问:
Number Groupings for Coding
On a coding assignment, I want my program to detect 3 groups of any number going from 1 to 6.
eg.
1
3
4
5
5
3
5
There is 1 group of three 5's but how would I specify that there are three fives? Also, how may pairs of 3 are there in the numbers going from one to 6?
eg. 1,2,3 or 4,5,6.</p>

答: You should post this on Stack Overflow, and not Mathematics... But anyway, you can create a 2nd array of size 6 (since we have 6 optional values) and then loop through the original array, and for each value in it, add one in the counter array (e.g. counterArr[value-1]++). Then all you need to do is just loop through counterArr and check for values >= 3.

Title: Number Groupings for Coding


问:
Error for model building
After building a model from historical data to create  prediction for example

$\hat{Y}=B_0+B_1X$

and now I want to calculate MSE or $R^2$ value, do I use same data that was used to create the model to calculate errors or do I need another data set with known target values ($Y$ values) that was not used in building the model.

答:
You shouldn't evaluate a model based on the samples it was built on. The goal of such metrics is to emulate how a model would perform in the real world, i.e. unseen data. You can either have two separate datasets, one for training and one for testing or you can use $k$-fold cross validation to not "waste" any data.

For example, let $k=10$ and suppose you have 1000 datapoints. You can partition your dataset into 10 sets of 100 samples, these sets are called folds. You can train 10 models using every 9-fold subset of the 10-folds and use the tenth fold to evaluate each model. For example, one model would be trained with folds 2-10 and tested with fold 1. Another one would be trained with folds 1,3,4,5,..,10 and tested with fold 2 and so on.

From those 10 evaluations, you get a sense of how good your model is for your data and you can get your final model by training it with all 10 folds that is expected to perform close to the average of your 10 evaluations.

ContentLicense: CC BY-SA 4.0

Title: Error for model building


问:
Question from $p$-adic HodgeTheory, linear algebra data
I want to understand the definition $2.1.4$ of $\text{Mod}_{\mathscr{O}}^{\varphi}$ category in the pages $4$-$5$ in the $p$-adic Hodge Theory survey paper here, by Brian Conrad.

The map $\varphi_{\mathcal{M}}: \mathcal{M} \to \mathcal{M}$ is $\varphi_{\mathscr{O}}$-semilinear and injective.

In page $5$, it says that " a $\varphi_{\mathscr{O}}$-semilinear operator $\varphi_{\mathcal{M}}: \mathcal{M} \to \mathcal{M}$ is injective if its $\mathscr{O}$-linearisation $$1 \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$ is injective.

My Question:

$(1)$ What does mean by $\mathscr{O}$-linearisation of $\varphi_{\mathcal{M}}$ ?

$(2)$ The map $1 \otimes \varphi_{\mathcal{M}}$ act as an identity on $\mathscr{O}$, why is so ?

$(3)$  Why we denote $\varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M}$ ?

please help me by explaining those

答:
(1) He means that map
$$1 \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$
which could also be written as
$$id_{\mathscr{O}} \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$

and explicitly is given by

$$\sum_i a_i \otimes m_i \mapsto \sum_i a_i \cdot \varphi_{\mathcal{M}}(m_i).$$

Note that this is well-defined because 1) in the tensor product, we have (EDIT: corrected) $a \otimes b m = a \cdot \varphi_{\mathscr{O}}(b)\otimes m $, and 2) we have $\varphi_{\mathcal{M}}(am) = \varphi_{\mathscr{O}}(a)\cdot \varphi_{\mathcal{M}}(m)$ because that is what $\varphi_{\mathcal{M}}$ being $\varphi_{\mathscr{O}}$-semilinear means.

Note that this map is indeed $\mathscr{O}$-linear in the sense that e.g.

$$(1 \otimes \varphi_{\mathcal{M}}) \left( b\cdot \sum_i a_i \otimes m_i\right) = b \cdot\left((1 \otimes \varphi_{\mathcal{M}}) ( \sum_i a_i \otimes m_i) \right)$$

so it is "a linear version of $\varphi_{\mathcal{M}}$". For example if $\mathcal{M}$ is a free module of finite rank, the matrices representing the semilinear map $\varphi_{\mathcal{M}}$ and the linear map $1 \otimes \varphi_{\mathcal{M}}$ are "the same".

(2) I do not know what you mean by that map "acting as the identity on $\mathscr{O}$" because it does not act on $\mathscr{O}$ or anything that canonically identifies with $\mathscr{O}$. If by $\mathscr{O}$ here you mean the first factor in the tensor product (which, however, is not a subset of the teesor product), then, well, it acts as identity there by definition.

(3) $\varphi_{\mathscr{O}}^{*} \mathcal{M}$ naturally identifies with the extension of scalars of the $\mathscr{O}$-module $\mathcal{M}$ via the ring homomorphism $\varphi_{\mathscr{O}}: \mathscr{O} \rightarrow \mathscr{O}$, and that is often described with that notation with the asterisk. I think this notation, in turn, was chosen to be reminiscent of pullbacks, see https://ncatlab.org/nlab/show/extension+of+scalars. I believe however that understanding the reasons for certain notations is secondary in understanding the actual theory here.


问:
Why is an element that belongs to a set that belongs to another set not an element of the second set?
If $A \in B$ and $a \in A$, does $a \in B$?
I've been told it does not.
Why does it not?

答:
Actually, from just knowing that $A\in B$ and $a\in A$, you cannot decide whether $a\in B$ or not. Both situations are possible.

For example, consider the case that $A=\{a\}$ and $B=\{A\}=\{\{a\}\}$. This could be depicted as follows:

Here the dots denote the sets, and the loops enclose their elements. As you can see, the set $B$ (blue) contains the set $A$ (red), but not the element $a$ (black), as $B$'s loop does not enclose the dot $a$.

On the other hand, consider $A=\{a\}$ and $B=\{a,A\}=\{a,\{a\}\}$. Then $B$ does contain both the set $A$ and its element $a$. This can be depicted as follows:

As you can see, now $B$'s loop not only encloses the set $A$, but also the element $a$ (which is also enclosed by $A$'s loop).

But in both cases you have $A\in B$ (the dot $A$ is enclosed in $B$'s loop) and $a\in A$ (the dot $a$ is enclosed by $A$'s loop).


问:
Set of elements in a Galois group after restriction
Let $K$ be a field and $\bar{K}$ be an algebraic closure of $K$. Let $L$ and $M$ be finite extensions of $K$.

Question: Is $\{\sigma|_M \, | \, \sigma \in \operatorname{Gal}(\bar{K}/L)\} = \operatorname{Gal}(M/M\cap L)$?

This seemed intuitive for me but I am not able to show it myself. I guess one must somehow use the Fundamental Theorem of Galois Theory (as always) but I have trouble doing that.

Any help or reference is highly appreciated!

答:
Assume $M/K$ is Galois. Then there is an isomorphism $$\operatorname{Gal}(LM/L)\to\operatorname{Gal}(M/M\cap L)$$ given by restricting to $M$ (this is Theorem 2.6 of https://kconrad.math.uconn.edu/blurbs/galoistheory/galoiscorrthms.pdf). Can you conclude from here?

EDIT: Now by transitivity of restriction, $$\operatorname{Gal}(\bar{K}/L)_{\mid_M}=(\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}})_{\mid_{M}},$$ and $$\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}}=\operatorname{Gal}(LM/L).$$ We conclude that $$\{\sigma_{\mid_M}\mid \sigma\in\operatorname{Gal}(\bar{K}/L)\}\overset{1}{=}\operatorname{Gal}(\bar{K}/L)_{\mid_M}\overset{2}{=}(\operatorname{Gal}(\bar{K}/L)_{\mid_{LM}})_{\mid_{M}}\overset{3}{=}(\operatorname{Gal}(LM/L))_{\mid_M}\overset{4}{=}\operatorname{Gal}(M/M\cap L).$$


答:
I always find this notation also a bit confusing and incomplete. Let me spell out, regarding 1. and 2., the construction of linearization explicitly:

$\varphi: M \rightarrow M$ is an $\varphi_O$-semilinear map, meaning that the compatibility with the scalar multiplication is somewhat "twisted", i.e. one has $\varphi(am)=\varphi(a)\cdot \varphi(m), \;a \in O, m \in M$ (rather than the "usual" linearity "$\varphi(am)=a\varphi(m)$"). This can cause some problems, for example, the image $\varphi(M)$ is not an $O$-submodule of $M$ in general.

For that reason, it is good to associate some "actually linear" map to $\varphi,$ hence the name linearization. The way to do it is as follows:

Consider $_O O_\varphi$, that is, $O$ as the $O$-bimodule whose left multiplication is the usual one but whose right multiplication is given by viewing $O$ as an $O$-algebra via the ring homomorphism $\varphi: O \rightarrow O$. That is, one has, for $x\in O$ and a scalar $a \in O,$ $a\cdot x=ax$ but $x \cdot a=x\varphi(a)$. Then there is an actually $O$-bilinear map
\begin{align*}
_O O_\varphi\times M & \longrightarrow M \\
(a, m) &\longmapsto a \varphi(m),
\end{align*}

and so this induces an $O$-linear (with respect to scalars acting on the left, i.e. utilizing the left action of $_O O_{\varphi}$) map $_O O_{\varphi} \otimes_{O}M \rightarrow M$, given by $a \otimes m \mapsto a \varphi(m)$. This is the linearization of $\varphi$. From the explicit description, you can kind of see why the map is described by $1\otimes \varphi_M$.

Title: Question from $p$-adic HodgeTheory, linear algebra data

Body:
I want to understand the definition $2.1.4$ of $\text{Mod}_{\mathscr{O}}^{\varphi}$ category in the pages $4$-$5$ in the $p$-adic Hodge Theory survey paper here, by Brian Conrad.

The map $\varphi_{\mathcal{M}}: \mathcal{M} \to \mathcal{M}$ is $\varphi_{\mathscr{O}}$-semilinear and injective.

In page $5$, it says that " a $\varphi_{\mathscr{O}}$-semilinear operator $\varphi_{\mathcal{M}}: \mathcal{M} \to \mathcal{M}$ is injective if its $\mathscr{O}$-linearisation $$1 \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$ is injective.

My Question:

$(1)$ What does mean by $\mathscr{O}$-linearisation of $\varphi_{\mathcal{M}}$ ?

$(2)$ The map $1 \otimes \varphi_{\mathcal{M}}$ act as an identity on $\mathscr{O}$, why is so ?

$(3)$  Why we denote $\varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M}$ ?

please help me by explaining those


问:
Show with the intermediate value theorem that f(x)=314 has a solution.
$$f(x)=\frac{1}{\sqrt{1-|\; x\;| }}+\frac{1}{x^{4}}$$

I was asked to show though the intermediate value theorem that $f(x)$ has at least one solution when $f(x)=314$. I found that $f(x)$ is continuous when $-1<x<0$ and $0<x<1$. The problem I'm having is that I thought that the theorem only worked for closed intervalls.
Any tips are greatly appreciated!

答: When $x=0.000001, f$ is huge.  When $x=0.5$, $f$ is small.  Somewhere in the middle is $314$.


问:
$z = \sum_{i = 1}^n x_i \otimes y_i$
Let $U$ and $V$ be $K$-vector spaces, with $\{u_i\}_{i \in I}$ base of $U$ and $\{v_j\}_{j \in J}$ base of $V$. If $z \in U \otimes V$, prove there exists $y_i \in V$ unique such that $z = \sum_{i \in I} u_i \otimes y_i$.

The same can be proved for $x_j \in J$ unique such that $z = \sum_{j \in J} x_j \otimes v_j$.

Lastly, prove there exists $x_1, \dots, x_n \in U$ and $y_i, \dots, y_n \in V$, linearly independent such that $z = \sum_{i = 1}^n x_i \otimes y_i$.

I guess the first and second are easy. I did

$$z = (\sum_{i \in I} a_iu_i) \otimes v = \sum_{i \in I} u_i \otimes a_iv$$

Then $y_i = a_iv$.
The same for the second question. I don’t know if this is correct.

The last I don’t know how to prove.

答:
Your proof for the first (and second) question essentially establishes existence (see my comment below), but not uniqueness. You need to establish that if we have
$$
\sum_{i} u_i \otimes y_i = \sum_{i} u_i \otimes y_i',
$$
then we have $y_i = y_i'$ for all $i \in I$. Equivalently, we are given that $\sum_{i}u_i \otimes (y_i - y_i') = 0$, and we want to deduce that $y_i - y_i' = 0$ for all $i \in I$.

As in my answer on your previous, I would recommend that you show that if $y_i - y_i' \neq 0$ for some $i$, then $\sum_{i}u_i \otimes (y_i - y_i') \neq 0$. Do this by constructing a suitable linear map and appealing to the universal definition of the tensor product

I find the last question to be very tricky. Things are a bit easier if we are given that either $U$ or $V$ is finite dimensional, but even this case is a bit tricky.

