问:
Problem with the proof of of $\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta$ in Simon, Blume 1994
In Simon, Blume 1994, p. 215, we find the following proof of the identity
$$
\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta
$$

Without loss of generality, we can work with $\mathbf{u}$ and $\mathbf{v}$ as vectors with tails at the origin $\mathbf{0} ;$ say $\mathbf{u}=\overrightarrow{O P}$ and $v=\overrightarrow{O Q}$. Let $\ell$ be the line through the vector $\mathbf{v}$, that is, the line through the points 0 and $Q$. Draw the perpendicular line segment $m$ from the point $P$ (the head of $\mathbf{u}$ ) to the line $\ell$, as in Figure 10.20 . Let $R$ be the point where $m$ meets $\ell$. Since $R$ lies on $\ell, \overrightarrow{O R}$ is a scalar multiple of $\mathbf{v}=\overrightarrow{O Q}$. Write $\overrightarrow{O R}=t v .$ Since $\mathbf{u}, t \mathbf{v},$ and the segment $m$ are the three sides of the right triangle $O P R,$ we can write $m$ as the vector $\mathbf{u}-t \mathbf{v}$. Since $\mathbf{u}$ is the hypotenuse of this right triangle,
$$
\cos \theta=\frac{\|t \mathbf{v}\|}{\|\mathbf{u}\|}=\frac{t\|\mathbf{v}\|}{\|\mathbf{u}\|}
$$
On the other hand, by the Pythagorean Theorem and Theorem $10.2,$ the square of the length of the hypotenuse is:
$\|\mathbf{u}\|^{2}=\|t \mathbf{v}\|^{2}+\|\mathbf{u}-t \mathbf{v}\|^{2}$
$$
\begin{array}{l}
=t^{2}\|\mathbf{v}\|^{2}+(\mathbf{u}-t \mathbf{v}) \cdot(\mathbf{u}-t \mathbf{v}) \\
=t^{2}\|\mathbf{v}\|^{2}+\mathbf{u} \cdot \mathbf{u}-2 \mathbf{u} \cdot(t \mathbf{v})+(t \mathbf{v}) \cdot(t \mathbf{v}) \\
=t^{2}\|\mathbf{v}\|^{2}+\|\mathbf{u}\|^{2}-2 t(\mathbf{u} \cdot \mathbf{v})+t^{2}\|\mathbf{v}\|^{2}
\end{array}
$$
or
$$
2 t(\mathbf{u} \cdot \mathbf{v})=2 t^{2}\|\mathbf{v}\|^{2}
$$
It follows that
$$
t=\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^{2}}
$$
Plugging equation (5) into equation (4) yields
$$
\cos \theta=\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}
$$

This identity is then used to prove the following theorem

Theorem 10.4 The angle between vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbf{R}^{\mathbf{n}}$ is
(a) acute, if $\mathbf{u} \cdot \mathbf{v}>0$
(b) obtuse, if $\mathbf{u} \cdot \mathbf{v}<0$,
(c) right, if $\mathbf{u} \cdot \mathbf{v}=0$.

However, I don't understand this proof fully since it seems that
$$
\cos \theta=\frac{\|t \mathbf{v}\|}{\|\mathbf{u}\|}=\frac{|t|\|\mathbf{v}\|}{\|\mathbf{u}\|}\neq\frac{t\|\mathbf{v}\|}{\|\mathbf{u}\|}
$$
Especially, this would imply that
$$
\cos \theta=\frac{|\mathbf{u} \cdot \mathbf{v}|}{\|\mathbf{u}\|\|\mathbf{v}\|}
$$
which would contradict the theorem 10.4. Am I missing something, or is there an error in this proof?

Thank you in advance for your help!

答: The authors have written too much. They should have gone directly from $\cos\theta$ to $\frac{t\|\mathbf v\|}{\|\mathbf u\|}$ – in the vectors' plane $t\mathbf v$ is still a signed quantity.


问:
Show that the following subspace is two dimensional
L= span {$e_{1}$, $e_{2}$} and K = span{$f_{1}$ $f_{2}$} are two disjoint subspaces in $\mathbb R^4$. Every 2x2 matrix A represents a linear transformation from L to K such as

A($x_{1}$$e_{1}$+$x_{2}$$e_{2}$)= $y_{1}$$f_{1}$+$y_{2}$$f_{2}$ where [$y_{1}$ $y_{2}$]$^{T}$=$\textbf{A}$ [$x_{1}$ $x_{2}$]$^{T}$

Show that the subspace $W_{A}$ of all vectors $x_{1}$$e_{1}$+$x_{2}$$e_{2}$+A($x_{1}$$e_{1}$+$x_{2}$$e_{2}$) such that $x_{1}$, $x_{2}$ $\in$ $\mathbb R^2$ is a two dimenisonal subspace in $\mathbb R^4$

The solution defines the transformation, ($x_{1}$, $x_{2}$) $\rightarrow$ $x_{1}$$e_{1}$+$x_{2}$$e_{2}$+A($x_{1}$$e_{1}$+$x_{2}$$e_{2}$), and shows that the null space is trivial, making the image space ($W_{A}$) two dimensional. However my attempt was to rewrite all the vectors in the
subspace $W_{A}$ to $x_{1}$$e_{1}$+$x_{2}$$e_{2}$+$y_{1}$$f_{1}$+$y_{2}$$f_{2}$, which is a linear combination of 4 linearly independent vectors (as L and K are disjoint subspaces), how can the subspace then be 2 dimensional?

$\textbf{EDIT}$: Why does that fact that the kernel of the transformation ($x_{1}$, $x_{2}$) $\rightarrow$ $x_{1}$$e_{1}$+$x_{2}$$e_{2}$+A($x_{1}$$e_{1}$+$x_{2}$$e_{2}$) is trivial, imply that the dimension of $W_{A}$ is 2?

答: The space $W_A$ is a subspace of$$W=\{x_1e_1+x_2e_2+y_1f_1+y_2f_2\mid x_1,x_2,y_1,y_2\in\Bbb R^4\}$$and this space is indeed $4$-dimensional. But $W_A\varsubsetneq W$, and so $\dim W_A<4$. And we have $W_A\varsubsetneq W$ because, when $x_1e_1+x_2e_2+y_1f_1+y_2f_2\in W_A$, the numbers $x_1$, $x_2$, $y_1$, and $y_2$ cannot be just any numbers; we must have$$\begin{bmatrix}y_1\\y_2\end{bmatrix}=A.\begin{bmatrix}x_1\\x_2\end{bmatrix}.$$

ContentLicense: CC BY-SA 4.0


答: The quoted proof was careless at one point. The generally correct expression is $t\Vert v\Vert/\Vert u\Vert$ rather than $|t|\Vert v\Vert/\Vert u\Vert$. The sign of $t$ boils down to whether the multiple of $v$ onto which $u$ is projected is parallel or antiparallel to $v$.

Title: Problem with the proof of of $\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta$ in Simon, Blume 1994

Body:
In Simon, Blume 1994, p. 215, we find the following proof of the identity
$$
\mathbf{u} \cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\| \cos \theta
$$

Without loss of generality, we can work with $\mathbf{u}$ and $\mathbf{v}$ as vectors with tails at the origin $\mathbf{0} ;$ say $\mathbf{u}=\overrightarrow{O P}$ and $v=\overrightarrow{O Q}$. Let $\ell$ be the line through the vector $\mathbf{v}$, that is, the line through the points 0 and $Q$. Draw the perpendicular line segment $m$ from the point $P$ (the head of $\mathbf{u}$ ) to the line $\ell$, as in Figure 10.20 . Let $R$ be the point where $m$ meets $\ell$. Since $R$ lies on $\ell, \overrightarrow{O R}$ is a scalar multiple of $\mathbf{v}=\overrightarrow{O Q}$. Write $\overrightarrow{O R}=t v .$ Since $\mathbf{u}, t \mathbf{v},$ and the segment $m$ are the three sides of the right triangle $O P R,$ we can write $m$ as the vector $\mathbf{u}-t \mathbf{v}$. Since $\mathbf{u}$ is the hypotenuse of this right triangle,
$$
\cos \theta=\frac{\|t \mathbf{v}\|}{\|\mathbf{u}\|}=\frac{t\|\mathbf{v}\|}{\|\mathbf{u}\|}
$$
On the other hand, by the Pythagorean Theorem and Theorem $10.2,$ the square of the length of the hypotenuse is:
$\|\mathbf{u}\|^{2}=\|t \mathbf{v}\|^{2}+\|\mathbf{u}-t \mathbf{v}\|^{2}$
$$
\begin{array}{l}
=t^{2}\|\mathbf{v}\|^{2}+(\mathbf{u}-t \mathbf{v}) \cdot(\mathbf{u}-t \mathbf{v}) \\
=t^{2}\|\mathbf{v}\|^{2}+\mathbf{u} \cdot \mathbf{u}-2 \mathbf{u} \cdot(t \mathbf{v})+(t \mathbf{v}) \cdot(t \mathbf{v}) \\
=t^{2}\|\mathbf{v}\|^{2}+\|\mathbf{u}\|^{2}-2 t(\mathbf{u} \cdot \mathbf{v})+t^{2}\|\mathbf{v}\|^{2}
\end{array}
$$
or
$$
2 t(\mathbf{u} \cdot \mathbf{v})=2 t^{2}\|\mathbf{v}\|^{2}
$$
It follows that
$$
t=\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^{2}}
$$
Plugging equation (5) into equation (4) yields
$$
\cos \theta=\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}
$$

This identity is then used to prove the following theorem

Theorem 10.4 The angle between vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbf{R}^{\mathbf{n}}$ is
(a) acute, if $\mathbf{u} \cdot \mathbf{v}>0$
(b) obtuse, if $\mathbf{u} \cdot \mathbf{v}<0$,
(c) right, if $\mathbf{u} \cdot \mathbf{v}=0$.

However, I don't understand this proof fully since it seems that
$$
\cos \theta=\frac{\|t \mathbf{v}\|}{\|\mathbf{u}\|}=\frac{|t|\|\mathbf{v}\|}{\|\mathbf{u}\|}\neq\frac{t\|\mathbf{v}\|}{\|\mathbf{u}\|}
$$
Especially, this would imply that
$$
\cos \theta=\frac{|\mathbf{u} \cdot \mathbf{v}|}{\|\mathbf{u}\|\|\mathbf{v}\|}
$$
which would contradict the theorem 10.4. Am I missing something, or is there an error in this proof?

Thank you in advance for your help!


问:
Why does $\log^{100}(n)$ grow asymptotically slower than $n^{7/8}$
More formally:
How do I compute the following limit:

$$
\lim _{x \rightarrow \infty} \frac{\log ^{100}(x)}{x^{7 / 8}}
$$

I would start using l'Hôpital's rule and notice the pattern, is there some other procedure? (without knowing it)

答: Squaring $x$ multiplies the numerator by $2^{100}$ and the denominator by $x^{7/8}$. Since $2^{100}$ is a constant but $x^{7/8}$ is unbounded, there exists an $x$ for which $x^{7/8}>2^{100}$, i.e. the denominator grows faster than the numerator. Thus the limit is $0$.


问:
Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2]
I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.

I tried multiple ways to understand why this assumption was made, but I can't figure it out. My first assumption was that since $m \in S$ it's true that:

$m > 3^{m/3}$

and by consequence:

$m^3 > 3^{m}$

but it still does not prove $3^m < m$. Any help is very appreciated.

答:
I already accepted @fleablood's answer. I'd just like to share my own proof after some work on it as further reference for other people working on this problem.

Theorem: $n \leq 3^{n/3}\text{ for }n \geq 0$

Proof:

Let C be the set of counterexamples to the theorem, namely
$$C ::= \{n \in \mathbb{N} | n > 3^{\frac{n}{3}}\}$$

For the purpose of a contradiction, assume $C$ is not empty. By the Well-ordering principle, there's a least element, $c \in C$. The $c$ element can't be:

1, because $1<3^\frac{1}{3}$;

2, because $2<3^\frac{2}{3}$;

3, because $3=3^\frac{3}{3}$;

4, because $4<4^\frac{4}{3}$.

Therefore, $c>4$.

If $c$ is the least counterexample for the theorem, $c-1$ must hold true, once it's less than $c$ itself. So:

$$
c-1 \leq 3^\frac{c-1}{3}\\
c-1^3 \leq 3^{c-1}\\
c-1^3 \leq (3^c)(3^{-1})\\
c-1^3 \leq \frac{3^c}{3}\\
3(c-1)^3 \leq 3^c\\
$$

$c \in C$, therefore, $c > 3^{\frac{c}{3}}$:
$$
c > 3^{\frac{c}{3}}\\
c^3 > 3^c\\
3^c < c^3
$$

So, we can state that $3(c-1)^3 \leq 3^c < c^3$ and that $3(c-1)^3 < c^3$. Now, by manipulation:

$$
3(c-1)^3 < c^3\\
3(c^3-3c^2+3c-1) < c^3\\
3c^3-9c^2+9c-3 < c^3\\
2c^3-9c^2+9c-3 < 0\\
2c^3+9c < 9c^2+3\\
$$
Dividing both sides by $c^2$, we get:
$$
2c+\frac{9}{c}<9+\frac{3}{c^2}\\
$$
As stated  before, $c>4$, so:
$$
c>4\\
c^2>16\\
\frac{1}{c^2}<\frac{1}{16}\\
\frac{3}{c^2}<\frac{3}{16}\\
9+\frac{3}{c^2}<9+\frac{3}{16}\\
9+\frac{3}{c^2}<9.1875<10\\
9+\frac{3}{c^2}<10
$$

Back to the main inequality:

$$
2c+\frac{9}{c}<9+\frac{3}{c^2}<10\\
$$
Since $c \in \mathbb{N}$, we can assume $2c < 2c+\frac{9}{c}$:
$$
2c<2c+\frac{9}{c}<9+\frac{3}{c^2}<10\\
2c<10\\
c<5
$$

$c \in \mathbb{N}$, $c>4$ and $c<5$ is a contradiction and therefore $C$ must be empty. The theorem is true.

