问:
Equation describing an elliptical orbit using time and angle
I'm working on a project where I'm trying to describe the orbits of the planets in the solar system using the polar equation of an ellipse.

Below is an equation I got from here.
$$
r = \frac{b^{2}}{a - c \cos \theta}
$$

Where
$a$ is the semi-major axis,
$b$ is the semi-minor axis,
$c$ is the distance between the center of the orbit and a focus point of the orbit $(c^2 = a^2 - b^2)$. The sun is at one of these foci, so $c$ is the distance between the center of the orbit/ellipse and the sun, the origin of the graph. This form of the equation also assumes no displacement of the ellipse in the $y$ direction from the origin.

This equation does relate the radius $r$ of a point on the ellipse as a function of the angle $\theta$ it makes with the origin.

However, I am trying to look for an equation that models the angle $\theta$ as a function of time $t$. For example, if $T$ was the period of one full orbit, then after $T$ seconds, the change in angle should be $2\pi$ radians. In other words, where if $f(t) = \theta$, $f(T) = 2\pi$ or $0$.

Considering that I have the values for $a$, $b$, and $c$, the orbital period $T$, and the linear speed $v$ of each planet, and since I cannot use the radius directly (because it is not constant across the ellipse), how could I derive the polar angle $\theta$ as a function of time $t$ without $r$?

答:
For the time dynamic the Newton gravity law additionally results in the second Kepler law
$$
r^2\dotθ=L=const.\tag1
$$
that allows to solve the angle-time problem via numerical integration.

One can also try to solve it symbolically, or at least reduce it to a simpler non-ODE form. Eliminating first $r$ gives
$$
L=\frac{b^4\dotθ}{(a-c\cosθ)^2}.\tag2
$$
For integrals of fractions in trigonometric functions the recommendation is to use  the half-angle tangent $u=\tan\fracθ2$ with $\dot u=\frac12(1+u^2)\dotθ$ and $\cos θ=\frac{1-u^2}{1+u^2}$. The equation simplifies to
$$
\frac{L}{b^4}=\frac{2(1+u^2)\dot u}{((a-c)+(a+c)u^2)^2}\tag3
$$
This form now strongly suggests to set (using again a half-angle in the tangent for symmetry)
\begin{align}
\tan\fracε2&=\sqrt{\frac{a+c}{a-c}}u\tag4
\\
\implies
(a-c)+(a+c)u^2&=(a-c)\sec^2\fracε2\tag{5a}\\
1+u^2&=\frac{(a+c)\cos^2\fracε2+(a-c)\sin^2\fracε2}{a+c}\sec^2\fracε2=\frac{a+c\cosε}{a+c}\sec^2\fracε2\tag{5b}
\\
1-u^2&=\frac{(a+c)\cos^2\fracε2-(a-c)\sin^2\fracε2}{a+c}\sec^2\fracε2=\frac{c+a\cosε}{a+c}\sec^2\fracε2\tag{5c}
\end{align}
With the derivative formulas
$$\begin{align}
\frac{\dotε}2\left(1+\frac{a+c}{a-c}u^2\right)&=\sqrt{\frac{a+c}{a-c}}\dot u\tag{6a}
\\\text{ or }~~
\dotε((a-c)+(a+c)u^2)&=2b\dot u\tag{6b}
\end{align}$$
one can insert and eliminate $u$ for $ε$, simplifying the equation to
$$\begin{align}
\frac{L}{b^3}&=\frac{(1+u^2) \dotε}{(a-c)+(a+c)u^2}
=\frac{a+c\cosε}{a^2-c^2}\dotε\tag{7a}
\\~\\
\frac{L}{b}&=(a+c\cosε)\dotε \tag{7b}
\end{align}$$
Now with an arbitrary change of the momentum constant to $L=ab\omega$, integration results in the Kepler equation
$$\tag8
ωt=ε+\frac{c}{a}\sinε, ~~~ \tan\fracθ2=\sqrt{\frac{a-c}{a+c}}\tan \fracε2
$$
The integration constant was left out which is correct if time $t=0$ corresponds to apoapsis with $θ=0$, where $r=\frac{b^2}{a-c}=a+c$ and thus also $ε=0$. In this context $ε$ is called* the eccentric anomaly (and $θ$ the true anomaly, $M=ωt$ the mean anomaly, there might have been something lost in the historic process, one would think that anomaly refers to the deviation from circular motion.)

For the Cartesian coordinates one gets from the half-angle formulas
\begin{alignat}{1}
x=r\cosθ&=\frac{b^2(1-u^2)}{(a-c)+(a+c)u^2}&=c+a\cosε\tag{9a}\\
y=r\sinθ&=\frac{b^2·2u}{(a-c)+(a+c)u^2}&=b\sinε\tag{9b}
\end{alignat}

* Note that this deviates from the common definitions in that the anomalies are the angles at the gravity center relative to the pericenter position, not as here to the apocenter position.


问:
is $\int^b_a |f| = |F(b)|-|F(a)|$?
sorry if this seems trivial or obvious. But is $\int^b_a |f| = |F(b)|-|F(a)|$? I'm currently trying to prove that $|F(b)-F(a)|\leq \int^b_a|f|$. Using the fundamental theorem of calculus, we have

$$\int^b_af=F(b)-F(a)\leq \int^b_a|f|$$
if $\int^b_a |f| = |F(b)|-|F(a)|$, then via the triangle inequality
$$\int^b_af=F(b)-F(a)\leq \int^b_a|f|=|F(b)|-|F(a)|\leq |F(b)-F(a)|$$
which is the opposite of what I want to prove, so I suppose this would suggest that $\int^b_a |f| \neq |F(b)|-|F(a)|$,

答: Take $a=-1,b=1, F(x)=x^{2}$ which gives $f(x)=2x$. Clearly, $0=F(b)-F(a)\neq \int_a^{b} |f(x)| dx$.

ContentLicense: CC BY-SA 4.0

Title: is $\int^b_a |f| = |F(b)|-|F(a)|$?


问:
(intergation) area of region bounded by a curve and a line with unknown a
Let $S1$ be the area of the region bounded by the curve $y = x^2$ and the lines $y = ax, (0 < a < 1)$, and let $S2$ be the area of the region bounded by the curve $y = x^2$, the lines $y = ax, (0 < a < 1)$ and the line $x = 1$, where $a$ is the same constant.

(a) Find the value of $a$ that minimizes the value of $S1 + S2$. What is the minimum value?

(b) Find the volume of the solid generated by revolving the region, that takes the minimum value of $S1 + S2$, about the x-axis.

I find $S1 = \displaystyle \frac{a^3}{6}$ and $S2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$, but how can I calculate the value of $a$?

答:
You have calculated value of $\small S_1$ and $\small S_2$ correctly. Now to find value of $a$ that minimizes $ \small S = S_1+S_2$, take the derivative of $S$ with respect to $a$ and equate to zero.

$\small S = S_1 + S_2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$

$\small S' = \displaystyle a^2 - \frac{1}{2} = 0 \implies a = \frac{1}{\sqrt2}$

That leads to minimum area of $\small S_1 + S_2 = \displaystyle \frac{\sqrt2 - 1}{3 \sqrt2}$

Now you need to find volume of solid generated by revolution of the region about x-axis $\big($for $a = \displaystyle \small \frac{1}{\sqrt2}\big)$. For $\small S_1$, both shell and washer methods are fine. For $\small S_2$, it is easier to apply washer method as the shell method will require you to split the integral into two.


答:
If by $F$, you mean the antiderivative of $f$, then no.

Take $f(x)=\sin(x), a=\pi, b=2\pi$. Then $F(x) = -\cos x$.

Then $$\int_a^b |f(x)|dx = \int_\pi^{2\pi}|\sin x| dx = -\int_\pi^{2\pi}\sin xdx = -\left(-\cos x|_\pi^{2\pi}\right) = -\left(-\cos 2\pi - (-\cos \pi)\right) = -(-1 - 1) = 2$$

while $|F(b)| - |F(a)| = 0$

Title: is $\int^b_a |f| = |F(b)|-|F(a)|$?

Body:
sorry if this seems trivial or obvious. But is $\int^b_a |f| = |F(b)|-|F(a)|$? I'm currently trying to prove that $|F(b)-F(a)|\leq \int^b_a|f|$. Using the fundamental theorem of calculus, we have

$$\int^b_af=F(b)-F(a)\leq \int^b_a|f|$$
if $\int^b_a |f| = |F(b)|-|F(a)|$, then via the triangle inequality
$$\int^b_af=F(b)-F(a)\leq \int^b_a|f|=|F(b)|-|F(a)|\leq |F(b)-F(a)|$$
which is the opposite of what I want to prove, so I suppose this would suggest that $\int^b_a |f| \neq |F(b)|-|F(a)|$,


问:
Prove that the equation $3x-\sin(2x)=0$ has only one solution
I`m trying to prove that this equation has only one solution$$3x-\sin(2x)=0$$
I need some advice how to do that, Thanks!

EDIT

答: Given the range of $\sin 2x$, it's easy to find an interval in which all the solutions must lie. In this interval, the function has a certain property which immediately tells you there can be at most one solution. By giving an explicit solution, you'll prove that there's at least one solution.</p>

ContentLicense: CC BY-SA 3.0


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.

Body:
First multiply through with $a^2b^2$ expand and homogenize the equation:

$$(a^2k^2+b^2h^2-a^2b^2)z^2-2a^2kyz-2b^2hxz+a^2y^2+b^2x^2=0$$

Then the system is

$$\begin{align}X-\lambda (2b^2p-2b^2hr)&=0\\
Y-\lambda (2a^2q-2a^2kr)&=0\\
Z-\lambda (2(a^2k^2+b^2h^2-a^2b^2)r-2a^2kq-2b^2hp)&=0\\
Xp+Yq+Zr&=0\end{align}$$

Solving explicitly is feasible by hand (showing the Sylvester matrix of which the resultant is the determinant):

resultant(X-l*(2*b^2*p-2*b^2*h*r),Y-l*(2*a^2*q-2*a^2*k*r),p);

$$\begin{pmatrix} 2\,a^2\,k
 \,l\,r-2\,a^2\,l\,q+Y\\\end{pmatrix}$$

resultant(X-l*(2*b^2*p-2*b^2*h*r),Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),p);

$$\begin{pmatrix} -2\,b^2\,
 l&2\,b^2\,h\,l\,r+X\\ 2\,b^2\,h\,l&\left(
 -2\,a^2\,k^2-2\,b^2\,h^2+2\,a^2\,b^2\right)\,l\,r+2\,a^2\,k\,l\,q+Z
 \\ \end{pmatrix}$$

resultant(Y-l*(2*a^2*q-2*a^2*k*r),Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),p);

$$\begin{pmatrix} 2\,a^2\,k
 \,l\,r-2\,a^2\,l\,q+Y\\\end{pmatrix}$$

resultant(X-l*(2*b^2*p-2*b^2*h*r),X*p+Y*q+Z*r,p);

$$\begin{pmatrix}-2\,b^2\,
 l&2\,b^2\,h\,l\,r+X\\ X&Z\,r+Y\,q\\
 \end{pmatrix}$$

resultant(Y-l*(2*a^2*q-2*a^2*k*r),X*p+Y*q+Z*r,p);

$$\begin{pmatrix} 2\,a^2\,k
 \,l\,r-2\,a^2\,l\,q+Y\\
 \end{pmatrix}$$

resultant(Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),X*p+Y*q+Z*r,p);

$$\begin{pmatrix} 2\,b^2\,h
 \,l&\left(-2\,a^2\,k^2-2\,b^2\,h^2+2\,a^2\,b^2\right)\,
 l\,r+2\,a^2\,k\,l\,q+Z\\ X&Z\,r+Y\,q\\
 \end{pmatrix}$$

Among these choose two and take resultant w.r.t $q$

resultant((2*X*a^2*k^2+2*X*b^2*h^2+2*Z*b^2*h-2*X*a^2*b^2)*l*r+(2*Y*b^2*h-2*X*a^2*k)*l*q-X*Z,2*a^2*k*l*r-2*a^2*l*q+Y,q);

$$\begin{pmatrix}\left(2\,
 Y\,b^2\,h-2\,X\,a^2\,k\right)\,l&\left(2\,X\,a^2\,k^2+2
 \,X\,b^2\,h^2+2\,Z\,b^2\,h-2\,X\,a^2\,b^2\right)\,l\,r-X\,Z\\ -2\,a
 ^2\,l&2\,a^2\,k\,l\,r+Y\\ \end{pmatrix}$$

resultant((-(2*X*b^2*h+2*Z*b^2)*l*r)-2*Y*b^2*l*q-X^2,2*a^2*k*l*r-2*a^2*l*q+Y,q);

$$\begin{pmatrix} -2\,Y\,b^
 2\,l&\left(-2\,X\,b^2\,h-2\,Z\,b^2\right)\,l\,r-X^2\cr
 -2\,a^2\,l&2\,a^2\,k\,l\,r+Y\\ \end{pmatrix}$$

