Mathematical identity in equation (3.7) of John Cardy's Scaling and Renormalization I know I'm at risk of appearing quite stupid, but can someone explain to me the following identity, appearing in equation 3.7 of Cardy's book.
$$
e^{Ks_3s_4} = \cosh(K)(1 + \tanh(K) s_3s_4) \tag{3.7}
$$
It is a factor appearing in a partition sum term of the one-dimensional classical Ising model. Cardy does not explain this identity. It baffles me, since the right hand side is linear in $s_1s_2$, while it cannot be a first-order Taylor expansion.
 A: I have not looked at the book. I am assuming that $(s_3,s_4)$ can take values only from $(+1,+1)$,$(+1,-1)$,$(-1,+1)$ and $(-1,-1)$. In that case, the product $s_3 s_4$ can take only two values : $+1$ and $-1$. For these values of $s_3s_4$, the right hand side and left hand side are equal because $\cosh K \tanh K = \sinh K$ and $\cosh K \pm \sinh K=\exp(\pm K)$.
(The expression does not look like a Taylor expansion, as the statement is true only when $s_3s_4=\pm 1$).
Another way to see it is that $s_3s_4$ satisfies the property that it squares to $+1$. As a result, the terms in the regular Taylor expansion of the exponential function simplify.
Setting $x=s_3s_4$,
$\exp(Kx)=1+Kx+K^2 x^2/2! + ...$
Since $x^2=1$, the even and odd powered terms can be combined to obtain Taylor expansions of $\sinh K $ and $\cosh K$.
$\exp(Kx)=(1+K^2 /2! + ...) + x(K + K^3/3!...)=\cosh K + x \sinh K$
Since this relies on $x^2$ being $1$, you can see that the statement should be true also when $s_3s_4$ is a Pauli matrix.
