问:
What is a counterexample to LaSalle invariance principle?
I realized I have never understood LaSalle's invariance principle because I have never seen a counterexample to it.

LaSalle's invariance principle is usually stated as,

Let $\Omega \subset D$ be a compact set that is positively invariant
with respect to $\dot x = f(x)$. Let $V: D \subset R$ be a
continuously differentiable function such that $\dot V(x) \leq 0$ in
$\Omega$. Let $E$ be the set of all points in $\Omega$ where $\dot
 V(x) = 0$. let $M$ be the largest invariant set in $E$. Then every
solution starting in $\Omega$ approaches $M$ as $t\to \infty$.

For example, is it possible to come up with an example where the largest invariant set in $E$ is the empty set, hence we do not have conclusion?

or as a corollary to it

Let $x = 0$ be an equilibrium point of $\dot x = f(x)$. Let $V : D ⊂ R ^n → R$ be a
continuously differentiable, positive definite function on $D$ such
that $\dot V (x) ≤ 0$ for all $x ∈ D$. Assume that no solution, other
than the trivial solution $x(t) ≡ 0$, can stay identically in $\{x ∈ D
| \dot V (x) = 0\}$. Then $x = 0$ is asymptotically stable.

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.711.2224&rep=rep1&type=pdf
http://www.cds.caltech.edu/archive/help/uploads/wiki/files/237/Lecture2_notes_CDS270.pdf

For example, the statement "Assume that no solution, other
than the trivial solution $x(t) ≡ 0$, can stay identically in $\{x ∈ D | \dot V (x) = 0\}$." is quite vague to me. Can someone come up with an example where there does exist solution, other than $x(t) = 0$ that stays identically in $\{x ∈ D | \dot V (x) = 0\}$, such that this results in a violation of the theorem?

If not, any example that violates LaSalle invariance principle helps!

答:
Consider the linear dynamics
$$\begin{aligned}
\dot{x}(t) &= -x,\\
\dot{y}(t) &= 0.\\
\end{aligned}
$$

You can verify that if we consider the compact set $\Omega = [-1, 1]^2$ that the Lyapunov candidate

$$V(x) := x^2 + y^2$$

satisfies

$$\dot{V}(x) = - 2 x^2 \leq 0.$$

Moreover, the largest invariant set in $\Omega$ is the $y$-axis,

$$M := \left\{ (0, y) \in \mathbb{R}^2 \colon (x,y) \in \Omega \right\}.$$

In this example we can only show solutions tend to $M$ and can't use the corollary simply because solutions on the $y$-axis don't tend to the origin. That is, there exists a non-trivial solution $\gamma(t) := (0, y_0)$ with $y_0 \neq 0$ that satisfies $\dot{V} \leq 0$ for all time. The crucial fact is that the dynamics on the largest invariant set aren't, themselves, asymptotically tending to the origin. This could be because, like in this example, the dynamics restricted on that set aren't asymptotically stable. But it can be much worse than that. Consider the following example:

$$\begin{aligned}
\dot{x}(t) &= -x (1 - x^2 - y^2)^2\\
\dot{y}(t) &= -y (1 - x^2 - y^2)^2
\end{aligned}$$

and of course consider a compact set $\Omega$ that contains the unit circle.
You can verify that, using the same Lyapunov candidate, you can only conclude solutions tend to the largest invariant set, which in this case is the union of the unit circle and the origin. To see why, observe with the same choice of candidate

$$\dot{V}(x,y) = -2 (x^2 + y^2) (x^2 + y^2 - 1)^2.$$

This is negative everywhere except when $1 = x^2 + y^2$ and at the origin. Solutions on the unit circle will stay there and can never reach the origin even though all the dynamics outside the unit circle and inside the unit circle have strictly decreasing energy. So hopefully that gives you a picture to draw when thinking about why we need to make assumptions of the behaviour of solutions in $E.$

However, what about breaking the premises of your first statement? This is actually pretty difficult to do besides in the naive way. Let us simply take $\dot{V} \leq 0$ for granted. By assuming $V$ is $C^1,$ you have that $\dot{V}$ is continuous. Since $\Omega$ is compact, $\dot{V}$ obtains a minimum and maximum. Suppose, by way of contradiction, that the maximum of $\dot{V}$ is not $0.$ That is,

$$\dot{V}(x') \leq -\varepsilon < 0$$

at some point $x'\in\Omega.$ This would be the case you are thinking about, that $E$ is empty. Except, here is the problem. Since the solution $x(t) \in \Omega$ for all time, as $\Omega$ is positively invariant, we know that

$$V(t) - V(0) = \int_{0}^t \dot{V}(x(t)) dt$$

is well-defined. We can bound this integral with

$$V(t) - V(0) \leq -\varepsilon t.$$

Now we construct a contradiction. Since $V$ is continuous on the compact set $\Omega$ it obtains a maximum $M$ and see that we have

$$V(t) \leq M - \varepsilon t.$$

Pick a sufficiently large value for time $t$ and see that $V$ is not positive definite on $\Omega.$

This demonstrates that $\dot{V}$ must actually attain a value of $0$ somewhere on $\Omega.$ Where? Well that you don't know. But that's enough to show that $E$ is non-empty.


问:
In how many ways 6 coins can be chosen out of 20 coins of type A, 10 of type of B and 7 of type C?
I know a similar question exists, but I am not satisfied with the explanation

The formula used is $\binom {n+r-1}{r-1}$, where $n=6$ and $r=3$

I don’t understand how the is the correct formula. It is the number of solutions of
$x_1+x_2+x_3..x_r=n$ where every solution is non negative.

It can also be described as ‘number of ways of distributing n identical objects among r groups’

Here, the objects aren’t identical, and neither are they being distributed into groups. Then how does using this make sense?

答:
Your main problem seems to stem from your observations below:

It can also be described as ‘number of ways of distributing n identical objects among r groups’

Here, the objects aren’t identical, and neither are they being distributed into groups. Then how does using this make sense?

Well, to understand why this formula applies, one way is to imagine that the coins are identical, and are distributed to three bins, $A,B,C$, where they magically transform into coins of type $A, B,$ and  $C$ respectively.

ContentLicense: CC BY-SA 4.0


问:
hard-sequence problem and limits
Let be $x_n$ a sequence such that $$2\le x_1\le 4$$ and  $$x_{n+1}=\frac{1}{x_n}+\frac{4}{2n+1}$$
Find the value of $$\lim _{n\to \infty }\left(n\left(x_n-1\right)\right)$$
I would like somebody to provide a complete proof showing that the sequence $x_n$ is decreasing and convergent to 1 since the book where this come from states that and overlooked the proof. I would also like to know what I have done wrong there.
I did the following using Stolz Cesaro lemma. Obvious the sequence $\frac{1}{\left(x_n-1\right)}$ is increasing to infinity.
$$\lim _{n\to \infty }\left(n\left(x_n-1\right)\right)=\lim _{n\to \infty }\left(\frac{n}{\frac{1}{\left(x_n-1\right)}}\right)=\lim _{n\to \infty }\left(\frac{n+1-n}{\frac{1}{\left(x_{n+1}-1\right)}-\frac{1}{\left(x_n-1\right)}}\right)=\lim _{n\to \infty }\left(\frac{1}{\frac{1}{\left(x_{n+1}-1\right)}-\frac{1}{\left(x_n-1\right)}}\right)=\lim _{n\to \infty }\left(\frac{\left(x_{n+1}-1\right)\left(x_n-1\right)}{x_n-x_{n+1}}\right)=\lim _{n\to \infty }\left(\frac{\left(\frac{1}{x_n}+\frac{4}{2n+1}-1\right)\left(x_n-1\right)}{x_n-\frac{1}{x_n}+\frac{4}{2n+1}}\right)=\lim _{n\to \infty }\left(\frac{\left(\frac{1}{x_n}-1\right)\left(x_n-1\right)}{x_n-\frac{1}{x_n}}\right)=\lim _{n\to \infty }\left(\frac{\left(1-x_n\right)\left(x_n-1\right)}{x_n^2-1}\right)=\lim _{n\to \infty }\left(\frac{\left(1-x_n\right)}{x_n+1}\right)=\lim _{x\to 1}\left(\left(\frac{\left(1-x\right)}{x+1}\right)\right)=0$$
The book states that the correct answer is 1. What I have done wrong and what is the correct solution?

答:
This is probably not the expected answer, but it is a solution, and it might be of some interest: the recurrence is equivalent to $$x_{n+1}\,x_n=1+\frac{4\,x_n}{2n+1}.$$ With $x_n=y_n/y_{n-1}$, this becomes
$$y_{n+1}=y_{n-1}+\frac4{2n+1}\,y_n.$$ Now $y_n=2n+1$ is a particular solution of this linear recurrence, so let $y_n=(2n+1)\,z_n$. This gives
$$(2n+3)\,z_{n+1}=4\,z_n+(2n-1)\,z_{n-1},$$ and after some elementary algebra,
$$(2n+3)(2n+1)(z_{n+1}-z_n)=-(2n+1)(2n-1)(z_n-z_{n-1}).$$ This means
$$(-1)^{n-1}(2n+1)(2n-1)(z_n-z_{n-1})$$ is constant, and thus
$$(-1)^{n-1}(2n+1)(2n-1)(z_n-z_{n-1})=3\,(z_1-z_0).$$ Obviously, we can assume $y_0=z_0=1$, implying $y_1=3\,z_1=x_1$ and $3\,(z_1-z_0)=x_1-3$. This gives
$$z_n=1+(x_1-3)\,\sum^n_{k=1}\frac{(-1)^{k-1}}{(2k-1)(2k+1)}.$$ It's clear that $z_n$ converges to some non-zero value $z_\infty$ for all $x_1$ except when
$$z_\infty=1+(x_1-3)\,\sum^\infty_{k=1}\frac{(-1)^{k-1}}{(2k-1)(2k+1)}=0,$$ i.e.
$$x_1=3-\frac4{\pi-2}$$ (since the sum of the infinite series is $\frac{\pi}4-\frac12$). From $z_n=z_\infty+O(n^{-2})$ immediately follows
$$\frac{z_n}{z_{n-1}}=1+O(n^{-2})$$ and $$n\,(x_n-1)=n\,\left(\frac{2n+1}{2n-1}\,\frac{z_n}{z_{n-1}}-1\right)=\frac{2n}{2n-1}+O(n^{-1}).$$

Title: hard-sequence problem and limits


问:
Derive Equation for Dual Curve to Ellipse not at Origin Given Implicitly
This is a second follow-up to this question. For the ellipse centered at $(h,k)$ defined implicitly as

$$
\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}-1=0 \tag{1}
$$

Please show the steps for calculating its dual curve and provide the equation for the dual curve. Do not solve by redefining the function parametrically. I know how to solve for parametric functions, but need to see the general algorithm for implicit functions.

Wikipedia provides a trivial example that does not help me.

My Attempt

At point $(p,q)$,

$$
X = \lambda \frac{2(p-h)}{a^2} \implies p = h + \left( \frac{a^2}{2\lambda} \right) X
$$

$$
Y = \lambda \frac{2(q-k)}{b^2} \implies q = k + \left( \frac{b^2}{2\lambda} \right) Y
$$

For incidence to occur, the inner product of the line $[X,Y]$ and point $(p,q)$ must be zero:

$$
\frac{a^2}{2\lambda}X^2 + hX + \frac{b^2}{2\lambda}Y^2 + kY = 0
$$

$$
a^2X^2 + b^2Y^2 + 2\lambda (hX + kY) = 0 \tag{2}
$$

How should $\lambda$ be eliminated at this point?

Thought 1:

Can we homogenize the linear terms by $\frac{\left( hX + kY \right)}{-2\lambda}$? If so, we get

$$
a^2X^2 + b^2Y^2 - (hX+kY)^2 = 0
$$

$$
a^2X^2 + b^2Y^2 - (h^2X^2 + 2hkXY + k^2Y^2) = 0
$$

$$
(a^2-h^2)X^2 + (b^2-k^2)Y^2 - 2hkXY = 0
$$

but I'm not sure if this is the correct equation for the dual. Is there a way to check?

Thought 2:

Alternatively, can I just say that $\lambda = 1$? In this case, I get:

$$
a^2X^2+b^2Y^2+2(hX+kY)=0
$$

However, now my answer in Thought 1 is different from Thought 2.

答:
The dual conics are for $ax^2+bxy+cy^2+dx+ey+f=0$ given by $$ (-\frac14e^2+cf)x^2+(\frac12de-bf)xy+(-\frac14d^2+af)y^2+(-cd+\frac12be)x+(\frac12bd-ae)y-\frac14b^2+ac=0$$ or $\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}a&\frac{b}{2}&\frac{d}{2}\\\frac{b}{2}&c&\frac{e}{2}\\\frac{d}{2}&\frac{e}{2}&f\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$ to $$\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}-\frac14e^2+cf&\frac{\frac12de-bf}{2}&\frac{-cd+\frac12be}{2}\\\frac{\frac12de-bf}{2}&-\frac14d^2+af&\frac{\frac12bd-ae}{2}\\\frac{-cd+\frac12be}{2}&\frac{\frac12bd-ae}{2}&-\frac14b^2+ac\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$$

from which you can see it's to do with the matrix of a conic and it's adjoint matrix which is the inverse when divided by the determinant.

For an explicit derivation see Perspectives on Projective Geometry by Jürgen Richter-Gebert:

A conic consists of all points p that satisfy an equation $p^TAp = 0.$ The set
of all tangents to this conic can be described as $\{Ap | p^TAp = 0\}.$ We can
describe this set of homogeneous coordinates for lines directly as a quadratic
form by the following observation:

$p^TAp = p^TAA^{−1}Ap = p^TA^TA^{−1}Ap = (Ap)^TA^{−1}(Ap).$

The right side of the equation explains how the set of tangent lines of a
conic may be directly interpreted as the zero set of a quadratic form with
matrix $A^{−1}$.

