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Activity 8.6.3

Avg difficulty 2.5

In this activity we apply the Lagrange Error Bound to quantify the accuracy of several different approximations.

  1. Use the degree 1010 Taylor polynomial (centered at a=0a = 0) of f(x)=exf(x) = e^x to estimate the value of e2e^2. What is the maximum error of your estimate, according to the Lagrange Error Bound? How does this compare to the actual error between e2e^2 and T10(2)T_{10}(2) as reported by a computer algebra system?

  2. Use a degree nn Taylor polynomial (centered at a=0a = 0) of f(x)=cos⁡(x)f(x) = \cos(x) to estimate the value of cos⁡(1)\cos(1) within an accuracy of 0.000000010.00000001. According to the Lagrange Error Bound, what value of nn is needed to achieve this accuracy? What is the resulting approximate value of cos⁡(1)\cos(1)?

  3. Recall that for f(x)=ln⁡(1+x)f(x) = \ln(1+x), its Taylor series centered at a=0a = 0 is given by ln⁡(1+x)=x−12x2+13x3−⋯+(−1)n−11nxn+⋯\begin{gathered} \ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots + (-1)^{n-1}\frac{1}{n}x^n + \cdots \end{gathered} and that the nthn^{\text{th}} derivative of f(x)=ln⁡(1+x)f(x) = \ln(1+x) is given by f(n)(x)=(−1)n−1(n−1)!(1+x)n\begin{gathered} f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n} \end{gathered}. If we want to estimate f(0.5)=ln⁡(1.5)f(0.5) = \ln(1.5) to within an accuracy of 0.00010.0001, what value of nn is needed to achieve this accuracy from computing Tn(0.5)T_n(0.5), according to the Lagrange Error Bound?

Boelkins, Active Calculus, §8.6 Quantifying the accuracy of approximations, Activity 8.6.3

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.