27_pdfsam_math 54 differential equation solutions odd

# 27_pdfsam_math 54 differential equation solutions odd -...

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Exercises 1.4 For approximation of φ ( t ) at the point t = 1 with N = 20 steps, we take the step size h = (1 t 0 ) / 20 = 0 . 05. Thus, the recursive formulas for Euler’s method are t n +1 = t n + 0 . 05 , x n +1 = x n + 0 . 05 ( 1 + x 2 n ) . Applying these formulas with n = 0 , 1 , . . . , 19, we obtain x 1 = x 0 + 0 . 05 ( 1 + x 2 0 ) = 0 . 05 , x 2 = x 1 + 0 . 05 ( 1 + x 2 1 ) = 0 . 05 + 0 . 05 ( 1 + 0 . 05 2 ) = 0 . 100125 , x 3 = x 2 + 0 . 05 ( 1 + x 2 2 ) = 0 . 100125 + 0 . 05 ( 1 + 0 . 100125 2 ) 0 . 150626 , . . . x 19 = x 18 + 0 . 05 ( 1 + x 2 18 ) 1 . 328148 , φ (1) x 20 = x 19 + 0 . 05 ( 1 + x 2 19 ) = 1 . 328148 + 0 . 05 ( 1 + 1 . 328148 2 ) 1 . 466347 , which is a good enough approximation to φ (1) = tan 1 1 . 557408. 13. From Problem 12, y n = (1 + 1 /n ) n and so lim n →∞ [( e y n ) / (1 /n )] is a 0 / 0 indeterminant. If we let h = 1 /n in y n and use L’Hospital’s rule, we get lim n →∞
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