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Unformatted text preview: 1.2 Application of the Linear Approximation: Newtons Method Consider the problem of finding the point where the curves x y = 1 and x y sin = intersect: This means solving the equation x x = 1 sin , or, equivalently, 1 sin = + x x . We refer to solving ) ( = x f as finding the roots of the equation ) ( = x f , or of finding the zeroes of ) ( x f . For most functions (like 1 sin + x x ), this is hard. However, for linear functions its easy, so we approximate! Example (from Newton himself): Find a root of the equation 5 2 3 = x x . Solution: Were going to use a linear approximation ( ) ( , 1 x P x ) to ) ( x f , and solve ) ( , 1 = x P x instead. This raises an immediate question, though: whats x ? If we want a good approximation, we need x to be close to the root, so we start with a guess. For continuous functions we can apply the Intermediate Value Theorem (IVT): Consider a few values of f : 16 ) 3 ( 1 ) 2 ( 6 ) 1 ( 5 ) ( = = = = f f f f Since f is continuous, there must be a root between 2 = x and 3 = x (by the IVT). It appears as though 2 = x is probably the better guess, so lets use this as our value for x ....
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This note was uploaded on 04/05/2010 for the course MATH 119 taught by Professor Harmsworth during the Spring '08 term at Waterloo.
 Spring '08
 HARMSWORTH
 Approximation

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