테일러 다항식과 오차

The third Maclaurin polynomial for $f(x)$ is

$$\begin{aligned} f(0) + f'(0)x +\frac{f''(0)}{2}\cdot x^2 + \frac{f'''(0)}{6}\cdot x^3 &=4+3x^2+\frac{1}{2}x^3. \end{aligned}$$

The coefficient of $x$ is $f'(0)$ on one side, and $0$ on the other, so $f'(0)=0$.

The coefficient of $x^2$ is $\dfrac{1}{2}f''(0)$ on one side, and $3$ on the other, so $f''(0)=6$.

The third Maclaurin polynomial for $h(x)$ is

$$\begin{aligned} h(0) + h'(0)x +\frac{h''(0)}{2}\cdot x^2 + \frac{h^{(3)}(0)}{3!}\cdot x^3 &= 1+4x-\frac{1}{3}x^2 + \frac{2}{3}x^3 \end{aligned}$$

The coefficient of $x^3$ is $\dfrac{1}{3!}h^{(3)}(0)$ on one side, and $\dfrac{2}{3}$ on the other. Thus $\dfrac{h^{(3)}(0)}{6}=\dfrac{2}{3}$, so $h^{(3)}(0)=6\cdot\dfrac{2}{3} = 4$.

The third order Taylor polynomial for $h(x)$ about $x=2$ is $$h(2) + h'(2)(x-2) +\frac{h''(2)}{2}\cdot (x-2)^2 + \frac{h'''(2)}{6}\cdot (x-2)^3$$ The coefficient of $(x-2)$ is $h'(2)$ in the definition, and $\dfrac{1}{2}$ in the given function, so $h'(2)=\dfrac{1}{2}$.

The coefficient of $(x-2)^2$ is $~\dfrac{1}{2}h''(2)~$ in the definition, and $0$ in the given function, so $h'(2)=0$.

(a) For $x$ near 3, $$f(x)\approx f(3)+f'(3)(x-3)=2+4(x-3)$$ In particular $$f(2.98)\approx 2+4(2.98-3)=2-0.08=1.92$$ (b) For $x$ near 3,

$$\begin{aligned} f(x)&\approx f(3)+f'(3)(x-3)+\frac{1}{2} f''(3)(x-3)^2\\ &=2+4(x-3)-\frac{1}{2} 10(x-3)^2 \\ \text{In particular} \\ f(2.98)&\approx 2+4(2.98-3)-5(2.98-3)^2\\ &=2-0.08-0.002={1.918} \end{aligned}$$

We use the function $f(x)=\sqrt{x}$ and point $a=1$ as the centre of our approximation, since we can easily calculate

$$\begin{aligned} f(a)&=f(1)=\sqrt{1}=1. \end{aligned}$$

We compute $f'(x)=\dfrac{1}{2\sqrt{x}}$, so

$$\begin{aligned} f'(1)&=\frac{1}{2\sqrt{1}}=\frac{1}{2}. \end{aligned}$$

So, a linear approximation of $\sqrt{2}=f(2)$ is

$$\begin{aligned} \sqrt{2}&\approx T_1(2)= f(1) + f'(1)\cdot (2-1)\\ &=1+\frac{1}{2}=\frac{3}{2}. \end{aligned}$$

We use the function $f(x)=x^{1/3}$ and point $a=27$ as the centre of our approximation since we can easily compute

$$\begin{aligned} f(27)&=3 \end{aligned}$$

We compute $f'(x)=\dfrac{1}{3} x^{-2/3}$, so

$$\begin{aligned} f'(27)&= \frac{1}{3} \cdot (27)^{-2/3} = \frac{1}{27} \end{aligned}$$

So, the linear approximation of $26^{1/3} = f(26)$ is

$$\begin{aligned} 26^{1/3}&\approx T_1\left(26\right) = f(27) + f'(27)\cdot (26-27) \\ &= 3 - \frac{1}{27} = \frac{80}{27} \end{aligned}$$

We use the function $f(x)=x^5$ and point $a=10$ as the centre of our approximation since we know that $f(a)=f(10)=10^5$.

Since $f'(x) = 5x^4$ we have $f'(10) = 50,000$.

So, a linear approximation of $10.1^5$ is

$$\begin{aligned} T_1(10.1)&= f(10) + f'(10)\cdot (10.1-10) \\ &= 100,000 + 50,000 \cdot 0.1 = 105,000\,. \end{aligned}$$

We use the function $f(x)=\sin(x)$ and point $a=\pi$ as the centre of our approximation since we know that

$$\begin{aligned} \sin(a)&=f(\pi)=\sin{\pi}=0. \end{aligned}$$

and $\pi$ is reasonably close to $\dfrac{101\pi}{100}$.

We compute $f'(x)=\cos(x)$, so

$$\begin{aligned} f'(\pi)&=\cos(\pi) = -1. \\ \text{So, the linear approximation of $\sin\left(\frac{101\pi}{100}\right) $ is} \\ f\left(\frac{101\pi}{100}\right)&\approx T_1\left(\frac{101\pi}{100}\right)\\ &= f(\pi) + f'(\pi)\cdot \left(\frac{101\pi}{100}-\pi\right) \\ &=0+(-1)\cdot\frac{\pi}{100}=-\frac{\pi}{100}. \end{aligned}$$

Set $f(x)=\arctan(x)$. Then $f'(x)=\dfrac{1}{1+x^2}$, so $f'(1)=\dfrac{1}{2}$ and $$f(1.1)\approx f(1)+f'(1)(1.1-1)=\frac{\pi}{4}+\frac{1}{20}$$

Set $f(x)=(2+x)^3$, so we are approximating $f(0.001)$.

The obvious choice of $a$ is $a=0$.

Then $f'(x)=3(2+x)^2$, so $$(2.001)^3=f(0.001)\approx f(0)+f'(0)(0.001-0) =8+\frac{12}{1000}=\frac{8012}{1000}$$

Remark: if we had chosen $f(x)=x^3$ and $a=2$, the result would have been exactly the same.

We set $f(x)=(8+x)^{2/3}$, and choose $a=0$ as our centre.

Then $f'(x)=\dfrac{2}{3}(8+x)^{-1/3}$, so that

$$\begin{aligned} (8.06)^{2/3}=f(0.06)&\approx f(0)+f'(0)\cdot 0.06\\ &=8^{2/3}+\frac{2}{3}8^{-1/3}\cdot 0.06\\ &=\sqrt[3]{8}^2+\frac{2}{3\sqrt[3]{8}}\cdot 0.06\\ &=2^2+\frac{2}{3\cdot 2}\cdot 0.06\\ &=4+\frac{1}{3}\cdot 0.06\\&=4.02=\frac{402}{100}\end{aligned}$$

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 $-\dfrac{1}{2}\cdot \dfrac{1}{10.3^2}(0.3)^2<-0.00042$.