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**Unformatted text preview: **. 25 ( x n + 1 − y n ) + (0 . 25) 2 2 ( y n − x n ) . By starting with x = 0 and y = 1 (the initial values for the problem), we ±nd y 1 = 1 + . 0625 2 ≈ 1 . 03125 . Plugging this value into the recursive formulas yields y 2 = 1 . 03125 + 0 . 25(0 . 25 + 1 − 1 . 03125) + ± . 0625 2 ² (1 . 03125 − . 25) ≈ 1 . 11035 . By continuing in this way, we can ±ll in the ±rst three columns in Table 3-H. 154...

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