问:
Controversial logic problem from brilliant. Misinterpreted logic or flawed question?
You are offered two cupcakes. One is poisoned and the other is safe to eat.

You happen to be in a village full of knights (who always tell the truth) and knaves (who always lie), but you can't tell which is which by their appearance.

You ask one of them, "Which cupcake is safe to eat?" To this he makes the following two statements,

"If I were a knave, I'd say the one on the right."

"But I'd say the one on the left, if I were a knight."
Which cupcake is safe to eat?

Assumption: The person you ask knows which cupcake is which.

This problem appears on brilliant in the practice section (truth tellers and liars level 3 challenges) and it's causing a bit of a heated discussion. The answer is supposedly the cupcake on the left. Is this question flawed or are people misinterpreting the logic?

答:
In mathematical logic, the statement "If $A$, then $B$" means a very precise thing, in a way that it is not always interpreted as in everyday conversation. It means no more and no less than the following statement: "It is not the case that $A$ is true and $B$ is false".

For instance, "If $2+2=5$, then $2+2=6$" is an example of a true implication. We're only committing to the second half of the statement if the first half is true, which it's not - so the statement remains true! An "if" statement is only false when the first part is true but the second part is false. See the Wikipedia article for a more in-depth treatment.

With that cleared up, let's take a look at this problem, and name our mystery person P. In fact, we only need to look at the second sentence.

P: "But I'd say the one on the left, if I were a knight."

They're saying "If (P is a knight), then (P would answer "the one on the left" to your question)".

When is this statement false? It's false when (P is a knight) is true, but (P would answer "the one on the left" to your question) is false. So this will be false only in those scenarios where P is a knight and would tell you the safe cupcake is on the right.

What can we conclude from this? Well, P can't be a knave, because then the statement would be false, but it can only be false when P is a knight! So P is a knight, and therefore the statement is true. So, we can conclude that P really would tell you "the one on the left", and because they're a knight, that answer would be the correct one.


问:
I can't seem to simplify this trigonometric differential into the required form
This has been bugging me all day. It seems like a very straight forward question but for whatever reason, I just cannot get my answer in the right format.

The question reads -

The parametric equations of a curve are

x = sin(2A) + 2sin(A)

y = cos(2A) - 2cos(A), where 0 < A < PI

Show that dy/dx = -(sinA) / (1 + cosA)

The differentiation is fine, I can do that without problems and I checked WolframAlpha to make sure I was correct in that but I just cannot simplify the dy/dx to the required format whatever I do. I keep getting very close in the different ways but never exactly what I need.
I would be very grateful if someone could help.

Tip: You need to find dx/dA and dy/dA first then invert dx/dA and multiply them together to get to dy/dx

答:
Dear Mark,
$$dy/dx = \frac{dy/dA}{dx/dA} = \frac{2\sin A - 2 \sin 2A}{2\cos A + 2\cos 2A} = \frac{2}{2}\cdot\frac{\sin A (1- 2\cos A)}{(1+\cos A)(-1+2\cos A)} = -\frac{\sin A}{1+\cos A}$$
The result may also be written as $-\tan(A/2)$ because $1+\cos A = 2\cos^2 (A/2)$ while $\sin A = 2\cos(A/2)\sin(A/2)$ and $2\cos(A/2)$ cancels. The procedure in the tip (which includes inversion of the function) is totally unnecessary.</p>

ContentLicense: CC BY-SA 3.0


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
That's right. Here is another proof:
$\forall n\in\mathbb{Z}$ if $n$ is odd, then $n^2$ is odd .
Let $m$ be an integer such that $m^2$ is even. Then $m$ must be even, because if $m$ is odd then $m^2$ must be odd, contrary to the hypothesis.</p>


问:
Proving Pointwise Convergence of a Sequence $\{f_n\}$ with a given $f_n$
I am trying to prove pointwise convergence of a sequence $\{f_n\}$ given $f_n:[0,\infty)\rightarrow \mathbb{R}$ with

$$f_n(x)=\frac{x^n}{1+x^{2n}}$$

I started by delineating each possible value for x, so I have that

$$f(x)=\begin{cases}0 & \mathrm{if} \space 0\le x \lt 1\\ \frac12 & \mathrm{if} \space x=1 \\ 0 & \mathrm{if} \space x \gt 1 \end{cases}$$

Now to prove. So I set out to show that $\lim_{n\rightarrow\infty}\frac{x^n}{1+x^{2n}}=0 \quad \forall x\in[0,1)\cup(1,\infty)$.

Now $\mathit{let} \, ε\gt0$ be given, choose any $x_0\in[0,1)\cup(1,\infty)$. Find $n^*$ such that $|f_n(x)-f(x)|\ltε\quad\forall n\ge n^*$.

$$\Rightarrow f_n(x_0)=\frac{x_0^n}{1+x_0^{2n}}$$

Since $f(x_0)=0$, then we have:

$$\left\lvert\frac{x_0^n}{1+x_0^{2n}}-0\right\rvert=\left\lvert\frac{x_0^n}{1+x_0^{2n}}\right\rvert=\frac{x_0^n}{1+x_0^{2n}}\ltε$$

since $x_0$ is positive or zero. Taking the natural log of both sides gives:

$$\ln\left({\frac{x_0^n}{1+x_0^{2n}}}\right)\lt\ln{ε}\quad\Rightarrow\quad n\ln({x_0})-n\ln({x_0})=0\ltε$$

But the problem is that now I no longer have $n$ in my equation so I cannot generalize in terms of $n\ge n^*$. Where did I go wrong, or where could I take a different approach?

答:
Set $g(t)=\frac{t}{1+t^2}$. Note that $f_n(x)=g(x^n)$.

If $x>1$ is fixed, then $x^n\to\infty$, as $n\to\infty$, so $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}g(x^n)=\lim_{t\to\infty}g(t)=0$. On the other hand, if $x\in[0,1)$ is fixed, then $x^n\to0$, as $n\to\infty$, so $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}g(x^n)=\lim_{t\to0}g(t)=0$. Finally, for $x=1$ we have that $f_n(1)=1/2$ for all $n$, so $f_n(1)\to1/2$.

So the sequence $(f_n)$ converges pointwise to the piecewise-defined function you specified.

ContentLicense: CC BY-SA 4.0


答:
Note that
$$0< \frac{x_0^n }{1+x_0^{2n} } <\frac{x_0^n }{x_0^{2n} } = x_0^{-n}$$
With the squeeze lemma the desired convergence follows for $x_0 >1$. For $x_0<1$ we simply use that the numerator converges to zero while the denominator converges to one.

Hope this helps you.

Title: Proving Pointwise Convergence of a Sequence $\{f_n\}$ with a given $f_n$

Body:
I am trying to prove pointwise convergence of a sequence $\{f_n\}$ given $f_n:[0,\infty)\rightarrow \mathbb{R}$ with

$$f_n(x)=\frac{x^n}{1+x^{2n}}$$

I started by delineating each possible value for x, so I have that

$$f(x)=\begin{cases}0 & \mathrm{if} \space 0\le x \lt 1\\ \frac12 & \mathrm{if} \space x=1 \\ 0 & \mathrm{if} \space x \gt 1 \end{cases}$$

Now to prove. So I set out to show that $\lim_{n\rightarrow\infty}\frac{x^n}{1+x^{2n}}=0 \quad \forall x\in[0,1)\cup(1,\infty)$.

Now $\mathit{let} \, ε\gt0$ be given, choose any $x_0\in[0,1)\cup(1,\infty)$. Find $n^*$ such that $|f_n(x)-f(x)|\ltε\quad\forall n\ge n^*$.

$$\Rightarrow f_n(x_0)=\frac{x_0^n}{1+x_0^{2n}}$$

Since $f(x_0)=0$, then we have:

$$\left\lvert\frac{x_0^n}{1+x_0^{2n}}-0\right\rvert=\left\lvert\frac{x_0^n}{1+x_0^{2n}}\right\rvert=\frac{x_0^n}{1+x_0^{2n}}\ltε$$

since $x_0$ is positive or zero. Taking the natural log of both sides gives:

$$\ln\left({\frac{x_0^n}{1+x_0^{2n}}}\right)\lt\ln{ε}\quad\Rightarrow\quad n\ln({x_0})-n\ln({x_0})=0\ltε$$

But the problem is that now I no longer have $n$ in my equation so I cannot generalize in terms of $n\ge n^*$. Where did I go wrong, or where could I take a different approach?


问:
How to derive corrected sum of squares and the corrected sum of cross products given LHS of the equations?
I am self-studying the book INTRODUCTION TO LINEAR REGRESSION ANALYSIS by Montgomery et al., where it is not clear to me how to reach the RHS of the equations (2.9) & (2.10) from the given LHS.
I will appreciate any help. Thanks

Equation (2.9) is as follows:
$$S_{xx}=\sum_{i=1}^{n}x_{i}^{2} - \frac{\left ( \sum_{i=1}^{n}x_{i} \right )^{2}}{n} = \sum_{i=1}^{n}(x_{i} -\overline{x})^2$$

and Equation (2.10) is as follows:
$$S_{xy}=\sum_{i=1}^{n}y_{i}x_{i}- \frac{\left ( \sum_{i=1}^{n}y_{i} \right )\left ( \sum_{i=1}^{n}x_{i} \right )}{n}=\sum_{i=1}^{n}y_i(x_{i} -\overline{x})$$

where, $$\overline{x}=\frac{1}{n}\sum_{i=1}^{n}{x_{i}}$$

Body:
I figured it out, the proofs are as follows,

For eqn. (2.9)

\begin{align*}
&S_{xx}=\sum_{i=1}^{n}x_{i}^{2} - \frac{\left ( \sum_{i=1}^{n}x_{i} \right )^{2}}{n} =\sum_{i=1}^{n}(x_{i} -\overline{x})^{2} \\
&where,  \overline{x}= \frac{1}{n}\sum_{i=1}^{n}{x}_i \\ \\
&RHS = \sum_{i=1}^{n}(x_{i} -\overline{x})^{2} \\
&= \sum_{i=1}^{n}(x_{i}^2 - 2x_{i}\overline{x} + \overline{x}^2)\\
&= \sum_{i=1}^{n}x_{i}^2 - \sum_{i=1}^{n}2x_{i}\overline{x} + \sum_{i=1}^{n}\overline{x}^2  \\
&\text{Since, } \overline{x}\text{ is a constant,} \sum_{i=1}^{n} \overline{x}^2=
\overline{x}^2\sum_{i=1}^{n}1=n\overline{x}^2 \\
&RHS = \sum_{i=1}^{n}x_{i}^2 - 2\overline{x}\sum_{i=1}^{n}x_{i} + n\overline{x}^2  \\
&RHS = \sum_{i=1}^{n}x_{i}^2 - 2n\overline{x}^2 + n\overline{x}^2  \\
&RHS = \sum_{i=1}^{n}x_{i}^2 - n\overline{x}^2   \\
&RHS = \sum_{i=1}^{n}x_{i}^2 - n(\frac{\sum_{i=1}^{n}x_{i}}{n})^2   \\
&RHS = \sum_{i=1}^{n}x_{i}^2 - \frac{(\sum_{i=1}^{n}x_{i})^2}{n} = LHS   \\
\end{align*}
####################################################################

and for Eqn (2.10)
\begin{align*}
&S_{xy}=\sum_{i=1}^{n}y_{i}x_{i} - \frac{\left ( \sum_{i=1}^{n}y_{i} \right )\left ( \sum_{i=1}^{n}x_{i} \right )}{n} = \sum_{i=1}^{n}y_{i}(x_{i} -\overline{x}) \\
&where,  \overline{x}= \frac{1}{n}\sum_{i=1}^{n}{x}_i \\ \\
&RHS =\sum_{i=1}^{n}y_{i}(x_{i} -\overline{x}) \\ \\
&= \sum_{i=1}^{n}(y_{i}x_{i}-y_{i}\overline{x})\\ \\
&= \sum_{i=1}^{n}(y_{i}x_{i})-\sum_{i=1}^{n}(y_{i}\overline{x})\\ \\
&= \sum_{i=1}^{n}(y_{i}x_{i})-\sum_{i=1}^{n}(y_{i}\frac{1}{n}\sum_{i=1}^{n}{x}_i) \\ \\
&= \sum_{i=1}^{n}(y_{i}x_{i})-\frac{1}{n}(\sum_{i=1}^{n}y_{i}\sum_{i=1}^{n}{x}_i)\\ \\
&= \sum_{i=1}^{n}y_{i}x_{i}-\frac{(\sum_{i=1}^{n}y_{i})(\sum_{i=1}^{n}{x}_i)}{n}\\ \\
&= LHS
\end{align*}


答:
I think it would be better to treat both cases separately. First, for $x\in [0,1) $, you have :

$$\forall n\geq0, |f_n(x)|\leq \frac{x^n}{2} $$

which goes to $0$ at infinity since $ 0\leq x<1 $.

Then, for $x\in (1,\infty)$, you can use the fact that $x^{2n}+1\sim x^{2n} $

So $f_n(x)\sim\frac{x^n}{x^{2n}}=x^{-n}\to_{\infty} 0 $


答:
There are several issues in your "proof".

