Horner's Method of Solving Irrational RootsDate: 01/29/98 at 05:56:15 From: migs chilo Subject: Horner's Method of Solving Irrational Roots Can you give a method, formula, and answer for this equation? x^3 - 3x^2 - 2x + 5 = 0 Date: 01/29/98 at 16:46:05 From: Doctor Rob Subject: Re: Horner's Method of Solving Irrational Roots Horner's Method for polynomials with leading coefficient 1 goes like this: 1. Guess a value A1 for the root. Divide f(x) by x - A1, getting remainder r0 and quotient q0(x). Divide q0(x) by x - A1, getting remainder r1 and quotient q1(x). Continue until the quotient is 1. Then the polynomial f1(x) = r0 + r1*x + r2*x^2 + ... + x^n has roots which are A1 less than those of the original polynomial. If your guess is good, r0 will be small compared to r1. 2. For your next guess, use A2 = -r0/r1. Repeat using f1(x), generating f2(x), a polynomial all of whose roots are A2 less than the roots of f1(x), and so A1 + A2 less than the roots of f(x). 3. Continue this process until your root has the desired accuracy, that is, until the guess An is smaller than the tolerable error. In this case, let's start with A1 = 1. Then x^2 - 2*x - 4, remainder 1 ----------------------- x - 1 ) x^3 - 3*x^2 - 2*x + 5 x - 1, remainder -5 --------------- x - 1 ) x^2 - 2*x - 4 1, remainder 0 ------- x - 1 ) x - 1 The polynomial 1 - 5*x + 0*x^2 + x^3 has roots 1 smaller than the roots of the original polynomial. Now guess 1/5 for the root of this polynomial. x^2 + 0.2 *x - 4.96, remainder 0.008 ----------------------------- x - 0.2 ) x^3 + 0*x^2 - 5.00*x + 1 x + 0.4, remainder -4.88 -------------------- x - 0.2 ) x^2 + 0.2*x - 4.96 1, remainder 0.6 --------- x - 0.2 ) x + 0.4 The polynomial 0.008 - 4.88*x + 0.6*x^2 + x^3 has roots 0.2 smaller than the roots of the preceding polynomial, and so 1.2 smaller than the roots of the original polynomial. For a new guess, use 0.008/4.88 = 0.00164. (remainder -0.000004818170944) x^2 + 0.60164*x - 4.8809866896 -------------------------------- x - 0.00164 ) x^3 + 0.6*x^2 - 4.88*x + 0.008 and so on! This is already accurate to 5 decimal places, and the root is 1.20164 to that accuracy. The next guess would be -.0000003242766. Once you have that root A, the quotient f(x)/(x-A) is used to find the next root, and so on. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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