# Asymptotic series as x approaches infinity

1. Nov 29, 2011

### rg2004

1. The problem statement, all variables and given/known data

Find the first 3 (non-zero) terms of the asymptotic series as x→∞ for ln(ex+1)

2. Relevant equations

ln(1+ε) ~ ε-ε2/2+ε3/3-O(ε4) (x→0)

3. The attempt at a solution

I found
x+ln(1+e-x) (x→∞) = ln(ex+1) (x→∞)​
ε=e-x
which allows me to use the Maclaurin series above (since now the ε→0 as x→∞) yielding
x+[e-x-e-2x/2+e-3x/3+O(e-4x)]

My question
Is there some expansion I should be using for e-x (x→∞) or some rearrangement of the ex (x→0) that I can use as replacement for e-x to find the next two terms of the series

Thanks
Clay

2. Nov 29, 2011

### susskind_leon

You could set x'=1/x to express ln(1+e^(-x)) in terms of x' for x'->0

3. Nov 29, 2011

### Mute

No. The asymptotic expansion of exp(-x) for large x is just exp(-x). The taylor series expansion is valid only for small x, so there's nothing else you can do. If you're looking for an expansion that looks like

$$\ln(1+e^x) \sim f(x)\left( 1 + \frac{a_1}{x} + \frac{a_2}{x^2} + \mathcal O \left(\frac{1}{x^3}\right) \right),$$

then I don't think that's possible. (Mathematica, seems to agree, suggesting there is no expansion of the function around x = infinity). I'd say the best you could do is define y = e^x and then modify the asymptotic series you already derived that's expressed as ln(y) + inverse powers of e^x = y.

Last edited: Nov 29, 2011