Taylor Series cubic approximation

**TAG16** · April 7th 2009, 02:30 PM

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 ?

**Danny** · April 7th 2009, 03:35 PM

Originally Posted by TAG16

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 ?

The derivatives are

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

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

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