Let tk(x)tk(x): be the taylor polynomial of degree k of the function f(x)=sin(x)f(x)=sin⁡(x) at a=0a=0.suppose you approximate f(x)f(x) by tk(x)tk(x), and if |x|≤1|x|≤1, how many terms do you need (that is, what is k) for you to have your error to be less than 1616 ? (hint: use the alternating series approximation.)

Ans : Note that: sin(x) = sum(n=0 to infinity) [(-1)^n * x^(2n + 1)]/(2n + 1)!. Then, since the series is alternating, the error in the approximation found by taking the first n terms of the series is no bigger than the n+1'th term. In other words: E ≤ a_n+1 = x^(2n + 3)/(2n + 3)!. (Note that a_n does not include (-1)^n, the alternating part) We need that 1/6 ≤ x^(2n + 3)/(2n + 3)!. Given |x| < 1, n = 2 will be the least integer solution. Thus, we need 2 + 1 = 3 terms.