问:
Finding the focus of an ellipse with ruler and compass

Given an ellipse E, find his focus with ruler and compass.

I tried to generalize the next theorem:

Given a circle C find his center. The idea for prove this is with an chord and it's bisector line (the ortogonal line in the midle point of the chord)

Ortogonality is fundamental for this theorem but in the ellipse ortogonality doesn't work.

Inspired by the equations and the linear algebra, an ellipse is a circle transformed by a linear transformation, but there are infinite linear transformations...the chord determines a linear transformation if we know the center of the elipse...

答:
Draw two parallel chords $AB$ and $CD$ and let $M$, $N$ be their midpoints. Line $MN$ intersects the ellipse at the endpoints of a diameter $PQ$, whose midpoint $O$ is the center of the ellipse. The diameter through $O$ parallel to $AB$ is the conjugate of $PQ$.

Finally, having a pair of conjugate diameters, you can construct the axes as explained in this old answer of mine and from them find the foci.


问:
Let $(x_n)$ and $(y_n)$ be sequences such that $(x_n)$ is a subsequence of $(y_n)$ and $(y_n)$ is a subsequence of $(x_n)$. $x_n=y_n$ for all $n$?

Let $(x_n)$ and $(y_n)$ be sequences such that $(x_n)$ is a subsequence of $(y_n)$ and
$(y_n)$ is a subsequence of $(x_n)$. Given that $x_n$ converges, does it follow that $x_n=y_n$ for all $n$?

I strongly suspect this is true. I have an idea but I'm not convinced its super tight.

By our subsequence fact, let $x_{a_n}=y_n$ and $y_{b_n}=x_n$. Since we have to fit these terms in the sequence and they are all in order, we must have $a_n\ge n$ and $b_n \ge n$ with equality if and only if the first $n-1$ terms are identical.
Assume BWOC $\exists k: x_k\neq y_k$ with $k$ minimal. Well $x_{a_k}=y_k$ with $a_k\gt k$, but $y_{b_{a_k}}=x_{a_k}, b_{a_k}\gt a_k \ (>k)$. In other words, $x_k \neq y_k \implies \text{$y_k$ appears later in the sequence ($x_n$)}\implies \text{another $y_k$ appears later in the sequence $(y_n)$}$ and this keeps going on back and forth. Since $(x_n)$ converges and $y_k$ appears infinitely many times in $(x_n)$, it must converge to $y_k$. But, by the same argument, we can show it converges to $x_k$. Since $x_k\neq y_k$, this is a contradiction.

After writing this out, I feel like my idea is right but I just have this doubt in my execution that I can't shake. Is my argument correct or is the proposition false after all?

答:
Your statement and the idea for the proof is correct. I would write is as follows, using the notation $a(n)$, $b(n)$, ... instead of $a_n$, $b_n$, ... to avoid deeply nested indices:

$(x_n)$ is a subsequence of $(y_n)$, so there is a strictly increasing function $a: \Bbb N \to \Bbb N$ with $x_{a(n)} = y_n$ for all $n$.

Similarly, $(y_n)$ is a subsequence of $(x_n)$, so there is a strictly  increasing function $b: \Bbb N \to \Bbb N$ with $y_{b(n)} = x_n$ for all $n$. In particular, $x_{a(b(n))} = x_n$ for all $n$.

Assume that the sequences are not identical, and define $k = \min \{ n \mid x_n \ne y_n \}$. Then $a(k) > k$ and $b(k) > k$, which implies $a(n) > n$ and $b(n) > n$ for all $n \ge k$.

Now define two functions $c, d: \Bbb N \to \Bbb N$ recursively as
$$
 c(0) = k \, , \, c(j+1) = a(b(c(j))) \, ,\\
 d(0) = a(k) \, , \, d(j+1) = a(b(d(j))) \, .
$$
Both functions are strictly increasing. Then
$$
 x_k = x_{c(0)} = x_{c(1)} = x_{c(2)} = \cdots
$$
and
$$
 y_k = x_{d(0)} = x_{d(1)} = x_{d(2)} = \cdots
$$
Since $(x_n)$ is convergent, all its subsequences have the same limit, and it follows that
$$
 x_k = \lim_{n \to \infty} x_n = y_k \, .
$$
That is a contradiction to the construction of $k$.

ContentLicense: CC BY-SA 4.0


问:
Does this proof make sense(The well-ordering principle for N)?
I need to explain the proof given by Des McHale in his "The well-ordering principle for N". I was hoping you could tell me if it makes sense and what else I could add to it. This is what I have so far:

"
In his paper, Des MacHale talks about the Well-Ordering Principle for $\mathbb{N}$. This is a different method to solving proofs. He says that this is a less talked about tool in the mathematicians belt, but is useful to know. He goes on to say that it "appears to be a more natural approach than induction." Using the Well Ordering Principle, also known as WOP, he goes on to solve and prove many theorems. One such theorem that he proves is that $\sqrt2\notin\mathbb{N}$. He uses WOP as a proof for this. In this paper we will attempt to use WOP as well to prove that $\sqrt3\notin\mathbb{Q}$. To put this in English, we want to prove that the square root of three is not a rational number.

Similar to the theorem proving that $\sqrt{2}$ is irrational, to prove that $\sqrt{3}$ is also irrational, it needs to be proven that there is no pair of positive integers (a,b) where $(a/b)^2 = 3$.

Set S can be a set of pairs of integers (a,b) where $(a/b)^2 = 3$

If S is non-empty, according to the Well Ordering Principle, S has to have at least one element c = a + b where $a^2 = 3b^2$.

When $a^2$ is divided by 3, and if $a = 3k + 1$ is odd, $a^2 = 9k^2 + 6k + 1$ which is odd, creating a contradiction.

Following, $a = 3k$ is even where $k\in\mathbb{N}$.

Next, $(3k)^2 = 9k^2 = 3b^2$ so $b^2 = 3k$ is even.

Following our argument before, $b = 2s$ is even too where $s\in\mathbb{N}$.

Because $a^2 = 3b^2$, we have $(3k)^2 = 3(2s)^2$ and then $k^2 = 2s^2$.

However, $a + b = 2(k+s)$, so $k + s < a + b$ which is a contradiction.

This contradiction shows that the set S is empty and therefore there is no pair of positive integers (a,b) where $(a/b)^2 = 3$.

This proves that $\sqrt{3}$ is irrational.

"

Thanks in advance for any help.

答:
I’m afraid that the argument has a number of problems. First of all, it does not in fact use the well-ordering principle at all. The well-ordering principle says that every non-empty set of non-negative integers has a smallest element; your $S$ is not a set of integers at all. But there are quite a few other problems, so I’ll go through it in detail.

Set S can be a set of pairs of integers (a,b) where $(a/b)^2 = 3$

If S is non-empty, according to the Well Ordering Principle, S has to have at least one element c = a + b where $a^2 = 3b^2$.

You defined $S$ to be a set of pairs of integers, and now you’re saying that a single integer $c$ is an element of $S$; this is impossible. You are actually dealing with two sets here. Let $P$ be the set of ordered pairs $\langle a,b\rangle$ of positive integers such that $(a/b)^2=3$. Then let $S=\{a+b:\langle a,b\rangle\in P\}$; now $S$ really is a set of positive integers, so we can apply the well-ordering principle to it. Specifically, if $P$ is not empty, then clearly $S$ is non-empty, and we can let $c$ be the smallest element of $S$. The definition of $S$ then ensures that there is a pair $\langle a,b\rangle\in P$ such that $c=a+b$.

When $a^2$ is divided by 3, and if $a = 3k + 1$ is odd, $a^2 = 9k^2 + 6k + 1$ which is odd, creating a contradiction.

This does not make sense as English. A sentence that starts When $a^2$ is divided by $3$ should continue by telling us what happens when $a^2$ is divided by $3$; yours does not. What you should be saying here is that since $a^2=3b^2$, we know that $a^2$ is divisible by $3$. What happens when we try to divide $a$ by $3$? There are $3$ possibilities: $a$ is divisible by $3$, in which case there is a positive integer $k$ such that $a=3k$; there is a remainder of $1$ when we divide $a$ by $3$, in which case $a=3k+1$ for some positive integer $k$; or there is a remainder of $2$ when we divide $a$ by $3$, in which case $a=3k+2$ for some positive integer $k$. Since your goal is to show that $a$ is a multiple of $3$, you need to eliminate both of the last two possibilities, not just one of them.

Whether or not $3k+1$ is odd depends entirely on $k$: $3k+1$ is odd if $k$ is even, and even if $k$ is odd. I suspect that you were following the proof for $\sqrt2$ without paying enough attention to what is actually going on. The even and odd cases in that proof correspond to the three cases $a=3k$, $a=3k+1$, and $a=3k+2$ in this proof.

What you mean is that if $a=3k+1$, then $$a^2=9k^2+6k+1=3(3k^2+2k)+1\,,$$ which is not divisible by $3$; this is impossible, so $a$ does not leave a remainder of $1$ when divided by $3$. Similarly, if $a=3k+2$, then

$$a^2=9k^2+12k+4=3(3k^2+4k+1)+1\,;$$

this is not divisible by $3$, so this case can also be ruled out, and we conclude that $a=3k$ for some positive integer $k$:

Following, $a = 3k$ is even where $k\in\mathbb{N}$.

Again, this has nothing to do with being even or odd; the point is that $a$ is divisible by $3$.

Next, $(3k)^2 = 9k^2 = 3b^2$ so $b^2 = 3k$ is even.

This has the same problem: $b^2$ may or may not be even, since it is even if and only if $k$ is even, and we know nothing about $k$. The point here is that $b^2$ is divisible by $3$, not by $2$ as in the proof for $\sqrt2$.

Following our argument before, $b = 2s$ is even too where $s\in\mathbb{N}$.

And here you’ve simply copied from the proof for $\sqrt2$ in a way that makes no sense and does you no good.

We just showed that if $a^2$ is divisible by $3$, so is $a$. It is that argument that we want to use again, this time to show that since $b^2$ is divisible by $3$, so is $b$. We’ve done it once, with $a$, so we can certainly do it again with $b$ to show that $b=\color{red}3s$ (not $2s$) for some positive integer $s$: it’s just a matter of changing every $a$ to $b$ and every $k$ to $s$. This is clear enough that we don’t need to write out the details.

Because $a^2 = 3b^2$, we have $(3k)^2 = 3(2s)^2$ and then $k^2 = 2s^2$.

No, because $a^2=3b^2$ we have $(3k)^2=3(\color{red}3s)^2$, and hence $k^2=\color{red}3s^2$.

However, $a + b = 2(k+s)$, so $k + s < a + b$ which is a contradiction.

Here again you’re copying from the $\sqrt2$ without paying enough attention to the algebra that you’re doing. In fact $a+b=\color{red}3k+\color{red}3s=\color{red}3(k+s)$. Of course this still lets you conclude that $k+s<a+b=c$. The pair $\langle k,s\rangle$ is in $P$, so $k+s\in S$, but we’ve just shown that $k+s<c$, where $c$ is the smallest member of $S$. This is plainly impossible, so we conclude that $S$ is empty, and therefore $P$ is also empty — i.e., $\sqrt3$ is irrational.


问:
Inner product on a vector space in a real-valued interval
If we have a vector space $V = C([a,b])$ where $ a,b \in \mathbb{R}$ and and $V$ contains all real-valued continuous functions, then I know that $
\langle f,g\rangle = \int_a^b f(t)g(t) dt$ is an inner product given that both f and g are in V. But my question is would $\langle f,g\rangle_2 = \int_a^c f(t)g(t) dt$ be and inner product given that $c \leq b$.

For example, if $V = C([0,1])$, then I know that $\langle f,g\rangle = \int_0^1 f(t)g(t) dt$ is an inner product of $V$. But would $\langle f,g\rangle_2 = \int_0^{1/2} f(t)g(t) dt$ be an inner product on $V$? My instinct is telling me that it is an inner product, but not on $V$ (I don't know if that even makes sense) but I don't know how to justify my answer.

Any help is appreciated! Thanks.

答: No it won't be, because $$\int_0^{1/2}f^2=0,$$ won't implies that $f=0$ on $[0,1]$. However, it will be a semi-inner product.


问:
find a group with the property :
a)find a nontrivial  group $G$ such that $G$ is isomorphic to $G \times G $

what i'm sure is that $G$ must be infinite ! but have now idea how to get or construct such group
i chose many $G$'s but all of the homomorphism was not injective

b) an infinite group in which every element has finite order but for each positive integer n there is an element of order n

the group  $G = (Z_1 \times Z_2 \times Z_3 \times Z_4 \times ...) $ satisfies the conditions except the one which says that every element have finite order .

how can we use this group to reach the asked group ?

答: I think for b you can consider $G=\mathbb Z(p^{\infty})$ where $p$ is a prime as well.</p>

ContentLicense: CC BY-SA 3.0

Title: find a group with the property :


问:
Matrix transformation $f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$
I am self-learning Linear Algebra (mainly through Khan Academy and Youtube). I encountered this linear transformation, and was wondering how would this be solved.

$f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$

$f \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $  = $ \begin{pmatrix}
4 & 1 & 3\\
2 &-1 & 3\\
2 & 2 & 0
\end{pmatrix} $ $\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $

Establish $x, y$ and $z$ such that,

$f\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $ = $\begin{pmatrix}
4\\
5\\
-1
\end{pmatrix} $

Do I just need to multiply the values of $f$ for the $3\times3$ matrix? What does this mean overall?

