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Here are two sums given:

  1. $$e^{-5}\approx \sum_{i=0}^9 \frac{(-1)^i 5^i}{i!}$$
  2. $$e^{-5}=1/e^5\approx \frac{1}{\sum_{i=0}^9 \frac{5^i}{i!}}$$

Using Matlab ;) I've got (2) to be closer. But computer-aside, is there anyway to say which one is closer than the other, say by thinking about the errors?

Mill
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  • You could polynomial divide (2), I get that may be unfeasible. You might also go about just comparing (1) and (2) as just a series of 2 or 3 terms and then maybe induction. – Zach466920 Jun 19 '15 at 20:59
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    I believe you want $5^i$ and not $5!$ in the denominator of (2). – Joel Jun 19 '15 at 21:00
  • No @Joel , i thought the same thing, but no. :( – Mill Jun 19 '15 at 21:06
  • @Joel is correct, you have miswritten the series in the denominator. The (partial) series for $e^5$ should be $\sum_{i=0}^9 \frac{5^i}{i!}$ – MPW Jun 19 '15 at 21:08
  • @MPW I get it, $e^x= \sum_{i=0}^\infty \frac{x^n}{n!}$. But the problem in front of me has approximated as I've written above. Perhaps the problem has some problem! typo :/ – Mill Jun 19 '15 at 21:12
  • Yes it must...${}{}$ – MPW Jun 19 '15 at 21:13
  • @MPW Just changed it. Now obviously (1) has a $(-1)^i$ which makes troubles, but what exactly? – Mill Jun 19 '15 at 21:19
  • At first I thought the alternating series should give the better approximation, but now I realize that what I was thinking would have made sense only with a number between $0$ and $1$, and $5$ is nowhere near that. ${}\qquad{}$ – Michael Hardy Jun 19 '15 at 21:34

2 Answers2

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The approximation $$ e^x\approx\sum_{i=0}^{n-1}\frac{x^i}{i!} $$ will have error term $\approx x^n/n!$ when $n\gg x$ since at this stage the terms are getting closer to zero. Thus, for $e^x$ and $e^{-x}$, the error term is the same order of magnitude. However, for $x>0$, the error term relative to the value is larger for $e^{-x}$, which is what matters when we estimate $e^{-x}$ using $1/e^x$.

Thus, for $x>0$, it is better to approximate $e^{-x}$ as $1/e^x$.

Einar Rødland
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The proportionate error is related to the value of the sum so far. The absolute value of the terms here is the same, but in the first case the sign alternates and the terms cancel (at least a bit) so each new term is greater in proportion to the sum than in the second case where there is no cancellation.

Mark Bennet
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    This should be a comment, you should back up your statements with proof... – Zach466920 Jun 19 '15 at 21:29
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    @Zach466920 The question is, is there a way to do this by thinking about the errors. This answer says "yes, and here is a way of thinking which works". I left the detail to be worked out by OP - but I think this is an answer to the question which was asked. I have just been reading Tom Körner's "Calculus for the ambitious" which deals with this kind of question in chapter 6, just to add a useful reference. – Mark Bennet Jun 19 '15 at 21:34