테일러 다항식과 오차

The $n^\text{th}$ derivative of $f(x)=e^x$ is bounded by $3$ on intervals containing $0$ and 1. Thus $|R_n(1)|\leq \frac{3}{(n+1)!}1^{(n+1)}$. When $n=7$, this is less than $0.0001$.

The $n^\text{th}$ derivative of $f(x)=\sqrt x$ is bounded by $0.1$ on intervals containing $3$ and $4$. Thus $|R_n(\pi)|\leq \frac{0.1}{(n+1)!}(1)^{(n+1)}$. When $n=4$, this is less than $0.0001$.

The $n^\text{th}$ derivative of $f(x)=\cos x$ is bounded by $1$ on intervals containing $0$ and $\pi/3$. Thus $|R_n(\pi/3)|\leq \frac{1}{(n+1)!}(\pi/3)^{(n+1)}$. When $n=7$, this is less than $0.0001$. Since the Maclaurin polynomial of $\cos x$ only uses even powers, we can actually just use $n=6$.

The $n^\text{th}$ derivative of $f(x)=\sin x$ is bounded by $1$ on intervals containing $0$ and $\pi$. Thus $|R_n(\pi)|\leq \frac{1}{(n+1)!}(\pi)^{(n+1)}$. When $n=12$, this is less than $0.0001$. Since the Maclaurin polynomial of $\sin x$ only uses odd powers, we can actually just use $n=11$.

The $n^\text{th}$ term is $\frac{1}{n!}x^n$.

The $n^\text{th}$ term is: when $n$ is even, $\frac{(-1)^{n/2}}{n!}x^n$; when $n$ is odd, $0$.

The $n^\text{th}$ term is: when $n$ even, 0; when $n$ is odd, $\frac{(-1)^{(n-1)/2}}{n!}x^n$.

The $n^\text{th}$ term is $x^n$.

The $n^\text{th}$ term is $(-1)^nx^n$.

The $n^\text{th}$ term is $(-1)^n\frac{(x-1)^n}{n}$.

$\displaystyle 1+x+\frac12x^2+\frac16x^3+\frac1{24}x^4$

$\displaystyle 3+15x+\frac{75}{2}x^2+\frac{375}{6}x^3+\frac{1875}{24}x^4$

$\displaystyle 1+2x-2x^2+4x^3-10x^4$

If $Q(x)$ is the quadratic approximation of $f$ about $3$, then $Q(3)=f(3)$, $Q'(3)=f'(3)$, and $Q''(3)=f''(3)$. There is no guarantee that $f(x)$ and $Q(x)$ share the same third derivative, though, so we do not have enough information to know $f'''(3)$.

$$\begin{aligned} f(3)&=-3^2+6(3)=9\\ f'(3)&=\left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{-x^2+6x\right\}\right|_{x=3}=\left.-2x+6\right|_{x=3}=0\\ f''(3)&=\left.\frac{\mathrm{d}^2}{\mathrm{d}{x}^2}\left\{-x^2+6x\right\}\right|_{x=3}= \left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{-2x+6\right\}\right|_{x=3}=-2 \end{aligned}$$

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

$$\begin{aligned} f(x) &\approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2 \\ \text{We subsitute $f(a)=2a+5$, $f'(a)=2$, and $f''(a)=0$:} \\ f(x) &\approx (2a+5)+2(x-a)=2x+5 \end{aligned}$$

So, our approximation is $f(x) \approx 2x+5$.

Remark: Our approximation is exact for every value of $x$. This will always happen with a quadratic approximation of a function that is quadratic, linear, or constant.

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 $0.04$.