答:
You are given that $\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4 & 1 & 3 \\ 2 & -1 & 3 \\ 2 & 2 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}$.

Of course, $\begin{pmatrix}4 & 1 & 3 \\ 2 & -1 & 3 \\ 2 & 2 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4x+ y+ 3z \\ 2x- y+ 3z \\ 2x+ 2y \end{pmatrix}$.

So we have $\begin{pmatrix}x \\ y \\ z \end{pmatrix}= \begin{pmatrix}4x+ y+ 3z \\ 2x- y+ 3z \\ 2x+ 2y \end{pmatrix}$.

That gives us three equations:

$$x= 4x+ y+ 3z\quad\text{which reduces to}\quad 3x+ y+ 3z= 0.$$
$$y= 2x- y+ 3z\quad\text{which reduces to}\quad 2x- 2y+ 3z= 0.$$
$$z= 2x+ 2y\quad\text{which reduces to}\quad 2x+ 2y- z= 0.$$

An obvious solution to that is $x= y= z= 0.$
Notice that the determinant is

$$\begin{align*}
\left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & -2 & 3 \\ 2 & 2 & -1 \end{array}\right|&= 3\left|\begin{array}{cc} -2 & 3 \\ 2 & -1\end{array}\right|- \left|\begin{array}{cc}2 & 3 \\ 2 & -1\end{array}\right|+ 3\left|\begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array}\right|\\[5pt]
&= 3(2- 6)- (-2- 6)+ 3(4+ 4)\\[5pt]
&= -12+ 8+ 24\\[5pt]
&= 20\\[5pt]
\end{align*}$$

Since that is not $0, x= y= z= 0$  is the only solution.


问:
How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$
let $a,b,c\in \mathbb{R}$, if such

$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$

Does this problem has nice methods?

My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks

答:
One approach uses factoring. Define rational functions
$$ R_1 := \frac1{b c-a a} + \frac1{c a-b b} + \frac1{a b-c c}, \tag{1}$$

$$ R_2 := \frac{a}{(b c-a a)^2} + \frac{b}{(c a-b b)^2} + \frac{c}{(a b-c c)^2}, \tag{2}$$
and polynomials
$$ P_1:=a+b+c,\qquad \qquad P_2:=a b+b c+c a,\\ P_3:=P_2\!-\!a a\!-\!b b\!-\!c c,\quad
P_4:=a b c P_1\!-\!a a b b\!-\!b b c c\!-\!c c a a,
\tag{3} $$
and product of polynomials
$$ D := (b c-a a)(c a-b b)(a b-c c). \tag{3} $$
The factorization of $\,R_1,R_2\,$ are
$$ R_1 = \frac{P_2P_3}D, \qquad R_2 = \frac{P_1 P_2 P_3 P_4}{D^2}. \tag{4} $$
The result is that $\,R_1 = 0\,$ if $\,P_2P_3=0\,$ and this also implies
that $\,R_2 = 0.$


问:
Trouble understanding Munkres' proof of the Urysohn metrization theorem
My trouble seems to be related to this question.

To summarize what I understand so far:

We're given a regular (Hausdorff) second-countable space $X$, and choose a countable base $\{U_n\}_{n\in\mathbb{N}}$ for the topology on $X$ such that for all open sets $U\subseteq X$ and all $x\in U$, there exists $n\in\mathbb{N}$ such that $x\in\overline{U_n}\subseteq U$.

Regular second-countable spaces are normal, allowing us to use Urysohn's lemma. For each pair of integers $m, n$ such that $\overline{U_m}\subseteq U_n$, choose a continuous function $f_{m,n} : X\to [0, 1]$ such that $f_{m,n}|_{\overline{U}_m}\equiv 1$ and $f_{m,n}|_{X\backslash U_n}\equiv 0$.

Let $F : X\to [0, 1]^{\mathbb{N}\times\mathbb{N}}\cong [0,1]^{\mathbb{N}}$ take $x\mapsto (f_{m,n}(x))_{m,n\in\mathbb{N}}$ (ranging only over the indices $m, n$ which are applicable; I've skipped the reindexing step because I prefer the $f_{m,n}$ notation). This function $F$ is a continuous bijection.

Now Munkres (supposedly) proves that $F$ is an open map. Letting $U\subseteq X$ be open, and taking an arbitrary $z_0\in F(U)$, we need to find an open set $W$ such that $z_0\in W\subset F(U)$.

To do this, Munkres first chooses $x_0$ such that $F(x_0) = z_0$, chooses indices $m, n$ such that $x_0\in \overline{U_m}\subseteq U_n\subseteq U$, so that $f_{m,n}(x_0) = 1$ and $f_{m,n}(X\backslash U_n)=\{0\}$. Next we take $V = \pi_{m,n}^{-1}((0, 1])$ (so $V = [0, 1]\times[0, 1]\times\cdots$ with a $(0,1]$ in the $(m,n)^\text{th}$ position). Finally we set $W = V\cap F(X)$. Then supposedly, $W\subset F(U)$.

My question is: how on earth can this be possible? This seems completely nonsensical to me. For instance, suppose there were base sets $U_j, U_k$ such that $\overline{U_j}\subseteq U_k\subset X\backslash U$. Then $f_{j,k}(U) = \{0\}$ since $f_{j,k}$ vanishes outside of $U_k$, and $U$ lives outside of $U_k$. Yet the $(j,k)^\text{th}$ factor in $W$ is $f_{j,k}(X)$, which contains $0$ and $1$, so clearly $f_{j,k}(X)\not\subseteq f_{j,k}(U)$, and thus $W\not\subseteq F(U)$.

EDIT: Let's check even further. If $W\subseteq F(U)$, then $\pi(W)\subseteq \pi(F(U))$ for any projection $\pi$. Let $j, k$ be the same indices chosen in the previous paragraph.
$$
\pi_{j,k}(W) = \pi_{j,k}(\pi_{m,n}^{-1}((0, 1])\cap F(X)) = \pi_{j,k}\left(F(X)\right) = f_{j,k}(X)\supset \{0, 1\}
$$
where we have assumed $(j,k)\neq (m,n)$. Next
$$
\pi_{j,k}(F(U)) = f_{j,k}(U) = \{0\}
$$
since $\overline{U_j}\subseteq U_k\subseteq X\backslash U$, so $U\subseteq X\backslash U_k$, implying $f_{j,k}(U)\subseteq f_{j,k}(X\backslash U_k) = \{0\}$.

If $W\subseteq F(U)$, then we should have $\pi_{j,k}(W)\subseteq \pi_{j,k}(F(U))$, but
$$
\pi_{j,k}(W)\supset\{0,1\}\not\subseteq \{0\}=\pi_{j,k}(F(X))
$$
so we have arrived at a contradiction.

We've made an assumption, that $(j, k)\neq (m, n)$. Recall $m, n$ were chosen such that $x_0\in \overline{U_m}\subseteq U_n\subseteq U$, and $U_k\subseteq X\backslash U$, so $U_n$ and $U_k$ are disjoint, implying $k\neq n$, so our choice of indices is valid.

EDIT 2: This is all assuming that such indices $j, k$ even exist, however it's not too hard to dream up examples where they do.

We can simplify my concerns a bit by first answering this question: does $F(U)$ even "look" like an open set in $F(X)$? By the definition of the product topology, were $F(U)$ open we would see that each projection should be open (with only finitely many projections being different from the entire base space), and yet as I've illustrated if we can pick indices $j, k$ such that $U_j\subseteq \overline{U_k}\subset X\backslash U$, then $\pi_{j,k}(F(U)) = f_{j,k}(U) = \{0\}$, which (presumably) isn't an open set in $f_{j,k}(X)$

Am I missing something?

答:
We have $z_0\in F[U]$, $x_0\in X$ such that $F(x_0)=z_0$, and $m,n\in\Bbb N$ such that $x_0\in\operatorname{cl}U_m\subseteq U_n\color{red}{\subseteq U}$. As you say, $f_{m,n}(x_0)=1$, but $f_{m,n}[U_n]$ is not $\{0\}$: $f_{m,n}[\color{red}{X\setminus U_n}]=\{0\}$. Now let $V=\pi_{m,n}^{-1}[(0,1]]$ and $W=V\cap F[X]$. If $x\in W$, then $f_{m,n}(x)\in(0,1]$, so $f_{m,n}(x)\ne 0$, and therefore $x\notin X\setminus U_n$. Of course that just means that $x\in U_n\subseteq U$, so $F[W]\subseteq U$, as desired.

There is no problem with a pair $\langle j,k\rangle$ such that $\operatorname{cl}U_j\subseteq U_k\subseteq X\setminus U$. As you say, in that case $f_{j,k}[U]=\{0\}$, and $f_{j,k}[X]\nsubseteq f_{j,k}[U]$, but that does not imply that $W\nsubseteq F[U]$: it just says that the projection of $F[U]$ onto the $\langle j,k\rangle$ factor is a proper subset of the projection of $F[X]$ onto that factor.

In your first edit you say that

$$\pi_{j,k}\big[\pi_{m,n}^{-1}[(0,1]]\cap F[X]\big]=\pi_{j,k}[F[X]]\,;$$

this is not true in general, as can easily be seen in a simpler example. Let $X=[0,1]$ and $F:X\to[0,1]\times[0,1]:x\mapsto\langle x,x\rangle$, and let $\pi_0$ and $\pi_1$ be the projections to the first and second factors, respectively. Then

$$\pi_1\big[\pi_0^{-1}[(0,1]]\cap F[X]\big]=(0,1]\ne[0,1]=\pi_1\big[F[X]\big]\,.$$

That error vitiates everything that follows.

Your second edit contains a serious misunderstanding of the product topology. It is true that if $F[U]$ is open, then every projection is open, since the projection maps are open maps, but it is not true that all but finitely many of the projections should be the whole space $[0,1]$: that is the case only for basic open sets in the product. Open sets in general are arbitrary unions of basic open sets and can be much more complicated.

As it happens, $V$ is a basic open set in the product: its $\langle m,n\rangle$ projection is $(0,1]$, and all of its other projections are $[0,1]$. The projections of $W$, however, depend very much on how $F[X]$ is situated in the product.


问:
Evaluating a definite integral by introducing change of variables
In a Physics class, I encountered the following integral: $$\int_{0}^{\infty} \left(\frac{x^2}{\exp\left(x-a\right)+1}-\frac{x^2}{\exp\left(x+a\right)+1}\right)\,\mathrm{d}x\text{,}$$

where $a$ is a constant. It is stated to yield: $$\int_{0}^{\infty}\left(\frac{x^2}{\exp\left(x-a\right)+1}-\frac{x^2}{\exp\left(x+a\right)+1}\right)\,\mathrm{d}x=\frac{1}{3}\left[\pi^2a+a^3\right]$$

I first introduced a change of variables: $$y:=x-a$$ $$z:=x^\prime+a$$ This gives: $$\int_{0}^{\infty}\frac{x^2}{\exp\left(x-a\right)+1}\,\mathrm{d}x-\int_{0}^{\infty}\frac{\left(x^\prime\right)^2}{\exp\left(x^\prime+a\right)+1}\,\mathrm{d}x^\prime=\int_{-a}^{\infty}\frac{\left(y+a\right)^2}{e^y + 1}\,\mathrm{d}y-\int_{a}^{\infty}\frac{\left(z-a\right)^2}{e^z+1}\,\mathrm{d}z$$ However, when introduced to a computer, the definite integrals are stated to have no solution. I actually expected to get integrals of the form: $$\int_{0}^{\infty}\frac{y}{e^y+1}\,\mathrm{d}y=\frac{\pi^2}{12}$$ What am I missing here?

答:
If you encountered this integral in a statistical mechanics class, perhaps they wanted you to use the integral of the Fermi-Dirac distribution and properties of the polylogarithm function.

Since $$-\operatorname{Li}_{3}(-x) = \frac{1}{2} \int_{0}^{\infty} \frac{t^{2}}{e^{t}/x+1} \, \mathrm dt, \, \quad x \ge -1,$$ and the inversion formula for the trilogarithm function states that $$\operatorname{Li}_{3}(-x) - \operatorname{Li}_{3} \left(- \frac{1}{x} \right) = - \frac{1}{6} \, \ln^{3}(x)  - \frac{\pi^{2}}{6} \ln x \, , \quad x \notin (-1,0], \tag{1}$$

we have $$ \begin{align} \int_{0}^{\infty} \left(\frac{t^2}{e^{t-a}+1}-\frac{t^2}{e^{t+a}+1}\right)\,\mathrm dt &= - 2\operatorname{Li}_{3}(-e^{a})+ 2 \operatorname{Li}_{3} \left(-\frac{1}{e^{a}}\right) \\ &= 2 \left( \frac{1}{6} \, \ln^{3}(e^{a}) + \frac{\pi^{2}}{6} \ln(e^{a}) \right) \\ &= 2 \left(\frac{a^{3}}{6}+ \frac{ \pi^{2}a }{6} \right) \\ &= \frac{a}{3} \left(a^{2} + \pi^{2} \right). \end{align}$$

$(1)$ The inversion formula for the trilogarithm can be derived from the inversion formula of the dilogarithm by dividing both sides of the equation by $x$ and integrating.


