)=p(x)t(x), but I'm not sure if this is the correct interpretation. Any bit of insight would be appreciated.

答:
Your thinking is right. We can see that if $1_F$ is the unit of $F$, then we have $1_Fp(x)=p(x)=r(x)s(x)$ so,
for $1_F$ is a constant polynomial (that's the crucial part), we do have that there exists a polynomial $g(x)\in F[x]$ such that $r(x)s(x)=p(x)g(x)$. So that means, by definition, that $p(x)|r(x)s(x)$.

Title: Conditions of polynomial irreducibility


问:
Closed subgroups of $GL(n, \mathbb{C})$
Let $M_{n}(\mathbb{C})$ be the space of all complex $n\times n$ matrices equipped with the operator norm:
$$||A|| := \sup_{||x||\le 1}||Ax|| \quad A \in M_{n}(\mathbb{C})$$

Let $GL(n, \mathbb{C})$ be the subspace of all invertible matrices of $M_{n}(\mathbb{C})$. I would like to prove that (sub)-groups such as $SL(n,\mathbb{C})$ (the set of all $n\times n$ matrices with determinant equals to one), $O(n)$ (set of all matrices $A\in M_{n}(\mathbb{C})$ such $A^{T}A = I$) and $SO(n)$ (subset of $O(n)$ of matrices with determinant one) are closed in $GL(n,\mathbb{C})$.

I suppose that the idea (for all proofs) is to take a sequence $\{A_{n}\}_{n\in \mathbb{N}}$ such that $A_{n}\to A$ and prove that $A$ is in the same set. But in order to prove that $A$ is an element of, say, $SL(n,\mathbb{C})$ or $SO(n)$, I have to prove that $\text{det}A = 1$ and I have no idea of how this is a conseguence of the convergence $A_{n}\to A$.

Any explanations and/or references are helpful!

答:
Actually, it's better to use this approach: $SL(n,\Bbb C)=\det^{-1}(\{1\})$ and so, since $\det$ is continuous and $\{1\}$ is closed, $SL(n,\Bbb C)$ is closed.

With $O(n)$ it's similar; you consider the map $f\colon GL(n,\Bbb C)\longrightarrow GL(n,\Bbb C)$ defined by $f(A)=A^TA$, which is continuous. Then $O(n)=f^{-1}(\{\operatorname{Id}_n\})$, and therefore it is closed.

Title: Closed subgroups of $GL(n, \mathbb{C})$


问:
How to find a function $f$ to transform a stochastic process $X$ to a martingale $f(X)$
Let's think about a property we want to show for a process $X=(X_t)_{t\geq0}$, which is very hard to show. But maybe we can show another property for a martingale $H$ very easy and the hope is that the desiered property can be derived from this. So the Idea is to find a function $f:\Bbb R\to\Bbb R$, so that $H_t:=f(X_t)$ is a martingale with respect to the original filtration.

An exemplary property for this could be e.g. the expected value is bounded in any kind. Let's assume we found such a function $f$ which is concave and strictly monotonically increasing. As $f(X)$ is a martingale we get for a constant $c=\Bbb Ef(X_t)\leq f(\Bbb EX_t)$ (by jensen's inequality) and thus ($f^{-1}$ is increasing as well) $f^{-1}(c)\leq \Bbb EX_t$.

As this question is only about the general procedure how to find such a function, we could do this e.g. with the process $X_t:=B_t+t$, whereas $B$ is a Brownian motion. I know the property 'bounded below' could be shown ways easier in this case, but it's only about the general procedure.

答:
The easiest solution of $f(X)$ to be a martingale is to choose $f$ as a constant function. But this is not what we want. So let's look for a better solution:

By the Itô-formula we get
$$f(B_t+t)=\int_0^t f'(B_s+s)\mathrm dB_s+\int_0^t f'(B_s+s)\mathrm ds+\frac{1}{2}\int_0^t f''(B_s+s)\mathrm ds.
$$
Since the first summand is already a local martingale (a stochastic integral w.r.t a local martingale is a local martingale again), we could try to find a function $g=f'$ so that $g(x)+\frac{1}{2}g'(x)=0$ for all $x$. This is satisfied for example by $g(x)=e^{-2x}$. Since
$$f(B_t+t)=-\frac{1}{2}e^{-2(B_t+t)}=\int_0^t e^{-2(B_s+s)}\mathrm dB_s
$$ is a local martingale we just have to check some conditions anymore that it is even a (true) martingale. One of them is to proof $\mathbb E\left[f(B+\mathrm{Id}_{\mathbb R})\right]_t<\infty$ for all $t\geq0$ ($[.]_t$ denotes the quadratic variation). Note that
$$\mathbb E\left[f(B+\mathrm{Id}_{\mathbb R})\right]_t=\mathbb E\int_0^t e^{-4(B_s+s)}\mathrm ds=\int_0^t e^{-4s} \mathbb Ee^{-4B_s}\mathrm ds.
$$
As $\mathbb Ee^{-4B_s}=e^{8s}$ is the moment-generating function of a $(0,s)$-normally distributed random variable evaluated at the point $-4$, we get $$\mathbb E\left[f(B+\mathrm{Id}_{\mathbb R})\right]_t=\int_0^t e^{4s} \mathrm ds=(e^{4t}-1)/4<\infty.$$


问:
Changing a simple, linear, second order PDE to a set of coupled ODEs
I'm solving:
$$\frac{\partial^2f}{\partial x_1^2}+\sum_{i=1}^{n}a_{i}\frac{\partial f}{\partial x_i}=0
\label{1}\tag{1}$$
wit $a_i$ some complicated functions of $x_i's$. If we didn't have the second order derivative, we could apply the method of characteristics and we would have the desired results. Instead, if $n=2$, we could apply the theory of ellipctic, hyperbolic and parabolic PDE's to proceed. However, in the general case of arbitrary $n$ it looks like the equation will be difficult to solve.

Questions

Can we classify this second order, linear, homogeneous PDE further?
EDIT: Yes, we could say it is a parabolic PDE.

What techniques can be used for solving such equations?

In the case of $a_i=const.$ the equation can be solved using the Laplace-Fourier transforms assuming we have some adequate boundary/initial conditions. This technique can be even used for certain simple $a_i$, but to my understanding it fails in the general case. The same applies to the Green's function method.

