How to find the number of rabbits after an elapsed time if they increase each time they mate? The problem is as follows:

Jenny is given by her dad as a birthday gift two female rabbits and
two male rabbits. Six months later each female rabbit gives birth
three female rabbits and three male rabbits. If we assume that the
same happens each six months. How many rabbits as maximum will she
have when it has elapsed two years after received them as birthday
present?

The alternatives in my book are as follows
$\begin{array}{ll}
1.&\textrm{1024}\\
2.&\textrm{2048}\\
3.&\textrm{512}\\
4.&\textrm{2192}\\
\end{array}$
I often get confused in these sorts of problems. Can someone help me?. Is this related with the famous Fibonnaci sequence
I think the method which would help me the most is an approach which develops from step by step. The part where I'm stuck is how to account for the female rabbits already born? do they mate as well?. This is kind of confusing. Is there powers involved?.
From looking at the alternatives I can see that there are suspiciously powers of two. Why? what's the reason for this?. Because of this part, it would be very helpful for easing my understanding, explaining the basis of the solution.
I think let's suppose if I have two rabbits, and each one makes two, then in the next generation I will have four, and in the next eight, but I don't have an idea how to use this into the problem. All and all, can someone help me to better understand this?

Every six months, the number of rabbits of either gender gets quadrupled. Hence, the number of rabbits of either gender is $2\cdot4^{n}$ at the end of $n$ half years. Total rabbits$=2\cdot2\cdot4^n=4^{n+1}$. For your question, $n=4$.

For this solution the following is true:

The population of male rabbits is always the same as the population of female rabbits. This is because every female rabbit gives birth to an equal amount of male and female rabbits and we start with an equal amount of males and females.
Males have no direct influence on the growth of the population. This is because males don't give birth
Rabbits never die and females give birth even after they've given it once before.

Since the population of female and male rabbits stays equal and males don't give birth we can just calculate the population of female rabbits and then double it at the end to get the complete population.
The key idea
In the problem it is stated that one female gives birth to 3 other females. This concludes that the the number of new borns is 3 times bigger then the population before 6 months.
Since the old population lives on and gives birth to new rabbits we have to add them to the new borns
Calculating the result
We can transform this into a recursive function $f(t)$ where every $t$ represents $6$ months.
($t=1 => 6months;$ $t=2 => 12months...$) :
$p(t+1) = p(t) + 3*p(t)$
The starting is two females so at $t=0$ the population is $2$:
$p(0) = 2$
This gives us the following results:
$p(1) = p(0) + 2*3     = 8$
$p(2) = p(1) + 8*3     = 32$
$p(3) = p(2) + 32*3   = 128$
$p(4) = p(3) +  128*3   = 512$
...
This however is only the population of female rabbits. As I've stated in the beginning we just have to double the population to also include the male rabbits.
This gives us:
$2* p(4) = 2 * 512 = 1024$
So after $4 * 6$ months or $2$ years the population of the rabbits is at $1024$ rabbits.
To explain explain why the population always seems to be a power of 2 we can simply derive the explicit function of the recursive function wich gives us:
$f(t)=2*2*4^t$ or $ f(t)=2*2*(2*2)^t$
Connection to the fibonacii sequence:
As seen in this solution this problem can be solved via a recursive function. The common expression of the FibonaciiSequence is also a recursive function. So they are related in that way.


Power Series Expansion $e^{2x}$ I'm currently finding the power series expansion centered at 0 for a bunch of functions. My answers are starting not to add up with the 'correct' ones and I'm looking for some assistance as to what I'm doing wrong, or if my answers are correct but just written differently...
For the problem: $$f(x) = e^{2x}$$
I derived it 4 times:
$$\frac{f^1(x)}{n!} = \frac{2e^{2x}}{1!}$$
$$\frac{f^2(x)}{n!} = \frac{4e^{2x}}{2!}$$
$$\frac{f^3(x)}{n!} = \frac{8e^{2x}}{3!}$$
$$\frac{f^4(x)}{n!} = \frac{16e^{2x}}{4!}$$
That was enough for me to "come up with the pattern." So I wrote for my power series expansion:
$$\sum_{n=0}^\infty \frac{2^ne^{2n}}{n!}$$
However, my professors answer is:
$$e^{2x} = \sum_{n=0}^\infty \frac{(2x)^n}{n!} = \sum_{n=0}^\infty (2)^n\frac{x^n}{n!}$$
I'm not really sure where the $x$ comes from in his solution, replacing $e$.
Thanks for your help

As you mentioned in your comment, and as others have pointed out, you know that the Taylor series about $0$, or the Maclaurin series for a function $f(x)$, is defined:

$f(x)=\displaystyle \sum_{n=0}^∞ \frac{x^n \cdot f^{(n)}(0)}{n!}$

Where $f^{(n)}(0)$ represents the $n$th derivative of $f$ evaluated at $0$.

You did right in noticing that the $n$th derivative of $f(x)=e^{2x}$ is:

$f^{(n)}(x)=2^n e^{2x}$

Now, notice, as you let $x=0$, the $n$th derivative evaluated at $0$ is:

$f^{(n)}(0)=2^n$

Now, using the definition you provided for the Maclaurin series, we have that:

$f(x)=e^{2x}=\displaystyle \sum_{n=0}^∞ \frac{2^{n} x^n}{n!}$

as req'd.

A power series expansion of $f$ about a point $x=x_0$ is an expression of the form

$$\sum\limits_{n=0}^{\infty} a_n\cdot (x-x_0)^n$$

If $f$ is analytic, then the coefficient $a_n$ is given by

$$a_n =\frac{ f^{(n)}(x_0)}{n!}$$

In your case, $x_0=0$ and $f(x)=e^{2x}$. You've calculated the first few derivatives of $f$, but you have not evaluated them at $x_0=0$ (to calculate the $a_n$).

A different approach, which I recommend, is the following. The power series of $e^x$ about $0$ is well known and given by

$$\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$$

In fact, if you have never before calculated this power series, I encourage you to do so using the observations above.

Now, $e^x$ is a very wel-behaved function, and we have that it actually is equal to its power series at every $x$. In particular, if you substitute $x$ by $2x$ in the power series expansion of $e^x$, you will obtain the power sereis expansion for $e^{2x}$:

$$\sum\limits_{n=0}^{\infty}\frac{(2x)^n}{n!}= \sum\limits_{n=0}^{\infty}2^n \cdot \frac{x^n}{n!}$$


Why do we denote the sequence space as $l^\infty$ My question is really simple. I didn't understand the call the sequence space as $l^\infty$. I know this is from the $l^p$ spaces denotation but I didn't understand the connection between both.

$l^p$ space

  Let $p\ge 1$ a fixed real number. By definition , each element in the
   space $l^p$ is a sequence $x=(x_j)=(x_1,x_2,\ldots)$ of numbers such
   that $|x_1|^p+|x_2|^p+\ldots$ converges and the metric is defined by

  $$d(x,y)=\bigg(\sum_{j=1}^{\infty}|x_j-y_j|^p\bigg)^{1/p}$$

$l^{\infty}$ space

  Let $X$ be a set of all bounded sequences of complex numbers; that is,
   every element of $X$ is a complex sequence $x=(x_1,x_2,\ldots)$ such
   for all $j=1,2\ldots$ we have $|x_j|\le c_x$, where $c_x$ is a real
   number. We choose the metric defined by

  $$d(x,y)=sup_{j\in \mathbb N}|x_j-y_j|$$

We denote it because $$\lim_{p\to\infty} \left(\sum_{j=1}^\infty |x_j-y_j|^p\right)^{\frac1p} = \sup_{j\in\mathbb N} |x_j-y_j|$$


How to show that the set of orthogonal n x n matrices forms a group under multiplication I am studying orthogonal matrices and I am not sure if to show if a set of orthogonal $n \times n$ matrices forms a group under multiplication. We must check each of the group axioms.
I found that the axioms are:

Closure
Associativity
Existence of identity matrix
Existence of the inverse matrix.

I edited my question, since I was able to find more information about this topic.
This group is called $O(n)$.
To check the four axioms I did:
Let $A \text{ and } B \in O(n)$, denoted as orthogonal matrices and assume that $C=AB$, then,
Closure :

To prove that $C \in  O(n)$ we must prove that $C$ is a real $n \times n$ orthogonal matrix with uni-modular determinant. Since A and B are real $n \times n$ matrices, $C$ is also a real $n \times n$ matrix so,
$C^TC=(AB)^T AB=B^T A^T AB = B^TB=I$

Associativity :

Matrix multiplication is associative, so the law holds for $O(n)$ group elements.

I am not sure if this is enough to prove associativity.
Identity element :

The $n \times n$ identity matrix $I_{n \times n}$ represents the identity element.

In this case I am not sure if this is enough to prove the identity element.
Inverse element :

Let $A^{-1}$ be the inverse of $A$, then we need to prove that  $A^{-1} \in  O(n)$ since $(A^{-1})^T=(A^T)^{-1}$. We have that:
$(A^{-1})^T A^{-1}=(A^T)^{-1} A ^{-1}=(AA^T)^{-1}=I^{-1}=I$

Can anyone check if what I did is correct? I also would like to know if I can prove the associativity and the identity element in a better way.
thanks

$(U_1 U_2)^T (U_1 U_2) = I$, hence $U_1 \circ U_2$ is orthogonal.

Associativity follows from associativity of matrix multiplication.

The matrix $I$ is an identity for matrix multiplication.

$U^T U = U U^T = I$, hence $U^{-1} = U^T$ is the required inverse.

Your doubt about the identity element is correct.

First you need to show that $I \in O(n)$ by showing that
$$I^TI=II^T=I,~~\text{since}~I^T=I.$$
Then
$$AI=IA=A, \forall A\in O(n).$$


Rational approximation for irrational number Sometimes I go into some subject in the class (high school level) and I have to explain to my students how approximate an irrational number by a sequence of rationals. The problem is that I should explain that in a high school level. What I usually do is take $\pi$ as a example and take the sequence:
\begin{align}
& 3,1=31/10\\
& 3,14=314/100 \\
& 3,141=3141/1000 \\
& 3,1415=31415/10000\\
&\vdots
\end{align}
I think that approach is intuitive and the students feel satisfacted with that. I was trying to figure out another way to explain the rational approximation but I coudn't find any. My questions is, does anyone know another way to explain that approximation in a high school level?

Thanks in advance.

Newton's method for square roots can be a good candidate due to its simplicity.

For example for approximating $\sqrt2$ in few iterations. You can ask your students where do they think $\sqrt2$ is located. Between $1.41$ and $1.42$, then you can start with $x_0=1.41$

$g(x) = x - \frac{(x^2 - 2)}{2x}$

$g(1.41) = 1.41 - \frac{(1.41^2 - 2)}{2\cdot1.41} = 1.4142198581...$

$g(1.4142198581) = 1.4142135623 \approx \sqrt2$

I think high schoolers can understand Farey sequences.  The $n$th Farey sequence is all the fractions between $0$ and $1$, inclusive, with denominators less than or equal to $n$.  $F_5 = \{0,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,1\}.$  If your irrational number is between $0$ and $1$, then it lies between two of those fractions.  By increasing $n$, there are more and more fractions and some of them get closer to your number.  Wiki has a page:

https://en.wikipedia.org/wiki/Farey_sequence

I don't think this is more intuitive, but it has the advantage of leading to a number of elementary exercises.


Circle $1$ is circumscribed about a square of side $8$ and circle $2$ is inscribed in the square Circle $1$ is circumscribed about a square of side $8$ and circle $2$
  is inscribed in the square. What is the ratio of the area of circle
  $1$ to circle $2$?

I know the formula of a circle is $A=\pi r^2$ but does anyone know how I can find the radius of circle $1$ and circle $2$?

The circle inscribed in the square has a radious $r_1=4$, because it touches the sides of the square at the middle points. While the circle ciscunscribed about the square has a radious of $r_2=4\sqrt{2}$ becouse it touches the vertex of the square (its radious is just half of the diagonal of the square). Finally, the ratio between the areas of both circles is just the difference
\begin{equation}
A_2/A_1=\frac{\pi r_{2}^2}{\pi r_{1}^2}=2
\end{equation}

Their radius are $4$ and $1$ respectively.
So the ratio should be
$$\frac{4^2}{1^2} = 16$$


We throw $5$ dice: What is the probability to have $4$ different numbers? We throw $5$ dice: What is the probability to have $4$ different numbers?

I know it is $$\frac{6\cdot 5\cdot 4\cdot 3}{6^5}.$$

I wanted to use an other argument, but it look to not work : I take $\binom{6}{4}$ numbers. Then I have $$\frac{6\cdot 5\cdot 4\cdot 3}{4!}$$
possibilities. Then I have to multiply this result by $4!$ and I don't understand why. Indeed, I would like to multiply by $5!$ since we can distribute the $5$ colors in e.g. $1;2;3;4;4$ in $5!$ different ways.

If I want all dice different, this argument works: I take $\binom{6}{5}$ number, then I can distribute the colors in $5!$ different way which give $\frac{5\cdot 5\cdot 4\cdot 3\cdot 2}{5!}5!=6\cdot 5\cdot 4\cdot 3\cdot 2$ possibilities, that is the correct answer.

