Find the real root of equation sinx-2x+1 using bisection method.
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How can I find a real root of the equation f(x)=x^3+x-1=0 by using the bisection and Newton-Raphson method correct to three decimal points?
Bisection method: This is a very simple method. Identify two points x = a and x = b such that f (a) and f (b) are having opposite signs. Let f (a) be negative and f (b) be positive. Then there will be a root of f (x) = 0 in between a and b. Let the first approximation be the mid point of the interval (a, b). i.e. x1 = (a + b)/2 If f (x1) = 0, then x1 is a root, other wise root lies between a and x1 or x1 and b according as f (x1) is positive or negative. Then again we bisect the interval and continue the process until the root is found to desired accuracy. Solution: let a = 0 and b = 1 x1 = 0.5, f(x1) = -ve between x1 = 0.5 and b = 1 x2 = 0.75 (+ve) between x1 and x2 x3 = 0.625 x4 = 0.6875 x5 = … (you get the gist) I think you should continue this about 10 times till you get x(n) = x(n+1) approximately to 3 decimal places. x(n) will be the answer. Newton-Raphson method: Let x0 be approximation for the root of f (x) = 0. Let x1 = x0 + h be the correct root so that f (x1…
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