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Expert-verified Found in: Page 435 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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.

Function is not analytic at x=0 .

See the step by step solution

## Step 1: Taylor series

For a function f(x) the Taylor series expansion about a point x0 is given by, $f\left(x-{x}_{0}\right)=f\left({x}_{0}\right)+f\text{'}\left({x}_{0}\right).\left(x-{x}_{0}\right)+f"\left({x}_{0}\right)$.$\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+f"\text{'}\left({x}_{0}\right).\frac{{\left(x-{x}_{0}\right)}^{3}}{3!}+....$

## 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, ### Want to see more solutions like these? 