问:
Clarification on the proof $A_n$ is simple
In Lang's Algebra book theorem 5.5 (which can be found here https://math24.files.wordpress.com/2013/02/algebra-serge-lang.pdf on page 33), I don't quite understand the last paragraph. When the author said $\sigma$ move $r,s$ other than $i,j,k$, does it mean that $\sigma$ contains the cycle $(i,j,k,r,s)$? If not, then how come $\sigma'(i) = i$?

答:
It doesn't quite mean that $\sigma$ contains the cycle $[ijkrs]$. It just means that $r$ and $s$ are also not fixed points of $\sigma$.  So the assumption is limited to $\sigma(i) = j$, $\sigma(j) = k$, $\sigma(r) \ne r$, and $\sigma(s) \ne s$ (with all $i,j,k,r,s$ distinct).  That can happen in one of several ways: it might be that $\sigma$ contains the cycle $[ijkrs]$, or it might mean $\sigma$ contains $[ijkr]$ and some other cycle permuting $s$, or it might be $\sigma$ contains $[ijk]$ and other cycle(s) permuting $r$ and $s$.

In any case, $\sigma'(j) = \tau(\sigma(\tau^{-1}(\sigma^{-1}(j))) = j$ so $j$ is a fixed point of $\sigma'$.  I'm not sure if I'm using the same multiplication convention (left-to-right vs right-to-left) as the text, so that might be why I have $j$ instead of $i$, but the argument is extremely similar no matter which convention one uses.  The key is that as long as there are two elements other than $i,j$ to work with, you can use those to form a $3$-cycle that induces an extra fixed-point after combining it with $\sigma$.

EDIT: One important detail that is understated by the text is that we must be sure that $\sigma$ and $\tau$ do not commute (else $\sigma’$ is the identity so its fixed points don’t cause a contradiction).  This can happen if we are careless with our choice of $r,s$, for instance if $\sigma$ is the $9$-cycle $[ij \,k \,xy \,r \,zw \,s]$.  However, it is easy to check that we can choose $r,s$ differently to fix this (in the case that $\sigma$ has such a large cycle, we could just pick the two elements after $k$).


问:
Closed sets in $~\mathbb R^3~$, topology
A singleton set is closed in $~\mathbb R~$ and $~\mathbb R^2~$, but it is not closed in $~\mathbb R^3~$. Why? We can say a singleton set is not closed in $~\mathbb R^3~$ as its complement is not open. But how do we prove this?
Also I can say since in $~\mathbb R~$ complement of every singleton set is open thus the singleton set is closed and same goes for  $~\mathbb R^2~$. (I hope I am correct). As in space $~\mathbb R~,~\mathbb R$ \ $\{x\}~$ can give union of two open sets.

答: $\mathbb R^n$ is a metric space, and any metric space is $T_1$.  That's equivalent to saying that points are closed.

ContentLicense: CC BY-SA 4.0


问:
Proving that $\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$ using geometry
I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.

$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$

My Attempt:

Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$.

I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this

答:
$\text{We are going to prove that}$
$\displaystyle \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}. \tag*{}$
$\textrm{Considering the roots of the polynomial equation }\\ \text{of degree 2m}$
$\displaystyle (z+1)^{2 m+1}-(z-1)^{2 m+1}=0 \cdots(*)\tag*{} $
$\displaystyle \left(\frac{z-1}{z+1}\right)^{2 m+1}=1 \tag*{}  \\$
$\displaystyle \frac{z-1}{z+1} =\cos \frac{2 k \pi}{2 m+1}+i \sin \frac{2 k \pi}{2 m+1} =e^{\frac{2 k \pi}{2 m+1} i}, $
$\text { where } k=1,2, \cdots , 2 m.$
$\textrm{Hence the 2m complex roots of (*) are}$

$ \displaystyle z =\frac{e^{\frac{2 k \pi}{2 m+1} i}+1}{1-e^{\frac{2 k \pi}{2 m+1} i}} =\frac{e^{\frac{k \pi}{2 m+1} i}-e^{-\frac{k \pi}{2 m+1} i}}{e^{-\frac{k\pi}{2 m+1} i}-e^{\frac{k \pi}{2 m+1} i}} \tag*{} $
$\displaystyle \qquad  =\frac{2 \cos \left(\frac{k \pi}{2 m+1}\right)}{-2 i\sin \left(\frac{k \pi}{2 m+1}\right)}  =i \cot \left(\frac{k \pi}{2 m+1}\right)\tag*{}  $

$\displaystyle \because \text { product of roots of }(*) =\frac{\text{constant in (*)}}{\text{leading coefficient of (*)}}\tag*{}  $
$\therefore \displaystyle \prod_{k=1} ^{2 m}\left[i \cot \left(\frac{k \pi}{2 m+1}\right)\right]=\frac{2}{2(2 m+1)} \tag*{} \\$
$\displaystyle i^{2 m} \prod_{k=1}^ {2 m}\cot \left(\frac{k \pi}{2 m+1}\right)=\frac{1}{ 2 m+1} \tag*{} $
$\because \displaystyle \cot \frac{(2 m+1-k)\pi}{2 m+1} =-\cot \left(\frac{k \pi}{2 m+1}\right)\tag*{} $
$\therefore\displaystyle (-1)^{m}(-1) ^m  \prod_{k=1}^ { m}\left[ \cot ^{2} \frac{k \pi}{2 m+1}\right]=\frac{1}{2 m+1} \tag*{} $
$\displaystyle \left(\prod_{k=1}^{m} \cot \frac{k \pi}{2 m+1}\right)^{2}=\frac{1}{2 m +1 } \tag*{}  \\$
$$\displaystyle \quad \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}$$


问:
Deriving Divergence in Spherical Coordinates (Infinitesimal Volume Method)
I am trying to derive the divergence operator in spherical coordinates using the 'cuboid' volume method, which is used in the book Div, Grad, Curl and All That by Schey, Problem II 21.

See:
Using Cylindrical Coordinates to Compute Curl
gradient and divergence using coordinate free del definition in cylindrical coordinate

I used the volume element $\Delta V = \Delta r (r \Delta \phi) (r \Delta \theta) = r^2 \Delta r \Delta \phi \Delta \theta$

($\theta$ is the azimuthal angle)

However, the result I got is the divergence operator in spherical coordinates without the $\sin \phi$.

What is the correct volume?

P.S. it seems $\Delta V = \Delta r (r \Delta \phi) (r \sin\phi \Delta \theta) = r^2 \sin\phi \Delta r \Delta \phi \Delta \theta$ is correct. But why?

Edit:
I think I found out why? $r \sin\phi$ is the projection of the radius 'vector' on the x-y plane, and $r \sin\phi \Delta \theta$ is the length of the arc swept across by $\Delta \theta$ (or rather approximately)?

答:
The answer you gave yourself in your edit is correct :)

It is not really a PROOF in the strict sense, because how do you know that approximation is sufficient? But in some ways it is better than a proof, because it gives you a good intuition of where the volume element comes from. It turns out that when $\Delta r,\Delta \phi,\Delta\theta$ are made small enough, the result limits to the right value using the approximation that the region has the shape of a rectangular prism.


问:
How to use the trace of matrix to prove this problem?

Let $A,B$ be a square matrix of order $n$ and $$A B - B A = A^m$$ where $m \geqslant 1$. Prove that $|A|=0$.

I got that

$$\mathrm{tr}(A^m)=\mathrm{tr}(AB-BA)=\mathrm{tr}(AB)-\mathrm{tr}(BA)=0$$

but I don't know the next steps.

答: If $A$ were invertible, you'd have $ABA^{-m} - BAA^{-m} = I$. But $\operatorname{tr} ABA^{-m} = \operatorname{tr} BA^{-m}A = \operatorname{tr} BAA^{-m}$, while $\operatorname{tr} I \ne 0$ (presuming the characteristic of the field is zero).


问:
Can a powerset be a member of the original set?
Within ZFC set theory, is the following statement or its negative provable?

There exists a set, $s$, whose powerset, $\mathcal{P}s$, is a member of $s$, i.e. $\mathcal{P}s \in s$.

答:
No: $s\in\wp(s)$, so if $\wp(s)\in s$, there would be an infinite descending $\in$-chain,

$$\ldots\in s\in\wp(s)\in s\in\wp(s)\in s\in\wp(s)\in s\,,$$

contradicting a consequence of the axiom of regularity.


问:
Proving that the inverse of a matrix squared is equal to first squaring the matrix then taking its inverse
If I have a matrix A, I want to show that $$(A^{-1})^2=(A^2)^{-1}$$

Sorry if it is an obvious/simple proof. Does this even require proving? This is my first time on this site. Hope someone can help

答:
Use the associativity of matrix multiplication
$$\begin{align}
(A^{-1})^2\cdot (A^2)
&= (A^{-1}A^{-1})(AA) \\
&= A^{-1}(A^{-1}A)A \\
&=A^{-1}IA \\
&= A^{-1}A \\
&= I
\end{align}$$
Verifiying that $(A^2)\cdot(A^{-1})^2 = I$ is almost the same.


问:
Prime number question
Can somebody please give me a hint on how to start this question:

Let $a$ and $n$ be two positive integers with $a,n ≥ 2$. Assume that $a^n−1$ is a prime number. Prove that $a = 2$ and $n$ is a prime number.

答:
$$
a^n - 1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-2}+\cdots+a^2+a^1+a^0).
$$
That's a factorization, showing that $a^n$ is not prime unless the first factor $a-1$ is $1$, so that $a=2$.

If $n$ is not prime, then
$$
a^n = a^{pq} = (a^p)^q,
$$
and $a^p$ is not $2$, so the same thing that happens when $a\ne 2$ happens when $a^p\ne 2$.

ContentLicense: CC BY-SA 3.0

Title: Prime number question


问:
Finding the second derivative
I have this question:
Show that
$$
\frac{\partial^2r}{\partial x^2} = \frac{y^2}{r^3}
$$
in two ways:
(1) find the $x$ derivative of
$$
\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}
$$
and (2) find the $x$ derivative of
$$
\frac{\partial r}{\partial x} = \frac{x}{r}
$$
by the chain rule.

I am quite lost.  I mean I can take the $x$ derivative of the two equations, but I don't seem to in any way connect it to
$$
\frac{\partial^2 r}{\partial x^2} = \frac{y^2}{r^2}.
$$

Can anybody help?

David

EDIT
It is supposed to be
$$
r^3
$$
in the denominator.

答:
(1)
$$
\begin{align}
\frac{\partial^2 r}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial r}{\partial x} \right) \\
&= \frac{\partial}{\partial x} \left( \frac{\partial}{\partial x} \sqrt{x^2 + y^2} \right) \\
&= \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2 + y^2}} \right) \\
&= \frac{\sqrt{x^2 + y^2} \cdot 1 - x \cdot \frac{x}{\sqrt{x^2 + y^2}}}{\left(\sqrt{x^2 + y^2}\right)^2} \\
&= \frac{(x^2 + y^2) - x^2}{\left(\sqrt{x^2 + y^2}\right)^3} \\
&= \frac{y^2}{r^3}
\end{align}
$$
(2)
$$
\begin{align}
\frac{\partial^2 r}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial r}{\partial x} \right) \\
&= \frac{\partial}{\partial x} \left( \frac{x}{r} \right) \\
&= \frac{r \cdot \frac{\partial x}{\partial x} - x \cdot \frac{\partial r}{\partial x}}{r^2} \\
&= \frac{r - x \cdot \frac{\partial r}{\partial x}}{r^2} \\
&= \frac{r - x \cdot \frac{x}{r}}{r^2} \\
&= \frac{r^2 - x^2}{r^3} \\
&= \frac{y^2}{r^3} \\
\end{align}
$$

Title: Finding the second derivative


问:
Why is the angle that maximizes $a \cos(\theta) + b \sin(\theta)$ in the direction of $(a, b)$?
I'm wondering if there is an intuitive explanation to the result above. Mathematically, it is easy to prove: If we let $x = a \cos(\theta) + b \sin(\theta)$, then setting $\frac{\mathbb{d}x}{\mathbb{d}\theta} = 0$ tells us that $\theta = \arctan(\frac{b}{a})$. Thus $\theta$ is the angle between the positive x axis and the vector $(a, b)$. Another way to show this is to notice that $x = (a, b) \cdot (\cos(\theta), \sin(\theta)) = ||(a,b)||\cos(\alpha)$. If $\cos(\alpha) = 1$, then $x$ will be maximized and will be in the direction of $(a, b)$, since $\alpha = 0$.

At the same time, the answer just seems so simple that I can't help but wonder whether there's a more obvious way to look at it. Trigonometry is not really inherent to the problem, $\cos(\theta)$ and $\sin(\theta)$ can also be replaced by the components of some vector $\vec{v}$.

答: $$\frac{x}{\sqrt{a^2+b^2}} = \frac{a}{\sqrt{a^2+b^2}} \cos \theta + \frac{b}{\sqrt{a^2+b^2}} \sin \theta,$$ so if we let $$\varphi = \arctan \frac{b}{a},$$ we have $$\cos \varphi = \frac{a}{\sqrt{a^2+b^2}}, \quad \sin \varphi = \frac{b}{\sqrt{a^2+b^2}},$$ hence $$x = \sqrt{a^2+b^2} ( \cos \varphi \cos \theta + \sin \varphi \sin \theta) = \sqrt{a^2 + b^2} \cos (\theta - \varphi).$$  Since $a, b$ are constants, it follows that $x$ is maximized whenever $\cos (\theta - \varphi) = 1$, or $\theta = \varphi + 2\pi k$ for some integer $k$.  This completes the proof.


问:
How Do Mathematicians ‘make up’ or perhaps more correctly ‘derive’ equations
I know this is definitely a thing otherwise we wouldn’t have equations.

What are the methods for making up an equation from scratch and from making up an equation from an existing equation?

