테일러 다항식과 오차

We will approximate $f(x)=x^{\tfrac{1}{7}}$ using a Taylor polynomial. Since $3^7=2187$, we will use $x=2187$ as our centre.

We need to figure out which order Taylor polynomial will result in a small enough error.

If we use the $n$th Taylor polynomial, our error will be

$$\begin{aligned} \left|f(2200)-T_n(2200)\right|&=\left|\frac{f^{(n+1)}(c)}{(n+1)!}(2200-2187)^{n+1}\right|\\ &=\left|f^{(n+1)}(c)\right|\cdot\frac{13^{n+1}}{(n+1)!} \\ \text{for some $c$ in $(2187,2200)$. In order for this to be less than 0.001, we need} \\ \left|f^{(n+1)}(c)\right|\cdot\frac{13^{n+1}}{(n+1)!}&<0.001\\ \left|f^{(n+1)}(c)\right|&<\frac{ (n+1)!}{1000\cdot13^{n+1}} \end{aligned}$$

It's a tricky thing to figure out which $n$ makes this true. Let's make a table. We won't show all the work of filling it in, but the work is standard.

$$\begin{array}{c|c|c|c} n & \dfrac{(n+1)!}{1000\cdot 13^{n+1}} & \left|f^{(n+1)}(c)\right| & \textup{Is }\left|f^{(n+1)}(c)\right|<\dfrac{ (n+1)!}{1000\cdot13^{n+1}}\textup{?} \\~&~&~\\ \hline ~&~&~\\ 0&\dfrac{1}{1000\cdot 13}&|f'(c)|=\dfrac{1}{7c^{6/7}}<\dfrac{1}{7\cdot 3^6}& \\~&~&~\\ \hline ~&~&~\\ 1&\dfrac{2}{1000\cdot 13^2}&|f''(c)|=\dfrac{6}{7^2\cdot c^{\tfrac{13}{7}}}<\dfrac{6}{7^2\cdot 3^{13}}&\textup{Yes!} \end{array}$$

That is: if we use the first order Taylor polynomial, then for some $c$ between $2187$ and $2200$,

$$\begin{aligned} |f(2200)-T_1(2200)|&=\left|f''(c)\right|\cdot\frac{13^2}{2!}\\ &=\frac{6}{7^2\cdot c^{\tfrac{13}{7}}}\cdot\frac{13^2}{2}\\ &<\dfrac{6}{7^2\cdot 3^{13}}\cdot\dfrac{13^2}{2}\\ &=\dfrac{3\cdot 13^2}{7^2 \cdot 3^{13}}\approx 0.0000065 \end{aligned}$$

So, actually, the linear Taylor polynomial (or any higher order Taylor polynomial) will result in an approximation that is much more accurate than required. (We don't know, however, that the constant approximation will be accurate enough–so we'd better stick with $n \geq 1$.)

Now that we know we can take the first order Taylor polynomial, let's compute $T_1(x)$. Recall we are taking the Taylor polynomial for $f(x)=x^{\tfrac{1}{7}}$ about $x=2187$.

$$\begin{aligned} f(2187)&=2187^{\tfrac{1}{7}}=3\\ f'(x)&=\frac{1}{7}x^{-\tfrac{6}{7}}\\ f'(2187)&=\frac{1}{7\sqrt[7]{2187}^6}=\frac{1}{7\cdot3^6}\\ T_1(x)&=f(2187)+f'(2187)(x-2187)\\ &=3+\frac{x-2187}{7\cdot3^6}\\ T_1(2200)&=3+\frac{2200-2187}{7\cdot3^6}\\ &=3+\frac{13}{7\cdot 3^6}\\ &\approx 3.00255 \end{aligned}$$

We conclude $\sqrt[7]{2200}\approx 3.00255$.