The easiest method I can think of is to exploit the canonical identification of $U \otimes V$ with $\operatorname{Hom}(U^*,V)$ (where $U^*$ denotes the dual of $U$).  Because $z \in U \otimes V$, we can write $z = \sum_{i \in I} p_i \otimes q_i$ for some $p_i,q_i$ and finite indexing set $I$. Let $U_0 \subset U$ denote the (finite dimensional!) subspace of $U$ spanned by $\{p_i : i \in I\}$. Let $\Phi: \operatorname{Hom}(U_0^*,V) \to U_0 \otimes V$ denote the canonical isomorphism defined such that
$$
\Phi(vf(\cdot)) = f \otimes v.
$$
Let $\phi = \Phi^{-1}(z)$. Note that the (finite rank) map $\phi:U_0^* \to V$ can be decomposed into a "rank-factorization" $\phi = \phi_2 \circ \phi_1$ where $r$ is the rank of $\phi$, $\phi_1:U_0^* \to \Bbb F^r$, and $\phi_2:\Bbb F^r \to W$.  If $u_j \in U,v_j \in V$ are defined such that
$$
\phi_1(f) = (f(x_1),\dots,f(x_r)), \quad \phi_2(c_1,\dots,c_r) = c_1 y_1 + \cdots + c_r y_r,
$$
then it follows that
$$
\phi(f) = \sum_{j=1}^r f(x_j) y_j, \quad
\Phi(\phi) = \sum_{j=1}^r x_j \otimes y_j.
$$

Title: $z = \sum_{i = 1}^n x_i \otimes y_i$


问:
Simplify $\frac{\sin t - \cos t}{\sin t + \cos t}$
$$\frac{\sin t - \cos t}{\sin t + \cos t}$$

I am to simplify this into $$\tan(t - \frac{\pi}{4})$$

I'm not sure how to carry on, though. Multiplying numerator and denominator by denominator (or numerator) doesn't get me anywhere.

答:
You know that you already need to use an angle addition identity, because the desired result $\tan (t - \pi/4)$ contains the difference of two angles.

So, which such identities do you know?  Specifically, do you know of the tangent angle addition identity $$\tan (a - b) = \frac{\sin (a - b)}{\cos (a - b)} = \frac{\sin a \cos b - \cos a \sin b}{\cos a \cos b + \sin a \sin b}?$$  Now, what happens if we let $a = t$ and $b = \pi/4$?


问:
Calculate the residue of $\frac{e^{2z}}{(e^{iz} +1)}$?
I need to calculate the residue of $\text{res}_{z=\pi}\dfrac{e^{2z}}{(e^{iz} +1)}$
The expected answer is $\dfrac{e^{2z}}{(1-i)}$. Im getting $\dfrac{e^{2z}}{(-i)}$ as the answer.</p>

答:
The pole is simple, thus

$$\text{res}_{z=\pi }(f)=\lim_{z\to \pi }(z-\pi )\frac{e^{2z}}{e^{iz}+1}\stackrel{\text{l'Hospital, for ex.}}=\frac{e^{2\pi }}{ie^{i\pi }}=\frac{e^{2\pi}}{-i}=ie^{2\pi} $$

ContentLicense: CC BY-SA 3.0


问:
What square matrices cannot be expressed as the sum of symmetric and skew-symmetric parts
I thought I knew that any square matrix can be written as the sum of symmetric and antisymmetric matrices since we have the property that any $n\times n$ matrix $A$ can be expressed as $A=\frac{1}{2}(A-A^T)+\frac{1}{2}(A+A^T)$. However, I have come across the following statement and was told this is not always true and asked if I could find a counterexample where this isn't true but I have been struggling to find a counterexample (perhaps when $\mathbb{F}=\mathbb{Z}_2$?). Here is the statement:

Let $\mathbb{F}$ be a field. Then any $A\in \mathbb{F}(n,n)$ may be written as a sum of a symmetric and an anti-symmetric matrices in $\mathbb{F}(n,n)$.

I would greatly appreciate if someone could show me a counterexample to this.

答:
Another point, in addition to the failure of the standard formula, which is not instantly a proof of non-existence of the decomposition in characteristic $2$ (though close to it), is to observe that in char $2$ skew-symmetry is symmetry. So the question would be whether every matrix is a sum of symmetric and symmetric, hence symmetric. And we easily respond that, no, not every matrix in char 2 (of size bigger than $1\times 1$) is symmetric.

So, again, not only does the expected formula fail, ...


问:
How to solve this equation (find a power series)
Please i have this equation

$$\sum_{n\geq0} n^2 a_n x^{n+1} +\sum_{n\geq0} (n-1)a_n x^n=0$$

i change $x^{n+1}$ to $x^n$ i found

$$\sum_{n\geq1} (n-1)^2 a_{n-1} x^{n} +\sum_{n\geq0} (n-1)a_n x^n=0$$

then $a_0=0$

i think that $a_1$ is trivial ?

so $\forall n\geq 2, a_{n-1}=-\frac{a_n}{n-1}$
How to continue  please ?  in order to find $a_n$

答:
Find the coefficients of the power series given $a_0=0$:

$$\sum_{n\geq0} n^2 a_n x^{n+1} +\sum_{n\geq0} (n-1)a_n x^n=0$$

Replace each occurrence of $n$ in the first sum with $n-1$.

$$\sum_{n\geq1} (n-1)^2 a_{n-1} x^{n} +\sum_{n\geq0} (n-1)a_n x^n=0$$

Since the first non-zero term of both sums is when $n=2$ the sums can be combined into a single sum.

$$\sum_{n\ge2} [(n-1)^2 a_{n-1} + (n-1)a_n] x^n=0$$

Thus each coefficient equals $0$.

$$(n-1)^2 a_{n-1} + (n-1)a_n=0$$
for $n\ge2$.

This gives a recurrence relation $a_n=-(n-1)a_{n-1}$ for $n\ge2$.

So we get $a_2=-a_1$, $\quad a_3=-2a_2=+2a_1$, $\quad a_4=-3a_3=-3\cdot2\cdot1a_1$

In general we get $a_n=(-1)^{n-1}(n-1)!a_1$ for $n\ge2$.


答:
Step-by-step

$$ \dfrac{\sin t - \cos t}{\sin t + \cos t}$$

$$= \dfrac{\dfrac{\sin t}{\cos t} - \dfrac{\cos t}{\cos t}}{\dfrac{\sin t}{\cos t} + \dfrac{\cos t}{\cos t}}$$

$$= \dfrac{\tan t - 1}{\tan t + 1}$$

$$= \dfrac{\tan t - 1}{1 + \tan t \color{blue}{\cdot 1}}$$

Substitute $\color{blue}{\tan \frac{\pi}{4} = 1}$

$$= \dfrac{\tan t - \tan \frac{\pi}{4}}{1 + \tan t \color{blue}{\tan \frac{\pi}{4}}}$$

$$= \tan (t - \frac{\pi}{4})$$

We have used the formula

$$ \tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B} $$

Title: Simplify $\frac{\sin t - \cos t}{\sin t + \cos t}$

Body:
$$\frac{\sin t - \cos t}{\sin t + \cos t}$$

I am to simplify this into $$\tan(t - \frac{\pi}{4})$$

I'm not sure how to carry on, though. Multiplying numerator and denominator by denominator (or numerator) doesn't get me anywhere.


问:
Intuition for dimensions
I got confused on the web and on my previous answers.

Sometimes a line got 1 dimension because you just need one point to define every part of it, sometimes it's in 2 dimensions if you consider it in a plan and sometimes it's in 3 dimensions if you plot it considering x, y and z. So how many dimensions has a line for example ?

Also, I found out that an inequality changes the number of dimensions, for example a line x+y=1 is in 2 dimensions if I plot it considering x and y but as soon as I write x+y>1 it's not a line anymore but a plan, so the number of dimensions increased by 1 so it is now a plan with 3 dimensions ? I think that is not correct but I do not know how..

Can you give me some intuitions or method to understand when I got how many dimensions considering an equation.

NB:

答:
The dimension of an object should be thought of as the answer to the following question:

If I were on the object, how many independent "axes of motion" could I move in?

For example, if you were an ant crawling along a line, there is only one axis of motion you can move in:  forward/backward.  This is what we mean when we say that a line is 1 dimensional.  It doesn't matter whether the line is considered by itself, or is drawn in a plane, or is part of a 3D environment:  on the line there is only one axis of motion, so the line is 1 dimensional.

If you were an ant crawling on a plane, or on the surface of a sphere, there are two axes of motion:  (1) forward/backward and (2) left/right.  These directions are perpendicular to each other, and of course you can move in a mixture of them as well.  But the fact that there are two independent axes of motion makes a plane (or the surface of a sphere) a 2-dimensonal object.

If, on the other hand, you were a bee buzzing around in mid-air, or in the interior of a spherical balloon, you would have three independent axes of motion:  (1) forward/backward, (2) left/right, and (3) up/down.  So those are 3-dimensional environments.

The number of dimensions an object has is an intrinsic propery of the object, and does not depend on where the object is.  We can put a 2-dimensional object in a 3-dimensional space (imagine a piece of paper blowing in the breeze).  We can also bound a 3-dimensional space by a 2-dimensional surface (think of the water contained inside a water balloon).  Similarly, a line (or indeed any curve, like the circumference of a circle) is 1-dimensional, regardless of whether it is contained by a 2-dimensional space or a 3-dimensional space, and regardless of whether it bounds a 2-dimensional region.

Edited to add:  The OP has added some clarification to the original question, and I now think I better understand the issue, so let me take a shot at addressing it.

What the OP wants to know is, "How do we think about the geometry of an equation?"  For example, how is it that the equation $x = 1$ defines a point on $\mathbb R$, a line in the plane $\mathbb R^2$, and a plane in $\mathbb R^3$?

The answer, I think, is that when we write an equation like
$$x = 1$$
we are abbreviating a slightly more complicated question.  The actual question is

Find the set of all solutions, in some "universe" $S$, to the equation $x = 1$.

We usually don't specify what the universe is, because context generally makes it clear.  However, when we move from one universe to another, suddenly the question "What kind of solutions are we looking for?" becomes very important.

Before I return to the question of dimension, let me illustrate the general point.  Consider the following equations:

$3x - 5 = 0$.  If we are searching for solutions in the set of natural numbers, $\mathbb N$, then this equation has no solutions.  On the other hand if we are searching in the set of rational numbers, $\mathbb Q$, then this equation has one solution.

$x^2 - 5 = 0$.  If we are working over $\mathbb Q$, this equation has no solutions.  On the other hand if we are working over $\mathbb R$, the set of real numbers, then it has two solutions.

$x^2 + 5 = 0$.  Over $\mathbb R$, this has no solutions; over $\mathbb C$, the complex numbers, it has two solutions.

So you can see that the solution set for a particular equation depends not only on the equation itself, but also on the "types" of solutions we are willing to admit.

Now let's return to the question in the OP.  Consider the following three similar-looking questions:

Find all $x \in \mathbb R$ such that $x = 5$.

Find all $(x, y) \in \mathbb R^2$ such that $x = 5$.

Find all $(x, y, z) \in \mathbb R^3$ such that $x = 5$.

The solution to the first problem is a point; the solution to the second problem is a line; and the solution to the third problem is a plane.  The equation is the same in all three cases, but the context is different.

To connect this to my earlier answer, let's consider the third case:  Suppose you are sitting at a point $(x, y, z) \in \mathbb R^3$ in the solution set of $x = 5$.  How much freedom of movement do you have, if you want to stay in that solution set?  You can't change your $x$-coordinate, as that has to stay equal to $5$.  But the equation says nothing about the values of $y$ or $z$; they are unconstrained by the equation.  So you can freely move in the $y$ direction (left/right motion) or in the $z$ direction (up/down motion).  Two axes of freedom, hence you are sitting in a 2-dimensional object.

On the other hand, if you are sitting at a point $(x, y) \in \mathbb R^2$ in the solution set of $x = 5$, the only freedom of motion you have is in the $y$ direction.  One axis of freedom, therefore a 1-dimensional object.

Very loosely speaking, an equation in $n$ variables imposes a constraint on the variables, which means you can choose $n-1$ of the variables to have any values you want, but then the final variable's value is determined by the constraint. The result is then (typically) a surface of $n-1$ dimensions.  If you have a system of two equations in $n$ variables, each equation imposes a constraint, so you have (again, typically) only $n-2$ free variables left, resulting in a surface of dimension $n-2$.

Title: Intuition for dimensions


问:
Consider 8 points on a circle of radius 1. Show that at least two points have a distance less than $\frac{9}{10}$ from each other.
Consider 8 points on a circle of radius 1. Show that at least two points have a
distance less than $\frac{9}{10}$ from each other.

What I know so far:

I think I start off with 8 points on the circumference which creates an octagon with side lengths $\frac{1}{2}$. Hence all points have at most a maximum distance of $\frac{1}{2}$.

However I have no idea what to do next. Any help would be great!

答:
We use the following claim:

Claim 1: Let $d_0,\ldots, d_m$ be $m$ real numbers. Then if $\sum_{i=0}^m d_i=K$ then there exists an $i$ such that $d_i \le \frac{K}{m+1}$

So now we use Claim 1. Write the points $x_0,\ldots, x_7$, in the order around the circle. Let us write as $d'(x_i,x_j)$ the length of the shortest walk between $x_i$ and $x_j$ on the circle.

The circumference of the circle is $2\pi$, so $\sum_{i=0}^7 d'(x_i,x_{i+1}) = 2\pi$ the circumference of the circle. By Claim 1,  there exists an $i$ such that $d'(x_i,x'_{i+1}) \le \frac{2\pi}{8} <.8 < \frac{9}{10}$.

As the euclidian distance between $x_i$ and $x_{i+1}$ is shorter than the length of the shortest segement on the circle between $x_i$ and $x_{i+1}$, if the inequality $d'(x_i,x'_{i+1}) \le \frac{2\pi}{8} < \frac{9}{10}$ is satisfied then the euclidena distance between $x_i$ and $x_{i+1}$, which is less than $d'(x_i,x_{i+1})$, must also be less than 9/10.


