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Fermat Numbers

Wednesday, September 28th, 2011 06:00 / Leave a Comment

Fermat Number, a class of numbers, is an integer of the form $F_n=2^{2^n} +1 \ \ n \ge 0$ .

For example: Putting $n := 0,1,2 \ldots$ in $F_n=2^{2^n}$ we get $F_0=3$ , $F_1=5$ , $F_2=17$ , $F_3=257$ etc.

Fermat observed that all the integers $F_0, F_1, F_2, F_3, \ldots$ were prime numbers and announced that $F_n$ is a prime for each natural value of $n$ .

In writing to Prof. Mersenne, Fermat confidently announced:

I have found that numbers of the form $2^{2^n}+1$ are always prime numbers and have long since signified to analysts the truth of this theorem.

However, he also accepted that he was unable to prove it theoretically. Euler in 1732 negated Fermat’s fact and told that $F_1 -F_4$ are primes but $F_5=2^{2^5} =4294967297$ is not a prime since it is divisible by 641.
Euler also stated that all Fermat numbers are not necessarily primes and the Fermat number which is a prime, might be called a Fermat Prime. Euler used division to prove the fact that $F_5$ is not a prime. The elementary proof of Euler’s negation is due to G. Bennett.

Theorem:

The Fermat number $F_5$ is divisible by $641$ i.e., $641|F_5$ .

Proof:

As defined $F_5 :=2^{2^5}+1=2^{32}+1 \ \ldots (1)$

Factorising $641$ in such a way that $641=640+1 =5 \times 128+1 \\ =5 \times 2^7 +1$
Assuming $a=5 \bigwedge b=2^7$ we have $ab+1=641$ .

Subtracting $a^4=5^4=625$ from 641, we get $ab+1-a^4=641-625=16=2^4 \ \ldots (2)$ .

Now again, equation (1) could be written as
$F_5=2^{32}+1 \\ \ =2^4 \times {(2^7)}^4+1 \\ \ =2^4 b^4 +1 \\ \ =(1+ab-a^4)b^4 +1 \\ \ =(1+ab)[a^4+(1-ab)(1+a^2b^2)] \\ \ =641 \times \mathrm{an \, Integer}$
Which gives that $641|F_n$ .

Mathematics is on its progression and well developed now but it is yet not confirmed that whether there are infinitely many Fermat primes or, for that matter, whether there is at least one Fermat prime beyond $F_4$ . The best guess is that all Fermat numbers $F_n>F_4$ are composite (non-prime).
A useful property of Fermat numbers is that they are relatively prime to each other; i.e., for Fermat numbers $F_n, F_m \ m > n \ge 0$ , $\mathrm{gcd}(F_m, F_n) =1$ .

Following two theorems are very useful in determining the primality of Fermat numbers:

Pepin Test:

For $n \ge 1$ , the Fermat number $F_n$ is prime $\iff 3^{(F_n-1)/2} \equiv -1 \pmod {F_n}$

Euler- Lucas Theorem

Any prime divisor $p$ of $F_n$ , where $n \ge 2$ , is of form $p=k \cdot 2^{n+2}+1$ .

Fermat numbers ( $F_n$ ) with $n=0, 1, 2, 3, 4$ are prime; with $n=5,6,7,8,9,10,11$ have completely been factored; with $n=12, 13, 15, 16, 18, 19, 25, 27, 30$ have two or more prime factors known; with $n=17, 21, 23, 26, 28, 29, 31, 32$ have only one prime factor known; with $n=14,20,22,24$ have no factors known but proved composites. $F_{33}$ has not yet been proved either prime or composite.