Finally the resultant w.r.t $r$

resultant(2*l*((2*Y*a^2*b^2*h*k+2*X*a^2*b^2*h^2+2*Z*a^2*b^2*h-2*X*a^4*b^2)*l*r-X*Y*a^2*k+Y^2*b^2*h-X*Z*a^2),-2*l*((b^2*(2*Y*a^2*k+2*Z*a^2)+2*X*a^2*b^2*h)*l*r+Y^2*b^2+X^2*a^2),r);

$$\begin{pmatrix} \left(4\,
 Y\,a^2\,b^2\,h\,k+4\,X\,a^2\,b^2\,h^2+4\,Z\,a^2\,b^2\,h-4\,X\,a^4\,b
 ^2\right)\,l^2&\left(-2\,X\,Y\,a^2\,k+2\,Y^2\,b^2\,h-2
 \,X\,Z\,a^2\right)\,l\cr \left(-4\,Y\,a^2\,b^2\,k-4\,X\,a^2\,b^2\,h-
 4\,Z\,a^2\,b^2\right)\,l^2&\left(-2\,Y^2\,b^2-2\,X^2\,a
 ^2\right)\,l\\ \end{pmatrix}$$

factor(%);
-8*X*a^4*b^2*(Y^2*k^2+2*X*Y*h*k+2*Y*Z*k+X^2*h^2+2*X*Z*h-Y^2*b^2-X^2*a^2+Z^2)*l^3

i.e. $Y^2k^2+2XYhk+2YZk+X^2h^2+2XZh-Y^2b^2-X^2a^2+Z^2=0.$

or in one fell swoop using the maxima CAS function eliminate

eliminate([X-l*(2*b^2*p-2*b^2*h*r),Y-l*(2*a^2*q-2*a^2*k*r),Z-l*(2*(a^2*k^2+b^2*h^2-a^2*b^2)*r-2*a^2*k*q-2*b^2*h*p),X*p+Y*q+Z*r],[p,q,r]);
[-16*Y*a^2*b^6*(Y^2*k^2+2*X*Y*h*k+2*Y*Z*k+X^2*h^2+2*X*Z*h-Y^2*b^2-X^2*a^2+Z^2)*l^4]

Setting $Z=1$ we have the dual by the equation in the standard affine in the dual projective plane

$$(h^2-a^2)X^2+2hkXY+(k^2-b^2)Y^2+2hX+2kY+1=0.$$


答:
Let's say we restrict to continuous $f$ for simplicity. A giveaway is the fact that the quantity $\lvert F(b)\rvert-\lvert F(a)\rvert$ depends on the choice of an antiderivative $F$ of $f$. For instance, if we had chosen an antiderivative $F$ such that $F(a)$ and $F(b)$ are both positive, then this quantity would acutally be $\int_a^bf(x)\,dx$, but if we had it such that one is negative and the other is positive, then $\lvert F(b)\rvert-\lvert F(a)\rvert=\begin{cases}F(b)+F(a)&\text{if }F(b)>F(a)\\ -F(b)-F(a)&\text{if }F(b)<F(a)\end{cases}$.

Therefore, your statement would be that for all continuous $f$, for all $a,b$ and for all antiderivatives $F$ of $f$ we have $\int_a^b\lvert f(x)\rvert\,dx=\lvert F(b)\rvert-\lvert F(a)\rvert$. Which doesn't make sense, because there are two antiderivatives $F_1$ and $F_2$ that result in different RHS.

Title: is $\int^b_a |f| = |F(b)|-|F(a)|$?


问:
why for $f(x) =1/(1+1/x)$ the curve pass through the origin
HI thank's for the future replies.
My question is simple : why for $f(x) =1/(1+1/x)$ the curve pass through the origin so it means $f(0)=0.$
And why $0^{-1}$ is infinite? it's what's telle geogebra application.

答:
If you write $f(x)=\frac{x}{x+1}$ then you should specify that this is written for $x \neq 0$ as orignal function is undefined at $0$ Or you can say It is out of domain of the function. We can use limits here as $x$ will never be equal to $0$ but approaching to $0$ in the limit process.
$$f(0)=\lim_{x \to 0} \frac{1}{1+\frac{1}{x}}$$ $\implies   f(0)\to 0$ as when $x\to 0$ , $\frac{1}{x} \to \infty$ and something over $\infty$ (very large number) is indeed $0$(Actually $\to 0$)


问:
Question on Ricci flow and Einstein Tensor
My knowledge on this topic is solely based on physics books, introductory texts on differential geometry and this one paper $[1]$.

One of the formulas of the paper shows what is called the Ricci soliton equation:

$R_{\mu\nu} -\frac{1}{2}\mathcal{L}_{X}g_{\mu\nu} - cg_{\mu\nu} = 0 \tag{1}$

In physics, a similar tensor field can be constructed: the Einstein tensor, given by:

$R_{\mu\nu} -\frac{1}{2}Rg_{\mu\nu} + \Lambda g_{\mu\nu} = 0 \tag{2}$

I would like to know: is it possible to have a situation where $\mathcal{L}_{X}g_{\mu\nu} =Rg_{\mu\nu}$, and $c=-\Lambda$ ?

$$* * *$$

$[1]$ https://arxiv.org/abs/0708.2144

答:
A vector field $X$ is said to be a conformal Killing vector field if
${\cal L}_Xg=\varphi g$ some function $\varphi$. In special case $\varphi=0$ it is called a Killing vector field.

Consider an Einstein Riemannian manifold $(M,g)$ of zero scalar curvature $R=0$ thus it is Ricci flat i.e. $Ric=R g=0$. If $(M,g)$ admit a Killing vector field $X$ (i.e. $\mathcal{L}_Xg=0$) then $(M, g, X)$ satisfies in your conditions. That is $0=\mathcal{L}_{X}g_{\mu\nu} =Rg_{\mu\nu}=0$ and $c=-\Lambda$ for zero Cosmological constant $\Lambda=0$.

For example $\Bbb R^n$ with Euclidean metric satisfies in your requirements. another example is torus $\Bbb S^1\times \Bbb S^1$ with flat metric because it admit Killing vector field.

All of these are correct if cosmological constant were zero, that I don't know what it is intuitively.


问:
Travelling Wave Solution For Fisher Equation
I am working with the Fisher equation which I have non-dimensionalised as
\begin{equation}
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} +u(1-u)
\end{equation}

I am looking for travelling wave solutions for the previous equation. I let $u(x,t)=U(z)$ with $z=x-ct$ so that $c$ is the speed of the travelling wave with $x$ from -100 to 100. and sudying for $t \geq 0$. I found the second order ODE for $U(z) as from the last eq, this being
\begin{align}
U'=V
\end{align}
\begin{align}
V' = -cV - U(1-U)
\end{align}

This is the part that I am having issues with. I am given the following
$$
u_0(x)=u(x,0)= \frac{1}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^2}
$$
For some positive parameter $\beta$. I must show that there is some unique $c$, and state it's value, such that $U(z)=u_0(z)$ for some arbitrary $\beta$.

I would appreciate any help with this problem or an approach that will lead me in the right direction. Thank you all in advance.

答:
I agree with what you found : $\quad\begin{cases}
U'=V\\
V' = -cV - U(1-U)
\end{cases}$
$$U''=-cU'-U(1-U)\tag 1$$
With initial condition :
$$
u_0(x)=u(x,0)= \frac{1}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^2}=U(x) \tag 2
$$
because this condition must satisfy the ODE $(1)$.

$$U'(x)=\frac{-2}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^3}\frac{\beta}{\sqrt{6}}e^{\frac{x}{\sqrt{6}}}\tag 3$$
$$U''(x)=\frac{1}{(1 + \beta e^{\frac{x}{\sqrt{6}}})^4}\beta^2e^{2\frac{x}{\sqrt{6}}})
-\frac{1}{3(1 + \beta e^{\frac{x}{\sqrt{6}}})^3}\beta e^{\frac{x}{\sqrt{6}}}\tag 4$$
Putting $(2),(3),(4)$ into equation $(1)$ after simplification leads to :
$$\frac{c\sqrt{3}-5}{3(1 + \beta e^{\frac{x}{\sqrt{6}}})^3}\beta e^{\frac{x}{\sqrt{6}}}=0$$
This can be satisfied any $\beta$ only with
$$c=\frac{5}{\sqrt{3}}$$


问:
Getting a normal distribution when sampling a uniform disturbition
I encountered some behavior that somehow makes sense to be but I don't exactly remember why, and I'll be glad to hear an explanation.

To simplify things up (it wasn't an intended experiment), I was drawing integers in the range [1, 10000], each number with the same probability (1/10000). I repeated this for like 200k times. For each number (from 1 to 10000), I wrote how many times it has been drawn. Then, I counted how many numbers appeared one time, how many numbers appeared 2 times, and so on. This gave me a normal distribution. Is that related to the original distribution being uniform? Or is it because of CLT? Should it happen in any kind of distribution?

Thanks.

答:
Sampling from a continuous uniform distribution, you will find that the Central Limit Theorem begins to 'converge' to normal for surprisingly small $n.$

If I sum $n=12$ independent observations from $\mathsf{Unif}(0,1)$ and subtract $6,$ the resulting
random variable will be very nearly standard normal:
$Z = \sum_{1=1}^{12} U_i - 6 \stackrel{aprx}{\sim}\mathsf{Norm}(0,1).$
Demonstration of a thousand such values using R:

```<code>set.seed(317)
z = replicate(1000, sum(runif(12))-6)
summary(z);  sd(z)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
-3.265696 -0.720931  0.006213  0.003641  0.704164  2.854923
 [1] 1.010645  # sample SD
</code>```

A Shapiro-Wilk tests doesn't detect a difference from normal.

```<code>shapiro.test(z)

        Shapiro-Wilk normality test

data:  z
W = 0.99823, p-value = 0.3946
</code>```

And a Kolmogorov-Smirnov test doesn't detect a difference from standard normal.

```<code>ks.test(z, pnorm)

        One-sample Kolmogorov-Smirnov test

data:  z
D = 0.020192, p-value = 0.8096
alternative hypothesis: two-sided
</code>```

A histogram of the 1000 values of $Z$ generated in this way shows a reasonably
good fit to a standard normal density curve.

```<code>hist(z, prob=T, col="skyblue2", main="Aprox NORM(0,1) from Sample of Uniform")
 curve(dnorm(x), add=T, col="orange", lwd=2)
</code>```

