테일러 다항식과 오차

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.

Let's start by taking the first two derivative of $f(x)$.

$$\begin{aligned} f(x)&=2\arcsin x&&\Rightarrow& f(0)&=2(0)=0\\ f'(x)&=\frac{2}{\sqrt{1-x^2}}&&\Rightarrow& f'(0)&=\frac{2}{1}=2\\ f''(x)&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{2(1-x^2)^{-\tfrac{1}{2}}\right\}\\ &=2\left(-\tfrac{1}{2}\right)(1-x^2)^{-\tfrac{3}{2}}(-2x) \qquad\text{(chain rule)}\\ &=\frac{2x}{\left(\sqrt{1-x^2}\right)^3}&&\Rightarrow& f''(0)&=0 \\ \text{Now, we can find the quadratic approximation about $x=0$.} \\ f(x)&\approx f(0)+f'(0)x+\frac{1}{2}f''(0)x^2\\ &=2x\\ f(1)& \approx 2 \end{aligned}$$

Our quadratic approximation gives $2\arcsin(1) \approx 2$. However, $2\arcsin(1)$ is exactly equal to $2\left(\dfrac{\pi}{2}\right)=\pi$. We've just made the rather unfortunate approximation $\pi \approx 2$.

From the text, the quadratic approximation of $e^x$ about $x=0$ is $$e^x \approx 1+x+\frac{1}{2}x^2$$ So, $$e=e^1 \approx 1+1+\frac{1}{2}=2.5$$ We estimate $e \approx 2.5$.

Remark: actually, $e \approx 2.718$.

Since $T_3(x)$ is the third order Taylor polynomial for $f(x)$ about $x=1$:

• $T_3(1)=f(1)$
• $T_3'(1)=f'(1)$
• $T_3''(1)=f''(1)$
• $T_3'''(1)=f'''(1)$

In particular, $f''(1)=T_3''(1)$.

$$\begin{aligned} T_3'(x)&=3x^2-10x+9\\ T_3''(x)&=6x-10\\ T_3''(1)&=6-10=-4 \end{aligned}$$

So, $f''(1)=-4$.

The fourth order Maclaurin polynomial for $f(x)$ is

$$\begin{aligned} T_4(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f^{(4)}(0)x^4 \\ \text{while the third order Maclaurin polynomial for $f(x)$ is} \\ T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3 \\ \text{So, we simply ``chop off" the part of $T_4(x)$ that includes $x^4$:} \\ T_3(x)&=-x^3+x^2-x+1 \end{aligned}$$

The $n$th order Taylor polynomial for $f(x)$ about $x=5$ is

$$\begin{aligned}T_n(x)&=\sum_{k=0}^{n} \frac{1}{k!}f^{(k)}(5)(x-5)^{k} \\ \text{We expand this somewhat:} \\ T_n(x)&=f(5)+f'(x-5) + \cdots + \boxed{\frac{1}{10!}f^{(10)}(5)(x-5)^{10}}+\cdots + \frac{1}{n!}f^{(n)}(5)(x-5)^n \\ \text{So, the coefficient of $(x-5)^{10}$ is $\dfrac{1}{10!}f^{(10)}(5)$. Expanding the given form of the Taylor polynomial:} \\ T_n(x)&=\sum_{k=0}^{n/2} \frac{2k+1}{3k-9}(x-5)^{2k}\\ &= \underbrace{\frac{1}{-9}}_{k=0}+ \underbrace{\frac{3}{-6}(x-5)^2}_{k=1}+\cdots+ \underbrace{\boxed{\frac{11}{6}(x-5)^{10}}}_{k=5}+\cdots+ \underbrace{\frac{n+1}{(3/2)n-9}(x-5)^n}_{k=n/2} \\ \text{Equating the coefficients of $(x-5)^{10}$ in the two expressions:} \\ \frac{1}{10!}f^{(10)}(5)&=\frac{11}{6}\\ f^{(10)}(5)&=\frac{11\cdot 10!}{6} \end{aligned}$$

Since $T_3(x)$ is the third order Taylor polynomial for $f(x)$ about $x=a$, we know the following things to be true:

• $f(a)=T_3(a)$
• $f'(a)=T'_3(a)$
• $f''(a)=T''_3(a)$
• $f'''(a)=T'''_3(a)$

But, some of these don't look super useful. For instance, if we try to use the first bullet, we get this equation: $$a^3\left[2\log a - \frac{11}{3}\right]=-\frac{2}{3}\sqrt{e^3}+3ea-6\sqrt{e}a^2+a^3$$ Solving this would be terrible. Instead, let's think about how the equations look when we move further down the list. Since $T_3(x)$ is a cubic equation, $T_3'''(x)$ is a constant (and so $T_3'''(a)$ does not depend on $a$). That sounds like it's probably the simplest option. Let's start differentiating. We'll need to know both $f'''(a)$ and $T_3'''(a)$.

$$\begin{aligned} f(x)&=x^3\left[2\log x - \frac{11}{3}\right]\\ f'(x)&=x^3\left[\frac{2}{x}\right]+3x^2\left[2\log x - \frac{11}{3}\right]=6x^2\log x -9x^2\\ f''(x)&=6x^2\cdot\frac{1}{x}+12x\log x - 18x=12x\log x - 12 x\\ f'''(x)&=12x\cdot\frac{1}{x}+12\log x - 12 = 12\log x\\ f'''(a)&=12\log a \\ \text{Now, let's move to the Taylor polynomial. Remember that $e$ is a constant.} \\ T_3(x)&=-\frac{2}{3}\sqrt{e^3}+3ex-6\sqrt{e}x^2+x^3\\ T_3'(x)&=3e-12\sqrt{e}x+3x^2\\ T_3''(x)&=-12\sqrt{e}+6x\\ T_3'''(x)&=6\\ T_3'''(a)&=6 \\ \text{The final bullet point gives us the equation:} \\ f'''(a)&=T_3'''(a)\\ 12\log a &= 6\\ \log a &= \frac{1}{2}\\ a&=e^{\tfrac{1}{2}} \end{aligned}$$