130k views
5 votes
Use Taylor's Theorem to obtain an upper bound for the error of

the approximation. Then calculate the value of the error. (Round
your answers to three significant figures.)
cos(0.7) ≈ 1 − (0.7)22!

1 Answer

5 votes

Final answer:

To obtain an upper bound for the error of the approximation of cos(0.7), we can use Taylor's Theorem. By taking the second degree Taylor polynomial of cos(x) centered at x=0 and evaluating the expression [(0.7^3)/(3!)] * 1, we can obtain an upper bound for the error to be approximately 0.033.

Step-by-step explanation:

To obtain an upper bound for the error of the approximation, we can use Taylor's Theorem. Taylor's Theorem states that for a function f(x) that is differentiable n+1 times on an interval containing x=a, we can approximate the value of f(x) near a using a Taylor polynomial of degree n. The error of the approximation is given by the remainder term which can be upper bounded by the (n+1)th derivative of f evaluated at some point c between a and x, multiplied by [(x-a)^(n+1)]/(n+1)!. In this case, we want to approximate cos(0.7) using a Taylor polynomial centered at x=0. The Taylor series expansion of cos(x) at x=0 is 1 - (x^2)/2!, and if we take the second degree (n=2) Taylor polynomial, the error is given by [(0.7-0)^3]/(3!) * cos(c) for some c between 0 and 0.7. Since cos(c) is always between -1 and 1, the error can be upper bounded by [(0.7^3)/(3!)] * 1. Evaluating this expression, we get an upper bound for the error of the approximation of cos(0.7) to be approximately 0.033.

User GabrielAnzaldo
by
8.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.