테일러·매클로린 급수

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).

(a) To sketch $y=f(x)$, we note the following:

• $f(x)$ is never negative.
• $\lim\limits_{x \to \pm \infty} f(x)=e^0=1$, so the curve has horizontal asymptotes in both directions at $y=1$.
• $\lim\limits_{x \to \pm 0} f(x)=\lim\limits_{x \to \pm 0} \frac{1}{e^{1/x^2}}=\lim\limits_{u \to +\infty}\frac{1}{e^u}=0=f(0)$, so the curve is continuous at $x=0$.
• For $x\neq 0$, $f'(x)=\frac{2}{x^3}e^{-1/x^2}$, so our curve is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$
• For $x\neq 0$, $f''(x)=2x^{-6}(2-3x^2)e^{-1/x^2}$, so our curve is concave up on $(-\sqrt{2/3},\sqrt{2/3})$, and concave down elsewhere.

(b) Since $f^{(n)}(0)=0$ for all whole $n$ (that is, the graph is really quite flat at the origin), and since $f(0)=0$, the Maclaurin series for $f(x)$ is $\displaystyle\sum_{n=0}^\infty \frac{0}{n!}x^n=0$. (c) The Maclaurin series converges for all real values of $x$ (to the constant 0). (d) Since $e^y>0$ for any real $y$, we see $f(x)=0$ only when $x=0$. So, $f(x)$ is only equal to its Maclaurin series at the single point $x=0$.

Remark: the function $f(x)$ is an example of a function whose Maclaurin series converges, but not to $f(x)$! To describe this behaviour, we say $f(x)$ is non-analytic.

• 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 $0$.