Newton's method is run on a polynomial with a root at of unknown multiplicity

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Numerical and computational mathematics · Numerical and computational mathematics

Mathematical Methods for Physics, Ch Root finding, interpolation, numerical linear algebra and quadrature

Problem

Newton's method is run on a polynomial p(x)p(x) with a root at x=2x = 2 of unknown multiplicity. The printed errors ∣xn−2∣|x_n - 2| are 0.1000.100, 0.07500.0750, 0.05630.0563, 0.04220.0422. ((a)) For f(x)=(x−2)mf(x) = (x-2)^m with an integer m≥2m \ge 2, show that Newton's method gives en+1=m−1mene_{n+1} = \frac{m-1}{m}e_n exactly, where en=xn−2e_n = x_n - 2, and explain from the error law en+1=f′′(ξ)2f′(xn)en2e_{n+1} = \frac{f''(\xi)}{2f'(x_n)}e_n^2 why the quadratic convergence of a simple root is lost. ((b)) Read the multiplicity of the root of pp from the printed errors, and find how many further steps plain Newton needs to bring the error from 0.04220.0422 below 10−1010^{-10}. ((c)) Show that the modified step xn+1=xn−m f(xn)/f′(xn)x_{n+1} = x_n - m\,f(x_n)/f'(x_n) lands on the root of (x−2)m(x-2)^m in one step. ((d)) A colleague proposes to switch to bisection for its guarantee. Decide whether bisection can be used on this root, and compare its rate with that of plain Newton here.

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