Homework Bob 27.1 and 27.2

Homework Bob 27.1 and 27.2 - f2 =-0.1600...

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For problems 27.1 and 27.2 Ben performed the calculations in the Command Window of MATLab and I performed the calculations by hand. Both forms of the evaluations are included for your review. 27.1) We are given the following table, and asked to use it in order to find forward, back, and central approximations of f’(1): t 0 .5 1.0 1.5 2.0 y 0 .19 .26 .29 .31 Thus, we have the following MatLab results from the computations: Forward Approximation: >> (.29 - .26)/(1.5 - 1) ans = 0.0600 Backward Approximation: >> (.26 - .19)/(1.0 - .5) ans = 0.1400 Central Difference Approximation: >> (.29 - .19)/(1.5 - .5) ans = 0.1000 Now, to find f’’(1) [f2], f’’’(1) [f3], and f (4) (1) [f4], we have the following: >> f2 = (.29 - 2*.26 + .19)/((.5)^2)
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Unformatted text preview: f2 =-0.1600 >> f3 = (.31 - 2*.29 + 2*.19 - 0)/(2*(.5^3)) f3 = 0.4400 >> f4 = (.31 - 4*.29 + 6*.26 - 4*.19 + 0)/((.5)^4) f4 =-0.8000 27.2) We are given the following matrix and asked to use it to find approximations of certain partial derivates: U = 5.1000 6.5000 7.5000 8.1000 8.4000 5.5000 6.8000 7.8000 8.3000 8.9000 5.5000 6.9000 9.0000 8.4000 9.1000 5.4000 9.6000 9.1000 8.6000 9.4000 ux(x2, y4): >> ux1 = (1/(2*h))*(U(3,4) - U(1,4)) ux1 = 1.5000 uxx(x3, y2): >> uxx1 = (1/(h^2))*(U(4,2) - 2*U(3,2) + U(2,2)) uxx1 = 260.0000 uyy(x3,y2): >> uyy1 = (1/(k^2))*(U(3,3) - 2*U(3,2) + U(3,1)) uyy1 = 2.8000 uxy(x2, y3): >> uxy1 = (1/(4*h*k))*(U(3,4) - U(3,2) - U(1,4) + U(1,2)) uxy1 =-0.5000...
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This note was uploaded on 01/15/2011 for the course MATH 345 taught by Professor Staff during the Spring '08 term at Ohio State.

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Homework Bob 27.1 and 27.2 - f2 =-0.1600...

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