Math 128A - HW1 Solutions.pdf

# A intergrate the maclaurin series for e x 2 to show

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a. Intergrate the Maclaurin series for e - x 2 to show that erf( x ) = 2 π k =0 ( - 1) k x 2 k +1 (2 k +1) k ! Solution: We know e y = k =0 y k /k !. Thus e - x 2 = k =0 ( - 1) k x 2 k /k !. Inte- grating term by term gives the desired result. c. Use the series in part (a) to approximate erf(1) to within 10 - 7 . Solution: When x = 1, the series in part (a) has the form k =0 ( - 1) k a k with a k positive and a k +1 < a k . (Note that the ratio a k +1 /a k = 2 k +1 2 k +3 · k k +1 · x 2 , which is less than 1.) The alternating series test guarantees that the error in truncating the series is bounded by the first omitted term. Using matlab (or any computer algebra system such as Wolfram alpha), we find that 2 π · 1 (2 k +1) k ! = 1 . 64 × 10 - 7 when k = 9 and 1 . 48 × 10 - 8 when k = 10. Thus, we can omit the terms k 10. Our approximation is then erf(1) 2 π 9 k =0 ( - 1) k (2 k +1) k ! = 0 . 842700779. (Including more terms yields 0 . 842700793). Alternative solution: We can use Taylor’s theorem with remainder to bound the error. Let ˜ f ( y ) = e y = ˜ P n ( y ) + ˜ R n ( y ), where ˜ P n ( y ) = n k =0 1 k ! y k and ˜ R n ( y ) = 1 ( n +1)! e ξ ( y ) y n +1 . Then f ( x ) = e - x 2 is equal to ˜ f ( - x 2 ). It follows that P 2 n +1 ( x ) = ˜ P n ( - x 2 ) and R 2 n +1 ( x ) = ˜ R n ( - x 2 ). Note that | R 2 n +1 ( x ) | = e ξ ( - x 2 ) ( n +1)! ( - 1) n +1 x 2 n +2 1 ( n +1)! x 2 n +2 , since e x is an increasing function and ξ ( - x 2 ) is between - x 2 and 0 (so e ξ ( - x 2 ) e 0 = 1.) Integrating from 0 to 1, we have erf(1) - 2 π n X k =0 ( - 1) k 1 2 k +1 (2 k + 1) k ! 2 π Z 1 0 | R 2 n +1 ( x ) | dx 2 π x 2 n +3 (2 n + 3)( n + 1)! 1 0 . The bound is identical to the first approach (i.e. n = 9 suffices).

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4 28(27). A function f : [ a, b ] R is said to satisfy a Lipschitz condition with Lipschitz constant L on an interval [ a, b ] if, for every x, y [ a, b ], we have | f ( x ) - f ( y ) | ≤ L | x - y | . a. Show that if f satisfies a Lipschitz condition with Lipschitz constant L on [ a, b ], then f C [ a, b ]. Solution: Increasing L if necessary, we may assume L > 0. Let x 0 [ a, b ] and > 0. Choose δ = L . Suppose x [ a, b ] with | x - x 0 | < δ . Then | f ( x ) - f ( x 0 ) | ≤ L | x - x 0 | < L · δ = . So f is continuous at x 0 , by definition. Since x 0 was arbitrary, f C [ a, b ]. b. Show that if f has a derivative that is bounded on [ a, b ] by L , then f satisfies a Lipschitz condition with Lipschitz constant L on [ a, b ]. Solution: Assume that | f 0 ( x ) | ≤ L for every x [ a, b ]. Let x, y [ a, b ]. Our assumptions imply that f is differentiable on ( a, b ) so by the mean value theorem there exists a constant c [ x, y ] such that f 0 ( c ) = f ( x ) - f ( y ) x - y . Rewriting this as | f ( x ) - f ( y ) | = f 0 ( c ) · | x - y | and applying the bounded assumption, we obtain | f ( x ) - f ( y ) | ≤ L | x - y | . So f satisfies a Lipschitz condition with Lipschitz constant L on [ a, b ]. c. Give an example of a function that is continuous on a closed interval but does not satisfy a Lipschitz condition on the interval.
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• Spring '08
• Rieffel
• Continuous function, Rolle, 3 k, 0 K, ξ, a. Intergrate

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