Question

Newton-Raphson scheme

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Calculus Suppose you want to find zeros of the function f(x)102212 and plan to use the Newton-Raphson scheme. (a) Write down the Newton-Raphson algorithm for this. That is, write down explicitly a formula for computing your (n+1)st guess Tn+1 given your nth guess rn for a root. In other words, deter- mine the recurrence relation resulting from using this particular function f. (b) Modifying Algorithm 2.2 as required, find the values through x7 if you choose an initial value x1 -2. c)Run your algorithm for the values r to r with the different choice of initial condition 0.75. What happens? Can you explain why this happens for this particular function with this particular choice of 1?

Answer 1

a. Algorithm for Newton Raphson method First we have to guess a root Lets say X_n. We have f(X)- 10XA3-9*X.*2-12*X+12 And then we have to find df/dx=f'(X)= 30X^2-18X-12 Now as per Newton Raphson method the (n+1)th root can be written as f (Xn) n+1 We have to iterate until convergence arrived For X1--2.0000 The value of X2 X1 f(X1)/df(X1) --2.0000 (10*-2.00)A3-9*(-2.00)-12*(-2.00)+12/(30*-2.00)A2-18*(-2.00)-12) -2.0000-(-80.0000)/((144.0000)) -1.444444 For X2 --1.4444 The value of X3 X2 f(X2)/df(X2) -1.4444 (10 (-1.44)A3-9*(-1.44)-12*(-1.44)+12)/(30*(-1.44)A2-18 (-1.44)-12) --1.4444-(-19.5816)/(76.5926)) 1.188785 For X3--1.1888 The value of X4 X3 f(X3)/df(X3) -1.1888 (10*(-1.19)A3-9*(-1.19)-12*(-1.19)+12)/(30*(-1.19)A2-18*(-1.19)-12) -1.1888-(-3.2535)/(51.7944)) --1.125969 For X4-1.1260 The value of X5 X4 f(X4)/df(X4) 1.1260 (10 (-1.13)A3-9*(-1.13)-12*-1.13)+12)/(30*(-1.13)A2-18 (-1.13)-12) -1.1260-(-0.1738)/((46.3017)) -1.122217 For X5 -1.1222 The value of X6 X5 f(X5)/dfX5) 1.1222 (10 (-1.12)A3-9*(-1.12)-12*(-1.12)+12)/(30*(-1.12)A2-18 -1.12)-12) --1.1222--0.0006)/((45.9810)) 1.122204 For X61.1222 The value of X7 X6 f(X6)/df(X6) --1.1222 (10 (-1.12)A3-9*(-1.12)-12*(-1.12)+12)/(30*(-1.12)A2-18 (-1.12)-12) --1.1222-(-0.0000)/((45.9799)) -1.122204

C. For X1 0.7500 The value of X2 X1 f(X1)/df(X1) 0.7500 (10 (0.75)A3-9*(0.75)-12* (0.75)+12)/(30* (0.75)2-18*(0.75)-12 = 0.7500-(2.1562)/((-8.6250)) -1.000000 For X2-1.0000 The value of X3 X2 - fX2)/df(X2) = 1.0000-(10"(1.00)^ 3-9 *(1.00)-12° (1.00)+12)/(30"(1.00)12-18"(100-12) 1.0000-(1.0000)/((0.0000)) =-inf After few iterations the slope becomes zero and due to that it provides inf value and never become convergent