The only way I have been able to do this is using dimensional analysis and I’d like to learn other ways, if you want to see an example of me using dimensional analysis to ‘make up an equation’ see my answer to this question.

How for example can you make up the following without experimentation:

Speed equation:
$$\text v=\frac{\text d}{\text t}$$

Density equation:
$$\text p=\frac{\text m}{\text v}$$

After seeing some answers, how do you make up an equation for a definition? @littleO tells me the velocity equation is a definition and I’d like to know how the guy who made the equation went about it.

Or how do you look at a square and come up with a quadratic equation?

Are there any equations you can make up without experimentation?

If this is too general, specific examples are welcome.

答:
You always need to remember, that "equations" are nothing more than formalized language. Before people "invented" equations, they argued with words, which today are a pain to read. Here is Newtons Second Law (summed up nicely in $\vec{F}=m\vec{a}$):

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

So math, as you may learned it in school may be terrifying because much teaching is done using the formal set of rules, rather than words. However, what is important are the ideas that are conveyed either with words, or with the equations. To go from one equation to another, we simply use clever tautologies, definitions of certain variables, that replace other variables (in the $v$ example, $\frac{d}{s}$)

Let's do a basic example: Let's assume all you own is a bag of apples, of unknown quantity. Now let's say I give you an apple. How many apples do you have?

Clearly, you have 1 apple, plus the amount of apples in the bag, no? So now let's define some variables:

$A$ is the amount of apples you have after I give you an apple

$b$ is the amount of apples there are in the bag.

Going off of these definitions, we can say the following:

$$A=1+b$$

This makes sense, no? You definitely have one apple, in addition to those that are in the bag. Furthermore, you cannot have more apples, because we assumed that you don't have anything else (note how we are assuming things). Et voilà, you have an equation with certain dimensional values!

Now let's get funkier. Let's say that the amount of apples in the bag is equal to the amount of apples my grandmother has. So if:

$g$ is the amount of apples my grandmother has

then you know

$$g=b$$

Now you see, that there is no need for you to know explicitly and only how many apples are in the bag. If you know the amount of apples my grandmother has, you know the amount of apples in your bag. To formalize this in equation form, this:
$$A=1+b$$

becomes this:
$$A=1+g$$

because $g=b$. Now you have another equation. Clearly, you can own trucks with apples, divisions of apples, square root of the amount of apples that the CEO of ExampleFirm makes in USD per year, per store location, etc. You can formalize what I'm saying, but as you can tell, just because you have an equation, doesn't mean it's correct. That's why you need to assume that certain things are the case, in order for the equation to make any real world sense.

You can graph virtually any equation. It might be a boring graph (all variables defined), but remember: Even a graph is just a tautology, i.e. a way to convey the information that is in the equation (which in turn, is the information that is given linguistically). So when you ask "How do I make up an equation", what you are... really asking... is "How do I form words" or "How do I make an argument" or "How do I say something that can be quantified", and the answer, as far as I can tell, is "You quantify, and then say what you are quantifying".


问:
Probability of rolling a 6 immediately after a 1 is rolled
Question

Ann and Bob take turns to roll a fair six-sided die. The winner is the first person to roll a six immediately after the other person has rolled a one. Ann will go first. Find the probability that Ann will win.

Answer

$\mathbb{P} (\mathrm {Ann\ wins}) = \frac {36} {73}$

I have thought long and hard about this question but I am unable to even start. I have tried considering cases, but got stuck along the way. For example, it is trivial to calculate the probability that Ann wins if there are only three rolls (which is the minimum number of rolls needed for Ann to win). However, the problem easily becomes very complicated when we consider five rolls and more.

The suggested solution by my professor uses first-step decomposition, but it is a new concept to me and I am struggling to understand it.

If anyone can provide a detailed and intuitive explanation as to how this problem should be solved, that will be greatly appreciated!

答:
This is my attempt at using Markov chains to solve the problem.

Following one of the answers posted, we have $4$ active states and $2$ absorbing states.

Let the $4$ active states be $(A, 0)$, $(A, 1)$, $(B, 0)$ and $(B, 1)$, where

$(A,0)$ is the state in which it is $A$'s turn and the prior flip was not a $1$,

$(A,1)$ is the state in which it is $A$'s turn and the prior flip was a $1$,

$(B,0)$ is the state in which it is $B$'s turn and the prior flip was not a $1$ and

$(B,1)$ is the state in which it is $B$'s turn and the prior flip was a $1$.

Let the $2$ absorbing states be $(A)$ and $(B)$, where

$(A)$ is the state in which $A$ wins and

$(B)$ is the state in which $B$ wins.

Note also that $(A, 0)$ shall be the state in which the game starts.

In order of $(A, 0)$, $(A, 1)$, $(B, 0)$, $(B, 1)$, $(A)$ and $(B)$, the corresponding transition matrix shall be

$$T = \begin{pmatrix}
0 & 0 & \frac 5 6 & \frac 1 6 & 0 & 0 \\
0 & 0 & \frac 4 6 & \frac 1 6 & \frac 1 6 & 0\\
\frac 5 6 & \frac 1 6 & 0 & 0 & 0 & 0 \\
\frac 4 6 & \frac 1 6 & 0 & 0 & 0 & \frac 1 6 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}.$$

Then, we are simply interested in $T^n_{1, 5}$ as $n \rightarrow \infty$, which can be easily computed to give approximately $0.493$, as required.

P.S. This is my very first time working with Markov chains and all of my learning regarding this concept is independent, so do feel free to provide any comments on how I can improve my working! I do believe I will be covering Markov chains soon, so this could be a nice head start!


问:
Is there a simple criterion for when $e^{x}+ ax^{2}+ bx+ c$ has a zero?

Is there a simple criterion for when
$$e^{x}+ ax^{2}+ bx+ c$$
has a zero ?

Source: MathOverflow/@MattF.

one simpler function $e^{x}+ ax+ b$ has a zero iff either or $- a+ a\!\ln\!\left ( -a \right )+ b\!\leq 0>\!a,$ or $a\!= 0>\!b,$ or $a> 0,$ where the last clause comes from using calculus to minimize the function.

First edit. It might be possible to extract something from Wilkie's approach in his

paper "Model completeness results for expansions of the ordered field of real numbers by restricted Pfaffian functions and the exponential function," which implies, among other things, that there is a system of exponential equations and inequalities that has a solution iff $e^{x}+ ax^{2}+ bx+ c$ doesn't.

Second edit. There is an algorithm to determine whether $e^{x}+ ax^{2}+ bx+ c$ has a zero :

simultaneously check for each positive rational $q$ whether $e^{q}+ aq^{2}+ bq+ c< 0,$ for each $n$ whether for all the positives $x$ so that $ax^{2}+ bx+ c+ \sum_{k= 0}^{n}\frac{x^{k}}{k!}\geq 0,$ and something similar for the negatives $x.$ The first piece comes from the proof of the transcendence of $e,$ like_ https://mathoverflow.net/a/378064/164469, and the second is standard Tarski-Seidenberg quantifier elimination. But this algorithm does not provide a time bound for the algorithm or a single set of discriminants which can be used for all $a, b, c.$

答:
You basically ask when

$f(x)=e^x+c$ and $g(x)=-ax^2-bx=-x(ax+b)$ cross each other.

Always when $a < 0$.

For $a \geq 0$ it is as follows.

The reason we have chosen the above two functions is that it is easier to visualize what is going on. $g(x)$ has one of the legs through the origin and $e^x$ is elevated by $c$. For such arrangement, there is one critical value $c_0$ when $f(x)$ and $g(x)$ are touching each other. Then, if $c > c_0$, $f(x)$ cannot cross $g(x)$ or if $c < c_0$ they will have two intersections.

When $f(x)$ is touching $g(x)$ it is at that point $f'(x_0)=g'(x_0)$ giving us the condition:

$$e^{x_0}=-2 a x_0 - b$$

which we can effectively solve and find $x_0$. Another condition we are getting from replacing $e^{x_0}$ into $f(x_0)=g(x_0)$ which is a common point.

$$-2 a x_0 - b+c_0=-ax_0^2-bx_0$$

giving

$$c_0=-ax_0^2-bx_0+2 a x_0 + b$$

So the condition is

$$c < c_0=-ax_0^2 +(2 a-b) x_0 + b \text{ for } x_0 = -W(\frac{e^{\frac{-b}{2a}}}{2a}) - \frac{b}{2 a}$$

If $a = 0$ we have always the solution if $b>0$ and otherwise

$$e^{x_0}= - b, x_0=\ln(-b)$$

$$c_0=b-bx_0$$

or $$c < b(1-\ln(-b)), b<0$$

If $a = b = 0$ we have the solution only if $c < 0$.


问:
How can I draw the link between negation inconsistency and this result?
If Γ ⊢ ∃x∃y (φ(x)∧φ(y)∧x≠y) , how does adding ∀x(x=c↔φ(x)) result in an inconsistent set?

I know that inconsistent has two variants, namely negation inconsistency and absolute inconsistency? How can I prove the former?

答:
The first formula asserts that there are two different objects that are $\varphi$.

The first formula implies that every object that is $\varphi$ is equal to $c$.

From the first one, by Existential instantiation,we get: $\varphi(c_1)$ and $\varphi(c_2)$, with $\lnot (c_1 = c_2)$.

Using the second one, by Universal instantiation, we get: $c_1=c$ and $c_2=c$, from which [transitivity of equality] we have: $c_1=c_2$, and thus the contradiction.


问:
Is there a non-standard definition of an absolute value?
So I was thinking about how we can define a number $x$ such that $|x|=-1$. This topic was already talked about so many times as we can see here or there. So I wanted to consider it from a different angle.

Firstly, I tried to define what an absolute value on $\mathbb{R}$ is. I want to say that the function $f: \mathbb{R} \to \mathbb{R}$ with the following properties is an absolute value function.

$\forall x\in \mathbb{R},f(x)\geq 0$

$\forall x\in \mathbb{R}, f(x)=0 \Leftrightarrow x=0$

$\forall x,y\in \mathbb{R}, f(xy)=f(x)f(y)$

$\forall x,y\in  \mathbb{R}, f(x+y) \leq f(x)+f(y)$

I would like to know if there is a function with those properties other than the standard $f(x)=|x|$. If there is, what condition do we need so that $|x|$ will be the only function with such properties? Do we need $f(f(x))=f(x)$ or $f(-x)=f(x)$?

答:
Let me say a few words in addition to the trivial counter example I gave in the comments. Let's make a list of observations based on the properties:

$f(0)=0$, by property 2.

$f(1)=[f(1)]^2$, and $f(1)\neq 0$ (by property 2) so $f(1)=1$.

Similarly, $1=f(1)=[f(-1)]^2$, so $f(-1)=1$ or $f(-1)=-1$, but by property 1, we need $f\geq 0$, so we must have $f(-1)=1$.

Hence, $f(-x)=f(-1)f(x)=f(x)$, so $f$ is even.

As a result, in order to understand the behavior of $f$, we may restrict attention to the interval $[0,\infty)$. As mentioned in the comments, it is possible to have a discontinuous $f$ which satisfies all four properties, and we know that the absolute value function is continuous, so continuity of $f$ is an extra assumption which we have to make.

Next, property (3) together with continuity, $f(0)=0$ and $f(1)=1$ is a very classical functional equation (I'm not an expert here so I'm not sure what it's called). But essentially, from this it follows that there is an $\alpha>0$ such that for all $x\geq 0$, $f(x)= x^{\alpha}$. Hence, by evenness of $f$, it follows that for all $x\in\Bbb{R}$, $f(x) = |x|^{\alpha}$.

We would like to somehow force $\alpha =1$. Well, observe that when restricted to $(0,\infty)$, we have

$f$ is strictly concave if $0<\alpha<1$, and hence $f(x+y)< f(x)+f(y)$ for all $x,y\in (0,\infty)$

$f$ is strictly convex if $1<\alpha$, and hence $f(x+y)> f(x)+f(y)$ for all $x,y\in (0,\infty)$.

So, if we want our $f$ to satisfy property (4), we must have $0<\alpha \leq 1$.

In other words, consider the following statements about a function $f:\Bbb{R}\to\Bbb{R}$

$f\geq 0$

for every $x\in\Bbb{R}$, $f(x)=0$ if and only if $x=0$.

for every $x,y\in\Bbb{R}$, $f(xy)=f(x)f(y)$.

for every $x,y\in\Bbb{R}$, $f(x+y)\leq f(x)+f(y)$.

$f$ is continuous.

What my discussion above shows is that these five conditions imply there exists a $0<\alpha \leq 1$ such that for all $x\in\Bbb{R}$, $f(x)=|x|^{\alpha}$. Conversely, given any $0<\alpha\leq 1$, the function $f(x):= |x|^{\alpha}$ satisfies these $5$ conditions. Because of this equivalence, these five properties are not enough to single out $\alpha =1$.

If you add in a sixth condition that

for every $x\in \Bbb{R}$, $f(f(x))=f(x)$,

then yes, this is necessary and sufficient to ensure $\alpha =1$. So, the sixth property is essential.

But of course, note that we can safely remove property (4), and replace it with $(6)$, because (1),(2),(3),(5) is equivalent to the existence of some $\alpha>0$ such that $f(x)=|x|^{\alpha}$. If we add in (6), this forces $\alpha=1$, so (4) is redundant as part of the assumptions, and it comes out for free.


问:
$\overline{A \cup B}$ is read as "neither $A$ nor $B$", how about $\overline{A \cap B}$?
$\overline{A \cup B}$ is read as "neither $A$ nor $B$", how about $\overline{A \cap B}$?

I know that $\overline{A \cap B}$ equals to $\overline{A}\cup \overline{B}$ which can be read as "not A or not B". But it is not what I am looking for.