问:
How to calculate percentile? Is it possible to get 100 percentile?
How do we calculate percentile? I think it should be calculated as:

```<code>P = Total number of candidates
L = Number of candidates whose marks are below yours
</code>```

Percentile = (L/P)*100

That means if you are the highest in the range, and there are 200,000 candidates and if your marks are same as 7 others then;

```<code>your percentile = (199992/200000)*100 = 99.996
</code>```

That means in any circumstances you can't have perfect 100 percentile.

Then how come this is possible, see links:-

http://www.zeenews.com/news680105.html

http://economictimes.indiatimes.com/cat/two-cat-aspirants-get-100-percentile/articleshow/2685824.cms

答: What happens in case of an exam where all candidates in an exam score the same marks does that mean they are all at 0% percentile as they have 0 candidates below them ?</p>


问:
What does it mean for media to play at "1.5x" or (Nx) speed?
Suppose I have a video that plays for 60 minutes at normal "1x" speed. I know that if I set the video to play at "2x" speed, then it should play for 30 minutes.

Now, what if the video is set to play at "1.5x" speed? Intuition leads me to two answers:

Since 1.5x is directly in the middle between 1x and 2x, then the video should play directly in the middle of 60 minutes and 30 minutes, meaning it should play for 45 minutes.

When playing at 2x speed, I can compute that

$$
60\ \text{real-time minutes} \times \frac{1\ \text{play-back minute}}{2\ \text{real-time minutes}} = 30 \ \text{play-back minutes}
$$
Analogously, for 1.5 speed, I can compute $60 \times \frac{1}{1.5} =$ 40.0 minutes.

Which one is the correct answer, 45 minutes or 40 minutes, and why? What if I try to play the video at $Nx$ speed, where $N > 0$, to indicate playing media at a faster speed or slower speed?

答: The second. Your first idea obtains the completion time from the playback rate given by a harmonic mean of $1\times,\,2\times$, viz. $\frac{2}{1+\tfrac12}=\tfrac43$. Similarly, $\tfrac{2}{\tfrac{1}{30}+\tfrac{1}{60}}=40$ is a harmonic mean of two completion times, obtained with the arithmetic-mean playback rate. Finally, a geometric-mean playback rate obtains a geometric-mean completion time.


问:
Probability question a bag with $6$ black and $3$ white balls

From an bag with $6$ black and $3$ white balls, a ball is drawn (with re-laying) repeatedly until a black one is drawn. Let $Y$ denote the number of total number of draws. What is $P (Y = 1)$?

Is the correct way of thinking and answer:

$6/ 9 = 0,666667 = 67 \%$ because $p(y=1)$ means the probability of selecting a black on the first try?

答: Assuming that the balls in the bag are all equally likely to be chosen, yes, the probability is indeed equal to $2/3 \approx 67\%$, for the very reason that you have described.


问:
Finding the distribution $(_=)$ of $k$ heads when throwing a die n times and only when 4 or 1 is rolled tossing a coin.
Player $A$ throws the die $n$ times. Whenever 1 or 4 is thrown, he tosses an unbiased coin. Let $X_n$ be the random variable which counts the number of heads from the Bernoulli experiment of tossing a coin. I am trying to find the distribution $P(X_n=k)$ for every $k$, its mean and variance (giving me a hint towards the distribution would be good enough though).

I noticed that the number of $4$'s and $1$'s could be counted using a binomial approach, so let $Y$ be a random variable which counts the number of throwing $4$ or $1$.

Since $X$ counts the number $k$ heads I have to throw at least $k$ times $4$ or $1$.

Giving us $P(Y \geq k) = \sum\limits_{i = k}^n {n \choose i} {2 \over 6}^i {4 \over 6}^{n-i}$

My next steps was thinking about how to count the number of heads.

The naive approach was just to multiplying $P(Y \geq k)$ with $1 \over 2$ but I can see no possibility to count heads now since I dont know whether a head or tails is flipped.

The next thing I came up with was combining this with binomial distribution again for counting k heads.

Something along this way: $P(X = k) =  \sum\limits_{i = k}^n {n \choose i} {2 \over 6}^i {4 \over 6}^{n-i} {n \choose k} {1 \over 2}^k {1 \over 2}^{i-k}$

I am not sure whether this is going into the right direction or not and it doesn't seem handy at all. As the next step I would try to use the binomial theorem, but I can't find a way to get this simpler along this way because I have to shift k indices.

答:
You can write
$$
X_n = \sum_{i=1}^n Y_i,
$$

where $Y_i=1$ when if you get a head in the $i$-th trial, $Y_i=0$ otherwise. The probability that $Y_i=1$ is $$P(Y_i=1) = P(Y_i=1|1\text{ or }4\text{ appears in the die throw})P(1\text{ or }4\text{ appears in the die throw}) = \frac{1\times 2}{2\times 6}.$$  Thus $\mathbb{E}(X_n)=n/6$. Now for the variance, if you assume the trials are independent for each $i$ then $X_n$ is clearly (sum of Bernoulli's) a binomial that is,
$$
P(X_n=k) = {n\choose k}\left(\frac{1}{6}\right)^{k}\left(\frac{5}{6}\right)^{n-k},
$$
for $k=0,1,\ldots,n$.
Therefore the variance is $\frac{n}{6}\times\frac{5}{6}$. If they are not independent then you need to compute the covariances depending on the problem.


问:
Show that exponential map between complex matrix and general linear group is surjective
$\textbf{Problem}$

Trying to that $\exp: M_n(\mathbb{C}) \rightarrow GL(n,\mathbb{C})$  is surjective.

$\textbf{Attempt}$

I started by considering a complex matrix A $\in M_n \mathbb{C}$, to this matrix I applied the Jordan normal form so $ A = \lambda I + N $ where $I$ is the identity matrix and $N$ is an upper triangular matrix.

Thus, using the exponential map I get:

$$ M_n \mathbb{C} \rightarrow GL(n \mathbb{C}) \\
A = (\lambda I + N) \mapsto exp(\lambda I + N)
$$

Expanding the exponential we get and using the following Jordan Formula:

$\textbf{Jordan Formula:}$

$$\mathbf J=\begin{pmatrix}\lambda&1&&\\&\lambda&\ddots&\\&&\ddots&1\\&&&\lambda\end{pmatrix}$$

$$f(\mathbf A)=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$

$$exp(\mathbf A)=\begin{pmatrix}exp(\lambda)&exp(\lambda)&\cdots&\frac{exp(\lambda)}{(n-1)!}\\&exp(\lambda)&\ddots&\vdots\\&&\ddots&exp(\lambda)\\&&&exp(\lambda)\end{pmatrix}$$

and I can identify this last matrix as an element of $GL(N, \mathbb C)$? If yes, is this enough to prove the surjectiveness of the map? What am I missing?

Thanks in advance!

Edit:

I used @BenGrossmann and @user1551 hint, so I defined:

$ M = log(\lambda I ) + N $ and $J = \lambda I + N $ then

$\exp(M) = \exp(\log ( \lambda I) + N) = \lambda I \exp(N) $ as N is nilpotent of degree 3 thus $exp(N) \approx I + N + \frac{1}{2} N^2$, I plug this in the above equation and I get:

$$ \exp(M) = \lambda I (\approx I + N + \frac{1}{2} N^2) $$

How can I prove that $\exp(M) = J$ from this?

答:
Let $A$ be any  unital, commutative algebra over any field with characteristic zero.  Denote by
$$
  N = \{x\in A: x^n=0,  \text{ for some } n>0\}
  $$
and
$$
  U = \{y\in A: (y-1)^n=0,  \text{ for some } n>0\}.
  $$
Since $A$  commutative, it easy easy to see that $N$ is a subalgebra (the fact that it is closed under addition follows
from  Newton's Binomial Theorem).
Observe also  that  $U=1+N$.

Next consider the functions
$$
  \ln:U\to N, \quad\text{and}\quad
  \exp:N\to U,
  $$
given by
$$
  \ln(y) = -\sum_{n=1}^\infty \frac{(1-y)^n}n, \quad\text{and}\quad
  \exp(x) = \sum_{n=0}^\infty   \frac{x^n}{n!}.
  $$

Despite the occurence of "$\infty $" as the upper limits of the above sums, only finitely many terms are nonzero because
$1-y$ and $x$ are  nilpotent elements.  Notice that
the fact that $N$ is an ideal guarantees that  $\ln$ is indeed  $N$-valued, while  $\exp$ is $U$-valued.

Although it is a bit of a pain to prove it, these maps are each others inverse.

Given an invertible complex matrix $y$ whose Jordan form consists of a single Jordan block,  we may write
$$
  y=\lambda I+x
  $$
where  $\lambda \neq 0$ and $x$ is nilpotent.  Rewritting $y$ as
$$
  y=\lambda (I+x/\lambda ),
  $$
writing $\lambda =\exp(z)$, for some complex number $z$, and  working within the unital commutative  algebra of matrices generated by $y$, we then have that
$$
  y=\lambda (I+x/\lambda ) = $$$$ =
  \exp(z) \exp\big (\ln(I+x/\lambda )\big ) = $$$$ =
  \exp\big (z+ \ln(I+x/\lambda )\big ).
  $$

For a general invertible matrix $y$, one may repeat the above procedure for each of its Jordan blocks.


问:
Radius of Convergence for $\sum_{n=1}^{\infty}n!(2x+1)^n$
I'm trying to find out what the radius of convergence is for the power series:
$$\sum_{n=1}^{\infty}n!(2x+1)^n$$

I figured I'd use the Ratio Test, but then I would just end up with $(2x+1)\lim_{n\to\infty}(n+1)$, so I thought the radius would be $$-1 \leq 2x+1 \leq 1$$
which would be $$-1 \leq x \leq 0$$

But I guess I got the idea wrong, would appreciate some help! Thanks in advance!

答:
Let $$u_n(x)=n!(2x+1)^n$$

If $ x=-\frac 12 $, the series converges.

If $ x\ne -\frac 12 $, then

$$\lim_{n\to+\infty}\Bigl|\frac{u_{n+1}(x)}{u_n(x)}\Bigr|=$$

$$\lim_{n\to+\infty}(n+1)|2x+1|=+\infty$$

$$\implies \sum u_n(x) \text{ diverges} $$

So the Radius of convergence is $ R=0$.


答: As in previous answer find the center of the ellipse $C$. Draw a circle with center at $C$ which intersects the ellipse in 4 points. The points form a rectangle. Draw through $C$ lines parallel to the sides of the rectangle. They are the main axes of the ellipse.

Title: Finding the focus of an ellipse with ruler and compass

Body:
Given an ellipse E, find his focus with ruler and compass.

I tried to generalize the next theorem:

Given a circle C find his center. The idea for prove this is with an chord and it's bisector line (the ortogonal line in the midle point of the chord)

Ortogonality is fundamental for this theorem but in the ellipse ortogonality doesn't work.

Inspired by the equations and the linear algebra, an ellipse is a circle transformed by a linear transformation, but there are infinite linear transformations...the chord determines a linear transformation if we know the center of the elipse...


问:
Let $ V $ be open in a metric space $ (X,d) $. Show that, for all $ A⊂X $, $ V∩\overline{A}⊂\overline{A∩V} $.

Definition: If $ X $ is a metric space, $ E \subset X $ and $ E '$ represents the set of all limit points of $ E $ in $ X $, then let us denote the closure of $ E $ as $ \overline {E} = E \cup E '. $

Let $ V $ be an open in a metric space $ (X,d) $. Show that for all $ A⊂X $ we have $ V∩\overline{A}⊂\overline{A∩V} $.

My attempt.
Let $ x \in V \cap \overline {A} $, then $ x \in V $ and $ x \in \overline {A} $. Let us consider two cases:

If $x\in A$ and $x\in V$, then $x\in V\cap A\subset \overline{A\cap V}$. Since for any set $ A $ we have $ A \subset \overline {A} $.

If $x\in A'$ and $x\in V$. Since $ x \in A '$, there exists $ q \neq x $ such that $ q \in A $. Since $ V $ is open, then all its points are limits, that is, $ q \in V $. Therefore, $ q \in V \cap A $, that is $ x \in \overline {V \cap A} $.

答:
No, the proof is not correct.

The division into cases is good and the proof for the first case is OK.

If $x\in A'$ and $x\in V$, then you want to show that $x\in(V\cap A)'$.

Take a neighborhood $U$ of $x$. Then also $U\cap V$ is a neighborhood of $x$ and, by definition of limit point, there is $y\in (U\cap V)\cap A=U\cap(V\cap A)$, $y\ne x$. Thus $y\in V\cap A$ and this proves that $x\in (V\cap A)'$. Note that, by definition, $(V\cap A)'\subseteq\overline{V\cap A}$, so the proof is complete.


