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anonymous

  • 5 years ago

Which of the following represents the Maclaurin series expansion of the functions y=e^(sin 2x) truncated at the third term

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  1. anonymous
    • 5 years ago
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    f(x) = e^sin 2x => f(0) = e^0 = 1 f'(x) = 2(cos 2x) e^sin2x => f'(0) = 2(1)e^0 = 2(1)(1) = 2 f''(x) = 2 cos 2x [2 (cos 2x)e^sin 2x] + e^sin 2x [(-4sin 2x)] = 4 (cos 2x)^2 e^sin 2x - 4 (sin 2x) e^sin 2x f''(0) = 4(1)(1) - 4(0)(1) = 4 The first three terms of the Maclaurin Series expansion are as follows: 1 + 2x + (4/2!) x^2 = 1 + 2x + 2x^2

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