Edit:

The $\lambda$-method works in Macaulay2:

```<code>R=QQ[a,b,c,d,e,f]
S=R[l]
T=S[p,q,r,X,Y,Z,MonomialOrder=>Lex]
I=ideal(X-l*(2*a*p+b*q+d*r),Y-l*(b*p+2*q*c+e*r),Z-l*(d*p+e*q+2*f*r),X*p+Y*q+Z*r)
gens gb I -- matrix {{(e^2-4*c*f)*X^2+(-2*d*e+4*b*f)*X*Y+(4*c*d-2*b*e)*X*Z+(d^2-4*a*f)*Y^2+(-2*b*d+4*a*e)*Y*Z+(b^2-4*a*c)*Z^2, (2*c*d^2-2*b*d*e+2*a*e^2+2*b^2*f-8*a*c*f)*l*r+(-2*c*d+b*e)*X+(b*d-2*a*e)*Y+(-b^2+4*a*c)*Z,(e^2-4*c*f)*l*r*X+(-d*e+2*b*f)*l*r*Y+(2*c*d-b*e)*l*r*Z-e*X*Y+2*c*X*Z+d*Y^2-b*Y*Z, (d*e-2*b*f)*l*r*X+(-d^2+4*a*f)*l*r*Y+(b*d-2*a*e)*l*r*Z-e*X^2+d*X*Y+b*X*Z-2*a*Y*Z,(2*c*d-b*e)*l*r*X+(-b*d+2*a*e)*l*r*Y+(b^2-4*a*c)*l*r*Z-2*c*X^2+2*b*X*Y-2*a*Y^2, (2*c*d-b*e)*l*q+(d*e-2*b*f)*l*r-d*Y+b*Z, (b*d-2*a*e)*l*q+(d^2-4*a*f)*l*r-d*X+2*a*Z, (b^2-4*a*c)*l*q+(b*d-2*a*e)*l*r-b*X+2*a*Y,(2*c*d^2-2*b*d*e+2*a*e^2)*q*Y^2+(2*b^2*e-8*a*c*e)*q*Y*Z+(-2*b^2*c+8*a*c^2)*q*Z^2+(d*e^2-4*c*d*f)*r*X*Y+(-b*e^2+4*b*c*f)*r*X*Z+(-d^2*e+4*a*e*f)*r*Y^2+(4*c*d^2-2*a*e^2-8*a*c*f)*r*Y*Z+(-4*b*c*d+b^2*e+4*a*c*e)*r*Z^2,(2*c*d-b*e)*q*X+(-b*d+2*a*e)*q*Y+(b^2-4*a*c)*q*Z+(d*e-2*b*f)*r*X+(-d^2+4*a*f)*r*Y+(b*d-2*a*e)*r*Z, e*l*q*X-d*l*q*Y+2*f*l*r*X-d*l*r*Z-X*Z, 2*c*l*q*X-b*l*q*Y+e*l*r*X-b*l*r*Z-X*Y, b*l*q*X-2*a*l*q*Y+d*l*r*X-2*a*l*r*Z-X^2,e*q*X*Y-2*c*q*X*Z-d*q*Y^2+b*q*Y*Z+2*f*r*X*Y-e*r*X*Z-d*r*Y*Z+b*r*Z^2, e*q*X^2-d*q*X*Y-b*q*X*Z+2*a*q*Y*Z+2*f*r*X^2-2*d*r*X*Z+2*a*r*Z^2, 2*c*q*X^2-2*b*q*X*Y+2*a*q*Y^2+e*r*X^2-d*r*X*Y-b*r*X*Z+2*a*r*Y*Z, d*l*p+e*l*q+2*f*l*r-Z,b*l*p+2*c*l*q+e*l*r-Y, 2*a*l*p+b*l*q+d*l*r-X, 2*a*p*Z-e*q*X+d*q*Y+b*q*Z-2*f*r*X+2*d*r*Z, d*p*Y-b*p*Z+e*q*Y-2*c*q*Z+2*f*r*Y-e*r*Z, 2*a*p*Y-2*c*q*X+2*b*q*Y-e*r*X+d*r*Y+b*r*Z, p*X+q*Y+r*Z}}
</code>```

$$(e^2-4cf)X^2+(-2de+4bf)XY+(4cd-2be)XZ+(d^2-4af)Y^2+(-2bd+4ae)YZ+(b^2-4ac)Z^2=0$$

It even gives you via

```<code>oo // (gens I)
</code>```

the combination of the generators that give this relation:

$$(X-\lambda (2ap+bq+dr))((e^2-4cf)X+(-de+2bf)Y+(2cd-be)Z)\\+(Y-\lambda (bp+2qc+er))((-de+2bf)X+(d^2-4af)Y+(-bd+2ae)Z)\\+(Z-\lambda (dp+eq+2fr))((2cd-be)X+(-bd+2ae)Y+(b^2-4ac)Z)\\ + (pX + qY + rZ)((2cd^2-2bde+2ae^2+2b^2f-8acf)\lambda)$$

Body:
The dual conics are for $ax^2+bxy+cy^2+dx+ey+f=0$ given by $$ (-\frac14e^2+cf)x^2+(\frac12de-bf)xy+(-\frac14d^2+af)y^2+(-cd+\frac12be)x+(\frac12bd-ae)y-\frac14b^2+ac=0$$ or $\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}a&\frac{b}{2}&\frac{d}{2}\\\frac{b}{2}&c&\frac{e}{2}\\\frac{d}{2}&\frac{e}{2}&f\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$ to $$\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}-\frac14e^2+cf&\frac{\frac12de-bf}{2}&\frac{-cd+\frac12be}{2}\\\frac{\frac12de-bf}{2}&-\frac14d^2+af&\frac{\frac12bd-ae}{2}\\\frac{-cd+\frac12be}{2}&\frac{\frac12bd-ae}{2}&-\frac14b^2+ac\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$$

from which you can see it's to do with the matrix of a conic and it's adjoint matrix which is the inverse when divided by the determinant.

For an explicit derivation see Perspectives on Projective Geometry by Jürgen Richter-Gebert:

A conic consists of all points p that satisfy an equation $p^TAp = 0.$ The set
of all tangents to this conic can be described as $\{Ap | p^TAp = 0\}.$ We can
describe this set of homogeneous coordinates for lines directly as a quadratic
form by the following observation:

$p^TAp = p^TAA^{−1}Ap = p^TA^TA^{−1}Ap = (Ap)^TA^{−1}(Ap).$

The right side of the equation explains how the set of tangent lines of a
conic may be directly interpreted as the zero set of a quadratic form with
matrix $A^{−1}$.

Edit:

The $\lambda$-method works in Macaulay2:

R=QQ[a,b,c,d,e,f]
S=R[l]
T=S[p,q,r,X,Y,Z,MonomialOrder=>Lex]
I=ideal(X-l*(2*a*p+b*q+d*r),Y-l*(b*p+2*q*c+e*r),Z-l*(d*p+e*q+2*f*r),X*p+Y*q+Z*r)
gens gb I -- matrix {{(e^2-4*c*f)*X^2+(-2*d*e+4*b*f)*X*Y+(4*c*d-2*b*e)*X*Z+(d^2-4*a*f)*Y^2+(-2*b*d+4*a*e)*Y*Z+(b^2-4*a*c)*Z^2, (2*c*d^2-2*b*d*e+2*a*e^2+2*b^2*f-8*a*c*f)*l*r+(-2*c*d+b*e)*X+(b*d-2*a*e)*Y+(-b^2+4*a*c)*Z,(e^2-4*c*f)*l*r*X+(-d*e+2*b*f)*l*r*Y+(2*c*d-b*e)*l*r*Z-e*X*Y+2*c*X*Z+d*Y^2-b*Y*Z, (d*e-2*b*f)*l*r*X+(-d^2+4*a*f)*l*r*Y+(b*d-2*a*e)*l*r*Z-e*X^2+d*X*Y+b*X*Z-2*a*Y*Z,(2*c*d-b*e)*l*r*X+(-b*d+2*a*e)*l*r*Y+(b^2-4*a*c)*l*r*Z-2*c*X^2+2*b*X*Y-2*a*Y^2, (2*c*d-b*e)*l*q+(d*e-2*b*f)*l*r-d*Y+b*Z, (b*d-2*a*e)*l*q+(d^2-4*a*f)*l*r-d*X+2*a*Z, (b^2-4*a*c)*l*q+(b*d-2*a*e)*l*r-b*X+2*a*Y,(2*c*d^2-2*b*d*e+2*a*e^2)*q*Y^2+(2*b^2*e-8*a*c*e)*q*Y*Z+(-2*b^2*c+8*a*c^2)*q*Z^2+(d*e^2-4*c*d*f)*r*X*Y+(-b*e^2+4*b*c*f)*r*X*Z+(-d^2*e+4*a*e*f)*r*Y^2+(4*c*d^2-2*a*e^2-8*a*c*f)*r*Y*Z+(-4*b*c*d+b^2*e+4*a*c*e)*r*Z^2,(2*c*d-b*e)*q*X+(-b*d+2*a*e)*q*Y+(b^2-4*a*c)*q*Z+(d*e-2*b*f)*r*X+(-d^2+4*a*f)*r*Y+(b*d-2*a*e)*r*Z, e*l*q*X-d*l*q*Y+2*f*l*r*X-d*l*r*Z-X*Z, 2*c*l*q*X-b*l*q*Y+e*l*r*X-b*l*r*Z-X*Y, b*l*q*X-2*a*l*q*Y+d*l*r*X-2*a*l*r*Z-X^2,e*q*X*Y-2*c*q*X*Z-d*q*Y^2+b*q*Y*Z+2*f*r*X*Y-e*r*X*Z-d*r*Y*Z+b*r*Z^2, e*q*X^2-d*q*X*Y-b*q*X*Z+2*a*q*Y*Z+2*f*r*X^2-2*d*r*X*Z+2*a*r*Z^2, 2*c*q*X^2-2*b*q*X*Y+2*a*q*Y^2+e*r*X^2-d*r*X*Y-b*r*X*Z+2*a*r*Y*Z, d*l*p+e*l*q+2*f*l*r-Z,b*l*p+2*c*l*q+e*l*r-Y, 2*a*l*p+b*l*q+d*l*r-X, 2*a*p*Z-e*q*X+d*q*Y+b*q*Z-2*f*r*X+2*d*r*Z, d*p*Y-b*p*Z+e*q*Y-2*c*q*Z+2*f*r*Y-e*r*Z, 2*a*p*Y-2*c*q*X+2*b*q*Y-e*r*X+d*r*Y+b*r*Z, p*X+q*Y+r*Z}}

$$(e^2-4cf)X^2+(-2de+4bf)XY+(4cd-2be)XZ+(d^2-4af)Y^2+(-2bd+4ae)YZ+(b^2-4ac)Z^2=0$$

It even gives you via

oo // (gens I)

the combination of the generators that give this relation:

$$(X-\lambda (2ap+bq+dr))((e^2-4cf)X+(-de+2bf)Y+(2cd-be)Z)\\+(Y-\lambda (bp+2qc+er))((-de+2bf)X+(d^2-4af)Y+(-bd+2ae)Z)\\+(Z-\lambda (dp+eq+2fr))((2cd-be)X+(-bd+2ae)Y+(b^2-4ac)Z)\\ + (pX + qY + rZ)((2cd^2-2bde+2ae^2+2b^2f-8acf)\lambda)$$


问:
Subgroup of order n of group of order 2n
There is a subgroup of order $n$ in a group of order $2n$. Prove, that all squares of elements lie in this subgroup.

I thought about some solution, with the usage of Lagrange theorem and normality of this subgroup, but I didn't come up with anything good.

答: The subgroup of order  $n $ is normal.  Thus $G/N $ is a group of order $2$.  There are two equivalence classes, $N $ and $xN $, for some $x\in G $.  If $g\in N $, then $g^2\in N $, since  $N $ is a subgroup.   If $g\notin N $, then  $g=xn $ for some  $n\in N$.  But then $g^2=x^2n'\in N $, since $xN $ has order two.


问:
Structure Theorem for non-abelian finite groups or rings
How many structure theorems do we have in Abstract Algebra for finite algebraic structures? I know some of the following theorems:

If $G$ is a finite abelian group, then $G$ is a product cyclic groups of prime power order.

If $R$ is a finite commutative ring, then $R$ is Artinian and hence a direct product of local rings.

I am wondering if there are other structure theorems for finite algebraic structures.
Is it possible to write a non-commutative or a non-abelian group in this way?

Why are the theorems available for abelian cases but not for non-abelian ones?

If someone can please help me find some references where I can get structure theorems for non-commutative or a non-abelian group/rings, I will be thankful.

Please help.

答:
Non-abelian finite groups are quite complicated, but we have completely classified the finite simple groups. Here's an overview from Wikipedia. Moreover, every finite group can be "built out of finite simple groups" in an essentially unique way, in the same sense that every nonzero integer can be written as a product of primes in an essentially unique way! If you want to know precisely what this means, this is the content of the Jordan–Hölder theorem (you will first need to learn what a composition series is). In contrast to prime factorizations, where we just multiply the primes to reproduce the original integer, there are many complicated ways that simple groups can be put together to form a non-simple group. The general problem of understanding the possible isomorphism types of a group from its simple factors is what's known as an "extension problem". In short, while we know all the finite simple groups, and we know  that all finite groups are built from these, we are not close to understanding all finite groups.

Non-commutative finite rings are quite complex! Wedderburn's little theorem says that a finite ring with no zero divisors is a field. Besides this, I don't know of any general classification results for finite rings. Even the finite commutative rings are quite difficult to classify: we easily reduce to a product of finite local rings, but those are complicated to describe. The good news is that, by considering the possible underlying abelian groups, it is not too computationally expensive to compute all isomorphism classes of finite commutative local rings of any given cardinality $n$.

I also feel obligated to mention the $5/8$ theorem:

Theorem ($5/8$ Theorem for Finite Groups) Let $G$ be a finite group such that the probability that two randomly chosen elements of $G$ commute is greater than $5/8$. Then $G$ is abelian.

Theorem ($5/8$ Theorem for Finite Rings) Let $G$ be a finite ring such that the probability that two randomly chosen elements of $G$ commute is greater than $5/8$. Then $G$ is commutative.

See here for a proof, or here for a generalization to compact topological groups.


问:
Can anyone please tell me why it is called Parametric equation?
Parametric representation of the parabola $y^2 = x$  is $(t^2 , t)$.

Can anyone please tell me why it is called Parametric representation ?

Can I say $(x , \pm \sqrt x)$  is a parametric representation of $y^2 = x$ ?

I know for a family of equations for every value of parameter we get a particular member  of that family.

$x^2+y^2 = a^2$  is the family of circles with center at origin. If we put $a= 1$ , we get a particular circle. $a$ is also varying but for a particular value of $a$ we are getting a member. That is why , this is little different from variable.

答:
It is called a parametric equation, because we can relate the values of x and y (or even z, if we consider a 3-D space) to a single parameter (let's call it t).

Consider the example you gave, of the parabola $y^{2}=x$. We can relate the values of x and y, by a single parameter t, which gives the value of x and y, as the parameter changes, both x and y change also, such that it satisfies the original equation.

And no, the example you gave isn't a parametric equation because it doesn't satisfy the original equation.


问:
A Linear Algebra Doubt
Suppose we have a $3$-dim vector space $V$ and we choose any non orthogonal
set of basis ${v_1,v_2,v_3}$. Now we consider the linear transformation that
projects any vector in $V$ to span$(v_1,v_2)$. Clearly the matrix representation for this transformation is
$$\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix}$$. The eigenvectors corresponding to eigen-value $1$
are ${v_1,v_2}$, and corresponding to $0$ is $v_3$. As the matrix is symmetric,
it should have orthogonal eigenvectors. But $\langle v_1,v_3\rangle$ should be non zero
as they are non orthogonal. Why is this?