答:
One thing that can cause trouble in understanding set theory (or algebra of sets) is not to distnguish rigorously enough sets (or set expressions) and propostions.

Set expressions are just symbols, they are not true or false (though they have a denotation, a reference). Propositions are true or false; they have meaning or sense.

The proposition $\neg (A\lor B)$ is read " it is not the case that (A is true or B is true)". (Note that , here, A and B are propositions themselves, not sets).

But the correspondng set expression is read " the complement of the union of set A and set B". That is: " the set of all x such that it is not the case that (x belongs to set A or x belongs to set B)". So, we use the NOR (logical) operator to define the complement of  a $\cup$ - set.

The set expression you are trying to read is " the complement of the intersection of set A and of set B".


问:
Determine if the vectors form a subspace of $\mathbb{R}^{4}$
I have an exercise about determining if the vectors form a subspace of $\mathbb{R}^{4}.$ I do not understand much about this. Could you please guide me on how to do it? Thank you so much, here is the exercise

答:
If $W\subset V$, and $V$ is a vector space, it is enough to check:

For every $w_1,w_2 \in W$, $w_1+w_2 \in W$

For every $w\in W$ and scalar $\alpha\in \Bbb F$, $\alpha w\in W$

for $W$ to be a subspace of $V$. Goes without saying, every subspace must contain $0\in V$. Can you take it from here?

To answer your particular question:

(a) Yes, this is a subspace. In fact, any equation of the form $ax_1 + bx_2 + cx_3 + dx_4 = 0$ defines a subspace of $V$, where $a,b,c,d\in\mathbb F$.

(b) Not a subspace! $x_1x_3 = x_2x_4$ is anyway a non-linear relation between the coordinates of $x$, so I wouldn't expect it to define a subspace. It suffices to find a counterexample, as done by @X.Li in the comments.

(c) Not a subspace! $x_3 \ge x_4 \implies -x_3 \le -x_4$, so if $z\in \{x\in\mathbb R^4: x_3\ge x_4\}$, then $-z\notin \{x\in\mathbb R^4: x_3\ge x_4\}$.


问:
References for Algebraic number theory
I am doing algebraic number theory first time. I have done all ring theory and field theory. I am interested in algebra , so also pretty much excited about algebraic number theory. I have a month's break before my semester begins. I want to self study some beginner's text covering my syllabus so that in class I can understand it much better, and may be it will turn up a research interest for me. Can somebody please suggest me some wonderful text for beginners in algebraic number theory, I do not want any number theory book using analytical methods. Strictly algebraic. Here are my contents...

Course Outline

Characteristic and minimal polynomial of an element relative to a
finite extension, Equivalent definitions of norm and trace, Algebraic
numbers, algebraic integers and their properties.

Integral bases, discriminant, Stickelberger’s theorem, Brille’s
theorem, description of integral basis of quadratic, cyclotomic and
special cubic fields.

Ideals in the ring of algebraic integers and their norm, factorization
of ideals into prime ideals, generalised Fermat’s theorem and Euler’s
theorem.

Dirichlet’s theorem on units, regulator of an algebraic number fields,
explicit computation of fundamental units in real quadratic fields.

Dedekind’s theorem for decomposion of rational primes in algebraic
number fields and its application, splitting of rational primes in
quadratic and cyclotomic fields.

Any book that covers all these with best conceptual approach. Thanks!

Anyways here are some suggested readings by my institute-

Saban Alaca and Kenneth Williams, Introductory Algebraic Number Theory, Cambridge University Press (2003).

M. Ram Murty and J. Esmonde Problems in Algebraic Numbers Theory, Springer-Verlag (2004).

Erich Hecke, Lectures on the Theory of Algebraic Numbers, Springer-Verlag (1981).

Paula Ribenboim, Algebraic Numbers, John Wiley & Sons (1972).

Harry Pollard and Harold Diamond, The Theory of Algebraic Numbers, Dover Publications (2010).

Which one I should go for among them,  or is there some other than these all, please suggest if any? Thanks!

答:
Here is a list of recommended texts from my institute
K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, 2nd Edition, Springer-Verlag, Berlin,
1990.S. Lang, Algebraic Number Theory, Addison- Wesley,
1970.D. A. Marcus, Number Fields, Springer-Verlag, 1977.</p>

Title: References for Algebraic number theory


问:
Range of $x$ for which the modulus inequality will be valid : $|x^2-2x-8| > 2x$ : Need clarification about solution set
$|x^2-2x-8| > 2x \\ $

For this problem I took two cases where in one case $ x^2-2x-8 \geq 0$ and in second case $ x^2-2x-8 \lt 0$.

$1st$ Case  when $ x^2-2x-8 \geq 0$ :-

$ x^2-2x-8 \gt 2x$

For this case I got the solution set as $x \in (-\infty,-2] \cup \biggl(2(1+\sqrt{3}),\infty\biggr)$ upon solviing the above two equations i.e. $ x^2-2x-8 \geq 0$ and $ x^2-2x-8 \gt 2x$ and then taking the common range of values of $x$ from the solutions of the two equations.

$2nd$ Case  when $ x^2-2x-8 \lt 0$ :-

$ -(x^2-2x-8) \gt 2x$

For this case I got the solution set as $x \in (-2,2\sqrt{2})$

I am getting confused how to decide the overall solution set for the original problem as a whole because if we plot the solution set on the number line for both the above cases then as per my understanding the common range from both the solution set should be the answer and if that's the case the answer should be $x \in (-2,2)$ as this is the range which is common to the two solution set but unfortunately the answer is different.

Can someone please help me on this and help with some clarification with in these cases how do you decide the solution set?

答:
Consider the first part. You wrote that you were dealing with the case in which $x^2-2x-8\geqslant0$. That means that $x\leqslant-2$ or that $x\geqslant4$. Then you solved the inequation $x^2-2x+8>2x$, and you should have got that $x>2+2\sqrt3$ or that $x<2-2\sqrt3$. So, the solution when you are in the first situation is that $x\in(-\infty,-2]\cup\left[2+2\sqrt3,\infty\right)$

Now, the second part. Asserting that $x^2-2x-8<0$ means that $x\in(-2,4)$. And asserting that $-(x^2-2x-8)>2x$ means that $x\in\left(-2\sqrt2,2\sqrt2\right)$. So, the solution when you are in the second situation is that $x\in\left(-2,2\sqrt2\right)$.

So, the global answer is that $x\in\left(-\infty,2\sqrt2\right]\cup\left[4,\infty\right)$.


答:
You should take the union of the two region as you want to collect all the possible $x$ for the conditions to be true.

If you do so, you get $(-\infty, 2\sqrt2) \cup (2(1+\sqrt{3}, \infty)$.

Title: Range of $x$ for which the modulus inequality will be valid : $|x^2-2x-8| > 2x$ : Need clarification about solution set

Body:
$|x^2-2x-8| > 2x \\ $

For this problem I took two cases where in one case $ x^2-2x-8 \geq 0$ and in second case $ x^2-2x-8 \lt 0$.

$1st$ Case  when $ x^2-2x-8 \geq 0$ :-

$ x^2-2x-8 \gt 2x$

For this case I got the solution set as $x \in (-\infty,-2] \cup \biggl(2(1+\sqrt{3}),\infty\biggr)$ upon solviing the above two equations i.e. $ x^2-2x-8 \geq 0$ and $ x^2-2x-8 \gt 2x$ and then taking the common range of values of $x$ from the solutions of the two equations.

$2nd$ Case  when $ x^2-2x-8 \lt 0$ :-

$ -(x^2-2x-8) \gt 2x$

For this case I got the solution set as $x \in (-2,2\sqrt{2})$

I am getting confused how to decide the overall solution set for the original problem as a whole because if we plot the solution set on the number line for both the above cases then as per my understanding the common range from both the solution set should be the answer and if that's the case the answer should be $x \in (-2,2)$ as this is the range which is common to the two solution set but unfortunately the answer is different.

Can someone please help me on this and help with some clarification with in these cases how do you decide the solution set?


问:
How to prove gamma coefficient is zero, when (X,Y) is independent?
$\pi_c=P(concordant)=2P(X_1>X_2,Y_1>Y_2)=2E{(X_1>X_2,Y_1>Y_2)}$
$=2\sum_{i=1}^{I}\sum_{j=1}^{J}P[X_1>X_2=i,Y_1>Y_2=j]\pi_{ij}$$=2\sum_{i=1}^{I}\sum_{j=1}^{J}(\sum_{h>i}^{}\sum_{k>j}^{}\pi_{hk})$

$\pi_d=P(disconcordant)=2P(X_1>X_2,Y_1<Y_2)=2\sum_{i=1}^{I}\sum_{j=1}^{J}(\sum_{h>i}^{}\sum_{k<j}^{}\pi_{hk})$

The gamma coefficient is $\gamma=\frac{\pi_c+\pi_d}{\pi_c-\pi_d}$

Why gamma coefficient $=\pi_c-\pi_d=0$, when (X,Y) is independent?

Please show me the process.

I appreciate your any help.

答:
Suppose $(X_1,Y_1)$ and $(X_2,Y_2)$ are i.i.d pairs of observations on $(X,Y)$.

Then probability of concordance and discordance is respectively

$$\pi_c=P\{(X_1-X_2)(Y_1-Y_2)>0\}$$

and $$\pi_d=P\{(X_1-X_2)(Y_1-Y_2)<0\}$$

If $X$ and $Y$ are independent and continuous, then $X_1$ and $X_2$ have the same distribution. Moreover, $X_1-X_2$ is independent of $Y_1-Y_2$.

Hence, under independence of continuous random variables,

\begin{align}
\pi_c&=P\{X_1>X_2\}P\{Y_1>Y_2\}+P\{X_1<X_2\}P\{Y_1<Y_2\}
\\&=P\{X_1<X_2\}P\{Y_1>Y_2\}+P\{X_1>X_2\}P\{Y_1<Y_2\}
\\&=P\{(X_1-X_2)(Y_1-Y_2)<0\}
\\&=\pi_d
\end{align}


问:
Find the rational points on $1 + 18 x + 81 x^2 + 44 x^3 = y^2$ with Sage
I'm trying to use Sage on-line,but I meet some trouble with the code of it.
I want to find the rational points on an ellipse curve,such as $$1 + 18 x + 81 x^2 + 44 x^3 = y^2,\tag1$$
I know that $(x,y)=(0,1)(1,\pm12)(-\frac{1}{11},0)$ are on the curve,but I don't know how to find more,so I try to solve it with the math-software Sage.

What should I input?Thanks in advance!

Edit: Multiply by $44^2$,we can difine the ellipse curve by this way,

E = EllipticCurve([0,81,0,44*18,44^2])

Then how to find the rational points on it?

答:
By the Mordell-Weil theorem, the group of rational points $E(\mathbb{Q})$ is finitely generated and therefore isomorphic to a group of the form $T\oplus \mathbb{Z}^R$, where $T$ is finite abelian formed by those points of finite order, and $R\geq 0$ is the rank of the elliptic curve, and it counts how many $\mathbb{Z}$-linearly independent points of infinite order there are.

With Sage, define your curve by

E=EllipticCurve([0,81,0,44*18,44^2]);

Now you can ask Sage to calculate the rank $R$ with

E.rank();

and the answer is $0$. So there are no points of infinite order, only finite order points. Let us calculate the torsion subgroup as follows

T=E.torsion_subgroup();

if you type ``T;'' and evaluate, Sage will tell you that $T\cong \mathbb{Z}/6\mathbb{Z}$. Finally, if you type

T.0;

Sage returns a generator of the torsion subgroup, namely $P=(44,528)$, which is a point of exact order $6$. Hence, $E(\mathbb{Q})=\langle P \rangle$ or
$$E(\mathbb{Q})=\langle P \rangle = \{ \mathcal{O}, (44,528),(0,44), (-4,0), (0,-44), (44,-528)\}.$$


答:
You seem to have taken the range common to the two solution sets. In other words, you have taken the intersection of the two sets, which is also done incorrectly because the intersection of the two cases you have taken is a null set because those two conditions cannot be true simultaneously. $x^2-2x-8 \geq 0 $ AND $x^2-2x-8 < 0 $ is not possible.
$$ \{x^2-2x-8 \geq 0 \} ~ \cap ~ \{x^2-2x-8 < 0 \}  = \phi$$
The final solution set that you are looking for is the union of the two sets because either $x^2-2x-8 \geq 0 $ OR $x^2-2x-8 < 0 $. OR signifies that you are supposed to take the union of the sets.
$$ \{x^2-2x-8 \geq 0 \} ~ \cup ~ \{x^2-2x-8 < 0 \}  = (-\infty, 2\sqrt2) \cup (2+2\sqrt{3}, \infty) $$


问:
Simplex Algorithm and Unboundedness
When maximizing an objective function with the simplex algorithm, if there exist a positive reduced cost with all negative entries in the column, we then know that the solution is unbounded. The question is, is there a way to sniff out possible unboundedness before even starting the simplex algorithm?

For example,

Producing X requires 3 of resource A, and creates 3 of resource B as a by-product.

Producing Y requires 3 if resource B, and creates 4 if resource A as a by-product.

This is not from a real scenario, merely an example of conditions that will allow us to create infinite amount of product X, and thus there is no optimal solution, the maximum objective value is unbounded.

答: One possibility is to look at the dual problem. If the dual is infeasible, and you know the primal problem is feasible, then the primal problem must have an unbounded optimum. However, this depends on whether it is easy to see the dual is infeasible. In general, it is as difficult detecting infeasibility as solving a linear program. So I guess, the answer to your question in general is no, unless the problem has some special structure.</p>

Title: Simplex Algorithm and Unboundedness


问:
Understanding Humphreys LIe algebras, page 32
I have two questions.

