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# goldch5 - Contents V Solving Polynomial Equations V.A V.B...

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Contents V Solving Polynomial Equations 1 V.A Newton’s Method in the Special Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 V.B General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

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V Solving Polynomial Equations V.1 V Solving Polynomial Equations Suppose P ( x ) is a polynomial. We wish to consider the following problems. Problem A . Find all the roots of P ( x ) in a given interval [ a, b ] to a given accuracy. Problem B . Find all the roots of P ( x ) to a given accuracy. First we show how to solve problem A in a very special case, namely when P 0 ( x ) has no roots in [ a, b ] and P ( a ) and P ( b ) are non-zero and have opposite signs. Then we show how to solve the problem in general. V.A Newton’s Method in the Special Case The algorithm we use in the special case is Newton’s method . The fundamental observation by Isaac Newton (1643–1727) was that if we know approximately where a root r of P is located, we may evaluate P at some point x 0 nearby. Usually, P ( x 0 ) 6 = 0, but from x 0 to the root r , the graph has to run from P ( x 0 ) to 0. As the tangent line to the graph of P at ( x 0 , P ( x 0 )) tends to point in the same direction as the graph, why not follow the tangent line until it intercepts the x –axis? The real stroke of genius was then to take x 1 , the x –intercept of the tangent at ( x 0 , P ( x 0 )), as the new guess and to repeat the step, following now the tangent to the graph of P at ( x 1 , P ( x 1 )) — and to keep going! x y (x 0 , P(x 0 )) (x 1 , P(x 1 )) x 0 x 1 x 2 r tangent through (x 0 , P(x 0 )) tangent through (x 1 , P(x 1 )) Now we make things precise. Theorem: Suppose that P ( x ) is a polynomial such that
V.A Newton’s Method in the Special Case V.2 (i) P 0 ( x ) has no root in [ a, b ]. (ii) P ( a ) and P ( b ) are non-zero and have opposite signs, that is, P ( a ) P ( b ) < 0. Then P ( x ) has a exactly one root r in ( a, b ) and there is an algorithm to approximate r to any desired accuracy. Proof: First we prove the assertion that P ( x ) has a single root. Since P ( a ) and P ( b ) have opposite signs, the IVT tells us that P ( x ) must vanish at some point r ( a, b ); i.e. there is at least one root. On the other hand, Rolle’s theorem says that two roots of P ( x ) are separated by a root of P 0 ( x ). Since P 0 ( x ) has no roots in [ a, b ], P ( x ) cannot have two roots in [ a, b ], it has at most one root. The algorithm to approximate the root, r , consists of the following steps. Step 1 (Bounding derivatives): Find constants M 1 and M 2 so that 0 < M 1 ≤ | P 0 ( x ) | for all x [ a, b ] and | P 00 ( x ) | ≤ M 2 for all x [ a, b ] . (Important Note: | P 0 ( x ) | has to be bounded from below , whereas | P 00 ( x ) | has to be bounded from above !) Let K = M 2 2 M 1 and choose a constant h > 0 so that Kh 1 . Step 2 (Shrinking the interval): Find inside [ a, b ] an interval [ u, v ] which contains the root, and such that v - u h . This can be done as follows: Method 1 (Bisection): if b - a h set u = a, v = b . (The original interval is already narrow enough.) otherwise evaluate P ( x ) at the midpoint of [ a, b ]. This divides [ a, b ] into two equal subintervals and from the sign of P ( a ), P ( b ) and P (midpoint) you can tell which subinterval contains the root.

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