§8.2 Taylor polynomials

Introduction

Taylor polynomials

For $f(x) = \sin(x)$, differentiating repeatedly cycles through $\sin$, $\cos$, $-\sin$, $-\cos$. The value $f'''(0)$ equals

For $f(x) = \cos(x)$, the derivative table through eighth order shows $f''(0)$ and $f^{(4)}(0)$ equal to

Using the coefficient rule $c_k = \frac{f^{(k)}(0)}{k!}$ for $f(x) = \cos(x)$, the polynomial $T_2(x)$ equals

Following the pattern of $T_2$, $T_4$, and $T_6$ for cosine, the polynomial $T_{10}(x)$ is

Matching $T_3(x) = c_0+c_1x+c_2x^2+c_3x^3$ to $\sin(x)$ at $0$ produces

For $f(x) = \cos(x)$ with $\Delta x = 0.2$ sampling from $-2$, the tolerance $|f-T_n| \lt 0.1$ holds roughly on

Using $T_3(x) = x-\frac{1}{3!}x^3$ for $\sin(x)$, estimate $\sin(0.5)$, rounded to three decimals.

Using $T_4(x) = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4$ for $\cos(x)$, estimate $\cos(1)$, rounded to three decimals.

The degree $5$ Taylor polynomial of $\ln(1+x)$ at $0$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5$.

Plotting $T_2$, $T_4$, and $T_6$ against $\cos(x)$ shows accuracy improving both at each fixed $x$ and across a wider interval as the degree grows.