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Fixed-Point Iteration  

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Carrera Falk Mr. Moore IB HL Math Methods January 3, 2003 IB HL Portfolio Assignment: Fixed-Point Iteration 1. i. A. Equation: f(x) = 0.25+6 Fixed point: x = 8 First "guess": xn-1 = 1 f(1) = 0.25(1) + 6 = 6.25 Second iteration: f(6.25) = 0.25(6.25) +6 = 7.5625 Third iteration: f(7.5625) = 0.25(7.5625) + 6 = 7.890625 Fourth iteration: f(7.890625) = 0.25(7.890625) + 6 = 7.97265625 B. - Copied axes located on page 6 Staircase diagram from right side of graph: First "guess": 20 f(20) = 0.25(20) + 6 = 11 Second iteration: f(11) = 0.25(11) + 6 = 8.75 Third iteration: f(8.75) = 0.25(8.75) + 6 = 8.1875 - New "staircase" diagram shown on axes - "Staircases" graphed on TI-83+ 2. A. Solve x = 1.25x - 2 to find a value of x at the fixed point: x= 1.25x -2 -.25x = -2 x = 8 B. Iterations of: f(x) = 1.25x...

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