Transcript Slide 1
Module Arch-Pi 1 Archimedes and Pi How can you compute p? What is p? The number p is defined as the ratio of the circumference of a circle to its diameter. Prepared for SSAC by Eric Gaze Alfred University, Alfred, NY Quantitative concepts and skills Estimation Ratios Limits Proportions Pythagorean Theorem Geometrical Reasoning Logic Functions, IF © The Washington Center for Improving the Quality of Undergraduate Education. All rights reserved. 2005 1 Overview of Module Archimedes of Syracuse, 287 – 212 BCE, was a brilliant Greek mathematician credited with the first sophisticated approximation of p. Ancient civilizations knew this ratio was slightly larger than 3, but the Greeks were the first to explore exactly how much larger it is. •Slides 3-4 introduce the approach Archimedes took to approximate pi by using circumscribed polygons. •Slides 5-7 give a first approximation using a circumscribed hexagon and explore the related problem of estimating square roots. •Slides 8-12 explore Archimedes’ iterative process of cascading right triangles that share the same vertex angle, which is bisected at every step. •Slide 13 creates the spreadsheet which gives the approximation of pi using circumscribed polygons. •Slide 14 gives the assignment to hand in. 2 To find p we need to find the circumference… π How can you measure the arc-length of a circle? The ingenious technique devised by the Greeks and adroitly exploited by Archimedes was to circumscribe a polygon about the circle. The perimeter of the polygon then approximates the circumference! Polygons are made up of right triangles and straight sides, perfect for a civilization that included Euclid and Pythagoras. Can you subdivide this hexagon into a collection of right triangles? 3 To approximate p we need to compute the ratio of the perimeter to the diameter… π Archimedes focused on a single right triangle and the associated ratio of the two legs of the right triangle. c b a a b Which ratio, or , is more useful b a Note that 2b polygon side 2a for computing perimeter ? How? diameter diameter Multiplying the numerator by the number of sides will give: (# sides) b perimeter π a diameter 4 Archimedes started with a circumscribed hexagon, giving a right triangle well known to the Greeks. What is special about this triangle? x ½x This triangle is a 30-60-90 right triangle, which is ½ of an equilateral 60-60-60 triangle. b c a 3 2 x Use the Pythagorean Theorem to compute the height of this triangle in terms of x. Thus the ratio of sides in a 30-6090 right triangle: a 3 b c 2 b 5 Archimedes now has a big problem: how to compute 3? We can create a spreadsheet to estimate 3 . We know this number is between 1 and 2 since 2 . 1 1 3 4 22 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 B C Start Value Increment 1 0.1 Value 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 (Value)^2 1 1.21 1.44 1.69 1.96 2.25 2.56 2.89 Too Big Too Big = Given Number = Formula Recreate this spreadsheet with appropriate formulas. To begin the spreadsheet, we must choose a starting value and an increment to increase it by. For our first iteration, we will use a starting value of 1 and an increment of 0.1. Cells in the first column simply add the increment stored at the top to the number in the row above the cell. Cells in the second column have two possible outputs: the square of “value” or “TOO BIG”. We use the IF logic function: =IF (B6^2 < 3, B6^2, “TOO BIG”) 6 Now we iterate Use the copy and paste commands to transfer your IF function to each successive, two-column block. Add four more columns to your spreadsheet to find 3 to 5 decimal places. 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 B C D E F G H I J K Start Value Increment 1 0.1 Start Value Increment 1.7 0.01 Start Value Increment 1.73 0.001 Start Value Increment 1.732 0.0001 Start Value Increment 1.7320 0.00001 Value 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 (Value)2 1.0 1.2 1.4 1.7 2.0 2.3 2.6 2.9 Too Big Too Big Value 1.7 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 (Value)2 2.89 2.92 2.96 2.99 Too Big Too Big Too Big Too Big Too Big Too Big Value 1.73 1.731 1.732 1.733 1.734 1.735 1.736 1.737 1.738 1.739 (Value)2 2.993 2.996 3.000 Too Big Too Big Too Big Too Big Too Big Too Big Too Big Value 1.732 1.7321 1.7322 1.7323 1.7324 1.7325 1.7326 1.7327 1.7328 1.7329 (Value)2 2.9998 Too Big Too Big Too Big Too Big Too Big Too Big Too Big Too Big Too Big Value 1.7320 1.73201 1.73202 1.73203 1.73204 1.73205 1.73206 1.73207 1.73208 1.73209 (Value)2 2.99982 2.99986 2.99989 2.99993 2.99996 3.00000 Too Big Too Big Too Big Too Big Choose your starting value based on the best estimate from the previous iteration. We get a very good estimate of the square root of 3 by using five decimal places in the iteration. 