Video: Documentary on Proof of Fermat’s Last Theorem (wpgaurav.wordpress.com)

Real and Complex projective n-spaces

Sunday, April 17th, 2011 07:47 / Leave a Comment

Real projective n-space

Real projective $n$ space, $\mathbb{R} P^n$ is defined to be the space of all lines through the origin in $\mathbb{R}^{n+1}$ . Each such kind is determined by a non-zero vector in $\mathbb{R}^{n+1}$ , unique up to scalar multiplication, and $\mathbb{R} P^n$ is topologized as the quotient space of $\mathbb{R}^{n+1}-\{0\}$ under the equivalence relation $v \sim \lambda v$ for scalars $\lambda \neq 0$ . We can restrict to vectors of length 1, so $\mathbb{R}P^n$ is also the quotient spaces $\mathbf{S}^n / (v \sim -v)$ , the sphere with antipodal points identified. This is equivalent to saying that $\mathbb{R}P^n$ is the quotient space of a hemisphere $\mathbf{D^n}$ with antipodal points of $\partial D^n$ identified. Since $\partial D^n$ with antipodal points identified is just $\mathbb{R}P^{n-1}$ , we see that $\mathbb{R}P^n$ is obtained from $\mathbb{R}P^{n-1}$ by attaching an $n$ -cell, with the quotient projection $S^{n-1} \to \mathbb{R}P^{n-1}$ as the attaching map. It follows by induction on $n$ that $\mathbb{R}P^n$ has a cell complex structure $e^0 \cup e^1 \cup ... \cup e^n$ with one cell $e^i$ in each dimension $i \leq n$ . Since $\mathbb{R}P^n$ is obtained from $\mathbb{R}P^{n-1}$ by attaching an $n$ -cell, the infinite union $\mathbb{R}P^{\infty}={\bigcup}_n \mathbb{R}P^n$ becomes a Cell Complex with one cell in each dimension. We can view $\mathbb{R}P^{\infty}$ as the space of lines through the origin in $\mathbb{R}^{\infty}={\bigcup}_n \mathbb{R}^n$ .

Complex Projective n-space

Complex projective n-space $\mathbb{C}P^n$ is space of complex lines through the origin in $\mathbb{C}P^{n+1}$ , that is, 1-dimensional vector subspaces of $\mathbb{C}^{n+1}$ . As in the case of Real projective n-space, each line is determined by a non-zero vector in $\mathbb{C}^{n+1}$ , unique up to scalar multiplication, and $\mathbb{C}P^n$ is topologized as the quotient space of $\mathbb{C}^{n+1}-0$ under the equivalence relation $v \sim \lambda v$ for $\lambda = 0$ .

Equivalently, this is the quotient of the unit Sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$ with $v \sim \lambda v$ for $| \lambda |=1$ .

It is also possible to obtain $\mathbb{C}P^n$ as a quotient space of disk $D^{2n}$ under the identifications $v \sim \lambda v$ for $\lambda v \in \partial D^{2n}$ , in the following way:

The vectors in $S^{2n+1} \subset \mathbb{C}^{n+1}$ with last coordinate real and nonnegative are precisely the vectors of the form $(\omega, \sqrt{1-{|\omega|}^2} ) \in \mathbb{C}^n \times \mathbb{C}$ with $|\omega| \leq 1$ . Such vectors form the of the function $\omega \to \sqrt{1-{|\omega|}^2}$ . This is a disk ${D_+}^{2n}$ bounded by the sphere $S^{2n-1} \subset S^{2n+1}$ consisting of vectors $(\omega, 0) \in \mathbb{C}^n \times \mathbb{C}$ with $|\omega|=1$ . Each vector in $S^{2n+1}$ is equivalent under the identifications $v \sim \lambda v$ to a vector in ${D_+}^{2n}$ , and the latter vector is unique if its last coordinate is non-zero. If the last coordinate is zero, we have just the identifications $v \sim \lambda v$ for $v \in S^{2n-1}$ .

From the description of $\mathbb{C}P^n$ as the quotient of ${D_+}^{2n}$ under the identifications $v \sim \lambda v$ for $v \in S^{2n-1}$ , it follows that $\mathbb{C}P^n$ is obtained from $\mathbb{C}P^{n-1}$ by attaching a cell $e^{2n}$ via the quotient map $S^{2n-1} \to \mathbb{C}P^{n-1}$ . So by induction on $n$ , we obtain a cell structure

$\mathbb{C}P^n=e^0 \cup e^2 \cup ...\cup e^{2n}$

with cells only in even dimensions. Similarly, $\mathbb{C}P^{\infty}$ has a cell structure with one cell in each even dimension.