So why doesn't it work with the previous situation ?

Notice that for a favorable outcome, exactly two of the five dice must display the same number, while each of the other three dice must display different numbers.

Suppose the dice are five different colors so that we can distinguish between them.  There are $\binom{5}{2}$ ways for two of the five dice to display the same number and $\binom{6}{1} = 6$ possible numbers for those two dice to display. There are five possible numbers remaining for the other three dice to display.  The number of possible ways those three dice can display the numbers is $5 \cdot 4 \cdot 3$ (where we list the outcomes as ordered triples in the order the colors appear in an alphabetical list).  Hence, the number of favorable outcomes is
$$\binom{5}{2} \cdot 6 \cdot 5 \cdot 4 \cdot 3$$
from which we obtain the probability
$$\frac{\binom{5}{2} \cdot 6 \cdot 5 \cdot 4 \cdot 3}{6^5}$$
that exactly four different numbers are displayed when five dice are rolled.

Suppose the output is in 5 boxes like [x][x][x][x][x]
For 4 definite different outcomes we have 6 x 5 x 4 x 3 (in any order)
For the remaining one, it can be any of the 6 digits so we have to choose 1 out of 5 boxes
and also the one box can contain any of the 6 digits i.e. 5c1 x 6

Therefore the probability should be (5c1 x 6 x 6 x 5 x 4 x 3)/6^5


gcd(a,b)=d then show that gcd(a/d,b/d)=1 (without euclids lemma or bezout's theorem) I was solving the problem $\gcd(a,b)=d$ then show that $\gcd\left(\frac{a}{d},\frac{b}{d}\right)=1$ (without Euclid's lemma or Bézout's theorem) then stumbled upon the fact that if we say
$$\gcd\left(\frac{a}{d},\frac{b}{d}\right)= k \tag{1}$$
then $a=kdn$ and $b=kdm$ then,
$$\gcd(a,b)=\gcd(m,n) \, kd \tag{2}$$
but I seem to not be able to prove that $(1)$ holds. If this holds then from this it’s clear that $d = \gcd(m,n)\,kd$ then $\gcd(m,n) = \frac{1}{k}$
Since $k$ is an integer and $\gcd(m,n)$ is also an integer then only value of $k$ that satisfies such a relation is $k = 1$ then by $(1)$ our question is answered.
So the question here is if two numbers $a = A \cdot B$ and $b = M \cdot N$, then
is it true that,
$$\gcd(a,b) = \gcd(A,M) \cdot \gcd(B,N)?$$

Let $a,b>0$, $d=(a,b)$. Let $a'=ad^{-1},b'=bd^{-1}$. Suppose that $e\mid a',b'$. Then $de\mid a,b$, so $de\mid (a,b)=d$. This means $e\mid 1$. It follows that $(a',b')=1$.

Proof: Before starting with the proof proper, we should observe that although a/d and b/d have the appearance of the functions, in fact, they are integers because d is a divisor both of a and of b. Now, knowing that gcd(a, b) = d, it is possible to find integers x and y such that d = ax + by. Upon dividing each side of this equation by d, we obtain the expression
1 = (a/d)x + (b/d)y
Because a/d and b/d are integers, an appeal to the theorem is legitimate. The conclusion is that a/d and b/d are relatively prime.   (Proved)


Function notation terminology Given the function $f:X\longrightarrow Y$, $X$ is called the domain while $Y$ is called the codomain. But what do you call $f(x)=x^2$ in this context, where $x\in X$? That is to say - what is the name for the $f(x)$ notation?

And while I'm here, what is the proper way to write a function like this? Would it be $f:\mathbb{R}\to\mathbb{R},\;f(x)=x^2$?

Edit:

I figured I'd add this to add a bit of context into why I'm asking. I'm writing a set of notes in LaTeX, and I'd like to use the correct terminology for the definition of a function.

  A function from set $A$ to set $B$, denoted by
  $$f:A\to B;x\mapsto f(x)$$
  is a mapping of elements from set $A$, (the $\textit{domain}$) to elements in set $B$ (the $\textit{codomain}$) using the $\color{blue}{\sf function}$ $f(x)$. The domain of a function is the set of all valid elements for a function to map from. The codomain of a function is the set of all possible values that an element from the domain can be mapped to. The $\textit{range}$ (sometimes called the $\textit{image}$) of a function is a subset of the codomain, and is the set of all elements that actually get mapped to by the function $f$.

Here I'm pretty sure the highlighted word "function" is not right.

I can remember to read this text, and being puzzled with the exact same question. From what I've learned from my teacher, you're right, writing down something as "the function $f(x)$..."  is sloppy notation. However, many books/people will use it this way.

If you're are very precise, $f(x)$ is not a function or an map. I don't know of a standard way to refer to $f(x)$, but here is some usage I found on the internet:

The output of a function $f$ corresponding to an input $x$ is denoted by $f(x)$.
Some would call "$f(x)=x^2$" the rule of the function $f$.
For each argument $x$, the corresponding unique $y$ in the codomain is called the function value at $x$ or the image of $x$ under $f$. It is written as $f(x)$.
If there is some relation specifying $f(x)$ in terms of $x$, then $f(x)$ is known as a dependent variable (and $x$ is an independent variable).

A correct way to notate your function $f$ is:
$$f:\Bbb{R}\to\Bbb{R}:x\mapsto f(x)=x^2$$

Note that $f(x)\in\Bbb{R}$ and $f\not\in\Bbb{R}$. But the function $f$ is an element of the set of continuous functions, and $f(x)$ isn't.

In some areas of math it is very important to notate a function/map specifying it's domain, codomain and function rule. However in for example calculus/physics, you'll see that many times only the function rule $f(x)$ is specified, as the reader is supposed to understand domain/codmain from the context.

You can also check those questions:

In written mathematics, is $f(x)$a function or a number?
What is the difference between writing f and f(x)?

From what I understand, $f$ is the function, so saying "a function $f(x)$" would
be wrong. Instead, you could say "a function $f$", or if you didn't want to assign it a name, "a function $x\mapsto x^2$", or if you want to specify the domain and codomain, "a function $f:X\to Y$".

If you just want to define the function's input-output relationship, $f(x)=x^2$ suffices, maybe with a $\forall x\in X$ at the beginning if you want to do it properly. That doesn't necessarily define the domain though, that set could just be a subset of the domain. $f(x)$ is really no different from $f(1)$, except that $x$ is a (qualified) variable. It's basically a rule defining the input and output for an improper subset of its domain.

The addition of "$:X\to Y$" is useful if you want to specify both the domain and codomain when defining the function. As far as I can tell, you can add it after a function like $f$ and the whole expression still refers to the function. You can't really use it with the $f(x)$ style, so I'd do something like

$f:\mathbb R\to\mathbb R=x\mapsto x^2$


Cantor's diagonalisation argument in proving cartesian product of infinitely many countable sets as uncountable There are lot of posts related to argument. But i have some confusion in the proof. My thinking :-
If cartesian product is represented as tuple $\langle s_1, s_2, s_3 .... s_n.....\rangle$
Then say $s_{ij}$ represent jth element of $s_i$ sequence. Then we have
$s_{11}, s_{12}, s_{13}, s_{14}, ...... \\ s_{21}, s_{22}, s_{23}, s_{24}, ...... \\............. \\.........$
now we can cross-out the elements diagonally(like in cantor diagonalisation argument) and say that bijection exists. But this type of argument is used in proving countably infinite union of countable sets is countable. Why the same can not be used in proving for cartesian product of  countably many countable sets? I am not able to perceive the difference. Please clarify

In union, each countable set can be represented as a sequence
In cartesian product can each element be considered a 1 element of product of sequences like 1st tuple as 1 element of S1, S2, S3 .....ie., s11, s21, s31, s41 .. second element as s12, s22, s32, s42 ....
(I am taking A as sequence s1, B as sequence s2, C as sequence s3 .... and not take cartesian product of AXBX... then elements will be product of sequences)

Both arguments can be visualized with an infinite matrix of elements.
For the Cantor argument, view the matrix a countable list of (countably) infinite sequences, then use diagonalization to build a SEQUENCE which does not occur as a row is the matrix. So the countable list of sequences (i.e. rows) is missing a sequence, so you conclude the set of all possible (infinite) sequences is UNCOUNTABLE.
For the countable union of countable sets, each row represents one of the sets in the union, and you do some sort of zig zag enumeration of the matrix entries to show the that the set of ENTRIES in the matrix (i.e. the union of the row sets) is COUNTABLE.
Both matrices have countably many entries, both matrices must be missing possible sequences that can be formed from the elements available. But the union is about the entries in the matrix, while the infinite product is about all possible sequences you could construct.

If each of your factor sets has at least teo elements, you can use the other diagonal argument to construct $t=\langle t_1,t_2,\ldots,t_m,\ldots\rangle$ where $t_m\ne s_{m,m}$ and hence $t\ne s_m$


How to find the equation of a line, tangent to a circle, that passes through a given external point I am given the equation of a circle: $(x + 2)^2 + (y + 7)^2 = 25$. The radius is $5$. Center of the circle: $(-2, -7)$.

Two lines tangent to this circle pass through point $(4, -3)$, which is outside of said circle. How would I go about finding one of the equations of the lines tangent to the circle?

I haven't started calculus, so I ask for advice fitting for someone starting a topic like this.

A useful technique for this situation which avoids using calculus goes like this:

Find the equation of a line through the point $(4,-3)$ with gradient $k$, and then use this (linear) equation to substitute into the equation of the circle to get a quadratic. The two roots of this quadratic would give the 2 points of intersection of the line with the circle, but in our case, we want to know the values of $k$ for which the quadratic has a double root (i.e. the 2 points of intersection are actually coincident). Use the standard condition ($b^2=4ac$) for a double root of a quadratic.

In your case, the equation of the line through $(4,-3)$ with gradient $k$ is $y+3 = k(x-4)$, or $y+7 = k(x-4)+4$. Substituting this into the equation of the circle gives:

$$(x+2)^2 + (k(x-4)+4))^2 = 25$$

So you need to rearrange this quadratic in $x$ and find out which vales of $k$ give double roots.

Let eq of tangents be y= mx+ c then it passes through point (4,-3)
So -3=4m+c let it put aside
Now tangent touches circle so radius = distance between centre and line
5 = |-2m+7+c|/(m^2+1)^1/2
By solving we get
11 m^2 +24m=9
We get two values of m
These are slopes of linesenter preformatted text here
We can equation of lines and angle between lines
This is my idea


Why in Rubik cube moves can be combined in any sequence? This is from visual group theory by Nathan Carter.

It mentions that moves can be combined in any sequence, where move is rotation by 90 degrees of face of rubik cube.

How is that $a(bc) = (ab)c ?$

Association of Rubik moves is so trivial it can be hard to spot.

Consider the move TLF (Top face a quarter turn, then Left face a quarter turn, then Front face a quarter turn). If we first do T, then do LF, we get the same result as we do when we first do TL, then do F. In other words, T(LF) = (TL)F. The Rubik's cube is associative.

As Arthur stated, its trivial and therefore strangely hard to spot; I came here with the same question. People seem to be misled when trying to think in terms of ordering compositions of transformations. Its not so simple as (F ∗ U) ∗ R where ∗ means "then perform"; because they then think logically to apply PEMDAS to F ∗ (U ∗ R) where parentheses also suggests "do this first, then perform...". It fails because the cube is obviously non-abelian, you can't "F then do UR". This points out the failure of good notation (and therefore explanation). (I think) its made most clear by Janet Chen in her writeup Group Theory and the Rubik’s Cube

  If C is an oriented cubie, we will write M(C) for the oriented cubicle
  that C ends up in after we apply the move M, with the faces of M(C)
  written in the same order as the faces of C. That is, the first face
  of C should end up in the first face of M(C), and so on. For example,
  the move R puts the ur cubie in the br cubicle, with the u face of the
  cubie lying in the b face of the cubicle and the r face of the cubie
  lying in the r face of the cubicle. Thus, we write R(ur) = br. First,
  let’s investigate what a sequence of two moves does to the cubie. If
  M1 and M2 are two moves, then M1 ∗ M2 is the move where we first do M1
  and then do M2. The move M1 moves C to the cubicle M1(C); the move M2
  then moves it to M2(M1(C)). Therefore, (M1 ∗ M2)(C) = M2(M1(C)). To
  show that ∗ is associative, we need to show that (M1 ∗ M2) ∗ M3 = M1 ∗
  (M2 ∗ M3) for any moves M1, M2, and M3. This is the same as showing
  that (M1 ∗ M2) ∗ M3 and M1 ∗ (M2 ∗ M3) do the same thing to every
  cubie. That is, we want to show that (M1 ∗ M2) ∗ M3 = M1 ∗ (M2 ∗
  M3) for any cubie C. We know from our above calculation that (M1
  ∗ M2) ∗ M3 = M3(M1 ∗ M2) = M3(M2(M1(C))). On the other hand,
  M1 ∗ (M2 ∗ M3) = (M2 ∗ M3)(M1(C)) = M3(M2(M1(C))). So, (M1 ∗ M2)
  ∗ M3 = M1 ∗ (M2 ∗ M3). Thus, ∗ is associative.

