问:
How would I take a given triple integral in rectangular coordinates and rewrite it in cylindrical and spherical coordinates?

This is the given integral, which is currently in rectangular coordinates, how would I rewrite this in spherical and cylindrical coordinate systems?

答:
Hint:

$\sqrt{x^2 + y^2} \leq z \leq 1$, $\, 0 \leq y \leq \sqrt{1 - x^2}, 0 \leq x \leq 1$

So it is a cone with vertex at the origin and $z = r$ (consider inverted cone). The upper bound of $z$ is $1$ and you see $x^2 + y^2 \leq 1$. You should also consider the lower bound of $x$ and $y$ which means you are only integrating over the region in first octant.

So the bounds in cylindrical coordinates, for example, will be straightforward

$r \leq z \leq 1$, $\, 0 \leq r \leq 1, 0 \leq \theta \leq \frac{\pi}{2}$

Can you now try and do the same for spherical coordinates? Just think what happens as you change the polar angle and what does it do to the limits of integration? Till certain polar angle, $\rho$ will be defined by the top of your cone and beyond that by the ray of the cone. $z = r$ will play an important role in defining the limits of polar angle and splitting your integral. Please sketch so it becomes clear.


问:
Transform roll, pitch, yaw from one coordinate system to another
I'm trying to figure out how to transform a pose given with Euler angles roll (righthanded around X axis), pitch (righthanded around Y axis), and yaw (lefthanded around Z axis) from the Unreal Engine into a pose in a purely right-handed North, West, Up coordinate frame.

I know the full rotation matrix is $R = R_z R_y R_x$, as Euler angles are applied in that order.

But, irritatingly, the Y axis switches direction, and the whole coordinate system is rotated.

How can I elegantly determine pose in my target reference frame? I'm open to quaternions.

答:
There's actually a really neat identity for this particular problem.

Yaw, pitch, and roll are applied in that order to recover a body frame from the world frame. If I have $yaw = \psi$, $pitch = \theta$ and $roll = \phi$ in the UE4 reference frame, then I can get to my NWU coordinate frame in two stages.

First let's transform to a frame I'm calling $C$.

Here the coordinate axes all still span the same dimensions, so we can find $R_{Cb} = R_z(-\psi)R_y(-\theta)R_x(\phi)$, the same ZYX ordering as before; we just have to negate the pitch and yaw angles because we've flipped the direction of $+Y$, which reverses the direction of rotation, and we choose to reverse the rotation around $Z$ to make $C$ entirely right-handed.

Note that the body frame also gets these changes, but its orientation of course doesn't change.

Now we can find $R_{NWUC} = R_z(-\frac{\pi}{2})$, because the $R_y$ and $R_x$ terms are $I$.

This time the body frame won't change at all, because it's already similar to the world frame.

In general we have to find

$$ R_{NWUb} = R_{NWUC}R_{Cb} = R_z(-\frac{\pi}{2})R_z(-\psi)R_y(-\theta)R_x(\phi) = R_z(-\frac{\pi}{2}-\psi)R_y(-\theta)R_x(\phi) $$

This is because to find the cumulative rotation matrix you right multiply and follow some canceling rules.

And in the end, since you don't have to change the ZYX order of application, you're left with a rotation matrix with Euler angles you can easily pick out: $yaw = -\frac{\pi}{2} - \psi$, $pitch = -\theta$, and $roll = \phi$.

