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Introduction

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: <my 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: <Same equation and fixed point as above> 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. ...read more.

Middle

/3 or 1/3x2 - 4/3 ii. Iterations on page 8 It is an attracting point, especially due to the fact that I used the number 1 as my first "guess" which made every other answer to the iteration "-1" because it is one of the numbers that solve the problem evenly. iii. Iterations on page 8 It is a repelling point, and maybe this is because the first "guess" number was a decimal, but also, the staircase here is found to be on the left side of the graph, whereas the above iteration was a right staircase. B. Consider the equation x = -0.5x + 2 Iterations on page 8 It is an attracting point, staircasing left of the fixed point. Again, here I noticed that I used a whole number as a first "guess" to be multiplied by a fraction of one in the equation. ...read more.

Conclusion

It is a repelling fixed point when examining the last graph where x = 1 and the first "guess" is x = 1.1. Once looking over these last equations, I am beginning to think that a greater meaning lies behind why some equations have repelling and some have attracting fixed points. E. Table of Attracting Fixed Point and Repelling Fixed Points Tables on page 10 i. Conjecture: The condition, which must be present for a fixed point to be attracting, is apparent to me that every one has a fractional slope of the tangent, or f ' (x). The repelling points also have a few fractional tangential slopes, but all of the attracting ones do as a trend. ii. Testing the conjecture: Unfortunately, this conjecture did not hold true for the equation f(x) = x^5 -2 when viewing its staircase from the left, as its fixed point seems to be centered around -3, and the tangential slope of that becomes a positive term, -405. (Iterations on page 10.) iii. Iterations on page 11 ...read more.

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