答:
Recently I have answered, in a different context, to a question involving the class to which this equation belongs thus, by using formerly produced materials, I think I can give a comprehensive answer to this one.

Can we classify this second order, linear, homogeneous PDE further?
EDIT: Yes, we could say it is a parabolic PDE.

Precisely, this is a degenerate parabolic equation, since the matrix of the coefficients of the second order derivatives is evidently positive semi-definite, as it is
$$
\mathbf{A}\triangleq(a_{ij})_{1,j=1,\ldots,n}=
\begin{pmatrix}
1 & 0 & \ldots & 0 \\
0 & 0 & \ldots & 0 \\
\vdots & \vdots &\ddots &\vdots\\
0 & 0 & \ldots & 0
\end{pmatrix}
$$
thus we have
$$
\sum_{i,j=1}^n a_{ij} \xi_i\xi_j \ge 0\label{2}\tag{2}
$$
for all real vectors $\boldsymbol{\xi}=(\xi_1, \ldots,\xi_n)\in\Bbb R^n.$ The degeneracy comes from the fact in \eqref{2} equality holds for at least one $\boldsymbol{\xi}\neq \mathbf 0$ (as a matter of fact, in our case this happens for all the vectors belonging to the hyperplane $\{\boldsymbol{\xi}\in\Bbb R^n: \xi_1=0\}$), thus it is degenerate as the Tricomi equation is.
 Equations of this kind are called equations with nonnegative characteristics, and their theory is comprehensively addressed in the monograph [1] by Oleĭnik and Radkevič.

What techniques can be used for solving such equations?

The main technique, which is de facto a bundle of techniques, is called elliptic regularization: a parameter-dependent, say $\varepsilon$-dependent, part is added to the the equation under analysis in order to transform the problem to an elliptic one, which can be studied with more or less standard techniques. The  solution of the original problem is then recovered by letting $\varepsilon \to 0$ in the family of solution to the (elliptic-)regularized problem, ad done in [1], §1.5, pp. 30-41. Specifically, in our case we may construct the solutions to \eqref{1} by using the solutions to the following elliptic differential equation:
$$
\frac{\partial^2f_\varepsilon}{\partial x_1^2}+ \varepsilon \sum_{i=2}^{n}\frac{\partial^2f_\varepsilon}{\partial x_i^2}+\sum_{i=1}^{n}a_{i}\frac{\partial f_\varepsilon}{\partial x_i}=0\quad \varepsilon >0 \label{3}\tag{3}
$$
However, despite the apparent simplicity of the methodology described above, the study of equations with non-positive characteristics offers some hidden difficulties: for example, when solving the boundary problem for \eqref{1} in a sufficiently regular domain $\Omega$, there are (possibly finite measure) subsets of the boundary where it is not necessary to specify the boundary condition at all. These subsets are defined by the so called Fichera's function ([1], §1.1, p. 17) (named after Gaetano Fichera) and for example they influence also the statement of the maximum principle ([1], §1.1, theorem 1.1.2, pp. 21-22) for degenerate equations of nonpositive characteristics as \eqref{1} is. Definitely, in order to analyze such kind of equations, reference [1] is worth a look.

References

[1] Ol’ga Arsen’evna Oleĭnik, Evgeniĭ Vladimirovich Radkevič, "Second order equations with nonnegative characteristic form", (Russian) Itogi Nauki i Tekhniki. Seriya "Matematicheskii Analiz" 1969, 7-252 (1971), Zbl 0217.41502. Translated as the book:

Ol’ga Arsen’evna Oleĭnik, Evgeniĭ Vladimirovich Radkevič, Second order equations with nonnegative characteristic form, Translated from the Russian by Paul C. Fife.  New York-London: Plenum Press, 1973, pp VII+259, ISBN: 0-306-30751-0, MR0457908.


问:
Find minimum volume bounded by three planes and another plane tangent to an ellipsoid.
I am having trouble with a problem that I found in a multivariate calculus book. The problem is :

Let $a$, $b$, and $c$ be nonzero real numbers. Find the minimum volume bounded by the planes given by $x=0$, $y=0$, $z=0$, and a plane that is tangent to the ellipsoid given
by $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$.

I suspect that solving this problem would involve the use of the Lagrange multiplier method.

In my solution so far I start with defining a function :
\begin{equation}
h(x,y,z) = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} - 1
\end{equation}
So that the ellipse (call it $E$) is given by :
\begin{equation}
E = \{ (x,y,z) \in \mathbb{R}^{3} \; : \; h(x,y,z) = 0 \}
\end{equation}
We have the partial derivatives of $h$ as :
\begin{align}
h_{x}(x,y,z) & = \frac{2x}{a^{2}}\\
h_{y}(x,y,z) & = \frac{2y}{b^{2}}\\
h_{z}(x,y,z) & = \frac{2z}{c^{2}}
\end{align}
Now let $(x_{0},y_{0},z_{0}) \in E$. The tangent plane to E at $(x_{0},y_{0},z_{0})$ can be defined implicitly using a function $T$ :
\begin{align}
T_{x_{0},y_{0},z_{0}}(x,y,z) & = h_{x}(x_{0},y_{0},z_{0})(x-x_{0}) + h_{y}(x_{0},y_{0},z_{0})(y-y_{0}) + h_{z}(x_{0},y_{0},z_{0})(z-z_{0}) \\
  & = \frac{2x_{0}(x-x_{0})}{a^{2}} + \frac{2y_{0}(y-y_{0})}{b^{2}} + \frac{2z_{0}(z-z_{0})}{c^{2}}\\
\end{align}
Where the tangent plane (say $Q(x_{0},y_{0},z_{0})$) is :
\begin{equation}
Q(x_{0},y_{0},z_{0}) = \{(x,y,z) \in \mathbb{R}^{3} \; : \; T_{x_{0},y_{0},z_{0}}(x,y,z) =0 \}
\end{equation}
So I suspect the point of the problem is to find the $(x_{0},y_{0},z_{0}) \in E$ so that
the volume bounded by the planes defined by $Q(x_{0},y_{0},z_{0})$ , $x = 0$, $y=0$, and $z = 0$ is minimized. The problem that I have at this point is how to get an expression for this volume.

Can anyone help with this ?

