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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: $$R_3(x) < \frac{1}{26244}$$

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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 $c=0$.

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