Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\sim$.

share|cite|improve this question
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 , for some of the notation.

share|cite|improve this answer
Thanks, understood it! – Jonathan May 13 '12 at 20:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.