Finally, a normal quantile plot is very nearly linear:

```<code>qqnorm(z, pch=20); qqline(z, col="green2")
</code>```

This method of generating (nearly) a standard normal distribution isn't perfect
(12 is a long way from infinity), but it was used to get approximately normal
distributions in the early days of computation because it involves only simple
arithmetic. Notice, however, that this method cannot give values outside
the interval $[-6, 6],$ while the standard normal distribution theoretically
takes values throughout the real line.

Body:
Sampling from a continuous uniform distribution, you will find that the Central Limit Theorem begins to 'converge' to normal for surprisingly small $n.$

If I sum $n=12$ independent observations from $\mathsf{Unif}(0,1)$ and subtract $6,$ the resulting
random variable will be very nearly standard normal:
$Z = \sum_{1=1}^{12} U_i - 6 \stackrel{aprx}{\sim}\mathsf{Norm}(0,1).$
Demonstration of a thousand such values using R:

set.seed(317)
z = replicate(1000, sum(runif(12))-6)
summary(z);  sd(z)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
-3.265696 -0.720931  0.006213  0.003641  0.704164  2.854923
 [1] 1.010645  # sample SD

A Shapiro-Wilk tests doesn't detect a difference from normal.

shapiro.test(z)

        Shapiro-Wilk normality test

data:  z
W = 0.99823, p-value = 0.3946

And a Kolmogorov-Smirnov test doesn't detect a difference from standard normal.

ks.test(z, pnorm)

        One-sample Kolmogorov-Smirnov test

data:  z
D = 0.020192, p-value = 0.8096
alternative hypothesis: two-sided

A histogram of the 1000 values of $Z$ generated in this way shows a reasonably
good fit to a standard normal density curve.

hist(z, prob=T, col="skyblue2", main="Aprox NORM(0,1) from Sample of Uniform")
 curve(dnorm(x), add=T, col="orange", lwd=2)

Finally, a normal quantile plot is very nearly linear:

qqnorm(z, pch=20); qqline(z, col="green2")

This method of generating (nearly) a standard normal distribution isn't perfect
(12 is a long way from infinity), but it was used to get approximately normal
distributions in the early days of computation because it involves only simple
arithmetic. Notice, however, that this method cannot give values outside
the interval $[-6, 6],$ while the standard normal distribution theoretically
takes values throughout the real line.


答:
Your function $f$ is not defined for $x \in \{-1,0\}$, because division by $0$ is not defined, so $1+(1+1/0)$ and $1+(1+1(-1))$ are not defined.
But for $x \in \{-1,0\}$ we have $f(x)=x/(1+x)$ in a neighborhood of $0$. Wich means that the graphs of $f$ and $x/(1+x)$ look almost similar and on a picture you may not see a difference. Mathematically this means $\lim_\limits{x \to 0} f(x)=0$.

Body:
HI thank's for the future replies.
My question is simple : why for $f(x) =1/(1+1/x)$ the curve pass through the origin so it means $f(0)=0.$
And why $0^{-1}$ is infinite? it's what's telle geogebra application.


问:
Do lifts of maps from extremally disconnected compact Hausdorff spaces into certain colimits exist?
Let $T$ be a extremally disconnected compact Hausdorff space. Further let $X$ and $Y$ be two ind-compact Hausdorff spaces, i.e. both are the filtered colimit of compact Hausdorff spaces with injective transition maps. Assume that $f \colon X \to Y$ is a continuous surjective map and $h \colon T \to Y$ is continuous.

My question is wether there exists a continuous lift $g \colon T \to X$ of $h$ such that $h= f \circ g$. I know that this is true if $X$ and $Y$ are compact Hausdorff spaces themselves since extremally disconnected compact Hausdorff spaces are the projective  objects in the category of compact Hausdorff spaces.

I might have to add, that $X$ and $Y$ share the same underlying filtrated category $\mathbb{R}_{>0}$ and whenever $f$ sends an element $x \in X_c$ to an element $f(x) \in Y_d$ then $d \leq c$.

答:
In general, the answer is No. The simplest counterexample is the following; I think it also satisfies your further assumptions. Take $Y=\{0,1,2,\ldots,\infty\}$, i.e. the one point compactification of $\mathbb N$, which is compact Hausdorff. Let $X_n=\{0,1,2,\ldots,n,\infty\}$, a finite set, which has a natural inclusion into $Y$; and let $X$ be the colimit of the $X_n$'s, so $X$ is just a discrete infinite set, the disjoint union of $\mathbb N$ and $\infty$. Then $f: X\to Y$ is a continuous surjection, but if $T\to Y$ is a surjection from some extremally disconnected $T$, then it cannot lift to $X$ as otherwise $T\to X$ would be a surjection from a compact space and hence $X$ would have to be compact.

In the positive direction, something useful to know is that if your filtered index category is given by $\mathbb R_{>0}$ (essentially equivalently, the cofinal subcategory $\mathbb N_{>0}$), then for any sequential colimit $Y=\mathrm{colim}_n Y_n$ of compact Hausdorff spaces along closed immersions, any map $T\to Y$ from a compact Hausdorff space $T$ to $Y$ factors over some $Y_n$ -- see, for example, Proposition A.15 (i) here. So if in your situation you further ask that all $X_c\to Y_c$ are surjective, then the answer is Yes.


问:
Is there a nice way to present solution to E(S₄) for Poisson process assuming I am correct
Let S4 the the time of the 4th event in a poisson process with rate $\lambda$
My intuition tells me the expected amount of time to 4th event is $(4/λ)$
But that is not satisfactory
$$E(S_4)=E(E(S_4|S_3))=E(E(E(S_4|S_3|S_2)))=E(E(E(E(S_4|S_3|S_2|S_1))))
=E(E(E((1/λ)+S_3|S_2|S_1)))=E(E((1/λ)+(1/λ)+S_2|S_1))= \text{etc}...$$

Is this usual notation or am I missing something?

答:
We can simplify the question if we write $S_n$ as a sum of interarrival times. If we let $T_1, T_2, \cdots $ be the interarrival times of each event, then $T_i\sim \text{Exp}(\lambda)$. Hence \begin{align*}
S_4&=T_1+T_2+T_3+T_4\\
\mathbb E(S_4)&=\mathbb E(T_1+T_2+T_3+T_4)=\mathbb E(T_1)+\mathbb E(T_2)+\mathbb E(T_3)+\mathbb E(T_4)=4\cdot \frac1\lambda =\frac4\lambda
\end{align*}


答:
To solve your actual problem:

I'm currently trying to prove that $|F(b)-F(a)|\leq \int^b_a|f|$

That is true because
$$
|F(b)-F(a)| = \left|\int_a^b f(x) \, dx \right| \leq \int^b_a|f(x)| \, dx
$$
if the fundamental theorem of calculus applies, e.g. if $f$ is continuous and $F' = f$.

Title: is $\int^b_a |f| = |F(b)|-|F(a)|$?


问:
What structure is preserved in a vector space isomorphism?
The definition of an isomorphism is "a structure-preserving mapping between two structures of the same type that can be reversed by an inverse mapping".

In my linear algebra class, an isomorphism is defined as "a bijective linear transformation." I understand some of the structure that is preserved between the spaces, but is there some exact fixed set of things that are preserved by a bijective linear transformation? (or an isomorphism in general).  Further, before we can say that a bijective linear transformation is an isomorphism, do we not need to prove that all of the structure is actually preserved? I ask this because we seem to just assume that a lot of things are preserved between different vector spaces, but never have we proven that these things are preserved between spaces. Below are a few things that I am a little confused about is they need to be proven or follow from the fact that $T$ is a bijective linear transformation.

Let $T: V \to W$ be an isomorphism.  I understand that since $T$ is linear, scalar multiplication and addition will be preserved. Must it be shown that if there is a linear transformation $A$ in $V$ such that for all $x \in V$, there exists a unique transformation $B$ in $W$ such that $A(x) = T^{-1}(B(T(x))$, or that if $S$ is a basis of $W$ of where $S = (S_1, \dots, S_n)$, then $(T^{-1}(S_1), \dots, T^{-1}(S_n))$ is a basis of $V$?

I'm not exactly sure where this preservation of structure stops, and when I am no longer using the fact that we have an isomorphism, and rather sort of just saying things that are not proven true. Maybe I am not entirely sure what constitutes as structure that needs to be preserved in terms of vector spaces.

Any help clearing up this idea of isomorphism is greatly appreciated.

答:
In general, what you really have is some many-sorted FOL (first-order logic) structure. For example, a vector space is nothing more than a two-sorted FOL structure, with one sort $V$ of vectors and the other sort $F$ of scalars, and with four operations $⊕,•,+,·$, and typically also with three constants $z,0,1$ such that:

$⊕ : V^2{→}V$.

$• : F{×}V{→}V$.

$(V,⊕,z)$ is a commutative group.

$(F,+,·,0,1)$ is a field.

$1•x = x$ for every $x{∈}V$.

$(a·b)•x = a•(b•x)$ for every $a,b{∈}F$ and $x{∈}V$.

$(a+b)•x = (a•x)⊕(b•x)$ for every $a,b{∈}F$ and $x{∈}V$.

$a•(x⊕y) = (a•x)⊕(a•y)$ for every $a{∈}F$ and $x,y{∈}V$.

An isomorphism between (many-sorted) FOL structures is just a bijective mapping of all the elements of each sort in one structure to the elements of the corresponding sort in the other structure such that the mapping commutes with the operations. As an example, an isomorphism between vector spaces $S = (V,F,⊕,•,+,·)$ and $S' = (V',F',⊕',•',+',·')$ is a pair $(i,j)$ such that:

$i$ bijects from $V$ to $V'$.

$j$ bijects from $F$ to $F'$.

$i(x⊕y) = i(x)⊕'i(y)$ for every $x,y{∈}V$.

$i(a•x) = j(a)•'i(x)$ for every $a{∈}F$ and $x{∈}V$.

$j(a+b) = j(a)+'j(b)$ for every $a,b{∈}F$.

$j(a·b) = j(a)·'j(b)$ for every $a,b{∈}F$.

That's precisely all there is to isomorphisms in general. Note that for vector spaces in particular, we typically focus on vector spaces over a fixed field $F$, and $F$ is often either $ℝ$ or $ℂ$, in which case we have strictly more than a field structure. In particular, $(ℝ,0,1,+,·)$ is a complete ordered field, and any isomorphism from a complete ordered field to itself is the identity, and $ℂ$ is built from $ℝ$, so when we talk about isomorphism between vector spaces $S$ and $S'$ where $F$ is either $ℝ$ or $ℂ$, we want the bijection $j$ to actually be the identity, and that is why we are left with just the criterion that $i$ commutes with the operations involving vectors:

$i(x⊕y) = i(x)⊕'i(y)$ for every $x,y{∈}V$.

$i(a•x) = a•'i(x)$ for every $a{∈}F$ and $x{∈}V$.

This in turn is equivalent to the isomorphism being a bijective linear transformation!

By the way, if you apply this general viewpoint based on many-sorted FOL structure axiomatization to other algebraic concepts as well, you will see that many concepts that appear different for different kinds of structures are actually identical once expressed from the logical viewpoint. For example, see this post about quotient structures and how to derive the criteria for 'quotienting'.


问:
Getting an improved upper bound for a finite geometric series
The title says it all.

For an infinite geometric series, we know that if $|x| < 1$, then

$$\sum_{i=0}^{\infty}{x^i} = \frac{1}{1 - x}.$$

On the other hand, if $|x| \geq 1$, then the the infinite geometric series

$$\sum_{i=0}^{\infty}{x^i} = 1 + x + x^2 + x^3 + \cdots$$

diverges, so that we consider the case of a finite geometric series.

We know that the equation

$$Y = \sum_{j=0}^{k}{y^j} = 1 + y + \cdots + y^k = \frac{y^{k + 1} - 1}{y - 1}$$

holds.

Traditionally, an upper bound for $Y$ is obtained via

$$Y = \frac{y^{k + 1} - 1}{y - 1} < \frac{y^{k + 1}}{y - 1}$$

so that we get

$$Z = \sum_{j=0}^{k}{y^{-j}} = 1 + \frac{1}{y} + \cdots + \frac{1}{y^k} = \frac{Y}{y^k} < \frac{y}{y - 1}.$$

My question is this:  Is it possible to improve on the upper bound $$\frac{y}{y - 1}$$ for $$Z = \sum_{j=0}^{k}{y^{-j}}?$$  Or is this already the least upper bound for $Z$?

Of course, I would have to qualify that I am interested in upper bounds for $Z$ that do not depend on $k$.

答:
(This is not really an answer to the original question that I had, just some thoughts that recently occurred to me while pondering a closely related problem.)

It is known that, for $q$ prime and $k$ a positive integer,
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{1 + q + \ldots + q^k}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)},$$
where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.

This then gives the (trivial) upper bound
$$I(q^k) < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$

We prove the following theorem in this "answer".

THEOREM
If $q$ is a prime number and $k$ is a positive integer, then
$$I(q^k) < \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$
Moreover, we have
$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) < \frac{q}{q - 1}.$$

PROOF

Let $q$ be a prime number, and let $k$ be a positive integer.

First, we want to show that
$$I(q^k) < \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$
Assume to the contrary that
$$\frac{q^{k+1} - 1}{q^k (q - 1)} = I(q^k) \geq \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$
This implies that
$$q^{2k+2} - 1 = (q^{k+1} - 1)(q^{k+1} + 1) \geq {q^k}\cdot{q}\cdot{q^{k+1}} = q^{2k+2},$$
which is a contradiction.  This proves the first part of the proposition.

Next, we have to prove that
$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) < \frac{q}{q - 1}.$$
Suppose to the contrary that
$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) \geq \frac{q}{q - 1}.$$
It follows that
$$q^{k+1} \geq q^{k+1} + 1,$$
which is a contradiction.  This proves the second part of the proposition, and we are done.


问:
A function in $\Bbb{Q}$ that is not bounded but is bounded in $\Bbb{R}$
I can not find an example of when a function is bounded in $\Bbb{R}$ but not in $\Bbb{Q}$. Can anybody help me?
$f$ is defined as $f:\{x \in \Bbb{Q}| a \leq x \leq b\} \to \Bbb{Q}.$

答: Let $A \subseteq B$ and $f: B \to \mathbb{R}$ a function. If $f$ is unbounded on $A$, then also on $B$. Hence, no such function exists (if I interpret the question correctly, otherwise let me know).


问:
Geometry - length of triangle question
[![enter image description here][1]][1]

Could anyone please help me with this question

I can't seem to solve for the ?.

The answer provided for ? is 1.

Thanks!

答:
There is not enough information to solve this problem, because you gave us just one side of the triangle without any angle, this leaves us with nothing to solve

But for the sake of a solution, let's assume the extra information is hidden in the triangle, maybe similarities, angles, length, etc.......... But when I checked, no lines were parallel, no smaller triangles were similar, some angles were equal and opposite but unknown

$(1)$.Then meaning the small triangle in the core must be an equilateral triangle, true but no helpful maths here, say $x $ is the length we are looking for

$(2)$.Or maybe the angle of the right vertex as is divided into $3$ parts, $\theta_{1}, \theta_{2}, \theta_{3}$ , the sub angles share eachother $\theta_{2} = 2 \theta_{1}$ and $\theta_{3} = 3 \theta_{1}$, so that the angle there is $6 \theta_{1}$, now If we assume the whole shape was an equilateral here, it contradicts
$$ A^2+A^2-2AA \cos(6 \theta ) = 3+x+2$$
$$ 2A^2-2A^2 \cos( 6 \theta ) = 5+x$$
$$ 2A^2( 1- \cos( 6 \theta )) = 5+x$$
$ A \ne 5+x$ , $ A > 5+x$,

If we assume the shape was not an equilateral triangle, and that condition holds, then
$$ A^2+B^2-2AB \cos(6 \theta ) = 3+x+2$$
In this case, remember the sector formula, which we can also use to approximate triangle $\theta \rightarrow 0$, $ s ≈ r \cdot \theta$, then $ x = B \cdot \theta $,
$B$ is the two length the bisects the angle $6 \theta$
$$ B^2+B^2-2BB \cos( \theta ) = x$$
$$2B^2( 1- \cos( \theta)) = x$$
$$ A^2+B^2-2AB \cos(3 \theta ) = 3$$
$$ B^2+C^2-2BC \cos(2 \theta ) = 2$$
$C$ is the last side, now it's easy to see why $x = 1$

(3).Or maybe the whole shape was an equilateral triangle without the angle of that vertex corresponding, check @Brightstart solution

Title: Geometry - length of triangle question


问:
Disjoint attractors have a disjoint basin of attraction
Let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. $A\subseteq\Omega$ is called stable if for every neighborhood $V$ of $A$, there is a neighborhood $U$ of $A$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag1.$$ $A$ is called attractor if $A$ is

forward invariant, i.e. $A=\tau(A)$;

stable; and

there is a neighborhood $U_0$ of $A$ with $U_0\subseteq E(A)$,

where $E(A)$ is the the basin of attraction for $A$, i.e. the set of all $x\in\Omega$ for which the orbit $$\operatorname{orb}x:\mathbb N_0\to\Omega\;,\;\;\;n\mapsto\tau^n(x)$$ is eventually in every neighborhood of $A$.

I've read that if $A_1,A_2\subseteq\Omega$ are different (maybe we need even to assume that they are disjoint) attractors, then $E(A_1)\cap E(A_2)=\emptyset$. How can we show this?

Moreover, I would like to conclude that in that case there cannot be a global attractor $G$, i.e. an attractor with $E(G)=\Omega$. This should immediately follow from the former result, if I'm not missing something.

If the topology on $\Omega$ is induced by a metric $d$, then $$E(A)\subseteq E_0(A):=\left\{x\in\Omega:d(\tau^n(x),A)\xrightarrow{n\to\infty}0\right\}$$ for all $A\subseteq\Omega$ (and "$=$" if $A$ is compact).

So, in that case, the desired claim would clearly follow if even $E_0(A_1)\cap E_0(A_2)=\emptyset$. Now it clearly holds that if $x\in E_0(A_1)\cap E_0(A_2)$, then $$d(A_1,A_2)\le d(\tau^n(x),A_1)+d(\tau^n(x),A_2)\xrightarrow{n\to\infty}0\tag1;$$ hence $d(A_1,A_2)=0$, but that doesn't yield a contradiction to $A_1\ne A_2$ (or $A_1\cap A_2=\emptyset$) in general ...

答:
Your definition of attractor lacks the following property that is usually attributed to attractors: An attractor has no true subset that is an attractor itself. One way to fix this is to require that $\mathop{\text{orb}} x$ in definition basin of attraction is not only eventually in every neighbourhood of $A$ but in every neighbourhood of every point of $A$.

Without that condition, you can have $A_1 \subset A_2$ (e.g., by $A_1$ containing one and $A_2$ containing both fixed points of a bistable system), in which case you certainly don’t have $E(A_1) \cap E(A_2) = \emptyset$.


问:
Limit $\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$
