numerical analysis in the interval 10 and the other is in th

in the interval [-1,0] and the other is in the [l.2] Manipulate the equation x^4 + 2x^2 - x - 3 = 0 in such a way that fixed point iteration works to find the root in [-1,0], and use this method to find the root accurate to 3 decimal places. If the same Hexed point iteration works to find the second root, use it; if not, find a different manipulation that allows you to find the root in [1,2].

Solution

The fixed point iteration method solves an equation f(x)=0 by finding a fixed point of the function g(x) such that the equation f(x)=0 may be re-written as x=g(x). This method iterates as x_n+1 = g(x_n), (n=0,1,2,3,...) on this new equation starting from some initial point x₀ in a given interval [a b], which contains the actual root of the given equation i.e., a fixed point of g(x). Though, the convergence of the method is not always guaranteed for all manipulations x=g(x) of f(x)=0. There is a simple way of checking: If g(x) is continuous on [a b] and absolute value of g^\'(x) < k <1 for all x in [a b], the method converges to a fixed point of g(x) in [a b], giving the solution of original equation. If this condition is not satisfied, the convergence can\'t be guaranteed.

The given equation, may be rewritten as x = x⁴ + 2 x² - 3, i.e., g(x) = x⁴ + 2 x² - 3 here, which is continuous on both the intervals given to contain the roots. But, the absolute value of g\' on these intervals is not <1 for all x in these intervals, so there is no guarantee of convergence.

In order to have higher chances of convergence, the initial point x₀ is chosen in such a way that its distance from the actual root is as small as possible. As for the given function, the roots lie in the intervals [-1 0] and [1 2]. Since these intervals are of small lengths, we may choose their middle points as the inital points for their respective iterations, i.e., for finding the first root take x₀ = -0.5, and put it in x_n+1 = g(x_n) i.e., in x_n+1 = x_n⁴ + 2 x_n² - 3 which for n=0, becomes x₁ = x₀⁴ + 2 x₀² - 3, and for x₀ = -0.5 gives x₁= -2.4375 and then on putting this value in the iterative equation for n=1 i.e. in x₂ = x₁⁴ + 2 x₁² - 3 we get a value for x_2. In this way iterations are continued till we reach a value of x for which there is no further improvement till the third place of decimal, over a few iterations. This value of x will give one root for the given equation. In case, such a value is not reached, the method does not converge.

Similarly, for finding the second root the iterations may be started with x₀ = 1.5.

The computational work is easy and may be performed with the help of a calculator.

In case of any further difficulty plz feel free to send the query.