答:
It is much, much easier to perform a coordinate transformation that takes the ellipse to the sphere:  $$(u, v, w) = (x/a, y/b, z/c)$$ and then observe that this transformation preserves tangency and scales volume proportionally by a factor of $1/(abc)$.

Then, since the sphere is symmetric, the minimum volume in the $(u,v,w)$ coordinate space is attained when the point of tangency is $u = v = w = 1/\sqrt{3}$ (why?).  Solve for the points where this plane intersects the $u$-, $v$-, and $w$-axes, compute the volume of the tetrahedron, and reverse the scaling transformation by multiplying by $abc$.


问:
Pollution of 3 lakes with differential equations
Consider three lakes of equal volumes connected to each other with a flow.

An accident results in the spillage of 300 000 kg of chemicals in Lake 1. Calculate the quantity of this chemical present, after the accident, in each of the lakes, assuming that it remains in one of the three lakes.
Then the problem gives me this suggestion:

Suggestion: construct the illustrated compartmental model in which $l_i(t)$ indicates the quantity of
pollutant in lake $i$ at time $t$, $V$ is the volume of each of the lakes and $r$ is the flow rate between 2 lakes

and deduce the equations
\begin{align}
    \frac{dl_1(t)}{dt}&= -r\frac{l_1}{V}+r\frac{l_3}{V}\nonumber\\
    \frac{dl_2(t)}{dt}&=-r\frac{l_2}{V}+r\frac{l_1}{V}\nonumber\\
    \frac{dl_3(t)}{dt}&=-r\frac{l_3}{V}+r\frac{l_2}{V}\nonumber
\end{align}

with the initial condition $l_1 (0) = 300000$ kg, $l_2 (0) = 0$ and $l_3 (0) = 0$. Indicate how to bring these three equations to the system of two equations that need to be resolved:
\begin{align}
    \frac{dl_1(t)}{dt}&= -r\frac{l_1}{V}+r\frac{300 000-l_1-l_2}{V}\nonumber\\
    \frac{dl_2(t)}{dt}&=-r\frac{l_2}{V}+r\frac{l_1}{V}\nonumber
\end{align}

I was wondering how can I transform the 3 differential equations into the 2 differential equations above? Thank you!

答:
Hint. Use linear algebra! Let $a = V/r \in \mathbb R$. Then we need to solve the following constant-coefficient linear system of ODE:
$$\begin{cases}
a\dot x(t) = - y(t) - z(t) \\
a\dot y(t) = - x(t) - z(t) \\
a\dot z(t) = - x(t) - y(t)
\end{cases} \iff a \dot{\mathbf x}(t)=a \begin{bmatrix} \dot x(t) \\ \dot y(t) \\ \dot z(t) \end{bmatrix} = -\begin{bmatrix} 0 & 1 & 1 \\  1 & 0 & 1\\  1 & 1 & 0\end{bmatrix}\begin{bmatrix}  x(t) \\  y(t) \\  z(t) \end{bmatrix} = \mathbf A \mathbf x(t).$$
Then you should try to diagonalize the matrix $\mathbf A = \mathbf M \mathbf J \mathbf M^{-1}$, change basis/coordinates from $\mathbf x$ to $\mathbf x':=\mathbf M \mathbf x$, solve the diagonalized system, revert to the original basis, and fix the constant by imposing the initial conditions.


问:
Prove for every $1 \leq n$, that $47\mid3^n4^{2n}-1$
I tried to solve with induction, base $n=1$ is correct such that, $\frac{3*16-1}{47}=1$.

Then assuming that for $n$ the statement is correct such that,  $3^n4^{2n} \equiv 0 \pmod {47}$.

We need prove that for $n+1$ is true such that, $\frac{(3*16*48^n)-1}{47}$ .

Then I tried to simplify it to $\frac{48}{47}*48^n-\frac{1}{47}= 1\frac{1}{47}*48^n-\frac{1}{47}=\frac{(3^n4^{2n})+47(3^n4^{2n})}{47}-\frac{1}{47}=\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$

We get that from $\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$, the expression  $\frac{(3^n4^{2n})} {47}-\frac{1}{47}$ is from the assumption so we are left with $3^n4^{2n}$ which is a natural number.

Is that proof enough to show that $47\mid3^n4^{2n}-1$?

答:
Hint:
No induction required. For a direct proof, use congruences $\bmod 47$ and observe that : $$3^n4^{2n}-1=(3\cdot 4^2)^n-1. $$


问:
Baire space is homeomorphic to countably many copies of itself
On wikipedia I found that the Baire space $\mathcal{N}$ is homeomorphic to the product of a countable number of copies of itself, however, I haven't been able to find a proof.

The Baire space is defined as $\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$ with the product topology, $\mathbb{N}$ being endowed with the discrete topology.

Can anyone help me, please?

答:
HINT: consider the left-shift map $$S:\mathcal{N}\rightarrow\mathcal{N}: f\mapsto \lambda x.f(x+1).$$

E.g. $S(0,1,2,3,4,5,...)=(1,2,3,4,5,...)$ and so on.

In more detail:

 For each $i\in\mathbb{N}$, let $\mathcal{N}_i=\{f\in\mathcal{N}: f(0)=i\}$. Now think about what $S\upharpoonright \mathcal{N}_i$ does, topologically.


Title: Prove for every $1 \leq n$, that $47\mid3^n4^{2n}-1$

Body:
I tried to solve with induction, base $n=1$ is correct such that, $\frac{3*16-1}{47}=1$.

Then assuming that for $n$ the statement is correct such that,  $3^n4^{2n} \equiv 0 \pmod {47}$.

We need prove that for $n+1$ is true such that, $\frac{(3*16*48^n)-1}{47}$ .

Then I tried to simplify it to $\frac{48}{47}*48^n-\frac{1}{47}= 1\frac{1}{47}*48^n-\frac{1}{47}=\frac{(3^n4^{2n})+47(3^n4^{2n})}{47}-\frac{1}{47}=\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$

We get that from $\frac{(3^n4^{2n})}{47}-\frac{1}{47}+3^n4^{2n}$, the expression  $\frac{(3^n4^{2n})} {47}-\frac{1}{47}$ is from the assumption so we are left with $3^n4^{2n}$ which is a natural number.

Is that proof enough to show that $47\mid3^n4^{2n}-1$?