问:
solve for x - algebra and log function
Have to solve this algebraic. I end up with the wrong result. I think I'm messing up with the log rules or something. Help please

$ 2 \cdot 1,3^x = 12 \cdot 0,9^x $

$ 1,3^x = 6  \cdot 0,9^x  $

$ x \cdot \log(1,3) = x  \cdot \log (6 \cdot 0,9  )$

Title: solve for x - algebra and log function

Body:
$ 2 \cdot 1,3^x = 12 \cdot 0,9^x $

$ 1,3^x = 6  \cdot 0,9^x  $

$ x \log 1,3 = \log 6+ x\log 0,9$

$x(\log 1,3-\log 0,9) =\log 6$

$$x=\frac{\log 6}{\log 1,3-\log 0,9}$$


问:
Let A be a Boolean algebra that has infinitely many atoms Let X be the set of atoms. Define the map from A to ℘(X) as follows, a → {x ∈ X : x ≤ a}.
Let $A$ be a Boolean algebra that has infinitely many atoms.

Let $X$ be the set of atoms.
Define the map from $A$ to $\wp(X)$ as follows, $a \mapsto \{x \in X : x \leq a\}$.

(a) $f$ is 1-1. (b) $f$ is an isomorphism (c) $f$ is surjective (d) None of the last

here is my thought process: I think the answer is (a) because the set $A$ is atomic but is not complete so is it not an isomorphism
since for a Boolean algebra $A$ and a power set algebra $\wp(At (A))$, $f : A \to \wp(At (A))$, where $f(b) = \{a \in At (A) | a \leq b\}$, is a homomorphism.

If $A$ is atomic, $f$ is an embedding and if $A$ is complete, $f$ is an epimorphism.

答:
Let $\mathbf A = \wp(Y) \times \mathbf B$, where $Y$ is any infinite set and $\mathbf B$ an atomless Boolean algebra.

Then $\mathbf A$ has infinitely many atoms: the elements of the form $(a,0)$ with $a$ an atom of $\wp(Y)$ (that is, a singleton of $Y$).

Now let $b,b' \in B \setminus \{0\}$.
Then there is no atom of $\mathbf A$ below either $(0,b)$ or $(0,b')$.

So $f(0,b) = f(0,b')=\varnothing$, and so the map is not injective.

Now, let $\mathbf A = FC(\mathbb N)$, the set of finite and co-finite subsets of $\mathbb N$.

Then $\mathbf A$ has infinitely many atoms: the singletons of $\mathbb N$ are the atoms of $\mathbf A$.

Let $B = \{\{2n\} : n \in \mathbb N\}$, a subset of the atoms of $\mathbf A$.

Suppose there exists $a \in A$ such that $f(a) \supseteq B$;
it follows that $\{2n\} \subseteq a$, for all $n \in \mathbb N$, and so $E \subseteq a$, where $E$ is the set of even numbers.

Now, it's clear that $E$ is not finite, and so $a$ can't be finite, and therefore, if $a$ is an element of $\mathbf A$, then $a$ must be co-finite, that is, $a = \mathbb N \setminus C$, where $C$ is a finite set of odd numbers.

But this means that $a$ contains odd numbers (indeed, infinitely many), and so $f(a) \neq B$.

Hence, the map is not surjective.


问:
$p$-adic closures of infinite sets
Let $S\subsetneq\mathbb{Z}$ be an infinite set.  Does there always exist a prime $p$ such that the closure of $S$ in the $p$-adic integers, $\mathbb{Z}_p$, contains a rational integer $n\notin S$?

Or, in elementary language, does there always exist a prime $p$ and $n\in \mathbb{Z}\setminus S$ such that for all $k$ there is an $s\in S$ such that $p^k\mid n-s$.

答: How about $S = \{ 0 \} \cup \{ n! : n \in \mathbb{N} \}$?</p>

Title: $p$-adic closures of infinite sets


问:
Is $\sum_{n=1}^{\infty} {\frac{(-1)^n}{x+2^n}} $ in $(-2,\infty)$ uniformly convergent?

Is $\sum_{n=1}^{\infty} {\frac{(-1)^n}{x+2^n}} $ in $(-2,\infty)$ uniformly convergent?

I started by checking if it is pointwise convergent, because if it wasn't then especially it is not uniformly conveergent. But by Leibnitz, it is convergent.

But I got stuck on proving that it is uniformly convergent. Any help would be appreciated!

答: Also a hint: Set $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{x+2^{n}}$. Now consider sequences $x_{k}=-2+\frac{1}{2^{k}}$ and $y_{k}=-2+\frac{1}{2^{k+1}}$. Obviously $|x_{k}-y_{k}|\to 0$  when $k\to\infty$. Now see what happens with $|f(x_{k})-f(y_{k})|=|x_{k}-y_{k}||\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(x_{k}+2^{n})(y_{k}+2^{n})}|$. More precisely, consider the first term in summation...</p>


答:
The ratio test actually gives the expression
$$
\lim_{n\to\infty}\Big[(2x+1)(n+1)\Big]
$$
And you want this limit to be finite (actually you want it to be between $-1$ and $1$, but that detail turns out to be irrelevant in this case). When can that happen? Exactly when $2x+1$ is zero. Never otherwise. So the series converges at $x=-\frac12$ and the radius of convergence is $0$.

Title: Radius of Convergence for $\sum_{n=1}^{\infty}n!(2x+1)^n$

Body:
I'm trying to find out what the radius of convergence is for the power series:
$$\sum_{n=1}^{\infty}n!(2x+1)^n$$

I figured I'd use the Ratio Test, but then I would just end up with $(2x+1)\lim_{n\to\infty}(n+1)$, so I thought the radius would be $$-1 \leq 2x+1 \leq 1$$
which would be $$-1 \leq x \leq 0$$

But I guess I got the idea wrong, would appreciate some help! Thanks in advance!


问:
Solving Equation With Two Equal Unknowns
In this question, I asked why it was not possible to solve the following three linear equations for $x_n$ when I knew the values of $s$, $h$, $r$ and knew that $k$ was the same for all three but did not have the value:

$$s_1 = k^{-1}(h_1 + rx_1)$$
$$s_2 = k^{-1}(h_2 + rx_2)$$
$$s_3 = k^{-1}(h_3 + rx_3)$$

The answer was that there are 4 unknowns and three equations, despite knowing that the unknown $k$ was the same across all three.

I have a follow up question now regarding the following equation:

$$s = k^{-1}(h + rx)$$

If I know $s$, $h$ and $r$, and I know that $k = x$ (but don't know either of their values), can this be solved?

Body:
$$s = k^{-1}(h + rx);\;k=x$$
$$s=\frac{h+rx}{x}$$
$$sx=h+rx$$
$$sx-rx=h\to x(s-r)=h$$
$$x=\frac{h}{s-r}$$


问:
$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$
I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$

applying Taylor's series.

$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}=    \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$

$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3-  x^5+ o(x^5))}=  \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 -  x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2}  \sim \frac{2}{3}$$

The suggested solution in my book is $2$. can someone indicate where I made mistakes?

答:
I think the following step is wrong:$$\lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln \sin 2x - \ln \cos 2x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln \sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}.$$

I think you have used the false identity $$ \ln\cos2x \equiv 2\ln\cos x - 2\ln\sin x. $$

It is false because $$ \ln\cos 2x = \ln(\cos^2x-\sin^2x) \neq\ln\cos^2x - \ln\sin^2x = 2\ln\cos x - 2\ln\sin^2x. $$


问:
Determine center of circle if radius and 2 tangent line segments are given
Given the radius and its $2$ tangent lines and their point of intersection of a circle.

A similar question is
How to calculate the two tangent points to a circle with radius R from two lines given by three points

But how do I find coordinates of the center of the circle.
I believe there must be $2$ solutions—each side of the intersection point.
Please let me know what I am doing wrong am here, or if I am missing something?

Thanks in advance.

答:
You have equation of two tangents.
STEP $1$:Draw perpendicular(normal) from centre to tangents it is equal to radius.since you have two tangents you will draw two perpendicular(normal).

STEP $2$ :so you will get two equation consists of coordinates of centre.

STEP $3$: Then solve those equation you will get coordinates of centre.If you dont know how to draw perpendicular let me know and give data I'll solve it.


答:
Perhaps it helps to change the scenario. Suppose a tortoise had to run a $60$ metre race. At normal speed, it does one metre per minute, and so the tortoise takes $60$ minutes. At $1.5\times$ speed, the tortoise travels $1.5$ metres per minute, and so after $40$ minutes the tortoise has travelled $1.5 \times 40 = 60$ metres.

The reason that you can't simply take the mean of $30$ and $60$ is actually quite subtle. If you rearrange the formula
$$
\text{speed}=\frac{\text{distance}}{\text{time}} \, ,
$$
you get
$$
\text{time}=\frac{\text{distance}}{\text{speed}} \, .
$$
Here, the distance is $60$. Hence,
$$
\text{time} = \frac{60}{\text{speed}} \, .
$$
If you plot this on a graph, you get this:

At first, the time taken is decreasing rapidly as your speed increases. However, the steepness of the curve then starts to become less apparent. This means that between $1\times$ and $1.5 \times$ speed, the change in time is larger than the change in time between $1.5 \times$ speed and $2 \times$ speed. Hence, the time goes all the way down to $40$ as you go from $1\times$ to $1.5\times$ speed, which is less than $45$. Most of the 'going down' has already occured when you get to the midpoint of $1$ and $2$, if that makes sense.

Title: What does it mean for media to play at "1.5x" or (Nx) speed?

Body:
Suppose I have a video that plays for 60 minutes at normal "1x" speed. I know that if I set the video to play at "2x" speed, then it should play for 30 minutes.

Now, what if the video is set to play at "1.5x" speed? Intuition leads me to two answers:

Since 1.5x is directly in the middle between 1x and 2x, then the video should play directly in the middle of 60 minutes and 30 minutes, meaning it should play for 45 minutes.

When playing at 2x speed, I can compute that

$$
60\ \text{real-time minutes} \times \frac{1\ \text{play-back minute}}{2\ \text{real-time minutes}} = 30 \ \text{play-back minutes}
$$
Analogously, for 1.5 speed, I can compute $60 \times \frac{1}{1.5} =$ 40.0 minutes.

Which one is the correct answer, 45 minutes or 40 minutes, and why? What if I try to play the video at $Nx$ speed, where $N > 0$, to indicate playing media at a faster speed or slower speed?


问:
How to find the probability of one score over other if they follow a normal distribution?
I am predicting scores over a golf tournament, what I want is to find the probability of one player bets another player on the same tournament. For example, the predicted score of player A is 63, while the predicted score of player B is 65, how can I convert this into probabilities that let me decide how likely is that player A win over player B? I know from the distribution of predicted and actual scores that the scores follow either a normal or Poisson distribution.</p>

答:
If $X_1$ and $X_2$ are independent normal RVs with means $\mu_1$ and $\mu_2$ and variances $\sigma_1^2$ and $\sigma_2^2$, then $X_1-X_2$ is normal with mean $\mu_1 - \mu_2$ and variance $\sigma_1^2 + \sigma_2^2$, so
$$P(X_1 < X_2) = P(X_1-X_2 < 0) = P\left(\frac{X_1-X_2 - (\mu_1 - \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} < - \frac{\mu_1-\mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}} \right) = \Phi\left(- \frac{\mu_1-\mu_2}{\sqrt{\sigma_1^2+\sigma_2^2}} \right).$$

The missing pieces of information you need are the variances of the two players' predicted scores, and the assertion that their scores are independent. If they are not independent but you know the correlation $\rho$ of their scores, then the variance of $X_1-X_2$ will $\sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2$ instead.


问:
Given 'n' number of unique positive integers, how many unique sum values can I make?
Suppose I have three integer values

```<code>[1,2,3]
</code>```

With above three values, I can form total 8 values ($2^3$).

```<code> 0                        (Pick 0 value)
 1, 2, 3,                 (Pick only 1 value)
 3[1+2], 4[1+3], 5[2+3],  (Pick 2 values)
 6[1+2+3]                 (Pick 3 values)
</code>```

However, Notice how there are duplicate outcomes.
3 and 3[1+2] ?

So I would say that using three numbers 1,2,3, I can form total 7 unique numbers.

But is there any formula I can apply to find out the cardinality of only the unique outcomes ?

