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

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!

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1 Answer 1

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

$$\frac{0.3^5}5=0.000486>0.0002\;,$$

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

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