答:
You're trying to invoke the following result:

Suppose $A$ is a symmetric real matrix (or a Hermitian complex matrix, if you prefer). Then the eigenspaces of $A$ are orthogonal.

This is true. The matrix you provided has orthogonal eigenspaces $\operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$ and $\operatorname{span}\{(0, 0, 1)\}$. Note that these are subspaces of $\Bbb{R}^3$, which may or may not have anything to do with the inner product space $V$. Thus, it is not correct to say that these are eigenspaces of your projection operator $P$.

If you want the equivalent to the above result for general inner product spaces, this is it:

Suppose $V$ is a finite-dimensional inner product space, and $T : V \to V$ is self-adjoint. Then the eigenspaces of $T$ are orthogonal.

In our case, the projection map
$$P : a_1 v_1 + a_2 v_2 + a_3 v_3 \mapsto a_1 v_1 + a_2 v_2$$
is not self-adjoint if $v_3$ is not orthogonal to $v_1$ and $v_2$. Let's suppose it's not orthogonal to $v_1$. Then
$$\langle Pv_1, v_3 \rangle = \langle v_1, v_3 \rangle \neq 0 = \langle v_1, 0 \rangle = \langle v_1, Pv_3\rangle.$$

So, as you've observed, it's perfectly possible to take a linear operator that is not self-adjoint to a symmetric (or Hermitian) matrix, by way of a non-orthonormal basis. Just be aware of which theorems about inner product spaces require bases to be orthonormal to preserve inner product space properties.

Title: A Linear Algebra Doubt


问:
Why is $\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx = \frac{\pi}{6}$?
Wolfram alpha computes that $\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx = \frac{\pi}{6}$.

Why is this true? I thought this looked like something which could be justified with contour integrals, but I wasn't able to find a contour (or at least an easy contour) which gives the result.

What's a technique that can be used to compute this integral?

答:
Using $2\cos x=e^{ix}+e^{-ix}$ the integral can be represented as:
$$\begin{align}
\int_{-\infty}^\infty f(x)dx,\quad\text{with}\quad f(x)=\frac{e^{ix}-1+\frac{x^2}2}{x^4}.
\end{align}$$
The integrated function $f(x)$ has a singularity (pole of third order) at $x=0$ and we are looking for the principal value of the integral. Completing the integration path in the upper complex half-plane with a semicircle of large radius $R$  and letting the radius tend to infinity we obtain:
$$
\int_{-\infty}^\infty f(x)dx=\oint_C f(z)dz=\frac12\times 2\pi i\operatorname{res}_{z=0}f(z)=\pi i\frac{-i}6=\frac\pi6.
$$


问:
Strictly nondecreasing permutations of numbers
Suppose I have a 6 digit lock that accepts 0-9 as inputs for the 6 slots.

And I was told that the numbers in order from left to right are never less than the number to their left.
so 0,0,0,0,0,0 would be valid but 0,1,2,3,2,9 would not.

I tried to brute force it by looking for a pattern in increasing the number of slots on the lock.
If it allowed a single digit, 10 passwords.
if it allowed 2 digits, 55 passwords (I think)

10 that start with 0,

9 that start with 1,

8 that start with 2,

ect.

10+9+8+...+3+2+1=55

3 gets a lot bigger, 100 that start with 0, then 9 that start with "1,1" 8 that start with "1,2" but from there I get lost with my counting and its going to be a huge nightmare when I get up to 6.

Please tell me there is a better way!
Thank you.

答:
This is a "stars and bars" problem.  If you know how many examples of each digit appear in your number, the non-decreasing condition forces a unique solution.  So you need to know how many solutions there are to:

$$\sum_{k=0}^9 x_k=n,$$

where $x_k$ is the number of times the digit $k$ appears in your number, and $n$ is the length of your number.  That number is $\binom{n+9}{9}$.


答: You can also use a well known property of the Laplace transform $$\int_{0}^{+\infty}f\left(x\right)g\left(x\right)dx=\int_{0}^{+\infty}\left(\mathcal{L}f\right)\left(y\right)\left(\mathcal{L}^{-1}g\right)\left(y\right)dy$$ where $\left(\mathcal{L}f\right),\,\left(\mathcal{L}^{-1}g\right)$ is the Laplace and the inverse Laplace transform, respectively. Hence $$\int_{0}^{+\infty}\frac{2\cos\left(x\right)-x^{2}+2}{x^{4}}dx=\frac{1}{3}\int_{0}^{+\infty}\frac{y^{3}}{y^{5}+y^{3}}dy=\frac{1}{3}\int_{0}^{+\infty}\frac{1}{y^{2}+1}dy=\color{red}{\frac{\pi}{6}}.$$

Title: Why is $\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx = \frac{\pi}{6}$?

Body:
Wolfram alpha computes that $\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx = \frac{\pi}{6}$.

Why is this true? I thought this looked like something which could be justified with contour integrals, but I wasn't able to find a contour (or at least an easy contour) which gives the result.

What's a technique that can be used to compute this integral?


问:
Solving system of ODE with elimination method for $x'-2x+y'=0$ and $x'-x+y'-y=0$
Suppose that a system of ODE is comprised of
$$x'-2x+y'=0$$
$$x'-x+y'-y=0$$

Naturally, I'll reformat this into the following:

$$(1) \ \ \ (D-2)x+(D)y=0$$
$$(2) \ \ \ (D-1)x+(D-1)y=0$$

And when I eliminate the y, I get $(-2D+2)x=0$ and solving for $D$ gets $D=1$ so one of my general solution is $x(t)=c_1e^x$ but this is where the problem happens. Say I plug that solution into $(1)$, then we'd get $-c_1e^x+(D)y=0$. How am I supposed to find the solution on the Y side? Or is there a better way of solving for this ODE?

答:
$$x'-2x+y'=0$$
$$x'-x+y'-y=0$$
Substract both DE:
$$-2x+x+y=0$$
$$\implies y=x$$
Substitute this in the first DE and solve.
$$x'-2x+y'=0$$
$$x'-2x+x'=0$$
$$x'-x=0$$
This is easy to solve then deduce $y$ since $y=x$.


问:
Basic doubt about stochastic integrals over general local martingales
Consider $M = (M_t)$ is a continuous square integrable local martingale and $$ \mathbb H ^2(M):= \left \{ \psi =(\psi_t)\  \text{is a real previsible process s.t.,} \forall t\geq 0, \ \mathbb E\left \{ \int _0 ^t  \psi _s ~d \langle M \rangle _s \right \}< + \infty \ \text {a.e.}  \right\}. $$
 If $ \phi =(\phi_t)   \in \mathbb H ^2(M)$ can we conclude that $I_t =\left(\int _0 ^ t \phi_s dM_s\right) $ is a martingale ?

My doubt comes from this problems also asked here.

Thank's in advance for your enlightenments.

答:
As I said in my comment, no you cannot conclude that $I_t$ is a martingale solely from the fact that the integral is well defined, since $1 \in \mathbb{H}^2(M)$.

The result you want to complete your answer to the other question uses the following characterization of martingales (Prop 2.9): A local martingale $X_t$ is a true martingale if and only if the collection $\chi_t = \{ T : T \mbox{ is a stopping time and }T \leq t \}$ is uniformly integrable.  A consequence of this is that if your local martingale is dominated by an integrable random variable, then it is a martingale.  For the local martingale in that problem, $1$ is a dominating function.

ContentLicense: CC BY-SA 3.0


问:
Convergence using the ratio test
I am trying to determine the smallest value for natural number $k$ such that $$\sum\limits_{n=1}^\infty\frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}$$ converges. I attempted to solve using the ratio test:

$$\text{If} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1, \text{then the series $\sum\limits_{n=1}^\infty a_n$ is absolutely convergent.}$$

Using $a_n = \frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}$ and $a_{n+1} = \frac{((n+1)!)^{k-1}(27(n+1))^{n+1}}{(k(n+1))!e^{n+1}}$, I tried to simplify the following ratio like this:

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim_{n\to\infty}\left|\frac{\frac{((n+1)!)^{k-1}(27(n+1))^{n+1}}{(k(n+1))!e^{n+1}}}{\frac{(n!)^{k-1}(27n)^n}{(kn)!e^n}}\right|$$

$$= \lim_{n\to\infty}\left|\frac{\frac{(n+1)^{k-1}(n!)^{k-1}(27)^{n+1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)(kn)!e^{n+1}}}{\frac{(n!)^{k-1}(27^n)n^n}{(kn)!e^n}}\right|$$

$$= \frac{27^{n+1}e^n}{27^ne^{n+1}}\lim_{n\to\infty}\left|\frac{\frac{(n+1)^{k-1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)}}{\frac{n^n}{1}}\right|$$

$$ = \frac{27}{e}\lim_{n\to\infty}\left|\frac{(n+1)^{k-1}(n+1)^{n+1}}{(kn+k)(kn+(k-1))\cdots(kn+1)n^n}\right|$$

From here, I consider the only the highest order term in the numerator and denominator, with there only being $k$ terms remaining from the factorial, yielding leading term $(kn)^k = k^kn^k$,

$$ = \frac{27}{e}\lim_{n\to\infty}\left|\frac{n^{k+n} + O(n^{k+n-1})}{k^kn^{k+n}+O(n^{k+n-1})}\right|$$

Solving the limit by dividing top and bottom by ${n^{k+n}}$ gives

$$ = \frac{27}{e}\frac{1}{k^k}$$

Which would be first less than 1 and therefore convergent at $k=3$, i.e. $\frac{27}{e}\frac{1}{3^3} = \frac{1}{e} < 1$. Then, by the ratio test, $k=3$ would be the first instance that this is true, because at $k=1$ we get $\frac{27}{e} > 1$ and at $k=2$ we get $\frac{27}{4e} > 1$, which diverge by the ratio test.

However, when I punch $\sum\limits_{n=1}^\infty\frac{(n!)^{3-1}(27n)^n}{(3n)!e^n}$ into Wolfram Alpha, it states that the series diverges, when the ratio test above shows that it converges for $k=3$. I'm not sure where my work is wrong and any pointers to the right answer of $k=4$ would be much appreciated. Thanks in advance!

答:
$$a_n=\frac{(n!)^{k-1}\,(27n)^n}{(kn)!\,e^n}\implies$$
$$\log(a_n)=(k-1)\log(n!)+n \log(27n)-n-\log((kn)!)$$ Using Stirling approximation
$$\log(a_n)=n (\log (27)-k \log (k))+\frac{1}{2} ((k-2) \log (2 \pi  n)-\log (k))+\frac{(k-1)
   k-1}{12 k n}+O\left(\frac{1}{n^3}\right)$$ Apply it twice and continue with Taylor series to obtain
$$\log(a_{n+1})-\log(a_n)=(\log (27)-k \log
   (k))+\frac{k-2}{2n}+O\left(\frac{1}{n^2}\right)$$
$$\frac{a_{n+1} } {a_n }=e^{\log(a_{n+1})-\log(a_n) }\sim 27 k^{-k}\, e^{\frac{k-2}{2n} }$$ So, you need $k>3$ for convergence.

If $k=3$, we have
$$\frac{a_{n+1} } {a_n }=\frac{9 n^{-n} (n+1)^{n+2}}{e (9 n (n+1)+2)}\to 1$$ then divergence

Title: Convergence using the ratio test


答:
A parameter is one or more numbers that is varied to give a collection of "things". Each value of the parameter(s) should give a single thing. For instance,

In $(t,t^2)$, the parameter $t$ can be varied to give a collection of points

In $x^2+y^2=a^2$, the parameter $a$ can be varied to give a collection of circles

In $\displaystyle\frac1{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}$, the two parameters $\sigma$ and $\mu$ can be varied to give different distribution functions for different normal distributions

In the case where the collection of things is a collection of points in the plane, or in higher dimensional space, then we call that a parametric representation of that collection of points. Because we use a parameter to represent them.

Title: Can anyone please tell me why it is called Parametric equation?

Body:
Parametric representation of the parabola $y^2 = x$  is $(t^2 , t)$.

Can anyone please tell me why it is called Parametric representation ?

Can I say $(x , \pm \sqrt x)$  is a parametric representation of $y^2 = x$ ?

I know for a family of equations for every value of parameter we get a particular member  of that family.

$x^2+y^2 = a^2$  is the family of circles with center at origin. If we put $a= 1$ , we get a particular circle. $a$ is also varying but for a particular value of $a$ we are getting a member. That is why , this is little different from variable.


答:
The idea with a parametric equation (with a single real parameter) is that it "draws the curve" as you vary the parameter.

For the parabola $y^2=x$, your attempt $(x,\pm\sqrt x)$ makes sense, but multi-valued expressions like $\pm...$ aren't allowed. This would "draw two curves at once" as $x$ moves through positive values (and draw nothing when $x$ is negative). In contrast, the parametrization $(t^2,t)$ gives a clean one-to-one correspondence (in fact a homeomorphism) between the parameter space $\Bbb R$ and the parabola.

$x^2+y^2=a^2$ is an equation with a parameter that describes a family of circles, and positive values of $a$ are in one-to-one correspondence with the members of the family. But since it's not a family of points, I probably wouldn't call it a parametric equation.


问:
Distance from the intersection point of two circles to a certain point on the circumference of the greater circle
On line segment $AB$ whose length is $6$, we take $AC = 3$ and $DB = 4$. Two circles are drawn taking $A$ and $B$ and centres and $AC$ and $BD$ as radii respectively. The circles intersect at points $P$ and $Q$. We join $A$ and $P$, and extend $AP$ so that it meets the circumference of the greater circle at $E$. How would one then find the length of $PE$?

答:
Continue the line $AB$ till the second  intersection with the greater circle. Let the new intersection point be $F$.

By the power of a point we have:
$$
AP\cdot AE=AD\cdot AF
$$
Can you take it from here?

$$PE=\frac{AD\cdot AF}{AP}-AP=\frac{2\cdot10}{3}-3=\frac{11}3.$$


问:
Easiest way to solve $\frac{2}{9}x=\sin(\frac{\pi}{9}x)$
We wanted to find the points of intersection between 2 curves, so we have to solve this equation $\frac{2}{9} x= \sin(\frac{\pi}{9}x)$.

I couldn't figure out an easy way to find the points of intersection, they are $-\frac{9}{2}$, $0$, and $\frac{9}{2}$.

Any ideas please on how to solve such equations?