Hope that helps.


periodic function, find g(6) If $f$ is periodic, $g$ is polynomial function, $f(g(x))$ is periodic, $g(2)=3$, and $g(4)=7$, then $g(6)$ is
A) 13
B) 15
C) 11
D) none of these

The answer is c) 11, but I did not understand how $g(x)$ was considered linear polynomial (because i got answer when $g(x)$ is $2n -1$), isn't there any $g(x)$ with degree greater than 1 make $f(g(x))$ periodic?
Why? How will you solve the problem?

Let $g(x)$ be a polynomial, that is: $$g(x)=\sum_{n=0}^\infty c_nx^n$$
with every $c_t$ a constant, note a polynomial of degree $k$ has $k=\max\{t|c_t\neq 0\}$
We take that:
$$\forall x\in\Bbb R,\forall t\in\Bbb Z; f(x)=f(x+t\alpha)$$
$$\forall x\in \Bbb R, \forall t\in\Bbb Z; f(g(x))=f(g(x+t\alpha))$$
Note $t$ is a constant independent of $x$.
We then insert our sum form of the polynomial.
$$f(\sum_{n=0}^\infty c_nx^n)=f(\sum_{n=0}^\infty c_n(x+\alpha)^n)$$
Expanding gives:
$$f(\color{red}{c_0}+\color{purple}{c_1x}+\color{blue}{c_2x^2}+\color{green}{c_3x^3}+...)=f(\color{red}{c_0}+\color{purple}{(c_1x+c_1\alpha)}+\color{blue}{(c_2x^2+2c_2x\alpha+c_2\alpha^2)}+\color{green}{(c_3x^3+3c_3x^2\alpha+3c_3x\alpha^2+c_3\alpha^3)}+...)$$
Notice when $k$ (as defined earlier as the degree of the polynomial) is at least $2$ the $t$ which satisfies $f(g(x))=f(g(x+t\alpha))$ is dependent on $x$. This is the same thing as saying that $(x+\alpha)^n-x^n-\alpha^n$ has terms for $n\geq 2$, which is trivial by the binomial theorem.
$$(x+\alpha)^n-x^n-\alpha^n=\sum_{r=1}^{n-1}\bigg[\binom nr x^{n-r}\alpha^r\bigg]$$

NB when $k=1$, you get:
$$f(c_0+c_1x)=f(c_0+c_1x+c_1\alpha)$$
so $t=c_1$, a non-zero constant independent of $x$

$g(x) = 2x-1$ is still a polynomial, it's just a little atypical in that it's a very simple polynomial, which has no degrees greater than 1. That still fits the definition, it's just not what you'd typically think of when you hear the term "polynomial".


How do I prove that this inequality result holds for all positive integers How I can prove that this:
$$ \frac{1}{2} < \frac{2}{3} < \frac{3}{4} < \frac{4}{5} < \frac{5}{6}<...<\frac{n+1}{m+1}<...$$
always holds when $n < m$ and $n$ and $m$ are positive integers?
I started writing this $\frac{n}{m}<\frac{n+1}{m+1}$ but then I don't know how to proceed.

This a high-school standard exercise on inequalities:
As all numbers are positive,
$$ \frac nm<\frac{n+1}{m+1}\iff n(\not m+1)<m(\not n+1)\iff n<m. $$


find common ratio in geometric sequence Please Help Me to answer this question

now im using this formula $$a_n = a_1r^{n-1}$$

now i got a question if the given term is only

$$a_n = 3,960,298,332$$
$$a_1 = 92$$
$$ratio = ??$$
$$n = 16 $$

but i tried $${3960298332 \over 92}$$ $$=43046721$$
$$\sqrt[15]{43046721}$$

and the result is $$3.227968874176$$

now is possible to get double in common ratio or its wrong?

. any answer is highly appreciated thank you


Topological characterization of the closed interval $[0, 1]$. I would like to learn purely topological characterizations of the closed real intervals (to justify the existence of algebraic topology). In particular, such a characterization should not use real numbers. I would like to see in what sense $[0, 1]$ is topologically more important than, for example, the set of rationals, or the complex unit disc. I am especially interested in why the unit interval is so important in the study of compact Hausdorff spaces (metrization, Urysohn's lemma).

I have come up with the following one.

Start with a definition of path connectedness that does not use $\mathbb R$: $x_1$ and $x_2$ are connected by a path in $X$ if for every Hausdorff compact $C$ and $a,b\in C$, there is a continuous $f\colon C\to X$ such that $f(a)=x_1$ and $f(b)=x_2$.  Now, if $X$ is a Hausdorff space and distinct points $x_1$ and $x_2$ are connected by a path in $X$, then there is a minimal subspace of $X$ in which $x_1$ and $x_2$ are still connected by a path, and every such subspace is homeomorphic to $[0, 1]$.

Roughly speaking, $([0, 1], 0, 1)$ is the minimal bi-pointed Hausdorff space such that every bi-pointed Hausdorff compact can be mapped into it with the distinguished points being sent onto the distinguished points.

Maybe in some sense it can be also said that $X = [0, 1]$ is the "minimal" Hausdorff space such that every Hausdorff compact embeds into $X^N$ for some $N$, but i do not know how to make this precise.

My question is: what are other "natural" topological characterizations of $[0, 1]$?

Update:  I have duplicated this question on MathOverflow.

It is the unique second countable continuum with precisely two non-cutpoints. This is due to Veblen, according to this overview.

A continuum is a connected and compact Hausdorff space, and a cutpoint (in a connected space) is a point that when removed leaves the remainder disconnected. The second countable is a (non-reals using) way of saying metrisable. One can prove the compactness and connectedness purely from order completeness of the order.

In my opinion all characterizations of the interval $[0,1]$ that are not using the totality of real numbers are far from being natural and are awkward at least.

It is very simple to define what a closed interval is when you have the notion of the real numbers: it is a compact and connected subspace. The other way, from the interval to the whole totality of real numbers, is unnatural.

You can of course argue that you do not need the totality of real numbers to define the closed interval. It is true. But you still need the concept of the totality of real numbers (at least on the intuitive, naive level) to convince yourself that the space you defined really is (isomorphic) to the closed interval. Without it (assume you never heard about the set of real numbers) you just have defined some species of topological space. You do not know apriori that it has all the properties you demand from the closed interval, since you don't know what a closed interval is. Therefore you do not know that it is useful from your perspective.

From the philosophical point of view such a definition is a classical vicious circle.

Do you really want to justify the existence of algebraic topology on the basis of vicious circle?

Please stay with what is simple, beautiful and not awkward.

Alexey, in your comment you asked about properties making $[0,1]$ more significant than other spaces. I believe that the properties making $[0,1]$ (and its higher dimensional analogues) so special are not of purely topological nature.

Algebraic topology is about functors from the topological world to the algebraic world. Those functors are used to extract topological knowledge from algebraic relations. From the topological perspective the functors of algebraic topology have to be computable to be of any value. For instance: you want to prove that two spaces are not homeomorphic. You are computing their fundamental groups (or other algebraic functors) and see that they are not isomorphic. The reason that you succeded was beacause you were able to compute those algebraic objects effectively. In this toy example I would like to point out that the crucial property of the functors of algebraic topology is computability.

To make my point of view more precise: usefulness and significance of the intervals in the realm of algebraic topology are not of purely topological nature.

They are of topological, algebraic and computational nature.

Just like the algebraic topology is about topology, algebra and computation of functors.

The intervals are so special because they can be used to construct functors that are "easily" computable. (I don't think that this reason can be dissected any further. Also, it does not exclude the possibility that other spaces have this property.)


Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$? Suppose that $f(x)$ is $L^1$ and R- integrable function, problem is to resolve if it is possible existence of such a $f(x)$ that:
$$\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx\neq 0$$

All math tricks acceptable.

For any $f \in L^1$ we have

$$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}$$

this result is known as Riemann-Lebesgue lemma.

As $f \in L^1$ we can choose a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions such that $f_n \stackrel{L^1}{\to} f$. Since $\|\hat{g}\|_{\infty} \leq \|g\|_{L^1}$ for any $g \in L^1$, we get by the triangle inequality

$$\begin{align*} |\hat{f}(\omega)|  &\leq |\hat{f}(\omega)-\hat{f}_n(\omega)|+|\hat{f}_n(\omega)| \leq \|f_n-f\|_{L^1} + |\hat{f}_n(\omega)|\end{align*}$$

for all $\omega \in \mathbb{R}$. By virtue of our choice, $\|f_n-f\|_{L^1} \to 0$ as $n \to \infty$. Therefore, we see that it suffices to prove $(1)$ for simple functions. So let

$$f(x) = \sum_{j=1}^n c_j \cdot 1_{[x_j,x_{j+1}]}$$

for some constants $c_j \in \mathbb{R}$, $x_j<x_{j+1}$, $j=1,\ldots,n$. Then,

$$\hat{f}(\omega) = \sum_{j=1}^n c_j \int_{x_j}^{x_{j+1}} e^{-\imath \, x \omega} \, dx = \sum_{j=1}^n \frac{c_j}{-\imath \omega} \bigg( e^{-\imath \, x_{j+1} \omega}-e^{-\imath \, x_j \omega} \bigg) \stackrel{\omega \to \infty}{\to} 0,$$

i.e. $(1)$ holds. This finishes the proof.

Correct answer is:

$f(x)\in L^1$ does not imply $\lim\limits_{\omega\to \infty}\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}dx=0$ even if $f$ is R-integrable.

Sample: $f(x)=\arcsin(x)$, where $f\colon [-1,1] \to \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.

Please pay attention that notation $\int f(x) dx$ determines using Riemann integral, if you want to use Lebesgue integral, you can just skip $dx$ or write $\mu(dx)$.

Below is a simple option of proving Riemann-Lebesgue Lemma:

Let $f(x)$ be numeric $L^1(\mathbb{R})$, measurable function determined for any $x\in \mathbb{R}$.

$$\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}=-\int\limits_{\mathbb{R}}f(x) e^{-i(x\omega+\pi)}=-\int\limits_{\mathbb{R}}f(x) e^{-i\omega\left(x+\frac{\pi}{\omega}\right)}=-\int\limits_{\mathbb{R}}f\left(x-\frac{\pi}{\omega}\right) e^{-i\omega x}$$

Note: I am using substitution $v=x+\frac{\pi}{\omega}$ with renaming $v$ into $x$ what is not changing a value of the integral.

So:

$$2\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}=-\int\limits_{\mathbb{R}}\left[f\left(x-\frac{\pi}{\omega}\right)-f(x) \right]e^{-i\omega x}
$$
$$2\left|\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}\right|= \left|\int\limits_{\mathbb{R}}\left[f\left(x-\frac{\pi}{\omega}\right)-f(x) \right]e^{-i\omega x}\right|\leqslant \left\|f\left(x-\frac{\pi}{\omega}\right)-f(x)\right\|_{L^1}$$
what implies:
$$\lim\limits_{\omega\to \infty}\left|\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}\right|\leqslant \frac{1}{2}\lim\limits_{\omega\to \infty}\left\|f\left(x-\frac{\pi}{\omega}\right)-f(x)\right\|_{L^1}=\frac{1}{2}\lim\limits_{h \to 0}\|f(x-h)-f(x)\|_{L^1}$$

Functions from $L^1(\mathbb{R})$ space keep continuity of translation in meaning of $L^1$ norm what is described here and here, hence:

$$\lim\limits_{\omega\to \infty}\left|\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}\right|\leqslant \frac{1}{2}\lim\limits_{h \to 0}\|f(x-h)-f(x)\|_{L^1}=\frac{1}{2}\cdot 0$$

Note: If I am asking a question I am not looking for something what is described in popular sources like wikipedia or wikiproof. You can assume that aforementioned places of the Internet are checked before asking the question and it is not bringing a suitable answer.
After analyzing of some users I feel happy that I am a disabled person who is treating mathematics as a hobby quite similar as an old man solitaire.

Points here means nothing in the real life.


Positive harmonic function on $\mathbb{R}^n$ is a constant? Is it true that a positive harmonic function on $\mathbb{R}^n$ must be a constant? How might we show this? The mean value property seems not to be the way...for that we would need boundedness.

Since it is not clear whether the Wikipedia proof uses boundedness or not, please allow me to give a detailed proof that only uses nonnegativity.

Let $u$ be a nonnegative harmonic function in $\mathbb{R}^n$, and let $x,y\in\mathbb{R}^n$. Denote by $B_R(y)$ the ball of radius $R>0$ centred at $y$, and similarly by $B_r(x)$ the ball of radius $r>0$ centred at $x$. We consider such balls with $R>0$ large and $r=R-|x-y|$, that is, $B_r(x)\subset B_R(y)$ and the boundary of $B_r(x)$ is tangent to the boundary of $B_R(y)$.
Then we have
$$
u(x) = \frac1{|B_r|}\int_{B_r(x)}u \leq \frac1{|B_r|}\int_{B_R(y)}u = \frac{|B_R|}{|B_r|}u(y),
$$
where we have used the mean value property in the first and last equalities, and the nonnegativity of $u$ in the second inequality. Now we send $R\to\infty$ and get
$$
u(x) \leq u(y).
$$
As $x$ and $y$ were arbitrary, we conclude that $u$ is constant.

