테일러·매클로린 급수

We use that

$$\log(1+y)= \sum_{n=1}^\infty (-1)^{n-1} \frac{y^n}{n} \qquad \text{for all } -1 < y \le 1$$

with $y=\dfrac{x-2}{2}$ to give

$$\begin{aligned} \log(x) & = \log(2 +x -2) =\log\left[2\left(1+\frac{x-2}{2}\right)\right]\\ &= \log 2 +\log\Big(1+\frac{x-2}{2}\Big) =\log 2 +\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\,2^n}(x-2)^n \end{aligned}$$

It converges when $-1<y\le 1$, or equivalently, $0<x\le 4$.

Expanding the exponential using its Maclaurin series,

$$\begin{aligned} I&=\int_0^1 x^4 e^{-x^2}\ \mathrm{d}x =\sum_{n=0}^\infty\int_0^1 x^4 \frac{{(-x^2)}^n}{n!}\ \mathrm{d}x =\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{2n+4} \ \mathrm{d}x \\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+5)} = \underbrace{\frac{1}{5}}_{n=0} -\underbrace{\frac{1}{7}}_{n=1} +\underbrace{\frac{1}{18}}_{n=2} -\underbrace{\frac{1}{3!(11)}}_{n=3} +\cdots \end{aligned}$$

The signs of successive terms in this series alternate. Futhermore the magnitude of the $n^{\rm th}$ term decreases with $n$. Hence, by the alternating series test, $I$ lies between $\frac{1}{5}-\frac{1}{7}+\frac{1}{18}$ and $\frac{1}{5}-\frac{1}{7}+\frac{1}{18}-\frac{1}{3!(11)}$. So

$$|I - a| \le \frac{1}{3!(11)} =\frac{1}{66}$$

Expanding the exponential using its Taylor series,

$$\begin{aligned} I&=\int_0^{\frac{1}{2}} x^2 e^{-x^2}\ \mathrm{d}x =\sum_{n=0}^\infty\int_0^{\frac{1}{2}} x^2 \frac{{(-x^2)}^n}{n!}\ \mathrm{d}x =\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^{\frac{1}{2}} x^{2n+2} \ \mathrm{d}x \\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}} \end{aligned}$$

The signs of successive terms in this series alternate. Futhermore the magnitude of the $n^{\rm th}$ term decreases with $n$. Hence, by the alternating series test, $I$ lies between $\sum\limits_{n=0}^N \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}}$ and $\sum\limits_{n=0}^{N+1} \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}}$, for every $N$. The first few terms are, to five decimal places,

$$\begin{array}{|c|c|c|c|c|}\hline n & \text{0} & \text{1} & \text{2} & \text{3} \\ \hline \dfrac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}} & \text{0.04167} & \text{-0.00625} & \text{0.00056} & \text{-0.00004} \\ \hline \end{array}$$

Allowing for a roundoff error of $0.000005$ in each of these, $I$ must be between

$$0.04167-0.00625+0.00056+0.000005\times 3=0.035995$$

and

$$0.04167-0.00625+0.00056-0.00004-0.000005\times 4=0.035920$$

where the multiples of 0.000005 are the maximum possible accumulated roundoff errors in the added terms.

(a) For Newton's method, recall we approximate a root of the function $g(x)$ in iterations: given an approximation $x_n$, our next approximation is $x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}$. In our case, $$x_{n+1}=x_n - \frac{x_n^3-2}{3x_n^2}=\frac23\left(x_n+\frac{1}{x_n^2}\right).$$

We want to start somewhere reasonably close to the actual root we want, so let's set $x_0=1$. (Your starting point may vary.)

$$\begin{aligned} x_0&=1 &&\implies x_1=\dfrac23\left(1+\frac11\right)=\dfrac{4}{3}&\approx&1.3333\\ x_1&=\dfrac43 &&\implies x_2=\dfrac23\left(\frac43+\frac{9}{16}\right)=\dfrac{91}{72}&\approx&1.2639\\ x_2&=\dfrac{91}{72} &&\implies x_3=\dfrac23\left(\dfrac{91}{72}+\dfrac{72^2}{91^2}\right)=\dfrac{1126819 }{894348 }&\approx&1.2599\\ x_3&=\dfrac{1126819 }{894348 } &&\implies x_4=\dfrac23\left(\dfrac{1126819 }{894348 }+\dfrac{894348^2}{1126819^2}\right)&\approx&1.2599 \end{aligned}$$

So, $\sqrt[3]{2}\approx 1.26$. (b) We'll evaluate the given series at $x=2$. This yields the series $$\sqrt[3]{2}=1+\frac{1}{6}+\sum_{n=2}^\infty(-1)^{n-1}\frac{(2)(5)(8)\cdots(3n-4)}{3^n\, n!}.$$ This series is alternating, so if we use the partial sum $S_N$, our absolute error is at most $$|a_{N+1}|=\frac{(2)(5)(8)\cdots(3N-1)}{3^{N+1}\, (N+1)!}$$ (if $N \ge 2$). We want to know which value of $N$ makes this at most 0.01. We test several values.