答:
What tools are you allowed to use to solve this problem? There are multiple ways to solve this based on your background. If it's just basic algebra and trig identities, list all the values of $x$ for which $\sin(\frac{\pi}{9}x) = c$ where $c$ are the known sin values for special angles. Then it's just a matter of plugging in those values for the LHS and see if it's equal. Now you know that sin only gives a value between -1 and 1, so your LHS will also be restricted so x values can be restricted very quickly.

So $|\frac{2}{9}x| \leq 1$, then $|x| \leq 9/2$. So you'll only have to test angles between $0$ and $\pm \frac{\pi}{2}$ since $\frac{\pi}{9}(\frac{9}{2})=\frac{\pi}{2}$


问:
For transformations from $F^n \rightarrow F^m$ does $T(u)=[T]_\beta^\gamma (u)$?
I am currently studying linear algebra and there is a proof that goes as follows:

Theorem 2.14. Let V and W be finite-dimensional vector spaces having ordered bases $\beta$ and $\gamma$, respectively, and let $T:V \rightarrow W$ be linear. Then, for each $u∈V$, we have

$$[T(u)]_\gamma=[T]_\beta^\gamma[u]_\beta$$

I am wondering if we are working with tuples, where the transformations are from  $F^n \rightarrow F^m$, can we extrapolate from the theorem that: $T(u)=[T]_\beta^\gamma *u$?

答:
No, not necessarily. We could take, for example,
\begin{align*}
T\begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} y \\ x \end{bmatrix} \\
\beta, \gamma &= \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix}\right) \\
u &= \begin{bmatrix} 1 \\ 1 \end{bmatrix}.
\end{align*}
Then,
\begin{align*}
T\begin{bmatrix} 1 \\ 0 \end{bmatrix} &= \begin{bmatrix} 0 \\ 1 \end{bmatrix} = -1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1\begin{bmatrix} 1 \\ 1 \end{bmatrix} \\
T\begin{bmatrix} 1 \\ 1 \end{bmatrix} &= \begin{bmatrix} 1 \\ 1 \end{bmatrix} = 0\begin{bmatrix} 1 \\ 0 \end{bmatrix} + 1\begin{bmatrix} 1 \\ 1 \end{bmatrix}.
\end{align*}
Thus,
$$[T]_\beta^\gamma = \begin{bmatrix} -1 & 0 \\ 1 & 1 \end{bmatrix}.$$
Now, we have
$$Tu = T\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix},$$
but
$$[T]_\beta^\gamma u = \begin{bmatrix} -1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix} \neq T(u).$$


问:
Consistency of linear equations in 3 variables
Which condition $p$, $q$ and $r$  should satisfy so that the following system of linear simultaneous equations has at least one solution, such that $p+q+r\neq0?$

$x+2y-3z=p$

$2x+6y-11z=q\\x-2y+7z=r$

I approached this problem using Gauss Elimination method by creating an augmented matrix and reducing it to row echelon form. I got something like this :-
$$
  \left[\begin{array}{rrr|r}
    1 & 2 & -3 & p \\
    2 & 6 & -11 & q \\
    1 & -2 & 7 & r
  \end{array}\right] =
  \left[\begin{array}{rrr|r}
    1 & 2 & -3 & p \\
    0 & 2 & -5 & q-2p \\
    0 & 0 & -11 & r+2q-5p
  \end{array}\right]
$$

Now we can see that the co-efficient matrix and augmented matrix, both has rank of 3 which equals the number of variables present i.e. $3$. So in this condition there must be unique solution to this system of simultaneous linear equations.

I am getting confused by the language of the question which says that there must be "at least one solution". So does this mean that this system of simultaneous linear equations should have more than 1 solutions? How can this be possible if the Gauss Elimination method is telling that there should be a unique solution?

Also what can we say about the relationship between $p$,$q$ and $r?$ As per my understanding, in order to keep the rank 3 because then only we will have a unique solution, so is the relationship between $p$,$q$ and $r$ will be $r+2q-5p = 0?$ My confusion here is that it is not necessary to have $r+2q-5p = 0$ because even if this is not equal to $0$ then also the rank will remain as $3$ as $-11$ will be there in the $3rd$ row.

Please clarify me on my doubts!

答:
Your restriction is on the last row. If $r + 2q - 5p \neq -11z$ there is no solution.

EDIT: Upon double checking, I believe you made an error on your Gaussian Elimination. The last row should have $0 = r + 2q - 5p$.  So that is your restriction.


问:
An uncountable set of sequence
Let $Z$ be a set of sequence.$\forall \{a_n\}, \exists \{b_n\} \in Z$, such that $lim_{n\rightarrow \infty} \frac{a_n}{b_n}=0$. Prove that $Z$ is uncountable.

We are asked to prove this using the Cantor's Diagonal Argument, but I don't see how it works. Any other approaches are acceptable as well.

Thanks!

答:
If all you know about the set $Z$ is the statement $$\forall \{a_n\}\in Z \exists \{b_n\}\in Z : \lim_{n\to\infty} \frac{a_n}{b_n} = 0,$$

then it is not possible to prove that $Z$ is uncountable. The set $\{\}$ (i.e., the empty set) satisfies the above condition, and is very much countable. Since you cannot prove that $Z$ is not the empty set, you also cannot prove that $Z$ is not countable.

Title: An uncountable set of sequence


问:
What can you say about groups if every homomorphism is trivial?
Suppose we have two groups $G,H$, and the premise is that every homomorphism from $G$ to $H$ is trivial.

This certainly can happen when either of $G,H$ is trivial, but is it possible that neither of $G,H$ are trivial?

答: There are other examples. If $G=\mathbb{Q}$ and $H=\mathbb{Z}$ are the additive group of rational numbers and the additive group of integers respectively, then every homomorphism from $G$ to $H$ is trivial.


答:
uhm the following way isn't that easy... you have to consider several properties of functions...

first off: $\sin(\dfrac{\pi x}{9}) = \dfrac{2x}{9} \iff \sin^{-1}(\dfrac{2x}{9}) = \dfrac{\pi x}{9}$

Let: $f(x) =\sin^{-1}(\dfrac{2x}{9}) - \dfrac{\pi x}{9} $

Our goal is to find where $f(x) = 0$

Notice $f(x) = -f(-x) \implies f(x)$ is an odd function. If an odd function is defined at zero, then its graph must pass through the origin. Proof. This means $x=0$ is a solution.

Now lets examine the domain of $f(x)$. Note $\sin^{-1}(x)$ is defined only when $-1\leq x\leq1$

This means domain of $f(x) =[-\dfrac{9}{2},\dfrac{9}{2}]$

Now lets examine the derivative of f(x):

$f'(x) = \dfrac{2}{9\sqrt{1-\dfrac{4x^2}{81}}} - \dfrac{\pi}{9}$

Now $f'(x) =0 \implies x = \pm \dfrac{9\sqrt{1-\dfrac{4}{\pi^2}}}{2} = \pm a, a>0$.
After drawing a sign diagram one can see:

f(x) is increasing $-\dfrac{9}{2}\leq x\leq -a$

f(x) is decreasing $-a\leq x\leq a$

f(x) is increasing $a\leq x\leq \dfrac{9}{2}$

Since f(x) is strictly decreasing in the interval $[-a,a]$ where $a>0$ and $f(-a) >0$ and $f(a)<0$, then there must be only one root from $[-a,a]$ and we already saw that $x = 0$ is the solution and now we can confirm that this is the only root in $[-a,a]$ Also notice the boundary points are $\pm \dfrac{9}{2}$ where the function is zero. With the same increasing/decreasing function logic we can confirm these two are also the only roots in their respective interval.

Title: Easiest way to solve $\frac{2}{9}x=\sin(\frac{\pi}{9}x)$

Body:
We wanted to find the points of intersection between 2 curves, so we have to solve this equation $\frac{2}{9} x= \sin(\frac{\pi}{9}x)$.

I couldn't figure out an easy way to find the points of intersection, they are $-\frac{9}{2}$, $0$, and $\frac{9}{2}$.

Any ideas please on how to solve such equations?


答:
Let's consider the function $f(x)=\frac{2 \exp(ix) + x^2 - 2}{x^4}dx$

$I=\int_0^\infty \frac{2 \cos(x) + x^2 - 2}{x^4}dx=\frac{1}{2}\Re\int_{-\infty}^\infty{f}(x)dx$

Lets consider the following contour, which includes integral along a big half-circle of radius $R\to\infty$ counter clockwise and a small half-circle of radius $r\to 0$ around $x=0$ clockkwise.

Integral around this contour $\oint{f}(x)dx=0$, because there are not poles inside the contour. On the other hand $\oint{f}(x)dx=2I+\int_R+\int_r\Rightarrow 2I=-\int_R-\int_r$

Integral along a big half-circle $\to0$ at $R\to\infty$ due to Jordan lemma.

$2I=-\Re\int_r\frac{2 \exp(ix) + x^2 - 2}{x^4}dx\Rightarrow$ $${I}=\frac{1}{2}\Re\lim_{r\to0}\int_0^{\pi}\frac{2 \exp(ire^{it}) + r^2e^{2it} - 2}{r^4e^{4it}}ire^{it}dt$$

Expanding exponent into a series $\exp(ire^{it})=1+ire^{it}-\frac{1}{2}r^2e^{2it}-\frac{i}{3!}r^3e^{3it}$ and noting that $\int_0^{\pi}\frac{r^2e^{2it}}{r^4e^{4it}}dt=0$ we get:

$${I}=\frac{1}{2}\Re\lim_{r\to0}\int_0^{\pi}\frac{ 2\,r^4e^{4it}+O(r^5)}{3!\,r^4e^{4it}}dt=\frac{\pi}{3!}$$


问:
setting up a dynamic programming problem with multiple states and controls
For an optimization problem with multiple states ($x$), controls ($y$), and random disturbances ($z$), the Euler equation for a stochastic dynamic programming problem is:

$D_yU(x,y,z)+\beta E D_xU(x,y,z)=0$

where $D_x$ and $D_y$ are the gradients with respect to the states and controls, respectively, $\beta$ is the discount factor, $U$ is the objective function, and $E$ is the expectation operator. This is easy enough to do in the single-state case, but I'm having problems getting a clear picture of the multivariate case, and none of the textbooks I have include very lucid examples.

My problem is:

$\max \quad E_0 \sum_{t=0}^\infty\beta^t u(c_t,N_t)$

subject to:

$c_t+K_{t+1}+G_t=f(K_t,N_t,\zeta_t)$

and

$N_t\leq1$

About the notation - $u$ is the objective function (a utility function), $K$ is the capital stock, $N$ is the proportion of available time supplied as labor, $\zeta$ is a random productivity disturbance, and $G$ is the government's budget (exogenous). $f$ is a production function, and total production depends on the two inputs ($N,K$), as well as the disturbance $\zeta$. Decisions are made after the disturbance is realized, so the budget restriction is always binding. $G$ and $\zeta$ both follow Markov processes. Lastly, capital completely depreciates every period, which is why $K_t$ doesn't show up on the left hand side of the constraint.

OK, with that out of the way, what I really want to know is how to set the problem up so that I can get first order conditions for a solution. As I see it, if I substitute the budget constraint into the objective function, the states for the problem are $\{K,G,\zeta\}$, the controls are $\{K^+,N\}$, and the random disturbances are $\{G,\zeta\}$. If I substitute these into the Euler equation I get:

$\left[\begin{matrix} U_{K^+} \\ U_N \end{matrix}\right]+\beta E(U_{K^+})|G,\zeta=0,$

where $U_{K^+}$ and $U_N$ are the partial derivatives of the objective function with respect to $K^+$ (i.e., $K_{t+1}$) and $N$ (i.e., $N_t$) respectively. Does this look correct, or am I missing something? My biggest worry is that the dimensions of the gradients don't match -- is that a problem? I'd appreciate any help at all.

答:
On closer inspection, it seems that the Euler equation at the bottom of the question is the correct expression for the necessary conditions for a solution. It's composed of two conditions – an intertemporal condition on the growth of the state variable, $K$:

$U_{K^+}+\beta E\left[(U_{K^+})|G,\zeta \right]=0$

In other words, the sacrifice in utility gained by increasing $K^+$ has to be compensated for by the increase in expected future utility that is gained. In other words, increasing $K^+$ this period decreases current utility, but increases the expected value of future utility flows, and an optimum is achieved (provided an interior solution exists) when the two of these exactly offset one another.

The second condition is the intratemporal condition on $N$:

$U_N=\frac{\partial u}{\partial c}\times\frac{\partial c}{\partial N} + \frac{\partial u}{\partial N}=0$

$u$ is increasing in $N$ in its first argument (higher $N$, more production), but decreasing in its second argument (lower $N$, more utility from leisure). Therefore, it's a necessary condition for an optimum that the marginal increase in utility gained by increasing $N$ be offset by the marginal decrease in utility lost.


问:
Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^2=4ax$ is the parabola $9y^2-4ax+32a^2=0.$
Show that the locus of the centroids of equilateral triangles inscribed in the parabola $y^2=4ax$ is the parabola $9y^2-4ax+32a^2=0.$

I tried to solve it.I took three coordinates of the equilateral triangle as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$

let the coordinates of the centroid be $(h,k)$.

$h=\frac{x_1+x_2+x_3}{3},k=\frac{y_1+y_2+y_3}{3}$

$h=\frac{y^2_1+y^2_2+y^2_3}{12a}$

But eliminating $y_1,y_2,y_3$ is difficult.Is my method correct or is there any other better method.Please help me.

答:
Writing the solution from Gelca,Andreescu Putnam and Beyond (2007), pp. 213-214 as Tony suggested:

Let us determine first some algebraic conditions that the coordinates $(x_i, y_i)$,
$i = 1, 2, 3$, of the vertices of a triangle should satisfy in order for the triangle to be equilateral. The equation of the median from $(x_3, y_3)$ is $$\dfrac{y-y_3}{x-x_3}=\dfrac{y_1+y_2-2y_3}{x_1+x_2-2x_3}$$
or $$(x_1-x_2)(x_1+x_2-2x_3)+(y_1-y_2)(y_1+y_2-2y_3)=0$$
So this relation along with the two obtained by circular permutations of the indices are necessary and sufficient conditions for the triangle to be equilateral. Of course, the third
condition is redundant. In the case of three points on the parabola, namely $\left(\dfrac{y_i^2}{4a},y_i\right)$ , $i=1,2,3$, after dividing by $y_1-y_2$, respectively, by $y_2-y_3$ (which are both nonzero), we obtain $$(y_1+y_2)(y_1^2+y_2^2-2y_3^2)+16a^2(y_1+y_2-2y_3)=0 \\ (y_2+y_3)(y_2^2+y_3^2-2y_1^2)+16a^2(y_2+y_3-2y_1)=0$$
Subtracting the two gives $$y_1^3-y_3^3+(y_1-y_3)(y_2^2-2y_1y_3)+48a^2(y_1-y_3)=0$$
Divide this by $y_1-y_3 \ne 0$ to transform it into $$y_1^2+y_2^2+y_3^2+3(y_1y_2+y_2y_3+y_3y_1)+48a^2=0$$
This is the condition by the $y$ -coordinates of the vertices of the triangle. Keeping in mind that the coordinates of the centroid of the triangle are $$x=\dfrac{y_1^2+y_2^2+y_3^2}{12a}, y=\dfrac{y_1+y_2+y_3}{3}$$ we rewrite the relations as $$-\dfrac{1}{2}(y_1^2+y_2^2+y_3^2)+\dfrac{3}{2}(y_1+y_2+y_3)+48a^3=0$$ then substitute $12ax=y_1^2+y_2^2+y_3^2$ and $3y=y_1+y_2+y_3$ to obtain the equation of the locus $$-6ax+\dfrac{27}{2}y^2+48a^2=0$$ or $$9y^2-4ax+32a^2=0$$


问:
My attempt at $S = \overline{S}$ iff $S$ is closed
$E$ is a metric space and $S$ a set in it.

