Normalisation and validity

First make the constant term inside the bracket equal to one. For \(A>0\), \((A+Bx)^\alpha=A^\alpha\left(1+\frac BAx\right)^\alpha\).

The expansion parameter is \(u=\frac BAx\), so impose \(\left|\frac BAx\right|<1\), rather than automatically writing \(|x|<1\).

For example, \((9+2x)^{\tfrac{1}{2}}=3(1+\frac{2x}{9})^{\tfrac{1}{2}}\), with \(-\frac92<x<\frac92\).

With a substitution such as \(u=-4x^2\), impose \(|-4x^2|<1\): this gives \(-\frac12<x<\frac12\). Powers of \(u\) produce powers \(x^2,x^4,x^6,\ldots\).

When multiplying two infinite expansions, use the intersection of their validity ranges. The domain of the original expression can be larger than the range for a particular series.

These notes use the standard open range \(|u|<1\). Endpoint behaviour needs a separate check and must not be inferred by replacing the strict inequality with an inclusive one.