# Error bounds on series approximation

## Homework Statement

I want to estimate $f(x)=\ln (\frac{1}{1+x})$ on the interval $(1/10,1)$ with the error on the approximation being no more than $0.1$.

## Homework Equations

http://en.wikipedia.org/wiki/Taylor's_theorem#Example

## The Attempt at a Solution

Following the example from Wikipedia, I first found that the $n^{\text{th}}$ derivative for $n \geq 1$ of $f(x)$ is:

$$f^{(n)}(x)=(-1)^n \frac{(n-1)!}{(1+x)^n}$$

Then the remainder term is:

$$R_{k}(x)=\frac{f^{(k+1)}(\xi_{L})}{(k+1)!}(x-a)^{k+1}=(-1)^{k+1}\frac{x^{k+1}}{(1+x)^{k+1} (k+1)}$$

Here's the sketchy part, I think I'm supposed to find an upper bound for $f(x)$ on the interval I want, but since $f(x)$ is a decreasing function, it's maximum value is going to be when $x=1/10$, in which case we have $f(1/10) \approx -0.0953$ so I guess that means I can say $f(x) \leq -0.09$?

Then:

$$|R_{k}(x)| \leq \Big|(-1)^{k+1}\frac{-0.09 x^{k+1}}{(1+x)^{k+1} (k+1)}\Big| = \Big|(-1)^{k+1}\Big| \frac{|-0.09| |x|^{k+1}}{|1+x|^{k+1} |k+1|} = 0.09 \frac{x^{k+1}}{(1+x)^{k+1} (k+1)} \leq \frac{0.09}{(1+x)^{k+1} (k+1)} \leq \frac{0.09}{(k+1)}$$

Finally, impose the condition on the maximum error:

$$\frac{0.09}{(k+1)} < 0.1 \implies k>-1/10 \implies k=0$$

That can't be right though. I think it might have something to do with that $\xi_{L}$ I disregarded.

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