Find the limit

$$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

Let $x=r\sin \theta$ and $y=r\cos \theta$.

$$
\begin{align}
\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2} \\
&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{r^4(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\lim_{r\to0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{0}{(0+2\sin^2\theta)^2}\\
&=0
\end{align}
$$

Does this logic follow? Can I ignore the case where $2\sin^2\theta=0$ meaning it is of the form $\frac{0}{0}$?
How else would I approach this limit.

答:
By AM-GM inequality we have that

$$x^4+y^2+y^2 \geq 3 x^{\frac{4}{3}}y^{\frac{4}{3}}$$

which we can use to say

$$ \frac{16|x^3y^3|}{(x^4+2y^2)^2}\leq \frac{16|x^3y^3|}{9x^{\frac{8}{3}}y^{\frac{8}{3}}} \leq 2\sqrt[3]{|xy|} $$

Thus the limit exists and equals $0$ by squeeze theorem.


问:
What is a proof of Euler's formula $e^{ix}=\cos(x)+i \sin(x)$ that doesn't use differentiation or De Moivre's theorem, if it exists at all?
The Taylor series expansions of $e^x$, $\sin(x)$ and $\cos(x)$ are the standard way to prove it, but does there exist an elementary way that doesn't use limits, calculus or De Moivre's theorem?

答:
If you take the trigonometric addition formulas for granted,

$$(\cos(n)+i\sin(n))(\cos(1)+i\sin(1))
\\=\cos(n)\cos(1)-\sin(n)\sin(1)+i(\cos(n)\sin(1)+\sin(n)\cos(1))
\\=\cos(n+1)+i\sin(n+1)$$

and by induction,

$$\cos(n)+i\sin(n)=(\cos(1)+i\sin(1))^n.$$

From this we have

$$\left(\cos\left(\frac nm\right)+i\sin\left(\frac nm\right)\right)^m=\cos(n)+i\sin(n)=(\cos(1)+i\sin(1))^n$$ and

$$\cos\left(\frac nm\right)+i\sin\left(\frac nm\right)=(\cos(1)+i\sin(1))^{n/m}$$ where we extract some to some $m^{\text{th}}$ root. In other words, with $q$ a rational number,

$$\cos(q)+i\sin(q)=(\cos(1)+i\sin(1))^q.$$

If we denote the complex number $\cos(1)+i\sin(1)$ as $e^i$, we have

$$\cos(q)+i\sin(q)=(e^i)^q.$$

This is by no means a proof (and if you wanted to generalize to the reals, you would need limits anyway), but it shows the connection between the addition formulas, which can be proven by geometry, and the powers of a complex.

Disclaimer: I didn't say that $(e^i)^q=e^{iq}$.


问:
Left invariant k-form
I am facing a question, have just started to read a chapter from Smoth Manifolds by Lee and having trouble to do this excercise

Let w be a left invariant k-form on a Lie group G. Show
that dw is a left invariant (k+1)-form.

I have re-read the definations but I can't proceed anywhere,
I am new members in this site,I don't know many rules, but if I violate any, please forgive me.

Gracias señor!

答:
If $\alpha$ is an invariant form and $g\in G$, denote $L_g$ the left translation,
$L_g^*\alpha=\alpha$ implies that $d(L_g^*\alpha)=L_g^*d\alpha=d\alpha$ since the exterior derivative commutes with pullbacks.

Proving that the pullback map commutes with the exterior derivative

Title: Left invariant k-form


问:
Basis of $V=\{a\cdot(1,2,3)^T\}$
I've a vector space

$V=\left\{a\left(\begin{array}
 {l}1 \\ 2 \\ 3\end{array}\right)\right\}$ $a$ is any real number.

Can I choose it's basis as $\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$.

I think that it spans the space but then I can also have it's basis the set of these three vectors
$\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$
$\left(\begin{array}{l}0 \\ 1\\ 0\end{array}\right)$$\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)$.

Something is wrong with my logic. Can anyone please point out my error.

Thank you

答:
Yes, the set $A$ containing only one element, $\begin{pmatrix}1\\2\\3\end{pmatrix}$, is indeed a basis for $V$. It is very easy to show that the set satisfies both conditions that are in the definition of basis:

It is true that the span of $A$ is equal to $V$, i.e. $\mathrm{span}(A)=V$.

It is true that the set $A$ is linearly independent.

On the other hand, the set $B=\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}$, containing the three vectors you list is not a basis for $V$, because the first condition fails. The span of $B$ includes the vector $\begin{pmatrix}1\\0\\0\end{pmatrix}$ which is not an element of $V$. This means that the span is not equal to $V$, so $B$ cannot be a basis for $V$.

Title: Basis of $V=\{a\cdot(1,2,3)^T\}$


问:
Roots of factorization with expressions for constants
When faced with the factorization $$(\alpha - 1 -\lambda)(\mu - 1 -\lambda)=0$$ the roots of $\lambda$ can simply be said to be $\alpha - 1$ and $\mu - 1$. Why is that answer different to the following working?
$$\begin{align*} \alpha\mu - \alpha - \alpha\lambda - \mu +1 + 2\lambda - \lambda\mu + \lambda^2 &=0 \\
\lambda^2 + (2-\alpha - \mu) \lambda + \alpha\mu - \alpha - \mu + 1 &=0
\end{align*}$$
$$\begin{align*}
\lambda &= \frac{\alpha + \mu - 2 \pm\sqrt{(2-\alpha-\mu)^2 - 4(\alpha\mu - \alpha -\mu + 1)}}2 \\
 &= \frac{\alpha + \mu - 2 \pm\sqrt{4 - 4\alpha -4\mu + \alpha^2 + 2\alpha\mu + \mu^2 - 4(\alpha\mu - \alpha -\mu + 1)}}2 \\
  &= \frac{\alpha + \mu - 2 \pm\sqrt{\alpha^2 - 2\alpha\mu + \mu^2}}2 \\
  &= \frac{\alpha + \mu - 2 \pm\sqrt{(\alpha - \mu)^2}}2\\
  &= \frac{-\alpha + 3\mu }2\qquad\text{or}\qquad \frac {3\alpha - \mu}2
\end{align*}$$

答:
$$\lambda_{1,2}=\frac{\alpha+\mu-2\pm\sqrt{(\alpha-\mu)^2}}{2}=\frac{\alpha+\mu-2\pm(\alpha-\mu)}{2}\\ \implies\lambda_1=\frac{2\alpha-2}2=\alpha-1\\
\implies\lambda_2=\frac{2\mu-2}2=\mu-1$$
What seems to be the problem?  I cannot be sure how you ended up with the solutions that you did.


问:
What is the maximum number of vertices of degree one that a binary tree with 10 vertices can have?
Full question as the title was a maximum of 150 characters.
A binary tree is a connected graph with no cycles, where each vertex has a degree less than or equal to 3. What is the maximum number of vertices of degree one that a binary tree with 10 vertices can have?

I am really struggling to understand this question and have not seen questions that are similar. I was hoping someone could push me into the right direction. As I understand the problem, we can only use 10 vertices. So would the maximum amount of vertices not be 10? I tried drawing it out and don't see how it can be any larger.

答:
For a binary tree (each node has at most two children) with $n$ vertices,

Let $f(n)$ be the maximum possible number of vertices of degree $1$.

Let $g(n)$ be the maximum possible number of leaves.$\\[4pt]$

Our goal is to compute $f(10)$.

The values of $f$ can be expressed in terms of the values of $g$ by
$$
f(n)=
\begin{cases}
0&\qquad\;\;\,\text{if}\;\,n\le 1\\[4pt]
\max\,\bigl\{g(n),1+g(n-1)\bigr\}&\qquad\;\;\,\text{if}\;\,n > 1\\[4pt]
\end{cases}
$$
and $g$ can be computed recursively by
$$
\;\;\;\;\;\,
g(n)=
\begin{cases}
0&\;\text{if}\;\,n=0\\[4pt]
1&\;\text{if}\;\,n=1\\[4pt]
\max\,\bigl\{g(a)+g(b){\,{\large{\mid}}\,}(a,b)\in X_n\bigr\}&\;\text{if}\;\,n > 1\;\;\;\;\;\\[4pt]
\end{cases}
$$
where for $n > 1$, we define $X_n$ as the set of all pairs $(a,b)$ of integers such that $0\le a\le b$ and $a+b=n-1$.

Computing some early values of $g$, it becomes evident that
$$
g(n)=\left\lceil\frac{n}{2}\right\rceil
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;
$$
for all $n$, which can be proved by strong induction on $n$.

It follows that for all $n > 1$ we have
$$
f(n)=\left\lceil\frac{n+1}{2}\right\rceil
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,
$$
In particular we have $f(10)=6$.


答:
When you see concurrent lines, Ceva's Theorem and Menelaus's Theorem are often the intended way.

Consider the grey triangle, applying Ceva's Theorem yields $$\frac{3}{x}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$

Again, consider the grey triangle with the black transversal line, applying Menelaus's Theorem yields $$\frac{5+x}{2}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$

Equating both equations gives $x=1$.

PS: I'm not sure why using other lines and triangles (for instance, the line between the line segments with lengths $3$ and $x$) doesn't work. If anyone would like to expand on this answer, please do. Thanks in advance!

Title: Geometry - length of triangle question

Body:
[![enter image description here][1]][1]

Could anyone please help me with this question

I can't seem to solve for the ?.

The answer provided for ? is 1.

Thanks!


问:
Set of Characters for $C([0,1])$
Let $\Omega(A)$ be the set of (characters )non-zero homomorphic functions from a unital Banach algebra $A$ to $\mathbb{C}$.

Show $\Omega(C[0,1])$ are the pointwise evaluations.

I'm not really sure why this is true by considering $\phi \in \Omega(C[0,1])$ and taking $f\in C[0,1]$.

Thank you in advance!

答:
Let $\phi:C([0,1])\to \mathbb{C}$ be a non-zero character. Note that $\phi$ is an algebra morphism and thus $\ker(\phi)$ is an ideal. By the first isomorphism theorem, $C([0,1])/\ker(\phi)\cong \mathbb{C}$ and thus $\ker(\phi)$ is a maximal ideal of $C([0,1])$.

For each $c\in \mathbb{C}$, let $I_c=\{f\in C([0,1])\mid f(c)=0\}$. You can easily verify that each $I_c$ is a maximal ideal. In this short note, the converse is proved as well, i.e. every maximal ideal is of this form.

Thus, there exists a $c\in \mathbb{C}$ such that $\ker(\phi)=I_c
$. Note that $C([0,1])\cong \ker(\phi)\oplus \mathbb{C}1$ where $1$ denotes the constant function with value $1$, indeed, every $f\in C([0,1])$ can be written as $f=g+h$ where $h=f(c)1$ and $g=f-h$ and clearly $g\in \ker(\phi)$. Thus, $\phi(f)=\phi(g+h)=\phi(h)=\phi(f(c)1)=f(c)\phi(1)=f(c)$.

Title: Set of Characters for $C([0,1])$


问:
How do you construct a bivariate normal pdf (or 2D Gaussian) using measure theory?
Suppose I have an experiment $\mathcal{E}$ and a corresponding probability space $(\Omega, \mathcal{F}, \mathbb{P})$ where:

$\Omega$ is the set of possible outcomes

$\mathcal{F}$ is a sigma-algebra on it

$\mathbb{P}: \mathcal{F}\to [0,1]$ is a probability measure on $(\Omega, \mathcal{F})$.

Now suppose there are two additional measurable spaces $(\mathsf{X}, \mathcal{X}) = (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ and $(\mathsf{Y}, \mathcal{Y})=(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ and we define two random variables
$$
X: \Omega \to \mathsf{X} \\
Y: \Omega \to \mathsf{Y}
$$
with probability distributions defined as push-forward measures
$$
\pi_X := X_*\mathbb{P}: \mathcal{X} \to [0, 1] \\
\pi_Y := Y_*\mathbb{P}: \mathcal{Y} \to [0, 1]
$$
which we assume to satisfy $\pi_X \ll \lambda$ and $\pi_Y \ll \lambda$ where $\lambda$ is the $1$-dimensional Lebesgue measure. Then their density functions are the Random-Nykodym derivatives
$$
f_X := \frac{d \pi_X}{d\lambda}:X \to [0, +\infty) \\
f_Y := \frac{d \pi_Y}{d\lambda}:Y \to [0, +\infty).
$$
Finally, we assume that their densities have the form of two standard normal distributions
$$
f_X(x) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{x^2}{2}\right) \\
f_Y(y) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{y^2}{2}\right)
$$
so that their CDFs become
$$
\begin{align*}
F_X(x)
&:= \pi_X((-\infty, x]) = \int_{(-\infty, x]} d\pi_X = \int_{(-\infty, x]} f_X d\lambda = \frac{1}{2\pi} \int_{(-\infty, x]} \exp\left(- \frac{t^2}{2}\right) dt \\
F_Y(y)
&:= \pi_Y((-\infty, y]) = \int_{(-\infty, y]} d\pi_Y = \int_{(-\infty, y]} f_Y d\lambda = \frac{1}{2\pi} \int_{(-\infty, y]} \exp\left(- \frac{t^2}{2}\right) dt
\end{align*}
$$
But then I am not sure about how to construct a general bivariate normal distribution with arbitrary mean and covariance matrix.

Following the Answer by Woolier

We collect together the two random variables into a $2D$ random vector
$$
(X, Y): \Omega \times \Omega \to \mathsf{X}\times \mathsf{Y}
$$
and define its probability distribution as
$$
\pi_{X\times Y} := \pi_X \times \pi_Y: \mathcal{X} \otimes \mathcal{Y} \to [0, 1]
$$
The probability density function is then given by the product
$$
f_{X\times Y} = \frac{d \pi_X}{d \lambda} \times \frac{d \pi_Y}{d \lambda}: X\times Y \to [0, +\infty)
$$
giving the following (we define $\boldsymbol{x} = (x, y)^\top \in \mathbb{R}^{2\times 1}$)
$$
\begin{align*}
f_{X\times Y}(x, y)
&= \left[\frac{1}{\sqrt{2\pi}} \exp\left(-\frac{x^2}{2}\right)\right]\left[\frac{1}{\sqrt{2\pi}} \exp\left(-\frac{y^2}{2}\right)\right] \\
&= \frac{1}{2\pi}\exp\left(-\frac{(x^2+y^2)}{2}\right) \\
&= \frac{1}{2\pi}\exp\left(-\frac{1}{2} (x, y) I \begin{pmatrix}
x \\
y
\end{pmatrix}
\right) \\
&= \frac{1}{2\pi} \exp\left(-\frac{1}{2} \boldsymbol{x}^\top I \boldsymbol{x}\right)
\end{align*}
$$
Then can then define a new set of random variables
$$
(X', Y') : \Omega\times \Omega \to \mathsf{X}\times\mathsf{Y}
$$
defined as follows. Let $\omega_1, \omega_2 \in \Omega$. Then if we let $L$ be the cholesky decomposition (lower-triangular) of a positive-definite matrix $\Sigma = LL^\top$, the values $X(\omega)$ and $Y(\omega)$ are mapped to
$$
\begin{pmatrix}
    X'(\omega) \\
    Y'(\omega)
\end{pmatrix} =
\begin{pmatrix}
l_{11} & l_{12} \\
l_{21} & l_{22}
\end{pmatrix}
\begin{pmatrix}
X(\omega) \\
Y(\omega)
\end{pmatrix}
+
\begin{pmatrix}
\mu_1 \\
\mu_2
\end{pmatrix} =:
L\boldsymbol{x} + \boldsymbol{\mu}
$$

and therefore we say that $(X', Y')$ follow a $\mathcal{N}(L\boldsymbol{0} + \boldsymbol{\mu}, LIL^\top) = \mathcal{N}(\boldsymbol{\mu}, \Sigma)$

答:
Hint:

If $(X,Y)\sim \mathcal{N}(\mu,\Sigma),$ then
$$
A\cdot (X,Y)^T+\mathbf{b}\sim \mathcal{N}(A\mu+\mathbf{b},A\Sigma A^T)
$$
For any matrix $A$ and vector $\mathbf{b}$ of suitable dimensions.


问:
Is the difference between stable nonlinear systems Lyapunov stable?
If there are two nonlinear systems with stable equilibria $x_1 = x_2 = 0$
$$\dot x_1 = f(x_1, u) \qquad \dot x_2 = g(x_2, u)$$
with identical inputs $u$, is the difference between the system states $x_1, x_2 \in \mathbb{R}^n$
$$e = x_1 - x_2$$
stable?

Trivial case

If the systems are LTI with identical system and input matrices $A$ and $B$
$$f(x_1 = x, u) = g(x_2 = x, u) = Ax + Bu,$$
the system dynamics is
$$\dot e = Ae.$$
A Lyapunov function
$$V(e) = e^T P e$$
results in
$$\dot V(e) = \dot e P e^T + e^T P \dot e = e^T (A^T P + P A) e.$$
If $A$ is Hurwitz, then $A^T P + P A = -Q$ with a positive definite real symmetric $P$ and a positive definite $Q$. As $\dot V(e) = -e^TQe < 0\ \forall x \neq 0$, the difference between the two asymptotically stable LTI systems is asymptotically stable.

My Question

Can this be generalized to arbitrary nonlinear globally/locally/asymptotically/... stable systems,

where $f = g$? (1)

where $f \neq g$? (2)

Thought experiment: An equilibrium is stable if for each $\epsilon > 0$, there is a $\delta$ such that with $|| x || < \delta$ at $t = t_0$ the state remains within $|| x || < \epsilon$. Wouldn't that imply that the vector $e$ connecting two $x_1, x_2$ remains within a hypersphere $\mathcal{B}$ that includes both $|| x_1 || < \epsilon_1$ and $|| x_2 || < \epsilon_2$? Then for $|| e || < \min(\delta_1, \delta_2)$ at $t = t_0$, it would hold $|| e || < 2r_\mathcal{B}$ with $r_\mathcal{B}$ radius of $\mathcal{B}$?

At the same time: Starting from
$$V(e) = \frac{1}{2} e^T e,$$
I quickly get stuck at
$$\dot V(e) = e^T \dot e = e^T \left(f\left(x_1, u\right) - g\left(x_2, u\right)\right) =\ ...?$$

Update

Arastas answer has a counterexample for $f \neq g$ (loosely quoting):

$\dot x_1 = u$ and $\dot x_2 = -u$ leads to $\dot e = 2u$. With $u=1$, we get $e(t) \rightarrow \infty$ for $t \rightarrow \infty$.

In this SISO example, the derivatives of $f$ and $g$ w.r.t. $u$ differ:
$$\frac{\partial}{\partial u} f(x_1, u) = 1 \neq -1 = \frac{\partial}{\partial u} g(x_2, u).$$

Speaking more generally, the (transposed) gradient matrices of $f$ and $g$ w.r.t. $u$ differ:
$$(\nabla_u f)^T = \frac{\partial}{\partial u} f(x_1, u) \neq \frac{\partial}{\partial u} g(x_2, u) = (\nabla_u g)^T.$$

What can we say about the stability of $e$ for arbitrary $f, g$, but allowing for restrictions on the gradients $\nabla_u f$, $\nabla_u g$? (3)

If the restriction were made $\nabla_u f = \nabla_u g$, (3) would capture case (1).

答:
Consider $f\ne g$. Let $\dot{x}_1=u$ and $\dot{x}_2=-u$ (Lyapunov stable). Then $\dot{e}=2u$ and for $u=1$ we have $e(t)\to \infty$ -- unstable. If you want your system to be assymptotically stable, then consider $\dot{x}_1=-x_1+u$ and $\dot{x}_2=-x_2-u$. Then for $u=t$ we have $e(t)\to \infty$ -- unstable.

Moreover, you write

the difference between the asymptotically stable LTI systems is (unsurprisingly) asymptotically stable.

Your example is not correct since the ass. stability of an LTI system is not equivalent to $A$ being negative definite. $A$ must be Hurwitz.


问:
Truth of a sentence in predicate logic
I'm working through the exercises in this Logic problem set and I'm struggling to understand ex. 7.5:

Determine whether the following sentence is logically true in predicate logic:

There is someone such that, if he or she is asleep, everyone is asleep.

The solutions section suggests this is True and gives a formal 'Natural Deduction' proof that I don't really understand (it probably doesn't help that I don't have a copy of the book the solutions keep referring to).

Is there another way to approach this?

答:
[...] It seems entirely trivial by comparison to the long chain of formal manipulations given as the solution, so I assumed it had to be faulty. [...]

The problem with your assumption is that formal proofs from scratch need to prove every single bit of the reasoning involved. In particular, how exactly do you justify that either "there is someone who is not asleep" or "there is not anyone who is not asleep"? Depending on your specific chosen formal system, this may be an axiom called LEM (law of excluded middle) or require a few lines of proof. Similarly, to get from the latter to "everyone is asleep" again may be allowed in one inference step or require some more lines of proof.

Here is a completely formal proof using a Fitch-style natural deduction system with LEM. The key to get an intuitive proof is to use LEM, so if the system does not support LEM then you simply prove the instance of LEM that you need! Shorter proofs in a formal system may be less intuitive.

