Limit of AreaDate: 03/01/98 at 15:44:59 From: Jason Subject: Geomety POW-Regarding Mathematical Limitsz Given a square with sides of length 5. Each side of the square is divided into three congruent parts, and a square is constructed outwardly on the middle part. The new figure has twenty line segments determining its boundary. Each of these is further divided into three congruent parts, and again a square is formed outwardly on each middle part. The process is continued infinitely. What is the limit approached by the area of this figure? I have tried looking up what limit means, but I don't get what the area is according to the limit. I know the number of squares multiplies by 5 each time. Can you please help me? Date: 03/02/98 at 17:30:20 From: Doctor Sam Subject: Re: Geomety POW-Regarding Mathematical Limits Jason, This is a complex problem. Let's hold off on the idea of "limit" and just compute some values. The first square has side length 5. The next square has side length 5/ 3. The next squares will be one-third smaller, or 5/9, and the next would be one-third of that, or 5/27. Do you see the pattern? The side lengths of the squares are the numbers 5, 5/3, 5/(3^2), 5/(3^3), ... These form what are called a geometric sequence. Each number is a multiple (in this case, 1/3) of the number that preceeds it. If you know the lengths of the sides, you can find the area of each size square 5^2, (5/3)^2, (5/(3^2))^2, (5/(3^3))^2, ..., which can be simplified to 5^2, 5^2/3^2, 5^2/3^4, 5^2/3^6 ... Okay so far? Now, you said that you know the number of squares increases by a factor of 5 each time. So you start with 1, then have an additional 4, then another 20 = 4*5, then another 100 = 4*5^2, etc. You can find the total area as you go along, adding up the areas at each stage of the construction: Total Area = 1 square with side length 5 + 4 squares with side length 5/3 + 4*5 squares with side length (5/3^2) + 4*5*5 squares with side length (5/3^3) + ... etc. Now let's write them all down: 1*5^2 + 4*5^2/3^2 + 4*5*5^2/3^4 + 4*(5^2)*(5^2)/3^6 + ... Simplify each fraction (but keep them with exponents to show the pattern): Total Area = 5^2 + 4*5^2/3^2 + 4*5^3/3^4 + 4*5^4/3^6 + ... The idea of a limit is that if you keep adding up more and more terms in this pattern, the numbers you get will all get closer and closer to some number (called the limit of this series of numbers). The limit can be computed because this pattern is also a geometric series, and there is a formula for the sum of such a pattern. The sum is: first term Sum = ---------------- 1 - common ratio To see that this is a geometric series, and to discover its common ratio, notice that after the first term every term is 5/9 = 5/3^2 times the preceeding term. So the common ratio is 5/9, but you need to skip the first area of 25 because it isn't multiplied by 4. So take the first term as 4*5. Total Area = 5^2 + 4*5^2/3^2 + 4*5^3/3^4 + 4*5^4/3^6 + ... = 25 + 20(5/9) + 20(5/9)^2 + 20(5/9)^3 + ... so the first term is 20(5/9) and the common ratio is 5/9. The total area ... the limit as you keep constructing more and more squares ... is: 20(5/9) 20(5/9) 20(5) 25 + ------- = 25 + ------- = 25 + ------- = 25 + 25 = 50 1 - 5/9 (4/9) 4 I hope that helps. -Doctor Sam, The Math Forum http://mathforum.org/dr.math/ |
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