hw3solns

hw3solns - [2 , 3]. Note that this was a pretty arbitrary...

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Math 151A HW3 Solutions Will Feldman 1. (a) This is a translation of ln x (b) Noting that f ( x ) = ln ( x + 2) is monotone and continuous we have that ln 2 x ln 4 for x [0 , 2] and by the intermediate value theorem every value in [ln 2 , ln 4] [0 , 2] is obtained by f on x [0 , 2]. (c) So now since f (0) - 0 0 and f (2) - 2 0 f must have a fixed point on [0 , 2]. (d) | f 0 ( x ) | = | 1 / ( x + 2) | ≤ 1 / 2 since f 0 is monotone decreasing. (e) The fixed point of f on [0 , 2] is unique and the fixed point iteration will converge. (f) Use matlab. 2. Think about whether the fixed point theorem applies to any interval containing the initial guess. The rate of convergence then depends on the maximum of | f 0 ( x ) | on this interval. 3. The main challenge here is to think of a good function to use for the fixed point iteration. Something like g ( x ) = x - ( x 3 - 25) / 30 on the interval [2 , 3] will work since g 0 ( x ) = 1 - x 2 / 10 so 3 / 5 g 0 ( x ) 1 / 10 on [2 , 3] so g is increasing and g (2) > 2 and g (3) < 3 so g ([2 , 3])
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Unformatted text preview: [2 , 3]. Note that this was a pretty arbitrary choice, the main goal was to make it easy to check the conditions of the xed point theorem. 4. For part (b) try something similar to the previous problem. 5. Just use the formula for Newtons method. Note that for f ( x ) =-x 3-cos x we have that f ( x ) =-3 x 2 + sin x so that f (0) = 0 so it will not be possible to carry out the Newton iteration if 0 is chosen as the initial point because carrying out the rst step of the iteration requires dividing by f (0) = 0. Geometrically Newtons method calculates the zero of the tangent line at the initial point. If f ( p ) = 0 then the tangent line at p is horizontal and does not have a zero. 1 6. See Burden and Faires for some pseudo - code for these methods. 2...
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hw3solns - [2 , 3]. Note that this was a pretty arbitrary...

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