Note that the Harnack inequality is lurking behind this proof. Also, it was crucial to consider balls with different radii in order to make use of nonnegativity. The Wikipedia proof uses balls of equal radii.

One can use Louisville's Theorem for harmonic functions to prove this. Note that if $u$ is a harmonic function then so is $e^{-u}$ and if $u$ is bounded below, then $e^{-u}$ is bounded and thus constant. This then forces $u$ to be constant. This shows the slightly stronger fact that if $u$ is a harmonic function bounded below, then $u$ is constant.


What exactly is circular reasoning? The way I used to be getting it was that circular reasoning occurs when a proof contains its thesis within its assumptions. Then, everything such a proof "proves" is that this particular statement entails itself; which is trivial since any statement entails itself.

But I witnessed a conversation that made me think I'm not getting this at all.

In short, Bob accused Alice of circular reasoning. But Alice responded in a way that perplexed me:

  Of course my proof contains its thesis within its assumptions. Each and every proof must be based on axioms, which are assumptions that are not to be proved. Thus each set of axioms implicitly contains all theses that can be proven from this set of axioms. As we know, each theorem in mathematics and logic is little more than a tautology: so is mine.

Not sure what should I think? On the one hand, Alice's reasoning seems correct. I, at least, can't find any error there. On the other hand, this entails that... Every valid proof must be circular! Which is absurd.

What is a circular proof? And what is wrong with the reasoning above?

Of course my proof contains its thesis within its assumptions. Each and every proof must be based on axioms, which are assumptions that are not to be proved.

Hold it right there, Alice.  These specific axioms are to be accepted without proof but nothing else is.  For anything that is true that is not one of these axioms, the role of proof must be to demonstrate that such a truth can be derived from these axioms and how it would be so derived.

  Thus each set of axioms implicite contains all thesis that can be proven from this set of axioms.

Implicit.  But the role of a proof is to make the implicit explicit.  I can claim that Fermat's last theorem is true.  That is a true statement.  But merely claiming it is not the same as a proof.  I can claim the axioms of mathematics imply Fermat's last theorem and that would be true.  But that's still not a proof.  To prove it, I must demonstrate how the axioms imply it.  And in doing so I can not base any of my demonstration implications upon the knowledge that I know it to be true.

  As we know, each theorem in mathematics and logics is little more than a tautology:

That's not actually what a tautology is.  But I'll assume you mean a true statement.

  so is mine.

No one cares if your statement is true.  We care if you can demonstrate how it is true.  You did not do that.

The apparent dilemma arises from two words which do not mean the same thing.

You are talking about proofs. Alice is talking about truth. These are not the same thing.

A statement which is always true is a tautology, so in a sense, every such statement, including a true theorem, is a tautology.

But truth is not a proof.

Proofs are simply the re-expression of statements as other statements without relying on other statements (i.e., no circular reasoning).

But the result of a correct proof is no more true than the original statements, so a proof is likewise not always the truth.


Prove that for every real number $x$, if $|x-3|>3$ then $x^2>6x$. Not a duplicate of
Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$.
This is exercise $3.5.10$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Prove that for every real number $x$, if $|x-3|>3$ then $x^2>6x$.
Here is my proof:
Let $x$ be an arbitrary element of $\Bbb R$. Suppose $|x-3|>3$. Now we consider two different cases.
Case $1.$ Suppose $x-3\geq0$. Ergo $x-3>3$ and so $x>6$. Multiplying both sides of $x>6$ by $x$ we obtain $x^2>6x$.
Case $2.$ Suppose $x-3<0$. Ergo $3-x>3$ and so $x<0$. Since $x<0$, then clearly $x<6$. Multiplying both sides of $x<6$ by $x$ we obtain $x^2>6x$.
Since the above cases are exhaustive, $x^2>6x$. Therefore if $|x-3|>3$ then $x^2>6x$. Since $x$ is arbitrary, $\forall x(|x-3|>3\rightarrow x^2>6)$. $Q.E.D.$
Is my proof valid$?$
One other question: In the above linked-post, there is a clear mistake in the proof:

Since by one of the cases we have $x^2 > 6x$ then $|x − 3| > 3$
$\Rightarrow$ $x^2 > 6x$.

But the cases must be exhaustive and from both of them we should reach the conclusion that $x^2>6x$. Is my reasoning correct$?$ If yes, then why no one mentioned it in the comments or the answers to the above post$?$
Thanks for your attention.

\begin{align}
|x-3|>3 & \Rightarrow (x-3)^2>9 \\
 & \Rightarrow x^2-6x+9>9 \\
 & \Rightarrow x^2-6x>0 \\
 & \Rightarrow x^2>6x \\
\end{align}

For the second case we can say that $x$ is negative,
$x^2$ is positive and $6x$ is negative so $x^2> 6x$


Understanding the meaning of a function I am confused as to what really a function is ?

Is a function the steps by which we get the output from an input ?
Or is it the name of the process by which we get the output from an input ?

Simply put, a function is basically the description or the rule of the dependence  of the output on the input. It may also be seen as a mapping of the set $X$ - the domain to the set $Y$- the codomain. Notation-aly $f: X\mapsto Y$.

Take the example of $f(x)=x^2$. Let $X$ be the set of Real Numbers denoted $\mathbb{R}$. Clearly you can observe from the graph of $y=x^2$ that $f$ maps the input that is the real line to the positive real plane including $y=0$. Again, Notation-aly $f(x)=x^2 \mid f:\mathbb{R}\mapsto \mathbb{R^+}\cup \{0\}$.

I would suggest you consult Wikipedia's page on Functions for deeper detail.

Consider the function $x\mapsto x^2$.  That describes the steps from input to output.  Now you may refer to that function by a name, say $f$, so you write $f\colon x\mapsto x^2$


find the rank of a linear mapping such that $T^2=0$ Let $T:\Bbb R^6\to\Bbb R^6$ be a linear mapping such that $T^2=0$.Then which one is true?

a)Rank$(T)$ is less than or equal to 3

b)Rank$(T)$ is greater than  3

c)Rank$(T)$ is  equal to 5

d)Rank$(T)$ is less than or equal to 6

  I think, Rank of $T^2$ and Rank of $T$ are less than 6 since $T^2=0$ ...
  So d) may be false...Now How to proceed?Someone please explain...

$T^2=0$ implies $\text{Im}(T)\subseteq\text{ker}(T)$. Note that $\text{dim Im}(T)+\text{dim ker}(T)=6$ and $\text{dim Im}(T)\leq \text{dim ker}(T)$, we have that $\text{dim Im}(T)\leq 3$.

It could clearly seen that in Transformation $T^2$ that the Range of $T$ acts as the Domain for the other $T$. It's given that $T^2=0$, hence the $\operatorname{Rank}(T) \subseteq \operatorname{NullSpace}(T)$. Thus we know that
$$\operatorname{Basis}\big(\operatorname{Rank}(T)\big) \subseteq \operatorname{Basis}\big(\operatorname{Nullspace}(T)\big)
$$
and hence $$\dim(\operatorname{Rank}(T)) \le \dim(\operatorname{Nullspace}(T))$$
And now by the Rank-Nulity Theorem we know that
$$
\begin{split}
\operatorname{Rank}(T)+&\operatorname{Nullspace}(T)=6\\
&\qquad\Downarrow\\
&\operatorname{Rank}(T) \le 3
\end{split}
$$


How to find a normal vector in 2D I hope this is not a stupid question.  I am having a hard time finding a normal in 2D.  I am working on a game project and I am trying to perform vector reflection during collision.  I know the rest of the math, but I am having a hard time creating the normal vector that is required.
Is there a way to build a normal vector from the point of collision?  What do I need to create the normal vector?  My position vector does not necessarily point in the direction of the object.  My position can be (100, 200) but the object is a vertical object for example.  What else can I do for this?

$\vec V_{reflected} =\vec V_{incident} -2\vec n(\vec V_{incident} \cdot \vec n)$

I also had this problem. you can do $\vec{v} \times \vec{p}$ and consider $x$ and $y$ coordinates only.

In fact for any $\vec{v}=(x,y,0)$ and $\vec{p}=(0,0,1)$ then $\vec{v} \times \vec{p}=\begin{bmatrix}\vec{i} & \vec{j} & \vec{k}\\x & y & z \\ 0 & 0 & 1\end{bmatrix} = y\vec{i}-x\vec{j}+0\vec{k}=(y,-x)$

That implies that the normal vector $\vec{n}=(\vec{V}y, -\vec{V}x)$


Shapes that can be made with a bandsaw Is anything known about surfaces immersed in $\mathbb{R}^3$ with the following property?

  $S$ has a definable interior and exterior, and any point on $S$ has a tangent line which can be continuously deformed (rotation, displacement) away to infinity while never passing through $S$

The set of objects that have no blind holes or blind furrows?

By the way, the line being a tangent is not strictly necessary.  It is not necessary for the surface to be continuous, as the theoretical band saw can be retracted and plunged in for each surface point separately.

It is sufficient that every point on the surface has at least one infinitely long line intersecting the surface at that point, such that the line be continuously moved to infinitely far from the surface without intersecting the volume enclosed by the surface.

Obviously, this line is the position the theoretical band saw could be used to cut that surface point (and possibly others) from a sufficiently larger volume.

Points on the surface that have a well-defined normal vector do have an interesting property: the local curvature is such that in at least one direction, the curvature is either parallel or away from that normal vector.

While this is may not be useful for mathematicians, for people considering whether some specific object can be wholly cut with a bandsaw-like tool, that rule gives a necessary but not sufficient condition for bandsawability.

That means a computer aided design tool (like anything from Fusion 360 to SketchUp) could easily provide a warning for objects that cannot be produced using bandsaw-like tool alone.


Proof that it always exists a prime between $\sqrt n$ and $n$ I'm looking for a proof that ensures that for every natural number $n > 1$ there is always a prime $p$ such that $\sqrt{n} \leq p \leq n$ holds.

I know you can deduce my previous claim as a corollary of some famous results like "Bertrand's postulate", but I was wondering if it does exist any easy self-contained proof.

This doesn't prove the claim for all $n>2$, but it does show that there are at most finitely many counterexamples. I assume the Prime Number Theorem is allowed here (I know you want a self-contained proof, but this is the best I could do), I'm not sure how we could really do much without it.

Assume there are infinitely many $n \in \mathbb{N}$ such that $\pi(n)= \pi(\sqrt{n})$ and let $(n_i)_{i \in \mathbb{N}}$ be the ordered sequence of these natural numbers together with their square roots, so that $\pi(n_i)=\pi(\sqrt{n_i})$ for each $i$. We have, from the prime number theorem: $\displaystyle \lim_{x \to \infty} \frac{\pi(x) \log(x)}{x}=1$

In particular, this implies the following sequence converges: $( \frac{\pi(n_i) \log(n_i)}{n_i})_{i \in \mathbb{N}}$ and thus it must be Cauchy. So for every $\epsilon>0$ there is an $N \in \mathbb{N}$ such that $\sqrt{n_i}=n_{k(i)}>N$ implies:

$| \frac{\pi(n_i) \log(n_i)}{n_i}- \frac{\pi(\sqrt{n_i}) \log(\sqrt{n_i})}{\sqrt{n_i}}|< \frac{\epsilon}{4}$

But $\pi(n_i)= \pi(\sqrt{n_i})$ for each $i$, so we can simplify this to:

$\pi(n_i)| \frac{(\sqrt{n_i}/\log(\sqrt{n_i}))-1}{n_i^{3/2}/ \frac{1}{2} \log(n_i)^2}|< \frac{\epsilon}{4}$

Further simplification yields the following:

$\frac{1}{2}(\frac{\log(n_i)}{\sqrt{n_i}})(\frac{\log(n_i) \pi(n_i)}{n_i}) |\frac{\sqrt{n_i}}{\log(\sqrt{n_i})}-1|< \frac{\epsilon}{4}$

Now, $\lim \frac{\log(n_i) \pi(n_i)}{n_i}=1$ again by the prime number theorem, so we can choose $N' \geq N$ sufficiently large such that $\frac{\log(n_i) \pi(n_i)}{n_i} \geq \frac{1}{2}$ for $n_i >N'$. This gives us, for $n_i> N'$:

$\frac{1}{4}(\frac{\log(n_i)}{\sqrt{n_i}})|\frac{\sqrt{n_i}}{\log(\sqrt{n_i})}-1|< \frac{\epsilon}{4}$

So there is an $N' \in \mathbb{N}$ such that $n_i>N'$ implies:

$|2-\frac{\log(n_i)}{\sqrt{n_i}}|< \epsilon$

Since $\epsilon>0$ was arbitrary, this by definition implies:

$\lim \frac{\log(n_i)}{\sqrt{n_i}}=2$

But in general: $\displaystyle \lim_{x \to \infty} \frac{\log(x)}{\sqrt{x}}=0$, so every sequence with $x_n \rightarrow \infty$ will satsify: $(\frac{\log(x_n)}{\sqrt{x_n}}) \rightarrow 0$, so our contradiction is obvious.