First question is about Remark in page 32. Let $x, y, h$ be a standard basis of $\mathfrak sl(2, F).\,V_\lambda$ is a weight space. We have the following lemma :

Lemma. If $v \in V_\lambda,$ then $x.v \in V_{\lambda+2}$ and $y.v \in V_{\lambda-2}.$

And here is the remark related to this lemma.

Remark. The lemma implies that $x, y$ are represented by nilpotent endomorphisms of $V.$

Here's my opinion.

For any $v \in V, v \in V_\lambda$ for some $\lambda.$ Then, $x.v \in V_{\lambda+2} \Rightarrow x^2.v \in V_{\lambda+4} \Rightarrow ...\Rightarrow x^n.v \in V_{\lambda+2n}.$ Since $V$ is finite dimensional, $V_{\lambda+2n}=0$ for some $n.$ Hence $x$ is nilpotent. Nilpotency of $y$ can be shown similarly.

Am I right ?

Second question is about the paragraph right next to the proof of lemma in the same page 32.
It says that

Thanks to formula (a), the nonzero $v_i$ are all linearly independent.

And again, here's what I thought.

Formula (a) is $h.v_i=(\lambda-2i)v_i.$ Suppose that $\sum c_iv_i=0.$ By (a), $0=h(0)=h(\sum c_iv_i)=\sum(\lambda-2i)c_iv_i.$
Again, $0=h(0)=h(\sum(\lambda-2i)c_iv_i)=\sum(\lambda-2i)^2c_iv_i.$ Repeating this process, we obtain

$$
    \begin{pmatrix}
    1 & 1 &\cdots& 1 \\
    \lambda-2 & \lambda-4 &\cdots& \lambda-2n \\
    \vdots & \vdots & \cdots & \vdots \\
    (\lambda-2)^{n-1} & (\lambda-4)^{n-1} & \cdots & (\lambda-2n)^{n-1} \\
    \end{pmatrix}  \begin{pmatrix} c_1v_1\\ c_2v_2\\ \vdots \\ c_nv_n \end{pmatrix}=0
$$

Since the left matrix is invertible (indeed, it is Vandermonde matrix), we have $c_iv_i=0$ for each $i.$ Thus $c_i=0$ for all $i$ and hence $v_i$'s are linearly independent.

Am I correct ? If I'm wrong, could you give me some hint ?
Thanks for reading the long questions and thanks for your answer in advance.

答: You are correct in both cases. But the second assertion is a particular case of the general Linear Algebra fact that eigenvectors with distinct eigenvalues are always linearly independent.</p>


问:
Find a value of $\;\lim\limits_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$

I'm going to give you an extremal problem.

Given $a_{n}:=\left ( \text{the solution for the equation}\,x^{n}= \cos x,\quad x> 0 \right ),$ find a value of
$$\lim_{n\rightarrow\infty}n\left ( 1- a_{n} \right )$$
Source: Fujino_Yusui

If you imagine the graph of $x^{n},$ it should intersect at the point where it increases rapidly by $1,$ which is about where $x^{n}= \cos 1\,(y$ is increasing rapidly, so if you look at $y,$ you should be able to approximate $x$ well enough$).\,n\left ( 1- \cos^{\frac{1}{n}}1 \right )$ looks good, and $\frac{1}{n}= h$ is the derivative of $h\rightarrow 0^{+}.$ I guess $-\ln\cos 1$ by definition.

答: Look at the function $f(x) = x^n - \cos x\implies f'(x) = nx^{n-1}+\sin x > 0$ on $(0,1)$ and $f(0)\cdot f(1) < 0 \implies \exists ! a_n\in (0,1): f(a_n) = 0\implies 0 < a_n < 1$. Next you show $\{a_n\}$ is an increasing sequence ( I leave this for OP to tackle : it will be a good wrestling match for ya ) and you have to use the $x^n = \cos x$ to do it !. After that then it has a limit $L$ and $L = \left(\cos L\right)^{0} = 1$. Thus: $n(1-a_n)= -n(-1+\sqrt[n]{\cos a_n})\to -\ln(\cos 1)$. Note that you can show: $n(\sqrt[n]{L} - 1) \to \ln L, L > 0$


问:
Absolutely Convergent Complex Power Series Implies Uniform Convergence, How about Not Absolute?
There is a theorem in complex analysis stated as "if a complex power series (centered at z_0) converges absolutely at z = z_1, then it converges uniformly in region: |z-z_0| <= |z_1-z_0|."

I know how to prove this theorem, but I wonder what will happen if we loosen the restriction in the statement - absolute convergence - to just convergence? Does that hold, as well?
If not, is there any counter example? Or if you could kindly help me to understand this statement better, I really appreciate that.

The statement just reminds me of the difference of "conditional convergence" and "absolute convergence" in real analysis, so quite naively, I bring this idea to complex analysis. Sorry if it's a dumb question.

答:
It is not true. For instance, the series $\sum_{n=1}^\infty\frac{(-1)^n}n$
converges, but the series $\sum_{n=1}^\infty\frac{z^n}n$ does not converge uniformly when $|z|\leqslant1$. Actually, it diverges when $z=1$.

But the absolute convergence still holds. In fact, it is enough to assume that the sequence $\left(a_n(z_1-z_0)^n\right)_{n\in\Bbb N}$ is bounded. If $B$ is such that $(\forall n\in\Bbb Z_+):\left|a_n(z_1-z_0)^n\right|\leqslant B$ and if $|z-z_0|<|z_1-z_0|$, then, if $n\in\Bbb N$,\begin{align}\sqrt[n]{\left|a_n(z-z_0)^n\right|}&=\sqrt[n]{\left|a_n(z_1-z_0)^n\left(\frac{z-z_0}{z_1-z_0}\right)^n\right|}\\&\leqslant\sqrt[n]B\left|\frac{z-z_0}{z_1-z_0}\right|\end{align}and so, since $\lim_{n\to\infty}\sqrt[n]B=1$, if you take some $c\in\left(\left|\frac{z-z_0}{z_1-z_0}\right|,1\right)$, you have $\sqrt[n]{\left|a_n(z-z_0)^n\right|}\leqslant c$ if $n$ is large enough. So, by the root test, your series converges absolutely.


问:
If a function is not defined at a point then will its inverse be defined at that point?
I have a function $y=\frac{x-2}{x-3}$ clearly the function is not defined at $x= 3$

But for its inverse $ \frac{3x-2}{x-1},$ can we say that it is defined at $x= 3$?

答: Sure. The domain of the inverse of $f$ (if it exists) is the range of $f$. And the fact that $3$ belongs or not to the domain of $f$ is independent of the fact that it belongs or not to the range of $f$.


答:
Yes your map in indeed defined at $x=3$.

In general, for $A \subseteq B$ and a map $f : A \to B$, there is no relationship between the fact that $f$ is defined at $b \in B \setminus A$ and the fact that $f^{-1}$ is defined at $b$.

$f^{-1}$ will be defined at $b \in B$ providing that:

$b$ belongs to $f(A)$.

The inverse image of $b$ under $f$ is unique.

Title: If a function is not defined at a point then will its inverse be defined at that point?

Body:
I have a function $y=\frac{x-2}{x-3}$ clearly the function is not defined at $x= 3$

But for its inverse $ \frac{3x-2}{x-1},$ can we say that it is defined at $x= 3$?


问:
Problem: Sum of absolute values of polynomial roots
Can you please give me some hints as to how I might approach this problem? Thanks!

Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$

Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$

答:
You can find such $a$ by inspection: Put $a=2$ and the roots are given by
$$x^3 - x^2 - x + 1 = 0 \Leftrightarrow (x-1)^2(x+1) = 0.$$</p>


答:
$$y=\frac{x-2}{x-3}$$

Let's study it's range.

$$y=\frac{x-2}{x-3}=\frac{x-3+1}{x-3}=1+\frac{1}{x-3}$$

Hence $1$ is not the domain of $f^{-1}$ but $3$ is.


问:
Why this function is not continuous at $z=0$?
Let $z=re^{i\theta}$, for $\theta \in (0,2 \pi]$ Let $$
f(z)=\left\{\begin{array}{ll}
0 & z=0 \\
\frac{r}{\theta} & z \neq 0
\end{array}\right.
$$

Why is it true that for every line going through the origin the function is continuous but at $z=0$ the function is not continuous?

I understand $z=0$ is problematic because this point is not defined well in the complex plane, but I still not understand how to show those arguments formally, by definition.

答:
A line going through the origin is defined by $\theta =a$ where $a$ is a constant. You can indeed verify that then as $r \to 0$, $f(z) \to 0$.

However for $z_n= \frac{1}{n} \left(\cos\left(\frac{1}{n^2}\right)+ i\sin\left(\frac{1}{n^2}\right)\right)$, you’ll see that $\lim\limits_{n \to \infty}f(z_n) =n$. So the  limit doesn’t exist.


问:
Find the least number which when divided leaves following remainders
The question is, find the least number when divided by $36, 48$ and $64$ leaves remainder $25, 37$ and $53$ respectively. The answer is said to be $($the $\operatorname{lcm}$ of the three numbers $-11),$ where $36- 25= 48- 37= 64- 53= 11.$

Can anyone please give me the logic for doing this ? I am not familiar with the modular operator method. So any answer not using that would be helpful. Thanks

答:
So let us assume that the required number is $x$.

Now by the given conditions,
\begin{align}
x\equiv 25\quad (mod\quad 36) \tag{1}\\
x\equiv 37 \quad(mod \quad 48)\tag{2}\\
x\equiv 53\quad (mod \quad 64)\tag{3}\\
\end{align}

$(1)\implies$ $x=36u+25$ for some $u\in\Bbb{Z}$ $\tag{4}$

Now   $x=36u+25$ satisfies $(2)$

So $(2)\implies$ \begin{align}
&36u+25\equiv37\quad (mod \quad 48)\\
\text{i.e} \quad & 36u\equiv 12 \quad(mod \quad 48)\\
\text{i.e} \quad & 3u\equiv 1 \quad(mod \quad 4)\\
\end{align}
Solution of the above congruence is $u\equiv -1 \quad(mod\quad 4)\implies u\equiv 3 \quad(mod\quad 4)$

So $u=4v+3$ for some $v\in\Bbb{Z}$.

Thus from $(4),\quad x=144v+133$, $\tag{5}$

Again $(5)$ satisfies $(3)$. So

\begin{align}
& 144v+133\equiv 53\quad (mod \quad 64)\\
\text{i.e} \quad & 144v\equiv -80 \quad(mod \quad 64)\\
\text{i.e} \quad & 9v\equiv -5 \quad(mod \quad 4)\\
\end{align}
Solution of the above congruence is $v\equiv -1 \quad(mod\quad 4)\implies v\equiv 3 \quad(mod\quad 4)$

So $v=4w+3$ for some $w\in\Bbb{Z}$.

Thus from $(5),$$
\quad x=576w+565 \tag{6}$

Hence the required solution is $x\equiv 565 \quad(mod \quad 576)$

Since the least integer (obviously positive) is required, so putting $w=-1$ in $(6)$ yields $x=11$

PS: You could also use Chinese Remainder Theorem to find this value.


问:
Is there a simple criterion for when $e^{x}+ ax^{2}+ bx+ c$ has a zero?

Is there a simple criterion for when
$$e^{x}+ ax^{2}+ bx+ c$$
has a zero ?

Source: MathOverflow/@MattF.

one simpler function $e^{x}+ ax+ b$ has a zero iff either or $- a+ a\!\ln\!\left ( -a \right )+ b\!\leq 0>\!a,$ or $a\!= 0>\!b,$ or $a> 0,$ where the last clause comes from using calculus to minimize the function.

First edit. It might be possible to extract something from Wilkie's approach in his

paper "Model completeness results for expansions of the ordered field of real numbers by restricted Pfaffian functions and the exponential function," which implies, among other things, that there is a system of exponential equations and inequalities that has a solution iff $e^{x}+ ax^{2}+ bx+ c$ doesn't.

Second edit. There is an algorithm to determine whether $e^{x}+ ax^{2}+ bx+ c$ has a zero :

simultaneously check for each positive rational $q$ whether $e^{q}+ aq^{2}+ bq+ c< 0,$ for each $n$ whether for all the positives $x$ so that $ax^{2}+ bx+ c+ \sum_{k= 0}^{n}\frac{x^{k}}{k!}\geq 0,$ and something similar for the negatives $x.$ The first piece comes from the proof of the transcendence of $e,$ like_ https://mathoverflow.net/a/378064/164469, and the second is standard Tarski-Seidenberg quantifier elimination. But this algorithm does not provide a time bound for the algorithm or a single set of discriminants which can be used for all $a, b, c.$

答:
It's one of my first answers, so my MathJax and my answer won't be perfect. I appreciate every help.

I am not sure if my answer is correct, because the first answer has many likes and an easier criterium. However, when graphing the function, it doesn't seem to work for me.

I've gone the same way. It is the question for which $a, b, c$ are the functions $f(x)=e^x$ and $g(x)=-ax^2 -bx -c$ intersect. $a<0$ is trivial, so we will assume $a>0.$

The function $g(x)$ is concave and the function $f(x)$ is convex, so we will look at the argument, where both functions have the same derivative. If $f=g$ at that point, there is exactly one intersection. If $f>g,$ there is no intersection and when $f<g$ there are two intersections.

Set $f'(k)=g'(k)$ to get the desired argument. If you solve $$e^k = -2ak-b$$ you get
$$k=-W(\frac{e^{- \frac{b}{2a} }}{2a})- \frac {b}{2a}$$

There are zeroes, when $e^k \leq -ak^2 -bk-c.\,k$ is the root of the above, so we can substitute $e^k.$

We get :
$$-2ak-b\leq -ak^2-bk-c$$

So the answer to your question is $:\quad h(x)=e^x+ax^2+bx+c$ has zeroes, iff $$ak^2+(b-2a)k-b+c\leq 0$$ or $$a<0$$ with $k$ being the term at the top. One zero with equality and two zeroes with strict inequality.


