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I need to evaluate $e^{1/3}$ to the nearest $0.0001$ using Maclaurin Series. However, each term in the series I used is positive. Due to this, even if the $n$th term is less than $0.0001$, the term after that may have an impact. For example, if you have a certain non-alternating, decreasing, series to $2$ terms, it may result in $4.40$. Let's say the third term is $0.04$, and the fourth term is $0.01$. While the $4.44$ doesn't change the nearest tenth ($4.4$), $4.45$ sure does ($4.5$).

Now I already have a solution to this problem which is to compare $e^x$ to a greater geometric series which you can evaluate. If sum of that geometric series is insignificant, then the remaining terms of the expansion will also be. However, there comes a problem when the geometric series is significant.

So my question is about how to bound the series better - which I think is done better through alternating series. Are there any alternating series I can use in this situation?

Yash Jain
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2 Answers2

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It's not at all clear to me what's going on when you say "If sum of that geometric series is insignificant, then the remaining terms of the expansion will also be. However, there comes a problem when the geometric series is significant." The error in approximating $e^x$ by $n$ terms is exactly $$E=\sum_{k=n+1}\frac{x^k}{k!}.$$So if $|x|<1$ then $$E<\frac1{(n+1)!}\sum_{k=n+1}^\infty x^k=\frac1{(n+1)!}\frac{x^{n+1}}{1-x}.$$So let $x=1/3$ and choose $n$ so that last term is smaller than $0.0001$ and you're done.

Note that $E>\frac{x^{n+1}}{(n+1)!}$, which says the upper bound is within a constant factor of being equality. This is to say you're not going to find a better estimate lying around...

quid
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David C. Ullrich
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As @David C. Ullrich answered, you need to find $n$ such that $$\frac1{(n+1)!}\frac{x^{n+1}}{1-x} \leq 10^{-k}\implies (n+1)! >\frac{x^{n+1}}{1-x}10^{k} >x^{n+1}\,10^k$$

If you look at this question of mine, you will find a superbe approximation from @rogjohn. Applied to your case, this will give $$n\sim e\, x\,\exp\Big[W\left(\frac{2k \log (10)-\log (2 \pi x)}{2 e x}\right)\Big]-\frac 32$$ where appears Lambert function. Evaluated for $x=\frac 13$ and $k=4$, this would give $$n \sim\frac e 3\,\exp\Big[W\left(\frac{3 }{2 e}\log \left(\frac{15\times 10^7}{\pi }\right)\right)\Big]-\frac 32\approx 3.61$$ that is to say $n=4$.

Without any approximation and using calculus, the exact solution of $$\frac{3^{-n}}{2 (n+1)!}=10^{-4}$$ is $n=3.76$; again $n=4$.

Claude Leibovici
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