테일러 다항식과 오차

We approximate the function $f(x)=\log x$ about the point $x=1$. We choose 1 because it is close to $0.93$, and we can evaluate $f(x)$ and its first two derivatives at $x=1$.

$$\begin{aligned} f(1)&=0\\ f'(x)&=\frac{1}{x}\qquad \Rightarrow \qquad f'(1)=1\\ f''(x)&=\frac{-1}{x^2} \qquad \Rightarrow \qquad f''(1)=-1 \\ \text{So,} \\ f(x) &\approx f(1)+f'(1)(x-1)+\frac{1}{2}f''(1)(x-1)^2\\ &=0+(x-1)-\frac{1}{2}(x-1)^2 \\ \text{When $x=0.93$:} \\ f(0.93) &\approx (0.93-1)-\frac{1}{2}(0.93-1)^2=-0.07-\frac{1}{2}(0.0049)=-0.07245 \end{aligned}$$

We estimate $\log(0.93) \approx -0.07245$.

Remark: a calculator approximates $\log(0.93) \approx -0.07257$. We're pretty close.

We approximate the function $f(x)=\cos x$. We can easily evaluate $\cos x$ and $\sin x$ ($\sin x$ will appear in the first derivative) at $x=0$, and 0 is quite close to $\dfrac{1}{15}$, so we centre our approximation about $x=0$.

$$\begin{aligned} f(0)&=1\\ f'(x)&=-\sin x\\ f'(0)&=-\sin(0)=0\\ f''(x)&=-\cos x\\ f''(0)&=-\cos(0)=-1 \\ \text{Using the quadratic approximation $f(x) \approx f(0)+f'(0)(x-0)+\frac{1}{2}f''(0)(x-0)^2$:} \\ f(x)&\approx 1 -\frac{1}{2}x^2\\ f\left(\frac{1}{15}\right)&\approx 1-\frac{1}{2\cdot 15^2}=\frac{449}{450} \end{aligned}$$

We approximate $\cos\left(\dfrac{1}{15}\right)\approx\dfrac{449}{450}$.

Remark: $\dfrac{449}{450}= 0.99\overline{77}$, while a calculator gives $\cos\left(\frac{1}{15}\right )\approx 0.9977786$. Our approximation has an error of about $0.000001$.

The quadratic approximation of a function $f(x)$ about $x=a$ is

$$\begin{aligned}f(x) &\approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2 \\ \text{We compute derivatives.} \\ f(0)&=e^0=1\\ f'(x)&=2e^{2x}\\ f'(0)&=2e^0=2\\ f''(x)&=4e^{2x}\\ f''(0)&=4e^0=4 \\ \text{Substituting:} \\ f(x) &\approx 1+2(x-0)+\frac{4}{2}(x-0)^2 \\f(x) &\approx 1+2x+2x^2 \end{aligned}$$

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 $0.04$.

For each of these, there are many solutions. We provide some below.

(a) $1+2+3+4+5 = \displaystyle \sum_{n=1}^5n$ (b) $2+4+6+8=\displaystyle \sum_{n=1}^4 2n$ (c) $3+5+7+9+11=\displaystyle \sum_{n=1}^{5}(2n+1)$ (d) $9+16+25+36+49=\displaystyle \sum_{n=3}^7 n^2$ (e) $9+4+16+5+25+6+36+7+49+8=\displaystyle \sum_{n=3}^7 (n^2+n+1)$ (f) $8+15+24+35+48= \displaystyle \sum_{n=3}^7 (n^2-1)$ (g) $3-6+9-12+15-18=\displaystyle \sum_{n=1}^6 (-1)^{n+1}3n$ Remark: if we had written $(-1)^n$ instead of $(-1)^{n+1}$, with everything else the same, the signs would have been reversed.

