Exercise 10

In Example, we found that $$\begin{gathered} f(x) = \ln(1+x) \approx T_5(x) = 1x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 \end{gathered}$$, where $T_5$ is the degree $5$ Taylor approximation of $\ln(1+x)$ centered at $a = 0$.

In Activity, we found that $$\begin{gathered} g(x) = \ln(x) \approx P_5(x) = 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 \end{gathered}$$, where $P_5$ is the degree $5$ Taylor approximation of $\ln(x)$ centered at $a = 1$. (Here we are using “$T_5$” and “$P_5$” to distinguish between these two degree $5$ polynomial approximations of the two different functions $f(x) = \ln(1+x)$ and $g(x) = \ln(x)$, centered at two different values.)

1. Note that $f(0.5) = \ln(1.5)$. Use $T_5(x)$ appropriately to find an estimate of $\ln(1.5)$.

2. Observe that $g(1.5) = \ln(1.5)$. Use $P_5(x)$ to estimate $\ln(1.5)$.

3. Are the estimates of $\ln(1.5)$ generated by $T_5(x)$ and $P_5(x)$ the same or different? Why do you think this happens?