Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ f(x) = \ln(1+x)$$

The previous part of this question required me to write down the remainder term for the taylor polynomial of order n.

My remainder term worked out to be:

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

I checked this manually, and it seemed to be correct.

For this question I am quite sure I need to solve for $n$, such that $R_n(x) < 0.0002$.

I have:

$$(-1)^n \frac{(0.3)^{n+1}}{n+1} < 0.0002 $$

(I have $0.3$ for $x$, since $\ln(1+0.3) = \ln(1.3)$.) Am I on the right track here? I am a little stuck with the algebra. I have:

$$ (-1)^n(0.3)^{(n+1)} < 0.0002(n+1) $$

I have tried to take the natural log of both sides, but it seems like I can't isolate the $n$.

Any help would be greatly appreciated!

share|cite|improve this question
up vote 1 down vote accepted

You actually want $|R_n(0.3)|<0.0002$, so you can ignore the factor of $(-1)^n$.

Here I think that your easiest bet is a little intelligently directed trial and error. Note that after you take logs, you have $(n+1)\ln 0.3<\ln 0.0002+\ln(n+1)$. Now $\ln 0.3$ is roughly $-1.2$, $\ln 0.0002$ is roughly $-8.5$, and $7\cdot12=84$, so a very rough first approximation would be to take $n+1=7$. Since $\ln 7$ is roughly $2$, in round numbers we have $-8.4$ on the left and $-6.5$ on the right, so $n=6$ is big enough: $-8.4<-6.5$. Now it only remains to be seen how much smaller you can go. By actual computation I get

$$\frac{0.3^6}6=0.0001215<0.0002$$ and


so it appears that $n=5$ is the best we can do.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.