So then by contradiction, there are at most finitely many $n \in \mathbb{N}$ such that $\pi(n)= \pi(\sqrt{n})$. $\Box$

No, there is no such a proof. Some years ago, I thought that there "must" be a proof of this fact, or to restate your proposition,
"There is always a prime between $n$ and $n^2$, when $n\geq 2$ "
should be proved using only combinatoric arguments and basic number theory. But I couldn't find such a proof and by the way, if you ever find such a proof, let me know, I will be very pleased too see it.


How do you distribute this negative? So I have $-(x - 2)^2$. Do I rewrite it as $-(x - 2)\cdot-(x - 2)$ and distribute the negative to the inside making it $(-x + 2)(-x + 2)$ or add the negative at the end of doing FOIL?

The minus sign is not in the scope of the exponent, so apply the minus AFTER squaring what's in parentheses.

To indicate the other, you would have to write $(-(x-2))^2$.

In the absence of inclusion symbols such as parentheses, brackets, braces, etc., exponents only apply to the immediately preceding factor that carries them.

Just square $(x-2)$ and negate it, i.e. multiply it by $-1$


If $L$ is a limit point of the sequence $(x^{(n)})^∞_{n=m}$, then $L$ is an adherent point of the set $\{x^{(n)}:n\ge m\}$. Let $(x^{(n)})^∞_{n=m}$ be a sequence of points in a metric space $(X, d)$, and let $L ∈ X$. Show that if $L$ is a limit point of the sequence $(x^{(n)})^∞_{n=m}$, then $L$ is an adherent point of the set $\{x^{(n)}:n\ge m\}$. Is the converse true?

Assume that $L$ is a limit point of  $(x^{(n)})^∞_{n=m}$

Then $\forall N\ge m$ and $\forall \epsilon>0$, $\exists n\ge N $ such that $d((x^{(n)}, L) \le \epsilon$

Closure definition of an adherent point: We say that $x_0$ is an adherent point of E if for every radius $r>0$, the ball $B(x_0,r)$ has a non-empty intersection with $E$.

It is also equivalent to say that $\exists$ a sequence $(x^{(n)})^∞_{n=m}$ in E which converges to $x_0$ with respect to the metric $d$.

In these defintions, are $x_0$ and L equivalent? Also could I get some help in formally writing this down?

Use the closure definition. Choose $\epsilon <r$. Then the point $x^{(n)}$ in the definition of limit point belongs to E and $B(L,r)$.

It seems to me the converse is not true. For a counterexample,  we can construct a sequence like {0, 1, -1, 1, -1, 1, -1, ... } in the metric space {R, d:|x-y|}, then 0 is the adhere point of the sequence set. However, 0 is not the limit point of this sequence because |1 - 0| is bigger than an arbitrary $\epsilon$ which is smaller than 1. Could you please help check whether there is anything wrong in this statement?


The limit $\lim_{n \to \infty}\left(2 + 1/n\right)^{n}$ How can one show that
$\displaystyle{\lim_{n \to \infty}\left(2 + {1 \over n}\right)^n}$ ?.

I am trying to shoehorn the definition of $e$ somewhere but fruitlessly so.

$$\displaystyle \lim_{n \to \infty} \left(2 + \frac 1n\right)^n\ge\lim_{n \to \infty} 2^n=+\infty$$

Applying L'Hopital $\lim(2+\frac{1}{n})^n=\lim(e^{\log(2+\frac{1}{n})n})=e^{\lim(\log(2+\frac{1}{n})/n^{-1})}$  the exponent will blow up.


Factorize expression $(4a+3)^2-a^4$ The question is to factorize $(4a+3)^2-a^4.$

The answer I ended up with was $(4a+3-a^2)(4a+3-a^2).$

I got this by difference of two squares $\cdots.$

I would like to know if this is correct.

SIMPLE

The question was to factorize $(4a+3)^2-a^4.$

IF you use difference of two squares we will get

LET $4a+3$  be $X$ and $a^2$ be $Y$

the $X^2-Y^2$

WE GET

$(4a+3-a^2)(4a+3+a^2).$

UNDERSTOOD

Further each of the quadratic equation can be factorised

$(x-1)(x-3)(x-/7+2)(x+/7+2)$

SO THE FINAL AWSER IS HERE


Solving $\int e^{|x|} dx$ Solving $\displaystyle\int e^{|x|} dx$.

I just solved a quite similar definite integral, and it wasn't hard, but it was because I could divide the integral in two integrals and get rid of modulus. Problem is I can't do that here, since it's indefinite now. I tried already substitution, parts, series and got nothing. How can I aproach such integral?

Thanks.

Edit: I'd like to obtain Wolfram's answer, not a function defined by parts.

Let us express the antiderivative as a definite integral plus a constant.

$$\int e^{|x|}dx=\int_0^x e^{|x|}dx+C.$$

Then for positive $x$,

$$e^x-1+C$$

and for negative $x$,

$$1-e^{-x}+C.$$

Hence

$$\text{sgn}(x)\left(e^{|x|}-1\right)+C.$$

As we can check, the derivative is $\text{sgn}^2(x)e^{|x|}$ (and $1$ for $x=0$).

Note that WA seemed to strive to avoid the absolute value in the exponent. We can achieve the same effect with

$$\text{sgn}(x)\frac{(1+\text{sgn}(x))(e^x-1)+(1-\text{sgn}(x))(e^{-x}-1)}{2}
=\frac{(\text{sgn}(x)+1)(e^x-1)-(\text{sgn}(x)-1)(e^x-1)e^{-x}}2
=e^{-x}\frac{\text{sgn}(x)(e^x-1)^2+e^{2x}-e^x-e^x-1}2.$$

This is ugly.

Write $e^{|x|}=e^{\operatorname{sgn}(x) x}$. Then

$$\int e^{|x|} dx = \int e^{\operatorname{sgn}(x)x} dx = \frac{e^{\operatorname{sgn}(x)x}}{\operatorname{sgn}(x)} + C = \operatorname{sgn}(x)e^{|x|} + C$$

edit: And well, this is broken.


When we say that $x^2$ = -4, would x = $\pm$2i or 2i? I saw a problem on Brilliant.org recently that sparked quite a heated discussion. The question was: Compute $\sqrt{-4} * \sqrt{-9}$. Now, If $\sqrt{-4}$ = 2i, the answer can only be -6. However, if $\sqrt{-4}$ = $\pm$2i, the answer can be $\pm$6. What would be the correct answer to this question?

Let me also say this: If we can split $\sqrt{-4}$ to $\sqrt{-1}*\sqrt{4}$, can we write the value of $\sqrt{-4}$ as $\pm$2i?

It's a silly thing to argue about because there's no standard definition of the square root for negative numbers. For nonnegative numbers there's no problem: we define $\sqrt{4}$ to be the positive number $x$ that solves $x^2=4$. This definition breaks down more generally because there's no natural extension of "positive" and "negative" to non-real numbers. If you like, you can create your own definition:

  For $a\in\mathbb R$, let $\sqrt{a}$ be the solution to $x^2=a$ whose real and imaginary parts are both nonnegative.

This looks reasonable to me; the problem is it's not a standard definition that everyone has accepted like in the $a\geq0$ case.

Edit: Another answer mentions the "principal branch" of the square-root function. The problem with this, and the reason there's confusion about the square root of negative numbers, is that the principal branch of $\sqrt{z}$ avoids the ray $(-\infty,0]$ in the complex plane. For any other complex number $z$, there's a standard choice of square root:
$$z^{1/2}=|z|^{1/2}e^{i\theta/2}$$
where $\theta\in(-\pi,\pi)$ is the angle $z$ makes with the positive real axis. Of course the negative real numbers correspond to values of $z$ with $\theta=\pm\pi$, which is outside of its domain. Depending on which sign of $\pi$ you use, you get different signs for $z^{1/2}$. The reason mathematicians decline to make such a choice is because the square root function would necessarily cease to be continuous (and in particular, complex analytic) at this branch cut.

There's a 3rd way to interpret $\sqrt{-4}*\sqrt{-9}$. Since $-4*-9=36$, couldn't one combine the values under the radicals to get $\sqrt{36}$? Then the answer is 6.The moral of the story is: it all depends on context and how you define your notation.


What is the quotient group of $O_2$/$SO_2$? I am working on this as a challenge problem, but I have never worked with a quotient group outside of modular values in the integers.

$SO_2$ is, by definition,  the kernel of the homomorphism $O_2\to \Bbb F^\times$,  $A\mapsto \det A$, which happens to be onto.


Does $\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 16}?$ How do we prove these two results?

$$\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 16}\tag1$$

$$\int_{0}^{\infty}{\sin(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 8\pi}\tag2$$

Try an approach to split out the form

$${\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 4}{\sin(16x^2\pi)-\sin(4x^2\pi)\over 2\sinh(x\pi)\cosh(x\pi)}\tag3$$

$$={1\over 4}{\sin(16x^2\pi)-4\sin(x^2\pi)\cos(x^2\pi)+8\sin^3(x^2\pi)\cos(x^2\pi)\over 2\sinh(x\pi)\cosh(x\pi)}\tag4$$
 Or we could write $(1)$ as

$$\int_{0}^{\infty}{({e^{i10x^2\pi}+e^{-i10x^2\pi}})({e^{i6x^2\pi}-e^{-i6x^2\pi}})\over4i \sinh^2(2x\pi)}\mathrm dx\tag5$$

$$\int_{0}^{\infty}{{e^{i16x^2\pi}-e^{-i16x^2\pi}}-({e^{i4x^2\pi}-e^{-i4x^2\pi}})\over4i \sinh^2(2x\pi)}\mathrm dx\tag6$$

$$\int_{0}^{\infty}{\sinh(i16x^2\pi)-\sinh(i4x^2\pi)\over2i \sinh^2(2x\pi)}\mathrm dx\tag7$$

Surely this is not the correct  approach here.

I estimated the closed form using wolfram integrator, not sure it is correct.

Maybe the most effective way to calculate these integrals is not to start from scratch and use residue theory, but to apply the general formula proved in Ramanujan's Lost Notebook, part IV, formula 14.4.14. In another words there is no need to reinvent the wheel. I will demonstrate the method and its effectiveness by calculating $(1)$. Note that
\begin{align}
I_1&=\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx\\
&=\int_{0}^{\infty}{\sin(16x^2\pi)-\sin(4x^2\pi)\over 2\sinh^2(2x\pi)}\mathrm dx\\
&=\int_{0}^{\infty}{32\pi x\cos(16x^2\pi)-8\pi x\cos(4x^2\pi)\over 4\pi}\left(\frac{1}{\tanh(2x\pi)}-1\right)\mathrm dx\\
&=\frac12\int_{0}^{\infty}{\left(4\cos(4x^2\pi)-\cos(x^2\pi)\right)}x\left(\frac{1}{\tanh(x\pi)}-1\right)~\mathrm dx
\end{align}

It is known that the following diverging integral can be regularized through the introduction of a regularization factor $e^{-\delta x^2},~\delta\to+0$
$$
\int_{0}^{\infty}x\cos\pi a x^2~dx=0\tag{*}.
$$

Ramanujan considers the function
$$
F_w(t)=\int_{0}^{\infty}{\sin(\pi t x)\over\tanh(x\pi)}e^{-\pi wx^2}~\mathrm dx
$$
and proves the following proposition

where the prime $'$ on the summation sign indicates that the terms with $j = 0, n$ are to be multiplied by $\frac{1}{2}$.

This formula allows one to calculate $F_w(t)$ when $w=-\eta_2ni/(\eta_1 m)$, because in that case $s=t$ and 14.4.14 can be solved for $F_w(t)$.

For example, for odd $m$ and $n$ 14.4.14 becomes:

Now divide (14.5.3) by $t$ and take the limit $t\to 0$. Of course the integral will be divergent, however it is made convergent by substracting $4\int_{0}^{\infty}x\cos(\pi x^2)~dx=0$. For example for $m=n=1$ one gets
$$
4\int_{0}^{\infty}{x\cos(x^2\pi)\left(\frac{1}{\tanh(2x\pi)}-1\right)}~\mathrm dx=\frac{1}{2}+\frac{-1}{2}(-1)+\left(\frac{-1}{2}\cdot \frac{1}{\sqrt{2}}+\frac{1}{2}\frac{-1}{\sqrt{2}}\right)-0\\
=1-\frac{1}{\sqrt{2}}.\tag{**}
$$

Numerical check confirms that (**) is correct.

When one of $m,n$ is even, then analog of $14.5.3$ will have $\sinh(mt)$ on the LHS instead of $\cosh(mt)$, and in this case one needs to divide by $t^2$ and apply Lopital's rule to the RHS. Of course there is no need to calculate everything by hand, because formulas $14.4.14$ can be plugged in Mathematica and closed forms evaluated automatically, thus saving a great deal of time and effort.

I've actually no idea how to proof it, but the question got me interested and I tried to do a few things. You could try to break them all down to exponentials. This is what I wrote on MathType for formula (1):

I would suggest to substitute the $e^{2\pi x}$ into a value $a$ and eventually you should get four integrals with something like
$ \left(\frac{ca^2}{a^2-1}\right) $. Evaluating this integral should give a logaritm and subsituting the value back could give the pretty number you were looking for. Good luck!


The definition "module of finite type". I know the definition of "module of finite type"

Is that different from finitely generated module ?

Thanks.