Body: Hint : $3 \cdot 4^2 = 48 \equiv 1 \pmod{47}$


问:
Exponential distribution for continous rv
The time one has to wait $T$ (in hours) between the telephone calls one recieves is random with function $f(t) = (3/2)e^{-\frac{3}{2}t}$ for $t>=0$.

Find the expected time between recieved calls and find the corresponding variance.

For two independent waiting-times, $T_1$ and $T_2$, find the probability that both are greater than 1.

(1)

We can see that $T$ has an exponential distiribution with parameter $\lambda = \frac{3}{2}$ for a $t >= 0$. The expected time and variace are given by:

\begin{align}\\
& E(X)=\mu _X=\alpha\beta=\frac{1}{\lambda}=\frac{1}{\frac{3}{2}}=\frac{2}{3} \\
& V(X)=\sigma ^2_X=\alpha\beta^2=\frac{1}{(\frac{3}{2})^2}=\frac{4}{9} \\
\end{align}

2

We have:

\begin{align}\\
& \lambda=\frac{3}{2} && F(x;\lambda)=1-e^{-\lambda x} && \text{for} x\geqslant0 \\
\end{align}

and:

\begin{align}\\
& P(X > x) = 1 - P(X\leqslant x) \\
& P(X > x) = 1 - F(x;\lambda )   \\
& P(X > x) = 1 - (1-e^{-\lambda x}) \\
& P(X > x) = e^{-\lambda x}
\end{align}

But I don't know how to continue if I even have understood it correctly so far..

答: Part (1) is correct.  For part (2), all you have to do is compute the probability that $T_1 > 1$, i.e. $$\Pr[T_1 > t] = e^{-3t/2},$$ and then observe that since the times are independent and identically distributed, the probability that both $T_1$ and $T_2$ are greater than $1$ is simply the square of this probability.


Body:
Transform ($3^n4^{2n}-1) \to$ ($3^n16^n-1) \to (48^n-1)$

For sum of geometric series

$1+x+x^2+x^3+...+x^{n-1}=\frac{x^n-1}{x-1}$

or

$x^n-1=(x-1)(1+x+x^2+x^3+...+x^{n-1})$

Set x=48,

$48^n-1=(48-1)(1+48+48^2+48^3+...+48^{n-1})$

=
$47 (1+48+48^2+48^3+...+48^{n-1}) \to  47\mid3^n4^{2n}-1$


问:
discrete math monotonic functions question
i had this question to answer:

how many monotonic function $f:[1...k]\to[1...n]$, $(i>j \implies f(i)\geqslant f(j))$ are there?

my answer is: $\sum_{r_1=1}^n\sum_{r_2=r_1}^n\cdots\sum_{r_k=r_{k-1}}^n1$

is this the right answer?

Thanks in advance,

Yaron

答:
I believe you need to switch your k's and n's.

You have k inputs that you need to choose outputs for and the largest each of these could be is n while the lowest they can be is the output of the previous input.


问:
Milnor's proof of Sard's theorem
What does the notation mean in the last step of milnor's proof for Sard's theorem

$f(x+h) = f(x) + R(x+h)$

where $||R(x+h)|| \leq c||h||^{k+1}$

What do the lines || mean here?

答: If ${\bf x} \in \mathbb{R}^d$ is vector in $\mathbb{R}^d$, we usually use $\|{\bf x}\|$ to represent the Euclidean norm (or "length") of ${\bf x}$. Mathematically, suppose ${\bf x} = (x_1,x_2,\ldots,x_d)$, then $\|{\bf x}\| = \sqrt{\sum_{k=1}^d x^2_k}$

Title: Milnor's proof of Sard's theorem


问:
If $a-\sqrt{2021a} = 1$, find $a-45\sqrt{a}$.
If a is a positive real number such that $$a-\sqrt{2021a} = 1$$ evaluate $$a-45\sqrt{a}$$
I've factored the $2021$ into $43\cdot 47$ and then $(45+2)(45-2)$ so $$a -\sqrt{(45^2-2^2)a} = 1$$ but I can't figure out what to do next.

答:
Let $t = \sqrt a$. Then, $$a - \sqrt{2021a} = 1$$
becomes $$t^2 -  \alpha t - 1 = 0$$
where $\alpha^2 = 45^2 - 4$. Solving this for $t$, we get
$$t  = \frac{\alpha  \pm \sqrt{\alpha^2 + 4}}{2} = \frac{\alpha \pm 45}{2} \tag{*}$$
We take only $"+"$ from $"\pm"$ since $t > 0$.

Now, we need to find the value of $t^2 - 45t$.

We subtract $45$ from both sides of $(*)$, to get $$t - 45 = \frac{\alpha - 45}{2}$$ Multiply by $t$:
$$t^2-45t = \frac{\alpha - 45}{2}t = \frac{\alpha - 45}{2}\frac{\alpha + 45}{2} = \frac{\alpha^2 - 45^2}{4} = \frac{-4}{4} = -1$$


答:
We have $3^n4^{2n}-1 = 48^n-1$.

Using induction:

$n=1: 48^1-1=47$ is divisible by $47$ and

$48^{n+1}-1 = 48(48^n-1)+47$ and $48^n-1$ is divisible by $47$ per induction hypothesis.


问:
How can one solve this optimization problem by rewriting the matrix?
Given this problem:

$$maximize \, \, x^T\begin{bmatrix}
3 & 4 \\
0 & 3
\end{bmatrix}x
$$

$$ subject \,to \, \, ||x||^2 = 1$$

I want to solve this using Langrange Conditions. What I wish to do is find the eigenvalues of a symmetric matrix $Q$, and the corresponding eigenvectors to maximize this problem.

In the solution material the textbook author rewrites the matrix: $\begin{bmatrix}
3 & 4 \\
0 & 3
\end{bmatrix} $ into $ Q = \begin{bmatrix}
3 & 2 \\
2 & 3
\end{bmatrix}$ , where $Q^T = Q$.  How and why can one do this?

(Afterwards one can calculate the eigenvalues of $Q$ and find the corresponding eigenvectors)

Thank you.