问:
a necessary and sufficient condition that homomorphic image of $R$ is a field
Let $R$ be a commutative ring with unity.Find a necessary and sufficient condition that homomorphic image of $R$ is a field

I am little bit confused about the question. By a necessary and sufficient condition what did they mean? condition on $R$ or the condition on the homomorphism.

If we can find a polynomial $f(x)$ which is irreducible on $R$ then $R/(f(x))$ is a field

答:
By the "first isomorphism theorem" for rings, a homomorphic image of $R$ via $\phi:R\to S$ is isomorphic to $R/\ker(\phi)$.

Since the ideals of $R/\ker(\phi)$ correspond to ideals of $R$ containing $\ker(\phi)$, the quotient will be a field (=have no nontrivial ideals) if and only if there are no ideals strictly between $\ker(\phi)$ and $R$. By definition, that happens exactly when $\ker(\phi)$ is a maximal ideal of $R$.

Since you included unity in your ring, there exists a maximal ideal $M$, and $R/M$ is a homomorphic image of $R$ that is a field. So, the answer to your question as literally written is "this is true for all (nonzero) rings." However, I'm betting you probably did not express your question as you really meant, and that the above paragraph is along the lines of the answer you need.

A comment on your idea for a solution: it looks like you are thinking of the polynomial ring $F[x]$ containing an $F$-irreducible polynomial $p(x)$ and then forming the quotient $F[x]/(p(x))$ to produce a field, which is a homomorphic image of $F[x]$.

What you have written, though, makes it look like you've confused the quotient $R[x]/(f(x))$ with "$R/(f(x))$". Even if you succeeded in finding a polynomial that generates a maximal ideal of $R[x]$, this only yields a maximal ideal of $R[x]$, and not $R$ itself.


问:
Is there an adjoint functor to the hom bifunctor in the category of $R$-modules, $R$ is commutative unital ring.
I've seen that for $W\in R-\mathrm{\!Mod}$ ,both $\mathrm{Hom}(W,-)$ and $\mathrm{Hom}(-,W)$  have adjoint functors.

adjoint of Hom(-,W)

this leaves me with one question:

$C:= R-\mathrm{\!Mod}$

does $\mathrm{Hom}(-,-):C^{op}\times C \to C$ have an adjoint functor?

答: A functor between additive categories which has either a left or a right adjoint must be additive, but $\text{Hom}$ is not; it is bilinear rather than linear. The tensor bifunctor also fails to have an adjoint for the same reason.


答:
As you mentioned you get an octagon (not necessarily regular) with points on the circle. Each chord (side of the octagon) is shorter than the corresponding arclength it subtends. All the arclengths add to $2\pi r = 2\pi$, which is the circumference of the circle. The chords add up to the perimeter of the octagon. Therefore the perimeter of the octagon is less than the circumference of the circle.

$\frac{2\pi}{8} = 0.785$

Assume all the sides of the octagon have length $\ge \frac{2\pi}{8}$. Then the perimeter of the octagon will be $\ge 8(\frac{2\pi}{8}) = 2\pi$. But this is impossible because we know the perimeter of the octagon is less than the circumference of the circle.

So at least one of the sides of the octagon has length $< \frac{2\pi}{8} = 0.785 < \frac{9}{10}$

Body:
Consider 8 points on a circle of radius 1. Show that at least two points have a
distance less than $\frac{9}{10}$ from each other.

What I know so far:

I think I start off with 8 points on the circumference which creates an octagon with side lengths $\frac{1}{2}$. Hence all points have at most a maximum distance of $\frac{1}{2}$.

However I have no idea what to do next. Any help would be great!


答: Below is a geometric proof by contradiction. The grey circle in the center has radius 1, while each of the red circles has radius 0.9 and its center marked with a red X. We begin by placing one point on the center circle's circumference, in this case, we start with the point on the right labeled 1. The next point, labelled 2, must be at a distance of at least 0.9 from point 1. To save space, we put it as close as possible, exactly 0.9 units from point 1. We do the same proceeding around the circle, putting each consecutive point exactly 0.9 units away from the previous one. By the time we get to point number 7, we find that there is no more room to place a point on the grey circle that is at least 0.9 units from both points 6 and 1. Therefore, it is not possible to put 8 points (or even 7 points) on a circle of radius 1, such that no two points are closer than 0.9 units apart.

Title: Consider 8 points on a circle of radius 1. Show that at least two points have a distance less than $\frac{9}{10}$ from each other.


问:
Can someone carry me through the process of sketching $f(x, y) = (x^2 + 1)(y^2 -1)$
The level lines are for $f(x,y) = (-2,-1,0,1,2)$. I've always struggled with this and I usually just do it with software, but I'd like to know your reasoning behind drawing this by hand

result

答:
$(x^2+1)(y^2-1)=2$ is a curve in the plane $z=2$ whose equation can be written as
$$y=\pm\sqrt{\frac{x^2+3}{x^2+1}}$$
Let's consider the positive one, the other one is the symmetric wrt the $x$-axis
$$y=\sqrt{\frac{x^2+3}{x^2+1}}$$
It is an even function, symmetric wrt $y$-axis

$$\underset{x\to \infty }{\text{lim}}\sqrt{\frac{x^2+3}{x^2+1}}=1$$
Which means there is an horizontal asymptote $y=1$

$$y'=-\frac{2 x}{\left(x^2+1\right)^{3/2} \sqrt{x^2+3}}$$

$y'=0$ for $x=0$ and $y=\sqrt{3}$. It's a maximum because $y'<0$ for $x>0$ and $y'>0$ for $x<0$.

These informations are enough to plot the graph below

setting $1,0,-1,-2$ you get the others. The level curves are the sections of the surface $f(x,y)=(x^2+1)(y^2-1)$ with a plane $z=k$


问:
Is force of mortality monotonic function?
I'm trying to do solve some question from the actuarial exam. The task is to map a function with a graph. In the graph the function is increasing on $[0,80]$ and decreasing on $[80,100]$. The options I can choose from are : $l_x,\ l_x^2,\ u_x,\ l_x u_x$. The first two can't be correct because they are decreasing everywhere. What can we tell about the $u_x$?

答:
The curve in question is

The choices are:

(A) $\mu(x)$

(B) $l_x \mu(x)$

(C) $l_x p_x$

(D) $l_x$

(E) $l_x^2$.

As you noted, we can immediately eliminate (D) and (E) because $l_x$ is monotonically decreasing.  We can eliminate (C) for a similar reason, since $p_x$, the survival function of a life aged $x$, is bounded between $0$ and $1$, and it tends to be the case that $p_x < p_y$ for $x > y$ in adulthood, because the probability of surviving one additional year is smaller as the age is greater.

We should also eliminate (A) because a typical force of mortality at the ultimate age should be higher than at younger ages.  This leaves (B) as the only candidate for this plot.

What does $l_x \mu(x)$ represent?  It is the product of the number of lives still living at age $x$, times the force of mortality experienced by those lives.  Consequently this product is proportional to the probability that a newborn life dies at age $x$ (with the constant of proportionality equal to the size of the cohort at birth, $l_0$).  So this explains why the plot is increasing up to age $80$ and then curves back down at around $x = 80$, because the number of lives that have died out by then makes $l_x$ small, thus making the product small even when the force of mortality is high.  This curve is also known as the curve of deaths.  It captures the likelihood of infant mortality, the low likelihood of young and middle-age adult mortality, the gradually increasing likelihood of death at older ages, and importantly, the relatively low likelihood of reaching extreme old age because most people don't make it that far.


问:
Find all complex solutions to $\sin z = i \sinh z$
I have tried the usual method of expanding into exponentials and letting $t:= e^{z}$ and then solving a polynomial equation. However, in the equation there are some $t^i$ that I don't know how to deal with.

I know that the problem has a solution as Mathematica lists this solutions ($k \in \mathbb{Z}$):

$(2+2i)\pi k$

$(1-i)(i\pi + 2 i \pi k)$

$(1-i)(-\frac{\pi}{2} + 2 \pi k)$

$(1-i)(\frac{\pi}{2} + 2\pi k)$

I have also proved that the question asked is equivalent to finding all complex solutions of $\sinh{iz}=\sinh{-z}$.

Body:
As $e^z=\cosh z+\sinh z$ and $e^{iz}=cos z+i\sin z$

$$2i(e^{iz}-e^{-iz})=-2(e^z-e^{-z})$$

Now if $e^{iz}=a, e^{-z}=b$

we have $$a-\dfrac1a=b-\dfrac1b \iff a^2b-b=ab^2-a\iff0=ab(a-b)+1(a-b)=(a-b)(ab+1)$$

$$a=b\implies e^{-z}=e^{iz}\iff e^{(1+i)z}=1 \implies(1+i)z=2m\pi i$$

$$ab+1=0\implies ab=-1\iff e^{(1-i)z}=-1\implies(1-i)z=(2n+1)\pi i$$


答: Note that $$\frac{\sin t - \cos t}{\sin t + \cos t} = \frac{\sqrt{2}\sin(t - \frac{\pi}{4})}{\sqrt{2}\sin(t + \frac{\pi}{4})} =  \frac{\sqrt{2}\sin(t - \frac{\pi}{4})}{\sqrt{2}\cos(t - \frac{\pi}{4})} = \tan(t- \frac{\pi}{4})$$


问:
Devise System of Equations to Solve Age Problem
Hi this may be simple silly problem but it is bugging me as I am not able to devise a system of equations to solve it.

My husband's age," remarked a lady the other day, "is represented by the
  figures of my own age reversed. He is my senior, and the difference between
  our ages is one-eleventh of their sum.

The answer is 54 and 45, but not able to find way to get it.

答:
Not sure if this totally solves what you want, but here's a way to get a stage where the numbers should be easier to guess and check. I assume the ages of the two people are two digit numbers. So we can represent the husband's age as $10x+y$, where $x$ and $y$ are the decimal digits, and the wife's age is then $10y+x$, with $10x+y\gt 10y+x$.

Since the difference of their ages is one eleventh of the sum, this translates to
$$
(10x+y)-(10y+x)=\frac{1}{11}(10x+y+10y+x)
$$
but this implies
$$
9x-9y=\frac{1}{11}(11x+11y)=x+y.
$$

So you have $9(x-y)=x+y$. Since $0\leq x,y\leq 9$, it's not hard to experiment with these numbers to find that $x=5$ and $y=4$.


答:
Since the denominator $g(z)=e^{iz}-1$ has a simple zero at $z_0=\pi$, and numerator $f(z)=e^{2z}$ never vanishes, you can apply the following formula
$$\operatorname{Res}\Big(\frac{f}{g},z_0\Big)=\frac{f(z_0)}{g'(z_0)}$$
giving in your case the following result
$$\operatorname{Res}\Big(\frac{e^{2z}}{e^{iz}+1},\pi\Big)=\frac{e^{2\pi}}{ie^{i\pi}}=\frac{e^{2\pi}}{-i}=e^{2\pi}i$$</p>

Title: Calculate the residue of $\frac{e^{2z}}{(e^{iz} +1)}$?

Body:
I need to calculate the residue of $\text{res}_{z=\pi}\dfrac{e^{2z}}{(e^{iz} +1)}$
The expected answer is $\dfrac{e^{2z}}{(1-i)}$. Im getting $\dfrac{e^{2z}}{(-i)}$ as the answer.</p>


答: note that $i\sin iz=\frac{e^{-z}-e^z}{2}=-\sinh z$ so $\sin iz=i \sinh z$ hence the equation is $\sin z=\sin iz$ or $2\sin ((z-iz)/2) \cos ((z+iz)/2)=0$ so either $z−iz=2k\pi$ or $z+iz=π+2k\pi$ and simplifying the cases reduces to $z=(1+i)k \pi$ or $z=(1-i)(\pi/2 +k\pi)$

Title: Find all complex solutions to $\sin z = i \sinh z$

Body:
I have tried the usual method of expanding into exponentials and letting $t:= e^{z}$ and then solving a polynomial equation. However, in the equation there are some $t^i$ that I don't know how to deal with.

I know that the problem has a solution as Mathematica lists this solutions ($k \in \mathbb{Z}$):

$(2+2i)\pi k$

$(1-i)(i\pi + 2 i \pi k)$

$(1-i)(-\frac{\pi}{2} + 2 \pi k)$

$(1-i)(\frac{\pi}{2} + 2\pi k)$

I have also proved that the question asked is equivalent to finding all complex solutions of $\sinh{iz}=\sinh{-z}$.


问:
Is this random bit generator broken?
Consider a following problem:

Suppose, a random bit generator of type $p$ was brought to a random bit generator repair station. Before starting repairing it, the workers from the repair station were completely unsure, whether it is really broken, decided to check that it is really broken by launching it several times. They know, that random bit generators of type $p$ output $1$ with probability $p$ and $0$ otherwise, independently every time they are launched. However, when such random bit generators are broken, they always output $1$. After launching the generator $t$ times and receiving only $1$ the workers concluded that it is most likely broken (with probability of it functioning correctly being less than $\epsilon$). What is the least possible value of $t$?

Suppose the prior probability of the generator being broken is $q$. Then it was launched and returned $1$. The conditional probability of that outcome was $1$ if the generator was indeed broken, and the total probability of this output was $q + (1-q)p = p + (1-p)q$. Then the posterior probability of it being broken is $\frac{q}{p + (1-p)q} = \frac{1}{1-p}(1 - \frac{p}{p + (1-p)q}$).

