Question

Question: Let

Show that fⁿ(0)=0for n=0,1,2....and hence that the Maclaurin series for f(x)is 0+0+0+...., which converges for all xbut is equal to f(x)only when x=0. This is an example of a function possessing derivatives of all orders (at x₀=0), whose Taylor series converges, but the Taylor series (about x₀ =0) does not converge to the original function! Consequently, this function is not analytic at x=0.

Short answer

Function is not analytic at x=0 .

Step-by-step solution

Taylor series

For a function f(x) the Taylor series expansion about a point x₀ is given by, $f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).(x-x_0)+f"\left(x_0\right)$.$\frac{{(x-x_0)}^2}{2!}+f"\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+....$

The derivatives of f

Let $x\ne 0$, then f(x)=e ${}^{-\frac{1}{x^2}}$.

For , the derivatives of f are:

f'(x)=e

f''(x)=e $\left(\frac{2}{x^6}-\frac{6}{x^4}\right)$

f'''(x)=e $\left(\frac{-12}{x^7}+\frac{24}{x^5}+\frac{4}{x^9}-\frac{6}{x^4}\right)$

Note that f⁽ⁿ⁾(x) will be of the form ep(x), where p(x) is some polynomial.

From the definition of f it follows that, f(0)=0 .

To calculate the derivative at x=0

We need to calculate the derivative at x=0 by definition,