```<code>If ∃x∈People:  [✰]
  ∃y∈People ( ¬Asleep(y) ) ∨ ¬∃y∈People ( ¬Asleep(y) ).  [LEM]
  If ∃y∈People ( ¬Asleep(y) ):
    Let c∈People such that ¬Asleep(c).
    If Asleep(c):
      Contradiction.
      ∀x∈People ( Asleep(x) ).
    Asleep(c) ⇒ ∀x∈People ( Asleep(x) ).
    ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
  If ¬∃y∈People ( ¬Asleep(y) ):
    Let d∈People.  [from ✰]
    If Asleep(d):
      Given any x∈People:
        If ¬Asleep(x):
          ∃y∈People ( ¬Asleep(y) ).
          Contradiction.
        Asleep(x).
      ∀x∈People ( Asleep(x) ).
    Asleep(d) ⇒ ∀x∈People ( Asleep(x) ).
    ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
  ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
</code>```

Important note: The statement in your question is actually not a logical tautology. As you should see from the above proof, it depends crucially on the assumption that there exists some person, otherwise it is trivially false.

~ ~ ~ ~ ~ ~ ~

Also, contrary to popular belief, the apparent paradox here is not really due to any mismatch between the English "if" and logical "if"! Rather, it is due to incorrect translation of the English sentence, due to the use of the generic present tense! Notice that the following sentence is actually true when interpreted according to standard English:

There is someone such that, if he or she is asleep at midnight on 16 Mar 2021, then everyone is asleep at midnight on 16 Mar 2021.

The problem with the original sentence was that because it used the present tense without a context specifying a single time, the time ended up being quantified outside the "if", and so it ended up being interpreted as:

There is a person A such that, at every time t, if A is asleep at time t, then everyone is asleep at time t.

Such implicit quantifiers do not go past explicit quantifiers, which is why it did not go all the way to the outside. Note that the logical tautology corresponds precisely to that:

At every time t, there is a person A such that, if A is asleep at time t, then everyone is asleep at time t.

I hope this partly linguistic explanation satisfies your inquiry into the apparent paradox, and yes as you guessed it is a quantifier swap, though not quite involving sleep-states but involving time. On the other hand, be aware that in general it is difficult to set down any rules to translate English into logical form. After all, there are still many people who insist on saying "All that glitters is not gold." instead of "Not all that glitters is gold."...

Title: Truth of a sentence in predicate logic

Body:
[...] It seems entirely trivial by comparison to the long chain of formal manipulations given as the solution, so I assumed it had to be faulty. [...]

The problem with your assumption is that formal proofs from scratch need to prove every single bit of the reasoning involved. In particular, how exactly do you justify that either "there is someone who is not asleep" or "there is not anyone who is not asleep"? Depending on your specific chosen formal system, this may be an axiom called LEM (law of excluded middle) or require a few lines of proof. Similarly, to get from the latter to "everyone is asleep" again may be allowed in one inference step or require some more lines of proof.

