Problem 4, HW 4. MAE 185.

Problem 4, HW 4. MAE 185.

Nasser M. Abbasi

May 24, 2003

problem Employee a two-through six-points Gauss-Legendre formulas to solve $∫ 3 −3 1+22x2dx$

Answer

$f(x)= --2- 1+2x2$ This function is an even function:

f(x)

hence

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First transform the integral to be inside the limits of $− 1$ to $+ 1$ . Let $x= k1y+ k2$

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Solve for $k1,k2$ . Adding the above 2 equations gives $2k2 = 3 =⇒$ _____ $k2 = 1.5$

Hence, $0= − k1+ 1.5 =⇒$ _____ $k1 = 1.5$

hence the transformation is_______________________ $x = 1.5+ 1.5y$ and ____________________ $dx = 1.5 dy$

Using the above transformation, the integral becomes

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Hence $---1---- g(y)= 5.5+9y+4.5y2$

2 points Gauss

$c0 = 1,c1 = 1$ , $x0 = − 1√3,x1 = +√13-$

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Hence

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3 points Gauss

$c = 5∕9, c = 8∕9, c = 5∕9,x = − √0.6, x = 0.0, x = √0.6 0 1 2 0 1 2$

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Hence

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4 points Gauss

$c0 = 0.3478548, c1 = 0.6521452, c2 = 0.6521452, c3 = 0.3478548$

$x0 = − 0.861136312, x1 = − 0.339981044, x2 = 0.339981044, x3 = 0.861136312$

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so

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5 points Gauss

$c0 = 0.2369269, c1 = 0.4786287, c2 = 0.568889, c3 = 0.4786287, c4 = 0.2369269$

$x0 = − 0.906179846, x1 = − 0.538469310, x2 = 0.0, x3 = 0.53846931, x4 = 0.906179846$

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Hence,

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6 points Gauss

$c0 = 0.171, c1 = 0.361, c2 = 0.468, c3 = 0.468, c4 = 0.3607616, c5 = 0.1713245$

$x = − 0.932, x = − 0.661, x = − 0.239, x = 0.239, x = 0.661x = 0.93 0 1 2 3 4 5$

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Hence

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summary: This table shows the relative error normalized to the true value for value of number of points n, starting with n=2 up to n=6. Analytically,

∫ √ -- ( √--) ---2--dx = 2arctan 2x 1 +2x2

hence

∫ 3 2 [√ -- (√ -- ) √ -- ( √-- ) ] Ianalytical = 2 -----2dx = 2 2arctan 2(3) − 2arctan 2 (0 ) = 2[1.894]= 3.788 0 1+ 2x

So $| | 𝜀t = |true−tarupperox|100 %$

|-|-----|----------------| | | | | |n I 𝜀t | |-|-----|----------------| | | ||3.788−3.818|| | |2 3.818|--3.788---=-0.792-| | | | | | |3 3.902 |3.7883.−7388.902| = 3.017 |-|-----|----------------| |43.799 ||3.788−3.780||= 0.287 | |-|-----|--3.788-----------| | | | | | |5 3.783 |3.7883.−7388.783|= 0.144| |-|-----|----------------| |63.787 ||3.788−3.787||= 0.035 | -----------3.788-----------|

As number of points increases, the acurracy increases.

observation: I am able to ﬁnd why the n=3 case seems to have some abnormality. $𝜀 t$ should decrease as n increases, but for n=3 it did not. Checked the calculations but not able to see if I did some arithmitic error.