in this sequence are ther 2 numbers that the ratio between them will equal PERFECTLY the devine ratio...if so plaese answer
2006-10-29 01:37:50 · 5 answers · asked by norm n 2 in Science & Mathematics ➔ Mathematics
The two people who answered before me had the right idea. There are no two Fibonacci numbers where a(n)/a(n-1)=phi (phi is the golden/divine ratio.) However, the ratio of any two consecutive numbers gets closer and closer to the fibonacci sequence as you go through the sequence. The mathematical way to say this is that the limit as n tends to infinity of a(n+1)/a(n) is equal to phi, or:
lim (a[n+1]/a[n]) = phi.
n-->inf.
where a[n] and a[n+1] are consecutive numbers in the fibonacci series.
2006-10-29 04:04:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
No. Proof by contradiction:
If there were two numbers in the sequence a_n and a_(n+1) such that a_(n+1) / a_n = phi (the golden ratio), then the ratio of a_n to the previous number in the sequence, a_(n-1) would be:
a_n / a_(n-1)
= a_n / (a_(n+1) - a_n) because a_(n+1) = a_(n-1) + a_n
= 1 / (a_(n+1)/a_n + a_n/a_n)
= 1 / (phi + 1)
= phi, by definition of the golden ratio.
So... If two consecutive numbers are exactly in a golden ratio, it follows that the previous pair is also in golden ratio. By induction, this means ALL previous pairs of consecutive numbers are in an exact golden ratio, including the first pair in the sequence. This is clearly not true of the first pair, 1 and 1. Contradiction.
2006-10-29 09:53:52 · answer #2 · answered by Anonymous · 0⤊ 0⤋
No, because the divine ratio is irrational. Every number in the sequence is an integer, so the ratio of two of them will be rational.
2006-10-29 10:01:48 · answer #3 · answered by Steven S 3 · 1⤊ 0⤋
There's loads on the internet - just Google "fibonacci sequence divine" and you'll find plenty of stuff at different levels.
2006-10-29 09:47:58 · answer #4 · answered by JJ 7 · 0⤊ 0⤋
How'd you think of the question?
2006-10-29 09:39:35 · answer #5 · answered by bebackhome_safe 2 · 0⤊ 0⤋