University:
Glendale Community College
Course:
MAT 140 | College Mathematics
Academic year:

2022
Views:

130

Pages:

15
Author:

Prince Sampson

Unit 2 15 Bysection Method:- →for finding the roots of the equation → It is based on repeated application of intermediate valve of propenty. Propenly. Q. find the real cool of x ^ 5 - x - 1 by byscition methal (x) = x ^ 3 - x - 1 x ^ 3 - x - 1 =0 rfloor Fon po f(x) - f(0) = 0 ^ 3 + 0 - 1 1 (-ne value) fon n=1 6(1) = 1 ^ 3 - 1 - 1 1-1-1-1 (-ve value for x_{1} = 2 6 (2) = 2 ^ 3 - 2 - 1 = 8 - 2 - 1 6-1-5 value) Let us fin a value as 1 of braue as 2 alpha = 1 , b = 2 so root lies b/w 1 and 2. 2 continuous name from the towe were taken betone. a <= a y(x) = - v_{c} * H(a) a = b t(x)=+ ve (t(b) Table tenm a b 1 2 -1 f(a) (b) n = (a + b)/2 5 + cx ) fut n=1.5 0.875 came the the positive be replaced by point) new the value (middle point value 115-1 0.875 -0.2968 1.25 (125)-1.25-1 1-25 5-1 1.375 -2968 0.875 (1335)-1375-1 0-2246 Q. V NEM to find real wat cos x - x * e ^ x = 0 comated to your fecund t(M) = cocx - x * e ^ x = 0 I(1) = cos 1 - |c ^ 1 - 2.177| * (- x) f(x) = cos 0e ^ 0 = 1 t ^ 2 * (x) * s = sin x - e ^ x * (x + 1) le con take no m (tre) sin^2 x - x * xi ^ 2 = - 8mx - e ^ x_{3} but when transitential c_{1} deg comes the better to trive f-ve point thee windpring of both the passes f-1 Thee, let x_{0} = 0.7 alpha n + 1 = alpha 4^ n f(n) f^ 1 (n+n) =x A = cos x n - x_{n} * e ^ x_{n} -(sin x x e^ x n (x+)) =n n =- (cos x n -x n e^ x n ) (sin x n + e ^ x_{n} * (x + 1) ) =x n + c cos x n - x_{n}[e ^ x_{n}) sin x n +e^ x n (x n )) Put noo n tf =n 0 + togx 0 -n e e^ x t sinx 0 +e^ x 0 (x t +1)) 0.5+ cos((0.5)-(0.5)e^ 0E 3[nu(0.5) - e ^ 10.5) * (0.4 + 1) =0.96. Put mol x 1 +x 1 +x 1 + (05x 1 -x 2 e^ 2 | 3i*x_{1} + e ^ 33 * (x_{1} + 1) x_{2} = 0.51% 33 Here the ny welve got repeated tory we got the wond the root is 0.518 Q. Apply ag 2(x - 3) = log_10(x) y(x) = 2x = 6 + log(10 ^ x) f(3) =6-log 10 ^ 3 b(4) =8.-6 log 10 ^ 4 -ve) (the) Average b - 0.49712 & 1.39794 m_{0} = 3.5 L47) f' (3,2- 0.4(343 m x n + 1 =x n - t langle x n rangle b^ prime langle x n rangle Put multipling x 1 =x 0 - 2x 0 ^ 2 -6-0.4343)20ge^ n 0 2 - (0.43/3)/x_{0} no in ban nye no 2по- в ко-чакраде 2no-0.4343 x_{1} = 3 * 25696 Bisection Method:- (Bolzano Method):- * Transcedental egh 18 the gh where you get log funet? or trigonmetrin on exponential etc. →Bayed on the nepeated applicath of intermediate medue propety. Let fen) be continuous bho a and b & let f(a) be-ve. 4fcb) be the Then the first appronimation of fint root is math, ffy)=0, then ry its root of &cm)=0. 2 Othemise the root lies between a & or 개 46 Positive or negative accordingly اله fcm) is Then we bisect the informal as before & continue the process until the root & found anurately.