The taylor series for $ln(x)$, centered at $x=1$, is $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n} $$ Let $f$ be the function given by the sum of the first three nonzero terms of this series. The maximum value of $|\ln(x)-f(x)|$ for $0.3\le\ x \le\ 1.7$ is?

When I look at this question, I instinctively think of alternating series error bound. Therefore the maximum error should be equal to the first omitted term $$= (-1)^{4+1}\frac{(x-1)^4}{4}$$ when we substitute in the endpoints of x, the results are the same $ =0.060025$

This solution is incorrect, but I do not understand why. The correct solution is the tedious way of actually calculating $$|\ln(0.3)-(\frac{x-1}{1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3})|$$ $=0.145$ (assuming $\ln(0.3)$ gives the largest answer). Why is this so? Why does using error bound give an incorrect answer?


Does your series really alternate?

Each term you add is negative for $x = 0.3$.

Addendum: When $n$ is an even number you multiply an even term by $(-1)^{(n+1)}$ to reach a negative number.

  • $\begingroup$ not when $n$ is an even number $\endgroup$ – Harrison Mar 10 '14 at 3:53
  • $\begingroup$ @Harrison: user1789954 is pointing out that for $x \lt 1$, your series is not alternating. All terms are negative. Your alternating series error bound will work for $x \gt 1$, but not for $x$ below that "center of the expansion". $\endgroup$ – hardmath Mar 10 '14 at 3:57
  • $\begingroup$ Ahhh..I see. Thank you! $\endgroup$ – Harrison Mar 10 '14 at 4:01
  • $\begingroup$ Here are the first few terms of the series: $-.7, -.245, -.114333, -.060025...$ $\endgroup$ – Brad Mar 10 '14 at 4:01

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