Numerics in General

3) the attached for reference.

Non-textbook question: Hands on to solve fixed-point iteration for parabola (30% pts): _{n+1}=a*{z_n (2_n}^2), pport "a" being a constant. 1. Start from z_0-0.5, and let a-2, do you observe certain convergence in should be large enough? If yes, what's the fixed point(s)? 2. Start from z_00.5, and let a-3.4, do iteration until 2.40. Asnis reaching 40, do you observe certain convergence? If yes, whats the fixed point(s)? 3. Let a=3.5, do the iteration again, use a large value for n. do you see convergence and fixed point(s)? 4. Let a-3.6, and iterate 5. Make discussion on your solutions in 1), 2), and 3, and how it changes with value of *a". Note: 1) If you are not sure on convergence, do the iteration with a large n, as large as you want 2) better use Excel or programming to compute. 3) read the figure as attached for reference. 143 3.7 Newton's Method (and Choos) The period 2,4.... is the number of 's in a cycle. 8 16 3.14 Fig. 3.24 Period doubling and chaies from iterating Fle) (stolen by special permission from Introduction to Applied Mathematics by Gilbert Strang. Wellesley Cambridge Press see only three's. Period 3 is followed by 6.12.24..... There is period doubling at the end of every window (including all the windows that are too small to sec). You can reproduce this figure by iterating - - . - from any , and plotting the results. CANTOR SETS AND FRACIALS I can't tell what happens at a 38. There may be a stable cycle of some long period. The's may come close to every point between 0 and 1. A third possibility is 10 approach a very thin limit set, which looks like the famous Castor set To construct the Cantor set, divide (0, 1) into three pieces and remove the open interval (4.3). Then remove (1.1) and 3) from what remains. At cach step Take out the middle thirds. The points that are left form the Cantor set All the endpoints, $. 3. .... are in the set. So is 2 (Problem 21. Nevertheless the lengths of the removed intervals add to 1 and the Cantor set has measure rero." What is especially striking is its self-similarity: Between 0 and you see the same Cantor ser three times smaller. From 0 to the Cantor set is there again, scaled down by 9. Every section, when blown up, copies the larger picture Fractals That self similarity is typical of a fractal. There is an infinite sequence of scales. A mathematical snowflake starts with a triangle and adds a bump in the middle of each side. At every step the bumps lengthen the sides by 43. The final boundary is self-similar, like an infinitely long coastline. The word "fractal" comes from fractional dimension. The snowflake boundary has dimension larger than 1 and smaller than 2. The Cantor set has dimension larger than 0 and smaller than 1. Covering an ordinary line segment with circles of radius would take clr circles. For fractals it takes c circles and is the dimension Non-textbook question: Hands on to solve fixed-point iteration for parabola (30% pts): _{n+1}=a*{z_n (2_n}^2), pport "a" being a constant. 1. Start from z_0-0.5, and let a-2, do you observe certain convergence in should be large enough? If yes, what's the fixed point(s)? 2. Start from z_00.5, and let a-3.4, do iteration until 2.40. Asnis reaching 40, do you observe certain convergence? If yes, whats the fixed point(s)? 3. Let a=3.5, do the iteration again, use a large value for n. do you see convergence and fixed point(s)? 4. Let a-3.6, and iterate 5. Make discussion on your solutions in 1), 2), and 3, and how it changes with value of *a". Note: 1) If you are not sure on convergence, do the iteration with a large n, as large as you want 2) better use Excel or programming to compute. 3) read the figure as attached for reference. 143 3.7 Newton's Method (and Choos) The period 2,4.... is the number of 's in a cycle. 8 16 3.14 Fig. 3.24 Period doubling and chaies from iterating Fle) (stolen by special permission from Introduction to Applied Mathematics by Gilbert Strang. Wellesley Cambridge Press see only three's. Period 3 is followed by 6.12.24..... There is period doubling at the end of every window (including all the windows that are too small to sec). You can reproduce this figure by iterating - - . - from any , and plotting the results. CANTOR SETS AND FRACIALS I can't tell what happens at a 38. There may be a stable cycle of some long period. The's may come close to every point between 0 and 1. A third possibility is 10 approach a very thin limit set, which looks like the famous Castor set To construct the Cantor set, divide (0, 1) into three pieces and remove the open interval (4.3). Then remove (1.1) and 3) from what remains. At cach step Take out the middle thirds. The points that are left form the Cantor set All the endpoints, $. 3. .... are in the set. So is 2 (Problem 21. Nevertheless the lengths of the removed intervals add to 1 and the Cantor set has measure rero." What is especially striking is its self-similarity: Between 0 and you see the same Cantor ser three times smaller. From 0 to the Cantor set is there again, scaled down by 9. Every section, when blown up, copies the larger picture Fractals That self similarity is typical of a fractal. There is an infinite sequence of scales. A mathematical snowflake starts with a triangle and adds a bump in the middle of each side. At every step the bumps lengthen the sides by 43. The final boundary is self-similar, like an infinitely long coastline. The word "fractal" comes from fractional dimension. The snowflake boundary has dimension larger than 1 and smaller than 2. The Cantor set has dimension larger than 0 and smaller than 1. Covering an ordinary line segment with circles of radius would take clr circles. For fractals it takes c circles and is the dimension