Some more example set of numbers to consider.

```<code>[1,2,3,4]
[1,2,3,4,5]
[1,3,5,7]
</code>```

答:
In the following list I wrote $(n,values)$ where $n$ is the number of different elements in the set and $values$ is the number of different values we can get taking $0$ elements, $1$ element, $2$ element, and so on.
$$(0,1),(1,2),(2,4),(3,7),(4,11),(5,16),(6,22),\ldots$$
$$values=\frac{1}{2} \left(n^2+n+2\right)$$
Hope this can help

Edit

Only works for lists of consecutive integers, like $\{1,2,\ldots,n\}$


问:
metric on clustering of correlation matrix using silhouette score
Given a correlation matrix $A_n$, a metric $\|\cdot \|_2$ and a clustering $k$, you can calculate the silhouette score (https://scikit-learn.org/stable/modules/generated/sklearn.metrics.silhouette_samples.html). Then define a $score = silhouette.mean/silhouette.std$. If you have two different clustering's $k_1$ and $k_1$ s.t. $score(k_1)$ < $score(k_2)$ will it be true that $score(k_1) < score(k2)$ when $A_n$ is embedded inside any $A_m$ where $n<m$?

答:
It is easy to se that - choose 4 points on the real line such that they form 2 clusters. In $\mathbb{R}^2$ choose a $y$ and choose an element in each of the clusters. Choose $-y$ for the other 2 elements. Make $y$ sufficiently large and you get the opposite clustering.

edit - proof:
Let the distance between cluster 1 and cluster 2 in $\mathbb{R}$ be $r$. Let $x_1$, $x_2$ belong to cluster 1.  Let $x_3$ and $x_4$ belong to cluster 2.

In $\mathbb{R}^2$ choose $y>r$ then also $2y > r$ and choose points $x_1'=(x_1, y)$, $x_2'=(x_2,-y)$, $x_3'=(x_3, y)$ and $x_4'=(x_4, -y)$
$\implies 0 < || x_1' - x_3' ||  < ||x_1' - x_2' || = || x_1' - x_3' + x_3' - x_2' || \le || x_1' - x_3'|| + || x_3' - x_2' ||  = r + 2y
\Leftrightarrow 0 < r < 2y \le r + 2y$
Also:
$0 < || x_1' - x_3' ||  < ||x_1' - x_4' || = || x_1' - x_3' + x_3' - x_4' || \le || x_1' - x_3'|| + || x_3' - x_4' || = r + 2y $


问:
What methods are used to look into $\pi^{\pi^{\pi^\pi}}$ and other stacked towers of repeating irrational numbers?
I recently saw a YouTube video here and it got me thinking, does anyone on this site have any insight on how we would tackle (change to base ten, evaluate, compute, or even put limits on approaching) some things like this?

examples:

$\displaystyle e^{e^{e^e}}$

$\displaystyle \pi^{\pi^{\pi^\pi}}$

$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}}$

$\ln(2)^{\ln(2)^{\ln(2)^{\ln(2)}}}$

What methods are proposed to even look at these things? Would it be a good idea to change to complex for pi? Do all towers of root 2 just simplify to 2?

note: Apologies for being new and not knowing how to format the text correctly.

Edit: Things I am thinking, for

$$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}} = \sqrt{2}^{\sqrt{2}^{\sqrt{2}\cdot \sqrt{2}}} = \sqrt{2}^{\sqrt{2}^{2}} = \sqrt{2}^{2} = 2 $$

This gets me thinking that regardless of the height (greater then 3) of the tower of it will always be 2.

答:
The question is ambiguous, but we could interpret it as one about computability and decidable equality. Let $\mathbb{Q}$ be the set of rational numbers, $\mathbb{Q}^+$ be the set of positive rational numbers, and $\mathbb{R}$ be the set of real numbers. A computable real number is a real number $x \in \mathbb{R}$ such that there exists a computable function $f_x : \mathbb{Q}^+ \rightarrow \mathbb{Q}$ such that

$$ |f_x(\varepsilon) - x| \leq \varepsilon $$

That is, $f_x$ is a computer program such that, for any desired rational error tolerance $\varepsilon > 0$, $f_x$ yields a rational number within $\varepsilon$ of $x$. For an implementation of this concept, see Toward an API for the real numbers (summary).

Now note that

\begin{align}
a^{a^{a^a}}
&= \exp a^{a^a} \log a \\
&= \exp (\exp a^a \log a) \log a \\
&= \exp (\exp (\exp a \log a) \log a) \log a
\end{align}

So we can reduce any power tower to exponentials and logarithms, which are computable. Finally, note that the constants $e, \pi, \sqrt{2}, \ln 2$ are computable. Therefore, all the numbers listed in the question are computable.

In fact, we have something even stronger:

Can we decide equality for recursive reals computed from integer constants by a combination of the following operations, which we will refer to as "calculator operations": (1) the four basic arithmetic operations, and square roots; (2) the sin, cos and tan trigonometric functions and their inverses; and (3) exponential and (natural) logarithm functions.

The answer turns out to be yes, as proven in the 1994 paper The identity problem for elementary functions and constants. Since $e = \exp 1$ and $\pi = 2 \sin^{-1} 1$, we can decide equality between any of the numbers listed in your question and any given integer. Therefore, we can decide whether any of the numbers listed in your question is an integer.

For example, to decide whether $x = \pi^{\pi^{\pi^\pi}}$ is an integer: Compute an integer upper bound $u \geq x$, say $u = 4^{4^{4^4}}$. Then, for all integers $k \in \{1, \dots, u\}$ (of which there are only finitely many), decide whether $x = k$. If equality holds for any $k$, $x$ is integer-valued. Otherwise, it is not integer-valued.


问:
Standard deviation of {1, -1, i, -i}
(just in case: $i^2 = -1$)

mean:
$$m = \frac{1 - 1 + i - i}{4} = 0$$

standard deviation:
$$\sigma = \sqrt{\frac{1^2 + (-1)^2 + i^2 + (-i)^2}{4}} = \sqrt{\frac{1 + 1 - 1 - 1}{4}} = 0$$

It feels wrong, but I don't understand how or why. Would anyone tell me why it's wrong?

答:
It's wrong because you need to use the magnitude of the each element not its square. This is because complex numbers ($a + bi$) are two dimensional vectors where their length is given by $\sqrt{a^2+b^2}$. This magnitude can also be found by multiplying by the complex conjugate (ie: $|a+bi|^2 = (a+bi)(a-bi)$)

This now will give the correct answer to

$$\sigma = \sqrt{\frac{|1|^2+|-1|^2+|i|^2+|-i|^2}{4}} = \sqrt{\frac{(1)(1)+(-1)(-1)+(i)(-i)+(-i)(i)}{4}} = 1 $$

Title: Standard deviation of {1, -1, i, -i}


问:
Metric space and closed sets
Let $\mathbb{R}^n$ have a Euclidian metric $d$ and $\emptyset \neq A \subseteq \mathbb{R}^n$ be a subset such that $(A, d_{|A \times A})$ is a metric space.

(a) Let $x \in A$ and consider the set $D := \{y \in A: d(x, y) \geq r\}$. Is $D$ closed in $A$?

(b) Let $x \in A$ and consider the set $E := \{y \in A: d(x, y) > r\}$. Is $D$ contained in the closure of $E$ in $A$?

Justify your answers.

I am reviewing solutions to various problems for an upcoming examination and this was a problem in a practice set but there are no solutions available. Any assistance with a solution is much appreciated.

答:
$D$ is closed in $A$. If $r = 0$, then $D = A$ which is closed by default. If $r > 0$, then we can prove $D$ is closed as follows. If $y_0 \in A$ is a limit point in $D$, then there is a sequence $(y_n)$ in $D$ which converges to $x_0$. If $\varepsilon > 0$ is small enough that $\varepsilon < r$, then we can choose sufficiently large $N$ so that $n \geq N$ implies that $d(y_n, y_0) < \varepsilon$. The condition on $\varepsilon$ ensures that $d(x,y_n) - d(y_n, y_0) > r - \varepsilon > 0$ so that the absolute value can be removed. Hence, by the reverse triangle inequality $$d(x, y_0) \geq |d(x, y_n) - d(y_n, y_0)| = d(x,y_n) - d(y_n, y_0) > r - \varepsilon$$ which proves that $d(x, y_0) \geq r$ so that $y_0 \in D$ (since $\varepsilon$ can be made arbitrarily small). Thus $D$ contains all its limit points so it is closed.

Since $D$ is closed, it is its own closure. Moreover, if $y \in A$ so that $d(x,y) > r$, then in particular $d(x,y) \geq r$ so that $y \in D$. Thus $A \subseteq D$ which proves (b).

Title: Metric space and closed sets


问:
I am confused about starting index of unit step function
Let $$u[n]=\begin{cases}
1&\text{if $n\geq0$}\\
0&\text{if $n<0$}
\end{cases}
$$
Then $$u[n-1]=\begin{cases}
1&\text{if $n\geq1$}\\
0&\text{if $n<1$}
\end{cases}.$$

If we have an equation that contains something like $$\displaystyle\frac{1}{2^{n}}u[n]-\frac{1}{3^{n}}u[n-1]$$
If we wish to get rid of the unit step function, we shall assume $n\geq1$. My question is, would this mean that $u[n]=0$? To clarify my point if we start at $n=1$ and upwards, does that mean $u[n]=0$ if $n=0$? since we are starting from $n=1$ and above

答:
The unit sep, $u(n)$ is equal to $1$ wheneve its argument is non-negative.  So, for $n\ge 1$, $u(n)=1$.  Moreover, we have

$$\frac{1}{2^n}u(n)-\frac{1}{3^n}u(n-1)=\begin{cases}\frac{1}{2^n}-\frac{1}{3^n}&,n\ge1\\\\
\frac1{2^n}&,1>n\ge 0\\\\0&,0>n\end{cases}$$


问:
Prove that $\exists c\in (a,b):f''(c) = 0$, if there is a secant line that intersects the graph of $f$ on $a,x_0, b$ where $a < x_0 < b$
Let $f: I \to \mathbb{R}$, where $I$ is an interval, a twice derivable function. Suppose that the secant line through the points $(a, f(a))$ and $(b, f(b))$ with $a,b \in I$ and $a < b$, intercepts the graph of $f$ in a point $(x_0, f(x_0))$, with $a < x_0 < b$. Show that $\exists c \in (a,b): f''(c) = 0$.

I think what should be done to prove this is showing that there is $x, y$ such that $f'(x) = f'(y)$, and then conclude that there is $f''(c) = 0$ through Rolle's theorem.

Since the secant line intercepts a point between $a,b$ and $f$ is continuous, it is easy to see that there is $a < x < y < b$ such that $f'(x) = f'(y)$, since it must mean that the graph "goes back in the direction where it came from" and then "heads towards its original direction" again.

I don't know how to write this down formally in a proof.

答:
The line segments from $(a, f(a))$ to $(x_0, f(x_0))$ and from $(x_0, f(x_0))$ to $(b, f(b))$ are two parts of the same secant line, and therefore have the same slope:
$$
\frac{f(x_0)-f(a)}{x_0-a} = \frac{f(b)-f(x_0)}{b-x_0} \, .
$$
According to the mean-value theorem, the fraction on the left is equal to $f'(x)$ for some $x \in (a, x_0)$, and the fraction on the right is equal to $f'(y)$ for some $y \in (x_0, b)$. So $f'(x) = f'(y)$ with $x < y$, and it follows that $f''(c) = 0$ for some $c \in (x, y) \subset (a, b)$.

You can also consider the function $g(x) = f(x) - l(x)$ where $l$ is the given secant line. Then $g(a) = g(x_0) = g(b) = 0$. Now apply Rolle's theorem repeatedly to conclude that $g'$ has (at least) two distinct zeros, and $g'' = f''$ has (at least) one zero.


问:
Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$
I'm struggling to grasp this particular question:

When a polynomial $f(x)$ is divided by $x+3$, the quotient is $2x^2-x+7$ and the remainder is $10$. What is $f(x)$?

This is what I did:

$$\begin{align}
f(x) &= (x+3)(2x^2-x+7) \\
&= 2x^3-x^2+7x+6x^2-3x+21 \\
&= 2x^3+5x^2+4x+21
\end{align}$$

Making sure it is correct:

Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that

答: Just add $10$ to your $f(x)$...it becomes the remainder when you do the division.


答:
In general, when you divide a function $p(x)$ by $q(x)$ and you get $f(x)$ as the quotient with a remainder of $r(x)$, you can alyways represent the answer as: $\frac{r(x)}{q(x)} + f(x)$

So you can rewrite your question algebraically as $\frac{10}{x+3} + 2x^2-x+7$ and then can simplify as follows:

$\frac{10}{x+3} + 2x^2-x+7 = \frac{10 + (x+3)(2x^2-x+7)}{x+3} = \frac{10 + 2x^3 -x^2+7x+6x^2-3x+21}{x+3} = \frac{2x^3 +5x^2 + 4x + 31}{x+3}$

Title: Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$

Body:
I'm struggling to grasp this particular question:

When a polynomial $f(x)$ is divided by $x+3$, the quotient is $2x^2-x+7$ and the remainder is $10$. What is $f(x)$?

This is what I did:

$$\begin{align}
f(x) &= (x+3)(2x^2-x+7) \\
&= 2x^3-x^2+7x+6x^2-3x+21 \\
&= 2x^3+5x^2+4x+21
\end{align}$$

Making sure it is correct:

Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that


答: Now @AdamRubinson has identified the error, note the limit is $L+2$ with$$L:=\lim_{x\to0^+}\frac{\ln\frac{1-\cos2x}{\tan^22x}}{\ln\tan2x}=\lim_{x\to0^+}\frac{\ln\tfrac12}{\ln2x}=\tfrac{-\ln2}{-\infty}=0.$$

Body:
I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$

applying Taylor's series.

$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}=    \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$

$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3-  x^5+ o(x^5))}=  \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 -  x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2}  \sim \frac{2}{3}$$

The suggested solution in my book is $2$. can someone indicate where I made mistakes?


问:
