91Ó°ÊÓ

Let's find the Maclaurin series (Taylor series centered at \(x=0\)) for the function \(f(x) = e^{-\frac{1}{x^2}}\). We will find its derivatives and evaluate them at \(x=0\): 1. \(f(x) = e^{-\frac{1}{x^2}}\) 2. \(f'(x) = \frac{d}{dx}(e^{-\frac{1}{x^2}}) = 2x e^{-\frac{1}{x^2}}\frac{1}{x^3} = \frac{2}{x} e^{-\frac{1}{x^2}}\) 3. \(f''(x) = -\frac{4}{x^3} e^{-\frac{1}{x^2}} + \frac{4} {x^5} e^{-\frac{1}{x^2}} = \frac{4}{x^3} e^{-\frac{1}{x^2}} \left( 1 - \frac{1}{x^2} \right)\) 4. And so on... Now evaluating each derivative at \(x=0\): 1. \(f(0) = e^{-\frac{1}{0^2}}\) is not defined. 2. \(f'(0) = \frac{2}{0} e^{-\frac{1}{0^2}}\) is not defined. 3. \(f''(0) = \frac{4}{0^3} e^{-\frac{1}{0^2}} \left( 1 - \frac{1}{0^2} \right)\) is not defined. 4. And so on... Since the derivatives of the given function are not defined at \(x=0\), we cannot construct a Taylor series representation centered at \(x=0\). This indicates that the Lagrange's power series representation fails for the given function.

The fact that the derivatives are not defined at \(x=0\) suggests that the function is not analytic at \(x=0\). A function is said to be analytic at a point if it can be represented by a convergent power series in a neighborhood of that point. Since there are no convergent power series representations centered at \(x=0\) for the given function, it is not analytic at \(x=0\). This is another reason why the Lagrange's power series representation fails for the given function.