To verify this (particularly the right-multiplying, which defies my intuition), I wrote a little script to visualize the example of a south-easterly-facing camera pitched downward at 20 degrees. This would have a $psi = +45^{\circ}$, $\theta = -20^{\circ}$, and $\phi = 0^{\circ}$ in UE4 coordinates:

```<code>D2R = numpy.pi/180
R2D = 180/numpy.pi

R_z = R_(-135*D2R, 'z') # implementation of the three 3D rotation matrices
R_y = R_(20*D2R, 'y')

RR = R_z.dot(R_y)
print(RR)

angles = recover_angles(RR) # https://www.gregslabaugh.net/publications/euler.pdf
for a in angles: print(a*R2D)

fig = pyplot.figure()
ax = fig.gca(projection='3d')
ax.plot([0,5],[0,0],[0,0],c='k')
ax.plot([0,0],[0,5],[0,0],c='k')
ax.plot([0,0],[0,0],[0,5],c='k')

ax.plot([0,RR[0,0]],[0,RR[1,0]],[0,RR[2,0]],c='r') # recall columns of R are
ax.plot([0,RR[0,1]],[0,RR[1,1]],[0,RR[2,1]],c='g') # unit coordinate vectors
ax.plot([0,RR[0,2]],[0,RR[1,2]],[0,RR[2,2]],c='b') # of frame b in world frame

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
pyplot.show()
</code>```

which returns:

That's the rotation matrix and recovered Euler angles (listed roll, pitch, yaw order) from an implementation of this function.

And here are the NWU axes in black with the body axes in color. This visualization is based on the fact that the columns of the rotation matrix are unit-length vectors pointing in the directions of the body frame's coordinate axes.

ContentLicense: CC BY-SA 4.0

Body:
There's actually a really neat identity for this particular problem.

Yaw, pitch, and roll are applied in that order to recover a body frame from the world frame. If I have $yaw = \psi$, $pitch = \theta$ and $roll = \phi$ in the UE4 reference frame, then I can get to my NWU coordinate frame in two stages.

First let's transform to a frame I'm calling $C$.

Here the coordinate axes all still span the same dimensions, so we can find $R_{Cb} = R_z(-\psi)R_y(-\theta)R_x(\phi)$, the same ZYX ordering as before; we just have to negate the pitch and yaw angles because we've flipped the direction of $+Y$, which reverses the direction of rotation, and we choose to reverse the rotation around $Z$ to make $C$ entirely right-handed.

Note that the body frame also gets these changes, but its orientation of course doesn't change.

Now we can find $R_{NWUC} = R_z(-\frac{\pi}{2})$, because the $R_y$ and $R_x$ terms are $I$.

This time the body frame won't change at all, because it's already similar to the world frame.

In general we have to find

$$ R_{NWUb} = R_{NWUC}R_{Cb} = R_z(-\frac{\pi}{2})R_z(-\psi)R_y(-\theta)R_x(\phi) = R_z(-\frac{\pi}{2}-\psi)R_y(-\theta)R_x(\phi) $$

This is because to find the cumulative rotation matrix you right multiply and follow some canceling rules.

And in the end, since you don't have to change the ZYX order of application, you're left with a rotation matrix with Euler angles you can easily pick out: $yaw = -\frac{\pi}{2} - \psi$, $pitch = -\theta$, and $roll = \phi$.

To verify this (particularly the right-multiplying, which defies my intuition), I wrote a little script to visualize the example of a south-easterly-facing camera pitched downward at 20 degrees. This would have a $psi = +45^{\circ}$, $\theta = -20^{\circ}$, and $\phi = 0^{\circ}$ in UE4 coordinates:

D2R = numpy.pi/180
R2D = 180/numpy.pi

R_z = R_(-135*D2R, 'z') # implementation of the three 3D rotation matrices
R_y = R_(20*D2R, 'y')

RR = R_z.dot(R_y)
print(RR)

angles = recover_angles(RR) # https://www.gregslabaugh.net/publications/euler.pdf
for a in angles: print(a*R2D)

fig = pyplot.figure()
ax = fig.gca(projection='3d')
ax.plot([0,5],[0,0],[0,0],c='k')
ax.plot([0,0],[0,5],[0,0],c='k')
ax.plot([0,0],[0,0],[0,5],c='k')

ax.plot([0,RR[0,0]],[0,RR[1,0]],[0,RR[2,0]],c='r') # recall columns of R are
ax.plot([0,RR[0,1]],[0,RR[1,1]],[0,RR[2,1]],c='g') # unit coordinate vectors
ax.plot([0,RR[0,2]],[0,RR[1,2]],[0,RR[2,2]],c='b') # of frame b in world frame

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
pyplot.show()

which returns:

That's the rotation matrix and recovered Euler angles (listed roll, pitch, yaw order) from an implementation of this function.

And here are the NWU axes in black with the body axes in color. This visualization is based on the fact that the columns of the rotation matrix are unit-length vectors pointing in the directions of the body frame's coordinate axes.


问:
Equivalency of Modulo function
Is it valid to take modulo of both sides of a complex eqn? I can't seem to pinpoint why, but say we want to find the solutions of this expression

$$2z^2=-4 \\ |2z^2|=|-4| \\ |z^2| = 2 \\ |z| = \sqrt{2}$$

However, it is fairly obvious there are many points on $|z|=\sqrt{2}$ that don't satisfy the first equation? $2(\sqrt{2})^2 \neq -4$

答: This is because you have solved the equation only for the amplitude of the complex variable $z$. Any complex variable also comes with a phase, so you also need to solve it for the phase of $z$. Just express $z$ as $z=re^{i\theta}$ where $r=\sqrt 2$ as you implied correctly and find the constraints on $\theta$.

Title: Equivalency of Modulo function


问:
A subspace of the $H_{0}^1(\Omega) \cap H^2(\Omega)$ is dense in a subspace of $L^2(\Omega)$?
Let $n \in \mathbb{N}$ and $\Omega \subset \mathbb{R}^n$ bounded open and of class $C^1$. Consider the spaces $L^2(\Omega)$, $H_0^1(\Omega)$ and $H^2(\Omega)$.

I know that $H_0^1(\Omega) \cap H^2(\Omega)$ is densely embedded in $L^2(\Omega)$ and $H_0^1(\Omega)$ is also densely embedded in $L^2(\Omega)$. Now, define
$$H_{0,m}^1(\Omega):=\left\{ f \in H_0^1(\Omega) \; ; \; \int_{\Omega} f(x) \;dx=0\right\} \quad \text{and} \quad L_{m}^2(\Omega):=\left\{ f \in L^2(\Omega) \; ; \; \int_{\Omega} f(x) \;dx=0\right\} .$$

I know that $H_{0,m}^1(\Omega)$ is closed subspace of the $H_0^1(\Omega)$, $L_{m}^2(\Omega)$ is a closed subspace of the $L^2(\Omega)$ and $H_{0,m}^1(\Omega) \cap H^2(\Omega)$ is a subspace of $L_{m}^2(\Omega)$ . Moreover $H_{0,m}^1(\Omega)$  continuously embedded in $L_{m}^2(\Omega)$ (right?).

Question. We have that $H_{0,m}^1(\Omega) \cap H^2(\Omega)$ is densely embedded in $L_{m}^2(\Omega)$? And $H_{0,m}^1(\Omega)$ is also densely embedded in $L^2_m(\Omega)$?

答:
Yes. Let $\psi \in C_c^\infty(\Omega)$ with $\int_\Omega \psi \, \mathrm{d}x = 1$ be given.

For any $f \in L^2_m(\Omega)$ take a sequence $(f_n) \subset C_c^\infty(\Omega)$
with $f_n \to f$ in $L^2(\Omega)$.
Then, $\int f_n \, \mathrm dx \to \int f \, \mathrm dx$. Thus, we can define
$$
g_n := f_n - \psi \, \int_\Omega f_n \, \mathrm dx$$
and obtain
$g_n \to f$ in $L^2(\Omega)$.


问:
Given $y'(t)=\frac{t}{4}-\frac{1}{y^2+1},y(0)=1$ and let $x(t)=1-\frac{t}{2}$ approx $|x(t)-y(t)|$
Given $y'(t)=\frac{t}{4}-\frac{1}{y^2+1},y(0)=1$ and let $x(t)=1-\frac{t}{2}$

Approximate $|x(t)-y(t)|,|t|<1/2$

I need to do this with $\epsilon$ solution which is corollary of Gronwall thm but I don't know how to use it.

答:
First let us find lipschitz constant    $|f(t,y)-f(t,z)|=|1/4-\frac{1}{y^2+1}-1/4+\frac{1}{z^2+1}|\leq|\frac{1}{(z^2+1)(y^2+1)}|\leq 1$ therefore we can choose $L=1$
by the $\epsilon$ solutions $|y(t)-x(t)|\leq|x_0-y_0|e^{L|t-t_0}+2\epsilon\frac{e^{L|t-t_0|}-1}{L}=2\epsilon\frac{e^{L|t-t_0|}-1}{L}\leq2\epsilon(e-1)$.

Let us find $\epsilon$ $|x'(t)-f(t,x(t))|=|1/2-t/4+\frac{1}{x^2}|\leq1/2+\frac{|t|}{4}+|\frac{1}{x^2+1}|\leq1/2+1/8+1=1.625$ therefore we overall got

$\forall t ,|t|<1/2,|y(t)-x(t)|\leq2*1.625(e-1)$.
if anyone can varify this?


问:
Suppose $u_{xx}+u_{yy} = 1$ has a radial solution on the annulus $ a < r < b$ and that $u(x,y)$ is zero at both $r=a$ and $r=b.$ Solve the PDE.
Using the polar form of the Laplace equation and the fact that the solution is radial I get that,

$$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} = 1.$$

From here, do we just solve this like a normal nonhomogenous ODE initial value problem? If so, would initial conditions be $u(a)=0$ and $u(b)=0$?

答:
$$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} = 1
$$
You can simply rewrite the DE as:
$$r\frac{\partial^2 u}{\partial r^2} + \frac{\partial u}{\partial r} = r$$
$$(ru')'= r$$
$$u'=\dfrac {r}2+\dfrac {C_1}r$$
$$u(r)=\dfrac {r^2}4+ C_1\ln r+C_2$$


问:
Closed curves in 3D
Is $R^3$ a union of disjoint closed curves? (Obviously $R^3$ minus a line is). Is this a classical problem?</p>

答:
Yes, $R^3$ is the union of disjoint circles. Not just homeomorphs of circles, but actual circles. The following simple construction is quoted from M. Jonsson and J. Wästlund, "Partitions of $R^3$ into curves", Math. Scand. 83 (1998), 192-204 (see Theorem 1.1), where it is attributed to A. Szulkin, "$R^3$ is the union of disjoint circles", Amer. Math. Monthly 90 (1983), 640-641. Here is a link to the Jonsson-Wästlund paper: http://www.mscand.dk/article.php?id=77 You can find out about other results like this, some of them depending on the axiom of choice, by reading that paper.

It is easy to see that it is possible to partition a two-punctured sphere into disjoint circles. Let $(x,y,z)$ be coordinates in $R^3$ and consider the union $C$ of the circles of radius $1$ in the $(x,y)$-plane centered at the points $(4k+1,0,0), k\in Z$. Then any sphere centered at the origin intersects $C$ in exactly two points. By covering each such sphere with disjoint circles, we obtain the desired partition.

Here is a picture: http://www.cut-the-knot.org/proofs/tessellation.shtml

I've been told that a partition of $R^3$ into circles was found independently and earlier, but not published, by William Thurston.

ContentLicense: CC BY-SA 3.0

Title: Closed curves in 3D


问:
Is this $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$?
Is this  $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$? Can you help please?

答:
Let $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$. Define the new polynomial $f(x+1)=x^6+7x^5+21x^4+35x^2+21x+7$, using the binomial theorem expand $f(x+1)$. This polynomial is irreducible over $\mathbb{Q}$ by Eisenstein criterion, for $p=7$

If $f(x)$ were reducible over the rationals then let $p(x), q(x)\in \mathbb{Q}[x]$ such that $$f(x)=p(x)q(x)$$ where $\deg(p)< 6$ and $\deg(q)<6$ We would then have $$f(x+1)=p(x+1)q(x+1)$$ but that's impossible! Hence $x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Q}[x]$


问:
Using Peano Axioms Prove That Greater Than Holds For Successor Functions
In first order logic, $\ge$ can be defined as $\exists x(u = x + w)$.

I am trying to formally deduce a proof for:

$\exists x(u = x + w) \vdash \exists x(s(u) = x + s(w))$

I can use peano axioms and basic formal deduction rules. I'm just having a hard time figuring out how to start the question and always get confused if I should use induction or not. Kinda new to this and just looking for a hint on how to start or general guidance on how to approach the question.

If I'm replacing the bound variables with free variables to start, which I believe I should, I know my final step would most likely be to add the $\exists x$ back to the LSH of the proof.

Any help appreciated.

Edit: Forgot to just say domain is natural numbers

答:
$∃x(u=x+w)$ ---premise

$u=a+w$ --- assumed [a] form 1) for $\exists$-elimination

$s(u)=s(a+w)$ --- from 2) by substitution axiom for equality

$s(u)=a+s(w)$ --- from 3) by axiom for $+$

$\exists x (s(u)=x+s(w))$ --- from 4) by $\exists$-introduction.

$\exists x (s(u)=x+s(w))$ --- from 2)-5) and 1) by $\exists$-elimination, discharging assumption [a].


问:
Zeros of polynomial $ f(z)= z^2+ az+ p^{11}$
Consider the polynomial $ f(z)= z^2+ az+ p^{11}$ where $ a \in \mathrm{Z}- \{0\}, \, p \geq 13$ is a prime. Suppose, $ a^2 \leq 4 p^{11},$ then which of the following is true?

$f$ has a zero on imaginary axis.

$f$ has a zero for which real and imaginary parts are equal.

$f$ has distinct roots.

$f$ has exactly one root.

For $a= 1$ and $p= 13$, I got two distinct roots for $f(z)$, so option $3$ is possibly true. Also $ f'(z)$ has a zero at $z= \frac{-a}{2}$, at which $f$ will have a root if $4p^{11} = a^2.$ How to approach for this question and discard other options?

答:
Answer: (3) is true i.e. $f$ has distinct roots.

Use Quadratic equation formula for the root of the polynomial. $$\therefore z = \frac{-a \pm \sqrt{a^2-4p^{11}}}{2}$$ Now Option (1),(2) and (4) can be eliminated by using $a=4p^{11}$


问:
Differentiating and integrating $r(t)=\langle 8t^4+\frac{5}{t} ,\sec^2(\frac{t\pi}{6},3e^t-2t\ln(t))\rangle $
I am asked to differentiate and integrate
$$r(t)=\langle 8t^4+\frac{5}{t} ,\sec^2(\frac{t\pi}{6},3e^t-2t\ln(t))\rangle $$

Am I supposed to differentiate/integrate each term independently?

For example, I have that $r'(t)$ of  $-8t^4 + 5/t$ is $-32t^3 - 5/x^2$.

答:
Yes, you differentiate/integrate each term independently. Try it out yourself, and click on the spoilers to check your answers

Here are the answers:

Differentiation:

 $r'(t)=\langle -32t^3-\frac{5}{t^2}, \frac{\pi}{3} \sec^2(\frac{\pi}{6}x) \tan(\frac{\pi}{6}x) , -2\ln (x)+3e^x-2 \rangle$

Integration:

 $R(t)=\langle -\frac{8}{5}x^5+5\ln(|x|) + C,\frac{6}{\pi}\tan(\frac{\pi}{6}x) + C, -x^2 \ln(x) + 3e^x + \frac{x^2}{2} + C \rangle$


问:
Symmetric Powers Commute with Pullback of (Locally Free) Sheaves
This question was asked here and is exercise II 5.16 e) in Hartshorne. It is a small remark after Theorem 16.3.7 in Vakil's FOAG, which the author says we should be able to prove as a Corollary immediately. Its proof is omitted on Stacks Project. This is self-study.

