Solution of nonlinear equations

Transcription

1 Chapter 1 Solution of nonlinear equations This chapter is devoted the problem of locating roots of equations (or zeros functions). The problem occurs frequently in scientific work. In this chapter we are interested in solving nonlinear equation finding x such that. The general problem posed in the simplest case of a real-value function of a real variable is this: Given a function find the values of x for which. We shall consider several standard procedures for solving this problem. Examples of nonlinear equations can be found in many applications. In the theory of diffraction of light we need the roots of the equation. In the calculation of planetary orbits we need the roots of Kepler s equation of and. for various values Since locating the zeroes of functions has been an active area of study for several hundred years numerous methods have been developed. In this chapter we begin with three simple methods that are quite useful: the bisection method Fixed-point method Newton s method and the Secant method. 1.1 Bisection Method If is a continuous function on the interval [a b] and if then must have a zero in. Since the function f changes sign on the interval [a b] and therefore it has at least one zero in the interval. This is a consequence of the Intermediate-Value Theorem. Theorem (Intermediate-Value Theorem) Let be a continuous function on the interval Then realizes every value between and. More precisely if is a number between and then there exists a number with such that 1

2 Theorem 1.2 Let be a continuous function on satisfying. Then has a root between and there exists a number satisfying and. The bisection method exploits this idea in the following way. If then we compute and test whether. If this is true the has zero in. So we start again with the new interval which is half as large as the original interval. If. Then. In this case we restart with the new interval. In either case a new interval containing a zero of f has been produced and the process can be repeated. Bisection method finds one zero but not all the zeros in the interval. Of course if then and a zero has been found. However it is quite unlikely that is exactly 0 in the computer because of round off errors. Thus the stop criterion should not be whether A reasonable tolerance must be allowed such as Bisection method is also known as the method of interval having. Example Use the bisection method to find the root of the equation closest to 0. Solution. If the graphs of and sinx are roughly plotted it becomes clear that there are no positive roots of 2

3 and that the first root to the left of 0 is in the interval [-4-3]. When the bisection algorithm was run on a machine similar to the marc-32 the following output was produced starting with the interval (-4-3). R c f(c) Bisection Method Given initial interval such that While if stop end if if else ; end if end while 3

4 Error Analysis (How accurate and how fast) To analyze the bisection method let us write the successive intervals that arise in the process [ ] and so on. Here are some observations about these numbers. ( ) Since the sequence is non-decreasing and bounded above it converse. Likewise converges. If we apply equation (1) repeatedly we find that Thus If we put then by taking a limit in the inequality we obtain where. Suppose that at a certain stage in the process the interval [a b] has just been defined. If the process in now stopped the root is certain to lie in this interval. The best estimate of the root at this stage is not or but the midpoint of the interval. The error is then bounded as follows. Summarizing this discussion we have the following theorem on the bisection method Theorem (Theorem on Bisection Method) If denote the intervals in the bisection method then the limits and exist equal and represent a zero of f. If and then. 4

5 Example Suppose that the bisection method is started with the interval [5063] how many steps should be taken to compute a root with relative error accuracy of on root in. Practical considerations First the midpoint c is computed as rather than as To adhere to the general stratagem that in numerical calculation it is best to compute a quantity by adding a small correction to a previous approximation. Forsythe Malcolm and Moler [1977] present an example in which the midpoint moves outside of the interval on a machine with limited precision. Second it is better to determine whether the function changes sign over the interval using Rather than Since the lather requires an unnecessary multiplication and could cause an underflow or overflow. Finally notice that three stopping criteria are presented in the following algorithm. First gives the maximum number of steps that the user permits. Such a safeguard should always be present to reduce the possibility of the computation going into an infinite loop. Next the calculation can be stopped when either the error is small or the value of is small enough. 5

6 The Parameter and control this. Example can be easily given in which one of the latter two stopping criteria is satisfied but the other is not. We choose to stop the algorithm when any one of the three stop criteria is established in the interest of having a robust code. Function [abuv]=bisection (fun a b M ) u=feval(fun a); v=feval(fun b); e=b-a; if (sign (u)>0 & sign(v)>0) (sign (u)<0 & sign(v)<0) end return ; for k=1:m e=e/2; c=a+e ; =feval (fun c); if ( e < w< ) return enf if (sign( ) = = sign(u)) a=c ; u= else b=c; v= end end 6