Solutions 8 MT1100 From Euclid to Mandelbrot 1. The graph of the function f intersects the line x = y at 7 points, so f has 7 fixed points. The gradient of f at these points is clearly always of absolute value 1 or more at 6 of these points, so these 6 points are unstable. The gradient of f at the fixed point at 0 is (just) less than 1, and so this point is stable. 2. (i) A fixed point c satisfies f(c) = c, and so c3 + ac = c. Therefore the fixed points are exactly the roots of c3 + (a -1)c = 0. Clearly c = 0 is always a fixed point. If c # 0, then we can divide by c to get c2 + (a - 1) = 0. When a > 1, this equation has no solutions, and so we just have a single fixed point at 0. When a = 1, the equation implies that c = 0 and so we also have just a single fixed point at 0 in this case. Finally, when a < 1 the equation has solutions c = #v1 - a and so we have 3 fixed points: 0, v/1 - a and -v1 - a. (ii) We have that f'(x) = 3x2 + a. A fixed point c is stable when |f'(c) | < 1 and is unstable otherwise. We have f'(0) = a and so the fixed point at 0 is stable when 0 ? a < 1 and is unstable when a > 1. When a < 1 we have two more fixed points to consider. In this case, we have f' (=v/1-a) = 3(1-a) + a = 3 -2a > 1 and so these two fixed points are unstable. (iii) The fixed point at 0 is superstable when a = 0, but is not superstable otherwise. The other fixed points are never superstable, as f' takes a value of 1 or more at these points (see part (ii)). (iv) The bifurcation diagram looks as follows: 1
1.04 0.5 0.5 1.0 1.5 2.0 -0.5 -1.0 3. (i) The fixed points of x satisfy f(x) = x. This holds exactly when 2- (x2 - 2x + 1) = x, in other words when 1 - 2 + 2x = x, and simplifying we have x2 - x -1 =0. The two solutions x1 = (1+v5)/2 and x2 = (1-v5)/2 to this quadratic equation are the two fixed points of f. (ii) We have f(2)(x) = 2-(2-(x-1)2-1)2 = 2- (2x-x2)2 = 2- x2(2-x)2. The fixed points of f(2) are the solutions of f(2) (x) = x. Using our formula for f(2)(x) and simplifying we find: 2 - x2(2 - x)2 = x (2 - x) - x2(2- x)2 = 0 (2-x) (1-x2(2-x)) = 0 (2-x)(x3- 2x2 +1) = 0. Both fixed points x1 and x2 of f must be solutions, so (x- x1) (x-x2) = x2 - x - 1 divides this equation.