QED


问:
Polya - Szego question
The following problem can be found in Polya and Szego book - exercise 19.2 on page 51. Prove that the sum is $\gamma$.

$\frac{1}{1}\\
-\frac{1}{2} - \frac{1}{3}\\
\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\\
-\frac{1}{5} - \frac{1}{6} - \frac{1}{7} - \frac{1}{8}...\\$

And so on...
Does anyone know how to help with this?

答:
Let $H_k$ be the harmonic numbers  https://en.wikipedia.org/wiki/Harmonic_number

The requested sum can be written as the limit for $n \to \infty$ of
$$S_n=\sum_{k=1}^n  \left(H_{(k-1)^2}-H_{k^2-1}+\frac{k^2-(k-1)^2}{k^2}\right)=
\sum_{k=1}^n  \left(H_{(k-1)^2}-H_{k^2}+\frac{2k}{k^2}\right).$$
Summing the telescoping series gives
$$S_n=-H_{n^2}+2H_n=-H_{n^2}+\ln(n^2)+ 2(H_n-\ln n) \to -\gamma+2\gamma=\gamma$$
as $n \to \infty$.

Title: Polya - Szego question


问:
Good book for analysis in particular sequences and series
I’m currently studying pure mathematics and I’m finding the analysis sections quite difficult to digest from the course text books. Could anybody recommend some good textbooks that explain sequence and series well in particular finding if they are convergent or divergent.

Thanks

答:
Two recommendations:

The highly readable classic Theory and Application of Infinite Series by Konrad Knopp is too good to be mentioned in the comment section only. I appreciate his style providing a wealth of information for the beginner in an elaborate, gentle wording but never being lengthy.

Another very nice book is Real Infinite Series by D.D. Bonar and M. J. Khoury. The authors provide besides all basic convergent tests also some more sophisticated techniques. They present in chapter 4 $107$ gems of interesting series and discuss in chapter 5 series problems and solutions from the Putnam competitions. I also like Appendix A with $101$ True or False questions and Appendix C containing many interesting references.


答:
The issue goes beyond carelessness with signs, as the SOHCAHTOA definition $$ \cos \theta = \frac{\|t \mathbf{v}\|}{\| \mathbf{u} \|}$$ is, strictly speaking, only valid when $0 < \theta < 90$ is an acute angle.

But even though this is a little sloppy, I can empathize with the textbook authors. Particularly when many novice students would need help with far more elementary concepts (and not necessarily even bother reading the proofs for themselves) I wouldn't have wanted to take a lengthy detour into the five cases that full formal rigor would require, either:

$\theta = 0$

$0 < \theta < 90$

$\theta = 90$

$90 < \theta < 180$

$\theta = 180$

None of these are especially hard, mind you, but if you have other material to get to, it's very excusable to handwave this low-level stuff away. In the interests of full disclosure, though, authors who do this should mention the cases that exist, say "We're going to show you one of these; the other cases are very similar," and perhaps leave them as challenge exercises at the very end of the section.


问:
Let $v_1 = (1, 0); v_2 = (1,-1) \space\text{and} \space v_3 = (0, 1).$
I am stuck on the following problem :

Let $v_1 = (1, 0); v_2 = (1,-1) \space\text{and} \space v_3 = (0, 1).$ How many linear transformations
  $T \colon \Bbb R^2 \to \Bbb R^2$ are there such that $Tv_1 = v_2; Tv_2 = v_3$ and $Tv_3 = v_1?$ The options are as follows:

(A) $3!$

  (B) $3$

  (C) $1$

  (D) $0$

What I observed that $v_2=v_1-v_3$ and so $T(v_2)=T(v_1)-T(v_3) \implies v_3=v_2-v_1$.

But I do not know how to progress from here. Any idea?

答:
Here's a straight forward solution. Every linear transformation $T\colon \mathbb R^2 \longrightarrow \mathbb R^2$ can be represented as a $2\times 2$ matrix, so you want to find $a_1$, $a_2$, $a_3$ and $a_4$ such that

\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_1 \\ a_3\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix}\]
and
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ -1\end{pmatrix} = \begin{pmatrix}0 \\ 1\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_1-a_2 \\ a_3-a_4\end{pmatrix} = \begin{pmatrix}0 \\ 1\end{pmatrix}\]
and
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}0 \\ 1\end{pmatrix} = \begin{pmatrix}1 \\ 0\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_2 \\ a_4\end{pmatrix} = \begin{pmatrix} 1 \\ 0\end{pmatrix}\]

Now it's obvious that there are no such matrix, and so the answer is (d), 0.

Actually you've the answer in front of you too. Assume that there is some such $T$. Since $v_2 = v_1+v_3$ we can use the linearity of the transformation to get $v_3 = v_2+v_1$. Combine these two equations to get $v_1 = 0$, which is a contradiction, and again the answer is (d), 0.

ContentLicense: CC BY-SA 3.0


问:
Given set of 1011 positive numbers prove there are 2 numbers that their sum or difference is a multiple of 2020
I'm struggling solving this problem :
I need to prove that given any set of 1011 positive integers, there are 2 numbers such that
$x_1 + x_2$  or $x_1 - x_2$ is a multiple of 2020.

basically I understand that I need to show that there are 2 numbers such that $ x_1 + x_2 = 0 \pmod{ 2020} $ or $ x_1 - x_2 = 0\pmod {2020}$ but can't find a way to solve that.

Hope to get a help.

edit - also able to make $x_1+x_1=0 (mod 2020)$ if necessary

答: Hint: There are 2020 possible mod-2020 residues. What if two if the numbers have the same residue? What if no two have the same residue?


问:
How to convert 3 numbers (or axes) into one number
I'm sorry if my explication is not quite the best but I'm not sure how to explain this in mathematical terms.

So I'm working on a Machine Learning project where I'm trying to get the emotions from one text. Each emotion will have a value from 0 to 1, where 0 means that emotion those not exist, and 1 means it's 100% that.

For example, I could have:

positive 0.8466, neutral 0.1458, negative 0.0076

This been my 3 base emotions that I am getting from any text.

Is there any way in which I can convert those numbers into a single one from a spectrum between -1 and 1. That would mean if the number is closer to 0 then the most powerful emotion from that text is negative, if it's closer to 0 then it's neutral, and also if it's closer to 1 that means it's positive.

Those any of you know how can I achieve something like this? I was thinking to consider each of the emotions as an axis into a 3d space, but I'm not sure how could I calculate that point in space.

答: A very natural solution is to weight the values. For example take $f(x)=1 \times$ positive $+ 0 \times $ neutral $ + (-1) \times $ negative. It's easy to show that $f(x) \in [-1,1]$ and why you can ignore neutral.


问:
Natural log of a negative number
My teacher told me that the natural logarithm of a negative number does not exist,
but

$$\ln(-1)=\ln (e^{i\pi})=i\pi$$

So, is it logical to have the natural logarithm of a negative number?

答:
$\log(-1)=i\pi$ is the principal value of the logarithm of $-1$. In general: the principal value of $\log re^{i\theta}$ is $\log r+i\theta$ where $\theta\in(-\pi,\pi]$, and adding integer multiples of $2i\pi$ produces all other possible logarithms.

Your teacher was correct... but only in the fully real setting.

Title: Natural log of a negative number


问:
Proving the linear combination of symmetric matrices is convex
I am trying to prove that the set is convex:
$$\{(x \in \mathbb{R}^n ) | \lambda_{\text{max}}(A(x))\le1, \boldsymbol{Tr}(B^TA(x))\ge 1\}$$
where $A(x)=A_0+x_1A_1+...+x_nA_n$ and $A_i \in \mathbb{S}^n$, $B \in \mathbb{R}^{n \times n}$

It's not very obvious to me which operation that preserves convexity to use for this proof. Any hints would be very helpful.

答:
Note that $\text{tr}(B^TA(x)) = \sum_{j=1} ^n x_j \cdot \text{tr}(B^TA_j) + \text{Tr}(B^TA_0)$ which is affine in $x$. Hence, the subset defined by the rightmost constraint only is certainly convex (it's just a shifted half-plane).

For the leftmost constraint, you can use that $\lambda_{\text{max}}(A+B) \leq \lambda_{\text{max}}(A) + \lambda_{\text{max}}(B)$ (this is kind of non-trivial, look up Weyl's inequality) and $\lambda_{\text{max}}(tA) = t\lambda_{\text{max}}(A)$. Then indeed, $$ \lambda_{\text{max}} \bigg(A(tx_1 + (1-t)x_2) \bigg) = \lambda_{\text{max}} \bigg(tA(x_1) + (1-t)A(x_2) \bigg) \leq t\lambda_{\text{max}}A(x_1) + (1-t)\lambda_{\text{max}}A(x_2) \leq 1.$$

Your set is the intersection of the two, and hence also convex.


答:
Context is important here. In the context of real numbers, negative numbers have no logarithms (and neither does $0$) because $\log(x)$ is a number $y$ such that $e^y=x$ and $e^y$ is always greater than $0$.

On the other hand, in the context of complex numbers, every complex number other than $0$ has logarithms. In fact, any such complex number has infinitely many logarithms! You are right when you claim that $i\pi$ is a logarithm of $-1$. However, every complex number of the form $\pi i+2\pi in$ (with $n\in\Bbb Z)$ is also a logarithm of $-1$, since$$e^{\pi i+2n\pi i}=e^{\pi i}e^{2\pi in}=(-1)\times1=-1.$$

Title: Natural log of a negative number

Body:
My teacher told me that the natural logarithm of a negative number does not exist,
but

$$\ln(-1)=\ln (e^{i\pi})=i\pi$$

So, is it logical to have the natural logarithm of a negative number?


答:
Let $u$ be the negative weight, $v$ the neutral weight, and $w$ the positive weight.  The score you want is of form
$$S = \frac{f(w) - h(u)}{f(w) + g(v) + h(u)}$$
where $f$, $g$, and $h$ are nonnegative weighting functions.

The simplest one is linear: $f(x) = x$, $g(x) = x$, $h(x) = x$, so
$$S = \frac{w - u}{w + v + u}$$

Almost as simple is a monomial form of some degree $p$, $f(x) = x^p$, $g(x) = x^p$, $h(x) = x^p$, so
$$S = \frac{w^p - u^p}{w^p + v^p + u^p}$$
where $p \gt 0$.  Obviously, for $p = 1$ it is the linear above.

Title: How to convert 3 numbers (or axes) into one number

Body:
I'm sorry if my explication is not quite the best but I'm not sure how to explain this in mathematical terms.

So I'm working on a Machine Learning project where I'm trying to get the emotions from one text. Each emotion will have a value from 0 to 1, where 0 means that emotion those not exist, and 1 means it's 100% that.

For example, I could have:

positive 0.8466, neutral 0.1458, negative 0.0076

This been my 3 base emotions that I am getting from any text.

Is there any way in which I can convert those numbers into a single one from a spectrum between -1 and 1. That would mean if the number is closer to 0 then the most powerful emotion from that text is negative, if it's closer to 0 then it's neutral, and also if it's closer to 1 that means it's positive.

Those any of you know how can I achieve something like this? I was thinking to consider each of the emotions as an axis into a 3d space, but I'm not sure how could I calculate that point in space.


问:
Single Jordan Block
Guys would you help me about what does "single jordan block" mean? Does single represent geometric multiplicity or algebraic multiplicity?</p>

答:
A Jordan block is a matrix of the form
$$
J = \pmatrix{\lambda & 1\\&\lambda&1\\&&\ddots&\ddots\\&&&\lambda&1\\&&&&\lambda}.
$$

Title: Single Jordan Block


问:
Diagonalizable matrices in $M_{2\times 2}(\mathbb{F}_2)$
List all diagonalizable $2\times 2$ matrices over the a field $F$ consisting of two elements $0$ and $1$.

I want to try and do this using C++, but perhaps this isn't the place to ask. I have an idea as to how I'd do it.

答:
I'll answer this for $n\times n$ matrices, which gives only slightly more of a challenge. A diagonalisable matrix $A$ is completely determined by its eigenspaces, which form a direct sum giving the whole space, and of cuorse $A$ determines those subspaces as well. Over $\Bbb F_2$ only two eigenspaces are possible, for eigenvalues $0$ and $1$. The matrix $A$ will be the projection onto its $1$-eigenspace parallel to its $0$-eigenspace; it will be idempotent ($A^2=A$) and conversely all idempotent matrices are diagonalisable.

The number of such matrices equals the number of ways to decompose the space $\Bbb F_2^n$ as an ordered direct sum of two subspaces (the eigenspaces of $0$ respectively for $1$; "ordered" just means that swapping the two summands is considered to give a different decomposition, as $A$ will differ). The set of such decompositions into subspaces of dimension $k$ and $n-k$ is the $GL(n,\Bbb F_2)$-orbit of the standard decompostion $\Bbb F_2^n=\Bbb F_2^k\oplus\Bbb F_2^{n-k}$, so by the orbit-stabiliser theorem the number of such decompositions is
$$
  \frac{\#GL(n,\Bbb F_2)}{\#GL(k,\Bbb F_2)\times\#GL(n-k,\Bbb F_2)}
 =2^{k(n-k)}\binom nk_2
$$
(the final factor is the Gaussian binomial coefficient $\binom nk_q$ evaluated at $q=2$), and so the total number of diagonalisable matrices $A$ is
$$
  \sum_{k=0}^n2^{k(n-k)}\binom nk_2
$$
which does not seem to simplify. The values for $n=0,1,2,3,4$ are respectively $1,2,8,58,802$, and having said that I can of course not fail to mention that this is OEIS A132186.


