Benedict Irwin edited untitled.tex  about 8 years ago

Commit id: 6774037760687602df3a33df8f3cd357771bca26

deletions | additions      

       

\begin{equation}  I[G(x)]=\frac{1}{1-G(x)}-1  \end{equation}  Let $I^{(n)}[G(x)]$ denote a repeated application of the invert transform $n$ times. Then \begin{equation}  I^{(2)}[G(x)]=\frac{1}{1-\frac{1}{1-G(x)}-1}-1\\  I^{(3)}[G(x)]=\frac{1}{1-\frac{1}{1-\frac{1}{1-G(x)}-1}-1}-1  \end{equation}  We then have \begin{equation}  I^{(0)}[P(x)]=2x+3x^2+5x^3+7x^4+11x^5+\cdots\\  I^{(1)}[P(x)]=2x+7x^2+25x^3+88x^4+311x^5+\cdots\\