Find the first three non-zero terms of the Taylor series for f(x) = tan(x).

We have that the Taylor series of a function infinitely differentiable at a x = a is given by the expansion: f(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)2/2! + f'''(a)(x - a)3/3! + f(4)(a)(x - a)4​​​​​​​/4! +... Thus we differentiate f(x) 5 times and evaluate at zero (as in this case a = 0) in order to obtain all our coefficients. f(x) = tan(x), f(0) = tan(0) = 0 f'(x) = sec2(x) = 1 + tan2(x) = 1 + f(x)2, thus f'(0) = 1 + f(0)2 = 1 [by writing f'(x) in terms of f(x), we can skip differentiating reciprocal trig functions and simply leave the derivates in terms of f(x) and its derivatives of lower order] f''(x) = 2f'(x)f(x), f''(0) = 0 f'''(x) = 2(f''(x)f(x) + f'(x)2), f'''(0) = 2 f(4)(x) = 2(f'''(x)f(x) + 3f''(x)f'(x)), f(4)(0) = 0 f(5)(x) = 2(f(4)(x)f(x) + 4f'''(x)f'(x) + 3f''(x)2), f(5)(0) = 16 Thus the first three non-zero terms of the Taylor series for tan(x) are: x + 2x3/3! + 16x5/5! = x + x3/3 + 2x5/15

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Answered by Ashwin S. Further Mathematics tutor

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