Now, as initially the workers were completely unsure, with prior probability being uniform, the posterior probability of it being broken $q_n$ after $n$ trials will satisfy the recurrent relation:

$$q_0 = \frac{1}{2}$$
$$q_{n+1} = \frac{1}{1-p}(1 - \frac{p}{p + (1-p)q_n})$$

Now all that remained is to find the minimal $n$ for which $q_n > 1 - \epsilon$.

The alternative task can be further "prettified" by rewriting it in terms of $y_n = (1-p)p_n$:

Suppose the sequence $y_n$ is defined by a recurrence:
$$y_0 = \frac{1-p}{2}$$
$$y_{n+1} = 1 - \frac{p}{p + y_n}$$
Find $\min\{t|y_t \geq (1-p)(1-\epsilon)\}$.

All that remains now is to find the closed form of $y_n$ and derive $t$ from it. But, unfortunately, I do not know how.

答:
Sketch for the solution: you can linearize the recursion noticing that
$$
q_{n+1}=\frac{q_n}{p+(1-p)q_n}=\frac1{\frac{p}{q_n}+1-p}\\
\therefore\quad \frac1{q_{n+1}}=\frac{p}{q_n}+1-p\implies r_{n+1}=r_np+(1-p)p
$$
for $r_n:=p/q_n$.

P.S.: sorry I'm too lazy to write a full answer :)

Title: Is this random bit generator broken?


问:
The logarithm of the p-quantile
Maybe somebody can explain how to get this result:

it's the logarithm of the p-quantile for the Weibull distribution. I actually know the formula of the Weibull distribution and that the Survival function has this form:

but still don't know how to get the result above.

Title: The logarithm of the p-quantile

Body:
$$p=1-e^{-(\frac{t}{\eta})^v}$$

... solve it in $log (t)$

$$1-p=e^{-\Big(\frac{t}{\eta}\Big)^v}$$

$$-log(1-p)=\Big(\frac{t}{\eta}\Big)^v$$

$$log(-log(1-p))=v[log(t)-log(\eta)]$$

$$log(t)=\frac{1}{v}[log(-log(1-p))]+log(\eta)$$

... as claimed


问:
Combinatorics Urn Problem
I have 208 unique marbles:

From this 208, I select 15 marbles without replacement and place them into an urn.

I do this again with a new urn, 3 more times, but using the original
208 marble set, so I apply replacement after each selection of 15.

I now have 4 urns of 15 marbles each, with no duplicate marbles within
a particular urn, but possibly duplicates across urns.

What is the probability that there are exactly 14 marbles that appear in more
than 1 urn $?$.

