Step 1

part (a) Here the ﬁxed parts \(e_{fixed},\theta _{fixed}\) are calculated using

\begin{align} e_{fixed} & =\frac {1}{4}\left ( \min \left ( O\left ( \infty \right ) ,2\right ) -\deg \left ( t\right ) -3\deg \left ( t_{1}\right ) \right ) \tag {5}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {t^{\prime }}{t}+3\frac {t_{1}^{\prime }}{t_{1}}\right ) \tag {6}\end{align}

For an example, lets say that \(r=\frac {16x-3}{16x^{2}}\). Hence \(t=16x^{2}=t_{1}t_{2}^{2}\) where \(t_{1}=16,t_{2}=x\). And \(O\left ( \infty \right ) =2-1=1\). Therefore \(\deg \left ( t\right ) =2,\deg \left ( t_{1}\right ) =0\) and \(t^{\prime }=\frac {d}{dx}16x^{2}=32x\) and \(t_{1}^{\prime }=0\). The above becomes

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 1,2\right ) -2-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 1-2\right ) \\ & =-\frac {1}{4}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {32x}{16x^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2x}\end{align*}

Here the values \(e_{i},\theta _{i}\) are found for \(i=1\cdots k_{2}\) where \(k_{2}\) is the number of roots of \(t_{2}\). In other words, the number of poles of \(r\) that are of order \(2\). These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). Label these poles \(c_{1},c_{2},\cdots ,c_{k_{2}}\). For each \(c_{i}\) then \(e_{i}=\sqrt {1+4b}\) where \(b\) is the coeﬃcient of \(\frac {1}{\left ( x-c_{i}\right ) ^{2}}\) in the partial fraction expansion of \(r\) and \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). For an example, if \(t=x^{2}\). Then \(t=t_{1}t_{2}^{2}\) and \(t_{1}=1,t_{2}=x\). Hence the zeros of \(t_{2}\) are \(c_{1}=0\). There is only one zero. Hence \(k_{2}=1\) and we only have one iteration to do. Hence \(b=1,e_{1}=\sqrt {1+4\left ( 1\right ) }=\sqrt {5}\) and \(\theta _{e}=\frac {\sqrt {5}}{x-0}=\frac {\sqrt {5}}{x}\).

This part applied only to case 1. It generates additional values of \(e_{i},\theta _{i}\) in addition to what was generated in part (b). This is done for poles of \(r\) of order \(4,6,8,\cdots ,M\) if any exist. These are the roots of \(t_{4},t_{6},\cdots ,t_{M}\). These poles are labeled \(c_{k_{2}+1},c_{k_{2}+2},\cdots ,c_{k}\). The labeling starts from \(k_{2}\) since for the pole of order 2 in part b we used \(c_{1},c_{2},\cdots ,c_{k_{2}}\) for its zeros. Now we iterate for \(i\) from \(k_{2}+1\) to \(k\). For each pole we ﬁnd its \(e_{i}\) and \(\theta _{i}\). These are found similar to original Kovacic paper. Examples below will illustrate better how this is done.

Note that if there are no poles of order \(4,6,\cdots \) then \(k=k_{2}=M.\) The value \(M\) is used below to generate \(s\) sequences in step 2. What this means is that for case 1, if there are no poles of order \(4,6,\cdots \), then \(M\) is just the number of poles of \(r\) of order 2. For cases other than case 1, \(M\) is always number of poles of \(r\) of order 2. The checking on poles of order \(4,6,\cdots \) is only done for case 1.

Now we need to ﬁnd \(e_{0},\theta _{0}\). If \(O\left ( \infty \right ) >2\) then \(e_{0}=1,\theta _{0}=0\). But if \(O\left ( \infty \right ) =2\) then \(\theta _{0}=0\) and \(e_{0}=\sqrt {1+4b}\) where \(b\) is the coeﬃcient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). This is the same as was done in original Kovacic algorithm. If \(O\left ( \infty \right ) \) is none of the above two cases, then case 1 is handled on its own and examples below show how this is done. Otherwise for all other cases \(e_{0}=0,e_{0}=0\).

The above complete step 1, which is to generates the candidate \(e^{\prime }s\) and \(\theta ^{\prime }s\). In step 2, these are used to generate trials \(d\) and \(\theta \) and ﬁnd from them \(P\left ( x\right ) \) polynomial if possible.

4.0.2 Step 1