We seek to show symmetric/exterior powers commute with pulling back locally free sheaves. More precisely, if $\mathcal F$ is a locally free sheaf on, say, scheme $Y$, and $\pi: X \to Y$ is a morphism of schemes, then

$$\pi^* \operatorname{Sym}^n \mathcal F \simeq \operatorname{Sym}^n \pi^* \mathcal F$$

where $\pi^*$ is the pullback functor and $\operatorname{Sym}^n$ means taking the tensor product $T^n$ and modding out by the two-sided ideal $I$ "where tensor multiplication is commutative."

Things I am allowed to assume: $T^n$ commutes with $\pi^*$, and (by induction), $n = 2$. I also can assume that $\pi^*$ is right-exact, so if we take the exact sequence

$$0 \to I \to T^2 \mathcal F \to \operatorname{Sym}^2 \mathcal F \to 0$$

we know that

$$\pi^*I \to \pi^*T^2 \mathcal F \simeq T^2 \pi^* \mathcal F \to \pi^*\operatorname{Sym}^2 \mathcal F \to 0$$

is exact. This leads me to believe we merely need to show that $\pi^*I$ is the kernel of $T^2 \pi^* \mathcal F \to \pi^*\operatorname{Sym}^2 \mathcal F$. I suspect I need to use the fact that $\mathcal F$ is locally free somehow, but am unsure how to do this.

I have found two completely unconvincing solutions in Hartshorne solution documents, one of which merely states that it is obvious that $\pi^*I$ is the required kernel, and the other of which just says the overall claim is true by universal properties. I would prefer to stay away from the latter-style argument, because in FOAG, we have thus far never established the universal properties of symmetric powers.

答: Since everything involved is local, we may replace $X$ and $Y$ by affine schemes $\operatorname{Spec} B$ and $\operatorname{Spec} A$ and $\mathcal{F}\cong \mathcal{O}_Y^n\cong \widetilde{A}^n$. Then $I$ is the sheaf corresponding to the $A$-module $J$ generated by $\{e_i\otimes e_j - e_j\otimes e_i\}$ as $e_i,e_j$ run over the standard basis vectors of $A^n$. The pullback of the sheaf $I$ to $X$ corresponds to taking the sheaf associated to $J\otimes_A B$, which is the $B$-module generated by $\{e_i\otimes e_j-e_j\otimes e_i\}$ as $e_i,e_j$ run over the standard basis vectors of $B^n$. But this is exactly the kernel of $T^2\mathcal{O}_X^n\to \operatorname{Sym}^2\mathcal{O}_X^n$, so you have that $\pi^*\operatorname{Sym}\mathcal{O}_Y^n \cong \operatorname{Sym}^2 \mathcal{O}_X^n \cong \operatorname{Sym}^2\pi^*\mathcal{O}_Y$.


问:
How to calculate Z-transform from a series
If we have the sequence $\{1,1,\dots\}$, how would I go about calculating the Z-transform? Such that we find $Z\{1,1 \dots\}(z)$.

Body:
$$Z\{1,1 \dots\}(z)=\dfrac {z}{z-\color{red}{1}z}=\dfrac {1}{1-\color{red}{1}z^{-1}}=\sum_{n=0}^{\infty}\color{red}{1^n}z^{-n}$$
$$Z\{1,2,4 \dots\}(z)=Z\{2^0,2^1,2^2 \dots\}(z)$$
$$Z\{2^0,2^1,2^2 \dots\}(z)=\dfrac {z}{z-\color{red}{2}z}=\dfrac {1}{1-\color{red}{2}z^{-1}}=\sum_{n=0}^{\infty}\color{red}{2^n}z^{-n}$$
$$Z\{2,2,2 \dots\}(z)=\dfrac {2z}{z-\color{red}{1}z}=\dfrac {2}{1-\color{red}{1}z^{-1}}=2\sum_{n=0}^{\infty}\color{red}{1^n}z^{-n}$$


问:
How to factor this polynomial?
I was trying to factor this polynomial:

$x^3 + x^2 - 16x + 20$

There are four options in this question:

(a) It could be factored in the following form: $(x^2 + b)(x+c)$;

(b) It could be factored in the following form: $(x+b)(x+c)(x+d)$, assuming that $b \neq c \neq d$

(c) It could not be factored.

(d) It could be factored in the following form: $(x+b)^2  (x+c) $

Here's how I've tried to do it:
I've tried to factor by grouping the x, therefore I've obtained:
$x(x^2 + x - 16) + 20$. Now, I've put the $x$ and the $20$ together:
$(x+20)(x^2 + x - 16)$.
Then, I've tried to factor the second term: $(x+20)(x-16)(x+1)$. So, the answer would be "b", according to this algorithm.

I've completed the test (it's a simulation for the admission test I'm going to do), I submit the answers, and I've noticed that this question isn't correct.

答:
As @Fernis pointed out in the comments,

You cannot factor out the $(x+20)$ as you have done. There is no common factor of $(x+20)$ between $x^2+x−16$ and $20$.

Using the Rational Root Theorem, you can know that the possible rational roots are $\pm 1, \pm2, \pm4, \pm5, \pm10, \pm20$.

Through inspection and polynomial/synthetic division, you can get $(x-2)^2(x-5)$, as @saulspatz said. Therefore, (d) is your answer.

Title: How to factor this polynomial?


问:
Concerning: presentations of rational numbers into sums
Problem: Prove that all positive rational numbers can be expressed as the finite sum of different numbers $\displaystyle \frac {1} {n}$ ($n$ is a natural number).

Example: $\displaystyle \frac {19}{16}=1+ \frac {1}{8} + \frac {1}{16}.$

*We cant sum numbers as $\displaystyle \frac {3}{16}$ (denominator > 1) but we can sum $\displaystyle \frac {1}{8}+ \frac {1}{16}.$

Any solutions? Suggestions?

答:
This is a putnam problem. For a complete solution please look here.

Take a look at this article as well: J.C.Owings, American Mathematical Monthly Vol. 75 (1968), Pages $777-778$.


问:
Looking for a QUBO (quadratic unconstrained binary optimization) problem with a known solution
Some quantum computers such as D-Wave claim that a QUBO problem with thousands of variables can be solved. The trouble is that it is very difficult for me to use a classical computer to verify if the global minimum has actually been reached. I mean, the whole point of using a quantum computer is to solve those problems that are too difficult for a classical computer.

So, I am thinking of using a QUBO problem with a known theoretical solution to verify if a quantum computer really works.

The factoring problem can be formulated as  $min(P - ab)^2$. For example, $min(15 - ab)^2$ has a solution $a=3,b=5$. When that is sent to the D-Wave quantum computer, it does produce the correct solution. Unfortunately, however, when P is a very large number, it does not work because the computer is not good at solving the problems whose coefficients have a wide range. $a=4x_2^2+2x_1+x_0, b=4x_5^2+2x_4+x_3$ If you expand the expression, you will get a simple polynomial with 6 variables. Image $a$ and $b$ are very large! The D-Wave computers cannot handle them.

Could you please let me know if there are any other very large QUBO problems that have known solutions?

答:
Many of the standard combinatorial optimization problems can be formulated as constraint-satisfaction problems, and then translated into a QUBO.  For example,
if you have a graph that you want to colour with $c$ colours, you can have $0-1$
variables $x_{ik}$ which you interpret as $1$ if vertex $i$ has colour $k$, $0$ otherwise.  The constraints are $\sum_{k=1}^c x_{ik} = 1$ for all $i$, and
$x_{ik} + x_{jk} \leq 1$ if $\{i,j\}$ is an edge of the graph.  It's easy to construct penalty
functions for these constraints.

Now if you want a colouring problem that looks difficult but has a known solution,
start by colouring your vertices randomly, and then randomly choose edges so that
every edge joins vertices with different colours.


问:
Bounding a polynomial function
Suppose $P(z)$ and $Q(z)$ are polynomials satisfying $deg(Q) \geq deg(P) +2$ and $\gamma_R$ is an arc parameterized by $z=Re^{i\theta}$ for $\theta \in [\alpha,\beta]$. Show that $ \lim_{R \to \infty}\oint_{\gamma_R} \frac{P(z)}{Q(z)} = 0$.

Right, so we have that since $P(z)$ and $Q(z)$ are polynomials, that $P(z) = a_0 +a_1z+....+a_nz^n$ and that $Q(z) = b_0 + b_1z + .....+b_mz^m$. From here we see that $\frac{P(z)}{Q(z)} = \frac{a_0 +a_1z+.....+a_nz^n}{b_0 + b_1z + .....+b_mz^m} = \frac{z^n(\frac{a_0}{z^n}+.....+an)}{z^m(\frac{b_0}{z^m}.....+bm)}=z^{n-m}\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}$. Since $\gamma_R$ is an arc parameterized by $z=Re^{i\theta}$ we have that $|\frac{P(z)}{Q(z)}| = |z^{n-m}| |\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}| \leq R^{n-m}|\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}|$.

From here I am unsure on how to bound the second term. I need to bound it so that I can use the $M-L$ estimate, if I can bound it I know where to go with it.

答: You don't need an exact bound. $\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}+.....+b_m}$ is bounded because it is continuous for $|z|\geq 1$ and tends to finite limit (namely $\frac {a_n} {b_m}$) as $|z| \to  \infty$. Any bound for this function is good enough to prove that the integral tends to $0$ as $R \to \infty$.

Title: Bounding a polynomial function


答:
Let $P(z) = a_n z^n + \cdots a_1 z + a_0$ and $Q(z) = b_m z^m + \cdots + b_1 z + b_0$ with $m \geq n + 2$ (and $a_n, b_m \neq 0$). Let $R > 0$.

$$\bigg|\int_{\gamma_R} \frac{P(z)}{Q(z)} dz \bigg| \leq \ell(\gamma_R) \max_{z \in \gamma_R} \frac{|a_n z^n + \cdots a_1 z + a_0|}{|b_m z^m + \cdots + b_1 z + b_0|}$$

We now use the regular triangle inequality in the numerator and the opposite triangle inequality in the denominator. We also assume that $R$ is so large that we can omit the absolute value sign in the denominator after applying the opposite triangle inequality. We get

$$\bigg|\int_{\gamma_R} \frac{P(z)}{Q(z)} dz \bigg| \leq R (\beta-\alpha)\frac{a_n R^n + \cdots a_1 R + a_0}{b_m R^m - \cdots - b_1 R - b_0}$$

Now, it's easy to see that this expression goes to $0$ as $R \rightarrow \infty$ since we have polynomial in $R$ of degree $n+1$ in the numerator and of degree $m$ in the denominator, and by assumption $m > n+1$.

Title: Bounding a polynomial function

Body:
Suppose $P(z)$ and $Q(z)$ are polynomials satisfying $deg(Q) \geq deg(P) +2$ and $\gamma_R$ is an arc parameterized by $z=Re^{i\theta}$ for $\theta \in [\alpha,\beta]$. Show that $ \lim_{R \to \infty}\oint_{\gamma_R} \frac{P(z)}{Q(z)} = 0$.

Right, so we have that since $P(z)$ and $Q(z)$ are polynomials, that $P(z) = a_0 +a_1z+....+a_nz^n$ and that $Q(z) = b_0 + b_1z + .....+b_mz^m$. From here we see that $\frac{P(z)}{Q(z)} = \frac{a_0 +a_1z+.....+a_nz^n}{b_0 + b_1z + .....+b_mz^m} = \frac{z^n(\frac{a_0}{z^n}+.....+an)}{z^m(\frac{b_0}{z^m}.....+bm)}=z^{n-m}\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}$. Since $\gamma_R$ is an arc parameterized by $z=Re^{i\theta}$ we have that $|\frac{P(z)}{Q(z)}| = |z^{n-m}| |\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}| \leq R^{n-m}|\frac{\frac{a_0}{z^n}+.....+an}{\frac{b_0}{z^m}.....+bm}|$.

From here I am unsure on how to bound the second term. I need to bound it so that I can use the $M-L$ estimate, if I can bound it I know where to go with it.


问:
Values of a parameter $x$ in an infinite series that makes it converge
I am required to find the values of $x$ in the following infinite series, which cause the series to converge.
$$\sum_{n=1}^\infty \frac{x^n}{\ln(n+1)}$$

I tried to use the ratio test, and found that the series converges when $x$ is in $(-1,1)$. However, this answer is not correct.

答:
From ratio test, as you have concluded, we need
$$\limsup_{n \to \infty} \left \vert\dfrac{x^{n+1}/\ln(n+2)}{x^n/\ln(n+1)}\right \vert < 1$$
Hence, we need
$$\vert x \vert \limsup_{n \to \infty} \left \vert\dfrac{\ln(n+1)}{\ln(n+2)}\right \vert < 1 \implies \vert x \vert < 1$$
We need to check the boundaries. For $x=-1$, this is an alternating series and hence converges. At the other boundary, i.e., $x=1$, it clearly diverges. Hence, converges for $x \in [-1,1)$. If you are considering $x \in \mathbb{C}$, we then have that the series converges for all $x$ such that $\vert x \vert \leq 1$ and $x \neq 1$. The convergence on the boundary (not including $1$), can be proved using the Dirichlet test/ generalized alternating test.</p>


问:
A question on floor function.
Prove that for $n \in N$,

$[\sqrt{n} + \frac{1}{2} ] = [\sqrt{n-\frac{3}{4}} + \frac{1}{2}]$

where [.] is the greatest integer function.

My attempt: k < $\sqrt{n} + \frac{1}{2} $ < k+1 .
Obviously, $\sqrt{n-\frac{3}{4}} + \frac{1}{2}$ < $\sqrt{n} + \frac{1}{2} $,

so it would suffice to show that $\sqrt{n-\frac{3}{4}} + \frac{1}{2}$> k , but it turns out that the inequlity is very weak so i am not able to prove it.

Is there any way to do this using induction? Or am i trying the right thing?

Any help is appreciated , thanks!

答:
We have, where $k$ is the integer part of $\sqrt{n} + \frac{1}{2}$,

$$\begin{equation}\begin{aligned}
k & \lt \sqrt{n} + \frac{1}{2} \\
k - \frac{1}{2} & \lt \sqrt{n} \\
k^2 - k + \frac{1}{4} & \lt n
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since $k$ and $n$ are positive integers, \eqref{eq1A} can actually be restated as being

$$\begin{equation}\begin{aligned}
k^2 - k + 1 & \le n \\
k^2 - k + \frac{1}{4} & \le n - \frac{3}{4} \\
\left(k - \frac{1}{2}\right)^2 & \le n - \frac{3}{4} \\
k - \frac{1}{2} & \le \sqrt{n - \frac{3}{4}} \\
k & \le \sqrt{n - \frac{3}{4}} + \frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since $\sqrt{n} + \frac{1}{2} \gt \sqrt{n - \frac{3}{4}} + \frac{1}{2}$, as you stated, this shows $k$ is also the integer part of $\sqrt{n - \frac{3}{4}} + \frac{1}{2}$, which means that

$$\left\lfloor \sqrt{n} + \frac{1}{2} \right\rfloor = \left\lfloor \sqrt{n - \frac{3}{4}} + \frac{1}{2} \right\rfloor \tag{3}\label{eq3A}$$

Title: A question on floor function.


答:
The easiest to try is (d), because it says that the polynomial has a double root. We will look for a root of the derivative and check if it cancels the polynomial.