答:
Following the advice from Quester I simulated the problem with Monte Carlo

```<code>import numpy as np
from tqdm import tqdm
import matplotlib.pyplot as plt
count = np.zeros(208)
rng = np.arange(208).astype(int)
urns=np.zeros([4,15]).astype(int)
exactly14 = 0
morethan14 = 0
lessthan14 = 0
history = []

for i in tqdm(range(1000000)):
    for j in range(4):
        urns[j, :] = np.random.choice(rng, 15, replace=False)

    n_unique=len(np.unique(np.ravel(urns)))
    n_multi=len(np.ravel(urns))-n_unique
    history.append(n_multi)
    if n_multi == 14:
        exactly14 += 1
    if n_multi > 14:
        morethan14 += 1
    else:
        lessthan14+=1

print(exactly14/1000000)
</code>```

0.000489

Title: Combinatorics Urn Problem

Body:
Following the advice from Quester I simulated the problem with Monte Carlo

import numpy as np
from tqdm import tqdm
import matplotlib.pyplot as plt
count = np.zeros(208)
rng = np.arange(208).astype(int)
urns=np.zeros([4,15]).astype(int)
exactly14 = 0
morethan14 = 0
lessthan14 = 0
history = []

for i in tqdm(range(1000000)):
    for j in range(4):
        urns[j, :] = np.random.choice(rng, 15, replace=False)

    n_unique=len(np.unique(np.ravel(urns)))
    n_multi=len(np.ravel(urns))-n_unique
    history.append(n_multi)
    if n_multi == 14:
        exactly14 += 1
    if n_multi > 14:
        morethan14 += 1
    else:
        lessthan14+=1

print(exactly14/1000000)

0.000489


问:
Exponential equation problem, where bases are not the same.
$(3/4)^{2x-1}= (4/3)^{x-2}$

How do i get the bases to be the same? My teacher just added a $-1$ and everything magically turned to
$(3/4)^{2-x}$, How did this happen?

答:
$(3/4)^{2x-1}= (4/3)^{x-2}$ and then as Tito mentioned we have $(3/4)^{2x-1}= (3/4)^{-x+2} $
so they have the same base and we should solve the equation for powers:
$2x-1=-x+2$ and then $3x = +3$ so $x=1$.


问:
prove the trigonometry identity:
prove the trigonometry identity:

$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$

of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.

答:
I'll begin by saying that when you want to prove an identity like $ L=R $  you cannot multiply both sides by any value, since then it wouldn't be an identity.

Anyway, Lets say that $L$ is LHS, and $R$ is RHS. Let $t=\tan(A)$. We can then say that $\frac{1}{t}=\cot(A)$.

Therefore, RHS becomes
$$ R = \frac{t^3 + \frac{1-t}{\frac{1}{t}}}{\frac{1-\frac{1}{t}}{t} + \frac{1}{t^3}} $$

Simplifying the top and bottom (separately) shows us that

$$ R = \frac{t^3-t^2+t}{\frac{t-1}{t^2} + \frac{1}{t^3}}$$

Multiply the top and bottom by $ t^3 $, then

$$ R = \frac{t^3(t^3-t^2+t)}{t(t-1)+1} = \frac{t^4(t^2-t+1)}{t^2-t+1} = t^4 $$

Then, substituting back $ t=\tan(A) $, we get

$$ R = \tan^4(A) = L $$

Like we wanted.

Title: prove the trigonometry identity:


问:
Why is an element that belongs to a set that belongs to another set not an element of the second set?
If $A \in B$ and $a \in A$, does $a \in B$?
I've been told it does not.
Why does it not?

答:
I found an intuitive solution that I'm happy with.
If you have a set of Animals={Dog, Cat, Elephant, Bird}
and you looked at Dog as a set of parts such as Dog={Ears, Legs, Tail, Eyes, Mouth}
it would not make sense to say that Ears ∈ Animals because Ears are not an animal.
Only when Ears are combined with Legs, Tail, Eyes, and Mouth do you get Dog.
The full set of elements that make up Dog are Necessary to make an animal.
Ears all on its own is not an Animal therefore it cannot be an element of the Set Animals.
Thanks for everyone's help!</p>


问:
Global Extrema on a Multivariable function
Say there is a function $f(x, y) = x^3 + e^{3 y} - 3 x e^y$ . How can I deduce which components are the dominating factors of $f(x,y)$ and how to deduce whether $f(x,y)$ may have maximum or minimum values by just observing the dominating factors of $f(x,y)$?

答:
For positive $y$ values the dominant term is $e^{3y}$ and there is no abs  max. For negative $y$ values the dominant term is $x^3$ and there no abs min.

There is a local minimum, though, at $(1,0)$

First derivatives are zero when
$$
\begin{cases}
3 x^2-3 e^y=0\\
3 e^{3 y}-3 x e^y=0\\
\end{cases}
$$

$$H(x,y)=\left(
\begin{array}{rr}
 6 x & -3 e^y \\
 -3 e^y & 9 e^{3 y}-3 x e^y \\
\end{array}
\right)
$$
$$\det(H(x,y))=-18 x^2 e^y+54 x e^{3 y}-9 e^{2 y};\;\det(H(1,0))=27>0$$
The determinant of the Hessian matrix is negative at $(1,0)$, therefore it is a local minimum.


答: The quantile $x(p)$ satisfies $$\Pr[T > x(p)] = 1-p \tag{1}$$ where $T$ is Weibull with survival function $$S_T(t) = e^{-(t/\eta)^\nu}. \tag{2}$$  But the survival function literally is $$S_T(t) = \Pr[T > t].$$  So combining $(1)$ and $(2)$ (where I have abbreviated $x(p)$ to $x$) yields $$e^{-(x/\eta)^\nu} = 1 - p,$$ and we need to solve for $x$:  $$\left(\frac{x}{\eta}\right)^\nu = -\log (1-p),$$ hence $$\nu \log \frac{x}{\eta} = \log (- \log(1-p) ),$$ and $$\log x - \log \eta = \frac{1}{\nu} \log ( - \log(1-p) ),$$ therefore the log-quantile $$\gamma(p; \boldsymbol \theta) = \log x(p) = \log \eta + \frac{1}{\nu} \log ( - \log(1-p) )$$ as claimed.

Title: The logarithm of the p-quantile

Body:
Maybe somebody can explain how to get this result:

it's the logarithm of the p-quantile for the Weibull distribution. I actually know the formula of the Weibull distribution and that the Survival function has this form:

but still don't know how to get the result above.


问:
Show if the following sets are open
I'm doing some problems and there are some to which I simply don't have an answer.

Are the following sets open or closed?

$(0,1)\text{ embedded in }\mathbb R^2\colon\{(x,y)\mid x\in(0,1),y=0\}\subset\mathbb R^2$

$\{(x,y,z)\mid 0\le x+y\le 1, z=0\}\subset\mathbb R^3$

$\{\frac{1}{n}\mid n\in\mathbb N\}$ as a subset of $\mathbb R_{++}$

The thing that confuses me is the fact that the set being embedded into a higher dimension while setting one of the variables constant.

For the set 3, would the answer differ if instead we had $\{\frac{1}{n}\mid n\in\mathbb N\}$ as a subset of $\mathbb R$? Do I understand correctly that the latter is neither closed nor open?

Thanks.

答:
is not closed, because for all $n$, the point $(1/n, 0)$ is in the set, but the sequence $(1/n, 0)$ tends to $(0,0)$ which is not in the set. It is neither open, because $(1/2,0)$ is in the set, but $(1/2, \varepsilon)$ is not in the set, for any $\varepsilon \neq 0$.

is closed, as the preimage of the closed subset $[0,1] \times \lbrace 0\rbrace$ of $\mathbb{R}^2$ by the application $f(x,y,z)=(x+y,z)$. You deduce by connectedness of $\mathbb{R}^3$ that it is not open.

is not open because $1$ is in the set but $1+\varepsilon$ is not in the set, for any $\varepsilon \neq 0$. It is closed because you can write its complement in $\mathbb{R}_{++}$ as
$$\bigcup_{k \in \mathbb{N}} \left(\frac{1}{k+1}, +\infty \right) \setminus \left\lbrace 1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{k}\right\rbrace$$
which is an union of open sets (each term of this union being open as an open set minus a finite number of elements).

Title: Show if the following sets are open


问:
Proving a product of functions is lebesgue integrable
I know that for measurable functions $f,g:X\rightarrow \mathbb{R}$ that if $f^2$ and $g^2$ are lebesgue integrable then $fg$ is lebesgue integrable (for a finite measure space $X$)

I understand the proof of that and so I’m trying to mimic the idea to prove $fg^2$ is lebesgue integrable provided $f^3$ and $g^5$ are lebesgue integrable.

From my understanding it will be sufficient to show $f^2$ and $g^4$ are lebesgue integrable but I’m not sure how that is done. Thus I’ve tried replicating the proof by using both $(f+g)^5$ and $(f-g)^5$ to show $|fg^2|$ is bounded but there’s more unwanted terms involved than in the proof so I’m not struggling to proceed.

答:
$f^3$ and $g^3$ need to be integrable. Since integrability of $fg^2$ is equivalent to integrability of $|fg^2|$, we can assume that $f$ and $g$ are nonnegative.
Then
$$
3fg^2 \le (f+g)^3
$$
and everything is fine.

You can use Hoelder inequality next time: in order that a product of $k$ functions is integrable, it suffices that the $k$-th power of each function is integrable.


答:
The set

$$ A:=\{x\in]0,1[, \;y=0\} \subseteq \Bbb R^2 $$

is neither open nor closed. In fact, if it was open, taking $z=(1/2,0)\in A$, there should be an open disk $B^0(z;r)$ with $r>0$ such that $B^0(z;r)\subseteq A$. This is absurd since for example $(1/2,r/2)\in B^0-A$.

If it was closed, it should contain the limit of the sequence $\big((1/n,0)\big)_{n=2,3,4,\ldots}$ of $A$, i.e. $(0,0)$, absurd.

The set

$$B:=\{0\leq x+y\leq1, \;z=0\} \subseteq \Bbb R^3 $$

is not open by a similar argument we used for $A$. It is closed: if $(x_n, y_n, 0)$ is a sequence of $B$ convergent to $(x, y, 0)$, then $x_n\to x, y_n\to y$, and hence

$$ 0\leq x_n+y_n\leq 1 \;\;\implies\;\; 0\leq x+y\leq1 $$

i.e. $(x,y,0)\in B$.

The set

$$ C:=\{1/n \;:\; n\in\Bbb N\} \subseteq\Bbb R_+ =\{x\geq 0\}$$

is neither open (see above argument) nor closed ($\lim 1/n=0\notin C$), and that  regardless of whether C is a subset of $\Bbb R_+$ or of $\;\Bbb R$.

If $\Bbb R_{++}$ denote the set $\{x>0\}$, $C$ is closed in $\Bbb R_{++}$ as proved in the answer of @TheSilverDoe.

Title: Show if the following sets are open

Body:
I'm doing some problems and there are some to which I simply don't have an answer.

Are the following sets open or closed?

$(0,1)\text{ embedded in }\mathbb R^2\colon\{(x,y)\mid x\in(0,1),y=0\}\subset\mathbb R^2$

$\{(x,y,z)\mid 0\le x+y\le 1, z=0\}\subset\mathbb R^3$

$\{\frac{1}{n}\mid n\in\mathbb N\}$ as a subset of $\mathbb R_{++}$

The thing that confuses me is the fact that the set being embedded into a higher dimension while setting one of the variables constant.

For the set 3, would the answer differ if instead we had $\{\frac{1}{n}\mid n\in\mathbb N\}$ as a subset of $\mathbb R$? Do I understand correctly that the latter is neither closed nor open?

Thanks.


问:
Solving a system of 3 equations with 2 variables
How would a system of 3 equations with 2 variables be solved to ensure that each variable is consistent throughout the system?

$$
a
\begin{bmatrix}
v1\\
v2\\
v3\\
\end{bmatrix}
+
b
\begin{bmatrix}
w1\\
w2\\
w3\\
\end{bmatrix}
=
\begin{bmatrix}
c1\\
c2\\
c3\\
\end{bmatrix}
$$

答:
The rank of the coefficient matrix must be equal to the rank of the augmented matrix. In your example the matrix of the coefficients is $3\times 2$ so the rank is $2$, while
augmented matrix is $3\times 3$. To have the same rank the augmented mtrix determinant must be zero.


答:
Here is an approach using the residue theorem.

The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are:
$$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy to obtain using power series; we keep in mind $\omega^3=1\implies 1/\omega=-1-\omega$).

Now consider $\int_{\Gamma}f(z)\,dz$ where $\Gamma$ is the boundary of the parallelogram

$\hspace{3cm}$
$$\big\{z : \max\{|\Re z|,|\Re(\omega z)|\}\leqslant N+1/2\big\}$$
and take $N\to\infty$. On $AB$ or $CD$ we have $z=\pm\frac1\omega\left(N+\frac12+it\right)$, with "$+$" on $CD$ and "$-$" on $AB$, and $t\in\mathbb{R}$; using $\lim\limits_{y\to-\infty}\cot(x+iy)=i$ (uniformly w.r.t. $x\in\mathbb{R}$), we see that both $\int_{AB}$ and $\int_{CD}$ tend to $$-\frac\pi\omega\int_{-\infty}^\infty\frac{dt}{\cosh^2\pi t}=-\frac2\omega$$ as $N\to\infty$. The integrals along $BC$ and $DA$ both tend to $0$, hence the residue theorem gives $$\frac{1-\omega}{3}+2(1-\omega)\sum_{n=1}^\infty\csc^2\omega\pi n=-\frac{2}{\omega\pi i}\implies\color{blue}{\sum_{n=1}^\infty\csc^2\omega\pi n=\frac{1}{\pi\sqrt3}-\frac16}.$$

Body:
$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $$ if $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ find $\frac{A^2}{B^2}$

My Attempt

$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \sum_{n=1}^\infty csch^2(i\omega\pi n)= 4\sum_{n=1}^\infty  \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2} $$ $$\sum_{n=1}^\infty  \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2}= \big(ie^{\pi\frac{\sqrt{3}}{2}}+ie^{-\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(-e^{\pi\sqrt{3}}+e^{-\pi\sqrt{3}}\big)^{-2} +\big(-ie^{3\pi\frac{\sqrt{3}}{2}}-ie^{-3\pi\frac{\sqrt{3}}{2}}\big)^{-2}+  \big(e^{2\pi\sqrt{3}}-e^{-2\pi\sqrt{3}}\big)^{-2} +...$$$$=  \sum_{n=0}^\infty  \big(ie^{(4n+1)\pi\frac{\sqrt{3}}{2}}+ie^{-(4n+1)\pi\frac{\sqrt{3}}{2}}\big)^{-2} +\sum_{n=0}^\infty  \big(-e^{(2n+1)π√3}+e^{-(2n+1)π√3}\big)^{-2} +\sum_{n=0}^\infty  \big(-ie^{(3+4n)\pi\frac{\sqrt{3}}{2}}+-ie^{-(4n+3)\pi\frac{\sqrt{3}}{2}}\big)^{-2}+
\sum_{n=0}^\infty  \big(e^{(2n)π√3}-e^{-(2n)π√3}\big)^{-2}  $$ $$\sum_{n=0}^\infty  \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty   \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty  \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty  \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1}  $$ I have found the sums numerically and $\sum_{n=0}^\infty  \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty   \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty  \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty  \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} \approx -0.00429$
How can I evaluate this analytically?


问:
determining the derivative of the following function
Let $q(x)$ be a differentiable function. Determine the derivatives of the following
functions. Each answer will have $q$ or $q'$ in it.

$$ f(x) = \sin\big(q(x)\big) $$

答:
By the chain rule,
$$
f'(x)=\cos (q(x))q'(x) \, \text{.}
$$


问:
If $(X_{t})_{t\in \mathbb{R}}$ has independent increments than $X_t$ and $X_s$ are independent if $t\neq s$?
The stochastic process $(X_{t})_{t\in \mathbb{R}}$ is said to have independent increments if for $n\in \mathbb{N}$ $t_1<t_2<\ldots < t_n$ I have that $X_{t_2}-X_{t_1}, X_{t_3}-X_{t_2},...$ are independent. But does that mean in particular that for $t\neq s \in \mathbb{R}$ we have that $X_t$ and $X_s$ are independent?