$E[X^2|\mathcal{G}]=2\int_0^{\infty}xP(X>x|\mathcal{G})dx$
Let $X$ a random variable on $(\Omega,\mathcal{F},P),\mathcal{G}$ a sub $\sigma$-algebra. Prove that, almost surely,

$$E[X^2|\mathcal{G}]=2\int_0^{\infty}xP(X>x|\mathcal{G})dx$$
The above formula was true for simple function $X,$ if we pick an increasing function $X_n$ to $X,$ there is a problem in applying the conditional monotone theorem in $\int_0^{\infty}xP(X_n>x)dx$ because $P(X_n>x|\mathcal{G}) \to P(X>x | \mathcal{G})$ on an event $E_x$ (which depends of $x>0$)such that $P(E_x)=1,$ so we can't change interchange limit-integral.

What do you suggest to solve this problem?

答:
I have got a slightly different answer.
Let $P_{\mathcal{G}}$ be your conditional measure (that is $P_{\mathcal{G}}(A)=P(A|\mathcal{G})$), then
$$
E[X^2|\mathcal{G}]= \int X^2 \,dP_{\mathcal{G}}= \int \int_{0}^{\infty}1(X^2>l)\,dl \,dP_{\mathcal{G}} = \int \int_{0}^{\infty}1(|X|>\sqrt{l})\,dl \,dP_{\mathcal{G}}.
$$
Now let $x=\sqrt{l}$ then
$$
E[X^2|\mathcal{G}]=2\int \int_{0}^{\infty}x1(|X|>x)\,dx \,dP_{\mathcal{G}}.
$$
Now use Tonelli's theorem and interchange the integral to get

$$
E[X^2|\mathcal{G}]=2\int_{0}^{\infty}x\int 1(|X|>x) \,dP_{\mathcal{G}}\,dx = 2 \int_{0}^{\infty}xP(|X|>x|\mathcal{G})\,dx.
$$


问:
Line-grid intersection in 2D
Say I have a line segment with known length L and a grid of pixels with dimension dX and dY along the X and Y axes. The line can begin from any arbitrary point within the grid and can have any arbitrary angle w.r.t the X and Y axes but its length is always the same (L). I would like to know what the maximum number of cells in the grid is that can be intersected with the line?

I was thinking something like this but am not sure:

ceil(L/dX) + ceil(L/dY) + 1

Numerical Simulations

In order to actually find what the correct answer is, I did a numerical simulation experiment where basically hundreds of millions of randomly generated lines with length L were started from a random point and fired randomly along various directions and the exact number of intersected grid pixels was counted to see what the max value is (since this was numerical simulations, I cannot say for sure that these values are correct):

L = 200, dX = 5, dY = 10 => Max no of intersected voxels = 47

L = 200, dX = 15, dY = 2 => Max no of intersected voxels = 103

L = 200, dX = 10, dY = 10 => Max no of intersected voxels = 31

答:
Let us switch to polar coordinates.

$$x=L\cdot \cos(\phi)-x_1\\y=L\cdot\sin(\phi)-y_1$$
where $(x_0,y_0)$ is a known end-point and where:
$$x_1 = \min(x_0,d_X-x_0)\\y_1 = \min(y_0,d_Y-y_0)$$

This corresponds to choosing the quandrant which points in direction closest to a corner.

We can now express the number as a function of $\phi\in[0,\pi/4]$: $$1+\left\lfloor\frac{x}{d_X}\right\rfloor+\left\lfloor\frac{ y}{d_Y}\right\rfloor=1+\left\lfloor\frac{L\cdot \cos(\phi)-x_1}{d_X}\right\rfloor+\left\lfloor\frac{L\cdot \sin(\phi)-y_1}{d_Y}\right\rfloor$$

More specifically, what we seek will be:

$$1+\max_{\phi}\left\{ \left\lfloor\frac{L\cdot \cos(\phi)-x_1}{d_X}\right\rfloor+\left\lfloor\frac{L\cdot \sin(\phi)-y_1}{d_Y}\right\rfloor \right\}$$

In most cases I suppose this will be related to the maximum Manhattan distance given a known Euclidean distance and a known distance between consecutive streets and avenues.

I think max shall be somehwere near $\phi \approx \arctan(d_Y/d_X)$.

In the special case if $d_X=d_Y$, then a tighter estimate shall be $\phi\approx$ 45 degrees and a factor $\sin(\pi/4)=\cos(\pi/4) = \frac{1}{\sqrt{2}}$ smaller than your estimate. There do as you mention exist isolated $\phi$ values for where the line passes precisely through integer coordinates on the lattice. But arbitrarily close to every such $\phi$ will be one with larger value.

But your estimate should work good enough as a not-as-tight upper bound.

Title: Line-grid intersection in 2D


问:
For n odd, skew symmetrical matrices in $M(n,\mathbb{R})$ are homeomorphic to orthogonal matrices for which $\det(A+I) \neq 0$?
I have a continuous mapping $A \mapsto (A-I)(A+I)^{-1}$ between the set of orthogonal matrices $A^{t}A=I$ s.t. $\det(A+I) \neq 0$ and the set of skew symmetrical matrices $A^{t}=-A$ in $M(n, \mathbb{R})$ with $n$ odd, so that $\det(A)=0$. How to prove it’s a homeomorphism? By artificially creating an inverse? By noting that the domain is open and proving that the image is so too, and then...?

Body: If $B = (A-I)(A+I)^{-1}$, then $A = (I+B)(I-B)^{-1}$.


问:
Prove that a sequence $(x_n)$ is Cauchy given $x_n$ is a point in $E_n$, where each $E_n$ is a nonempty bounded sequence of $\mathbb{R}^d$
Suppose that $E_1 \supseteq E_2 \supseteq ....$, where each $E_n$ is a nonempty bounded subset of $\mathbb{R}^d$, and suppose also that $\lim_{n\to\infty} \mbox{diam}(E_n) =0$. For each $n \in \mathbb{N}$, let $x_n$ be one point in $E_n$. Prove that $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}^d$.

I am not sure at all how to do this. I know a sequence is Cauchy if for every $\epsilon >0$, there exists an $N>0$ such that $m,n \ge N$ implies $|x_m-x_n|\le \epsilon$.

答:
For each $m, n, k\in\Bbb N$ with $m, n\geqslant k$, we have from our hypotheses that $x_m, x_n \in E_k$.
This implies that $\lVert x_m - x_n\rVert \leqslant \mbox{diam}(E_k)$.
Can you use the fact that $\lim_{n\to\infty} \mbox{diam}(E_n) =0$ to conclude?


问:
What are the Laws of Rational Exponents?
On Math SE, I've seen several questions which relate to the following. By abusing the laws of exponents for rational exponents, one can come up with any number of apparent paradoxes, in which a number seems to be shown as equal to its opposite (negative). Possibly the most concise example:

$-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1$

Of the seven equalities in this statement, I'm embarrassed to say that I'm not totally sure which one is incorrect. Restricting the discussion to real numbers and rational exponents, we can look at some college algebra/precalculus books and find definitions like the following (here, Ratti & McWaters, Precalculus: a right triangle approach, section P.6):

The thing that looks the most suspect in my example above is the 4th equality, $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$, which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... but technically, that translation from rational exponent to radical expression was not used at this point. Rather, we're still only manipulating rational exponents, which seems fully compliant with Ratti's 2nd property: $(a^r)^s = a^{rs}$, where indeed "all of the expressions used are defined". The rational-exponent-to-radical-expression switch (via the rational exponent definition) doesn't actually happen until the 6th equality, $(1)^\frac{1}{2} = \sqrt{1}$, and that seems to undeniably be a true statement. So I'm a bit stumped at exactly where the falsehood lies.

We can find effectively identical definitions in other books. For example, in Sullivan's College Algebra, his definition is (sec. R.8): "If $a$ is a real number and $m$ and $n$ are integers containing no common factors, with $n \ge 2$, then: $a^\frac{m}{n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$, provided that $\sqrt[n]{a}$ exists"; and he briefly states that "the Laws of Exponents hold for rational exponents", but all examples are restricted to positive variables only. OpenStax College Algebra does the same (sec. 1.3): "In these cases, the exponent must be a fraction in lowest terms... All of the properties of exponents that we learned for integer exponents also hold for rational exponents."

So what exactly are the restrictions on the Laws of Exponents in the real-number context, with rational exponents? As one example, is there a reason missing from the texts above why $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$ is a false statement, or is it one of the other equalities that fails?

Edit: Some literature that discusses this issue:

Goel, Sudhir K., and Michael S. Robillard. "The Equation: $-2 = (-8)^\frac{1}{3} = (-8)^\frac{2}{6} = [(-8)^2]^\frac{1}{6} = 2$." Educational Studies in Mathematics 33.3 (1997): 319-320.

Tirosh, Dina, and Ruhama Even. "To define or not to define: The case of $(-8)^\frac{1}{3}$." Educational Studies in Mathematics 33.3 (1997): 321-330.

Choi, Younggi, and Jonghoon Do. "Equality Involved in 0.999... and $(-8)^\frac{1}{3}$" For the Learning of Mathematics 25.3 (2005): 13-36.

Woo, Jeongho, and Jaehoon Yim. "Revisiting 0.999... and $(-8)^\frac{1}{3}$ in School Mathematics from the Perspective of the Algebraic Permanence Principle." For the Learning of Mathematics 28.2 (2008): 11-16.

Gómez, Bernardo, and Carmen Buhlea. "The ambiguity of the sign √." Proceedings of the Sixth Congress of the European Society for Research in Mathematics Education. 2009.

Gómez, Bernardo. "Historical conflicts and subtleties with the √ sign in textbooks." 6th European Summer University on the History and Epistemology in Mathematics Education. HPM: Vienna University of Technology, Vienna, Austria (2010).

答: Graphically the results look very different. Reinforcing the differences.


问:
Show two topologies coincide on the unit ball.
Consider the following lemma from "Lectures on von Neumann algebras":

I understand the proof of $(i)$ and $(ii)$. However, the proof says that $(iii)$ and $(iv)$ follow immediately from $(i)$ and $(ii)$. I can't see this. Can someone please explain this?

答:
Since $\sigma(\mathscr E,\mathscr F)\subset \sigma(\mathscr E,\tilde{\mathscr F})$, the latter topology is stronger. Now suppose that $x_j\to x$ in $\sigma(\mathscr E,\mathscr F)$ and $x,x_j\in\mathscr E_1$ for all $j$. For any  $\varphi\in\tilde{\mathscr F}$, by $(ii)$, $\varphi$ is $\sigma(\mathscr E,\mathscr F)$-continuous, so $\varphi(x_j)\to\varphi(x)$. Thus $x_j\to x$ in $\sigma(\mathscr E,\bar{\mathscr F})$.

If $\mathscr F$ is closed and $\varphi$ is $\sigma(\mathscr E,\mathscr F)$-continuous on $\mathscr E_1$, by $(ii)$ you get that $\varphi\in\mathscr F$. By $(i)$, $\varphi$ is $\sigma(\mathscr E,\mathscr F)$-continuous.


答:
Welcome to StackExchange. There seems to be quite a few questions that you might be asking. First, using $I$ to represent an irrational number, let's look at converting $I^{I^{I^I}}$ to its decimal representation.

$$
I^{I^{I^I}} = I^{\left(I^{\left(I^I\right)}\right)}
$$

As with all algebra, we need to work on the inside-most parentheses. Add elements of the Taylor ($I^1 = I)$ or MacLaurin ($I^0=1$) Series in order until they are smaller than the precision of your desired answer. Repeat as you work your way outwards. This, of course, is a naive implementation, but it will work. Taylor Series for $e^x$ are better known and easier to work with, so let's check that out:

$$
\begin{align*}
e^x &= a^b \\
x &= \ln\left(a^b\right) = b \ln (a)\\
I^I &= e^{I \ln(I)} \text{ giving you more rational factors}\\
I^{\left(I^I\right)} &= e^{e^{I \ln(I)} \ln (I)}\\
I^{\left(I^{\left(I^I\right)}\right)} &= e^{e^{e^{I \ln(I)} \ln (I)}\ln(I)}
\end{align*}
$$

Still, to do a numerical approximation, there's a lot of work to use a Taylor Series. Another option would be to use Newton's method since you know the values at $0$ and $1$ and finding derivatives of the function seems doable. I believe this would be a much faster implementation than the naive Taylor. A more sophisticated Taylor is likely also available in which you find an approximation of the entire beast instead of each step individually. It sounds hard.

Next, can these towers have rational representations? It's possible with probability zero. There are a infinitely more irrational numbers than than rationals. Last, it appears your example with $\sqrt 2$ misunderstands exponentiation. $a^{b^c} = a^{\left(b^c\right)} \neq \left(a^b\right)^c \neq a^{(b\cdot c)}$.

Title: What methods are used to look into $\pi^{\pi^{\pi^\pi}}$ and other stacked towers of repeating irrational numbers?

Body:
I recently saw a YouTube video here and it got me thinking, does anyone on this site have any insight on how we would tackle (change to base ten, evaluate, compute, or even put limits on approaching) some things like this?

examples:

$\displaystyle e^{e^{e^e}}$

$\displaystyle \pi^{\pi^{\pi^\pi}}$

$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}}$

$\ln(2)^{\ln(2)^{\ln(2)^{\ln(2)}}}$

What methods are proposed to even look at these things? Would it be a good idea to change to complex for pi? Do all towers of root 2 just simplify to 2?

