테일러 다항식과 오차

We begin by finding the derivatives of $f$ at $x=0$.

$$\begin{aligned} f(x)&=(1 - 3x)^{-1/3} & f(0) & = 1 \\ f'(x)&=(-3)\frac{-1}{3}(1 - 3x)^{-4/3}=(1-3x)^{-4/3} & f'(0) & = 1 \\ f''(x)&=(-3)\frac{-4}{3}(1-3x)^{-7/3}={4}(1 - 3x)^{-7/3} & f''(0) & = 4 \\ f^{(3)}(x)&=(-3)(4)\frac{-7}{3}(1-3x)^{-10/3}=28(1 - 3x)^{-10/3} & f^{(3)}(0) & = 28 \\ \end{aligned}$$

Plugging these into the definition of a Taylor Polynomial, we find that the third–order Taylor polynomial for $f$ around $x = 0$ is

$$\begin{aligned} T_3(x)&=1+x+\frac{4}{2!}x^2+\frac{28}{3!}x^3\\ &=1 + x +2x^2+\frac{14}{3}x^3 \end{aligned}$$

The fourth order Maclaurin polynomial for $f(x)$ is

$$\begin{aligned} T_4(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f^{(4)}(0)x^4 \\ \text{while the third order Maclaurin polynomial for $f(x)$ is} \\ T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3 \\ \text{So, we simply ``chop off" the part of $T_4(x)$ that includes $x^4$. Since that's already 0, in this case $T_3(x)=T_4(x)$.} \\ T_3(x)&=5x^2-9 \end{aligned}$$

Let $f(x)=\log x$ and $x_0=10$. Then

$$\begin{aligned} f(x)&=\log x & f'(x)&=\frac{1}{x} & f''(x)&=-\frac{1}{x^2} \\ f(10)&=\log 10\approx 2.30259 & f'(10)&=\frac{1}{10} & f''(10)&=-\frac{1}{100} \end{aligned}$$

so that, with $x=10.3$, $$\log 10.3=f(10.3)\approx f(10)+f'(10)(10.3-10) =2.30259+\frac{0.3}{10} =2.33259$$ The error in this approximation (excluding the error in the given data $\log 10\approx 2.30259$) is $\dfrac{1}{2} f''(z)(10.3-10)^2$ for some $z$ between $10$ and $10.3$. Because $f''(z)=-\dfrac{1}{z^2}$ increases as $z$ increases, it must be between $-\dfrac{1}{10^2}$ and $-\dfrac{1}{10.3^2}$. This forces $\dfrac{1}{2} f''(z)(10.3-10)^2$ to be between $-\dfrac{1}{2}\cdot \dfrac{1}{10^2}(0.3)^2=-0.00045$ and $-\dfrac{1}{2}\cdot \dfrac{1}{10.3^2}(0.3)^2<-0.00042$.