5333-Solns-Assignment9-Sp13 - 5333 Analysis Solutions Assignment 9 Section 28 1(a True(b False pn has the same values at f AT x0 pn approximates the

# 5333-Solns-Assignment9-Sp13 - 5333 Analysis Solutions...

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5333: Analysis Solutions, Assignment # 9 Section 28. 1. (a) True. (b) False. p n has the same values at f AT x 0 ; p n approximates the values of f at points “near” x 0 . 4. By Taylor’s Theorem, | e x - p n ( x ) | ≤ max | f ( n +1) ( t ) | ( n + 1)! | x | n +1 e 2 ( n + 1)! 2 n +1 < 9 · 2 n +1 ( n + 1)! We want, 9 · 2 n +1 ( n + 1)! < 0 . 2 = 1 5 , which is equivalent to: ( n + 1)! > 45 · 2 n +1 . The least n which satisfies this inequality is n = 7. p 7 ( x ) = 1 + x + x 2 2! + x 3 3! + · · · + x 7 7! is the Taylor polynomial of least degree that approximates e x to within 0.2 on [ - 2 , 2] 7. f ( x ) = cos x, x 0 = 0. p 5 ( x ) = 1 - x 2 2! + x 4 4! . p 5 (1) = 1 - 1 2 + 1 24 = 0 . 541666667 error = | cos 1 - p 5 (1) | ≤ max | f (6) | 6! 1 6 1 6! 0 . 001389 Calculator value: cos 1 0 . 540302306 8. f ( x ) = x, x 0 = 9. p 2 ( x ) = 3 + 1 6 ( x - 9) - 1 216 ( x - 9) 2 . p 2 (8 . 8) = 3 - 1 6 (0 . 2) - 1 216 (0 . 2) 2 2 . 96648148. error = | 8 . 8 - p 2 (8 . 8) | ≤ max | f (3) | 3! | - 0 . 2 | 3 1 8 · 3 4 3! · (0 . 2) 3 0 . 000002058 Calculator value: 8 . 8 2 . 966479395 1
Section 29 2. (a) True. If integraldisplay b a f ( x ) dx exists, then integraldisplay b a f ( x ) dx = L ( f ) = U ( f ) and for any partition P , L ( { , P ) ≤ L ( { ) = U ( { ) ≤ U ( { , P ). (b) True. For any partitions P and Q , P ∪ Q is a partition of P and of Q . Therefore, (c) True. Let epsilon1 > 0. Since L ( f ) = sup Q ( f, Q ) : Q is a partition of [ a, b ], there is a partition P such that L ( f ) - epsilon1 < L ( f, P ) ≤ L ( { ). Since f is integrable on [ a, b ], L ( f ) = U ( f ) and the result follows. 4. (a) L ( f, P ) = f (2) · 1 + f (3) · 1 = 1 2 + 1 3 = 5 6 U ( f, P ) = f (1) · 1 + f (2) · 1 = 1 + 1 2 = 3 2 (b) L ( f, P ) = f (1 . 5) · 1 2 + f (2) · 1 2 + f (2 . 5) · 1 2 + f (3) · 1 2 = 0 . 95 U ( f, P ) = f (1) · 1 2 + f (1 . 5) · 1 2 + f (2) · 1 2 + f (2 . 5) · 1 2 = 0 . 1 . 283 (c) integraldisplay 3 1 1 x dx = ln x bracketrightbig 3 1 = ln 3 1 . 0986 5. Set f ( x ) = x and let P = { 0 , b/n, 2 b/n, 3 b/n, · · · , ( n - 1) b/n, nb/n } be the parti- tion of [0 , b ] into n subintervals of equal length. Then L ( f, P ) = 0 · b n + b n · b n + 2 b n · b n + · · · + ( n - 1) b n · b n = b 2 n 2 (1 + 2 + 3 + · · · + ( n - 1)) = b 2 n 2 · ( n - 1) n 2 1 2 as n → ∞ U ( f, P ) = b n

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