note: Apologies for being new and not knowing how to format the text correctly.

Edit: Things I am thinking, for

$$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^\sqrt{2}}} = \sqrt{2}^{\sqrt{2}^{\sqrt{2}\cdot \sqrt{2}}} = \sqrt{2}^{\sqrt{2}^{2}} = \sqrt{2}^{2} = 2 $$

This gets me thinking that regardless of the height (greater then 3) of the tower of it will always be 2.


问:
Example of a graded ring with respect to any completely additive arithmetic function - is it isomorphic to anything?
Let $\varphi : \Bbb{N} \to \Bbb{N}$ be a completely additive arithmetic function, i.e. $\varphi(nm) = \varphi(n) + \varphi(m)$ for all $n, m \in \Bbb{N}$.

For example, $\Omega(n) = $ the total number of prime factors of $n$ including multiplicities.

Let $U \subset \Bbb{Z}^{\Bbb{N}}$ be a subset of integer sequences defined by $U = \{ u \in \Bbb{Z}^{\Bbb{N}} : \sum_{n \geq 1} u_n n = 0, \text{supp}(u) \lt \infty\}$.  That is to say vectors in $U$ represent $\Bbb{Z}$-linear relationships among some finite subset of natural numbers.  $U$ is clearly an abelian group, as adding $0$ to $0$ equals $0$ or another relation among the integers.  The set of linear relationships is in bijection with $U$.

Let $(u + v)_n = u_n + v_n$ be defined componentwise.  However, instead of componentwise multiplication, we define a convolution-type product - this is why $U$ may be isomorphic to something else in arithmetic function theory.

To make an elegant definition of multiplication, we first define the abelian groups $A_i = \{ u \in U : u_n = 0, \ \forall n \in \Bbb{N}$ such that $\varphi(n) \neq i \}$.

Then such subsets of $U$ are indeed abelian groups since the condition $\varphi(n) \neq i \implies u_n = 0$ works strictly with the indices.  So the values of the sequence are free to be any value $u_n \in \Bbb{Z}$.

Also notice that these abelian groups have pairwise intersection $A_i \cap A_j = 0 \in U$ for all $i \neq j$.  This is because of the fact that two distinct inverse images of a function are disjoint, here $\varphi^{-1}(i) \cap \varphi^{-1}(j) = \varnothing$ and so form a partition of $\Bbb{N}$.  In the case of $\varphi = \Omega$, you can write this as $\Bbb{N} = \{1\} \uplus \Bbb{P} \uplus \Bbb{P}^{2} \uplus \dots$ where $\Bbb{P} \subset \Bbb{N}$ is the set of prime numbers and $\Bbb{P}^k = \{ q_1 \cdots q_k : q_j \in \Bbb{P}\}$ is the set of products of $k$ primes.

Now, since we're defining a graded ring structure any two general elements $x, y \in U$ that are nonzero belong to a unique $A_i, A_j$, respectively.

Thus let $x, y \in U$ and $x \in A_i, y \in A_j$.  Then define the convolution product for the ring as:

$$
(xy)_n = \sum_{ab = n \\ \varphi(a) = i \\ \varphi(b) = j} x_a y_b
$$

This is essentially derived from multiplying two linear relations and taking their coefficients:

$$
0 = (\sum_{n \geq 1} x_n n)(\sum_{n \geq 1} y_n n)
$$

For example: multiplying $\sum_{p \text{ prime}} x_p p = 0$ and $\sum_{p, q \text{ prime}} y_p pq = 0$, corresponding to elements $x \in A_1, y \in A_2$, will clearly yield a relation corresponding to an element of $A_3$, however we have to keep careful track of where the coefficients end up and that is the purpose of the above formula.

You can see by looking at the formulas for $+$ and $\cdot$ that the distributive law should hold.  Associativity of $\cdot$ is brought over from associativity in $\Bbb{Z}$, when we multiply two linear relations and take their coefficients.

Thus, we have that for all $i,j \in \Bbb{N} \cup 0$, $A_i A_j \subset A_{i + j}$, each $A_i$ is an abelian group and $A_i \cap A_j = 0$.  That is precisely the definition of a graded ring:

$$
R_{\varphi} = \bigoplus_{n \geq 0 } A_n
$$

Note that $A_0 := \Bbb{Z}$ must be defined.  See arguments in comments of an answer below.  This ensures unitality of the ring.

Thus, let $R_{\varphi}$ be called the graded ring associated with the completely additive arithmetic function $\varphi$.

Note that for two completely additive arithmetic functions $\varphi, \psi$, $R_{\varphi}$ need not be isomorphic to $R_{\psi}$ as rings, though their underlying additive abelian groups $U$ are equal.

My question is, can you see whether $R_{\varphi}$ is isomorphic to some ring already known about?

My next question is: how would you generalize this, such as replacing the coefficients with $\Bbb{Q}$ or some integral domain, replacing $\Bbb{N}$ with $\Bbb{Z}$ or a group, so that $\varphi$ becomes an additive group character, etc. ?

Note that any such $\varphi : \Bbb{N} \to \Bbb{N} \cup 0$ extends nicely to all of $\Bbb{Z}$ by setting $\varphi(-n) = \varphi(|n|)$ and then it extends nicely to all of $\Bbb{Q}^{\times}$ via $\varphi(x/y) := \varphi(x) - \varphi(y)$ where it becomes an abelian group homomorphism onto $(\Bbb{Z}, +)$.

答:
Without the $U$ condition your ring is that of Dirichlet polynomials with integer coefficients (more generally formal Dirichlet series), decomposed (graded by the values taken by $\varphi$) as $$\sum_{n\ge 1} a_n n^{-s}= \sum_j \sum_{n\ge 1,\varphi(n)=j} a_n n^{-s}$$

This ring is $\Bbb{Z}[2^{-s},3^{-s},5^{-s},\ldots]\cong \Bbb{Z}[X_1,X_2,X_3,\ldots]$ (for the formal Dirichlet series take the formal series in the $X_j$).

Don't forget that the simplest additive function (and the natural grading on Dirichlet series) is $\log n$.

With the $U$ condition then $1$ is not in your ring and everything is not quite natural (the Dirichlet polynomials whose every graded piece vanish at $-1$..).

Grading by $\Omega(n)$ is somewhat natural in $\zeta(s)=\exp(\sum_{p^k} p^{-sk}/k)$.


问:
Rate of convergence of an inductively defined sequence $x_{n+1}=f(x_n)$
Let $x_0\in(0,1)$, $\mu>1$ and $f(x)=x-x^\mu$. Define $x_{n+1}=f(x_n)$. I need help proving that there exist positive $c_1, c_2$ such that
\begin{equation}
x_n\leq c_1 n^{-c_2}.
\end{equation}

答:
It's easy to prove that $x_n \in (0,1)$. Besides, the sequence $x_n$ is decreasing, hence this sequence converges to the solution of $x = x-x^\mu$ (which is $0$). In other word, $$x_n \xrightarrow{n\rightarrow +\infty} 0 \tag{1}$$

Now, we study the sequence $x_n^{1-\mu}$, we have
\begin{align}
x_{n+1}^{1-\mu} &= (x_n-x_n^{\mu})^{1-\mu} \\
&= x_n^{1-\mu}(1-x_n^{\mu-1})^{1-\mu} \\
&= x_n^{1-\mu}\left(1+(1-\mu) x_n^{\mu-1}-\frac{(1-\mu)\mu}{2} x_n^{2(\mu-1)}+ \mathcal{O}(x_n^{3(\mu-1)})   \right) \\
&= x_n^{1-\mu}+(1-\mu) -\frac{(1-\mu)\mu}{2} x_n^{(\mu-1)}+ \mathcal{O}(x_n^{2(\mu-1)}) \tag{2}   \\
\end{align}

From (1), we deduce that the terms $-\frac{(1-\mu)\mu}{2} x_n^{(\mu-1)}+ \mathcal{O}(x_n^{2(\mu-1)}) \xrightarrow{n\rightarrow +\infty} 0$, so from (2), we have
$$x_{n+1}^{1-\mu}-x_n^{1-\mu} \xrightarrow{n\rightarrow +\infty}(1-\mu) \tag{3}$$

According to the Stolz-Cesano theorem for the case $a_n = x_n^{1-\mu}$ and $b_n=n$, we can deduce from (3) that
$$\frac{x_n^{1-\mu}}{n}\xrightarrow{n\rightarrow +\infty} (1-\mu)\iff \frac{x_n}{n^{-\frac{1}{\mu-1}}}\xrightarrow{n\rightarrow +\infty} (1-\mu)^{-\frac{1}{\mu-1}} \tag{4}$$

From (4), from the definition of limit, for all $\epsilon >0$, there exists $N_0(\epsilon)$ such that for all $n>N_0(\epsilon)$, we have
$$\left|\frac{x_n}{n^{-\frac{1}{\mu-1}}} - (1-\mu)^{-\frac{1}{\mu-1}} \right| \le \epsilon$$
In particular, for all $n>N_0(\epsilon)$, we have
$$x_n \le ((1-\mu)^{-\frac{1}{\mu-1}}+\epsilon)n^{-\frac{1}{\mu-1}}$$

Hence, we can define

$$c_2=\frac{1}{\mu-1}$$
$$c_1 = \max\left(1-\mu)^{-\frac{1}{\mu-1}}+\epsilon;  \max_{i=1,\ldots,N_0(\epsilon)}\left(\frac{x_i}{i^{-c_2}}\right) \right)$$

and then we can conclude that for all $n \ge 1$
$$x_n \le c_1 n^{-c_2}$$


问:
How to prove $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$?
I have a question regarding Functional Analysis:

Assume that $X$ is a Banach space and that $P \in B(X)$ is a projection with $ran P \neq {[0]}$.

Then I have easily shown that $\lambda = 0$ and $\lambda = 1$ are eigenvalues of $P$.

I have troubles proving that if $\lambda \notin\{0,1\}$, then it follows that we have $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$.

Any help would be grateful.

答:
You can derive the inverse for $|\lambda| > 1$ using a geometric series:
$$
       (\lambda I-P)^{-1}=\frac{1}{\lambda}(I-\frac{1}{\lambda}P)^{-1}=\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^n}P^n=\frac{1}{\lambda}(I+\sum_{n=1}^{\infty}\frac{1}{\lambda^n}P) \\
       = \frac{1}{\lambda}(I+\frac{1/\lambda}{1-1/\lambda}P) = \frac{1}{\lambda}I+\frac{1}{\lambda(\lambda-1)}P.
$$
Both sides are holomorphic in $\lambda$ except at $\lambda=0,1$. So the expressions are equal for all such $\lambda$.


问:
Why is this functional norm-continuous?
Consider the following fragment from the book "Lectures on von Neumann algebras":

Why is the marked line true? I'm probably missing some basic functional analytic fact here.

答:
With a stronger topology, continuity is easier because it implies the convergence in the weaker topology.

If $x_j\to0$ in norm, then $\psi(x_j)\to0$ for all $\psi\in\mathscr F$, so $x_j\to0$ in $\sigma(\mathscr E,\mathscr F)$ and then $\varphi(x_j)\to0$.

Title: Why is this functional norm-continuous?


问:
$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$
I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$

applying Taylor's series.

$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}=    \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}=  \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$

$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3-  x^5+ o(x^5))}=  \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 -  x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2}  \sim \frac{2}{3}$$

The suggested solution in my book is $2$. can someone indicate where I made mistakes?

答:
The OP's error has been pegged to an error with the logarithm of a difference being equated to the difference of logarithms. An alternative approach, which avoids all this, is to note that

$${\ln(1-\cos2x)\over\ln\tan2x}={\ln(2\sin^2x)\over\ln\displaystyle\left({2\tan x\over1-\tan^2x}\right)}={2\ln\sin x+\ln2\over\ln\sin x-\ln\cos x+\ln2-\ln(1-\tan^2x)}$$

Since $\ln\sin x\to-\infty$ as $x\to0^+$ while all the other terms tend to finite limits (i.e. $\ln2\to\ln2$, $\ln\cos x\to\ln\cos0=\ln1=0$, and $\ln(1-\tan^2x)\to\ln(1-\tan^20)=\ln1=0$), the requested limit is easily seen to equal $2$.


问:
Is the cardinality of a set necessarily a natural number?
I've never seen phrases like "$\sqrt{5}$ people" or "a set with $\pi$ many elements". Are there sets with cardinality, say, $\frac{1}{2}$?

Edit: As Brian M. Scott pointed out, the only real numbers that are cardinalities of sets are the non-negative integers. Could you explain why is it so?

答:
Fundamentally, cardinals and real numbers are different things.  You can think of the cardinality of a set as some abstract object ("cardinal") assigned to it, in such a way that two sets get assigned the same cardinal if and only if there is a bijection between them.

But there is a natural way to identify the finite cardinals with the natural numbers: namely, identify a cardinal $a$ with the natural number $n_a$ such that the set $\{1, 2, \dots, n_a\}$ has cardinality $a$ (i.e. any other set with cardinality $a$ has a bijection with $\{1,2,\dots, n_a\}$ and any other set with cardinality $a$).  (In this discussion, "natural numbers" includes 0.)  This is a nice identification because it makes cardinal arithmetic match up with the arithmetic of natural numbers.  For instance:

Ordering: $n_a \le n_b$ if and only if any set of cardinality $a$ has an injection into any set of cardinality $b$

Addition: If $A,B$ are disjoint and have cardinalities $a,b$ respectively, then the cardinality $c$ of their union $C = A \cup B$ has $n_c = n_a + n_b$.

Multiplication: If $A,B$ have cardinalities $a,b$ and $C = A \times B$ has cardinality $c$, then $n_c = n_a n_b$.

Exponents: If $A,B$ have cardinalities $a,b$, and $C = A^B$ is the set of all functions from $B$ to $A$, then $n_c = n_a^{n_b}$.

One could imagine a system that identified other (infinite) cardinals with real numbers other than the natural numbers.  For instance, nobody could stop me from proposing a system in which the cardinality $\aleph_0$ of the set of integers is identified with the real number $22/7$.  But this system wouldn't have the properties listed above.  For instance, $22/7 \le 4$, but there is no injection from $\mathbb{Z}$ to $\{1,2,3,4\}$, and $\mathbb{Z} \times \{1,2,3,4,5,6,7\}$ definitely does not have the same cardinality as $\{1,2,\dots, 21,22\}$.  In fact, it's not hard to see that there would be no way of identifying infinite cardinals with real numbers that would preserve the above list of properties.