Here is a completely formal proof using a Fitch-style natural deduction system with LEM. The key to get an intuitive proof is to use LEM, so if the system does not support LEM then you simply prove the instance of LEM that you need! Shorter proofs in a formal system may be less intuitive.

If ∃x∈People:  [✰]
  ∃y∈People ( ¬Asleep(y) ) ∨ ¬∃y∈People ( ¬Asleep(y) ).  [LEM]
  If ∃y∈People ( ¬Asleep(y) ):
    Let c∈People such that ¬Asleep(c).
    If Asleep(c):
      Contradiction.
      ∀x∈People ( Asleep(x) ).
    Asleep(c) ⇒ ∀x∈People ( Asleep(x) ).
    ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
  If ¬∃y∈People ( ¬Asleep(y) ):
    Let d∈People.  [from ✰]
    If Asleep(d):
      Given any x∈People:
        If ¬Asleep(x):
          ∃y∈People ( ¬Asleep(y) ).
          Contradiction.
        Asleep(x).
      ∀x∈People ( Asleep(x) ).
    Asleep(d) ⇒ ∀x∈People ( Asleep(x) ).
    ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).
  ∃a∈People ( Asleep(a) ⇒ ∀x∈People ( Asleep(x) ) ).

Important note: The statement in your question is actually not a logical tautology. As you should see from the above proof, it depends crucially on the assumption that there exists some person, otherwise it is trivially false.

~ ~ ~ ~ ~ ~ ~

Also, contrary to popular belief, the apparent paradox here is not really due to any mismatch between the English "if" and logical "if"! Rather, it is due to incorrect translation of the English sentence, due to the use of the generic present tense! Notice that the following sentence is actually true when interpreted according to standard English:

There is someone such that, if he or she is asleep at midnight on 16 Mar 2021, then everyone is asleep at midnight on 16 Mar 2021.

The problem with the original sentence was that because it used the present tense without a context specifying a single time, the time ended up being quantified outside the "if", and so it ended up being interpreted as:

There is a person A such that, at every time t, if A is asleep at time t, then everyone is asleep at time t.

Such implicit quantifiers do not go past explicit quantifiers, which is why it did not go all the way to the outside. Note that the logical tautology corresponds precisely to that:

At every time t, there is a person A such that, if A is asleep at time t, then everyone is asleep at time t.

I hope this partly linguistic explanation satisfies your inquiry into the apparent paradox, and yes as you guessed it is a quantifier swap, though not quite involving sleep-states but involving time. On the other hand, be aware that in general it is difficult to set down any rules to translate English into logical form. After all, there are still many people who insist on saying "All that glitters is not gold." instead of "Not all that glitters is gold."...


问:
Is there any way to circumference of a circle with radius but without pi?
Please let me know if anybody knows how to calculate the circumference of a circle with radius but without pi?</p>

答:
It is a secret of euclidean geometry (not valid, e.g.,  in spherical geometry) that all figures can be linearly scaled by arbitrary real factors $\lambda>0$, whereby the lengths of all segments and curves are multiplied by $\lambda$ and the areas of nice domains are multiplied by $\lambda^2$. From this it follows that for circles there has to be a formula
$${\rm circumference}=C\cdot {\rm diameter}\ ,$$
whereby $C$ is a "world constant". This is known to humanity since thousands of years. Later the constant of value slightly larger than $3$ has been denoted by $\pi$, and only in the last years of the $19^{\rm th}$ century it has been proven that this $\pi$ is very irrational, and not even expressible in terms of square roots or similar. But $\pi$ is an ordinary real number and has its "infinite precision" like every such number, e.g. $7$.


问:
Upper bound on arc length by chord length
Consider an arc and the corresponding chord of a circle, as in https://mathworld.wolfram.com/CircularSegment.html. Then is it possible to bound the arc length by the length of the corresponding chord? Specifically, can we have the following inequality,
$$
s\leq Ca
$$
for some universal constant $C>0$.

答:
In the unit circle, we have $a = 2\sin \left({\frac {s}{2}}\right)$, and so
$$
\frac s a = \frac{s}{2\sin(s/2)} \le \frac \pi 2
$$
for $s \in [0,\pi]$.


问:
A surjective linear mapping $L$ from $X$ to $Y$ is open iff there is a constant $c$ for each $y=Lx$ satisfying $\|x\| \leq c \|y\|$
So I have to show both directions. For the if direction, I began as below.

Suppose such a $c$ exists for each pair of $x \in X$ and its corresponding $y = Lx \in Y$. I have to show that the image of an open set containing $x$ by $L$ is an open set in $Y$. Let $x \in U$ where $U$ is an open set in $X$. Also, since $Y$ is subjective, there is a $y \in Y$ such that $y=Lx$. I have to find an open ball in $Y$ containing $y$. Such a ball would be of the form $\{y' \in Y, \|y'-y\|<r \}$ for some $r>0$. Plus, again due to subjectivity, every $y'=Lx'$ for some $x'\in X$. Thus, $$r>\|y'-y\|=\|Lx'-Lx\| = \|L(x'-x)\| > \frac{\|x-x'\|}{c}.$$ Which gives me $\|x-x'\|< rc$. Am I on the right track for this part? How can I conclude from this that $L$ maps each open ball of radius $rc$ in $X$ into another open ball of radius $r$ in $Y$? And is it so?

As for the other direction, I guess that I should somehow use the fact that the image of the open unit ball under any open mapping contains another open ball. That is $B_Y(0,\rho) \subset L\Big( B_X(0,1)  \Big)$ for some $\rho >0$.

I really appreciate any help on both sides (the opposite direction seems more challenging!) Thanks.
Edit: $X$ and $Y$ are not necessarily Banach spaces.

答:
For the other direction, if $L$ is an open mapping, then we can find     $ \delta>0 $ such that  $ L(B_X) \supset \delta B_Y $, where $B_X$ and $B_Y$ are the open unit balls of $X$ and $Y$ resp.

Let  $ y \in Y  $.  If $y=0$, we have nothing to prove. Therefore, we may assume that  $ y \neq 0 $. Note that  $ \frac{\delta }{2} \frac{y}{\ ||{y}||} \in L(B_x) $ and so there exists  $ z \in B_X $ such that  $ \frac{\delta }{2} \frac{y}{||{y}||}=L(z)$. In other words, $  y=L(\frac{\delta }{2} ||y|| z  ) $. If we let  $ x =\frac{\delta }{2} ||y||  z     $, then  $ ||x|| \leq \frac{\delta}{2} ||y||$.  Take $c=\delta/2$.

Edit: For the original direction, notice that it suffices to show that there exists $\delta >0$ such that $ L(Β_X) \supset \delta B_Y $ (which follows immediately from the hypothesis). Indeed, suppose such a $\delta$ exists. Let $U \subset X$ be open and let $y_0 \in L(U)$. Then, $y_0 = Lx_0 $ for some $x_0 \in U$. Since $U$ is open, there exists  $r>0$ such that $B_X(x_0,r) \subset U$, or equivalently, $x_0 +B_Χ(0,r) \subset U$. Hence,
$$ y_0 +L(B_X(0,r)) \subset L(U).$$
Since   $ L(B_X(0,r)) \supset  \delta  Β_Y (0,r) $ we have that
$$ B_Y(y_0,\delta r) \subset L(U),$$
that is, $L(U)$ is open in $Y$.


问:
What are the steps to find the oblique asymptote of this function?
The oblique asymptote of $\frac{x^2}{2x - 5}$ is $\frac{x}{2}+\frac{5}{4}$. What steps do I need to take in order to get from the original function to the oblique asymptote?

答:
You can do polynomial division to get:
$$\frac{x^2}{2x-5} = \frac{x}{2}+\frac{5}{4}+\frac{25}{8x-20}$$ Now the last part will “disappear” as $x \to \infty$ and so you will be left with the oblique asymptote.


问:
Solutions to Laplace's Equation in the plane in polar coordinates
I'm trying to solve Laplace's equation in the disc $r\leq a$ using separation of variables (so relatively simple stuff, compared to what's often on here), and I've proceeded like so thus far:

Let $T(r,\theta)$ be a solution to $\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2}=0$.
Substituting $T(r,\theta)=F(r)G(\theta)$ into this equation, and equating both functions to a constant, say, $\lambda$, gives:

(1) $r^2F''(r)+rF'(r)=\lambda F(r)$

(2) $G''(\theta)=-\lambda G(\theta)$

First considering (2), we know that $G(\theta)$ must be $2\pi$ periodic, which is only possible if either $\lambda=n^2$ for $n\in\mathbb{N}$, in which case $G(\theta)=A\sin(n\theta)+B\cos(n\theta)$, or $\lambda=0$, giving $G(\theta)=c$ for some constant $c$.

Now looking back at (1), $\lambda=0$ gives $F(r)=A + B\log{r}$, but what about when $\lambda=n^2$?

I have in my online notes that the solution is a linear combination of $r^n$ and $r^{-n}$, and I can see that these fit, but apart from getting that answer by a lucky guess, I don't see how one would come about it. The notes say that this equation is of 'Euler's type', but I'm not too sure what that means either, and if I search for it on the internet I just come across a lot of other unrelated equations that Euler had something to do with.

Thanks in advance, and this is my first question, so sorry if I've done anything wrong!

答: Too late to help I imagine but the $\lambda=n^2$ eigenvalues arise because we want a single-valued solution, so $G(\theta+2\pi)=G(\theta)$. Having found the possible values for $\lambda$, the equation for $F$ is equidimensional (each time we differentiate we multiply by $r$, and $r \frac{d}{dr} r^k=kr^k$), so we try $F=r^k$ and find the values of $k$ that work.</p>


问:
Connectedness for subsets of $\mathbb{R}^2$
I have some confusion on my understanding of connectedness of subsets of $\mathbb{R}^2$. For example I was given a practice problem to explain why the comb space $S=[0,1] \times \{0\} \cup \{(0,1)\} \cup (\cup_{n=1}^{\infty}\{\frac{1}{n}\} \times [0,1])$ is connected, and the answer says "Since any ball around the point $(0,1)$ contains some points on the spikes(near $(0,1)$ , we cannot find an open set separating $(0,1)$ from the rest of $S$".

Now I am seeing a practice problem in topology by Franzosa which says the union of the two subsets $y=0$ and $y=e^x$ is not connected. But then I think about the explanation given for the comb space, and it seems I can always find an open ball of any given radius around some point arbitrarily far down the negative $x-$axis that will intersect $y=e^x$.I do not need rigorous answers for this, I just need a conceptual understanding. In what way do these situations differ, so that the comb space is connected, but not the latter?

Also, what would be the best way(without proof) to determine subsets of $\mathbb{R}^2$ are connected or not? Path connected?

答:
"Since any ball around the point $(0,1)$ contains some points on the spikes near $(0,1)$, we cannot find an open set separating $(0,1)$ from the rest of $S$."

If that should be the only argument to prove the connectivity of $S$, then it is sheer nonsense. To see this, consider
$$S'  = \{(0,1)\} \cup (\bigcup_{n=1}^{\infty}\{\frac{1}{n}\} \times [0,1]) .$$
This space is certainly not connected (each of the sets $\{\frac{1}{n}\} \times [0,1])$ is open and closed in $S'$), but we can correctly argue

"Since any ball around the point $(0,1)$ contains some points on the spikes near $(0,1)$, we cannot find an open set separating $(0,1)$ from the rest of $S'$."

To give a complete proof, observe that the space
$$T  = [0,1] \times \{0\}  \cup (\bigcup_{n=1}^{\infty}\{\frac{1}{n}\} \times [0,1])$$
is path connected, hence connected. Now assume that $S = U \cup V$ with disjoint nonempty open subsets $U, V$ of $S$. W.l.o.g. $(1,0) \in U$. But now $(1,0) \in B_r(1,0) \cap S \subset U$ for some $r > 0$. Your argument says that $B_r(1,0)$ must contain a point of $T$, thus $U \cap T  \ne \emptyset$. But then $T =  U' \cup V$ with $U' = U \cap T$. The sets $U',V$ are disjoint nonempty open subsets of $T$, a contradiction since $T$ is connected.


问:
Is it possible to show that no multiset may exist to satisfy a given constraint without knowing the multiplicity of the set's elements?
I'd like to preface my question by stating that I have no education in mathematics beyond A levels (UK), which was 7 years ago, so I apologise if I'm asking a silly question or if I misuse any terms.

In short, is there a way to find a multiset, that satisfies a given value constraint, without knowing the multiplicity of each element within the set, and without knowing the cardinality of the multiset? In my scenario, you would know the maximum possible multiplicity of each element, and you would have a maximum and minimum cardinality for the multiset.

I'm not even sure if that question makes sense.

I'm wondering if it is possible to arrive at X with any combination of elements within a set, where each element can be repeated, but the number of times any particular element can be repeated varies from element to element.

To give some context, imagine you have in front of you a list of daily total sales from a restaurant and you want to determine if it is possible for a sum of items on the menu to equal the daily total figure, assuming no discounts or human error. The restaurant provides no additional information.

Let's assume the restaurant declared £500 total sales on Monday, and we know the menu has 10 possible items. We can determine how many of any particular item could have been sold if we divide £500 by the item's cost (and this would assume no sales from any of the other items), and thus give us a maximum possible multiplicity for each element. We can calculate the minimum cardinality (I'm not even sure that's an actual concept) of the multiset by assuming the multiset contains only the most expensive item sold, and thus use its maximum multiplicity; the reverse can be done for the maximum cardinality of the set, by assuming only the cheapest item was sold, and thus use its maximum multiplicity.

