The power series representation of $e^x = \sum \limits_{k=0}^{\infty} \frac{x^k}{k!}$. Can I use this approximation for $e^{-x} = 1/e^x = 1/\sum \limits_{k=0}^{\infty} \frac{x^k}{k!}$ instead of $e^{-x} = \sum \limits_{k=0}^{\infty} \frac{(-1)^k x^k}{k!}$. What is the difference between two approximation?
Power series representation of $e^x$ and $e^{-x}$
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approximation
power-series
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0Note: The power series representations are not approximations of $e^x$ or $e^{-x}$. They are the functions themselves... – 2011-11-13
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0You could. The series $\sum_{k=0}^\infty {(-1)^kx^k\over k!}$ should have a faster rate of convergence, since it is an alternating series. – 2011-11-13
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0You can use either, but I would recommend using the alternating series, because then you have the standard upper bound for the cut-off error, so less work to do. – 2011-11-13
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1However, for practical computational work, using the alternating series is a bad idea, as it is quite prone to subtractive cancellation. You eventually reach the point where you're adding up terms of (nearly) the same magnitude but of opposite sign. – 2011-11-13
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1I have not done any thinking about it, but in in some cases at least the $1/(1+x+\cdots)$ approach seems to give a better approximation for given number of terms. It would be interesting to know something general. Nice question! – 2011-11-13
1 Answers
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The two are equivalent. If you know about power series multiplication, you can see that
$$ \left(\sum_{k=0}^\infty \frac{x^k}{k!}\right)\cdot \left(\sum_{k=0}^\infty \frac{(-1)^kx^k}{k!}\right)=1. $$
If you expand the product of the two power series, terms will cancel, leaving you with $1$.