By Nathan Jacobson
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Example text
The bisection method is a simple but slowly convergent method. 2 Solve the equation x3 – 9x + 1 = 0 for the root lying between 2 and 3, correct to three significant figures. Solution We have f(x) = x3 – 9x + 1, f(2) = –9, f(3) = 1 ∴ f(2) f(3) < 0. 003 2 f(xn + 1) In the 8th step an, bn and xn + 1 are equal up to three significant figures. 94 as a root up to three significant figures. 94. 3 Compute one root of ex – 3x = 0 correct to two decimal places. 6. 00184 24 NUMERICAL ANALYSIS In the 4th step an, bn and xn + 1 are equal up to two decimal places.
2565) 3 − sin ( −1. 2565) + 1 3 ( −1. 2565) 2 − cos ( −1. 249. 14 Solve x4 – 5x3 + 20x2 – 40x + 60 = 0, by Newton–Raphson method given that all the roots of the given equation are complex. Solution Let f(x) = x4 – 5x3 + 20x2 – 40x + 60 ∴ So that bg f ′ x = 4 x 3 − 15x 2 + 40 x − 40 The given equation is f(x) = 0 (1) Using Newton–Raphson Method. We obtain x n + 1 = xn − = xn − b g b g f xn f ′ xn xn4 − 5xn3 + 20 xn2 − 40 xn + 60 4 xn3 − 15xn2 + 40 xn − 40 = 3xn4 − 10 xn3 + 20 xn2 − 60 4 xn3 − 15xn2 + 40 xn − 40 x1 = 3x04 − 10 x03 + 20 x02 − 60 4 x03 − 15x02 + 40 x0 − 40 (2) Putting n = 0 and, Taking x0 = 2(1 + i) by trial, we get b g − 10b2 + 2ig + 20b2 + 2ig − 60 4b2 + 2i g − 15b2 + 2i g + 40b2 + 2i g − 40 .
12 Find the Newton’s method, the root of the ex = 4x, which is approximately 2, correct to three places of decimals. 085537 = + ve ∴ f(2) f(3) < 0 ∴ f(x) = 0 has a root between 2 and 3. 23383 . 16617. 1561 (approximately). 6106516. 1532. 13 Find the root of the equation sin x = 1 + x3, between –2 and –1 correct to 3 decimal place by Newton Rappon method. 0907 f(–1) f (–2) < 0 f(x) = 0 has a root between –2 and –1. x0 = –1 : be the initial approximation of the root. 3421 2. 4597 38 NUMERICAL ANALYSIS The second approximation to the root is x2 = x1 − f ( x1 ) f ′ ( x1 ) = −1.