答:
That manipulation is valid since $$x^T\begin{bmatrix}3&4\\0&3\end{bmatrix}x = 3x_1^2+4x_1x_2+3x_2^2 = 3x_1^2+2x_1x_2+2x_2x_1+3x_2^2 = x^T\begin{bmatrix}3&2\\2&3\end{bmatrix}x.$$

In general, if $A \in \mathbb{R}^{n \times n}$ and $x \in \mathbb{R}^n$, then $x^TAx$ is a scalar. Hence, $x^TAx = (x^TAx)^T = x^TA^Tx$, and thus, $$x^TAx = \dfrac{1}{2}x^TAx+\dfrac{1}{2}x^TA^Tx = x^T\left(\dfrac{A+A^T}{2}\right)x.$$


问:
Random projection a fixing vector is equivalent to projecting a random vector
In the book Vershynin the following exercise is presented:

Let $P$ be a projection in $\mathbb{R}^n$ onto a random $m$-dimensional subspace $E \sim \mbox{Unif}(G_{n,m})$, where $G_{n,m}$ is the Grassmanian, the set of $m$-dimensional linear subspaces of $\mathbb{R}^n$. Let $Q$ be a $m \times n$ matrix obtained by choosing the first $m$ rows of a random $n \times n$ matrix $\sim \mbox{Unif}( O(n) )$, namely drawn uniformly from the orthogonal group.

(a) show that for any fixed point $x \in \mathbb{R}^n$, $\| P x \|_2$ and $\|Q x\|_2$ have the same distribution,

(b) show that, for any fixed point $z \in \mathbb{S}^{m-1}$,
$$
Q^T z \sim \mbox{Unif} ( \mathbb{S}^{n-1}),
$$
where $\mathbb{S}^{n-1} \subset \mathbb{R}^n$ is the $n-1$ dimensional unit sphere and here Unif means uniform distribution in $\mathbb{S}^{n-1}$.

I think that one should use the singular value decomposition for $P$ and make use of the rotational invariance but I am not able to figure our how. Does anyone have a suggestion?

答:
For (a), one approach is to show that the distribution of $\|Px\|_2$ where $P$ is random and $x$ is fixed is the same as the distribution of $\|Px\|_2$ when $P$ is fixed and $x$ is random. All this really uses is rotational invariance (i.e., invariance under orthonormal changes of basis).

Formally, we proceed as follows: Let $S\in O(n)$ be fixed, and let $P$ be drawn randomly from the Grassmanian manifold $G_{n,m}$. Because multiplication by $S$ is an orthonormal change of basis and the Grassmanian does not depend on choice of basis,  $P\sim S^\top P S$. Because this is true for any fixed $S$, if we take $T\sim \mathrm{Unif}(O(n))$, then $P\sim T^\top P T$ as well. So
$$ \|Px\|_2\sim \|T^\top PT x\|_2 = (PTx)^\top TT^\top (P T x) = \|PTx\|_2. $$
Let $y = Tx$. I claim $\|Py\|_2\sim\|\pi y\|_2$ for any fixed $P\in G_{n,m}$, where $\pi\in G_{n,m}$ is the projection onto the first $m$ coordinates. Since $P$ is a projection, it has singular value decomposition $P =  S^\top\pi S$ for $S\in O(n)$. It follows that $\|Py\|_2=\|\pi S y\|_2\sim\|\pi y\|_2$, where we use the fact that $S T \sim T$ because $\mathrm{Unif}(O(n))$ is invariant under orthonormal changes of basis. To finish, observe that we can write $Q = \pi T$, so $Qx\sim \pi Tx = \pi y$.

For (b), we have $Q^\top z\sim (QT)^\top z = T^\top (Q^\top z)$, for $T\sim\mathrm{Unif}(O(n))$, using similar reasoning as above.


问:
Is there a general summation formula for the polygamma function at $z=1/2$? i.e $\psi^{(s)}(\frac{1}{2})$ for all s.
For $s>0$ one has $ \psi^{(s)}(\frac{1}{2}) =  s! \cdot \zeta(s+1, \frac{1}{2}) \cdot (-1)^{s+1} $.

E.g.

$ \psi^{(1)}(\frac{1}{2}) = 3 \cdot \zeta(2) $

$ \psi^{(2)}(\frac{1}{2}) = -14 \cdot \zeta(3) $

$ \psi^{(3)}(\frac{1}{2}) = 90 \cdot \zeta(4) $

etc.

However, Mathematica gives

$ \psi^{(-1)}(\frac{1}{2}) = \frac{\ln(\pi)}{2} $

$ \psi^{(-2)}(\frac{1}{2}) = \ln(A^\frac{3}{2}2^\frac{5}{24}\pi^\frac{1}{4}), $ where $A$ is the Glaisher-Kinkelin Constant

$ \psi^{(-3)}(\frac{1}{2}) = \ln(A^\frac{1}{2}2^\frac{1}{16}\pi^\frac{1}{16}) + \frac{7 \zeta(3)}{32\pi^2} $

and I'm not sure how these are obtained since negative integers plugged into the formula above yields wonky results.

答:
First off we have for reference this formula which states for all $n\in\Bbb N$
$$
\psi ^{(n)}\left(\tfrac{1}{2}\right)=(-1)^{n+1} \left(2^{n+1}-1\right) n! \zeta (n+1).
$$
Note that this is not the same formula as given in your question statement. As for negative values, I was not able to find a closed-form for all $-n\in\Bbb N$ at $z=1/2$ but there is one for $-2n$ with $n\in\Bbb N$ and $z=1/2$ given here.
Furthermore, we do have a formula for general $-n$ and $z$ given here which can be evaluated at $z=1/2$. With a little extra work you could further simplify this to the special cases you obtained using Mathematica using the well-known properties of the polygamma function.


问:
Duality and Concreteness in Category Theory
I'm sorry for the silly question, but is it true that the opposite of a concrete category is also concrete?

Moreover, if I have a category and a faithful functor from it to a concrete category, may I deduce by composing the faithful functors that even the starting category is concrete?

答: There is small subtlety to pay attention to here. If by $C$ is concrete you mean that there exists a faithful functor to sets, then your question is answered in the comments. However, if by $C$ is concrete you mean "I have chosen a faithful functor to sets", then things are different. The question is what do we take the category of concrete categories to be. Are its objects concrete categories but with no specified functor to sets, or is it concrete categories with a choice of such a functor. In the latter case, the opposite is not concrete since there is no canonical functor $\mathbf {Set}^\mathrm {op}\to \mathbf {Set}$. In that case the opposite of a concrete category of $\mathbf {Set} $ is a concrete category over $\mathbf {Set}^\mathrm {op}$.


问:
Explore the conditions under which the measures of two angles in a triangle are kept to be constants
Given A,C are two points on DP and DQ of  $\triangle DPQ$, B is the intercept of AQ and CP, PQ=AP+CQ,  $\angle \alpha+\angle \beta=180^o$.  If also given  AP=CQ, it can be proven $\angle PDQ$ = 60$^o$ and $\angle PBQ$=120$^o$.

Question:

If the requirement of  AP=CQ removed, are there any other conditions that   can lead to the same conclusion  $\angle PDQ$ = 60$^o$ and $\angle PBQ$=120$^o$?

It seems that we need to construct something to link PQ,AP and CQ together, need help to continue.

答:
The suggested solution assumes some basic knowledge.

Let me know if this is what you have in mind and if all is clear.


问:
Prove that the $\lim _\limits{x \to 1} (4/(3x+2))= 4/5$.
Use the definition of a limit to prove that the  $\lim _\limits{x \to 1} \frac 4 {3x+2}= \frac 4 5$.
For my scratch work I have:
$$|4/(3x+2) - 4/5| = |20/5(3x+2)-4(3x+2)/5(3x+2)| = |-12x-12/15x+10|$$

This is where I am stuck. I know somehow I need to get it so that $(x-1)$ is in there somehow but if I factor at a $12$ or $-12$ I get $12(-x-1)$ or $-12(x+1).$

I know I need to get $(x-1)$ somewhere so I can suppose that $|x-1| < 1$ so I can find the values that the other factor can fall into which I can then use to set my $\gamma = \min \{1,\_\_ \}.$

答:
hint

Given $ \epsilon >0$, we look for small $ \eta>0 $, smaller than $ 1$ by adding the condition

$$|x-1|<1 \iff 0<x<2$$
$$\iff \frac{1}{40}\frac{1}{5(3x+2)}<\frac{1}{10}$$

such that

$$|\frac{12}{5(3x+2)}|<\frac 65$$


问:
Showing a collection of sets forms a semiring

Definition: A nonempty collection $S$ of subsets of $X$ is a semiring if for any two sets
$A,B\in S$,
$$A\cap B\in S\tag{a}$$
and there is a finite disjoint collection $\{s_k\}_{k=1}^{n}$ with $s_k\in S$, so that
$$A\setminus B = \bigcup_{k=1}^{n}s_k\tag{b}$$

Question: Consider the collection $S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $\mathbb{R}$. I want to show that $S$ forms a semiring.

For property $(a)$; $\emptyset$ intersected with any set in $S$ is again $\emptyset$ which is in $S$, $[0,1]$ intersected with any set in $S$ is either the $\emptyset$ or $[0,1]$ both are in $S$, $[0,3]$ intersected with any set in $S$ is either the $\emptyset$ or $[0,1]$ or $[2,3]$ all in $S$, and lasty $[2,3]$ intersected with any set in $S$ is either the $\emptyset$ or $[2,3]$ all in $S$. Therefore, property $(a)$ holds.

For property $(b)$; $\emptyset$ set-minus any set in $S$ is empty hence the finite disjoint collection
$\{\emptyset\}$ works, $[0,1]$ set-minus any set in $S$ is either $\emptyset$ or $[0,1]$ hence the finite disjoint collection $\{\emptyset\}$ or $\{[0,1]\}$ works respectively, $[0,3]$ set-minus any set in $S$ is either $[0,3]$ or $[2,3]$ or $[0,1]$ hence the finite disjoint collection $\{[0,3]\}$ or $\{[2,3]\}$ or $\{[0,1]\}$ works respectively, and lastly $[2,3]$ set-minus any set in $S$ is either $[2,3]$ or $[0,1]$ hence the finite disjoint collection $\{[2,3]\}$ or $\{[0,1]\}$ works respectively.

My question is simply, is this correct?

答:
$S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $\mathbb{R}$ is not a semiring.

Note that $[0,3] \setminus [0,1] = (1,3]$ and $(1,3]$  can not be written as a finite union of elements in $S$.

Also, $[0,3] \setminus [2,3] = [0,2)$ and $[0,2)$  can not be written as a finite union of elements in $S$.

Regarding your argument:

You wrote "$[0,3]$ set-minus any set in $S$ is either $[0,3]$ or $[2,3]$ or $[0,1]$". This is not true.

You also wrote "$[2,3]$ set-minus any set in $S$ is either $[2,3]$ or $[0,1]$. That is incorrect. The right statement would be "$[2,3]$ set-minus any set in $S$ is either $\emptyset$ or $[2,3]$.


问:
Derivation of Poisson PDF from Binomial: questions on a Wiki proof
Wiki has a proof and I have questions regarding it.

a)
$$
\lim_{n\rightarrow\infty} {n \choose k} p^k_n(1-p_n)^{n-k} = \lim_{n \rightarrow \infty} \frac {n(n - 1)(n - 2) \cdots (n - k + 1)}{k!} \left ( \frac \lambda n\right)^k \left(1 - \frac \lambda n \right)^{n-k}
$$
Question 1 how was the $(n -k +1)$ term in the numerator derived?

b)
$$
= \lim_{n\rightarrow \infty} \frac {n^k + O(n^{k-1})}{k!} \frac {\lambda^k}{n^k} \left(1 - \frac \lambda n \right)^{n-k}
$$
c)
$$
= \lim_{n\rightarrow\infty} \frac {\lambda^k}{k!} \left(1 - \frac \lambda n\right)^{n-k}
$$

Question 2 how does form (b) become (c)?  Does $\frac {n^k + O(n^{k-1})}{n^k} = 1$ somehow?  If big-O $O(n^{k-1}) = (\lim_{n\rightarrow\infty} n^{k-1}) \le C n^{k-1}$, then (b) becomes $\frac {n^k + Cn^{k-1}}{n^k} = 1 + C n^{-1}$.  I'm confused here.

答:
Question 1: See big O notation.

