6.2.1.1 step 1

Assuming that the necessary conditions for case one are satisfied and \(z''=r z, r=\frac {s}{t}\). Let \(\Gamma \) be the set of all poles of \(r\). For each pole \(c\) in this set, three quantities are calculated: Rational function \(\left [ \sqrt {r}\right ]_{c}\) and two complex numbers \(\alpha _{c}^{+},\alpha _{c}^{-}\).

How this is done depends on the order of the pole as described below. If the set \(\Gamma \) is empty (when there are no poles), then this part is skipped.

1. If the pole \(c\) has order \(1\) then

   \begin{align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =1\\ \alpha _{c}^{-} & =1 \end{align*}

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

   \begin{align*} \left [ \sqrt {r}\right ]_{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end{align*}

   Where \(b\) is the coefficient of \(\frac {1}{(x-c)^{2}}\) in the partial fraction decomposition of \(r\).

3. If the pole is of order \(\{4,6,8,\dots \}\) (poles must be all even from the conditions of case one), then the computation is more involved. Let \(2v\) be the order of the pole. Hence if the pole was order 4, then \(v=2\). Let \(\left [ \sqrt {r}\right ]_{c}\) be the sum of terms involving \(\frac {1}{(x-c)^i}\) for \(2\leq i\leq v\) in the Laurent series expansion of \(\sqrt {r}\) (not \(r\)) at \(c\). Therefore

   \begin{align*} \left [ \sqrt {r}\right ] _{c} &=\sum _{i=2}^{v}\frac {a_{i}}{\left (x-c\right ) ^{i}} \\ &= \frac {a_{2}}{\left ( x-c\right ) ^{2}}+\frac {a_{3}}{\left ( x-c\right ) ^{3}}+\cdots +\frac {a_{v}}{\left ( x-c\right ) ^{v}}\tag {1} \end{align*}

   \(\alpha _{c}^{+},\alpha _{c}^{-}\) are found using

   \begin{align*} \alpha _{c}^{+} & =\frac {1}{2}\left ( \frac {b}{a_v}+v\right )\\ \alpha _{c}^{-} & =\frac {1}{2}\left ( -\frac {b}{a_v}+v\right ) \end{align*}

   Where in the above \(a_v\) is the coefficient of the term \(\frac {a_v}{(x-c)^v}\) in (1) and \(b\) is the coefficient of the term \(\frac {1}{(x-c)^{v+1}}\) in \(r\) itself (found from the partial fraction decomposition), minus the coefficient of same term in the Laurent series expansion of \(\sqrt {r}\) at \(c\).

   The coefficients in the Laurent series can be obtained as follows. Given \(r(x)\) with a pole of finite order \(N\) at \(x=c\), then its Laurent series expansion at \(c\) is given by the sum of the analytic part and the principal part of the of the Laurent series. The coefficients \(b_n\) are contained in the principal part of the series.

   \begin{align} r\left ( x\right ) & =\sum _{n=0}^{\infty }a_{n}\left ( x-c\right )^{n}+ \sum _{n=1}^{N}\frac {b_{n}}{\left ( x-c\right ) ^{n}}\tag {2}\\ & =\sum _{n=0}^{\infty }a_{n}\left ( x-c\right ) ^{n}+\frac {b_{1}}{\left ( x-c\right ) }+\frac {b_{2}}{\left ( x-c\right ) ^{2}}+\frac {b_{3}}{\left ( x-c\right ) ^{3}} +\cdots +\frac {b_{N}}{\left ( x-c\right )^{N}}\nonumber \end{align}

   To obtain \(b_{1}\) (which is the residue of \(r\left ( x\right ) \) at \(c\)), both sides of the above are multiplied by \(\left (x-c\right )^{N}\) which gives