$\overline{S}$ denotes the closure of $S$. The definition for closure that I am using is:

The closure of a set $A$ is the intersection of all closed sets containing $A$. More explicitly:

$\displaystyle \overline{A} = \bigcap_{i \in I} C_i $ , where for each $i \in I$ the set $C_i$ is closed in $E$ and $S \subset C_i$.

$I$ is an index set.

Here is my write-up of a proof for the statement in the title. It seems a bit too facile and terse to me, so I wanted to ask for some feedback on its correctness.

Proof:

Suppose $S = \overline{S}$. The set $\overline{S}$ is an intersection of closed sets, and such intersections are always themselves closed. Therefore $S$ is closed.

Suppose now $S$ is closed. Since $S$ contains itself, it qualifies as $S = C_j$ for some $j \in I$.

$S$ is a subset of each $C_i$ so $S$ is necessarily in the intersection of all such sets, that is $S \subset \overline{S}$.

Next, for each $x \in \overline{S}$:

$x \in \overline{S} = \displaystyle \bigcap_{i \in I} C_i$ means $x \in C_i$ for each $i \in I$. In particular, $x \in C_j = S$. Therefore $\overline{S} \subset S$.

Hence $ \overline{S} = S$. We are done. $\blacksquare$.

Any thoughts or comments appreciated! Peace!

EDIT:

After José's feedback, I suppose the definition below is incorrect since it does not capture every possible closed set containing $S$:

More explicitly:

$\displaystyle \overline{A} = \bigcap_{i \in I} C_i $ , where for each $i \in I$ the set $C_i$ is closed in $E$ and $S \subset C_i$.

$I$ is an index set.

The index set is not specified, so if there are infinitely many closed sets containing $S$ but $I$ is finite, we're missing some sets.

I think this ought to be the correct way to state it explicitly:

$\overline{S} = \displaystyle \bigcap_{C \in A} C$

where $ C = \{  X \subset E: X \text{ is closed and } S \subset X \} $

答:
Having the definition

$$\overline{S} = \bigcap \{C\mid C \text{ closed }, S \subseteq C\}$$

we see that 1. $\overline{S}$ is closed, as an intersection of closed sets and 2. $S \subseteq \overline{S}$ as all sets in the intersection contain $S$, hence so does their intersection.

If now $S=\overline{S}$, the first fact immediately implies $S$ is closed.

If $S$ is closed, $S$ is one of the sets that is in the defining intersection, so $\overline{S} \subseteq S$ and together with fact 2 we have $S=\overline{S}$ (having two inclusions).


问:
Extension operator $P: W^{k, p}(\Omega) \to W^{k, p}(\mathbb{R})$.
In my functional analysis course, we are studying the extension operator and there is something I really don't understand. Let me explain my problem for the case $\Omega = \mathbb R^n_+ = \{x \in \mathbb R^n~|~x_n > 0\}$ and $k = 1$.

For such an open set, there is a linear continuous operator $P: W^{1, p}(\mathbb R^n_+) \to W^{1, p}(\mathbb R^n)$ such that $(Pu)|_{\mathbb R^n_+} = u$. To construct such an operator, we use the approximation of $W^{1, p}(\mathbb R^n_+)$ by $C^\infty (\overline{\mathbb R^n_+})\cap W^{1, p}(\mathbb R^n_+)$ to define
$$
Pu(x',x_n) = \begin{cases}
u(x', x_n) & \text{if }~ x_n \ge 0,\\
u(x', -x_n) & \text{if }~ x_n < 0,
\end{cases}
$$
Such that $Pu \in W^{1, p}(\mathbb R^n)$ and we can conclude by density. It is the same kind of argument on $\Omega$ smooth and bounded.

My question is the following: why do we need to construct the extension like that ? It seems much more natural to define
$$
Pu(x',x_n) = \begin{cases}
u(x', x_n) & \text{if }~ x_n \ge 0,\\
0 & \text{if }~ x_n < 0.
\end{cases}
$$
What could possibly goes wrong with the extension by zero of functions of $W^{1, p}(\Omega)$ ?

答:
Your problem is the jump of the derivative of the function at the boundary!

Indeed, for a function in $L^2(\Omega)$, it is natural to extend by $0$ outside. However, for a function $u \in H^1(\Omega)$, if you extend by $0$ outside $\Omega$, you might have that $u$ is discontinuous at $\partial \Omega$, and therefore nothing ensures you that

$$\nabla u \in L^2(\mathbb{R}^n).$$

However, if $u \in H^1_0(\Omega)$, you can extend by $0$ without any problem.


问:
Prove $p^2=p$ and $qp=0$
I am not really aware what's going on in this question. I appreciate your help.

Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$.

Prove $p^2=p$ and $qp=0$.

$$p(p+q) = p(\text{id}_U) \Rightarrow p^2+pq=p \Rightarrow p^2 =p.$$ I actually first found this by letting $(\text{id}_U) =1$, although probably a wrong to do it.

Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$. Again, I'm not sure if this actually is the right way to do it.

Then we have $$q(p+q) = q(\text{id}_U) \Rightarrow q^2+qp=q \Rightarrow q^2+ qp= q^2 \Rightarrow qp=0.$$

Prove $K=\text{im}(q)$.

For this question, I honestly do not know how to tackle it from the following definitions.

$K = \ker(p)=\left\{u \in U \ |\ p(u)=0_U\right\}$ and $\text{im}(q)=\left\{q(u) \ | \ u \in U \right\}$.

What's the correct way of doing these questions? Thank you for your time.

答:
You did the questions correctly (see below). It might be slightly more common to capitalize $P$ and $Q$, and it would be useful to see these as projections onto orthogonal subspaces to ge geometric intution. For instance, you could take $P : R^2 \rightarrow R^2$ to be $P(x,y) = (x,0)$ and $Q(x,y) = (0,y)$. In this case, the vector space is finite dimensional and the operators are defined on a basis. I find the best way to convince myself of the kind of operations you're carrying out is to say, "if $P$ and $Q$  both do the same thing to any $u \in U$, then $P = Q$."

To finish your question, we show $im(Q) = K = ker(P)$.

i) $im(Q) \subset ker(P)$. Let $v \in im(Q)$. Then $ v = Qw$ for some $w \in U$. Then $Pv = PQw = 0w = 0.$

ii) $ker(P) \subset im(Q)$. Let $v \in ker(P)$. Then $v = (P+Q)v = Pv + Qv = 0 + Qv$, so in fact $v = Qv$.

To address something mentioned in the comments: you might be treating the linear maps as matrices and implicitly using the fact that matrix multiplication corresponds to operator composition. In other words, I can define an operator $P$ on $\mathbb{R}^2$ that, say, rotates clockwise by $\pi/4.$ This is a linear map on $\mathbb{R}^2$, so if I pick a basis $\mathcal{B}$, I can represent $P$ by its standard matrix $[P]_\mathcal{B}$. Then if I have any other operator $Q$, $(P \circ Q)(v) = [P]_\mathcal{B}[B]_\mathcal{B}v_\mathcal{B}$.

Title: Prove $p^2=p$ and $qp=0$


问:
Dimension of $\Bbb{R^{\infty}}$
Prove $\dim(\Bbb{R^{\infty}})$=$\infty$.  Suppose $\Bbb{R^{\infty}}$ has finite dimension, a number $n\in\Bbb{N}$. Define the generic basis vector $e_i$ in the usual fashion. Then the set containing $n+1$ such vectors is independent ( by induction I guess ). But why would $\Bbb{R^{\infty}}$ contain $n+1$ such vectors?

答:
$X:=\mathbb{R}^{\infty}$ contains the infinite vectors (or sequences, to be more precise) $e_i$, whose member $j$ is $(e_i)_j=\delta_{ij}$.

Those vectors are linearly independent, by definition. Indeed, just take a finite linear combination of them, suppose it is the $0$ vector in $X$, and reason component-wise to deduce the coefficients of the linear combination must be all $0$.

Hence $X$ cannot have finite dimension, because we found an infinite set of independent vectors inside him.

Addendum.

$\mathbb{R}^{\infty}$ is technically the set of all functions $x: \mathbb{N} \rightarrow \mathbb{R}$. These functions are in turn, by definition, the real sequences, and we just denote $x(n)$ by $x_n$ to make this connection more explicit.

Then, a sequence $x$ can be represented as an "infinite vector" $x"="(x_1,x_2,...)$.

Title: Dimension of $\Bbb{R^{\infty}}$


问:
Is there an easy/fast method for calculating $\text{P}(X_i = \text{max}(X_1, \dots, X_n)), \forall i$ where $X_i \sim \text{N}(\mu_i, \sigma_i)$?
For $X_i \sim \text{N}(\mu_i, \sigma_i)$,  is there a fast method of (roughly) computing

$$\text{P}\left(X_i = \text{max}(X_1, \dots, X_n)\right), \forall i$$

for a set of $n$ tuples of $(\mu_i, \sigma_i)$, $i \in 1, \dots, n$? Or in words, for all $i$, what is the probability that $X_i$ is the maximum in the sample of $\{X_1, \dots, X_N\}$

I believe the exact answer is given by

$$
\text{P}\left(X_i = \text{max}(X_i, \dots, X_n)\right) =
  \frac{1}{\sqrt{2\pi\sigma_i}}
  \int_{-\infty}^{\infty} \exp\left(\frac{-(x-\mu_i)^2}{2\sigma_i}\right)
  \prod_{k \neq i} \Phi\left(\frac{x-\mu_k}{\sigma_k}\right)
  \text{dx}
$$

where $\Phi(x)$ is the cumulative normal distribution function.

I need to repeat this calculation for $M$ different sets of $n$ tuples. I am wondering if there is perhaps a formula which could give an approximation that could possibly be vectorised for fast computation when $n$ is known in advance. Ideally, avoiding having to perform $M\times n$ numerical integrations.

One thought I had was to train a model.

Given that I can calculate the exact values in advance for a set of $n$ tuples, perhaps I could train a machine learning model which might be able to give me a fast but adequate approximation when $n$ is fixed? I could set it up as a multiple response model where the input data consists of rows of vectors made from concatenating $n$ randomly generated $(\mu_i, \sigma_i)$ tuples and the responses are calculated using the formula above. I can generate as much data as I need.

答:
We have

\begin{align}
P\left(X_i = \max(X_1, \dots, X_n)\right) &= E\left(\mathbb{I}_{\{X_i = \max(X_1, \dots, X_n)\}} \right)\\
&= E\left(E(\mathbb{I}_{\{X_i = \max(X_1, \dots, X_n)\}}|X_i) \right)\\
&= E\left(P(X_i = \max(X_1, \dots, X_n)|X_i) \right)\\
&= E\left(P(\bigcap_{k \ne i} \{X_k \le X_i\} |X_i) \right)\\
&= E\left( \prod_{k \ne i} P(X_k \le X_i |X_i) \right) \tag{1}\\
&= E\left( \prod_{k \ne i} \Phi\left(\frac{X_i-\mu_k}{\sigma_k}\right) \right) \\
&= \frac{1}{\sqrt{2\pi\sigma_i}}\int_\mathbb{R}\left(\exp\left(\frac{-(x-\mu_i)^2}{2\sigma_i}\right)  \prod_{k \neq i} \Phi\left(\frac{x-\mu_k}{\sigma_k}\right)\right)dx
\end{align}

At $(1)$, we need the assumption that $(X_i)_{i=1,...,n}$ are independant.


问:
Using dimensional analysis to ‘get’ planck’s ${\lambda}$ equation with avogadro’s constant
This is for when you calculate bond energy in: ${j\cdot mol(n)^{-1}}$ and want to use planck’s equation for photon energy to calculate ${\lambda}$

Planck’s equation for photon energy:
$${\lambda=\frac{h\cdot c}{E}}$$

Planck’s equation for photon energy with avogadro’s constant:

$${\lambda=\frac{h\cdot c\cdot N_A}{E}}$$

I don’t understand how ${N_A}$ (${\frac{1}{n}}$) is at the top as:

${n=\frac{g}{g\cdot n^{-1}}\Rightarrow{n=\frac{1}{n}}}$

Energy in this case:

${E=\frac{j}{mol\ or\ n \ (1/n)}}$

I can’t rearrange ${1/n}$ to get it to the to the top in order to get rid of the ${n}$ at the bottom so that:

${E=j}$

Can someone show me verbosely with dimensional analysis how to do this?

Units:

${\lambda=m}$

${E=j}$

${h=js}$

${c=ms^{-1}}$

${N_A=n^{-1}}$

Dimensional analysis:

${\lambda=js\cdot ms^{-1}\cdot j^{-1}(1^{-1}n^{-2})}$

${\lambda=\frac{j\cdot s\cdot m}{j\cdot s(\frac{1}{n})}}\Rightarrow{\lambda=\frac{j\cdot s\cdot m}{\dfrac {j\cdot s}{n}}}\Rightarrow{\lambda=\frac{m}{\dfrac{1}{n}}}$

Then I can’t get ${\lambda=m}$

Is this correct, and if not then why?:

${\frac{j}{mol(1/n)}=\frac{j}{\dfrac{1}{n}}}$

Bonus question:

I have been using dimensional analysis to ‘get’ equations before I even knew that word, as my memory is bad and I couldn’t remember equations, I used to think it was a trick

Are there more/better ways to ‘get’ equations and cases where dimensional analysis won’t work?

I really need help here and would appreciate it very much

答:
Maybe I wrongly understand the question but:
$$
[\text m]=[\lambda]\stackrel{?}=\left[\frac{hc N_A}{E}\right]=\frac{[\text J\cdot\text s][\text m\cdot \text s^{-1}][\text{mol}^{-1}]}{[\text J\cdot\text{mol}^{-1}]}=[\text m].
$$
so that there is no inconsistency in this equation.

