급수 심화 전체 문제

(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.