$$3x^2+2x-16=0\iff x=2\text{ or }x=-\dfrac83.$$

Now $p(2)=0$, bingo !

Title: How to factor this polynomial?

Body:
I was trying to factor this polynomial:

$x^3 + x^2 - 16x + 20$

There are four options in this question:

(a) It could be factored in the following form: $(x^2 + b)(x+c)$;

(b) It could be factored in the following form: $(x+b)(x+c)(x+d)$, assuming that $b \neq c \neq d$

(c) It could not be factored.

(d) It could be factored in the following form: $(x+b)^2  (x+c) $

Here's how I've tried to do it:
I've tried to factor by grouping the x, therefore I've obtained:
$x(x^2 + x - 16) + 20$. Now, I've put the $x$ and the $20$ together:
$(x+20)(x^2 + x - 16)$.
Then, I've tried to factor the second term: $(x+20)(x-16)(x+1)$. So, the answer would be "b", according to this algorithm.

I've completed the test (it's a simulation for the admission test I'm going to do), I submit the answers, and I've noticed that this question isn't correct.


问:
How to count the total unique numbers in this set?
If we have the following set:

$\Big\{ \frac{1}{1},\frac{2}{1},...,\frac{k}{1},\frac{1}{2},...,\frac{k}{2},...,\frac{1}{k},...,\frac{k}{k} \Big\}$

it is very clear that there are some doublets of numbers, such as $\frac{1}{2}=\frac{2}{4}$. My question is, how do I compute the total number of unique numbers in this set? Is there any theorem or formula I could potentially use? I started to enumerate all the possibilities as:

$1,2,...,k,\frac{1}{2},\frac{1}{3},...,\frac{1}{k},\frac{3}{2},\frac{5}{2},\frac{7}{2},...$

but it just seems way too tedious.

答:
This is OEIS A018805. A formula is given there:
$$f(k)=\left(2\sum_{j=1}^k\varphi(j)\right)-1$$
where $\varphi$ is the totient function.


问:
How could one say that it doesnt exist while still its having a value?
The doubt araised when I was in my class studying about iterated limits.

$$\lim_{y\to 0}\lim_{x\to 0}y\sin\big(\frac{1}{x}\big) $$

In the above limit, it's taught that the first limit do not exist thus the whole will not. But what my doubt is, when the first limit is operated we are getting $ y*k$ where $k$ is some value between $-1$ and $1$ ... so we have a value for that limit but don't know what it is. Then if we apply the next limit... It becomes zero... Can someone pls explain why we cannot do like as I thought

答:
You have to do things step by step, and the order matters very much. limit in $x$ first, then the limit in $y$, because that's your question. Because for any $y\neq 0$, $\lim_{x\to 0}y\sin\left(\frac{1}{x}\right)$ doesn't exist, neither does $\lim_{y\to 0}\lim_{x\to 0}y\sin\left(\frac{1}{x}\right)$.

On the other hand, if you asked: "what is $\lim\limits_{x\to 0} \lim\limits_{y\to 0}y\sin\left(\frac{1}{x}\right)$", then the answer is $0$, because we first fix a value of $x\neq 0$. Then, we are looking at $\lim\limits_{y\to 0}y\sin\left(\frac{1}{x}\right)$. THis is zero, because $-1\leq \sin\left(\frac{1}{x}\right)\leq 1$, so if we multiply by $y$, and let $y\to 0$ the result is $0$.

So,
\begin{align}
\lim_{x\to 0}\lim_{y\to 0}y\sin\left(\frac{1}{x}\right) = \lim_{x\to 0}(0) = 0.
\end{align}

So, your reasoning is actually addressing a different question. And probably the reason you're confused is because intuitively, you may expect that $\lim_{y\to 0}\lim_{x\to 0}$ is the same thing as $\lim_{x\to 0}\lim_{y\to 0}$. Unfortunately, IT IS NOT (and this example shows you exactly why the order matters)! Welcome to calculus/analysis; you'll encounter many such results which show you that you have to be very very careful with limits, the order of limits etc.


问:
prove or disprove $\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\left(\frac{k}{p}\right)\equiv 0\pmod{ p^2}$
let $p=8k+1$ prime,prove or disprove
$$\sum_{k=1}^{p-1}\dfrac{(-1)^k}{k}\left(\dfrac{k}{p}\right)\equiv  0\pmod {p^2}\tag{1}$$
where $\left(\dfrac{k}{p}\right)$ is Legendre symbol

It seem this Notes on Wolstenholme’s Theorem  How prove this $(p-1)!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}\right)\equiv 0\pmod{p^2}$ I  can't prove $(1)$

答:
This is what I thought of. It is not finished but it is a start.

Observe that $$\frac{1}{p}\binom{p}{k}=\frac{1}{p}\cdot\frac{p!}{k!\cdot(p-k)!}\equiv\frac{(p-1)!}{k!\cdot-k\cdot...\cdot-(p-1)}\equiv\frac{(p-1)!}{(p-1)!\cdot k}\cdot(-1)^{p-k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}$$

So $$\frac{1}{p}\binom{p}{k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}\Rightarrow\binom{p}{k}\equiv\frac{p\cdot(-1)^{k+1}}{k}\pmod{p^2}$$

To prove $$\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\bigg(\frac{k}{p}\bigg)$$ is divisible by $p^2$, it is enough to prove that $p^2$ divides $$\frac{1}{p}\bigg(\sum_{k=1}^{p-1}\frac{p\cdot(-1)^k}{k}\bigg(\frac{k}{p}\bigg)\bigg)\equiv\frac{1}{p}\bigg(\sum_{k=1}^{p-1}\binom{p}{k}\cdot\bigg(\frac{k}{p}\bigg)\bigg)\pmod{p^2}$$ But $$\sum_{k=1}^{p-1}\binom{p}{k}\cdot\bigg(\frac{k}{p}\bigg)=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-\sum_{k=1}^{p-1}\binom{p}{k}=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-2^p+2$$ so we want to prove that $$\frac{1}{p}\bigg(2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-2^p+2\bigg)$$ is divible by $p^2$, in other words $$\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}+1\equiv2^{p-1}\pmod{p^3}$$


答:
You are largely right when you say that the series converges for all $x$ in the interval $(-1,1)$. However, the series also converges for $x=-1$, because then it is an alternating series.

The series diverges at $x=1$, and also whenver $|x|\gt 1$.

Title: Values of a parameter $x$ in an infinite series that makes it converge

Body:
I am required to find the values of $x$ in the following infinite series, which cause the series to converge.
$$\sum_{n=1}^\infty \frac{x^n}{\ln(n+1)}$$

I tried to use the ratio test, and found that the series converges when $x$ is in $(-1,1)$. However, this answer is not correct.


问:
How to find the local maximum and minimum of $f(x)=x+\sin(2x)$ bounded by the interval $-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$?

Find the local maximum and minimum of $y=x+\sin(2x)$ between
$-\frac{2\pi}{3}$  and $\frac{2\pi}{3}$.

What I have done so far:

$f'(x) = 1+2\cos(2x)$  and $f''(x)=-4\sin(2x)$.

答:
First, find where the derivative is zero ($y'=1+2\cos(2x)=0$).

$$1+2\cos(2x)=0$$
$$2\cos(2x)=-1$$
$$\cos(2x)=-\frac{1}{2}$$
$$2x=-\frac{2\pi}{3}, \frac{2\pi}{3}$$
(within the bounds)
$$x=-\frac{\pi}{3}, \frac{\pi}{3}$$
Checking the second derivative, you can find that $x=-\frac{\pi}{3}$ is local minimum and $x=\frac{\pi}{3}$ is local maximum.


问:
For $N\geq 5$. Which one of the following inequality holds?
For $N\geq 5$. Which one of the following inequality holds?

$\sum _{n=1}^N\frac{1}{n}\le 1+\log \left(N\right)$

$\sum _{n=1}^N\frac{1}{n}< 1+\log \left(N\right)$

$\sum _{n=1}^N\frac{1}{n}\geq \log \left(N\right)$

$\sum _{n=1}^N\frac{1}{n} \leq \log \left(N\right)$

My Attempt

When I substitute $N=5$, I am getting Option(1), Option(2) and Option(3) are correct. How do I do it formally?

答: $\frac 1  {k+1} <\int_k^{k+1} \frac  1 x dx < \frac  1k$. Sum this w.r.t. $k$ from $k=1$ to $k=N-1$. You will see that 1),2) and 3) are all true. You can see that 4) is false by taking $N=2$.


问:
Quick question on taking the gradient of a 3-variable function defined by norm
Let $g(x) : \mathbb{R}^3 \to \mathbb{R}^2$, and $c = (c_1,c_2)$.

We define $f(x) = ||c - g(x)||^2$, mapping from $\mathbb{R}^3 \to \mathbb{R}^2$. How do we find $\nabla f$(x)?

答: Chain rule, bearing in mind that if $Df(x):\Bbb R^3\to \Bbb R$ is the total derivative of $f$ (a linear map), then $\nabla f(x)$ is the unique vector in $\Bbb R^3$ with $Df(x)(v) = \langle \nabla f(x),v\rangle$, and $Dg(x):\Bbb R^3 \to \Bbb R^2$ is the linear map whose matrix relative to the standard bases of $\Bbb R^3$ and $\Bbb R^2$ is the Jacobian of $g$, we have $$f(x) = \|g(x)-c\|^2 \implies Df(x)(v) = 2\langle g(x)-c, Dg(x)(v)\rangle = \langle Dg(x)^*(g(x)-c),v\rangle,$$where $Dg(x)^*:\Bbb R^2 \to \Bbb R^3$ is the adjoint of $Dg(x)$, represented by the transpose of the Jacobian of $g$. This means that $$\nabla f(x) = Dg(x)^*(g(x)-c).$$In words, the gradient of $f$ at $x$ is obtained by multiplying the transpose of the Jacobian of $g$ at $x$ (a $3\times 2$ matrix) by the column vector $g(x)-c$ (a $2\times 1$ matrix) --- the result is a column vector ($3\times 1$ matrix), as expected.


答:
This one is easy to factor. Let's see how.

Start by plugging in $x=0,1,-1,2$ and so on.

You will find by inspection that $x=2$ is a zero of the polynomial. Therefore,$(x-2)$ is its factor.

Now factor the polynomial in such a way that $(x-2)$ gets common.

$$x^3+x^2-16x+20$$
$$=x^3-2x^2+3x^2-6x-10x+20$$
$$=x^2(x-2)+3x(x-2)-10(x-2)$$
$$=(x-2)(x^2+3x-10)$$
$$=(x-2)(x^2+5x-2x-10)$$
$$=(x-2)[x(x+5)-2(x+5)]$$
$$=(x-2)^2(x+5)$$

Title: How to factor this polynomial?


问:
Helpful to review certain calculus topics before first real analysis course?
This is my first time posting, so I apologize in advance if my question is inappropriate here.  I wanted to know if it would be beneficial for me to review certain calculus topics before I take my first real analysis course.  I have noticed that some calculus topics are involved in real analysis courses (e.g. sequences, series, definition of limit).  If this is true, then what calculus topics would be helpful to review?  If this is not the case, what self-study methods would most likely benefit me before I journey into proof-based math?

EDIT:  I am completely new to proof-based math, so I am looking for a couple of self-study suggestions that would likely prepare me well for the rigor of real analysis. By self-study suggestions, I mean reviewing certain calculus topics (if it would be helpful), and/or certain titles of books, etc.

答:
Certainly, a solid understanding of limits would help.

By this, I don't just mean that you can look at $\lim_{x\rightarrow 1}(x^2-1)/(x-1)$ and be able to compute the answer.  I mean that you have a picture in mind, understand how the $\varepsilon\mbox{-}\delta$ language relates to that picture and (ideally) are able to write down the $\varepsilon\mbox{-}\delta$ proof.

Beyond that, maybe the way you think about calculus is more important than the topics (which are pretty standardized) that you know.  For example, when you look at the equation $x^3-x-1=0$ do you think purely algebraically?  Or do you think in terms of a picture? Can you tie that picture to theorems, such as the intermediate value theorem?


问:
Icosahedral symmetry as permutation group
Hopefully an easy question: the icosahedral group of order 60 (orientation preserving symmetries of a regular icosahedron) is isomorphic to the alternating group on 5 points.  In terms of the icosahedron, what are the 5 "points"?

It would be ideal if the "points" were actually points.  The wikipedia article mentions some compounds of inscribed solids, and I think I'd need a physical demonstration to see that.  However, according to this question and this question, we should be able to give just a few points, so that the stabilizer of those points has order 12.

However, I am not sure what the points would be.  Maybe they are the vertices of an associated tetrahedron.  It would be nice if it was easy to describe those vertices, as I certainly don't see any tetrahedra myself, but I can imagine a collection of 4 vertices easily enough.

答:
The icosahedral group $I$ permutes the 5 colors in this coloring of the icosahedron. A rotation of the icosahedron about a vertex gives a 5 cycle permutation of the colors. A rotation of the icosahedron about a triangle face is a 3 cycle permutation of colors. A rotation about an edge gives (ab)(cd) signature.

As @draks says I think the action is more obvious that this coloring is preserved if you think of inscribed tetrahedra in the dodecahedron, but this is the icosahedron equivalent. The corners of those tetrahedra partition the vertices of the dodecahedron, corresponding to this partition of faces in the icosahedron.

Proof $I$ acts on the coloring as $A_5$:

Up to reflection this coloring is unique in having the properties that: each color has 4 faces, no pair of faces sharing an edge or vertex are the same color, and every vertex is incident to all 5 colors. Clearly those properties are preserved by $I$. I don't have a good uniqueness argument other than after you assign the colors around a vertex, you have a single choice you can make before all other colors are forced by the non-adjacency conditions. Therefore $I$ must preserve this coloring.

Once you convince yourself that this coloring is preserved, then the action of the icosahedral group on the set of colors must be $A_5$, since $A_5$ is simple. Any kernel of the action must be trivial.


问:
How to find basis for this set?
let $A$ be a $5 \times 7$ matrix

and let
$$K_A  = \left\{ X \in \mathbb{R}^7 | AX = [0, 0, 0, 0, 0 ]^T \right\}$$

If I am given the matrix $A$, how do I find the basis for $K_A$?

答:
Find an orthonormal basis $b_1, \dots, b_k$ of $K_A^\perp$. To do this, notice that the rows of $A$ span $K_A^\perp$. Thus, we can apply Gram-schmidt process to them to get such a basis.

Extend $b_1, \dots, b_k$ to a basis $b_1, \dots, b_k, c_1, \dots, c_r$ of $\mathbb{R}^7$, for example by appending random vectors $c_1, \dots, c_r$ sampled from the standard Gaussian distribution.

