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Screwdriver

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## Homework Statement

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

## 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 [itex]n^{\text{th}}[/itex] derivative for [itex]n \geq 1[/itex] of [itex]f(x)[/itex] is:

[tex]f^{(n)}(x)=(-1)^n \frac{(n-1)!}{(1+x)^n}[/tex]

Then the remainder term is:

[tex]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)}[/tex]

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

Then:

[tex]|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)} [/tex]

Finally, impose the condition on the maximum error:

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

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

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