   \begin{equation} \left ( x-c\right ) ^{N}r\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}\left ( x-c\right ) ^{n+N}+b_{1}\left ( x-c\right ) ^{N-1}+b_{2}\left ( x-c\right ) ^{N-2}+\cdots +b_{N}\tag {3}\end{equation}
   Differentiating both sides of (3) \((N-1)\) times w.r.t. \(x\) gives
   \[ \frac {d^{N-1}}{dx^{\left ( N-1\right ) }}\left ( \left ( x-c\right )^{N}f\left ( x\right ) \right ) =\sum _{n=0}^{\infty }\frac {d^{N-1}}{dx^{\left (N-1\right ) }}\left ( a_{n}\left ( x-c\right ) ^{n+N}\right ) +b_{1}\left ( N-1\right ) ! \]
   Evaluating the above at \(x=c\) gives
   \[ b_{1}=\frac {\lim _{x\rightarrow c}\frac {d^{N-1}}{dx^{\left ( N-1\right ) }}\left ( \left ( x-c\right ) ^{N}r(x) \right ) }{\left ( N-1\right ) !}\]
   To find the next coefficient \(b_{2}\), both sides of (3) are differentiated \((N-2)\) times
   \[ \frac {d^{N-2}}{dx^{\left ( N-2\right ) }}\left ( \left ( x-c\right ) ^{N}r\left ( x\right ) \right ) =\sum _{n=0}^{\infty }\frac {d^{N-2}}{dx^{\left ( N-2\right ) }}\left ( a_{n}\left ( x-c\right ) ^{n+N}\right ) +b_{1}\left ( N-1\right ) !\left ( x-c\right ) +b_{2}\left ( N-2\right ) ! \]
   Evaluating the above at \(x=c\) gives
   \[ b_{2}=\frac {\lim _{x\rightarrow c}\frac {d^{N-2}}{dx^{\left ( N-2\right ) }}\left ( \left ( x-c\right ) ^{N}r\left ( x\right ) \right ) }{\left ( N-2\right ) !}\]
   The above is repeated to find \(b_{3},b_{4},\cdots ,b_{N}\). The general formula for find coefficient \(b_n\) is therefore
   \begin{align*} b_{n}=\frac {\lim _{x\rightarrow c}\frac {d^{N-n}}{dx^{\left ( N-n\right ) }}\left ( \left ( x-c\right ) ^{N}r\left ( x\right ) \right ) }{\left ( N-n\right ) !}\tag {4} \end{align*}

   For the special case of the last term \(b_{N}\) the above simplifies to

   \begin{align*} b_{N}=\lim _{x\rightarrow c}(x-c)^{N} r(x) \tag {5} \end{align*}

   The above is implemented in the function laurent_coeff() in the Kovacic class.

   This completes finding all the quantities \(\left \{\left [ \sqrt {r}\right ] _{c}, \alpha _{c}^{+},\alpha _{c}^{+}\right \}\) for each pole in the set \(\Gamma \) for case one.

The next step calculates the following three quantities for \(\mathcal {O}(\infty )\).

1. If \(\mathcal {O}(\infty )\leq 0\), which must be even, then let \(-2 v =\mathcal {O}(\infty )\) and \(\left [ \sqrt {r}\right ] _{\infty }\) is then the sum of all terms \(x^i\) for for \(0\leq i\leq v\) in the Laurent series expansion of \(\sqrt {r}\) at \(\infty \).

   \begin{align*} \left [ \sqrt {r}\right ] _{\infty }&= \sum _{i=0}^{v} a_i x^v = a_0 + a_1 x + a_2 x^2 \cdots + a_{v} x^v \tag {6} \end{align*}

   The coefficients \(a_i\) are found by setting \(x=\frac {1}{y}\) in \(r\) and then finding the Laurent series of \(\left [ \sqrt {r(y)}\right ]\) expanded around \(y=zero\). The process for finding the coefficient is the same one used as described earlier where now the limit is taken as \(y\) approaches zero from the right. This gives all the terms of (6). This is implemented in the function laurent_coeff() in the Kovacic class.

   The corresponding \(\{\alpha _{\infty }^{+},\alpha _{\infty }^{-}\}\) are given by

   \begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a_v}-v\right ) \\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a_v}-v\right ) \end{align*}

   Where \(a_v\) is coefficient of \(x^v\) in (6) and \(b\) is the coefficient of \(x^{v-1}\) in \(r\) itself (found using long division) minus the coefficient of \(x^{v-1}\) in \(\left (\left [ \sqrt {r}\right ] _{\infty }\right )^2\).

2. If \(\mathcal {O}(\infty )=2\) then \(\left [ \sqrt {r}\right ] _{\infty }=0\). The corresponding \(\{\alpha _{\infty }^{+},\alpha _{\infty }^{-}\}\) are given by

   \begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}\end{align*}

   Here \(b=\frac {\operatorname {lcoef}(s)}{\operatorname {lcoeff}(t)}\) where \(r=\frac {s}{t}\). \(\operatorname {lcoef}(s)\) is the leading coefficient of \(s\) and similarly, \(\operatorname {lcoef}(t)\) is the leading coefficient of \(t\).

3. If \(\mathcal {O}(\infty )>2\) then

   \begin{align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =0\\ \alpha _{\infty }^{-} & =1 \end{align*}