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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

Expert verified

Function is not analytic at x=0 .

Step by step solution

01

Taylor series

For a function f(x) the Taylor series expansion about a point x0 is given by, f(x-x0)=f(x0)+f'(x0).(x-x0)+f"(x0).(x-x0)22!+f"'(x0).(x-x0)33!+....

02

The derivatives of f

Let x≠0, then f(x)=e -1x2.

For , the derivatives of f are:

f'(x)=e

f''(x)=e 2x6-6x4

f'''(x)=e -12x7+24x5+4x9-6x4

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 .

03

To calculate the derivative at x=0

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

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