And of course the inconsistency will appear if $N_A$ is absent or wrongly  placed.


答:
Let $(z_n)_n$ be a sequence of all sequences in $Z$ for $n\in\mathbb{N}$.
Then consider the following sequence $a_n$ defined by the following equation.
$$a_n = z_{\operatorname{ord}_2(n)+1}[n]$$
This sequence infinitely often coincides with every element in $(z_n)_n$.

There needs to be a sequence $(b_n)_n \in Z$ such that $\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = 0$. This sequence $(b_n)_n\in Z$ has to be $z_m$ for some $m\in\mathbb{N}$, but $(a_n)_n$ is equal to $z_m$ infinitely often at positions $2^{m-1}\mathbb{N}$. so $\big(\frac{a_n}{b_n}\big)$ can't converge to $0$ given that a subsequence of $\big(\frac{a_n}{b_n}\big)$ converges to $1$.

This contradiction shows that the assumption that $Z$ is countable is false.

Title: An uncountable set of sequence

Body:
Let $Z$ be a set of sequence.$\forall \{a_n\}, \exists \{b_n\} \in Z$, such that $lim_{n\rightarrow \infty} \frac{a_n}{b_n}=0$. Prove that $Z$ is uncountable.

We are asked to prove this using the Cantor's Diagonal Argument, but I don't see how it works. Any other approaches are acceptable as well.

Thanks!


问:
Using modular arithmetic to find the remainder of $12^{157}$ when divided by $10$
So, I've figured out that:
$12 \equiv 2 \mod 10$

And hence,

$$ 12^{157} \equiv 2^{157}  \mod 10$$

I thought of expanding out $2^{157} = 2^{156} \cdot 2 = 4^{78} \cdot 2 = 16^{39}
 \cdot 2$, then I could write:
$$ 12^{157} \equiv 2^{157} \equiv 16^{39} \cdot 2 \equiv  6^{39} \cdot 2 \equiv 36^{19}\cdot 12\equiv 6^{19} \cdot 2  \equiv 36^9 \cdot 12 \equiv 6^9 \cdot 2 \equiv 36^4 \cdot 12 \equiv 6^4 \cdot 2  \equiv 72 \cdot 36 \equiv 2 \cdot 6 \equiv 2 $$

This is quite a long process. I found that you could use fermats little theorem to speed up the case when the number we taking mod with is prime, however are there any theorems/methods which can be used to speed up calculations when we take a composite numbers?

答:
You could use Fermat`s Little Theorem and Chinese Remainder Theorem together.

Note that

$12^{157} \equiv 0 \pmod{2} \Rightarrow 12^{157}=2k$

$12^4 \equiv 1 \pmod{5} \Rightarrow 12^{157} \equiv 2 \pmod{5}$

$2k \equiv 2\pmod{5} \Rightarrow k \equiv 1 \pmod{5}$

$k \equiv 1 \pmod{5} \Rightarrow k=5x+1 $ so $12^{157}=2k=10x+2 \equiv 2 \pmod{10}$


问:
convolution of exponential distribution and uniform distribution
Given $X$ an exponentially distributed random variable with parameter $\lambda$ and $Y$ a uniformly distributed random variable between $-C$ and $C$. $X$ and $Y$ are independent. I'm supposed to calculate the distribution of $X + Y$ using convolution.

Does anyone have any tips on how to do this?

I understand that the convolution is represented by
$$\int_{-\infty}^{+\infty} f_1(x) \cdot f_2(z-x)dx\tag {1}$$

and so given $$f_1(x) = \begin{cases}
\lambda e^{-\lambda x} &, x\ge 0\\ 0 &, x < 0 \end{cases}$$

and $$f_2(y) = \begin{cases} \frac{1}{2C} &, y \in [-C,C]\\ 0 & , \text{otherwise}\end{cases}$$

(1) becomes: $$\int_{-\infty}^{+\infty} \lambda e^{-\lambda x}\cdot \frac{1}{2C} dx$$

but I don't know how to procede from here. How do I choose which intervals of $z$ to integrate?

答:
Define $Z=X+Y$. We need to find the CDF/PDF of $Z$. By definition, we have:

$P(Z\le z) = P(X+Y\le z)=\int_{-\infty}^{+ \infty}P(X \le z-y) f(y) dy$

Now, for $z-y < 0$

$$P(X \le z-y) = 0$$

and for $z-y \ge 0$

$$P(X \le z-y) = 1-e^{-\lambda(z-y)}$$

and $f(y) = \frac{1}{2C}$

Therefore, the above integral reduces to:

$P(Z\le z) = \int_{-\infty}^{+ z} (1-e^{-\lambda(z-y)}) \frac{1}{2C} dy$

Therefore, the pdf of $Z$ is given by:

$$\int_{-\infty}^{+ z}  \frac{\lambda e^{-\lambda(z-y)}}{2C} dy$$


问:
Calculation of simplex coordinates that are bound by vectors and the simplex has to contain a target coordinate
Notation:

$p_{i, j, k}$; the $k^{th}$ component of the $j^{th}$ boundary coordinate on the $i^{th}$ boundary

Finding a weight in 2D

I have 2 sets of 2 boundary coordinates in $\mathbb{R}^2$ space.
The boundaries and their coordinates are indicated in green in Figure 1 below.
Corresponding coordinates in each set are tied together by a line (a tie-line or tie-1-simplex).
The tie-1-simplices are indicated in solid red in Figure 1.

<img src="https://i.stack.imgur.com/caRXM.png" alt="Figure 1: Two sets of 2 boundary coordinates in $\mathbb{R}^2$" />

In each set ($i$), I can create a vector; $\bar{v}_i = p_{i, 2} - p_{i, 1} \quad :i \in \mathbb{N} [1, 2]$

A point on a vector can be described by the two coordinates and a weight; $r_i(p_{i, 1}, p_{i, 2}, \beta) = p_{i,1} + \beta \bar{v}_i = p_{i,1}+ \beta(p_{i,2} - p_{i,1}) \quad :\beta \in \mathbb{R} [0, 1]$

Within the convex hull of these coordinates I have another target coordinate; $p_t$, indicated in blue in Figure 1.

A simplex of $\mathbb{R}^1$ (a line), indicated in hashed red in Figure 1, exist in this $\mathbb{R}^2$ space that contains the target coordinate and whose vertices are found on each of the vectors where the weight $\beta$ is the same for both vectors.

Given the sets of boundary coordinates and target coordinate, how can I explicitly calculate $\beta$. An iterative method is to be avoided as this calculation is critical to the performance of an algorithm.

Finding a weight in 3D. Case 1

This is very similar to the 2D case.

I have 3 sets of 2 boundary coordinates in $\mathbb{R}^3$ space.
The boundaries and their coordinates are indicated in green in Figure 2 below.
Corresponding coordinates in each set are tied together by a triangle (a tie-triangle or tie-2-simplex).
The tie-2-simplices are indicated in solid red in Figure 2.

<img src="https://i.stack.imgur.com/vSzZ6.png" alt="Figure 2: Three sets of 2 boundary coordinates in $\mathbb{R}^3$" />

In each set ($i$), I can create a vector; $\bar{v}_i = p_{i, 2} - p_{i, 1} \quad :i \in \mathbb{N} [1, 3]$

A point on a vector can be described by the two coordinates and a weight; $r_i(p_{i, 1}, p_{i, 2}, \beta) = p_{i,1} + \beta \bar{v}_i = p_{i,1}+ \beta(p_{i,2} - p_{i,1}) \quad :\beta \in \mathbb{R} [0, 1]$

Within the convex hull of these coordinates I have another target coordinate; $p_t$, indicated in blue in Figure 2.

A simplex of $\mathbb{R}^2$ (a triangle), indicated in hashed red in Figure 2, exist in this $\mathbb{R}^3$ space that contains the target coordinate and whose vertices are found on each of the vectors where the weight $\beta$ is the same for both vectors.

Given the sets of boundary coordinates and target coordinate, how can I explicitly calculate $\beta$. An iterative method is to be avoided as this calculation is critical to the performance of an algorithm.

Finding a weight in 3D. Case 2

Similar to the first 3D case where the boundaries were lines (one vector per boundary), in this case the boundaries are surfaces defined by three coordinates (two vectors per boundary).

I now have 2 sets of 3 boundary coordinates in $\mathbb{R}^3$ space.
The boundaries and their coordinates are indicated in green in Figure 3 below.
Corresponding coordinates in each set are tied together by a line (a tie-line or tie-1-simplex).
The tie-1-simplices are indicated in solid red in Figure 3.

<img src="https://i.stack.imgur.com/3VIxA.png" alt="Figure 3: Two sets of 3 boundary coordinates in $\mathbb{R}^3$" />

In each set ($i$), I can create two vectors, each originating from the same coordinate (coordinate 1 for instance); $\bar{v}_{i, 12}  = p_{i, 2} - p_{i, 1}, \bar{v}_{i, 13}  = p_{i, 3} - p_{i, 1} \quad :i \in \mathbb{N} [1, 2]$

With these two vectors, I can describe a coordinate on the surface created between them.
A point on the surface can be described by the three coordinates and two weights; $r_i(p_{i, 1}, p_{i, 2}, p_{i, 3}, \beta_1, \beta_2) = p_{i,1} + \beta_1 \bar{v}_{i, 12} + \beta_2 \bar{v}_{i, 13} = p_{i,1}+ \beta_1(p_{i,2} - p_{i,1}) + \beta_2(p_{i,3} - p_{i,1}) \quad :\beta_i \in \mathbb{R} [0, 1]; \beta_i >= 0; \Sigma{\beta_i} <= 1$

Within the convex hull of these coordinates I have another target coordinate; $p_t$, , indicated in blue in Figure 3.

A simplex of $\mathbb{R}^1$ (a line), indicated in hashed red in Figure 3, exist in this $\mathbb{R}^3$ space that contains the target coordinate and whose vertices are found on each of the surfaces described by the two sets of vectors where the weights $\beta_i$ and $\beta_2$ are the same for each surface.

Given the sets of boundary coordinates and target coordinate, how can I explicitly calculate $\beta_1$ and $\beta_2$. An iterative method is to be avoided as this calculation is critical to the performance of an algorithm.

Finding a weight in generic case

I would ultimately like to get to the point where I can calculate any number of $\beta$ weights in $\mathbb{R}^n$.

Is there someone that has solved a similar problem before or has a suggestion on how this can be attempted? I would appreciate it.

答:
I have found an expression that can be used to determine the $\beta$ values, but unfortunately it is not an explicit solution but can be solved with non-linear methods. I have shown below how I derived the solution for each case, referring to some expressions in the question. This was done to show where I got the pattern shown in the generic solution.

Notation:

$B$; the number of boundaries

$C$; the number of boundary coordinates a boundary is composed of

$N$; the dimension where the coordinates are found in

$p_{i, j, k}$; the $k^{th}$ component of the $j^{th}$ boundary coordinate on the $i^{th}$ boundary

Finding a weight in 2D

As indicated, a point on each boundary can be represented by the vector that connects the boundary coordinates and a weight, $\beta$;

$r_i(p_{i, 1}, p_{i, 2}, \beta) = p_{i,1} + \beta \bar{v}_i = p_{i,1}+ \beta(p_{i,2} - p_{i,1}) \quad :\beta \in \mathbb{R} [0, 1]$

This was expanded and factorised so that each boundary coordinate was expressed individually together with a weight;

$r_i(p_{i, 1}, p_{i, 2}, \beta) = p_{i,1}(1 - \beta) + p_{i,2}(\beta) \quad :\beta \in \mathbb{R} [0, 1]$

With the $\beta$ value equal for both boundaries, the two coordinates required to create the tie-simplex that contains the target coordinate, are known, albeit in terms of an unknown variable.
These two coordinates can be used to create another vector, $\bar{t}$, that represents the tie-simplex;

$\bar{s} = r_{2} - r_{1}$

This vector $\bar{s}$, together with a tie-simplex weight, $\alpha$, can be used to express the target coordinate;

$t(r_{1}, r_{2}, \alpha) = r_{1} + \alpha \bar{s} = r_{1} + \alpha(r_{2} - r_{1}) \quad :\alpha \in \mathbb{R} [0, 1]$.

This was expanded and factorised so that each of the coordinates could be expressed individually together with a weight;

$t(r_{1}, r_{2}, \alpha) = r_{1}(1 - \alpha) + r_{2}(\alpha) \quad :\alpha \in \mathbb{R} [0, 1]$.

After substituting the expressions for $r_{1}$ and $r_{2}$, the target coordinate $t$, can be expressed in terms of the known boundary coordinates, the boundary weight $\beta$, and tie-simplex weight $\alpha$;

$t = r_{1}(1 - \alpha) + r_{2}(\alpha) = [p_{1,1}(1 - \beta) + p_{1,2}(\beta)](1 - \alpha) + [p_{2,1}(1 - \beta) + p_{2,2}(\beta)](\alpha) \quad :\alpha, \beta \in \mathbb{R} [0, 1]$

After some expanding and factorising, this can be written in matrix and vector notation as;

$\bar{t} = \bf{P}\bar{\omega}$

where

$\bar{t} = \begin{bmatrix}
 t_{x}\\
    t_{y} \\
\end{bmatrix}$

$\bf{P} = \begin{bmatrix}
 p_{1, 1, x} & p_{1, 1, y} \\
    p_{1, 2, x} & p_{1, 2, y} \\
    p_{2, 1, x} & p_{2, 1, y} \\
    p_{2, 2, x} & p_{2, 2, y} \\
\end{bmatrix} ^T$

$\bar{\omega} = \begin{bmatrix}
 (1 - \beta)(1 - \alpha) \\
    (\beta)(1 - \alpha)\\
    (1 - \beta)(\alpha) \\
    (\beta)(\alpha)\\
\end{bmatrix} $

With the target coordinate known as well as all the boundary coordinates, this system of equations can be solved to obtain the boundary weight $\beta$.
This however cannot be done with linear algebra seeing that this is a non-linear expression.

Finding a weight in 3D, case 1

As with the 2D case, a coordinate can be expressed with the following expression because there are only two coordinates that creates the 1D boundary;

$r_i(p_{i, 1}, p_{i, 2}, \beta) = p_{i,1}(1 - \beta) + p_{i,2}(\beta) \quad :\beta
\in \mathbb{R} [0, 1]$

In this case however, there are three boundaries, and therefore three coordinates that creates the tie-simplex that contains the target coordinate.
Two vectors can be created from one of the coordinates to the others;

$\bar{s}_1 = r_{2} - r_{1}$, and $\bar{s}_2 = r_{3} - r_{1}$

These vectors $\bar{s}_1$ and $\bar{s}_2$, together with two tie-simplex weights, $\alpha_1$ and $\alpha_2$, can be used to express the target coordinate;

$t(r_{1}, r_{2}, r_{3}, \alpha_1, \alpha_2) = r_{1} + \alpha_1 \bar{s}_1 + \alpha_2 \bar{s}_2 = r_{1} + \alpha_1(r_{2} - r_{1}) + \alpha_2(r_{3} - r_{1}) \quad :\alpha_1, \alpha_2 \in \mathbb{R} [0, 1]$.