This term is sometimes used to mean a finitely generated module. That is to say, if someone says ``an $R$-module of finite type," he or she definitely means finitely generated as a module. I think that nowadays it is more common to say "finitely generated $R$-module," or, especially in commutative algebra, "finite $R$-module." I prefer this terminology. Of course the latter could lead to confusion in principle, but not usually in practice. In the contexts of $R$-algebras, an $R$-algebra of finite type is definitely not the same thing as an $R$-algebra that is finitely generated as an $R$-module (often just called a finite $R$-algebra).

finite type means  finitely generated $R$- algebra
finite means finitely generated $R$-module
so in Atiyah-Macdonald's famous book this proposition

results


If $f(a_n)\rightarrow L$ then $a_{n}\rightarrow L$ Let $\{a_{k}\}$ be a non-negative, increasing sequence. Consider

$$S_{n} = \frac{1}{n}\sum_{k=1}^{n}a_{k}.$$

I would like to show that if $$\lim_{n\rightarrow\infty}S_{n} = L,$$ then $$\lim_{k\rightarrow\infty}a_{k} = L.$$

I don't really have any clue how I could solve this problem. By definition, there exists $N$ such that

$$\lvert S_{n} -L\rvert <\epsilon$$

or

$$\frac{1}{n}\left\lvert \sum_{k=1}^{n}a_{k} - nL\right\rvert< \epsilon$$

for all $\epsilon>0$ whenever $n>N$ and

$$\frac{1}{n}\left\lvert \sum_{k=1}^{n}a_{k} - nL\right\rvert \leq \frac{1}{n}\sum_{k=1}^{n}\lvert a_{k}-L \rvert.$$

Now, if we define a new sequence $\{\alpha_{k}\}$ by

$$\alpha_{k} := \lvert a_{k} -L\rvert,$$

then I am pretty sure there is a theorem from harmonic analysis that states something like $\sum_{k=1}^{n}{\alpha_{k}}$ converges if and only if $(1/n)\sum_{k=1}^{n}{\alpha_{k}}$ converges too.

From here, by the Divergence Theorem, I would get that necessarily $\alpha_{k} \rightarrow L$ for $\sum{\alpha_{k}}$ to converge and hence $(1/n)\sum \alpha_{k}$ would also require $\alpha_{k}\rightarrow L$ to converge.

However, I can neither find a reference to this theorem I only vaguely remember and this still doesn't get me anywhere because I don't think the convergence of $(1/n)\sum \alpha_{k}$ is helpful anyway.

I have also tried proving this by contradiction too, but I would really prefer a direct proof.

Since the sequence is increasing $a_n \rightarrow \sup \{a_k : k \in \mathbb {N}\}$

Let prove that $L = \sup  \{a_k : k \in \mathbb {N}\}$. First for every $n$, $a_n \le L$. Indeed if there is $N$ that $a_N > L+\epsilon$ for $\epsilon > 0$. Since the sequence is increasing then for every $n \ge N$, $a_n > L+\epsilon$. Since the sequence is non negative, then $S_n > \frac{n - N}{n} (L+\epsilon)$ after tending $n$ to $\infty$ we have $L \ge L + \epsilon$ which is absurde.

On the other hand if for all $n$, $a_n < L - \epsilon$ for $\epsilon > 0$ then $S_n < L-\epsilon$. Tending $n$ to infinity $L \le L- \epsilon$ which is also absurde.

If $a_n$ tends to infinity them $L$ is infinity and if $a_n$ tends to a finite limit a then $L=a$. This is standard and the proof is elementary. The result follows immediately by considering these two cases.


Calculating integral with different boundary I am confused about calculting such integral with two ways of choosing boundary:
$$I_1 = \int_0^1 \frac 1{x^2}dx,\quad \text{for } 0 \le x \le 1$$
$$I_2 = \int_0^1 \frac 1{x^2}dx, \quad \text{for } 0 < x \le 1$$

What is the difference between $I_1$ and $I_2$ when we calculate these integral including the case $x=0$ and not? Thank you for your answers.

The expression $$I = \int_0^1 \frac{1}{x^2}\text{d}x$$ would contain enough information to unambiguously evaluate the integral, if it converged. This is because integrals over single points evaluate to 0.

To see this clearly, consider splitting the expression:
$$\left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 \le x \le 1 \right] = \left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 < x \le 1 \right] + \int_0^0 f(x) \, \text{d}x$$

However, both diverge when $f(x) = \frac{1}{x^2}$ because the closure of the interval of integration contains an essential singularity of the integrand.


The cardinal of finite field I know that the cardinal of a finite field is $p^n$ for $p$ a prime. The proof what I found is: if $K$ is a finite field then there is a  sub-field  $\Bbb F_p$ with cardinal $p$ and then we can see $K$ as a vector space over $\Bbb F_p$ so its cardinal is $p^n$ for some $n$. The problem with this proof is that I can't use the vector spaces so my question: is there another proof?

Suppose the field has an order divisible by two primes, $p\neq q$. The additive group of the field has elements $a$ of order $p$ and $b$ of order $q$ (this is group theory). The field also, therefore, contains the non-zero elements $pb=b+b+\dots +b\neq 0$ where $b$ is taken $p$ times, and this is non-zero because $b$ has order $q$, and also $qa\neq 0$.

Now $pb\cdot qa=pa\cdot qb=0$, contradicting the fact that we have a field.

So there can only be one prime dividing the order of the field, which must therefore be a power of that prime.

The fact that $K$ is a vector space over $\mathbb{F}_p$ is the core of the result so you're not going to find a truly different proof. If you don't want to use the language of vector spaces, you can imitate the proof as follows.  Since the field $K$ is finite, it contains $\mathbb{F}_p$ for some prime $p$.  Let $K_0 = \mathbb{F}_p$.  If $K_0 = K$ we are done. If $K \ne K_0$, choose some element $x \in K - K_0$ and let $K_1 = K_0[x]$.  Now prove by hand that $K_1$ is a field and its number of elements is a power of $p$.  If $K_1 = K$ we are done.  If not, choose another element and add it to $K_1$ to get a field $K_2$.  Prove that $|K_2|$ is still a power of $p$.  If $K_2 = K$ we are done, else repeat again to $K_3$, etc.  Since $K$ is finite, this algorithm eventually terminates and proves that $|K|$ is a power of $p$.


For $m \in \text{max} \ R[X]$ and $f \in R[X] \setminus m, \ R[X]_f$ is not a local ring Let $R$ be a commutative ring with $1$ and $m$ be a maximal ideal of $R[X]$. Let $f \in R[X]$ such that $f \notin m$. Then I want to show that $R[X]_f$ is not a local ring, where $R[X]_f$ is the localization of $R[X]$ with respect to the multiplicative closed set $\{ 1 , f ,f ^{2}, \ldots \}$.

My effort: On the contrary if we assume that $R[X]_f$ is local then clearly $R[X]_f \cong R[X]_{m}.$

But then I couldn't draw any contradiction. Please don't give me a reference where it was proved that there is another maximal ideal $n$ of $R[X]$ different from $m$ such that $f \notin n$ as well. To be honest I want to prove the previously stated fact in a different way.

We refer to the following commutative diagram:

Let $\text{loc}_f:R[X]\to R[X]_f$ be the canonical localization and $\eta:R[X]\to R[X]/m$ be the canonical quotient. Note that $R[X]/m$ is a field with $\eta(f)\neq 0$ and thus by the universal property of localization that there exists a unique ring morphism $k:R[X]_f\to R[X]/m$ with $\eta=k\circ \text{loc}$. Clearly $\text{ker}(k)\subsetneq R[X]_f$ is maximal.

Now denote by $\pi:R[X]\to R[X]$ the unique automorphism fixing $R\subset R[X]$ and taking $x\mapsto x+1$ and by $\pi^{-1}$ its inverse. From the U.P. of localization this induces an automorphism $\text{loc}(\pi):R[X]_f\to R[X]_f$ such that the "obvious" square $\text{loc}_f\circ \pi= \text{loc}(\pi)\circ \text{loc}_f$ commutes.

Observe that $\text{ker}(k\circ \text{loc}(\pi)^{-1})\subsetneq R[X]_f$ is like $\text{ker}(k)$ maximal. But $$k\circ\text{loc}_f=\eta=k\circ \text{loc}(\pi)^{-1}\circ \text{loc}_f\circ \pi$$ From which it follows that $$\text{loc}_f(p)\in \text{ker}(k)\iff p\in \text{ker}(\eta)\iff \text{loc}_f(p+1)\in \text{ker}(k\circ \text{loc}(\pi)^{-1})$$ And thus that $\text{ker}(k)$ and $\text{ker}(k\circ \text{loc}(\pi)^{-1})$ are disjoint maximal ideals of $R[X]_f$, as desired $\blacksquare$

We have $R[X]_{m} /(m\cdot R[X]_m)$ is a field. Whereas,
$B = (R[x]_f)/(m \cdot R[x]_f) = (R/m) [X]_f$ is not a field as the image $\bar{f}$, of $f$ in $(R/m) [X]$ which is a principal ideal domain with infinitely many prime ideals, has only finitely many prime
divisors and all the prime ideal to which $\bar{f}$ does not belong survive in $B$.


Classification of isolated singularity I have found the Laurent Series of $\displaystyle \frac{\sin3z}{z^5}$  about $z=0$ to be:

$$ \sum_{n=-2}^{\infty} \frac{(-1)^n3^{2n+5}z^{2n}}{(2n+5)!}$$

I now have to show that $\displaystyle \frac{\sin3z}{z^5}$ has a pole of order $4$ at $z=0.
$

How do I go about classifying this seeing as $z$ is to the power of $2n$ as opposed to the usual $n$?

We don't have to use the series expansion to see that $\sin(3z)/z^5$ has a pole of order four. From Calculus, we learned that
$$
\lim_{z\to 0}\frac{\sin(z)}{z} = 1
$$
and diverges to infinity when the degree of the denominator is positive and odd and DNE for positive even degrees.
For a residue at $z=0$ for $\sin(3z)/z^5$, we have
$$
\lim_{z\to 0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\frac{\sin(3z)}{z^5}(z-z_j)^n =
\lim_{z\to 0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\frac{\sin(3z)}{z^5}z^n
$$
where $z_j = 0$. It should be pretty apparent now what order the pole at $z=0$ is.

If you are interested in poles at $z=0$, you only need the terms in the series with a negative exponent of $z$.
Write down the first few terms and you'll see what happens.

Also, you can define
$$
a_n
=
\frac{(-1)^{n/2}3^{n+5}}{(n+5)!}
$$
for even $n$ and $a_n=0$ for odd $n$.
Then your series is $\sum_{n=-4}^\infty a_nz^n$, if that form is what you are after.


Find the derivative of y with respect to the given independent variable Find the derivative of y with respect to the given independent variable:

$y = 3^{-x} \stackrel{D}{\longrightarrow} y' = 3^{-x} \cdot (-1) \cdot \ln 3 $

This is my teacher's solution. I don't understand what he did. Can someone explain to me?

Your teacher implicitly wrote the function $y$ as
\begin{eqnarray}
y &=& 3^{-x} \\
  &=& e^{-x \ln 3}
\end{eqnarray}
From which the derivative is straightforward
\begin{eqnarray}
\frac{dy}{dx} &=& - \ln 3 . e^{-x \ln 3} \\
              &=& (-1).3^{-x}.\ln 3
\end{eqnarray}

use this standard result$$\frac{d a^x}{dx}= a^xln(a)$$
and apply chain rule


Prove the derivative of $\sin(1/x)$ exists How do I prove the derivative of $$\sin(1/x)=-\frac{1}{x^2}\cos(1/x)$$? I understand that you use $$f'(x_0) = \lim_{x \to x_0} \frac{\sin(1/x) - \sin(1/x_0)}{x-x_0} = -\frac{1}{x_0^2}\cos(1/x_0)$$ somehow but am not sure how to show it directly.

Any guidance would be greatly appreciated!

You could try implicit differentiation to solve this. Let $y = \sin\left(\frac{1}{x} \right)$. Then $\arcsin(y) = \frac{1}{x}$ and $$\frac{dy}{dx}\arcsin(y)=\frac{dy}{dx}\frac{1}{x} \\ \implies \frac{y'}{\sqrt{1-y^2}}=-\frac{1}{x^2} \\ \implies y' = -\frac{\sqrt{1-y^2}}{x^2}$$ Now plug back in $y = \sin\left(\frac{1}{x}\right)$ to get $$y'=-\frac{\sqrt{1-\sin^2\left(\frac{1}{x}\right)}}{x^2}$$ Then use the Pythagorean identity $$1-\sin^2\left(\frac{1}{x}\right)=\cos^2\left(\frac{1}{x}\right)$$ to conclude that $$y'=-\frac{\sqrt{\cos^2\left(\frac{1}{x}\right)}}{x^2} \\ =-\frac{\cos\left(\frac{1}{x}\right)}{x^2}  $$

\begin{align*}
\lim_{h\rightarrow 0}\frac{\sin(1/(x+h))-\sin(1/x)}{h}&=\lim_{h\rightarrow 0}\frac{\sin(1/x-h/x(x+h))-\sin(1/x)}{h}
\end{align*}
Now use $\sin(A-B)=\dots$ to reduce the problem to showing the following:
\begin{align*}
\lim_{h\rightarrow 0}\frac{\sin(h/x(x+h))}{h}&=\frac{1}{x^2}\\
\lim_{h\rightarrow 0}\frac{\cos(h/x(x+h))-1}{h}&=0
\end{align*}
(Use the substitution $\alpha=h/x(x+h),$ noting that $\alpha\rightarrow 0$ as $h\rightarrow 0$ ).


