Explanation The five-point midpoint formula is given by, f ′ ( x 0 ) = 1 12 h [ f ( x 0 − 2 h ) − 8 f ( x 0 − h ) + 8 f ( x 0 + h ) − f ( x 0 + 2 h ) ] When x 0 = 2.0 and h = 0.1 , the above equation becomes, f ′ ( x 0 ) = 1 12 × 0.1 [ f ( 2 − 0.2 ) − 8 f ( 2 − 0.1 ) + 8 f ( 2 + 0.1 ) − f ( 2 + 0.2 ) ] = 1 1.2 [ f ( 1.8 ) − 8 f ( 1.9 ) + 8 f ( 2.1 ) − f ( 2.2 ) ] = 1 1.2 [ ( 1.8 ) e 1.8 − 8 ( 1.9 ) e 1.9 + 8 ( 2.1 ) e 2.1 − ( 2.2 ) e 2.2 ] = 1 1.2 [ 10.889365435 − 8 ( 12.703199440 ) + 8 ( 17.148956816 ) − ( 19.855029698 ) ] ≈ 22.166995 Thus, the formula in Eq. (4.6) applied to f ( x ) = x e x at x 0 = 2

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Chapter 4.2, Problem 7ES

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To show: The formula in Eq. (4.6) applied to f(x)=xex at x0=2.0 gives N2(0.2) when h=0.1 and gives N2(0.1) when h=0.05.

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Chapter 4.2, Problem 6ES

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Numerical Analysis

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The formula in Eq. (4.6) applied to f ( x ) = x e x at x 0 = 2.0 gives N 2 ( 0.2 ) when h = 0.1 and gives N 2 ( 0.1 ) when h = 0.05 .

Solution Summary: The author explains the five-point midpoint formula in Eq. (4.6) applied to f(x)=x

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To show: The formula in Eq. (4.6) applied to f(x)=xex at x0=2.0 gives N2(0.2) when h=0.1 and gives N2(0.1) when h=0.05.

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Chapter 4 Solutions