问:
Quick ways to _verify_ determinant, minimal polynomial, characteristic polynomial, eigenvalues, eigenvectors ...
What are easy and quick ways to verify determinant, minimal polynomial, characteristic polynomial, eigenvalues, eigenvectors after calculating them?

So if I calculated determinant, minimal polynomial, characteristic polynomial, eigenvalues, eigenvectors, what are ways to be sure that I didn't do a major mistake? I don't want to verify my solutions all the way through, I just want a quick way which gives me that it is highly likely that the calculated determinant is right etc.

Let $A$ be a matrix $A \in \operatorname{Mat}(n, \mathbb{C})$,

let $\det(A)$ be the determinant of matrix $A$,

let $v_1, v_2, ..., v_k$ be eigenvectors of matrix $A$,

let $\lambda_1, \lambda_2, ..., \lambda_n$ be eigenvalues of matrix $A$,

let $\chi_A(t) = t^n + a_{n-1}t^{n-1}+\cdots + a_0 = (t-\lambda_1)\cdots(t-\lambda_n)$ be the characteristic polynomial of matrix $A$,

let $\mu_A(t)$ be the minimal polynomial of matrix $A$.

Verifications suggested so far:

eigenvectors / eigenvalues

$\det(A) = \lambda_1^{m_1} \lambda_2^{m_2} \cdots \lambda_n^{m_l}$ where $m_i$ is the multiplicity of the corresponding eigenvalue

$a_0 = (-1)^n\lambda_1\cdots\lambda_n$

eigenvectors can be verified by multiplying with the matrix; the eigenvalues can be verified at the same time; i.e. $A v_i = \lambda_i v_i$

determinant

$\det(A) = \lambda_1^{m_1} \lambda_2^{m_2} \cdots \lambda_l^{m_l}$ where $m_i$ is the multiplicity of the corresponding eigenvalue

characteristic / minimal polynomial

$a_0 = (-1)^n\lambda_1\cdots\lambda_n$

$\mu_A(A) = 0$ and $\chi_A(A) = 0$

$\mu_A(t) \mid \chi_A(t)$

答:
You can use the Caley-Hamilton theorem. Put your matrix in the characteristic polynomial and you should find $0$. Explicitly, $\chi_A(A) = A^n + a_{n-1}A^{n-1}+\cdots + a_0I=0$. This checks the

characteristic polynomial, the minimal polynomial and the eigenvalues. The only thing not checked is the dimension of the eigenspace, and the eigenvectors.

Of course this checking is simple, if your software computes matrix algebra.


问:
How many "distinct" polynomials are there mod $n$?
(This sounds like a fairly natural question, but I couldn't find it asked on this site before…)

Fix an integer $n$ as the modulus, and consider polynomials (with integer coefficients) modulo $n$.

For any polynomial $f$ and any fixed $k$, we have $f(x) + nx^k \equiv f(x) \pmod n$ for all integers $x$; thus any two polynomials with some coefficient differing by a multiple of $n$ are equivalent in the sense that they give the same value mod $n$ for any integer $x$ as input. So we can assume that each coefficient of $f$ is one of $\{0, 1, 2, \dots, n-1\}$.

When $n$ is prime, we know (from Fermat's little theorem) that $x^n \equiv x \pmod n$ for all integers $x$, which means any polynomial of degree $n$ or greater is equivalent (in the same sense as above) to a polynomial of lower degree. Even when $n$ is not prime, some such relationship probably exists; for example when $n=4$, we can see that $x^4 \equiv x^2 \pmod 4$. (I guess, though this may be wrong and I haven't checked it, that if the prime factorization of $n$ is $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$, then $x^n \equiv x^{n/(p_1p_2\cdots p_r)} \pmod n$.)

In any case, whenever $x \equiv y \pmod n$, we have $f(x) \equiv f(y) \pmod n$, so it is enough to consider the polynomial's values at $n$ consecutive integers say $\{0, 1, 2, \dots, n-1\}$. As there are at most $n$ distinct values it can have mod $n$ for each input $n$, there are at most $n^n$ distinct polynomials in this sense.

The question: How many distinct polynomials with integer coefficients are there mod $n$? Here, the meaning of "distinct" is that polynomials $f$ and $g$ are "distinct" if there exists at least one integer $x$ at which $f(x) \not\equiv g(x) \pmod n$. Or, using the terminology from what I found later (see below), we want the number of polynomial functions from $\mathbb{Z}$ to $\mathbb{Z}/n\mathbb{Z}$.

That is the question, but of course before asking it here I tried a few things to find the answer myself, including writing the following program.

Define the "signature" (just a name I made up for writing this program: is there already a standard term for this?) of a polynomial $f$ as the tuple $(f(0), f(1), f(2), …, f(n-1))$ of its values at all $n$ points. We can count the number of distinct polynomials mod $n$ by keeping track of all the encountered signatures, as we start with the zero polynomial and successively add $ax^k$ for each $a$ from $0$ to $n-1$, for increasing $k$ until $x^k$ has the same signature as some already seen polynomial.

```<code>#!/usr/bin/env python3

# Adding the signatures of two polynomials mod n
def add(t1, t2):
    n = len(t1)
    assert n == len(t2)
    return tuple((t1[i] + t2[i]) % n for i in range(n))

def count(n):
    szero = tuple([0]*n)  # The signature of the zero polynomial
    seen = {szero}  # Only the zero polynomial has been seen
    k = 0
    sxk = tuple([1]*n)  # The signature of x^k
    while True:
        news = set()  # New signatures, by adding ax^k for some a
        saxk = szero
        for a in range(1, n):
            saxk = add(saxk, sxk)  # The signature of a*(x^k)
            for s in seen:
                newp = add(saxk, s)  # a*(x^k) + (smaller polynomial)
                if newp not in seen: news.add(newp)
        seen.update(news)
        # Now done with all polynomials of degree k
        k += 1
        sxk = tuple((sxk[i]*i) % n for i in range(n))
        if sxk in seen: break
    return len(seen)

print('n', '\t', 'count(n)')
for n in range(1, 11):  # Gets very slow for n=11
    print(n, '\t', count(n))
</code>```

This prints:

```<code>n    count(n)
1    1
2    4
3    27
4    64
5    3125
6    108
7    823543
8    1024
9    19683
10   12500
</code>```

The good thing about computing these small values is that there is the OEIS, using which you can, to quote Donald Knuth, "compute your way into the literature". And accordingly, searching the OEIS for this sequence $1$, $4$, $27$, $64$, $3125$, $108$, $823543$, $1024$, $19683$ gives A058067 described as "Number of polynomial functions from Z to Z/nZ" which is exactly our question. We also learn from that page that there's a formula

$$
\prod_{k=0}^{n-1} \frac{n}{\gcd(n, k!)}
$$

that fits the numbers we computed and (per that page) was first proved by Aubrey J. Kempner in 1921 in a two-part paper called "Polynomials and their residue systems" (1, 2), and there's what appears to be a nice paper in 2000 by Manjul Bhargava called "The Factorial Function and Generalizations" PDF that generalizes it (and maybe gives a shorter proof).

But I haven't read either of these papers as I ran out of my time budget for this (well, I exceeded it by writing up this question), and I hope someone can give a clear proof/derivation of the formula either using those papers or independently.

Edit: I read the latter paper and understood it enough to answer this question (left an answer below), but I have a follow-up question: Smallest number of residue classes for a polynomial function mod $n$ that separates $0$.

答:
Here is a proof based on the article by Manjul Bhargava (2000): this article is nicely written and won a Merten M. Hasse Prize ("for a noteworthy expository paper appearing in an Association [MAA] publication, at least one of whose authors is a younger mathematician"). Though the paper is about a more general setting, we can restrict to the case of integers (undo the generalization!), to get a nice and illuminating proof.

Lemma 1: Any integer polynomial $f$ can be written in the "falling factorial basis" instead of the monomial basis. In other words, if $f(x) = \sum\limits_{k=0}^d a_k x^k$ where each $a_k$ is an integer, then we can also write $$f(x) = \sum_{k=0}^d c_k x^{\underline k}$$ for some integer coefficients $c_k$, where $x^{\underline k}$ means $x(x-1)\cdots(x-k+1)$.

Proof: We have the expression $x^n = \sum_k {n \brace k} x^{\underline k}$ where ${n \brace k}$ is a Stirling subset number ("Stirling number of the second kind"), etc. See for instance Chapter 6 of the book Concrete Mathematics, or Wikipedia. $\blacksquare$

So from now on we'll write all polynomials in this basis.

Lemma 2: The polynomial $c_k x^{\underline k}$ vanishes identically mod $n$ (meaning that $c_k x^{\underline k} \equiv 0 \pmod n$ for all integers $x$) if and only if $c_k k!$ is divisible by $n$.

Proof: If $c_k x^{\underline k} \equiv 0 \pmod n$ for all $x$, then taking $x = k$ in particular gives $c_k k! \equiv 0 \pmod n$. For the other direction, if $c_k k!$ is divisible by $n$, then (for any integer $x$) note that $x^{\underline k}$ is a product of $k$ consecutive integers and is thus divisible by $k!$, which means that $c_k x^{\underline k}$ is divisible by $c_k k!$ which in turn is divisible by $n$. $\blacksquare$

Lemma 3: Let $f(x) = \sum_{k=0}^d c_k x^{\underline k}$. Then $f$ vanishes identically mod $n$ (meaning that $f(x) \equiv 0 \pmod n$ for all integer $x$) if and only if each term $c_k x^{\underline k}$ vanishes identically mod $n$.

Proof: One direction is obvious: if each $c_k x^{\underline k}$ vanishes identically mod $n$, then so does their sum $f$. For the other direction, suppose $f \equiv 0 \pmod n$, and let $j$ be the smallest index such that $c_j x^{\underline j} \not\equiv 0 \pmod n$. Then,

for $k < j$ we have $c_k x^{\underline k} \equiv 0 \pmod n$ by definition of $j$,

for $k > j$, by taking $x = j$ we see that $c_k j^{\underline k}$ includes a factor $0$ so $c_k j^{\underline k} \equiv 0 \pmod n$.

As $f(x) \equiv 0 \pmod n$, this means (still looking at $x=j$) that the only remaining term $c_j j^{\underline j} \equiv 0 \pmod n$ as well, i.e. $c_j j!$ is divisible by $n$. This from Lemma 2 means that $c_j x^{\underline j} \equiv 0 \pmod n$ for all $x$. $\blacksquare$

Lemma 4: $c_k k!$ is divisible by $n$ if and only if $\frac{n}{\gcd(n, k!)}$ divides $c_k$.

Proof: A special case of "If $a$ divides $bc$, then $a/gcd(a,b)$ divides $c$", which can be proved in many ways, say using the unique prime factorization.$\blacksquare$

This leads to a theorem characterizing all the polynomial functions mod $n$:

Theorem: Any polynomial function $f: \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ can be expressed uniquely in the form $$f(x) = \sum_k c_k x^{\underline k} \qquad\text{ where }0 \le c_k < \frac{n}{\gcd(n, k!)}.$$

Proof: By the previous lemmas, note that changing one of the coefficients $c_k$ by a multiple of $\frac{n}{\gcd(n, k!)}$ means adding a term which vanishes identically mod $n$, and thus does not change the function $f$. This means that each of the coefficients $c_k$ is determined only modulo $\frac{n}{\gcd(n, k!)}$. $\blacksquare$

(Note that this is essentially a "structure theorem" for the abelian group (under addition) of polynomial functions mod $n$.)

Finally, as there are $\frac{n}{\gcd(n, k!)}$ choices for each coefficient $c_k$, the number of distinct polynomial functions modulo $n$ is $$\prod_k \frac{n}{\gcd(n, k!)}$$

Edit. Here's another proof I came up with, that's more computational and less "magic". It came up while trying to enumerate all possible "signatures" $(f(0), f(1), \ldots, f(n-1))$.

Specifically, suppose we are given a "signature" $(y_0, y_1, \ldots y_{n-1})$ and want to find a polynomial (or all polynomials) $f$ such that $f(0) \equiv y_0$ and $f(1) \equiv y_1$ and so on (all congruences are modulo $n$). By Lemma 1 from the proof above, we can suppose that the polynomial we are trying to find has been written in the basis of "falling factorial powers" $x^{\underline k}$, where $x^{\underline k}$ means $x(x-1)(x-2)\cdots(x-k+1)$. That is, we can suppose that the polynomial we want is $$f(x) = c_0 + c_1 x + c_2 x^{\underline 2} + c_3 x^{\underline 3} + \ldots,$$ where we are trying to solve for the coefficients $c_0, c_1, \dots$.

This we can do iteratively, by considering the given values of $f(0), f(1), \dots$ in order:

$f(0) = y_0$ is the same as $c_0 \equiv y_0$, which can always be "solved" for $c_0$, i.e. just set $c_0 \equiv y_0$.

Having thus determined $c_0$, the next equation we were given, namely $f(1) = y_1$, is now the same as $c_0 + c_1 \equiv y_1$ or equivalently $c_1 \equiv y_1 - c_0$ which can always be solved for $c_1$, as $c_0$ is known.

Next, $f(2) = y_2$ is the same as $c_0 + 2c_1 + 2c_2 \equiv y_2$ or equivalently $2c_2 \equiv y_2 - c_0 - 2c_1$ which (now that we have determined $c_0$ and $c_1$) can be solved for $c_2$ except in one case: if $n$ is even and the RHS is odd, then our congruence $2c_2 = \text{(an odd number)} \pmod n$ is impossible. So we need to impose a condition, namely: if $n$ is even then $y_2$ must be such that the RHS is also even. When $n$ is odd, any value mod $n$ will do for $y_2$, but when $n$ is even only half the values mod $n$ will do for $y_2$. In short, there are $\dfrac{n}{\gcd(n, 2)}$ values for $y_2$ for which we can find $c_2$ to satisfy $f(2) = y_2$.