答: No, certainly not.  For example, let $X_0$ be a non-trivial random variable, and let $X_t = X_0$ for all $t$.  Then the increments are all zero, and thus independent.  But $X_0$ is not independent of itself.


问:
Groups with only two subgroups
I want to find all groups $G$ which have only two subgroups $G$ and $\{e\}$.

I think there can be some connection with simple groups, but it's only my intuition. I would like also to prove why these groups have only such two subgroups.

答:
First, notice that for $G=\lbrace e \rbrace$, then $G$ has only one subgroup, so it is not solution.

So let's suppose that $G\neq\lbrace e \rbrace$ and $G$ has only two subgroups. Then take $x \in G \setminus \lbrace e \rbrace$. The subgroup generated by $x$ must be different from $\lbrace e \rbrace$, therefore it must be equal to $G$. This means that $G$ is generated by one element : if $G$ is infinite, then $G$ is isomorphic to $\mathbb{Z}$, and $\mathbb{Z}$ has a lot of subgroups, so $G$ is not a solution. So $G$ is finite, so $G$ is cyclic. If its order is not a prime number, you can easily find several subgroups ; whereas if its order is a prime number, you cannot.

So the solutions are the cyclic groups of prime orders.

Title: Groups with only two subgroups


问:
Computing Probability given CDF
Given $$F(x) =\begin{cases}(x-1)^2/4,& 1 < x < 3\\0,& x\le 1\\ 1,& x\ge3\end{cases}$$

Is $P(2 < X < 5) = P(2 < X)$ since $F(x) = 1$ when $x\ge 3$? Also would $P(X = 1.5) = f(1.5) = F'(1.5)$?

答:
We have $$P(2<X<5) = F_x(5)-F_x(2) = 1 - F_x(2)  $$
On the other hand, $$P(2<X) = 1-P(X<2) = 1 - F_x(X<2)$$
So the first answer is yes.

But when we are talking about one specific point in a continuous distribution, its probability considered zero as its occurrence will be almost impossible. Assume we have
$$
P(X=1.5) = P(1.5-\epsilon < X < 1.5+\epsilon) = F_x(1.5+\epsilon) - F_x(1.5-\epsilon)
$$
where $\epsilon$ is very small number. Based on Taylor series expansion $F_x(1.5+\epsilon)$ could be written as
$$
F_x(1.5+\epsilon) = F_x(1.5) + f_x(1.5)(\epsilon) = F_x(1.5) + \epsilon
$$
and $F_x(1.5-\epsilon) = F_x(1.5) - \epsilon $ in the same way. Therefore,
$$
P(X=1.5) =  F_x(1.5) + \epsilon - F_x(1.5) + \epsilon = \epsilon \approx 0
$$
I hope it helps.

Title: Computing Probability given CDF


问:
Question on the demonstration of Morse theorem
We have theorem of Morse

and this is the proof

i dont understand this :

"$(c_i)$ has no cluster point since each $M^a=f^{-1}]-\infty,a]$ is compact "

Thank you.

答: Nondegenerate critical points must be isolated.</p>


问:
Why is $H^n(\mathbb{R}P^2;\mathbb{Z}) \otimes H^n(S^1;\mathbb{Z}) \not =0$ in degree $n = 1,2$?

Compute $$H^*(\mathbb{R}P^2\times \mathbb{R}P^2, \mathbb{Z}).$$

Hint: Consider the Mayer-Vietoris sequence of suitable open sets of the form $\mathbb{R}P^2\times U$ and $\mathbb{R}P^2\times V$ where
$U$ is a small disk in $\mathbb{R}P^2$.

So i was trying to understand the proposed solution to this exercise. But building the Mayer-Vietoris sequence goes wrong since i somehow end up getting $$H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z}) = 0$$

which is apparently wrong.

The solutions begins with:

Viewing $\mathbb{R}P^2$ as $D^2/{\sim}$ where we identify $p \in S^1 =
 \partial D^2$ with $−p$, we set $U$ to be a small disk around $0 \in
 D^2$ and $V = \mathbb{R}P^2 − \{0\}$. Thus $U$ is contractible and $V$
deformation retracts onto the subspace $S^1/{\sim}$, which is
homeomorphic to $S^1$ through a loop joining $p$ and $−p$.

We consider the Mayer-Vietoris sequence of the cover $\mathbb{R}P^2 \times U, \mathbb{R}P^2 \times V$. Note that the intersection is given by $\mathbb{R}P^2 \times (U \cap V )$. As $U$ is
contractible $\mathbb{R}P^2 \times U$ is homotopy equivalent to $\mathbb{R}P^2$.

That's all fine. But my issue begins now by plugging in the cohomologies into the Mayer-Vietoris sequence which is given by

$$\cdots \to H^1(X;\mathbb{Z}) \to H^1(A;\mathbb{Z})\oplus H^1(B;\mathbb{Z}) \to H^1(A\cap B;\mathbb{Z}) \to H^2(X;\mathbb{Z})\to \cdots$$

Let's ignore the degree $0$ stuff for now.

I know that $$H^*(\mathbb{R}P^2,\mathbb{Z}) = \begin{cases} \mathbb{Z}/2\mathbb{Z} & * = 2 \\ 0 & \text{else} \end{cases}$$
and clearly $$H^*(S^1,\mathbb{Z}) = \begin{cases} \mathbb{Z} & * = 1 \\ 0 & \text{else} \end{cases}$$

Since $A:= \mathbb{R}P^2 \times U$ is homotopy equivalent to $\mathbb{R}P^2$, we have $$H^*(A;\mathbb{Z}) = H^*(\mathbb{R}P^2\times U;\mathbb{Z}) \cong H^*(\mathbb{R}P^2;\mathbb{Z})$$

$B:= \mathbb{R}P^2 \times V$ is homotopy equivalent to $\mathbb{R}P^2\times S^1$ and since $S^1$ is finitely generated free abelian, by the Künneth formula we get

$$ H^*(\mathbb{R}P^2\times V;\mathbb{Z}) \cong H^*(\mathbb{R}P^2\times S^1;\mathbb{Z}) = H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z})$$

Here is where i have issues:

Since $H^n(\mathbb{R}P^2;\mathbb{Z})$ vanishes in degree other than $n=2$ and $H^*(S^1;\mathbb{Z})$ vanishes in degree other than $n=1$, the tensor product $H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z})$ should give me always $0$, shouldn't it?

Somehow this seems to be wrong but i don't know why.

答:
$H^n(\mathbb{R}P^2;\mathbb{Z})$ vanishes in degree other than $n=2$ and $H^*(S^1;\mathbb{Z})$ vanishes in degree other than $n=1$

This is only true for reduced cohomology.  For unreduced cohomology, both of them have $\mathbb{Z}$ in $H^0$ as well.

In addition, note that $H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z})$ does not mean you are taking the tensor product $H^n(\mathbb{R}P^2;\mathbb{Z}) \otimes H^n(S^1;\mathbb{Z})$ separately in each degree.  Rather, it means you are taking the tensor product of the entire cohomology rings (all degrees together).  This means that the degree $n$ part of the tensor product will be the direct sum of all ways to tensor parts whose degrees add up to $n$, i.e. $\bigoplus_{p+q=n}H^p(\mathbb{R}P^2;\mathbb{Z}) \otimes H^q(S^1;\mathbb{Z})$.  So for instance, the nontrivial $H^2(\mathbb{R}P^2;\mathbb{Z})$ and $H^1(S^1;\mathbb{Z})$ will get tensored together to give something nontrivial in $H^3$.  (Note that this additionally means that even if you don't care about $H^0$ in the end, you still have to take into account the unreduced $H^0$ of each factor, since terms like $H^0\otimes H^n$ and $H^n\otimes H^0$ will appear in the $n$th degree of the tensor product.)


问:
Why is the $\lim_{x \to \infty} \frac{(-1)^x}{x}= 0$?
I know that the answer is $0$ by using Wolfram. But I'd appreciate it if someone shows the solution to this problem. If I changed it to $\lim_{n \to \infty} \frac{(-1)^n}{n}= 0$ where $n$ takes on positive integers including $0$, would the answer change?

答:
$(-1)^x$ for non-integer $x$ is "multi-valued" in the complex plane.  But all values have the property
$$
\big|(-1)^x\big| = |(-1)|^x = 1.
$$
so, no matter what values you choose, for $x>0$ we have
$$
\left|\frac{(-1)^x}{x}\right| = \frac{1}{x}
$$


问:
What square matrices cannot be expressed as the sum of symmetric and skew-symmetric parts
I thought I knew that any square matrix can be written as the sum of symmetric and antisymmetric matrices since we have the property that any $n\times n$ matrix $A$ can be expressed as $A=\frac{1}{2}(A-A^T)+\frac{1}{2}(A+A^T)$. However, I have come across the following statement and was told this is not always true and asked if I could find a counterexample where this isn't true but I have been struggling to find a counterexample (perhaps when $\mathbb{F}=\mathbb{Z}_2$?). Here is the statement:

Let $\mathbb{F}$ be a field. Then any $A\in \mathbb{F}(n,n)$ may be written as a sum of a symmetric and an anti-symmetric matrices in $\mathbb{F}(n,n)$.

I would greatly appreciate if someone could show me a counterexample to this.

答:
Another way of understanding why this isn't always true, in lieu of a counterexample, is via a dimension argument.  $n\times n$ matrices may be viewed as living in a vector space $V = \mathbb F^{n\times n}$.  This space has dimension $n^2$.   (Use standard basis vectors of the form $\mathbf e_k\mathbf e_j^T$ and count.)

For fields with characteristic $\neq 2$ then we have two subspaces $S_y$ and $S_k$ consisting of symmetric and skew symmetric matrices respectively. By inspection:

$\dim\big(S_y\big) = \binom{n}{2} +n$

$\dim\big(S_k\big) = \binom{n}{2}$

so if
$V=S_y + S_k$ then $V=S_y \oplus S_k $

because there is a trivial intersection between $S_y$ and $S_k$

i.e. if we find some $X$ such that $X\in S_y$ and $X\in S_k$, then $X=X^T$ because it is symmetric, and $-X=X^T$ because it is skew symmetric.  So $X=-X$ or $2X=\mathbf 0\implies X=\mathbf 0$.

To confirm $V=S_y \oplus S_k $, check the dimensions.  i.e. if the RHS, a subspace of $V$, provides a basis with $n^2$ elements then it must be consist of all of $V$. To confirm:

$n^2 =\dim\big(V\big) =  \dim\big(S_y\big) + \dim\big(S_k\big)=\big(n + \binom{n}{2}\big) + \binom{n}{2}=n^2$

However in fields of characteristic 2 you have $S_k\subseteq S_y$,

if $V=S_y + S_k$ then  $V=S_y + S_k = S_y $  which further implies

$n^2= \dim\big(V\big)=\dim\big(S_y + S_k\big)=\dim\big(S_y \big) =\big(n + \binom{n}{2}\big) \lt n^2  $

which is impossible

note:

it is not necessarily true that $S_y\subseteq S_k$ in characteristic 2.  Many texts will explicitly define skew symmetry such that the matrix has all zeros on the diagonal hence $I_n$ is in $S_y$ but not $S_k$.  Conventions vary.  That said this convention is quite useful-- e.g. when working with skew symmetric bilinear forms and showing congruence of invertible skew symmetric matrices to the symplectic matrix over any field.


问:
How Do I Find the Value of t*, the Critical Value of a t Distribution, Given the Sample Size and the Probability of Being Greater than t*?
I am trying to find the value of t*, the critical value of the t distribution for a sample of size 6, such that the probability of being greater than t* is 1%.

I see that $t(a/2,n-1) = t*$, however, I am unsure how to use that formula in a logical way. I'm guessing that it would be $t(0.005,5)$, but I am unsure what to do with those numbers.

Does anyone know how to solve this problem?

Thanks!

Body:
P(t>t*) = 1% = 0.01
P(t<t*) = 1-0.01 = 0.99
t* = T.INV(0.99,6-1)


答:
"The thing that confuses me is the fact that the set being embedded into a higher dimension while setting one of the variables constant."

EXACTLY!!!!!

In general that is dead giveaway they can not be open-- any neighborhood will have points pertruding out in the higher dimensions.  For example in question 1) with a neighborhood with radius $\epsilon$ around point $(x,0)$ will contain points $(w,\delta)$ where $\delta\ne  0$ so $(w,\delta)$ is not in the set so the set is not open.

Same thing with question 2).

For the set 3, would the answer differ if instead we had {1n∣n∈N} as a subset of R

It makes all the difference in the world!

$0\in \mathbb R$ and $0\not \in \mathbb R_+$.

In $\mathbb R$ we have $\lim_{n\to \infty} \frac 1n = 0$ so $0$ is a limit point of the set.  So the set is not closed.

But in $\mathbb R_+$ we have $\lim_{n\to \infty}$ non-existent. We show that for any $x \in \mathbb R$ that $x$ is not a limit point of the set.  so the set has no limit points and therefore is closed.

Title: Show if the following sets are open


问:
Probability of duration of a phone calls like a draw made random with replacement
The durations of phone calls taken by the receptionist at an office are like draws made at random with replacement from a list that has an average of $8.5$ minutes (that's $8$ minutes and $30$ seconds) and an SD of $3$ minutes. Approximately what is the chance that the total duration of the next $100$ calls is more than $15$ hours?

First convert everything into seconds. The mean is $8.5$ minutes. $8.5$ minutes is $510$ seconds for one call,

so it's $100$ times that amount for $100$ calls. $510\times 100=51000$ seconds

The SD is $3$ minutes. $3\times 60=180$ seconds for one call,

the variance is the SD squared. $\sigma=180^2=32,400$ seconds.

Multiply the variance by $100$, then take the square root of it for the SD for $100$ calls.

$SD=\sqrt{32400\times 100}=1,800$ seconds.

$x$ is the time for $100$ calls. The mean is the mean for $100$ calls, SD is the SD for $100$ calls. You'll use the $z$ score to find the probability:

$15$  hours$\times60$  minutes$\times 60$  seconds$=54,000$ seconds

$$z=\frac{x−\bar{x}}{SD}= \frac{54,000-51,000}{1,800}=1.67$$

Area under normal curve Lower end $-5$

and upper end $1.67=95.22\%= 0.9522$ THEREFORE $P(t\gt54,000)=1-0.9522=0.0482$

答:
So you have 100 random variables, each with average $\mu_0=8.5$ minutes and standard deviation $\sigma_0=3$ minutes. The central limit theorem thus tells you that the sum of these random variables is approximately normally distributed with $\mu=100\mu_0=850$ minutes and standard deviation $\sigma=\sqrt{100}\sigma_0 = 30$ minutes.

You're looking for the probability that such a standard distribution yields a value greater than 15 hours, or equivalently greater than $15\cdot 60 = 900$ minutes. If a random variable $X$ is normally distributed with average $\mu$ and standard deviation $\sigma$, then the probability that $X \leq a$ (or equivalently $X < a$, since the normal distribution is continuous) is $$
  \mathbb{P}(X \leq a) = \Phi\left(\frac{a - \mu}{\sigma}\right) \text{,}
$$
where $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution, i.e. the one with average 0 and standard deviation $1$. You're interested in the probability that $X = X_1+\ldots+X_{100}$ is greater than 900 minutes, and must thuse compute $$
  \mathbb{P}(X > 900) = 1 - \mathbb{P}(X \leq 900) = 1 - \Phi\left(\frac{900 - 850}{30}\right) = 1 - \Phi\left(\frac{5}{3}\right) \text{.}
$$

Now you just need to use a table (or software) to evaluate $\Phi\left(\frac{5}{3}\right) \approx 0.952$ and get $$
   \mathbb{P}(X > 900) \approx 0.048 \text{.}
$$

So you simply forgot to invert the probability you got at the end, i.e. your value (which is $\Phi\left(\frac{5}{3}\right)$) is the probability that the calls take less than 15 hours in total.

