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# MATH-M343/S343 Section 2.7 Notes MATH-S343

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This 2 page Class Notes was uploaded by Kathryn Brinser on Wednesday September 21, 2016. The Class Notes belongs to MATH-S343 at Indiana University taught by Michael Jolly in Fall 2016. Since its upload, it has received 2 views. For similar materials see Honors Differential Equations in Mathematics at Indiana University.

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Date Created: 09/21/16

S343 Section 2.7 Notes- Euler’s Method of Numerical Approximation 9-13-16 Used for equations that cannot be solved without computer- give numerical approximation of ???? ???? at given ???? values Known as first-order method/tangent line method- error at each step (iteration) decreases like ℎ decreases ???????? ???? ????+ℎ −???? ????) Recall???????? = ????(????,???? ????)) = ℎ→0 ( ℎ ) where ℎ = uniform step size between ???? and ????????+1 (ie. ???? = ???? + ℎ) ????+1 ???? ???? ????+ℎ −????(????) o ???? (,???? ???? )) ≈ ℎ ℎ(???? (,???? ???? ))≈ ???? ???? + ℎ − ???? ???? ( ) ???? ???? + ℎ ≈ ???? ???? + ℎ(???? ????,????(???? )( )) o If ???? 0 = ???? 0 ???? 0 + ℎ ≈ ???? +0ℎ ???? (,????( 0)) ???? ℎ = ???? ≈ ???? + ℎ ???? 0(???? ( )) 1 0 0 o Let ????????≈ ???? ???? ???? = ???? ????ℎ : ???? = ???? + ℎ ????(???? ,???? )) ????+1 ???? ???? ???? ′ Ex. Approximate ???? 0.1 and ???? 0.2 for ???? = 2???? − 1 with ???? 0 = 1 and ℎ = 0.1 and compare the results to the exact values. ( ) o ???? 1 ???? +0ℎ ???? ( ,????0 0 ) = 1 + 0.1 ( 0,1 )) = 1 + 0.1 2 1 − 1 ) = 1 + 0.1 1 ) = 1.1 ≈ ???? 0.1 ) o ???? = ???? + ℎ ???? ( ,???? )) 2 1 1 1 = 1.1 + 0.1(???? 0.1,1.1 )) = 1.1 + 0.1 2 1.1 − 1 ) ( ) = 1.1 + 0.1 2.2 − 1 = 1.1 + 0.1 1.2) = 1.1 + 0.12 = 1.22 ≈ ???? 0.2 ) ′ o ???? = 2???? − 1 ???? − 2???? = −1 Let ???? = ????∫ −2????????= ???? −2???? ′ −2???? −2???? −2???? ???? ???? − 2???????? = −???? ????(???????? −2????) = −???? −2???? ???????? ∫ ???? (???????? −2????)= ∫ −???? −2???????????? ???????? 1 ???????? −2???? = ???? −2????+ ???? 1 2 ???? = + ???????? 2???? 2 o Given ???? 0 = 1: 1 = + ???????? 0 1 1 = + ???? 1 2 2 = ???? 1 2???? 1 ???? ???? = ????2 + 2 1 2 0.1 1 o ???? 0.1 = ???? 2 + 2 1.1107 o ???? 0.2 = ???? 1 2 0.2 + = 1.2459 2 2 o Both approximations too low, which makes sense with concavity of graph of ???? vs. ????: Ex. Approximate ???? 0.1 and ???? 0.2 for ???? = 2???? − 1 with ???? 0 = 1 and ℎ = 0.05. o ???? = 1 + 0.05 ????(0,1 )) 1 = 1 + 0.05 2 1 − 1 ) = 1 + 0.05 = 1.05 ≈ ???? 0.05 ) ( ) o ????2= 1.05 + 0.05 ????(0.1,1.05 ) = 1.05 + 0.05 2 1.05 − 1 ) = 1.05 + 0.05 2.1 − 1 ) = 1.05 + 0.05 1.1 ) = 1.05 + 0.055 = 1.105 ≈ ???? 0.1 ) o Noticeably better approximation at ???? = 0.1 using smaller step between iterations o From above, ???? 0.1 = 1.1107 At ℎ = 0.1: 1.1107 − 1.1000 = 0.0107 At ℎ = 0.05: 1.1107 − 1.0500 = 0.00570 Error decreased by approx. ½ when ℎ halved; expect same effect if ℎ = 0.025 ( ) ( ) ( ) ′ ???? ( ) Ex. Approximate ???? 0.1 , ???? 0.2 , and ???? 0.3 for ???? = w????th ???? 0 = 1 and ℎ = 0.1. o ????1= 1 + 0.1 ( 0,1 )) 0 = 1 + 0.1( 1 = 1 ≈ ???? 0.1 ) o ????2= 1 + 0.1 ( 0.1,1 )) = 1 + 0.1( ).1 1 = 1 + 0.01 = 1.01 ≈ ???? 0.2 ) ( ) o ????3= 1.01 + 0.1 ( 0.2,1.01 ) = 1.01 + 0.1( 0.2) 0.02 1.01 = 1.01 + 1.01 = 1.0298 ≈ ???? 0.3 )

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