Showing $\left\lbrace x: \lim f_n(x) \text{ exists} \right\rbrace$ is measurable Assuming $\left\lbrace f_n\right\rbrace$ is measurable for all n, I was able to show using the fact that sequences of real numbers converge iff they are Cauchy that:

$\left\lbrace x:  \lim f_n(x) \text{ exists} \right\rbrace =\bigcap_{k=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{m=N}^{\infty} \bigcap_{n=N}^{\infty} \left\lbrace x:  f_n(x)-f_m(x) < \frac{1}{k} \right\rbrace \cap \left\lbrace x:  f_m(x)-f_n(x) < \frac{1}{k} \right\rbrace$

Since $f_n$ is measurble for all n, $\left\lbrace x:  f_n(x)-f_m(x) < \frac{1}{k} \right\rbrace $ should be a measurable set for all $n,m,$ and $k$, and then since there's a bunch of countable intersections and unions of all these sets, I should have that $\left\lbrace x:  \lim f_n(x) exists \right\rbrace$ is measurable as well. What I don't understand is why $f_n$ being measurable immediately implies that $\left\lbrace x:  f_n(x)-f_m(x) < \frac{1}{k}\right\rbrace $ is measurable. We just started learning about measurable functions and I know that $f:X \rightarrow Y$ is $M,N-$measurable if $f^{-1}(E) \in M$ for all $E \in N$ where $M$ is the sigma-algebra on $X$ and $N$ the sigma algebra on $Y$, but I don't see how to apply that definition here.

For sake of completness I shall give a sketch for another proof.

$1)$ Show that the maps $x \mapsto \lim \inf f_n(x)$ and $x \mapsto \lim \sup f_n(x)$ are measurables.

$2)$ Show that the set if $f, g$ are measurable functions, then the set $\{x : f(x) = g(x)\}$ is measurable.
To do this, use that since $\overline{\mathbb{R}}$ is Hausdorff, then the diagonal $\Delta := \{(x,x) : x \in \mathbb{R}\}$ is closed. Further, use thtat $f\times g(x) := (f(x),g(x))$.

$3)$ Recall that a $\lim f_n(x)$ exists if, and only if, $\lim \sup f_n(x) = \lim \inf f_n(x).$

What is the advantage of such approach? It is because you can change the counterdomain to any second countable Hausdorff space with Borel sigma algebra.


Prove that $A ∩ (B △ C) = (A △ B) ∩ (A △ C)$ I have to prove that for arbitrary sets $A, B, C$ the following statement is true:

$$A ∩ (B △ C) = (A △ B) ∩ (A △ C).$$

I'm familiar with the property $A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)$ which seems very close to the original problem and yet I can't figure out how to solve it. I'm trying to

Edit: The operator $△$ is defined as $A △ B = (A \setminus B) ∪ (B \setminus A)$. I'm trying to prove the problem with any known properties of sets and axioms from set theory.

Any suggestions and ideas will be greatly appreciated.
Thank you!

Your intended identity is false. Let's use this diagram:

So $A=\{1,2,4,5\}$, $B=\{2,3,5,6\}$, and $C=\{4,5,6,7\}$. Substituting into your "identity",

$$A\cap(B\Delta C)=(A\Delta B)\cap(A\Delta C)$$
$$\{1,2,4,5\}\cap(\{2,3,5,6\}\Delta \{4,5,6,7\})
=(\{1,2,4,5\}\Delta \{2,3,5,6\})\cap(\{1,2,4,5\}\Delta \{4,5,6,7\})$$
$$\{1,2,4,5\}\cap\{2,3,4,7\}
=\{1,3,4,6\}\cap \{1,2,6,7\}$$
$$\{2,4\}=\{1,6\}$$

which is clearly false.

I would love to use a shaded Venn diagram, but I'm not sure how to do that in Geogebra. Remember, Venn diagrams are your friends! They can be used to check "identities" and to give ideas on how to prove them.

Hint: $A=B$ means $A\subseteq B$ and $B\subseteq A$.

Also, $A\subseteq B$ means $x\in A\Rightarrow x\in B$.

Let $x\in A\cap (B\triangle C)$.

Then $x\in A\cap (B\setminus C\cup C\setminus B)$

$x\in (A\cap B\setminus C)\cup (A\cap C\setminus B)$

$x\in ((A\cap B\setminus C)\cup A)\cap ((A\cap B\setminus C)\cup C\setminus B)$

and so on.


How to prove $a_{n} < 2$ if $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}$ Let $(a_{n})_{n \geq1}$ be a real sequence such that $a_{1}=a_{2}=1$ and $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}, n\geq 1$.

  Prove that $a_{n} < 2, \forall n \geq 1.$

I write $$\sum a_{k+2}-a_{k+1}=\sum \frac{a_{k}}{3}$$
and I obtained :

$$3a_{n+2}=a_{1}+\ldots+a_{n}+3$$

or

$$a_{n}=\frac{a_{1}+\ldots+a_{n-2}+3}{3} < 2$$

And what remains to prove it is :

$$a_{1}+\ldots +a_{n-2} < 3,$$ but from this point I don't know how I have to do.

I need a proof without derivatives.

Thanks :)

if we assume there is a generating function so

$$ f(x)= \sum_{n=0}^{\infty}a_{n}x^{n} $$

then $ f(x) $ satisfy the functional equation

$$ f(x)-a_{0}-a_{1}x=f(x)x-a_{0}x+x^{2}f(x/3) $$

from this i think you could obtaien the derivatives so $$ n!f^{(n)}(0)= a_{n} $$


Simplifying trig (sin and cos) expression with identities I was working on a trig exercise and I'm not exactly sure how to simplify these expressions. I know the identities are needed to solve the problem but I've never seen angles like these before:
$\cos(\frac{4\pi}{9})$
and
$\sin(\frac{\pi}{55})$

These angles are in radians
I much prefer degrees to radians, so let's do the conversion! $\frac{4}{9} \cdot pi=80^\circ$. pi radians is equal to 180 degrees.
So we would like to find cos 80 degrees. You can use your calculator to solve that one.
sin(pi/55) is similar.
Hope this helped!


Trigonometry- Double Angle Trigonometric Equations $$\tan(3x)=((\tan x(3-\tan^2x))/(1-3\tan^2x))$$ I'm just having a hard time seeing where the double angle formula fits in with verifying this equation.

Using a Adicional formula: $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y},$$ we have:

$$\tan(2x+x)=\frac{\tan 2x+\tan x}{1-\tan 2x\tan x}=\frac{\tan (x+x)+\tan x}{1-\tan (x+x) \tan x}=\frac{\frac{\tan x + \tan x}{1-\tan x \tan x } +\tan x}{1-\frac{\tan x + \tan x}{1-\tan x \tan x } \tan x}$$
$$=\frac{\tan x +\frac{2\tan x}{1-\tan^2 x}}{1-\tan x \frac{2\tan x}{1-\tan^2 x}}=\frac{\frac{\tan x -\tan^3 x + 2\tan x}{1-\tan^2 x}}{\frac{1-\tan^2 x-2\tan^2 x}{1-\tan^2 x}}$$
$$=\frac{{\tan x -\tan^3 x + 2\tan x}}{{1-\tan^2 x-2\tan^2 x}}=\frac{{3\tan x -\tan^3 x }}{{1-3\tan^2 x}}=\frac{\tan x (3-\tan^2 x)}{1-3\tan^2 x}$$


How do I solve this Differential Algebraic Equation? I am trying to model a physical system which is goverened by the equation:

$$ (r(t))^2 \cdot \frac{d^2 r(t)}{dt^2} = c q(t)$$

I am a junior electrical engineering student, so I have had courses in linear algebra, ordinary differential equations, and LTI system theory + Fourier/Laplace/DFT/Z transforms.

Based on what I can tell about this equation, it is a Differential Algebraic Equation, which I am unfamiliar with how to solve. I would like to solve for $r(t)$ given an arbitrary function of $q(t)$ - say a sinusoidal, exponential, or polynomial function.

From what I've read online, an analytic solution may not be possible, so numerical solutions are fine too.

Also, all I can find online is how to solve a SYSTEM of DAEs, not simply a singular DAE equation.

What I would like to do is to be able to predict the output $r(t)$ given some input signal $q(t)$

As Yves mentioned your problem is an ordinary differential equation, however it is implicit (the highest derivative is not isolated). If you are expecting the solution cross the zero line the reformulation into an explicit system by diving by $r(t)^2$ is inappropiate. I will focus my answer on the implicit formulation.
Please note that $\dfrac{q(t)}{r^2(t)}$ is not Lipschitz-continous (has unbounded derivative) meaning it has likely no solution for all points in time/value space. Solvers might fail to solve it over the complete time interval you want it to be solved. You can solve implicit ODEs numerically using an DAE solver.
The idea is that DAE solver tries to find an du such that the residual function returns 0 and then step forward using that du.
Here is how you would do it in a Julia command line:
]add DifferentialEquations #installs a kitchen sink of tools
]add Plots # if you need plots

function residual(res,du,u,p,t)
   res[1] = du[1] - u[2]
   res[2] = du[2]*u[1]*u[1] - c*q(t)
end

c = 3 # when changed just rerun solve
q(t) = sin(t) # when changed just rerun solve

tspan = (0.0,1.0) #solve for virtual 1.0 time units
u0 = [1.,2.] # initial conditions for r(t) and s(t)

prob = DAEProblem(DAEFunction(residual),u0,u0,tspan) # If you change u0, or tspan rerun this too

sol = solve(prob, DABDF2())

sol(0.8) #interpolated solution at time 0.8
Array(sol) #solution without interpolation

plot(sol) ## savefig(plot(sol), "name.png") to save in working directory, you can use a path too

Another reformulation would be using a mass matrix. The idea would be that you bring your equation into this form.
$$M(\vec{x},t) \dfrac{d\vec{x}}{dt} = f(\vec{x},t)$$
However whether you have problems at zero crossings (where the $M^{-1}$ is illdefined) depends on the solver in question, i think. If you want to see code for that too, please add a comment saying so. I could also show how to solve the explicit formulation of Yves in Julia.

You can use a Runge-Kutta solver, for instance, but as your equation is of the second order, you will turn it to a system anyway:

$$\begin{cases}\dfrac{dr(t)}{dt}=s(t),\\\dfrac{ds(t)}{dt}=c\dfrac{q(t)}{r^2(t)}.\end{cases}$$


Why we assume henselian ring is local? Henselian ring is defined as local ring in which hensel lemma holds.
Why do we assume local ring?
What is wrong with the definition that 'henselian ring is defined as hensel lemma holds' ?

The version of Hensel's lemma I'm familiar with says that in $\mathbb{Z}_p$, if you have a simple root of a polynomial mod the maximal ideal then it lifts to a simple root. How would you generalize this if there is more than one maximal ideal?


Different meanings of $\int_0^T X(t) dt$, and its meaning in Ito isometry? Given a stochastic process $X: [0,T] \times \Omega \to \mathbb R$, I realized there are different meanings of $\int_0^T X(t) dt$.

$\int_0^T X(t, \omega) dt$, $\forall \omega \in \Omega$ or a.e., where the integral is Lebesgue integral wrt the Lebesgue measure on $[0,T]$. (This is the definition that seems most natural.)
$\int_0^T X(t, \omega) dt$, $\forall \omega \in \Omega$ or a.e., where the integral is Riemann integral over $[0,T]$. (This is the definition similar to Ito integral except the integrator $W(t)$ replaced with $t$.)
Given a partition $\mathcal P$ of $[0,T]$ into $0=t_0 < t_1 < t_2 < \dots < t_n=T$, define $\delta(\mathcal P) := \max_i (t_{i+1} - t_i)$. For each $i \leq n-1$, suppose that $t_i \leq \psi_i \leq t_{i+1}$. Define Riemann sum as $S_{\mathcal P} := \sum_i X(\psi_i) (t_{i+1} - t_i)$. If $S(\mathcal P)$ converges in $L^2$ norm as $\delta(\mathcal P)$ goes to $0$, then define $\int_0^T X(t) dt$ to be the limit. (This is the definition taken from Lamperti's Stochastic Processes.)

My questions are:

Is $\int_0^T X(t) dt$ always a random variable under each of the three definitions?
For a stochastic process $X$, does existence of one definition imply existence of another definition?
What if the first two definitions are relaxed to allow $\int_0^T X(t, \omega) dt$ exist a.e.?
In Ito isometry $E([\int_0^T X(t) dW(t)]^2) = E(\int_0^T X(t)^2 dt)$, how is $\int_0^T X(t)^2 dt$ on RHS defined?

Thanks and regards

I am in the middle of taking a stochastic calculus course, so I can't guarantee that all my answers are 100% correct, but I can share some ideas that my instructor mentioned related to the first question:

$ \int_0^T X(t)dt $ is non-random since this is an integral with respect to T, and therefore we are not looking at the randomness of X(t) but rather we are looking at the entire path, and integrating over this given path.