This was expanded and factorised so that each of the coordinates could be expressed individually together with a weight;

$t(r_{1}, r_{2}, r_{3}, \alpha_1, \alpha_2) = r_{1}(1 - \alpha_1 - \alpha_2) + r_{2}(\alpha_1) + r_{3}(\alpha_2) \quad :\alpha_1, \alpha_2 \in \mathbb{R} [0, 1]$.

After substituting the expressions for $r_{1}$, $r_{2}$ and $r_{2}$, the target coordinate $t$, can be expressed in terms of the known boundary coordinates, the boundary weight $\beta$, and tie-simplex weights $\alpha_1$ and $\alpha_2$;

\begin{align}
t &= r_{1}(1 - \alpha_1 - \alpha_2) + r_{2}(\alpha_1) + r_{3}(\alpha_2) \\
&= [p_{1,1}(1 - \beta) + p_{1,2}(\beta)](1 - \alpha_1 - \alpha_2) + [p_{2,1}(1 - \beta) + p_{2,2}(\beta)](\alpha_1) + [p_{3,1}(1 - \beta) + p_{3,2}(\beta)](\alpha_2) \\
& \alpha_1, \alpha_2, \beta \in \mathbb{R} [0, 1]
\end{align}

After some expanding and factorising, this can be written in matrix and vector notation as;

$\bar{t} = \bf{P}\bar{\omega}$

where

$\bar{t} = \begin{bmatrix}
 t_{x} \\
    t_{y} \\
    t_{z} \\
\end{bmatrix}$

$\bf{P} = \begin{bmatrix}
 p_{1, 1, x} & p_{1, 1, y} & p_{1, 1, z}\\
    p_{1, 2, x} & p_{1, 2, y} & p_{1, 2, z}\\
    p_{2, 1, x} & p_{2, 1, y} & p_{2, 1, z}\\
    p_{2, 2, x} & p_{2, 2, y} & p_{2, 2, z}\\
    p_{3, 1, x} & p_{3, 1, y} & p_{3, 1, z}\\
    p_{3, 2, x} & p_{3, 2, y} & p_{3, 2, z}\\
\end{bmatrix} ^T$

$\bar{\omega} = \begin{bmatrix}
 (1 - \beta)(1 - \alpha_1 - \alpha_2) \\
    (\beta)(1 - \alpha_1 - \alpha_2)\\
    (1 - \beta)(\alpha_1) \\
    (\beta)(\alpha_1)\\
    (1 - \beta)(\alpha_2) \\
    (\beta)(\alpha_2)\\
\end{bmatrix} $

With the target coordinate known as well as all the boundary coordinates, this system of equations can be solved to obtain the boundary weight $\beta$.
This however cannot be done with linear algebra seeing that this is a non-linear expression.

Finding a weight in 3D, case 2

In the first case of 3D, the boundaries were 1D with two boundary coordinates each.
In this case, the boundaries are 2D with three boundary coordinates each.

As indicated, a point on each boundary can be represented by two vectors (with the same origin) and two boundary weights, $\beta_1$ and $\beta_2$;

$r_i(p_{i, 1}, p_{i, 2}, p_{i, 3}, \beta_1, \beta_2) = p_{i,1} + \beta_1 \bar{v}_{i, 12} + \beta_2 \bar{v}_{i, 13} = p_{i,1}+ \beta_1(p_{i,2} - p_{i,1}) + \beta_2(p_{i,3} - p_{i,1}) \quad :\beta_i \in \mathbb{R} [0, 1]$

This was expanded and factorised so that each of the coordinates could be expressed individually together with a weight;

$r_i(p_{i, 1}, p_{i, 2}, p_{i, 3}, \beta_1, \beta_2) = p_{i,1}(1 - \beta_1 - \beta_2) + p_{i,2}(\beta_1) + p_{i,3}(\beta_2) \quad :\beta_i \in \mathbb{R} [0, 1]$

Because there are only two boundaries in this case, the tie-simplex between the two boundaries can be expressed by a vector and single tie-simplex weight, $\alpha$, as in the 2D case (shown in expanded and factorised format);

$t(r_{1}, r_{2}, \alpha) = r_{1}(1 - \alpha) + r_{2}(\alpha) \quad :\alpha \in \mathbb{R} [0, 1]$.

After substituting the expressions for $r_{1}$ and $r_{2}$, the target coordinate $t$, can be expressed in terms of the known boundary coordinates, the boundary weights $\beta_1$ and $\beta_2$, and tie-simplex weight $\alpha$;

\begin{align}
t &= r_{1}(1 - \alpha) + r_{2}(\alpha) \\
&= [p_{1,1}(1 - \beta_1 - \beta_2) + p_{1,2}(\beta_1) + p_{1,3}(\beta_2)](1 - \alpha) + [p_{2,1}(1 - \beta_1 - \beta_2) + p_{2,2}(\beta_1) + p_{2,3}(\beta_2)](\alpha) \\
&\alpha, \beta_1, \beta_2 \in \mathbb{R} [0, 1]
\end{align}

After some expanding and factorising, this can be written in matrix and vector notation as;

$\bar{t} = \bf{P}\bar{\omega}$

where

$\bar{t} = \begin{bmatrix}
 t_{x} \\
    t_{y} \\
    t_{z} \\
\end{bmatrix}$

$\bf{P} = \begin{bmatrix}
 p_{1, 1, x} & p_{1, 1, y} & p_{1, 1, z}\\
    p_{1, 2, x} & p_{1, 2, y} & p_{1, 2, z}\\
    p_{1, 3, x} & p_{1, 3, y} & p_{1, 3, z}\\
    p_{2, 1, x} & p_{2, 1, y} & p_{2, 1, z}\\
    p_{2, 2, x} & p_{2, 2, y} & p_{2, 2, z}\\
    p_{2, 3, x} & p_{2, 3, y} & p_{2, 3, z}\\
\end{bmatrix} ^T$

$\bar{\omega} = \begin{bmatrix}
 (1 - \beta_1 - \beta_2)(1 - \alpha) \\
    (\beta_1)(1 - \alpha)\\
    (\beta_2)(1 - \alpha)\\
    (1 - \beta_1 - \beta_2)(\alpha) \\
    (\beta_1)(\alpha)\\
    (\beta_2)(\alpha)\\
\end{bmatrix} $

With the target coordinate known as well as all the boundary coordinates, this system of equations can be solved to obtain the boundary weight $\beta$.
This however cannot be done with linear algebra seeing that this is a non-linear expression.

Finding weights in generic case

The boundaries and the tie-simplex should always be simplices.
The tie-simplex can be of a different order as the boundaries, but all the boundaries should be of the same order.

Given the generic case where $B$ number of boundaries are found and is expressed by $C$ number of boundary coordinates each, in $N$ dimensions.
When $C$ number of boundary coordinates are present, the number of boundary weights $\beta$ is equal to $C - 1$.
If $B$ number of boundaries are present, the number of tie-simplex weights $\alpha$ is equal to $B - 1$.

$\bar{t} = \begin{bmatrix}
 t_{x} \\
    t_{y} \\
    t_{z} \\
    \vdots \\
    t_{N} \\
\end{bmatrix}$

$\bf{P} = \begin{bmatrix}
 p_{1, 1, x} & p_{1, 1, y} & p_{1, 1, z} & \dots & p_{1, 1, N}\\
    p_{1, 2, x} & p_{1, 2, y} & p_{1, 2, z} & \dots & p_{1, 2, N}\\
    \vdots & \vdots & \vdots & \vdots & \vdots \\
    p_{1, C, x} & p_{1, C, y} & p_{1, C, z} & \dots & p_{1, C, N}\\
 p_{2, 1, x} & p_{2, 1, y} & p_{2, 1, z} & \dots & p_{2, 1, N}\\
    p_{2, 2, x} & p_{2, 2, y} & p_{2, 2, z} & \dots & p_{2, 2, N}\\
    \vdots & \vdots & \vdots & \vdots & \vdots \\
    p_{2, C, x} & p_{2, C, y} & p_{2, C, z} & \dots & p_{2, C, N}\\
    \vdots & \vdots & \vdots & \vdots & \vdots \\
    p_{B, C, x} & p_{B, C, y} & p_{B, C, z} & \dots & p_{B, C, N}\\
\end{bmatrix} ^T$

$\bar{\omega} = \begin{bmatrix}
 (1 - \Sigma \beta)(1 - \Sigma \alpha) \\
    (\beta_1)(1 - \Sigma \alpha)\\
    (\beta_2)(1 - \Sigma \alpha)\\
    \vdots \\
    (\beta_{C-1})(1 - \Sigma \alpha)\\
 (1 - \Sigma \beta)(\alpha_1) \\
    (\beta_1)(\alpha_1)\\
    (\beta_2)(\alpha_1)\\
    \vdots \\
    (\beta_{C-1})(\alpha_1)\\
 (1 - \Sigma \beta)(\alpha_2) \\
    (\beta_1)(\alpha_2)\\
    (\beta_2)(\alpha_2)\\
    \vdots \\
    (\beta_{C-1})(\alpha_2)\\
    \vdots \\
 (1 - \Sigma \beta)(\alpha_{B-1}) \\
    (\beta_1)(\alpha_{B-1})\\
    (\beta_2)(\alpha_{B-1})\\
    \vdots \\
    (\beta_{C-1})(\alpha_{B-1})\\
\end{bmatrix} $

$\Sigma \alpha = \sum\limits_{i=1}^{B-1} \alpha_i \quad : \alpha_i \in \mathbb{R}[0, 1] $

$\Sigma \beta = \sum\limits_{j=1}^{C-1} \beta_j \quad : \beta_j \in \mathbb{R}[0, 1] $

With the target coordinate known as well as all the boundary coordinates, this system of equations can be solved to obtain the boundary weights $\beta$.
This however cannot be done with linear algebra seeing that this is a non-linear expression.

If any $\alpha_i$ < 0, $\alpha_i$ > 1 or $\beta_j$ < 0, $\beta_j$ > 1 the target coordinate is outside the boundaries and tie-simplices connecting the sets of boundary coordinates.


问:
Numerical solution to differential equation
I want to find a numerical solution to this differential equation:

$$x''-\frac{2kqQ}{m\left(x^2+(\frac{ab}{2})^2\right)}=0$$

$k,q,Q,a,b,m$ are constants.

Can anyone recommend a program or a method for solving this? I have mathematica but I have no idea how to solve this in mathematica.

答:
Just another way (leading to the same result as @uranix.

To make notations simpler, let
$$\alpha=\frac{2kqQ}{m}\qquad \text{and} \qquad \beta=\frac{a b}{2}$$ making the equation to be
$$x''-\frac{\alpha}{x^2+\beta^2}=0$$ Switching variables
$$\frac{t''}{[t']^3}+\frac{\alpha}{x^2+\beta^2}=0$$ Reduction of order $p=t'$ gives
$$p=\pm \frac{\sqrt{\beta }}{\sqrt{2 \alpha  \tan ^{-1}\left(\frac{x}{\beta }\right)+ c_1}}$$ No way to go further even if $c_1=0$.


问:
Minkowski's upper bound
Show that every ideal class of K contains an integral ideal $\mathfrak{b}$ such that

$N\mathfrak{b}\leq \frac{n!}{n^n}(\frac{4}{\pi})^s\sqrt{|d_K|}$

I couldn't find a proper proof of the statement. I can prove that every non-zero integral idea $\mathfrak{a}$ in K contains non-zero element $\alpha$ such that

$N_{K/\mathbb{Q}}(\alpha)\leq \frac{n!}{n^n}(\frac{4}{\pi})^s\sqrt{|d_K|}N\mathfrak{a}$.

But was not able to proceed from this.

答: Hint: given $\mathfrak b$, take an integral ideal $\mathfrak a$ such that $\mathfrak a \mathfrak b$ is principal.

Title: Minkowski's upper bound


问:
Convergence of a series of a given metric..
I'm trouble with a metric defined over a given set: Consider $\mathcal{P}=C_p^\infty([-\pi, \pi])$, that is, $\mathcal{P}$ is the set of all infinitely differentiable functions $f:\mathbb R\rightarrow \mathbb C$ which are $2\pi$-periodic ($f(x+2\pi)=f(x)$, for all $x\in \mathbb R$). We can introduce a metric in this set setting $$\displaystyle d(f, g)=\sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f^{(k)}-g^{(k)}\|_\infty}{1+\|f^{(k)}-g^{(k)}\|_\infty},$$ where $\displaystyle \|f\|_\infty=\sup_{x\in J}|f(x)|$ where $J\subset \mathbb R$ is an interval of length $2\pi$. The problem is: why does the series above converge? Also, how can I prove $f_n\to f$ in $\mathcal{P}$ if and only if $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k\in\mathbb N$?</p>

答:
Each term in the series is bounded by $\frac{1}{2^k}$, since $\frac{x}{1+x}<1$ for all $x\ge 0$. Thus we have
$$\sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f^{(k)}-g^{(k)}\|_\infty}{1+\|f^{(k)}-g^{(k)}\|_\infty}\leq \sum_{k=0}^\infty \frac{1}{2^k}=\frac{1}{1-\frac12}=2.$$

I suspect you wish to show that $f_n\to f$ in $\mathcal P$ if and only if $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k$. One direction (that if $f_n\to f$ in $\mathcal P$, $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k$) is easy, and I will leave to you. For the other direction, suppose $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k$ and let $\epsilon>0$. Let $i$ be such that $\frac{1}{2^i}<\frac{\epsilon}{2}$ and $N$ be such that, for all $k\leq i$,
$$n\ge N\implies \|f_n^{(i)}-f^{(i)}\|_\infty < \frac{\epsilon}{2(i+1)}.$$
Then for $n\ge N$ we have
$$\begin{align}
d(f_n,f) &= \sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{1+\|f_n^{(k)}-f^{(k)}\|_\infty}\\
&= \sum_{k=0}^i \frac{1}{2^k}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{1+\|f_n^{(k)}-f^{(k)}\|_\infty}+\sum_{k=i+1}^\infty \frac{1}{2^k}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{1+\|f_n^{(k)}-f^{(k)}\|_\infty}\\
&\leq \sum_{k=0}^i \frac{1}{2^k}\|f_n^{(k)}-f^{(k)}\|_\infty+\sum_{k=i+1}^\infty \frac{1}{2^k}\\
&\leq \sum_{k=0}^i \frac{1}{2^k}\frac{\epsilon}{2(i+1)}+\frac{1}{2^i}\\
&\leq \sum_{k=0}^i \frac{\epsilon}{2(i+1)}+\frac{\epsilon}{2}=\epsilon\\
\end{align}$$
hence $f_n\to f$ in $\mathcal P$.


