Does Taylor's Theorem with Remainder guarantee that the second Taylor polynomial of f(x)=2ex at x=1 has an error less than 0.01 in the estimate of 2e1.2 ?

Taylor's Theorem with Remainder states that the error R_n of a Taylor polynomial approximation P_n(x) of a function f(x) is bounded by |R_n(x)| ≤ M / (n+1)! * |x - a|^(n+1), where M is the maximum value of the |f^(n+1)(t)| on the interval between x and a.

The second Taylor polynomial for f(x) = 2e^x at x = 1 is P_2(x) = 2e + 2e(x - 1) + e(x - 1)^2.

We want to estimate the error at x = 1.2, so |x - a| = |1.2 - 1| = 0.2.

The third derivative of f(x) = 2e^x is f'''(x) = 2e^x, which is always increasing. Therefore, the maximum value of |f'''(t)| on the interval from 1 to 1.2 is M = |f'''(1.2)| = 2e^1.2.

Substituting these values into the error bound, we get |R_2(1.2)| ≤ 2e^1.2 / 3! * 0.2^3 = 2e^1.2 / 6 * 0.008 = e^1.2 * 0.00267.

Since e^1.2 < e, the error is less than e * 0.00267, which is less than 0.01.

Therefore, Taylor's Theorem with Remainder guarantees that the second Taylor polynomial of f(x) = 2e^x at x = 1 has an error less than 0.01 in the estimate of 2e^1.2.