Apply Gram-Schmidt process to the basis $b_1, \dots, b_k, c_1, \dots, c_r$. In the resulting list of vectors, the last $r$ vectors will constitute an orthonormal basis of $K_A$.

Title: How to find basis for this set?


问:
Direct sum of the group of $n$-chains $S_n(X)$ for all $n$ induces the direct sum of the chain complex $S_{\ast}(X)$?
Let $(X,A)$ be a topological space pair in $(\mathbf{Top}^2)$, i.e $A$ is a subspace of $X$. For any $q \geq 0$, we have the decomposition
$$
S_q(X) = S_q(A) \oplus S_q(X,A). \quad \quad (\star)
$$

Reasons: Since any $q$-simplex $\sigma: \Delta_q \rightarrow A$ of $A$ is a $q$-simplex of $X$, and hence the basis of $S_q(A)$ (which is exactly the $q$-simplexes of $A$) is a subset of the basis of $S_q(X)$ (which consists all $q$-simplexes of $X$). So $S_q(A)$ is a direct summand of $S_q(X)$, and hence we obtain $(\star)$.

My main question: Does $(\star)$ means that we have the following decomposition $(\star\star)$ of chain complexes?
$$
S_{\ast}(X) = S_{\ast}(A) \oplus S_{\ast}(X,A) \quad \quad (\star\star)
$$

I shall explain my question further as follows:

1. My attempts - a counterexample: This should be incorrect, since if $(\star\star)$ holds, we shall apply the homology functor $H_{\ast}$ on both sides and obtain
$$
H_{\ast}(X) = H_{\ast}(A) \oplus H_{\ast}(X,A) \quad\quad(\star\star\star)
$$
However, let $X=\mathbb{D}^n$,$A=\mathbb{S}^{n-1}$ and $\ast = n$, then
$$
\mathrm{LHS} = H_{n}(X) = H_{n}(\mathbb{D}^n) = 0,
$$
while
$$
\mathrm{RHS} = H_{n}(A) \oplus H_{n}(X,A) = H_{n}(\mathbb{S}^{n-1}) \oplus H_{n}(\mathbb{D}^n, \mathbb{S}^{n-1}) = \mathbb{Z},
$$
which yields $\mathrm{LHS} \neq \mathrm{RHS}$.

2. My questions: From the counterexample above, $(\star\star)$ should be incorrect, but I don't know why $(\star)$ DOES NOT imply $(\star\star)$? OR is my argument in "Reasons" incorrect, which means that even $(\star)$ does not holds in general?

3. Wild guess: Does these things related to the contractiablity of $A$ as a subspace of $X$. Since once $A$ is a retract of $X$, we shall obtain $r: X \rightarrow A$, satisfying
$$\iota \circ r = \mathrm{id}_{A} \quad\quad (\ast)$$
where $\iota: A \rightarrow X$ is the embedding. Applying the functor $S_{\ast}$ on both sides of $(\ast)$, we see that the exact sequence of chain complexes
$$
0 \rightarrow S_{\ast}(A) \rightarrow S_{\ast}(X) \rightarrow S_{\ast}(X,A) \rightarrow 0
$$
splits, and hence we obtain $(\star\star)$ again, then we get $(\star\star\star)$. So the counterexample shows that $\mathbb{S}^{n-1}$ is not a retract of $\mathbb{D}^n$.

4. Further question: Is my wild guess in "3" correct?

Sorry for such a long post and thank you all for commenting and answering!

答:
The short answer to your main question is "no". If $A_*$, $B_*$, and $C_*$ are chain complexes, having $C_q = A_q \oplus B_q$ for each $q$ in some sense doesn't guarantee that $C_* = A_* \oplus B_*$ as chain complexes. The relevant maps between $C_q$ and $A_q \oplus B_q$ need to also respect the boundary maps in the chain complexes in order for that to be true.

The counterexample you gave in "1" is essentially correct.

You know $(\star \star \star)$ doesn't hold, which leads you to conclude that $(\star \star)$ doesn't hold and the question you're asking in "2" is essentially "how come?". One way to arrive at an answer to that is to be more explicit about what's actually going on with $(\star \star)$, and $(\star)$ in your counterexample.

Concretize, concretize, concretize

For concreteness you might as well fix $n = 1$. So $X = [0,1]$, and $A = \{0, 1\}$. The relevant homology groups are trivial except $H_0(A) = \mathbb Z^2$, $H_0(X) = \mathbb Z$, and $H_1(X, A) = \mathbb Z$. In particular, we have $H_0(X) \neq H_0(A) \oplus H_0(X,A)$. What gives? Let's work out what these things should be from first principles.

You've got $H_0(A) = S_0(A) / \partial S_1(A).$ In a rare case of sanity, we literally just have $S_0(A) = \mathbb Z^2$ and $\partial S_1(A) = 0$, so we can be really concrete about this.

You've got $H_0(X) = S_0(X) / \partial S_1(X).$ We have $S_0(X)$ is the free abelian group generated by points in $X$, and $\partial S_1(X)$ is generated by differences between arbitrary pairs of those generators. This is messy in that the two spaces are infinite-dimensional, but still pretty concrete, and it's hopefully clear why the quotient is indeed just $\mathbb Z$.

Now what about $H_0(X,A)$, which is $S_0(X,A) / \partial S_1(X,A)$? We have conflicting information about this, according to one argument it's trivial, and according to another argument it should be $\mathbb Z$. The claim that it's trivial just means every element of $S_0(X,A)$ is a boundary of some element of $S_1(X,A)$. This is something that you can dig into "by hand". Here are the two arguments when you unwrap them.

(1) It's trivial. (You can get this by concretizing whatever abstract argument you have for this, e.g. the argument in terms of the long exact sequence.) We have $S_0(X,A) = S_0(X) / S_0(A)$, so an element of $S_0(X,A)$ is an abstract sum of points in X, except two abstract sums are considered the same if their difference is a sum of points in A. We have $S_1(X,A) = S_1(X) / S_1(A)$, so an element of $S_1(X, A)$ is an abstract sum of paths in $X$, except two abstract sums are considered the same if their difference is paths that live entirely in $A$. The boundary map between them is "the obvious thing" (take the signed boundary of each path), though it takes a bit of thought to see that it's well-defined. Is it true that every abstract sum of points in X (modulo points in A) is the boundary of some abstract sum of paths in X (modulo paths in A)? Yes. Given an abstract sum of points in X, pick a representative with total signed sum of points equal to 0 (since you're free to "add or remove points in A"). Then pair up the positive points with the negative points and draw a path for each pair. The abstract sum of these paths hits the element of $S_0(X, A)$ that you started with.

(2) It's not trivial. We have $S_0(X) = S_0(A) \oplus S_0(X, A)$. Since $S_0(X)$ is abstract sums of points in $X$, and $S_0(A)$ is abstract sums of points in $A$, we have that $S_0(X, A)$ is abstract sums of points in $X$ which aren't in $A$. Similarly, $S_1(X, A)$ is abstract sums of paths in $X$ which aren't (entirely) in $A$. Is it true that every element of $S_0(X, A)$ is a boundary of some element of $S_1(X, A)$? No, the only way an abstract sum of points can be a boundary of an abstract sum of paths is if the total signed sum of all the points is 0.

So which of these is right? Is the boundary map $\partial: S_1(X, A) \to S_0(X, A)$ surjective or isn't it? Even more concretely, is there an element of $S_1(X, A)$ which maps, via the boundary map, to the element of $S_0(X, A)$ corresponding to just the midpoint $1/2 \in X$? Argument (1) constructs one, e.g. a path in $X$ from $0 \in A \subseteq X$ to $1/2 \in X$ maps to that element of $S_0(X, A)$ according to argument (1). But not according to argument (2).

Well what does that path map to according to argument (2)? It should map to the abstract sum of positive copy of $1/2 \in X$ and one negative copy of $0 \in X$. Oops, according to argument (2), that's not even an element of $S_0(X, A)$ since it uses points in $A$! So if we believe argument (2) then we've run into worse issues; the boundary map from $S_1(X, A)$ to $S_0(X, A)$ doesn't even make sense!

Then abstract

Hopefully that's enough exposition to show that just by concretizing things as much as possible, you're forced to realize that there's something pretty sketchy about argument (2), and that this element of $S_1(X, A)$ really should have some image with the boundary map.

One way to phrase this is that your mistake was to implicitly treat $S_q(X, A)$ as contained in $S_q(X)$ when in fact it is a quotient of $S_q(X)$. You've perhaps hidden more than you realized in the claim $S_q(X) = S_q(A) \oplus S_q(A,X).$

Specifically, it's true that for each $q$ the exact sequence $$0 \to S_q(A) \to S_q(X) \to S_q(A,X) \to 0$$ splits, but the way you phrased it brushes the role of the splitting map $S_q(A, X) \to S_q(X)$ under the carpet. It turns out the splitting map you have in mind does not play well with boundaries, so it doesn't correspond to a chain complex map, which means it doesn't give you a split exact sequence of chain complexes $$0 \to S_*(A) \to S_*(X) \to S_*(A,X) \to 0.$$


问:
How do you find the common ratio of a geometric sequence if not given the first term?
The only given values are the sum of an infinite geometric series which is equal to 9/2, and the second term which is equal to -2. How do I find the common ratio here?</p>

答:
Assume the geometric sequence to be

$$a,ar,ar^2...$$

(where $r$ is the common ratio)

Given that $ar=-2$ and $\frac{a}{1-r}=\frac92$

Two equations and two variables. I bet you can solve it now.


问:
The Euler's phi function for interval
I've read about Euler's phi function. I know that we use it to calculate the number of co-prime numbers with n that are smaller than n. So, I want to know can we somehow use this function, to calculate the number of co-prime numbers with n in any interval? For example, if I want to know the number of three-digit numbers, that are co-prime with 60. I will be glad if you can help me with this!

答:
Any $n$ consecutive numbers form a complete residue system modulo $n$. Hence within these numbers, there are exactly $\phi (n)$ numbers that are coprime to $n$.

In your case, the set of 3-digit numbers is actually $900$ consecutive numbers. Among every $60$ consecutive numbers, there are $\phi (60)$ numbers that are coprime to $60$.

Since $900$ is a multiple of $60$, there are exactly $\left(\dfrac {900}{60} \phi(60)\right)$ 3-digit numbers coprime to $60$.

Title: The Euler's phi function for interval


问:
Question about residually finite dimensional algebras. What is this construction doing?
Let $A$ be a separable C*-algebra and $\{\pi_n\}_{n\in\mathbb N}$ be a sequence of finite dimensional representations of $A$ such that for every $a\neq 0$ there is $n$ such that $\pi_n(a)\neq 0$. In other words, $\bigoplus_n \pi_n $ is faithful. (such $A$ is said to be residually finite dimensional.)

Let $B$ be unital separable and $\{\pi_n\}_{n\in\mathbb N}$ be such sequence. Assume every $\pi_n$ is irreducible and has dimension $k(n)$. Also assume each representation in $\{\pi_n\}_{n\in\mathbb N}$ repeats infinitely many times.

Suppose $l:\mathbb N\to \mathbb N$ is an increasing sequence.

For each $n$, define $\phi_n:M_{k(n)}\to M_{k(n)}(B)$ by viewing $M_{k(n)}$ as a subalgebra of $M_{k(n)}(B)$, since $B$ is unital. Then define $\psi_n^1=\phi_n\circ \pi_n:B\to M_{k(n)}(B)$.

By defining $h_1:b\mapsto \text{diag}(b,\psi_1^1(b),...,\psi_{l(1)}^1(b))$
, $h_1$ is a homomorphism from $B=M_1(B)=M_{I(1)}(B)$ to $M_{I(2)}(B)$, where $I(1)=1$ and $I(2)=1+k(1)+...+k(l(1))$.

Suppose $h_m:M_{I(1)}(B)\to M_{I(2)}(B)$ and $\psi_n^m:M_{I(m)}(B)\to M_{I(m)k(n)}(B)$ are defined. Define $\psi_n^{m+1}=\psi_n^1\otimes 1_{I(m+1)}:M_{I(m+1)}(B)\to M_{k(n)I(m+1)}(B)$ and define $h_{m+1}:b\mapsto \text{diag}(b,\psi_1^{m+1}(b),...,\psi_{l(m+1)}^{m+1}(b))$ as homomorphism from $M_{I(m+1)}(B)$ to $M_{I(m+2)}(B)$, where $I(m+2)=I(m+1)(1+k(1)+...+k(l(m+1)))$.

Then we have unital homomorphisms $h_m:M_{I(m)}(B)\to M_{I(m+1)}(B)$ and therefore we can define the inductive limit C*-algebra $A=\lim_{n\to\infty}M_{I(m)}(B)$.

Proposition. $A$ is simple.

This is from An introduction to the classification of amenable C*-algebras, p158. This definition is too complicated for me to understand.

Since $h_m$ maps $b$ to the matrix with $b$ on its diagonal, $h_m$ is a norm preserving injection. Therefore there are $M_{I(n)}(B)\simeq B_n\leq A$ with $\mathbf{closure}[\bigcup B_n]=A$. It only needs to show if ideal $I$ intersects with large enough $M_{I(n)}(B)$ then $M_{I(n)}(B)\subseteq I$. However, the definition is so complicated that I don't even know where to start.

As in the proof the book gives, it says,

For $0\neq y\in M_{l(n)}$, there is $\pi_m$ such that $\pi_m\otimes \text{id}_{M_{I(n)}}(y)\neq 0$. Let $n+l\geq m$, by considering $h_{n+l}\circ h_{n+l-1}\circ...\circ h_{n+1}$ we may write $h_{n,n+l}(y)=H(y)\oplus (\pi_m\otimes \text{id}_{M_{I(n)}}\otimes \text{id}_{M_J})$ for some positive integer $J$ and some homomorphism $H$. Since $\psi_m^1\otimes \text{id}_{M_{I(n)}}\otimes \text{id}_{M_J}(y)$ is a nonzero element in $M_{I(n+l)}$, the ideal generated by $\psi_m^1\otimes \text{id}_{M_{I(n)}}\otimes \text{id}_{M_J}(y)$ contains $M_{I(n+l)}$.

I do not understand it at all.

答:
I get it now.

The point is, if $y\in M_{n}(B)$ contains nonzero constants among its elements, the ideal generated by $y$ is the whole $M_{n}(B)$.

Suppose $y\in M_{I(n)}(B)$ is nonzero, by the assumption, there is a sufficiently large $m$ such that $\pi_m\otimes 1_{I(n)}(y)\neq 0$, which means $\psi_m^n(y)$ has nonzero constant elements. Then $h_{m-1}\circ ...\circ h_{n}(y)$ therefore generates $M_{I(m)}(B)\supseteq M_n(B)$.

As simple as it is.


问:
find maximum of $|e^{-z^2}|$ for $|z|\geq5$ and $0\leq Arg(z)\leq\pi/8$
Goal: find maximum of $m=|e^{-z^2}|$ for $|z|\geq5$ and $0\leq Arg(z)\leq\pi/8$

