121616949-math.143 - case we can test the result for...

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6.3 Newton’s Method 129 1 2 .................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 6.17 Newton’s method. and in general from x i we compute x i +1 : x i +1 = x i f ± ( x i ) - f ( x i ) f ± ( x i ) = x i - f ( x i ) f ± ( x i ) . EXAMPLE 6.20 Returning to the previous example, f ( x ) = x 2 - 3, f ± ( x ) = 2 x , and the formula becomes x i +1 = x i - ( x 2 i - 3) / (2 x i ) = ( x 2 i + 3) / (2 x i ), as before. In practice, which is to say, if you need to approximate a value in the course of designing a bridge or a building or an airframe, you will need to have some confidence that the approximation you settle on is accurate enough. As a rule of thumb, once a certain number of decimal places stop changing from one approximation to the next it is likely that those decimal places are correct. Still, this may not be enough assurance, in which
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Unformatted text preview: case we can test the result for accuracy. EXAMPLE 6.21 Find the x coordinate of the intersection of the curves y = 2 x and y = tan x , accurate to three decimal places. To put this in the context of Newton’s method, we note that we want to know where 2 x = tan x or tan x-2 x = 0. We compute f ± ( x ) = sec 2 x-2 and set up the formula: x i +1 = x i-tan x i-2 x i sec 2 x i-2 . From the graph in figure 6.18 we guess x = 1 as a starting point, then using the formula we compute x 1 = 1 . 310478030, x 2 = 1 . 223929096, x 3 = 1 . 176050900, x 4 = 1 . 165926508,...
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