Next, $f(3) = y_3$ is the same as $c_0 + 3c_1 + 6c_2 + 6c_3 \equiv y_3$ or equivalently $6c_3 \equiv y_3 - …$ which can be solved for $c_3$, for $\dfrac{n}{\gcd(n, 6)}$ values of $y_3$.

And so on: in general $f(k) = y_k$ is the same as $k!c_k \equiv y_k - …$ which can be solved (for $c_k$) for $\dfrac{n}{\gcd(n, k!)}$ values of $y_k$.

So the number of $(y_0, y_1, \ldots)$ that can possibly be the signatures of a polynomial (and thus also the number of polynomial functions) is:

$$\prod_{k \ge 0} \frac{n}{\gcd(n, k!)}$$

(Deep down there are of course some similarities between this proof and the previous one, but at least this version is something I could imagine having come up with by myself, with only the falling factorial basis idea.)


问:
How to prove that this series which is summation of rational polynomial fractions is absolutely convergent?
Consider the sequence $a_n= \frac{n^3+2n^2+2n+4}{n^5+n^4+7n^2+1}$,$n \geq0$.

Prove that the series $\sum_{n \geq0}{a_n}$ is absolutely convergent.

I try to prove it using ratio test, but when I try to calculate the limit $\lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|$, I find that it equals 1, and I can't use the ratio test.

Can someone give me some hints?

Now I try to use compare test, however, I have no idea about finding the right series to compare to. I really want to know if there are some principles for finding such series. I think that exercises of limits and series are too tricky for me, they are too flexible, many times when I meet such exercise I have no direction to work in.

答: Note, that $|a_n|\leq c/n^2$ for all $n$ where $c$ is suitable large. This may be seen by writing $a_n=\frac{1+2/n+2/n^2+4/n^3}{n^2(1+1/n+7/n^3+1/n^5)}$.


问:
Find $a_n$ for the fourier series $\cos(3x) =\sum_{n=0}^{\infty}a_n\cos(2n+1)x$
Given $\cos(3x) = u(x,0)=\sum_{n=0}^{\infty}a_n\cos(2n+1)x,\quad\forall 0<x<\frac{\pi}{2}$

Hello !

So I have been a solving a PDE question and this is the final step, where I have applied initial condition to the general solution and I am face to face with the Fourier series. I have to find the the value of term $a_n.$ The book gives the answer that $a_1=1$ and $a_n=0,\quad\forall n = 0,2,3,4,5\ldots$

Following is my calculations so far :

$\cos(3x) =a_0\cos(x)+\sum_{n=1}^{\infty}a_n\cos(2n+1)x$

Since the series is Fourier cosine with $p=\frac{\pi}{2}$ we have $a_0, a_n$ are given by :

$a_0=\frac{1}{p}\int_{0}^{p} \cos(3x){\rm d}x$

$a_n=\frac{2}{p}\int_{0}^{p} \cos(2n+1)\cos(3x){\rm d}x$

I am solving both but answer for $a_0=-\frac{2}{3}\pi$ and I didn't solve for $a_n$ since $a_0$ is not equal to zero required as per answer in the book.
Can any one guide me where I am making the mistake ?

答:
By Werner's identities
$$ \begin{split}
2\cos (\alpha) \cos (\beta)&=\cos (\alpha-\beta)+\cos (\alpha+\beta),\\
2\sin (\alpha) \cos (\beta)&=\sin (\alpha-\beta)+\sin (\alpha+\beta),
\end{split}$$
and the related prosthaphaeresis formulae
$$\sin(\alpha\pm\beta)=\sin(\alpha)\cos(\beta)\pm\sin(\beta)\cos(\alpha),  $$
we find
$$\begin{split}
\frac 4 \pi \int_0^{\pi/2} \cos(3x)\cos(kx)dx&= \frac 2 {\pi} \int_0^{\pi/2} \cos((3-k)x)dx+\frac 1 \pi\int_0^{\pi/2}\cos((3+k)x)dx \\
&=\frac 2 {\pi} \left[\frac 1 {k-3}\int_0^{\pi(k-3)/2}\cos(u)du+\frac 1{k+3}\int_0^{\pi(k+3)/2}\cos(u)du \right]\\
&=\frac 2 {\pi} \left[\frac {1} {k-3} \sin \left(\frac{\pi(k-3)}{2} \right)+\frac {1} {k+3} \sin \left(\frac{\pi(k+3)}{2} \right)\right]\\
&= \frac{2}{\pi} \left[\frac{2k \sin(k\pi/2)\cos(3\pi/2) - 6\sin(3\pi/2)\cos(k\pi/2)}{k^2-9} \right]\\
&= \frac{12}\pi \frac{1}{k^2-9} \cos\left(\frac{\pi k}{2} \right).
\end{split}$$
Your book's notation is retrieved by setting $k=(2n+1)$; thus, the case $n=0$ corresponds to $k=1$, where the integral evaluates to $12/\pi \cdot \cos(\pi/2)/(-8) = 0$.

Observe that the singularity at $k=3$ is removable: by continuity, you can insert the value
$$\lim_{k\to 3} \frac{12}{\pi} \frac{1}{k^2-9} \cos\left(\frac{\pi k}{2} \right) = 1. $$
This corresponds to the fact that
$$\frac 4 \pi \int_0^{\pi/2} \cos^2(3x) dx = 1. $$


问:
Question about integration
I am currently revising some calculus notes, and I got a question I hope somebody could help me with. I seen the following done various places (not in my own notes, but in other proofs)

$\int \text{some-expression-in-terms-of-y  }\frac{dy}{dx} dx = \int \text{some-expression-in-terms-of-y  } dy$

I was told never to think of $\frac{dy}{dx}$ as a fraction, but rather as a symbol. So what exactly is happening there? Its like part of the integral is evaluated, changing the integral so its now with respect to $dy$.

答:
It's not a fraction, and you are right not to think of it as a fraction. What you've seen is the substitution rule (the opposite of the chain rule you're used to), which states that
$$\int{f(u(x))\cdot u'(x)} dx=\int{f(u)} du$$
You're not "cancelling" the dx you were integrating against before with the dx on the bottom - it's more like you're changing to a more convenient coordinate system.

The idea that you can "cross out" the dx's in $\frac{dy}{dx}dx$ to get dy is an abuse of notation that many nonrigorous, pedagogical texts adopt so their students can remember the rule better.

Title: Question about integration


问:
metrics of Product space
Let for each $i\in\mathbb{N}$ $(X_{i},\rho_{i})$ be discrete metric spaces, the metrics defined on product space $\prod_{i\in\mathbb{N}}X_{i}$ \
$d(x,y)=\sum_{i=1}^{\infty}\frac{\rho_{i}(x_{i},y_{i})}{2^{i}}$ and $d^*(x,y)=\frac{1}{n+1}$ with $n=min\{i:x(i)\neq y(i)\}$ are equivalent
I tried so hard for Days but I could not proceed it, i guess special choices is required, any hint or help would be appreciated

答:
In this answer I gave full details as to why the sum metric $d(x,y)=\sum_{i=1}^\infty \frac{\rho_i(x_i,y_i)}{2^i}$ induces the product topology on $\prod_i X_i$ (under the mild assumption that the $\rho_i$ are bounded by $1$ to ensure convergence of the series; this is true for all metric spaces generally).

It's also not too hard to see that $d^\ast(x,y)=\frac{1}{n+1}$ with $n = \min\{i\mid x_i \neq y_i\}$  induces the product topology too if each $X_i$ has the discrete topology, as is the case here:

Let $O$ be product open and $x \in O$. We need to show that $O$ contains a $d^\ast$-ball around $x$, so that $O$ is open in the $d^\ast$-metric too.

First we find a basic open set of the product topology, which is of the form $\prod_i O_i$ where all $O_i$ are open in $X_i$ and $F = \{i\mid O_i \neq X_i\}$ is finite, such that $x \in \prod_i O_i \subseteq O$. This is just by definition of the product topology.

Let $n_0 = \max F \in \Bbb N$. Note that for any $y \in \prod_i X_i$: if $d^\ast(x,y) < \frac{1}{n_0+1}$ then we know that $y_i = x_i$ for all $i \le n_0$ (otherwise $i_0 := \min\{i\mid x_i \neq y_i\} \le n_0$ and $d^\ast(x,y)=\frac{1}{i_0+1} \ge \frac{1}{n_0+1}$, which is not the case). And this in turn means that $x_i = y_i$ on all $i \in F$, so:

$$x \in B_{d^\ast}(x,\frac{1}{n_0+1}) \subseteq \prod_i O_i \subseteq O$$

which thus shows that $O$ is $d^\ast$-open.

If $O$ is $d^\ast$-open in $\prod_i X_i$ we need to show it is product open as well: let $x \in O$ and find $r>0$ such that $B_{d^\ast}(x,r) \subseteq O$. Also find $n \in \Bbb N$ such that $\frac{1}{n} < r$. Now define $O'=\prod_i O_i$ by $O_i = X_i$ when $i > n$ and $O_i = \{x_i\}$ for $i \le n$. This is a basic open subset of the product topology on $\prod_i X_i$ when all $X_i$ are discrete:  all singleton sets in $X_i$ are indeed open and all but finitely many coordinates have trivial coordinate set $X_i$.

Now, as before, if $x_i = y_i$ for $i \le n$ we know that $i_0 = \min\{i\mid x_i \neq y_i\} > n$ and so $d^\ast(x,y) = \frac{1}{i_0+1} \le \frac{1}{n} < r$, hence:

$$x \in O' \subseteq  B_{d^\ast}(x,r) \subseteq O$$

which shows that $x$ is an interior point of $O$ in the product topology and as $x$ was arbitrary, $O$ is product open.

So we have $\mathcal{T}_{d^\ast} = \mathcal{T}_{\text{prod}} = \mathcal{T}_d$ and both metrics induce the same topology, namely the natural product topology on $\prod_i X_i$.

Title: metrics of Product space


问:
The condition for $\lim(a_nb_n)=\lim(a_n)\lim(b_n)$
I have seen many proofs here for
$(n!)^{1/n}$ being divergent.
But suppose, if we consider
$$(n!)^{1/n}=(1\cdot 2\cdots n)^{1/n}
       =(1)^{1/n}\cdots(n)^{1/n}$$

Here each individual term converges to 1.

I know this is not true as $(n!)^{1/n}$ is divergent. So, is there a condition to $\lim(a_n b_n)=\lim(a_n)\lim(b_n)$?

答:
If the sequences $\{a_n\}$ and $\{b_n\}$ converge, i.e. $$\lim a_n = A, \lim b_n = B$$ then
$$\lim a_nb_n = \lim a_n \lim b_n = AB$$
You can extend this to the case when the number of sequences involved is finite, by induction.

The issue with your example is that it doesn't fit in the above category, i.e. since the number of product terms are not finite as $n\to\infty$. Do you see the problem?


问:
How should I solve for $y$ in the following exponential fraction
How to solve for $y$ without complex solutions. That is how to algebraically find $y$ in the following equation:

$$x=\frac{2^{y}+2^{-y}}{2}$$

答:
Let $z=2^y$.

Your equation is then equivalent to
$$2x = z + \frac{1}{z} \iff z^2 - 2xz + 1 = 0.$$

By the Quadratic Formula, you get
$$z = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}.$$

(Note that $x \geq 1$ by the Arithmetic Mean-Geometric Mean Inequality.)

Therefore, you have
$$y = \log_{2}\bigg(x \pm \sqrt{x^2 - 1}\bigg)$$


问:
How to Factorise these expressions?
I'm dong some Factorisation revision and I'm a bit stuck. I've never done this before and I'm stuck a bit. First of all I know how to do small factorisations and I know you have to find the Highest Common Factor and that it's the opposite of expanding but these expressions are scaring me a bit.

e.g. how would you do this $x^2 + 8x + 16$

or this $9m^2 - 24mn + 16n^2$

答:
Given:

$\boxed{x^2 + 8x + 16}$

The first question is best solved by guessing.  We need to find a number that multiplies to positive $16$ and adds to positive $8$.

Two numbers which fit this are: positive $4$ and positive $4$. Because $\sqrt{16}$ = $4$, we can call this a perfect square.

Therefore, we can factor this as: $$(x+4)(x+4)$$

Which is the same as: $$(x+4)^2$$

We can check this by using the FOIL method to expand the factoring:

$$x^2+4x+4x+16=x^2+8x+16$$

The next question is a little more difficult and so we will make use of a slightly different strategy.

Given:

$\boxed{9m^2 - 24mn + 16n^2}$

We can use a method called grouping to factor this equation.

We start by multiping $9m^2\cdot16n^2$ which equals $144m^2n^2$. Similarily to the last question we want to find two numbers which multiply to $144m^2n^2$ and add up to $-24mn$.

In this case, $-12mn$ and $-12mn$ are two numbers which multiply to positive $144m^2n^2$ and add to $-24mn.$ We now want to split the $-24mn$ and use grouping to factor this.

$$(9m^2-12mn)(-12mn+16n^2)$$

Notice how in both cases we can factor $3m$ out:

$$3m(9m^2-12mn)-3m(-12mn+16n^2)$$

Leaving us with:

$$3m(3m-4n)-4n(3m-4n)$$

We now have our final factored form:

$$(3m-4n) (3m-4n)$$

Which can also be written as:

$$(3m-4n)^2$$

This is also a perfect square.

Once again, we can check this is the correct factoring by simply expanding it using the FOIL method.

$$(3m-4n) (3m-4n) = 9m^2-12mn-12mn+16n^2\\=9m^2-24mn+16n^2$$

Title: How to Factorise these expressions?