First of all, you don't need the redundant subscript in $x_0$, which makes your writing unnecessarily busy.

The first $\Rightarrow$ does not make sense: not even wrong. There is no implication there.

Second, the estimate you want is
$$
\left|\frac{x^n}{1+x^{2n}}-0\right|<\epsilon\tag{0}
$$

You seem to apply the following incorrect formula when taking logarithm (also with a missing factor of $2$ when taking $\ln(x_0^2)$):
$$
\ln(1+y)\stackrel{!}{=} \ln(1)+\ln(y)
$$

If $x\in[0,1)$, then
$$
\lim_{n\to\infty}f_n(x)=\frac{\lim_n x^n}{1+\lim_n (x^2)^n}=\frac{0}{1+0}=0\tag{1}
$$

If $x>1$, then
$$
\lim_{n\to\infty}f_n(x)=\lim_n\frac{(\frac{1}{x})^n}{(\frac{1}{x^2})^n+1}
=\frac{\lim_n(\frac{1}{x})^n}{\lim_n(\frac{1}{x^2})^n+1}
=\frac{0}{0+1}=0\tag{2}
$$

If you do want an $\epsilon$-$N$ proof, you can simply prove that
$\lim_{n\to\infty} r^n=0$ for $0\le|r|<1$ and then apply the limit laws to get (1) and (2).


问:
How do you find the second constant in a parabolic pde solution?
Okay, so I am given the following Parabolic PDE:

$y^2u_{xx}-2xyu_{xy}+x^2u_{yy}=x^{-1}y^2u_x+y^{-1}x^2u_y$

I find the characteristics to be:

$\frac{dy}{dx} = \frac{-x}{y}$ and therefore $\psi(x,y) = x^2+y^2$

Now, I have been told that I need to choose the other constant, $\eta$. In the answer for the question, the lecturer decides to choose $\eta(x,y) = y^2$ but gives no reason as to why he does. I decided to choose $\eta(x,y) = x^2$ but this does not give the same answer.

Is there a method on how to know what value to choose for the other constant?

答:
Your equation is $Du = 0$ where
$$
D = \left(y^2\partial_{x}^2 - 2xy\partial_{x}\partial_{y} + x^2\partial_y^2 - \frac{y^2}x\partial_x - \frac{x^2}y\partial_y\right).
$$
With some work, we can factor $D$:
\begin{align*}
D & = (y\partial_x - x\partial_y)^2 + \left(\frac yx - \frac xy\right)x\partial_y + \left(\frac xy - \frac yx\right)y\partial_x \\
& = (y\partial_x - x\partial_y)^2 + \left(\frac xy - \frac yx\right)(y\partial_x - x\partial_y) \\
& = \left(y\partial_x - x\partial_y + \frac xy - \frac yx\right)(y\partial_x - x\partial_y).
\end{align*}
This suggests solving the PDE in two steps. The first one is to solve for $v$ from
$$
\left(y\partial_x - x\partial_y + \frac xy - \frac yx\right)v = 0.
$$
The second one is to solve for $u$ from
$$
\left(y\partial_x - x\partial_y\right)u = v.
$$
In both steps, the characteristics are defined by $\frac{\partial y}{\partial s} = -x$ and $\frac{\partial x}{\partial s} = y$ where $s$ is the curve parameter (or in your notation, $\frac{dy}{dx} = -\frac{x}y$, and the curve parameter is $x$). The equation for $v$ is $\frac{\partial v}{\partial s} = -\left(\frac xy - \frac yx\right)v$ while the equation for $u$ is $\frac{\partial u}{\partial s} = v$.

To solve for characteristics, differentiate equations for $y$ and $x$ once more to get $\frac{\partial^2 y}{\partial s^2} = -y$ and $\frac{\partial^2 x}{\partial s^2} = -x$. The solution is $x = c_1(r)\cos(s) + c_2(r)\sin(s)$, $y = -c_1(r)\sin(s) + c_2(r)\cos(s)$, where $r$ is the index of the characteristic curve. (Graphically, characteristics are circles, as you have seen from $x^2 + y^2 = \psi$. Your $\psi$ is the same as $c_1(r)^2 + c_2(r)^2$ in my notation.)

With two arbitrary functions of $r$, you do have some freedom here. You get to choose where each curve starts, which is the point on the circle that $s = 0$ corresponds to. Usually, you use the initial condition to decide, because making $s = 0$ correspond to the initial solution curve will simplify a lot of calculation. Anyway, since the initial condition is not given, I'll go ahead and do some computation.

Your lecturer's choice makes the positive $y$-axis correspond to $s = 0$ (so $c_1(r) = 0$), while your choice makes the positive $x$-axis correspond to $s = 0$ (so $c_2(r) = 0$). Your lecturer's choice leads to something like
\begin{align*}
x(r, s) & = r\sin s \\
y(r, s) & = r\cos s
\end{align*}
while your choice leads to something like
\begin{align*}
x(r, s) & = r\cos s \\
y(r, s) & = -r\sin s.
\end{align*}
It should not matter which one you choose. Note that with your lecturer's choice, you have
\begin{align*}
\frac yx - \frac xy & = \cot s - \tan s \\
& = 2\cot(2s)
\end{align*}
and with your choice, you have
\begin{align*}
\frac yx - \frac xy & = -\tan s + \cot s \\
& = 2\cot(2s).
\end{align*}
In both cases, $\frac yx - \frac xy$ becomes the same $2\cot(2s)$, and that is the expression you'll need to solve for $v$ and $u$:
\begin{align*}
\frac{\partial v}{\partial s} & = \left(\frac yx - \frac xy\right)v \\
& = 2\cot(2s)v\\
\therefore v & = C(r)\sin(2s) \\
\text{and from} \frac{\partial u}{\partial s} = v \quad \Longrightarrow \quad
u & = -\frac 12C(r)\cos(2s) + D(r) \\
\therefore u & = \tilde C(x^2 + y^2) \cdot (x^2 - y^2) + \tilde D(x^2 + y^2).
\end{align*}
Although $\cos(2s)$ becomes different expressions in different parametrizations, the final expression is the same because the sign difference is subsumed in $\tilde C$. I did get sloppy and ignored the case $\sin(2s) < 0$. However, the final solution works out if we keep in mind that $\tilde C$ and $\tilde D$, once solved from the initial condition, are valid only in the quadrant that contains the initial solution curve. It is natural to expect that the solution does not emanate past the $x$-axis or the $y$-axis because we have $x$ and $y$ in the denominators in the original PDE.


问:
On a definition of Spivak's fuzzy set
In the paper "Metric Realization of Fuzzy Simplicial Sets"  of David Spivak it takes $I=(0,1]$ as  poset and consider it as a category. He gives it a Grothendieck topology induce it from consider $I$ as topological space with topology induced from sets of the form $(0,a)$ with $a\in I$.

In the paper, he defines the notation $S((0,a))=S^{\geq a}$ for a sheaf $S$. So he also defines:

$S(a)=S^{\geq a}-colim_{b>a}S^{\geq a}$

So my question is how to understand the definition of $S(a)$. From a cetegorial point of view the colimit is unique up to isomorphism.So diferente sets makes the same colimit, in fact the colimit is determined by its cardanility, as we are working in the category of sets. But this difference depends of the choice of the specific set used to represent the colimit.

答:
(Co)limits aren't just objects in a category: they're really (co)cones, which is an object equipped maps to or from other objects, which is universal with respect to this property.

So $\operatorname{colim}_{b > a} S^{\geq b}$ isn't just a set (which as you point out would be determined up to isomorphism by its cardinality).  It is a set such that given a compatible system of maps $S^{\geq b} \to T$ for any other object $T$, there exists a unique map $\operatorname{colim}_{b > a} S^{\geq b} \to T$.

In particular we may take $T = S^{\geq a}$, and as Spivak explained there is a compatible system of maps $S^{\geq b} \to S^{\geq a}$ for all $b > a$.  So by the universal property of the colimit there is a unique map $\phi: \operatorname{colim}_{b > a} S^{\geq b} \to S^{\geq a}$, and by $S^{\geq a} \setminus \operatorname{colim}_{b > a} S^{\geq b}$ Spivak presumably means the complement of $\operatorname{colim}_{b > a} S^{\geq b}$ in $S^{\geq a}$, where $\operatorname{colim}_{b > a} S^{\geq b}$ sits inside $S^{\geq a}$ as the image of $\phi$.

Title: On a definition of Spivak's fuzzy set


问:
Mis-statement of First Principle of Mathematical Induction?
Maybe my problem is the English, but I cannot see why Gallian’s statement of the First Principle of Mathematical Induction is true.  In the 7th edition of Contemporary Abstract Algebra (p13) he states:

Let $S$ be a set of integers containing $a$.  Suppose $S$ has the property that whenever some integer $n$ greater than or equal to $a$ belongs to $S$, then the integer $n+1$ also belongs to $S$. Then $S$ contains every integer greater than or equal to $a$.

Let $S=\{1,5,7,12,13,14,15,...\}$ and let $a=7$. The integers $(n)$ will be the integers $12$ and above. This set $S$ meets the criterion since the inclusion of every such $n$ implies $n+1$ is also in $S$.  But clearly $S$ does not contain every integer greater than or equal to $a$.

答:
Your confusion seems to be the fact that you think you can choose both $a$ and the starting point for $n$ independently. (you chose $a=7$ but your $n$ starts at 12$).

