Exercise 7

In this exercise, we extend our work in Exercise. We continue to consider the function $f(x) = \frac{1}{3}x^3 + \frac{1}{4}x^2 - 2x - 1$, but now build the cubic (degree $3$) approximation to $f$ near $a = 0$.

1. Let $T_3(x) = k_0 + k_1x + k_2x^2 + k_3x^3$ and determine $T_3'(x)$, $T_3''(x)$, and $T_3'''(x)$. In addition, determine $f'''(x)$. Record your results below, along with the values of each of these functions at $a = 0$.

   $$\begin{aligned} f(x) &= x^3/3 + x^2/4 - 2x - 1 \\ f'(x) &= \underline{\hspace{2em}} \\ f''(x) &= \underline{\hspace{2em}} \\ f'''(x) &= \underline{\hspace{2em}} \\ & \\ f(0) &= \underline{\hspace{2em}} \\ f'(0) &= \underline{\hspace{2em}} \\ f''(0) &= \underline{\hspace{2em}} \\ f'''(0) &= \underline{\hspace{2em}} \end{aligned}$$

   $$\begin{aligned} T_3(x) &= k_0 + k_1 x + k_2 x^2 + k_3 x^3 \\ T_3'(x) &= \underline{\hspace{2em}} \\ T_3''(x) &= \underline{\hspace{2em}} \\ T_3'''(x) &= \underline{\hspace{2em}} \\ & \\ T_3(0) &= \underline{\hspace{2em}} \\ T_3'(0) &= \underline{\hspace{2em}} \\ T_3''(0) &= \underline{\hspace{2em}} \\ T_3'''(0) &= \underline{\hspace{2em}} \end{aligned}$$

2. Use your work in above to find the formula for $T_3(x)$ that satisfies $T_3(0) = f(0)$, $T_3'(0) = f'(0)$, $T_3''(0) = f''(0)$, and $T_3'''(0) = f'''(0)$.

3. Plot $f(x)$, $T_1(x)$, $T_2(x)$, and $T_3(x)$ on the same axes, centered at $a = 0$. What do you notice?

4. What do you expect the degree $4$ approximation to $f(x)$, $T_4(x)$ to be? Why?

5. What do you expect will happen if we find the degree $6$ approximation to any function $f(x)$ that is itself a degree $6$ polynomial?