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Q-37E

Expert-verifiedFound in: Page 435

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**Question: Let **

**Show that f ^{n}(0)=0**

Function is not analytic at x=0 .

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

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 .

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

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