But the following sais that you have to check for all $n$'s starting at $n=a$

```<code>"whenever some integer <span class="math-container">$n$</span> greater than or equal to <span class="math-container">$a$</span> belongs to S, then the integer n+1 also belongs to S."
</code>```

Body:
Your confusion seems to be the fact that you think you can choose both $a$ and the starting point for $n$ independently. (you chose $a=7$ but your $n$ starts at 12$).

But the following sais that you have to check for all $n$'s starting at $n=a$

"whenever some integer $n$ greater than or equal to $a$ belongs to S, then the integer n+1 also belongs to S."


答: The definition is not wrong. The thing you are missing is that if $S$ is our set containing $a$, the condition is that for every $n\geq a$, if $n \in S$ then $n+1\in S$. So in your example while it is true that for some $n \geq a$, (particularly $n\geq 12$), it is not true that every integer $\geq a$ satisfies the property. For example since $a=7$ and $n=7\geq 7$, the fact that $7 \in S$ should imply that $8\in S$ but this is clearly not the case.

Title: Mis-statement of First Principle of Mathematical Induction?

Body:
Maybe my problem is the English, but I cannot see why Gallian’s statement of the First Principle of Mathematical Induction is true.  In the 7th edition of Contemporary Abstract Algebra (p13) he states:

Let $S$ be a set of integers containing $a$.  Suppose $S$ has the property that whenever some integer $n$ greater than or equal to $a$ belongs to $S$, then the integer $n+1$ also belongs to $S$. Then $S$ contains every integer greater than or equal to $a$.

Let $S=\{1,5,7,12,13,14,15,...\}$ and let $a=7$. The integers $(n)$ will be the integers $12$ and above. This set $S$ meets the criterion since the inclusion of every such $n$ implies $n+1$ is also in $S$.  But clearly $S$ does not contain every integer greater than or equal to $a$.


答:
I think you have overlooked the fact that if $a = 7,$ then $7$ is an integer greater than or equal to $a.$

If $a = 7$ and if $S$ had the property that whenever some integer $n$ greater than or equal to $a$ belongs to $S$, then the integer $n+1$ also belongs to $S$,
we could set $n = 7$, and then the fact that $n \geq a$ and $n \in S$
would imply that $n + 1 = 8$ also belonged to $S.$

Since $8$ is not in $S$, $S$ does not have the property that the principle of induction requires.

The whole point of the principle of induction is that it applies to all integers $n$ that are greater than or equal to $a$, including $a$ itself.
So first you set $n = a$ and therefore $n+1 = a+1$ must be in the set.
Then you set $n = a+1$ and therefore $n+1 = a+2$ must be in the set.
And so forth ad infinitum.


答:
The statement is meant to be read as $$\forall S\subseteq \Bbb Z, ((a\in S\land \forall n\ge a, (n\in S\Rightarrow n+1\in S))\Rightarrow (\forall m\ge a, m\in S))$$

Id est, that if $S$ is a subset of $\Bbb Z$ containing $a$ and such that, for all $n\ge a$, if $n\in S$ then $n+1\in S$, then $S$ contains all the integers greater or equal to $a$. It's a cumbersome statement to make in natural language, all the more if you try to squeeze it inside a single sentence. In my opinion the statement you are quoting, as an English sentence, is rather clumsy and imprecise, but I think that even there you could try and grasp where the quantifiers are supposed to be.


问:
Monic polynomial with certain properties is the characteristic polynomial?
I'm trying to determine whether this statement is true or false:

Assume that $dim V = n$ and let $T \in L(V)$. Let $f(z) \in P_n(F)$ be a
monic polynomial of degree
$n$ such that $f(T) = 0$. Then $f(z)$ is the characteristic polynomial of $T$.

I know that because $f(T) = 0$, the minimal polynomial of T divides it and that the minimal polynomial of T divides the characteristic polynomial of T as well, but I can't seem to put these together in a way that conclusively establishes the truth or falsehood of the above statement.

If anyone could shed some insight on the matter, it would be greatly appreciated. Thanks!

答: Hint Take $T$ to be the $0$ operator. Then $f(T)=0$ for all $f(z)$ divisible by $z$.


问:
Alternative Methods for Trigonometry Proofs (Varying Angles)
Consider all the ways to prove $\sin(2x)=2\sin(x)\cos(x)$. There are many!

I personally would use Euler's formula involving complex numbers due to it being extremely simple and straightforward. We know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. The LH and RH sides become $\frac{e^{i2x}-e^{-i2x}}{2i}=2(\frac{e^{ix}-e^{-ix}}{2i})(\frac{e^{ix}+e^{-ix}}{2})$. And now this proof can be done with some basic algebra!

At the same time, I have always been bothered by rewriting $\sin(x)$ and $\cos(x)$ this way. To rewrite everything in terms of $e$ feels like an utter avoidance to use the words "sine" and "cosine." It also seems like an avoidance to represent such formulas along the unit circle to prove them. The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

This got me thinking... is there a dependable way to prove results like these using the unit circle? Or is there a way to rewrite everything in terms of sine and cosine another way?

I have been curious about this for a long time, and I would be curious as to what other dependable ways are used to prove trigonometric results such as these without $i$ and infinite sequences. What do other people do?

答: A rather twisted idea: $y = \sin 2x$ and $y = 2\sin x\cos x$ are solutions of the ODE $y'' + 4y = 0$ with initial conditions $y(0) = 0$, $y'(0) = 2$. Now, apply unicity...


问:
Relation between Halting problem and Godel's incompleteness Theorem
I've heard that these are closely related to each other.
However, while the former looks totally trivial to me, the latter still puzzles me even after I read the proof of it. So I made up a scenario to check my understanding.

Let's suppose that I believe Goldbach's conjecture to be false for some reason.
And I'm running a simple python program to find the counterexample of it on my desktop.
The program checks each case in increasing order. When it finds a counterexample, it returns the number and stops.

Then one day, someone proved that Goldbach's conjecture is unprovable.
From my understanding, it doesn't mean that I should stop the program right then.
What is proven is that there is no proof stating that my program will run forever, even if Goldbach's conjecture is true and my program actually run forever.
So it's reasonable to keep the program running and expect the program to find the counterexample even tomorrow.
Also, I can say that there is still a chance to disprove Goldbach's conjecture.

Is that correct?

答:
I'm not sure what your example has to do with the Incompleteness Theorem. If the Goldbach Conjecture is false, then yes, your program will eventually find a counterexample and thus prove it to be false. But if it is true, we might still be able to prove it true as well.

For example, suppose I have a computer program that tests the hypothesis that every even number greater than $2$ is the sum of two odd numbers by going through all such even numbers one by one, just like your program.  Now, we know this hypothesis to be true, and so your computer program will run forever. But, does this mean that we can't prove the hypothesis?  Of course not .. and the proof is elementary.


问:
Prove that $s_{n+1}=s_n^2-2$ diverges.
Let $s_1=5$ and for $n\geq 1$, $s_{n+1}=s_n^2-2$. Do the following steps to show that $(s_n)$ diverges.

1.If $\lim s_n = a$ then $a^2 -2 = a$.

$a=\lim s_{n+1}=\lim {s_n^2-2}=\lim {s_n^2}-\lim (2)=(\lim {s_n})^2-\lim (2)=a^2-2$

2. If $\lim s_n = a$ then a has to be either $−1$ or $2$.

From (1), we have

$a=a^2-2$

$a^2-a-2=0$

$a=−1, 2$

3. Show that $\lim s_n$ can neither be $−1$ nor $2$. (Hint: write down the first few terms of $(s_n)$ to get some insights.)

$s_n\leq s_{n+1}$ for all $n$

Proof, by induction:

$P_1$ is true; at $n=1$, $5=s_1\leq s_2=5^2-2=23$

Suppose that $s_n\leq s_{n+1}$ for some $n$

By the assumption we have $s_{n+1}=s_n^2-2\leq s_{n+1}^2-2=s_{n+2}$

As a result, $s_n$ is increasing, so $s_n\geq 5$ for all $n$. Now since $s_n\geq 5$, this implies that $\lim s_n \geq 5$ which implies that it cannot be $-1$ or $2$.

4. Combine these steps to show that $\lim s_n$ does not exist.

Is there any further step to do here or just combine the 2 and 3?

Are my answers correct, please? Thank you in advance.

答:
More convenient to begin with an $s_0.$

Suppose we find the real number $A>1$ for which
$$ s_0 = A + \frac{1}{A}  $$
From $s_{n+1} = s_n^2 - 2$  we find
$$ s_1 = A^2 + \frac{1}{A^2} \; , \; \;   $$
$$ s_2 = A^4 + \frac{1}{A^4} \; , \; \;   $$
$$ s_3 = A^8 + \frac{1}{A^8} \; , \; \;   $$
generally
$$ s_n = A^{2^n} + \frac{1}{A^{2^n}} \; . \; \;   $$
This gets bigger and bigger, because $A> 1.$

With $ s_0 = 5 = A + \frac{1}{A}  $ we find $A = \frac{5 + \sqrt{21}}{2} \approx 4.79.$

If we were to change the $s_0$ to something between $0$ and $2,$  most things would be the same but $A = \frac{s_0 + \sqrt{s_0^2 -4}}{2}$  is now a  complex number of norm exactly one. So $s_n,$ which remains real, oscillates but always has absolute value less that two. Indeed, we have $s_n = 2 \cos \left(2^n T \right)$  where $T = \arctan \frac{\sqrt{4-s_0^2 }}{s_0} $

Title: Prove that $s_{n+1}=s_n^2-2$ diverges.


问:
Sufficient condition for biconnectivity
Assume we have a planar graph  $G$ with $2n+1$ vertices and $3n$ edges, such that it is bipartite $G=A\cup B$ and every vertex from $A$ is of degree $\leq 3$. Does it suffice to conclude that $G$ is a biconnected graph?

or guide me to the conditions from literature so that I can write what I have tried.

答:
Such a graph does not even need to be connected. For example, the disjoint union $K_{4,2} \sqcup K_{5,2}$ satisfies your hypotheses: it's planar, it has $13=2\cdot 6+1$ vertices and $18 = 3\cdot 6$ edges, and if we put the larger half of each component in $A$, then each vertex in $A$ has degree $2$.

(We can add an edge between an $A$-vertex in $K_{4,2}$ and a $B$-vertex in $K_{5,2}$, and delete an old edge, to get a counterexample that's connected but not biconnected.)

Title: Sufficient condition for biconnectivity


答:
If $n$ is odd say $n=2k+1$ then
$$n^2=(2k+1)^2=(2k+1)\times (2k+1)=2\times(k(2k+2))+1\quad\text{is odd}$$
hence necessary $n$ is even</p>

Title: Prove if $n^2$ is even, then $n$ is even.

Body:
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?


问:
Example of Separation of Variables

Seek separable solutions, $u(x,t)=X(x)T(t)$ of the equation, $u_{tt}=c^{2}\dfrac{1}{x}\dfrac{\partial}{\partial x}\left(x\dfrac{\partial u}{\partial x}\right)$. Find the general solution assuming the separation constant is zero.

I am only able to get as far as $X(x)T''(t)=c^{2}\dfrac{1}{x}\dfrac{\partial}{\partial x}\left(x\dfrac{\partial u}{\partial x}\right)$, in which I'm unsure on how to proceed afterwards. When I referred back to an example of applying the separation of variables method to the diffusion equation, it was possible to express the equation $u(x,t)=X(x)T(t)$ in the form $\dfrac{X''(x)}{X(x)}=\dfrac{T'(t)}{DT(t)}=k$ where $D$ is the diffusion constant, and $k$ an arbitrary constant. I wonder if this is also possible for this specific problem, or am I supposed to follow another procedure? I appreciate any pointers I can get on this, thanks!

答:
$$\frac{1}{x} \frac{\partial}{\partial x} \left( x \frac{\partial u}{\partial x} \right) = \frac{1}{x} \left( \frac{\partial u}{\partial x} + x \frac{\partial^2 u}{\partial x^2} \right)$$
Therefore the equation becomes
$$X(x) T''(t) = c^2 \left(\frac{X'(x) T(t)}{x} + X''(x) T(t) \right)$$
You can divide both sides of this equation by $X(x) T(t)$ to separate variables.

Title: Example of Separation of Variables


答:
We have $$c^2\frac{1}{x}\frac{\partial }{\partial x}(x\frac{\partial u}{\partial x})=c^2\frac{1}{x}\big(\frac{\partial u}{\partial x}+x\frac{\partial^2 u}{\partial x^2}\big)=c^2T(t)\frac{1}{x}\big(X'(x)+xX''(x)\big)$$

So the partial differential equation can be rewitten as

$$X(x)T''(t)=c^2T(t)\frac{1}{x}\big(X'(x)+xX''(x)\big)$$

$$\frac{1}{c^2}\frac{T''(t)}{T(t)}=\frac{1}{xX(x)}\big(X'(x)+xX''(x)\big)=K$$

where $K$ is the separation constant. But we assume that $K=0$ so we have

$$T''(t)=0 \space\space\space\text{and}\space\space\space xX''(x)+X'(x)=(xX'(x))'=0$$

That is $T(t)=a+bt$ and $X(x)=c_1+c_2\ln(x)$ for constants $a,b,c_1,c_2$.

Title: Example of Separation of Variables

Body:
Seek separable solutions, $u(x,t)=X(x)T(t)$ of the equation, $u_{tt}=c^{2}\dfrac{1}{x}\dfrac{\partial}{\partial x}\left(x\dfrac{\partial u}{\partial x}\right)$. Find the general solution assuming the separation constant is zero.

