Using a 3rd order Maclaurin polynomial, find an upper bound on the error when log(1+x) is approximated by a 3rd order polynomial for |x|<= 0.1. For some reason I keep getting a different answer from my book. Can someone check where I have gone wrong
$$ f(x) = log(x+1) $$
$$ f'(x) = \frac{1}{x+1} $$
$$ f''(x) = \frac{-1}{(x+1)^2} $$
$$ f'''(x) = \frac{2}{(x+1)^3} $$
$$ f(0) = 0, f'(0)=1, f''(0)=-1, f'''(0) = 2$$
$$ f(x) \approx x – \frac{x^2}{2} + \frac{x^3}{3} $$
$$Error:$$
$$ f''''(x) = \frac{-6}{(x+1)^4} $$
$$ f''''(c) = \frac{-6}{(c+1)^4} $$
$$R_3(x) = |\frac{x^4}{4(1+c)^4} |$$
From here on I thought, to find the upper bound on the error, x^4 must be a maximum, so x = 0.1, and 4(1+c)^4 must be a minimum. Therefore I let x = 0.1, and I let c = 0, since 0<=c<=0.1.
So I got the upperbound of the error to be:
$$R_3(x) < \frac{1}{40000}$$
For some reason, the solutions in my book states the result is:
$$R_3(x) < \frac{1}{26244}$$
Best Answer
The calculation is basically right. However, the range is supposed to be $|x|\le 0.1$. The "worst case" bound for the fourth derivative should be obtained by setting $c=-0.1$, and not $c=0$.