Fibonacci Sequence and PHI
The Fibonacci sequence is a famous sequence of numbers starting with 1,1 where each term (from the third onwards) is the sum of the previous two terms. The first few numbers in the sequence are as follows:
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,
46368,75025,121393,196418,317811,514229,832040,1346269….
which quickly become very large.
The Fibonacci numbers, fr can be defined by
f1 = f2 = 1
and
fk = fk-1 + fk-2 for k≥3
PHI the golden ratio
PHI or Φ is said to be the golden ratio since so many things in nature seem to naturally arrange themselves in this ratio. It is approximately equal to 1.618033988749895.
Phi is also the positive solution to the equation
x2 = 1 + x
which has the irrational solution
phi = (1+√5)/2
Interestingly an irrational number is one which cant be written as the ratio of two integers so the golden ratio is not in fact a ratio meaning ratios in nature can only become very close to it but cant actually equal it.
PHI and Division of Fibonacci numbers
It is interesting to calculate the ratio of consecutive numbers in the Fibonacci sequence. These ratios quickly approach the golden ratio know as PHI or Φ.
The start of list of value obtained from these divisions are:
1
2
1.5
1.6666666666666667
1.6
1.625
1.6153846153846154
1.619047619047619
1.6176470588235294
1.6181818181818182
1.6179775280898876
1.6180555555555556
1.6180257510729614
1.6180371352785146
Calculation of PHI from limit of Fibonacci divisions
If we suppose that the ratio of consecutive terms in the Fibonacci sequence do approach a limit we can use this to find the value of phi.
IF we denote the nth Fibonacci number by fn and the nth ratio as rnthen
rn = fn+1 / fn
But using the definition of a Fibonacci number:
fn+1 = fn + fn-1
then the ratio is
rn = (fn + fn-1 )
fn
Which can be simplified to be
rn = 1 + fn-1/fn
but fn-1/ f = 1/rn-1
so
rn = 1 + 1/rn-1
Now supposing that the ratios tend to a limit p as n tends to infinity then p is the solution of the equation
p = 1 + 1/p
which can be re-arranged to give
p2 – p – 1 = 0
and can be solved using the quadratic equation to give
p = (1 + √5)/2
and
p = (1 – √5)/2
The first of these solutions happens to be the golden ratio PHI or Φ
If anyone has a proof that these ratios do infact approach a limit please include it in the comments of this post.
Comments Left
I suppose it is important to note that
the second option for the solution of p
p = (1 – √5)/2
is going to be a negative value.
We can eliminate this as a solution
entirely since the entire problem is based
on positive integers and their ratios. The
limit cannot possibly be negative.
As for a proof that the limit exists… Im
not entirely sure. For each n, increasing
incrementally one unit at a time, the
difference between consecutive ratios
decreases significantly. I perceive the
convergence as intuitively obvious, but
that doesnt constitute a proof.
@CogitoErgoCogitoSum
particularly your comment that the other solution is negative and cannot therefore be the limit,
thans for taking the time to comment and read
david
Great article! Here’s how to prove your sequence of the Fibonacci ratios has a limit (that is, it converges).
NOTE: Because I can’t use subscripts in these comments, let a_n denote the nth term of the sequence called a.
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As shown in the article, the sequence a_n = f_(n+1)/f_n can be defined recursively as:
a_(n+1) = 1 + 1/a_n
It can be shown using induction that
a_2(n+1) =1
That is, all of our even terms of the sequence decrease.
It can also be shown by induction that:
a_2n > 1 for all n >= 1
We have all the even terms of our sequence are bounded below.
Because our sequence of even terms is both decreasing and bounded below, it must converge, approaching a limit which is greater than or equal to our lower bound.
It can then be shown by writing out the defenition of our sequence in terms of only even terms that this limit must be equal to phi:
a_2(n+1) = 1 + 1/( 1 + 1/a_2n )
By replacing the symbols a_2(n+1) and a_2n with our limit L, we can solve for L in the above expression using the quadratic formula (with L >= 1 > 0) to obtain:
lim n->inf. a_2n = L = ( 1 + 5^(1/2) )/2 = phi
We can then define the sequence of odd terms of a using our even terms:
a_(2n+1) = 1 + 1/a_2n
Taking the limit as n approaches infinity again yeilds phi, because we have already found the limit of a_2n.
Now for any positive integer k, we have that
a_k = a_2n -OR- a_(2n+1) for some other integer n.
Because a_2n > a_(2n+1), we have
a_2n >= a_k >= a_(2n+1)
Taking limits as k (and therefore n) approach infinity yields:
phi >= lim k->inf. a_k >= phi
(Squeeze theorem) Therefore:
lim k->inf. a_k = phi
Q.E.D.
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