问:
Consider the set of points in $\operatorname{PG}(3,q)$ given by $X_0^2+X_1^2+X_2^2+X_3^2=0$. Show that no plane is disjoint to this set.
My attempt: call this set $\Omega$. I feel like we will have to look at $q$ even and $q$ odd separately. Suppose that $q$ is even, then (since the characteristic of the underlying field is 2), we can rewrite the equation for $\Omega$ as $\sum X_i = 0$. So, $\Omega$ is a plane in $\operatorname{PG}(3,q)$. Since the projective space has dimension three, we know that every other plane will intersect $\Omega$ in at least line, proving the statement for even $q$'s.

Now, I don't really know how to handle the case for odd $q$. When we intersect $\Omega$ with a plane $\pi$, we will end up with a quadratic in this plane. Since the field now has odd characteristic, and the matrix of such a quadratic is the identity (indeed, take f.ex. $\pi: X_3 = 0$, then $\Omega\cap \pi: (X_0, X_1,X_2) I_3(X_0,X_1,X_2)^t=X_3=0$), we could consider the corresponding orthogonal polarity* $\beta$ such that $\Omega$ is the set of absolute points of $\beta$. This is where the inspiration ends for me.

$\ast$: a polarity $\beta$ is an anti-automorphism of the projective space for which $\beta^2=1$. An orthogonal polarity in a plane over a field of odd characteristic can be represented as $(a_0,a_1,a_2)^t = A (x_0,x_1,x_2)^t$, where $A$ is symmetric.

Does anyone know how to build on from my reasoning? Or is there another way to tackle this problem?

Thanks!

答: As you wrote, the intersection is always projectively equivalent to $$x_1^2+x_2^2+x_3^2=0,$$ from here we get that $$x_1^2+x_2^2=-x_3^2.$$ It always has a solution, because in a finite field of odd order every element is the sum of two square-elements. (So for all $-x_3^2$ we can get $x_1$ and $x_2$ this way.)


答:
Let set $f(x)=\dfrac{\ln(x)^{100}}{x^{\frac 78}}$ it is not to difficult to see that $f\searrow$ at infinity.

Indeed $f'(x)=-\dfrac{\overbrace{\ln(x)^{99}}^{>0}(\overbrace{7\ln(x)-800}^{>0})}{\underbrace{x^{\frac{15}8}}_{>0}}<0\quad$ for $x$ large enough.

In particular since $f$ is continuous and $f(x)>0$ for $x>1$ we can conclude that $f$ is bounded. $$\exists M>0\text{ : }|f(x)|<M\quad\forall x>1$$

Now we just use apply to $x=u^2$ and use the logarithm multiplicative to additive property to show the denominator is negligible.

$$|f(x)|=|f(u^2)|=\left|\dfrac{2^{100}\ln(u)^{100}}{u^{\frac 78}u^{\frac 78}}\right|=\left|\dfrac{2^{100}f(u)}{u^{\frac 78}}\right|\le2^{100}M\dfrac 1{|u|^{\frac 78}}\to 0$$

Naturally, you can easily transpose the proof for any suitable $\, \alpha,\beta\, $ and $\, \dfrac{\ln(x)^\alpha}{x^\beta}$

Title: Why does $\log^{100}(n)$ grow asymptotically slower than $n^{7/8}$

Body:
More formally:
How do I compute the following limit:

$$
\lim _{x \rightarrow \infty} \frac{\log ^{100}(x)}{x^{7 / 8}}
$$

I would start using l'Hôpital's rule and notice the pattern, is there some other procedure? (without knowing it)


问:
Hawaiian Earrings are covering space for shrinking wedge of circles

for any $P=(x,y) \in \mathbb{R}^2$,  Let $C_{P,n}$ denote a circle centred at $\big(x+\frac1n, y \big)$ of radius $\frac1n$. Let $W_{P}$ denote the shrinking wedge of circle $\bigcup_{n \in \mathbb{Z}} W_{(0,0)} $..

Define the space
$$X=\{0\} \times \mathbb{R} \cup \big(\bigcup_{i \in \mathbb{Z}}  W_{(0,4i)} \big)$$.
Show that X is a covering space for $W_{(0,0)}$.

Hatcher claims that $X$ is a covering space for $W_{(0,0)}$ but he did not provide any hint. I tried some maps but I wonder if they will be covering space. I was thinking that the number of circles in $X$ are countable and this is the case with $W_{(0,0)}$. Hence there is a bijection between them. Hence a circle in $X$ would map to a circle in $W_{(0,0)}$, but then I doubt this would be continous. Please help me with this problems.

答: You map the copy of $\mathbb{R}$ in the usual winding fashion onto the outermost circle. In your case, this should be $(0, y) \mapsto \left(\cos\left(\frac{\pi}{2}y\right), \sin\left(\frac{\pi}{2} y\right)\right)$. As for the copies of the earrings, you map them to the earrings in the base space after skipping the outer most circle. So, $C_{P, n} \subset W_{(0, 4i)}$ in the total space gets mapped to $C_{P, n+1}$ in the base.


问:
Rudin Theorem 1.17
1.17 THEOREM

Let $f: X \to [0, \infty)$ be measurable. There exist simple measurable functions $s_n$ on $X$ such that

(a) $0 \le s_1 \le s_2 \le \dots \le f.$

(b) $s_n(x) \to f(x)$ as $n \to \infty \forall x$

PROOF Put $\delta_n = 2^{-n}$. To each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k \delta_n \le t < (k + 1) \delta_n$. Define

$
\phi_n(t)=
\begin{cases}
k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n \\
n \space \space \text{if} \space n \le t < \infty \\
\end{cases}
$

Each $\phi_n$ is then a Borel function on $[0,\infty)$,

$t~\delta_n< \phi_n(t)\le t \space \space \space \space \text{if} \space \space 0 \le t <n$

$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le  t$ and $\phi_n(t) \to t$ as $n \to \infty$ It follows that the functions

$s_n = \phi_n (f)$

satisfy (a) and (b); they are measurable, by Theorem 1.12(d). $\square$

$\textbf{Question}$: Why can't we set $\phi_n(t)= n$ ? What would we lose?

$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le  t$ is only true for $t<n$. I just don't see what we gain by having $k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n $

答: The construction partitions a subset of the range of $f$, i.e. the values of $f$ that fall in the interval $[0,n]$.  The definition of $\phi_{n}$ ensures that the interval goes to $[0,\infty)$ (i.e. covers the entire range of $f$) while the requirement $\phi$ be equal to $k_{n}(t)\delta_{n}$ ensures that the range is partitioned finer and finer (in particular, at a "faster rate" than the interval is being extended).  What you lose by just setting $\phi$ equal to $n$ is the approximation of $f$!</p>

Title: Rudin Theorem 1.17


问:
Show that $f(A)^c\subseteq f(A^c)$ if and only if $f$ is injective
Let $E_i$ be a set, $f:E_1\to E_2$ and $A\subseteq E_1$.

It is easy to show (see, for example, https://math.stackexchange.com/a/1447826/47771) that $f(A)^c\subseteq f(A^c)$ if and only if $f$ is surjective.

Are we able to show that

$f(A^c)\subseteq f(A)^c$ if and only if $f$ is injective

as well? (It's quite easy to prove $f(A^c)=f(A)^c$ if $f$ is bijective; see, for example, https://math.stackexchange.com/a/3301866/47771.)

答:
I will denote $X = E_1$ and $Y = E_2$ and $f:X \rightarrow Y$.

Suppose $f$ is injective. Let $y \in f[X\setminus A]$ be arbitrary. Then, there exists $x_1 \in X\setminus A$, equivalently $x_1 \not\in A$, such that $f(x_1) = y$. Now, suppose there is $x_2 \in A$ such that $f(x_2) = y$. Then, $x_1 \neq x_2$ and this means $f(x_1) \neq f(x_2)$ by injectivity, which is a contradiction. So, for all $x \in A$, $f(x) \neq y$, which means $y \not\in f[A]$, equivalently $y \in Y\setminus f[A]$.

Suppose $f[X\setminus A] \subseteq Y\setminus f[A]$. Let $x_1 \neq x_2$ be elements of $X$ and assume $X$ has at least two elements. Then, there is a subset $A$ of $X$ such that $x_1\in A$ and $x_2\not\in A$. Then, we immediately have $f(x_1) \in f[A]$. $x_2 \in X\setminus A$ implies $f(x_2) \in f[X\setminus A] \subseteq Y\setminus f[A]$, that is, $f(x_2) \not\in f[A]$. Therefore, $f(x_1) \neq f(x_2)$.


问:
How my professor made the following conclusion?
Update: The suggested Answer didn't solve the problem

Note: This question can be solved without knowing probability at all. (all you need is the bold text)

Today my lecturer wrote the following on board:

where the text in red specefies for which values the multiplication under integral sign isn't zero.

I understand that we need to split into conditions:

where $\max\{0,z+a\}=0$ which means: $z<-a$

where $\max\{0,z+a\}=z+a$ which means: $z>-a$

but in the first condition where did we get $-b<=z$ from? I understand that without it something will be wrong since for all values the integral will not be 0 (while we know from the text in red that for some it's zero for sure) but I don't understand where it specifically came from...

I have been thinking on this for hours.

答:
It looks like $Y \sim \text{Exp}(\lambda)$ and $X \sim \text{Unif}(a,b)$. In that case $Y \geq 0$ and $a \leq X \leq b$. So $Y-X \geq -b$. Thus $f_{Y-X}(z)=0$ when $z <-b$.

Another way to think about it is combining the inequality $y \geq 0$ with the inequality $z+b \geq y$ gives $z+b \geq 0$, so $z \geq -b$.


问:
I solved the problem. But I have some doubts on the solution.
Given the sequence $a_n = \sin(n)$, where $n$ is an integer. Prove that limit superior is $1$.

My solution: by previous theorems we know that if $x,y \in \Bbb R$ and the ratio of $x$ and y is irrational, then there exists the integers $a$ and $b$ such that $$\forall \epsilon, \quad  0<ax+by<\epsilon$$ So we can put integrs $m$ ands $n$ such that $$0<\frac{n}{m}-\frac{\pi}{2}<\epsilon$$ And we are done.

But here m should be in form of $m=4k+1,\, k\in \Bbb Z$.

My question is, how we can be sure that we can find such $m$?

答:
What you actually need is, given $\epsilon>0$, to show there exist integers $n,k$ such that
$$| (4k+1)\pi/2-n |=  |2k\pi+\pi/2-n|<\epsilon \,.$$
This follows from the density of the orbit of irrational rotations
https://en.wikipedia.org/wiki/Irrational_rotation

Prove that the orbit of an iterated rotation of 0 (by (A)(Pi), A irrational) around a circle centered at the origin is dense in the circle.

Irrational angle rotation argument.

Dense set in the unit circle- reference needed


问:
Prove that $\int^1_0|f'(x)|dx \le 2\int^1_0 |f(x)|dx+\int^1_0|f''(x)|dx.$
$f(x),f'(x),f''(x)$ are  continuous functions, $f(0)f(1) \ge 0$, prove that $$\int^1_0|f'(x)|dx \le 2\int^1_0 |f(x)|dx+\int^1_0|f''(x)|dx.$$

This problem comes from here.

答:
Replacing $f$ with $-f$, we may assume that $f(0)\geq 0$.

If $f'$ has a root $a\in[0,1]$, then write $$\int_0^a|f'|=\int_0^a\left|\int_a^xf''(t)\,dt\right|dx\leq\int_0^a\int_x^a|f''(t)|\,dtdx=\int_0^a\int_0^t|f''(t)|\,dxdt\leq\int_0^a|f''|,$$

and similarly $\int_a^1|f'|\leq\int_a^1|f''|$, so the estimate holds.

If $f'$ does not have a root in $[0,1]$, then it has a fixed sign; replacing $f(x)$ with $f(1-x)$, we may assume that $f'>0$ in $[0,1]$. So, $f\geq 0$ in $(0,1)$.

We claim that there exists $\xi\in (0,1)$ such that $f'(\xi)\leq 2\int_0^1f$: if not, then $f'(x)>2\int_0^1f$ for all $x\in(0,1)$. Integrating in $(0,x)$ for $x\in(0,1)$, we then have $$f(x)-f(0)=\int_0^xf'>2x\int_0^1f,$$ therefore $$\int_0^1f(x)\,dx-f(0)=\int_0^1(f(x)-f(0))\,dx>\int_0^1f\int_0^12x\,dx=\int_0^1f(x)\,dx,$$ which implies that $f(0)<0$, a contradiction with $f\geq 0$ in $(0,1)$. So, for some $\xi\in (0,1)$, we have $f'(\xi)\leq 2\int_0^1f$. Then, \begin{align*}\int_0^1|f'(x)|\,dx&\leq\int_0^1|f'(x)-f'(\xi)|\,dx+|f'(\xi)|\\
&\leq\int_0^1\int_{\xi}^x|f''(t)|\,dtdx+2\int_0^1f,\end{align*} which implies the estimate.


问:
Obtain the conditional distribution of $X$ given $X^2=t$
Consider the model $X \sim N(\theta, 1)$ for $\theta \in \mathbb R$. Obtain the conditional distribution of $X$ given $X^2=t$ for $t\ge 0$.

Hint: This is a discrete distribution.

Current work:
$f(X=x|X^2=t)=\displaystyle \frac{f(X=x, X^2=t)}{f(X^2=t)}=\frac{f(X=x, X^2=t)}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}=\frac{f(X=\sqrt t)}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}=\frac{\frac{1}{\sqrt {2\pi}}e^{-(\sqrt t-\theta)^2/2}}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}$? But this can't be right because there's no $x$ in the answer. What does one do here?

答:
Disclaimer: I'm not an expert, so please correct me if there are serious flaws in my argument.

Hand-waving/guessing

First observe that if $X^2=t$ then $X$ is either $\sqrt{t}$ or $-\sqrt{t}$, so this is a discrete distribution, and to specify the distribution you only need to compute $P(X=\sqrt{t} \mid X^2=t)$ and $P(X=-\sqrt{t} \mid X^2=t)$.

