# Approximation problem

1. Aug 7, 2008

### Narcol2000

I'm having problems understanding how

$$\frac{e^{-\hbar \omega / 2k_BT}}{1-e^{-\hbar \omega / k_BT}}$$

approximates to

$$k_BT/ \hbar\omega$$

when

$$T >> \hbar\omega/k_B$$

Seems like it should be simple but don't quite see how to arrive at this result.

*update*

I have tried using taylor expansions of $$exp(-x)$$ and $$1-exp(-x)$$ and just using the first expansion term since if $$T>>\hbar\omega/k_B$$ then $$\hbar\omega/k_BT$$ should be small. This seems to give the right answer but i'd be interested in knowing if indeed my method is ok and if there are alternate methods.

Last edited: Aug 7, 2008
2. Aug 7, 2008

### lemma28

If you call
$$x = - {{\hbar \omega } \over {2 k_B T}}$$
(and $$x \to 0$$ when $$T >> \hbar\omega/k_B$$)

then your expression is equivalent to
$${{e^x } \over {1 - e^{2x} }}$$

Utilizing the known limit
$$\mathop {\lim }\limits_{x \to 0} {{e^x -1} \over x} = 1$$

you can write
$$\mathop {\lim }\limits_{x \to 0} {{e^x } \over {1 - e^{2x} }} = \mathop {\lim }\limits_{x \to 0} {{e^x } \over {\left( {1 - e^x } \right)\left( {1 + e^x } \right)}}\left( {{{e^x - 1 } \over x}} \right) \to -{1 \over {2x}}$$

So the expression near zero goes like
$$-{1 \over {2x}}$$
that means that the original expression goes like
$${{k_B T} \over {\hbar \omega }}$$

3. Aug 8, 2008

### Narcol2000

Thanks, the way i did it was equivalent it seems, but yours was a lot more clearer..

thanks again.