Fibonacci number: Difference between revisions

From Citizendium
Jump to navigation Jump to search
imported>Aleksander Stos
(→‎Direct formula: straightforward claim of the formula)
imported>Wlodzimierz Holsztynski
(→‎Properties: derivations + more (also -- a clean up). Formatting (the look) is not satisfactory, mea culpa.)
Line 16: Line 16:
==Properties==
==Properties==


*The quotient of two consecutive fibonacci numbers converges to the [[golden ratio]]: <math>\lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\varphi</math>
*The quotient of two consecutive fibonacci numbers converges to the [[golden ratio]]:
*If <math>\scriptstyle m > 2\ </math> divides <math>\scriptstyle n\ </math> then <math>\scriptstyle F_m\ </math> divides <math>\scriptstyle F_n\ </math>
 
*If <math>\scriptstyle p \ge 5</math> and <math>\scriptstyle F_p\ </math> is a prime number then <math>p</math> is prime. (The converse is false.)  
:::<math>\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi</math>
*<math>\operatorname{gcd}(f_m, f_n) = f_{\operatorname{gcd}(m,n)} </math>
 
*<math>F_0^2 + F_1^2 + F_2^2 + ... + F_n^2 = \sum_{i=0}^n F_i^2 = F_n \cdot F_{n+1}</math>
Below, we will use the following, simple observation:
 
if three integers <math>\ a,b,c,</math>&nbsp; satisfy equality <math>\ c = a+b,</math>&nbsp; then
 
::<math>\ \gcd(a,b)\ =\ \gcd(a,c)=\gcd(b,c)=1.</math>
 
 
* <math>\gcd\left(F_n,F_{n+1}\right)\ =\ \gcd\left(F_n,F_{n+2}\right)\ =\ 1</math>
 
Indeed,
 
::<math>\gcd\left(F_0,F_1\right)\ =\ \gcd\left(F_0,F_2\right)\ =\ 1</math>
 
and the rest is an easy induction.
 
 
* <math>F_n\ =\ F_{k+1}\cdot F_{n-k} + F_k\cdot F_{n-k-1}</math>
 
:for all integers <math>\ k,n,</math>&nbsp; such that <math>\ 0\le k < n.</math>
 
 
Indeed, the equality holds for <math>\ k=0,</math>&nbsp; and the rest is a routine induction on <math>\ k.</math>
 
Since <math>\gcd\left(F_k,F_{k+1}\right) = 1</math>,&nbsp; the above equality implies:
 
::: <math>\gcd\left(F_k,F_n\right)\ =\ \gcd\left(F_k,F_{n-k}\right)</math>
 
which, via Euclid algorithm, leads to:
 
 
*<math>\gcd(F_m, F_n)\ =\ F_{\gcd(m,n)} </math>
 
Let's note the two instant corollaries of the above statement:
 
 
*If <math>\ m</math>&nbsp; divides <math>\ n\ </math> then <math>\ F_m\ </math> divides <math>\ F_n\ </math>
 
*If <math>\ F_p\ </math>&nbsp; is a prime number then <math>\ p</math>&nbsp; is prime. (The converse is false.)
 
 
*<math>\sum_{i=0}^n F_i^2 = F_n \cdot F_{n+1}</math>


== Direct formula ==
== Direct formula ==

Revision as of 18:06, 29 December 2007

This article is a stub and thus not approved.
Main Article
Discussion
Related Articles  [?]
Bibliography  [?]
External Links  [?]
Citable Version  [?]
 
This editable Main Article is under development and subject to a disclaimer.

In mathematics, the Fibonacci numbers form a sequence defined by the following recurrence relation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n := \begin{cases} 0 & \mbox{if } n = 0; \\ 1 & \mbox{if } n = 1; \\ F_{n-1}+F_{n-2} & \mbox{if } n > 1. \\ \end{cases} }

The sequence of fibonacci numbers start: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...

Properties

  • The quotient of two consecutive fibonacci numbers converges to the golden ratio:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi}

Below, we will use the following, simple observation:

if three integers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ a,b,c,}   satisfy equality Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ c = a+b,}   then


Indeed,

and the rest is an easy induction.


for all integers   such that


Indeed, the equality holds for   and the rest is a routine induction on

Since ,  the above equality implies:

which, via Euclid algorithm, leads to:


Let's note the two instant corollaries of the above statement:


  • If   divides then divides
  • If   is a prime number then   is prime. (The converse is false.)


Direct formula

We have

for every .

Indeed, let    and   .  Let

Then:

  •     and    
  •     hence    
  •     hence    

for every . Thus   for every and the formula is proved.

Furthermore, we have:

  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A\cdot a = -1\ }
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A > 1\ }
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1 < a < 0\ }
  • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}\ >\ \left|\frac{1}{\sqrt{5}}\cdot a^n\right|\quad\rightarrow\quad 0}

It follows that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_n\ }   is the nearest integer to  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{5}}\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n}

for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ n=0,1,\dots} . It follows that  Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty}\frac{F(n+1)}{F(n)}=A} ;  thus the value of the golden ratio is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \varphi\ =\ A\ =\ \frac{1+\sqrt{5}}{2}} .

Further reading

Applications

The sequence of Fibonacci numbers was first used to represent the growth of a colony of rabbits, starting with one pair of rabbits.