Question 2:

We have
$$\frac {n!}{k!(n -k)!}=\frac {n(n-1)(n-2)(n-3)...(n-k+1)}{k!}$$

This really is obvious ($n!$ is series of products from $n$ to one and $(n-k)!$ a series from $n-k$ to one), but we can also use induction to prove it. If $k=1$, then

$$\frac {n!}{k!(n -k)!}=\frac {n!}{1!(n -1)!}=\frac {n (n-1)!}{(n -1)!}=n=\frac{(n-k+1)}{k!}$$

Let now the relation hold for $k$, and we will show that this holds for $k+1$. We have

$$\frac {n!}{(k+1)!(n -(k+1))!}$$
$$=\frac {(n-k)n!}{(k+1)k!(n -k)!}$$
$$=\frac {n (n-1)(n-2)...(n-(k+1)+1)}{(k+1)!}$$

This gives the claim.


问:
A question about the following theorem: $c(A\cup B)=c(A)\cup c(B)$.
Let there be a point $p\in(A\cup B)'$ such that some open sets containing it contain points in $A\setminus B$, and other open  sets containing it contain points in $B\setminus A$.

$A\setminus B=A-(A\cap B)$

Then $p$ is not a limit point of either $A$ or $B$. However, it is a limit point of $A\cup B$!! Is the theorem $c(A\cup B)=c(A)\cup c(B)$ untrue then?

EDIT: $c(A)$ is the closure of set $A$ in a topological space. $A'$ is the complement of set $A$ in the topological space

答:
Let $X$ be our topological space. Any point $p\in X$ whatsoever has the property that

some open sets containing it contain points in $A\setminus B$, and other open sets containing it contain points in $B\setminus A$

because $X$ is an open set, and contains every point of the topological space.


问:
Expression for $\{B,e^A\}$ where $\{[B,A],A\}=0$
I found this problem in a quantum field theory class, talking about fermionic modes. $B$ and $A$ are operators that obey anti-commutation relations. I have tried to solve it doing the following:
\begin{align*}
\{B,e^A\}=\{B,\sum_n \frac{A^n}{n!}\}=\sum_n\frac{1}{n!}\{B,A^n\}
\end{align*}
So the problem is about calculating $\{B,A^n\}$ under the condition $\{[B,A],A\}=0$.

答:
A sketch:

We have $[B,\,A^2]=0$. Use $\{B,\,UV\}=[B,\,U]V+U\{B,\,V\}$, so $\{B,\,A^2V\}=A^2\{B,\,V\}$ and $\{B,\,A^{2n}V\}=A^{2n}\{B,\,V\}$, to show$$\{B,\,e^A\}=(\cosh A)\underbrace{2B}_{\{B,\,I\}}+(\operatorname{sinch}A)\{B,\,A\},$$provided we have the necessary convergence conditions to suitably manipulate infinite series.


问:
Path lifting lemma - proof clarification
So i was studying the proof of the 'path lifting lemma' on my book and I got stuck.
Here $ S^1 $ is identified with complex numbers of norm 1 and $ E: \mathbb{R} \rightarrow S^1 $ is defined by $ E(x)= e^{2\pi ix}, x\in \mathbb{R}$.

Lemma: If $ \alpha $ is a path in $ S^1 $ with starting point  1, there exist a unique path $\alpha '$ in $ \mathbb{R} $ with starting point $0$ such that $ E \alpha ' = \alpha $.

Proof: Because I is compact and $ \alpha $ is continuous, we can choose subintervals of  I, $ [t_0,t_1],...[t_{n-1},t_n],\  0=t_0 < t_1$ $ <...< t_n=1$ such that $$|\alpha (t) - \alpha (t')| < 1 \; \forall t,t' \in [t_{i-1},t_i] \;\; \forall i=1...n
\; \; \; \; \; \; \;\; \; \; \; \; \; \;\; \; \; \; \; \; \;
\textbf{(1)} $$
and so such that $$ \alpha ([t_{i-1},t_i])\mbox{ is contained in } S^1 \backslash \{ - \alpha (t_{i-1}) \} \;\; \forall i=1...n
\; \; \; \; \; \; \;\; \; \; \; \
\textbf{ (2) }
$$...(the proof goes on)

I'm ok with the rest of the proof (that I didn't write) but I don't understand this beginning: why can we choose such intervals? Where the compactness of I and the continuity of $ \alpha $ are used? Why (1) implies (2) ?

Thanks to everyone.

答:
Fix a path $\alpha: I\to S^1$. Note that you can reformulate the statement as follows: "It is possible to subdivide the interval into pieces $[0,1] = [0,t_1]\cup [t_1,t_2]\cup...\cup[t_{n-1},t_n]$  such that the image $\alpha([t_i,t_{i+1}])$ of any piece is contained in an open ball of radius say $1/4$. " My statement is stronger than the one from the book.

Here is how to prove this. Let $\mathcal U$ be the open cover of $S^1$ consisting of all open balls of radius $1/4$. Set $\mathcal V = \{\gamma^{-1}(U)\,|\, U\in \mathcal U\}$. Note that $\mathcal V$ is an open cover of the interval $I$. The result now follows from the Lebegue-number lemma: Explanations of Lebesgue number lemma. There is a number $1/n >0$ such that each set of diameter at most $1/n$ is contained in a set from the cover $\mathcal V$. The partition $[0,1/n] \cup [1/n,2/n]\cup...$ will do the job. $[t_i,t_{i+1}] \subset \gamma^{-1}(U)$ for some $U \in \mathcal U$, hence $\gamma([t_i,t_{i+1}])\subset U$ and $U$ is an open ball of radius $1/4$.


问:
Is this series well defined?
Let's consider functional following:

$$f : l^2 \ni(x_n)_{n=1}^\infty \rightarrow \sum_{n=1}^\infty\frac{x_n}{\sqrt{n}}$$

I'm trying to prove or disprove fact - if this series well defined ?

Because $x_n \in l^2$, then:

$$\sqrt{x_1^2+x_2^2+x_3^2+...} <\infty$$

I wasn't sure how to judge on definition of $f$. It's quite obvious when it's in $l^1$ - we just need to consider absolut convergence:

$$\sum_n |\frac{x_n}{\sqrt{n}}| \le \sum_n|x_n| \;\;\;\; \text{And this would be convergent if $x_n \in l^1$}$$

However, we have $l^2$ space and I don't see how we can use this fact. Could you please give me a hand doing so ?