Btw, there's an easy way to see that your answer cannot be correct. You're asking from the probability that the value is greater than some value which itself is greater than the mean of the distribution. Since the normal distribution is symmetric, the mean is also the median, and thus the probability cannot be larger than 0.5, i.e. 50%. And since the distance from the mean isn't particularly small compared to the standard deviation (your distance is 50, the standard deviation is 30), it'll have to be substantially smaller than 50%.


问:
Probability puzzle for a fair die
What is the expected number of times we have to roll a fair six-sided die to obtain that the number of times any two faces appear are equal (after the first rolling)?</p>

答:
These are just some back-of-the-envelope calculations, but I think they'll convince you that the probability that the event occurs is less than $1$, so that the expectation is infinite.

What is the probability that each number occurs $k$ times in $6k$ rolls?  There are $\frac{(6k)!}{(k!)^6}$ acceptable sequences of rolls out of $6^{6k}$ possible sequences so that the probability is $$\frac1{6^{6k}}\frac{(6k)!}{(k!)^6}$$  Stirling's approximation gives $$\frac1{6^{6k}}\frac{(6k)!}{(k!)^6}\sim\frac1{6^{6k}}\frac{(6k)^{6k}e^{-6k}\sqrt{12\pi k}}{(k^ke^{-k}\sqrt{2\pi k})^6}=\frac{\sqrt3}4(\pi k)^{-5/2}$$

Numerical calculation shows that $$\sum_{k=1}^\infty\frac{\sqrt3}4(\pi k)^{-5/2}\approx\frac1{30}$$

an this sequence overstates the probability, since it would count sequence where the phenomenon more than once multiple times.

This should convince you that doing the calculations carefully, if it were thought worthwhile, would demonstrate that the phenomenon is not very likely to occur, and that the expected waiting time until it does is infinite.

Title: Probability puzzle for a fair die


问:
Basic question on local compactness
Let $X$ be a topological space.

Let $A \subset F \subset X$.

Let $F$ be a subspace of $X$.

Let $A$ be a subspace of $F$.

Suppose that under this subspace topology inherited from $F$, $A$ forms a locally compact topological space.

Now let $A$ is a subspace of $X$.

Then how can I show that under the subspace topology inherited from $X$, $A$ is a locally compact topological space?

答: Because the topology that $A$ inherits from $F$ is the same topology that it inherits from $X$ (since the topology of $F$ is the one that it inherits from $X$). That's so because, for both topologies, a subset $O$ of $A$ is open if and only if $O=O^\star\cap A$, for some open subset $O^\star$ of $X$.

Title: Basic question on local compactness


问:
Clarity on Big-O
so I just had this question, I know that for example $g(n) = 100$, then $O\big(g(n)\big) = O(1)$. Let us take $a(n)$ as some function like $n\log n$, does this mean that $a(n)\in O\big(g(n)\big)$?

答:
So you have $g(n) = 100$, in which case $g(n)$ is $O(1)$. You then define $a(n) = n\log n$ and ask if $a(n)$ is $O(g(n))=O(1)$.

The answer is no.

For $a(n)$ to be $O(1)$, it would have to be true that $n\log n$ was bounded by a constant from some point and onwards, but $n\log n$ grows to infinity, making such a bound impossible.

Title: Clarity on Big-O


问:
Complex numbers and absolute values
If i have equation:

\begin{align}
P = \left|\psi\right|^2
\end{align}

where $P$ is a probability and we know there is no negative probability. This means $P$ must belong to $\mathbb{R}$. If i want to calculate $|\psi|$ i can do it simply by sq. rt. the equation:

\begin{align}
\left|\psi\right| = \sqrt{P}
\end{align}

Is $\left|\psi\right|$ a real number? What about $\psi$ by itself? Please explain.

答: The absolute value of a complex number (sometimes called its modulus) is real by construction. If $\psi=\alpha+i\beta$ with $\alpha$ and $\beta$ real, then $|\psi|=\sqrt{\alpha^2+\beta^2}$.

Title: Complex numbers and absolute values


问:
Why is the expectation of $Y=X^2$ more than $\textbf{E}[X^2]$?
Given $X$ is an RV with pdf $f(x)=2x$ where $0<x<1$. Then the $$\textbf{E}[X^2]=\int_0^12x^3 = \frac{1}{2}$$
Given $Y=X^2$, then the pdf of $Y$ is $g(x)=f(x)f(x)=4x^2$ giving us the expectation, $$\textbf{E}[Y] = \int_0^1 4x^3dx = 1.$$

Why is the expectation of $Y=X^2$ more than $\textbf{E}[X^2]$?

Because it seems like $\textbf{E}[Y]=\textbf{E}[X^2]$.

答: Hint: The pdf of $Y$ is not given by $f(x)\cdot f(x)$, as is evidenced by the fact that integrating that function from $0$ to $1$ does not give a total probability of $1$.


问:
For a Brownian motion $B(t)$, show $0=\mathbb E[ B(\tau)]$ where $\tau := \max \{\tau_a, \tau_b\}$ is the latter hitting time of the levels $a<0<b$.
Let $B(t)$ be a Brownian motion. Let $a<0<b$ and define corresponding hitting times of those levels $\tau_a$ and $\tau_b$. Put $$\tau := \max \{\tau_a,  \tau_b\}.$$ We know $\tau$ is also a stopping time. I want to show that $$0=\mathbb E[ B(\tau)].$$

It is well known that the above equality holds for the stopping time $\tau_a \wedge \tau_b$ and I tried mimicking the proof using martingales, but it does not seem to work since $\mathbb E \tau =\infty$.

Any help is appreciated.

答: By the (a.s.) continuity of the sample paths of Brownian motion, we have $B_{\tau} \in \{a,b\}$. If we set $\sigma:=\min\{\tau_a,\tau_b\}$, then $$B(\tau,\omega) = a \iff \tau_a(\omega)>\tau_b(\omega) \iff  B(\sigma,\omega)=b$$ and, analogously, $$B(\tau,\omega)=b \iff B(\sigma,\omega)=a.$$ Consequently, $$\mathbb{E}(B_{\tau}) = a \mathbb{P}(B_{\sigma}=b) + b \mathbb{P}(B_{\sigma}=a).$$ Since $$\mathbb{P}(B_{\sigma}=a) = \frac{b}{b-a} \qquad \mathbb{P}(B_{\sigma}=b) = \frac{-a}{b-a}$$ (see e.g. this question), we get $$\mathbb{E}(B_{\tau}) =- \frac{a^2}{b-a} +\frac{b^2}{b-a}= \frac{(b-a)(b+a)}{b-a} = b+a.$$


问:
Extending lines from Cyclic Quadilateral
Take a cyclic quadilateral, ABCD. Line BD bisects angle ABC. Extend the side BA beyond A to a point H. Show that DH = DB only if AH = BC.

I've attempted to draw this out, but anything I've drawn is a dead end. Any guidance or solutions are appreciated, thanks.

答:
$CD\cong AD$ because angles $CBD$ and $ABD$ are congruent as $BD$ is bisector.

$BC\cong AH$ by hypothesis.

$BCD$ and $DAB$ are supplementary because $ABCD$ is cyclic, thus $BCD\cong DAH$.

Therefore triangles $CBD$ and $DAH$ are congruent and $BD\cong DH$.


问:
What is the probability of getting red marbles from bag B?
So there are 3 bags: bag A, bag B, and bag C.

Bag A contains 2 red marbles and 3 white marbles.

Bag B contains 3 red marbles and 5 white marbles.

Bag C contains 4 red marbles and 6 white marbles.

Suppose a bag was randomly selected and marble was taken out.
What is the probability of taking a red marble from bag B?

So I've tried the conditional probability
https://i.stack.imgur.com/YSEky.png, but kinda confused.
If we make it like $P(A)$ is the probability of bag B chosen out of 3 bags and $P(A ∩ B)$ is the probability of getting red marbles out of marbles in bag B, making $P(A | B)$ the result.
But if we make it like $P(A)$ is the probability of red marbles get chosen out of all marbles, and $P(A | B)$ is the probability of only red marbles from bag B get chosen out of all marbles and making $P(A ∩ B)$ the result, it's going to be different. Which one is the right calculation (or maybe both are wrong...)?

答:
This is a funny situation.  Considering my comment, while the OP has clearly shown work, I am still not sure which question is being asked.

So I will answer each question separately.

Bag A contains 2 red marbles and 3 white marbles.

Bag B contains 3 red marbles and 5 white marbles.

Bag C contains 4 red marbles and 6 white marbles.

Let $E_1$ be the event that bag B is chosen.

Let $E_2$ be the event that a red marble is chosen.

Q1: Compute $p(E_2 | E_1)$.

Q2: Compute $p(E_1 ~\text{and} ~E_2 ~\text{both occurring})$.

Q3: Compute $p(E_1 | E_2)$.

Answer-1:

Given that bag B is chosen, the answer is immediately seen to be $\frac{3}{8}$.

Answer-2:

$p(E_1 ~\text{and} ~E_2 ~\text{both occurring})$ is (using Answer-1)

$$p(E_1) \times p(E_2|E_1) = \frac{1}{3} \times \frac{3}{8} = \frac{1}{8}.$$

Answer-3:

$$p(E_2) = \left[\frac{1}{3} \times \frac{2}{5}\right]
+ \left[\frac{1}{3} \times \frac{3}{8}\right]
+ \left[\frac{1}{3} \times \frac{4}{10}\right]
= \frac{47}{120}.
$$

Using Answer-2,

$$p(E_1 | E_2) = \frac{p(E_1 ~\text{and} ~E_2 ~\text{both occurring})}{p(E_2)}$$

$$= \frac{(1/8)}{(47/120)} = \frac{15}{47}.$$


问:
Coloring the line
You probably know the Hadwiger-Nelson problem, which states:

What is the minimal number $k$ such that there is a $k$-coloring of the plane such that no two points at distance $1$ have the same color?

This problem is open. There are known upper ($k=7$) and lower ($k=4$) bounds.

My question is: is the same problem for the line solved? I can see the obvious bound below given by $2$ and $3$ as a bound above, given by the coloring by intervals of length $\frac{2}{3}$. But is there a $2$-coloring of the line solving the problem?

答: Sure! For $x, y \in \mathbb{R}$, consider a relation $a \sim b \iff a - b \in \mathbb{Z}$ -- that is, two numbers are in the relation if they differ by an integer. This is an equivalence relation, and it's easy to see that its equivalence classes are of the form $[x] = \{ x + k: k \in \mathbb{Z}\}$. Consider the set of equivalence classes $A = \mathbb{R}/\sim$. Pick a choice function $f: A \to \mathbb{R}$ that from each equivalence class picks a representative, i.e. satisfies $x = [f(x)]$: it exists by the axiom of choice. Now we'll construct a coloring, that is, a function $g: \mathbb{R} \to \{0, 1\}$: for $a \in \mathbb{R}$, consider $x = f([a])$. Since $[x] = [a]$, $a = x + k$ for some integer $k$. Put $g(a) = k \mod 2$. I'll leave it for you to show that $g$ is actually the coloring we want.</p>

Title: Coloring the line


问:
Is $f(x) \in \mathbb{R}$ if $x \in \mathbb{R}$?
Let $f:\Omega\subset\mathbb{C}\rightarrow\mathbb{C}$ be holomorphic with $\Omega$ open. Assume there is an infinite sequence $(x_n)_n$ in $\Omega \cap \mathbb{R}$ that converges to $x_0 \in \Omega \cap \mathbb{R}$ and assume that for every $n$, $f(x_n) \in \mathbb{R}$. Let $D_r$ be the open disk of radius $r>0$ and centre $x_0$ such that $D_r \subseteq \Omega$. Is it true that $f(x) \in \mathbb{R}$ for every $x \in D_r \cap \mathbb{R}$?

I tried proving this by constructing a holomorphic function $g$ such that $g(z)$ is real whenever $z$ is real and $g(x_n)=f(x_n)$. Then by the identity theorem it follows that $g=f$. Unfortunately, I was unable to find such $g$.

Any help would be welcome. Thank you in advance.

答: Let $f=u+iv.$ Then $v$ is real analytic on $I=(x_0-r,x_0+r).$ Furthermore $v=0$ on the set $E=\{x_n: x_n\in I\}\cup \{x_0\}.$ Since $E$ is a subset of $I$ with limit point in $I,$ we have $v=0$ on $I$ by the identity principle for real analytic functions. Thus $f(x)$ is real for all $x\in I.$


问:
Fermi-Dirac integral
I'm trying to do some integral calculation for Fermi-Dirac distribution, specifically for:

$$
\int_{0}^{\infty}{E^{2} \over
1 + \exp\left(E - \mu \over k_{B}\,T\right)} \, dE
$$

I know that it can be only solved numerically, but I got information that when we set $T = 0$ it can be solved analytically.

And here is my question: When we will do that directly, all is left is $0$. So, how can I approach such sort of problem $?$.

答:
You have

$$I(T)=\int\limits_0^\infty {E^2\over 1+\exp{E-\mu\over k_BT}}\mathrm{d}E = -\operatorname{Li}_3\left(-e^{\mu\over k_BT}\right)$$

in general. However, for your case of interest where $T\rightarrow 0$, you will have a Dirac impulse for $E=\mu$ and the solution is just $I(0)=\mu^3$.

Title: Fermi-Dirac integral


问:
Can someone guide me through the proof for this? Combining inverse functions and intersection is confusing me
Let f be a function that maps X into Y (f: X $\to$Y). Let {$G_\alpha$| $\alpha$ $\in$ A} be an indexed family of subsets of Y.
 Show f$^{-1}$ $\bigl($$\bigcap_{\alpha\in A}$ G$_\alpha$ $\bigl)$ = $\bigcap_{\alpha\in A}$ f $^{-1}$ $\bigl(G_\alpha\bigl)$

答:
Can you proof the simpler version for finite intersections?

$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$.

The general version is just a slight variation of this proof.

We have to proof equality of sets.
For that we proof $\subseteq$ and $\supseteq$.

To proof "$\supseteq$", we take $x\in f^{-1}(A\cap B)=\text{recall definition of $f^{-1}(\text{set})$}$. We have to show that $x\in f^{-1}(A)\cap f^{-1}(B)$

What can you conclude?
All you have to do is basically that you write carefully down how the 'signs' are defined.  There are no tricks involved.

Then "$\supseteq$" is proved similarly, or you can just 'reverse' your proof of $\subseteq$.

Once you finished this proof, the other one should be problem.

Give it a try, and feel free to ask if you get stuck.


问:
Linear algebra, Approximation and fundamental theorem of linear algebra
For a given $m\times n$ matrix $A$ and $\vec{x} \in \mathbb{R}^n$ find $\vec{y} \in \mathbb{R}^n$ which is a solution to $A\vec{y}=0$ and is a minimum distance from $\vec{x}$.

Show that for $\vec{y}$ such that $A\vec{y}=0$ and $\left \| \vec{x}-\vec{y} \right \|$ is as small as possible, then $\vec{x}-\vec{y}$ is in the $Row(A)$.

I am very close to this, can someone help me clear my thoughts a bit?
Here is my process,
because $A\vec{y}=0$ and $\vec{y}$ is in $R^n$  so $\vec{y}$ is in the Null(A),
x-ay=x-projection of x onto Col(A), which is equal to perp ColA (x) – Logan 6 mins ago

So $\vec{x}-A\vec{y}$ is in the complement of Row space of A, using "the fundamental theorem of linear algebra" again.
How do you go from there? Please help me!!!!

答:
There are some mistakes in what you've written, but you essentially have the idea for the second question.

We are looking for the point $\vec y$ in the nullspace of $A$ that is as close to $\vec x$ as possible, which is in other words the projection of $\vec x$ onto the nullspace of $A$. By the properties of orthogonal projection, such a $\vec y$ has the property that $\vec x - \vec y$ lies in the orthogonal complement to the nullspace of $A$.