Note: The Sphere $S^n$ has the structure of a cell complex with just two cells, $e^0$ and $e^n$ , the n-cell being attached by the constant map $S^{n-1} \to e^0$ . This is equivalent to regarding $S^n$ as the quotient space $\dfrac{D^n}{\partial D^n}$ .

How to Draw a Cell Complex (or CW Complex)

Let we try to construct a space by following procedure:

Start with a discrete set ${X^0}$ , whose points are regarded as $\mathbf{0}$ -cells.
Inductively, form the $\textbf{n}$ -skeleton $X^n$ from $X^{n-1}$ by attaching $n$ -cells $e^n_{\alpha}$ via maps $\Phi_{\alpha} : S^{n-1} \rightarrow X^{n-1}$ . This means that $X^n$ is the quotient space of the disjoint union $X^{n-1} \, \mathbf{\sqcup_{\alpha}} D^n_{\alpha}$ of $X^{n-1}$ with a collection of n-disks $D^n_\alpha$ under the identifications $x \sim \Phi_{\alpha} (x)$ for $x \in \partial D^n_\alpha$ . Thus as a set

$X^n=X^{n-1} \mathbf{\sqcup_{\alpha}} e^n_\alpha$

where each $e^n_\alpha$ is an open $n$ disk.

One can either stop this inductive process at a finite stage, setting $X=X^n$ for some $n < \infty$ , one can continue indefinitely, setting $X=\bigcup_n X^n$ .

A space $X$ constructed in this way is called a CELL COMPLEX or CW COMPLEX.

Reference:

ALGEBRAIC TOPOLOGY

Allen Hatcher

http://www.math.cornell.edu/~hatcher/#ATI

D’ ALEMBERT’s Test of Convergence of Series

Saturday, April 9th, 2011 11:36 / 1 Comment

Statement

A series $\sum {u_n}$ of positive terms is convergent if from and after some fixed term $\dfrac {u_{n+1}} {u_n} < r < {1}$ , where r is a fixed number. The series is divergent if $\dfrac{u_{n+1}} {u_n} > 1$ from and after some fixed term.

D’ Alembert’s Test is also known as ratio test of convergency of a series.

Definitions for Generally Interested Readers

(Definition 1) An infinite series $\sum {u_n}$ i.e. $\mathbf {u_1+u_2+u_3+....+u_n}$ is said to be convergent if $S_n$ , the sum of its first $n$ terms, tends to a finite limit $S$ as n tends to infinity.
We call $S$ the sum of the series, and write $S=\displaystyle {\lim_{n \to \infty} } S_n$ .
Thus an infinite series $\sum {u_n}$ converges to a sum S, if for any given positive number $\epsilon$ , however small, there exists a positive integer $n_0$ such that
$|S_n-S| < \epsilon$ for all $n \ge n_0$ .
(Definition 2)
If $S_n \to \pm \infty$ as $n \to \infty$ , the series is said to be divergent.
Thus, $\sum {u_n}$ is said to be divergent if for every given positive number $\lambda$ , however large, there exists a positive integer $n_0$ such that $|S_n|>\lambda$ for all $n \ge n_0$ .
(Definition 3)
If $S_n$ does not tends to a finite limit, or to plus or minus infinity, the series is called Oscillatory

Discussions

Let a series be $\mathbf {u_1+u_2+u_3+.......}$ . We assume that the above inequalities are true.

