Solution to HW7

# Solution to HW7 - Solution to HW#7 EGM6341 Spring 2009 pp...

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Unformatted text preview: Solution to HW#7 EGM6341 Spring 2009 pp. 323-329 #1 Write a program to evaluate ( ) b a I f x dx = using the trapezoidal rule with n subdivisions, calling the result I n . Use the program to calculate the following integrals with n=2,4,8,16,…,512 a) dx x I ) exp( 2 1- ∫ = n I n Error Rn 2 0.731370251828563 0.015453881 4 0.742984097800381 0.003840035 8 0.745865614845695 0.000958518 4.030462 16 0.746584596788222 0.000239536 4.007774 32 0.746764254652294 5.98782E-05 4.001951 64 0.746809163637828 1.49692E-05 4.000488 128 0.746820390541618 3.74227E-06 4.000122 256 0.746823197246153 9.35566E-07 4.000031 512 0.746823898920948 2.33891E-07 4.000008 1024 0.746824074339563 5.84729E-08 4.000002 I_extrap 0.746824132812397 2.95319E-14 Note: Error is calculated using |I exact-I n |; I exact = 0.746824132812427 based on 4-point Gauss-Legendre quadrature with 1024 intervals. Rn = (I n-1-I n-2 )/(I n-I n-1 ) I_extrap = (R(n)*I n-I n-1 )/(R(n)-1) = 0.746824132812397 for n=1024 Observation: * R(n) ~ 4 is close to theoretical estimate for a 2 nd order convergence * Error_extrap(n=1024) <<Error(n=1024) extrapolation is very powerful b) dx x I 5 . 2 1 ∫ = n I n Error Rn 2 0.338388347648318 0.052674062 4 0.298791496231346 0.013077211 8 0.288974739670143 0.003260454 4.033598 16 0.286528567896037 0.000814282 4.013110 32 0.285917779698734 0.000203494 4.004943 64 0.285765152250462 5.08665E-05 4.001824 128 0.285727001721098 1.2716E-05 4.000664 256 0.285717464659795 3.17895E-06 4.000240 512 0.285715080445649 7.94731E-07 4.000086 1024 0.285714484396680 1.98682E-07 4.000031 I_extrap 0.285714285715720 1.43469E-12 Note: Error is calculated using |I exact-I(n)|; I exact =1/3.5=0.285714285714285714285714… Rn = (I n-1-I n-2 )/(I n-I n-1 ) I_extrap = (R(n)*I n-I n-1 )/(R(n)-1)= 0.285714285715720 for n=1024 Observation: * R(n) ~ 4 is close to theoretical estimate for a 2 nd order convergence * Error_extrap(n=1024) <<Error(n=1024) extrapolation is very powerful * Is there a “Singularity” at x=0? T.E. = - 2 12 1 h [ f ' (b)- f ' (a)] Or T.E. = - 2 12 1 h ( b- a ) f " ( η ), 5 . 1 ~ ) ( ' x x f , 5 . ~ ) ( " x x f are both continuous near x=0. Thus, for trapezoidal rule, 5 . 2 ) ( x x f = is NOT singular. Thus, R(n) = 4 as predicted for continuous integrand. c) dx x I 2 4 4 1 1 + ∫ =- n I n Error Rn 2 4.23529411764706 1.58365879 4 2.91764705882353 0.266011731 8 2.65882352941177 0.007188202 5.090909 16 2.65050680499416 0.001128522 31.12085 32 2.65134716346583 0.000288164-9.89664 64 2.65156325136377 7.2076E-05 3.888966 128 2.65161730615210 1.80212E-05 3.997572 256 2.65163082190308 4.50543E-06 3.999392 512 2.65163420096925 1.12637E-06 3.999848 1024 2.65163504574383 2.81592E-07 3.999962 I_extrap 2.65163532733893 2.86349E-12 Note: Error is calculated using |I exact-I(n)|; I exact = 2tan-1 (4)= 2.65163532733607 Rn = (I n-1-I n-2 )/(I n-I n-1 ) I_extrap = (R(n)*I n-I n-1 )/(R(n)-1)= 2.65163532733893 for n=1024 Observation: * R(n) ~ 4 is close to theoretical estimate for a 2...
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## This note was uploaded on 04/23/2009 for the course EGM 6341 taught by Professor Mei during the Spring '09 term at University of Florida.

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Solution to HW7 - Solution to HW#7 EGM6341 Spring 2009 pp...

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