Taylor polynomials

In Exercise, we found that the degree 2 Taylor polynomial centered at $a = 0$ of a quadratic function is the quadratic function itself. In this exercise, we explore how changing the center of the approximation offers additional insight into the function.

Let $f(x) = \frac{1}{2}x^2 - 2x + 5$, and let $a = 2$ be the center at which we will find a degree $2$ Taylor polynomial approximation of $f$.

1. By finding $f'(x)$, $f''(x)$, $f(2)$, $f'(2)$, and $f''(2)$, determine $T_2(x)$, the degr…

Recall that we found in Preview and subsequent work that $\sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7$, which is the degree $7$ Taylor approximation centered at $0$. And in Activity, we found that the degree $6$ Taylor approximation centered at $a = 0$ for $\cos(x)$ is $\cos(x) \approx 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6$.

In this exercise, we investigate Taylor polynomial approximations of $f(x) = \sin(x)$ centered at $a = \frac{\pi}{2}$.

1. By finding the appropriate derivatives of $f(x) = \sin(x)$ and evaluating them at $a = \frac{\pi}{2}$, determine the degree $6$ Taylor polynom…

In Example, we found that , where $T_5$ is the degree $5$ Taylor approximation of $\ln(1+x)$ centered at $a = 0$.

In Activity, we found that , 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…