테일러 다항식과 오차

There are a few functions we could choose to approximate. For example:

• $f(x)=x^{4/3}$. In this case, we would probably choose to approximate $f(x)$ about $x=8$ (since 8 is a cube, $8^{4/3}=2^4=16$ is something we can evaluate) or $x=1$.
• $f(x)=5^{x}$. We can easily figure out $f(x)$ when $x$ is a whole number, so we would want to centre our approximation around some whole number $x=a$, but then $f'(a)=5^a\log(5)$ gives us a problem: without a calculator, it's hard to know what $\log(5)$ is.
• Since $5^{4/3}=5\sqrt[3]{5}$, we can use $f(x)=5\sqrt[3]{x}$. As in the first bullet, we would centre about $x=8$, or $x=1$.

There isn't much difference between the first and third bullets. We'll go with $f(x)=5\sqrt[3]{x}$, centred about $x=8$.

$$\begin{aligned} f(x)&=5x^{\frac{1}{3}}&&\Rightarrow& f(8)&=5\cdot 2 = 10\\ f'(x)&=\frac{5}{3}x^{-\frac{2}{3}}&&\Rightarrow& f'(8)&=\frac{5}{3}\left(2^{-2}\right)=\frac{5}{12}\\ f''(x)&=\frac{5}{3}\left(-\frac{2}{3}\right)x^{-\frac{5}{3}}=-\frac{10}{9}x^{-\frac{5}{3}}&&\Rightarrow& f''(8)&=-\frac{10}{9}\left(2^{-5}\right)=-\frac{5}{144} \\ \text{Using the quadratic approximation $f(x) \approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2$:} \\ f(x)&\approx 10+\frac{5}{12}(x-8)-\frac{5}{288}(x-8)^2\\ f(5)&\approx 10+\frac{5}{12}(-3)-\frac{5}{288}(9)=\frac{275}{32} \end{aligned}$$

We estimate $5^{4/3}\approx\dfrac{275}{32}$

Remark: $\dfrac{275}{32} = 8.59375$, and a calculator gives $5^{4/3}\approx 8.5499$. Although 5 and 8 are somewhat far apart, our estimate is only off by about $0.04$.