I am only able to get as far as $X(x)T''(t)=c^{2}\dfrac{1}{x}\dfrac{\partial}{\partial x}\left(x\dfrac{\partial u}{\partial x}\right)$, in which I'm unsure on how to proceed afterwards. When I referred back to an example of applying the separation of variables method to the diffusion equation, it was possible to express the equation $u(x,t)=X(x)T(t)$ in the form $\dfrac{X''(x)}{X(x)}=\dfrac{T'(t)}{DT(t)}=k$ where $D$ is the diffusion constant, and $k$ an arbitrary constant. I wonder if this is also possible for this specific problem, or am I supposed to follow another procedure? I appreciate any pointers I can get on this, thanks!


问:
For a given $f \in \text{Hom}(V,W)$ there exist bases $B$ and $C$ of $V$ and $W$ such that $M_C^B(f) = \begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix}$
Let $V$ and $W$ be vector spaces and $f \in \text{Hom}(V,W)$ is a homomorphism with $\text{rank}(f) = r.$

Prove that there exist bases $B$ and $C$ of $V$ and $W$, respectively, such that
$$M_C^B(f) = \begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix},$$
where $I_r$ is an $r \times r$ identity matrix.

My attempt: Let $\{u_1, \dots, u_{n-r}\}$ be a basis of $\text{Ker}(f)$ (where $n$ denotes $\text{dim}V$) and add $r$ vectors $\{v_1, \dots, v_r\}$ to form the base for $V$. Denote
$$B = \{u_1^*, \dots, u_n^*\} = \{v_1, \dots, v_r, u_1, \dots, u_{n-r}\}.$$
I thought that this would get me somewhere, but I can't seem to obtain $C$ with this idea.

Another approach: If $\{u_1, \dots, u_n\}$ is some basis for $V$, then there exist $\{u_{i_1}, \dots, u_{i_r}\} \subseteq \{u_1, \dots, u_n\}$ such that $\{f(u_{i_1}), \dots, f(u_{i_r})\}$ is a linear independent set. Renumerate the basis,
$$B = \{u_{i_1}, \dots, u_{i_r}, u_{i_{r+1}}, \dots, u_{i_n}\}$$
where the ordering of last $n-r$ vectors is not important. Then I've managed to get the first part of $C$ and not the second part. Should I choose specific $u_{i_{r+1}}, \dots, u_{i_n}$?

Any hints would be appreciated!

答:
Your first approach will work:

The key is that by construction $(f(v_1),\dots,f(v_r))$ are linearly independent. (Prove it.)

So, $B=(v_1,\dots,v_r,u_1,\dots,u_{n-r})$ is a basis of $V$ with $(u_1,\dots,u_{n-r})$ being a basis of $\ker f$.

And let $C$ be any extension of $(f(v_1),\dots,f(v_r))$ to a basis of $W$.


问:
Integral involving exponentials of cosh functions
I recently encountered the following integral

$$I_p\equiv \int\limits_0^{+\infty}\,\int\limits_{-\infty}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}x\,\mathrm{d}c\,,$$

with $p>0$. Can $I_p$ be calculated in terms of elementary functions?

Update:

Using the following integral representation for the Bessel function

$$
\int\limits_0^{+\infty}e^{-a\cosh x}\;\mathrm{d}x=K_0(a)\,,
$$

one can bring $I_p$ to the following form:

$$
I_p=2 \int\limits_0^{+\infty}e^{-c\,p}\;K_0\left[\left(c^2+1\right) p\right]\;\mathrm{d}c\,.
$$

Which shows that $I_p$ can be seen as a Laplace transform. However, I haven't managed to do the last integral either.

答:
By Fubinis theorem:
$$I_p = \int\limits_0^{+\infty}\,\int\limits_{-\infty}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}x\,\mathrm{d}c =\int\limits_{-\infty}^{+\infty}\,\int\limits_{0}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}c\,\mathrm{d}x.$$
We can solve the integral with respect to $c$ analytically:
$$\int\limits_{0}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}c = -\dfrac{\sqrt{{\pi}}\mathrm{e}^{\frac{p}{4\cosh\left(x\right)}-p\cosh\left(x\right)}\left(\operatorname{erf}\left(\frac{\sqrt{p}}{2\sqrt{\cosh\left(x\right)}}\right)-1\right)}{2\sqrt{p\cosh\left(x\right)}}$$
I am afraid that we can't solve the second integral analytically.


问:
Does the Laplace transform of $f(t)=(1-H_{a} (t)) \exp{[(-1)^b (t-a)^2+(t-a)^{2021}]}$ exist?
Let $a,b\in \mathbb{N}$, $a>b$. Does the Laplace transform of the following exist? $$f(t)=(1-H_{a} (t)) \exp{[(-1)^b (t-a)^2+(t-a)^{2021}]}$$

Should Ι distinguish the cases for $b$?

$b$ is odd

$b$ is even

答:
$$\mathcal{L} (f(t)) = \int_0^a e^{(-1)^b (t - a)^2 + (t - a)^{2021}} \cdot e^{-st} \: dt$$
This is an integral of a continuous function over the closed interval $[0, a]$. So it exists. The fun part is evaluating it in closed form.


问:
Existence of proof of consistency of stronger axiomatic systems
Assume $\mathsf{ZFC}$ is consistent. Are there axioms $\mathsf{P}$ and $\mathsf{Q}$ independent from $\mathsf{ZFC}$ such that?:

$\mathsf{Q}$ is a stronger axiom than $\mathsf{P}$

There is a proof under $\mathsf{ZFC}$ that $\mathsf{ZFC} + \mathsf{P}$ is consistent

There is no proof under $\mathsf{ZFC}$ that $\mathsf{ZFC} + \mathsf{Q}$ is consistent

There is a proof under $\mathsf{ZFC} + \mathsf{P}$ that $\mathsf{ZFC} + \mathsf{Q}$ is consistent

This question arose while developing a new logic system. This logic system deals with undecidable statements by assigning them a truth value other than true and false. They are the elements of a $\sigma$-algebra on $\mathsf{ZFC}$'s house.

<s>If such $\mathsf{P}$ and $\mathsf{Q}$ don't exist, it would imply that all nonempty subhouses (which all have uncountably many residents) have the trivial topology, which is a very disappointing result.</s> Well, it turns out I asked the wrong question. This question is essentially equivalent to verifying that $X = \overline{Y} \land Y = \overline{Z} → X = \overline{Z}$, where overlines mean topological closure. I'll try to formulate the original question I had, but that should be another post.

答:
Well right off the bat, there's a problem: if $\mathsf{ZFC}$ proves that $\mathsf{ZFC}+P$ is consistent, then $\mathsf{ZFC}$ is inconsistent right off the bat per Godel's second incompleteness theorem! I was silly and missed this initially.

We can rephrase this by using conditional consistency claims, as follows:

Can there be $P,Q$ such that:

$\mathsf{ZFC}\vdash Q\rightarrow P$,

$\mathsf{ZFC}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC}+P)$,

$\mathsf{ZFC}+P\vdash Con(\mathsf{ZFC}+P)\rightarrow Con(\mathsf{ZFC}+Q)$, but

$\mathsf{ZFC}\not\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC}+Q)$?

However, this changes the picture drastically. Take $P=\neg Con(\mathsf{ZFC})$. Then bulletpoint $(3)$ holds trivially for every $Q$, because $\mathsf{ZFC}+P\vdash\neg Con(\mathsf{ZFC}+P)$. So just pick any $Q$ such that $(1)$ and $(4)$ hold.


问:
Proof of uniqueness of Pythagorean triples generated by Euclid's formula
How it can be proven that Euclid's formula for generating Pythagorean triples obtains each$$a=m^2-n^2,\,b=2mn,\,c=m^2+n^2$$at unique $m,\,n$?

Thanks in advance.

答:
A right triangle with side lengths $a,b,c$ where $c$ is the hypotenuse and using integers $m, n $ where $ m > n$, we can find Euclid's Formula.

First, given a right triangle with an hypotenuse $\sqrt{c}$ we know that the side lengths must be m and n telling  us  $c = m^2 + n^2$.

Using this we now can find the integer values of a and b.

$(m^2 + n^2)^2 = c^2$

$m^4 + 2m^2n^2 + n^4 = c^2$

$(m - n)^2 + (2mn)^2 = c^2$

Which results in Euclid's formula:

$(2mn, m^2 - n^2, m^2 + n^2)$


问:
MSC of Point Set Topology (Mathematical Subject Classification)
In which lower level class belongs Point Set Topology?

[1]: https://zbmath.org/static/msc2020.pdf

Is it part of MSC03E-Set-theory, or part of MSC-54-General-topology?

Do you have a collection of relating keywords to MSC-classes?

It helps to see the connections.

Thanks in advance.

答:
It depends on what you are doing. I assume by lower level you really mean high level, or general, or 2-digit class. In that case, 54 is general topology, 26 is real functions, 03 is mathematical logic and foundations. "Point-set topology" most likely refers to the stuff in 54, or to the theory of Baire functions, as in 26A21, or to descriptive set theory, as in 03E15. That said, the latter two are also considered in 54 (perhaps with different emphasis).

MathSciNet allows you to search for reviews containing specific words or sentences. If you have access to the database, this may help you identify the appropriate classes you are interested in.


问:
Homotopy groups of non-contractible manifolds
Motivated by a proof in a differential geometry book and so far my lack of knowledge in algebraic topology I would like to know the following :

Is it possible to have a compact non-contractible manifold $M$ with finite fundamental group and trivial homotopy groups $\pi_k(M)$ for $k\geq 2$?

Any help or reference is appreciated, thanks in advance.

答: This is impossible even without assuming compactness (assuming you want $M$ to be connected).  Suppose you had such a manifold $M$.  Then $M$ would be a $K(G,1)$ space for the finite nontrivial group $G=\pi_1(M)$.  However, any nontrivial finite group has infinite cohomological dimension, so $M$ would have nontrivial cohomology (with appropriate coefficients) in infinitely many degrees.  This is a contradiction, since if $M$ is a manifold of dimension $n$ it can't have nontrivial cohomology above degree $n$.


问:
Shifrin Differential Geometry Exercise $1.2.27$ -- A Differential Equation For Bikes

The Question

Suppose the front wheel of a bicycle follows the arclength-parametrized plane curve $\vec{\alpha}$. Determine the path $\vec{\beta}$ of the rear wheel, $1$ unit away. As the hint explains, the goal is a differential equation involving $\theta$, the angle of the front wheel with the axle of the bike, and $\kappa$, the curvature of $\vec{\alpha}$.

This question is very interesting to me, and I haven't seen a solution written up anywhere. This is sort of shocking to me as it seems like it should be a very relevant problem for e.g. autonomous driving.

What I've Tried

I've only been able to make minimal progress.

The hint tells us to write $\vec{\alpha} - \vec{\beta}$ in terms of $\theta$, $\vec{T}$ (i.e. $\vec{\alpha}'$), and $\vec{N}$ (i.e. $\frac{\vec{\alpha}''}{\kappa}$). We obviously have $\| \vec{\alpha} - \vec{\beta}\|^2 = 1$. Differentiating, we obtain
$$
(\vec{\alpha}' - \vec{\beta}') \cdot (\vec{\alpha} - \vec{\beta}) = 0
$$
That is:
$$
(\vec{T} - \vec{\beta}') \cdot (\vec{\alpha} - \vec{\beta}) = 0
$$
Differentiating again, we obtain
$$
(\kappa \vec{N} - \vec{\beta}'') \cdot (\vec{\alpha} - \vec{\beta}) + (\vec{T} - \vec{\beta}') \cdot (\vec{T} - \vec{\beta}') = 0
$$
Now it seems to me we should have
$$
\vec{T} \cdot \vec{\beta}' = \|\vec{\beta}'\| \cos \theta
$$
So we can expand
$$
(\vec{T} - \vec{\beta}') \cdot (\vec{T} - \vec{\beta}') = 1 - 2\|\beta'\| \cos\theta + \|\vec{\beta}'\|^2
$$
And that's as far as I've gotten. It seems like I'm going about this all wrong. In particular, I have no idea what to do with the derivatives of $\vec{\beta}$. If only $\vec{\beta}$ were arclength parametrized I feel like I could make some progress, but I don't think there is any reason it should be. The only thing I can think is that we should have
$$
\vec{\alpha} - \vec{\beta} = \lambda \vec{\beta}'
$$
for some $\lambda$ that could depend on the arclength of $\vec{\alpha}$. I didn't push too far in this direction, though, since it required introducing yet another unknown.

I thought of yet another line of attack. Since $\|\vec{\alpha} - \vec{\beta}\| = 1$, we can say that $\|\vec{\alpha} - \vec{\beta}\|$ is just $\vec{T}$ rotated by $\theta$, i.e.
$$
\vec{\alpha} - \vec{\beta} = \cos (\theta) \vec{T} + \sin(\theta) \vec{N}
$$
With this expression I've gone as far as the hint suggests, but I don't see what to do next.

What am I missing?

If this post summons Ted Shifrin, and he'd rather answers to his textbook questions not be given out, I'd be happy to delete this question and post it as a reference request instead. I really am shocked I haven't been able to find this problem written about anywhere. I'm guessing it's because I'm bad at searching the literature, not because it actually hasn't been written about.

答:
With help from Ted Shifrin, I was able to solve the problem. Here's a sketch of the solution.

The first observation is that $\|\vec{\alpha} - \vec{\beta}\| = 1$ since the bike has length $1$. Since $\vec{\alpha}$ is arc-length parametrized, $\|\vec{T}\| = \|\vec{\alpha}\| = 1$, so $\vec{T}$ and $\vec{\alpha} - \vec{\beta}$ must be related by a rotation. The angle of this rotation is exactly the angle $\theta$ that the front wheel makes with the frame of the bike (up to a sign depending on how you define things). Applying the rotation to $\vec{T}$ gives you an expression for $\vec{\alpha} - \vec{\beta}$ in the $\{\vec{T} , \vec{N}\}$ basis in terms of $\theta$. Differentiate this expression for $\vec{\alpha}- \vec{\beta}$ to obtain an expression for $\vec{\beta}'$ in the $\{\vec{T}, \vec{N}\}$ basis in terms of $\theta$ and $\kappa$.

On the other hand, $\vec{\beta}'$ must point in the same direction as $\vec{\alpha} - \vec{\beta}$ since the rear wheel is fixed to point in the same direction as the frame of the bike. This allows us to write
$$
\vec{\beta}' = \lambda(\vec{\alpha} - \vec{\beta})
$$
from which it easily follows that
$$