问:
This random variable $Z$ seems to have the same distribution as $\min{X,Y}$ as $|X-Y|$, for $X,Y \sim \operatorname{Unif}(0,1)$.
Based on these (but I hope this is self-contained):

Why does $\min(X,Y)$ and $|X-Y|$ have the same distribution when $X,Y\sim U(0,1)$?

Random points on a circle

For $X,Y \sim \operatorname{Unif}(0,1)$, $\min\{X,Y\}$ and $|X-Y|$ have the same distribution, as explained intuitively in the 1st question. Apparently, for this random variable $Z$ described in the 2nd question, $Z$ also has the same distribution as $\min\{X,Y\}$ and $|X-Y|$. $Z$ is described as

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $Z$ be the random variable denoting the distance of the point $B$ to its neighbour in the anti-clockwise direction.

Of course the precise answer to this is that they all have pdf $f_Z(z) = 2(1-z)1_{z \in (0,1)}$. But I'm wondering here if $Z$ is actually the absolute difference or minimum or 2 iid uniform(0,1) or if $Z$ is something else altogether.

答:
I believe $Z=\min\{X,Y\}$, where $X$ and $Y$ are described in David K's answer:

Let $X$ be the anticlockwise distance along the circle's circumference from $B$ to $A$.
Let $Y$ be the anticlockwise distance from $B$ to $C$.

Since the three points are independently uniformly distributed along the circumference of the circle, $X$ and $Y$ are iid variables with uniform distributions on $[0,1).$


问:
An uncountable set of sequence
Let $Z$ be a set of sequence.$\forall \{a_n\}, \exists \{b_n\} \in Z$, such that $lim_{n\rightarrow \infty} \frac{a_n}{b_n}=0$. Prove that $Z$ is uncountable.

We are asked to prove this using the Cantor's Diagonal Argument, but I don't see how it works. Any other approaches are acceptable as well.

Thanks!

答:
If $Z$ were countable, we could label its elements $\{z_n^1\}, \{z_n^2\}, \ldots$ (using superscripts because subscripts are taken already). Define a new sequence $\{a_n\}$ by $$a_n = \max\limits_{i,j \leq n}z^i_j.$$

Note that for all $m \geq n$, we have $a_m \geq z^n_m$, so $a_m/z^n_m \geq 1$, so we cannot have $a_m/z^n_m\to 0$, a contradiction.

Title: An uncountable set of sequence


问:
Show that $ab \mod m = ((a \bmod m)(b \bmod m)) \bmod m.$
With modular definition in mind that two integers $a,b \in \mathbb{Z}$, are congruent module $m$ meaning that $a\equiv b \pmod{m}$, where $m$ is a positive integer.

Details:

$x \equiv y \pmod{m}$ is by definition equivalent to  $m|(x−y)$.

$x \equiv y \pmod{m}$ , $x,y\in \mathbb{Z}, m \in \mathbb{Z^+}$ iff $a \pmod {m} = b \pmod{m}$.

if $a | b$ and $a|c$, then $a|(b+c)$.

if $a | b$ and $b|c$, then $a|c$.

if $a | b$ then $a|bc$, for all integers $c$.

Question: Show that $ab \pmod{m} = ((a \pmod{m} )(b \pmod{m})) \mod m.$

答:
Suppose $a = mq + r, b = mq' + r':$

So $ a \mod m = r$ and $b \mod m = r'$

$ab = mq(mq') + rmq' +r'(mq) + rr'$

All terms are multiples of $m$ except perhaps $rr'$ So to get modulo $m$ of $ab$, you simply get modulo $m$ of $rr'$.

Hence $ab \mod m = rr' \mod m $ and $rr' = (a \mod m)(b \mod m)$ , yielding the result.


问:
If $A$ is an $n \times n$ matrix such that $A^3 = O_{3}$, show that $I - A$ is invertible with inverse $I + A + A^2$
So this question is basically a proof.

If $A$ is an $n \times n$ matrix (so square) which satisfies the condition $A^3 = O_{3}$ ($A^{3}$ gives the $3 \times 3$ zero matrix), then show that $(I - A)$ is invertible with inverse $(I + A + A^2)$.

I have no idea where to start, all help welcome.

答: Hint: for any invertible matrix $A$ with inverse $A^{-1}$, we have $AA^{-1} = I$. Try multiplying out $(I-A)(I+A+A^{2})$. You should get some cancellation and then use your condition to conclude something...</p>


问:
Prove that the polyhedron has at least $2^n$ vertices
Let A be the n x n identity matrix and let b be the n x 1 vector with every entry equal to one. Prove that the polyhedron P(A, b) has at least 2$^n$ vertices.

What I have so far:

P(A, b)= set of x $\in$ $\mathbb{R}$ so that

$\begin{bmatrix} -I_n \\A\end{bmatrix}$ x $\leq$ $\begin{bmatrix} **0** \\**b**\end{bmatrix}$

is satisfied. The 0 is the zero vector.

Therefore we will have the constraints x$_1$, x$_2$, ..., x$_n$ $\geq$ 0 and x$_1$ $\leq$ 1, x$_n$ $\leq$ 1, ..., x$_n$ $\leq$ 1. We will have 2n constraints. To make the proof a little less wordy, I've called the first set of constraints "0 constraints" and the second set of constraints "1 constraints" - this is just for my draft proof.

The number of vertices will be: 1 + n - 1 + n$^2$

The 1 comes from the origin always being a vertex as it is where the 0 constraints all intersect

The n - 1 comes from there being an intersection between every 1 constraint with every other 1 constraint except when the constraints are the same

The n$^2$ comes from there being an intersection between every 0 constraint and every 1 constraint - since we have n of each constraints, we get n$^2$.

That's where I get stuck. I think I'm heading in the correct direction but my counting or something is off.

答:
For every variable, $x_i$, you can pick it to be $0$ or $1$ to make it a vertex, that is for each variable, you have two choices.

Since there are $n$ variables, you have $2^n$ choices.


答:
First treating id$_U$=1 is correct. Because when you multiply a matrix by identity matrix, you got the matrix. So you can just ignore identity matrix during matrix multiplication, as you ignore 1 during multiplication.

Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$.

But I believe the reasoning is the following computation similar to the previous one. The claim you made "because $q$ is also a linear map" justified the commutativity of addition for linear maps, a.k.a $p+q = q+p$:

$$q(q+p) = q(p+q) = q(\text{id}_U) \Rightarrow qp + q^2=q \Rightarrow q^2 =q.$$

Then following your own computation we're done.

Title: Prove $p^2=p$ and $qp=0$

Body:
I am not really aware what's going on in this question. I appreciate your help.

Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$.

Prove $p^2=p$ and $qp=0$.

$$p(p+q) = p(\text{id}_U) \Rightarrow p^2+pq=p \Rightarrow p^2 =p.$$ I actually first found this by letting $(\text{id}_U) =1$, although probably a wrong to do it.

Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$. Again, I'm not sure if this actually is the right way to do it.

Then we have $$q(p+q) = q(\text{id}_U) \Rightarrow q^2+qp=q \Rightarrow q^2+ qp= q^2 \Rightarrow qp=0.$$

Prove $K=\text{im}(q)$.

For this question, I honestly do not know how to tackle it from the following definitions.

$K = \ker(p)=\left\{u \in U \ |\ p(u)=0_U\right\}$ and $\text{im}(q)=\left\{q(u) \ | \ u \in U \right\}$.

What's the correct way of doing these questions? Thank you for your time.


答:
What you want to show is that you can multiply by representatives thus showing that the relation $a\equiv b\pmod {m}$ is a congruence relation. So following @Karl's comment suppose that $a\equiv a'\pmod m$ and $b\equiv b'\pmod m$, i.e. $m|a-a'$ and $m|b-b'$, then
$$ab-a'b'=(a-a')b'+a(b-b')$$
and since $a-a',b-b'$are divisble by $m$ it follows that $ab-a'b'$ is divisble by $m$ and thus
$$ab\equiv a'b'\pmod m.$$

Title: Show that $ab \mod m = ((a \bmod m)(b \bmod m)) \bmod m.$

Body:
With modular definition in mind that two integers $a,b \in \mathbb{Z}$, are congruent module $m$ meaning that $a\equiv b \pmod{m}$, where $m$ is a positive integer.

Details:

$x \equiv y \pmod{m}$ is by definition equivalent to  $m|(x−y)$.

$x \equiv y \pmod{m}$ , $x,y\in \mathbb{Z}, m \in \mathbb{Z^+}$ iff $a \pmod {m} = b \pmod{m}$.

if $a | b$ and $a|c$, then $a|(b+c)$.

if $a | b$ and $b|c$, then $a|c$.

if $a | b$ then $a|bc$, for all integers $c$.

Question: Show that $ab \pmod{m} = ((a \pmod{m} )(b \pmod{m})) \mod m.$


问:
Find the c.d.f. from the p.m.f
Verify that the following p.m.f. is valid and if it is, then find its c.d.f.

$$\mathbb{P}(X=n)=\frac{1}{n(n+1)}\;\forall\;n\in\mathbb{N}^+$$

My Attempt
Verifying that it is valid is easy. I just need to check whether or not it sums to $1$. So
$$\sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$
We know that this infinite series is true, so our p.m.f. is valid. Now to find the c.d.f.. Since this p.m.f. is discrete, I will use the following formula:
$$F(t)=\sum_{k\in X, k\leq t}p(k)$$
My question is how should I divide the intervals so that I can write the c.d.f.? It is a piecewise function that changes with each $n$ so would my c.d.f. be
\begin{equation}
F(t)=
\begin{cases}
0 & \text{if } t\leq0\\
\frac{1}{1(1+1)} &\text{if } 0<t\leq1\\
\frac{1}{2(2+1)} &\text{if } 1<t\leq2\\
...
\end{cases}
\end{equation}

This doesn't seem like a very good way to write the c.d.f. so I am wondering if there are other ways to express it.

答:
You can use induction  as pointed out in the hint, or you can see it as follows:

$$
\mathbb{P}(X\leq n) = \sum_{i=1}^n \frac{1}{i(i+1)}=\sum_{i=1}^n \frac{(i+1)-i}{i(i+1)} = \sum_{i=1}^n \left(\frac{1}{i} -\frac{1}{i+1} \right)= 1-\frac{1}{n+1}.
$$

Title: Find the c.d.f. from the p.m.f


问:
Infinite type algebra
Maybe this question is trivial : most of the algebras I encounter (issued from invariant theory !) are of finite type. Does someone have a 'simple' example of an algebra/subalgebra of infinite type. In fact, I don't see immediately how to show that some algebra is of infinite type.

For instance, if we take the subalgebra
$$\mathbb{R}[x^{k}+k,\quad k\geq 2]$$
it seems to me to be of infinite type...

Same for
$$\mathbb{R}[x^{k_1}y^{k_2},k_1\geq 1,k_2\geq 2]$$

Is it obvious ?

答:
Any finite type $K$-algebra is isomorphic to $K[X_1,\ldots,X_n]/I$, for some $n\geq 0$ and some ideal $I$. In particular, a finite type $K$-algebra is a Noetherian ring.

Hence any non Noetherian ring containing a field $K$ as a subring will give you a counterexample. The simplest one is $K[X_m, m\geq 0]$.

If you want an example of a finite type algebra having a subalgebra not of finite type, you can take $K[X,Y]$ and $K[XY^k,k\geq 0]$ , since the second one is not Noetherian (see  How to prove that $k[x, xy, xy^2, \dotsc]$ is not noetherian?)

Title: Infinite type algebra


问:
$F(x)$ versus $F[x]$ where $F$ is a field
What is the difference, if there is any, between $F(x)$ and $F[x]$, where $F$ is a field and $x$ is an element of a field extension $E/F$ that is not in $F$? I have some guesses, but I was never sure about this. For example, for one specific usage in a textbook, I see $\mathbb{Q}[i]$ and $\mathbb{Q}(\pi)$. And maybe a more general question: what happens when $F$ is just a ring? For example, for the ring of Gaussian integers, the notation I usually see is $\mathbb{Z}[i]$, with brackets instead of open parentheses. I guess I never properly learned the notation.

答:
Well, $F[x]$ means ring adjunction of $x$, while $F(x)$ means field adjunction of $x$.
More concretely,

$F(x) = \left\{\frac{f(x)}{g(x)}\mid f(x),g(x)\in F[x], g(x)\ne 0\right\}$
is the field of quotients of $F[x]$.

$\Bbb Q(i) = \{a+ib\mid a,b\in\Bbb Q\}$ with $i^2=-1$ is the field adjunction of $i$ to $\Bbb Q$. It's a field and the same as the ring adjunction $\Bbb Q[i]$ (as its also a field).

$\Bbb Z[i] = \{a+ib\mid a,b\in\Bbb Z\}$ with $i^2=-1$ is the ring adjungation of $i$ to $\Bbb Q$. Its's a ring.

Beware of $\Bbb Q(\pi)$. As $\pi$ is tranzendential over $\Bbb Q$, this field extension of $\Bbb Q$ is infinite.

You need to learn about ring and field adjunction.


答: $F[x]$ is the ring of all polynomials in $x$ with coefficients from $F$. In contrast, $F(x)$ is the smallest field which extends $F$ and contains $x$: in other words, it contains all rational functions (i.e. quotients of polynomials) in $x$ with coefficients from $F$, rather than just all polynomials. Note that sometimes these can be the same, e.g. $\mathbb{Q}\left[\sqrt{2}\right]$ is the same as $\mathbb{Q}\left(\sqrt{2}\right)$, since any polynomial in $\sqrt{2}$ can be written in the form $a+b\sqrt{2}$, and if $a,b,c,d \in \mathbb{Q}$ then $$\frac{a+b\sqrt{2}}{c+d\sqrt{2}} = \frac{ac-2bd}{c^2-2d^2} + \frac{bc-ad}{c^2-2d^2}$$ where the coefficients are still in $\mathbb{Q}$.

Title: $F(x)$ versus $F[x]$ where $F$ is a field

Body: What is the difference, if there is any, between $F(x)$ and $F[x]$, where $F$ is a field and $x$ is an element of a field extension $E/F$ that is not in $F$? I have some guesses, but I was never sure about this. For example, for one specific usage in a textbook, I see $\mathbb{Q}[i]$ and $\mathbb{Q}(\pi)$. And maybe a more general question: what happens when $F$ is just a ring? For example, for the ring of Gaussian integers, the notation I usually see is $\mathbb{Z}[i]$, with brackets instead of open parentheses. I guess I never properly learned the notation.