For Octave users, OctSymPy is a symbolic package which try to be compatible with SMT unless otherwise specified. It also claims to work on Matlab, but I don't use Matlab, so I cannot test it. (Please file a bug if it doesn't!)

To run your example, simply type (This should also work on SMT):

eig(sym([8 -2 1 -1; 23 -6 3 -3; -4 1 1 1; -4 2 0 1]))

A screenshot showing the result:

Feel free to ask if you don't understand something.

You're working with floating point arithmetic, so unless you use a symbolic package, there's going to be some incorrect digits.  You could set a tolerance and round to the nearest whole number, but that may impact your computation of other eigenvalues.  In Matlab, you could round to the second digit by way of:

round(1.003 + 0.0031i,2)

Edit: why the downvote, was something I said incorrect?


Prove that the number $ \underbrace{11 \ldots 1}_{(p-1) \mathrm{l}^{\prime} \mathrm{s}} $ is divisible by $p$ Question -

Example 1.30. Let $p \geq 7$ be a prime. Prove that the number
$
\underbrace{11 \ldots 1}_{(p-1) \mathrm{l}^{\prime} \mathrm{s}}
$
is divisible by $p$

Proof: We have
$
\underbrace{11 \ldots 1}_{p-1^{\prime} \mathrm{s}}=\frac{10^{p-1}-1}{9}
$
and the conclusion follows from Fermat's little theorem.

But by FLT we get ${10^{p-1}-1} \equiv 0 \pmod{p}$ how we will get that $\frac{10^{p-1}-1}{9} \equiv 0 \pmod{p}$?

Since $p$ is not divisible by $3$, modulo $p$ we have that division by $9$ corresponds to multiplication by some constant. Which constant will depend on $p$, but for instance, for $p=7$ we have $\frac19\equiv 4$, and for $p=11$ we have $\frac19\equiv 5$.

This gives us $10^{p-1}-1\equiv 0\implies \frac{10^{p-1}-1}{9}\equiv 0$.


If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable. I'm attempting to understand a proof presented on this Wikipedia page: https://en.wikipedia.org/wiki/Diagonalizable_matrix

The claim is as follows:

  Let $A$ be a matrix over $F$. If $A$ is diagonalizable, then so is any power of it. Conversely, if $A$ is invertible, $F$ is algebraically closed, and $A^n$ is diagonalizable for some $n$ that is not an integer multiple of the characteristic of $F$, then $A$ is diagonalizable.

This is the provided proof:

If $A^n$ is diagonalizable, then $A$ is annihilated by some polynomial ${\displaystyle \left(x^{n}-\lambda _{1}\right)\cdots \left(x^{n}-\lambda _{k}\right)}$, which has no multiple root (since ${\displaystyle \lambda _{j}\neq 0})$ and is divided by the minimal polynomial of $A$.

Here are my questions:

(a) Why is it clear that $A$ is annihilated by that polynomial? $A$ certainly is annihilated by $(x-\lambda_1)\cdots(x-\lambda_k)$ and $A^n$ is annihilated by $(x-\lambda_1^n)\cdots(x-\lambda_k^n)$ (as powers of matrices have powers of eigenvalues as their eigenvalues), but I don't see the connection to Wikipedia's claim.

(b) What tells us that there is no multiple root? It's clear to me that $\lambda_j\neq0$ ($A$ is invertible implies $A^n$ is invertible implies $A^n$ does not have $0$ as an eigenvalue), but why can't two of the $\lambda_j$'s be equal? What says that we have distinct eigenvalues?

(c) Any polynomial that annihilates a matrix certainly is a multiple of the minimal polynomial...but why does this tell us that $A$ is diagonalizable?

(d) Earlier in the article, the following is claimed: A matrix or linear map is diagonalizable over the field $F$ if and only if its minimal polynomial is a product of distinct linear factors over $F$. If questions (a) and (b) are resolved, I can see how this would imply (c), but why is this claim true?

Here is another approach to this problem, but this one seems to be more complicated than what is presented on Wikipedia. Positive power of an invertible matrix with complex entries is diagonalizable only if the matrix itself is diagonalizable.

That passage was written by a user with a Japanese user name "TakuyaMurata". I suppose what he meant was actually this: suppose $F$ is an algebraically closed field and $n>0$ is not an integer multiple of $\operatorname{char}(F)$. If $A$ is invertible and $A^n$ is diagonalisable over $F$, then:

$f(x)=(x^n-\lambda_1)\cdots(x^n-\lambda_k)$ annihilates $A$, where $\lambda_1,\ldots,\lambda_k$ are the distinct eigenvalues of $A^n$.
Since the $\lambda_i$s are distinct, $x^n-\lambda_i$ and $x^n-\lambda_j$ have not any common root when $i\ne j$. As $g(x):=x^n-\lambda_i$ and $g'(x)=nx^{n-1}$ also have no common roots (because $\lambda_i\ne0$ and $n\ne0$ --- we have used the assumptions that $A$ is invertible and $n$ is not an integer multiple of $\operatorname{char} F$ here), we see that $f$ has not any multiple roots.
As the minimal polynomial $m(x)$ of $A$ divides $f(x)$, $m$ also has not any multiple roots.
Therefore, $m$ is a product of distinct linear factors (note that $m$ splits because $F$ by assumption is algebraically closed). Hence $A$ is diagonalisable over $F$.

https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)

quote: An endomorphism φ of a finite dimensional vector space over a field F is diagonalizable if and only if its minimal polynomial factors completely over F into distinct linear factors. The fact that there is only one factor X − λ for every eigenvalue λ means that the generalized eigenspace for λ is the same as the eigenspace for λ: every Jordan block has size 1.

Some quick justifications here: https://www.maths.ed.ac.uk/~tl/minimal.pdf

So since we supposed $A^n$ diagonalisable, it has such a minimal polynomial $\mu(x)=\prod(x-\lambda_i)$ with distinct $\lambda_i$ that satisfies $\mu(A^n)=0$

But $\mu(A^n)=0$ also means that $M(A)=0$ where $M(x)=\prod(x^n-\lambda_i)$ or equivalently $M(x)=\mu(x^n)$.

Now if the $\lambda_i=r_ie^{i\theta_i}$ are distinct then so are their k-roots.

Indeed $(\lambda_1)^{1/n}=(\lambda_2)^{1/n}\implies\begin{cases} r_1=r_2\\ \frac{\theta_1+2n_1\pi}n=\frac{\theta_2+2n_2\pi}n+2m\pi\end{cases}\implies\begin{cases} r_1=r_2\\ \theta_1=\theta_2\pmod{2\pi}\end{cases}\implies \lambda_1=\lambda_2$

(We get the result by contraposition).

Considering now the minimal polynomial of $A$, it has to divide $M(x)$, but we have just seen that because of the hypothesis made on $F$, that $M$ is completely split, and all the linear factors are distinct.

Coming back to the quote at the beginning, it means that $A$ itself should be diagonalisable.


If I know values for 0% and 100%, how can I calculate the value at 90%? If I have 2 values, one representing 0% of a range (lets say 1.0), and another value representing 100% (lets say 20.0), how would I work out what value 90% would be? A formula would be handy if poss. Thanks.

So, I take this to mean you have an interval of values, $[a,b]$ where $a$ is the min and $b$ is the max and you wish to find a point $90\%$ of the way along. We can write $x\in[a,b]$ as $x=a+(b-a)t$ where $t\in[0,1]$. It turns out that the value of $t$ in this formula is the percentage. At $t=0$ we get $a$, at $t=1$ we get $b$, and at $t=0.9$ we get $0.9b+0.1a$.

Again, the general formula (with $t$ as the percentage) is:

$$x(t)=a+(b-a)t=a(1-t)+bt$$

$1=\frac {100}{100}=100$%.

$0=\frac {0}{100}=0$%

$0.9=\frac {90}{100}=90$%

So in your example $90$%$=\frac {9}{10} \cdot (20-1)=17.1$, in a sense that we interpret this as $90$% of interval from $1$ to $20$, which we could write as $[1,20]$, is $17.1$.

In general you get $a$% buy multiplying $\frac {a}{100}$ with the value of which you want to take $a$%.


Structure of inflation/deflation on Penrose Tilings? Thinking about P2/P3 type penrose tilings (kites/darts or rhombs - for this question they should be equivalent?) we know we can "inflate"/"deflate" any tiling of the plane to get another. We also know that the number of Penrose tilings is uncountably infinite. Do we know what structure the "inflation/deflation" procedure bestows upon the set of penrose tilings?

To be more precise, for example we know that the 'sun' and the 'star' penrose tilings are dual under inflation/deflation. Are there any other 'cycles' of penrose tilings? In my mind, the inflation/deflation introduces a kind of 'local linear order' on the set of penrose tilings - is it known how to describe this structure more carefully?

Unsure of where to go for answers - I've been trying to find Penrose's paper “The Role of Aesthetics in Pure and Applied Mathematical Research.” in Bulletin of the Institute of Mathematics and Its Applications, 10, 1974. pp. 266–271. to no avail

The short answer is that there are infinitely many 'cycles' with arbitrarily long lengths. We would call them periodic orbits under the substitution action. (Not to be confused with periodic orbits under the translation action which obviously do not exist for aperiodic tiling spaces).

Periodic points for the substitution action on the tiling space are encoded in the dynamical zeta function which can be computed via the Lefschetz fixed point theorem and using inverse limit techniques from work of Anderson and Putnam.

https://www.cambridge.org/core/journals/ergodic-theory-and-dynamical-systems/article/topological-invariants-for-substitution-tilings-and-their-associated-castalgebras/5D7DB90543165287CA5C4CBF012323F5

There are many other ways of calculating the zeta function (And the associated cohomology groups), for example via Barge-Diamond complexes or the (dual) Walton homology groups of a tiling.

The minimal embedding of the rhombic version of the Penrose tiling, if to be constructed by cut and Project, is known to be the root lattice $A_4$.

All details of the according construction, lifting of tiling space paths to lattice paths (or re-projecting these to perp space), esp. the structure of the according acceptance domain, the "origins" of the local tiling configurations in the language of acceptance domain sub-regions, the rules of its in-/deflation symmetries, the consideration of isomorphism classes of tilings, etc. all were contained already within the very first according publication. Cf.:

M. Baake, P. Kramer, M. Schlottmann, D. Zeidler, "Planar Patterns with fivefold symmetry as sections of periodic structures in 4-space", Int. J. Mod. Phys. B4 (1990) 2217-2268.

--- rk


$k[x]/(x^2)$ has only one prime ideal Let $k$ be a field. Then $k[x]/(x^2)$ has only one prime ideal.

It is clear that $(X)$ is the maximal ideal thus prime where $X$ is the class elements $x$. How do I know there aren't other prime ideals? I tried to define homomorphism $\phi':k[x]/(x^2)\to k$ by evaluation but it seems there is no other way to define the map well defined other than evaluating at 0 due to $\phi'([f])=\phi'([g])$ not equal for different representatives.

Suppose $\mathfrak{p}$ is a prime ideal. Then $x\cdot x = 0\in \mathfrak{p}$, so $x\in \mathfrak{p}$, and $(x)\subseteq \mathfrak{p}$. Since, as you note, $(x)$ is maximal, $(x) = \mathfrak{p}$.

Don't forget nontrivial assumption. Anyway suppose the representative ax+b is contained. Then as will (1/b×x)(ax+b). This tells you the class of 1 is there too after subtraction of ax and division by b. That then says everything is. Notice b was nonzero for this. That just leaves you with (x) as you already have.


Proof verification: Compact set has sup and inf I was reading this post compact set always contains its supremum and infimum

There was an answer reposted as follows:

  As $K$ is compact, we have that $K$ is bounded. So $\sup K$ and $\inf
 K$ exists. By definition $\sup K$, for every $n \in N$ exists $x_n \in
 K$ such that $\sup K- x_n<1/n$ then $\sup K = \lim x_n$ with $x_n \in
 K$, as K is closed follows that $\sup K \in K$. To inf is analogous.
  Ps: Compact ⇒ closed and bounded.

This proof is quite attractive but I do not get the idea that just because $\sup K$ exists, therefore there exists a sequence approaching it.

Can someone please explain this part to clarify why that $\sup K$ must exist for a compact set $K$?

Here is why there always exists a sequence in $K$ converging to $\sup K$, provided its existence (which is guaranteed by the fact that bounded sets always have a least upper bound in $\Bbb R$).

Let $A\subseteq \Bbb R$ and let $s=\sup A<+\infty$.

If there were a $n>0$ such that $\left(s-\frac1n,+\infty\right)\cap A=\emptyset$, then $s-\frac1{2n}$ would be an upper bound for $A$, which would contradict the fact that $s$ is the least upper bound of $A$. Therefore, for all $n$ there must exist $x_n\in A$ such that $x_n> s-\frac1n$.

But since $s$ is an upper bound of $A$ and $x_n\in A$, it must hold $x_n\le s$ as well.

Therefore $s-\frac1n < x_n\le s$. By the squeeze therem, $x_n\to s$.

Compact is equivalent to closed + bounded in R. Bounded sets in R have sup and inf.