\lambda = \vec{\beta}' \cdot (\vec{\alpha} - \vec{\beta})
$$
We have expressions for both vectors on the RHS in the $\{\vec{T}, \vec{N}\}$ basis, so we can easily calculate this dot product in terms of $\theta$ and $\kappa$.

Knowing $\lambda$, we have yet another expression for $\vec{\beta}'$. Since $\{\vec{T}, \vec{N}\}$ is a basis, we may equate the components of each expression. After just a little algebra, you should obtain
$$
\kappa = \theta ' + \sin\theta
$$
an expression for $\kappa$ entirely in terms of $\theta$ and its derivatives.


问:
Compact map of a closed convex set
Question: $X$ is a $B^*$ space and $C$ is a closed convex set of $X$. Let $T: C\to C$ be a compact map (defined below). Prove that $T(C)$ is sequentially pre-compact (That is, any sequence in $T(C)$ has a Cauchy subsequence).

(The real goal is to show that $T$ has a fixed point in $C$, but showing $T(C)$ is sequentially pre-compact is good enough because of Schauder's fixed point theorem)

Compact map: Given $X$ a $B^*$ space and $E \subset X$, Let $T:E \to X$. We call $T$ a compact map if $T$ is continuous and for all $K\subset E$ with $K$ bounded, we have that $T(K)$ is sequentially compact. (That is, every sequence in $T(K)$ has a convergent subsequence with limits in $T(K)$.).

My attempt: If $C$ has only one element there is nothing to prove. So assume there are at least two element in $C$. I am able to show that the following is true,
$$\Lambda =  \bigcup_{\substack{K \subseteq C\\ K \text{ bounded}}}K = C$$
(The forward inclusion is trivial. For the reverse inclusion, if we assume $x \in C\setminus\Lambda$, since there is some other $y$ also in $C$; we can look at the convex hull of $\{x,y\} \subseteq C$ which is totally bounded and therefore bounded and thus is $\subseteq \Lambda$, which puts $x\in \Lambda$ which is a contradiction!)

This also gives me
$$T(C) = \bigcup_{\substack{K \subseteq C\\ K \text{ bounded}}} T(K)$$

But I am sort of lost here trying to show that any sequence in $T(C)$ has a Cauchy subsequence. Intuitively, since the union is arbitrary, it seems to me that it should be possible to create a sequence such that for any $N \in \mathbb{N}$ and for any $K \subseteq C$ bounded, there is some $k > N$ such that $x_k \not\in T(K)$.

Correction/Edit: The question (which originally appeared in my homework set) has been updated to include the additional hypothesis that $C$ is bounded.

My solution: Now the entire question is trivial since we have the bounded subset $C\subseteq C$ such that $T(C)$ is sequentially compact. Finally, in order to use Schauder's fixed point theorem, we need the set to be complete; but this is achieved easily. First we notice that the sequential compactness of $T(C)$ ensures the compact-ness of $T(C)$. So if we consider a Cauchy sequence $\{x_k\} \subset C$ that doesn't coverge in $C$, by the continuity of $T$, we obtain a Cauchy sequence $\{T(x_k)\} \subset T(C)$ that does not converge in $T(C)$ - but this contradicts the compactness of $T(C)$! So, $C$ is complete and we can use Schauder; and so we win!

Further question: I am still curious to know if the statement still holds if we remove the bounded assumption on $C$. According to the answer by @Kavi Rama Murthy, it does look like we don't always have sequential pre-compactness of $T(C)$ in this case; but the question about the fixed point itself remains elusive (as noted in my reply comment).

答:
The definition of a compact map has been revised after I posted this answer.

$T(C)$ need not be sequentially compact: Let $C=X$ and $Tx=f(x)x_0$ where $x_0$ is a fixed nonzero elmeent of $X$ and $f$ is a fixed non-zero continuous linear functional on $X$. Then the hypothesis is satisfed but $T(C)$ is the span of $x_0$ which is not pre-compact.

Take $X=C=\mathbb R$ and $Tx=x+1$ to see that $T$ may not have any fixed point.

Title: Compact map of a closed convex set


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
Of course, you can always use the fact that if $p$ is a prime which divides a product $ab$ ,then $p | a$ or $p |b$. But, I'm not sure if you have this result, and the question suggests that you don't. If you want to prove it, you can argue along the following lines--if you know Bezout's identity.

Suppose that $p$ divides $ab$ ,but not $a$. Then, it folows that $gcd(a,p)=1$. Hence, we may express $1=np+ma$. Or $b=nbp+mba$. Now, since we know that $p|ab$, we have that $ab=\alpha p$, and so $a=\frac{\alpha p}{b}$. Hence, by substitution, we have that
$b=nbp+m\alpha p=p(nb+\alpha m)$. Hence, b is divisible by p.

For this particular problem, a=b=n.

Note, that if you do not know Bezout's identity, then you will have no idea how I was able to write 1 as a linear combo of p and a. First note that 1 is the gcd of $p$ and $a$. Hence, we can obtain one by iterating the Euclidian algorithm again and again. Then, we just backwards substitute from the last equation upwards to get the gcd as a linear combo of the two numbers involved. Try it with 5 and 17, for instance. This is Bezout's idenitity.


问:
What is the solution of limit $\lim_{x\to\infty} \frac{x}{x} $?
What is the solution of this limit: $$\lim_{x\to\infty} \frac{x}{x} $$ I'm sure it's pretty basic but somehow I can't figure out how to solve it? Is it just $1$?

答:
Yes. You can just divide
$$\lim_{x\to \infty}\frac{x}{x}=\lim_{x\to\infty}1.$$
The limit of a constant is always the constant itself:
$$\lim_{x\to p}\mathrm{const.}=\mathrm{const.}$$


问:
How to find matrix of linear map sample of vectors and their projection?
I would like to find matrix of linear transformation $f$, which projects vectors like this:

$$f((x, y, z)^T)= (a, b, c, d, e)^T$$

How Can I find this matrix ? I am confused that vectors have different amount of elements.

Can you provide please minimal explanation / proof why your solution works ?

Thank you for help

答:
As @Lukas suggested the matrix can be found using the bases, or direct computation. Let's assume that the matrix that we are looking for is M then, M is 5x3 (since it transforms a 3x1 vector to 5x1) and let assume that

$$
M=\begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix}
$$

where $c_i$ is the columns of M and each of them is 5x1

$$
M \begin{pmatrix} 1 \\ 0 \\0 \\ \end{pmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 0 \\0 \\ \end{pmatrix} = c_1 = \begin{pmatrix} 3 \\ 5 \\7 \\ 9 \\ 6\end{pmatrix}
$$

$$
M \begin{pmatrix} 1 \\ 2 \\0 \\ \end{pmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 2 \\0 \\ \end{pmatrix} = c_1 + 2 c_2 =\begin{pmatrix} 2 \\ 3 \\4 \\ 5 \\ 1\end{pmatrix}
$$

Hence,

$$
c_2 = 1/2\left( \begin{pmatrix} 2 \\ 3 \\4 \\ 5 \\ 1\end{pmatrix} -  \begin{pmatrix} 3 \\ 5 \\7 \\ 9 \\ 6\end{pmatrix}\right) = \begin{pmatrix} -.5 \\ -1 \\-1.5 \\ -2 \\ -2.5\end{pmatrix}
$$

$$
M \begin{pmatrix} 1 \\ 2 \\3 \\ \end{pmatrix} = \begin{bmatrix} c_1 & c_2 & c_3 \\ \end{bmatrix} \begin{pmatrix} 1 \\ 2 \\3 \\ \end{pmatrix} =c_1 + 2c_2 + 3c_3 = \begin{pmatrix} 1\\2 \\ 3 \\4 \\ 5 \end{pmatrix}
$$

Hence,

$$
c_3 = 1/3\left( \begin{pmatrix} 1 \\ 2 \\ 3 \\4 \\ 5 \end{pmatrix} -  \begin{pmatrix} 2 \\ 3 \\4 \\ 5 \\ 1\end{pmatrix}\right) = \begin{pmatrix} -1/3 \\ -1/3 \\-1/3 \\ -1/3 \\ 4/3\end{pmatrix}
$$


问:
Collecting math websites
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.

答:
ProofWiki is an online collection of proofs that I have found to be useful at times. Each page contains links to definitions and lemmas that are used in the proof, as well as a list of external sources on the topic. More importantly, there is a list of math jokes ;)

Another useful site is Wolfram MathWorld. It contains articles on many topics in mathematics, with links to related topics and references. It too has a few jokes.

Title: Collecting math websites


问:
Formula to get the exact number of primes below $n$
Total noob question so please bear with me.

There seems to be several formulae which estimate the number of primes below $n$, but none which give the exact number (short of computing the primes directly).

Is that because it has been proven that such a formula does not exist? Or because it hasn't been discovered yet?

ie. A formula akin to area = πr² that gives an instant answer rather than using ∑

Thanks!

答:
In other words - Is there an explicit analytical formula for the prime-counting function $\pi(n)$?

Yes there are many. If it's about computation time one would use boundaries, resulting from the prime number theorem.

If it's about easyness one may use Wilsons theorem to get to an anaytical expression or other theorems, for example
$$\pi(n) = \sum_{j=1}^{n}\left\lfloor\cos^2\pi\frac{(n-1)!+1}{n}\right\rfloor,$$
where $\lfloor\cdot\rfloor$ is the floor function. Basically the whole function inside the sum is an indicator function, that returns $1$ if $n$ is prime and $0$ otherwise.

If it's about beauty, than it's this formula:

$$\pi(x)=R(x)-\sum_{\rho}R(x^\rho),$$
where $$R(x) := \sum_{n=1}^{\infty}\frac{\mu(n)}{n}\operatorname{li}(x^{1/n}),$$
$\mu$ the Möbius-function, $\operatorname{li}$ the logarithmic integral and $\rho$ are all the zeros of the $\zeta$-function.


问:
What happens if we remove requirement for cofinite vanishing in simplical chains
Let $\mathcal{K}$ be a simplical complex. Define (simplical) $n$-chains as maps from $n$-simplices in $\mathcal{K}$ to $\mathbb{Z}$ such that they vanish cofinitely many times. We then get a basis for a free abelian group "for free" since every chain is a sum of characteristic functions over the support for some coefficients and this sum is guaranteed to be finite by the cofinite vanishing condition.

What if we remove this requirement? Could we fix it by using $\mathbb{Z}_2$? I can't find anything about it and all the references I looked at (Munkres, Hatcher, Rotman, Hocking & Young) do it either with this requirement or even define them as finite formal sums straight away. I verified that $\partial$ behaves as expected. Where does this approach exactly fail, apart from that it doesn't benefit from "niceness" of having a free abelian group as the output.

答:
I think there's not any interesting mathematical content here... more that things simply won't work right for not-cofinitely-vanishing.

For example, consider a $0$-simplex $z_o$ with (countably) infinitely-many $1$-oriented simplices $t_1, t_2, t_3, ...$ having boundaries $\partial t_i=z_o-z_i$ with $0$-simplices $z_i$. (Can do this over $\mathbb Z/2$, as well.) Then the boundary of $\sum_i t_i$ is $\sum_{i=1}^\infty z_o-\sum_{j\not=0} z_j$. The latter sum might be ok, but what could the former mean? Doing things mod $2$ does not seem to help.

So, unless we enlarge the possible meanings of these sums (which is not out of the question), it's simply unclear what the meaning would be, in the context of boundaries and so on.


问:
4 Color Theorem - What am I not seeing??
Let me say first that I am in no way a mathematician. Just slightly interested in mathematics. I think I may have found an exception to the 4 color theorem.
I don't claim to be smarter than those who proved the theorem, and I'll assume I'm wrong.
What am I missing here?
(forgive my quickly drawn graph!)

(I created this account just to ask this question, so the website made me embed the image instead of just adding it straight to the question.)

I may be missing some fundamental rule in setting up my problem. But I think my map fits the desired intent of the "map" motif of the theorems original question.

答: I've put numbers instead of colours (I left yours because whatever).


答: Here is your map colored with only four colours.

Title: 4 Color Theorem - What am I not seeing??

Body:
Let me say first that I am in no way a mathematician. Just slightly interested in mathematics. I think I may have found an exception to the 4 color theorem.
I don't claim to be smarter than those who proved the theorem, and I'll assume I'm wrong.
What am I missing here?
(forgive my quickly drawn graph!)

(I created this account just to ask this question, so the website made me embed the image instead of just adding it straight to the question.)

I may be missing some fundamental rule in setting up my problem. But I think my map fits the desired intent of the "map" motif of the theorems original question.


问:
Chain rule for functional derivates with the same variables
I am following an online course where we should sometimes compute functional derivatives. The scope of the course is density functional theory, so we have the density $n(r)$ and we have functionals of the form $F[n(r)]$.

$$n(r) = \sum_i \Psi_i^*(r)\Psi_i(r)$$

In the course we need to perform derivatives with respect to $\Psi_i^*(r)$ not with respect to $n(r)$. One quick result is that:

$$\frac{\delta n(r)}{\delta \Psi_i^*(r)} = \frac{\delta \Psi_i^*(r)}{\delta \Psi_i^*(r)} \Psi_i(r) = \Psi_i(r)$$

One of the students used the chain rules for functional derivatives:

