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ive got to write about it so plllllzzzzzzz help!

2006-11-20 06:04:04 · 5 answers · asked by Anonymous in Arts & Humanities History

5 answers

Have a look at this website, it think it might be helpful to you:

http://www.educ.queensu.ca/~fmc/may2002/RabFib.htm

2006-11-20 06:06:38 · answer #1 · answered by Funky Little Spacegirl 6 · 0 0

Easy. Add the numbers 0 1 1 2 3 5 8 13 21 34 55 etc just add the previous 2 numbers to get the answer of the next Anyway you may wanna check wikipedia for the history of how Mr. Fibonacci found the sequence.

2016-03-14 10:26:23 · answer #2 · answered by Anonymous · 0 0

The Fibonacci numbers first appear, under the name maatraameru (mountain of cadence), in the work of the Sanskrit grammarian Pingala (Chhandah-shāstra, the Art of Prosody, 450 or 200 BC). Prosody was important in ancient Indian ritual because of an emphasis on the purity of utterance. The Indian mathematician Virahanka (6th century AD) showed how the Fibonacci sequence arose in the analysis of metres with long and short syllables. Subsequently, the Jain philosopher Hemachandra (c.1150) composed a well known text on these. A commentary on Virahanka by Gopala in the 12th c. also revisits the problem in some detail.

Sanskrit vowel sounds can be long (L) or short (S), and Virahanka's analysis, which came to be known as mAtrA-vritta wishes to compute how many metres (mAtrAs) of a given overall length can be composed of these syllables. If the long syllable is twice as long as the short, the solutions are:

1 mora: S (1 pattern)
2 morae: SS; L (2)
3 morae: SSS, SL; LS (3)
4 morae: SSSS, SSL, SLS; LSS, LL (5)
5 morae: SSSSS, SSSL, SSLS, SLSS, SLL; LSSS, LSL, LLS (8)

2006-11-20 08:29:37 · answer #3 · answered by Anonymous · 2 0

In the West, the sequence was first studied by Leonardo of Pisa, known as Fibonacci (1202). He considers the growth of an idealised (biologically unrealistic) rabbit population, assuming that:

* in the first month there is just one newly-born pair,
* new-born pairs become fertile from their second month on
* each month every fertile pair begets a new pair, and
* the rabbits never die

Let the population at month n be F(n). At this time, only rabbits who were alive at month n−2 are fertile and produce offspring, so F(n−2) pairs are added to the current population of F(n−1). Thus the total is F(n) = F(n−1) + F(n−2)

2006-11-20 06:15:53 · answer #4 · answered by Steve-o87 2 · 0 0

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2015-01-26 00:55:15 · answer #5 · answered by Anonymous · 0 0

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2016-05-01 15:44:23 · answer #6 · answered by ? 3 · 0 0

he looked at rabbit populations. start with 1 male 1 female. they have 1 baby per month who takes 1 month to become able to have offspring, so on and so on

1, 1, (1+1) 2, (2+1) 3, (3+2) 5 etc

2006-11-20 06:08:12 · answer #7 · answered by Anonymous · 0 0

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2016-06-04 00:53:45 · answer #8 · answered by Anonymous · 0 0

1

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2014-12-19 03:40:56 · answer #10 · answered by Anonymous · 0 0

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