Q-37E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Let   

Show that fⁿ(0)=0 for n=0,1,2....  and hence that the Maclaurin series for f(x)  is 0+0+0+.... , which converges for all x but 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.

Step-by-Step Solution

Answer

Function is not analytic at x=0 .

 

Step 1: Taylor series

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

Step 2: 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 e p(x), where p(x) is some polynomial.

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

Step 3: To calculate the derivative at x=0

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