§8.1 Extending local linearization (continued)

As the degree of the approximation increases

Summary

For $f(x) = e^x$ and $T_2(x) = 1+x+\frac{1}{2}x^2$, the absolute error at $x = 1$ is approximately

For the cubic $T_3(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3$, the absolute error $|f(-1)-T_3(-1)|$ is approximately

When the spreadsheet window widens to something like $-2 \lt x \lt 2$, the polynomial approximations show that

Scanning the spreadsheet columns at a fixed $x$ while the degree climbs from $1$ to $4$ shows the error

For $f(x) = \frac{1}{3}x^3+\frac{1}{4}x^2-2x-1$ near $a = 0$, the quadratic approximation is

For $T_2(x) = 1+x+\frac{1}{2}x^2$ approximating $e^x$, the absolute error at $x = -1$ is about $0.13212$. Report this error.

For $T_4(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4$, the absolute error at $x = 1$ is approximately $0.00995$. Report this value.

For $f(x) = \frac{1}{3}x^3+\frac{1}{4}x^2-2x-1$, the linear approximation at $a=0$ is $T_1(x) = -1-2x$. Compute $T_1(0.5)$.

For the cubic function $f(x) = \frac{1}{3}x^3+\frac{1}{4}x^2-2x-1$, the degree $3$ Taylor polynomial at $a = 0$ equals $f$ itself.