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I was reading some stuff on asymptotic analysis, but how do you get from the 1st line to the 2nd line?

$y \sim \frac{1+x}{2\lambda}\exp\left(\frac{\lambda x}{1+x}\right) - \frac{1+x}{2\lambda}\exp\left(\frac{-\lambda x}{1+x}\right)\\y(1) \sim\frac{1}{\lambda}\exp\left(\frac{\lambda}{2}\right) \text{as } \lambda \rightarrow \infty$

I see they substituted $x=1$, but where does the $- \frac{1+x}{2\lambda}\exp\left(\frac{-\lambda x}{1+x}\right)$ term go?

What does $y(1) \sim\frac{1}{\lambda}\exp\left(\frac{\lambda}{2}\right) \text{as } \lambda \rightarrow \infty$ actually mean? The main part im not sure about is the $\sim$.

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1 Answer 1

up vote 3 down vote accepted

$$f\sim g \quad (\text{as }x\rightarrow\infty)$$

reads "$f$ is asymptotic to $g$ as $x$ goes to infinity" . This means basically that $\lim_{x\rightarrow \infty} \frac{f}{g} = 1$ .

Because $e^x$ dominates $e^{-x}$ as $x$ becomes large (i.e. $e^{-x} \in \mathcal{o}(e^{x})$) they neglected the term with the negative exponent in the second line.

See also http://en.wikipedia.org/wiki/Asymptotics , http://en.wikipedia.org/wiki/Big_O_notation for some of the notation.

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Thanks, understood it! –  Jonathan May 13 '12 at 20:22

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