In analogy with the discrete case, you might rewrite $\{X^2 = t\} = \{X=\sqrt{t}\} \cup \{X = -\sqrt{t}\}$ and guess
$$P(X=\sqrt{t} \mid X^2=t) = \frac{f_X(\sqrt{t})}{f_X(\sqrt{t}) + f_X(-\sqrt{t})}$$
$$P(X=-\sqrt{t} \mid X^2=t) = \frac{f_X(-\sqrt{t})}{f_X(\sqrt{t}) + f_X(-\sqrt{t})}$$

This would certainly be a probability distribution since the probabilities sum to one. However it is not clear why we can add/divide the densities in this way, since the density values are not really probabilities. The discussion below gives a little more background on why this ends up being valid. However, in my experience undergraduate courses are content to sweep this under the rug since this discussion would involve some rather technical measure theory concepts.

Before I turn to the more rigorous discussion, I want to comment that I noticed your denominator is the density of $f_{X^2}$ (which you correctly noted to be $f_{X^2}(t) = \frac{1}{2\sqrt{t}}(f_X(\sqrt{t}) + f_X(-\sqrt{t}))$. However, you used $f_X(\sqrt{t})$ as the numerator. Due to the change of variables, the $\frac{1}{2\sqrt{t}}$ scaling factor prevents your attempt from being a probability distribution (in that $\frac{f_X(\sqrt{t})}{f_{X^2}(t)} + \frac{f_X(-\sqrt{t})}{f_{X^2}(t)} \ne 1$). This is one reason why it can be dangerous to manipulate densities as you would with discrete probabilities, even if sometimes you do get the correct answer.

Verifying guess with slightly more technical stuff

Conditioning on events of measure zero is a bit of a thorny discussion. One framework for handling this situation is the notion of regular conditional probabilities. The definition translates to the following. The map $(t, A) \mapsto \nu(t, A)$ is the regular conditional probability of $X$ given $X^2$ if

$\nu(t, \cdot)$ is a probability measure for each $t$

$\nu(\cdot, A)$ is measurable for each measurable set $A$

$P(\{X \in A\} \cap \{X^2 \in B\}) = \int_B \nu(t, A) f_{X^2}(t) \, dt, \qquad \text{for all measurable sets $A,B$}.$

(Note that the right-hand side of the third condition can be written as $E[\nu(X^2, A) \mathbf{1}_{X^2 \in B}]$.)

Then $\nu(t, A)$ is what we would usually denote by $P(X \in A \mid X^2 = t)$.

Let's try the guess we made above. That is, let's try
$$\nu(t, A) =
\frac{f_X(\sqrt{t})}{f_X(\sqrt{t}) + f_X(-\sqrt{t})} \mathbf{1}_A(\sqrt{t}) +
\frac{f_X(-\sqrt{t})}{f_X(\sqrt{t}) + f_X(-\sqrt{t})}\mathbf{1}_A(-\sqrt{t}).$$
This satisfies the first two conditions. It remains to check the third condition.

Notice that for $t \ge 0$, we have $f_{X^2}(t) = \frac{d}{dt} P(X^2 \le t) = \frac{d}{dt} \int_{-\sqrt{t}}^\sqrt{t} f_X(s) \, ds = \frac{1}{2\sqrt{t}} (f_X(\sqrt{t}) + f_X(-\sqrt{t}))$. (This is in the denominator of your attempt.)
Let $B_+ := B \cap [0, \infty)$. Then,
\begin{align}
&\int_{B} \nu(t, A) f_{X^2}(t) \, dt
\\
&= \int_{B_+} \frac{1}{2\sqrt{t}} f_X(\sqrt{t}) \mathbf{1}_A(\sqrt{t}) \, dt
+ \int_{B_+} \frac{1}{2\sqrt{t}} f_X(-\sqrt{t}) \mathbf{1}_A(-\sqrt{t}) \, dt
\\
&= \int_0^\infty f_X(u) \mathbf{1}_A(u) \mathbf{1}_{B_+}(u^2) \, du + \int_0^\infty f_X(-u) \mathbf{1}_A(-u) \mathbf{1}_{B_+}((-u)^2) \, du
\\
&= \int_{-\infty}^\infty f_X(v) \mathbf{1}_A(v) \mathbf{1}_{B_+}(v^2) \, dv
\\
&= P(\{X \in A\} \cap \{X^2 \in B\}).
\end{align}

Alternate definition involving a limit of events with vanshing probability

Teresa Lisbon and Matthew Pilling consider an alternate way to define conditioning on a zero-probability event by approximating it with events with probability approaching zero. This is discussed on the same Wikipedia page. In our situation, it would be

$$P(X \in A \mid X^2 = t ) = \lim_{\epsilon \to 0} \frac{P(\{X \in A\} \cap \{t-\epsilon < X^2 < t + \epsilon\})}{P(t-\epsilon < X^2 < t + \epsilon)}.$$

I have not checked whether this alternate definition yields the same answer as above, but I want to note that this is not always considered a "good" way to define conditional probability. Indeed there is a warning on that Wikipedia section with a link to a discussion that contains a simple example for where this definition leads to an irregular conditional probability distribution.

Consider $U \sim \text{Uniform}(0,1)$. For any $\delta > 0$, we have
\begin{align}
&P(0.5-\delta < U < 0.5+\delta \mid U=0.5)
\\
&= \lim_{\epsilon \to 0} \frac{P(\{0.5-\delta < U < 0.5+\delta\} \cap  \{0.5-\epsilon < U < 0.5+\epsilon\})}{P(0.5-\epsilon < U < 0.5+\epsilon)}
\\
&= 1\end{align}
but
$$P(U=0.5 \mid U=0.5) = \lim_{\epsilon \to 0} \frac{P(\{U=0.5\} \cap  \{0.5-\epsilon < U < 0.5+\epsilon\})}{P(0.5-\epsilon < U < 0.5+\epsilon)} = 0.$$

Appendix

My understanding of conditional probabilities may be shaky so please correct any misunderstandings in my answer. I found the Wikipedia pages useful (linked above) as well as this answer by Stefan Hansen.


问:
Burnside's lemma for colourings on a triangle-square
Suppose that we have $n$ colours available and we paint the edges and vertices of an equilateral triangle (resp. square) and we want to count how many colourings there are, up to equivalence, where two colourings are equivalent if one can be transformed into the other by applying a symmetry of the triangle (resp. square). How many colourings are there?

I know how to use Burnside's lemma to count the colourings when we only paint either the edges or the vertices of a regular $n$-gon, but not both. In the above problem, I initally thought that the case of painting the vertices and edges of a triangle was exactly the same as painting just the vertices of a regular hexagon, and similarly I considered working with a regular octagon in the place of the vertices and edges of a square. It seems intuitive, but something is not convincing me: Is the group of symmetries of the vertices and edges of a regular $n$-gon isomorphic to the group of symmetries of just the edges of a regular $2n$-gon (the dihedral group of $4n$ elements). Another possibility that passed through my mind was: just multiply the orbits of the action of $D_3$ by $2$ to get the colourings when we paint the vertices and edges.

Is there any merit in my thoughts? If so, please point out my mistakes. Thanks in advance, and sorry for the long text!

答:
There are  two possibilities  here, rotational symmetry  (necklace) or
dihedral  symmetry (bracelet).  For the  first one  we have  the cycle
index of the cyclic group:

$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$

For second one we have the cycle index of the dihedral group

$$Z(D_n) =
\frac{1}{2} Z(C_n) +
\begin{cases}
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$

With rotations the cycle structure induced by the vertices on the edges
is the same for vertices and edges  and with reflections the two types of
reflections are swapped when $n$  is even (use $a$  for vertices and $b$
for edges; we use the variable $n$ for the number  of vertices and $M$
for the number of colors). We thus obtain

$$Z(C'_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d} b_d^{n/d}.$$

and for the dihedral group

$$Z(D'_n) =
\frac{1}{2} Z(C'_n) +
\begin{cases}
\frac{1}{2} a_1 a_2^{(n-1)/2} b_1 b_2^{(n-1)/2}
& n \text{ odd} \\
\frac{1}{4}
\left( a_1^2 a_2^{n/2-1}  b_2^{n/2}
+ a_2^{n/2} b_1^2 b_2^{n/2-1} \right)
& n \text{ even.}
\end{cases}$$

Thus we have by Burnside for colorings with at most $M$ colors
(for all permutations of the group in question to fix a coloring it must be constant on each cycle so we have $M$ choices for the equal color of all slots on the cycle giving the substitutions $a_d = b_d = M$):

$$Z(C'_n; a_1 = M, b_1 = M, \ldots)
= \frac{1}{n} \sum_{d|n} \varphi(d) M^{2n/d}$$

and for the dihedral group

$$Z(D'_n; a_1 = M, b_1 = M, \ldots) =
\frac{1}{2} Z(C'_n; a_1 = M, b_2 = M, \ldots)
+ \frac{1}{2} M^{n+1}
\\ = \frac{1}{2n} \sum_{d|n} \varphi(d) M^{2n/d}
+ \frac{1}{2} M^{n+1}
.$$

This gives with $M=2$ (two colors) the sequence

$$4, 9, 20, 51, 136, 414, 1300, 4371, 15084, 53508, \ldots$$

which points us to OEIS A161221, where these
data are confirmed. We also obtain with $n=3$ (triangle)

$$\frac{1}{6} (M^6 + 2 M^2) + \frac{1}{2} M^4.$$

This factors as

$$\frac{1}{6} M^2 (M^2+1) (M^2+2)$$

as pointed out in the accepted answer. We are asked to do squares
as well and obtain

$$\frac{1}{8} (M^8 + M^4 + 2 M^2) + \frac{1}{2} M^5
= \frac{1}{8} (M^8 + 4 M^5 + M^4 + 2 M^2).$$

N.B. the two types of  reflections when $n$ is even are first, those that
fix two opposite  vertices and second, those that fix two opposite edges.

There is more on this at
MSE 3216181.


问:
is a relation R total/linear/well-order
Let $\mathcal{R}$ be a relation on $\mathbb{N}\times \mathbb{N}$ i.e $\mathcal{R}\subseteq(\mathbb{N}\times \mathbb{N})\times (\mathbb{N}\times \mathbb{N})$ s.t $(x,y)\mathcal{R}(z,w)$ iff $x<z$ or $x=z \wedge y\le w$.

a)is $\mathcal{R}$ total order/linear order?

b) does the greatest and the smallest element exist?

c) is $\mathcal{R}$ a well order?

My solution

reflexivity:

$(x,y)\mathcal{R}(x,y)$ as $x=x$ and $y\le y$ $\forall x,y\in \mathbb{N}\times \mathbb{N}$

antisymmetry

$(x,y)\mathcal{R}(y,x)$ as $x<y\vee x=y \wedge y\le x$ I am not sure about this one, I considered two examples:

$(1,2)\mathcal{R}(2,1)$  then $1<2$ so it is okay but

$(2,1)not\mathcal{R}(1,2)$ as $2\not< 1$ futhermore $2\neq 1$
so $\mathcal{R}$ is not antisymmetric? thus not total/linear order and not well-order?

答:
You have shown that the relation is reflexive, and that's fine. But the others you haven't done right.$\newcommand{\R}{\mathrel{\mathcal{R}}}$

For antisymmetry, for example, you need to assume that $(x,y)\R(z,w)$ and $(z,w)\R(x,y)$. From that assumption you need to show that $(x,y)=(z,w)$. This condition needs to be shown for every $(x,y),(z,w)\in\Bbb{N\times N}$. One example is not enough!

You also have to show that this is a transitive relation. For the totality you also made a mistake. You have shown that there are two pairs such that $\R$ doesn't hold in one direction. But remember the definition of a total order:

$(A,\preceq)$ is totally ordered if for every $a,b\in A$ it holds $a\preceq b$ or $b\preceq a$.

So the fact that $(2,1)\R(1,2)$ is false doesn't prove anything. You need to show that $(1,2)\R(2,1)$ is also false -- but you can't, because $(1,2)\R(2,1)$ is in fact true.

Finally, your title asks whether or not it is a total order, or a linear order, or a well-order. It is unclear whether or not you think that total and linear orders are different things; or that you think that linear orders and well-orders are the same thing. In either case it is wrong. All total orders are linear orders, and not all linear orders are well-orders.

Now, let me reveal you a little secret, this is indeed a well-order (so it is a total order as well). But I am going to leave you with the task of proving that.

Title: is a relation R total/linear/well-order


问:
Expectation of Cumulative Gaussian Function
If M(x) is the cumulative Gaussian function and X is N(0,1) then what is E[M(X)]?

Thus if X ~ N(0,1):

$M(x) = P(X \leq x)=\Phi(x)$

The answer given is for x in (0,1):

\begin{align} P(M(X) \leq x) &= P(\Phi(X) \leq x) \\ &=P(X \leq \Phi^{-1}(x)) \\ &=x  \end{align}

Hence M(X) has a uniform distribution on (0,1) and hence an expectation of 0.5.

My understanding:

$\Phi(x)$ is not a r.v. it returns for a given x the Gaussian cdf for that value (so it's range is between 0 and 1).

In the second line the inverse Gaussian cdf is applied, thus on the right side of the inequality you have the value of the RV distributed N(0,1) corresponding to each value of x, where x is restricted to be between 0 and 1. The whole of line 2 is the probability that X is less than or equal to this value.

Is my understanding of the first two lines correct?

What I don't understand how you end up with a uniformly distributed variable is the third line.
Algebraically I can see that you are applying $\Phi(\Phi^{-1}(x))$ so you get x which s a value between zero and one but how do we know that these values are uniformly distributed?.

Thanks

答:
If you know that $P(X \le t) = \Phi(t)$, then the last line is simply this equation applied with $t := \Phi^{-1}(x)$, i.e.
$$P(X \le \Phi^{-1}(x)) = \Phi(\Phi^{-1}(x)) = x.$$
To conclude, you have shown the CDF of $M(X)$ is $P(M(X) \le x)=x$ for $x \in [0,1]$, which is the CDF of the uniform distribution.


