the golden ratio-II

…well lets leave the “golden ratio” aside for some time n come to something even more amazing..

Here’s a very interesting question..

“A man put a pair of rabbits in a place surrounded on all sides by a wall.

How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes productive?”

Anything familiar? C’mon all you maths freaks…get ur pencil n paper…this shudnt b al that difficult..

If I tell u the person who posed the question the answers gonna b on the tip of ur tongue!

Lets play with the problem a little more…

Let fn denote the number of pairs of rabbits after n months. The key fact is that the number of rabbits at the end of a month is the number at the beginning of the month plus the number of births produced by the mature pairs:

fn = fn1 + fn2

The initial conditions are that in the first month there is one pair of rabbits and in the second there are two pairs:

f1 = 1; f2 = 2

now in the 3^rd month there are 3 pairs (3 and not 4!!!, read the last line carefully, the new pair becomes productive every second month)

thus we get

f3=3

proceeding in this manner , we get a recurrence relation

f_n = f_{n –1} + f_{n – 2}

to give

f12=233

(work it out if u please…or read on further to know how!)

If we place the above recurrence relation in the form of an equation, we can expect solutions of the form

fn= cρⁿ

the recurrence relation becomes

ρ­­²= ρ+ 1

we have come across this equation before…rewriting it

ρ­­² – ρ – 1 = 0

this equation has 2 possible solutions , φ and φ—1, so the general solution is

fn = c₁ φ­­­ⁿ +c₂ (φ—1)ⁿ

the constants can be determined by the initial conditions of the problem

c₁ = φ/(2 φ—1 )

and

c₂ = - (1—φ)/( 2 φ—1 )

inserting these in the general solution gives

fn = ( φ­­­ⁿ⁺¹ + (φ—1)ⁿ⁺¹ )/( 2 φ—1 ) …(1)

if one decides to solve the equation, u wil find that the right hand side involves powers and quotients of irrational numbers, but the result is a sequence of integers!!

The n-1 th term of the equation is

f_n-1 = ( φ­­­ⁿ + (φ—1)ⁿ )/ ( 2 φ—1 ) …(2)

now divide (1) by (2) and take limit as nà ∞

the result…… ‘φ’

what does this imply? It simply means that the population of rabbits in the pen doesn’t double every month but it is multiplied by the golden ratio every month.

Now if we use the general solution above, n substitute n =12 and round of the answers to integers, the sequence obtained…

1,1,2,3,5,8,13,21,34,55,89,144,233

…..the sequence is…the “Fibonacci Sequence”