6.2.3.1 step 1

Assuming the necessary conditions for case three are satisfied and \(z''=r z, r=\frac {s}{t}\). Let \(\Gamma \) be the set of all poles of \(r\). Recall that case three can have either a pole of order 1 or order 2 only. For each pole \(c\) in this set, \(E_c\) is found as follows

1. If the pole \(c\) has order \(1\) then \(E_c=\{12\}\).

2. If the pole \(c\) is of order \(2\) then

   \begin{alignat*}{2} E_c &=\left \{ 6 + \frac {12 k}{n} \sqrt {1+ 4 b} \right \} \qquad \text {for} \quad k &&=-\frac {n}{2}\cdots \frac {n}{2} \tag {1} \end{alignat*}

   Where \(k\) is incremented by \(1\) each time, and \(n\) is any of \(\{4,6,12\}\) and \(b\) is the coefficient of \(\frac {1}{(x-c)^2}\) in the partial fraction decomposition of \(r\). In the above set \(E_c\), only integer values are kept. For an example, when \(n=4\) then \(k=\{-2,-1,0,1,2\}\) and \(E_c =\{ 6 -6 \sqrt {1+ 4 b},6 -3 \sqrt {1+ 4 b},6,6 + 3 \sqrt {1+ 4 b},6 + 6 \sqrt {1+ 4 b}\}\) and similarly for \(n=6\) and \(n=12\).

The next step determines \(E_{\infty }\). This is found using same formula as (1) but \(b\) is calculated differently using \(b=\frac {\operatorname {lcoef}(s)}{\operatorname {lcoeff}(t)}\) where \(r=\frac {s}{t}\). \(\operatorname {lcoef}(s)\) is the leading coefficient of \(s\) and \(\operatorname {lcoef}(t)\) is the leading coefficient of \(t\).