Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the magnitude of the error in the approximation when lxl<0.1 ?

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- Apr 7th 2009, 02:30 PMTAG16Taylor Series cubic approximation
Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the magnitude of the error in the approximation when lxl

__<__0.1 ? - Apr 7th 2009, 03:35 PMJester
The derivatives are

$\displaystyle f'(x) = \frac{1}{(1-x)^2},\;\; f'(0) = 1$

$\displaystyle f''(x) = \frac{2}{(1-x)^3},\;\; f''(0) = 2!$

$\displaystyle f'''(x) = \frac{3!}{(1-x)^4},\;\; f'''(0) = 3!$

$\displaystyle f^{(4)}(x) = \frac{4!}{(1-x)^5},\;\; \text{for the remainder}$

The Taylor polynomial $\displaystyle P_3(x) = 1 + x + x^2 + x^3$ and the Taylor remainder $\displaystyle R_3(x) = f^{(4)}(c) \frac{x^4}{4!} = \frac{x^4}{(1-c)^5} $ where c is between 0 and $\displaystyle x$. Since $\displaystyle |x| < 0.1 $ then we are trying to determine the maximum of $\displaystyle |R_3(x)| = \left|\frac{x^4}{(1-c)^5}\right|$ on the domain $\displaystyle [-0.1,0.1]\;\times\;[-0.1,0.1]$ and so this would occur at $\displaystyle x = \pm 0.1,\;\;c = 0.1$ and the maximum error would be

$\displaystyle |R_3(x)| \le 0.00006209$.