问:
Distance to a curve via Newton's Method
So I wanted a quick confirmation if my thinking is correct.
The particular problem I'm working on is a $2$D Bezier Curve, with at most $4$ control points and at a minimum of $2$ control points.
The more general question (also I'm going to use the distance squared because it is cleaner to look at);

If given an arbitrary coordinate, $(x,y)$, and $2$ arbitrary functions, $F_x(t)$ and $F_y(t)$, with the distance squared being

$D(t)=(x-F_x(t))^2+(y-F_y(t))^2$

The first order derivative of that equation with respect to $t$ being

$D'(t)=2*(x-F_x(t))(-F_x'(t))+2*(y-Fy(t))(-Fy'(t))$

The second order derivative being;
$D''(t)=2*(x-F_x(t))(-F''_x(t))+2*F'_x(t)^2+2*(y-F_y(t))(-F_y''(t))+2*F'_y(t)^2$

I should be able to use Newton's method of

$t_{n+1}=t_n-D'(t_n)/D''(t_n)$

to approximate the t with the minimum distance to $F_x(t)$ and $F_y(t)$, correct?

(Clamping on t as well from the range of $t\in\{0,1\}$)

答: At the closest point $P$ on the curve, the tangent vector is perpendicular to the vector from $P$ to the reference point $(x,y)$. So, the dot product of these two vectors is zero. This gives you a polynomial equation of degree 5 in the curve parameter, $t$, which you can solve using your favorite root finder (Newton or otherwise).

Title: Distance to a curve via Newton's Method


问:
A question about a statement in the proof of Riemann Mapping Theorem from Gamelin
I am reading the proof of the Riemann Mapping theorem from Gamelin's Complex Analysis,
in his proof, let $D$ be a simply connected domain where $D \neq \mathbb{C}$, ($D$ can be unbounded), fix a point $z_0 \in D$,

Gamelin defined $\mathcal{F}$ to be the collection of all univalent function defined on $D$, such that $|f(z)| < 1$ and $f(z_0) = 0$, and Gamelin claims that this collection is nonempty.

I have no idea why this is true, Gamelin's argument makes no sense to me.
His original proof is, $\mathcal{F}$ is nonempty, since if $h(z)$ maps $D$ conformally onto a bounded domain, then the function $f(z) = \epsilon(h(z) -h(z_0))$ is in $\mathcal{F}$. But he did not define what is $h(z)$.

Can somebody fill in the gap for me please?

答:
The existence of a univalent holomorphic function from $D$ onto a bounded domain is indeed an essential and non-trivial part in the proof of the Riemann mapping theorem. It is also the part where the condition that $D$ is simply connected plays a crucial role.

The proof can be found in Alhfors' “Complex Analysis“ or Rudin's “Real and Complex Analysis” books, and also in the Wikipedia article.

The argument goes roughly as follows:

Choose $w_0 \in \Bbb C \setminus D$. Then $z-w_0$ is non-zero on the simply connected domain $D$, so that is has a holomorphic square root, i.e. a function $\phi: D \to \Bbb C$ with $\phi(z)^2 = z-w_0$.

If $w_1 \in \phi(D)$ and $w_2 \in \phi(D)$ then $w_1 \ne -w_2$.

Choose a disk $\overline{B_r(a)} \subset \phi(D)$ with $r < |a|$. Then $\overline{B_r(-a)} \cap D = \emptyset$.

$h(z) = r/(\phi(z)+a)$ is holomorphic and injective in $D$ with $|h(z)| < 1$.


答:
Let denote some-expression-in-terms-of-$y $ by $f(y)$ where $y$ is a function that depend to $x$. Now since $dy=y'(x) dx=\frac{dy}{dx}dx$ then  we have
$$\int f(y) y' dx=\int f(y) dy$$</p>

Title: Question about integration

Body:
I am currently revising some calculus notes, and I got a question I hope somebody could help me with. I seen the following done various places (not in my own notes, but in other proofs)

$\int \text{some-expression-in-terms-of-y  }\frac{dy}{dx} dx = \int \text{some-expression-in-terms-of-y  } dy$

I was told never to think of $\frac{dy}{dx}$ as a fraction, but rather as a symbol. So what exactly is happening there? Its like part of the integral is evaluated, changing the integral so its now with respect to $dy$.


问:
How to prove $A=\mathcal{O}_n$, use trace of matrix
Let $A=B+C$, and $B=B^T$, $C=-C^T$, $BC=\mathcal{O}_n$. Prove that if $A^2=\mathcal{O}_n$, then $A=\mathcal{O}_n$.
I want to prove $B=C=
\mathcal{O}_n$, and I got that
$$A^2=B^2+CB+BC+C^2\implies0=\mathrm{tr}(A^2)=\mathrm{tr}(B^2)+\mathrm{tr}(C^2)$$
So$$\mathrm{tr}(BB^T)=\mathrm{tr}(CC^T)\geqslant0.$$
But I don't know how to go on.

答:
Hint

You also have $\left(A’\right)^2=0$.

Expend this equality in term of $B’,C’$.

Then prove that $B=C=0$.


问:
$|G|=22$ abelian group , prove exist only $1$ element of order $2$ and order $11$
I'm trying to solve this question for a while and I think I'm missing something.

if you can please help me without using "Cauchy's theorem".

My try

By Lagrange's theorem we know that for each $g \in G : o(g)||G| \Rightarrow o(g)|22$

So we have only $4$ options : $o(g)=1 , o(g)=2 , o(g)=11 , o(g)=22$

Let's cancel the option $o(g)=1$ because then $g=e$ and it won't help us.

if $o(g)=22 \rightarrow G$ is cyclic so we can pick $o(g^2)=11$ and $o(g^{11})=2$ so we are done.

I'm stuck in this phase of if $o(g)=2$ or $o(g)=11$

I tried proving by contradiction that it can't be that exist more than $1$ elemet of order $2$ or $11$

I know that the amount of elements of order $2$ is odd (since $22$ is even).

I know that $2$ and $11$ is prime.

I noticed I didn't use the fact that $G$ is abelian so I guess I need to use it somehow yet I don't know how.

Any clues would help.

Have a nice day and thank you!

答:
For an odd prime $p$, an abelian group of order $2p$ is cyclic, so we have $G\cong C_{2p}$. A cyclic group of order $n$ has a unique subgroup of order $d$ for each divisor $d\mid n$.

So there is a unique subgroup of order $2$ resp. $11$ for $G$ with $|G|=2p=2\cdot 11$. However, the number of elements of order $d$ in $G$ is given by $\phi(d)$.
We have $\phi(2)=1$ and $\phi(11)=10$, so that there is a unique element of order $2$ in $G$, but certainly not a unique element of order $11$.

References:

Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .

Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$


问:
Is $\mathbb{Q} \times \mathbb{Q}$ a $G_\delta$ set?
I can prove that $\mathbb{Q}$ is not a $G_\delta$ set in $\mathbb{R}$. I was applying the same Baire space argument to show that $\mathbb{Q} \times \mathbb{Q}$ is not a $G_\delta$ set.

I was thinking like this:

We can write $(\mathbb{Q} \times \mathbb{Q})^c= \displaystyle{\bigcap_{(p,q)\in \mathbb{Q} \times \mathbb{Q}}\{(p,q)\}^c}= \displaystyle{\bigcap_{n=1}^{\infty}V_n}$ where each $V_n$ is open in $\mathbb{R}^2$. If we can write $\mathbb{Q} \times \mathbb{Q}= \displaystyle{\bigcap_{n=1}^{\infty}U_n}$ where each $U_n$ is open in $\mathbb{R}^2$. Then $\displaystyle{\bigcap_{n=1}^{\infty}(U_n \cap V_n)}= \phi$ is dense in $\mathbb{R}^2$-----which is a contradiction.!

Am I thinking correctly ?

Please help. Thanks in advance.

答:
The argument for $\Bbb Q \times \Bbb Q$ in $\Bbb R^2$ is exactly the same as for $\Bbb Q$ in $\Bbb R$, in essence:

If $(X,d)$ is a complete metric space with no isolated points then any countable dense set $D$ is not a $G_\delta$:

Suppose $D=\bigcap_n O_n$, where all $O_n$ are open. It follows that the $O_n$ are dense too as they all contain the dense set $D$.

Also, $X\setminus \{d\}$ is open and dense for any $d \in D$ as there are no isolated points.

But then $$\emptyset= D \cap D^\complement = \bigcap_n O_n \cap \bigcap_{d \in D} (X\setminus \{d\}$$

contradicts the Baire theorem for the complete $(X,d)$: a countable intersection of dense open sets with empty (not dense!) intersection.

$\Bbb Q \times \Bbb Q$ is still countable and dense and $\Bbb R^2$ is complete metric without isolated points so the same proof applies.


问:
Probability of some dice rolls
Question

A fair six-sided die is rolled $12$ times. Find the probability that each of the six possible outcomes ($1$, $2$, $3$, $4$, $5$ and $6$) come up at least once.

My working

Let $X_n$ be the random variable denoting the number of times $n$ appears, out of $12$ rolls. For example, $X_1$ will be the random variable denoting the number of times $1$ appears, out of $12$ rolls.

Then, $$X_n \sim B(12, \frac 1 6)$$ and the required probability is

\begin{align}
P(X_1 \geq 1)P(X_2 \geq 1)P(X_3 \geq 1)P(X_4 \geq 1)P(X_5 \geq 1)P(X_6 \geq 1) & = [P(X \geq 1)]^6
\\[5 mm] & =
[1 - P(X = 0)]^6
\\[5 mm] & =
\left[1 - \left(\frac 5 6\right)^{12}\right]^6
\\[5 mm] & \approx
0.490
\end{align}

Answer

The solution by my professor gives the required probability as $$1 -
\left[\binom 6 1 \left(\frac 5 6\right)^{12} - \binom 6 2 \left(\frac 4 6\right)^{12} + \binom 6 3 \left(\frac 3 6\right)^{12} - \binom 6 4 \left(\frac 2 6\right)^{12} + \binom 6 5 \left(\frac 1 6\right)^{12}\right] \approx 0.438.$$

The solution does come with an explanation, but I am very confused by it, so I am hoping to find more intuitive suggestions on why my working is wrong as well as how to go about approaching the problem.

答:
The issue is that $X_1, X_2, X_3, X_4, X_5, X_6$ as you defined are NOT independent random variables, due to the constraint that
$$\sum_{i=1}^6 X_i = 12.$$
Therefore, the probability that each of the six possible outcomes (1, 2, 3, 4, 5 and 6) come up at least once, i.e. $P(X_1 \geq 1, X_2 \geq 1, X_3 \geq 1, X_4 \geq 1, X_5 \geq 1, X_6 \geq 1)$, cannot be factorized as the product of $P(X_i \geq 1), i = 1,\dots,6$.

Title: Probability of some dice rolls


问:
Why is the angle that maximizes $a \cos(\theta) + b \sin(\theta)$ in the direction of $(a, b)$?
I'm wondering if there is an intuitive explanation to the result above. Mathematically, it is easy to prove: If we let $x = a \cos(\theta) + b \sin(\theta)$, then setting $\frac{\mathbb{d}x}{\mathbb{d}\theta} = 0$ tells us that $\theta = \arctan(\frac{b}{a})$. Thus $\theta$ is the angle between the positive x axis and the vector $(a, b)$. Another way to show this is to notice that $x = (a, b) \cdot (\cos(\theta), \sin(\theta)) = ||(a,b)||\cos(\alpha)$. If $\cos(\alpha) = 1$, then $x$ will be maximized and will be in the direction of $(a, b)$, since $\alpha = 0$.

At the same time, the answer just seems so simple that I can't help but wonder whether there's a more obvious way to look at it. Trigonometry is not really inherent to the problem, $\cos(\theta)$ and $\sin(\theta)$ can also be replaced by the components of some vector $\vec{v}$.

答:
Here is a geometric-intuitive way but also easy to verify analytically.

If you interpret $z:=(a,b) = a+ib$ as complex number, then the expression

$$a\cos \theta +b\sin \theta = \Re \left( (\cos \theta -i\sin \theta\right)\cdot (a+ib)) =\Re \left(e^{-i\theta}z\right)$$

is the real part of $e^{-i\theta}z$ after rotating $z$ counterclockwise by $\theta$.

Of course, you maximize the real part of $e^{-i\theta}z$ by rotating $z$ back to the positive real axis. Analytically this is clear because

$$e^{-i\theta}z=x+iy\Rightarrow x^2+y^2=|z|^2 = a^2+b^2$$

So, analytically, you maximize $x^2$ by minimizing $y^2$, which means that the imaginary part $y$ of $e^{-i\theta}z$ should become zero and the real part $x$ should be positive.

With $z=|z|e^{i\phi}$ you get

$$e^{-i\theta}z = e^{-i\theta}\cdot|z|e^{i\phi}=|z|e^{i(\phi-\theta)}=|z| \text{ for } \boxed{\theta = \phi}$$


问:
Units in $\mathbb{Q}[x]/(x^2-1)$
I know for integral domain $R$, units of $R[x]$ are units of $R$.

How about if we introduce a quotient ring? i.e., suppose I have an ideal $(x^2-1) \in \mathbb{Q}[x]$ and quotients by this ideal. Note since $x^2-1 = (x-1) (x+1) $ and $-1,1\in \mathbb{Q}$ $x^2-1$ is reducible so $(x^2-1)$ is not a maximal ideal.

I wonder what is units of $\mathbb{Q}[x]/(x^2-1)$.

Let $\bar{a} \in \mathbb{Q}[x]/(x^2-1)$ be units. then $\exists \bar b$ such that $\overline{ab}=1$

\begin{align}
&\bar{a} = a + (x^2-1), \quad \bar{b} = b+ (x^2-1) \\
&\Rightarrow \quad \overline{ab} = ab + (x^2-1) = (x^2-1)
\end{align}
then $ab\in(x^2-1)$, so $a$ can be either $c, c(x+1), c(x-1)$ where $c \in \mathbb{Q}$

Is this approach correct? How one can compute the units of $\mathbb{Q}[x]/(x^2-1)$ in a formal way?

