# Coth approximation

## Main Question or Discussion Point

In a QM textbook (Newton), I found the below expression for large x:

$coth(x)\cong 1+2e^{-2x}$

I tried

$coth(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}$

George Jones
Staff Emeritus
Gold Member
Write $z = e^{-2x}$, so that $z$ is small when $x$ is large. Then,

$$\coth x = \left(1+z\right)\left(1-z\right)^{-1}.$$

Do a power series expansion of of this, or do a power series expansion of just $\left(1-z\right)^{-1}$.

Is it correct to do like this:
for large x; ex=1+ex.
So on mere substitution in to the eqn.
$$coth(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}$$
one will get
$$coth(x)\cong 1+2e^{-2x}$$
at low temperature the weighing factor is (1/E).
$$x=E/(2k_{\rm B}T)$$.

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