Question

NOTE: Algebraic expressions follow FORTRAN conventions. Use full calculator precision for intermediate values Use Newton's method with the function defined by: F(X) 0.85"X 3.64* sin(0.89 *X) = - Use the ( X, F (X) ) starting point: (2.12, -1.65777) new approximation to root after ONE iteration is new approximation to root after TWO iterations is new approximation to root after THREE iterations is If X satisfies the convergence criterior If (X) 0.00001, then the root X is 4 ANSNER SECTION: SELECTIONS FOR BLANK NUMBER 1 2.95675 2.77238 0.342821 -8.45434 3.01282 (b) SELECTIONS FOR BLANK NUMBER 2 2.75062 2.76117 2.77397 (c) 6.48334 2.06117 SELECTIONS FOR BLANK NUMBER 3 2.74735 -4.49634 2.74973 4.52735 2.74773 (c) SELECTIONS FOR BLANK NUMBER 4 -899.777 2.74729 5.18729 4.57729 5.79729

Answer 1

Given f(x) = 0.85 lambda - 3.64 sin(0.89 lambda) then f�(lambda) = 0.85 - 3.64 times 0.89 times cos (089 lambda) = 0.85 - 3.2396 cos (0.89 lambda) By Newton�s iteration formula, lambda_n + 1 = lambda_n - f(x_n)/f�(lambda_n) = x_n- 0.85 - 3.64 sin(0.89x_n)/0.85 - 3.2396 cos (0.89 x_n)/0.85 -3.2396 cos(0.89 lambda_n)