$$\frac{\delta F[n(r)]}{\delta \Psi_i^*(r')} = \int \frac{\delta F[n(r)]}{\delta n(r)}\frac{\delta n(r)}{\delta \Psi_i^*(r')}dr = \int \frac{\delta F[n(r)]}{\delta n(r)}\delta(r - r')\Psi_i(r) dr = \frac{\delta F[n(r)]}{\delta n(r')}\Psi_i(r')$$

Which is a really helpful result because it allows to compute derivatives directly with respect to $n(r)$ and still get what we want by just multiplying by $\Psi_i(r')$.

Now the problem I have, is that, in this line he introduced a new variable $r'$ while it shouldn't, the derivative should be with one variable $r$ everywhere, and I believe that this is not valid when the variables are the same (the integral doesn't vanish.).

What I did is that I used the functional derivative definition of composed function $F[g(f(r))]$ and adapted to each functional I needed to derive which lead to the same results with more work

While I acknowledge that the first solution is faster, I wouldn't be able to find it by myself because it doesn't make so much sense to me.

Any thoughts?

答:
Some of your expressions are problematic, if not meaningless, and maybe that's at the heart of your confusion. Fair warning: I would use the mainstream definition of the functional derivative used in physics, very often not math, as, e.g. in Largangian mechanics. Drop the different indices i, and the summation over them, since they don't enter the problem and just contribute extraneous functions which act like numerical coefficients in ordinary calculus.

You then pretend n(r) itself is a functional(which depends on some "parameter" r, even though not on the argument r' entering in the variation of the function you are about to differentiate it with respect to!), which you may differentiate unambiguously, to obtain the canonical expression,
$$  n(r)= \int\!\! dr' ~n(r') \delta(r'-r) ~~~\leadsto  \\
\frac{\delta n(r)}{\delta  \Psi^*(r')} = \int\!\! dr' ~  \Psi^*(r') \Psi(r') \delta(r'-r)   = \Psi (r')  \delta(r-r'),  \tag{1}$$
a function of r and r'.

Now, switch gears and think of n(r) as a function, again, entering in a functional F[n]. Since the functional  F does not depend on the argument of this function, we skip the "dummy" argument r of it.  Then, the functional derivative of it is the following  function of r',

$$\frac{\delta F[n]}{\delta \Psi^*(r')} = \int \frac{\delta F[n]}{\delta n(r)}\frac{\delta n(r)}{\delta \Psi^*(r')}dr = \int \frac{\delta F[n]}{\delta n(r)}\delta(r - r')\Psi(r') dr = \frac{\delta F[n]}{\delta n(r')}\Psi(r')~.\tag{2}$$
You might think of this as the directional functional derivative of F in the direction of $\frac{\delta n(r)}{\delta \Psi^*(r')}$.

Test-drive it with an overly silly easy functional, such as
$$
F[n]=  \int\!\! dr   ~ n(r)^5 =  \int\!\! dr   ~ \Psi^*(r)^5 \Psi(r) ^5 .
$$
Directly, you have the function
$$
\frac{\delta F[n]}{\delta \Psi^*(r')} = 5\Psi^*(r')^4 \Psi(r') ^5 ,
$$
but, by use of (2), you also get the same function
$$
\frac{\delta F[n]}{\delta \Psi^*(r')} = \int\!\!dr~ 5n(r)^4  \delta(r - r') \Psi(r')  =   5\Psi^*(r')^4 \Psi(r') ^5.
$$

When in doubt, use finite-dimensional vectors, and take  vector derivatives
of scalar functions thereof, and only at the very end take the number of components j to go to infinity and utilize a continuous variable r for them.
So, (2) is something like $$\partial F(y)/\partial x_i^* =\sum_j \partial F /\partial y_i~~ \partial y_j/\partial x_i^* = \sum_j \partial F /\partial y_i~~   x_i \delta_{ij} ,$$ which might illustrate the necessity of two variables before their ultimate merge.


问:
How to prove statistical hypothesis?
I developed a caching method. I took 100 experiments and got that hit ratio is not less than 75%. Now, I want to prove that my method with some probability gives hit ratio not less than 75%. How should I make this?</p>

答:
You can calculate the chance of a type I or type II error in your hypothesis, which is essentially what you are looking for, the probability that your hypothesis is incorrect:

If you are running a Bernoulli trial, which I'm assuming you are, your hypothesis is that p>=0.75 where P(x=0)=1-p P(x=1)=p

type I error occurs when your hypothesis was actually correct but you rejected it based on your expirements. Since the criteria you are using to determine if you reject or accept the hypothesis is whether or not your expirements gave a successful result in over 75% of the time, the chance for type I error is:

P(sum(1 through 100)(x)<75 | p>0.75)

type II error occurs when your hypothesis was incorrect, but you accepted it:

P(sum(1 through 100)(x)>75 | p<0.75)

The sum of the results of Bernoulli trials have a binomial distribution, so you can calculate these probabilities using the CDF/PDF of the binomial distribution, which can be found here:

http://en.wikipedia.org/wiki/Binomial_distribution

Title: How to prove statistical hypothesis?


问:
Show whether the vector projection is idempotent
A projection is a mapping of a mathematical structure into a substructure such that the mapping is idempotent. In a familiar sense, 'structures' could be sets, topological spaces, metric spaces, vector spaces, inner product spaces, groups, graphs, hypergraphs, ..., and whatever else.

I'm interested in a classic of linear algebra: vector projection.

$$\vec{a_1} = \frac{\vec{a} \cdot \vec{a}}{||\vec{b}||^2}\vec{b} = \frac{\vec{a} \cdot \vec{a}}{\vec{b} \cdot \vec{b}}\vec{b} = \left( \vec{a} \cdot \hat{b}\right) \hat{b}$$

It is natural to me that a projection of two vectors in $\mathbb{R}$ is a vector within the original space. What confused me is how to think about the projection as idempotent. The operation maps two vectors into one vector, so how could I apply the projection a second time if the second application of the map needs two vectors but the last projection only 'produced' one?

Perhaps I've confused two uses of the word 'projection', or that I'm just looking at the composition of functions the wrong way. Please show whether the vector projection is a projection in the sense defined above.

答:
You're right, vector projection is a special case of "abstract" projection:

Let $V$ be a vector space. In linear algebra, a linear map $P: V \to V$ is a projection if it satisfies the equation $P^2 = P$. (This is your first usage of the word.)

Now, fix a vector $\mathbf b \in V$ and consider the linear map $\text{proj}_{\mathbf b} : V \to V$ given by the formula $\mathbf a \mapsto \text{proj}_{\mathbf b} \mathbf a$. In your question you were considering the case of one vectors undergoing projection; here, we're projecting the entire space onto $\mathbf b$. We now calculate that this map is in fact a projection, i.e. that it's idempotent:
$$\text{proj}_{\mathbf b}(\text{proj}_{\mathbf b} \mathbf a) = (\text{proj}_{\mathbf b} \mathbf a \cdot \hat{\mathbf b})\hat{\mathbf b} = [(\mathbf a \cdot \hat{\mathbf b})\hat{\mathbf b}\cdot \hat{\mathbf b}]\hat{\mathbf b} = (\mathbf a \cdot \hat{\mathbf b})(\hat{\mathbf b} \cdot \hat{\mathbf b})\hat{\mathbf b} = (\mathbf a \cdot \hat{\mathbf b})\hat{\mathbf b} = \text{proj}_{\mathbf b} \mathbf a. $$

As a bonus, you could choose a basis for $V$ and write down the matrix $P$ of $\text{proj}_{\mathbf b}$ in that basis. You'll find that $P^2 = P$ holds.


问:
Every affine k-scheme can be embedded into an affine space?
While reading through Algebraic Geometry I by Görtz and Wedhorn (2nd version) I came across the following remark on p. 352 (between Proposition 12.66 and Corollary 12.67):

If $Y$ is an affine $k$-scheme, we can embed it into affine space $\mathbb{A}^{(I)}_k$ ($I$ some index set).

This remark is not further explained. Why is it true? For me it seems non-trivial, if $Y$ is not of finite type.

答: An affine $k$-scheme $X$ is given by the spectrum of a $k$-algebra $A$. We can pick generators $\{a_i\}_{i\in I}$ for $A$ as a $k$-algebra, which gives a surjective morphism from the polynomial ring $k[x_i]_{i\in I}\to A$ given by sending $x_i\mapsto a_i$. The spectrum of this map is exactly the closed immersion of $X$ in to $\Bbb A^{(I)}_k$. This proof does not care whether $I$ is finite or not.


问:
Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$?

Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$, where $\mathfrak{ p,q,r,s}$ are cardinal numbers?

Is it possible to prove in $\mathsf{ZFC}$ that there is a counterexample?

答:
Andres Caicedo asked whether or not we can prove in $\sf ZF$ that there exists a counterexample (i.e. a situation where inequality holds in the assumption, but the exponents are equal).

In $\sf ZFC$ we know that there exists such counterexample, so let us assume $\sf ZF+\lnot AC$. Let $P$ be a set which cannot be well-ordered, and let $\newcommand{\fp}{\mathfrak p}\fp$ denote its cardinal. We make the following assumptions:

$\fp^\omega=\fp$, otherwise replace $P$ by $P^\omega$. From this assumption we can conclude that: $\fp=\fp+\fp=\fp\cdot\fp$, and from those we can deduce that $\fp^\fp=2^\fp$.

If $\kappa$ is the least ordinal not less or equal than $\fp$, then $\kappa<2^\fp$. If this is not true we can replace $\fp$ by $2^\fp$ (because $\kappa$ is the least ordinal with that property for $2^\fp$ as well) or by $2^{2^{\fp}}$ if needed. Since we know that $\kappa<2^{2^{2^\fp}}$, one of the three options must have the wanted property.

Note that the above properties are preserved by taking powers, so replacing $\fp$ by its power set (once or twice) would not change the first assumption.

Other important properties following from the first property are: $$2^\fp=(2^\fp)^\fp=2^\fp+2^\fp=2^\fp\cdot2^\fp.$$

We know by a lemma of Tarski that if $\lambda$ is an $\aleph$ and $\mathfrak m$ is a cardinal such that $\frak\lambda+m=\lambda\cdot m$, then $\lambda$ and $\frak m$ are comparable. Because we took $\kappa$ to be incomparable with $\fp$ we know that $\fp+\kappa<\fp\cdot\kappa$.

Note that $\fp(\fp+\kappa)=\kappa(\fp+\kappa)=(\fp+\kappa)(\fp\cdot\kappa)=\fp\cdot\kappa$ by the properties of $\fp$ and $\kappa$.

Case I: $2^\kappa\nleq2^\fp$.

We observe that $2^\fp<2^{\fp+\kappa}$, and now we calculate:
$$\begin{align}
  &(2^\fp)^{\fp+\kappa}=2^{\fp(\fp+\kappa)}=2^{\fp\cdot\kappa}&\tag{1}\\
  &(2^{\fp+\kappa})^{\fp\cdot\kappa}=2^{(\fp+\kappa)\fp\cdot\kappa}=2^{\fp\cdot\kappa}&\tag{2}
  \end{align}$$
And it is not hard to see that the inequalities are satisfied, but the exponentiation ends up equal.

Case II: $2^\kappa\leq2^\fp$.

We note that $2^\fp=2^{\fp+\kappa}$, and we make the following calculations:
$$\begin{align}
  & 2^\fp=2^{\fp+\kappa}\leq\fp^{\fp+\kappa}\leq(2^\fp)^{\fp+\kappa}=2^{\fp\cdot\kappa}=(2^\kappa)^\fp\leq(2^\fp)^\fp=2^\fp&\tag{3}\\
  & 2^\fp\leq\kappa^\fp\leq(2^\fp)^\fp=2^\fp &\tag{4}
  \end{align}$$
And in this case we see that $\kappa<2^\fp$ and that $\fp<\fp+\kappa$, but the exponentiation is again equal.


问:
Constants in the sigma notation for sums
This really is a question about the semantics of the sigma notation for writing long sums in a concise way. Let's say we have a sum given by the following notation,

$$\sum_{i = 0}^{n} (\frac{1}{n} + i^2)$$

My question now is rather simple. For a given $n$, let's say $n = 3$, would this sum written out look like this,

$$\sum_{i = 0}^{3} (\frac{1}{3} + i^2) = (\frac{1}{3} + 0^2) + (\frac{1}{3} + 1^2) + (\frac{1}{3} + 2^2) + (\frac{1}{3} + 3^2)$$

or like this,

$$\sum_{i = 0}^{3} (\frac{1}{3} + i^2) = \frac{1}{3} + 0^2 + 1^2 + 2^2 + 3^2$$

Basically, my question is, whether the constant term $\frac{1}{n}$ is only counted once in the sum or for every iteration. I realize this may seem like a stupid or obvious question to some, which is why its so difficult finding anything related to this specific thing.

答:
As summation is linear, we can actually think of the summation of a sum as a sum of summations: $$\sum\left(a+b\right)=\sum a+\sum b.$$

For your summation, we would be able to write $$\sum_{i=0}^n\left(\frac1n+i^2\right)=\sum_{i=0}^n\frac1n+\sum_{i=0}^ni^2.$$

As such, for $n=3$, we have $$\sum_{i=0}^3\frac13+\sum_{i=0}^3i^2=\underbrace{\left(\frac13+\frac13+\frac13+\frac13\right)}_{\text{one for each index number}}+\left(0^2+1^2+2^2+3^2\right)=\frac43+14=\frac{46}3.$$


问:
$\lim_{\overline{x} \to \infty} \left [ 1 + Ae^{-\overline{x}} + (1 - A)e^{\overline{x}}\right ] = \lim_{x \to 0}\left [ e^{x}\right] \implies A = 1$?
Why does $\lim_{\overline{x} \to \infty} \left [ 1 + Ae^{-\overline{x}} + (1 - A)e^{\overline{x}}\right ] = \lim_{x \to 0}\left [ e^{x}\right] \implies A = 1$ ?

My book says:

where $$y_0 = e^{x}$$
and the notation $Y_0(\infty)$ mean $lim_{\overline{x} \to \infty}Y_0$ and $y_0(0)$ means $\lim_{x \to 0}y_0$.

How is that from the above equation of limits they deduce that $A = 1$? I get that the right hand side limit is equal to $1$, but the limit on the left goes to infinity because of the $e^{\overline{x}}$ term. Can someone please explain what is going on?