问:
Deriving a basis from a generating subset and a free subset

Let $K$ be a field and $E$ a $K$-vector space. Let $S$ be a generating
system $E$ and $L$ a free subset of $E$ such that $L\subset S$. Then
there exists a basis of $E$ such that $L\subset B\subset S$.

Let $\mathcal{L}$ be the set of free subsets of $E$ contained in $S$. This set is inductive. Then the set $\mathcal{M}$ of free subsets of $E$ containing $L$ and contained in $S$ is also inductive. By Zorn's lemma, there exists a maximal $B$ in $\mathcal{M}$. We want to show that $E\subset\text{span}(B)$. Suppose not: let $y\in E$ such that $y\not\in\text{span}(B)$. Then $B\cup\{y\}$ is a free subset of $E$ and $L\subset B\cup\{y\}$. I still need to show that $B\cup\{y\}\subset S$ in order to get a contradiction. But why should $y\in S$?

答: It doesn't have to be true. But instead of taking some $y\in E$ such that $y\notin\operatorname{span}(B)$, take $y\in S$ such that $y\notin\operatorname{span}(B)$. Such a $y$ must exist, because otherwise $S\subset\operatorname{span}(B)$, and therefore $\operatorname{span}(B)$ would be the whole space.


问:
Diagonal entries of symmetric matrices and eigenvalues
suppose I have a symmetric matrix $S$ with diagonal entries $s_{jj}$. I'm trying to prove that none of these diagonal entries $s_{jj}$ can be be smaller than all the eigenvalues of $S$. My textbook tells me that if one of these diagonal entries were small than all the eigenvalues, then $S-s_{jj}I$ would have wholly positive eigenvalues and be positive definite - however that couldn't be the case because $S-s_{jj}I$ would have a zero on the main diagonal.

What I don't understand is how one can skip straight to $S-s_{jj}I$ having wholly positive eigenvalues.

Help will be appreciated, thanks!

答: In general, the eigenvalues of $S - \mu I$ are $\lambda_1 - \mu, \lambda_2 - \mu, \ldots, \lambda_n - \mu$ where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $S$. You can check this by noting that the eigenvectors of $S$ are eigenvectors of $S-\mu I$ and vice versa.


问:
Divergence theorem is not working for this example?
I'm trying to compute the flux of $F=(x+y,y,z^2)$ across the surface $S$ of equation $z=x^2+y^2$ between the two planes $z=1$ and $z=2$ with the plain definition $\int \int_{S} F \cdot n dS$ and with the divergence theorem

With the definition of surface integral

I write the parametrization of the lateral surface $S_1$ with $\sigma(z,\theta)=(z \cos(\theta), z \sin(\theta),z)$, where $(z,\theta) \in [1,2] \times [0,2 \pi)$.

$\int_{S_1}F \cdot n dS = \int_{0}^{2 \pi} \int_{1}^2 (z(\cos(\theta) + \sin(\theta)),z  \sin(\theta),z^2) \cdot (z \cos(\theta), z \sin(\theta),-z) dzd \theta = \frac{14}{3} \pi - \frac{15}{2} \pi$

$\sigma$ is only a parametrization for the lateral surface of $S$: I need to fill also the two cuts on the upper part $S_2$ and lower part $S_3$.

$S_1$ it's parametrized by $(\rho,\theta)$ polar coordinates with $\rho \in(0,2)$ and $\theta \in [0,2 \pi)$.

By integration, I obtain that $$\int \int_{S_2}F \cdot n dS = 16\pi$$ while $$\int \int_{S_3}F \cdot n dS = \pi$$

So the net flux is $$\frac{14}{3} \pi - \frac{15}{2} \pi + 16 \pi + \pi$$

Using the divergence theorem

Here the volume $V$ is described by $V=\{x^2+y^2 \leq z^2, z \in [1,2] \}$. By divergence thm, we know that $$\int_{S} F \cdot n dS = \int \int \int_{V} \text{div}(F) dxdydz = \int \int \int_V 2+2z = 2 \text{Vol(V)}- 2 \int \int \int_V z dxdydz$$.

The volume is $\text{Vol(V)}= \int_1^2 \int_{B_{\sqrt{z}}(0,0)}dxdy dz = \frac{7}{3} \pi$

The other term is $2 \int \int \int_V z dxdydz = 15 \pi $ so the net flux is $$\frac{14}{3} \pi -15 \pi$$

which is different from the previous one (and also negative). I can't spot any error in my computations/reasoning: where am I wrong?

答:
The theorem works: as @Ian wrote in his comment, the flux across the lower surface must be negative. The right computation is $$\int \int_{S_3}(x+y,y,z^2) \cdot n dS  = \int_0^{2 \pi} \int_0^1(*,*,1) \cdot(0,0,-\rho)d\rho d\theta = - \pi$$  since $S_3$ has radius $1$. Notice that the outward normal is pointing out, while yours is probably pointing inside the volume $V$ defined by your surface.
The other flux across $S_1$ is $16 \pi$, which can be done as the one above. That said, your net flux using the definition is $\frac{14}{3} \pi - \frac{15}{2} \pi + 16 \pi - \pi = \frac{14}{3} \pi + \frac{15}{2}\pi$

There's another problem in your divergence: that integral is $$\int \int \int_V 2 + 2z dxdydz = \int_1^2 \int_{B_z(0,0)}2+2z dxdydz =[\ldots ] = \frac{14}{3} \pi +\frac{15}{2} \pi$$
which confirms the previous result.


问:
Find the smallest possible integer that satisfies both modular equations
Find the smallest positive integer that satisfies both. x ≡ 4 (mod 9) and x ≡ 7 (mod 8) Explain how you calculated this answer.

I am taking a math for teachers course in university, so I'm worried about guessing too much in fear that I will teach myself incorrectly. Our textbook (Pearson Custom Math Text) is absolutely terrible! It skips briefly over certain elements.

If I had to guess how to do this question, it would be to convert them to x-4 and x-7 respectively then use multiples of 9 and 8 to see if I could come up with a positive integer which left the respective remainders... Is there an easier way to solve this? Am I doing it correctly?

答:
Apply the Chinese Remainder Theorem (the following is just applying the proof of the theorem to this case):

$$1=1\cdot 9+(-1)\cdot 8\implies x:=(4)(-1)\cdot8+7\cdot 1\cdot9=31$$


问:
Does $M = \oplus_i M_i = \sum_j M'_j$ with $M_i, M'_j$ simples implies $M_i \simeq M'_j$ for some i,j
Let $M$ be a $R$-module.
We assume that there exists two family $(M_i)_i$ and $(M'_j)_j$ of simple submodules of $M$ such that
$$ M = \bigoplus_i M_i = \sum_j M'_j. $$
Is there some $i,j$ such that $M_i \simeq M'_j$?

答:
For notational convenience I will let $I$ and $J$ be index sets for the $M_i$ and $M_j'$.

The answer is to your question is yes, and in fact for any $j\in J$ we can find $i\in I$ with $M_i\cong M_j'$. To see this, let $f:M_j'\hookrightarrow M$ be the inclusion map, and define $f_i=\pi_i\circ f$ for every $i\in I$, where $\pi_i:M\to M_i$ is the projection map. We cannot have every $f_i$ identically zero, or else $f$ would be identically zero, contradicting that $M_j'$ is simple. Hence there is some $i$ with $f_i$ non-zero. But any non-zero map between simple modules is an isomorphism, so $f_i$ is in fact an isomorphism $M_j'\cong M_i$, as desired.

In fact, a similar statement holds for $I$ instead of $J$: for every $i\in I$, we can find $j$ with $M_i\cong M_j'$. This follows from (the proof of) lemma 1 here; indeed, since $M=\sum_{j\in J}M'_j$, and each $M'_j$ is simple, there is some $J'\subseteq J$ with $M=\bigoplus _{j\in J'}M_j'$. Now we are in a position to apply exactly the same argument as above, by considering the compositions of the projections $\pi_j:M\to M'_j$ (for all $j\in J'$) with the inclusion $M_i\hookrightarrow M$.


答:
As $S$ is a symmetric matrix it is diagonalizable and we can write $D = U^{-1} S U$ with $D$ a diagonal matrix and $U$ an invertible matrix. Now you can transform
$$ S- s_{jj} I \to U^{-1} (S- s_{jj} I) U = D-s_{jj} I $$
which has the eigenvalues $\lambda_i$ of $S$ minus $s_{jj}$ which are positive per definition.

Title: Diagonal entries of symmetric matrices and eigenvalues

Body:
suppose I have a symmetric matrix $S$ with diagonal entries $s_{jj}$. I'm trying to prove that none of these diagonal entries $s_{jj}$ can be be smaller than all the eigenvalues of $S$. My textbook tells me that if one of these diagonal entries were small than all the eigenvalues, then $S-s_{jj}I$ would have wholly positive eigenvalues and be positive definite - however that couldn't be the case because $S-s_{jj}I$ would have a zero on the main diagonal.

What I don't understand is how one can skip straight to $S-s_{jj}I$ having wholly positive eigenvalues.

Help will be appreciated, thanks!


问:
Uniqueness of MLE for Poisson Regression
The maximum likelihood-based parameter estimation for the Poisson Regression is shown on various occasions, but I could not find the fact if the MLE that one can find via Gradient descend or newton-Raphson method is unique or not?</p>

答:
It is indeed unique, because the negative log-likelihood is strictly convex. I will show that here.

On a probability space $(\Omega, \mathscr{F}, P)$ let $y_i : \Omega \to \mathbb{N}$ be $n$ independent Poisson random-variables given some unknown parameters $\theta \in \mathbb{R}^m$ and corresponding "feature" vectors $b_i \in \mathbb{R}^m$. Assume that $m$ of the $n$ feature vectors are linearly-independent (this is almost always the case in practice as $n \gg m$).

The likelihood model for basic Poisson regression is:
\begin{align}
p\big{(}y_1,y_2,\ldots,y_n|\theta\big{)} &= \prod_{i=1}^n \text{Pois}\big{(}y_i;\lambda_i=e^{b_i^\intercal \theta}\big{)}\\[3pt]
&= \prod_{i=1}^n \frac{\big{(}e^{b_i^\intercal \theta}\big{)}^{y_i} e^{-\big{(}e^{b_i^\intercal \theta}\big{)}}}{y_i!}
\end{align}

The maximizing $\theta$ must be a stationary point:
\begin{align}
\frac{d}{d\theta}\log p\big{(}y_1,y_2,\ldots,y_n|\theta\big{)} &= 0\\[3pt]
\frac{d}{d\theta} \sum_{i=1}^n \log e^{y_i b_i^\intercal \theta} + \log e^{-e^{b_i^\intercal \theta}} - \log y_i! &= 0\\[3pt]
\sum_{i=1}^n \frac{d}{d\theta} y_i b_i^\intercal \theta - \frac{d}{d\theta} e^{b_i^\intercal \theta} - \frac{d}{d\theta} \log y_i! &= 0\\[3pt]
\sum_{i=1}^n y_i b_i^\intercal - e^{b_i^\intercal \theta}b_i^\intercal &= 0\\[3pt]
\sum_{i=1}^n \big{(}y_i - e^{b_i^\intercal \theta}\big{)}b_i &= 0\\[3pt]
\end{align}

For a method like gradient ascent or Newton's to maximize uniquely, we must have strict concavity:
\begin{align}
\frac{d^2}{d\theta^2}\log p\big{(}y_1,y_2,\ldots,y_n|\theta\big{)} &\overset{?}{<} 0\\[3pt]
\frac{d}{d\theta} \sum_{i=1}^n y_i b_i^\intercal - e^{b_i^\intercal \theta}b_i^\intercal &\overset{?}{<} 0\\[3pt]
- \sum_{i=1}^n e^{b_i^\intercal \theta}b_i b_i^\intercal &< 0\\[3pt]
\end{align}

We do, because the real exponential is positive-definite, $b_ib_i^\intercal$ is positive-semidefinite, and the sum contains $m$ linearly-independent rank-1 terms. Note: even if that last condition isn't met, we still have concavity ($\small{\implies}$ global maximum), just not strict concavity ($\small{\implies}$ unique global maximum).


问:
Can a primitive root of a polynomial over $GF(2)$ ever not generate a multiplicative group?
Can a primitive root of a polynomial over $GF(2)$ ever not generate a multiplicative group?

I have some notes from my review of finite field extensions a while ago that I've been rereading. It's the last statement that's throwing me.  I've included some preceding notes for context.

If $p(x)$ is an irreducible polynomial
  of degree $n$, then adjoining a root
  of $p$ to $GF(2)$ generates an
  extension of degree $n$, which is
  necessarily a field, $E$, with $2^n$
  elements.

The multiplicative group of nonzero
  elements in $E$ has order $2^n - 1$.

  Thus by Lagrange's Theorem, every
  nonzero element $a$ of $E$ satisfies
  $a^{2^n - 1} = 1$.  Thus every element
  $a$ in $E$ is a root of $g(X) =
> X^{2^n} - X$.

In other words, $E$ is exactly the set
  of all roots of $g(X)$. Now the roots
  of the original $p(x)$ are also roots
  of $g(X)$, and so $p(x)$ divides
  $g(X)$ (after making the variables the
  same).

Conversely, if $f(x)$ is any
  polynomial that divides $g(x)$, then
  the roots of $f(x)$ lie in $E$, so
  they generate a subfield of $E$. If
  they generate all of $E$, and if
  $f(x)$ is irreducible, then they must
  have degree $n$.

Now, let $\rho$ be a primitive root of
  $f$, where f is irreducible. So $\rho$
  will generate $E$ as a field, but not
  necessarily generate $E-\{0\}$ as a
  multiplicative group.

