테일러 다항식과 오차

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 $T_3'''(a)$.

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

So, $a=\sqrt{e}$.

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.

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