答: Taking limit for each term we get $1+0+\lim_{\overline x \to \infty} (1-A)e^{\overline x}=1$. So $\lim_{\overline x \to \infty} (1-A)e^{\overline x}=0$.  This also implies $\lim_{\overline x \to \infty} |1-A|e^{\overline x}=0$. If $A \neq 1$ then the limit is  $+\infty$ leading to a contradiction. Hence, $A=1$.


问:
How do I prove that $x,y$ and $z$ lie on the same line if they are related to the same parameter?
Related to this question

How exactly does that work? I would be grateful for a proof but barring that just the name of the theorem, so I can look it up myself.

My work/thoughts so far:

If we start with a system like the one below:

$$y+az=b$$
$$x+fz=g$$

I guess I could solve both sides for z and get something like:

$$z=\frac{b-y}{a}$$
$$z=\frac{g-x}{f}$$

Which I guess implies that:
$$\frac{b-y}{a}=\frac{g-x}{f} $$
$$y=\frac{a(g-x)}{f}+b $$

Which is really just another way of writing $y=kx+m$, so for every value of z, the relationship between x and y can be written as a line. Since there is an infinite number of z, that means we can vizualize all the possible lines as an infinite plane.

If we could define a vectpr z we could restrict the possible solutions to the values in the plane that satisfiy both $y=\frac{a(g-x)}{f}+b $ and the equation for z. It just so happens we have two expressions for z, $z=\frac{b-y}{a}$ and $z=\frac{g-x}{f}$.

I dont' know, where do I go from here?

答:
I’m awfully sorry, but your question is imprecise, especially since I don’t see mention of any parameter anywhere in the title or body. So my attempt at an answer may not satisfy you at all. Nonetheless:

It seems to me that $x$, $y$, and $z$ are to be the coordinates of points in $3$-space $\Bbb R^3$. And that $a$, $b$, $f$, and $g$ are parameters which define the problem. You pretty clearly are of the same opinion.

Furthermore, you have correctly solved for $z$ in the two cases, $z=-\frac1ay+\frac ba$, $z=-\frac1fx+\frac gf\,$. These say, separately, that a choice of $x$ and $y$ gives you $z$. But not necessarily the same $z$ in the two cases.
For instance, say $x=1$ and $y=2$. Then you get, in the first case, the point $(1,2,-\frac2a+\frac ba$) and in the second case $(1,2,\frac1f+\frac gf)$. As you see, the $z$-values need not be the same.

There’s where you went wrong, when you said, “Which I guess implies that…”

You want to think of the situation geometrically, where the first equation defines a particular plane in space (depending on the values of $a$ and $b$), and the second equation defines some other plane in space. (They will not be parallel planes in this case, because in the two cases, the vectors perpendicular are $(0,1,a)$ and $(1,0,f)$, definitely not parallel.) Since the planes are not parallel, they have a line in common. (If your geometric intuition is still in its infancy, think of the east wall of your room and the south wall, which intersect in a vertical line, namely the SE seam between the two walls.)

If you want a parametric representation of this line, the work is done for you. You have the three parametric formulas, in terms of a parameter $t$:

\begin{align}
x&=g-ft\\
y&=b-at\\
z&=t
\end{align}


问:
Proving a set is open iff its complement is closed from scratch
We have that a set $F$ is closed if every limit point of $F$ is a point of $F$ and we say that $p$ is a limit point of a set $E$ is every neighborhood $N$ of $p$ contains a point $q \neq p$. In other words, if $N \cap E \neq \varnothing$.

${\bf TASK:}$ Prove that $U$ is open iff $U^c$ is closed.

TRY: Let $U$ be a open set. Let $x$ be a limit point of $U^c$. We need to show that $x \in U^c$. If $x \in U$, then there is a neighboorhood $N$ of $U$ such that $N \cap U = \varnothing$ so that $x$ is not limit point. Therefore, $x$ better be in $U^c$, as desired.

On the other hand, say $U^c$ is closed. Let $x \in U$. We must find a neighborhood of $x$ that is contained in $U$.

If we can show that $x$ is not  limit point of $U^c$, then we will be able to have a neightborhood $N$ that is completely outside $U^c$ (that is, inside of $U$). So if $x$ is a limit point of $U^c$, then $x \in U^c$ by assumption and this contradiction proves that $x$ is not a limit point of $U^c$.

Is this a correct proof?

Thanks

答:
Your last paragraph isn't quite correct.

By assumption, $U^c$ is closed, so if $x \in U$, then because $U^c$ contains all of its limit points and $x \notin U^c$, we have that $x$ is not a limit point of $U^c$.  That means there is some neighborhood $N$ of $x$ that is disjoint from $U^c$, and a neighborhood that is disjoint from $U^c$ is necessarily contained within $U$.  Thus, $N$ is an open neighborhood of $x$ that is contained within $U$, and since $x$ was an arbitrary point of $U$, that means that $U$ is open.


问:
A question on logic - where intuition can fail
Suppose I have two predicates $P(x)$ and $Q(x)$, such that $\overline{P(x)\wedge Q(x)}$ holds for all $x$.

Now, if $\displaystyle \bigwedge_{x\in A}P(x)$ for a set $A$, it must be certainly true, that $\displaystyle \overline{\bigwedge_{x\in A}Q(x)}$ by intuition. But suppose, that $A = \{\;\;\}$, then obviously both $\displaystyle \bigwedge_{x\in A}P(x)$ and $\displaystyle\bigwedge_{x\in A}Q(x)$.

Where is my mistake? I guess it is in the assumption that
$$\bigwedge_x\overline{P(x)\wedge Q(x)} \rightarrow \left(\bigwedge_{x\in A}P(x) \rightarrow \overline{\bigwedge_{x\in A}Q(x)}\right)?$$

*Note: $\displaystyle \overline{P(x)\wedge Q(x)} \equiv \lnot(P(x)\land Q(x))$

$\displaystyle \quad\quad\quad\quad \overline{\bigwedge_{x\in A}Q(x)} \equiv \lnot \left(\bigwedge_{x\in A}Q(x)\right)$

答:
When I TA'd a course which covered similar topics we specifically required for the structure to be nonempty.

This is in order to avoid the very problem that you refer to in your question.

In fact, the assumption that you made is very true if you require the structure to have a nonempty universe. Let us prove it quickly:

Suppose that for all $x\in A$ we have $\overline{P(x)\land Q(x)}$, and that for all $x\in A,\ P(x)$ holds.

Let $x$ be an arbitrary element of $A$, then $\overline{P(x)\land Q(x)}$ which by DeMorgan is equivalent to $\overline{P(x)}\lor\overline{Q(x)}$. We assume that $P(x)$ is also true, therefore it must be that $\overline{Q(x)}$.

Since this $x$ was arbitrary it holds for all $x\in A$, i.e. $\bigwedge_{x\in A}\overline{Q(x)}$.


问:
A numerical coincidence with continued fractions
My brother built a garage that measures 45 feet by 30 feet.  To make sure the right angles were accurate, he measured the two diagonals of the rectangle to see that they were equal.  In  inches,
\begin{align}
& \sqrt{540^2+360^2} \approx 648.999229548\text{ inches} \\[6pt]
= {} & 54\text{ feet}+1\text{ inch} - \text{less than $0.001$ inches}.
\end{align}
It's a bit odd to come within a thousandth of an inch when rounding to the nearest inch, <s>but there's more:
$$
\sqrt{540^2+360^2} = 649 - \cfrac{1}{1298+\cfrac{1}{24073+\cdots}}.
$$
One part in twenty-four thousand??</s>

Did I just happen to be there when someone rolled boxcars two dozen times in a row, or is there more to be said?

(Maybe I should add that $540^2+360^2 = 649^2-1$.)

PS:
$$
\sqrt{540^2+360^2} = 648 + \cfrac{1}{1+\cfrac{1}{1297+\cfrac{1}{25700+\cdots}}} \\
\text{(This part is mistaken; see below.)}
$$

Later edit: A calculator gave me the results above repeatedly; later another calculator disagreed, just as persistently, and I figured out what the truth is.

答:
Since $540^2+360^2+1^2=649^2$, we have $\sqrt{540^2+360^2} = \sqrt{649^2-1}$, and notice that since $2\cdot649=1298$, we get that
$$-649+\sqrt{649^2-1}=\frac{-1298+\sqrt{1298^2-4}}{2}\tag{1}$$
is a solution to
$$
x^2+1298x+1=0.
$$
Rearrange the quadratic equation:
$$
x = \frac{-1}{1298+x}
$$
and then substituting the expression on the right for $x$ within that very expression gives us
$$
x=\cfrac{-1}{1298-\cfrac{1}{1298-\cfrac{1}{1298-\cfrac{1}{1298-\cdots\cdots\cdots}}}}
$$
and from $(1)$ we have
$$
\sqrt{649^2-1} = 649+x.
$$
That's ONE WAY OF LOOKING AT IT, and at one level it explains it and at another it doesn't.  But it proves that this expansion is right.</p>


问:
Continuous function related to group of homeomorphisms acting properly discontinuously must be a homeomorphism
I'm trying to solve the following exercise:

Let $X$ be a connected space and $G$ a properly discontinuous group of homeomorphisms of $X$. Suppose that a continous map $f: X \to X$ has the following property: for every $x \in X$, there exists $g \in G$ such that $f(x) = gx$. Prove that $f$ is a homeomorphism.

I was able to show $f$ must be injective: if $f(x) = gx = hy = f(y)$ then $x = g^{-1}hy$, and since the action is properly discontinuous, it follows that $g = h$ and hence that $x = y$. I know I also have to show $f$ is surjective, takes open sets to open sets and the inverse image of open sets via $f$ is open, but I've had a hard time with that. I'd appreciate any help!

答:
Since you have shown that $f$ is injective, it is enough to show that $f$ is a local homeomorphism and surjective.

Let $x\in X$, and $g$ such that $f(x)=g(x)$. There exists an open subset $U$ containing $x$ such that for every $h\in G$, the intersection $h(U)\cap U$ is not empty implies that $h=e$ the neutral element since the action is properly discontinuous. Write $V=g(U)$. Suppose there exists $h\in G, y\in U$ such that $h(y)\in V$, we have $g^{-1}h(y)\in U$ implies that $g^{-1}h=e$ and $g=h$.

The set $\{y\in X:f(y)=g(x)\}$ is closed, the previous argument shows that it is open since $X$ is connected, it is $X$.

$f^{-1}(V)\cap U=W$ is an open subset. Let $z\in W$, $f(z)=g_z(z), g_z\in G$, since $z\in W\subset U, g_z(z)\in V$, the previous remark implies that $g_z=g$. We deduce that the restriction of $f$ and $g$ to $W$ coincide and $f$ is a local homeomorphism.


问:
What is the Haar measure on the unit sphere?
I try to understand the proof of Lemma 4.2. in the paper 'The Euler equations as a differential inclusion' by De Lellis and Székelyhidi. In the proof they use the Haar measure on the unit sphere $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$. How is the Haar measure defined here?

答:
See Section 2.1 in these lecture notes and especially Remark 2.6. One can construct the Haar measure as the normalized area of a subset of $\mathbb{S}^{n-1}$.

To randomly choose a point from a distribution that is uniform on $\mathbb{S}^{n-1}$ according to the Haar measure, it suffices to take $n$ normally-distributed random variables $(X_1,\cdots,X_n)$ and choose the point to correspond to the normalized vector $$\frac{(X_1,\cdots,X_n)}{\sqrt{\sum_{i=1}^n X_i^2}}$$


问:
Trying to figure out formula to find the angle to rotate a pivot so that an attached point with a direction is '"facing" a third point
Sorry for the title, I am finding it hard to describe the problem I am trying to solve...

I am trying to figure out the angle I need to rotate a point in 2d space which is currently "facing" another point in the distance so that another point which is offset from the pivot position can now face the point in the distance. Hopefully the below image explains this better than words:

I am trying to figure out angle (?) which the pivot (a) has to rotate by so that point (c) is now facing point (b).

答:
You're almost there. Here's the right-hand side of your diagram with a few added markings:

The labels $a$ and $b$ are restored to the pivot position and the point in the distance;
the label $x$ is restored to the segment whose length represents the offset distance, and the label $z$ is restored so that it indicates the distance from $a$ to $b$.

Note that the line $ab$ crosses two parallel lines and therefore makes the same angle $B$ with each line.
The angle $B$ at the distant point $b$ is one angle of a right triangle whose hypotenuse is $z$ and whose opposite leg is $x$;
therefore
$$ \sin B = \frac xz. $$

This can be inverted by using the arc sine function (the inverse of the sine function):

$$ B = \arcsin\left(\frac xz\right) . $$

Now as we have already noted, both angles labeled $B$ have the same angular measure;
the angle $B$ between the lines meeting at $a$ is the angle you want.

Note that the distance $y$ in the original figure is irrelevant as long as $y^2 < z^2 - x^2.$
If $y^2 = z^2 - x^2$ then the circle passes through $b$ and if
$y^2 > z^2 - x^2$ then $b$ is inside the circle, so it is not possible in those cases for $c$ to "face" $b$ as shown in the figure.


问:
Dartboard paradox and understanding independence

By definition, events $A$ and $B$ are independent if $$P(A \cap
    B) = P(A)\:P(B).$$ Thus if an event $A$ happens almost never
$\left(P(A)=0\right),$ then
$A$ is independent of all events, including itself. So, hitting the
exact centre of a dartboard (happens almost never) and hitting within its inner ring are independent events.

On the other hand, it is standard to characterise
independence as follows:

Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

Since hitting the exact centre of a dartboard guarantees hitting
within its inner ring (which is otherwise not guaranteed), the
two events are dependent.

Isn't this a contradiction?? If yes, then do we accept that the definition of independence is not meant to fully correspond to its verbal/intuitive characterisation?

Is there any semantic difference between the following two versions?

$(i)$ Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

$(ii)$ Two events are independent if the occurrence of one does not
affect the occurrence of the other.

ADDENDUM

On further pondering, I have resolved both questions:

2.

Consider this experiment: flip two fair coins, letting $H_1$ be the event that the first coin lands on Heads, and $X$ be the event that the coins land on different sides. Then $$ P\left(H_1 \cap X\right)=\frac14=P(H_1)\:P(X);$$ i.e., $H_1$ and $X$ are independent events.

Knowledge that $H_1$ happens reduces the possible number of ways that $X$ can eventuate—from $2$ (outcomes HT and TH) to $1$ (outcome HT)—but does not change the probability $\left(\frac12\right)$ of $X.$

1.

The following revision characterises pairwise independence more clearly and accurately:

Let $P(A)\neq0.$

Events $A$ and $B$ are independent iff. the knowledge that $A$ happens doesn't change $B$'s probability.

In this informal characterisation, almost-never events are now excluded from being conditioned on. (What does it even mean to say that an almost-never event has happened: in what sense have I hit the exact centre of a dartboard?) It motivates the definition of pairwise independence, which does allow both events to be impossible.

答:
Events $A$ and $B$ are indeed defined to be independent if $P[AB] = P[A]P[B]$.  This definition holds for all cases, including cases when $P[A]=0$ and/or $P[B]=0$. The definition leads to your desired interpretation "knowing if $B$ occurred does not affect the likelihood of $A$"  in the special case $P[B]>0$, since it means $P[A|B] = P[A]$ in that case. It does not allow your interpretation in the case when $P[B]=0$, since we cannot condition on a zero-probability event.  So indeed, there are peculiarities when $P[B]=0$ since intuition fails in that case. Nevertheless, your intuition is still useful since it applies when $P[B]>0$.

It may also be helpful to observe that if $P[B]=0$ then $P[A|B^c] = P[A]$, which is intuitively what we would want when $A$ and $B^c$ are independent.

A situation about darts similar to the one you posed can be addressed by the definitions of conditioning with respect to a random variable. For a dartboard, we can imagine $(X,Y)$ uniformly distributed over a 2-d ball of radius 1.  Then define $R = \sqrt{X^2+Y^2}$ as the distance from the center.  Consider the events:
$$ A = \{R \leq 1/10\}, B = \{R = 1/10\}$$
Now $P[B]=0$ so $A$ and $B$ are independent.  Nevertheless this is similar to the situation you describe because $B \subseteq A$, so "intuitively" we would want $P[A|B]=1$. This holds if we use:
$$ P[A|R=r] = \frac{f_{R|A}(r)P[A]}{f_R(r)} $$
which shall be applied when $f_R$ and $f_{R|A}$ are continuous at point $r$ and $f_R(r)\neq 0$,
and we observe:
\begin{align}
f_R(r) &= \left\{\begin{array}{cc}
2r  &  \mbox{ if $r \in [0,1]$}\\
0 & \mbox{else}
\end{array}\right.\\
f_{R|A}(r) &= \left\{\begin{array}{cc}
\frac{2r}{P[A]} & \mbox{ if $r \in [0,1/10]$} \\
0 & \mbox{ else}
\end{array}\right.
\end{align}
so
$$ P[A|R=r] = \left\{\begin{array}{cc}
0 & \mbox{if $1/10 < r<1$} \\
1 & \mbox{ if $0<r\leq 1/10$}
\end{array}\right.$$

Our expression $P[A|R=r]$ cannot be viewed as an instance of the definition of conditional probability $P[A|B]$ for events $A$ and $B$, since the event $\{R=r\}$ has probability 0.  Rather, it is an expression related to the definition of conditioning on a random variable $R$. Formally, you can say the random variable $Z$ defined by
$$ Z = \left\{\begin{array}{cc}
0 & \mbox{if $R>1/10$} \\
1 & \mbox{ if $0\leq R \leq 1/10$}
\end{array}\right.$$
is a "version" of the conditional expectation $E[1_A|R]$.  As such, it is almost surely the same as any other version. Thus, while this version takes the liberty to define $Z=1$ when $R=0$, and $Z=0$ when $R\geq 1$, this is fine because it only affects a set of measure 0.


问:
Square root of $8^3$
I'm only in intermediate algebra. I know that $\sqrt{8^3}$ is equal to $16\sqrt{2}$ but could you simply explain the process on how to get to that? </p>

答:
Using the standard rules of algebra, we compute:

$$\sqrt{8^{3}} = \sqrt{8^2 \cdot8} = \sqrt{8^2}\cdot\sqrt{8} = 8\cdot\sqrt{4\cdot2} = 8\cdot\sqrt{2^2}\cdot\sqrt{2} = 16\sqrt{2}$$

Title: Square root of $8^3$


问:
Triple quotient product of a triangle in the Nine Point Circle
Show that the 9-point Circle of a triangle satisfies the cyclic ordered product of segments

$$ \dfrac{BP}{PC}\times \dfrac{CQ} {QA}\times \dfrac{AR}{RB}=1 $$

Show that it could be included in a generalization Menelaus's theorem as a special case of a proposition where the product of ratios is any real number.

Context was yesterday's  attempt to find  such triple ratio product locus with my following logic for this particular case.

The circle has equal segments on its sides,  satisfying the generalization

At foot of perpendicular through Ortho-center the opposite sides are partitioned as:

$$ \dfrac{a \cos B}{b \cos A}\cdot\dfrac{b \cos C}{c \cos B}\cdot \dfrac{c \cos A}{a \cos C} =1$$

which also satisfies the generalization.

答: This has nothing to do with the nine-point circle. The three points in the sides are the feet of the altitudes. The altitidues are concurrent and, as such, your triple product is a special case of Ceva's theorem.</p>


问:
Linear Algebra Done Right: Notation 1.23
I recently started reading Linear Algebra Done Right by Axler, and I find it great up until Notation 1.23, where the first bullet point states:

If $S$ is a set then $F^S$ denotes the set of functions from $S$ to $F$

I would really like to know what this means; in other words, does this mean that for all $f \in S$, then $f \in F$, but the only difference is that $F$ is a vector field?

答: It means that if $f\in \mathbb{F} ^S$ then $f$ is a function from the set $S$ to a field (think $\mathbb{R}$ or $\mathbb{C}$ for example), $f:S\to\mathbb{F} $. Also, $\mathbb{F} ^S$, with the sum (sum of functions) and product by scalar (scalar times function) as defined in 1.23 is a vector space.


问:
Response of a specific LTI system to an exponential
I am taking a Signals & Systems course in my Telecommunications Engineering degree, and now I am trying to put together everything I have seen.
I came across this basic problem of determining the output of a system whose transfer function is $H(s)=\frac{1}{s+2}$ when it is fed with a signal $x(t)=e^{2jt}$ (where $j^2=-1$).
As every other student of S&S I know that exponentials are eigenfunctions to LTI systems, hence I tried this
$$y(t)=H(s_0)e^{s_0t}=H(2j)e^{2jt}=\frac{1}{2j+2}e^{2jt}=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}e^{2jt}$$
"Nice!" I said. But then I wanted to test the convolution theorem, and so I happily did
$$Y(s) = H(s)X(s) = \frac{1}{s+2}\frac{1}{s-2j}=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(\frac{1}{s-2j}-\frac{1}{s+2}\right)$$
Seeing this, and that its time-domain counterpart would be $y(t)=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(e^{2jt}-e^{-2t}\right)$ I thought I made a mistake, but Wolfram Alpha spat the same. Where does the $e^{-2t}$ term come from?. My first idea was that maybe I was messing it up with the ROC's or something, but all the present signals are right-sided.

Then I tried the convolution with the impulse response, which is precisely $h(t)=\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}$.
$$(x\ast h)(t)=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau=\int_{0}^{\infty}e^{-2\tau}e^{2j(t-\tau)}d\tau=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}e^{2jt}$$
I calmed down a bit when I saw that something matched what was predicted by the eigenfunction property.
After this I attempted the same by applying the commutativity of the convolution:
$$(h\ast x)(t)=\int_{0}^{\infty}e^{2j\tau}e^{-2(t-\tau)}d\tau$$
And it diverges, but I found out that
$$(h\ast x)(t)=\int_{0}^{t}e^{2j\tau}e^{-2(t-\tau)}d\tau=\int_{0}^{t}e^{-2\tau}e^{2j(t-\tau)}d\tau=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(e^{2jt}-e^{-2t}\right)$$

I must admit that I do not quite understand the difference between integrating from $0$ to $\infty$ and from $0$ to $t$ in the convolution, and therefore I am unable to explain why with the integration interval $[0,\infty)$ the convolution gives the same as the inverse Laplace transform. And I definitely cannot understand why all this methods do not give the same answer.

On the other hand, I tried feeding the system with $e^{-2t}$. The convolution with the interval of integration $[0,\infty)$ is divergent, but the following is certainly true:
$$\int_{0}^{t}e^{-2\tau}e^{-2(t-\tau)}d\tau = te^{-2t} = \mathcal{L}^{-1}\left\{\frac{1}{(s+2)^2}\right\}$$
Beside, a (very simple) "tool" I created in geogebra which shows graphically the convolution between two signals showed the same function (and it has worked very well for every pair of real signals I have tried).

But I don't understand how to apply the property of eigenfunction here since $H(-2)=\frac{1}{0}$.

May someone help me understand this?

答:
There is no problem with the computations above, but there is definitely a misunderstanding in the way they are interpreted.

When one considers the response $\Psi(s)=H(s)X(s)$ of the system , you are essentially saying that the system is satisfying a differential equation with the unperturbed initial condition $y(0)=0$, in your case the equation is $y'(t)+2y(t)=x(t)~,~ y(0)=0$. Laplace transforming this equation gives the exact expression above. The solution you have found is correct and satisfies this problem, and you can check that explicitly.

On the other hand saying that the solution of this problem is $x(t)/\lambda$ where $\lambda$ is the eigenvalue of the function $x(t)$ with respect to the linear operator $T=\frac{d}{dx}+2$ is NOT correct. What this function represents is a particular  solution to the inhomogeneous ODE, and in no way a complete solution to the problem.  In your case it can be shown that the full solution to the ODE mentioned above is

$$y(t)=Ce^{-2t}+\frac{e^{2jt}}{2j+2}\equiv y_0(t)+y_p(t)~~,~C\in\mathbb{R}$$

where $C$ is an arbitrary constant which is determined by the initial conditions. The only way $y_p(t)$ can be a solution by itself is if there is an initial condition that sets $C=0$. In this case no real number can achieve this, so the inhomogeneous solution will always be accompanied by some sort of term $\propto e^{-2t}$.

As far as the convolution integrals are concerned, since all signals are right-sided one can show that the following three expressions are equivalent:

$$\Psi(s)=H(s)X(s)\iff y(t)=\int_0^{t}dt' h(t')x(t-t')\iff y'(t)+2y(t)=x(t)~,y(0)=0$$

which you have shown to be true. Exchanging the order of $h,x$ does not change anything.

However, the interpretation of the same convolution integral from $(0,\infty)$ is completely different. To obtain that integral you secretly assume that $x(t)$ is not right-sided, and instead has preexisted for all times $t\in \mathbb{R}$. This means that the system started getting kicked in the infinite past, which means that it must have reached the equilibrium solution at any finite time (all transients have died since the system is causal) and that is the reason why you obtain the equilibrium solution when you perform the integral in $(0, \infty)$ instead. Indeed you can demonstrate this using the ODE representation:

$$y'(t)+2y(t)=x(t)~,~ y(t_0)=y_0 \Rightarrow y(t)=e^{2t_0}\left(y_0-\frac{e^{2jt_0}}{2j+2}\right)e^{-2t}+\frac{e^{2jt}}{2j+2}$$
Taking the limit $t_0\to -\infty$ requires the transients to vanish
$$\lim_{t_0\to-\infty}y(t)=\frac{e^{2jt}}{2j+2}$$

and only the particular solution remains. It is not fair to ask both convolution integrals to give you the same answer, because they are designed to answer different problems. The method of particular solutions-eigenfunctions yields the long-time behavior of a causal system, the Laplace transform is useful for studying the full response of a system.

Finally, in the last example the method of eigenfunctions simply does not apply, because $e^{-2t}$ is a solution to the homogeneous ODE $y'+2y=0$, which doesn't allow the application of the method. Regardless of that subtlety, if we follow the same procedure we used above, we see that in this problem the initial condition $y(0)=0$ sets the arbitrary constant to zero, and thus renders the two solutions equal, but this is a coincidence owing to the fact that the particular solution $y_p(t)=t e^{-2t}$ has $y_p(0)=0$.