Math 231 Worksheet 5 1 Right Hand Approximation Let Rnf be the Right Hand Approximation to cid 82 b a Give a formula for R2f R3f and Rnf Solution cid 18 cid 18 cid 20 cid 18 n cid 88 f 2 cid 19 b a cid 19 b a cid 19 cid 18 cid 19 cid 19 cid 18 b a b a 2 b a 3 b a n f 3 a cid 18 cid 19 f a 2 2 b a 3 f b b a cid 18 cid 19 n cid 88 b a 2 n a R2f f a R3f f a Rnf a f x dx over n equal partitions b a f b b a cid 16 3 b a a 3 cid 17 cid 21 b a f n n b a n b Explain why it is that if f cid 48 x K1 for all x in a b then for all x in a b n f a i x x with x b a a i 1 i 1 n n f i f x f a K1 x a di ers from cid 82 b D cid 82 b Solution Solution By the Mean Value Theorem there exists a c in a b such that f cid 48 c f b f a There exists a K1 0 such that f cid 48 x K1 for all x in a b Then f x f a K1 x a c Use your result from b to show that if f cid 48 x K1 for all x in a b then R1f b a a f x dx by at most K1 b a 2 2 Di erence between integral and approximation a f x dx f b b a cid 82 b a f x dx cid 82 b a f b dx cid 82 b By MVT f x f b f cid 48 c x b for some c in b x The derivative is assumed bounded f cid 48 x K1 for all x a b Then a f x f b dx cid 12 cid 12 cid 12 cid 12 cid 90 b a cid 12 cid 12 cid 12 cid 12 cid 90 b a cid 90 b a f x f b dx f x f b dx f cid 48 c x b K1 x b dx cid 90 b cid 90 b a a K1 b x dx f x dx R1 xi 1 xi cid 12 cid 12 cid 12 cid 12 2 2 2 2 a cid 17 b b 2 b a 2 b a 2 K1 b x 2 Then cid 105 b K1 b a 2 cid 104 K1 cid 16 K1 cid 17 cid 16 K1 cid 82 b a K1 b x dx K1 for all x in a b then Rnf approximates cid 82 b d Use your result from part c and your formula in part a to show that if f cid 48 x a f x dx to an error no more than the points as a x0 x1 xn b then Rn a b cid 80 n have cid 82 b cid 12 cid 12 cid 12 cid 12 cid 12 n cid 88 cid 12 cid 12 cid 12 cid 12 cid 90 b Solution The de nition of Rn is to subdivide the interval a b into n equal pieces If we label i 1 R1 xi 1 xi We also f x dx Thus by the triangle inequality we have that f x dx Rn a b cid 12 cid 12 cid 12 cid 12 cid 90 xi cid 90 xi R1 xi 1 xi xi 1 xi 1 2n a i 1 i 1 xi 1 cid 82 xi a f x dx cid 80 n cid 12 cid 12 cid 12 cid 12 cid 12 n cid 88 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 90 xi i 1 xi 1 By part c we know that f x dx R1 xi 1 xi xi xi 1 2 and since we cut the interval a b into n equal pieces each xi xi 1 b a i 1 f x dx n cid 88 cid 12 cid 12 cid 12 cid 12 K1 cid 19 2 cid 18 b a 2 b a 2 n2 K1 2 n and so b a 2 n i 1 i 1 xi 1 K1 2 n K1 2 f x dx R1 xi 1 xi cid 12 cid 12 cid 12 cid 12 n cid 88 cid 12 cid 12 cid 12 cid 12 cid 90 xi n cid 88 2 The Midpoint Approximation Mn can be used to approximate cid 82 b K2 for all x on a b then M1 approximates cid 82 b cid 12 cid 12 cid 12 cid 12 K2 f x dx Mn b a 3 24n2 cid 12 cid 12 cid 12 cid 12 cid 90 b n a this is shown using the Mean Value Theorem Use this to show more generally that when f 2 x K2 for all x on a b then a f x dx If f 2 x b a 3 a f x dx to an error no more than K2 24 Solution 2 Following the argument from 2 d f x dx n cid 88 i 1 M1 xi 1 xi f x dx M1 xi 1 xi cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 a f x dx Mn a b cid 12 cid 12 cid 12 cid 12 cid 12 n cid 88 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 cid 90 b cid 90 xi cid 12 cid 12 cid 12 cid 12 cid 90 xi triangle inequality n cid 88 error estimation n cid 88 cid 0 b a n cid 88 K2 xi 1 xi 1 i 1 i 1 i 1 equal partitions K2 xi xi 1 3 24 cid 1 3 n 24 i 1 n K2 b a 3 24n3 K2 b a 3 24n2 3 Mary has a fast process to approximate cid 82 b that if f 6 x K6 for all x in a b then P f approximates cid 82 b a f x dx which she calls P f She knows a f x dx to an error no more than K6 b a 7 Mary wants to use her process to numerically approximate an integral by subdividing an interval into n equal pieces and applying P f to each of the smaller intervals and then adding up the result much like the Left Right Midpoint rules do If Mary calls her approximation Pnf what is an upper bound for the error 48 of Pnf to approximate cid 82 b a f x dx Why Solution Following the solutions to 1 d and 2 f x …
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