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 ﬁxed point theorem. 4. For part (b) try something similar to the previous problem. 5. Just use the formula for Newton’s 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 ﬁrst step of the iteration requires dividing by f (0) = 0. Geometrically Newton’s 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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## This note was uploaded on 01/30/2012 for the course MATH 131a taught by Professor Hitrik during the Spring '08 term at UCLA.

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

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