答:
No, let $x_n = \dfrac{1}{\sqrt n \ln(n+1)}.$ Then $(x_n)\in l^2.$ But

$$\sum_{n=1}^{\infty}\frac{x_n}{\sqrt n}=\sum_{n=1}^{\infty}\frac{1}{n\ln(n+1)} = \infty.$$

Title: Is this series well defined?


答:
There is no such point: Let $p \in c(A \cup B)$, then any open set containing $p$ contains points of $A$ or $B$. Suppose there is an open set $U$ such that $p \in U$, $U \cap A = \emptyset$ and another open $V$ such that $p \in V$, $V \cap B\ne\emptyset$. Then $U \cap V$ is open $p \in U \cap V$, but $(U \cap V) \cap (A \cup B) = \emptyset$, contradicting $p \in c(A \cup B)$.

Hence, a point as you want, cannot exists, therefore $c(A \cup B) = c(A) \cup c(B)$.

Title: A question about the following theorem: $c(A\cup B)=c(A)\cup c(B)$.

Body:
Let there be a point $p\in(A\cup B)'$ such that some open sets containing it contain points in $A\setminus B$, and other open  sets containing it contain points in $B\setminus A$.

$A\setminus B=A-(A\cap B)$

Then $p$ is not a limit point of either $A$ or $B$. However, it is a limit point of $A\cup B$!! Is the theorem $c(A\cup B)=c(A)\cup c(B)$ untrue then?

EDIT: $c(A)$ is the closure of set $A$ in a topological space. $A'$ is the complement of set $A$ in the topological space


问:
Rewriting function
Given function
$$
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
$$
and
$$
g(x) = \frac{2}{1+\sqrt{x+1}}
$$
Prove that $f(x) = g(x)$ where $x\neq 0$.

This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows:
$$
f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2x}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2}{\sqrt{x+1}}
$$

However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.

EDIT:
Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.

答:
Hint: consider the following identity
$$
(\sqrt{x+1} + 1)(\sqrt{x+1} - 1) = (\sqrt{x+1})^2 - 1^2 = x + 1 - 1 = x
$$

Title: Rewriting function


问:
Closed form solution for $a^x - b^x = 1$
I'm looking for a closed form solution for $a^x - b^x = 1$ where $a,b\in \mathbb{R}$ are known and $x$ is unknown. There is a similar problem with positive sign instead of minus. Here are my attempts:

Let $b = a+\epsilon $ where $\epsilon = b-a$ and use Newton's generalized binomial theorem to get $$\sum_{i=1}^\infty\binom{x}{i}\epsilon^ia^{x-i} = -1$$but I don't know how to proceed. Also I think for $a\gt b \gt 0$ there is a unique solution and for $0\lt a\lt b$ there is no solution since $a^x$ is an increasing function. We can rewrite the equation as $$a^{-x}(1+b^x) = 1 \implies -x\ln a + \ln(1+b^x) = 0 \implies x =\frac{\ln(1+b^x)}{\ln a}$$Assuming that we can use Taylor series for $\ln(1+b^x)$, we have $$x = \frac{1}{\ln a}(b^x-\frac{b^{2x}}{2}+\frac{b^{3x}}{3}-\frac{b^{4x}}{4}+\cdots)$$and this seems useless.

答:
Add $b^x$ to get $$a^x = 1+b^x$$ then divide by $a^{x}$ to get $$1 = \left(\frac{1}{a}\right)^x+\left(\frac{b}{a}\right)^x$$

Assuming that $a, b>0$, if $\frac{1}{a}, \frac{b}{a}<1$, there is a solution. If only one out of $\frac{1}{a}, \frac{b}{a}$ is less than $1$, there will be no solution, and if both are greater than $1$, the solution will be the negative of the solution for $1=a^x + \left(\frac{a}{b}\right)^x$.

This question is now the same as the linked question. Let's say we are trying to solve $$1=a^x + \left(\frac{a}{b}\right)^x$$

Using this answer, if $b < 1$, let $p = \frac{\ln(a/b)}{\ln(a)}$ and $u = a^x$. Otherwise, let $p = \frac{\ln(a)}{\ln(a/b)}$ and $u = \left(\frac{a}{b}\right)^x$. Then $$u = \sum_{k=0}^\infty \frac{\Gamma(pk+1)(-1)^k}{\Gamma((p-1)k+2) k!}$$


问:
Probability of exactly 1 given everything appears
I'm trying to figure out the following question, and I'm stuck! It comes from an old PUMaC problem with a partial solution (there's a typo in the one provided on Stack Exchange as well):

Stack Post:

Collecting cards

Original Problem and Solution (problem 4):

https://static1.squarespace.com/static/570450471d07c094a39efaed/t/5de43032c573c502d7fe2f7d/1575236145801/2019Combo_A_Sols.pdf

If I'm not mistaken, I think what the problem is saying is that if you have cards $\{A,B,C,D,E,F\}$, drawn independently and with equal probability 1/6 (uniformly), what is (1 minus) the probability that you obtain exactly one A, given that you stop drawing when you've drawn every card? (We can choose A w.l.o.g. by law of total probability).

e.g. $(A,B,C,D,E,E,F)$, $(B,B,C,C,D,F,F,E,E,A)$ <-- these draws work.

According to their partial solution, it seems not so much of a coincidence that we see the harmonic series as would be in the case if we found E(number of draws until all 6 appear) using linearity of expectation. But, beyond that I'm not sure how they achieved this answer.

When I tried to go about it, it seemed like it would take significant case work (dependent on the number of draws and whether or not A is the last to appear).

答:
Consider pulling cards until you pull $A$ twice. Note that the order of $A$ is equally likely to be the first, second, ..., sixth. Now, condition on the order of $A$, let it be the $n$th unique card. Let us assume that your position $n$ is not equal to $6$. Now, $A$ is again equally likely to be the last card among the remaining $(6 - n)$ cards, giving a probability of $A$ being last $1/(6-n + 1)$.

For $A$ to be drawn once in your scenario, $A$ had to be the last card in the first or the second draw (do you see why?), giving the final answer $(1/6) + (1/6) \sum_{n=1}^5 1/(6-n + 1)$, which matches the formula they gave. I believe they used a different counting argument, but since their solution is incomplete, and this one is simpler, you might as well use this.

As an important side note, I think the better way to think about this problem to understand all the symmetries and independences is by considering a Poisson splitting (that is how I solved it). It is a hard problem without Poisson processes for sure.