By the fundamental theorem, this is the same as saying that $\vec x - \vec y$ lies in the row space of $A$.

As for how to compute this vector: note that the matrix that projects onto the row space of $A$ is given by $A^T(AA^T)^{-1}A$ (assuming the rows of $A$ are linearly independent). It follows that the projection onto the nullspace has the matrix
$$
P = I - A^T(AA^T)^{-1}A.
$$
The vector that we are after is the projection of $\vec x$ onto the nullspace, which will simply be $\vec y = P \vec x$.

If the rows of $A$ are not linearly independent, we can proceed as follows. The projection of $\vec x$ onto the row-space of $A$ is the vector $A^T \vec z$, where $\vec z$  is the least squares solution to the equation $A^T \vec z = \vec x$. In other words, if $\vec z$ is any solution to the equation
$$
AA^T \vec z = A \vec x,
$$
then $A^T \vec z$ is the projection of $\vec x$ onto the row space. It follows that the desired vector $\vec y$ is equal to $\vec y = \vec x - A^T\vec z$.


问:
Find generating function expressing number of rectangles
I have a 3 x n rectangle. I need to find a generating function expressing the number of little 1 x 3 rectangles inside this bigger one. How can I do this? I have no idea how to even begin.</p>

答:
Hint: $a_{n+3} = a_n+a_{n+2}$.

Expanded: $a_n$ obviously, is the number of tilings of a $3\times n$ rectangle. Now, consider a $3\times (n+3)$ rectangle. In any tiling, the bottom left corner is either tiled by a vertical tile, or a horizontal tile. In the first case, the rest of the is $3 \times (n+2)$ and can be tiled $a_{n+2}$ ways. Otherwise, the cell just above the bottom left has to be tiled by a horizontal tile, and same for the one above it. In which case the rest is a $3 \times n$ rectangle, which can be tiled $a_n$ ways. These cases are obviously mutually exclusive.

The answer seems to be $$\sum_{n=0}^\infty a_n x^n = \dfrac{1}{1-x-x^3}$$.


问:
In relation to classic semantics, let $ \alpha $ be an arbitrary of $ \mathcal L_1 ^ {=} $ - closed formula ...
In relation to classic semantics, let $ \alpha $ be an arbitrary of $ \mathcal L_1 ^ {=} $ - closed formula, such that $ \vDash \alpha $ and $ \Delta $ a set of $ \mathcal L_1 ^ { =} $ - closed formula, such that the formula $ \lnot \alpha $ belongs to $ \Delta $, so $\Delta $$\vDash $$\alpha $ is valid?

Is the answer no, right?

Because let's assume that $ \Delta $ = {$ \beta, \phi, P \to Q, \lnot \alpha, .... $}
this means that all formulas are true, so $ \lnot \alpha $ is true, so if $ \alpha $ is true and is in $ \Delta $, then $ \lnot \alpha $ cannot be in $ \Delta $, but by chance $ \lnot \alpha $ is in $ \Delta $, so $ \alpha $ is not a semantic consequence (sorry I don't know how to write the symbol of this).

答:
Your intuition that "$\neg \alpha$ cannot be in $\Delta$" goes in the right direction: If $\alpha$ is valid, then any set of formulas containing $\neg \alpha$ is unsatisfiable (aka contradictory or inconsistent), i.e. there can be no interpretation that makes all formulas in $\Delta$ true.

But precisely because $\Delta$ is unsatisfiable, it entails $\alpha$: Remember that $\Delta \vDash \alpha$ iff every interpretation that satisfies all formulas in $\Delta$ also satisfies $\alpha$, i.o.w., there is no interpretation that satisfies all formulas in $\Delta$ but fails $\alpha$. Now since there is no interpretation that can satisfy $\Delta$ to begin with, then in particular there can be no such counter interpretation which invalidates the entailment: $\Delta \vDash \alpha$ is vacuously valid because $\Delta$ is unsatisfiable.

The inference is actually valid for a second reason: There can be no interpretation that satisfies all formulas in $\Delta$ and simulateneously does not satisfy $\alpha$, because there is no interpretation at all that would not satisfy $\alpha$ -- after all, we assumed that $\alpha$ is valid. In classical logic, adding additional premises to an already valid formula never invalidates the entailment: $\Delta \vDash \alpha$ is trivially valid because $\alpha$ is valid.

In general, we have that

$\Gamma \vDash \phi$ for arbitary formulas $\phi$, if $\Gamma$ is unsatisfiable, and

$\Gamma \vDash \phi$ for arbitrary sets of formulas $\Gamma$, if $\phi$ is valid.


问:
Prove that $12(ab+ba+ac) <7a^2+15b^2+18c^2$ holds for all positive numbers.
Prove that $12(ab+ba+ac) < 7a^2+15b^2+18c^2$ holds for all positive numbers.

I tried completing the square, but that solution would suggest that inequality holds for all real numbers. Inequalities between means did not work for me either.

$$12(ab+ba+ac) < 7a^2+15b^2+18c^2$$
$$(2a-3b)^2+(2b-3c)^2+(2a-3c)^2+2b^2-a^2>0$$

答:
We write $f(a,b,c)=7a^2+15b^2+18c^2-12(ab+bc+ca)$ as a quadratic form with matrix
$$\begin{bmatrix}
7&-6&-6\\
-6&15&-6\\
-6&-6&18\end{bmatrix}$$
This can be verified to be a positive-definite matrix, so $f(a,b,c)\ge0$ for all $a,b,c\in\mathbb R$. In particular, if $a,b,c>0$ then $f(a,b,c)>0$.


Title: Prove that $12(ab+ba+ac) <7a^2+15b^2+18c^2$ holds for all positive numbers.

Body:
Prove that $12(ab+ba+ac) < 7a^2+15b^2+18c^2$ holds for all positive numbers.

I tried completing the square, but that solution would suggest that inequality holds for all real numbers. Inequalities between means did not work for me either.

$$12(ab+ba+ac) < 7a^2+15b^2+18c^2$$
$$(2a-3b)^2+(2b-3c)^2+(2a-3c)^2+2b^2-a^2>0$$

Body:
The completing to square works:

We need to prove that:
$$18c^2-12(a+b)c+7a^2+15b^2-12ab>0$$ or
$$9c^2-6(a+b)c+\frac{7}{2}a^2+\frac{15}{2}b^2-6ab>0$$ or
$$(3c-a-b)^2+\frac{5}{2}a^2+\frac{13}{2}b^2-8ab>0,$$
which is true by AM-GM:
$$\frac{5}{2}a^2+\frac{13}{2}b^2-8ab\geq2\sqrt{\frac{5}{2}a^2\cdot\frac{13}{2}b^2}-8ab=\left(\sqrt{65}-8\right)ab>0$$


答:
I'm assuming you meant to write that $\lnot\alpha$ is in $\Delta$, right?

The statement is true. Since $\models\alpha$, every structure satisfies $\alpha$. Thus every model of $\Delta$ is a model of $\alpha$, which is the definition of $\Delta\models\alpha$.

The fact that $\lnot\alpha\in \Delta$ is irrelevant here - but it also allows us to conclude that $\Delta$ has no models at all!

Title: In relation to classic semantics, let $ \alpha $ be an arbitrary of $ \mathcal L_1 ^ {=} $ - closed formula ...

Body:
In relation to classic semantics, let $ \alpha $ be an arbitrary of $ \mathcal L_1 ^ {=} $ - closed formula, such that $ \vDash \alpha $ and $ \Delta $ a set of $ \mathcal L_1 ^ { =} $ - closed formula, such that the formula $ \lnot \alpha $ belongs to $ \Delta $, so $\Delta $$\vDash $$\alpha $ is valid?

Is the answer no, right?

Because let's assume that $ \Delta $ = {$ \beta, \phi, P \to Q, \lnot \alpha, .... $}
this means that all formulas are true, so $ \lnot \alpha $ is true, so if $ \alpha $ is true and is in $ \Delta $, then $ \lnot \alpha $ cannot be in $ \Delta $, but by chance $ \lnot \alpha $ is in $ \Delta $, so $ \alpha $ is not a semantic consequence (sorry I don't know how to write the symbol of this).


答:
By AM-GM, or completing the square,

$(p-q)^2 \ge 0 \Rightarrow p^2 + q^2 \ge 2pq$

we have

$$\color{blue}{(4a^2 + 9b^2)} + \color{red}{(3a^2 + 12c^2)} + \color{green}{(6b^2 + 6c^2)} \ge \color{blue}{12ab} + \color{red}{12ca} + \color{green}{12bc}$$

with equality for $2a=3b$, $a=2c$, $b=c$ whose simultaneous solution is $(a,b,c)=(0,0,0)$

For $a,b,c > 0$, we have strict inequality.


问:
Inverse properties
Suppose we fave a continuous function $f:X\rightarrow Y$ and $\mathbb{E}\subseteq X$ where $\mathbb{E}$ is connected and there exist $A,B$ open sets such that
$f(\mathbb{E})\subseteq f(A)\cup f(B)$ and
$f(\mathbb{E})\cap f(A)\cap f(B)= \varnothing $

is it possible to do :$(f^{-1}f(\mathbb{E}))\subseteq f^{-1}(f(A)\cup f(B))\subseteq f^{-1}(f(A))\cup f^{-1}(f(B))$

$f^{-1}(f(\mathbb{E}))\cap f^{-1}(f(A)\cap f(B))= \varnothing \Rightarrow $ $(\mathbb{E})\cap f^{-1}(f(A))\cap f^{-1}(f(B))=\varnothing$

i am trying to understand the proof that $f(\mathbb{E})$ is also connected and i just wonder why we didn't just do the above.I suppose that its not true but for what exactly reson is my question.

thank you.

答:
We suppose that
$$ f(E) = U_1 \cup U_2,$$
and
$$ U_1 \cap U_2 = \emptyset.$$
$U_1$ and $U_2$ are open sets of $Y.$
Then
$$f^{-1}(f(E))=f^{-1}(U_1) \cup f^{-1}(U_2),$$
and
$$E= f^{-1}(U_1) \cup f^{-1}(U_2).$$
$f$ is continuous then $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open.
$E$ is connected. So
$f^{-1}(U_1)=E$ or $f^{-1}(U_2)=E.$
And because that
$$U_1 \cap U_2 = \emptyset$$ we have
$$f^{-1}(U_1) \cap f^{-1}(u_2)=\emptyset$$  and consequently

$$f^{-1}(U_2) = \emptyset.$$
Hence $f(E)$ is connected.

Title: Inverse properties


问:
Find the greatest and least values for $|z - w|$ for points on these loci.
So I'm given that
$$ |z - (-i)| = 1$$
and $$arg(w - 2) = \frac{3\pi}{4}$$

I have plotted the loci. The first is the circumference of a circle with center -i and radius 1, and the second is a half-line starting at -2 and making the given angle with the positive real axis.

Now, my next question is:
Find the greatest and least values for $|z - w|$ for points on these loci.

I am having trouble understanding what $|z - w|$ means. So z is an arbitrary complex number on the circle, right? And w is an arbitrary complex number on the half-line?

If that is so, does $|z - w|$ mean the "magnitude of of z minus the magnitude of w"?

And if that is so, isn't the least value for   $|z - w|$ going to be from the origin to -2? And what of the greatest?

I wish I could attach a sketch or something, but I can't (I also hope my half-line is correct -- it starts at $-2$).

答:
You're close. If you have points $z$ and $w$, then $z-w$ is the vector pointing from $w$ to $z$. This makes $|z-w|$ the length of the line segment between those two points. Your guess of "the magnitude of $z$ minus the magnitude of $w$" would be $|z|-|w|$, which is a different animal. Note that since $|\cdot|$ is always non-negative, the least value of $|z-w|$ cannot be $-2$.

In particular, given your two curves, the smallest value of $|z-w|$ would be the length of the smallest line segment between those curves.

Side comment: here are a couple of terminology tips that will save you a few words. Instead of "circumference of a circle", you can just call it a circle. A "filled circle" would be a disc. Second, a "half-line" is a ray.


问:
Application of Fatou's lemma to show that a function is the characteristic of some measurable set
I need some help solving this problem:

Let $(X,\mathcal{M},\mu)$ be a measure space and $f:X\mapsto[0,\infty]$ be a $\mu-$measurable function such that there exists $C\geq0$ such that $$\int_X(f(x))^nd\mu(x)=C,\;\text{for each }n\geq1$$
I'm asked to show that there exist a measurable set $E$ such that $f=\chi_E$ almost everywhere.

My idea was to show that the sets $A=\{x\in X|f(x)>1\}$ and $B=\{x\in X|f(x)\in(0,1)\}$ have measure 0, so the function only takes values 0 and 1 a.e. and then I could take $E=f^{-1}(1)$ which will be measurable because $f$ is measurable and $f=\chi_E$ a.e.

We're told to use Fatou's lemma but I'm a bit lost. I think that we can say that $\mu(A)=0$ because for each $x\in A$ the sequence $(f(x)^n)$ is not bounded so the integral cannot be finite. And $\mu(B)=0$ because for $x\in B$ $(f(x)^n)$ tends to 0 so the function would be end up being 0 in $B$ because $f\geq 0$. But I don't know if it is a good start and how to prove it rigorously. Any hint?

Thanks for your help.

答:
Define $A_k := \{ x \in X \mid f(x) \geq 1+1/k\}$, then by assumption

$$C = \int_X f(x)^n \, \mu(dx) \geq \int_{A_k} f(x)^n \, \mu(dx) \geq \left(1+ \frac{1}{k} \right)^n \mu(A_k)$$

for any $n \in \mathbb{N}$. When we let $n \to \infty$, then the right-hand side converges to $\infty$ unless $\mu(A_k)=0$. Since we know that the right-hand side is bounded in $n$, we obtain that $\mu(A_k)=0$ for all $k \geq 1$. Hence, $\mu(\{f>1\}) = \mu(\bigcup_k A_k)=0$.

It remains to show that $\mu(\{0<f<1\})=0$.

Possibility 1: Since $f(x)^n \to 0$ for $x \in \{0<f<1\}$, it follows from the dominated convergence theorem that

$$\mu(\{f =1 \}) = \int_{0 \leq f \leq 1} \lim_{n \to \infty} f(x)^n \, \mu(dx) = \lim_{n \to \infty} \int f(x)^n \, \mu(dx)=C,$$

where we used that we know from our previous consideration that $f \in [0,1]$ almost everywhere. Thus $$C = \int f(x) = \underbrace{\int_{f=1}\, d\mu}_{=\mu(f=1)=C} + \int_{0<f<1} f(x) \, \mu(dx),$$ i.e. $\int_{0<f<1} f(x)\,\mu(dx)=0$, implying $\mu(\{0<f<1\})=0$.

Possibility 2: We prove this by contradiction. If the set $\{x \in X \mid f(x) \in (0,1)\}$ had strictly positve measure, then

$$\int f(x) \, \mu(dx) = \int_{0<f<1} f(x)\, \mu(dx) + \mu(\{f=1\})$$

would be strictly larger than

$$\int f(x)^2 \, \mu(dx) = \int_{0<f<1} f(x)^2 \, \mu(dx) + \mu(\{f=1\}),$$

contradicting the assumption that both integrals equal the constant $C$.


问:
Loss of "significance" in $x-y$, $x=0.3721478693, y =0.3720230572$
If we perform the calculations in a computer with a five digit mantissa then I get:

$$\begin{aligned}
\mathrm{fl}(x) &=0.37215 \\
\mathrm{fl}(y) &=0.37202 \\
\mathrm{fl}(x)-\mathrm{fl}(y) &=0.00013
\end{aligned}$$

The exact calculation are: $$\begin{aligned}
x &=0.3721478693 \\
y &=0.3720230572 \\
x-y &=0.0001248121
\end{aligned}$$

So the relative error will be:
$$\left|\frac{x-y-[\mathrm{fl}(x)-\mathrm{fl}(y)]}{x-y}\right|=\left|\frac{0.0001248121-0.00013}{0.0001248121}\right| \approx 4 \%$$

One say that a loss of significance has occurred. But what have been lost? Is it numbers of significance digits in the result $x-y$? If that is correct, can one say that a loss of 5-6 significances digits has been lost?
Also, how many have been lost: 5 or 6 significant digits?

答: What has been lost is potential significant figures: it could have been seven significant figures if we did it at full precision, but we only got two. Five significant figures have been lost here.