From the first part of the statement:
$\dfrac {u_2}{u_1} < r$ , $\dfrac {u_3}{u_2} < r$ ……… where r <1.
Therefore $\mathbf {{u_1+u_2+u_3+....}= u_1 {(1+\frac{u_2}{u_1}+\frac{u_3}{u_1}+....)}}$
$=\mathbf {u_1{(1+\frac{u_2}{u_1}+\frac{u_3}{u_2} \times \frac{u_2}{u_1}+....)}}$
$< \mathbf {u_1(1+r+r^2+.....)}$
Therefore, $\sum{u_n} < u_1 (1+r+r^2+.....)$
or, $\sum{u_n} < \displaystyle{\lim_{n \to \infty}} \dfrac {u_1 (1-r^n)} {1-r}$
Since r<1, therefore as $n \to \infty , \ r^n \to 0$
therefore $\sum{u_n} < \dfrac{u_1} {1-r}$ =k say, where k is a fixed number.
Therefore $\sum{u_n}$ is convergent.
Since, $\dfrac{u_{n+1}}{u_n} > 1$ then, $\dfrac{u_2}{u_1} > 1$ , $\dfrac{u_3}{u_2} > 1$ …….
Therefore $u_2 > u_1, \ u_3 >u_2>u_1, \ u_4 >u_3 > u_2 >u_1$ and so on.
Therefore $\sum {u_n}=u_1+u_2+u_3+....+u_n$ > $nu_1$ . By taking n sufficiently large, we see that $nu_1$ can be made greater than any fixed quantity.
Hence the series is divergent.

Comments

When $\dfrac {u_{n+1}} {u_n}=1$ , the test fails.

Another form of the test–

The series $\sum {u_n}$ of positive terms is convergent if $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ >1 and divergent if $\displaystyle{\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ <1.
One should use this form of the test in the practical applications.

A Problem:
Verify whether the infinite series $\dfrac{x}{1.2} + \dfrac {x^2} {2.3} + \dfrac {x^3} {3.4} +....$ is convergent or divergent.

Solution

We have $u_{n+1}= \dfrac {x^{n+1}}{(n+1)(n+2)}$ and $u_n= \dfrac {x^n} {n(n+1)}$
Therefore $\displaystyle {\lim_{n \to \infty}} \dfrac{u_n} {u_{n+1}} = \displaystyle{\lim_{n \to \infty}} (1+\frac{2}{n}) \frac{1}{x} = \frac{1}{x}$
Hence, when 1/x >1 , i.e., x <1, the series is convergent and when x >1 the series is divergent.
When x=1, $u_n=\dfrac{1} {n(n+1)}=\dfrac {1}{n^2} {(1+1/n)}^{-1}$
or, $u_n=\dfrac{1}{n^2}(1-\frac{1}{n}+ \frac {1}{n^2}-.....)$
Take $\dfrac{1}{n^2}=v_n$ Now $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{v_n}=1$ , a non-zero finite quantity.
But $\sum {v_n}=\sum {\frac{1}{n^2}}$ is convergent.
Hence, $\sum {u_n}$ is also Convergent.

MY DIGITAL NOTEBOOK

Tag Archives: Theorems

Fermat Numbers

Theorem:

Proof:

Pepin Test:

Euler- Lucas Theorem

Related articles

Real and Complex projective n-spaces

Real projective n-space

Complex Projective n-space

How to Draw a Cell Complex (or CW Complex)

D’ ALEMBERT’s Test of Convergence of Series

Statement

Definitions for Generally Interested Readers

Discussions

Comments

Another form of the test–

Solution

Follow “MY DIGITAL NOTEBOOK”

MY DIGITAL NOTEBOOK

Tag Archives: Theorems

Fermat Numbers

Theorem:

Proof:

Pepin Test:

Euler- Lucas Theorem

Related articles

Better You Share!

Like this:

Real and Complex projective n-spaces

Real projective n-space

Complex Projective n-space

How to Draw a Cell Complex (or CW Complex)

Better You Share!

Like this:

D’ ALEMBERT’s Test of Convergence of Series

Statement

Definitions for Generally Interested Readers

Discussions

Comments

Another form of the test–

Solution

Better You Share!

Like this:

Follow “MY DIGITAL NOTEBOOK”