Determine the degree of the Maclaurin polynomial required for the error in the approximation of the value sin(2) to be less than 0.001 according to Taylor's Theorem.
A. 8
B. 9
C. 10
D. 11
E. 12
To use Taylor's Theorem, we need to find the Maclaurin polynomial of sin(x) and the corresponding error term.
The Maclaurin polynomial for sin(x) is given by:
P(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
The error term for the Maclaurin polynomial is given by:
R(x) = (f^(n+1))(c) * (x^(n+1))/(n+1)!
where f^(n+1) represents the (n+1)th derivative of f(x) and c lies between 0 and x.
To ensure that the error in the approximation of sin(2) is less than 0.001, we need to find the smallest value of n such that the error term R(x) is less than 0.001 when x = 2.
Let's calculate the error term for x = 2 and different values of n until we find a value that yields an error less than 0.001:
For n = 8:
R(2) = (f^(9))(c) * (2^9)/(9!)
= (cos(c)) * (512)/(362880)
= (cos(c)) * 0.00141093474
For n = 9:
R(2) = (f^(10))(c) * (2^10)/(10!)
= (-sin(c)) * (1024)/(3628800)
= (-sin(c)) * 0.0002821860889
For n = 10:
R(2) = (f^(11))(c) * (2^11)/(11!)
= (-cos(c)) * (2048)/(39916800)
= (-cos(c)) * 5.129844e-05
From the above calculations, we see that for n = 10, the error term is already less than 0.001. Therefore, the degree of the Maclaurin polynomial required for the error in the approximation of sin(2) to be less than 0.001 is 10.
The correct answer is C. 10.