7 Archimedes is now ready to estimate p using a hexagon. Recreate this spreadsheet. Refer back to slide 4 for the approximation formula. Use your work from slides 5 and 6 to estimate p. 2 3 B Iteration 0 C a/b (sqrt(3)) 1.73205 D c/b E F # sides Pi ~ # sides(b/a) 2 6 3.4641032303 Archimedes recognized that this estimate was not very close, and he also knew that to get a better approximation he would need to increase the number of sides. His genius was figuring out how to compute the crucial ratio for polygons which have more sides and are not composed of 30-60-90 right triangles. Better Approximation! 8 The approach... Archimedes’ approach is to bisect the vertex angle of the right triangle at the center of the polygon thereby creating a new right triangle. This new right triangle belongs to a new polygon with more sides. c b1 b How many sides does the new polygon have? c2 b2 a Proposition 3 from Book VI of Euclid’s Elements: Archimedes needs to find the new ratio: a b2 c b1 a b2 (for the case where the angle opposite b1 is the same as the angle opposite b2) For more information on this equation, click here 9 Archimedes used Euclid’s Proposition VI to arrive at the desired ratio as follows: To get to c b1 b a b2 Start with Proposition VI from Book 3 of Euclid’s Elements: c b1 a b2 c2 b2 a c a b1 b2 ca b a a b2 b2 a b2 First, Add 1 to Both Sides! What do we need to multiply both sides by to get the ratio a b2 ? π in the sky 10 Continuing the calculation from the previous slide we arrive at a as follows: b2 Proposition 3 from Book VI of Euclid’s Elements: c b1 b c2 c b1 a b2 b b22 a ca b a ca a b2 b b b2 a a c a How is related to and ? b2 b b 11 We are now ready to estimate p using a dodecagon. B Iteration 2 3 4 C a/b 0 1 1.73205 3.73205 E F # sides Pi ~ # sides(b/a ) 2 6 3.4641032303 12 3.2153910049 c b1 b D c/b Recreate this spreadsheet c2 b2 a The ratio c/b is not necessary for estimating p. If we knew the ratio, c2/b2 for the smaller triangle, however, we could bisect the angle again and repeat the process for a 24-sided polygon, and then again and again for polygons with more and more sides! Excel excels at this type of iterative process. 12 Use the formula you found for c2/b2 to recreate this spreadsheet. Use the Pythagorean Theorem for the smaller green triangle and divide through by b2. Solve for c2/b2. 2 3 4 B Iteration C a/b 0 1 D c/b E F # sides Pi ~ # sides(b/a ) 1.73205 2 6 3.4641032303 3.73205 3.863703 12 3.2153910049 c b1 b c2 b2 a We now have a huge advantage over Archimedes, who had to do each calculation one at a time by hand. We can fill the formulas down the columns and estimate p using polygons with as many sides as we please! 13 Expand your spreadsheet from the previous slide. 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 B Iteration C a/b 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1.73205 3.73205 7.595752525 15.25704853 30.54683367 61.11003127 122.2282439 244.4605785 488.9232024 977.8474274 1955.695366 3911.390988 7822.782104 15645.56427 31291.12857 62582.25717 125164.5143 250329.0287 D c/b 2 3.863703 7.661296 15.28979 30.5632 61.11821 122.2323 244.4626 488.9242 977.8479 1955.696 3911.391 7822.782 15645.56 31291.13 62582.26 125164.5 250329 E F # sides Pi ~ # sides(b/a ) 6 3.4641032303 12 3.2153910049 24 3.1596606025 48 3.1460868671 96 3.1427152495 192 3.1418736993 384 3.1416633963 768 3.1416108258 1536 3.1415976835 3072 3.1415943979 6144 3.1415935765 12288 3.1415933712 24576 3.1415933199 49152 3.1415933070 98304 3.1415933038 196608 3.1415933030 393216 3.1415933028 786432 3.1415933028 π 3.14159265 358979 Hi π 14 End of Module Assignment Goodbye π 1. Write out your calculations from slide 5 involving the Pythagorean Theorem. Verify the ratios for a/b and c/b. 2. Submit an electronic copy of your spreadsheet from Slide 6 on Blackboard. 3. Write out your answer to the question from Slide 8 with clear explanation and a picture. 4. Write out your answers to the questions on Slides 9 and 10 with clear explanations. 5. Write out your calculations from Slide 12 involving the Pythagorean Theorem. 6. Submit an electronic copy of your spreadsheet from Slide 13 on Blackboard. 7. Use your spreadsheet to determine how many decimal places of accuracy 265 Archimedes was able to get for a 96-sided polygon using 153 to approximate the square root of 3. We do not know how Archimedes arrived at this fraction. 8. Determine how many decimal places of accuracy you need for the square root of 3 to get 9 decimal places of accuracy for p. 15 Euclid’s Elements: Proposition 3, Book VI What it says: c b1 a b2 If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle; and, if segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the point of section bisects the angle of the triangle. This statement comes from http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html, which also contains further information about the proposition, including a proof. Back to Slide 9 16