Views
8 months ago

# numerical3j

## {24} • Numerical

{24} • Numerical Methods 1.4: x n approaches the solution x = 1. The outcome seems to switch somewhere between the initial estimates x 1 = 1.44 and x 1 = 1.45. With some more computations, we can narrow down further and find that the switch happens somewhere between x 1 = 1.447 and x 1 = 1.448. The computations (to 3 decimal places) from these initial values are shown in the table below. So, if we try x 1 = 1.4475, will the x n converge to the solution x = 1 or x = 0? The answer is neither: as shown in the table, it converges to the solution x = 2! x 1 1.4 1.44 1.447 1.4475 1.448 1.45 1.5 x 2 0.754 0.594 0.554 0.551 0.548 0.536 0 x 3 1.036 1.266 1.440 1.458 1.477 1.567 0 x 4 1.000 0.952 0.596 0.484 0.317 -9.156 0 x 5 1.000 1.000 1.260 2.370 -0.598 -5.793 0 x 6 1.000 1.000 0.956 2.111 -0.225 -3.562 0 x 7 1.000 1.000 1.000 2.015 -0.050 -2.091 0 x 8 1.000 1.000 1.000 2.000 -0.003 -1.135 0 x 9 1.000 1.000 1.000 2.000 0.000 -0.536 0 x 10 1.000 1.000 1.000 2.000 0.000 -0.192 0 x 11 1.000 1.000 1.000 2.000 0.000 -0.038 0 x 12 1.000 1.000 1.000 2.000 0.000 -0.002 0 Furthermore, if we try x 1 = 1 + 1 5 ∼ 1.447214, we find different behaviour again! As it turns out, x 2 = 1 − 1 5 and x 3 = 1 + 1 5 so x 1 = x 3 and the x n alternate between these two values. Exercise 12 Verify that if f (x) = 2x − 3x 2 + x 3 = x(x − 1)(x − 2) and x 1 = 1 + 1 , then x 5 2 = 1 − 1 and 5 x 3 = 1 + 1 = x 5 1 . So there is no simple transition from those values of x 1 which lead to the solution x = 1 and x = 0. The behaviour of Newton’s method depends extremely sensitively on the initial x 1 . Indeed, we have only tried only a handful of values for x 1 , and have only seen the tip of the iceberg of this behaviour.

A guide for teachers – Years 11 and 12 • {25} In the Links Forward section we examine this behaviour further, showing some pictures of this sensitive dependence on initial conditions, which is indicative of mathematical chaos. Pictures illustrating the behaviour of Newton’s method show the intricate detail of a fractal. Getting Newton’s method to work In general, it is difficult to state precisely when Newton’s method will provide good approximations to a solution of f (x) = 0. But we can make the following notes, summarising our previous observations. • If f ′ (x n ) = 0 then Newton’s method immediately fails, as it attempts to divide by zero. • The “approximations" x 1 , x 2 , x 3 ,... can cycle, rather than converging to a solution. • The “approximate solutions" x 1 , x 2 , x 3 ,... can even diverge to infinity. This problem happens when f ′ (x n ) is small, but not zero. • The behaviour of Newton’s method can depend extremely sensitively on the choice of x 1 , and lead to chaotic dynamics. However, a useful rule of thumb is as follows: • Suppose you can sketch a graph of y = f (x), and you can see the approximate location of a solution, and f ′ (x) is not too small near that solution. Then taking x 1 near that solution, and away from other solutions or critical points, Newton’s method will tend to produce x n which converge rapidly to the solution. Links Forward Everything we know is only some kind of approximation, because we know that we do not know all the laws yet. Therefore, things must be learned only to be unlearned again or, more likely, to be corrected. – Richard Feynman Speed of the bisection algorithm As we saw previously, one nice aspect of the bisection algorithm is that it always works. We now consider its accuracy more precisely, and how long it takes to work. In order to examine the accuracy of the method, we find the length of each [a n ,b n ]. Exercise 2 asked you to find the length of each [a n ,b n ] in terms of the original interval