With all of that, is it possible to determine if a multiset exists to satisfy the value of £500? Or rather, is it possible to show that no multiset exists to satisfy the value of £500? Because if the target value of £500 can be divided without remainders by any of the items, then of course at least one multiset exists.

Hope that's clear enough to understand.

Edit: Just editing to see if the questions will show in 'Latest'

答:
It is certainly possible: as you have observed, there is a maximum number of sales per item, which leaves only finitely many options. If there are $K$ items, and the cheapest item could have been sold $M$ times, there are at most $(M+1)^K$ possible multisets, and you could just check each of them to see if it works.

So the question remains: is there a fast way to do it? Essentially, your problem is a version of the subset sum problem (can we find a sub(multi)set of the items whose price adds to the total?), and this problem is NP-complete. In general, you should not expect such problems to be efficiently solvable.

However, there are special cases which are solvable quickly (by a computer): in case the total sum is not too large, there is a fairly efficient solution described for example on the Wikipedia page.


问:
arithmetic operations on alephs in WolframAlpha
Can someone please enlighten me if or how both of these facts can be true? I entered $\aleph_0-\aleph_0$ into WolframAlpha, which gave the result as undefined. But it also plotted the result on a number line at zero. Similarly, $\frac{\aleph_0}{\aleph_0}$ is 1 on the number line, but undefined. Is this just a bug? </p>

答:
Cardinal addition and multiplication (at least for $\aleph$ numbers, which in $\sf ZFC$ means all cardinal arithmetics) has the following property:$$\aleph_\alpha+\aleph_\beta=\aleph_\alpha\cdot\aleph_\beta=\max\{\aleph_\alpha,\aleph_\beta\}=\aleph_{\max\{\alpha,\beta\}}.$$

When an operation can be reversed we usually expect the answer to be uniquely determined. But $\aleph_0+\aleph_0=\aleph_0\cdot\aleph_0=\aleph_0$, and so if we could cancel either of the operations we would get:
$$\begin{align}
&\aleph_0+0=\aleph_0\\&\aleph_0+\aleph_0=\aleph_0\\&\aleph_0+\diagup\hspace{-.7em}\aleph_0=\diagup\hspace{-.7em}\aleph_0&
\implies\aleph_0=0.\end{align}$$

Or similarly, for multiplication:

$$\begin{align}&\aleph_0\cdot1=\aleph_0\\&\aleph_0\cdot\aleph_0=\aleph_0\\&\aleph_0\cdot\diagup\hspace{-.7em}\aleph_0=\diagup\hspace{-.7em}\aleph_0&\implies\aleph_0=1.\end{align}$$

Either result is absurd, of course.


问:
Find the value of $VN$ in Hexagon

He asked for $ VN $ and said that you have $ 2 $ squares and a regular hexagon, in addition to the hexagon side being $ 1 $.

The only thing I found out here was that the triangles are equilateral $ \frac {VNS} {SUM} $

What can I do?

答:
Since you have spotted the equilateral triangles, note that:
$$\measuredangle UMB=\measuredangle MBC \implies \bigtriangleup MUR \sim \bigtriangleup BCR$$
Both of these triangles are $120°-30°-30°$ triangles, so the longer side is $√3$ times the shorter sides.

You can find $MR$ via:
$$MR=AB-BR$$
Applying similarity:

$$(1-1/√3)/(1/√3)=UM/1 \implies UM=√3-1$$

Since both are equilateral triangles:

$$VN=1-(√3-1)=2-√3$$

Title: Find the value of $VN$ in Hexagon


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

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

Drawbacks:

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

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

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

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

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

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

答:
Technically they belong to the same optimization method: namely that of steepest descent or gradient descent.

The strategy in line search is to evaluate $f(\mathbf{x}-\alpha\nabla_x f(\mathbf{x})$ for several values of $\alpha$ and choose the one that results in the smallest objective function value, while the more popular approach is to set $\alpha$ to a small constant.

Line search and using a (constant) learning rate are thus two different strategies but belong to the same optimization method.

Title: Drawbacks of gradient descent


问:
Limit $\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$
Find the limit

$$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

Let $x=r\sin \theta$ and $y=r\cos \theta$.

$$
\begin{align}
\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2} \\
&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{r^4(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\lim_{r\to0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{0}{(0+2\sin^2\theta)^2}\\
&=0
\end{align}
$$

Does this logic follow? Can I ignore the case where $2\sin^2\theta=0$ meaning it is of the form $\frac{0}{0}$?
How else would I approach this limit.

答:
Another approach is to homogenize degrees with $y=ux^2$ ($u$ variable).

After substitution $x^8$ can be simplified and we get $f(x,y)=\underbrace{\dfrac{-16u^3}{(1+2u^2)^2}}_{g(u)}\times x$

It is not difficult to see that $g$ is bounded, indeed $1+2u^2\ge 1>0$ so the denominator never annulates so $g$ is continuous on $\mathbb R$.

Also $|g(u)|\sim \frac{16|u|^3}{4|u|^4}=\frac 4{|u|}\to 0$ at infinity, therefore $g$ is bounded by some $M$.

We conclude that $|f(x,y)|\le |g(u)||x|\le M|x|\to 0$


问:
Consider the inequality $9-x^2>|x-a|$ where $a$ is real, then find set of values for $a$ such that at least one negative solution exists
For $x>a$

$$9-x^2 >x-a$$
$$x^2 + x -(9+a)<0$$

So
$$1+4(9+a)<0$$
$$a<\frac{-37}{4}$$

Similarly for $x<a$
$$x^2-x+a-9<0$$
So $a>\frac{37}{4}$

Now in both cases, we need at least one $x<0$

In case one, $a<0$ and $x>a$ so $x$ can be both positive and negative, so I don’t know how to deal with that

Similar problem in case 2.

Another inequality I found was $a>-9$, obtained by solving the quadratic for first case, giving $x\in (\frac{1-\sqrt {37+4a}}{2}, \frac{1+\sqrt {37+4a}}{2})$, and then setting the lower limit to $<0$

I can also find another inequality from the second case, but there are way too many inequalities to choose from and the condition that $x>a$ and $x<a$ is throwing me off.

Also is a downward parabola even possible for these cases? The equation demands it to be downward if $f(x)<0$ for all $x$, but the coefficient of $x^2$ is positive here.

答:
No, you started good, then went off the rails, by trying to take the shortcut that the discriminant must be negative.

$\underline{\text{Case 1}:~~ (x-a) \geq 0}$

$$9 - x^2 > x - a. \tag1$$

This implies that

$\displaystyle x^2 + x - (a + 9) < 0 \implies$

$$\left(x + \frac{1}{2}\right)^2 - \left(a + 9 + \frac{1}{4}\right) < 0. \tag2 $$

Examining inequality (2) at the critical value $\left(x = -\frac{1}{2}\right)$:

$\left(a + 9 + \frac{1}{4}\right)$ must be $> 0$, which  implies that the lower bound for $a$ in Case 1 is $a > \frac{-37}{4}.$

As $a$ approaches $0$ from below, an $x$ will always be available such that $a \leq x < 0$.  In this circumstance, it is clear that the LHS of inequality (1) above will be large enough to satisfy the inequality.

Therefore, the upper bound for $a$ in Case 1 is $(a < 0)$.

$\underline{\text{Case 2}:~~ (x-a) < 0}$

$$9 - x^2 > a - x. \tag3$$

This implies that

$\displaystyle x^2 - x + (a - 9) < 0 \implies$

$$\left[\left(x - \frac{1}{2}\right)^2 -
\left(\frac{1}{4}\right)\right] + (a - 9)  < 0. \tag4 $$

With $x$ constrained by $x < 0$,

$\displaystyle \left[\left(x - \frac{1}{2}\right)^2 -
\left(\frac{1}{4}\right)\right]$ will approach $0$ from above, as $x$ approaches $0$ from below.

Therefore, in Case 2, an upper bound for $a$ is $a < 9.$

So the final question is : what is the lower bound for $a$ in Case 2.

Consider the inequality :

$$x^2 - x + (x - 9) < 0\tag5$$

which is what happens when $x=a$.

Clearly $x$ can never be $\leq -3$, or inequality (5) will be violated.  To prove that $(a > -3)$ is a lower bound for $a$ in Case 2, assume that

$\exists ~\delta > 0$ such that $a = (-3 + \delta).$

Then inequality (4) above may be manipulated to

$$\left[\left(x - \frac{1}{2}\right)^2 -
\left(\frac{1}{4}\right)\right] + (a - 9)  < 0 \implies $$

$$\left(x - \frac{1}{2}\right)^2 < \left(\frac{1}{4}\right)
+ (9 - [-3 + \delta]) =
(12) + \left(\frac{1}{4}\right) - \delta.\tag6$$

Regardless of how small the fixed size of $\delta$ is, as long as $\delta > 0$, you will be able to find a value of $x > -3$, so that inequality (6) above is satisfied.

Therefore, in Case 2, $a$ is constrained by $-3 < a < 9.$

In summary:

Case 1 bounds on $a$ : $~ -\frac{37}{4} < a < 0$.

Case 2 bounds on $a$ : $~ -3 < a < 9$.

Therefore, the overall constraints on $a$ are $-\frac{37}{4} < a < 9.$


问:
Proving $\binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n}$ algebraically
I am working through the exercises and have spent half a day on one problem so I decided to get some help because I can't figure it out.

Show that if $n$ is a positive integer at most equal to $m$, then
$$
\binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n}
$$
I expand the sum of two binomial coefficients and get
$$
\frac{m!}{n!(m-n)!} + \frac{m!}{(n-1)!(m-n+1)!}
$$
Now, I have trouble summing up the two fractions, specifically, finding the common denominator of two fractions. I tried all sorts of things I could think of but can't seem to get to $n!(m-n+n)!$ which is in the solutions at the back of the book.

My last try
$$
\frac{m!(n!(m-n)!) + m!((n-1)!(m-n+1)!)}{n!(m-n)!(m-n+1)!(n-1)!} =
$$
$$
= \frac{m!(n!(m-n)! + (n-1)!(m-n+1)!)}{n!(m-n+1)!(m-n)!(n-1)!} =
$$
$$
= \frac{m!(mn!-nn!+((mn-n^2+n)!-(m-n+1)!)!}{n!(m-n+1)!(m-n)!(n-1)!}
$$

And so on, I feel like I am missing something as I don't see any opportunity to simplify this mess.

Body:
Proof using formula:
$$
{{m\choose n} := {m(m-1)\ldots(m-n+1)\over n!}}
$$

$$
\eqalign{{m\choose n}+{m\choose n-1}&=
{m(m-1)\ldots(m-n+1)\over n!}+{m(m-1)\ldots(m-(n-1)+1)\over (n-1)!}\cr
&={m(m-1)\ldots(m-n+2)\cdot(m-n+1)\over n!}+{m(m-1)\ldots(m-n+2)\over (n-1)!} \cdot {n \over n}\cr
&={m(m-1)\ldots(m-n+2)\cdot[(m-n+1)+n]\over n!}\cr
&={m(m-1)\ldots(m-n+2)(m+1)\over n!}\cr
&={(m+1)m(m-1)\ldots[(m+1)-n+1)]\over n!}\cr
&={m+1\choose n}.
}
$$


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
Well, of course you're free to adopt the Pythagorean as an axiom in your system. But that doesn't remove the burden of proof; it merely transforms it, as now you have to prove that a) your system as stated is consistent; and, almost as important, b) that your new axiom is not redundant i.e. could not itself be derived from other axioms in that system. That is, you'll have to prove that you can neither prove nor disprove the Pythagorean using the other axioms of your system. If I was your teacher I'd definitely take you up on that. But if I was you, I might not want to go there. Unless you really are game.

Basically, axioms aka postulates aren't universal truths; they're Lego bricks. At the end of the day it doesn't matter what bricks you have or decide to use, so much as what you can build with them, how that relates to other builds, and occasionally whether it's useful for anything.

I could adopt as my only axiom that I'm God or Chuck Norris, and hence everything I say is true just because I say so - or; it is true that I said it, since "things Jesper said" is the only universe of which this system is valid. And even then fairly useless. You can't build things with it. Even I can't build things with it. You may note, say, that Jesper said A, and then he said not-A. But you can't even take that and derive a proper contradiction. Systems with that property are called dialethian - and they may not all be as useless as this.

Or I might adopt as axiomatic that only those propositions that have been proved or disproved are allowed to have a truth value, so that there will be propostions that have none; and consequently the classical dictum, known as the law of the excluded middle, that a proposition is either true or false, is itself false - in this kind of system, called intuitionistic or constructivist. And that has some very interesting consequences, or builds, or meta-builds known as category theory and topos theory.

All I'm saying is, you're on the right track. Challenge everything up to and including your ability to challenge everything. But derive the consequences. Cuz that's where the fun is. You're absolutely right that you can take anything at all to be your axioms in your system. The challenge is still to do something interesting with it, and hopefully tell us something of how your system relates to everything else.

But right now you're at school. And school tells you what system to use. Use it, then. It doesn't have to become part of your identity.

You can prove the Pythagorean theorem given Euclid's postulates. Once you've done that, you'll probably just go right on using it wherever those postulates hold, without even bothering to remember or recount your proof. And hence it is, in a sense, axiomatic for your new build.

