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Unformatted text preview: five times as many curves, so there are a total of 4 Â· 5 n 1 segments in the n th curve in the series. The segments in each curve are onethird the length of those in the previous one, so the length of each segment in the n th curve is (1 / 3) n 1 . Thus the total length is 4 Â· (5 / 3) n 1 . (b) What is the area enclosed by the n th curve in this series? (Note that the n th curve in this series is obtained from the ( n 1)st curve by adding a large number of small squares. For example, we obtain the second curve from the first one by adding four squares, each of which has side length 1 / 3 and thus area 1 / 9. So the area enclosed by the second curve is 1 + 4 / 9.) Your answer should involve summing a geometric series. Solution. To obtain the n th curve from the ( n 1)st curve, note that the ( n 1)st curve consists of 4 Â· 5 n 2 line segments of length (1 / 3) n 2 . On each of these line segments we will place a square of side length (1 / 3) n 1 , and thus of area (1 / 9) n 1 . So to get from the ( n 1)st curve to the n th curve, we add squares having total area 4 Â· 5 n 2 Â· (1 / 9) n 1 ; we can rearrange this to get (4 / 5) Â· (5 / 9) n 1 . The area enclosed by the n th curve in this series (for n â‰¥ 2) is therefore 1 + (4 / 5)(5 / 9) + (4 / 5)(5 / 9) 2 + Â·Â·Â· + (4 / 5)(5 / 9) n 1 and all the terms but the first make up a geometric series. We can rewrite this as 1 + h (4 / 9) + (4 / 9)(5 / 9) + (4 / 9)(5 / 9) 2 + Â·Â·Â· + (4 / 9)(5 / 9) n 2 i and recall that a + ar + ... + ar k = a (1 r k +1 ) / (1 r ). With a = 4 / 9 ,r = 5 / 9 ,k = n 2, the area enclosed is therefore 1 + 4 / 9 1 (5 / 9) n 1 1 5 / 9 ! This can be simplified to give 1 + (1 (5 / 9) n 1 ) and if n is very large this is very nearly 2. In fact, the area enclosed by this curve is very close to being a square of side length âˆš 2. 6. Start with the point 0. Flip a coin. If it comes up heads, move 2 / 3 of the way toward 1, if it comes up tails, move 2 / 3 of the way to 0 from wherever you are at the time. Repeat forever. The points you find are drawing a picture of the Cantor set. Verify that any point in the Cantor set will move to another point in the Cantor set under the coinflippingandmoving process. (B+S 6.3.26) Solution. Moving twothirds of the way towards 1 takes the point x to (2 + x ) / 3. To see this, note that the distance from x to 1 is 1 x ; the distance from (2 + x ) / 3 to 1 is 1 2 + x 3 = 1 x 3 . 4 Moving twothirds of the way towards 0 takes the point x to x/ 3. So we need to show that if x is in the Cantor set, then so are (2 + x ) / 3 and x/ 3. But we did this in class, by looking at the ternary expansion of x . 5...
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 Summer '09
 Lugo
 Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

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