答:
Hint: The elements of the quotient ring are of the form $a+bx$ with $a,b\in\Bbb Q$ and  $x^2=1$.

Consider the equation
$(a+bx)(c+dx)=1$, when $a+bx$ is given.

Then $(ac+bd) + (ad+bc)x=1$ and so $ac+bd=1$ and $ad+bc=0$.

This gives you two equation with two unknowns $c,d$. Solve them.

Title: Units in $\mathbb{Q}[x]/(x^2-1)$


问:
Solve one system of equations
Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$

I tried solving this using cross multiplication method  but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$

答:
As @Aderinsola Joshua commented, use successive eliminations

From $(1)$, you have $y=-\frac{a x+c z}{b}$.

Plug in $(2)$ and then $z=\frac{c x \left(a^2-b^2\right)}{a \left(b^2-c^2\right)}$

Plug in $(3)$ and simplify to get
$$\frac{b c  (a-b) (a+b) (a-c) (a+c) \left(a^2
   \left(b^2-c^2\right)^2-x^2\right)x}{a^2 \left(b^2-c^2\right)^2}=0$$
So, assuming $a\neq b \neq c$, you are left with
$$\left(a^2
   \left(b^2-c^2\right)^2-x^2\right)x=0$$

Title: Solve one system of equations


问:
Nonlogical axioms and their validity
I am reading A Friendly Introduction to Mathematical Logic by Leary and Kristansen. The authors show that the logical axioms are valid. Since the logical axioms are independent of the language, they are true in any $\mathcal{L}$-structure and any first order language $\mathcal{L}$.

This gets confusing when it comes to nonlogical axioms. Peano Axioms $N$ (see Example 2.8.3) are nonlogical axioms in the language of number theory $\mathcal{L}_{NT}$. Here's my question: If $\mathfrak{N}$ is the standard structure of the number theory, is it correct to say that $\mathfrak{N}\models N$? Is it simply true by definition? Or do we accept nonlogical axioms to be true in any structure?

Likewise, if we are in the language of set theory, do we call the structure $\mathfrak{S}$ to be "standard" where ZF is accepted as nonlogical axioms?

Also, the authors prove that there is no way that a set of $\mathcal{L}_{NT}$ sentence can ever characterise the standard structure, i.e., even if there was some other structure such that $\mathfrak{A}$ where the Peano axioms $N$ was true then it would not hold that $\mathfrak{A}\simeq\mathfrak{N}$.

答:
If $\mathfrak N$ is the standard structure of the number theory, is it correct to say that $\mathfrak {N} \vDash \mathsf {PA}$ [where $\mathsf {PA}$ are the Peano Axioms]?

Exactly; we assume that Peano Axioms are true in the standard model.

And no; non-logical axioms are not valid, i.e. true in every structure. The axiom $¬∃x(0=s(x))$ does not hold in a domain with only one object.

And yes, there are non-standard models of arithmetic, i.e. structures satisfying first-order Peano arithmetic that contains non-standard numbers.

Not every collection of non-logical axioms have a "standard" model; see Group theory: every structure satisfying the group axioms is called a group, but there are many "varieties".

In the same way, there are many different structures called vector spaces: everyone satisfies the axioms for vector space.

Title: Nonlogical axioms and their validity


答:
You are assuming that the random variables $X_i$'s are independent but this is not the case.

Let $A_i$ be the event that $i$ occur, we want to compute $$P\left(\bigcap_{i=1}^6 A_i\right)=1-P\left( \bigcup_{i=1}^6 A_i^c\right)$$

We then compute $P\left( \bigcup_{i=1}^6 A_i^c\right)$ by inclusion-exclusion principle.

Notice that we have

$$P\left(\bigcap_{i=1}^m A_i^c\right)=\left( \frac{6-m}{6}\right)^{12}$$

By symmetry:

$$P\left( \bigcup_{i=1}^6 A_i^c\right)=\sum_{m=1}^6 (-1)^{m+1}\binom{6}{m}P\left(\bigcap_{i=1}^m A_i^c\right)$$

Title: Probability of some dice rolls

Body:
Question

A fair six-sided die is rolled $12$ times. Find the probability that each of the six possible outcomes ($1$, $2$, $3$, $4$, $5$ and $6$) come up at least once.

My working

Let $X_n$ be the random variable denoting the number of times $n$ appears, out of $12$ rolls. For example, $X_1$ will be the random variable denoting the number of times $1$ appears, out of $12$ rolls.

Then, $$X_n \sim B(12, \frac 1 6)$$ and the required probability is

\begin{align}
P(X_1 \geq 1)P(X_2 \geq 1)P(X_3 \geq 1)P(X_4 \geq 1)P(X_5 \geq 1)P(X_6 \geq 1) & = [P(X \geq 1)]^6
\\[5 mm] & =
[1 - P(X = 0)]^6
\\[5 mm] & =
\left[1 - \left(\frac 5 6\right)^{12}\right]^6
\\[5 mm] & \approx
0.490
\end{align}

Answer

The solution by my professor gives the required probability as $$1 -
\left[\binom 6 1 \left(\frac 5 6\right)^{12} - \binom 6 2 \left(\frac 4 6\right)^{12} + \binom 6 3 \left(\frac 3 6\right)^{12} - \binom 6 4 \left(\frac 2 6\right)^{12} + \binom 6 5 \left(\frac 1 6\right)^{12}\right] \approx 0.438.$$

The solution does come with an explanation, but I am very confused by it, so I am hoping to find more intuitive suggestions on why my working is wrong as well as how to go about approaching the problem.


问:
Evaluating this Riemann sum
$$\lim_{x\rightarrow\infty}\sum_{k=1}^{n}\frac{6(k+1)^2}{n^3}\sqrt{1+\frac{2(k+1)^3}{n^3}}$$

How do we determine if it is a Right/Left or midpoint Riemann Sum?

How do we find the values a and b?

Is $$f(x)=\frac{3x^2}{2}\sqrt{1+\frac{x^3}{4}}$$ possible? if $$x=\frac{2(k-1)}{n}$$

But then again how do I find the interval for which I have to integrate?

答:
$$
\sum_{k=1}^{n}\frac{6\left(k+1\right)^2}{n^3}\sqrt{1+\frac{2\left(k+1\right)^3}{n^3}}=\sum_{k=2}^{n+1}6\frac{k^2}{n^3}\sqrt{1+2\frac{k^3}{n^3}}
$$
You need to have only an expression with $\frac{k}{n}$ to idenfity with $f\left(\frac{k}{n}\right)$. So you write it as
$$
\frac{1}{n}\sum_{k=2}^{n+1}6\frac{k^2}{n^2}\sqrt{1+2\frac{k^3}{n^3}}
$$

Can you identify $a$, $b$ and $f$ now ? (notice also that's it starts at $k=2$)

Title: Evaluating this Riemann sum


Title: Solve one system of equations

Body:
Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$

I tried solving this using cross multiplication method  but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$

Body:
$$x = a(c-b)(c+b)$$
$$y = b(a-c)(a+c)$$
$$z = c(b-a)(b+a)$$


问:
Show that analytic $f_n$ on $D_1(0)$ have an uniformly convergent subsequence, given that $\int_0^{2\pi} |f_n(e^{i\theta})|d\theta< 1$
I am preparing for my qualifying exam in complex analysis, and I'm working on the following problem.

Let $f_n = u_n + iv_n$ be a family of functions that are analytic on $D_1(0)$ and continuous up to $\overline{D_1(0)}$. Furthermore, assume that, for each $n$, we have $\int_0^{2\pi} |f_n(e^{i\theta})|d\theta < 1.$ Prove that $\{f_n\}$ has a subsequence that converges uniformly on compact subsets of $D_1(0)$.

I'm pretty sure that I need to apply Montel's Theorem, which says that if a family of analytic functions is uniformly bounded on compact subsets of the domain, then that family has a uniformly convergent subsequence.

Here's my attempt at a solution. I'd like to get some feeback about whether I'm on the right track. Let $K \subset D_1(0)$ be compact, and set $D = \operatorname{dist}(K, \{z : |z| =1 \}) > 0$. Since $f_n$ is analytic on $D_1(0)$ and continuous up to $\overline{D_1(0)}$, both $u_n$ and $v_n$ are harmonic on $D_1(0)$ and continuous up to $\overline{D_1(0)}$. Thus $u_n$ and $v_n$ satisfy the Poisson integral formula. That is, for $a \in K$, we have

$$u_n(a) = \int_0^{2\pi} P(a, \theta)u_n(e^{i\theta})d\theta,$$

$$v_n(a) = \int_0^{2\pi} P(a, \theta)v_n(e^{i\theta})d\theta, $$

where $P(a,\theta)$ is the Poisson kernel given by $P(a, \theta) = \frac{1}{2\pi} \frac{1- |a|^2}{|e^{i\theta} - a|^2}$.

Noting that, $|P(a,\theta)| \le \frac{1}{\pi D^2}$, we get

$$|f_n(a)| \le |u_n(a)| + |v_n(a)| \le \frac{1}{\pi D^2}(\int_0^{2\pi} |u_n(e^{i\theta})|d\theta + \int_0^{2\pi} |v_n(e^{i\theta})|d\theta) \le \frac{2}{\pi D^2}\int_0^{2\pi} |f_n(e^{i\theta})|d\theta < \frac{2}{\pi D^2},  $$

for all $n$ and $a \in K$. Thus the $f_n$ are uniformly bounded on $K$, and Montel's Theorem gives us what we want.

Am I on the right track here with exploiting the fact that $u_n$, $v_n$ are harmonic?

答:
Yes, your approach is fine. Notice however that you can simplify it by using Cauchy's integral formula (which is more or less equivalent to what you did): </p>

<p>$$
f(z)=\frac{1}{2\pi i} \int_{\partial D_1} \frac{f(\zeta)}{\zeta-z} d \zeta
$$
from which you obtain $|f(z)| < \text{dist}(z, \partial D_1)^{-1}$. In fact taking derivatives on both sides you get $|f'(z)|<\text{dist}(z, \partial D_1)^{-2}$ from which Arzela-Ascoli gives convergence in compact subsets.</p>

Body:
Yes, your approach is fine. Notice however that you can simplify it by using Cauchy's integral formula (which is more or less equivalent to what you did):

$$
f(z)=\frac{1}{2\pi i} \int_{\partial D_1} \frac{f(\zeta)}{\zeta-z} d \zeta
$$
from which you obtain $|f(z)| < \text{dist}(z, \partial D_1)^{-1}$. In fact taking derivatives on both sides you get $|f'(z)|<\text{dist}(z, \partial D_1)^{-2}$ from which Arzela-Ascoli gives convergence in compact subsets.


答:
Here is a systematic method that brings back your issue to solving a linear system.

I assume $abc \ne 0$. Setting

$$X=x/a, Y=y/b, Z=z/c \ \ \ \text{and} \ \ \ A=a^2,B=b^2,C=c^2,$$

then replacing $x,y,z$ by $aX,bY,cZ$ resp., and inverting equ. 1, with equ. 2, your system can be written under the form:

$$\begin{pmatrix}1&1&1\\A&B&C\\A^2&B^2&C^2\end{pmatrix}\begin{pmatrix}X\\Y\\Z\end{pmatrix}=\begin{pmatrix}0\\0\\-XYZ\end{pmatrix}$$

Otherwise said:

$$\begin{pmatrix}1&1&1\\A&B&C\\A^2&B^2&C^2\end{pmatrix}\begin{pmatrix}1/(YZ)\\1/(XZ)\\1/(XY)\end{pmatrix}=\begin{pmatrix}0\\0\\-1\end{pmatrix}$$

It remains to solve a simple linear system (with a so-called "Vandermonde matrix", always invertible, because its determinant is $(B-A)(C-A)(C-B) \ne 0$) then obtain $X,Y,Z$ out of the result.

Title: Solve one system of equations


问:
Show that $\frac{\arctan(x^2)}{\ln(1+x^2)}$ is bounded by $\frac{1+\sqrt{2}}{2}$
Given $F(x) = \frac{\arctan(x^2)}{\ln(1+x^2)},$ which is defined for $x>0.$ I would like to show that it is bounded above by $\frac{1+\sqrt{2}}{2}$.

I tried to differentiate the function in order to find maximal points however I've reached an equation which I don't know how to solve :

$$F'(x) = \frac{\frac{2x}{1+x^4}\ln(1+x^2)-\arctan(x^2)\frac{2x}{1+x^2}}{\ln^2(1+x^2)}\overset{{\rm ?}}{=}0$$

Any help would be appreciated.

答: We can simplify the problem by treating the function in terms of $x^2$. That is, the maximum value of $\arctan(x^2)/\log(1+x^2)$ is the same as that of $\arctan x/\log(1+x)$ which has derivative $$\frac1{\log^2(1+x)}\left(\frac{\log(1+x)}{1+x^2}-\frac{\arctan x}{1+x}\right).$$ Setting this to zero yields an expression for the stationary point $x_s$ $$\frac{\log(1+x_s)}{1+x_s^2}=\frac{\arctan x_s}{1+x_s}\implies\frac{\arctan x_s}{\log(1+x_s)}=\frac{1+x_s}{1+x_s^2}.$$ To find the maximum value of $(1+x)/(1+x^2)$, we set its derivative $(1-2x-x^2)/(1+x^2)^2$ to zero which yields $x=-1\pm\sqrt2$. We know that $x_s>0$ (since the original function is a function of $x^2$) so it follows that $$\frac{\arctan x^2}{\log(1+x^2)}\le\frac{1+(-1+\sqrt2)}{1+(-1+\sqrt2)^2}=\frac{1+\sqrt2}2$$ as desired.