You can also quite easily disprove the Pythagorean: just do your geometry on a sphere or a saddle (hyperboloid). In non-Eucledian geometry, while still incredibly important, the Pythagorean is only true in the limit of infinitesimal triangles. And may that be a lesson to us all: today's universal truth is likely tomorrows special case. Maybe.

Title: What is a proof?


问:
Eigenvalues of a 2x2 stochastic Matrix
In one of my lecture notes it is stated that one of the Eigenvalues (EV) of a stochastic Matrix can be calculated by $P_{11} - P_{21}$

I.e I have the following matrix :
$$\begin{matrix}
    0.1 & 0.9 \\
    0.5 & 0.5 \\
\end{matrix}$$
Obviously the EV are 1 and -0.4

Okay lets look at some more random Numbers :
$$\begin{matrix}
    0.42653 & 0.57347 \\
    0.2956 & 0.7044 \\
\end{matrix}$$   $P_{11} - P_{21}$ = 0.42653 - 0.2956 = 0.13093 which is its EV. The other is obviously 1

Is that always right for these kind of stochastic matrices?

答:
When $P$ is a $2\times2$ row-stochastic matrix, one of its eigenvalues is $1$. Let the other eigenvalue be $\lambda$. Then
$$
\lambda+1=\operatorname{tr}(P)=p_{11}+p_{22}=p_{11}+(1-p_{21}).
$$
Therefore $\lambda=p_{11}-p_{21}$.

Alternatively, let
$$
P=\pmatrix{p_{11}&p_{12}\\ p_{21}&p_{22}}=\pmatrix{a&1-a\\ b&1-b}.
$$
Then
$$
\pmatrix{1&-1}P=\pmatrix{1&-1}\pmatrix{a&1-a\\ b&1-b}=(a-b)\pmatrix{1&-1}.
$$
Therefore $a-b=p_{11}-p_{21}=p_{22}-p_{12}$ is an eigenvalue of $P$.

Title: Eigenvalues of a 2x2 stochastic Matrix


答:
You already got the answer, in finding limits just avoid the intermidate
$$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$
You
let $x=r\sin \theta$ and $y=r\cos \theta$.

$$
\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}  = \lim_{r\to  0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2} = \frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2} = \frac{0}{(0+2\sin^2\theta)^2}$$

Remember that $\theta $ is also a variable $x=r\sin \theta$, so you can ignore the case when $2\sin^2 \theta$ is $0$, If fact we can deceive the equation by saying $\lim_{(\theta ,r) \to (0,0)}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2}$, since $x$ and $y$ are approaching same objective $$\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2} = \lim_{x \to 0} \lim_{y \to x}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

$$\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2} = \lim_{x \to 0} \frac{-16x^3x^3}{(x^4+2x^2)^2} = \lim_{x \to 0} \frac{-16x^6}{(x^4+2x^2)^2}$$
As $x$ gets smaller the numerator shrinkes rapidly compared to the denominator, when you check L'hospitals rule after repeated sections the numerator would vanish meaning the limit is $0$

Title: Limit $\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$

Body:
Find the limit

$$\displaystyle\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}$$

Let $x=r\sin \theta$ and $y=r\cos \theta$.

$$
\begin{align}
\lim_{(x,y)\to(0,0)}\frac{-16x^3y^3}{(x^4+2y^2)^2}&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{(r^4\cos^4\theta+2r^2\sin^2\theta)^2} \\
&=\lim_{r\to0}\frac{-16r^3\cos ^3 \theta r^3\sin ^3\theta}{r^4(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\lim_{r\to0}\frac{-16r^2\cos ^3 \theta \sin ^3\theta}{(r^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{-16(0)^2\cos ^3 \theta \sin ^3\theta}{((0)^2\cos^4\theta+2\sin^2\theta)^2}\\
&=\frac{0}{(0+2\sin^2\theta)^2}\\
&=0
\end{align}
$$

Does this logic follow? Can I ignore the case where $2\sin^2\theta=0$ meaning it is of the form $\frac{0}{0}$?
How else would I approach this limit.


答:
Multiply top and bottom by $2x + 5$ to complete the square. Then we have:

$$\frac{x^2(2x+5)}{(2x)^2 - 5^2} = \frac{2x^3+5x^2}{4x^2 - 25}$$

and dividing by $x^2$, we get:

$$\frac{2x+5}{4-25/x^2}$$

As $x$ tends to infinity, $25/x^2$ tends to $0$, so we are left with $\frac{2x+5}{4} = \frac{x}{2} + \frac{5}{4}$.

Title: What are the steps to find the oblique asymptote of this function?

Body: The oblique asymptote of $\frac{x^2}{2x - 5}$ is $\frac{x}{2}+\frac{5}{4}$. What steps do I need to take in order to get from the original function to the oblique asymptote?


问:
Limit exists with definition but not with polar coordinates
I would like to know why if I try to prove with the delta epsilon definition that the limit as $(x,y)$ tends to $(0,0)$ is $0$ of this function:
$$\frac{x^2 +y}{\sqrt{x^2+y^2}}$$
I get a positive result: for $\delta = \varepsilon -1$, I get that $|f(x,y)-0|$ is less than $\varepsilon$.

But if you evaluate the limit with polar coordinates, you get that the limit depends on the path, and thus it doesn't exist.

答:
For the epsilon-delta definition you need to prove that for every positive $\epsilon$ there exists a positive $\delta$ such that...

So you can't take $\delta=\epsilon-1$, since if $0<\epsilon\leq 1$, this won't give you a positive $\delta$.


问:
Confirming convergence of a sequence and finding its limit.
The problem I'm working on (from Rudin) is the following.

If $s_1=\sqrt{2}$ and $$s_{n+1} = \sqrt{2+\sqrt{s_n}} \quad\text{      } (n=1,2,3,\cdots)$$

prove that $\{s_n\}$ converges and that ${s_n}<2
$ for $(n=1,2,3,\cdots)$.

The following is what I tried to do.

a), I can see that the sequence is monotonically increasing. So if I could see if it is bounded, then the sequence must be convergent. The problem is, I cannot confirm the part that it is bounded. I though about the sequence $$\{t_n\} = \{ \sqrt{2},\sqrt{4},\sqrt{6},\cdots\}$$ which is clearly "larger" than $\{s_n\}$ , but this is not bounded. Can someone help me out ?

b), I am also trying to find the limit. Assuming that convergence were confirmed, can someone tell me how to work on this ? The following is what I tried but it seems to not work well...

$$x = \sqrt{2+\sqrt[4]{2+\sqrt[4]{2+\cdots}}} $$
$$x^2 - 2 = \sqrt[4]{2+\sqrt[4]{2+\sqrt[4]{2+\cdots}}}$$
$$x^2 - 2 = \sqrt{x}$$
$$(x-1)(x^3+x^2-3x-4)=0$$

It's clear why $x=1$ is not the solution here, so $x$ must be one of the three solutions that the cubic has as a root. Besides the fact that I cannot do anything simple to find the solution of the cubic since it is irrational, did I at least proceed in a right way?

How would you solve this problem ?

The book implies that the upper bound of $\{s_n\}$ is $2$, but how is it confirmed ?

Edit:
Thanks everyone, for the help. I just need someone to confirm how to find the limit.

答:
Induction. $n=1$ is obvious.
$$
s_{n+1} = \sqrt{2+\sqrt{s_n}} < 2 \iff 2+\sqrt{s_n} < 4 \iff s_n<4
$$

Then your bounded + monotonic $\implies$ convergent logic is good.


答:
A general way to find an asymptote (at $x \to \infty$) is to use the derivative (you can consider the asymptote as the tangent of the curve at $\infty$):

$\begin{align*}
  f(x)
    &= \frac{x^2}{2 x - 5} \\
  f'(x)
    &= \frac{2 x^2  - 10}{(2 x - 5)^2} \\
  \lim_{x \to \infty} f'(x)
    &= \frac{1}{2}
\end{align*}$

This gives the slope $m$. To get the intercept, just compute;

$\begin{align*}
   \lim_{x \to \infty} f(x) - m x
     &= \lim_{x \to \infty} \frac{x^2}{2 x - 5} - \frac{x}{2} \\
     &= \frac{5}{4}
\end{align*}$

Thus the asymptote is:

$\begin{align*}
  \frac{x}{2} + \frac{5}{4}
\end{align*}$


答:
You cannot take $\delta=\varepsilon-1$, since you are after a $\delta>0$.

And you don't need polar coordinates to show that that limit does not exist. Just use the fact that, if $f(x,y)=\frac{x^2+y}{\sqrt{x^2+y^2}}$, then

if $t>0$, $f(0,t)=1$, and therefore the limit, if it exists, can only be $1$;

if $y=-x^2$, $f(x,y)=0$, and therefore the limt, if it exists, can only be $0$.

Title: Limit exists with definition but not with polar coordinates

Body:
I would like to know why if I try to prove with the delta epsilon definition that the limit as $(x,y)$ tends to $(0,0)$ is $0$ of this function:
$$\frac{x^2 +y}{\sqrt{x^2+y^2}}$$
I get a positive result: for $\delta = \varepsilon -1$, I get that $|f(x,y)-0|$ is less than $\varepsilon$.

But if you evaluate the limit with polar coordinates, you get that the limit depends on the path, and thus it doesn't exist.


答:
By Induction you can show that $s_n<2,\forall n\in N$.

Hint: $s_{n+1}=\sqrt{2+\sqrt{s_n}}\le \sqrt{2+\sqrt{2}}<\sqrt{2+2}=2$

Title: Confirming convergence of a sequence and finding its limit.

Body:
The problem I'm working on (from Rudin) is the following.

If $s_1=\sqrt{2}$ and $$s_{n+1} = \sqrt{2+\sqrt{s_n}} \quad\text{      } (n=1,2,3,\cdots)$$

prove that $\{s_n\}$ converges and that ${s_n}<2
$ for $(n=1,2,3,\cdots)$.

The following is what I tried to do.

a), I can see that the sequence is monotonically increasing. So if I could see if it is bounded, then the sequence must be convergent. The problem is, I cannot confirm the part that it is bounded. I though about the sequence $$\{t_n\} = \{ \sqrt{2},\sqrt{4},\sqrt{6},\cdots\}$$ which is clearly "larger" than $\{s_n\}$ , but this is not bounded. Can someone help me out ?

b), I am also trying to find the limit. Assuming that convergence were confirmed, can someone tell me how to work on this ? The following is what I tried but it seems to not work well...

$$x = \sqrt{2+\sqrt[4]{2+\sqrt[4]{2+\cdots}}} $$
$$x^2 - 2 = \sqrt[4]{2+\sqrt[4]{2+\sqrt[4]{2+\cdots}}}$$
$$x^2 - 2 = \sqrt{x}$$
$$(x-1)(x^3+x^2-3x-4)=0$$

It's clear why $x=1$ is not the solution here, so $x$ must be one of the three solutions that the cubic has as a root. Besides the fact that I cannot do anything simple to find the solution of the cubic since it is irrational, did I at least proceed in a right way?

How would you solve this problem ?

The book implies that the upper bound of $\{s_n\}$ is $2$, but how is it confirmed ?

Edit:
Thanks everyone, for the help. I just need someone to confirm how to find the limit.


问:
In definite integration, do I have to watch out for the behaviour of a function between the limits?

Calculate -
$$\int^{1}_{-1}\frac{e^\frac{-1}{x}dx}{x^2(1+e^\frac{-2}{x})}$$

This is my approach-

Substitute $t=e^\frac{-1}{x}$

$$\therefore \int^{\frac{1}{e}}_{e}\frac{dt}{1+t^2}$$

which gives us $\frac{\pi}{2}- 2\arctan{e}$, but the answer given is $\pi- 2\arctan{e}$. I'm guessing that my wrong answer has something to do with the exponential term.

答:
You're pretty much correct. The usual theorem for integration by substitution says that we can use a substitution $t=\phi(x)$ if $\phi$ is $C^1$ (continuously differentiable) in the interval of integration. (This is just a sufficient condition of course).

But your substitution $t=\phi(x)=e^{-\frac 1x}$ is not even continuous at $x=0$. In fact
$$
\lim_{x\to0^-} \phi(x) = \infty
$$
$$
\lim_{x\to0^+} \phi(x) = 0
$$
So we have to split the integral up like this:
$$
\int_{-1}^0 \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})} dx
= \int_e^\infty \frac{1}{1+t^2} dt
= \arctan{\infty} - \arctan e
= \frac{\pi}{2} - \arctan e
$$
and
$$
\int_0^1 \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})} dx
= \int_0^{e^{-1}} \frac{1}{1+t^2} dt
= \arctan{e^{-1}} - \arctan 0
= \frac{\pi}{2} - \arctan e
$$
so
$$
\int_{-1}^1 \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})} dx
= \pi - 2\arctan e
$$

Bottom line: If $\phi$ is not continuous (or $C^1$, but that would be rarer) on the interval, then split the interval up.

BONUS: In this case we could also have taken a common short cut. We can notice that $f(x) = \frac{e^{-\frac{1}{x}}}{x^2(1+e^{-\frac{2}{x}})}$ is an even function (i.e. $f(x)= f(-x)$). Therefore we can also write
$$
\int_{-1}^1f(x)dx = 2\int_{0}^1 f(x)dx
$$
and then the substitution is legal again.