答: I think this is an example of what you're after. The polynomial $x^4+x^3+x^2+x+1$ is irreducible over the field of 2 elements, so any root $\alpha$ generates the field of 16 elements. But $\alpha^5=1$ (note that the given polynomial is a factor of $x^5-1$), so $\alpha$ doesn't generate the 15-element multiplicative subgroup as a group. </p>


问:
Why does $\log^{100}(n)$ grow asymptotically slower than $n^{7/8}$
More formally:
How do I compute the following limit:

$$
\lim _{x \rightarrow \infty} \frac{\log ^{100}(x)}{x^{7 / 8}}
$$

I would start using l'Hôpital's rule and notice the pattern, is there some other procedure? (without knowing it)

答:
Taking the logarithm of the numerator and denominator gives

$$ \frac{\log \left( \,(\log x)^{100}\,\right)}{\log \left(x^{\frac{7}{8}}\right)} \;\;= \;\; \frac{100 \log \log x}{\frac{7}{8}\log x} \;\; = \;\; \frac{800}{7} \cdot \frac{\log \log x}{\log x} \;\; \rightarrow \;\; 0, $$

which suggests the limit will be $0.$ However, this process can change the value of a limit, as can be seen by considering $\lim\limits_{x \rightarrow \infty} \frac{2x + 42}{3x + 42}$ — the limit is $\frac{2}{3}$ before taking logarithms of the numerator and denominator, and the limit is $\frac{\log 2}{\log 3}$ after taking logarithms of the numerator and denominator. Although I believe this process will NOT change the value of the limit (for positive numerators and positive $\neq 1$ denominators) if the original limit is zero or infinite, we can circumvent this issue by rewriting in such a way that the original numerator and denominator are not changed, but still have present the simplifying aspect of taking a logarithm. This can be done by applying TWO successive operations, taking a logarithm followed by an exponentiation:

$$ \frac{(\log x)^{100}}{x^{\frac{7}{8}}} \;\;= \;\; \frac{\exp \left[ \log \left( \,(\log x)^{100}\,\right)\right]}{\exp \left[\log \left(x^{\frac{7}{8}}\right)\right]} \;\;= \;\; \frac{\exp \left[100 \log \log x \right]}{\exp \left[ \frac{7}{8}\log x \right]}$$

$$ = \;\; \exp \left[100\log \log x \; - \; \frac{7}{8}\log x  \right] $$

Since we clearly have $\;100\log \log x - \frac{7}{8}\log x \; \rightarrow \; -\infty,\;$ and $\exp(\rightarrow -\infty) \rightarrow 0,$ it follows that the original quotient approaches $0.$


问:
Definition of the determinant as of a function $d:\operatorname{End}(V)\rightarrow\mathbb{R}$
I'm trying to characterize the determinant in the following way. Consider a finite-dimensional vector space $V$ over $\mathbb{R}$ and the natural ring of its endomorphisms  $\operatorname{End}(V)$. Given the determinant, we may define a function $d:\operatorname{End}(V)\rightarrow\mathbb{R}$ with the following properties:

$d(1)=1$

$d(fg)=d(f)d(g)$

$d(f)\neq0\iff f$ has a two-sided inverse (i.e. $f$ is an automorphism)

Can we complete this set of conditions in such a way that the determinant could be defined. Is there some category theoretical point of view?

答: If you add the condition that $d$ is an alternating multilinear map of $f(e_1), \dots, f(e_n)$ where $\{e_1, \dots, e_n\}$ is the canonical basis of $V$ then you’ll get that $d$ is the determinant.


问:
Reference request on spherical harmonics
I'd like to find a (hopefully modern, mathematician-friendly) reference which proves that homogeneous harmonic polynomials restrict to an orthonormal basis for $L^2$ functions on the sphere $S^n$. Particularly I'd like to understand why we should expect that harmonic functions on $\mathbb{R}^{n+1}$ restrict to eigenfunctions of the Laplacian on the sphere.

答: Chapter 5 of Harmonic Function Theory by Axler, Bourdon and Ramey. As with any Axler book, it's a very thorough treatment of harmonic polynomials with a fair bit of motivation.</p>


问:
Orthogonal transformations and metric tensor: prove that $U^{T} g U=g$
I'm attempting to prove that, if $U=(U^i_j)_{n \times n}$ is the matrix associated to an orthogonal transformation and $g=(g_{ij})_{n \times n}$ is the metric tensor, then

$$U^{T} g U=g \tag{1}$$

My attempt

Applying the condition that an orthogonal transformation must preserve the scalar product, $\big(U(\boldsymbol{x}), U(\boldsymbol{y})\big) = (\boldsymbol{x}, \boldsymbol{y})$, I have found that that

$$ \sum ^n _{i=1} \sum ^n _{j=1} \sum ^n _{k=1} \sum ^n _{l=1} g_{i j} U_{k}^{i} U_{l}^{j} x^{k} y^{l}= \sum ^n _{i=1}\sum ^n _{j=1} g_{i j} x^{i} y^{j} \tag{2}$$

Using the Einstein summation convention and changing in the RHS the indexes $i \rightarrow k, j \rightarrow l$

$$g_{i j} U_{k}^{i} U_{l}^{j} x^{k} y^{l} = g_{kl} x^{k} y^{l} \tag{3}$$

Since the terms $x^{k} y^{l}$ appears at both sides, if we could identify $g_{i j} U_{k}^{i} U_{l}^{j}=g_{k l}$, this would correspond to the matrix product $U^{T} g U=g$, and the proof would be complete.

But I'm stuck here, because I don't see how we can match parts of different sums... Could someone tell me how could it be justified from $(3)$ that $g_{i j} U_{k}^{i} U_{l}^{j}=g_{k l}$ (if it is actually possible to do this)?

答:
As already said, the key is that the equality must be satisfied for any values of $\boldsymbol{x}=(x^1,...,x^n)$ and $\boldsymbol{y}=(y^1,...,y^n)$. Also notice that in the LHS, once the sum is performed over $i$ and $j$, there is a term that only depends on $k$ and $l$. This is clear by reordering equation $(3)$ as follows:

$$ \sum ^n _{k=1} \sum ^n _{l=1} x^{k} y^{l} \bigg(\sum ^n _{i=1} \sum ^n _{j=1} g_{i j} U_{k}^{i} U_{l}^{j} \bigg) = \sum ^n _{k=1}\sum ^n _{l=1} x^{k} y^{l}g_{kl} $$

$$ \sum ^n _{k=1} \sum ^n _{l=1} x^{k} y^{l} h_{kl} = \sum ^n _{k=1}\sum ^n _{l=1} x^{k} y^{l}g_{kl} $$

So $h_{kl}=g_{i j} U_{k}^{i} U_{l}^{j}=g_{kl}$.


问:
$n$th root of $\alpha$
Let $n$ be a positive integer and $K$ be a field of positive characteristic coprime to $n$. Show that for any $a \in K^\times$, $[K(a^{1/n}):K]$ is a divisor of $n$. We are not assuming $\zeta_n\in K$.

I know a special case of this to be true like if $K=\mathbb{F}_p$, but how to show this in general, if $K$ is not finite?

答:
With $a^{1/n}$ I mean a root of $x^n-a\in K[x]$ minimizing $[K(a^{1/n}):K]$. Since $$[K(a^{1/{pn}}):K]=[K((a^{1/n})^{1/p}):K(a^{1/n})][K(a^{1/{n}}):K]$$
It will suffice to check the case $F(b^{1/p})/F$ with $p$ prime and $b \in F$.

The $F(\zeta_p)$ minimal polynomial of $b^{1/p}$ divides $x^p-b\in F(\zeta_p)[x]$, it is $$f(x)=\prod_{j=1}^d (x-\zeta_p^{c_j} b^{1/p}), \qquad d=[F(\zeta_p,b^{1/p}):F(\zeta_p)]\ \le p$$ $$f(0)=\zeta_p^t b^{d/p}\in F(\zeta_p)$$

Thus, either $d=p$ or $b^{1/p}\in F(\zeta_p),d=1$.

If $d=p$ then $x^p-b\in F[x]$ is irreducible.

Otherwise for $b^{1/p}\in F(\zeta_p),d=1$ then $$N_{F(\zeta_p)/F}(b^{1/p})=\zeta_p^s b^{r/p}\in F,\qquad p\nmid r=[F(\zeta_p):F]$$ Whence $\zeta_p^{s r^{-1}} b^{1/p}\in F$ and $F$ contains a root of $x^p-b$. Since we chose a root of $x^p-b$ minimizing $[F(b^{1/p}):F]$ it means that $[F(b^{1/p}):F]=1$.

All this proves that $$[F(b^{1/p}):F]\ |\ p \qquad \implies\qquad [K(a^{1/n}):K]\ | \ n$$

Title: $n$th root of $\alpha$


问:
Gradient of a norm with power
Can someone suggest me the approach to find the gradient of 4-norm ?  eg:(||x||) where it is not a one norm but 4 norm to the power of 4.

I need to solve this question for my practice for the exams. Any input would be helpful. We will consider 4th root for the norm here.

答: The norm itself must have a 4th root. But in the picture, the function $f$ has a power 4. So we can look at the function simply as $f(x) = x_1^4 + x_2^4 + ... +x_d^4$, where $x_i$s are the components of $x$. Therefore the gradient is simply $$\nabla f = 4 (x_1^3, x_2^3, ..., x_d^3 ) \in \mathbb{R}^d$$

Title: Gradient of a norm with power


问:
How to solve 700 = 7x + 6y for a range of values?
Last time I did any maths was A-Level (some time ago!).

This is a programming/layout problem where I need to display 7 items across a page, with 6 gaps between them. I have a fixed width, but need to determine values of x and y that are integers where x > y.

This 'feels' like something I could plot on a graph, hence solve with calculus but I need a pointer in the right direction. Ideally I'd like to solve it programatically but I need an understanding of how to solve it first. Further down the line, I'd like to be able to vary the number of items to get different values.

Thanks

答:
$$700 = 7x + 6y\implies y = \frac{-7(x - 100)}{6}$$
Experimentation in a spreadsheet shows that [integer] $x,y$ values increase/decrease by amounts corresponding to their opposite coefficients. For example, a valid $x$-value occurs only every $6$ integers and a resulting $y$-value occurs every $7$ integers. Here is a sample to show the effect.

$$(-14,133)\quad (-8,126)\quad (-2,119)\quad (4,112)\quad (10,105)\quad (16,98)\quad $$

A little math shows the predictable values and relationships to be as follows.

$$x = 6 n + 4 \qquad y = 112 - 7 n \qquad  n \in\mathbb{Z}$$

Now, if we have a known $x$-range, we can find $n$ by plugging $x$-lo and $x$-hi into
$$n = \frac{(x - 4)}{6}$$
and rounding up or down to get the integer values desired and needed.

ContentLicense: CC BY-SA 2.5


问:
Infinite strictly decreasing sequence of ordinals
I know that on a linearly ordered set $X$, $\leq$ is a well order if and only if $X$ contains no strictly decreasing sequence. However this conflicts in my head with the following problematic 'intuition' which I do not know where it fails.

Let $\alpha$ be a limit ordinal such that $\alpha\geq \omega_1$. All ordinals strictly less than $\alpha$ are a proper set and therefore we should not be able to find an infinite strictly decreasing sequence of ordinals. However since there uncountably many ordinals lesser than $\alpha$, it seems like we should be able to choose a strictly decreasing sequence in some complicated way. Aside from this proposition, can someone explain to me why a decreasing sequence must stablize?

I know there is probably some simple point which has not fully clicked for me.

答:
This is a direct consequence of the definition of a well-ordered set and the definition of an ordinal number.

A linearly ordered set $(X, \leq)$ is well-ordered if and only if every nonempty subset $S \subseteq X$ has a least element.

Contrapositively, if we could come up with some infinite decreasing sequence $$x_1 > x_2 > x_3 > ... $$ of elements of $X$, then this would correspond to a nonempty, infinite subset $$S := \{ x_1, x_2, x_3, ... \} \subseteq X$$ with no least element, so $X$ would not be well-ordered.

Ordinal numbers, meanwhile, are typically defined to be certain well-ordered sets. In Von Neumann's formulation, for instance, any given ordinal is defined to be the well-ordered set of all smaller ordinals:

A set S is an ordinal if and only if S is strictly well-ordered with respect to set membership and every element of S is also a subset of S.

So if I try to build my infinite descending sequence of ordinals starting from some infinite--even some uncountably infinite--ordinal number $\alpha$, I can't do it, because that sequence necessarily lives inside the well-ordered set $\alpha$, and as a subset of a well-ordered set, it has a least element.


问:
Show that $\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})\mathbf{Y}^T]$
Let $\mathbf{Y}$ be a $p$-dimensional random vector with $\mathbb{E}|\mathbf{Y}|^2<\infty$. Show that $$\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})\mathbf{Y}^T]$$

I tried using the fact that $$\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})(\mathbf{Y}-\mathbb{E}\mathbf{Y})^T]$$

and then expanding it and simplifying it, but I couldn't derive the former.

How do I go about with this?

答: The difference between the two terms is $E[(Y-EY) (EY)^\top]$. Since $(EY)^\top$ is not random, you can pull it out to obtain $E[(Y-EY)] (EY)^\top$. Can you conclude from here?


答: if $F(z)=\frac{az+b}{cz+d}$ and $ad-bc=1$ then upper and lower half plane are  mapped to themselves only</p>

Title: $H^+\to L^-$ and $H^-\to L^+$?

Body:
Define

$H^{+}=\{z:y>0\}$

$H^{-}=\{z:y<0\}$

$L^{+}=\{z:x>0\}$

$L^{-}=\{z:x<0\}$

$f(z)=\frac{2z+1}{5z+3}$  maps

$1.$ $H^+\to H^+$ and $H^-\to H^-$

$2$. $H^+\to H^-$ and $H^-\to H^+$

$3.$ $H^+\to L^-$ and $H^-\to L^+$

$4.$ $H^+\to L^+$ and $H^-\to L^-$

here  $\to$ means "onto", well, I have no idea how to solve this one. please help.
$X$ axis is mapped to the curve ${2x+1\over 5x+3}$


问:
How many words can be created by using the letters of "MATHEMATIK" when no same letters can be next to each other?
The word MATHEMATIK (in german) has 10 letters.

