Note if m 0 the graph is a horizontal line so if f a

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Unformatted text preview: e it explicitly. We have 0 = f (c) = f (a) + m(c − a). This implies that −f (a) = m(c − a). If m = 0, we can divide both sides of the equation by m to get −f (a) = c − a. m Finally, adding a to both sides of the equation yields f (a) m f (a) = a− f (a) c = a− Therefore, if f (x) = f (a) + m(x − a) and m = 0, we can easily solve the equation f (x) = 0. Note: If m = 0, the graph is a horizontal line, so if f (a) = 0, it does NOT cross the x-axis and NO such c exists. What do we do if the graph of f (x) is not a line? In this case we can use tangent line approximation (this is formally called linear approximation ). The following steps outline a general method for using tangent lines to approximate the point c for f (c) = 0. Step 1: Pick a point x1 that is reasonably close to a point c with f (c) = 0. (The IVT or a rough plot might be helpful in finding such an x1 .) Step 2: If f (x) is differentiable at x = x1 , then we can approximate f (x) by the function Lx1 (x) = f (x1 ) + f (x1 )(x − x1 ) The graph of Lx1 (x) is the tangent line to f (x) at x = x1 . Since, for x values near x1 , f (x) ∼ Lx1 (x) = it would make sense that the graphs of f (x) and Lx1 (x) cross the x-axis at roughly the same place. vii Copyrighted by B.A. Forrest (bforrest@sympatico.ca) Newton’s Method in Maple Recall from above that Lx1 (x) is a linear function. Therefore, if f (x1 ) = 0, the graph of Lx1 (x) crosses the x-axis at the point x2 where x2 = x1 − f (x1 ) . f (x1 ) Step 3: We now repeat the above procedure, replacing x1 by x2 and using the linear approximation at x2 , to get a new approximation x3 = x2 − f (x2 ) f (x2 ) for c. The diagram shows that in this example, x3 is very close to c. Continuing in this manner, we get a recursively defined sequence xn+1 = xn − f (xn ) f (xn ) where xn+1 is simply the point at which the tangent line to the graph of f (x) through (xn , f (xn )) crosses the x-axis. It can be shown that for most nice functions and reasonable choices of x1 , the sequence {xn } converges very rapidly to a number c with f (c) = 0. Still, it makes sense for us to ask: How do we know how good our approximation is? If we want to approximate c to k decimal places of accuracy, at least k decimal places must be carried throughout our calculations. The procedure stops when two successive terms, xn and xn+1 , agree to k decimal places. In most case...
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