$$\begin{array}{|c|c|}\hline N & |a_{N+1}| \\ \hline \text{3} & \dfrac{(2)(5)(8)}{3^4\cdot 4!}\approx 0.04 \\ \hline \text{4} & \dfrac{(2)(5)(8)(11)}{3^5\, 5!}\approx 0.03 \\ \hline \text{5} & \dfrac{(2)(5)(8)(11)(14)}{3^6\, 6!}\approx 0.023 \\ \hline \text{6} & \dfrac{(2)(5)(8)(11)(14)(17)}{3^7\, 7!}\approx 0.019 \\ \hline \text{7} & \dfrac{(2)(5)(8)(11)(14)(17)(20)}{3^8\, 8!}\approx 0.016 \\ \hline \text{8} & \dfrac{(2)(5)(8)(11)(14)(17)(20)(23)}{3^9\, 9!}\approx 0.013 \\ \hline \text{9} & \dfrac{(2)(5)(8)(11)(14)(17)(20)(23)(26)}{3^{10}\, 10!}\approx 0.012 \\ \hline \text{10} & \dfrac{(2)(5)(8)(11)(14)(17)(20)(23)(26)(29)}{3^{11}\, 11!}\approx 0.0103 \\ \hline \text{11} & \dfrac{(2)(5)(8)(11)(14)(17)(20)(23)(26)(29)(32)}{3^{12}\, 12!}\approx 0.009 \\ \hline \end{array}$$

So, the approximation $S_{11}$ has a sufficiently small error. That is, we would add up the first twelve terms.

Our plan is as follows:

• Make a Taylor series for $f(x)$
• Calculate the tenth derivative of the Taylor series of $f(x)$.
• Decide how many terms we need to add to achieve the desired accuracy.
• Approximate $f^{(10)}\left(\frac15\right)$ with a partial sum.

We know that the Taylor series for $\arctan x$ is $\displaystyle\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$, which converges for $-1\le x\le1$. So, the Taylor series for $\arctan(x^3)$ is $$f(x)=\arctan(x^3)=\sum_{n=0}^\infty (-1)^n\frac{(x^3)^{2n+1}}{2n+1} =\sum_{n=0}^\infty (-1)^n\frac{x^{6n+3}}{2n+1}$$ It is much easier to differentiate this series many times than it is to differentiate $\arctan(x^3)$ directly many times.

$$\begin{aligned} f'(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)x^{6n+2}}{2n+1}\\ f''(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)(6n+2)x^{6n+1}}{2n+1}\\ f'''(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)(6n+2)(6n+1)x^{6n}}{2n+1}\\ \vdots&\\ f^{(10)}(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)(6n+2)(6n+1)\cdots(6n-6)x^{6n-7}}{2n+1}\\ &=\sum_{n=2}^\infty (-1)^n\frac{(6n+3)(6n+2)(6n+1)\cdots(6n-6)x^{6n-7}}{2n+1}\\ &=\sum_{n=2}^\infty (-1)^n\frac{(6n+3)!}{(2n+1)(6n-7)!}x^{6n-7}\\ f^{(10)}\left(\frac{1}{5}\right)&=\sum_{n=2}^\infty (-1)^n\frac{(6n+3)!}{(2n+1)(6n-7)!\cdot 5^{6n-7}} \end{aligned}$$

(Notice, after ten differentiations, the terms $a_0$ and $a_1$ are both zero.)

Since this is an alternating series, the absolute error involved in using the approximation $S_N$ is at most $$|a_{N+1}|=\frac{(6N+9)!}{(2N+3)(6N-1)!\cdot 5^{6N-1}}$$ By testing a few values of $N$, we find $$|a_6|=|a_{5+1}|=\frac{39!}{(13)(29!)\cdot 5^{29}}\approx 0.00000095<10^{-6}$$ So, $S_5$ is a sufficient approximation. That is,

$$\begin{aligned} f^{(10)}\left(\frac15\right)&\approx \sum_{n=2}^5 (-1)^n\frac{(6n+3)!}{(2n+1)(6n-7)!\cdot 5^{6n-7}}\\ &=(-1)^2\frac{15!}{5\cdot 5!\cdot 5^{5}}+(-1)^3\frac{21!}{7!\cdot 11!\cdot 5^{11}}+ (-1)^4\frac{27!}{9!\cdot 17!\cdot 5^{17}}+ (-1)^5\frac{33!}{11!\cdot 23!\cdot 5^{23}}\\ &=\frac{15!}{5!\cdot5^6}-\frac{21!}{7!\cdot 11!\cdot 5^{11}}+\frac{27!}{9!\cdot 17!\cdot 5^{17}}-\frac{33!}{11!\cdot 23!\cdot 5^{23}} \end{aligned}$$

Remark: if we had calculated $f^{(10)}(1/5)$ directly, using derivative rules instead of series, we would have found an exact value; however, our value here is easier to find, and is highly accurate (if not exact).

• Solution 1: Since $f(x)$ is odd, $f(-x)=-f(x)$ for all $x$ in its domain. We plug this into our power series, then consider the even-indexed terms and the odd-indexed terms separately.

$$\begin{aligned} f(-x)&=-f(x)\\ \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}(-x)^n&=-\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\\ \sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}(-x)^{2n+1}+ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}(-x)^{2n}&= -\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1} -\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\\ -\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}+ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&= -\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1} -\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\\ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&= -\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}\\ 2 \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&=0\\ \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}&=0 \end{aligned}$$

• Solution 2: Alternately, we could note the following:

• Since all derivative of $f(x)$ exist, all its derivatives are continuous.

• The derivative of an odd function is even, and the derivative of an even function is odd.

• So, the even-indexed derivatives of $f(x)$ are continuous, odd functions.

• Every continuous, odd function passes through the origin. That is, $f^{(2n)}(0)=0$.

• So, every term in the series is $0$.