my understanding is after expanding with $z = re^{i\theta}, m=e^{-r^2cos\theta}$,$cos\theta>0\therefore argmax_r\ m=0$. But the given answer says the maximum is $e^{-25/\sqrt{2}}$ what have I done wrong?

答: $|e^{-z^{2}}|=e^{-r^{2} \cos (2 \theta)}$ where $z=re^{i\theta}$. The minimum vaue of $\cos (2 \theta)$ for $ 0 \leq \theta \leq \frac {\pi} 8$ is $\cos (\frac {\pi} 4)=\frac  1 {\sqrt 2}$ since $\cos$ is decreasing in $[0,\frac {\pi} 2]$.   Hence $|e^{-z^{2}}| \leq e^{-25 /\sqrt 2}$ and this value is attained when  $z=5e^{i\frac {\pi} 8}$.


答:
For multiples of $60$, it is easy: the number of natural numbers less than $60m$ that is coprime to $60$ is just $m\cdot\phi(60)$. That's because the pattern of coprimes and non-coprimes of 60 repeats with a period of 60.

In fact, it repeats with a period of 30 (30 is the result of multiplying all the unique primes in the factorisation of 60), so multiples of 30 are easy: the number of natural numbers less than $30n$ that is coprime to $60$ is just $n\cdot\frac{\phi(60)}{2}$. In fact, the same is true for any interval of $30n$ consecutive integers.

As the set of 3-digit numbers is an interval with $30\cdot30$ integers, we get that there are $30\cdot 8=240$ three-digit numbers that are coprime to 60.

When the period length doesn't go nicely into the length of the interval as discussed above, things get a bit more complicated. Your best bet is probably to find the closest whole multiple of the period length, then check each of the remaining numbers by hand. For instance, three-digit numbers whose first digit is 1 gives an interval of length 100. Use the above approach on the first 90 (up to and including 189), then check 190 up to 199 one by one.

Title: The Euler's phi function for interval

Body: I've read about Euler's phi function. I know that we use it to calculate the number of co-prime numbers with n that are smaller than n. So, I want to know can we somehow use this function, to calculate the number of co-prime numbers with n in any interval? For example, if I want to know the number of three-digit numbers, that are co-prime with 60. I will be glad if you can help me with this!


问:
What is the relationship between $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ and $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$?
Is $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ > or < or = $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, given $x_1$, $x_2$ and $x_3$ are all positive?

I know there exists a real number $a\in(1,2)$ such that $\frac{1}{3}x=x^{\frac {1}{3}}$, but there are three variables even though they are all positive.

Also I know how to simplify $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3-x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, but then I get stuck too.

I also tried to relate this to the roots of polynomial, but I don't quite know how.

Could anyone help please?

答:
As you said, $x_1$,$x_2$,$x_3$ are positive reals. Furthermore,
by AM-GM inequality  $\frac{x_1 + x_2 + x_3}{3} \geq  (x_1)^\frac{1}{3}(x_2)^\frac{1}{3}(x_3)^\frac{1}{3}$

so the proof is using just AM-GM inequality.


答:
Here is a neater version. From the formula:

$$a = \frac{9}{2}(1-r)$$
$$ar = -2= \frac{9}{2}(1-r)r$$

and remember that $|r| < 1$ for the series to converge.

Title: How do you find the common ratio of a geometric sequence if not given the first term?

Body: The only given values are the sum of an infinite geometric series which is equal to 9/2, and the second term which is equal to -2. How do I find the common ratio here?</p>


问:
Calculation: all alphanumeric passcode combinations
This is a statement from the Apple Platform Security document:

The iteration count is calibrated so that one attempt takes approximately 80 milliseconds. This means it would take more than five and one-half years to try all combinations of a six-character alphanumeric passcode with lowercase letters and numbers.

Question: What calculation was used to determine that the duration is five and one-half years?

What would the time duration be for a seven, eight, nine and ten-character alphanumeric password with lowercase letters and numbers?

答: There are 36 possible characters if you use lowercase and numbers 0-9, this means in every digit you input there are 36 times more new different combinations, this means you can describe the number of possible combinations with $36^n$ where n denotes the number of digits the password has. If you want to work out how long it would take to brute force this password using the 80ms you have given you would simply have to times the 80 by the number of combinations.


问:
Asking for clarification as to ( inductive) proof methods in metalogic. ( Comparing The Logic Book's template to a proof by Shapiro).

I'm currently trying to make my way through the basic metatheory of FOL reading Shapiro's entry on Classical Logic in SEP.

Since I encounter difficulties in understanding proof procedures, I took a look at The Logic Book ( by Bergmann) which is recommended in the Bibliography ( at the end of Shapiro's article).

Note : Here is an example of the kind of proof I can find in Shapiro's text ( https://plato.stanford.edu/entries/logic-classical/#FeatSynt)

Lemma 2. Each formula consists of a string of zero or more unary
markers followed by either an atomic formula or a formula produced
using a binary connective, via one of clauses (3)–(5).

Proof: We proceed by induction on the complexity of the formula or, in
other words, on the number of formation rules that are applied. The
Lemma clearly holds for atomic formulas. Let n be a natural number,
and suppose that the Lemma holds for any formula constructed from n or
fewer instances of clauses (2)–(7). Let θ be a formula constructed
from n+1 instances. The Lemma holds if the last clause used to
construct θ was either (3), (4), or (5). If the last clause used to
construct θ was (2), then θ is ¬ψ. Since ψ was constructed with n
instances of the rule, the Lemma holds for ψ (by the induction
hypothesis), and so it holds for θ. Similar reasoning shows the Lemma
to hold for θ if the last clause was (6) or (7). By clause (8), this
exhausts the cases, and so the Lemma holds for θ, by induction.

In The Logic Book, I can read this

" We may now generally charaterize arguments by mathematical induction. In such an argument, we arrange the items about which we want to prove some thesis in a series of groups . In our example we arranged the sentences of Sentential Logic into the series : all sentences  contaning 0 occurrence of connectives, all sentences containing 1 occurrences..., all sentences containing 2 occurrence...
Having arranged the items in such a series , an argument by mathematical induction then takes the following form:

(1) the thesis holds for every member of the first group of the series

(2) For each group of the series, if the thesis holds for every member
of every prior group, then it holds for every member of that group as
well.

(3) Therefore, the thesis holds for every member of every group in the
series."

My question(s) :

(1) is this the method applied by Shapiro in his proof of " lemma 2" ( reproduced above)?

(2) In The Logic Book, the method is termed " strong induction" ; does Shapiro use rather structural induction? ( Is " induction on the compelxity of a formula" structural inducton or strong induction, or ordinary induction? )

(3) In the inductive hypothesis of Shapiro's proof, how to understand the " OR FEWER" part? If Shapiro applied stong induction, and if - in this method -  the inductive hypothesis is that the thesis ( or property) holds for every group prior to the $n^{th}$ group, shouldn't we find instead " for any formula constructed from $n$ AND fewer instances of clauses (2)–(7)"

(4) also, induction hypothesis in the template ( in he Logic Book) seems to go from group $1$, $2$, $3$ , $n-1$ to group $n$  , while Shapiro goes from  $1$, $2$, $3$ , $n$ to $n+1$.

(5) In case you had any reading recommendation regarding the uses of induction ( strong? structural? ) and proof methods in the metalogic of FOL , I'd be very pleased.

答:
(1) Yes. Induction on "complexity" means to apply induction on the number of connectives.
Base case: $0$ connectives, i.e. an atom.

Induction step: assume that the property to be proved holds for formulas $\varphi, \psi$ of "complexity" $n$ and consider a formula of complexity $n+1$, i.e. $\lnot \varphi$, $\varphi \lor \psi$.

(2) We use the term structural induction when we apply induction to objects that are not numbers, i.e. to some sort of recursively defined structure, such as formulas, lists, or trees.

(3) IMO it is only a question of wording (and I'm not an English native...) If we write "for any formula constructed from $n$ and fewer instances of clauses..." we may allude to the fact that a single formula $\varphi$ is built up applying more than one clause, which contradicts the unique readibility lemma.

(4) No issue here: if you set $k=n-1$, then $k+1=n$ and you can reconcile the two.

(5) Most mathematical logic textbooks; e.g. D.van dalen, Logic and Structure.


答:
Alphanumeric = $26$ lowercase letters and $10$ numbers, so there are $36$ choices for each character.

Since there are $6$ characters, and each character has the same number of choices, there are in total $36^6 = 36 \times 36 \times 36 \times 36 \times 36 \times 36$ possibilities, which is roughly $2.177 \cdot 10^9$.

So this would take $(2.177 \cdot 10^9) \cdot (80 \cdot 10^{-3})$ seconds. Converting this into years (seconds, minutes, hours, days, years), we have:

$$(2.177 \cdot 10^9) \cdot (80 \cdot 10^{-3}) \cdot \frac{1}{60} \cdot \frac{1}{60} \cdot \frac{1}{24} \cdot \frac{1}{365} \approx 5.52 \text{ years}$$

For any other number of characters, say $n$, repeat this process but with $36^n$ instead of $36^6$.

If you increase the number of characters by $1$, you add an extra factor of $36$ into the calculation without changing anything else, so the decoding time increases exponentially.

Title: Calculation: all alphanumeric passcode combinations

Body:
This is a statement from the Apple Platform Security document:

The iteration count is calibrated so that one attempt takes approximately 80 milliseconds. This means it would take more than five and one-half years to try all combinations of a six-character alphanumeric passcode with lowercase letters and numbers.

Question: What calculation was used to determine that the duration is five and one-half years?

What would the time duration be for a seven, eight, nine and ten-character alphanumeric password with lowercase letters and numbers?


问:
For periodical smooth $f$, is the sequence $(f^{(n)})_{n\in \mathbb{N}}$ bounded?
For any infinitely differentiable function $f: \mathbb{R}\to \mathbb{R}$ and positive integer $k\in\mathbb{N}$, let $f^{(k)}$ denote the $k$-th derivative of $f$.

It is easy to see that for any periodical function $f\in C^\infty$ the maximum $\max(\text{im}(f))$ exists.

Is $$\big\{\max\big(\text{im}(f^{(n)})\big):n\in\mathbb{N}\big\}$$ always bounded for $f\in C^\infty$ with $f$ periodical?

答: For $f(x)=\sin (2x)$ we have $max (Im (f^{(n)}))=2^{n} \to \infty$.


问:
Computing $\pi_1(f(\mathbf{S}^1),1)$
Given a map $f:\mathbf{S}^1 \to \mathbf{S}^1$ such that $f(1)=1$, find $\pi_1(f(\mathbf{S}^1),1)$
By definition $\pi_1(f(\mathbf{S}^1),1)=\{[\alpha]_h: \alpha\in \pi(f(\mathbf{S^1}),1)\}$ and I only know that $\pi_1(\mathbf{S}^1,1)$ is generated by $[\omega_1]_h$, $\omega_1(s)=e^{2 \pi i s}$.
Well, $[f \circ \omega_1]_h \in \pi_1(f(\mathbf{S}^1),1)$ then I have the idea that $\pi_1(f(\mathbf{S}^1),1)$ is generated by $[f \circ \omega_1]_h$.

答:
If $f$ is surjective then $f(S^1)=S^1$ and thus $\pi_1(f(S^1))=\pi_1(S^1)\simeq\mathbb{Z}$.

On the other hand if $f$ is not surjective, then the image $f(S^1)$ is a subset of $S^1\backslash\{p\}$ for some point $p\in S^1$. And the latter space is homeomorphic to $\mathbb{R}$. And thus $\pi_1(f(S^1))$ has to be trivial, since path connected subsets of $\mathbb{R}$ are precisely intervals and so the path connected component of any point in $f(S^1)$ is an interval. And intervals are contractible.

In particular what I'm saying here is that any subset of $\mathbb{R}$ has trivial fundamental group (or any homotopy group), regardless of the choice of base point.

Note that I didn't even assume that $f$ is continuous (you didn't as well, although it might be implicit). If it is then "not surjective" case simplifies since $f(S^1)$ is homeomorphic either to $[0,1]$ or a singleton $\{*\}$ as a compact and path connected subset of $\mathbb{R}$.

Title: Computing $\pi_1(f(\mathbf{S}^1),1)$


问:
Let $(X,d)$ be a compact metric space. Then which of the following statements are true?
Let $(X,d)$ be a compact metric space. Let $T:X \to X $ be a continuous function satisfying $\inf_{n\in \mathbb N}d(T^n(x),T^n(y))\neq 0$ for every distinct $x$ and $y$. Then which of the following statements are true?

$T:X \to X $ is one-one.

$T:X \to X $ is not one-one.

Image of $T:X \to X $ is closed.

If $X$ is finite, then $T$ is onto.

My Try

$\inf_{n\in \mathbb N}d(T^n(x),T^n(y))\neq 0$. So, $n=1$ $x\neq y, d(T^1(x),T^1(y))\neq 0$. Hence, $T$ is one-one. So, prtion 1 is true. option 2 is false. Let $f:\mathbb [0,1]\to [0,1]$, $f(x)=x$, which indicates that image $T$ is closed. If $X$ is finite, If $T$ is injective, then $T$ must be onto. So, option (1),(3),(4) are correct. How do I give formal proof?