I want to know how many different words can be created by using the letters of this word. However, the same letter can never be next to each other.

So for example, MAATHEMTIK is not allowed.

I did the following:

$n = 10$

$k_1 = 2$ (M)

$k_2 = 2$ (A)

$k_3 = 2$ (T)

$k_4 = 1$ (H)

$k_5 = 1$ (E)

$k_6 = 1$ (I)

$k_7 = 1$ (K)

I use the formula $\frac{n!}{k_1! \cdot k_2! ... \cdot k_n! }$ and get:

$$\frac{13!}{2! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1! \cdot 1!} = 778377600$$

This is the number of all possible words (permutations). Now we need to subtract the number of words which can be created by having the same letter next to each other, i.e.

MAATHEMTIK

AAMTHEMTIK

MTAAHEMTIK

etc.

But I don't know how to figure that out.

答:
Someone mentioned another solution involving the permutations of "Mississippi," but it makes things unnecessarily complicated.

Simple solution:

Assume initially that all the letters of 'mathematik' are distinct, subscripting them if necessary. So the first $m$ is labelled $m_1$ and the second is $m_2$. Then with 10 "distinct" letters (not really distinct, just different b/c of the subscripts) there are $10!$ permutations.

Now to deal with the overcounting due to the subscripts: for each subscripted letter divide by the factorial of its count. For example, since $m$ appears twice, divide by 2!. Likewise divide by 2! because of the 'a' appearing twice, and again by 2! because of the 't' appearing twice.

So the final answer is ${10! \over {2! 2!2!}}$.

In general, if a letter appears 3 times, divide by 3! to compensate for overcounting, etc.

[edit]: I forgot to incorporate the fact that a letter can't appear next to the same letter. Please see Aryan's answer below.


Body:
Using
$$ \ln(1+u)<u, u>0$$
one has, for $x>1$,
$$ 0<\frac{\ln^{100}x}{x^{7/8}}=\bigg(\frac{\ln^{\frac12}x}{x^{175}}\bigg)^{200}=\bigg(\frac{\ln^{\frac12}[1+(x-1)]}{x^{175}}\bigg)^{200}<\bigg(\frac{(x-1)^{\frac12}}{x^{175}}\bigg)^{200}<\frac1{x^{174\cdot200}}$$
from which it is easy to obtain
$$ \lim_{x\to\infty}\frac{\ln^{100}x}{x^{7/8}}=0.$$


问:
Quickly finding $\theta$ from $\sin\theta=\frac{3}{\sqrt{73}}$. (In particular, why can we say that $\theta<37^\circ$?)

If it is given to you that
$$\sin\theta = \frac{3}{\sqrt{73}}$$
then how will you find this angle quickly?

So you can imagine that real diagram is like this

The thing is to find that angle $\theta$.

Also, my teacher assumed it intuitively that $\theta$ is less than $37^\circ$. What is the way to do that?

答: You're evaluating $\arctan\tfrac38\approx20.56^\circ$, which needs a calculator, but $\arctan\tfrac{1}{\sqrt{3}}=30^\circ$ is a tighter upper bound than your teacher requested. @Arthur obtained it another way.


答: You can even safely assume it's under 30 degrees, way less than 30 in fact, since at 30 degrees, the opposing side reaches half of hypothenuse, and 3 is way less than $\frac{\sqrt{73}}{2}$, so your angle is below $30 ^ \circ$.

Title: Quickly finding $\theta$ from $\sin\theta=\frac{3}{\sqrt{73}}$. (In particular, why can we say that $\theta<37^\circ$?)

Body:
If it is given to you that
$$\sin\theta = \frac{3}{\sqrt{73}}$$
then how will you find this angle quickly?

So you can imagine that real diagram is like this

The thing is to find that angle $\theta$.

Also, my teacher assumed it intuitively that $\theta$ is less than $37^\circ$. What is the way to do that?


问:
Equivalent definition of fractional ideal on ring of integers of a number field
I know the following definition on a Dedekind domain:

Let $D$ be a Dedekind domain and $K$ its field of fractions, then we say that $F$ is a fractional ideal of $D$ if it is a $D$-submodule of $K$ and there exists a $c\in D \setminus \{0_D\}$ such that $cF \subset D$

Now, if we have that $K$ is a number field and $R$ is its ring of integers, as $R$ is a Dedekind domain we can define as before the fractional ideal of $R$ but, in the book Problems in algebraic number theory by M. Ray Murty and Jody Esmode, the following definition is given:

Given $K$ a fractional ideal and $R$ is its ring of integers, we say that $F$ is a fractional ideal of $R$ if it is a $R$-submodule of $K$ and there exists a $m \in \mathbb{Z} \setminus \{0\}$ such that $mF \subset R$

It is clear that the general definition implies this one as $\mathbb{Z} \subset R$ but, is this definition in fact equivalent to the general one?

答:
In the case where $R$ is the ring of integers (or more general an order) of a number field $K$ they are equivalent. Suppose that $F$ is a fractional ideal of $R$ in the sense that there is some $s\in R\setminus\{0\}$ with $sF\subseteq R$. Then we can take the norm $m:=N_{K/\Bbb Q}(s)\in \Bbb Z\setminus\{0\}$. Since $s\mid m$ in $R$ we still have $mF\subseteq R$.

A counterexample for the general case of arbitrary Dedekind domains $D$:
Take $D=k[x]$ and $F=\frac{1}{x}D$ where $k$ is any field.


答:
The only thing you have to remind is $E[E[Y]] = E[Y]$.

By definition,
$$
\begin{align}
\text{Cov}(Y)
&= E[(Y-E[Y])(Y-E[Y])^\top]\\
&= E[YY^\top - YE[Y]^\top - E[Y] Y^\top +E[Y]E[Y]^\top]\\
&= E[YY^\top]-E[Y]E[Y]^\top
\end{align}
$$

The form you wrote is
$$
E[(Y - E[Y])Y^\top] = E[YY^\top - E[Y]Y^\top] = E[YY^\top] - E[Y]E[Y]^\top
$$
which gives what you want.

Title: Show that $\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})\mathbf{Y}^T]$

Body:
Let $\mathbf{Y}$ be a $p$-dimensional random vector with $\mathbb{E}|\mathbf{Y}|^2<\infty$. Show that $$\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})\mathbf{Y}^T]$$

I tried using the fact that $$\text{Cov}(\mathbf{Y})=\mathbb{E}[(\mathbf{Y}-\mathbb{E}\mathbf{Y})(\mathbf{Y}-\mathbb{E}\mathbf{Y})^T]$$

and then expanding it and simplifying it, but I couldn't derive the former.

How do I go about with this?


问:
Projection on Banach spaces
Let $X$ be a Banach space, $P: X\rightarrow X$ with $P^2=P$, linear. Prove that $P$ is bounded $\iff X=\ker P \bigoplus PX$.

Context: exercise $5.14$ in Bollobas' Linear Analysis.

My try: since $P^2=P$, then $\exists x$ with $ Px=y \iff Py=y$, so $PX$ are the fixed points of $P$. Also, suppose $x \in PX\cap\ker P$. Then $x=Px=0$, therefore $PX\cap\ker P=\{0\}$.

Now, let $x \in X$. Then $P(x-Px)=Px-Px=0$, so that $PX+\ker P=X\implies X=\ker P \bigoplus PX$.

However I didn't use that $X$ is Banach, and neither I had to suppose that $X$ is continuous. In fact, I didn't use at all the topological structure of $X$. What is wrong in my reasoning?

答:
By definition of the projection $P$, the projection $p$ on $PX$ is equal to $P$. Therefore $p$ is continuous if and only if $P$ is bounded. As the projection $q$ on $\ker P$ is equal to $Id -p$, $q$ is also continuous if and only if $P$ is bounded as the identity is bounded.

Therefore $P$ bounded is equivalent to $X = \ker P \oplus PX$ where $\oplus$ stands for the topological sum.

Title: Projection on Banach spaces


问:
Measure of the union of the sets of a numerable family of measurable sets
I'm working with some measure exercises . I've tried to get a formula to know the measure of the union of n sets ( finite number ), I think I have it, proved by induction, I wonder if there is also a formula to measure the infinite union of all the sets in a numerable family or if that can't be simply measured or if there doesn't exist a formula.
Thanks.</p>

答:
Welcome to MSE!

I'm going to link a related question that doesn't look like it ever got a full answer, in case it's of interest to you.

The result you're looking for is called inclusion-exclusion, and because the relevant sum is alternating it gives rise to a family of inequalities called
Bonferroni Inequalities. This is another place where knowing the name of something lets you find endless information about it. I suggest you google "countable inclusion exclusion principle" or something similar now that you know!

For convenience, though, here is one result in this vein, which you can find in the paper "Exact Conditions for Countable Inclusion-Exclusion Identity and Extensions" by Friedland and Krop (see here).

$$\mu \left ( \bigcup_{i \in \mathbb{N}} A_i \right ) = \sum_{k \in \mathbb{N}} (-1)^{k-1} \sum_{1 \leq i_1 \leq \ldots \leq i_k} \mu \left ( A_{i_1} \cap \ldots \cap A_{i_k} \right )$$

Then the authors give necessary conditions for this to converge (after all, it's an infinite sum). First let

$$ S_k \triangleq \sum_{1 \leq i_1 \leq \ldots \leq i_k} \mu \left ( A_{i_1} \cap \ldots \cap A_{i_k} \right )$$

Then the paper says

[The result] holds if $S_k$, $k \in \mathbb{N}$ is a sequence of nonnegative numbers that converge exponentially to zero:

$$ S_k \in [0, \infty) \text{ for } k \in \mathbb{N} \text{ and } \limsup_{k \to \infty} S_k^{\frac{1}{k}} < 1$$

You can check out the original paper for more information.

I hope this helps ^_^


答:
First of all the letters are :

2×A,2×M,2×T,H,E,I,K.

Now we will do as the following first of all it is clear that there is $7!$ different ways of making a word just by using the letters A,M,T,H,E,I,K.

note that there are $8$ different places for putting a letter in this word (one before the first letter one between the first and second letters one between the second and third and so on) so if we want to place a second letter A there are $8-2$ different possible places since there are two places adjacent to the first A and we should not place our second A in those places. So there is $7!×6$ different methods of putting 2×A,M,T,H,E,I,K such that no two same letters are adjacent.

Now in the same way there are 9 different places for putting the second M but two of them are out of access since the second M should not be adjacent to the first letter M. So there is $7!×6×7$ different methods of putting 2×A,2×M,T,H,E,I,K such that no two same letters are adjacent.

And finally for putting the second T there are 10 different places two of which or not allowed to have another T in them. So at last there are $7!×6×7×8$ different words having the same letters as MATEMATIKS such that no two same letters are adjacent.

Now note that in this counting, each valid word will be counted exactly 8 times and this is the reason:

If for a 10-letter word $W$ made of the letters of MATEMATIKS, we choose one A, one M and one T and the H,E,I,K as the initial letters, then by removing the other 3 letters there will be 7 initial letters, using which, by going through the foresaid procedure $W$ can be built. So technically each word is produced 8 times each time by a different initial 7-letter. (That's clearly because there are 2×2×2 choices for choosing one A, one M and one T to be the initial ones out of two A, two M and two T. Also note that because $W$ is a word in which no two same letters are adjacent, all of these initial 7-letters that produce $W$ are pairwisely different)

Hence any word is counted 8 times and the answer is $7!×6×7×8÷8=7!×6×7$

Title: How many words can be created by using the letters of "MATHEMATIK" when no same letters can be next to each other?

Body:
The word MATHEMATIK (in german) has 10 letters.

I want to know how many different words can be created by using the letters of this word. However, the same letter can never be next to each other.

So for example, MAATHEMTIK is not allowed.

I did the following:

$n = 10$

$k_1 = 2$ (M)

$k_2 = 2$ (A)

$k_3 = 2$ (T)

$k_4 = 1$ (H)

$k_5 = 1$ (E)

$k_6 = 1$ (I)

$k_7 = 1$ (K)

I use the formula $\frac{n!}{k_1! \cdot k_2! ... \cdot k_n! }$ and get:

$$\frac{13!}{2! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1! \cdot 1!} = 778377600$$

This is the number of all possible words (permutations). Now we need to subtract the number of words which can be created by having the same letter next to each other, i.e.

MAATHEMTIK

AAMTHEMTIK

MTAAHEMTIK

etc.

But I don't know how to figure that out.


问:
sum of polynoms of given property
I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.

I have to prove that:

$f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.

I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.

答:
Because $f$ is a nonzero polynomial, it has only finitely many zeros $x_1< x_2 < \dots < x_r$. Moreover, $f(x) \underset{x \to \pm \infty}{\sim} P(x)$ so $P(x) \geq 0$ on $(- \infty,x_1]$ and $[x_r,+ \infty)$. Notice also that the sign of $f$ is constant on $[x_i,x_{i+1}]$.

Let $1 \leq i \leq r-1$. Because $f(x_i)=f(x_{i+1})=0$ there exists $y_i \in (x_i,x_{i+1})$ such that $f'(y_i)=0$ by mean value theorem. But $f(y_i)=P(y_i)+f'(y_i) \geq 0$. Therefore, $f$ is non-negative on $[x_i,x_{i+1}]$.

Finally, you deduce that $f$ is non-negative.

Title: sum of polynoms of given property


问:
Name of distribution with pdf $r\exp(-r^2/2)1_{[0,\,\infty)}(r)$?
Is there a name for the distribution with pdf $$f(r) = r\exp(-r^2/2), \ \ r\ge0$$

?

