Infinite Geometric Progression
Date: 10/02/97 at 20:56:26 From: Franklin Lum Subject: Repetend or infinite repeating decimal Hi Dr. Math, I teach eightt grade math. I can convert infinite repeating numbers from fractions to decimals with no trouble, e.g. 3/11 = .2727 or .27 bar on the top 27. My problem is how do you go about solving the .27 bar on both numbers into fractions? How do you do that? Could you tell me the steps? Please help. Judiana
Date: 10/03/97 at 08:42:59 From: Doctor Jerry Subject: Re: Repetend or infinite repeating decimal The idea is to use the formula for the sum of an infinite geometric progression. Let me first remind you of that. A geometric progression is one that generates each new term by multiplying by a fixed ratio, usually named r. If the first term is a, then the second term is a*r, the third term is a*r*r = a*r^2, and so on. Let S be the sum a+a*r+a*r^2+a*r^3+... Unless a happens to be zero, the sum is infinite or not well defined if the absolute value of r is 1 or bigger: a = 1, r = 2: 1+1*2+1*2^2+1*2^3+... = 1+2+4+8+... a = 1, r = -2: 1-1*2+2*2^2-1*2^3+... = 1-2+4-8+... Suppose a is not zero and |r|<1 (this is the case in the repeating decimals). Okay, the sum a+a*r+a*r^2+a*r^3+... is given by S = a/(1-r). This is easily seen. Just write S = a+a*r+a*r^2+a*r^3+... Now multiply both sides by r r*S = a*r+a*r^2+a*r^3+... Now substract, S-r*S = a. So, S = a/(1-r). Now for 0.272727... This (and other such repeating decimals) can be written as 27/100+(27/100)*(1/100)+(27/100)(1/100)^2+... So, 0.272727... = a/(1-r) = (27/100)/(1-1/100) = 27/99 = 3/11. Neat, huh? I'm not patting myself on the back. The above is just my memory of some earlier person's work. I enjoy doing the demo, though. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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