答: Your answers for 1), 20 and 4) are correct but you cannot prove that 3) is true by just giving an example. Since $X$ is compact and $T$ is continuous the image of $T$ is compact, hence closed in $X$.


答: For what it's worth, you can prove this inequality without knowing the more general result: write $x_i=y_i^3$ so$$\frac{y_1^3+y_2^3+y_3^3}{3}-y_1y_2y_3=\tfrac16(y_1+y_2+y_3)((y_1-y_2)^2+(y_2-y_3)^2+(y_3-y_1)^2).$$

Title: What is the relationship between $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ and $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$?

Body:
Is $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ > or < or = $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, given $x_1$, $x_2$ and $x_3$ are all positive?

I know there exists a real number $a\in(1,2)$ such that $\frac{1}{3}x=x^{\frac {1}{3}}$, but there are three variables even though they are all positive.

Also I know how to simplify $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3-x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, but then I get stuck too.

I also tried to relate this to the roots of polynomial, but I don't quite know how.

Could anyone help please?


问:
Show that there two nodes with $k$ edge-disjoint paths connecting them
I do not know how to solve this exercise:

Let $G(V,E)$ be a simple undirected graph with $\lvert V(G)\rvert  \ge 2$ and $deg(v) \ge k$ for all $v \in V(G)$. Show that there are two nodes $s$ and $t$ in $G$ such that $s$ and $t$ are connected by at least $k$ edge-disjoint paths. Is the claim also true if $G$ has exactly one node $w$ with $deg(w) < k$.

We have recently done Gomory-Hu trees in class, so I suppose this should be used here, but I do not see how. Could you give me a hint?

答:
By Menger's theorem, your problem is equivalent to finding two vertices $s$ and $t$ such that the size $\lambda_{s,t}$ of a minimum $s-t$ cut is at least $k$. Let $s$ be a leaf in a Gomory-Hu tree of $G$ and let $t$ be its unique neighbour in the tree. Then by the properties of Gomory-Hu trees, $\lambda_{s,t}$ is the number of edges crossing the cut $(\{s\}, V-\{s\})$; but this is just $\deg(s) \geq k$.

Since a tree has at least two leaves, the same proof works in the case when there is a single vertex with degree less than $k$.


问:
How to find the local maximum and minimum of $f(x)=x+\sin(2x)$ bounded by the interval $-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$?

Find the local maximum and minimum of $y=x+\sin(2x)$ between
$-\frac{2\pi}{3}$  and $\frac{2\pi}{3}$.

What I have done so far:

$f'(x) = 1+2\cos(2x)$  and $f''(x)=-4\sin(2x)$.

答:
Setting first derivative to zero and solving for $x$
$$1 + 2 \cos 2x =0 \to x =\pm \frac {\pi}{3},\pm \frac {2\pi}{3}...$$

So sine sine wave roots are placed with constant period

The second derivative has negative,positive, negative, positive, negative signs alternating signs so that from $ x=0$ you get successively maximum, minimum, maximum, minimum, maximum...extrema after plugging in roots as shown.


问:
Solve a cubic equation?
Need help with solving an equation:

Solve the equation $5x^3 - 24x^2 + 9x + 54 = 0$ given that two of it's roots are equal.

Any help would be greatly appreciated.
Thanks!

答:
If two roots are equal, then the derivative will have that as a root.

If you can't use derivatives, you can set $P(x) = 5(x-a)^2(x-b)$ and solve for $a$ and $b$, by comparing coefficients.

Title: Solve a cubic equation?


问:
If $u\in H^s(\mathbb R^d),\,s>d/2$, then $\|u\|_{L^\infty(\mathbb R^d)}\leq C\|u\|_{H^s(\mathbb R^d)}$ via Fourier transform
Problem: Use the Fourier transform to prove that if $u\in H^s(\mathbb R^d)$ for $s>d/2$, then $u\in L^\infty(\mathbb R^d)$, with the bound
$$\|u\|_{L^\infty(\mathbb R^d)}\leq C\|u\|_{H^s(\mathbb R^d)}$$
for some constant $C$ depending only on $s$ and $d$.

My Thoughts: We will use the Fourier characterization of the Sobolev space $H^s(\mathbb R^d)$, where we define the Fourier tranform by
$$\widehat{u}(\xi)=\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}u(x)e^{-ix\cdot\xi}\,dx.$$
Let $x\in\mathbb R^d$ be arbitrary but fixed. Then using the Fourier inversion theorem and the Cauchy-Schwarz inequality we get that
\begin{align*}
\vert u(x)\vert&=\frac{1}{(2\pi)^{d/2}}\left\vert\int_{\mathbb R^d}e^{ix\cdot \xi}\widehat{u}(\xi)\,d\xi\right\vert\\
&\leq\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}\vert \widehat{u}(\xi)\vert\,d\xi\\
&=\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}(1+\vert\xi\vert^2)^{s/2}\vert \widehat{u}(\xi)\vert(1+\vert\xi\vert^2)^{-s/2}\,d\xi\\
&\leq\frac{1}{(2\pi)^{d/2}}\left[\int_{\mathbb R^d}(1+\vert\xi\vert^2)^s\vert\widehat{u}(\xi)\vert^2\,d\xi\right]^{1/2}\left[\int_{\mathbb R^d}(1+\vert\xi\vert^2)^{-s}\,d\xi\right]^{1/2}\\
&= C\|u\|_{H^s(\mathbb R^d)},
\end{align*}
where
$$C=\frac{1}{(2\pi)^{d/2}}\int_{\mathbb R^d}(1+\vert\xi\vert^2)^{-s}\,d\xi,$$
where the latter integral, using polar integration, can be shown to be finite if and only if $s>d/2,$ hence the dependance of $C$ on $d$ and $s$.

Since $x\in\mathbb R^d$ above was arbitrary, it follows that
$$\|u\|_{L^\infty(\mathbb R^d)}\leq C\|u\|_{H^s(\mathbb R^d)}$$
and the proof is complete.

Do you agree with my proof above? Any feedback is most welcomed and appreciated.

Thank you for your time.

答: Yes, that's exactly it. Technically this proof establishes the estimate on a dense subspace of $H^s$ such as Schwartz space, because you are assuming that you can write $u$ as the inverse Fourier transform of an $L^1$ function, whereas the Fourier transform on $L^2$ (which the Fourier characterization of $H^s$ depends on) is defined through a limiting process. So you might also have to provide the standard density argument showing how this estimate extends to arbitrary $H^s$ functions, depending on what sort of details you're willing to sweep under the rug.


问:
How to prove a subset of L(V,W) is a subspace- how to prove not empty
Does the set have to contain the zero transformation i.e. $T$ such that $T(v)=0$ for all $v \in V$ to be non empty?

答: If $L(V,W)$ is the set of all linear mappings $T:V \to W$, then the zero transformation is linear (check this !), hence the zero transformation is in $L(V,W)$.


答:
No, a general nonempty subset doesn't have to contain the zero transformation.

But as I understand, you ask specifically about subspaces. If a subset is a subspace it indeed must contain the zero element. In general, to show a subset $U$ is a subspace you need to show the following:

$1$. $U$ is nonempty.

$2$. $U$ is closed under addition and scalar multiplication.

From these two conditions it automatically follows that the zero element is in $U$. So to prove the first condition you can show it is nonempty in any way, just show it contains any element. It's just that most of the time the easiest way is to show that specifically the zero element is in $U$. (but it isn't a must)

Title: How to prove a subset of L(V,W) is a subspace- how to prove not empty

Body: Does the set have to contain the zero transformation i.e. $T$ such that $T(v)=0$ for all $v \in V$ to be non empty?


问:
Cardinality of the set of prime numbers
It was proved by Euclid that there are infinitely many primes. But what is the cardinality of the set of prime numbers ?

Cantor showed that the sets $\mathbb{Q}$ and $\mathbb{Z}$ have the same cardinality as the natural numbers $\mathbb{N}$ by constructing a pairing of the two sets, or a bijective function $ \pi_{\mathbb{Z}} : \mathbb{N} \rightarrow \mathbb{Z}$ and $ \pi_{\mathbb{Q}} : \mathbb{N} \rightarrow \mathbb{Q}$.

Let $\mathbb{P}$ denote the set of prime numbers. Is it possible to construct such a pairing function, $ \pi_{\mathbb{P}} : \mathbb{N} \rightarrow \mathbb{P}$ ?

It's clear that $|\mathbb{P}| \leq |\mathbb{N}| = \aleph_0$ since $\mathbb{P} \subset \mathbb{N}$. Is it possible to show that $|\mathbb{P}| \geq |\mathbb{N}|$, or do we have $|\mathbb{P}| < |\mathbb{N}|$ ?

答: $\aleph_0$ is the smallest infinite cardinal. Therefore, since $\mathbb{P}$ is infinite, $|\mathbb{P}| \geq \aleph_0$. Finally, $|\mathbb{P}|= \aleph_0$.</p>

Title: Cardinality of the set of prime numbers


问:
Clock angle problem from simulation of planetary orbits

Description

I'm trying to reproduce a Solar System mathematically using ellipses. The initial conditions are:

All ellipses will have a common foci $(0,0)$, but different centers

A planet orbiting an ellipse will have a period $T$

The position of a planet in the ellipse will be the reflected point (from the center of the ellipse) given the angle $\theta$ at time $t$ (*)

Every ellipse can have a different initial phase angle $\theta_0$

Every ellipse can be rotated about the common foci $(0,0)$

To simplify the problem. We will assume conditions $4.$ and $5.$ are nonexistent. Here's a simple representation of a single ellipse:

where

$(j,k)$ is the center of the ellipse

$(x,y)$ is the result of the
ellipse formula at angle $\theta$

$(x',y')$ is the reflected
point using the center as a mirror

$\omega$ is the angle at point $(x',y')$

*: This is needed to simulate Kepler's second law

Problem

Given two ellipses (whose variables are denoted by subindex $_1$ and $_2$ in the following formulas), I'd like to find a formula that describes at which times $t$, $\omega_1$ and $\omega_2$ are equal. In other words, find the equation of $\omega$ in relation to time $t$. Here's a gif that shows the specified system of two orbits:

where

The green dot is the center of the large ellipse

The red dot is the center of the smaller ellipse

The period of the large ellipse is $T_1 = 6$s

The period of the smaller ellipse is $T_2 = 14$s

The eccentricity of the large ellipse is 0.77

The eccentricity of the smaller ellipse is 0.75

The directrix of both ellipses is 2

So far, I've been able to figure out at which times $t$ both $\theta$ are equal:

We know how $\theta$ varies in relation to $t$,
$$\theta = \frac{2\pi t}{T} + \theta_0$$

Since we have stablished that $\theta_0$ doesn't exist,
$$\theta = \frac{2\pi t}{T}$$

We know that at time $t$, both $\theta_1$ and $\theta_2$ must be equal:
$$\theta_2 = \theta_1 + 2\pi k\ \ ,\  k \in \mathbb{Z}$$

where $2\pi k$ represents the periodicity of the orbit. If we substitute:
$$\frac{2\pi t}{T_2} = \frac{2\pi t}{T_1} + 2\pi k\ \ ,\  k \in \mathbb{Z}$$

Solve for t:
$$t = \frac{k}{\frac{1}{T_2}-\frac{1}{T_1}}$$

This formula gives me at which time $t$ both ellipses will have the same $\theta$. If we substitute $T_1$ and $T_2$ with the periods of the ellipses in the example gif, we got for $k=1$, $t=10.5$s. If you look at the gif, after 10.5s, both red lines will be aligned, meaning $\theta_1$ and $\theta_2$ will be equal. Another way of finding this formula is using the polar expression of an ellipse:

We know that the angle $\theta$ will be
$$\theta = \arctan{\frac{y}{x}}$$

And we know that the point $(x,y)$ in polar coordinates is represented as:
$$(x,y) = (r\cos{\theta}, r\sin{\theta})$$

Where $r$ is the distance from the center to the point $(x,y)$, and this distance is given by the polar formula of an ellipse:
$$r=\frac{ed}{1+e\cos{\theta}}$$
$$(x,y) = \left(\frac{ed}{1+e\cos{\theta}}\cos{\theta}, \frac{ed}{1+e\cos{\theta}}\sin{\theta}\right)$$

Going back to the formula described at step $1.$, we have
$$\theta = \arctan{\frac{y}{x}} = \arctan{\frac{\frac{ed}{1+e\cos{\theta}}\sin{\theta}}{\frac{ed}{1+e\cos{\theta}}\cos{\theta}}}$$

Simplifying, we have
$$\theta = \arctan{\frac{y}{x}} = \arctan{\frac{\sin{\theta}}{\cos{\theta}}} = \arctan{(tan{\theta})}$$

Again, we want both $\theta_1$ and $\theta_2$ to be equal:
$$\arctan{\left(tan{\frac{2\pi t}{T_2}}\right)} = \arctan{\left(tan{\frac{2\pi t}{T_1}}\right)}$$

Which is equivalent to
$$\frac{2\pi t}{T_2} = \frac{2\pi t}{T_1} + 2\pi k\ \ ,\  k \in \mathbb{Z}$$

So far so good. Now we want to find at which time $t$, both $\omega_1$ and $\omega_2$ are equal. We can use the second approach:

We know that the point (x',y') is
$$(x',y')=2(j,k)-(x,y) = (2j-x,2k-y)$$

And we already know how to represent $x$ and $y$ in relation to time:
$$(x,y) = \left(\frac{ed}{1+e\cos{\theta}}\cos{\theta}, \frac{ed}{1+e\cos{\theta}}\sin{\theta}\right)$$

This means that $(x',y')$ will be
$$(x',y') = \left(2j-\frac{ed\cos{\theta}}{1+e\cos{\theta}}, 2k-\frac{ed\sin{\theta}}{1+e\cos{\theta}}\right)$$

Which means that the angle $\omega$ will be
$$\omega = \arctan{\left(\frac{2k-\frac{ed\sin{\theta}}{1+e\cos{\theta}}}{2j-\frac{ed\cos{\theta}}{1+e\cos{\theta}}}\right)}$$

Since we want $\omega_1$ and $\omega_2$ to be equal
$$\omega_2 = \omega_1 + 2\pi k$$
$$\arctan{\left(\frac{2k_2-\frac{e_2d_2\sin{\theta_2}}{1+e_2\cos{\theta_2}}}{2j_2-\frac{e_2d_2\cos{\theta_2}}{1+e_2\cos{\theta_2}}}\right)} = \arctan{\left(\frac{2k_1-\frac{e_1d_1\sin{\theta_1}}{1+e_1\cos{\theta_1}}}{2j_1-\frac{e_1d_1\cos{\theta_1}}{1+e_1\cos{\theta_1}}}\right)} + 2\pi k$$

and we know that $k_2=k_1=0$ since condition $5.$ doesn't apply:
$$\arctan{\left(\frac{-\frac{e_2d_2\sin{\theta_2}}{1+e_2\cos{\theta_2}}}{2j_2-\frac{e_2d_2\cos{\theta_2}}{1+e_2\cos{\theta_2}}}\right)} = \arctan{\left(\frac{-\frac{e_1d_1\sin{\theta_1}}{1+e_1\cos{\theta_1}}}{2j_1-\frac{e_1d_1\cos{\theta_1}}{1+e_1\cos{\theta_1}}}\right)} + 2\pi k$$

