$U = \frac{h\omega}{e^{hw/KT}-1}\approx KT - \frac{h\omega}{2}+....O(\frac{h\omega}{KT})$

Question

$$U = \frac{h\omega}{e^{hw/KT}-1}\approx KT - \frac{h\omega}{2}+....O(\frac{h\omega}{KT})$$

If have to prove this for $KT\gg h\omega$

I dont understand what the O in the equation means.
can someone explain what the O is.
There was also a hint saying to take $x= h\omega = KT$
and to then taylor expand around x=0 i dont understand that because that would mean $e^0 -1 = 0$ and that would lead to dividing by zero so how does that work?

Answer 1

HINTS

We say $f(x) = O(g(x))$ when $\exists c, N \in \mathbb{R}$ such that $f(x) \le cg(x)$ whenever $x > N$. In other words, $O(\cdot)$ denotes an upper bound on something.

As for your exact problem, recall that the first term of Taylor expansion allows to approximate $$ e^x \approx 1+x $$ for small $x$, and so $$ e^{h\omega/KT}-1 \approx \left(1 + \frac{h\omega}{KT}\right)-1 = \frac{h\omega}{KT} $$ Can you now do your problem?

Answer 2

The problem should be stated as $\frac{\hbar\omega}{e^{\hbar\omega/(kT)}-1}\approx kT-\frac{\hbar\omega}{2}$. Dividing by $kT$, it's just the $x:=\hbar\omega/(kT)$ special case of $\frac{x}{e^x-1}\approx 1-\frac{x}{2}$, which follows from $\frac{e^x-1}{x}\approx 1+\frac{x}{2}$. This requires $e^x\approx 1+x+\frac12 x^2$, not just a linear approximation. We can state the approximation more fully as $\frac{x}{e^x-1}\approx 1-\frac{x}{2}+O(x^2)$ or $\frac{x}{e^x-1}\approx 1-\frac{x}{2}+o(x)$, to indicate the error term is approximately proportional to $x^2$ if $x$ is small.