If we were to find the 16th order Maclaurin polynomial for a generic function, we might expect to have to differentiate 16 times (ugh). But, we know that the derivatives of sines and cosines repeat themselves. So, it's enough to figure out the pattern:

$$\begin{aligned} f(x)&=\sin x + \cos x & f(0)&=1\\ f'(x)&=\cos x - \sin x & f'(0)&=1\\ f''(x)&=-\sin x - \cos x & f''(0)&=-1\\ f'''(x)&=-\cos x +\sin x & f'''(0)&=-1\\ f^{(4)}&=\sin x + \cos x & f^{(4)}(0)&=1 \end{aligned}$$

Since $f^{(4)}(x)=f(x)$, our derivatives repeat from here. They follow the pattern: $+1$, $+1$, $-1$, $-1$.

$$\begin{aligned} T_{16}(x)&=1 +x -\frac{1}{2}x^2 -\frac{1}{3!}x^3 +\frac{1}{4!}x^4 +\frac{1}{5!}x^5 -\frac{1}{6!}x^6 -\frac{1}{7!}x^7 +\frac{1}{8!}x^8 +\frac{1}{9!}x^9 -\frac{1}{10!}x^{10} -\frac{1}{11!}x^{11}\\ &\qquad +\frac{1}{12!}x^{12} +\frac{1}{13!}x^{13} -\frac{1}{14!}x^{14} -\frac{1}{15!}x^{15} +\frac{1}{16!}x^{16} \end{aligned}$$

A Taylor polynomial gives a polynomial approximation for a function $s(t)$. Since $s(t)$ is itself a polynomial, any $n$th order Taylor polynomial, with $n$ greater than or equal to the degree of $s(t)$, will simply give $s(t)$. So, in our case, $T_{100}(t)=s(t)=4.9t^2-t+10$.

If that isn't satisfying, you can go through the normal method of calculating $T_{100}(t)$.

$$\begin{aligned} s(t)&=4.9t^2-t+10 & s(5)&=4.9(25)-5+10=127.5\\ s'(t)&=9.8t-1 & s'(5)&=9.8(5)-1=48\\ s''(t)&=9.8 & s''(5)&=9.8 \end{aligned}$$

The rest of the derivatives of $s(t)$ are identically zero, so they are (in particular) zero when $t=5$. Therefore,

$$\begin{aligned} T_{100}(t)&=127.5+48(t-5)+\frac{1}{2}9.8(t-5)^2\\ &=127.5+48(t-5)+4.9(t-5)^2 \end{aligned}$$

We can now check that $T_{100}(t)$ really is the same as $s(t)$.

$$\begin{aligned} T_{100}(t)&=127.5+48(t-5)+4.9(t-5)^2\\ &=127.5+48(t-5)+4.9(t^2-10t+25)\\ &=[127.5+48(-5)+4.9(25)]+[48-4.9(10)]t+4.9t^2\\ &=10-t+4.9t^2=s(t) \end{aligned}$$

as expected.

Let's start by differentiating $f(x)$ and looking for a pattern. Remember that $\log 2 = \log_e 2$ is a constant number.

$$\begin{aligned} f(x)&=2^x\\ f'(x)&=2^x\log 2\\ f''(x)&=2^x\left(\log 2\right)^2\\ f^{(3)}(x)&=2^x\left(\log 2\right)^3\\ f^{(4)}(x)&=2^x\left(\log 2\right)^4\\ f^{(5)}(x)&=2^x\left(\log 2\right)^5 \\ \text{So, in general,} \\ f^{(k)}(x)&=2^x\left(\log 2\right)^k \\ \text{We notice that this formula works even when $k=0$ and $k=1$. When $x=1$,} \\ f^{(k)}(1)&=2\left(\log 2\right)^k \\ \text{The $n$th order Taylor polynomial of $f(x)$ about $x=1$ is} \\ T_{n}(x)&=\sum_{k=0}^n \frac{f^{(k)}(1)}{k!}(x-1)^k\\ &=\sum_{k=0}^n \frac{2(\log 2)^k}{k!}(x-1)^k \end{aligned}$$

We need to know the first six derivatives of $f(x)$ at $x=1$. Let's get started.

$$\begin{aligned} f(x)&=x^2\log x + 2x^2+5 & f(1)&=7\\ f'(x)&=(x^2)\frac{1}{x}+(2x)\log x + 4x \\ &=2x\log x + 5x& f'(1)&=5\\ f''(x)&=(2x)\frac{1}{x}+(2)\log x + 5\\ &=2\log x + 7& f''(1)&=7\\ f'''(x)&=2x^{-1}& f'''(1)&=2\\ f^{(4)}&=-2x^{-2}& f^{(4)}(1)&=-2\\ f^{(5)}&=4x^{-3}& f^{(5)}(1)&=4\\ f^{(6)}&=-12x^{-4}& f^{(6)}(1)&=-12 \end{aligned}$$