1.4 Newton-Raphson and Secant Methods 1.4.1 Slope Methods for Finding Roots
1.4 Newton-Raphson and Secant Methods 1.4.1 Slope Methods for Finding Roots
Theorem 1.5(Newton-Raphson Theorem). Assume that f E C a, b and there exists a number p e Ja, bl, where f(p)=0. If f(p)#0, then there exists a 8>0 such that the sequence pk igso defined by the iteration Dk=9h-1)=D,f=1) f(pk-1) for k= 1.2 (1.40) will converge to p for any initial approximation po E, p+8 Remark. The function g(a) defined by formula 9(x)=x
Corollary 1. 2(Newtons Iteration for Finding Square Roots). Assume that A>O is a real number and let po>0 be an initial approximation to vA.Define the sequence pr g-o using the recursive rule Pk-1+ pk-1 for k=1.2 1.47) 2 Then the sequence (pk )k-o converges to VA; that is, limn-oo Pk =VA
Example 1.11. Use Newton's square-root algorithm to findv5 Starting with po=2 and using formula(1.47), we compute 2+5/ 2.25 2.25+5/2.25 p =2.236111111 2236111111+5/2.23611111 =2.236067978 236067978+5/223606797 p4 =2.236067978
0=ft)=(Cn2+3221-0)-32Ct r=r()=C2(1-e1)
Example 1. 12. A projectile is fired with an angle of elevation bo= 450, Wu=Ur= 160ft/sec, and C=10. Find the elapsed time until impact and find the range Using formulas(1.51)and (1.52), the equations of motion are y=f(t)=4800(1 31.534367, we will use the initial guess Po =8. The derivative is f'(t)=480 e-4/10 W e-t10)-320 t and a=r(t)=16001-c-10). Since f(8)=83.22972andf(9) 320, and its value f(po)=f(8)=-104.3220972 is used in formula(1.40)to get 83.22097200 p1 =8.797731010 104.3220972 A summary of the calculation is given in Table 1.4 The value pa has eight decimal places of accuracy, and the time until impact is C8.74217466 seconds. The range can now be computed using r(t); and we get r(8.74217466)=1601-c 0.847217466 =932.4986302ft
Table 1. 4 Finding the Time When the Height f(t)Is Zero k Time, Pk Pk+1-Pk Height, f(pk 0|8.00000797310183.2297200 1|8.7973101-0.05530160-6.68369700 28.74242941-0.00025475-0.03050700 38.74217467-0.0000001-0.000100 48.742174660.00000000000
1.4.2 The division -by- Zero Error Definition 1.4(Order of a Root ). Assume that f(a) and its derivatives f(a) (M) (ar)are defined and continuous on an interval about a= p. We say that f(a)=0 has a root of order M at 2 =p if and only if f(p)=0,f(p), f(-(p)=0,andf(0(p)≠0
1.4.2 The Division-by-Zero Error
Lemma 1. 1. If the equation f(a)=0 has a root of order M at a= p, then there exists a continuous function h(a) so that f(a) can be expressed as the product f(ar)=(a-p)h(),where h(p)+0 1.54
Example 1. 13. The function f(a)=xs-3.c+2 has a simple root at p=-2 and a double root at p=l. This can be verified by considering the derivatives f(a)=352-3 and f"(a)=6. At th ne value P=-2, we have f(2)=0 and f(-2)=9, So M=1 in Definition 1.4; hence p=-2 is a simple root. For the value p= l, we have f(1)=0,f(1)=0, and f"(1)=6, so M=2 in Definition 1.4; hence p= l is a double root. Also, notice that f(a)has the factorization f( )=(c+2)(a-1)2