The properties above are very useful, and so in order to preserve them, we generally do not attempt to identify any other real numbers with cardinals.  In principle, there could be another system that, although it didn't satisfy the above properties, had some other useful properties.  But I've never heard of one that was useful enough to attract much attention.


问:
Proof of Lagrange theorem - Order of a subgroup divides order of the group
The Lagrange theorem states:

If $G$ is a finite group, and $H$ a subgroup of $G$, then the order of $H$ will divide the order of $G$. More precisely, $|G| = |H| \cdot (\text{number of left cosets of }H)$:

$$|G| = |H| \cdot (G:H)$$

The proof I have in my notes says:

$G$ consists of $\{G:H\}$ cosets, each of them consists of $|H|$ elements, the cosets are disjoint.

That's it. How does this prove the theorem?

答:
Given a group $G$ and $H\le G$:

$x\sim y \stackrel{(def.)}{\iff} xy^{-1}\in H$ is an equivalence relation on $G$;

$[x]_\sim = Hx$;

the map $h\mapsto hx$ is a bijection from $H$ to $Hx$.

If $G$ is finite, from 1 to 3 follows that $|H|\mid |G|$.


问:
Inscribed circle of an isosceles triangle
In a triangle $ABC$, $AC = BC = 24$ and a circle with center $J$ is inscribed. If $CH$ is altitude $(CH\perp AB,H\in AB)$ and $CJ:CH=12:17$, then find the length of $AB$.

Triangle $ABC$ is isosceles so $H$ is also the midpoint of $AB.$ So we can just find $AH$ (or $BH$) and double it to find $AB$. Let $CJ=12x, CH=17x\Rightarrow JH=5x$. I am not sure what to do from here. Thank you in advance!

答: Let $JH = r$, and $CH = h$.  Then there are two formulas for the triangle's area:  $$|\triangle ABC| = \frac{1}{2}ch = rs,$$ where $AB = c$ and $s = (a+b+c)/2$ is the semiperimeter.  Since we know $a = b = 24$, it follows that $$ch = r(48+c).$$  But we are given that $CJ:CH = 12:17$, or $$\frac{h-r}{h} = 1 - \frac{r}{h} = \frac{12}{17}.$$   Thus $$c = \frac{r}{h}\left(48 + c\right) = \frac{5}{17}\left(48 + c\right),$$ which allows us to solve for $c$.


问:
Permutations with No letters-digits repetitions
A car license plate consists of 3 capital letters of the English alphabet in the first 3 positions of the license plate followed by 4 digits from 0 to 9.  How many different plates can we have if neither the same letter nor the same digit on the plate is allowed to be repeated?

I was wondering if the right answer is $26! \cdot 10!$, based on permutations?

I would appreciate for your guidance, as I face great difficulty in it.
Thank you very much in advance.

答:
You have not taken the number of letters or digits into account.

Answer the following questions, keeping in mind that neither a letter nor a digit may be repeated.

In how many ways can the first position be filled with a letter?

Given that the first position has been filled with a letter, in how many ways can the second position be filled with a letter?

Given that the first two positions have been filled with letters, in how many ways can the third position be filled with a letter?

In how many ways can the fourth position be filled with a digit?

Given that the fourth position has been filled with a digit, in how many ways can the fifth position be filled with a digit?

Given that the fourth and fifth positions have been filled with digits, in how many ways can the sixth position be filled with a digit?

Given that the fourth, fifth, and sixth positions have been filled with digits, in how many ways can the seventh position be filled with a digit?

Use the Multiplication Principle to conclude.

  There are $26$ ways to fill the first position with a letter, $25$ ways to fill the second position with a different letter, $24$ ways to fill the third position with a letter that has not yet been used, $10$ ways to fill the fourth position with a digit, $9$ ways to fill the fifth position with a different digit, $8$ ways to fill the sixth position with a digit that has not yet been used, and $7$ ways to fill the seventh position with a digit that has not yet been used.  Hence, by the Multiplication Principle, there are $$26 \cdot 25 \cdot 24 \cdot 10 \cdot 9 \cdot 8 \cdot 7$$ license plates in which the first three characters are distinct letters of the English alphabet and the last four characters are distinct decimal digits.


问:
How to isolate j?
can anyone explain me how to isolate the j variable please?

$$q = \frac{1 - (1 + j)^{-n}}{j} p $$

TIA

答: This can be solved exactly ("isolating $j$") only under very special circumstances. If $n > 4$ it is probably hopeless (and for $n = 3$ or 4 the exact solution will turn out to be a horrible mess).</p>

Title: How to isolate j?


问:
Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$
I'm struggling to grasp this particular question:

When a polynomial $f(x)$ is divided by $x+3$, the quotient is $2x^2-x+7$ and the remainder is $10$. What is $f(x)$?

This is what I did:

$$\begin{align}
f(x) &= (x+3)(2x^2-x+7) \\
&= 2x^3-x^2+7x+6x^2-3x+21 \\
&= 2x^3+5x^2+4x+21
\end{align}$$

Making sure it is correct:

Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that

答:
"Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that"

Don't you?

If $f(x) = (x+3)(2x^2 -x + 7)$ then

$f(x) + r = (x+3)(2x^2-x +7) + r$

And $f(x) + 10 = (x+3)(2x^2-x +7) +10$.

And doesn't that mean the remainder of $f(x)\div (x+3) = 10$?

After all if we divide $(x+3)(2x^2 -x +7) + 10$ we'd get $\frac {(x+3)(2x^2 - x+ 7)}{x+3} = 2x^2 -x + 7$ and then .... we'd be left with the remainder of $10$.

So if you are told $f(x)$ divided by $d(x)$ has a quotient of $q(x)$ and a remainder of $r$ then you can just do $f(x) = d(x)q(x) + r$.

And so $g(x) = (x+3)(2x^2 - x+7) + 10=$

$(2x^3+5x^2+4x+21) + 10=$

$2x^3 + 5x^2 +4 + 31$

has to do it and if we did that we'd have:


问:
A projection is a fibration
I recently learned about the concept of a Hurewicz fibration. I tried to prove that for a product of topological spaces $X \times Y$, the projection $p: X \times Y \rightarrow X$ is a Hurewicz fibration, something that on wikipedia is described as 'very easily seen'. I thought that maybe I should somehow use the universal propery of the product, but I still don't see how to proceed after setting up the commutative diagram

$$
\require{AMScd}
\begin{CD}
Z @>g>> X \times Y\\
@V{i_0}VV @VVpV \\
Z \times I @>>h> X
\end{CD}
$$

in order to obtain a lifting morphism $d: Z \times I \rightarrow X \times Y$. Could someone give me some suggestions, or a reference to a resource where this fact is proven?

答:
Let $g: Z \to X \times Y$ be a given map, with a specified homotopy $h: Z \times I \to X$ such that $h(z,0) = p(g(z))$ for all $z$.

Then, we can lift the homotopy to $d: Z \times I \to X \times Y$ by $d(z,t) := (h(z,t), \pi_2(g(z)))$.

Title: A projection is a fibration


答: There are $\frac{26!}{23!}$ ways to select the letter portion of the plate and $\frac{10!}{6!}$ ways to select the number portion.  The product is the total number of possible plates with these contraints.

Title: Permutations with No letters-digits repetitions

Body:
A car license plate consists of 3 capital letters of the English alphabet in the first 3 positions of the license plate followed by 4 digits from 0 to 9.  How many different plates can we have if neither the same letter nor the same digit on the plate is allowed to be repeated?

I was wondering if the right answer is $26! \cdot 10!$, based on permutations?

I would appreciate for your guidance, as I face great difficulty in it.
Thank you very much in advance.


问:
Matrix transformation $f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$
I am self-learning Linear Algebra (mainly through Khan Academy and Youtube). I encountered this linear transformation, and was wondering how would this be solved.

$f:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$

$f \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $  = $ \begin{pmatrix}
4 & 1 & 3\\
2 &-1 & 3\\
2 & 2 & 0
\end{pmatrix} $ $\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $

Establish $x, y$ and $z$ such that,

$f\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} $ = $\begin{pmatrix}
4\\
5\\
-1
\end{pmatrix} $

Do I just need to multiply the values of $f$ for the $3\times3$ matrix? What does this mean overall?

答:
You need to solve the non-homogeneous system

$$ \begin{pmatrix}
4 & 1 & 3\\
2 &-1 & 3\\
2 & 2 & 0
\end{pmatrix} \begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =  \begin{pmatrix}
4\\
5\\
-1
\end{pmatrix}$$

Form then the augmented coefficients matrix and reduce it by rows:

$$ \begin{pmatrix}
4 & 1 & 3&:&4\\
2 &-1 & 3&:&5\\
2 & 2 & 0&:&-1
\end{pmatrix}\stackrel{R_1\leftrightarrow R_3\cdot\left(\frac12\right)}\longrightarrow\begin{pmatrix}
1 & 1 & 0&:&-1/2\\
2 &-1 & 3&:&5\\
4 & 1 & 3&:&4\\
 \end{pmatrix}\stackrel{R_2-2R_1\\R_3-4R_1}\longrightarrow$$

$$\begin{pmatrix}
1 & 1 & 0&:&-1/2\\
0 &-3 & 3&:&6\\
0 & -3 & 3&:&6\\
 \end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1 & 1 & 0&:&-1/2\\
0 &-3 & 3&:&6\\
0 & 0 & 0&:&0\\
 \end{pmatrix}$$

Well, there are infinite solutions to the above system: begin with row $3$ , and remember the columns represent $\;x, y, z\;$ from left to right, so:

$$R_2:\;\;-3y+3z=6\implies y=z-2\\R_1:\;\;x+y=-\frac12\implies x=-\frac12-y=-\frac12-z+2=\frac32-z$$

Now just choose a nice $\;z\;$ , say $\;z=0\implies y=-2\,,\,\,x=\cfrac32\;$ , and one solution to your problem is

$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
3/2\\
-2\\
0
\end{pmatrix}$$

as you can easily check.


问:
oddness and evenness of sum of three square integers
I am trying to prove that $\forall x, y ,z, v \in \mathbb{N}, x^2+y^2+z^2=v^2,$ $v$ is even if and only if $x,y,$ and $z$ are even. So far, after having remarked that studying the evenness of $v^2$ is  the same as studying the evenness of $v^2$ ($v^2\equiv v \mod 2$), I've been trying to separate it into four main cases : either there is none of $x,y,z$ that is even, or only one, or only two, or all three. However I find that if there are two that are odd, then their individual squares are odd, so the sum of these squares will be even, which eventually makes it so that $v^2$ is even with only $x,y$ odd. Contradicting what I have to prove. Do you have any ideea where I might have gone wrong ?

答:
A square is always congruent to $1$ or $0$ $\pmod{4}$. Hence if you assume that $x,y$ are odd they'll be congruent to $1 \pmod{4}$. Thus, if $z$ is odd we have $x^2+y^2+z^2 \equiv v^2 \equiv 2 \pmod{4}$ which is absurd.
One possible way of doing the problem is the following one :
Firstly it is easy to prove that if $x,y,z$ are even so is $v$. Now lets prove that if $v$ is even so are $x,y,z$.
Suppose that among t $x,y,z$ two of them have different parities. As we already showed that the case two odds/one even doesnt work lets consider the case where we have two even numbers and one odd. The sum of the square of two even numbers is congruent to $0 \pmod{4}$. Adding the square of the one odd number we have we find that $v^2 \equiv 1\pmod{4}$ which is absurd as we supposed $v$ even.
Hope this'll help you.
Hubert


问:
Difference between prime counting functions(Resolved)
Good afternoon everyone! Given $N$ and $k$ are natural numbers, with $k<N$, and $\pi(x)$ counts the number of primes less than x or equal to x(edited), can someone give me some tips for simplifying(perhaps removing the dependency on N entirely) the following expression. $\pi(N) - \pi(N-k)$, it seems like it should be simpler than I'm making it. Thank you in advance!

答:
By the prime number theorem $\pi(n) \approx x / \log n$, and thus
$$ \pi(N + k) - \pi(N) \approx \frac{N+k}{\log(N + k)} - \frac{N}{\log N},$$
and if $N$ is much larger than $k$ then this is well-approximated by
$$ \frac{N + k}{\log N} - \frac{N}{\log N} = \frac{k}{\log N}.$$

I'll note that this is the same as you would get from the prime number theorem probability statement that the probability that a random integer of size $N$ is prime is about $\frac{1}{\log N}$, and thus taking $k$ such integers would give you approximately $k / \log N$ primes.

Finally, I'll also note that this heuristic shows that you absolutely cannot hope to get a bound independent of $N$.