Musique & Mathématique

14 septembre 2009

Continued fraction

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Continued fractions

Introduction

The decomposition in continued fractions provides an approximation of a real by a series of fractions more and more precise.

Procedure

For example, starting from the real S = log(2)/log(3/2) = 1,70951129 encounted in the Score article.
The suite of the fractions for the S approximation is :
1/1, 2/1, 5/3, 12/7, 41/24, 53/31, 306/179, 665/389, …

Ok, but how to do that ??
First, we define the suite un as :

u0 = S = 1,70951129…
un+1 = 1 / fract(un) , where fract(un) is the fractionnal part of un.

Then, we defined the suite vn = ent(un), integer part of un.

With a simple calculator (four operations) we get the un and vn values:

u0 = 1,709511291 v0 = 1
u1 = 1,40942084 v1 = 1
u2 = 2,442474596 v2 = 2
u3 = 2,260016753 v3 = 2
u4 = 3,84590604 v4 = 3
u5 = 1,1821644 v5 = 1
u6 = 5,489547 v6 = 5
u7 = 2,04270 v7 = 2

After the vn suite, we can get the pn and qn suite defined as:

p0 = 1
p1 = ent(S) , integer part of S
pn+1 = vn * pn + pn-1

q0 = 0
q1 = 1
qn+1 = vn * qn + qn-1

La n-th approximation of S is given by the pn/qn fraction.

In our case, S = 1,70951129), we get :

n vn pn qn pn/qn
0 1 1 0 infinite
1 1 ent(S) = 1 1 1/1
2 2 1+1*1=2 0+1*1=1 2/1
3 2 1+2*2=5 1+2*1=3 5/3
4 3 2+2*5=12 1+2*3=7 12/7
5 1 5+3*12=41 3+3*7=24 41/24
6 5 12+1*41=53 7+1*24=31 53/31
7 2 41+5*53=306 24+5*31=179 306/179
8 53+2*306=665 31+2*179=389 665/389

La last row contains the result: the suite of the fractions for the S approximation.

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