Show that $f(x) = 0$ if $x = 0$ $f(x) = \text{exp}(-1/x^{2})$ if $x \neq 0$ is infinitely differentiable but not analytic at $x_{0} = 0$ `

Question

So taking the derivative of the latter half we get $$\left(\frac{2}{x^{3}}\right)e^{\frac{-1}{x^{2}}}$$ by chain rule. Im not sure how to find the derivative at x=0, and show that it equals 0 as it only produces more '1/0' expressions. Additionally I assume to show it is not analytic at $x_0=0$, we use the taylor series for $e^{x}$ which is $$1+x+\frac{x^2}{2}+\cdots+\frac{x^{n}}{n!}$$ Where we substitute $\frac{-1}{x^{2}}$ for x to get $$e^{\frac{-1}{x^{2}}}=1-\frac{1}{x^{2}}+\frac{1}{x^{4}2!}-\frac{1}{x^{6}3!}\cdots$$ So when we plug in x=0 this is invalid and thus is not analytic.

Is my latter attempt sufficient or do I need to add more? And how do I reach the conclusion that the function is differentiable at x=0.

Answer 1

To prove that the function is infinitely differentiable you need some formula for the n-th derivative at points $x>0$. The fact that $f$ is not analytic is immediate from the fact that nonzero analytic functions cannot vanish on a set with limit points.

Hints: prove by a simple induction argument that $f^{(n)} (x)=p_n(\frac 1 x) f(x)$ for some polynomial $p_n$. Then change $x$ to $\frac 1 x$ and use the fact that $\frac {e^{x}} {p(x)} \to \infty$ as $x \to \infty$ for any polynomial $p$. [ This last fact follows by repeated application of L'Hopital's Rule].

Answer 2

Hint: You should be able to show the derivatives at $0$ are all $0$. Thus the Taylor series at $0$ would be $0$.

Answer 3

I expect this is a duplicate of an existing question.

But if not, some hints are:

(1) Show by induction that, for all $n \in \mathbb N$ there is a rational function $r_n(x)$ such that $$ f^{(n)}(x) = r_n(x)\exp(-1/x^q),\quad x \ne 0. $$

(2) if $r(x)$ is a ratonal function, then $$ \lim_{x \to 0} r(x)\exp(-1/x^2) = 0. $$

Answer 4

Every Real -analytic function $f(x)$ is the Real part of a complex function $F(z)=U+iV$. Since f is compact and continuous, it is bounded. Since $f$ , the Real part ,is bounded, so is $F(z)$ , its Complex extension. Then $F(z)$ is an entire, bounded function, and so must be constant, which it is not -- a contradiction.