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Laguerre's method may even converge to a complex root of the polynomial, because the radicand of the square root may be of a negative number, in the formula for the correction, , given above – manageable so long as complex numbers can be conveniently accommodated for the calculation. This may be considered an advantage or a liability ...
If the numbers , …, are pairwise different, then the polynomials in the terms of the right hand side form a basis of the n-dimensional space [] of polynomials with maximal degree n − 1. Thus a solution w → {\displaystyle {\vec {w}}} to the increment equation exists in this case.
Points are colored according to the final point of the Bairstow iteration, black points indicate divergent behavior. The first image is a demonstration of the single real root case. The second indicates that one can remedy the divergent behavior by introducing an additional real root, at the cost of slowing down the speed of convergence.
Given n + 1 points, there is a unique polynomial of degree ≤ n which goes through the given points. Neville's algorithm evaluates this polynomial. Neville's algorithm evaluates this polynomial. Neville's algorithm is based on the Newton form of the interpolating polynomial and the recursion relation for the divided differences .
Muller's method fits a parabola, i.e. a second-order polynomial, to the last three obtained points f(x k-1), f(x k-2) and f(x k-3) in each iteration. One can generalize this and fit a polynomial p k,m (x) of degree m to the last m+1 points in the k th iteration. Our parabola y k is written as p k,2 in this notation. The degree m must be 1 or ...
Choosing the points of intersection as interpolation nodes we obtain the interpolating polynomial coinciding with the best approximation polynomial. The defect of this method, however, is that interpolation nodes should be calculated anew for each new function f ( x ), but the algorithm is hard to be implemented numerically.
This polynomial is further reduced to = + + which is shown in blue and yields a zero of −5. The final root of the original polynomial may be found by either using the final zero as an initial guess for Newton's method, or by reducing () and solving the linear equation. As can be seen, the expected roots of −8, −5, −3, 2, 3, and 7 were ...
Finding the root of a linear polynomial (degree one) is easy and needs only one division: the general equation + = has solution = /. For quadratic polynomials (degree two), the quadratic formula produces a solution, but its numerical evaluation may require some care for ensuring numerical stability.