Fibonacci Algorithm and Egyptian FractionsDate: 04/25/99 at 19:06:37 From: liz Subject: Egyptian fractions I would like to know how to do 4 rounds in the Fibonacci algorithm. I have been able to get 3 rounds (521/1050 = 1/3 + 1/7 + 150 ) - could you please show me how to do 4 rounds with 2 examples? Also, I would like to know whether it is always possible to change any fraction into an Egyptian fraction using the decomposition method. Thanks. Liz Date: 04/26/99 at 16:30:42 From: Doctor Rick Subject: Re: Egyptian fractions Hi, Liz, welcome to Ask Dr. Math! The number of terms in an Egyptian fraction found by the Fibonacci algorithm depends only on what fraction you start with. There are fractions that lead to 4 or more terms using this method; one of these (the one with the smallest denominator) is 8/11. Try 16/17 too. You can easily write an Egyptian fraction as long as you want. For instance, any unit fraction 1/n in your expansion can be replaced by 1/n = 1/(2n) + 1/(3n) + 1/(6n) In your example, 521/1050 = 1/3 + 1/7 + 1/50 = (1/6 + 1/9 + 1/18) + (1/14 + 1/21 + 1/42) + (1/100 + 1/150 + 1/300) There, we just multiplied the number of terms by 3, and you can multiply it by 3 again by replacing each of these terms, and so on forever. So finding long Egyptian fractions isn't a challenge; but finding the SHORTEST Egyptian fraction expansion for any given fraction is a challenge. The Fibonacci algorithm isn't guaranteed to find the shortest. Take a look at this interesting Web site: An Introduction to Egyptian Fractions (Ron Knott) http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html It has lots of interesting information about Egyptian fractions, including the outline of a proof that the Fibonacci algorithm will produce an Egyptian fraction of finite length for any fraction. I don't know what you mean by the "decomposition method" so I can't say whether it will always work, if it's different from the Fibonacci method (and there are lots of methods). - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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