It's not quite Exponential, not quite Normal, nor Half-Normal, I can't find any distribution I know of that this could be. Does it have some standard name?

Context:
I'm working through a question in probability that gives iids $X,Y \thicksim N(0,1)$, and $R,\,\Theta$ as the corresponding polar coordinates for $R \in [0, \infty), \Theta \in [0, 2\pi)$, so

$$ X = R\cos(\Theta), \ Y = R\sin(\Theta)$$

Working out the joint pdf of R and Theta, I get

$$f_{\Theta}(t) = 1/(2\pi) $$
$$f_R(r) = r\exp(-r^2/2)$$
In their respective domains

Which both seem like valid pdfs, give $R$ and $\Theta$ independent and $\Theta \thicksim U[0, 2\pi)$

答: A $\sigma=1$ Rayleigh distribution.

Body: A $\sigma=1$ Rayleigh distribution.


问:
Strong convexity and the composition of convex functions
My question concerns strong convexity and the composition of functions. Let $f:\Omega \subseteq \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous differentiable function. Recall that $f$ is convex if its epigraph $\{ (\mathbf{x}, \mu) \in \Omega \times \mathbb{R} \colon \mu \geq f(\mathbf{x}) \}$ is a convex set. Furthermore, $f$ is strongly convex if it holds $$( \nabla_{\mathbf{x}} f -  \nabla_{\mathbf{y}}f )^T (\mathbf{x} - \mathbf{y})\geq m \| \mathbf{x} - \mathbf{y} \|_2^2$$ for a constant $m > 0$. Note that strong convexity is a strictly stronger definition than convexity.

It is well-known that if $f$ is convex and $g$ is convex non-decreasing over an univariate domain, then the function $g \circ f$ is also convex. Does this property extends to strong convexity? Specifically, if $f$ is a strongly convex function as described above and $g$ is convex non-decreasing, is the function $g \circ f$ also strongly convex?

Progress I made on this question: It can be easily proven that $f$ is strong convex iff. the function $f(\boldsymbol{x}) - \frac{m}{2} \| \boldsymbol{x} \|_2^2$ is convex. Hence, to answer this question it is sufficient to prove that $(g \circ f)(\boldsymbol{x}) - \frac{\beta}{2} \| \boldsymbol{x} \|_2^2$ is a convex function for some parameter $\beta > 0$, given that $f$ is strongly convex and $g$ is convex non-decreasing.

答:
No; here is a counterexample:

Let $f=\|\cdot\|^2$ which is strongly convex. However, if we let $g$ be the zero function, then $g\circ f$ is also the zero function, which is convex, but not strongly convex.


问:
Inverse sine simplifying problem
It's a simple but not popular problem and there isn't a lot to say here, the problem is:

simplify the following function:

\begin{align}
 \arcsin(b\sin(x))
\end{align}

And for example, if b=2 what is the resulting simplified function?

答: You don't just simplify $\sin^{-1}(b \sin(x))$, unless $|b|<1 $... and even if you only consider $-1<b<1$, I doubt you'll get anything pretty...

Title: Inverse sine simplifying problem


问:
Problem getting to and understanding the answer of $\lim \limits_{x \to 0^+} \frac{(\sin{x})^x}{x}$
$$\lim \limits_{x \to 0^+} \frac{(\sin{x})^x}{x}$$

Somehow right limit ($0^+$) is $+\infty$ but left ($0^-$) one is undefined. I wrote limit like $ \lim \limits_{x \to 0^+} \frac{\sin{x}}{x} * \lim \limits_{x \to 0^+} (\sin{x})^{x-1}$ and the first one is 1 so it leaves us with just $\lim \limits_{x \to 0^+} (\sin{x})^{x-1}$. So if we put the values of the limit in: $\sin{x}$ when $x$ is converging to $0$ is $0$ and we get $0^{0-1}$ or $\frac{1}{0}$. So if the right limit is $\frac{1}{0^+} = +\infty$ why isn't $\frac{1}{0^-} = -\infty$?

答: Hint: use that $\sin x\geq\frac{x}{2}$ for $x$ close to $0$.


问:
Find min of $IA + IB + IC +ID$ in tetrahedron $ABCD$
Let the point $I$ in tetrahedron $ABCD$. Find $\min\{IA + IB + IC + ID\}$.

I can't solve this problem, even in the case ABCD regular. Please help

答:
I only can solve regular case, here is the solution:

$H$ and $F$ are one the plane $BCD$, $H,F$ is the foot of $A,I$,

$G$ is on $AH$ and $GH=IF=q,IG=FH=p,\angle FHB=\theta,BH=HD=HB=b,AH=h $

$AI=\sqrt{p^2+(h-q)^2},BF^2=p^2+b^2-2pbcos(\theta),FD^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta),FC^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)$

$AI+BI+CI+DI=\sqrt{p^2+(h-q)^2}+\sqrt{p^2-2pbcos(\theta)+b^2+q^2}+\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta)+q^2}+\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)+q^2}=f(p,q,\theta)$

we note that $p,q$ is independent, so we take $p$ first to let $f_{p}=f_{p_{min}}$

$f'_{p}=\dfrac{2p}{\sqrt{p^2+(h-q)^2}}+\dfrac{2p-2bcos(\theta)}{\sqrt{p^2-2pbcos(\theta)+b^2+q^2}}+\dfrac{2p-2bcos(\dfrac{2\pi}{3}-\theta)}{\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta)}}+\dfrac{2p-2bcos(\dfrac{2\pi}{3}+\theta)}{\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)+q^2}}=0$

here we proof that $g(x)=\dfrac{2x-2m}{\sqrt{x^2-2mx+m^2+q^2}}$ will be mono increasing function. because:

$g'(x)=\dfrac{q^2}{(x^2-2mx+m^2+q^2)^{\frac{3}{2}}}>0 \to f'_{p}$is mono increasing function also . luckily $f'_{p}(0)=0$, so we have only one root. it is trivial that $\sqrt{p^2-tp+s}$ will be increasing function when $p$ is big, so $ p=0 \to f_{p_{min}}$  and again we are lucky to clean $\theta$ also.
now $f=3\sqrt{b^2+q^2}+h-q$, $b,h$ is known so you can find final result.


问:
Uniform convergence sequence of positive reals
Suppose $f_n$ for $n ∈ \Bbb N$ and $f$ are functions on $J$. If there exists a sequence $(a_n)$ of positive reals satisfying $a_n→0$ as $n→∞$ and
$$|f_n(x)−f(x)|≤a_n, \quad  ∀n∈\Bbb N, ∀x∈J$$
then $(f_n)$ converges uniformly to $f$.

Can someone help me with the proof of this theorem?

答:
Let $\epsilon>0.$ Let $N$ be large enough so that $n>N$ implies $a_n<\epsilon.$ Then for $n>N$,
$$
|f_n(x)-f(x)|\leq a_n < \epsilon.
$$


问:
Localization property of spherical harmonics on incomplete spheres
This question follows from another question I asked here.

I am currently trying to read this paper and I am having difficulty in understanding the interpretations of eq. 7 which are given in lines immediately after the equation. It would be very helpful if you can provide hints on what they mean and also the calculation of the $v_i$ and $\lambda_i$.

As I understand, the point that the author is trying to make is that eigenvectors are localized in either the partial sky or the sky which is masked. But I am unable to follow how the inference is derived.

Any help in this regards would be appreciated.

Update 1 Jan 2021
So, here is the math that I did.The integral in equation 7 of the paper that I linked to goes like this,
$$ \int_{S'} v_i(\pmb{\hat{r}}')\delta(\pmb{\hat{r}}-\pmb{\hat{r}}')d\Omega' = \lambda_iv_i(\pmb{\hat{r}})$$
So, the unit vector $\pmb{\hat{r}}$ can point to any direction (in the partial sky $S'$ covered or in the cut sky $S-S'$. The paragraph immediately after this equation mentions that for a given eigenvector, this equation is true for any direction $\pmb{\hat{r}}$. Now, if we take that the $\pmb{\hat{r}}$ is in the partial sky $S'$, then the integral (due to the dirac delta function would just be 1 at the location when $\pmb{\hat{r}}$=$\pmb{\hat{r}}'$ and zero at all other directions which just reduces the equation to, $$v_i(\pmb{\hat{r}}) = \lambda_iv_i(\pmb{\hat{r}})$$, which implies that $v_i(\pmb{\hat{r}}) \neq 0$ but $\lambda_i = 1$. On the other hand, when $\pmb{\hat{r}}$ \neq $\pmb{\hat{r}}'$ then the dirac delta on the left hand side evaluates to zero, hence $$\lambda_iv_i(\pmb{\hat{r}}) = 0$$
In the first case, it is rather easy to see that $\lambda_i$ is 1. But,  I am unable to understand all the other results and constraints of $v_i(\pmb{\hat{r}})$. I cant use $v_i(\pmb{\hat{r}})$ for the two directions together because they are in principle different directions. So, how do I get the result of the paper that, for cut region ($S-S'$) $v_i(\pmb{\hat{r}}) = 0$ and $\lambda_i = 1$ and in the partial covered region, $S'$, $v_i(\pmb{\hat{r}}) = 0$ and $\lambda_i = 0$.

答:
Due to the Dirac delta, the evaluation of the integral separates into the two cases, depending on whether $\hat{}$ is in the partial sky or not:
$$
\lambda_i v_i(\hat{})
= \int_{S'} v_i(\hat{}') \, \delta(\hat{} - \hat{}') \, \mathrm{d}\Omega'
= \begin{cases}
    v_i(\hat{}) & \text{if $\hat{} \in S'$} \\
    0           & \text{if $\hat{} \notin S'.$}
  \end{cases}
$$
The key, as the paper suggests, is that this equation holds for every $\hat{}$ in the full sky $S$, as $v_i$ is supposed to be an eigenfunction of the operator with eigenvalue $\lambda_i$. The $\hat{} \in S'$ case of the equation, which can be written perhaps more clearly as $(\lambda_i - 1) v_i(\hat{}) = 0$, implies that either

(a) $\lambda_i = 1$

or

(b) $v_i = 0$ for all $\hat{} \in S'$,

while the $\hat{} \notin S'$ case implies either

(c) $\lambda_i = 0$

or

(d) $v_i = 0$ for all $\hat{} \notin S'$.

Since the zero function is not considered an eigenfunction, the only compatible pairs of possibilities are (a)+(d), or (b)+(c), as claimed.


问:
GRE quant question
If $a > 0$ and $b < 0$, which of the following statements are true about the values of $x$ that solve the equation $x^2-ax+ b =0$?

Indicate all such statements.

A) They have opposite signs

B) Their sum is greater than zero.

C) Their product equals $-b$.

While solving some GRE practice problems, I came across the problem above. the question seemed easy enough at first sight, but after solving it and looking at the solution. the solution states that only choices A and B are correct. why is this true? why is choice C incorrect? I googled the question but I was not convinced by any of the answers. while glancing at the question again I also realized that the first statement seems to contradict the second. a>0 but x^2 - ax + b =0. how can "a" be positive but at the same time be negative in the quad. equation? "a" can be produced by the sum of the two roots of the equation. and how is choice C incorrect? the product of a positive and negative integer will produce a negative integer?  I've been looking at this question for about 30 mins and i cannot help but feel that this is a poorly written gottca question. A good explanation is appreciated thanks.

答:
Here is a concrete example of how this might come about.

Suppose $a = 2$. This satisfies the condition $a > 0.$

Suppose $b = -3.$ This satisfies the condition $b < 0.$
It also implies that $-b = -(-3) = 3.$

Now consider the equation $x^2-ax+ b =0$.
Making the substitutions for $a$ and $b$ with the values we have supposed,
the equation is literally

$$ x^2 - 2x - 3 = 0. $$

Now $a$ is still $2,$ and $2$ is still a positive number. Of course that also means $-2$, which is the coefficient of $x$ in this equation, is a negative number.
But the only way $-2$ (or $-a$) could be negative is if $2$ (or $a$) is positive.
The statement "$a$ is positive" doesn't contradict the statement "$-a$ is negative"; in order for the second statement to be true, the first statement has to be true.

The roots of this equation are $-1, 3.$
The have opposite signs and their sum is greater than zero.
But the product of the roots is $-1 \times 3 = -3,$
whereas $-b$ is $3.$
Without doubt, $-3 \neq 3.$

Note: This is not a proof of the correct responses to the GRE. Look in the other answer of this question to find a proof of those responses.
This answer is a disproof of the mistaken notions in the question.

Title: GRE quant question


问:
Continuously extend this function to the unit sphere.
Given a nonzero vector $u$, I can define the map
$$
f_{u} : S^2 \to \mathbb{R}^3 \\
f_{u}(x) = \frac{u-(u\cdot x)x}{({(u \cdot x)^2}/(u\cdot u))-1}
$$
which is well-defined everywhere except when $u$ and $x$ are parallel. So, I have two questions:

Does $f_{u}$ have limits at those two points where it is not well-defined, and what are those limits?

Is there a simple geometric interpretation of what $f_{u}$ does to the unit sphere?

答:
Note that $f_{cu}(x)=cf_u(x)$ so you may as well let $u\in S^2$.

To see what the function is doing, take the great circle $C$ containing $\frac{u}{|u|},x\in S^2$

Set the coordinates $u=(|u|,0),, x=(\cos\theta,\sin\theta)$ so that $u\cdot x=|u|\cos\theta$.

Then

\begin{eqnarray}
f_u(x)&=\frac{(1,0)-\cos\theta(\cos\theta,\sin\theta)}{\cos^2\theta-1}\cdot |u|\\
      &=\left(-1,\cot\theta\right)\cdot|u|
\end{eqnarray}

Which is a 'vertical' line. So the function maps $S^2$ onto a plane.

Here is a geometrical illustration of how the mapping works.

Addendum: Note that $f$ is discontinuous at $(\pm1,0)$.