Math 1206 Calculus Sec. 4.9: Newton s Method I. Introduction For linear and quadratic equations there are simple formulas for solving for the roots. For third- and fourth-degree equations there are also formulas for the roots but they are more complicated. If f is a polynomial of degree 5 or higher, there is no such formula. (In 1824, Neils Abel proved that no general formula can be given for the roots of a 5 th -degree equation in terms of radicals. Evariste Galois later proved that it is impossible to find a general formula for roots of an n th -degree equation if n is any integer >4.) Likewise, there is no formula that will enable us to find the exact roots of a transcendental equation. When exact formulas for solving an equation f(x)=0 are not available, we will have to approximate the root. One way for us to approximate a solution to these equations is to graph the function and zoom in until we can identify the root to a specific decimal place. Another method is to use a numerical technique from calculus, such as Newton s method or as it is more accurately called the Newton-Raphson method. A faster alternative is to use a numerical rootfinder on a calculator or computer software, most of which use of Newton s Method. II. The Theory A. Background Linearization is the key to solving these problems. Newton s Method is based on the idea of using tangent lines to replace the graph of y=f(x) near the points where f is zero,i.e., since the tangent line is close to the curve, its x-intercept is close to the x-intercept of the curve. B. Theory Newton s Method begins by obtaining the initial approximation x 1 by either guessing or from a rough graph. The method then uses the tangent line to the curve y=f(x) at the point (x 1,f(x 1 )) to approximate the curve; the x-intercept of this tangent line, let s call it x 2, is usually a better approximation to the solution than x 1 was. The value x 3 where the tangent line to the curve y=f(x) at the point (x 2,f(x 2 )) crosses the x-axis is the next approximation in the method. We continue this process until we obtain a sequence of approximations x 1, x 2,x 3, If the numbers x n become close to the root r as n becomes large, we say that the sequence converges to r, i.e., lim x x n = r. When Newton s method converges to a root, it may not always be the root you have in mind. The sequence of successive approximations usually converges to the desired root, in certain circumstances the sequence may not converge. It could occur when f (x 1 )=0 or is close to 0. An approximation may also fall outside the domain of f. Newton s method fails and a better initial approximation x 1 should be chosen. C. Derivation of the formula Given the approximation x n, the point-slope equation for the tangent line to the curve y=f(x) at the point (x n,f(x n )) is y f ( x n ) = f ( x n ) ( x x n ). We can find where the tangent line crosses the x-axis setting y=0 in the above equation and solving for x.
( ) = f ( x n ) ( x x n ) f ( x n ) = f ( x n ) x f ( x n ) x n f ( x ( ) x = f ( x n ) x n f ( x n ) n ) x n f ( x n ) x = f ( x n ) ( ) f ( x n ) 0 f x n f x n x = x n f x n The value of x is the next approximation, x n+1. D. Strategy for Newton s Method 1. Guess a first approximation to a root of the equation f(x)=0. A graph of y=f(x) will help. 2. Determine f (x). 3. Use the formula x n+1 = x n f x n f x n to find the next approximation. ( ) ( ), f ( x n) 0 where f (x n ) is the derivative of f at x n 4. Continue the method until 2 approximations are accurate to the desired decimal place. III. EXAMPLES A. Graphical Examples 1. Use a geometric approach to illustrate Newton's Method to find the negative root of f x ( ) = x 4 + 32x 8 f ( x) = 4x 3 + 32 Starting with x 1 = 3, draw the tangent line to the curve at x=-3 From the graph and a program written in Mathematica, we find that the root for the tangent line gives us x 2 3.30263. Tangent at (-3,f(-3)) is y=-76 x-251 Using x 2 3.30263, the tangent line through that point and the program written in Mathematica, we find that the root for the tangent line gives us x 3 3.25547 Tangent at (x 2,f(x 2 )) is y= - 112.092 x -364.912
Using x 3 3.25547 the tangent line through that point and the program written in Mathematica, we find that the root for the tangent line gives us x 4 3.25411 Tangent at (x 3,f(x 3 )) is y = - 106.007 x-344.958 Using x 4 3.25411 the tangent line through that point and the program written in Mathematica, we find that the root for the tangent line gives us x 5 3.25411 Therefore, the root is approximately 3.25411 2. What happens if we use the same function, f ( x) = x 4 + 32x 8with an initial guess of x=-2? f (-2)=0 at x=-2 => A horizontal tangent was encountered at x =- 2. Therefore, Newton s Method fails and a new initial value should be chosen. B. Algebraic Examples 1. Use Newton s Method to approximate 2 to 5 decimal places.
2. Find the x-coordinate of the point where the curve y=x 3 -x crosses the horizontal line y=1. C. Additional Geometric Example The root for h(x)=x-cos(2x) cannot be approximated by Newton's Method using x 1 =-1. If you start Newton s Method with x 1 =-1, then x 2-1.71324, x 3-1.23085,
x 4-2.99144, x 5-0.51178, The process produces answers that continually waver back and forth around x= -1.