Final answer:
The Taylor series for sin(x) around x = 0 includes the terms x, -x³/3!, and x⁵/5!, with all even powers having a coefficient of zero. The general form includes terms (-1)ⁿn x¹⁴ⁿ÷(2n+1)! for n going from 0 to infinity. Evaluating the series requires substituting the specific value of x into the given terms.
Step-by-step explanation:
The Taylor series for sin(x) expanded around x = 0 up to order x⁶ is:
- sin(x) = x - ⅓x³ + ⅛x⁵ + ...
(a) The first six terms of the Taylor series for sin(x) are actually only three non-zero terms: x, -⅓x³, and ⅛x⁵, since the coefficients for even powers of x are zero in the sin(x) series.
(b) The general form of the Taylor series for sin(x) is:
- sin(x) = ∑((-1)ⁿn x¹⁴ⁿ÷(2n+1)!), where n ranges from 0 to ∞.
(c) The coefficient of x⁴ in the series is 0, because sin(x) only has odd-powered terms in its Taylor expansion.
(d) To evaluate the series for a specific value of x, substitute x into the series and calculate the sum of the terms.