Review : Taylor Series — Question 6

Question 6 – Solution

Strategy. Separate the existence of every derivative from the limiting behavior of the Taylor remainders.

Step 1: Exponential decay beats every power. Put $u=1/x^2\to\infty$. Choose an integer $k>m/2$. Since $e^u\ge u^k/k!$, $$|x|^{-m}e^{-1/x^2}=u^{m/2}e^{-u} \le k!u^{m/2-k}\longrightarrow 0.$$

Step 2: Prove smoothness, including the center. For $x\ne 0$, induction gives $f^{(n)}(x)=P_n(1/x)e^{-1/x^2}$, where $P_0(z)=1$ and $$P_{n+1}(z)=-z^2P_n'(z)+2z^3P_n(z).$$ Define each candidate derivative to be zero at $0$. Step 1 proves its continuity there. Its difference quotient at $0$ is $P_n(1/h)e^{-1/h^2}/h\to 0$, again by Step 1. Inductively each candidate is the derivative of the preceding one, so $f\in C^\infty(\mathbb R)$ and $f^{(n)}(0)=0$ for every $n\ge 0$.

Step 3: Compare the series with the function. Every Taylor polynomial is zero. Thus the Maclaurin series is $\boxed{\sum_{n=0}^{\infty}0x^n=0}$, with infinite radius. It equals $f$ only at $x=0$, since $f(x)>0$ for every nonzero $x$. Infinite radius does not by itself identify the represented function.

Step 4: Locate the missing hypothesis. Taylor’s finite theorem remains valid on the segment from $0$ to any fixed $x$. But $\boxed{R_n(x)=f(x)-T_n(x)=e^{-1/x^2}>0}$ for $x\ne 0$, independently of $n$. Thus $R_n(x)\not\to 0$. Smoothness provides each finite remainder formula, not a derivative-growth bound forcing remainders to vanish. The graph and the zero Taylor sum touch to every order at $0$ without agreeing nearby.