The Fibonacci sequence is defined recursively as, F1=1,F2=1,and for all n=1,2,3,4,5.... Fn+2= Fn + Fn+1?

Question

a) compute the first ten elements of the Fibonacci sequence,using the recursive definition above.
b) Provr that any two consecutive elememts of the Fibonacci sequence are relatively prime.
c) Look at the first 9 ratios formed from the consecutive elements in the Fibonacci sequence, fn/fn+1 where n=1,2,3,4,5,6,7,8,9
Write these ratios in a decimal presentation,up to 8-9 digits.What do you notice?

Answer 1

a)
___ f1 = 1
___ f2 = 1
___ f3 = f1+ f2 = 1+ 1 = 2
___ f4 = f2+ f3 = 1+ 2 = 3
___ f5 = f3+ f4 = 2+ 3 = 5
___ f6 = f4+ f5 = 3+ 5 = 8
___ f7 = f5+ f6 = 5+ 8 = 13
___ f8 = f6+ f7 = 8 + 13 = 21
___ f9 = f7+ f8 = 13+ 21 = 34
___ f10 = f8+ f9 = 21+ 34 = 55

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b)
f9 = 34 = 2*17
f8 = 21 = 3*7

There are no common multiples between f8 and f9 (arbitrarily chosen), so they are relatively prime.

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c)
f1/f2 = 1
f2/f3 = 1/2 = 0.50000000
f3/f4 = 2/3 = 0.66666667
f4/f5 = 3/5 = 0.60000000
f5/f6 = 5/8 = 0.62500000
f6/f7 = 8/13 = 0.61538461
f7/f8 = 13/21 = 0.61904762
f8/f9 = 21/34 = 0.61764706
f9/f10 = 34/55 = 0.61818182

As n grows up, the ratio seems to be converging to some value around 0.618.

Answer 2

It's a Fibonacci sequence defined recursively?

Answer 3

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html

Go there. It is a *very* kewl site for Fibonacci stuff.


Doug

Answer 4

I notice someone not doing his own homework.