Simplifying...
$$\tan{\left(\frac{2\pi t}{T_1}\right)}e_2d_2(4j_1-e_1d_1) = tan{\left(\frac{2\pi t}{T_2}\right)}e_1d_1(4j_2-e_2d_2)$$

At which point, I got stuck. WolframAlpha does not give me a clear answer for $t$, and if I substitute the example gif variables in this formula, WolframAlpha gives 5 possible $t$, of which 3 contain an imaginary number $i$. Which doesn't make sense since time doesn't have 2 dimensions.

Questions

Am I doing something wrong?

Is there a more elegant solution to this problem that I'm not aware of?

Why is WolframAlpha giving me imaginary numbers?

答:
With respect to ellipse center the angles using standard notation $(a,b,c,e)$ major, minor axes, semi inter focal length distance and eccentricity respectively are found as:

$$ (x,y)= \dfrac{a(1-e^2)}{1+e\cos \theta} (\cos \theta, \sin \theta)$$

$$ (x',y')=  \dfrac{-a(1-e^2)}{1+e\cos \theta} (\cos \theta, \sin \theta)$$

$$ \omega= \theta + \tan^{-1}\dfrac{y}{c-x}+ \tan^{-1}\dfrac{y}{2c-x}; \tag 1 $$

$$  \theta_1 + \tan^{-1}\dfrac{y_1}{c_1-x_1}+ \tan^{-1}\dfrac{y_1}{2c_1-x_1}$$

$$=\theta_2 + \tan^{-1}\dfrac{y_2}{c_2-x_2}+ \tan^{-1}\dfrac{y_2}{2c_2-x_2};\tag 2  $$

Eccentric anomaly $E$ at center is connected to the $\theta's$ from Kepler's Equation:

$$  y_1/x_1=\tan E_1=\sqrt{\dfrac{1+e_1}{1-e_1}}\tan \theta_1;\;$$
$$  y_2/x_2=\tan E_2=\sqrt{\dfrac{1+e_2}{1-e_2}}\tan \theta_2 ;\tag 3$$

From  Kepler's third law

$$ \frac{T_1^2}{T_2^2}= \frac {a_1^3}{a_2^3} \tag4 $$

Time dependence is from the classical Newton planetary orbits ( tangential direction & Coriolis accelerations and normal direction the  gravity & centrifugal accelerations ) withgiven  perigee start boundary conditions. we can calculate all time dependent dynamical quantities either using elliptic integrals in closed form or numerically. Finding roots of angles establishes times of clock angles.

Fuller answer could follow after OP's clarifying reply to my comments.


问:
How do I get the normal domain from an inequality condition (application of Fubini's Theorem)?
I want to compute the volume of the 3-dimensional unit ball.

$A := \{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + z^2 \leq 1$

From this I want to get the normal domain as:

$A = \{(x,y,z): x \in [-1,1], y \in [-\sqrt{1-x^2}, \sqrt{1-x^2}], z \in [-\sqrt{1-x^2-y^2},\sqrt{1-x^2-y^2}]\}$

The question is, how do I get the x, y, z intervals from the inequality?

The more concrete (but less 'exact') question is: Is it ok to 'leave out' z in the definition of the y interval, i. e. disregard z?

答:
I found the answer (any corrections or improvements appreciated):

$$
x^2 + y^2 + z^2 \leq 1
$$

Because $x^2 \geq 0$, $y^2 \geq 0$, $z^2 \geq 0$:

$$
0 \leq x^2 + y^2 + z^2 \leq 1
$$

From that we can say, because $z^2 \geq 0$ and $0 \leq x^2 + y^2 + z^2 \leq 1$ that leaving out $z$ will not change the in equality and get

$$
0 \leq x^2 + y^2 \leq 1
$$

From this we can split off

$$
x^2 + y^2 \leq 1
$$

which can be changed to express $y$ with $x$.

The same holds for expressing $z$ with $x$ and $y$.


答:
Since $\Bbb P \subset \Bbb N$, we have a natural ordering inherited on $\Bbb P$. Since it is a subset of a well-ordering, the induced ordering is again a well-ordering.

Thus we have that $(\Bbb P, <)$ is an infinite well-ordering; it is clear that it contains no limits.

Therefore, the order-type of $(\Bbb P, <)$ must be $\omega$, meaning that it can be put in direct bijective correspondence with $\Bbb N$.

Effectively, this proof (which hinges on the theorem "every well-ordering is order-isomorphic to a unique ordinal") gives a bijection by the following recursive definition:

$$f: \Bbb N \to \Bbb P: f(n) := \begin{cases}\inf \Bbb P &: n = 0\\\inf(\Bbb P \setminus\{f(1)\ldots f(m)\}) &: n = m+1\end{cases}$$

That is, "$f(n)$ is the $n+1$st smallest prime number".

Title: Cardinality of the set of prime numbers

Body:
It was proved by Euclid that there are infinitely many primes. But what is the cardinality of the set of prime numbers ?

Cantor showed that the sets $\mathbb{Q}$ and $\mathbb{Z}$ have the same cardinality as the natural numbers $\mathbb{N}$ by constructing a pairing of the two sets, or a bijective function $ \pi_{\mathbb{Z}} : \mathbb{N} \rightarrow \mathbb{Z}$ and $ \pi_{\mathbb{Q}} : \mathbb{N} \rightarrow \mathbb{Q}$.

Let $\mathbb{P}$ denote the set of prime numbers. Is it possible to construct such a pairing function, $ \pi_{\mathbb{P}} : \mathbb{N} \rightarrow \mathbb{P}$ ?

It's clear that $|\mathbb{P}| \leq |\mathbb{N}| = \aleph_0$ since $\mathbb{P} \subset \mathbb{N}$. Is it possible to show that $|\mathbb{P}| \geq |\mathbb{N}|$, or do we have $|\mathbb{P}| < |\mathbb{N}|$ ?


问:
Explain into disjunctive normal form
Can anyone tell me how to Express the following formula into disjunctive normal form ⌐ (p V q) ↔ (p ^ q). I have done few steps but I need your help.</p>

答:
A DNF is an OR of ANDS.

$\neg (p\lor q) \leftrightarrow (p\land q)$

$(\neg p\land \neg q) \leftrightarrow (p \land q)$

$((\neg p\land \neg q) \to (p \land q))\land ((p \land q) \to (\neg p\land \neg q))$

$(\neg(\neg p\land \neg q) \lor (p \land q))\land (\neg(p \land q) \lor (\neg p\land \neg q))$

$(p\lor q) \lor (p \land q)\land ((\neg p\lor \neg q) \lor (\neg p\land \neg q))$

Now, just distribute and simplify.

Title: Explain into disjunctive normal form


问:
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

Solve:   $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

My Try

$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!

答:
Let $t = \tan (x/2)$.  Then $\sin x = 2t/(1+t^2)$ and $\cos x = (1-t^2)/(1+t^2)$.  Your equation looks quartic after multiplying both sides by $(1+t^2)^2$.
$$12t(1-t^2)+4((1-t^2)^2-4t^2)-(2-2t^2-12t)(1+t^2)-6(1+t^2)^2$$
but it simplifies to $-4(3t-1)^2=0$, so $t = 1/3$, giving $x = 2(\tan^{-1} t + n\pi) = 2(\tan^{-1}{\frac13}+n\pi)$ for any integer $n$, since $\tan$ has period $\pi$.

The above $t$-substitution omits the possibility that $x = (2n+1) \pi$, which is actually a solution: cross out the $\sin$ in the equation and observe that $4(1)-2(-1)-6 = 0$.


问:
Rotations by degrees other than $90, 180,$ and $270$.
Say I have a triangle with vertices $(0,0), (2,4), (4,0)$ that I want to rotate along the origin. Rotation by multiples of $90^{\circ}$ is simple. However, I want to rotate by something a bit more complicated, such as $54^{\circ}$. How do I figure out where the vertices would be then?</p>

答: One way is to use complex numbers.  Multiplying by $\cos\theta+i\sin\theta$ rotates by $\theta$ about 0, so you could multiply $(2+4i)(\cos 54^\circ+i\sin 54^\circ)$ to get the rotation image of (2,4).</p>

ContentLicense: CC BY-SA 2.5


问:
integral transforms: why do roots in frequency domain correspond to eigenvalues in time domain (and how does it help solve differential equations)?
In Wikipedia you can read about integral transforms, esp. the Laplace transform which maps a differential equation in the time domain into a polynomial equation in the complex frequency domain:

roots of the polynomial equations in the complex frequency domain correspond to eigenvalues in the time domain

Can anybody give a simple example and foremost give an intuition why this is so - and why does this help solving the differential equation? I don't see the big picture.

答:
I'm going to give an example of how it works. I don't know why it works. The example is the simplest case from electric circuit theory, the discharge of a capacitor. You know the solution f(t) is just an exponential decay, truncated to the left at t=0. The Fourier Transform of f(t) is F(w) = 1/1+jw. Now observe:

F(w) + jw*F(w) = 1/1+jw  +  jw/1+jw  =  1

Now remember some properties of the Fourier transform: multiplication by jw in the frequency domain is equivalent to differentiation in the time domain. And the Fourier transform of a constant function in one domain is the delta function in the other domain. Using only these properties we can translate the above equation into words:

"A function plus its derivative is equal to the delta function".

This sentence is exactly the differential equation of a capactor which has a unit of charge dumped into it at t=0.


问:
A question in Ring theory
R be a commutative ring with unity and it has exactly one maximal ideal. Then prove that the equation  $x^2 =x$  has exactly two solutions.

Show me the right way to solbe this one.

答: Let $\mathfrak{m}$ be the unique maximal ideal of $R$. If $x^2=x$, then $0=x^2-x=x(x-1)$. If $x\in\mathfrak{m}$, then $1-x\in R^\times$ since $x\in\mathcal{J}(R)=\mathfrak{m}$, hence not a zero-divisor. If $x\notin\mathfrak{m}$, then $x\in R^\times$, hence not a zero-divisor. So $x=0,1$ are the only solutions.</p>

Title: A question in Ring theory


Body:
Solve:   $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

My Try

$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!

Body:
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

$6\sin{x}\cos{x} -9\sin^2{x} - \cos^2{x} -2\cos{x} + 6\sin{x} - 1=0$

$-(3 \sin x - \cos x)^2 + 2(3 \sin x - \cos x) - 1= 0$

$(3 \sin x - \cos x)^2 - 2(3 \sin x - \cos x) + 1= 0$


答: $x=2y$  we get after some manipulation $$-4\cos^4 y-36\cos^2y\sin^2 y+24\cos^3 y\sin y=0$$ $$-4\cos^2 y{(\cos y-3\sin y)}^2=0$$ can you end it??

Title: Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$


问:
Outer Product Definition
In my quantum mechanics course, the lecturer do the following definition for outer product, then equate it a matrix. Then, she want us to show this equality.
$$\sum \limits_{i}^{n}\sum \limits_{j}^{n}
\text{out}(\vec e_{i},\vec e_{j}) \equiv E_{ij}
$$

However, when I perform the calculations, I found a unit matrix as you can see from the following steps. For the sake of simplicity, I did not write the summation signs.
$$\begin{equation*}
\begin{split}
 &= \vec e_{i} \, \vec e_{j}\,^{T} \\
 &= \vec e_{i} \, \vec e_{i}\,^{T} \delta_{ij} \\
 &= \vec e_{i} \, \vec e_{i}\,^{T} \\
 &= I
\end{split}
\end{equation*}$$

I is an nxn unit matrix. I want to also mention that $e_i,e_j$ are linearly independent unit vectors.

Where is my mistake?

答:
Consider this example:
$$v = (1,2,3),\ w= (4,5,6)^\top$$
Then the outer product would be
$$ \begin{pmatrix}
1*4 & 1*5 & 1*6\\
2*4 & 2*5 & 2*6 \\
3*4 & 3*5 & 3*6
\end{pmatrix} $$

Considering $e_1 =(1,0,0), e_3 = (0,0,1)$ you will get
$$E_{13} = \begin{pmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix} $$
$E_{13}$ is the matrix which just zeros expect at the position $(1,3)$

In general you get $1$ for the outer product of two unity vectors when $i=j$ and $0$ if $i\neq j$

Title: Outer Product Definition


问:
How to get the marginal density function of dependent random variables
I am unsure how to get the marginal probability density function if the two random variables are dependent.

In general for given $f_{XY}(x,y)$ we get $$f_X(x)=\int_{-\infty}^{+\infty}f_{X,Y}(x,y)dy\tag1$$and $$f_Y(y)=\int_{-\infty}^{+\infty}f_{X,Y}(x,y)dx\tag2$$
However I am unsure what happens if e.g. $x$ depends on $y$. Consider this example
$$ f_{XY}(x,y) =
\begin{cases}
x+y& \text{ for } y\geq x\geq0, y\geq0  \\
0 & \text{ otherwise }
\end{cases} $$
Then we get
$$f_Y(y)=\int_{0}^{+\infty}(x+y)dx = \frac{x^{2}}{2}+xy$$
but what will I get for $f_{X}(x)$ ?

Edit: To clarify this question, would the "easiest" solution
$$f_X(x)=\int_{0}^{y}(x+y)dy $$
be a valid solution?

答:
Consider the following example

$$ f_{XY}(x,y) =
\begin{cases}
2(x+y)& \text{ for } 1\geq y\geq x\geq0 \\
0 & \text{ otherwise }
\end{cases} $$

this is similar to yours but now $f_{XY}(x,y)$ is a nice density...

to obtain the marginals let's focus on the ${X,Y}$ domain

and consider that  the rv $X$ takes values in $(0;y)$

$$\underbrace{0<x<y}_{\text{X-integration bounds}}<1$$

while the rv $Y$ takes values in $(x;1)$

$$0<\underbrace{x<y<1}_{\text{Y-integration bounds}}$$

thus to calculate the marginals you can do

$$f_X(x)=\int_x^1 f(x,y)dy$$

$$f_Y(y)=\int_0^y f(x,y)dx$$


问:
what is $\lim_{z\rightarrow 0}{(\cos(z))^{1/z^2}}$
The suggestion was to use L'Hospital's rule but I couldn't see how to apply it.

I tried manipulating the term to $\lim_{z\rightarrow 0} \exp(1/z^2\ln|º\cos(z)|+i\arg(\cos(z)))=\lim_{z\rightarrow 0} \exp(i\arg(\cos(z))$ and I also don't know how to proceed.

答:
$$(\cos z)^{1/z^2}=[1+(\cos z-1)]^{1/z^2}=\left\{[1+(\cos z-1)]^{\frac{1}{\cos z-1}}\right\}^{\frac{\cos z-1}{z^2}}.
$$

The par between $\{\cdot\}$ goes to $e$, since $\cos z-1\to0$ ($z\to0$). So you can apply L'Hôpital to the exponent $\frac{\cos z-1}{z^2}$.

But if you remember that $1-\cos z\sim \frac{z^2}{2}$ as $z\to 0$, then $$\lim_{z\to0}\frac{\cos z-1}{z^2}=\lim_{z\to0} \frac{-z^2/2}{z^2}=-\frac{1}{2},$$
so your limit is $$\lim_{z\to0}(\cos z)^{1/z^2}=e^{-1/2}=\frac{1}{\sqrt e}.$$


答:
If you want to use the fact that two roots are equal, then we have that if $a$ is a double root of $f(x)$, then $a$ is a root of $f'(x)$. Use this to narrow down the roots.

You could also use the rational root test to see if there are some trivial roots, we can conclude directly.

Title: Solve a cubic equation?

Body:
Need help with solving an equation:

Solve the equation $5x^3 - 24x^2 + 9x + 54 = 0$ given that two of it's roots are equal.

Any help would be greatly appreciated.
Thanks!