University:
California State University, Northridge
Course:
MATH 331 | Mathematical Explorations
Academic year:

2023
Views:

369

Pages:

34
Author:

Yung D

UNIT-I SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS method of False position (os) Regula Falsi method (1) Lincar interpolation method. If (24) f(a) 20, then as lies between 2 and a af(21) -2, fla) +(x)-(a) Find the positive root of x²-22-5=0 by the Ragula falsi method. Let f(x)=-2x-5=0 There is only poistive reet by Descarte's rule of sigus f(0) = -5 = -ve f(1) = 1-2-5-6-ve f(2)= 8-4-5 = -1=-ve f(2)=27-6-5=16= +ve Therefore the positive root les between 2 and 3. It is closer to 2 also. af (5)-b.b(0) f(5)-(a) 2f(3)-38(2) 6(3)-6(2) =2(16)-3(-1) 16-(-1) 32 +3 17 35 17 2.0588 [ correct to 4 decimal places] 3 6(11)=f(2.0588) = (2.0588)-2(2.0588)-5 8.7265-4-1176-5 F-0.3911 The root lies between 2.0588 and 3 2.05884(3)-34(20588) (3)-(2.0588) 2.0588 (16)-3(-0.3911) 16-(-0.3991). 32.9408 +1.1733 34-1141 16.3911 16.39 11 = 2.0813 f(2)= (2.0813) = (2.0813) - 2 (2.0813)-5 =9-0158-4-1626-5 = -0.1468 The root lies between 2.0813 and 3 αι 2.08136(3)-34(2.0813) f(3) -12.0813) 2.0813(16)-3(-0.1468) 16-(-0.1468) 33.3008 +0.4404= 33.7412 16.1468 16.1468 2.08965 = 2.0897 (four decimal places) f (23) = f(2.0897) = (2.0897)-2(2-0897)-5 = -0.054-ve The root lies between 2.0897 and 3 24 = (2.0897) (3)-36(2.0897) f(3)-(2-0897) =(2.0897)(16)-3(-0.054) 16-(-0.054) - 33.4352 +0.162 16.054 33.5972 16.054 2.0928 3 +(2.0928) = (2.0928) - 2 (2.0928)-5 6(x4) = f( = 9.1661 -4-1856-5 -0.0195 =-ve The reet les between 2. 0928 and 3 D= 2.0928 (3)-36(2-0928) (3)f (2.0928) (2-0928) (16)-3(-0.0195) 16-(-0.0195) 33.4848 + 0.0585 16.0195 33-5433 16-995 = 2.0939 *(x5) = (2.0939) = (2.0939)-2(2.0939)-5 9-1805-4-1878-5 =-0.0073 The 96 root lies between 2.0939 And 3 (2.0939) f (3)-36 (2.0939) (3)-(2.0939) =(2.093円) (16)-3(-0.0073) 16-(-0.0073) 33.5024 +0.0219 33-5243 16.0073 16.0073 =2.0943 6(76)=4(2.0943) = (2.0943)3-2(2.0943)-5 9.1868-4-1886-5 -0.0028 = "The root lies between 2.0943 and 3 = (2.0943) f (3) - 84(2.0943) (3)-(2.0943) " (2.0943) (16)-3(-0.0028) 16-(-0.0028) 33.5088 + 0.0084 16.0028 33.51722.0945 16.0028 6(7) = f(2.0945) = (2.0945)-2(2.0945)-5 =9-1884-4-189-5 -0.0006- ve The root lies between 2.0945 and 3 2.0945 6-(3)-36(2.0945) 6(3)-(2.0945) " (2.0945) (16)-3(-0.0006) 16-(-0.0006) 33-512 + 0.0018 16.0006 33.5138 16.0006 = 2.0945 6/6)=4(2.0943) = (2.0943)-2(2.0943)-5 9.1868-4-1886-5 = -0.0028 The root lies between 2.0943 and 3 7= (2.0943) f (3) - 86 (2.0943) (3)+(2.0943) = (2.0943) (16) -3(-0.0028) 16-(-0-0028) 33.5088 + 0.0084 16.0028 " 33.51722.0945 16.0028 f(x) = f(2.0945)= (2.0945)-2(2.0945)-5 9-1884-4-189-5 -0.0006 - ve . The root lies between 2.0945 and 3 8 = 2.0945 6-(3)-36(2.0945) f (3)-(2-0945) (2.0945) (16)-3(-0.0006) 16-1-0-0006) = 33.512 + 0.0018 16.0006 33.5138 16.0006 " 2.0945 we observe that xy= x8 = 2.0945 Correct to 4 places. of decimals. Hence, the required root correct to four places of decimals is 2.0945 below. The results of the Complete working are tabulated 2 using method of false position find a x²-32-5=0 rect of the equation Iterationes Formula 2s = aftb)-bf1a) f(5)-f(a) we observe that xy = x2=2-0945 Hence the required rest is 2-0945. -Given f(x)=-3x-5 10)= 0-0-5=-5=-ve ELD = 1-3-5-1-8-7=-ve f(2)=8-6-58-11--3=-ve 6(3)=27-9-5-27-14-13= +ve