Home » Posts tagged 'Theorems'

Tag Archives: Theorems

Fermat Numbers

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.

Related articles

Real and Complex projective n-spaces

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:

  1. Start with a discrete set {X^0}, whose points are regarded as \mathbf{0}-cells.
  2. 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

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.

Follow

Get every new post delivered to your Inbox.

Join 755 other followers

%d bloggers like this: