Solutions 7 MT1100 From Euclid to Mandelbrot 1. (a) The Mathematica command PolarPlot [Sqrt [Cos [2*t]] , {t, 0, 2*Pi}] generates the following output: 0.3 0.2 0.1 -1.0 -0.5 0.5 -0.1 1.0 -0.2 -0.3 (b) There are (many) values of 0 that do not give points on the curve. [This is true whenever cos(20) < 0. So when T/4 < 0 < 3T/4 and 5/4 < 0 < 7T/4 we don't get a point on the curve. Can you see this from the Mathematica output?] There are no values of 0 that give two or more points on the curve. [Since r cannot be negative, r is uniquely determined as the positive square root of a cos 20. Can you see this from the Mathematica output?] (c) Multiplying by r2, we see that rt = a2r2 cos 20 = a2r2(cos2 0 - sin2 0) = a2(x2 - y2). But r2 = r2 (cos2 0 + sin2 0) = x2 + y2, and so r4 = (x2 + y2)2. Hence (x2 + y2)2 - a2(x2-y2) = r2 - r2 = 0. Multiplying out the left hand side gives us the equation we want. 1
2. The following Mathematica commands f [x_] := (2 x^2 - x + 0.6) /x points = Table [{t, Nest [f, 1, t]}, {t, 0, 10}] ListPlot [points, Joined -> True, Mesh -> All] produce this output: 400 300 200 100 2 4 6 8 10 The time series plot indicates that lim -.. f(n) (1) = oo. [Note that f(x) ~ 2x when x is large. We can prove that f(x) > (3/2)x when x > 1, and so we can prove by induction on n that f(n) (1) > (3/2)" > .] 3. (i) For any positive integer n, let P(n) be the statement that " f(n) (x) = ax + 1-a 1-a" b ". We find that P(1) is true, since the left hand side is f(1)(x) = f(x), and the right hand side is a1x + 1ª b = ax + b. Suppose, as an inductive hypothesis, that P(k) is true for some integer k. Then 1 - ak 1 - a b. f 2
So f(k+1) (x) = f(f(k) (x) 0 1 - ak 0 = f akx + 1 - a b 1 - ak b + b =a ax+ 0 1-a + 1 - a = ak+1x + b 1 - a = ak+1x + 1 - @k+1 b. 1 - a and so P(k + 1) holds whenever P(k) holds. The result now follows, by the Principle of Mathematical Induction. (ii) We have that f(x)=x+ b, so f(2)(x)=f(f(x))=f(x+b) =x+b+b = x + 2b, and f f x In general, f(n)(x)=x + nb. 4. (i) We have f(u) = u exactly when u2 - u + (1/2) = u and so u is a fixed point exactly when it is a solution to u2-2u+(1/2) = 0. Hence u = p 2+ (-2)2 2 4 x (1/2) And'so we have a