§8.1 Extending local linearization

Introduction

Finding a quadratic approximation

Over and over again

For $f(x) = e^x$ near $a = 0$, using $f(0) = 1$ and $f'(0) = 1$, the tangent line approximation is

For the quadratic template $T_2(x) = b_0+b_1x+b_2x^2$ with constants $b_0$, $b_1$, $b_2$, the second derivative is

For $f(x) = e^x$, differentiating repeatedly to fill the table gives $f'(x)$ and $f''(x)$ as

To make $T_2$ match $f(x) = e^x$ up to second order at $0$, the condition $T_2''(0) = f''(0)$ gives

Comparing $T_2(x) = 1+x+\frac{1}{2}x^2$ with $T_1(x) = 1+x$ as approximations of $e^x$ near $0$ shows that

Matching the cubic $T_3(x) = c_0+c_1x+c_2x^2+c_3x^3$ to $e^x$ at $0$ through third order forces

Following the pattern $1$, $1$, $\frac{1}{2}$, $\frac{1}{6}$, the quartic coefficient $d_4$ in $T_4$ for $e^x$ should be

Using $T_2(x) = 1+x+\frac{1}{2}x^2$ to approximate $e^x$, compute $T_2(1)$ as a decimal.

Using $T_4(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4$, compute $T_4(1)$, rounded to three decimals.

The coefficient $c_3$ of the cubic Taylor polynomial for $e^x$ at $0$ satisfies $6c_3 = 1$. Report $c_3$, rounded to three decimals.

The difference $e^x-(1+x)$ equals $0$ at $x=0$ and grows as $|x|$ increases, reaching about $0.718$ at $x=1$.

For $T_3(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3$, the tolerance $|f-T_3| \lt 0.1$ holds on roughly $-1.3 \lt x \lt 1.1$, wider than the quadratic's window.